# Foundation of Mathematical Economics Solutions

July 30, 2017 | Author: Misbahul Islam | Category: Mathematical Concepts, Mathematical Analysis, Mathematical Objects, Mathematical Structures, Space

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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

Chapter 1: Sets and Spaces 1.1 { 1, 3, 5, 7 . . . } or { ð â ð : ð is odd } 1.2 Every ð¥ â ðŽ also belongs to ðµ. Every ð¥ â ðµ also belongs to ðŽ. Hence ðŽ, ðµ have precisely the same elements. 1.3 Examples of ï¬nite sets are â the letters of the alphabet { A, B, C, . . . , Z } â the set of consumers in an economy â the set of goods in an economy â the set of players in a game. Examples of inï¬nite sets are â the real numbers â â the natural numbers ð â the set of all possible colors â the set of possible prices of copper on the world market â the set of possible temperatures of liquid water. 1.4 ð = { 1, 2, 3, 4, 5, 6 }, ðž = { 2, 4, 6 }. 1.5 The player set is ð = { Jenny, Chris }. Their action spaces are ðŽð = { Rock, Scissors, Paper }

ð = Jenny, Chris

1.6 The set of players is ð = {1, 2, . . . , ð }. The strategy space of each player is the set of feasible outputs ðŽð = { ðð â â+ : ðð â€ ðð } where ðð is the output of dam ð. 1.7 The player set is ð = {1, 2, 3}. There are 23 = 8 coalitions, namely ð«(ð ) = {â, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}} There are 2

10

coalitions in a ten player game.

/ ð âª ð . This implies ð¥ â / ð and ð¥ â / ð, 1.8 Assume that ð¥ â (ð âª ð )ð . That is ð¥ â or ð¥ â ð ð and ð¥ â ð ð. Consequently, ð¥ â ð ð â© ð ð . Conversely, assume ð¥ â ð ð â© ð ð . This implies that ð¥ â ð ð and ð¥ â ð ð . Consequently ð¥ â / ð and ð¥ â / ð and therefore ð¥â / ð âª ð . This implies that ð¥ â (ð âª ð )ð . The other identity is proved similarly. 1.9

âª

ð=ð

ðâð

ð=â

ðâð

1

Solutions for Foundations of Mathematical Economics

1

-1

ð¥2

0

1

ð¥1

-1 Figure 1.1: The relation { (ð¥, ðŠ) : ð¥2 + ðŠ 2 = 1 } 1.10 The sample space of a single coin toss is { ð», ð }. The set of possible outcomes in three tosses is the product { {ð», ð } Ã {ð», ð } Ã {ð», ð } = (ð», ð», ð»), (ð», ð», ð ), (ð», ð, ð»), } (ð», ð, ð ), (ð, ð», ð»), (ð, ð», ð ), (ð, ð, ð»), (ð, ð, ð ) A typical outcome is the sequence (ð», ð», ð ) of two heads followed by a tail. 1.11 ð â© âð+ = {0} where 0 = (0, 0, . . . , 0) is the production plan using no inputs and producing no outputs. To see this, ï¬rst note that 0 is a feasible production plan. Therefore, 0 â ð . Also, 0 â âð+ and therefore 0 â ð â© âð+ . To show that there is no other feasible production plan in âð+ , we assume the contrary. That is, we assume there is some feasible production plan y â âð+ â {0}. This implies the existence of a plan producing a positive output with no inputs. This technological infeasible, so that ðŠ â / ð. 1.12

1. Let x â ð (ðŠ). This implies that (ðŠ, âx) â ð . Let xâ² â¥ x. Then (ðŠ, âxâ² ) â€ (ðŠ, âx) and free disposability implies that (ðŠ, âxâ² ) â ð . Therefore xâ² â ð (ðŠ).

2. Again assume x â ð (ðŠ). This implies that (ðŠ, âx) â ð . By free disposal, (ðŠ â² , âx) â ð for every ðŠ â² â€ ðŠ, which implies that x â ð (ðŠ â² ). ð (ðŠ â² ) â ð (ðŠ). 1.13 The domain of â ðŠ1 or ð¥1 = ðŠ1 and ð¥2 > ðŠ2 Since all elements ð¥ â â2 are comparable, â» is complete; it is a total order. 1.35 Assume â¿ð is complete for every ð. Then for every ð¥, ðŠ â ð and for all ð = 1, 2, . . . , ð, either ð¥ð â¿ð ðŠð or ðŠð â¿ð ð¥ð or both. Either ð¥ð âŒð ðŠð for all ð Then deï¬ne ð¥ âŒ ðŠ. ð¥ð ââŒð ðŠð for some ð Let ð be the ï¬rst individual with a strict preference, that is ð = minð (ð¥ð ââŒ ðŠð ). (Completeness of â¿ð ensures that ð is deï¬ned). Then deï¬ne ð¥ â» ðŠ if ð¥ð â»ð ðŠð ðŠ â» ð¥ otherwise 1.36 Let ð, ð and ð be subsets of a ï¬nite set ð. Set inclusion â is reï¬exive since ð â ð. transitive since ð â ð and ð â ð implies ð â ð . anti-symmetric since ð â ð and ð â ð implies ð = ð Therefore â is a partial order. 1.37 Assume ð¥ and ðŠ are both least upper bounds of ðŽ. That is ð¥ â¿ ð for all ð â ðŽ and ðŠ â¿ ð for all ð â ðŽ. Further, if ð¥ is a least upper bound, ðŠ â¿ ð¥. If ðŠ is a least upper bound, ð¥ â¿ ðŠ. By anti-symmetry, ð¥ = ðŠ. 1.38 ð¥ âŒ ðŠ =â ð¥ â¿ ðŠ and ðŠ â¿ ð¥ which implies that ð¥ = ðŠ by antisymmetry. Each equivalence class âŒ (ð¥) = { ðŠ â ð : ðŠ âŒ ð¥ } comprises just a single element ð¥. 1.39 max ð«(ð) = ð and min ð«(ð) = â. 1.40 The subset {2, 4, 8} forms a chain. More generally, the set of integer powers of a given number { ð, ð2 , ð3 , . . . } forms a chain. 1.41 Assume ð¥ and ðŠ are maximal elements of the chain ðŽ. Then ð¥ â¿ ð for all ð â ðŽ and in particular ð¥ â¿ ðŠ. Similarly, ðŠ â¿ ð for all ð â ðŽ and in particular ðŠ â¿ ð¥. Since â¿ is anti-symmetric, ð¥ = ðŠ. 1.42

1. By assumption, for every ð¡ â ð â ð , âº(ð¡) is a nonempty ï¬nite chain. Hence, it has a unique maximal element, ð(ð¡).

2. Let ð¡ be any node. Either ð¡ is an initial node or ð¡ has a unique predecessor ð(ð¡). Either ð(ð¡) is an initial node, or it has a unique predecessor ð(ð(ð¡)). Continuing in this way, we trace out a unique path from ð¡ back to an initial node. We can be sure of eventually reaching an initial node since ð is ï¬nite. 1.43 (1, 2) âš (3, 1) = (3, 2) and (1, 2) â§ (3, 2) = (1, 2) 6

Solutions for Foundations of Mathematical Economics 1.44

1. ð¥âšðŠ is an upper bound for { ð¥, ðŠ }, that is xâšy â¿ ð¥ and xâšy â¿ ðŠ. Similarly, ð¥ âš ðŠ is a lower bound for { ð¥, ðŠ }.

2. Assume ð¥ â¿ ðŠ. Then ð¥ is an upper bound for { ð¥, ðŠ }, that is ð¥ â¿ ð¥ âš ðŠ. If ð is any upper bound for { ð¥, ðŠ }, then ð â¿ ð¥. Therefore, ð¥ is the least upper bound for { ð¥, ðŠ }. Similarly, ðŠ is a lower bound for { ð¥, ðŠ }, and is greater than any other lower bound. Conversely, assume ð¥ âš ðŠ = ð¥. Then ð¥ is an upper bound for { ð¥, ðŠ }, that is ð¥ â¿ ðŠ. 3. Using the preceding equivalence ð¥ â¿ ð¥ â§ ðŠ =â ð¥ âš (ð¥ â§ ðŠ) = ð¥ ð¥ âš ðŠ â¿ ð¥ =â (ð¥ âš ðŠ) â§ ð¥ = ð¥ 1.45 A chain ð is a complete partially ordered set. For every ð¥, ðŠ â ð with ð¥ â= ðŠ, either ð¥ â» ðŠ or ðŠ â» ð¥. Therefore, deï¬ne the meet and join by { ðŠ if ð¥ â» ðŠ ð¥â§ðŠ = ð¥ if ðŠ â» ð¥ { ð¥ if ð¥ â» ðŠ ð¥âšðŠ = ðŠ if ðŠ â» ð¥ ð is a lattice with these operations. 1.46 Assume ð1 and ð2 are lattices, and let ð = ð1 Ã ð2 . Consider any two elements x = (ð¥1 , ð¥2 ) and y = (ðŠ1 , ðŠ2 ) in ð. Since ð1 and ð2 are lattices, ð1 = ð¥1 âš ðŠ1 â ð1 and ð2 = ð¥2 âš ðŠ2 â ð2 , so that b = (ð1 , ð2 ) = (ð¥1 âš ðŠ1 , ð¥2 âš ðŠ2 ) â ð. Furthermore b â¿ x and b â¿ y in the natural product order, so that b is an upper bound for the Ë = (Ëð1 , Ëð2 ) of {x, y} must have ðð â¿ð ð¥ð and ðð â¿ð ðŠð , {x, y}. Every upper bound b Ë â¿ b. Therefore, b is the least upper bound of {x, y}, that is b = x âš y. so that b Similarly, x â§ y = (ð¥1 â§ ðŠ1 , ð¥2 â§ ðŠ2 ). 1.47 Let ð be a subset of ð and let ð â = { ð¥ â ð : ð¥ â¿ ð  for every ð  â ð } be the set of upper bounds of ð. Then ð¥â â ð â â= â. By assumption, ð â has a greatest lower bound ð. Since every ð  â ð is a lower bound of ð â , ð â¿ ð  for every ð  â ð. Therefore ð is an upper bound of ð. Furthermore, ð is the least upper bound of ð, since ð âŸ ð¥ for every ð¥ â ð â . This establishes that every subset of ð also has a least upper bound. In particular, every pair of elements has a least upper and a greatest lower bound. Consequently ð is a complete lattice. 1.48 Without loss of generality, we will prove the closed interval case. Let [ð, ð] be an interval in a lattice ð¿. Recall that ð = inf[ð, ð] and ð = sup[ð, ð]. Choose any ð¥, ðŠ in [ð, ð] â ð¿. Since ð¿ is a lattice, ð¥ âš ðŠ â ð¿ and ð¥ âš ðŠ = sup{ ð¥, ðŠ } âŸ ð Therefore ð¥ âš ðŠ â [ð, ð]. Similarly, ð¥ â§ ðŠ â [ð, ð]. [ð, ð] is a lattice. Similarly, for any subset ð â [ð, ð] â ð¿, sup ð â ð¿ if ð¿ is complete. Also, sup ð âŸ ð = sup[ð, ð]. Therefore sup ð â [ð, ð]. Similarly inf ð â [ð, ð] so that [ð, ð] is complete.

7

Solutions for Foundations of Mathematical Economics 1.49

1. The strong set order â¿ð is antisymmetric Let ð1 , ð2 â ð with ð1 â¿ð ð2 and ð2 â¿ð ð1 . Choose ð¥1 â ð1 and ð¥2 â ð2 . Since ð1 â¿ð ð2 , ð¥1 âš ð¥2 â ð1 and ð¥1 â§ ð¥2 â ð2 . On the other hand, since ð2 â¿ ð1 , ð¥1 = (ð¥1 âš (ð¥1 â§ ð¥2 ) â ð2 and ð¥2 = ð¥2 â§ (ð¥1 âš ð¥2 ) â ð1 (Exercise 1.44. Therefore ð1 = ð2 and â¿ð is antisymmetric. transitive Let ð1 , ð2 , ð3 â ð with ð1 â¿ð ð2 and ð2 â¿ð ð3 . Choose ð¥1 â ð1 , ð¥2 â ð2 and ð¥3 â ð3 . Since ð1 â¿ð ð2 and ð2 â¿ð ð3 , ð¥1 âš ð¥2 and ð¥2 â§ ð¥3 are in ð2 . Therefore ðŠ2 = ð¥1 âš (ð¥2 â§ ð¥3 ) â ð2 which implies ) ( ð¥1 âš ð¥3 = ð¥1 âš (ð¥2 â§ ð¥3 ) âš ð¥3 ( ) = ð¥1 âš (ð¥2 â§ ð¥3 ) âš ð¥3 = ðŠ2 âš ð¥3 â ð3 since ð2 â¿ð ð3 . Similarly ð§2 = (ð¥1 âš ð¥2 ) â§ ð¥3 â ð2 and ( ) ð¥1 â§ ð¥3 = ð¥1 â§ (ð¥1 âš ð¥2 ) â§ ð¥3 ) ( = ð¥1 â§ (ð¥1 âš ð¥2 ) â§ ð¥3 = ð¥1 â§ ð§2 â ð1 Therefore, ð1 â¿ð ð3 .

2. ð â¿ð ð if and only if, for every ð¥1 , ð¥2 â ð, ð¥1 âš ð¥2 â ð and ð¥1 â§ ð¥2 â ð, which is the case if and only if ð is a sublattice. 3. Let ð¿(ð) denote the set of all sublattices of ð. We have shown that â¿ð is reï¬exive, transitive and antisymmetric on ð¿(ð). Hence, it is a partial order on ð¿(ð). 1.50 Assume ð1 â¿ð ð2 . For any ð¥1 â ð1 and ð¥2 â ð2 , ð¥1 âš ð¥2 â ð1 and ð¥1 â§ ð¥2 â ð2 . Therefore sup ð1 â¿ ð¥1 âš ð¥2 â¿ ð¥2

for every ð¥2 â ð2

which implies that sup ð1 â¿ sup ð2 . Similarly inf ð2 âŸ ð¥1 â§ ð¥2 âŸ ð¥1

for every ð¥1 â ð1

which implies that inf ð2 âŸ inf ð1 . Note that completeness ensures the existence of sup ð and inf ð respectively. 1.51 An argument analogous to the preceding exercise establishes =â . (Completeness is not required, since for any interval ð = inf[ð, ð] and ð = sup[ð, ð]). To establish the converse, assume that ð1 = [ð1 , ð1 ] and ð2 = [ð2 , ð2 ]. Consider any ð¥1 â ð1 and ð¥2 â ð2 . There are two cases. Case 1. ð¥1 â¿ ð¥2 Since ð is a chain, ð¥1 âš ð¥2 = ð¥1 â ð1 . ð¥1 â§ ð¥2 = ð¥2 â ð2 . Case 2. ð¥1 âº ð¥2 Since ð is a chain, ð¥1 âš ð¥2 = ð¥2 . Now ð1 âŸ ð¥1 âº ð¥2 âŸ ð2 âŸ ð2 . Therefore, ð¥2 = ð¥1 âš ð¥2 â ð1 . Similarly ð2 âŸ ð1 âŸ ð¥1 âº ð¥2 âŸ ð2 . Therefore ð¥1 â§ ð¥2 = ð¥1 â ð2 . We have shown that ð1 â¿ð ð2 in both cases. 1.52 Assume that â¿ is a complete relation on ð. This means that for every ð¥, ðŠ â ð, either ð¥ â¿ ðŠ or ðŠ â¿ ð¥. In particular, letting ð¥ = ðŠ, ð¥ â¿ ð¥ for ð¥ â ð. â¿ is reï¬exive. 8

Solutions for Foundations of Mathematical Economics

1.53 Anti-symmetry implies that each indiï¬erence class contains a single element. If the consumerâs preference relation was anti-symmetric, there would be no baskets of goods between which the consumer was indiï¬erent. Each indiï¬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 ð¥ is maximal in the weakly ordered set ð. We have to show that ð¥ â¿ ðŠ for all ðŠ â ð. Assume otherwise, that is assume there is some ðŠ â ð for which ð¥ ââ¿ ðŠ. Since â¿ is complete, this implies that ðŠ â» ð¥ which contradicts the assumption that ð¥ is maximal. Hence we conclude that ð¥ â¿ ðŠ for ðŠ â ð and ð¥ 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, ð and ð are both best elements in the weakly ordered set {ð âŒ ð â» ð}. 1.56

1. For every ð¥ â ð, either ð¥ â¿ ðŠ =â ð¥ â â¿(ðŠ) or ðŠ â¿ ð¥ =â ð¥ â âŸ(ðŠ) since â¿ is complete. Consequently, â¿(ðŠ) âª âº(ðŠ) = ð If ð¥ â â¿(ðŠ) â© âŸ(ðŠ), then ð¥ â¿ ðŠ and ðŠ â¿ ð¥ so that ð¥ âŒ ðŠ and ð¥ â ðŒðŠ .

2. For every ð¥ â ð, either ð¥ â¿ ðŠ =â ð¥ â â¿(ðŠ) or ðŠ â» ð¥ =â ð¥ â âº(ðŠ) since â¿ is complete. Consequently, â¿(ðŠ) âª âº(ðŠ) = ð and â¿(ðŠ) â© âº(ðŠ) = â. 3. For every ðŠ â ð, â»(ðŠ) and ðŒðŠ partition â¿(ðŠ) and therefore â»(ðŠ), ðŒðŠ and âº(ðŠ) partition ð. 1.57 Assume ð¥ â¿ ðŠ and ð§ â â¿(ð¥). Then ð§ â¿ ð¥ â¿ ðŠ by transitivity. Therefore ð§ â â¿(ðŠ). This shows that â¿(ð¥) â â¿(ðŠ). Similarly, assume ð¥ â» ðŠ and ð§ â â»(ð¥). Then ð§ â» ð¥ â» ðŠ by transitivity. Therefore ð§ â â»(ðŠ). This shows that â¿(ð¥) â â¿(ðŠ). To show that â¿(ð¥) â= â¿(ðŠ), observe that ð¥ â â»(ðŠ) but that ð¥ â / â»(ð¥) 1.58 Every ï¬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 deï¬ning the core (Chapter 13) but the strong Pareto order in the Nash bargaining solution (Chapter 15). 1.60 Assume that a group ð is decisive over ð¥, ðŠ â ð. Let ð, ð â ð be two other states. We have to show that ð is decisive over ð and ð. Without loss of generality, assume for all individuals ð â¿ð ð¥ and ðŠ â¿ð ð. Then, the Pareto order implies that ð â» ð¥ and ðŠ â» ð. Assume that for every ð â ð, ð¥ â¿ð ðŠ. Since ð is decisive over ð¥ and ðŠ, the social order ranks ð¥ â¿ ðŠ. By transitivity, ð â¿ ð. By IIA, this holds irrespective of individual preferences on other alternatives. Hence, ð is decisive over ð and ð. 1.61 Assume that ð is decisive. Let ð¥, ðŠ and ð§ be any three alternatives and assume ð¥ â¿ ðŠ for every ð â ð. Partition ð into two subgroups ð1 and ð2 so that ð¥ â¿ð ð§ for every ð â ð1 and ð§ â¿ð ðŠ for every ð â ð2 Since ð is decisive, ð¥ â¿ ðŠ. By completeness, either ð¥ â¿ ð§ in which case ð1 is decisive over ð¥ and ð§. By the ï¬eld expansion lemma (Exercise 1.60), ð1 is decisive. 9

Solutions for Foundations of Mathematical Economics

ð§ â» ð¥ which implies ð§ â¿ ðŠ. In this case, ð2 is decisive over ðŠ and ð§, and therefore (Exercise 1.60) decisive. 1.62 Assume â» is a social order which is Pareto and satisï¬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 ðŽ is decisive over ð¥ and ðŠ and ðµ is decisive over ð€ and ð§. That is, assume ð¥ â»ðŽ ðŠ =â ð¥ â» ðŠ ð€ â»ðµ ð§ =â ð€ â» ð§ Also assume ðŠ â¿ð ð€

for every ð

ð§ â¿ð ð¥

for every ð

This implies that ðŠ â¿ ð€ and ð§ â¿ ð¥ (Pareto principle). Combining these preferences, transitivity implies that ð¥â»ðŠâ¿ð€â»ð§ which contradicts the assumption that ð§ â¿ ð¥. Therefore, the implied social ordering is intransitive. 1.64 Assume ð¥ â core. In particular this implies that there does not exist any ðŠ â ð (ð ) such that ðŠ â» ð¥. Therefore ð¥ â Pareto. 1.65 No state will accept a cost share which exceeds what it can achieve on its own, so that if ð¥ â core then ð¥ðŽð â€ 1870 ð¥ð ð â€ 5330 ð¥ðŽð â€ 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 ð¥ðŽð + ð¥ð ð â€ 6990 Similarly ð¥ðŽð + ð¥ðŸð â€ 1960 ð¥ð ð + ð¥ðŸð â€ 5020 Finally, the sum of the shares should equal the total cost ð¥ðŽð + ð¥ð ð + ð¥ðŸð = 6530 The core is the set of all allocations of the total cost which satisfy the preceding inequalities. For example, the allocation (ð¥ðŽð = 1500, ð¥ð ð = 5000, ð¥ðŸð = 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 (ð¥ðŽð = 1510, ð¥ð ð = 5000, ð¥ðŸð = 20), which therefore belongs to the core. 10

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 ð€(ð ). 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 ð€(ðŽð, ð ð ). By similar calculations, we can compute the worth of each coalition, namely ð€(ðŽð ) = 0 ð€(ð ð ) = 0

ð€(ðŽð, ð ð ) = 210 ð€(ðŽð, ðŸð ) = 770

ð€(ðŸð ) = 0

ð€(ðŸð, ð ð ) = 1170

ð€(ð ) = 1530

An outcome in this game is an allocation of the total cost savings ð€(ð ) = 1530 amongst the three players. This can be translated into ï¬nal cost shares by subtracting each players share of the cost savings from their standalone cost. For example, a speciï¬c outcome in this game is (ð¥ðŽð = 370, ð¥ð ð = 330, ð¥ðŸð = 830), which corresponds to ï¬nal cost shares of 1500 for AP, 5000 for TN and 30 for KM. 1.67 Let ð¶ = {x â ð :

â

ð¥ð â¥ ð€(ð) for every ð â ð }

ðâð

/ core. This implies there exists some 1. ð¶ â core Assume that x â ð¶. Suppose x â coalition ð and outcome y â ð€(ð) such that y â»ð x for every ð â ð. â â y â ð€(ð) implies ðâð ðŠð â€ ð€(ð) while â y â»ð x for every ð â ð implies ðŠð > ð¥ð for every ð â ð. Summing, this implies â â ðŠð > ð¥ð â¥ ð€(ð) ðâð

ðâð

This contradiction establishes that x â core. 2. core â ð¶ Assume that x â core. Suppose x â / ð¶. This implies there exists some â â coalition ð such that ðâð ð¥ð < ð€(ð). Let ð = ð€(ð) â ðâð ð¥ð and consider the allocation y obtained by reallocating ð from ð ð to ð, that is { ð¥ð + ð/ð  ðâð ðŠð = ð¥ð â ð/(ð â ð ) ð â /ð where ð  = â£ðâ£ is the number of players in ð and ð = â£ð â£ is the number in ð . Then â ðŠð > ð¥ð forâ every ð â ð so that y â»ð x for every ð â ð. Further, y â ð€(ð) since ðâð ðŠð = ðâð ð¥ð + ð = ð€(ð) and y â ð since â ðâð

ðŠð =

â ðâð

(ð¥ð + ð/ð ) +

â

(ð¥ð â ð/(ð â ð )) =

ðâð /

â

ð¥ð = ð€(ð )

ðâð

This contradicts our assumption that x â / core, establishing that x â ð¶.

11

Solutions for Foundations of Mathematical Economics

1.68 The 7 unanimity games for the player set ð = {1, 2, 3} are { 1 S = {1}, {1,2}, {1,3}, N ð¢{1} (ð) = 0 otherwise { 1 S = {2}, {1,2}, {2,3}, N ð¢{2} (ð) = 0 otherwise { 1 S = {3}, {1,3}, {2,3}, N ð¢{3} (ð) = 0 otherwise { 1 S = {1,2}, N ð¢{1,2} (ð) = 0 otherwise { 1 S = {1,3}, N ð¢{1,3} (ð) = 0 otherwise { 1 S = {2,3}, N ð¢{2,3} (ð) = 0 otherwise { 1 S=N ð¢ð (ð) = 0 otherwise 1.69 Firstly, consider a simple game which is a unanimity game with essential coalition ð and let ð¥ be an outcome in which ð¥ð â¥ 0

for every ð â ð

ð¥ð = 0

for every ð â /ð

and â

ð¥ð = 1

ðâð

We claim that ð¥ â core. Winning coalitions If ð is winning coalition, then ð€(ð) = 1. Furthermore, if it is a winning coalition, it must contain ð , that is ð â ð and â â ð¥ð â¥ ð¥ð = 1 = ð€(ð) ðâð

ðâð

Losing coalitions If ð is a losing coalition, ð€(ð) = 0 and â ð¥ð â¥ 0 = ð€(ð) ðâð

Therefore ð¥ â core and so core â= â. Conversely, consider a simple game which is not a unanimity game. Suppose there exists an outcome ð¥ â core. Then â ð¥ð ð€(ð ) = 1 (1.15) ðâð

12

Solutions for Foundations of Mathematical Economics

Since there are no veto players (ð = â), ð€(ð â {ð}) = 1 for every player ð â ð and â ð¥ð â¥ ð€(ð â {ð}) = 1 ðâ=ð

which implies that ð¥ð = 0 for every ð â ð contradicting (1.15). Thus we conclude that core = â. 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 ð(x1 ) = (â180, â180, â365, â365, â395, â955) and ð(x2 ) = (â160, â200, â370, â380, â380, â950) d(x1 ) âºð¿ d(x2 ) which implies x1 â»ð x2 . 1.71 It is a weak order on ð, that is â¿ is reï¬exive, transitive and complete. Reï¬exivity ð and transitivity ï¬ow from the corresponding properties of â¿ð¿ on â2 . Similarly, for ð any x, y â ð, either d(x) âŸð¿ d(y) or d(y) âŸð¿ d(x) since â¿ð¿ is complete on â2 . Consequently either x â¿ y or y â¿ x (or both). â¿ is not a partial order since it is not antisymmetric d(x) âŸð¿ d(y) and d(y) âŸð¿ d(x) does not imply x = y 1.72 ð(ð, x) = ð€(ð) â

â

ð¥ð

ðâð

so that ð(ð, x) â€ 0 ââ

â

ð¥ð â¥ ð€(ð)

ðâð

1.73 Assume to the contrary that x â Nu but that x â / core. Then, there exists a coalition ð with a positive deï¬cit ð(ð, x) > 0. Since core â= â, there exists some y â ð such that ð(ð, y) â€ 0 for every ð â Nu. Consequently, d(y) âº d(x) and y â» x, so that x â / Nu. This contradiction establishes that Nu â core. 1.74 For player 1, ðŽ1 = {ð¶, ð } and (ð¶, ð¶) â¿1 (ð¶, ð¶) (ð¶, ð¶) â¿1 (ð, ð¶) Similarly for player 2 (ð¶, ð¶) â¿2 (ð¶, ð¶) (ð¶, ð¶) â¿2 (ð¶, ð ) Therefore, (ð¶, ð¶) satisï¬es the requirements of the deï¬nition of a Nash equilibrium (Example 1.51). 13

Solutions for Foundations of Mathematical Economics

1.75 If aâð is the best element in (ðŽð , â¿â²ð ) for every player ð, then (ðâð , aâð ) â»ð (ðð , aâð ) for every ðð â ðŽð and aâð â ðŽâð for every ð â ð . Therefore, aâ is a Nash equilibrium. Â¯ is another Nash equilibrium. Then for every To show that it is unique, assume that a player ð â ð Â¯âð ) â¿ð (ðð , a Â¯âð ) for every ðð â ðŽð (Â¯ ðð , a Â¯ is a maximal element of â¿â²ð . To see this, assume not. That is, which implies that a assume that there exists some ðËð â ðŽð such that ðËð â»â²ð ð Â¯ð which implies ðð , aâð ) for every aâð â ðŽâð (Ë ðð , aâð ) â»ð (Â¯ In particular Â¯âð ) â»ð (ðâð , aââð ) (Ë ðð , a Â¯ is anwhich contradicts the assumption that aâ is a Nash equilibrium. Therefore, a other Nash equilibrium, then ð Â¯ð is maximal in â¿â²ð and hence also a best element of â¿â²ð (Exercise 1.54), which contradicts the assumption that ðâð is the unique best element. Consequently, we conclude that aâ is the unique Nash equilibrium of the game. 1.76 We show that ð(ð¥, ðŠ) = â£ð¥ â ðŠâ£ satisï¬es the requirements of a metric, namely 1. â£ð¥ â ðŠâ£ â¥ 0. 2. â£ð¥ â ðŠâ£ = 0 if and only if ð¥ = ðŠ. 3. â£ð¥ â ðŠâ£ = â£ðŠ â ð¥â£. To establish the triangle inequality, we can consider various cases. For example, if ð¥â€ðŠâ€ð§ â£ð¥ â ð§â£ + â£ð§ â ðŠâ£ â¥ â£ð¥ â ð§â£ = ð§ â ð¥ â¥ ðŠ â ð¥ = â£ð¥ â ðŠâ£ If ð¥ â€ ð§ â€ ðŠ â£ð¥ â ð§â£ + â£ð§ â ðŠâ£ = ð§ â ð¥ + ðŠ â ð§ = ðŠ â ð¥ = â£ð¥ â ðŠâ£ and so on. 1.77 We show that ðâ ð¥, ðŠ = maxðð=1 â£ð¥ð â ðŠð â£ satisï¬es the requirements of a metric, namely 1. maxðð=1 â£ð¥ð â ðŠð â£ â¥ 0 2. maxðð=1 â£ð¥ð â ðŠð â£ = 0 if and only if ð¥ð = ðŠð for all ð. 3. maxðð=1 â£ð¥ð â ðŠð â£ = maxðð=1 â£ðŠð â ð¥ð â£ 4. For every ð, â£ð¥ð â ðŠð â£ â€ â£ð¥ð â ð§ð â£ + â£ð§ð â ðŠð â£ from previous exercise. Therefore max â£ð¥ð â ðŠð â£ â€ max (â£ð¥ð â ð§ð â£ + â£ð§ð â ðŠð â£) â€ max â£ð¥ð â ð§ð â£ + max â£ð§ð â ðŠð â£ 1.78 For any ð, any neighborhood of 1/ð contains points of ð (namely 1/ð) and points not in ð (1/ð + ð). Hence every point in ð is a boundary point. Also, 0 is a boundary point. Therefore b(ð) = ð âª {0}. Note that ð â b(ð). Therefore, ð has no interior points. 14

Solutions for Foundations of Mathematical Economics 1.79

1. Let ð¥ â int ð. Thus ð is a neighborhood of ð¥. Therefore, ð â ð is a neighborhood of ð¥, so that ð¥ is an interior point of ð .

2. Clearly, if ð¥ â ð, then ð¥ â ð â ð . Therefore, assume ð¥ â ð â ð which implies that ð¥ is a boundary point of ð. Every neighborhood of ð¥ contains other points of ð â ð . Hence ð¥ â ð . 1.80 Assume that ð is open. Every ð¥ â ð has a neighborhood which is disjoint from ð ð . Hence no ð¥ â ð is a closure point of ð ð . ð ð contains all its closure points and is therefore closed. Conversely, assume that ð is closed. Let ð¥ be a point its complement ð ð . Since ð is closed and ð¥ â / ð, ð¥ is not a boundary point of ð. This implies that ð¥ has a neighborhood ð which is disjoint from ð, that is ð â ð ð . Hence, ð¥ is an interior point of ð ð . This implies that ð ð contains only interior points, and hence is open. 1.81 Clearly ð¥ is a neighborhood of every point ð¥ â ð, since ðµð (ð¥) â ð for every ð > 0. Hence, every point ð¥ â ð is an interior point of ð¥. Similarly, every point ð¥ â â is an interior point (there are none). Since ð¥ and â are open, there complements â and ð¥ are closed. Alternatively, â has no boundary points, and is therefore is open. Trivialy, on the other hand, â contains all its boundary points, and is therefore closed. 1.82 Let ð be a metric space. Assume ð is the union of two disjoint closed sets ðŽ and ðµ, that is ð =ðŽâªðµ

Then ðŽ = ðµ ð is open as is ðµ = ðŽð . Therefore ð is not connected. Conversely, assume that ð is not connected. Then there exist disjoint open sets ðŽ and ðµ such that ð = ðŽ âª ðµ. But ðŽ = ðµ ð is also closed as is ðµ = ðŽð . Therefore ð is the union of two disjoint closed sets. 1.83 Assume ð is both open and closed, â â ð â ð. We show that we can represent ð as the union of two disjoint open sets, ð and ð ð . For any ð â ð, ð = ð âª ð ð and ð â© ð ð = â. ð is open by assumption. It complement ð ð is open since ð is closed. Therefore, ð is not connected. Conversely, assume that ð is not connected. That is, there exists two disjoint open sets ð and ð such that ð = ð âª ð . Now ð = ð ð , which implies that ð is closed since ð is open. Therefore ð is both open and closed. 1.84 Assume that ð is both open and closed. Then so is ð ð and ð is the disjoint union of two closed sets ð¥ = ð âª ðð so that b(ð) = ð â© ð ð = ð â© ð ð = â Conversely, assume that b(ð) = ð â© ð ð = â. This implies that Consider any ð¥ â ð. Since ð â© ð ð = â, ð¥ â / ð ð . A fortiori, x â / ð ð which implies that ð¥ â ð and therefore ð â ð. ð is closed. Similarly we can show that ð ð â ð ð so that ð ð is closed and therefore ð is open. ð is both open and closed.

15

Solutions for Foundations of Mathematical Economics 1.85

1. Let {ðºð } be a (possibly inï¬nite) collection of open sets. Let ðº = âªð ðºð . Let ð¥ be a point in ðº. Then there exists some particular ðºð which contains ð¥. Since ðºð is open, ðºð is a neighborhood of ð¥. Since ðºð â ðº, ð¥ is an interior point of ðº. Since ð¥ is an arbitrary point in ðº, we have shown that every ð¥ â ðº is an interior point. Hence, ðº is open. What happens if every ðºð is empty? In this case, ðº = â and is open (Exercise 1.81). The other possibility is that the collection {ðºð } is empty. Again ðº = â which is open. Suppose { ðº1 , ðº2 , . . . , ðºð } is a ï¬nite collection of open sets. Let ðº = â©ð ðºð . If ðº = â, then it is trivially open. Otherwise, let ð¥ be a point in ðº. Then ð¥ â ðºð for all ð = 1, 2, . . . , ð. Since the sets ðºð are open, for every ð, there exists an open ball ðµ(ð¥, ðð ) â ðºð about ð¥. Let ð be the smallest radius of these open balls, that is ð = min{ ð1 , ð2 , . . . , ðð }. Then ðµð (ð¥) â ðµ(ð¥, ðð ), so that ðµð (ð¥) â ðºð for all i. Hence ðµð (ð¥) â ðº. ð¥ is an interior point of ðº and ðº is open. To complete the proof, we need to deal with the trivial case in which the collection is empty. In that case, ðº = â©ð ðºð = ð and hence is open.

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

1. Let ð¥0 be an interior point of ð. This implies there exists an open ball ðµ â ð about ð¥0 . Every ð¥ â ðµ is an interior point of ð. Hence ðµ â int ð. ð¥0 is an interior point of int ð which is therefore open. Let ðº be any open subset of ð and ð¥ be a point in ðº. ðº is neighborhood of ð¥, which implies that ð â ðº is also neighborhood of ð¥. Therefore ð¥ is an interior point of ð. Therefore int ð contains every open subset ðº â ð, and hence is the largest open set in ð.

2. Let ð denote the closure of the set ð. Clearly, ð â ð. To show the converse, let ð¥ be a closure point of ð and let ð be a neighborhood of ð¥. Then ð contains some other point ð¥â² â= ð which is a closure point of ð. ð is a neighborhood of ð¥â² which intersects ð. Hence ð¥ is a closure point of ð. Consequently ð = ð which implies that ð is closed. Assume ð¹ is a closed subset of containing ð. Then ðâð¹ =ð¹ since ð¹ is closed. Hence, ð is a subset of every closed set containing ð. 1.87 Every ð¥ â ð is either an interior point or a boundary point. Consequently, the interior of ð is the set of all ð¥ â ð which are not boundary points int ð = ð â b(ð) 1.88 Assume that ð is closed, that is ð = ð âª b(ð) = ð This implies that b(ð) â ð. ð contains its boundary. Assume that ð contains its boundary, that is ð â b(ð). Then ð = ð âª b(ð) = ð ð is closed. 16

Solutions for Foundations of Mathematical Economics

1.89 Assume ð is bounded, and let ð = ð(ð). Choose any ð¥ â ð. For all ðŠ â ð, ð(ð¥, ðŠ) â€ ð < ð + 1. Therefore, ðŠ â ðµ(ð¥, ð + 1). ð is contained in the open ball ðµ(ð¥, ð + 1). Conversely, assume ð is contained in the open ball ðµð (ð¥). Then for any ðŠ, ð§ â ð ð(ðŠ, ð§) â€ ð(ðŠ, ð¥) + ð(ð¥, ð§) < 2ð by the triangle inequality. Therefore ð(ð) < 2ð and the set is bounded. 1.90 Let ðŠ â ð â© ðµð (ð¥0 ). For every ð¥ â ð, ð(ð¥, ðŠ) < ð and therefore ð(ð¥, ð¥0 ) â€ ð(ð¥, ðŠ) + ð(ðŠ, ð¥0 ) < ð + ð = 2ð so that ð¥ â ðµ2ð (ð¥0 ). 1.91 Let y0 â ð . For any ð > 0, let yâ² = y â ð be the production plan which is ð units less in every commodity. Then, for any y â ðµð (yâ² ) ðŠð â ðŠðâ² â€ ðâ (y, yâ² ) < ð

for every ð

and therefore y < y0 . Thus ðµð (yâ² ) â ð and so yâ² â int ð â= â. 1.92 For any ð¥ â ð1 ðð¥ = ð(ð¥, ð2 ) > 0 Similarly, for every ðŠ â ð2 ððŠ = ð(ðŠ, ð1 ) > 0 Let ð1 =

âª

ðµðð¥ /2 (ð¥)

ð¥âð1

ð2 =

âª

ðµððŠ /2 (ð¥)

ðŠâð2

Then ð1 and ð2 are open sets containing ð1 and ð2 respectively. To show that ð1 and ð2 are disjoint, suppose to the contrary that ð§ â ð1 â© ð2 . Then, there exist points ð¥ â ð1 and ðŠ â ð2 such that ð(ð¥, ð§) < ðð¥ /2,

ð(ðŠ, ð§) < ððŠ /2

Without loss of generality, suppose that ðð¥ â€ ððŠ and therefore ð(ð¥, ðŠ) â€ ð(ð¥, ð§) + ð(ðŠ, ð§) < ðð¥ /2 + ððŠ /2 â€ ððŠ which contradicts the deï¬nition of ððŠ and shows that ð1 â© ð2 = â. 1.93 By Exercise 1.92, there exist disjoint open sets ð1 and ð2 such that ð1 â ð1 and ð2 â ð2 . Since ð2 â ð2 , ð2 â© ð2ð = â. ð2ð is a closed set which contains ð1 , and therefore ð2 â© ð1 = â. ð = ð1 is the desired set. 1.94 See Figure 1.2.

17

Solutions for Foundations of Mathematical Economics

ð

1

ðµ1/2 ((2, 0)) 1

2

Solutions for Foundations of Mathematical Economics ðµ(ð¥, ð)

ð

ðµ(ðŠ, ð)

ð

ð

ð¥

ðŠ

Figure 1.3: A convergent sequence cannot have two distinct limits 1.98 The share ð ð of the ðth guest is ð ð =

1ð 2

lim ð ð = 0 However, ð ð > 0 for all ð. 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 ð¥ð â ð¥. That is, there exists some ð such that ð(ð¥ð , ð¥) < ð/2 for all ð â¥ ð . Then, for all ð, ð â¥ ð ð(ð¥ð , ð¥ð ) â€ ð(ð¥ð , ð¥) + ð(ð¥, ð¥ð ) < ð/2 + ð/2 = ð 1.100 Let (ð¥ð ) be a Cauchy sequence. There exists some ð such that ð(ð¥ð â ð¥ð ) < 1 for all ð â¥ ð . Let ð = max{ ð(ð¥1 â ð¥ð ), ð(ð¥2 â ð¥ð ), . . . , ð(ð¥ð â1 â ð¥ð ), 1 } Every ð¥ð belongs to ðµ(ð¥ð , ð + 1), the ball of radius ð + 1 centered on ð¥ð . 1.101 Let (ð¥ð ) be a bounded increasing sequence in â and let ð = { ð¥ð } be the set of elements of (ð¥ð ). Let ð be the least upper bound of ð. We show that ð¥ð â ð. First observe that ð¥ð â€ ð for every ð (since ð is an upper bound). Since ð is the least upper bound, for every ð > 0 there exists some element ð¥ð such that ð¥ð > ð â ð. Since (ð¥ð ) is increasing, we must have ð â ð < ð¥ð â€ ð for every ð â¥ ð That is, for every ð > 0 there exists an ð such that ð(ð¥ð , ð¥) < ð for every ð â¥ ð ð¥ð â ð. 1.102 If ðœ > 1, the sequence ðœ, ðœ 2 , ðœ 3 , . . . is unbounded. Otherwise, if ðœ â€ 1, ðœ ð â€ ðœ ðâ1 and the sequence is decreasing and bounded by ðœ â€ 1. Therefore the sequence converges (Exercise 1.101). Let ð¥ = limðââ . Then ðœ ð+1 = ðœðœ ð and therefore ð¥ = lim ðœ ð+1 = ðœ lim ðœ ð = ðœð¥ ðââ

ðââ

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

Solutions for Foundations of Mathematical Economics â ðœ = 1, in which case ð¥ = lim 1ð = 1 â ð¥ = 0 when 0 â€ ðœ < 1 Therefore ðœ ð â 0 ââ ðœ < 1 1.103

1. For every ð¥ â â (ð¥ â

Expanding

â 2 2) â¥ 0

â ð¥2 â 2 2ð¥ + 2 â¥ 0

â ð¥2 + 2 â¥ 2 2ð¥

Dividing by ð¥ â 2 â¥2 2 ð¥

ð¥+ for every ð¥ > 0. Therefore 1 2

(

2 ð¥+ ð¥

) â¥

â 2

2. Let (ð¥ð ) be the sequence deï¬ned in Example 1.64. That is ) ( 1 2 ð¥ð = ð¥ðâ1 + ðâ1 2 ð¥ Starting from ð¥0 = 2, it is clear that ð¥ð â¥ 0 for all ð. Substituting in ( ) â 1 2 ð¥+ â¥ 2 2 ð¥ 1 ð¥ = 2 ð

That is ð¥ð â¥

( ð¥ðâ1 +

2 ð¥ðâ1

) â¥

â 2

â 2 for every ð. Therefore for every ð ) ( 1 2 ð¥ð â ð¥ð+1 = ð¥ð â ð¥ð + ð 2 ð¥ ) ( 1 2 = ð¥ð â ð 2 ð¥ ( ) 1 2 â¥ ð¥ð â â 2 2 â ð =ð¥ â 2 â¥0

â This implies that ð¥ð+1 â€ ð¥ð . Consequently 2 â€ ð¥ð â€ 2 for every ð. (ð¥ð ) is a bounded monotone sequence. By Exercise 1.101, ð¥ð â ð¥. The limit ð¥ satisï¬es the equation ( ) 1 2 ð¥= ð¥+ 2 ð¥ â Solving, this implies ð¥2 = 2 or ð¥ = 2 as required. 20

Solutions for Foundations of Mathematical Economics

1.104 The following sequence approximates the square root of any positive number ð ð¥1 = ð 1( ð ) ð¥ð+1 = ð¥ð + ð 2 ð¥ 1.105 Let ð¥ â ð. If ð¥ â ð, then ð¥ is the limit of the sequence (ð¥, ð¥, ð¥, . . . ). If ð¥ â / ð, then ð¥ is a boundary point of ð. For every ð, the ball ðµ(ð¥, 1/ð) contains a point ð¥ð â ð. From the sequence of open balls ðµ(ð¥, 1/ð) for ð = 1, 2, 3, . . . , we can generate of a sequence of points ð¥ð which converges to ð¥. Conversely, assume that ð¥ is the limit of a sequence (ð¥ð ) of points in ð. Either ð¥ â ð and therefore ð¥ â ð. Or ð¥ â / ð. Since ð¥ð â ð¥, every neighborhood of ð¥ contains points ð ð¥ of the sequence. Hence, ð¥ is a boundary point of ð and ð¥ â ð. 1.106 ð is closed if and only if ð = ð. The result follows from Exercise 1.105. 1.107 Let ð be a closed subset of a complete metric space ð. Let (ð¥ð ) be a Cauchy sequence in ð. Since ð is complete, ð¥ð â ð¥ â ð. Since ð is closed, ð¥ â ð (Exercise 1.106). 1.108 Since ð(ð ð ) â 0, ð cannot contain more than one point. Therefore, it suï¬ces to show that ð is nonempty. Choose some ð¥ð from each ð ð . Since ð(ð ð ) â 0, (ð¥ð ) is a Cauchy sequence. Since ð is complete, there exists some ð¥ â ð such that ð¥ð â ð¥. Choose some ð. Since the sets are nested, the subsequence { ð¥ð : ð â¥ ð } â ð ð . Since ð ð is closed, ð¥ â ð ð (Exercise 1.106). Since ð¥ â ð ð for every ð ð¥â

ðð

ð=1

1.109 If player 1 picks closed balls whose radius decreases by at least half after each pair of moves, then { ð 1 , ð 3 , ð 5 , . . . } is a nested sequence of closed sets which has a nonempty intersection (Exercise 1.108). 1.110 Let (ð¥ð ) be a sequence in ð â ð with ð closed and ð compact. Since ð is compact, there exists a convergent subsequence ð¥ð â ð¥ â ð . Since ð is closed, we must have ð¥ â ð (Exercise 1.106). Therefore (ð¥ð ) contains a subsequence which converges in ð, so that ð is compact. 1.111 Let (ð¥ð ) be a Cauchy sequence in a metric space. For every ð > 0, there exists ð such that ð(ð¥ð , ð¥ð ) < ð/2 for all ð, ð â¥ ð Trivially, if (ð¥ð ) converges, it has a convergent subsequence (the whole sequence). Conversely, assume that (ð¥ð ) has a subsequence (ð¥ð ) which converges to ð¥. That is, there exists some ð such that ð(ð¥ð , ð¥) < ð/2 for all ð â¥ ð Therefore, by the triangle inequality ð(ð¥ð , ð¥) â€ ð(ð¥ð , ð¥ð ) + ð(ð¥ð , ð¥) < ð/2 + ð/2 = ð for all ð â¥ max ð, ð

21

Solutions for Foundations of Mathematical Economics

1.112 We proceed sequentially as follows. Choose any ð¥1 in ð. If the open ball ðµ(ð¥1 , ð) contains ð, we are done. Otherwise, choose some ð¥2 â / ðµ(ð¥1 , ð) and consider the set âª2 ðµ(ð¥ð , ð). If this set contains ð, we are done. Otherwise, choose some ð¥3 â / âªð=1 âª3 2 ðµ(ð¥ , ð) and consider ðµ(ð¥ , ð) ð ð ð=1 ð=1 The process must terminate with a ï¬nite number of open balls. Otherwise, if the process could be continued indeï¬nitely, we could construct an inï¬nite sequence (ð¥1 , ð¥2 , ð¥3 , . . . ) which had no convergent subsequence. The would contradict the compactness of ð. 1.113 Assume ð is compact. The previous exercise showed that ð is totally bounded. Further, since every sequence has a convergent subsequence, every Cauchy sequence converges (Exercise 1.111). Therefore ð is complete. Conversely, assume that ð is complete and totally bounded and let ð1 = { ð¥11 , ð¥21 , ð¥31 , . . . } be an inï¬nite sequence of points in ð. Since ð is totally bounded, it is covered by a ï¬nite collection of open balls of radius 1/2. ð1 has a subsequence ð2 = { ð¥12 , ð¥22 , ð¥32 , . . . } all of whose points lie in one of the open balls. Similarly, ð2 has a subsequence ð3 = { ð¥13 , ð¥23 , ð¥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 âdiagonalâ subsequence { ð¥11 , ð¥22 , ð¥33 , . . . } get closer and closer together. That is, ð is a Cauchy sequence. Since ð is complete, ð converges in ð and ð1 has a convergent subsequence ð. Hence, ð is compact. 1.114

1. Every big set ð â â¬ has a least two distinct points. 0 for every ð â â¬.

Hence ð(ð ) >

2. Otherwise, there exists ð such that ð(ð ) â¥ 1/ð for every ð â â¬ and therefore ð¿ = inf ð ââ¬ ð(ð ) â¥ 1/ð > 0. 3. Choose a point ð¥ð in each ðð . Since ð is compact, the sequence (ð¥ð ) has a convergent subsequence (ð¥ð ) which converges to some point ð¥0 â ð. 4. The point ð¥0 belongs to at least one ð0 in the open cover ð. Since ð0 is open, there exists some open ball ðµð (ð¥0 ) â ð0 . 5. Consider the concentric ball ðµð/2 (ð¥0 ). Since (ð¥ð ) is a convergent subsequence, there exists some ð such that ð¥ð â ðµð/2 (ð¥) for every ð â¥ ð . 6. Choose some ð0 â¥ min{ ð, 2/ð }. Then 1/ð0 < ð/2 and ð(ðð0 ) < 1/ð0 < ð/2. ð¥ð0 â ðð0 â© ðµð/2 (ð¥) and therefore (Exercise 1.90) ðð0 â ðµð (ð¥) â ð 0 . This contradicts the assumption that ðð is a big set. Therefore, we conclude that ð¿ > 0. 1.115

1. ð is totally bounded (Exercise 1.112). Therefore, for every ð > 0, there exists a ï¬nite number of open balls ðµð (ð¥ð ) such that ð=

ð âª

ðµð (ð¥ð )

ð=1

2. ð(ðµð (ð¥ð )) = 2ð < ð¿. By deï¬nition of the Lebesgue number, every ðµð (ð¥ð ) is contained in some ðð â ð. 3. The collection of open balls {ðµð (ð¥ð )} covers ð. Therefore, fore every ð¥ â ð, there exists ð such that ð¥ â ðµð (ð¥ð ) â ðð 22

Solutions for Foundations of Mathematical Economics Therefore, the ï¬nite collection ð1 , ð2 , . . . , ðð covers ð. 1.116 For any family of subsets ð â© âª ð = â ââ ðð = ð ðâð

ðâð

Suppose to the contrary that ð is a collection of closed sets with the ï¬nite intersection â© property, but that ðâð ð = â. Then { ð ð : ð â ð } is a open cover of ð which does not have a ï¬nite subcover. Consequently ð cannot be compact. Conversely, assume every collection of closed sets with the ï¬nite intersection property has a nonempty intersection. Let â¬ be an open cover of ð. Let ð = { ð â ð : ðð â â¬ } That is

âª

ð ð = ð which implies

ðâð

ð=â

ðâð

Consequently, ð does not have the ï¬nite intersection property. There exists a ï¬nite subcollection { ð1 , ð2 , . . . , ðð } such that ð â©

ðð = â

ð=1

which implies that ð âª

ððð = ð

ð=1

{ ð1ð , ð2ð , . . . , ððð } is a ï¬nite subcover of ð. Thus, ð is compact. 1.117 Every ï¬nite collection of nested (nonempty) sets has the ï¬nite intersection property. By Exercise 1.116, the sequence has a non-empty intersection. (Note: every set ðð is a subset of the compact set ð1 .) 1.118 (1) =â (2) Exercises 1.114 and 1.115. (2) =â (3) Exercise 1.116 (3) =â (1) Let ð be a metric space in which every collection of closed subsets with the ï¬nite intersection property has a ï¬nite intersection. Let (ð¥ð ) be a sequence in ð. For any ð, let ðð be the tail of the sequence minus the ï¬rst ð terms, that is ðð = { ð¥ð : ð = ð + 1, ð + 2, . . . } The collection (ðð ) has the ï¬nite intersection property since, for any ï¬nite set of integers { ð1 , ð2 , . . . , ðð } ð â©

ððð â ððŸ â= â

ð=1

where ðŸ = max{ ð1 , ð2 , . . . , ðð }. Therefore â â© ð=1

23

ðð â= â

Solutions for Foundations of Mathematical Economics

âªâ Choose any ð¥ â ð=1 ðð . That is, ð¥ â ðð for each ð = 1, 2, . . . . Thus, for every ð > 0 and ð = 1, 2, . . . , there exists some ð¥ð â ðµð (ð¥) â© ðð We construct a subsequence as follows. For ð = 1, 2, . . . , let ð¥ð be the ï¬rst term in ðð which belongs to ðµ1/ð (ð¥). Then, (ð¥ð ) is a subsequence of (ð¥ð ) which converges to ð¥. We conclude that every sequence has a convergent subsequence. 1.119 Assume (ð¥ð ) is a bounded sequence in â. Without loss of generality, we can assume that { ð¥ð } â [0, 1]. Divide ðŒ 0 = [0, 1] into two sub-intervals [0, 1/2] and [1/2, 1]. At least one of the sub-intervals must contain an inï¬nite number of terms of the sequence. Call this interval ðŒ 1 . Continuing this process of subdivision, we obtain a nested sequence of intervals ðŒ0 â ðŒ1 â ðŒ2 â . . . each of which contains an inï¬nite number of terms of the sequence. Consequently, we can construct a subsequence (ð¥ð ) with ð¥ð â ðŒ ð . Furthermore, the intervals get smaller and smaller with ð(ðŒ ð ) â 0, so that (ð¥ð ) is a Cauchy sequence. Since â is complete, the subsequence (ð¥ð ) converges to ð¥ â â. 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 (ð¥ð ) be a Cauchy sequence in â. That is, for every ð > 0, there exists ð such that â£ð¥ð â ð¥ð â£ < ð for all ð, ð â¥ ð . (ð¥ð ) is bounded (Exercise 1.100) and hence by the Bolzano-Weierstrass theorem, it has a convergent subsequence (ð¥ð ) with ð¥ð â ð¥ â â. Choose ð¥ð from the convergent subsequence such that ð â¥ ð and â£ð¥ð â ð¥â£ < ð/2. By the triangle inequality â£ð¥ð â ð¥â£ â€ â£ð¥ð â ð¥ð â£ + â£ð¥ð â ð¥â£ < ð/2 + ð/2 = ð Hence the sequence (ð¥ð ) converges to ð¥ â â. 1.121 Since ð1 and ð2 are linear spaces, x1 + y1 â ð1 and x2 + y2 â ð2 , so that (x1 + y1 , x2 + y2 ) â ð1 Ã ð2 . Similarly (ðŒx1 , ðŒx2 ) â ð1 Ã ð2 for every (x1 , x2 ) â ð1 Ã ð2 . Hence, ð = ð1 Ã ð2 is closed under addition and scalar multiplication. With addition and scalar multiplication deï¬ned component-wise, ð inherits the arithmetic properties (like associativity) of its constituent spaces. Verifying this would proceed identically as for âð . It is straightforward though tedious. The zero element in ð is 0 = (01 , 02 ) where 01 is the zero element in ð1 and 02 is the zero element in ð2 . Similarly, the inverse of x = (x1 , x2 ) is âx = (âx1 , âx2 ). 1.122

1. x+y =x+z âx + (x + y) = âx + (x + z) (âx + x) + y = (âx + x) + z 0+y =0+z y=z

24

Solutions for Foundations of Mathematical Economics

2. ðŒx = ðŒy 1 1 (ðŒx) = (ðŒy) ðŒ ( ) (ðŒ ) 1 1 ðŒ x= ðŒ y ðŒ ðŒ x=y 3. ðŒx = ðœx implies (ðŒ â ðœ)x = ðŒx â ðœx = 0 Provided x = 0, we must have (ðŒ â ðœ)x = 0x That is ðŒ â ðœ = 0 which implies ðŒ = ðœ. 4. (ðŒ â ðœ)x = (ðŒ + (âðœ))x = ðŒx + (âðœ)x = ðŒx â ðœx 5. ðŒ(x â y) = ðŒ(x + (â1)y) = ðŒx + ðŒ(â1)y = ðŒx â ðŒy 6. ðŒ0 = ðŒ(x + (âx)) = ðŒx + ðŒ(âx) = ðŒx â ðŒx =0 1.123 The linear hull of the vectors {(1, 0), (0, 2)} is { ( ) ( )} 1 0 lin {(1, 0), (0, 2)} = ðŒ1 + ðŒ2 0 2 ) ( { ðŒ1 } = ðŒ2 = â2 The linear hull of the vectors {(1, 0), (0, 2)} is the whole plane â2 . Figure 1.4 illustrates how any vector in â2 can be obtained as a linear combination of {(1, 0), (0, 2)}. 1.124

1. From the deï¬nition of ðŒ, ðŒð = ð€(ð) â

â ð âð

25

ðŒð

Solutions for Foundations of Mathematical Economics

(â2, 3)

3 2 1

-2

-1

0

1

Figure 1.4: Illustrating the span of { (1, 0), (0, 2) }. for every ð â ð . Rearranging ð€(ð) = ðŒð + =

â

â

ðŒð +

ð =ð

=

â

ðŒð

ð âð

â

ðŒð

ð âð

ðŒð

ð âð

2.

â

ðŒð ð€ð (ð) =

ð âð

â

ðŒð ð€ð (ð) +

ð âð

=

â

ðŒð 1 +

ð âð

=

â

â

â

ðŒð ð€ð (ð)

ð ââð

ðŒð 0

ð ââð

ðŒð 1

ð âð

= ð€(ð) 1.125

1. Choose any x â ð. By homogeneity 0x = ð â ð.

2. For every x â ð, âx = (â1)x â ð by homogeneity. 1.126 Examples of subspaces in âð include: 1. The set containing just the null vector {0} is subspace. 2. Let x be any element in âð and let ð be the set of all scalar multiples of x ð = { ðŒx : ðŒ â â } ð is a line through the origin in âð and is a subspace. 3. Let ð be the set of all ð-tuples with zero ï¬rst coordinate, that is ð = { (ð¥1 , ð¥2 , . . . , ð¥ð ) : ð¥1 = 0, ð¥ð â â, ð â= 1 } For any x, y â ð x + y = (0, ð¥2 , ð¥3 , . . . , ð¥ð ) + (0, ðŠ2 , ðŠ3 , . . . , ðŠð ) = (0, ð¥2 + ðŠ2 , ð¥3 + ðŠ3 , . . . , ð¥ð + ðŠð ) â ð 26

Solutions for Foundations of Mathematical Economics

Similarly ðŒx = ðŒ(0, ð¥2 , ð¥3 , . . . , ð¥ð ) = (0, ðŒð¥2 , ðŒð¥3 , . . . , ðŒð¥ð ) â ð Therefore ð is a subspace of âð . Generalizing, any set of vectors with one or more coordinates identically zero is a subspace of âð . 4. We will meet some more complicated subspaces in Chapter 2. 1.127 No, âx â / âð+ if x â âð+ unless x = 0. âð+ is an example of a cone (Section 1.4.5). 1.128 lin ð is a subspace Let x, y be two elements in lin ð. x is a linear combination of elements of ð, that is x = ðŒ1 ð¥1 + ðŒ2 ð¥2 + . . . ðŒð ð¥ð Similarly y = ðœ1 ð¥1 + ðœ2 ð¥2 + . . . ðœð ð¥ð and x + y = (ðŒ1 + ðœ1 )ð¥1 + (ðŒ2 + ðœ2 )ð¥2 + â â â + (ðŒð + ðœð )ð¥ð â lin ð and ðŒx = ðŒðŒ1 ð¥1 + ðŒðŒ2 ð¥2 + â â â + ðŒðŒð ð¥ð â lin ð This shows that lin ð is closed under addition and scalar multiplication and hence is a subspace. lin ð is the smallest subspace containing ð Let ð be any subspace containing ð. Then ð contains all linear combinations of elements in ð, so that lin ð â ð . Hence lin ð is the smallest subspace containing S. 1.129 The previous exercise showed that lin ð is a subspace. Therefore, if ð = lin ð, ð is a subspace. Conversely, assume that ð is a subspace. Then ð is the smallest subspace containing ð, and therefore ð = lin ð (again by the previous exercise). 1.130 Let x, y â ð = ð1 â© ð2 . Hence x, y â ð1 and for any ðŒ, ðœ â â, ðŒx + ðœy â ð1 . Similarly ðŒx + ðœy â ð2 and therefore ðŒx + ðœy â ð. ð is a subspace. 1.131 Let ð = ð1 + ð2 . First note that 0 = 0 + 0 â ð. Suppose x, y belong to ð. Then there exist s1 , t1 â ð1 and s2 , t2 â ð2 such that x = s1 + s2 and y = t1 + t2 . For any ðŒ, ðœ â â, ðŒx + ðœy = ðŒ(s1 + s2 ) + ðœ(t1 + t2 ) = (ðŒs1 + ðœt1 ) + (ðŒs2 + ðœt2 ) â ð since ðŒs1 + ðœt1 â ð1 and ðŒs2 + ðœt2 â ð2 . 1.132 Let ð1 = { ðŒ(1, 0) : ðŒ â â } ð2 = { ðŒ(0, 1) : ðŒ â â } 27

Solutions for Foundations of Mathematical Economics

ð1 and ð2 are respectively the horizontal and vertical axes in â2 . Their union is not a subspace, since for example ( ) ( ) ( ) 1 1 0 = + â / ð1 âª ð2 1 0 1 However, any vector in â2 can be written as the sum of an element of ð1 and an element of ð2 . Therefore, their sum is the whole space â2 , that is ð 1 + ð 2 = â2 1.133 Assume that ð is linearly dependent, that is there exists x1 , . . . , xð â ð and ðŒ2 , . . . , ðŒð â ð such that x1 = ðŒ2 x2 + ðŒ3 x3 + . . . , ðŒð xð Rearranging, this implies 1x1 â ðŒ2 x2 â ðŒ3 x3 â . . . ðŒð xð = 0 Conversely, assume there exist x1 , x2 , . . . , xð â x and ðŒ1 , ðŒ2 , . . . , ðŒð â â such that ðŒ1 x1 + ðŒ2 x2 . . . + ðŒð xð = 0 Assume without loss of generality that ðŒ1 â= 0. Then x1 = â

ðŒ2 ðŒ3 ðŒð x2 â x3 â . . . â xð ðŒ1 ðŒ1 ðŒ1

which shows that x1 â lin ð â {x1 } 1.134 Assume {(1, 1, 1), (0, 1, 1), (0, 0, 1)} are linearly dependent. Then there exists ðŒ1 , ðŒ2 , ðŒ3 such that â â â â â â â â 1 0 0 0 ðŒ1 â1â  + ðŒ2 â1â  + ðŒ3 â0â  = â0â  1 1 1 0 or equivalently ðŒ1 = 0 ðŒ1 + ðŒ2 = 0 ðŒ1 + ðŒ2 + ðŒ3 = 0 which imply that ðŒ1 = ðŒ2 = ðŒ3 = 0 Therefore {(1, 1, 1), (0, 1, 1), (0, 0, 1)} are linearly independent. 1.135 Suppose on the contrary that ð is linearly dependent. That is, there exists a set of games { ð¢ð1 , ð¢ð2 , . . . , ð¢ðð } and nonzero coeï¬cients (ðŒ1 , ðŒ2 , . . . , ðŒð ) such that (Exercise 1.133) ðŒ1 ð¢ð1 + ðŒ2 ð¢ð2 + . . . + ðŒð ð¢ðð = 0 28

(1.16)

Solutions for Foundations of Mathematical Economics

Assume that the coalitions are ordered so that ð1 has the smallest number of players of any of the coalitions ð1 , ð2 , . . . , ðð . This implies that no coalition ð2 , ð3 , . . . , ðð is a subset of ð1 and ð¢ðð (ð1 ) = 0

for every ð = 2, 3, . . . , ð

(1.17)

Using (1.39), ð¢ð1 can be expressed as a linear combination of the other games, ð¢ð1 = â1/ðŒ1

ð â

ðŒð ð¢ðð

(1.18)

ð=2

Substituting (1.40) this implies that ð¢ð1 (ð1 ) = 0 whereas ð¢ð (ð ) = 1

for every ð

by deï¬nition. This contradiction establishes that the set ð is linearly independent. 1.136 If ð is a subspace, then 0 â ð and ðŒx1 = 0 with ðŒ â= 0 and x1 = 0 (Exercise 1.122). Therefore ð is linearly dependent (Exercise 1.133). 1.137 Suppose x has two representations, that is x = ðŒ1 x1 + ðŒ2 x2 + . . . + ðŒð xð x = ðœ1 x1 + ðœ2 x2 + . . . + ðœð xð Subtracting 0 = (ðŒ1 â ðœ1 )x1 + (ðŒ2 â ðœ2 )x2 + . . . + (ðŒð â ðœð )xð

(1.19)

Since {x1 , x2 , . . . , , xð } is linearly independent, (1.19) implies that ðŒð = ðœð = 0 for all ð (Exercise 1.133) 1.138 Let ð be the set of all linearly independent subsets of a linear space ð. ð is partially ordered by inclusion. Every chain ð¶ = {ððŒ } â ð has an upper bound, namely âª ð. By Zornâs lemma, ð has a maximal element ðµ. We show that ðµ is a basis ðâð¶ for ð. ðµ is linearly independent since ðµ â ð . Suppose that ðµ does not span ð so that lin ðµ â ð. Then there exists some x â ð â lin ðµ. The set ðµ âª {x} is a linearly independent and contains ðµ, which contradicts the assumption that ðµ is the maximal element of ð . Consequently, we conclude that ðµ spans ð and hence is a basis. 1.139 Exercise 1.134 established that the set ðµ = { (1, 1, 1), (0, 1, 1), (0, 0, 1)} is linearly independent. Since dim ð3 = 3, any other vectors must be linearly dependent on ðµ. That is lin ðµ = â3 . ð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.

29

Solutions for Foundations of Mathematical Economics

1.140 Let ðŽ = {a1 , a2 , . . . , að } and ðµ = {b1 , b2 , . . . , bð } be two bases for a linear space ð. Let ð1 = {ð1 } âª ðŽ = {b1 , a1 , a2 , . . . , að } ð is linearly dependent (since ð1 â lin ðŽ) and spans ð. ðŒ1 , ðŒ2 , . . . , ðŒð and ðœ1 such that

Therefore, there exists

ðœ1 b1 + ðŒ1 a1 + ðŒ2 a2 + . . . + ðŒð að = 0 At least one ðŒð â= 0. Deleting the corresponding element að , we obtain another set ð1â² of ð elements ð1â² = {b1 , a1 , a2 , . . . , aðâ1 , að+1 , . . . , að } which is also spans ð. Adding the second element from ðµ, we obtain the ð + 1 element set ð2 = {b1 , b2 , a1 , a2 , . . . , aðâ1 , að+1 , . . . , að } which again is linearly dependent and spans ð. Continuing in this way, we can replace ð vectors in ðŽ with the ð vectors from ðµ while maintaining a spanning set. This process cannot eliminate all the vectors in ðŽ, because this would imply that ðµ was linearly dependent. (Otherwise, the remaining bð would be linear combinations of preceding elements of ðµ.) We conclude that necessarily ð â€ ð. Reversing the process and replacing elements of ðµ with elements of ðŽ establishes that ð â€ ð. Together these inequalities imply that ð = ð and ðŽ and ðµ have the same number of elements. 1.141 Suppose that the coalitions are ordered in some way, so that ð«(ð ) = {ð0 , ð1 , ð2 , . . . , ð2ð â1 } with ð0 = â. There are 2ð coalitions. Each game ðº â ð¢ ð corresponds to a unique list of length 2ð of coalitional worths v = (ð£0 , ð£1 , ð£2 , . . . , ð£2ð â1 ) ð

with ð£0 = 0. That is, each game deï¬nes a vector ð£ = (0, ð£1 , . . . , ð£2ð â1 ) â â2 and ð conversely each vector ð£ â â2 (with ð£0 = 0) deï¬nes a game. Therefore, the space of ð all games ð¢ ð is formally identical to the subspace of â2 in which the ï¬rst component ð is identically zero, which in turn is equivalent to the space â2 â1 . Thus, ð¢ ð is a 2ð â 1-dimensional linear space. 1.142 For illustrative purposes, we present two proofs, depending upon whether the linear space is assumed to be ï¬nite dimensional or not. In the ï¬nite dimensional case, a constructive proof is possible, which forms the basis for practical algorithms for constructing a basis. Let ð be a linearly independent set in a linear space ð. ð is ï¬nite dimensional Let ð = dim ð. Assume ð has ð elements and denote it ðð . If lin ðð = ð, then ðð is a basis and we are done. Otherwise, there exists some xð+1 â ð â lin ðð . Adding xð+1 to ðð gives a new set of ð + 1 elements ðð+1 = ðð âª { xð+1 } 30

Solutions for Foundations of Mathematical Economics

which is also linearly independent ( since xð+1 â / lin ðð ). If lin ðð+1 = ð, then ðð+1 is a basis and we are done. Otherwise, there exists some xð+2 â ð â lin ðð+1 . Adding xð+2 to ðð+1 gives a new set of ð + 2 elements ðð+2 = ðð+1 âª { xð+2 } which is also linearly independent ( since xð+2 â / lin ðð+2 ). Repeating this process, we can construct a sequence of linearly independent sets ðð , ðð+1 , ðð+2 . . . such that lin ðð â« lin ðð+1 â« lin ðð+2 â â â â ð. Eventually, we will reach a set which spans ð and hence is a basis. ð is possibly inï¬nite dimensional For the general case, we can adapt the proof of the existence of a basis (Exercise 1.138), restricting ð to be the class of all linearly independent subsets of ð containing ð. 1.143 Otherwise (if a set of ð + 1 elements was linearly independent), it could be extended to basis at least ð + 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 ðµ = {x1 , x2 , . . . , xð } be a set of linearly independent elements in an ð-dimensional linear space ð. We have to show that lin ðµ = ð. Take any x â ð. The set ðµ âª {x} = {x1 , x2 , . . . , xð , x } must be linearly dependent (Exercise 1.143). That is there exists numbers ðŒ1 , ðŒ2 , . . . , ðŒð , ðŒ, not all zero, such that ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð + ðŒx = 0

(1.20)

Furthermore, it must be the case that ðŒ â= 0 since otherwise ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð = 0 which contradicts the linear independence of ðŽ. Solving (1.20) for x, we obtain x=

1 ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð ðŒ

Since x was an arbitrary element of ð, we conclude that ðµ spans ð and hence ðµ is a basis. 1.145 A basis spans ð. To establish the converse, assume that ðµ = {x1 , x2 , . . . , xð } is a set of ð elements which span ð. If ð is linearly dependent, then one element is linearly dependent on the other elements. Without loss of generality, assume that x1 â lin ðµ â {x1 }. Deleting x1 the set ðµ â {x1 } = {x2 , x3 , . . . , xð } also spans ð. Continuing in this fashion by eliminating dependent elements, we ï¬nish with a linearly independent set of ð < ð elements which spans ð. That is, we can ï¬nd a basis of ð < ð elements, which contradicts the assumption that the dimension of ð is ð (Exercise 1.140). Thus any set of ð vectors which spans ð must be linearly independent and hence a basis. 31

Solutions for Foundations of Mathematical Economics

1.146 We have previously shown â that the set ð is linearly independent (Exercise 1.135). â the space ð¢ ð has dimension 2ðâ1 (Exercise 1.141). There are 2ðâ1 distinct T-unanimity games ð¢ð in ð . Hence ð spans the 2ðâ1 space ð¢ ð . Alternatively, note that any game ð€ â ð¢ ð can be written as a linear combination of T-unanimity games (Exercise 1.75). 1.147 Let ðµ = {x1 , x2 , . . . , xð } be a basis for ð. Since ðµ is linearly independent, ð â€ ð (Exercise 1.143). There are two possibilities. Case 1: ð = ð. ðµ is a set of ð linearly independent elements in an ð-dimensional space ð. Hence ðµ is a basis for ð and ð = lin ðµ = ð. Case 2: ð < ð. Since ðµ is linearly independent but cannot be a basis for the ðdimensional space ð, we must have ð = lin ðµ â ð. Therefore, we conclude that if ð â ð is a proper subspace, it has a lower dimension than ð. 1.148 Let ðŒ1 , ðŒ2 , ðŒ3 be the coordinates of (1, 1, 1) for the basis {(1, 1, 1), (0, 1, 1), (0, 0, 1)}. That is â â â â â â â â 1 0 0 1 â1â  = ðŒ1 â1â  + ðŒ2 â1â  + ðŒ3 â0â  1 1 1 1 which implies that ðŒ1 = 1, ðŒ2 = ðŒ3 = 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 ð of a linear space ð is a subspace of ð if ðŒx + ðœy â ð for every x, y â ð and for every ðŒ, ðœ â â Letting ðœ = 1 â ðŒ, this implies that ðŒx + (1 â ðŒ)y â ð

for every x, y â ð and ðŒ â â

ð is an aï¬ne set. Conversely, suppose that ð is an aï¬ne set containing 0, that is ðŒx + (1 â ðŒ)y â ð

for every x, y â ð and ðŒ â â

Letting y = 0, this implies that ðŒx â ð

for every x â ð and ðŒ â â

so that ð is homogeneous. Now letting ðŒ = 12 , for every x and y in ð, 1 1 x+ y âð 2 2 and homogeneity implies

( x+y =2

1 1 x+ y 2 2

ð is also additive. Hence ð is subspace. 32

) âð

Solutions for Foundations of Mathematical Economics

1.150 For any x â ð, let ð = ð âx = {v â ð : v +x â ð } ð is an aï¬ne set For any v1 , v2 â ð , there exist corresponding s1 , s2 â ð such that v1 = s1 â x and v2 = s2 â x and therefore ðŒv1 + (1 â ðŒ)v2 = ðŒ(s1 â x) + (1 â ðŒ)(s1 â x) = ðŒs1 + (1 â ðŒ)s2 â ðŒx + (1 â ðŒ)x =sâx where s = ðŒð 1 + (1 â ðŒ)ð 2 â ð. There ð is an aï¬ne set. ð is a subspace Since x â ð, 0 = x â x â ð . Therefore ð is a subspace (Exercise 1.149). ð is unique Suppose that there are two subspaces ð 1 and ð 2 such that ð = ð 1 + x1 and ð = ð 2 + x2 . Then ð1 + x1 = ð2 + x2 ð1 = ð2 + (x2 â x1 ) = ð2 + x where x = x2 â x1 â ð. Therefore ð1 is parallel to ð2 . Since ð1 is a subspace, 0 â ð1 which implies that âx â ð2 . Since ð2 is a subspace, this implies that x â ð2 and ð2 + x â ð2 . Therefore ð1 = ð2 + x â ð2 . Similarly, ð2 â ð1 and hence ð1 = ð2 . Therefore the subspace ð is unique. 1.151 Let ð â¥ ð denote the relation ð is parallel to ð , that is ð â¥ ð ââ ð = ð + x for some x â ð The relation â¥ is reï¬exive ð â¥ ð since ð = ð + 0 transitive Assume ð = ð + x and ð = ð + y. Then ð = ð + (x + y) symmetric ð = ð + x =â ð = ð + (âx) Therefore â¥ is an equivalence relation. 1.152 See exercises 1.130 and 1.162. 1.153

1. Exercise 1.150

2. Assume x0 â ð . For every x â ð» x = x0 + v = w â ð which implies that ð» â ð . Conversely, assume ð» = ð . Then x0 = 0 â ð since ð is a subspace. 3. By deï¬nition, ð» â ð. Therefore ð = ð» â x â ð. / ð . Suppose to the contrary 4. Let x1 â lin {x1 , ð } = ð â² â ð

33

Solutions for Foundations of Mathematical Economics

Then ð» â² = x0 + ð â² is an aï¬ne set (Exercise 1.150) which strictly contains ð». This contradicts the deï¬nition of ð» as a maximal proper aï¬ne set. 5. Let x1 â / ð . By the previous part, x â lin {x1 , ð }. That is, there exists ðŒ â â such that x = ðŒx1 + v for some v â ð To see that ðŒ is unique, suppose that there exists ðœ â â such that x = ðœx1 + vâ² for some vâ² â ð Subtracting 0 = (ðŒ â ðœ)x1 + (v â vâ² ) / ð. which implies that ðŒ = ðœ since x1 â 1.154 Assume x, y â ð. That is, x, y â âð and â â ð¥ð = ðŠð = ð€(ð ) ðâð

ðâð

ð

For any ðŒ â â, ðŒx + (1 â ðŒ)y â â and â â â ðŒð¥ð + (1 â ðŒ)ðŠð = ðŒ ð¥ð + (1 â ðŒ) ðŠð ðâð

ðâð

ðâð

= ðŒð€(ð ) + (1 â ðŒ)ð€(ð ) = ð€(ð ) Hence ð is an aï¬ne subset of âð . 1.155 See Exercise 1.129. 1.156 No. A straight line through any two points in âð+ extends outside âð+ . Put diï¬erently, the aï¬ne hull of âð+ is the whole space âð . 1.157 Let ð = aï¬ ð â x1 = aï¬ {0, x2 â x1 , x3 â x1 , . . . , xð â x1 } ð is a subspace (0 â ð ) and aï¬ ð = ð + x1 and dim aï¬ ð = dim ð Note that the choice of x1 is arbitrary. ð is aï¬nely dependent if and only if there exists some xð â ð such that xð â â x1 . aï¬ (ð â {xð }). Since the choice of x1 is arbitrary, we assume that xð = xð â aï¬ (ð â {xð }) ââ xð â (ð + x1 ) â {xð } ââ xð â x1 â ð â {xð â x1 } ââ xð â x1 â lin {x2 â x1 , x3 â x1 , . . . , xðâ1 â x1 , . . . , xð+1 â x1 , . . . , xð â x1 } 34

Solutions for Foundations of Mathematical Economics

Therefore, ð is aï¬nely dependent if and only if {x2 â x1 , x3 â x1 , . . . , xð â x1 } is linearly independent. 1.158 By the previous exercise, the set ð = {x1 , x2 , . . . , xð } is aï¬nely dependent if and only if the set {x2 â x1 , x3 â x1 , . . . , xð â x1 } is linearly dependent, so that there exist numbers ðŒ2 , ðŒ3 , . . . , ðŒð , not all zero, such that ðŒ2 (x2 â x1 ) + ðŒ3 (x3 â x1 ) + â â â + ðŒð (xð â x1 ) = 0 or ðŒ2 x2 + ðŒ3 x3 + â â â + ðŒð xð â

ð â

ðŒð x1 = 0

ð=2

Let ðŒ1 = â

âð

ð=2

ðŒð . Then ðŒ1 x1 + ðŒ2 x2 + . . . + ðŒð xð = 0

and ðŒ1 + ðŒ2 + . . . + ðŒð = 0 as required. 1.159 Let ð = aï¬ ð â x1 = aï¬ { 0, x2 â x1 , x3 â x1 , . . . , xð â x1 } Then aï¬ ð = x1 + ð If ð is aï¬nely independent, every x â aï¬ ð has a unique representation as x = x1 + v,

vâð

with v = ðŒ2 (x2 â x1 ) + ðŒ3 (x3 â x1 ) + â â â + ðŒð (xð â x1 ) so that x = x1 + ðŒ2 (x2 â x1 ) + ðŒ3 (x3 â x1 ) + â â â + ðŒð (xð â x1 ) âð Deï¬ne ðŒ1 = 1 â ð=2 ðŒð . Then x = ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð with ðŒ1 + ðŒ2 + â â â + ðŒð = 1 x is a unique aï¬ne combination of the elements of ð. 1.160 Assume that ð¥, ðŠ â (ð, ð) â â. This means that ð < ð¥ < ð and ð < ðŠ < ð. For every 0 â€ ðŒ â€ 1 ðŒð¥ + (1 â ðŒ)ðŠ > ðŒð + (1 â ðŒ)ð 35

Solutions for Foundations of Mathematical Economics

and ðŒð¥ + (1 â ðŒ)ðŠ < ðŒð + (1 â ðŒ)ð Therefore ð < ðŒð¥+(1âðŒ)ðŠ < ð and ðŒð¥+(1âðŒ)ðŠ â (ð, ð). (ð, ð) is convex. Substituting â€ for < demonstrates that [ð, ð] is convex. Let ð be an arbitrary convex set in â. Assume that ð is not an interval. This implies that there exist numbers ð¥, ðŠ, ð§ such that ð¥ < ðŠ < ð§ and ð¥, ð§ â ð while ðŠ â / ð. Deï¬ne ðŒ=

ð§âðŠ ð§âð¥

so that 1âðŒ=

ðŠâð¥ ð§âð¥

Note that 0 â€ ðŒ â€ 1 and that ðŒð¥ + (1 â ðŒ)ð§ =

ðŠâð¥ ð§âðŠ ð¥+ ð§=ðŠâ /ð ð§âð¥ ð§âð¥

which contradicts the assumption that ð is convex. We conclude that every convex set in â is an interval. Note that ð may be a hybrid interval such (ð, ð] or [ð, ð) as well as an open (ð, ð) or closed [ð, ð] interval. 1.161 Let (ð, ð€) be a TP-coalitional game. If core(ð, ð€) = â then it is trivially convex. Otherwise, assume core(ð, ð€) is nonempty and let x1 and x2 belong to core(ð, ð€). That is â ð¥1ð â¥ ð€(ð) for every ð â ð ðâð

â

ð¥1ð = ð€(ð )

ðâð

and therefore for any 0 â€ ðŒ â€ 1 â ðŒð¥1ð â¥ ðŒð€(ð)

for every ð â ð

ðâð

â

ðŒð¥1ð = ðŒð€(ð )

ðâð

Similarly â

(1 â ðŒ)ð¥2ð â¥ (1 â ðŒ)ð€(ð)

for every ð â ð

ðâð

â

(1 â ðŒ)ð¥2ð = (1 â ðŒ)ð€(ð )

ðâð

Summing these two systems â ðŒð¥1ð + (1 â ðŒ)ð¥2ð â¥ ðŒð€(ð) + (1 â ðŒ)ð€(ð) = ð€(ð) ðâð

â

ðŒð¥1ð + (1 â ðŒ)ð¥2ð = ðŒð€(ð ) + (1 â ðŒ)ð€(ð ) = ð€(ð )

ðâð

That is, ðŒð¥1ð + (1 â ðŒ)ð¥2ð belongs to core(ð, ð€). 36

for every ð â ð

Solutions for Foundations of Mathematical Economics

â© 1.162 Let â­ be a collection of convex sets and let x, y belong to ðââ­ ð. for every ð â â­, x, y â ð and therefore â© ðŒx + (1 â ðŒ)y â ð for all 0 â€ ðŒ â€ 1 (since ð is convex). Therefore ðŒx + (1 â ðŒ)y â ðââ­ ð. 1.163 Fix some output ðŠ. Assume that x1 , x2 â ð (ðŠ). This implies that both (ðŠ, âx1 ) and (ðŠ, âx2 ) belong to the production possibility set ð . If ð is convex ðŒ(ðŠ, âx1 ) + (1 â ðŒ)(ðŠ, âx2 ) = (ðŒðŠ + (1 â ðŒ)ðŠ, ðŒx1 + (1 â ðŒ)x2 ) = (ðŠ, ðŒx1 + (1 â ðŒ)x2 ) â ð for every ðŒ â [0, 1]. This implies that ðŒx1 + (1 â ðŒ)x2 â ð (ðŠ). Since the choice of ðŠ was arbitrary, this implies that ð (ðŠ) is convex for every ðŠ. 1.164 Assume ð1 and ð2 are convex sets. Let ð = ð1 + ð2 . Suppose x, y belong to ð. Then there exist s1 , t1 â ð1 and s2 , t2 â ð2 such that x = s1 + s2 and y = t1 + t2 . For any ðŒ â [0, 1] ðŒx + (1 â ðŒ)y = ðŒs1 + s2 + (1 â ðŒ)t1 + t2 = ðŒs1 + (1 â ðŒ)t1 + ðŒs2 + (1 â ðŒ)t2 â ð since ðŒs1 + (1 â ðŒ)t1 â ð1 and ðŒs2 + (1 â ðŒ)t2 â ð2 . The argument readily extends to any ï¬nite number of sets. 1.165 Without loss of generality, assume that ð = 2. Let ð = ð1 Ã ð2 â ð = ð1 Ã ð2 . Suppose x = (ð¥1 , ð¥2 ) and y = (ðŠ1 , ðŠ2 ) belong to ð. Then ðŒx + (1 â ðŒ)y = ðŒ(ð¥1 , ð¥2 ) + (1 â ðŒ)(ðŠ1 , ðŠ2 ) = (ðŒð¥1 , ðŒð¥2 ) + ((1 â ðŒ)ðŠ1 , (1 â ðŒ)ðŠ2 ) = (ðŒð¥1 + (1 â ðŒ)ðŠ1 , ðŒð¥2 + (1 â ðŒ)ðŠ2 ) â ð 1.166 Let ðŒx, ðŒy be points in ðŒð so that x, y â ð. Since ð is convex, ðœx+ (1 â ðœ)y â ð for every 0 â€ ðœ â€ 1. Multiplying by ðŒ ðŒ(ðœx + (1 â ðœ)y) = ðœ(ðŒx) + (1 â ðœ)(ðŒy) â ðŒð Therefore, ðŒð is convex. 1.167 Combine Exercises 1.164 and 1.166. 1.168 The inclusion ð â ðŒð + (1 â ðŒ)ð is true for any set (whether convex or not), since for every x â ð x = ðŒx + (1 â ðŒ)x â ðŒð + (1 â ðŒ)ð The reverse inclusion ðŒð +(1âðŒ)ð â ð follows directly from the deï¬nition of convexity. 1.169 Given any two convex sets ð and ð in a linear space, the largest convex set contained in both is ð â© ð ; the smallest convex set containing both is conv ð âª ð . Therefore, the set of all convex sets is a lattice with ð â§ð =ð â©ð ð âš ð = conv ð âª ð The lattice is complete since every collection {ðð } has a least upper bound conv âª ðð and a greatest lower bound â©ðð . 37

Solutions for Foundations of Mathematical Economics

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 ð is convex. Let x be a convex combination of elements in ð, that is let x = ðŒ1 x1 + ðŒ2 x2 + . . . + ðŒð xð where x1 , x2 , . . . , xð â ð and ðŒ1 , ðŒ2 , . . . , ðŒð â â+ with ðŒ1 + ðŒ2 + . . . + ðŒð = 1. We need to show that x â ð. We proceed by induction of the number of points ð. Clearly, x â ð if ð = 1 or ð = 2. To show that it is true for ð = 3, let x = ðŒ1 x1 + ðŒ2 x2 + ðŒ3 x3 where x1 , x2 , x3 â ð and ðŒ1 , ðŒ2 , ðŒ3 â â+ with ðŒ1 + ðŒ2 + ðŒ3 = 1. Assume that ðŒð > 0 for all ð (otherwise ð = 1 or ð = 2) so that ðŒ1 < 1. Rewriting x = ðŒ1 x1 + ðŒ2 x2 + ðŒ3 x3 ( ) ðŒ2 ðŒ2 = ðŒ1 x1 + (1 â ðŒ1 ) x2 + x3 1 â ðŒ1 1 â ðŒ1 = ðŒ1 x1 + (1 â ðŒ1 )y where ( y=

ðŒ2 ðŒ2 x2 + x3 1 â ðŒ1 1 â ðŒ1

)

y is a convex combination of two elements x2 and x3 since ðŒ2 ðŒ2 ðŒ2 + ðŒ3 + = =1 1 â ðŒ1 1 â ðŒ1 1 â ðŒ1 and ðŒ2 + ðŒ3 = 1 â ðŒ1 . Hence y â ð. Therefore x is a convex combination of two elements x1 and ðŠ and is also in ð. Proceeding in this fashion, we can show that every convex combination belongs to ð, that is conv ð â ð. 1.171 This is precisely analogous to Exercise 1.128. We observe that 1. conv ð is a convex set. 2. if ð¶ is any convex set containing ð, then conv ð â ð¶. Therefore, conv ð is the smallest convex set containing S. 1.172 Note ï¬rst that ð â conv ð for any set ð. The converse for convex sets follows from Exercise 1.170. 1.173 Assume x â conv (ð1 + ð2 ). Then, there exist numbers ðŒ1 , ðŒ2 , . . . , ðŒð and vectors x1 , x2 , . . . , xð in ð1 + ð2 such that x = ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð For every xð , there exists x1ð â ð1 and x2ð â ð2 such that xð = x1ð + x2ð

38

Solutions for Foundations of Mathematical Economics

and therefore ð â

x=

ðŒð x1ð +

ð=1

ð â

ðŒð x2ð

ð=1

= x1 + x2 âð âð where x1 = ð=1 ðŒð x1ð â ð1 and x2 = ð=1 ðŒð x2ð â ð2 . Therefore x â conv ð1 + conv ð2 . Conversely, assume that x â conv ð1 + conv ð2 . Then x = x1 + x2 , where x1 =

ð â

ðŒð ð¥1ð ,

x1ð â ð1

ðœð ð¥2ð ,

x2ð â ð2

ð=1

x2 =

ð â ð=1

and x = x1 + x2 =

ð â

ðŒð ð¥1ð +

ð=1

ð â

ðœð ð¥2ð â conv (ð1 + ð2 )

ð=1

since x1ð , x2ð â ð1 + ð2 for every ð and ð. 1.174 The dimension of the input requirement set ð (ðŠ) is ð. Its aï¬ne hull is âð . 1.175

1. Let x = ðŒ1 x1 + ðŒ2 x2 + . . . + ðŒð xð

(1.21)

If ð > dim ð +1, the elements x1 , x2 , . . . , xð â ð are aï¬nely dependent (Exercise 1.157 and therefore there exist numbers ðœ1 , ðœ2 , . . . , ðœð , not all zero, such that (Exercise 1.158) ðœ1 x1 + ðœ2 x2 + . . . + ðœð xð = 0

(1.22)

and ðœ1 + ðœ2 + . . . + ðœð = 0 2. Combining (1.21) and (1.22) x = x â ð¡0 ð ð â â ðŒð xð â ð¡ ðœð xð = ð=1

=

ð â

ð=1

(ðŒð â ð¡ðœð )xð

ð=1

for any ð¡ â â. } { 3. Let ð¡ = minð ðŒðœðð : ðœð > 0 =

ðŒð ðœð

We note that â ð¡ > 0 since ðŒð > 0 for every ð. 39

(1.23)

Solutions for Foundations of Mathematical Economics

â If ðœð > 0, then ðŒð /ðœð â¥ ðŒð /ðœð â¥ ð¡ and therefore ðŒð â ð¡ðœð â¥ 0 â If ðœð â€ 0 then ðŒð â ð¡ðœð > 0 for every ð¡ > 0. â Therefore ðŒð â ð¡ðœð â¥ 0 for every ð¡ and â ðŒð â ð¡ðœð = 0 for ð = ð. Therefore, (1.23) represents x as a convex combination of only ð â 1 points. 4. This process can be repeated until x is represented as a convex combination of at most dim ð + 1 elements. 1.176 Assume x is not an extreme point of ð. Then there exists distinct x1 and x2 in S such that x = ðŒx1 + (1 â ðŒ)x2 Without loss of generality, assume ðŒ â€ 1/2 and let y = x2 â x. Then x + y = x2 â ð. Furthermore x â y = x â x2 + x = 2x â x2 = 2(ðŒx1 + (1 â ðŒ)x2 ) â x2 = 2ðŒx1 + (1 â 2ðŒ)x2 â ð since ðŒ â€ 1/2. 1.177

1. For any x = (ð¥1 , ð¥2 ) â ð¶2 , there exists some ðŒ1 â [0, 1] such that ð¥1 = ðŒ1 ð + (1 â ðŒ1 )(âð) = (2ðŒ1 â 1)ð In fact, ðŒ1 is deï¬ned by ðŒ1 = Therefore (see Figure 1.5) ) ( ( ð¥1 = ðŒ1 ð ( ) ( ð¥1 = ðŒ1 âð

ð¥1 + ð 2ð

) âð + (1 â ðŒ1 ) ð ( ) ) âð ð + (1 â ðŒ1 ) âð âð ð ð

)

(

Similarly ð¥2 = ðŒ2 ð + (1 â ðŒ2 )(âð) where ðŒ2 =

ð¥2 + ð 2ð

Therefore, for any x â ð¶2 , ( ( ) ) ) ( ð¥1 ð¥1 ð¥1 x= + (1 â ðŒ2 ) = ðŒ2 ð¥2 ð âð ( ) ( ) ð âð = ðŒ1 ðŒ2 + (1 â ðŒ1 )ðŒ2 ð ð ( ) ( ) ð âð + ðŒ1 (1 â ðŒ2 ) + (1 â ðŒ1 )(1 â ðŒ2 ) âð âð ( ) ( ) ( ) ( ) ð âð ð âð = ðœ1 + ðœ2 + ðœ3 + ðœ4 ð ð âð âð 40

Solutions for Foundations of Mathematical Economics

(ð¥1 , c) x ð¥1

(ð¥1 , -c) Figure 1.5: A cube in â2 where 0 â€ ðœð â€ 1 and ðœ1 + ðœ2 + ðœ3 + ðœ4 = ðŒ1 ðŒ2 + (1 â ðŒ1 )ðŒ2 + ðŒ1 (1 â ðŒ2 ) + (1 â ðŒ1 )(1 â ðŒ2 ) = ðŒ1 ðŒ2 + ðŒ2 â ðŒ1 ðŒ2 + ðŒ1 â ðŒ1 ðŒ2 + 1 â ðŒ1 â ðŒ2 + ðŒ1 ðŒ2 =1 That is

{( ð¥ â conv

ð ð

) ( ) ( ) ( )} âð ð âð , , , ð âð âð

2. (a) For any point (ð¥1 , ð¥2 , . . . , ð¥ðâ1 , ð) which lies on face of the cube ð¶ð , (ð¥1 , ð¥2 , . . . , ð¥ðâ1 ) â ð¶ðâ1 and therefore (ð¥1 , ð¥2 , . . . , ð¥ðâ1 ) â conv { Â±ð, Â±ð, . . . , Â±ð) } â âðâ1 so that x â conv { (Â±ð, Â±ð, . . . , Â±ð, ð) } â âð Similarly, any point (ð¥1 , ð¥2 , . . . , ð¥ðâ1 , âð) on the opposite face lies in the convex hull of the points { (Â±ð, Â±ð, . . . , Â±ð, âð) }. (b) For any other point x = (ð¥1 , ð¥2 , . . . , ð¥ð ) â ð¶ð , let ðŒð =

ð¥ð + ð 2ð

so that ð¥ð = ðŒð ð + (1 â ðŒð )(âð) Then

â

â â â â â ð¥1 ð¥1 ð¥1 â ð¥2 â â ð¥2 â â ð¥2 â â â â â â â â â â â â x = â . . . â = ðŒð â . . . â + (1 â ðŒð ) â â ... â âð¥ðâ1 â  âð¥ðâ1 â  âð¥ðâ1 â  ð¥ð ð âð 41

Solutions for Foundations of Mathematical Economics Hence

x â conv { (Â±ð, Â±ð, . . . , Â±ð) } â âð In other words ð¶ð â conv { (Â±ð, Â±ð, . . . , Â±ð) } â âð 3. Let ðž denote the set of points of the form { (Â±ð, Â±ð, . . . , Â±ð) } â âð . Clearly, every point in ðž is an extreme point of ð¶ð . Conversely, we have shown that ð¶ð â conv ðž. Therefore, no point x â ð¶ ð â ðž can be an extreme point of ð¶ ð . ðž is the set of extreme points of ð¶ ð . 4. Since ð¶ ð is convex, and ðž â ð¶ð , conv ðž â ð¶ ð . Consequently, ð¶ ð = conv ðž. 1.178 Let x, y belong to ð â ð¹ is convex. For any ðŒ â [0, 1] â ðŒx + (1 â ðŒ)y â ð since ð convex â ðŒx + (1 â ðŒ)y â / ð¹ since ð¹ is a face Thus ðŒx + (1 â ðŒ)y â ð â ð¹ which is convex. 1.179

1. Trivial.

âª 2. Let {ð¹ð } be a collection of faces of ð and let ð¹ = ð¹ð . Choose any x, y â ð. If the line segment between x and y intersects ð¹ , then âª it intersects some face ð¹ð which implies that x, y â ð¹ð . Therefore, x, y â ð¹ = ð¹ð . â© 3. Let {ð¹ð } be a collection of faces of ð and let ð¹ = ð¹ð . Choose any x, y â ð. if the line segment between x and y intersects ð¹ , then it intersects âª every face ð¹ð which implies that x, y â ð¹ð for every ð. Therefore, x, y â ð¹ = ð¹ð . 4. Let ð be the collection of all faces of ð. This is partially ordered by inclusion. By â© the previous result, every nonempty subcollection ð has a least upper bound ( ð¹ âð ð¹ ). Hence ð is a complete lattice (Exercise 1.47).

1.180 Let ð be a polytope. Then ð = conv { x1 , x2 , . . . , xð }. Note that every extreme point belongs to { x1 , x2 , . . . , xð }. Now choose the smallest subset whose convex hull is still ð, that is delete elements which can be written as convex combinations of other elements. Suppose the minimal subset is { x1 , x2 , . . . , xð }. We claim that each of these elements is an extreme point of ð, that is { x1 , x2 , . . . , xð } = ðž. Assume not, that is assume that xð is not an extreme point so that there exists x, y â ð with xð = ðŒx + (1 â ðŒ)y

with 0 < ðŒ < 1

(1.24)

Since x, y â conv {x1 , x2 , . . . , xð } x=

ð â

ðŒð xð

y=

ð=1

ð â

ðœxð

ð=1

Substituting in (1.24), we can write xð as a convex combination of {x1 , x2 , . . . , xð }. xð =

ð ð â â ( ) ðŒðŒð + (1 â ðŒ)ðœð xð = ðŸð xð ð=1

ð=1

where ðŸð = ðŒðŒð + (1 â ðŒ)ðœð 42

Solutions for Foundations of Mathematical Economics

Note that 0 â€ ðŸð â€ 1, so that either ðŸð < 1 or ðŸð = 1. We show that both cases lead to a contradiction. â ðŸð < 1. Then ðâ1 â( ) 1 ðŒðŒð + (1 â ðŒ)ðœð xð 1 â ðŸð ð=1

xð =

which contradicts the minimality of the set {x1 , x2 , . . . , xð }. â ðŸð = 1. Then ðŸð = 0 for every ð â= ð. That is ðŒðŒð + (1 â ðŒ)ðœð = 0 which implies that ðŒð = ðœð

for every ð â= ð

for every ð â= ð and therefore x = y.

Therefore, if {x1 , x2 , . . . , xð } 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 â ð and 0 < ðŒ < 1 such that x1 = ðŒy + (1 â ðŒ)z

(1.25)

Now, since y â ð, there exist ðœ1 , ðœ2 , . . . , ðœð such that y=

ð â

ðœð xð ,

ð=1

ð â

ðœð = 1

ð=1

Similarly, there exist ð¿1 , ð¿2 , . . . , ð¿ð such that z=

ð â

ð â

ð¿ð xð ,

ð=1

ð¿ð = 1

ð=1

Substituting in (1.25) x1 = ðŒ =

ð â

ðœð xð + (1 â ðŒ)

ð=1 ð â

ð â

ð¿ð xð

ð=1

( ) ðŒðœð + (1 â ðŒ)ð¿ð xð

ð=1

Since

âð

ð=1

( ) â â ðŒðœð + (1 â ðŒ)ð¿ð = ðŒ ðð=1 ðœð + (1 â ðŒ) ð=1 ð¿ð = 1 x1 =

ð â ( ) ðŒðœð + (1 â ðŒ)ð¿ð xð ð=1

Subtracting, this implies 0=

ð â ( ) ðŒðœð + (1 â ðŒ)ð¿ð (xð â x1 ) ð=2

This establishes that the set {x2 â x1 , x3 â x1 , . . . , xð â x1 } is linearly dependent and therefore ðž = {x1 , x2 , . . . , xð } is aï¬nely dependent (Exercise 1.157). This contradicts the assumption that ð is a simplex. 43

Solutions for Foundations of Mathematical Economics

1.182 Let ð be the dimension of a convex set ð in a linear space ð. Then ð = dim aï¬ ð and there exists a set { x1 , x2 , . . . , xð+1 } of aï¬nely independent points in ð. Deï¬ne ð â² = conv { x1 , x2 , . . . , xð+1 } Then ð â² is an ð-dimensional simplex contained in ð. 1.183 Let w = (ð€({1}), ð€({2}), . . . , ð€({ð})) denote the vector of individual worths and let ð  denote the surplus to be distributed, that is â ð  = ð€(ð ) â ð€({ð}) ðâð

ð  > 0 if the game is essential. For each player ð = 1, 2, . . . , ð, let yð = w + ð eð be the outcome in which player ð receives the entire surplus. (eð is the ðth unit vector.) Note that { ð€({ð}) + ð  ð = ð ð ðŠð = ð€({ð}) ð â= ð Each yð is an imputation since ðŠðð â¥ ð€({ð}) and â â ðŠðð = ð€({ð}) + ð  = ð€(ð ) ðâð

ðâð

Therefore {y1 , y2 , . . . , yð } â ðŒ. Since ðŒ is convex (why ?), ð = conv {y1 , y2 , . . . , yð } â ðŒ. Further, for every ð, ð â ð the vectors yð â yð = ð (eð â eð ) are linearly independent. Therefore ð is an ð â 1-dimensional simplex in âð . For any x â ðŒ deï¬ne ðŒð =

ð¥ð â ð€({ð}) ð

so that ð¥ð = ð€({ð}) + ðŒð ð  Since x is an imputation â ðŒð â¥ 0 (â ) â â â ðâð ðŒð = ðâð ð¥ð â ðâð ð€({ð}) /ð  = 1 â We claim that x = ðâð ðŒð yð since for each ð = 1, 2, . . . , ð â â ðŒð ðŠðð = ðŒð ð€({ð}) + ðŒð ð  ðâð

ðâð

= ð€({ð}) + ðŒð ð  = ð¥ð Therefore x â conv {y1 , y2 , . . . , yð } = ð, that is ðŒ â ð. Since we previously showed that ð â ðŒ, we have established that ðŒ = ð, which is an ð â 1 dimensional simplex in âð . 44

Solutions for Foundations of Mathematical Economics

ð¥2

ð¥2

ð¥2

ð¥1

ð¥1

ð¥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 = (ð¥1 , ð¥2 , . . . , ð¥ð ) belong to âð+ , which means that ð¥ð â¥ 0 for every ð. For every ðŒ > 0 ðŒx = (ðŒx1 , ðŒx2 , . . . , ðŒxð ) and ðŒð¥ð â¥ 0 for every ð. Therefore ðŒx â âð+ . âð+ is a cone in âð . 1.186 Assume ðŒx + ðœy â ð for every x, y â ð and ðŒ, ðœ â â+

(1.26)

Letting ðœ = 0, this implies that ðŒx â ð for every x â ð and ðŒ â â+ so that ð is a cone. To show that ð is convex, let x and y be any two elements in ð. For any ðŒ â [0, 1], (1.26) implies that ðŒx + (1 â ðŒ)y â ð Therefore ð is convex. Conversely, assume that ð is a convex cone. For any ðŒ, ðœ â â+ and x, y â ð ðœ ðŒ x+ yâð ðŒ+ðœ ðŒ+ðœ and therefore ðŒx + ðœy â ð 1.187 Assume ð satisï¬es 1. ðŒð â ð for every ðŒ â¥ 0 2. ð + ð â ð

45

Solutions for Foundations of Mathematical Economics

By (1), ð is a cone. To show that it is convex, let x and y belong to ð. By (1), ðŒx and (1 â ðŒ)y belong to ð, and therefore ðŒx + (1 â ðŒ)y belongs to ð by (2). ð is convex. Conversely, assume that ð is a convex cone. Then ðŒð â ð

for every ðŒ â¥ 0

Let x and y be any two elements in ð. Since ð is convex, ð§ = ðŒ 12 x + (1 â ðŒ) 12 y â ð and since it is a cone, 2ð§ = x + y â ð. Therefore ð +ð âð 1.188 We have to show that ð is convex cone. By assumption, ð is convex. To show that ð is a cone, let y be any production plan in ð . By convexity ðŒy = ðŒy + (1 â ðŒ)0 â ð for every 0 â€ ðŒ â€ 1 Repeated use of additivity ensures that ðŒy â ð for every ðŒ = 1, 2, . . . Combining these two conclusions implies that ðŒy â ð for every ðŒ â¥ 0 1.189 Let ð® â ð¢ ð denote the set of all superadditive games. Let ð€1 , ð€2 â ð be two superadditive games. Then, for all distinct coalitions ð, ð â ð with ð â© ð = â ð€1 (ð âª ð ) â¥ ð€1 (ð) + ð€1 (ð ) ð€2 (ð âª ð ) â¥ ð€2 (ð) + ð€2 (ð ) Adding (ð€1 + ð€2 )(ð âª ð ) = ð€1 (ð âª ð ) + ð€2 (ð âª ð ) â¥ ð€1 (ð) + ð€2 (ð) + ð€1 (ð ) + ð€2 (ð ) = (ð€1 + ð€2 )(ð) + (ð€1 + ð€2 )(ð ) so that ð€1 + ð€2 is superadditive. Similarly, we can show that ðŒð€1 is superadditive for all ðŒ â â+ . Hence ð® is a convex cone in ð¢ ð . â©ð 1.190 Let x belong to ð=1 ðð . Then x â ð â©ð for every ð. Since each ðð is a cone, ðŒx â ðð for every ðŒ â¥ 0 and therefore ðŒx â ðð=1 ðð . Let ð = ð1 + ð2 + â â â + ðð and assume x belongs to ð. Then there exist xð â ðð , ð = 1, 2, . . . , ð such that x = x1 + x2 + â â â + xð For any ðŒ â¥ 0 ðŒx = ðŒ(x1 + x2 + â â â + xð ) = ðŒx1 + ðŒx2 + â â â + ðŒxð â ð since ðŒxð â ðð for every ð.

46

Solutions for Foundations of Mathematical Economics 1.191

1. Suppose that y â ð . Then, there exist ðŒ, ðŒ2 , . . . , ðŒ8 â¥ 0 such that y=

8 â

ðŒð yð

ð=1

and for the ï¬rst commodity 8 â

y1 =

ðŒð ðŠð1

ð=1

If y â= 0, at least one of the ðŒð > 0 and hence y1 < 0 since ðŠð1 < 0 for ð = 1, 2, . . . , 8. 2. Free disposal requires that y â ð, yâ² â€ y =â yâ² â ð . Consider the production plan yâ² = (â2, â2, â2, â2). Note that yâ² â€ y3 and yâ² â€ y6 . Suppose that yâ² â ð . Then there exist ðŒ1 , ðŒ2 , . . . , ðŒ8 â¥ 0 such that y=

8 â

ðŒð yð

ð=1

For the third commodity (component), we have 4ðŒ1 + 3ðŒ2 + 3ðŒ3 + 3ðŒ4 + 12ðŒ5 â 2ðŒ6 + 5ðŒ8 = â2

(1.27)

and for the fourth commodity 2ðŒ2 â 1ðŒ3 + 1ðŒ4 + 5ðŒ6 + 10ðŒ7 â 2ðŒ8 = â2

(1.28)

Adding to (1.28) to (1.27) gives 4ðŒ1 + 5ðŒ2 + 2ðŒ3 + 4ðŒ4 + 12ðŒ5 + 3ðŒ6 + 10ðŒ7 + 3ðŒ8 = â4 / ð. which is impossible given that ðŒð â¥ 0. Therefore, we conclude that yâ² â 3. y2 = (â7, â9, 3, 2) â¥ (â8, â13, 3, 1) = y4 3y1 = (â9, â18, 12, 0) â¥ (â11, â19, 12, 0) = y5 y7 = (â8, â5, 0, 10) â¥ (â8, â6, â4, 10) = 2y6 2y3 = (â2, â4, 6, â2) â¥ (â2, â4, 5, â2) = y8 4. 2y3 + y7 = (â2, â4, 6, â2) + (â8, â5, 0, 10) = (â10, â9, 6, 8) â¥ (â14, â18, 6, 4) = 2y2 20y3 + 2y7 = 20(â1, â2, 3, â1) + 2(â8, â5, 0, 10) = (â20, â40, 60, â20) + (â16, â10, 0, 20) = (â36, â50, 60, 0) â¥ (â45, â90, 60, 0) = 15y1 47

Solutions for Foundations of Mathematical Economics

5. Eï¬(ð ) = cone { y3 , y7 }. 1.192 This is precisely analogous to Exercise 1.128. We observe that 1. cone ð is a convex cone. 2. if ð¶ is any convex cone containing ð, then conv ð â ð¶. Therefore, cone ð is the smallest convex cone containing S. 1.193 For any set ð, ð â cone ð. If ð is a convex cone, Exercise 1.186 implies that cone ð â ð. 1.194

1. If ð > ð = dim cone ð = dim lin ð, the elements x1 , x2 , . . . , xð â ð are linearly dependent and therefore there exist numbers ðœ1 , ðœ2 , . . . , ðœð , not all zero, such that (Exercise 1.134) ðœ1 x1 + ðœ2 x2 + . . . + ðœð xð = 0

(1.29)

2. Combining (1.14) and (1.29) x = x â ð¡0 ð ð â â = ðŒð xð â ð¡ ðœð xð ð=1

=

ð â

ð=1

(ðŒð â ð¡ðœð )xð

(1.30)

ð=1

for any ð¡ â â. { } 3. Let ð¡ = minð ðŒðœðð : ðœð > 0 =

ðŒð ðœð

We note that â ð¡ > 0 since ðŒð > 0 for every ð. â If ðœð > 0, then ðŒð /ðœð â¥ ðŒð /ðœð â¥ ð¡ and therefore ðŒð â ð¡ðœð â¥ 0. â If ðœð â€ 0 then ðŒð â ð¡ðœð > 0 for every ð¡ > 0. â Therefore ðŒð â ð¡ðœð â¥ 0 for every ð¡ and â ðŒð â ð¡ðœð = 0 for ð = ð. Therefore, (1.30) represents x as a nonnegative combination of only ð â 1 points. 4. This process can be repeated until x is represented as a convex combination of at most ð points. 1.195

1. The aï¬ne hull of ðË is parallel to the aï¬ne hull of ð. Therefore ( Since

0 0

)

dim ð = dim aï¬ ð = dim aï¬ ðË Ë â / aï¬ ð, dim cone ðË = dim aï¬ ðË + 1 = dim ð + 1

48

Solutions for Foundations of Mathematical Economics

( ) x 2. For every x â conv ð, â conv ðË and there exist (Exercise 1.194) ð + 1 1 ( ) xð points â ðË such that 1 (

x 1

)

( â conv {

x1 1

) ( ) ( ) x2 xð+1 , ,... } 1 1

This implies that x â conv { x1 , x2 , . . . , xð+1 } with x1 , x2 , . . . , xð+1 â ð. 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, . . . , ð. Since it has ð + 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. ð(x, y) = â¥x â yâ¥ â¥ 0.

2. ð(x, y) = â¥x â yâ¥ = 0 if and only if x â y = 0, that is x = y. 3. Property (3) ensures that â¥âxâ¥ = â¥xâ¥ and therefore â¥x â yâ¥ = â¥y â xâ¥ so that ð(x, y) = â¥x â yâ¥ = â¥y â xâ¥ = ð(y, x) 4. For any z â ð ð(x, y) = â¥x â yâ¥ = â¥x â z + z â yâ¥ â€ â¥x â zâ¥ + â¥z â yâ¥ = ð(x, z) + ð(z, y) Therefore ð(x, y) = â¥x â yâ¥ satisï¬es the properties required of a metric. 1.198 Clearly â¥xâ¥â â¥ 0 and â¥xâ¥â = 0 if and only if x = 0. Thirdly ð

ð

ð=1

ð=1

â¥ðŒxâ¥ = max â£ðŒð¥ð â£ = â£ðŒâ£ max â£ð¥ð â£ = â£ðŒâ£ â¥xâ¥ To prove the triangle inequality, we note that for any ð¥ð , ðŠð â â max(ð¥ð + ðŠð ) â€ max ð¥ð + max ðŠð Therefore ð

ð

ð

ð=1

ð=1

ð=1

â¥xâ¥ = max(ð¥ð + ðŠð ) â€ max ð¥ð + max ðŠð = â¥xâ¥ + â¥yâ¥ 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, â2, â3) and the average net output, â2, is negative. A norm is required to be nonnegative. Moreover, theâquantities of inputs ð and outputs may balance out yielding a zero average. That is, ( ð=1 ðŠð )/ð = 0 does not imply that ðŠð = 0 for all ð.

49

Solutions for Foundations of Mathematical Economics 1.200 â¥xâ¥ â â¥yâ¥ = â¥x â y + yâ¥ â â¥yâ¥ â€ â¥x â yâ¥ + â¥yâ¥ â â¥yâ¥ = â¥x â yâ¥ 1.201 Using the previous exercise â¥xð â¥ â â¥xâ¥ â€ â¥xð â xâ¥ â 0

1.202 First note that each term xð + yð â ð by linearity. Similarly, x + y â ð. Fix some ð > 0. There exists some ðx such that â¥xð â xâ¥ < ð for all ð â¥ ðx . Similarly, there exists some ðy such that â¥yð â yâ¥ < ð for all ð â¥ ðy . For all ð â¥ max{ ðx , ðy }, â¥(xð + yð ) â (x + y)â¥ = â¥(xð â x) + (yð â y)â¥ â€ â¥xð â xâ¥ + â¥yð â yâ¥ 0, there exists some ð such that â¥zð â zð â¥ = max{ â¥xð â xð â¥ , â¥yð â yð â¥ } < ð for every ð, ð â¥ ð . This implies that (xð ) and (yð ) are Cauchy sequences in ð and ð respectively. Since ð and ð are complete, both sequences converge. That is, there exists x â ð and y â ð such that â¥xð â xâ¥ â 0 and â¥yð â yâ¥ â 0. In other words, given ð > 0 there exists ð such that â¥xð â xâ¥ < ð and â¥yð â yâ¥ < ð for every ð â¥ ð . Let z = (x, y). Then, for every ð â¥ ð â¥zð â zâ¥ = max{ â¥xð â xâ¥ , â¥yð â yâ¥ } < ð zð â z. 1.210

1. By assumption, for every ð = 1, 2, . . . , there exists a point yð such that ( ð ) 1 â â¥yâ¥ < â£ðŒð â£ ð ð=1 where y = ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð Let ð ð =

âð

ð=1

â£ðŒð â£. By assumption ð ð > ð â¥yð â¥ â¥ 0. Deï¬ne xð =

1 ð y ð ð

Then xð = ðœ1ð x1 + ðœ2ð x2 + â â â + ðœðð xð âð 1 ð ð ð where ðœðð = ðŒð ð /ð  , ð=1 â£ðœð â£ = 1 and â¥x â¥ < ð for every ð = 1, 2, . . . . ð Consequently â¥x â¥ â 0. 51

Solutions for Foundations of Mathematical Economics

âð 2. Since ð=1 â£ðœðð â£ = 1, â£ðœðð â£ â€ 1 for every ð. Consequently, for every coordinate ð, the sequence (ðœðð ) is bounded. By the Bolzano-Weierstrass theorem (Exercise 1.119), the sequence (ðœ1ð ) has a convergent subsequence with ðœ1ð â ðœ1 . Let xð,1 denote the corresponding subsequence of xð . Similarly, ðœ2ð,1 has a subsequence converging to ðœ2 . Let (xð,2 ) denote the corresponding subsequence of (xð ). Proceeding coordinate by coordinate, we obtain a subsequence (xð,ð ) where each term is xð,ð = ðœ ð,ð x1 + ðœ ð,ð x2 + â â â + ðœ ð,ð xð and each coeï¬cient converges ðœðð,ð â ðœð . Let x = ðœ1 x1 + ðœ2 x2 + â â â + ðœ2 xð Then xð,ð â x (Exercise 1.202). âð âð ð 3. Since ð=1 â£ðœð â£ = 1 for every ð, ð=1 â£ðœð â£ = 1. Consequently, at least one of the coeï¬cients ðœð â= 0. Since x1 , x2 , . . . , xð are linearly independent, x â= 0 (Exercise 1.133) and therefore â¥xâ¥ â= 0. But (xð,ð ) is a subsequence of (xð ). This contradicts the earlier conclusion (part 1) that â¥xð â¥ â 0. 1.211

1. Let ð be a normed linear space ð of dimension ð and let { x1 , x2 , . . . , xð } be a basis for ð. Let (xð ) be a Cauchy sequence in ð. Each term xð has a unique representation ð ð xð = ðŒð 1 x1 + ðŒ2 x2 + â â â + ðŒð xð

We show that each of the sequences ðŒð ð is a Cauchy sequence in â. Since xð is a Cauchy sequence, for every ð > 0 there exists an ð such that â¥xð â xð â¥ < ð for all ð, ð â¥ ð . Using Lemma 1.1, there exists ð > 0 such that   ð ð  â â   ð ð ð ð ð â£ðŒð â ðŒð â£ â€  (ðŒð â ðŒð )xð  = â¥xð â xð â¥ < ð   ð=1

ð=1

for all ð, ð â¥ ð . Dividing by ð > 0 we get for every ð ð â£ðŒð ð â ðŒð â£ â€

ð â

ð â£ðŒð ð â ðŒð â£ <

ð=1

ð ð

ðŒð ð

is a Cauchy sequence in â. Since â is complete, each Thus each sequence sequence converges to some limit ðŒð â â. 2. Let x = ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð Then x â ð and

  ð ð  â â   â¥xð â xâ¥ =  (ðŒð â£ðŒð ð â ðŒð )xð  â€ ð â ðŒð â£ â¥xð â¥   ð=1

ð=1

ð ð Since ðŒð ð â ðŒð for every ð, â¥x â xâ¥ â 0 which implies that x â x.

3. Since (xð ) was an arbitrary Cauchy sequence, we have shown that every Cauchy sequence in ð converges. Hence ð is complete. 52

Solutions for Foundations of Mathematical Economics

1.212 Let ð be an open set according to the â¥ââ¥ð and let x0 be a point in ð. Since ð is open, it contains an open ball in the â¥ââ¥ð topology about x0 , namely ðµð (x0 , ð) = { x â ð : â¥x â x0 â¥ð < ð } â ð Let ðµð (x0 , ð) = { x â ð : â¥x â x0 â¥ð < ð } be the open ball about x0 in the â¥ââ¥ð topology. The condition (1.15) implies that ðµð (x0 , ð) â ðµð (x0 , ð) â ð and therefore x0 â ðµð (x0 , ð) â ð ð is open in the â¥ââ¥ð topology. Similarly, any ð open in the â¥ââ¥ð topology is open in the â¥ââ¥ð topology. 1.213 Let ð be a normed linear space of dimension ð. and let { x1 , x2 , . . . , xð } be a basis for ð. Let â¥ââ¥ð and â¥ââ¥ð be two norms on ð. Every x â ð has a unique representation x = ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð Repeated application of the triangle inequality gives â¥xâ¥ð = â¥ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð â¥ð ð â â€ â¥ðŒð xð â¥ð ð=1

=

ð â

â£ðŒð â£ â¥xð â¥ð

ð=1 ð â

â€ð

â£ðŒð â£

ð=1

where ð = maxð â¥xð â¥. By Lemma 1.1, there is a positive constant ð such that ð â ð=1

â£ðŒð â£ â€ â¥xâ¥ð /ð

Combining these two inequalities, we have â¥xâ¥ð â€ ð â¥xâ¥ð /ð Setting ðŽ = ð/ð > 0, we have shown ðŽ â¥xâ¥ð â€ â¥xâ¥ð The other inequality in (1.15) is obtained by interchanging the roles of â¥ââ¥ð and â¥ââ¥ð . 1.214 Assume xð â x = (ð¥1 , ð¥2 , . . . , ð¥ð ). Then, for every ð > 0, there exists some ð such that â¥xð â xâ¥â < ð. Therefore, for ð = 1, 2, . . . , ð â£ð¥ðð â ð¥ð â£ â€ max â£ð¥ðð â ð¥ð â£ = â¥xð â xâ¥â < ð ð

Therefore ð¥ðð â xð .

53

Solutions for Foundations of Mathematical Economics

Conversely, assume that (xð ) is a sequence in âð with ð¥ðð â ð¥ð for every coordinate ð. Choose some ð > 0. For every ð, there exists some integer ðð such that â£ð¥ðð â ð¥ð â£ < ð for every ð â¥ ðð Let ð = maxð { ð1 , ð2 , . . . , ðð }. Then â£ð¥ðð â ð¥ð â£ < ð for every ð â¥ ð and â¥xð â xâ¥â = max â£ð¥ðð â ð¥ð â£ < ð for every ð â¥ ð ð

ð

That is, x â x. A similar proof can be given using the Euclidean norm â¥ââ¥2 , but it is slightly more complicated. This illustrates an instance where the sup norm is more tractable. 1.215

1. Let ð â ð be closed and bounded and let xð be a sequence in ð. Every term xð has a representation ð â

xð =

ðŒð ð xð

ð=1

Since ð is bounded, so is xð . That is, there exists ð such that â¥xð â¥ â€ ð for all ð. Applying Lemma 1.1, there is a positive constant ð such that ð

ð â

â£ðŒð â£ â€ â¥xð â¥ â€ ð

ð=1

Hence, for every ð, the sequence of scalars ðŒðð is bounded. 2. By the Bolzano-Weierstrass theorem (Exercise 1.119), the sequence ðŒð 1 has a convergent subsequence with limit ðŒ1 . Let ð¥ð (1) be the corresponding subsequence of xð . ð 3. Similarly, ð¥ð (1) has a subsequence for which the corresponding scalars ðŒ2 converge to ðŒ2 . Repeating this process ð times (this is were ï¬niteness is important), we deduce the existence of a subsequence ð¥ð (ð) whose scalars converge to (ðŒ1 , ðŒ2 , . . . , ðŒð ).

4. Let x=

ð â

ðŒð xð

ð=1 ð ð Since ðŒð ð â ðŒð for every ð, â¥x â xâ¥ â 0 which implies that x â x.

5. Since ð is closed, x â ð. 6. We have shown that every sequence in ð has a subsequence which converges in ð. ð is compact. 1.216 Let x and y belong to ðµ = { x : â¥xâ¥ < 1 }, the unit ball in the normed linear space ð. Then â¥xâ¥ , â¥yâ¥ < 1. By the triangle inequality â¥ðŒx + (1 â ðŒ)yâ¥ â€ ðŒ â¥xâ¥ + (1 â ðŒ) â¥yâ¥ â€ ðŒ + (1 â ðŒ) = 1 Hence ðŒx + (1 â ðŒ)y â ðµ. 54

Solutions for Foundations of Mathematical Economics

1.217 If int ð is empty, it is trivially convex. Therefore, assume int ð â= â and let x, y â int ð. We must show that z = ðŒx + (1 â ðŒ)y â int ð. Since x, y â int ð, there exists some ð > 0 such that the open balls ðµ(x, ð) and ðµ(y, ð) are both contained in int ð. Let w be any vector with â¥wâ¥ < ð. The point z + w = ðŒ(x + w) + (1 â ðŒ)(y + w) â ð since x + w â ðµ(x, ð) â ð and y + w â ðµ(y, ð) â ð and ð is convex. Hence z is an interior point of ð. Similarly, if ð is empty, it is trivially convex. Therefore, assume ð â= â and let x, y â ð. Choose some ðŒ. We must show that ð§ = ðŒx + (1 â ðŒ)y â ð. There exist sequences (xð ) and (yð ) in ð which converge to x and y respectively (Exercise 1.105). Since ð is convex, the sequence (ðŒxð + (1 â ðŒ)yð ) lies in ð and moreover converges to ðŒx + (1 â ðŒ)y = z (Exercise 1.202). Therefore ð§ is the limit of a sequence in ð and hence ð§ â ð. Therefore, ð is convex. Â¯ = ðŒx1 + (1 â ðŒ)x2 for some ðŒ â (0, 1). Since x1 â ð, 1.218 Let x x1 â ð + ððµ ðŒx1 â ðŒ(ð + ððµ) Â¯ of radius ð is The open ball about x Â¯ + ððµ ðµ(Â¯ x, ð) = x = ðŒx1 + (1 â ðŒ)x2 + ððµ â ðŒ(ð + ððµ) + (1 â ðŒ)x2 + ððµ = ðŒð + (1 â ðŒ)x2 + (1 + ðŒ)ððµ ( ) 1+ðŒ = ðŒð + (1 â ðŒ) x2 + ððµ 1âðŒ Since x2 â int ð x2 +

( ) 1+ðŒ 1+ðŒ ððµ = ðµ x2 , ð âð 1âðŒ 1âðŒ

for suï¬ciently small ð. For such ð ðµ(Â¯ x, ð) â ðŒð + (1 â ðŒ)ð =ð Â¯ â int ð. by Exercise 1.168. Therefore x 1.219 It is easy to show that ðâ

ðð

ðâðŒ

To show the converse, choose any x â ð and let x0 â ðð for every ð â ðŒ. By Exercise 1.218, ðŒx + (1 â ðŒ)x0 â ðð for all 0 < ðŒ < 1. This implies that ðŒx + (1 â ðŒ)x0 â â©ðâðŒ ðð = ð for all 0 < ðŒ < 1, and therefore that x0 = limðŒâ0 ðŒx + (1 â ðŒ)x0 â ð. 1.220 Assume that x â int ð. Then, there exists some ð such that ðµ(x, ð) = x + ððµ â ð 55

Solutions for Foundations of Mathematical Economics Let y be any element in the unit ball ðµ. Then ây â ðµ and x1 = x + ðy â ð x2 = x â ðy â ð 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(ð) â b(ð). To show the converse, assume that x is a boundary point which is not an extreme point. That is, there exist x1 , x2 â ð such that x = ðŒx1 + (1 â ðŒ)x2

0 0 and ð¥ð < 1 1.229 Let ð = dim ð. By Exercise 1.182, ð contains a simplex ð ð of the same dimension. That is, there exist ð vertices v1 , v2 , . . . , vð such that ð ð = conv { v1 , v2 , . . . , vð } { = ðŒ1 v1 + ðŒ2 v2 + â â â + ðŒð vð : ðŒ1 , ðŒ2 , . . . , ðŒð â¥ 0, ðŒ1 + ðŒ2 + . . . + ðŒð = 1

}

Analogous to the previous part, the relative interior of ð ð is ri ð ð = conv { v1 , v2 , . . . , vð } { = ðŒ1 v1 + ðŒ2 v2 + â â â + ðŒð vð : ðŒ1 , ðŒ2 , . . . , ðŒð > 0, ðŒ1 + ðŒ2 + . . . + ðŒð = 1

}

which is nonempty. Note, the proposition is trivially true for a set containing a single point (ð = 0), since this point is the whole aï¬ne space. 1.230 If int ð â= â, then aï¬ ð = ð and ri ð = int ð. The converse follows from Exercise 1.229. 1.231 Since ð > inf

xâð

ð â ð=1

58

ðð ð¥ð

Solutions for Foundations of Mathematical Economics there exists some x â ð such that ð â

ðð ð¥ð â€ ð

ð=1

Therefore x â ð(p, ð) which is nonempty. Let ðË = minð ðð be the lowest price of the ð goods. Then ð(p, ð) â ðµ(0, ð/Ë ð) and so is bounded. (That is, no component of an aï¬ordable bundle can contain more than ð/Ë ð units.) To show that ð(p, ð) is closed, let (xð ) be a sequence of consumption bundles in ð(p, ð). Since ð(p, ð) is bounded, xð â x â ð. Furthermore ð1 ð¥ð1 + ð2 ð¥ð2 + â â â + ðð ð¥ðð â€ ð for every ð This implies that ð1 ð¥1 + ð2 ð¥2 + â â â + ðð ð¥ð â€ ð ð

so that x â x â ð(p, ð). Therefore ð(p, ð) is closed. We have shown that ð(p, ð) is a closed and bounded subset of âð . Hence it is compact (Proposition 1.4). 1.232 Let x, y â ð(p, ð). That is ð â

ðð ð¥ð â€ ð

ð=1 ð â

ðð ðŠ ð â€ ð

ð=1

For any ðŒ â [0, 1], the cost of the weighted average bundle z = ðŒx + (1 â ðŒ)y (where each component ð§ð = ðŒð¥ð + (1 â ðŒ)ðŠð ) is ð â ð=1

ðð ð§ ð =

ð â

ðð (ðŒð¥ð ð=1 ð â

=ðŒ

+ (1 â ðŒ)ðŠð

ðð ð¥ð + (1 â ðŒ)

ð=1

ð â

ðð ðŠ ð

ð=1

â€ ðŒð + (1 â ðŒ)ð =ð Therefore z â ð(p, ð). The budget set ð(p, ð) is convex. 1.233

1. Assume that â» is strongly monotone. Let x, y â ð with x â¥ y. Either x â© y so that x â» y by strong monotonicity or x = y so that x â¿ y by reï¬exivity. In either case, x â¿ y so that â¿ is weakly monotonic.

2. Again, assume that â¿ is strongly monotonic and let y â ð. ð is open (relative to itself). Therefore, there exists some ð such that ðµ(y, ð) = y + ððµ â ð Let x = y + ðe1 be the consumption bundle containing ð more units of good 1. Then e1 â ðµ, x â ðµ(y, ð) and therefore â¥x â yâ¥ < ð. Furthermore, x â© y and therefore x â» y. 59

Solutions for Foundations of Mathematical Economics

3. Assume â¿ is locally nonsatiated. Then, for every x â ð, there exists some y â ð such that y â» x. Therefore, there is no best element. 1.234 is assume that xâ â¿ x for every x â ðµ(p, ð) but that âð Assume otherwise, thatâ ð ð=1 ðð ð¥ð < ð. Let ð = ð â ð=1 ðð ð¥ð be the unspent income. Spending the residual on good 1, the commodity bundle x = xâ + ðð1 e1 is aï¬ordable ð â

ðð ð¥ð =

ð=1

ð â

ðð ð¥âð + ð1

ð=1

ð =ð ð1

Moreover, since x â© xâ , x â» xâ , which contradicts the assumption that xâ is the best element in ð(p, ð). 1.235 otherwise, that is assume that xâ â¿ x for every x â ðµ(p, ð) but that âð Assume â â ð=1 ðð ð¥ð < ð. This implies that x â int ð(p, ð). Therefore, there exists a neighâ borhood ð of x with ð â ð(p, ð). Within this neighborhood, there exists some x â ð â ð(p, ð) with x â» xâ , which contradicts the assumption that xâ is the best element in ð(p, ð). 1.236

1. Assume â¿ is continuous. Choose some y â ð. For any x0 in â»(y), x0 â» y and (since â¿ is continuous) there exists some neighborhood ð(x0 ) such that x â» y for every x â ð(x0 ). That is, ð(x0 ) â â»(y) and â»(y) is open. Similarly, for any x0 â âº(y), x0 âº y and there exists some neighborhood ð(x0 ) such that x âº y for every x â ð(x0 ). Thus ð(x0 ) â âº(y) and âº(y) is open.

2. Conversely, assume that the sets â»(y) = { x : x â» y } and âº(y) = { x : x âº y } are open in x. Assume x0 â» y0 . (a) Suppose there exists some y such that x0 â» y â» z0 . Then x0 â â»(y), which is open by assumption. That is, â»(y) is an open neighborhood of x0 and x â» y for every x â â»(y). Similarly, âº(y) is an open neighborhood of z0 for which y â» z for every z â âº(y). Therefore ð(x0 ) = â»(y) and ð(z0 ) = âº(y) are the required neighborhoods of x0 and z0 respectively such that xâ»yâ»z

for every x â ð(x0 ) and y â ð(z0 )

(b) Suppose there is no y such that x0 â» y â» z0 . i. By assumption â â»(z0 ) is open â x0 â» z0 which implies x0 â â»(z0 ), Therefore â»(z0 ) is an open neighborhood of x0 . ii. Since â¿ is complete, either y âº x0 or y â¿ x0 for every y â ð (Exercise 1.56. Since there is no y such that x0 â» y â» z0 y â» z0 =â y ââº x0 =â y â¿ x0 Therefore â»(z0 ) = â¿(x0 ). iii. Since x â¿ x0 â» z0 for every x â â¿(x0 ) = â»(z0 ) x â» z0 for every x â â»(z0 )

60

Solutions for Foundations of Mathematical Economics

iv. Therefore ð(x0 ) = â»(z0 ) is an open neighborhood of x0 such that x â» z0 for every x â ð(x0 ) Similarly, ð(z0 ) = âº(x0 ) is an open neighborhood of z0 such that z âº x0 for every z â ð(z0 ). Consequently xâ»z

for every x â ð(x0 ) and z â ð(z0 )

( )ð 3. â¿(y) = âº(y) (Exercise 1.56). Therefore, â¿(y) is closed if and only if âº(y) is open (Exercise 1.80). Similarly, âŸ(y) is closed if and only if â»(y) is open. 1.237

1. Let ð¹ = { (x, y) â ð Ãð : x â¿ y }. Let ((xð , yð )) be a sequence in ð¹ which converges to (x, y). Since (xð , yð ) â ð¹ , xð â¿ yð for every ð. By assumption, x â¿ y. Therefore, (x, y) â ð¹ which establishes that ð¹ is closed (Exercise 1.106) Conversely, assume that ð¹ is closed and let ((xð , yð )) be a sequence converging to (x, y) with xð â¿ yð for every ð. Then ((xð , yð )) â ð¹ which implies that (x, y) â ð¹ . Therefore x â¿ y.

2. Yes. Setting yð = y for every ð, their deï¬nition implies that for every sequence (xð ) in ð with xð â¿ y, x = lim xð â¿ y. That is, the upper contour set â¿(y) = { x : x â¿ y } is closed. Similarly, the lower contour set âŸ(y) is closed. Conversely, assume that the preference relation is continuous (in our deï¬nition). We show that the set ðº = { (x, y) : x âº y } is open. Let (x0 , y0 ) â ðº. Then x0 âº y0 . By continuity, there exists neighborhoods ð(x0 ) and ð(y0 ) of x0 and y0 such that x âº y for every x â ð(x0 ) and y â ð(y0 ). Hence, for every (x, y) â ð = ð(x0 ) Ã ð(y0 ), x âº y. Therefore ð â ðº which implies that ðº is open. Consequently ðºð = { (x, y) : x â¿ y } is closed. 1.238 Assume the contrary. That is, assume there is no y with x â» y â» z. Since â¿ is complete, either y âº x0 or y â¿ x0 for every y â ð (Exercise 1.56). Since there is no y such that x0 â» y â» z0 y â» z0 =â y ââº x0 =â y â¿ x0 Therefore â»(z0 ) = â¿(x0 ). By continuity, â»(z0 ) is open and â¿(x0 ) is closed. Hence â»(z0 ) = â¿(x0 ) is both open and closed (Exercise 1.83). Alternatively, â¿(x0 ) and âŸ(z0 ) are both open sets which partition ð. This contradicts the assumption that ð is connected. 1.239 Let ð â denote the set of best elements. As demonstrated in the preceding proof â© ðâ = â¿(yð ) yâð

Therefore ð â is closed (Exercise 1.85) and hence compact (Exercise 1.110). 1.240 Assume for simplicity that ð1 = ð2 = 1 and that ð = 1. Then, the budget set is ðµ(1, 1) = { x â â2++ : ð¥1 + ð¥2 â€ 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) â / ð.

61

Solutions for Foundations of Mathematical Economics

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 aï¬ordable alternative in her budget set. The theory would be empty if there was no such optimal choice. 1.242

1. Let ð 0 = ð â© âð+ . Then ð 0 is compact and ð 1 â ð 0 . Deï¬ne the order x â¿1 y if and only if ð1 (x) â€ ð1 (y). Then â¿1 is continuous on ð and ð 1 = { x â ð : ð1 (x) â€ ð1 (y) for every y â ð } is the set of best elements in ð with respect to the order â¿1 . By Exercise 1.239, ð 1 is nonempty and compact.

2. Assume ð ðâ1 is compact. Deï¬ne the order x â¿ð y if and only if ðð (x) â€ ðð (y). Then â¿ð is continuous on ð ðâ1 and ð ð = { x â ð ðâ1 : ðð (x) â€ ðð (y) for every y â ð ðâ1 } is the set of best elements in ð ðâ1 with respect to the order â¿ð . By Exercise 1.239, ð ð is nonempty and compact. 3. Assume x â Nu. Then x â¿ y for every y â ð d(x) âŸð¿ d(y) for every y â ð For every ð = 1, 2, . . . , 2ð dð (x) â€ dð (y) for every y â ð ð

ð

which implies x â ð ð . In particular x â ð 2 . Therefore Nu â ð 2 . ð

ð

ð

Suppose Nu â ð 2 . Then there exists some x, y â ð, x â / ð 2 and y â ð 2 such ð / ð ð . Then ðð (x) > ðð (y). that x â¿ y. Let ð be the smallest integer such that x â ð But x â ð for every ð < ð and therefore ðð (x) = ðð (y) for ð = 1, 2, . . . , ð â 1. This means that d(y) âºð¿ d(x) so that x âºð y. This contradiction establishes ð that Nu = ð 2 . 1.243 Assume â¿ is convex. Choose any y â ð and let x â â¿(y). Then x â¿ y. Since â¿ is convex, this implies that ðŒx + (1 â ðŒ)y â¿ y

for every 0 â€ ðŒ â€ 1

and therefore ðŒx + (1 â ðŒ)y â â¿(y)

for every 0 â€ ðŒ â€ 1

Therefore â¿(y) is convex. To show the converse, assume that â¿(y) is convex for every y â ð. Choose x, y â ð. Interchanging x and y if necessary, we can assume that x â¿ y so that x â â¿(y). Of course, y â â¿(y). Since â¿(y) is convex ðŒx + (1 â ðŒ)y â â¿(y)

for every 0 â€ ðŒ â€ 1

which implies ðŒx + (1 â ðŒ)y â¿ y

for every 0 â€ ðŒ â€ 1 62

Solutions for Foundations of Mathematical Economics

1.244 ð â may be empty, in which case it is trivially convex. Otherwise, let xâ â ð â . For every x â ð â x â¿ xâ which implies x â â¿(xâ ) Therefore ð â â â¿(xâ ). Conversely, by transitivity x â¿ xâ â¿ y for every y â ð for every x â â¿(xâ ) which implies â¿(xâ ) â ð â . Therefore, ð â = â¿(xâ ) which is convex. 1.245 To show that â¿ð is strictly convex, assume that x, y â ð are such d(x) = d(y) with x â= y. Suppose ) ( d(x) = ð(ð1 , x), ð(ð2 , x) . . . , ð(ð2ð , x) In the order ð1 , ð2 , . . . , ð2ð , let ðð be the ï¬rst coalition for which ð(ðð , x) â= ð(ðð , y). That is ð(ðð , x) = ð(ðð , y) for every ð < ð

(1.32)

Since ð(ðð , x) â= ð(ðð , y) and d(x) is listed in descending order, we must have ð(ðð , x) > ð(ðð , y)

(1.33)

ð(ðð , x) â¥ ð(ðð , y) for every ð > ð

(1.34)

and

Choose 0 < ðŒ < 1 and let z = ðŒx + (1 â ðŒ)y. For any coalition ð â ð(ð, z) = ð€(ð) â ð§ð ðâð

â( ) = ð€(ð) â ðŒð¥ð + (1 â ðŒ)ðŠð ðâð

= ð€(ð) â ðŒ (

â

ð¥ð â (1 â ðŒ)

ðâð

= ðŒ ð€(ð) â

â

ðâð

) ð¥ð

â

ðŠð (

+ (1 â ðŒ) ð€(ð) â

ðâð

â

) ðŠð

ðâð

= ðŒð(ð, x) + (1 â ðŒ)ð(ð, y) Using (1.55) to (1.57), this implies that ð(ðð , z) = ð(ðð , x),

ðð

for every 0 < ðŒ < 1, Therefore d(z) âºð¿ d(x). Thus z â»ð x, which establishes that â¿ is strictly convex. The set of feasible outcomes is convex. Assume x, y â Nu â ð, x â= y. Then d(x) = d(y) and z = ðŒx + (1 â ðŒ)y â»ð x for every 0 < ðŒ < 1 which contradicts the assumption that x â Nu. We conclude that the nucleolus contains only one element. 63

Solutions for Foundations of Mathematical Economics 1.246

1. (a) Clearly âº(x0 ) â âŸ(x0 ) and â»(y0 ) â â¿(y0 ). Consequently âº(x0 ) âª â»(y0 ) â âŸ(x0 ) âª â¿(y0 ). We claim that these sets are in fact equal. Let z â âŸ(x0 ) âª â¿(y0 ). Suppose that z â âŸ(x0 ) but z â / âº(x0 ). Then z â¿ x0 . By transitivity, z â¿ x0 â» y0 which implies that z â â»(y0 ). Similarly z â â¿(y0 ) â â»(y0 ) implies z â âº(x0 ). Therefore âº(x0 ) âª â»(y0 ) = âŸ(x0 ) âª â¿(y0 ) (b) By continuity, âº(x0 ) âª â»(y0 ) is open and âŸ(x0 ) âª â¿(y0 ) = âº(x0 ) âª â»(y0 ) is closed. Further x0 â» y0 implies that x0 â â»(y0 ) so that âº(x0 ) âª â»(y0 ) â= â. We have established that âº(x0 ) âª â»(y0 ) is a nonempty subset of ð which is both open and closed. Since ð is connected, this implies (Exercise 1.83) that âº(x0 ) âª â»(y0 ) = ð

Solutions for Foundations of Mathematical Economics

1.248 For every x â ð, there exists some z such that z â» x (Nonsatiation). For any ð, choose some ðŒ â (0, ð/ â¥x â zâ¥) and let y = ðŒz + (1 â ðŒ)x. Then â¥x â yâ¥ = ðŒ â¥x â zâ¥ < ð Moreover, since â¿ is strictly convex, y = ðŒz + (1 â ðŒ)x â» x Thus, â¿ 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 eï¬cient. That is, there exist allocation y such that y â¿ð x for all ð and some individual ð for which y â»ð x. Take 1 â ð¡ percent of ðâs consumption and distribute it equally to the other participants. By continuity, 1âð¡ there exists some ð¡ such that ð¡y â»ð x. The other agents receive yð + ðâ1 yð which, by monotonicity, they strictly prefer to xð . 1.250 Assume that (pâ , xâ ) is a competitive equilibrium of an exchange economy, but that xâ does not belong to the core of the corresponding market game. Then there exists some coalition ð and allocation y ââ ð (ð) such that yð â»ð xâð for every ð â ð. â Since y â ð (ð), we must have ðâð yð = ðâð wð . Since xâ is a competitive equilibrium and yð â»ð xâð for every ð â ð, y must be unaffordable, that is ð â

ðð ðŠðð >

ð=1

ð â

ðð wðð for every ð â ð

ð=1

and therefore ð ââ ðâð ð=1

ðð ðŠðð >

ð ââ

ðð wðð

ðâð ð=1

which contradicts the assumption that ðŠ â ð (ð). 1.251 Combining the previous exercise with Exercise 1.64 xâ â core â Pareto

65

Solutions for Foundations of Mathematical Economics

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 oneâs birthday. 2.2 The origin 0 is ï¬xed point for every ð. Furthermore, when ð = 0, ð is an identity function and every point is a ï¬xed point. 2.3 For every ð¥ â ð, there exists some ðŠ â ð such that ð (ð¥) = ðŠ, whence ð¥ â ð â1 (ðŠ). Therefore, every ð¥ belongs to some contour. To show that distinct contours are disjoint, assume ð¥ â ð â1 (ðŠ1 ) â© ð â1 (ðŠ2 ). Then ð (ð¥) = ðŠ1 and also ð (ð¥) = ðŠ2 . Since ð is a function, this implies that ðŠ1 = ðŠ2 . 2.4 Assume ð is one-to-one and onto. Then, for every ðŠ â ð , there exists ð¥ â ð such that ð (ð¥) = ðŠ. That is, ð â1 (ðŠ) â= â for every ðŠ â ð . If ð is one to one, ð (ð¥) = ðŠ = ð (ð¥â² ) implies ð¥ = ð¥â² . Therefore, ð â1 (ðŠ) consists of a single element. Therefore, the inverse function ð â1 exists. Conversely, assume that ð : ð â ð has an inverse ð â1 . As ð â1 is a function mapping ð to ð, it must be deï¬ned for every ðŠ â ð . Therefore ð is onto. Assume there exists ð¥, ð¥â² â ð and ðŠ â ð such that ð (ð¥) = ðŠ = ð (ð¥â² ). Then ð â1 (ðŠ) = ð¥ and also ð â1 (ðŠ) = ð¥â² . Since ð â1 is a function, this implies that ð¥ = ð¥â² . Therefore ð is one-to-one. 2.5 Choose any ð¥ â ð and let ðŠ = ð (ð¥). Since ð is one-to-one, ð¥ = ð â1 (ðŠ) = ð â1 (ð (ð¥)). The second identity is proved similarly. 2.6 (2.2) implies for every ð¥ â â ðð¥ ðâð¥ = ð0 = 1 and therefore ðâð¥ =

1 ðð¥

For every ð¥ â¥ 0 ðð¥ = 1 +

ð¥3 ð¥ ð¥2 + + + âââ > 0 1 2 6

and therefore by (2.28) ðð¥ > 0 for every ð¥ â â. For every ð¥ â¥ 1 ðð¥ = 1 +

ð¥3 ð¥ ð¥2 + + + â â â â¥ 1 + ð¥ â â as ð¥ â â 1 2 6

and therefore ðð¥ â â as ð¥ â â. By (2.28) ðð¥ â 0 as ð¥ â ââ. 2.7 ðð¥/2 ðð¥/2 ðð¥ = ð¥ 2 ð¥2 ( ð¥/2 ) 1 ð = ðð¥/2 â â as ð¥ â â 2 ð¥/2 66

(2.28)

Solutions for Foundations of Mathematical Economics

since the term in brackets is strictly greater than 1 for any ð¥ > 0. Similarly ðð¥ (ðð¥/(ð+1) )ð ðð¥/(ð+1) = ð¥ ð ð¥ (ð + 1)ð ( ð+1 ) ( ð¥/(ð+1) )ð 1 ð = ðð¥/(ð+1) â â (ð + 1)ð ð¥/(ð + 1) 2.8 Assume that ð â â is compact. Then ð is bounded (Proposition 1.1), and there exists ð such that â£ð¥â£ â€ ð for every ð¥ â ð. For all ð â¥ ð â¥ 2ð     ð  â ð¥ð   ð¥ð+1 ðâð â ( ð¥ )ð     â£ðð (ð¥) â ðð (ð¥)â£ =  â€   ð!   (ð + 1)! ð  ð=ð+1 ð=0    ð ð+1 ðâð â ( ð )ð   â€   (ð + 1)! ð  ð=0 ( ( )ðâð ) ð ð+1 1 1 1 â€ 1 + + + âââ + (ð + 1)! 2 4 2 ( ( )ð ) ð+1 ð 1 ð ð+1 â€2 â€2 â€ (ð + 1)! ð+1 2 by Exercise 1.206. Therefore ðð converges to ð for all ð¥ â ð. 2.9 This is a special case of Example 2.8. For any ð, ð â ð¹ (ð), deï¬ne (ð + ð) = ð (ð¥) + ð(ð¥) (ðŒð )(ð¥) = ðŒð (ð¥) With these deï¬nitions ð + ð and ðŒð also map ð to â. Hence ð¹ (ð) is closed under addition and scalar multiplication. It is straightforward but tedious to verify that ð¹ (ð) satisï¬es the other requirements of a linear space. 2.10 The zero element in ð¹ (ð) is the constant function ð (ð¥) = 0 for every ð¥ â ð. 2.11

1. From the deï¬nition of â¥ð â¥ it is clear that â â¥ð â¥ â¥ 0. â â¥ð â¥ = 0 of and only ð is the zero functional. â â¥ðŒð â¥ = â£ðŒâ£ â¥ð â¥ since supð¥âð â£ðŒð (ð¥)â£ = â£ðŒâ£ supð¥âð â£ð (ð¥)â£ It remains to verify the triangle inequality, namely â¥ð + ðâ¥ = sup â£(ð + ð)(ð¥)â£ ð¥âð

= sup â£ð (ð¥) + ð(ð¥)â£ ð¥âð { } â€ sup â£ð (ð¥)â£ + â£ð(ð¥)â£ ð¥âð

â€ sup â£(ð (ð¥)â£ + sup â£ð(ð¥)â£ ð¥âð

ð¥âð

= â¥ð â¥ + â¥ðâ¥ 2. Consequently, for any ð â ðµ(ð), ðŒð (ð¥) â€ â£ðŒâ£ â¥ð â¥ for every ð¥ â ð and therefore ðŒð â ðµ(ð). Similarly, for any ð, ð â ðµ(ð), (ð + ð)(ð¥) â€ â¥ð â¥ + â¥ðâ¥ for every 67

Solutions for Foundations of Mathematical Economics

ð¥ â ð and therefore ð + ð â ðµ(ð). Hence, ðµ(ð) is closed under addition and scalar multiplication; it is a subspace of the linear space ð¹ (ð). We conclude that ðµ(ð) is a normed linear space. 3. To show that ðµ(ð) is complete, assume that (ð ð ) is a Cauchy sequence in ðµ(ð). For every ð¥ â ð â£ð ð (ð¥) â ð ð (ð¥)â£ â€ â¥ð ð â ð ð â¥ â 0 Therefore, for ð¥ â ð, ð ð (ð¥) is a Cauchy sequence of real numbers. Since â is complete, this sequence converges. Deï¬ne the function ð (ð¥) = lim ð ð (ð¥) ðââ

We need to show â â¥ð ð â ð â¥ â 0 and â ð â ðµ(ð) ð

(ð ) is a Cauchy sequence. For given ð > 0, choose ð such that â¥ð ð â ð ð â¥ < ð/2 for very ð, ð â¥ ð . For any ð¥ â ð and ð â¥ ð , â£ð ð (ð¥) â ð (ð¥)â£ â€ â£ð ð (ð¥) â ð ð (ð¥)â£ + â£ð ð (ð¥) â ð (ð¥)â£ â€ â¥ð ð â ð ð â¥ + â£ð ð (ð¥) â ð (ð¥)â£ By suitable choice of ð (which may depend upon ð¥), each term on the right can be made smaller than ð/2 and therefore â£ð ð (ð¥) â ð (ð¥)â£ < ð for every ð¥ â ð and ð â¥ ð . â¥ð ð â ð â¥ = sup â£ð ð (ð¥) â ð (ð¥)â£ â€ ð ð¥âð

Finally, this implies â¥ð â¥ = limðââ â¥ð ð â¥. Therefore ð â ðµ(ð). 2.12 If the die is fair, the probability of the elementary outcomes is ð ({1}) = ð ({2}) = ð ({3}) = ð ({4}) = ð ({5}) = ð ({6}) = 1/6 By Condition 3 ð ({2, 4, 6}) = ð ({2}) + ð ({4}) + ð ({6}) = 1/2 2.13 The proï¬t maximization problem of a competitive single-output ï¬rm is to choose the combination of inputs x â âð+ and scale of production ðŠ to maximize net proï¬t. This is summarized in the constrained maximization problem max ððŠ â x,ðŠ

âð

ð â

ð€ð ð¥ð

ð=1

subject to x â ð (ðŠ)

where ððŠ is total revenue and ð=1 ð€ð ð¥ð total cost. The proï¬t function, which depends upon both ð and w, is deï¬ned by Î (ð, w) =

max ððŠ â

ðŠ,xâð (ðŠ)

68

ð â ð=1

ð€ð ð¥ð

Solutions for Foundations of Mathematical Economics

For analysis, it is convenient to represent the technology ð (ðŠ) by a production function (Example 2.24). The ï¬rmâs optimization can then be expressed as maxð ðð (x) â

xââ+

ð â

ð€ð ð¥ð

ð=1

and the proï¬t function as Î (ð, w) = maxð ðð (x) â xââ+

2.14

ð â

ð€ð ð¥ð

ð=1

1. Assume that production is proï¬table at p. That is, there exists some y â ð such that ð (y, p) > 0. Since the technology exhibits constant returns to scale, ð is a cone (Example 1.101). Therefore ðŒy â ð for every ðŒ > 0 and â â ðð (ðŒðŠð ) = ðŒ ðð ðŠð = ðŒð (y, p) ð (ðŒy, p) = ð

ð

Therefore { ð (ðŒy, p) : ðŒ > 0 } is unbounded and Î (p) = sup ð (y, p) â¥ sup ð (ðŒy, p) = +â ðŒ>0

yâð

2. Assume to the contrary that there exists p â âð+ with Î (p) = ð â / { 0, +â, ââ }. There are two possible cases. (a) 0 < ð < +â. Since ð = supðŠâð ð (y, p) > 0, there exists y â ð such that ð (y, p) > 0 The previous part implies Î (p) = +â. (b) ââ < ð < 0. Then there exists y â ð such that ð (y, p) < 0 By a similar argument to the previous part, this implies Î (p) = ââ. 2.15 Assume xâ is a solution to (2.4). ð (xâ , ðœ) â¥ ð (x, ðœ) for every x â ðº(ðœ) and therefore ð (xâ , ðœ) â¥ sup ð (x, ðœ) = ð£(ðœ) xâðº(ðœ)

On the other hand xâ â ðº(ðœ) and therefore ð£(ðœ) = sup ð (x, ðœ) â¥ ð (xâ , ðœ) xâðº(ðœ)

Therefore, xâ satisï¬es (2.5). Conversely, assume xâ â ðº(ðœ) satisï¬es (2.5). Then ð (xâ , ðœ) = ð£(ðœ) = sup ð (x, ðœ) â¥ ð (x, ðœ) for every x â ðº(ðœ) xâðº(ðœ)

xâ solve (2.4). 2.16 The assumption that ðº(ð¥) â= â for every ð¥ â ð implies Î(ð¥0 ) â= â for every ð¥0 â ð. There always exist feasible plans from any starting point. Since ð¢ is bounded, there exists ð such that â£ð (ð¥ð¡ , ð¥ð¡+1 )â£ â€ ð for every x â Î(ð¥0 ). Consequently, for every x â Î(ð¥0 ), ð (x) â â and â  â â â  â â ð   â£ð (x)â£ =  ðœ ð¡ ð (ð¥ð¡ , ð¥ð¡+1 ) â€ ðœ ð¡ â£ð (ð¥ð¡ , ð¥ð¡+1 )â£ â€ ðœð¡ð =   1âðœ ð¡=0 ð¡=0 ð¡=0 69

Solutions for Foundations of Mathematical Economics

using the formula for a geometric series (Exercise 1.108). Therefore ð£(ð¥0 ) =

sup ð (x) â€

xâÎ(ð¥0 )

ð 1âðœ

and ð£ â ðµ(ð). Next, we note that for every feasible plan x â Î(ð¥0 ) ð (x) =

â â

ðœ ð¡ ð (ð¥ð¡ , ð¥ð¡+1 )

ð¡=0

= ð (ð¥0 , ð¥1 ) + ðœ

â â

ðœ ð¡ ð (ð¥ð¡+1 , ð¥ð¡+2 )

ð¡=0

= ð (ð¥0 , ð¥1 ) + ðœð (xâ² )

(2.29)

where xâ² = (ð¥1 , ð¥2 , . . . ) is the continuation of the plan x starting at ð¥1 . For any ð¥0 â ð and ð > 0, there exists a feasible plan x â Î(ð¥0 ) such that ð (x) â¥ ð£(ð¥0 ) â ð Let xâ² = (ð¥1 , ð¥2 , . . . ) be the continuation of the plan x starting at ð¥1 . Using (2.29) and the fact that ð (xâ² ) â€ ð£(ð¥1 ), we conclude that ð£(ð¥0 ) â ð â€ ð (x) = ð (ð¥0 , ð¥1 ) + ðœð (xâ² ) â€ ð (ð¥0 , ð¥1 ) + ðœð£(ð¥1 ) â€ sup { ð (ð¥0 , ðŠ) + ðœð£(ðŠ) } ðŠâðº(ð¥)

Since this is true for every ð > 0, we must have ð£(ð¥0 ) â€ sup { ð (ð¥0 , ðŠ) + ðœð£(ðŠ) } ðŠâðº(ð¥)

On the other hand, choose any ð¥1 â ðº(ð¥0 ) â ð. Since ð£(ð¥1 ) =

sup ð (x)

xâÎ(ð¥1 )

there exists a feasible plan xâ² = (ð¥1 , ð¥2 , . . . ) starting at ð¥1 such that ð (xâ² ) â¥ ð£(ð¥1 ) â ð Moreover, the plan x = (ð¥0 , ð¥1 , ð¥2 , . . . ) is feasible from ð¥0 and ð£(ð¥0 ) â¥ ð (x) = ð (ð¥0 , ð¥1 ) + ðœð (xâ² ) â¥ ð (ð¥0 , ð¥1 ) + ðœð£(ð¥1 ) â ðœð Since this is true for every ð > 0 and ð¥1 â ðº(ð¥0 ), we conclude that ð£(ð¥0 ) â¥ sup { ð (ð¥0 , ðŠ) + ðœð£(ðŠ) } ðŠâðº(ð¥)

Together with (2.30) this establishes the required result, namely ð£(ð¥0 ) = sup { ð (ð¥0 , ðŠ) + ðœð£(ðŠ) } ðŠâðº(ð¥)

70

(2.30)

Solutions for Foundations of Mathematical Economics

2.17 Assume x is optimal, so that ð (xâ ) â¥ ð (x) for every x â Î(ð¥0 ) This implies (using (2.39)) ð (ð¥0 , ð¥â1 ) + ðœð (xâ â² ) â¥ ð (ð¥0 , ð¥1 ) + ðœð (xâ² ) where xâ² = (ð¥1 , ð¥2 , . . . ) is the continuation of the plan x starting at ð¥1 and xâ â² = (ð¥â1 , ð¥â2 , . . . ) is the continuation of the plan xâ . In particular, this is true for every plan x â Î(ð¥0 ) with ð¥1 = ð¥â1 and therefore ð (ð¥0 , ð¥â1 ) + ðœð (xâ â² ) â¥ ð (ð¥0 , ð¥â1 ) + ðœð (xâ² ) for every xâ² â Î(ð¥â1 ) which implies that ð (xâ â² ) â¥ ð (xâ² ) for every xâ² â Î(ð¥â1 ) That is, xâ â² is optimal starting at ð¥â1 and therefore ð (xâ â² ) = ð£(ð¥â1 ) (Exercise 2.15). Consequently ð£(ð¥0 ) = ð (xâ ) = ð (ð¥0 , ð¥â1 ) + ðœð (xâ â² ) = ð (ð¥0 , ð¥â1 ) + ðœð£(ð¥â1 ) This veriï¬es (2.13) for ð¡ = 0. A similar argument veriï¬es (2.13) for any period ð¡. To show the converse, assume that xâ = (ð¥0 , ð¥â1 , ð¥â2 , . . . ) â Î(ð¥0 ) satisï¬es (2.13). Successively using (2.13) ð£(ð¥0 ) = ð (ð¥0 , ð¥â1 ) + ðœð£(ð¥â1 ) = ð (ð¥0 , ð¥â1 ) + ðœð (ð¥â1 , ð¥â2 ) + ðœ 2 ð£(ð¥â1 ) =

1 â

ðœ ð¡ ð (ð¥âð¡ , ð¥âð¡+1 ) + ðœ 2 ð£(ð¥â2 )

ð¡=0

=

2 â

ðœ ð¡ ð (ð¥âð¡ , ð¥âð¡+1 ) + ðœ 3 ð£(ð¥â3 )

ð¡=0

=

ð â1 â

ðœ ð¡ ð (ð¥ð¡ , ð¥ð¡ + 1) + ðœ ð ð£(ð¥âð )

ð¡=0

for any ð = 1, 2, . . . . Since ð£ is bounded (Exercise 2.16), ðœ ð ð£(ð¥âð ) â 0 as ð â â and therefore ð£(ð¥0 ) =

â â

ðœ ð¡ ð (ð¥ð¡ , ð¥ð¡+1 ) = ð (xâ )

ð¡=0

Again using Exercise 2.15, xâ is optimal. 2.18 We have to show that â for any ð£ â ðµ(ð), ð ð£ is a functional on ð. â ð ð£ is bounded. Since ð¹ â ðµ(ð Ã ð), there exists ð1 < â such that â£ð (ð¥, ðŠ)â£ â€ ð1 for every (ð¥, ðŠ) â ð Ã ð. Similarly, for any ð£ â ðµ(ð), there exists ð2 < â such that â£ð£(ð¥)â£ â€ ð2 for every ð¥ â ð. Consequently for every (ð¥, ðŠ) â ð Ã ð and ð£ â ðµ(ð) â£ð (ð¥, ðŠ) + ðœð£(ðŠ)â£ â€ â£ð (ð¥, ðŠ)â£ + ðœ â£ð£(ðŠ)â£ â€ ð1 + ðœð2 < â 71

(2.31)

Solutions for Foundations of Mathematical Economics For each ð¥ â ð, the set ðð¥ =

{

ð (ð¥, ðŠ) + ðœð£(ðŠ) : ðŠ â ðº(ð¥)

}

is a nonempty bounded subset of â, which has least upper bound. Therefore (ð ð£)(ð¥) = sup ðð¥ = sup ð (ð¥, ðŠ) + ðœð£(ðŠ) ðŠâðº(ð¥)

deï¬nes a functional on ð. Moreover by (2.31) â£(ð ð£)(ð)â£ â€ ð1 + ðœð2 < â Therefore ð ð£ â ðµ(ð). 2.19 Let ð = {1, 2, 3}. Any individual is powerless so that ð€({ð}) = 0

ð = 1, 2, 3

Any two players can allocate the \$1 to between themselves, leaving the other player out. Therefore ð€({ð, ð}) = 1

ð, ð â ð, ð â= ð

The best that the three players can achieve is to allocate the \$1 amongst themselves, so that ð€(ð ) = 1 2.20 If the consumers preferences are continuous and strictly convex, she has a unique optimal choice xâ for every set of prices p and income ð in ð (Example 1.116). Therefore, the demand correspondence is single valued. 2.21 Assume ð âð â ðµ(sâ ) for every ð â ð . Then for every player ð â ð (ð ð , sâð ) â¿ð (ð â²ð , sâð ) for every ð â²ð â ðð sâ = (ð â1 , ð â2 , . . . , ð âð ) is a Nash equilibrium. Conversely, assume sâ = (ð â1 , ð â2 , . . . , ð âð ) is a Nash equilibrium. Then for every player ð â ð (ð ð , sâð ) â¿ð (ð â²ð , sâð ) for every ð â²ð â ðð which implies that ð âð â ðµ(sâ ) for every ð â ð 2.22 For any nonempty compact set ð â ð, ðµ(ð ) is nonempty and compact provided â¿ð is continuous (Proposition 1.5) and ðµ(ð ) â ð . Therefore ðµð1 â ðµð2 â ðµð3 . . . is a nested sequence ofâ©nonempty compact sets. By the nested intersection theorem â (Exercise 1.117), ðð = ð=0 ðµðð â= â. 2.23 If sâ is a Nash equilibrium, ð ð â ðµðð for every ð.

72

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2.24 For any ðœ, let xâ â ð(ðœ). Then ð (xâ , ðœ) â¥ ð (x, ðœ)

for every x â ðº(ðœ)

Therefore ð (xâ , ðœ) â¥ ð£(ðœ) = sup ð (x, ðœ) xâðº(ðœ)

Conversely ð£(ðœ) = sup ð (x, ðœ) â¥ sup ð (x, ðœ) â¥ ð (xâ , ðœ) for every xâ â ð(ðœ) xâðº(ðœ)

xâðº(ðœ)

Consequently ð£(ðœ) = ð (xâ , ðœ) for any xâ â ð(ðœ) 2.25 The graph of ð is graph(ð ) = { (ðŠ, x) â â+ Ã âð+ : x â ð (ðŠ) } while the production possibility set ð is ð = { (ðŠ, âx) â â+ Ã âð+ : ð¥ â ð (ðŠ) } Assume that ð is convex and let (ðŠ ð , xð ) â graph(ð ), ð = 1, 2. This means that (ðŠ 1 , âx1 ) â ð and (ðŠ 2 , âx2 ) â ð Let Â¯ = ðŒx1 + (1 â ðŒ)x2 ðŠÂ¯ = ðŒðŠ 1 + (1 â ðŒ)ðŠ 2 and x for some 0 â€ ðŒ â€ 1. Since ð is convex (Â¯ ðŠ , âÂ¯ x) = ðŒ(ðŠ 1 , âx1 ) + (1 â ðŒ)(ðŠ 2 , âx2 ) â ð Â¯ â ð (Â¯ Â¯ ) â graph(ð ). That is graph(ð ) is convex. and therefore x ðŠ ) so that (Â¯ ðŠ, x Conversely, assuming graph(ð ) is convex, if (ðŠ ð , âxð ) â ð , ð = 1, 2, then (ðŠ ð , xð ) â graph(ð ) and therefore Â¯ ) â graph(ð ) =â x Â¯ â ð (Â¯ (Â¯ ðŠ, x ðŠ ) =â (Â¯ ðŠ, âÂ¯ x) â ð so that ð is convex. 2.26 The graph of ð is graph(ðº) = { (ðœ, x) â Î Ã ð : x â ðº(ðœ) } Assume that (ðœð , xð ) â graph(ðº), ð = 1, 2. This means that xð â ðº(ðœ) and therefore ð ð (x, ðœ) â€ ðð for every ð and ð = 1, 2. Since ð ð is convex ð(ðŒx1 + (1 â ðŒ)x2 , ðŒðœ1 + (1 â ðŒ)ðœ 2 ) â¥ ðŒð(x1 , ðœ 1 ) + (1 â ðŒ)ð(x2 , ðœ2 ) â¥ ðð Therefore ðŒx1 +(1âðŒ)x2 â ðº(ðŒðœ 1 +(1âðŒ)ðœ2 ) and (ðŒðœ 1 +(1âðŒ)ðœ 2 , ðŒx1 +(1âðŒ)x2 ) â graph(ðº). ðº is convex. 73

Solutions for Foundations of Mathematical Economics

2.27 The identity function ðŒð : ð â ð is deï¬ned by ðŒð (ð¥) = ð¥ for every ð¥ â ð. Therefore ð¥2 â»ð ð¥1 =â ðŒð (ð¥2 ) = ð¥2 â»ð ð¥1 = ðŒð (ð¥1 ) 2.28 Assume that ð and ð are increasing. Then, for every ð¥1 , ð¥2 â ð with ð¥2 â¿ð ð¥1 ð (ð¥2 ) â¿ð ð (ð¥1 ) =â ð(ð (ð¥2 )) â¿ð ð(ð (ð¥1 )) ð â ð is also increasing. Similarly, if ð and ð are strictly increasing, ð¥2 â»ð ð¥1 =â ð (ð¥2 ) â»ð ð (ð¥1 ) =â ð(ð (ð¥2 )) â»ð ð(ð (ð¥1 )) 2.29 For every ðŠ â ð (ð), there exists a unique ð¥ â ð such that ð (ð¥) = y. (For if ð¥1 , ð¥2 are such that ð (ð¥1 ) = ð (ð¥2 ), then ð¥1 = ð¥2 .) Therefore, ð is one-to-one and onto ð (ð), and so has an inverse (Exercise 2.4). Further ð¥2 > ð¥2 ââ ð (ð¥2 ) > ð (ð¥1 ) Therefore ð â1 is strictly increasing. 2.30 Assume ð : ð â â is increasing. Then, for every ð¥2 â¿ ð¥1 , ð (ð¥2 ) â¥ ð (ð¥1 ) which implies that âð (ð¥2 ) â€ âð (ð¥1 ). âð is decreasing. 2.31 For every ð¥2 â¿ ð¥1 . ð (ð¥2 ) â¥ ð (ð¥1 ) ð(ð¥2 ) â¥ ð(ð¥1 ) Adding (ð + ð)(ð¥2 ) = ð (ð¥2 ) + ð(ð¥2 ) â¥ ð (ð¥1 ) + ð (ð¥1 ) = (ð + ð)(ð¥1 ) That is, ð + ð is increasing. Similarly for every ðŒ â¥ 0 ðŒð (ð¥2 ) â¥ ðŒð (ð¥1 ) and therefore ðŒð is increasing. By Exercise 1.186, the set of all increasing functionals is a convex cone in ð¹ (ð). If ð is strictly increasing, then for every ð¥2 â» ð¥1 , ð (ð¥2 ) > ð (ð¥1 ) ð(ð¥2 ) â¥ ð(ð¥1 ) Adding (ð + ð)(ð¥2 ) = ð (ð¥2 ) + ð(ð¥2 ) > ð (ð¥1 ) + ð(ð¥1 ) = (ð + ð)(ð¥1 ) ð + ð is strictly increasing. Similarly for every ðŒ > 0 ðŒð (ð¥2 ) > ðŒð (ð¥1 ) ðŒð is strictly increasing. 2.32 For every ð¥2 â» ð¥1 . ð (ð¥2 ) > ð (ð¥1 ) > 0 ð(ð¥2 ) > ð(ð¥1 ) > 0 and therefore (ð ð)(ð¥2 ) = ð (ð¥2 )ð(ð¥2 ) > ð (ð¥2 )ð(ð¥1 ) > ð (ð¥1 )ð(ð¥1 ) = (ð ð)(ð¥1 ) using Exercise 2.31. 74

Solutions for Foundations of Mathematical Economics

2.33 By Exercise 2.31 and Example 2.53, each ðð is strictly increasing on â+ . That is ð¥1 < ð¥2 =â ðð (ð¥1 ) < ðð (ð¥2 ) for every ð

(2.32)

and therefore lim ðð (ð¥1 ) â€ lim ðð (ð¥2 )

ðââ

ðââ

This suï¬ces to show that ð(ð¥) = limðââ ðð (ð¥) is increasing (not strictly increasing). However, 1 + ð¥ is strictly increasing, and therefore by Exercise 2.31 ðð¥ = 1 + ð¥ + ð(ð¥) is strictly increasing on â+ . While it is the case that ð = lim ðð is strictly increasing on â+ , (2.32) does not suï¬ce to show this. 2.34 For every ð > 0, ð log ð¥ is strictly increasing (Exercise 2.32) and therefore ðð log ð¥ is strictly increasing (Exercise 2.28). For every ð < 0, âð log ð¥ is strictly increasing and therefore (Exercise 2.30 ð log ð¥ is strictly decreasing. Therefore ðð log ð¥ is strictly decreasing (Exercise 2.28). 2.35 Apply Exercises 2.31 and 2.28 to Example 2.56. 2.36 ð¢ is (strictly) increasing so that ð¥2 â¿ ð¥1 =â ð¢(ð¥2 ) â¥ ð¢(ð¥1 ) To show the converse, assume that ð¥1 , ð¥2 â ð with ð¢(ð¥2 ) â¥ ð¢(ð¥1 ). Since â¿ is complete, either ð¥2 â¿ ð¥1 or ð¥1 â» ð¥2 . However, the second possibility cannot occur since ð¢ is strictly increasing and therefore ð¥1 â» ð¥2 =â ð¢(ð¥1 ) > ð¢(ð¥2 ) contradicting the hypothesis that ð¢(ð¥2 ) â¥ ð¢(ð¥1 ). We conclude that ð¢(ð¥2 ) â¥ ð¢(ð¥1 ) =â ð¥2 â¿ ð¥1 2.37 Assume ð¢ represents the preference ordering â¿ on ð and let ð : â â â be strictly increasing. Then composition ð â ð¢ : ð â â is strictly increasing (Exercise 2.28). Therefore ð â ð¢ is a utility function (Example 2.58). Since ð is strictly increasing ð(ð¢(ð¥2 )) â¥ ð(ð¢(ð¥1 )) ââ ð¢(ð¥2 ) â¥ ð¢(ð¥1 ) ââ ð¥2 â¿ ð¥1 for every ð¥1 , ð¥2 â ð Therefore, ð â ð¢ also represents â¿. 2.38

Â¯ = ð§Â¯1 â¿ x and therefore zÂ¯ â ðx+ . Similarly, 1. (a) Let ð§Â¯ = maxðð=1 ð¥ð . Then z ð let ð§ = minð=1 ð¥ð . Then z = ð§1 â ðxâ . Therefore, ðx+ and ðxâ are both nonempty. By continuity, the upper and lower contour sets â¿(x) and âŸ(x) are closed. ð is a closed cone. Since ðx+ = â¿(x) â© ð and ðxâ = âŸ(x) â© ð ðx+ and ðxâ are closed. (b) By completeness, ðx+ âª ðxâ = ð. Since ð is connected, ðx+ â© ðxâ â= â. (Otherwise, ð is the union of two disjoint closed sets and hence the union of two disjoint open sets.) (c) Let zx â ðx+ â© ðxâ . Then z â¿ x and also z âŸ x. That is, z âŒ x. 75

Solutions for Foundations of Mathematical Economics

(d) Suppose x âŒ z1x and x âŒ z2x with z1x â= z2x . Then either z1x > z2x or z1x < z2x . Without loss of generality, assume z2x > z1x . Then monotonicity and transitivity imply x âŒ z2x â» z1x âŒ x which is a contradiction. Therefore zx is unique. Let ð§x denote the scale of zx , that is zx = ð§x 1. For every x â âð+ , there is a unique zx âŒ x and the function ð¢ : âð+ â â given by ð¢(x) = ð§x is well-deï¬ned. Moreover x2 â¿ x1 ââ zx2 â¿ zx1 ââ ð§x2 â¥ ð§x1 ââ ð¢(x2 ) â¥ ð¢(x1 ) ð¢ represents the preference order â¿. 2.39

1. For every ð¥1 â â, (ð¥1 , 2) â»ð¿ (ð¥1 , 1) in the lexicographic order. If ð¢ represents â¿ð¿ , ð¢ is strictly increasing and therefore ð¢(ð¥1 , 2) > ð¢(ð¥1 , 1). There exists a rational number ð(ð¥1 ) such that ð¢(ð¥1 , 2) > ð(ð¥1 ) > ð¢(ð¥1 , 1).

2. The preceding construction associates a rational number with every real number ð¥1 â â. Hence ð is a function from â to the set of rational numbers ð. For any ð¥11 , ð¥21 â â with ð¥21 > ð¥11 ð(ð¥21 ) > ð¢(ð¥21 , 1) > ð¢(ð¥11 , 2) > ð(ð¥11 ) Therefore ð¥21 > ð¥11 =â ð(ð¥21 ) > ð(ð¥11 ) ð is strictly increasing. 3. By Exercise 2.29, ð has an inverse. This implies that ð is one-to-one and onto, which is impossible since ð is countable and â is uncountable (Example 2.16). This contradiction establishes that â¿ð¿ has no such representation ð¢. 2.40 Let a1 , a2 â ðŽ with a1 â¿2 a2 . Since the game is strictly competitive, a2 â¿1 a1 . Since ð¢1 represents â¿1 , ð¢1 (a2 ) â¥ ð¢1 (a1 ) which implies that âð¢1 (a2 ) â€ âð¢1 (a1 ), that is ð¢2 (a1 ) â¥ ð¢2 (a2 ) where ð¢2 = âð¢1 . Similarly ð¢2 (a1 ) â¥ ð¢2 (a2 ) =â ð¢1 (a1 ) â€ ð¢1 (a2 ) ââ a1 âŸ1 a2 =â a1 â¿2 a2 Therefore ð¢2 = âð¢1 represents â¿2 and ð¢1 (a) + ð¢2 (a) = 0 for every a â ðŽ 2.41 Assume ð â« ð . By superadditivity ð€(ð ) â¥ ð€(ð) + ð€(ð â ð) â¥ ð€(ð) 2.42 Assume ð£, ð€ â ðµ(ð) with ð€(ðŠ) â¥ ð£(ðŠ) for every ðŠ â ð. Then for any ð¥ â ð ð (ð¥, ðŠ) + ðœð€(ðŠ) â¥ ð (ð¥, ðŠ) + ðœð£(ðŠ) for every ðŠ â ð and therefore (ð ð€)(ð¥) = sup {ð (ð¥, ðŠ) + ðœð€(ðŠ)} â¥ sup {ð (ð¥, ðŠ) + ðœð£(ðŠ)} = (ð ð£)(ð¥) ðŠâðº(ð¥)

ðŠâðº(ð¥)

T is increasing. 76

Solutions for Foundations of Mathematical Economics

2.43 For every ð2 â¥ ð1 â Î, if ð¥1 â ðº(ð1 ) and ð¥2 â ðº(ð2 ), then ð¥1 â§ ð¥2 â€ ð¥1 and therefore ð¥1 â§ ð¥2 â ðº(ð1 ). If ð¥1 â¥ ð¥2 , then ð¥1 âš ð¥2 = ð¥1 â€ ð(ð1 ) â€ ð(ð2 ) and therefore ð¥1 âš ð¥2 â ðº(ð2 ). On the other hand, if ð¥1 â€ ð¥2 , then ð¥1 âš ð¥2 = ð¥2 â ðº(ð2 ). 2.44 Assume ð is increasing, and let ð¥1 , ð¥2 â ð with ð¥2 â¿ ð¥1 . Let ðŠ1 â ð(ð¥1 ). Choose any ðŠ â² â ð(ð¥2 ). Since ð is increasing, ð(ð¥2 ) â¿ð ð(ð¥1 ) and therefore ðŠ2 = ðŠ1 âš ðŠ â² â ð(ð¥2 ). ðŠ2 â¿ ðŠ1 as required. Similarly, for every ðŠ2 â ð(ð¥2 ), there exists some ðŠ â² â ð(ð¥2 ) such that ðŠ1 = ðŠ â² â§ ðŠ2 â ð(ð¥1 ) with ðŠ2 â¿ ðŠ1 . 2.45 Since ð(ð¥) is a sublattice, sup ð(ð¥) â ð(ð¥) for every ð¥. Therefore, the function ð (ð¥) = sup ð(ð¥) is a selection. Similarly ð(ð¥) = inf ð(ð¥) is a selection. Both ð and ð are increasing (Exercise 1.50).

â 2.46 Let ð¥1 , ð¥2 belong to ðâ with ð¥2 â¿ ð¥1 . Choose y1 = (ðŠ11 , ðŠ21 , . . . , ðŠð1 ) â ð ðð (ð¥1 ) and y2 = (ðŠ12 , ðŠ22 , . . . , ðŠð2 ) â ð ðð (ð¥2 ). Then, for each ð = 1, 2, . . . , ð, ðŠð1 â ðð (ð¥1 ) and (ð¥1 ) and ðŠð1 âš ðŠð2 ââðð (ð¥2 ) for ðŠð2 â ðð (ð¥2 ). Since each ðð isâincreasing, ðŠð1 â§ ðŠð2 â ððâ 1 2 1 1 2 each ð. Therefore y â§ y â ð ðð (ð¥ ) and y âš y â ð ðð (ð¥2 ). ð(ð¥) = ð ðð (ð¥) is increasing. â© â© 2.47 Let ð¥1 , ð¥2 belong to ð with ð¥2 â¿ ð¥1 . Choose ðŠ 1 â ð ðð (ð¥1 ) and ðŠ 2 â ð ðð (ð¥2 ). Then ðŠ 1 â ðð (ð¥1 ) and ðŠ 2 â ðð (ð¥2 ) for each ð = 1, 2, . . . , ð. Since each ððâ©is increasing, ðð (ð¥1 ) and ðŠ 1 âšâ©ðŠ 2 â ðð (ð¥2 ) for each ð. Therefore ðŠ 1 â§ ðŠ 2 â ð ðð (ð¥1 ) and ðŠ1 â§ ðŠ2 â â© 1 2 ðŠ âš ðŠ â ð ðð (ð¥2 ). ð = ð ð is increasing. 2.48 Let ð be a selection from an always increasing correspondence ð : ð â ð . For every ð¥1 , ð¥2 â ð, ð (ð¥1 ) â ð(ð¥1 ) and ð (ð¥2 ) â ð(ð¥2 ). Since ð is always increasing ð¥1 â¿ð ð¥2 =â ð (ð¥1 ) â¿ð ð (ð¥2 ) ð is increasing. Conversely, assume every selection ð â ð is increasing. Choose any ð¥1 , ð¥2 â ð with ð¥1 â¿ ð¥2 . For every ðŠ1 â ð(ð¥1 ) and ðŠ2 â ð(ð¥2 ), there exists a selection ð with ðŠð = ð(ð¥ð ), ð = 1, 2. Since ð is increasing, ð¥1 â¿ð ð¥2 =â ðŠ1 â¿ð ðŠ2 ð is increasing. 2.49 Let ð¥1 , ð¥2 â ð. If ð is a chain, either ð¥1 â¿ ð¥2 or ð¥2 â¿ ð¥1 . Without loss of generality , assume ð¥2 â¿ ð¥1 . Then ð¥1 âš ð¥2 = ð¥2 and ð¥1 â§ ð¥2 = ð¥1 and (2.17) is satisï¬ed as an identity. 2.50 (ð + ð)(ð¥1 âš ð¥2 ) + (ð + ð)(ð¥1 â§ ð¥2 ) = ð (ð¥1 âš ð¥2 ) + ð(ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) + ð(ð¥1 â§ ð¥2 ) = ð (ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) + ð(ð¥1 âš ð¥2 ) + ð(ð¥1 â§ ð¥2 ) â¥ ð (ð¥1 ) + ð (ð¥2 ) + ð(ð¥1 ) + ð(ð¥2 ) = (ð + ð)(ð¥1 ) + (ð + ð)(ð¥2 ) Similarly ð (ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) â¥ ð (ð¥1 ) + ð (ð¥2 ) 77

Solutions for Foundations of Mathematical Economics

implies ðŒð (ð¥1 âš ð¥2 ) + ðŒð (ð¥1 â§ ð¥2 ) â¥ ðŒð (ð¥1 ) + ðŒð (ð¥2 ) for all ðŒ â¥ 0. By Exercise 1.186, the set of all supermodular functions is a convex cone in ð¹ (ð). 2.51 Since ð is supermodular and ð is nonnegative deï¬nite, ( ) ð (ð¥1 âš ð¥2 )ð(ð¥1 âš ð¥2 ) â¥ ð (ð¥1 ) + ð (ð¥2 ) â ð (ð¥1 â§ ð¥2 ) ð(ð¥1 âš ð¥2 ) ( ) = ð (ð¥2 )ð(ð¥1 âš ð¥2 ) + ð (ð¥1 ) â ð (ð¥1 â§ ð¥2 ) ð(ð¥1 âš ð¥2 ) for any ð¥1 , ð¥2 â ð. Since ð and ð are increasing, this implies ( ) ð (ð¥1 âš ð¥2 )ð(ð¥1 âš ð¥2 ) â¥ ð (ð¥2 )ð(ð¥1 âš ð¥2 ) + ð (ð¥1 ) â ð (ð¥1 â§ ð¥2 ) ð(ð¥1 )

(2.33)

Similarly, since ð is nonnegative deï¬nite, ð supermodular, and ð and ð increasing ( ) ð (ð¥2 )ð(ð¥1 âš ð¥2 ) â¥ ð (ð¥2 ) ð(ð¥1 ) + ð(ð¥2 ) â ð(ð¥1 â§ ð¥2 ) ( ) = ð (ð¥2 )ð(ð¥2 ) + ð (ð¥2 ) ð(ð¥1 ) â ð(ð¥1 â§ ð¥2 ) ( ) â¥ ð (ð¥2 )ð(ð¥2 ) + ð (ð¥1 â§ ð¥2 ) ð(ð¥1 ) â ð(ð¥1 â§ ð¥2 ) Combining this inequality with (2.33) gives ( ) ð (ð¥1 âš ð¥2 )ð(ð¥1 âš ð¥2 ) â¥ ð (ð¥2 )ð(ð¥2 ) + ð (ð¥1 â§ ð¥2 ) ð(ð¥1 ) â ð(ð¥1 â§ ð¥2 ) ( ) + ð (ð¥1 ) â ð (ð¥1 â§ ð¥2 ) ð(ð¥1 ) = ð (ð¥2 )ð(ð¥2 ) + ð (ð¥1 â§ ð¥2 )ð(ð¥1 ) â ð (ð¥1 â§ ð¥2 )ð(ð¥1 â§ ð¥2 ) + ð (ð¥1 )ð(ð¥1 ) â ð (ð¥1 â§ ð¥2 )ð(ð¥1 ) = ð (ð¥2 )ð(ð¥2 ) â ð (ð¥1 â§ ð¥2 )ð(ð¥1 â§ ð¥2) + ð (ð¥1 )ð(ð¥1 ) or ð ð(ð¥1 âš ð¥2 ) + ð ð(ð¥1 â§ ð¥2 ) â¥ ð ð(ð¥1 ) + ð ð(ð¥2 ) ð ð 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 ï¬rm produces two products. For every production plan y = (ðŠ1 , ðŠ2 ), y = (ðŠ1 , 0) âš (0, ðŠ2 ) 0 = (ðŠ1 , 0) â§ (0, ðŠ2 ) If ð is strictly submodular ð(w, y) + ð(w, 0) < ð(w, (ðŠ1 , 0)) + ð(w, (0, ðŠ2 )) Since ð(w, 0) = 0 ð(w, y) < ð(w, (ðŠ1 , 0)) + ð(w, (0, ðŠ2 )) The technology displays economies of scope.

78

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2.54 Assume (ð, ð€) is convex, that is ð€(ð âª ð ) + ð€(ð â© ð ) â¥ ð€(ð) + ð€(ð ) for every ð, ð â ð For all disjoint coalitions ð â© ð = â ð€(ð âª ð ) â¥ ð€(ð) + ð€(ð ) ð€ is superadditive. 2.55 Rewriting (2.18), this implies ð€(ð âª ð ) â ð€(ð ) â¥ ð€(ð) â ð€(ð â© ð ) for every ð, ð â ð

(2.34)

Let ð â ð â ð â {ð} and let ð â² = ð âª {ð}. Substituting in (2.34) ð€(ð â² âª ð ) â ð€(ð ) â¥ ð€(ð â² ) â ð€(ð â² â© ð ) Since ð â ð ð â² âª ð = (ð âª {ð}) âª ð = ð âª {ð} ð â² â© ð = (ð âª {ð}) â© ð = ð Substituting in the previous equation gives the required result, namely ð€(ð âª {ð}) â ð€(ð ) â¥ ð€(ð âª {ð}) â ð€(ð) Conversely, assume that ð€(ð âª {ð}) â ð€(ð ) â¥ ð€(ð âª {ð}) â ð€(ð)

(2.35)

for every ð â ð â ð â {ð}. Let ð and ð be arbitrary coalitions. Assume ð â© ð â ð and ð â© ð â ð (otherwise (2.18) is trivially satisï¬ed). This implies that ð â ð â= â. Assume these players are labelled 1, 2, . . . , ð, that is ð â ð = {1, 2, . . . , ð}. By (2.35) ð€(ð âª {1}) â ð€(ð) â¥ ð€((ð â© ð ) âª {1}) â ð€(ð â© ð )

(2.36)

Successively adding the remaining players in ð â ð ð€(ð âª {1, 2}) â ð€(ð âª {1}) â¥ ð€((ð â© ð ) âª {1, 2}) â ð€((ð â© ð ) âª {1}) .. . ( ) ð€(ð âª (ð â ð)) â ð€(ð âª {1, 2, . . . , ð â 1}) â¥ ð€ ð â© ð ) âª (ð â ð) â ð€((ð â© ð ) âª {1, 2, . . . , ð â 1}) Adding these inequalities to (2.36), we get ( ) ð€(ð âª (ð â ð)) â ð€(ð) â¥ ð€ ð â© ð ) âª (ð â ð) â ð€(ð â© ð ) This simpliï¬es to ð€(ð âª ð ) â ð€(ð) â¥ ð(ð ) â ð€(ð â© ð ) which can be arranged to give (2.18).

79

Solutions for Foundations of Mathematical Economics

2.56 The cost allocation game is not convex. Let ð = {ðŽð, ðŸð }, ð = {ðŸð, ð ð }. Then ð âª ð = {ðŽð, ðŸð, ð ð } = ð and ð â© ð = {ðŸð } and ð€(ð âª ð ) + ð€(ð â© ð ) = 1530 < 1940 = 770 + 1170 = ð€(ð) + ð€(ð ) Alternatively, observe that TNâs marginal contribution to coalition {ðŸð, ð ð } is 1170, which is greater than its marginal contribution to the grand coalition {ðŽð, ðŸð, ð ð } (1530 â 770 = 760). 2.57 ð is supermodular if ð (ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) â¥ ð (ð¥1 ) + ð (ð¥2 ) which can be rearranged to give ð (ð¥1 âš ð¥2 ) â ð (ð¥2 ) â¥ ð (ð¥1 ) â ð (ð¥1 â§ ð¥2 ) If the right hand side of this inequality is nonnegative, then so a fortiori is the left hand side, that is ð (ð¥1 ) â¥ ð (ð¥1 â§ ð¥2 ) =â ð (ð¥1 âš ð¥2 ) â¥ ð (ð¥2 ) If the right hand side is strictly positive, so must be the left hand side ð (ð¥1 ) > ð (ð¥1 â§ ð¥2 ) =â ð (ð¥1 âš ð¥2 ) > ð (ð¥2 ) 2.58 Assume ð¥2 â¿ ð¥1 â ð and ðŠ2 â¿ð ðŠ2 â ð . Assume that ð displays increasing diï¬erences in (ð¥, ðŠ), that is ð (ð¥2 , ðŠ2 ) â ð (ð¥1 , ðŠ2 ) â¥ ð (ð¥2 , ðŠ1 ) â ð (ð¥1 , ðŠ1 )

(2.37)

ð (ð¥2 , ðŠ2 ) â ð (ð¥2 , ðŠ1 ) â¥ ð (ð¥1 , ðŠ2 ) â ð (ð¥1 , ðŠ1 )

(2.38)

Rearranging

Conversely, (2.38) implies (2.37) . 2.59 We showed in the text that supermodularity implies increasing diï¬erences. To show that reverse, assume that ð : ð Ã ð â â displays increasing diï¬erences in (ð¥, ðŠ). Choose any (ð¥1 , ðŠ1 ), (ð¥2 , ðŠ2 ) â ð Ã ð . If (ð¥1 , ðŠ1 ), (ð¥2 , ðŠ2 ) are comparable, so that either (ð¥1 , ðŠ1 ) â¿ (ð¥2 , ðŠ2 ) or (ð¥1 , ðŠ1 ) âŸ (ð¥2 , ðŠ2 ), then (2.17) holds has an equality. Therefore assume that (ð¥1 , ðŠ1 ), (ð¥2 , ðŠ2 ) are incomparable. Without loss of generality, assume that ð¥1 âŸ ð¥2 while ðŠ1 â¿ ðŠ2 . (This is where we require that ð and ð be chains). This implies (ð¥1 , ðŠ1 ) â§ (ð¥2 , ðŠ2 ) = (ð¥1 , ðŠ2 ) and (ð¥1 , ðŠ1 ) âš (ð¥2 , ðŠ2 ) = (ð¥2 , ðŠ1 ) Increasing diï¬erences implies that ð (ð¥2 , ðŠ1 ) â ð (ð¥1 , ðŠ1 ) â¥ ð (ð¥2 , ðŠ2 ) â ð (ð¥1 , ðŠ2 ) which can be rewritten as ð (ð¥2 , ðŠ1 ) + ð (ð¥1 , ðŠ2 ) â¥ ð (ð¥1 , ðŠ1 ) + ð (ð¥2 , ðŠ2 ) Substituting (2.39) ( ) ( ) ð (ð¥1 , ðŠ1 ) âš (ð¥2 , ðŠ2 ) + ð (ð¥1 , ðŠ1 ) â§ (ð¥2 , ðŠ2 ) â¥ ð (ð¥1 , ðŠ1 ) + ð (ð¥2 , ðŠ2 ) which establishes the supermodularity of ð on ð Ã ð (2.17). 80

(2.39)

Solutions for Foundations of Mathematical Economics

2.60 In the standard Bertrand model of oligopoly â the strategy space of each ï¬rm is â+ , a lattice. â ð¢ð (ðð , pâð ) is supermodular in ðð (Exercise 2.51). â If the other ï¬rmâs increase their prices from p1âð to p2âð , the eï¬ect on the demand for ï¬rm ðâs product is â ð (ðð , p2âð ) â ð (ðð , p1âð ) = ððð (ð2ð â ð1ð ) ðâ=ð

If the goods are gross substitutes, demand for ï¬rm ð increases and the amount of the increase is independent of ðð . Consequently, the eï¬ect on proï¬t will be increasing in ðð . That is the payoï¬ function (net revenue) has increasing diï¬erences in (ðð , pâð ). Speciï¬cally, â ð¢(ðð , p2âð ) â ð¢(ðð , p1âð ) = ððð (ðð â ðÂ¯ð )(ð2ð â ð1ð ) ðâ=ð

For any price increase p2âð â© p1âð , the change in proï¬t ð¢(ðð , p2âð ) â ð¢(ðð , p1âð ) is increasing in ðð . Hence, the Bertrand oligopoly model is a supermodular game. 2.61 Suppose ð displays increasing diï¬erences so that for all ð¥2 â¿ ð¥1 and ðŠ2 â¿ ðŠ1 ð (ð¥2 , ðŠ2 ) â ð (ð¥1 , ðŠ2 ) â¥ ð (ð¥2 , ðŠ1 ) â ð (ð¥1 , ðŠ1 ) Then ð (ð¥2 , ðŠ1 ) â ð (ð¥1 , ðŠ1 ) â¥ 0 =â ð (ð¥2 , ðŠ2 ) â ð (ð¥1 , ðŠ2 ) â¥ 0 and ð (ð¥2 , ðŠ1 ) â ð (ð¥1 , ðŠ1 ) > 0 =â ð (ð¥2 , ðŠ2 ) â ð (ð¥1 , ðŠ2 ) > 0 2.62 For any ðœ â Îâ , let x1 , x2 â ð(ðœ). Supermodularity implies ð (x1 âš x2 , ðœ) + ð (x1 â§ x2 , ðœ) â¥ ð (x1 , ðœ) + ð (x2 , ðœ) which can be rearranged to give ð (x1 âš x2 , ðœ) â ð (x2 , ðœ) â¥ ð (x1 , ðœ) â ð (x1 â§ x2 , ðœ)

(2.40)

However x1 and x2 are both maximal in ðº(ðœ). ð (x2 , ðœ) â¥ ð (x1 âš x2 , ðœ) =â ð (x1 âš x2 , ðœ) â ð (x2 , ðœ) â€ 0 ð (x1 , ðœ) â¥ ð (x1 â§ x2 , ðœ) =â ð (x1 , ðœ) â ð (x1 â§ x2 , ðœ) â¥ 0 Substituting in (2.40), we conclude 0 â¥ ð (x1 âš x2 , ðœ) â ð (x2 , ðœ) â¥ ð (x1 , ðœ) â ð (x1 â§ x2 , ðœ) â¥ 0 This inequality must be satisï¬ed as an equality with ð (x1 âš x2 , ðœ) = ð (x2 , ðœ) ð (x1 â§ x2 , ðœ) = ð (x1 , ðœ) That is x1 âš x2 â ð(ðœ) and x1 â§ x2 â ð(ðœ). By Exercise 2.45, ð has an increasing selection. 81

Solutions for Foundations of Mathematical Economics

2.63 As in the proof of the theorem, let ðœ1 , ðœ 2 belong to Î with ðœ 2 â¿ ðœ1 . Choose any optimal solutions x1 â ð(ðœ 1 ) and x2 â ð(ðœ2 ). We claim that x2 â¿ð x1 . Assume otherwise, that is assume x2 ââ¿ð x1 . This implies (Exercise 1.44) that x1 â§ x2 â= x1 . Since x1 â¿ x1 â§ x2 , we must have x1 â» x1 â§ x2 . Strictly increasing diï¬erences implies ð (x1 , ðœ 2 ) â ð (x1 , ðœ1 ) > ð (x1 â§ x2 , ðœ2 ) â ð (x1 â§ x2 , ðœ 1 ) which can be rearranged to give ð (x1 , ðœ 2 ) â ð (x1 â§ x2 , ðœ2 ) > ð (x1 , ðœ1 ) â ð (x1 â§ x2 , ðœ 1 )

(2.41)

Supermodularity implies ð (x1 âš x2 , ðœ2 ) + ð (x1 â§ x2 , ðœ2 ) â¥ ð (x1 , ðœ2 ) + ð (x2 , ðœ 2 ) which can be rearranged to give ð (x1 âš x2 , ðœ2 ) â ð (x2 , ðœ2 ) â¥ ð (x1 , ðœ2 ) â ð (x1 â§ x2 , ðœ 2 ) Combining this inequality with (2.41) gives ð (x1 âš x2 , ðœ2 ) â ð (x2 , ðœ2 ) > ð (x1 , ðœ1 ) â ð (x1 â§ x2 , ðœ 1 )

(2.42)

However x1 and x2 are optimal for their respective parameter values, that is ð (x2 , ðœ2 ) â¥ ð (x1 âš x2 , ðœ 2 ) =â ð (x1 âš x2 , ðœ 2 ) â ð (x2 , ðœ2 ) â€ 0 ð (x1 , ðœ1 ) â¥ ð (x1 â§ x2 , ðœ 1 ) =â ð (x1 , ðœ 1 ) â ð (x1 â§ x2 , ðœ1 ) â¥ 0 Substituting in (2.42), we conclude 0 â¥ ð (x1 âš x2 , ðœ2 ) â ð (x2 , ðœ2 ) > ð (x1 , ðœ1 ) â ð (x1 â§ x2 , ðœ 1 ) â¥ 0 This contradiction implies that our assumption that x2 ââ¿ð x1 is false. x2 â¿ð x1 as required. ð is always increasing. 2.64 The budget correspondence is descending in p and therefore ascending in âp. Consequently, the indirect utility function ð£(p, ð) =

sup

xâð(p,ð)

ð¢(x)

is increasing in âp, that is decreasing in p. 2.65 â= Let ðœ2 â¿ ðœ 1 and ðº2 â¿ð ðº1 . Select x1 â ð(ðœ 1 , ðº1 ) and x2 â ð(ðœ 2 , ðº2 ). Since ðº2 â¿ð ðº1 , x1 â§ x2 â ðº1 . Since x1 is optimal (x1 â ð(ðœ1 , ðº1 )), ð (x1 , ðœ1 ) â¥ ð (x1 â§ x2 , ðœ1 ). Quasisupermodularity implies ð (x1 âš x2 , ðœ 1 ) â¥ ð (x2 , ðœ1 ). By the single crossing condition ð (x1 âš x2 , ðœ2 ) â¥ ð (x2 , ðœ 2 ). Therefore x1 âš x2 â ð(ðœ2 , ðº2 ). Similarly, since ðº2 â¿ð ðº1 , x1 âš x2 â ðº(ðœ 2 ). But x2 is optimal, which implies that ð (x2 , ðœ2 ) â¥ ð (x1 âš x2 , ðœ2 ) or ð (x1 âš x2 , ðœ 2 ) â€ ð (x2 , ðœ2 ). The single crossing condition implies that a similar inequality holds at ðœ 1 , that is ð (x1 âš x2 , ðœ1 ) â€ ð (x2 , ðœ1 ). Quasisupermodularity implies that ð (x1 , ðœ 1 ) â€ ð (x1 â§x2 , ðœ 1 ). Therefore x1 â§x2 â ð(ðœ 1 , ðº1 ). Since x1 âš x2 â ð(ðœ 2 , ðº2 ) and x1 â§ x2 â ð(ðœ 1 , ðº1 ), ð is increasing in (ðœ, ðº). =â To show that ð is quasisupermodular, suppose that ðœ is ï¬xed. Choose any x1 , x2 â ð. Let ðº1 = {x1 , x1 â§ x2 } and ðº2 = {x2 , x1 âš x2 }. Then ðº2 â¿ð ðº1 . Assume that ð (x1 , ðœ) â¥ ð (x1 â§x2 , ðœ). Then x1 â ð(ðœ, ðº1 ) which implies that x1 âšx2 â ð(ðœ, ðº2 ). (If x2 â ð(ðœ, ðº2 ), then also x1 âš x2 â ð(ðœ, ðº2 ) since ð is increasing in (ðœ, ðº)). But this implies that ð (x1 âš x2 , ðœ) â¥ ð (x2 , ðœ). ð is quasisupermodular in ð. 82

Solutions for Foundations of Mathematical Economics

To show that ð satisï¬es the single crossing condition, choose any x2 â¿ x1 and let ðº = {x1 , x2 }. Assume that ð (x2 , ðœ1 ) â¥ ð (x1 , ðœ 1 ). Then x2 â ð(ðœ1 , ðº) which implies that x2 â ð(ðœ 2 , ðº) for any ðœ 2 â¿ ðœ1 . (If x1 â ð(ðœ 2 , ðº), then also x1 âšx2 = x2 â ð(ðœ 2 , ðº) since ð is increasing in (ðœ, ðº).) But this implies that ð (x2 , ðœ2 ) â¥ ð (x1 , ðœ2 ). ð satisï¬es the single crossing condition. 2.66 First, assume that ð is continuous. Let ð be an open subset in ð and ð = ð â1 (ð ). If ð = â, it is open. Otherwise, choose ð¥0 â ð and let ðŠ0 = ð (ð¥0 ) â ð . Since ð is open, there exists a neighborhood ð (ðŠ0 ) â ð . Since ð is continuous, there exists a corresponding neighborhood ð (ð¥0 ) with ð (ð (ð¥0 )) â ð (ð (ð¥0 )). Since ð (ð (ð¥0 )) â ð , ð (ð¥0 ) â ð. This establishes that for every ð¥0 â ð there exist a neighborhood ð (ð¥0 ) contained in ð. That is, ð is open in ð. Conversely, assume that the inverse image of every open set in ð is open in ð. Choose some ð¥0 â ð and let ðŠ0 = ð (ð¥0 ). Let ð â ð be a neighborhood of ðŠ0 . ð contains an open ball ðµð (ðŠ0 ) about ðŠ0 . By hypothesis, the inverse image ð = ð â1 (ðµð (ðŠ0 )) is open in ð. Therefore, there exists a neighborhood ð (ð¥0 ) â ð â1 (ðµð (ðŠ0 )). Since ðµð (ðŠ0 ) â ð , ð (ð (ð¥0 )) â ð . Since the choice of ð¥0 was arbitrary, we conclude that ð is continuous. 2.67 Assume ð is continuous. Let ð be a closed set in ð and let ð = ð â1 (ð ). Then, ð ð is open. By the previous exercise, ð â1 (ð ð ) = ð ð is open and therefore ð is closed. Conversely, for every open set ð â ð , ð ð is closed. By hypothesis, ð ð = ð â1 (ð ð ) is closed and therefore ð = ð â1 (ð ) is open. ð is continuous by the previous exercise. 2.68 Assume ð is continuous. Let ð¥ð be a sequence converging to ð¥ Let ð be a neighborhood of ð (ð¥). Since ð is continuous, there exists a neighborhood ð â ð¥ such that ð (ð) â ð . Since ð¥ð converges to ð¥, there exists some ð such that ð¥ð â ð for all ð â¥ ð . Consequently ð (ð¥ð ) â ð for every ð â¥ ð . This establishes that ð (ð¥ð ) â ð (ð¥). Conversely, assume that for every sequence ð¥ð â ð¥, ð (ð¥ð ) â ð (ð¥). We show that if ð were not continuous, it would be possible to construct a sequence which violates this hypothesis. Suppose then that ð is not continuous. Then there exists a neighborhood / ð . In ð of ð (ð¥) such that for every neighborhood ð of ð¥, there is ð¥â² â ð with ð (ð¥â² ) â particular, consider the sequence of open balls ðµ1/ð (ð¥). For every ð, choose a point / ð . Then ð¥ð â ð¥ but ð (ð¥ð ) does not converge to ð (ð¥). ð¥ð â ðµ1/ð (ð¥) with ð (ð¥ð ) â This contradicts the assumption. We conclude that ð must be continuous. 2.69 Since ð is one-to-one and onto, it has an inverse ð = ð â1 which maps ð onto ð. Let ð be an open set in ð. Since ð is open, ð = ð â1 (ð) = ð (ð) is open in ð . Therefore ð = ð â1 is continuous. 2.70 Assume ð is continuous. Let (ð¥ð , ðŠ ð ) be a sequence of points in graph(ð ) converging to (ð¥, ðŠ). Then ðŠ ð = ð (ð¥ð ) and ð¥ð â ð¥. Since ð is continuous, ðŠ = ð (ð¥) = limðââ ð (ð¥ð ) = limðââ ðŠ ð . Therefore (ð¥, ðŠ) â graph(ð ) which is therefore closed. 2.71 By the previous exercise, ð continuous implies graph(ð ) closed. Conversely, suppose graph(ð ) is closed and let ð¥ð be a sequence converging to ð¥. Then (ð¥ð , ð (ð¥ð ))) is a sequence in graph(ð ). Since ð is compact, ð (ð¥ð ) contains a subsequence which converges ðŠ. Since graph(ð ) is closed, (ð¥, ðŠ) â graph(ð ) and therefore ðŠ = ð (ð¥) and ð (ð¥ð ) â ð (ð¥). 2.72 Let ð be an open set in ð. Since ð and ð are continuous, ð â1 (ð ) is open in ð and ð â1 (ð â1 (ð )) is open in ð. But ð â1 (ð â1 (ð )) = (ð â ð)â1 (ð ). Therefore ð â ð is continuous. 2.73 Exercises 1.201 and 2.68.

83

Solutions for Foundations of Mathematical Economics

2.74 Let ð¢ be deï¬ned as in Exercise 2.38. Let (xð ) be a sequence converging to x. Let ð§ ð = ð¢(xð ) and ð§ = ð¢(x). We need to show that ð§ ð â ð§. (ð§ ð ) has a convergent subsequence. Let ð§Â¯ = maxð ð¥ð and ð§ = minð ð¥ð . Then ð§ â [ð§, ð§Â¯]. Fix some ð > 0. Since xð â x, there exists some ð such that â¥xð â xâ¥â < ð for every ð â¥ ð . Consequently, for all ð â¥ ð , the terms of the sequence (ð§ ð ) lie in the compact set [ð§ â ð, ð§Â¯ + ð]. Hence, (ð§ ð ) has a convergent subsequence (ð§ ð ). Every convergent subsequence (ð§ ð ) converges to ð§. Suppose not. That is, suppose there exists a convergent subsequence which converges to ð§ â² . Without loss of generality, assume ð§ â² > ð§. Let ð§Ë = 12 (ð§ + ð§ â² ) and let z = ð§1, zâ² = ð§ â² 1, Ëz = ð§Ë1 be the corresponding commodity bundles (see Exercise 2.38). Since ð§ ð â ð§ â² > ð§Ë, there exists some ð such that ð§ ð > ð§Ë for every ð â¥ ð . This implies that xð âŒ zð â» zË for every ð â¥ ð by monotonicity. Now xð â x and continuity of preferences implies that x â¿ Ëz. However x âŒ z which implies that z â¿ zË which contradicts monotonicity, since Ë z > z. Consequently, every convergent subsequence (ð§ ð ) converges to ð§. 2.75 Assume ð is compact. Let ðŠ ð be a sequence in ð (ð). There exists a sequence ð¥ð in ð with ðŠ ð = ð (ð¥ð ). Since ð is compact, it contains a convergent subsequence ð¥ð â ð¥. If ð is continuous, the subsequence ðŠ ð = ð (ð¥ð ) converges in ð (ð) (Exercise 2.68). Therefore ð (ð) is compact. Assume ð is connected but âª ð (ð) is not. This means â© there exists open subsets ðº and ð» in ð such that ð (ð) â ðº ð» and (ðº â© ð (ð)) (ð» â© ð (ð)) = â. This implies that ð = ð â1 (ðº) âª ð â1 (ð») is a disconnection of ð, which contradicts the connectedness of ð. 2.76 Let ð be any open set in ð. Its complement ð ð is closed and therefore compact. Consequently, ð (ð ð ) is compact (Exercise 2.3) and hence closed. Since ð is one-to-one and onto, ð (ð) is the complement of ð (ð ð ), and thus open in ð . Therefore, ð is an open mapping. By Exercise 2.69, ð â1 is continuous and ð is a homeomorphism. 2.77 Assume ð continuous. The sets {ð (ð¥) â¥ ð} and {ð (ð¥) â€ ð} are closed subsets of the â and hence â¿(ð) = ð â1 {ð (ð¥) â¥ ð} and âŸ(ð) = ð â1 {ð (ð¥) â€ ð} are closed subsets of ð (Exercise 2.67). Conversely, assume that all upper â¿(ð) and lower âŸ(ð) contour sets are closed. This implies that the sets â»(ð) and âº(ð) are open. Let ðŽ be an open set in â. Then for every ð â ðŽ, there exists an open ball ðµðð (ð) â ðŽ âª ðŽ= ðµðð (ð) ðâðŽ

For every ð â ðŽ, ðµðð (ð) = (ð â ðð , ð + ðð ) and ð â1 (ðµðð (ð)) = â»(ð â ðð ) â© âº(ð + ðð ) which is open. Consequently âª âª ( ) â»(ð â ðð ) â© âº(ð + ðð ) ð â1 (ðµðð (ð)) = ð â1 (ðŽ) = ðâðŽ

ðâðŽ

is open. ð is continuous by Exercise 2.66. 84

Solutions for Foundations of Mathematical Economics

2.78 Choose any ð¥0 â ð and ð > 0. Since ð is continuous, there exists ð¿1 such that ð(ð¥, ð¥0 ) < ð¿1 =â â£ð (ð¥) â ð (ð¥0 )â£ < ð/2 Similarly, there exists ð¿2 such that ð(ð¥, ð¥0 ) < ð¿2 =â â£ð(ð¥) â ð(ð¥0 )â£ < ð/2 Let ð¿ = min{ð¿1 , ð¿2 }. Then, provided ð(ð¥, ð¥0 ) < ð¿ â£(ð + ð)(ð¥) â (ð + ð)(ð¥0 )â£ = â£ð (ð¥) + ð(ð¥) â ð (ð¥0 ) â ð(ð¥0 )â£ â€ â£ð (ð¥) â ð (ð¥0 )â£ + â£ð(ð¥) â ð(ð¥0 )â£ 0 such that â£ð (ð¥) â ð (ð¥0 )â£ < ð and â£ð(ð¥) â ð(ð¥0 )â£ < ð whenever ð(ð¥, ð¥0 ) < ð¿. Consequently, while ð(ð¥, ð¥0 ) < ð¿ â£ð (ð¥)â£ â€ â£ð (ð¥) â ð (ð¥0 )â£ + â£ð (ð¥0 )â£ < ð + â£ð (ð¥0 )â£ â€ 1 + â£ð (ð¥0 )â£ and â£(ð ð)(ð¥) â (ð ð)(ð¥0 )â£ = â£ð (ð¥)ð(ð¥) â ð (ð¥0 )ð(ð¥0 )â£ = â£ð (ð¥)(ð(ð¥) â ð(ð¥0 )) + ð(ð¥0 )(ð (ð¥) â ð (ð¥0 ))â£ â€ â£ð (ð¥)â£ â£ð(ð¥) â ð(ð¥0 )â£ + â£ð(ð¥0 )â£ â£ð (ð¥) â ð (ð¥0 )â£ < ð(1 + â£ð (ð¥0 )â£ + â£ð(ð¥0 )â£) Given ð > 0, let ð = min{1, ð/(1 + â£ð (ð¥0 )â£ + â£ð(ð¥0 )â£)}. Then, we have shown that there exists ð¿ > 0 such that ð(ð¥, ð¥0 ) < ð¿ =â â£(ð ð)(ð¥) â (ð ð)(ð¥0 )â£ < ð Therefore, ð ð is continuous at ð¥0 . 2.80 Apply Exercises 2.78 and 2.72. 2.81 For any ð â â, the upper and lower contour sets of ð âš ð, namely { ð¥ : max{ð (ð¥), ð(ð¥)} â¥ ð} = {ð¥ : ð (ð¥) â¥ ð } âª { ð¥ : ð(ð¥) â¥ ð } { ð¥ : max{ð (ð¥), ð(ð¥)} â€ ð} = {ð¥ : ð (ð¥) â€ ð } â© { ð¥ : ð(ð¥) â€ ð } are closed. Therefore ð âš ð is continuous (Exercise 2.77). Similarly for ð â§ ð. 2.82 The set ð = ð (ð) is compact (Proposition 2.3). We want to show that ð has both largest and smallest elements. Assume otherwise, that is assume that ð has no largest element. Then, the set of intervals {(ââ, ð¡) : ð¡ â ð } forms an open covering of ð . Since ð is compact, there exists a ï¬nite subcollection of intervals {(ââ, ð¡1 ), (ââ, ð¡2 ), . . . , (ââ, ð¡ð )} which covers ð . Let ð¡â be the largest of these ð¡ð . Then ð¡â does not belong to any of the intervals {(ââ, ð¡1 ), (ââ, ð¡2 ), . . . , (ââ, ð¡ð )}, contrary to the fact that they cover ð . This contradiction shows that, contrary to our assumption, there must exist a largest element ð¡â â ð , that is ð¡â â¥ ð¡ for all ð¡ â ð . Let ð¥â â ð â1 (ð¡â ). Then ð¡â = ð (ð¥â ) â¥ ð (ð¥) for all ð¥ â ð. The existence of a smallest element is proved analogously. 85

Solutions for Foundations of Mathematical Economics

2.83 By Proposition 2.3, ð (ð) is connected and hence an interval (Exercise 1.95). 2.84 The range ð (ð) is a compact subset of â (Proposition 2.3). Therefore ð is bounded (Proposition 1.1). Ë 2.85 Let ð¶(ð) denote the set of all continuous (not necessarily bounded) functionals on ð. Then Ë ð¶(ð) = ðµ(ð) â© ð¶(ð) Ë ðµ(ð), ð¶(ð) are a linear subspaces of the set of all functionals ð¹ (ð) (Exercises 2.11, Ë 2.78 respectively). Therefore ð¶(ð) = ðµ(ð) â© ð¶(ð) is a subspace of ð¹ (ð) (Exercise 1.130). Clearly ð¶(ð) â ðµ(ð). Therefore ð¶(ð) is a linear subspace of ðµ(ð). Let ð be a bounded function in the closure of ð¶(ð), that is ð â ð¶(ð). We show that ð is continuous. For any ð > 0, there exists ð0 â ð¶(ð) such that â¥ð â ð0 â¥ < ð/3. Therefore â£ð (ð¥) â ð0 (ð¥)â£ < ð/3 for every ð¥ â ð. Choose some ð¥0 â ð. Since ð0 is continuous, there exists ð¿ > 0 such that ð(ð¥, ð¥0 ) < ð¿ =â â£ð0 (ð¥) â ð0 (ð¥0 )â£ < ð/3 Therefore, for every ð¥ â ð such that ð(ð¥, ð¥0 ) < ð¿ â£ð (ð¥) â ð (ð¥0 )â£ = â£ð (ð¥) â ð0 (ð¥) + ð0 (ð¥) â ð0 (ð¥0 ) + ð0 (ð¥0 ) â ð (ð¥0 )â£ â€ â£ð (ð¥) â ð0 (ð¥)â£ + â£ð0 (ð¥) â ð0 (ð¥0 )â£ + â£ð0 (ð¥0 ) â ð (ð¥0 )â£ < ð/3 + ð/3 + ð/3 = ð Therefore ð is continuous at ð¥0 . Since ð¥0 was arbitrary, we conclude that is continuous everywhere, that is ð â ð¶(ð). Therefore ð¶(ð) = ð¶(ð) and ð¶(ð) is closed in ðµ(ð). Since ðµ(ð) is complete (Exercise 2.11), we conclude that ð¶(ð) is complete (Exercise 1.107). Therefore ð¶(ð) is a Banach space. 2.86 For every ðŒ â â, { ð¥ : ð (ð¥) â¥ ðŒ } = {ð¥ : âð (ð¥) â€ âðŒ } and therefore { ð¥ : ð (ð¥) â¥ ðŒ } is closed ââ {ð¥ : âð (ð¥) â€ âðŒ } is closed 2.87 Exercise 2.77. 2.88 1 implies 2 Suppose ð is upper semi-continuous. Let ð¥ð be a sequence converging to ð¥0 . Assume ð (ð¥ð ) â ð. For every ðŒ < ð, there exists some ð such that ð (ð¥ð ) > ðŒ for every ð â¥ ð . Hence ð¥0 â { ð¥ : ð (ð¥) â¥ ðŒ } = { ð¥ : ð (ð¥) â¥ ðŒ } since ð is upper semi-continuous. That is, ð (ð¥0 ) â¥ ðŒ for every ðŒ < ð. Hence ð (ð¥0 ) â¥ ð = limðââ ð (ð¥ð ). 2 implies 3 Let (ð¥ð , ðŠ ð ) be a sequence in hypo ð which converges to (ð¥, ðŠ). That is, ð¥ð â ð¥, ðŠ ð â ðŠ and ðŠ ð â€ ð (ð¥ð ). Condition 2 implies that ð (ð¥) â¥ ðŠ. Hence, (ð¥, ðŠ) â hypo ð . Therefore hypo ð is closed. 3 implies 1 For ï¬xed ðŒ â â, let ð¥ð be a sequence in { ð¥ : ð (ð¥) â¥ ðŒ }. Suppose ð¥ð â ð¥0 . Then, the sequence (ð¥ð , ðŒ) converges to (ð¥0 , ðŒ) â hypo ð . Hence ð (ð¥0 ) â¥ ðŒ and ð¥0 â { ð¥ : ð (ð¥) â¥ ðŒ }, which is therefore closed (Exercise 1.106). 86

Solutions for Foundations of Mathematical Economics 2.89 Let ð = supð¥âð ð (ð¥), so that ð (ð¥) â€ ð for every ð¥ â ð

(2.43)

There exists a sequence ð¥ð in ð with ð (ð¥ð ) â ð . Since ð is compact, there exists a convergent subsequence ð¥ð â ð¥â and ð (ð¥ð ) â ð . However, since ð is upper semi-continuous, ð (ð¥â ) â¥ lim ð (ð¥ð ) = ð . Combined with (2.43), we conclude that ð (ð¥â ) = ð . 2.90 Choose some ð > 0. Since ð is uniformly continuous, there exists some ð¿ > 0 such that ð(ð (ð¥ð ), ð (ð¥ð )) < ð for every ð¥ð , ð¥ð â ð such that ð(ð¥ð , ð¥ð ) < ð¿. Let (ð¥ð ) be a Cauchy sequence in ð. There exists some ð such that ð(ð¥ð , ð¥ð ) < ð¿ for every ð, ð â¥ ð . Uniform continuity implies that ð(ð (ð¥ð ), ð (ð¥ð )) < ð for every ð, ð â¥ ð . (ð (ð¥ð )) is a Cauchy sequence. 2.91 Suppose not. That is, suppose ð is continuous but not uniformly continuous. Then there exists some ð > 0 such that for ð = 1, 2, . . . , there exist points ð¥1ð , ð¥2ð such that ð(ð¥1ð , ð¥2ð ) < 1/ð but ð(ð (ð¥1ð ), ð (ð¥2ð )) â¥ ð

(2.44)

Since ð is compact, (ð¥1ð ) has a subsequence (ð¥1ð ) converging to some ð¥ â ð. By construction (ð(ð¥1ð , ð¥2ð ) < 1/ð), the sequence (ð¥2ð ) also converges to ð¥ and by continuity lim ð (ð¥1ð ) = lim ð (ð¥2ð )

ðââ

ðââ

which contradicts (2.44). 2.92 Assume ð is Lipschitz with constant ðœ. For any ð > 0, let ð¿ = ð/2ðœ. Then, provided ð(ð¥, ð¥0 ) â€ ð¿ ð(ð (ð¥), ð (ð¥0 )) â€ ðœð(ð¥, ð¥0 ) = ðœð¿ = ðœ

ð ð = 0. ð¹ is totally bounded (Exercise 1.113), so that there exist ï¬nite set of functions {ð1 , ð2 , . . . , ðð } in F such that ð

min â¥ð â ðð â¥ â€ ð/3 ð=1

Each ðð is uniformly continuous (Exercise 2.91), so that there exists ð¿ð > 0 such that ð(ð¥, ð¥0 ) â€ ð¿ =â ð(ðð (ð¥), ðð (ð¥0 ) < ð/3 Let ð¿ = min{ð¿1 , ð¿2 , . . . , ð¿ð }. Given any ð â ð¹ , let ð be such that â¥ð â ðð â¥ < ð/3. Then for any ð¥, ð¥0 â ð, ð(ð¥, ð¥0 ) â€ ð¿ implies ð(ð (ð¥), ð (ð¥0 ) â€ ð(ð (ð¥), ðð (ð¥)) + ð(ðð (ð¥), ðð (ð¥0 )) + ð(ðð (ð¥0 ), ð (ð¥0 )) <

ð ð ð + + =ð 3 3 3

for every ð â ð¹ . Therefore, ð¹ is equicontinuous. Conversely, assume that ð¹ â ð¶(ð) is closed, bounded and equicontinuous. Let (ðð ) be a bounded equicontinuous sequence of functions in ð¹ . We show that (ðð ) has a convergent subsequence. 1. First, we show that for any ð > 0, there is exists a subsequence (ðð ) such that â¥ðð â ððâ² â¥ < ð for every ðð , ððâ² in the subsequence. Since the functions are equicontinuous, there exists ð¿ > 0 such that ð(ðð (ð¥) â ðð (ð¥0 ) <

ð 3

for every ð¥, ð¥0 in ð with ð(ð¥, ð¥0 ) â€ ð¿. Since ð is compact, it is totally bounded (Exercise 1.113). That is, there exist a ï¬nite number of open balls ðµð¿ (ð¥ð ), ð = 1, 2 . . . , ð which cover ð. The sequence (ðð (ð¥1 ), ðð (ð¥2 , . . . , ðð (ð¥ð )) is a bounded sequence in âð . By the Bolzano-Weierstrass theorem (Exercise 1.119), this sequence has a convergent subsequence (ðð (ð¥1 ), ðð (ð¥2 ), . . . , ðð (ð¥ð )) such that ðð (ð¥ð ) â ððâ² (ð¥ð ) < ð/3 for ð and every ðð , ððâ² in the subsequence. Consequently, for any ð¥ â ð, there exists ð such that ð(ðð (ð¥), ððâ² (ð¥) â€ ð(ðð (ð¥), ðð (ð¥ð )) + ð(ðð (ð¥ð ), ððâ² (ð¥ð )) + ð(ððâ² (ð¥ð ), ððâ² (ð¥)) ð ð ð < + + =ð 3 3 3 That is, â¥ðð â ððâ² â¥ < ð for every ðð , ððâ² in the subsequence. 2. Choose a ball ðµ1 of radius 1 in ð¶(ð) which contains inï¬nitely many elements of (ðð ). Applying step 1, there exists a ball ðµ2 of radius 1/2 containing inï¬nitely many elements of (ðð ). Proceeding in this fashion, we obtain a nested sequence ðµ1 â ðµ2 â . . . of balls in ð¶(ð) such that (a) ð(ðµð ) â 0 and (b) each ðµð contains inï¬nitely many terms of (ðð ). Choosing ððð â ðµð gives a convergent subsequence. 88

Solutions for Foundations of Mathematical Economics

2.96 Let ð â ð¹ . Then for every ð > 0 there exists ð¿ > 0 and ð â ð¹ such that â¥ð â ðâ¥ < ð/3 and ð(ð¥, ð¥0 ) â€ ð¿ =â ð(ð (ð¥), ð (ð¥0 ) < ð/3 so that if ð(ð¥, ð¥0 ) â€ ð¿ â¥ð(ð¥) â ð(ð¥0 )â¥ â€ â¥ð (ð¥) â ð(ð¥)â¥ + â¥ð (ð¥) â ð (ð¥0 )â¥ + â¥ð (ð¥0 ) â ð(ð¥0 )â¥ <

ð ð ð + + =ð 3 3 3

2.97 For every ð â ð ðâ (ð ð ) = { ð¥ â ð : ð(ð¥) â© ð ð â= â } ð+ (ð ) = { ð¥ â ð : ð(ð¥) â ð } For every x â ð either ð(ð¥) â ð or ð(ð¥) â© ð ð â= â but not both. Therefore ð+ (ð ) âª ðâ (ð ð ) = ð ð+ (ð ) â© ðâ (ð ð ) = â That is

( )ð ð+ (ð ) = ðâ (ð ð )

2.98 Assume ð¥ â ð(ð )â1 . Then ð(ð¥) = ð , ð(ð¥) â ð and ð¥ â ð+ (ð ). Now assume ð¥ â ð+ (ð ) so that ð(ð¥) â ð . Consequently, ð(ð¥) â© ð = ð(ð¥) â= â and ð¥ â ðâ (ð ). 2.99 The respective inverses are: {ð¡1 } {ð¡2 } {ð¡1 , ð¡2 } {ð¡2 , ð¡3 } {ð¡1 , ð¡2 , ð¡3 }

ðâ1 2 â â {ð 1 } {ð 2 } â

ð+ 2 â â {ð 1 } {ð 2 } {ð 1 , ð 2 }

ðâ 2 {ð 1 } {ð 1 , ð 2 } {ð 1 , ð 2 } {ð 1 , ð 2 } {ð 1 , ð 2 }

2.100 Let ð be an open interval meeting ð(1), that is ð(1) â© ð â= â. Since ð(1) = {1}, we must have 1 â ð and therefore ð(ð¥) â© ð â= â for every ð¥ â ð. Therefore ð is lhc at ð¥ = 1. On the other hand, the open interval ð = (1/2, 3/2) contains ð(1) but it does not contain ð(ð¥) for any ð¥ > 1. Therefore, ð is not uhc at ð¥ = 1. 2.101 Choose any open set ð â ð and ð¥ â ð. Since ð(ð¥) = ðŸ = ð(ð¥â² ) for every ð¥, ð¥â² â ð â ð(ð¥) â ð if and only if ð(ð¥â² ) â ð for every ð¥, ð¥â² â ð â ð(ð¥) â© ð â= â if and only if ð(ð¥â² ) â© ð â= â for every ð¥, ð¥â² â ð. Consequently, ð is both uhc and lhc at all ð¥ â ð. 2.102 First assume that the ð is uhc. Let ð be any open subset in ð and ð = ð+ (ð ). If ð = â, it is open. Otherwise, choose ð¥0 â ð so that ð(ð¥0 ) â ð . Since ð is uhc, there exists a neighborhood ð(ð¥0 ) such that ð(ð¥) â ð for every ð¥ â ð(ð¥0 ). That is, ð(ð¥0 ) â ð+ (ð ) = ð. This establishes that for every ð¥0 â ð there exist a neighborhood ð(ð¥0 ) contained in ð. That is, ð is open in ð. Conversely, assume that the upper inverse of every open set in ð is open in ð. Choose some ð¥0 â ð and let ð be an open set containing ð(ð¥0 ). Let ð = ð+ (ð ). ð is an open set containing ð¥0 . That is, ð is a neighborhood of ð¥0 with ð(ð¥) â ð for every ð¥ â ð. Since the choice of ð¥0 was arbitrary, we conclude that ð is uhc. The lhc case is analogous. 89

Solutions for Foundations of Mathematical Economics

Solutions for Foundations of Mathematical Economics 2.106

1. Assume ð is closed. For any ð¥ â ð, let (ðŠ ð ) be a sequence in ð(ð¥). Since ð is closed, ðŠ ð â ðŠ â ð(ð¥). Therefore ð(ð¥) is closed.

2. Assume ð is closed-valued and uhc. Choose any (ð¥, ðŠ) â / graph(ð). Since ð(ð¥) is closed, there exist disjoint open sets ð1 and ð2 in ð such that ðŠ â ð1 and ð(ð¥) â ð2 (Exercise 1.93). Since ð is uhc, ð+ (ð2 ) is a neighborhood of ð¥. Therefore ð+ (ð2 ) Ã ð1 is a neighborhood of (ð¥, ðŠ) disjoint from graph(ð). Therefore the complement of graph(ð) is open, which implies that graph(ð) is closed. 3. Since ð is closed and ð compact, ð is compact-valued. Let (ð¥ð ) â ð¥ be a sequence in ð and (ðŠ ð ) a sequence in ð with ðŠ ð â ð(ð¥ð ). Since ð is compact, there exists a subsequence ðŠ ð â ðŠ. Since ð is closed, ðŠ â ð(ð¥). Therefore, by Exercise 2.104, ð is uhc. 2.107 Assume ð is closed-valued and uhc. Then ð is closed (Exercise 2.106). Conversely, if ð is closed, then ð(ð¥) is closed for every ð¥ (Exercise 2.106). If ð is compact, then ð is compact-valued (Exercise 1.110). By Exercise 2.104, ð is uhc. 2.108 ð1 is closed-valued (Exercise 2.106). Similarly, ð2 is closed-valued (Proposition 1.1). Therefore, for every ð¥ â ð, ð(ð¥) = ð1 (ð¥) â© ð2 (ð¥) is closed (Exercise 1.85) and hence compact (Exercise 1.110). Hence ð is compact-valued. Now, for any ð¥0 â ð, let ð be an open neighborhood of ð(ð¥0 ). We need to show that there is a neighborhood ð of ð¥0 such that ð(ð) â ð . Case 1 ð â ð2 (ð¥0 ): Since ð2 is uhc, there exists a neighborhood such that ð â ð¥0 such that ð2 (ð) â ð which implies that ð(ð) â ð2 (ð) â ð Case 2 ð ââ ð2 (ð¥0 ): Let ðŸ = ð2 (ð¥0 ) â ð â= â. For every ðŠ â ðŸ, there exist neighborhoods ððŠ (ð¥0 ) and ð (ðŠ) such that ð1 (ððŠ (ð¥0 )) â© ð (ðŠ) = â (Exercise 1.93). The sets ð (ðŠ) constitute an open covering of ðŸ. Since ðŸ is compact, there exists a ï¬nite subcover, that is there exists a ï¬nite number of elements ðŠ1 , ðŠ2 , . . . ðŠð such that ð âª

ðŸâ

ð (ðŠð )

ð=1

âªð Let ð (ðŸ) denote ð=1 ð (ðŠð ). Note that ð âªð (ðŸ) is an open set containing ð2 (ð¥0 ). Since ð2 is uhc, there exists a neighborhood ð â² (ð¥0 ) such that ð2 (ð â² (ð¥0 )) â ð âª ð (ðŸ). Let ð(ð¥0 ) =

ððŠð (ð¥0 ) â© ð â² (ð¥0 )

ð=1

ð(ð¥0 ) is an open neighborhood of ð¥0 for which ð1 (ð(ð¥0 )) â© ð (ðŸ) = â and ð2 (ð(ð¥0 )) â ð âª ð (ðŸ) from which we conclude that ð(ð(ð¥0 )) = ð1 (ð(ð¥0 )) â© ð2 (ð(ð¥0 )) â ð âð 2.109 1. Let x â ð(p, ð) â© ð . Then x â ð(p, ð) and ð=1 ðð ð¥ð â€ ð. Since ð is Ë = ðŒx â ð and open, there exists ðŒ < 1 such that x ð â ð=1

ðð ð¥Ëð = ðŒ

ð â

ðð ð¥ð <

ð=1

91

ð â ð=1

ðð ð¥ð â€ ð

Solutions for Foundations of Mathematical Economics

2. (a) Suppose that ð(p, ð) is not lhc. Then for every neighborhood ð of (p, ð), there exists (pâ² , ðâ² ) â ð such that ð(pâ² , ðâ² ) â© ð = â. In particular, for every open ball ðµð (p, ð), there exists a point (pð , ðð ) â ðµð (p, ð) such that ð(pð , ðð ) â© ð = â. ((pð , ðð )) is the required sequence. (b) By construction, â¥pð â pâ¥ < 1/ð â 0 which implies that ððð â ðð for every ð. Therefore (Exercise 1.202) â â Ëð â ðð ð¥Ëð < ð and ðð â ð ððð ð¥ and therefore there exists ð such that â Ë ð < ðð ðð ð ð¥ which implies that Ë â ð(pð , ðð ) x (c) Also by construction ð(pð , ðð ) â© ð = â which implies ð(pð , ðð ) â ð ð and therefore Ë â ð(pð , ðð ) =â x Ëâ x /ð Ë â The assumption that ð(p, ð) is not lhc at (p, ð) implies that x / ð , contraË â ð. dicting the conclusion in part 1 that x 3. This contradiction establishes that (p, ð) is lhc at (p, ð). Since the choice of (p, ð) was arbitrary, we conclude that the budget correspondence ð(p, ð) is lhc for all (p, ð) â ð (assuming ð = âð+ ). 4. In the previous example (Example 2.89), we have shown that ð(p, ð) is uhc. Hence, â the budget correspondence is continuous for all (p, ð) such that ð > ð inf xâð ð=1 ðð ð¥ð . 2.110 We give two alternative proofs. Proof 1 Let ð = {ð} be an open cover of ð(ðŸ). For every ð¥ â ðŸ, ð(ð¥) â ð(ðŸ) is compact and hence can be covered by a ï¬nite number of the sets ð â ð. Let ðð¥ denote the union of the ï¬nite cover of ð(ð¥). Since ð is uhc, every ð+ (ðð¥ ) is open in ð. Therefore { ð+ (ðð¥ ) : ð¥ â ðŸ } is an open covering of ðŸ. If ðŸ is compact, it contains an ï¬nite covering { ð+ (ðð¥1 ), ð+ (ðð¥2 ), . . . , ð+ (ðð¥ð ) }. The sets ðð¥1 , ðð¥2 , . . . , ðð¥ð are a ï¬nite subcovering of ð(ðŸ). Proof 2 Let (ðŠ ð ) be a sequence in ð(ðŸ). We have to show that (ðŠ ð ) has a convergent subsequence with a limit in ð(ðŸ). For every ðŠ ð , there is an ð¥ð with ðŠ ð â ð(ð¥ð ). Since ðŸ is compact, the sequence (ð¥ð ) has a convergent subsequence ð¥ð â ð¥ â ðŸ. Since ð is uhc, the sequence (ðŠ ð ) has a subsequence (ðŠ ð ) which converges to ðŠ â ð(ð¥) â ð(ðŸ). Hence the original sequence (ðŠ ð ) has a convergent subsequence. 2.111 The sets ð, ð(ð), ð2 (ð), . . . form a sequence of nonempty compact sets. Since ð(ð) â ð, ð2 (ð) â ð(ð) and so on, the sequence of sets ðð ð is decreasing. Let ðŸ=

ðð (ð)

ð=1

By the nested intersection theorem (Exercise 1.117), ðŸ â= â. Since ðŸ â ððâ1 (ð), ð(ðŸ) â ðð (ð) for every ð, which implies that ð(ðŸ) â ðŸ. 92

Solutions for Foundations of Mathematical Economics To show that ðŸ that ðŠ â ð(ð¥ð ). ð¥ð â ðð (ð) for (Exercise 2.107),

â ð(ðŸ), let ðŠ â ðŸ. For every ð there exists an ð¥ð â ðð (ð) such Since ð is compact, there exists a subsequence ð¥ð â ð¥0 . Since every ð, ð¥0 â ðŸ. The sequence (ð¥ð , ðŠ) â (ð¥0 , ðŠ). Since ð is closed ðŠ â ð(ð¥0 ). Therefore ðŠ â ð(ðŸ) which implies that ðŸ â ð(ðŸ).

2.112 ð(ð¥) is compact for every ð¥ â ð by Tychonoï¬âs theorem (Proposition 1.2). Let ð¥ð â ð¥ be a sequence in ð and let ðŠ ð = (ðŠ1ð , ðŠ2ð , . . . , ðŠðð ) with ðŠðð â ð(ð¥ð ) be a corresponding sequence of points in ð . For each ðŠðð , ð = 1, 2, . . . , ð, there exists a â² subsequence ðŠðð â ðŠð with ðŠð â ðð (ð¥) (Exercise 2.104). Therefore ðŠ = (ðŠ1 , ðŠ2 , . . . , ðŠð ) â ð(ð¥) which implies that ð is uhc. 2.113 Let ð£ â ð¶(ð). For every x â ð, the maximand ð (ð¥, ðŠ) + ðœð£(ðŠ) is a continuous function on a compact set ðº(ð¥). Therefore the supremum is attained, and max can replace sup in the deï¬nition of the operator ð (Theorem 2.2). ð ð£ is the value function for the constrained optimization problem max { ð (ð¥, ðŠ) + ðœð£(ðŠ) }

ðŠâðº(ð¥)

satisfying the requirements of the continuous maximum theorem (Theorem 2.3), which ensures that ð ð£ is continuous on ð. We have previously shown that ð ð£ is bounded (Exercise 2.18). Therefore ð ð£ â ð¶(ð). 2.114

1. ð has a least upper bound since ð is a complete lattice. Let ð â = sup ð. Then ð â = â¿(ð â ) is a complete sublattice of ð (Exercise 1.48).

2. For every ð  â ð, ð  âŸ ð â and since ð is increasing and ð  is a ï¬xed point ð  = ð (ð ) âŸ ð (ð â ) Therefore ð (ð â ) â ð â . (ð (ð â ) is an upper bound of ð). Again, since ð is increasing, this implies that ð (ð¥) â¿ ð (ð â ) for every ð¥ â ð â . Therefore ð (ð â ) â ð â . 3. Let ð be the restriction of ð to the sublattice ð â . Since ð (ð â ) â ð â , ð is an increasing function on a complete lattice. Applying Theorem 2.4, ð has a smallest ï¬xed point ð¥ Ë. 4. ð¥ Ë is a ï¬xed point of ð , that is ð¥ Ë â ðž. Furthermore, ð¥Ë â ð â . Therefore ð¥Ë is an upper bound for ð in ðž. Moreover, ð¥ Ë is the smallest ï¬xed point of ð in ð â . Therefore, ð¥ Ë is the least upper bound of ð in ðž. 5. By Exercise 1.47, this implies that ðž is a complete lattice. In Example 2.91, if ð = {(2, 1), (1, 2)}, ð â = {(2, 2), (3, 2), (2, 3), (3, 3)} and ð¥Ë = (3, 3). 2.115

1. For every ð¥ â ð , there exists some ðŠð¥ â ð(ð¥) such that ðŠð¥ âŸ ð¥. Moreover, ð¥) such that since ð is increasing and ð¥ Ë âŸ ð¥, there exists some ð§ð¥ â ð(Ë ð§ð¥ âŸ ðŠð¥ âŸ ð¥ for every ð¥ â ð

2. Let ð§Ë = inf{ð§ð¥ } Ë. (a) Since ð§ð¥ âŸ ð¥ for every ð¥ â ð , ð§Ë = inf{ð§ð¥ } âŸ inf{ð¥} = ð¥ (b) Since ð(Ë ð¥) is a complete sublattice of ð, ð§Ë = inf{ð§ð¥ } â ð(Ë ð¥). 3. Therefore, ð¥ Ë â ð. 4. Since ð§Ë âŸ ð¥Ë and ð is increasing, there exists some ðŠ â ð(Ë ð§ ) such that ðŠ âŸ ð§Ë â ð(Ë ð¥) Hence ð§Ë â ð . 93

Solutions for Foundations of Mathematical Economics

5. This implies that ð¥ Ë âŸ ð§Ë. Therefore ð¥ Ë = ð§Ë â ð(Ë ð¥) ð¥ Ë is a ï¬xed point of ð. 6. Since ðž â ð , ð¥Ë = inf ð is the least ï¬xed point of ð. 2.116

1. Let ð â ðž and ð â = sup ð. For every ð¥ â ð, ð¥ â ð(ð¥). Since ð is increasing, there exists some ð§ð¥ â ð(ð â ) such that ð§ð¥ â¿ ð¥.

2. Let ð§ â = sup ð§ð¥ . Then (a) Since ð§ð¥ â¿ ð¥ for every ð¥ â ð, ð§ â = sup ð§ð¥ â¿ sup ð¥ = ð â (b) ð§ â â ð(ð â ) since ð(ð â ) is a complete sublattice. 3. Deï¬ne ð â = { ð¥ â ð : ð¥ â¿ ð  for every ð  â ð } ð â is the set of all upper bounds of ð in ð. Then ð â is a complete lattice, since ð â = â¿(ð â ) 4. Let ð : ð â â ð â be the correspondence ð(ð¥) = ð(ð¥) â© ð(ð¥) where ð : ð â â ð â is the constant correspondence deï¬ned by ð(ð¥) = ð â for every ð¥ â ð â . Then (a) Since ð is increasing, for every ð¥ â¿ ð â , there exists some ðŠð¥ â ð(ð¥) such that ðŠð¥ â¿ ð â . Therefore ð(ð¥) â= â for every ð¥ â ð â . (b) Both ð(ð¥) and ð(ð¥) are complete sublattices for every ð¥ â ð â . Therefore ð(ð¥) is a complete sublattice for every ð¥ â ð â . (c) Since both ð and ð are increasing on ð â , ð is increasing on ð â (Exercise 2.47). 5. By the previous exercise, ð has a least ï¬xed point ð¥Ë. 6. ð¥ Ë â ð â is an upper bound of ð. Therefore ð¥ Ë is the least upper bound of ð in ðž. 7. By the previous exercise, ðž has a least element. Since we have shown every subset ð â ðž has a least upper bound, this establishes that ðž is complete lattice (Exercise 1.47). 2.117 For any ð, let a1âð , a2âð â ðŽâð with a2âð â¿ a1âð . Let ð Â¯1ð = ð (a1âð ) and ð Â¯2ð = ð (a2âð ). 2 1 1 1 We want to show that ð Â¯ð â¿ ð Â¯ð . Since ð Â¯ð â ðµ(aâð ) and ðµ(aâð ) is increasing, there Â¯1ð . (Exercise 2.44). Therefore exists some ðð â ðµ(a2âð ) such that ðð â¿ ð sup ðµ(aâð ) = ð Â¯2ð â¿ ðð â¿ ð Â¯1ð ðÂ¯ð is increasing. 2.118 For any player ð, their best response correspondence ðµð (aâð ) is 1. increasing by the monotone maximum theorem (Theorem 2.1). 2. a complete sublattice of ðŽð for every aâð â ðŽâð (Corollary 2.1.1). 94

Solutions for Foundations of Mathematical Economics The joint best response correspondence

ðµ(a) = ðµ1 (aâ1 ) Ã ðµ2 (aâ2 ) Ã â â â Ã ðµð (aâð ) is also 1. increasing (Exercise 2.46) 2. a complete sublattice of ðŽ for every a â ðŽ Therefore, the best response correspondence ðµ(a) satisï¬es the conditions of Zhouâs theorem, which implies that the set ðž of ï¬xed points of ðµ is a nonempty complete lattice. ðž is precisely the set of Nash equilibria of the game. 2.119 In proving the theorem, we showed that ð(ð¥ð , ð¥ð+ð ) â€

ðœð ð(ð¥0 , ð¥1 ) 1âðœ

for every ð, ð â¥ 0. Letting ð â â, ð¥ð+ð â ð¥ and therefore ð(ð¥ð , ð¥) â€

ðœð ð(ð¥0 , ð¥1 ) 1âðœ

Similarly, for every ð, ð â¥ 0 ð(ð¥ð , ð¥ð+ð ) â€ ð(ð¥ð , ð¥ð+1 ) + ð(ð¥ð+1 , ð¥ð+2 ) + â â â + ð(ð¥ð+ðâ1 , ð¥ð+ð ) â€ (ðœ + ðœ 2 + â â â + ðœ ð )ð(ð¥ðâ1 , ð¥ð ) â€

ðœ(1 â ðœ ð ) ð(ð¥ðâ1 , ð¥ð ) 1âðœ

Letting ð â â, ð¥ð+ð â ð¥ and ðœ ð â 0 so that ð(ð¥ð , ð¥) â€

ðœ ð(ð¥ðâ1 , ð¥ð ) 1âðœ

2.120 First observe that ð (ð¥) â¥ 1 for every ð¥ â¥ 1. Therefore ð : ð â ð. For any ð¥, ð§ â ð ð¥ â ðŠ + ð¥2 â ð (ð¥) â ð (ðŠ) = ð¥âðŠ 2(ð¥ â ðŠ) Since

1 ð¥ðŠ

=

1 1 â 2 ð¥ðŠ

â€ 1 for all ð¥, ðŠ â ð â

so that

2 ðŠ

ð (ð¥) â ð (ðŠ) 1 1 â€ â€ 2 ð¥âðŠ 2

   ð (ð¥) â ð (ðŠ)  â£ð (ð¥) â ð (ðŠ)â£ 1  = â€  ð¥âðŠ  â£ð¥ â ðŠâ£ 2

or â£ð (ð¥) â ð (ðŠ)â£ â€ ð is a contraction on ð with modulus 1/2. 95

1 â£ð¥ â ðŠâ£ 2

Solutions for Foundations of Mathematical Economics

ð is closed and hence complete (Exercise 1.107). Therefore, ð has a ï¬xed point. That is, there exists ð¥0 â ð such that ð¥0 = ð (ð¥0 ) =

1 2 (ð¥0 + ) 2 ð¥0

Rearranging

so that ð¥0 =

2ð¥20 = ð¥20 + 2 =â ð¥20 = 2

â 2.

Letting ð¥0 = 2 ð¥1 =

3 1 (2 + 1) = 2 2

Using the error bounds in Corollary 2.5.1, â ðœð ð(ð¥ð , 2) â€ ð(ð¥0 , ð¥1 ) 1âðœ (1/2)ð = 1/2 1/2 1 = ð 2 1 < 0.001 = 1024 when ð = 10. Therefore, we conclude that 10 iterations are ample to reduce the error below 0.001. Actually, with experience, we can reï¬ne this a priori estimate. In Example 1.64, we calculated the ï¬rst ï¬ve terms of the sequence to be (2, 1.5, 1.416666666666667, 1.41421568627451, 1.41421356237469) We observe that ð(ð¥3 , ð¥4 ) = 1.41421568627451 â 1.41421356237469) = 0.0000212389982 so that using the second inequality of Corollary 2.5.1 ð(ð¥4 ,

â

2) â€

1/2 0.0000212389982 < 0.001 1/2

ð¥4 = 1.41421356237469 is the desired approximation. 2.121 Choose any ð¥0 â ð. Deï¬ne the sequence ð¥ð = ð (ð¥ð ) = ð ð (ð¥0 ). Then (ð¥ð ) is a Cauchy sequence in ð converging to ð¥. Since ð is closed, ð¥ â ð. 2.122 By the Banach ï¬xed point theorem, ð ð has a unique ï¬xed point ð¥. Let ðœ be the Lipschitz constant of ð ð . We have to show ð¥ is a ï¬xed point of ð ð(ð (ð¥), ð¥) = ð(ð (ð ð (ð¥), ð ð (ð¥)) = ð(ð ð (ð (ð¥), ð ð (ð¥)) â€ ðœð(ð (ð¥), ð¥) Since ðœ < 1, this implies that ð(ð (ð¥), ð¥) = 0 or ð (ð¥) = ð¥. ð¥ is the only ï¬xed point of ð Suppose ð§ = ð (ð§) is another ï¬xed point of ð . Then ð§ is a ï¬xed point of ð ð and ð(ð¥, ð§) = ð(ð ð (ð¥), ð ð (ð§)) â€ ðœð(ð¥, ð§) which implies that ð¥ = ð§. 96

Solutions for Foundations of Mathematical Economics

2.123 By the Banach ï¬xed point theorem, for every ð â Î, there exists ð¥ð â ð such that ðð (ð¥ð ) = ð¥ð . Choose any ð0 â Î. ð(ð¥ð , ð¥ð0 ) = ð(ðð (ð¥ð ), ðð0 (ð¥ð0 )) â€ ð(ðð (ð¥ð ), ðð (ð¥ð0 )) + ð(ðð (ð¥ð0 ), ðð0 (ð¥ð0 )) â€ ðœð(ð¥ð , ð¥ð0 ) + ð(ðð (ð¥ð0 ), ðð0 (ð¥ð0 )) (1 â ðœ)ð(ð¥ð , ð¥ð0 ) â€ ð(ðð (ð¥ð0 ), ðð0 (ð¥ð0 )) ð(ð¥ð , ð¥ð0 ) â€

ð(ðð (ð¥ð0 ), ðð0 (ð¥ð0 )) â0 (1 â ðœ)

as ð â ð0 . Therefore ð¥ð â ð¥ð0 . 2.124

1. Let x be a ï¬xed point of ð . Then x satisï¬es x = (ðŒ â ðŽ)x + c = x â ðŽx + ð which implies that ðŽx = ð.

2. For any x1 , x2 â ð     ð (x1 ) â ð (x2 ) = (ðŒ â ðŽ)(x1 â x2 )   â€ â¥ðŒ â ðŽâ¥ x1 â x2  Since ððð = 1, the norm of ðŒ â ðŽ is â¥ðŒ â ðŽâ¥ = max ð

â

â£ððð â£ = ð

ðâ=ð

and     ð (x1 ) â ð (x2 ) â€ ð x1 â x2  By the assumption of strict diagonal dominance, ð < 1. Therefore ð is a contraction and has a unique ï¬xed point x. 2.125

1. ð(ð¥) = { ðŠ â â ðº(ð¥) : ð (ð¥, ðŠ â ) + ðœð£(ðŠ â ) = ð£(ð¥) } = {ðŠ â â ðº(ð¥) : ð (ð¥, ðŠ â ) + ðœð£(ðŠ â ) = sup {ð (ð¥, ðŠ) + ðœð£(ðŠ)}} ðŠâðº(ð¥)

â

â

â

= {ðŠ â ðº(ð¥) : ð (ð¥, ðŠ ) + ðœð£(ðŠ ) â¥ ð (ð¥, ðŠ) + ðœð£(ðŠ) for every ðŠ â ðº(ð¥)} = arg max {ð (ð¥, ðŠ) + ðœð£(ðŠ)} ðŠâðº(ð¥)

2. ð(ð¥) is the solution correspondence of a standard constrained maximization problem, with ð¥ as parameter and ðŠ the decision variable. By assumption the maximand ð (ð¥, ðŠ) = ð (ð¥, ðŠ) + ðœð£(ðŠ) is continuous and the constraint correspondence ðº(ð¥) is continuous and compact-valued. Applying the continuous maximum theorem (Theorem 2.3), ð is nonempty, compact-valued and uhc. 3. We have just shown that ð(ð¥) is nonempty for every ð¥ â ð. Starting at ð¥0 , choose some ð¥â1 â ð(ð¥0 ). Then choose ð¥â2 â ð(ð¥â1 ). Proceeding in this way, we can construct a plan xâ = ð¥0 , ð¥â1 , ð¥â2 , . . . such that ð¥âð¡+1 â ð(ð¥âð¡ ) for every ð¡ = 0, 1, 2, . . . .

97

Solutions for Foundations of Mathematical Economics

4. Since ð¥âð¡+1 â ð(ð¥âð¡ ) for every ð¡, x satisï¬es Bellmanâs equation, that is ð£(ð¥âð¡ ) = ð (ð¥âð¡ , ð¥âð¡+1 ) + ðœð£(ð¥âð¡+1 ),

ð¡ = 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 ð of solutions to Bellmanâs equation (Exercise 2.17) is the solution correspondence of the constrained maximization problem ð(ð¥) = arg max { ð (ð¥, ðŠ) + ðœð£(ðŠ) } ðŠâðº(ð¥)

This problem satisï¬es the requirements of the monotone maximum theorem (Theorem 2.1), since the objective function ð (ð¥, ðŠ) + ðœð£(ðŠ) â supermodular in ðŠ â displays strictly increasing diï¬erences in (ð¥, ðŠ) since for every ð¥2 â¥ ð¥1 ð (ð¥2 , ðŠ) + ðœð£(ðŠ) â ð (ð¥1 , ðŠ) + ðœð£(ðŠ) = ð (ð¥2 , ðŠ) â ð (ð¥1 , ðŠ) â ðº(ð¥) is increasing. By Corollary 2.1.2, ð(ð¥) is always increasing. 2. Let xâ = (ð¥0 , ð¥â1 , ð¥â2 , . . . ) be an optimal plan. Then (Exercise 2.17) ð¥âð¡+1 â ð(ð¥âð¡ ),

ð¡ = 0, 1, 2, . . .

Since ð is always increasing ð¥âð¡ â¥ ð¥âð¡â1 =â ð¥âð¡+1 â¥ ð¥âð¡ for every ð¡ = 1, 2, . . . . Similarly ð¥âð¡ â€ ð¥âð¡â1 =â ð¥âð¡+1 â€ ð¥âð¡ xâ = (ð¥0 , ð¥â1 , ð¥â2 , . . . ) is a monotone sequence. 2.127 Let ð(ð¥) = ð (ð¥) â ð¥. ð is continuous (Exercise 2.78) with ð(0) â¥ 0 and ð(1) â€ 0 By the intermediate value theorem (Exercise 2.83), there exists some point ð¥ â [0, 1] with ð(ð¥) = 0 which implies that ð (ð¥) = ð¥. 2.128

1. To show that a label min{ ð : ðœð â€ ðŒð â= 0 } exists for every x â ð, assume to the contrary that, for some x â ð, ðœð > ðŒð for every ð = 0, 1, . . . , ð. This implies ð â

ðœð >

ð=0

ð â

ðŒð = 1

ð=0

contradicting the requirement that ð â

ðœð = 1 for every ð (x) â ð

ð=0

98

Solutions for Foundations of Mathematical Economics

2. The barycentric coordinates of vertex xð are ðŒð = 1 with ðŒð = 0 for every ð â= ð. Therefore the rule assigns vertex xð the label ð. 3. Similarly, if x belongs to a proper face of ð, 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 â conv {x1 , x2 , x4 }. Then x = ðŒ1 x1 + ðŒ2 x2 + ðŒ4 x4 ,

ðŒ1 + ðŒ2 + ðŒ4 = 1

/ {1, 2, 4}. Therefore and ðŒð = 0 for ð â x +ââ min{ ð : ðœð â€ ðŒð â= 0 } â {1, 2, 4} 2.129

1. Since ð is compact, it is bounded (Proposition 1.1) and therefore it is contained in a suï¬ciently large simplex ð .

2. By Exercise 3.74, there exists a continuous retraction ð : ð â ð. The composition ð â ð : ð â ð â ð . Furthermore as the composition of continuous functions, ð â ð is continuous (Exercise 2.72). Therefore ð â ð has a ï¬xed point xâ â ð , that is ð â ð(xâ ) = xâ . 3. Since ð â ð(x) â ð for every x â ð , we must have ð â ð(xâ ) = xâ â ð. Therefore, ð(xâ ) = xâ which implies that ð (xâ ) = xâ . That is, xâ is a ï¬xed point of ð . 2.130 Convexity of ð is required to ensure that there is a continuous retraction of the simplex onto ð. 2.131

1. ð (ð¥) = ð¥2 on ð = (0, 1) or ð (ð¥) = ð¥ + 1 on ð = â+ .

2. ð (ð¥) = 1 â ð¥ on ð = [0, 1/3] âª [2/3, 1]. 3. Let ð = [0, 1] and deï¬ne

{ ð (ð¥) =

1 0

0 â€ ð¥ < 1/2 otherwise

2.132 Suppose such a function exists. Deï¬ne ð (x) = âð(x). Then ð : ðµ â ðµ continously, and has no ï¬xed point since for â x â ð, ð (x) = âð(x) = âx â= x â x â ðµ â ð, ð (x) â / ðµ â ð and thereforeð (x) â= x Therefore ð has no ï¬xed point contradicting Brouwerâs theorem. 2.133 Suppose to the contrary that ð has no ï¬xed point. For every x â ðµ, let ð(z) denote the point where the line segment from ð (x) through x intersects the boundary ð of ðµ. Since ð is continuous and ð (x) â= x, ð is a continuous function from ðµ to its boundary, that is a retraction, contradicting Exercise 2.132. We conclude that ð must have a ï¬xed point. 2.134 No-retraction =â Brouwer Note ï¬rst that the no-retraction theorem (Exercise 2.132) generalizes immediately to a closed ball about 0 of arbitrary radius. Assume that ð is a continuous operator on a compact, convex set ð in a ï¬nite dimensional normed linear space. There exists a closed ball ðµ containing ð (Proposition 1.1). Deï¬ne ð : ðµ â ð by ð(y) = { x â ð : x is closest to y } As in Exercise 2.129, ð is well-deï¬ned, continuous and ð(x) = x for every x â ð. ð â ð : ðµ â ð â ðµ and has a ï¬xed point xâ = ð (ð(xâ )) by Exercise 2.133. Since 99

Solutions for Foundations of Mathematical Economics

ð â ð(x) â ð for every x â ðµ, we must have ð â ð(xâ ) = xâ â ð. Therefore, ð(xâ ) = xâ which implies that ð (xâ ) = xâ . That is, xâ is a ï¬xed point of ð . Brouwer =â no-retraction Exercise 2.132. 2.135 Let Îð , ð = 1, 2, . . . be a sequence of simplicial partitions of ð in which the maximum diameter of the subsimplices tend to zero as ð â â. By Spernerâs lemma (Proposition 1.3), every partition Îð has a completely labeled subsimplex with vertices xð0 , xð1 , . . . , xðð . By construction of an admissible labeling, each xðð belongs to a face containing xð , that is xðð â conv {xð , . . . } and therefore xðð â ðŽð ,

ð = 0, 1, . . . , ð â²

Since ð is compact, each sequence xðð has a convergent subsequence xðð . Moreover, since the diameters of the subsimplices converge to zero, these subsequences must converge to the same point, say xâ . That is, â²

lim xðð = xâ ,

ð = 0, 1, . . . , ð

ðâ² ââ

Since the sets ðŽð are closed, xâ â ðŽð for every ð and therefore ð â©

xâ â

ðŽð â= â

ð=0

2.136

=â Let ð : ð â ð be a continuous operator on an ð-dimensional simplex ð with vertices x0 , x1 , . . . , xð . For ð = 0, 1, . . . , ð, let ðŽð = { x â ð : ðœð â€ ðŒð } where ðŒ0 , ðŒ1 , . . . , ðŒð and ðœ0 , ðœ1 , . . . , ðœð are the barycentric coordinates of x and ð (x) respectively. Then â ð continuous =â ðŽð closed for every ð = 0, 1, . . . , ð (Exercise 1.106) â Let x â conv { xð : ð â ðŒ } for some ðŒ â { 0, 1, . . . , ð }. Then â

ðŒð = 1 =

ð â

ðœð

ð=0

ðâðŒ

which implies that ðœð â€ ðŒð for some ð â ðŒ, so that x â ðŽð . Therefore âª ðŽð conv { xð : ð â ðŒ } â ðâðŒ

Therefore the collection ðŽ0 , ðŽ1 , . . . , ðŽð satisï¬es the hypotheses of the K-K-M theorem and their intersection is nonempty. That is, there exists xâ â

ðŽð â= â with ðœðâ â€ ðŒâð ,

ð = 0, 1, . . . , ð

ð=0

where â ðŒâ and â ðœ â are the barycentric coordinates of xâ and ð (xâ ) respectively. â Since ðœð = ðŒâð = 1, this implies that ðœðâ = ðŒâð

ð = 0, 1, . . . , ð

In other words, ð (xâ ) = xâ . 100

Solutions for Foundations of Mathematical Economics â=

Let ðŽ0 , ðŽ1 , . . . , ðŽð be closed subsets of an ð dimensional simplex ð with vertices x0 , x1 , . . . , xð such that âª conv { xð : ð â ðŒ } â ðŽð ðâðŒ

for every ðŒ â { 0, 1, . . . , ð }. For ð = 0, 1, . . . , ð, let ðð (x) = ð(x, ðŽð ) For any x â ð with barycentric coordinates ðŒ0 , ðŒ1 , . . . , ðŒð , deï¬ne ð (x) = ðœ0 x0 + ðœ1 x1 + â â â + ðœð xð where ðœð =

ðŒð + ðð (x) â 1 + ðð=0 ðð (x)

ð = 0, 1, . . . , ð

(2.45)

â By construction ðœð â¥ 0 and ðð=0 ðœð = 1. Therefore ð (x) â ð. That is, ð : ð â ð. Furthermore ð is continuous. By Brouwerâs theorem, there exists a ï¬xed point ð¥â with ð (xâ ) = xâ . That is ðœðâ = ðŒâð for ð = 0, 1, . . . , ð. Now, since the collection ðŽ0 , ðŽ1 , . . . , ðŽð covers ð, there exists some ð for which ð(xâ , ðŽð ) = 0. Substituting ðœðâ = ðŒâð in (2.45) we have ðŒâð =

1+

ðŒâ âð ð

ð=0

ðð (xâ )

which implies that ðð (xâ ) = 0 for every ð. Since the ðŽð are closed, xâ â ðŽð for every ð and therefore xâ â

ðŽð â= â

ð=0

( ) 2.137 To simplify the notation, let ð§ð+ (p) = max 0, zð (p) . Assume pâ is a ï¬xed point of ð. Then for every ð = 1, 2, . . . , ð ðâð =

ðð + ð§ð+ (pâ ) âð 1 + ð=1 ð§ð+ (pâ )

Cross-multiplying ðâð + ðâð

ð â ð=1

ð§ð+ (p) = ðâð + ð§ð+ (pâ )

or ðâð

ð â ð=1

ð§ð+ (p) = ð§ð+ (pâ )

ð = 1, 2, . . . ð

Multiplying each equation by ð§ð (p) we get ðâð ð§ð (pâ )

ð â ð=1

ð§ð+ (p) = ð§ð (pâ )ð§ð+ (pâ ) 101

ð = 1, 2, . . . ð

Solutions for Foundations of Mathematical Economics Summing over ð ð â

Since

âð

ð=1

ðâð ð§ð (pâ )

ð â ð=1

ð=1

ð§ð+ (p) =

ð â ð=1

ð§ð (pâ )ð§ð+ (pâ )

ðâð ð§ð (pâ ) = 0 this implies that ð â ð=1

ð§ð (pâ )ð§ð+ (pâ ) = 0

( )2 Each term of this sum is nonnegative, since it is either 0 or ð§ð (pâ ) . Consequently, every term must be zero which implies that ð§ð (pâ) â€ 0 for every ð = 1, 2, . . . , ð. In other words, z(pâ ) â€ 0. 2.138 Every individual demand function xð (p, ð) is continuous (Example 2.90) in p and ð. For given endowment ð ð ðð =

ð â

ðð ð ðð

ð=1

is continuous in p (Exercise 2.78). Therefore the excess demand function zð (p) = xð (p, ð) â ð ð is continuous for every consumer ð and hence the aggregate excess demand function is continuous. Similarly, the consumerâs demand function xð (p, ð) is homogeneous of degree 0 in p and ð. For given endowment ð ð , the consumerâs wealth is homogeneous of degree 1 in p and therefore the consumerâs excess demand function zð (p) is homogeneous of degree 0. So therefore is the aggregate excess demand function z(p). 2.139 z(p) = =

ð â ð=1 ð â

zð (p) ( ) xð (p, ð) â ð ð

ð=1

and therefore pð z(p) =

ð â

pð xð (p, ð) â

ð=1

ð â

pð ð ð

ð=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 pð xð (p, ð) = pð ðð for every ð = 1, 2, . . . , ð Therefore in aggregate pð z(p) =

ð â

pð xð (p, ð) â

ð=1

ð â ð=1

for every p. 102

pð ð ð = 0

Solutions for Foundations of Mathematical Economics 2.140 Assume there exists pâ such that z(pâ ) â€ 0. That is z(pâ ) =

ð â

zð (p) =

ð=1

ð ð ð â â ( ) â xð (p, ð) â ðð = xð (p, ð) â ðð â€ 0 ð=1

or

ð=1

â ðâð

xð â€

â

ð=1

ðð

ðâð

Aggregate demand is less or equal to available supply. âð Let ðâð = ð=1 ðâð ð ðð denote the wealth of consumer ð when the price system is pâ and let xâð = x(pâ , ðâ ) be his chosen consumption bundle. Then xâð â¿ xð for every xð â ð(pâ , ðð ) Let xâ = (xâ1 , xâ2 , . . . , xâð ) be the allocation comprising these optimal bundles. The pair (pâ , xâ ) is a competitive equilibrium. 2.141 For each xð , let ð ð denote the subsimplex of Îð which contains xð and let xð0 , xð1 , . . . , xðð denote the vertices of ð ð . Let ðŒð0 , ðŒð1 , . . . , ðŒðð denote the barycentric coordinates (Exercise 1.159) of x with respect to the vertices of ð ð and let yðð = ð ð (xðð ), ð = 0, 1, . . . , ð, denote the images of the vertices. Since ð is compact, there exists â² â² â² subsequences xðð , yðð and ðŒð such that xðð â xâð

yðð â yðâ and ðŒðð â ðŒâð

ð = 0, 1, . . . , ð

Furthermore, ðŒâð â¥ 0 and ðŒâ0 +ðŒâ1 +â â â+ðŒâð = 1. Since the diameters of the subsimplices converge to zero, their vertices must converge to the same point. That is, we must have xâ0 = xâ1 = â â â = xâð = xâ By deï¬nition of ð ð ð ð (xð ) = ðŒð0 ð (xð0 ) + ðŒð1 ð (xð1 ) + â â â + ðŒðð ð (xðð ) Substituting yðð = ð ð (xðð ), ð = 0, 1, . . . , ð and recognizing that xð is a ï¬xed point of ð ð , we have ð¥ð = ð ð (xð ) = ðŒð0 y0ð + ðŒð1 y1ð + â â â + ðŒðð yðð Taking limits xâ = ðŒâ0 y0â + ðŒâ1 y1â + â â â + ðŒâð yðâ

(2.46)

For each coordinate ð, (xðð , yðð ) â graph(ð) for every ð = 0, 1, . . . . Since ð is closed, (xâð , yðâ ) â graph(ð). That is, yðâ â ð(xâð ) = ð(xâ ) for every ð = 0, 1, . . . , ð. Therefore, (2.46) implies xâ â conv ð(xâ ) Since ð is convex valued, xâ â ð(xâ )

103

Solutions for Foundations of Mathematical Economics

2.142 Analogous to Exercise 2.129, there exists a simplex ð containing ð and a retraction of ð onto ð, that is a continuous function ð : ð â ð with ð(x) = x for every x â ð. Then ð â ð : ð â ð â ð is closed-valued (Exercise 2.106) and uhc (Exercise 2.103). By the argument in the proof, there exists a point xâ â ð such that xâ â ð â ð(xâ ). However, since ð â ð(xâ ) â ð, we must have xâ â ð and therefore ð(xâ ) = xâ . This implies xâ â ð(xâ ). That is, xâ is a ï¬xed point of ð. 2.143 ðµ = ðµ1 Ã ðµ2 Ã . . . Ã ðµð is the Cartesian product of uhc, compact- and convexvalued correspondences. Therefore ðµ is also compact-valued and uhc (Exercise 2.112 and also convex-valued (Exercise 1.165). By Exercise 2.106, ðµ is closed. 2.144 Strict quasiconcavity ensures that the best response correspondence is in fact a function ðµ : ð â ð. Since the hypotheses of Example 2.96 apply, there exists at least one equilibrium. Suppose that there are two Nash equilibria s and sâ² . Since ðµ is a contraction, ð(ðµ(s), ðµ(sâ² ) â€ ðœð(s, sâ² ) for some ðœ < 1. However ðµ(s) = s and ðµ(sâ² ) = sâ² and (2.46) implies that ð(s, sâ² ) â€ ðœð(s, sâ² ) which is possible if and only if s = sâ² . This implies that the equilibrium must be unique. 2.145 Since ðŸ is compact, it is totally bounded (Exercise 1.112). There exists a ï¬nite set of points x1 , x2 , . . . , xð such that ð â©

ðŸâ

ðµð (xð )

ð=1

Let ð = conv {x1 , x2 , . . . , xð }. For ð = 1, 2, . . . , ð and x â ð, deï¬ne ðŒð (x) = max{0, ð â â¥x â xð â¥} Then for every x â ðŸ, 0 â€ ðŒð (x) â€ ð,

ð = 1, 2, . . . , ð

and ðŒð (x) > 0 ââ x â ðµð (xð ) Note that ðŒð (x) > 0 for some ð. Deï¬ne

â ðŒð (x)xð â(x) = â ðŒð (x)

Then â(x) â ð and therefore â : ðŸ â ð. Furthermore, â is continuous and  â   ðŒð (x)xð  â â x â¥â(x) â xâ¥ =    ðŒð (x)  â  ðŒð (x)(xð â x)   â =   ðŒð (x) â ðŒð (x) â¥xð â xâ¥ â = ðŒð (x) â ðŒð (x)ð â€ â =ð ðŒð (x) since ðŒð (x) > 0 ââ â¥xð â xâ¥ â€ ð. 104

Solutions for Foundations of Mathematical Economics 2.146

( ) 1. For every x â ð ð , ð (x) â ð and therefore ð ð (x) = âð ð (x) â ð ð .

2. For any x â ð ð , let y = ð (x) â ð (ð) and therefore   ð â (y) â y < 1 ð which implies   ð ð (x) â ð (x) â€ 1 for every x â ð ð ð 2.147 By the Triangle inequality  ð      x â ð (x) â€ ð ð (xð ) â ð (xð ) + ð (xð ) â ð (x) As shown in the previous exercise   ð ð ð (x ) â ð (xð ) â€ 1 â 0 ð as ð â â. Also since ð is continuous   ð (xð ) â ð (x) â 0 Therefore  ð  x â ð (x) â 0 =â x = ð (x) x is a ï¬xed point of ð . 2.148 ð (ð¹ ) is bounded and equicontinuous and so therefore is ð (ð¹ ) (Exercise 2.96). By Ascoliâs theorem (Exercise 2.95), ð (ð¹ ) is compact. Therefore ð is a compact operator. Applying Corollary 2.8.1, ð has a ï¬xed point.

105

Solutions for Foundations of Mathematical Economics

Chapter 3: Linear Functions 3.1 Let x1 , x2 â ð and ðŒ1 , ðŒ2 â â. Homogeneity implies that ð (ðŒ1 x1 ) = ðŒ1 ð (ð¥1 ) ð (ðŒ2 x2 ) = ðŒ2 ð (ð¥2 ) and additivity implies ð (ðŒ1 x1 + ðŒ2 x2 ) = ðŒ1 ð (x1 ) + ðŒ2 ð (x2 ) Conversely, assume ð (ðŒ1 x1 + ðŒ2 x2 ) = ðŒ1 ð (x1 ) + ðŒ2 ð (x2 ) for all x1 , x2 â ð and ðŒ1 , ðŒ2 â â. Letting ðŒ1 = ðŒ2 = 1 implies ð (x1 + x2 ) = ð (x1 ) + ð (x2 ) while setting x2 = 0 implies ð (ðŒ1 x1 ) = ðŒ1 ð (x1 ) 3.2 Assume ð1 , ð2 â ð¿(ð, ð ). Deï¬ne the mapping ð1 + ð2 : ð â ð by (ð1 + ð2 )(x) = ð1 (x) + ð2 (x) We have to conï¬rm that ð1 + ð2 is linear, that is (ð1 + ð2 )(x1 + x2 ) = ð1 (x1 + x2 ) + ð2 (x1 + x2 ) = ð1 (x1 ) + ð1 (x2 ) + ð2 (x1 ) + ð2 (x2 ) = ð1 (x1 ) + ð2 (x1 ) + ð1 (x1 ) + ð2 (x2 ) = (ð1 + ð2 )(x1 ) + (ð1 + ð2 )(x2 ) and (ð1 + ð2 )(ðŒx) = ð1 (ðŒx) + ð2 (ðŒx) = ðŒ(ð1 (x) + ð2 (x)) = ðŒ(ð1 + ð2 )(x) Similarly let ð â ð¿(ð, ð ) and deï¬ne ðŒð : ð â ð by (ðŒð )(x) = ðŒð (x) ðŒð is also linear, since (ðŒð )(ðœx) = ðŒð (ðœx) = ðŒðœð (x) = ðœðŒð (x) = ðœ(ðŒð )(x) 106

Solutions for Foundations of Mathematical Economics

3.3 Let x, x1 , x2 â â2 . Then ð (x1 + x2 ) = ð (ð¥11 + ð¥21 , ð¥12 + ð¥22 ) ) ( = (ð¥11 + ð¥21 ) cos ð â (ð¥12 + ð¥22 ) sin ð, (ð¥11 + ð¥21 ) sin ð â (ð¥12 + ð¥22 ) cos ð ) ( = (ð¥11 cos ð â ð¥12 sin ð) + (ð¥21 cos ð â ð¥22 sin ð), (ð¥11 sin ð + ð¥12 cos ð) + (ð¥21 sin ð â ð¥22 cos ð) ( ) = (ð¥11 cos ð â ð¥12 sin ð, ð¥11 sin ð + ð¥12 cos ð) + (ð¥21 cos ð â ð¥22 sin ð, ð¥21 sin ð â ð¥22 cos ð = ð (ð¥11 , ð¥12 ) + ð (ð¥21 , ð¥22 ) = ð (x1 ) + ð (x2 ) and ð (ðŒx) = ð (ðŒð¥1 , ðŒð¥2 ) = (ðŒð¥1 cos ð â ðŒð¥2 sin ð, ðŒð¥1 sin ð + ðŒð¥2 cos ð) = ðŒ (ð¥1 cos ð â ð¥1 sin ð, ð¥1 sin ð + ð¥2 cos ð) = ðŒð (ð¥1 , ð¥2 ) = ðŒð (x) 3.4 Let x, x1 , x2 â â3 . ð (x1 + x2 ) = ð (ð¥11 + ð¥2 , ð¥12 + ð¥22 , ð¥13 + ð¥23 ) = (ð¥11 + ð¥21 , ð¥12 + ð¥22 , 0) = (ð¥11 , ð¥12 , 0) + (ð¥21 , ð¥22 , 0) = ð (ð¥11 , ð¥12 , ð¥13 ) + ð (ð¥21 , ð¥22 , ð¥23 ) = ð (x1 ) + ð (x2 ) and ð (ðŒx) = ð (ðŒð¥1 , ðŒð¥2 , ðŒð¥3 ) = (ðŒð¥1 , ðŒð¥2 , 0) = ðŒ(ð¥1 , ð¥2 , 0) = ðŒð (ð¥1 , ð¥2 , ð¥3 ) = ðŒð (x) This mapping is the projection of 3-dimensional space onto the (2-dimensional) plane. 3.5 Applying the deï¬nition

( )( ) 0 1 ð¥1 ð (ð¥1 , ð¥2 ) = 1 0 ð¥2 = (ð¥2 , ð¥1 )

This function interchanges the two coordinates of any point in the plane â2 . Its action corresponds to reï¬ection about the line ð¥1 = ð¥2 ( 45 degree diagonal). 3.6 Assume (ð, ð€) and (ð, ð€â² ) are two games in ð¢ ð . For any coalition ð â ð (ð€ + ð€â² )(ð) â (ð€ + ð€â² )(ð â {ð}) = ð€(ð) + ð€(ð â² ) â ð€(ð â {ð}) â ð€â² (ð â {ð}) = (ð€(ð) â ð€(ð â {ð})) + (ð€â² (ð) â ð€â² (ð â {ð})) = ðð (ð€) + ðð (ð€â² ) 107

Solutions for Foundations of Mathematical Economics 3.7 The characteristic function of cost allocation game is ð€(ðŽð ) = 0 ð€(ð ð ) = 0

ð€(ðŽð, ð ð ) = 210 ð€(ðŽð, ðŸð ) = 770

ð€(ðŸð ) = 0

ð€(ðŸð, ð ð ) = 1170

ð€(ð ) = 1530

The following table details the computation of the Shapley value for player ðŽð . ð ðŽð ðŽð, ð ð ðŽð, ðŸð ðŽð, ð ð, ðŸð ðð (ð€)

ðŸð 1/3 1/6 1/6 1/3

ð€(ð) 0 210 770 1530

ð€(ð â {ð}) 0 0 0 1170

ðŸð (ð€(ð) â ð€(ð â {ð})) 0 35 128 1/3 120 283 1/3

Thus ððŽð ð€ = 283 1/3. Similarly, we can calculate that ðð ð ð€ = 483 1/3 and ððŸð ð€ = 763 1/3. 3.8 â

ðð ð€ =

ðâð

=

â

( â

ðâð

ðâð

â

(

ðâð

=

â

) ðŸð (ð€(ð) â ð€(ð â {ð})) ) ðŸð (ð€(ð) â ð€(ð â {ð}))

ðâð

ââ

ðŸð ð€(ð) â

ðâð ðâð

=

â â

ðŸð ð€(ð â {ð})

ðâð ðâð

ð  Ã ðŸð ð€(ð) â

ðâð

=

ââ

(

â

ðŸð

ðâð

ð  Ã ðŸð ð€(ð) â

â

â

) ð€(ð â {ð})

ðâð

ð  Ã ðŸð ð€(ð)

ðâð

ðâð

= ð Ã ðŸð ð€(ð ) = ð€(ð ) 3.9 If ð, ð â ð ð€(ð â {ð}) = ð€(ð â {ð, ð} âª {ð}) = ð€(ð â {ð, ð} âª {ð}) = ð€(ð â {ð}) ðð (ð€) =

â

ðŸð (ð€(ð) â ð€(ð â {ð}))

ðâð

=

â

ðŸð (ð€(ð) â ð€(ð â {ð})) +

ðâð,ð

=

â

â

ðŸð (ð€(ð) â ð€(ð â {ð})) +

=

â

ðŸð (ð€(ð âª {ð}) â ð€(ð))

ðââð,ð

ðŸð (ð€(ð) â ð€(ð â {ð})) +

â

ðŸð â² (ð€(ð â² âª {ð}) â ð€(ð â² ))

ð â² ââð,ð

ðâð,ð

â

ðŸð (ð€(ð) â ð€(ð â {ð}))

ðâð,ðââð

ðâð,ð

=

â

ðŸð (ð€(ð) â ð€(ð â {ð})) +

ðâð,ð

â

ðââð,ðâð

= ðð (ð€) 108

ðŸð (ð€(ð) â ð€(ð â {ð}))

Solutions for Foundations of Mathematical Economics

3.10 For any null player ð€(ð) â ð€(ð â {ð}) = 0 for every ð â ð . Consequently

â

ðð (ð€) =

ðŸð (ð€(ð) â ð€(ð â {ð})) = 0

ðâð

3.11 Every ð â / ð is a null player, so that ðð (ð¢ð ) = 0 Feasibility requires that

â

for every ð â /ð â

ðð (ð¢ð ) =

ðâð

ðð (ð¢ð ) = 1

ðâð

Further, any two players in ð are substitutes, so that symmetry requires that ðð (ð¢ð ) = ðð (ð¢ð )

for every ð, ð â ð

Together, these conditions require that ðð (ð¢ð ) =

1 ð¡

for every ð â ð

The Shapley value of the a T-unanimity game is { 1 ðâð ðð (ð¢ð ) = ð¡ 0 ðâ /ð where ð¡ = â£ð â£. 3.12 Any coalitional game can be represented as a linear combination of unanimity games ð¢ð (Example 1.75) â ð€= ðŒð ð¢ð ð

By linearity, the Shapley value is

â

â

ðð€ = ð â =

â

â ðŒð ð¢ð â

ð âð

ðŒð ðð¢ð

ð âð

and therefore for player ð ðð ð€ =

â

ðŒð ðð ð¢ð

ð âð

=

â 1 ðŒð ð¡

ð âð ð âð

â 1 â 1 ðŒð â ðŒð = ð¡ ð¡ ð âð

ð âð ðâð /

= ð (ð, ð€) â ð (ð â {ð}, ð€) 109

Solutions for Foundations of Mathematical Economics Using Exercise 3.8 ð€(ð ) =

â

ðð ð€

ðâð

=

â(

) ð (ð, ð€) â ð (ð â {ð}, ð£)

ðâð

= ðð (ð, ð€) â

â

ð (ð â {ð}, ð£)

ðâð

which implies that 1 ð (ð, ð€) = ð

( ð€(ð ) â

â

) ð (ð â {ð}, ð£)

ðâð

3.13 Choose any x â= 0 â ð. 0ð = x â x and by additivity ð (0ð ) = ð (x â x) = ð (x) â ð (x) = 0ð 3.14 Let x1 , x2 belong to ð. Then ð â ð (x1 + x2 ) = ð â ð (x1 + x2 ) ) ( = ð ð (x1 ) + ð (x2 ) ) ( ) ( = ð ð (x1 ) + ð ð (x2 ) = ð â ð (x1 ) + ð â ð (x2 ) and ð â ð (ðŒx) = ð (ð (ðŒx)) = ð (ðŒð (x)) = ðŒð (ð (x)) = ðŒð â ð (x) Therefore ð â ð is linear. 3.15 Let ð be a subspace of ð and let y1 , y2 belong to ð (ð). Choose any x1 â ð â1 (y1 ) and x2 â ð â1 (y2 ). Then for ðŒ1 , ðŒ2 â â ðŒ1 x1 + ðŒ2 x2 â ð Since ð is linear (Exercise 3.1) ðŒ1 y1 + ðŒ2 y2 = ðŒ1 ð (x1 ) + ðŒ2 ð (x2 ) = ð (ðŒ1 x1 + ðŒ2 x2 ) â ð (ð) ð (ð) is a subspace. Let ð be a subspace of ð and let x1 , x2 belong to ð â1 (ð ). Let y1 = ð (x1 ) and y2 = ð (x2 ). Then y1 , y2 â ð . For every ðŒ1 , ðŒ2 â â ðŒ1 y1 + ðŒ2 y2 = ðŒ1 ð (x1 ) + ðŒ2 ð (x2 ) â ð 110

Solutions for Foundations of Mathematical Economics

Since ð is linear, this implies that ð (ðŒ1 x1 + ðŒ2 x2 ) = ðŒ1 ð (x1 ) + ðŒ2 ð (x2 ) â ð Therefore ðŒ1 x1 + ðŒ2 x2 â ð â1 (ð ) We conclude that ð â1 (ð ) is a subspace. 3.16 ð (ð) is a subspace of ð . rank ð (ð) = rank ð implies that ð (ð) = ð . ð is onto. 3.17 This is a special case of the previous exercise, since {0ð } is a subspace of ð . 3.18 Assume not. That is, assume that there exist two distinct elements x1 and x2 with ð (x1 ) = ð (x2 ). Then x1 â x2 â= 0ð but ð (x1 â x2 ) = ð (x1 ) â ð (x2 ) = 0ð so that x1 â x2 â kernel ð which contradicts the assumption that kernel ð = {0}. 3.19 If ð has an inverse, then it is one-to-one and onto (Exercise 2.4), that is ð â1 (0) = 0 and ð (ð) = ð . Conversely, if kernel ð = {0} then ð is one-to-one by the previous exercise. If furthermore ð (ð) = ð , then ð is one-to-one and onto, and therefore has an inverse (Exercise 2.4). 3.20 Let ð be a nonsingular linear function from ð to ð with inverse ð â1 . Choose y1 , y2 â ð and let x1 = ð â1 (y1 ) x2 = ð â1 (y2 ) so that y1 = ð (x1 ) y2 = ð (x2 ) Since ð is linear ð (x1 + x2 ) = ð (x1 ) + ð (x2 ) = y1 + y2 which implies that ð â1 (y1 + y2 ) = x1 + x2 = ð â1 (y1 ) + ð â1 (y2 ) The homogeneity of ð â1 can be demonstrated similarly. 3.21 Assume that ð : ð â ð and ð : ð â ð are nonsingular. Then (Exercise 3.19) â ð (ð) = ð and ð(ð ) = ð â kernel ð = {0ð } and kernel ð = {0ð } We have previously shown (Exercise 3.14) that â = ð â ð : ð â ð is linear. To show that â is nonsingular, we note that â â(ð) = ð â ð (ð) = ð(ð ) = ð

111

Solutions for Foundations of Mathematical Economics

â If x â kernel (â) then â(x) = ð (ð (x)) = 0 and ð (x) â kernel ð = {0ð }. Therefore ð (x) = 0ð which implies that x = 0ð . Thus kernel â = {0ð }. We conclude that â is nonsingular. Finally, let z be any point in ð and let x1 = ââ1 (z) = (ð â ð )â1 (z) y = ð â1 (z) x2 = ð â1 (y) Then z = â(x1 ) = ð â ð (x1 ) z = ð(y) = ð â ð (x2 ) which implies that x1 = x2 . 3.22 Suppose ð were one-to-one. Then kernel ð = {0} â kernel â and ð = â â ð â1 is a well-deï¬ned linear function mapping ð (ð) to ð with ( ) ð â ð = â â ð â1 â ð = â We need to show that this still holds if ð is not one-to-one. In this case, for arbitrary y â ð (ð), ð â1 (y) may contain more than one element. Suppose x1 and x2 are distinct elements in ð â1 (y). Then ð (x1 â x2 ) = ð (x1 ) â ð (x2 ) = y â y = 0 so that x1 â x2 â kernel ð â kernel â (by assumption). Therefore â(x1 ) â â(x2 ) = â(x1 â x2 ) = 0 which implies that â(x1 ) = â(x2 ) for all x1 , x2 â ð â1 (y). Thus ð = ââ ð â1 : ð (ð) â ð is well deï¬ned even if ð is many-to-one. To show that ð is linear, choose y1 , y2 in ð (ð) and let x1 â ð â1 (y1 ) x2 â ð â1 (y2 ) Since ð (x1 + x2 ) = ð (x1 ) + ð (x2 ) = y1 + y2 x1 + x2 â ð â1 (y1 + y2 ) and ð(y1 + y2 ) = â(x1 + x2 ) Therefore ð(y1 ) + ð(y2 ) = â(x1 ) + â(x2 ) = â(x1 + x2 ) = ð(y1 + y2 ) 112

Solutions for Foundations of Mathematical Economics

Similarly ðŒx1 â ð â1 (ðŒy1 ) and ð(ðŒy1 ) = â(ðŒx1 ) = ðŒâ(x1 ) = ðŒð(y1 ) We conclude that ð = â â ð â1 is a linear function mapping ð (ð) to ð with â = ð â ð . 3.23 Let y be an arbitrary element of ð (ð) with x â ð â1 (y). Since B is a basis for ð, x can be represented as a linear combination of elements of ðµ, that is there exists x1 , x2 , .., xð â ðµ and ðŒ1 , ..., ðŒð â ð such that x=

ð â

ðŒð xð

ð=1

y = ð (x) ) ( â ðŒð xð =ð =

â

ð

ðŒð ð (xð )

ð

Since ð (xð ) â ð (ðµ), we have shown that y can be written as a linear combination of elements of ð (ðµ), that is y â lin ðµ Since the choice of y was arbitrary, ð (ðµ) spans ð (ð), that is lin ðµ = ð (ð) 3.24 Let ð = dim ð and ð = dim kernel ð . Let x1 , . . . , xð be a basis for the kernel of ð . This can be extended (Exercise 1.142) to a basis ðµ for ð. Exercise 3.23 showed lin ðµ = ð (ð) Since x1 , x2 , . . . , xð â kernel ð , ð (xð ) = 0 for ð = 1, 2, . . . , ð. This implies that {ð (xð+1 ), . . . , ð (xð )} spans ð (ð), that is lin {(xð+1 ), ..., ð (xð )} = ð (ð) To show that dim ð (ð) = ð â ð, we have to show that {ð (ð¥ð+1 ), ð (ð¥ð+2 ), . . . , ð (ð¥ð )} is linearly independent. Assume not. That is, assume there exist ðŒð+1 , ðŒð+2 , ..., ðŒð â ð such that ð â

ðŒð ð (xð ) = 0

ð=ð+1

This implies that

( ð

)

ð â

ðŒð xð

=0

ð=ð+1

or x=

ð â

ðŒð ð¥ð â kernel ð

ð=ð+1

113

Solutions for Foundations of Mathematical Economics

This implies that x can also be expressed as a linear combination of elements in {x1 , ð¥2 , ..., xð }, that is there exist scalars ðŒ1 , ðŒ2 , . . . , ðŒð such that x=

ð â

ðŒð xð

ð=1

or x=

ð â

ð â

ðŒð xð =

ð=1

ðŒð xð

ð=ð+1

which contradicts the assumption that ðµ is a basis for ð. Therefore {ð (xð+1 ), . . . , ð (xð )} is a basis for ð (ð) and therefore dim ð (ð¥) = ð â ð. We conclude that dim kernel ð + dim ð (ð) = ð = dim ð 3.25 Equation (3.2) implies that nullity ð = 0, and therefore ð is one-to-one (Exercise 3.18). 3.26 Choose some x = (ð¥1 , ð¥2 , . . . , ð¥ð ) â ð. x has a unique representation in terms of the standard basis (Example 1.79) x=

ð â

ð¥ð eð

ð=1

Let y = ð (x). Since ð is linear

â

y = ð (x) = ð â

â

ð â

ð¥ð eð â  =

ð=1

ð â

xð ð (eð )

ð=1

Each ð (eð ) has a unique representation of the form ð (eð ) =

ð â

ððð eð

ð=1

so that y = ð (x) =

ð â ð=1

=

ð â ð=1

(

ð¥ð

ð â

) ððð eð

ð=1

â â ð â â ððð ð¥ð â  eð ð=1

â â âð ð1ð ð¥ð âð=1 ð â ð=1 ð2ð ð¥ð â â â =â â .. â  â âð . ð ð¥ ð=1 ðð ð = ðŽx where

â

â ð11 ð12 . . . ð1ð â ð21 ð22 . . . ð2ð â â ðŽ=â â . . . . . . . . . . . . . . . . . . . . .â  ðð1 ðð2 . . . ððð 114

Solutions for Foundations of Mathematical Economics 3.27

( 1 0 0 1

) 0 0

3.28 We must specify bases for each space. The most convenient basis for ðºð is the T-unanimity games. We adopt the standard basis for âð . With respect to these bases, the Shapley value ð is represented by the 2ðâ1 Ãð matrix where each row is the Shapley value of the corresponding T-unanimity game. For three player games (ð = 3), the matrix is â 1 0 â0 1 â â0 0 â1 1 â â 21 2 â â 2 01 â0 2 1 3

1 3

â 0 0â â 1â â 0â â 1â 2â 1â  2 1 3

3.29 Clearly, if ð is continuous, ð is continuous at 0. To show the converse, assume that ð : ð â ð is continuous at 0. Let (xð ) be a sequence which converges to x â ð. Then the sequence (xð â x) converges to 0ð and therefore ð (xð âx) â 0ð by continuity (Exercise 2.68). By linearity, ð (xð )âð (x) = ð (xð âx) â 0ð and therefore ð (xð ) converges to ð (x). We conclude that ð is continuous at x. 3.30 Assume that ð is bounded, that is â¥ð (x)â¥ â€ ð â¥xâ¥ for every x â ð Then ð is Lipschitz at 0 (with Lipschitz constant ð ) and hence continuous (by the previous exercise). Conversely, assume ð is continuous but not bounded. Then, for every positive integer ð, there exists some xð â ð such that â¥ð (xð )â¥ > ð â¥xð â¥ which implies that  ( )   xð ð >1  ð â¥xð â¥  Deï¬ne yð =

xð ð â¥xð â¥

Then yð â 0 but ð (yð ) ââ 0. This implies that ð is not continuous at the origin, contradicting our hypothesis. 3.31 Let {x1 , x2 , . . . , xð } be a basis for ð. For every x â ð, there exists numbers ðŒ1 , ðŒ2 , . . . , ðŒð such that x=

ð â

ðŒð xð

ð=1

115

Solutions for Foundations of Mathematical Economics and ð (x) =

ð â

ðŒð ð (xð )

ð=1 ð â

 â¥ð (x)â¥ =   â€

ð=1

ð â

   ðŒð ð (xð ) 

â£ðŒð â£ â¥ð (xð )â¥

ð=1

ð )â ( ð â£ðŒð â£ â€ max â¥ð (xð )â¥ ð=1

ð=1

By Lemma 1.1, there exists a constant ð such that   ð ð  1 â 1  â â£ðŒð â£ â€  ðŒð xð  = â¥xâ¥  ð  ð ð=1 ð=1 Combining these two inequalities â¥ð (x)â¥ â€ ð â¥xâ¥ where ð = maxðð=1 â¥ð (xð )â¥ /ð. 3.32 For any x â ð, let ð = â¥xâ¥ and deï¬ne y = x/ð. Linearity implies that â¥ð (x)â¥ = sup â¥ð (x/ð)â¥ = sup â¥ð (y)â¥ ð xâ=0 xâ=0 â¥yâ¥=1

â¥ð â¥ = sup

3.33 â¥ð â¥ is a norm Let ð â ðµð¿(ð, ð ). Clearly â¥ð â¥ = sup â¥ð (x)â¥ â¥ 0 â¥xâ¥=1

Further, for every ðŒ â â, â¥ðŒð â¥ = sup â¥ðŒð (x)â¥ = â£ðŒâ£ â¥ð â¥ â¥xâ¥=1

Finally, for every ð â ðµð¿(ð, ð ), â¥ð + ðâ¥ = sup â¥ð (x) + ð(x)â¥ â€ sup â¥ð (x)â¥ + sup â¥ð(x)â¥ â€ â¥ð â¥ + â¥ðâ¥ â¥xâ¥=1

â¥xâ¥=1

â¥xâ¥=1

verifying the triangle inequality. There â¥ð â¥ is a norm. ðµð¿(ð, ð ) is a linear space Let ð, ð â ðµð¿(ð, ð ). Since ðµð¿(ð, ð ) â ð¿(ð, ð ), ð + ð is linear, that is ð +ð â ð¿(ð, ð ) (Exercise 3.2). Similarly, ðŒð â ð¿(ð, ð ) for every ðŒ â â. Further, by the triangle inequality â¥ð + ðâ¥ â€ â¥ð â¥ + â¥ðâ¥ and therefore for every x â ð â¥(ð + ð)(x)â¥ â€ â¥ð + ðâ¥ â¥xâ¥ â€ (â¥ð â¥ + â¥ðâ¥) â¥xâ¥ Therefore ð + ð â ðµð¿(ð, ð ). Similarly â¥(ðŒð )(x)â¥ â€ (â£ðŒâ£ â¥ð â¥) â¥xâ¥ so that ðŒð â ðµð¿(ð, ð ). 116

Solutions for Foundations of Mathematical Economics

ðµð¿(ð, ð ) is complete with this norm Let (ð ð ) be a Cauchy sequence in ðµð¿(ð, ð ). For every x â ð â¥ð ð (x) â ð ð (x)â¥ â€ â¥ð ð â ð ð â¥ â¥xâ¥ Therefore (ð ð (x)) is a Cauchy sequence in ð , which converges since ð is complete. Deï¬ne the function ð : ð â ð by ð (x) = limðââ ð ð (x). ð is linear since ð (x1 + x2 ) = lim ð ð (x1 + x2 ) = lim ð ð (x1 ) + lim ð ð (x2 ) = ð (x1 ) + ð (x2 ) and ð (ðŒx) = lim ð ð (ðŒx) = ðŒ lim ð ð (x) = ðŒð (x) To show that ð is bounded, we observe that     â¥ð (x)â¥ = lim ð ð (x) = lim â¥ð ð (x)â¥ â€ sup â¥ð ð (x)â¥ â€ sup â¥ð ð â¥ â¥xâ¥ ð

ð

ð

ð

Since (ð ð ) is a Cauchy sequence, (ð ð ) is bounded (Exercise 1.100), that is there exists ð such that â¥ð ð â¥ â€ ð . This implies â¥ð (x)â¥ â€ sup â¥ð ð â¥ â¥xâ¥ â€ ð â¥xâ¥ ð

Thus, ð is bounded. To complete the proof, we must show ð ð â ð , that is â¥ð ð â ð â¥ â 0. Since (ð ð ) is a Cauchy sequence, for every ð > 0, there exists ð such that â¥ð ð â ð ð â¥ â€ ð for every ð, ð â¥ ð and consequently â¥ð ð (x) â ð ð (x)â¥ = â¥(ð ð â ð ð )(x)â¥ â€ ð â¥xâ¥ Letting ð go to inï¬nity, â¥ð ð (x) â ð (x)â¥ = â¥(ð ð â ð )(x)â¥ â€ ð â¥xâ¥ for every x â ð and ð â¥ ð and therefore â¥ð ð â ð â¥ = sup {ð ð â ð )(x)} â€ ð â¥xâ¥=1

for every ð â¥ ð . 3.34

1. Since ð is ï¬nite-dimensional, ð is compact (Proposition 1.4). Since ð is continuous, ð (ð) is a compact set in ð (Exercise 2.3). Since 0ð â / ð, 0ð = ð (0ð ) â / ð (ð). ( )ð 2. Consequently, ð (ð) is an open set containing 0ð . It contains an open ball ( )ð ð â ð (ð) around 0ð . 3. Let y â ð and choose any x â ð â1 (y) and consider y/ â¥xâ¥. Since ð is linear, ( ) x ð (x) y = =ð â ð (ð) â¥xâ¥ â¥xâ¥ â¥xâ¥ and therefore y/ â¥xâ¥ â / ð since ð â© ð (ð) = â. 117

Solutions for Foundations of Mathematical Economics Suppose that y â / ð (ðµ). Then â¥xâ¥ â¥ 1 and therefore y â ð =â

y âð â¥xâ¥

since ð is convex. This contradiction establishes that y â ð (ðµ) and therefore ð â ð (ðµ). We conclude that ð (ðµ) contains an open ball around 0ð . 4. Let ð be any open set in ð. We need to show that ð (ð) is open in ð . Choose any y â ð (ð) and x â ð â1 (y). Then x â ð and, since ð is open, there exists some ð > 0 such that ðµð (x) â ð. Now ðµð (x) = x + ððµ and ð (ðµð (x)) = y + ðð (ðµ) â ð (ð) by linearity. As we have just shown, there exists an open ball T about 0ð such that ð â ð (ðµ). Let ð (x) = y + ðð . ð (x) is an open ball about y. Since ð â ð (ðµ), ð (x) = y + ðð â ð (ðµð (x)) â ð (ð). This implies that ð (ð) is open. Since ð was an arbitrary open set, ð is an open map. 5. Exercise 2.69. 3.35 ð is linear ð (ðŒ + ðœ) =

ð â

(ðŒð + ðœð )xð =

ð=1

ð â

ðŒð xð +

ð=1

ð â

ðœð xð = ð (ðŒ) + ð (ðœ)

ð=1

Similarly for every ð¡ â â ð (ð¡ðŒ) = ð¡

ð â

ðŒð xð = ð¡ð (ð¡ðŒ)

ð=1

ð is one-to-one Exercise 1.137. ð is onto By deï¬nition of a basis lin {x1 , x2 , . . . , xð } = ð ð is continuous Exercise 3.31 ð is an open map Proposition 3.2 3.36 ð is bounded and therefore there exists ð such that â¥ð (x)â¥ â€ ð â¥xâ¥. Similarly, ð â1 is bounded and therefore there exists ð such that for every x ð â1 (y) â€

1 â¥yâ¥ ð

where y = ð (x). This implies ð â¥xâ¥ â€ â¥ð (x)â¥ and therefore for every x â ð. ð â¥xâ¥ â€ â¥ð (x)â¥ â€ ð â¥xâ¥ By the linearity of ð , ð â¥x1 â x2 â¥ â€ â¥ð (x1 â x2 )â¥ = â¥ð (x1 ) â ð (x2 )â¥ â€ ð â¥x1 â x2 â¥

118

Solutions for Foundations of Mathematical Economics

3.37 For any function, continuity implies closed graph (Exercise 2.70). To show the converse, assume that ðº = graph(ð ) is closed. ð Ãð with norm â¥(x, y)â¥ = max{â¥xâ¥ , â¥yâ¥} is a Banach space (Exercise 1.209). Since ðº is closed, ðº is complete. Also, ðº is a subspace of ð Ã ð . Consequently, ðº is a Banach space in its own right. Consider the projection â : ðº â ð deï¬ned by â(x, ð (x)) = x. Clearly â is linear, one-to-one and onto with ââ1 (x) = (x, ð (x)) It is also bounded since â¥â(x, ð (x))â¥ = â¥xâ¥ â€ â¥(x, ð (x)â¥ By the open mapping theorem, ââ1 is bounded. For every x â ð     â¥ð (x)â¥ â€ â¥(x, ð (x))â¥ = ââ1 (x) â€ ââ1  â¥xâ¥ We conclude that ð is bounded and hence continuous. 3.38 ð (1) = 5, ð (2) = 7 but ð (1 + 2) = ð (3) = 9 â= ð (1) + ð (2) Similarly ð (3 Ã 2) = ð (6) = 15 â= 3 Ã ð (2) 3.39 Assume ð is aï¬ne. Let y = ð (0) and deï¬ne ð(x) = ð (x) â y ð is homogeneous since for every ðŒ â â ð(ðŒx) = ð(ðŒx + (1 â ðŒ)0) = ð (ðŒx + (1 â ðŒ)0) â y = ðŒð (x) + (1 â ðŒ)ð (0) â y = ðŒð (x) + (1 â ðŒ)y â y = ðŒð (x) â ðŒy = ðŒ(ð (ð¥) â y) = ðŒð(x) Similarly for any x1 , x2 â ð ð(ðŒx1 + (1 â ðŒ)x2 ) = ð (ðŒx1 + (1 â ðŒ)x2 ) â ðŠ = ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) â ðŠ Therefore, for ðŒ = 1/2 1 1 1 1 ð( x1 + x2 ) = ð (x1 ) + ð (x2 ) â ðŠ 2 2 2 2 1 1 = (ð (x1 ) â ðŠ) + (ð (x2 ) â ðŠ) 2 2 1 1 = ð(x1 ) + ð(x2 ) 2 2 119

Solutions for Foundations of Mathematical Economics

Since ð is homogeneous ð(x1 + x2 ) = ð(x1 ) + ð(x2 ) which shows that ð is additive and hence linear. Conversely if ð (x) = ð(x) + y with ð linear ð (ðŒx1 + (1 â ðŒ)x2 ) = ðŒð(x1 ) + (1 â ðŒ)ð(x2 ) + ðŠ = ðŒð(x1 ) + ðŠ + (1 â ðŒ)ð(x2 ) + ðŠ = ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) 3.40 Let ð be an aï¬ne subset of ð and let y1 , y2 belong to ð (ð). Choose any x1 â ð â1 (y1 ) and x2 â ð â1 (y2 ). Then for any ðŒ â â ðŒx1 + (1 â ðŒ)x2 â ð Since ð is aï¬ne ðŒy1 + (1 â ðŒ)y2 = ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) = ð (ðŒx1 + (1 â ðŒ)x2 ) â ð (ð) ð (ð) is an aï¬ne set. Let ð be an aï¬ne subset of ð and let x1 , x2 belong to ð â1 (ð ). Let y1 = ð (x1 ) and y2 = ð (x2 ). Then y1 , y2 â ð . For every ðŒ â â ðŒy1 + (1 â ðŒ)y2 = ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) â ð Since ð is aï¬ne, this implies that ð (ðŒx1 + (1 â ðŒ)x2 ) = ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) â ð Therefore ðŒx1 + (1 â ðŒ)x2 â ð â1 (ð ) We conclude that ð â1 (ð ) is an aï¬ne set. 3.41 For any y1 , y2 â ð (ð), choose x1 , x2 â ð such that yð = ð (xð ). Since ð is convex, ðŒx1 + (1 â ðŒ)x2 â ð and therefore ð (ðŒx1 + (1 â ðŒ)x2 ) = ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) = ðŒy1 + (1 â ðŒ)y2 â ð (ð) Therefore ð (ð) is convex. 3.42 Suppose otherwise that y is not eï¬cient. Then there exists another production plan yâ² â ð such that yâ² â¥ y. Since p > 0, this implies that pyâ² > py, contradicting the assumption that y maximizes proï¬t. 3.43 The random variable ð can be represented as the sum â ð(ð )ð{ð } ð= ð âð

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where ð{ð } is the indicator function of the set {ð }. Since ðž is linear ðž(ð) =

â

ð(ð )ðž(ð{ð } )

ð âð

=

â

ðð ð(ð )

ð âð

since ðž(ð{ð } = ð ({ð }) = ðð  â¥ 0. For the random variable ð = 1, ð(ð ) = 1 for every ð  â ð and â ðð = 1 ðž(1) = ð âð

3.44 Let ð¥1 , ð¥2 â ð¶[0, 1]. Recall that addition in C[0,1] is deï¬ned by (ð¥1 + ð¥2 )(ð¡) = ð¥1 (ð¡) + ð¥2 (ð¡) Therefore ð (ð¥1 + ð¥2 ) = (ð¥1 + ð¥2 )(1/2) = ð¥1 (1/2) + ð¥2 (1/2) = ð (ð¥1 ) + ð (ð¥2 ) Similarly ð (ðŒð¥1 ) = (ðŒð¥1 )(1/2) = ðŒð¥1 (1/2) = ðŒð (ð¥1 ) 3.45 Assume that xâ = xâ1 + xâ2 + â â â + xâð maximizes ð over ð. Suppose to the contrary that there exists yð â ðð such that ð (yð ) > ð (xâð ). Then y = xâ1 + xâ2 + â â â + yð + â â â + xâð â ð and â â ð (y) = ð (xâð ) + ð (yð ) > ð (xâð ) = ð (xâ ) ð

ðâ=ð

contradicting the assumption at ð is maximized at xâ . Conversely, assume ð (xâð ) â¥ ð (xð ) for every xð â ðð for every ð = 1, 2, . . . , ð. Summing â â â â ð (xâ ) = ð ( xâð ) = ð (ð¥âð ) â¥ ð (xð ) = ð ( xð ) = ð (x) for every x â ð xâ = xâ1 + xâ2 + â â â + xâð maximizes ð over ð. 3.46

1. Assume (ð¥ð¡ ) is a sequence in ð1 with ð  = the sequence of partial sums ð ð¡ =

ð¡ â

ââ

ð¡=1

â£ð¥ð â£ < â. Let (ð ð¡ ) denote

â£ð¥ð â£

ð=1

Then (ð ð¡ ) is a bounded monotone sequence in âð which converges to ð . Consequently, (ð ð¡ ) is a Cauchy sequence. For every ð > 0 there exists an ð such that ð+ð â

â£ð¥ð¡ â£ < ð

ð=ð

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for every ð â¥ ð and ð â¥ 0. Letting ð = 0 â£ð¥ð¡ â£ < ð for every ð â¥ ð We conclude that ð¥ð¡ â 0 so that (ð¥ð¡ ) â ð0 . This establishes ð1 â ð0 . To see that the inclusion is strict, that is ð1 â ð0 , we observe that the sequence (1/ð) = (1, 1/2, 1/3, . . . ) converges to zero but that since â   â 1   = 1 + 1 + 1+ = â ð 2 3 ð=1 (1/ð) â / ð1 . Every convergent sequence is bounded (Exercise 1.97). Therefore ð0 â ðâ . 2. Clearly, every sequence (ðð¡ ) â ð1 deï¬nes a linear functional ð â ðâ²0 given by ð (x) =

â â

ðð¡ ð¥ð¡

ð=1

for every x = (ð¥ð¡ ) â ð0 . To show that ð is bounded we observe that every (ð¥ð¡ ) â ð0 is bounded and consequently â£ð (x)â£ â€

â â ð=1

â£ðð¡ â£ â£ð¥ð¡ â£ â€ â¥(ð¥ð¡ )â¥â

â â ð=1

â£ðð¡ â£ = â¥(ðð¡ )â¥1 â¥(ð¥ð¡ )â¥â

Therefore ð â ðâ0 . To show the converse, let eð¡ denote the unit sequences e1 = (1, 0, 0, . . . ) e2 = (0, 1, 0, . . . ) e3 = (0, 0, 1, . . . ) {e1 , e2 , e3 , . . . , } form a basis for ð0 . Then every sequence (ð¥ð¡ ) â ð0 has a unique representation (ð¥ð¡ ) =

â â

ð¥ð¡ eð¡

ð=1

Let ð â ðâ0 be a continuous linear functional on ð0 . By continuity and linearity ð (x) =

â â

ð¥ð¡ ð (eð¡ )

ð=1

Let ðð¡ = ð (eð¡ ) so that ð (x) =

â â

ðð¡ ð¥ð¡

ð=1

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

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We need to show that the sequence (ðð¡ ) â ð1 . For any ð , consider the sequence xð¡ = (ð¥1 , ð¥2 , . . . , ð¥ð¡ , 0, 0, . . . ) where â§ âš0 ðð¡ = 0 or ð â¥ ð ð¥ð¡ = â£ðð¡ â£ â© otherwise ðð¡ Then (xð¡ ) â ð0 , â¥xð¡ â¥â = 1 and ð (xð¡ ) =

ð¡ â

ðð¡ ð¥ð¡ =

ð=1

ð¡ â

â£ðð¡ â£

ð=1

Since ð â ðâ0 , ð is bounded and therefore ð (xð¡ ) â€ â¥ð â¥ â¥xð¡ â¥ = â¥ð â¥ < â and therefore ð¡ â

â£ðð¡ â£ < â for every ð = 1, 2, . . .

ð=1

Consequently â â

â£ðð¡ â£ = sup

ð¡ â

ð ð=1

ð=1

â£ðð¡ â£ â€ â¥ð â¥ < â

We conclude that (ðð¡ ) â ð1 and therefore ðâ0 = ð1 3. Similarly, every sequence (ðð¡ ) â ðâ deï¬nes a linear functional ð on ð1 given by ð (x) =

â â

ðð¡ ð¥ð¡

ð=1

for every x = (ð¥ð¡ ) â ð1 . Moreover ð is bounded since â£ð (x)â£ â€

â â

â£ðð¡ â£ â£ð¥ð¡ â£ â€ â¥(ðð¡ )â¥

ð=1

â â

â£ð¥ð¡ â£ < â

ð=1

for every x = (ð¥ð¡ ) â ð1 Again, given any linear functional ð â ð1â , let ðð¡ = ð (eð¡ ) where eð¡ is the ð unit sequence. Then ð has the representation ð (x) =

â â

ðð¡ ð¥ð¡

ð=1

To show that (ðð¡ ) â ðâ , for ð = 1, 2, . . . , consider the sequence xð¡ = (0, 0, . . . , ð¥ð¡ , 0, 0, . . . ) where â§ âš â£ðð¡ â£ ð = ð and ð â= 0 ð¡ ðð¡ ð¥ð¡ = â© 0 otherwise Then xð¡ â ð1 , â¥xð¡ â¥1 = 1 and ð (xð¡ ) = â£ðð¡ â£ Since ð â

ð1â ,

ð is bounded and therefore  ð ð  = ð (xð ) â€ â¥ð â¥ â¥xð â¥ = â¥ð â¥

for every ð . Consequently (ðð ) â ðâ . We conclude that ð1â = ðâ 123

Solutions for Foundations of Mathematical Economics 3.47 By linearity ð(ð¥, ð¡) = ð(ð¥, 0) + ð(0, ð¡) = ð(ð¥, 0) + ð(0, 1)ð¡

Considered as a function of ð¥, ð(ð¥, 0) is a linear functional on ð. Deï¬ne ð(ð¥) = ð(ð¥, 0) ðŒ = ð(0, 1) Then ð(ð¥, ð¡) = ð(ð¥) + ðŒð¡ 3.48 Suppose ð â©

kernel ðð â kernel ð

ð=1

Deï¬ne the function ðº : ð â âð by ðº(x) = (ð1 (x), ð2 (x), . . . , ðð (x)) Then kernel ðº = { x â ð : ðð (x) = 0, ð = 1, 2, . . . ð } ð â© = kernel ðð ð=1

â kernel ð ð : ð â â and ðº : ð â âð . By Exercise 3.22, there exists a linear function ð» : âð â â such that ð = ð» â ðº. That is, for every ð¥ â ð ð (x) = ð» â ðº(x) = ð»(ð1 (x), ð2 (x), . . . , ðð (x)) Let ðŒð = ð»(eð ) where eð is the ð-th unit vector in âð . Since every linear mapping is determined by its action on a basis, we must have ð (x) = ðŒ1 ð1 (x) + ðŒ2 ð2 (x) + â â â + ðŒð ðð (x)

for every ð¥ â ð

That is ð â lin ð1 , ð2 , . . . , ðð Conversely, suppose ð â lin ð1 , ð2 , . . . , ðð That is ð (x) = ðŒ1 ð1 (x) + ðŒ2 ð2 (x) + â â â + ðŒð ðð (x) for every ð¥ â ð â©ð For every x â ð=1 kernel ðð , ðð (x) = 0, ð = 1, 2, . . . , ð and therefore ð (x) = 0. Therefore ð¥ â kernel ð . That is ð â©

kernel ðð â kernel ð

ð=1

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3.49 Let ð» be a hyperplane in ð. Then there exists a unique subspace ð such that ð» = x0 + ð for some x0 â ð» (Exercise 1.153). There are two cases to consider. Case 1: x0 â / ð . For every x â ð, there exists unique ðŒx â â such x = ðŒx x0 + ð£ for some ð£ â ð Deï¬ne ð (x) = ðŒx . Then ð : ð â â. It is straightforward to show that ð is linear. Since ð» = x0 + ð , ðŒx = 1 if and only if x â ð». Therefore ð» = { x â ð : ð (x) = 1 } Case 2: x0 â ð . In this case, choose some x1 â / ð . Again, for every x â ð, there exists a unique ðŒx â â such x = ðŒx x1 + ð£ for some ð£ â ð and ð (x) = ðŒx is a linear functional on ð. Furthermore x0 â ð implies ð» = ð (Exercise 1.153) and therefore ð (x) = 0 if and only if x â ð». Therefore ð» = { x â ð : ð (x) = 0 } Conversely, let ð be a nonzero linear functional in ð â² . Let ð = kernel ð and choose x0 â ð â1 (1). (This is why we require ð â= 0). For any x â ð ð (x â ð (x)x0 ) = ð (x) â ð (x) Ã 1 = 0 so that x â ð (x)x0 â ð . That is, x = ð (x)x0 + ð£ for some ð£ â ð . Therefore, ð = lin (x0 , ð ) so that ð is a maximal proper subspace. For any ð â â, let x1 â ð â1 (ð). Then, for every x â ð â1 (ð), ð (x â x1 ) = 0 and { x : ð (x) = ð} = {x : ð (x â x1 ) = 0 } = x1 + ð which is a hyperplane. 3.50 By the previous exercise, there exists a linear functional ð such that ð» = { ð¥ â ð : ð (ð¥) = ð } for some ð â â. Since 0 â / ð», ð â= 0. Without loss of generality, we can assume that ð = 1. (Otherwise, take the linear functional 1ð ð ). To show that ð is unique, assume that ð is another linear functional with ð» = { x : ð (ð¥) = 1} = {x : ð(ð¥) = 1 } Then ð» â { x : ð (ð¥) â ð(ð¥) = 0 } Since ð» is a maximal subset, ð is the smallest subspace containing ð». Therefore ð (ð¥) = ð(ð¥) for every ð¥ â ð.

125

Solutions for Foundations of Mathematical Economics

3.51 By Exercise 3.49, there exists a linear functional ð such that ð» = { ð¥ â ð : ð (ð¥) = 0 } Since x0 â / ð», ð (x0 ) â= 0. Without loss of generality, we can normalize so that ð (x0 ) = 1. (If ð (x0 ) = ð â= 1, then the linear functional ð â² = 1/cð has ð â² (x0 ) = 1 and kernel ð â² = ð».) To show that ð is unique, suppose that ð is another linear functional with kernel ð = ð» and ð(x0 ) = 1. For any x â ð, there exists ðŒ â â such that x = ðŒx0 + v with ð£ â ð» (Exercise 1.153). Since ð (v) = ð(v) = 0 and ð (x0 ) = ð(x0 ) = 1 ð(x) = ð(ðŒx0 + v) = ðŒð(ð¥0 ) = ðŒð (x0 ) = ð (ðŒx0 + v) = ð (x) 3.52 Assume ð = ðð, ð â= 0. Then ð (ð¥) = 0 ââ ð(ð¥) = 0 Conversely, let ð» = ð â1 (0) = ð â1 (0). If ð» = ð, then ð = ð = 0. Otherwise, ð» is a hyperplane containing 0. Choose some x0 â / ð». Every x â ð has a unique representation x = ðŒx0 + v with v â ð» (Exercise 1.153) and ð (x) = ðŒð (x0 ) ð(x) = ðŒð(x0 ) Let ð = ð (x0 )/ð(x0 ) so that ð (x0 ) = ðð(x0 ). Substituting ð (x) = ðŒð (x0 ) = ðŒðð(x0 ) = ðð(x) 3.53 ð continuous implies that the set { ð¥ â ð : ð (ð¥) = ð } = ð â1 (ð) is closed for every ð â â (Exercise 2.67). Conversely, let ð = 0 and assume that ð» = { ð¥ â ð : ð (ð¥) = 0 } is closed. There exists x0 â= 0 such that ð = lin {ð¥0 , ð»} (Exercise 1.153). Let xð â x be a convergent sequence in ð. Then there exist ðŒð , ðŒ â â and vð , ð£ â ð» such that xð = ðŒð x0 + vð , x = ðŒx0 + v and â¥xð â xâ¥ = â¥ðŒð x0 + vð â ðŒx0 + vâ¥ = â¥ðŒð x0 â ðŒx0 + vð â vâ¥ â€ â£ðŒð â ðŒâ£ â¥x0 â¥ + â¥vð â vâ¥ â0 which implies that ðŒð â ðŒ. By linearity ð (xð ) = ðŒð ð (x0 ) + ð (vð ) = ðŒð ð (x0 ) since vð â ð» and therefore ð (xð ) = ðŒð ð (x0 ) â ðŒð (x0 ) = ð (x) ð is continuous.

126

Solutions for Foundations of Mathematical Economics 3.54 ð (x + xâ² , y) =

ð â ð â

ððð (ð¥ð + ð¥â²ð )ðŠð

ð=1 ð=1

=

ð â ð â

ððð ð¥ð ðŠð +

ð=1 ð=1

ð â ð â

ððð ð¥â²ð ðŠð

ð=1 ð=1

= ð (x, y) + ð (xâ² , y) Similarly, we can show that ð (x, y + yâ² ) = ð (x, y) + ð (x, yâ² ) and ð (ðŒx, y) = ðŒð (x, y) = ð (x, ðŒy) for every ðŒ â â 3.55 Let x1 , x2 , . . . , xð be a basis for ð and y1 , y2 , . . . , yð be a basis for ð . Let the numbers ððð represent the action of ð on these bases, that is ððð = ð (xð , yð )

ð = 1, 2, . . . , ð, ð = 1, 2, . . . , ð

and let ðŽ be the ð Ã ð matrix of numbers ððð . Choose any x â ð and y â ð and let their representations in terms of the bases be x=

ð â

ðŒð xð and y =

ð=1

ð â

ðœð yð

ð=1

respectively. By the bilinearity of ð â â ð (x, y) = ð ( ðŒð xð , ðŒð yð ) =

â

ð

ðŒð ð (xð ,

ð

=

â

ðŒð

ð

=

â

ð

â

â

ðŒð yð )

ð

ðŒð ð (xð , yð )

ð

ðŒð

â

ð

=

â

ðŒð ððð

ð

ðŒð ðŽy

ð â²

= x ðŽy 3.56 Every y â ð â² is a linear functional on ð. Hence y(x + xâ² ) = y(x) + y(xâ² ) y(ðŒx) = ðŒy(x) and therefore ð (x + xâ² , y) = y(x + xâ² ) = y(x) + y(xâ² ) = ð (x, y) + ð (xâ² , y) ð (ðŒx, y) = y(ðŒx) = ðŒy(x) = ðŒð (x, y) 127

Solutions for Foundations of Mathematical Economics

In the dual space ð â² (y + yâ² )(x) â¡ y(x) + yâ² (x) (ðŒy)(x) â¡ ðŒy(x) and therefore ð (x, y + yâ² ) = (y + yâ² )(x) = y(x) + yâ² (x) = ð (x, y) + ð (x, yâ² ) ð (x, ðŒy) = (ðŒy)(x) = ðŒy(x) = ðŒð (x, y) 3.57 Assume ð1 , ð2 â ðµðð¿(ð Ã ð, ð). Deï¬ne the mapping ð1 + ð2 : ð Ã ð â ð by (ð1 + ð2 )(x, y) = ð1 (x, y) + ð2 (x, y) We have to conï¬rm that ð1 + ð2 is bilinear, that is (ð1 + ð2 )(x1 + x2 , y) = ð1 (x1 + x2 , y) + ð2 (x1 + x2 , y) = ð1 (x1 , y) + ð1 (x2 , y) + ð2 (x1 , y) + ð2 (x2 , y) = ð1 (x1 , y) + ð2 (x1 , y) + ð1 (x1 , y) + ð2 (x2 , y) = (ð1 + ð2 )(x1 , y) + (ð1 + ð2 )(x2 , y) Similarly, we can show that (ð1 + ð2 )(x, y1 + y2 ) = (ð1 + ð2 )(x, y1 ) + (ð1 + ð2 )(x, y2 ) and (ð1 + ð2 )(ðŒx, y) = ðŒ(ð1 + ð2 )(x, y) = (ð1 + ð2 )(x, ðŒy) For every ð â ðµðð¿(ð Ã ð, ð) deï¬ne the function ðŒð : ð Ã ð â ð by (ðŒð )(x, y) = ðŒð (x, y) ðŒð is also bilinear, since (ðŒð )(x1 + x2 , y) = ðŒð (x1 + x2 , y) = ðŒð (x1 , y) + ðŒð (x2 , y) = (ðŒð )(x1 , y) + (ðŒð (x2 , y) Similarly (ðŒð )(x, y1 + y2 ) = (ðŒð )(x, y1 ) + (ðŒð )(x, y2 ) (ðŒð )(ðœx, y) = ðœ(ðŒð )(x, y) = (ðŒð )(x, ðœy) Analogous to (Exercise 2.78), ð1 + ð2 and ðŒð are also continuous 3.58

1. ðµð¿(ð, ð) is a linear space and therefore so is ðµð¿(ð, ðµð¿(ð, ð)) (Exercise 3.33).

128

Solutions for Foundations of Mathematical Economics

2. ðx is linear and therefore ð (x, y1 + y2 ) = ð(x)(y1 + y2 ) = ð(x)(y1 ) + ð(x)(y2 ) = ð (x, y1 ) + ð (x, y2 ) and ð (x, ðŒy) = ð(x)(ðŒy) = ðŒð(x)(y) = ðŒð (x, y) Similarly, ð is linear and therefore ð (x1 + x2 , y) = ðx1 +x2 (y) = ðx1 (y) + ðx2 (y) = ð (x1 , y) + ð (x2 , y) and ð (ðŒx, y) = ððŒx (y) = ðŒðx (y) = ðŒð (x, y) ð is bilinear 3. Let ð â ðµðð¿(ð Ã ð, ð). For every x â ð, the partial function ðx : ð â ð is linear. Therefore ðx â ðµð¿(ð, ð) and ð â ðµð¿(ð, ðµð¿(ð, ð)). 3.59 Bilinearity and symmetry imply ð (x â ðŒy, x â ðŒy) = ð (x, x â ðŒy) â ðŒð (y, x â ðŒy) = ð (x, x) â ðŒð (x, y) â ðŒð (y, x) + ðŒ2 ð (y, y) = ð (x, x) â 2ðŒð (x, y) + ðŒ2 ð (y, y) Nonnegativity implies ð (x â ðŒy, x â ðŒy) = ð (x, x) â 2ðŒð (x, y) + ðŒ2 ð (y, y) â¥ 0

(3.38)

for every x, y â ð and ðŒ â â Case 1 ð (x, x) = ð (y, y) = 0 Then (3.38) becomes â2ðŒð (x, y) â¥ 0 Setting ðŒ = ð (x, y) generates ( )2 â2 ð (x, y) â¥ 0 which implies that ð (x, y) = 0 Case 2 Either ð (x, x) > 0 or ð (y, y) > 0. Without loss of generality, assume ð (y, y) > 0 and set ðŒ = ð (x, y)/ð (y, y) in (3.38). That is ( ) ( )2 ð (x, y) ð (x, y) ð (x, x) â 2 ð (x, y) + ð (y, y) â¥ 0 ð (y, y) ð (y, y) or ð (x, x) â

ð (x, y)2 â¥0 ð (y, y)

which implies ( )2 ð (x, y) â€ ð (x, x)ð (y, y) for every x, y â ð 129

Solutions for Foundations of Mathematical Economics

3.60 A Euclidean space is a ï¬nite-dimensional normed space, which is complete (Proposition 1.4). 3.61 ð (x, y) = xð y satisï¬es the requirements of Exercise 3.59 and therefore (xð y)2 â€ (xð x)(yð y) Taking square roots  ð  x y â€ â¥xâ¥ â¥yâ¥ 3.62 By deï¬nition, the inner product is a bilinear functional. To show that it is continuous, let ð be an inner product space with inner product denote by xð y. Let xð â x and yð â y be sequences in ð.  ð ð ð    (x ) y â xð y = (xð )ð yð â (xð )ð y + (xð )ð y â xð y     â€ (xð )ð yð â (xð )ð y + (xð )ð y â xð y     â€ (xð )ð (yð â y) + (xð â x)ð y Applying the Cauchy-Schwartz inequality   ð ð ð (x ) y â xð y â€ â¥xð â¥ â¥yð â yâ¥ + â¥xð â xâ¥ â¥yâ¥ Since the sequence xð converges, it is bounded, that is there exists ð such that â¥xð â¥ â€ ð for every ð. Therefore  ð ð ð  (x ) y â xð y â€ â¥xð â¥ â¥yð â yâ¥ + â¥xð â xâ¥ â¥yâ¥ â€ ð â¥yð â yâ¥ + â¥xð â xâ¥ â¥yâ¥ â 0 3.63 Applying the properties of the inner product â â â¥xâ¥ = xð x â¥ 0 â â â¥xâ¥ = xð x = 0 if and only if x = 0 â â â â¥ðŒxâ¥ = (ðŒx)ð (ðŒx) = ðŒ2 xð x = â£ðŒâ£ â¥xâ¥ To prove the triangle inequality, observe that bilinearity and symmetry imply 2

â¥x + yâ¥ = (x + y)ð (x + y) = xð x + xð y + yð x + zð z = xð x + 2xð y + yð y 2

2

= â¥xâ¥ + 2xð y + â¥yâ¥   â€ â¥xâ¥2 + 2 xð y + â¥yâ¥2 Applying the Cauchy-Schwartz inequality 2

2

â¥x + yâ¥ â€ â¥xâ¥ + 2 â¥xâ¥ â¥yâ¥ + â¥yâ¥

2

= (â¥xâ¥ + â¥yâ¥)2 3.64 For every y â ð, the partial function ðy (x) = xð y is a linear functional on ð (since xð y is bilinear). Continuity follows from the Cauchy-Schwartz inequality, since for every x â ð   â£ðy (x)â£ = xð y â€ â¥yâ¥ â¥xâ¥ 130

Solutions for Foundations of Mathematical Economics which shows that â¥ðy â¥ â€ â¥yâ¥. In fact, â¥ðy â¥ = â¥yâ¥ since â¥ðy â¥ = sup â£ðy (x)â£ â¥xâ¥=1

 ( )  y  â¥ ðy â¥yâ¥  (   y )ð    = y  â¥yâ¥  =

1 ð y y = â¥yâ¥ â¥yâ¥

3.65 By the Weierstrass Theorem (Theorem 2.2), the continuous function ð(x) = â¥xâ¥ attains a maximum on the compact set ð at some point x0 . We claim that x0 is an extreme point. Suppose not. Then, there exist x1 , x2 â ð such that x0 = ðŒx1 + (1 â ðŒ)x2 = x2 + ðŒ(x1 â x2 ) Since x0 maximizes â¥xâ¥ on ð

( )ð ( ) 2 2 x2 + ðŒ(x1 â x2 ) â¥x2 â¥ â€ â¥x0 â¥ = x2 + ðŒ(x1 â x2 ) 2

= â¥x2 â¥ + 2ðŒxð2 (x1 â x2 ) + ðŒ2 â¥x1 â x2 â¥

2

or 2

2xð2 (x1 â x2 ) + ðŒ â¥x1 â x2 â¥ â¥ 0

(3.39)

Similarly, interchanging the role of x1 and x2 2

2xð1 (x2 â x1 ) + ðŒ â¥x2 â x1 â¥ â¥ 0 or â2xð1 (x1 â x2 ) + ðŒ â¥x1 â x2 â¥2 â¥ 0

(3.40)

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

2(x2 â x1 )ð (x1 â x2 ) + 2ðŒ â¥x1 â x2 â¥ â¥ 0 or 2(x2 â x1 )ð (x2 â x1 ) = â2(x2 â x1 )ð (x1 â x2 ) â€ 2ðŒ â¥x1 â x2 â¥2 and therefore â¥x2 â x1 â¥ â€ ðŒ â¥x2 â x1 â¥ Since 0 < ðŒ < 1, this implies that â¥x1 â x2 â¥ = 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 â¥x + yâ¥2 + â¥x â yâ¥2 = (x + y)ð (x + y) + (x â y)ð (x â y) = xð x + xð y + yð x + yð y + xð x â xð y â yð x + yð y = 2xð x + 2yð y 2

= 2 â¥xâ¥ + 2 â¥yâ¥ 131

2

Solutions for Foundations of Mathematical Economics 3.67 Note that â¥ð¥â¥ = â¥ðŠâ¥ = 1 and â¥ð¥ + ðŠâ¥ = sup

( ) ð¥(ð¡) + ðŠ(ð¡) = sup (1 + ð¡) = 2

â¥ð¥ â ðŠâ¥ = sup

( ) ð¥(ð¡) â ðŠ(ð¡) = sup (1 â ð¡) = 1

0â€ð¡â€1 0â€ð¡â€1

0â€ð¡â€1 0â€ð¡â€1

so that 2

2

2

â¥ð¥ + ðŠâ¥ + â¥ð¥ â ðŠâ¥ = 5 â= 2 â¥ð¥â¥ + 2 â¥ð¥â¥

2

Since ð¥ and ðŠ do not satisfy the parallelogram law (Exercise 3.66), ð¶(ð) cannot be an inner product space. 3.68 Let {x1 , x2 , . . . , xð } be a set of pairwise orthogonal vectors. Assume 0 = ðŒx1 + ðŒ2 x2 + â â â + ðŒð xð Using bilinearity, this implies 0 = 0ð xð =

ð â

ðŒð xðð xð = ðŒð â¥xð â¥

ð=1

for every ð = 1, 2, . . . , ð. Since xð â= 0, this implies ðŒð = 0 for every ð = 1, 2, . . . , ð. We conclude that the set {x1 , x2 , . . . , xð } is linearly independent (Exercise 1.133). 3.69 Let x1 , x2 , . . . , xð be a orthonormal basis for ð. Since ðŽ represents ð ð (xð ) =

ð â

ððð xð

ð=1

for ð = 1, 2, . . . , ð. Taking the inner product with xð , ( ð ) ð â â ð ð xð ð (xð ) = xð ððð xð = ððð xðð xð ð=1

ð=1

Since {x1 , x2 , . . . , xð } is orthonormal { xðð xð

=

1 0

if ð = ð otherwise

so that the last sum simpliï¬es to xðð ð (xð ) = ððð for every ð, ð 3.70

1. By the Cauchy-Schwartz inequality  ð  x y â€ â¥xâ¥ â¥xâ¥ for every x and y, so that  ð   x y  â€1 â£cos ðâ£ =  â¥xâ¥ â¥yâ¥  which implies â1 â€ cos ð â€ 1 132

Solutions for Foundations of Mathematical Economics

2. Since cos 90 = 0, ð = 90 implies that xð y = 0 or x â¥ y. Conversely, if x â¥ y, xð y = 0 and cos ð = 0 which implies ð = 90 degrees. 3.71 By bilinearity 2

2

â¥x + yâ¥ = (x + y)ð (x + y) = â¥xâ¥ + xð y + yð x + â¥yâ¥

2

If x â¥ y, xð y = yð x = 0 and 2

2

â¥x + yâ¥ = â¥xâ¥ + â¥yâ¥ 3.72

2

Ë â ð and let ðË be the set of all x â ð which are closer to y 1. Choose some x Ë , that is than x ðË = { x â ð : â¥x â yâ¥ â€ â¥Ë x â yâ¥ } ðË is compact (Proposition 1.4). By the Weierstrass theorem (Theorem 2.2), the continuous function ð(x) = â¥xyâ¥ Ë That is attains a minimum on ðË at some point x0 â ð. â¥x0 â yâ¥ â€ â¥x â yâ¥ for every x â ðË A fortiori â¥x0 â yâ¥ â€ â¥x â yâ¥ for every x â ð

2. Suppose there exists some x1 â ð such that â¥x1 â yâ¥ = â¥x0 â yâ¥ = ð¿ By the parallelogram law (Exercise 3.66) 2

â¥x0 â x1 â¥ = â¥x0 â y + y â x1 â¥ 2

2 2

= 2 â¥x0 â yâ¥ + 2 â¥x1 â yâ¥ â â¥(x0 â y) â (y â x1 )â¥  2   2 2 2 1 = 2 â¥x0 â yâ¥ + 2 â¥x1 â yâ¥ â 2  (x0 + x1 ) â y  2  2 1   = 2ð¿ 2 + 2ð¿ 2 â 22   2 (x0 + x1 ) â y

2

  since 12 (x0 + x1 ) â ð and therefore  12 (x0 + x1 ) â y â¥ ð¿ so that 2

â¥x0 â x1 â¥ â€ 2ð¿ 2 + 2ð¿ 2 â 4ð¿ 2 = 0 which implies that x1 = x0 . 3. Let x â ð. Since ð is convex, the line segment ðŒx+(1âðŒ)x0 = x0 +ðŒ(xâx0 ) â ð and therefore (since x0 is the closest point) ( 2 ) â¥x0 â yâ¥2 â€  x0 + ðŒ(x â x0 ) â y 2

= â¥(x0 â y) + ðŒ(x â x0 )â¥ ( )ð ( ) = (x0 â y) + ðŒ(x â x0 ) (x0 â y) + ðŒ(x â x0 ) 2

= â¥x0 â yâ¥ + 2ðŒ(x0 â y)ð (x â x0 ) + ðŒ2 â¥x â x0 â¥ 133

2

Solutions for Foundations of Mathematical Economics which implies that 2

2ðŒ(x0 â y)ð (x â x0 ) + ðŒ2 â¥x â x0 â¥ â¥ 0 Dividing through by ðŒ 2

2(x0 â y)ð (x â x0 ) + ðŒ â¥x â x0 â¥ â¥ 0 which inequality must hold for every 0 < ðŒ < 1. Letting ðŒ â 0, we must have (x0 â y)ð (x â x0 ) â¥ 0 as required. 3.73

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

â¥xð â xð â¥ = â¥(xð â y) + (y â xð )â¥

2

2

2

2

= 2 â¥xð â yâ¥ + 2 â¥y â xð â¥ â 2 â¥xð + xð â¥

for every ð, ð. Since ð is convex, (xð +xð )/2 â ð and therefore â¥xð + xð â¥ â¥ 2ð. Therefore 2

2

2

â¥xð â xð â¥ = 2 â¥xð â yâ¥ + 2 â¥y â xð â¥ â 4ð2 Since â¥xð â yâ¥ â ð and â¥xð â yâ¥ â ð as ð, ð â â, we conclude that 2 â¥xð â xð â¥ â 0. That is, (xð ) is a Cauchy sequence. 2. Since ð is a closed subspace of complete space, there exists x0 â ð such that xð â x0 . By continuity of the norm â¥x0 â yâ¥ = lim â¥xð â yâ¥ = ð ðââ

Therefore â¥x0 â yâ¥ â€ â¥x â yâ¥ for every x â ð Uniqueness follows in the same manner as the ï¬nite-dimensional case. 3.74 Deï¬ne ð : ð â ð by ð(y) = { x â ð : x is closest to y } The function ð is well-deï¬ned since for every y â ð there exists a unique point x â ð which is closest to y (Exercise 3.72). Clearly, for every x â ð, x is the closest point to x. Therefore ð(x) = x for every x â ð. To show that ð is continuous, choose any y1 and y2 in ð x1 = ð(y1 ) and x2 = ð(y2 )

134

Solutions for Foundations of Mathematical Economics

be the corresponding closest points in ð. Then ( )ð ( ) 2 â¥(y1 â y2 ) â (x1 â x2 )â¥ = (y1 â y2 ) â (x1 â x2 ) (y1 â y1 ) â (x1 â x2 ) = (y1 â y2 )ð (y1 â y2 ) + (x1 â x2 )ð (x1 â x2 ) â 2(y1 â y2 )ð (x1 â x2 ) = â¥y1 â y2 â¥2 + â¥x1 â x2 â¥2 â 2(y1 â y2 )ð (x1 â x2 ) 2

2

= â¥y1 â y2 â¥ + â¥x1 â x2 â¥ â 2(y1 â y2 )ð (x1 â x2 ) 2

â 2 â¥x1 â x2 â¥ + 2(x1 â x2 )ð (x1 â x2 ) 2

2

= â¥y1 â y2 â¥ â â¥x1 â x2 â¥ ( )ð + 2 (x1 â x2 ) â (y1 â y2 ) (x1 â x2 ) = â¥y1 â y2 â¥2 â â¥x1 â x2 â¥2 + 2(x1 â y1 )ð (x1 â x2 ) â 2(x2 â y2 )ð (x1 â x2 ) 2

2

= â¥y1 â y2 â¥ â â¥x1 â x2 â¥

â 2(x1 â y1 )ð (x2 â x1 ) â 2(x2 â y2 )ð (x1 â x2 ) so that 2

2

â¥y1 â y2 â¥ â â¥x1 â x2 â¥ = â¥(y1 â y2 ) â (x1 â x2 )â¥

2

+ 2(x1 â y1 )ð (x2 â x1 ) + 2(x2 â y2 )ð (x1 â ð¥2 ) Using Exercise 3.72 (x1 â y1 )ð (x2 â x1 ) â¥ 0 and (x2 â y2 )ð (x1 â x2 ) â¥ 0 which implies that the left-hand side â¥y1 â y2 â¥2 â â¥x1 â x2 â¥2 â¥ 0 or â¥x1 â x2 â¥ = â¥ð(y1 ) â ð(y2 )â¥ â€ â¥y1 â y2 â¥ ð is Lipschitz continuous. 3.75 Let ð = kernel ð . Then ð is a closed subspace of ð. If ð = ð, then ð is the zero functional and y = 0 is the required element. Otherwise chose any y â / ð and let x0 be the closest point in ð (Exercise 3.72). Deï¬ne z = x0 â y. Then z â= 0 and zð x â¥ 0 for every x â ð Since ð is subspace, this implies that zð x = 0 for every x â ð that is z is orthogonal to ð. Let ðË be the subset of ð deï¬ned by ðË = { ð (x)z â ð (z)x : x â ð } For every x â ðË

( ) ð (x) = ð ð (x)z â ð (z)x = ð (x)ð (z) â ð (z)ð (x) = 0 135

Solutions for Foundations of Mathematical Economics Therefore ðË â ð. For every x â ð

( )ð ð (x)z â ð (z)x z = ð (x)zð z â ð (z)xð z = 0 since z â ð â¥ . Therefore ð (x) =

ð (z) â¥zâ¥

(

ð 2x z

= xð

zð (z) â¥zâ¥

2

) = xð y

where y=

zð (z) â¥zâ¥

2

3.76 ð â 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 ð is a Hilbert space. By the Riesz representation theorem (Exercise 3.75), for every ð â ð â there exists yð â ð such that ð (x) =< x, yð > for every x â ð Furthermore, if yð represents ð and yð represents ð â ð â , then yð + yð represents ð + ð and ðŒyð represents ðŒð since (ð + ð)(x) = ð (x) + ð(x) =< x, yð > + < x, yð >=< x, yð + yð > (ðŒð )(x) = ðŒð (x) = ðŒ < x, yð >=< x, ðŒyð > Deï¬ne an inner product on ð â by < ð, ð >=< yð , yð > We show that it satisï¬es the properties of an inner product, namely symmetry < ð, ð >=< yð , yð >=< yð , yð >=< ð, ð > additivity < ð1 + ð2 , ð >=< yð , yð1 +ð2 >=< yð , yð1 + yð2 >=< ð1 , ð > + < ð2 , ð > homogeneity < ðŒð, ð >=< yð , ðŒyð >= ðŒ < yð , yð >= ðŒ < ð, ð > positive deï¬niteness < ð, ð >=< yð , yð >â¥ 0 and < ð, ð >=< yð , yð >= 0 if and only if ð = ð. Therefore, ð â is a complete inner product space, that is a Hilbert space. 3.77 Let ð be a Hilbert space. Applying the previous exercise a second time, ð ââ is also a Hilbert space. Let ð¹ be an arbitrary functional in ð ââ . By the Riesz representation theorem, there exists ð â ð â such that ð¹ (ð ) =< ð, ð > for every ð â ð â Again by the Riesz representation theorem, there exists xð (representing ð ) and xð¹ (representing ð) in ð such that ð¹ (ð ) =< ð, ð >=< xð¹ , xð > and ð (x) =< x, xð > 136

Solutions for Foundations of Mathematical Economics In particular, ð (xð¹ ) =< xð¹ , xð >= ð¹ (ð )

That is, for every ð¹ â ð ââ , there exists an element xð¹ â ð such that ð¹ (ð ) = ð (xð¹ ) ð is reï¬exive. 3.78

1. Adapt Exercise 3.64.

2. By Exercise 3.75, there exists unique xâ â ð such that ðy (x) = xð xâ 3. Substituting ð (x)ð y = ðy (x) = xð xâ = ð¥ð ð â (y) 4. For every y1 , y2 â ð ( ( ) ) xð ð â (y1 + y2 ) = ð (x)ð y1 + y2 = ð (x)ð y1 + ð (x)ð y1 = xð ð â (y1 ) + xð ð â (y1 ) and for every y â ð xð ð â (ðŒy) = ð (x)ð ðŒy = ðŒð (x)ð y = ðŒxð ð â (y) = xð ðŒð â (y) 3.79 The zero element 0ð is a ï¬xed point of every linear operator (Exercise 3.13). 3.80 ðŽðŽâ1 = ðŒ so that det(ðŽ) det(ðŽâ1 ) = det(ðŒ) = 1 3.81 Expanding along the ðth row using (3.8) det(ð¶) =

ð â

(â1)ð+ð (ðŒððð + ðœððð ) det(ð¶ðð )

ð=1 ð â

=ðŒ

(â1)ð+ð ððð det(ð¶ðð ) + ðœ

ð=1

ð â

(â1)ð+ð ððð det(ð¶ðð )

ð=1

But the matrices diï¬er only in the ðth row and therefore ðŽðð = ðµðð = ð¶ðð ,

ð = 1, 2, . . . ð

so that det(ð¶) = ðŒ

ð â

(â1)ð+ð ððð det(ðŽðð ) + ðœ

ð=1

ð â

(â1)ð+ð ððð det(ðµðð )

ð=1

= ðŒ det(ðŽ) + ðœ det(ðµ) 3.82 Suppose that x1 and x2 are eigenvectors corresponding to the eigenvalue ð. By linearity ð (x1 + x2 ) = ð (x1 ) + ð (x2 ) = ðx1 + ðx2 = ð(x1 + x2 ) and ð (ðŒx1 ) = ðŒð (x1 ) = ðŒðx Therefore x1 + x2 and ðŒx1 are also eigenvectors. 137

Solutions for Foundations of Mathematical Economics

3.83 Suppose ð is singular. Then there exists x â= 0 such that ð (x) = 0. Therefore x is an eigenvector with eigenvalue 0. Conversely, if 0 is an eigenvalue ð (x) = 0x = 0 for any x â= 0. Therefore ð is singular. 3.84 Since ð (x) = ðx ð (x)ð x = ðxð x = ðxð x 3.85 By Exercise 3.69 ððð = xðð ð (xð ) ððð = xðð ð (xð ) = ð (xð )ð xð and therefore ððð = ððð ââ xðð ð (xð ) = ð (xð )ð xð 3.86 By bilinearity xð1 ð (x2 ) = xð1 ð2 x2 = ð2 xð1 x2 ð (x1 )ð x2 = ð1 xð1 x2 = ð1 xð1 x2 Since ð is symmetric, this implies (ð1 â ð2 )xð1 x2 = 0 and ð1 â= ð2 implies xð1 x2 = 0. 3.87

1. Since ð compact and ð is continuous (Exercises 3.31, 3.62), the maximum is attained at some x0 â ð (Theorem 2.2), that is ð = ð (x0 )ð x0 â¥ ð (x)ð x for every x â ð Hence ( )ð ð(x, y) = ðx â ð (x) y is well-deï¬ned.

2. For any x â ð ( )ð ð(x, x) = ðx â ð (x) x = ðxð x â ð (x)ð x = ð â¥xâ¥2 â ð (x)ð x ( 2

2

= ð â¥xâ¥ â â¥xâ¥ ð

x

)ð ( 2

â¥xâ¥ ) 2( ð = â¥xâ¥ ð â ð (z) z â¥ 0

since z = x/ â¥xâ¥ â ð. 138

x â¥xâ¥

) 2

Solutions for Foundations of Mathematical Economics

3. Since ð is symmetric ( )ð ð(y, x) = ðy â ð (y) x = ðyð x â ð (y)ð x = ðxð y â ð (x)ð ðŠ ( )ð = ðx â ð (x) y = ð(x, y) 4. ð satisï¬es the conditions of Exercise 3.59 and therefore (ð(x, y))2 â€ ð(x, x)ð(y, y) for every x, y â ð

(3.41)

By deï¬nition ð(x0 , x0 ) = 0 and (3.41) implies that ð(x0 , y) = 0 for every y â ð That is ( )ð ð(x0 , y) = ðx0 â ð (x0 ) y = 0 for every ðŠ â ð and therefore ðx0 â ð (x0 ) = 0 or ð (x0 ) = ðx0 In other words, x0 is an eigenvector. By construction, â¥x0 â¥ = 1. 3.88

1. Suppose x2 , x3 â ð. Then ( )ð ðŒx2 + ðœx3 x1 = ðŒxð2 x1 + ðœxð3 x1 = 0 so that ðŒx2 + ðœx3 â ð. ð is a subspace. Let {x1 , x2 , . . . , xð } be a basis for ð (Exercise 1.142). For x â ð, there exists (Exercise 1.137) unique ðŒ1 , ðŒ2 , . . . , ðŒð such that x = ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð If x â ð xð x1 = ðŒ1 xð1 x1 = 0 which implies that ðŒ1 = 0. dim ð = ð â 1.

Therefore, x2 , x3 , . . . , xð span ð and therefore

2. For every x â ð, ð (x)ð x0 = xð ð (x0 ) = xð ðx0 = ðxð x0 = 0 since ð is symmetric. Therefore ð (x) â {x0 }â¥ = ð. 3.89 Let ð be a symmetric operator. By the spectral theorem (Proposition 3.6), there exists a diagonal matrix ðŽ which represents ð . The elements of ðŽ are the eigenvalues of ð . By Proposition 3.5, the determinant of ðŽ is the product of these diagonal elements. 139

Solutions for Foundations of Mathematical Economics 3.90 By linearity ð (x) =

â

ð¥ð ð (xð )

ð

ð deï¬nes a quadratic form since â ( )ð â â ââ ââ â â ð(x) = xð ð (x) = ð¥ð xð ð¥ð ð (xð )â  = ð¥ð ð¥ð xðð ð (xð ) = ððð ð¥ð ð¥ð ð

ð

ð

ð

ð

ð

by Exercise 3.69. 3.91 Let ð be the symmetric linear operator deï¬ning ð ð(x) = xð ð (x) By the spectral theorem (Proposition 3.6), there exists an orthonormal basis x1 , x2 , . . . , xð comprising the eigenvectors of ð . Let ð1 , ð2 , . . . , ðð be the corresponding eigenvalues, that is ð (xð ) = ðð xð

ð = 1, 2 . . . , ð

Then for x = ð¥1 x1 + ð¥2 x2 + â â â + ð¥ð xð ð(x) = xð ð (x) = (ð¥1 x1 + ð¥2 x2 + â â â + ð¥ð xð )ð ð (ð¥1 x1 + ð¥2 x2 + â â â + ð¥ð xð ) = (ð¥1 x1 + ð¥2 x2 + â â â + ð¥ð xð )ð (ð¥1 ð (x1 ) + ð¥2 ð (x2 ) + â â â + ð¥ð ð (xð )) = (ð¥1 x1 + ð¥2 x2 + â â â + ð¥ð xð )ð (ð¥1 ð1 x1 + ð¥2 ð2 x2 + â â â + ð¥ð ðð xð ) = ð¥1 ð1 ð¥1 + ð¥2 ð2 ð¥2 + â â â + ð¥ð ðð ð¥ð = ð1 ð¥21 + ð2 ð¥22 + â â â + ðð ð¥2ð 3.92

1. Assuming that ð11 â= 0, the quadratic form can be rewritten as follows ð(ð¥1 , ð¥2 ) = ð11 ð¥21 + 2ð12 ð¥1 ð¥2 + ð22 ð¥22 = ð11 ð¥21 + 2ð12 ð¥1 ð¥2 +

ð212 2 ð212 2 ð¥ â ð¥ + ð22 ð¥22 ð11 2 ð11 2

(

( )2 ) ( ) ð ð ð212 12 12 2 = ð11 ð¥1 + 2 ð¥1 ð¥2 + ð¥2 + ð22 â ð¥22 ð11 ð11 ð11 ( )2 ( ) ð11 ð22 â ð212 ð12 = ð11 ð¥1 + ð¥2 + ð¥22 ð11 ð11 2. We observe that ð must be positive for every ð¥1 and ð¥2 provided ð11 > 0 and ð11 ð22 â ð212 > 0. Similarly ð must be negative for every ð¥1 and ð¥2 if ð11 > 0 and ð11 ð22 â ð212 > 0. Otherwise, we can choose values for ð¥1 and ð¥2 which make ð both positive and negative. Note that the condition ð11 ð22 > ð212 > 0 implies that ð11 and ð12 must have the same sign. 3. If ð11 = ð22 = 0, then ð is indeï¬nite. Otherwise, if ð11 = 0 but ð22 â= 0, then the ð can we can âcomplete the squareâ using ð22 and deduce { } { } nonnegative ð11 , ð22 â¥ 0 ð is deï¬nite if and only if and ð11 ð22 â¥ ð212 nonpositive ð11 , ð22 â€ 0 140

Solutions for Foundations of Mathematical Economics

3.93 Let ð : ð â â be a quadratic form on ð. Then there exists a linear operator ð such that ð(x) = xð ð (x) and (Exercise 3.13) ð(0) = 0ð ð (0) = 0 3.94 Suppose to the contrary that the positive (negative) deï¬nite matrix ðŽ is singular. Then there exists x â= 0 such that ðŽx = 0 and therefore xâ² ðŽx = 0 contradicting the deï¬niteness of ðŽ. 3.95 Let e1 , e2 , . . . , eð be the standard basis for âð (Example 1.79). Then for every ð eâ²ð ðŽeð = ððð > 0 3.96 Let ð be the quadratic form deï¬ned by ðŽ. By Exercise 3.91, there exists an orthonormal basis such that ð(x) = ð1 x21 + ð2 x22 + â â â + ðð x2ð where ð1 , ð2 , . . . , ðð are the eigenvalues of ðŽ. This implies â§ â§ â« â« ðð > 0ï£Ž ð(x) > 0ï£Ž ï£Ž ï£Ž ï£Ž ï£Ž ï£Ž ï£Ž âš âš â¬ â¬ ð(x) â¥ 0 ðð â¥ 0 ââ ð = 1, 2, . . . , ð ï£Žðð < 0ï£Ž ï£Žð(x) < 0ï£Ž ï£Ž ï£Ž ï£Ž ï£Ž â© â­ â© â­ ð(x) â€ 0 ðð â€ 0 3.97 Let ð1 , ð2 , . . . , ðð be the eigenvalues of ðŽ. By Exercise 3.89 det(ðŽ) = ð1 ð2 . . . ðð By Exercise 3.96, ðð â¥ 0 for every ð and therefore det(ðŽ) â¥ 0. We conclude that det(ðŽ) > 0 ââ ðð > 0 for every ð ââ ðŽ is positive deï¬nite by Exercise 3.96. 3.98

1. ðŽ0 = 0. Therefore, 0 is always a solution.

2. Assume x1 and x2 are solutions, that is ðŽx1 = 0 and ðŽx2 = 0 Then ðŽ(x1 + x2 ) = ðŽx1 + ðŽx2 = 0 x1 + x2 is also a solution. 3. Let ð be the linear function deï¬ned by ð (x) = ðŽx The system of equations ðŽx = 0 has a nontrivial solution if and only if kernel ð â= {0} ââ nullity ð > 0 By the rank theorem (Exercise 3.24) rankð + nullityð = dim ð so that nullity ð > 0 ââ rankð < dim ð = ð 141

Solutions for Foundations of Mathematical Economics 3.99

1. Assume x1 and x2 are solutions of (3.16). That is ðŽx1 = c and ðŽx2 = c Subtracting ðŽx1 â ðŽx2 = ðŽ(x1 â x2 ) = 0

2. Assume xð solves (3.16) while x is any solution to (3.17). That is ðŽxð = c and ðŽx = 0 Adding ðŽxð + ðŽx = ðŽ(xð + x) = c We conclude that xð + x solves (3.16) for every x â ðŸ. 3. If 0 is the only solution of (3.17), ðŸ = {0}. Assume x1 and x2 are solutions of (3.16). Then x1 â x2 â ðŸ = {0} which implies x1 = x2 . 3.100 Let ð = { x : ðŽx = ð }. For every x, y â ð and ðŒ â â ðŽðŒx + (1 â ðŒ)y = ðŒðŽx + (1 â ðŒ)ðŽy = ðŒc + (1 â ðŒ)c = ð Therefore, z = ðŒx + (1 â ðŒ)y â ð. ð is aï¬ne. 3.101 Let ð â= â be an aï¬ne set âð . Then there exists a unique subspace ð such that ð = x0 + ð for some x0 â ð (Exercise 1.150). The orthogonal complement of ð is ð â¥ = { a â ð : ax = 0 for every x â ð } Let (a1 , a2 , . . . , að ) be a basis for ð â¥ . Then ð = (ð â¥ )â¥ = {x : að x = 0,

ð = 1, 2, . . . ð}

Let ðŽ be the ðÃð matrix whose rows are a1 , a2 , . . . , að . Then ð is the set of solutions to the homogeneous linear system ðŽx = 0, that is ð = { ð¥ : ðŽx = 0 } Therefore ð = x0 + ð = x0 + { x : ðŽx = 0 } = { x : ðŽ(x â x0 ) = 0 } = { x : ðŽx = c } where c = ðŽx0 . 3.102 Consider corresponding homogeneous system ð¥1 + 3ð¥2 = 0 ð¥1 â ð¥2 = 0 142

Solutions for Foundations of Mathematical Economics Multiplying the second equation by 3 ð¥1 + 6ð¥2 = 0 3ð¥1 â 3ð¥2 = 0 and adding yields 4ð¥1 = 0

for which the only solution is ð¥1 = 0. Substituting in the ï¬rst equation implies ð¥2 = 0. The kernel of ð = ðŽx is {0}. Therefore dim ð (â2 ) = 2, and the system ðŽx = ð has a unique solution for every ð1 , ð2 . 3.103 We can write the system ðŽx = c in the form â â â â â â â â ð11 ð1ð ð1ð ð1 â .. â â .. â â .. â â .. â ð¥1 â . â  + â â â + ð¥ð â . â  + â â â + ð¥ð â . â  = â . â  ðð1

ððð

ððð

ðð

Subtracting c from the ðth column gives â â â â â â ð11 ð1ð ð¥ð ð1ð â ð1 â â â â . â â .. ð¥1 â ... â  + â â â + â â  + â â â + ð¥ð â .. â  = 0 . ð¥ð ððð â ðð

ðð1

so that the columns of the matrix â ð11 . . . â .. ð¶=â . ...

ðð1

ððð

(ð¥ð ð1ð â ð1 ) .. .

(ð¥ð ððð â ðð )

... ...

â ð1ð .. â . â

ððð

are linearly dependent (Exercise 1.133). Therefore det(ð¶) = 0. Let ðµð denote the matrix obtained from ðŽ by replacing the ðth column with c. Then ðŽ, ðµð and ð¶ diï¬er only in the ðth column, with the ðth column of ð¶ being a linear combination of the ðth columns of ðŽ and ðµð . â â â â â â ð1ð ð1ð ð1ð â .. â â .. â â .. â = ð¥ â â . â  â . â  ðâ . â  ððð

ððð

ððð

By Exercise 3.81 det(ð¶) = xð det(ðŽ) â det(ðµð ) = 0 and therefore ð¥ð =

det(ðµð ) det(ðŽ)

as required. 3.104 Let ( ð ð

)â1 ( ð ðŽ = ð ð¶ 143

ðµ ð·

)

Solutions for Foundations of Mathematical Economics The inverse satisï¬es the equation ( )( ð ð ðŽ ð ð ð¶ In particular, this means that ðŽ and ( ð ð By Cramerâs rule (Exercise 3.103)

where Î = ðð â ðð. Similarly

ðµ ð·

)

( 1 = 0

) 0 1

ð¶ satisfy the equation )( ) ( ) ð ðŽ 1 = ð ð¶ 0

  1 ð    0 ð ð = ðŽ =   Î ð ð   ð ð

  ð 1    ð 0 âð  = ð¶ =   Î ð ð   ð ð

ðµ and ð· are determined analogously. 3.105 A portfolio is duplicable if and only if there is a diï¬erent portfolio y â= x such that ðx = ðy or ð(x â y) = 0 There exists a duplicable portfolio if and only if this homogeneous system has a nontrivial solution, that is if rank ð < ðŽ. 3.106 State ð Â¯ is insurable if there is a solution to the linear system ðx = eð Â¯

(3.42)

where eð Â¯ is the ð Â¯-th unit vector (the ð Â¯ Arrow-Debreu security). (3.42) has a solution for every state ð  if and only if ð (âðŽ ) = âð , that is rank ð = ð. 3.107 Figure 3.1. 3.108 Let ð be an aï¬ne subset of âð . Then there exists (Exercise 3.101) a system of linear equations ðŽx = c such that ð = { x : ðŽx = c } Let að denote the ð-th row of ðŽ. Then ð = { ð¥ : að x = ðð , ð = 1, 2, . . . , ð } =

{ x : að x = ðð }

ð=1

where each { x : að x = ðð } is a hyperplane in âð (Example 3.21). 144

Solutions for Foundations of Mathematical Economics

Figure 3.1: The solutions of three equations in two unknowns 3.109 Let ð = { x : ðŽx â€ c }. For every x, y â ð and 0 â€ ðŒ â€ 1 ðŽx â€ c ðŽy â€ c and therefore ðŽðŒx + (1 â ðŒ)y = ðŒðŽx + (1 â ðŒ)ðŽy â€ ðŒc + (1 â ðŒ)c = ð Therefore, z = ðŒx + (1 â ðŒ)y â ð. ð is a convex set. 3.110 We have already seen that ð = { x : ðŽx â€ 0 } is convex. To show that it is a cone, let x â ð. Then ðŽx â€ 0 ðŽðŒx â€ 0 so that ðŒx â ð. ð is a convex cone. 3.111

1. Each column ðŽð is a vector in âð . If the set {ðŽ1 , ðŽ2 , . . . , ðŽð } is linearly independent, it has at most ð elements, that is ð â€ ð and x is a basic feasible solution.

2. (a) Assume {ðŽ1 , ðŽ2 , . . . , ðŽð } are linearly dependent. Then (Exercise 1.133) there exist numbers ðŠ1 , ðŠ2 , . . . , ðŠð , not all zero, such that ðŠ1 ðŽ1 + ðŠ2 ðŽ2 + â â â + ðŠð ðŽð = 0 y = (ðŠ1 , ðŠ2 , . . . , ðŠð ) is a nontrivial solution to the homogeneous system. (b) For every ð¡ â â, âð¡y â kernel ð = ðŽx and xâ² = x â ð¡y is a solution of the corresponding nonhomogeneous system ðŽx = c. To see this directly, subtract ðŽð¡y = 0 145

Solutions for Foundations of Mathematical Economics

from ðŽx = c to give ðŽxâ² = ðŽ(x â ð¡y) = c (c) Note that x > 0 and therefore ð¡Ë > 0 which implies that ð¥ Ëð > 0 for every ðŠð â€ 0. For every ðŠð > 0, ð¥ð /ðŠð â¥ ð¡Ë, which implies that ð¥ð â¥ ð¡ËðŠð , so that ð¥Ëð â¥ ð¥ð â ð¡ËðŠð â¥ 0 Ë is a feasible solution. Therefore, x (d) There exists some coordinate â such that ð¡Ë = ð¥â /ðŠâ so that ð¥ Ëâ = ð¥â â ð¡ËðŠâ = 0 so that ð â

c=

ð¥Ëð ðŽðœ

ð =1

ðâ=â

ð¥ Ë 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 ð, there exists ðð elements xðð and coeï¬cients ððð > 0 such that xð =

ðð â

ððð xðð

ð=1

and

â ðð

ð=1

ððð = 1. Hence x=

ð â

xð =

ð=1

ð â ðð â

ððð xðð

ð=1 ð=1

3. Direct computation. 4. Regarding the ððð as âvariablesâ and the points ð§ðð as coeï¬cents, z=

ðð ð â â

ððð zðð

ð=1 ð=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

Solutions for Foundations of Mathematical Economics

a basic feasible solution. That is, there exists coeï¬cients ððð â¥ 0 and ððð > 0 for at most (ð + ð) components such that z=

ðð ð â â

ððð zðð

(3.43)

ð=1 ð=1

Decomposing, (3.43) implies x=

ðð ð â â

ððð xðð

ð=1 ð=1

and ðð â

ððð = 1

for every ð

ð=1

5. (3.43) implies that at least one ððð > 0 for every ð. This accounts for at least ð of the positive ððð . Since there are at most (ð + ð) coeï¬cients ððð which are strictly positive, there are at most ð indices ð which have more than one positive coeï¬cient ððð . For the remaining ð â ð indices, xð = xðð for some ð; that is xð â ðð . 3.113

1. Since ðŽ is productive, there exists x â¥ 0 such that ðŽx > 0. Consider any z for which ðŽz â¥ 0. For every ðŒ > 0 ðŽ(x + ðŒz) = ðŽx + ðŒðŽz > 0

(3.44)

Suppose to the contrary that z ââ¥ 0. That is, there exists some component ð§ð < 0. Let ð§ð ðŒ = max{â } ð¥ð Without loss of generality, ð§1 attains this maximum, that is assume ðŒ = ð§1 /ð¥1 . Then ð¥1 + ðŒð§1 = 0 and ð¥ð + ðŒð§ð â¥ 0 for every ð. Now consider the matrix ðµ = ðŒ â ðŽ. By the assumptions of the Leontief model (Example 3.35), the matrix ðŽ has 1 along the diagonal and negative oï¬-diagonal elements. That is ððð = 1

ð = 1, 2, . . . , ð

ððð â€ 0

ð, ð = 1, 2, . . . , ð,

Therefore ðµ =ðŒ âðŽâ¥0 147

ð â= ð

Solutions for Foundations of Mathematical Economics

That is, every element of ðµ is nonnegative. Consequently since x + ðŒz â¥ 0 ðµ(x + ðŒz) â¥ 0

(3.45)

On the other hand, substituting ðŽ = ðŒ â ðµ in (3.45) (ðŒ â ðµ)(x + ðŒz) > 0 x + ðŒz > ðµ(x + ðŒz) which implies that the ï¬rst component of ðµ(x + ðŒz) is negative, contradicting (3.45). This contradiction establishes that z â¥ 0. Suppose ðŽx = 0. A fortiori ðŽx â¥ 0. By the previous part this implies x â¥ 0. On the other hand, it also implies that âðŽx = ðŽ(âx) = 0 so that âx â¥ 0. We conclude that x = 0 is the only solution to ðŽx = 0. ðŽ is nonsingular. Since ðŽ is nonsingular, the system ðŽx = y has a unique solution x for any y â¥ 0. By the ï¬rst part, x â¥ 0. 3.114 Suppose ðŽ is productive. By the previous exercise, ðŽ is nonsingular with inverse ðŽâ1 . Let eð be the ðth unit vector. Since eð â¥ 0, there exists xð â¥ 0 such that ðŽxð = eð Multiplying by ðŽâ1 xð = ðŽâ1 ðŽxð = ðŽâ1 eð = ðŽâ1 ð where ðŽâ1 is the ð column of ðŽâ1 . Since xð â¥ 0 for every ð, we conclude that ðŽâ1 â¥ 0. ð Conversely, assume that ðŽâ1 â¥ 0. Let 1 = (1, 1, . . . , 1) denote a net output of 1 for each commodity. Then x = ðŽâ1 1 â¥ 0 and ðŽx = 1 > 0 ðŽ is productive. 3.115 Takayama 1985, p.383, Theorem 4.C.4. 3.116 Let a0 = (ð01 , ð02 , . . . , ð0ð ) be the vector of labour requirements and ð€ the wage rate. The unit proï¬t of industry ð is â ðð = ðð + ððð ðð â ð€ð0 ðâ=ð

Recall that ððð â€ 0 for ð â= ð. The vector of unit proï¬ts for all industries is Î  = ðŽp â ð€ð0 Proï¬ts will be zero in all industries if there exists a price system p such that Î  = ðŽp â ð€ð0 = 0 or ðŽp = ð€ð0

(3.46)

By the previous results, (3.46) has a unique nonnegative solution p = ðŽâ1 ð€ð0 if the technology ðŽ is productive. Furthermore, ðŽâ1 is nonnegative. Since ð0 > 0, so is p > 0. 148

Solutions for Foundations of Mathematical Economics

3.117 Let ð¢ðµ denote the steady state unemployment rate for blacks. Then ð¢ðµ satisï¬es the equation ð¢ðµ = 0.0038(1 â ð¢ðµ ) + 0.8975ð¢ðµ which implies that ð¢ðµ = 0.036. That is, the data implies an unemployment rate of 3.6 percent for blacks. Similarly, the unemployment rate for white males ð¢ð satisï¬es the equation ð¢ð = 0.0022(1 â ð¢ð ) + 0.8614ð¢ð which implies that ð¢ð = 0.016 or 1.6 percent. 3.118 The transition matrix is

( ð =

) .6 .25 .4 .75

If the current state vector is x0 = (.4, .6), the state vector after a single mailing will be x1 = ð 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 ð (ð¥) = ð¥2 . For every ð¥1 , ð¥2 â â and 0 â€ ðŒ â€ 1 ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) = (ðŒð¥1 + (1 â ðŒ)ð¥2 )2 = (ðŒð¥1 + (1 â ðŒ)ð¥2 )(ðŒð¥1 + (1 â ðŒ)ð¥2 ) = ðŒ2 ð¥21 + 2ðŒ(1 â ðŒ)ð¥1 ð¥2 + (1 â ðŒ)2 ð¥22 = ðŒð¥21 + (1 â ðŒ)ð¥22 â ðŒð¥21 â (1 â ðŒ)ð¥22 + ðŒ2 ð¥21 + 2ðŒ(1 â ðŒ)ð¥1 ð¥2 + (1 â ðŒ)2 ð¥22 ) ( = ðŒð¥21 + (1 â ðŒ)ð¥22 â ðŒ(1 â ðŒ)ð¥21 â 2ðŒ(1 â ðŒ)ð¥1 ð¥2 + ðŒ(1 â ðŒ)ð¥22 = ðŒð¥21 + (1 â ðŒ)ð¥22 â ðŒ(1 â ðŒ)(ð¥1 â ð¥2 )2 â€ ðŒð¥21 + (1 â ðŒ)ð¥22 = ðŒð (ð¥1 ) + (1 â ðŒ)(ð¥2 ) 3.120 ð (ð¥) = ð¥ is linear and therefore convex. In the previous exercise we showed that ð¥2 is convex. Therefore ð (ð¥) = ð¥ð is convex for ð = 1, 2. Assume that ð is convex for

149

Solutions for Foundations of Mathematical Economics

ð â 1. Then ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) = (ðŒð¥1 + (1 â ðŒ)ð¥2 )ð = (ðŒð¥1 + (1 â ðŒ)ð¥2 )(ðŒð¥1 + (1 â ðŒ)ð¥2 )ðâ1 â€ (ðŒð¥1 + (1 â ðŒ)ð¥2 )(ðŒð¥ðâ1 + (1 â ðŒ)ð¥ðâ1 ) 1 2

(since ð¥ðâ1 is convex)

= ðŒ2 ð¥ð1 + ðŒ(1 â ðŒ)ð¥ðâ1 ð¥2 + ðŒ(1 â ðŒ)ð¥1 ð¥ðâ1 + (1 â ðŒ)2 ð¥ð2 1 2

= ðŒð¥ð1 + (1 â ðŒ)ð¥ð2 â ðŒð¥ð1 â (1 â ðŒ)ð¥ð2

+ ðŒ2 ð¥ð1 + ðŒ(1 â ðŒ)ð¥ðâ1 ð¥2 + ðŒ(1 â ðŒ)ð¥1 ð¥ðâ1 + (1 â ðŒ)2 ð¥ð2 1 2 ) ( ðâ1 ð = ðŒð¥ð1 + (1 â ðŒ)ð¥ð2 â ðŒ(1 â ðŒ) ð¥ð1 â ð¥1 ð¥ðâ1 â ð¥ ð¥ + ð¥ 2 2 2 1 ( ) ðâ1 ðâ1 ð ð = ðŒð¥1 + (1 â ðŒ)ð¥2 â ðŒ(1 â ðŒ) ð¥1 (ð¥1 â ð¥2 ) â ð¥2 (ð¥1 â ð¥2 ) ( ) ðâ1 â ð¥ ) = ðŒð¥ð1 + (1 â ðŒ)ð¥ð2 â ðŒ(1 â ðŒ) (ð¥1 â ð¥2 )(ð¥ðâ1 1 2 Since ð¥ð is monotonic (Example 2.53) ð¥ðâ1 â ð¥ðâ1 â¥ 0 ââ ð¥1 â ð¥2 â¥ 0 1 2 and therefore (ð¥1 â ð¥2 )(ð¥ðâ1 â ð¥ðâ1 )â¥0 1 2 We conclude that ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) â€ ðŒð¥ð1 + (1 â ðŒ)ð¥ð2 = ðŒð (ð¥1 ) + (1 â ðŒ)(ð¥2 ) ð is convex for all ð = 1, 2, . . . . 3.121 For given x1 , x2 â ð, deï¬ne ð : [0, 1] â ð by ð(ð¡) = (1 â ð¡)x1 + ð¡x2 Then ð(0) = x1 , ð(1) = x2 and â = ð â ð . Assume ð is convex. For any ð¡1 , ð¡2 â [0, 1], let Â¯ 1 and ð(ð¡2 ) = x Â¯2 ð(ð¡1 ) = x For any ðŒ â [0, 1]

) ( x1 + (1 â ðŒ)Â¯ ð ðŒð¡1 + (1 â ðŒ)ð¡2 = ðŒÂ¯ x2 ) ( ) ( x1 + (1 â ðŒ)Â¯ â ðŒð¡1 + (1 â ðŒ)ð¡2 = ð ðŒÂ¯ x2 â€ ðŒð (Â¯ x1 ) + (1 â ðŒ)ð (Â¯ x2 ) â€ ðŒâ(ð¡1 ) + (1 â ðŒ)ð¡2 )

â is convex. Conversely, assume â is convex for any x1 , x2 â ð. For any ðŒ â [0, 1] ð(ðŒ) = ðŒx1 + (1 â ðŒ)x2 and

) ( ð ðŒx1 + (1 â ðŒ)x2 = â(ðŒ) â€ ðŒâ(0) + (1 â ðŒ)â(1) = ðŒð (x1 ) + (1 â ðŒ)ð (x2 )

Since this is true for any x1 , x2 â ð, we conclude that ð is convex. 150

Solutions for Foundations of Mathematical Economics

3.122 Assume ð is convex which implies epi ð is convex. The points (xð , ð (xð )) â epi ð . Since epi ð is convex ðŒ1 (x1 , ð (x1 )) + ðŒ2 (x1 , ð (x1 )) + â â â + (xð , ð (xð )) â epi ð that is ð (ðŒ1 x1 + ðŒ2 x2 + â â â + ðŒð xð ) â€ ðŒ1 ð (x1 ) + ðŒ2 ð (x1 ) + â â â + ðŒð ð (xð )) Conversely, letting ð = 2 and ðŒ = ðŒ1 , (3.25) implies that ð (ðŒx1 + (1 â ðŒ)x2 ) â€ ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) Jensenâs inequality can also be proved by induction from the deï¬nition of a convex function (see for example Sydsaeter + Hammond 1995; p.624). 3.123 For each ð, let ðŠð = log ð¥ð so that ð¥ð = ððŠð ðŒð ðŠð ð ð¥ðŒ ð = ð

Since ðð¥ is convex (Example 3.41) ð¥ð1 1 ð¥ð2 2 . . . ð¥ððð

ðð > 0 =

â

exp(ðŒð ðŠð ) = exp

(â

) â â ðŒð ððŠð = ðŒð ð¥ð ðŒð ðŠð â€

by Jensenâs inequality. Setting ðŒð = 1/ð, we have (ð¥1 ð¥2 . . . ð¥ð )1/ð â€

ð

1â ð¥ð ð ð=1

as required. 3.124 Assume ð is concave. That is for every x1 , x2 â ð and 0 â€ ðŒ â€ 1 ð (ðŒx1 + (1 â ðŒ)x2 ) â¥ ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) Multiplying through by â1 reverses the inequality so that âð (ðŒx1 + (1 â ðŒ)x2 ) â€ âðŒð (x1 ) + (1 â ðŒ)ð (x2 ) = ðŒ â ð (x1 ) + (1 â ðŒ) â ð (x2 ) which shows that âð is concave. The converse follows analogously. 3.125 Assume that ð is concave. Then âð is convex and by Theorem 3.7 epi â ð = { (ð¥, ðŠ) â ð Ã â : ðŠ â¥ âð (ð¥), ð¥ â ð } is convex. But epi â ð = { (ð¥, ðŠ) â ð Ã â : ðŠ â¥ âð (ð¥), ð¥ â ð } = { (ð¥, ðŠ) â ð Ã â : ðŠ â€ ð (ð¥), ð¥ â ð } = hypo ð Therefore hypo ð is convex. Conversely, if hypo ð is convex, epi â ð is convex which implies that âð is convex and hence ð is concave.

151

Solutions for Foundations of Mathematical Economics

3.126 Suppose that x1 minimizes the cost of producing ðŠ at input prices w1 while x2 Â¯ be the weighted average price, that is minimizes cost at w2 . For some ðŒ â [0, 1], let w Â¯ = ðŒw1 + (1 â ðŒ)w2 w Â¯ minimizes cost at w. Â¯ Then and suppose that x Â¯ ðŠ) = wÂ¯ Â¯x ð(w, x = (ðŒw1 + (1 â ðŒ)w2 )Â¯ Â¯ + (1 â ðŒ)w2 x Â¯ = ðŒw1 x But since x1 and x2 minimize cost at w1 and w2 respectively Â¯ â¥ ðŒw1 x1 = ðŒð(w1 , ðŠ) ðŒw1 x Â¯ â¥ (1 â ðŒ)w2 x2 = (1 â ðŒ)ð(w2 , ðŠ) (1 â ðŒ)w2 x so that Â¯ + (1 â ðŒ)w2 x Â¯ â¥ ðŒð(w1 , ðŠ) + (1 â ðŒ)ðw2 , ðŠ) Â¯ ðŠ) = ð(ðŒw1 + (1 â ðŒ)w2 , ðŠ) = ðŒw1 x ð(w, This establishes that the cost function ð is concave in w. 3.127 Since ð¢ is concave, Jensenâs inequality implies ( ð ) ð ð â1 â 1 1â ðð¡ â¥ ð¢(ðð¡ ) = ð¢ ð¢(ðð¡ ) ð ð ð ð¡=1 ð¡=1 ð¡=1 for any consumption stream ð1 , ð2 , . . . , ðð so that ( ð ) ð â â1 ðð¡ = ð ð¢(Â¯ ð= ð¢(ðð¡ ) â€ ð ð¢ ð) ð ð¡=1 ð¡=1 It is impossible to do better than consume a constant fraction ðÂ¯ = ð€/ð of wealth in each period. 3.128 If ð¥1 = ð¥3 , the inequality is trivially satisï¬ed. Now assume ð¥1 â= ð¥3 . Since ð¥2 â [ð¥1 , ð¥3 ], there exists ðŒ â [0, 1] such that ð¥2 = ðŒð¥1 + (1 â ðŒ)ð¥2 Let ð¥ Â¯ = ð¥1 â ð¥2 + ð¥3 . Then ð¥ Â¯ â [ð¥1 , ð¥3 ] and there exists ðœ â [0, 1] such that ð¥Â¯ = ðœð¥1 + (1 â ðœ)ð¥2 Adding

( ) ð¥Â¯ + ð¥2 = (ðŒ + ðœ)ð¥1 + (1 â ðŒ) + (1 â ðœ) ð¥3

or ð¥1 â ð¥3 = (ðŒ + ðœ)(ð¥3 â ð¥1 ) which implies that ðŒ + ðœ = 1 and therefore ðœ = 1 â ðŒ. Since ð is convex ð (ð¥2 ) â€ ðŒð (ð¥1 ) + (1 â ðŒ)ð (ð¥2 ð (Â¯ ð¥) â€ ðœð (ð¥1 ) + (1 â ðœ)ð (ð¥2 ) = (1 â ðŒ)ð (ð¥1 ) + ðŒð (ð¥3 ) Adding ð (Â¯ ð¥) + ð (ð¥2 ) â€ ð (ð¥1 ) + ð (ð¥3 ) 152

Solutions for Foundations of Mathematical Economics

3.129 Let ð¥1 , ð¥2 , ðŠ1 , ðŠ2 â â with ð¥1 < ð¥2 and ðŠ1 < ðŠ2 . Note that ð¥1 â ðŠ2 â€ ð¥2 â ðŠ2 â€ ð¥2 â ðŠ1 and therefore (Exercise 3.128) ( ) ð ð¥1 â ðŠ2 ) â (ð¥2 â ðŠ2 ) + (ð¥2 â ðŠ1 ) > ð (ð¥1 â ðŠ2 ) â ð (ð¥2 â ðŠ2 ) + ð (ð¥2 â ðŠ1 ) That is ð (ð¥1 â ðŠ1 ) > ð (ð¥1 â ðŠ2 ) â ð (ð¥2 â ðŠ2 ) + ð (ð¥2 â ðŠ1 ) Rearranging ð (ð¥2 â ðŠ2 ) â ð (ð¥1 â ðŠ2 ) > ð (ð¥2 â ðŠ1 ) â ð (ð¥1 â ðŠ1 ) as required. 3.130 A functional is aï¬ne if and only if inequalities (3.24) and (3.26) are satisï¬ed as equalities. 3.131 Since ð and ð are convex on ð ð (ðœx1 + (1 â ðœ)x2 ) â€ ðœð (x1 ) + (1 â ðœ)ð (x2 ) 1

2

1

2

ð(ðœx + (1 â ðœ)x ) â€ ðœð(x ) + (1 â ðœ)ð(x )

(3.47) (3.48)

for every x1 , x2 â ð and ðœ â [0, 1]. Adding (ð + ð)(ðœx1 + (1 â ðœ)x2 ) â€ ðœ(ð + ð)(x1 ) + (1 â ðœ)ð (x2 ) ð + ð is convex. Multiplying (3.47) by ðŒ â¥ 0 ðŒð (ðœx1 + (1 â ðœ)x2 ) â€ ðŒ(ðœð (x1 ) + (1 â ðœ)ð (x2 )) = (ðœðŒð (x1 ) + (1 â ðœ)ðŒð (x2 )) ðŒð is convex. Moreover, if ð is strictly convex, ð (ðœx1 + (1 â ðœ)x2 ) < ðœð (x1 ) + (1 â ðœ)ð (x2 )

(3.49)

for every x1 , x2 â ð, x1 â= x2 and ðœ â (0, 1). Adding this to (3.48) (ð + ð)(ðœx1 + (1 â ðœ)x2 ) < ðœ(ð + ð)(x1 ) + (1 â ðœ)ð (x2 ) so that ð + ð is strictly convex. Multiplying (3.49) by ðŒ > 0 ðŒð (ðœx1 + (1 â ðœ)x2 ) < ðŒ(ðœð (x1 ) + (1 â ðœ)ð (x2 )) = (ðœðŒð (x1 ) + (1 â ðœ)ðŒð (x2 )) ðŒð is strictly convex. 3.132 x â epi (ð âš ð) ââ x â epi ð and x â epi ð That is epi (ð âš ð) = epi ð â© epi ð Therefore epi ð âš ð is convex (Exercise 1.162) and therefore ð is convex (Proposition 3.7). 153

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3.133 If ð is convex ð (ðŒx1 + (1 â ðŒ)x2 ) â€ ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) Since ð is increasing ( ) ( ) ð ð (ðŒx1 + (1 â ðŒ)x2 ) â€ ð ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) ) ( ) ( â€ ðŒð ð (x1 ) + (1 â ðŒ)ð ð (x2 ) since ð is also convex. The concave case is proved similarly. 3.134 Let ð¹ = log ð . If ð¹ is convex, ð (x) = ðð¹ (x) is an increasing convex function of a convex function and is therefore convex (Exercise 3.133). 3.135 If ð is positive and concave, then log ð is concave (Exercise 3.51). Therefore log

1 = log 1 â log ð = â log ð ð

is convex. By the previous exercise (Exercise 3.134), this implies that 1/ð is convex. If ð is negative and convex, then âð is positive and concave, 1/ â ð is convex, and therefore 1/ð is concave. 3.136 Consider the identity ) ( ) ( ) ( ) ( ð ð (ð¥1 âš ð¥2 ) + ð ð (ð¥1 â§ ð¥2 ) â ð ð (ð¥1 ) â ð ð (ð¥2 ) ) ( ) ( ) ( ) ) ) ( = ð ð (ð¥1 âš ð¥2 ) + ð ð (ð¥1 â§ ð¥2 ) â ð ð (ð¥1 ) â ð ð (ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) â ð (ð¥1 ) ( ) ( ) + ð ð (ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) â ð (ð¥1 ) â ð ð (ð¥2 ) (3.50) Deï¬ne

( ) ( ) ( ) ( ) ð(ð¥1 , ð¥2 ) = ð ð (ð¥1 âš ð¥2 ) + ð ð (ð¥1 â§ ð¥2 ) â ð ð (ð¥1 ) â ð ð (ð¥2 )

Then ð â ð is supermodular if ð is nonnegative deï¬nite and submodular if ð is nonpositive deï¬nite. Using the identity (3.50), ð can be decomposed into two components ð(ð¥1 , ð¥2 ) = ð1 (ð¥1 , ð¥2 ) + ð2 (ð¥1 , ð¥2 ) ( ) ( ) ( ) ð1 (ð¥1 , ð¥2 ) = ð ð (ð¥1 âš ð¥2 ) + ð ð (ð¥1 â§ ð¥2 ) â ð ð (ð¥1 ) ( ) â ð ð (ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) â ð (ð¥1 ) ( ) ( ) ð2 (ð¥1 , ð¥2 ) = ð ð (ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) â ð (ð¥1 ) â ð ð (ð¥2 )

(3.51)

ð will deï¬nite if both components are deï¬nite. For any ð¥1 , ð¥2 â ð¥1 , let ð = ð (ð¥1 â§ ð¥2 ), ð = ð (ð¥1 ) and ð = ð (ð¥1 âš ð¥2 ). Provided ð is monotone, ð lies between ð and ð. Substituting in (3.51) ð1 (ð¥1 , ð¥2 ) = ð(ð) + ð(ð) â ð(ð) â ð(ð + ð â ð) and Exercise 3.128 implies

{

ð1 (ð¥1 , ð¥2 ) = ð(ð) + ð(ð) â ð(ð) â ð(ð + ð â ð) Now consider ð2 .

} { } â¥ð convex if ð is â€0 concave

{ } { } â¥ supermodular ð (ð¥1 âš ð¥2 ) + ð (ð¥1 â§ ð¥2 ) â ð (ð¥1 ) is ð (ð¥2 ) if ð is â€ submodular 154

(3.52)

Solutions for Foundations of Mathematical Economics and therefore since ð is increasing

{

ð2 (ð¥1 , ð¥2 ) =

â¥ 0 if ð is supermodular â€ 0 if ð is submodular

(3.53)

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

1. Assume that ð is bounded above in a neighborhood of x0 . Then there exists a ball ðµ(ð¥0 ) and constant ð such that ð (x) â€ ð for every x â ðµ(ð¥0 ) Since ð is convex ð (ðŒx + (1 â ðŒ)x0 ) â€ ðŒð (x) + (1 â ðŒ)ð (x0 ) â€ ðŒð + (1 â ðŒ)ð (x0 )

(3.54)

2. Given x â ðµ(ð¥0 ) and ðŒ â [0, 1] let z = ðŒx + (1 â ðŒ)x0

(3.55)

Subtracting ð (x0 ) from (3.54) gives ð (z) â ð (x0 ) â€ ðŒ(ð â ð (x0 )) Rewriting (3.55) (1 â ðŒ)x0 = z â ðŒx (1 + ðŒ)x0 = z + ðŒ(2x0 â x) ðŒ 1 z+ (2x0 â x) x0 = 1+ðŒ 1+ðŒ 3. Note that (2x0 â x) = x0 â (x â x0 ) â ðµ(x0 ) so that ð (2x0 â x) â€ ð and therefore ð (x0 ) â€

ðŒ ðŒ 1 1 ð (z) + ð (2x0 â x) â€ ð (z) + ð 1+ðŒ 1+ðŒ 1+ðŒ 1+ðŒ

which implies (1 + ðŒ)ð (x0 ) â€ ð (z) + ðŒð ðŒ(ð (x0 ) â ð ) â€ ð (z) â ð (x0 ) 4. Combined with (3.56) we have ðŒ(ð (x0 ) â ð ) â€ ð (z) â ð (x0 ) â€ ðŒ(ð â ð (x0 )) or â£ð (z) â ð (x0 )â£ â€ ðŒ(ð â ð (x0 )) and therefore ð (z) â ð (x0 ) as z â x0 . ð is continuous. 155

(3.56)

Solutions for Foundations of Mathematical Economics 3.138

1. Since ð is open, there exists a ball ðµð (x1 ) â ð. Let ð¡ = 1 + ð2 . Then x0 + ð¡(x1 â x0 ) â ðµð (ð¥1 ) â ð.

2. Let ð  = ð¡â1 ð¡ ð. The open ball ðµð  (x1 ) of radius ð  centered on x1 is contained in ð . Therefore ð is a neighborhood of x1 . 3. Since ð is convex, for every y â ð ð (y) â€ (1 â ðŒ)ð (x) + ðŒð (z) â€ (1 â ðŒ)ð + ðŒð (z) â€ ð + ð (z) Therefore ð is bounded on ð . 3.139 The previous exercise showed that ð is locally bounded from above for every x â ð. To show that it is also locally bounded from below, choose some x0 â ð. There exists some ðµ(x0 and ð such that ð (x) â€ ð for every x â ðµ(x0 ) Choose some ð¥1 â ðµ(x0 ) and let x2 = 2x0 â x1 . Then x2 = 2x0 â x1 = x0 â (x1 â x0 ) â ðµ(x0 ) and ð (x2 ) â€ ð . Since ð¹ is convex ð (x) â€

1 1 ð (x1 ) + ð (x2 ) 2 2

and therefore ð (x1 ) â¥ 2ð (x) â ð (x2 ) Since ð (x2 ) â€ ð , âð (x2 ) â¥ âð and therefore ð (x1 ) â¥ 2ð (x) â ð so that ð is bounded from below. 3.140 Let ð be a convex function deï¬ned on an open convex set ð in a normed linear space, which is bounded from above in a neighborhood of a single point x0 â ð. By Exercise 3.138, ð is bounded above at every x â ð. This implies (Exercise 3.137) that ð is continuous at every x â ð. 3.141 Without loss of generality, assume 0 â ð. Assume ð has dimension ð and let x1 , x2 , . . . , xð be a basis for the subspace containing ð. Choose some ð > 0 small enough so that ð = conv {0, ðx1 , ðx2 , . . . , ðð¥ð } â ð Any x â ð is a convexâ combination of the points 0, x1 , x2 , . . . , xð and so there exists ðŒ0 , ðŒ1 , ðŒ2 , . . . , ðŒð â¥ 0, ðŒð = 1 such that x = ðŒ0 0 + ðŒ1 x1 + â â â + ðŒð xð . By Jensenâs inequality ð (x) = ð (ðŒ0 0 + ðŒ1 x1 + â â â + ðŒð xð ) â€ ðŒ0 ð (0) + ðŒ1 ð (x1 ) + â â â + ðŒð ð (xð ) â€ max{ ð (0), ð (x1 ), . . . , ð (xð ) } Therefore, ð is bounded above on a neighbourhood of some x0 â int ð (which is nonempty by Exercise 1.229). By Proposition 3.8, ð is continuous on ð.

156

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3.142 Clearly, if ð is convex, it is locally convex at every x â ð, where ð is the required neighborhood. To prove the converse, assume to the contrary that ð is locally convex at every x â ð but it is not globally convex. That is, there exists x1 , x2 â ð such that ð (ðŒx1 + (1 â ðŒ)x2 ) > ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) Let ) ( â(ð¡) = ð ð¡x1 + (1 â ð¡)x2 Local convexity implies that ð is continuous at every x â ð (Corollary 3.8.1), and therefore continuous on ð. Therefore, â is continuous on [0, 1]. By the continuous maximum theorem (Theorem 2.3), ð = arg max â(ð¡) xâ[x1 ,x2 ]

is nonempty and compact. Let ð¡0 = max ð . For every ð > 0, â(ð¡0 â ð) â€ â(ð¡0 ) and â(ð¡0 + ð) < â(ð¡0 ) Let x0 = ð¡0 x1 + (1 â ð¡0 )x2 and xð = (ð¡0 + ð)x1 + (1 â ð¡0 â ð)x2 Every neighborhood ð of x0 contains xâð , xð â [x1 , x2 ] with 1 1 1 1 ð (xâð ) + ð (xð ) = â(ð¡0 â ð) + â(ð¡0 + ð) < â(ð¡0 ) = ð (x0 ) = ð 2 2 2 2

(

1 1 xâð + xð 2 2

)

contradicting the local convexity of ð at x0 . 3.143 Assume ð is quasiconcave. That is for every x1 , x2 â ð and 0 â€ ðŒ â€ 1 ð (ðŒx1 + (1 â ðŒ)x2 ) â¥ min{ð (x1 ), (x2 )} Multiplying through by â1 reverses the inequality so that âð (ðŒx1 + (1 â ðŒ)x2 ) â€ â min{ð (x1 ), ð (x2 )} = max{âð (x1 ), âð (x2 )} which shows that âð is quasiconvex. The converse follows analogously. 3.144 Assume ð is concave, that is ð (ðŒx1 + (1 â ðŒ)x2 ) â¥ ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) for every x1 , x2 â ð and 0 â€ ðŒ â€ 1 Without loss of generality assume that ð (x1 ) â€ ð (x2 ). Then ð (ðŒx1 + (1 â ðŒ)x2 ) â¥ ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) â¥ ðŒð (x1 ) + (1 â ðŒ)ð (x1 ) = ð (x1 ) = min{ð (x1 ), ð (x2 )} ð is quasiconcave. 3.145 Let ð : â â â. Choose any ð¥1 , ð¥2 in â with ð¥1 < ð¥2 . If ð is increasing, then ð (ð¥1 ) â€ ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) â€ ð (ð¥2 ) for every 0 â€ ðŒ â€ 1. The ï¬rst inequality implies that ð (ð¥1 ) = min{ð (ð¥1 ), ð (ð¥2 )} â€ ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) 157

Solutions for Foundations of Mathematical Economics

so that ð is quasiconcave. The second inequality implies that ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) â€ max{ð (ð¥1 ), ð (ð¥2 )} = ð (ð¥2 ) so that ð is also quasiconvex. Conversely, if ð is decreasing ð (ð¥1 ) â¥ ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) â¥ ð (ð¥2 ) for every 0 â€ ðŒ â€ 1. The ï¬rst inequality implies that ð (ð¥1 ) = max{ð (ð¥1 ), ð (ð¥2 )} â¥ ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) so that ð is quasiconvex. The second inequality implies that ð (ðŒð¥1 + (1 â ðŒ)ð¥2 ) â€ max{ð (ð¥1 ), ð (ð¥2 )} = ð (ð¥2 ) so that ð is also quasiconcave. 3.146 âŸð (ð) = { x â ð : ð (x) â€ ð } = {x â ð : âð (x) â¥ âð} = â¿âð (âð) 3.147 For given ð and ð, choose any p1 and p2 in âŸð£ (ð). For any 0 â€ ðŒ â€ 1, let Â¯ = ðŒp1 + (1 â ðŒ)p2 . The key step is to show that any commodity bundle x which is p Â¯ is also aï¬ordable at either p1 or p2 . Assume that x is aï¬ordable at p Â¯, aï¬ordable at p that is x is in the budget set Â¯x â€ ð } x â ð(Â¯ p, ð) = { x : p To show that x is aï¬ordable at either p1 or p2 , that is x â ð(p1 , ð) or x â ð(p2 , ð) assume to the contrary that xâ / ð(p1 , ð) and x â / ð(p2 , ð) This implies that p1 x > ð and p2 x > ð so that ðŒp1 x > ðŒð and (1 â ðŒ)p2 > (1 â ðŒ)ð Summing these two inequalities Â¯ x = (ðŒp1 + (1 â ðŒ)p2 )x > ð p contradicting the assumption that x â ð(Â¯ p, ð). We conclude that ð(Â¯ p, ð) â ð(p1 , ð) âª ð(p2 , ð) Now ð£(Â¯ ð, ð) = sup{ ð¢(x) : x â ð(Â¯ p, ð) } â€ sup{ ð¢(x) : x â ð(p1 , ð) âª ð(p2 , ð) } â€ð Â¯ â âŸð£ (ð) for every 0 â€ ðŒ â€ 1. Thus, âŸð£ (ð) is convex and so ð£ is quasiconvex Therefore p (Exercise 3.146). 158

Solutions for Foundations of Mathematical Economics

3.148 Since ð is quasiconcave ð (ðŒx1 + (1 â ðŒ)x2 ) â¥ min{ð (x1 ), ð (x2 )} for every x1 , x2 â ð and 0 â€ ðŒ â€ 1 Since ð is increasing ) ( ) ( ) ( ) ( ð ð (ðŒx1 + (1 â ðŒ)x2 ) â¥ ð( min{ð (x1 ), ð (x2 )}) â¥ min{ð ð (x1 ) , ð ð (x2 ) } ð â ð is quasiconcave. 3.149 When ð â¥ 1, the function â(x) = ðŒ1 ð¥ð1 + ðŒ2 ð¥ð2 + . . . ðŒð ð¥ðð is convex (Example 3.58) as is ðŠ 1/ð . Therefore ð (x) = (â(x))

1/ð

is an increasing convex function of a convex function and is therefore convex (Exercise 3.133). 3.150 ð is a monotonic transformation of the concave function â(x) = x. 3.151 By Exercise 3.39, there exist linear functionals ðË and ðË and scalars ð and ð such that ð (x) = ðË(x) + ð and ð(x) = ðË(x) + ð The upper contour set â¿â (ð) = { ð¥ â ð : â(x) â¥ ð } ðË(ð¥) + ð â¥ ð} = {ð¥ â ð : ðË(ð¥) + ð = { ð¥ â âð : ðË(x) + ð â¥ ðË ð(x) + ðð } +

ð (x) â¥ ð â ðð } = { ð¥ â âð+ : ðË(x) â ðË which is a halfspace in ð and therefore convex. Similarly, the lower contour set âŸâ (ð) = { ð¥ â ð : â(x) â¥ ð } is also a halfspace and hence convex. Therefore â is both quasiconcave and quasiconvex. 3.152 For ð â€ 0 â¿(ð) = { ð¥ â ð : â(x) â¥ 0 } = ð which is convex. For ð > 0 â¿â (ð) = { ð¥ â ð : â(x) â¥ ð } ð (x) â¥ ð} ð(x) = { ð¥ â ð : ð (x) â¥ ðð(x) } = {ð¥ â ð :

= { ð¥ â ð : ð (x) â ðð(x) â¥ 0 } is convex since ð â ðð = ð + ð(âð) is concave (Exercises 3.124 and 3.131). Since â¿â (ð) is convex for every ð, â is quasiconcave. 159

Solutions for Foundations of Mathematical Economics

3.153 ð (x) ðË(x)

â(x) =

where ðË = 1/ð is positive and convex by Exercise 3.135. By the previous exercise, â is quasiconcave. 3.154 Let ð¹ = log ð . If ð¹ is concave, ð (x) = ðð¹ (x) is an increasing function of (quasi)concave function, and hence is quasiconcave (Exercise 3.148). 3.155 Let ð¹ (x) = log ð (x) =

ð â

ðŒð log ðð (x)

ð=1

As the sum of concave functions, ð¹ is concave (Exercise 3.131). By the previous exercise, ð is quasiconcave. Â¯ = ðŒðœ1 + (1 â ðŒ)ðœ 2 Â¯ are optimal solutions for ðœ1 , ðœ2 and ðœ 3.156 Assume x1 , x2 and x respectively. That is ð (x1 , ðœ 1 ) = ð£(ðœ 1 ) ð (x2 , ðœ 2 ) = ð£(ðœ 2 ) Â¯ = ð£(ðœ) Â¯ ð (Â¯ x, ðœ) Since ð is convex in ðœ Â¯ = ð (Â¯ Â¯ ð£(ðœ) x, ðœ) = ð (Â¯ x, ðŒðœ 1 + (1 â ðŒ)ðœ 2 ) â€ ðŒð (Â¯ x, ðœ 1 ) + (1 â ðŒ)ð (xâ , ðœ2 ) â€ ðŒð (x1 , ðœ1 ) + (1 â ðŒ)ð (x2 , ðœ 2 ) = ðŒð£(ðœ 1 ) + (1 â ðŒ)ð£(ðœ 2 ) ð£ is convex. 3.157 Assume to the contrary that x1 and x2 are distinct optimal solutions, that is â x2 , for some ðœ â Îâ , so that x1 , x2 â ð(ðœ), x1 = ð (x1 , ðœ) = ð (x2 , ðœ) = ð£(ðœ) â¥ ð (x, ðœ) for every x â ðº(ðœ) Â¯ = ðŒx1 + (1 â ðŒ)x2 for ðŒ â (0, 1). Since ðº(ðœ) is convex, x Â¯ is feasible. Since ð is Let x strictly quasiconcave ð (Â¯ x, ðœ) > min{ ð (x1 , ðœ), ð (x2 , ðœ) } = ð£(ðœ) contradicting the optimality of x1 and x2 . We conclude that ð(ðœ) is single-valued for every ðœ â Îâ . In other words, ð is a function. 3.158

1. The value function is ð£(ð¥0 ) =

sup ð (x)

xâÎ(ð¥0 )

where ð (x) =

â â ð¡=0

160

ðœ ð¡ ð (ð¥ð¡ , ð¥ð¡+1 )

Solutions for Foundations of Mathematical Economics

and Î(ð¥0 ) = {x â ð â : ð¥ð¡+1 â ðº(ð¥ð¡ ), ð¡ = 0, 1, 2, . . . } Since an optimal policy exists (Exercise 2.125), the maximum is attained and ð£(ð¥0 ) = max ð (x) xâÎ(ð¥0 )

(3.57)

It is straightforward to show that â ð (x) is strictly concave and â Î(ð¥0 ) is convex Applying the Concave Maximum Theorem (Theorem 3.1) to (3.57), we conclude that the value function ð£ is strictly concave. 2. Assume to the contrary that xâ² and xâ²â² are distinct optimal plans, so that ð£(ð¥0 ) = ð (xâ² ) = ð (xâ²â² ) Â¯ is feasible and Â¯ = ðŒxâ² + (1 â ðŒ)xâ²â² . Since Î(ð¥0 ) is convex, x Let x ð (Â¯ x) > ðŒð (xâ² ) + (1 â ðŒ)ð (xâ²â² ) = ð (xâ² ) which contradicts the optimality of xâ² . We conclude that the optimal plan is unique. 3.159 We observe that â ð¢(ð¹ (ð) â ðŠ) is supermodular in ðŠ (Exercise 2.51) â ð¢(ð¹ (ð) â ðŠ) displays strictly increasing diï¬erences in (ð, ðŠ) (Exercise 3.129) â ðº(ð) = [0, ð¹ (ð)] is increasing. Applying Exercise 2.126, we can conclude that the optimal policy (ð0 , ð1â , ð2â , . . . ) is a monotone sequence. Since ð is compact, kâ is a bounded monotone sequence, which converges monotonically to some steady state ð â (Exercise 1.101). 3.160 Suppose there exists (xâ , yâ ) â ð Ã ð such that ð (x, yâ ) â€ ð (xâ , yâ ) â€ ð (xâ , y) for every x â ð and y â ð Let ð£ = ð (xâ , yâ ). Since ð (x, yâ ) â€ ð£ for every x â ð max ð (ð¥, yâ ) â€ ð£ xâð

and therefore min max ð (x, y) â€ max ð (x, yâ ) â€ ð£

yâð xâð

xâð

Similarly max min ð (x, y) â¥ ð£ xâð yâðŠ

Combining the last two inequalities, we have max min ð (x, y) â¥ ð£ â¥ min max ð (ð¥, y) xâð yâðŠ

yâð xâð

161

Solutions for Foundations of Mathematical Economics

Together with (3.28), this implies equality max min ð (x, y) = min max ð (x, y) xâð yâð

yâð xâð

Conversely, suppose that max min ð (x, y) = ð£ = min max ð (x, y) xâð yâð

yâð xâð

The function ð(x) = min ð (x, y) yâð

is a continuous function (Theorem 2.3) on a compact set ð. By the Weierstrass theorem (Theorem 2.2), there exists xâ which maximizes ð on ð, that is ð(xâ ) = min ð (xâ , y) = max ð(x) = max min ð (x, y) = ð£ yâð

xâð

xâð yâð

which implies that ð (xâ , y) â¥ ð£ for every y â ð Similarly, there exists y â ð such that ð (x, yâ ) â€ ð£ for every x â ð Combining these inequalities, we have ð (x, yâ ) â€ ð£ â€ ð (xâ , y) for every x â ð and y â ð In particular, we have ð (xâ , yâ ) â€ ð£ â€ ð (xâ , yâ ) so that ð£ = ð (xâ , yâ ) as required. 3.161 For any ð¥ â ð and ðŠ â ð , let ð(ð¥) = min ð (ð¥, ðŠ) and â(ðŠ) = max ð (ð¥, ðŠ) ðŠâð

ð¥âð

Then ð(ð¥) = min ð (ð¥, ðŠ) â€ max ð (ð¥, ðŠ) = â(ðŠ) ðŠâð

ð¥âð

and therefore max ð(ð¥) â€ max â(ðŠ) ð¥âð

ðŠâð

That is max min ð (ð¥, ðŠ) â€ ððŠ max ð (ð¥, ðŠ) ð¥

ðŠ

ð¥

162

Solutions for Foundations of Mathematical Economics

3.162 Clearly ð (ð¥) = ð¥ð his homogeneous of degree ð. Conversely assume ð is homogeneous of degree ð, that is ð (ð¡ð¥) = ð¡ð ð (ð¥) Letting ð¥ = 1 ð (ð¡) = ð¡ð ð (1) Setting ð (1) = ðŽ â â and interchanging ð¥ and ð¡ yields the result. 3.163 1/ð

ð (ð¡x) = (ð1 (ð¡ð¥1 )ð + ð2 (ð¡ð¥2 )ð + . . . ðð (ð¡ð¥ð )ð ) 1/ð

= ð¡ (ð1 ð¥ð1 + ð2 ð¥ð2 + . . . ðð ð¥ðð ) = ð¡ð (x) 3.164 For ðœ â â++

â(ðœð¡) = ð (ðœð¡x0 ) = ðœ ð ð (ð¡x0 ) = ðœ ð â(ð¡) 3.165 Suppose that xâ minimizes the cost of producing output ðŠ at prices w. That is wð xâ â€ wð x

for every x â ð (ðŠ)

It follows that ð¡wð xâ â€ ð¡wð x

for every x â ð (ðŠ)

for every ð¡ > 0, verifying that xâ minimizes the cost of producing ðŠ at prices ð¡w. Therefore ð(ð¡w, ðŠ) = (ð¡w)xâ = ð¡(wð xâ ) = ð¡ð(w, ðŠ) ð(w, ðŠ) homogeneous of degree one in input prices w. 3.166 For given prices w, let xâ minimize the cost of producing one unit of output, so that ð(w, 1) = wð xâ . Clearly ð (xâ ) = 1 where ð is the production function. Now consider any output ðŠ. Since ð is homogeneous ð (ðŠxâ ) = ðŠð (xâ ) = ðŠ Therefore ðŠxâ is suï¬cient to produce ðŠ, so that ð(w, ðŠ) â€ wð (ðŠxâ ) = ðŠwð xâ = ðŠð(w, 1) Suppose that ð(w, ðŠ) < wð (ðŠxâ ) = ðŠð(w, 1) Then there exists xâ² such that ð (xâ² ) = ðŠ and wð xâ² < wð (ðŠxâ ) which implies that w

ð

(

xâ² ðŠ

)

< wð xâ = ð(w, 1) 163

Solutions for Foundations of Mathematical Economics Since ð is homogeneous

( ð

xâ² ðŠ

) =

1 ð (xâ² ) = 1 ðŠ

Therefore, xâ² is a lower cost method of producing one unit of output, contradicting the deï¬nition of xâ . We conclude that ð(w, ðŠ) = ðŠð(w, 1) ð(w, ðŠ) is homogeneous of degree one in ðŠ. 3.167 If the consumerâs demand is invariant to proportionate changes in all prices and income, so also will the derived utility. More formally, suppose that xâ maximizes utility at prices p and income ð, that is xâ â¿ x

for every x â ð(p, ð)

Then ð£(p, ð) = ð¢(xâ ) Since ð(ð¡p, ð¡ð) = ð(p, ð) xâ â¿ x

for every x â ð(ð¡p, ð¡ð)

and ð£(ð¡p, ð¡ð) = ð¢(xâ ) = ð£(p, ð) 3.168 Assume ð is homogeneous of degree one, so that ð (ð¡x) = ð¡ð (x)

for every ð¡ > 0

Let (x, ðŠ) â epi ð , so that ð (x) â€ ðŠ For any ð¡ > 0 ð (ð¡x) = ð¡ð (x) â€ ð¡ðŠ which implies that (ð¡x, ð¡ðŠ) â epi ð . Therefore epi ð is a cone. Conversely assume epi ð is a cone. Let x â ð and deï¬ne ðŠ = ð (x). Then (x, ðŠ) â epi ð and therefore (ð¡x, ð¡ðŠ) â epi ð so ð (ð¡x) â€ ð¡ðŠ Now suppose to the contrary that ð (ð¡x) = ð§ < ð¡ðŠ = ð¡ð (x)

(3.58)

Then (ð¡x, ð§) â epi ð . Since epi ð is a cone, we must have (x, ð§/ð¡) â epi ð so that ð (x) â€

ð§ ð¡

and ð¡ð (x) â€ ð§ = ð (ð¡x) contradicting (3.58). We conclude that ð (ð¡x) = ð¡ð (x) for every ð¡ > 0 164

Solutions for Foundations of Mathematical Economics

3.169 Take any x1 and x2 in ð and let ðŠ1 = ð (x1 ) > 0 and ðŠ2 = ð (x2 ) > 0 Since ð is homogeneous of degree one, ( ) ( ) x1 x2 ð =ð =1 ðŠ1 ðŠ2 Since ð is quasiconcave

( ) x2 x1 + (1 â ðŒ) ð ðŒ â¥1 ðŠ1 ðŠ2

for every 0 â€ ðŒ â€ 1. Choose ðŒ = ðŠ1 /(ðŠ1 + ðŠ2 ) so that (1 â ðŒ) = ðŠ2 /(ðŠ1 + ðŠ2 ). Then ( ) x1 x2 ð + â¥1 ðŠ1 + ðŠ2 ðŠ1 + ðŠ2 Again using the homogeneity of ð , this implies ð (x1 + x2 ) â¥ ðŠ1 + ðŠ2 = ð (x1 ) + ð (x2 ) 3.170 Let ð â ð¹ (ð) be a strictly positive deï¬nite, quasiconcave functional which is homogeneous of degree one. For any x1 , x2 in ð and 0 â€ ðŒ â€ 1 ðŒx1 , (1 â ðŒ)x2 in ð and therefore ð (ðŒx1 + (1 â ðŒ)x2 ) â¥ ð (ðŒx1 ) + ð ((1 â ðŒ)x2 ) since ð is superadditive (Exercise 3.169). But ð (ðŒx1 ) = ðŒð (x1 ) ð ((1 â ðŒ)x2 ) = (1 â ðŒ)ð (x2 ) by homogeneity. Substituting in (3.58), we conclude that ð (ðŒx1 + (1 â ðŒ)x2 ) â¥ ðŒð (x1 ) + (1 â ðŒ)ð ((1 â ðŒ)x2 ) ð is concave. 3.171 Assume that ð is strictly positive deï¬nite, quasiconcave and homogeneous of degree ð, 0 < ð < 1. Deï¬ne â(x) = (ð (x))

1/ð

Then â is quasiconcave (Exercise 3.148. Further, for every ð¡ > 0 1/ð

â(ð¡x) = (ð (ð¡x)) ( )1/ð = ð¡ð ð (x) = ð¡ (ð (x))

1/ð

= ð¡â(x) so that â is homogeneous of degree 1. By Exercise 3.170, â is concave. ð (x) = (â(x))

ð

That is ð = ð â â where ð(ðŠ) = ðŠ ð is monotone and concave provided ð â€ 1. By Exercise 3.133, ð = ð â â is concave. 165

Solutions for Foundations of Mathematical Economics

3.172 Continuity is a necessary and suï¬cient condition for the existence of a utility function representing â¿ (Remark 2.9). Suppose ð¢ represents the homothetic preference relation â¿. For any x1 , x2 â ð ð¢(x1 ) = ð¢(x2 ) =â x1 âŒ x2 =â ð¡x1 âŒ ð¡x2 =â ð¢(ð¡x1 ) = ð¢(ð¡x2 ) for every ð¡ > 0 Conversely, if ð¢ is a homothetic functional, x1 âŒ x2 =â ð¢(x1 ) = ð¢(x2 ) =â ð¢(ð¡x1 ) = ð¢(ð¡x2 ) =â ð¡x1 âŒ ð¡x2 for every ð¡ > 0 3.173 Suppose that ð = ð â â where ð is strictly increasing and â is homogeneous of degree ð. Then ( )1/ð Ë â(x) = â(x) Ë where is homogeneous of degree one and ð = ðË â â ( ) ðË(ðŠ) = ð ðŠ ð ) is increasing. 3.174 Assume x1 , x2 â ð with ð (x1 ) = ð(â(x1 )) = ð(âx2 )) = ð (x2 ) Since ð is strictly increasing, this implies that â(x1 ) = â(x2 ) Since â is homogeneous â(ð¡x1 ) = ð¡ð â(x1 ) = ð¡ð â(x2 ) = â(ð¡x2 ) for some ð. Therefore ð (ð¡x1 ) = ð(â(ð¡x1 )) = ð(â(ð¡x2 )) = ð (ð¡x2 ) 3.175 Let x0 â= 0 be any point in ð, and deï¬ne ð : â â â by ð(ðŒ) = ð (ðŒx0 ) Since ð is strictly increasing, so is ð and therefore ð has a strictly increasing inverse ð â1 . Let â = ð â1 â ð so that ð = ð â â. We need to show that â is homogeneous. For any x â ð, there exists ðŒ such that ð(ðŒ) = ð (ðŒx0 ) = ð (x) that is ðŒ = â(x) = ð â1 (ð (x)). Since ð is homothetic ð(ð¡ðŒ) = ð (ð¡ðŒx0 )ð (ð¡x) for every ð¡ > 0 and therefore â(ð¡x) = ð â1 (ð (ð¡x)) = ð â1 (ð (ð¡ðŒx0 )) = ð â1 ð(ð¡ðŒ) = ð¡ðŒ = ð¡â(x) â is homogeneous of degree one. 166

Solutions for Foundations of Mathematical Economics

3.176 Let ð be the production function. If ð is homothetic, there exists (Exercise 3.175) a linearly homogeneous function â and strictly increasing function ð such that ð = ð ââ. ð(w, ðŠ) = min{ wð x : ð (x) â¥ ðŠ } x

= min{ wð x : ð(â(x)) â¥ ðŠ } x

= min{ wð x : â(x) â¥ ð â1 (ðŠ) } x

= ð â1 (ðŠ)ð(w, 1) by Exercise 3.166. 3.177 Let ð : ð â â be positive, strictly increasing, homothetic and quasiconcave. By Exercise 3.175, there exists a linearly homogeneous function â : ð â â and strictly increasing function ð â ð¹ (ð) such that ð = ð â â. â = ð â1 â ð is positive, quasiconcave (Exercise 3.148) and homogeneous of degree one. By Proposition 3.12, â is concave and therefore ð = ð â â is concaviï¬able. 3.178 Since ð»ð (ð) is a supporting hyperplane to ð at x0 , then ð (x0 ) = ð and either ð (x) â¥ ð = ð (x0 ) for every x â ð or ð (x) â€ ð = ð (x0 ) for every x â ð 3.179 Suppose to the contrary that y = (â, ð) â int ðŽ â© ðµ. Then y â¿ yâ . By strict convexity yðŒ = ðŒy + (1 â ðŒ)yâ â» yâ for every ðŒ â (0, 1) Since y â int ðŽ, yðŒ â ðŽ for ðŒ suï¬ciently small. That is, there exists some ðŒ such that yðŒ is feasible and yðŒ â» yâ , contradicting the optimality of yâ . 3.180 For notational simplicity, let ð be the linear functional which separates ðŽ and ðµ in Example 3.77. ð (y) measure the cost of the plan y = (â, ð), that is ð (y) = ð€â + ðð. Assume to the contrary there exists a preferred lifestyle in ð, that is there exists some y = (â, ð) â ð such that y â» yâ = (ââ , ð â ). Since y â ðµ, ð (y) â¥ ð (yâ ) by (3.29). On the other hand, y â ð which implies that ð (y) â€ ð (yâ ). Consequently, ð (y) = ð (yâ ). By continuity, there exists some ðŒ < 1 such that ðŒy â» yâ which implies that ðŒy â ðµ. By linearity ð (ðŒy) = ðŒð (y) < ð (y) = ð (yâ ) = ðŒ contrary to (3.29). This contradiction establishes that yâ is the best choice in budget set ð. 3.181 By Proposition 3.7, epi ð is a convex set in ð Ã â with (x0 , ð (x0 )) a point on its boundary. By Corollary 3.2.2 of the Separating Hyperplane Theorem, there exists linear a functional ð â (ð Ã â)â² such that ð(x, ðŠ) â¥ ð(x0 , ð (x0 )) for every (x, ðŠ) â epi ð 167

(3.59)

Solutions for Foundations of Mathematical Economics

ð can be decomposed into two components (Exercise 3.47) ð(x, ðŠ) = âð(x) + ðŒðŠ The assumption that x0 â int ð ensures that ðŒ > 0 and we can normalize so that ðŒ = 1. Substituting in (3.59) âð(x) + ð (x) â¥ âð(x0 ) + ð (x0 ) ð (x) â¥ ð (x0 ) + ð(x â x0 ) for every x â ð. 3.182 By Exercise 3.72, there exists a unique point x0 â ð such that (x0 â y)ð (x â x0 ) â¥ 0 for every x â ð Deï¬ne the linear functional (Exercise 3.64) ð (x) = (x0 â y)ð x and let ð = ð (x0 ). For all x â ð ð (x) â ð (x0 ) = ð (x â x0 ) = (x0 â y)ð (x â x0 ) â¥ 0 and therefore ð (x) â¥ ð (x0 ) = ð for every x â ð Furthermore 2

ð (x0 ) â ð (y) = ð (x0 â y) = (x0 â y)ð (x0 â y) = â¥x0 â yâ¥ > 0 since y â= x0 . Therefore ð (x0 ) > ð (y) and ð (y) < ð â€ ð (x) for every x â ð 3.183 If y â b(ð), y â ð ð and there exists a sequence of points {yð } â ð ð converging to y (Exercise 1.105). That is, there exists a sequence of nonboundary points {yð } â /ð converging to y. For every point yð , there is a linear functional ð ð â ð â and ðð such that ð ð (yð ) < ðð â€ ð ð (x)

for every x â ð

Deï¬ne ð ð = ð ð / â¥ð ð â¥. By construction, the sequence of linear functionals ð ð belong to the unit ball in ð â (since â¥ð â¥ = 1). Since ð â is ï¬nite dimensional, the unit ball is compact as so ð ð has a convergent subsequence with limit ð such that ð (y) â€ ð (x)

for every ð¥ â ð

ð (y) â€ ð (x)

for every ð¥ â ð

A fortiori

3.184 There are two possible cases. yâ / ð By Exercise 3.182, there exists a hyperplane which separates y and ð which a fortiori separates y and ð, that is ð (y) â€ ð (x)

for every x â ð 168

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y â ð Since y â / ð, y must be a boundary point of ð. By the previous exercise, there exists a supporting hyperplane at y, that is there exists a continuous linear functional ð â ð â such that ð (y) â€ ð (x) 3.185

for every x â ð

1. ð (ð) â â.

2. ð (ð) is convex and hence an interval (Exercise 1.160. 3. ð (ð) is open in â (Proposition 3.2). 3.186 ð is nonempty and convex and 0 â / ð. (Otherwise, there exists x â ðŽ and y â ðµ such that 0 = y + (âx) which implies that x = y contradicting the assumption that ðŽ â© ðµ = â.) Thus there exists a continuous linear functional ð â ð â such that ð (y â x) â¥ ð (0) = 0

for every x â ðŽ, y â ðµ

so that ð (x) â€ ð (y) for every x â ðŽ, y â ðµ Let ð = supxâðŽ ð (x). Then ð (x) â€ ð â€ ð (y) for every x â ðŽ, y â ðµ By Exercise 3.185, ð (int ðŽ) is an open interval in (ââ, ð], hence ð (int ðŽ) â (ââ, ð), so that ð (x) < ð for every x â int ðŽ. Similarly, ð (int ðµ) > ð and ð (x) < ð < ð (y) for every x â int ðŽ, y â int ðµ 3.187 Since int ðŽ â© ðµ = â, int ðŽ and ðµ can be separated. That is, there exists a continuous linear functional ð â ð â and a number ð such that ð (x) â€ ð â€ ð (y)

for every x â ðŽ, y â int ðµ

which implies that ð (x) â€ ð â€ ð (y)

for every x â ðŽ, y â ðµ

since ðâ€

inf

yâint ðµ

ð (y) = inf ð (y) yâðµ

Conversely, suppose that ðŽ and ðµ can be separated. That is, there exists ð â ð â such that ð (x) â€ ð â€ ð (y)

for every x â ðŽ, y â ðµ

Then ð (int ðŽ) is an open interval in [ð, â), which is disjoint from the interval ð (ðµ) â (ââ, ð]. This implies that int ðŽ â© ðµ = â. 3.188 Since x0 â b(ð), {x0 } â© int ð = â and int ð â= â. By Corollary 3.2.1, {x0 } and ð can be separated, that is there exist ð â ð â such that ð (x0 ) â€ ð (x) for every x â ð

169

Solutions for Foundations of Mathematical Economics

3.189 Let x â ð¶. Since ð¶ is a cone, ðx â ð¶ for every ð â¥ 0 and therefore ð (ðx) â¥ ð or ð (x) â¥ ð/ð

for every ð â¥ 0

Taking the limit as ð â â implies that ð (x) â¥ 0

for every x â ð¶

3.190 First note that 0 â ð and therefore ð (0) = 0 â€ ð so that ð â¥ 0. Suppose that there exists some z â ð for which ð (z) = ð â= 0. By linearity, this implies ð(

2ð 2ð z) = ð (z) = 2ð > ð ð ð

which contradicts the requirement ð (z) â€ ð for every z â ð 3.191 By Corollary 3.2.1, there exists ð â ð â such that ð (z) â€ ð â€ ð (x)

for every x â ð, z â ð

By Exercise 3.190 ð (z) = 0

for every z â ð

ð (x) â¥ 0

for every x â ð

and therefore

Therefore ð is contained in the hyperplane ð»ð (0) which separates ð from ð. 3.192 Combining Theorem 3.2 and Corollary 3.2.1, there exists a hyperplane ð»ð (ð) such that ð (x) â€ ð â€ ð (y)

for every x â ðŽ, y â ðµ

and such that ð (x) < ð â€ ð (y)

for every x â int ðŽ, y â ðµ

Since int ðŽ â= â, there exists some x â int ðŽ with ð (x) < ð. Hence ðŽ â ð â1 (ð) = ð»ð (ð). 3.193 Follows directly from the basic separation theorem, since ðŽ = int ðŽ and ðµ = int ðµ. 3.194 Let ð = ðµ â ðŽ. Then 1. ð is a nonempty, closed, convex set (Exercise 1.203). 2. 0 â / ð. There exists a continuous linear functional ð â ð â such that ð (x) â¥ ð > ð (0) = 0 170

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ðŽ

ðµ

Figure 3.2: ðŽ and ðµ cannot be strongly separated. for every z â ð (Exercise 3.182). For every x â ðŽ, y â ðµ, z = y â x â ð and ð (z) = ð (y) â ð (x) â¥ ð > 0 or ð (x) + ð â€ ð (y) which implies that sup ð (x) + ð â€ inf ð (y) yâðµ

xâðŽ

and sup ð (x) < inf ð (y)

xâðŽ

yâðµ

3.195 No. See Figure 3.2. 3.196

1. Assume that there exists a convex neighborhood ð â 0 such that (ðŽ + ð ) â© ðµ = â Then (ðŽ + ð ) is convex and ðŽ â int (ðŽ + ð ) â= â and int (ðŽ + ð ) â© ðµ = â. By Corollary 3.2.1, there exists continuous linear functional such that ð (x + u) â€ ð (y)

for every x â ðŽ, u â ð, y â ðµ

Since ð (ð ) is an open interval containing 0, there exists some u0 with ð (u0 ) = ð > 0. ð (x) + ð â€ ð (y)

for every x â ðŽ, y â ðµ

which implies that sup ð (x) < inf ð (y) yâðµ

xâðŽ

Conversely, assume that ðŽ and ðµ can be strongly separated. That is, there exists a continuous linear functional ð â ð â and number ð > 0 such that ð (x) â€ ð â ð < ð + ð â€ ð (y) for every x â ðŽ, y â ðµ Let ð = { ð¥ â ð : â£ð (ð¥)â£ < ð }. ð is a convex neighborhood of 0 such that (ðŽ + ð ) â© ðµ = â. 171

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2. Let ðŽ and ðµ be nonempty, disjoint, convex subsets in a normed linear space ð with ðŽ compact and ðµ closed. By Exercise 1.208, there exists a convex neighborhood ð â 0 such that (ðŽ + ð ) â© ðµ = â. By the previous part, ðŽ and ðµ can be strongly separated. 3.197 Assume ð(ðŽ, ðµ) = inf{ â¥x â yâ¥ : x â ðŽ, y â ðµ } = 2ð > 0. Let ð = ðµð (0) be the open ball around 0 of radius ð. For every x â ðŽ, u â ð, y â ðµ â¥x + (âu) â yâ¥ = â¥x â y â uâ¥ â¥ â¥x â yâ¥ â â¥uâ¥ so that ð(ðŽ + ð, ðµ) = inf â¥x + (âu) â yâ¥ â¥ inf (â¥x â yâ¥ â â¥uâ¥) x,u,y

x,u,y

â¥ inf â¥x â yâ¥ â sup â¥uâ¥) x,y

u

= 2ð â ð =ð>0 Therefore (ðŽ + ð ) â© ðµ = â and so ðŽ and ðµ can be strongly separated. Conversely, assume that ðŽ and ðµ can be strongly separated, so that there exists a convex neighborhood ð of 0 such that (ðŽ + ð ) â© ðµ = â. Therefore, there exists ð > 0 such that ðµð (0) â ð and ðŽ + ðµð â© ðµ = â This implies that ð(ðŽ, ðµ) = inf{ â¥x â yâ¥ : x â ðŽ, y â ðµ } > ð > 0 3.198 Take ðŽ = {y} and ðµ = ð in Proposition 3.14. There exists ð â ð â such that ð (y) < ð â€ ð (x)

for every x â ð

By Corollary 3.2.3, ð = 0. 3.199

1. Consider the set ð = { ð (ð¥), âð1 (ð¥), âð2 (ð¥), . . . , âðð (ð¥) : ð¥ â ð } ð is the image of a linear mapping from ð to ð = âð+1 and hence is a subspace of âð+1 .

2. By hypothesis, the point e0 = (1, 0, 0, . . . , 0) â âð+1 does not belong to ð. Otherwise, we have an ð¥ â ð such that ðð (ð¥) = 0 for every ð but ð (ð¥) = 1. 3. By the previous exercise, there exists a linear functional ð â ð â such that ð(e0 ) > 0 ð(z) = 0

for every z â ð

4. In other words, there exists a vector ð = (ð0 , ð1 , . . . , ðð ) â ð = â(ð+1)â such that ðe0 > 0 ðz = 0

(3.60) for every z â ð 172

(3.61)

Solutions for Foundations of Mathematical Economics Equation (3.61) states that ðz = ð0 z0 + ð1 z1 + â â â + ðð zð = 0

for every z â ð

That is, for every ð¥ â ð, ð0 ð (x) â ð1 ð1 (x) â ð2 ð2 (x) â . . . â ðð ðð (x) = 0 5. Inequality (3.60) establishes that ð0 > 0. Without loss of generality we can normalize so that ð0 = 1. 6. Therefore ð (ð¥) =

ð â

ðð ðð (ð¥)

ð=1

3.200 For every x â ð, ðð (x) = 0, ð = 1, 2 . . . ð and therefore ð (x) =

ð â

ðð ðð (x) = 0

ð=1

3.201 The set ð = { ð1 (ð¥), ð2 (ð¥), . . . , ðð (ð¥) : ð¥ â ð } is a closed subspace in âð . If the system is inconsistent, c = (ð1 , ð2 , . . . , ðð ) â / ð. By Exercise 3.198, there exists a linear functional ð on âð such ð(z) = 0 for every z â ð ð(c) > 0 That is, there exist numbers ð1 , ð2 , . . . , ðð such that ð â

ðð ðð (x) = 0

ð=1

and ð â

ðð ðð > 0

ð=1

which contradicts the hypothesis ð â

ðð ðð = 0 =â

ð=1

ð â

ðð ðð = 0

ð=1

Conversely, if for some x â ð ðð (x) = ðð

ð = 1, 2, . . . , ð

then ð â

ðð ðð (x) =

ð=1

ð â

ðð ðð

ð=1

and ð â

ðð ðð = 0 =â

ð=1

ð â ð=1

173

ðð ðð = 0

Solutions for Foundations of Mathematical Economics

Ë = { x â ðŸ : â¥xâ¥ = 1 } is 3.202 The set ðŸ 1 â compact (the unit ball is compact if and only if ð is ï¬nite-dimensional) â convex (which is why we need the 1 norm) By Proposition 3.14, there exists a linear functional ð â ð â such that Ë ËâðŸ for every x for every x â ð

ð (Ë x) > 0 ð (x) = 0

Ë Then Ë = x/ â¥xâ¥1 â ðŸ. For any x â ðŸ, x â= 0, deï¬ne x Ë ) = â¥xâ¥1 ð (Ë ð (x) = ð (â¥xâ¥1 x x) > 0 3.203

1. Let ðŽ = { (x, ðŠ) : ðŠ â¥ ð(x), x â ð } ðµ = { (x, ðŠ) : ðŠ = ð0 (x), x â ð } ðŽ is the epigraph of a convex functional and hence convex. ðµ is a subspace of ð = ð Ã â and also convex.

2. Since ð is convex, int ðŽ â= â. Furthermore ð0 (x) â€ ð(x) =â int ðŽ â© ðµ = â 3. By Exercise 3.2.3, there exists linear functional ð â ð â such that ð(x, ðŠ) â¥ 0 ð(x, ðŠ) = 0

for every (x, ðŠ) â ðŽ for every (x, ðŠ) â ðµ

There exists ðŠ such that ðŠ > ð(0) and therefore (0, ðŠ) â int ðŽ and ð(0, ðŠ) > 0. Therefore ð(0, 1) =

1 ð(0, 1) > 0 ðŠ

4. Let ð â ð â be deï¬ned by 1 ð (x) = â ð(x, 0) ð where ð = ð(0, 1). Since ð(x, 0) = ð(x, ðŠ) â ð(0, ðŠ) = ð(x, ðŠ) â ððŠ 1 1 ð (x) = â (ð(x, ðŠ) â ððŠ) = â ð(x, ðŠ) + ðŠ ð ð for every ðŠ â â 5. For every x â ð 1 ð (x) = â ð(x, ð0 (x)) + ð0 (x) ð = ð0 (x) since ð(x, ð0 (x)) = 0 for every x â ð. Thus ð is an extension of ð0 . 174

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6. For any x â ð, let ðŠ = ð(x). Then (x, ðŠ) â ðŽ and ð(x, ðŠ) â¥ 0. Therefore 1 ð (x) = â ð(x, ðŠ) + ðŠ ð 1 = â ð(x, ðŠ) + ð(x) ð â€ ð(x) Therefore ð is bounded by ð as required. 3.204 Let ð â ð â be deï¬ned by ð(x) = â¥ð0 â¥ð â¥xâ¥ Then ð0 (x) â€ ð(x) for all x â ð. By the Hahn-Banach theorem (Exercise 3.15), there exists an extension ð â ð â such that ð (x) â€ ð(x) = â¥ð0 â¥ð â¥xâ¥ Therefore â¥ð â¥ð = sup â¥ð (x)â¥ = â¥ð0 â¥ð â¥xâ¥=1

3.205 If x0 = 0, any bounded linear functional will do. Therefore, assume x0 â= 0. On the subspace lin {x0 } = {ðŒx0 : ðŒ â â}, deï¬ne the function ð0 (ðŒx0 ) = ðŒ â¥x0 â¥ ð0 is a bounded linear functional on lin {x0 } with norm 1. By the previous part, ð0 can be extended to a bounded linear functional ð â ð â with the same norm, that is â¥ð â¥ = 1 and ð (x0 ) = â¥x0 â¥. 3.206 Since x1 â= x2 , x1 â x2 â= 0. There exists a bounded linear functional such that ð (x1 â x2 ) = â¥x1 â x2 â¥ â= 0 so that ð (x1 ) â= ð (x2 ) 3.207

1.

â ð is a complete lattice (Exercise 1.179).

â The intersection of any chain is â nonempty (since ð is compact) â a face (Exercise 1.179) Hence every chain has a minimal element. â By Zornâs lemma (Remark 1.5), ð has a minimal element ð¹0 . 2. Assume to the contrary that ð¹0 contains two distinct elements x1 , x2 . Then (Exercise 3.206) there exists a continuous linear functional ð â ð â such that ð (x1 ) â= ð (x2 ) Let ð be in the minimum value of ð (x) on ð¹0 and let ð¹1 be the set on which it attains this minimum. (Since ð¹0 is compact, ð is well-deï¬ned and ð¹1 is nonempty. That is ð = min{ ð (ð¥) : x â ð¹0 } ð¹1 = { x â ð : ð (x) = ð } 175

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Now ð¹1 â ð¹0 since ð (x1 ) â= ð (x2 ). To show that ð¹1 is a face of ð¹0 , assume that ðŒx+(1âðŒ)y â ð¹1 for some x, y â ð¹0 . Then ð = ð (ðŒx + (1 â ðŒ)y) = ðŒð (x) + (1 â ðŒ)ð (y) = ð. Since x, y â ð¹0 , this implies that ð (x) = ð (y) = ð so that x, y â ð¹1 . Therefore ð¹1 is a face. We have shown that, if ð¹0 contains two distinct elements, there exists a smaller face ð¹1 â ð¹0 , contradicting the minimality of ð¹0 . We conclude that ð¹0 comprises a single element x0 . 3. ð¹0 = {x0 } which is an extreme point of ð. 3.208 Let ð» = ð»ð (ð) be a supporting hyperplane to ð. Without loss of generality assume ð (x) â€ ð for every x â ð

(3.62)

and there exists some xâ â ð such that ð (xâ ) = ð That is ð is maximized at xâ . Version 1 By the previous exercise, ð achieves its maximum at an extreme point. That is, there exists an extreme point x0 â ð such that ð (x0 ) â¥ ð (x) for every x â ð In particular, ð (x0 ) â¥ ð (xâ ) = ð. But (3.62) implies ð (x0 ) â€ ð. Therefore, we conclude that ð (x0 ) = ð and therefore x0 â ð». Version 2 The set ð» â© ð is a nonempty, compact, convex subset of a linear space. Hence, by Exercise 3.207, ð» â© ð contains an extreme point, say x0 . We show that x0 is an extreme point of ð. Assume not, that is assume that there exists x1 , x2 â ð such that x0 = ðŒx1 + (1 â ðŒ)x2 for some ðŒ â (0, 1). Since x0 is an extreme point of ð» â© ð, at least one of the points x1 , x2 must lie outside ð». Assume x1 â / ð» which implies that ð (x1 ) < ð. Since ð (x2 ) â€ ð ð (x0 ) = ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) < ð

(3.63)

However, since x0 â ð» â© ð, we must have ð (x0 ) = ð which contradicts (3.63). Therefore x0 is an extreme point of ð. In fact, we have shown that every extreme point of ð» â© ð must be an extreme point of ð. 3.209 Let ðË denote the closed, convex hull of the extreme points of ð. (The closed, convex hull of a set is simply the closure of the convex hull.) Clearly ðË â ð and it remains to show that ðË contains all of ð. Ë By the Strong Separation Assume not. That is, assume ðË â ð and let x0 â ð â ð. Theorem, there exists a linear functional ð â ð â such that ð (x0 ) > ð (x) for every x â ðË 176

(3.64)

Solutions for Foundations of Mathematical Economics

On the other hand, by Exercise 3.16, ð attains its maximum at an extreme point of ð. That is, there exists x1 â ðË such that ð (x1 ) â¥ ð (x) for every x â ð In particular ð (x1 ) â¥ ð (x0 ) Ë since x0 â ðË â ð. This contradicts (3.64) since x1 â ð. Thus our assumption that ð â ðË yields a contradiction. We conclude that ð = ðË 3.210

1. (a) ð is compact and convex, since it is the product of compact, convex sets (Proposition 1.2, Exercise 1.165). âð âð (b) Since x â ð=1 conv ðð , there exist xð â conv ðð such that x = ð=1 xð . (x1 , x2 , . . . , xð ) â ð (x) so that ð (x) â= â. (c) By the Krein-Millman theorem (or Exercise 3.207), ð (x) has an extreme point z = (z1 , z2 , . . . , zð ) such that â zð â conv ðð for every ð âð â ð=1 zð = x. since z â ð (x).

2. (a) Exercise 1.176 (b) Since ð > ð = dim ð, the vectors y1 , y2 , . . . , yð are linearly dependent (Exercise 1.143). Consequently, there exists numbers ðŒâ²1 , ðŒâ²2 , . . . , ðŒâ²ð , not all zero, such that ðŒâ²1 y1 + ðŒâ²2 y2 + â â â + ðŒâ²ð yð = 0 (Exercise 1.133). Let ðŒð =

ðŒâ²ð maxð â£ðŒð â£

Then â£ðŒð â£ â€ 1 for every ð and ðŒ1 y1 + ðŒ2 y2 + â â â + ðŒð yð = 0 (c) Since â£ðŒð â£ â€ 1, zð + ðŒð yð â conv ðð for every ð = 1, 2, . . . , ð. Furthermore ð â ð=1

z+ ð =

ð â

zð +

ð=1

ð â

ðŒð yð =

ð=1

ð â

zð = x

ð=1

Therefore, z+ â ð (x). Similarly, zâ â ð (x). (d) By direct computation z=

1 + 1 â z + z 2 2

which implies that z is not an extreme point of ð (x), contrary to our assumption. This establishes that at least ð â ð zð are extreme points of the corresponding conv ðð . 177

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4 3 (0, 2.5)

conv ð2 2

(.5, 2) P(x)

1 0

1

2

3

4

conv ð1 Figure 3.3: Illustrating the proof of the Shapley Folkman theorem. 3. Every extreme point of conv ðð is an element of ðð . 3.211 See Figure 3.3. 3.212 Let {ð1 , ð2 , . . . , ðð } be a â collection of subsets of an ð-dimensional ânonempty ð ð linear space and let x â â conv ð = conv ð . That is, there exists xð â ð ð ð=1 ð=1 conv ðð such that x = ðð=1 xð . By CarathÂŽeodoryâs theorem, there exists for every xð a ï¬nite number of points xð1 , xð2 , . . . , xððð such that xð â conv {xð1 , xð2 , . . . , xððð }. For every ð = 1, 2, . . . , ð, let ðËð = { xðð : ð = 1, 2, . . . , ðð } Then x=

ð â

xð ,

xð â conv ðËð

ð=1

â âË ðð . Moreover, the sets ðð are compact (in fact That is, x â conv ðËð = conv ï¬nite). By the previous exercise, there exists ð points zð â ðËð such that x=

ð â

zð ,

zð â conv ðËð

ð=1

and moreover zð â ðËð â ðð for at least ð â ð indices ð. 3.213 Let ð be a closed convex set in a normed linear space. Clearly, ð is contained in the intersection of all the closed halfspaces which contain ð. For any y â / ð, there exists a hyperplane which strongly separates {y} and ð. One of its closed halfspaces contains ð but not y. Consequently, y does not belong to the intersection of all the closed halfspaces containing ð. 3.214

1. Since ð â (ðŠ) is the intersection of closed, convex sets, it is closed and convex. Assume x is feasible, that is x â ð (ðŠ). Then wð x â€ (ðw, ðŠ) and x â ð â (ðŠ). That is, ð (ðŠ) â ð â (ðŠ).

178

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2. Assume ð (ðŠ) is convex. For any x0 â / ð (ðŠ) there exists w such that wð x0 <

inf wð x = ð(w, ðŠ)

xâð (ðŠ)

by the Strong Separation Theorem. Monotonicity ensures that w â¥ 0 and hence x0 â / ð â (ðŠ). 3.215 Assume x â ð (ðŠ) = ð â (ðŠ). That is ð(w) for every x wð x â¥ ðŠË Therefore, for any ð¡ â â+ ð¡wð x â¥ ð¡ðŠð(w) for every x which implies that ð¡x â ð â (ðŠ) = ð (ðŠ). 3.216 A polyhedron ð = { ð¥ â ð : ðð (ð¥) â€ ðð , ð = 1, 2, . . . , ð } ð â© = { x â ð : ðð (x) â€ ðð } ð=1

is the intersection of a ï¬nite number of closed convex sets. 3.217 Each row að = (ðð1 , ðð2 , . . . ððð ) of ðŽ deï¬nes a linear functional ðð (x) = ðð1 ð¥1 + ðð2 ð¥2 + â â â + ððð ð¥ð on âð . The set ð of solutions to ðŽx â€ c is ð = { ð¥ â ð : ðð (x) â€ ðð , ð = 1, 2, . . . , ð } is a polyhedron. 3.218 For simplicity, we assume that the game is superadditive, so that ð€(ð) â¥ 0 for every ð. Consequently, in every core allocation x, 0 â€ ð¥ð â€ ð€(ð ) and core â [0, ð€(ð )] Ã [0, ð€(ð )] Ã â â â Ã [0, ð€(ð )] â âð 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 =â polyhedron Assume that ð is a polytope generated by the points { x1 , x2 , . . . , xð } and let ð¹1 , ð¹2 , . . . , ð¹ð denote the proper faces of ð . For each ð = 1, 2, . . . , ð, let ð»ð denote the hyperplane containing ð¹ð so that ð¹ð = ð â© ð»ð . For every such hyperplane, there exists a nonzero linear functional ðð and constant ðð such that ðð (x) = ðð for every x â ð»ð . Furthermore, every such hyperplane is a bounding hyperplane of ð . Without loss of generality, we can assume that ðð (x) â€ ð for every x â ð . Let ð = { x â ð : ðð (x) â€ ðð , ð = 1, 2, . . . , ð } Clearly ð â ð. To show that ð â ð , assume not. That is, assume that there exists y â ð âð and let x â ri ð . (ri ð is nonempty by exercise 1.229). Since ð is Â¯ = ðŒx+(1âðŒ)y belongs closed (Exercise 1.227), there exists a some ðŒ such that x Â¯ â ð¹ð â ð»ð . to the relative boundary of ð , and there exists some ð such that x Let ð»ð+ = { x â ð : ðð (x) â€ ðð } denote the closed half-space bounded by ð»ð and Â¯ = ðŒx+(1âðŒ)y, which implies that containing ð . ð»ð is a face of ð»ð+ containing x x, y â ð»ð . This in turn implies that x â ð¹ð , which contradicts the assumption that x â ri ð . We conclude that ð = ð . 179

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polyhedron =â polytope Conversely, assume ð is a nonempty compact polyhedral set in a normed linear space. Then, there exist linear functionals ð1 , ð2 , . . . , ðð in ð â and numbers ð1 , ð2 , . . . , ðð such that ð = { x â ð : ðð (x) â€ ðð , ð = 1, 2, . . . , ð }. We show that ð has a ï¬nite number of extreme points. Let ð denote the dimension of ð. If ð = 1, ð is either a single point or closed line segment (since ð is compact), and therefore has a ï¬nite number of extreme points (that is, 1 or 2). Now assume that every compact polyhedral set of dimension ð â 1 has a ï¬nite number of extreme points. Let ð»ð , ð = ð = 1, 2, . . . , ð denote the hyperplanes associated with the linear functionals ðð deï¬ning ð (Exercise 3.49). Let x be an extreme point of ð. Then ð is a boundary point of ð (Exercise 1.220) and therefore belongs to some ð»ð . We claim that x is also an extreme point of the set ð â© ð»ð . To see this, assume otherwise. That is, assume that x is not an extreme point of ð â©ð»ð . Then, there exists x1 , x2 â ð â©ð»ð such that x = ðŒx1 + (1 â ðŒ)x2 . But then x1 , x2 â ð and x is not an extreme point of ð. Therefore, every extreme point of ð is an extreme point of some ð â© ð»ð , which is a compact polyhedral set of dimension ð â 1. By hypothesis, each ð â© ð»ð has a ï¬nite number of extreme points. Since there are only ð such hyperplanes ð»ð , ð has a ï¬nite number of extreme points. By the Krein-Milman theorem (Exercise 3.209), ð is the closed convex hull of its extreme points. Since there are only ï¬nite extreme points, ð is a polytope. 3.220

1. Let ð, ð â ð â so that ð (x) â€ 0 and ð(x) â€ 0 for every x â ð. For every ðŒ, ðœ â¥ 0 ðŒð (x) + ðœð (x) â€ 0 for every ð¥ â ð. This shows that ðŒð + ðœð â ð â . ð â is a convex cone. To show that ð â is closed, let ð be the limit of a sequence (ðð ) of functionals in ð â . Then, for every x â ð, ðð (x) â€ 0 so that ð (x) = lim ðð (x) â€ 0

2. Let x, y â ð ââ . Then, for every ð â ð â ð (x) â€ 0 and ð (y) â€ 0 and therefore ð (ðŒx + ðœy) = ðŒð (x) + ðœð (y) â€ 0 for every ðŒ, ðœ â¥ 0. There ðŒx + ðœy â ð ââ . ð ââ is a convex cone. To show that ð ââ is closed, let xð be a sequence of points in ð ââ converging to ð¥. For every ð = 1, 2, . . . ð (xð ) â€ 0 for every ð â ð â By continuity ð (x) = lim ð (xð ) â€ 0 for every ð â ð â Consequently x â ð ââ which is therefore closed. 180

Solutions for Foundations of Mathematical Economics

3. Let x â ð. Then ð (x) â€ 0 for every ð â ð â so that x â ð ââ . 4. Exercise 1.79. 3.221 Let ð â ð2â . Then ð (x) â€ 0 for every x â ð2 . A fortiori, since ð1 â ð2 , ð (x) â€ 0 for every x â ð1 . Therefore ð â ð1â . 3.222 Exercise 3.220 showed that ð â ð ââ . To show the converse, let y â / ð. By Proposition 3.14, there exists some ð â ð â and ð such that ð (y) > ð ð (x) < ð

for every x â ð

Since ð is a cone, 0 â ð and ð (0) = 0 < ð. Since ðŒð = ð for every ðŒ > 0 then ð (x) < 0

for every x â ð

/ ð ââ . That is so that ð â ð â . ð (y) > 0, y â yâ / ð =â y â / ð ââ from which we conclude that ð ââ â ð. 3.223 Let ðŸ = cone {ð1 , ð2 , . . . , ðð } ð â = { ð â ðâ : ð = ðð ðð , ðð â¥ 0 } ð=1

be the set of all nonnegative linear combinations of the linear functionals ðð . ðŸ is a closed convex cone. Suppose that ð â / cone {ð1 , ð2 , . . . , ðð }, that is assume that ð â / ðŸ. Then {ð } is a compact convex set disjoint from ðŸ. By Proposition 3.14, there exists a continuous linear functional ð and number ð such that sup ð(ð) < ð < ð(ð )

ðâðŸ

Since 0 â ðŸ, ð â¥ 0 and so ð(ð ) > 0. Further, for every ð â ðº ð â ðð ðð ) ð(ð) = ð( ð=1

=

ð â

ðð ð(ðð ) < ð for every ðð â¥ 0

ð=1

Since ðð can be made arbitrarily large, this last inequality implies that ð(ðð ) â€ 0

ð = 1, 2, . . . , ð

By the Riesz representation theorem (Exercise 3.75), there exists x â ð ð(ðð ) = ðð (x) and ð(ð ) = ð (x) Since ð(ðð ) = ðð (x) â€ 0 181

Solutions for Foundations of Mathematical Economics

x â ð. By hypothesis ð (x) = ð(ð ) â€ 0 contradicting the conclusion that ð(ð ) > 0. This contradiction establishes that ð â ðŸ, that is ð (ð¥) =

ð â

ðð ðð (ð¥),

ðð â¥ 0

ð=1

3.224 Let a1 , a2 , . . . , að denote the rows of ðŽ and deï¬ne the linear functional ð, ð1 , ð2 , . . . , ðð by ð (x) = cx ðð (x) = að x ð = 1, 2, . . . , ð Assume cx â€ 0 for every x satisfying ðŽx â€ 0, that is ð (x) â€ 0 for every x â ð where ð = { x â ð : ðð (x) â€ 0, ð = 1, 2, . . . , ð } By Proposition 3.18, there exists y â âð + such that ð (x) =

ð â

ðŠð ðð (x)

ð=1

or c=

ð â

ðŠð að = ðŽð y

ð=1

Conversely, assume that c = ðŽð y =

ð â

ðŠð að

ð=1

Then ðŽx â€ 0 =â að x â€ 0 for every ð =â cx â€ 0 3.225 Let ð = âð+ denote the positive orthant of âð . ð is a convex set (indeed cone) with a nonempty interior. By Corollary 3.2.1, there exists a hyperplane ð»p (ð) such that pð x â€ ð â€ py

for every x â ð, y â ð

Since 0 â ð p0 = 0 â¥ ð which implies that ð â€ 0 and pð x â€ ð â€ 0

for every ð¥ â ð

To show that p is nonnegative, let e1 , e2 , . . . , eð denote the standard basis for âð . Each eð belongs to ð so that peð = ðð â¥ 0 182

for every ð

Solutions for Foundations of Mathematical Economics

3.226 Assume yâ is an eï¬cient production plan in ð and let ð = ð â ðŠ â . ð is convex. We claim that ð â©âð++ = â. Otherwise, if there exists some z â ð â©âð++ , let yâ² = yâ +z â z â ð implies yâ² â ð while â z â âð++ implies yâ² > yâ contradicting the eï¬ciency of yâ . Therefore, ð is a convex set which contains no interior points of the nonnegative orthant âð+ . By Exercise 3.225, there exists a price system p such that pð x â€ 0 for every x â ð Since ð = ð â ðŠ â , this implies p(y â yâ ) â€ 0 for every y â ð or pyâ â¥ py for every y â ð ðŠ â maximizes the producerâs proï¬t at prices p. 3.227 Consider the set ð â = { x â âð : âx â ð }. ð â© int âðâ = â =â ð â â© int âð+ = â From the previous exercise, there exists a hyperplane with nonnegative normal p â© 0 such that pð x â€ 0

for every x â ð â

pð x â¥ 0

for every x â ð

Since p â© 0, this implies

3.228

1. Suppose x â â¿(xâ ). Then, there exists an allocation (x1 , x2 , . . . , xð ) such that x=

ð â

xð

ð=1

where xð â â¿(xâð ) for every ð = 1, 2, . . . , ð. Conversely, if (x1â , x2 , . . . , xð ) is an ð allocation with xð â â¿(xâð ) for every ð = 1, 2, . . . , ð, then x = ð=1 xð â â¿(xâ ). 2. For every agent ð, xâð â â¿(xâð ), which implies that xâ =

ð â

xâð â â¿(xâ )

ð=1

and therefore 0 â ð = â¿(xâ ) â xâ â= â Since individual preferences are convex, â¿(xâð ) is convex for each ð and therefore â â â â ð = â¿(x ) â x = ð â¿(xð ) â xâ is convex (Exercise 1.164).

183

Solutions for Foundations of Mathematical Economics

Assume to the contrary that ð â© int âðâ â= â. That is, there exists some z â ð with z < 0. This implies that there exists some allocation (x1 , x2 , . . . , xð ) such that â xð â xâ < 0 z= ð

xâð

and xð â¿ for every ð â ð . Distribute z equally to all the consumers. That is, consider the allocation yð = xð + z/ð By strict monotonicity, yð â» xð â¿ xâð for every ð â ð . Since â â â yð = xð + z = xâ = xâð ð

ð

ð

(y1 , y2 , . . . , yð ) is a reallocation of the original allocation xâ which is strictly preferred by all consumers. This contradicts the assumed Pareto eï¬ciency of xâ . We conclude that ð â© int âðâ â= â 3. Applying Exercise 3.227, there exists a hyperplane with nonnegative normal pâ â© 0 such that pâ z â¥ 0 for every z â ð That is pâ (x â xâ ) â¥ 0 or pâ x â¥ pâ xâ for every x â â¿(xâ )

(3.65)

4. Consider any allocation which is strictly preferred to xâ by consumer ð, that is xð â â»ð (xâð ). Construct another allocation y by taking ð > 0 of each commodity away from agent ð and distributing amongst the other agents to give yð = (1 â ð)xð ð yð = xâð + xð , ðâ1

ð â= ð

By continuity, there exists some ð > 0 such that yð = (1 â ð)xð â»ð xâð . By monotonicity, yð â»ð xâð for every ð â= ð. We have constructed â an allocation y which is strictly preferred to xâ by all the agents, so that y = ð yð â â¿(xâ ). (3.65) implies that py â¥ pxâ That is â

â( xâð + p â(1 â ð)xð + ðâ=ð

â â â â â ) â â ð xð â  = p âxð + xâð â  â¥ p âxâð + xâð â  ðâ1 ðâ=ð

ðâ=ð

which implies that pxð â¥ pxâð

for every xð â â»(xâð ) 184

(3.66)

Solutions for Foundations of Mathematical Economics

5. Trivially, xâ is a feasible allocation with endowments wð = xâð and ðð = pâ wð = pâ xâð . To show that (pâ , xâ ) is a competitive equilibrium, we have to show that xâð is the best allocation in the budget set ðð (p, ðð ) for each consumer ð. Suppose to the contrary there exists some consumer ð and allocation yð such that yð â» xð and pyð â€ ðð = pxâð . By continuity, there exists some ðŒ â (0, 1) such that ðŒyð â»ð xâð and p(ðŒyð ) = ðŒpyð < pyð â€ pxâ contradicting (3.66). We conclude that xâð â¿ð xð for every x â ð(pâ , ðð ) for every consumer ð. (pâ , xâ ) is a competitive equilibrium. 3.229 By the previous exercise, there exists a price system pâ such that xâð is optimal for each consumer ð in the budget set ð(pâ , pâ xâð ), that is xâð â¿ð xð for every xð â ð(pâ , pâ xâð )

(3.67)

For each consumer, let ð¡ð be the diï¬erence between her endowed wealth pâ wð and her required wealth pâ xâð . That is, deï¬ne ð¡ð = pâ xâð â pâ wð = pâ (xâð â wð ) Then pâ xâð = pâ + wð

(3.68)

â

By assumption x is feasible, so that â â â xâð â wð = (xâð â wð ) = 0 ð

so that

ð

â ð

ð¡ð = pâ

ð

â

(xâð â wð ) = 0

ð

â

Furthermore, for ðð = ð wð + ð¡ð , (3.68) implies ð(pâ , ðð ) = { xð : pâ xð â€ pâ wð + ð¡ð } = { xð : pâ xð â€ pâ xâð } = ð(pâ , pâ xâð ) for each consumer ð. Using (3.67) we conclude that xâð â¿ð xð for every xð â ð(pâ , ðð ) for every agent ð. (pâ , xâ ) is a competitive equilibrium where each consumerâs after-tax wealth is ðð = pwð + ð¡ð 3.230 Apply Exercise 3.202 with ðŸ = âð+ . 3.231 ðŸ â = { p : pð x â€ 0 for every x â ðŸ } No such hyperplane exists if and only if ðŸ â â© âð++ = â. Assume this is the case. By Exercise 3.225, there exists x â© 0 such that xp = pð x â€ 0 for every p â ðŸ â In other words, x â ðŸ ââ . By the duality theorem ðŸ ââ = ðŸ which implies that x â ðŸ as well as âð+ , contrary to the hypothesis that ðŸ â© âð+ = {0}. This contradiction establishes that ðŸ â â© âð++ â= â. 185

Solutions for Foundations of Mathematical Economics

3.232 Given a set of ï¬nancial assets with prices p and payoï¬ matrix ð, let ð = { (âpx, ðð¥) : x â âð } ð is the set of all possible (cost, payoï¬) pairs. It is a subspace of âð+1 . Let ð be the nonnegative orthant in âð+1 . The no arbitrage condition ðx â¥ 0 =â pð x â¥ 0 implies that ð â© ð = {0}. By Exercise 3.230, there exists a hyperplane with positive normal ð = ð0 , ð1 , . . . , ðð such that ðz = 0

for every z â ð

ðz > 0

â {0} for every z â âð+1 +

That is for every x â âð

âð0 px + ððx = 0 or pð x = ð/ð0 ðx

for every x â âð

ð/ð0 is required state price vector. Conversely, if a state price vector exists ðð =

ð â

ððð  ðð

ð=1

then clearly ðx â¥ 0 =â pð x â¥ 0 No arbitrage portfolios exist. 3.233 Apply the Farkas lemma to the system âðŽx â€ 0 âcð x > 0 3.234 The inequality system ðŽð y â¥ c has a nonnegative solution if and only if the corresponding system of equations ðŽð y â z = c ð has a nonnegative solution y â âð + , z â â+ . This is equivalent to the system ( ) y â² ðµ =c z

(3.69)

where ðµ â² = (ðŽð , âðŒð ) and ðŒð is the ð Ã ð identity matrix. By the Farkas lemma, system (3.69) has no solution if and only if the system ðµx â€ 0 and cð x > 0 ( ) ðŽ has a solution x â âð . Since ðµ = , ðµx â€ 0 implies âðŒ ðŽx â€ 0 and â ðŒx â€ 0 and the latter inequality implies x â âð+ . Thus we have established that the system ðŽð y â¥ c has no nonnegative solution if and only if ðŽx â€ 0 and cð x > 0 for some x â âð+ 186

Solutions for Foundations of Mathematical Economics

Ë â âð+ such that 3.235 Assume system I has a solution, that is there exists x Ëâ¥0 ðŽË x = 0, cË x > 0, x Ë /cË Then x = x x satisï¬es the system ðŽx = 0, cx = 1, x â¥ 0

(3.70)

xâ² ðŽð = 0, xc = 1, x â¥ 0

(3.71)

which is equivalent to

Suppose y â âð satisï¬es ðŽy â¥ c Multiplying by x â¥ 0 gives xâ² ðŽð y â¥ xc Substituting (3.71), this implies the contradiction 0â¥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 ( ) ( ) ðŽ 0 x= c 1 or ðµx = b (3.72) ) ( ) ( âðŽ 0 is (ð + 1) Ã ð and b = â âð+1 . If (3.72) has no solution, where ðµ = c 1 there exists (by the Farkas alternative) some z â âð+1 such that ðµ â² z â€ 0 and bz > 0 Decompose z into z = (y, ð§) with y â âð and ð§ â â. The second inequality implies that (0, 1)â² (y, ð§) = 0y + ð§ = ð§ > 0 Without loss of generality, we can normalize so that ð§ = 1 and z = (y, 1). Now ðµ â² = (âðŽð , c) and so the ï¬rst inequality implies that ( ) y ð (âðŽ , c) = âðŽð y + c â€ 0 1 or ðŽð y â¥ c We conclude that II has a solution. 187

Solutions for Foundations of Mathematical Economics

3.236 For every linear functional ðð , there exists a vector a â âð such that ðð (x) = að xË (Proposition 3.11). Let ðŽð be the matrix whose rows are að , that is â 1â a â a2 â â ðŽ=â â. . .â  að Then, the system of inequalities (3.31) is ðŽð x â¥ c where c = (ð1 , ð2 , . . . , ðð ). By the preceding exercise, this system is consistent if and only there is no solution to the system ðŽð = 0

cð > 0

ðâ¥0

Now ð â

ðŽð = 0 ââ

ðð ðð = 0

ð = 1, 2, . . . , ð

ð=1

Therefore, the inequalities (3.31) is consistent if an only if ð â

ðð ðð = 0 =â

ð=1

ð â

ðð ðð â€ 0

ð=1

for every set of nonnegative numbers ð1 , ð2 , . . . , ðð . 3.237 Let ðµ be the 2ð Ã ð matrix comprising ðŽ and âðŽ as follows ( ) ðŽ ðµ= âðŽ Then the Fredholm alternative I cð x = 1

ðŽx = 0 is equivalent to the system ðµx â€ 0

cx > 0

(3.73)

2ð By the Farkas alternative theorem, either (3.73) has a solution or there exists ð â ð+ such that

ðµâ²ð = c

(3.74)

Decompose ð into two ð-vectors ð ð = (ð, ð¿), ð, ð¿ â ð+

so that (3.74) can be rewritten as ðµ â² ð = ðŽð ð â ðŽð ð¿ = ðŽð (ð â ð¿) = c Deï¬ne y = ð â ð¿ â âð We have established that either (3.73) has a solution or there exists a vector y â âð such that ðŽð y = c 188

Solutions for Foundations of Mathematical Economics

3.238 Let að , ð = 1, 2, . . . , ð denote the rows of ðŽ. Each að deï¬nes linear functional ðð (ð¥) = að ð¥ on âð , and c deï¬nes another linear functional ð (ð¥) = cð x. Assume that ð (ð¥) = cð x = 0 for every x â ð where ð = { x : ðð (x) = að x = 0, ð = 1, 2, . . . , ð } Then the system ðŽð¥ = 0 has no solution satisfying the constraint cð x > 0. By Exercise 3.20, there exists scalars ðŠ1 , ðŠ2 , . . . , ðŠð such that ð (x)=

ð â

ðŠð ðð (x)

ð=1

or c=

ð â

ðŠð ðð = ðŽð y

ð=1

That is y = (ðŠ1 , ðŠ2 , . . . , ðŠð ) solves the related nonhomogeneous system ðŽð y = c Conversely, assume that ðŽð y = c for some ðŠ â âð . Then cð x = ðŠðŽð¥ = 0 for all ð¥ such that ðŽð¥ = 0 and therefore there is no solution satisfying the constraint cð x = 1. 3.239 Let ð = { z : z = ðŽx, x â â } the image of ð. ð is a subspace. Assume that system I has no solution, that is ð â© âð ++ = â By Exercise 3.225, there exists y â âð + â {0} such that yz = 0 for every z â ð That is yðŽx = 0 for every x â âð Letting x = ðŽð y, we have yðŽðŽð y = 0 which implies that ðŽð y = 0 System II has a solution y. Ë is a solution to I. Suppose to the contrary there also exists Conversely, assume that x Ë to II. Then, since ðŽË Ë â© 0, we must have y Ë ðŽË Ë ðŽð y Ë > 0. a solution y x > 0 and y x=x ðË ð ËðŽ y Ë = 0, a contradiction. Hence, we On the other hand, ðŽ y = 0 which implies x conclude that II cannot have a solution if I has a solution. 189

Solutions for Foundations of Mathematical Economics

3.240 We have already shown (Exercise 3.239) that the alternatives I and II are mutually incompatible. If Gordanâs system II ðŽð y = 0 has a semipositive solution y â© 0, then we can normalize y such that 1y = 1 and the system ðŽð y = 0 1y = 1 has a nonnegative solution. Conversely, if Gordanâs system II has no solution, the system ðµâ²y = c (

) ðŽð where ðµ = and c = (0, 1) = (0, 0, . . . , 0, 1), 0 â âð , is the (ð + 1)st unit 1 vector has no solution y â¥ 0. By the Farkas lemma, there exists z â âð+1 such that â²

ðµz â¥ 0 cz < 0 Decompose z into z = (x, ð¥) with x â âð . The second inequality implies that ð¥ < 0 since cz = (0, 1)â² (x, ð¥) = ð¥ < 0 Since ðµ = (ðŽ, 1), the ï¬rst inequality implies that ðµz = (ðŽ, 1)(x, ð¥) = ðŽx + 1ð¥ â¥ 0 or ðŽx â¥ â1ð¥ > 0 x solves Gordanâs system I. 3.241 Let a1 , a2 , . . . , að be a basis for ð. Let ðŽ = (a1 , a2 , . . . , að ) be the matrix whose columns are að . To say that ð contains no positive vector means that the system ðŽx > 0 has no solution. By Gordanâs theorem, there exists some y â© 0 such that ðŽð y = 0 that is að y = yað = 0, ð = 1, 2, . . . , ð so that y â ð â¥ . 190

Solutions for Foundations of Mathematical Economics

3.242 Let ð be the subspace ð = { z : ðŽx : x â âð }. System I has no solution ðŽx â© 0 if and only if ð has no nonnegative vector z â© 0. By the previous exercise, ð â¥ contains a positive vector y > 0 such that yz = 0 for every z â ð Letting x = ðŽð y, we have yðŽðŽð y = 0 which implies that ðŽð y = 0 System II has a solution y. 3.243 Let ð = { z : z = ðŽx, x â â } the image of ð. ð is a subspace. Assume that system I has no solution, that is ð â© âð + = {0} By Exercise 3.230, there exists y â âð ++ such that yz = 0 for every z â ð That is yðŽx = 0 for every x â âð Letting x = ðŽð y, we have yðŽðŽð y = 0 which implies that ðŽð y = 0 System II has a solution y. Ë is a solution to I. Suppose to the contrary there also exists Conversely, assume that x Ë > 0. Ë to II. Then, since ðŽË Ë > 0, we must have y Ë ðŽË Ë ðŽð y a solution y x â© 0 and y x=x ðË ð ËðŽ y Ë = 0, a contradiction. Hence, we On the other hand, ðŽ y = 0 which implies x conclude that II cannot have a solution if I has a solution. 3.244 The inequality system ðŽð y â€ 0 has a nonnegative solution if and only if the corresponding system of equations ðŽð y + z = 0 ð has a nonnegative solution y â âð + , z â â+ . This is equivalent to the system ( ) y â² ðµ =0 z

(3.75)

where ðµ â² = (ðŽð , ðŒð ) and ðŒð is the ð Ã ð identity matrix. By Gordanâs theorem, system (3.75) has no solution if and only if the system ( has a solution x â âð . Since ðµ =

ðµx > 0 ) ðŽ , ðµx > 0 implies ðŒ

ðŽx > 0 and ðŒx > 0 and the latter inequality implies x â âð++ . Thus we have established that the system ðŽð y â€ 0 has no nonnegative solution if and only if ðŽx > 0 for some x â âð++ 191

Solutions for Foundations of Mathematical Economics

3.245 Assume system II has no solution, that is there is no y â âð such that ðŽy â€ 0, y â© 0 This implies that the system âðŽy â¥ 0 1y â¥ 1 ( ) âðŽ ð â² has no solution y â â+ . Deï¬ning ðµ = , the latter can be written as 1â² ðµ â² y â¥ âeð+1

(3.76)

where âeð+1 = (0, 1), 0 â âð . By the Gale alternative (Exercise 3.234), if system (3.76) has no solution, the alternative system ðµz â€ 0, âeð+1 z > 0 ð has a nonnegative solution z â âð+1 + . Decompose z into z = (x, ð§) where x â â+ and ð§ â â+ . The second inequality implies ð§ > 0 since eð+1 z = ð§.

ðµ = (âðŽð , 1) and the ï¬rst inequality implies ( ) x ðµz = (âðŽð , 1) = âðŽð x + 1ð§ â€ 0 ð§ or ðŽð x â¥ 1ð§ > 0 Thus system I has a solution x â âð+ . Since x = 0 implies ðŽx = 0, we conclude that x â© 0. Conversely, assume that II has a solution y â© 0 such that ðŽy â€ 0. Then, for every x â âð+ xðŽð y = yâ² ðŽð x â€ 0 Since y â© 0, this implies ðŽð x â€ 0 for every x â âð+ which contradicts I. 3.246 We give a constructive proof, by proposing an algorithm which will generate the desired decomposition. Assume that x satisï¬es ðŽx â© 0. Arrange the rows of ðŽ such that the positive elements of ðŽx are listed ï¬rst. That is, decompose ðŽ into two submatrices such that ðµ1x > 0 ð¶1x = 0 Either Case 1 ð¶ 1 x â© 0 has no solution and the result is proved or 192

Solutions for Foundations of Mathematical Economics

Case 2 ð¶ 1 x â© 0 has a solution xâ² . Â¯ be a linear combination of x and xâ² . Speciï¬cally, deï¬ne Let x Â¯ = ðŒx + xâ² x where ðŒ > max

âbð x bð xâ²

where bð is the ðth row of ðµ 1 . ðŒ is chosen so that ðŒðµ 1 x > ðµ 1 xâ² By direct computation Â¯ = ðŒðµ 1 x + ðµ 1 xâ² > 0 ðµ1x Â¯ = ðŒð¶ 1 x + ð¶ 1 xâ² â© 0 ð¶1x Â¯ is another solution to ðŽx â© 0 since ð¶ 1 x = 0 and ð¶ 1 xâ² â© 0. By construction, x such that ðŽÂ¯ x has more positive components than ðŽx. Again, collect all the positive components together, decomposing ðŽ into two submatrices such that Â¯>0 ðµ2x Â¯=0 ð¶2x Either Case 1 ð¶ 2 x â© 0 has no solution and the result is proved or Case 2 ð¶ 2 x â© 0 has a solution xâ²â² . In the second case, we can repeat the previous procedure, generating another decomposition ðµ 3 , ð¶ 3 and so on. At each stage ð, the matrix ðµ ð get larger and ð¶ ð smaller. The algorithm must terminate before ðµ ð equals ðŽ, since we began with the assumption that ðŽx > 0 has no solution. 3.247 There are three possible cases to consider. Case 1: y = 0 is the only solution of ðŽð y = 0. Then ðŽx > 0 has a solution xâ² by Gordanâs theorem and ðŽxâ² + 0 > 0 Case 2: ðŽð y = 0 has a positive solution y > 0 Then 0 is the only solution ðŽx â¥ 0 by Stiemkeâs theorem and ðŽ0 + y > 0 Case 3 ðŽð y = 0 has a solution y â© 0 but y â> 0. By Gordanâs theorem ðŽx > 0 has no solution. By the previous exercise, ðŽ can be decomposed into two consistent subsystems ðµx > 0 ð¶x = 0 193

Solutions for Foundations of Mathematical Economics

such that ð¶x â© 0 has no solution. Assume that ðµ is ð Ã ð and ð¶ is ð Ã ð where ð = ð â ð. Applying Stiemkeâs theorem to ð¶, there exists z > 0, z â âð . Deï¬ne y â âð + by { 0 ð = 1, 2, . . . , ð ðŠð = ðŠð = ð§ðâð ð = ð + 1, ð + 2, . . . , ð Then x, y is the desired solution since for every ð, ð = 1, 2, . . . , ð either ðŠð > 0 or (ðŽx)ð = (ðµx)ð > 0. 3.248 Consider the dual pair ( ) ( ) y ðŽ = 0, y â¥ 0, z â¥ 0 x â¥ 0 and (ðŽð , ðŒ) z ðŒ By Tuckerâs theorem, this has a solution xâ , yâ , zâ such that ðŽxâ â¥ 0, xâ â¥ 0, ðŽð yâ + zâ = 0, yâ â¥ 0, zâ â¥ 0 ðŽx + y > 0 ðŒxâ + ðŒz > 0 Substituting zâ = âðŽð yâ implies ðŽð y â€ 0 and x â ðŽð yâ > 0 3.249 Consider the dual pair ðŽx â¥ 0 and ðŽð y = 0, y â¥ 0 where ðŽ is an ð Ã ð matrix. By Tuckerâs theorem, there exists a pair of solutions xâ â âð and yâ â âð such that ðŽxâ + yâ > 0

(3.77)

Assume that ðŽx > 0 has no solution (Gordan I). Then there exists some ð such that (ðŽxâ )ð = 0 and (3.77) implies that ðŠðâ > 0. Therefore yâ â© 0 and solves Gordan II. Conversely, assume that ðŽð y = 0 has no solution y > 0 (Stiemke II). Then, there exists some ð such that ðŠðâ = 0 and (3.77) implies that (ðŽxâ )ð > 0). Therefore xâ solves ðŽx â© 0 (Stiemke I). 3.250 We have already shown that Farkas I and II are mutually inconsistent. Assume that Farkas system I ðŽx â¥ 0, cð x < 0 ( has no solution. Deï¬ne the (ð + 1) Ã ð matrix ðµ = the system ðµx â¥ 0 194

ðŽ âcâ²

) . Our assumption is that

Solutions for Foundations of Mathematical Economics

has no solution with (ðµx)ð+1 = âcx > 0. By Tuckerâs theorem, the dual system ðµâ²z = 0 has a solution z â âð+1 with zð+1 > 0. Without loss of generality, we can normalize + â² ð so that zð+1 = 1. Decompose z into z = (y, 1) with y â âð + . Since ðµ = (ðŽ , âc), â² ðµ z = 0 implies ðµ â² z = (ðŽð , âc)(y, 1) = ðŽð y â c = 0 or ðŽð y = c y â âð + solves Farkas II. 3.251 If x â¥ 0 solves I, then xâ² (ðŽð y1 + ðµ â² y2 + ð¶ â² y3 ) = xâ² ðŽð y1 + xâ² ðµ â² y2 + xâ² ð¶ â² y3 ) > 0 since xâ² ðŽð y1 = y1 ðŽx > 0, xâ² ðµ â² y2 = y2 ðµx â¥ 0 and xâ² ð¶ â² y3 = y3 ð¶x = 0 which contradicts II. The equation ð¶x = 0 is equivalent to the pair of inequalities ð¶x â¥ 0, âð¶x â¥ 0. By Tuckerâs theorem the dual pair ðŽð y1 + ðµ â² y2 + ð¶ â² y3 â ð¶ â² y4 = 0

ðŽx â¥ 0 ðµx â¥ 0 ð¶x â¥ 0 âð¶x â¥ 0

has solutions ð¥ â âð , y1 â âð1 , y2 â âð2 , u3 , v3 â âð3 such that y1 â¥ 0

ðŽx + y1 > 0

y2 â¥ 0 u3 â¥ 0

ðµx + y2 > 0 ð¶x + u3 > 0

v3 â¥ 0

âð¶x + v3 > 0

Assume Motzkin I has no solution. That is, there is y1 â© 0. Deï¬ne y3 = u3 â v3 . Then y1 , y2 , y3 satisï¬es Motzkin II. 3.252

1. For every a â ð, let ðaâ be the polar set ðaâ = { x â âð : â¥xâ¥ = 1, xa â¥ 0 } ðaâ is nonempty since 0 â ðaâ . Let x be the limit of a sequence xð of points in ðaâ . Since xð a â¥ 0 for every ð, xa â¥ 0 so that x â ðaâ . Hence ðaâ is a closed subset of ðµ = { x â âð : â¥xâ¥ = 1 }.

2. Let {a1 , a2 , . . . , að } be any ï¬nite set of points in ð. Since 0 â / ð, the system ð â

ðŠð að = 0,

ð=1

ð â

ðŠð = 1, ðŠð â¥ 0

ð=1

has no solution. A fortiori, the system ð â

ðŠð að = 0

ð=1

195

Solutions for Foundations of Mathematical Economics

has no solution ðŠ â âð + . If ðŽ is the ðÃn matrix whose rows are að , the latter system can be written as ðŽð y = 0 3. By Gordanâs theorem, the system ðŽx > 0

(3.78)

Â¯ â= 0. has a solution x 4. Without loss of generality, we can take â¥Â¯ xâ¥ = 1. (3.78) implies that Â¯=x Â¯ að > 0 að x Â¯ â ðaâð . Hence for every ð = 1, 2 . . . , ð so that x ð â©

Â¯â x

ð=1

ðaâð

â©ð 5. We have shown that for every ï¬nite set {a1 , a2 , . . . , að } â ð, ð=1 ðaâð is nonempty closed subset of the compact set ðµ = {ð¥ â âð : â¥xâ¥ = 1}. By the Finite intersection property (Exercise 1.116) â© ðaâ â= â aâð

6. For every p â

ðaâ pa â¥ 0 for every a â ð

p deï¬nes a hyperplane ð (a) = pa which separates ð from 0. 3.253 The expected outcome if player 1 adopts the mixed strategy p = (ð1 , ð2 , . . . , ðð ) and player 2 plays her ð pure strategy is ð¢(p, ð) =

ð â

ðð ððð = pað

ð=1

where að is the ðth column of ðŽ. The expected payoï¬ to 1 for all possible responses of player 2 is the vector (pðŽ)â² = ðŽð p. The mixed strategy p ensures player 1 a nonnegative security level provided ðŽð p â¥ 0. Similarly, if 2 adopts the mixed strategy q = (ð1 , ð2 , . . . , ðð ), the expected payoï¬ to 2 if 1 plays his ð strategy is að q where að is the ðth row of ðŽ. The expected outcome for all the possible responses of player 1 is the vector ðŽq. The mixed strategy q ensures player 2 a nonpositive security level provided ðŽq â€ 0. By the von Neumann alternative theorem (Exercise 3.245), at least one of these alternatives must be true. That is, either Either I ðŽð p > 0, p â© 0 for some p â âð or II ðŽq â€ 0, q â© 0 for some q â âð Since p â© 0 and q â© 0, we can normalize so that p â Îðâ1 and q â Îðâ1 . At least one of the players has a strategy which guarantees she cannot lose. 196

Solutions for Foundations of Mathematical Economics 3.254

1. For any ð â â, deï¬ne the game ð¢Ë(a1 , a2 ) = ð¢(a1 , a2 ) â ð with Ë(p, ð) = max min ð¢(p, ð) â ð = ð£1 â ð ð£Ë1 = max min ð¢ p

p

ð

ð

ð£Ë2 = min max ð¢ Ë(ð, q) = min max ð¢(ð, q) â ð = ð£2 â ð q

q

ð

ð

By the previous exercise, Either ð£Ë1 â¥ 0 or ð£ËðŠ â€ 0 That is Either ð£1 â¥ ð or ð£2 â€ ð 2. Since this applies for arbitrary ð â â, it implies that while ð£1 â€ ð£2 and there is no ð such that ð£1 < ð < ð£2 Therefore, we conclude that ð£1 = ð£2 as required. 3.255

1. The mixed strategies p of player 1 are elements of the simplex Îðâ1 , which is compact (Example 1.110). Since ð£1 (p) = minðð=1 ð¢(p, ð) is continuous (Maximum theorem 2.3), ð£1 (p) achieves its maximum on Îðâ1 (Weierstrass theorem 2.2). That is, there exists pâ â Îðâ1 such that ð£1 = ð£1 (pâ ) = max ð£1 (p) p

Similarly, there exists qâ â Îðâ1 such that ð£2 = ð£2 (qâ ) = min ð£2 (q) q

2. Let ð¢(p, q) denote the expected outcome when player 1 adopts mixed strategy p and player 2 plays q. That is ð¢(p, q) =

ð â ð â

ðð ðð ððð

ð=1 ð=1

Then ð£ = ð¢(pâ , qâ ) = max ð¢(ð, qâ ) â¥ ð

â

ðð ð¢(ð, qâ ) = ð¢(p, qâ ) for every p â Îðâ1

ð

Similarly ð£ = ð¢(pâ , qâ ) = min ð¢(pâ , ð) â€ ð

â

ðð ð¢(pâ , ð) = ð¢(pâ , q) for every q â Îðâ1

ð

(pâ , qâ ) is a Nash equilibrium. 197

Solutions for Foundations of Mathematical Economics

3.256 By the Minimax theorem, every ï¬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 ð¡1 ðp (ð¡1 ) = âð1 + 2ð2 < 0 if ð1 > 2ð2 If player 2 adopts the strategy ð¡5 ðp (ð¡5 ) = ð1 â 2ð2 < 0 if ð1 < 2ð2 Therefore ð£1 (p) = min ðp (z) â€ min{ðp (ð¡1 ), ðp (ð¡5 )} < 0 ð§âð

for every p such that ð1 â= ð2 . Since ð1 + ð2 = 1, we conclude that { = 0 p = pâ = ( 2/3, 1/3) ð£1 (p) < 0 otherwise We conclude that ð£1 = max ð£1 (p) = 0 p

which is attained at pâ = ( 2/3, 1/3). 3.258

1. ð

ð£2 = min max ð§ð zâð ð=1

Since ð is compact, ð£2 = 0 implies there exists zÂ¯ â ð such that ð

max ð§Â¯ð = 0 ð=1

which implies that Â¯ z â€ 0. Consequently ð â© âðâ â= â. 2. Assume to the contrary that there exists z â ð â© int âðâ That is, there exists some strategy q â Îðâ1 such that ðŽq < 0 and therefore ð£2 < 0, contrary to the hypothesis. 3. There exists a hyperplane with nonnegative normal separating ð from âðâ (Exercise 3.227). That is, there exists pâ â âð+ , pâ â= 0 such that ðpâ (z) â¥ 0 for every z â ð and therefore ð£1 (pâ ) = min ðpâ (z) â¥ 0 zâð

Without loss of generality, we can normalize so that pâ â Îðâ1 .

198

âð

ð=1

ðâð = 1 and therefore

Solutions for Foundations of Mathematical Economics

4. Consequently ð£1 = max ð£1 (p) â¥ ð£1 (pâ ) â¥ 0 p

On the other hand, we know that ð contains a point zÂ¯ â€ 0. For every p â¥ 0 ðp (Â¯z) â€ 0 and therefore ð§) â€ 0 ð£1 (p) = min ðp (z) â€ ðp (Â¯ zâð

so that ð£1 = max ð£1 (p) â€ 0 p

We conclude that ð£1 = 0 = ð£2 3.259 Consider the game with the same strategies and the payoï¬ function ð¢ Ë(a1 , a2 ) = ð¢(a1 , a2 ) â ð The expected value to player 2 is Ë(ð, q) = min max ð¢(ð, q) â ð = ð£2 â ð = 0 ð£Ë2 = min max ð¢ q

q

ð

ð

By the previous exercise ð£Ë1 = ð£Ë2 = 0 and ð£1 = max min ð¢(p, ð) = max min ð¢ Ë(p, ð) + ð = ð£Ë1 + ð = ð = ð£2 p

q

ð

ð

3.260 Assume that p1 and p2 are both optimal strategies for player 1. Then ð¢(p1 , q) â¥ ð£ for every q â Îðâ1 ð¢(p2 , q) â¥ ð£ for every q â Îðâ1 Â¯ = ðŒp1 , p2 + (1 â ðŒ). Since ð¢ is bilinear Let p ð¢(Â¯ p, q) = ðŒð¢(p1 , q) + (1 â ðŒ)ð¢(p2 , q) â¥ ð£ for every q â Îðâ1 Â¯ is also an optimal strategy for player 1. Consequently, p 3.261 ð is the payoï¬ function of some 2 person zero-sum game in which the players have ð + 1 and ð + 1 strategies respectively. The result follows from the Minimax Theorem. 3.262

1. The possible partitions of ð = {1, 2, 3} are: {1}, {2}, {3} {ð, ð}, {ð},

ð, ð, ð =â ð, ð â= ð â= ð

{1, 2, 3} In any partition, at most one coalition can have two or more players, and therefore ðŸ â

ð€(ðð ) â€ 1

ð=1

199

Solutions for Foundations of Mathematical Economics

2. Assume x = (ð¥1 , ð¥2 , ð¥3 ) â core. Then x must satisfy the following system of inequalities ð¥1 + ð¥2 â¥ 1 = ð€({1, 2}) ð¥1 + ð¥3 â¥ 1 = ð€({1, 3}) ð¥2 + ð¥3 â¥ 1 = ð€({2, 3}) which can be summed to yield 2(ð¥1 + ð¥2 + ð¥3 ) â¥ 3 or ð¥1 + ð¥2 + ð¥3 â¥ 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 (ð, ð€) is not cohesive. Then there exists a partition {ð1 , ð2 , . . . , ððŸ } of ð such that ð€(ð ) <

ðŸ â

ð€(ðð )

ð=1

Assume x â core. Then

â

ð¥ð â¥ ð€(ðð )

ð = 1, 2, . . . , ðŸ

ðâðð

Since {ð1 , ð2 , . . . , ððŸ } is a partition â

ð¥ð =

ðâð

ðŸ â â

ð¥ð â¥

ð=1 ðâðð

ð â

ð€(ðð ) > ð€(ð )

ð=1

which contradicts the assumption that x â core. This establishes that cohesivity is necessary for the existence of the core. To show that cohesivity is not suï¬cient, 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. â¬ = {ð } with weights { ð€(ð) =

1 0

ð=ð otherwise

2. â¬ = {{1}, {2}, {3}} with weights ð€(ð) = 1 for every ð â â¬ 3. â¬ = {{ð}, {ð, ð}}, ð, ð, ð â â¬, ð â= ð â= ð with weights ð€(ð) = 1 for every ð â â¬ 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}

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 eð =

â

ðð eð

ðââ¬

are vectors, with each component corresponding to a particular player. For player ð, the ðð¡ â component of eð is 1 and the ðð¡ â component of eð is 1 if ð â ð and 0 otherwise. Therefore, for each player ð, the preceding expression can be written â ðð = 1 ðââ¬â£ðâð

For each coalition ð, the share of the coalition ð at the allocation x is â ðð (x) = ð â ðð¥ð = eð xË The condition ðð =

â

(3.79)

ðð ðð

ðââ¬

means that for every x â ð ðð (x) =

â

ðð ðð (x)

ðââ¬

Substituting (3.79) eð xË =

â

ðð ðð xË

ðââ¬

which is equivalent to the condition â

ðð eð = eð

ðââ¬

3.267 By construction, ð â¥ 0. If ð = 0, â ðð ðð â ððð = 0 ðâð

implies that ðð = 0 for all ð and consequently â ðð ð€(ð) â ðð€(ð ) â€ 0 ðâð

is trivially satisï¬ed. On the other hand, if ð > 0, we can divide both conditions by ð.)

201

Solutions for Foundations of Mathematical Economics

3.268 Let (ð, ð€1 ) and (ð, ð€2 ) be balanced games. By the Bondareva-Shapley theorem, they have nonempty cores. Let x1 â core(ð, ð€1 ) and x2 â core(ð, ð€2 ). That is, ðð (x1 ) â¥ ð€1 (ð) for every ð â ð ðð (x2 ) â¥ ð€2 (ð) for every ð â ð Adding, we have ðð (x1 ) + ðð (x2 ) = ðð (x1 + x2 ) â¥ ð€1 (ð) + ð€2 (ð) for every ð â ð which implies that x1 + x2 belongs to core(ð, ð€1 + ð€2 ). Therefore (ð, ð€1 + ð€2 ) is balanced. Similarly, if x â core(ð, ð€), then ðŒx belongs to core(ð, ðŒð€) for every ðŒ â â+ . That is (ð, ðŒð€) is balanced for every ðŒ â â+ . 3.269

1. Assume otherwise. That is assume there exists some y â ðŽ â© ðµ. Taking the ï¬rst ð components, this implies that â eð = ðð  eð ðâð

for some (ðð â¥ 0 : ð â ð ). Let â¬ = {ð â ð â£ ðð > 0} be the set of coalitions with strictly positive weights. Then â¬ is a balanced family of coalitions with weights ðð (Exercise 3.266). However, looking at the last coordinate, y â ðŽ â© ðµ implies â ðð  ð€(ð) = ð€(ð ) + ð > ð€(ð ) ðââ¬

which contradicts the assumption that the game is balanced. We conclude that ðŽ and ðµ are disjoint if the game is balanced. 2. (a) Substituting y = (eâ , 0) in (3.36) gives (z, ð§0 )â² (0, 0) = 0 â¥ ð which implies that ð â€ 0. NOTE We still have to show that ð â¥ 0. (b) Substituting (eð , ð€ðŠ(ð )) in (3.36) gives ð§eð + ð§0 ð€(ð ) > ð§eð + ð§0 ð€(ð ) + ð§0 ð for all ð > 0, which implies that ð§0 < 0. 3. Without loss of generality, we can normalize so that ð§0 = â1. Then the separating hyperplane conditions become (z, â1)â² y â¥ 0 â²

(z, â1) (eð , ð€(ð ) + ð) < 0

for every y â ðŽ

(3.80)

for every ð > 0

(3.81)

For any ð â ð , (eð , ð€(ð)) â ðŽ. Substituting y = (eð , ð€(ð)) in (3.80) gives eâ²ð z â ð€(ð) â¥ 0 that is ðð (z) = eâ²ð z =â¥ ð€(ð) 202

Solutions for Foundations of Mathematical Economics while (3.81) implies ðð (z) = eâ²ð z > ð€(ð ) + ð

for every ð > 0

This establishes that z belongs to the core. Hence the core is nonempty. â 3.270 1. Let ðŒ = ð€(ð ) â ðâð ð€ð > 0 since (ð, ð€) is essential. For every ð â ð , deï¬ne ( ) â 1 0 ð€(ð) â ð€ð ð€ (ð) = ðŒ ðâð

Then ð€0 ({ð}) = 0 for every ð â ð ð€0 (ð ) = 1 ð€0 is 0â1 normalized. 2. Let y â core(ð, ð€0 ). Then for every ð â ð â ðŠð â¥ ð€0 (ð)

(3.82)

ðâð

â

ðŠð = 1

(3.83)

ðâð

Let w = (ð€1 , ð€2 , . . . , ð€ð ) where ð€ð = ð€({ð}). Let x = ðŒy + w. Using (3.82) and (3.83) â â ð¥ð = (ðŒðŠð + ð€ð ) ðâð

ðâð

=ðŒ

â

ðŠð +

ðâð

â ðâð

ð€ð

ðâð

â¥ ðŒð€0 (ð) + 1 =ðŒ ðŒ

â â

ð€ð

ðâð

(

ð€(ð) â

â ðâð

) ð€ð

+

â

ð€ð

ðâð

= ð€(ð) â ð¥ð = (ðŒðŠð + ð€ð ) ðâð

=ðŒ+

â

ð€ð

ðâð

= ð€(ð ) Therefore, x = ðŒy + w â core(ð, ð€). Similarly, we can show that x â core(ð, ð€) =â y =

1 (x â w) â core(ð, ð€0 ) ðŒ

and therefore core(ð, ð€) = ðŒcore(ð, ð€0 ) + w 203

Solutions for Foundations of Mathematical Economics 3. This immediately implies

core(ð, ð€) = â ââ core(ð, ð€0 ) = â 3.271 (ð, ð€) is 0â1 normalized, that is ð€({ð} = 0 for every ð â ð ð€(ð ) = 1 Consequently, x belongs to the core of (ð, ð€) if and only if â

ð¥ð â¥ ð€ð = 0

(3.84)

ð¥ð = ð€(ð ) = 1

(3.85)

ð¥ð â¥ ð€(ð) for every ð â ð

(3.86)

ðâð

â ðâð

(3.84) and (3.85) ensure that x = (ð¥1 , ð¥2 , . . . , ð¥ð ) is a mixed strategy for player 1 in the two-person zero-sum game. Using this mixed strategy, the expected payoï¬ to player I for any strategy ð of player II is ð¢(x, ð) =

â

ð¥ð ð¢(ð, ð) =

ðâð

â ðâð

ð¥ð

1 ð€(ð)

(3.86) implies ð¢(x, ð) =

â ðâð

ð¥ð

1 â¥ 1 for every ð â ð ð€(ð)

That is any x â core(ð, ð€) provides a mixed strategy for player I which ensures a payoï¬ at least 1. That is core(ð, ð€) â= â =â ð¿ â¥ 1 Conversely, if the ð¿ < 1, there is no mixed strategy for player I which satisï¬es (3.86) and consequently no x which satisï¬es (3.84), (3.85) and (3.86). In other words, core(ð, ð€) = â. 3.272 If ð¿ is the value of ðº, there exists a mixed strategy which will guarantee that II pays no more than ð¿. That is, there exists numbers ðŠð â¥ 0 for every coalition ð â ð such that â ðŠð = 1 ðâð

and â

ðŠð ð¢(ð, ð) â€ ð¿

for every ð â ð

ðâð

that is â ðâð

ðŠð

1 â€ð¿ ð€(ð)

for every ð â ð

ðâð

204

Solutions for Foundations of Mathematical Economics

or â ðâð

ðŠð â€1 ð¿ð€(ð)

for every ð â ð

(3.87)

ðâð

For each coalition ð â ð let ðð =

ðŠð ð¿ð€(ð)

in (3.87) â

ðð â€ 1

ðâð ðâð

Augment the collection ð with the single-player coalitions to form the collection â¬ = ð âª { {ð} : ð â ð } and with weights { ðð : ð â ð } and â

ð{ð} = 1 â

ðð

ðâð

Then â¬ is a balanced collection. Since the game (ð, ð€) is balanced 1 = ð€(ð ) â¥

â

ðð ð€(ð)

ðââ¬

=

â

ðð ð€(ð)

ðâð

â

ðŠð ð€(ð) ð¿ð€(ð) ðââ¬ 1â = ðŠð ð¿ =

ðââ¬

1 = ð¿ that is 1â¥

1 ð¿

Â¯ = (1/ð, 1/ð, . . . , 1/ð), the payoï¬ is If I plays the mixed strategy x ð¢(Â¯ x, ð) =

â ðâð

1 1 = > 0 for every ð â ð ðð€(ð) ð€(ð)

Therefore ð¿ > 0 and (3.88) implies that ð¿â¥1

205

(3.88)

Solutions for Foundations of Mathematical Economics 3.273 Assume core(ð, ð€) â= â and let ð¥ â core(ð, ð€). Then ðð (x) â¥ ð€(ð) for every ð â ð where ðð =

â ðâð

(3.89)

ð¥ð measures the share coalition ð at the allocation x.

Let â¬ be a balanced family of coalitions with weights ðð . For every ð â â¬, (3.89) implies ðð ðð (x) â¥ ðð ð€(ð) Summing over all ð â â¬ â

â

ðð ðð (x) â¥

ðââ¬

ðð ð€(ð)

ðââ¬

Evaluating the left hand side of this inequality â â â ðð ðð (x) = ð ð¥ð ðââ¬

ðââ¬

=

ðâð

ââ

ðð¥ð

ðâð ðââ¬

=

â

ðâð

ð¥ð

ðâð

=

â

â

ðââ¬ ðâð

ð¥ð

ðâð

= ð€(ð ) Substituting this in (3.90) gives ð€(ð ) â¥

â ðââ¬

The game is balanced.

206

ðð ð€(ð)

ð

(3.90)

Solutions for Foundations of Mathematical Economics

Chapter 4: Smooth Functions 4.1 Along the demand curve, price and quantity are related according to the equation ð = 10 â ð¥ This is called the inverse demand function. Total revenue ð(ð¥) (price times quantity) is given by ð(ð¥) = ðð¥ = (10 â ð¥)ð¥ = 10ð¥ â ð¥2 = ð (ð¥) ð(ð¥) can be rewritten as ð(ð¥) = 21 + 4(ð¥ â 3) At ð¥ = 3, the price is 7 but the marginal revenue of an additional unit is only 4. The function ð decomposes (approximately) the total revenue into two components â the revenue from the sale of 3 units (21 = 3 Ã 7) plus the marginal revenue from the sale of additional units (4(ð¥ â 3)). 4.2 If your answer is 5 per cent, obtained by subtracting the inï¬ation rate from the growth rate of nominal GDP, you are implicitly using a linear approximation. To see this, let ð ð ðð ðð

= 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, (ð + ðð)(ð + ðð), equals 1.10 times nominal GDP at the beginning of the year, ðð. That is (ð + ðð)(ð + ðð) = 1.10ðð

(4.42)

Furthermore, the price level at the end of the year, ð + ðð equals 1.05 times the price level of the start of year, ð: ð + ðð = 1.05ð Substituting this in equation (4.38) yields 1.05ð(ð + ðð) = 1.10ðð which can be solved to give ðð = (

1.10 â 1)ð = 0.0476 1.05

The growth rate of real GDP (ðð/ð) is equal to 4.76 per cent. 207

Solutions for Foundations of Mathematical Economics

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. (ð + ðð)(ð + ðð) = ðð + ððð + ððð + ðððð Dividing by ðð (ð + ðð)(ð + ðð) ðð ðð ðððð =1+ + + ðð ð ð ðð The growth rate of nominal GDP is (ð + ðð)(ð + ðð) â ðð (ð + ðð)(ð + ðð) = â1 ðð ðð ðð ðð ðððð = + + ð ðð ðð = Growth rate of output + Inï¬ation rate + Error term For small changes, the error term ðððð/ðð is insigniï¬cant, and we can approximate the growth rate of output according to the sum Growth rate of nominal GDP = Growth rate of output + Inï¬ation rate This is a linear approximation since it approximates the function (ð + ðð)(ð + ðð) by the linear function ðð + ððð + ððð. In eï¬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) â¥xâ¥ ð(x) = ð (x0 + x) â ð (x0 ) â ð(x) and therefore ð(x) =

ð (x0 + x) â ð (x0 ) â ð(x) â¥xâ¥

ð(x) â 0ð as x â 0ð can be expressed as lim ð(x) = 0ð

xâ0ð

4.4 Suppose not. That is, there exist two linear maps such that ð (x0 + x) = ð (x0 ) + ð1 (x) + â¥xâ¥ ð1 (x) ð (x0 + x) = ð (x0 ) + ð2 (x) + â¥xâ¥ ð2 (x) with lim ðð (x) = 0,

xâ0

208

ð = 1, 2

Solutions for Foundations of Mathematical Economics Subtracting we have ð¿1 (x) â ð¿2 (x) = â¥xâ¥ (ð1 (x) â ð2 (x)) and lim

xâ0

ð1 (x) â ð2 (x) =0 â¥xâ¥

Since ð1 â ð2 is linear, (4) implies that ð1 (x) = ð2 (x) for all x â ð. To see this, we proceed by contradiction. Again, suppose not. That is, suppose there exists some x â ð such that ð1 (x) â= ð2 (x) For this x, let ð=

ð1 (x) â ð2 (x) â¥xâ¥

By linearity, ð1 (ð¡x) â ð2 (ð¡x) = ð for every âð¡ > 0 â¥ð¡xâ¥ and therefore lim

ð¡â0

ð1 (ð¡x) â ð2 (ð¡x) = ð â= 0 â¥ð¡xâ¥

which contradicts (4). Therefore ð1 (x) = ð2 (x) for all x â ð. 4.5 If ð : ð â ð is diï¬erentiable at x0 , then ð (x0 + x) = ð (x0 ) + ð(x) + ð(x) â¥xâ¥ where ð(x) â 0ð as x â 0ð . Since ð is a continuous linear function, ð(x) â 0ð as x â 0ð . Therefore lim ð (x0 + x) = lim ð (x0 ) + lim ð(x) + lim ð(x) â¥xâ¥

xâ0

xâ0

xâ0

xâ0

= ð (x0 ) ð is continuous. 4.6 4.7 4.8 The approximation error at the point (2, 16) is ð (2, 16) ð(2, 16) Absolute error Percentage error Relative error

=8.0000 =11.3333 =-3.3333 =-41.6667 =-4.1667

By contrast, â(2, 16) = 8 = ð (2, 16). Table 4.1 shows that â gives a good approximation to ð in the neighborhood of (2, 16). 209

Solutions for Foundations of Mathematical Economics

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

ð (x0 + x)

â(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

210

Solutions for Foundations of Mathematical Economics

4.9 To show that ð is nonlinear, consider ð((1, 2, 3, 4, 5) + (66, 55, 75, 81, 63)) = ð(67, 57, 78, 85, 68) = (85, 78, 68, 67, 58) â= (5, 4, 3, 2, 1) + (81, 75, 67, 63, 55) To show that ð is diï¬erentiable, consider a particular point, say (66, 55, 75, 81, 63). Consider the permutation ð : âð â âð deï¬ned by ð(ð¥1 , ð¥2 , . . . , ð¥5 ) = (ð¥4 , ð¥3 , ð¥1 , ð¥5 , ð¥2 ) ð is linear and ð(66, 55, 75, 81, 63) = (81, 75, 67, 63, 55) = ð(66, 55, 75, 81, 63) Furthermore, ð(x) = ð(x) for all x close to (66, 55, 75, 81, 63). Hence, ð(x) approximates ð(x) in a neighborhood of (66, 55, 75, 81, 63) and so ð is diï¬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 xð â= xð . In summary, each application of ð involves a permutation, although the particular permutation depends upon the argument, x. However, for any given x0 with x0ð â= x0ð , 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 ð at x0 . 4.10 Using (4.3), we have for any x ð (x0 + ð¡x) â ð (x0 ) â ð·ð [x0 ](ð¡x) =0 ð¡xâ0 â¥ð¡xâ¥ lim

or ð (x0 + ð¡x) â ð (x0 ) â ð¡ð·ð [x0 ](x) =0 ð¡â0 ð¡ â¥xâ¥ lim

For â¥xâ¥ = 1, this implies ð¡ð·ð [x0 ](x) ð (x0 + ð¡x) â ð (x0 ) = ð¡â0 ð¡ ð¡ lim

that is 0 0 â x ð [x0 ] = lim ð (x + ð¡x) â ð (x ) = ð·ð [x0 ](x) ð· ð¡â0 ð¡

4.11 By direct calculation â(x0ð + ð¡) â â(x0ð ) ð¡â0 ð¡ ð (x01 , x02 , . . . , x0ð + ð¡, . . . , x0ð ) â ð (x01 , x02 , . . . , x0ð , . . . , x0ð ) = lim ð¡â0 ð¡ ð (x0 + ð¡eð ) â ð (x0 ) = lim ð¡â0 ð¡ 0 â = ð·eð ð [x ]

ð·ð¥ð ð [x0 ] = lim

211

Solutions for Foundations of Mathematical Economics

4.12 Deï¬ne the function ( ) â(ð¡) = ð (8, 8) + ð¡(1, 1) = (8 + ð¡)1/3 (8 + ð¡)2/3 =8+ð¡ The directional derivative of ð in the direction (1, 1) is â (1,1) ð (8, 8) = lim â(ð¡) â â(0) ð· ð¡â0 ð¡ =1 Generalization of this example reveals that the directional derivative of ð along any â x0 ð [x0 ] = 1 for every x0 . Economically, ray through the origin equals 1, that is ð· 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 = âð (x0 ). Each component of p represents the action of the derivative on an element of the standard basis {e1 , e2 , . . . , eð }(see proof of Theorem 3.4) ðð = ð·ð [x0 ](eð ) Since â¥eð â¥ = 1, ð·ð [x0 ](eð ) is the directional derivative at x0 in the direction eð (Exercise 4.10) â eð (x0 ) ðð = ð·ð [x0 ](eð ) = ð· But this is simply the ð partial derivative of ð (Exercise 4.11) â eð (x0 ) = ð·ð¥ð ð (x0 ) ðð = ð·ð [x0 ](eð ) = ð· 4.14 Using the standard inner product on âð (Example 3.26) and Exercise 4.13 < âð (x0 ), x >=

ð â

ð·ð¥ð ð [x0 ]xð = ð·ð [x0 ](x)

ð=1

4.15 Since ð is diï¬erentiable ð (x1 + ð¡x) = ð (x1 ) + âð (x0 )ð ð¡x + ð(ð¡x) â¥ð¡xâ¥ with ð(ð¡x) â 0 as ð¡x â 0. If ð is increasing, ð (x1 + ð¡x) â¥ ð (x1 ) for every x â¥ 0 and ð¡ > 0. Therefore âð (x0 )ð ð¡x + ð(ð¡x) â¥ð¡xâ¥ = ð¡âð (x0 )ð x + ð¡ð(ð¡x) â¥xâ¥ â¥ 0 Dividing by ð¡ and letting ð¡ â 0 âð (x0 )ð x â¥ 0 for every x â¥ 0 In particular, this applies for unit vectors eð . Therefore ð·ð¥ð ð (x1 ) â¥ 0,

ð = 1, 2, . . . , ð

212

Solutions for Foundations of Mathematical Economics

â x ð (x0 ) measures the rate of increase of ð in the di4.16 The directional derivative ð· rection x. Using Exercises 4.10, 4.14 and 3.61, assuming x has unit norm,   â x ð (x0 ) = ð·ð [x0 ](x) =< âð (x0 ), x >â€ âð (x0 ) ð·   This bound is attained when x = âð (x0 )/ âð (x0 ) since   âð (x0 )2   âð (x0 ) 0 0 â ð·x ð (x ) =< âð (x ), >= = âð (x0 ) 0 0 â¥âð (x )â¥ â¥âð (x )â¥ The directional derivative is maximized when âð (x0 ) and x are aligned. 4.17 Using Exercise 4.14 ð» = { x â ð :< âð [x0 ], x >= 0 } 4.18 Assume each ðð is diï¬erentiable at x0 and let ð·ð [x0 ] = (ð·ð1 [x0 ], ð·ð2 [x0 ], . . . , ð·ðð [x0 ]) Then

â â â f (x0 + x) â f [x0 ] â ð·f [x0 ]x = â â

ð1 (x0 + x) â ð1 [x0 ] â ð·ð1 [x0 ]x ð2 (x0 + x) â ð2 [x0 ] â ð·ð2 [x0 ]x .. .

â â â â â

ðð (x0 + x) â ðð (x0 ) â ð·ðð [x0 ]x and ðð (x0 + x) â ðð (x0 ) â ð·ðð [x0 ]x â 0 as â¥xâ¥ â 0 â¥xâ¥ for every ð implies f (x0 + x) â f (x0 ) â ð·f [x0 ](x) â 0 as â¥xâ¥ â 0 â¥xâ¥

(4.43)

Therefore f is diï¬erentiable with derivative ð·f [x0 ] = ð¿ = (ð·ð1 (x0 ), ð·ð2 [x0 ], . . . , ð·ðð [x0 ]) Each ð·ðð [x0 ] is represented by the gradient âðð [x0 ] (Exercise 4.13) and therefore ð·ð [x0 ] is represented by the matrix â â â â âð1 [x0 ] ð·ð¥1 ð1 [x0 ] ð·ð¥2 ð1 [x0 ] . . . ð·ð¥ð ð1 [x0 ] â âð2 [x0 ] â â ð·ð¥1 ð2 [x0 ] ð·ð¥2 ð2 [x0 ] . . . ð·ð¥ð ð2 [x0 ] â â â â â ðœ =â â=â â .. .. .. .. .. â â  â â  . . . . . âðð [x0 ]

ð·ð¥1 ðð [x0 ] ð·ð¥2 ðð [x0 ] . . .

ð·ð¥ð ðð [x0 ]

Conversely, if f is diï¬erentiable, its derivative ð·f [x0 ] : âð â âð be decomposed into ð component ð·ð1 [x0 ], ð·ð2 [x0 ], . . . , ð·ðð [x0 ] functionals such that â â ð1 (x0 + x) â ð1 (x0 ) â ð·ð1 [x0 ]x â ð2 (x0 + x) â ð2 (x0 ) â ð·ð2 [x0 ]x â â â f (x0 + x) â f (x0 ) â ð·f [x0 ]x = â â .. â  â . ðð (x0 + x) â ðð (x0 ) â ð·ðð [x0 ]x

213

Solutions for Foundations of Mathematical Economics

(4.43) implies that ðð (x0 + x) â ðð (x0 ) â ð·ðð [x0 ]x â 0 as â¥xâ¥ â 0 â¥xâ¥ for every ð. 4.19 If ð·ð [x0 ] has full rank, then it is one-to-one (Exercise 3.25) and onto (Exercise 3.16). Therefore ð·ð [x0 ] is nonsingular. The Jacobian ðœð (x0 ) represents ð·ð [x0 ], which is therefore nonsingular if and only if det ðœð (x0 ) â= 0. 4.20 When ð is a functional, rank ð â¥ ððððð = 1. If ð·ð [x0 ] has full rank (1), then ð·ð [x0 ] maps ð onto â (Exercise 3.16), which requires that âð (x0 ) â= 0. 4.21 4.23 If ð : ð Ã ð â ð is bilinear ð (x0 + x, y0 + y) = ð (x0 , y0 ) + ð (x0 , y) + ð (x, y0 ) + ð (x, y) Deï¬ning ð·ð [x0 , y0 ](x, y) = ð (x0 , y) + ð (x, y0 ) ð (x0 + x, y0 + y) = ð (x0 , y0 ) + ð·ð [x0 , y0 ](x, y) + ð (x, y) Since ð is continuous, there exists ð such that ð (x, y) â€ ð â¥xâ¥ â¥yâ¥

for every x â ð and y â ð

and therefore NOTE This is not quite right. See Spivak p. 23. Avez (Tilburg) has ( )2 â¥ð (x, y)â¥ â€ ð â¥xâ¥ â¥yâ¥ â€ ð â¥xâ¥ + â¥yâ¥ â€ ð â¥(x, y)â¥2 which implies that â¥ð (x, y)â¥ â 0 as (x, y) â 0 â¥(x, y)â¥

lim

x1 ,x2 â0

ð (x1 , x2 ) =0 â¥x1 â¥ â¥x2 â¥

Therefore ð is diï¬erentiable with derivative ð·ð [x0 , y0 ] = ð (x0 , y) + ð (x, y0 ) 4.24 Deï¬ne ð : â2 â â by ð(ð§1 , ð§2 ) = ð§1 ð§2 Then ð is bilinear (Example 3.23) and continuous (Exercise 2.79) and therefore diï¬erentiable (Exercise 4.23) with derivative ð·ð[ð§1 , z2 ] = ð(z1 , â) + ð(â, z2 ) 214

Solutions for Foundations of Mathematical Economics

The function ð ð is the composition of ð with ð and ð, ð ð(x, y) = ð(ð (x), ð(y)) By the chain rule, the derivative of ð ð is ( ) ð·ð ð[x, y] = ð·ð[ð§1 , z2 ] ð·ð [x], ð·ð[x] = ð(z1 , ð·ð[y]) + ð(ð·ð [x], z2 ) = ð [x]ð·ð[y]) + ð(y)ð·ð [x] where z1 = ð (x) and z2 = ð(y). 4.25 For ð = 1, ð (ð¥) = ð¥ is linear and therefore (Exercise 4.6) ð·ð [ð¥] = 1 (ð·ð [ð¥](ð¥) = ð¥). For ð = 2, let ð(ð¥) = ð¥ so that ð (ð¥) = ð¥2 = ð(ð¥)ð(ð¥). Using the product rule ð·ð [x] = ð(ð¥)ð·ð(ð¥) + ð(ð¥)ð·ð(ð¥) = 2ð¥ Now assume it is true for ð â 1 and let ð(ð¥) = ð¥ðâ1 , so that ð (x) = ð¥ð(ð¥). By the product rule ð·ð [x] = ð¥ð·ð[ð¥] + ð(ð¥)1 By assumption ð·ð[ð¥] = (ð â 1)ð¥ðâ2 and therefore ð·ð [x] = ð¥ð·ð[ð¥] + ð(ð¥)1 = ð¥(ð â 1)ð¥ðâ2 + ð¥ðâ1 = ðð¥ðâ1 4.26 Using the product rule (Exercise 4.24) ð·ð¥ ð(ð¥0 ) = ð (ð¥0 )ð·ð¥ ð¥ + ð¥0 ð·ð¥ ð (ð¥0 ) = ð0 + ð¥0 ð·ð¥ ð (ð¥0 ) where ð0 = ð (ð¥0 ). Marginal revenue equals one unit at the current price minus the reduction in revenue caused by reducing the price on existing sales. ( )â1 4.27 Fix some x0 and let ð = ð·ð [x0 ] . Let y0 = ð (x0 ). For any y, let x = â1 0 â1 0 0 ð (y + y) â ð (y ) so that ð(y) = ð (x + x) â ð (x) and    â1 0 ( ) ð (y + y) â ð â1 (y0 ) â ð(y) = (x â ð ð (x0 + x) â ð (x0 ))  Since ð is diï¬erentiable at x0 with ð·ð [x0 ] = ð â1 ð (x0 + x) â ð (x0 ) = ð â1 (x) + ð(x) â¥xâ¥ Substituting ( )  â1 0    ð (y + y) â ð â1 (y0 ) â ð(y) =  x â ð ð â1 (x) + ð(x) â¥xâ¥   ( )   = ð ð(x) â¥xâ¥   ( )   = â¥xâ¥ ð ð(x)  ( ) with ð(x) â 0ð as x â 0ð . Since ð â1 and ð are continuous, ð ð(x) â 0ð as y â 0. )â1 ( . We conclude that ð â1 is diï¬erentiable with derivative ð = ð·ð [x0 ]

215

Solutions for Foundations of Mathematical Economics 4.28 log ð (ð¥) = ð¥ log ð and therefore

( ) ð (ð¥) = exp log ð (ð¥) = ðð¥ log ð

By the Chain Rule, ð is diï¬erentiable with derivative ð·ð¥ ð (ð¥) = ðð¥ log ð log ð = ðð¥ log ð 4.29 By Exercise 4.15, the function ð : â â â deï¬ned by ð(ðŠ) = tiable with derivative 1 ð·ðŠ ð[ðŠ] = âðŠ â2 = â 2 ðŠ

1 ðŠ

= ðŠ â1 is diï¬eren-

Applying the Chain Rule, 1/ð = ð â ð is diï¬erentiable with derivative 1 ð·ð [x] ð· [x] = ð·ð[ð (x)]ð·ð [x] = â ( )2 ð ð (x) 4.30 Applying the Product Rule to ð Ã (1/ð) 1 1 ð ð·ð [x] ð· [x, y] = ð (x)ð· [y] + ð ð ð(y) ð·ð[y] 1 = âð (x) ( ð·ð [x] )2 + ð(y) ð(y) ð(y)ð·ð [x] â ð (x)ð·ð[y] = ( )2 ð(y) 4.31 In the particular case where 1/3 2/3

ð (x1 , x2 ) = x1 x2 the partial derivatives at the point (8, 8) are ð·ð¥1 ð [(8, 8)] =

2 1 and ð·ð¥2 ð¹ [(8, 8)] = 3 3

4.32 The partial derivatives of ð (x) are from Table 4.4 ð·ð¥ð ð [x] = ð¥ð1 1 ð¥ð2 2 . . . ðð ð¥ððð â1 . . . ð¥ððð = ðð so that the gradient is

( âð (x) =

ð (x) ð¥ð ð1 ð2 ðð , ,..., ð¥1 ð¥2 ð¥ð

) ð (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 ð·ð¥ð ð [x] =

1 1 â1 (ð1 ð¥ð1 + ð2 ð¥ð2 + â â â + ðð ð¥ðð ) ð ðð ðð¥ðâ1 ð ð

= ðð ð¥ðâ1 (ð1 ð¥ð1 + ð2 ð¥ð2 + â â â + ðð ð¥ðð ) ð ( )1âð ð (x) = ðð ð¥ð 216

1âð ð

Solutions for Foundations of Mathematical Economics

4.34 Deï¬ne â(ð¥) = ð (ð¥) â

ð (ð) â ð (ð) (ð¥ â ð) ðâð

Then â is continuous on [ð, ð] and diï¬erentiable on (ð, ð) with â(ð) = ð (ð) â

ð (ð) â ð (ð) (ð â ð)ð (ð) = â(ð) ðâð

By Rolleâs theorem (Exercise 5.8), there exists ð¥ â (ð, ð) such that ââ² (ð¥) = ð â² (ð¥) â

ð (ð) â ð (ð) =0 ðâð

4.35 Assume âð (x) â¥ 0 for every x â ð. By the mean value theorem, for any x2 â¥ x1 Â¯ â (x1 , x2 ) such that in ð, there exists x ð (x2 ) = ð (x1 ) + ð·ð [Â¯ x](x2 â x1 ) Using (4.6) ð (x2 ) = ð (x1 ) +

ð â

ð·ð¥ð ð (Â¯ x)(ð¥2ð â ð¥1ð )

(4.44)

ð=1

âð (Â¯ x) â¥ 0 and x2 â¥ x1 implies that ð â

ð·ð¥ð ð (Â¯ x)(ð¥2ð â ð¥1ð ) â¥ 0

ð=1

and therefore ð (x2 ) â¥ ð (x1 ). ð is increasing. The converse was established in Exercise 4.15 4.36 âð (Â¯ x) > 0 and x2 â¥ x1 implies that ð â

ð·ð¥ð ð (Â¯ x)(ð¥2ð â ð¥1ð ) > 0

ð=1

Substituting in (4.44) ð (x2 ) = ð (x1 ) +

ð â

ð·ð¥ð ð (Â¯ x)(ð¥2ð â ð¥1ð ) > ð (x1 )

ð=1

ð is strictly increasing. 4.37 Diï¬erentiability implies the existence of the gradient and hence the partial derivatives of ð (Exercise 4.13). Continuity of ð·ð [x] implies the continuity of the partial derivatives. To prove the converse, choose some x0 â ð and deï¬ne for the partial functions âð (ð¡) = ð (ð¥01 , ð¥02 , . . . , ð¥0ðâ1 , ð¡, ð¥0ð+1 + ð¥ð+1 , . . . , ð¥0ð + ð¥ð )

ð = 1, 2, . . . , ð

so that ââ²ð (ð¡) = ð·ð¥ð ð (xð ) where xð = (ð¥01 , ð¥02 , . . . , ð¥0ð , ð¡, ð¥0ð+1 + ð¥ð+1 , . . . , ð¥0ð + ð¥ð ). Further, â1 (ð¥01 + ð¥1 ) = ð (x0 + x), âð (ð¥0ð ) = ð (x0 ), and âð (ð¥0ð + ð¥ð ) = âðâ1 (ð¥0ð ) so that ð (x0 + x) â ð (x0 ) =

ð â ( ) âð (ð¥0ð + ð¥ð ) â âð (ð¥0ð ) ð=1

217

Solutions for Foundations of Mathematical Economics

By the mean value theorem, there exists, for each ð, ð¡Â¯ð between ð¥0ð + ð¥ð and ð¥ð such that âð (ð¥0ð + ð¥ð ) â âð (ð¥ð ) = ð·ð¥ð ð (Â¯ xð )ð¥ð Â¯ ð = (ð¥01 , ð¥02 , . . . , ð¥0ð , ð¡Â¯, ð¥0ð+1 + ð¥ð+1 , . . . , ð¥0ð + ð¥ð ). Therefore where x ð (x0 + x) â ð (x0 ) =

ð â

ð·ð¥ð ð (Â¯ xð )ð¥ð

ð=1

Deï¬ne the linear functional ð(x) =

ð â

ð·ð¥ð ð (x0 )ð¥ð

ð=1

Then ð (x0 + x) â ð (x0 ) â ð(x) =

ð ( ) â ð·ð¥ð ð (Â¯ xð ) â ð·ð¥ð ð (x0 ) ð¥ð ð=1

and ð   â   ð (x0 + x) â ð (x0 ) â ð(x) â€ (ð·ð¥ð ð (Â¯ xð ) â ð·ð¥ð ð (x0 ) â£ð¥ð â£ ð=1

so that

  ð ð (x0 + x) â ð (x0 ) â ð(x) â   â£ð¥ð â£ (ð·ð¥ð ð (Â¯ â€ lim xð ) â ð·ð¥ð ð (x0 ) xâ0 â¥xâ¥ â¥xâ¥ ð=1 â€

ð â   (ð·ð¥ð ð (Â¯ xð ) â ð·ð¥ð ð (x0 ) ð=1

=0 since the partial derivatives ð·ð¥ð ð (x) are continuous. Therefore ð is diï¬erentiable with derivative ð(x) =

ð â

ð·ð¥ð ð [x0 ]ð¥ð

ð=1

4.38 For every x1 , x2 â ð â¥ð (x1 ) â ð (x2 )â¥ â€

sup

xâ[x1 ,x2 ]

â¥ð·ð (x)â¥ â¥x1 â x2 â¥

by Corollary 4.1.1. If ð·ð [x] = 0 for every x â ð, then â¥ð (x1 ) â ð (x2 )â¥ = 0 which implies that ð (x1 ) = ð (x2 ). We conclude that ð is constant on ð. The converse was established in Exercise 4.7. 4.39 For any x0 â ð, let ðµ â ð be an open ball of radius of radius ð centered on x0 . Applying the mean value inequality (Corollary 4.1.1) to ðð â ðð we have  ( ) ðð (x) â ðð (x) â ðð (x0 ) â ðð (x0 )  â€ sup â¥ð·ðð [Â¯ x] â ð·ðð [Â¯ x]â¥ â¥x â x0 â¥ Â¯ âðµ x

â€ ð sup â¥ð·ðð [Â¯ x] â ð·ðð [Â¯ x]â¥ Â¯ âðµ x

218

Solutions for Foundations of Mathematical Economics

for every x â ðµ. Given ð > 0, there exists ð such that for every ð, ð > ð â¥ð·ðð â ð·ðð â¥ < ð/ð and â¥ð·ðð â ðâ¥ < ð Letting ð â â

 ( ) ðð (x) â ð (x) â ðð (x0 ) â ð (x0 )  â€ ð â¥x â x0 â¥

(4.45)

for ð â¥ ð and x â ðµ. Applying the mean value inequality to ðð , there exists ð¿ such that â¥ðð (x) â ðð (x0 )â¥ â€ ð â¥x â x0 â¥

(4.46)

Using (4.45) and (4.46) and the fact that â¥ð·ðð â ðâ¥ < ð we deduce that â¥ð (x) â ð (x0 ) â ð(x0 )â¥ â€ 3ð â¥x â x0 â¥ ð is diï¬erentiable with derivative ð. 4.40 Deï¬ne ð (ð¥) =

ðð¥+ðŠ ððŠ

By the chain rule (Exercise 4.22) ð â² (ð¥) =

ðð¥+ðŠ = ð (ð¥) ððŠ

which implies (Example 4.21) that ð (ð¥) =

ðð¥+ðŠ = ðŽðð¥ for some ðŽ â â ððŠ

Evaluating at ð¥ = 0 using ð0 = 1 gives ð (0) =

ððŠ = ðŽ for some ðŽ â â ððŠ

so that ð (ð¥) =

ððŠ ð¥ ðð¥+ðŠ = ð ððŠ ððŠ

which implies that ðð¥+ðŠ = ðð¥ ððŠ 4.41 If ð = ðŽð¥ð , ð â² (ð¥) = ððŽð¥ðâ1 and ðž(ð¥) = ð¥

ððŽð¥ðâ1 =ð ðŽð¥ð

To show that this is the only function with constant elasticity, deï¬ne ð(ð¥) =

ð (ð¥) ð¥ð

ð is diï¬erentiable (Exercise 4.30) with derivative ð â² (ð¥) =

ð¥ð ð â² (ð¥) â ð (ð¥)ðð¥ðâ1 ð¥ð â² (ð¥) â ðð (ð¥) = ð¥2ð ð¥ð+1 219

(4.47)

Solutions for Foundations of Mathematical Economics If ðž(ð¥) = ð¥

ð â² (ð¥) =ð ð (ð¥)

then ð¥ð â² (ð¥) = ðð (ð¥) Substituting in (4.47) ð â² (ð¥) =

ð¥ð â² (ð¥) â ðð (ð¥) = 0 for every ð¥ â â ð¥ð+1

Therefore, ð is a constant function (Exercise 4.38). That is, there exists ðŽ â â such that ð(ð¥) =

ð (ð¥) = ðŽ or ð (ð¥) = ðŽð¥ð ð¥ð

4.42 Deï¬ne ð : ð â ð by ð(x) = ð (x) â ð·ð [x0 ](x) ð is diï¬erentiable with ð·ð[x] = ð·ð [x] â ð·ð [x0 ] Applying Corollary 4.1.1 to ð, â¥ð(x1 ) â ð(x2 )â¥ â€

sup

xâ[x1 ,x2 ]

â¥ð·ð[x]â¥ â¥x1 â x2 â¥

for every x1 , x2 â ð. Substituting for ð and ð·ð â¥ð (x1 ) â ð·ð [x0 ](x1 ) â ð (x2 ) + ð·ð [x0 ](x2 )â¥ = â¥ð (x1 ) â ð (x2 ) â ð·ð [x0 ](x1 â x2 )â¥ â€

sup

xâ[x1 ,x2 ]

â¥ð·ð [x] â ð·ð [x0 ]â¥ â¥x1 â x2 â¥

4.43 Since ð·ð is continuous, there exists a neighborhood ð of x0 such that â¥ð·ð [x] â ð·ð [x0 ]â¥ < ð for every x â ð and therefore for every x1 , x2 â ð sup

xâ[x1 ,x2 ]

â¥ð·ð [x] â ð·ð [x0 ]â¥ < ð

By the previous exercise (Exercise 4.42) â¥ð (x1 ) â ð (x2 ) â ð·ð [x0 ](x1 â x2 )â¥ â€ ð â¥x1 â x2 â¥ 4.44 By the previous exercise (Exercise 4.43), there exists a neighborhood such that â¥ð (x1 ) â ð (x2 ) â ð·ð [x0 ](x1 â x2 )â¥ â€ ð â¥x1 â x2 â¥ The Triangle Inequality (Exercise 1.200) implies â¥ð (x1 ) â ð (x2 )â¥ â â¥ð·ð [x0 ](x1 â x2 )â¥ â€ â¥ð (x1 ) â ð (x2 ) â ð·ð [x0 ](x1 â x2 )â¥ â€ ð â¥x1 â x2 â¥ and therefore â¥ð (x1 ) â ð (x2 )â¥ â€ â¥ð·ð [x0 ](x1 â x2 )â¥ + ð â¥x1 â x2 â¥ â€ â¥ð·ð [x0 ] + ðâ¥ â¥x1 â x2 â¥ 220

Solutions for Foundations of Mathematical Economics 4.45 Assume not. That is, assume that y = ð (x1 ) â ð (x2 ) ââ conv ðŽ

Then by the (strong) separating hyperplane theorem (Proposition 3.14) there exists a linear functional ð on ð such that ð(y) > ð(a)

for every a â ðŽ

(4.48)

where ð(ðŠ) = ð(ð (x1 ) â ð (x2 )) = ð(ð (x1 )) â ð(ð (x2 )) ðð is a functional on ð. By the mean value theorem (Theorem 4.1), there exists some Â¯ â [x1 , x2 ] such that x x](x1 â x2 ) = ð â ð·ð [Â¯ x](x â x2 ) = ð(ð) ð â ð (x1 ) â ð â ð (x2 )) = ð·(ð â ð )[Â¯ for some ð â ðŽ contradicting (4.44). 4.46 Deï¬ne â : [ð, ð] â â by

( ) ( ) â(ð¥) = ð (ð) â ð (ð) ð(ð¥) â ð(ð) â ð(ð) ð (ð¥)

â â ð¶[ð, ð] and is diï¬erentiable on ð, ð) with ( ) ( ) â(ð) = ð (ð) â ð (ð) ð(ð) â ð(ð) â ð(ð) ð (ð) = ð (ð)ð(ð) â ð (ð)ð(ð) = â(ð) By Rolleâs theorem (Exercise 5.8), there exists ð¥ â (ð, ð) such that ( ) ( ) ââ² (ð¥) = ð (ð) â ð (ð) ð â² (ð¥) â ð(ð) â ð(ð) ð â² (ð¥) = 0 4.47 The hypothesis that limð¥âð ð·ð (ð¥)/ð·ð(ð¥) exists contains two implicit assumptions, namely â ð and ð are diï¬erentiable on a neighborhood ð of ð (except perhaps at ð) â ð â² (ð¥) â= 0 in this neighborhood (except perhaps at ð). Applying the Cauchy mean value theorem, for every ð¥ â ð, there exists some ðŠð¥ â (ð, ð¥) such that ð â² (ðŠð¥ ) ð (ð¥) â ð (ð) ð (ð¥) = = ð â² (ðŠð¥ ) ð(ð¥) â ð(ð) ð(ð¥) and therefore ð (ð¥) ð â² (ðŠð¥ ) ð â² (ð¥) = lim â² = lim â² ð¥âð ð(ð¥) ð¥âð ð (ðŠð¥ ) ð¥âð ð (ð¥) lim

4.48 Let ðŽ = ð1 + ð2 + â â â + ðð â= 1. Then from (4.12) ð1 log ð¥1 + ð2 log ð¥2 + . . . ðð log ð¥ð ðŽ ð2 ðð ð1 log ð¥1 + log ð¥2 + . . . log ð¥ð = ðŽ ðŽ ðŽ

lim ð(ð) =

ðâ0

and therefore lim log ð (ð, x) =

ðâ0

ð1 ð2 ðð log ð¥1 + log ð¥2 + . . . log ð¥ð ðŽ ðŽ ðŽ

so that ð1

ð1

ð1

lim ð (ð, x) = ð¥1ðŽ ð¥2ðŽ . . . ð¥ððŽ

ðâ0

which is homogeneous of degree one. 221

Solutions for Foundations of Mathematical Economics

4.49 Average cost is given by ð(ðŠ)/ðŠ which is undeï¬ned at ðŠ = 0. We seek limðŠâ0 ð(ðŠ)/ðŠ. By LâHË opitalâs rule ð(ðŠ) ðâ² (ðŠ) = lim ðŠâ0 ðŠ ðŠâ0 1 = ðâ² (0) lim

which is marginal cost at zero output. 4.50

1. Since limð¥ââ ð â² (ð¥)/ð â² (ð¥) = k, for every ð > 0 there exists ð such that  â²   ð (Â¯   ð¥) â ð  < ð/2 for every ð¥ Â¯>ð (4.49)  ð â² (Â¯  ð¥) For every ð¥ > ð, there exists (Exercise 4.46) ð¥ Â¯ â (ð, ð¥) such that ð (ð¥) â ð (ð) ð â² (Â¯ ð¥) = â² ð(ð¥) â ð(ð) ð (Â¯ ð¥) and therefore by (4.49)    ð (ð¥) â ð (ð)    < ð/2 for every ð¥ > ð â ð  ð(ð¥) â ð(ð) 

2. ð (ð¥) ð (ð¥) â ð (ð) ð (ð¥) ð(ð¥) â ð(ð) = Ã Ã ð(ð¥) ð(ð¥) â ð(ð) ð (ð¥) â ð (ð) ð(ð¥) ð (ð¥) â ð (ð) 1 â Ã = ð(ð¥) â ð(ð) 1â

ð(ð) ð(ð¥) ð (ð) ð (ð¥)

For ï¬xed ð lim

1â

ð¥ââ

1â

ð(ð) ð(ð¥) ð (ð) ð (ð¥)

=1

and therefore there exists ð2 such that 1â 1â

ð(ð) ð(ð¥) ð (ð) ð (ð¥)

< 2 for every ð¥ > ð2

which implies that    ð  ð (ð¥)    ð(ð¥) â ð  < 2 Ã 2 for every ð¥ > ð = max{ð1 , ð2 } 4.51 We know that the result holds for ð = 1 (Exercise 4.22). Assume that the result holds for ð â 1. By the chain rule ð·(ð â ð )[x] = ð·ð[ð (x)] â ð·ð [x] If ð, ð â ð¶ ð , the ð·ð, ð·ð â ð¶ ðâ1 and therefore (by assumption) ð·(ð â ð ) â ð¶ ðâ1 , which implies that ð â ð â ð¶ ð . 222

Solutions for Foundations of Mathematical Economics 4.52 The partial derivatives of the quadratic function are ð·1 ð = 2ðð¥1 + 2ðð¥2 ð·2 ð = 2ðð¥1 + 2ðð¥2 The second-order partial derivatives are ð·11 ð = 2ð

ð·21 ð = 2ð

ð·12 ð = 2ð

ð·22 ð = 2ð

4.53 Apply Exercise 4.37 to each partial derivative ð·ð ð [x]. 4.54 ð»(x0 ) =

(

ð·11 ð ð [x0 ] ð·12 ð ð [x0 ] ð·21 ð ð [x0 ] ð·22 ð ð [x0 ]

)

( =2

ð ð

ð ð

)

4.55 4.56 For any ð¥1 â ð, deï¬ne ð : ð â â by ð(ð¡) = ð (ð¡) + ð â² [ð¡](ð¥1 â ð¡) + ð2 (ð¥1 â ð¡)2 ð is diï¬erentiable on ð with ðâ² (ð¡) = ð â² [ð¡] â ð â² [ð¡] + ð â²â² [ð¡](ð¥1 â ð¡) â 2ð2 (ð¥1 â ð¡) = ð â²â² [ð¡](ð¥1 â ð¡) â 2ð2 (ð¥1 â ð¡) Note that ð(ð¥1 ) = ð (ð¥1 ) and ð(ð¥0 ) = ð (ð¥0 ) + ð â² (ð¥0 )(ð¥1 â ð¥0 ) + ð2 (ð¥1 â ð¥0 )2

(4.50)

is a quadratic approximation for ð near ð¥0 . If we require that this be exact at ð¥1 â= ð¥0 , then ð(ð¥0 ) = ð (ð¥1 ) = ð(ð¥1 ). By the mean value theorem (Theorem 4.1), there exists some ð¥ Â¯ between ð¥0 and ð¥1 such that ð(ð¥1 ) â ð(ð¥0 ) = ðâ² (Â¯ ð¥)(ð¥1 â ð¥0 ) = ð â²â² (Â¯ ð¥)(ð¥1 â ð¥0 ) â 2ð2 (ð¥1 â ð¡) = 0 which implies that ð2 =

1 â²â² ð (Â¯ ð¥) 2

Setting ð¥ = ð¥1 â ð¥0 in (4.50) gives the required result. 4.57 For any ð¥1 â ð, deï¬ne ð : ð â â by 1 1 ð(ð¡) = ð (ð¡) + ð â² [ð¡](ð¥1 â ð¡) + ð â²â² [ð¡](ð¥1 â ð¡)2 + ð (3) [ð¡](ð¥1 â ð¡)3 + . . . 2 3! 1 (ð) ð ð+1 + ð [ð¡](ð¥1 â ð¡) + ðð+1 (ð¥1 â ð¡) ð! ð is diï¬erentiable on ð with 1 1 ð â² (ð¡) = ð â² [ð¡] â ð â² [ð¡] + ð â²â² [ð¡](ð¥1 â ð¡) â ð â²â² [ð¡](ð¥1 â ð¡) + ð (3) [ð¡](ð¥1 â ð¡)2 â ð (3) [ð¡](ð¥1 â ð¥0 )2 + . . . 2 2 1 1 (ð+1) (ð) ðâ1 ð ð [ð¡](ð¥1 â ð¡) + + ð [ð¡](ð¥1 â ð¡) â (ð + 1)ðð+1 (ð¥1 â ð¡)ð (ð â 1)! ð! All but the last two terms cancel, so that 1 ð (ð¡) = ð (ð+1) [ð¡](ð¥1 â ð¡)ð â (ð + 1)ðð+1 (ð¥1 â ð¡)ð = ð! â²

223

(

) 1 (ð+1) ð [ð¡] â (ð + 1)ðð+1 (ð¥1 â ð¡)ð ð!

Solutions for Foundations of Mathematical Economics

Note that ð(ð¥1 ) = ð (ð¥1 ) and 1 1 ð(ð¥0 ) = ð (ð¥0 ) + ð â² [ð¥0 ](ð¥1 â ð¥0 ) + ð â²â² [ð¥0 ](ð¥1 â ð¥0 )2 + ð (3) [ð¥0 ](ð¥1 â ð¥0 )3 + . . . 2 3! 1 + ð (ð+1) [ð¥0 ](ð¥1 â ð¥0 )ð + ðð+1 (ð¥1 â ð¥0 )ð+1 (4.51) ð! is a polynomial approximation for ð near ð¥0 . If we require that ðð+1 be such that ð(ð¥0 ) = ð (ð¥1 ) = ð(ð¥1 ), there exists (Theorem 4.1) some ð¥ Â¯ between ð¥0 and ð¥1 such that ð¥)(ð¥1 â ð¥0 ) = 0 ð(ð¥1 ) â ð(ð¥0 ) = ð â² (Â¯ which for ð¥1 â= ð¥0 implies that ð â² (Â¯ ð¥) =

1 ð+1 [Â¯ ð¥] â (ð + 1)ðð+1 = 0 ð ð!

or ðð+1 =

1 ð ð+1 [Â¯ ð¥] (ð + 1)!

Setting ð¥ = ð¥1 â ð¥0 in (4.51) gives the required result. 4.58 By Taylorâs theorem (Exercise 4.57), for every ð¥ â ð â ð¥0 , there exists ð¥ Â¯ between 0 and ð¥ such that 1 ð (ð¥0 + ð¥) = ð (ð¥0 ) + ð â² [ð¥0 ]ð¥ + ð â²â² [ð¥0 ]ð¥2 + ð(ð¥) 2 where ð(ð¥) =

1 (3) ð [Â¯ ð¥]ð¥3 3!

and 1 ð(ð¥) = ð (3) [Â¯ ð¥](ð¥) ð¥2 3! ð¥] is bounded on [0, ð¥] and therefore Since ð â ð¶ 3 , ð (3) [Â¯ lim â£

ð¥â0

ð(ð¥) 1 â£ = lim â£ð (3) [Â¯ ð¥](ð¥)â£ = 0 ð¥â0 3! ð¥2

4.59 The function ð : â â ð deï¬ned by ð(ð¡) = ð¡x0 + (1 â ð¡)x ð is ð¶ â with ð·ð[ð¡] = x and ð·ð ð(ð¡) = 0 for ð = 2, 3, . . . . By Exercise 4.51, the composite function â = ð â ð is ð¶ ð+1 . By the Chain rule ââ² (ð¡) = ð·ð [ð(ð¡)] â ð·ð[ð¡] = ð·ð [ð(ð¡)](x) Similarly

( ) ââ²â² (ð¡) = ð· ð·ð [ð(ð¡)](x) = ð·2 ð [ð(ð¡)] â ð·ð[ð¡](x â x0 ) = ð·2 ð [ð(ð¡)](x)(2) 224

Solutions for Foundations of Mathematical Economics and for all 1 â€ ð â€ ð + 1

) ( â(ð) (ð¡) = ð· ð·(ðâ1) ð [ð(ð¡)](x)(ðâ1) = ð·ð ð [ð(ð¡)] â ð·ð[ð¡](x â x0 )(ðâ1) = ð·ð ð [ð(ð¡)](x)(ð)

4.60 From Exercise 4.54, the Hessian of ð is ( ð ð»(x) = 2 ð

ð ð

)

and the gradient of ð is

( ) âð (x) = (2ðð¥1 , 2ðð¥2 ) with âð (0, 0) = 0

so that the second order Taylor series at (0, 0) is 1 ð (x) = ð (0, 0) + âð (0, 0)x + 2xð 2

(

ð ð ð ð

) x

= ðð¥21 + 2ðð¥1 ð¥2 + ðð¥22 Not surprisingly, we conclude that the best quadratic approximation of a quadratic function is the function itself. 4.61

1. Since ð·ð [x0 ] is continuous and one-to-one (Exercise 3.36), there exists a constant ð such that ð â¥x1 â x2 â¥ â€ â¥ð·ð [x0 ](x1 â x2 )â¥

(4.52)

Let ð = ð/2. By Exercise 4.43, there exists a neighborhood ð such that â¥ð·ð [x0 ](x1 â x2 ) â (ð (x1 ) â ð (x2 ))â¥ = â¥ð (x1 ) â ð (x2 ) â ð·ð [x0 ](x1 â x2 )â¥ â€ ð â¥x1 â x2 â¥ for every x1 , x2 â ð. The Triangle Inequality (Exercise 1.200) implies â¥ð·ð [x0 ](x1 â x2 )â¥ â â¥(ð (x1 ) â ð (x2 ))â¥ â€ ð â¥x1 â x2 â¥ Substituting (4.52) 2ð â¥x1 â x2 â¥ â â¥(ð (x1 ) â ð (x2 ))â¥ â€ ð â¥x1 â x2 â¥ That is ð â¥x1 â x2 â¥ â€ â¥(ð (x1 ) â ð (x2 ))â¥

(4.53)

and therefore ð (x1 ) = ð (x2 ) =â x1 = x2 2. Let ð = ð (ð). Since the restriction of ð to ð is one-to-one and onto, and therefore there exists an inverse ð â1 : ð â ð. For any y1 , y2 â ð , let x1 = ð â1 (y1 ) and x2 = ð â1 (y2 ). Substituting in (4.53)   ð ð â1 (y1 ) â ð â1 (y2 ) â€ â¥y1 â y2 â¥ so that   â1 ð (y1 ) â ð â1 (y2 ) â€ 1 â¥y1 â y2 â¥ ð ð â1 is continuous. 225

Solutions for Foundations of Mathematical Economics

3. Since ð is open, ð = ð â1 (ð) is open. Therefore, ð = ð (ð) is a neighborhood of ð (x0 ). Therefore, ð is locally onto. 4.62 Assume to the contrary that there exists x0 â= x1 â ð with ð (x0 ) = ð (x1 ). Let x = x1 â x0 . Deï¬ne ð : [0, 1] â ð by ð(ð¡) = (1 â ð¡)x0 + ð¡x1 = x0 + ð¡x. Then ð(0) = x0 Deï¬ne

ð(1) = x1

ð â² (ð¡) = x

( ( ) ) â(ð¡) = xð ð ð(ð¡) â ð (x0 )

Then â(0) = 0 = â(1) By the mean value theorem (Mean value theorem), there exists 0 < ðŒ < 1 such that ð(ðŒ) â ð and ââ² (ðŒ) = xð ð·ð [ð(ðŒ)]x = xð ðœð (ð(ðŒ))x = 0 which contradicts the deï¬niteness of ðœð . 4.63 Substituting the linear functions in (4.35) and (4.35), the IS-LM model can be expressed as (1 â ð¶ðŠ )ðŠ â ðŒð ð = ð¶0 + ðŒ0 + ðº â ð¶ðŠ ð ð¿ðŠ ðŠ + ð¿ð ð = ð/ð which can be rewritten in matrix form as )( ) ( ) ( ðŠ ð â ð¶ðŠ ð 1 â ð¶ðŠ ðŒð = ð¿ðŠ ð¿ð ð ð/ð where ð = ð¶0 + ðŒ0 + ðº. Provided the system is nonsingular, that is    1 â ð¶ðŠ ðŒð    â= 0 ð·= ð¿ðŠ ð¿ð  the system can be solved using Cramerâs rule (Exercise 3.103) to yield (1 â ð¶ðŠ )ð/ð â ð¿ðŠ (ð â ð¶ðŠ ð ) ð· ð¿ð (ð â ð¶ðŠ )ð â ðŒð ð/ð ðŠ= ð· ð=

4.64 The kernel kernel ð·ð¹ [(x0 , ðœ0 )] = { (x, ðœ) : ð·ð¹ [(x0 , ðœ0 )](x, ðœ) = 0 } is the set of solutions to the equation ( ) ( ) ( ) x ð·x ð (x0 , ðœ0 )x + ð·ðœ ð (x0 , ðœ0 )ðœ 0 = ð·ð¹ [x0 , ðœ0 ] = ðœ ðœ 0 Only ðœ = 0 satisï¬es this equation. Substituting ðœ = 0, the equation reduces to ð·x ð (x0 , ðœ0 )x = 0 which has a unique solution x = 0 since ð·x ð [x0 , ðœ0 ] is nonsingular. Therefore the kernel of ð·ð¹ [x0 , ðœ0 ] consists of the single point (0, 0) which implies that ð·ð¹ [x0 , ðœ 0 ] is nonsingular (Exercise 3.19). 226

Solutions for Foundations of Mathematical Economics

4.65 The IS curve is horizontal if its slope is zero, that is ð·ðŠ ð = â

1 â ð·ðŠ ð¶ âð·ð ðŒ

This requires either 1. unit marginal propensity to consume (ð·ðŠ ð¶ = 1) 2. inï¬nite interest elasticity of investment (ð·ð ðŒ = â) 4.66 The LM curve ð = â(ðŠ) is implicitly deï¬ned by the equation ð (ð, ðŠ; ðº, ð, ð ) = ð¿(ðŠ, ð) â ð/ð = 0 the slope of which is given by ð·ðŠ ð ð·ð ð ð·ðŠ ð¿ =â ð·ð ð¿

ð·ðŠ â = â

Economic considerations dictate that the numerator (ð·ðŠ ð ) is positive while the denominator (ð·ð ð¿) is negative. Preceded by a negative sign, the slope of the LM curve is positive. The LM curve would be vertical (inï¬nite slope) if the interest elasticity of the demand for money was zero (ð·ð ð¿ = 0). 4.67 Suppose ð is convex. For any x, x0 â ð let ) ( â(ð¡) = ð ð¡x + (1 â ð¡)x0 â€ ð¡ð (x) + (1 â ð¡)ð (x0 ) for 0 < ð¡ < 1. Subtracting â(0) = ð (x0 ) â(ð¡) â â(0) â€ ð¡ð (x) â ð¡ð (x0 ) and therefore ð (x) â ð (x0 ) â¥

â(ð¡) â â(0) ð¡

Using Exercise 4.10 ð (x) â ð (x0 ) â¥ lim

ð¡â0

â(ð¡) â â(0) â x ð [x0 ] = ð·ð [x0 ](x â x0 ) =ð· ð¡

Conversely, let x0 = ðŒx1 + (1 â ðŒ)x2 for any x1 , x2 â ð. If ð satisï¬es (4.29) on ð, then ð (x1 ) â¥ ð (x0 ) + ð·ð [x0 ](x1 â x0 ) ð (x2 ) â¥ ð (x0 ) + ð·ð [x0 ](x2 â x0 ) and therefore for any 0 â€ ðŒ â€ 1 ðŒð (x1 ) â¥ ðŒð (x0 + ðŒð·ð [x0 ](x1 â x0 ) (1 â ðŒ)ð (x2 ) â¥ (1 â ðŒ)ð (x0 + (1 â ðŒ)ð·ð [x0 ](x2 â x0 ) Adding and using the linearity of ð·ð (Exercise 4.21) ðŒð (x1 ) + (1 â ðŒ)ð (x2 ) â¥ ð (x0 ) + ð·ð [x0 ](ðŒx1 + (1 â ðŒ)x2 â x0 ) = ð (x0 ) = ð (ðŒx1 + (1 â ðŒ)x2 ) That is, ð is convex. If (4.29) is strict, so is (4.54). 227

(4.54)

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4.68 Since â is convex, it has a subgradient ð â ð â (Exercise 3.181) such that â(x) â¥ â(x0 ) + ð(x â x0 ) for every x â ð (4.31) implies that ð is also a subgradient of ð on ð ð (x) â¥ ð (ð¥0 ) + ð(x â x0 ) for every x â ð Since ð is diï¬erentiable, this implies that ð is unique (Remark 4.14) and equal to the derivative of ð . Hence â is diï¬erentiable at x0 with ð·â[x0 ] = ð·ð [x0 ]. 4.69 Assume ð is convex. For every x, x0 â ð, Exercise 4.67 implies ( ) ð (x) â¥ ð (x0 ) + âð (x0 )ð x â x0 ) ( ð (x0 ) â¥ ð (x) + âð (x)ð x0 â x Adding ) ( ( ) ð (x) + ð (x0 ) â¥ ð (x) + ð (x0 ) + âð (x)ð x0 â x + âð (x0 )ð x â x0 or ( ) ( ) âð (x)ð x â x0 â¥ âð (x0 )ð x â x0 and therefore âð (x) â âð (x0 )ð x â x0 â¥ 0 When ð is strictly convex, the inequalities are strict. Conversely, assume (4.32). By the mean value theorem (Theorem 4.1), there exists Â¯ â (x, x0 ) such that x x)ð x â x0 ð (x) â ð (x0 ) = âð (Â¯ By assumption Â¯ â x0 â¥ 0 âð (Â¯ x) â âð (x0 )ð x But Â¯ â x0 = ðŒx0 + (1 â ðŒ)x â x0 = (1 â ðŒ)(x â x0 ) x and therefore (1 â ðŒ)âð (Â¯ x) â âð (x0 )ð x â x0 â¥ 0 so that âð (Â¯ x)ð x â x0 â¥ âð (x0 )ð x â x0 â¥ 0 and therefore ð (x) â ð (x0 ) = âð (Â¯ x)ð x â x0 â¥ âð (x0 )ð x â x0 Therefore ð is convex by Exercise 4.67. 228

Solutions for Foundations of Mathematical Economics 4.70 For ð â â, âð (ð¥) = ð â² (ð¥) and (4.32) becomes (ð â² (ð¥2 ) â ð â² (ð¥1 )(ð¥2 â ð¥1 ) â¥ 0 for every ð¥1 , ð¥2 â ð. This is equivalent to ð â² (ð¥2 )(ð¥2 â ð¥1 ) â¥ ð â² (ð¥1 )(ð¥2 â ð¥1 )0 or ð¥2 > ð¥1 =â ð â² (ð¥2 ) â¥ ð â² (ð¥1 ) ð is strictly convex if and only if the inequalities are strict.

4.71 ð â² is increasing if and only if ð â²â² = ð·ð â² â¥ 0 (Exercise 4.35). ð â² is strictly increasing if ð â²â² = ð·ð â² > 0 (Exercise 4.36). 4.72 Adapting the previous example

â§ ï£Ž âš= 0 ð â²â² (ð¥) = ð(ð â 1)ð¥ð â 2 = â¥ 0 ï£Ž â© indeterminate

if ð = 1 if ð = 2, 4, 6, ððð¡ð  otherwise

Therefore, the power function is convex if ð is even, and neither convex if ð â¥ 3 is odd. It is both convex and concave when ð = 1. 4.73 Assume ð is quasiconcave, and ð (x) â¥ ð (x0 ). Diï¬erentiability at x0 implies for all 0 < ð¡ < 1 ð (x0 + ð¡(x â x0 ) = ð (x0 ) + âð (x0 )ð¡(x â x0 ) + ð(ð¡) â¥ð¡(x â x0 )â¥ where ð(ð¡) â 0 and ð¡ â 0. Quasiconcavity implies ð (x0 + ð¡(x â x0 ) â¥ ð (x0 ) and therefore âð (x0 )ð¡(x â x0 ) + ð(ð¡) â¥ð¡(x â x0 )â¥ â¥ 0 Dividing by ð¡ and letting ð¡ â 0, we get âð (x0 )(x â x0 ) â¥ 0 Conversely, assume ð is a diï¬erentiable functional satisfying (4.36). For any x1 , x2 â ð with ð (x1 ) â¥ ð (x2 for every x, x0 â ð), deï¬ne â : [0, 1] â â by ( ) ) ( â(ð¡) = ð (1 â ð¡)x1 + ð¡x2 = ð x1 + ð¡(x2 â x1 ) We need to show that â(ð¡) â¥ â(1) for every ð¡ â (0, 1). Suppose to the contrary that â(ð¡1 ) < â(1). Then (see below) there exists ð¡0 with â(ð¡0 ) < â(1) and ââ² (ð¡0 ) < 0. By the Chain Rule, this implies ââ² (ð¡0 ) = âð (x0 )(x2 â x1 ) < 0 critical where x0 = x1 + ð¡(x2 â x1 ). Since x2 â x0 = (1 â ð¡)(x2 â x1 ) this implies that ââ² (ð¡0 ) =

1 âð (x0 )(x2 â x0 ) 1âð¡ 229

(4.55)

Solutions for Foundations of Mathematical Economics

On the other hand, since ð (x0 ) â¥ ð (x2 ), (4.36) implies âð (x0 )(x2 â x0 ) â¥ 0 contradicting (4.55). To show that there exists ð¡0 with â(ð¡0 ) < â(1) and ââ² (ð¡0 ) < 0: Since ð is continuous, there exists an open interval (ð, ð) with ð < ð¡1 < ð with â(ð) = â(ð) = â(1) and â(ð¡) < â(1) for every ð¡ â (ð, ð). By the Mean Value Theorem, there exist ð¡0 â (ð, ð¡1 ) such that 0 < â(ð¡1 ) â â(ð) = ââ² (ð¡0 )(ð¡1 â ð) which implies that ââ² (ð¡0 ) > 0. 4.74 Suppose to the contrary that ð (x) > ð (x0 ) and âð (x0 )(x â x0 ) â€ 0 critical Let x1 = ââð (x0 ) â= 0. For every ð¡ â â+ âð (x0 )(x + ð¡x1 â x0 ) = âð (x0 )ð¡x1 + âð (x0 )(x â x0 ) â€ ð¡âð (x0 )x1 2

= âð¡ â¥âð (x0 )â¥ < 0 Since ð is continuous, there exists ð¡ > 0 such that ð (x + ð¡x1 ) > ð (x0 ) and âð (x0 )(x + ð¡x1 â x0 ) < 0 contradicting the quasiconcavity of ð (4.36). 4.75 Suppose ð (x) < ð (x0 ) =â âð (x0 )(x â x0 ) < 0 This implies that âð (x) > âð (x0 ) =â â â ð (x0 )(x â x0 ) > 0 and âð is pseudoconcave. 4.76

1. If ð â ð¹ [ð] is concave (and diï¬erentiable) ð (x) â€ ð (x0 ) + âð (x0 )ð (x â x0 ) for every x, x0 â ð(equation 4.30). Therefore ð (x) > ð (x0 ) =â âð (x0 )ð (x â x0 ) > 0 ð is pseudoconcave.

2. Assume to the contrary that ð is pseudoconcave but not quasiconcave. Then, Â¯ = ðŒx1 + (1 â ðŒ)x2 , x1 , x2 â ð such that there exists x ð (Â¯ x) < min{ð (x1 ), ð (x2 )} Assume without loss of generality that ð (Â¯ x) < ð (x1 ) â€ ð (x2 ) 230

(4.56)

Solutions for Foundations of Mathematical Economics

Pseudoconcavity (4.38) implies Â¯) > 0 âð (Â¯ x)(x2 â x

(4.57)

x â (1 â ðŒ)x2 )/ðŒ Since x1 = (Â¯ Â¯= x1 â x

) 1( 1âðŒ Â¯ â (1 â ðŒ)x2 â ðŒÂ¯ Â¯) x (x2 â x x =â ðŒ ðŒ

Substituting in (4.57) gives Â¯) < 0 âð (Â¯ x)(x1 â x which by pseudoconcavity implies ð (x1 ) â€ ð (Â¯ x) contradicting our assumption (4.56) . 3. Exercise 4.74. 4.77 The CES function is quasiconcave provided ð â€ 1 (Exercise 3.58). Since ð·ð¥ð ð (x) > 0 for all x â âð+ +, the CES function with ð â€ 1 is pseudoconcave on âð++ . 4.78 Assume that ð : ð â â is homogeneous of degree ð, so that for every x â ð ð (ð¡x) = ð¡ð ð (x) for every ð¡ > 0 Diï¬erentiating both sides of this identity with respect to ð¥ð ð·ð¥ð ð (ð¡x)ð¡ = ð¡ð ð·ð¥ð ð (x) and dividing by ð¡ > 0 ð·ð¥ð ð (ð¡x) = ð¡ðâ1 ð·ð¥ð ð (x) 4.79 If ð is homogeneous of degree ð â x ð (x) = lim ð (x + ð¡x) â ð (x) ð· ð¡â0 ð¡ ð ((1 + ð¡)x) â ð (x) = lim ð¡â0 ð¡ (1 + ð¡)ð ð (x) â ð (x) = lim ð¡â0 ð¡ (1 + ð¡)ð â 1 = lim ð (x) ð¡â0 ð¡ Applying LâHËopitalâs Rule (Exercise 4.47) (1 + ð¡)ðâ1 ð(1 + ð¡)ðâ1 ð (x) = lim =ð ð¡â0 ð¡â0 ð¡ 1 lim

and therefore â x ð (x) = ðð (x) ð·

(4.58)

4.80 For ï¬xed x, deï¬ne â(ð¡) = ð (ð¡x) By the Chain Rule ââ² (ð¡) = ð¡ð·ð [ð¡x](x) = ð¡ðð (ð¡x) = ð¡ðâ(ð¡) 231

(4.59)

Solutions for Foundations of Mathematical Economics

using(4.40). Diï¬erentiating the product â(ð¡)ð¡âð ( ) ( ) ð·ð¡ â(ð¡)ð¡âð = âðâ(ð¡)ð¡âðâ1 + ð¡âð ââ² (ð¡) = ð¡âð ââ² (ð¡) â ðð¡â(ð¡) = 0 from (4.59). Since this holds for every ð¡, â(ð¡)ð¡âð must be constant (Exercise 4.38), that is there exists ð â â such that â(ð¡)ð¡âð = ð =â â(ð¡) = ðð¡ð Evaluating at ð¡ = 1, â(1) = ð and therefore â(ð¡) = ð¡ð â(1) Since â(ð¡) = ð (ð¡x) and â(1) = ð (x), this implies ð (ð¡x) = ð¡ð ð (x) for every x and ð¡ > 0 ð is homogeneous of degree ð. 4.81 If ð is linearly homogeneous and quasiconcave, then ð is concave (Proposition 3.12). Therefore, its Hessian is nonpositive deï¬nite (Proposition 4.1). and its diagonal elements ð·ð¥2 ð ð¥ð ð (x) are nonpositive (Exercise 3.95). By Wicksellâs law, ð·ð¥2 ð ð¥ð ð (x) is nonnegative. 4.82 Assume ð is homogeneous of degree ð, that is ð (ð¡x) = ð¡ð ð (x) for every x â ð and ð¡ > 0 By Eulerâs theorem ð·ð¡ ð [ð¡x](ð¡x) = ðð (ð¡x) and therefore the elasticity of scale is

  ð¡ ð¡ ð·ð¡ ð (ð¡x) ðð (ð¡x) = ð = ðž(x) = ð (ð¡x) ð (ð¡x) ð¡=1

Conversely, assume that

  ð¡ ðž(x) = ð·ð¡ ð (ð¡x) =ð ð (ð¡x) ð¡=1

that is ð·ð¡ ð (ð¡x) = ðð (ð¡x) By Eulerâs theorem, ð is homogeneous of degree ð. 4.83 Assume ð â ð¹ (ð) is diï¬erentiable and homogeneous of degree ð â= 0. By Eulerâs theorem ð·ð [x](x) = ðð (x) â= 0 for every x â ð such that ð (x) â= 0. 4.84 ð satisï¬es Eulerâs theorem ðð (x) =

ð â

ð·ð ð (x)ð¥ð

ð=1

232

Solutions for Foundations of Mathematical Economics

Diï¬erentiating with respect to ð¥ð ðð·ð ð (x) =

ð â

ð·ðð ð (x)ð¥ð + ð·ð ð (x)

ð=1

or (ð â 1)ð·ð ð (x) =

ð â

ð·ðð ð (x)ð¥ð

ð = 1, 2, . . . , ð

ð=1

Multiplying each equation by ð¥ð and summing (ð â 1)

ð â

ð·ð ð (x)ð¥ð =

ð=1

ð â ð â

ð·ðð ð (x)ð¥ð ð¥ð = xâ² ð»x

ð=1 ð=1

By Eulerâs theorem, the left hand side is (ð â 1)ðð (x) = xâ² ð»x 4.85 If ð is homothetic, there exists strictly increasing ð and linearly homogeneous â such that ð = ð â â (Exercise 3.175). Using the Chain Rule and Exercise 4.78 ð·ð¥ð ð (ð¡x) = ð â² (ð (ð¡x))ð·ð¥ð â(ð¡x) = ð¡ð â² (ð (ð¡x))ð·ð¥ð â(x) and therefore ð·ð¥ð ð (ð¡x) ð¡ð â² (ð (ð¡x))ð·ð¥ð â(x) = ð·ð¥ð ð (ð¡x) ð·ð¥ð ð¡ð â² (ð (ð¡x))ð·ð¥ð â(x) ð·ð¥ð â(x) = ð·ð¥ð â(x) ð·ð¥ð ð â² (ð (x)ð·ð¥ð â(x) = ð·ð¥ð ð â² (ð (x))ð·ð¥ð â(x) ð·ð¥ð ð (x) = ð·ð¥ð ð (x) ð·ð¥ð

233

Solutions for Foundations of Mathematical Economics

Chapter 5: Optimization 5.1 As stated, this problem has no optimal solution. Revenue ð (ð¥) increases without bound as the rate of exploitation ð¥ gets smaller and smaller. Given any positive exploitation rate ð¥0 , a smaller rate will increase total revenue. Nonexistence arises from inadequacy in modeling the island leadersâ 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 â Current revenues can be invested to provide a future return. There is an opportunity cost (the interest foregone) to delaying extraction and sale. â 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-oï¬ of lower total return against reduced risk. 5.2 Suppose that xâ is a local optimum which is not a global optimum. That is, there exists a neighborhood ð of xâ such that ð (xâ , ðœ) â¥ ð (x, ðœ) for every x â ð â© ðº(ðœ) and also another point xââ â ðº(ðœ) such that ð (xââ , ðœ) > ð (xâ , ðœ) Since ðº(ðœ) is convex, there exists ðŒ â (0, 1) such that ðŒxâ + (1 â ðŒ)xââ â ð â© ðº(ðœ) By concavity of ð ð (ðŒxâ + (1 â ðŒ)xââ , ðœ) â¥ ðŒð (xâ , ðœ) + (1 â ðŒ)ð (xââ , ðœ) > ð (xâ , ðœ) contradicting the assumption that xâ is a local optimum. 5.3 Suppose that xâ is a local optimum which is not a global optimum. That is, there exists a neighborhood ð of xâ such that ð (xâ , ðœ) â¥ ð (x, ðœ) for every x â ð â© ðº(ðœ) and also another point xââ â ðº(ðœ) such that ð (xââ , ðœ) > ð (xâ , ðœ) Since ðº(ðœ) is convex, there exists ðŒ â (0, 1) such that ðŒxâ + (1 â ðŒ)xââ â ð â© ðº(ðœ)

234

Solutions for Foundations of Mathematical Economics

By strict quasiconcavity of ð ð (ðŒxâ + (1 â ðŒ)xââ , ðœ) > min{ ð (xâ , ðœ), ð (xââ , ðœ) } > ð (xâ , ðœ) contradicting the assumption that xâ is a local optimum. Therefore, if xâ is local optimum, it must be a global optimum. Now suppose that xâ is a weak global optimum, that is ð (xâ , ðœ) â¥ ð (x, ðœ) for every x â ð but there another point xââ â ð such that ð (xââ , ðœ) = ð (xâ , ðœ) Since ðº(ðœ) is convex, there exists ðŒ â (0, 1) such that ðŒxâ + (1 â ðŒ)xââ â ð â© ðº(ðœ) By strict quasiconcavity of ð ð (ðŒxâ + (1 â ðŒ)xââ , ðœ) > min{ ð (xâ , ðœ), ð (xââ , ðœ) } = ð (xâ , ðœ) contradicting the assumption that xâ is a global optimum. We conclude that every optimum is a strict global optimum and hence unique. 5.4 Suppose that xâ is a local optimum of (5.3) in ð, so that ð (xâ ) â¥ ð (x)

(5.80)

for every x in a neighborhood ð of xâ . If ð is diï¬erentiable, ð (x) = ð (xâ ) + ð·ð [xâ ](x â xâ ) + ð(x) â¥x â xâ â¥ where ð(x) â 0 as x â xâ . (5.80) implies that there exists a ball ðµð (xâ ) such that ð·ð [xâ ](x â xâ ) + ð(x) â¥x â xâ â¥ â€ 0 for every x â ðµð (xâ ). Letting x â xâ , we conclude that ð·ð [xâ ](x â xâ ) â€ 0 for every x â ðµð (xâ ). Suppose there exists x â ðµð (xâ ) such that ð·ð [xâ ](x â xâ ) = ðŠ < 0 Let dx = x â xâ so that x = xâ + dx. Then xâ â dx â ðµð (xâ ). Since ð·ð [xâ ] is linear, ð·ð [xâ ](âdx) = âð·ð [xâ ](dx) = âðŠ > 0 contradicting (5.80). Therefore ð·ð [xâ ](x â xâ ) = 0 for every x â ðµð (xâ ).

235

Solutions for Foundations of Mathematical Economics

5.5 We apply the reasoning of Example 5.5 to each component. Formally, for each ð, let ðËð be the projection of ð along the ðð¡â axis ðËð (ð¥ð ) = ð (ð¥â1 , ð¥â2 , . . . , ð¥âðâ1 , ð¥ð , ð¥âð+1 , . . . , ð¥âð ) ð¥âð maximizes ðËð (ð¥ð ) over â+ , for which it is necessary that ð·ð¥ð ðËð (ð¥âð ) â€ 0

ð¥âð â¥ 0

ð¥âð ð·ð¥ð ðËð (ð¥âð ) = 0

Substituting ð·ð¥ð ðËð (ð¥âð ) = ð·ð¥ð ð [xâ ] yields ð·ð¥ð ð [xâ ] â€ 0

ð¥âð â¥ 0

ð¥âð ð·ð¥ð ð [xâ ] = 0

5.6 By Taylorâs Theorem (Example 4.33) 1 2 ð (xâ + dx) = ð (xâ ) + âð (xâ )dx + dxð ð»ð (xâ )dx + ð(dx) â¥dxâ¥ 2 with ð(dx) â 0 as dx â 0. Given 1. âð (xâ ) = 0 and 2. ð»ð (xâ ) is negative deï¬nite and letting dx â 0, we conclude that ð (xâ + dx) < ð (xâ ) for small dx. xâ is a strict local maximum. 5.7 If xâ is a local minimum of ð (x), it is necessary that ð (xâ ) â€ ð (x) for every x in a neighborhood ð of xâ . Assuming that ð is ð¶ 2 , ð (x) can be approximated by 1 ð (x) â ð (xâ ) + âð (xâ )dx + dxð ð»ð (xâ )dx 2 where dx = x â xâ . If xâ is a local minimum, then there exists a ball ðµð (xâ ) such that 1 ð (xâ ) â€ ð (xâ ) + âð (xâ )dx + dxð ð»ð (xâ )dx 2 or 1 âð (xâ )dx + dxð ð»ð (xâ )dx â¥ 0 2 for every dx â ðµð (xâ ). To satisfy this inequality for all small dx requires that the ï¬rst term be zero and the second term nonnegative. In other words, for a point xâ to be a local minimum of a function ð , it is necessary that the gradient be zero and the Hessian be nonnegative deï¬nite at xâ . Furthermore, by Taylorâs Theorem 1 2 ð (xâ + dx) = ð (xâ ) + âð (xâ )dx + dxð ð»ð (xâ )dx + ð(dx) â¥dxâ¥ 2 with ð(dx) â 0 as dx â 0. Given 236

Solutions for Foundations of Mathematical Economics

(1,2,3)

3 2 ð¥ 2

0

1

1 ð¥1

2

Figure 5.1: The strictly concave function ð (ð¥1 , ð¥2 ) = ð¥1 ð¥2 + 3ð¥2 â ð¥21 â ð¥22 has a unique global maximum. 1. âð (xâ ) = 0 and 2. ð»ð (xâ ) is positive deï¬nite and letting dx â 0, we conclude that ð (xâ + dx) > ð (xâ ) for small dx. xâ is a strict local minimum. 5.8 By the Weierstrass theorem (Theorem 2.2), ð has a maximum ð¥â and a minimum ð¥â on [ð, ð]. Either â ð¥â â (ð, ð) and ð â² (ð¥â ) = 0 (Theorem 5.1) or â ð¥â â (ð, ð) and ð â² (ð¥â ) = 0 (Exercise 5.7) or â Both maxima and minima are boundary points, that is ð¥â , ð¥â â {ð, ð} which implies that ð is constant on [ð, ð] and therefore ð â² (ð¥) = 0 for every ð¥ â (ð, ð) (Exercise 4.7). 5.9 The ï¬rst-order conditions for a maximum are ð·ð¥1 ð (ð¥1 , ð¥2 ) = ð¥2 â 2ð¥1 = 0 ð·ð¥2 ð (ð¥1 , ð¥2 ) = ð¥1 + 3 â 2ð¥2 = 0 which have the unique solution ð¥â1 = 1, ð¥â2 = 2. (1, 2) is the only stationary point of ð and hence the only possible candidate for a maximum. To verify that (1, 2) satisï¬es the second-order condition for a maximum, we compute the Hessian of ð ) ( â2 1 ð»(x) = 1 â2 which is negative deï¬nite everywhere. Therefore (1, 2) is a strict local maximum of ð . Further, since ð is strictly concave (Proposition 4.1), we conclude that (1, 2) is a strict global maximum of ð (Exercise 5.2), where it attains its maximum value ð (1, 2) = 3 (Figure 5.1). 237

Solutions for Foundations of Mathematical Economics

5.10 The ï¬rst-order conditions for a maximum (or minimum) are ð·1 ð (ð¥) = 2ð¥1 = 0 ð·2 ð (ð¥) = 2ð¥2 = 0 which have a unique solution ð¥1 = ð¥2 = 0. This is the only stationary point of ð . Since the Hessian of ð ( ) 2 0 ð»= 0 2 is positive deï¬nite, we deduce (0, 0) is a strict global minimum of ð (Proposition 4.1, Exercise 5.2). 5.11 The average ï¬rmâs proï¬t function is 1 1 Î (ð, ð) = ðŠ â ð â ð â 2 6 and the ï¬rmâs proï¬t maximization problem is 1 1 max Î (ð, ð) = ð 1/6 ð1/3 â ð â ð â ð,ð 2 6 A necessary condition for a proï¬t maximum is that the proï¬t function be stationary, that is 1 â5/6 1/3 1 ð ð â =0 6 2 1 1/6 â2/3 â1=0 ð·ð Î (ð, ð) = ð ð 3

ð·ð Î (ð, ð) =

which can be solved to yield ð=ð=

1 9

The ï¬rmâs output is ðŠ=

1 1/6 1 1/3 1 = 9 9 3

and its proï¬t is 1 11 1 1 1 1 â â =0 Î ( , ) = â 3 3 3 29 9 6 5.12 By the Chain Rule ð·x (â â ð )[xâ ] = ð·â â ð·x ð [xâ ] = 0 Since ð·â > 0 ð·x (â â ð )[ð¥â ] = 0 ââ ð·x ð [xâ ] = 0 â â ð has the same stationary points as ð .

238

Solutions for Foundations of Mathematical Economics

5.13 Since the log function is monotonic, ï¬nding the maximum likelihood estimators is equivalent to solving the maximization problem ( Exercise 5.12) max log ð¿(ð, ð) = â ð,ð

ð 1 â ð log 2ð â ð log ð â 2 (ð¥ð¡ â ð)2 2 2ð ð¡=1

For (Ë ð, ð Ë 2 ) to solve this problem, it is necessary that log ð¿ be stationary at (Ë ð, ð Ë 2 ), that is ð, ð Ë2) = ð·ð log ð¿(Ë

ð 1 â (ð¥ð¡ â ð Ë) = 0 ð Ë 2 ð¡=1

ð·ð log ð¿(Ë ð, ð Ë2) = â

ð 1 â ð + 3 (ð¥ð¡ â ð Ë)2 = 0 ð Ë ð Ë ð¡=1

which can be solved to yield ð Ë=ð¥ Â¯= ð Ë2 =

ð 1 â ð¥ð¡ ð ð¡=1

ð 1â (ð¥ð¡ â ð¥ Â¯)2 ð ð¡=1

5.14 The gradient of the objective function is ) ( â2(ð¥1 â 1) âð (x) = â2(ð¥2 â 1) while that of the constraint is

( âð(x) =

2ð¥1 2ð¥2

)

A necessary condition for the optimal solution is that these be proportional that is ) ( ( ) 2ð¥1 â2(ð¥1 â 1) =ð âð (ð¥) = = âð(x) â2(ð¥2 â 1) 2ð¥2 which can be solved to yield ð¥1 = ð¥2 =

1 1+ð

which includes an unknown constant of proportionality ð. However, any solution must also satisfy the constraint ( )2 1 ð(ð¥1 , ð¥2 ) = 2 =1 1+ð This can be solved for ð ð=

â 2â1

and substituted into (5.80) 1 ð¥1 = ð¥2 = â 2 239

Solutions for Foundations of Mathematical Economics 5.15 The consumerâs problem is max ð¢(x) = ð¥1 + ð log ð¥2 xâ¥0

subject to ð(x) = ð¥1 + ð2 ð¥2 â ð = 0 The ï¬rst-order conditions for a (local) optimum are ð·ð¥1 ð¢(xâ ) = 1 â€ ð = ð·ð¥1 ð(ð¥â )

ð¥1 â¥ 0

ð â€ ðð2 = ð·ð¥2 ð(ð¥â ) ð·ð¥2 ð¢(xâ ) = ð¥2

ð¥2 â¥ 0

( ð¥2

ð¥1 (1 â ð) = 0 ) =0

ð â ðð2 ð¥2

(5.81) (5.82)

We can distinguish two cases: Case 1 ð¥1 = 0 in which case the budget constraint implies that ð¥2 = ð/ð2 . Case 2 ð¥1 > 0 In this case, (5.81) implies that ð = 1. Consequently, the ï¬rst inequality of (5.82) implies that ð¥2 > 0 and therefore the last equation implies ð¥2 = ð/ð2 with ð¥1 = ð â ð. We deduce that the consumer ï¬rst spends portion ð of her income on good 2 and the remainder on good 1. 5.16 Suppose without loss of generality that the ï¬rst ð components of yâ are strictly positive while the remaining components are zero. That is ðŠðâ > 0 ðŠðâ = 0

ð = 1, 2, . . . , ð ð = ð + 1, ð + 2, . . . , ð

(xâ , yâ ) solves the problem max ð (x) subject to g(x) = 0 ð = ð + 1, ð + 2, . . . , ð ðŠð = 0 By Theorem 5.2, there exist multipliers ð1 , ð2 , . . . , ðð and ðð+1 , ðð+2 , . . . , ðð such that ð·x ð [xâ , yâ ] = ð·y ð [xâ , yâ ] =

ð â ð=1 ð â

ðð ð·x ðð [xâ , yâ ] ðð ð·y ðð [xâ , yâ ] +

ð=1

ð â

ðð ðŠ ð

ð=ð+1

Furthermore, ðð â¥ 0 for every ð so that ð·y ð [xâ , yâ ] â€

ð â

ðð ð·y ðð [xâ , yâ ]

ð=1

with ð·ðŠð ð [xâ , yâ ] =

ð â

ðð ð·ðŠð ðð [xâ , yâ ] if ðŠð > 0

ð=1

5.17 Assume that xâ = (ð¥â1 , ð¥â2 ) solves max ð (ð¥1 , ð¥2 )

ð¥1 ,ð¥2

240

Solutions for Foundations of Mathematical Economics

subject to ð(ð¥1 , ð¥2 ) = 0 By the implicit function theorem, there exists a function â : â â â such that ð¥1 = â(ð¥2 )

(5.83)

and ð(â(ð¥2 ), ð¥2 ) = 0 for ð¥2 in a neighborhood of ð¥â2 . Furthermore ð·â[ð¥â2 ] = â

ð·ð¥1 ð[xâ ] ð·ð¥2 ð[xâ ]

(5.84)

Using (5.41), we can convert the original problem into the unconstrained maximization of a function of a single variable max ð (â(ð¥2 ), ð¥2 ) ð¥2

If ð¥â2 maximizes this function, it must satisfy the ï¬rst-order condition (applying the Chain Rule) ð·ð¥1 ð [ð¥â] â ð·â[ð¥â2 ] + ð·ð¥2 ð [xâ ] = 0 Substituting (5.42) yields ( ) ð·ð¥1 ð[xâ ] ð·ð¥1 ð [ð¥â] â + ð·ð¥2 ð [xâ ] = 0 ð·ð¥2 ð[xâ ] or ð·ð¥1 ð [ð¥â] ð·ð¥1 ð[xâ ] = ð·ð¥2 ð [xâ ] ð·ð¥2 ð[xâ ] 5.18 The consumerâs problem is max ð¢(x) xâð

subject to pð x = ð Solving for ð¥1 from the budget constraint yields âð ð â ð=2 ðð ð¥ð ð¥1 = ð1 Substituting this in the utility function, the aï¬ordable utility levels are â ( ) ð â ðð=2 ðð ð¥ð , ð¥2 , ð¥3 , . . . , ð¥ð ð¢ Ë(ð¥2 , ð¥3 , . . . , ð¥ð ) = ð¢ ð1

(5.85)

and the consumerâs problem is to choose (ð¥2 , ð¥3 , . . . , ð¥ð ) to maximize (5.85). The ï¬rst-order conditions are that ð¢ Ë(ð¥2 , ð¥3 , . . . , ð¥ð ) be stationary, that is for every good ð = 2, 3, . . . , ð âð ( ) ð â ð=2 ðð ð¥ð â ð·ð¥ð ð¢ Ë(ð¥2 , ð¥3 , . . . , ð¥ð ) = ð·ð¥1 ð¢(x )ð·ð¥ð + ð·ð¥ð ð¢(xâ ) = 0 ð1 241

Solutions for Foundations of Mathematical Economics

which reduces to ð·ð¥1 ð¢(xâ )(â

ð1 ) + ð·ð¥ð ð¢(xâ ) = 0 ðð

or ð1 ð·ð¥1 ð¢(xâ ) = â ð·ð¥ð ð¢(x ) ðð

ð = 2, 3, . . . , ð

This is the familiar equality between the marginal rate of substitution and the price ratio (Example 5.15). Since our selection of ð¥1 was arbitrary, this applies between any two goods. 5.19 Adapt Exercise 5.6. 5.20 Corollary 5.1.2 implies that xâ is a global maximum of ð¿(x, ð), that is ð¿(xâ , ð) â¥ ð¿(x, ð) for every x â ð which implies ð (xâ ) â

â

ðð ðð (xâ ) â¥ ð (x) â

â

ðð ðð (x) for every x â ð

Since g(xâ ) = 0 this implies ð (xâ ) â¥ ð (x) â

â

ðð ðð (x) for every x â ð

A fortiori ð (xâ ) â¥ ð (x) for every x â ðº = { x â ð : g(x) = 0 } 5.21 Suppose that xâ is a local maximum of ð on ðº. That is, there exists a neighborhood ð such that ð (xâ ) â¥ ð (x) for every x â ð â© ðº But for every x â ðº, ðð (x) = 0 for every ð and â ð¿(x) = ð (x) + ðð ðð (x) = ð (x) and therefore ð¿(xâ ) â¥ ð¿(x) for every x â ð â© ðº 5.22 The area of the base is Base = ð€2 = ðŽ/3 and the four sides Sides = 4ð€â â â ðŽ ðŽ =4 3 12 4ðŽ = 16 2ðŽ = 3 242

Solutions for Foundations of Mathematical Economics

5.23 Let the dimensions of the vat be ð€ Ã ð Ã â. We wish to min Surface area = ðŽ = ð€ Ã ð + 2ð€â + 2ðâ

ð€,ð,â

subject to ð€ Ã ð Ã â = 32 The Lagrangean is ð¿(ð€, ð, â, ð) = ð€ð + 2ð€â + 2ðâ â ðð€ðâ. The ï¬rst-order conditions for a maximum are ð·ð€ ð¿ = ð + 2â â ððâ = 0

(5.86)

ð·ð ð¿ = ð€ + 2â â ðð€â = 0

(5.87)

ð·â ð¿ = 2ð€ + 2ð â ðð€ð = 0 ð€ðâ = 32

(5.88)

Subtracting (5.45) from (5.44) ð â ð€ = ð(ð â ð€)â This equation has two possible solutions. Either ð=

1 or ð = ð€ â

But if ð = 1/â, (5.44) implies that ð = 0 and the volume is zero. Therefore, we conclude that ð€ = ð. Substituting ð€ = ð in (5.45) and (5.46) gives ð€ + 2â = ðð€â 4ð€ = ðð€2 from which we deduce that ð=

4 ð€

Substituting in (5.45) ð€ + 2â =

4 ð€â = 4â ð€

which implies that ð€ = 2â or â =

1 ð€ 2

To achieve the required volume of 32 cubic metres requires that 1 ð€ Ã ð Ã â = ð€ Ã ð€ Ã ð€ = 32 2 so that the dimensions of the vat are ð€=4

ð=4

â=2

The area of sheet metal required is ðŽ = ð€ð + 2ð€â + 2ðâ = 48 243

Solutions for Foundations of Mathematical Economics 5.24 The Lagrangean for this problem is

ð¿(x, ð) = ð¥21 + ð¥22 + ð¥23 â ð(2ð¥1 â 3ð¥2 + 5ð¥3 â 19) A necessary condition for xâ to solve the problem is that the Lagrangean be stationary at xâ , that is ð·ð¥1 ð¿ = 2ð¥â1 â 2ð = 0 ð·ð¥2 ð¿ = 2ð¥â2 + 3ð = 0 ð·ð¥3 ð¿ = 2ð¥â3 â 5ð = 0

which implies 3 ð¥â2 = â ð 2

ð¥â1 = ð

ð¥â2 =

5 ð 2

(5.89)

It is also necessary that the solution satisfy the constraint, that is 2ð¥â1 â 3ð¥â2 + 5ð¥â3 = 19 Substituting (5.89) into the constraint we get 9 25 2ð + ð + ð = 19ð = 19 2 2 which implies ð = 1. Substituting in (5.89), the solution is xâ = (1, â 32 , 52 ). Since the constraint is aï¬ne and the objective (âð ) is concave, stationarity of the Lagrangean is also suï¬cient for global optimum (Corollary 5.2.4). 5.25 The Lagrangean is 1âðŒ ð¿(ð¥1 , ð¥2 , ð) = ð¥ðŒ â ð(ð1 ð¥1 + ð2 ð¥2 â ð) 1 ð¥2

The Lagrangean is stationary where ð·ð¥1 ð¿ = ðŒð¥ðŒâ1 ð¥1âðŒ â ðð1 = 0 1 2 1âðŒâ1 ð·ð¥2 ð¿ = 1 â ðŒð¥ðŒ â ðð2 = 0 1 ð¥2

Therefore the ï¬rst-order conditions for a maximum are 1â

ðŒð¥ðŒâ1 ð¥1âðŒ = ðð1 1 2

(5.90)

1âðŒâ1 ðŒð¥ðŒ 1 ð¥2

(5.91)

= ðð2 ð1 ð¥1 + ð2 ð¥2 â ð

Dividing (5.48) by (5.49) gives ðŒð¥ðŒâ1 ð¥1âðŒ 1 2

(1âðŒ)â1

1 â ðŒð¥ðŒ 1 ð¥2

= ð1 ð2

which simpliï¬es to ð1 ðŒð¥2 = (1 â ðŒ)ð¥1 ð2

or

ð2 ð¥2 =

(1 â ðŒ) ð1 ð¥1 ðŒ

Substituting in the budget constraint (5.50) (1 â ðŒ) ð1 ð¥1 = ð ðŒ ðŒ + (1 â ðŒ) ð1 ð¥1 = ð ðŒ

ð1 ð¥1 +

244

(5.92)

Solutions for Foundations of Mathematical Economics so that ð¥â1 =

ð ðŒ ðŒ + (1 â ðŒ) ð1

From the budget constraint (5.92) ð¥â2 =

(1 â ðŒ) ð ðŒ + (1 â ðŒ) ð2

5.26 The Lagrangean is ðŒð 1 ðŒ2 ð¿(x, ð) = ð¥ðŒ 1 ð¥2 . . . ð¥ð â ð(ð1 ð¥1 + ð2 ð¥2 + ... + ðð ð¥ð )

The ï¬rst-order conditions for a maximum are ðŒð â1 1 ðŒ2 ð ð·ð¥ð ð¿ = ðŒð ð¥ðŒ . . . ð¥ðŒ ð â ððð = 1 ð¥2 . . . ð¥ð

ðŒð ð¢(ð¥) â ððð = 0 ð¥ð

or ðŒð ð¢(ð¥) = ðð ð¥ð ð

ð = 1, 2, . . . , ð

Summing over all goods and using the budget constraint ð â ðŒð ð¢(ð¥) ð=1

Letting

âð

ð=1

ð

ð

=

ð

â ð¢(ð¥) â ðŒð = ðð ð¥ð = ð ð ð=1 ð=1

ðŒð = ðŒ, this implies ð ð¢(x) = ð ðŒ

Substituting in (5.93) ðð ð¥ð =

ðŒð ð ðŒ

or ð¥âð =

ðŒð ð ðŒ ðð

ð = 1, 2, . . . , ð

5.27 The Lagrangean is ð¿(x, ð) = ð€1 ð¥1 + ð€2 ð¥2 â ð(ð¥ð1 + ð¥ð2 â ðŠ ð ). The necessary conditions for stationarity are ð·ð¥1 ð¿(x, ð) = ð€1 â ððð¥ðâ1 =0 1 ð·ð¥2 ð¿(x, ð) = ð€2 â ððð¥ðâ1 =0 2 or ð€1 = ððð¥ðâ1 1 ð€2 = ððð¥ðâ1 2

245

(5.93)

Solutions for Foundations of Mathematical Economics which reduce to ð¥ðâ1 ð€1 = 1ðâ1 ð€2 ð¥ 2 ð€2 ðâ1 ðâ1 ð¥2 = ð¥ ð€1 1 ð ( ) ðâ1 ð€2 ð¥ð2 = ð¥ð1 ð€1 Substituting in the production constraint (

) ð ð€2 ðâ1 ð + ð¥1 = ðŠ ð ð€1 ( ð ) ( ) ðâ1 ð€2 1+ ð¥ð1 = ðŠ ð ð€1 ð¥ð1

we can solve ð¥1 ( ð¥ð1

=

(

1+ (

ð¥1 =

(

1+

ð€2 ð€1 ð€2 ð€1

ð )â1 ) ðâ1

ðŠð

ð )â1/ð ) ðâ1

ðŠ

Similarly ( ð¥2 =

( 1+

ð€1 ð€2

ð )â1/ð ) ðâ1

ðŠ

5.28 Example 5.27 is ï¬awed. The optimum of the constrained maximization problem (â = ð€/2) is in fact a saddle point of the Lagrangean. It maximizes the Lagrangean in the feasible set, but not globally. The net beneï¬t approach to the Lagrange multiplier method is really only applicable when the Lagrangean (net beneï¬t function) is concave, so that every stationary point is a global maximum. This requirement is satisï¬ed in many standard examples, such as the consumerâ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 ð¿(x, ð) =

ð â

( ðð (ð¥ð ) + ð ð· â

ð=1

ð â

) ð¥ð

(5.94)

ð=1

can be rewritten as ð¿(x, ð) = â

ð â ) ( ðð¥ð â ðð (ð¥ð ) + ðð·

(5.95)

ð=1

The ðth term in the sum is the net proï¬t of plant ð if its output is valued at ð. Therefore, if the company undertakes to buy electricity from its plants at the price ð and instructs each plant manager to produce so as to maximize the plantâs net proï¬t, each manager 246

Solutions for Foundations of Mathematical Economics

will be induced to choose an output level which maximizes the proï¬t of the company as a whole. This is the case whether the price ð 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 ð·. In this way, the shadow price ð can be used to decentralize the production decision. 5.30 The Lagrangean for this problem is ð¿(ð¥1 , ð¥2 , ð1 , ð2 ) = ð¥1 ð¥2 â ð1 (ð¥21 + 2ð¥22 â 3) â ð2 (2ð¥21 + ð¥22 â 3) The ï¬rst-order conditions for stationarity ð·ð¥1 ð¿ = ð¥2 â 2ð1 ð¥1 â 4ð2 ð¥1 = 0 ð·ð¥2 ð¿ = ð¥1 â 4ð1 ð¥2 â 2ð2 ð¥2 = 0 can be written as ð¥2 = 2(ð1 + 2ð2 )ð¥1

(5.96)

ð¥1 = 2(2ð1 + ð2 )ð¥2

(5.97)

which must be satisï¬ed along with the complementary slackness conditions ð¥21 + 2ð¥22 â 3 â€ 0

ð1 â¥ 0

ð1 (ð¥21 + 2ð¥22 â 3) = 0

2ð¥21 + ð¥22 â 3 â€ 0

ð2 â¥ 0

ð2 (2ð¥21 + ð¥22 â 3) = 0

First suppose that both constraints are slack so that ð1 = ð2 = 0. Then the ï¬rst-order conditions (5.96) and (5.97) imply that ð¥1 = ð¥2 = 0. (0, 0) satisï¬es the Kuhn-Tucker conditions. Next suppose that the ï¬rst constraint is binding while the second constraint have two solutions, is slack â (ð2 = 0).â The ï¬rst-order â and (5.97) â â conditions (5.96) â ð¥1 = 3/2, ð¥2 = 3/2, ð = 1/(2 2) and ð¥1 = â 3/2, ð¥2 = â 3/2, ð = 1/(2 2), but these violate the second constraint. Similarly, there is no solution in which the ï¬rst constraint is slack and the second constraint binding. Finally, assume that the both constraints are binding. This implies that ð¥1 = ð¥2 = 1 or ð¥1 = ð¥2 = â1, which points satisfy the ï¬rst-order conditions (5.96) and (5.97) with ð1 = ð2 = 1/6. We conclude that three points satisfy the Kuhn-Tucker conditions, namely (0, 0), (1, 1) and (â1, â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 (â1, â1), both of which achieve the same level of the objective function. 5.31 Dividing the ï¬rst-order conditions, we obtain ð ð(ð  â ð) ð·ð ð(ð, ð) = â ð·ð ð(ð, ð) ð€ (1 â ð)ð€ Using the revenue function ð(ð, ð) = ð(ð (ð, ð))ð (ð, ð) the marginal revenue products of capital and labor are ð·ð ð(ð, ð) = ð·ðŠ ð(ðŠ)ð·ð ð (ð, ð) ð·ð ð(ð, ð) = ð·ðŠ ð(ðŠ)ð·ð ð (ð, ð) so that their ratio is equal to the ratio of the marginal products ð·ð ð (ð, ð) ð·ð ð(ð, ð) = ð·ð ð(ð, ð) ð·ð ð (ð, ð 247

Solutions for Foundations of Mathematical Economics

The necessary condition for optimality can be expressed as ð ð ð âð ð·ð ð (ð, ð) = â ð·ð ð (ð, ð) ð€ 1âð ð€ whereas the necessary condition for cost minimization is (Example 5.16) ð·ð ð (ð, ð) ð = ð·ð ð (ð, ð) ð€ The regulated ï¬rm does not use the cost-minimizing combination of inputs. 5.32 The general constrained optimization problem max ð (x) x

subject to g(x) â€ 0 can be transformed into an equivalent equality constrained problem max ð (x) x,s

subject to g(x) + s = 0 and s â¥ 0 Ë (x, s) = g(x) + s, the through the addition of nonnegative slack variables s. Letting g ï¬rst-order conditions a local optimum are (Exercise 5.16) â â ð·x ð (xâ ) = ðð ð·x ðËð (xâ , sâ ) = ðð ð·x ðð (xâ ) â ðð ð·s ðËð (x, s) = ð (5.98) 0 = ð·s ð (xâ ) â€ ðð s = 0

sâ¥0

(5.99)

Condition (5.98) implies that ðð â¥ 0 for every ð. Furthermore, rewriting the constraint as s = âg(x) the complementary slackness condition (5.99) becomes ðð g(x) = 0

g(x) â€ 0

This establishes the necessary conditions of Theorem 5.3. 5.33 The equality constrained maximization problem max ð (x) x

subject to g(x) = 0 is equivalent to the problem max ð (x) x

subject to g(x) â€ 0 âg(x) â€ â0 By the Kuhn-Tucker theorem (Theorem 5.3), there exists nonnegative multipliers + â â + â ð+ 1 , ð2 , . . . , ðð and ð1 , ð2 , . . . , ðð such that â â â â ð·ð (xâ ) = ð+ ðâ (5.100) ð ð·ðð [x ] â ð ð·ðð [x ] = 0 248

Solutions for Foundations of Mathematical Economics with â ð+ ð ðð (x) = 0 and ðð ðð (x) = 0

ð = 1, 2, . . . , ð

â Deï¬ning ðð = ð+ ð â ðð , (5.100) can be written as

ð·ð (xâ ) =

â

ðð ð·ðð [xâ ]

which is the ï¬rst-order condition for an equality constrained problem. Furthermore, if xâ satisï¬es the inequality constraints ð(xâ ) â€ 0 and ð(xâ ) â¥ 0 it satisï¬es the equality ð(xâ ) = 0 5.34 Suppose that xâ solves the problem max cð x subject to ðŽx â€ 0 x

with Lagrangean ð¿ = cð x â ðð ðŽx Then there exists ð â¥ 0 such that ð·x ð¿ = cð â ðð ðŽ = 0 that is, ðŽð ð = c. Conversely, if there is no solution, there exists x such that ðŽx â€ 0 and cð x > cð 0 = 0 5.35 There are two binding constraints at (4, 0), namely ð(ð¥1 , ð¥2 ) = ð¥1 + ð¥2 â€ 4 â(ð¥1 , ð¥2 ) = âð¥2 â€ 0 with gradients âð(4, 0) = (1, 1) ââ(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 ð¿(x, ð) = ð¢(x) â ð(pð x â ð) and the ï¬rst-order (Kuhn-Tucker) conditions are (Corollary 5.3.2) ð·ð¥ð ð¿[xâ , ð] = ð·ð¥ð ð¢[xâ ] â ððð â€ 0 ð

â

p x â€ð

xâð â¥ 0 ðâ¥0

ð¥âð (ð·ð¥ð ð¢[xâ ] â ððð ) = 0 ð

â

ð(p x â ð) = 0

for every good ð = 1, 2, . . . , ð. Two cases must be distinguished. 249

(5.101) (5.102)

Solutions for Foundations of Mathematical Economics

Case 1 ð > 0 This implies that pð x = ð, the consumer spends all her income. Condition (5.101) implies ð·ð¥ð ð¢[xâ ] â€ ððð for every ð with ð·ð¥ð ð¢[xâ ] = ððð for every ð for which ð¥ð > 0 This case was analyzed in Example 5.17. Case 2 ð = 0 This allows the possibility that the consumer does not spend all her income. Substituting ð = 0 in (5.101) we have ð·ð¥ð ð¢[xâ ] = 0 for every ð. At the optimal consumption bundle xâ , 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 xâ , either â the consumer is satiated (ð·ð¥ð ð¢[xâ ] = 0 for every ð) or â the consumer consumes only those goods whose marginal utility exceeds the threshold ð·ð¥ð ð¢[xâ ] â¥ ððð and adjusts consumption so that the marginal utility is proportional to price for all consumed goods. 5.37 Assume x â ð·(xâ ). Then there exists ðŒ Â¯ â â such that xâ + ðŒx â ð for every 0 â€ ðŒ â€ ðŒ. Â¯ Deï¬ne ð â ð¹ ([0, ðŒ Â¯ ]) by ð(ðŒ) = ð (xâ + ðŒx). If xâ is a local maximum, ð has a local maximum at 0, and therefore ð â² (0) â€ 0 (Theorem 5.1). By the chain rule (Exercise 4.22), this implies ð â² (0) = ð·ð [xâ ](x) â€ 0 and therefore x â / ð» + (xâ ). 5.38 If x is a tangent vector, so is ðœx for any nonnegative ðœ (replace 1/ðŒð by ðœ/ðŒð in the preceding deï¬nition. Also, trivially, x = 0 is a tangent vector (with xð = xâ and ðŒð = 1 for all ð). The set ð of all vectors tangent to ð at xâ is therefore a nonempty cone, which is called the cone of tangents to ð at xâ . To show that ð is closed, let xð be a sequence in ð converging to some x â âð . We need to show that x â ð . Since xð â ð , there exist feasible points xðð â ð and ðŒðð such that (xðð â xâ )/ðŒðð â xð as ð â â For any ð choose ð such that â¥xð â xâ¥ â€

1 ð 2

and then choose ð such that â¥xðð â xâ â¥ â€ ð and â¥(xðð â xâ )/ðŒðð â xð â¥ â€

1 ð 2

Relabeling xðð as xð and ðŒðð as ðŒð we have we have constructed a sequence xð in S such that   ð x â xâ  â€ ð and

 ð    (x â xâ )/ðŒð â x â€ (xð â xâ )/ðŒð â xð  + â¥xð â xâ¥ â€ 1 ð 1

Letting ð â â, xð converges to xâ and (xð â xâ )/ðŒð converges to x, which proves that x â ð as required. 250

Solutions for Foundations of Mathematical Economics

5.39 Assume x â ð·(xâ ). That is, there exists ðŒ Â¯ such that xâ + ðŒx â ð for every ðŒ â [0, ðŒ Â¯ ]. For ð = 1, 2, . . . , let ðŒð = ðŒ Â¯ /ð. Then xð = xâ + ðŒð x â ð, xð â xâ and ð â â â (x â x )/ðŒð = (x + ðŒð x â x )/ðŒð = x. Therefore, x â ð (xâ ). 5.40 Let dx â ð (xâ ). Then there exists a feasible sequence {xð } converging to xâ and a sequence {ðŒð } of nonnegative scalars such that the sequence {(xð â xâ )/ðŒð } converges to dx. For any ð â ðµ(xâ ), ðð (xâ ) = 0 and   ðð (xð ) = ð·ðð [xâ ](xð â xâ ) + ðð xð â xâ  where ðð â 0 as k â â. This implies   1 1 ðð (xð ) = ð ð·ðð [xâ ](xð â xâ ) + ðð (xð â xâ )/ðŒð  ð ðŒ ðŒ Since xð is feasible 1 ðð (xð ) â€ 0 ðŒð and therefore   ð·ðð [xâ ]((xð â xâ )/ðŒð ) + ð ð (xð â xâ )/ðŒð  â€ 0 Letting ð â â we conclude that ð·ðð [xâ ](dx) â€ 0 That is, dx â ð¿. 5.41 ð¿0 â ð¿1 by deï¬nition. Assume dx â ð¿1 . That is ð·ðð [xâ ](dx) < 0 â

ð·ðð [x ](dx) â€ 0

for every ð â ðµ ð (xâ ) ð¶

â

for every ð â ðµ (x )

(5.103) (5.104)

where ðµ ð¶ (xâ ) = ðµ(xâ ) â ðµ ð (xâ ) is the set of concave binding constraints at xâ . By concavity (Exercise 4.67), (5.104) implies that ðð (xâ + ðŒdx) â€ ðð (xâ ) = 0 for every ðŒ â¥ 0 and ð â ðµ ð¶ (xâ ) From (5.103) there exists some ðŒð such that ðð (xâ + ðŒdx) < 0 for every ðŒ â [0, ðŒð ] and ð â ðµ ð (xâ ) Furthermore, since ðð (xâ ) < 0 for all ð â ð(xâ ), there exists some ðŒð > 0 such that ðð (xâ + ðŒdx) < 0 for every ðŒ â [0, ðŒð ] and ð â ð(xâ ) Setting ðŒ Â¯ = min{ðŒð , ðŒð } we have ðð (xâ + ðŒdx) â€ 0 for every ðŒ â [0, ðŒ Â¯ ] and ð = 1, 2, . . . , ð or xâ + ðŒdx â ðº = { x : ðð (x) â€ 0, ð = 1, 2, . . . , ð } for every ðŒ â [0, ðŒ Â¯] Therefore dx â ð·. We have previously shown (Exercises 5.39 and 5.40) that ð· â ð â ð¿. 251

Solutions for Foundations of Mathematical Economics

5.42 Assume that g satisï¬es the Quasiconvex CQ condition at xâ . That is, for every Ë such that ðð (Ë ð â ðµ(xâ ), ðð is quasiconvex, âðð (xâ ) â= 0 and there exists x x) < 0. Ë â xâ . Quasiconvexity and regularity implies that Consider the perturbation dx = x for every binding constraint ð â ðµ(xâ ) (Exercises 4.74 and 4.75) ðð (Ë x) < ðð (xâ ) =â âðð (xâ )ð (Ë x â xâ ) = âðð (xâ )ð dx < 0 That is ð·ðð [xâ ](dx) < 0 Therefore, dx â ð¿0 (xâ ) â= â and g satisï¬es the Cottle constraint qualiï¬cation condition. 5.43 If the binding constraints ðµ(xâ ) are regular at xâ , their gradients are linearly independent. That is, there exists no ðð â= 0, ð â ðµ(xâ ) such that â ðð âðð [xâ ] = 0 ðâðµ(xâ )

By Gordanâs theorem (Exercise 3.239), there exists dx â âð such that âðð [xâ ]ð dx < 0 for every ð â ðµ(xâ ) Therefore dx â ð¿0 (xâ ) â= â. 5.44 If ðð concave, ðµ ð (xâ ) = â, and AHUCQ is trivially satisï¬ed (with dx = 0 â ð¿1 ). For every ð, let ðð = { dx : ð·ðð [xâ ](dx) < 0 } Then

â ð¿1 (xâ ) = â

â

ðð â

â

â

ðâðµ ð (xâ )

â ðð â

ðâðµ ð¶ (xâ )

where ðµ ð¶ (xâ ) and ðµ ð (xâ ) are respectively the concave and nonconcave constraints binding at xâ . If ðð satisï¬es the AHUCQ condition, ð¿1 (xâ ) â= â and Exercise 1.219 implies that â â â â â© â© â© ð¿1 = â ðð â  â ðð â  ðâðµ ð (xâ )

ðâðµ ð¶ (xâ )

Now ðð = { dx : ð·ðð [xâ ](dx) â€ 0 } and therefore

ð¿1 =

ðð = ð¿

ðâðµ(xâ )

Since (Exercise 5.41) ð¿1 â ð â ð¿ and ð is closed (Exercise 5.38), we have ð¿ = ð¿1 â ð â ð¿ which implies that ð = ð¿. 252

Solutions for Foundations of Mathematical Economics

5.45 For each ð = 1, 2, . . . , ð, either ðð (xâ ) < 0 which implies that ðð = 0 and therefore ðð ð·ðð [xâ ](x â xâ ) = 0 or â

ðð (x ) = 0 Since ðð is quasiconvex and ðð (x) â€ 0 = ð(xâ ), Exercise 4.73 implies that ð·ðð [xâ ](x â xâ ) â€ 0. Since ðð â¥ 0, this implies that ðð ð·ðð [xâ ](x â xâ ) â€ 0. We have shown that for every ð, ðð ð·ðð [xâ ](x â xâ ) â€ 0. The ï¬rst-order condition implies that â ðð ð·ðð [xâ ](x â xâ ) â€ 0 ð·ð [xâ ](x â xâ ) = ð

If âð (xâ ) â€

â

ðð âðð (xâ )

xâ â¥ 0

( )ð âð (xâ ) â ðð âðð (xâ ) xâ = 0

The ï¬rst-order conditions imply that for every x â ðº, x â¥ 0 and (

)ð âð (xâ ) â ðð âðð (xâ ) x â€ 0

and therefore ( )ð âð (xâ ) â ðð âðð (xâ ) (x â xâ ) â€ 0 or âð (xâ )ð (x â xâ ) â€

â

ðð âðð (xâ )ð (x â xâ ) â€ 0

5.46 Assuming ð¥ð = ð¥ð = 0, the constraints become 2ð¥ð â€ 30 2ð¥ð â€ 25 ð¥ð â€ 20 The ï¬rst and third conditions are redundant, which implies that ðð = ðð = 0. Complementary slackness requires that, if ð¥ð > 0, ð·ð¥ð ð¿ = 1 â 2ðð â 2ðð â ðð = 0 or ðð = 12 . Evaluating the Lagrangean at (0, 1/2, 0) yields )) ( ( 1 = 3ð¥ð + ð¥ð + 3ð¥ð ð¿ x, 0, , 0 2 1 â (ð¥ð + 2ð¥ð + 3ð¥ð â 25) 2 25 5 3 = + ð¥ð + ð¥ð 2 2 2 This basic feasible solution is clearly not optimal, since proï¬t would be increased by increasing either ð¥ð or ð¥ð .

253

Solutions for Foundations of Mathematical Economics

Following the hint, we allow ð¥ð > 0, retaining the assumption that ð¥ð = 0. We must be alert to the possibility that ð¥ð = 0. With ð¥ð = 0, the constraints become 2ð¥ð + ð¥ð â€ 30 2ð¥ð + 3ð¥ð â€ 25 ð¥ð + ð¥ð â€ 20 The ï¬rst constraint is redundant, which implies that ðð = 0. If ð¥ð > 0, complementary slackness requires that ð·ð¥ð ð¿ = 3 â 3ðð â ðð = 0 or ðð = 3(1 â ðð )

(5.105)

The requirement that ðð â¥ 0 implies that ðð â€ 1. Substituting (5.105) in the second ï¬rst-order condition ð·ð¥ð ð¿ = 1 â 2ðð â ðð = 1 â 2ðð â 3(1 â ðð ) = â2 + ðð implies that ð·ð¥ð ð¿ = â2 + ðð < 0

for every ðð â€ 1

Complementary slackness then requires implies that ð¥ð = 0. The constraints now become ð¥ð â€ 30 3ð¥ð â€ 25 ð¥ð â€ 20 The ï¬rst and third are redundant, so that ðð and ðð = 0. Equation (5.105) implies that ðð = 1. Evaluating the Lagrangean at this point (ð = 0, 1, 0), we have ð¿(ð¥, (0, 1, 0)) = 3ð¥ð + ð¥ð + 3ð¥ð â (ð¥ð + 2ð¥ð + 3ð¥ð â 25) = 25 + 2ð¥ð â ð¥ð Clearly this is not an optimal solution, An increase in ð¥ð is indicated. This leads us to the hypothesis ð¥ð > 0, ð¥ð > 0, ð¥ð = 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 ð¥ð > 0, ð¥ð â¥ 0, ð¥ð = 0, the constraints become 2ð¥ð + 2ð¥ð â€ 30 ð¥ð + 2ð¥ð â€ 25 2ð¥ð + ð¥ð â€ 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 ï¬rst constraint is equal to the sum of 254

Solutions for Foundations of Mathematical Economics

the second and third constraints. The feasible solution ð¥ð = 0, ð¥ð = 5, ð¥ð = 10, where the constraints are linearly dependent, is known as a degenerate solution. Degeneracy is a signiï¬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 ðð = 0 and proceeding as before, complementary slackness (ð¥ð > 0) requires that ð·ð¥ð ð¿ = 3 â 2ðð â ðð = 0 or ðð = 3 â 2ðð

(5.106)

Nonnegativity of ðð implies that ðð â€ 32 . Substituting (5.106) in the second ï¬rst-order condition yields ð·ð¥ð ð¿ = 1 â 2ðð â 2ðð = 1 â 2ðð â 2(3 â 2ðð ) = â5 + 2ðð < 0 for every ðð â€

3 2

Complementary slackness therefore implies that ð¥ð = 0, which takes us back to the starting point of the presentation in the text, where ð¥ð > 0, ð¥ð = ð¥ð = 0. 5.48 Assume that (c1 , ð§1 ) and (c2 , ð§2 ) belong to ðµ. That is ð§1 â¥ ð§ â ð§2 â¥ ð§ â

c1 â€ 0 c2 â€ 0 For any ðŒ â (0, 1),

ð§Â¯ = ðŒð§1 + (1 â ðŒ)ð§2 â€ ð§ â

Â¯ c = ðŒc1 + (1 â ðŒ)c2 â€ 0

and therefore (Â¯ c, ð§Â¯) â ðµ. This shows that ðµ is convex. Let 1 = (1, 1, . . . , 1) â âð . Then (c â 1, ð§ + 1) â int ðµ â= â. There ðµ has a nonempty interior. 5.49 Let (c, ð§) â int ðµ. This implies that c < 0 and ð§ > ð§ â . Since ð£ is monotone ð£(c) â€ ð£(0) = zâ < ð§ which implies that (c, ð§) â / ðŽ. 5.50 The linear functional ð¿ can be decomposed into separate components, so that there exists (Exercise 3.47) ð â ð â and ðŒ â â such that ð¿(c, ð§) = ðŒð§ â ð(c) Assuming ð â âð , there exists (Proposition 3.4) ð â âð such that ð(c) = ðð c and therefore ð¿(c, ð§) = ðŒð§ â ðð c The point (0, ð§ â + 1) belongs to ðµ. Therefore, by (5.75), ð¿(0, ð§ â ) â€ ð¿(0, ð§ â + 1) 255

Solutions for Foundations of Mathematical Economics which implies that ðŒð§ â â ðð 0 â€ ðŒ(ð§ â + 1) â ðð 0

or ðŒ â¥ 0. Similarly, let { e1 , e2 , . . . , eð } denote the standard basis for âð (Example 1.79). For any ð = 1, 2, . . . , ð, the point (0 â eð , ð§ â ) (which corresponds to decreasing resource ð by one unit) belongs to ðµ and therefore (from (5.75)) ð§ â â ðð (0 â eð ) = ð§ â + ðð â¥ ð§ â â ðð 0 = ð§ â which implies that ðð â¥ 0. 5.51 Let cË = ð(Ë x) < 0 and ð§Ë = ð (Ë x) Suppose ðŒ = 0. Then, since ð¿ is nonzero, at least one component of ð must be nonzero. That is, ð â© 0 and therefore ðð ðË < 0

(5.107)

But (Ë c, ð§Ë) â ðŽ and (5.74) implies ðŒË ð§ â ðð cË â€ ðŒð§ â â ðð 0 and therefore ðŒ = 0 implies ðð ðË â¥ 0 contradicting (5.107). Therefore, we conclude that ðŒ > 0. 5.52 The utilityâs optimization problem is max ð(ðŠ, ð ) =

ðŠ,ð â¥0

ð â« â ð=1

ðŠð 0

(ðð (ð ) â ðð )ðð â ð0 ð

subject to ðð (y, ð ) = ðŠð â ð â€ 0

ð = 1, 2, . . . , ð

The demand independence assumption ensures that the objective function ð is concave, since its Hessian â â ð·ð1 0 . . . 0 0 â 0 ð·ð2 . . . 0 0â â ð»ð = â â 0 ... ð·ðð 0â  0 ... 0 0 is nonpositive deï¬nite (Exercise 3.96). The constraints are linear and hence convex. Moreover, there exists a point (0, 1) such that for every ð = 1, 2, . . . , ð ðð (0, 1) = 0 â 1 < 0 Therefore the problem satisï¬es the conditions of Theorem 5.6. The optimal solution (yâ , ð â ) satisï¬es the Kuhn-Tucker conditions, that is there exist multipliers ð1 , ð2 , . . . , ðð such that for every period ð = 1, 2, . . . , ð ð·ðŠð ð¿ = ðð (ðŠð ) â ðð â ðð â€ 0

ðŠð â¥ 0

ðŠð (ðð (ðŠð ) â ðð â ðð ) = 0

ðŠð â€ ð

ðð â¥ 0

ð(ð â ðŠð ) = 0

256

(5.108)

Solutions for Foundations of Mathematical Economics and that capacity be chosen such that ð· ð ð¿ = ð0 â

ð â

( ðð â€ 0

ð â¥0

ð

ð0 â

ð=1

ð â

) ðð

=0

(5.109)

ð=1

where ð¿ is the Lagrangean ð¿(ðŠ, ð, ð) =

ð â« â ð=1

0

ðŠð

(ðð (ð ) â ðð )ðð â ð0 ð â

ð â

ðð (ðŠð â ð )

ð=1

In oï¬-peak periods (ðŠð < ð ), complementary slackness requires that ðð = 0 and therefore from (5.108) ðð (ðŠð ) = ðð assuming ðŠð > 0. In peak periods (ðŠð = ð ) ðð (ðŠð ) = ðð + ðð We conclude that it is optimal to price at marginal cost in oï¬-peak periods and charge a premium during peak periods. Furthermore, (5.109) implies that the total premium is equal to the marginal capacity cost ð â

ðð = ð0

ð=1

Furthermore, note that ð â

ðð ðŠð =

ð=1

â Peak

=

=

ðð ðŠð

Oï¬-peak

â

ðð ðŠð +

ðŠð =ð

=

â

ðð ðŠð +

â

â

ðð ðŠð

ðð =0

ðð ð

ðŠð =ð ð â

ðð ð = ð0 ð

ð=1

Therefore, the utilityâs total revenue is ð(ðŠ, ð ) = = = =

ð â ð=1 ð â ð=1 ð â ð=1 ð â

ðð (ðŠð )ðŠð (ðð + ðð )ðŠð ðð ðŠ ð +

ð â

ðð ðŠð

ð=1

ðð ðŠð + ð0 ð = ð(ðŠ, ð )

ð=1

Under the optimal pricing policy, revenue equals cost and the utility breaks even. 257

Solutions for Foundations of Mathematical Economics

Chapter 6: Comparative Statics 6.1 The Jacobian is

( ð»ð¿ ðœg

ðœ=

ðœgð 0

)

where ð»ð¿ is the Hessian of the Lagrangean. We note that â ð»ð¿ (x0 ) is negative deï¬nite in the subspace ð = { x : ðœg x = 0 } (since x0 satisï¬es the conditions for a strict local maximum) â ðœg has rank ð (since the constraints are regular). Consider the system of equations ( ð»ð¿ ðœg

ðœgð 0

)( ) ( ) x 0 = y 0

(6.28)

where x â âð and y â âð . It can be decomposed into ð»ð¿ x + ðœgð y = 0

(6.29)

ðœg x = 0

(6.30)

Suppose x solves (6.30). Multiplying (6.29) by xð gives xð ð»ð¿ x + xð ðœgð y = xð ð»ð¿ x + (ðœg x)ð y = 0 But (6.30) implies that the second term is 0 and therefore xð ð»ð¿ x = 0. Since ð»ð¿ is positive deï¬nite on ð = { x : ðœg x = 0 }, we must have x = 0. Then (6.29) reduces to ðœgð y = 0 Since ðœg has rank ð, 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 ðœ is nonsingular. 6.2 The Lagrangean for this problem is

( ) ð¿ = ð (x) â ðð g(x) â c

By Corollary 6.1.1 âð£(c) = ð·c ð¿ = ð 6.3 Optimality implies ð (x1 , ðœ1 ) â¥ ð (x, ðœ 1 ) and ð (x2 , ðœ 2 ) â¥ ð (x, ðœ2 ) for every x â ð In particular ð (x1 , ðœ1 ) â¥ ð (x2 , ðœ1 ) and ð (x2 , ðœ2 ) â¥ ð (x1 , ðœ2 ) Adding these inequalities ð (x1 , ðœ1 ) + ð (x2 , ðœ2 ) â¥ ð (x2 , ðœ1 ) + ð (x1 , ðœ2 ) 258

Solutions for Foundations of Mathematical Economics

Rearranging and using the bilinearity of ð gives ð (x1 â x2 , ðœ1 ) â¥ ð (x1 â x2 , ðœ2 ) and ð (x1 â x2 , ðœ 1 â ðœ 2 ) â¥ 0 6.4 Let ð1 denote the proï¬t maximizing price with the cost function ð1 (ðŠ) and let ðŠ1 be the corresponding output. Similarly let ð2 and ðŠ2 be the proï¬t maximizing price and output when the costs are given by ð2 (ðŠ). With cost function ð1 , the ï¬rms proï¬t is Î  = ððŠ â ð1 (ðŠ) Since this is maximised at ð1 and ðŠ1 (although the monopolist could have sold ðŠ2 at price ð2 ) ð1 ðŠ1 â ð1 (ðŠ1 ) â¥ ð2 ðŠ2 â ð1 (ðŠ2 ) Rearranging ð1 ðŠ1 â ð2 ðŠ2 â¥ ð1 (ðŠ1 ) â ð1 (ðŠ2 )

(6.31)

The increase in revenue in moving from ðŠ2 to ðŠ1 is greater than the increase in cost. Similarly ð2 ðŠ2 â ð2 (ðŠ2 ) â¥ ð1 ðŠ1 â ð2 (ðŠ1 ) which can be rearranged to yield ð2 (ðŠ1 ) â ð2 (ðŠ2 ) â¥ ð1 ðŠ1 â ð2 ðŠ2 Combining the previous inequality with (6.31) yields ð2 (ðŠ1 ) â ð2 (ðŠ2 ) â¥ ð1 (ðŠ1 ) â ð1 (ðŠ2 )

(6.32)

6.5 By Theorem 6.2 ð·w Î [w, ð] = âxâ and ð·ð Î [w, ð] = ðŠ â and therefore 2 Î (ð, w) â¥ 0 ð·ð ðŠ(ð, w) = ð·ðð 2 Î (ð, w) â€ 0 ð·ð€ð ð¥ð (ð, w) = âð·ð€ ð ð€ð 2 Î (ð, w) = ð·ð€ð ð¥ð (ð, w) ð·ð€ð ð¥ð (ð, w) = âð·ð€ ð ð€ð 2 Î (ð, w) = âð·ð€ð ðŠ(ð, w) ð·ð ð¥ð (ð, w) = âð·ð€ ðð

since Î  is convex and therefore ð»Î  (w, ð) is symmetric (Theorem 4.2) and nonnegative deï¬nite (Proposition 4.1). 6.6 By Shephardâs lemma (6.17) ð¥ð (ð€, ðŠ) = ð·ð€ð ð(ð€, ðŠ) 259

Solutions for Foundations of Mathematical Economics Using Youngâs theorem (Theorem 4.2), 2 ð·ðŠ ð¥ð [w, ðŠ] = ð·ð€ ð[w, ðŠ] ððŠ 2 ð[w, ðŠ] = ð·ðŠð€ ð

= ð·ð€ð ð·ðŠ ð[w, ðŠ] Therefore ð·ðŠ ð¥ð [w, ðŠ] â¥ 0 ââ ð·ð€ð ð·ðŠ ð[w, ðŠ] â¥ 0 6.7 The demand functions must satisfy the budget contraint identically, that is ð â

ðð ð¥ð (p, ð) = ð for every p and ð

ð=1

Diï¬erentiating with respect to m ð â

ðð ð·ð ð¥ð [p, ð] = 1

ð=1

This is the Engel aggregation condition, which simply states that any additional income be spent on some goods. Multiplying each term by ð¥ð ð/(ð¥ð ð) ð â ðð ð¥ð ð=1

ð ð·ð ð¥ð [p, ð] = 1 ð ð¥ð (p, ð)

the Engel aggregation condition can be written in elasticity form ð â

ðŒð ðð = 1

ð=1

where ðŒð = ðð ð¥ð /ð is the budget share of good ð. On average, goods must have unit income elasticities. Diï¬erentiating the budget constraint with respect to ðð ð â

ðð ð·ðð ð¥ð [p, ð] + ð¥ð (ð, ð) = 0

ð=1

This is the Cournot aggregation condition, which implies that an increase in the price of ðð is equivalent to a decrease in real income of ð¥ð ððð . Multiplying each term in the sum by ð¥ð /ð¥ð gives ð â ðð ð¥ð ð=1

ð¥ð

ð·ðð ð¥ð [p, ð] = âð¥ð

Multiplying through by ðð /ð ð â ðð ð¥ð ðð ð=1

ð ð¥ð

ð·ðð ð¥ð [p, ð] = â

ð â

ðŒð ððð = âðŒð

ð=1

260

ðð ð¥ð ð

Solutions for Foundations of Mathematical Economics

6.8 Supermodularity of Î (x, ð, âw) follows from Exercises 2.50 and 2.51. To show strictly increasing diï¬erences, consider two price vectors w2 â¥ w1 Î (x, ð, âw1 ) â Î (x, ð, âw2 ) =

ð â

(âð€ð1 )ð¥ð â

ð=1

=

ð â

ð â

(âð€ð2 )ð¥ð

ð=1

(ð€ð2 â ð€ð1 )ð¥ð

ð=1

âð

Since w2 â¥ w1 , w2 â w1 â¥ 0 and

2 ð=1 (ð€ð

â ð€ð1 )ð¥ð is strictly increasing in x.

6.9 For any ð2 â¥( ð1 , ðŠ 2 = ð (ð2 ) â€ ð (ð1 ) = ðŠ 1 and ð(ðŠ 1 , ð) â ð(ðŠ 2 , ð) is increasing in ð and therefore â ð(ð (ð2 ), ð) â ð(ð (ð1 ), ð)) is increasing in ð. 6.10 The ï¬rmâs optimization problem is max ðððŠ â ð(ðŠ)

ðŠââ+

The objective function ð (ðŠ, ð, ð) = ðððŠ â ð(ðŠ) is â supermodular in ðŠ (Exercise 2.49) â displays strictly increasing diï¬erences in (ðŠ, ð) since ( ) ð (ðŠ 2 , ð, ð) â ð (ðŠ 1 , ð, ð) = ðð(ðŠ 2 â ðŠ 1 ) â ð(ðŠ 2 ) â ð(ðŠ 1 ) is strictly increasing in ð for ðŠ 2 > ðŠ 1 . Therefore (Corollary 2.1.2), the ï¬rmâs output correspondence is strongly increasing and every selection is increasing (Exercise 2.45). Therefore, the ï¬rmâ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 ( ð11 ð»ð = ð21

ð12 ð22

)

Therefore, its inverse is (Exercise 3.104) ð»ðâ1

1 = Î

(

ð22 âð21

âð12 ð11

)

where Î = ð11 ð22 â ð12 ð21 â¥ 0 by the second-order condition. Therefore, the Jacobian of the demand functions is ( ) ) ( 1 1 â1 ð·ð€1 ð¥1 ð·ð€2 ð¥1 ð22 âð12 = ð»ð = ð·ð€1 ð¥2 ð·ð€2 ð¥2 ð11 ð ðÎ âð21 Therefore ð21 ð·ð€1 ð¥2 = â ðÎ

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