Foundation of Mathematical Economics Solutions

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Foundation of Mathematical Economics by Michael Carter Solution manual The solutions manual contains detailed answers t...

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Solutions Manual Foundations of Mathematical Economics Michael Carter November 15, 2002

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Chapter 1: Sets and Spaces 1.1 { 1, 3, 5, 7 . . . } or { š‘› āˆˆ š‘ : š‘› is odd } 1.2 Every š‘„ āˆˆ š“ also belongs to šµ. Every š‘„ āˆˆ šµ also belongs to š“. Hence š“, šµ have precisely the same elements. 1.3 Examples of ļ¬nite sets are āˆ™ the letters of the alphabet { A, B, C, . . . , Z } āˆ™ the set of consumers in an economy āˆ™ the set of goods in an economy āˆ™ the set of players in a game. Examples of inļ¬nite sets are āˆ™ the real numbers ā„œ āˆ™ the natural numbers š”‘ āˆ™ the set of all possible colors āˆ™ the set of possible prices of copper on the world market āˆ™ the set of possible temperatures of liquid water. 1.4 š‘† = { 1, 2, 3, 4, 5, 6 }, šø = { 2, 4, 6 }. 1.5 The player set is š‘ = { Jenny, Chris }. Their action spaces are š“š‘– = { Rock, Scissors, Paper }

š‘– = Jenny, Chris

1.6 The set of players is š‘ = {1, 2, . . . , š‘› }. The strategy space of each player is the set of feasible outputs š“š‘– = { š‘žš‘– āˆˆ ā„œ+ : š‘žš‘– ā‰¤ š‘„š‘– } where š‘žš‘– is the output of dam š‘–. 1.7 The player set is š‘ = {1, 2, 3}. There are 23 = 8 coalitions, namely š’«(š‘ ) = {āˆ…, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} There are 2

10

coalitions in a ten player game.

/ š‘† āˆŖ š‘‡ . This implies š‘„ āˆˆ / š‘† and š‘„ āˆˆ / š‘‡, 1.8 Assume that š‘„ āˆˆ (š‘† āˆŖ š‘‡ )š‘ . That is š‘„ āˆˆ or š‘„ āˆˆ š‘† š‘ and š‘„ āˆˆ š‘‡ š‘. Consequently, š‘„ āˆˆ š‘† š‘ āˆ© š‘‡ š‘ . Conversely, assume š‘„ āˆˆ š‘† š‘ āˆ© š‘‡ š‘ . This implies that š‘„ āˆˆ š‘† š‘ and š‘„ āˆˆ š‘‡ š‘ . Consequently š‘„ āˆˆ / š‘† and š‘„ āˆˆ / š‘‡ and therefore š‘„āˆˆ / š‘† āˆŖ š‘‡ . This implies that š‘„ āˆˆ (š‘† āˆŖ š‘‡ )š‘ . The other identity is proved similarly. 1.9

āˆŖ

š‘†=š‘

š‘†āˆˆš’ž

āˆ©

š‘†=āˆ…

š‘†āˆˆš’ž

1

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

1

-1

š‘„2

0

1

š‘„1

-1 Figure 1.1: The relation { (š‘„, š‘¦) : š‘„2 + š‘¦ 2 = 1 } 1.10 The sample space of a single coin toss is { š», š‘‡ }. The set of possible outcomes in three tosses is the product { {š», š‘‡ } Ɨ {š», š‘‡ } Ɨ {š», š‘‡ } = (š», š», š»), (š», š», š‘‡ ), (š», š‘‡, š»), } (š», š‘‡, š‘‡ ), (š‘‡, š», š»), (š‘‡, š», š‘‡ ), (š‘‡, š‘‡, š»), (š‘‡, š‘‡, š‘‡ ) A typical outcome is the sequence (š», š», š‘‡ ) of two heads followed by a tail. 1.11 š‘Œ āˆ© ā„œš‘›+ = {0} where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no outputs. To see this, ļ¬rst note that 0 is a feasible production plan. Therefore, 0 āˆˆ š‘Œ . Also, 0 āˆˆ ā„œš‘›+ and therefore 0 āˆˆ š‘Œ āˆ© ā„œš‘›+ . To show that there is no other feasible production plan in ā„œš‘›+ , we assume the contrary. That is, we assume there is some feasible production plan y āˆˆ ā„œš‘›+ āˆ– {0}. This implies the existence of a plan producing a positive output with no inputs. This technological infeasible, so that š‘¦ āˆˆ / š‘Œ. 1.12

1. Let x āˆˆ š‘‰ (š‘¦). This implies that (š‘¦, āˆ’x) āˆˆ š‘Œ . Let xā€² ā‰„ x. Then (š‘¦, āˆ’xā€² ) ā‰¤ (š‘¦, āˆ’x) and free disposability implies that (š‘¦, āˆ’xā€² ) āˆˆ š‘Œ . Therefore xā€² āˆˆ š‘‰ (š‘¦).

2. Again assume x āˆˆ š‘‰ (š‘¦). This implies that (š‘¦, āˆ’x) āˆˆ š‘Œ . By free disposal, (š‘¦ ā€² , āˆ’x) āˆˆ š‘Œ for every š‘¦ ā€² ā‰¤ š‘¦, which implies that x āˆˆ š‘‰ (š‘¦ ā€² ). š‘‰ (š‘¦ ā€² ) āŠ‡ š‘‰ (š‘¦). 1.13 The domain of ā€œ š‘¦1 or š‘„1 = š‘¦1 and š‘„2 > š‘¦2 Since all elements š‘„ āˆˆ ā„œ2 are comparable, ā‰» is complete; it is a total order. 1.35 Assume ā‰æš‘– is complete for every š‘–. Then for every š‘„, š‘¦ āˆˆ š‘‹ and for all š‘– = 1, 2, . . . , š‘›, either š‘„š‘– ā‰æš‘– š‘¦š‘– or š‘¦š‘– ā‰æš‘– š‘„š‘– or both. Either š‘„š‘– āˆ¼š‘– š‘¦š‘– for all š‘– Then deļ¬ne š‘„ āˆ¼ š‘¦. š‘„š‘– āˆ•āˆ¼š‘– š‘¦š‘– for some š‘– Let š‘˜ be the ļ¬rst individual with a strict preference, that is š‘˜ = minš‘– (š‘„š‘– āˆ•āˆ¼ š‘¦š‘– ). (Completeness of ā‰æš‘– ensures that š‘˜ is deļ¬ned). Then deļ¬ne š‘„ ā‰» š‘¦ if š‘„š‘˜ ā‰»š‘– š‘¦š‘˜ š‘¦ ā‰» š‘„ otherwise 1.36 Let š‘†, š‘‡ and š‘ˆ be subsets of a ļ¬nite set š‘‹. Set inclusion āŠ† is reļ¬‚exive since š‘† āŠ† š‘†. transitive since š‘† āŠ† š‘‡ and š‘‡ āŠ† š‘ˆ implies š‘† āŠ† š‘ˆ . anti-symmetric since š‘† āŠ† š‘‡ and š‘‡ āŠ† š‘† implies š‘† = š‘‡ Therefore āŠ† is a partial order. 1.37 Assume š‘„ and š‘¦ are both least upper bounds of š“. That is š‘„ ā‰æ š‘Ž for all š‘Ž āˆˆ š“ and š‘¦ ā‰æ š‘Ž for all š‘Ž āˆˆ š“. Further, if š‘„ is a least upper bound, š‘¦ ā‰æ š‘„. If š‘¦ is a least upper bound, š‘„ ā‰æ š‘¦. By anti-symmetry, š‘„ = š‘¦. 1.38 š‘„ āˆ¼ š‘¦ =ā‡’ š‘„ ā‰æ š‘¦ and š‘¦ ā‰æ š‘„ which implies that š‘„ = š‘¦ by antisymmetry. Each equivalence class āˆ¼ (š‘„) = { š‘¦ āˆˆ š‘‹ : š‘¦ āˆ¼ š‘„ } comprises just a single element š‘„. 1.39 max š’«(š‘‹) = š‘‹ and min š’«(š‘‹) = āˆ…. 1.40 The subset {2, 4, 8} forms a chain. More generally, the set of integer powers of a given number { š‘›, š‘›2 , š‘›3 , . . . } forms a chain. 1.41 Assume š‘„ and š‘¦ are maximal elements of the chain š“. Then š‘„ ā‰æ š‘Ž for all š‘Ž āˆˆ š“ and in particular š‘„ ā‰æ š‘¦. Similarly, š‘¦ ā‰æ š‘Ž for all š‘Ž āˆˆ š“ and in particular š‘¦ ā‰æ š‘„. Since ā‰æ is anti-symmetric, š‘„ = š‘¦. 1.42

1. By assumption, for every š‘” āˆˆ š‘‡ āˆ– š‘Š , ā‰ŗ(š‘”) is a nonempty ļ¬nite chain. Hence, it has a unique maximal element, š‘(š‘”).

2. Let š‘” be any node. Either š‘” is an initial node or š‘” has a unique predecessor š‘(š‘”). Either š‘(š‘”) is an initial node, or it has a unique predecessor š‘(š‘(š‘”)). Continuing in this way, we trace out a unique path from š‘” back to an initial node. We can be sure of eventually reaching an initial node since š‘‡ is ļ¬nite. 1.43 (1, 2) āˆØ (3, 1) = (3, 2) and (1, 2) āˆ§ (3, 2) = (1, 2) 6

Solutions for Foundations of Mathematical Economics 1.44

c 2001 Michael Carter āƒ All rights reserved

1. š‘„āˆØš‘¦ is an upper bound for { š‘„, š‘¦ }, that is xāˆØy ā‰æ š‘„ and xāˆØy ā‰æ š‘¦. Similarly, š‘„ āˆØ š‘¦ is a lower bound for { š‘„, š‘¦ }.

2. Assume š‘„ ā‰æ š‘¦. Then š‘„ is an upper bound for { š‘„, š‘¦ }, that is š‘„ ā‰æ š‘„ āˆØ š‘¦. If š‘ is any upper bound for { š‘„, š‘¦ }, then š‘ ā‰æ š‘„. Therefore, š‘„ is the least upper bound for { š‘„, š‘¦ }. Similarly, š‘¦ is a lower bound for { š‘„, š‘¦ }, and is greater than any other lower bound. Conversely, assume š‘„ āˆØ š‘¦ = š‘„. Then š‘„ is an upper bound for { š‘„, š‘¦ }, that is š‘„ ā‰æ š‘¦. 3. Using the preceding equivalence š‘„ ā‰æ š‘„ āˆ§ š‘¦ =ā‡’ š‘„ āˆØ (š‘„ āˆ§ š‘¦) = š‘„ š‘„ āˆØ š‘¦ ā‰æ š‘„ =ā‡’ (š‘„ āˆØ š‘¦) āˆ§ š‘„ = š‘„ 1.45 A chain š‘‹ is a complete partially ordered set. For every š‘„, š‘¦ āˆˆ š‘‹ with š‘„ āˆ•= š‘¦, either š‘„ ā‰» š‘¦ or š‘¦ ā‰» š‘„. Therefore, deļ¬ne the meet and join by { š‘¦ if š‘„ ā‰» š‘¦ š‘„āˆ§š‘¦ = š‘„ if š‘¦ ā‰» š‘„ { š‘„ if š‘„ ā‰» š‘¦ š‘„āˆØš‘¦ = š‘¦ if š‘¦ ā‰» š‘„ š‘‹ is a lattice with these operations. 1.46 Assume š‘‹1 and š‘‹2 are lattices, and let š‘‹ = š‘‹1 Ɨ š‘‹2 . Consider any two elements x = (š‘„1 , š‘„2 ) and y = (š‘¦1 , š‘¦2 ) in š‘‹. Since š‘‹1 and š‘‹2 are lattices, š‘1 = š‘„1 āˆØ š‘¦1 āˆˆ š‘‹1 and š‘2 = š‘„2 āˆØ š‘¦2 āˆˆ š‘‹2 , so that b = (š‘1 , š‘2 ) = (š‘„1 āˆØ š‘¦1 , š‘„2 āˆØ š‘¦2 ) āˆˆ š‘‹. Furthermore b ā‰æ x and b ā‰æ y in the natural product order, so that b is an upper bound for the Ė† = (Ė†š‘1 , Ė†š‘2 ) of {x, y} must have š‘š‘– ā‰æš‘– š‘„š‘– and š‘š‘– ā‰æš‘– š‘¦š‘– , {x, y}. Every upper bound b Ė† ā‰æ b. Therefore, b is the least upper bound of {x, y}, that is b = x āˆØ y. so that b Similarly, x āˆ§ y = (š‘„1 āˆ§ š‘¦1 , š‘„2 āˆ§ š‘¦2 ). 1.47 Let š‘† be a subset of š‘‹ and let š‘† āˆ— = { š‘„ āˆˆ š‘‹ : š‘„ ā‰æ š‘  for every š‘  āˆˆ š‘† } be the set of upper bounds of š‘†. Then š‘„āˆ— āˆˆ š‘† āˆ— āˆ•= āˆ…. By assumption, š‘† āˆ— has a greatest lower bound š‘. Since every š‘  āˆˆ š‘† is a lower bound of š‘† āˆ— , š‘ ā‰æ š‘  for every š‘  āˆˆ š‘†. Therefore š‘ is an upper bound of š‘†. Furthermore, š‘ is the least upper bound of š‘†, since š‘ ā‰¾ š‘„ for every š‘„ āˆˆ š‘† āˆ— . This establishes that every subset of š‘‹ also has a least upper bound. In particular, every pair of elements has a least upper and a greatest lower bound. Consequently š‘‹ is a complete lattice. 1.48 Without loss of generality, we will prove the closed interval case. Let [š‘Ž, š‘] be an interval in a lattice šæ. Recall that š‘Ž = inf[š‘Ž, š‘] and š‘ = sup[š‘Ž, š‘]. Choose any š‘„, š‘¦ in [š‘Ž, š‘] āŠ† šæ. Since šæ is a lattice, š‘„ āˆØ š‘¦ āˆˆ šæ and š‘„ āˆØ š‘¦ = sup{ š‘„, š‘¦ } ā‰¾ š‘ Therefore š‘„ āˆØ š‘¦ āˆˆ [š‘Ž, š‘]. Similarly, š‘„ āˆ§ š‘¦ āˆˆ [š‘Ž, š‘]. [š‘Ž, š‘] is a lattice. Similarly, for any subset š‘† āŠ† [š‘Ž, š‘] āŠ† šæ, sup š‘† āˆˆ šæ if šæ is complete. Also, sup š‘† ā‰¾ š‘ = sup[š‘Ž, š‘]. Therefore sup š‘† āˆˆ [š‘Ž, š‘]. Similarly inf š‘† āˆˆ [š‘Ž, š‘] so that [š‘Ž, š‘] is complete.

7

Solutions for Foundations of Mathematical Economics 1.49

c 2001 Michael Carter āƒ All rights reserved

1. The strong set order ā‰æš‘† is antisymmetric Let š‘†1 , š‘†2 āŠ† š‘‹ with š‘†1 ā‰æš‘† š‘†2 and š‘†2 ā‰æš‘† š‘†1 . Choose š‘„1 āˆˆ š‘†1 and š‘„2 āˆˆ š‘†2 . Since š‘†1 ā‰æš‘† š‘†2 , š‘„1 āˆØ š‘„2 āˆˆ š‘†1 and š‘„1 āˆ§ š‘„2 āˆˆ š‘†2 . On the other hand, since š‘†2 ā‰æ š‘†1 , š‘„1 = (š‘„1 āˆØ (š‘„1 āˆ§ š‘„2 ) āˆˆ š‘†2 and š‘„2 = š‘„2 āˆ§ (š‘„1 āˆØ š‘„2 ) āˆˆ š‘†1 (Exercise 1.44. Therefore š‘†1 = š‘†2 and ā‰æš‘† is antisymmetric. transitive Let š‘†1 , š‘†2 , š‘†3 āŠ† š‘‹ with š‘†1 ā‰æš‘† š‘†2 and š‘†2 ā‰æš‘† š‘†3 . Choose š‘„1 āˆˆ š‘†1 , š‘„2 āˆˆ š‘†2 and š‘„3 āˆˆ š‘†3 . Since š‘†1 ā‰æš‘† š‘†2 and š‘†2 ā‰æš‘† š‘†3 , š‘„1 āˆØ š‘„2 and š‘„2 āˆ§ š‘„3 are in š‘†2 . Therefore š‘¦2 = š‘„1 āˆØ (š‘„2 āˆ§ š‘„3 ) āˆˆ š‘†2 which implies ) ( š‘„1 āˆØ š‘„3 = š‘„1 āˆØ (š‘„2 āˆ§ š‘„3 ) āˆØ š‘„3 ( ) = š‘„1 āˆØ (š‘„2 āˆ§ š‘„3 ) āˆØ š‘„3 = š‘¦2 āˆØ š‘„3 āˆˆ š‘†3 since š‘†2 ā‰æš‘† š‘†3 . Similarly š‘§2 = (š‘„1 āˆØ š‘„2 ) āˆ§ š‘„3 āˆˆ š‘†2 and ( ) š‘„1 āˆ§ š‘„3 = š‘„1 āˆ§ (š‘„1 āˆØ š‘„2 ) āˆ§ š‘„3 ) ( = š‘„1 āˆ§ (š‘„1 āˆØ š‘„2 ) āˆ§ š‘„3 = š‘„1 āˆ§ š‘§2 āˆˆ š‘†1 Therefore, š‘†1 ā‰æš‘† š‘†3 .

2. š‘† ā‰æš‘† š‘† if and only if, for every š‘„1 , š‘„2 āˆˆ š‘†, š‘„1 āˆØ š‘„2 āˆˆ š‘† and š‘„1 āˆ§ š‘„2 āˆˆ š‘†, which is the case if and only if š‘† is a sublattice. 3. Let šæ(š‘‹) denote the set of all sublattices of š‘‹. We have shown that ā‰æš‘† is reļ¬‚exive, transitive and antisymmetric on šæ(š‘‹). Hence, it is a partial order on šæ(š‘‹). 1.50 Assume š‘†1 ā‰æš‘† š‘†2 . For any š‘„1 āˆˆ š‘†1 and š‘„2 āˆˆ š‘†2 , š‘„1 āˆØ š‘„2 āˆˆ š‘†1 and š‘„1 āˆ§ š‘„2 āˆˆ š‘†2 . Therefore sup š‘†1 ā‰æ š‘„1 āˆØ š‘„2 ā‰æ š‘„2

for every š‘„2 āˆˆ š‘†2

which implies that sup š‘†1 ā‰æ sup š‘†2 . Similarly inf š‘†2 ā‰¾ š‘„1 āˆ§ š‘„2 ā‰¾ š‘„1

for every š‘„1 āˆˆ š‘†1

which implies that inf š‘†2 ā‰¾ inf š‘†1 . Note that completeness ensures the existence of sup š‘† and inf š‘† respectively. 1.51 An argument analogous to the preceding exercise establishes =ā‡’ . (Completeness is not required, since for any interval š‘Ž = inf[š‘Ž, š‘] and š‘ = sup[š‘Ž, š‘]). To establish the converse, assume that š‘†1 = [š‘Ž1 , š‘1 ] and š‘†2 = [š‘Ž2 , š‘2 ]. Consider any š‘„1 āˆˆ š‘†1 and š‘„2 āˆˆ š‘†2 . There are two cases. Case 1. š‘„1 ā‰æ š‘„2 Since š‘‹ is a chain, š‘„1 āˆØ š‘„2 = š‘„1 āˆˆ š‘†1 . š‘„1 āˆ§ š‘„2 = š‘„2 āˆˆ š‘†2 . Case 2. š‘„1 ā‰ŗ š‘„2 Since š‘‹ is a chain, š‘„1 āˆØ š‘„2 = š‘„2 . Now š‘Ž1 ā‰¾ š‘„1 ā‰ŗ š‘„2 ā‰¾ š‘2 ā‰¾ š‘2 . Therefore, š‘„2 = š‘„1 āˆØ š‘„2 āˆˆ š‘†1 . Similarly š‘Ž2 ā‰¾ š‘Ž1 ā‰¾ š‘„1 ā‰ŗ š‘„2 ā‰¾ š‘2 . Therefore š‘„1 āˆ§ š‘„2 = š‘„1 āˆˆ š‘†2 . We have shown that š‘†1 ā‰æš‘† š‘†2 in both cases. 1.52 Assume that ā‰æ is a complete relation on š‘‹. This means that for every š‘„, š‘¦ āˆˆ š‘‹, either š‘„ ā‰æ š‘¦ or š‘¦ ā‰æ š‘„. In particular, letting š‘„ = š‘¦, š‘„ ā‰æ š‘„ for š‘„ āˆˆ š‘‹. ā‰æ is reļ¬‚exive. 8

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.53 Anti-symmetry implies that each indiļ¬€erence class contains a single element. If the consumerā€™s preference relation was anti-symmetric, there would be no baskets of goods between which the consumer was indiļ¬€erent. Each indiļ¬€erence curve which consist a single point. 1.54 We previously showed (Exercise 1.27) that every best element is maximal. To prove the converse, assume that š‘„ is maximal in the weakly ordered set š‘‹. We have to show that š‘„ ā‰æ š‘¦ for all š‘¦ āˆˆ š‘‹. Assume otherwise, that is assume there is some š‘¦ āˆˆ š‘‹ for which š‘„ āˆ•ā‰æ š‘¦. Since ā‰æ is complete, this implies that š‘¦ ā‰» š‘„ which contradicts the assumption that š‘„ is maximal. Hence we conclude that š‘„ ā‰æ š‘¦ for š‘¦ āˆˆ š‘‹ and š‘„ is a best element. 1.55 False. A chain has at most one maximal element (Exercise 1.41). Here, uniqueness is ensured by anti-symmetry. A weakly ordered set in which the order is not antisymmetric may have multiple maximal and best elements. For example, š‘Ž and š‘ are both best elements in the weakly ordered set {š‘Ž āˆ¼ š‘ ā‰» š‘}. 1.56

1. For every š‘„ āˆˆ š‘‹, either š‘„ ā‰æ š‘¦ =ā‡’ š‘„ āˆˆ ā‰æ(š‘¦) or š‘¦ ā‰æ š‘„ =ā‡’ š‘„ āˆˆ ā‰¾(š‘¦) since ā‰æ is complete. Consequently, ā‰æ(š‘¦) āˆŖ ā‰ŗ(š‘¦) = š‘‹ If š‘„ āˆˆ ā‰æ(š‘¦) āˆ© ā‰¾(š‘¦), then š‘„ ā‰æ š‘¦ and š‘¦ ā‰æ š‘„ so that š‘„ āˆ¼ š‘¦ and š‘„ āˆˆ š¼š‘¦ .

2. For every š‘„ āˆˆ š‘‹, either š‘„ ā‰æ š‘¦ =ā‡’ š‘„ āˆˆ ā‰æ(š‘¦) or š‘¦ ā‰» š‘„ =ā‡’ š‘„ āˆˆ ā‰ŗ(š‘¦) since ā‰æ is complete. Consequently, ā‰æ(š‘¦) āˆŖ ā‰ŗ(š‘¦) = š‘‹ and ā‰æ(š‘¦) āˆ© ā‰ŗ(š‘¦) = āˆ…. 3. For every š‘¦ āˆˆ š‘‹, ā‰»(š‘¦) and š¼š‘¦ partition ā‰æ(š‘¦) and therefore ā‰»(š‘¦), š¼š‘¦ and ā‰ŗ(š‘¦) partition š‘‹. 1.57 Assume š‘„ ā‰æ š‘¦ and š‘§ āˆˆ ā‰æ(š‘„). Then š‘§ ā‰æ š‘„ ā‰æ š‘¦ by transitivity. Therefore š‘§ āˆˆ ā‰æ(š‘¦). This shows that ā‰æ(š‘„) āŠ† ā‰æ(š‘¦). Similarly, assume š‘„ ā‰» š‘¦ and š‘§ āˆˆ ā‰»(š‘„). Then š‘§ ā‰» š‘„ ā‰» š‘¦ by transitivity. Therefore š‘§ āˆˆ ā‰»(š‘¦). This shows that ā‰æ(š‘„) āŠ† ā‰æ(š‘¦). To show that ā‰æ(š‘„) āˆ•= ā‰æ(š‘¦), observe that š‘„ āˆˆ ā‰»(š‘¦) but that š‘„ āˆˆ / ā‰»(š‘„) 1.58 Every ļ¬nite ordered set has a least one maximal element (Exercise 1.28). 1.59 Kreps (1990, p.323), Luenberger (1995, p.170) and Mas-Colell et al. (1995, p.313) adopt the weak Pareto order, whereas Varian (1992, p.323) distinguishes the two orders. Osborne and Rubinstein (1994, p.7) also distinguish the two orders, utilizing the weak order in deļ¬ning the core (Chapter 13) but the strong Pareto order in the Nash bargaining solution (Chapter 15). 1.60 Assume that a group š‘† is decisive over š‘„, š‘¦ āˆˆ š‘‹. Let š‘Ž, š‘ āˆˆ š‘‹ be two other states. We have to show that š‘† is decisive over š‘Ž and š‘. Without loss of generality, assume for all individuals š‘Ž ā‰æš‘– š‘„ and š‘¦ ā‰æš‘– š‘. Then, the Pareto order implies that š‘Ž ā‰» š‘„ and š‘¦ ā‰» š‘. Assume that for every š‘– āˆˆ š‘†, š‘„ ā‰æš‘– š‘¦. Since š‘† is decisive over š‘„ and š‘¦, the social order ranks š‘„ ā‰æ š‘¦. By transitivity, š‘Ž ā‰æ š‘. By IIA, this holds irrespective of individual preferences on other alternatives. Hence, š‘† is decisive over š‘Ž and š‘. 1.61 Assume that š‘† is decisive. Let š‘„, š‘¦ and š‘§ be any three alternatives and assume š‘„ ā‰æ š‘¦ for every š‘– āˆˆ š‘†. Partition š‘† into two subgroups š‘†1 and š‘†2 so that š‘„ ā‰æš‘– š‘§ for every š‘– āˆˆ š‘†1 and š‘§ ā‰æš‘– š‘¦ for every š‘– āˆˆ š‘†2 Since š‘† is decisive, š‘„ ā‰æ š‘¦. By completeness, either š‘„ ā‰æ š‘§ in which case š‘†1 is decisive over š‘„ and š‘§. By the ļ¬eld expansion lemma (Exercise 1.60), š‘†1 is decisive. 9

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

š‘§ ā‰» š‘„ which implies š‘§ ā‰æ š‘¦. In this case, š‘†2 is decisive over š‘¦ and š‘§, and therefore (Exercise 1.60) decisive. 1.62 Assume ā‰» is a social order which is Pareto and satisļ¬es Independence of Irrelevant Alternatives. By the Pareto principle, the whole group is decisive over any pair of alternatives. By the previous exercise, some proper subgroup is decisive. Continuing in this way, we eventually arrive at a decisive subgroup of one individual. By the Field Expansion Lemma (Exercise 1.60), that individual is decisive over every pair of alternatives. That is, the individual is a dictator. 1.63 Assume š“ is decisive over š‘„ and š‘¦ and šµ is decisive over š‘¤ and š‘§. That is, assume š‘„ ā‰»š“ š‘¦ =ā‡’ š‘„ ā‰» š‘¦ š‘¤ ā‰»šµ š‘§ =ā‡’ š‘¤ ā‰» š‘§ Also assume š‘¦ ā‰æš‘– š‘¤

for every š‘–

š‘§ ā‰æš‘– š‘„

for every š‘–

This implies that š‘¦ ā‰æ š‘¤ and š‘§ ā‰æ š‘„ (Pareto principle). Combining these preferences, transitivity implies that š‘„ā‰»š‘¦ā‰æš‘¤ā‰»š‘§ which contradicts the assumption that š‘§ ā‰æ š‘„. Therefore, the implied social ordering is intransitive. 1.64 Assume š‘„ āˆˆ core. In particular this implies that there does not exist any š‘¦ āˆˆ š‘Š (š‘ ) such that š‘¦ ā‰» š‘„. Therefore š‘„ āˆˆ Pareto. 1.65 No state will accept a cost share which exceeds what it can achieve on its own, so that if š‘„ āˆˆ core then š‘„š“š‘ƒ ā‰¤ 1870 š‘„š‘‡ š‘ ā‰¤ 5330 š‘„š“š‘ƒ ā‰¤ 860 Similarly, the combined share of the two states AP and TN should not exceed 6990, which they could achieve by proceeding without KM, that is š‘„š“š‘ƒ + š‘„š‘‡ š‘ ā‰¤ 6990 Similarly š‘„š“š‘ƒ + š‘„š¾š‘€ ā‰¤ 1960 š‘„š‘‡ š‘ + š‘„š¾š‘€ ā‰¤ 5020 Finally, the sum of the shares should equal the total cost š‘„š“š‘ƒ + š‘„š‘‡ š‘ + š‘„š¾š‘€ = 6530 The core is the set of all allocations of the total cost which satisfy the preceding inequalities. For example, the allocation (š‘„š“š‘ƒ = 1500, š‘„š‘‡ š‘ = 5000, š‘„š¾š‘€ = 30) does not belong to the core, since TN and KM will object to their combined share of 5030; since they can meet their needs jointly at a total cost of 5020. One the other hand, no group can object to the allocation (š‘„š“š‘ƒ = 1510, š‘„š‘‡ š‘ = 5000, š‘„š¾š‘€ = 20), which therefore belongs to the core. 10

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

1.66 The usual way to model a cost allocation problem as a TP-coalitional game is to regard the potential cost savings from cooperation as the sum to be allocated. In this example, the total joint cost of 6530 represents a potential saving of 1530 over the aggregate cost of 8060 if each region goes its own way. This potential saving of 1530 measures š‘¤(š‘ ). Similarly, undertaking a joint development, AP and TN could satisfy their combined requirements at a total cost of 6890. This compares with the standalone costs of 7100 (= 1870 (AP) + 5330 (TN)). Hence, the potential cost savings from their collaboration are 210 (= 7100 - 6890), which measures š‘¤(š“š‘ƒ, š‘‡ š‘ ). By similar calculations, we can compute the worth of each coalition, namely š‘¤(š“š‘ƒ ) = 0 š‘¤(š‘‡ š‘ ) = 0

š‘¤(š“š‘ƒ, š‘‡ š‘ ) = 210 š‘¤(š“š‘ƒ, š¾š‘€ ) = 770

š‘¤(š¾š‘€ ) = 0

š‘¤(š¾š‘€, š‘‡ š‘ ) = 1170

š‘¤(š‘ ) = 1530

An outcome in this game is an allocation of the total cost savings š‘¤(š‘ ) = 1530 amongst the three players. This can be translated into ļ¬nal cost shares by subtracting each players share of the cost savings from their standalone cost. For example, a speciļ¬c outcome in this game is (š‘„š“š‘ƒ = 370, š‘„š‘‡ š‘ = 330, š‘„š¾š‘€ = 830), which corresponds to ļ¬nal cost shares of 1500 for AP, 5000 for TN and 30 for KM. 1.67 Let š¶ = {x āˆˆ š‘‹ :

āˆ‘

š‘„š‘– ā‰„ š‘¤(š‘†) for every š‘† āŠ† š‘ }

š‘–āˆˆš‘†

/ core. This implies there exists some 1. š¶ āŠ† core Assume that x āˆˆ š¶. Suppose x āˆˆ coalition š‘† and outcome y āˆˆ š‘¤(š‘†) such that y ā‰»š‘– x for every š‘– āˆˆ š‘†. āˆ‘ āˆ™ y āˆˆ š‘¤(š‘†) implies š‘–āˆˆš‘† š‘¦š‘– ā‰¤ š‘¤(š‘†) while āˆ™ y ā‰»š‘– x for every š‘– āˆˆ š‘† implies š‘¦š‘– > š‘„š‘– for every š‘– āˆˆ š‘†. Summing, this implies āˆ‘ āˆ‘ š‘¦š‘– > š‘„š‘– ā‰„ š‘¤(š‘†) š‘–āˆˆš‘†

š‘–āˆˆš‘†

This contradiction establishes that x āˆˆ core. 2. core āŠ† š¶ Assume that x āˆˆ core. Suppose x āˆˆ / š¶. This implies there exists some āˆ‘ āˆ‘ coalition š‘† such that š‘–āˆˆš‘† š‘„š‘– < š‘¤(š‘†). Let š‘‘ = š‘¤(š‘†) āˆ’ š‘–āˆˆš‘† š‘„š‘– and consider the allocation y obtained by reallocating š‘‘ from š‘† š‘ to š‘†, that is { š‘„š‘– + š‘‘/š‘  š‘–āˆˆš‘† š‘¦š‘– = š‘„š‘– āˆ’ š‘‘/(š‘› āˆ’ š‘ ) š‘– āˆˆ /š‘† where š‘  = āˆ£š‘†āˆ£ is the number of players in š‘† and š‘› = āˆ£š‘ āˆ£ is the number in š‘ . Then āˆ‘ š‘¦š‘– > š‘„š‘– forāˆ‘ every š‘– āˆˆ š‘† so that y ā‰»š‘– x for every š‘– āˆˆ š‘†. Further, y āˆˆ š‘¤(š‘†) since š‘–āˆˆš‘† š‘¦š‘– = š‘–āˆˆš‘† š‘„š‘– + š‘‘ = š‘¤(š‘†) and y āˆˆ š‘‹ since āˆ‘ š‘–āˆˆš‘

š‘¦š‘– =

āˆ‘ š‘–āˆˆš‘†

(š‘„š‘– + š‘‘/š‘ ) +

āˆ‘

(š‘„š‘– āˆ’ š‘‘/(š‘› āˆ’ š‘ )) =

š‘–āˆˆš‘† /

āˆ‘

š‘„š‘– = š‘¤(š‘ )

š‘–āˆˆš‘

This contradicts our assumption that x āˆˆ / core, establishing that x āˆˆ š¶.

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.68 The 7 unanimity games for the player set š‘ = {1, 2, 3} are { 1 S = {1}, {1,2}, {1,3}, N š‘¢{1} (š‘†) = 0 otherwise { 1 S = {2}, {1,2}, {2,3}, N š‘¢{2} (š‘†) = 0 otherwise { 1 S = {3}, {1,3}, {2,3}, N š‘¢{3} (š‘†) = 0 otherwise { 1 S = {1,2}, N š‘¢{1,2} (š‘†) = 0 otherwise { 1 S = {1,3}, N š‘¢{1,3} (š‘†) = 0 otherwise { 1 S = {2,3}, N š‘¢{2,3} (š‘†) = 0 otherwise { 1 S=N š‘¢š‘ (š‘†) = 0 otherwise 1.69 Firstly, consider a simple game which is a unanimity game with essential coalition š‘‡ and let š‘„ be an outcome in which š‘„š‘– ā‰„ 0

for every š‘– āˆˆ š‘‡

š‘„š‘– = 0

for every š‘– āˆˆ /š‘‡

and āˆ‘

š‘„š‘– = 1

š‘–āˆˆš‘

We claim that š‘„ āˆˆ core. Winning coalitions If š‘† is winning coalition, then š‘¤(š‘†) = 1. Furthermore, if it is a winning coalition, it must contain š‘‡ , that is š‘‡ āŠ† š‘† and āˆ‘ āˆ‘ š‘„š‘– ā‰„ š‘„š‘– = 1 = š‘¤(š‘†) š‘–āˆˆš‘†

š‘–āˆˆš‘‡

Losing coalitions If š‘† is a losing coalition, š‘¤(š‘†) = 0 and āˆ‘ š‘„š‘– ā‰„ 0 = š‘¤(š‘†) š‘–āˆˆš‘†

Therefore š‘„ āˆˆ core and so core āˆ•= āˆ…. Conversely, consider a simple game which is not a unanimity game. Suppose there exists an outcome š‘„ āˆˆ core. Then āˆ‘ š‘„š‘– š‘¤(š‘ ) = 1 (1.15) š‘–āˆˆš‘

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Since there are no veto players (š‘‡ = āˆ…), š‘¤(š‘ āˆ– {š‘–}) = 1 for every player š‘– āˆˆ š‘ and āˆ‘ š‘„š‘— ā‰„ š‘¤(š‘ āˆ– {š‘–}) = 1 š‘—āˆ•=š‘–

which implies that š‘„š‘– = 0 for every š‘– āˆˆ š‘ contradicting (1.15). Thus we conclude that core = āˆ…. 1.70 The excesses of the proper coalitions at x1 and x2 are x1 -180 -955 -395 -365 -365 -180

{AP} {KM} {TN} {AP, KM} {AP, TN} {KM, TN}

x2 -200 -950 -380 -380 -370 -160

Therefore š‘‘(x1 ) = (āˆ’180, āˆ’180, āˆ’365, āˆ’365, āˆ’395, āˆ’955) and š‘‘(x2 ) = (āˆ’160, āˆ’200, āˆ’370, āˆ’380, āˆ’380, āˆ’950) d(x1 ) ā‰ŗšæ d(x2 ) which implies x1 ā‰»š‘‘ x2 . 1.71 It is a weak order on š‘‹, that is ā‰æ is reļ¬‚exive, transitive and complete. Reļ¬‚exivity š‘› and transitivity ļ¬‚ow from the corresponding properties of ā‰æšæ on ā„œ2 . Similarly, for š‘› any x, y āˆˆ š‘‹, either d(x) ā‰¾šæ d(y) or d(y) ā‰¾šæ d(x) since ā‰æšæ is complete on ā„œ2 . Consequently either x ā‰æ y or y ā‰æ x (or both). ā‰æ is not a partial order since it is not antisymmetric d(x) ā‰¾šæ d(y) and d(y) ā‰¾šæ d(x) does not imply x = y 1.72 š‘‘(š‘†, x) = š‘¤(š‘†) āˆ’

āˆ‘

š‘„š‘–

š‘–āˆˆš‘†

so that š‘‘(š‘†, x) ā‰¤ 0 ā‡ā‡’

āˆ‘

š‘„š‘– ā‰„ š‘¤(š‘†)

š‘–āˆˆš‘†

1.73 Assume to the contrary that x āˆˆ Nu but that x āˆˆ / core. Then, there exists a coalition š‘‡ with a positive deļ¬cit š‘‘(š‘‡, x) > 0. Since core āˆ•= āˆ…, there exists some y āˆˆ š‘‹ such that š‘‘(š‘†, y) ā‰¤ 0 for every š‘† āŠ† Nu. Consequently, d(y) ā‰ŗ d(x) and y ā‰» x, so that x āˆˆ / Nu. This contradiction establishes that Nu āŠ† core. 1.74 For player 1, š“1 = {š¶, š‘ } and (š¶, š¶) ā‰æ1 (š¶, š¶) (š¶, š¶) ā‰æ1 (š‘, š¶) Similarly for player 2 (š¶, š¶) ā‰æ2 (š¶, š¶) (š¶, š¶) ā‰æ2 (š¶, š‘ ) Therefore, (š¶, š¶) satisļ¬es the requirements of the deļ¬nition of a Nash equilibrium (Example 1.51). 13

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.75 If aāˆ—š‘– is the best element in (š“š‘– , ā‰æā€²š‘– ) for every player š‘–, then (š‘Žāˆ—š‘– , aāˆ’š‘– ) ā‰»š‘– (š‘Žš‘– , aāˆ’š‘– ) for every š‘Žš‘– āˆˆ š“š‘– and aāˆ’š‘– āˆˆ š“āˆ’š‘– for every š‘– āˆˆ š‘ . Therefore, aāˆ— is a Nash equilibrium. ĀÆ is another Nash equilibrium. Then for every To show that it is unique, assume that a player š‘– āˆˆ š‘ ĀÆāˆ’š‘– ) ā‰æš‘– (š‘Žš‘– , a ĀÆāˆ’š‘– ) for every š‘Žš‘– āˆˆ š“š‘– (ĀÆ š‘Žš‘– , a ĀÆ is a maximal element of ā‰æā€²š‘– . To see this, assume not. That is, which implies that a assume that there exists some š‘ŽĖœš‘– āˆˆ š“š‘– such that š‘ŽĖœš‘– ā‰»ā€²š‘– š‘Ž ĀÆš‘– which implies š‘Žš‘– , aāˆ’š‘– ) for every aāˆ’š‘– āˆˆ š“āˆ’š‘– (Ėœ š‘Žš‘– , aāˆ’š‘– ) ā‰»š‘– (ĀÆ In particular ĀÆāˆ’š‘– ) ā‰»š‘– (š‘Žāˆ—š‘– , aāˆ—āˆ’š‘– ) (Ėœ š‘Žš‘– , a ĀÆ is anwhich contradicts the assumption that aāˆ— is a Nash equilibrium. Therefore, a other Nash equilibrium, then š‘Ž ĀÆš‘– is maximal in ā‰æā€²š‘– and hence also a best element of ā‰æā€²š‘– (Exercise 1.54), which contradicts the assumption that š‘Žāˆ—š‘– is the unique best element. Consequently, we conclude that aāˆ— is the unique Nash equilibrium of the game. 1.76 We show that šœŒ(š‘„, š‘¦) = āˆ£š‘„ āˆ’ š‘¦āˆ£ satisļ¬es the requirements of a metric, namely 1. āˆ£š‘„ āˆ’ š‘¦āˆ£ ā‰„ 0. 2. āˆ£š‘„ āˆ’ š‘¦āˆ£ = 0 if and only if š‘„ = š‘¦. 3. āˆ£š‘„ āˆ’ š‘¦āˆ£ = āˆ£š‘¦ āˆ’ š‘„āˆ£. To establish the triangle inequality, we can consider various cases. For example, if š‘„ā‰¤š‘¦ā‰¤š‘§ āˆ£š‘„ āˆ’ š‘§āˆ£ + āˆ£š‘§ āˆ’ š‘¦āˆ£ ā‰„ āˆ£š‘„ āˆ’ š‘§āˆ£ = š‘§ āˆ’ š‘„ ā‰„ š‘¦ āˆ’ š‘„ = āˆ£š‘„ āˆ’ š‘¦āˆ£ If š‘„ ā‰¤ š‘§ ā‰¤ š‘¦ āˆ£š‘„ āˆ’ š‘§āˆ£ + āˆ£š‘§ āˆ’ š‘¦āˆ£ = š‘§ āˆ’ š‘„ + š‘¦ āˆ’ š‘§ = š‘¦ āˆ’ š‘„ = āˆ£š‘„ āˆ’ š‘¦āˆ£ and so on. 1.77 We show that šœŒāˆž š‘„, š‘¦ = maxš‘›š‘–=1 āˆ£š‘„š‘– āˆ’ š‘¦š‘– āˆ£ satisļ¬es the requirements of a metric, namely 1. maxš‘›š‘–=1 āˆ£š‘„š‘– āˆ’ š‘¦š‘– āˆ£ ā‰„ 0 2. maxš‘›š‘–=1 āˆ£š‘„š‘– āˆ’ š‘¦š‘– āˆ£ = 0 if and only if š‘„š‘– = š‘¦š‘– for all š‘–. 3. maxš‘›š‘–=1 āˆ£š‘„š‘– āˆ’ š‘¦š‘– āˆ£ = maxš‘›š‘–=1 āˆ£š‘¦š‘– āˆ’ š‘„š‘– āˆ£ 4. For every š‘–, āˆ£š‘„š‘– āˆ’ š‘¦š‘– āˆ£ ā‰¤ āˆ£š‘„š‘– āˆ’ š‘§š‘– āˆ£ + āˆ£š‘§š‘– āˆ’ š‘¦š‘– āˆ£ from previous exercise. Therefore max āˆ£š‘„š‘– āˆ’ š‘¦š‘– āˆ£ ā‰¤ max (āˆ£š‘„š‘– āˆ’ š‘§š‘– āˆ£ + āˆ£š‘§š‘– āˆ’ š‘¦š‘– āˆ£) ā‰¤ max āˆ£š‘„š‘– āˆ’ š‘§š‘– āˆ£ + max āˆ£š‘§š‘– āˆ’ š‘¦š‘– āˆ£ 1.78 For any š‘›, any neighborhood of 1/š‘› contains points of š‘† (namely 1/š‘›) and points not in š‘† (1/š‘› + šœ–). Hence every point in š‘† is a boundary point. Also, 0 is a boundary point. Therefore b(š‘†) = š‘† āˆŖ {0}. Note that š‘† āŠ‚ b(š‘†). Therefore, š‘† has no interior points. 14

Solutions for Foundations of Mathematical Economics 1.79

c 2001 Michael Carter āƒ All rights reserved

1. Let š‘„ āˆˆ int š‘†. Thus š‘† is a neighborhood of š‘„. Therefore, š‘‡ āŠ‡ š‘† is a neighborhood of š‘„, so that š‘„ is an interior point of š‘‡ .

2. Clearly, if š‘„ āˆˆ š‘†, then š‘„ āˆˆ š‘‡ āŠ† š‘‡ . Therefore, assume š‘„ āˆˆ š‘† āˆ– š‘† which implies that š‘„ is a boundary point of š‘†. Every neighborhood of š‘„ contains other points of š‘† āŠ† š‘‡ . Hence š‘„ āˆˆ š‘‡ . 1.80 Assume that š‘† is open. Every š‘„ āˆˆ š‘† has a neighborhood which is disjoint from š‘† š‘ . Hence no š‘„ āˆˆ š‘† is a closure point of š‘† š‘ . š‘† š‘ contains all its closure points and is therefore closed. Conversely, assume that š‘† is closed. Let š‘„ be a point its complement š‘† š‘ . Since š‘† is closed and š‘„ āˆˆ / š‘†, š‘„ is not a boundary point of š‘†. This implies that š‘„ has a neighborhood š‘ which is disjoint from š‘†, that is š‘ āŠ† š‘† š‘ . Hence, š‘„ is an interior point of š‘† š‘ . This implies that š‘† š‘ contains only interior points, and hence is open. 1.81 Clearly š‘„ is a neighborhood of every point š‘„ āˆˆ š‘‹, since šµš‘Ÿ (š‘„) āŠ† š‘‹ for every š‘Ÿ > 0. Hence, every point š‘„ āˆˆ š‘‹ is an interior point of š‘„. Similarly, every point š‘„ āˆˆ āˆ… is an interior point (there are none). Since š‘„ and āˆ… are open, there complements āˆ… and š‘„ are closed. Alternatively, āˆ… has no boundary points, and is therefore is open. Trivialy, on the other hand, āˆ… contains all its boundary points, and is therefore closed. 1.82 Let š‘‹ be a metric space. Assume š‘‹ is the union of two disjoint closed sets š“ and šµ, that is š‘‹ =š“āˆŖšµ

š“āˆ©šµ =āˆ…

Then š“ = šµ š‘ is open as is šµ = š“š‘ . Therefore š‘‹ is not connected. Conversely, assume that š‘‹ is not connected. Then there exist disjoint open sets š“ and šµ such that š‘‹ = š“ āˆŖ šµ. But š“ = šµ š‘ is also closed as is šµ = š“š‘ . Therefore š‘‹ is the union of two disjoint closed sets. 1.83 Assume š‘† is both open and closed, āˆ… āŠ‚ š‘† āŠ‚ š‘‹. We show that we can represent š‘‹ as the union of two disjoint open sets, š‘† and š‘† š‘ . For any š‘† āŠ‚ š‘‹, š‘‹ = š‘† āˆŖ š‘† š‘ and š‘† āˆ© š‘† š‘ = āˆ…. š‘† is open by assumption. It complement š‘† š‘ is open since š‘† is closed. Therefore, š‘‹ is not connected. Conversely, assume that š‘† is not connected. That is, there exists two disjoint open sets š‘† and š‘‡ such that š‘‹ = š‘† āˆŖ š‘‡ . Now š‘† = š‘‡ š‘ , which implies that š‘† is closed since š‘‡ is open. Therefore š‘† is both open and closed. 1.84 Assume that š‘† is both open and closed. Then so is š‘† š‘ and š‘‹ is the disjoint union of two closed sets š‘„ = š‘† āˆŖ š‘†š‘ so that b(š‘†) = š‘† āˆ© š‘† š‘ = š‘† āˆ© š‘† š‘ = āˆ… Conversely, assume that b(š‘†) = š‘† āˆ© š‘† š‘ = āˆ…. This implies that Consider any š‘„ āˆˆ š‘†. Since š‘† āˆ© š‘† š‘ = āˆ…, š‘„ āˆˆ / š‘† š‘ . A fortiori, x āˆˆ / š‘† š‘ which implies that š‘„ āˆˆ š‘† and therefore š‘† āŠ† š‘†. š‘† is closed. Similarly we can show that š‘† š‘ āŠ† š‘† š‘ so that š‘† š‘ is closed and therefore š‘† is open. š‘† is both open and closed.

15

Solutions for Foundations of Mathematical Economics 1.85

c 2001 Michael Carter āƒ All rights reserved

1. Let {šŗš‘– } be a (possibly inļ¬nite) collection of open sets. Let šŗ = āˆŖš‘– šŗš‘– . Let š‘„ be a point in šŗ. Then there exists some particular šŗš‘— which contains š‘„. Since šŗš‘— is open, šŗš‘— is a neighborhood of š‘„. Since šŗš‘— āŠ† šŗ, š‘„ is an interior point of šŗ. Since š‘„ is an arbitrary point in šŗ, we have shown that every š‘„ āˆˆ šŗ is an interior point. Hence, šŗ is open. What happens if every šŗš‘– is empty? In this case, šŗ = āˆ… and is open (Exercise 1.81). The other possibility is that the collection {šŗš‘– } is empty. Again šŗ = āˆ… which is open. Suppose { šŗ1 , šŗ2 , . . . , šŗš‘› } is a ļ¬nite collection of open sets. Let šŗ = āˆ©š‘– šŗš‘– . If šŗ = āˆ…, then it is trivially open. Otherwise, let š‘„ be a point in šŗ. Then š‘„ āˆˆ šŗš‘– for all š‘– = 1, 2, . . . , š‘›. Since the sets šŗš‘– are open, for every š‘–, there exists an open ball šµ(š‘„, š‘Ÿš‘– ) āŠ† šŗš‘– about š‘„. Let š‘Ÿ be the smallest radius of these open balls, that is š‘Ÿ = min{ š‘Ÿ1 , š‘Ÿ2 , . . . , š‘Ÿš‘› }. Then šµš‘Ÿ (š‘„) āŠ† šµ(š‘„, š‘Ÿš‘– ), so that šµš‘Ÿ (š‘„) āŠ† šŗš‘– for all i. Hence šµš‘Ÿ (š‘„) āŠ† šŗ. š‘„ is an interior point of šŗ and šŗ is open. To complete the proof, we need to deal with the trivial case in which the collection is empty. In that case, šŗ = āˆ©š‘– šŗš‘– = š‘‹ and hence is open.

2. The corresponding properties of closed sets are established analogously. 1.86

1. Let š‘„0 be an interior point of š‘†. This implies there exists an open ball šµ āŠ† š‘† about š‘„0 . Every š‘„ āˆˆ šµ is an interior point of š‘†. Hence šµ āŠ† int š‘†. š‘„0 is an interior point of int š‘† which is therefore open. Let šŗ be any open subset of š‘† and š‘„ be a point in šŗ. šŗ is neighborhood of š‘„, which implies that š‘† āŠ‡ šŗ is also neighborhood of š‘„. Therefore š‘„ is an interior point of š‘†. Therefore int š‘† contains every open subset šŗ āŠ† š‘†, and hence is the largest open set in š‘†.

2. Let š‘† denote the closure of the set š‘†. Clearly, š‘† āŠ† š‘†. To show the converse, let š‘„ be a closure point of š‘† and let š‘ be a neighborhood of š‘„. Then š‘ contains some other point š‘„ā€² āˆ•= š‘‹ which is a closure point of š‘†. š‘ is a neighborhood of š‘„ā€² which intersects š‘†. Hence š‘„ is a closure point of š‘†. Consequently š‘† = š‘† which implies that š‘† is closed. Assume š¹ is a closed subset of containing š‘†. Then š‘†āŠ†š¹ =š¹ since š¹ is closed. Hence, š‘† is a subset of every closed set containing š‘†. 1.87 Every š‘„ āˆˆ š‘† is either an interior point or a boundary point. Consequently, the interior of š‘† is the set of all š‘„ āˆˆ š‘† which are not boundary points int š‘† = š‘† āˆ– b(š‘†) 1.88 Assume that š‘† is closed, that is š‘† = š‘† āˆŖ b(š‘†) = š‘† This implies that b(š‘†) āŠ† š‘†. š‘† contains its boundary. Assume that š‘† contains its boundary, that is š‘† āŠ‡ b(š‘†). Then š‘† = š‘† āˆŖ b(š‘†) = š‘† š‘† is closed. 16

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.89 Assume š‘† is bounded, and let š‘‘ = š‘‘(š‘†). Choose any š‘„ āˆˆ š‘†. For all š‘¦ āˆˆ š‘†, šœŒ(š‘„, š‘¦) ā‰¤ š‘‘ < š‘‘ + 1. Therefore, š‘¦ āˆˆ šµ(š‘„, š‘‘ + 1). š‘† is contained in the open ball šµ(š‘„, š‘‘ + 1). Conversely, assume š‘† is contained in the open ball šµš‘Ÿ (š‘„). Then for any š‘¦, š‘§ āˆˆ š‘† šœŒ(š‘¦, š‘§) ā‰¤ šœŒ(š‘¦, š‘„) + šœŒ(š‘„, š‘§) < 2š‘Ÿ by the triangle inequality. Therefore š‘‘(š‘†) < 2š‘Ÿ and the set is bounded. 1.90 Let š‘¦ āˆˆ š‘† āˆ© šµš‘Ÿ (š‘„0 ). For every š‘„ āˆˆ š‘†, šœŒ(š‘„, š‘¦) < š‘Ÿ and therefore šœŒ(š‘„, š‘„0 ) ā‰¤ šœŒ(š‘„, š‘¦) + šœŒ(š‘¦, š‘„0 ) < š‘Ÿ + š‘Ÿ = 2š‘Ÿ so that š‘„ āˆˆ šµ2š‘Ÿ (š‘„0 ). 1.91 Let y0 āˆˆ š‘Œ . For any š‘Ÿ > 0, let yā€² = y āˆ’ š‘Ÿ be the production plan which is š‘Ÿ units less in every commodity. Then, for any y āˆˆ šµš‘Ÿ (yā€² ) š‘¦š‘– āˆ’ š‘¦š‘–ā€² ā‰¤ šœŒāˆž (y, yā€² ) < š‘Ÿ

for every š‘–

and therefore y < y0 . Thus šµš‘Ÿ (yā€² ) āŠ‚ š‘Œ and so yā€² āˆˆ int š‘Œ āˆ•= āˆ…. 1.92 For any š‘„ āˆˆ š‘†1 šœŒš‘„ = šœŒ(š‘„, š‘†2 ) > 0 Similarly, for every š‘¦ āˆˆ š‘†2 šœŒš‘¦ = šœŒ(š‘¦, š‘†1 ) > 0 Let š‘‡1 =

āˆŖ

šµšœŒš‘„ /2 (š‘„)

š‘„āˆˆš‘†1

š‘‡2 =

āˆŖ

šµšœŒš‘¦ /2 (š‘„)

š‘¦āˆˆš‘†2

Then š‘‡1 and š‘‡2 are open sets containing š‘†1 and š‘†2 respectively. To show that š‘‡1 and š‘‡2 are disjoint, suppose to the contrary that š‘§ āˆˆ š‘‡1 āˆ© š‘‡2 . Then, there exist points š‘„ āˆˆ š‘†1 and š‘¦ āˆˆ š‘†2 such that šœŒ(š‘„, š‘§) < šœŒš‘„ /2,

šœŒ(š‘¦, š‘§) < šœŒš‘¦ /2

Without loss of generality, suppose that šœŒš‘„ ā‰¤ šœŒš‘¦ and therefore šœŒ(š‘„, š‘¦) ā‰¤ šœŒ(š‘„, š‘§) + šœŒ(š‘¦, š‘§) < šœŒš‘„ /2 + šœŒš‘¦ /2 ā‰¤ šœŒš‘¦ which contradicts the deļ¬nition of šœŒš‘¦ and shows that š‘‡1 āˆ© š‘‡2 = āˆ…. 1.93 By Exercise 1.92, there exist disjoint open sets š‘‡1 and š‘‡2 such that š‘†1 āŠ† š‘‡1 and š‘†2 āŠ† š‘‡2 . Since š‘†2 āŠ† š‘‡2 , š‘†2 āˆ© š‘‡2š‘ = āˆ…. š‘‡2š‘ is a closed set which contains š‘‡1 , and therefore š‘†2 āˆ© š‘‡1 = āˆ…. š‘‡ = š‘‡1 is the desired set. 1.94 See Figure 1.2.

17

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

š‘†

1

šµ1/2 ((2, 0)) 1

2

Figure 1.2: Open ball about (2, 0) relative to š‘‹ 1.95 Assume š‘† is connected. Suppose š‘† is not an interval. This implies that there exists numbers š‘„, š‘¦, š‘§ such that š‘„ < š‘¦ < š‘§ and š‘„, š‘§ āˆˆ š‘† while š‘¦ āˆˆ / š‘†. Then š‘† = (š‘† āˆ© (āˆ’āˆž, š‘¦)) āˆŖ (š‘† āˆ© (š‘¦, āˆž)) represents š‘† as the union of two disjoint open sets (relative to š‘†), contradicting the assumption that š‘† is connected. Conversely, assume that š‘† is an interval. Suppose that š‘† is not connected. That is, š‘† = š“ āˆŖ šµ where š“ and šµ are nonempty disjoint closed sets. Choose š‘„ āˆˆ š“ and š‘§ āˆˆ šµ. Since š“ and šµ are disjoint, š‘„ āˆ•= š‘§. Without loss of generality, we may assume š‘„ < š‘§. Since š‘† is an interval, [š‘„, š‘§] āŠ† š‘† = š“ āˆŖ šµ. Let š‘¦ = sup{ [š‘„, š‘§] āˆ© š‘† } Clearly š‘„ ā‰¤ š‘¦ ā‰¤ š‘§ so that š‘¦ āˆˆ š‘†. Now š‘¦ belongs to either š“ or šµ. Since š“ is closed in š‘†, [š‘„, š‘§] āˆ© š“ is closed and š‘¦ = sup{ [š‘„, š‘§] āˆ© š‘† } āˆˆ š“. This implies the š‘¦ < š‘§. Consequently, š‘¦ + šœ– āˆˆ šµ for every šœ– > 0 such that š‘¦ + šœ– ā‰¤ š‘§. Since šµ is closed, š‘¦ āˆˆ šµ. This implies that š‘¦ belongs to both š“ and šµ contradicting the assumption that š“ āˆ© šµ = āˆ…. We conclude that š‘† must be connected. 1.96 Assume š‘„š‘› ā†’ š‘„ and also š‘„š‘› ā†’ š‘¦. We have to show that š‘„ = š‘¦. Suppose not, that is suppose š‘„ āˆ•= š‘¦ (see Figure 1.3). Then šœŒ(š‘„, š‘¦) = š‘… > 0. Let š‘Ÿ = š‘…/3 > 0. Since š‘„š‘› ā†’ š‘„, there exists some š‘š‘„ such that š‘„š‘› āˆˆ šµš‘Ÿ (š‘„) for all š‘› ā‰„ š‘š‘„ . Since š‘„š‘› ā†’ š‘¦, there exists some š‘š‘¦ such that š‘„š‘› āˆˆ šµš‘Ÿ (š‘¦) for all š‘› ā‰„ š‘š‘¦ . But these statements are contradictory since šµš‘Ÿ (š‘„) āˆ© šµ(š‘¦, š‘Ÿ) = āˆ…. We conclude that the successive terms of a convergent sequence cannot get arbitrarily close to two distinct points, so that the limit a convergent sequence is unique. 1.97 Let (š‘„š‘› ) be a sequence which converges to š‘„. There exists some š‘ such that šœŒ(š‘„š‘› āˆ’ š‘„) < 1 for all š‘› ā‰„ š‘ . Let š‘… = max{ šœŒ(š‘„1 āˆ’ š‘„), šœŒ(š‘„2 āˆ’ š‘„), . . . , šœŒ(š‘„š‘ āˆ’1 āˆ’ š‘„), 1 } Then for all š‘›, šœŒ(š‘„š‘› āˆ’š‘„) ā‰¤ š‘…. That is every element š‘„š‘› in the sequence (š‘„š‘› ) belongs to šµ(š‘„, š‘… + 1), the open ball about š‘„ of radius š‘… + 1. Therefore the sequence is bounded. 18

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics šµ(š‘„, š‘Ÿ)

š‘…

šµ(š‘¦, š‘Ÿ)

š‘Ÿ

š‘Ÿ

š‘„

š‘¦

Figure 1.3: A convergent sequence cannot have two distinct limits 1.98 The share š‘ š‘› of the š‘›th guest is š‘ š‘› =

1š‘› 2

lim š‘ š‘› = 0 However, š‘ š‘› > 0 for all š‘›. There is no limit to the number of guests who will get a share of the cake, although the shares will get vanishingly small for large parties. 1.99 Suppose š‘„š‘› ā†’ š‘„. That is, there exists some š‘ such that šœŒ(š‘„š‘› , š‘„) < šœ–/2 for all š‘› ā‰„ š‘ . Then, for all š‘š, š‘› ā‰„ š‘ šœŒ(š‘„š‘š , š‘„š‘› ) ā‰¤ šœŒ(š‘„š‘š , š‘„) + šœŒ(š‘„, š‘„š‘› ) < šœ–/2 + šœ–/2 = šœ– 1.100 Let (š‘„š‘› ) be a Cauchy sequence. There exists some š‘ such that šœŒ(š‘„š‘› āˆ’ š‘„š‘ ) < 1 for all š‘› ā‰„ š‘ . Let š‘… = max{ šœŒ(š‘„1 āˆ’ š‘„š‘ ), šœŒ(š‘„2 āˆ’ š‘„š‘ ), . . . , šœŒ(š‘„š‘ āˆ’1 āˆ’ š‘„š‘ ), 1 } Every š‘„š‘› belongs to šµ(š‘„š‘ , š‘… + 1), the ball of radius š‘… + 1 centered on š‘„š‘ . 1.101 Let (š‘„š‘› ) be a bounded increasing sequence in ā„œ and let š‘† = { š‘„š‘› } be the set of elements of (š‘„š‘› ). Let š‘ be the least upper bound of š‘†. We show that š‘„š‘› ā†’ š‘. First observe that š‘„š‘› ā‰¤ š‘ for every š‘› (since š‘ is an upper bound). Since š‘ is the least upper bound, for every šœ– > 0 there exists some element š‘„š‘ such that š‘„š‘ > š‘ āˆ’ šœ–. Since (š‘„š‘› ) is increasing, we must have š‘ āˆ’ šœ– < š‘„š‘› ā‰¤ š‘ for every š‘› ā‰„ š‘ That is, for every šœ– > 0 there exists an š‘ such that šœŒ(š‘„š‘› , š‘„) < šœ– for every š‘› ā‰„ š‘ š‘„š‘› ā†’ š‘. 1.102 If š›½ > 1, the sequence š›½, š›½ 2 , š›½ 3 , . . . is unbounded. Otherwise, if š›½ ā‰¤ 1, š›½ š‘› ā‰¤ š›½ š‘›āˆ’1 and the sequence is decreasing and bounded by š›½ ā‰¤ 1. Therefore the sequence converges (Exercise 1.101). Let š‘„ = limš‘›ā†’āˆž . Then š›½ š‘›+1 = š›½š›½ š‘› and therefore š‘„ = lim š›½ š‘›+1 = š›½ lim š›½ š‘› = š›½š‘„ š‘›ā†’āˆž

š‘›ā†’āˆž

which can be satisļ¬ed if and only if 19

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics āˆ™ š›½ = 1, in which case š‘„ = lim 1š‘› = 1 āˆ™ š‘„ = 0 when 0 ā‰¤ š›½ < 1 Therefore š›½ š‘› ā†’ 0 ā‡ā‡’ š›½ < 1 1.103

1. For every š‘„ āˆˆ ā„œ (š‘„ āˆ’

Expanding

āˆš 2 2) ā‰„ 0

āˆš š‘„2 āˆ’ 2 2š‘„ + 2 ā‰„ 0

āˆš š‘„2 + 2 ā‰„ 2 2š‘„

Dividing by š‘„ āˆš 2 ā‰„2 2 š‘„

š‘„+ for every š‘„ > 0. Therefore 1 2

(

2 š‘„+ š‘„

) ā‰„

āˆš 2

2. Let (š‘„š‘› ) be the sequence deļ¬ned in Example 1.64. That is ) ( 1 2 š‘„š‘› = š‘„š‘›āˆ’1 + š‘›āˆ’1 2 š‘„ Starting from š‘„0 = 2, it is clear that š‘„š‘› ā‰„ 0 for all š‘›. Substituting in ( ) āˆš 1 2 š‘„+ ā‰„ 2 2 š‘„ 1 š‘„ = 2 š‘›

That is š‘„š‘› ā‰„

( š‘„š‘›āˆ’1 +

2 š‘„š‘›āˆ’1

) ā‰„

āˆš 2

āˆš 2 for every š‘›. Therefore for every š‘› ) ( 1 2 š‘„š‘› āˆ’ š‘„š‘›+1 = š‘„š‘› āˆ’ š‘„š‘› + š‘› 2 š‘„ ) ( 1 2 = š‘„š‘› āˆ’ š‘› 2 š‘„ ( ) 1 2 ā‰„ š‘„š‘› āˆ’ āˆš 2 2 āˆš š‘› =š‘„ āˆ’ 2 ā‰„0

āˆš This implies that š‘„š‘›+1 ā‰¤ š‘„š‘› . Consequently 2 ā‰¤ š‘„š‘› ā‰¤ 2 for every š‘›. (š‘„š‘› ) is a bounded monotone sequence. By Exercise 1.101, š‘„š‘› ā†’ š‘„. The limit š‘„ satisļ¬es the equation ( ) 1 2 š‘„= š‘„+ 2 š‘„ āˆš Solving, this implies š‘„2 = 2 or š‘„ = 2 as required. 20

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1.104 The following sequence approximates the square root of any positive number š‘Ž š‘„1 = š‘Ž 1( š‘Ž ) š‘„š‘›+1 = š‘„š‘› + š‘› 2 š‘„ 1.105 Let š‘„ āˆˆ š‘†. If š‘„ āˆˆ š‘†, then š‘„ is the limit of the sequence (š‘„, š‘„, š‘„, . . . ). If š‘„ āˆˆ / š‘†, then š‘„ is a boundary point of š‘†. For every š‘›, the ball šµ(š‘„, 1/š‘›) contains a point š‘„š‘› āˆˆ š‘†. From the sequence of open balls šµ(š‘„, 1/š‘›) for š‘› = 1, 2, 3, . . . , we can generate of a sequence of points š‘„š‘› which converges to š‘„. Conversely, assume that š‘„ is the limit of a sequence (š‘„š‘› ) of points in š‘†. Either š‘„ āˆˆ š‘† and therefore š‘„ āˆˆ š‘†. Or š‘„ āˆˆ / š‘†. Since š‘„š‘› ā†’ š‘„, every neighborhood of š‘„ contains points š‘› š‘„ of the sequence. Hence, š‘„ is a boundary point of š‘† and š‘„ āˆˆ š‘†. 1.106 š‘† is closed if and only if š‘† = š‘†. The result follows from Exercise 1.105. 1.107 Let š‘† be a closed subset of a complete metric space š‘‹. Let (š‘„š‘› ) be a Cauchy sequence in š‘†. Since š‘‹ is complete, š‘„š‘› ā†’ š‘„ āˆˆ š‘‹. Since š‘† is closed, š‘„ āˆˆ š‘† (Exercise 1.106). 1.108 Since š‘‘(š‘† š‘› ) ā†’ 0, š‘† cannot contain more than one point. Therefore, it suļ¬ƒces to show that š‘† is nonempty. Choose some š‘„š‘› from each š‘† š‘› . Since š‘‘(š‘† š‘› ) ā†’ 0, (š‘„š‘› ) is a Cauchy sequence. Since š‘‹ is complete, there exists some š‘„ āˆˆ š‘‹ such that š‘„š‘› ā†’ š‘„. Choose some š‘š. Since the sets are nested, the subsequence { š‘„š‘› : š‘› ā‰„ š‘š } āŠ† š‘† š‘š . Since š‘† š‘š is closed, š‘„ āˆˆ š‘† š‘š (Exercise 1.106). Since š‘„ āˆˆ š‘† š‘š for every š‘š š‘„āˆˆ

āˆž āˆ©

š‘†š‘š

š‘š=1

1.109 If player 1 picks closed balls whose radius decreases by at least half after each pair of moves, then { š‘† 1 , š‘† 3 , š‘† 5 , . . . } is a nested sequence of closed sets which has a nonempty intersection (Exercise 1.108). 1.110 Let (š‘„š‘› ) be a sequence in š‘† āŠ† š‘‡ with š‘† closed and š‘‡ compact. Since š‘‡ is compact, there exists a convergent subsequence š‘„š‘š ā†’ š‘„ āˆˆ š‘‡ . Since š‘† is closed, we must have š‘„ āˆˆ š‘† (Exercise 1.106). Therefore (š‘„š‘› ) contains a subsequence which converges in š‘†, so that š‘† is compact. 1.111 Let (š‘„š‘› ) be a Cauchy sequence in a metric space. For every šœ– > 0, there exists š‘ such that šœŒ(š‘„š‘š , š‘„š‘› ) < šœ–/2 for all š‘š, š‘› ā‰„ š‘ Trivially, if (š‘„š‘› ) converges, it has a convergent subsequence (the whole sequence). Conversely, assume that (š‘„š‘› ) has a subsequence (š‘„š‘š ) which converges to š‘„. That is, there exists some š‘€ such that šœŒ(š‘„š‘š , š‘„) < šœ–/2 for all š‘š ā‰„ š‘€ Therefore, by the triangle inequality šœŒ(š‘„š‘› , š‘„) ā‰¤ šœŒ(š‘„š‘› , š‘„š‘€ ) + šœŒ(š‘„š‘€ , š‘„) < šœ–/2 + šœ–/2 = šœ– for all š‘› ā‰„ max š‘€, š‘

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1.112 We proceed sequentially as follows. Choose any š‘„1 in š‘‹. If the open ball šµ(š‘„1 , š‘Ÿ) contains š‘‹, we are done. Otherwise, choose some š‘„2 āˆˆ / šµ(š‘„1 , š‘Ÿ) and consider the set āˆŖ2 šµ(š‘„š‘– , š‘Ÿ). If this set contains š‘‹, we are done. Otherwise, choose some š‘„3 āˆˆ / āˆŖš‘–=1 āˆŖ3 2 šµ(š‘„ , š‘Ÿ) and consider šµ(š‘„ , š‘Ÿ) š‘– š‘– š‘–=1 š‘–=1 The process must terminate with a ļ¬nite number of open balls. Otherwise, if the process could be continued indeļ¬nitely, we could construct an inļ¬nite sequence (š‘„1 , š‘„2 , š‘„3 , . . . ) which had no convergent subsequence. The would contradict the compactness of š‘‹. 1.113 Assume š‘‹ is compact. The previous exercise showed that š‘‹ is totally bounded. Further, since every sequence has a convergent subsequence, every Cauchy sequence converges (Exercise 1.111). Therefore š‘‹ is complete. Conversely, assume that š‘‹ is complete and totally bounded and let š‘†1 = { š‘„11 , š‘„21 , š‘„31 , . . . } be an inļ¬nite sequence of points in š‘‹. Since š‘‹ is totally bounded, it is covered by a ļ¬nite collection of open balls of radius 1/2. š‘†1 has a subsequence š‘†2 = { š‘„12 , š‘„22 , š‘„32 , . . . } all of whose points lie in one of the open balls. Similarly, š‘†2 has a subsequence š‘†3 = { š‘„13 , š‘„23 , š‘„33 , . . . } all of whose points lie in an open ball of radius 1/3. Continuing in this fashion, we construct a sequence of subsequences, each of which lies in a ball of smaller and smaller radius. Consequently, successive terms of the ā€œdiagonalā€ subsequence { š‘„11 , š‘„22 , š‘„33 , . . . } get closer and closer together. That is, š‘† is a Cauchy sequence. Since š‘‹ is complete, š‘† converges in š‘‹ and š‘†1 has a convergent subsequence š‘†. Hence, š‘‹ is compact. 1.114

1. Every big set š‘‡ āˆˆ ā„¬ has a least two distinct points. 0 for every š‘‡ āˆˆ ā„¬.

Hence š‘‘(š‘‡ ) >

2. Otherwise, there exists š‘› such that š‘‘(š‘‡ ) ā‰„ 1/š‘› for every š‘‡ āˆˆ ā„¬ and therefore š›æ = inf š‘‡ āˆˆā„¬ š‘‘(š‘‡ ) ā‰„ 1/š‘› > 0. 3. Choose a point š‘„š‘› in each š‘‡š‘› . Since š‘‹ is compact, the sequence (š‘„š‘› ) has a convergent subsequence (š‘„š‘š ) which converges to some point š‘„0 āˆˆ š‘‹. 4. The point š‘„0 belongs to at least one š‘†0 in the open cover š’ž. Since š‘†0 is open, there exists some open ball šµš‘Ÿ (š‘„0 ) āŠ† š‘†0 . 5. Consider the concentric ball šµš‘Ÿ/2 (š‘„0 ). Since (š‘„š‘š ) is a convergent subsequence, there exists some š‘€ such that š‘„š‘š āˆˆ šµš‘Ÿ/2 (š‘„) for every š‘š ā‰„ š‘€ . 6. Choose some š‘›0 ā‰„ min{ š‘€, 2/š‘Ÿ }. Then 1/š‘›0 < š‘Ÿ/2 and š‘‘(š‘‡š‘›0 ) < 1/š‘›0 < š‘Ÿ/2. š‘„š‘›0 āˆˆ š‘‡š‘›0 āˆ© šµš‘Ÿ/2 (š‘„) and therefore (Exercise 1.90) š‘‡š‘›0 āŠ† šµš‘Ÿ (š‘„) āŠ† š‘† 0 . This contradicts the assumption that š‘‡š‘› is a big set. Therefore, we conclude that š›æ > 0. 1.115

1. š‘‹ is totally bounded (Exercise 1.112). Therefore, for every š‘Ÿ > 0, there exists a ļ¬nite number of open balls šµš‘Ÿ (š‘„š‘› ) such that š‘‹=

š‘› āˆŖ

šµš‘Ÿ (š‘„š‘– )

š‘–=1

2. š‘‘(šµš‘Ÿ (š‘„š‘– )) = 2š‘Ÿ < š›æ. By deļ¬nition of the Lebesgue number, every šµš‘Ÿ (š‘„š‘– ) is contained in some š‘†š‘– āˆˆ š’ž. 3. The collection of open balls {šµš‘Ÿ (š‘„š‘– )} covers š‘‹. Therefore, fore every š‘„ āˆˆ š‘‹, there exists š‘– such that š‘„ āˆˆ šµš‘Ÿ (š‘„š‘– ) āŠ† š‘†š‘– 22

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Solutions for Foundations of Mathematical Economics Therefore, the ļ¬nite collection š‘†1 , š‘†2 , . . . , š‘†š‘› covers š‘‹. 1.116 For any family of subsets š’ž āˆ© āˆŖ š‘† = āˆ… ā‡ā‡’ š‘†š‘ = š‘‹ š‘†āˆˆš’ž

š‘†āˆˆš’ž

Suppose to the contrary that š’ž is a collection of closed sets with the ļ¬nite intersection āˆ© property, but that š‘†āˆˆš’ž š‘† = āˆ…. Then { š‘† š‘ : š‘† āˆˆ š’ž } is a open cover of š‘‹ which does not have a ļ¬nite subcover. Consequently š‘‹ cannot be compact. Conversely, assume every collection of closed sets with the ļ¬nite intersection property has a nonempty intersection. Let ā„¬ be an open cover of š‘‹. Let š’ž = { š‘† āŠ† š‘‹ : š‘†š‘ āˆˆ ā„¬ } That is

āˆŖ

š‘† š‘ = š‘‹ which implies

š‘†āˆˆš’ž

āˆ©

š‘†=āˆ…

š‘†āˆˆš’ž

Consequently, š’ž does not have the ļ¬nite intersection property. There exists a ļ¬nite subcollection { š‘†1 , š‘†2 , . . . , š‘†š‘› } such that š‘› āˆ©

š‘†š‘– = āˆ…

š‘–=1

which implies that š‘› āˆŖ

š‘†š‘–š‘ = š‘‹

š‘–=1

{ š‘†1š‘ , š‘†2š‘ , . . . , š‘†š‘›š‘ } is a ļ¬nite subcover of š‘‹. Thus, š‘‹ is compact. 1.117 Every ļ¬nite collection of nested (nonempty) sets has the ļ¬nite intersection property. By Exercise 1.116, the sequence has a non-empty intersection. (Note: every set š‘†š‘– is a subset of the compact set š‘†1 .) 1.118 (1) =ā‡’ (2) Exercises 1.114 and 1.115. (2) =ā‡’ (3) Exercise 1.116 (3) =ā‡’ (1) Let š‘‹ be a metric space in which every collection of closed subsets with the ļ¬nite intersection property has a ļ¬nite intersection. Let (š‘„š‘› ) be a sequence in š‘‹. For any š‘›, let š‘†š‘› be the tail of the sequence minus the ļ¬rst š‘› terms, that is š‘†š‘› = { š‘„š‘š : š‘š = š‘› + 1, š‘› + 2, . . . } The collection (š‘†š‘› ) has the ļ¬nite intersection property since, for any ļ¬nite set of integers { š‘›1 , š‘›2 , . . . , š‘›š‘˜ } š‘˜ āˆ©

š‘†š‘›š‘— āŠ† š‘†š¾ āˆ•= āˆ…

š‘—=1

where š¾ = max{ š‘›1 , š‘›2 , . . . , š‘›š‘˜ }. Therefore āˆž āˆ© š‘›=1

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āˆŖāˆž Choose any š‘„ āˆˆ š‘›=1 š‘†š‘› . That is, š‘„ āˆˆ š‘†š‘› for each š‘› = 1, 2, . . . . Thus, for every š‘Ÿ > 0 and š‘› = 1, 2, . . . , there exists some š‘„š‘› āˆˆ šµš‘Ÿ (š‘„) āˆ© š‘†š‘› We construct a subsequence as follows. For š‘˜ = 1, 2, . . . , let š‘„š‘˜ be the ļ¬rst term in š‘†š‘˜ which belongs to šµ1/š‘˜ (š‘„). Then, (š‘„š‘˜ ) is a subsequence of (š‘„š‘› ) which converges to š‘„. We conclude that every sequence has a convergent subsequence. 1.119 Assume (š‘„š‘› ) is a bounded sequence in ā„œ. Without loss of generality, we can assume that { š‘„š‘› } āŠ‚ [0, 1]. Divide š¼ 0 = [0, 1] into two sub-intervals [0, 1/2] and [1/2, 1]. At least one of the sub-intervals must contain an inļ¬nite number of terms of the sequence. Call this interval š¼ 1 . Continuing this process of subdivision, we obtain a nested sequence of intervals š¼0 āŠƒ š¼1 āŠƒ š¼2 āŠƒ . . . each of which contains an inļ¬nite number of terms of the sequence. Consequently, we can construct a subsequence (š‘„š‘š ) with š‘„š‘š āˆˆ š¼ š‘š . Furthermore, the intervals get smaller and smaller with š‘‘(š¼ š‘› ) ā†’ 0, so that (š‘„š‘š ) is a Cauchy sequence. Since ā„œ is complete, the subsequence (š‘„š‘š ) converges to š‘„ āˆˆ ā„œ. Note how we implicitly called on the Axiom of Choice (Remark 1.5) in choosing a subsequence from the nested sequence of intervals. 1.120 Let (š‘„š‘› ) be a Cauchy sequence in ā„œ. That is, for every šœ– > 0, there exists š‘ such that āˆ£š‘„š‘› āˆ’ š‘„š‘š āˆ£ < šœ– for all š‘š, š‘› ā‰„ š‘ . (š‘„š‘› ) is bounded (Exercise 1.100) and hence by the Bolzano-Weierstrass theorem, it has a convergent subsequence (š‘„š‘š ) with š‘„š‘š ā†’ š‘„ āˆˆ ā„œ. Choose š‘„š‘Ÿ from the convergent subsequence such that š‘Ÿ ā‰„ š‘ and āˆ£š‘„š‘Ÿ āˆ’ š‘„āˆ£ < šœ–/2. By the triangle inequality āˆ£š‘„š‘› āˆ’ š‘„āˆ£ ā‰¤ āˆ£š‘„š‘› āˆ’ š‘„š‘Ÿ āˆ£ + āˆ£š‘„š‘Ÿ āˆ’ š‘„āˆ£ < šœ–/2 + šœ–/2 = šœ– Hence the sequence (š‘„š‘› ) converges to š‘„ āˆˆ ā„œ. 1.121 Since š‘‹1 and š‘‹2 are linear spaces, x1 + y1 āˆˆ š‘‹1 and x2 + y2 āˆˆ š‘‹2 , so that (x1 + y1 , x2 + y2 ) āˆˆ š‘‹1 Ɨ š‘‹2 . Similarly (š›¼x1 , š›¼x2 ) āˆˆ š‘‹1 Ɨ š‘‹2 for every (x1 , x2 ) āˆˆ š‘‹1 Ɨ š‘‹2 . Hence, š‘‹ = š‘‹1 Ɨ š‘‹2 is closed under addition and scalar multiplication. With addition and scalar multiplication deļ¬ned component-wise, š‘‹ inherits the arithmetic properties (like associativity) of its constituent spaces. Verifying this would proceed identically as for ā„œš‘› . It is straightforward though tedious. The zero element in š‘‹ is 0 = (01 , 02 ) where 01 is the zero element in š‘‹1 and 02 is the zero element in š‘‹2 . Similarly, the inverse of x = (x1 , x2 ) is āˆ’x = (āˆ’x1 , āˆ’x2 ). 1.122

1. x+y =x+z āˆ’x + (x + y) = āˆ’x + (x + z) (āˆ’x + x) + y = (āˆ’x + x) + z 0+y =0+z y=z

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2. š›¼x = š›¼y 1 1 (š›¼x) = (š›¼y) š›¼ ( ) (š›¼ ) 1 1 š›¼ x= š›¼ y š›¼ š›¼ x=y 3. š›¼x = š›½x implies (š›¼ āˆ’ š›½)x = š›¼x āˆ’ š›½x = 0 Provided x = 0, we must have (š›¼ āˆ’ š›½)x = 0x That is š›¼ āˆ’ š›½ = 0 which implies š›¼ = š›½. 4. (š›¼ āˆ’ š›½)x = (š›¼ + (āˆ’š›½))x = š›¼x + (āˆ’š›½)x = š›¼x āˆ’ š›½x 5. š›¼(x āˆ’ y) = š›¼(x + (āˆ’1)y) = š›¼x + š›¼(āˆ’1)y = š›¼x āˆ’ š›¼y 6. š›¼0 = š›¼(x + (āˆ’x)) = š›¼x + š›¼(āˆ’x) = š›¼x āˆ’ š›¼x =0 1.123 The linear hull of the vectors {(1, 0), (0, 2)} is { ( ) ( )} 1 0 lin {(1, 0), (0, 2)} = š›¼1 + š›¼2 0 2 ) ( { š›¼1 } = š›¼2 = ā„œ2 The linear hull of the vectors {(1, 0), (0, 2)} is the whole plane ā„œ2 . Figure 1.4 illustrates how any vector in ā„œ2 can be obtained as a linear combination of {(1, 0), (0, 2)}. 1.124

1. From the deļ¬nition of š›¼, š›¼š‘† = š‘¤(š‘†) āˆ’

āˆ‘ š‘‡ āŠŠš‘†

25

š›¼š‘‡

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

(āˆ’2, 3)

3 2 1

-2

-1

0

1

Figure 1.4: Illustrating the span of { (1, 0), (0, 2) }. for every š‘† āŠ† š‘ . Rearranging š‘¤(š‘†) = š›¼š‘† + =

āˆ‘

āˆ‘

š›¼š‘‡ +

š‘‡ =š‘†

=

āˆ‘

š›¼š‘‡

š‘‡ āŠŠš‘†

āˆ‘

š›¼š‘‡

š‘‡ āŠŠš‘†

š›¼š‘‡

š‘‡ āŠ†š‘†

2.

āˆ‘

š›¼š‘‡ š‘¤š‘‡ (š‘†) =

š‘‡ āŠ†š‘

āˆ‘

š›¼š‘‡ š‘¤š‘‡ (š‘†) +

š‘‡ āŠ†š‘†

=

āˆ‘

š›¼š‘‡ 1 +

š‘‡ āŠ†š‘†

=

āˆ‘

āˆ‘

āˆ‘

š›¼š‘‡ š‘¤š‘‡ (š‘†)

š‘‡ āˆ•āŠ†š‘†

š›¼š‘‡ 0

š‘‡ āˆ•āŠ†š‘†

š›¼š‘‡ 1

š‘‡ āŠ†š‘†

= š‘¤(š‘†) 1.125

1. Choose any x āˆˆ š‘†. By homogeneity 0x = šœƒ āˆˆ š‘†.

2. For every x āˆˆ š‘†, āˆ’x = (āˆ’1)x āˆˆ š‘† by homogeneity. 1.126 Examples of subspaces in ā„œš‘› include: 1. The set containing just the null vector {0} is subspace. 2. Let x be any element in ā„œš‘› and let š‘‡ be the set of all scalar multiples of x š‘‡ = { š›¼x : š›¼ āˆˆ ā„œ } š‘‡ is a line through the origin in ā„œš‘› and is a subspace. 3. Let š‘† be the set of all š‘›-tuples with zero ļ¬rst coordinate, that is š‘† = { (š‘„1 , š‘„2 , . . . , š‘„š‘› ) : š‘„1 = 0, š‘„š‘— āˆˆ ā„œ, š‘— āˆ•= 1 } For any x, y āˆˆ š‘† x + y = (0, š‘„2 , š‘„3 , . . . , š‘„š‘› ) + (0, š‘¦2 , š‘¦3 , . . . , š‘¦š‘› ) = (0, š‘„2 + š‘¦2 , š‘„3 + š‘¦3 , . . . , š‘„š‘› + š‘¦š‘› ) āˆˆ š‘† 26

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Similarly š›¼x = š›¼(0, š‘„2 , š‘„3 , . . . , š‘„š‘› ) = (0, š›¼š‘„2 , š›¼š‘„3 , . . . , š›¼š‘„š‘› ) āˆˆ š‘† Therefore š‘† is a subspace of ā„œš‘› . Generalizing, any set of vectors with one or more coordinates identically zero is a subspace of ā„œš‘› . 4. We will meet some more complicated subspaces in Chapter 2. 1.127 No, āˆ’x āˆˆ / ā„œš‘›+ if x āˆˆ ā„œš‘›+ unless x = 0. ā„œš‘›+ is an example of a cone (Section 1.4.5). 1.128 lin š‘† is a subspace Let x, y be two elements in lin š‘†. x is a linear combination of elements of š‘†, that is x = š›¼1 š‘„1 + š›¼2 š‘„2 + . . . š›¼š‘› š‘„š‘› Similarly y = š›½1 š‘„1 + š›½2 š‘„2 + . . . š›½š‘› š‘„š‘› and x + y = (š›¼1 + š›½1 )š‘„1 + (š›¼2 + š›½2 )š‘„2 + ā‹… ā‹… ā‹… + (š›¼š‘› + š›½š‘› )š‘„š‘› āˆˆ lin š‘† and š›¼x = š›¼š›¼1 š‘„1 + š›¼š›¼2 š‘„2 + ā‹… ā‹… ā‹… + š›¼š›¼š‘› š‘„š‘› āˆˆ lin š‘† This shows that lin š‘† is closed under addition and scalar multiplication and hence is a subspace. lin š‘† is the smallest subspace containing š‘† Let š‘‡ be any subspace containing š‘†. Then š‘‡ contains all linear combinations of elements in š‘†, so that lin š‘† āŠ‚ š‘‡ . Hence lin š‘† is the smallest subspace containing S. 1.129 The previous exercise showed that lin š‘† is a subspace. Therefore, if š‘† = lin š‘†, š‘† is a subspace. Conversely, assume that š‘† is a subspace. Then š‘† is the smallest subspace containing š‘†, and therefore š‘† = lin š‘† (again by the previous exercise). 1.130 Let x, y āˆˆ š‘† = š‘†1 āˆ© š‘†2 . Hence x, y āˆˆ š‘†1 and for any š›¼, š›½ āˆˆ ā„œ, š›¼x + š›½y āˆˆ š‘†1 . Similarly š›¼x + š›½y āˆˆ š‘†2 and therefore š›¼x + š›½y āˆˆ š‘†. š‘† is a subspace. 1.131 Let š‘† = š‘†1 + š‘†2 . First note that 0 = 0 + 0 āˆˆ š‘†. Suppose x, y belong to š‘†. Then there exist s1 , t1 āˆˆ š‘†1 and s2 , t2 āˆˆ š‘†2 such that x = s1 + s2 and y = t1 + t2 . For any š›¼, š›½ āˆˆ ā„œ, š›¼x + š›½y = š›¼(s1 + s2 ) + š›½(t1 + t2 ) = (š›¼s1 + š›½t1 ) + (š›¼s2 + š›½t2 ) āˆˆ š‘† since š›¼s1 + š›½t1 āˆˆ š‘†1 and š›¼s2 + š›½t2 āˆˆ š‘†2 . 1.132 Let š‘†1 = { š›¼(1, 0) : š›¼ āˆˆ ā„œ } š‘†2 = { š›¼(0, 1) : š›¼ āˆˆ ā„œ } 27

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

š‘†1 and š‘†2 are respectively the horizontal and vertical axes in ā„œ2 . Their union is not a subspace, since for example ( ) ( ) ( ) 1 1 0 = + āˆˆ / š‘†1 āˆŖ š‘†2 1 0 1 However, any vector in ā„œ2 can be written as the sum of an element of š‘†1 and an element of š‘†2 . Therefore, their sum is the whole space ā„œ2 , that is š‘† 1 + š‘† 2 = ā„œ2 1.133 Assume that š‘† is linearly dependent, that is there exists x1 , . . . , xš‘› āˆˆ š‘† and š›¼2 , . . . , š›¼š‘› āˆˆ š‘… such that x1 = š›¼2 x2 + š›¼3 x3 + . . . , š›¼š‘› xš‘› Rearranging, this implies 1x1 āˆ’ š›¼2 x2 āˆ’ š›¼3 x3 āˆ’ . . . š›¼š‘› xš‘› = 0 Conversely, assume there exist x1 , x2 , . . . , xš‘› āˆˆ x and š›¼1 , š›¼2 , . . . , š›¼š‘› āˆˆ ā„œ such that š›¼1 x1 + š›¼2 x2 . . . + š›¼š‘› xš‘› = 0 Assume without loss of generality that š›¼1 āˆ•= 0. Then x1 = āˆ’

š›¼2 š›¼3 š›¼š‘› x2 āˆ’ x3 āˆ’ . . . āˆ’ xš‘› š›¼1 š›¼1 š›¼1

which shows that x1 āˆˆ lin š‘† āˆ– {x1 } 1.134 Assume {(1, 1, 1), (0, 1, 1), (0, 0, 1)} are linearly dependent. Then there exists š›¼1 , š›¼2 , š›¼3 such that āŽ› āŽž āŽ› āŽž āŽ› āŽž āŽ› āŽž 1 0 0 0 š›¼1 āŽ1āŽ  + š›¼2 āŽ1āŽ  + š›¼3 āŽ0āŽ  = āŽ0āŽ  1 1 1 0 or equivalently š›¼1 = 0 š›¼1 + š›¼2 = 0 š›¼1 + š›¼2 + š›¼3 = 0 which imply that š›¼1 = š›¼2 = š›¼3 = 0 Therefore {(1, 1, 1), (0, 1, 1), (0, 0, 1)} are linearly independent. 1.135 Suppose on the contrary that š‘ˆ is linearly dependent. That is, there exists a set of games { š‘¢š‘‡1 , š‘¢š‘‡2 , . . . , š‘¢š‘‡š‘š } and nonzero coeļ¬ƒcients (š›¼1 , š›¼2 , . . . , š›¼š‘š ) such that (Exercise 1.133) š›¼1 š‘¢š‘‡1 + š›¼2 š‘¢š‘‡2 + . . . + š›¼š‘š š‘¢š‘‡š‘š = 0 28

(1.16)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Assume that the coalitions are ordered so that š‘‡1 has the smallest number of players of any of the coalitions š‘‡1 , š‘‡2 , . . . , š‘‡š‘š . This implies that no coalition š‘‡2 , š‘‡3 , . . . , š‘‡š‘š is a subset of š‘‡1 and š‘¢š‘‡š‘— (š‘‡1 ) = 0

for every š‘— = 2, 3, . . . , š‘›

(1.17)

Using (1.39), š‘¢š‘‡1 can be expressed as a linear combination of the other games, š‘¢š‘‡1 = āˆ’1/š›¼1

š‘š āˆ‘

š›¼š‘— š‘¢š‘‡š‘—

(1.18)

š‘—=2

Substituting (1.40) this implies that š‘¢š‘‡1 (š‘‡1 ) = 0 whereas š‘¢š‘‡ (š‘‡ ) = 1

for every š‘‡

by deļ¬nition. This contradiction establishes that the set š‘ˆ is linearly independent. 1.136 If š‘† is a subspace, then 0 āˆˆ š‘† and š›¼x1 = 0 with š›¼ āˆ•= 0 and x1 = 0 (Exercise 1.122). Therefore š‘† is linearly dependent (Exercise 1.133). 1.137 Suppose x has two representations, that is x = š›¼1 x1 + š›¼2 x2 + . . . + š›¼š‘› xš‘› x = š›½1 x1 + š›½2 x2 + . . . + š›½š‘› xš‘› Subtracting 0 = (š›¼1 āˆ’ š›½1 )x1 + (š›¼2 āˆ’ š›½2 )x2 + . . . + (š›¼š‘› āˆ’ š›½š‘› )xš‘›

(1.19)

Since {x1 , x2 , . . . , , xš‘› } is linearly independent, (1.19) implies that š›¼š‘– = š›½š‘– = 0 for all š‘– (Exercise 1.133) 1.138 Let š‘ƒ be the set of all linearly independent subsets of a linear space š‘‹. š‘ƒ is partially ordered by inclusion. Every chain š¶ = {š‘†š›¼ } āŠ† š‘ƒ has an upper bound, namely āˆŖ š‘†. By Zornā€™s lemma, š‘ƒ has a maximal element šµ. We show that šµ is a basis š‘†āˆˆš¶ for š‘‹. šµ is linearly independent since šµ āˆˆ š‘ƒ . Suppose that šµ does not span š‘‹ so that lin šµ āŠ‚ š‘‹. Then there exists some x āˆˆ š‘‹ āˆ– lin šµ. The set šµ āˆŖ {x} is a linearly independent and contains šµ, which contradicts the assumption that šµ is the maximal element of š‘ƒ . Consequently, we conclude that šµ spans š‘‹ and hence is a basis. 1.139 Exercise 1.134 established that the set šµ = { (1, 1, 1), (0, 1, 1), (0, 0, 1)} is linearly independent. Since dim š‘…3 = 3, any other vectors must be linearly dependent on šµ. That is lin šµ = ā„œ3 . šµ is a basis. By a similar argument to exercise 1.134, it is readily seen that {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is linearly independent and hence constitutes a basis.

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.140 Let š“ = {a1 , a2 , . . . , aš‘› } and šµ = {b1 , b2 , . . . , bš‘š } be two bases for a linear space š‘‹. Let š‘†1 = {š‘1 } āˆŖ š“ = {b1 , a1 , a2 , . . . , aš‘› } š‘† is linearly dependent (since š‘1 āˆˆ lin š“) and spans š‘‹. š›¼1 , š›¼2 , . . . , š›¼š‘› and š›½1 such that

Therefore, there exists

š›½1 b1 + š›¼1 a1 + š›¼2 a2 + . . . + š›¼š‘› aš‘› = 0 At least one š›¼š‘– āˆ•= 0. Deleting the corresponding element aš‘– , we obtain another set š‘†1ā€² of š‘› elements š‘†1ā€² = {b1 , a1 , a2 , . . . , aš‘–āˆ’1 , aš‘–+1 , . . . , aš‘› } which is also spans š‘‹. Adding the second element from šµ, we obtain the š‘› + 1 element set š‘†2 = {b1 , b2 , a1 , a2 , . . . , aš‘–āˆ’1 , aš‘–+1 , . . . , aš‘› } which again is linearly dependent and spans š‘‹. Continuing in this way, we can replace š‘š vectors in š“ with the š‘š vectors from šµ while maintaining a spanning set. This process cannot eliminate all the vectors in š“, because this would imply that šµ was linearly dependent. (Otherwise, the remaining bš‘– would be linear combinations of preceding elements of šµ.) We conclude that necessarily š‘š ā‰¤ š‘›. Reversing the process and replacing elements of šµ with elements of š“ establishes that š‘› ā‰¤ š‘š. Together these inequalities imply that š‘› = š‘š and š“ and šµ have the same number of elements. 1.141 Suppose that the coalitions are ordered in some way, so that š’«(š‘ ) = {š‘†0 , š‘†1 , š‘†2 , . . . , š‘†2š‘› āˆ’1 } with š‘†0 = āˆ…. There are 2š‘› coalitions. Each game šŗ āˆˆ š’¢ š‘ corresponds to a unique list of length 2š‘› of coalitional worths v = (š‘£0 , š‘£1 , š‘£2 , . . . , š‘£2š‘› āˆ’1 ) š‘›

with š‘£0 = 0. That is, each game deļ¬nes a vector š‘£ = (0, š‘£1 , . . . , š‘£2š‘› āˆ’1 ) āˆˆ ā„œ2 and š‘› conversely each vector š‘£ āˆˆ ā„œ2 (with š‘£0 = 0) deļ¬nes a game. Therefore, the space of š‘› all games š’¢ š‘ is formally identical to the subspace of ā„œ2 in which the ļ¬rst component š‘› is identically zero, which in turn is equivalent to the space ā„œ2 āˆ’1 . Thus, š’¢ š‘ is a 2š‘› āˆ’ 1-dimensional linear space. 1.142 For illustrative purposes, we present two proofs, depending upon whether the linear space is assumed to be ļ¬nite dimensional or not. In the ļ¬nite dimensional case, a constructive proof is possible, which forms the basis for practical algorithms for constructing a basis. Let š‘† be a linearly independent set in a linear space š‘‹. š‘‹ is ļ¬nite dimensional Let š‘› = dim š‘‹. Assume š‘† has š‘š elements and denote it š‘†š‘š . If lin š‘†š‘š = š‘‹, then š‘†š‘š is a basis and we are done. Otherwise, there exists some xš‘š+1 āˆˆ š‘‹ āˆ– lin š‘†š‘š . Adding xš‘š+1 to š‘†š‘š gives a new set of š‘š + 1 elements š‘†š‘š+1 = š‘†š‘š āˆŖ { xš‘š+1 } 30

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

which is also linearly independent ( since xš‘š+1 āˆˆ / lin š‘†š‘š ). If lin š‘†š‘š+1 = š‘‹, then š‘†š‘š+1 is a basis and we are done. Otherwise, there exists some xš‘š+2 āˆˆ š‘‹ āˆ– lin š‘†š‘š+1 . Adding xš‘š+2 to š‘†š‘š+1 gives a new set of š‘š + 2 elements š‘†š‘š+2 = š‘†š‘š+1 āˆŖ { xš‘š+2 } which is also linearly independent ( since xš‘š+2 āˆˆ / lin š‘†š‘š+2 ). Repeating this process, we can construct a sequence of linearly independent sets š‘†š‘š , š‘†š‘š+1 , š‘†š‘š+2 . . . such that lin š‘†š‘š ā«‹ lin š‘†š‘š+1 ā«‹ lin š‘†š‘š+2 ā‹… ā‹… ā‹… āŠ† š‘‹. Eventually, we will reach a set which spans š‘‹ and hence is a basis. š‘‹ is possibly inļ¬nite dimensional For the general case, we can adapt the proof of the existence of a basis (Exercise 1.138), restricting š‘ƒ to be the class of all linearly independent subsets of š‘‹ containing š‘†. 1.143 Otherwise (if a set of š‘› + 1 elements was linearly independent), it could be extended to basis at least š‘› + 1 elements (exercise 1.142). This would contradict the fundamental result that all bases have the same number of elements (Exercise 1.140). 1.144 Every basis is linearly independent. Conversely, let šµ = {x1 , x2 , . . . , xš‘› } be a set of linearly independent elements in an š‘›-dimensional linear space š‘‹. We have to show that lin šµ = š‘‹. Take any x āˆˆ š‘‹. The set šµ āˆŖ {x} = {x1 , x2 , . . . , xš‘› , x } must be linearly dependent (Exercise 1.143). That is there exists numbers š›¼1 , š›¼2 , . . . , š›¼š‘› , š›¼, not all zero, such that š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› + š›¼x = 0

(1.20)

Furthermore, it must be the case that š›¼ āˆ•= 0 since otherwise š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› = 0 which contradicts the linear independence of š“. Solving (1.20) for x, we obtain x=

1 š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› š›¼

Since x was an arbitrary element of š‘‹, we conclude that šµ spans š‘‹ and hence šµ is a basis. 1.145 A basis spans š‘‹. To establish the converse, assume that šµ = {x1 , x2 , . . . , xš‘› } is a set of š‘› elements which span š‘‹. If š‘† is linearly dependent, then one element is linearly dependent on the other elements. Without loss of generality, assume that x1 āˆˆ lin šµ āˆ– {x1 }. Deleting x1 the set šµ āˆ– {x1 } = {x2 , x3 , . . . , xš‘› } also spans š‘‹. Continuing in this fashion by eliminating dependent elements, we ļ¬nish with a linearly independent set of š‘š < š‘› elements which spans š‘‹. That is, we can ļ¬nd a basis of š‘š < š‘› elements, which contradicts the assumption that the dimension of š‘‹ is š‘› (Exercise 1.140). Thus any set of š‘› vectors which spans š‘‹ must be linearly independent and hence a basis. 31

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.146 We have previously shown āˆ™ that the set š‘ˆ is linearly independent (Exercise 1.135). āˆ™ the space š’¢ š‘ has dimension 2š‘›āˆ’1 (Exercise 1.141). There are 2š‘›āˆ’1 distinct T-unanimity games š‘¢š‘‡ in š‘ˆ . Hence š‘ˆ spans the 2š‘›āˆ’1 space š’¢ š‘ . Alternatively, note that any game š‘¤ āˆˆ š’¢ š‘ can be written as a linear combination of T-unanimity games (Exercise 1.75). 1.147 Let šµ = {x1 , x2 , . . . , xš‘š } be a basis for š‘†. Since šµ is linearly independent, š‘š ā‰¤ š‘› (Exercise 1.143). There are two possibilities. Case 1: š‘š = š‘›. šµ is a set of š‘› linearly independent elements in an š‘›-dimensional space š‘‹. Hence šµ is a basis for š‘‹ and š‘† = lin šµ = š‘‹. Case 2: š‘š < š‘›. Since šµ is linearly independent but cannot be a basis for the š‘›dimensional space š‘‹, we must have š‘† = lin šµ āŠ‚ š‘‹. Therefore, we conclude that if š‘† āŠ‚ š‘‹ is a proper subspace, it has a lower dimension than š‘‹. 1.148 Let š›¼1 , š›¼2 , š›¼3 be the coordinates of (1, 1, 1) for the basis {(1, 1, 1), (0, 1, 1), (0, 0, 1)}. That is āŽ› āŽž āŽ› āŽž āŽ› āŽž āŽ› āŽž 1 0 0 1 āŽ1āŽ  = š›¼1 āŽ1āŽ  + š›¼2 āŽ1āŽ  + š›¼3 āŽ0āŽ  1 1 1 1 which implies that š›¼1 = 1, š›¼2 = š›¼3 = 0. Therefore (1, 0, 0) are the required coordinates of the (1, 1, 1) with respect to the basis {(1, 1, 1), (0, 1, 1), (0, 0, 1)}. (1, 1, 1) are the coordinates of the vector (1, 1, 1) with respect to the standard basis. 1.149 A subset š‘† of a linear space š‘‹ is a subspace of š‘‹ if š›¼x + š›½y āˆˆ š‘† for every x, y āˆˆ š‘† and for every š›¼, š›½ āˆˆ ā„œ Letting š›½ = 1 āˆ’ š›¼, this implies that š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘†

for every x, y āˆˆ š‘† and š›¼ āˆˆ ā„œ

š‘† is an aļ¬ƒne set. Conversely, suppose that š‘† is an aļ¬ƒne set containing 0, that is š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘†

for every x, y āˆˆ š‘† and š›¼ āˆˆ ā„œ

Letting y = 0, this implies that š›¼x āˆˆ š‘†

for every x āˆˆ š‘† and š›¼ āˆˆ ā„œ

so that š‘† is homogeneous. Now letting š›¼ = 12 , for every x and y in š‘†, 1 1 x+ y āˆˆš‘† 2 2 and homogeneity implies

( x+y =2

1 1 x+ y 2 2

š‘† is also additive. Hence š‘† is subspace. 32

) āˆˆš‘†

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.150 For any x āˆˆ š‘†, let š‘‰ = š‘† āˆ’x = {v āˆˆ š‘‹ : v +x āˆˆ š‘† } š‘‰ is an aļ¬ƒne set For any v1 , v2 āˆˆ š‘‰ , there exist corresponding s1 , s2 āˆˆ š‘† such that v1 = s1 āˆ’ x and v2 = s2 āˆ’ x and therefore š›¼v1 + (1 āˆ’ š›¼)v2 = š›¼(s1 āˆ’ x) + (1 āˆ’ š›¼)(s1 āˆ’ x) = š›¼s1 + (1 āˆ’ š›¼)s2 āˆ’ š›¼x + (1 āˆ’ š›¼)x =sāˆ’x where s = š›¼š‘ 1 + (1 āˆ’ š›¼)š‘ 2 āˆˆ š‘†. There š‘‰ is an aļ¬ƒne set. š‘‰ is a subspace Since x āˆˆ š‘†, 0 = x āˆ’ x āˆˆ š‘‰ . Therefore š‘‰ is a subspace (Exercise 1.149). š‘‰ is unique Suppose that there are two subspaces š‘‰ 1 and š‘‰ 2 such that š‘† = š‘‰ 1 + x1 and š‘† = š‘‰ 2 + x2 . Then š‘‰1 + x1 = š‘‰2 + x2 š‘‰1 = š‘‰2 + (x2 āˆ’ x1 ) = š‘‰2 + x where x = x2 āˆ’ x1 āˆˆ š‘‹. Therefore š‘‰1 is parallel to š‘‰2 . Since š‘‰1 is a subspace, 0 āˆˆ š‘‰1 which implies that āˆ’x āˆˆ š‘‰2 . Since š‘‰2 is a subspace, this implies that x āˆˆ š‘‰2 and š‘‰2 + x āŠ† š‘‰2 . Therefore š‘‰1 = š‘‰2 + x āŠ† š‘‰2 . Similarly, š‘‰2 āŠ† š‘‰1 and hence š‘‰1 = š‘‰2 . Therefore the subspace š‘‰ is unique. 1.151 Let š‘† āˆ„ š‘‡ denote the relation š‘† is parallel to š‘‡ , that is š‘† āˆ„ š‘‡ ā‡ā‡’ š‘† = š‘‡ + x for some x āˆˆ š‘‹ The relation āˆ„ is reļ¬‚exive š‘† āˆ„ š‘† since š‘† = š‘† + 0 transitive Assume š‘† = š‘‡ + x and š‘‡ = š‘ˆ + y. Then š‘† = š‘ˆ + (x + y) symmetric š‘† = š‘‡ + x =ā‡’ š‘‡ = š‘† + (āˆ’x) Therefore āˆ„ is an equivalence relation. 1.152 See exercises 1.130 and 1.162. 1.153

1. Exercise 1.150

2. Assume x0 āˆˆ š‘‰ . For every x āˆˆ š» x = x0 + v = w āˆˆ š‘‰ which implies that š» āŠ† š‘‰ . Conversely, assume š» = š‘‰ . Then x0 = 0 āˆˆ š‘‰ since š‘‰ is a subspace. 3. By deļ¬nition, š» āŠ‚ š‘‹. Therefore š‘‰ = š» āˆ’ x āŠ‚ š‘‹. / š‘‰ . Suppose to the contrary 4. Let x1 āˆˆ lin {x1 , š‘‰ } = š‘‰ ā€² āŠ‚ š‘‹

33

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Then š» ā€² = x0 + š‘‰ ā€² is an aļ¬ƒne set (Exercise 1.150) which strictly contains š». This contradicts the deļ¬nition of š» as a maximal proper aļ¬ƒne set. 5. Let x1 āˆˆ / š‘‰ . By the previous part, x āˆˆ lin {x1 , š‘‰ }. That is, there exists š›¼ āˆˆ ā„œ such that x = š›¼x1 + v for some v āˆˆ š‘‰ To see that š›¼ is unique, suppose that there exists š›½ āˆˆ ā„œ such that x = š›½x1 + vā€² for some vā€² āˆˆ š‘‰ Subtracting 0 = (š›¼ āˆ’ š›½)x1 + (v āˆ’ vā€² ) / š‘‰. which implies that š›¼ = š›½ since x1 āˆˆ 1.154 Assume x, y āˆˆ š‘‹. That is, x, y āˆˆ ā„œš‘› and āˆ‘ āˆ‘ š‘„š‘– = š‘¦š‘– = š‘¤(š‘ ) š‘–āˆˆš‘

š‘–āˆˆš‘

š‘›

For any š›¼ āˆˆ ā„œ, š›¼x + (1 āˆ’ š›¼)y āˆˆ ā„œ and āˆ‘ āˆ‘ āˆ‘ š›¼š‘„š‘– + (1 āˆ’ š›¼)š‘¦š‘– = š›¼ š‘„š‘– + (1 āˆ’ š›¼) š‘¦š‘– š‘–āˆˆš‘

š‘–āˆˆš‘

š‘–āˆˆš‘

= š›¼š‘¤(š‘ ) + (1 āˆ’ š›¼)š‘¤(š‘ ) = š‘¤(š‘ ) Hence š‘‹ is an aļ¬ƒne subset of ā„œš‘› . 1.155 See Exercise 1.129. 1.156 No. A straight line through any two points in ā„œš‘›+ extends outside ā„œš‘›+ . Put diļ¬€erently, the aļ¬ƒne hull of ā„œš‘›+ is the whole space ā„œš‘› . 1.157 Let š‘‰ = aļ¬€ š‘† āˆ’ x1 = aļ¬€ {0, x2 āˆ’ x1 , x3 āˆ’ x1 , . . . , xš‘› āˆ’ x1 } š‘‰ is a subspace (0 āˆˆ š‘‰ ) and aļ¬€ š‘† = š‘‰ + x1 and dim aļ¬€ š‘† = dim š‘‰ Note that the choice of x1 is arbitrary. š‘† is aļ¬ƒnely dependent if and only if there exists some xš‘˜ āˆˆ š‘† such that xš‘˜ āˆˆ āˆ• x1 . aļ¬€ (š‘† āˆ– {xš‘˜ }). Since the choice of x1 is arbitrary, we assume that xš‘˜ = xš‘˜ āˆˆ aļ¬€ (š‘† āˆ– {xš‘˜ }) ā‡ā‡’ xš‘˜ āˆˆ (š‘‰ + x1 ) āˆ– {xš‘˜ } ā‡ā‡’ xš‘˜ āˆ’ x1 āˆˆ š‘‰ āˆ– {xš‘˜ āˆ’ x1 } ā‡ā‡’ xš‘˜ āˆ’ x1 āˆˆ lin {x2 āˆ’ x1 , x3 āˆ’ x1 , . . . , xš‘˜āˆ’1 āˆ’ x1 , . . . , xš‘˜+1 āˆ’ x1 , . . . , xš‘› āˆ’ x1 } 34

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

Therefore, š‘† is aļ¬ƒnely dependent if and only if {x2 āˆ’ x1 , x3 āˆ’ x1 , . . . , xš‘› āˆ’ x1 } is linearly independent. 1.158 By the previous exercise, the set š‘† = {x1 , x2 , . . . , xš‘› } is aļ¬ƒnely dependent if and only if the set {x2 āˆ’ x1 , x3 āˆ’ x1 , . . . , xš‘› āˆ’ x1 } is linearly dependent, so that there exist numbers š›¼2 , š›¼3 , . . . , š›¼š‘› , not all zero, such that š›¼2 (x2 āˆ’ x1 ) + š›¼3 (x3 āˆ’ x1 ) + ā‹… ā‹… ā‹… + š›¼š‘› (xš‘› āˆ’ x1 ) = 0 or š›¼2 x2 + š›¼3 x3 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› āˆ’

š‘› āˆ‘

š›¼š‘– x1 = 0

š‘–=2

Let š›¼1 = āˆ’

āˆ‘š‘›

š‘–=2

š›¼š‘– . Then š›¼1 x1 + š›¼2 x2 + . . . + š›¼š‘› xš‘› = 0

and š›¼1 + š›¼2 + . . . + š›¼š‘› = 0 as required. 1.159 Let š‘‰ = aļ¬€ š‘† āˆ’ x1 = aļ¬€ { 0, x2 āˆ’ x1 , x3 āˆ’ x1 , . . . , xš‘› āˆ’ x1 } Then aļ¬€ š‘† = x1 + š‘‰ If š‘† is aļ¬ƒnely independent, every x āˆˆ aļ¬€ š‘† has a unique representation as x = x1 + v,

vāˆˆš‘‰

with v = š›¼2 (x2 āˆ’ x1 ) + š›¼3 (x3 āˆ’ x1 ) + ā‹… ā‹… ā‹… + š›¼š‘› (xš‘› āˆ’ x1 ) so that x = x1 + š›¼2 (x2 āˆ’ x1 ) + š›¼3 (x3 āˆ’ x1 ) + ā‹… ā‹… ā‹… + š›¼š‘› (xš‘› āˆ’ x1 ) āˆ‘š‘› Deļ¬ne š›¼1 = 1 āˆ’ š‘–=2 š›¼š‘– . Then x = š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› with š›¼1 + š›¼2 + ā‹… ā‹… ā‹… + š›¼š‘› = 1 x is a unique aļ¬ƒne combination of the elements of š‘†. 1.160 Assume that š‘„, š‘¦ āˆˆ (š‘Ž, š‘) āŠ† ā„œ. This means that š‘Ž < š‘„ < š‘ and š‘Ž < š‘¦ < š‘. For every 0 ā‰¤ š›¼ ā‰¤ 1 š›¼š‘„ + (1 āˆ’ š›¼)š‘¦ > š›¼š‘Ž + (1 āˆ’ š›¼)š‘Ž 35

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

and š›¼š‘„ + (1 āˆ’ š›¼)š‘¦ < š›¼š‘ + (1 āˆ’ š›¼)š‘ Therefore š‘Ž < š›¼š‘„+(1āˆ’š›¼)š‘¦ < š‘ and š›¼š‘„+(1āˆ’š›¼)š‘¦ āˆˆ (š‘Ž, š‘). (š‘Ž, š‘) is convex. Substituting ā‰¤ for < demonstrates that [š‘Ž, š‘] is convex. Let š‘† be an arbitrary convex set in ā„œ. Assume that š‘† is not an interval. This implies that there exist numbers š‘„, š‘¦, š‘§ such that š‘„ < š‘¦ < š‘§ and š‘„, š‘§ āˆˆ š‘† while š‘¦ āˆˆ / š‘†. Deļ¬ne š›¼=

š‘§āˆ’š‘¦ š‘§āˆ’š‘„

so that 1āˆ’š›¼=

š‘¦āˆ’š‘„ š‘§āˆ’š‘„

Note that 0 ā‰¤ š›¼ ā‰¤ 1 and that š›¼š‘„ + (1 āˆ’ š›¼)š‘§ =

š‘¦āˆ’š‘„ š‘§āˆ’š‘¦ š‘„+ š‘§=š‘¦āˆˆ /š‘† š‘§āˆ’š‘„ š‘§āˆ’š‘„

which contradicts the assumption that š‘† is convex. We conclude that every convex set in ā„œ is an interval. Note that š‘† may be a hybrid interval such (š‘Ž, š‘] or [š‘Ž, š‘) as well as an open (š‘Ž, š‘) or closed [š‘Ž, š‘] interval. 1.161 Let (š‘, š‘¤) be a TP-coalitional game. If core(š‘, š‘¤) = āˆ… then it is trivially convex. Otherwise, assume core(š‘, š‘¤) is nonempty and let x1 and x2 belong to core(š‘, š‘¤). That is āˆ‘ š‘„1š‘– ā‰„ š‘¤(š‘†) for every š‘† āŠ† š‘ š‘–āˆˆš‘†

āˆ‘

š‘„1š‘– = š‘¤(š‘ )

š‘–āˆˆš‘

and therefore for any 0 ā‰¤ š›¼ ā‰¤ 1 āˆ‘ š›¼š‘„1š‘– ā‰„ š›¼š‘¤(š‘†)

for every š‘† āŠ† š‘

š‘–āˆˆš‘†

āˆ‘

š›¼š‘„1š‘– = š›¼š‘¤(š‘ )

š‘–āˆˆš‘

Similarly āˆ‘

(1 āˆ’ š›¼)š‘„2š‘– ā‰„ (1 āˆ’ š›¼)š‘¤(š‘†)

for every š‘† āŠ† š‘

š‘–āˆˆš‘†

āˆ‘

(1 āˆ’ š›¼)š‘„2š‘– = (1 āˆ’ š›¼)š‘¤(š‘ )

š‘–āˆˆš‘

Summing these two systems āˆ‘ š›¼š‘„1š‘– + (1 āˆ’ š›¼)š‘„2š‘– ā‰„ š›¼š‘¤(š‘†) + (1 āˆ’ š›¼)š‘¤(š‘†) = š‘¤(š‘†) š‘–āˆˆš‘†

āˆ‘

š›¼š‘„1š‘– + (1 āˆ’ š›¼)š‘„2š‘– = š›¼š‘¤(š‘ ) + (1 āˆ’ š›¼)š‘¤(š‘ ) = š‘¤(š‘ )

š‘–āˆˆš‘

That is, š›¼š‘„1š‘– + (1 āˆ’ š›¼)š‘„2š‘– belongs to core(š‘, š‘¤). 36

for every š‘† āŠ† š‘

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

āˆ© 1.162 Let ā„­ be a collection of convex sets and let x, y belong to š‘†āˆˆā„­ š‘†. for every š‘† āˆˆ ā„­, x, y āˆˆ š‘† and therefore āˆ© š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘† for all 0 ā‰¤ š›¼ ā‰¤ 1 (since š‘† is convex). Therefore š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘†āˆˆā„­ š‘†. 1.163 Fix some output š‘¦. Assume that x1 , x2 āˆˆ š‘‰ (š‘¦). This implies that both (š‘¦, āˆ’x1 ) and (š‘¦, āˆ’x2 ) belong to the production possibility set š‘Œ . If š‘Œ is convex š›¼(š‘¦, āˆ’x1 ) + (1 āˆ’ š›¼)(š‘¦, āˆ’x2 ) = (š›¼š‘¦ + (1 āˆ’ š›¼)š‘¦, š›¼x1 + (1 āˆ’ š›¼)x2 ) = (š‘¦, š›¼x1 + (1 āˆ’ š›¼)x2 ) āˆˆ š‘Œ for every š›¼ āˆˆ [0, 1]. This implies that š›¼x1 + (1 āˆ’ š›¼)x2 āˆˆ š‘‰ (š‘¦). Since the choice of š‘¦ was arbitrary, this implies that š‘‰ (š‘¦) is convex for every š‘¦. 1.164 Assume š‘†1 and š‘†2 are convex sets. Let š‘† = š‘†1 + š‘†2 . Suppose x, y belong to š‘†. Then there exist s1 , t1 āˆˆ š‘†1 and s2 , t2 āˆˆ š‘†2 such that x = s1 + s2 and y = t1 + t2 . For any š›¼ āˆˆ [0, 1] š›¼x + (1 āˆ’ š›¼)y = š›¼s1 + s2 + (1 āˆ’ š›¼)t1 + t2 = š›¼s1 + (1 āˆ’ š›¼)t1 + š›¼s2 + (1 āˆ’ š›¼)t2 āˆˆ š‘† since š›¼s1 + (1 āˆ’ š›¼)t1 āˆˆ š‘†1 and š›¼s2 + (1 āˆ’ š›¼)t2 āˆˆ š‘†2 . The argument readily extends to any ļ¬nite number of sets. 1.165 Without loss of generality, assume that š‘› = 2. Let š‘† = š‘†1 Ɨ š‘†2 āŠ† š‘‹ = š‘‹1 Ɨ š‘‹2 . Suppose x = (š‘„1 , š‘„2 ) and y = (š‘¦1 , š‘¦2 ) belong to š‘†. Then š›¼x + (1 āˆ’ š›¼)y = š›¼(š‘„1 , š‘„2 ) + (1 āˆ’ š›¼)(š‘¦1 , š‘¦2 ) = (š›¼š‘„1 , š›¼š‘„2 ) + ((1 āˆ’ š›¼)š‘¦1 , (1 āˆ’ š›¼)š‘¦2 ) = (š›¼š‘„1 + (1 āˆ’ š›¼)š‘¦1 , š›¼š‘„2 + (1 āˆ’ š›¼)š‘¦2 ) āˆˆ š‘† 1.166 Let š›¼x, š›¼y be points in š›¼š‘† so that x, y āˆˆ š‘†. Since š‘† is convex, š›½x+ (1 āˆ’ š›½)y āˆˆ š‘† for every 0 ā‰¤ š›½ ā‰¤ 1. Multiplying by š›¼ š›¼(š›½x + (1 āˆ’ š›½)y) = š›½(š›¼x) + (1 āˆ’ š›½)(š›¼y) āˆˆ š›¼š‘† Therefore, š›¼š‘† is convex. 1.167 Combine Exercises 1.164 and 1.166. 1.168 The inclusion š‘† āŠ† š›¼š‘† + (1 āˆ’ š›¼)š‘† is true for any set (whether convex or not), since for every x āˆˆ š‘† x = š›¼x + (1 āˆ’ š›¼)x āˆˆ š›¼š‘† + (1 āˆ’ š›¼)š‘† The reverse inclusion š›¼š‘† +(1āˆ’š›¼)š‘† āŠ† š‘† follows directly from the deļ¬nition of convexity. 1.169 Given any two convex sets š‘† and š‘‡ in a linear space, the largest convex set contained in both is š‘† āˆ© š‘‡ ; the smallest convex set containing both is conv š‘† āˆŖ š‘‡ . Therefore, the set of all convex sets is a lattice with š‘† āˆ§š‘‡ =š‘† āˆ©š‘‡ š‘† āˆØ š‘‡ = conv š‘† āˆŖ š‘‡ The lattice is complete since every collection {š‘†š‘– } has a least upper bound conv āˆŖ š‘†š‘– and a greatest lower bound āˆ©š‘†š‘– . 37

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.170 If a set contains all convex combinations of its elements, it contains all convex combinations of any two points, and hence is convex. Conversely, assume that š‘† is convex. Let x be a convex combination of elements in š‘†, that is let x = š›¼1 x1 + š›¼2 x2 + . . . + š›¼š‘› xš‘› where x1 , x2 , . . . , xš‘› āˆˆ š‘† and š›¼1 , š›¼2 , . . . , š›¼š‘› āˆˆ ā„œ+ with š›¼1 + š›¼2 + . . . + š›¼š‘› = 1. We need to show that x āˆˆ š‘†. We proceed by induction of the number of points š‘›. Clearly, x āˆˆ š‘† if š‘› = 1 or š‘› = 2. To show that it is true for š‘› = 3, let x = š›¼1 x1 + š›¼2 x2 + š›¼3 x3 where x1 , x2 , x3 āˆˆ š‘† and š›¼1 , š›¼2 , š›¼3 āˆˆ ā„œ+ with š›¼1 + š›¼2 + š›¼3 = 1. Assume that š›¼š‘– > 0 for all š‘– (otherwise š‘› = 1 or š‘› = 2) so that š›¼1 < 1. Rewriting x = š›¼1 x1 + š›¼2 x2 + š›¼3 x3 ( ) š›¼2 š›¼2 = š›¼1 x1 + (1 āˆ’ š›¼1 ) x2 + x3 1 āˆ’ š›¼1 1 āˆ’ š›¼1 = š›¼1 x1 + (1 āˆ’ š›¼1 )y where ( y=

š›¼2 š›¼2 x2 + x3 1 āˆ’ š›¼1 1 āˆ’ š›¼1

)

y is a convex combination of two elements x2 and x3 since š›¼2 š›¼2 š›¼2 + š›¼3 + = =1 1 āˆ’ š›¼1 1 āˆ’ š›¼1 1 āˆ’ š›¼1 and š›¼2 + š›¼3 = 1 āˆ’ š›¼1 . Hence y āˆˆ š‘†. Therefore x is a convex combination of two elements x1 and š‘¦ and is also in š‘†. Proceeding in this fashion, we can show that every convex combination belongs to š‘†, that is conv š‘† āŠ† š‘†. 1.171 This is precisely analogous to Exercise 1.128. We observe that 1. conv š‘† is a convex set. 2. if š¶ is any convex set containing š‘†, then conv š‘† āŠ† š¶. Therefore, conv š‘† is the smallest convex set containing S. 1.172 Note ļ¬rst that š‘† āŠ† conv š‘† for any set š‘†. The converse for convex sets follows from Exercise 1.170. 1.173 Assume x āˆˆ conv (š‘†1 + š‘†2 ). Then, there exist numbers š›¼1 , š›¼2 , . . . , š›¼š‘› and vectors x1 , x2 , . . . , xš‘› in š‘†1 + š‘†2 such that x = š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› For every xš‘– , there exists x1š‘– āˆˆ š‘†1 and x2š‘– āˆˆ š‘†2 such that xš‘– = x1š‘– + x2š‘–

38

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

and therefore š‘› āˆ‘

x=

š›¼š‘– x1š‘– +

š‘–=1

š‘› āˆ‘

š›¼š‘– x2š‘–

š‘–=1

= x1 + x2 āˆ‘š‘› āˆ‘š‘› where x1 = š‘–=1 š›¼š‘– x1š‘– āˆˆ š‘†1 and x2 = š‘–=1 š›¼š‘– x2š‘– āˆˆ š‘†2 . Therefore x āˆˆ conv š‘†1 + conv š‘†2 . Conversely, assume that x āˆˆ conv š‘†1 + conv š‘†2 . Then x = x1 + x2 , where x1 =

š‘› āˆ‘

š›¼š‘– š‘„1š‘– ,

x1š‘– āˆˆ š‘†1

š›½š‘— š‘„2š‘— ,

x2š‘– āˆˆ š‘†2

š‘–=1

x2 =

š‘š āˆ‘ š‘—=1

and x = x1 + x2 =

š‘› āˆ‘

š›¼š‘– š‘„1š‘– +

š‘–=1

š‘š āˆ‘

š›½š‘— š‘„2š‘— āˆˆ conv (š‘†1 + š‘†2 )

š‘—=1

since x1š‘– , x2š‘— āˆˆ š‘†1 + š‘†2 for every š‘– and š‘—. 1.174 The dimension of the input requirement set š‘‰ (š‘¦) is š‘›. Its aļ¬ƒne hull is ā„œš‘› . 1.175

1. Let x = š›¼1 x1 + š›¼2 x2 + . . . + š›¼š‘› xš‘›

(1.21)

If š‘› > dim š‘† +1, the elements x1 , x2 , . . . , xš‘› āˆˆ š‘† are aļ¬ƒnely dependent (Exercise 1.157 and therefore there exist numbers š›½1 , š›½2 , . . . , š›½š‘› , not all zero, such that (Exercise 1.158) š›½1 x1 + š›½2 x2 + . . . + š›½š‘› xš‘› = 0

(1.22)

and š›½1 + š›½2 + . . . + š›½š‘› = 0 2. Combining (1.21) and (1.22) x = x āˆ’ š‘”0 š‘› š‘› āˆ‘ āˆ‘ š›¼š‘– xš‘– āˆ’ š‘” š›½š‘– xš‘– = š‘–=1

=

š‘› āˆ‘

š‘–=1

(š›¼š‘– āˆ’ š‘”š›½š‘– )xš‘–

š‘–=1

for any š‘” āˆˆ ā„œ. } { 3. Let š‘” = minš‘– š›¼š›½š‘–š‘– : š›½š‘– > 0 =

š›¼š‘— š›½š‘—

We note that āˆ™ š‘” > 0 since š›¼š‘– > 0 for every š‘–. 39

(1.23)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

āˆ™ If š›½š‘– > 0, then š›¼š‘– /š›½š‘– ā‰„ š›¼š‘— /š›½š‘— ā‰„ š‘” and therefore š›¼š‘– āˆ’ š‘”š›½š‘– ā‰„ 0 āˆ™ If š›½š‘– ā‰¤ 0 then š›¼š‘– āˆ’ š‘”š›½š‘– > 0 for every š‘” > 0. āˆ™ Therefore š›¼š‘– āˆ’ š‘”š›½š‘– ā‰„ 0 for every š‘” and āˆ™ š›¼š‘– āˆ’ š‘”š›½š‘– = 0 for š‘– = š‘—. Therefore, (1.23) represents x as a convex combination of only š‘› āˆ’ 1 points. 4. This process can be repeated until x is represented as a convex combination of at most dim š‘† + 1 elements. 1.176 Assume x is not an extreme point of š‘†. Then there exists distinct x1 and x2 in S such that x = š›¼x1 + (1 āˆ’ š›¼)x2 Without loss of generality, assume š›¼ ā‰¤ 1/2 and let y = x2 āˆ’ x. Then x + y = x2 āˆˆ š‘†. Furthermore x āˆ’ y = x āˆ’ x2 + x = 2x āˆ’ x2 = 2(š›¼x1 + (1 āˆ’ š›¼)x2 ) āˆ’ x2 = 2š›¼x1 + (1 āˆ’ 2š›¼)x2 āˆˆ š‘† since š›¼ ā‰¤ 1/2. 1.177

1. For any x = (š‘„1 , š‘„2 ) āˆˆ š¶2 , there exists some š›¼1 āˆˆ [0, 1] such that š‘„1 = š›¼1 š‘ + (1 āˆ’ š›¼1 )(āˆ’š‘) = (2š›¼1 āˆ’ 1)š‘ In fact, š›¼1 is deļ¬ned by š›¼1 = Therefore (see Figure 1.5) ) ( ( š‘„1 = š›¼1 š‘ ( ) ( š‘„1 = š›¼1 āˆ’š‘

š‘„1 + š‘ 2š‘

) āˆ’š‘ + (1 āˆ’ š›¼1 ) š‘ ( ) ) āˆ’š‘ š‘ + (1 āˆ’ š›¼1 ) āˆ’š‘ āˆ’š‘ š‘ š‘

)

(

Similarly š‘„2 = š›¼2 š‘ + (1 āˆ’ š›¼2 )(āˆ’š‘) where š›¼2 =

š‘„2 + š‘ 2š‘

Therefore, for any x āˆˆ š¶2 , ( ( ) ) ) ( š‘„1 š‘„1 š‘„1 x= + (1 āˆ’ š›¼2 ) = š›¼2 š‘„2 š‘ āˆ’š‘ ( ) ( ) š‘ āˆ’š‘ = š›¼1 š›¼2 + (1 āˆ’ š›¼1 )š›¼2 š‘ š‘ ( ) ( ) š‘ āˆ’š‘ + š›¼1 (1 āˆ’ š›¼2 ) + (1 āˆ’ š›¼1 )(1 āˆ’ š›¼2 ) āˆ’š‘ āˆ’š‘ ( ) ( ) ( ) ( ) š‘ āˆ’š‘ š‘ āˆ’š‘ = š›½1 + š›½2 + š›½3 + š›½4 š‘ š‘ āˆ’š‘ āˆ’š‘ 40

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

(š‘„1 , c) x š‘„1

(š‘„1 , -c) Figure 1.5: A cube in ā„œ2 where 0 ā‰¤ š›½š‘– ā‰¤ 1 and š›½1 + š›½2 + š›½3 + š›½4 = š›¼1 š›¼2 + (1 āˆ’ š›¼1 )š›¼2 + š›¼1 (1 āˆ’ š›¼2 ) + (1 āˆ’ š›¼1 )(1 āˆ’ š›¼2 ) = š›¼1 š›¼2 + š›¼2 āˆ’ š›¼1 š›¼2 + š›¼1 āˆ’ š›¼1 š›¼2 + 1 āˆ’ š›¼1 āˆ’ š›¼2 + š›¼1 š›¼2 =1 That is

{( š‘„ āˆˆ conv

š‘ š‘

) ( ) ( ) ( )} āˆ’š‘ š‘ āˆ’š‘ , , , š‘ āˆ’š‘ āˆ’š‘

2. (a) For any point (š‘„1 , š‘„2 , . . . , š‘„š‘›āˆ’1 , š‘) which lies on face of the cube š¶š‘› , (š‘„1 , š‘„2 , . . . , š‘„š‘›āˆ’1 ) āˆˆ š¶š‘›āˆ’1 and therefore (š‘„1 , š‘„2 , . . . , š‘„š‘›āˆ’1 ) āˆˆ conv { Ā±š‘, Ā±š‘, . . . , Ā±š‘) } āŠ† ā„œš‘›āˆ’1 so that x āˆˆ conv { (Ā±š‘, Ā±š‘, . . . , Ā±š‘, š‘) } āŠ† ā„œš‘› Similarly, any point (š‘„1 , š‘„2 , . . . , š‘„š‘›āˆ’1 , āˆ’š‘) on the opposite face lies in the convex hull of the points { (Ā±š‘, Ā±š‘, . . . , Ā±š‘, āˆ’š‘) }. (b) For any other point x = (š‘„1 , š‘„2 , . . . , š‘„š‘› ) āˆˆ š¶š‘› , let š›¼š‘› =

š‘„š‘› + š‘ 2š‘

so that š‘„š‘› = š›¼š‘› š‘ + (1 āˆ’ š›¼š‘› )(āˆ’š‘) Then

āŽ›

āŽž āŽ› āŽ› āŽž āŽž š‘„1 š‘„1 š‘„1 āŽœ š‘„2 āŽŸ āŽœ š‘„2 āŽŸ āŽœ š‘„2 āŽŸ āŽœ āŽŸ āŽœ āŽœ āŽŸ āŽŸ āŽœ āŽŸ āŽœ āŽŸ āŽŸ x = āŽœ . . . āŽŸ = š›¼š‘› āŽœ . . . āŽŸ + (1 āˆ’ š›¼š‘› ) āŽœ āŽœ ... āŽŸ āŽš‘„š‘›āˆ’1 āŽ  āŽš‘„š‘›āˆ’1 āŽ  āŽš‘„š‘›āˆ’1 āŽ  š‘„š‘› š‘ āˆ’š‘ 41

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Hence

x āˆˆ conv { (Ā±š‘, Ā±š‘, . . . , Ā±š‘) } āŠ‚ ā„œš‘› In other words š¶š‘› āŠ† conv { (Ā±š‘, Ā±š‘, . . . , Ā±š‘) } āŠ‚ ā„œš‘› 3. Let šø denote the set of points of the form { (Ā±š‘, Ā±š‘, . . . , Ā±š‘) } āŠ† ā„œš‘› . Clearly, every point in šø is an extreme point of š¶š‘› . Conversely, we have shown that š¶š‘› āŠ† conv šø. Therefore, no point x āˆˆ š¶ š‘› āˆ– šø can be an extreme point of š¶ š‘› . šø is the set of extreme points of š¶ š‘› . 4. Since š¶ š‘› is convex, and šø āŠ‚ š¶š‘› , conv šø āŠ† š¶ š‘› . Consequently, š¶ š‘› = conv šø. 1.178 Let x, y belong to š‘† āˆ– š¹ is convex. For any š›¼ āˆˆ [0, 1] āˆ™ š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘† since š‘† convex āˆ™ š›¼x + (1 āˆ’ š›¼)y āˆˆ / š¹ since š¹ is a face Thus š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘† āˆ– š¹ which is convex. 1.179

1. Trivial.

āˆŖ 2. Let {š¹š‘– } be a collection of faces of š‘† and let š¹ = š¹š‘– . Choose any x, y āˆˆ š‘†. If the line segment between x and y intersects š¹ , then āˆŖ it intersects some face š¹š‘– which implies that x, y āˆˆ š¹š‘– . Therefore, x, y āˆˆ š¹ = š¹š‘– . āˆ© 3. Let {š¹š‘– } be a collection of faces of š‘† and let š¹ = š¹š‘– . Choose any x, y āˆˆ š‘†. if the line segment between x and y intersects š¹ , then it intersects āˆŖ every face š¹š‘– which implies that x, y āˆˆ š¹š‘– for every š‘–. Therefore, x, y āˆˆ š¹ = š¹š‘– . 4. Let š”‰ be the collection of all faces of š‘†. This is partially ordered by inclusion. By āˆ© the previous result, every nonempty subcollection š”Š has a least upper bound ( š¹ āˆˆš”Š š¹ ). Hence š”‰ is a complete lattice (Exercise 1.47).

1.180 Let š‘† be a polytope. Then š‘† = conv { x1 , x2 , . . . , xš‘› }. Note that every extreme point belongs to { x1 , x2 , . . . , xš‘› }. Now choose the smallest subset whose convex hull is still š‘†, that is delete elements which can be written as convex combinations of other elements. Suppose the minimal subset is { x1 , x2 , . . . , xš‘š }. We claim that each of these elements is an extreme point of š‘†, that is { x1 , x2 , . . . , xš‘š } = šø. Assume not, that is assume that xš‘š is not an extreme point so that there exists x, y āˆˆ š‘† with xš‘š = š›¼x + (1 āˆ’ š›¼)y

with 0 < š›¼ < 1

(1.24)

Since x, y āˆˆ conv {x1 , x2 , . . . , xš‘š } x=

š‘š āˆ‘

š›¼š‘– xš‘–

y=

š‘–=1

š‘š āˆ‘

š›½xš‘–

š‘–=1

Substituting in (1.24), we can write xš‘š as a convex combination of {x1 , x2 , . . . , xš‘š }. xš‘š =

š‘š š‘š āˆ‘ āˆ‘ ( ) š›¼š›¼š‘– + (1 āˆ’ š›¼)š›½š‘– xš‘– = š›¾š‘– xš‘– š‘–=1

š‘–=1

where š›¾š‘– = š›¼š›¼š‘– + (1 āˆ’ š›¼)š›½š‘– 42

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

Note that 0 ā‰¤ š›¾š‘– ā‰¤ 1, so that either š›¾š‘š < 1 or š›¾š‘š = 1. We show that both cases lead to a contradiction. āˆ™ š›¾š‘š < 1. Then š‘šāˆ’1 āˆ‘( ) 1 š›¼š›¼š‘– + (1 āˆ’ š›¼)š›½š‘– xš‘– 1 āˆ’ š›¾š‘š š‘–=1

xš‘š =

which contradicts the minimality of the set {x1 , x2 , . . . , xš‘š }. āˆ™ š›¾š‘š = 1. Then š›¾š‘– = 0 for every š‘– āˆ•= š‘š. That is š›¼š›¼š‘– + (1 āˆ’ š›¼)š›½š‘– = 0 which implies that š›¼š‘– = š›½š‘–

for every š‘– āˆ•= š‘š

for every š‘– āˆ•= š‘š and therefore x = y.

Therefore, if {x1 , x2 , . . . , xš‘š } is a minimal spanning set, every point must be an extreme point. 1.181 Assume to the contrary that one of the vertices is not an extreme point of the simplex. Without loss of generality, assume this is x1 . Then, there exist distinct y, z āˆˆ š‘† and 0 < š›¼ < 1 such that x1 = š›¼y + (1 āˆ’ š›¼)z

(1.25)

Now, since y āˆˆ š‘†, there exist š›½1 , š›½2 , . . . , š›½š‘› such that y=

š‘› āˆ‘

š›½š‘– xš‘– ,

š‘–=1

š‘› āˆ‘

š›½š‘– = 1

š‘–=1

Similarly, there exist š›æ1 , š›æ2 , . . . , š›æš‘› such that z=

š‘› āˆ‘

š‘› āˆ‘

š›æš‘– xš‘– ,

š‘–=1

š›æš‘– = 1

š‘–=1

Substituting in (1.25) x1 = š›¼ =

š‘› āˆ‘

š›½š‘– xš‘– + (1 āˆ’ š›¼)

š‘–=1 š‘› āˆ‘

š‘› āˆ‘

š›æš‘– xš‘–

š‘–=1

( ) š›¼š›½š‘– + (1 āˆ’ š›¼)š›æš‘– xš‘–

š‘–=1

Since

āˆ‘š‘›

š‘–=1

( ) āˆ‘ āˆ‘ š›¼š›½š‘– + (1 āˆ’ š›¼)š›æš‘– = š›¼ š‘›š‘–=1 š›½š‘– + (1 āˆ’ š›¼) š‘–=1 š›æš‘– = 1 x1 =

š‘› āˆ‘ ( ) š›¼š›½š‘– + (1 āˆ’ š›¼)š›æš‘– xš‘– š‘–=1

Subtracting, this implies 0=

š‘› āˆ‘ ( ) š›¼š›½š‘– + (1 āˆ’ š›¼)š›æš‘– (xš‘– āˆ’ x1 ) š‘–=2

This establishes that the set {x2 āˆ’ x1 , x3 āˆ’ x1 , . . . , xš‘› āˆ’ x1 } is linearly dependent and therefore šø = {x1 , x2 , . . . , xš‘› } is aļ¬ƒnely dependent (Exercise 1.157). This contradicts the assumption that š‘† is a simplex. 43

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.182 Let š‘› be the dimension of a convex set š‘† in a linear space š‘‹. Then š‘› = dim aļ¬€ š‘† and there exists a set { x1 , x2 , . . . , xš‘›+1 } of aļ¬ƒnely independent points in š‘†. Deļ¬ne š‘† ā€² = conv { x1 , x2 , . . . , xš‘›+1 } Then š‘† ā€² is an š‘›-dimensional simplex contained in š‘†. 1.183 Let w = (š‘¤({1}), š‘¤({2}), . . . , š‘¤({š‘›})) denote the vector of individual worths and let š‘  denote the surplus to be distributed, that is āˆ‘ š‘  = š‘¤(š‘ ) āˆ’ š‘¤({š‘–}) š‘–āˆˆš‘

š‘  > 0 if the game is essential. For each player š‘– = 1, 2, . . . , š‘›, let yš‘– = w + š‘ eš‘– be the outcome in which player š‘– receives the entire surplus. (eš‘– is the š‘–th unit vector.) Note that { š‘¤({š‘–}) + š‘  š‘— = š‘– š‘– š‘¦š‘— = š‘¤({š‘–}) š‘— āˆ•= š‘– Each yš‘– is an imputation since š‘¦š‘—š‘– ā‰„ š‘¤({š‘—}) and āˆ‘ āˆ‘ š‘¦š‘—š‘– = š‘¤({š‘—}) + š‘  = š‘¤(š‘ ) š‘—āˆˆš‘

š‘—āˆˆš‘

Therefore {y1 , y2 , . . . , yš‘› } āŠ† š¼. Since š¼ is convex (why ?), š‘† = conv {y1 , y2 , . . . , yš‘› } āŠ† š¼. Further, for every š‘–, š‘— āˆˆ š‘ the vectors yš‘– āˆ’ yš‘— = š‘ (eš‘– āˆ’ eš‘— ) are linearly independent. Therefore š‘† is an š‘› āˆ’ 1-dimensional simplex in ā„œš‘› . For any x āˆˆ š¼ deļ¬ne š›¼š‘– =

š‘„š‘– āˆ’ š‘¤({š‘–}) š‘ 

so that š‘„š‘— = š‘¤({š‘—}) + š›¼š‘— š‘  Since x is an imputation āˆ™ š›¼š‘– ā‰„ 0 (āˆ‘ ) āˆ‘ āˆ‘ āˆ™ š‘–āˆˆš‘ š›¼š‘– = š‘–āˆˆš‘ š‘„š‘– āˆ’ š‘–āˆˆš‘ š‘¤({š‘–}) /š‘  = 1 āˆ‘ We claim that x = š‘–āˆˆš‘ š›¼š‘– yš‘– since for each š‘— = 1, 2, . . . , š‘› āˆ‘ āˆ‘ š›¼š‘– š‘¦š‘—š‘– = š›¼š‘– š‘¤({š‘—}) + š›¼š‘— š‘  š‘–āˆˆš‘

š‘–āˆˆš‘

= š‘¤({š‘—}) + š›¼š‘— š‘  = š‘„š‘— Therefore x āˆˆ conv {y1 , y2 , . . . , yš‘› } = š‘†, that is š¼ āŠ† š‘†. Since we previously showed that š‘† āŠ† š¼, we have established that š¼ = š‘†, which is an š‘› āˆ’ 1 dimensional simplex in ā„œš‘› . 44

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

š‘„2

š‘„2

š‘„2

š‘„1

š‘„1

š‘„1

1. A non-convex cone

2. A convex set

3. A convex cone

Figure 1.6: A cone which is not convex, a convex set and a convex cone 1.184 See Figure 1.6. 1.185 Let x = (š‘„1 , š‘„2 , . . . , š‘„š‘› ) belong to ā„œš‘›+ , which means that š‘„š‘– ā‰„ 0 for every š‘–. For every š›¼ > 0 š›¼x = (š›¼x1 , š›¼x2 , . . . , š›¼xš‘› ) and š›¼š‘„š‘– ā‰„ 0 for every š‘–. Therefore š›¼x āˆˆ ā„œš‘›+ . ā„œš‘›+ is a cone in ā„œš‘› . 1.186 Assume š›¼x + š›½y āˆˆ š‘† for every x, y āˆˆ š‘† and š›¼, š›½ āˆˆ ā„œ+

(1.26)

Letting š›½ = 0, this implies that š›¼x āˆˆ š‘† for every x āˆˆ š‘† and š›¼ āˆˆ ā„œ+ so that š‘† is a cone. To show that š‘† is convex, let x and y be any two elements in š‘†. For any š›¼ āˆˆ [0, 1], (1.26) implies that š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘† Therefore š‘† is convex. Conversely, assume that š‘† is a convex cone. For any š›¼, š›½ āˆˆ ā„œ+ and x, y āˆˆ š‘† š›½ š›¼ x+ yāˆˆš‘† š›¼+š›½ š›¼+š›½ and therefore š›¼x + š›½y āˆˆ š‘† 1.187 Assume š‘† satisļ¬es 1. š›¼š‘† āŠ† š‘† for every š›¼ ā‰„ 0 2. š‘† + š‘† āŠ† š‘†

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

By (1), š‘† is a cone. To show that it is convex, let x and y belong to š‘†. By (1), š›¼x and (1 āˆ’ š›¼)y belong to š‘†, and therefore š›¼x + (1 āˆ’ š›¼)y belongs to š‘† by (2). š‘† is convex. Conversely, assume that š‘† is a convex cone. Then š›¼š‘† āŠ† š‘†

for every š›¼ ā‰„ 0

Let x and y be any two elements in š‘†. Since š‘† is convex, š‘§ = š›¼ 12 x + (1 āˆ’ š›¼) 12 y āˆˆ š‘† and since it is a cone, 2š‘§ = x + y āˆˆ š‘†. Therefore š‘† +š‘† āŠ†š‘† 1.188 We have to show that š‘Œ is convex cone. By assumption, š‘Œ is convex. To show that š‘Œ is a cone, let y be any production plan in š‘Œ . By convexity š›¼y = š›¼y + (1 āˆ’ š›¼)0 āˆˆ š‘Œ for every 0 ā‰¤ š›¼ ā‰¤ 1 Repeated use of additivity ensures that š›¼y āˆˆ š‘Œ for every š›¼ = 1, 2, . . . Combining these two conclusions implies that š›¼y āˆˆ š‘Œ for every š›¼ ā‰„ 0 1.189 Let š’® āŠ‚ š’¢ š‘ denote the set of all superadditive games. Let š‘¤1 , š‘¤2 āˆˆ š‘† be two superadditive games. Then, for all distinct coalitions š‘†, š‘‡ āŠ‚ š‘ with š‘† āˆ© š‘‡ = āˆ… š‘¤1 (š‘† āˆŖ š‘‡ ) ā‰„ š‘¤1 (š‘†) + š‘¤1 (š‘‡ ) š‘¤2 (š‘† āˆŖ š‘‡ ) ā‰„ š‘¤2 (š‘†) + š‘¤2 (š‘‡ ) Adding (š‘¤1 + š‘¤2 )(š‘† āˆŖ š‘‡ ) = š‘¤1 (š‘† āˆŖ š‘‡ ) + š‘¤2 (š‘† āˆŖ š‘‡ ) ā‰„ š‘¤1 (š‘†) + š‘¤2 (š‘†) + š‘¤1 (š‘‡ ) + š‘¤2 (š‘‡ ) = (š‘¤1 + š‘¤2 )(š‘†) + (š‘¤1 + š‘¤2 )(š‘‡ ) so that š‘¤1 + š‘¤2 is superadditive. Similarly, we can show that š›¼š‘¤1 is superadditive for all š›¼ āˆˆ ā„œ+ . Hence š’® is a convex cone in š’¢ š‘ . āˆ©š‘› 1.190 Let x belong to š‘–=1 š‘†š‘– . Then x āˆˆ š‘† āˆ©š‘– for every š‘–. Since each š‘†š‘– is a cone, š›¼x āˆˆ š‘†š‘– for every š›¼ ā‰„ 0 and therefore š›¼x āˆˆ š‘›š‘–=1 š‘†š‘– . Let š‘† = š‘†1 + š‘†2 + ā‹… ā‹… ā‹… + š‘†š‘› and assume x belongs to š‘†. Then there exist xš‘– āˆˆ š‘†š‘– , š‘– = 1, 2, . . . , š‘› such that x = x1 + x2 + ā‹… ā‹… ā‹… + xš‘› For any š›¼ ā‰„ 0 š›¼x = š›¼(x1 + x2 + ā‹… ā‹… ā‹… + xš‘› ) = š›¼x1 + š›¼x2 + ā‹… ā‹… ā‹… + š›¼xš‘› āˆˆ š‘† since š›¼xš‘– āˆˆ š‘†š‘– for every š‘–.

46

Solutions for Foundations of Mathematical Economics 1.191

c 2001 Michael Carter āƒ All rights reserved

1. Suppose that y āˆˆ š‘Œ . Then, there exist š›¼, š›¼2 , . . . , š›¼8 ā‰„ 0 such that y=

8 āˆ‘

š›¼š‘– yš‘–

š‘–=1

and for the ļ¬rst commodity 8 āˆ‘

y1 =

š›¼š‘– š‘¦š‘–1

š‘–=1

If y āˆ•= 0, at least one of the š›¼š‘– > 0 and hence y1 < 0 since š‘¦š‘–1 < 0 for š‘– = 1, 2, . . . , 8. 2. Free disposal requires that y āˆˆ š‘Œ, yā€² ā‰¤ y =ā‡’ yā€² āˆˆ š‘Œ . Consider the production plan yā€² = (āˆ’2, āˆ’2, āˆ’2, āˆ’2). Note that yā€² ā‰¤ y3 and yā€² ā‰¤ y6 . Suppose that yā€² āˆˆ š‘Œ . Then there exist š›¼1 , š›¼2 , . . . , š›¼8 ā‰„ 0 such that y=

8 āˆ‘

š›¼š‘– yš‘–

š‘–=1

For the third commodity (component), we have 4š›¼1 + 3š›¼2 + 3š›¼3 + 3š›¼4 + 12š›¼5 āˆ’ 2š›¼6 + 5š›¼8 = āˆ’2

(1.27)

and for the fourth commodity 2š›¼2 āˆ’ 1š›¼3 + 1š›¼4 + 5š›¼6 + 10š›¼7 āˆ’ 2š›¼8 = āˆ’2

(1.28)

Adding to (1.28) to (1.27) gives 4š›¼1 + 5š›¼2 + 2š›¼3 + 4š›¼4 + 12š›¼5 + 3š›¼6 + 10š›¼7 + 3š›¼8 = āˆ’4 / š‘Œ. which is impossible given that š›¼š‘– ā‰„ 0. Therefore, we conclude that yā€² āˆˆ 3. y2 = (āˆ’7, āˆ’9, 3, 2) ā‰„ (āˆ’8, āˆ’13, 3, 1) = y4 3y1 = (āˆ’9, āˆ’18, 12, 0) ā‰„ (āˆ’11, āˆ’19, 12, 0) = y5 y7 = (āˆ’8, āˆ’5, 0, 10) ā‰„ (āˆ’8, āˆ’6, āˆ’4, 10) = 2y6 2y3 = (āˆ’2, āˆ’4, 6, āˆ’2) ā‰„ (āˆ’2, āˆ’4, 5, āˆ’2) = y8 4. 2y3 + y7 = (āˆ’2, āˆ’4, 6, āˆ’2) + (āˆ’8, āˆ’5, 0, 10) = (āˆ’10, āˆ’9, 6, 8) ā‰„ (āˆ’14, āˆ’18, 6, 4) = 2y2 20y3 + 2y7 = 20(āˆ’1, āˆ’2, 3, āˆ’1) + 2(āˆ’8, āˆ’5, 0, 10) = (āˆ’20, āˆ’40, 60, āˆ’20) + (āˆ’16, āˆ’10, 0, 20) = (āˆ’36, āˆ’50, 60, 0) ā‰„ (āˆ’45, āˆ’90, 60, 0) = 15y1 47

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

5. Eļ¬€(š‘Œ ) = cone { y3 , y7 }. 1.192 This is precisely analogous to Exercise 1.128. We observe that 1. cone š‘† is a convex cone. 2. if š¶ is any convex cone containing š‘†, then conv š‘† āŠ† š¶. Therefore, cone š‘† is the smallest convex cone containing S. 1.193 For any set š‘†, š‘† āŠ† cone š‘†. If š‘† is a convex cone, Exercise 1.186 implies that cone š‘† āŠ† š‘†. 1.194

1. If š‘› > š‘š = dim cone š‘† = dim lin š‘†, the elements x1 , x2 , . . . , xš‘› āˆˆ š‘† are linearly dependent and therefore there exist numbers š›½1 , š›½2 , . . . , š›½š‘› , not all zero, such that (Exercise 1.134) š›½1 x1 + š›½2 x2 + . . . + š›½š‘› xš‘› = 0

(1.29)

2. Combining (1.14) and (1.29) x = x āˆ’ š‘”0 š‘› š‘› āˆ‘ āˆ‘ = š›¼š‘– xš‘– āˆ’ š‘” š›½š‘– xš‘– š‘–=1

=

š‘› āˆ‘

š‘–=1

(š›¼š‘– āˆ’ š‘”š›½š‘– )xš‘–

(1.30)

š‘–=1

for any š‘” āˆˆ ā„œ. { } 3. Let š‘” = minš‘– š›¼š›½š‘–š‘– : š›½š‘– > 0 =

š›¼š‘— š›½š‘—

We note that āˆ™ š‘” > 0 since š›¼š‘– > 0 for every š‘–. āˆ™ If š›½š‘– > 0, then š›¼š‘– /š›½š‘– ā‰„ š›¼š‘— /š›½š‘— ā‰„ š‘” and therefore š›¼š‘– āˆ’ š‘”š›½š‘– ā‰„ 0. āˆ™ If š›½š‘– ā‰¤ 0 then š›¼š‘– āˆ’ š‘”š›½š‘– > 0 for every š‘” > 0. āˆ™ Therefore š›¼š‘– āˆ’ š‘”š›½š‘– ā‰„ 0 for every š‘” and āˆ™ š›¼š‘– āˆ’ š‘”š›½š‘– = 0 for š‘– = š‘—. Therefore, (1.30) represents x as a nonnegative combination of only š‘› āˆ’ 1 points. 4. This process can be repeated until x is represented as a convex combination of at most š‘š points. 1.195

1. The aļ¬ƒne hull of š‘†Ėœ is parallel to the aļ¬ƒne hull of š‘†. Therefore ( Since

0 0

)

dim š‘† = dim aļ¬€ š‘† = dim aļ¬€ š‘†Ėœ Ėœ āˆˆ / aļ¬€ š‘†, dim cone š‘†Ėœ = dim aļ¬€ š‘†Ėœ + 1 = dim š‘† + 1

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

( ) x 2. For every x āˆˆ conv š‘†, āˆˆ conv š‘†Ėœ and there exist (Exercise 1.194) š‘š + 1 1 ( ) xš‘– points āˆˆ š‘†Ėœ such that 1 (

x 1

)

( āˆˆ conv {

x1 1

) ( ) ( ) x2 xš‘š+1 , ,... } 1 1

This implies that x āˆˆ conv { x1 , x2 , . . . , xš‘š+1 } with x1 , x2 , . . . , xš‘š+1 āˆˆ š‘†. 1.196 A subsimplex with precisely one distinguished face is completely labeled. Suppose a subsimplex has more than one distinguished face. This means that it has vertices labeled 1, 2, . . . , š‘›. Since it has š‘› + 1 vertices, one of these labels must be repeated (twice). The distinguished faces lie opposite the repeated vertices. There are precisely two distinguished faces. 1.197

1. šœŒ(x, y) = āˆ„x āˆ’ yāˆ„ ā‰„ 0.

2. šœŒ(x, y) = āˆ„x āˆ’ yāˆ„ = 0 if and only if x āˆ’ y = 0, that is x = y. 3. Property (3) ensures that āˆ„āˆ’xāˆ„ = āˆ„xāˆ„ and therefore āˆ„x āˆ’ yāˆ„ = āˆ„y āˆ’ xāˆ„ so that šœŒ(x, y) = āˆ„x āˆ’ yāˆ„ = āˆ„y āˆ’ xāˆ„ = šœŒ(y, x) 4. For any z āˆˆ š‘‹ šœŒ(x, y) = āˆ„x āˆ’ yāˆ„ = āˆ„x āˆ’ z + z āˆ’ yāˆ„ ā‰¤ āˆ„x āˆ’ zāˆ„ + āˆ„z āˆ’ yāˆ„ = šœŒ(x, z) + šœŒ(z, y) Therefore šœŒ(x, y) = āˆ„x āˆ’ yāˆ„ satisļ¬es the properties required of a metric. 1.198 Clearly āˆ„xāˆ„āˆž ā‰„ 0 and āˆ„xāˆ„āˆž = 0 if and only if x = 0. Thirdly š‘›

š‘›

š‘–=1

š‘–=1

āˆ„š›¼xāˆ„ = max āˆ£š›¼š‘„š‘– āˆ£ = āˆ£š›¼āˆ£ max āˆ£š‘„š‘– āˆ£ = āˆ£š›¼āˆ£ āˆ„xāˆ„ To prove the triangle inequality, we note that for any š‘„š‘– , š‘¦š‘– āˆˆ ā„œ max(š‘„š‘– + š‘¦š‘– ) ā‰¤ max š‘„š‘– + max š‘¦š‘– Therefore š‘›

š‘›

š‘›

š‘–=1

š‘–=1

š‘–=1

āˆ„xāˆ„ = max(š‘„š‘– + š‘¦š‘– ) ā‰¤ max š‘„š‘– + max š‘¦š‘– = āˆ„xāˆ„ + āˆ„yāˆ„ 1.199 Suppose that producing one unit of good 1 requires two units of good 2 and three units of good 3. The production plan is (1, āˆ’2, āˆ’3) and the average net output, āˆ’2, is negative. A norm is required to be nonnegative. Moreover, theāˆ‘quantities of inputs š‘› and outputs may balance out yielding a zero average. That is, ( š‘–=1 š‘¦š‘– )/š‘› = 0 does not imply that š‘¦š‘– = 0 for all š‘–.

49

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 1.200 āˆ„xāˆ„ āˆ’ āˆ„yāˆ„ = āˆ„x āˆ’ y + yāˆ„ āˆ’ āˆ„yāˆ„ ā‰¤ āˆ„x āˆ’ yāˆ„ + āˆ„yāˆ„ āˆ’ āˆ„yāˆ„ = āˆ„x āˆ’ yāˆ„ 1.201 Using the previous exercise āˆ„xš‘› āˆ„ āˆ’ āˆ„xāˆ„ ā‰¤ āˆ„xš‘› āˆ’ xāˆ„ ā†’ 0

1.202 First note that each term xš‘› + yš‘› āˆˆ š‘‹ by linearity. Similarly, x + y āˆˆ š‘‹. Fix some šœ– > 0. There exists some š‘x such that āˆ„xš‘› āˆ’ xāˆ„ < šœ– for all š‘› ā‰„ š‘x . Similarly, there exists some š‘y such that āˆ„yš‘› āˆ’ yāˆ„ < šœ– for all š‘› ā‰„ š‘y . For all š‘› ā‰„ max{ š‘x , š‘y }, āˆ„(xš‘› + yš‘› ) āˆ’ (x + y)āˆ„ = āˆ„(xš‘› āˆ’ x) + (yš‘› āˆ’ y)āˆ„ ā‰¤ āˆ„xš‘› āˆ’ xāˆ„ + āˆ„yš‘› āˆ’ yāˆ„ 0, there exists some š‘ such that āˆ„zš‘› āˆ’ zš‘š āˆ„ = max{ āˆ„xš‘› āˆ’ xš‘š āˆ„ , āˆ„yš‘› āˆ’ yš‘š āˆ„ } < šœ– for every š‘›, š‘š ā‰„ š‘ . This implies that (xš‘› ) and (yš‘› ) are Cauchy sequences in š‘‹ and š‘Œ respectively. Since š‘‹ and š‘Œ are complete, both sequences converge. That is, there exists x āˆˆ š‘‹ and y āˆˆ š‘Œ such that āˆ„xš‘› āˆ’ xāˆ„ ā†’ 0 and āˆ„yš‘› āˆ’ yāˆ„ ā†’ 0. In other words, given šœ– > 0 there exists š‘ such that āˆ„xš‘› āˆ’ xāˆ„ < šœ– and āˆ„yš‘› āˆ’ yāˆ„ < šœ– for every š‘› ā‰„ š‘ . Let z = (x, y). Then, for every š‘› ā‰„ š‘ āˆ„zš‘› āˆ’ zāˆ„ = max{ āˆ„xš‘› āˆ’ xāˆ„ , āˆ„yš‘› āˆ’ yāˆ„ } < šœ– zš‘› ā†’ z. 1.210

1. By assumption, for every š‘š = 1, 2, . . . , there exists a point yš‘š such that ( š‘› ) 1 āˆ‘ āˆ„yāˆ„ < āˆ£š›¼š‘– āˆ£ š‘š š‘–=1 where y = š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› Let š‘ š‘š =

āˆ‘š‘›

š‘–=1

āˆ£š›¼š‘– āˆ£. By assumption š‘ š‘š > š‘š āˆ„yš‘š āˆ„ ā‰„ 0. Deļ¬ne xš‘š =

1 š‘š y š‘ š‘š

Then xš‘š = š›½1š‘š x1 + š›½2š‘š x2 + ā‹… ā‹… ā‹… + š›½š‘›š‘š xš‘› āˆ‘š‘› 1 š‘š š‘š š‘š where š›½š‘–š‘š = š›¼š‘š š‘– /š‘  , š‘–=1 āˆ£š›½š‘– āˆ£ = 1 and āˆ„x āˆ„ < š‘š for every š‘› = 1, 2, . . . . š‘š Consequently āˆ„x āˆ„ ā†’ 0. 51

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

āˆ‘š‘› 2. Since š‘–=1 āˆ£š›½š‘–š‘š āˆ£ = 1, āˆ£š›½š‘–š‘š āˆ£ ā‰¤ 1 for every š‘–. Consequently, for every coordinate š‘–, the sequence (š›½š‘–š‘š ) is bounded. By the Bolzano-Weierstrass theorem (Exercise 1.119), the sequence (š›½1š‘š ) has a convergent subsequence with š›½1š‘š ā†’ š›½1 . Let xš‘š,1 denote the corresponding subsequence of xš‘š . Similarly, š›½2š‘š,1 has a subsequence converging to š›½2 . Let (xš‘š,2 ) denote the corresponding subsequence of (xš‘š ). Proceeding coordinate by coordinate, we obtain a subsequence (xš‘š,š‘› ) where each term is xš‘š,š‘› = š›½ š‘š,š‘› x1 + š›½ š‘š,š‘› x2 + ā‹… ā‹… ā‹… + š›½ š‘š,š‘› xš‘› and each coeļ¬ƒcient converges š›½š‘–š‘š,š‘› ā†’ š›½š‘– . Let x = š›½1 x1 + š›½2 x2 + ā‹… ā‹… ā‹… + š›½2 xš‘› Then xš‘š,š‘› ā†’ x (Exercise 1.202). āˆ‘š‘› āˆ‘š‘› š‘š 3. Since š‘–=1 āˆ£š›½š‘– āˆ£ = 1 for every š‘š, š‘–=1 āˆ£š›½š‘– āˆ£ = 1. Consequently, at least one of the coeļ¬ƒcients š›½š‘– āˆ•= 0. Since x1 , x2 , . . . , xš‘› are linearly independent, x āˆ•= 0 (Exercise 1.133) and therefore āˆ„xāˆ„ āˆ•= 0. But (xš‘š,š‘› ) is a subsequence of (xš‘š ). This contradicts the earlier conclusion (part 1) that āˆ„xš‘š āˆ„ ā†’ 0. 1.211

1. Let š‘‹ be a normed linear space š‘‹ of dimension š‘› and let { x1 , x2 , . . . , xš‘› } be a basis for š‘‹. Let (xš‘š ) be a Cauchy sequence in š‘‹. Each term xš‘š has a unique representation š‘š š‘š xš‘š = š›¼š‘š 1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘›

We show that each of the sequences š›¼š‘š š‘– is a Cauchy sequence in ā„œ. Since xš‘š is a Cauchy sequence, for every šœ– > 0 there exists an š‘ such that āˆ„xš‘š āˆ’ xš‘Ÿ āˆ„ < šœ– for all š‘š, š‘Ÿ ā‰„ š‘ . Using Lemma 1.1, there exists š‘ > 0 such that   š‘› š‘›  āˆ‘ āˆ‘   š‘š š‘Ÿ š‘š š‘Ÿ š‘ āˆ£š›¼š‘– āˆ’ š›¼š‘– āˆ£ ā‰¤  (š›¼š‘– āˆ’ š›¼š‘– )xš‘–  = āˆ„xš‘š āˆ’ xš‘Ÿ āˆ„ < šœ–   š‘–=1

š‘–=1

for all š‘š, š‘Ÿ ā‰„ š‘ . Dividing by š‘ > 0 we get for every š‘– š‘Ÿ āˆ£š›¼š‘š š‘– āˆ’ š›¼š‘– āˆ£ ā‰¤

š‘› āˆ‘

š‘Ÿ āˆ£š›¼š‘š š‘– āˆ’ š›¼š‘– āˆ£ <

š‘–=1

šœ– š‘

š›¼š‘š š‘–

is a Cauchy sequence in ā„œ. Since ā„œ is complete, each Thus each sequence sequence converges to some limit š›¼š‘– āˆˆ ā„œ. 2. Let x = š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› Then x āˆˆ š‘‹ and

  š‘› š‘›  āˆ‘ āˆ‘   āˆ„xš‘š āˆ’ xāˆ„ =  (š›¼š‘š āˆ£š›¼š‘š š‘– āˆ’ š›¼š‘– )xš‘–  ā‰¤ š‘– āˆ’ š›¼š‘– āˆ£ āˆ„xš‘– āˆ„   š‘–=1

š‘–=1

š‘š š‘š Since š›¼š‘š š‘– ā†’ š›¼š‘– for every š‘–, āˆ„x āˆ’ xāˆ„ ā†’ 0 which implies that x ā†’ x.

3. Since (xš‘š ) was an arbitrary Cauchy sequence, we have shown that every Cauchy sequence in š‘‹ converges. Hence š‘‹ is complete. 52

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.212 Let š‘† be an open set according to the āˆ„ā‹…āˆ„š‘Ž and let x0 be a point in š‘†. Since š‘† is open, it contains an open ball in the āˆ„ā‹…āˆ„š‘Ž topology about x0 , namely šµš‘Ž (x0 , š‘Ÿ) = { x āˆˆ š‘‹ : āˆ„x āˆ’ x0 āˆ„š‘Ž < š‘Ÿ } āŠ† š‘† Let šµš‘ (x0 , š‘Ÿ) = { x āˆˆ š‘‹ : āˆ„x āˆ’ x0 āˆ„š‘ < š‘Ÿ } be the open ball about x0 in the āˆ„ā‹…āˆ„š‘ topology. The condition (1.15) implies that šµš‘ (x0 , š‘Ÿ) āŠ† šµš‘Ž (x0 , š‘Ÿ) āŠ† š‘† and therefore x0 āˆˆ šµš‘ (x0 , š‘Ÿ) āŠ‚ š‘† š‘† is open in the āˆ„ā‹…āˆ„š‘ topology. Similarly, any š‘† open in the āˆ„ā‹…āˆ„š‘ topology is open in the āˆ„ā‹…āˆ„š‘Ž topology. 1.213 Let š‘‹ be a normed linear space of dimension š‘›. and let { x1 , x2 , . . . , xš‘› } be a basis for š‘‹. Let āˆ„ā‹…āˆ„š‘Ž and āˆ„ā‹…āˆ„š‘ be two norms on š‘‹. Every x āˆˆ š‘‹ has a unique representation x = š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› Repeated application of the triangle inequality gives āˆ„xāˆ„š‘Ž = āˆ„š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› āˆ„š‘Ž š‘› āˆ‘ ā‰¤ āˆ„š›¼š‘– xš‘– āˆ„š‘Ž š‘–=1

=

š‘› āˆ‘

āˆ£š›¼š‘– āˆ£ āˆ„xš‘– āˆ„š‘Ž

š‘–=1 š‘› āˆ‘

ā‰¤š‘˜

āˆ£š›¼š‘– āˆ£

š‘–=1

where š‘˜ = maxš‘– āˆ„xš‘– āˆ„. By Lemma 1.1, there is a positive constant š‘ such that š‘› āˆ‘ š‘–=1

āˆ£š›¼š‘– āˆ£ ā‰¤ āˆ„xāˆ„š‘ /š‘

Combining these two inequalities, we have āˆ„xāˆ„š‘Ž ā‰¤ š‘˜ āˆ„xāˆ„š‘ /š‘ Setting š“ = š‘/š‘˜ > 0, we have shown š“ āˆ„xāˆ„š‘Ž ā‰¤ āˆ„xāˆ„š‘ The other inequality in (1.15) is obtained by interchanging the roles of āˆ„ā‹…āˆ„š‘Ž and āˆ„ā‹…āˆ„š‘ . 1.214 Assume xš‘› ā†’ x = (š‘„1 , š‘„2 , . . . , š‘„š‘› ). Then, for every šœ– > 0, there exists some š‘ such that āˆ„xš‘› āˆ’ xāˆ„āˆž < šœ–. Therefore, for š‘– = 1, 2, . . . , š‘› āˆ£š‘„š‘›š‘– āˆ’ š‘„š‘– āˆ£ ā‰¤ max āˆ£š‘„š‘›š‘– āˆ’ š‘„š‘– āˆ£ = āˆ„xš‘› āˆ’ xāˆ„āˆž < šœ– š‘–

Therefore š‘„š‘›š‘– ā†’ xš‘– .

53

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Conversely, assume that (xš‘› ) is a sequence in ā„œš‘› with š‘„š‘›š‘– ā†’ š‘„š‘– for every coordinate š‘–. Choose some šœ– > 0. For every š‘–, there exists some integer š‘š‘– such that āˆ£š‘„š‘›š‘– āˆ’ š‘„š‘– āˆ£ < šœ– for every š‘› ā‰„ š‘š‘– Let š‘ = maxš‘– { š‘1 , š‘2 , . . . , š‘š‘› }. Then āˆ£š‘„š‘›š‘– āˆ’ š‘„š‘– āˆ£ < šœ– for every š‘› ā‰„ š‘ and āˆ„xš‘› āˆ’ xāˆ„āˆž = max āˆ£š‘„š‘›š‘– āˆ’ š‘„š‘– āˆ£ < šœ– for every š‘› ā‰„ š‘ š‘–

š‘›

That is, x ā†’ x. A similar proof can be given using the Euclidean norm āˆ„ā‹…āˆ„2 , but it is slightly more complicated. This illustrates an instance where the sup norm is more tractable. 1.215

1. Let š‘† āŠ† š‘‹ be closed and bounded and let xš‘š be a sequence in š‘†. Every term xš‘š has a representation š‘› āˆ‘

xš‘š =

š›¼š‘š š‘– xš‘–

š‘–=1

Since š‘† is bounded, so is xš‘š . That is, there exists š‘˜ such that āˆ„xš‘š āˆ„ ā‰¤ š‘˜ for all š‘š. Applying Lemma 1.1, there is a positive constant š‘ such that š‘

š‘› āˆ‘

āˆ£š›¼š‘– āˆ£ ā‰¤ āˆ„xš‘š āˆ„ ā‰¤ š‘˜

š‘–=1

Hence, for every š‘–, the sequence of scalars š›¼š‘›š‘– is bounded. 2. By the Bolzano-Weierstrass theorem (Exercise 1.119), the sequence š›¼š‘š 1 has a convergent subsequence with limit š›¼1 . Let š‘„š‘š (1) be the corresponding subsequence of xš‘š . š‘š 3. Similarly, š‘„š‘š (1) has a subsequence for which the corresponding scalars š›¼2 converge to š›¼2 . Repeating this process š‘› times (this is were ļ¬niteness is important), we deduce the existence of a subsequence š‘„š‘š (š‘›) whose scalars converge to (š›¼1 , š›¼2 , . . . , š›¼š‘› ).

4. Let x=

š‘› āˆ‘

š›¼š‘– xš‘–

š‘–=1 š‘š š‘š Since š›¼š‘š š‘– ā†’ š›¼š‘– for every š‘–, āˆ„x āˆ’ xāˆ„ ā†’ 0 which implies that x ā†’ x.

5. Since š‘† is closed, x āˆˆ š‘†. 6. We have shown that every sequence in š‘† has a subsequence which converges in š‘†. š‘† is compact. 1.216 Let x and y belong to šµ = { x : āˆ„xāˆ„ < 1 }, the unit ball in the normed linear space š‘‹. Then āˆ„xāˆ„ , āˆ„yāˆ„ < 1. By the triangle inequality āˆ„š›¼x + (1 āˆ’ š›¼)yāˆ„ ā‰¤ š›¼ āˆ„xāˆ„ + (1 āˆ’ š›¼) āˆ„yāˆ„ ā‰¤ š›¼ + (1 āˆ’ š›¼) = 1 Hence š›¼x + (1 āˆ’ š›¼)y āˆˆ šµ. 54

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.217 If int š‘† is empty, it is trivially convex. Therefore, assume int š‘† āˆ•= āˆ… and let x, y āˆˆ int š‘†. We must show that z = š›¼x + (1 āˆ’ š›¼)y āˆˆ int š‘†. Since x, y āˆˆ int š‘†, there exists some š‘Ÿ > 0 such that the open balls šµ(x, š‘Ÿ) and šµ(y, š‘Ÿ) are both contained in int š‘†. Let w be any vector with āˆ„wāˆ„ < š‘Ÿ. The point z + w = š›¼(x + w) + (1 āˆ’ š›¼)(y + w) āˆˆ š‘† since x + w āˆˆ šµ(x, š‘Ÿ) āŠ‚ š‘† and y + w āˆˆ šµ(y, š‘Ÿ) āŠ‚ š‘† and š‘† is convex. Hence z is an interior point of š‘†. Similarly, if š‘† is empty, it is trivially convex. Therefore, assume š‘† āˆ•= āˆ… and let x, y āˆˆ š‘†. Choose some š›¼. We must show that š‘§ = š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘†. There exist sequences (xš‘› ) and (yš‘› ) in š‘† which converge to x and y respectively (Exercise 1.105). Since š‘† is convex, the sequence (š›¼xš‘› + (1 āˆ’ š›¼)yš‘› ) lies in š‘† and moreover converges to š›¼x + (1 āˆ’ š›¼)y = z (Exercise 1.202). Therefore š‘§ is the limit of a sequence in š‘† and hence š‘§ āˆˆ š‘†. Therefore, š‘† is convex. ĀÆ = š›¼x1 + (1 āˆ’ š›¼)x2 for some š›¼ āˆˆ (0, 1). Since x1 āˆˆ š‘†, 1.218 Let x x1 āˆˆ š‘† + š‘Ÿšµ š›¼x1 āˆˆ š›¼(š‘† + š‘Ÿšµ) ĀÆ of radius š‘Ÿ is The open ball about x ĀÆ + š‘Ÿšµ šµ(ĀÆ x, š‘Ÿ) = x = š›¼x1 + (1 āˆ’ š›¼)x2 + š‘Ÿšµ āŠ† š›¼(š‘† + š‘Ÿšµ) + (1 āˆ’ š›¼)x2 + š‘Ÿšµ = š›¼š‘† + (1 āˆ’ š›¼)x2 + (1 + š›¼)š‘Ÿšµ ( ) 1+š›¼ = š›¼š‘† + (1 āˆ’ š›¼) x2 + š‘Ÿšµ 1āˆ’š›¼ Since x2 āˆˆ int š‘† x2 +

( ) 1+š›¼ 1+š›¼ š‘Ÿšµ = šµ x2 , š‘Ÿ āŠ†š‘† 1āˆ’š›¼ 1āˆ’š›¼

for suļ¬ƒciently small š‘Ÿ. For such š‘Ÿ šµ(ĀÆ x, š‘Ÿ) āŠ† š›¼š‘† + (1 āˆ’ š›¼)š‘† =š‘† ĀÆ āˆˆ int š‘†. by Exercise 1.168. Therefore x 1.219 It is easy to show that š‘†āŠ†

āˆ©

š‘†š‘–

š‘–āˆˆš¼

To show the converse, choose any x āˆˆ š‘† and let x0 āˆˆ š‘†š‘– for every š‘– āˆˆ š¼. By Exercise 1.218, š›¼x + (1 āˆ’ š›¼)x0 āˆˆ š‘†š‘– for all 0 < š›¼ < 1. This implies that š›¼x + (1 āˆ’ š›¼)x0 āˆˆ āˆ©š‘–āˆˆš¼ š‘†š‘– = š‘† for all 0 < š›¼ < 1, and therefore that x0 = limš›¼ā†’0 š›¼x + (1 āˆ’ š›¼)x0 āˆˆ š‘†. 1.220 Assume that x āˆˆ int š‘†. Then, there exists some š‘Ÿ such that šµ(x, š‘Ÿ) = x + š‘Ÿšµ āŠ† š‘† 55

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Let y be any element in the unit ball šµ. Then āˆ’y āˆˆ šµ and x1 = x + š‘Ÿy āˆˆ š‘† x2 = x āˆ’ š‘Ÿy āˆˆ š‘† so that x=

1 1 x1 + x2 2 2

x is not an extreme point. We have shown that no interior point is an extreme point; hence every extreme point must be a boundary point. 1.221 We showed in Exercise 1.220 that ext(š‘†) āŠ† b(š‘†). To show the converse, assume that x is a boundary point which is not an extreme point. That is, there exist x1 , x2 āˆˆ š‘† such that x = š›¼x1 + (1 āˆ’ š›¼)x2

0 0 and š‘„š‘– < 1 1.229 Let š‘› = dim š‘†. By Exercise 1.182, š‘† contains a simplex š‘† š‘› of the same dimension. That is, there exist š‘› vertices v1 , v2 , . . . , vš‘› such that š‘† š‘› = conv { v1 , v2 , . . . , vš‘› } { = š›¼1 v1 + š›¼2 v2 + ā‹… ā‹… ā‹… + š›¼š‘› vš‘› : š›¼1 , š›¼2 , . . . , š›¼š‘› ā‰„ 0, š›¼1 + š›¼2 + . . . + š›¼š‘› = 1

}

Analogous to the previous part, the relative interior of š‘† š‘› is ri š‘† š‘› = conv { v1 , v2 , . . . , vš‘› } { = š›¼1 v1 + š›¼2 v2 + ā‹… ā‹… ā‹… + š›¼š‘› vš‘› : š›¼1 , š›¼2 , . . . , š›¼š‘› > 0, š›¼1 + š›¼2 + . . . + š›¼š‘› = 1

}

which is nonempty. Note, the proposition is trivially true for a set containing a single point (š‘› = 0), since this point is the whole aļ¬ƒne space. 1.230 If int š‘† āˆ•= āˆ…, then aļ¬€ š‘† = š‘‹ and ri š‘† = int š‘†. The converse follows from Exercise 1.229. 1.231 Since š‘š > inf

xāˆˆš‘‹

š‘› āˆ‘ š‘–=1

58

š‘š‘– š‘„š‘–

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics there exists some x āˆˆ š‘‹ such that š‘› āˆ‘

š‘š‘– š‘„š‘– ā‰¤ š‘š

š‘–=1

Therefore x āˆˆ š‘‹(p, š‘š) which is nonempty. Let š‘Ė‡ = minš‘– š‘š‘– be the lowest price of the š‘› goods. Then š‘‹(p, š‘š) āŠ† šµ(0, š‘š/Ė‡ š‘) and so is bounded. (That is, no component of an aļ¬€ordable bundle can contain more than š‘š/Ė‡ š‘ units.) To show that š‘‹(p, š‘š) is closed, let (xš‘› ) be a sequence of consumption bundles in š‘‹(p, š‘š). Since š‘‹(p, š‘š) is bounded, xš‘› ā†’ x āˆˆ š‘‹. Furthermore š‘1 š‘„š‘›1 + š‘2 š‘„š‘›2 + ā‹… ā‹… ā‹… + š‘š‘› š‘„š‘›š‘› ā‰¤ š‘š for every š‘› This implies that š‘1 š‘„1 + š‘2 š‘„2 + ā‹… ā‹… ā‹… + š‘š‘› š‘„š‘› ā‰¤ š‘š š‘›

so that x ā†’ x āˆˆ š‘‹(p, š‘š). Therefore š‘‹(p, š‘š) is closed. We have shown that š‘‹(p, š‘š) is a closed and bounded subset of ā„œš‘› . Hence it is compact (Proposition 1.4). 1.232 Let x, y āˆˆ š‘‹(p, š‘š). That is š‘› āˆ‘

š‘š‘– š‘„š‘– ā‰¤ š‘š

š‘–=1 š‘› āˆ‘

š‘š‘– š‘¦ š‘– ā‰¤ š‘š

š‘–=1

For any š›¼ āˆˆ [0, 1], the cost of the weighted average bundle z = š›¼x + (1 āˆ’ š›¼)y (where each component š‘§š‘– = š›¼š‘„š‘– + (1 āˆ’ š›¼)š‘¦š‘– ) is š‘› āˆ‘ š‘–=1

š‘š‘– š‘§ š‘– =

š‘› āˆ‘

š‘š‘– (š›¼š‘„š‘– š‘–=1 š‘› āˆ‘

=š›¼

+ (1 āˆ’ š›¼)š‘¦š‘–

š‘š‘– š‘„š‘– + (1 āˆ’ š›¼)

š‘–=1

š‘› āˆ‘

š‘š‘– š‘¦ š‘–

š‘–=1

ā‰¤ š›¼š‘š + (1 āˆ’ š›¼)š‘š =š‘š Therefore z āˆˆ š‘‹(p, š‘š). The budget set š‘‹(p, š‘š) is convex. 1.233

1. Assume that ā‰» is strongly monotone. Let x, y āˆˆ š‘‹ with x ā‰„ y. Either x ā‰© y so that x ā‰» y by strong monotonicity or x = y so that x ā‰æ y by reļ¬‚exivity. In either case, x ā‰æ y so that ā‰æ is weakly monotonic.

2. Again, assume that ā‰æ is strongly monotonic and let y āˆˆ š‘‹. š‘‹ is open (relative to itself). Therefore, there exists some š‘Ÿ such that šµ(y, š‘Ÿ) = y + š‘Ÿšµ āŠ† š‘‹ Let x = y + š‘Ÿe1 be the consumption bundle containing š‘Ÿ more units of good 1. Then e1 āˆˆ šµ, x āˆˆ šµ(y, š‘Ÿ) and therefore āˆ„x āˆ’ yāˆ„ < š‘Ÿ. Furthermore, x ā‰© y and therefore x ā‰» y. 59

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c 2001 Michael Carter āƒ All rights reserved

3. Assume ā‰æ is locally nonsatiated. Then, for every x āˆˆ š‘‹, there exists some y āˆˆ š‘‹ such that y ā‰» x. Therefore, there is no best element. 1.234 is assume that xāˆ— ā‰æ x for every x āˆˆ šµ(p, š‘š) but that āˆ‘š‘› Assume otherwise, thatāˆ‘ š‘› š‘–=1 š‘š‘– š‘„š‘– < š‘š. Let š‘Ÿ = š‘š āˆ’ š‘–=1 š‘š‘– š‘„š‘– be the unspent income. Spending the residual on good 1, the commodity bundle x = xāˆ— + š‘š‘Ÿ1 e1 is aļ¬€ordable š‘› āˆ‘

š‘š‘– š‘„š‘– =

š‘–=1

š‘› āˆ‘

š‘š‘– š‘„āˆ—š‘– + š‘1

š‘–=1

š‘Ÿ =š‘š š‘1

Moreover, since x ā‰© xāˆ— , x ā‰» xāˆ— , which contradicts the assumption that xāˆ— is the best element in š‘‹(p, š‘š). 1.235 otherwise, that is assume that xāˆ— ā‰æ x for every x āˆˆ šµ(p, š‘š) but that āˆ‘š‘› Assume āˆ— āˆ— š‘–=1 š‘š‘– š‘„š‘– < š‘š. This implies that x āˆˆ int š‘‹(p, š‘š). Therefore, there exists a neighāˆ— borhood š‘ of x with š‘ āŠ† š‘‹(p, š‘š). Within this neighborhood, there exists some x āˆˆ š‘ āŠ† š‘‹(p, š‘š) with x ā‰» xāˆ— , which contradicts the assumption that xāˆ— is the best element in š‘‹(p, š‘š). 1.236

1. Assume ā‰æ is continuous. Choose some y āˆˆ š‘‹. For any x0 in ā‰»(y), x0 ā‰» y and (since ā‰æ is continuous) there exists some neighborhood š‘†(x0 ) such that x ā‰» y for every x āˆˆ š‘†(x0 ). That is, š‘†(x0 ) āŠ† ā‰»(y) and ā‰»(y) is open. Similarly, for any x0 āˆˆ ā‰ŗ(y), x0 ā‰ŗ y and there exists some neighborhood š‘†(x0 ) such that x ā‰ŗ y for every x āˆˆ š‘†(x0 ). Thus š‘†(x0 ) āŠ† ā‰ŗ(y) and ā‰ŗ(y) is open.

2. Conversely, assume that the sets ā‰»(y) = { x : x ā‰» y } and ā‰ŗ(y) = { x : x ā‰ŗ y } are open in x. Assume x0 ā‰» y0 . (a) Suppose there exists some y such that x0 ā‰» y ā‰» z0 . Then x0 āˆˆ ā‰»(y), which is open by assumption. That is, ā‰»(y) is an open neighborhood of x0 and x ā‰» y for every x āˆˆ ā‰»(y). Similarly, ā‰ŗ(y) is an open neighborhood of z0 for which y ā‰» z for every z āˆˆ ā‰ŗ(y). Therefore š‘†(x0 ) = ā‰»(y) and š‘†(z0 ) = ā‰ŗ(y) are the required neighborhoods of x0 and z0 respectively such that xā‰»yā‰»z

for every x āˆˆ š‘†(x0 ) and y āˆˆ š‘†(z0 )

(b) Suppose there is no y such that x0 ā‰» y ā‰» z0 . i. By assumption āˆ™ ā‰»(z0 ) is open āˆ™ x0 ā‰» z0 which implies x0 āˆˆ ā‰»(z0 ), Therefore ā‰»(z0 ) is an open neighborhood of x0 . ii. Since ā‰æ is complete, either y ā‰ŗ x0 or y ā‰æ x0 for every y āˆˆ š‘‹ (Exercise 1.56. Since there is no y such that x0 ā‰» y ā‰» z0 y ā‰» z0 =ā‡’ y āˆ•ā‰ŗ x0 =ā‡’ y ā‰æ x0 Therefore ā‰»(z0 ) = ā‰æ(x0 ). iii. Since x ā‰æ x0 ā‰» z0 for every x āˆˆ ā‰æ(x0 ) = ā‰»(z0 ) x ā‰» z0 for every x āˆˆ ā‰»(z0 )

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c 2001 Michael Carter āƒ All rights reserved

iv. Therefore š‘†(x0 ) = ā‰»(z0 ) is an open neighborhood of x0 such that x ā‰» z0 for every x āˆˆ š‘†(x0 ) Similarly, š‘†(z0 ) = ā‰ŗ(x0 ) is an open neighborhood of z0 such that z ā‰ŗ x0 for every z āˆˆ š‘†(z0 ). Consequently xā‰»z

for every x āˆˆ š‘†(x0 ) and z āˆˆ š‘†(z0 )

( )š‘ 3. ā‰æ(y) = ā‰ŗ(y) (Exercise 1.56). Therefore, ā‰æ(y) is closed if and only if ā‰ŗ(y) is open (Exercise 1.80). Similarly, ā‰¾(y) is closed if and only if ā‰»(y) is open. 1.237

1. Let š¹ = { (x, y) āˆˆ š‘‹ Ɨš‘‹ : x ā‰æ y }. Let ((xš‘› , yš‘› )) be a sequence in š¹ which converges to (x, y). Since (xš‘› , yš‘› ) āˆˆ š¹ , xš‘› ā‰æ yš‘› for every š‘›. By assumption, x ā‰æ y. Therefore, (x, y) āˆˆ š¹ which establishes that š¹ is closed (Exercise 1.106) Conversely, assume that š¹ is closed and let ((xš‘› , yš‘› )) be a sequence converging to (x, y) with xš‘› ā‰æ yš‘› for every š‘›. Then ((xš‘› , yš‘› )) āˆˆ š¹ which implies that (x, y) āˆˆ š¹ . Therefore x ā‰æ y.

2. Yes. Setting yš‘› = y for every š‘›, their deļ¬nition implies that for every sequence (xš‘› ) in š‘‹ with xš‘› ā‰æ y, x = lim xš‘› ā‰æ y. That is, the upper contour set ā‰æ(y) = { x : x ā‰æ y } is closed. Similarly, the lower contour set ā‰¾(y) is closed. Conversely, assume that the preference relation is continuous (in our deļ¬nition). We show that the set šŗ = { (x, y) : x ā‰ŗ y } is open. Let (x0 , y0 ) āˆˆ šŗ. Then x0 ā‰ŗ y0 . By continuity, there exists neighborhoods š‘†(x0 ) and š‘†(y0 ) of x0 and y0 such that x ā‰ŗ y for every x āˆˆ š‘†(x0 ) and y āˆˆ š‘†(y0 ). Hence, for every (x, y) āˆˆ š‘ = š‘†(x0 ) Ɨ š‘†(y0 ), x ā‰ŗ y. Therefore š‘ āŠ† šŗ which implies that šŗ is open. Consequently šŗš‘ = { (x, y) : x ā‰æ y } is closed. 1.238 Assume the contrary. That is, assume there is no y with x ā‰» y ā‰» z. Since ā‰æ is complete, either y ā‰ŗ x0 or y ā‰æ x0 for every y āˆˆ š‘‹ (Exercise 1.56). Since there is no y such that x0 ā‰» y ā‰» z0 y ā‰» z0 =ā‡’ y āˆ•ā‰ŗ x0 =ā‡’ y ā‰æ x0 Therefore ā‰»(z0 ) = ā‰æ(x0 ). By continuity, ā‰»(z0 ) is open and ā‰æ(x0 ) is closed. Hence ā‰»(z0 ) = ā‰æ(x0 ) is both open and closed (Exercise 1.83). Alternatively, ā‰æ(x0 ) and ā‰¾(z0 ) are both open sets which partition š‘‹. This contradicts the assumption that š‘‹ is connected. 1.239 Let š‘‹ āˆ— denote the set of best elements. As demonstrated in the preceding proof āˆ© š‘‹āˆ— = ā‰æ(yš‘– ) yāˆˆš‘‹

Therefore š‘‹ āˆ— is closed (Exercise 1.85) and hence compact (Exercise 1.110). 1.240 Assume for simplicity that š‘1 = š‘2 = 1 and that š‘š = 1. Then, the budget set is šµ(1, 1) = { x āˆˆ ā„œ2++ : š‘„1 + š‘„2 ā‰¤ 1 } The consumer would like to spend as much as possible of her income on good 1. However, the point (1, 0) is not feasible, since (1, 0) āˆˆ / š‘‹.

61

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.241 Essentially, consumer theory (in economics) is concerned with predicting the way in which consumer purchases vary with changes in observable parameters such as prices and incomes. Predictions are deduced by assuming that the consumer will consistently choose the best aļ¬€ordable alternative in her budget set. The theory would be empty if there was no such optimal choice. 1.242

1. Let š‘‹ 0 = š‘‹ āˆ© ā„œš‘›+ . Then š‘‹ 0 is compact and š‘‹ 1 āŠ† š‘‹ 0 . Deļ¬ne the order x ā‰æ1 y if and only if š‘‘1 (x) ā‰¤ š‘‘1 (y). Then ā‰æ1 is continuous on š‘‹ and š‘‹ 1 = { x āˆˆ š‘‹ : š‘‘1 (x) ā‰¤ š‘‘1 (y) for every y āˆˆ š‘‹ } is the set of best elements in š‘‹ with respect to the order ā‰æ1 . By Exercise 1.239, š‘‹ 1 is nonempty and compact.

2. Assume š‘‹ š‘˜āˆ’1 is compact. Deļ¬ne the order x ā‰æš‘˜ y if and only if š‘‘š‘˜ (x) ā‰¤ š‘‘š‘˜ (y). Then ā‰æš‘˜ is continuous on š‘‹ š‘˜āˆ’1 and š‘‹ š‘˜ = { x āˆˆ š‘‹ š‘˜āˆ’1 : š‘‘š‘˜ (x) ā‰¤ š‘‘š‘˜ (y) for every y āˆˆ š‘‹ š‘˜āˆ’1 } is the set of best elements in š‘‹ š‘˜āˆ’1 with respect to the order ā‰æš‘˜ . By Exercise 1.239, š‘‹ š‘˜ is nonempty and compact. 3. Assume x āˆˆ Nu. Then x ā‰æ y for every y āˆˆ š‘‹ d(x) ā‰¾šæ d(y) for every y āˆˆ š‘‹ For every š‘˜ = 1, 2, . . . , 2š‘› dš‘˜ (x) ā‰¤ dš‘˜ (y) for every y āˆˆ š‘‹ š‘›

š‘›

which implies x āˆˆ š‘‹ š‘˜ . In particular x āˆˆ š‘‹ 2 . Therefore Nu āŠ† š‘‹ 2 . š‘›

š‘›

š‘›

Suppose Nu āŠ‚ š‘‹ 2 . Then there exists some x, y āˆˆ š‘‹, x āˆˆ / š‘‹ 2 and y āˆˆ š‘‹ 2 such š‘‘ / š‘‹ š‘˜ . Then š‘‘š‘˜ (x) > š‘‘š‘˜ (y). that x ā‰æ y. Let š‘˜ be the smallest integer such that x āˆˆ š‘™ But x āˆˆ š‘‹ for every š‘™ < š‘˜ and therefore š‘‘š‘™ (x) = š‘‘š‘™ (y) for š‘™ = 1, 2, . . . , š‘˜ āˆ’ 1. This means that d(y) ā‰ŗšæ d(x) so that x ā‰ŗš‘‘ y. This contradiction establishes š‘› that Nu = š‘‹ 2 . 1.243 Assume ā‰æ is convex. Choose any y āˆˆ š‘‹ and let x āˆˆ ā‰æ(y). Then x ā‰æ y. Since ā‰æ is convex, this implies that š›¼x + (1 āˆ’ š›¼)y ā‰æ y

for every 0 ā‰¤ š›¼ ā‰¤ 1

and therefore š›¼x + (1 āˆ’ š›¼)y āˆˆ ā‰æ(y)

for every 0 ā‰¤ š›¼ ā‰¤ 1

Therefore ā‰æ(y) is convex. To show the converse, assume that ā‰æ(y) is convex for every y āˆˆ š‘‹. Choose x, y āˆˆ š‘‹. Interchanging x and y if necessary, we can assume that x ā‰æ y so that x āˆˆ ā‰æ(y). Of course, y āˆˆ ā‰æ(y). Since ā‰æ(y) is convex š›¼x + (1 āˆ’ š›¼)y āˆˆ ā‰æ(y)

for every 0 ā‰¤ š›¼ ā‰¤ 1

which implies š›¼x + (1 āˆ’ š›¼)y ā‰æ y

for every 0 ā‰¤ š›¼ ā‰¤ 1 62

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.244 š‘‹ āˆ— may be empty, in which case it is trivially convex. Otherwise, let xāˆ— āˆˆ š‘‹ āˆ— . For every x āˆˆ š‘‹ āˆ— x ā‰æ xāˆ— which implies x āˆˆ ā‰æ(xāˆ— ) Therefore š‘‹ āˆ— āŠ† ā‰æ(xāˆ— ). Conversely, by transitivity x ā‰æ xāˆ— ā‰æ y for every y āˆˆ š‘‹ for every x āˆˆ ā‰æ(xāˆ— ) which implies ā‰æ(xāˆ— ) āŠ† š‘‹ āˆ— . Therefore, š‘‹ āˆ— = ā‰æ(xāˆ— ) which is convex. 1.245 To show that ā‰æš‘‘ is strictly convex, assume that x, y āˆˆ š‘‹ are such d(x) = d(y) with x āˆ•= y. Suppose ) ( d(x) = š‘‘(š‘†1 , x), š‘‘(š‘†2 , x) . . . , š‘‘(š‘†2š‘› , x) In the order š‘†1 , š‘†2 , . . . , š‘†2š‘› , let š‘†š‘˜ be the ļ¬rst coalition for which š‘‘(š‘†š‘˜ , x) āˆ•= š‘‘(š‘†š‘˜ , y). That is š‘‘(š‘†š‘— , x) = š‘‘(š‘†š‘— , y) for every š‘— < š‘˜

(1.32)

Since š‘‘(š‘†š‘˜ , x) āˆ•= š‘‘(š‘†š‘˜ , y) and d(x) is listed in descending order, we must have š‘‘(š‘†š‘˜ , x) > š‘‘(š‘†š‘˜ , y)

(1.33)

š‘‘(š‘†š‘˜ , x) ā‰„ š‘‘(š‘†š‘— , y) for every š‘— > š‘˜

(1.34)

and

Choose 0 < š›¼ < 1 and let z = š›¼x + (1 āˆ’ š›¼)y. For any coalition š‘† āˆ‘ š‘‘(š‘†, z) = š‘¤(š‘†) āˆ’ š‘§š‘– š‘–āˆˆš‘†

āˆ‘( ) = š‘¤(š‘†) āˆ’ š›¼š‘„š‘– + (1 āˆ’ š›¼)š‘¦š‘– š‘–āˆˆš‘†

= š‘¤(š‘†) āˆ’ š›¼ (

āˆ‘

š‘„š‘– āˆ’ (1 āˆ’ š›¼)

š‘–āˆˆš‘†

= š›¼ š‘¤(š‘†) āˆ’

āˆ‘

š‘–āˆˆš‘†

) š‘„š‘–

āˆ‘

š‘¦š‘– (

+ (1 āˆ’ š›¼) š‘¤(š‘†) āˆ’

š‘–āˆˆš‘†

āˆ‘

) š‘¦š‘–

š‘–āˆˆš‘†

= š›¼š‘‘(š‘†, x) + (1 āˆ’ š›¼)š‘‘(š‘†, y) Using (1.55) to (1.57), this implies that š‘‘(š‘†š‘— , z) = š‘‘(š‘†š‘— , x),

š‘—š‘˜

for every 0 < š›¼ < 1, Therefore d(z) ā‰ŗšæ d(x). Thus z ā‰»š‘‘ x, which establishes that ā‰æ is strictly convex. The set of feasible outcomes is convex. Assume x, y āˆˆ Nu āŠ† š‘‹, x āˆ•= y. Then d(x) = d(y) and z = š›¼x + (1 āˆ’ š›¼)y ā‰»š‘‘ x for every 0 < š›¼ < 1 which contradicts the assumption that x āˆˆ Nu. We conclude that the nucleolus contains only one element. 63

Solutions for Foundations of Mathematical Economics 1.246

c 2001 Michael Carter āƒ All rights reserved

1. (a) Clearly ā‰ŗ(x0 ) āŠ† ā‰¾(x0 ) and ā‰»(y0 ) āŠ† ā‰æ(y0 ). Consequently ā‰ŗ(x0 ) āˆŖ ā‰»(y0 ) āŠ† ā‰¾(x0 ) āˆŖ ā‰æ(y0 ). We claim that these sets are in fact equal. Let z āˆˆ ā‰¾(x0 ) āˆŖ ā‰æ(y0 ). Suppose that z āˆˆ ā‰¾(x0 ) but z āˆˆ / ā‰ŗ(x0 ). Then z ā‰æ x0 . By transitivity, z ā‰æ x0 ā‰» y0 which implies that z āˆˆ ā‰»(y0 ). Similarly z āˆˆ ā‰æ(y0 ) āˆ– ā‰»(y0 ) implies z āˆˆ ā‰ŗ(x0 ). Therefore ā‰ŗ(x0 ) āˆŖ ā‰»(y0 ) = ā‰¾(x0 ) āˆŖ ā‰æ(y0 ) (b) By continuity, ā‰ŗ(x0 ) āˆŖ ā‰»(y0 ) is open and ā‰¾(x0 ) āˆŖ ā‰æ(y0 ) = ā‰ŗ(x0 ) āˆŖ ā‰»(y0 ) is closed. Further x0 ā‰» y0 implies that x0 āˆˆ ā‰»(y0 ) so that ā‰ŗ(x0 ) āˆŖ ā‰»(y0 ) āˆ•= āˆ…. We have established that ā‰ŗ(x0 ) āˆŖ ā‰»(y0 ) is a nonempty subset of š‘‹ which is both open and closed. Since š‘‹ is connected, this implies (Exercise 1.83) that ā‰ŗ(x0 ) āˆŖ ā‰»(y0 ) = š‘‹

2. (a) By deļ¬nition, x āˆˆ / ā‰ŗ(x). So ā‰ŗ(x) āˆ© ā‰ŗ(y) = š‘‹ implies x āˆˆ ā‰»(y), that is x ā‰æ y contradicting the noncomparability of x and y. Therefore ā‰ŗ(x) āˆ© ā‰ŗ(y) āˆ•= š‘‹ (b) By assumption, there exists at least one pair x0 , y0 such that x0 ā‰» y0 . By the previous part ā‰ŗ(x0 ) āˆŖ ā‰»(y0 ) = š‘‹ This implies either x ā‰ŗ x0 or x ā‰» y0 . Without loss of generality, assume x ā‰» y0 . Again using the previous part, we have ā‰ŗ(x) āˆŖ ā‰»(y0 ) = š‘‹ Since x and y are not comparable, y āˆˆ / ā‰ŗ(x) which implies that y āˆˆ ā‰»(y0 ). Therefore x ā‰» y0 and y ā‰» y0 or alternatively y0 āˆˆ ā‰ŗ(x0 ) āˆ© ā‰»(y0 ) āˆ•= āˆ… (c) Clearly ā‰ŗ(x) āŠ† ā‰¾(x) and ā‰»(y) āŠ† ā‰æ(y). Consequently ā‰ŗ(x) āˆ© ā‰ŗ(y) āŠ† ā‰¾(x) āˆ© ā‰¾(y) Let z āˆˆ ā‰¾(x) āˆ© ā‰¾(y). That is, z ā‰¾ x. If x ā‰¾ z, then transitivity implies x ā‰¾ z ā‰¾ y, which contradicts the noncomparability of x and y. Consequently x āˆ•ā‰¾ z which implies z ā‰ŗ x and z āˆˆ ā‰ŗ(x). Similarly z āˆˆ ā‰ŗ(y) and therefore ā‰ŗ(x) āˆ© ā‰ŗ(y) = ā‰¾(x) āˆ© ā‰¾(y) 3. If x and y are noncomparable, ā‰ŗ(x)āˆ©ā‰ŗ(y) is a nonempty proper subset of š‘‹. By continuity ā‰ŗ(x) āˆ© ā‰ŗ(y) = ā‰¾(x) āˆ© ā‰¾(y) is both open and closed which contradicts the assumption that š‘‹ is connected (Exercise 1.83). We conclude that ā‰æ must be complete. 1.247 Assume x ā‰» y. Then x āˆˆ ā‰»(y). Since ā‰»(y) is open, x āˆˆ int ā‰»(y). Also y āˆˆ ā‰»(y). By Exercise 1.218, š›¼x + (1 āˆ’ š›¼)y āˆˆ int ā‰»(y) for every 0 < š›¼ < 1, which implies š›¼x + (1 āˆ’ š›¼)y ā‰» y for every 0 < š›¼ < 1 64

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

1.248 For every x āˆˆ š‘‹, there exists some z such that z ā‰» x (Nonsatiation). For any š‘Ÿ, choose some š›¼ āˆˆ (0, š‘Ÿ/ āˆ„x āˆ’ zāˆ„) and let y = š›¼z + (1 āˆ’ š›¼)x. Then āˆ„x āˆ’ yāˆ„ = š›¼ āˆ„x āˆ’ zāˆ„ < š‘Ÿ Moreover, since ā‰æ is strictly convex, y = š›¼z + (1 āˆ’ š›¼)x ā‰» x Thus, ā‰æ is locally nonsatiated. We have previously shown that local nonsatiation implies nonsatiation (Exercise 1.233). Consequently, these two properties are equivalent for strictly convex preferences. 1.249 Assume that x is not strongly Pareto eļ¬ƒcient. That is, there exist allocation y such that y ā‰æš‘– x for all š‘– and some individual š‘— for which y ā‰»š‘— x. Take 1 āˆ’ š‘” percent of š‘—ā€™s consumption and distribute it equally to the other participants. By continuity, 1āˆ’š‘” there exists some š‘” such that š‘”y ā‰»š‘— x. The other agents receive yš‘– + š‘›āˆ’1 yš‘— which, by monotonicity, they strictly prefer to xš‘– . 1.250 Assume that (pāˆ— , xāˆ— ) is a competitive equilibrium of an exchange economy, but that xāˆ— does not belong to the core of the corresponding market game. Then there exists some coalition š‘† and allocation y āˆˆāˆ‘ š‘Š (š‘†) such that yš‘– ā‰»š‘– xāˆ—š‘– for every š‘– āˆˆ š‘†. āˆ‘ Since y āˆˆ š‘Š (š‘†), we must have š‘–āˆˆš‘† yš‘– = š‘–āˆˆš‘† wš‘– . Since xāˆ— is a competitive equilibrium and yš‘– ā‰»š‘– xāˆ—š‘– for every š‘– āˆˆ š‘†, y must be unaffordable, that is š‘™ āˆ‘

š‘š‘— š‘¦š‘–š‘— >

š‘—=1

š‘™ āˆ‘

š‘š‘— wš‘–š‘— for every š‘– āˆˆ š‘†

š‘—=1

and therefore š‘™ āˆ‘āˆ‘ š‘–āˆˆš‘† š‘—=1

š‘š‘— š‘¦š‘–š‘— >

š‘™ āˆ‘āˆ‘

š‘š‘— wš‘–š‘—

š‘–āˆˆš‘† š‘—=1

which contradicts the assumption that š‘¦ āˆˆ š‘Š (š‘†). 1.251 Combining the previous exercise with Exercise 1.64 xāˆ— āˆˆ core āŠ† Pareto

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Chapter 2: Functions 2.1 In general, the birthday mapping is not one-to-one since two individuals may have the same birthday. It is not onto since some days may be no oneā€™s birthday. 2.2 The origin 0 is ļ¬xed point for every šœƒ. Furthermore, when šœƒ = 0, š‘“ is an identity function and every point is a ļ¬xed point. 2.3 For every š‘„ āˆˆ š‘‹, there exists some š‘¦ āˆˆ š‘Œ such that š‘“ (š‘„) = š‘¦, whence š‘„ āˆˆ š‘“ āˆ’1 (š‘¦). Therefore, every š‘„ belongs to some contour. To show that distinct contours are disjoint, assume š‘„ āˆˆ š‘“ āˆ’1 (š‘¦1 ) āˆ© š‘“ āˆ’1 (š‘¦2 ). Then š‘“ (š‘„) = š‘¦1 and also š‘“ (š‘„) = š‘¦2 . Since š‘“ is a function, this implies that š‘¦1 = š‘¦2 . 2.4 Assume š‘“ is one-to-one and onto. Then, for every š‘¦ āˆˆ š‘Œ , there exists š‘„ āˆˆ š‘‹ such that š‘“ (š‘„) = š‘¦. That is, š‘“ āˆ’1 (š‘¦) āˆ•= āˆ… for every š‘¦ āˆˆ š‘Œ . If š‘“ is one to one, š‘“ (š‘„) = š‘¦ = š‘“ (š‘„ā€² ) implies š‘„ = š‘„ā€² . Therefore, š‘“ āˆ’1 (š‘¦) consists of a single element. Therefore, the inverse function š‘“ āˆ’1 exists. Conversely, assume that š‘“ : š‘‹ ā†’ š‘Œ has an inverse š‘“ āˆ’1 . As š‘“ āˆ’1 is a function mapping š‘Œ to š‘‹, it must be deļ¬ned for every š‘¦ āˆˆ š‘Œ . Therefore š‘“ is onto. Assume there exists š‘„, š‘„ā€² āˆˆ š‘‹ and š‘¦ āˆˆ š‘Œ such that š‘“ (š‘„) = š‘¦ = š‘“ (š‘„ā€² ). Then š‘“ āˆ’1 (š‘¦) = š‘„ and also š‘“ āˆ’1 (š‘¦) = š‘„ā€² . Since š‘“ āˆ’1 is a function, this implies that š‘„ = š‘„ā€² . Therefore š‘“ is one-to-one. 2.5 Choose any š‘„ āˆˆ š‘‹ and let š‘¦ = š‘“ (š‘„). Since š‘“ is one-to-one, š‘„ = š‘“ āˆ’1 (š‘¦) = š‘“ āˆ’1 (š‘“ (š‘„)). The second identity is proved similarly. 2.6 (2.2) implies for every š‘„ āˆˆ ā„œ š‘’š‘„ š‘’āˆ’š‘„ = š‘’0 = 1 and therefore š‘’āˆ’š‘„ =

1 š‘’š‘„

For every š‘„ ā‰„ 0 š‘’š‘„ = 1 +

š‘„3 š‘„ š‘„2 + + + ā‹…ā‹…ā‹… > 0 1 2 6

and therefore by (2.28) š‘’š‘„ > 0 for every š‘„ āˆˆ ā„œ. For every š‘„ ā‰„ 1 š‘’š‘„ = 1 +

š‘„3 š‘„ š‘„2 + + + ā‹… ā‹… ā‹… ā‰„ 1 + š‘„ ā†’ āˆž as š‘„ ā†’ āˆž 1 2 6

and therefore š‘’š‘„ ā†’ āˆž as š‘„ ā†’ āˆž. By (2.28) š‘’š‘„ ā†’ 0 as š‘„ ā†’ āˆ’āˆž. 2.7 š‘’š‘„/2 š‘’š‘„/2 š‘’š‘„ = š‘„ 2 š‘„2 ( š‘„/2 ) 1 š‘’ = š‘’š‘„/2 ā†’ āˆž as š‘„ ā†’ āˆž 2 š‘„/2 66

(2.28)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

since the term in brackets is strictly greater than 1 for any š‘„ > 0. Similarly š‘’š‘„ (š‘’š‘„/(š‘›+1) )š‘› š‘’š‘„/(š‘›+1) = š‘„ š‘› š‘„ (š‘› + 1)š‘› ( š‘›+1 ) ( š‘„/(š‘›+1) )š‘› 1 š‘’ = š‘’š‘„/(š‘›+1) ā†’ āˆž (š‘› + 1)š‘› š‘„/(š‘› + 1) 2.8 Assume that š‘† āŠ† ā„œ is compact. Then š‘† is bounded (Proposition 1.1), and there exists š‘€ such that āˆ£š‘„āˆ£ ā‰¤ š‘€ for every š‘„ āˆˆ š‘†. For all š‘› ā‰„ š‘š ā‰„ 2š‘€     š‘›  āˆ‘ š‘„š‘˜   š‘„š‘š+1 š‘›āˆ’š‘š āˆ‘ ( š‘„ )š‘˜     āˆ£š‘“š‘› (š‘„) āˆ’ š‘“š‘š (š‘„)āˆ£ =  ā‰¤   š‘˜!   (š‘š + 1)! š‘š  š‘˜=š‘š+1 š‘˜=0    š‘€ š‘š+1 š‘›āˆ’š‘š āˆ‘ ( š‘€ )š‘˜   ā‰¤   (š‘š + 1)! š‘š  š‘˜=0 ( ( )š‘›āˆ’š‘š ) š‘€ š‘š+1 1 1 1 ā‰¤ 1 + + + ā‹…ā‹…ā‹… + (š‘š + 1)! 2 4 2 ( ( )š‘š ) š‘š+1 š‘€ 1 š‘€ š‘š+1 ā‰¤2 ā‰¤2 ā‰¤ (š‘š + 1)! š‘š+1 2 by Exercise 1.206. Therefore š‘“š‘› converges to š‘“ for all š‘„ āˆˆ š‘†. 2.9 This is a special case of Example 2.8. For any š‘“, š‘” āˆˆ š¹ (š‘‹), deļ¬ne (š‘“ + š‘”) = š‘“ (š‘„) + š‘”(š‘„) (š›¼š‘“ )(š‘„) = š›¼š‘“ (š‘„) With these deļ¬nitions š‘“ + š‘” and š›¼š‘“ also map š‘‹ to ā„œ. Hence š¹ (š‘‹) is closed under addition and scalar multiplication. It is straightforward but tedious to verify that š¹ (š‘‹) satisļ¬es the other requirements of a linear space. 2.10 The zero element in š¹ (š‘‹) is the constant function š‘“ (š‘„) = 0 for every š‘„ āˆˆ š‘‹. 2.11

1. From the deļ¬nition of āˆ„š‘“ āˆ„ it is clear that āˆ™ āˆ„š‘“ āˆ„ ā‰„ 0. āˆ™ āˆ„š‘“ āˆ„ = 0 of and only š‘“ is the zero functional. āˆ™ āˆ„š›¼š‘“ āˆ„ = āˆ£š›¼āˆ£ āˆ„š‘“ āˆ„ since supš‘„āˆˆš‘‹ āˆ£š›¼š‘“ (š‘„)āˆ£ = āˆ£š›¼āˆ£ supš‘„āˆˆš‘‹ āˆ£š‘“ (š‘„)āˆ£ It remains to verify the triangle inequality, namely āˆ„š‘“ + š‘”āˆ„ = sup āˆ£(š‘“ + š‘”)(š‘„)āˆ£ š‘„āˆˆš‘‹

= sup āˆ£š‘“ (š‘„) + š‘”(š‘„)āˆ£ š‘„āˆˆš‘‹ { } ā‰¤ sup āˆ£š‘“ (š‘„)āˆ£ + āˆ£š‘”(š‘„)āˆ£ š‘„āˆˆš‘‹

ā‰¤ sup āˆ£(š‘“ (š‘„)āˆ£ + sup āˆ£š‘”(š‘„)āˆ£ š‘„āˆˆš‘‹

š‘„āˆˆš‘‹

= āˆ„š‘“ āˆ„ + āˆ„š‘”āˆ„ 2. Consequently, for any š‘“ āˆˆ šµ(š‘‹), š›¼š‘“ (š‘„) ā‰¤ āˆ£š›¼āˆ£ āˆ„š‘“ āˆ„ for every š‘„ āˆˆ š‘‹ and therefore š›¼š‘“ āˆˆ šµ(š‘‹). Similarly, for any š‘“, š‘” āˆˆ šµ(š‘‹), (š‘“ + š‘”)(š‘„) ā‰¤ āˆ„š‘“ āˆ„ + āˆ„š‘”āˆ„ for every 67

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

š‘„ āˆˆ š‘‹ and therefore š‘“ + š‘” āˆˆ šµ(š‘‹). Hence, šµ(š‘‹) is closed under addition and scalar multiplication; it is a subspace of the linear space š¹ (š‘‹). We conclude that šµ(š‘‹) is a normed linear space. 3. To show that šµ(š‘‹) is complete, assume that (š‘“ š‘› ) is a Cauchy sequence in šµ(š‘‹). For every š‘„ āˆˆ š‘‹ āˆ£š‘“ š‘› (š‘„) āˆ’ š‘“ š‘š (š‘„)āˆ£ ā‰¤ āˆ„š‘“ š‘› āˆ’ š‘“ š‘š āˆ„ ā†’ 0 Therefore, for š‘„ āˆˆ š‘‹, š‘“ š‘› (š‘„) is a Cauchy sequence of real numbers. Since ā„œ is complete, this sequence converges. Deļ¬ne the function š‘“ (š‘„) = lim š‘“ š‘› (š‘„) š‘›ā†’āˆž

We need to show āˆ™ āˆ„š‘“ š‘› āˆ’ š‘“ āˆ„ ā†’ 0 and āˆ™ š‘“ āˆˆ šµ(š‘‹) š‘›

(š‘“ ) is a Cauchy sequence. For given šœ– > 0, choose š‘ such that āˆ„š‘“ š‘› āˆ’ š‘“ š‘š āˆ„ < šœ–/2 for very š‘š, š‘› ā‰„ š‘ . For any š‘„ āˆˆ š‘‹ and š‘› ā‰„ š‘ , āˆ£š‘“ š‘› (š‘„) āˆ’ š‘“ (š‘„)āˆ£ ā‰¤ āˆ£š‘“ š‘› (š‘„) āˆ’ š‘“ š‘š (š‘„)āˆ£ + āˆ£š‘“ š‘š (š‘„) āˆ’ š‘“ (š‘„)āˆ£ ā‰¤ āˆ„š‘“ š‘› āˆ’ š‘“ š‘š āˆ„ + āˆ£š‘“ š‘š (š‘„) āˆ’ š‘“ (š‘„)āˆ£ By suitable choice of š‘š (which may depend upon š‘„), each term on the right can be made smaller than šœ–/2 and therefore āˆ£š‘“ š‘› (š‘„) āˆ’ š‘“ (š‘„)āˆ£ < šœ– for every š‘„ āˆˆ š‘‹ and š‘› ā‰„ š‘ . āˆ„š‘“ š‘› āˆ’ š‘“ āˆ„ = sup āˆ£š‘“ š‘› (š‘„) āˆ’ š‘“ (š‘„)āˆ£ ā‰¤ šœ– š‘„āˆˆš‘‹

Finally, this implies āˆ„š‘“ āˆ„ = limš‘›ā†’āˆž āˆ„š‘“ š‘› āˆ„. Therefore š‘“ āˆˆ šµ(š‘‹). 2.12 If the die is fair, the probability of the elementary outcomes is š‘ƒ ({1}) = š‘ƒ ({2}) = š‘ƒ ({3}) = š‘ƒ ({4}) = š‘ƒ ({5}) = š‘ƒ ({6}) = 1/6 By Condition 3 š‘ƒ ({2, 4, 6}) = š‘ƒ ({2}) + š‘ƒ ({4}) + š‘ƒ ({6}) = 1/2 2.13 The proļ¬t maximization problem of a competitive single-output ļ¬rm is to choose the combination of inputs x āˆˆ ā„œš‘›+ and scale of production š‘¦ to maximize net proļ¬t. This is summarized in the constrained maximization problem max š‘š‘¦ āˆ’ x,š‘¦

āˆ‘š‘›

š‘› āˆ‘

š‘¤š‘– š‘„š‘–

š‘–=1

subject to x āˆˆ š‘‰ (š‘¦)

where š‘š‘¦ is total revenue and š‘–=1 š‘¤š‘– š‘„š‘– total cost. The proļ¬t function, which depends upon both š‘ and w, is deļ¬ned by Ī (š‘, w) =

max š‘š‘¦ āˆ’

š‘¦,xāˆˆš‘‰ (š‘¦)

68

š‘› āˆ‘ š‘–=1

š‘¤š‘– š‘„š‘–

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

For analysis, it is convenient to represent the technology š‘‰ (š‘¦) by a production function (Example 2.24). The ļ¬rmā€™s optimization can then be expressed as maxš‘› š‘š‘“ (x) āˆ’

xāˆˆā„œ+

š‘› āˆ‘

š‘¤š‘– š‘„š‘–

š‘–=1

and the proļ¬t function as Ī (š‘, w) = maxš‘› š‘š‘“ (x) āˆ’ xāˆˆā„œ+

2.14

š‘› āˆ‘

š‘¤š‘– š‘„š‘–

š‘–=1

1. Assume that production is proļ¬table at p. That is, there exists some y āˆˆ š‘Œ such that š‘“ (y, p) > 0. Since the technology exhibits constant returns to scale, š‘Œ is a cone (Example 1.101). Therefore š›¼y āˆˆ š‘Œ for every š›¼ > 0 and āˆ‘ āˆ‘ š‘š‘– (š›¼š‘¦š‘– ) = š›¼ š‘š‘– š‘¦š‘– = š›¼š‘“ (y, p) š‘“ (š›¼y, p) = š‘–

š‘–

Therefore { š‘“ (š›¼y, p) : š›¼ > 0 } is unbounded and Ī (p) = sup š‘“ (y, p) ā‰„ sup š‘“ (š›¼y, p) = +āˆž š›¼>0

yāˆˆš‘Œ

2. Assume to the contrary that there exists p āˆˆ ā„œš‘›+ with Ī (p) = šœ‹ āˆˆ / { 0, +āˆž, āˆ’āˆž }. There are two possible cases. (a) 0 < šœ‹ < +āˆž. Since šœ‹ = supš‘¦āˆˆš‘Œ š‘“ (y, p) > 0, there exists y āˆˆ š‘Œ such that š‘“ (y, p) > 0 The previous part implies Ī (p) = +āˆž. (b) āˆ’āˆž < šœ‹ < 0. Then there exists y āˆˆ š‘Œ such that š‘“ (y, p) < 0 By a similar argument to the previous part, this implies Ī (p) = āˆ’āˆž. 2.15 Assume xāˆ— is a solution to (2.4). š‘“ (xāˆ— , šœ½) ā‰„ š‘“ (x, šœ½) for every x āˆˆ šŗ(šœ½) and therefore š‘“ (xāˆ— , šœ½) ā‰„ sup š‘“ (x, šœ½) = š‘£(šœ½) xāˆˆšŗ(šœ½)

On the other hand xāˆ— āˆˆ šŗ(šœ½) and therefore š‘£(šœ½) = sup š‘“ (x, šœ½) ā‰„ š‘“ (xāˆ— , šœ½) xāˆˆšŗ(šœ½)

Therefore, xāˆ— satisļ¬es (2.5). Conversely, assume xāˆ— āˆˆ šŗ(šœ½) satisļ¬es (2.5). Then š‘“ (xāˆ— , šœ½) = š‘£(šœ½) = sup š‘“ (x, šœ½) ā‰„ š‘“ (x, šœ½) for every x āˆˆ šŗ(šœ½) xāˆˆšŗ(šœ½)

xāˆ— solve (2.4). 2.16 The assumption that šŗ(š‘„) āˆ•= āˆ… for every š‘„ āˆˆ š‘‹ implies Ī“(š‘„0 ) āˆ•= āˆ… for every š‘„0 āˆˆ š‘‹. There always exist feasible plans from any starting point. Since š‘¢ is bounded, there exists š‘€ such that āˆ£š‘“ (š‘„š‘” , š‘„š‘”+1 )āˆ£ ā‰¤ š‘€ for every x āˆˆ Ī“(š‘„0 ). Consequently, for every x āˆˆ Ī“(š‘„0 ), š‘ˆ (x) āˆˆ ā„œ and āˆž  āˆž āˆž āˆ‘  āˆ‘ āˆ‘ š‘€   āˆ£š‘ˆ (x)āˆ£ =  š›½ š‘” š‘“ (š‘„š‘” , š‘„š‘”+1 ) ā‰¤ š›½ š‘” āˆ£š‘“ (š‘„š‘” , š‘„š‘”+1 )āˆ£ ā‰¤ š›½š‘”š‘€ =   1āˆ’š›½ š‘”=0 š‘”=0 š‘”=0 69

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

using the formula for a geometric series (Exercise 1.108). Therefore š‘£(š‘„0 ) =

sup š‘ˆ (x) ā‰¤

xāˆˆĪ“(š‘„0 )

š‘€ 1āˆ’š›½

and š‘£ āˆˆ šµ(š‘‹). Next, we note that for every feasible plan x āˆˆ Ī“(š‘„0 ) š‘ˆ (x) =

āˆž āˆ‘

š›½ š‘” š‘“ (š‘„š‘” , š‘„š‘”+1 )

š‘”=0

= š‘“ (š‘„0 , š‘„1 ) + š›½

āˆž āˆ‘

š›½ š‘” š‘“ (š‘„š‘”+1 , š‘„š‘”+2 )

š‘”=0

= š‘“ (š‘„0 , š‘„1 ) + š›½š‘ˆ (xā€² )

(2.29)

where xā€² = (š‘„1 , š‘„2 , . . . ) is the continuation of the plan x starting at š‘„1 . For any š‘„0 āˆˆ š‘‹ and šœ– > 0, there exists a feasible plan x āˆˆ Ī“(š‘„0 ) such that š‘ˆ (x) ā‰„ š‘£(š‘„0 ) āˆ’ šœ– Let xā€² = (š‘„1 , š‘„2 , . . . ) be the continuation of the plan x starting at š‘„1 . Using (2.29) and the fact that š‘ˆ (xā€² ) ā‰¤ š‘£(š‘„1 ), we conclude that š‘£(š‘„0 ) āˆ’ šœ– ā‰¤ š‘ˆ (x) = š‘“ (š‘„0 , š‘„1 ) + š›½š‘ˆ (xā€² ) ā‰¤ š‘“ (š‘„0 , š‘„1 ) + š›½š‘£(š‘„1 ) ā‰¤ sup { š‘“ (š‘„0 , š‘¦) + š›½š‘£(š‘¦) } š‘¦āˆˆšŗ(š‘„)

Since this is true for every šœ– > 0, we must have š‘£(š‘„0 ) ā‰¤ sup { š‘“ (š‘„0 , š‘¦) + š›½š‘£(š‘¦) } š‘¦āˆˆšŗ(š‘„)

On the other hand, choose any š‘„1 āˆˆ šŗ(š‘„0 ) āŠ† š‘‹. Since š‘£(š‘„1 ) =

sup š‘ˆ (x)

xāˆˆĪ“(š‘„1 )

there exists a feasible plan xā€² = (š‘„1 , š‘„2 , . . . ) starting at š‘„1 such that š‘ˆ (xā€² ) ā‰„ š‘£(š‘„1 ) āˆ’ šœ– Moreover, the plan x = (š‘„0 , š‘„1 , š‘„2 , . . . ) is feasible from š‘„0 and š‘£(š‘„0 ) ā‰„ š‘ˆ (x) = š‘“ (š‘„0 , š‘„1 ) + š›½š‘ˆ (xā€² ) ā‰„ š‘“ (š‘„0 , š‘„1 ) + š›½š‘£(š‘„1 ) āˆ’ š›½šœ– Since this is true for every šœ– > 0 and š‘„1 āˆˆ šŗ(š‘„0 ), we conclude that š‘£(š‘„0 ) ā‰„ sup { š‘“ (š‘„0 , š‘¦) + š›½š‘£(š‘¦) } š‘¦āˆˆšŗ(š‘„)

Together with (2.30) this establishes the required result, namely š‘£(š‘„0 ) = sup { š‘“ (š‘„0 , š‘¦) + š›½š‘£(š‘¦) } š‘¦āˆˆšŗ(š‘„)

70

(2.30)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.17 Assume x is optimal, so that š‘ˆ (xāˆ— ) ā‰„ š‘ˆ (x) for every x āˆˆ Ī“(š‘„0 ) This implies (using (2.39)) š‘“ (š‘„0 , š‘„āˆ—1 ) + š›½š‘ˆ (xāˆ— ā€² ) ā‰„ š‘“ (š‘„0 , š‘„1 ) + š›½š‘ˆ (xā€² ) where xā€² = (š‘„1 , š‘„2 , . . . ) is the continuation of the plan x starting at š‘„1 and xāˆ— ā€² = (š‘„āˆ—1 , š‘„āˆ—2 , . . . ) is the continuation of the plan xāˆ— . In particular, this is true for every plan x āˆˆ Ī“(š‘„0 ) with š‘„1 = š‘„āˆ—1 and therefore š‘“ (š‘„0 , š‘„āˆ—1 ) + š›½š‘ˆ (xāˆ— ā€² ) ā‰„ š‘“ (š‘„0 , š‘„āˆ—1 ) + š›½š‘ˆ (xā€² ) for every xā€² āˆˆ Ī“(š‘„āˆ—1 ) which implies that š‘ˆ (xāˆ— ā€² ) ā‰„ š‘ˆ (xā€² ) for every xā€² āˆˆ Ī“(š‘„āˆ—1 ) That is, xāˆ— ā€² is optimal starting at š‘„āˆ—1 and therefore š‘ˆ (xāˆ— ā€² ) = š‘£(š‘„āˆ—1 ) (Exercise 2.15). Consequently š‘£(š‘„0 ) = š‘ˆ (xāˆ— ) = š‘“ (š‘„0 , š‘„āˆ—1 ) + š›½š‘ˆ (xāˆ— ā€² ) = š‘“ (š‘„0 , š‘„āˆ—1 ) + š›½š‘£(š‘„āˆ—1 ) This veriļ¬es (2.13) for š‘” = 0. A similar argument veriļ¬es (2.13) for any period š‘”. To show the converse, assume that xāˆ— = (š‘„0 , š‘„āˆ—1 , š‘„āˆ—2 , . . . ) āˆˆ Ī“(š‘„0 ) satisļ¬es (2.13). Successively using (2.13) š‘£(š‘„0 ) = š‘“ (š‘„0 , š‘„āˆ—1 ) + š›½š‘£(š‘„āˆ—1 ) = š‘“ (š‘„0 , š‘„āˆ—1 ) + š›½š‘“ (š‘„āˆ—1 , š‘„āˆ—2 ) + š›½ 2 š‘£(š‘„āˆ—1 ) =

1 āˆ‘

š›½ š‘” š‘“ (š‘„āˆ—š‘” , š‘„āˆ—š‘”+1 ) + š›½ 2 š‘£(š‘„āˆ—2 )

š‘”=0

=

2 āˆ‘

š›½ š‘” š‘“ (š‘„āˆ—š‘” , š‘„āˆ—š‘”+1 ) + š›½ 3 š‘£(š‘„āˆ—3 )

š‘”=0

=

š‘‡ āˆ’1 āˆ‘

š›½ š‘” š‘“ (š‘„š‘” , š‘„š‘” + 1) + š›½ š‘‡ š‘£(š‘„āˆ—š‘‡ )

š‘”=0

for any š‘‡ = 1, 2, . . . . Since š‘£ is bounded (Exercise 2.16), š›½ š‘‡ š‘£(š‘„āˆ—š‘‡ ) ā†’ 0 as š‘‡ ā†’ āˆž and therefore š‘£(š‘„0 ) =

āˆž āˆ‘

š›½ š‘” š‘“ (š‘„š‘” , š‘„š‘”+1 ) = š‘ˆ (xāˆ— )

š‘”=0

Again using Exercise 2.15, xāˆ— is optimal. 2.18 We have to show that āˆ™ for any š‘£ āˆˆ šµ(š‘‹), š‘‡ š‘£ is a functional on š‘‹. āˆ™ š‘‡ š‘£ is bounded. Since š¹ āˆˆ šµ(š‘‹ Ɨ š‘‹), there exists š‘€1 < āˆž such that āˆ£š‘“ (š‘„, š‘¦)āˆ£ ā‰¤ š‘€1 for every (š‘„, š‘¦) āˆˆ š‘‹ Ɨ š‘‹. Similarly, for any š‘£ āˆˆ šµ(š‘‹), there exists š‘€2 < āˆž such that āˆ£š‘£(š‘„)āˆ£ ā‰¤ š‘€2 for every š‘„ āˆˆ š‘‹. Consequently for every (š‘„, š‘¦) āˆˆ š‘‹ Ɨ š‘‹ and š‘£ āˆˆ šµ(š‘‹) āˆ£š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦)āˆ£ ā‰¤ āˆ£š‘“ (š‘„, š‘¦)āˆ£ + š›½ āˆ£š‘£(š‘¦)āˆ£ ā‰¤ š‘€1 + š›½š‘€2 < āˆž 71

(2.31)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics For each š‘„ āˆˆ š‘‹, the set š‘†š‘„ =

{

š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) : š‘¦ āˆˆ šŗ(š‘„)

}

is a nonempty bounded subset of ā„œ, which has least upper bound. Therefore (š‘‡ š‘£)(š‘„) = sup š‘†š‘„ = sup š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) š‘¦āˆˆšŗ(š‘„)

deļ¬nes a functional on š‘‹. Moreover by (2.31) āˆ£(š‘‡ š‘£)(š‘˜)āˆ£ ā‰¤ š‘€1 + š›½š‘€2 < āˆž Therefore š‘‡ š‘£ āˆˆ šµ(š‘‹). 2.19 Let š‘ = {1, 2, 3}. Any individual is powerless so that š‘¤({š‘–}) = 0

š‘– = 1, 2, 3

Any two players can allocate the $1 to between themselves, leaving the other player out. Therefore š‘¤({š‘–, š‘—}) = 1

š‘–, š‘— āˆˆ š‘, š‘– āˆ•= š‘—

The best that the three players can achieve is to allocate the $1 amongst themselves, so that š‘¤(š‘ ) = 1 2.20 If the consumers preferences are continuous and strictly convex, she has a unique optimal choice xāˆ— for every set of prices p and income š‘š in š‘ƒ (Example 1.116). Therefore, the demand correspondence is single valued. 2.21 Assume š‘ āˆ—š‘– āˆˆ šµ(sāˆ— ) for every š‘– āˆˆ š‘ . Then for every player š‘– āˆˆ š‘ (š‘ š‘– , sāˆ’š‘– ) ā‰æš‘– (š‘ ā€²š‘– , sāˆ’š‘– ) for every š‘ ā€²š‘– āˆˆ š‘†š‘– sāˆ— = (š‘ āˆ—1 , š‘ āˆ—2 , . . . , š‘ āˆ—š‘› ) is a Nash equilibrium. Conversely, assume sāˆ— = (š‘ āˆ—1 , š‘ āˆ—2 , . . . , š‘ āˆ—š‘› ) is a Nash equilibrium. Then for every player š‘– āˆˆ š‘ (š‘ š‘– , sāˆ’š‘– ) ā‰æš‘– (š‘ ā€²š‘– , sāˆ’š‘– ) for every š‘ ā€²š‘– āˆˆ š‘†š‘– which implies that š‘ āˆ—š‘– āˆˆ šµ(sāˆ— ) for every š‘– āˆˆ š‘ 2.22 For any nonempty compact set š‘‡ āŠ† š‘†, šµ(š‘‡ ) is nonempty and compact provided ā‰æš‘– is continuous (Proposition 1.5) and šµ(š‘‡ ) āŠ† š‘‡ . Therefore šµš‘–1 āŠ‡ šµš‘–2 āŠ‡ šµš‘–3 . . . is a nested sequence ofāˆ©nonempty compact sets. By the nested intersection theorem āˆž (Exercise 1.117), š‘…š‘– = š‘›=0 šµš‘–š‘› āˆ•= āˆ…. 2.23 If sāˆ— is a Nash equilibrium, š‘ š‘– āˆˆ šµš‘–š‘› for every š‘›.

72

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.24 For any šœ½, let xāˆ— āˆˆ šœ‘(šœ½). Then š‘“ (xāˆ— , šœ½) ā‰„ š‘“ (x, šœ½)

for every x āˆˆ šŗ(šœ½)

Therefore š‘“ (xāˆ— , šœ½) ā‰„ š‘£(šœ½) = sup š‘“ (x, šœ½) xāˆˆšŗ(šœ½)

Conversely š‘£(šœ½) = sup š‘“ (x, šœ½) ā‰„ sup š‘“ (x, šœ½) ā‰„ š‘“ (xāˆ— , šœ½) for every xāˆ— āˆˆ šœ‘(šœ½) xāˆˆšŗ(šœ½)

xāˆˆšŗ(šœ½)

Consequently š‘£(šœ½) = š‘“ (xāˆ— , šœ½) for any xāˆ— āˆˆ šœ‘(šœ½) 2.25 The graph of š‘‰ is graph(š‘‰ ) = { (š‘¦, x) āˆˆ ā„œ+ Ɨ ā„œš‘›+ : x āˆˆ š‘‰ (š‘¦) } while the production possibility set š‘Œ is š‘Œ = { (š‘¦, āˆ’x) āˆˆ ā„œ+ Ɨ ā„œš‘›+ : š‘„ āˆˆ š‘‰ (š‘¦) } Assume that š‘Œ is convex and let (š‘¦ š‘– , xš‘– ) āˆˆ graph(š‘‰ ), š‘– = 1, 2. This means that (š‘¦ 1 , āˆ’x1 ) āˆˆ š‘Œ and (š‘¦ 2 , āˆ’x2 ) āˆˆ š‘Œ Let ĀÆ = š›¼x1 + (1 āˆ’ š›¼)x2 š‘¦ĀÆ = š›¼š‘¦ 1 + (1 āˆ’ š›¼)š‘¦ 2 and x for some 0 ā‰¤ š›¼ ā‰¤ 1. Since š‘Œ is convex (ĀÆ š‘¦ , āˆ’ĀÆ x) = š›¼(š‘¦ 1 , āˆ’x1 ) + (1 āˆ’ š›¼)(š‘¦ 2 , āˆ’x2 ) āˆˆ š‘Œ ĀÆ āˆˆ š‘‰ (ĀÆ ĀÆ ) āˆˆ graph(š‘‰ ). That is graph(š‘‰ ) is convex. and therefore x š‘¦ ) so that (ĀÆ š‘¦, x Conversely, assuming graph(š‘‰ ) is convex, if (š‘¦ š‘– , āˆ’xš‘– ) āˆˆ š‘Œ , š‘– = 1, 2, then (š‘¦ š‘– , xš‘– ) āˆˆ graph(š‘‰ ) and therefore ĀÆ ) āˆˆ graph(š‘‰ ) =ā‡’ x ĀÆ āˆˆ š‘‰ (ĀÆ (ĀÆ š‘¦, x š‘¦ ) =ā‡’ (ĀÆ š‘¦, āˆ’ĀÆ x) āˆˆ š‘Œ so that š‘Œ is convex. 2.26 The graph of šœ‘ is graph(šŗ) = { (šœ½, x) āˆˆ Ī˜ Ɨ š‘‹ : x āˆˆ šŗ(šœ½) } Assume that (šœ½š‘– , xš‘– ) āˆˆ graph(šŗ), š‘– = 1, 2. This means that xš‘– āˆˆ šŗ(šœ½) and therefore š‘” š‘— (x, šœ½) ā‰¤ š‘š‘— for every š‘— and š‘– = 1, 2. Since š‘” š‘— is convex š‘”(š›¼x1 + (1 āˆ’ š›¼)x2 , š›¼šœ½1 + (1 āˆ’ š›¼)šœ½ 2 ) ā‰„ š›¼š‘”(x1 , šœ½ 1 ) + (1 āˆ’ š›¼)š‘”(x2 , šœ½2 ) ā‰„ š‘š‘— Therefore š›¼x1 +(1āˆ’š›¼)x2 āˆˆ šŗ(š›¼šœ½ 1 +(1āˆ’š›¼)šœ½2 ) and (š›¼šœ½ 1 +(1āˆ’š›¼)šœ½ 2 , š›¼x1 +(1āˆ’š›¼)x2 ) āˆˆ graph(šŗ). šŗ is convex. 73

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.27 The identity function š¼š‘‹ : š‘‹ ā†’ š‘‹ is deļ¬ned by š¼š‘‹ (š‘„) = š‘„ for every š‘„ āˆˆ š‘‹. Therefore š‘„2 ā‰»š‘‹ š‘„1 =ā‡’ š¼š‘‹ (š‘„2 ) = š‘„2 ā‰»š‘‹ š‘„1 = š¼š‘‹ (š‘„1 ) 2.28 Assume that š‘“ and š‘” are increasing. Then, for every š‘„1 , š‘„2 āˆˆ š‘‹ with š‘„2 ā‰æš‘‹ š‘„1 š‘“ (š‘„2 ) ā‰æš‘Œ š‘“ (š‘„1 ) =ā‡’ š‘”(š‘“ (š‘„2 )) ā‰æš‘ š‘”(š‘“ (š‘„1 )) š‘” āˆ˜ š‘“ is also increasing. Similarly, if š‘“ and š‘” are strictly increasing, š‘„2 ā‰»š‘‹ š‘„1 =ā‡’ š‘“ (š‘„2 ) ā‰»š‘Œ š‘“ (š‘„1 ) =ā‡’ š‘”(š‘“ (š‘„2 )) ā‰»š‘ š‘”(š‘“ (š‘„1 )) 2.29 For every š‘¦ āˆˆ š‘“ (š‘‹), there exists a unique š‘„ āˆˆ š‘‹ such that š‘“ (š‘„) = y. (For if š‘„1 , š‘„2 are such that š‘“ (š‘„1 ) = š‘“ (š‘„2 ), then š‘„1 = š‘„2 .) Therefore, š‘“ is one-to-one and onto š‘“ (š‘‹), and so has an inverse (Exercise 2.4). Further š‘„2 > š‘„2 ā‡ā‡’ š‘“ (š‘„2 ) > š‘“ (š‘„1 ) Therefore š‘“ āˆ’1 is strictly increasing. 2.30 Assume š‘“ : š‘‹ ā†’ ā„œ is increasing. Then, for every š‘„2 ā‰æ š‘„1 , š‘“ (š‘„2 ) ā‰„ š‘“ (š‘„1 ) which implies that āˆ’š‘“ (š‘„2 ) ā‰¤ āˆ’š‘“ (š‘„1 ). āˆ’š‘“ is decreasing. 2.31 For every š‘„2 ā‰æ š‘„1 . š‘“ (š‘„2 ) ā‰„ š‘“ (š‘„1 ) š‘”(š‘„2 ) ā‰„ š‘”(š‘„1 ) Adding (š‘“ + š‘”)(š‘„2 ) = š‘“ (š‘„2 ) + š‘”(š‘„2 ) ā‰„ š‘“ (š‘„1 ) + š‘“ (š‘„1 ) = (š‘“ + š‘”)(š‘„1 ) That is, š‘“ + š‘” is increasing. Similarly for every š›¼ ā‰„ 0 š›¼š‘“ (š‘„2 ) ā‰„ š›¼š‘“ (š‘„1 ) and therefore š›¼š‘“ is increasing. By Exercise 1.186, the set of all increasing functionals is a convex cone in š¹ (š‘‹). If š‘“ is strictly increasing, then for every š‘„2 ā‰» š‘„1 , š‘“ (š‘„2 ) > š‘“ (š‘„1 ) š‘”(š‘„2 ) ā‰„ š‘”(š‘„1 ) Adding (š‘“ + š‘”)(š‘„2 ) = š‘“ (š‘„2 ) + š‘”(š‘„2 ) > š‘“ (š‘„1 ) + š‘”(š‘„1 ) = (š‘“ + š‘”)(š‘„1 ) š‘“ + š‘” is strictly increasing. Similarly for every š›¼ > 0 š›¼š‘“ (š‘„2 ) > š›¼š‘“ (š‘„1 ) š›¼š‘“ is strictly increasing. 2.32 For every š‘„2 ā‰» š‘„1 . š‘“ (š‘„2 ) > š‘“ (š‘„1 ) > 0 š‘”(š‘„2 ) > š‘”(š‘„1 ) > 0 and therefore (š‘“ š‘”)(š‘„2 ) = š‘“ (š‘„2 )š‘”(š‘„2 ) > š‘“ (š‘„2 )š‘”(š‘„1 ) > š‘“ (š‘„1 )š‘”(š‘„1 ) = (š‘“ š‘”)(š‘„1 ) using Exercise 2.31. 74

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.33 By Exercise 2.31 and Example 2.53, each š‘”š‘› is strictly increasing on ā„œ+ . That is š‘„1 < š‘„2 =ā‡’ š‘”š‘› (š‘„1 ) < š‘”š‘› (š‘„2 ) for every š‘›

(2.32)

and therefore lim š‘”š‘› (š‘„1 ) ā‰¤ lim š‘”š‘› (š‘„2 )

š‘›ā†’āˆž

š‘›ā†’āˆž

This suļ¬ƒces to show that š‘”(š‘„) = limš‘›ā†’āˆž š‘”š‘› (š‘„) is increasing (not strictly increasing). However, 1 + š‘„ is strictly increasing, and therefore by Exercise 2.31 š‘’š‘„ = 1 + š‘„ + š‘”(š‘„) is strictly increasing on ā„œ+ . While it is the case that š‘” = lim š‘”š‘› is strictly increasing on ā„œ+ , (2.32) does not suļ¬ƒce to show this. 2.34 For every š‘Ž > 0, š‘Ž log š‘„ is strictly increasing (Exercise 2.32) and therefore š‘’š‘Ž log š‘„ is strictly increasing (Exercise 2.28). For every š‘Ž < 0, āˆ’š‘Ž log š‘„ is strictly increasing and therefore (Exercise 2.30 š‘Ž log š‘„ is strictly decreasing. Therefore š‘’š‘Ž log š‘„ is strictly decreasing (Exercise 2.28). 2.35 Apply Exercises 2.31 and 2.28 to Example 2.56. 2.36 š‘¢ is (strictly) increasing so that š‘„2 ā‰æ š‘„1 =ā‡’ š‘¢(š‘„2 ) ā‰„ š‘¢(š‘„1 ) To show the converse, assume that š‘„1 , š‘„2 āˆˆ š‘‹ with š‘¢(š‘„2 ) ā‰„ š‘¢(š‘„1 ). Since ā‰æ is complete, either š‘„2 ā‰æ š‘„1 or š‘„1 ā‰» š‘„2 . However, the second possibility cannot occur since š‘¢ is strictly increasing and therefore š‘„1 ā‰» š‘„2 =ā‡’ š‘¢(š‘„1 ) > š‘¢(š‘„2 ) contradicting the hypothesis that š‘¢(š‘„2 ) ā‰„ š‘¢(š‘„1 ). We conclude that š‘¢(š‘„2 ) ā‰„ š‘¢(š‘„1 ) =ā‡’ š‘„2 ā‰æ š‘„1 2.37 Assume š‘¢ represents the preference ordering ā‰æ on š‘‹ and let š‘” : ā„œ ā†’ ā„œ be strictly increasing. Then composition š‘” āˆ˜ š‘¢ : š‘‹ ā†’ ā„œ is strictly increasing (Exercise 2.28). Therefore š‘” āˆ˜ š‘¢ is a utility function (Example 2.58). Since š‘” is strictly increasing š‘”(š‘¢(š‘„2 )) ā‰„ š‘”(š‘¢(š‘„1 )) ā‡ā‡’ š‘¢(š‘„2 ) ā‰„ š‘¢(š‘„1 ) ā‡ā‡’ š‘„2 ā‰æ š‘„1 for every š‘„1 , š‘„2 āˆˆ š‘‹ Therefore, š‘” āˆ˜ š‘¢ also represents ā‰æ. 2.38

ĀÆ = š‘§ĀÆ1 ā‰æ x and therefore zĀÆ āˆˆ š‘x+ . Similarly, 1. (a) Let š‘§ĀÆ = maxš‘›š‘–=1 š‘„š‘– . Then z š‘› let š‘§ = minš‘–=1 š‘„š‘– . Then z = š‘§1 āˆˆ š‘xāˆ’ . Therefore, š‘x+ and š‘xāˆ’ are both nonempty. By continuity, the upper and lower contour sets ā‰æ(x) and ā‰¾(x) are closed. š‘ is a closed cone. Since š‘x+ = ā‰æ(x) āˆ© š‘ and š‘xāˆ’ = ā‰¾(x) āˆ© š‘ š‘x+ and š‘xāˆ’ are closed. (b) By completeness, š‘x+ āˆŖ š‘xāˆ’ = š‘. Since š‘ is connected, š‘x+ āˆ© š‘xāˆ’ āˆ•= āˆ…. (Otherwise, š‘ is the union of two disjoint closed sets and hence the union of two disjoint open sets.) (c) Let zx āˆˆ š‘x+ āˆ© š‘xāˆ’ . Then z ā‰æ x and also z ā‰¾ x. That is, z āˆ¼ x. 75

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c 2001 Michael Carter āƒ All rights reserved

(d) Suppose x āˆ¼ z1x and x āˆ¼ z2x with z1x āˆ•= z2x . Then either z1x > z2x or z1x < z2x . Without loss of generality, assume z2x > z1x . Then monotonicity and transitivity imply x āˆ¼ z2x ā‰» z1x āˆ¼ x which is a contradiction. Therefore zx is unique. Let š‘§x denote the scale of zx , that is zx = š‘§x 1. For every x āˆˆ ā„œš‘›+ , there is a unique zx āˆ¼ x and the function š‘¢ : ā„œš‘›+ ā†’ ā„œ given by š‘¢(x) = š‘§x is well-deļ¬ned. Moreover x2 ā‰æ x1 ā‡ā‡’ zx2 ā‰æ zx1 ā‡ā‡’ š‘§x2 ā‰„ š‘§x1 ā‡ā‡’ š‘¢(x2 ) ā‰„ š‘¢(x1 ) š‘¢ represents the preference order ā‰æ. 2.39

1. For every š‘„1 āˆˆ ā„œ, (š‘„1 , 2) ā‰»šæ (š‘„1 , 1) in the lexicographic order. If š‘¢ represents ā‰æšæ , š‘¢ is strictly increasing and therefore š‘¢(š‘„1 , 2) > š‘¢(š‘„1 , 1). There exists a rational number š‘Ÿ(š‘„1 ) such that š‘¢(š‘„1 , 2) > š‘Ÿ(š‘„1 ) > š‘¢(š‘„1 , 1).

2. The preceding construction associates a rational number with every real number š‘„1 āˆˆ ā„œ. Hence š‘Ÿ is a function from ā„œ to the set of rational numbers š‘„. For any š‘„11 , š‘„21 āˆˆ ā„œ with š‘„21 > š‘„11 š‘Ÿ(š‘„21 ) > š‘¢(š‘„21 , 1) > š‘¢(š‘„11 , 2) > š‘Ÿ(š‘„11 ) Therefore š‘„21 > š‘„11 =ā‡’ š‘Ÿ(š‘„21 ) > š‘Ÿ(š‘„11 ) š‘Ÿ is strictly increasing. 3. By Exercise 2.29, š‘Ÿ has an inverse. This implies that š‘Ÿ is one-to-one and onto, which is impossible since š‘„ is countable and ā„œ is uncountable (Example 2.16). This contradiction establishes that ā‰æšæ has no such representation š‘¢. 2.40 Let a1 , a2 āˆˆ š“ with a1 ā‰æ2 a2 . Since the game is strictly competitive, a2 ā‰æ1 a1 . Since š‘¢1 represents ā‰æ1 , š‘¢1 (a2 ) ā‰„ š‘¢1 (a1 ) which implies that āˆ’š‘¢1 (a2 ) ā‰¤ āˆ’š‘¢1 (a1 ), that is š‘¢2 (a1 ) ā‰„ š‘¢2 (a2 ) where š‘¢2 = āˆ’š‘¢1 . Similarly š‘¢2 (a1 ) ā‰„ š‘¢2 (a2 ) =ā‡’ š‘¢1 (a1 ) ā‰¤ š‘¢1 (a2 ) ā‡ā‡’ a1 ā‰¾1 a2 =ā‡’ a1 ā‰æ2 a2 Therefore š‘¢2 = āˆ’š‘¢1 represents ā‰æ2 and š‘¢1 (a) + š‘¢2 (a) = 0 for every a āˆˆ š“ 2.41 Assume š‘† ā«‹ š‘‡ . By superadditivity š‘¤(š‘‡ ) ā‰„ š‘¤(š‘†) + š‘¤(š‘‡ āˆ– š‘†) ā‰„ š‘¤(š‘†) 2.42 Assume š‘£, š‘¤ āˆˆ šµ(š‘‹) with š‘¤(š‘¦) ā‰„ š‘£(š‘¦) for every š‘¦ āˆˆ š‘‹. Then for any š‘„ āˆˆ š‘‹ š‘“ (š‘„, š‘¦) + š›½š‘¤(š‘¦) ā‰„ š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) for every š‘¦ āˆˆ š‘‹ and therefore (š‘‡ š‘¤)(š‘„) = sup {š‘“ (š‘„, š‘¦) + š›½š‘¤(š‘¦)} ā‰„ sup {š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦)} = (š‘‡ š‘£)(š‘„) š‘¦āˆˆšŗ(š‘„)

š‘¦āˆˆšŗ(š‘„)

T is increasing. 76

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.43 For every šœƒ2 ā‰„ šœƒ1 āˆˆ Ī˜, if š‘„1 āˆˆ šŗ(šœƒ1 ) and š‘„2 āˆˆ šŗ(šœƒ2 ), then š‘„1 āˆ§ š‘„2 ā‰¤ š‘„1 and therefore š‘„1 āˆ§ š‘„2 āˆˆ šŗ(šœƒ1 ). If š‘„1 ā‰„ š‘„2 , then š‘„1 āˆØ š‘„2 = š‘„1 ā‰¤ š‘”(šœƒ1 ) ā‰¤ š‘”(šœƒ2 ) and therefore š‘„1 āˆØ š‘„2 āˆˆ šŗ(šœƒ2 ). On the other hand, if š‘„1 ā‰¤ š‘„2 , then š‘„1 āˆØ š‘„2 = š‘„2 āˆˆ šŗ(šœƒ2 ). 2.44 Assume šœ‘ is increasing, and let š‘„1 , š‘„2 āˆˆ š‘‹ with š‘„2 ā‰æ š‘„1 . Let š‘¦1 āˆˆ šœ‘(š‘„1 ). Choose any š‘¦ ā€² āˆˆ šœ‘(š‘„2 ). Since šœ‘ is increasing, šœ‘(š‘„2 ) ā‰æš‘† šœ‘(š‘„1 ) and therefore š‘¦2 = š‘¦1 āˆØ š‘¦ ā€² āˆˆ šœ‘(š‘„2 ). š‘¦2 ā‰æ š‘¦1 as required. Similarly, for every š‘¦2 āˆˆ šœ‘(š‘„2 ), there exists some š‘¦ ā€² āˆˆ šœ‘(š‘„2 ) such that š‘¦1 = š‘¦ ā€² āˆ§ š‘¦2 āˆˆ šœ‘(š‘„1 ) with š‘¦2 ā‰æ š‘¦1 . 2.45 Since šœ‘(š‘„) is a sublattice, sup šœ‘(š‘„) āˆˆ šœ‘(š‘„) for every š‘„. Therefore, the function š‘“ (š‘„) = sup šœ‘(š‘„) is a selection. Similarly š‘”(š‘„) = inf šœ‘(š‘„) is a selection. Both š‘“ and š‘” are increasing (Exercise 1.50).

āˆ 2.46 Let š‘„1 , š‘„2 belong to š‘‹āˆ with š‘„2 ā‰æ š‘„1 . Choose y1 = (š‘¦11 , š‘¦21 , . . . , š‘¦š‘›1 ) āˆˆ š‘– šœ‘š‘– (š‘„1 ) and y2 = (š‘¦12 , š‘¦22 , . . . , š‘¦š‘›2 ) āˆˆ š‘– šœ‘š‘– (š‘„2 ). Then, for each š‘– = 1, 2, . . . , š‘›, š‘¦š‘–1 āˆˆ šœ‘š‘– (š‘„1 ) and (š‘„1 ) and š‘¦š‘–1 āˆØ š‘¦š‘–2 āˆˆāˆšœ‘š‘– (š‘„2 ) for š‘¦š‘–2 āˆˆ šœ‘š‘– (š‘„2 ). Since each šœ‘š‘– isāˆincreasing, š‘¦š‘–1 āˆ§ š‘¦š‘–2 āˆˆ šœ‘š‘–āˆ 1 2 1 1 2 each š‘–. Therefore y āˆ§ y āˆˆ š‘– šœ‘š‘– (š‘„ ) and y āˆØ y āˆˆ š‘– šœ‘š‘– (š‘„2 ). šœ‘(š‘„) = š‘– šœ‘š‘– (š‘„) is increasing. āˆ© āˆ© 2.47 Let š‘„1 , š‘„2 belong to š‘‹ with š‘„2 ā‰æ š‘„1 . Choose š‘¦ 1 āˆˆ š‘– šœ‘š‘– (š‘„1 ) and š‘¦ 2 āˆˆ š‘– šœ‘š‘– (š‘„2 ). Then š‘¦ 1 āˆˆ šœ‘š‘– (š‘„1 ) and š‘¦ 2 āˆˆ šœ‘š‘– (š‘„2 ) for each š‘– = 1, 2, . . . , š‘›. Since each šœ‘š‘–āˆ©is increasing, šœ‘š‘– (š‘„1 ) and š‘¦ 1 āˆØāˆ©š‘¦ 2 āˆˆ šœ‘š‘– (š‘„2 ) for each š‘–. Therefore š‘¦ 1 āˆ§ š‘¦ 2 āˆˆ š‘– šœ‘š‘– (š‘„1 ) and š‘¦1 āˆ§ š‘¦2 āˆˆ āˆ© 1 2 š‘¦ āˆØ š‘¦ āˆˆ š‘– šœ‘š‘– (š‘„2 ). šœ‘ = š‘– šœ‘ is increasing. 2.48 Let š‘“ be a selection from an always increasing correspondence šœ‘ : š‘‹ ā‡‰ š‘Œ . For every š‘„1 , š‘„2 āˆˆ š‘‹, š‘“ (š‘„1 ) āˆˆ šœ‘(š‘„1 ) and š‘“ (š‘„2 ) āˆˆ šœ‘(š‘„2 ). Since šœ‘ is always increasing š‘„1 ā‰æš‘‹ š‘„2 =ā‡’ š‘“ (š‘„1 ) ā‰æš‘Œ š‘“ (š‘„2 ) š‘“ is increasing. Conversely, assume every selection š‘“ āˆˆ šœ‘ is increasing. Choose any š‘„1 , š‘„2 āˆˆ š‘‹ with š‘„1 ā‰æ š‘„2 . For every š‘¦1 āˆˆ šœ‘(š‘„1 ) and š‘¦2 āˆˆ šœ‘(š‘„2 ), there exists a selection š‘“ with š‘¦š‘– = šœ‘(š‘„š‘– ), š‘– = 1, 2. Since š‘“ is increasing, š‘„1 ā‰æš‘‹ š‘„2 =ā‡’ š‘¦1 ā‰æš‘Œ š‘¦2 šœ‘ is increasing. 2.49 Let š‘„1 , š‘„2 āˆˆ š‘‹. If š‘‹ is a chain, either š‘„1 ā‰æ š‘„2 or š‘„2 ā‰æ š‘„1 . Without loss of generality , assume š‘„2 ā‰æ š‘„1 . Then š‘„1 āˆØ š‘„2 = š‘„2 and š‘„1 āˆ§ š‘„2 = š‘„1 and (2.17) is satisļ¬ed as an identity. 2.50 (š‘“ + š‘”)(š‘„1 āˆØ š‘„2 ) + (š‘“ + š‘”)(š‘„1 āˆ§ š‘„2 ) = š‘“ (š‘„1 āˆØ š‘„2 ) + š‘”(š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) + š‘”(š‘„1 āˆ§ š‘„2 ) = š‘“ (š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) + š‘”(š‘„1 āˆØ š‘„2 ) + š‘”(š‘„1 āˆ§ š‘„2 ) ā‰„ š‘“ (š‘„1 ) + š‘“ (š‘„2 ) + š‘”(š‘„1 ) + š‘”(š‘„2 ) = (š‘“ + š‘”)(š‘„1 ) + (š‘“ + š‘”)(š‘„2 ) Similarly š‘“ (š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) ā‰„ š‘“ (š‘„1 ) + š‘“ (š‘„2 ) 77

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implies š›¼š‘“ (š‘„1 āˆØ š‘„2 ) + š›¼š‘“ (š‘„1 āˆ§ š‘„2 ) ā‰„ š›¼š‘“ (š‘„1 ) + š›¼š‘“ (š‘„2 ) for all š›¼ ā‰„ 0. By Exercise 1.186, the set of all supermodular functions is a convex cone in š¹ (š‘‹). 2.51 Since š‘“ is supermodular and š‘” is nonnegative deļ¬nite, ( ) š‘“ (š‘„1 āˆØ š‘„2 )š‘”(š‘„1 āˆØ š‘„2 ) ā‰„ š‘“ (š‘„1 ) + š‘“ (š‘„2 ) āˆ’ š‘“ (š‘„1 āˆ§ š‘„2 ) š‘”(š‘„1 āˆØ š‘„2 ) ( ) = š‘“ (š‘„2 )š‘”(š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 ) āˆ’ š‘“ (š‘„1 āˆ§ š‘„2 ) š‘”(š‘„1 āˆØ š‘„2 ) for any š‘„1 , š‘„2 āˆˆ š‘‹. Since š‘“ and š‘” are increasing, this implies ( ) š‘“ (š‘„1 āˆØ š‘„2 )š‘”(š‘„1 āˆØ š‘„2 ) ā‰„ š‘“ (š‘„2 )š‘”(š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 ) āˆ’ š‘“ (š‘„1 āˆ§ š‘„2 ) š‘”(š‘„1 )

(2.33)

Similarly, since š‘“ is nonnegative deļ¬nite, š‘” supermodular, and š‘“ and š‘” increasing ( ) š‘“ (š‘„2 )š‘”(š‘„1 āˆØ š‘„2 ) ā‰„ š‘“ (š‘„2 ) š‘”(š‘„1 ) + š‘”(š‘„2 ) āˆ’ š‘”(š‘„1 āˆ§ š‘„2 ) ( ) = š‘“ (š‘„2 )š‘”(š‘„2 ) + š‘“ (š‘„2 ) š‘”(š‘„1 ) āˆ’ š‘”(š‘„1 āˆ§ š‘„2 ) ( ) ā‰„ š‘“ (š‘„2 )š‘”(š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) š‘”(š‘„1 ) āˆ’ š‘”(š‘„1 āˆ§ š‘„2 ) Combining this inequality with (2.33) gives ( ) š‘“ (š‘„1 āˆØ š‘„2 )š‘”(š‘„1 āˆØ š‘„2 ) ā‰„ š‘“ (š‘„2 )š‘”(š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) š‘”(š‘„1 ) āˆ’ š‘”(š‘„1 āˆ§ š‘„2 ) ( ) + š‘“ (š‘„1 ) āˆ’ š‘“ (š‘„1 āˆ§ š‘„2 ) š‘”(š‘„1 ) = š‘“ (š‘„2 )š‘”(š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 )š‘”(š‘„1 ) āˆ’ š‘“ (š‘„1 āˆ§ š‘„2 )š‘”(š‘„1 āˆ§ š‘„2 ) + š‘“ (š‘„1 )š‘”(š‘„1 ) āˆ’ š‘“ (š‘„1 āˆ§ š‘„2 )š‘”(š‘„1 ) = š‘“ (š‘„2 )š‘”(š‘„2 ) āˆ’ š‘“ (š‘„1 āˆ§ š‘„2 )š‘”(š‘„1 āˆ§ š‘„2) + š‘“ (š‘„1 )š‘”(š‘„1 ) or š‘“ š‘”(š‘„1 āˆØ š‘„2 ) + š‘“ š‘”(š‘„1 āˆ§ š‘„2 ) ā‰„ š‘“ š‘”(š‘„1 ) + š‘“ š‘”(š‘„2 ) š‘“ š‘” is supermodular. (I acknowledge the help of Don Topkis in formulating this proof.) 2.52 Exercises 2.49 and 2.50. 2.53 For simplicity, assume that the ļ¬rm produces two products. For every production plan y = (š‘¦1 , š‘¦2 ), y = (š‘¦1 , 0) āˆØ (0, š‘¦2 ) 0 = (š‘¦1 , 0) āˆ§ (0, š‘¦2 ) If š‘ is strictly submodular š‘(w, y) + š‘(w, 0) < š‘(w, (š‘¦1 , 0)) + š‘(w, (0, š‘¦2 )) Since š‘(w, 0) = 0 š‘(w, y) < š‘(w, (š‘¦1 , 0)) + š‘(w, (0, š‘¦2 )) The technology displays economies of scope.

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2.54 Assume (š‘, š‘¤) is convex, that is š‘¤(š‘† āˆŖ š‘‡ ) + š‘¤(š‘† āˆ© š‘‡ ) ā‰„ š‘¤(š‘†) + š‘¤(š‘‡ ) for every š‘†, š‘‡ āŠ† š‘ For all disjoint coalitions š‘† āˆ© š‘‡ = āˆ… š‘¤(š‘† āˆŖ š‘‡ ) ā‰„ š‘¤(š‘†) + š‘¤(š‘‡ ) š‘¤ is superadditive. 2.55 Rewriting (2.18), this implies š‘¤(š‘† āˆŖ š‘‡ ) āˆ’ š‘¤(š‘‡ ) ā‰„ š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ© š‘‡ ) for every š‘†, š‘‡ āŠ† š‘

(2.34)

Let š‘† āŠ‚ š‘‡ āŠ‚ š‘ āˆ– {š‘–} and let š‘† ā€² = š‘† āˆŖ {š‘–}. Substituting in (2.34) š‘¤(š‘† ā€² āˆŖ š‘‡ ) āˆ’ š‘¤(š‘‡ ) ā‰„ š‘¤(š‘† ā€² ) āˆ’ š‘¤(š‘† ā€² āˆ© š‘‡ ) Since š‘† āŠ‚ š‘‡ š‘† ā€² āˆŖ š‘‡ = (š‘† āˆŖ {š‘–}) āˆŖ š‘‡ = š‘‡ āˆŖ {š‘–} š‘† ā€² āˆ© š‘‡ = (š‘† āˆŖ {š‘–}) āˆ© š‘‡ = š‘† Substituting in the previous equation gives the required result, namely š‘¤(š‘‡ āˆŖ {š‘–}) āˆ’ š‘¤(š‘‡ ) ā‰„ š‘¤(š‘† āˆŖ {š‘–}) āˆ’ š‘¤(š‘†) Conversely, assume that š‘¤(š‘‡ āˆŖ {š‘–}) āˆ’ š‘¤(š‘‡ ) ā‰„ š‘¤(š‘† āˆŖ {š‘–}) āˆ’ š‘¤(š‘†)

(2.35)

for every š‘† āŠ‚ š‘‡ āŠ‚ š‘ āˆ– {š‘–}. Let š‘† and š‘‡ be arbitrary coalitions. Assume š‘† āˆ© š‘‡ āŠ‚ š‘† and š‘† āˆ© š‘‡ āŠ‚ š‘‡ (otherwise (2.18) is trivially satisļ¬ed). This implies that š‘‡ āˆ– š‘† āˆ•= āˆ…. Assume these players are labelled 1, 2, . . . , š‘š, that is š‘‡ āˆ– š‘† = {1, 2, . . . , š‘š}. By (2.35) š‘¤(š‘† āˆŖ {1}) āˆ’ š‘¤(š‘†) ā‰„ š‘¤((š‘† āˆ© š‘‡ ) āˆŖ {1}) āˆ’ š‘¤(š‘† āˆ© š‘‡ )

(2.36)

Successively adding the remaining players in š‘‡ āˆ– š‘† š‘¤(š‘† āˆŖ {1, 2}) āˆ’ š‘¤(š‘† āˆŖ {1}) ā‰„ š‘¤((š‘† āˆ© š‘‡ ) āˆŖ {1, 2}) āˆ’ š‘¤((š‘† āˆ© š‘‡ ) āˆŖ {1}) .. . ( ) š‘¤(š‘† āˆŖ (š‘‡ āˆ– š‘†)) āˆ’ š‘¤(š‘† āˆŖ {1, 2, . . . , š‘š āˆ’ 1}) ā‰„ š‘¤ š‘† āˆ© š‘‡ ) āˆŖ (š‘‡ āˆ– š‘†) āˆ’ š‘¤((š‘† āˆ© š‘‡ ) āˆŖ {1, 2, . . . , š‘š āˆ’ 1}) Adding these inequalities to (2.36), we get ( ) š‘¤(š‘† āˆŖ (š‘‡ āˆ– š‘†)) āˆ’ š‘¤(š‘†) ā‰„ š‘¤ š‘† āˆ© š‘‡ ) āˆŖ (š‘‡ āˆ– š‘†) āˆ’ š‘¤(š‘† āˆ© š‘‡ ) This simpliļ¬es to š‘¤(š‘† āˆŖ š‘‡ ) āˆ’ š‘¤(š‘†) ā‰„ š‘†(š‘‡ ) āˆ’ š‘¤(š‘† āˆ© š‘‡ ) which can be arranged to give (2.18).

79

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.56 The cost allocation game is not convex. Let š‘† = {š“š‘ƒ, š¾š‘€ }, š‘‡ = {š¾š‘€, š‘‡ š‘ }. Then š‘† āˆŖ š‘‡ = {š“š‘ƒ, š¾š‘€, š‘‡ š‘ } = š‘ and š‘† āˆ© š‘‡ = {š¾š‘€ } and š‘¤(š‘† āˆŖ š‘‡ ) + š‘¤(š‘† āˆ© š‘‡ ) = 1530 < 1940 = 770 + 1170 = š‘¤(š‘†) + š‘¤(š‘‡ ) Alternatively, observe that TNā€™s marginal contribution to coalition {š¾š‘€, š‘‡ š‘ } is 1170, which is greater than its marginal contribution to the grand coalition {š“š‘ƒ, š¾š‘€, š‘‡ š‘ } (1530 āˆ’ 770 = 760). 2.57 š‘“ is supermodular if š‘“ (š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) ā‰„ š‘“ (š‘„1 ) + š‘“ (š‘„2 ) which can be rearranged to give š‘“ (š‘„1 āˆØ š‘„2 ) āˆ’ š‘“ (š‘„2 ) ā‰„ š‘“ (š‘„1 ) āˆ’ š‘“ (š‘„1 āˆ§ š‘„2 ) If the right hand side of this inequality is nonnegative, then so a fortiori is the left hand side, that is š‘“ (š‘„1 ) ā‰„ š‘“ (š‘„1 āˆ§ š‘„2 ) =ā‡’ š‘“ (š‘„1 āˆØ š‘„2 ) ā‰„ š‘“ (š‘„2 ) If the right hand side is strictly positive, so must be the left hand side š‘“ (š‘„1 ) > š‘“ (š‘„1 āˆ§ š‘„2 ) =ā‡’ š‘“ (š‘„1 āˆØ š‘„2 ) > š‘“ (š‘„2 ) 2.58 Assume š‘„2 ā‰æ š‘„1 āˆˆ š‘‹ and š‘¦2 ā‰æš‘Œ š‘¦2 āˆˆ š‘Œ . Assume that š‘“ displays increasing diļ¬€erences in (š‘„, š‘¦), that is š‘“ (š‘„2 , š‘¦2 ) āˆ’ š‘“ (š‘„1 , š‘¦2 ) ā‰„ š‘“ (š‘„2 , š‘¦1 ) āˆ’ š‘“ (š‘„1 , š‘¦1 )

(2.37)

š‘“ (š‘„2 , š‘¦2 ) āˆ’ š‘“ (š‘„2 , š‘¦1 ) ā‰„ š‘“ (š‘„1 , š‘¦2 ) āˆ’ š‘“ (š‘„1 , š‘¦1 )

(2.38)

Rearranging

Conversely, (2.38) implies (2.37) . 2.59 We showed in the text that supermodularity implies increasing diļ¬€erences. To show that reverse, assume that š‘“ : š‘‹ Ɨ š‘Œ ā†’ ā„œ displays increasing diļ¬€erences in (š‘„, š‘¦). Choose any (š‘„1 , š‘¦1 ), (š‘„2 , š‘¦2 ) āˆˆ š‘‹ Ɨ š‘Œ . If (š‘„1 , š‘¦1 ), (š‘„2 , š‘¦2 ) are comparable, so that either (š‘„1 , š‘¦1 ) ā‰æ (š‘„2 , š‘¦2 ) or (š‘„1 , š‘¦1 ) ā‰¾ (š‘„2 , š‘¦2 ), then (2.17) holds has an equality. Therefore assume that (š‘„1 , š‘¦1 ), (š‘„2 , š‘¦2 ) are incomparable. Without loss of generality, assume that š‘„1 ā‰¾ š‘„2 while š‘¦1 ā‰æ š‘¦2 . (This is where we require that š‘‹ and š‘Œ be chains). This implies (š‘„1 , š‘¦1 ) āˆ§ (š‘„2 , š‘¦2 ) = (š‘„1 , š‘¦2 ) and (š‘„1 , š‘¦1 ) āˆØ (š‘„2 , š‘¦2 ) = (š‘„2 , š‘¦1 ) Increasing diļ¬€erences implies that š‘“ (š‘„2 , š‘¦1 ) āˆ’ š‘“ (š‘„1 , š‘¦1 ) ā‰„ š‘“ (š‘„2 , š‘¦2 ) āˆ’ š‘“ (š‘„1 , š‘¦2 ) which can be rewritten as š‘“ (š‘„2 , š‘¦1 ) + š‘“ (š‘„1 , š‘¦2 ) ā‰„ š‘“ (š‘„1 , š‘¦1 ) + š‘“ (š‘„2 , š‘¦2 ) Substituting (2.39) ( ) ( ) š‘“ (š‘„1 , š‘¦1 ) āˆØ (š‘„2 , š‘¦2 ) + š‘“ (š‘„1 , š‘¦1 ) āˆ§ (š‘„2 , š‘¦2 ) ā‰„ š‘“ (š‘„1 , š‘¦1 ) + š‘“ (š‘„2 , š‘¦2 ) which establishes the supermodularity of š‘“ on š‘‹ Ɨ š‘Œ (2.17). 80

(2.39)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.60 In the standard Bertrand model of oligopoly āˆ™ the strategy space of each ļ¬rm is ā„œ+ , a lattice. āˆ™ š‘¢š‘– (š‘š‘– , pāˆ’š‘– ) is supermodular in š‘š‘– (Exercise 2.51). āˆ™ If the other ļ¬rmā€™s increase their prices from p1āˆ’š‘– to p2āˆ’š‘– , the eļ¬€ect on the demand for ļ¬rm š‘–ā€™s product is āˆ‘ š‘“ (š‘š‘– , p2āˆ’š‘– ) āˆ’ š‘“ (š‘š‘– , p1āˆ’š‘– ) = š‘‘š‘–š‘— (š‘2š‘— āˆ’ š‘1š‘— ) š‘–āˆ•=š‘—

If the goods are gross substitutes, demand for ļ¬rm š‘– increases and the amount of the increase is independent of š‘š‘– . Consequently, the eļ¬€ect on proļ¬t will be increasing in š‘š‘– . That is the payoļ¬€ function (net revenue) has increasing diļ¬€erences in (š‘š‘– , pāˆ’š‘– ). Speciļ¬cally, āˆ‘ š‘¢(š‘š‘– , p2āˆ’š‘– ) āˆ’ š‘¢(š‘š‘– , p1āˆ’š‘– ) = š‘‘š‘–š‘— (š‘š‘– āˆ’ š‘ĀÆš‘– )(š‘2š‘— āˆ’ š‘1š‘— ) š‘–āˆ•=š‘—

For any price increase p2āˆ’š‘– ā‰© p1āˆ’š‘– , the change in proļ¬t š‘¢(š‘š‘– , p2āˆ’š‘– ) āˆ’ š‘¢(š‘š‘– , p1āˆ’š‘– ) is increasing in š‘š‘– . Hence, the Bertrand oligopoly model is a supermodular game. 2.61 Suppose š‘“ displays increasing diļ¬€erences so that for all š‘„2 ā‰æ š‘„1 and š‘¦2 ā‰æ š‘¦1 š‘“ (š‘„2 , š‘¦2 ) āˆ’ š‘“ (š‘„1 , š‘¦2 ) ā‰„ š‘“ (š‘„2 , š‘¦1 ) āˆ’ š‘“ (š‘„1 , š‘¦1 ) Then š‘“ (š‘„2 , š‘¦1 ) āˆ’ š‘“ (š‘„1 , š‘¦1 ) ā‰„ 0 =ā‡’ š‘“ (š‘„2 , š‘¦2 ) āˆ’ š‘“ (š‘„1 , š‘¦2 ) ā‰„ 0 and š‘“ (š‘„2 , š‘¦1 ) āˆ’ š‘“ (š‘„1 , š‘¦1 ) > 0 =ā‡’ š‘“ (š‘„2 , š‘¦2 ) āˆ’ š‘“ (š‘„1 , š‘¦2 ) > 0 2.62 For any šœ½ āˆˆ Ī˜āˆ— , let x1 , x2 āˆˆ šœ‘(šœ½). Supermodularity implies š‘“ (x1 āˆØ x2 , šœ½) + š‘“ (x1 āˆ§ x2 , šœ½) ā‰„ š‘“ (x1 , šœ½) + š‘“ (x2 , šœ½) which can be rearranged to give š‘“ (x1 āˆØ x2 , šœ½) āˆ’ š‘“ (x2 , šœ½) ā‰„ š‘“ (x1 , šœ½) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½)

(2.40)

However x1 and x2 are both maximal in šŗ(šœ½). š‘“ (x2 , šœ½) ā‰„ š‘“ (x1 āˆØ x2 , šœ½) =ā‡’ š‘“ (x1 āˆØ x2 , šœ½) āˆ’ š‘“ (x2 , šœ½) ā‰¤ 0 š‘“ (x1 , šœ½) ā‰„ š‘“ (x1 āˆ§ x2 , šœ½) =ā‡’ š‘“ (x1 , šœ½) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½) ā‰„ 0 Substituting in (2.40), we conclude 0 ā‰„ š‘“ (x1 āˆØ x2 , šœ½) āˆ’ š‘“ (x2 , šœ½) ā‰„ š‘“ (x1 , šœ½) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½) ā‰„ 0 This inequality must be satisļ¬ed as an equality with š‘“ (x1 āˆØ x2 , šœ½) = š‘“ (x2 , šœ½) š‘“ (x1 āˆ§ x2 , šœ½) = š‘“ (x1 , šœ½) That is x1 āˆØ x2 āˆˆ šœ‘(šœ½) and x1 āˆ§ x2 āˆˆ šœ‘(šœ½). By Exercise 2.45, šœ‘ has an increasing selection. 81

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.63 As in the proof of the theorem, let šœ½1 , šœ½ 2 belong to Ī˜ with šœ½ 2 ā‰æ šœ½1 . Choose any optimal solutions x1 āˆˆ šœ‘(šœ½ 1 ) and x2 āˆˆ šœ‘(šœ½2 ). We claim that x2 ā‰æš‘‹ x1 . Assume otherwise, that is assume x2 āˆ•ā‰æš‘‹ x1 . This implies (Exercise 1.44) that x1 āˆ§ x2 āˆ•= x1 . Since x1 ā‰æ x1 āˆ§ x2 , we must have x1 ā‰» x1 āˆ§ x2 . Strictly increasing diļ¬€erences implies š‘“ (x1 , šœ½ 2 ) āˆ’ š‘“ (x1 , šœ½1 ) > š‘“ (x1 āˆ§ x2 , šœ½2 ) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½ 1 ) which can be rearranged to give š‘“ (x1 , šœ½ 2 ) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½2 ) > š‘“ (x1 , šœ½1 ) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½ 1 )

(2.41)

Supermodularity implies š‘“ (x1 āˆØ x2 , šœ½2 ) + š‘“ (x1 āˆ§ x2 , šœ½2 ) ā‰„ š‘“ (x1 , šœ½2 ) + š‘“ (x2 , šœ½ 2 ) which can be rearranged to give š‘“ (x1 āˆØ x2 , šœ½2 ) āˆ’ š‘“ (x2 , šœ½2 ) ā‰„ š‘“ (x1 , šœ½2 ) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½ 2 ) Combining this inequality with (2.41) gives š‘“ (x1 āˆØ x2 , šœ½2 ) āˆ’ š‘“ (x2 , šœ½2 ) > š‘“ (x1 , šœ½1 ) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½ 1 )

(2.42)

However x1 and x2 are optimal for their respective parameter values, that is š‘“ (x2 , šœ½2 ) ā‰„ š‘“ (x1 āˆØ x2 , šœ½ 2 ) =ā‡’ š‘“ (x1 āˆØ x2 , šœ½ 2 ) āˆ’ š‘“ (x2 , šœ½2 ) ā‰¤ 0 š‘“ (x1 , šœ½1 ) ā‰„ š‘“ (x1 āˆ§ x2 , šœ½ 1 ) =ā‡’ š‘“ (x1 , šœ½ 1 ) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½1 ) ā‰„ 0 Substituting in (2.42), we conclude 0 ā‰„ š‘“ (x1 āˆØ x2 , šœ½2 ) āˆ’ š‘“ (x2 , šœ½2 ) > š‘“ (x1 , šœ½1 ) āˆ’ š‘“ (x1 āˆ§ x2 , šœ½ 1 ) ā‰„ 0 This contradiction implies that our assumption that x2 āˆ•ā‰æš‘‹ x1 is false. x2 ā‰æš‘‹ x1 as required. šœ‘ is always increasing. 2.64 The budget correspondence is descending in p and therefore ascending in āˆ’p. Consequently, the indirect utility function š‘£(p, š‘š) =

sup

xāˆˆš‘‹(p,š‘š)

š‘¢(x)

is increasing in āˆ’p, that is decreasing in p. 2.65 ā‡= Let šœ½2 ā‰æ šœ½ 1 and šŗ2 ā‰æš‘† šŗ1 . Select x1 āˆˆ šœ‘(šœ½ 1 , šŗ1 ) and x2 āˆˆ šœ‘(šœ½ 2 , šŗ2 ). Since šŗ2 ā‰æš‘† šŗ1 , x1 āˆ§ x2 āˆˆ šŗ1 . Since x1 is optimal (x1 āˆˆ šœ‘(šœ½1 , šŗ1 )), š‘“ (x1 , šœ½1 ) ā‰„ š‘“ (x1 āˆ§ x2 , šœ½1 ). Quasisupermodularity implies š‘“ (x1 āˆØ x2 , šœ½ 1 ) ā‰„ š‘“ (x2 , šœ½1 ). By the single crossing condition š‘“ (x1 āˆØ x2 , šœ½2 ) ā‰„ š‘“ (x2 , šœ½ 2 ). Therefore x1 āˆØ x2 āˆˆ šœ‘(šœ½2 , šŗ2 ). Similarly, since šŗ2 ā‰æš‘† šŗ1 , x1 āˆØ x2 āˆˆ šŗ(šœ½ 2 ). But x2 is optimal, which implies that š‘“ (x2 , šœ½2 ) ā‰„ š‘“ (x1 āˆØ x2 , šœ½2 ) or š‘“ (x1 āˆØ x2 , šœ½ 2 ) ā‰¤ š‘“ (x2 , šœ½2 ). The single crossing condition implies that a similar inequality holds at šœ½ 1 , that is š‘“ (x1 āˆØ x2 , šœ½1 ) ā‰¤ š‘“ (x2 , šœ½1 ). Quasisupermodularity implies that š‘“ (x1 , šœ½ 1 ) ā‰¤ š‘“ (x1 āˆ§x2 , šœ½ 1 ). Therefore x1 āˆ§x2 āˆˆ šœ‘(šœ½ 1 , šŗ1 ). Since x1 āˆØ x2 āˆˆ šœ‘(šœ½ 2 , šŗ2 ) and x1 āˆ§ x2 āˆˆ šœ‘(šœ½ 1 , šŗ1 ), šœ‘ is increasing in (šœ½, šŗ). =ā‡’ To show that š‘“ is quasisupermodular, suppose that šœ½ is ļ¬xed. Choose any x1 , x2 āˆˆ š‘‹. Let šŗ1 = {x1 , x1 āˆ§ x2 } and šŗ2 = {x2 , x1 āˆØ x2 }. Then šŗ2 ā‰æš‘† šŗ1 . Assume that š‘“ (x1 , šœ½) ā‰„ š‘“ (x1 āˆ§x2 , šœ½). Then x1 āˆˆ šœ‘(šœ½, šŗ1 ) which implies that x1 āˆØx2 āˆˆ šœ‘(šœ½, šŗ2 ). (If x2 āˆˆ šœ‘(šœ½, šŗ2 ), then also x1 āˆØ x2 āˆˆ šœ‘(šœ½, šŗ2 ) since šœ‘ is increasing in (šœ½, šŗ)). But this implies that š‘“ (x1 āˆØ x2 , šœ½) ā‰„ š‘“ (x2 , šœ½). š‘“ is quasisupermodular in š‘‹. 82

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

To show that š‘“ satisļ¬es the single crossing condition, choose any x2 ā‰æ x1 and let šŗ = {x1 , x2 }. Assume that š‘“ (x2 , šœ½1 ) ā‰„ š‘“ (x1 , šœ½ 1 ). Then x2 āˆˆ šœ‘(šœ½1 , šŗ) which implies that x2 āˆˆ šœ‘(šœ½ 2 , šŗ) for any šœ½ 2 ā‰æ šœ½1 . (If x1 āˆˆ šœ‘(šœ½ 2 , šŗ), then also x1 āˆØx2 = x2 āˆˆ šœ‘(šœ½ 2 , šŗ) since šœ‘ is increasing in (šœ½, šŗ).) But this implies that š‘“ (x2 , šœ½2 ) ā‰„ š‘“ (x1 , šœ½2 ). š‘“ satisļ¬es the single crossing condition. 2.66 First, assume that š‘“ is continuous. Let š‘‡ be an open subset in š‘Œ and š‘† = š‘“ āˆ’1 (š‘‡ ). If š‘† = āˆ…, it is open. Otherwise, choose š‘„0 āˆˆ š‘† and let š‘¦0 = š‘“ (š‘„0 ) āˆˆ š‘‡ . Since š‘‡ is open, there exists a neighborhood š‘ (š‘¦0 ) āŠ† š‘‡ . Since š‘“ is continuous, there exists a corresponding neighborhood š‘ (š‘„0 ) with š‘“ (š‘ (š‘„0 )) āŠ† š‘ (š‘“ (š‘„0 )). Since š‘ (š‘“ (š‘„0 )) āŠ† š‘‡ , š‘ (š‘„0 ) āŠ† š‘†. This establishes that for every š‘„0 āˆˆ š‘† there exist a neighborhood š‘ (š‘„0 ) contained in š‘†. That is, š‘† is open in š‘‹. Conversely, assume that the inverse image of every open set in š‘Œ is open in š‘‹. Choose some š‘„0 āˆˆ š‘‹ and let š‘¦0 = š‘“ (š‘„0 ). Let š‘‡ āŠ‚ š‘Œ be a neighborhood of š‘¦0 . š‘‡ contains an open ball šµš‘Ÿ (š‘¦0 ) about š‘¦0 . By hypothesis, the inverse image š‘† = š‘“ āˆ’1 (šµš‘Ÿ (š‘¦0 )) is open in š‘‹. Therefore, there exists a neighborhood š‘ (š‘„0 ) āŠ† š‘“ āˆ’1 (šµš‘Ÿ (š‘¦0 )). Since šµš‘Ÿ (š‘¦0 ) āŠ† š‘‡ , š‘“ (š‘ (š‘„0 )) āŠ† š‘‡ . Since the choice of š‘„0 was arbitrary, we conclude that š‘“ is continuous. 2.67 Assume š‘“ is continuous. Let š‘‡ be a closed set in š‘Œ and let š‘† = š‘“ āˆ’1 (š‘‡ ). Then, š‘‡ š‘ is open. By the previous exercise, š‘“ āˆ’1 (š‘‡ š‘ ) = š‘† š‘ is open and therefore š‘† is closed. Conversely, for every open set š‘‡ āŠ† š‘Œ , š‘‡ š‘ is closed. By hypothesis, š‘† š‘ = š‘“ āˆ’1 (š‘‡ š‘ ) is closed and therefore š‘† = š‘“ āˆ’1 (š‘‡ ) is open. š‘“ is continuous by the previous exercise. 2.68 Assume š‘“ is continuous. Let š‘„š‘› be a sequence converging to š‘„ Let š‘‡ be a neighborhood of š‘“ (š‘„). Since š‘“ is continuous, there exists a neighborhood š‘† āˆ‹ š‘„ such that š‘“ (š‘†) āŠ† š‘‡ . Since š‘„š‘› converges to š‘„, there exists some š‘ such that š‘„š‘› āˆˆ š‘† for all š‘› ā‰„ š‘ . Consequently š‘“ (š‘„š‘› ) āˆˆ š‘‡ for every š‘› ā‰„ š‘ . This establishes that š‘“ (š‘„š‘› ) ā†’ š‘“ (š‘„). Conversely, assume that for every sequence š‘„š‘› ā†’ š‘„, š‘“ (š‘„š‘› ) ā†’ š‘“ (š‘„). We show that if š‘“ were not continuous, it would be possible to construct a sequence which violates this hypothesis. Suppose then that š‘“ is not continuous. Then there exists a neighborhood / š‘‡ . In š‘‡ of š‘“ (š‘„) such that for every neighborhood š‘† of š‘„, there is š‘„ā€² āˆˆ š‘† with š‘“ (š‘„ā€² ) āˆˆ particular, consider the sequence of open balls šµ1/š‘› (š‘„). For every š‘›, choose a point / š‘‡ . Then š‘„š‘› ā†’ š‘„ but š‘“ (š‘„š‘› ) does not converge to š‘“ (š‘„). š‘„š‘› āˆˆ šµ1/š‘› (š‘„) with š‘“ (š‘„š‘› ) āˆˆ This contradicts the assumption. We conclude that š‘“ must be continuous. 2.69 Since š‘“ is one-to-one and onto, it has an inverse š‘” = š‘“ āˆ’1 which maps š‘Œ onto š‘‹. Let š‘† be an open set in š‘‹. Since š‘“ is open, š‘‡ = š‘” āˆ’1 (š‘†) = š‘“ (š‘†) is open in š‘Œ . Therefore š‘” = š‘“ āˆ’1 is continuous. 2.70 Assume š‘“ is continuous. Let (š‘„š‘› , š‘¦ š‘› ) be a sequence of points in graph(š‘“ ) converging to (š‘„, š‘¦). Then š‘¦ š‘› = š‘“ (š‘„š‘› ) and š‘„š‘› ā†’ š‘„. Since š‘“ is continuous, š‘¦ = š‘“ (š‘„) = limš‘›ā†’āˆž š‘“ (š‘„š‘› ) = limš‘›ā†’āˆž š‘¦ š‘› . Therefore (š‘„, š‘¦) āˆˆ graph(š‘“ ) which is therefore closed. 2.71 By the previous exercise, š‘“ continuous implies graph(š‘“ ) closed. Conversely, suppose graph(š‘“ ) is closed and let š‘„š‘› be a sequence converging to š‘„. Then (š‘„š‘› , š‘“ (š‘„š‘› ))) is a sequence in graph(š‘“ ). Since š‘Œ is compact, š‘“ (š‘„š‘› ) contains a subsequence which converges š‘¦. Since graph(š‘“ ) is closed, (š‘„, š‘¦) āˆˆ graph(š‘“ ) and therefore š‘¦ = š‘“ (š‘„) and š‘“ (š‘„š‘› ) ā†’ š‘“ (š‘„). 2.72 Let š‘‡ be an open set in š‘. Since š‘“ and š‘” are continuous, š‘” āˆ’1 (š‘‡ ) is open in š‘Œ and š‘“ āˆ’1 (š‘” āˆ’1 (š‘‡ )) is open in š‘‹. But š‘“ āˆ’1 (š‘” āˆ’1 (š‘‡ )) = (š‘“ āˆ˜ š‘”)āˆ’1 (š‘‡ ). Therefore š‘“ āˆ˜ š‘” is continuous. 2.73 Exercises 1.201 and 2.68.

83

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.74 Let š‘¢ be deļ¬ned as in Exercise 2.38. Let (xš‘› ) be a sequence converging to x. Let š‘§ š‘› = š‘¢(xš‘› ) and š‘§ = š‘¢(x). We need to show that š‘§ š‘› ā†’ š‘§. (š‘§ š‘› ) has a convergent subsequence. Let š‘§ĀÆ = maxš‘– š‘„š‘– and š‘§ = minš‘– š‘„š‘– . Then š‘§ āˆˆ [š‘§, š‘§ĀÆ]. Fix some šœ– > 0. Since xš‘› ā†’ x, there exists some š‘ such that āˆ„xš‘› āˆ’ xāˆ„āˆž < šœ– for every š‘› ā‰„ š‘ . Consequently, for all š‘› ā‰„ š‘ , the terms of the sequence (š‘§ š‘› ) lie in the compact set [š‘§ āˆ’ šœ–, š‘§ĀÆ + šœ–]. Hence, (š‘§ š‘› ) has a convergent subsequence (š‘§ š‘š ). Every convergent subsequence (š‘§ š‘š ) converges to š‘§. Suppose not. That is, suppose there exists a convergent subsequence which converges to š‘§ ā€² . Without loss of generality, assume š‘§ ā€² > š‘§. Let š‘§Ė† = 12 (š‘§ + š‘§ ā€² ) and let z = š‘§1, zā€² = š‘§ ā€² 1, Ė†z = š‘§Ė†1 be the corresponding commodity bundles (see Exercise 2.38). Since š‘§ š‘š ā†’ š‘§ ā€² > š‘§Ė†, there exists some š‘€ such that š‘§ š‘š > š‘§Ė† for every š‘š ā‰„ š‘€ . This implies that xš‘š āˆ¼ zš‘š ā‰» zĖ† for every š‘š ā‰„ š‘€ by monotonicity. Now xš‘š ā†’ x and continuity of preferences implies that x ā‰æ Ė†z. However x āˆ¼ z which implies that z ā‰æ zĖ† which contradicts monotonicity, since Ė† z > z. Consequently, every convergent subsequence (š‘§ š‘š ) converges to š‘§. 2.75 Assume š‘‹ is compact. Let š‘¦ š‘› be a sequence in š‘“ (š‘‹). There exists a sequence š‘„š‘› in š‘‹ with š‘¦ š‘› = š‘“ (š‘„š‘› ). Since š‘‹ is compact, it contains a convergent subsequence š‘„š‘š ā†’ š‘„. If š‘“ is continuous, the subsequence š‘¦ š‘š = š‘“ (š‘„š‘š ) converges in š‘“ (š‘‹) (Exercise 2.68). Therefore š‘“ (š‘‹) is compact. Assume š‘‹ is connected but āˆŖ š‘“ (š‘‹) is not. This means āˆ© there exists open subsets šŗ and š» in š‘Œ such that š‘“ (š‘‹) āŠ‚ šŗ š» and (šŗ āˆ© š‘“ (š‘‹)) (š» āˆ© š‘“ (š‘‹)) = āˆ…. This implies that š‘‹ = š‘“ āˆ’1 (šŗ) āˆŖ š‘“ āˆ’1 (š») is a disconnection of š‘‹, which contradicts the connectedness of š‘‹. 2.76 Let š‘† be any open set in š‘‹. Its complement š‘† š‘ is closed and therefore compact. Consequently, š‘“ (š‘† š‘ ) is compact (Exercise 2.3) and hence closed. Since š‘“ is one-to-one and onto, š‘“ (š‘†) is the complement of š‘“ (š‘† š‘ ), and thus open in š‘Œ . Therefore, š‘“ is an open mapping. By Exercise 2.69, š‘“ āˆ’1 is continuous and š‘“ is a homeomorphism. 2.77 Assume š‘“ continuous. The sets {š‘“ (š‘„) ā‰„ š‘Ž} and {š‘“ (š‘„) ā‰¤ š‘Ž} are closed subsets of the ā„œ and hence ā‰æ(š‘Ž) = š‘“ āˆ’1 {š‘“ (š‘„) ā‰„ š‘Ž} and ā‰¾(š‘Ž) = š‘“ āˆ’1 {š‘“ (š‘„) ā‰¤ š‘Ž} are closed subsets of š‘‹ (Exercise 2.67). Conversely, assume that all upper ā‰æ(š‘Ž) and lower ā‰¾(š‘Ž) contour sets are closed. This implies that the sets ā‰»(š‘Ž) and ā‰ŗ(š‘Ž) are open. Let š“ be an open set in ā„œ. Then for every š‘Ž āˆˆ š“, there exists an open ball šµš‘Ÿš‘Ž (š‘Ž) āŠ† š“ āˆŖ š“= šµš‘Ÿš‘Ž (š‘Ž) š‘Žāˆˆš“

For every š‘Ž āˆˆ š“, šµš‘Ÿš‘Ž (š‘Ž) = (š‘Ž āˆ’ š‘Ÿš‘Ž , š‘Ž + š‘Ÿš‘Ž ) and š‘“ āˆ’1 (šµš‘Ÿš‘Ž (š‘Ž)) = ā‰»(š‘Ž āˆ’ š‘Ÿš‘Ž ) āˆ© ā‰ŗ(š‘Ž + š‘Ÿš‘Ž ) which is open. Consequently āˆŖ āˆŖ ( ) ā‰»(š‘Ž āˆ’ š‘Ÿš‘Ž ) āˆ© ā‰ŗ(š‘Ž + š‘Ÿš‘Ž ) š‘“ āˆ’1 (šµš‘Ÿš‘Ž (š‘Ž)) = š‘“ āˆ’1 (š“) = š‘Žāˆˆš“

š‘Žāˆˆš“

is open. š‘“ is continuous by Exercise 2.66. 84

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.78 Choose any š‘„0 āˆˆ š‘‹ and šœ– > 0. Since š‘“ is continuous, there exists š›æ1 such that šœŒ(š‘„, š‘„0 ) < š›æ1 =ā‡’ āˆ£š‘“ (š‘„) āˆ’ š‘“ (š‘„0 )āˆ£ < šœ–/2 Similarly, there exists š›æ2 such that šœŒ(š‘„, š‘„0 ) < š›æ2 =ā‡’ āˆ£š‘”(š‘„) āˆ’ š‘”(š‘„0 )āˆ£ < šœ–/2 Let š›æ = min{š›æ1 , š›æ2 }. Then, provided šœŒ(š‘„, š‘„0 ) < š›æ āˆ£(š‘“ + š‘”)(š‘„) āˆ’ (š‘“ + š‘”)(š‘„0 )āˆ£ = āˆ£š‘“ (š‘„) + š‘”(š‘„) āˆ’ š‘“ (š‘„0 ) āˆ’ š‘”(š‘„0 )āˆ£ ā‰¤ āˆ£š‘“ (š‘„) āˆ’ š‘“ (š‘„0 )āˆ£ + āˆ£š‘”(š‘„) āˆ’ š‘”(š‘„0 )āˆ£ 0 such that āˆ£š‘“ (š‘„) āˆ’ š‘“ (š‘„0 )āˆ£ < šœ‚ and āˆ£š‘”(š‘„) āˆ’ š‘”(š‘„0 )āˆ£ < šœ‚ whenever šœŒ(š‘„, š‘„0 ) < š›æ. Consequently, while šœŒ(š‘„, š‘„0 ) < š›æ āˆ£š‘“ (š‘„)āˆ£ ā‰¤ āˆ£š‘“ (š‘„) āˆ’ š‘“ (š‘„0 )āˆ£ + āˆ£š‘“ (š‘„0 )āˆ£ < šœ‚ + āˆ£š‘“ (š‘„0 )āˆ£ ā‰¤ 1 + āˆ£š‘“ (š‘„0 )āˆ£ and āˆ£(š‘“ š‘”)(š‘„) āˆ’ (š‘“ š‘”)(š‘„0 )āˆ£ = āˆ£š‘“ (š‘„)š‘”(š‘„) āˆ’ š‘“ (š‘„0 )š‘”(š‘„0 )āˆ£ = āˆ£š‘“ (š‘„)(š‘”(š‘„) āˆ’ š‘”(š‘„0 )) + š‘”(š‘„0 )(š‘“ (š‘„) āˆ’ š‘“ (š‘„0 ))āˆ£ ā‰¤ āˆ£š‘“ (š‘„)āˆ£ āˆ£š‘”(š‘„) āˆ’ š‘”(š‘„0 )āˆ£ + āˆ£š‘”(š‘„0 )āˆ£ āˆ£š‘“ (š‘„) āˆ’ š‘“ (š‘„0 )āˆ£ < šœ‚(1 + āˆ£š‘“ (š‘„0 )āˆ£ + āˆ£š‘”(š‘„0 )āˆ£) Given šœ– > 0, let šœ‚ = min{1, šœ–/(1 + āˆ£š‘“ (š‘„0 )āˆ£ + āˆ£š‘”(š‘„0 )āˆ£)}. Then, we have shown that there exists š›æ > 0 such that šœŒ(š‘„, š‘„0 ) < š›æ =ā‡’ āˆ£(š‘“ š‘”)(š‘„) āˆ’ (š‘“ š‘”)(š‘„0 )āˆ£ < šœ– Therefore, š‘“ š‘” is continuous at š‘„0 . 2.80 Apply Exercises 2.78 and 2.72. 2.81 For any š‘Ž āˆˆ ā„œ, the upper and lower contour sets of š‘“ āˆØ š‘”, namely { š‘„ : max{š‘“ (š‘„), š‘”(š‘„)} ā‰„ š‘Ž} = {š‘„ : š‘“ (š‘„) ā‰„ š‘Ž } āˆŖ { š‘„ : š‘”(š‘„) ā‰„ š‘Ž } { š‘„ : max{š‘“ (š‘„), š‘”(š‘„)} ā‰¤ š‘Ž} = {š‘„ : š‘“ (š‘„) ā‰¤ š‘Ž } āˆ© { š‘„ : š‘”(š‘„) ā‰¤ š‘Ž } are closed. Therefore š‘“ āˆØ š‘” is continuous (Exercise 2.77). Similarly for š‘“ āˆ§ š‘”. 2.82 The set š‘‡ = š‘“ (š‘‹) is compact (Proposition 2.3). We want to show that š‘‡ has both largest and smallest elements. Assume otherwise, that is assume that š‘‡ has no largest element. Then, the set of intervals {(āˆ’āˆž, š‘”) : š‘” āˆˆ š‘‡ } forms an open covering of š‘‡ . Since š‘‡ is compact, there exists a ļ¬nite subcollection of intervals {(āˆ’āˆž, š‘”1 ), (āˆ’āˆž, š‘”2 ), . . . , (āˆ’āˆž, š‘”š‘› )} which covers š‘‡ . Let š‘”āˆ— be the largest of these š‘”š‘– . Then š‘”āˆ— does not belong to any of the intervals {(āˆ’āˆž, š‘”1 ), (āˆ’āˆž, š‘”2 ), . . . , (āˆ’āˆž, š‘”š‘› )}, contrary to the fact that they cover š‘‡ . This contradiction shows that, contrary to our assumption, there must exist a largest element š‘”āˆ— āˆˆ š‘‡ , that is š‘”āˆ— ā‰„ š‘” for all š‘” āˆˆ š‘‡ . Let š‘„āˆ— āˆˆ š‘“ āˆ’1 (š‘”āˆ— ). Then š‘”āˆ— = š‘“ (š‘„āˆ— ) ā‰„ š‘“ (š‘„) for all š‘„ āˆˆ š‘‹. The existence of a smallest element is proved analogously. 85

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.83 By Proposition 2.3, š‘“ (š‘‹) is connected and hence an interval (Exercise 1.95). 2.84 The range š‘“ (š‘‹) is a compact subset of ā„œ (Proposition 2.3). Therefore š‘“ is bounded (Proposition 1.1). Ėœ 2.85 Let š¶(š‘‹) denote the set of all continuous (not necessarily bounded) functionals on š‘‹. Then Ėœ š¶(š‘‹) = šµ(š‘‹) āˆ© š¶(š‘‹) Ėœ šµ(š‘‹), š¶(š‘‹) are a linear subspaces of the set of all functionals š¹ (š‘‹) (Exercises 2.11, Ėœ 2.78 respectively). Therefore š¶(š‘‹) = šµ(š‘‹) āˆ© š¶(š‘‹) is a subspace of š¹ (š‘‹) (Exercise 1.130). Clearly š¶(š‘‹) āŠ† šµ(š‘‹). Therefore š¶(š‘‹) is a linear subspace of šµ(š‘‹). Let š‘“ be a bounded function in the closure of š¶(š‘‹), that is š‘“ āˆˆ š¶(š‘‹). We show that š‘“ is continuous. For any šœ– > 0, there exists š‘“0 āˆˆ š¶(š‘‹) such that āˆ„š‘“ āˆ’ š‘“0 āˆ„ < šœ–/3. Therefore āˆ£š‘“ (š‘„) āˆ’ š‘“0 (š‘„)āˆ£ < šœ–/3 for every š‘„ āˆˆ š‘‹. Choose some š‘„0 āˆˆ š‘‹. Since š‘“0 is continuous, there exists š›æ > 0 such that šœŒ(š‘„, š‘„0 ) < š›æ =ā‡’ āˆ£š‘“0 (š‘„) āˆ’ š‘“0 (š‘„0 )āˆ£ < šœ–/3 Therefore, for every š‘„ āˆˆ š‘‹ such that šœŒ(š‘„, š‘„0 ) < š›æ āˆ£š‘“ (š‘„) āˆ’ š‘“ (š‘„0 )āˆ£ = āˆ£š‘“ (š‘„) āˆ’ š‘“0 (š‘„) + š‘“0 (š‘„) āˆ’ š‘“0 (š‘„0 ) + š‘“0 (š‘„0 ) āˆ’ š‘“ (š‘„0 )āˆ£ ā‰¤ āˆ£š‘“ (š‘„) āˆ’ š‘“0 (š‘„)āˆ£ + āˆ£š‘“0 (š‘„) āˆ’ š‘“0 (š‘„0 )āˆ£ + āˆ£š‘“0 (š‘„0 ) āˆ’ š‘“ (š‘„0 )āˆ£ < šœ–/3 + šœ–/3 + šœ–/3 = šœ– Therefore š‘“ is continuous at š‘„0 . Since š‘„0 was arbitrary, we conclude that is continuous everywhere, that is š‘“ āˆˆ š¶(š‘‹). Therefore š¶(š‘‹) = š¶(š‘‹) and š¶(š‘‹) is closed in šµ(š‘‹). Since šµ(š‘‹) is complete (Exercise 2.11), we conclude that š¶(š‘‹) is complete (Exercise 1.107). Therefore š¶(š‘‹) is a Banach space. 2.86 For every š›¼ āˆˆ ā„œ, { š‘„ : š‘“ (š‘„) ā‰„ š›¼ } = {š‘„ : āˆ’š‘“ (š‘„) ā‰¤ āˆ’š›¼ } and therefore { š‘„ : š‘“ (š‘„) ā‰„ š›¼ } is closed ā‡ā‡’ {š‘„ : āˆ’š‘“ (š‘„) ā‰¤ āˆ’š›¼ } is closed 2.87 Exercise 2.77. 2.88 1 implies 2 Suppose š‘“ is upper semi-continuous. Let š‘„š‘› be a sequence converging to š‘„0 . Assume š‘“ (š‘„š‘› ) ā†’ šœ‡. For every š›¼ < šœ‡, there exists some š‘ such that š‘“ (š‘„š‘› ) > š›¼ for every š‘› ā‰„ š‘ . Hence š‘„0 āˆˆ { š‘„ : š‘“ (š‘„) ā‰„ š›¼ } = { š‘„ : š‘“ (š‘„) ā‰„ š›¼ } since š‘“ is upper semi-continuous. That is, š‘“ (š‘„0 ) ā‰„ š›¼ for every š›¼ < šœ‡. Hence š‘“ (š‘„0 ) ā‰„ šœ‡ = limš‘›ā†’āˆž š‘“ (š‘„š‘› ). 2 implies 3 Let (š‘„š‘› , š‘¦ š‘› ) be a sequence in hypo š‘“ which converges to (š‘„, š‘¦). That is, š‘„š‘› ā†’ š‘„, š‘¦ š‘› ā†’ š‘¦ and š‘¦ š‘› ā‰¤ š‘“ (š‘„š‘› ). Condition 2 implies that š‘“ (š‘„) ā‰„ š‘¦. Hence, (š‘„, š‘¦) āˆˆ hypo š‘“ . Therefore hypo š‘“ is closed. 3 implies 1 For ļ¬xed š›¼ āˆˆ ā„œ, let š‘„š‘› be a sequence in { š‘„ : š‘“ (š‘„) ā‰„ š›¼ }. Suppose š‘„š‘› ā†’ š‘„0 . Then, the sequence (š‘„š‘› , š›¼) converges to (š‘„0 , š›¼) āˆˆ hypo š‘“ . Hence š‘“ (š‘„0 ) ā‰„ š›¼ and š‘„0 āˆˆ { š‘„ : š‘“ (š‘„) ā‰„ š›¼ }, which is therefore closed (Exercise 1.106). 86

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 2.89 Let š‘€ = supš‘„āˆˆš‘‹ š‘“ (š‘„), so that š‘“ (š‘„) ā‰¤ š‘€ for every š‘„ āˆˆ š‘‹

(2.43)

There exists a sequence š‘„š‘› in š‘‹ with š‘“ (š‘„š‘› ) ā†’ š‘€ . Since š‘‹ is compact, there exists a convergent subsequence š‘„š‘š ā†’ š‘„āˆ— and š‘“ (š‘„š‘š ) ā†’ š‘€ . However, since š‘“ is upper semi-continuous, š‘“ (š‘„āˆ— ) ā‰„ lim š‘“ (š‘„š‘š ) = š‘€ . Combined with (2.43), we conclude that š‘“ (š‘„āˆ— ) = š‘€ . 2.90 Choose some šœ– > 0. Since š‘“ is uniformly continuous, there exists some š›æ > 0 such that šœŒ(š‘“ (š‘„š‘š ), š‘“ (š‘„š‘› )) < šœ– for every š‘„š‘š , š‘„š‘› āˆˆ š‘‹ such that šœŒ(š‘„š‘š , š‘„š‘› ) < š›æ. Let (š‘„š‘› ) be a Cauchy sequence in š‘‹. There exists some š‘ such that šœŒ(š‘„š‘š , š‘„š‘› ) < š›æ for every š‘š, š‘› ā‰„ š‘ . Uniform continuity implies that šœŒ(š‘“ (š‘„š‘š ), š‘“ (š‘„š‘› )) < šœ– for every š‘š, š‘› ā‰„ š‘ . (š‘“ (š‘„š‘› )) is a Cauchy sequence. 2.91 Suppose not. That is, suppose š‘“ is continuous but not uniformly continuous. Then there exists some šœ– > 0 such that for š‘› = 1, 2, . . . , there exist points š‘„1š‘› , š‘„2š‘› such that šœŒ(š‘„1š‘› , š‘„2š‘› ) < 1/š‘› but šœŒ(š‘“ (š‘„1š‘› ), š‘“ (š‘„2š‘› )) ā‰„ šœ–

(2.44)

Since š‘‹ is compact, (š‘„1š‘› ) has a subsequence (š‘„1š‘š ) converging to some š‘„ āˆˆ š‘‹. By construction (šœŒ(š‘„1š‘› , š‘„2š‘› ) < 1/š‘›), the sequence (š‘„2š‘š ) also converges to š‘„ and by continuity lim š‘“ (š‘„1š‘š ) = lim š‘“ (š‘„2š‘š )

š‘šā†’āˆž

š‘šā†’āˆž

which contradicts (2.44). 2.92 Assume š‘“ is Lipschitz with constant š›½. For any šœ– > 0, let š›æ = šœ–/2š›½. Then, provided šœŒ(š‘„, š‘„0 ) ā‰¤ š›æ šœŒ(š‘“ (š‘„), š‘“ (š‘„0 )) ā‰¤ š›½šœŒ(š‘„, š‘„0 ) = š›½š›æ = š›½

šœ– šœ– = 0. š¹ is totally bounded (Exercise 1.113), so that there exist ļ¬nite set of functions {š‘“1 , š‘“2 , . . . , š‘“š‘› } in F such that š‘›

min āˆ„š‘“ āˆ’ š‘“š‘˜ āˆ„ ā‰¤ šœ–/3 š‘˜=1

Each š‘“š‘˜ is uniformly continuous (Exercise 2.91), so that there exists š›æš‘˜ > 0 such that šœŒ(š‘„, š‘„0 ) ā‰¤ š›æ =ā‡’ šœŒ(š‘“š‘˜ (š‘„), š‘“š‘˜ (š‘„0 ) < šœ–/3 Let š›æ = min{š›æ1 , š›æ2 , . . . , š›æš‘˜ }. Given any š‘“ āˆˆ š¹ , let š‘˜ be such that āˆ„š‘“ āˆ’ š‘“š‘˜ āˆ„ < šœ–/3. Then for any š‘„, š‘„0 āˆˆ š‘‹, šœŒ(š‘„, š‘„0 ) ā‰¤ š›æ implies šœŒ(š‘“ (š‘„), š‘“ (š‘„0 ) ā‰¤ šœŒ(š‘“ (š‘„), š‘“š‘˜ (š‘„)) + šœŒ(š‘“š‘˜ (š‘„), š‘“š‘˜ (š‘„0 )) + šœŒ(š‘“š‘˜ (š‘„0 ), š‘“ (š‘„0 )) <

šœ– šœ– šœ– + + =šœ– 3 3 3

for every š‘“ āˆˆ š¹ . Therefore, š¹ is equicontinuous. Conversely, assume that š¹ āŠ† š¶(š‘‹) is closed, bounded and equicontinuous. Let (š‘“š‘› ) be a bounded equicontinuous sequence of functions in š¹ . We show that (š‘“š‘› ) has a convergent subsequence. 1. First, we show that for any šœ– > 0, there is exists a subsequence (š‘“š‘š ) such that āˆ„š‘“š‘š āˆ’ š‘“š‘šā€² āˆ„ < šœ– for every š‘“š‘š , š‘“š‘šā€² in the subsequence. Since the functions are equicontinuous, there exists š›æ > 0 such that šœŒ(š‘“š‘› (š‘„) āˆ’ š‘“š‘› (š‘„0 ) <

šœ– 3

for every š‘„, š‘„0 in š‘‹ with šœŒ(š‘„, š‘„0 ) ā‰¤ š›æ. Since š‘‹ is compact, it is totally bounded (Exercise 1.113). That is, there exist a ļ¬nite number of open balls šµš›æ (š‘„š‘– ), š‘– = 1, 2 . . . , š‘˜ which cover š‘‹. The sequence (š‘“š‘› (š‘„1 ), š‘“š‘› (š‘„2 , . . . , š‘“š‘› (š‘„š‘˜ )) is a bounded sequence in ā„œš‘› . By the Bolzano-Weierstrass theorem (Exercise 1.119), this sequence has a convergent subsequence (š‘“š‘š (š‘„1 ), š‘“š‘š (š‘„2 ), . . . , š‘“š‘š (š‘„š‘˜ )) such that š‘“š‘š (š‘„š‘– ) āˆ’ š‘“š‘šā€² (š‘„š‘– ) < šœ–/3 for š‘– and every š‘“š‘š , š‘“š‘šā€² in the subsequence. Consequently, for any š‘„ āˆˆ š‘‹, there exists š‘– such that šœŒ(š‘“š‘š (š‘„), š‘“š‘šā€² (š‘„) ā‰¤ šœŒ(š‘“š‘š (š‘„), š‘“š‘š (š‘„š‘– )) + šœŒ(š‘“š‘š (š‘„š‘– ), š‘“š‘šā€² (š‘„š‘– )) + šœŒ(š‘“š‘šā€² (š‘„š‘– ), š‘“š‘šā€² (š‘„)) šœ– šœ– šœ– < + + =šœ– 3 3 3 That is, āˆ„š‘“š‘š āˆ’ š‘“š‘šā€² āˆ„ < šœ– for every š‘“š‘š , š‘“š‘šā€² in the subsequence. 2. Choose a ball šµ1 of radius 1 in š¶(š‘‹) which contains inļ¬nitely many elements of (š‘“š‘› ). Applying step 1, there exists a ball šµ2 of radius 1/2 containing inļ¬nitely many elements of (š‘“š‘› ). Proceeding in this fashion, we obtain a nested sequence šµ1 āŠ‡ šµ2 āŠ‡ . . . of balls in š¶(š‘‹) such that (a) š‘‘(šµš‘– ) ā†’ 0 and (b) each šµš‘– contains inļ¬nitely many terms of (š‘“š‘› ). Choosing š‘“š‘›š‘– āˆˆ šµš‘– gives a convergent subsequence. 88

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

2.96 Let š‘” āˆˆ š¹ . Then for every šœ– > 0 there exists š›æ > 0 and š‘“ āˆˆ š¹ such that āˆ„š‘“ āˆ’ š‘”āˆ„ < šœ–/3 and šœŒ(š‘„, š‘„0 ) ā‰¤ š›æ =ā‡’ šœŒ(š‘“ (š‘„), š‘“ (š‘„0 ) < šœ–/3 so that if šœŒ(š‘„, š‘„0 ) ā‰¤ š›æ āˆ„š‘”(š‘„) āˆ’ š‘”(š‘„0 )āˆ„ ā‰¤ āˆ„š‘“ (š‘„) āˆ’ š‘”(š‘„)āˆ„ + āˆ„š‘“ (š‘„) āˆ’ š‘“ (š‘„0 )āˆ„ + āˆ„š‘“ (š‘„0 ) āˆ’ š‘”(š‘„0 )āˆ„ <

šœ– šœ– šœ– + + =šœ– 3 3 3

2.97 For every š‘‡ āŠ† š‘Œ šœ‘āˆ’ (š‘‡ š‘ ) = { š‘„ āˆˆ š‘‹ : šœ‘(š‘„) āˆ© š‘‡ š‘ āˆ•= āˆ… } šœ‘+ (š‘‡ ) = { š‘„ āˆˆ š‘‹ : šœ‘(š‘„) āŠ† š‘‡ } For every x āˆˆ š‘‹ either šœ‘(š‘„) āŠ† š‘‡ or šœ‘(š‘„) āˆ© š‘‡ š‘ āˆ•= āˆ… but not both. Therefore šœ‘+ (š‘‡ ) āˆŖ šœ‘āˆ’ (š‘‡ š‘ ) = š‘‹ šœ‘+ (š‘‡ ) āˆ© šœ‘āˆ’ (š‘‡ š‘ ) = āˆ… That is

( )š‘ šœ‘+ (š‘‡ ) = šœ‘āˆ’ (š‘‡ š‘ )

2.98 Assume š‘„ āˆˆ šœ‘(š‘‡ )āˆ’1 . Then šœ‘(š‘„) = š‘‡ , šœ‘(š‘„) āŠ† š‘‡ and š‘„ āˆˆ šœ‘+ (š‘‡ ). Now assume š‘„ āˆˆ šœ‘+ (š‘‡ ) so that šœ‘(š‘„) āŠ† š‘‡ . Consequently, šœ‘(š‘„) āˆ© š‘‡ = šœ‘(š‘„) āˆ•= āˆ… and š‘„ āˆˆ šœ‘āˆ’ (š‘‡ ). 2.99 The respective inverses are: {š‘”1 } {š‘”2 } {š‘”1 , š‘”2 } {š‘”2 , š‘”3 } {š‘”1 , š‘”2 , š‘”3 }

šœ‘āˆ’1 2 āˆ… āˆ… {š‘ 1 } {š‘ 2 } āˆ…

šœ‘+ 2 āˆ… āˆ… {š‘ 1 } {š‘ 2 } {š‘ 1 , š‘ 2 }

šœ‘āˆ’ 2 {š‘ 1 } {š‘ 1 , š‘ 2 } {š‘ 1 , š‘ 2 } {š‘ 1 , š‘ 2 } {š‘ 1 , š‘ 2 }

2.100 Let š‘‡ be an open interval meeting šœ‘(1), that is šœ‘(1) āˆ© š‘‡ āˆ•= āˆ…. Since šœ‘(1) = {1}, we must have 1 āˆˆ š‘‡ and therefore šœ‘(š‘„) āˆ© š‘‡ āˆ•= āˆ… for every š‘„ āˆˆ š‘‹. Therefore šœ‘ is lhc at š‘„ = 1. On the other hand, the open interval š‘‡ = (1/2, 3/2) contains šœ‘(1) but it does not contain šœ‘(š‘„) for any š‘„ > 1. Therefore, šœ‘ is not uhc at š‘„ = 1. 2.101 Choose any open set š‘‡ āŠ† š‘Œ and š‘„ āˆˆ š‘‹. Since šœ‘(š‘„) = š¾ = šœ‘(š‘„ā€² ) for every š‘„, š‘„ā€² āˆˆ š‘‹ āˆ™ šœ‘(š‘„) āŠ† š‘‡ if and only if šœ‘(š‘„ā€² ) āŠ† š‘‡ for every š‘„, š‘„ā€² āˆˆ š‘‹ āˆ™ šœ‘(š‘„) āˆ© š‘‡ āˆ•= āˆ… if and only if šœ‘(š‘„ā€² ) āˆ© š‘‡ āˆ•= āˆ… for every š‘„, š‘„ā€² āˆˆ š‘‹. Consequently, šœ‘ is both uhc and lhc at all š‘„ āˆˆ š‘‹. 2.102 First assume that the šœ‘ is uhc. Let š‘‡ be any open subset in š‘Œ and š‘† = šœ‘+ (š‘‡ ). If š‘† = āˆ…, it is open. Otherwise, choose š‘„0 āˆˆ š‘† so that šœ‘(š‘„0 ) āŠ† š‘‡ . Since šœ‘ is uhc, there exists a neighborhood š‘†(š‘„0 ) such that šœ‘(š‘„) āŠ† š‘‡ for every š‘„ āˆˆ š‘†(š‘„0 ). That is, š‘†(š‘„0 ) āŠ† šœ‘+ (š‘‡ ) = š‘†. This establishes that for every š‘„0 āˆˆ š‘† there exist a neighborhood š‘†(š‘„0 ) contained in š‘†. That is, š‘† is open in š‘‹. Conversely, assume that the upper inverse of every open set in š‘Œ is open in š‘‹. Choose some š‘„0 āˆˆ š‘‹ and let š‘‡ be an open set containing šœ‘(š‘„0 ). Let š‘† = šœ‘+ (š‘‡ ). š‘† is an open set containing š‘„0 . That is, š‘† is a neighborhood of š‘„0 with šœ‘(š‘„) āŠ† š‘‡ for every š‘„ āˆˆ š‘†. Since the choice of š‘„0 was arbitrary, we conclude that šœ‘ is uhc. The lhc case is analogous. 89

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.103 Assume šœ‘ is uhc and š‘‡ be any closed set in š‘Œ . By Exercise 2.97 [ ] šœ‘āˆ’ (š‘‡ ) = šœ‘+ (š‘‡ š‘ ) š‘‡ š‘ is open. By the previous exercise, šœ‘+ (š‘‡ š‘ ) is open which implies that šœ‘āˆ’ (š‘‡ ) is closed. Conversely, assume šœ‘āˆ’ (š‘‡ ) is closed for every closed set š‘‡ . Let š‘‡ be an open subset of š‘Œ so that š‘‡ š‘ is closed. Again by Exercise 2.97, [ ] šœ‘+ (š‘‡ ) = šœ‘āˆ’ (š‘‡ š‘ ) By assumption šœ‘āˆ’ (š‘‡ š‘ ) is closed and therefore šœ‘+ (š‘‡ ) is open. By the previous exercise, šœ‘ is uhc. The lhc case is analogous. 2.104 Assume that šœ‘ is uhc at š‘„0 . We ļ¬rst show that (š‘¦ š‘› ) is bounded and hence has a convergent subsequence. Since šœ‘(š‘„0 ) is compact, there exists a bounded open set š‘‡ containing šœ‘(š‘„0 ). Since šœ‘ is uhc, there exists a neighborhood š‘† of š‘„0 such that šœ‘(š‘„) āŠ† š‘‡ for š‘„ āˆˆ š‘†. Since š‘„š‘› ā†’ š‘„0 , there exists some š‘ such that š‘„š‘› āˆˆ š‘† for every š‘› ā‰„ š‘ . Consequently, šœ‘(š‘„š‘› ) āŠ† š‘‡ for every š‘› ā‰„ š‘ and therefore š‘¦ š‘› āˆˆ š‘‡ for every š‘› ā‰„ š‘ . The sequence š‘¦ š‘› is bounded and hence has a convergent subsequence š‘¦ š‘š ā†’ š‘¦0 . To complete the proof, we have to show that š‘¦0 āˆˆ šœ‘(š‘„0 ). Assume not, assume that / šœ‘(š‘„0 ). Then, there exists an open set š‘‡ containing šœ‘(š‘„0 ) such that š‘¦0 āˆˆ / š‘‡ š‘¦0 āˆˆ (Exercise 1.93). Since šœ‘ is uhc, there exists š‘ such that šœ‘(š‘„š‘› ) āŠ† š‘‡ for every š‘› ā‰„ š‘ . This implies that š‘¦ š‘š āˆˆ š‘‡ for every š‘š ā‰„ š‘ . Since š‘¦ š‘š ā†’ š‘¦0 , we conclude that š‘¦0 āˆˆ š‘‡ , contradicting the speciļ¬cation of š‘‡ . Conversely, suppose that for every sequence š‘„š‘› ā†’ š‘„0 , š‘¦ š‘› āˆˆ šœ‘(š‘„š‘› ), there is a subsequence of š‘¦ š‘š ā†’ š‘¦0 āˆˆ šœ‘(š‘„0 ). Suppose that šœ‘ is not uhc at š‘„0 . That is, there exists an open set š‘‡ āŠ‡ šœ‘(š‘„0 ) such that every neighborhood contains some š‘„ with šœ‘(š‘„) āˆ•āŠ† š‘‡ . From the sequence of neighborhoods šµ1/š‘› (š‘„0 ), we can construct a sequence š‘„š‘› ā†’ š‘„ and š‘¦ š‘› āˆˆ šœ‘(š‘„š‘› ) but š‘¦ š‘› āˆˆ / š‘‡ . Such a sequence cannot have a subsequence which converges to š‘¦ 0 āˆˆ šœ‘(š‘„) , contradicting the hypothesis. We conclude that šœ‘ must be uhc at š‘„0 . 2.105 Assume that šœ‘ is lhc. Let š‘„š‘› be a sequence converging to š‘„0 and š‘¦0 āˆˆ šœ‘(š‘„0 ). Consider the sequence of open balls šµ1/š‘š (š‘¦0 ), š‘š = 1, 2, . . . . Note that every šµ1/š‘š (š‘¦0 ) meets šœ‘(š‘„0 ). Since šœ‘ is lhc, there exists a sequence (š‘† š‘š ) of neighborhoods of š‘„0 such that šœ‘(š‘„) āˆ© šµ1/š‘š āˆ•= āˆ… for every š‘„ āˆˆ š‘† š‘š . Since š‘„š‘› ā†’ š‘„, for every š‘š, there exists some š‘š‘š such that š‘„š‘› āˆˆ š‘†š‘š for every š‘› ā‰„ š‘š‘š . Without loss of generality, we can assume that š‘1 < š‘2 < š‘3 . . . . We can now construct the desired sequence š‘¦ š‘› . For each š‘› = 1, 2, . . . , choose š‘¦ š‘› in the set šœ‘(š‘„š‘› ) āˆ© šµ 1/m where š‘š‘š ā‰¤ š‘› ā‰¤ š‘š‘š+1 since š‘› ā‰„ š‘š‘š =ā‡’ š‘„š‘› āˆˆ š‘†š‘š =ā‡’ šœ‘(š‘„š‘› ) āˆ© šµ1/š‘š āˆ•= āˆ… Since š‘¦ š‘› āˆˆ šµ 1/m (š‘¦0 ), the sequence (š‘¦ š‘› ) converges to š‘¦0 and š‘› ā†’ āˆž. Conversely, assume that šœ‘ is not lhc at š‘„0 , that is there exists an open set š‘‡ with š‘‡ āˆ© šœ‘(š‘„0 ) āˆ•= āˆ… such that every neighborhood š‘† āˆ‹ š‘„0 contains some š‘„ with šœ‘(š‘„) āˆ© š‘‡ = āˆ…. Therefore, there exists a sequence š‘„š‘› ā†’ š‘„ with šœ‘(š‘„)āˆ©š‘‡ = āˆ…. Choose any š‘¦0 āˆˆ šœ‘(š‘„0 )āˆ©š‘‡ . By assumption, there exists a sequence š‘¦ š‘› ā†’ š‘¦ with š‘¦ š‘› āˆˆ šœ‘(š‘„š‘› ). Since š‘‡ is open and š‘¦0 āˆˆ š‘‡ , there exists some š‘ such that š‘¦ š‘› āˆˆ š‘‡ for all š‘› ā‰„ š‘ , for which šœ‘(š‘¦ š‘› ) āˆ© š‘‡ āˆ•= āˆ…. This contradiction establishes that šœ‘ is lhc at š‘„0 . 90

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 2.106

1. Assume šœ‘ is closed. For any š‘„ āˆˆ š‘‹, let (š‘¦ š‘› ) be a sequence in šœ‘(š‘„). Since šœ‘ is closed, š‘¦ š‘› ā†’ š‘¦ āˆˆ šœ‘(š‘„). Therefore šœ‘(š‘„) is closed.

2. Assume šœ‘ is closed-valued and uhc. Choose any (š‘„, š‘¦) āˆˆ / graph(šœ‘). Since šœ‘(š‘„) is closed, there exist disjoint open sets š‘‡1 and š‘‡2 in š‘Œ such that š‘¦ āˆˆ š‘‡1 and šœ‘(š‘„) āŠ† š‘‡2 (Exercise 1.93). Since šœ‘ is uhc, šœ‘+ (š‘‡2 ) is a neighborhood of š‘„. Therefore šœ‘+ (š‘‡2 ) Ɨ š‘‡1 is a neighborhood of (š‘„, š‘¦) disjoint from graph(šœ‘). Therefore the complement of graph(šœ‘) is open, which implies that graph(šœ‘) is closed. 3. Since šœ‘ is closed and š‘Œ compact, šœ‘ is compact-valued. Let (š‘„š‘› ) ā†’ š‘„ be a sequence in š‘‹ and (š‘¦ š‘› ) a sequence in š‘Œ with š‘¦ š‘› āˆˆ šœ‘(š‘„š‘› ). Since š‘Œ is compact, there exists a subsequence š‘¦ š‘š ā†’ š‘¦. Since šœ‘ is closed, š‘¦ āˆˆ šœ‘(š‘„). Therefore, by Exercise 2.104, šœ‘ is uhc. 2.107 Assume šœ‘ is closed-valued and uhc. Then šœ‘ is closed (Exercise 2.106). Conversely, if šœ‘ is closed, then šœ‘(š‘„) is closed for every š‘„ (Exercise 2.106). If š‘Œ is compact, then šœ‘ is compact-valued (Exercise 1.110). By Exercise 2.104, šœ‘ is uhc. 2.108 šœ‘1 is closed-valued (Exercise 2.106). Similarly, šœ‘2 is closed-valued (Proposition 1.1). Therefore, for every š‘„ āˆˆ š‘‹, šœ‘(š‘„) = šœ‘1 (š‘„) āˆ© šœ‘2 (š‘„) is closed (Exercise 1.85) and hence compact (Exercise 1.110). Hence šœ‘ is compact-valued. Now, for any š‘„0 āˆˆ š‘‹, let š‘‡ be an open neighborhood of šœ‘(š‘„0 ). We need to show that there is a neighborhood š‘† of š‘„0 such that šœ‘(š‘†) āŠ† š‘‡ . Case 1 š‘‡ āŠ‡ šœ‘2 (š‘„0 ): Since šœ‘2 is uhc, there exists a neighborhood such that š‘† āˆ‹ š‘„0 such that šœ‘2 (š‘†) āŠ† š‘‡ which implies that šœ‘(š‘†) āŠ† šœ‘2 (š‘†) āŠ† š‘‡ Case 2 š‘‡ āˆ•āŠ‡ šœ‘2 (š‘„0 ): Let š¾ = šœ‘2 (š‘„0 ) āˆ– š‘‡ āˆ•= āˆ…. For every š‘¦ āˆˆ š¾, there exist neighborhoods š‘†š‘¦ (š‘„0 ) and š‘‡ (š‘¦) such that šœ‘1 (š‘†š‘¦ (š‘„0 )) āˆ© š‘‡ (š‘¦) = āˆ… (Exercise 1.93). The sets š‘‡ (š‘¦) constitute an open covering of š¾. Since š¾ is compact, there exists a ļ¬nite subcover, that is there exists a ļ¬nite number of elements š‘¦1 , š‘¦2 , . . . š‘¦š‘› such that š‘› āˆŖ

š¾āŠ†

š‘‡ (š‘¦š‘– )

š‘–=1

āˆŖš‘› Let š‘‡ (š¾) denote š‘–=1 š‘‡ (š‘¦š‘– ). Note that š‘‡ āˆŖš‘‡ (š¾) is an open set containing šœ‘2 (š‘„0 ). Since šœ‘2 is uhc, there exists a neighborhood š‘† ā€² (š‘„0 ) such that šœ‘2 (š‘† ā€² (š‘„0 )) āŠ† š‘‡ āˆŖ š‘‡ (š¾). Let š‘†(š‘„0 ) =

š‘› āˆ©

š‘†š‘¦š‘– (š‘„0 ) āˆ© š‘† ā€² (š‘„0 )

š‘–=1

š‘†(š‘„0 ) is an open neighborhood of š‘„0 for which šœ‘1 (š‘†(š‘„0 )) āˆ© š‘‡ (š¾) = āˆ… and šœ‘2 (š‘†(š‘„0 )) āŠ† š‘‡ āˆŖ š‘‡ (š¾) from which we conclude that šœ‘(š‘†(š‘„0 )) = šœ‘1 (š‘†(š‘„0 )) āˆ© šœ‘2 (š‘†(š‘„0 )) āŠ† š‘‡ āˆ‘š‘› 2.109 1. Let x āˆˆ š‘‹(p, š‘š) āˆ© š‘‡ . Then x āˆˆ š‘‹(p, š‘š) and š‘–=1 š‘š‘– š‘„š‘– ā‰¤ š‘š. Since š‘‡ is Ėœ = š›¼x āˆˆ š‘‡ and open, there exists š›¼ < 1 such that x š‘› āˆ‘ š‘–=1

š‘š‘– š‘„Ėœš‘– = š›¼

š‘› āˆ‘

š‘š‘– š‘„š‘– <

š‘–=1

91

š‘› āˆ‘ š‘–=1

š‘š‘– š‘„š‘– ā‰¤ š‘š

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2. (a) Suppose that š‘‹(p, š‘š) is not lhc. Then for every neighborhood š‘† of (p, š‘š), there exists (pā€² , š‘šā€² ) āˆˆ š‘† such that š‘‹(pā€² , š‘šā€² ) āˆ© š‘‡ = āˆ…. In particular, for every open ball šµš‘› (p, š‘š), there exists a point (pš‘› , š‘šš‘› ) āˆˆ šµš‘› (p, š‘š) such that š‘‹(pš‘› , š‘šš‘› ) āˆ© š‘‡ = āˆ…. ((pš‘› , š‘šš‘› )) is the required sequence. (b) By construction, āˆ„pš‘› āˆ’ pāˆ„ < 1/š‘› ā†’ 0 which implies that š‘š‘›š‘– ā†’ š‘š‘– for every š‘–. Therefore (Exercise 1.202) āˆ‘ āˆ‘ Ėœš‘– ā†’ š‘š‘– š‘„Ėœš‘– < š‘š and š‘šš‘› ā†’ š‘š š‘š‘›š‘– š‘„ and therefore there exists š‘ such that āˆ‘ Ėœ š‘– < š‘šš‘ š‘š‘ š‘– š‘„ which implies that Ėœ āˆˆ š‘‹(pš‘ , š‘šš‘ ) x (c) Also by construction š‘‹(pš‘ , š‘šš‘ ) āˆ© š‘‡ = āˆ… which implies š‘‹(pš‘ , š‘šš‘ ) āŠ† š‘‡ š‘ and therefore Ėœ āˆˆ š‘‹(pš‘› , š‘šš‘› ) =ā‡’ x Ėœāˆˆ x /š‘‡ Ėœ āˆˆ The assumption that š‘‹(p, š‘š) is not lhc at (p, š‘š) implies that x / š‘‡ , contraĖœ āˆˆ š‘‡. dicting the conclusion in part 1 that x 3. This contradiction establishes that (p, š‘š) is lhc at (p, š‘š). Since the choice of (p, š‘š) was arbitrary, we conclude that the budget correspondence š‘‹(p, š‘š) is lhc for all (p, š‘š) āˆˆ š‘ƒ (assuming š‘‹ = ā„œš‘›+ ). 4. In the previous example (Example 2.89), we have shown that š‘‹(p, š‘š) is uhc. Hence, āˆ‘ the budget correspondence is continuous for all (p, š‘š) such that š‘š > š‘š inf xāˆˆš‘‹ š‘–=1 š‘š‘– š‘„š‘– . 2.110 We give two alternative proofs. Proof 1 Let š’ž = {š‘†} be an open cover of šœ‘(š¾). For every š‘„ āˆˆ š¾, šœ‘(š‘„) āŠ† šœ‘(š¾) is compact and hence can be covered by a ļ¬nite number of the sets š‘† āˆˆ š’ž. Let š‘†š‘„ denote the union of the ļ¬nite cover of šœ‘(š‘„). Since šœ‘ is uhc, every šœ‘+ (š‘†š‘„ ) is open in š‘‹. Therefore { šœ‘+ (š‘†š‘„ ) : š‘„ āˆˆ š¾ } is an open covering of š¾. If š¾ is compact, it contains an ļ¬nite covering { šœ‘+ (š‘†š‘„1 ), šœ‘+ (š‘†š‘„2 ), . . . , šœ‘+ (š‘†š‘„š‘› ) }. The sets š‘†š‘„1 , š‘†š‘„2 , . . . , š‘†š‘„š‘› are a ļ¬nite subcovering of šœ‘(š¾). Proof 2 Let (š‘¦ š‘› ) be a sequence in šœ‘(š¾). We have to show that (š‘¦ š‘› ) has a convergent subsequence with a limit in šœ‘(š¾). For every š‘¦ š‘› , there is an š‘„š‘› with š‘¦ š‘› āˆˆ šœ‘(š‘„š‘› ). Since š¾ is compact, the sequence (š‘„š‘› ) has a convergent subsequence š‘„š‘š ā†’ š‘„ āˆˆ š¾. Since šœ‘ is uhc, the sequence (š‘¦ š‘š ) has a subsequence (š‘¦ š‘ ) which converges to š‘¦ āˆˆ šœ‘(š‘„) āŠ† šœ‘(š¾). Hence the original sequence (š‘¦ š‘› ) has a convergent subsequence. 2.111 The sets š‘‹, šœ‘(š‘‹), šœ‘2 (š‘‹), . . . form a sequence of nonempty compact sets. Since šœ‘(š‘‹) āŠ† š‘‹, šœ‘2 (š‘‹) āŠ† šœ‘(š‘‹) and so on, the sequence of sets šœ‘š‘› š‘‹ is decreasing. Let š¾=

āˆž āˆ©

šœ‘š‘› (š‘‹)

š‘›=1

By the nested intersection theorem (Exercise 1.117), š¾ āˆ•= āˆ…. Since š¾ āŠ† šœ‘š‘›āˆ’1 (š‘‹), šœ‘(š¾) āŠ† šœ‘š‘› (š‘‹) for every š‘›, which implies that šœ‘(š¾) āŠ† š¾. 92

Solutions for Foundations of Mathematical Economics To show that š¾ that š‘¦ āˆˆ šœ‘(š‘„š‘› ). š‘„š‘š āˆˆ šœ‘š‘š (š‘‹) for (Exercise 2.107),

c 2001 Michael Carter āƒ All rights reserved

āŠ† šœ‘(š¾), let š‘¦ āˆˆ š¾. For every š‘› there exists an š‘„š‘› āˆˆ šœ‘š‘› (š‘‹) such Since š‘‹ is compact, there exists a subsequence š‘„š‘š ā†’ š‘„0 . Since every š‘š, š‘„0 āˆˆ š¾. The sequence (š‘„š‘š , š‘¦) ā†’ (š‘„0 , š‘¦). Since šœ‘ is closed š‘¦ āˆˆ šœ‘(š‘„0 ). Therefore š‘¦ āˆˆ šœ‘(š¾) which implies that š¾ āŠ† šœ‘(š¾).

2.112 šœ‘(š‘„) is compact for every š‘„ āˆˆ š‘‹ by Tychonoļ¬€ā€™s theorem (Proposition 1.2). Let š‘„š‘˜ ā†’ š‘„ be a sequence in š‘‹ and let š‘¦ š‘˜ = (š‘¦1š‘˜ , š‘¦2š‘˜ , . . . , š‘¦š‘›š‘˜ ) with š‘¦š‘–š‘˜ āˆˆ šœ‘(š‘„š‘˜ ) be a corresponding sequence of points in š‘Œ . For each š‘¦š‘–š‘˜ , š‘– = 1, 2, . . . , š‘›, there exists a ā€² subsequence š‘¦š‘–š‘˜ ā†’ š‘¦š‘– with š‘¦š‘– āˆˆ šœ‘š‘– (š‘„) (Exercise 2.104). Therefore š‘¦ = (š‘¦1 , š‘¦2 , . . . , š‘¦š‘› ) āˆˆ šœ‘(š‘„) which implies that šœ‘ is uhc. 2.113 Let š‘£ āˆˆ š¶(š‘‹). For every x āˆˆ š‘‹, the maximand š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) is a continuous function on a compact set šŗ(š‘„). Therefore the supremum is attained, and max can replace sup in the deļ¬nition of the operator š‘‡ (Theorem 2.2). š‘‡ š‘£ is the value function for the constrained optimization problem max { š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) }

š‘¦āˆˆšŗ(š‘„)

satisfying the requirements of the continuous maximum theorem (Theorem 2.3), which ensures that š‘‡ š‘£ is continuous on š‘‹. We have previously shown that š‘‡ š‘£ is bounded (Exercise 2.18). Therefore š‘‡ š‘£ āˆˆ š¶(š‘‹). 2.114

1. š‘† has a least upper bound since š‘‹ is a complete lattice. Let š‘ āˆ— = sup š‘†. Then š‘† āˆ— = ā‰æ(š‘ āˆ— ) is a complete sublattice of š‘‹ (Exercise 1.48).

2. For every š‘  āˆˆ š‘†, š‘  ā‰¾ š‘ āˆ— and since š‘“ is increasing and š‘  is a ļ¬xed point š‘  = š‘“ (š‘ ) ā‰¾ š‘“ (š‘ āˆ— ) Therefore š‘“ (š‘ āˆ— ) āˆˆ š‘† āˆ— . (š‘“ (š‘ āˆ— ) is an upper bound of š‘†). Again, since š‘“ is increasing, this implies that š‘“ (š‘„) ā‰æ š‘“ (š‘ āˆ— ) for every š‘„ āˆˆ š‘† āˆ— . Therefore š‘“ (š‘† āˆ— ) āŠ† š‘† āˆ— . 3. Let š‘” be the restriction of š‘“ to the sublattice š‘† āˆ— . Since š‘“ (š‘† āˆ— ) āŠ† š‘† āˆ— , š‘” is an increasing function on a complete lattice. Applying Theorem 2.4, š‘” has a smallest ļ¬xed point š‘„ Ėœ. 4. š‘„ Ėœ is a ļ¬xed point of š‘“ , that is š‘„ Ėœ āˆˆ šø. Furthermore, š‘„Ėœ āˆˆ š‘† āˆ— . Therefore š‘„Ėœ is an upper bound for š‘† in šø. Moreover, š‘„ Ėœ is the smallest ļ¬xed point of š‘“ in š‘† āˆ— . Therefore, š‘„ Ėœ is the least upper bound of š‘† in šø. 5. By Exercise 1.47, this implies that šø is a complete lattice. In Example 2.91, if š‘† = {(2, 1), (1, 2)}, š‘† āˆ— = {(2, 2), (3, 2), (2, 3), (3, 3)} and š‘„Ėœ = (3, 3). 2.115

1. For every š‘„ āˆˆ š‘€ , there exists some š‘¦š‘„ āˆˆ šœ‘(š‘„) such that š‘¦š‘„ ā‰¾ š‘„. Moreover, š‘„) such that since šœ‘ is increasing and š‘„ Ėœ ā‰¾ š‘„, there exists some š‘§š‘„ āˆˆ šœ‘(Ėœ š‘§š‘„ ā‰¾ š‘¦š‘„ ā‰¾ š‘„ for every š‘„ āˆˆ š‘€

2. Let š‘§Ėœ = inf{š‘§š‘„ } Ėœ. (a) Since š‘§š‘„ ā‰¾ š‘„ for every š‘„ āˆˆ š‘€ , š‘§Ėœ = inf{š‘§š‘„ } ā‰¾ inf{š‘„} = š‘„ (b) Since šœ‘(Ėœ š‘„) is a complete sublattice of š‘‹, š‘§Ėœ = inf{š‘§š‘„ } āˆˆ šœ‘(Ėœ š‘„). 3. Therefore, š‘„ Ėœ āˆˆ š‘€. 4. Since š‘§Ėœ ā‰¾ š‘„Ėœ and šœ‘ is increasing, there exists some š‘¦ āˆˆ šœ‘(Ėœ š‘§ ) such that š‘¦ ā‰¾ š‘§Ėœ āˆˆ šœ‘(Ėœ š‘„) Hence š‘§Ėœ āˆˆ š‘€ . 93

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

5. This implies that š‘„ Ėœ ā‰¾ š‘§Ėœ. Therefore š‘„ Ėœ = š‘§Ėœ āˆˆ šœ‘(Ėœ š‘„) š‘„ Ėœ is a ļ¬xed point of šœ‘. 6. Since šø āŠ† š‘€ , š‘„Ėœ = inf š‘€ is the least ļ¬xed point of šœ‘. 2.116

1. Let š‘† āŠ† šø and š‘ āˆ— = sup š‘†. For every š‘„ āˆˆ š‘†, š‘„ āˆˆ šœ‘(š‘„). Since šœ‘ is increasing, there exists some š‘§š‘„ āˆˆ šœ‘(š‘ āˆ— ) such that š‘§š‘„ ā‰æ š‘„.

2. Let š‘§ āˆ— = sup š‘§š‘„ . Then (a) Since š‘§š‘„ ā‰æ š‘„ for every š‘„ āˆˆ š‘†, š‘§ āˆ— = sup š‘§š‘„ ā‰æ sup š‘„ = š‘ āˆ— (b) š‘§ āˆ— āˆˆ šœ‘(š‘ āˆ— ) since šœ‘(š‘ āˆ— ) is a complete sublattice. 3. Deļ¬ne š‘† āˆ— = { š‘„ āˆˆ š‘‹ : š‘„ ā‰æ š‘  for every š‘  āˆˆ š‘† } š‘† āˆ— is the set of all upper bounds of š‘† in š‘‹. Then š‘† āˆ— is a complete lattice, since š‘† āˆ— = ā‰æ(š‘ āˆ— ) 4. Let šœ‡ : š‘† āˆ— ā‡‰ š‘† āˆ— be the correspondence šœ‡(š‘„) = šœ‘(š‘„) āˆ© šœ“(š‘„) where šœ“ : š‘† āˆ— ā‡‰ š‘† āˆ— is the constant correspondence deļ¬ned by šœ“(š‘„) = š‘† āˆ— for every š‘„ āˆˆ š‘† āˆ— . Then (a) Since šœ‘ is increasing, for every š‘„ ā‰æ š‘ āˆ— , there exists some š‘¦š‘„ āˆˆ šœ‘(š‘„) such that š‘¦š‘„ ā‰æ š‘ āˆ— . Therefore šœ‡(š‘„) āˆ•= āˆ… for every š‘„ āˆˆ š‘† āˆ— . (b) Both šœ‘(š‘„) and šœ“(š‘„) are complete sublattices for every š‘„ āˆˆ š‘† āˆ— . Therefore šœ‡(š‘„) is a complete sublattice for every š‘„ āˆˆ š‘† āˆ— . (c) Since both šœ‘ and šœ“ are increasing on š‘† āˆ— , šœ‡ is increasing on š‘† āˆ— (Exercise 2.47). 5. By the previous exercise, šœ‡ has a least ļ¬xed point š‘„Ėœ. 6. š‘„ Ėœ āˆˆ š‘† āˆ— is an upper bound of š‘†. Therefore š‘„ Ėœ is the least upper bound of š‘† in šø. 7. By the previous exercise, šø has a least element. Since we have shown every subset š‘† āŠ† šø has a least upper bound, this establishes that šø is complete lattice (Exercise 1.47). 2.117 For any š‘–, let a1āˆ’š‘– , a2āˆ’š‘– āˆˆ š“āˆ’š‘– with a2āˆ’š‘– ā‰æ a1āˆ’š‘– . Let š‘Ž ĀÆ1š‘– = š‘“ (a1āˆ’š‘– ) and š‘Ž ĀÆ2š‘– = š‘“ (a2āˆ’š‘– ). 2 1 1 1 We want to show that š‘Ž ĀÆš‘– ā‰æ š‘Ž ĀÆš‘– . Since š‘Ž ĀÆš‘– āˆˆ šµ(aāˆ’š‘– ) and šµ(aāˆ’š‘– ) is increasing, there ĀÆ1š‘– . (Exercise 2.44). Therefore exists some š‘Žš‘– āˆˆ šµ(a2āˆ’š‘– ) such that š‘Žš‘– ā‰æ š‘Ž sup šµ(aāˆ’š‘– ) = š‘Ž ĀÆ2š‘– ā‰æ š‘Žš‘– ā‰æ š‘Ž ĀÆ1š‘– š‘“ĀÆš‘– is increasing. 2.118 For any player š‘–, their best response correspondence šµš‘– (aāˆ’š‘– ) is 1. increasing by the monotone maximum theorem (Theorem 2.1). 2. a complete sublattice of š“š‘– for every aāˆ’š‘– āˆˆ š“āˆ’š‘– (Corollary 2.1.1). 94

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Solutions for Foundations of Mathematical Economics The joint best response correspondence

šµ(a) = šµ1 (aāˆ’1 ) Ɨ šµ2 (aāˆ’2 ) Ɨ ā‹… ā‹… ā‹… Ɨ šµš‘› (aāˆ’š‘› ) is also 1. increasing (Exercise 2.46) 2. a complete sublattice of š“ for every a āˆˆ š“ Therefore, the best response correspondence šµ(a) satisļ¬es the conditions of Zhouā€™s theorem, which implies that the set šø of ļ¬xed points of šµ is a nonempty complete lattice. šø is precisely the set of Nash equilibria of the game. 2.119 In proving the theorem, we showed that šœŒ(š‘„š‘› , š‘„š‘›+š‘š ) ā‰¤

š›½š‘› šœŒ(š‘„0 , š‘„1 ) 1āˆ’š›½

for every š‘š, š‘› ā‰„ 0. Letting š‘š ā†’ āˆž, š‘„š‘›+š‘š ā†’ š‘„ and therefore šœŒ(š‘„š‘› , š‘„) ā‰¤

š›½š‘› šœŒ(š‘„0 , š‘„1 ) 1āˆ’š›½

Similarly, for every š‘›, š‘š ā‰„ 0 šœŒ(š‘„š‘› , š‘„š‘›+š‘š ) ā‰¤ šœŒ(š‘„š‘› , š‘„š‘›+1 ) + šœŒ(š‘„š‘›+1 , š‘„š‘›+2 ) + ā‹… ā‹… ā‹… + šœŒ(š‘„š‘›+š‘šāˆ’1 , š‘„š‘›+š‘š ) ā‰¤ (š›½ + š›½ 2 + ā‹… ā‹… ā‹… + š›½ š‘š )šœŒ(š‘„š‘›āˆ’1 , š‘„š‘› ) ā‰¤

š›½(1 āˆ’ š›½ š‘š ) šœŒ(š‘„š‘›āˆ’1 , š‘„š‘› ) 1āˆ’š›½

Letting š‘š ā†’ āˆž, š‘„š‘›+š‘š ā†’ š‘„ and š›½ š‘š ā†’ 0 so that šœŒ(š‘„š‘› , š‘„) ā‰¤

š›½ šœŒ(š‘„š‘›āˆ’1 , š‘„š‘› ) 1āˆ’š›½

2.120 First observe that š‘“ (š‘„) ā‰„ 1 for every š‘„ ā‰„ 1. Therefore š‘“ : š‘‹ ā†’ š‘‹. For any š‘„, š‘§ āˆˆ š‘‹ š‘„ āˆ’ š‘¦ + š‘„2 āˆ’ š‘“ (š‘„) āˆ’ š‘“ (š‘¦) = š‘„āˆ’š‘¦ 2(š‘„ āˆ’ š‘¦) Since

1 š‘„š‘¦

=

1 1 āˆ’ 2 š‘„š‘¦

ā‰¤ 1 for all š‘„, š‘¦ āˆˆ š‘‹ āˆ’

so that

2 š‘¦

š‘“ (š‘„) āˆ’ š‘“ (š‘¦) 1 1 ā‰¤ ā‰¤ 2 š‘„āˆ’š‘¦ 2

   š‘“ (š‘„) āˆ’ š‘“ (š‘¦)  āˆ£š‘“ (š‘„) āˆ’ š‘“ (š‘¦)āˆ£ 1  = ā‰¤  š‘„āˆ’š‘¦  āˆ£š‘„ āˆ’ š‘¦āˆ£ 2

or āˆ£š‘“ (š‘„) āˆ’ š‘“ (š‘¦)āˆ£ ā‰¤ š‘“ is a contraction on š‘‹ with modulus 1/2. 95

1 āˆ£š‘„ āˆ’ š‘¦āˆ£ 2

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

š‘‹ is closed and hence complete (Exercise 1.107). Therefore, š‘“ has a ļ¬xed point. That is, there exists š‘„0 āˆˆ š‘‹ such that š‘„0 = š‘“ (š‘„0 ) =

1 2 (š‘„0 + ) 2 š‘„0

Rearranging

so that š‘„0 =

2š‘„20 = š‘„20 + 2 =ā‡’ š‘„20 = 2

āˆš 2.

Letting š‘„0 = 2 š‘„1 =

3 1 (2 + 1) = 2 2

Using the error bounds in Corollary 2.5.1, āˆš š›½š‘› šœŒ(š‘„š‘› , 2) ā‰¤ šœŒ(š‘„0 , š‘„1 ) 1āˆ’š›½ (1/2)š‘› = 1/2 1/2 1 = š‘› 2 1 < 0.001 = 1024 when š‘› = 10. Therefore, we conclude that 10 iterations are ample to reduce the error below 0.001. Actually, with experience, we can reļ¬ne this a priori estimate. In Example 1.64, we calculated the ļ¬rst ļ¬ve terms of the sequence to be (2, 1.5, 1.416666666666667, 1.41421568627451, 1.41421356237469) We observe that šœŒ(š‘„3 , š‘„4 ) = 1.41421568627451 āˆ’ 1.41421356237469) = 0.0000212389982 so that using the second inequality of Corollary 2.5.1 šœŒ(š‘„4 ,

āˆš

2) ā‰¤

1/2 0.0000212389982 < 0.001 1/2

š‘„4 = 1.41421356237469 is the desired approximation. 2.121 Choose any š‘„0 āˆˆ š‘†. Deļ¬ne the sequence š‘„š‘› = š‘“ (š‘„š‘› ) = š‘“ š‘› (š‘„0 ). Then (š‘„š‘› ) is a Cauchy sequence in š‘† converging to š‘„. Since š‘† is closed, š‘„ āˆˆ š‘†. 2.122 By the Banach ļ¬xed point theorem, š‘“ š‘ has a unique ļ¬xed point š‘„. Let š›½ be the Lipschitz constant of š‘“ š‘ . We have to show š‘„ is a ļ¬xed point of š‘“ šœŒ(š‘“ (š‘„), š‘„) = šœŒ(š‘“ (š‘“ š‘ (š‘„), š‘“ š‘ (š‘„)) = šœŒ(š‘“ š‘ (š‘“ (š‘„), š‘“ š‘ (š‘„)) ā‰¤ š›½šœŒ(š‘“ (š‘„), š‘„) Since š›½ < 1, this implies that šœŒ(š‘“ (š‘„), š‘„) = 0 or š‘“ (š‘„) = š‘„. š‘„ is the only ļ¬xed point of š‘“ Suppose š‘§ = š‘“ (š‘§) is another ļ¬xed point of š‘“ . Then š‘§ is a ļ¬xed point of š‘“ š‘ and šœŒ(š‘„, š‘§) = šœŒ(š‘“ š‘ (š‘„), š‘“ š‘ (š‘§)) ā‰¤ š›½šœŒ(š‘„, š‘§) which implies that š‘„ = š‘§. 96

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2.123 By the Banach ļ¬xed point theorem, for every šœƒ āˆˆ Ī˜, there exists š‘„šœƒ āˆˆ š‘‹ such that š‘“šœƒ (š‘„šœƒ ) = š‘„šœƒ . Choose any šœƒ0 āˆˆ Ī˜. šœŒ(š‘„šœƒ , š‘„šœƒ0 ) = šœŒ(š‘“šœƒ (š‘„šœƒ ), š‘“šœƒ0 (š‘„šœƒ0 )) ā‰¤ šœŒ(š‘“šœƒ (š‘„šœƒ ), š‘“šœƒ (š‘„šœƒ0 )) + šœŒ(š‘“šœƒ (š‘„šœƒ0 ), š‘“šœƒ0 (š‘„šœƒ0 )) ā‰¤ š›½šœŒ(š‘„šœƒ , š‘„šœƒ0 ) + šœŒ(š‘“šœƒ (š‘„šœƒ0 ), š‘“šœƒ0 (š‘„šœƒ0 )) (1 āˆ’ š›½)šœŒ(š‘„šœƒ , š‘„šœƒ0 ) ā‰¤ šœŒ(š‘“šœƒ (š‘„šœƒ0 ), š‘“šœƒ0 (š‘„šœƒ0 )) šœŒ(š‘„šœƒ , š‘„šœƒ0 ) ā‰¤

šœŒ(š‘“šœƒ (š‘„šœƒ0 ), š‘“šœƒ0 (š‘„šœƒ0 )) ā†’0 (1 āˆ’ š›½)

as šœƒ ā†’ šœƒ0 . Therefore š‘„šœƒ ā†’ š‘„šœƒ0 . 2.124

1. Let x be a ļ¬xed point of š‘“ . Then x satisļ¬es x = (š¼ āˆ’ š“)x + c = x āˆ’ š“x + š‘ which implies that š“x = š‘.

2. For any x1 , x2 āˆˆ š‘‹     š‘“ (x1 ) āˆ’ š‘“ (x2 ) = (š¼ āˆ’ š“)(x1 āˆ’ x2 )   ā‰¤ āˆ„š¼ āˆ’ š“āˆ„ x1 āˆ’ x2  Since š‘Žš‘–š‘– = 1, the norm of š¼ āˆ’ š“ is āˆ„š¼ āˆ’ š“āˆ„ = max š‘–

āˆ‘

āˆ£š‘Žš‘–š‘— āˆ£ = š‘˜

š‘—āˆ•=š‘–

and     š‘“ (x1 ) āˆ’ š‘“ (x2 ) ā‰¤ š‘˜ x1 āˆ’ x2  By the assumption of strict diagonal dominance, š‘˜ < 1. Therefore š‘“ is a contraction and has a unique ļ¬xed point x. 2.125

1. šœ‘(š‘„) = { š‘¦ āˆ— āˆˆ šŗ(š‘„) : š‘“ (š‘„, š‘¦ āˆ— ) + š›½š‘£(š‘¦ āˆ— ) = š‘£(š‘„) } = {š‘¦ āˆ— āˆˆ šŗ(š‘„) : š‘“ (š‘„, š‘¦ āˆ— ) + š›½š‘£(š‘¦ āˆ— ) = sup {š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦)}} š‘¦āˆˆšŗ(š‘„)

āˆ—

āˆ—

āˆ—

= {š‘¦ āˆˆ šŗ(š‘„) : š‘“ (š‘„, š‘¦ ) + š›½š‘£(š‘¦ ) ā‰„ š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) for every š‘¦ āˆˆ šŗ(š‘„)} = arg max {š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦)} š‘¦āˆˆšŗ(š‘„)

2. šœ‘(š‘„) is the solution correspondence of a standard constrained maximization problem, with š‘„ as parameter and š‘¦ the decision variable. By assumption the maximand š‘“ (š‘„, š‘¦) = š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) is continuous and the constraint correspondence šŗ(š‘„) is continuous and compact-valued. Applying the continuous maximum theorem (Theorem 2.3), šœ‘ is nonempty, compact-valued and uhc. 3. We have just shown that šœ‘(š‘„) is nonempty for every š‘„ āˆˆ š‘‹. Starting at š‘„0 , choose some š‘„āˆ—1 āˆˆ šœ‘(š‘„0 ). Then choose š‘„āˆ—2 āˆˆ šœ‘(š‘„āˆ—1 ). Proceeding in this way, we can construct a plan xāˆ— = š‘„0 , š‘„āˆ—1 , š‘„āˆ—2 , . . . such that š‘„āˆ—š‘”+1 āˆˆ šœ‘(š‘„āˆ—š‘” ) for every š‘” = 0, 1, 2, . . . .

97

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Solutions for Foundations of Mathematical Economics

4. Since š‘„āˆ—š‘”+1 āˆˆ šœ‘(š‘„āˆ—š‘” ) for every š‘”, x satisļ¬es Bellmanā€™s equation, that is š‘£(š‘„āˆ—š‘” ) = š‘“ (š‘„āˆ—š‘” , š‘„āˆ—š‘”+1 ) + š›½š‘£(š‘„āˆ—š‘”+1 ),

š‘” = 0, 1, 2, . . .

Therefore x is optimal (Exercise 2.17). 2.126

1. In the previous exercise (Exercise 2.125) we showed that the set šœ‘ of solutions to Bellmanā€™s equation (Exercise 2.17) is the solution correspondence of the constrained maximization problem šœ‘(š‘„) = arg max { š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) } š‘¦āˆˆšŗ(š‘„)

This problem satisļ¬es the requirements of the monotone maximum theorem (Theorem 2.1), since the objective function š‘“ (š‘„, š‘¦) + š›½š‘£(š‘¦) āˆ™ supermodular in š‘¦ āˆ™ displays strictly increasing diļ¬€erences in (š‘„, š‘¦) since for every š‘„2 ā‰„ š‘„1 š‘“ (š‘„2 , š‘¦) + š›½š‘£(š‘¦) āˆ’ š‘“ (š‘„1 , š‘¦) + š›½š‘£(š‘¦) = š‘“ (š‘„2 , š‘¦) āˆ’ š‘“ (š‘„1 , š‘¦) āˆ™ šŗ(š‘„) is increasing. By Corollary 2.1.2, šœ‘(š‘„) is always increasing. 2. Let xāˆ— = (š‘„0 , š‘„āˆ—1 , š‘„āˆ—2 , . . . ) be an optimal plan. Then (Exercise 2.17) š‘„āˆ—š‘”+1 āˆˆ šœ‘(š‘„āˆ—š‘” ),

š‘” = 0, 1, 2, . . .

Since šœ‘ is always increasing š‘„āˆ—š‘” ā‰„ š‘„āˆ—š‘”āˆ’1 =ā‡’ š‘„āˆ—š‘”+1 ā‰„ š‘„āˆ—š‘” for every š‘” = 1, 2, . . . . Similarly š‘„āˆ—š‘” ā‰¤ š‘„āˆ—š‘”āˆ’1 =ā‡’ š‘„āˆ—š‘”+1 ā‰¤ š‘„āˆ—š‘” xāˆ— = (š‘„0 , š‘„āˆ—1 , š‘„āˆ—2 , . . . ) is a monotone sequence. 2.127 Let š‘”(š‘„) = š‘“ (š‘„) āˆ’ š‘„. š‘” is continuous (Exercise 2.78) with š‘”(0) ā‰„ 0 and š‘”(1) ā‰¤ 0 By the intermediate value theorem (Exercise 2.83), there exists some point š‘„ āˆˆ [0, 1] with š‘”(š‘„) = 0 which implies that š‘“ (š‘„) = š‘„. 2.128

1. To show that a label min{ š‘– : š›½š‘– ā‰¤ š›¼š‘– āˆ•= 0 } exists for every x āˆˆ š‘†, assume to the contrary that, for some x āˆˆ š‘†, š›½š‘– > š›¼š‘– for every š‘– = 0, 1, . . . , š‘›. This implies š‘› āˆ‘

š›½š‘– >

š‘–=0

š‘› āˆ‘

š›¼š‘– = 1

š‘–=0

contradicting the requirement that š‘› āˆ‘

š›½š‘– = 1 for every š‘“ (x) āˆˆ š‘†

š‘–=0

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2. The barycentric coordinates of vertex xš‘– are š›¼š‘– = 1 with š›¼š‘— = 0 for every š‘— āˆ•= š‘–. Therefore the rule assigns vertex xš‘– the label š‘–. 3. Similarly, if x belongs to a proper face of š‘†, it coordinates relative to the vertices not in that face are 0, and it cannot be assigned a label corresponding to a vertex not in the face. To be concrete, suppose that x āˆˆ conv {x1 , x2 , x4 }. Then x = š›¼1 x1 + š›¼2 x2 + š›¼4 x4 ,

š›¼1 + š›¼2 + š›¼4 = 1

/ {1, 2, 4}. Therefore and š›¼š‘– = 0 for š‘– āˆˆ x +āˆ’ā†’ min{ š‘– : š›½š‘– ā‰¤ š›¼š‘– āˆ•= 0 } āˆˆ {1, 2, 4} 2.129

1. Since š‘† is compact, it is bounded (Proposition 1.1) and therefore it is contained in a suļ¬ƒciently large simplex š‘‡ .

2. By Exercise 3.74, there exists a continuous retraction š‘Ÿ : š‘‡ ā†’ š‘†. The composition š‘“ āˆ˜ š‘Ÿ : š‘‡ ā†’ š‘† āŠ† š‘‡ . Furthermore as the composition of continuous functions, š‘“ āˆ˜ š‘Ÿ is continuous (Exercise 2.72). Therefore š‘“ āˆ˜ š‘Ÿ has a ļ¬xed point xāˆ— āˆˆ š‘‡ , that is š‘“ āˆ˜ š‘Ÿ(xāˆ— ) = xāˆ— . 3. Since š‘“ āˆ˜ š‘Ÿ(x) āˆˆ š‘† for every x āˆˆ š‘‡ , we must have š‘“ āˆ˜ š‘Ÿ(xāˆ— ) = xāˆ— āˆˆ š‘†. Therefore, š‘Ÿ(xāˆ— ) = xāˆ— which implies that š‘“ (xāˆ— ) = xāˆ— . That is, xāˆ— is a ļ¬xed point of š‘“ . 2.130 Convexity of š‘† is required to ensure that there is a continuous retraction of the simplex onto š‘†. 2.131

1. š‘“ (š‘„) = š‘„2 on š‘† = (0, 1) or š‘“ (š‘„) = š‘„ + 1 on š‘† = ā„œ+ .

2. š‘“ (š‘„) = 1 āˆ’ š‘„ on š‘† = [0, 1/3] āˆŖ [2/3, 1]. 3. Let š‘† = [0, 1] and deļ¬ne

{ š‘“ (š‘„) =

1 0

0 ā‰¤ š‘„ < 1/2 otherwise

2.132 Suppose such a function exists. Deļ¬ne š‘“ (x) = āˆ’š‘Ÿ(x). Then š‘“ : šµ ā†’ šµ continously, and has no ļ¬xed point since for āˆ™ x āˆˆ š‘†, š‘“ (x) = āˆ’š‘Ÿ(x) = āˆ’x āˆ•= x āˆ™ x āˆˆ šµ āˆ– š‘†, š‘“ (x) āˆˆ / šµ āˆ– š‘† and thereforeš‘“ (x) āˆ•= x Therefore š‘“ has no ļ¬xed point contradicting Brouwerā€™s theorem. 2.133 Suppose to the contrary that š‘“ has no ļ¬xed point. For every x āˆˆ šµ, let š‘Ÿ(z) denote the point where the line segment from š‘“ (x) through x intersects the boundary š‘† of šµ. Since š‘“ is continuous and š‘“ (x) āˆ•= x, š‘Ÿ is a continuous function from šµ to its boundary, that is a retraction, contradicting Exercise 2.132. We conclude that š‘“ must have a ļ¬xed point. 2.134 No-retraction =ā‡’ Brouwer Note ļ¬rst that the no-retraction theorem (Exercise 2.132) generalizes immediately to a closed ball about 0 of arbitrary radius. Assume that š‘“ is a continuous operator on a compact, convex set š‘† in a ļ¬nite dimensional normed linear space. There exists a closed ball šµ containing š‘† (Proposition 1.1). Deļ¬ne š‘” : šµ ā†’ š‘† by š‘”(y) = { x āˆˆ š‘† : x is closest to y } As in Exercise 2.129, š‘” is well-deļ¬ned, continuous and š‘”(x) = x for every x āˆˆ š‘†. š‘“ āˆ˜ š‘” : šµ ā†’ š‘† āŠ† šµ and has a ļ¬xed point xāˆ— = š‘“ (š‘”(xāˆ— )) by Exercise 2.133. Since 99

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

š‘“ āˆ˜ š‘”(x) āˆˆ š‘† for every x āˆˆ šµ, we must have š‘“ āˆ˜ š‘”(xāˆ— ) = xāˆ— āˆˆ š‘†. Therefore, š‘”(xāˆ— ) = xāˆ— which implies that š‘“ (xāˆ— ) = xāˆ— . That is, xāˆ— is a ļ¬xed point of š‘“ . Brouwer =ā‡’ no-retraction Exercise 2.132. 2.135 Let Ī›š‘˜ , š‘˜ = 1, 2, . . . be a sequence of simplicial partitions of š‘† in which the maximum diameter of the subsimplices tend to zero as š‘˜ ā†’ āˆž. By Spernerā€™s lemma (Proposition 1.3), every partition Ī›š‘˜ has a completely labeled subsimplex with vertices xš‘˜0 , xš‘˜1 , . . . , xš‘˜š‘› . By construction of an admissible labeling, each xš‘˜š‘– belongs to a face containing xš‘– , that is xš‘˜š‘– āˆˆ conv {xš‘– , . . . } and therefore xš‘˜š‘– āˆˆ š“š‘– ,

š‘– = 0, 1, . . . , š‘› ā€²

Since š‘† is compact, each sequence xš‘˜š‘– has a convergent subsequence xš‘˜š‘– . Moreover, since the diameters of the subsimplices converge to zero, these subsequences must converge to the same point, say xāˆ— . That is, ā€²

lim xš‘˜š‘– = xāˆ— ,

š‘– = 0, 1, . . . , š‘›

š‘˜ā€² ā†’āˆž

Since the sets š“š‘– are closed, xāˆ— āˆˆ š“š‘– for every š‘– and therefore š‘› āˆ©

xāˆ— āˆˆ

š“š‘– āˆ•= āˆ…

š‘–=0

2.136

=ā‡’ Let š‘“ : š‘† ā†’ š‘† be a continuous operator on an š‘›-dimensional simplex š‘† with vertices x0 , x1 , . . . , xš‘› . For š‘– = 0, 1, . . . , š‘›, let š“š‘– = { x āˆˆ š‘† : š›½š‘– ā‰¤ š›¼š‘– } where š›¼0 , š›¼1 , . . . , š›¼š‘› and š›½0 , š›½1 , . . . , š›½š‘› are the barycentric coordinates of x and š‘“ (x) respectively. Then āˆ™ š‘“ continuous =ā‡’ š“š‘– closed for every š‘– = 0, 1, . . . , š‘› (Exercise 1.106) āˆ™ Let x āˆˆ conv { xš‘– : š‘– āˆˆ š¼ } for some š¼ āŠ† { 0, 1, . . . , š‘› }. Then āˆ‘

š›¼š‘– = 1 =

š‘› āˆ‘

š›½š‘–

š‘–=0

š‘–āˆˆš¼

which implies that š›½š‘– ā‰¤ š›¼š‘– for some š‘– āˆˆ š¼, so that x āˆˆ š“š‘– . Therefore āˆŖ š“š‘– conv { xš‘– : š‘– āˆˆ š¼ } āŠ† š‘–āˆˆš¼

Therefore the collection š“0 , š“1 , . . . , š“š‘› satisļ¬es the hypotheses of the K-K-M theorem and their intersection is nonempty. That is, there exists xāˆ— āˆˆ

š‘› āˆ©

š“š‘– āˆ•= āˆ… with š›½š‘–āˆ— ā‰¤ š›¼āˆ—š‘– ,

š‘– = 0, 1, . . . , š‘›

š‘–=0

where āˆ‘ š›¼āˆ— and āˆ‘ š›½ āˆ— are the barycentric coordinates of xāˆ— and š‘“ (xāˆ— ) respectively. āˆ— Since š›½š‘– = š›¼āˆ—š‘– = 1, this implies that š›½š‘–āˆ— = š›¼āˆ—š‘–

š‘– = 0, 1, . . . , š‘›

In other words, š‘“ (xāˆ— ) = xāˆ— . 100

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Solutions for Foundations of Mathematical Economics ā‡=

Let š“0 , š“1 , . . . , š“š‘› be closed subsets of an š‘› dimensional simplex š‘† with vertices x0 , x1 , . . . , xš‘› such that āˆŖ conv { xš‘– : š‘– āˆˆ š¼ } āŠ† š“š‘– š‘–āˆˆš¼

for every š¼ āŠ† { 0, 1, . . . , š‘› }. For š‘– = 0, 1, . . . , š‘›, let š‘”š‘– (x) = šœŒ(x, š“š‘– ) For any x āˆˆ š‘† with barycentric coordinates š›¼0 , š›¼1 , . . . , š›¼š‘› , deļ¬ne š‘“ (x) = š›½0 x0 + š›½1 x1 + ā‹… ā‹… ā‹… + š›½š‘› xš‘› where š›½š‘– =

š›¼š‘– + š‘”š‘– (x) āˆ‘ 1 + š‘›š‘—=0 š‘”š‘— (x)

š‘– = 0, 1, . . . , š‘›

(2.45)

āˆ‘ By construction š›½š‘– ā‰„ 0 and š‘›š‘–=0 š›½š‘– = 1. Therefore š‘“ (x) āˆˆ š‘†. That is, š‘“ : š‘† ā†’ š‘†. Furthermore š‘“ is continuous. By Brouwerā€™s theorem, there exists a ļ¬xed point š‘„āˆ— with š‘“ (xāˆ— ) = xāˆ— . That is š›½š‘–āˆ— = š›¼āˆ—š‘– for š‘– = 0, 1, . . . , š‘›. Now, since the collection š“0 , š“1 , . . . , š“š‘› covers š‘†, there exists some š‘– for which šœŒ(xāˆ— , š“š‘– ) = 0. Substituting š›½š‘–āˆ— = š›¼āˆ—š‘– in (2.45) we have š›¼āˆ—š‘– =

1+

š›¼āˆ— āˆ‘š‘› š‘–

š‘—=0

š‘”š‘— (xāˆ— )

which implies that š‘”š‘— (xāˆ— ) = 0 for every š‘—. Since the š“š‘– are closed, xāˆ— āˆˆ š“š‘– for every š‘– and therefore xāˆ— āˆˆ

š‘› āˆ©

š“š‘– āˆ•= āˆ…

š‘–=0

( ) 2.137 To simplify the notation, let š‘§š‘˜+ (p) = max 0, zš‘– (p) . Assume pāˆ— is a ļ¬xed point of š‘”. Then for every š‘˜ = 1, 2, . . . , š‘› š‘āˆ—š‘˜ =

š‘š‘˜ + š‘§š‘˜+ (pāˆ— ) āˆ‘š‘› 1 + š‘—=1 š‘§š‘—+ (pāˆ— )

Cross-multiplying š‘āˆ—š‘˜ + š‘āˆ—š‘˜

š‘› āˆ‘ š‘—=1

š‘§š‘—+ (p) = š‘āˆ—š‘˜ + š‘§š‘˜+ (pāˆ— )

or š‘āˆ—š‘˜

š‘› āˆ‘ š‘—=1

š‘§š‘—+ (p) = š‘§š‘˜+ (pāˆ— )

š‘˜ = 1, 2, . . . š‘›

Multiplying each equation by š‘§š‘˜ (p) we get š‘āˆ—š‘˜ š‘§š‘˜ (pāˆ— )

š‘› āˆ‘ š‘—=1

š‘§š‘–+ (p) = š‘§š‘˜ (pāˆ— )š‘§š‘˜+ (pāˆ— ) 101

š‘˜ = 1, 2, . . . š‘›

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Summing over š‘˜ š‘› āˆ‘

Since

āˆ‘š‘›

š‘˜=1

š‘āˆ—š‘˜ š‘§š‘˜ (pāˆ— )

š‘› āˆ‘ š‘—=1

š‘˜=1

š‘§š‘–+ (p) =

š‘› āˆ‘ š‘˜=1

š‘§š‘˜ (pāˆ— )š‘§š‘˜+ (pāˆ— )

š‘āˆ—š‘˜ š‘§š‘˜ (pāˆ— ) = 0 this implies that š‘› āˆ‘ š‘˜=1

š‘§š‘˜ (pāˆ— )š‘§š‘˜+ (pāˆ— ) = 0

( )2 Each term of this sum is nonnegative, since it is either 0 or š‘§š‘˜ (pāˆ— ) . Consequently, every term must be zero which implies that š‘§š‘˜ (pāˆ—) ā‰¤ 0 for every š‘˜ = 1, 2, . . . , š‘™. In other words, z(pāˆ— ) ā‰¤ 0. 2.138 Every individual demand function xš‘– (p, š‘š) is continuous (Example 2.90) in p and š‘š. For given endowment šŽ š‘– š‘šš‘– =

š‘™ āˆ‘

š‘š‘— šŽ š‘–š‘—

š‘—=1

is continuous in p (Exercise 2.78). Therefore the excess demand function zš‘– (p) = xš‘– (p, š‘š) āˆ’ šŽ š‘– is continuous for every consumer š‘– and hence the aggregate excess demand function is continuous. Similarly, the consumerā€™s demand function xš‘– (p, š‘š) is homogeneous of degree 0 in p and š‘š. For given endowment šŽ š‘– , the consumerā€™s wealth is homogeneous of degree 1 in p and therefore the consumerā€™s excess demand function zš‘– (p) is homogeneous of degree 0. So therefore is the aggregate excess demand function z(p). 2.139 z(p) = =

š‘› āˆ‘ š‘–=1 š‘› āˆ‘

zš‘– (p) ( ) xš‘– (p, š‘š) āˆ’ šŽ š‘–

š‘–=1

and therefore pš‘‡ z(p) =

š‘› āˆ‘

pš‘‡ xš‘– (p, š‘š) āˆ’

š‘–=1

š‘› āˆ‘

pš‘‡ šŽ š‘–

š‘–=1

Since preferences are nonsatiated and strictly convex, they are locally nonsatiated (Exercise 1.248) which implies (Exercise 1.235) that every consumer must satisfy his budget constraint pš‘‡ xš‘– (p, š‘š) = pš‘‡ šŽš‘– for every š‘– = 1, 2, . . . , š‘› Therefore in aggregate pš‘‡ z(p) =

š‘› āˆ‘

pš‘‡ xš‘– (p, š‘š) āˆ’

š‘–=1

š‘› āˆ‘ š‘–=1

for every p. 102

pš‘‡ šŽ š‘– = 0

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 2.140 Assume there exists pāˆ— such that z(pāˆ— ) ā‰¤ 0. That is z(pāˆ— ) =

š‘› āˆ‘

zš‘– (p) =

š‘–=1

š‘› š‘› š‘› āˆ‘ āˆ‘ ( ) āˆ‘ xš‘– (p, š‘š) āˆ’ šŽš‘– = xš‘– (p, š‘š) āˆ’ šŽš‘– ā‰¤ 0 š‘–=1

or

š‘–=1

āˆ‘ š‘–āˆˆš‘

xš‘– ā‰¤

āˆ‘

š‘–=1

šŽš‘–

š‘–āˆˆš‘

Aggregate demand is less or equal to available supply. āˆ‘š‘™ Let š‘šāˆ—š‘– = š‘—=1 š‘āˆ—š‘— šŽ š‘–š‘— denote the wealth of consumer š‘– when the price system is pāˆ— and let xāˆ—š‘– = x(pāˆ— , š‘šāˆ— ) be his chosen consumption bundle. Then xāˆ—š‘– ā‰æ xš‘– for every xš‘– āˆˆ š‘‹(pāˆ— , š‘šš‘– ) Let xāˆ— = (xāˆ—1 , xāˆ—2 , . . . , xāˆ—š‘› ) be the allocation comprising these optimal bundles. The pair (pāˆ— , xāˆ— ) is a competitive equilibrium. 2.141 For each xš‘˜ , let š‘† š‘˜ denote the subsimplex of Ī›š‘˜ which contains xš‘˜ and let xš‘˜0 , xš‘˜1 , . . . , xš‘˜š‘› denote the vertices of š‘† š‘˜ . Let š›¼š‘˜0 , š›¼š‘˜1 , . . . , š›¼š‘˜š‘› denote the barycentric coordinates (Exercise 1.159) of x with respect to the vertices of š‘† š‘˜ and let yš‘–š‘˜ = š‘“ š‘˜ (xš‘˜š‘– ), š‘– = 0, 1, . . . , š‘›, denote the images of the vertices. Since š‘† is compact, there exists ā€² ā€² ā€² subsequences xš‘˜š‘– , yš‘–š‘˜ and š›¼š‘˜ such that xš‘˜š‘– ā†’ xāˆ—š‘–

yš‘–š‘˜ ā†’ yš‘–āˆ— and š›¼š‘˜š‘– ā†’ š›¼āˆ—š‘–

š‘– = 0, 1, . . . , š‘›

Furthermore, š›¼āˆ—š‘– ā‰„ 0 and š›¼āˆ—0 +š›¼āˆ—1 +ā‹… ā‹… ā‹…+š›¼āˆ—š‘› = 1. Since the diameters of the subsimplices converge to zero, their vertices must converge to the same point. That is, we must have xāˆ—0 = xāˆ—1 = ā‹… ā‹… ā‹… = xāˆ—š‘› = xāˆ— By deļ¬nition of š‘“ š‘˜ š‘“ š‘˜ (xš‘˜ ) = š›¼š‘˜0 š‘“ (xš‘˜0 ) + š›¼š‘˜1 š‘“ (xš‘˜1 ) + ā‹… ā‹… ā‹… + š›¼š‘˜š‘› š‘“ (xš‘˜š‘› ) Substituting yš‘–š‘˜ = š‘“ š‘˜ (xš‘˜š‘– ), š‘– = 0, 1, . . . , š‘› and recognizing that xš‘˜ is a ļ¬xed point of š‘“ š‘˜ , we have š‘„š‘˜ = š‘“ š‘˜ (xš‘˜ ) = š›¼š‘˜0 y0š‘˜ + š›¼š‘˜1 y1š‘˜ + ā‹… ā‹… ā‹… + š›¼š‘˜š‘› yš‘›š‘˜ Taking limits xāˆ— = š›¼āˆ—0 y0āˆ— + š›¼āˆ—1 y1āˆ— + ā‹… ā‹… ā‹… + š›¼āˆ—š‘› yš‘›āˆ—

(2.46)

For each coordinate š‘–, (xš‘˜š‘– , yš‘–š‘˜ ) āˆˆ graph(šœ‘) for every š‘˜ = 0, 1, . . . . Since šœ‘ is closed, (xāˆ—š‘– , yš‘–āˆ— ) āˆˆ graph(šœ‘). That is, yš‘–āˆ— āˆˆ šœ‘(xāˆ—š‘– ) = šœ‘(xāˆ— ) for every š‘– = 0, 1, . . . , š‘›. Therefore, (2.46) implies xāˆ— āˆˆ conv šœ‘(xāˆ— ) Since šœ‘ is convex valued, xāˆ— āˆˆ šœ‘(xāˆ— )

103

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2.142 Analogous to Exercise 2.129, there exists a simplex š‘‡ containing š‘† and a retraction of š‘‡ onto š‘†, that is a continuous function š‘” : š‘‡ ā†’ š‘† with š‘”(x) = x for every x āˆˆ š‘†. Then šœ‘ āˆ˜ š‘” : š‘‡ ā‡‰ š‘† āŠ‚ š‘‡ is closed-valued (Exercise 2.106) and uhc (Exercise 2.103). By the argument in the proof, there exists a point xāˆ— āˆˆ š‘‡ such that xāˆ— āˆˆ šœ‘ āˆ˜ š‘”(xāˆ— ). However, since šœ‘ āˆ˜ š‘”(xāˆ— ) āŠ† š‘†, we must have xāˆ— āˆˆ š‘† and therefore š‘”(xāˆ— ) = xāˆ— . This implies xāˆ— āˆˆ šœ‘(xāˆ— ). That is, xāˆ— is a ļ¬xed point of šœ‘. 2.143 šµ = šµ1 Ɨ šµ2 Ɨ . . . Ɨ šµš‘› is the Cartesian product of uhc, compact- and convexvalued correspondences. Therefore šµ is also compact-valued and uhc (Exercise 2.112 and also convex-valued (Exercise 1.165). By Exercise 2.106, šµ is closed. 2.144 Strict quasiconcavity ensures that the best response correspondence is in fact a function šµ : š‘† ā†’ š‘†. Since the hypotheses of Example 2.96 apply, there exists at least one equilibrium. Suppose that there are two Nash equilibria s and sā€² . Since šµ is a contraction, šœŒ(šµ(s), šµ(sā€² ) ā‰¤ š›½šœŒ(s, sā€² ) for some š›½ < 1. However šµ(s) = s and šµ(sā€² ) = sā€² and (2.46) implies that šœŒ(s, sā€² ) ā‰¤ š›½šœŒ(s, sā€² ) which is possible if and only if s = sā€² . This implies that the equilibrium must be unique. 2.145 Since š¾ is compact, it is totally bounded (Exercise 1.112). There exists a ļ¬nite set of points x1 , x2 , . . . , xš‘› such that š‘› āˆ©

š¾āŠ†

šµšœ– (xš‘– )

š‘–=1

Let š‘† = conv {x1 , x2 , . . . , xš‘› }. For š‘– = 1, 2, . . . , š‘› and x āˆˆ š‘‹, deļ¬ne š›¼š‘– (x) = max{0, šœ– āˆ’ āˆ„x āˆ’ xš‘– āˆ„} Then for every x āˆˆ š¾, 0 ā‰¤ š›¼š‘– (x) ā‰¤ šœ–,

š‘– = 1, 2, . . . , š‘›

and š›¼š‘– (x) > 0 ā‡ā‡’ x āˆˆ šµšœ– (xš‘– ) Note that š›¼š‘– (x) > 0 for some š‘–. Deļ¬ne

āˆ‘ š›¼š‘– (x)xš‘– ā„Ž(x) = āˆ‘ š›¼š‘– (x)

Then ā„Ž(x) āˆˆ š‘† and therefore ā„Ž : š¾ ā†’ š‘†. Furthermore, ā„Ž is continuous and  āˆ‘   š›¼š‘– (x)xš‘–  āˆ‘ āˆ’ x āˆ„ā„Ž(x) āˆ’ xāˆ„ =    š›¼š‘– (x)  āˆ‘  š›¼š‘– (x)(xš‘– āˆ’ x)   āˆ‘ =   š›¼š‘– (x) āˆ‘ š›¼š‘– (x) āˆ„xš‘– āˆ’ xāˆ„ āˆ‘ = š›¼š‘– (x) āˆ‘ š›¼š‘– (x)šœ– ā‰¤ āˆ‘ =šœ– š›¼š‘– (x) since š›¼š‘– (x) > 0 ā‡ā‡’ āˆ„xš‘– āˆ’ xāˆ„ ā‰¤ šœ–. 104

Solutions for Foundations of Mathematical Economics 2.146

c 2001 Michael Carter āƒ All rights reserved

( ) 1. For every x āˆˆ š‘† š‘˜ , š‘“ (x) āˆˆ š‘† and therefore š‘” š‘˜ (x) = ā„Žš‘˜ š‘“ (x) āˆˆ š‘† š‘˜ .

2. For any x āˆˆ š‘† š‘˜ , let y = š‘“ (x) āˆˆ š‘“ (š‘†) and therefore   š‘˜ ā„Ž (y) āˆ’ y < 1 š‘˜ which implies   š‘˜ š‘” (x) āˆ’ š‘“ (x) ā‰¤ 1 for every x āˆˆ š‘† š‘˜ š‘˜ 2.147 By the Triangle inequality  š‘˜      x āˆ’ š‘“ (x) ā‰¤ š‘” š‘˜ (xš‘˜ ) āˆ’ š‘“ (xš‘˜ ) + š‘“ (xš‘˜ ) āˆ’ š‘“ (x) As shown in the previous exercise   š‘˜ š‘˜ š‘” (x ) āˆ’ š‘“ (xš‘˜ ) ā‰¤ 1 ā†’ 0 š‘˜ as š‘˜ ā†’ āˆž. Also since š‘“ is continuous   š‘“ (xš‘˜ ) āˆ’ š‘“ (x) ā†’ 0 Therefore  š‘˜  x āˆ’ š‘“ (x) ā†’ 0 =ā‡’ x = š‘“ (x) x is a ļ¬xed point of š‘“ . 2.148 š‘‡ (š¹ ) is bounded and equicontinuous and so therefore is š‘‡ (š¹ ) (Exercise 2.96). By Ascoliā€™s theorem (Exercise 2.95), š‘‡ (š¹ ) is compact. Therefore š‘‡ is a compact operator. Applying Corollary 2.8.1, š‘‡ has a ļ¬xed point.

105

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Chapter 3: Linear Functions 3.1 Let x1 , x2 āˆˆ š‘‹ and š›¼1 , š›¼2 āˆˆ ā„œ. Homogeneity implies that š‘“ (š›¼1 x1 ) = š›¼1 š‘“ (š‘„1 ) š‘“ (š›¼2 x2 ) = š›¼2 š‘“ (š‘„2 ) and additivity implies š‘“ (š›¼1 x1 + š›¼2 x2 ) = š›¼1 š‘“ (x1 ) + š›¼2 š‘“ (x2 ) Conversely, assume š‘“ (š›¼1 x1 + š›¼2 x2 ) = š›¼1 š‘“ (x1 ) + š›¼2 š‘“ (x2 ) for all x1 , x2 āˆˆ š‘‹ and š›¼1 , š›¼2 āˆˆ ā„œ. Letting š›¼1 = š›¼2 = 1 implies š‘“ (x1 + x2 ) = š‘“ (x1 ) + š‘“ (x2 ) while setting x2 = 0 implies š‘“ (š›¼1 x1 ) = š›¼1 š‘“ (x1 ) 3.2 Assume š‘“1 , š‘“2 āˆˆ šæ(š‘‹, š‘Œ ). Deļ¬ne the mapping š‘“1 + š‘“2 : š‘‹ ā†’ š‘Œ by (š‘“1 + š‘“2 )(x) = š‘“1 (x) + š‘“2 (x) We have to conļ¬rm that š‘“1 + š‘“2 is linear, that is (š‘“1 + š‘“2 )(x1 + x2 ) = š‘“1 (x1 + x2 ) + š‘“2 (x1 + x2 ) = š‘“1 (x1 ) + š‘“1 (x2 ) + š‘“2 (x1 ) + š‘“2 (x2 ) = š‘“1 (x1 ) + š‘“2 (x1 ) + š‘“1 (x1 ) + š‘“2 (x2 ) = (š‘“1 + š‘“2 )(x1 ) + (š‘“1 + š‘“2 )(x2 ) and (š‘“1 + š‘“2 )(š›¼x) = š‘“1 (š›¼x) + š‘“2 (š›¼x) = š›¼(š‘“1 (x) + š‘“2 (x)) = š›¼(š‘“1 + š‘“2 )(x) Similarly let š‘“ āˆˆ šæ(š‘‹, š‘Œ ) and deļ¬ne š›¼š‘“ : š‘‹ ā†’ š‘Œ by (š›¼š‘“ )(x) = š›¼š‘“ (x) š›¼š‘“ is also linear, since (š›¼š‘“ )(š›½x) = š›¼š‘“ (š›½x) = š›¼š›½š‘“ (x) = š›½š›¼š‘“ (x) = š›½(š›¼š‘“ )(x) 106

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

3.3 Let x, x1 , x2 āˆˆ ā„œ2 . Then š‘“ (x1 + x2 ) = š‘“ (š‘„11 + š‘„21 , š‘„12 + š‘„22 ) ) ( = (š‘„11 + š‘„21 ) cos šœƒ āˆ’ (š‘„12 + š‘„22 ) sin šœƒ, (š‘„11 + š‘„21 ) sin šœƒ āˆ’ (š‘„12 + š‘„22 ) cos šœƒ ) ( = (š‘„11 cos šœƒ āˆ’ š‘„12 sin šœƒ) + (š‘„21 cos šœƒ āˆ’ š‘„22 sin šœƒ), (š‘„11 sin šœƒ + š‘„12 cos šœƒ) + (š‘„21 sin šœƒ āˆ’ š‘„22 cos šœƒ) ( ) = (š‘„11 cos šœƒ āˆ’ š‘„12 sin šœƒ, š‘„11 sin šœƒ + š‘„12 cos šœƒ) + (š‘„21 cos šœƒ āˆ’ š‘„22 sin šœƒ, š‘„21 sin šœƒ āˆ’ š‘„22 cos šœƒ = š‘“ (š‘„11 , š‘„12 ) + š‘“ (š‘„21 , š‘„22 ) = š‘“ (x1 ) + š‘“ (x2 ) and š‘“ (š›¼x) = š‘“ (š›¼š‘„1 , š›¼š‘„2 ) = (š›¼š‘„1 cos šœƒ āˆ’ š›¼š‘„2 sin šœƒ, š›¼š‘„1 sin šœƒ + š›¼š‘„2 cos šœƒ) = š›¼ (š‘„1 cos šœƒ āˆ’ š‘„1 sin šœƒ, š‘„1 sin šœƒ + š‘„2 cos šœƒ) = š›¼š‘“ (š‘„1 , š‘„2 ) = š›¼š‘“ (x) 3.4 Let x, x1 , x2 āˆˆ ā„œ3 . š‘“ (x1 + x2 ) = š‘“ (š‘„11 + š‘„2 , š‘„12 + š‘„22 , š‘„13 + š‘„23 ) = (š‘„11 + š‘„21 , š‘„12 + š‘„22 , 0) = (š‘„11 , š‘„12 , 0) + (š‘„21 , š‘„22 , 0) = š‘“ (š‘„11 , š‘„12 , š‘„13 ) + š‘“ (š‘„21 , š‘„22 , š‘„23 ) = š‘“ (x1 ) + š‘“ (x2 ) and š‘“ (š›¼x) = š‘“ (š›¼š‘„1 , š›¼š‘„2 , š›¼š‘„3 ) = (š›¼š‘„1 , š›¼š‘„2 , 0) = š›¼(š‘„1 , š‘„2 , 0) = š›¼š‘“ (š‘„1 , š‘„2 , š‘„3 ) = š›¼š‘“ (x) This mapping is the projection of 3-dimensional space onto the (2-dimensional) plane. 3.5 Applying the deļ¬nition

( )( ) 0 1 š‘„1 š‘“ (š‘„1 , š‘„2 ) = 1 0 š‘„2 = (š‘„2 , š‘„1 )

This function interchanges the two coordinates of any point in the plane ā„œ2 . Its action corresponds to reļ¬‚ection about the line š‘„1 = š‘„2 ( 45 degree diagonal). 3.6 Assume (š‘, š‘¤) and (š‘, š‘¤ā€² ) are two games in š’¢ š‘ . For any coalition š‘† āŠ† š‘ (š‘¤ + š‘¤ā€² )(š‘†) āˆ’ (š‘¤ + š‘¤ā€² )(š‘† āˆ– {š‘–}) = š‘¤(š‘†) + š‘¤(š‘† ā€² ) āˆ’ š‘¤(š‘† āˆ– {š‘–}) āˆ’ š‘¤ā€² (š‘† āˆ– {š‘–}) = (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–})) + (š‘¤ā€² (š‘†) āˆ’ š‘¤ā€² (š‘† āˆ– {š‘–})) = šœ‘š‘– (š‘¤) + šœ‘š‘– (š‘¤ā€² ) 107

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 3.7 The characteristic function of cost allocation game is š‘¤(š“š‘ƒ ) = 0 š‘¤(š‘‡ š‘ ) = 0

š‘¤(š“š‘ƒ, š‘‡ š‘ ) = 210 š‘¤(š“š‘ƒ, š¾š‘€ ) = 770

š‘¤(š¾š‘€ ) = 0

š‘¤(š¾š‘€, š‘‡ š‘ ) = 1170

š‘¤(š‘ ) = 1530

The following table details the computation of the Shapley value for player š“š‘ƒ . š‘† š“š‘ƒ š“š‘ƒ, š‘‡ š‘ š“š‘ƒ, š¾š‘€ š“š‘ƒ, š‘‡ š‘, š¾š‘€ šœ‘š‘“ (š‘¤)

š›¾š‘† 1/3 1/6 1/6 1/3

š‘¤(š‘†) 0 210 770 1530

š‘¤(š‘† āˆ– {š‘–}) 0 0 0 1170

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–})) 0 35 128 1/3 120 283 1/3

Thus šœ‘š“š‘ƒ š‘¤ = 283 1/3. Similarly, we can calculate that šœ‘š‘‡ š‘ š‘¤ = 483 1/3 and šœ‘š¾š‘€ š‘¤ = 763 1/3. 3.8 āˆ‘

šœ‘š‘– š‘¤ =

š‘–āˆˆš‘

=

āˆ‘

( āˆ‘

š‘–āˆˆš‘

š‘†āˆ‹š‘–

āˆ‘

(

š‘†āˆ‹š‘–

=

āˆ‘

) š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–})) ) š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–}))

š‘–āˆˆš‘

āˆ‘āˆ‘

š›¾š‘† š‘¤(š‘†) āˆ’

š‘†āŠ†š‘ š‘–āˆˆš‘†

=

āˆ‘ āˆ‘

š›¾š‘† š‘¤(š‘† āˆ– {š‘–})

š‘†āŠ†š‘ š‘–āˆˆš‘†

š‘  Ɨ š›¾š‘† š‘¤(š‘†) āˆ’

š‘†āŠ†š‘

=

āˆ‘āˆ‘

(

āˆ‘

š›¾š‘†

š‘†āŠ†š‘

š‘  Ɨ š›¾š‘† š‘¤(š‘†) āˆ’

āˆ‘

āˆ‘

) š‘¤(š‘† āˆ– {š‘–})

š‘–āˆˆš‘†

š‘  Ɨ š›¾š‘† š‘¤(š‘†)

š‘†āŠ‚š‘

š‘†āŠ†š‘

= š‘› Ɨ š›¾š‘ š‘¤(š‘ ) = š‘¤(š‘ ) 3.9 If š‘–, š‘— āˆˆ š‘† š‘¤(š‘† āˆ– {š‘–}) = š‘¤(š‘† āˆ– {š‘–, š‘—} āˆŖ {š‘—}) = š‘¤(š‘† āˆ– {š‘–, š‘—} āˆŖ {š‘–}) = š‘¤(š‘† āˆ– {š‘–}) šœ‘š‘– (š‘¤) =

āˆ‘

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–}))

š‘†āˆ‹š‘–

=

āˆ‘

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–})) +

š‘†āˆ‹š‘–,š‘—

=

āˆ‘

āˆ‘

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘—})) +

=

āˆ‘

š›¾š‘† (š‘¤(š‘† āˆŖ {š‘–}) āˆ’ š‘¤(š‘†))

š‘†āˆ•āˆ‹š‘–,š‘—

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘—})) +

āˆ‘

š›¾š‘† ā€² (š‘¤(š‘† ā€² āˆŖ {š‘—}) āˆ’ š‘¤(š‘† ā€² ))

š‘† ā€² āˆ•āˆ‹š‘–,š‘—

š‘†āˆ‹š‘–,š‘—

āˆ‘

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–}))

š‘†āˆ‹š‘–,š‘†āˆ•āˆ‹š‘—

š‘†āˆ‹š‘–,š‘—

=

āˆ‘

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘—})) +

š‘†āˆ‹š‘–,š‘—

āˆ‘

š‘†āˆ•āˆ‹š‘–,š‘†āˆ‹š‘—

= šœ‘š‘— (š‘¤) 108

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘—}))

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

3.10 For any null player š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–}) = 0 for every š‘† āŠ† š‘ . Consequently

āˆ‘

šœ‘š‘– (š‘¤) =

š›¾š‘† (š‘¤(š‘†) āˆ’ š‘¤(š‘† āˆ– {š‘–})) = 0

š‘†āŠ†š‘

3.11 Every š‘– āˆˆ / š‘‡ is a null player, so that šœ‘š‘– (š‘¢š‘‡ ) = 0 Feasibility requires that

āˆ‘

for every š‘– āˆˆ /š‘‡ āˆ‘

šœ‘š‘– (š‘¢š‘‡ ) =

š‘–āˆˆš‘‡

šœ‘š‘– (š‘¢š‘‡ ) = 1

š‘–āˆˆš‘

Further, any two players in š‘‡ are substitutes, so that symmetry requires that šœ‘š‘– (š‘¢š‘‡ ) = šœ‘š‘— (š‘¢š‘‡ )

for every š‘–, š‘— āˆˆ š‘‡

Together, these conditions require that šœ‘š‘– (š‘¢š‘‡ ) =

1 š‘”

for every š‘– āˆˆ š‘‡

The Shapley value of the a T-unanimity game is { 1 š‘–āˆˆš‘‡ šœ‘š‘– (š‘¢š‘‡ ) = š‘” 0 š‘–āˆˆ /š‘‡ where š‘” = āˆ£š‘‡ āˆ£. 3.12 Any coalitional game can be represented as a linear combination of unanimity games š‘¢š‘‡ (Example 1.75) āˆ‘ š‘¤= š›¼š‘‡ š‘¢š‘‡ š‘‡

By linearity, the Shapley value is

āŽ›

āˆ‘

šœ‘š‘¤ = šœ‘ āŽ =

āˆ‘

āŽž š›¼š‘‡ š‘¢š‘‡ āŽ 

š‘‡ āŠ†š‘

š›¼š‘‡ šœ‘š‘¢š‘‡

š‘‡ āŠ†š‘

and therefore for player š‘– šœ‘š‘– š‘¤ =

āˆ‘

š›¼š‘‡ šœ‘š‘– š‘¢š‘‡

š‘‡ āŠ†š‘

=

āˆ‘ 1 š›¼š‘‡ š‘”

š‘‡ āŠ†š‘ š‘‡ āˆ‹š‘–

āˆ‘ 1 āˆ‘ 1 š›¼š‘‡ āˆ’ š›¼š‘‡ = š‘” š‘” š‘‡ āŠ†š‘

š‘‡ āŠ†š‘ š‘–āˆˆš‘‡ /

= š‘ƒ (š‘, š‘¤) āˆ’ š‘ƒ (š‘ āˆ– {š‘–}, š‘¤) 109

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Using Exercise 3.8 š‘¤(š‘ ) =

āˆ‘

šœ‘š‘– š‘¤

š‘–āˆˆš‘

=

āˆ‘(

) š‘ƒ (š‘, š‘¤) āˆ’ š‘ƒ (š‘ āˆ– {š‘–}, š‘£)

š‘–āˆˆš‘

= š‘›š‘ƒ (š‘, š‘¤) āˆ’

āˆ‘

š‘ƒ (š‘ āˆ– {š‘–}, š‘£)

š‘–āˆˆš‘

which implies that 1 š‘ƒ (š‘, š‘¤) = š‘›

( š‘¤(š‘ ) āˆ’

āˆ‘

) š‘ƒ (š‘ āˆ– {š‘–}, š‘£)

š‘–āˆˆš‘

3.13 Choose any x āˆ•= 0 āˆˆ š‘‹. 0š‘‹ = x āˆ’ x and by additivity š‘“ (0š‘‹ ) = š‘“ (x āˆ’ x) = š‘“ (x) āˆ’ š‘“ (x) = 0š‘Œ 3.14 Let x1 , x2 belong to š‘‹. Then š‘” āˆ˜ š‘“ (x1 + x2 ) = š‘” āˆ˜ š‘“ (x1 + x2 ) ) ( = š‘” š‘“ (x1 ) + š‘“ (x2 ) ) ( ) ( = š‘” š‘“ (x1 ) + š‘” š‘“ (x2 ) = š‘” āˆ˜ š‘“ (x1 ) + š‘” āˆ˜ š‘“ (x2 ) and š‘” āˆ˜ š‘“ (š›¼x) = š‘” (š‘“ (š›¼x)) = š‘” (š›¼š‘“ (x)) = š›¼š‘” (š‘“ (x)) = š›¼š‘” āˆ˜ š‘“ (x) Therefore š‘” āˆ˜ š‘“ is linear. 3.15 Let š‘† be a subspace of š‘‹ and let y1 , y2 belong to š‘“ (š‘†). Choose any x1 āˆˆ š‘“ āˆ’1 (y1 ) and x2 āˆˆ š‘“ āˆ’1 (y2 ). Then for š›¼1 , š›¼2 āˆˆ ā„œ š›¼1 x1 + š›¼2 x2 āˆˆ š‘† Since š‘“ is linear (Exercise 3.1) š›¼1 y1 + š›¼2 y2 = š›¼1 š‘“ (x1 ) + š›¼2 š‘“ (x2 ) = š‘“ (š›¼1 x1 + š›¼2 x2 ) āˆˆ š‘“ (š‘†) š‘“ (š‘†) is a subspace. Let š‘‡ be a subspace of š‘Œ and let x1 , x2 belong to š‘“ āˆ’1 (š‘‡ ). Let y1 = š‘“ (x1 ) and y2 = š‘“ (x2 ). Then y1 , y2 āˆˆ š‘‡ . For every š›¼1 , š›¼2 āˆˆ ā„œ š›¼1 y1 + š›¼2 y2 = š›¼1 š‘“ (x1 ) + š›¼2 š‘“ (x2 ) āˆˆ š‘‡ 110

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Since š‘“ is linear, this implies that š‘“ (š›¼1 x1 + š›¼2 x2 ) = š›¼1 š‘“ (x1 ) + š›¼2 š‘“ (x2 ) āˆˆ š‘‡ Therefore š›¼1 x1 + š›¼2 x2 āˆˆ š‘“ āˆ’1 (š‘‡ ) We conclude that š‘“ āˆ’1 (š‘‡ ) is a subspace. 3.16 š‘“ (š‘‹) is a subspace of š‘Œ . rank š‘“ (š‘‹) = rank š‘Œ implies that š‘“ (š‘‹) = š‘Œ . š‘“ is onto. 3.17 This is a special case of the previous exercise, since {0š‘Œ } is a subspace of š‘Œ . 3.18 Assume not. That is, assume that there exist two distinct elements x1 and x2 with š‘“ (x1 ) = š‘“ (x2 ). Then x1 āˆ’ x2 āˆ•= 0š‘‹ but š‘“ (x1 āˆ’ x2 ) = š‘“ (x1 ) āˆ’ š‘“ (x2 ) = 0š‘Œ so that x1 āˆ’ x2 āˆˆ kernel š‘“ which contradicts the assumption that kernel š‘“ = {0}. 3.19 If š‘“ has an inverse, then it is one-to-one and onto (Exercise 2.4), that is š‘“ āˆ’1 (0) = 0 and š‘“ (š‘‹) = š‘Œ . Conversely, if kernel š‘“ = {0} then š‘“ is one-to-one by the previous exercise. If furthermore š‘“ (š‘‹) = š‘Œ , then š‘“ is one-to-one and onto, and therefore has an inverse (Exercise 2.4). 3.20 Let š‘“ be a nonsingular linear function from š‘‹ to š‘Œ with inverse š‘“ āˆ’1 . Choose y1 , y2 āˆˆ š‘Œ and let x1 = š‘“ āˆ’1 (y1 ) x2 = š‘“ āˆ’1 (y2 ) so that y1 = š‘“ (x1 ) y2 = š‘“ (x2 ) Since š‘“ is linear š‘“ (x1 + x2 ) = š‘“ (x1 ) + š‘“ (x2 ) = y1 + y2 which implies that š‘“ āˆ’1 (y1 + y2 ) = x1 + x2 = š‘“ āˆ’1 (y1 ) + š‘“ āˆ’1 (y2 ) The homogeneity of š‘“ āˆ’1 can be demonstrated similarly. 3.21 Assume that š‘“ : š‘‹ ā†’ š‘Œ and š‘” : š‘Œ ā†’ š‘ are nonsingular. Then (Exercise 3.19) āˆ™ š‘“ (š‘‹) = š‘Œ and š‘”(š‘Œ ) = š‘ āˆ™ kernel š‘“ = {0š‘‹ } and kernel š‘” = {0š‘Œ } We have previously shown (Exercise 3.14) that ā„Ž = š‘” āˆ˜ š‘“ : š‘‹ ā†’ š‘ is linear. To show that ā„Ž is nonsingular, we note that āˆ™ ā„Ž(š‘‹) = š‘” āˆ˜ š‘“ (š‘‹) = š‘”(š‘Œ ) = š‘

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āˆ™ If x āˆˆ kernel (ā„Ž) then ā„Ž(x) = š‘” (š‘“ (x)) = 0 and š‘“ (x) āˆˆ kernel š‘” = {0š‘Œ }. Therefore š‘“ (x) = 0š‘Œ which implies that x = 0š‘‹ . Thus kernel ā„Ž = {0š‘‹ }. We conclude that ā„Ž is nonsingular. Finally, let z be any point in š‘ and let x1 = ā„Žāˆ’1 (z) = (š‘” āˆ˜ š‘“ )āˆ’1 (z) y = š‘” āˆ’1 (z) x2 = š‘“ āˆ’1 (y) Then z = ā„Ž(x1 ) = š‘” āˆ˜ š‘“ (x1 ) z = š‘”(y) = š‘” āˆ˜ š‘“ (x2 ) which implies that x1 = x2 . 3.22 Suppose š‘“ were one-to-one. Then kernel š‘“ = {0} āŠ† kernel ā„Ž and š‘” = ā„Ž āˆ˜ š‘“ āˆ’1 is a well-deļ¬ned linear function mapping š‘“ (š‘‹) to š‘Œ with ( ) š‘” āˆ˜ š‘“ = ā„Ž āˆ˜ š‘“ āˆ’1 āˆ˜ š‘“ = ā„Ž We need to show that this still holds if š‘“ is not one-to-one. In this case, for arbitrary y āˆˆ š‘“ (š‘‹), š‘“ āˆ’1 (y) may contain more than one element. Suppose x1 and x2 are distinct elements in š‘“ āˆ’1 (y). Then š‘“ (x1 āˆ’ x2 ) = š‘“ (x1 ) āˆ’ š‘“ (x2 ) = y āˆ’ y = 0 so that x1 āˆ’ x2 āˆˆ kernel š‘“ āŠ† kernel ā„Ž (by assumption). Therefore ā„Ž(x1 ) āˆ’ ā„Ž(x2 ) = ā„Ž(x1 āˆ’ x2 ) = 0 which implies that ā„Ž(x1 ) = ā„Ž(x2 ) for all x1 , x2 āˆˆ š‘“ āˆ’1 (y). Thus š‘” = ā„Žāˆ˜ š‘“ āˆ’1 : š‘“ (š‘‹) ā†’ š‘ is well deļ¬ned even if š‘“ is many-to-one. To show that š‘” is linear, choose y1 , y2 in š‘“ (š‘‹) and let x1 āˆˆ š‘“ āˆ’1 (y1 ) x2 āˆˆ š‘“ āˆ’1 (y2 ) Since š‘“ (x1 + x2 ) = š‘“ (x1 ) + š‘“ (x2 ) = y1 + y2 x1 + x2 āˆˆ š‘“ āˆ’1 (y1 + y2 ) and š‘”(y1 + y2 ) = ā„Ž(x1 + x2 ) Therefore š‘”(y1 ) + š‘”(y2 ) = ā„Ž(x1 ) + ā„Ž(x2 ) = ā„Ž(x1 + x2 ) = š‘”(y1 + y2 ) 112

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Similarly š›¼x1 āˆˆ š‘“ āˆ’1 (š›¼y1 ) and š‘”(š›¼y1 ) = ā„Ž(š›¼x1 ) = š›¼ā„Ž(x1 ) = š›¼š‘”(y1 ) We conclude that š‘” = ā„Ž āˆ˜ š‘“ āˆ’1 is a linear function mapping š‘“ (š‘‹) to š‘ with ā„Ž = š‘” āˆ˜ š‘“ . 3.23 Let y be an arbitrary element of š‘“ (š‘‹) with x āˆˆ š‘“ āˆ’1 (y). Since B is a basis for š‘‹, x can be represented as a linear combination of elements of šµ, that is there exists x1 , x2 , .., xš‘š āˆˆ šµ and š›¼1 , ..., š›¼š‘š āˆˆ š‘… such that x=

š‘š āˆ‘

š›¼š‘– xš‘–

š‘–=1

y = š‘“ (x) ) ( āˆ‘ š›¼š‘– xš‘– =š‘“ =

āˆ‘

š‘–

š›¼š‘– š‘“ (xš‘– )

š‘–

Since š‘“ (xš‘– ) āˆˆ š‘“ (šµ), we have shown that y can be written as a linear combination of elements of š‘“ (šµ), that is y āˆˆ lin šµ Since the choice of y was arbitrary, š‘“ (šµ) spans š‘“ (š‘‹), that is lin šµ = š‘“ (š‘‹) 3.24 Let š‘› = dim š‘‹ and š‘˜ = dim kernel š‘“ . Let x1 , . . . , xš‘˜ be a basis for the kernel of š‘“ . This can be extended (Exercise 1.142) to a basis šµ for š‘‹. Exercise 3.23 showed lin šµ = š‘“ (š‘‹) Since x1 , x2 , . . . , xš‘˜ āˆˆ kernel š‘“ , š‘“ (xš‘– ) = 0 for š‘– = 1, 2, . . . , š‘˜. This implies that {š‘“ (xš‘˜+1 ), . . . , š‘“ (xš‘› )} spans š‘“ (š‘‹), that is lin {(xš‘˜+1 ), ..., š‘“ (xš‘› )} = š‘“ (š‘‹) To show that dim š‘“ (š‘‹) = š‘› āˆ’ š‘˜, we have to show that {š‘“ (š‘„š‘˜+1 ), š‘“ (š‘„š‘˜+2 ), . . . , š‘“ (š‘„š‘› )} is linearly independent. Assume not. That is, assume there exist š›¼š‘˜+1 , š›¼š‘˜+2 , ..., š›¼š‘› āˆˆ š‘… such that š‘› āˆ‘

š›¼š‘– š‘“ (xš‘– ) = 0

š‘–=š‘˜+1

This implies that

( š‘“

)

š‘› āˆ‘

š›¼š‘– xš‘–

=0

š‘–=š‘˜+1

or x=

š‘› āˆ‘

š›¼š‘– š‘„š‘– āˆˆ kernel š‘“

š‘–=š‘˜+1

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Solutions for Foundations of Mathematical Economics

This implies that x can also be expressed as a linear combination of elements in {x1 , š‘„2 , ..., xš‘˜ }, that is there exist scalars š›¼1 , š›¼2 , . . . , š›¼š‘˜ such that x=

š‘˜ āˆ‘

š›¼š‘– xš‘–

š‘–=1

or x=

š‘˜ āˆ‘

š‘› āˆ‘

š›¼š‘– xš‘– =

š‘–=1

š›¼š‘– xš‘–

š‘–=š‘˜+1

which contradicts the assumption that šµ is a basis for š‘‹. Therefore {š‘“ (xš‘˜+1 ), . . . , š‘“ (xš‘› )} is a basis for š‘“ (š‘‹) and therefore dim š‘“ (š‘„) = š‘› āˆ’ š‘˜. We conclude that dim kernel š‘“ + dim š‘“ (š‘‹) = š‘› = dim š‘‹ 3.25 Equation (3.2) implies that nullity š‘“ = 0, and therefore š‘“ is one-to-one (Exercise 3.18). 3.26 Choose some x = (š‘„1 , š‘„2 , . . . , š‘„š‘› ) āˆˆ š‘‹. x has a unique representation in terms of the standard basis (Example 1.79) x=

š‘› āˆ‘

š‘„š‘— eš‘—

š‘—=1

Let y = š‘“ (x). Since š‘“ is linear

āŽ›

y = š‘“ (x) = š‘“ āŽ

āŽž

š‘› āˆ‘

š‘„š‘— eš‘— āŽ  =

š‘—=1

š‘› āˆ‘

xš‘— š‘“ (eš‘— )

š‘—=1

Each š‘“ (eš‘— ) has a unique representation of the form š‘“ (eš‘— ) =

š‘š āˆ‘

š‘Žš‘–š‘— eš‘–

š‘–=1

so that y = š‘“ (x) =

š‘› āˆ‘ š‘—=1

=

š‘š āˆ‘ š‘–=1

(

š‘„š‘—

š‘š āˆ‘

) š‘Žš‘–š‘— eš‘–

š‘–=1

āŽ› āŽž š‘› āˆ‘ āŽ š‘Žš‘–š‘— š‘„š‘— āŽ  eš‘– š‘—=1

āŽž āŽ› āˆ‘š‘› š‘Ž1š‘— š‘„š‘— āˆ‘š‘—=1 š‘› āŽœ š‘—=1 š‘Ž2š‘— š‘„š‘— āŽŸ āŽŸ āŽœ =āŽœ āŽŸ .. āŽ  āŽ āˆ‘š‘› . š‘Ž š‘„ š‘—=1 š‘šš‘— š‘— = š“x where

āŽ›

āŽž š‘Ž11 š‘Ž12 . . . š‘Ž1š‘› āŽœ š‘Ž21 š‘Ž22 . . . š‘Ž2š‘› āŽŸ āŽŸ š“=āŽœ āŽ . . . . . . . . . . . . . . . . . . . . .āŽ  š‘Žš‘š1 š‘Žš‘š2 . . . š‘Žš‘šš‘› 114

Solutions for Foundations of Mathematical Economics 3.27

( 1 0 0 1

c 2001 Michael Carter āƒ All rights reserved

) 0 0

3.28 We must specify bases for each space. The most convenient basis for šŗš‘ is the T-unanimity games. We adopt the standard basis for ā„œš‘› . With respect to these bases, the Shapley value šœ‘ is represented by the 2š‘›āˆ’1 Ɨš‘› matrix where each row is the Shapley value of the corresponding T-unanimity game. For three player games (š‘› = 3), the matrix is āŽ› 1 0 āŽœ0 1 āŽœ āŽœ0 0 āŽœ1 1 āŽœ āŽœ 21 2 āŽœ āŽœ 2 01 āŽ0 2 1 3

1 3

āŽž 0 0āŽŸ āŽŸ 1āŽŸ āŽŸ 0āŽŸ āŽŸ 1āŽŸ 2āŽŸ 1āŽ  2 1 3

3.29 Clearly, if š‘“ is continuous, š‘“ is continuous at 0. To show the converse, assume that š‘“ : š‘‹ ā†’ š‘Œ is continuous at 0. Let (xš‘› ) be a sequence which converges to x āˆˆ š‘‹. Then the sequence (xš‘› āˆ’ x) converges to 0š‘‹ and therefore š‘“ (xš‘› āˆ’x) ā†’ 0š‘Œ by continuity (Exercise 2.68). By linearity, š‘“ (xš‘› )āˆ’š‘“ (x) = š‘“ (xš‘› āˆ’x) ā†’ 0š‘Œ and therefore š‘“ (xš‘› ) converges to š‘“ (x). We conclude that š‘“ is continuous at x. 3.30 Assume that š‘“ is bounded, that is āˆ„š‘“ (x)āˆ„ ā‰¤ š‘€ āˆ„xāˆ„ for every x āˆˆ š‘‹ Then š‘“ is Lipschitz at 0 (with Lipschitz constant š‘€ ) and hence continuous (by the previous exercise). Conversely, assume š‘“ is continuous but not bounded. Then, for every positive integer š‘›, there exists some xš‘› āˆˆ š‘‹ such that āˆ„š‘“ (xš‘› )āˆ„ > š‘› āˆ„xš‘› āˆ„ which implies that  ( )   xš‘› š‘“ >1  š‘› āˆ„xš‘› āˆ„  Deļ¬ne yš‘› =

xš‘› š‘› āˆ„xš‘› āˆ„

Then yš‘› ā†’ 0 but š‘“ (yš‘› ) āˆ•ā†’ 0. This implies that š‘“ is not continuous at the origin, contradicting our hypothesis. 3.31 Let {x1 , x2 , . . . , xš‘› } be a basis for š‘‹. For every x āˆˆ š‘‹, there exists numbers š›¼1 , š›¼2 , . . . , š›¼š‘› such that x=

š‘› āˆ‘

š›¼š‘– xš‘–

š‘–=1

115

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics and š‘“ (x) =

š‘› āˆ‘

š›¼š‘– š‘“ (xš‘– )

š‘–=1 š‘› āˆ‘

 āˆ„š‘“ (x)āˆ„ =   ā‰¤

š‘–=1

š‘› āˆ‘

   š›¼š‘– š‘“ (xš‘– ) 

āˆ£š›¼š‘– āˆ£ āˆ„š‘“ (xš‘– )āˆ„

š‘–=1

š‘› )āˆ‘ ( š‘› āˆ£š›¼š‘– āˆ£ ā‰¤ max āˆ„š‘“ (xš‘– )āˆ„ š‘–=1

š‘–=1

By Lemma 1.1, there exists a constant š‘ such that   š‘› š‘›  1 āˆ‘ 1  āˆ‘ āˆ£š›¼š‘– āˆ£ ā‰¤  š›¼š‘– xš‘–  = āˆ„xāˆ„  š‘  š‘ š‘–=1 š‘–=1 Combining these two inequalities āˆ„š‘“ (x)āˆ„ ā‰¤ š‘€ āˆ„xāˆ„ where š‘€ = maxš‘›š‘–=1 āˆ„š‘“ (xš‘– )āˆ„ /š‘. 3.32 For any x āˆˆ š‘‹, let š‘Ž = āˆ„xāˆ„ and deļ¬ne y = x/š‘Ž. Linearity implies that āˆ„š‘“ (x)āˆ„ = sup āˆ„š‘“ (x/š‘Ž)āˆ„ = sup āˆ„š‘“ (y)āˆ„ š‘Ž xāˆ•=0 xāˆ•=0 āˆ„yāˆ„=1

āˆ„š‘“ āˆ„ = sup

3.33 āˆ„š‘“ āˆ„ is a norm Let š‘“ āˆˆ šµšæ(š‘‹, š‘Œ ). Clearly āˆ„š‘“ āˆ„ = sup āˆ„š‘“ (x)āˆ„ ā‰„ 0 āˆ„xāˆ„=1

Further, for every š›¼ āˆˆ ā„œ, āˆ„š›¼š‘“ āˆ„ = sup āˆ„š›¼š‘“ (x)āˆ„ = āˆ£š›¼āˆ£ āˆ„š‘“ āˆ„ āˆ„xāˆ„=1

Finally, for every š‘” āˆˆ šµšæ(š‘‹, š‘Œ ), āˆ„š‘“ + š‘”āˆ„ = sup āˆ„š‘“ (x) + š‘”(x)āˆ„ ā‰¤ sup āˆ„š‘“ (x)āˆ„ + sup āˆ„š‘”(x)āˆ„ ā‰¤ āˆ„š‘“ āˆ„ + āˆ„š‘”āˆ„ āˆ„xāˆ„=1

āˆ„xāˆ„=1

āˆ„xāˆ„=1

verifying the triangle inequality. There āˆ„š‘“ āˆ„ is a norm. šµšæ(š‘‹, š‘Œ ) is a linear space Let š‘“, š‘” āˆˆ šµšæ(š‘‹, š‘Œ ). Since šµšæ(š‘‹, š‘Œ ) āŠ† šæ(š‘‹, š‘Œ ), š‘“ + š‘” is linear, that is š‘“ +š‘” āˆˆ šæ(š‘‹, š‘Œ ) (Exercise 3.2). Similarly, š›¼š‘“ āˆˆ šæ(š‘‹, š‘Œ ) for every š›¼ āˆˆ ā„œ. Further, by the triangle inequality āˆ„š‘“ + š‘”āˆ„ ā‰¤ āˆ„š‘“ āˆ„ + āˆ„š‘”āˆ„ and therefore for every x āˆˆ š‘‹ āˆ„(š‘“ + š‘”)(x)āˆ„ ā‰¤ āˆ„š‘“ + š‘”āˆ„ āˆ„xāˆ„ ā‰¤ (āˆ„š‘“ āˆ„ + āˆ„š‘”āˆ„) āˆ„xāˆ„ Therefore š‘“ + š‘” āˆˆ šµšæ(š‘‹, š‘Œ ). Similarly āˆ„(š›¼š‘“ )(x)āˆ„ ā‰¤ (āˆ£š›¼āˆ£ āˆ„š‘“ āˆ„) āˆ„xāˆ„ so that š›¼š‘“ āˆˆ šµšæ(š‘‹, š‘Œ ). 116

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

šµšæ(š‘‹, š‘Œ ) is complete with this norm Let (š‘“ š‘› ) be a Cauchy sequence in šµšæ(š‘‹, š‘Œ ). For every x āˆˆ š‘‹ āˆ„š‘“ š‘› (x) āˆ’ š‘“ š‘š (x)āˆ„ ā‰¤ āˆ„š‘“ š‘› āˆ’ š‘“ š‘š āˆ„ āˆ„xāˆ„ Therefore (š‘“ š‘› (x)) is a Cauchy sequence in š‘Œ , which converges since š‘Œ is complete. Deļ¬ne the function š‘“ : š‘‹ ā†’ š‘Œ by š‘“ (x) = limš‘›ā†’āˆž š‘“ š‘› (x). š‘“ is linear since š‘“ (x1 + x2 ) = lim š‘“ š‘› (x1 + x2 ) = lim š‘“ š‘› (x1 ) + lim š‘“ š‘› (x2 ) = š‘“ (x1 ) + š‘“ (x2 ) and š‘“ (š›¼x) = lim š‘“ š‘› (š›¼x) = š›¼ lim š‘“ š‘› (x) = š›¼š‘“ (x) To show that š‘“ is bounded, we observe that     āˆ„š‘“ (x)āˆ„ = lim š‘“ š‘› (x) = lim āˆ„š‘“ š‘› (x)āˆ„ ā‰¤ sup āˆ„š‘“ š‘› (x)āˆ„ ā‰¤ sup āˆ„š‘“ š‘› āˆ„ āˆ„xāˆ„ š‘›

š‘›

š‘›

š‘›

Since (š‘“ š‘› ) is a Cauchy sequence, (š‘“ š‘› ) is bounded (Exercise 1.100), that is there exists š‘€ such that āˆ„š‘“ š‘› āˆ„ ā‰¤ š‘€ . This implies āˆ„š‘“ (x)āˆ„ ā‰¤ sup āˆ„š‘“ š‘› āˆ„ āˆ„xāˆ„ ā‰¤ š‘€ āˆ„xāˆ„ š‘›

Thus, š‘“ is bounded. To complete the proof, we must show š‘“ š‘› ā†’ š‘“ , that is āˆ„š‘“ š‘› āˆ’ š‘“ āˆ„ ā†’ 0. Since (š‘“ š‘› ) is a Cauchy sequence, for every šœ– > 0, there exists š‘ such that āˆ„š‘“ š‘› āˆ’ š‘“ š‘š āˆ„ ā‰¤ šœ– for every š‘›, š‘š ā‰„ š‘ and consequently āˆ„š‘“ š‘› (x) āˆ’ š‘“ š‘š (x)āˆ„ = āˆ„(š‘“ š‘› āˆ’ š‘“ š‘š )(x)āˆ„ ā‰¤ šœ– āˆ„xāˆ„ Letting š‘š go to inļ¬nity, āˆ„š‘“ š‘› (x) āˆ’ š‘“ (x)āˆ„ = āˆ„(š‘“ š‘› āˆ’ š‘“ )(x)āˆ„ ā‰¤ šœ– āˆ„xāˆ„ for every x āˆˆ š‘‹ and š‘› ā‰„ š‘ and therefore āˆ„š‘“ š‘› āˆ’ š‘“ āˆ„ = sup {š‘“ š‘› āˆ’ š‘“ )(x)} ā‰¤ šœ– āˆ„xāˆ„=1

for every š‘› ā‰„ š‘ . 3.34

1. Since š‘‹ is ļ¬nite-dimensional, š‘† is compact (Proposition 1.4). Since š‘“ is continuous, š‘“ (š‘†) is a compact set in š‘Œ (Exercise 2.3). Since 0š‘‹ āˆˆ / š‘†, 0š‘Œ = š‘“ (0š‘‹ ) āˆˆ / š‘“ (š‘†). ( )š‘ 2. Consequently, š‘“ (š‘†) is an open set containing 0š‘Œ . It contains an open ball ( )š‘ š‘‡ āŠ† š‘“ (š‘†) around 0š‘Œ . 3. Let y āˆˆ š‘‡ and choose any x āˆˆ š‘“ āˆ’1 (y) and consider y/ āˆ„xāˆ„. Since š‘“ is linear, ( ) x š‘“ (x) y = =š‘“ āˆˆ š‘“ (š‘†) āˆ„xāˆ„ āˆ„xāˆ„ āˆ„xāˆ„ and therefore y/ āˆ„xāˆ„ āˆˆ / š‘‡ since š‘‡ āˆ© š‘“ (š‘†) = āˆ…. 117

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Suppose that y āˆˆ / š‘“ (šµ). Then āˆ„xāˆ„ ā‰„ 1 and therefore y āˆˆ š‘‡ =ā‡’

y āˆˆš‘‡ āˆ„xāˆ„

since š‘‡ is convex. This contradiction establishes that y āˆˆ š‘“ (šµ) and therefore š‘‡ āŠ† š‘“ (šµ). We conclude that š‘“ (šµ) contains an open ball around 0š‘Œ . 4. Let š‘† be any open set in š‘‹. We need to show that š‘“ (š‘†) is open in š‘Œ . Choose any y āˆˆ š‘“ (š‘†) and x āˆˆ š‘“ āˆ’1 (y). Then x āˆˆ š‘† and, since š‘† is open, there exists some š‘Ÿ > 0 such that šµš‘Ÿ (x) āŠ† š‘†. Now šµš‘Ÿ (x) = x + š‘Ÿšµ and š‘“ (šµš‘Ÿ (x)) = y + š‘Ÿš‘“ (šµ) āŠ† š‘“ (š‘†) by linearity. As we have just shown, there exists an open ball T about 0š‘Œ such that š‘‡ āŠ† š‘“ (šµ). Let š‘‡ (x) = y + š‘Ÿš‘‡ . š‘‡ (x) is an open ball about y. Since š‘‡ āŠ† š‘“ (šµ), š‘‡ (x) = y + š‘Ÿš‘‡ āŠ† š‘“ (šµš‘Ÿ (x)) āŠ† š‘“ (š‘†). This implies that š‘“ (š‘†) is open. Since š‘† was an arbitrary open set, š‘“ is an open map. 5. Exercise 2.69. 3.35 š‘“ is linear š‘“ (š›¼ + š›½) =

š‘› āˆ‘

(š›¼š‘– + š›½š‘– )xš‘– =

š‘–=1

š‘› āˆ‘

š›¼š‘– xš‘– +

š‘–=1

š‘› āˆ‘

š›½š‘– xš‘– = š‘“ (š›¼) + š‘“ (š›½)

š‘–=1

Similarly for every š‘” āˆˆ ā„œ š‘“ (š‘”š›¼) = š‘”

š‘› āˆ‘

š›¼š‘– xš‘– = š‘”š‘“ (š‘”š›¼)

š‘–=1

š‘“ is one-to-one Exercise 1.137. š‘“ is onto By deļ¬nition of a basis lin {x1 , x2 , . . . , xš‘› } = š‘‹ š‘“ is continuous Exercise 3.31 š‘“ is an open map Proposition 3.2 3.36 š‘“ is bounded and therefore there exists š‘€ such that āˆ„š‘“ (x)āˆ„ ā‰¤ š‘€ āˆ„xāˆ„. Similarly, š‘“ āˆ’1 is bounded and therefore there exists š‘š such that for every x š‘“ āˆ’1 (y) ā‰¤

1 āˆ„yāˆ„ š‘š

where y = š‘“ (x). This implies š‘š āˆ„xāˆ„ ā‰¤ āˆ„š‘“ (x)āˆ„ and therefore for every x āˆˆ š‘‹. š‘š āˆ„xāˆ„ ā‰¤ āˆ„š‘“ (x)āˆ„ ā‰¤ š‘€ āˆ„xāˆ„ By the linearity of š‘“ , š‘š āˆ„x1 āˆ’ x2 āˆ„ ā‰¤ āˆ„š‘“ (x1 āˆ’ x2 )āˆ„ = āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 )āˆ„ ā‰¤ š‘€ āˆ„x1 āˆ’ x2 āˆ„

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

3.37 For any function, continuity implies closed graph (Exercise 2.70). To show the converse, assume that šŗ = graph(š‘“ ) is closed. š‘‹ Ɨš‘Œ with norm āˆ„(x, y)āˆ„ = max{āˆ„xāˆ„ , āˆ„yāˆ„} is a Banach space (Exercise 1.209). Since šŗ is closed, šŗ is complete. Also, šŗ is a subspace of š‘‹ Ɨ š‘Œ . Consequently, šŗ is a Banach space in its own right. Consider the projection ā„Ž : šŗ ā†’ š‘‹ deļ¬ned by ā„Ž(x, š‘“ (x)) = x. Clearly ā„Ž is linear, one-to-one and onto with ā„Žāˆ’1 (x) = (x, š‘“ (x)) It is also bounded since āˆ„ā„Ž(x, š‘“ (x))āˆ„ = āˆ„xāˆ„ ā‰¤ āˆ„(x, š‘“ (x)āˆ„ By the open mapping theorem, ā„Žāˆ’1 is bounded. For every x āˆˆ š‘‹     āˆ„š‘“ (x)āˆ„ ā‰¤ āˆ„(x, š‘“ (x))āˆ„ = ā„Žāˆ’1 (x) ā‰¤ ā„Žāˆ’1  āˆ„xāˆ„ We conclude that š‘“ is bounded and hence continuous. 3.38 š‘“ (1) = 5, š‘“ (2) = 7 but š‘“ (1 + 2) = š‘“ (3) = 9 āˆ•= š‘“ (1) + š‘“ (2) Similarly š‘“ (3 Ɨ 2) = š‘“ (6) = 15 āˆ•= 3 Ɨ š‘“ (2) 3.39 Assume š‘“ is aļ¬ƒne. Let y = š‘“ (0) and deļ¬ne š‘”(x) = š‘“ (x) āˆ’ y š‘” is homogeneous since for every š›¼ āˆˆ ā„œ š‘”(š›¼x) = š‘”(š›¼x + (1 āˆ’ š›¼)0) = š‘“ (š›¼x + (1 āˆ’ š›¼)0) āˆ’ y = š›¼š‘“ (x) + (1 āˆ’ š›¼)š‘“ (0) āˆ’ y = š›¼š‘“ (x) + (1 āˆ’ š›¼)y āˆ’ y = š›¼š‘“ (x) āˆ’ š›¼y = š›¼(š‘“ (š‘„) āˆ’ y) = š›¼š‘”(x) Similarly for any x1 , x2 āˆˆ š‘‹ š‘”(š›¼x1 + (1 āˆ’ š›¼)x2 ) = š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) āˆ’ š‘¦ = š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) āˆ’ š‘¦ Therefore, for š›¼ = 1/2 1 1 1 1 š‘”( x1 + x2 ) = š‘“ (x1 ) + š‘“ (x2 ) āˆ’ š‘¦ 2 2 2 2 1 1 = (š‘“ (x1 ) āˆ’ š‘¦) + (š‘“ (x2 ) āˆ’ š‘¦) 2 2 1 1 = š‘”(x1 ) + š‘”(x2 ) 2 2 119

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Since š‘” is homogeneous š‘”(x1 + x2 ) = š‘”(x1 ) + š‘”(x2 ) which shows that š‘” is additive and hence linear. Conversely if š‘“ (x) = š‘”(x) + y with š‘” linear š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) = š›¼š‘”(x1 ) + (1 āˆ’ š›¼)š‘”(x2 ) + š‘¦ = š›¼š‘”(x1 ) + š‘¦ + (1 āˆ’ š›¼)š‘”(x2 ) + š‘¦ = š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) 3.40 Let š‘† be an aļ¬ƒne subset of š‘‹ and let y1 , y2 belong to š‘“ (š‘†). Choose any x1 āˆˆ š‘“ āˆ’1 (y1 ) and x2 āˆˆ š‘“ āˆ’1 (y2 ). Then for any š›¼ āˆˆ ā„œ š›¼x1 + (1 āˆ’ š›¼)x2 āˆˆ š‘† Since š‘“ is aļ¬ƒne š›¼y1 + (1 āˆ’ š›¼)y2 = š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) = š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) āˆˆ š‘“ (š‘†) š‘“ (š‘†) is an aļ¬ƒne set. Let š‘‡ be an aļ¬ƒne subset of š‘Œ and let x1 , x2 belong to š‘“ āˆ’1 (š‘‡ ). Let y1 = š‘“ (x1 ) and y2 = š‘“ (x2 ). Then y1 , y2 āˆˆ š‘‡ . For every š›¼ āˆˆ ā„œ š›¼y1 + (1 āˆ’ š›¼)y2 = š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) āˆˆ š‘‡ Since š‘“ is aļ¬ƒne, this implies that š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) = š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) āˆˆ š‘‡ Therefore š›¼x1 + (1 āˆ’ š›¼)x2 āˆˆ š‘“ āˆ’1 (š‘‡ ) We conclude that š‘“ āˆ’1 (š‘‡ ) is an aļ¬ƒne set. 3.41 For any y1 , y2 āˆˆ š‘“ (š‘†), choose x1 , x2 āˆˆ š‘† such that yš‘– = š‘“ (xš‘– ). Since š‘† is convex, š›¼x1 + (1 āˆ’ š›¼)x2 āˆˆ š‘† and therefore š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) = š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) = š›¼y1 + (1 āˆ’ š›¼)y2 āˆˆ š‘“ (š‘†) Therefore š‘“ (š‘†) is convex. 3.42 Suppose otherwise that y is not eļ¬ƒcient. Then there exists another production plan yā€² āˆˆ š‘Œ such that yā€² ā‰„ y. Since p > 0, this implies that pyā€² > py, contradicting the assumption that y maximizes proļ¬t. 3.43 The random variable š‘‹ can be represented as the sum āˆ‘ š‘‹(š‘ )šœ’{š‘ } š‘‹= š‘ āˆˆš‘†

120

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

where šœ’{š‘ } is the indicator function of the set {š‘ }. Since šø is linear šø(š‘‹) =

āˆ‘

š‘‹(š‘ )šø(šœ’{š‘ } )

š‘ āˆˆš‘†

=

āˆ‘

š‘š‘† š‘‹(š‘ )

š‘ āˆˆš‘†

since šø(šœ’{š‘ } = š‘ƒ ({š‘ }) = š‘š‘  ā‰„ 0. For the random variable š‘‹ = 1, š‘‹(š‘ ) = 1 for every š‘  āˆˆ š‘† and āˆ‘ š‘š‘† = 1 šø(1) = š‘ āˆˆš‘†

3.44 Let š‘„1 , š‘„2 āˆˆ š¶[0, 1]. Recall that addition in C[0,1] is deļ¬ned by (š‘„1 + š‘„2 )(š‘”) = š‘„1 (š‘”) + š‘„2 (š‘”) Therefore š‘“ (š‘„1 + š‘„2 ) = (š‘„1 + š‘„2 )(1/2) = š‘„1 (1/2) + š‘„2 (1/2) = š‘“ (š‘„1 ) + š‘“ (š‘„2 ) Similarly š‘“ (š›¼š‘„1 ) = (š›¼š‘„1 )(1/2) = š›¼š‘„1 (1/2) = š›¼š‘“ (š‘„1 ) 3.45 Assume that xāˆ— = xāˆ—1 + xāˆ—2 + ā‹… ā‹… ā‹… + xāˆ—š‘› maximizes š‘“ over š‘†. Suppose to the contrary that there exists yš‘— āˆˆ š‘†š‘— such that š‘“ (yš‘— ) > š‘“ (xāˆ—š‘— ). Then y = xāˆ—1 + xāˆ—2 + ā‹… ā‹… ā‹… + yš‘— + ā‹… ā‹… ā‹… + xāˆ—š‘› āˆˆ š‘† and āˆ‘ āˆ‘ š‘“ (y) = š‘“ (xāˆ—š‘– ) + š‘“ (yš‘– ) > š‘“ (xāˆ—š‘– ) = š‘“ (xāˆ— ) š‘–

š‘–āˆ•=š‘—

contradicting the assumption at š‘“ is maximized at xāˆ— . Conversely, assume š‘“ (xāˆ—š‘– ) ā‰„ š‘“ (xš‘– ) for every xš‘– āˆˆ š‘†š‘– for every š‘– = 1, 2, . . . , š‘›. Summing āˆ‘ āˆ‘ āˆ‘ āˆ‘ š‘“ (xāˆ— ) = š‘“ ( xāˆ—š‘– ) = š‘“ (š‘„āˆ—š‘– ) ā‰„ š‘“ (xš‘– ) = š‘“ ( xš‘– ) = š‘“ (x) for every x āˆˆ š‘† xāˆ— = xāˆ—1 + xāˆ—2 + ā‹… ā‹… ā‹… + xāˆ—š‘› maximizes š‘“ over š‘†. 3.46

1. Assume (š‘„š‘” ) is a sequence in š‘™1 with š‘  = the sequence of partial sums š‘ š‘” =

š‘” āˆ‘

āˆ‘āˆž

š‘”=1

āˆ£š‘„š‘— āˆ£ < āˆž. Let (š‘ š‘” ) denote

āˆ£š‘„š‘— āˆ£

š‘—=1

Then (š‘ š‘” ) is a bounded monotone sequence in ā„œš‘› which converges to š‘ . Consequently, (š‘ š‘” ) is a Cauchy sequence. For every šœ– > 0 there exists an š‘ such that š‘š+š‘˜ āˆ‘

āˆ£š‘„š‘” āˆ£ < šœ–

š‘›=š‘š

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

for every š‘š ā‰„ š‘ and š‘˜ ā‰„ 0. Letting š‘˜ = 0 āˆ£š‘„š‘” āˆ£ < šœ– for every š‘› ā‰„ š‘ We conclude that š‘„š‘” ā†’ 0 so that (š‘„š‘” ) āˆˆ š‘0 . This establishes š‘™1 āŠ† š‘0 . To see that the inclusion is strict, that is š‘™1 āŠ‚ š‘0 , we observe that the sequence (1/š‘›) = (1, 1/2, 1/3, . . . ) converges to zero but that since āˆž   āˆ‘ 1   = 1 + 1 + 1+ = āˆž š‘› 2 3 š‘›=1 (1/š‘›) āˆˆ / š‘™1 . Every convergent sequence is bounded (Exercise 1.97). Therefore š‘0 āŠ‚ š‘™āˆž . 2. Clearly, every sequence (š‘š‘” ) āˆˆ š‘™1 deļ¬nes a linear functional š‘“ āˆˆ š‘ā€²0 given by š‘“ (x) =

āˆž āˆ‘

š‘š‘” š‘„š‘”

š‘›=1

for every x = (š‘„š‘” ) āˆˆ š‘0 . To show that š‘“ is bounded we observe that every (š‘„š‘” ) āˆˆ š‘0 is bounded and consequently āˆ£š‘“ (x)āˆ£ ā‰¤

āˆž āˆ‘ š‘›=1

āˆ£š‘š‘” āˆ£ āˆ£š‘„š‘” āˆ£ ā‰¤ āˆ„(š‘„š‘” )āˆ„āˆž

āˆž āˆ‘ š‘›=1

āˆ£š‘š‘” āˆ£ = āˆ„(š‘š‘” )āˆ„1 āˆ„(š‘„š‘” )āˆ„āˆž

Therefore š‘“ āˆˆ š‘āˆ—0 . To show the converse, let eš‘” denote the unit sequences e1 = (1, 0, 0, . . . ) e2 = (0, 1, 0, . . . ) e3 = (0, 0, 1, . . . ) {e1 , e2 , e3 , . . . , } form a basis for š‘0 . Then every sequence (š‘„š‘” ) āˆˆ š‘0 has a unique representation (š‘„š‘” ) =

āˆž āˆ‘

š‘„š‘” eš‘”

š‘›=1

Let š‘“ āˆˆ š‘āˆ—0 be a continuous linear functional on š‘0 . By continuity and linearity š‘“ (x) =

āˆž āˆ‘

š‘„š‘” š‘“ (eš‘” )

š‘›=1

Let š‘š‘” = š‘“ (eš‘” ) so that š‘“ (x) =

āˆž āˆ‘

š‘š‘” š‘„š‘”

š‘›=1

Every linear function is determined by its action on a basis (Exercise 3.23). 122

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Solutions for Foundations of Mathematical Economics

We need to show that the sequence (š‘š‘” ) āˆˆ š‘™1 . For any š‘ , consider the sequence xš‘” = (š‘„1 , š‘„2 , . . . , š‘„š‘” , 0, 0, . . . ) where āŽ§ āŽØ0 š‘š‘” = 0 or š‘› ā‰„ š‘ š‘„š‘” = āˆ£š‘š‘” āˆ£ āŽ© otherwise š‘š‘” Then (xš‘” ) āˆˆ š‘0 , āˆ„xš‘” āˆ„āˆž = 1 and š‘“ (xš‘” ) =

š‘” āˆ‘

š‘š‘” š‘„š‘” =

š‘›=1

š‘” āˆ‘

āˆ£š‘š‘” āˆ£

š‘›=1

Since š‘“ āˆˆ š‘āˆ—0 , š‘“ is bounded and therefore š‘“ (xš‘” ) ā‰¤ āˆ„š‘“ āˆ„ āˆ„xš‘” āˆ„ = āˆ„š‘“ āˆ„ < āˆž and therefore š‘” āˆ‘

āˆ£š‘š‘” āˆ£ < āˆž for every š‘ = 1, 2, . . .

š‘›=1

Consequently āˆž āˆ‘

āˆ£š‘š‘” āˆ£ = sup

š‘” āˆ‘

š‘ š‘›=1

š‘›=1

āˆ£š‘š‘” āˆ£ ā‰¤ āˆ„š‘“ āˆ„ < āˆž

We conclude that (š‘š‘” ) āˆˆ š‘™1 and therefore š‘āˆ—0 = š‘™1 3. Similarly, every sequence (š‘š‘” ) āˆˆ š‘™āˆž deļ¬nes a linear functional š‘“ on š‘™1 given by š‘“ (x) =

āˆž āˆ‘

š‘š‘” š‘„š‘”

š‘›=1

for every x = (š‘„š‘” ) āˆˆ š‘™1 . Moreover š‘“ is bounded since āˆ£š‘“ (x)āˆ£ ā‰¤

āˆž āˆ‘

āˆ£š‘š‘” āˆ£ āˆ£š‘„š‘” āˆ£ ā‰¤ āˆ„(š‘š‘” )āˆ„

š‘›=1

āˆž āˆ‘

āˆ£š‘„š‘” āˆ£ < āˆž

š‘›=1

for every x = (š‘„š‘” ) āˆˆ š‘™1 Again, given any linear functional š‘“ āˆˆ š‘™1āˆ— , let š‘š‘” = š‘“ (eš‘” ) where eš‘” is the š‘› unit sequence. Then š‘“ has the representation š‘“ (x) =

āˆž āˆ‘

š‘š‘” š‘„š‘”

š‘›=1

To show that (š‘š‘” ) āˆˆ š‘™āˆž , for š‘ = 1, 2, . . . , consider the sequence xš‘” = (0, 0, . . . , š‘„š‘” , 0, 0, . . . ) where āŽ§ āŽØ āˆ£š‘š‘” āˆ£ š‘› = š‘ and š‘ āˆ•= 0 š‘” š‘š‘” š‘„š‘” = āŽ© 0 otherwise Then xš‘” āˆˆ š‘™1 , āˆ„xš‘” āˆ„1 = 1 and š‘“ (xš‘” ) = āˆ£š‘š‘” āˆ£ Since š‘“ āˆˆ

š‘™1āˆ— ,

š‘“ is bounded and therefore  š‘ š‘  = š‘“ (xš‘ ) ā‰¤ āˆ„š‘“ āˆ„ āˆ„xš‘› āˆ„ = āˆ„š‘“ āˆ„

for every š‘ . Consequently (š‘š‘ ) āˆˆ š‘™āˆž . We conclude that š‘™1āˆ— = š‘™āˆž 123

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 3.47 By linearity šœ‘(š‘„, š‘”) = šœ‘(š‘„, 0) + šœ‘(0, š‘”) = šœ‘(š‘„, 0) + šœ‘(0, 1)š‘”

Considered as a function of š‘„, šœ‘(š‘„, 0) is a linear functional on š‘‹. Deļ¬ne š‘”(š‘„) = šœ‘(š‘„, 0) š›¼ = šœ‘(0, 1) Then šœ‘(š‘„, š‘”) = š‘”(š‘„) + š›¼š‘” 3.48 Suppose š‘š āˆ©

kernel š‘”š‘— āŠ† kernel š‘“

š‘—=1

Deļ¬ne the function šŗ : š‘‹ ā†’ ā„œš‘› by šŗ(x) = (š‘”1 (x), š‘”2 (x), . . . , š‘”š‘š (x)) Then kernel šŗ = { x āˆˆ š‘‹ : š‘”š‘— (x) = 0, š‘— = 1, 2, . . . š‘š } š‘š āˆ© = kernel š‘”š‘— š‘—=1

āŠ† kernel š‘“ š‘“ : š‘‹ ā†’ ā„œ and šŗ : š‘‹ ā†’ ā„œš‘› . By Exercise 3.22, there exists a linear function š» : ā„œš‘› ā†’ ā„œ such that š‘“ = š» āˆ˜ šŗ. That is, for every š‘„ āˆˆ š‘‹ š‘“ (x) = š» āˆ˜ šŗ(x) = š»(š‘”1 (x), š‘”2 (x), . . . , š‘”š‘š (x)) Let š›¼š‘— = š»(eš‘— ) where eš‘— is the š‘—-th unit vector in ā„œš‘š . Since every linear mapping is determined by its action on a basis, we must have š‘“ (x) = š›¼1 š‘”1 (x) + š›¼2 š‘”2 (x) + ā‹… ā‹… ā‹… + š›¼š‘š š‘”š‘š (x)

for every š‘„ āˆˆ š‘‹

That is š‘“ āˆˆ lin š‘”1 , š‘”2 , . . . , š‘”š‘š Conversely, suppose š‘“ āˆˆ lin š‘”1 , š‘”2 , . . . , š‘”š‘š That is š‘“ (x) = š›¼1 š‘”1 (x) + š›¼2 š‘”2 (x) + ā‹… ā‹… ā‹… + š›¼š‘š š‘”š‘š (x) for every š‘„ āˆˆ š‘‹ āˆ©š‘š For every x āˆˆ š‘—=1 kernel š‘”š‘— , š‘”š‘— (x) = 0, š‘— = 1, 2, . . . , š‘š and therefore š‘“ (x) = 0. Therefore š‘„ āˆˆ kernel š‘“ . That is š‘š āˆ©

kernel š‘”š‘— āŠ† kernel š‘“

š‘—=1

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3.49 Let š» be a hyperplane in š‘‹. Then there exists a unique subspace š‘‰ such that š» = x0 + š‘‰ for some x0 āˆˆ š» (Exercise 1.153). There are two cases to consider. Case 1: x0 āˆˆ / š‘‰ . For every x āˆˆ š‘‹, there exists unique š›¼x āˆˆ ā„œ such x = š›¼x x0 + š‘£ for some š‘£ āˆˆ š‘‰ Deļ¬ne š‘“ (x) = š›¼x . Then š‘“ : š‘‹ ā†’ ā„œ. It is straightforward to show that š‘“ is linear. Since š» = x0 + š‘‰ , š›¼x = 1 if and only if x āˆˆ š». Therefore š» = { x āˆˆ š‘‹ : š‘“ (x) = 1 } Case 2: x0 āˆˆ š‘‰ . In this case, choose some x1 āˆˆ / š‘‰ . Again, for every x āˆˆ š‘‹, there exists a unique š›¼x āˆˆ ā„œ such x = š›¼x x1 + š‘£ for some š‘£ āˆˆ š‘‰ and š‘“ (x) = š›¼x is a linear functional on š‘‹. Furthermore x0 āˆˆ š‘‰ implies š» = š‘‰ (Exercise 1.153) and therefore š‘“ (x) = 0 if and only if x āˆˆ š». Therefore š» = { x āˆˆ š‘‹ : š‘“ (x) = 0 } Conversely, let š‘“ be a nonzero linear functional in š‘‹ ā€² . Let š‘‰ = kernel š‘“ and choose x0 āˆˆ š‘“ āˆ’1 (1). (This is why we require š‘“ āˆ•= 0). For any x āˆˆ š‘‹ š‘“ (x āˆ’ š‘“ (x)x0 ) = š‘“ (x) āˆ’ š‘“ (x) Ɨ 1 = 0 so that x āˆ’ š‘“ (x)x0 āˆˆ š‘‰ . That is, x = š‘“ (x)x0 + š‘£ for some š‘£ āˆˆ š‘‰ . Therefore, š‘‹ = lin (x0 , š‘‰ ) so that š‘‰ is a maximal proper subspace. For any š‘ āˆˆ ā„œ, let x1 āˆˆ š‘“ āˆ’1 (š‘). Then, for every x āˆˆ š‘“ āˆ’1 (š‘), š‘“ (x āˆ’ x1 ) = 0 and { x : š‘“ (x) = š‘} = {x : š‘“ (x āˆ’ x1 ) = 0 } = x1 + š‘‰ which is a hyperplane. 3.50 By the previous exercise, there exists a linear functional š‘” such that š» = { š‘„ āˆˆ š‘‹ : š‘“ (š‘„) = š‘ } for some š‘ āˆˆ ā„œ. Since 0 āˆˆ / š», š‘ āˆ•= 0. Without loss of generality, we can assume that š‘ = 1. (Otherwise, take the linear functional 1š‘ š‘“ ). To show that š‘“ is unique, assume that š‘” is another linear functional with š» = { x : š‘“ (š‘„) = 1} = {x : š‘”(š‘„) = 1 } Then š» āŠ† { x : š‘“ (š‘„) āˆ’ š‘”(š‘„) = 0 } Since š» is a maximal subset, š‘‹ is the smallest subspace containing š». Therefore š‘“ (š‘„) = š‘”(š‘„) for every š‘„ āˆˆ š‘‹.

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3.51 By Exercise 3.49, there exists a linear functional š‘“ such that š» = { š‘„ āˆˆ š‘‹ : š‘“ (š‘„) = 0 } Since x0 āˆˆ / š», š‘“ (x0 ) āˆ•= 0. Without loss of generality, we can normalize so that š‘“ (x0 ) = 1. (If š‘“ (x0 ) = š‘ āˆ•= 1, then the linear functional š‘“ ā€² = 1/cš‘“ has š‘“ ā€² (x0 ) = 1 and kernel š‘“ ā€² = š».) To show that š‘“ is unique, suppose that š‘” is another linear functional with kernel š‘” = š» and š‘”(x0 ) = 1. For any x āˆˆ š‘‹, there exists š›¼ āˆˆ ā„œ such that x = š›¼x0 + v with š‘£ āˆˆ š» (Exercise 1.153). Since š‘“ (v) = š‘”(v) = 0 and š‘“ (x0 ) = š‘”(x0 ) = 1 š‘”(x) = š‘”(š›¼x0 + v) = š›¼š‘”(š‘„0 ) = š›¼š‘“ (x0 ) = š‘“ (š›¼x0 + v) = š‘“ (x) 3.52 Assume š‘“ = šœ†š‘”, šœ† āˆ•= 0. Then š‘“ (š‘„) = 0 ā‡ā‡’ š‘”(š‘„) = 0 Conversely, let š» = š‘“ āˆ’1 (0) = š‘” āˆ’1 (0). If š» = š‘‹, then š‘“ = š‘” = 0. Otherwise, š» is a hyperplane containing 0. Choose some x0 āˆˆ / š». Every x āˆˆ š‘‹ has a unique representation x = š›¼x0 + v with v āˆˆ š» (Exercise 1.153) and š‘“ (x) = š›¼š‘“ (x0 ) š‘”(x) = š›¼š‘”(x0 ) Let šœ† = š‘“ (x0 )/š‘”(x0 ) so that š‘“ (x0 ) = šœ†š‘”(x0 ). Substituting š‘“ (x) = š›¼š‘“ (x0 ) = š›¼šœ†š‘”(x0 ) = šœ†š‘”(x) 3.53 š‘“ continuous implies that the set { š‘„ āˆˆ š‘‹ : š‘“ (š‘„) = š‘ } = š‘“ āˆ’1 (š‘) is closed for every š‘ āˆˆ ā„œ (Exercise 2.67). Conversely, let š‘ = 0 and assume that š» = { š‘„ āˆˆ š‘‹ : š‘“ (š‘„) = 0 } is closed. There exists x0 āˆ•= 0 such that š‘‹ = lin {š‘„0 , š»} (Exercise 1.153). Let xš‘› ā†’ x be a convergent sequence in š‘‹. Then there exist š›¼š‘› , š›¼ āˆˆ ā„œ and vš‘› , š‘£ āˆˆ š» such that xš‘› = š›¼š‘› x0 + vš‘› , x = š›¼x0 + v and āˆ„xš‘› āˆ’ xāˆ„ = āˆ„š›¼š‘› x0 + vš‘› āˆ’ š›¼x0 + vāˆ„ = āˆ„š›¼š‘› x0 āˆ’ š›¼x0 + vš‘› āˆ’ vāˆ„ ā‰¤ āˆ£š›¼š‘› āˆ’ š›¼āˆ£ āˆ„x0 āˆ„ + āˆ„vš‘› āˆ’ vāˆ„ ā†’0 which implies that š›¼š‘› ā†’ š›¼. By linearity š‘“ (xš‘› ) = š›¼š‘› š‘“ (x0 ) + š‘“ (vš‘› ) = š›¼š‘› š‘“ (x0 ) since vš‘› āˆˆ š» and therefore š‘“ (xš‘› ) = š›¼š‘› š‘“ (x0 ) ā†’ š›¼š‘“ (x0 ) = š‘“ (x) š‘“ is continuous.

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Solutions for Foundations of Mathematical Economics 3.54 š‘“ (x + xā€² , y) =

š‘š āˆ‘ š‘› āˆ‘

š‘Žš‘–š‘— (š‘„š‘– + š‘„ā€²š‘– )š‘¦š‘—

š‘–=1 š‘—=1

=

š‘š āˆ‘ š‘› āˆ‘

š‘Žš‘–š‘— š‘„š‘– š‘¦š‘— +

š‘–=1 š‘—=1

š‘š āˆ‘ š‘› āˆ‘

š‘Žš‘–š‘— š‘„ā€²š‘– š‘¦š‘—

š‘–=1 š‘—=1

= š‘“ (x, y) + š‘“ (xā€² , y) Similarly, we can show that š‘“ (x, y + yā€² ) = š‘“ (x, y) + š‘“ (x, yā€² ) and š‘“ (š›¼x, y) = š›¼š‘“ (x, y) = š‘“ (x, š›¼y) for every š›¼ āˆˆ ā„œ 3.55 Let x1 , x2 , . . . , xš‘š be a basis for š‘‹ and y1 , y2 , . . . , yš‘› be a basis for š‘Œ . Let the numbers š‘Žš‘–š‘— represent the action of š‘“ on these bases, that is š‘Žš‘–š‘— = š‘“ (xš‘– , yš‘— )

š‘– = 1, 2, . . . , š‘š, š‘— = 1, 2, . . . , š‘›

and let š“ be the š‘š Ɨ š‘› matrix of numbers š‘Žš‘–š‘— . Choose any x āˆˆ š‘‹ and y āˆˆ š‘Œ and let their representations in terms of the bases be x=

š‘š āˆ‘

š›¼š‘– xš‘– and y =

š‘–=1

š‘› āˆ‘

š›½š‘– yš‘—

š‘—=1

respectively. By the bilinearity of š‘“ āˆ‘ āˆ‘ š‘“ (x, y) = š‘“ ( š›¼š‘– xš‘– , š›¼š‘— yš‘— ) =

āˆ‘

š‘–

š›¼š‘– š‘“ (xš‘– ,

š‘–

=

āˆ‘

š›¼š‘–

š‘–

=

āˆ‘

š‘—

āˆ‘

āˆ‘

š›¼š‘— yš‘— )

š‘—

š›¼š‘— š‘“ (xš‘– , yš‘— )

š‘—

š›¼š‘–

āˆ‘

š‘–

=

āˆ‘

š›¼š‘— š‘Žš‘–š‘—

š‘—

š›¼š‘– š“y

š‘– ā€²

= x š“y 3.56 Every y āˆˆ š‘‹ ā€² is a linear functional on š‘‹. Hence y(x + xā€² ) = y(x) + y(xā€² ) y(š›¼x) = š›¼y(x) and therefore š‘“ (x + xā€² , y) = y(x + xā€² ) = y(x) + y(xā€² ) = š‘“ (x, y) + š‘“ (xā€² , y) š‘“ (š›¼x, y) = y(š›¼x) = š›¼y(x) = š›¼š‘“ (x, y) 127

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In the dual space š‘‹ ā€² (y + yā€² )(x) ā‰” y(x) + yā€² (x) (š›¼y)(x) ā‰” š›¼y(x) and therefore š‘“ (x, y + yā€² ) = (y + yā€² )(x) = y(x) + yā€² (x) = š‘“ (x, y) + š‘“ (x, yā€² ) š‘“ (x, š›¼y) = (š›¼y)(x) = š›¼y(x) = š›¼š‘“ (x, y) 3.57 Assume š‘“1 , š‘“2 āˆˆ šµš‘–šæ(š‘‹ Ɨ š‘Œ, š‘). Deļ¬ne the mapping š‘“1 + š‘“2 : š‘‹ Ɨ š‘Œ ā†’ š‘ by (š‘“1 + š‘“2 )(x, y) = š‘“1 (x, y) + š‘“2 (x, y) We have to conļ¬rm that š‘“1 + š‘“2 is bilinear, that is (š‘“1 + š‘“2 )(x1 + x2 , y) = š‘“1 (x1 + x2 , y) + š‘“2 (x1 + x2 , y) = š‘“1 (x1 , y) + š‘“1 (x2 , y) + š‘“2 (x1 , y) + š‘“2 (x2 , y) = š‘“1 (x1 , y) + š‘“2 (x1 , y) + š‘“1 (x1 , y) + š‘“2 (x2 , y) = (š‘“1 + š‘“2 )(x1 , y) + (š‘“1 + š‘“2 )(x2 , y) Similarly, we can show that (š‘“1 + š‘“2 )(x, y1 + y2 ) = (š‘“1 + š‘“2 )(x, y1 ) + (š‘“1 + š‘“2 )(x, y2 ) and (š‘“1 + š‘“2 )(š›¼x, y) = š›¼(š‘“1 + š‘“2 )(x, y) = (š‘“1 + š‘“2 )(x, š›¼y) For every š‘“ āˆˆ šµš‘–šæ(š‘‹ Ɨ š‘Œ, š‘) deļ¬ne the function š›¼š‘“ : š‘‹ Ɨ š‘Œ ā†’ š‘ by (š›¼š‘“ )(x, y) = š›¼š‘“ (x, y) š›¼š‘“ is also bilinear, since (š›¼š‘“ )(x1 + x2 , y) = š›¼š‘“ (x1 + x2 , y) = š›¼š‘“ (x1 , y) + š›¼š‘“ (x2 , y) = (š›¼š‘“ )(x1 , y) + (š›¼š‘“ (x2 , y) Similarly (š›¼š‘“ )(x, y1 + y2 ) = (š›¼š‘“ )(x, y1 ) + (š›¼š‘“ )(x, y2 ) (š›¼š‘“ )(š›½x, y) = š›½(š›¼š‘“ )(x, y) = (š›¼š‘“ )(x, š›½y) Analogous to (Exercise 2.78), š‘“1 + š‘“2 and š›¼š‘“ are also continuous 3.58

1. šµšæ(š‘Œ, š‘) is a linear space and therefore so is šµšæ(š‘‹, šµšæ(š‘Œ, š‘)) (Exercise 3.33).

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2. šœ‘x is linear and therefore š‘“ (x, y1 + y2 ) = šœ‘(x)(y1 + y2 ) = šœ‘(x)(y1 ) + šœ‘(x)(y2 ) = š‘“ (x, y1 ) + š‘“ (x, y2 ) and š‘“ (x, š›¼y) = šœ‘(x)(š›¼y) = š›¼šœ‘(x)(y) = š›¼š‘“ (x, y) Similarly, šœ‘ is linear and therefore š‘“ (x1 + x2 , y) = šœ‘x1 +x2 (y) = šœ‘x1 (y) + šœ‘x2 (y) = š‘“ (x1 , y) + š‘“ (x2 , y) and š‘“ (š›¼x, y) = šœ‘š›¼x (y) = š›¼šœ‘x (y) = š›¼š‘“ (x, y) š‘“ is bilinear 3. Let š‘“ āˆˆ šµš‘–šæ(š‘‹ Ɨ š‘Œ, š‘). For every x āˆˆ š‘‹, the partial function š‘“x : š‘Œ ā†’ š‘ is linear. Therefore š‘“x āˆˆ šµšæ(š‘Œ, š‘) and šœ‘ āˆˆ šµšæ(š‘‹, šµšæ(š‘Œ, š‘)). 3.59 Bilinearity and symmetry imply š‘“ (x āˆ’ š›¼y, x āˆ’ š›¼y) = š‘“ (x, x āˆ’ š›¼y) āˆ’ š›¼š‘“ (y, x āˆ’ š›¼y) = š‘“ (x, x) āˆ’ š›¼š‘“ (x, y) āˆ’ š›¼š‘“ (y, x) + š›¼2 š‘“ (y, y) = š‘“ (x, x) āˆ’ 2š›¼š‘“ (x, y) + š›¼2 š‘“ (y, y) Nonnegativity implies š‘“ (x āˆ’ š›¼y, x āˆ’ š›¼y) = š‘“ (x, x) āˆ’ 2š›¼š‘“ (x, y) + š›¼2 š‘“ (y, y) ā‰„ 0

(3.38)

for every x, y āˆˆ š‘‹ and š›¼ āˆˆ ā„œ Case 1 š‘“ (x, x) = š‘“ (y, y) = 0 Then (3.38) becomes āˆ’2š›¼š‘“ (x, y) ā‰„ 0 Setting š›¼ = š‘“ (x, y) generates ( )2 āˆ’2 š‘“ (x, y) ā‰„ 0 which implies that š‘“ (x, y) = 0 Case 2 Either š‘“ (x, x) > 0 or š‘“ (y, y) > 0. Without loss of generality, assume š‘“ (y, y) > 0 and set š›¼ = š‘“ (x, y)/š‘“ (y, y) in (3.38). That is ( ) ( )2 š‘“ (x, y) š‘“ (x, y) š‘“ (x, x) āˆ’ 2 š‘“ (x, y) + š‘“ (y, y) ā‰„ 0 š‘“ (y, y) š‘“ (y, y) or š‘“ (x, x) āˆ’

š‘“ (x, y)2 ā‰„0 š‘“ (y, y)

which implies ( )2 š‘“ (x, y) ā‰¤ š‘“ (x, x)š‘“ (y, y) for every x, y āˆˆ š‘‹ 129

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Solutions for Foundations of Mathematical Economics

3.60 A Euclidean space is a ļ¬nite-dimensional normed space, which is complete (Proposition 1.4). 3.61 š‘“ (x, y) = xš‘‡ y satisļ¬es the requirements of Exercise 3.59 and therefore (xš‘‡ y)2 ā‰¤ (xš‘‡ x)(yš‘‡ y) Taking square roots  š‘‡  x y ā‰¤ āˆ„xāˆ„ āˆ„yāˆ„ 3.62 By deļ¬nition, the inner product is a bilinear functional. To show that it is continuous, let š‘‹ be an inner product space with inner product denote by xš‘‡ y. Let xš‘› ā†’ x and yš‘› ā†’ y be sequences in š‘‹.  š‘› š‘‡ š‘›    (x ) y āˆ’ xš‘‡ y = (xš‘› )š‘‡ yš‘› āˆ’ (xš‘› )š‘‡ y + (xš‘› )š‘‡ y āˆ’ xš‘‡ y     ā‰¤ (xš‘› )š‘‡ yš‘› āˆ’ (xš‘› )š‘‡ y + (xš‘› )š‘‡ y āˆ’ xš‘‡ y     ā‰¤ (xš‘› )š‘‡ (yš‘› āˆ’ y) + (xš‘› āˆ’ x)š‘‡ y Applying the Cauchy-Schwartz inequality   š‘› š‘‡ š‘› (x ) y āˆ’ xš‘‡ y ā‰¤ āˆ„xš‘› āˆ„ āˆ„yš‘› āˆ’ yāˆ„ + āˆ„xš‘› āˆ’ xāˆ„ āˆ„yāˆ„ Since the sequence xš‘› converges, it is bounded, that is there exists š‘€ such that āˆ„xš‘› āˆ„ ā‰¤ š‘€ for every š‘›. Therefore  š‘› š‘‡ š‘›  (x ) y āˆ’ xš‘‡ y ā‰¤ āˆ„xš‘› āˆ„ āˆ„yš‘› āˆ’ yāˆ„ + āˆ„xš‘› āˆ’ xāˆ„ āˆ„yāˆ„ ā‰¤ š‘€ āˆ„yš‘› āˆ’ yāˆ„ + āˆ„xš‘› āˆ’ xāˆ„ āˆ„yāˆ„ ā†’ 0 3.63 Applying the properties of the inner product āˆš āˆ™ āˆ„xāˆ„ = xš‘‡ x ā‰„ 0 āˆš āˆ™ āˆ„xāˆ„ = xš‘‡ x = 0 if and only if x = 0 āˆš āˆš āˆ™ āˆ„š›¼xāˆ„ = (š›¼x)š‘‡ (š›¼x) = š›¼2 xš‘‡ x = āˆ£š›¼āˆ£ āˆ„xāˆ„ To prove the triangle inequality, observe that bilinearity and symmetry imply 2

āˆ„x + yāˆ„ = (x + y)š‘‡ (x + y) = xš‘‡ x + xš‘‡ y + yš‘‡ x + zš‘‡ z = xš‘‡ x + 2xš‘‡ y + yš‘‡ y 2

2

= āˆ„xāˆ„ + 2xš‘‡ y + āˆ„yāˆ„   ā‰¤ āˆ„xāˆ„2 + 2 xš‘‡ y + āˆ„yāˆ„2 Applying the Cauchy-Schwartz inequality 2

2

āˆ„x + yāˆ„ ā‰¤ āˆ„xāˆ„ + 2 āˆ„xāˆ„ āˆ„yāˆ„ + āˆ„yāˆ„

2

= (āˆ„xāˆ„ + āˆ„yāˆ„)2 3.64 For every y āˆˆ š‘‹, the partial function š‘“y (x) = xš‘‡ y is a linear functional on š‘‹ (since xš‘‡ y is bilinear). Continuity follows from the Cauchy-Schwartz inequality, since for every x āˆˆ š‘‹   āˆ£š‘“y (x)āˆ£ = xš‘‡ y ā‰¤ āˆ„yāˆ„ āˆ„xāˆ„ 130

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics which shows that āˆ„š‘“y āˆ„ ā‰¤ āˆ„yāˆ„. In fact, āˆ„š‘“y āˆ„ = āˆ„yāˆ„ since āˆ„š‘“y āˆ„ = sup āˆ£š‘“y (x)āˆ£ āˆ„xāˆ„=1

 ( )  y  ā‰„ š‘“y āˆ„yāˆ„  (   y )š‘‡    = y  āˆ„yāˆ„  =

1 š‘‡ y y = āˆ„yāˆ„ āˆ„yāˆ„

3.65 By the Weierstrass Theorem (Theorem 2.2), the continuous function š‘”(x) = āˆ„xāˆ„ attains a maximum on the compact set š‘† at some point x0 . We claim that x0 is an extreme point. Suppose not. Then, there exist x1 , x2 āˆˆ š‘† such that x0 = š›¼x1 + (1 āˆ’ š›¼)x2 = x2 + š›¼(x1 āˆ’ x2 ) Since x0 maximizes āˆ„xāˆ„ on š‘†

( )š‘‡ ( ) 2 2 x2 + š›¼(x1 āˆ’ x2 ) āˆ„x2 āˆ„ ā‰¤ āˆ„x0 āˆ„ = x2 + š›¼(x1 āˆ’ x2 ) 2

= āˆ„x2 āˆ„ + 2š›¼xš‘‡2 (x1 āˆ’ x2 ) + š›¼2 āˆ„x1 āˆ’ x2 āˆ„

2

or 2

2xš‘‡2 (x1 āˆ’ x2 ) + š›¼ āˆ„x1 āˆ’ x2 āˆ„ ā‰„ 0

(3.39)

Similarly, interchanging the role of x1 and x2 2

2xš‘‡1 (x2 āˆ’ x1 ) + š›¼ āˆ„x2 āˆ’ x1 āˆ„ ā‰„ 0 or āˆ’2xš‘‡1 (x1 āˆ’ x2 ) + š›¼ āˆ„x1 āˆ’ x2 āˆ„2 ā‰„ 0

(3.40)

Adding the inequalities (3.39) and (3.40) yields 2

2(x2 āˆ’ x1 )š‘‡ (x1 āˆ’ x2 ) + 2š›¼ āˆ„x1 āˆ’ x2 āˆ„ ā‰„ 0 or 2(x2 āˆ’ x1 )š‘‡ (x2 āˆ’ x1 ) = āˆ’2(x2 āˆ’ x1 )š‘‡ (x1 āˆ’ x2 ) ā‰¤ 2š›¼ āˆ„x1 āˆ’ x2 āˆ„2 and therefore āˆ„x2 āˆ’ x1 āˆ„ ā‰¤ š›¼ āˆ„x2 āˆ’ x1 āˆ„ Since 0 < š›¼ < 1, this implies that āˆ„x1 āˆ’ x2 āˆ„ = 0 or x1 = x2 which contradicts our premise that x0 is not an extreme point. 3.66 Using bilinearity and symmetry of the inner product āˆ„x + yāˆ„2 + āˆ„x āˆ’ yāˆ„2 = (x + y)š‘‡ (x + y) + (x āˆ’ y)š‘‡ (x āˆ’ y) = xš‘‡ x + xš‘‡ y + yš‘‡ x + yš‘‡ y + xš‘‡ x āˆ’ xš‘‡ y āˆ’ yš‘‡ x + yš‘‡ y = 2xš‘‡ x + 2yš‘‡ y 2

= 2 āˆ„xāˆ„ + 2 āˆ„yāˆ„ 131

2

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Solutions for Foundations of Mathematical Economics 3.67 Note that āˆ„š‘„āˆ„ = āˆ„š‘¦āˆ„ = 1 and āˆ„š‘„ + š‘¦āˆ„ = sup

( ) š‘„(š‘”) + š‘¦(š‘”) = sup (1 + š‘”) = 2

āˆ„š‘„ āˆ’ š‘¦āˆ„ = sup

( ) š‘„(š‘”) āˆ’ š‘¦(š‘”) = sup (1 āˆ’ š‘”) = 1

0ā‰¤š‘”ā‰¤1 0ā‰¤š‘”ā‰¤1

0ā‰¤š‘”ā‰¤1 0ā‰¤š‘”ā‰¤1

so that 2

2

2

āˆ„š‘„ + š‘¦āˆ„ + āˆ„š‘„ āˆ’ š‘¦āˆ„ = 5 āˆ•= 2 āˆ„š‘„āˆ„ + 2 āˆ„š‘„āˆ„

2

Since š‘„ and š‘¦ do not satisfy the parallelogram law (Exercise 3.66), š¶(š‘‹) cannot be an inner product space. 3.68 Let {x1 , x2 , . . . , xš‘› } be a set of pairwise orthogonal vectors. Assume 0 = š›¼x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› Using bilinearity, this implies 0 = 0š‘‡ xš‘— =

š‘› āˆ‘

š›¼š‘– xš‘‡š‘– xš‘— = š›¼š‘— āˆ„xš‘— āˆ„

š‘–=1

for every š‘— = 1, 2, . . . , š‘›. Since xš‘— āˆ•= 0, this implies š›¼š‘— = 0 for every š‘— = 1, 2, . . . , š‘›. We conclude that the set {x1 , x2 , . . . , xš‘› } is linearly independent (Exercise 1.133). 3.69 Let x1 , x2 , . . . , xš‘› be a orthonormal basis for š‘‹. Since š“ represents š‘“ š‘“ (xš‘— ) =

š‘› āˆ‘

š‘Žš‘–š‘— xš‘–

š‘–=1

for š‘— = 1, 2, . . . , š‘›. Taking the inner product with xš‘– , ( š‘› ) š‘› āˆ‘ āˆ‘ š‘‡ š‘‡ xš‘– š‘“ (xš‘— ) = xš‘– š‘Žš‘–š‘— xš‘– = š‘Žš‘–š‘— xš‘‡š‘– xš‘— š‘–=1

š‘–=1

Since {x1 , x2 , . . . , xš‘› } is orthonormal { xš‘‡š‘˜ xš‘—

=

1 0

if š‘– = š‘— otherwise

so that the last sum simpliļ¬es to xš‘‡š‘– š‘“ (xš‘— ) = š‘Žš‘–š‘— for every š‘–, š‘— 3.70

1. By the Cauchy-Schwartz inequality  š‘‡  x y ā‰¤ āˆ„xāˆ„ āˆ„xāˆ„ for every x and y, so that  š‘‡   x y  ā‰¤1 āˆ£cos šœƒāˆ£ =  āˆ„xāˆ„ āˆ„yāˆ„  which implies āˆ’1 ā‰¤ cos šœƒ ā‰¤ 1 132

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Solutions for Foundations of Mathematical Economics

2. Since cos 90 = 0, šœƒ = 90 implies that xš‘‡ y = 0 or x āŠ„ y. Conversely, if x āŠ„ y, xš‘‡ y = 0 and cos šœƒ = 0 which implies šœƒ = 90 degrees. 3.71 By bilinearity 2

2

āˆ„x + yāˆ„ = (x + y)š‘‡ (x + y) = āˆ„xāˆ„ + xš‘‡ y + yš‘‡ x + āˆ„yāˆ„

2

If x āŠ„ y, xš‘‡ y = yš‘‡ x = 0 and 2

2

āˆ„x + yāˆ„ = āˆ„xāˆ„ + āˆ„yāˆ„ 3.72

2

Ė† āˆˆ š‘† and let š‘†Ė† be the set of all x āˆˆ š‘† which are closer to y 1. Choose some x Ė† , that is than x š‘†Ė† = { x āˆˆ š‘† : āˆ„x āˆ’ yāˆ„ ā‰¤ āˆ„Ė† x āˆ’ yāˆ„ } š‘†Ė† is compact (Proposition 1.4). By the Weierstrass theorem (Theorem 2.2), the continuous function š‘”(x) = āˆ„xyāˆ„ Ė† That is attains a minimum on š‘†Ė† at some point x0 āˆˆ š‘†. āˆ„x0 āˆ’ yāˆ„ ā‰¤ āˆ„x āˆ’ yāˆ„ for every x āˆˆ š‘†Ė† A fortiori āˆ„x0 āˆ’ yāˆ„ ā‰¤ āˆ„x āˆ’ yāˆ„ for every x āˆˆ š‘†

2. Suppose there exists some x1 āˆˆ š‘† such that āˆ„x1 āˆ’ yāˆ„ = āˆ„x0 āˆ’ yāˆ„ = š›æ By the parallelogram law (Exercise 3.66) 2

āˆ„x0 āˆ’ x1 āˆ„ = āˆ„x0 āˆ’ y + y āˆ’ x1 āˆ„ 2

2 2

= 2 āˆ„x0 āˆ’ yāˆ„ + 2 āˆ„x1 āˆ’ yāˆ„ āˆ’ āˆ„(x0 āˆ’ y) āˆ’ (y āˆ’ x1 )āˆ„  2   2 2 2 1 = 2 āˆ„x0 āˆ’ yāˆ„ + 2 āˆ„x1 āˆ’ yāˆ„ āˆ’ 2  (x0 + x1 ) āˆ’ y  2  2 1   = 2š›æ 2 + 2š›æ 2 āˆ’ 22   2 (x0 + x1 ) āˆ’ y

2

  since 12 (x0 + x1 ) āˆˆ š‘† and therefore  12 (x0 + x1 ) āˆ’ y ā‰„ š›æ so that 2

āˆ„x0 āˆ’ x1 āˆ„ ā‰¤ 2š›æ 2 + 2š›æ 2 āˆ’ 4š›æ 2 = 0 which implies that x1 = x0 . 3. Let x āˆˆ š‘†. Since š‘† is convex, the line segment š›¼x+(1āˆ’š›¼)x0 = x0 +š›¼(xāˆ’x0 ) āˆˆ š‘† and therefore (since x0 is the closest point) ( 2 ) āˆ„x0 āˆ’ yāˆ„2 ā‰¤  x0 + š›¼(x āˆ’ x0 ) āˆ’ y 2

= āˆ„(x0 āˆ’ y) + š›¼(x āˆ’ x0 )āˆ„ ( )š‘‡ ( ) = (x0 āˆ’ y) + š›¼(x āˆ’ x0 ) (x0 āˆ’ y) + š›¼(x āˆ’ x0 ) 2

= āˆ„x0 āˆ’ yāˆ„ + 2š›¼(x0 āˆ’ y)š‘‡ (x āˆ’ x0 ) + š›¼2 āˆ„x āˆ’ x0 āˆ„ 133

2

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Solutions for Foundations of Mathematical Economics which implies that 2

2š›¼(x0 āˆ’ y)š‘‡ (x āˆ’ x0 ) + š›¼2 āˆ„x āˆ’ x0 āˆ„ ā‰„ 0 Dividing through by š›¼ 2

2(x0 āˆ’ y)š‘‡ (x āˆ’ x0 ) + š›¼ āˆ„x āˆ’ x0 āˆ„ ā‰„ 0 which inequality must hold for every 0 < š›¼ < 1. Letting š›¼ ā†’ 0, we must have (x0 āˆ’ y)š‘‡ (x āˆ’ x0 ) ā‰„ 0 as required. 3.73

1. Using the parallelogram law (Exercise 3.66), 2

āˆ„xš‘š āˆ’ xš‘› āˆ„ = āˆ„(xš‘š āˆ’ y) + (y āˆ’ xš‘› )āˆ„

2

2

2

2

= 2 āˆ„xš‘š āˆ’ yāˆ„ + 2 āˆ„y āˆ’ xš‘› āˆ„ āˆ’ 2 āˆ„xš‘š + xš‘› āˆ„

for every š‘š, š‘›. Since š‘† is convex, (xš‘š +xš‘› )/2 āˆˆ š‘† and therefore āˆ„xš‘š + xš‘› āˆ„ ā‰„ 2š‘‘. Therefore 2

2

2

āˆ„xš‘š āˆ’ xš‘› āˆ„ = 2 āˆ„xš‘š āˆ’ yāˆ„ + 2 āˆ„y āˆ’ xš‘› āˆ„ āˆ’ 4š‘‘2 Since āˆ„xš‘š āˆ’ yāˆ„ ā†’ š‘‘ and āˆ„xš‘› āˆ’ yāˆ„ ā†’ š‘‘ as š‘š, š‘› ā†’ āˆž, we conclude that 2 āˆ„xš‘š āˆ’ xš‘› āˆ„ ā†’ 0. That is, (xš‘› ) is a Cauchy sequence. 2. Since š‘† is a closed subspace of complete space, there exists x0 āˆˆ š‘† such that xš‘› ā†’ x0 . By continuity of the norm āˆ„x0 āˆ’ yāˆ„ = lim āˆ„xš‘› āˆ’ yāˆ„ = š‘‘ š‘›ā†’āˆž

Therefore āˆ„x0 āˆ’ yāˆ„ ā‰¤ āˆ„x āˆ’ yāˆ„ for every x āˆˆ š‘† Uniqueness follows in the same manner as the ļ¬nite-dimensional case. 3.74 Deļ¬ne š‘” : š‘‡ ā†’ š‘† by š‘”(y) = { x āˆˆ š‘† : x is closest to y } The function š‘” is well-deļ¬ned since for every y āˆˆ š‘‡ there exists a unique point x āˆˆ š‘† which is closest to y (Exercise 3.72). Clearly, for every x āˆˆ š‘†, x is the closest point to x. Therefore š‘”(x) = x for every x āˆˆ š‘†. To show that š‘” is continuous, choose any y1 and y2 in š‘‡ x1 = š‘”(y1 ) and x2 = š‘”(y2 )

134

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Solutions for Foundations of Mathematical Economics

be the corresponding closest points in š‘†. Then ( )š‘‡ ( ) 2 āˆ„(y1 āˆ’ y2 ) āˆ’ (x1 āˆ’ x2 )āˆ„ = (y1 āˆ’ y2 ) āˆ’ (x1 āˆ’ x2 ) (y1 āˆ’ y1 ) āˆ’ (x1 āˆ’ x2 ) = (y1 āˆ’ y2 )š‘‡ (y1 āˆ’ y2 ) + (x1 āˆ’ x2 )š‘‡ (x1 āˆ’ x2 ) āˆ’ 2(y1 āˆ’ y2 )š‘‡ (x1 āˆ’ x2 ) = āˆ„y1 āˆ’ y2 āˆ„2 + āˆ„x1 āˆ’ x2 āˆ„2 āˆ’ 2(y1 āˆ’ y2 )š‘‡ (x1 āˆ’ x2 ) 2

2

= āˆ„y1 āˆ’ y2 āˆ„ + āˆ„x1 āˆ’ x2 āˆ„ āˆ’ 2(y1 āˆ’ y2 )š‘‡ (x1 āˆ’ x2 ) 2

āˆ’ 2 āˆ„x1 āˆ’ x2 āˆ„ + 2(x1 āˆ’ x2 )š‘‡ (x1 āˆ’ x2 ) 2

2

= āˆ„y1 āˆ’ y2 āˆ„ āˆ’ āˆ„x1 āˆ’ x2 āˆ„ ( )š‘‡ + 2 (x1 āˆ’ x2 ) āˆ’ (y1 āˆ’ y2 ) (x1 āˆ’ x2 ) = āˆ„y1 āˆ’ y2 āˆ„2 āˆ’ āˆ„x1 āˆ’ x2 āˆ„2 + 2(x1 āˆ’ y1 )š‘‡ (x1 āˆ’ x2 ) āˆ’ 2(x2 āˆ’ y2 )š‘‡ (x1 āˆ’ x2 ) 2

2

= āˆ„y1 āˆ’ y2 āˆ„ āˆ’ āˆ„x1 āˆ’ x2 āˆ„

āˆ’ 2(x1 āˆ’ y1 )š‘‡ (x2 āˆ’ x1 ) āˆ’ 2(x2 āˆ’ y2 )š‘‡ (x1 āˆ’ x2 ) so that 2

2

āˆ„y1 āˆ’ y2 āˆ„ āˆ’ āˆ„x1 āˆ’ x2 āˆ„ = āˆ„(y1 āˆ’ y2 ) āˆ’ (x1 āˆ’ x2 )āˆ„

2

+ 2(x1 āˆ’ y1 )š‘‡ (x2 āˆ’ x1 ) + 2(x2 āˆ’ y2 )š‘‡ (x1 āˆ’ š‘„2 ) Using Exercise 3.72 (x1 āˆ’ y1 )š‘‡ (x2 āˆ’ x1 ) ā‰„ 0 and (x2 āˆ’ y2 )š‘‡ (x1 āˆ’ x2 ) ā‰„ 0 which implies that the left-hand side āˆ„y1 āˆ’ y2 āˆ„2 āˆ’ āˆ„x1 āˆ’ x2 āˆ„2 ā‰„ 0 or āˆ„x1 āˆ’ x2 āˆ„ = āˆ„š‘”(y1 ) āˆ’ š‘”(y2 )āˆ„ ā‰¤ āˆ„y1 āˆ’ y2 āˆ„ š‘” is Lipschitz continuous. 3.75 Let š‘† = kernel š‘“ . Then š‘† is a closed subspace of š‘‹. If š‘† = š‘‹, then š‘“ is the zero functional and y = 0 is the required element. Otherwise chose any y āˆˆ / š‘† and let x0 be the closest point in š‘† (Exercise 3.72). Deļ¬ne z = x0 āˆ’ y. Then z āˆ•= 0 and zš‘‡ x ā‰„ 0 for every x āˆˆ š‘† Since š‘† is subspace, this implies that zš‘‡ x = 0 for every x āˆˆ š‘† that is z is orthogonal to š‘†. Let š‘†Ė† be the subset of š‘‹ deļ¬ned by š‘†Ė† = { š‘“ (x)z āˆ’ š‘“ (z)x : x āˆˆ š‘‹ } For every x āˆˆ š‘†Ė†

( ) š‘“ (x) = š‘“ š‘“ (x)z āˆ’ š‘“ (z)x = š‘“ (x)š‘“ (z) āˆ’ š‘“ (z)š‘“ (x) = 0 135

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Therefore š‘†Ė† āŠ† š‘†. For every x āˆˆ š‘‹

( )š‘‡ š‘“ (x)z āˆ’ š‘“ (z)x z = š‘“ (x)zš‘‡ z āˆ’ š‘“ (z)xš‘‡ z = 0 since z āˆˆ š‘† āŠ„ . Therefore š‘“ (x) =

š‘“ (z) āˆ„zāˆ„

(

š‘‡ 2x z

= xš‘‡

zš‘“ (z) āˆ„zāˆ„

2

) = xš‘‡ y

where y=

zš‘“ (z) āˆ„zāˆ„

2

3.76 š‘‹ āˆ— is always complete (Proposition 3.3). To show that it is a Hilbert space, we have to that it has an inner product. For this purpose, it will be clearer if we use an alternative notation < x, y > to denote the inner product of x and y. Assume š‘‹ is a Hilbert space. By the Riesz representation theorem (Exercise 3.75), for every š‘“ āˆˆ š‘‹ āˆ— there exists yš‘“ āˆˆ š‘‹ such that š‘“ (x) =< x, yš‘“ > for every x āˆˆ š‘‹ Furthermore, if yš‘“ represents š‘“ and yš‘” represents š‘” āˆˆ š‘‹ āˆ— , then yš‘“ + yš‘” represents š‘“ + š‘” and š›¼yš‘“ represents š›¼š‘“ since (š‘“ + š‘”)(x) = š‘“ (x) + š‘”(x) =< x, yš‘“ > + < x, yš‘” >=< x, yš‘“ + yš‘” > (š›¼š‘“ )(x) = š›¼š‘“ (x) = š›¼ < x, yš‘“ >=< x, š›¼yš‘“ > Deļ¬ne an inner product on š‘‹ āˆ— by < š‘“, š‘” >=< yš‘” , yš‘“ > We show that it satisļ¬es the properties of an inner product, namely symmetry < š‘“, š‘” >=< yš‘” , yš‘“ >=< yš‘“ , yš‘” >=< š‘”, š‘“ > additivity < š‘“1 + š‘“2 , š‘” >=< yš‘” , yš‘“1 +š‘“2 >=< yš‘” , yš‘“1 + yš‘“2 >=< š‘“1 , š‘” > + < š‘“2 , š‘” > homogeneity < š›¼š‘“, š‘” >=< yš‘” , š›¼yš‘“ >= š›¼ < yš‘” , yš‘“ >= š›¼ < š‘“, š‘” > positive deļ¬niteness < š‘“, š‘” >=< yš‘” , yš‘“ >ā‰„ 0 and < š‘“, š‘” >=< yš‘” , yš‘“ >= 0 if and only if š‘“ = š‘”. Therefore, š‘‹ āˆ— is a complete inner product space, that is a Hilbert space. 3.77 Let š‘‹ be a Hilbert space. Applying the previous exercise a second time, š‘‹ āˆ—āˆ— is also a Hilbert space. Let š¹ be an arbitrary functional in š‘‹ āˆ—āˆ— . By the Riesz representation theorem, there exists š‘” āˆˆ š‘‹ āˆ— such that š¹ (š‘“ ) =< š‘“, š‘” > for every š‘“ āˆˆ š‘‹ āˆ— Again by the Riesz representation theorem, there exists xš‘“ (representing š‘“ ) and xš¹ (representing š‘”) in š‘‹ such that š¹ (š‘“ ) =< š‘“, š‘” >=< xš¹ , xš‘“ > and š‘“ (x) =< x, xš‘“ > 136

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Solutions for Foundations of Mathematical Economics In particular, š‘“ (xš¹ ) =< xš¹ , xš‘“ >= š¹ (š‘“ )

That is, for every š¹ āˆˆ š‘‹ āˆ—āˆ— , there exists an element xš¹ āˆˆ š‘‹ such that š¹ (š‘“ ) = š‘“ (xš¹ ) š‘‹ is reļ¬‚exive. 3.78

1. Adapt Exercise 3.64.

2. By Exercise 3.75, there exists unique xāˆ— āˆˆ š‘‹ such that š‘“y (x) = xš‘‡ xāˆ— 3. Substituting š‘“ (x)š‘‡ y = š‘“y (x) = xš‘‡ xāˆ— = š‘„š‘‡ š‘“ āˆ— (y) 4. For every y1 , y2 āˆˆ š‘Œ ( ( ) ) xš‘‡ š‘“ āˆ— (y1 + y2 ) = š‘“ (x)š‘‡ y1 + y2 = š‘“ (x)š‘‡ y1 + š‘“ (x)š‘‡ y1 = xš‘‡ š‘“ āˆ— (y1 ) + xš‘‡ š‘“ āˆ— (y1 ) and for every y āˆˆ š‘Œ xš‘‡ š‘“ āˆ— (š›¼y) = š‘“ (x)š‘‡ š›¼y = š›¼š‘“ (x)š‘‡ y = š›¼xš‘‡ š‘“ āˆ— (y) = xš‘‡ š›¼š‘“ āˆ— (y) 3.79 The zero element 0š‘‹ is a ļ¬xed point of every linear operator (Exercise 3.13). 3.80 š“š“āˆ’1 = š¼ so that det(š“) det(š“āˆ’1 ) = det(š¼) = 1 3.81 Expanding along the š‘–th row using (3.8) det(š¶) =

š‘› āˆ‘

(āˆ’1)š‘–+š‘— (š›¼š‘Žš‘–š‘— + š›½š‘š‘–š‘— ) det(š¶š‘–š‘— )

š‘—=1 š‘› āˆ‘

=š›¼

(āˆ’1)š‘–+š‘— š‘Žš‘–š‘— det(š¶š‘–š‘— ) + š›½

š‘—=1

š‘› āˆ‘

(āˆ’1)š‘–+š‘— š‘š‘–š‘— det(š¶š‘–š‘— )

š‘—=1

But the matrices diļ¬€er only in the š‘–th row and therefore š“š‘–š‘— = šµš‘–š‘— = š¶š‘–š‘— ,

š‘— = 1, 2, . . . š‘›

so that det(š¶) = š›¼

š‘› āˆ‘

(āˆ’1)š‘–+š‘— š‘Žš‘–š‘— det(š“š‘–š‘— ) + š›½

š‘—=1

š‘› āˆ‘

(āˆ’1)š‘–+š‘— š‘š‘–š‘— det(šµš‘–š‘— )

š‘—=1

= š›¼ det(š“) + š›½ det(šµ) 3.82 Suppose that x1 and x2 are eigenvectors corresponding to the eigenvalue šœ†. By linearity š‘“ (x1 + x2 ) = š‘“ (x1 ) + š‘“ (x2 ) = šœ†x1 + šœ†x2 = šœ†(x1 + x2 ) and š‘“ (š›¼x1 ) = š›¼š‘“ (x1 ) = š›¼šœ†x Therefore x1 + x2 and š›¼x1 are also eigenvectors. 137

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Solutions for Foundations of Mathematical Economics

3.83 Suppose š‘“ is singular. Then there exists x āˆ•= 0 such that š‘“ (x) = 0. Therefore x is an eigenvector with eigenvalue 0. Conversely, if 0 is an eigenvalue š‘“ (x) = 0x = 0 for any x āˆ•= 0. Therefore š‘“ is singular. 3.84 Since š‘“ (x) = šœ†x š‘“ (x)š‘‡ x = šœ†xš‘‡ x = šœ†xš‘‡ x 3.85 By Exercise 3.69 š‘Žš‘–š‘— = xš‘‡š‘– š‘“ (xš‘— ) š‘Žš‘—š‘– = xš‘‡š‘— š‘“ (xš‘– ) = š‘“ (xš‘– )š‘‡ xš‘— and therefore š‘Žš‘–š‘— = š‘Žš‘—š‘– ā‡ā‡’ xš‘‡š‘– š‘“ (xš‘— ) = š‘“ (xš‘– )š‘‡ xš‘— 3.86 By bilinearity xš‘‡1 š‘“ (x2 ) = xš‘‡1 šœ†2 x2 = šœ†2 xš‘‡1 x2 š‘“ (x1 )š‘‡ x2 = šœ†1 xš‘‡1 x2 = šœ†1 xš‘‡1 x2 Since š‘“ is symmetric, this implies (šœ†1 āˆ’ šœ†2 )xš‘‡1 x2 = 0 and šœ†1 āˆ•= šœ†2 implies xš‘‡1 x2 = 0. 3.87

1. Since š‘† compact and š‘“ is continuous (Exercises 3.31, 3.62), the maximum is attained at some x0 āˆˆ š‘† (Theorem 2.2), that is šœ† = š‘“ (x0 )š‘‡ x0 ā‰„ š‘“ (x)š‘‡ x for every x āˆˆ š‘† Hence ( )š‘‡ š‘”(x, y) = šœ†x āˆ’ š‘“ (x) y is well-deļ¬ned.

2. For any x āˆˆ š‘‹ ( )š‘‡ š‘”(x, x) = šœ†x āˆ’ š‘“ (x) x = šœ†xš‘‡ x āˆ’ š‘“ (x)š‘‡ x = šœ† āˆ„xāˆ„2 āˆ’ š‘“ (x)š‘‡ x ( 2

2

= šœ† āˆ„xāˆ„ āˆ’ āˆ„xāˆ„ š‘“

x

)š‘‡ ( 2

āˆ„xāˆ„ ) 2( š‘‡ = āˆ„xāˆ„ šœ† āˆ’ š‘“ (z) z ā‰„ 0

since z = x/ āˆ„xāˆ„ āˆˆ š‘†. 138

x āˆ„xāˆ„

) 2

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

3. Since š‘“ is symmetric ( )š‘‡ š‘”(y, x) = šœ†y āˆ’ š‘“ (y) x = šœ†yš‘‡ x āˆ’ š‘“ (y)š‘‡ x = šœ†xš‘‡ y āˆ’ š‘“ (x)š‘‡ š‘¦ ( )š‘‡ = šœ†x āˆ’ š‘“ (x) y = š‘”(x, y) 4. š‘” satisļ¬es the conditions of Exercise 3.59 and therefore (š‘”(x, y))2 ā‰¤ š‘”(x, x)š‘”(y, y) for every x, y āˆˆ š‘‹

(3.41)

By deļ¬nition š‘”(x0 , x0 ) = 0 and (3.41) implies that š‘”(x0 , y) = 0 for every y āˆˆ š‘‹ That is ( )š‘‡ š‘”(x0 , y) = šœ†x0 āˆ’ š‘“ (x0 ) y = 0 for every š‘¦ āˆˆ š‘‹ and therefore šœ†x0 āˆ’ š‘“ (x0 ) = 0 or š‘“ (x0 ) = šœ†x0 In other words, x0 is an eigenvector. By construction, āˆ„x0 āˆ„ = 1. 3.88

1. Suppose x2 , x3 āˆˆ š‘†. Then ( )š‘‡ š›¼x2 + š›½x3 x1 = š›¼xš‘‡2 x1 + š›½xš‘‡3 x1 = 0 so that š›¼x2 + š›½x3 āˆˆ š‘†. š‘† is a subspace. Let {x1 , x2 , . . . , xš‘› } be a basis for š‘‹ (Exercise 1.142). For x āˆˆ š‘‹, there exists (Exercise 1.137) unique š›¼1 , š›¼2 , . . . , š›¼š‘› such that x = š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› If x āˆˆ š‘† xš‘‡ x1 = š›¼1 xš‘‡1 x1 = 0 which implies that š›¼1 = 0. dim š‘† = š‘› āˆ’ 1.

Therefore, x2 , x3 , . . . , xš‘› span š‘† and therefore

2. For every x āˆˆ š‘†, š‘“ (x)š‘‡ x0 = xš‘‡ š‘“ (x0 ) = xš‘‡ šœ†x0 = šœ†xš‘‡ x0 = 0 since š‘“ is symmetric. Therefore š‘“ (x) āˆˆ {x0 }āŠ„ = š‘†. 3.89 Let š‘“ be a symmetric operator. By the spectral theorem (Proposition 3.6), there exists a diagonal matrix š“ which represents š‘“ . The elements of š“ are the eigenvalues of š‘“ . By Proposition 3.5, the determinant of š“ is the product of these diagonal elements. 139

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Solutions for Foundations of Mathematical Economics 3.90 By linearity š‘“ (x) =

āˆ‘

š‘„š‘— š‘“ (xš‘— )

š‘—

š‘„ deļ¬nes a quadratic form since āŽž ( )š‘‡ āŽ› āˆ‘ āˆ‘āˆ‘ āˆ‘āˆ‘ āˆ‘ āŽ š‘„(x) = xš‘‡ š‘“ (x) = š‘„š‘– xš‘– š‘„š‘— š‘“ (xš‘— )āŽ  = š‘„š‘– š‘„š‘— xš‘‡š‘– š‘“ (xš‘— ) = š‘Žš‘–š‘— š‘„š‘– š‘„š‘— š‘–

š‘—

š‘–

š‘—

š‘–

š‘—

by Exercise 3.69. 3.91 Let š‘“ be the symmetric linear operator deļ¬ning š‘„ š‘„(x) = xš‘‡ š‘“ (x) By the spectral theorem (Proposition 3.6), there exists an orthonormal basis x1 , x2 , . . . , xš‘› comprising the eigenvectors of š‘“ . Let šœ†1 , šœ†2 , . . . , šœ†š‘› be the corresponding eigenvalues, that is š‘“ (xš‘– ) = šœ†š‘– xš‘–

š‘– = 1, 2 . . . , š‘›

Then for x = š‘„1 x1 + š‘„2 x2 + ā‹… ā‹… ā‹… + š‘„š‘› xš‘› š‘„(x) = xš‘‡ š‘“ (x) = (š‘„1 x1 + š‘„2 x2 + ā‹… ā‹… ā‹… + š‘„š‘› xš‘› )š‘‡ š‘“ (š‘„1 x1 + š‘„2 x2 + ā‹… ā‹… ā‹… + š‘„š‘› xš‘› ) = (š‘„1 x1 + š‘„2 x2 + ā‹… ā‹… ā‹… + š‘„š‘› xš‘› )š‘‡ (š‘„1 š‘“ (x1 ) + š‘„2 š‘“ (x2 ) + ā‹… ā‹… ā‹… + š‘„š‘› š‘“ (xš‘› )) = (š‘„1 x1 + š‘„2 x2 + ā‹… ā‹… ā‹… + š‘„š‘› xš‘› )š‘‡ (š‘„1 šœ†1 x1 + š‘„2 šœ†2 x2 + ā‹… ā‹… ā‹… + š‘„š‘› šœ†š‘› xš‘› ) = š‘„1 šœ†1 š‘„1 + š‘„2 šœ†2 š‘„2 + ā‹… ā‹… ā‹… + š‘„š‘› šœ†š‘› š‘„š‘› = šœ†1 š‘„21 + šœ†2 š‘„22 + ā‹… ā‹… ā‹… + šœ†š‘› š‘„2š‘› 3.92

1. Assuming that š‘Ž11 āˆ•= 0, the quadratic form can be rewritten as follows š‘„(š‘„1 , š‘„2 ) = š‘Ž11 š‘„21 + 2š‘Ž12 š‘„1 š‘„2 + š‘Ž22 š‘„22 = š‘Ž11 š‘„21 + 2š‘Ž12 š‘„1 š‘„2 +

š‘Ž212 2 š‘Ž212 2 š‘„ āˆ’ š‘„ + š‘Ž22 š‘„22 š‘Ž11 2 š‘Ž11 2

(

( )2 ) ( ) š‘Ž š‘Ž š‘Ž212 12 12 2 = š‘Ž11 š‘„1 + 2 š‘„1 š‘„2 + š‘„2 + š‘Ž22 āˆ’ š‘„22 š‘Ž11 š‘Ž11 š‘Ž11 ( )2 ( ) š‘Ž11 š‘Ž22 āˆ’ š‘Ž212 š‘Ž12 = š‘Ž11 š‘„1 + š‘„2 + š‘„22 š‘Ž11 š‘Ž11 2. We observe that š‘ž must be positive for every š‘„1 and š‘„2 provided š‘Ž11 > 0 and š‘Ž11 š‘Ž22 āˆ’ š‘Ž212 > 0. Similarly š‘ž must be negative for every š‘„1 and š‘„2 if š‘Ž11 > 0 and š‘Ž11 š‘Ž22 āˆ’ š‘Ž212 > 0. Otherwise, we can choose values for š‘„1 and š‘„2 which make š‘ž both positive and negative. Note that the condition š‘Ž11 š‘Ž22 > š‘Ž212 > 0 implies that š‘Ž11 and š‘Ž12 must have the same sign. 3. If š‘Ž11 = š‘Ž22 = 0, then š‘ž is indeļ¬nite. Otherwise, if š‘Ž11 = 0 but š‘Ž22 āˆ•= 0, then the š‘ž can we can ā€œcomplete the squareā€ using š‘Ž22 and deduce { } { } nonnegative š‘Ž11 , š‘Ž22 ā‰„ 0 š‘ž is deļ¬nite if and only if and š‘Ž11 š‘Ž22 ā‰„ š‘Ž212 nonpositive š‘Ž11 , š‘Ž22 ā‰¤ 0 140

Solutions for Foundations of Mathematical Economics

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3.93 Let š‘„ : š‘‹ ā†’ ā„œ be a quadratic form on š‘‹. Then there exists a linear operator š‘“ such that š‘„(x) = xš‘‡ š‘“ (x) and (Exercise 3.13) š‘„(0) = 0š‘‡ š‘“ (0) = 0 3.94 Suppose to the contrary that the positive (negative) deļ¬nite matrix š“ is singular. Then there exists x āˆ•= 0 such that š“x = 0 and therefore xā€² š“x = 0 contradicting the deļ¬niteness of š“. 3.95 Let e1 , e2 , . . . , eš‘› be the standard basis for ā„œš‘› (Example 1.79). Then for every š‘– eā€²š‘– š“eš‘– = š‘Žš‘–š‘– > 0 3.96 Let š‘„ be the quadratic form deļ¬ned by š“. By Exercise 3.91, there exists an orthonormal basis such that š‘„(x) = šœ†1 x21 + šœ†2 x22 + ā‹… ā‹… ā‹… + šœ†š‘› x2š‘› where šœ†1 , šœ†2 , . . . , šœ†š‘› are the eigenvalues of š“. This implies āŽ§ āŽ§ āŽ« āŽ« šœ†š‘– > 0ļ£“ š‘„(x) > 0ļ£“ ļ£“ ļ£“ ļ£“ ļ£“ ļ£“ ļ£“ āŽØ āŽØ āŽ¬ āŽ¬ š‘„(x) ā‰„ 0 šœ†š‘– ā‰„ 0 ā‡ā‡’ š‘– = 1, 2, . . . , š‘› ļ£“šœ†š‘– < 0ļ£“ ļ£“š‘„(x) < 0ļ£“ ļ£“ ļ£“ ļ£“ ļ£“ āŽ© āŽ­ āŽ© āŽ­ š‘„(x) ā‰¤ 0 šœ†š‘– ā‰¤ 0 3.97 Let šœ†1 , šœ†2 , . . . , šœ†š‘› be the eigenvalues of š“. By Exercise 3.89 det(š“) = šœ†1 šœ†2 . . . šœ†š‘› By Exercise 3.96, šœ†š‘– ā‰„ 0 for every š‘– and therefore det(š“) ā‰„ 0. We conclude that det(š“) > 0 ā‡ā‡’ šœ†š‘– > 0 for every š‘– ā‡ā‡’ š“ is positive deļ¬nite by Exercise 3.96. 3.98

1. š“0 = 0. Therefore, 0 is always a solution.

2. Assume x1 and x2 are solutions, that is š“x1 = 0 and š“x2 = 0 Then š“(x1 + x2 ) = š“x1 + š“x2 = 0 x1 + x2 is also a solution. 3. Let š‘“ be the linear function deļ¬ned by š‘“ (x) = š“x The system of equations š“x = 0 has a nontrivial solution if and only if kernel š‘“ āˆ•= {0} ā‡ā‡’ nullity š‘“ > 0 By the rank theorem (Exercise 3.24) rankš‘“ + nullityš‘“ = dim š‘‹ so that nullity š‘“ > 0 ā‡ā‡’ rankš‘“ < dim š‘‹ = š‘› 141

Solutions for Foundations of Mathematical Economics 3.99

c 2001 Michael Carter āƒ All rights reserved

1. Assume x1 and x2 are solutions of (3.16). That is š“x1 = c and š“x2 = c Subtracting š“x1 āˆ’ š“x2 = š“(x1 āˆ’ x2 ) = 0

2. Assume xš‘ solves (3.16) while x is any solution to (3.17). That is š“xš‘ = c and š“x = 0 Adding š“xš‘ + š“x = š“(xš‘ + x) = c We conclude that xš‘ + x solves (3.16) for every x āˆˆ š¾. 3. If 0 is the only solution of (3.17), š¾ = {0}. Assume x1 and x2 are solutions of (3.16). Then x1 āˆ’ x2 āˆˆ š¾ = {0} which implies x1 = x2 . 3.100 Let š‘† = { x : š“x = š‘ }. For every x, y āˆˆ š‘† and š›¼ āˆˆ ā„œ š“š›¼x + (1 āˆ’ š›¼)y = š›¼š“x + (1 āˆ’ š›¼)š“y = š›¼c + (1 āˆ’ š›¼)c = š‘ Therefore, z = š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘†. š‘† is aļ¬ƒne. 3.101 Let š‘† āˆ•= āˆ… be an aļ¬ƒne set ā„œš‘› . Then there exists a unique subspace š‘‰ such that š‘† = x0 + š‘‰ for some x0 āˆˆ š‘† (Exercise 1.150). The orthogonal complement of š‘‰ is š‘‰ āŠ„ = { a āˆˆ š‘‹ : ax = 0 for every x āˆˆ š‘‰ } Let (a1 , a2 , . . . , aš‘š ) be a basis for š‘‰ āŠ„ . Then š‘‰ = (š‘‰ āŠ„ )āŠ„ = {x : aš‘– x = 0,

š‘– = 1, 2, . . . š‘š}

Let š“ be the š‘šĆ—š‘› matrix whose rows are a1 , a2 , . . . , aš‘š . Then š‘‰ is the set of solutions to the homogeneous linear system š“x = 0, that is š‘‰ = { š‘„ : š“x = 0 } Therefore š‘† = x0 + š‘‰ = x0 + { x : š“x = 0 } = { x : š“(x āˆ’ x0 ) = 0 } = { x : š“x = c } where c = š“x0 . 3.102 Consider corresponding homogeneous system š‘„1 + 3š‘„2 = 0 š‘„1 āˆ’ š‘„2 = 0 142

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Multiplying the second equation by 3 š‘„1 + 6š‘„2 = 0 3š‘„1 āˆ’ 3š‘„2 = 0 and adding yields 4š‘„1 = 0

for which the only solution is š‘„1 = 0. Substituting in the ļ¬rst equation implies š‘„2 = 0. The kernel of š‘“ = š“x is {0}. Therefore dim š‘“ (ā„œ2 ) = 2, and the system š“x = š‘ has a unique solution for every š‘1 , š‘2 . 3.103 We can write the system š“x = c in the form āŽ› āŽž āŽ› āŽž āŽ› āŽž āŽ› āŽž š‘Ž11 š‘Ž1š‘— š‘Ž1š‘› š‘1 āŽœ .. āŽŸ āŽœ .. āŽŸ āŽœ .. āŽŸ āŽœ .. āŽŸ š‘„1 āŽ . āŽ  + ā‹… ā‹… ā‹… + š‘„š‘— āŽ . āŽ  + ā‹… ā‹… ā‹… + š‘„š‘› āŽ . āŽ  = āŽ . āŽ  š‘Žš‘›1

š‘Žš‘›š‘—

š‘Žš‘›š‘›

š‘š‘›

Subtracting c from the š‘—th column gives āŽ› āŽž āŽž āŽ› āŽž āŽ› š‘Ž11 š‘Ž1š‘› š‘„š‘— š‘Ž1š‘— āˆ’ š‘1 āŽœ āŽŸ āŽŸ āŽœ . āŽŸ āŽœ .. š‘„1 āŽ ... āŽ  + ā‹… ā‹… ā‹… + āŽ āŽ  + ā‹… ā‹… ā‹… + š‘„š‘› āŽ .. āŽ  = 0 . š‘„š‘— š‘Žš‘›š‘— āˆ’ š‘š‘›

š‘Žš‘›1

so that the columns of the matrix āŽ› š‘Ž11 . . . āŽœ .. š¶=āŽ . ...

š‘Žš‘›1

š‘Žš‘›š‘›

(š‘„š‘— š‘Ž1š‘— āˆ’ š‘1 ) .. .

(š‘„š‘— š‘Žš‘›š‘— āˆ’ š‘š‘› )

... ...

āŽž š‘Ž1š‘› .. āŽŸ . āŽ 

š‘Žš‘›š‘›

are linearly dependent (Exercise 1.133). Therefore det(š¶) = 0. Let šµš‘— denote the matrix obtained from š“ by replacing the š‘—th column with c. Then š“, šµš‘— and š¶ diļ¬€er only in the š‘—th column, with the š‘—th column of š¶ being a linear combination of the š‘—th columns of š“ and šµš‘— . āŽ› āŽž āŽ› āŽž āŽ› āŽž š‘1š‘— š‘Ž1š‘— š‘1š‘— āŽœ .. āŽŸ āŽœ .. āŽŸ āŽœ .. āŽŸ = š‘„ āˆ’ āŽ . āŽ  āŽ . āŽ  š‘—āŽ . āŽ  š‘š‘›š‘—

š‘Žš‘›š‘—

š‘š‘›š‘—

By Exercise 3.81 det(š¶) = xš‘— det(š“) āˆ’ det(šµš‘— ) = 0 and therefore š‘„š‘— =

det(šµš‘— ) det(š“)

as required. 3.104 Let ( š‘Ž š‘

)āˆ’1 ( š‘ š“ = š‘‘ š¶ 143

šµ š·

)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics The inverse satisļ¬es the equation ( )( š‘Ž š‘ š“ š‘ š‘‘ š¶ In particular, this means that š“ and ( š‘Ž š‘ By Cramerā€™s rule (Exercise 3.103)

where Ī” = š‘Žš‘‘ āˆ’ š‘š‘. Similarly

šµ š·

)

( 1 = 0

) 0 1

š¶ satisfy the equation )( ) ( ) š‘ š“ 1 = š‘‘ š¶ 0

  1 š‘    0 š‘‘ š‘‘ = š“ =   Ī” š‘Ž š‘   š‘ š‘‘

  š‘Ž 1    š‘ 0 āˆ’š‘  = š¶ =   Ī” š‘Ž š‘   š‘ š‘‘

šµ and š· are determined analogously. 3.105 A portfolio is duplicable if and only if there is a diļ¬€erent portfolio y āˆ•= x such that š‘…x = š‘…y or š‘…(x āˆ’ y) = 0 There exists a duplicable portfolio if and only if this homogeneous system has a nontrivial solution, that is if rank š‘… < š“. 3.106 State š‘ ĀÆ is insurable if there is a solution to the linear system š‘…x = eš‘ ĀÆ

(3.42)

where eš‘ ĀÆ is the š‘ ĀÆ-th unit vector (the š‘ ĀÆ Arrow-Debreu security). (3.42) has a solution for every state š‘  if and only if š‘“ (ā„œš“ ) = ā„œš‘† , that is rank š‘… = š‘†. 3.107 Figure 3.1. 3.108 Let š‘† be an aļ¬ƒne subset of ā„œš‘› . Then there exists (Exercise 3.101) a system of linear equations š“x = c such that š‘† = { x : š“x = c } Let aš‘– denote the š‘–-th row of š“. Then š‘† = { š‘„ : aš‘– x = š‘š‘– , š‘– = 1, 2, . . . , š‘› } =

š‘› āˆ©

{ x : aš‘– x = š‘š‘– }

š‘–=1

where each { x : aš‘– x = š‘š‘– } is a hyperplane in ā„œš‘› (Example 3.21). 144

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Figure 3.1: The solutions of three equations in two unknowns 3.109 Let š‘† = { x : š“x ā‰¤ c }. For every x, y āˆˆ š‘† and 0 ā‰¤ š›¼ ā‰¤ 1 š“x ā‰¤ c š“y ā‰¤ c and therefore š“š›¼x + (1 āˆ’ š›¼)y = š›¼š“x + (1 āˆ’ š›¼)š“y ā‰¤ š›¼c + (1 āˆ’ š›¼)c = š‘ Therefore, z = š›¼x + (1 āˆ’ š›¼)y āˆˆ š‘†. š‘† is a convex set. 3.110 We have already seen that š‘† = { x : š“x ā‰¤ 0 } is convex. To show that it is a cone, let x āˆˆ š‘†. Then š“x ā‰¤ 0 š“š›¼x ā‰¤ 0 so that š›¼x āˆˆ š‘†. š‘† is a convex cone. 3.111

1. Each column š“š‘— is a vector in ā„œš‘š . If the set {š“1 , š“2 , . . . , š“š‘˜ } is linearly independent, it has at most š‘š elements, that is š‘˜ ā‰¤ š‘š and x is a basic feasible solution.

2. (a) Assume {š“1 , š“2 , . . . , š“š‘˜ } are linearly dependent. Then (Exercise 1.133) there exist numbers š‘¦1 , š‘¦2 , . . . , š‘¦š‘˜ , not all zero, such that š‘¦1 š“1 + š‘¦2 š“2 + ā‹… ā‹… ā‹… + š‘¦š‘˜ š“š‘˜ = 0 y = (š‘¦1 , š‘¦2 , . . . , š‘¦š‘˜ ) is a nontrivial solution to the homogeneous system. (b) For every š‘” āˆˆ ā„œ, āˆ’š‘”y āˆˆ kernel š‘“ = š“x and xā€² = x āˆ’ š‘”y is a solution of the corresponding nonhomogeneous system š“x = c. To see this directly, subtract š“š‘”y = 0 145

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from š“x = c to give š“xā€² = š“(x āˆ’ š‘”y) = c (c) Note that x > 0 and therefore š‘”Ė† > 0 which implies that š‘„ Ė†š‘— > 0 for every š‘¦š‘— ā‰¤ 0. For every š‘¦š‘— > 0, š‘„š‘— /š‘¦š‘— ā‰„ š‘”Ė†, which implies that š‘„š‘— ā‰„ š‘”Ė†š‘¦š‘— , so that š‘„Ė†š‘— ā‰„ š‘„š‘— āˆ’ š‘”Ė†š‘¦š‘— ā‰„ 0 Ė† is a feasible solution. Therefore, x (d) There exists some coordinate ā„Ž such that š‘”Ė† = š‘„ā„Ž /š‘¦ā„Ž so that š‘„ Ė†ā„Ž = š‘„ā„Ž āˆ’ š‘”Ė†š‘¦ā„Ž = 0 so that š‘˜ āˆ‘

c=

š‘„Ė†š‘— š“š½

š‘— =1

š‘—āˆ•=ā„Ž

š‘„ Ė† is a feasible solution with one less positive component. 3. Starting with any nonbasic feasible solution, this elimination technique can be repeated until the remaining vectors are linearly independent and a basic feasible solution is obtained. 3.112

1. Exercise 1.173.

2. For each š‘–, there exists š‘™š‘– elements xš‘–š‘— and coeļ¬ƒcients š‘Žš‘–š‘— > 0 such that xš‘– =

š‘™š‘– āˆ‘

š‘Žš‘–š‘— xš‘–š‘—

š‘–=1

and

āˆ‘ š‘™š‘–

š‘—=1

š‘Žš‘–š‘— = 1. Hence x=

š‘› āˆ‘

xš‘– =

š‘–=1

š‘› āˆ‘ š‘™š‘– āˆ‘

š‘Žš‘–š‘— xš‘–š‘—

š‘–=1 š‘—=1

3. Direct computation. 4. Regarding the š‘Žš‘–š‘— as ā€œvariablesā€ and the points š‘§š‘–š‘— as coeļ¬ƒcents, z=

š‘™š‘– š‘› āˆ‘ āˆ‘

š‘Žš‘–š‘— zš‘–š‘—

š‘–=1 š‘—=1

is a linear equation system in which variables are restricted to be nonnegative. By the fundamental theorem of linear programming (Exercise 3.111), there exists

146

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Solutions for Foundations of Mathematical Economics

a basic feasible solution. That is, there exists coeļ¬ƒcients š‘š‘–š‘— ā‰„ 0 and š‘š‘–š‘— > 0 for at most (š‘š + š‘›) components such that z=

š‘™š‘– š‘› āˆ‘ āˆ‘

š‘š‘–š‘— zš‘–š‘—

(3.43)

š‘–=1 š‘—=1

Decomposing, (3.43) implies x=

š‘™š‘– š‘› āˆ‘ āˆ‘

š‘š‘–š‘— xš‘–š‘—

š‘–=1 š‘—=1

and š‘™š‘– āˆ‘

š‘š‘–š‘— = 1

for every š‘–

š‘—=1

5. (3.43) implies that at least one š‘š‘–š‘— > 0 for every š‘–. This accounts for at least š‘› of the positive š‘š‘–š‘— . Since there are at most (š‘š + š‘›) coeļ¬ƒcients š‘š‘–š‘— which are strictly positive, there are at most š‘š indices š‘– which have more than one positive coeļ¬ƒcient š‘š‘–š‘— . For the remaining š‘š āˆ’ š‘› indices, xš‘– = xš‘–š‘— for some š‘—; that is xš‘– āˆˆ š‘†š‘– . 3.113

1. Since š“ is productive, there exists x ā‰„ 0 such that š“x > 0. Consider any z for which š“z ā‰„ 0. For every š›¼ > 0 š“(x + š›¼z) = š“x + š›¼š“z > 0

(3.44)

Suppose to the contrary that z āˆ•ā‰„ 0. That is, there exists some component š‘§š‘– < 0. Let š‘§š‘– š›¼ = max{āˆ’ } š‘„š‘– Without loss of generality, š‘§1 attains this maximum, that is assume š›¼ = š‘§1 /š‘„1 . Then š‘„1 + š›¼š‘§1 = 0 and š‘„š‘– + š›¼š‘§š‘– ā‰„ 0 for every š‘–. Now consider the matrix šµ = š¼ āˆ’ š“. By the assumptions of the Leontief model (Example 3.35), the matrix š“ has 1 along the diagonal and negative oļ¬€-diagonal elements. That is š‘Žš‘–š‘– = 1

š‘– = 1, 2, . . . , š‘›

š‘Žš‘–š‘— ā‰¤ 0

š‘–, š‘— = 1, 2, . . . , š‘›,

Therefore šµ =š¼ āˆ’š“ā‰„0 147

š‘— āˆ•= š‘—

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That is, every element of šµ is nonnegative. Consequently since x + š›¼z ā‰„ 0 šµ(x + š›¼z) ā‰„ 0

(3.45)

On the other hand, substituting š“ = š¼ āˆ’ šµ in (3.45) (š¼ āˆ’ šµ)(x + š›¼z) > 0 x + š›¼z > šµ(x + š›¼z) which implies that the ļ¬rst component of šµ(x + š›¼z) is negative, contradicting (3.45). This contradiction establishes that z ā‰„ 0. Suppose š“x = 0. A fortiori š“x ā‰„ 0. By the previous part this implies x ā‰„ 0. On the other hand, it also implies that āˆ’š“x = š“(āˆ’x) = 0 so that āˆ’x ā‰„ 0. We conclude that x = 0 is the only solution to š“x = 0. š“ is nonsingular. Since š“ is nonsingular, the system š“x = y has a unique solution x for any y ā‰„ 0. By the ļ¬rst part, x ā‰„ 0. 3.114 Suppose š“ is productive. By the previous exercise, š“ is nonsingular with inverse š“āˆ’1 . Let eš‘– be the š‘–th unit vector. Since eš‘– ā‰„ 0, there exists xš‘– ā‰„ 0 such that š“xš‘– = eš‘– Multiplying by š“āˆ’1 xš‘– = š“āˆ’1 š“xš‘– = š“āˆ’1 eš‘– = š“āˆ’1 š‘– where š“āˆ’1 is the š‘– column of š“āˆ’1 . Since xš‘– ā‰„ 0 for every š‘–, we conclude that š“āˆ’1 ā‰„ 0. š‘– Conversely, assume that š“āˆ’1 ā‰„ 0. Let 1 = (1, 1, . . . , 1) denote a net output of 1 for each commodity. Then x = š“āˆ’1 1 ā‰„ 0 and š“x = 1 > 0 š“ is productive. 3.115 Takayama 1985, p.383, Theorem 4.C.4. 3.116 Let a0 = (š‘Ž01 , š‘Ž02 , . . . , š‘Ž0š‘› ) be the vector of labour requirements and š‘¤ the wage rate. The unit proļ¬t of industry š‘– is āˆ‘ šœ‹š‘– = š‘š‘– + š‘Žš‘–š‘— š‘š‘— āˆ’ š‘¤š‘Ž0 š‘—āˆ•=š‘–

Recall that š‘Žš‘–š‘— ā‰¤ 0 for š‘— āˆ•= š‘–. The vector of unit proļ¬ts for all industries is Ī  = š“p āˆ’ š‘¤š‘Ž0 Proļ¬ts will be zero in all industries if there exists a price system p such that Ī  = š“p āˆ’ š‘¤š‘Ž0 = 0 or š“p = š‘¤š‘Ž0

(3.46)

By the previous results, (3.46) has a unique nonnegative solution p = š“āˆ’1 š‘¤š‘Ž0 if the technology š“ is productive. Furthermore, š“āˆ’1 is nonnegative. Since š‘Ž0 > 0, so is p > 0. 148

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3.117 Let š‘¢šµ denote the steady state unemployment rate for blacks. Then š‘¢šµ satisļ¬es the equation š‘¢šµ = 0.0038(1 āˆ’ š‘¢šµ ) + 0.8975š‘¢šµ which implies that š‘¢šµ = 0.036. That is, the data implies an unemployment rate of 3.6 percent for blacks. Similarly, the unemployment rate for white males š‘¢š‘Š satisļ¬es the equation š‘¢š‘Š = 0.0022(1 āˆ’ š‘¢š‘Š ) + 0.8614š‘¢š‘Š which implies that š‘¢š‘Š = 0.016 or 1.6 percent. 3.118 The transition matrix is

( š‘‡ =

) .6 .25 .4 .75

If the current state vector is x0 = (.4, .6), the state vector after a single mailing will be x1 = š‘‡ x0 ( )( ) .6 .25 .4 = .4 .75 .6 ( ) 0.39 = .61 Following a single mailing, the number of subscribers will drop to 30 percent of the mailing list, comprising 24 percent from renewals and 15 percent new subscriptions. 3.119 Let š‘“ (š‘„) = š‘„2 . For every š‘„1 , š‘„2 āˆˆ ā„œ and 0 ā‰¤ š›¼ ā‰¤ 1 š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) = (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 )2 = (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 )(š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) = š›¼2 š‘„21 + 2š›¼(1 āˆ’ š›¼)š‘„1 š‘„2 + (1 āˆ’ š›¼)2 š‘„22 = š›¼š‘„21 + (1 āˆ’ š›¼)š‘„22 āˆ’ š›¼š‘„21 āˆ’ (1 āˆ’ š›¼)š‘„22 + š›¼2 š‘„21 + 2š›¼(1 āˆ’ š›¼)š‘„1 š‘„2 + (1 āˆ’ š›¼)2 š‘„22 ) ( = š›¼š‘„21 + (1 āˆ’ š›¼)š‘„22 āˆ’ š›¼(1 āˆ’ š›¼)š‘„21 āˆ’ 2š›¼(1 āˆ’ š›¼)š‘„1 š‘„2 + š›¼(1 āˆ’ š›¼)š‘„22 = š›¼š‘„21 + (1 āˆ’ š›¼)š‘„22 āˆ’ š›¼(1 āˆ’ š›¼)(š‘„1 āˆ’ š‘„2 )2 ā‰¤ š›¼š‘„21 + (1 āˆ’ š›¼)š‘„22 = š›¼š‘“ (š‘„1 ) + (1 āˆ’ š›¼)(š‘„2 ) 3.120 š‘“ (š‘„) = š‘„ is linear and therefore convex. In the previous exercise we showed that š‘„2 is convex. Therefore š‘“ (š‘„) = š‘„š‘› is convex for š‘› = 1, 2. Assume that š‘“ is convex for

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š‘› āˆ’ 1. Then š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) = (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 )š‘› = (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 )(š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 )š‘›āˆ’1 ā‰¤ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 )(š›¼š‘„š‘›āˆ’1 + (1 āˆ’ š›¼)š‘„š‘›āˆ’1 ) 1 2

(since š‘„š‘›āˆ’1 is convex)

= š›¼2 š‘„š‘›1 + š›¼(1 āˆ’ š›¼)š‘„š‘›āˆ’1 š‘„2 + š›¼(1 āˆ’ š›¼)š‘„1 š‘„š‘›āˆ’1 + (1 āˆ’ š›¼)2 š‘„š‘›2 1 2

= š›¼š‘„š‘›1 + (1 āˆ’ š›¼)š‘„š‘›2 āˆ’ š›¼š‘„š‘›1 āˆ’ (1 āˆ’ š›¼)š‘„š‘›2

+ š›¼2 š‘„š‘›1 + š›¼(1 āˆ’ š›¼)š‘„š‘›āˆ’1 š‘„2 + š›¼(1 āˆ’ š›¼)š‘„1 š‘„š‘›āˆ’1 + (1 āˆ’ š›¼)2 š‘„š‘›2 1 2 ) ( š‘›āˆ’1 š‘› = š›¼š‘„š‘›1 + (1 āˆ’ š›¼)š‘„š‘›2 āˆ’ š›¼(1 āˆ’ š›¼) š‘„š‘›1 āˆ’ š‘„1 š‘„š‘›āˆ’1 āˆ’ š‘„ š‘„ + š‘„ 2 2 2 1 ( ) š‘›āˆ’1 š‘›āˆ’1 š‘› š‘› = š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 āˆ’ š›¼(1 āˆ’ š›¼) š‘„1 (š‘„1 āˆ’ š‘„2 ) āˆ’ š‘„2 (š‘„1 āˆ’ š‘„2 ) ( ) š‘›āˆ’1 āˆ’ š‘„ ) = š›¼š‘„š‘›1 + (1 āˆ’ š›¼)š‘„š‘›2 āˆ’ š›¼(1 āˆ’ š›¼) (š‘„1 āˆ’ š‘„2 )(š‘„š‘›āˆ’1 1 2 Since š‘„š‘š is monotonic (Example 2.53) š‘„š‘›āˆ’1 āˆ’ š‘„š‘›āˆ’1 ā‰„ 0 ā‡ā‡’ š‘„1 āˆ’ š‘„2 ā‰„ 0 1 2 and therefore (š‘„1 āˆ’ š‘„2 )(š‘„š‘›āˆ’1 āˆ’ š‘„š‘›āˆ’1 )ā‰„0 1 2 We conclude that š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) ā‰¤ š›¼š‘„š‘›1 + (1 āˆ’ š›¼)š‘„š‘›2 = š›¼š‘“ (š‘„1 ) + (1 āˆ’ š›¼)(š‘„2 ) š‘“ is convex for all š‘› = 1, 2, . . . . 3.121 For given x1 , x2 āˆˆ š‘†, deļ¬ne š‘” : [0, 1] ā†’ š‘† by š‘”(š‘”) = (1 āˆ’ š‘”)x1 + š‘”x2 Then š‘”(0) = x1 , š‘”(1) = x2 and ā„Ž = š‘” āˆ˜ š‘“ . Assume š‘“ is convex. For any š‘”1 , š‘”2 āˆˆ [0, 1], let ĀÆ 1 and š‘”(š‘”2 ) = x ĀÆ2 š‘”(š‘”1 ) = x For any š›¼ āˆˆ [0, 1]

) ( x1 + (1 āˆ’ š›¼)ĀÆ š‘” š›¼š‘”1 + (1 āˆ’ š›¼)š‘”2 = š›¼ĀÆ x2 ) ( ) ( x1 + (1 āˆ’ š›¼)ĀÆ ā„Ž š›¼š‘”1 + (1 āˆ’ š›¼)š‘”2 = š‘“ š›¼ĀÆ x2 ā‰¤ š›¼š‘“ (ĀÆ x1 ) + (1 āˆ’ š›¼)š‘“ (ĀÆ x2 ) ā‰¤ š›¼ā„Ž(š‘”1 ) + (1 āˆ’ š›¼)š‘”2 )

ā„Ž is convex. Conversely, assume ā„Ž is convex for any x1 , x2 āˆˆ š‘†. For any š›¼ āˆˆ [0, 1] š‘”(š›¼) = š›¼x1 + (1 āˆ’ š›¼)x2 and

) ( š‘“ š›¼x1 + (1 āˆ’ š›¼)x2 = ā„Ž(š›¼) ā‰¤ š›¼ā„Ž(0) + (1 āˆ’ š›¼)ā„Ž(1) = š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 )

Since this is true for any x1 , x2 āˆˆ š‘†, we conclude that š‘“ is convex. 150

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Solutions for Foundations of Mathematical Economics

3.122 Assume š‘“ is convex which implies epi š‘“ is convex. The points (xš‘– , š‘“ (xš‘– )) āˆˆ epi š‘“ . Since epi š‘“ is convex š›¼1 (x1 , š‘“ (x1 )) + š›¼2 (x1 , š‘“ (x1 )) + ā‹… ā‹… ā‹… + (xš‘› , š‘“ (xš‘› )) āˆˆ epi š‘“ that is š‘“ (š›¼1 x1 + š›¼2 x2 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› ) ā‰¤ š›¼1 š‘“ (x1 ) + š›¼2 š‘“ (x1 ) + ā‹… ā‹… ā‹… + š›¼š‘› š‘“ (xš‘› )) Conversely, letting š‘› = 2 and š›¼ = š›¼1 , (3.25) implies that š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰¤ š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) Jensenā€™s inequality can also be proved by induction from the deļ¬nition of a convex function (see for example Sydsaeter + Hammond 1995; p.624). 3.123 For each š‘–, let š‘¦š‘– = log š‘„š‘– so that š‘„š‘– = š‘’š‘¦š‘– š›¼š‘– š‘¦š‘– š‘– š‘„š›¼ š‘– = š‘’

Since š‘’š‘„ is convex (Example 3.41) š‘„š‘Ž1 1 š‘„š‘Ž2 2 . . . š‘„š‘Žš‘›š‘›

š‘Žš‘– > 0 =

āˆ

exp(š›¼š‘– š‘¦š‘– ) = exp

(āˆ‘

) āˆ‘ āˆ‘ š›¼š‘– š‘’š‘¦š‘– = š›¼š‘– š‘„š‘– š›¼š‘– š‘¦š‘– ā‰¤

by Jensenā€™s inequality. Setting š›¼š‘– = 1/š‘›, we have (š‘„1 š‘„2 . . . š‘„š‘› )1/š‘› ā‰¤

š‘›

1āˆ‘ š‘„š‘– š‘› š‘–=1

as required. 3.124 Assume š‘“ is concave. That is for every x1 , x2 āˆˆ š‘† and 0 ā‰¤ š›¼ ā‰¤ 1 š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰„ š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) Multiplying through by āˆ’1 reverses the inequality so that āˆ’š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰¤ āˆ’š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) = š›¼ āˆ’ š‘“ (x1 ) + (1 āˆ’ š›¼) āˆ’ š‘“ (x2 ) which shows that āˆ’š‘“ is concave. The converse follows analogously. 3.125 Assume that š‘“ is concave. Then āˆ’š‘“ is convex and by Theorem 3.7 epi āˆ’ š‘“ = { (š‘„, š‘¦) āˆˆ š‘‹ Ɨ ā„œ : š‘¦ ā‰„ āˆ’š‘“ (š‘„), š‘„ āˆˆ š‘‹ } is convex. But epi āˆ’ š‘“ = { (š‘„, š‘¦) āˆˆ š‘‹ Ɨ ā„œ : š‘¦ ā‰„ āˆ’š‘“ (š‘„), š‘„ āˆˆ š‘‹ } = { (š‘„, š‘¦) āˆˆ š‘‹ Ɨ ā„œ : š‘¦ ā‰¤ š‘“ (š‘„), š‘„ āˆˆ š‘‹ } = hypo š‘“ Therefore hypo š‘“ is convex. Conversely, if hypo š‘“ is convex, epi āˆ’ š‘“ is convex which implies that āˆ’š‘“ is convex and hence š‘“ is concave.

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3.126 Suppose that x1 minimizes the cost of producing š‘¦ at input prices w1 while x2 ĀÆ be the weighted average price, that is minimizes cost at w2 . For some š›¼ āˆˆ [0, 1], let w ĀÆ = š›¼w1 + (1 āˆ’ š›¼)w2 w ĀÆ minimizes cost at w. ĀÆ Then and suppose that x ĀÆ š‘¦) = wĀÆ ĀÆx š‘(w, x = (š›¼w1 + (1 āˆ’ š›¼)w2 )ĀÆ ĀÆ + (1 āˆ’ š›¼)w2 x ĀÆ = š›¼w1 x But since x1 and x2 minimize cost at w1 and w2 respectively ĀÆ ā‰„ š›¼w1 x1 = š›¼š‘(w1 , š‘¦) š›¼w1 x ĀÆ ā‰„ (1 āˆ’ š›¼)w2 x2 = (1 āˆ’ š›¼)š‘(w2 , š‘¦) (1 āˆ’ š›¼)w2 x so that ĀÆ + (1 āˆ’ š›¼)w2 x ĀÆ ā‰„ š›¼š‘(w1 , š‘¦) + (1 āˆ’ š›¼)š‘w2 , š‘¦) ĀÆ š‘¦) = š‘(š›¼w1 + (1 āˆ’ š›¼)w2 , š‘¦) = š›¼w1 x š‘(w, This establishes that the cost function š‘ is concave in w. 3.127 Since š‘¢ is concave, Jensenā€™s inequality implies ( š‘‡ ) š‘‡ š‘‡ āˆ‘1 āˆ‘ 1 1āˆ‘ š‘š‘” ā‰„ š‘¢(š‘š‘” ) = š‘¢ š‘¢(š‘š‘” ) š‘‡ š‘‡ š‘‡ š‘”=1 š‘”=1 š‘”=1 for any consumption stream š‘1 , š‘2 , . . . , š‘š‘‡ so that ( š‘‡ ) š‘‡ āˆ‘ āˆ‘1 š‘š‘” = š‘‡ š‘¢(ĀÆ š‘ˆ= š‘¢(š‘š‘” ) ā‰¤ š‘‡ š‘¢ š‘) š‘‡ š‘”=1 š‘”=1 It is impossible to do better than consume a constant fraction š‘ĀÆ = š‘¤/š‘‡ of wealth in each period. 3.128 If š‘„1 = š‘„3 , the inequality is trivially satisļ¬ed. Now assume š‘„1 āˆ•= š‘„3 . Since š‘„2 āˆˆ [š‘„1 , š‘„3 ], there exists š›¼ āˆˆ [0, 1] such that š‘„2 = š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 Let š‘„ ĀÆ = š‘„1 āˆ’ š‘„2 + š‘„3 . Then š‘„ ĀÆ āˆˆ [š‘„1 , š‘„3 ] and there exists š›½ āˆˆ [0, 1] such that š‘„ĀÆ = š›½š‘„1 + (1 āˆ’ š›½)š‘„2 Adding

( ) š‘„ĀÆ + š‘„2 = (š›¼ + š›½)š‘„1 + (1 āˆ’ š›¼) + (1 āˆ’ š›½) š‘„3

or š‘„1 āˆ’ š‘„3 = (š›¼ + š›½)(š‘„3 āˆ’ š‘„1 ) which implies that š›¼ + š›½ = 1 and therefore š›½ = 1 āˆ’ š›¼. Since š‘“ is convex š‘“ (š‘„2 ) ā‰¤ š›¼š‘“ (š‘„1 ) + (1 āˆ’ š›¼)š‘“ (š‘„2 š‘“ (ĀÆ š‘„) ā‰¤ š›½š‘“ (š‘„1 ) + (1 āˆ’ š›½)š‘“ (š‘„2 ) = (1 āˆ’ š›¼)š‘“ (š‘„1 ) + š›¼š‘“ (š‘„3 ) Adding š‘“ (ĀÆ š‘„) + š‘“ (š‘„2 ) ā‰¤ š‘“ (š‘„1 ) + š‘“ (š‘„3 ) 152

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3.129 Let š‘„1 , š‘„2 , š‘¦1 , š‘¦2 āˆˆ ā„œ with š‘„1 < š‘„2 and š‘¦1 < š‘¦2 . Note that š‘„1 āˆ’ š‘¦2 ā‰¤ š‘„2 āˆ’ š‘¦2 ā‰¤ š‘„2 āˆ’ š‘¦1 and therefore (Exercise 3.128) ( ) š‘“ š‘„1 āˆ’ š‘¦2 ) āˆ’ (š‘„2 āˆ’ š‘¦2 ) + (š‘„2 āˆ’ š‘¦1 ) > š‘“ (š‘„1 āˆ’ š‘¦2 ) āˆ’ š‘“ (š‘„2 āˆ’ š‘¦2 ) + š‘“ (š‘„2 āˆ’ š‘¦1 ) That is š‘“ (š‘„1 āˆ’ š‘¦1 ) > š‘“ (š‘„1 āˆ’ š‘¦2 ) āˆ’ š‘“ (š‘„2 āˆ’ š‘¦2 ) + š‘“ (š‘„2 āˆ’ š‘¦1 ) Rearranging š‘“ (š‘„2 āˆ’ š‘¦2 ) āˆ’ š‘“ (š‘„1 āˆ’ š‘¦2 ) > š‘“ (š‘„2 āˆ’ š‘¦1 ) āˆ’ š‘“ (š‘„1 āˆ’ š‘¦1 ) as required. 3.130 A functional is aļ¬ƒne if and only if inequalities (3.24) and (3.26) are satisļ¬ed as equalities. 3.131 Since š‘“ and š‘” are convex on š‘† š‘“ (š›½x1 + (1 āˆ’ š›½)x2 ) ā‰¤ š›½š‘“ (x1 ) + (1 āˆ’ š›½)š‘“ (x2 ) 1

2

1

2

š‘”(š›½x + (1 āˆ’ š›½)x ) ā‰¤ š›½š‘”(x ) + (1 āˆ’ š›½)š‘”(x )

(3.47) (3.48)

for every x1 , x2 āˆˆ š‘† and š›½ āˆˆ [0, 1]. Adding (š‘“ + š‘”)(š›½x1 + (1 āˆ’ š›½)x2 ) ā‰¤ š›½(š‘“ + š‘”)(x1 ) + (1 āˆ’ š›½)š‘“ (x2 ) š‘“ + š‘” is convex. Multiplying (3.47) by š›¼ ā‰„ 0 š›¼š‘“ (š›½x1 + (1 āˆ’ š›½)x2 ) ā‰¤ š›¼(š›½š‘“ (x1 ) + (1 āˆ’ š›½)š‘“ (x2 )) = (š›½š›¼š‘“ (x1 ) + (1 āˆ’ š›½)š›¼š‘“ (x2 )) š›¼š‘“ is convex. Moreover, if š‘“ is strictly convex, š‘“ (š›½x1 + (1 āˆ’ š›½)x2 ) < š›½š‘“ (x1 ) + (1 āˆ’ š›½)š‘“ (x2 )

(3.49)

for every x1 , x2 āˆˆ š‘†, x1 āˆ•= x2 and š›½ āˆˆ (0, 1). Adding this to (3.48) (š‘“ + š‘”)(š›½x1 + (1 āˆ’ š›½)x2 ) < š›½(š‘“ + š‘”)(x1 ) + (1 āˆ’ š›½)š‘“ (x2 ) so that š‘“ + š‘” is strictly convex. Multiplying (3.49) by š›¼ > 0 š›¼š‘“ (š›½x1 + (1 āˆ’ š›½)x2 ) < š›¼(š›½š‘“ (x1 ) + (1 āˆ’ š›½)š‘“ (x2 )) = (š›½š›¼š‘“ (x1 ) + (1 āˆ’ š›½)š›¼š‘“ (x2 )) š›¼š‘“ is strictly convex. 3.132 x āˆˆ epi (š‘“ āˆØ š‘”) ā‡ā‡’ x āˆˆ epi š‘“ and x āˆˆ epi š‘” That is epi (š‘“ āˆØ š‘”) = epi š‘“ āˆ© epi š‘” Therefore epi š‘“ āˆØ š‘” is convex (Exercise 1.162) and therefore š‘“ is convex (Proposition 3.7). 153

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3.133 If š‘“ is convex š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰¤ š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) Since š‘” is increasing ( ) ( ) š‘” š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰¤ š‘” š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) ) ( ) ( ā‰¤ š›¼š‘” š‘“ (x1 ) + (1 āˆ’ š›¼)š‘” š‘“ (x2 ) since š‘” is also convex. The concave case is proved similarly. 3.134 Let š¹ = log š‘“ . If š¹ is convex, š‘“ (x) = š‘’š¹ (x) is an increasing convex function of a convex function and is therefore convex (Exercise 3.133). 3.135 If š‘“ is positive and concave, then log š‘“ is concave (Exercise 3.51). Therefore log

1 = log 1 āˆ’ log š‘“ = āˆ’ log š‘“ š‘“

is convex. By the previous exercise (Exercise 3.134), this implies that 1/š‘“ is convex. If š‘“ is negative and convex, then āˆ’š‘“ is positive and concave, 1/ āˆ’ š‘“ is convex, and therefore 1/š‘“ is concave. 3.136 Consider the identity ) ( ) ( ) ( ) ( š‘” š‘“ (š‘„1 āˆØ š‘„2 ) + š‘” š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘” š‘“ (š‘„1 ) āˆ’ š‘” š‘“ (š‘„2 ) ) ( ) ( ) ( ) ) ) ( = š‘” š‘“ (š‘„1 āˆØ š‘„2 ) + š‘” š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘” š‘“ (š‘„1 ) āˆ’ š‘” š‘“ (š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘“ (š‘„1 ) ( ) ( ) + š‘” š‘“ (š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘“ (š‘„1 ) āˆ’ š‘” š‘“ (š‘„2 ) (3.50) Deļ¬ne

( ) ( ) ( ) ( ) šœ‘(š‘„1 , š‘„2 ) = š‘” š‘“ (š‘„1 āˆØ š‘„2 ) + š‘” š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘” š‘“ (š‘„1 ) āˆ’ š‘” š‘“ (š‘„2 )

Then š‘” āˆ˜ š‘“ is supermodular if šœ‘ is nonnegative deļ¬nite and submodular if šœ‘ is nonpositive deļ¬nite. Using the identity (3.50), šœ‘ can be decomposed into two components šœ‘(š‘„1 , š‘„2 ) = šœ‘1 (š‘„1 , š‘„2 ) + šœ‘2 (š‘„1 , š‘„2 ) ( ) ( ) ( ) šœ‘1 (š‘„1 , š‘„2 ) = š‘” š‘“ (š‘„1 āˆØ š‘„2 ) + š‘” š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘” š‘“ (š‘„1 ) ( ) āˆ’ š‘” š‘“ (š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘“ (š‘„1 ) ( ) ( ) šœ‘2 (š‘„1 , š‘„2 ) = š‘” š‘“ (š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘“ (š‘„1 ) āˆ’ š‘” š‘“ (š‘„2 )

(3.51)

šœ‘ will deļ¬nite if both components are deļ¬nite. For any š‘„1 , š‘„2 āˆˆ š‘„1 , let š‘Ž = š‘“ (š‘„1 āˆ§ š‘„2 ), š‘ = š‘“ (š‘„1 ) and š‘ = š‘“ (š‘„1 āˆØ š‘„2 ). Provided š‘“ is monotone, š‘ lies between š‘Ž and š‘. Substituting in (3.51) šœ‘1 (š‘„1 , š‘„2 ) = š‘”(š‘) + š‘”(š‘Ž) āˆ’ š‘”(š‘) āˆ’ š‘”(š‘ + š‘Ž āˆ’ š‘) and Exercise 3.128 implies

{

šœ‘1 (š‘„1 , š‘„2 ) = š‘”(š‘) + š‘”(š‘Ž) āˆ’ š‘”(š‘) āˆ’ š‘”(š‘ + š‘Ž āˆ’ š‘) Now consider šœ‘2 .

} { } ā‰„š‘‚ convex if š‘” is ā‰¤0 concave

{ } { } ā‰„ supermodular š‘“ (š‘„1 āˆØ š‘„2 ) + š‘“ (š‘„1 āˆ§ š‘„2 ) āˆ’ š‘“ (š‘„1 ) is š‘“ (š‘„2 ) if š‘“ is ā‰¤ submodular 154

(3.52)

Solutions for Foundations of Mathematical Economics and therefore since š‘” is increasing

{

šœ‘2 (š‘„1 , š‘„2 ) =

c 2001 Michael Carter āƒ All rights reserved

ā‰„ 0 if š‘“ is supermodular ā‰¤ 0 if š‘“ is submodular

(3.53)

Together (3.52) and (3.53) gives the desired result. 3.137

1. Assume that š‘“ is bounded above in a neighborhood of x0 . Then there exists a ball šµ(š‘„0 ) and constant š‘€ such that š‘“ (x) ā‰¤ š‘€ for every x āˆˆ šµ(š‘„0 ) Since š‘“ is convex š‘“ (š›¼x + (1 āˆ’ š›¼)x0 ) ā‰¤ š›¼š‘“ (x) + (1 āˆ’ š›¼)š‘“ (x0 ) ā‰¤ š›¼š‘€ + (1 āˆ’ š›¼)š‘“ (x0 )

(3.54)

2. Given x āˆˆ šµ(š‘„0 ) and š›¼ āˆˆ [0, 1] let z = š›¼x + (1 āˆ’ š›¼)x0

(3.55)

Subtracting š‘“ (x0 ) from (3.54) gives š‘“ (z) āˆ’ š‘“ (x0 ) ā‰¤ š›¼(š‘€ āˆ’ š‘“ (x0 )) Rewriting (3.55) (1 āˆ’ š›¼)x0 = z āˆ’ š›¼x (1 + š›¼)x0 = z + š›¼(2x0 āˆ’ x) š›¼ 1 z+ (2x0 āˆ’ x) x0 = 1+š›¼ 1+š›¼ 3. Note that (2x0 āˆ’ x) = x0 āˆ’ (x āˆ’ x0 ) āˆˆ šµ(x0 ) so that š‘“ (2x0 āˆ’ x) ā‰¤ š‘€ and therefore š‘“ (x0 ) ā‰¤

š›¼ š›¼ 1 1 š‘“ (z) + š‘“ (2x0 āˆ’ x) ā‰¤ š‘“ (z) + š‘€ 1+š›¼ 1+š›¼ 1+š›¼ 1+š›¼

which implies (1 + š›¼)š‘“ (x0 ) ā‰¤ š‘“ (z) + š›¼š‘€ š›¼(š‘“ (x0 ) āˆ’ š‘€ ) ā‰¤ š‘“ (z) āˆ’ š‘“ (x0 ) 4. Combined with (3.56) we have š›¼(š‘“ (x0 ) āˆ’ š‘€ ) ā‰¤ š‘“ (z) āˆ’ š‘“ (x0 ) ā‰¤ š›¼(š‘€ āˆ’ š‘“ (x0 )) or āˆ£š‘“ (z) āˆ’ š‘“ (x0 )āˆ£ ā‰¤ š›¼(š‘€ āˆ’ š‘“ (x0 )) and therefore š‘“ (z) ā†’ š‘“ (x0 ) as z ā†’ x0 . š‘“ is continuous. 155

(3.56)

Solutions for Foundations of Mathematical Economics 3.138

c 2001 Michael Carter āƒ All rights reserved

1. Since š‘† is open, there exists a ball šµš‘Ÿ (x1 ) āŠ† š‘†. Let š‘” = 1 + š‘Ÿ2 . Then x0 + š‘”(x1 āˆ’ x0 ) āˆˆ šµš‘Ÿ (š‘„1 ) āŠ† š‘†.

2. Let š‘  = š‘”āˆ’1 š‘” š‘Ÿ. The open ball šµš‘  (x1 ) of radius š‘  centered on x1 is contained in š‘‡ . Therefore š‘‡ is a neighborhood of x1 . 3. Since š‘“ is convex, for every y āˆˆ š‘‡ š‘“ (y) ā‰¤ (1 āˆ’ š›¼)š‘“ (x) + š›¼š‘“ (z) ā‰¤ (1 āˆ’ š›¼)š‘€ + š›¼š‘“ (z) ā‰¤ š‘€ + š‘“ (z) Therefore š‘“ is bounded on š‘‡ . 3.139 The previous exercise showed that š‘“ is locally bounded from above for every x āˆˆ š‘†. To show that it is also locally bounded from below, choose some x0 āˆˆ š‘†. There exists some šµ(x0 and š‘€ such that š‘“ (x) ā‰¤ š‘€ for every x āˆˆ šµ(x0 ) Choose some š‘„1 āˆˆ šµ(x0 ) and let x2 = 2x0 āˆ’ x1 . Then x2 = 2x0 āˆ’ x1 = x0 āˆ’ (x1 āˆ’ x0 ) āˆˆ šµ(x0 ) and š‘“ (x2 ) ā‰¤ š‘€ . Since š¹ is convex š‘“ (x) ā‰¤

1 1 š‘“ (x1 ) + š‘“ (x2 ) 2 2

and therefore š‘“ (x1 ) ā‰„ 2š‘“ (x) āˆ’ š‘“ (x2 ) Since š‘“ (x2 ) ā‰¤ š‘€ , āˆ’š‘“ (x2 ) ā‰„ āˆ’š‘€ and therefore š‘“ (x1 ) ā‰„ 2š‘“ (x) āˆ’ š‘€ so that š‘“ is bounded from below. 3.140 Let š‘“ be a convex function deļ¬ned on an open convex set š‘† in a normed linear space, which is bounded from above in a neighborhood of a single point x0 āˆˆ š‘†. By Exercise 3.138, š‘“ is bounded above at every x āˆˆ š‘†. This implies (Exercise 3.137) that š‘“ is continuous at every x āˆˆ š‘†. 3.141 Without loss of generality, assume 0 āˆˆ š‘†. Assume š‘† has dimension š‘› and let x1 , x2 , . . . , xš‘› be a basis for the subspace containing š‘†. Choose some šœ† > 0 small enough so that š‘ˆ = conv {0, šœ†x1 , šœ†x2 , . . . , šœ†š‘„š‘› } āŠ† š‘† Any x āˆˆ š‘ˆ is a convexāˆ‘ combination of the points 0, x1 , x2 , . . . , xš‘› and so there exists š›¼0 , š›¼1 , š›¼2 , . . . , š›¼š‘› ā‰„ 0, š›¼š‘– = 1 such that x = š›¼0 0 + š›¼1 x1 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› . By Jensenā€™s inequality š‘“ (x) = š‘“ (š›¼0 0 + š›¼1 x1 + ā‹… ā‹… ā‹… + š›¼š‘› xš‘› ) ā‰¤ š›¼0 š‘“ (0) + š›¼1 š‘“ (x1 ) + ā‹… ā‹… ā‹… + š›¼š‘› š‘“ (xš‘› ) ā‰¤ max{ š‘“ (0), š‘“ (x1 ), . . . , š‘“ (xš‘› ) } Therefore, š‘“ is bounded above on a neighbourhood of some x0 āˆˆ int š‘ˆ (which is nonempty by Exercise 1.229). By Proposition 3.8, š‘“ is continuous on š‘†.

156

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c 2001 Michael Carter āƒ All rights reserved

3.142 Clearly, if š‘“ is convex, it is locally convex at every x āˆˆ š‘†, where š‘† is the required neighborhood. To prove the converse, assume to the contrary that š‘“ is locally convex at every x āˆˆ š‘† but it is not globally convex. That is, there exists x1 , x2 āˆˆ š‘† such that š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) > š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) Let ) ( ā„Ž(š‘”) = š‘“ š‘”x1 + (1 āˆ’ š‘”)x2 Local convexity implies that š‘“ is continuous at every x āˆˆ š‘† (Corollary 3.8.1), and therefore continuous on š‘†. Therefore, ā„Ž is continuous on [0, 1]. By the continuous maximum theorem (Theorem 2.3), š‘‡ = arg max ā„Ž(š‘”) xāˆˆ[x1 ,x2 ]

is nonempty and compact. Let š‘”0 = max š‘‡ . For every šœ– > 0, ā„Ž(š‘”0 āˆ’ šœ–) ā‰¤ ā„Ž(š‘”0 ) and ā„Ž(š‘”0 + šœ–) < ā„Ž(š‘”0 ) Let x0 = š‘”0 x1 + (1 āˆ’ š‘”0 )x2 and xšœ– = (š‘”0 + šœ–)x1 + (1 āˆ’ š‘”0 āˆ’ šœ–)x2 Every neighborhood š‘‰ of x0 contains xāˆ’šœ– , xšœ– āˆˆ [x1 , x2 ] with 1 1 1 1 š‘“ (xāˆ’šœ– ) + š‘“ (xšœ– ) = ā„Ž(š‘”0 āˆ’ šœ–) + ā„Ž(š‘”0 + šœ–) < ā„Ž(š‘”0 ) = š‘“ (x0 ) = š‘“ 2 2 2 2

(

1 1 xāˆ’šœ– + xšœ– 2 2

)

contradicting the local convexity of š‘“ at x0 . 3.143 Assume š‘“ is quasiconcave. That is for every x1 , x2 āˆˆ š‘† and 0 ā‰¤ š›¼ ā‰¤ 1 š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰„ min{š‘“ (x1 ), (x2 )} Multiplying through by āˆ’1 reverses the inequality so that āˆ’š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰¤ āˆ’ min{š‘“ (x1 ), š‘“ (x2 )} = max{āˆ’š‘“ (x1 ), āˆ’š‘“ (x2 )} which shows that āˆ’š‘“ is quasiconvex. The converse follows analogously. 3.144 Assume š‘“ is concave, that is š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰„ š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) for every x1 , x2 āˆˆ š‘† and 0 ā‰¤ š›¼ ā‰¤ 1 Without loss of generality assume that š‘“ (x1 ) ā‰¤ š‘“ (x2 ). Then š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰„ š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) ā‰„ š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x1 ) = š‘“ (x1 ) = min{š‘“ (x1 ), š‘“ (x2 )} š‘“ is quasiconcave. 3.145 Let š‘“ : ā„œ ā†’ ā„œ. Choose any š‘„1 , š‘„2 in ā„œ with š‘„1 < š‘„2 . If š‘“ is increasing, then š‘“ (š‘„1 ) ā‰¤ š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) ā‰¤ š‘“ (š‘„2 ) for every 0 ā‰¤ š›¼ ā‰¤ 1. The ļ¬rst inequality implies that š‘“ (š‘„1 ) = min{š‘“ (š‘„1 ), š‘“ (š‘„2 )} ā‰¤ š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) 157

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

so that š‘“ is quasiconcave. The second inequality implies that š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) ā‰¤ max{š‘“ (š‘„1 ), š‘“ (š‘„2 )} = š‘“ (š‘„2 ) so that š‘“ is also quasiconvex. Conversely, if š‘“ is decreasing š‘“ (š‘„1 ) ā‰„ š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) ā‰„ š‘“ (š‘„2 ) for every 0 ā‰¤ š›¼ ā‰¤ 1. The ļ¬rst inequality implies that š‘“ (š‘„1 ) = max{š‘“ (š‘„1 ), š‘“ (š‘„2 )} ā‰„ š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) so that š‘“ is quasiconvex. The second inequality implies that š‘“ (š›¼š‘„1 + (1 āˆ’ š›¼)š‘„2 ) ā‰¤ max{š‘“ (š‘„1 ), š‘“ (š‘„2 )} = š‘“ (š‘„2 ) so that š‘“ is also quasiconcave. 3.146 ā‰¾š‘“ (š‘) = { x āˆˆ š‘‹ : š‘“ (x) ā‰¤ š‘Ž } = {x āˆˆ š‘‹ : āˆ’š‘“ (x) ā‰„ āˆ’š‘} = ā‰æāˆ’š‘“ (āˆ’š‘) 3.147 For given š‘ and š‘š, choose any p1 and p2 in ā‰¾š‘£ (š‘). For any 0 ā‰¤ š›¼ ā‰¤ 1, let ĀÆ = š›¼p1 + (1 āˆ’ š›¼)p2 . The key step is to show that any commodity bundle x which is p ĀÆ is also aļ¬€ordable at either p1 or p2 . Assume that x is aļ¬€ordable at p ĀÆ, aļ¬€ordable at p that is x is in the budget set ĀÆx ā‰¤ š‘š } x āˆˆ š‘‹(ĀÆ p, š‘š) = { x : p To show that x is aļ¬€ordable at either p1 or p2 , that is x āˆˆ š‘‹(p1 , š‘š) or x āˆˆ š‘‹(p2 , š‘š) assume to the contrary that xāˆˆ / š‘‹(p1 , š‘š) and x āˆˆ / š‘‹(p2 , š‘š) This implies that p1 x > š‘š and p2 x > š‘š so that š›¼p1 x > š›¼š‘š and (1 āˆ’ š›¼)p2 > (1 āˆ’ š›¼)š‘š Summing these two inequalities ĀÆ x = (š›¼p1 + (1 āˆ’ š›¼)p2 )x > š‘š p contradicting the assumption that x āˆˆ š‘‹(ĀÆ p, š‘š). We conclude that š‘‹(ĀÆ p, š‘š) āŠ† š‘‹(p1 , š‘š) āˆŖ š‘‹(p2 , š‘š) Now š‘£(ĀÆ š‘, š‘š) = sup{ š‘¢(x) : x āˆˆ š‘‹(ĀÆ p, š‘š) } ā‰¤ sup{ š‘¢(x) : x āˆˆ š‘‹(p1 , š‘š) āˆŖ š‘‹(p2 , š‘š) } ā‰¤š‘ ĀÆ āˆˆ ā‰¾š‘£ (š‘) for every 0 ā‰¤ š›¼ ā‰¤ 1. Thus, ā‰¾š‘£ (š‘) is convex and so š‘£ is quasiconvex Therefore p (Exercise 3.146). 158

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c 2001 Michael Carter āƒ All rights reserved

3.148 Since š‘“ is quasiconcave š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰„ min{š‘“ (x1 ), š‘“ (x2 )} for every x1 , x2 āˆˆ š‘† and 0 ā‰¤ š›¼ ā‰¤ 1 Since š‘” is increasing ) ( ) ( ) ( ) ( š‘” š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰„ š‘”( min{š‘“ (x1 ), š‘“ (x2 )}) ā‰„ min{š‘” š‘“ (x1 ) , š‘” š‘“ (x2 ) } š‘” āˆ˜ š‘“ is quasiconcave. 3.149 When šœŒ ā‰„ 1, the function ā„Ž(x) = š›¼1 š‘„šœŒ1 + š›¼2 š‘„šœŒ2 + . . . š›¼š‘› š‘„šœŒš‘› is convex (Example 3.58) as is š‘¦ 1/šœŒ . Therefore š‘“ (x) = (ā„Ž(x))

1/šœŒ

is an increasing convex function of a convex function and is therefore convex (Exercise 3.133). 3.150 š‘“ is a monotonic transformation of the concave function ā„Ž(x) = x. 3.151 By Exercise 3.39, there exist linear functionals š‘“Ė† and š‘”Ė† and scalars š‘ and š‘ such that š‘“ (x) = š‘“Ė†(x) + š‘ and š‘”(x) = š‘”Ė†(x) + š‘ The upper contour set ā‰æā„Ž (š‘Ž) = { š‘„ āˆˆ š‘† : ā„Ž(x) ā‰„ š‘Ž } š‘“Ė†(š‘„) + š‘ ā‰„ š‘Ž} = {š‘„ āˆˆ š‘† : š‘”Ė†(š‘„) + š‘ = { š‘„ āˆˆ ā„œš‘› : š‘“Ė†(x) + š‘ ā‰„ š‘ŽĖ† š‘”(x) + š‘Žš‘ } +

š‘” (x) ā‰„ š‘ āˆ’ š‘Žš‘ } = { š‘„ āˆˆ ā„œš‘›+ : š‘“Ė†(x) āˆ’ š‘ŽĖ† which is a halfspace in š‘‹ and therefore convex. Similarly, the lower contour set ā‰¾ā„Ž (š‘Ž) = { š‘„ āˆˆ š‘† : ā„Ž(x) ā‰„ š‘Ž } is also a halfspace and hence convex. Therefore ā„Ž is both quasiconcave and quasiconvex. 3.152 For š‘Ž ā‰¤ 0 ā‰æ(š‘Ž) = { š‘„ āˆˆ š‘† : ā„Ž(x) ā‰„ 0 } = š‘† which is convex. For š‘Ž > 0 ā‰æā„Ž (š‘Ž) = { š‘„ āˆˆ š‘† : ā„Ž(x) ā‰„ š‘Ž } š‘“ (x) ā‰„ š‘Ž} š‘”(x) = { š‘„ āˆˆ š‘† : š‘“ (x) ā‰„ š‘Žš‘”(x) } = {š‘„ āˆˆ š‘† :

= { š‘„ āˆˆ š‘† : š‘“ (x) āˆ’ š‘Žš‘”(x) ā‰„ 0 } is convex since š‘“ āˆ’ š‘Žš‘” = š‘“ + š‘Ž(āˆ’š‘”) is concave (Exercises 3.124 and 3.131). Since ā‰æā„Ž (š‘Ž) is convex for every š‘Ž, ā„Ž is quasiconcave. 159

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3.153 š‘“ (x) š‘”Ė†(x)

ā„Ž(x) =

where š‘”Ė† = 1/š‘” is positive and convex by Exercise 3.135. By the previous exercise, ā„Ž is quasiconcave. 3.154 Let š¹ = log š‘“ . If š¹ is concave, š‘“ (x) = š‘’š¹ (x) is an increasing function of (quasi)concave function, and hence is quasiconcave (Exercise 3.148). 3.155 Let š¹ (x) = log š‘“ (x) =

š‘› āˆ‘

š›¼š‘– log š‘“š‘– (x)

š‘–=1

As the sum of concave functions, š¹ is concave (Exercise 3.131). By the previous exercise, š‘“ is quasiconcave. ĀÆ = š›¼šœ½1 + (1 āˆ’ š›¼)šœ½ 2 ĀÆ are optimal solutions for šœ½1 , šœ½2 and šœ½ 3.156 Assume x1 , x2 and x respectively. That is š‘“ (x1 , šœ½ 1 ) = š‘£(šœ½ 1 ) š‘“ (x2 , šœ½ 2 ) = š‘£(šœ½ 2 ) ĀÆ = š‘£(šœ½) ĀÆ š‘“ (ĀÆ x, šœ½) Since š‘“ is convex in šœ½ ĀÆ = š‘“ (ĀÆ ĀÆ š‘£(šœ½) x, šœ½) = š‘“ (ĀÆ x, š›¼šœ½ 1 + (1 āˆ’ š›¼)šœ½ 2 ) ā‰¤ š›¼š‘“ (ĀÆ x, šœ½ 1 ) + (1 āˆ’ š›¼)š‘“ (xāˆ— , šœ½2 ) ā‰¤ š›¼š‘“ (x1 , šœ½1 ) + (1 āˆ’ š›¼)š‘“ (x2 , šœ½ 2 ) = š›¼š‘£(šœ½ 1 ) + (1 āˆ’ š›¼)š‘£(šœ½ 2 ) š‘£ is convex. 3.157 Assume to the contrary that x1 and x2 are distinct optimal solutions, that is āˆ• x2 , for some šœ½ āˆˆ Ī˜āˆ— , so that x1 , x2 āˆˆ šœ‘(šœ½), x1 = š‘“ (x1 , šœ½) = š‘“ (x2 , šœ½) = š‘£(šœ½) ā‰„ š‘“ (x, šœ½) for every x āˆˆ šŗ(šœ½) ĀÆ = š›¼x1 + (1 āˆ’ š›¼)x2 for š›¼ āˆˆ (0, 1). Since šŗ(šœ½) is convex, x ĀÆ is feasible. Since š‘“ is Let x strictly quasiconcave š‘“ (ĀÆ x, šœ½) > min{ š‘“ (x1 , šœ½), š‘“ (x2 , šœ½) } = š‘£(šœ½) contradicting the optimality of x1 and x2 . We conclude that šœ‘(šœ½) is single-valued for every šœ½ āˆˆ Ī˜āˆ— . In other words, šœ‘ is a function. 3.158

1. The value function is š‘£(š‘„0 ) =

sup š‘ˆ (x)

xāˆˆĪ“(š‘„0 )

where š‘ˆ (x) =

āˆž āˆ‘ š‘”=0

160

š›½ š‘” š‘“ (š‘„š‘” , š‘„š‘”+1 )

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c 2001 Michael Carter āƒ All rights reserved

and Ī“(š‘„0 ) = {x āˆˆ š‘‹ āˆž : š‘„š‘”+1 āˆˆ šŗ(š‘„š‘” ), š‘” = 0, 1, 2, . . . } Since an optimal policy exists (Exercise 2.125), the maximum is attained and š‘£(š‘„0 ) = max š‘ˆ (x) xāˆˆĪ“(š‘„0 )

(3.57)

It is straightforward to show that āˆ™ š‘ˆ (x) is strictly concave and āˆ™ Ī“(š‘„0 ) is convex Applying the Concave Maximum Theorem (Theorem 3.1) to (3.57), we conclude that the value function š‘£ is strictly concave. 2. Assume to the contrary that xā€² and xā€²ā€² are distinct optimal plans, so that š‘£(š‘„0 ) = š‘ˆ (xā€² ) = š‘ˆ (xā€²ā€² ) ĀÆ is feasible and ĀÆ = š›¼xā€² + (1 āˆ’ š›¼)xā€²ā€² . Since Ī“(š‘„0 ) is convex, x Let x š‘ˆ (ĀÆ x) > š›¼š‘ˆ (xā€² ) + (1 āˆ’ š›¼)š‘ˆ (xā€²ā€² ) = š‘ˆ (xā€² ) which contradicts the optimality of xā€² . We conclude that the optimal plan is unique. 3.159 We observe that āˆ™ š‘¢(š¹ (š‘˜) āˆ’ š‘¦) is supermodular in š‘¦ (Exercise 2.51) āˆ™ š‘¢(š¹ (š‘˜) āˆ’ š‘¦) displays strictly increasing diļ¬€erences in (š‘˜, š‘¦) (Exercise 3.129) āˆ™ šŗ(š‘˜) = [0, š¹ (š‘˜)] is increasing. Applying Exercise 2.126, we can conclude that the optimal policy (š‘˜0 , š‘˜1āˆ— , š‘˜2āˆ— , . . . ) is a monotone sequence. Since š‘‹ is compact, kāˆ— is a bounded monotone sequence, which converges monotonically to some steady state š‘˜ āˆ— (Exercise 1.101). 3.160 Suppose there exists (xāˆ— , yāˆ— ) āˆˆ š‘‹ Ɨ š‘Œ such that š‘“ (x, yāˆ— ) ā‰¤ š‘“ (xāˆ— , yāˆ— ) ā‰¤ š‘“ (xāˆ— , y) for every x āˆˆ š‘‹ and y āˆˆ š‘Œ Let š‘£ = š‘“ (xāˆ— , yāˆ— ). Since š‘“ (x, yāˆ— ) ā‰¤ š‘£ for every x āˆˆ š‘‹ max š‘“ (š‘„, yāˆ— ) ā‰¤ š‘£ xāˆˆš‘‹

and therefore min max š‘“ (x, y) ā‰¤ max š‘“ (x, yāˆ— ) ā‰¤ š‘£

yāˆˆš‘Œ xāˆˆš‘‹

xāˆˆš‘‹

Similarly max min š‘“ (x, y) ā‰„ š‘£ xāˆˆš‘‹ yāˆˆš‘¦

Combining the last two inequalities, we have max min š‘“ (x, y) ā‰„ š‘£ ā‰„ min max š‘“ (š‘„, y) xāˆˆš‘‹ yāˆˆš‘¦

yāˆˆš‘Œ xāˆˆš‘‹

161

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Together with (3.28), this implies equality max min š‘“ (x, y) = min max š‘“ (x, y) xāˆˆš‘‹ yāˆˆš‘Œ

yāˆˆš‘Œ xāˆˆš‘‹

Conversely, suppose that max min š‘“ (x, y) = š‘£ = min max š‘“ (x, y) xāˆˆš‘‹ yāˆˆš‘Œ

yāˆˆš‘Œ xāˆˆš‘‹

The function š‘”(x) = min š‘“ (x, y) yāˆˆš‘Œ

is a continuous function (Theorem 2.3) on a compact set š‘‹. By the Weierstrass theorem (Theorem 2.2), there exists xāˆ— which maximizes š‘” on š‘‹, that is š‘”(xāˆ— ) = min š‘“ (xāˆ— , y) = max š‘”(x) = max min š‘“ (x, y) = š‘£ yāˆˆš‘Œ

xāˆˆš‘‹

xāˆˆš‘‹ yāˆˆš‘Œ

which implies that š‘“ (xāˆ— , y) ā‰„ š‘£ for every y āˆˆ š‘Œ Similarly, there exists y āˆˆ š‘Œ such that š‘“ (x, yāˆ— ) ā‰¤ š‘£ for every x āˆˆ š‘‹ Combining these inequalities, we have š‘“ (x, yāˆ— ) ā‰¤ š‘£ ā‰¤ š‘“ (xāˆ— , y) for every x āˆˆ š‘‹ and y āˆˆ š‘Œ In particular, we have š‘“ (xāˆ— , yāˆ— ) ā‰¤ š‘£ ā‰¤ š‘“ (xāˆ— , yāˆ— ) so that š‘£ = š‘“ (xāˆ— , yāˆ— ) as required. 3.161 For any š‘„ āˆˆ š‘‹ and š‘¦ āˆˆ š‘Œ , let š‘”(š‘„) = min š‘“ (š‘„, š‘¦) and ā„Ž(š‘¦) = max š‘“ (š‘„, š‘¦) š‘¦āˆˆš‘Œ

š‘„āˆˆš‘‹

Then š‘”(š‘„) = min š‘“ (š‘„, š‘¦) ā‰¤ max š‘“ (š‘„, š‘¦) = ā„Ž(š‘¦) š‘¦āˆˆš‘Œ

š‘„āˆˆš‘‹

and therefore max š‘”(š‘„) ā‰¤ max ā„Ž(š‘¦) š‘„āˆˆš‘‹

š‘¦āˆˆš‘Œ

That is max min š‘“ (š‘„, š‘¦) ā‰¤ š‘“š‘¦ max š‘“ (š‘„, š‘¦) š‘„

š‘¦

š‘„

162

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3.162 Clearly š‘“ (š‘„) = š‘„š‘Ž his homogeneous of degree š‘Ž. Conversely assume š‘“ is homogeneous of degree š‘Ž, that is š‘“ (š‘”š‘„) = š‘”š‘Ž š‘“ (š‘„) Letting š‘„ = 1 š‘“ (š‘”) = š‘”š‘Ž š‘“ (1) Setting š‘“ (1) = š“ āˆˆ ā„œ and interchanging š‘„ and š‘” yields the result. 3.163 1/šœŒ

š‘“ (š‘”x) = (š‘Ž1 (š‘”š‘„1 )šœŒ + š‘Ž2 (š‘”š‘„2 )šœŒ + . . . š‘Žš‘› (š‘”š‘„š‘› )šœŒ ) 1/šœŒ

= š‘” (š‘Ž1 š‘„šœŒ1 + š‘Ž2 š‘„šœŒ2 + . . . š‘Žš‘› š‘„šœŒš‘› ) = š‘”š‘“ (x) 3.164 For š›½ āˆˆ ā„œ++

ā„Ž(š›½š‘”) = š‘“ (š›½š‘”x0 ) = š›½ š‘˜ š‘“ (š‘”x0 ) = š›½ š‘˜ ā„Ž(š‘”) 3.165 Suppose that xāˆ— minimizes the cost of producing output š‘¦ at prices w. That is wš‘‡ xāˆ— ā‰¤ wš‘‡ x

for every x āˆˆ š‘‰ (š‘¦)

It follows that š‘”wš‘‡ xāˆ— ā‰¤ š‘”wš‘‡ x

for every x āˆˆ š‘‰ (š‘¦)

for every š‘” > 0, verifying that xāˆ— minimizes the cost of producing š‘¦ at prices š‘”w. Therefore š‘(š‘”w, š‘¦) = (š‘”w)xāˆ— = š‘”(wš‘‡ xāˆ— ) = š‘”š‘(w, š‘¦) š‘(w, š‘¦) homogeneous of degree one in input prices w. 3.166 For given prices w, let xāˆ— minimize the cost of producing one unit of output, so that š‘(w, 1) = wš‘‡ xāˆ— . Clearly š‘“ (xāˆ— ) = 1 where š‘“ is the production function. Now consider any output š‘¦. Since š‘“ is homogeneous š‘“ (š‘¦xāˆ— ) = š‘¦š‘“ (xāˆ— ) = š‘¦ Therefore š‘¦xāˆ— is suļ¬ƒcient to produce š‘¦, so that š‘(w, š‘¦) ā‰¤ wš‘‡ (š‘¦xāˆ— ) = š‘¦wš‘‡ xāˆ— = š‘¦š‘(w, 1) Suppose that š‘(w, š‘¦) < wš‘‡ (š‘¦xāˆ— ) = š‘¦š‘(w, 1) Then there exists xā€² such that š‘“ (xā€² ) = š‘¦ and wš‘‡ xā€² < wš‘‡ (š‘¦xāˆ— ) which implies that w

š‘‡

(

xā€² š‘¦

)

< wš‘‡ xāˆ— = š‘(w, 1) 163

Solutions for Foundations of Mathematical Economics Since š‘“ is homogeneous

( š‘“

xā€² š‘¦

) =

c 2001 Michael Carter āƒ All rights reserved

1 š‘“ (xā€² ) = 1 š‘¦

Therefore, xā€² is a lower cost method of producing one unit of output, contradicting the deļ¬nition of xāˆ— . We conclude that š‘(w, š‘¦) = š‘¦š‘(w, 1) š‘(w, š‘¦) is homogeneous of degree one in š‘¦. 3.167 If the consumerā€™s demand is invariant to proportionate changes in all prices and income, so also will the derived utility. More formally, suppose that xāˆ— maximizes utility at prices p and income š‘š, that is xāˆ— ā‰æ x

for every x āˆˆ š‘‹(p, š‘š)

Then š‘£(p, š‘š) = š‘¢(xāˆ— ) Since š‘‹(š‘”p, š‘”š‘š) = š‘‹(p, š‘š) xāˆ— ā‰æ x

for every x āˆˆ š‘‹(š‘”p, š‘”š‘š)

and š‘£(š‘”p, š‘”š‘š) = š‘¢(xāˆ— ) = š‘£(p, š‘š) 3.168 Assume š‘“ is homogeneous of degree one, so that š‘“ (š‘”x) = š‘”š‘“ (x)

for every š‘” > 0

Let (x, š‘¦) āˆˆ epi š‘“ , so that š‘“ (x) ā‰¤ š‘¦ For any š‘” > 0 š‘“ (š‘”x) = š‘”š‘“ (x) ā‰¤ š‘”š‘¦ which implies that (š‘”x, š‘”š‘¦) āˆˆ epi š‘“ . Therefore epi š‘“ is a cone. Conversely assume epi š‘“ is a cone. Let x āˆˆ š‘† and deļ¬ne š‘¦ = š‘“ (x). Then (x, š‘¦) āˆˆ epi š‘“ and therefore (š‘”x, š‘”š‘¦) āˆˆ epi š‘“ so š‘“ (š‘”x) ā‰¤ š‘”š‘¦ Now suppose to the contrary that š‘“ (š‘”x) = š‘§ < š‘”š‘¦ = š‘”š‘“ (x)

(3.58)

Then (š‘”x, š‘§) āˆˆ epi š‘“ . Since epi š‘“ is a cone, we must have (x, š‘§/š‘”) āˆˆ epi š‘“ so that š‘“ (x) ā‰¤

š‘§ š‘”

and š‘”š‘“ (x) ā‰¤ š‘§ = š‘“ (š‘”x) contradicting (3.58). We conclude that š‘“ (š‘”x) = š‘”š‘“ (x) for every š‘” > 0 164

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3.169 Take any x1 and x2 in š‘† and let š‘¦1 = š‘“ (x1 ) > 0 and š‘¦2 = š‘“ (x2 ) > 0 Since š‘“ is homogeneous of degree one, ( ) ( ) x1 x2 š‘“ =š‘“ =1 š‘¦1 š‘¦2 Since š‘“ is quasiconcave

( ) x2 x1 + (1 āˆ’ š›¼) š‘“ š›¼ ā‰„1 š‘¦1 š‘¦2

for every 0 ā‰¤ š›¼ ā‰¤ 1. Choose š›¼ = š‘¦1 /(š‘¦1 + š‘¦2 ) so that (1 āˆ’ š›¼) = š‘¦2 /(š‘¦1 + š‘¦2 ). Then ( ) x1 x2 š‘“ + ā‰„1 š‘¦1 + š‘¦2 š‘¦1 + š‘¦2 Again using the homogeneity of š‘“ , this implies š‘“ (x1 + x2 ) ā‰„ š‘¦1 + š‘¦2 = š‘“ (x1 ) + š‘“ (x2 ) 3.170 Let š‘“ āˆˆ š¹ (š‘†) be a strictly positive deļ¬nite, quasiconcave functional which is homogeneous of degree one. For any x1 , x2 in š‘† and 0 ā‰¤ š›¼ ā‰¤ 1 š›¼x1 , (1 āˆ’ š›¼)x2 in š‘† and therefore š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰„ š‘“ (š›¼x1 ) + š‘“ ((1 āˆ’ š›¼)x2 ) since š‘“ is superadditive (Exercise 3.169). But š‘“ (š›¼x1 ) = š›¼š‘“ (x1 ) š‘“ ((1 āˆ’ š›¼)x2 ) = (1 āˆ’ š›¼)š‘“ (x2 ) by homogeneity. Substituting in (3.58), we conclude that š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) ā‰„ š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ ((1 āˆ’ š›¼)x2 ) š‘“ is concave. 3.171 Assume that š‘“ is strictly positive deļ¬nite, quasiconcave and homogeneous of degree š‘˜, 0 < š‘˜ < 1. Deļ¬ne ā„Ž(x) = (š‘“ (x))

1/š‘˜

Then ā„Ž is quasiconcave (Exercise 3.148. Further, for every š‘” > 0 1/š‘˜

ā„Ž(š‘”x) = (š‘“ (š‘”x)) ( )1/š‘˜ = š‘”š‘˜ š‘“ (x) = š‘” (š‘“ (x))

1/š‘˜

= š‘”ā„Ž(x) so that ā„Ž is homogeneous of degree 1. By Exercise 3.170, ā„Ž is concave. š‘“ (x) = (ā„Ž(x))

š‘˜

That is š‘“ = š‘” āˆ˜ ā„Ž where š‘”(š‘¦) = š‘¦ š‘˜ is monotone and concave provided š‘˜ ā‰¤ 1. By Exercise 3.133, š‘“ = š‘” āˆ˜ ā„Ž is concave. 165

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

3.172 Continuity is a necessary and suļ¬ƒcient condition for the existence of a utility function representing ā‰æ (Remark 2.9). Suppose š‘¢ represents the homothetic preference relation ā‰æ. For any x1 , x2 āˆˆ š‘† š‘¢(x1 ) = š‘¢(x2 ) =ā‡’ x1 āˆ¼ x2 =ā‡’ š‘”x1 āˆ¼ š‘”x2 =ā‡’ š‘¢(š‘”x1 ) = š‘¢(š‘”x2 ) for every š‘” > 0 Conversely, if š‘¢ is a homothetic functional, x1 āˆ¼ x2 =ā‡’ š‘¢(x1 ) = š‘¢(x2 ) =ā‡’ š‘¢(š‘”x1 ) = š‘¢(š‘”x2 ) =ā‡’ š‘”x1 āˆ¼ š‘”x2 for every š‘” > 0 3.173 Suppose that š‘“ = š‘” āˆ˜ ā„Ž where š‘” is strictly increasing and ā„Ž is homogeneous of degree š‘˜. Then ( )1/š‘˜ Ė† ā„Ž(x) = ā„Ž(x) Ė† where is homogeneous of degree one and š‘“ = š‘”Ė† āˆ˜ ā„Ž ( ) š‘”Ė†(š‘¦) = š‘” š‘¦ š‘˜ ) is increasing. 3.174 Assume x1 , x2 āˆˆ š‘† with š‘“ (x1 ) = š‘”(ā„Ž(x1 )) = š‘”(ā„Žx2 )) = š‘“ (x2 ) Since š‘” is strictly increasing, this implies that ā„Ž(x1 ) = ā„Ž(x2 ) Since ā„Ž is homogeneous ā„Ž(š‘”x1 ) = š‘”š‘˜ ā„Ž(x1 ) = š‘”š‘˜ ā„Ž(x2 ) = ā„Ž(š‘”x2 ) for some š‘˜. Therefore š‘“ (š‘”x1 ) = š‘”(ā„Ž(š‘”x1 )) = š‘”(ā„Ž(š‘”x2 )) = š‘“ (š‘”x2 ) 3.175 Let x0 āˆ•= 0 be any point in š‘†, and deļ¬ne š‘” : ā„œ ā†’ ā„œ by š‘”(š›¼) = š‘“ (š›¼x0 ) Since š‘“ is strictly increasing, so is š‘” and therefore š‘” has a strictly increasing inverse š‘” āˆ’1 . Let ā„Ž = š‘” āˆ’1 āˆ˜ š‘“ so that š‘“ = š‘” āˆ˜ ā„Ž. We need to show that ā„Ž is homogeneous. For any x āˆˆ š‘†, there exists š›¼ such that š‘”(š›¼) = š‘“ (š›¼x0 ) = š‘“ (x) that is š›¼ = ā„Ž(x) = š‘” āˆ’1 (š‘“ (x)). Since š‘“ is homothetic š‘”(š‘”š›¼) = š‘“ (š‘”š›¼x0 )š‘“ (š‘”x) for every š‘” > 0 and therefore ā„Ž(š‘”x) = š‘” āˆ’1 (š‘“ (š‘”x)) = š‘” āˆ’1 (š‘“ (š‘”š›¼x0 )) = š‘” āˆ’1 š‘”(š‘”š›¼) = š‘”š›¼ = š‘”ā„Ž(x) ā„Ž is homogeneous of degree one. 166

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

3.176 Let š‘“ be the production function. If š‘“ is homothetic, there exists (Exercise 3.175) a linearly homogeneous function ā„Ž and strictly increasing function š‘” such that š‘“ = š‘” āˆ˜ā„Ž. š‘(w, š‘¦) = min{ wš‘‡ x : š‘“ (x) ā‰„ š‘¦ } x

= min{ wš‘‡ x : š‘”(ā„Ž(x)) ā‰„ š‘¦ } x

= min{ wš‘‡ x : ā„Ž(x) ā‰„ š‘” āˆ’1 (š‘¦) } x

= š‘” āˆ’1 (š‘¦)š‘(w, 1) by Exercise 3.166. 3.177 Let š‘“ : š‘† ā†’ ā„œ be positive, strictly increasing, homothetic and quasiconcave. By Exercise 3.175, there exists a linearly homogeneous function ā„Ž : š‘† ā†’ ā„œ and strictly increasing function š‘” āˆˆ š¹ (š‘…) such that š‘“ = š‘” āˆ˜ ā„Ž. ā„Ž = š‘” āˆ’1 āˆ˜ š‘“ is positive, quasiconcave (Exercise 3.148) and homogeneous of degree one. By Proposition 3.12, ā„Ž is concave and therefore š‘“ = š‘” āˆ˜ ā„Ž is concaviļ¬able. 3.178 Since š»š‘“ (š‘) is a supporting hyperplane to š‘† at x0 , then š‘“ (x0 ) = š‘ and either š‘“ (x) ā‰„ š‘ = š‘“ (x0 ) for every x āˆˆ š‘† or š‘“ (x) ā‰¤ š‘ = š‘“ (x0 ) for every x āˆˆ š‘† 3.179 Suppose to the contrary that y = (ā„Ž, š‘ž) āˆˆ int š“ āˆ© šµ. Then y ā‰æ yāˆ— . By strict convexity yš›¼ = š›¼y + (1 āˆ’ š›¼)yāˆ— ā‰» yāˆ— for every š›¼ āˆˆ (0, 1) Since y āˆˆ int š“, yš›¼ āˆˆ š“ for š›¼ suļ¬ƒciently small. That is, there exists some š›¼ such that yš›¼ is feasible and yš›¼ ā‰» yāˆ— , contradicting the optimality of yāˆ— . 3.180 For notational simplicity, let š‘“ be the linear functional which separates š“ and šµ in Example 3.77. š‘“ (y) measure the cost of the plan y = (ā„Ž, š‘ž), that is š‘“ (y) = š‘¤ā„Ž + š‘š‘ž. Assume to the contrary there exists a preferred lifestyle in š‘‹, that is there exists some y = (ā„Ž, š‘ž) āˆˆ š‘‹ such that y ā‰» yāˆ— = (ā„Žāˆ— , š‘ž āˆ— ). Since y āˆˆ šµ, š‘“ (y) ā‰„ š‘“ (yāˆ— ) by (3.29). On the other hand, y āˆˆ š‘‹ which implies that š‘“ (y) ā‰¤ š‘“ (yāˆ— ). Consequently, š‘“ (y) = š‘“ (yāˆ— ). By continuity, there exists some š›¼ < 1 such that š›¼y ā‰» yāˆ— which implies that š›¼y āˆˆ šµ. By linearity š‘“ (š›¼y) = š›¼š‘“ (y) < š‘“ (y) = š‘“ (yāˆ— ) = š›¼ contrary to (3.29). This contradiction establishes that yāˆ— is the best choice in budget set š‘‹. 3.181 By Proposition 3.7, epi š‘“ is a convex set in š‘‹ Ɨ ā„œ with (x0 , š‘“ (x0 )) a point on its boundary. By Corollary 3.2.2 of the Separating Hyperplane Theorem, there exists linear a functional šœ‘ āˆˆ (š‘‹ Ɨ ā„œ)ā€² such that šœ‘(x, š‘¦) ā‰„ šœ‘(x0 , š‘“ (x0 )) for every (x, š‘¦) āˆˆ epi š‘“ 167

(3.59)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

šœ‘ can be decomposed into two components (Exercise 3.47) šœ‘(x, š‘¦) = āˆ’š‘”(x) + š›¼š‘¦ The assumption that x0 āˆˆ int š‘† ensures that š›¼ > 0 and we can normalize so that š›¼ = 1. Substituting in (3.59) āˆ’š‘”(x) + š‘“ (x) ā‰„ āˆ’š‘”(x0 ) + š‘“ (x0 ) š‘“ (x) ā‰„ š‘“ (x0 ) + š‘”(x āˆ’ x0 ) for every x āˆˆ š‘†. 3.182 By Exercise 3.72, there exists a unique point x0 āˆˆ š‘† such that (x0 āˆ’ y)š‘‡ (x āˆ’ x0 ) ā‰„ 0 for every x āˆˆ š‘† Deļ¬ne the linear functional (Exercise 3.64) š‘“ (x) = (x0 āˆ’ y)š‘‡ x and let š‘ = š‘“ (x0 ). For all x āˆˆ š‘† š‘“ (x) āˆ’ š‘“ (x0 ) = š‘“ (x āˆ’ x0 ) = (x0 āˆ’ y)š‘‡ (x āˆ’ x0 ) ā‰„ 0 and therefore š‘“ (x) ā‰„ š‘“ (x0 ) = š‘ for every x āˆˆ š‘† Furthermore 2

š‘“ (x0 ) āˆ’ š‘“ (y) = š‘“ (x0 āˆ’ y) = (x0 āˆ’ y)š‘‡ (x0 āˆ’ y) = āˆ„x0 āˆ’ yāˆ„ > 0 since y āˆ•= x0 . Therefore š‘“ (x0 ) > š‘“ (y) and š‘“ (y) < š‘ ā‰¤ š‘“ (x) for every x āˆˆ š‘† 3.183 If y āˆˆ b(š‘†), y āˆˆ š‘† š‘ and there exists a sequence of points {yš‘› } āˆˆ š‘† š‘ converging to y (Exercise 1.105). That is, there exists a sequence of nonboundary points {yš‘› } āˆˆ /š‘† converging to y. For every point yš‘› , there is a linear functional š‘” š‘› āˆˆ š‘‹ āˆ— and š‘š‘› such that š‘” š‘› (yš‘› ) < š‘š‘› ā‰¤ š‘” š‘› (x)

for every x āˆˆ š‘†

Deļ¬ne š‘“ š‘› = š‘” š‘› / āˆ„š‘” š‘› āˆ„. By construction, the sequence of linear functionals š‘“ š‘› belong to the unit ball in š‘‹ āˆ— (since āˆ„š‘“ āˆ„ = 1). Since š‘‹ āˆ— is ļ¬nite dimensional, the unit ball is compact as so š‘“ š‘› has a convergent subsequence with limit š‘“ such that š‘“ (y) ā‰¤ š‘“ (x)

for every š‘„ āˆˆ š‘†

š‘“ (y) ā‰¤ š‘“ (x)

for every š‘„ āˆˆ š‘†

A fortiori

3.184 There are two possible cases. yāˆˆ / š‘† By Exercise 3.182, there exists a hyperplane which separates y and š‘† which a fortiori separates y and š‘†, that is š‘“ (y) ā‰¤ š‘“ (x)

for every x āˆˆ š‘† 168

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

y āˆˆ š‘† Since y āˆˆ / š‘†, y must be a boundary point of š‘†. By the previous exercise, there exists a supporting hyperplane at y, that is there exists a continuous linear functional š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (y) ā‰¤ š‘“ (x) 3.185

for every x āˆˆ š‘†

1. š‘“ (š‘†) āŠ† ā„œ.

2. š‘“ (š‘†) is convex and hence an interval (Exercise 1.160. 3. š‘“ (š‘†) is open in ā„œ (Proposition 3.2). 3.186 š‘† is nonempty and convex and 0 āˆˆ / š‘†. (Otherwise, there exists x āˆˆ š“ and y āˆˆ šµ such that 0 = y + (āˆ’x) which implies that x = y contradicting the assumption that š“ āˆ© šµ = āˆ….) Thus there exists a continuous linear functional š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (y āˆ’ x) ā‰„ š‘“ (0) = 0

for every x āˆˆ š“, y āˆˆ šµ

so that š‘“ (x) ā‰¤ š‘“ (y) for every x āˆˆ š“, y āˆˆ šµ Let š‘ = supxāˆˆš“ š‘“ (x). Then š‘“ (x) ā‰¤ š‘ ā‰¤ š‘“ (y) for every x āˆˆ š“, y āˆˆ šµ By Exercise 3.185, š‘“ (int š“) is an open interval in (āˆ’āˆž, š‘], hence š‘“ (int š“) āŠ† (āˆ’āˆž, š‘), so that š‘“ (x) < š‘ for every x āˆˆ int š“. Similarly, š‘“ (int šµ) > š‘ and š‘“ (x) < š‘ < š‘“ (y) for every x āˆˆ int š“, y āˆˆ int šµ 3.187 Since int š“ āˆ© šµ = āˆ…, int š“ and šµ can be separated. That is, there exists a continuous linear functional š‘“ āˆˆ š‘‹ āˆ— and a number š‘ such that š‘“ (x) ā‰¤ š‘ ā‰¤ š‘“ (y)

for every x āˆˆ š“, y āˆˆ int šµ

which implies that š‘“ (x) ā‰¤ š‘ ā‰¤ š‘“ (y)

for every x āˆˆ š“, y āˆˆ šµ

since š‘ā‰¤

inf

yāˆˆint šµ

š‘“ (y) = inf š‘“ (y) yāˆˆšµ

Conversely, suppose that š“ and šµ can be separated. That is, there exists š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (x) ā‰¤ š‘ ā‰¤ š‘“ (y)

for every x āˆˆ š“, y āˆˆ šµ

Then š‘“ (int š“) is an open interval in [š‘, āˆž), which is disjoint from the interval š‘“ (šµ) āŠ† (āˆ’āˆž, š‘]. This implies that int š“ āˆ© šµ = āˆ…. 3.188 Since x0 āˆˆ b(š‘†), {x0 } āˆ© int š‘† = āˆ… and int š‘† āˆ•= āˆ…. By Corollary 3.2.1, {x0 } and š‘† can be separated, that is there exist š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (x0 ) ā‰¤ š‘“ (x) for every x āˆˆ š‘†

169

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c 2001 Michael Carter āƒ All rights reserved

3.189 Let x āˆˆ š¶. Since š¶ is a cone, šœ†x āˆˆ š¶ for every šœ† ā‰„ 0 and therefore š‘“ (šœ†x) ā‰„ š‘ or š‘“ (x) ā‰„ š‘/šœ†

for every šœ† ā‰„ 0

Taking the limit as šœ† ā†’ āˆž implies that š‘“ (x) ā‰„ 0

for every x āˆˆ š¶

3.190 First note that 0 āˆˆ š‘ and therefore š‘“ (0) = 0 ā‰¤ š‘ so that š‘ ā‰„ 0. Suppose that there exists some z āˆˆ š‘ for which š‘“ (z) = šœ– āˆ•= 0. By linearity, this implies š‘“(

2š‘ 2š‘ z) = š‘“ (z) = 2š‘ > š‘ šœ– šœ–

which contradicts the requirement š‘“ (z) ā‰¤ š‘ for every z āˆˆ š‘ 3.191 By Corollary 3.2.1, there exists š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (z) ā‰¤ š‘ ā‰¤ š‘“ (x)

for every x āˆˆ š‘†, z āˆˆ š‘

By Exercise 3.190 š‘“ (z) = 0

for every z āˆˆ š‘

š‘“ (x) ā‰„ 0

for every x āˆˆ š‘†

and therefore

Therefore š‘ is contained in the hyperplane š»š‘“ (0) which separates š‘† from š‘. 3.192 Combining Theorem 3.2 and Corollary 3.2.1, there exists a hyperplane š»š‘“ (š‘) such that š‘“ (x) ā‰¤ š‘ ā‰¤ š‘“ (y)

for every x āˆˆ š“, y āˆˆ šµ

and such that š‘“ (x) < š‘ ā‰¤ š‘“ (y)

for every x āˆˆ int š“, y āˆˆ šµ

Since int š“ āˆ•= āˆ…, there exists some x āˆˆ int š“ with š‘“ (x) < š‘. Hence š“ āŠˆ š‘“ āˆ’1 (š‘) = š»š‘“ (š‘). 3.193 Follows directly from the basic separation theorem, since š“ = int š“ and šµ = int šµ. 3.194 Let š‘† = šµ āˆ’ š“. Then 1. š‘† is a nonempty, closed, convex set (Exercise 1.203). 2. 0 āˆˆ / š‘†. There exists a continuous linear functional š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (x) ā‰„ š‘ > š‘“ (0) = 0 170

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c 2001 Michael Carter āƒ All rights reserved

š“

šµ

Figure 3.2: š“ and šµ cannot be strongly separated. for every z āˆˆ š‘† (Exercise 3.182). For every x āˆˆ š“, y āˆˆ šµ, z = y āˆ’ x āˆˆ š‘† and š‘“ (z) = š‘“ (y) āˆ’ š‘“ (x) ā‰„ š‘ > 0 or š‘“ (x) + š‘ ā‰¤ š‘“ (y) which implies that sup š‘“ (x) + š‘ ā‰¤ inf š‘“ (y) yāˆˆšµ

xāˆˆš“

and sup š‘“ (x) < inf š‘“ (y)

xāˆˆš“

yāˆˆšµ

3.195 No. See Figure 3.2. 3.196

1. Assume that there exists a convex neighborhood š‘ˆ āˆ‹ 0 such that (š“ + š‘ˆ ) āˆ© šµ = āˆ… Then (š“ + š‘ˆ ) is convex and š“ āŠ‚ int (š“ + š‘ˆ ) āˆ•= āˆ… and int (š“ + š‘ˆ ) āˆ© šµ = āˆ…. By Corollary 3.2.1, there exists continuous linear functional such that š‘“ (x + u) ā‰¤ š‘“ (y)

for every x āˆˆ š“, u āˆˆ š‘ˆ, y āˆˆ šµ

Since š‘“ (š‘ˆ ) is an open interval containing 0, there exists some u0 with š‘“ (u0 ) = šœ– > 0. š‘“ (x) + šœ– ā‰¤ š‘“ (y)

for every x āˆˆ š“, y āˆˆ šµ

which implies that sup š‘“ (x) < inf š‘“ (y) yāˆˆšµ

xāˆˆš“

Conversely, assume that š“ and šµ can be strongly separated. That is, there exists a continuous linear functional š‘“ āˆˆ š‘‹ āˆ— and number šœ– > 0 such that š‘“ (x) ā‰¤ š‘ āˆ’ šœ– < š‘ + šœ– ā‰¤ š‘“ (y) for every x āˆˆ š“, y āˆˆ šµ Let š‘ˆ = { š‘„ āˆˆ š‘‹ : āˆ£š‘“ (š‘„)āˆ£ < šœ– }. š‘ˆ is a convex neighborhood of 0 such that (š“ + š‘ˆ ) āˆ© šµ = āˆ…. 171

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c 2001 Michael Carter āƒ All rights reserved

2. Let š“ and šµ be nonempty, disjoint, convex subsets in a normed linear space š‘‹ with š“ compact and šµ closed. By Exercise 1.208, there exists a convex neighborhood š‘ˆ āˆ‹ 0 such that (š“ + š‘ˆ ) āˆ© šµ = āˆ…. By the previous part, š“ and šµ can be strongly separated. 3.197 Assume šœŒ(š“, šµ) = inf{ āˆ„x āˆ’ yāˆ„ : x āˆˆ š“, y āˆˆ šµ } = 2šœ– > 0. Let š‘ˆ = šµšœ– (0) be the open ball around 0 of radius šœ–. For every x āˆˆ š“, u āˆˆ š‘ˆ, y āˆˆ šµ āˆ„x + (āˆ’u) āˆ’ yāˆ„ = āˆ„x āˆ’ y āˆ’ uāˆ„ ā‰„ āˆ„x āˆ’ yāˆ„ āˆ’ āˆ„uāˆ„ so that šœŒ(š“ + š‘ˆ, šµ) = inf āˆ„x + (āˆ’u) āˆ’ yāˆ„ ā‰„ inf (āˆ„x āˆ’ yāˆ„ āˆ’ āˆ„uāˆ„) x,u,y

x,u,y

ā‰„ inf āˆ„x āˆ’ yāˆ„ āˆ’ sup āˆ„uāˆ„) x,y

u

= 2šœ– āˆ’ šœ– =šœ–>0 Therefore (š“ + š‘ˆ ) āˆ© šµ = āˆ… and so š“ and šµ can be strongly separated. Conversely, assume that š“ and šµ can be strongly separated, so that there exists a convex neighborhood š‘ˆ of 0 such that (š“ + š‘ˆ ) āˆ© šµ = āˆ…. Therefore, there exists šœ– > 0 such that šµšœ– (0) āŠ† š‘ˆ and š“ + šµšœ– āˆ© šµ = āˆ… This implies that šœŒ(š“, šµ) = inf{ āˆ„x āˆ’ yāˆ„ : x āˆˆ š“, y āˆˆ šµ } > šœ– > 0 3.198 Take š“ = {y} and šµ = š‘€ in Proposition 3.14. There exists š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (y) < š‘ ā‰¤ š‘“ (x)

for every x āˆˆ š‘€

By Corollary 3.2.3, š‘ = 0. 3.199

1. Consider the set š‘ = { š‘“ (š‘„), āˆ’š‘”1 (š‘„), āˆ’š‘”2 (š‘„), . . . , āˆ’š‘”š‘š (š‘„) : š‘„ āˆˆ š‘‹ } š‘ is the image of a linear mapping from š‘‹ to š‘Œ = ā„œš‘š+1 and hence is a subspace of ā„œš‘š+1 .

2. By hypothesis, the point e0 = (1, 0, 0, . . . , 0) āˆˆ ā„œš‘š+1 does not belong to š‘. Otherwise, we have an š‘„ āˆˆ š‘‹ such that š‘”š‘– (š‘„) = 0 for every š‘– but š‘“ (š‘„) = 1. 3. By the previous exercise, there exists a linear functional šœ‘ āˆˆ š‘Œ āˆ— such that šœ‘(e0 ) > 0 šœ‘(z) = 0

for every z āˆˆ š‘

4. In other words, there exists a vector šœ† = (šœ†0 , šœ†1 , . . . , šœ†š‘š ) āˆˆ š‘Œ = ā„œ(š‘š+1)āˆ— such that šœ†e0 > 0 šœ†z = 0

(3.60) for every z āˆˆ š‘ 172

(3.61)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Equation (3.61) states that šœ†z = šœ†0 z0 + šœ†1 z1 + ā‹… ā‹… ā‹… + šœ†š‘š zš‘š = 0

for every z āˆˆ š‘

That is, for every š‘„ āˆˆ š‘‹, šœ†0 š‘“ (x) āˆ’ šœ†1 š‘”1 (x) āˆ’ šœ†2 š‘”2 (x) āˆ’ . . . āˆ’ šœ†š‘š š‘”š‘š (x) = 0 5. Inequality (3.60) establishes that šœ†0 > 0. Without loss of generality we can normalize so that šœ†0 = 1. 6. Therefore š‘“ (š‘„) =

š‘š āˆ‘

šœ†š‘– š‘”š‘– (š‘„)

š‘–=1

3.200 For every x āˆˆ š‘†, š‘”š‘— (x) = 0, š‘— = 1, 2 . . . š‘š and therefore š‘“ (x) =

š‘š āˆ‘

šœ†š‘– š‘”š‘– (x) = 0

š‘–=1

3.201 The set š‘ = { š‘”1 (š‘„), š‘”2 (š‘„), . . . , š‘”š‘š (š‘„) : š‘„ āˆˆ š‘‹ } is a closed subspace in ā„œš‘š . If the system is inconsistent, c = (š‘1 , š‘2 , . . . , š‘š‘š ) āˆˆ / š‘. By Exercise 3.198, there exists a linear functional šœ‘ on ā„œš‘š such šœ‘(z) = 0 for every z āˆˆ š‘ šœ‘(c) > 0 That is, there exist numbers šœ†1 , šœ†2 , . . . , šœ†š‘š such that š‘š āˆ‘

šœ†š‘— š‘”š‘— (x) = 0

š‘—=1

and š‘š āˆ‘

šœ†š‘— š‘š‘— > 0

š‘—=1

which contradicts the hypothesis š‘š āˆ‘

šœ†š‘— š‘”š‘— = 0 =ā‡’

š‘—=1

š‘š āˆ‘

šœ†š‘— š‘š‘— = 0

š‘—=1

Conversely, if for some x āˆˆ š‘‹ š‘”š‘— (x) = š‘š‘—

š‘— = 1, 2, . . . , š‘š

then š‘š āˆ‘

šœ†š‘— š‘”š‘— (x) =

š‘—=1

š‘š āˆ‘

šœ†š‘— š‘š‘—

š‘—=1

and š‘š āˆ‘

šœ†š‘— š‘”š‘— = 0 =ā‡’

š‘—=1

š‘š āˆ‘ š‘—=1

173

šœ†š‘— š‘š‘— = 0

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Ė† = { x āˆˆ š¾ : āˆ„xāˆ„ = 1 } is 3.202 The set š¾ 1 āˆ™ compact (the unit ball is compact if and only if š‘‹ is ļ¬nite-dimensional) āˆ™ convex (which is why we need the 1 norm) By Proposition 3.14, there exists a linear functional š‘“ āˆˆ š‘‹ āˆ— such that Ė† Ė†āˆˆš¾ for every x for every x āˆˆ š‘€

š‘“ (Ė† x) > 0 š‘“ (x) = 0

Ė† Then Ė† = x/ āˆ„xāˆ„1 āˆˆ š¾. For any x āˆˆ š¾, x āˆ•= 0, deļ¬ne x Ė† ) = āˆ„xāˆ„1 š‘“ (Ė† š‘“ (x) = š‘“ (āˆ„xāˆ„1 x x) > 0 3.203

1. Let š“ = { (x, š‘¦) : š‘¦ ā‰„ š‘”(x), x āˆˆ š‘‹ } šµ = { (x, š‘¦) : š‘¦ = š‘“0 (x), x āˆˆ š‘ } š“ is the epigraph of a convex functional and hence convex. šµ is a subspace of š‘Œ = š‘‹ Ɨ ā„œ and also convex.

2. Since š‘” is convex, int š“ āˆ•= āˆ…. Furthermore š‘“0 (x) ā‰¤ š‘”(x) =ā‡’ int š“ āˆ© šµ = āˆ… 3. By Exercise 3.2.3, there exists linear functional šœ‘ āˆˆ š‘Œ āˆ— such that šœ‘(x, š‘¦) ā‰„ 0 šœ‘(x, š‘¦) = 0

for every (x, š‘¦) āˆˆ š“ for every (x, š‘¦) āˆˆ šµ

There exists š‘¦ such that š‘¦ > š‘”(0) and therefore (0, š‘¦) āˆˆ int š“ and šœ‘(0, š‘¦) > 0. Therefore šœ‘(0, 1) =

1 šœ‘(0, 1) > 0 š‘¦

4. Let š‘“ āˆˆ š‘‹ āˆ— be deļ¬ned by 1 š‘“ (x) = āˆ’ šœ‘(x, 0) š‘ where š‘ = šœ‘(0, 1). Since šœ‘(x, 0) = šœ‘(x, š‘¦) āˆ’ šœ‘(0, š‘¦) = šœ‘(x, š‘¦) āˆ’ š‘š‘¦ 1 1 š‘“ (x) = āˆ’ (šœ‘(x, š‘¦) āˆ’ š‘š‘¦) = āˆ’ šœ‘(x, š‘¦) + š‘¦ š‘ š‘ for every š‘¦ āˆˆ ā„œ 5. For every x āˆˆ š‘ 1 š‘“ (x) = āˆ’ šœ‘(x, š‘“0 (x)) + š‘“0 (x) š‘ = š‘“0 (x) since šœ‘(x, š‘“0 (x)) = 0 for every x āˆˆ š‘. Thus š‘“ is an extension of š‘“0 . 174

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

6. For any x āˆˆ š‘‹, let š‘¦ = š‘”(x). Then (x, š‘¦) āˆˆ š“ and šœ‘(x, š‘¦) ā‰„ 0. Therefore 1 š‘“ (x) = āˆ’ šœ‘(x, š‘¦) + š‘¦ š‘ 1 = āˆ’ šœ‘(x, š‘¦) + š‘”(x) š‘ ā‰¤ š‘”(x) Therefore š‘“ is bounded by š‘” as required. 3.204 Let š‘” āˆˆ š‘‹ āˆ— be deļ¬ned by š‘”(x) = āˆ„š‘“0 āˆ„š‘ āˆ„xāˆ„ Then š‘“0 (x) ā‰¤ š‘”(x) for all x āˆˆ š‘. By the Hahn-Banach theorem (Exercise 3.15), there exists an extension š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (x) ā‰¤ š‘”(x) = āˆ„š‘“0 āˆ„š‘ āˆ„xāˆ„ Therefore āˆ„š‘“ āˆ„š‘‹ = sup āˆ„š‘“ (x)āˆ„ = āˆ„š‘“0 āˆ„š‘ āˆ„xāˆ„=1

3.205 If x0 = 0, any bounded linear functional will do. Therefore, assume x0 āˆ•= 0. On the subspace lin {x0 } = {š›¼x0 : š›¼ āˆˆ ā„œ}, deļ¬ne the function š‘“0 (š›¼x0 ) = š›¼ āˆ„x0 āˆ„ š‘“0 is a bounded linear functional on lin {x0 } with norm 1. By the previous part, š‘“0 can be extended to a bounded linear functional š‘“ āˆˆ š‘‹ āˆ— with the same norm, that is āˆ„š‘“ āˆ„ = 1 and š‘“ (x0 ) = āˆ„x0 āˆ„. 3.206 Since x1 āˆ•= x2 , x1 āˆ’ x2 āˆ•= 0. There exists a bounded linear functional such that š‘“ (x1 āˆ’ x2 ) = āˆ„x1 āˆ’ x2 āˆ„ āˆ•= 0 so that š‘“ (x1 ) āˆ•= š‘“ (x2 ) 3.207

1.

āˆ™ š”‰ is a complete lattice (Exercise 1.179).

āˆ™ The intersection of any chain is ā€“ nonempty (since š‘† is compact) ā€“ a face (Exercise 1.179) Hence every chain has a minimal element. āˆ™ By Zornā€™s lemma (Remark 1.5), š”‰ has a minimal element š¹0 . 2. Assume to the contrary that š¹0 contains two distinct elements x1 , x2 . Then (Exercise 3.206) there exists a continuous linear functional š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (x1 ) āˆ•= š‘“ (x2 ) Let š‘ be in the minimum value of š‘“ (x) on š¹0 and let š¹1 be the set on which it attains this minimum. (Since š¹0 is compact, š‘ is well-deļ¬ned and š¹1 is nonempty. That is š‘ = min{ š‘“ (š‘„) : x āˆˆ š¹0 } š¹1 = { x āˆˆ š‘† : š‘“ (x) = š‘ } 175

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Now š¹1 āŠ‚ š¹0 since š‘“ (x1 ) āˆ•= š‘“ (x2 ). To show that š¹1 is a face of š¹0 , assume that š›¼x+(1āˆ’š›¼)y āˆˆ š¹1 for some x, y āˆˆ š¹0 . Then š‘ = š‘“ (š›¼x + (1 āˆ’ š›¼)y) = š›¼š‘“ (x) + (1 āˆ’ š›¼)š‘“ (y) = š‘. Since x, y āˆˆ š¹0 , this implies that š‘“ (x) = š‘“ (y) = š‘ so that x, y āˆˆ š¹1 . Therefore š¹1 is a face. We have shown that, if š¹0 contains two distinct elements, there exists a smaller face š¹1 āŠ‚ š¹0 , contradicting the minimality of š¹0 . We conclude that š¹0 comprises a single element x0 . 3. š¹0 = {x0 } which is an extreme point of š‘†. 3.208 Let š» = š»š‘“ (š‘) be a supporting hyperplane to š‘†. Without loss of generality assume š‘“ (x) ā‰¤ š‘ for every x āˆˆ š‘†

(3.62)

and there exists some xāˆ— āˆˆ š‘† such that š‘“ (xāˆ— ) = š‘ That is š‘“ is maximized at xāˆ— . Version 1 By the previous exercise, š‘“ achieves its maximum at an extreme point. That is, there exists an extreme point x0 āˆˆ š‘† such that š‘“ (x0 ) ā‰„ š‘“ (x) for every x āˆˆ š‘† In particular, š‘“ (x0 ) ā‰„ š‘“ (xāˆ— ) = š‘. But (3.62) implies š‘“ (x0 ) ā‰¤ š‘. Therefore, we conclude that š‘“ (x0 ) = š‘ and therefore x0 āˆˆ š». Version 2 The set š» āˆ© š‘† is a nonempty, compact, convex subset of a linear space. Hence, by Exercise 3.207, š» āˆ© š‘† contains an extreme point, say x0 . We show that x0 is an extreme point of š‘†. Assume not, that is assume that there exists x1 , x2 āˆˆ š‘† such that x0 = š›¼x1 + (1 āˆ’ š›¼)x2 for some š›¼ āˆˆ (0, 1). Since x0 is an extreme point of š» āˆ© š‘†, at least one of the points x1 , x2 must lie outside š». Assume x1 āˆˆ / š» which implies that š‘“ (x1 ) < š‘. Since š‘“ (x2 ) ā‰¤ š‘ š‘“ (x0 ) = š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) < š‘

(3.63)

However, since x0 āˆˆ š» āˆ© š‘†, we must have š‘“ (x0 ) = š‘ which contradicts (3.63). Therefore x0 is an extreme point of š‘†. In fact, we have shown that every extreme point of š» āˆ© š‘† must be an extreme point of š‘†. 3.209 Let š‘†Ė† denote the closed, convex hull of the extreme points of š‘†. (The closed, convex hull of a set is simply the closure of the convex hull.) Clearly š‘†Ė† āŠ‚ š‘† and it remains to show that š‘†Ė† contains all of š‘†. Ė† By the Strong Separation Assume not. That is, assume š‘†Ė† āŠŠ š‘† and let x0 āˆˆ š‘† āˆ– š‘†. Theorem, there exists a linear functional š‘“ āˆˆ š‘‹ āˆ— such that š‘“ (x0 ) > š‘“ (x) for every x āˆˆ š‘†Ė† 176

(3.64)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

On the other hand, by Exercise 3.16, š‘“ attains its maximum at an extreme point of š‘†. That is, there exists x1 āˆˆ š‘†Ė† such that š‘“ (x1 ) ā‰„ š‘“ (x) for every x āˆˆ š‘† In particular š‘“ (x1 ) ā‰„ š‘“ (x0 ) Ė† since x0 āˆˆ š‘†Ė† āŠ‚ š‘†. This contradicts (3.64) since x1 āˆˆ š‘†. Thus our assumption that š‘† āŠŠ š‘†Ė† yields a contradiction. We conclude that š‘† = š‘†Ė† 3.210

1. (a) š‘ƒ is compact and convex, since it is the product of compact, convex sets (Proposition 1.2, Exercise 1.165). āˆ‘š‘› āˆ‘š‘› (b) Since x āˆˆ š‘–=1 conv š‘†š‘– , there exist xš‘– āˆˆ conv š‘†š‘– such that x = š‘–=1 xš‘– . (x1 , x2 , . . . , xš‘› ) āˆˆ š‘ƒ (x) so that š‘ƒ (x) āˆ•= āˆ…. (c) By the Krein-Millman theorem (or Exercise 3.207), š‘ƒ (x) has an extreme point z = (z1 , z2 , . . . , zš‘› ) such that āˆ™ zš‘– āˆˆ conv š‘†š‘– for every š‘– āˆ‘š‘› āˆ™ š‘–=1 zš‘– = x. since z āˆˆ š‘ƒ (x).

2. (a) Exercise 1.176 (b) Since š‘™ > š‘š = dim š‘‹, the vectors y1 , y2 , . . . , yš‘™ are linearly dependent (Exercise 1.143). Consequently, there exists numbers š›¼ā€²1 , š›¼ā€²2 , . . . , š›¼ā€²š‘™ , not all zero, such that š›¼ā€²1 y1 + š›¼ā€²2 y2 + ā‹… ā‹… ā‹… + š›¼ā€²š‘™ yš‘™ = 0 (Exercise 1.133). Let š›¼š‘– =

š›¼ā€²š‘– maxš‘– āˆ£š›¼š‘– āˆ£

Then āˆ£š›¼š‘– āˆ£ ā‰¤ 1 for every š‘– and š›¼1 y1 + š›¼2 y2 + ā‹… ā‹… ā‹… + š›¼š‘™ yš‘™ = 0 (c) Since āˆ£š›¼š‘– āˆ£ ā‰¤ 1, zš‘– + š›¼š‘– yš‘– āˆˆ conv š‘†š‘– for every š‘– = 1, 2, . . . , š‘™. Furthermore š‘› āˆ‘ š‘–=1

z+ š‘– =

š‘› āˆ‘

zš‘– +

š‘–=1

š‘™ āˆ‘

š›¼š‘– yš‘– =

š‘–=1

š‘› āˆ‘

zš‘– = x

š‘–=1

Therefore, z+ āˆˆ š‘ƒ (x). Similarly, zāˆ’ āˆˆ š‘ƒ (x). (d) By direct computation z=

1 + 1 āˆ’ z + z 2 2

which implies that z is not an extreme point of š‘ƒ (x), contrary to our assumption. This establishes that at least š‘› āˆ’ š‘š zš‘– are extreme points of the corresponding conv š‘†š‘– . 177

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Solutions for Foundations of Mathematical Economics

4 3 (0, 2.5)

conv š‘†2 2

(.5, 2) P(x)

1 0

1

2

3

4

conv š‘†1 Figure 3.3: Illustrating the proof of the Shapley Folkman theorem. 3. Every extreme point of conv š‘†š‘– is an element of š‘†š‘– . 3.211 See Figure 3.3. 3.212 Let {š‘†1 , š‘†2 , . . . , š‘†š‘› } be a āˆ‘ collection of subsets of an š‘š-dimensional āˆ‘nonempty š‘› š‘› linear space and let x āˆ‘ āˆˆ conv š‘† = conv š‘† . That is, there exists xš‘– āˆˆ š‘– š‘– š‘–=1 š‘–=1 conv š‘†š‘– such that x = š‘›š‘–=1 xš‘– . By CarathĀ“eodoryā€™s theorem, there exists for every xš‘– a ļ¬nite number of points xš‘–1 , xš‘–2 , . . . , xš‘–š‘™š‘– such that xš‘– āˆˆ conv {xš‘–1 , xš‘–2 , . . . , xš‘–š‘™š‘– }. For every š‘– = 1, 2, . . . , š‘›, let š‘†Ėœš‘– = { xš‘–š‘— : š‘— = 1, 2, . . . , š‘™š‘– } Then x=

š‘› āˆ‘

xš‘– ,

xš‘– āˆˆ conv š‘†Ėœš‘–

š‘–=1

āˆ‘ āˆ‘Ėœ š‘†š‘– . Moreover, the sets š‘†š‘– are compact (in fact That is, x āˆˆ conv š‘†Ėœš‘– = conv ļ¬nite). By the previous exercise, there exists š‘› points zš‘– āˆˆ š‘†Ėœš‘– such that x=

š‘› āˆ‘

zš‘– ,

zš‘– āˆˆ conv š‘†Ėœš‘–

š‘–=1

and moreover zš‘– āˆˆ š‘†Ėœš‘– āŠ† š‘†š‘– for at least š‘› āˆ’ š‘š indices š‘–. 3.213 Let š‘† be a closed convex set in a normed linear space. Clearly, š‘† is contained in the intersection of all the closed halfspaces which contain š‘†. For any y āˆˆ / š‘†, there exists a hyperplane which strongly separates {y} and š‘†. One of its closed halfspaces contains š‘† but not y. Consequently, y does not belong to the intersection of all the closed halfspaces containing š‘†. 3.214

1. Since š‘‰ āˆ— (š‘¦) is the intersection of closed, convex sets, it is closed and convex. Assume x is feasible, that is x āˆˆ š‘‰ (š‘¦). Then wš‘‡ x ā‰¤ (š‘w, š‘¦) and x āˆˆ š‘‰ āˆ— (š‘¦). That is, š‘‰ (š‘¦) āŠ† š‘‰ āˆ— (š‘¦).

178

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

2. Assume š‘‰ (š‘¦) is convex. For any x0 āˆˆ / š‘‰ (š‘¦) there exists w such that wš‘‡ x0 <

inf wš‘‡ x = š‘(w, š‘¦)

xāˆˆš‘‰ (š‘¦)

by the Strong Separation Theorem. Monotonicity ensures that w ā‰„ 0 and hence x0 āˆˆ / š‘‰ āˆ— (š‘¦). 3.215 Assume x āˆˆ š‘‰ (š‘¦) = š‘‰ āˆ— (š‘¦). That is š‘(w) for every x wš‘‡ x ā‰„ š‘¦Ė† Therefore, for any š‘” āˆˆ ā„œ+ š‘”wš‘‡ x ā‰„ š‘”š‘¦š‘(w) for every x which implies that š‘”x āˆˆ š‘‰ āˆ— (š‘¦) = š‘‰ (š‘¦). 3.216 A polyhedron š‘† = { š‘„ āˆˆ š‘‹ : š‘”š‘– (š‘„) ā‰¤ š‘š‘– , š‘– = 1, 2, . . . , š‘š } š‘š āˆ© = { x āˆˆ š‘‹ : š‘”š‘– (x) ā‰¤ š‘š‘– } š‘–=1

is the intersection of a ļ¬nite number of closed convex sets. 3.217 Each row aš‘– = (š‘Žš‘–1 , š‘Žš‘–2 , . . . š‘Žš‘–š‘› ) of š“ deļ¬nes a linear functional š‘”š‘– (x) = š‘Žš‘–1 š‘„1 + š‘Žš‘–2 š‘„2 + ā‹… ā‹… ā‹… + š‘Žš‘–š‘› š‘„š‘› on ā„œš‘› . The set š‘† of solutions to š“x ā‰¤ c is š‘† = { š‘„ āˆˆ š‘‹ : š‘”š‘– (x) ā‰¤ š‘š‘– , š‘– = 1, 2, . . . , š‘š } is a polyhedron. 3.218 For simplicity, we assume that the game is superadditive, so that š‘¤(š‘–) ā‰„ 0 for every š‘–. Consequently, in every core allocation x, 0 ā‰¤ š‘„š‘– ā‰¤ š‘¤(š‘ ) and core āŠ† [0, š‘¤(š‘ )] Ɨ [0, š‘¤(š‘ )] Ɨ ā‹… ā‹… ā‹… Ɨ [0, š‘¤(š‘ )] āŠ‚ ā„œš‘› Thus, the core is bounded. Since it is the intersection of closed halfspaces, the core is also closed. By Proposition 1.1, the core is compact. 3.219 polytope =ā‡’ polyhedron Assume that š‘ƒ is a polytope generated by the points { x1 , x2 , . . . , xš‘š } and let š¹1 , š¹2 , . . . , š¹š‘˜ denote the proper faces of š‘ƒ . For each š‘– = 1, 2, . . . , š‘˜, let š»š‘– denote the hyperplane containing š¹š‘– so that š¹š‘– = š‘ƒ āˆ© š»š‘– . For every such hyperplane, there exists a nonzero linear functional š‘”š‘– and constant š‘š‘– such that š‘”š‘– (x) = š‘š‘– for every x āˆˆ š»š‘– . Furthermore, every such hyperplane is a bounding hyperplane of š‘ƒ . Without loss of generality, we can assume that š‘”š‘– (x) ā‰¤ š‘ for every x āˆˆ š‘ƒ . Let š‘† = { x āˆˆ š‘‹ : š‘”š‘– (x) ā‰¤ š‘š‘– , š‘– = 1, 2, . . . , š‘š } Clearly š‘ƒ āŠ† š‘†. To show that š‘† āŠ† š‘ƒ , assume not. That is, assume that there exists y āˆˆ š‘† āˆ–š‘ƒ and let x āˆˆ ri š‘ƒ . (ri š‘ƒ is nonempty by exercise 1.229). Since š‘ƒ is ĀÆ = š›¼x+(1āˆ’š›¼)y belongs closed (Exercise 1.227), there exists a some š›¼ such that x ĀÆ āˆˆ š¹š‘– āŠ† š»š‘– . to the relative boundary of š‘ƒ , and there exists some š‘– such that x Let š»š‘–+ = { x āˆˆ š‘‹ : š‘”š‘– (x) ā‰¤ š‘š‘– } denote the closed half-space bounded by š»š‘– and ĀÆ = š›¼x+(1āˆ’š›¼)y, which implies that containing š‘ƒ . š»š‘– is a face of š»š‘–+ containing x x, y āˆˆ š»š‘– . This in turn implies that x āˆˆ š¹š‘– , which contradicts the assumption that x āˆˆ ri š‘ƒ . We conclude that š‘† = š‘ƒ . 179

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polyhedron =ā‡’ polytope Conversely, assume š‘† is a nonempty compact polyhedral set in a normed linear space. Then, there exist linear functionals š‘”1 , š‘”2 , . . . , š‘”š‘š in š‘‹ āˆ— and numbers š‘1 , š‘2 , . . . , š‘š‘š such that š‘† = { x āˆˆ š‘‹ : š‘”š‘– (x) ā‰¤ š‘š‘– , š‘– = 1, 2, . . . , š‘š }. We show that š‘† has a ļ¬nite number of extreme points. Let š‘› denote the dimension of š‘†. If š‘› = 1, š‘† is either a single point or closed line segment (since š‘† is compact), and therefore has a ļ¬nite number of extreme points (that is, 1 or 2). Now assume that every compact polyhedral set of dimension š‘› āˆ’ 1 has a ļ¬nite number of extreme points. Let š»š‘– , š‘– = š‘– = 1, 2, . . . , š‘š denote the hyperplanes associated with the linear functionals š‘”š‘– deļ¬ning š‘† (Exercise 3.49). Let x be an extreme point of š‘†. Then š‘† is a boundary point of š‘† (Exercise 1.220) and therefore belongs to some š»š‘— . We claim that x is also an extreme point of the set š‘† āˆ© š»š‘— . To see this, assume otherwise. That is, assume that x is not an extreme point of š‘† āˆ©š»š‘— . Then, there exists x1 , x2 āˆˆ š‘† āˆ©š»š‘— such that x = š›¼x1 + (1 āˆ’ š›¼)x2 . But then x1 , x2 āˆˆ š‘† and x is not an extreme point of š‘†. Therefore, every extreme point of š‘† is an extreme point of some š‘† āˆ© š»š‘– , which is a compact polyhedral set of dimension š‘› āˆ’ 1. By hypothesis, each š‘† āˆ© š»š‘– has a ļ¬nite number of extreme points. Since there are only š‘š such hyperplanes š»š‘– , š‘† has a ļ¬nite number of extreme points. By the Krein-Milman theorem (Exercise 3.209), š‘† is the closed convex hull of its extreme points. Since there are only ļ¬nite extreme points, š‘† is a polytope. 3.220

1. Let š‘“, š‘” āˆˆ š‘† āˆ— so that š‘“ (x) ā‰¤ 0 and š‘”(x) ā‰¤ 0 for every x āˆˆ š‘†. For every š›¼, š›½ ā‰„ 0 š›¼š‘“ (x) + š›½š‘“ (x) ā‰¤ 0 for every š‘„ āˆˆ š‘†. This shows that š›¼š‘“ + š›½š‘” āˆˆ š‘† āˆ— . š‘† āˆ— is a convex cone. To show that š‘† āˆ— is closed, let š‘“ be the limit of a sequence (š‘“š‘› ) of functionals in š‘† āˆ— . Then, for every x āˆˆ š‘†, š‘“š‘› (x) ā‰¤ 0 so that š‘“ (x) = lim š‘“š‘› (x) ā‰¤ 0

2. Let x, y āˆˆ š‘† āˆ—āˆ— . Then, for every š‘“ āˆˆ š‘† āˆ— š‘“ (x) ā‰¤ 0 and š‘“ (y) ā‰¤ 0 and therefore š‘“ (š›¼x + š›½y) = š›¼š‘“ (x) + š›½š‘“ (y) ā‰¤ 0 for every š›¼, š›½ ā‰„ 0. There š›¼x + š›½y āˆˆ š‘† āˆ—āˆ— . š‘† āˆ—āˆ— is a convex cone. To show that š‘† āˆ—āˆ— is closed, let xš‘› be a sequence of points in š‘† āˆ—āˆ— converging to š‘„. For every š‘› = 1, 2, . . . š‘“ (xš‘› ) ā‰¤ 0 for every š‘“ āˆˆ š‘† āˆ— By continuity š‘“ (x) = lim š‘“ (xš‘› ) ā‰¤ 0 for every š‘“ āˆˆ š‘† āˆ— Consequently x āˆˆ š‘† āˆ—āˆ— which is therefore closed. 180

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3. Let x āˆˆ š‘†. Then š‘“ (x) ā‰¤ 0 for every š‘“ āˆˆ š‘† āˆ— so that x āˆˆ š‘† āˆ—āˆ— . 4. Exercise 1.79. 3.221 Let š‘“ āˆˆ š‘†2āˆ— . Then š‘“ (x) ā‰¤ 0 for every x āˆˆ š‘†2 . A fortiori, since š‘†1 āŠ† š‘†2 , š‘“ (x) ā‰¤ 0 for every x āˆˆ š‘†1 . Therefore š‘“ āˆˆ š‘†1āˆ— . 3.222 Exercise 3.220 showed that š‘† āŠ† š‘† āˆ—āˆ— . To show the converse, let y āˆˆ / š‘†. By Proposition 3.14, there exists some š‘“ āˆˆ š‘‹ āˆ— and š‘ such that š‘“ (y) > š‘ š‘“ (x) < š‘

for every x āˆˆ š‘†

Since š‘† is a cone, 0 āˆˆ š‘† and š‘“ (0) = 0 < š‘. Since š›¼š‘† = š‘† for every š›¼ > 0 then š‘“ (x) < 0

for every x āˆˆ š‘†

/ š‘† āˆ—āˆ— . That is so that š‘“ āˆˆ š‘† āˆ— . š‘“ (y) > 0, y āˆˆ yāˆˆ / š‘† =ā‡’ y āˆˆ / š‘† āˆ—āˆ— from which we conclude that š‘† āˆ—āˆ— āŠ† š‘†. 3.223 Let š¾ = cone {š‘”1 , š‘”2 , . . . , š‘”š‘š } š‘š āˆ‘ = { š‘” āˆˆ š‘‹āˆ— : š‘” = šœ†š‘— š‘”š‘— , šœ†š‘— ā‰„ 0 } š‘—=1

be the set of all nonnegative linear combinations of the linear functionals š‘”š‘— . š¾ is a closed convex cone. Suppose that š‘“ āˆˆ / cone {š‘”1 , š‘”2 , . . . , š‘”š‘š }, that is assume that š‘“ āˆˆ / š¾. Then {š‘“ } is a compact convex set disjoint from š¾. By Proposition 3.14, there exists a continuous linear functional šœ‘ and number š‘ such that sup šœ‘(š‘”) < š‘ < šœ‘(š‘“ )

š‘”āˆˆš¾

Since 0 āˆˆ š¾, š‘ ā‰„ 0 and so šœ‘(š‘“ ) > 0. Further, for every š‘” āˆˆ šŗ š‘š āˆ‘ šœ†š‘— š‘”š‘— ) šœ‘(š‘”) = šœ‘( š‘—=1

=

š‘š āˆ‘

šœ†š‘— šœ‘(š‘”š‘— ) < š‘ for every šœ†š‘— ā‰„ 0

š‘—=1

Since šœ†š‘— can be made arbitrarily large, this last inequality implies that šœ‘(š‘”š‘— ) ā‰¤ 0

š‘— = 1, 2, . . . , š‘š

By the Riesz representation theorem (Exercise 3.75), there exists x āˆˆ š‘‹ šœ‘(š‘”š‘— ) = š‘”š‘— (x) and šœ‘(š‘“ ) = š‘“ (x) Since šœ‘(š‘”š‘— ) = š‘”š‘— (x) ā‰¤ 0 181

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x āˆˆ š‘†. By hypothesis š‘“ (x) = šœ‘(š‘“ ) ā‰¤ 0 contradicting the conclusion that šœ‘(š‘“ ) > 0. This contradiction establishes that š‘“ āˆˆ š¾, that is š‘“ (š‘„) =

š‘š āˆ‘

šœ†š‘— š‘”š‘— (š‘„),

šœ†š‘— ā‰„ 0

š‘—=1

3.224 Let a1 , a2 , . . . , aš‘š denote the rows of š“ and deļ¬ne the linear functional š‘“, š‘”1 , š‘”2 , . . . , š‘”š‘š by š‘“ (x) = cx š‘”š‘— (x) = aš‘— x š‘— = 1, 2, . . . , š‘š Assume cx ā‰¤ 0 for every x satisfying š“x ā‰¤ 0, that is š‘“ (x) ā‰¤ 0 for every x āˆˆ š‘† where š‘† = { x āˆˆ š‘‹ : š‘”š‘— (x) ā‰¤ 0, š‘— = 1, 2, . . . , š‘š } By Proposition 3.18, there exists y āˆˆ ā„œš‘š + such that š‘“ (x) =

š‘š āˆ‘

š‘¦š‘— š‘”š‘— (x)

š‘—=1

or c=

š‘š āˆ‘

š‘¦š‘— aš‘— = š“š‘‡ y

š‘—=1

Conversely, assume that c = š“š‘‡ y =

š‘š āˆ‘

š‘¦š‘— aš‘—

š‘—=1

Then š“x ā‰¤ 0 =ā‡’ aš‘— x ā‰¤ 0 for every š‘— =ā‡’ cx ā‰¤ 0 3.225 Let š‘ = ā„œš‘›+ denote the positive orthant of ā„œš‘› . š‘ is a convex set (indeed cone) with a nonempty interior. By Corollary 3.2.1, there exists a hyperplane š»p (š‘) such that pš‘‡ x ā‰¤ š‘ ā‰¤ py

for every x āˆˆ š‘†, y āˆˆ š‘

Since 0 āˆˆ š‘ p0 = 0 ā‰„ š‘ which implies that š‘ ā‰¤ 0 and pš‘‡ x ā‰¤ š‘ ā‰¤ 0

for every š‘„ āˆˆ š‘†

To show that p is nonnegative, let e1 , e2 , . . . , eš‘› denote the standard basis for ā„œš‘› . Each eš‘– belongs to š‘ so that peš‘– = š‘š‘– ā‰„ 0 182

for every š‘–

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3.226 Assume yāˆ— is an eļ¬ƒcient production plan in š‘Œ and let š‘† = š‘Œ āˆ’ š‘¦ āˆ— . š‘† is convex. We claim that š‘† āˆ©ā„œš‘›++ = āˆ…. Otherwise, if there exists some z āˆˆ š‘† āˆ©ā„œš‘›++ , let yā€² = yāˆ— +z āˆ™ z āˆˆ š‘† implies yā€² āˆˆ š‘Œ while āˆ™ z āˆˆ ā„œš‘›++ implies yā€² > yāˆ— contradicting the eļ¬ƒciency of yāˆ— . Therefore, š‘† is a convex set which contains no interior points of the nonnegative orthant ā„œš‘›+ . By Exercise 3.225, there exists a price system p such that pš‘‡ x ā‰¤ 0 for every x āˆˆ š‘† Since š‘† = š‘Œ āˆ’ š‘¦ āˆ— , this implies p(y āˆ’ yāˆ— ) ā‰¤ 0 for every y āˆˆ š‘Œ or pyāˆ— ā‰„ py for every y āˆˆ š‘Œ š‘¦ āˆ— maximizes the producerā€™s proļ¬t at prices p. 3.227 Consider the set š‘† āˆ’ = { x āˆˆ ā„œš‘› : āˆ’x āˆˆ š‘† }. š‘† āˆ© int ā„œš‘›āˆ’ = āˆ… =ā‡’ š‘† āˆ’ āˆ© int ā„œš‘›+ = āˆ… From the previous exercise, there exists a hyperplane with nonnegative normal p ā‰© 0 such that pš‘‡ x ā‰¤ 0

for every x āˆˆ š‘† āˆ’

pš‘‡ x ā‰„ 0

for every x āˆˆ š‘†

Since p ā‰© 0, this implies

3.228

1. Suppose x āˆˆ ā‰æ(xāˆ— ). Then, there exists an allocation (x1 , x2 , . . . , xš‘› ) such that x=

š‘› āˆ‘

xš‘–

š‘–=1

where xš‘– āˆˆ ā‰æ(xāˆ—š‘– ) for every š‘– = 1, 2, . . . , š‘›. Conversely, if (x1āˆ‘ , x2 , . . . , xš‘› ) is an š‘› allocation with xš‘– āˆˆ ā‰æ(xāˆ—š‘– ) for every š‘– = 1, 2, . . . , š‘›, then x = š‘–=1 xš‘– āˆˆ ā‰æ(xāˆ— ). 2. For every agent š‘–, xāˆ—š‘– āˆˆ ā‰æ(xāˆ—š‘– ), which implies that xāˆ— =

š‘› āˆ‘

xāˆ—š‘– āˆˆ ā‰æ(xāˆ— )

š‘–=1

and therefore 0 āˆˆ š‘† = ā‰æ(xāˆ— ) āˆ’ xāˆ— āˆ•= āˆ… Since individual preferences are convex, ā‰æ(xāˆ—š‘– ) is convex for each š‘– and therefore āˆ‘ āˆ— āˆ— āˆ— š‘† = ā‰æ(x ) āˆ’ x = š‘– ā‰æ(xš‘– ) āˆ’ xāˆ— is convex (Exercise 1.164).

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Assume to the contrary that š‘† āˆ© int ā„œš‘™āˆ’ āˆ•= āˆ…. That is, there exists some z āˆˆ š‘† with z < 0. This implies that there exists some allocation (x1 , x2 , . . . , xš‘› ) such that āˆ‘ xš‘– āˆ’ xāˆ— < 0 z= š‘–

xāˆ—š‘–

and xš‘– ā‰æ for every š‘– āˆˆ š‘ . Distribute z equally to all the consumers. That is, consider the allocation yš‘– = xš‘– + z/š‘› By strict monotonicity, yš‘– ā‰» xš‘– ā‰æ xāˆ—š‘– for every š‘– āˆˆ š‘ . Since āˆ‘ āˆ‘ āˆ‘ yš‘– = xš‘– + z = xāˆ— = xāˆ—š‘– š‘–

š‘–

š‘–

(y1 , y2 , . . . , yš‘› ) is a reallocation of the original allocation xāˆ— which is strictly preferred by all consumers. This contradicts the assumed Pareto eļ¬ƒciency of xāˆ— . We conclude that š‘† āˆ© int ā„œš‘™āˆ’ āˆ•= āˆ… 3. Applying Exercise 3.227, there exists a hyperplane with nonnegative normal pāˆ— ā‰© 0 such that pāˆ— z ā‰„ 0 for every z āˆˆ š‘† That is pāˆ— (x āˆ’ xāˆ— ) ā‰„ 0 or pāˆ— x ā‰„ pāˆ— xāˆ— for every x āˆˆ ā‰æ(xāˆ— )

(3.65)

4. Consider any allocation which is strictly preferred to xāˆ— by consumer š‘—, that is xš‘— āˆˆ ā‰»š‘— (xāˆ—š‘— ). Construct another allocation y by taking šœ– > 0 of each commodity away from agent š‘— and distributing amongst the other agents to give yš‘— = (1 āˆ’ šœ–)xš‘— šœ– yš‘– = xāˆ—š‘– + xš‘— , š‘›āˆ’1

š‘– āˆ•= š‘—

By continuity, there exists some šœ– > 0 such that yš‘— = (1 āˆ’ šœ–)xš‘— ā‰»š‘— xāˆ—š‘— . By monotonicity, yš‘– ā‰»š‘– xāˆ—š‘– for every š‘– āˆ•= š‘—. We have constructed āˆ‘ an allocation y which is strictly preferred to xāˆ— by all the agents, so that y = š‘– yš‘– āˆˆ ā‰æ(xāˆ— ). (3.65) implies that py ā‰„ pxāˆ— That is āŽ›

āˆ‘( xāˆ—š‘– + p āŽ(1 āˆ’ šœ–)xš‘— + š‘–āˆ•=š‘—

āŽž āŽž āŽž āŽ› āŽ› ) āˆ‘ āˆ‘ šœ– xš‘— āŽ  = p āŽxš‘— + xāˆ—š‘– āŽ  ā‰„ p āŽxāˆ—š‘— + xāˆ—š‘– āŽ  š‘›āˆ’1 š‘–āˆ•=š‘—

š‘–āˆ•=š‘—

which implies that pxš‘— ā‰„ pxāˆ—š‘—

for every xš‘— āˆˆ ā‰»(xāˆ—š‘— ) 184

(3.66)

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c 2001 Michael Carter āƒ All rights reserved

5. Trivially, xāˆ— is a feasible allocation with endowments wš‘– = xāˆ—š‘– and š‘šš‘– = pāˆ— wš‘– = pāˆ— xāˆ—š‘– . To show that (pāˆ— , xāˆ— ) is a competitive equilibrium, we have to show that xāˆ—š‘– is the best allocation in the budget set š‘‹š‘– (p, š‘šš‘– ) for each consumer š‘–. Suppose to the contrary there exists some consumer š‘— and allocation yš‘— such that yš‘— ā‰» xš‘— and pyš‘— ā‰¤ š‘šš‘— = pxāˆ—š‘— . By continuity, there exists some š›¼ āˆˆ (0, 1) such that š›¼yš‘— ā‰»š‘– xāˆ—š‘— and p(š›¼yš‘— ) = š›¼pyš‘— < pyš‘— ā‰¤ pxāˆ— contradicting (3.66). We conclude that xāˆ—š‘– ā‰æš‘– xš‘– for every x āˆˆ š‘‹(pāˆ— , š‘šš‘– ) for every consumer š‘–. (pāˆ— , xāˆ— ) is a competitive equilibrium. 3.229 By the previous exercise, there exists a price system pāˆ— such that xāˆ—š‘– is optimal for each consumer š‘– in the budget set š‘‹(pāˆ— , pāˆ— xāˆ—š‘– ), that is xāˆ—š‘– ā‰æš‘– xš‘– for every xš‘– āˆˆ š‘‹(pāˆ— , pāˆ— xāˆ—š‘– )

(3.67)

For each consumer, let š‘”š‘– be the diļ¬€erence between her endowed wealth pāˆ— wš‘– and her required wealth pāˆ— xāˆ—š‘– . That is, deļ¬ne š‘”š‘– = pāˆ— xāˆ—š‘– āˆ’ pāˆ— wš‘– = pāˆ— (xāˆ—š‘– āˆ’ wš‘– ) Then pāˆ— xāˆ—š‘– = pāˆ— + wš‘–

(3.68)

āˆ—

By assumption x is feasible, so that āˆ‘ āˆ‘ āˆ‘ xāˆ—š‘– āˆ’ wš‘– = (xāˆ—š‘– āˆ’ wš‘– ) = 0 š‘–

so that

š‘–

āˆ‘ š‘–

š‘”š‘– = pāˆ—

š‘–

āˆ‘

(xāˆ—š‘– āˆ’ wš‘– ) = 0

š‘–

āˆ—

Furthermore, for š‘šš‘– = š‘ wš‘– + š‘”š‘– , (3.68) implies š‘‹(pāˆ— , š‘šš‘– ) = { xš‘– : pāˆ— xš‘– ā‰¤ pāˆ— wš‘– + š‘”š‘– } = { xš‘– : pāˆ— xš‘– ā‰¤ pāˆ— xāˆ—š‘– } = š‘‹(pāˆ— , pāˆ— xāˆ—š‘– ) for each consumer š‘–. Using (3.67) we conclude that xāˆ—š‘– ā‰æš‘– xš‘– for every xš‘– āˆˆ š‘‹(pāˆ— , š‘šš‘– ) for every agent š‘–. (pāˆ— , xāˆ— ) is a competitive equilibrium where each consumerā€™s after-tax wealth is š‘šš‘– = pwš‘– + š‘”š‘– 3.230 Apply Exercise 3.202 with š¾ = ā„œš‘›+ . 3.231 š¾ āˆ— = { p : pš‘‡ x ā‰¤ 0 for every x āˆˆ š¾ } No such hyperplane exists if and only if š¾ āˆ— āˆ© ā„œš‘›++ = āˆ…. Assume this is the case. By Exercise 3.225, there exists x ā‰© 0 such that xp = pš‘‡ x ā‰¤ 0 for every p āˆˆ š¾ āˆ— In other words, x āˆˆ š¾ āˆ—āˆ— . By the duality theorem š¾ āˆ—āˆ— = š¾ which implies that x āˆˆ š¾ as well as ā„œš‘›+ , contrary to the hypothesis that š¾ āˆ© ā„œš‘›+ = {0}. This contradiction establishes that š¾ āˆ— āˆ© ā„œš‘›++ āˆ•= āˆ…. 185

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3.232 Given a set of ļ¬nancial assets with prices p and payoļ¬€ matrix š‘…, let š‘ = { (āˆ’px, š‘…š‘„) : x āˆˆ ā„œš‘› } š‘ is the set of all possible (cost, payoļ¬€) pairs. It is a subspace of ā„œš‘†+1 . Let š‘ be the nonnegative orthant in ā„œš‘†+1 . The no arbitrage condition š‘…x ā‰„ 0 =ā‡’ pš‘‡ x ā‰„ 0 implies that š‘ āˆ© š‘ = {0}. By Exercise 3.230, there exists a hyperplane with positive normal šœ† = šœ†0 , šœ†1 , . . . , šœ†š‘† such that šœ†z = 0

for every z āˆˆ š‘

šœ†z > 0

āˆ– {0} for every z āˆˆ ā„œš‘†+1 +

That is for every x āˆˆ ā„œš‘›

āˆ’šœ†0 px + šœ†š‘…x = 0 or pš‘‡ x = šœ†/šœ†0 š‘…x

for every x āˆˆ ā„œš‘›

šœ†/šœ†0 is required state price vector. Conversely, if a state price vector exists š‘š‘Ž =

š‘† āˆ‘

š‘…š‘Žš‘  šœ‹š‘ 

š‘Ž=1

then clearly š‘…x ā‰„ 0 =ā‡’ pš‘‡ x ā‰„ 0 No arbitrage portfolios exist. 3.233 Apply the Farkas lemma to the system āˆ’š“x ā‰¤ 0 āˆ’cš‘‡ x > 0 3.234 The inequality system š“š‘‡ y ā‰„ c has a nonnegative solution if and only if the corresponding system of equations š“š‘‡ y āˆ’ z = c š‘› has a nonnegative solution y āˆˆ ā„œš‘š + , z āˆˆ ā„œ+ . This is equivalent to the system ( ) y ā€² šµ =c z

(3.69)

where šµ ā€² = (š“š‘‡ , āˆ’š¼š‘› ) and š¼š‘› is the š‘› Ɨ š‘› identity matrix. By the Farkas lemma, system (3.69) has no solution if and only if the system šµx ā‰¤ 0 and cš‘‡ x > 0 ( ) š“ has a solution x āˆˆ ā„œš‘› . Since šµ = , šµx ā‰¤ 0 implies āˆ’š¼ š“x ā‰¤ 0 and āˆ’ š¼x ā‰¤ 0 and the latter inequality implies x āˆˆ ā„œš‘›+ . Thus we have established that the system š“š‘‡ y ā‰„ c has no nonnegative solution if and only if š“x ā‰¤ 0 and cš‘‡ x > 0 for some x āˆˆ ā„œš‘›+ 186

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Ė† āˆˆ ā„œš‘›+ such that 3.235 Assume system I has a solution, that is there exists x Ė†ā‰„0 š“Ė† x = 0, cĖ† x > 0, x Ė† /cĖ† Then x = x x satisļ¬es the system š“x = 0, cx = 1, x ā‰„ 0

(3.70)

xā€² š“š‘‡ = 0, xc = 1, x ā‰„ 0

(3.71)

which is equivalent to

Suppose y āˆˆ ā„œš‘š satisļ¬es š“y ā‰„ c Multiplying by x ā‰„ 0 gives xā€² š“š‘‡ y ā‰„ xc Substituting (3.71), this implies the contradiction 0ā‰„1 We conclude that system II cannot have a solution if I has a solution. Now, assume system I has no solution. System I is equivalent to (3.70) which in turn is equivalent to the system ( ) ( ) š“ 0 x= c 1 or šµx = b (3.72) ) ( ) ( āˆ’š“ 0 is (š‘š + 1) Ɨ š‘› and b = āˆˆ ā„œš‘š+1 . If (3.72) has no solution, where šµ = c 1 there exists (by the Farkas alternative) some z āˆˆ ā„œš‘š+1 such that šµ ā€² z ā‰¤ 0 and bz > 0 Decompose z into z = (y, š‘§) with y āˆˆ ā„œš‘š and š‘§ āˆˆ ā„œ. The second inequality implies that (0, 1)ā€² (y, š‘§) = 0y + š‘§ = š‘§ > 0 Without loss of generality, we can normalize so that š‘§ = 1 and z = (y, 1). Now šµ ā€² = (āˆ’š“š‘‡ , c) and so the ļ¬rst inequality implies that ( ) y š‘‡ (āˆ’š“ , c) = āˆ’š“š‘‡ y + c ā‰¤ 0 1 or š“š‘‡ y ā‰„ c We conclude that II has a solution. 187

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3.236 For every linear functional š‘”š‘— , there exists a vector a āˆˆ ā„œš‘› such that š‘”š‘— (x) = aš‘— xĖ™ (Proposition 3.11). Let š“š‘‡ be the matrix whose rows are aš‘— , that is āŽ› 1āŽž a āŽœ a2 āŽŸ āŽŸ š“=āŽœ āŽ. . .āŽ  aš‘š Then, the system of inequalities (3.31) is š“š‘‡ x ā‰„ c where c = (š‘1 , š‘2 , . . . , š‘š‘š ). By the preceding exercise, this system is consistent if and only there is no solution to the system š“šœ† = 0

cšœ† > 0

šœ†ā‰„0

Now š‘š āˆ‘

š“šœ† = 0 ā‡ā‡’

šœ†š‘— š‘”š‘– = 0

š‘– = 1, 2, . . . , š‘š

š‘—=1

Therefore, the inequalities (3.31) is consistent if an only if š‘š āˆ‘

šœ†š‘— š‘”š‘– = 0 =ā‡’

š‘—=1

š‘š āˆ‘

šœ†š‘— š‘š‘— ā‰¤ 0

š‘—=1

for every set of nonnegative numbers šœ†1 , šœ†2 , . . . , šœ†š‘š . 3.237 Let šµ be the 2š‘š Ɨ š‘› matrix comprising š“ and āˆ’š“ as follows ( ) š“ šµ= āˆ’š“ Then the Fredholm alternative I cš‘‡ x = 1

š“x = 0 is equivalent to the system šµx ā‰¤ 0

cx > 0

(3.73)

2š‘š By the Farkas alternative theorem, either (3.73) has a solution or there exists šœ† āˆˆ š‘…+ such that

šµā€²šœ† = c

(3.74)

Decompose šœ† into two š‘š-vectors š‘š šœ† = (šœ‡, š›æ), šœ‡, š›æ āˆˆ š‘…+

so that (3.74) can be rewritten as šµ ā€² šœ† = š“š‘‡ šœ‡ āˆ’ š“š‘‡ š›æ = š“š‘‡ (šœ‡ āˆ’ š›æ) = c Deļ¬ne y = šœ‡ āˆ’ š›æ āˆˆ ā„œš‘š We have established that either (3.73) has a solution or there exists a vector y āˆˆ ā„œš‘š such that š“š‘‡ y = c 188

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3.238 Let aš‘— , š‘— = 1, 2, . . . , š‘š denote the rows of š“. Each aš‘– deļ¬nes linear functional š‘”š‘— (š‘„) = aš‘— š‘„ on ā„œš‘› , and c deļ¬nes another linear functional š‘“ (š‘„) = cš‘‡ x. Assume that š‘“ (š‘„) = cš‘‡ x = 0 for every x āˆˆ š‘† where š‘† = { x : š‘”š‘— (x) = aš‘– x = 0, š‘— = 1, 2, . . . , š‘š } Then the system š“š‘„ = 0 has no solution satisfying the constraint cš‘‡ x > 0. By Exercise 3.20, there exists scalars š‘¦1 , š‘¦2 , . . . , š‘¦š‘š such that š‘“ (x)=

š‘š āˆ‘

š‘¦š‘— š‘”š‘— (x)

š‘—=1

or c=

š‘š āˆ‘

š‘¦š‘— š‘Žš‘— = š“š‘‡ y

š‘—=1

That is y = (š‘¦1 , š‘¦2 , . . . , š‘¦š‘š ) solves the related nonhomogeneous system š“š‘‡ y = c Conversely, assume that š“š‘‡ y = c for some š‘¦ āˆˆ ā„œš‘š . Then cš‘‡ x = š‘¦š“š‘„ = 0 for all š‘„ such that š“š‘„ = 0 and therefore there is no solution satisfying the constraint cš‘‡ x = 1. 3.239 Let š‘† = { z : z = š“x, x āˆˆ ā„œ } the image of š‘†. š‘† is a subspace. Assume that system I has no solution, that is š‘† āˆ© ā„œš‘š ++ = āˆ… By Exercise 3.225, there exists y āˆˆ ā„œš‘š + āˆ– {0} such that yz = 0 for every z āˆˆ š‘† That is yš“x = 0 for every x āˆˆ ā„œš‘› Letting x = š“š‘‡ y, we have yš“š“š‘‡ y = 0 which implies that š“š‘‡ y = 0 System II has a solution y. Ė† is a solution to I. Suppose to the contrary there also exists Conversely, assume that x Ė† to II. Then, since š“Ė† Ė† ā‰© 0, we must have y Ė† š“Ė† Ė† š“š‘‡ y Ė† > 0. a solution y x > 0 and y x=x š‘‡Ė† š‘‡ Ė†š“ y Ė† = 0, a contradiction. Hence, we On the other hand, š“ y = 0 which implies x conclude that II cannot have a solution if I has a solution. 189

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3.240 We have already shown (Exercise 3.239) that the alternatives I and II are mutually incompatible. If Gordanā€™s system II š“š‘‡ y = 0 has a semipositive solution y ā‰© 0, then we can normalize y such that 1y = 1 and the system š“š‘‡ y = 0 1y = 1 has a nonnegative solution. Conversely, if Gordanā€™s system II has no solution, the system šµā€²y = c (

) š“š‘‡ where šµ = and c = (0, 1) = (0, 0, . . . , 0, 1), 0 āˆˆ ā„œš‘š , is the (š‘š + 1)st unit 1 vector has no solution y ā‰„ 0. By the Farkas lemma, there exists z āˆˆ ā„œš‘›+1 such that ā€²

šµz ā‰„ 0 cz < 0 Decompose z into z = (x, š‘„) with x āˆˆ ā„œš‘› . The second inequality implies that š‘„ < 0 since cz = (0, 1)ā€² (x, š‘„) = š‘„ < 0 Since šµ = (š“, 1), the ļ¬rst inequality implies that šµz = (š“, 1)(x, š‘„) = š“x + 1š‘„ ā‰„ 0 or š“x ā‰„ āˆ’1š‘„ > 0 x solves Gordanā€™s system I. 3.241 Let a1 , a2 , . . . , aš‘š be a basis for š‘†. Let š“ = (a1 , a2 , . . . , aš‘š ) be the matrix whose columns are aš‘— . To say that š‘† contains no positive vector means that the system š“x > 0 has no solution. By Gordanā€™s theorem, there exists some y ā‰© 0 such that š“š‘‡ y = 0 that is aš‘— y = yaš‘— = 0, š‘— = 1, 2, . . . , š‘š so that y āˆˆ š‘† āŠ„ . 190

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3.242 Let š‘ be the subspace š‘ = { z : š“x : x āˆˆ ā„œš‘› }. System I has no solution š“x ā‰© 0 if and only if š‘ has no nonnegative vector z ā‰© 0. By the previous exercise, š‘ āŠ„ contains a positive vector y > 0 such that yz = 0 for every z āˆˆ š‘ Letting x = š“š‘‡ y, we have yš“š“š‘‡ y = 0 which implies that š“š‘‡ y = 0 System II has a solution y. 3.243 Let š‘† = { z : z = š“x, x āˆˆ ā„œ } the image of š‘†. š‘† is a subspace. Assume that system I has no solution, that is š‘† āˆ© ā„œš‘š + = {0} By Exercise 3.230, there exists y āˆˆ ā„œš‘š ++ such that yz = 0 for every z āˆˆ š‘† That is yš“x = 0 for every x āˆˆ ā„œš‘› Letting x = š“š‘‡ y, we have yš“š“š‘‡ y = 0 which implies that š“š‘‡ y = 0 System II has a solution y. Ė† is a solution to I. Suppose to the contrary there also exists Conversely, assume that x Ė† > 0. Ė† to II. Then, since š“Ė† Ė† > 0, we must have y Ė† š“Ė† Ė† š“š‘‡ y a solution y x ā‰© 0 and y x=x š‘‡Ė† š‘‡ Ė†š“ y Ė† = 0, a contradiction. Hence, we On the other hand, š“ y = 0 which implies x conclude that II cannot have a solution if I has a solution. 3.244 The inequality system š“š‘‡ y ā‰¤ 0 has a nonnegative solution if and only if the corresponding system of equations š“š‘‡ y + z = 0 š‘› has a nonnegative solution y āˆˆ ā„œš‘š + , z āˆˆ ā„œ+ . This is equivalent to the system ( ) y ā€² šµ =0 z

(3.75)

where šµ ā€² = (š“š‘‡ , š¼š‘› ) and š¼š‘› is the š‘› Ɨ š‘› identity matrix. By Gordanā€™s theorem, system (3.75) has no solution if and only if the system ( has a solution x āˆˆ ā„œš‘› . Since šµ =

šµx > 0 ) š“ , šµx > 0 implies š¼

š“x > 0 and š¼x > 0 and the latter inequality implies x āˆˆ ā„œš‘›++ . Thus we have established that the system š“š‘‡ y ā‰¤ 0 has no nonnegative solution if and only if š“x > 0 for some x āˆˆ ā„œš‘›++ 191

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3.245 Assume system II has no solution, that is there is no y āˆˆ ā„œš‘› such that š“y ā‰¤ 0, y ā‰© 0 This implies that the system āˆ’š“y ā‰„ 0 1y ā‰„ 1 ( ) āˆ’š“ š‘š ā€² has no solution y āˆˆ ā„œ+ . Deļ¬ning šµ = , the latter can be written as 1ā€² šµ ā€² y ā‰„ āˆ’eš‘š+1

(3.76)

where āˆ’eš‘š+1 = (0, 1), 0 āˆˆ ā„œš‘š . By the Gale alternative (Exercise 3.234), if system (3.76) has no solution, the alternative system šµz ā‰¤ 0, āˆ’eš‘š+1 z > 0 š‘› has a nonnegative solution z āˆˆ ā„œš‘›+1 + . Decompose z into z = (x, š‘§) where x āˆˆ ā„œ+ and š‘§ āˆˆ ā„œ+ . The second inequality implies š‘§ > 0 since eš‘š+1 z = š‘§.

šµ = (āˆ’š“š‘‡ , 1) and the ļ¬rst inequality implies ( ) x šµz = (āˆ’š“š‘‡ , 1) = āˆ’š“š‘‡ x + 1š‘§ ā‰¤ 0 š‘§ or š“š‘‡ x ā‰„ 1š‘§ > 0 Thus system I has a solution x āˆˆ ā„œš‘›+ . Since x = 0 implies š“x = 0, we conclude that x ā‰© 0. Conversely, assume that II has a solution y ā‰© 0 such that š“y ā‰¤ 0. Then, for every x āˆˆ ā„œš‘›+ xš“š‘‡ y = yā€² š“š‘‡ x ā‰¤ 0 Since y ā‰© 0, this implies š“š‘‡ x ā‰¤ 0 for every x āˆˆ ā„œš‘›+ which contradicts I. 3.246 We give a constructive proof, by proposing an algorithm which will generate the desired decomposition. Assume that x satisļ¬es š“x ā‰© 0. Arrange the rows of š“ such that the positive elements of š“x are listed ļ¬rst. That is, decompose š“ into two submatrices such that šµ1x > 0 š¶1x = 0 Either Case 1 š¶ 1 x ā‰© 0 has no solution and the result is proved or 192

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Case 2 š¶ 1 x ā‰© 0 has a solution xā€² . ĀÆ be a linear combination of x and xā€² . Speciļ¬cally, deļ¬ne Let x ĀÆ = š›¼x + xā€² x where š›¼ > max

āˆ’bš‘— x bš‘— xā€²

where bš‘— is the š‘—th row of šµ 1 . š›¼ is chosen so that š›¼šµ 1 x > šµ 1 xā€² By direct computation ĀÆ = š›¼šµ 1 x + šµ 1 xā€² > 0 šµ1x ĀÆ = š›¼š¶ 1 x + š¶ 1 xā€² ā‰© 0 š¶1x ĀÆ is another solution to š“x ā‰© 0 since š¶ 1 x = 0 and š¶ 1 xā€² ā‰© 0. By construction, x such that š“ĀÆ x has more positive components than š“x. Again, collect all the positive components together, decomposing š“ into two submatrices such that ĀÆ>0 šµ2x ĀÆ=0 š¶2x Either Case 1 š¶ 2 x ā‰© 0 has no solution and the result is proved or Case 2 š¶ 2 x ā‰© 0 has a solution xā€²ā€² . In the second case, we can repeat the previous procedure, generating another decomposition šµ 3 , š¶ 3 and so on. At each stage š‘˜, the matrix šµ š‘˜ get larger and š¶ š‘˜ smaller. The algorithm must terminate before šµ š‘˜ equals š“, since we began with the assumption that š“x > 0 has no solution. 3.247 There are three possible cases to consider. Case 1: y = 0 is the only solution of š“š‘‡ y = 0. Then š“x > 0 has a solution xā€² by Gordanā€™s theorem and š“xā€² + 0 > 0 Case 2: š“š‘‡ y = 0 has a positive solution y > 0 Then 0 is the only solution š“x ā‰„ 0 by Stiemkeā€™s theorem and š“0 + y > 0 Case 3 š“š‘‡ y = 0 has a solution y ā‰© 0 but y āˆ•> 0. By Gordanā€™s theorem š“x > 0 has no solution. By the previous exercise, š“ can be decomposed into two consistent subsystems šµx > 0 š¶x = 0 193

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

such that š¶x ā‰© 0 has no solution. Assume that šµ is š‘˜ Ɨ š‘› and š¶ is š‘™ Ɨ š‘› where š‘™ = š‘š āˆ’ š‘˜. Applying Stiemkeā€™s theorem to š¶, there exists z > 0, z āˆˆ ā„œš‘™ . Deļ¬ne y āˆˆ ā„œš‘š + by { 0 š‘— = 1, 2, . . . , š‘˜ š‘¦š‘— = š‘¦š‘— = š‘§š‘—āˆ’š‘˜ š‘— = š‘˜ + 1, š‘˜ + 2, . . . , š‘š Then x, y is the desired solution since for every š‘—, š‘— = 1, 2, . . . , š‘š either š‘¦š‘— > 0 or (š“x)š‘— = (šµx)š‘— > 0. 3.248 Consider the dual pair ( ) ( ) y š“ = 0, y ā‰„ 0, z ā‰„ 0 x ā‰„ 0 and (š“š‘‡ , š¼) z š¼ By Tuckerā€™s theorem, this has a solution xāˆ— , yāˆ— , zāˆ— such that š“xāˆ— ā‰„ 0, xāˆ— ā‰„ 0, š“š‘‡ yāˆ— + zāˆ— = 0, yāˆ— ā‰„ 0, zāˆ— ā‰„ 0 š“x + y > 0 š¼xāˆ— + š¼z > 0 Substituting zāˆ— = āˆ’š“š‘‡ yāˆ— implies š“š‘‡ y ā‰¤ 0 and x āˆ’ š“š‘‡ yāˆ— > 0 3.249 Consider the dual pair š“x ā‰„ 0 and š“š‘‡ y = 0, y ā‰„ 0 where š“ is an š‘š Ɨ š‘› matrix. By Tuckerā€™s theorem, there exists a pair of solutions xāˆ— āˆˆ ā„œš‘› and yāˆ— āˆˆ ā„œš‘š such that š“xāˆ— + yāˆ— > 0

(3.77)

Assume that š“x > 0 has no solution (Gordan I). Then there exists some š‘— such that (š“xāˆ— )š‘— = 0 and (3.77) implies that š‘¦š‘—āˆ— > 0. Therefore yāˆ— ā‰© 0 and solves Gordan II. Conversely, assume that š“š‘‡ y = 0 has no solution y > 0 (Stiemke II). Then, there exists some š‘— such that š‘¦š‘—āˆ— = 0 and (3.77) implies that (š“xāˆ— )š‘— > 0). Therefore xāˆ— solves š“x ā‰© 0 (Stiemke I). 3.250 We have already shown that Farkas I and II are mutually inconsistent. Assume that Farkas system I š“x ā‰„ 0, cš‘‡ x < 0 ( has no solution. Deļ¬ne the (š‘š + 1) Ɨ š‘› matrix šµ = the system šµx ā‰„ 0 194

š“ āˆ’cā€²

) . Our assumption is that

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

has no solution with (šµx)š‘š+1 = āˆ’cx > 0. By Tuckerā€™s theorem, the dual system šµā€²z = 0 has a solution z āˆˆ ā„œš‘š+1 with zš‘š+1 > 0. Without loss of generality, we can normalize + ā€² š‘‡ so that zš‘š+1 = 1. Decompose z into z = (y, 1) with y āˆˆ ā„œš‘š + . Since šµ = (š“ , āˆ’c), ā€² šµ z = 0 implies šµ ā€² z = (š“š‘‡ , āˆ’c)(y, 1) = š“š‘‡ y āˆ’ c = 0 or š“š‘‡ y = c y āˆˆ ā„œš‘š + solves Farkas II. 3.251 If x ā‰„ 0 solves I, then xā€² (š“š‘‡ y1 + šµ ā€² y2 + š¶ ā€² y3 ) = xā€² š“š‘‡ y1 + xā€² šµ ā€² y2 + xā€² š¶ ā€² y3 ) > 0 since xā€² š“š‘‡ y1 = y1 š“x > 0, xā€² šµ ā€² y2 = y2 šµx ā‰„ 0 and xā€² š¶ ā€² y3 = y3 š¶x = 0 which contradicts II. The equation š¶x = 0 is equivalent to the pair of inequalities š¶x ā‰„ 0, āˆ’š¶x ā‰„ 0. By Tuckerā€™s theorem the dual pair š“š‘‡ y1 + šµ ā€² y2 + š¶ ā€² y3 āˆ’ š¶ ā€² y4 = 0

š“x ā‰„ 0 šµx ā‰„ 0 š¶x ā‰„ 0 āˆ’š¶x ā‰„ 0

has solutions š‘„ āˆˆ ā„œš‘› , y1 āˆˆ ā„œš‘š1 , y2 āˆˆ ā„œš‘š2 , u3 , v3 āˆˆ ā„œš‘š3 such that y1 ā‰„ 0

š“x + y1 > 0

y2 ā‰„ 0 u3 ā‰„ 0

šµx + y2 > 0 š¶x + u3 > 0

v3 ā‰„ 0

āˆ’š¶x + v3 > 0

Assume Motzkin I has no solution. That is, there is y1 ā‰© 0. Deļ¬ne y3 = u3 āˆ’ v3 . Then y1 , y2 , y3 satisļ¬es Motzkin II. 3.252

1. For every a āˆˆ š‘†, let š‘†aāˆ— be the polar set š‘†aāˆ— = { x āˆˆ ā„œš‘› : āˆ„xāˆ„ = 1, xa ā‰„ 0 } š‘†aāˆ— is nonempty since 0 āˆˆ š‘†aāˆ— . Let x be the limit of a sequence xš‘› of points in š‘†aāˆ— . Since xš‘› a ā‰„ 0 for every š‘›, xa ā‰„ 0 so that x āˆˆ š‘†aāˆ— . Hence š‘†aāˆ— is a closed subset of šµ = { x āˆˆ ā„œš‘› : āˆ„xāˆ„ = 1 }.

2. Let {a1 , a2 , . . . , aš‘š } be any ļ¬nite set of points in š‘†. Since 0 āˆˆ / š‘†, the system š‘š āˆ‘

š‘¦š‘– aš‘– = 0,

š‘–=1

š‘š āˆ‘

š‘¦š‘– = 1, š‘¦š‘– ā‰„ 0

š‘–=1

has no solution. A fortiori, the system š‘š āˆ‘

š‘¦š‘– aš‘– = 0

š‘–=1

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has no solution š‘¦ āˆˆ ā„œš‘š + . If š“ is the š‘šĆ—n matrix whose rows are aš‘– , the latter system can be written as š“š‘‡ y = 0 3. By Gordanā€™s theorem, the system š“x > 0

(3.78)

ĀÆ āˆ•= 0. has a solution x 4. Without loss of generality, we can take āˆ„ĀÆ xāˆ„ = 1. (3.78) implies that ĀÆ=x ĀÆ aš‘– > 0 aš‘– x ĀÆ āˆˆ š‘†aāˆ—š‘– . Hence for every š‘– = 1, 2 . . . , š‘š so that x š‘š āˆ©

ĀÆāˆˆ x

š‘–=1

š‘†aāˆ—š‘–

āˆ©š‘š 5. We have shown that for every ļ¬nite set {a1 , a2 , . . . , aš‘š } āŠ† š‘†, š‘–=1 š‘†aāˆ—š‘– is nonempty closed subset of the compact set šµ = {š‘„ āˆˆ ā„œš‘› : āˆ„xāˆ„ = 1}. By the Finite intersection property (Exercise 1.116) āˆ© š‘†aāˆ— āˆ•= āˆ… aāˆˆš‘†

6. For every p āˆˆ

āˆ© aāˆˆš‘†

š‘†aāˆ— pa ā‰„ 0 for every a āˆˆ š‘†

p deļ¬nes a hyperplane š‘“ (a) = pa which separates š‘† from 0. 3.253 The expected outcome if player 1 adopts the mixed strategy p = (š‘1 , š‘2 , . . . , š‘š‘š ) and player 2 plays her š‘— pure strategy is š‘¢(p, š‘—) =

š‘š āˆ‘

š‘š‘– š‘Žš‘–š‘— = paš‘—

š‘–=1

where aš‘— is the š‘—th column of š“. The expected payoļ¬€ to 1 for all possible responses of player 2 is the vector (pš“)ā€² = š“š‘‡ p. The mixed strategy p ensures player 1 a nonnegative security level provided š“š‘‡ p ā‰„ 0. Similarly, if 2 adopts the mixed strategy q = (š‘ž1 , š‘ž2 , . . . , š‘žš‘› ), the expected payoļ¬€ to 2 if 1 plays his š‘– strategy is aš‘– q where aš‘– is the š‘–th row of š“. The expected outcome for all the possible responses of player 1 is the vector š“q. The mixed strategy q ensures player 2 a nonpositive security level provided š“q ā‰¤ 0. By the von Neumann alternative theorem (Exercise 3.245), at least one of these alternatives must be true. That is, either Either I š“š‘‡ p > 0, p ā‰© 0 for some p āˆˆ ā„œš‘š or II š“q ā‰¤ 0, q ā‰© 0 for some q āˆˆ ā„œš‘› Since p ā‰© 0 and q ā‰© 0, we can normalize so that p āˆˆ Ī”š‘šāˆ’1 and q āˆˆ Ī”š‘›āˆ’1 . At least one of the players has a strategy which guarantees she cannot lose. 196

Solutions for Foundations of Mathematical Economics 3.254

c 2001 Michael Carter āƒ All rights reserved

1. For any š‘ āˆˆ ā„œ, deļ¬ne the game š‘¢Ė†(a1 , a2 ) = š‘¢(a1 , a2 ) āˆ’ š‘ with Ė†(p, š‘—) = max min š‘¢(p, š‘—) āˆ’ š‘ = š‘£1 āˆ’ š‘ š‘£Ė†1 = max min š‘¢ p

p

š‘—

š‘—

š‘£Ė†2 = min max š‘¢ Ė†(š‘–, q) = min max š‘¢(š‘–, q) āˆ’ š‘ = š‘£2 āˆ’ š‘ q

q

š‘–

š‘–

By the previous exercise, Either š‘£Ė†1 ā‰„ 0 or š‘£Ė†š‘¦ ā‰¤ 0 That is Either š‘£1 ā‰„ š‘ or š‘£2 ā‰¤ š‘ 2. Since this applies for arbitrary š‘ āˆˆ ā„œ, it implies that while š‘£1 ā‰¤ š‘£2 and there is no š‘ such that š‘£1 < š‘ < š‘£2 Therefore, we conclude that š‘£1 = š‘£2 as required. 3.255

1. The mixed strategies p of player 1 are elements of the simplex Ī”š‘šāˆ’1 , which is compact (Example 1.110). Since š‘£1 (p) = minš‘›š‘—=1 š‘¢(p, š‘—) is continuous (Maximum theorem 2.3), š‘£1 (p) achieves its maximum on Ī”š‘šāˆ’1 (Weierstrass theorem 2.2). That is, there exists pāˆ— āˆˆ Ī”š‘šāˆ’1 such that š‘£1 = š‘£1 (pāˆ— ) = max š‘£1 (p) p

Similarly, there exists qāˆ— āˆˆ Ī”š‘›āˆ’1 such that š‘£2 = š‘£2 (qāˆ— ) = min š‘£2 (q) q

2. Let š‘¢(p, q) denote the expected outcome when player 1 adopts mixed strategy p and player 2 plays q. That is š‘¢(p, q) =

š‘š āˆ‘ š‘› āˆ‘

š‘š‘– š‘žš‘– š‘Žš‘–š‘—

š‘–=1 š‘—=1

Then š‘£ = š‘¢(pāˆ— , qāˆ— ) = max š‘¢(š‘–, qāˆ— ) ā‰„ š‘–

āˆ‘

š‘š‘– š‘¢(š‘–, qāˆ— ) = š‘¢(p, qāˆ— ) for every p āˆˆ Ī”š‘šāˆ’1

š‘–

Similarly š‘£ = š‘¢(pāˆ— , qāˆ— ) = min š‘¢(pāˆ— , š‘—) ā‰¤ š‘—

āˆ‘

š‘žš‘— š‘¢(pāˆ— , š‘—) = š‘¢(pāˆ— , q) for every q āˆˆ Ī”š‘›āˆ’1

š‘—

(pāˆ— , qāˆ— ) is a Nash equilibrium. 197

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Solutions for Foundations of Mathematical Economics

3.256 By the Minimax theorem, every ļ¬nite two person zero-sum game has a value. The previous result shows that this is attained at a Nash equilibrium. 3.257 If player 2 adopts the strategy š‘”1 š‘“p (š‘”1 ) = āˆ’š‘1 + 2š‘2 < 0 if š‘1 > 2š‘2 If player 2 adopts the strategy š‘”5 š‘“p (š‘”5 ) = š‘1 āˆ’ 2š‘2 < 0 if š‘1 < 2š‘2 Therefore š‘£1 (p) = min š‘“p (z) ā‰¤ min{š‘“p (š‘”1 ), š‘“p (š‘”5 )} < 0 š‘§āˆˆš‘

for every p such that š‘1 āˆ•= š‘2 . Since š‘1 + š‘2 = 1, we conclude that { = 0 p = pāˆ— = ( 2/3, 1/3) š‘£1 (p) < 0 otherwise We conclude that š‘£1 = max š‘£1 (p) = 0 p

which is attained at pāˆ— = ( 2/3, 1/3). 3.258

1. š‘š

š‘£2 = min max š‘§š‘– zāˆˆš‘ š‘–=1

Since š‘ is compact, š‘£2 = 0 implies there exists zĀÆ āˆˆ š‘ such that š‘š

max š‘§ĀÆš‘– = 0 š‘–=1

which implies that ĀÆ z ā‰¤ 0. Consequently š‘ āˆ© ā„œš‘›āˆ’ āˆ•= āˆ…. 2. Assume to the contrary that there exists z āˆˆ š‘ āˆ© int ā„œš‘›āˆ’ That is, there exists some strategy q āˆˆ Ī”š‘›āˆ’1 such that š“q < 0 and therefore š‘£2 < 0, contrary to the hypothesis. 3. There exists a hyperplane with nonnegative normal separating š‘ from ā„œš‘›āˆ’ (Exercise 3.227). That is, there exists pāˆ— āˆˆ ā„œš‘›+ , pāˆ— āˆ•= 0 such that š‘“pāˆ— (z) ā‰„ 0 for every z āˆˆ š‘ and therefore š‘£1 (pāˆ— ) = min š‘“pāˆ— (z) ā‰„ 0 zāˆˆš‘

Without loss of generality, we can normalize so that pāˆ— āˆˆ Ī”š‘šāˆ’1 .

198

āˆ‘š‘›

š‘–=1

š‘āˆ—š‘– = 1 and therefore

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

4. Consequently š‘£1 = max š‘£1 (p) ā‰„ š‘£1 (pāˆ— ) ā‰„ 0 p

On the other hand, we know that š‘ contains a point zĀÆ ā‰¤ 0. For every p ā‰„ 0 š‘“p (ĀÆz) ā‰¤ 0 and therefore š‘§) ā‰¤ 0 š‘£1 (p) = min š‘“p (z) ā‰¤ š‘“p (ĀÆ zāˆˆš‘

so that š‘£1 = max š‘£1 (p) ā‰¤ 0 p

We conclude that š‘£1 = 0 = š‘£2 3.259 Consider the game with the same strategies and the payoļ¬€ function š‘¢ Ė†(a1 , a2 ) = š‘¢(a1 , a2 ) āˆ’ š‘ The expected value to player 2 is Ė†(š‘–, q) = min max š‘¢(š‘–, q) āˆ’ š‘ = š‘£2 āˆ’ š‘ = 0 š‘£Ė†2 = min max š‘¢ q

q

š‘–

š‘–

By the previous exercise š‘£Ė†1 = š‘£Ė†2 = 0 and š‘£1 = max min š‘¢(p, š‘—) = max min š‘¢ Ė†(p, š‘—) + š‘ = š‘£Ė†1 + š‘ = š‘ = š‘£2 p

q

š‘—

š‘—

3.260 Assume that p1 and p2 are both optimal strategies for player 1. Then š‘¢(p1 , q) ā‰„ š‘£ for every q āˆˆ Ī”š‘›āˆ’1 š‘¢(p2 , q) ā‰„ š‘£ for every q āˆˆ Ī”š‘›āˆ’1 ĀÆ = š›¼p1 , p2 + (1 āˆ’ š›¼). Since š‘¢ is bilinear Let p š‘¢(ĀÆ p, q) = š›¼š‘¢(p1 , q) + (1 āˆ’ š›¼)š‘¢(p2 , q) ā‰„ š‘£ for every q āˆˆ Ī”š‘›āˆ’1 ĀÆ is also an optimal strategy for player 1. Consequently, p 3.261 š‘“ is the payoļ¬€ function of some 2 person zero-sum game in which the players have š‘š + 1 and š‘› + 1 strategies respectively. The result follows from the Minimax Theorem. 3.262

1. The possible partitions of š‘ = {1, 2, 3} are: {1}, {2}, {3} {š‘–, š‘—}, {š‘˜},

š‘–, š‘—, š‘˜ =āˆˆ š‘, š‘– āˆ•= š‘— āˆ•= š‘˜

{1, 2, 3} In any partition, at most one coalition can have two or more players, and therefore š¾ āˆ‘

š‘¤(š‘†š‘˜ ) ā‰¤ 1

š‘˜=1

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c 2001 Michael Carter āƒ All rights reserved

2. Assume x = (š‘„1 , š‘„2 , š‘„3 ) āˆˆ core. Then x must satisfy the following system of inequalities š‘„1 + š‘„2 ā‰„ 1 = š‘¤({1, 2}) š‘„1 + š‘„3 ā‰„ 1 = š‘¤({1, 3}) š‘„2 + š‘„3 ā‰„ 1 = š‘¤({2, 3}) which can be summed to yield 2(š‘„1 + š‘„2 + š‘„3 ) ā‰„ 3 or š‘„1 + š‘„2 + š‘„3 ā‰„ 3/2 which implies that x exceeds the sum available. This contradiction establishes that the core is empty. Alternatively, observe that the three person majority game is a simple game with no veto players. By Exercise 1.69, its core is empty. 3.263 Assume that the game (š‘, š‘¤) is not cohesive. Then there exists a partition {š‘†1 , š‘†2 , . . . , š‘†š¾ } of š‘ such that š‘¤(š‘ ) <

š¾ āˆ‘

š‘¤(š‘†š‘˜ )

š‘˜=1

Assume x āˆˆ core. Then

āˆ‘

š‘„š‘– ā‰„ š‘¤(š‘†š‘˜ )

š‘˜ = 1, 2, . . . , š¾

š‘–āˆˆš‘†š‘˜

Since {š‘†1 , š‘†2 , . . . , š‘†š¾ } is a partition āˆ‘

š‘„š‘– =

š‘–āˆˆš‘

š¾ āˆ‘ āˆ‘

š‘„š‘– ā‰„

š‘˜=1 š‘–āˆˆš‘†š‘˜

š‘ āˆ‘

š‘¤(š‘†š‘˜ ) > š‘¤(š‘ )

š‘˜=1

which contradicts the assumption that x āˆˆ core. This establishes that cohesivity is necessary for the existence of the core. To show that cohesivity is not suļ¬ƒcient, we observe that the three person majority game is cohesive, but its core is empty. 3.264 The other balanced families of coalitions in a three player game are 1. ā„¬ = {š‘ } with weights { š‘¤(š‘†) =

1 0

š‘†=š‘ otherwise

2. ā„¬ = {{1}, {2}, {3}} with weights š‘¤(š‘†) = 1 for every š‘† āˆˆ ā„¬ 3. ā„¬ = {{š‘–}, {š‘—, š‘˜}}, š‘–, š‘—, š‘˜ āˆˆ ā„¬, š‘– āˆ•= š‘— āˆ•= š‘˜ with weights š‘¤(š‘†) = 1 for every š‘† āˆˆ ā„¬ 3.265 The following table lists some nontrivial balanced families of coalitions for a four player game. Other balanced families can be obtained by permutation of the players. 200

Solutions for Foundations of Mathematical Economics

{123}, {124}, {34} {12}, {13}, {23}, {4} {123}, {14}, {24}, {3} {123}, {14}, {24}, {34} {123}, {124}, {134}, {234}

c 2001 Michael Carter āƒ All rights reserved

Weights 1/2, 1/2, 1/2 1/2, 1/2, 1/2, 1 1/2, 1/2, 1/2, 1/2 2/3, 1/3, 1/3, 1/3 1/3, 1/3, 1/3, 1/3

3.266 Both sides of the expression eš‘ =

āˆ‘

šœ†š‘† eš‘†

š‘†āˆˆā„¬

are vectors, with each component corresponding to a particular player. For player š‘–, the š‘–š‘” ā„Ž component of eš‘ is 1 and the š‘–š‘” ā„Ž component of eš‘† is 1 if š‘– āˆˆ š‘† and 0 otherwise. Therefore, for each player š‘–, the preceding expression can be written āˆ‘ šœ†š‘† = 1 š‘†āˆˆā„¬āˆ£š‘†āˆ‹š‘–

For each coalition š‘†, the share of the coalition š‘† at the allocation x is āˆ‘ š‘”š‘† (x) = š‘– āˆˆ š‘†š‘„š‘– = eš‘† xĖ™ The condition š‘”š‘ =

āˆ‘

(3.79)

šœ†š‘† š‘”š‘†

š‘†āˆˆā„¬

means that for every x āˆˆ š‘‹ š‘”š‘ (x) =

āˆ‘

šœ†š‘† š‘”š‘† (x)

š‘†āˆˆā„¬

Substituting (3.79) eš‘ xĖ™ =

āˆ‘

šœ†š‘† š‘’š‘† xĖ™

š‘†āˆˆā„¬

which is equivalent to the condition āˆ‘

šœ†š‘† eš‘† = eš‘

š‘†āˆˆā„¬

3.267 By construction, šœ‡ ā‰„ 0. If šœ‡ = 0, āˆ‘ šœ†š‘† š‘”š‘† āˆ’ šœ‡š‘”š‘ = 0 š‘†āŠ†š‘

implies that šœ†š‘† = 0 for all š‘† and consequently āˆ‘ šœ†š‘† š‘¤(š‘†) āˆ’ šœ‡š‘¤(š‘ ) ā‰¤ 0 š‘†āŠ†š‘

is trivially satisļ¬ed. On the other hand, if šœ‡ > 0, we can divide both conditions by šœ‡.)

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Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

3.268 Let (š‘, š‘¤1 ) and (š‘, š‘¤2 ) be balanced games. By the Bondareva-Shapley theorem, they have nonempty cores. Let x1 āˆˆ core(š‘, š‘¤1 ) and x2 āˆˆ core(š‘, š‘¤2 ). That is, š‘”š‘† (x1 ) ā‰„ š‘¤1 (š‘†) for every š‘† āŠ† š‘ š‘”š‘† (x2 ) ā‰„ š‘¤2 (š‘†) for every š‘† āŠ† š‘ Adding, we have š‘”š‘† (x1 ) + š‘”š‘† (x2 ) = š‘”š‘† (x1 + x2 ) ā‰„ š‘¤1 (š‘†) + š‘¤2 (š‘†) for every š‘† āŠ† š‘ which implies that x1 + x2 belongs to core(š‘, š‘¤1 + š‘¤2 ). Therefore (š‘, š‘¤1 + š‘¤2 ) is balanced. Similarly, if x āˆˆ core(š‘, š‘¤), then š›¼x belongs to core(š‘, š›¼š‘¤) for every š›¼ āˆˆ ā„œ+ . That is (š‘, š›¼š‘¤) is balanced for every š›¼ āˆˆ ā„œ+ . 3.269

1. Assume otherwise. That is assume there exists some y āˆˆ š“ āˆ© šµ. Taking the ļ¬rst š‘› components, this implies that āˆ‘ eš‘ = šœ†š‘  eš‘† š‘†āŠ†š‘

for some (šœ†š‘† ā‰„ 0 : š‘† āŠ† š‘ ). Let ā„¬ = {š‘† āŠ‚ š‘ āˆ£ šœ†š‘† > 0} be the set of coalitions with strictly positive weights. Then ā„¬ is a balanced family of coalitions with weights šœ†š‘† (Exercise 3.266). However, looking at the last coordinate, y āˆˆ š“ āˆ© šµ implies āˆ‘ šœ†š‘  š‘¤(š‘†) = š‘¤(š‘ ) + šœ– > š‘¤(š‘ ) š‘†āˆˆā„¬

which contradicts the assumption that the game is balanced. We conclude that š“ and šµ are disjoint if the game is balanced. 2. (a) Substituting y = (eāˆ… , 0) in (3.36) gives (z, š‘§0 )ā€² (0, 0) = 0 ā‰„ š‘ which implies that š‘ ā‰¤ 0. NOTE We still have to show that š‘ ā‰„ 0. (b) Substituting (eš‘ , š‘¤š‘¦(š‘ )) in (3.36) gives š‘§eš‘ + š‘§0 š‘¤(š‘ ) > š‘§eš‘ + š‘§0 š‘¤(š‘ ) + š‘§0 šœ– for all šœ– > 0, which implies that š‘§0 < 0. 3. Without loss of generality, we can normalize so that š‘§0 = āˆ’1. Then the separating hyperplane conditions become (z, āˆ’1)ā€² y ā‰„ 0 ā€²

(z, āˆ’1) (eš‘ , š‘¤(š‘ ) + šœ–) < 0

for every y āˆˆ š“

(3.80)

for every šœ– > 0

(3.81)

For any š‘† āŠ† š‘ , (eš‘† , š‘¤(š‘†)) āˆˆ š“. Substituting y = (eš‘† , š‘¤(š‘†)) in (3.80) gives eā€²š‘† z āˆ’ š‘¤(š‘†) ā‰„ 0 that is š‘”š‘† (z) = eā€²š‘† z =ā‰„ š‘¤(š‘†) 202

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics while (3.81) implies š‘”š‘ (z) = eā€²š‘ z > š‘¤(š‘ ) + šœ–

for every šœ– > 0

This establishes that z belongs to the core. Hence the core is nonempty. āˆ‘ 3.270 1. Let š›¼ = š‘¤(š‘ ) āˆ’ š‘–āˆˆš‘ š‘¤š‘– > 0 since (š‘, š‘¤) is essential. For every š‘† āŠ† š‘ , deļ¬ne ( ) āˆ‘ 1 0 š‘¤(š‘†) āˆ’ š‘¤š‘– š‘¤ (š‘†) = š›¼ š‘–āˆˆš‘†

Then š‘¤0 ({š‘–}) = 0 for every š‘– āˆˆ š‘ š‘¤0 (š‘ ) = 1 š‘¤0 is 0ā€“1 normalized. 2. Let y āˆˆ core(š‘, š‘¤0 ). Then for every š‘† āŠ† š‘ āˆ‘ š‘¦š‘– ā‰„ š‘¤0 (š‘†)

(3.82)

š‘–āˆˆš‘†

āˆ‘

š‘¦š‘– = 1

(3.83)

š‘–āˆˆš‘

Let w = (š‘¤1 , š‘¤2 , . . . , š‘¤š‘› ) where š‘¤š‘– = š‘¤({š‘–}). Let x = š›¼y + w. Using (3.82) and (3.83) āˆ‘ āˆ‘ š‘„š‘– = (š›¼š‘¦š‘– + š‘¤š‘– ) š‘–āˆˆš‘†

š‘–āˆˆš‘†

=š›¼

āˆ‘

š‘¦š‘– +

š‘–āˆˆš‘†

āˆ‘ š‘–āˆˆš‘

š‘¤š‘–

š‘–āˆˆš‘†

ā‰„ š›¼š‘¤0 (š‘†) + 1 =š›¼ š›¼

āˆ‘ āˆ‘

š‘¤š‘–

š‘–āˆˆš‘†

(

š‘¤(š‘†) āˆ’

āˆ‘ š‘–āˆˆš‘†

) š‘¤š‘–

+

āˆ‘

š‘¤š‘–

š‘–āˆˆš‘†

= š‘¤(š‘†) āˆ‘ š‘„š‘– = (š›¼š‘¦š‘– + š‘¤š‘– ) š‘–āˆˆš‘

=š›¼+

āˆ‘

š‘¤š‘–

š‘–āˆˆš‘

= š‘¤(š‘ ) Therefore, x = š›¼y + w āˆˆ core(š‘, š‘¤). Similarly, we can show that x āˆˆ core(š‘, š‘¤) =ā‡’ y =

1 (x āˆ’ w) āˆˆ core(š‘, š‘¤0 ) š›¼

and therefore core(š‘, š‘¤) = š›¼core(š‘, š‘¤0 ) + w 203

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 3. This immediately implies

core(š‘, š‘¤) = āˆ… ā‡ā‡’ core(š‘, š‘¤0 ) = āˆ… 3.271 (š‘, š‘¤) is 0ā€“1 normalized, that is š‘¤({š‘–} = 0 for every š‘– āˆˆ š‘ š‘¤(š‘ ) = 1 Consequently, x belongs to the core of (š‘, š‘¤) if and only if āˆ‘

š‘„š‘– ā‰„ š‘¤š‘– = 0

(3.84)

š‘„š‘– = š‘¤(š‘ ) = 1

(3.85)

š‘„š‘– ā‰„ š‘¤(š‘†) for every š‘† āˆˆ š’œ

(3.86)

š‘–āˆˆš‘

āˆ‘ š‘–āˆˆš‘†

(3.84) and (3.85) ensure that x = (š‘„1 , š‘„2 , . . . , š‘„š‘› ) is a mixed strategy for player 1 in the two-person zero-sum game. Using this mixed strategy, the expected payoļ¬€ to player I for any strategy š‘† of player II is š‘¢(x, š‘†) =

āˆ‘

š‘„š‘– š‘¢(š‘–, š‘†) =

š‘–āˆˆš‘

āˆ‘ š‘–āˆˆš‘†

š‘„š‘–

1 š‘¤(š‘†)

(3.86) implies š‘¢(x, š‘†) =

āˆ‘ š‘–āˆˆš‘†

š‘„š‘–

1 ā‰„ 1 for every š‘† āˆˆ š’œ š‘¤(š‘†)

That is any x āˆˆ core(š‘, š‘¤) provides a mixed strategy for player I which ensures a payoļ¬€ at least 1. That is core(š‘, š‘¤) āˆ•= āˆ… =ā‡’ š›æ ā‰„ 1 Conversely, if the š›æ < 1, there is no mixed strategy for player I which satisļ¬es (3.86) and consequently no x which satisļ¬es (3.84), (3.85) and (3.86). In other words, core(š‘, š‘¤) = āˆ…. 3.272 If š›æ is the value of šŗ, there exists a mixed strategy which will guarantee that II pays no more than š›æ. That is, there exists numbers š‘¦š‘† ā‰„ 0 for every coalition š‘† āˆˆ š’œ such that āˆ‘ š‘¦š‘† = 1 š‘†āˆˆš’œ

and āˆ‘

š‘¦š‘† š‘¢(š‘–, š‘†) ā‰¤ š›æ

for every š‘– āˆˆ š‘

š‘†āˆˆš’œ

that is āˆ‘ š‘†āˆˆš’œ

š‘¦š‘†

1 ā‰¤š›æ š‘¤(š‘†)

for every š‘– āˆˆ š‘

š‘†āˆ‹š‘–

204

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

or āˆ‘ š‘†āˆˆš’œ

š‘¦š‘† ā‰¤1 š›æš‘¤(š‘†)

for every š‘– āˆˆ š‘

(3.87)

š‘†āˆ‹š‘–

For each coalition š‘† āˆˆ š’œ let šœ†š‘† =

š‘¦š‘† š›æš‘¤(š‘†)

in (3.87) āˆ‘

šœ†š‘† ā‰¤ 1

š‘†āˆˆš’œ š‘†āˆ‹š‘–

Augment the collection š’œ with the single-player coalitions to form the collection ā„¬ = š’œ āˆŖ { {š‘–} : š‘– āˆˆ š‘ } and with weights { šœ†š‘† : š‘† āˆˆ š’œ } and āˆ‘

šœ†{š‘–} = 1 āˆ’

šœ†š‘†

š‘†āˆˆš’œ

Then ā„¬ is a balanced collection. Since the game (š‘, š‘¤) is balanced 1 = š‘¤(š‘ ) ā‰„

āˆ‘

šœ†š‘† š‘¤(š‘†)

š‘†āˆˆā„¬

=

āˆ‘

šœ†š‘† š‘¤(š‘†)

š‘†āˆˆš’œ

āˆ‘

š‘¦š‘† š‘¤(š‘†) š›æš‘¤(š‘†) š‘†āˆˆā„¬ 1āˆ‘ = š‘¦š‘† š›æ =

š‘†āˆˆā„¬

1 = š›æ that is 1ā‰„

1 š›æ

ĀÆ = (1/š‘›, 1/š‘›, . . . , 1/š‘›), the payoļ¬€ is If I plays the mixed strategy x š‘¢(ĀÆ x, š‘†) =

āˆ‘ š‘–āˆˆš‘

1 1 = > 0 for every š‘† āŠ† š’œ š‘›š‘¤(š‘†) š‘¤(š‘†)

Therefore š›æ > 0 and (3.88) implies that š›æā‰„1

205

(3.88)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 3.273 Assume core(š‘, š‘¤) āˆ•= āˆ… and let š‘„ āˆˆ core(š‘, š‘¤). Then š‘”š‘† (x) ā‰„ š‘¤(š‘†) for every š‘† āŠ† š‘ where š‘”š‘† =

āˆ‘ š‘–āˆˆš‘†

(3.89)

š‘„š‘– measures the share coalition š‘† at the allocation x.

Let ā„¬ be a balanced family of coalitions with weights šœ†š‘† . For every š‘† āˆˆ ā„¬, (3.89) implies šœ†š‘† š‘”š‘† (x) ā‰„ šœ†š‘† š‘¤(š‘†) Summing over all š‘† āˆˆ ā„¬ āˆ‘

āˆ‘

šœ†š‘† š‘”š‘† (x) ā‰„

š‘†āˆˆā„¬

šœ†š‘† š‘¤(š‘†)

š‘†āˆˆā„¬

Evaluating the left hand side of this inequality āˆ‘ āˆ‘ āˆ‘ šœ†š‘† š‘”š‘† (x) = šœ† š‘„š‘– š‘†āˆˆā„¬

š‘†āˆˆā„¬

=

š‘–āˆˆš‘†

āˆ‘āˆ‘

šœ†š‘„š‘–

š‘–āˆˆš‘ š‘†āˆˆā„¬

=

āˆ‘

š‘†āˆ‹š‘–

š‘„š‘–

š‘–āˆˆš‘

=

āˆ‘

āˆ‘

š‘†āˆˆā„¬ š‘†āˆ‹š‘–

š‘„š‘–

š‘–āˆˆš‘

= š‘¤(š‘ ) Substituting this in (3.90) gives š‘¤(š‘ ) ā‰„

āˆ‘ š‘†āˆˆā„¬

The game is balanced.

206

šœ†š‘† š‘¤(š‘†)

šœ†

(3.90)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Chapter 4: Smooth Functions 4.1 Along the demand curve, price and quantity are related according to the equation š‘ = 10 āˆ’ š‘„ This is called the inverse demand function. Total revenue š‘…(š‘„) (price times quantity) is given by š‘…(š‘„) = š‘š‘„ = (10 āˆ’ š‘„)š‘„ = 10š‘„ āˆ’ š‘„2 = š‘“ (š‘„) š‘”(š‘„) can be rewritten as š‘”(š‘„) = 21 + 4(š‘„ āˆ’ 3) At š‘„ = 3, the price is 7 but the marginal revenue of an additional unit is only 4. The function š‘” decomposes (approximately) the total revenue into two components ā€” the revenue from the sale of 3 units (21 = 3 Ɨ 7) plus the marginal revenue from the sale of additional units (4(š‘„ āˆ’ 3)). 4.2 If your answer is 5 per cent, obtained by subtracting the inļ¬‚ation rate from the growth rate of nominal GDP, you are implicitly using a linear approximation. To see this, let š‘ š‘ž š‘‘š‘ š‘‘š‘ž

= price level at the beginning of the year = real GDP at the beginning of the year = change in prices during year = change in output during year

We are told that nominal GDP at the end of the year, (š‘ + š‘‘š‘)(š‘ž + š‘‘š‘ž), equals 1.10 times nominal GDP at the beginning of the year, š‘š‘ž. That is (š‘ + š‘‘š‘)(š‘ž + š‘‘š‘ž) = 1.10š‘š‘ž

(4.42)

Furthermore, the price level at the end of the year, š‘ + š‘‘š‘ equals 1.05 times the price level of the start of year, š‘: š‘ + š‘‘š‘ = 1.05š‘ Substituting this in equation (4.38) yields 1.05š‘(š‘ž + š‘‘š‘ž) = 1.10š‘š‘ž which can be solved to give š‘‘š‘ž = (

1.10 āˆ’ 1)š‘ž = 0.0476 1.05

The growth rate of real GDP (š‘‘š‘ž/š‘ž) is equal to 4.76 per cent. 207

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

To show how the estimate of 5 per cent involves a linear approximation, we expand the expression for real GDP at the end of the year. (š‘ + š‘‘š‘)(š‘ž + š‘‘š‘ž) = š‘š‘ž + š‘š‘‘š‘ž + š‘žš‘‘š‘ + š‘‘š‘š‘‘š‘ž Dividing by š‘š‘ž (š‘ + š‘‘š‘)(š‘ž + š‘‘š‘ž) š‘‘š‘ž š‘‘š‘ š‘‘š‘š‘‘š‘ž =1+ + + š‘š‘ž š‘ž š‘ š‘š‘ž The growth rate of nominal GDP is (š‘ + š‘‘š‘)(š‘ž + š‘‘š‘ž) āˆ’ š‘š‘ž (š‘ + š‘‘š‘)(š‘ž + š‘‘š‘ž) = āˆ’1 š‘š‘ž š‘š‘ž š‘‘š‘ž š‘‘š‘ š‘‘š‘š‘‘š‘ž = + + š‘ž š‘‘š‘ š‘š‘ž = Growth rate of output + Inļ¬‚ation rate + Error term For small changes, the error term š‘‘š‘š‘‘š‘ž/š‘š‘ž is insigniļ¬cant, and we can approximate the growth rate of output according to the sum Growth rate of nominal GDP = Growth rate of output + Inļ¬‚ation rate This is a linear approximation since it approximates the function (š‘ + š‘‘š‘)(š‘ž + š‘‘š‘ž) by the linear function š‘š‘ž + š‘š‘‘š‘ž + š‘žš‘‘š‘. In eļ¬€ect, we are evaluating the change output at the old prices, and the change in prices at the old output, and ignoring in interaction between changes in prices and changes in quantities. The use of linear approximation in growth rates is extremely common in practice. 4.3 From (4.2) āˆ„xāˆ„ šœ‚(x) = š‘“ (x0 + x) āˆ’ š‘“ (x0 ) āˆ’ š‘”(x) and therefore šœ‚(x) =

š‘“ (x0 + x) āˆ’ š‘“ (x0 ) āˆ’ š‘”(x) āˆ„xāˆ„

šœ‚(x) ā†’ 0š‘Œ as x ā†’ 0š‘‹ can be expressed as lim šœ‚(x) = 0š‘Œ

xā†’0š‘‹

4.4 Suppose not. That is, there exist two linear maps such that š‘“ (x0 + x) = š‘“ (x0 ) + š‘”1 (x) + āˆ„xāˆ„ šœ‚1 (x) š‘“ (x0 + x) = š‘“ (x0 ) + š‘”2 (x) + āˆ„xāˆ„ šœ‚2 (x) with lim šœ‚š‘– (x) = 0,

xā†’0

208

š‘– = 1, 2

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Subtracting we have šæ1 (x) āˆ’ šæ2 (x) = āˆ„xāˆ„ (šœ‚1 (x) āˆ’ šœ‚2 (x)) and lim

xā†’0

š‘”1 (x) āˆ’ š‘”2 (x) =0 āˆ„xāˆ„

Since š‘”1 āˆ’ š‘”2 is linear, (4) implies that š‘”1 (x) = š‘”2 (x) for all x āˆˆ š‘‹. To see this, we proceed by contradiction. Again, suppose not. That is, suppose there exists some x āˆˆ š‘‹ such that š‘”1 (x) āˆ•= š‘”2 (x) For this x, let šœ‚=

š‘”1 (x) āˆ’ š‘”2 (x) āˆ„xāˆ„

By linearity, š‘”1 (š‘”x) āˆ’ š‘”2 (š‘”x) = šœ‚ for every āˆ€š‘” > 0 āˆ„š‘”xāˆ„ and therefore lim

š‘”ā†’0

š‘”1 (š‘”x) āˆ’ š‘”2 (š‘”x) = šœ‚ āˆ•= 0 āˆ„š‘”xāˆ„

which contradicts (4). Therefore š‘”1 (x) = š‘”2 (x) for all x āˆˆ š‘‹. 4.5 If š‘“ : š‘‹ ā†’ š‘Œ is diļ¬€erentiable at x0 , then š‘“ (x0 + x) = š‘“ (x0 ) + š‘”(x) + šœ‚(x) āˆ„xāˆ„ where šœ‚(x) ā†’ 0š‘Œ as x ā†’ 0š‘‹ . Since š‘” is a continuous linear function, š‘”(x) ā†’ 0š‘Œ as x ā†’ 0š‘‹ . Therefore lim š‘“ (x0 + x) = lim š‘“ (x0 ) + lim š‘”(x) + lim šœ‚(x) āˆ„xāˆ„

xā†’0

xā†’0

xā†’0

xā†’0

= š‘“ (x0 ) š‘“ is continuous. 4.6 4.7 4.8 The approximation error at the point (2, 16) is š‘“ (2, 16) š‘”(2, 16) Absolute error Percentage error Relative error

=8.0000 =11.3333 =-3.3333 =-41.6667 =-4.1667

By contrast, ā„Ž(2, 16) = 8 = š‘“ (2, 16). Table 4.1 shows that ā„Ž gives a good approximation to š‘“ in the neighborhood of (2, 16). 209

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Table 4.1: Approximating the Cobb-Douglas function at (2, 16) x x0 + x At their intersection: (0.0, 0.0) (2.0, 16.0)

Approximation Error Percentage Relative

š‘“ (x0 + x)

ā„Ž(x0 + x)

8.0000

8.0000

0.0000

NIL

Around the unit circle: (1.0, 0.0) (3.0, 16.0) (0.7, 0.7) (2.7, 16.7) (0.0, 1.0) (2.0, 17.0) (-0.7, 0.7) (1.3, 16.7) (-1.0, 0.0) (1.0, 16.0) (-0.7, -0.7) (1.3, 15.3) (0.0, -1.0) (2.0, 15.0) (0.7, -0.7) (2.7, 15.3)

9.1577 9.1083 8.3300 7.1196 6.3496 6.7119 7.6631 8.5867

9.3333 9.1785 8.3333 7.2929 6.6667 6.8215 7.6667 8.7071

-1.9177 -0.7712 -0.0406 -2.4342 -4.9934 -1.6323 -0.0466 -1.4018

-0.1756 -0.0702 -0.0034 -0.1733 -0.3171 -0.1096 -0.0036 -0.1204

Around a smaller circle: (0.10, 0.00) (2.1, 16.0) (0.07, 0.07) (2.1, 16.1) (0.00, 0.10) (2.0, 16.1) (-0.07, 0.07) (1.9, 16.1) (-0.10, 0.00) (1.9, 16.0) (-0.07, -0.07) (1.9, 15.9) (0.00, -0.10) (2.0, 15.9) (0.07, -0.07) (2.1, 15.9)

8.1312 8.1170 8.0333 7.9279 7.8644 7.8813 7.9666 8.0693

8.1333 8.1179 8.0333 7.9293 7.8667 7.8821 7.9667 8.0707

-0.0266 -0.0103 -0.0004 -0.0181 -0.0291 -0.0110 -0.0004 -0.0171

-0.0216 -0.0083 -0.0003 -0.0143 -0.0229 -0.0087 -0.0003 -0.0138

Parallel to the (-2.0, 0.0) (-1.0, 0.0) (-0.5, 0.0) (-0.1, 0.0) (0.0, 0.0) (0.1, 0.0) (0.5, 0.0) (1.0, 0.0) (2.0, 0.0) (4.0, 0.0)

x1 axis: (0.0, 16.0) (1.0, 16.0) (1.5, 16.0) (1.9, 16.0) (2.0, 16.0) (2.1, 16.0) (2.5, 16.0) (3.0, 16.0) (4.0, 16.0) (6.0, 16.0)

0.0000 6.3496 7.2685 7.8644 8.0000 8.1312 8.6177 9.1577 10.0794 11.5380

5.3333 6.6667 7.3333 7.8667 8.0000 8.1333 8.6667 9.3333 10.6667 13.3333

NIL -4.9934 -0.8922 -0.0291 0.0000 -0.0266 -0.5678 -1.9177 -5.8267 -15.5602

-2.6667 -0.3171 -0.1297 -0.0229 NIL -0.0216 -0.0979 -0.1756 -0.2936 -0.4488

Parallel to the (0.0, -4.0) (0.0, -2.0) (0.0, -1.0) (0.0, -0.5) (0.0, -0.1) (0.0, 0.0) (0.0, 0.1) (0.0, 0.5) (0.0, 1.0) (0.0, 2.0) (0.0, 4.0)

x2 axis: (2.0, 12.0) (2.0, 14.0) (2.0, 15.0) (2.0, 15.5) (2.0, 15.9) (2.0, 16.0) (2.0, 16.1) (2.0, 16.5) (2.0, 17.0) (2.0, 18.0) (2.0, 20.0)

6.6039 7.3186 7.6631 7.8325 7.9666 8.0000 8.0333 8.1658 8.3300 8.6535 9.2832

6.6667 7.3333 7.6667 7.8333 7.9667 8.0000 8.0333 8.1667 8.3333 8.6667 9.3333

-0.9511 -0.2012 -0.0466 -0.0112 -0.0004 0.0000 -0.0004 -0.0105 -0.0406 -0.1522 -0.5403

-0.0157 -0.0074 -0.0036 -0.0018 -0.0003 NIL -0.0003 -0.0017 -0.0034 -0.0066 -0.0125

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4.9 To show that š‘Ÿ is nonlinear, consider š‘Ÿ((1, 2, 3, 4, 5) + (66, 55, 75, 81, 63)) = š‘Ÿ(67, 57, 78, 85, 68) = (85, 78, 68, 67, 58) āˆ•= (5, 4, 3, 2, 1) + (81, 75, 67, 63, 55) To show that š‘Ÿ is diļ¬€erentiable, consider a particular point, say (66, 55, 75, 81, 63). Consider the permutation š‘” : ā„œš‘› ā†’ ā„œš‘› deļ¬ned by š‘”(š‘„1 , š‘„2 , . . . , š‘„5 ) = (š‘„4 , š‘„3 , š‘„1 , š‘„5 , š‘„2 ) š‘” is linear and š‘”(66, 55, 75, 81, 63) = (81, 75, 67, 63, 55) = š‘Ÿ(66, 55, 75, 81, 63) Furthermore, š‘”(x) = š‘Ÿ(x) for all x close to (66, 55, 75, 81, 63). Hence, š‘”(x) approximates š‘Ÿ(x) in a neighborhood of (66, 55, 75, 81, 63) and so š‘Ÿ is diļ¬€erentiable at (66, 55, 75, 81, 63). The choice of (66, 55, 75, 81, 63) was arbitrary, and the argument applies at every x such that xš‘– āˆ•= xš‘— . In summary, each application of š‘Ÿ involves a permutation, although the particular permutation depends upon the argument, x. However, for any given x0 with x0š‘– āˆ•= x0š‘— , the same permutation applies to all x in the neighborhood of x0 , so that the permutation (which is a linear function) is the derivative of š‘Ÿ at x0 . 4.10 Using (4.3), we have for any x š‘“ (x0 + š‘”x) āˆ’ š‘“ (x0 ) āˆ’ š·š‘“ [x0 ](š‘”x) =0 š‘”xā†’0 āˆ„š‘”xāˆ„ lim

or š‘“ (x0 + š‘”x) āˆ’ š‘“ (x0 ) āˆ’ š‘”š·š‘“ [x0 ](x) =0 š‘”ā†’0 š‘” āˆ„xāˆ„ lim

For āˆ„xāˆ„ = 1, this implies š‘”š·š‘“ [x0 ](x) š‘“ (x0 + š‘”x) āˆ’ š‘“ (x0 ) = š‘”ā†’0 š‘” š‘” lim

that is 0 0 āƒ— x š‘“ [x0 ] = lim š‘“ (x + š‘”x) āˆ’ š‘“ (x ) = š·š‘“ [x0 ](x) š· š‘”ā†’0 š‘”

4.11 By direct calculation ā„Ž(x0š‘– + š‘”) āˆ’ ā„Ž(x0š‘– ) š‘”ā†’0 š‘” š‘“ (x01 , x02 , . . . , x0š‘– + š‘”, . . . , x0š‘› ) āˆ’ š‘“ (x01 , x02 , . . . , x0š‘– , . . . , x0š‘› ) = lim š‘”ā†’0 š‘” š‘“ (x0 + š‘”eš‘– ) āˆ’ š‘“ (x0 ) = lim š‘”ā†’0 š‘” 0 āƒ— = š·eš‘– š‘“ [x ]

š·š‘„š‘– š‘“ [x0 ] = lim

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c 2001 Michael Carter āƒ All rights reserved

4.12 Deļ¬ne the function ( ) ā„Ž(š‘”) = š‘“ (8, 8) + š‘”(1, 1) = (8 + š‘”)1/3 (8 + š‘”)2/3 =8+š‘” The directional derivative of š‘“ in the direction (1, 1) is āƒ— (1,1) š‘“ (8, 8) = lim ā„Ž(š‘”) āˆ’ ā„Ž(0) š· š‘”ā†’0 š‘” =1 Generalization of this example reveals that the directional derivative of š‘“ along any āƒ— x0 š‘“ [x0 ] = 1 for every x0 . Economically, ray through the origin equals 1, that is š· this means that increasing inputs in the same proportions leads to a proportionate increase in output, which is the property of constant returns to scale. We will study this property of homogeneity is some depth in Section 4.6. 4.13 Let p = āˆ‡š‘“ (x0 ). Each component of p represents the action of the derivative on an element of the standard basis {e1 , e2 , . . . , eš‘› }(see proof of Theorem 3.4) š‘š‘– = š·š‘“ [x0 ](eš‘– ) Since āˆ„eš‘– āˆ„ = 1, š·š‘“ [x0 ](eš‘– ) is the directional derivative at x0 in the direction eš‘– (Exercise 4.10) āƒ— eš‘– (x0 ) š‘š‘– = š·š‘“ [x0 ](eš‘– ) = š· But this is simply the š‘– partial derivative of š‘“ (Exercise 4.11) āƒ— eš‘– (x0 ) = š·š‘„š‘– š‘“ (x0 ) š‘š‘– = š·š‘“ [x0 ](eš‘– ) = š· 4.14 Using the standard inner product on ā„œš‘› (Example 3.26) and Exercise 4.13 < āˆ‡š‘“ (x0 ), x >=

š‘› āˆ‘

š·š‘„š‘– š‘“ [x0 ]xš‘– = š·š‘“ [x0 ](x)

š‘–=1

4.15 Since š‘“ is diļ¬€erentiable š‘“ (x1 + š‘”x) = š‘“ (x1 ) + āˆ‡š‘“ (x0 )š‘‡ š‘”x + šœ‚(š‘”x) āˆ„š‘”xāˆ„ with šœ‚(š‘”x) ā†’ 0 as š‘”x ā†’ 0. If š‘“ is increasing, š‘“ (x1 + š‘”x) ā‰„ š‘“ (x1 ) for every x ā‰„ 0 and š‘” > 0. Therefore āˆ‡š‘“ (x0 )š‘‡ š‘”x + šœ‚(š‘”x) āˆ„š‘”xāˆ„ = š‘”āˆ‡š‘“ (x0 )š‘‡ x + š‘”šœ‚(š‘”x) āˆ„xāˆ„ ā‰„ 0 Dividing by š‘” and letting š‘” ā†’ 0 āˆ‡š‘“ (x0 )š‘‡ x ā‰„ 0 for every x ā‰„ 0 In particular, this applies for unit vectors eš‘– . Therefore š·š‘„š‘– š‘“ (x1 ) ā‰„ 0,

š‘– = 1, 2, . . . , š‘›

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āƒ— x š‘“ (x0 ) measures the rate of increase of š‘“ in the di4.16 The directional derivative š· rection x. Using Exercises 4.10, 4.14 and 3.61, assuming x has unit norm,   āƒ— x š‘“ (x0 ) = š·š‘“ [x0 ](x) =< āˆ‡š‘“ (x0 ), x >ā‰¤ āˆ‡š‘“ (x0 ) š·   This bound is attained when x = āˆ‡š‘“ (x0 )/ āˆ‡š‘“ (x0 ) since   āˆ‡š‘“ (x0 )2   āˆ‡š‘“ (x0 ) 0 0 āƒ— š·x š‘“ (x ) =< āˆ‡š‘“ (x ), >= = āˆ‡š‘“ (x0 ) 0 0 āˆ„āˆ‡š‘“ (x )āˆ„ āˆ„āˆ‡š‘“ (x )āˆ„ The directional derivative is maximized when āˆ‡š‘“ (x0 ) and x are aligned. 4.17 Using Exercise 4.14 š» = { x āˆˆ š‘‹ :< āˆ‡š‘“ [x0 ], x >= 0 } 4.18 Assume each š‘“š‘— is diļ¬€erentiable at x0 and let š·š‘“ [x0 ] = (š·š‘“1 [x0 ], š·š‘“2 [x0 ], . . . , š·š‘“š‘š [x0 ]) Then

āŽ› āŽœ āŽœ f (x0 + x) āˆ’ f [x0 ] āˆ’ š·f [x0 ]x = āŽœ āŽ

š‘“1 (x0 + x) āˆ’ š‘“1 [x0 ] āˆ’ š·š‘“1 [x0 ]x š‘“2 (x0 + x) āˆ’ š‘“2 [x0 ] āˆ’ š·š‘“2 [x0 ]x .. .

āŽž āŽŸ āŽŸ āŽŸ āŽ 

š‘“š‘š (x0 + x) āˆ’ š‘“š‘š (x0 ) āˆ’ š·š‘“š‘š [x0 ]x and š‘“š‘— (x0 + x) āˆ’ š‘“š‘— (x0 ) āˆ’ š·š‘“š‘— [x0 ]x ā†’ 0 as āˆ„xāˆ„ ā†’ 0 āˆ„xāˆ„ for every š‘— implies f (x0 + x) āˆ’ f (x0 ) āˆ’ š·f [x0 ](x) ā†’ 0 as āˆ„xāˆ„ ā†’ 0 āˆ„xāˆ„

(4.43)

Therefore f is diļ¬€erentiable with derivative š·f [x0 ] = šæ = (š·š‘“1 (x0 ), š·š‘“2 [x0 ], . . . , š·š‘“š‘š [x0 ]) Each š·š‘“š‘— [x0 ] is represented by the gradient āˆ‡š‘“š‘— [x0 ] (Exercise 4.13) and therefore š·š‘“ [x0 ] is represented by the matrix āŽ› āŽž āŽ› āŽž āˆ‡š‘“1 [x0 ] š·š‘„1 š‘“1 [x0 ] š·š‘„2 š‘“1 [x0 ] . . . š·š‘„š‘› š‘“1 [x0 ] āŽœ āˆ‡š‘“2 [x0 ] āŽŸ āŽœ š·š‘„1 š‘“2 [x0 ] š·š‘„2 š‘“2 [x0 ] . . . š·š‘„š‘› š‘“2 [x0 ] āŽŸ āŽœ āŽŸ āŽœ āŽŸ š½ =āŽœ āŽŸ=āŽœ āŽŸ .. .. .. .. .. āŽ āŽ  āŽ āŽ  . . . . . āˆ‡š‘“š‘š [x0 ]

š·š‘„1 š‘“š‘š [x0 ] š·š‘„2 š‘“š‘š [x0 ] . . .

š·š‘„š‘› š‘“š‘š [x0 ]

Conversely, if f is diļ¬€erentiable, its derivative š·f [x0 ] : ā„œš‘› ā†’ ā„œš‘š be decomposed into š‘š component š·š‘“1 [x0 ], š·š‘“2 [x0 ], . . . , š·š‘“š‘š [x0 ] functionals such that āŽž āŽ› š‘“1 (x0 + x) āˆ’ š‘“1 (x0 ) āˆ’ š·š‘“1 [x0 ]x āŽœ š‘“2 (x0 + x) āˆ’ š‘“2 (x0 ) āˆ’ š·š‘“2 [x0 ]x āŽŸ āŽŸ āŽœ f (x0 + x) āˆ’ f (x0 ) āˆ’ š·f [x0 ]x = āŽœ āŽŸ .. āŽ  āŽ . š‘“š‘š (x0 + x) āˆ’ š‘“š‘š (x0 ) āˆ’ š·š‘“š‘š [x0 ]x

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(4.43) implies that š‘“š‘— (x0 + x) āˆ’ š‘“š‘— (x0 ) āˆ’ š·š‘“š‘— [x0 ]x ā†’ 0 as āˆ„xāˆ„ ā†’ 0 āˆ„xāˆ„ for every š‘—. 4.19 If š·š‘“ [x0 ] has full rank, then it is one-to-one (Exercise 3.25) and onto (Exercise 3.16). Therefore š·š‘“ [x0 ] is nonsingular. The Jacobian š½š‘“ (x0 ) represents š·š‘“ [x0 ], which is therefore nonsingular if and only if det š½š‘“ (x0 ) āˆ•= 0. 4.20 When š‘“ is a functional, rank š‘‹ ā‰„ š‘Ÿš‘Žš‘›š‘˜š‘Œ = 1. If š·š‘“ [x0 ] has full rank (1), then š·š‘“ [x0 ] maps š‘‹ onto ā„œ (Exercise 3.16), which requires that āˆ‡š‘“ (x0 ) āˆ•= 0. 4.21 4.23 If š‘“ : š‘‹ Ɨ š‘Œ ā†’ š‘ is bilinear š‘“ (x0 + x, y0 + y) = š‘“ (x0 , y0 ) + š‘“ (x0 , y) + š‘“ (x, y0 ) + š‘“ (x, y) Deļ¬ning š·š‘“ [x0 , y0 ](x, y) = š‘“ (x0 , y) + š‘“ (x, y0 ) š‘“ (x0 + x, y0 + y) = š‘“ (x0 , y0 ) + š·š‘“ [x0 , y0 ](x, y) + š‘“ (x, y) Since š‘“ is continuous, there exists š‘€ such that š‘“ (x, y) ā‰¤ š‘€ āˆ„xāˆ„ āˆ„yāˆ„

for every x āˆˆ š‘‹ and y āˆˆ š‘Œ

and therefore NOTE This is not quite right. See Spivak p. 23. Avez (Tilburg) has ( )2 āˆ„š‘“ (x, y)āˆ„ ā‰¤ š‘€ āˆ„xāˆ„ āˆ„yāˆ„ ā‰¤ š‘€ āˆ„xāˆ„ + āˆ„yāˆ„ ā‰¤ š‘€ āˆ„(x, y)āˆ„2 which implies that āˆ„š‘“ (x, y)āˆ„ ā†’ 0 as (x, y) ā†’ 0 āˆ„(x, y)āˆ„

lim

x1 ,x2 ā†’0

š‘“ (x1 , x2 ) =0 āˆ„x1 āˆ„ āˆ„x2 āˆ„

Therefore š‘“ is diļ¬€erentiable with derivative š·š‘“ [x0 , y0 ] = š‘“ (x0 , y) + š‘“ (x, y0 ) 4.24 Deļ¬ne š‘š : ā„œ2 ā†’ ā„œ by š‘š(š‘§1 , š‘§2 ) = š‘§1 š‘§2 Then š‘š is bilinear (Example 3.23) and continuous (Exercise 2.79) and therefore diļ¬€erentiable (Exercise 4.23) with derivative š·š‘š[š‘§1 , z2 ] = š‘š(z1 , ā‹…) + š‘š(ā‹…, z2 ) 214

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

The function š‘“ š‘” is the composition of š‘š with š‘“ and š‘”, š‘“ š‘”(x, y) = š‘š(š‘“ (x), š‘”(y)) By the chain rule, the derivative of š‘“ š‘” is ( ) š·š‘“ š‘”[x, y] = š·š‘š[š‘§1 , z2 ] š·š‘“ [x], š·š‘”[x] = š‘š(z1 , š·š‘”[y]) + š‘š(š·š‘“ [x], z2 ) = š‘“ [x]š·š‘”[y]) + š‘”(y)š·š‘“ [x] where z1 = š‘“ (x) and z2 = š‘”(y). 4.25 For š‘› = 1, š‘“ (š‘„) = š‘„ is linear and therefore (Exercise 4.6) š·š‘“ [š‘„] = 1 (š·š‘“ [š‘„](š‘„) = š‘„). For š‘› = 2, let š‘”(š‘„) = š‘„ so that š‘“ (š‘„) = š‘„2 = š‘”(š‘„)š‘”(š‘„). Using the product rule š·š‘“ [x] = š‘”(š‘„)š·š‘”(š‘„) + š‘”(š‘„)š·š‘”(š‘„) = 2š‘„ Now assume it is true for š‘› āˆ’ 1 and let š‘”(š‘„) = š‘„š‘›āˆ’1 , so that š‘“ (x) = š‘„š‘”(š‘„). By the product rule š·š‘“ [x] = š‘„š·š‘”[š‘„] + š‘”(š‘„)1 By assumption š·š‘”[š‘„] = (š‘› āˆ’ 1)š‘„š‘›āˆ’2 and therefore š·š‘“ [x] = š‘„š·š‘”[š‘„] + š‘”(š‘„)1 = š‘„(š‘› āˆ’ 1)š‘„š‘›āˆ’2 + š‘„š‘›āˆ’1 = š‘›š‘„š‘›āˆ’1 4.26 Using the product rule (Exercise 4.24) š·š‘„ š‘…(š‘„0 ) = š‘“ (š‘„0 )š·š‘„ š‘„ + š‘„0 š·š‘„ š‘“ (š‘„0 ) = š‘0 + š‘„0 š·š‘„ š‘“ (š‘„0 ) where š‘0 = š‘“ (š‘„0 ). Marginal revenue equals one unit at the current price minus the reduction in revenue caused by reducing the price on existing sales. ( )āˆ’1 4.27 Fix some x0 and let š‘” = š·š‘“ [x0 ] . Let y0 = š‘“ (x0 ). For any y, let x = āˆ’1 0 āˆ’1 0 0 š‘“ (y + y) āˆ’ š‘“ (y ) so that š‘”(y) = š‘“ (x + x) āˆ’ š‘“ (x) and    āˆ’1 0 ( ) š‘“ (y + y) āˆ’ š‘“ āˆ’1 (y0 ) āˆ’ š‘”(y) = (x āˆ’ š‘” š‘“ (x0 + x) āˆ’ š‘“ (x0 ))  Since š‘“ is diļ¬€erentiable at x0 with š·š‘“ [x0 ] = š‘” āˆ’1 š‘“ (x0 + x) āˆ’ š‘“ (x0 ) = š‘” āˆ’1 (x) + šœ‚(x) āˆ„xāˆ„ Substituting ( )  āˆ’1 0    š‘“ (y + y) āˆ’ š‘“ āˆ’1 (y0 ) āˆ’ š‘”(y) =  x āˆ’ š‘” š‘” āˆ’1 (x) + šœ‚(x) āˆ„xāˆ„   ( )   = š‘” šœ‚(x) āˆ„xāˆ„   ( )   = āˆ„xāˆ„ š‘” šœ‚(x)  ( ) with šœ‚(x) ā†’ 0š‘Œ as x ā†’ 0š‘‹ . Since š‘“ āˆ’1 and š‘” are continuous, š‘” šœ‚(x) ā†’ 0š‘‹ as y ā†’ 0. )āˆ’1 ( . We conclude that š‘“ āˆ’1 is diļ¬€erentiable with derivative š‘” = š·š‘“ [x0 ]

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Solutions for Foundations of Mathematical Economics 4.28 log š‘“ (š‘„) = š‘„ log š‘Ž and therefore

( ) š‘“ (š‘„) = exp log š‘“ (š‘„) = š‘’š‘„ log š‘Ž

By the Chain Rule, š‘“ is diļ¬€erentiable with derivative š·š‘„ š‘“ (š‘„) = š‘’š‘„ log š‘Ž log š‘Ž = š‘Žš‘„ log š‘Ž 4.29 By Exercise 4.15, the function š‘” : ā„œ ā†’ ā„œ deļ¬ned by š‘”(š‘¦) = tiable with derivative 1 š·š‘¦ š‘”[š‘¦] = āˆ’š‘¦ āˆ’2 = āˆ’ 2 š‘¦

1 š‘¦

= š‘¦ āˆ’1 is diļ¬€eren-

Applying the Chain Rule, 1/š‘“ = š‘” āˆ˜ š‘“ is diļ¬€erentiable with derivative 1 š·š‘“ [x] š· [x] = š·š‘”[š‘“ (x)]š·š‘“ [x] = āˆ’ ( )2 š‘“ š‘“ (x) 4.30 Applying the Product Rule to š‘“ Ɨ (1/š‘”) 1 1 š‘“ š·š‘“ [x] š· [x, y] = š‘“ (x)š· [y] + š‘” š‘” š‘”(y) š·š‘”[y] 1 = āˆ’š‘“ (x) ( š·š‘“ [x] )2 + š‘”(y) š‘”(y) š‘”(y)š·š‘“ [x] āˆ’ š‘“ (x)š·š‘”[y] = ( )2 š‘”(y) 4.31 In the particular case where 1/3 2/3

š‘“ (x1 , x2 ) = x1 x2 the partial derivatives at the point (8, 8) are š·š‘„1 š‘“ [(8, 8)] =

2 1 and š·š‘„2 š¹ [(8, 8)] = 3 3

4.32 The partial derivatives of š‘“ (x) are from Table 4.4 š·š‘„š‘– š‘“ [x] = š‘„š‘Ž1 1 š‘„š‘Ž2 2 . . . š‘Žš‘– š‘„š‘–š‘Žš‘– āˆ’1 . . . š‘„š‘Žš‘›š‘› = š‘Žš‘– so that the gradient is

( āˆ‡š‘“ (x) =

š‘“ (x) š‘„š‘– š‘Ž1 š‘Ž2 š‘Žš‘› , ,..., š‘„1 š‘„2 š‘„š‘›

) š‘“ (x)

4.33 Applying the chain rule (Exercise 4.22) to general power function (Example 4.15), the partial derivatives of the CES function are š·š‘„š‘– š‘“ [x] =

1 1 āˆ’1 (š‘Ž1 š‘„šœŒ1 + š‘Ž2 š‘„šœŒ2 + ā‹… ā‹… ā‹… + š‘Žš‘› š‘„šœŒš‘› ) šœŒ š‘Žš‘– šœŒš‘„šœŒāˆ’1 š‘– šœŒ

= š‘Žš‘– š‘„šœŒāˆ’1 (š‘Ž1 š‘„šœŒ1 + š‘Ž2 š‘„šœŒ2 + ā‹… ā‹… ā‹… + š‘Žš‘› š‘„šœŒš‘› ) š‘– ( )1āˆ’šœŒ š‘“ (x) = š‘Žš‘– š‘„š‘– 216

1āˆ’šœŒ šœŒ

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

4.34 Deļ¬ne ā„Ž(š‘„) = š‘“ (š‘„) āˆ’

š‘“ (š‘) āˆ’ š‘“ (š‘Ž) (š‘„ āˆ’ š‘Ž) š‘āˆ’š‘Ž

Then ā„Ž is continuous on [š‘Ž, š‘] and diļ¬€erentiable on (š‘Ž, š‘) with ā„Ž(š‘) = š‘“ (š‘) āˆ’

š‘“ (š‘) āˆ’ š‘“ (š‘Ž) (š‘ āˆ’ š‘Ž)š‘“ (š‘Ž) = ā„Ž(š‘Ž) š‘āˆ’š‘Ž

By Rolleā€™s theorem (Exercise 5.8), there exists š‘„ āˆˆ (š‘Ž, š‘) such that ā„Žā€² (š‘„) = š‘“ ā€² (š‘„) āˆ’

š‘“ (š‘) āˆ’ š‘“ (š‘Ž) =0 š‘āˆ’š‘Ž

4.35 Assume āˆ‡š‘“ (x) ā‰„ 0 for every x āˆˆ š‘‹. By the mean value theorem, for any x2 ā‰„ x1 ĀÆ āˆˆ (x1 , x2 ) such that in š‘‹, there exists x š‘“ (x2 ) = š‘“ (x1 ) + š·š‘“ [ĀÆ x](x2 āˆ’ x1 ) Using (4.6) š‘“ (x2 ) = š‘“ (x1 ) +

š‘› āˆ‘

š·š‘„š‘– š‘“ (ĀÆ x)(š‘„2š‘– āˆ’ š‘„1š‘– )

(4.44)

š‘–=1

āˆ‡š‘“ (ĀÆ x) ā‰„ 0 and x2 ā‰„ x1 implies that š‘› āˆ‘

š·š‘„š‘– š‘“ (ĀÆ x)(š‘„2š‘– āˆ’ š‘„1š‘– ) ā‰„ 0

š‘–=1

and therefore š‘“ (x2 ) ā‰„ š‘“ (x1 ). š‘“ is increasing. The converse was established in Exercise 4.15 4.36 āˆ‡š‘“ (ĀÆ x) > 0 and x2 ā‰„ x1 implies that š‘› āˆ‘

š·š‘„š‘– š‘“ (ĀÆ x)(š‘„2š‘– āˆ’ š‘„1š‘– ) > 0

š‘–=1

Substituting in (4.44) š‘“ (x2 ) = š‘“ (x1 ) +

š‘› āˆ‘

š·š‘„š‘– š‘“ (ĀÆ x)(š‘„2š‘– āˆ’ š‘„1š‘– ) > š‘“ (x1 )

š‘–=1

š‘“ is strictly increasing. 4.37 Diļ¬€erentiability implies the existence of the gradient and hence the partial derivatives of š‘“ (Exercise 4.13). Continuity of š·š‘“ [x] implies the continuity of the partial derivatives. To prove the converse, choose some x0 āˆˆ š‘† and deļ¬ne for the partial functions ā„Žš‘– (š‘”) = š‘“ (š‘„01 , š‘„02 , . . . , š‘„0š‘–āˆ’1 , š‘”, š‘„0š‘–+1 + š‘„š‘–+1 , . . . , š‘„0š‘› + š‘„š‘› )

š‘– = 1, 2, . . . , š‘›

so that ā„Žā€²š‘– (š‘”) = š·š‘„š‘– š‘“ (xš‘– ) where xš‘– = (š‘„01 , š‘„02 , . . . , š‘„0š‘– , š‘”, š‘„0š‘–+1 + š‘„š‘–+1 , . . . , š‘„0š‘› + š‘„š‘› ). Further, ā„Ž1 (š‘„01 + š‘„1 ) = š‘“ (x0 + x), ā„Žš‘› (š‘„0š‘› ) = š‘“ (x0 ), and ā„Žš‘– (š‘„0š‘– + š‘„š‘– ) = ā„Žš‘–āˆ’1 (š‘„0š‘– ) so that š‘“ (x0 + x) āˆ’ š‘“ (x0 ) =

š‘› āˆ‘ ( ) ā„Žš‘– (š‘„0š‘– + š‘„š‘– ) āˆ’ ā„Žš‘– (š‘„0š‘– ) š‘–=1

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By the mean value theorem, there exists, for each š‘–, š‘”ĀÆš‘– between š‘„0š‘– + š‘„š‘– and š‘„š‘– such that ā„Žš‘– (š‘„0š‘– + š‘„š‘– ) āˆ’ ā„Žš‘– (š‘„š‘– ) = š·š‘„š‘– š‘“ (ĀÆ xš‘– )š‘„š‘– ĀÆ š‘– = (š‘„01 , š‘„02 , . . . , š‘„0š‘– , š‘”ĀÆ, š‘„0š‘–+1 + š‘„š‘–+1 , . . . , š‘„0š‘› + š‘„š‘› ). Therefore where x š‘“ (x0 + x) āˆ’ š‘“ (x0 ) =

š‘› āˆ‘

š·š‘„š‘– š‘“ (ĀÆ xš‘– )š‘„š‘–

š‘–=1

Deļ¬ne the linear functional š‘”(x) =

š‘› āˆ‘

š·š‘„š‘– š‘“ (x0 )š‘„š‘–

š‘–=1

Then š‘“ (x0 + x) āˆ’ š‘“ (x0 ) āˆ’ š‘”(x) =

š‘› ( ) āˆ‘ š·š‘„š‘– š‘“ (ĀÆ xš‘– ) āˆ’ š·š‘„š‘– š‘“ (x0 ) š‘„š‘– š‘–=1

and š‘›   āˆ‘   š‘“ (x0 + x) āˆ’ š‘“ (x0 ) āˆ’ š‘”(x) ā‰¤ (š·š‘„š‘– š‘“ (ĀÆ xš‘– ) āˆ’ š·š‘„š‘– š‘“ (x0 ) āˆ£š‘„š‘– āˆ£ š‘–=1

so that

  š‘› š‘“ (x0 + x) āˆ’ š‘“ (x0 ) āˆ’ š‘”(x) āˆ‘   āˆ£š‘„š‘– āˆ£ (š·š‘„š‘– š‘“ (ĀÆ ā‰¤ lim xš‘– ) āˆ’ š·š‘„š‘– š‘“ (x0 ) xā†’0 āˆ„xāˆ„ āˆ„xāˆ„ š‘–=1 ā‰¤

š‘› āˆ‘   (š·š‘„š‘– š‘“ (ĀÆ xš‘– ) āˆ’ š·š‘„š‘– š‘“ (x0 ) š‘–=1

=0 since the partial derivatives š·š‘„š‘– š‘“ (x) are continuous. Therefore š‘“ is diļ¬€erentiable with derivative š‘”(x) =

š‘› āˆ‘

š·š‘„š‘– š‘“ [x0 ]š‘„š‘–

š‘–=1

4.38 For every x1 , x2 āˆˆ š‘† āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 )āˆ„ ā‰¤

sup

xāˆˆ[x1 ,x2 ]

āˆ„š·š‘“ (x)āˆ„ āˆ„x1 āˆ’ x2 āˆ„

by Corollary 4.1.1. If š·š‘“ [x] = 0 for every x āˆˆ š‘‹, then āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 )āˆ„ = 0 which implies that š‘“ (x1 ) = š‘“ (x2 ). We conclude that š‘“ is constant on š‘†. The converse was established in Exercise 4.7. 4.39 For any x0 āˆˆ š‘†, let šµ āŠ† š‘† be an open ball of radius of radius š‘Ÿ centered on x0 . Applying the mean value inequality (Corollary 4.1.1) to š‘“š‘› āˆ’ š‘“š‘š we have  ( ) š‘“š‘› (x) āˆ’ š‘“š‘š (x) āˆ’ š‘“š‘› (x0 ) āˆ’ š‘“š‘š (x0 )  ā‰¤ sup āˆ„š·š‘“š‘› [ĀÆ x] āˆ’ š·š‘“š‘š [ĀÆ x]āˆ„ āˆ„x āˆ’ x0 āˆ„ ĀÆ āˆˆšµ x

ā‰¤ š‘Ÿ sup āˆ„š·š‘“š‘› [ĀÆ x] āˆ’ š·š‘“š‘š [ĀÆ x]āˆ„ ĀÆ āˆˆšµ x

218

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for every x āˆˆ šµ. Given šœ– > 0, there exists š‘ such that for every š‘š, š‘› > š‘ āˆ„š·š‘“š‘› āˆ’ š·š‘“š‘š āˆ„ < šœ–/š‘Ÿ and āˆ„š·š‘“š‘› āˆ’ š‘”āˆ„ < šœ– Letting š‘š ā†’ āˆž

 ( ) š‘“š‘› (x) āˆ’ š‘“ (x) āˆ’ š‘“š‘› (x0 ) āˆ’ š‘“ (x0 )  ā‰¤ šœ– āˆ„x āˆ’ x0 āˆ„

(4.45)

for š‘› ā‰„ š‘ and x āˆˆ šµ. Applying the mean value inequality to š‘“š‘› , there exists š›æ such that āˆ„š‘“š‘› (x) āˆ’ š‘“š‘› (x0 )āˆ„ ā‰¤ šœ– āˆ„x āˆ’ x0 āˆ„

(4.46)

Using (4.45) and (4.46) and the fact that āˆ„š·š‘“š‘› āˆ’ š‘”āˆ„ < šœ– we deduce that āˆ„š‘“ (x) āˆ’ š‘“ (x0 ) āˆ’ š‘”(x0 )āˆ„ ā‰¤ 3šœ– āˆ„x āˆ’ x0 āˆ„ š‘“ is diļ¬€erentiable with derivative š‘”. 4.40 Deļ¬ne š‘“ (š‘„) =

š‘’š‘„+š‘¦ š‘’š‘¦

By the chain rule (Exercise 4.22) š‘“ ā€² (š‘„) =

š‘’š‘„+š‘¦ = š‘“ (š‘„) š‘’š‘¦

which implies (Example 4.21) that š‘“ (š‘„) =

š‘’š‘„+š‘¦ = š“š‘’š‘„ for some š“ āˆˆ ā„œ š‘’š‘¦

Evaluating at š‘„ = 0 using š‘’0 = 1 gives š‘“ (0) =

š‘’š‘¦ = š“ for some š“ āˆˆ ā„œ š‘’š‘¦

so that š‘“ (š‘„) =

š‘’š‘¦ š‘„ š‘’š‘„+š‘¦ = š‘’ š‘’š‘¦ š‘’š‘¦

which implies that š‘’š‘„+š‘¦ = š‘’š‘„ š‘’š‘¦ 4.41 If š‘“ = š“š‘„š‘Ž , š‘“ ā€² (š‘„) = š‘Žš“š‘„š‘Žāˆ’1 and šø(š‘„) = š‘„

š‘Žš“š‘„š‘Žāˆ’1 =š‘Ž š“š‘„š‘Ž

To show that this is the only function with constant elasticity, deļ¬ne š‘”(š‘„) =

š‘“ (š‘„) š‘„š‘Ž

š‘” is diļ¬€erentiable (Exercise 4.30) with derivative š‘” ā€² (š‘„) =

š‘„š‘Ž š‘“ ā€² (š‘„) āˆ’ š‘“ (š‘„)š‘Žš‘„š‘Žāˆ’1 š‘„š‘“ ā€² (š‘„) āˆ’ š‘Žš‘“ (š‘„) = š‘„2š‘Ž š‘„š‘Ž+1 219

(4.47)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics If šø(š‘„) = š‘„

š‘“ ā€² (š‘„) =š‘Ž š‘“ (š‘„)

then š‘„š‘“ ā€² (š‘„) = š‘Žš‘“ (š‘„) Substituting in (4.47) š‘” ā€² (š‘„) =

š‘„š‘“ ā€² (š‘„) āˆ’ š‘Žš‘“ (š‘„) = 0 for every š‘„ āˆˆ ā„œ š‘„š‘Ž+1

Therefore, š‘” is a constant function (Exercise 4.38). That is, there exists š“ āˆˆ ā„œ such that š‘”(š‘„) =

š‘“ (š‘„) = š“ or š‘“ (š‘„) = š“š‘„š‘Ž š‘„š‘Ž

4.42 Deļ¬ne š‘” : š‘† ā†’ š‘Œ by š‘”(x) = š‘“ (x) āˆ’ š·š‘“ [x0 ](x) š‘” is diļ¬€erentiable with š·š‘”[x] = š·š‘“ [x] āˆ’ š·š‘“ [x0 ] Applying Corollary 4.1.1 to š‘”, āˆ„š‘”(x1 ) āˆ’ š‘”(x2 )āˆ„ ā‰¤

sup

xāˆˆ[x1 ,x2 ]

āˆ„š·š‘”[x]āˆ„ āˆ„x1 āˆ’ x2 āˆ„

for every x1 , x2 āˆˆ š‘†. Substituting for š‘” and š·š‘” āˆ„š‘“ (x1 ) āˆ’ š·š‘“ [x0 ](x1 ) āˆ’ š‘“ (x2 ) + š·š‘“ [x0 ](x2 )āˆ„ = āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 ) āˆ’ š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„ ā‰¤

sup

xāˆˆ[x1 ,x2 ]

āˆ„š·š‘“ [x] āˆ’ š·š‘“ [x0 ]āˆ„ āˆ„x1 āˆ’ x2 āˆ„

4.43 Since š·š‘“ is continuous, there exists a neighborhood š‘† of x0 such that āˆ„š·š‘“ [x] āˆ’ š·š‘“ [x0 ]āˆ„ < šœ– for every x āˆˆ š‘† and therefore for every x1 , x2 āˆˆ š‘† sup

xāˆˆ[x1 ,x2 ]

āˆ„š·š‘“ [x] āˆ’ š·š‘“ [x0 ]āˆ„ < šœ–

By the previous exercise (Exercise 4.42) āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 ) āˆ’ š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„ ā‰¤ šœ– āˆ„x1 āˆ’ x2 āˆ„ 4.44 By the previous exercise (Exercise 4.43), there exists a neighborhood such that āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 ) āˆ’ š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„ ā‰¤ šœ– āˆ„x1 āˆ’ x2 āˆ„ The Triangle Inequality (Exercise 1.200) implies āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 )āˆ„ āˆ’ āˆ„š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„ ā‰¤ āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 ) āˆ’ š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„ ā‰¤ šœ– āˆ„x1 āˆ’ x2 āˆ„ and therefore āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 )āˆ„ ā‰¤ āˆ„š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„ + šœ– āˆ„x1 āˆ’ x2 āˆ„ ā‰¤ āˆ„š·š‘“ [x0 ] + šœ–āˆ„ āˆ„x1 āˆ’ x2 āˆ„ 220

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 4.45 Assume not. That is, assume that y = š‘“ (x1 ) āˆ’ š‘“ (x2 ) āˆ•āˆˆ conv š“

Then by the (strong) separating hyperplane theorem (Proposition 3.14) there exists a linear functional šœ‘ on š‘Œ such that šœ‘(y) > šœ‘(a)

for every a āˆˆ š“

(4.48)

where šœ‘(š‘¦) = šœ‘(š‘“ (x1 ) āˆ’ š‘“ (x2 )) = šœ‘(š‘“ (x1 )) āˆ’ šœ‘(š‘“ (x2 )) šœ‘š‘“ is a functional on š‘†. By the mean value theorem (Theorem 4.1), there exists some ĀÆ āˆˆ [x1 , x2 ] such that x x](x1 āˆ’ x2 ) = šœ‘ āˆ˜ š·š‘“ [ĀÆ x](x āˆ’ x2 ) = šœ‘(š‘Ž) šœ‘ āˆ˜ š‘“ (x1 ) āˆ’ šœ‘ āˆ˜ š‘“ (x2 )) = š·(šœ‘ āˆ˜ š‘“ )[ĀÆ for some š‘Ž āˆˆ š“ contradicting (4.44). 4.46 Deļ¬ne ā„Ž : [š‘Ž, š‘] ā†’ ā„œ by

( ) ( ) ā„Ž(š‘„) = š‘“ (š‘) āˆ’ š‘“ (š‘Ž) š‘”(š‘„) āˆ’ š‘”(š‘) āˆ’ š‘”(š‘Ž) š‘“ (š‘„)

ā„Ž āˆˆ š¶[š‘Ž, š‘] and is diļ¬€erentiable on š‘Ž, š‘) with ( ) ( ) ā„Ž(š‘Ž) = š‘“ (š‘) āˆ’ š‘“ (š‘Ž) š‘”(š‘Ž) āˆ’ š‘”(š‘) āˆ’ š‘”(š‘Ž) š‘“ (š‘Ž) = š‘“ (š‘)š‘”(š‘Ž) āˆ’ š‘“ (š‘Ž)š‘”(š‘) = ā„Ž(š‘) By Rolleā€™s theorem (Exercise 5.8), there exists š‘„ āˆˆ (š‘Ž, š‘) such that ( ) ( ) ā„Žā€² (š‘„) = š‘“ (š‘) āˆ’ š‘“ (š‘Ž) š‘” ā€² (š‘„) āˆ’ š‘”(š‘) āˆ’ š‘”(š‘Ž) š‘“ ā€² (š‘„) = 0 4.47 The hypothesis that limš‘„ā†’š‘Ž š·š‘“ (š‘„)/š·š‘”(š‘„) exists contains two implicit assumptions, namely āˆ™ š‘“ and š‘” are diļ¬€erentiable on a neighborhood š‘† of š‘Ž (except perhaps at š‘Ž) āˆ™ š‘” ā€² (š‘„) āˆ•= 0 in this neighborhood (except perhaps at š‘Ž). Applying the Cauchy mean value theorem, for every š‘„ āˆˆ š‘†, there exists some š‘¦š‘„ āˆˆ (š‘Ž, š‘„) such that š‘“ ā€² (š‘¦š‘„ ) š‘“ (š‘„) āˆ’ š‘“ (š‘Ž) š‘“ (š‘„) = = š‘” ā€² (š‘¦š‘„ ) š‘”(š‘„) āˆ’ š‘”(š‘Ž) š‘”(š‘„) and therefore š‘“ (š‘„) š‘“ ā€² (š‘¦š‘„ ) š‘“ ā€² (š‘„) = lim ā€² = lim ā€² š‘„ā†’š‘Ž š‘”(š‘„) š‘„ā†’š‘Ž š‘” (š‘¦š‘„ ) š‘„ā†’š‘Ž š‘” (š‘„) lim

4.48 Let š“ = š‘Ž1 + š‘Ž2 + ā‹… ā‹… ā‹… + š‘Žš‘› āˆ•= 1. Then from (4.12) š‘Ž1 log š‘„1 + š‘Ž2 log š‘„2 + . . . š‘Žš‘› log š‘„š‘› š“ š‘Ž2 š‘Žš‘› š‘Ž1 log š‘„1 + log š‘„2 + . . . log š‘„š‘› = š“ š“ š“

lim š‘”(šœŒ) =

šœŒā†’0

and therefore lim log š‘“ (šœŒ, x) =

šœŒā†’0

š‘Ž1 š‘Ž2 š‘Žš‘› log š‘„1 + log š‘„2 + . . . log š‘„š‘› š“ š“ š“

so that š‘Ž1

š‘Ž1

š‘Ž1

lim š‘“ (šœŒ, x) = š‘„1š“ š‘„2š“ . . . š‘„š‘›š“

šœŒā†’0

which is homogeneous of degree one. 221

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

4.49 Average cost is given by š‘(š‘¦)/š‘¦ which is undeļ¬ned at š‘¦ = 0. We seek limš‘¦ā†’0 š‘(š‘¦)/š‘¦. By Lā€™HĖ† opitalā€™s rule š‘(š‘¦) š‘ā€² (š‘¦) = lim š‘¦ā†’0 š‘¦ š‘¦ā†’0 1 = š‘ā€² (0) lim

which is marginal cost at zero output. 4.50

1. Since limš‘„ā†’āˆž š‘“ ā€² (š‘„)/š‘” ā€² (š‘„) = k, for every šœ– > 0 there exists š‘Ž such that  ā€²   š‘“ (ĀÆ   š‘„) āˆ’ š‘˜  < šœ–/2 for every š‘„ ĀÆ>š‘Ž (4.49)  š‘” ā€² (ĀÆ  š‘„) For every š‘„ > š‘Ž, there exists (Exercise 4.46) š‘„ ĀÆ āˆˆ (š‘Ž, š‘„) such that š‘“ (š‘„) āˆ’ š‘“ (š‘Ž) š‘“ ā€² (ĀÆ š‘„) = ā€² š‘”(š‘„) āˆ’ š‘”(š‘Ž) š‘” (ĀÆ š‘„) and therefore by (4.49)    š‘“ (š‘„) āˆ’ š‘“ (š‘Ž)    < šœ–/2 for every š‘„ > š‘Ž āˆ’ š‘˜  š‘”(š‘„) āˆ’ š‘”(š‘Ž) 

2. š‘“ (š‘„) š‘“ (š‘„) āˆ’ š‘“ (š‘Ž) š‘“ (š‘„) š‘”(š‘„) āˆ’ š‘”(š‘Ž) = Ɨ Ɨ š‘”(š‘„) š‘”(š‘„) āˆ’ š‘”(š‘Ž) š‘“ (š‘„) āˆ’ š‘“ (š‘Ž) š‘”(š‘„) š‘“ (š‘„) āˆ’ š‘“ (š‘Ž) 1 āˆ’ Ɨ = š‘”(š‘„) āˆ’ š‘”(š‘Ž) 1āˆ’

š‘”(š‘Ž) š‘”(š‘„) š‘“ (š‘Ž) š‘“ (š‘„)

For ļ¬xed š‘Ž lim

1āˆ’

š‘„ā†’āˆž

1āˆ’

š‘”(š‘Ž) š‘”(š‘„) š‘“ (š‘Ž) š‘“ (š‘„)

=1

and therefore there exists š‘Ž2 such that 1āˆ’ 1āˆ’

š‘”(š‘Ž) š‘”(š‘„) š‘“ (š‘Ž) š‘“ (š‘„)

< 2 for every š‘„ > š‘Ž2

which implies that    šœ–  š‘“ (š‘„)    š‘”(š‘„) āˆ’ š‘˜  < 2 Ɨ 2 for every š‘„ > š‘Ž = max{š‘Ž1 , š‘Ž2 } 4.51 We know that the result holds for š‘› = 1 (Exercise 4.22). Assume that the result holds for š‘› āˆ’ 1. By the chain rule š·(š‘” āˆ˜ š‘“ )[x] = š·š‘”[š‘“ (x)] āˆ˜ š·š‘“ [x] If š‘“, š‘” āˆˆ š¶ š‘› , the š·š‘“, š·š‘” āˆˆ š¶ š‘›āˆ’1 and therefore (by assumption) š·(š‘” āˆ˜ š‘“ ) āˆˆ š¶ š‘›āˆ’1 , which implies that š‘” āˆ˜ š‘“ āˆˆ š¶ š‘› . 222

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 4.52 The partial derivatives of the quadratic function are š·1 š‘“ = 2š‘Žš‘„1 + 2š‘š‘„2 š·2 š‘“ = 2š‘š‘„1 + 2š‘š‘„2 The second-order partial derivatives are š·11 š‘“ = 2š‘Ž

š·21 š‘“ = 2š‘

š·12 š‘“ = 2š‘

š·22 š‘“ = 2š‘

4.53 Apply Exercise 4.37 to each partial derivative š·š‘– š‘“ [x]. 4.54 š»(x0 ) =

(

š·11 š‘“ š‘“ [x0 ] š·12 š‘“ š‘“ [x0 ] š·21 š‘“ š‘“ [x0 ] š·22 š‘“ š‘“ [x0 ]

)

( =2

š‘Ž š‘

š‘ š‘‘

)

4.55 4.56 For any š‘„1 āˆˆ š‘†, deļ¬ne š‘” : š‘† ā†’ ā„œ by š‘”(š‘”) = š‘“ (š‘”) + š‘“ ā€² [š‘”](š‘„1 āˆ’ š‘”) + š‘Ž2 (š‘„1 āˆ’ š‘”)2 š‘” is diļ¬€erentiable on š‘† with š‘ā€² (š‘”) = š‘“ ā€² [š‘”] āˆ’ š‘“ ā€² [š‘”] + š‘“ ā€²ā€² [š‘”](š‘„1 āˆ’ š‘”) āˆ’ 2š‘Ž2 (š‘„1 āˆ’ š‘”) = š‘“ ā€²ā€² [š‘”](š‘„1 āˆ’ š‘”) āˆ’ 2š‘Ž2 (š‘„1 āˆ’ š‘”) Note that š‘”(š‘„1 ) = š‘“ (š‘„1 ) and š‘”(š‘„0 ) = š‘“ (š‘„0 ) + š‘“ ā€² (š‘„0 )(š‘„1 āˆ’ š‘„0 ) + š‘Ž2 (š‘„1 āˆ’ š‘„0 )2

(4.50)

is a quadratic approximation for š‘“ near š‘„0 . If we require that this be exact at š‘„1 āˆ•= š‘„0 , then š‘”(š‘„0 ) = š‘“ (š‘„1 ) = š‘”(š‘„1 ). By the mean value theorem (Theorem 4.1), there exists some š‘„ ĀÆ between š‘„0 and š‘„1 such that š‘”(š‘„1 ) āˆ’ š‘”(š‘„0 ) = š‘ā€² (ĀÆ š‘„)(š‘„1 āˆ’ š‘„0 ) = š‘“ ā€²ā€² (ĀÆ š‘„)(š‘„1 āˆ’ š‘„0 ) āˆ’ 2š‘Ž2 (š‘„1 āˆ’ š‘”) = 0 which implies that š‘Ž2 =

1 ā€²ā€² š‘“ (ĀÆ š‘„) 2

Setting š‘„ = š‘„1 āˆ’ š‘„0 in (4.50) gives the required result. 4.57 For any š‘„1 āˆˆ š‘†, deļ¬ne š‘” : š‘† ā†’ ā„œ by 1 1 š‘”(š‘”) = š‘“ (š‘”) + š‘“ ā€² [š‘”](š‘„1 āˆ’ š‘”) + š‘“ ā€²ā€² [š‘”](š‘„1 āˆ’ š‘”)2 + š‘“ (3) [š‘”](š‘„1 āˆ’ š‘”)3 + . . . 2 3! 1 (š‘›) š‘› š‘›+1 + š‘“ [š‘”](š‘„1 āˆ’ š‘”) + š‘Žš‘›+1 (š‘„1 āˆ’ š‘”) š‘›! š‘” is diļ¬€erentiable on š‘† with 1 1 š‘” ā€² (š‘”) = š‘“ ā€² [š‘”] āˆ’ š‘“ ā€² [š‘”] + š‘“ ā€²ā€² [š‘”](š‘„1 āˆ’ š‘”) āˆ’ š‘“ ā€²ā€² [š‘”](š‘„1 āˆ’ š‘”) + š‘“ (3) [š‘”](š‘„1 āˆ’ š‘”)2 āˆ’ š‘“ (3) [š‘”](š‘„1 āˆ’ š‘„0 )2 + . . . 2 2 1 1 (š‘›+1) (š‘›) š‘›āˆ’1 š‘› š‘“ [š‘”](š‘„1 āˆ’ š‘”) + + š‘“ [š‘”](š‘„1 āˆ’ š‘”) āˆ’ (š‘› + 1)š‘Žš‘›+1 (š‘„1 āˆ’ š‘”)š‘› (š‘› āˆ’ 1)! š‘›! All but the last two terms cancel, so that 1 š‘” (š‘”) = š‘“ (š‘›+1) [š‘”](š‘„1 āˆ’ š‘”)š‘› āˆ’ (š‘› + 1)š‘Žš‘›+1 (š‘„1 āˆ’ š‘”)š‘› = š‘›! ā€²

223

(

) 1 (š‘›+1) š‘“ [š‘”] āˆ’ (š‘› + 1)š‘Žš‘›+1 (š‘„1 āˆ’ š‘”)š‘› š‘›!

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Note that š‘”(š‘„1 ) = š‘“ (š‘„1 ) and 1 1 š‘”(š‘„0 ) = š‘“ (š‘„0 ) + š‘“ ā€² [š‘„0 ](š‘„1 āˆ’ š‘„0 ) + š‘“ ā€²ā€² [š‘„0 ](š‘„1 āˆ’ š‘„0 )2 + š‘“ (3) [š‘„0 ](š‘„1 āˆ’ š‘„0 )3 + . . . 2 3! 1 + š‘“ (š‘›+1) [š‘„0 ](š‘„1 āˆ’ š‘„0 )š‘› + š‘Žš‘›+1 (š‘„1 āˆ’ š‘„0 )š‘›+1 (4.51) š‘›! is a polynomial approximation for š‘“ near š‘„0 . If we require that š‘Žš‘›+1 be such that š‘”(š‘„0 ) = š‘“ (š‘„1 ) = š‘”(š‘„1 ), there exists (Theorem 4.1) some š‘„ ĀÆ between š‘„0 and š‘„1 such that š‘„)(š‘„1 āˆ’ š‘„0 ) = 0 š‘”(š‘„1 ) āˆ’ š‘”(š‘„0 ) = š‘” ā€² (ĀÆ which for š‘„1 āˆ•= š‘„0 implies that š‘” ā€² (ĀÆ š‘„) =

1 š‘›+1 [ĀÆ š‘„] āˆ’ (š‘› + 1)š‘Žš‘›+1 = 0 š‘“ š‘›!

or š‘Žš‘›+1 =

1 š‘“ š‘›+1 [ĀÆ š‘„] (š‘› + 1)!

Setting š‘„ = š‘„1 āˆ’ š‘„0 in (4.51) gives the required result. 4.58 By Taylorā€™s theorem (Exercise 4.57), for every š‘„ āˆˆ š‘† āˆ’ š‘„0 , there exists š‘„ ĀÆ between 0 and š‘„ such that 1 š‘“ (š‘„0 + š‘„) = š‘“ (š‘„0 ) + š‘“ ā€² [š‘„0 ]š‘„ + š‘“ ā€²ā€² [š‘„0 ]š‘„2 + šœ–(š‘„) 2 where šœ–(š‘„) =

1 (3) š‘“ [ĀÆ š‘„]š‘„3 3!

and 1 šœ–(š‘„) = š‘“ (3) [ĀÆ š‘„](š‘„) š‘„2 3! š‘„] is bounded on [0, š‘„] and therefore Since š‘“ āˆˆ š¶ 3 , š‘“ (3) [ĀÆ lim āˆ£

š‘„ā†’0

š‘’(š‘„) 1 āˆ£ = lim āˆ£š‘“ (3) [ĀÆ š‘„](š‘„)āˆ£ = 0 š‘„ā†’0 3! š‘„2

4.59 The function š‘” : ā„œ ā†’ š‘† deļ¬ned by š‘”(š‘”) = š‘”x0 + (1 āˆ’ š‘”)x š‘” is š¶ āˆž with š·š‘”[š‘”] = x and š·š‘˜ š‘”(š‘”) = 0 for š‘˜ = 2, 3, . . . . By Exercise 4.51, the composite function ā„Ž = š‘“ āˆ˜ š‘” is š¶ š‘›+1 . By the Chain rule ā„Žā€² (š‘”) = š·š‘“ [š‘”(š‘”)] āˆ˜ š·š‘”[š‘”] = š·š‘“ [š‘”(š‘”)](x) Similarly

( ) ā„Žā€²ā€² (š‘”) = š· š·š‘“ [š‘”(š‘”)](x) = š·2 š‘“ [š‘”(š‘”)] āˆ˜ š·š‘”[š‘”](x āˆ’ x0 ) = š·2 š‘“ [š‘”(š‘”)](x)(2) 224

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics and for all 1 ā‰¤ š‘˜ ā‰¤ š‘› + 1

) ( ā„Ž(š‘˜) (š‘”) = š· š·(š‘˜āˆ’1) š‘“ [š‘”(š‘”)](x)(š‘˜āˆ’1) = š·š‘˜ š‘“ [š‘”(š‘”)] āˆ˜ š·š‘”[š‘”](x āˆ’ x0 )(š‘˜āˆ’1) = š·š‘˜ š‘“ [š‘”(š‘”)](x)(š‘˜)

4.60 From Exercise 4.54, the Hessian of š‘“ is ( š‘Ž š»(x) = 2 š‘

š‘ š‘‘

)

and the gradient of š‘“ is

( ) āˆ‡š‘“ (x) = (2š‘Žš‘„1 , 2š‘š‘„2 ) with āˆ‡š‘“ (0, 0) = 0

so that the second order Taylor series at (0, 0) is 1 š‘“ (x) = š‘“ (0, 0) + āˆ‡š‘“ (0, 0)x + 2xš‘‡ 2

(

š‘Ž š‘ š‘ š‘‘

) x

= š‘Žš‘„21 + 2š‘š‘„1 š‘„2 + š‘š‘„22 Not surprisingly, we conclude that the best quadratic approximation of a quadratic function is the function itself. 4.61

1. Since š·š‘“ [x0 ] is continuous and one-to-one (Exercise 3.36), there exists a constant š‘š such that š‘š āˆ„x1 āˆ’ x2 āˆ„ ā‰¤ āˆ„š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„

(4.52)

Let šœ– = š‘š/2. By Exercise 4.43, there exists a neighborhood š‘† such that āˆ„š·š‘“ [x0 ](x1 āˆ’ x2 ) āˆ’ (š‘“ (x1 ) āˆ’ š‘“ (x2 ))āˆ„ = āˆ„š‘“ (x1 ) āˆ’ š‘“ (x2 ) āˆ’ š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„ ā‰¤ šœ– āˆ„x1 āˆ’ x2 āˆ„ for every x1 , x2 āˆˆ š‘†. The Triangle Inequality (Exercise 1.200) implies āˆ„š·š‘“ [x0 ](x1 āˆ’ x2 )āˆ„ āˆ’ āˆ„(š‘“ (x1 ) āˆ’ š‘“ (x2 ))āˆ„ ā‰¤ šœ– āˆ„x1 āˆ’ x2 āˆ„ Substituting (4.52) 2šœ– āˆ„x1 āˆ’ x2 āˆ„ āˆ’ āˆ„(š‘“ (x1 ) āˆ’ š‘“ (x2 ))āˆ„ ā‰¤ šœ– āˆ„x1 āˆ’ x2 āˆ„ That is šœ– āˆ„x1 āˆ’ x2 āˆ„ ā‰¤ āˆ„(š‘“ (x1 ) āˆ’ š‘“ (x2 ))āˆ„

(4.53)

and therefore š‘“ (x1 ) = š‘“ (x2 ) =ā‡’ x1 = x2 2. Let š‘‡ = š‘“ (š‘†). Since the restriction of š‘“ to š‘† is one-to-one and onto, and therefore there exists an inverse š‘“ āˆ’1 : š‘‡ ā†’ š‘†. For any y1 , y2 āˆˆ š‘‡ , let x1 = š‘“ āˆ’1 (y1 ) and x2 = š‘“ āˆ’1 (y2 ). Substituting in (4.53)   šœ– š‘“ āˆ’1 (y1 ) āˆ’ š‘“ āˆ’1 (y2 ) ā‰¤ āˆ„y1 āˆ’ y2 āˆ„ so that   āˆ’1 š‘“ (y1 ) āˆ’ š‘“ āˆ’1 (y2 ) ā‰¤ 1 āˆ„y1 āˆ’ y2 āˆ„ šœ– š‘“ āˆ’1 is continuous. 225

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

3. Since š‘† is open, š‘‡ = š‘“ āˆ’1 (š‘†) is open. Therefore, š‘‡ = š‘“ (š‘†) is a neighborhood of š‘“ (x0 ). Therefore, š‘“ is locally onto. 4.62 Assume to the contrary that there exists x0 āˆ•= x1 āˆˆ š‘† with š‘“ (x0 ) = š‘“ (x1 ). Let x = x1 āˆ’ x0 . Deļ¬ne š‘” : [0, 1] ā†’ š‘† by š‘”(š‘”) = (1 āˆ’ š‘”)x0 + š‘”x1 = x0 + š‘”x. Then š‘”(0) = x0 Deļ¬ne

š‘”(1) = x1

š‘” ā€² (š‘”) = x

( ( ) ) ā„Ž(š‘”) = xš‘‡ š‘“ š‘”(š‘”) āˆ’ š‘“ (x0 )

Then ā„Ž(0) = 0 = ā„Ž(1) By the mean value theorem (Mean value theorem), there exists 0 < š›¼ < 1 such that š‘”(š›¼) āˆˆ š‘† and ā„Žā€² (š›¼) = xš‘‡ š·š‘“ [š‘”(š›¼)]x = xš‘‡ š½š‘“ (š‘”(š›¼))x = 0 which contradicts the deļ¬niteness of š½š‘“ . 4.63 Substituting the linear functions in (4.35) and (4.35), the IS-LM model can be expressed as (1 āˆ’ š¶š‘¦ )š‘¦ āˆ’ š¼š‘Ÿ š‘Ÿ = š¶0 + š¼0 + šŗ āˆ’ š¶š‘¦ š‘‡ šæš‘¦ š‘¦ + šæš‘Ÿ š‘Ÿ = š‘€/š‘ƒ which can be rewritten in matrix form as )( ) ( ) ( š‘¦ š‘ āˆ’ š¶š‘¦ š‘‡ 1 āˆ’ š¶š‘¦ š¼š‘Ÿ = šæš‘¦ šæš‘Ÿ š‘Ÿ š‘€/š‘ƒ where š‘ = š¶0 + š¼0 + šŗ. Provided the system is nonsingular, that is    1 āˆ’ š¶š‘¦ š¼š‘Ÿ    āˆ•= 0 š·= šæš‘¦ šæš‘Ÿ  the system can be solved using Cramerā€™s rule (Exercise 3.103) to yield (1 āˆ’ š¶š‘¦ )š‘€/š‘ƒ āˆ’ šæš‘¦ (š‘ āˆ’ š¶š‘¦ š‘‡ ) š· šæš‘Ÿ (š‘ āˆ’ š¶š‘¦ )š‘‡ āˆ’ š¼š‘Ÿ š‘€/š‘ƒ š‘¦= š· š‘Ÿ=

4.64 The kernel kernel š·š¹ [(x0 , šœ½0 )] = { (x, šœ½) : š·š¹ [(x0 , šœ½0 )](x, šœ½) = 0 } is the set of solutions to the equation ( ) ( ) ( ) x š·x š‘“ (x0 , šœ½0 )x + š·šœ½ š‘“ (x0 , šœ½0 )šœ½ 0 = š·š¹ [x0 , šœ½0 ] = šœ½ šœ½ 0 Only šœ½ = 0 satisļ¬es this equation. Substituting šœ½ = 0, the equation reduces to š·x š‘“ (x0 , šœ½0 )x = 0 which has a unique solution x = 0 since š·x š‘“ [x0 , šœ½0 ] is nonsingular. Therefore the kernel of š·š¹ [x0 , šœ½0 ] consists of the single point (0, 0) which implies that š·š¹ [x0 , šœ½ 0 ] is nonsingular (Exercise 3.19). 226

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4.65 The IS curve is horizontal if its slope is zero, that is š·š‘¦ š‘” = āˆ’

1 āˆ’ š·š‘¦ š¶ āˆ’š·š‘Ÿ š¼

This requires either 1. unit marginal propensity to consume (š·š‘¦ š¶ = 1) 2. inļ¬nite interest elasticity of investment (š·š‘Ÿ š¼ = āˆž) 4.66 The LM curve š‘Ÿ = ā„Ž(š‘¦) is implicitly deļ¬ned by the equation š‘“ (š‘Ÿ, š‘¦; šŗ, š‘‡, š‘€ ) = šæ(š‘¦, š‘Ÿ) āˆ’ š‘€/š‘ƒ = 0 the slope of which is given by š·š‘¦ š‘“ š·š‘Ÿ š‘“ š·š‘¦ šæ =āˆ’ š·š‘Ÿ šæ

š·š‘¦ ā„Ž = āˆ’

Economic considerations dictate that the numerator (š·š‘¦ š‘“ ) is positive while the denominator (š·š‘Ÿ šæ) is negative. Preceded by a negative sign, the slope of the LM curve is positive. The LM curve would be vertical (inļ¬nite slope) if the interest elasticity of the demand for money was zero (š·š‘Ÿ šæ = 0). 4.67 Suppose š‘“ is convex. For any x, x0 āˆˆ š‘† let ) ( ā„Ž(š‘”) = š‘“ š‘”x + (1 āˆ’ š‘”)x0 ā‰¤ š‘”š‘“ (x) + (1 āˆ’ š‘”)š‘“ (x0 ) for 0 < š‘” < 1. Subtracting ā„Ž(0) = š‘“ (x0 ) ā„Ž(š‘”) āˆ’ ā„Ž(0) ā‰¤ š‘”š‘“ (x) āˆ’ š‘”š‘“ (x0 ) and therefore š‘“ (x) āˆ’ š‘“ (x0 ) ā‰„

ā„Ž(š‘”) āˆ’ ā„Ž(0) š‘”

Using Exercise 4.10 š‘“ (x) āˆ’ š‘“ (x0 ) ā‰„ lim

š‘”ā†’0

ā„Ž(š‘”) āˆ’ ā„Ž(0) āƒ— x š‘“ [x0 ] = š·š‘“ [x0 ](x āˆ’ x0 ) =š· š‘”

Conversely, let x0 = š›¼x1 + (1 āˆ’ š›¼)x2 for any x1 , x2 āˆˆ š‘†. If š‘“ satisļ¬es (4.29) on š‘†, then š‘“ (x1 ) ā‰„ š‘“ (x0 ) + š·š‘“ [x0 ](x1 āˆ’ x0 ) š‘“ (x2 ) ā‰„ š‘“ (x0 ) + š·š‘“ [x0 ](x2 āˆ’ x0 ) and therefore for any 0 ā‰¤ š›¼ ā‰¤ 1 š›¼š‘“ (x1 ) ā‰„ š›¼š‘“ (x0 + š›¼š·š‘“ [x0 ](x1 āˆ’ x0 ) (1 āˆ’ š›¼)š‘“ (x2 ) ā‰„ (1 āˆ’ š›¼)š‘“ (x0 + (1 āˆ’ š›¼)š·š‘“ [x0 ](x2 āˆ’ x0 ) Adding and using the linearity of š·š‘“ (Exercise 4.21) š›¼š‘“ (x1 ) + (1 āˆ’ š›¼)š‘“ (x2 ) ā‰„ š‘“ (x0 ) + š·š‘“ [x0 ](š›¼x1 + (1 āˆ’ š›¼)x2 āˆ’ x0 ) = š‘“ (x0 ) = š‘“ (š›¼x1 + (1 āˆ’ š›¼)x2 ) That is, š‘“ is convex. If (4.29) is strict, so is (4.54). 227

(4.54)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

4.68 Since ā„Ž is convex, it has a subgradient š‘” āˆˆ š‘‹ āˆ— (Exercise 3.181) such that ā„Ž(x) ā‰„ ā„Ž(x0 ) + š‘”(x āˆ’ x0 ) for every x āˆˆ š‘‹ (4.31) implies that š‘” is also a subgradient of š‘“ on š‘† š‘“ (x) ā‰„ š‘“ (š‘„0 ) + š‘”(x āˆ’ x0 ) for every x āˆˆ š‘† Since š‘“ is diļ¬€erentiable, this implies that š‘” is unique (Remark 4.14) and equal to the derivative of š‘“ . Hence ā„Ž is diļ¬€erentiable at x0 with š·ā„Ž[x0 ] = š·š‘“ [x0 ]. 4.69 Assume š‘“ is convex. For every x, x0 āˆˆ š‘†, Exercise 4.67 implies ( ) š‘“ (x) ā‰„ š‘“ (x0 ) + āˆ‡š‘“ (x0 )š‘‡ x āˆ’ x0 ) ( š‘“ (x0 ) ā‰„ š‘“ (x) + āˆ‡š‘“ (x)š‘‡ x0 āˆ’ x Adding ) ( ( ) š‘“ (x) + š‘“ (x0 ) ā‰„ š‘“ (x) + š‘“ (x0 ) + āˆ‡š‘“ (x)š‘‡ x0 āˆ’ x + āˆ‡š‘“ (x0 )š‘‡ x āˆ’ x0 or ( ) ( ) āˆ‡š‘“ (x)š‘‡ x āˆ’ x0 ā‰„ āˆ‡š‘“ (x0 )š‘‡ x āˆ’ x0 and therefore āˆ‡š‘“ (x) āˆ’ āˆ‡š‘“ (x0 )š‘‡ x āˆ’ x0 ā‰„ 0 When š‘“ is strictly convex, the inequalities are strict. Conversely, assume (4.32). By the mean value theorem (Theorem 4.1), there exists ĀÆ āˆˆ (x, x0 ) such that x x)š‘‡ x āˆ’ x0 š‘“ (x) āˆ’ š‘“ (x0 ) = āˆ‡š‘“ (ĀÆ By assumption ĀÆ āˆ’ x0 ā‰„ 0 āˆ‡š‘“ (ĀÆ x) āˆ’ āˆ‡š‘“ (x0 )š‘‡ x But ĀÆ āˆ’ x0 = š›¼x0 + (1 āˆ’ š›¼)x āˆ’ x0 = (1 āˆ’ š›¼)(x āˆ’ x0 ) x and therefore (1 āˆ’ š›¼)āˆ‡š‘“ (ĀÆ x) āˆ’ āˆ‡š‘“ (x0 )š‘‡ x āˆ’ x0 ā‰„ 0 so that āˆ‡š‘“ (ĀÆ x)š‘‡ x āˆ’ x0 ā‰„ āˆ‡š‘“ (x0 )š‘‡ x āˆ’ x0 ā‰„ 0 and therefore š‘“ (x) āˆ’ š‘“ (x0 ) = āˆ‡š‘“ (ĀÆ x)š‘‡ x āˆ’ x0 ā‰„ āˆ‡š‘“ (x0 )š‘‡ x āˆ’ x0 Therefore š‘“ is convex by Exercise 4.67. 228

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 4.70 For š‘† āŠ† ā„œ, āˆ‡š‘“ (š‘„) = š‘“ ā€² (š‘„) and (4.32) becomes (š‘“ ā€² (š‘„2 ) āˆ’ š‘“ ā€² (š‘„1 )(š‘„2 āˆ’ š‘„1 ) ā‰„ 0 for every š‘„1 , š‘„2 āˆˆ š‘†. This is equivalent to š‘“ ā€² (š‘„2 )(š‘„2 āˆ’ š‘„1 ) ā‰„ š‘“ ā€² (š‘„1 )(š‘„2 āˆ’ š‘„1 )0 or š‘„2 > š‘„1 =ā‡’ š‘“ ā€² (š‘„2 ) ā‰„ š‘“ ā€² (š‘„1 ) š‘“ is strictly convex if and only if the inequalities are strict.

4.71 š‘“ ā€² is increasing if and only if š‘“ ā€²ā€² = š·š‘“ ā€² ā‰„ 0 (Exercise 4.35). š‘“ ā€² is strictly increasing if š‘“ ā€²ā€² = š·š‘“ ā€² > 0 (Exercise 4.36). 4.72 Adapting the previous example

āŽ§ ļ£“ āŽØ= 0 š‘“ ā€²ā€² (š‘„) = š‘›(š‘› āˆ’ 1)š‘„š‘› āˆ’ 2 = ā‰„ 0 ļ£“ āŽ© indeterminate

if š‘› = 1 if š‘› = 2, 4, 6, š‘‘š‘œš‘”š‘  otherwise

Therefore, the power function is convex if š‘› is even, and neither convex if š‘› ā‰„ 3 is odd. It is both convex and concave when š‘› = 1. 4.73 Assume š‘“ is quasiconcave, and š‘“ (x) ā‰„ š‘“ (x0 ). Diļ¬€erentiability at x0 implies for all 0 < š‘” < 1 š‘“ (x0 + š‘”(x āˆ’ x0 ) = š‘“ (x0 ) + āˆ‡š‘“ (x0 )š‘”(x āˆ’ x0 ) + šœ‚(š‘”) āˆ„š‘”(x āˆ’ x0 )āˆ„ where šœ‚(š‘”) ā†’ 0 and š‘” ā†’ 0. Quasiconcavity implies š‘“ (x0 + š‘”(x āˆ’ x0 ) ā‰„ š‘“ (x0 ) and therefore āˆ‡š‘“ (x0 )š‘”(x āˆ’ x0 ) + šœ‚(š‘”) āˆ„š‘”(x āˆ’ x0 )āˆ„ ā‰„ 0 Dividing by š‘” and letting š‘” ā†’ 0, we get āˆ‡š‘“ (x0 )(x āˆ’ x0 ) ā‰„ 0 Conversely, assume š‘“ is a diļ¬€erentiable functional satisfying (4.36). For any x1 , x2 āˆˆ š‘† with š‘“ (x1 ) ā‰„ š‘“ (x2 for every x, x0 āˆˆ š‘†), deļ¬ne ā„Ž : [0, 1] ā†’ ā„œ by ( ) ) ( ā„Ž(š‘”) = š‘“ (1 āˆ’ š‘”)x1 + š‘”x2 = š‘“ x1 + š‘”(x2 āˆ’ x1 ) We need to show that ā„Ž(š‘”) ā‰„ ā„Ž(1) for every š‘” āˆˆ (0, 1). Suppose to the contrary that ā„Ž(š‘”1 ) < ā„Ž(1). Then (see below) there exists š‘”0 with ā„Ž(š‘”0 ) < ā„Ž(1) and ā„Žā€² (š‘”0 ) < 0. By the Chain Rule, this implies ā„Žā€² (š‘”0 ) = āˆ‡š‘“ (x0 )(x2 āˆ’ x1 ) < 0 critical where x0 = x1 + š‘”(x2 āˆ’ x1 ). Since x2 āˆ’ x0 = (1 āˆ’ š‘”)(x2 āˆ’ x1 ) this implies that ā„Žā€² (š‘”0 ) =

1 āˆ‡š‘“ (x0 )(x2 āˆ’ x0 ) 1āˆ’š‘” 229

(4.55)

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c 2001 Michael Carter āƒ All rights reserved

On the other hand, since š‘“ (x0 ) ā‰„ š‘“ (x2 ), (4.36) implies āˆ‡š‘“ (x0 )(x2 āˆ’ x0 ) ā‰„ 0 contradicting (4.55). To show that there exists š‘”0 with ā„Ž(š‘”0 ) < ā„Ž(1) and ā„Žā€² (š‘”0 ) < 0: Since š‘“ is continuous, there exists an open interval (š‘Ž, š‘) with š‘Ž < š‘”1 < š‘ with ā„Ž(š‘Ž) = ā„Ž(š‘) = ā„Ž(1) and ā„Ž(š‘”) < ā„Ž(1) for every š‘” āˆˆ (š‘Ž, š‘). By the Mean Value Theorem, there exist š‘”0 āˆˆ (š‘Ž, š‘”1 ) such that 0 < ā„Ž(š‘”1 ) āˆ’ ā„Ž(š‘Ž) = ā„Žā€² (š‘”0 )(š‘”1 āˆ’ š‘Ž) which implies that ā„Žā€² (š‘”0 ) > 0. 4.74 Suppose to the contrary that š‘“ (x) > š‘“ (x0 ) and āˆ‡š‘“ (x0 )(x āˆ’ x0 ) ā‰¤ 0 critical Let x1 = āˆ’āˆ‡š‘“ (x0 ) āˆ•= 0. For every š‘” āˆˆ ā„œ+ āˆ‡š‘“ (x0 )(x + š‘”x1 āˆ’ x0 ) = āˆ‡š‘“ (x0 )š‘”x1 + āˆ‡š‘“ (x0 )(x āˆ’ x0 ) ā‰¤ š‘”āˆ‡š‘“ (x0 )x1 2

= āˆ’š‘” āˆ„āˆ‡š‘“ (x0 )āˆ„ < 0 Since š‘“ is continuous, there exists š‘” > 0 such that š‘“ (x + š‘”x1 ) > š‘“ (x0 ) and āˆ‡š‘“ (x0 )(x + š‘”x1 āˆ’ x0 ) < 0 contradicting the quasiconcavity of š‘“ (4.36). 4.75 Suppose š‘“ (x) < š‘“ (x0 ) =ā‡’ āˆ‡š‘“ (x0 )(x āˆ’ x0 ) < 0 This implies that āˆ’š‘“ (x) > āˆ’š‘“ (x0 ) =ā‡’ āˆ‡ āˆ’ š‘“ (x0 )(x āˆ’ x0 ) > 0 and āˆ’š‘“ is pseudoconcave. 4.76

1. If š‘“ āˆˆ š¹ [š‘†] is concave (and diļ¬€erentiable) š‘“ (x) ā‰¤ š‘“ (x0 ) + āˆ‡š‘“ (x0 )š‘‡ (x āˆ’ x0 ) for every x, x0 āˆˆ š‘†(equation 4.30). Therefore š‘“ (x) > š‘“ (x0 ) =ā‡’ āˆ‡š‘“ (x0 )š‘‡ (x āˆ’ x0 ) > 0 š‘“ is pseudoconcave.

2. Assume to the contrary that š‘“ is pseudoconcave but not quasiconcave. Then, ĀÆ = š›¼x1 + (1 āˆ’ š›¼)x2 , x1 , x2 āˆˆ š‘† such that there exists x š‘“ (ĀÆ x) < min{š‘“ (x1 ), š‘“ (x2 )} Assume without loss of generality that š‘“ (ĀÆ x) < š‘“ (x1 ) ā‰¤ š‘“ (x2 ) 230

(4.56)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Pseudoconcavity (4.38) implies ĀÆ) > 0 āˆ‡š‘“ (ĀÆ x)(x2 āˆ’ x

(4.57)

x āˆ’ (1 āˆ’ š›¼)x2 )/š›¼ Since x1 = (ĀÆ ĀÆ= x1 āˆ’ x

) 1( 1āˆ’š›¼ ĀÆ āˆ’ (1 āˆ’ š›¼)x2 āˆ’ š›¼ĀÆ ĀÆ) x (x2 āˆ’ x x =āˆ’ š›¼ š›¼

Substituting in (4.57) gives ĀÆ) < 0 āˆ‡š‘“ (ĀÆ x)(x1 āˆ’ x which by pseudoconcavity implies š‘“ (x1 ) ā‰¤ š‘“ (ĀÆ x) contradicting our assumption (4.56) . 3. Exercise 4.74. 4.77 The CES function is quasiconcave provided šœŒ ā‰¤ 1 (Exercise 3.58). Since š·š‘„š‘– š‘“ (x) > 0 for all x āˆˆ ā„œš‘›+ +, the CES function with šœŒ ā‰¤ 1 is pseudoconcave on ā„œš‘›++ . 4.78 Assume that š‘“ : š‘† ā†’ ā„œ is homogeneous of degree š‘˜, so that for every x āˆˆ š‘† š‘“ (š‘”x) = š‘”š‘› š‘“ (x) for every š‘” > 0 Diļ¬€erentiating both sides of this identity with respect to š‘„š‘– š·š‘„š‘– š‘“ (š‘”x)š‘” = š‘”š‘› š·š‘„š‘– š‘“ (x) and dividing by š‘” > 0 š·š‘„š‘– š‘“ (š‘”x) = š‘”š‘˜āˆ’1 š·š‘„š‘– š‘“ (x) 4.79 If š‘“ is homogeneous of degree š‘˜ āƒ— x š‘“ (x) = lim š‘“ (x + š‘”x) āˆ’ š‘“ (x) š· š‘”ā†’0 š‘” š‘“ ((1 + š‘”)x) āˆ’ š‘“ (x) = lim š‘”ā†’0 š‘” (1 + š‘”)š‘› š‘“ (x) āˆ’ š‘“ (x) = lim š‘”ā†’0 š‘” (1 + š‘”)š‘› āˆ’ 1 = lim š‘“ (x) š‘”ā†’0 š‘” Applying Lā€™HĖ†opitalā€™s Rule (Exercise 4.47) (1 + š‘”)š‘˜āˆ’1 š‘˜(1 + š‘”)š‘˜āˆ’1 š‘“ (x) = lim =š‘˜ š‘”ā†’0 š‘”ā†’0 š‘” 1 lim

and therefore āƒ— x š‘“ (x) = š‘˜š‘“ (x) š·

(4.58)

4.80 For ļ¬xed x, deļ¬ne ā„Ž(š‘”) = š‘“ (š‘”x) By the Chain Rule ā„Žā€² (š‘”) = š‘”š·š‘“ [š‘”x](x) = š‘”š‘˜š‘“ (š‘”x) = š‘”š‘˜ā„Ž(š‘”) 231

(4.59)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

using(4.40). Diļ¬€erentiating the product ā„Ž(š‘”)š‘”āˆ’š‘˜ ( ) ( ) š·š‘” ā„Ž(š‘”)š‘”āˆ’š‘˜ = āˆ’š‘˜ā„Ž(š‘”)š‘”āˆ’š‘˜āˆ’1 + š‘”āˆ’š‘˜ ā„Žā€² (š‘”) = š‘”āˆ’š‘˜ ā„Žā€² (š‘”) āˆ’ š‘˜š‘”ā„Ž(š‘”) = 0 from (4.59). Since this holds for every š‘”, ā„Ž(š‘”)š‘”āˆ’š‘˜ must be constant (Exercise 4.38), that is there exists š‘ āˆˆ ā„œ such that ā„Ž(š‘”)š‘”āˆ’š‘˜ = š‘ =ā‡’ ā„Ž(š‘”) = š‘š‘”š‘˜ Evaluating at š‘” = 1, ā„Ž(1) = š‘ and therefore ā„Ž(š‘”) = š‘”š‘˜ ā„Ž(1) Since ā„Ž(š‘”) = š‘“ (š‘”x) and ā„Ž(1) = š‘“ (x), this implies š‘“ (š‘”x) = š‘”š‘˜ š‘“ (x) for every x and š‘” > 0 š‘“ is homogeneous of degree š‘˜. 4.81 If š‘“ is linearly homogeneous and quasiconcave, then š‘“ is concave (Proposition 3.12). Therefore, its Hessian is nonpositive deļ¬nite (Proposition 4.1). and its diagonal elements š·š‘„2 š‘– š‘„š‘– š‘“ (x) are nonpositive (Exercise 3.95). By Wicksellā€™s law, š·š‘„2 š‘– š‘„š‘— š‘“ (x) is nonnegative. 4.82 Assume š‘“ is homogeneous of degree š‘˜, that is š‘“ (š‘”x) = š‘”š‘˜ š‘“ (x) for every x āˆˆ š‘† and š‘” > 0 By Eulerā€™s theorem š·š‘” š‘“ [š‘”x](š‘”x) = š‘˜š‘“ (š‘”x) and therefore the elasticity of scale is

  š‘” š‘” š·š‘” š‘“ (š‘”x) š‘˜š‘“ (š‘”x) = š‘˜ = šø(x) = š‘“ (š‘”x) š‘“ (š‘”x) š‘”=1

Conversely, assume that

  š‘” šø(x) = š·š‘” š‘“ (š‘”x) =š‘˜ š‘“ (š‘”x) š‘”=1

that is š·š‘” š‘“ (š‘”x) = š‘˜š‘“ (š‘”x) By Eulerā€™s theorem, š‘“ is homogeneous of degree š‘˜. 4.83 Assume š‘“ āˆˆ š¹ (š‘†) is diļ¬€erentiable and homogeneous of degree š‘˜ āˆ•= 0. By Eulerā€™s theorem š·š‘“ [x](x) = š‘˜š‘“ (x) āˆ•= 0 for every x āˆˆ š‘† such that š‘“ (x) āˆ•= 0. 4.84 š‘“ satisļ¬es Eulerā€™s theorem š‘˜š‘“ (x) =

š‘› āˆ‘

š·š‘– š‘“ (x)š‘„š‘–

š‘–=1

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Diļ¬€erentiating with respect to š‘„š‘— š‘˜š·š‘— š‘“ (x) =

š‘› āˆ‘

š·š‘–š‘— š‘“ (x)š‘„š‘– + š·š‘— š‘“ (x)

š‘–=1

or (š‘˜ āˆ’ 1)š·š‘— š‘“ (x) =

š‘› āˆ‘

š·š‘–š‘— š‘“ (x)š‘„š‘–

š‘— = 1, 2, . . . , š‘›

š‘–=1

Multiplying each equation by š‘„š‘— and summing (š‘˜ āˆ’ 1)

š‘› āˆ‘

š·š‘— š‘“ (x)š‘„š‘— =

š‘—=1

š‘› āˆ‘ š‘› āˆ‘

š·š‘–š‘— š‘“ (x)š‘„š‘– š‘„š‘— = xā€² š»x

š‘—=1 š‘–=1

By Eulerā€™s theorem, the left hand side is (š‘˜ āˆ’ 1)š‘˜š‘“ (x) = xā€² š»x 4.85 If š‘“ is homothetic, there exists strictly increasing š‘” and linearly homogeneous ā„Ž such that š‘“ = š‘” āˆ˜ ā„Ž (Exercise 3.175). Using the Chain Rule and Exercise 4.78 š·š‘„š‘– š‘“ (š‘”x) = š‘” ā€² (š‘“ (š‘”x))š·š‘„š‘– ā„Ž(š‘”x) = š‘”š‘” ā€² (š‘“ (š‘”x))š·š‘„š‘– ā„Ž(x) and therefore š·š‘„š‘– š‘“ (š‘”x) š‘”š‘” ā€² (š‘“ (š‘”x))š·š‘„š‘– ā„Ž(x) = š·š‘„š‘— š‘“ (š‘”x) š·š‘„š‘— š‘”š‘” ā€² (š‘“ (š‘”x))š·š‘„š‘— ā„Ž(x) š·š‘„š‘– ā„Ž(x) = š·š‘„š‘— ā„Ž(x) š·š‘„š‘— š‘” ā€² (š‘“ (x)š·š‘„š‘– ā„Ž(x) = š·š‘„š‘— š‘” ā€² (š‘“ (x))š·š‘„š‘— ā„Ž(x) š·š‘„š‘– š‘“ (x) = š·š‘„š‘— š‘“ (x) š·š‘„š‘—

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Chapter 5: Optimization 5.1 As stated, this problem has no optimal solution. Revenue š‘“ (š‘„) increases without bound as the rate of exploitation š‘„ gets smaller and smaller. Given any positive exploitation rate š‘„0 , a smaller rate will increase total revenue. Nonexistence arises from inadequacy in modeling the island leadersā€™ problem. For example, the model ignores any costs of extraction and sale. Realistically, we would expect per-unit costs to decrease with volume (increasing returns to scale) at least over lower outputs. Extraction and transaction costs should make vanishingly small rates of output prohibitively expensive and encourage faster utilization. Secondly, even if the government weights future generations equally with the current generation, it would be rational to value current revenue more highly than future revenue and discount future returns. Discounting is appropriate for two reasons āˆ™ Current revenues can be invested to provide a future return. There is an opportunity cost (the interest foregone) to delaying extraction and sale. āˆ™ Innovation may create substitutes which reduce the future demand for the fertilizer. If the government is risk averse, it has an incentive to accelerate exploitation, trading-oļ¬€ of lower total return against reduced risk. 5.2 Suppose that xāˆ— is a local optimum which is not a global optimum. That is, there exists a neighborhood š‘† of xāˆ— such that š‘“ (xāˆ— , šœ½) ā‰„ š‘“ (x, šœ½) for every x āˆˆ š‘† āˆ© šŗ(šœ½) and also another point xāˆ—āˆ— āˆˆ šŗ(šœ½) such that š‘“ (xāˆ—āˆ— , šœ½) > š‘“ (xāˆ— , šœ½) Since šŗ(šœ½) is convex, there exists š›¼ āˆˆ (0, 1) such that š›¼xāˆ— + (1 āˆ’ š›¼)xāˆ—āˆ— āˆˆ š‘† āˆ© šŗ(šœ½) By concavity of š‘“ š‘“ (š›¼xāˆ— + (1 āˆ’ š›¼)xāˆ—āˆ— , šœ½) ā‰„ š›¼š‘“ (xāˆ— , šœ½) + (1 āˆ’ š›¼)š‘“ (xāˆ—āˆ— , šœ½) > š‘“ (xāˆ— , šœ½) contradicting the assumption that xāˆ— is a local optimum. 5.3 Suppose that xāˆ— is a local optimum which is not a global optimum. That is, there exists a neighborhood š‘† of xāˆ— such that š‘“ (xāˆ— , šœ½) ā‰„ š‘“ (x, šœ½) for every x āˆˆ š‘† āˆ© šŗ(šœ½) and also another point xāˆ—āˆ— āˆˆ šŗ(šœ½) such that š‘“ (xāˆ—āˆ— , šœ½) > š‘“ (xāˆ— , šœ½) Since šŗ(šœ½) is convex, there exists š›¼ āˆˆ (0, 1) such that š›¼xāˆ— + (1 āˆ’ š›¼)xāˆ—āˆ— āˆˆ š‘† āˆ© šŗ(šœ½)

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By strict quasiconcavity of š‘“ š‘“ (š›¼xāˆ— + (1 āˆ’ š›¼)xāˆ—āˆ— , šœ½) > min{ š‘“ (xāˆ— , šœ½), š‘“ (xāˆ—āˆ— , šœ½) } > š‘“ (xāˆ— , šœ½) contradicting the assumption that xāˆ— is a local optimum. Therefore, if xāˆ— is local optimum, it must be a global optimum. Now suppose that xāˆ— is a weak global optimum, that is š‘“ (xāˆ— , šœ½) ā‰„ š‘“ (x, šœ½) for every x āˆˆ š‘† but there another point xāˆ—āˆ— āˆˆ š‘† such that š‘“ (xāˆ—āˆ— , šœ½) = š‘“ (xāˆ— , šœ½) Since šŗ(šœ½) is convex, there exists š›¼ āˆˆ (0, 1) such that š›¼xāˆ— + (1 āˆ’ š›¼)xāˆ—āˆ— āˆˆ š‘† āˆ© šŗ(šœ½) By strict quasiconcavity of š‘“ š‘“ (š›¼xāˆ— + (1 āˆ’ š›¼)xāˆ—āˆ— , šœ½) > min{ š‘“ (xāˆ— , šœ½), š‘“ (xāˆ—āˆ— , šœ½) } = š‘“ (xāˆ— , šœ½) contradicting the assumption that xāˆ— is a global optimum. We conclude that every optimum is a strict global optimum and hence unique. 5.4 Suppose that xāˆ— is a local optimum of (5.3) in š‘‹, so that š‘“ (xāˆ— ) ā‰„ š‘“ (x)

(5.80)

for every x in a neighborhood š‘† of xāˆ— . If š‘“ is diļ¬€erentiable, š‘“ (x) = š‘“ (xāˆ— ) + š·š‘“ [xāˆ— ](x āˆ’ xāˆ— ) + šœ‚(x) āˆ„x āˆ’ xāˆ— āˆ„ where šœ‚(x) ā†’ 0 as x ā†’ xāˆ— . (5.80) implies that there exists a ball šµš‘Ÿ (xāˆ— ) such that š·š‘“ [xāˆ— ](x āˆ’ xāˆ— ) + šœ‚(x) āˆ„x āˆ’ xāˆ— āˆ„ ā‰¤ 0 for every x āˆˆ šµš‘Ÿ (xāˆ— ). Letting x ā†’ xāˆ— , we conclude that š·š‘“ [xāˆ— ](x āˆ’ xāˆ— ) ā‰¤ 0 for every x āˆˆ šµš‘Ÿ (xāˆ— ). Suppose there exists x āˆˆ šµš‘Ÿ (xāˆ— ) such that š·š‘“ [xāˆ— ](x āˆ’ xāˆ— ) = š‘¦ < 0 Let dx = x āˆ’ xāˆ— so that x = xāˆ— + dx. Then xāˆ— āˆ’ dx āˆˆ šµš‘Ÿ (xāˆ— ). Since š·š‘“ [xāˆ— ] is linear, š·š‘“ [xāˆ— ](āˆ’dx) = āˆ’š·š‘“ [xāˆ— ](dx) = āˆ’š‘¦ > 0 contradicting (5.80). Therefore š·š‘“ [xāˆ— ](x āˆ’ xāˆ— ) = 0 for every x āˆˆ šµš‘Ÿ (xāˆ— ).

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5.5 We apply the reasoning of Example 5.5 to each component. Formally, for each š‘–, let š‘“Ė†š‘– be the projection of š‘“ along the š‘–š‘”ā„Ž axis š‘“Ė†š‘– (š‘„š‘– ) = š‘“ (š‘„āˆ—1 , š‘„āˆ—2 , . . . , š‘„āˆ—š‘–āˆ’1 , š‘„š‘– , š‘„āˆ—š‘–+1 , . . . , š‘„āˆ—š‘› ) š‘„āˆ—š‘– maximizes š‘“Ė†š‘– (š‘„š‘– ) over ā„œ+ , for which it is necessary that š·š‘„š‘– š‘“Ė†š‘– (š‘„āˆ—š‘– ) ā‰¤ 0

š‘„āˆ—š‘– ā‰„ 0

š‘„āˆ—š‘– š·š‘„š‘– š‘“Ė†š‘– (š‘„āˆ—š‘– ) = 0

Substituting š·š‘„š‘– š‘“Ė†š‘– (š‘„āˆ—š‘– ) = š·š‘„š‘– š‘“ [xāˆ— ] yields š·š‘„š‘– š‘“ [xāˆ— ] ā‰¤ 0

š‘„āˆ—š‘– ā‰„ 0

š‘„āˆ—š‘– š·š‘„š‘– š‘“ [xāˆ— ] = 0

5.6 By Taylorā€™s Theorem (Example 4.33) 1 2 š‘“ (xāˆ— + dx) = š‘“ (xāˆ— ) + āˆ‡š‘“ (xāˆ— )dx + dxš‘‡ š»š‘“ (xāˆ— )dx + šœ‚(dx) āˆ„dxāˆ„ 2 with šœ‚(dx) ā†’ 0 as dx ā†’ 0. Given 1. āˆ‡š‘“ (xāˆ— ) = 0 and 2. š»š‘“ (xāˆ— ) is negative deļ¬nite and letting dx ā†’ 0, we conclude that š‘“ (xāˆ— + dx) < š‘“ (xāˆ— ) for small dx. xāˆ— is a strict local maximum. 5.7 If xāˆ— is a local minimum of š‘“ (x), it is necessary that š‘“ (xāˆ— ) ā‰¤ š‘“ (x) for every x in a neighborhood š‘† of xāˆ— . Assuming that š‘“ is š¶ 2 , š‘“ (x) can be approximated by 1 š‘“ (x) ā‰ˆ š‘“ (xāˆ— ) + āˆ‡š‘“ (xāˆ— )dx + dxš‘‡ š»š‘“ (xāˆ— )dx 2 where dx = x āˆ’ xāˆ— . If xāˆ— is a local minimum, then there exists a ball šµš‘Ÿ (xāˆ— ) such that 1 š‘“ (xāˆ— ) ā‰¤ š‘“ (xāˆ— ) + āˆ‡š‘“ (xāˆ— )dx + dxš‘‡ š»š‘“ (xāˆ— )dx 2 or 1 āˆ‡š‘“ (xāˆ— )dx + dxš‘‡ š»š‘“ (xāˆ— )dx ā‰„ 0 2 for every dx āˆˆ šµš‘Ÿ (xāˆ— ). To satisfy this inequality for all small dx requires that the ļ¬rst term be zero and the second term nonnegative. In other words, for a point xāˆ— to be a local minimum of a function š‘“ , it is necessary that the gradient be zero and the Hessian be nonnegative deļ¬nite at xāˆ— . Furthermore, by Taylorā€™s Theorem 1 2 š‘“ (xāˆ— + dx) = š‘“ (xāˆ— ) + āˆ‡š‘“ (xāˆ— )dx + dxš‘‡ š»š‘“ (xāˆ— )dx + šœ‚(dx) āˆ„dxāˆ„ 2 with šœ‚(dx) ā†’ 0 as dx ā†’ 0. Given 236

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(1,2,3)

3 2 š‘„ 2

0

1

1 š‘„1

2

Figure 5.1: The strictly concave function š‘“ (š‘„1 , š‘„2 ) = š‘„1 š‘„2 + 3š‘„2 āˆ’ š‘„21 āˆ’ š‘„22 has a unique global maximum. 1. āˆ‡š‘“ (xāˆ— ) = 0 and 2. š»š‘“ (xāˆ— ) is positive deļ¬nite and letting dx ā†’ 0, we conclude that š‘“ (xāˆ— + dx) > š‘“ (xāˆ— ) for small dx. xāˆ— is a strict local minimum. 5.8 By the Weierstrass theorem (Theorem 2.2), š‘“ has a maximum š‘„āˆ— and a minimum š‘„āˆ— on [š‘Ž, š‘]. Either āˆ™ š‘„āˆ— āˆˆ (š‘Ž, š‘) and š‘“ ā€² (š‘„āˆ— ) = 0 (Theorem 5.1) or āˆ™ š‘„āˆ— āˆˆ (š‘Ž, š‘) and š‘“ ā€² (š‘„āˆ— ) = 0 (Exercise 5.7) or āˆ™ Both maxima and minima are boundary points, that is š‘„āˆ— , š‘„āˆ— āˆˆ {š‘Ž, š‘} which implies that š‘“ is constant on [š‘Ž, š‘] and therefore š‘“ ā€² (š‘„) = 0 for every š‘„ āˆˆ (š‘Ž, š‘) (Exercise 4.7). 5.9 The ļ¬rst-order conditions for a maximum are š·š‘„1 š‘“ (š‘„1 , š‘„2 ) = š‘„2 āˆ’ 2š‘„1 = 0 š·š‘„2 š‘“ (š‘„1 , š‘„2 ) = š‘„1 + 3 āˆ’ 2š‘„2 = 0 which have the unique solution š‘„āˆ—1 = 1, š‘„āˆ—2 = 2. (1, 2) is the only stationary point of š‘“ and hence the only possible candidate for a maximum. To verify that (1, 2) satisļ¬es the second-order condition for a maximum, we compute the Hessian of š‘“ ) ( āˆ’2 1 š»(x) = 1 āˆ’2 which is negative deļ¬nite everywhere. Therefore (1, 2) is a strict local maximum of š‘“ . Further, since š‘“ is strictly concave (Proposition 4.1), we conclude that (1, 2) is a strict global maximum of š‘“ (Exercise 5.2), where it attains its maximum value š‘“ (1, 2) = 3 (Figure 5.1). 237

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5.10 The ļ¬rst-order conditions for a maximum (or minimum) are š·1 š‘“ (š‘„) = 2š‘„1 = 0 š·2 š‘“ (š‘„) = 2š‘„2 = 0 which have a unique solution š‘„1 = š‘„2 = 0. This is the only stationary point of š‘“ . Since the Hessian of š‘“ ( ) 2 0 š»= 0 2 is positive deļ¬nite, we deduce (0, 0) is a strict global minimum of š‘“ (Proposition 4.1, Exercise 5.2). 5.11 The average ļ¬rmā€™s proļ¬t function is 1 1 Ī (š‘˜, š‘™) = š‘¦ āˆ’ š‘˜ āˆ’ š‘™ āˆ’ 2 6 and the ļ¬rmā€™s proļ¬t maximization problem is 1 1 max Ī (š‘˜, š‘™) = š‘˜ 1/6 š‘™1/3 āˆ’ š‘˜ āˆ’ š‘™ āˆ’ š‘˜,š‘™ 2 6 A necessary condition for a proļ¬t maximum is that the proļ¬t function be stationary, that is 1 āˆ’5/6 1/3 1 š‘˜ š‘™ āˆ’ =0 6 2 1 1/6 āˆ’2/3 āˆ’1=0 š·š‘™ Ī (š‘˜, š‘™) = š‘˜ š‘™ 3

š·š‘˜ Ī (š‘˜, š‘™) =

which can be solved to yield š‘˜=š‘™=

1 9

The ļ¬rmā€™s output is š‘¦=

1 1/6 1 1/3 1 = 9 9 3

and its proļ¬t is 1 11 1 1 1 1 āˆ’ āˆ’ =0 Ī ( , ) = āˆ’ 3 3 3 29 9 6 5.12 By the Chain Rule š·x (ā„Ž āˆ˜ š‘“ )[xāˆ— ] = š·ā„Ž āˆ˜ š·x š‘“ [xāˆ— ] = 0 Since š·ā„Ž > 0 š·x (ā„Ž āˆ˜ š‘“ )[š‘„āˆ— ] = 0 ā‡ā‡’ š·x š‘“ [xāˆ— ] = 0 ā„Ž āˆ˜ š‘“ has the same stationary points as š‘“ .

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5.13 Since the log function is monotonic, ļ¬nding the maximum likelihood estimators is equivalent to solving the maximization problem ( Exercise 5.12) max log šæ(šœ‡, šœŽ) = āˆ’ šœ‡,šœŽ

š‘‡ 1 āˆ‘ š‘‡ log 2šœ‹ āˆ’ š‘‡ log šœŽ āˆ’ 2 (š‘„š‘” āˆ’ šœ‡)2 2 2šœŽ š‘”=1

For (Ė† šœ‡, šœŽ Ė† 2 ) to solve this problem, it is necessary that log šæ be stationary at (Ė† šœ‡, šœŽ Ė† 2 ), that is šœ‡, šœŽ Ė†2) = š·šœ‡ log šæ(Ė†

š‘‡ 1 āˆ‘ (š‘„š‘” āˆ’ šœ‡ Ė†) = 0 šœŽ Ė† 2 š‘”=1

š·šœŽ log šæ(Ė† šœ‡, šœŽ Ė†2) = āˆ’

š‘‡ 1 āˆ‘ š‘‡ + 3 (š‘„š‘” āˆ’ šœ‡ Ė†)2 = 0 šœŽ Ė† šœŽ Ė† š‘”=1

which can be solved to yield šœ‡ Ė†=š‘„ ĀÆ= šœŽ Ė†2 =

š‘‡ 1 āˆ‘ š‘„š‘” š‘‡ š‘”=1

š‘‡ 1āˆ‘ (š‘„š‘” āˆ’ š‘„ ĀÆ)2 š‘‡ š‘”=1

5.14 The gradient of the objective function is ) ( āˆ’2(š‘„1 āˆ’ 1) āˆ‡š‘“ (x) = āˆ’2(š‘„2 āˆ’ 1) while that of the constraint is

( āˆ‡š‘”(x) =

2š‘„1 2š‘„2

)

A necessary condition for the optimal solution is that these be proportional that is ) ( ( ) 2š‘„1 āˆ’2(š‘„1 āˆ’ 1) =šœ† āˆ‡š‘“ (š‘„) = = āˆ‡š‘”(x) āˆ’2(š‘„2 āˆ’ 1) 2š‘„2 which can be solved to yield š‘„1 = š‘„2 =

1 1+šœ†

which includes an unknown constant of proportionality šœ†. However, any solution must also satisfy the constraint ( )2 1 š‘”(š‘„1 , š‘„2 ) = 2 =1 1+šœ† This can be solved for šœ† šœ†=

āˆš 2āˆ’1

and substituted into (5.80) 1 š‘„1 = š‘„2 = āˆš 2 239

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Solutions for Foundations of Mathematical Economics 5.15 The consumerā€™s problem is max š‘¢(x) = š‘„1 + š‘Ž log š‘„2 xā‰„0

subject to š‘”(x) = š‘„1 + š‘2 š‘„2 āˆ’ š‘š = 0 The ļ¬rst-order conditions for a (local) optimum are š·š‘„1 š‘¢(xāˆ— ) = 1 ā‰¤ šœ† = š·š‘„1 š‘”(š‘„āˆ— )

š‘„1 ā‰„ 0

š‘Ž ā‰¤ šœ†š‘2 = š·š‘„2 š‘”(š‘„āˆ— ) š·š‘„2 š‘¢(xāˆ— ) = š‘„2

š‘„2 ā‰„ 0

( š‘„2

š‘„1 (1 āˆ’ šœ†) = 0 ) =0

š‘Ž āˆ’ šœ†š‘2 š‘„2

(5.81) (5.82)

We can distinguish two cases: Case 1 š‘„1 = 0 in which case the budget constraint implies that š‘„2 = š‘š/š‘2 . Case 2 š‘„1 > 0 In this case, (5.81) implies that šœ† = 1. Consequently, the ļ¬rst inequality of (5.82) implies that š‘„2 > 0 and therefore the last equation implies š‘„2 = š‘Ž/š‘2 with š‘„1 = š‘š āˆ’ š‘Ž. We deduce that the consumer ļ¬rst spends portion š‘Ž of her income on good 2 and the remainder on good 1. 5.16 Suppose without loss of generality that the ļ¬rst š‘˜ components of yāˆ— are strictly positive while the remaining components are zero. That is š‘¦š‘–āˆ— > 0 š‘¦š‘–āˆ— = 0

š‘– = 1, 2, . . . , š‘˜ š‘– = š‘˜ + 1, š‘˜ + 2, . . . , š‘›

(xāˆ— , yāˆ— ) solves the problem max š‘“ (x) subject to g(x) = 0 š‘– = š‘˜ + 1, š‘˜ + 2, . . . , š‘› š‘¦š‘– = 0 By Theorem 5.2, there exist multipliers šœ†1 , šœ†2 , . . . , šœ†š‘š and šœ‡š‘˜+1 , šœ‡š‘˜+2 , . . . , šœ‡š‘› such that š·x š‘“ [xāˆ— , yāˆ— ] = š·y š‘“ [xāˆ— , yāˆ— ] =

š‘š āˆ‘ š‘—=1 š‘š āˆ‘

šœ†š‘— š·x š‘”š‘— [xāˆ— , yāˆ— ] šœ†š‘— š·y š‘”š‘— [xāˆ— , yāˆ— ] +

š‘—=1

š‘› āˆ‘

šœ‡š‘– š‘¦ š‘–

š‘–=š‘˜+1

Furthermore, šœ‡š‘– ā‰„ 0 for every š‘– so that š·y š‘“ [xāˆ— , yāˆ— ] ā‰¤

š‘š āˆ‘

šœ†š‘— š·y š‘”š‘— [xāˆ— , yāˆ— ]

š‘—=1

with š·š‘¦š‘– š‘“ [xāˆ— , yāˆ— ] =

š‘š āˆ‘

šœ†š‘— š·š‘¦š‘– š‘”š‘— [xāˆ— , yāˆ— ] if š‘¦š‘– > 0

š‘—=1

5.17 Assume that xāˆ— = (š‘„āˆ—1 , š‘„āˆ—2 ) solves max š‘“ (š‘„1 , š‘„2 )

š‘„1 ,š‘„2

240

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subject to š‘”(š‘„1 , š‘„2 ) = 0 By the implicit function theorem, there exists a function ā„Ž : ā„œ ā†’ ā„œ such that š‘„1 = ā„Ž(š‘„2 )

(5.83)

and š‘”(ā„Ž(š‘„2 ), š‘„2 ) = 0 for š‘„2 in a neighborhood of š‘„āˆ—2 . Furthermore š·ā„Ž[š‘„āˆ—2 ] = āˆ’

š·š‘„1 š‘”[xāˆ— ] š·š‘„2 š‘”[xāˆ— ]

(5.84)

Using (5.41), we can convert the original problem into the unconstrained maximization of a function of a single variable max š‘“ (ā„Ž(š‘„2 ), š‘„2 ) š‘„2

If š‘„āˆ—2 maximizes this function, it must satisfy the ļ¬rst-order condition (applying the Chain Rule) š·š‘„1 š‘“ [š‘„āˆ—] āˆ˜ š·ā„Ž[š‘„āˆ—2 ] + š·š‘„2 š‘“ [xāˆ— ] = 0 Substituting (5.42) yields ( ) š·š‘„1 š‘”[xāˆ— ] š·š‘„1 š‘“ [š‘„āˆ—] āˆ’ + š·š‘„2 š‘“ [xāˆ— ] = 0 š·š‘„2 š‘”[xāˆ— ] or š·š‘„1 š‘“ [š‘„āˆ—] š·š‘„1 š‘”[xāˆ— ] = š·š‘„2 š‘“ [xāˆ— ] š·š‘„2 š‘”[xāˆ— ] 5.18 The consumerā€™s problem is max š‘¢(x) xāˆˆš‘‹

subject to pš‘‡ x = š‘š Solving for š‘„1 from the budget constraint yields āˆ‘š‘› š‘š āˆ’ š‘–=2 š‘š‘– š‘„š‘– š‘„1 = š‘1 Substituting this in the utility function, the aļ¬€ordable utility levels are āˆ‘ ( ) š‘š āˆ’ š‘›š‘–=2 š‘š‘– š‘„š‘– , š‘„2 , š‘„3 , . . . , š‘„š‘› š‘¢ Ė†(š‘„2 , š‘„3 , . . . , š‘„š‘› ) = š‘¢ š‘1

(5.85)

and the consumerā€™s problem is to choose (š‘„2 , š‘„3 , . . . , š‘„š‘› ) to maximize (5.85). The ļ¬rst-order conditions are that š‘¢ Ė†(š‘„2 , š‘„3 , . . . , š‘„š‘› ) be stationary, that is for every good š‘— = 2, 3, . . . , š‘› āˆ‘š‘› ( ) š‘š āˆ’ š‘–=2 š‘š‘– š‘„š‘– āˆ— š·š‘„š‘— š‘¢ Ė†(š‘„2 , š‘„3 , . . . , š‘„š‘› ) = š·š‘„1 š‘¢(x )š·š‘„š‘— + š·š‘„š‘— š‘¢(xāˆ— ) = 0 š‘1 241

Solutions for Foundations of Mathematical Economics

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which reduces to š·š‘„1 š‘¢(xāˆ— )(āˆ’

š‘1 ) + š·š‘„š‘— š‘¢(xāˆ— ) = 0 š‘š‘—

or š‘1 š·š‘„1 š‘¢(xāˆ— ) = āˆ— š·š‘„š‘— š‘¢(x ) š‘š‘—

š‘— = 2, 3, . . . , š‘›

This is the familiar equality between the marginal rate of substitution and the price ratio (Example 5.15). Since our selection of š‘„1 was arbitrary, this applies between any two goods. 5.19 Adapt Exercise 5.6. 5.20 Corollary 5.1.2 implies that xāˆ— is a global maximum of šæ(x, š€), that is šæ(xāˆ— , š€) ā‰„ šæ(x, š€) for every x āˆˆ š‘‹ which implies š‘“ (xāˆ— ) āˆ’

āˆ‘

šœ†š‘— š‘”š‘— (xāˆ— ) ā‰„ š‘“ (x) āˆ’

āˆ‘

šœ†š‘— š‘”š‘— (x) for every x āˆˆ š‘‹

Since g(xāˆ— ) = 0 this implies š‘“ (xāˆ— ) ā‰„ š‘“ (x) āˆ’

āˆ‘

šœ†š‘— š‘”š‘— (x) for every x āˆˆ š‘‹

A fortiori š‘“ (xāˆ— ) ā‰„ š‘“ (x) for every x āˆˆ šŗ = { x āˆˆ š‘‹ : g(x) = 0 } 5.21 Suppose that xāˆ— is a local maximum of š‘“ on šŗ. That is, there exists a neighborhood š‘† such that š‘“ (xāˆ— ) ā‰„ š‘“ (x) for every x āˆˆ š‘† āˆ© šŗ But for every x āˆˆ šŗ, š‘”š‘— (x) = 0 for every š‘— and āˆ‘ šæ(x) = š‘“ (x) + šœ†š‘— š‘”š‘— (x) = š‘“ (x) and therefore šæ(xāˆ— ) ā‰„ šæ(x) for every x āˆˆ š‘† āˆ© šŗ 5.22 The area of the base is Base = š‘¤2 = š“/3 and the four sides Sides = 4š‘¤ā„Ž āˆš āˆš š“ š“ =4 3 12 4š“ = 16 2š“ = 3 242

Solutions for Foundations of Mathematical Economics

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5.23 Let the dimensions of the vat be š‘¤ Ɨ š‘™ Ɨ ā„Ž. We wish to min Surface area = š“ = š‘¤ Ɨ š‘™ + 2š‘¤ā„Ž + 2š‘™ā„Ž

š‘¤,š‘™,ā„Ž

subject to š‘¤ Ɨ š‘™ Ɨ ā„Ž = 32 The Lagrangean is šæ(š‘¤, š‘™, ā„Ž, šœ†) = š‘¤š‘™ + 2š‘¤ā„Ž + 2š‘™ā„Ž āˆ’ šœ†š‘¤š‘™ā„Ž. The ļ¬rst-order conditions for a maximum are š·š‘¤ šæ = š‘™ + 2ā„Ž āˆ’ šœ†š‘™ā„Ž = 0

(5.86)

š·š‘™ šæ = š‘¤ + 2ā„Ž āˆ’ šœ†š‘¤ā„Ž = 0

(5.87)

š·ā„Ž šæ = 2š‘¤ + 2š‘™ āˆ’ šœ†š‘¤š‘™ = 0 š‘¤š‘™ā„Ž = 32

(5.88)

Subtracting (5.45) from (5.44) š‘™ āˆ’ š‘¤ = šœ†(š‘™ āˆ’ š‘¤)ā„Ž This equation has two possible solutions. Either šœ†=

1 or š‘™ = š‘¤ ā„Ž

But if šœ† = 1/ā„Ž, (5.44) implies that š‘™ = 0 and the volume is zero. Therefore, we conclude that š‘¤ = š‘™. Substituting š‘¤ = š‘™ in (5.45) and (5.46) gives š‘¤ + 2ā„Ž = šœ†š‘¤ā„Ž 4š‘¤ = šœ†š‘¤2 from which we deduce that šœ†=

4 š‘¤

Substituting in (5.45) š‘¤ + 2ā„Ž =

4 š‘¤ā„Ž = 4ā„Ž š‘¤

which implies that š‘¤ = 2ā„Ž or ā„Ž =

1 š‘¤ 2

To achieve the required volume of 32 cubic metres requires that 1 š‘¤ Ɨ š‘™ Ɨ ā„Ž = š‘¤ Ɨ š‘¤ Ɨ š‘¤ = 32 2 so that the dimensions of the vat are š‘¤=4

š‘™=4

ā„Ž=2

The area of sheet metal required is š“ = š‘¤š‘™ + 2š‘¤ā„Ž + 2š‘™ā„Ž = 48 243

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics 5.24 The Lagrangean for this problem is

šæ(x, šœ†) = š‘„21 + š‘„22 + š‘„23 āˆ’ šœ†(2š‘„1 āˆ’ 3š‘„2 + 5š‘„3 āˆ’ 19) A necessary condition for xāˆ— to solve the problem is that the Lagrangean be stationary at xāˆ— , that is š·š‘„1 šæ = 2š‘„āˆ—1 āˆ’ 2šœ† = 0 š·š‘„2 šæ = 2š‘„āˆ—2 + 3šœ† = 0 š·š‘„3 šæ = 2š‘„āˆ—3 āˆ’ 5šœ† = 0

which implies 3 š‘„āˆ—2 = āˆ’ šœ† 2

š‘„āˆ—1 = šœ†

š‘„āˆ—2 =

5 šœ† 2

(5.89)

It is also necessary that the solution satisfy the constraint, that is 2š‘„āˆ—1 āˆ’ 3š‘„āˆ—2 + 5š‘„āˆ—3 = 19 Substituting (5.89) into the constraint we get 9 25 2šœ† + šœ† + šœ† = 19šœ† = 19 2 2 which implies šœ† = 1. Substituting in (5.89), the solution is xāˆ— = (1, āˆ’ 32 , 52 ). Since the constraint is aļ¬ƒne and the objective (āˆ’š‘“ ) is concave, stationarity of the Lagrangean is also suļ¬ƒcient for global optimum (Corollary 5.2.4). 5.25 The Lagrangean is 1āˆ’š›¼ šæ(š‘„1 , š‘„2 , šœ†) = š‘„š›¼ āˆ’ šœ†(š‘1 š‘„1 + š‘2 š‘„2 āˆ’ š‘š) 1 š‘„2

The Lagrangean is stationary where š·š‘„1 šæ = š›¼š‘„š›¼āˆ’1 š‘„1āˆ’š›¼ āˆ’ šœ†š‘1 = 0 1 2 1āˆ’š›¼āˆ’1 š·š‘„2 šæ = 1 āˆ’ š›¼š‘„š›¼ āˆ’ šœ†š‘2 = 0 1 š‘„2

Therefore the ļ¬rst-order conditions for a maximum are 1āˆ’

š›¼š‘„š›¼āˆ’1 š‘„1āˆ’š›¼ = šœ†š‘1 1 2

(5.90)

1āˆ’š›¼āˆ’1 š›¼š‘„š›¼ 1 š‘„2

(5.91)

= šœ†š‘2 š‘1 š‘„1 + š‘2 š‘„2 āˆ’ š‘š

Dividing (5.48) by (5.49) gives š›¼š‘„š›¼āˆ’1 š‘„1āˆ’š›¼ 1 2

(1āˆ’š›¼)āˆ’1

1 āˆ’ š›¼š‘„š›¼ 1 š‘„2

= š‘1 š‘2

which simpliļ¬es to š‘1 š›¼š‘„2 = (1 āˆ’ š›¼)š‘„1 š‘2

or

š‘2 š‘„2 =

(1 āˆ’ š›¼) š‘1 š‘„1 š›¼

Substituting in the budget constraint (5.50) (1 āˆ’ š›¼) š‘1 š‘„1 = š‘š š›¼ š›¼ + (1 āˆ’ š›¼) š‘1 š‘„1 = š‘š š›¼

š‘1 š‘„1 +

244

(5.92)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics so that š‘„āˆ—1 =

š‘š š›¼ š›¼ + (1 āˆ’ š›¼) š‘1

From the budget constraint (5.92) š‘„āˆ—2 =

(1 āˆ’ š›¼) š‘š š›¼ + (1 āˆ’ š›¼) š‘2

5.26 The Lagrangean is š›¼š‘› 1 š›¼2 šæ(x, šœ†) = š‘„š›¼ 1 š‘„2 . . . š‘„š‘› āˆ’ šœ†(š‘1 š‘„1 + š‘2 š‘„2 + ... + š‘š‘› š‘„š‘› )

The ļ¬rst-order conditions for a maximum are š›¼š‘– āˆ’1 1 š›¼2 š‘› š·š‘„š‘– šæ = š›¼š‘– š‘„š›¼ . . . š‘„š›¼ š‘› āˆ’ šœ†š‘š‘– = 1 š‘„2 . . . š‘„š‘–

š›¼š‘– š‘¢(š‘„) āˆ’ šœ†š‘š‘– = 0 š‘„š‘–

or š›¼š‘– š‘¢(š‘„) = š‘š‘– š‘„š‘– šœ†

š‘– = 1, 2, . . . , š‘›

Summing over all goods and using the budget constraint š‘› āˆ‘ š›¼š‘– š‘¢(š‘„) š‘–=1

Letting

āˆ‘š‘›

š‘–=1

šœ†

š‘›

=

š‘›

āˆ‘ š‘¢(š‘„) āˆ‘ š›¼š‘– = š‘š‘– š‘„š‘– = š‘š šœ† š‘–=1 š‘–=1

š›¼š‘– = š›¼, this implies š‘š š‘¢(x) = šœ† š›¼

Substituting in (5.93) š‘š‘– š‘„š‘– =

š›¼š‘– š‘š š›¼

or š‘„āˆ—š‘– =

š›¼š‘– š‘š š›¼ š‘š‘–

š‘– = 1, 2, . . . , š‘›

5.27 The Lagrangean is šæ(x, šœ†) = š‘¤1 š‘„1 + š‘¤2 š‘„2 āˆ’ šœ†(š‘„šœŒ1 + š‘„šœŒ2 āˆ’ š‘¦ šœŒ ). The necessary conditions for stationarity are š·š‘„1 šæ(x, šœ†) = š‘¤1 āˆ’ šœ†šœŒš‘„šœŒāˆ’1 =0 1 š·š‘„2 šæ(x, šœ†) = š‘¤2 āˆ’ šœ†šœŒš‘„šœŒāˆ’1 =0 2 or š‘¤1 = šœ†šœŒš‘„šœŒāˆ’1 1 š‘¤2 = šœ†šœŒš‘„šœŒāˆ’1 2

245

(5.93)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics which reduce to š‘„šœŒāˆ’1 š‘¤1 = 1šœŒāˆ’1 š‘¤2 š‘„ 2 š‘¤2 šœŒāˆ’1 šœŒāˆ’1 š‘„2 = š‘„ š‘¤1 1 šœŒ ( ) šœŒāˆ’1 š‘¤2 š‘„šœŒ2 = š‘„šœŒ1 š‘¤1 Substituting in the production constraint (

) šœŒ š‘¤2 šœŒāˆ’1 šœŒ + š‘„1 = š‘¦ šœŒ š‘¤1 ( šœŒ ) ( ) šœŒāˆ’1 š‘¤2 1+ š‘„šœŒ1 = š‘¦ šœŒ š‘¤1 š‘„šœŒ1

we can solve š‘„1 ( š‘„šœŒ1

=

(

1+ (

š‘„1 =

(

1+

š‘¤2 š‘¤1 š‘¤2 š‘¤1

šœŒ )āˆ’1 ) šœŒāˆ’1

š‘¦šœŒ

šœŒ )āˆ’1/š‘ ) šœŒāˆ’1

š‘¦

Similarly ( š‘„2 =

( 1+

š‘¤1 š‘¤2

šœŒ )āˆ’1/š‘ ) šœŒāˆ’1

š‘¦

5.28 Example 5.27 is ļ¬‚awed. The optimum of the constrained maximization problem (ā„Ž = š‘¤/2) is in fact a saddle point of the Lagrangean. It maximizes the Lagrangean in the feasible set, but not globally. The net beneļ¬t approach to the Lagrange multiplier method is really only applicable when the Lagrangean (net beneļ¬t function) is concave, so that every stationary point is a global maximum. This requirement is satisļ¬ed in many standard examples, such as the consumerā€™s problem (Example 5.21) and cost minimization (Example 5.28). It is also met in Example 5.29. The requirement of concavity is not recognized in the text, and Section 5.3.6 should be amended accordingly. 5.29 The Lagrangean šæ(x, šœ†) =

š‘› āˆ‘

( š‘š‘– (š‘„š‘– ) + šœ† š· āˆ’

š‘–=1

š‘› āˆ‘

) š‘„š‘–

(5.94)

š‘–=1

can be rewritten as šæ(x, šœ†) = āˆ’

š‘› āˆ‘ ) ( šœ†š‘„š‘– āˆ’ š‘š‘– (š‘„š‘– ) + šœ†š·

(5.95)

š‘–=1

The š‘–th term in the sum is the net proļ¬t of plant š‘– if its output is valued at šœ†. Therefore, if the company undertakes to buy electricity from its plants at the price šœ† and instructs each plant manager to produce so as to maximize the plantā€™s net proļ¬t, each manager 246

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c 2001 Michael Carter āƒ All rights reserved

will be induced to choose an output level which maximizes the proļ¬t of the company as a whole. This is the case whether the price šœ† is the market price at which the company can buy electricity from external suppliers or the shadow price determined by the need to satisfy the total demand š·. In this way, the shadow price šœ† can be used to decentralize the production decision. 5.30 The Lagrangean for this problem is šæ(š‘„1 , š‘„2 , šœ†1 , šœ†2 ) = š‘„1 š‘„2 āˆ’ šœ†1 (š‘„21 + 2š‘„22 āˆ’ 3) āˆ’ šœ†2 (2š‘„21 + š‘„22 āˆ’ 3) The ļ¬rst-order conditions for stationarity š·š‘„1 šæ = š‘„2 āˆ’ 2šœ†1 š‘„1 āˆ’ 4šœ†2 š‘„1 = 0 š·š‘„2 šæ = š‘„1 āˆ’ 4šœ†1 š‘„2 āˆ’ 2šœ†2 š‘„2 = 0 can be written as š‘„2 = 2(šœ†1 + 2šœ†2 )š‘„1

(5.96)

š‘„1 = 2(2šœ†1 + šœ†2 )š‘„2

(5.97)

which must be satisļ¬ed along with the complementary slackness conditions š‘„21 + 2š‘„22 āˆ’ 3 ā‰¤ 0

šœ†1 ā‰„ 0

šœ†1 (š‘„21 + 2š‘„22 āˆ’ 3) = 0

2š‘„21 + š‘„22 āˆ’ 3 ā‰¤ 0

šœ†2 ā‰„ 0

šœ†2 (2š‘„21 + š‘„22 āˆ’ 3) = 0

First suppose that both constraints are slack so that šœ†1 = šœ†2 = 0. Then the ļ¬rst-order conditions (5.96) and (5.97) imply that š‘„1 = š‘„2 = 0. (0, 0) satisļ¬es the Kuhn-Tucker conditions. Next suppose that the ļ¬rst constraint is binding while the second constraint have two solutions, is slack āˆš (šœ†2 = 0).āˆš The ļ¬rst-order āˆš and (5.97) āˆš āˆš conditions (5.96) āˆš š‘„1 = 3/2, š‘„2 = 3/2, šœ† = 1/(2 2) and š‘„1 = āˆ’ 3/2, š‘„2 = āˆ’ 3/2, šœ† = 1/(2 2), but these violate the second constraint. Similarly, there is no solution in which the ļ¬rst constraint is slack and the second constraint binding. Finally, assume that the both constraints are binding. This implies that š‘„1 = š‘„2 = 1 or š‘„1 = š‘„2 = āˆ’1, which points satisfy the ļ¬rst-order conditions (5.96) and (5.97) with šœ†1 = šœ†2 = 1/6. We conclude that three points satisfy the Kuhn-Tucker conditions, namely (0, 0), (1, 1) and (āˆ’1, āˆ’1). Noting the objective function, we observe that (0, 0) in fact minimizes the objective. We conclude that there are two local maxima, (1, 1) and (āˆ’1, āˆ’1), both of which achieve the same level of the objective function. 5.31 Dividing the ļ¬rst-order conditions, we obtain š‘Ÿ šœ†(š‘  āˆ’ š‘Ÿ) š·š‘˜ š‘…(š‘˜, š‘™) = āˆ’ š·š‘™ š‘…(š‘˜, š‘™) š‘¤ (1 āˆ’ šœ†)š‘¤ Using the revenue function š‘…(š‘˜, š‘™) = š‘(š‘“ (š‘˜, š‘™))š‘“ (š‘˜, š‘™) the marginal revenue products of capital and labor are š·š‘˜ š‘…(š‘˜, š‘™) = š·š‘¦ š‘(š‘¦)š·š‘˜ š‘“ (š‘˜, š‘™) š·š‘™ š‘…(š‘˜, š‘™) = š·š‘¦ š‘(š‘¦)š·š‘™ š‘“ (š‘˜, š‘™) so that their ratio is equal to the ratio of the marginal products š·š‘˜ š‘“ (š‘˜, š‘™) š·š‘˜ š‘…(š‘˜, š‘™) = š·š‘™ š‘…(š‘˜, š‘™) š·š‘™ š‘“ (š‘˜, š‘™ 247

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

The necessary condition for optimality can be expressed as š‘Ÿ šœ† š‘ āˆ’š‘Ÿ š·š‘˜ š‘“ (š‘˜, š‘™) = āˆ’ š·š‘™ š‘“ (š‘˜, š‘™) š‘¤ 1āˆ’šœ† š‘¤ whereas the necessary condition for cost minimization is (Example 5.16) š·š‘˜ š‘“ (š‘˜, š‘™) š‘Ÿ = š·š‘™ š‘“ (š‘˜, š‘™) š‘¤ The regulated ļ¬rm does not use the cost-minimizing combination of inputs. 5.32 The general constrained optimization problem max š‘“ (x) x

subject to g(x) ā‰¤ 0 can be transformed into an equivalent equality constrained problem max š‘“ (x) x,s

subject to g(x) + s = 0 and s ā‰„ 0 Ė† (x, s) = g(x) + s, the through the addition of nonnegative slack variables s. Letting g ļ¬rst-order conditions a local optimum are (Exercise 5.16) āˆ‘ āˆ‘ š·x š‘“ (xāˆ— ) = šœ†š‘— š·x š‘”Ė†š‘— (xāˆ— , sāˆ— ) = šœ†š‘— š·x š‘”š‘— (xāˆ— ) āˆ‘ šœ†š‘— š·s š‘”Ė†š‘— (x, s) = š€ (5.98) 0 = š·s š‘“ (xāˆ— ) ā‰¤ š€š‘‡ s = 0

sā‰„0

(5.99)

Condition (5.98) implies that šœ†š‘— ā‰„ 0 for every š‘—. Furthermore, rewriting the constraint as s = āˆ’g(x) the complementary slackness condition (5.99) becomes š€š‘‡ g(x) = 0

g(x) ā‰¤ 0

This establishes the necessary conditions of Theorem 5.3. 5.33 The equality constrained maximization problem max š‘“ (x) x

subject to g(x) = 0 is equivalent to the problem max š‘“ (x) x

subject to g(x) ā‰¤ 0 āˆ’g(x) ā‰¤ āˆ’0 By the Kuhn-Tucker theorem (Theorem 5.3), there exists nonnegative multipliers + āˆ’ āˆ’ + āˆ’ šœ†+ 1 , šœ†2 , . . . , šœ†š‘š and šœ†1 , šœ†2 , . . . , šœ†š‘š such that āˆ‘ āˆ‘ āˆ— āˆ— š·š‘“ (xāˆ— ) = šœ†+ šœ†āˆ’ (5.100) š‘— š·š‘”š‘— [x ] āˆ’ š‘— š·š‘”š‘— [x ] = 0 248

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics with āˆ’ šœ†+ š‘— š‘”š‘— (x) = 0 and šœ†š‘— š‘”š‘— (x) = 0

š‘— = 1, 2, . . . , š‘š

āˆ’ Deļ¬ning šœ†š‘— = šœ†+ š‘— āˆ’ šœ†š‘— , (5.100) can be written as

š·š‘“ (xāˆ— ) =

āˆ‘

šœ†š‘— š·š‘”š‘— [xāˆ— ]

which is the ļ¬rst-order condition for an equality constrained problem. Furthermore, if xāˆ— satisļ¬es the inequality constraints š‘”(xāˆ— ) ā‰¤ 0 and š‘”(xāˆ— ) ā‰„ 0 it satisļ¬es the equality š‘”(xāˆ— ) = 0 5.34 Suppose that xāˆ— solves the problem max cš‘‡ x subject to š“x ā‰¤ 0 x

with Lagrangean šæ = cš‘‡ x āˆ’ š€š‘‡ š“x Then there exists š€ ā‰„ 0 such that š·x šæ = cš‘‡ āˆ’ š€š‘‡ š“ = 0 that is, š“š‘‡ š€ = c. Conversely, if there is no solution, there exists x such that š“x ā‰¤ 0 and cš‘‡ x > cš‘‡ 0 = 0 5.35 There are two binding constraints at (4, 0), namely š‘”(š‘„1 , š‘„2 ) = š‘„1 + š‘„2 ā‰¤ 4 ā„Ž(š‘„1 , š‘„2 ) = āˆ’š‘„2 ā‰¤ 0 with gradients āˆ‡š‘”(4, 0) = (1, 1) āˆ‡ā„Ž(4, 0) = (0, 1) which are linearly independent. Therefore the binding constraints are regular at (0, 4). 5.36 The Lagrangean for this problem is šæ(x, šœ†) = š‘¢(x) āˆ’ šœ†(pš‘‡ x āˆ’ š‘š) and the ļ¬rst-order (Kuhn-Tucker) conditions are (Corollary 5.3.2) š·š‘„š‘– šæ[xāˆ— , šœ†] = š·š‘„š‘– š‘¢[xāˆ— ] āˆ’ šœ†š‘š‘– ā‰¤ 0 š‘‡

āˆ—

p x ā‰¤š‘š

xāˆ—š‘– ā‰„ 0 šœ†ā‰„0

š‘„āˆ—š‘– (š·š‘„š‘– š‘¢[xāˆ— ] āˆ’ šœ†š‘š‘– ) = 0 š‘‡

āˆ—

šœ†(p x āˆ’ š‘š) = 0

for every good š‘– = 1, 2, . . . , š‘š. Two cases must be distinguished. 249

(5.101) (5.102)

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Case 1 šœ† > 0 This implies that pš‘‡ x = š‘š, the consumer spends all her income. Condition (5.101) implies š·š‘„š‘– š‘¢[xāˆ— ] ā‰¤ šœ†š‘š‘– for every š‘– with š·š‘„š‘– š‘¢[xāˆ— ] = šœ†š‘š‘– for every š‘– for which š‘„š‘– > 0 This case was analyzed in Example 5.17. Case 2 šœ† = 0 This allows the possibility that the consumer does not spend all her income. Substituting šœ† = 0 in (5.101) we have š·š‘„š‘– š‘¢[xāˆ— ] = 0 for every š‘–. At the optimal consumption bundle xāˆ— , the marginal utility of every good is zero. The consumer is satiated, that is no additional consumption can increase satisfaction. This case was analyzed in Example 5.31. In summary, at the optimal consumption bundle xāˆ— , either āˆ™ the consumer is satiated (š·š‘„š‘– š‘¢[xāˆ— ] = 0 for every š‘–) or āˆ™ the consumer consumes only those goods whose marginal utility exceeds the threshold š·š‘„š‘– š‘¢[xāˆ— ] ā‰„ šœ†š‘š‘– and adjusts consumption so that the marginal utility is proportional to price for all consumed goods. 5.37 Assume x āˆˆ š·(xāˆ— ). Then there exists š›¼ ĀÆ āˆˆ ā„œ such that xāˆ— + š›¼x āˆˆ š‘† for every 0 ā‰¤ š›¼ ā‰¤ š›¼. ĀÆ Deļ¬ne š‘” āˆˆ š¹ ([0, š›¼ ĀÆ ]) by š‘”(š›¼) = š‘“ (xāˆ— + š›¼x). If xāˆ— is a local maximum, š‘” has a local maximum at 0, and therefore š‘” ā€² (0) ā‰¤ 0 (Theorem 5.1). By the chain rule (Exercise 4.22), this implies š‘” ā€² (0) = š·š‘“ [xāˆ— ](x) ā‰¤ 0 and therefore x āˆˆ / š» + (xāˆ— ). 5.38 If x is a tangent vector, so is š›½x for any nonnegative š›½ (replace 1/š›¼š‘˜ by š›½/š›¼š‘˜ in the preceding deļ¬nition. Also, trivially, x = 0 is a tangent vector (with xš‘˜ = xāˆ— and š›¼š‘˜ = 1 for all š‘˜). The set š‘‡ of all vectors tangent to š‘† at xāˆ— is therefore a nonempty cone, which is called the cone of tangents to š‘† at xāˆ— . To show that š‘‡ is closed, let xš‘› be a sequence in š‘‡ converging to some x āˆˆ ā„œš‘› . We need to show that x āˆˆ š‘‡ . Since xš‘› āˆˆ š‘‡ , there exist feasible points xš‘šš‘› āˆˆ š‘† and š›¼š‘šš‘› such that (xš‘šš‘› āˆ’ xāˆ— )/š›¼š‘šš‘› ā†’ xš‘› as š‘š ā†’ āˆž For any š‘ choose š‘› such that āˆ„xš‘› āˆ’ xāˆ„ ā‰¤

1 š‘ 2

and then choose š‘š such that āˆ„xš‘šš‘› āˆ’ xāˆ— āˆ„ ā‰¤ š‘ and āˆ„(xš‘šš‘› āˆ’ xāˆ— )/š›¼š‘šš‘› āˆ’ xš‘› āˆ„ ā‰¤

1 š‘ 2

Relabeling xš‘šš‘› as xš‘ and š›¼š‘šš‘› as š›¼š‘ we have we have constructed a sequence xš‘ in S such that   š‘ x āˆ’ xāˆ—  ā‰¤ š‘ and

 š‘    (x āˆ’ xāˆ— )/š›¼š‘ āˆ’ x ā‰¤ (xš‘ āˆ’ xāˆ— )/š›¼š‘ āˆ’ xš‘›  + āˆ„xš‘› āˆ’ xāˆ„ ā‰¤ 1 š‘ 1

Letting š‘ ā†’ āˆž, xš‘ converges to xāˆ— and (xš‘ āˆ’ xāˆ— )/š›¼š‘ converges to x, which proves that x āˆˆ š‘‡ as required. 250

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Solutions for Foundations of Mathematical Economics

5.39 Assume x āˆˆ š·(xāˆ— ). That is, there exists š›¼ ĀÆ such that xāˆ— + š›¼x āˆˆ š‘† for every š›¼ āˆˆ [0, š›¼ ĀÆ ]. For š‘˜ = 1, 2, . . . , let š›¼š‘˜ = š›¼ ĀÆ /š‘˜. Then xš‘˜ = xāˆ— + š›¼š‘˜ x āˆˆ š‘†, xš‘˜ ā†’ xāˆ— and š‘˜ āˆ— āˆ— āˆ— (x āˆ’ x )/š›¼š‘˜ = (x + š›¼š‘˜ x āˆ’ x )/š›¼š‘˜ = x. Therefore, x āˆˆ š‘‡ (xāˆ— ). 5.40 Let dx āˆˆ š‘‡ (xāˆ— ). Then there exists a feasible sequence {xš‘˜ } converging to xāˆ— and a sequence {š›¼š‘˜ } of nonnegative scalars such that the sequence {(xš‘˜ āˆ’ xāˆ— )/š›¼š‘˜ } converges to dx. For any š‘— āˆˆ šµ(xāˆ— ), š‘”š‘— (xāˆ— ) = 0 and   š‘”š‘— (xš‘˜ ) = š·š‘”š‘— [xāˆ— ](xš‘˜ āˆ’ xāˆ— ) + šœ‚š‘— xš‘˜ āˆ’ xāˆ—  where šœ‚š‘— ā†’ 0 as k ā†’ āˆž. This implies   1 1 š‘”š‘— (xš‘˜ ) = š‘˜ š·š‘”š‘— [xāˆ— ](xš‘˜ āˆ’ xāˆ— ) + šœ‚š‘— (xš‘˜ āˆ’ xāˆ— )/š›¼š‘˜  š‘˜ š›¼ š›¼ Since xš‘˜ is feasible 1 š‘”š‘— (xš‘˜ ) ā‰¤ 0 š›¼š‘˜ and therefore   š·š‘”š‘— [xāˆ— ]((xš‘˜ āˆ’ xāˆ— )/š›¼š‘˜ ) + šœ‚ š‘– (xš‘˜ āˆ’ xāˆ— )/š›¼š‘˜  ā‰¤ 0 Letting š‘˜ ā†’ āˆž we conclude that š·š‘”š‘— [xāˆ— ](dx) ā‰¤ 0 That is, dx āˆˆ šæ. 5.41 šæ0 āŠ† šæ1 by deļ¬nition. Assume dx āˆˆ šæ1 . That is š·š‘”š‘— [xāˆ— ](dx) < 0 āˆ—

š·š‘”š‘— [x ](dx) ā‰¤ 0

for every š‘— āˆˆ šµ š‘ (xāˆ— ) š¶

āˆ—

for every š‘— āˆˆ šµ (x )

(5.103) (5.104)

where šµ š¶ (xāˆ— ) = šµ(xāˆ— ) āˆ’ šµ š‘ (xāˆ— ) is the set of concave binding constraints at xāˆ— . By concavity (Exercise 4.67), (5.104) implies that š‘”š‘— (xāˆ— + š›¼dx) ā‰¤ š‘”š‘— (xāˆ— ) = 0 for every š›¼ ā‰„ 0 and š‘— āˆˆ šµ š¶ (xāˆ— ) From (5.103) there exists some š›¼š‘ such that š‘”š‘— (xāˆ— + š›¼dx) < 0 for every š›¼ āˆˆ [0, š›¼š‘ ] and š‘— āˆˆ šµ š‘ (xāˆ— ) Furthermore, since š‘”š‘— (xāˆ— ) < 0 for all š‘— āˆˆ š‘†(xāˆ— ), there exists some š›¼š‘† > 0 such that š‘”š‘— (xāˆ— + š›¼dx) < 0 for every š›¼ āˆˆ [0, š›¼š‘† ] and š‘— āˆˆ š‘†(xāˆ— ) Setting š›¼ ĀÆ = min{š›¼š‘ , š›¼š‘† } we have š‘”š‘— (xāˆ— + š›¼dx) ā‰¤ 0 for every š›¼ āˆˆ [0, š›¼ ĀÆ ] and š‘— = 1, 2, . . . , š‘š or xāˆ— + š›¼dx āˆˆ šŗ = { x : š‘”š‘— (x) ā‰¤ 0, š‘— = 1, 2, . . . , š‘š } for every š›¼ āˆˆ [0, š›¼ ĀÆ] Therefore dx āˆˆ š·. We have previously shown (Exercises 5.39 and 5.40) that š· āŠ‚ š‘‡ āŠ‚ šæ. 251

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Solutions for Foundations of Mathematical Economics

5.42 Assume that g satisļ¬es the Quasiconvex CQ condition at xāˆ— . That is, for every Ė† such that š‘”š‘— (Ė† š‘— āˆˆ šµ(xāˆ— ), š‘”š‘— is quasiconvex, āˆ‡š‘”š‘— (xāˆ— ) āˆ•= 0 and there exists x x) < 0. Ė† āˆ’ xāˆ— . Quasiconvexity and regularity implies that Consider the perturbation dx = x for every binding constraint š‘— āˆˆ šµ(xāˆ— ) (Exercises 4.74 and 4.75) š‘”š‘— (Ė† x) < š‘”š‘— (xāˆ— ) =ā‡’ āˆ‡š‘”š‘— (xāˆ— )š‘‡ (Ė† x āˆ’ xāˆ— ) = āˆ‡š‘”š‘— (xāˆ— )š‘‡ dx < 0 That is š·š‘”š‘— [xāˆ— ](dx) < 0 Therefore, dx āˆˆ šæ0 (xāˆ— ) āˆ•= āˆ… and g satisļ¬es the Cottle constraint qualiļ¬cation condition. 5.43 If the binding constraints šµ(xāˆ— ) are regular at xāˆ— , their gradients are linearly independent. That is, there exists no šœ†š‘— āˆ•= 0, š‘— āˆˆ šµ(xāˆ— ) such that āˆ‘ šœ†š‘— āˆ‡š‘”š‘— [xāˆ— ] = 0 š‘—āˆˆšµ(xāˆ— )

By Gordanā€™s theorem (Exercise 3.239), there exists dx āˆˆ ā„œš‘› such that āˆ‡š‘”š‘— [xāˆ— ]š‘‡ dx < 0 for every š‘— āˆˆ šµ(xāˆ— ) Therefore dx āˆˆ šæ0 (xāˆ— ) āˆ•= āˆ…. 5.44 If š‘”š‘— concave, šµ š‘ (xāˆ— ) = āˆ…, and AHUCQ is trivially satisļ¬ed (with dx = 0 āˆˆ šæ1 ). For every š‘—, let š‘†š‘— = { dx : š·š‘”š‘— [xāˆ— ](dx) < 0 } Then

āŽ› šæ1 (xāˆ— ) = āŽ

āŽž

āˆ©

š‘†š‘– āŽ 

āˆ©

āŽ›

āˆ©

āŽ

š‘–āˆˆšµ š‘ (xāˆ— )

āŽž š‘†š‘– āŽ 

š‘–āˆˆšµ š¶ (xāˆ— )

where šµ š¶ (xāˆ— ) and šµ š‘ (xāˆ— ) are respectively the concave and nonconcave constraints binding at xāˆ— . If š‘”š‘— satisļ¬es the AHUCQ condition, šæ1 (xāˆ— ) āˆ•= āˆ… and Exercise 1.219 implies that āŽž āŽ› āŽž āŽ› āˆ© āˆ© āˆ© šæ1 = āŽ š‘†š‘– āŽ  āŽ š‘†š‘– āŽ  š‘–āˆˆšµ š‘ (xāˆ— )

š‘–āˆˆšµ š¶ (xāˆ— )

Now š‘†š‘– = { dx : š·š‘”š‘— [xāˆ— ](dx) ā‰¤ 0 } and therefore

āˆ©

šæ1 =

š‘†š‘– = šæ

š‘—āˆˆšµ(xāˆ— )

Since (Exercise 5.41) šæ1 āŠ† š‘‡ āŠ† šæ and š‘‡ is closed (Exercise 5.38), we have šæ = šæ1 āŠ† š‘‡ āŠ† šæ which implies that š‘‡ = šæ. 252

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c 2001 Michael Carter āƒ All rights reserved

5.45 For each š‘— = 1, 2, . . . , š‘š, either š‘”š‘— (xāˆ— ) < 0 which implies that šœ†š‘— = 0 and therefore šœ†š‘— š·š‘”š‘— [xāˆ— ](x āˆ’ xāˆ— ) = 0 or āˆ—

š‘”š‘— (x ) = 0 Since š‘”š‘— is quasiconvex and š‘”š‘— (x) ā‰¤ 0 = š‘”(xāˆ— ), Exercise 4.73 implies that š·š‘”š‘— [xāˆ— ](x āˆ’ xāˆ— ) ā‰¤ 0. Since šœ†š‘— ā‰„ 0, this implies that šœ†š‘— š·š‘”š‘— [xāˆ— ](x āˆ’ xāˆ— ) ā‰¤ 0. We have shown that for every š‘—, šœ†š‘— š·š‘”š‘— [xāˆ— ](x āˆ’ xāˆ— ) ā‰¤ 0. The ļ¬rst-order condition implies that āˆ‘ šœ†š‘— š·š‘”š‘— [xāˆ— ](x āˆ’ xāˆ— ) ā‰¤ 0 š·š‘“ [xāˆ— ](x āˆ’ xāˆ— ) = š‘—

If āˆ‡š‘“ (xāˆ— ) ā‰¤

āˆ‘

šœ†š‘— āˆ‡š‘”š‘— (xāˆ— )

xāˆ— ā‰„ 0

( )š‘‡ āˆ‡š‘“ (xāˆ— ) āˆ’ šœ†š‘— āˆ‡š‘”š‘— (xāˆ— ) xāˆ— = 0

The ļ¬rst-order conditions imply that for every x āˆˆ šŗ, x ā‰„ 0 and (

)š‘‡ āˆ‡š‘“ (xāˆ— ) āˆ’ šœ†š‘— āˆ‡š‘”š‘— (xāˆ— ) x ā‰¤ 0

and therefore ( )š‘‡ āˆ‡š‘“ (xāˆ— ) āˆ’ šœ†š‘— āˆ‡š‘”š‘— (xāˆ— ) (x āˆ’ xāˆ— ) ā‰¤ 0 or āˆ‡š‘“ (xāˆ— )š‘‡ (x āˆ’ xāˆ— ) ā‰¤

āˆ‘

šœ†š‘— āˆ‡š‘”š‘— (xāˆ— )š‘‡ (x āˆ’ xāˆ— ) ā‰¤ 0

5.46 Assuming š‘„š‘‘ = š‘„š‘‘ = 0, the constraints become 2š‘„š‘ ā‰¤ 30 2š‘„š‘ ā‰¤ 25 š‘„š‘ ā‰¤ 20 The ļ¬rst and third conditions are redundant, which implies that šœ†š‘“ = šœ†š‘š = 0. Complementary slackness requires that, if š‘„š‘ > 0, š·š‘„š‘ šæ = 1 āˆ’ 2šœ†š‘“ āˆ’ 2šœ†š‘™ āˆ’ šœ†š‘š = 0 or šœ†š‘™ = 12 . Evaluating the Lagrangean at (0, 1/2, 0) yields )) ( ( 1 = 3š‘„š‘ + š‘„š‘ + 3š‘„š‘‘ šæ x, 0, , 0 2 1 āˆ’ (š‘„š‘ + 2š‘„š‘ + 3š‘„š‘‘ āˆ’ 25) 2 25 5 3 = + š‘„š‘ + š‘„š‘‘ 2 2 2 This basic feasible solution is clearly not optimal, since proļ¬t would be increased by increasing either š‘„š‘ or š‘„š‘‘ .

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Following the hint, we allow š‘„š‘‘ > 0, retaining the assumption that š‘„š‘ = 0. We must be alert to the possibility that š‘„š‘ = 0. With š‘„š‘ = 0, the constraints become 2š‘„š‘ + š‘„š‘‘ ā‰¤ 30 2š‘„š‘ + 3š‘„š‘‘ ā‰¤ 25 š‘„š‘ + š‘„š‘‘ ā‰¤ 20 The ļ¬rst constraint is redundant, which implies that šœ†š‘“ = 0. If š‘„š‘‘ > 0, complementary slackness requires that š·š‘„š‘‘ šæ = 3 āˆ’ 3šœ†š‘™ āˆ’ šœ†š‘š = 0 or šœ†š‘š = 3(1 āˆ’ šœ†š‘™ )

(5.105)

The requirement that šœ†š‘š ā‰„ 0 implies that šœ†š‘™ ā‰¤ 1. Substituting (5.105) in the second ļ¬rst-order condition š·š‘„š‘ šæ = 1 āˆ’ 2šœ†š‘™ āˆ’ šœ†š‘š = 1 āˆ’ 2šœ†š‘™ āˆ’ 3(1 āˆ’ šœ†š‘™ ) = āˆ’2 + šœ†š‘™ implies that š·š‘„š‘ šæ = āˆ’2 + šœ†š‘™ < 0

for every šœ†š‘™ ā‰¤ 1

Complementary slackness then requires implies that š‘„š‘ = 0. The constraints now become š‘„š‘‘ ā‰¤ 30 3š‘„š‘‘ ā‰¤ 25 š‘„š‘‘ ā‰¤ 20 The ļ¬rst and third are redundant, so that šœ†š‘“ and šœ†š‘š = 0. Equation (5.105) implies that šœ†š‘™ = 1. Evaluating the Lagrangean at this point (šœ† = 0, 1, 0), we have šæ(š‘„, (0, 1, 0)) = 3š‘„š‘ + š‘„š‘ + 3š‘„š‘‘ āˆ’ (š‘„š‘ + 2š‘„š‘ + 3š‘„š‘‘ āˆ’ 25) = 25 + 2š‘„š‘ āˆ’ š‘„š‘ Clearly this is not an optimal solution, An increase in š‘„š‘ is indicated. This leads us to the hypothesis š‘„š‘ > 0, š‘„š‘‘ > 0, š‘„š‘ = 0 which was evaluated in the text, and in fact lead to the optimal solution. 5.47 If we ignore the hint and consider solutions with š‘„š‘ > 0, š‘„š‘ ā‰„ 0, š‘„š‘‘ = 0, the constraints become 2š‘„š‘ + 2š‘„š‘ ā‰¤ 30 š‘„š‘ + 2š‘„š‘ ā‰¤ 25 2š‘„š‘ + š‘„š‘ ā‰¤ 20 These three constraints are linearly dependent, so that any one of them is redundant and can be eliminated. For example, 3/2 times the ļ¬rst constraint is equal to the sum of 254

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c 2001 Michael Carter āƒ All rights reserved

the second and third constraints. The feasible solution š‘„š‘ = 0, š‘„š‘ = 5, š‘„š‘‘ = 10, where the constraints are linearly dependent, is known as a degenerate solution. Degeneracy is a signiļ¬cant feature of linear programming, allowing the theoretical possibility of a breakdown in the simplex algorithm. Fortunately, such breakdown seems very rare in practice. Degeneracy at the optimal solution indicates multiple optima. One way to proceed in this example is to arbitrarily designate one constraint as redundant, assuming the corresponding multiplier is zero. Arbitrarily choosing šœ†š‘š = 0 and proceeding as before, complementary slackness (š‘„š‘‘ > 0) requires that š·š‘„š‘‘ šæ = 3 āˆ’ 2šœ†š‘“ āˆ’ šœ†š‘™ = 0 or šœ†š‘™ = 3 āˆ’ 2šœ†š‘“

(5.106)

Nonnegativity of šœ†š‘™ implies that šœ†š‘“ ā‰¤ 32 . Substituting (5.106) in the second ļ¬rst-order condition yields š·š‘„š‘ šæ = 1 āˆ’ 2šœ†š‘“ āˆ’ 2šœ†š‘™ = 1 āˆ’ 2šœ†š‘“ āˆ’ 2(3 āˆ’ 2šœ†š‘“ ) = āˆ’5 + 2šœ†š‘“ < 0 for every šœ†š‘“ ā‰¤

3 2

Complementary slackness therefore implies that š‘„š‘ = 0, which takes us back to the starting point of the presentation in the text, where š‘„š‘ > 0, š‘„š‘ = š‘„š‘‘ = 0. 5.48 Assume that (c1 , š‘§1 ) and (c2 , š‘§2 ) belong to šµ. That is š‘§1 ā‰„ š‘§ āˆ— š‘§2 ā‰„ š‘§ āˆ—

c1 ā‰¤ 0 c2 ā‰¤ 0 For any š›¼ āˆˆ (0, 1),

š‘§ĀÆ = š›¼š‘§1 + (1 āˆ’ š›¼)š‘§2 ā‰¤ š‘§ āˆ—

ĀÆ c = š›¼c1 + (1 āˆ’ š›¼)c2 ā‰¤ 0

and therefore (ĀÆ c, š‘§ĀÆ) āˆˆ šµ. This shows that šµ is convex. Let 1 = (1, 1, . . . , 1) āˆˆ ā„œš‘š . Then (c āˆ’ 1, š‘§ + 1) āˆˆ int šµ āˆ•= āˆ…. There šµ has a nonempty interior. 5.49 Let (c, š‘§) āˆˆ int šµ. This implies that c < 0 and š‘§ > š‘§ āˆ— . Since š‘£ is monotone š‘£(c) ā‰¤ š‘£(0) = zāˆ— < š‘§ which implies that (c, š‘§) āˆˆ / š“. 5.50 The linear functional šæ can be decomposed into separate components, so that there exists (Exercise 3.47) šœ‘ āˆˆ š‘Œ āˆ— and š›¼ āˆˆ ā„œ such that šæ(c, š‘§) = š›¼š‘§ āˆ’ šœ‘(c) Assuming š‘Œ āŠ† ā„œš‘š , there exists (Proposition 3.4) š€ āˆˆ ā„œš‘š such that šœ‘(c) = š€š‘‡ c and therefore šæ(c, š‘§) = š›¼š‘§ āˆ’ š€š‘‡ c The point (0, š‘§ āˆ— + 1) belongs to šµ. Therefore, by (5.75), šæ(0, š‘§ āˆ— ) ā‰¤ šæ(0, š‘§ āˆ— + 1) 255

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics which implies that š›¼š‘§ āˆ— āˆ’ š€š‘‡ 0 ā‰¤ š›¼(š‘§ āˆ— + 1) āˆ’ š€š‘‡ 0

or š›¼ ā‰„ 0. Similarly, let { e1 , e2 , . . . , eš‘š } denote the standard basis for ā„œš‘š (Example 1.79). For any š‘— = 1, 2, . . . , š‘š, the point (0 āˆ’ eš‘— , š‘§ āˆ— ) (which corresponds to decreasing resource š‘— by one unit) belongs to šµ and therefore (from (5.75)) š‘§ āˆ— āˆ’ š€š‘‡ (0 āˆ’ eš‘— ) = š‘§ āˆ— + šœ†š‘— ā‰„ š‘§ āˆ— āˆ’ š€š‘‡ 0 = š‘§ āˆ— which implies that šœ†š‘— ā‰„ 0. 5.51 Let cĖ† = š‘”(Ė† x) < 0 and š‘§Ė† = š‘“ (Ė† x) Suppose š›¼ = 0. Then, since šæ is nonzero, at least one component of š€ must be nonzero. That is, š€ ā‰© 0 and therefore š€š‘‡ š‘Ė† < 0

(5.107)

But (Ė† c, š‘§Ė†) āˆˆ š“ and (5.74) implies š›¼Ė† š‘§ āˆ’ š€š‘‡ cĖ† ā‰¤ š›¼š‘§ āˆ— āˆ’ š€š‘‡ 0 and therefore š›¼ = 0 implies š€š‘‡ š‘Ė† ā‰„ 0 contradicting (5.107). Therefore, we conclude that š›¼ > 0. 5.52 The utilityā€™s optimization problem is max š‘†(š‘¦, š‘Œ ) =

š‘¦,š‘Œ ā‰„0

š‘› āˆ« āˆ‘ š‘–=1

š‘¦š‘– 0

(š‘š‘– (šœ ) āˆ’ š‘š‘– )š‘‘šœ āˆ’ š‘0 š‘Œ

subject to š‘”š‘– (y, š‘Œ ) = š‘¦š‘– āˆ’ š‘Œ ā‰¤ 0

š‘– = 1, 2, . . . , š‘›

The demand independence assumption ensures that the objective function š‘† is concave, since its Hessian āŽ› āŽž š·š‘1 0 . . . 0 0 āŽœ 0 š·š‘2 . . . 0 0āŽŸ āŽŸ š»š‘† = āŽœ āŽ 0 ... š·š‘š‘› 0āŽ  0 ... 0 0 is nonpositive deļ¬nite (Exercise 3.96). The constraints are linear and hence convex. Moreover, there exists a point (0, 1) such that for every š‘– = 1, 2, . . . , š‘› š‘”š‘– (0, 1) = 0 āˆ’ 1 < 0 Therefore the problem satisļ¬es the conditions of Theorem 5.6. The optimal solution (yāˆ— , š‘Œ āˆ— ) satisļ¬es the Kuhn-Tucker conditions, that is there exist multipliers šœ†1 , šœ†2 , . . . , šœ†š‘š such that for every period š‘– = 1, 2, . . . , š‘› š·š‘¦š‘– šæ = š‘š‘– (š‘¦š‘– ) āˆ’ š‘š‘– āˆ’ šœ†š‘– ā‰¤ 0

š‘¦š‘– ā‰„ 0

š‘¦š‘– (š‘š‘– (š‘¦š‘– ) āˆ’ š‘š‘– āˆ’ šœ†š‘– ) = 0

š‘¦š‘– ā‰¤ š‘Œ

šœ†š‘– ā‰„ 0

šœ†(š‘Œ āˆ’ š‘¦š‘– ) = 0

256

(5.108)

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics and that capacity be chosen such that š· š‘Œ šæ = š‘0 āˆ’

š‘› āˆ‘

( šœ†š‘– ā‰¤ 0

š‘Œ ā‰„0

š‘Œ

š‘0 āˆ’

š‘–=1

š‘› āˆ‘

) šœ†š‘–

=0

(5.109)

š‘–=1

where šæ is the Lagrangean šæ(š‘¦, š‘Œ, šœ†) =

š‘› āˆ« āˆ‘ š‘–=1

0

š‘¦š‘–

(š‘š‘– (šœ ) āˆ’ š‘š‘– )š‘‘šœ āˆ’ š‘0 š‘Œ āˆ’

š‘› āˆ‘

šœ†š‘– (š‘¦š‘– āˆ’ š‘Œ )

š‘–=1

In oļ¬€-peak periods (š‘¦š‘– < š‘Œ ), complementary slackness requires that šœ†š‘– = 0 and therefore from (5.108) š‘š‘– (š‘¦š‘– ) = š‘š‘– assuming š‘¦š‘– > 0. In peak periods (š‘¦š‘– = š‘Œ ) š‘š‘– (š‘¦š‘– ) = š‘š‘– + šœ†š‘– We conclude that it is optimal to price at marginal cost in oļ¬€-peak periods and charge a premium during peak periods. Furthermore, (5.109) implies that the total premium is equal to the marginal capacity cost š‘› āˆ‘

šœ†š‘– = š‘0

š‘–=1

Furthermore, note that š‘› āˆ‘

šœ†š‘– š‘¦š‘– =

š‘–=1

āˆ‘ Peak

=

=

šœ†š‘– š‘¦š‘–

Oļ¬€-peak

āˆ‘

šœ†š‘– š‘¦š‘– +

š‘¦š‘– =š‘Œ

=

āˆ‘

šœ†š‘– š‘¦š‘– +

āˆ‘

āˆ‘

šœ†š‘– š‘¦š‘–

šœ†š‘– =0

šœ†š‘– š‘Œ

š‘¦š‘– =š‘Œ š‘› āˆ‘

šœ†š‘– š‘Œ = š‘0 š‘Œ

š‘–=1

Therefore, the utilityā€™s total revenue is š‘…(š‘¦, š‘Œ ) = = = =

š‘› āˆ‘ š‘–=1 š‘› āˆ‘ š‘–=1 š‘› āˆ‘ š‘–=1 š‘› āˆ‘

š‘š‘– (š‘¦š‘– )š‘¦š‘– (š‘š‘– + šœ†š‘– )š‘¦š‘– š‘š‘– š‘¦ š‘– +

š‘› āˆ‘

šœ†š‘– š‘¦š‘–

š‘–=1

š‘š‘– š‘¦š‘– + š‘0 š‘Œ = š‘(š‘¦, š‘Œ )

š‘–=1

Under the optimal pricing policy, revenue equals cost and the utility breaks even. 257

Solutions for Foundations of Mathematical Economics

c 2001 Michael Carter āƒ All rights reserved

Chapter 6: Comparative Statics 6.1 The Jacobian is

( š»šæ š½g

š½=

š½gš‘‡ 0

)

where š»šæ is the Hessian of the Lagrangean. We note that āˆ™ š»šæ (x0 ) is negative deļ¬nite in the subspace š‘‡ = { x : š½g x = 0 } (since x0 satisļ¬es the conditions for a strict local maximum) āˆ™ š½g has rank š‘š (since the constraints are regular). Consider the system of equations ( š»šæ š½g

š½gš‘‡ 0

)( ) ( ) x 0 = y 0

(6.28)

where x āˆˆ ā„œš‘› and y āˆˆ ā„œš‘š . It can be decomposed into š»šæ x + š½gš‘‡ y = 0

(6.29)

š½g x = 0

(6.30)

Suppose x solves (6.30). Multiplying (6.29) by xš‘‡ gives xš‘‡ š»šæ x + xš‘‡ š½gš‘‡ y = xš‘‡ š»šæ x + (š½g x)š‘‡ y = 0 But (6.30) implies that the second term is 0 and therefore xš‘‡ š»šæ x = 0. Since š»šæ is positive deļ¬nite on š‘‡ = { x : š½g x = 0 }, we must have x = 0. Then (6.29) reduces to š½gš‘‡ y = 0 Since š½g has rank š‘š, this has only the trivial solution y = 0 (Section 3.6.1). We have shown that the system (6.38) has only the trivial solution (0, 0). This implies that the matrix š½ is nonsingular. 6.2 The Lagrangean for this problem is

( ) šæ = š‘“ (x) āˆ’ š€š‘‡ g(x) āˆ’ c

By Corollary 6.1.1 āˆ‡š‘£(c) = š·c šæ = š€ 6.3 Optimality implies š‘“ (x1 , šœ½1 ) ā‰„ š‘“ (x, šœ½ 1 ) and š‘“ (x2 , šœ½ 2 ) ā‰„ š‘“ (x, šœ½2 ) for every x āˆˆ š‘‹ In particular š‘“ (x1 , šœ½1 ) ā‰„ š‘“ (x2 , šœ½1 ) and š‘“ (x2 , šœ½2 ) ā‰„ š‘“ (x1 , šœ½2 ) Adding these inequalities š‘“ (x1 , šœ½1 ) + š‘“ (x2 , šœ½2 ) ā‰„ š‘“ (x2 , šœ½1 ) + š‘“ (x1 , šœ½2 ) 258

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c 2001 Michael Carter āƒ All rights reserved

Rearranging and using the bilinearity of š‘“ gives š‘“ (x1 āˆ’ x2 , šœ½1 ) ā‰„ š‘“ (x1 āˆ’ x2 , šœ½2 ) and š‘“ (x1 āˆ’ x2 , šœ½ 1 āˆ’ šœ½ 2 ) ā‰„ 0 6.4 Let š‘1 denote the proļ¬t maximizing price with the cost function š‘1 (š‘¦) and let š‘¦1 be the corresponding output. Similarly let š‘2 and š‘¦2 be the proļ¬t maximizing price and output when the costs are given by š‘2 (š‘¦). With cost function š‘1 , the ļ¬rms proļ¬t is Ī  = š‘š‘¦ āˆ’ š‘1 (š‘¦) Since this is maximised at š‘1 and š‘¦1 (although the monopolist could have sold š‘¦2 at price š‘2 ) š‘1 š‘¦1 āˆ’ š‘1 (š‘¦1 ) ā‰„ š‘2 š‘¦2 āˆ’ š‘1 (š‘¦2 ) Rearranging š‘1 š‘¦1 āˆ’ š‘2 š‘¦2 ā‰„ š‘1 (š‘¦1 ) āˆ’ š‘1 (š‘¦2 )

(6.31)

The increase in revenue in moving from š‘¦2 to š‘¦1 is greater than the increase in cost. Similarly š‘2 š‘¦2 āˆ’ š‘2 (š‘¦2 ) ā‰„ š‘1 š‘¦1 āˆ’ š‘2 (š‘¦1 ) which can be rearranged to yield š‘2 (š‘¦1 ) āˆ’ š‘2 (š‘¦2 ) ā‰„ š‘1 š‘¦1 āˆ’ š‘2 š‘¦2 Combining the previous inequality with (6.31) yields š‘2 (š‘¦1 ) āˆ’ š‘2 (š‘¦2 ) ā‰„ š‘1 (š‘¦1 ) āˆ’ š‘1 (š‘¦2 )

(6.32)

6.5 By Theorem 6.2 š·w Ī [w, š‘] = āˆ’xāˆ— and š·š‘ Ī [w, š‘] = š‘¦ āˆ— and therefore 2 Ī (š‘, w) ā‰„ 0 š·š‘ š‘¦(š‘, w) = š·š‘š‘ 2 Ī (š‘, w) ā‰¤ 0 š·š‘¤š‘– š‘„š‘– (š‘, w) = āˆ’š·š‘¤ š‘– š‘¤š‘– 2 Ī (š‘, w) = š·š‘¤š‘– š‘„š‘— (š‘, w) š·š‘¤š‘— š‘„š‘– (š‘, w) = āˆ’š·š‘¤ š‘– š‘¤š‘— 2 Ī (š‘, w) = āˆ’š·š‘¤š‘– š‘¦(š‘, w) š·š‘ š‘„š‘– (š‘, w) = āˆ’š·š‘¤ š‘–š‘

since Ī  is convex and therefore š»Ī  (w, š‘) is symmetric (Theorem 4.2) and nonnegative deļ¬nite (Proposition 4.1). 6.6 By Shephardā€™s lemma (6.17) š‘„š‘– (š‘¤, š‘¦) = š·š‘¤š‘– š‘(š‘¤, š‘¦) 259

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics Using Youngā€™s theorem (Theorem 4.2), 2 š·š‘¦ š‘„š‘– [w, š‘¦] = š·š‘¤ š‘[w, š‘¦] š‘–š‘¦ 2 š‘[w, š‘¦] = š·š‘¦š‘¤ š‘–

= š·š‘¤š‘– š·š‘¦ š‘[w, š‘¦] Therefore š·š‘¦ š‘„š‘– [w, š‘¦] ā‰„ 0 ā‡ā‡’ š·š‘¤š‘– š·š‘¦ š‘[w, š‘¦] ā‰„ 0 6.7 The demand functions must satisfy the budget contraint identically, that is š‘› āˆ‘

š‘š‘– š‘„š‘– (p, š‘š) = š‘š for every p and š‘š

š‘–=1

Diļ¬€erentiating with respect to m š‘› āˆ‘

š‘š‘– š·š‘š š‘„š‘– [p, š‘š] = 1

š‘–=1

This is the Engel aggregation condition, which simply states that any additional income be spent on some goods. Multiplying each term by š‘„š‘– š‘š/(š‘„š‘– š‘š) š‘› āˆ‘ š‘š‘– š‘„š‘– š‘–=1

š‘š š·š‘š š‘„š‘– [p, š‘š] = 1 š‘š š‘„š‘– (p, š‘š)

the Engel aggregation condition can be written in elasticity form š‘› āˆ‘

š›¼š‘– šœ‚š‘– = 1

š‘–=1

where š›¼š‘– = š‘š‘– š‘„š‘– /š‘š is the budget share of good š‘–. On average, goods must have unit income elasticities. Diļ¬€erentiating the budget constraint with respect to š‘š‘— š‘› āˆ‘

š‘š‘– š·š‘š‘— š‘„š‘– [p, š‘š] + š‘„š‘— (š‘, š‘š) = 0

š‘–=1

This is the Cournot aggregation condition, which implies that an increase in the price of š‘š‘— is equivalent to a decrease in real income of š‘„š‘— š‘‘š‘š‘— . Multiplying each term in the sum by š‘„š‘– /š‘„š‘– gives š‘› āˆ‘ š‘š‘– š‘„š‘– š‘–=1

š‘„š‘–

š·š‘š‘— š‘„š‘– [p, š‘š] = āˆ’š‘„š‘—

Multiplying through by š‘š‘— /š‘š š‘› āˆ‘ š‘š‘– š‘„š‘– š‘š‘— š‘–=1

š‘š š‘„š‘–

š·š‘š‘— š‘„š‘– [p, š‘š] = āˆ’

š‘› āˆ‘

š›¼š‘– šœ–š‘–š‘— = āˆ’š›¼š‘—

š‘–=1

260

š‘š‘— š‘„š‘— š‘š

c 2001 Michael Carter āƒ All rights reserved

Solutions for Foundations of Mathematical Economics

6.8 Supermodularity of Ī (x, š‘, āˆ’w) follows from Exercises 2.50 and 2.51. To show strictly increasing diļ¬€erences, consider two price vectors w2 ā‰„ w1 Ī (x, š‘, āˆ’w1 ) āˆ’ Ī (x, š‘, āˆ’w2 ) =

š‘› āˆ‘

(āˆ’š‘¤š‘–1 )š‘„š‘– āˆ’

š‘–=1

=

š‘› āˆ‘

š‘› āˆ‘

(āˆ’š‘¤š‘–2 )š‘„š‘–

š‘–=1

(š‘¤š‘–2 āˆ’ š‘¤š‘–1 )š‘„š‘–

š‘–=1

āˆ‘š‘›

Since w2 ā‰„ w1 , w2 āˆ’ w1 ā‰„ 0 and

2 š‘–=1 (š‘¤š‘–

āˆ’ š‘¤š‘–1 )š‘„š‘– is strictly increasing in x.

6.9 For any š‘2 ā‰„( š‘1 , š‘¦ 2 = š‘“ (š‘2 ) ā‰¤ š‘“ (š‘1 ) = š‘¦ 1 and š‘(š‘¦ 1 , šœƒ) āˆ’ š‘(š‘¦ 2 , šœƒ) is increasing in šœƒ and therefore āˆ’ š‘(š‘“ (š‘2 ), šœƒ) āˆ’ š‘(š‘“ (š‘1 ), šœƒ)) is increasing in šœƒ. 6.10 The ļ¬rmā€™s optimization problem is max šœƒš‘š‘¦ āˆ’ š‘(š‘¦)

š‘¦āˆˆā„œ+

The objective function š‘“ (š‘¦, š‘, šœƒ) = šœƒš‘š‘¦ āˆ’ š‘(š‘¦) is āˆ™ supermodular in š‘¦ (Exercise 2.49) āˆ™ displays strictly increasing diļ¬€erences in (š‘¦, šœƒ) since ( ) š‘“ (š‘¦ 2 , š‘, šœƒ) āˆ’ š‘“ (š‘¦ 1 , š‘, šœƒ) = šœƒš‘(š‘¦ 2 āˆ’ š‘¦ 1 ) āˆ’ š‘(š‘¦ 2 ) āˆ’ š‘(š‘¦ 1 ) is strictly increasing in šœƒ for š‘¦ 2 > š‘¦ 1 . Therefore (Corollary 2.1.2), the ļ¬rmā€™s output correspondence is strongly increasing and every selection is increasing (Exercise 2.45). Therefore, the ļ¬rmā€™s output increases as the yield increases. It is analogous to an increase in the exogenous price. 6.11 With two factors, the Hessian is ( š‘“11 š»š‘“ = š‘“21

š‘“12 š‘“22

)

Therefore, its inverse is (Exercise 3.104) š»š‘“āˆ’1

1 = Ī”

(

š‘“22 āˆ’š‘“21

āˆ’š‘“12 š‘“11

)

where Ī” = š‘“11 š‘“22 āˆ’ š‘“12 š‘“21 ā‰„ 0 by the second-order condition. Therefore, the Jacobian of the demand functions is ( ) ) ( 1 1 āˆ’1 š·š‘¤1 š‘„1 š·š‘¤2 š‘„1 š‘“22 āˆ’š‘“12 = š»š‘“ = š·š‘¤1 š‘„2 š·š‘¤2 š‘„2 š‘“11 š‘ š‘Ī” āˆ’š‘“21 Therefore š‘“21 š·š‘¤1 š‘„2 = āˆ’ š‘Ī”

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