Measures, Integrals, and Martingales - Solutions Manual
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Measures, Integrals & Martingales (2nd edition) Cambridge University Press, Cambridge 2017 ISBN: 978–1–316–62024–3
Solution Manual René L. Schilling Dresden, May 2017
R.L. Schilling: Measures, Integrals & Martingales
Acknowledgement. I am grateful for the help of Dr. Franziska Kühn in the preparation of this solution manual. Dr. Björn Böttcher, Dr. Julian Hollender and Dr. Franziska Kühn contributed problems and solutions to this collection. Dresden, Mai 2017
2
René Schilling
Contents 1 Prologue. Solutions to Problems 1.1–1.5
7
2 The pleasures of counting. Solutions to Problems 2.1–2.22 3
9
-Algebras. Solutions to Problems 3.1–3.16
21
4 Measures. Solutions to Problems 4.1–4.22
31
5 Uniqueness of measures. Solutions to Problems 5.1–5.13
49
6 Existence of measures. Solutions to Problems 6.1–6.14
59
7 Measurable mappings. Solutions to Problems 7.1–7.13
71
8 Measurable functions. Solutions to Problems 8.1–8.26
79
9 Integration of positive functions. Solutions to Problems 9.1–9.14
93
10 Integrals of measurable functions. Solutions to Problems 10.1–10.9
101
11 Null sets and the ‘almost everywhere’. Solutions to Problems 11.1–11.12
109
12 Convergence theorems and their applications. Solutions to Problems 12.1–12.37
119
3
R.L. Schilling: Measures, Integrals & Martingales 13 The function spaces Lp . Solutions to Problems 13.1–13.26
147
14 Product measures and Fubini’s theorem. Solutions to Problems 14.1–14.20
165
15 Integrals with respect to image measures. Solutions to Problems 15.1–15.16
185
16 Integrals of images and Jacobi’s transformation rule. Solutions to Problems 16.1–16.12
197
17 Dense and determining sets. Solutions to Problems 17.1–17.9
209
18 Hausdorff measure. Solutions to Problems 18.1–18.7
219
19 The Fourier transform. Solutions to Problems 19.1–19.9
223
20 The Radon–Nikodým theorem. Solutions to Problems 20.1–20.9
233
21 Riesz representation theorems. Solutions to Problems 21.1–21.7
241
22 Uniform integrability and Vitali’s convergence theorem. Solutions to Problems 22.1–22.17
253
23 Martingales. Solutions to Problems 23.1–23.16
267
24 Martingale convergence theorems. Solutions to Problems 24.1–24.9
275
25 Martingales in action. Solutions to Problems 25.1–25.15
283
26 Abstract Hilbert space. Solutions to Problems 26.1–26.19
293
27 Conditional expectations. Solutions to Problems 27.1–27.19
4
311
Solution Manual. Chapter 1–13. Last update 23rd August 2017 28 Orthonormal systems and their convergence behaviour. Solutions to Problems 28.1–28.11
327
5
1 Prologue. Solutions to Problems 1.1–1.5 Problem 1.1 Solution: We have to calculate the area of an isosceles triangle of side-length r, base b, height h and opening angle
:= 2⇡_2j . From elementary geometry we know that cos
so that Since we have lim
2
area (triangle) = ô0
sin
= 1 we find
=
and
h r
sin
1 hb = r2 cos 2
2
2
b 2r
=
sin
2
=
r2 sin . 2
r2 sin 2⇡ 2j jôÿ 2 sin 2⇡ j 2 = 2r ⇡ lim 2⇡2
area (circle) = lim 2j
jôÿ
2j
= 2r2 ⇡ just as we had expected. ∑∑ Problem 1.2 Solution: By construction, Cn+1 = [0, 1] ‰
Hn+1 Õ
I
Õ
i=1 t1 ,…,ti À{0,2}
It1 ,…,ti
and each interval It1 ,…,ti has length 2*i . We have used this when calculating l(Cn+1 ) l(Cn+1 ) = l[0, 1] * 20 ù
1 1 1 * 21 ù 2 * 5 * 2n ù n+1 1 3 3 3
(note that we have removed 2n intervals of length 3*n*1 ). If we let n ô ÿ, we get for all removed intervals
l
Hÿ Õ
Õ
i=1 t1 ,…,ti À{0,2}
I It1 ,…,ti
=
ÿ … i=1
2i*1 ù
1 = 1. 3i
The last line requires -additivity. (Just in case: you will see in the next chapter that the number of removed intervals is indeed countable).
∑∑
7
R.L. Schilling: Measures, Integrals & Martingales Problem 1.3 Solution: We record the lenghts of the removed pieces in each step 1. In Step 1 we remove one (= 20 ) piece of length 12 r;
2. In Step 2 we remove two (= 21 ) pieces, each of length 18 r; 3. In Step 3 we remove four (= 22 ) pieces, each of length n. In Step n we remove 2n pieces, each of length
1 r; 32
1 r; 22n*1
In each step we remove 2n ù 2*2n+1 ù r = 2*n+1 r units of length, i.e. we remove ÿ … r = r. n*1 2 n=1
Thus, l(I) = l[0, 1] * r = 1 * r. This means that the modified Cantor set does have a length! Consequently it cannot be empty. ∑∑ Problem 1.4 Solution: In each step the total length is increased by the factor 4_3, since we remove the middle interval (relative length 1_3) and replace it by two copies constituting the sides of an equilateral triangle (relative length 2_3). Thus, l(Kn ) =
⇠ ⇡n ⇠ ⇡n 4 4 4 ù l(Kn*1 ) = 5 = l(K0 ) = . 3 3 3
In particular, limnôÿ l(Kn ) = ÿ.
Again -additivity comes in in the form of a limit (compare with Problem 1.2). ∑∑ Problem 1.5 Solution: In each step the total area is decreased by the factor 3_4, since we remove the middle triangle (relative area 1_4). Thus,
˘ ⇠ ⇡n ⇠ ⇡n 3 3 3 3 area(Sn ) = ù area(Sn*1 ) = 5 = area(S0 ) = . 4 4 4 4
In particular, area(S) = limnôÿ area(Sn ) = 0.
Again -additivity comes in in the form of a limit (compare with Problem 1.2). Notice that S is not empty as it contains the vertices of all black triangles (see figure) of each stage.
∑∑
8
2 The pleasures of counting. Solutions to Problems 2.1–2.22 Problem 2.1 Solution: (i) We have x À A ‰ B ◊Ÿ x À A and x à B
◊Ÿ x À A and x À B c ◊Ÿ x À A „ B c .
(ii) Using (i) and de Morgan’s laws (*) yields (i)
(A ‰ B) ‰ C = (A „ B c ) „ C c = A „ B c „ C c ( shows that A = B, hence that
is injective.
(B), which is also a contradiction. This, finally
On the other hand, each natural number can be expressed in terms of finite sums of powers of base-2, so that
is also surjective.
Thus, #F = #N. ∑∑ Problem 2.20 Solution: (Let F be as in the previous exercise.) Observe that the infinite sets from P(N), I := P(N) ‰ F can be surjectively mapped onto {0, 1}N : if {a1 , a2 , a3 , …} = A œ N, then define an infinite 0-1-sequence (b1 , b2 , b3 , …) by setting bj = 0 or bj = 1 according to whether
aj is even or odd. This is a surjection of P(N) onto {0, 1}N and so #P(N) Π#{0, 1}N . Call this map
and consider the family
*1 (s),
s À {0, 1}N in I, consisting of obviously disjoint infinite
subsets of N which lead to the same 0-1-sequence s. Now choose from each family
*1 (s)
a
representative, call it r(s) À I. Then the map s ;ô r(s) is a bijection between {0, 1}N and a subset of I, the set of all representatives. Hence, I has at least the same cardinality as {0, 1}N and as such a bigger cardinality than N.
∑∑ Problem 2.21 Solution: Denote by ⇥ the map P(N) Œ A ;ô 1A À {0, 1}N . Let = (d1 , d2 , d3 , …) À {0, 1}N and define A( ) := {j À N : dj = 1}. Then ive.
= (1A( ) (j))jÀN showing that ⇥ is surject-
On the other hand, 1A = 1B ◊Ÿ 1A (j) = 1B (j) ≈j À N ◊Ÿ A = B. 19
R.L. Schilling: Measures, Integrals & Martingales This shows the injectivity of ⇥, and #P(N) = #{0, 1}N follows. ∑∑ Problem 2.22 Solution: Since for A, A® , B, B ® œ X we have the ‘multiplication rule’ (A „ B) ‰ (A® „ B ® ) = (A ‰ A® ) „ (A ‰ B ® ) „ (B ‰ A® ) „ (B ‰ B ® ) and since this rule carries over to the infinite case, we get the formula from the problem by ‘multiplying out’ the countable union
(A01 „ A11 ) ‰ (A02 „ A12 ) ‰ (A03 „ A13 ) ‰ (A04 „ A14 ) ‰ 5 . More formally, one argues as follows: xÀ
Õ nÀN
(*)
(A0n „ A1n ) ◊Ÿ « n0 : x À A0n „ A1n 0
0
while xÀ
Ã
Õ
i=(i(k))kÀN À{0,1}N kÀN
Ai(k) k
◊Ÿ ≈i = (i(k))kÀN À {0, 1}N : x À
Õ kÀN
Ai(k) k
N
i(k0 )
◊Ÿ ≈i = (i(k))kÀN À {0, 1} « k0 À N : x À Ak
0
(**)
Clearly, (*) implies (**). On the other hand, assume that (**) holds but that (*) is wrong, i.e.
suppose that for every n we have that either x À A0n or x À A1n or x is in neither of A0n , A1n . Thus we can construct a uniquely defined sequence i(n) À {0, 1}, n À N, by setting h n0 n i(n) = l1 n n0 j
if x À A0n ; if x À A1n ;
if x à A0n and x à A1n .
Define by i® (n) := 1 * i(n) the ‘complementary’ 0-1-sequence. Then xÀ
Õ n
Ai(n) but x à n
Õ n
®
Ain (n)
contradicting our assumption (**). ∑∑
20
3
-Algebras.
Solutions to Problems 3.1–3.16 Problem 3.1 Solution: (i) It is clearly enough to show that A, B À A
-Ÿ A „ B À A, because the case of N sets
follows from this by induction, the induction step being
A1 „ … „ AN „AN+1 = B „ AN+1 À A. ´≠≠≠≠≠≠Ø≠≠≠≠≠≠¨ =:BÀA
Let A, B À A. Then, by (⌃2 ) also Ac , B c À A and, by (⌃3 ) and (⌃2 ) A „ B = (Ac ‰ B c )c = (Ac ‰ B c ‰ Á ‰ Á ‰ …)c À A. Alternative: Of course, the last argument also goes through for N sets: A1 „ A2 „ … „ AN = (Ac1 ‰ Ac2 ‰ … ‰ AcN )c = (Ac1 ‰ … ‰ AcN ‰ Á ‰ Á ‰ …)c À A. (ii) By (⌃2 ) we have A À A -Ÿ Ac À A. Use Ac instead of A and observe that (Ac )c = A to see the claim.
(iii) Clearly Ac , B c À A and so, by part (i), A ‰ B = A „ B c À A as well as A Ã B = (A ‰ B) ‰ (B ‰ A) À A. ∑∑ Problem 3.2 Solution:
(iv) Let us assume that B ë Á and B ë X. Then B c à {Á, B, X}. Since with B also B c must be contained in a -algebra, the family {Á, B, X} cannot be one.
(vi) Set AE := {E „ A : A À A}. The key observation is that all set operations in AE are now relative to E and not to X. This concerns mainly the complementation of sets! Let us check (⌃1 )–(⌃3 ).
Clearly Á = E „ Á À AE . If B À A, then B = E „ A for some A À A and the complement of B relative to E is E ‰ B = E „ B c = E „ (E „ A)c = E „ (E c ‰ Ac ) = E „ Ac À AE as
Ac À A. Finally, let (Bj )jÀN œ AE . Then there are (Aj )jÀN œ A such that Bj = E „ Aj . ∑ ∑ ∑ ∑ Since A = jÀN Aj À A we get jÀN Bj = jÀN (E „Aj ) = E „ jÀN Aj = E „A À AE . 21
R.L. Schilling: Measures, Integrals & Martingales (vii) Note that f *1 interchanges with all set operations. Let A, Aj , j À N be sets in A. We know
that then A = f *1 (A® ), Aj = f *1 (A®j ) for suitable A, A®j À A ® . Since A ® is, by assumption a -algebra, we have
as Á À A ®
Á = f *1 (Á) À A c
= f *1 (A® ) À A H I Õ Õ Õ Aj = f *1 (A®j ) = f *1 A®j À A Ac = f *1 (A® )
jÀN
jÀN
which proves (⌃1 )–(⌃3 ) for A.
c
jÀN
as A® À A ® c
as
Õ jÀN
A®j À A ®
∑∑ Problem 3.3 Solution:
Denote by ⌃ =
({x}, x À R). Let A be the -algebra defined in Ex-
ample 3.3(v). It is clear that {x} À A, and so ⌃ œ A. On the other hand, if A À A, then either A
or Ac is countable. Wlog assume that A is countable. Then A is a countable union of singletons, as such A À ⌃ as well as Ac À ⌃. This means A œ ⌃.
∑∑ Problem 3.4 Solution: (i) Since G is a -algebra, G ‘competes’ in the intersection of all -algebras C – G appearing in the definition of A in the proof of Theorem 3.4(ii). Thus, G – (G) while G œ (G) is always true.
(ii) Without loss of generality we can assume that Á ë A ë X since this would simplify the
problem. Clearly {Á, A, Ac , X} is a -algebra containing A and no element can be removed without losing this property. Thus {Á, A, Ac , X} is minimal and, therefore, = ({A}).
(iii) Assume that F œ G. Then we have F œ G œ
(G). Now C :=
(G) is a potential
‘competitor’ in the intersection appearing in the proof of Theorem 3.4(ii), and as such C – (F), i.e. (G) – (F).
∑∑ Problem 3.5 Solution: (i) {Á, (0, 12 ), {0} ‰ [ 12 , 1], [0, 1]}.
We have 2 atoms (see the explanations below): (0, 12 ), (0, 12 )c .
(ii) {Á, [0, 14 ), [ 14 , 34 ], ( 34 , 1], [0, 34 ], [ 14 , 1], [0, 14 ) ‰ ( 34 , 1], [0, 1]}. We have 3 atoms (see below): [0, 14 ), [ 14 , 34 ], ( 34 , 1].
(iii) —same solution as (ii)—
Parts (ii) and (iii) are quite tedious to do and they illustrate how difficult it can be to find a -algebra
containing two distinct sets.... imagine how to deal with something that is generated by 10, 20,
or infinitely many sets. Instead of giving a particular answer, let us describe the method to find ({A, B}) practically, and then we are going to prove it.
22
Solution Manual. Chapter 1–13. Last update 23rd August 2017 1. Start with trivial sets and given sets: Á, X, A, B. 2. now add their complements: Ac , B c
3. now add their unions and intersections and differences: A ‰ B, A „ B, A ‰ B, B ‰ A 4. now add the complements of the sets in 3.: Ac „ B c , Ac ‰ B c , (A ‰ B)c , (B ‰ A)c
5. finally, add unions of differences and their complements: (A‰B)‰(B ‰A), (A‰B)c „(B ‰A)c .
All in all one should have 16 sets (some of them could be empty or X or appear several times,
depending on how much A differs from B). That’s it, but the trouble is: is this construction correct? Here is a somewhat more systematic procedure:
Definition: An atom of a -algebra A is a non-void set Á ë A À A that contains no other set of A. Since A is stable under intersections, it is also clear that all atoms are disjoint sets! Now we can make up every set from A as union (finite or countable) of such atoms. The task at hand is to find atoms if A, B are given. This is easy: the atoms of our future -algebra must be: A ‰ B,
B ‰ A, A „ B, (A ‰ B)c . (Test it: if you make a picture, this is a tesselation of our space X using
disjoint sets and we can get back A, B as union! It is also minimal, since these sets must appear in ({A, B}) anyway.) The crucial point is now:
Theorem. If A is a -algebra with N atoms (finitely many!), then A consists of exactly 2N elements. Proof. The question is how many different unions we can make out of N sets. Simple answer: we find
, 0 Õ j Õ N different unions involving exactly j sets (j = 0 will, of course, produce ≥ N the empty set) and they are all different as the atoms were disjoint. Thus, we get N = j=0 j N j
(1 + 1)N = 2N different sets.
It is clear that they constitute a -algebra. This answers the above question. The number of atoms depends obviously on the relative position of A, B: do they intersect, are they disjoint etc. Have fun with the exercises and do not try to find
-algebras generated by three or more sets..... (By the way: can you think of a situation in [0, 1]
with two subsets given and exactly four atoms? Can there be more?)
∑∑ Problem 3.6 Solution: (i) See the solution to Problem 3.5.
(ii) If A1 , … , AN œ X are given, there are at most 2N atoms. This can be seen by induction. If
N = 1, then there are #{A, Ac } = 2 atoms. If we add a further set AN+1 , then the worst case
would be that AN+1 intersects with each of the 2N atoms, thus splitting each atom into two sets which amounts to saying that there are 2 2N = 2N+1 atoms.
23
R.L. Schilling: Measures, Integrals & Martingales ∑∑ Problem 3.7 Solution: We follow the hint. Since #A = #N, the following set is a countable intersection of measurable sets, hence itself in A:
≈x À X : A(x) :=
Ã
A À A.
AÀA,AŒx
(*)
Write A0 for the atoms of A. Then • A(x) À A is an atom which contains x.
Indeed: Otherwise, there is some B œ A(x) such that B À A, B ë Á, B ë A(x). We can
assume that x À B, or we would take B ® := A(x) ‰ B instead of B. Since x À B, B is part of the intersection appearing in (*) so that B – A(x), hence B = A(x), which is impossible.
• Every atom A ë Á of A is of the form (*).
Indeed: By assumption, x0 À A so that A = A(x0 ). • A has #N many atoms. Indeed: Since #A = #N, there are countably infinitely many disjoint sets in A, thus the procedure (*) yields at least #N many atoms. On the other hand, there cannot be more atoms than members of A, and the claim follows.
Since A contains all countable unions of sets from A0 , and since there are more than countably many such unions, it is clear that #A > #N.
Remark: A -algebra may have no non-empty atoms at all! Here is an example (which I learned from Julian Hollender). Let I be an uncountable set, e.g. I = [0, 1], and consider ⌦ = {0, 1}I . We can construct a -algebra on ⌦ in the following way: Let K œ I and define PK : {0, 1}I ô {0, 1}K
the coordinate projection. A cylinder set or finitely based set with basis K œ I is a set of the form PK*1 (B) where #K < ÿ and B œ {0, 1}K . Now consider the -algebra A := ({cylinder sets})
on {0, 1}I . Intuitively, A À A is of the form PL*1 (B) where L is countable. (The proof as such is not obvious, a possible source is Lemma 4.5 in Schilling & Partzsch: Brownian Motion. De
Gruyter, Berlin 2012.) Assume that A0 À A were an atom. Then A0 has the basis L. Take i À I ‰ L, consider L® = L ‰ {i} and construct a set PL*1® (B ® ) where B ® = B ù {0}, say. Then PL*1® (B ® ) œ A0 and PL*1® (B ® ) À A.
∑∑ Problem 3.8 Solution: We begin with an example: Let X = (0, 1] and A = B(0, 1] be the Borel sets. Define
An :=
((j * 1)2*n , j2*n ] , j = 1, 2, … , 2n
the dyadic -algebra of step 2*n . Clearly, #An = 2n . Moreover, An ◊ An+1
24
and Aÿ :=
Õ n
An .
Solution Manual. Chapter 1–13. Last update 23rd August 2017 However, Aÿ is NOT a -algebra. Argument 1: I À Aÿ ◊Ÿ I À An for some n, i.e. I is a finite union of intervals with dyadic end-points. (More precisely: the topological boundary I ‰ I ˝ consists of dyadic points). On the other hand, every open set (a, b) œ [0, 1] is a countable union of sets from Aÿ : Õ
(a, b) =
I
IÀAÿ ,Iœ(a,b)
which follows from the fact that the dyadic numbers are dense in (0, 1]. (If you want it more
elementary, then approximate a and b from the right and left, respectively, by dyadic numbers and construct the approximating intervals by hand....). If, for example, a and b are irrational, then (a, b) Ã Aÿ . This shows that Aÿ cannot be a -algebra. In fact, our argument shows that (Aÿ ) = B(0, 1]. Argument 2: Since #An = 2n we see that #Aÿ = #N. But Problem 3.7 tells us that Aÿ can’t be a -algebra.
Let us now turn to the general case. We follow the note by A. Broughton and B.W. Huff: A comment on unions of sigma-fields. Am. Math. Monthly 84 (1977) 553–554.
Since the An are strictly increasing, we may assume that A1 ë {Á, X}. Recall also the notion of a trace -Algebra
B „ An := {B „ A : A À An }. Step 1. Claim: There exists a set E À A1 such that (E „ An+1 ) ‰ (E „ An ) ë Á for infinitely many n À N.
To see this, assume – to the contrary – that for some n and some B À A1 we have B „ An = B „ An+1
and
B c „ An = B c „ An+1 .
If U À An+1 ‰ An , then U=
(B „ U ) ´Ø¨
ÀB„An+1 =B„An œAn
‰
(B c „ U ) ´≠Ø≠¨
ÀB c „An+1 =B c „An œAn
leading to the contradiction U À An . Thus the claim holds with either E = B or E = B c . Step 2. Let E be the set from Step 1 and denote by n1 , n2 , … a sequence for which the assertion in Step 1 holds. Then
Fk := E „ Ank ,
kÀN 25
R.L. Schilling: Measures, Integrals & Martingales is a strictly increasing sequence of -Algebras over the set E. Again we may assume that F1 ë {Á, E} As in Step 1, we find some E1 À F1 such that E1 is not trivial (i.e. E1 ë Á and E1 ë E) and (E1 „ Fk+1 ) ‰ (E1 „ Fk ) ë Á holds for infinitely many k.
Step 3. Now we repeat Step 2 and construct recursively a sequence of -algebras Ai1 œ Ai2 œ Ai3 … and a sequence of sets E1 – E2 – E3 … such that and
Ek À Aik
Ek+1 À (Ek „ Aik+1 ) ‰ (Ek „ Aik ).
Step 4. The sets Fk := Ek ‰ Ek+1 have the property that they are disjoint and Fk À Aik+1 ‰ Aik . Since the -algebras are increasing, we have Õ nÀN
An =
Õ kÀN
Aik
which means that we can restrict ourselves to a subsequence. This means that we can assume that ik = k. Step 5. Without loss of generality we can identify Fk with {k} and assume that the An are algebras on N such that {k} À Ak+1 ‰ Ak . Let Bn the smallest set in An such that n À Bn . Then n À Bn œ {n, n + 1, n + 2, …} and Bn ë {n}. Moreover m À Bn -Ÿ Bm œ Bn
since m À Bn „ Bm À Am .
Now define n1 = 1 and pick nk+1 recursively: nk+1 À Bnk such that nk+1 ë nk . Then Bn1 – Bn2 –
… . Set E = {n2 , n4 , n6 , …}. If Aÿ were a -algebra, then E À An for some n, thus E À An2k for
some k. Then {n2k , n2k+2 , …} À An2k and thus Bn2k œ {n2k , n2k+2 , …}. This contradicts the fact n2k+1 À Bn2k .
∑∑ Problem 3.9 Solution: O1 Since Á contains no element, every element x À Á admits certainly some neighbourhood B (x) and so Á À O. Since for all x À Rn also B (x) œ Rn , Rn is clearly open.
O2 Let U , V À O. If U „ V = Á, we are done. Else, we find some x À U „ V . Since U , V are open, we find some h := min{ 1 ,
2}
1, 2
> 0 such that B 1 (x) œ U and B 2 (x) œ V . But then we can take
> 0 and find
Bh (x) œ B 1 (x) „ B 2 (x) œ U „ V , i.e. U „ V À O. For finitely many, say N, sets, the same argument works. Notice that already for countably many sets we will get a problem as the radius h := min{ necessarily any longer > 0. 26
j
: j À N} is not
Solution Manual. Chapter 1–13. Last update 23rd August 2017 O2 Let I be any (finite, countable, not countable) index set and (Ui )iÀI œ O be a family of open ∑ sets. Set U := iÀI Ui . For x À U we find some j À I with x À Uj , and since Uj was open, ∑ we find some j > 0 such that B j (x) œ Uj . But then, trivially, B j (x) œ Uj œ iÀI Ui = U proving that U is open.
The family O n cannot be a -algebra since the complement of an open set U ë Á, ë Rn is closed. ∑∑ Problem 3.10 Solution: Let X = R and set Uk := (* k1 , k1 ) which is an open set. Then {0} but a singleton like {0} is closed and not open.
∂
kÀN Uk
=
∑∑ Problem 3.11 Solution: We know already that the Borel sets B = B(R) are generated by any of the following systems:
{[a, b) : a, b À Q}, {[a, b) : a, b À R},
{(a, b) : a, b À Q}, {(a, b) : a, b À R}, O 1 , or C 1 Here is just an example (with the dense set D = Q) how to solve the problem. Let b > a. Since (*ÿ, b) ‰ (*ÿ, a) = [a, b) we get that
{[a, b) : a, b À Q} œ ({(*ÿ, c) : c À Q}) -Ÿ B = ({[a, b) : a, b À Q}) œ ({(*ÿ, c) : c À Q}). On the other hand we find that (*ÿ, a) =
∑
kÀN [*k, a)
proving that
{(*ÿ, a) : a À Q} œ ({[c, d) : c, d À Q}) = B -Ÿ
({(*ÿ, a) : a À Q}) œ B
and we get equality. The other cases are similar. ∑∑ Problem 3.12 Solution: Let B := {Br (x) : x À Rn , r > 0} and let B® := {Br (x) : x À Qn , r À Q+ }. Clearly,
B® œ B œ O n -Ÿ
(B® ) œ (B) œ (O n ) = B(Rn ).
On the other hand, any open set U À O n can be represented by U= Indeed, U –
∑
BÀB® , BœU
Õ BÀB® , BœU
B.
( 0 was arbitrary, we can make ✏ ô 0 and the claim follows. ∑ (iii) Since 0ÕxÕ1 {x} is a disjoint union, only the countability assumption is violated. Let’s see what happens if we could use ‘ -additivity’ for such non-countable unions: 0 1 … … Õ 0= 0= ({x}) = {x} = ([0, 1]) = 1 0ÕxÕ1
0ÕxÕ1
0ÕxÕ1
which is impossible. Problem 4.14 Solution: Without loss of generality we may assume that a ë b; set
∑∑ :=
a
+
b.
Then (B) = 0 if, and only if, a à B and b à B. Since {a}, {b} and {a, b} are Borel sets, all null sets of
are given by
N = B ‰ {a, b} : B À B(R) . (This shows that, in some sense, null sets can be fairly large!). ∑∑ Problem 4.15 Solution: Let us write N for the family of all (proper and improper) subsets of
sets. We note that sets in N can be measurable (that is: N À A) but need not be measurable.
null
(i) Since Á À N, we find that A = A ‰ Á À A for every A À A; thus, A œ A. Let us check that A is a -algebra.
(⌃1 ) Since Á À A œ A, we have Á À A. (⌃2 ) Let A< À A. Then A< = A ‰ N for A À A and N À N. By definition, N œ M À A where (M) = 0. Now
A< c = (A ‰ N)c = Ac „ N c = Ac „ N c „ (M c ‰ M) = (Ac „ N c „ M c ) ‰ (Ac „ N c „ M) 40
Solution Manual. Chapter 1–13. Last update 23rd August 2017 = (Ac „ M c ) ‰ (Ac „ N c „ M) where we use that N œ M, hence M c œ N c , hence M c „ N c = M c . But now we see
that Ac „ M c À A and Ac „ N c „ M À N since Ac „ N c „ M œ M and M À A is a null set: (M) = 0.
(⌃3 ) Let (A 0 there is some G À G such that (A Ã G) Õ ✏. From
Ac à Gc = (Gc ‰ Ac ) ‰ (Ac ‰ Gc ) = (Gc „ A) ‰ (Ac „ G) = (A ‰ G) ‰ (G ‰ A) = AÃG we conclude that (Ac à Gc ) Õ ✏; consequently, Ac À D (observe that Gc À G!). (D3 ) Let (Aj )jÀN œ D be a sequence of mutually disjoint sets and ✏ > 0. Since measure, we get
… jÀN
H (Aj ) =
” jÀN
is a finite
I < ÿ,
Aj
and, in particular, we can pick N À N so large, that ÿ …
(Aj ) Õ ✏.
j=N+1
For j À {1, … , N} there is some Gj À G such that (Aj à Gj ) Õ ✏. Thus, G := ∑N j=1 Gj À G satisfies H
”
jÀN
I
Aj
H
‰G =
”
jÀN
H =
”
jÀN
54
I
„ Gc
Aj I Aj
„
HN Ã j=1
I Gjc
Solution Manual. Chapter 1–13. Last update 23rd August 2017 H
”
=
Aj „
jÀN
N Ã k=1
I Gkc
ÿ ” ” (Aj „ Gjc ) ‰ Aj . N
œ
j=1
In the same way we get
H G‰
j=N+1
I
Õ jÀN
HN ”
œG‰
Aj
j=1
=G„
N Ã
I Aj
Acj
j=1
N Õ (Gj „ Acj ).
œ
j=1
Thus,
HH
”
jÀN
I Aj
I ÃG
HN I ÿ Õ ” (Aj à Gj ) ‰ Aj
Õ
j=1
Õ
N
… j=1
j=N+1
H
(Aj à Gj ) +
ÿ …
j=N+1
I Aj
Õ N✏ + ✏. Since ✏ > 0 is arbitrary, we conclude that
Ω
jÀN Aj
À D.
Obviously, G œ D (take G = A À G). Since G is „-stable, we get A = (G) = (G) œ D. (ii) Using the family D ® := {A À A : ≈ ✏ > 0 « G À G : (A à G) Õ ✏, ⌫(A à G) Õ ✏}, we find, just as in (i), that D ® is a Dynkin system. The rest of the proof is as before. ∑ ∑ (iii) “◊”: Let A À A such that A œ nÀN In and nÀN In Õ ✏. Because of the monotonicity of measures we get
H (A) Õ
and so (A) = 0.
Õ nÀN
I
“Ÿ”: Set K := {A œ Rn : «(Ik )kÀN œ J : A = I ÀKŸ
Ic
À K. Define, furthermore,
Õ ✏,
In ∑
k Ik
or Ac =
D := {A œ Rn : ≈ ✏« J , K À K, J œ A œ K,
∑
k Ik }
and observe that
(K ‰ J ) Õ ✏}.
We claim that D is a Dynkin system. 55
R.L. Schilling: Measures, Integrals & Martingales (D1 ) Clearly, X = Rn À D (take J = K = Rn ). (D2 ) Pick A À D and ✏ > 0. Then there are J , K À K such that J œ A œ K and (K ‰ J ) Õ ✏. From J c , K c À K, (K c ‰ J c ) = (J ‰ K) Õ ✏ and J c – Ac – K c we
get immediately Ac À D.
(D3 ) Let (Aj )jÀN œ D be a sequence of mutually disjoint sets and ✏ > 0. Pick Jj À K and Kj À K such that Jj œ Aj œ Kj , (Kj ‰ Jj ) Õ ✏2*j and set Õ
J :=
jÀN
K :=
Aj
Õ jÀN
Kj .
Since K is stable under countable unions, we get J À K, K À K. Moreover, J œ Ω j Aj œ K and HH
(K ‰ J ) =
I
Õ
HH = HL = H Õ Õ
…
Õ
jÀN
Ã
jÀN
H
jÀN
Õ
H „
Kj
Õ
jÀN
I
Õ
jÀN
„
Kj
jÀN
H
Kj „
à kÀN
Ic I
Jj II Jjc IMI
Jkc
I
(Kj „
Jjc )
(Kj „ Jjc ) jÀN ´≠≠≠Ø≠≠≠¨
(Kj ‰Jj )Õ✏2*j
Õ ✏. Thus,
Ω j
Aj À D.
Finally, J œ D entails that B(Rn ) = (J) œ D. Now let A be a set satisfying (A) = 0. Therefore, for every ✏ > 0 there is a set K✏ = K À K ∑ ∑ such that A œ K and (K) < ✏. If K = i Ii , we are done. If K c = i Ii we have to argue like this: Let J := JR := [*R, R)d À J. Then K=
à i
Iic
and J „ K =
à i
Iic
„J =
à i
J ‰ Ii =
k ÃÃ k i=1
J ‰ Ii
and each set J ‰ Ii is a finite union of sets from J (since J is a semiring), hence
∂k
i=1 J
‰ Ii is a
finite union of sets from J. Since (J „ K) Õ (K) Õ ✏, a continuity-of-measure argument shows ∂ ∂ that there exists some k such that J „ K œ ki=1 J ‰ Ii and ( ki=1 J ‰ Ii ) Õ 2✏.
If we pick ✏ = ✏_2R , we see that we can cover A „ [*R, R)d by a countable union of J-sets, call their union UR , such that (UR ) Õ ✏_2R . Finally, (A) Õ
… RÀN
56
(UR ) Õ ✏
Solution Manual. Chapter 1–13. Last update 23rd August 2017 and we can combine all covers which make up the UR , R À N. ∑∑ Problem 5.13 Solution: (i) mind the misprint: we also need stability of M under finite intersections. Clearly, any
-algebra is also a monotone class. Conversely, if M is a monotone class such that M À
M -Ÿ M c À M, then the condition (⌃2 ) holds, while (⌃1 ) is satisfied by the very definition of a monotone class. If M is also stable under finite intersections, we get M, N À M -Ÿ
M ‰ N = (M c „ N c )c À M, so (⌃3 ) follows from the stability under finite unions and the stability of monotone classes under increasing limits of sets.
(ii) Since (G) is a monotone class containing G, we have – by minimality – that m(G) œ (G). On the other hand, by the monotone class theorem, we get G œ m(G) -Ÿ which means that m(G) = (G).
(G) œ m(G) ∑∑
57
6 Existence of measures. Solutions to Problems 6.1–6.14 Problem 6.1 Solution: (i) Monotonicity: If x Õ 0 Õ y, then F (x) Õ 0 Õ F (y).
If 0 < x Õ y, we have [0, x) œ [0, y) and so 0 Õ F (x) = [0, x) Õ [0, y) = F (y).
If x Õ y < 0, we have [y, 0) œ [x, 0) and so 0 Õ *F (y) = [y, 0) Õ [x, 0) = *F (x), i.e. F (x) Õ F (y) Õ 0. Left-continuity: Let us deal with the case x Œ 0 only, the case x < 0 is analogous (and
even easier). Assume first that x > 0. Take any sequence xk < x and xk ~ x as k ô ÿ. Without loss of generality we can assume that 0 < xk < x. Then [0, xk ) ~ [0, x) and using Proposition 4.3 (continuity of measures) implies
lim F (xk ) = lim [0, xk ) = [0, x) = F (x).
kôÿ
kôÿ
If x = 0 we must take a sequence xk < 0 and we have then [xk , 0) ö [0, 0) = Á. Again by Proposition 4.3, now (iii® ), we get
lim F (xk ) = * lim [xk , 0) = (Á) = 0 = F (0).
kôÿ
kôÿ
which shows left-continuity at this point, too.
We remark that, since for a sequence yk ö y, yk > y we have [0, yk ) ö [0, y], and not [0, y), we cannot expect right-continuity in general.
(ii) Since J = {[a, b), a Õ b} is a semi-ring (cf. the remark preceding Proposition 6.3 or Proposition 6.5) it is enough to check that ⌫F is a premeasure on J. This again amounts to showing (M1 ) and (M2 ) relative to J (mind you: ⌫F is not a measure as J is not a -algebra....). We do this in the equivalent form of Lemma 4.9:
(i) ⌫F (Á) = ⌫F [a, a) = F (a) * F (a) = 0 for any a.
(ii) Let a Õ b Õ c so that [a, b), [b, c) À J are disjoint sets and [a, c) = [a, b) ⁄ [b, c) À J (the latter is crucial). Then we have
⌫F [a, b) + ⌫F [b, c) = F (b) * F (a) + F (c) * F (b) = F (c) * F (a) = ⌫F [a, c) = ⌫F [a, b) ⁄ [b, c) .
59
R.L. Schilling: Measures, Integrals & Martingales (iii® ) (Sufficient since ⌫F is finite for every set [a, b)). Now take a sequence of intervals
[ak , bk ) which decreases towards some [a, b) À J. This means that ak ~ a, ak Õ a and
bk ö b, bk Œ b because the intervals are nested (gives increasing-decreasing sequences). If bk > b for infinitely many k, this would mean that [ak , bk ) ô [a, b] à J since
b À [ak , bk ) for all k. Since we are only interested in sequences whose limits stay in J, the sequence bk must reach b after finitely many steps and stay there to give [a, b). Thus, we may assume directly that we have only [ak , b) ö [a, b) with ak ~ a, ak Õ a. But then we can use left-continuity and get
lim ⌫F [ak , b) = lim (F (b) * F (ak )) = F (b) * F (a)
kôÿ
kôÿ
= ⌫F [a, b). Note that ⌫F takes on only positive values because F increases.
This means that we find at least one extension. Uniqueness follows since ⌫F [*k, k) = F (k) * F (*k) < ÿ (iii) Now let
and [*k, k) ~ R.
be a measure with [*n, n) < ÿ. The latter means that the function F (x),
as defined in part (i), is finite for every x À R. Now take this F and define, as in (ii) a (uniquely defined) measure ⌫F . Let us see that
= ⌫F . For this, it is enough to show
equality on the sets of type [a, b) (since such sets generate the Borel sets and the uniqueness theorem applies....) If 0 Õ a Õ b, ⌫F [a, b) = F (b) * F (a) = [0, b) * [0, a) =
[0, b) ‰ [0, a)
= [a, b)
Ç
If a Õ b Õ 0, ⌫F [a, b) = F (b) * F (a) = * [b, 0) * (* [a, 0)) = [a, 0) * [b, 0) =
[a, 0) ‰ [b, 0)
= [a, b)
Ç
If a Õ 0 Õ b, ⌫F [a, b) = F (b) * F (a) = [0, b)) * (* [a, 0)) = [a, 0)) + [0, b) =
[a, 0) ⁄ [0, b)
= [a, b)
60
Ç
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (iv) F : R ô R with F (x) = x, since [a, b) = b * a = F (b) * F (a). h n0, x Õ 0 (v) F : R ô R, with, say, F (x) = l = 1(0,ÿ) (x) since 0 [a, b) = 0 whenever n1, x > 0 j a, b < 0 or a, b > 0. This means that F must be constant on (*ÿ, 0) and (0, ÿ) If a Õ 0 < b we have, however,
= 1 which indicates that F (x) must jump by 1 at the point 0.
0 [a, b)
Given the fact that F must be left-continuous, it is clear that it has, in principle, the above
form. The only ambiguity is, that if F (x) does the job, so does c + F (x) for any constant c À R. (vi) Assume that F is continuous at the point x. Then 0 ⇡1 Ã⌧ 1 ({x}) = x, x + k kÀN
⇡⇡ x, x + k1 kôÿ ⇠ ⇠ ⇡ ⇡ def = lim F x + k1 * F (x) kôÿ ⇠ ⇡ = lim F x + k1 * F (x) 4.3
⇠⌧
= lim
( 0 a ∑ œ A such that B ✏ := j Bj✏ – Q and, because of -subadditivity, < (Q)
(B ✏ ) *
<
(Q) Õ
… j
Set B :=
∂
kB
1_k
(Bj✏ ) *
À A. Then B – Q and (B) =
<
< (B)
(Q) Õ ✏.
=
< (Q).
Now let N À A and N œ B ‰ Q. Then B ‰ N – B ‰ (B ‰ Q) = B „ [(B „ Qc )c ] = B „ [B c ‰ Q] =B„Q 62
Solution Manual. Chapter 1–13. Last update 23rd August 2017 = Q. So,
<
<
(Q) * (N) = (B) * (N) = (B ‰ N) =
(B ‰ N) Œ
<
(Q)
which means that (N) = 0. If
< (Q)
= ÿ, we take the exhausting sequence (Ak )kÀN œ A with Ak ~ X and (Ak ) < ÿ
and set Qk := Ak „ Q for every k À N. By the first part we can find sets Bk À A with Bk – Qk , (Bk ) =
< (Q ) k
and (N) = 0 for all N À A with N œ Bk ‰ Qk . Without loss
of generality we can assume that Bk œ Ak , otherwise we replace Bk by Ak „ Bk . Indeed, Bk „ Ak – Qk , Bk „ Ak À A, <
<
(Qk ) = (Bk ) Œ (Ak „ Bk ) Œ
(Qk )
and Bk ‰ Qk – (Bk „ Ak ) ‰ Qk , i.e. we have again that all measurable N œ (Bk „ Ak ) ‰ Qk satisfy (N) = 0.
Assume now that N œ B ‰ Q, B = ∑ have N = k Nk as well as
∑
k Bk
and N À A. Then Nk := N „ Bk À A and we
Nk = N „ Bk œ (B ‰ Q) „ Bk = Bk ‰ Q = Bk ‰ Qk . Thus (Nk ) = 0 and, by -subadditivity, (N) Õ
(ii) Define Ñ :=
≥ÿ
0. 71
R.L. Schilling: Measures, Integrals & Martingales If 2n À T *1 (A), n > 0, then we see that 2n + 2 = 2(n + 1) À A. As A À A this yields 2n + 3 À A and so 2n + 1 = T *1 (2n + 3) À T *1 (A). Therefore, T is measurable.
On the other hand, T *1 is not measurable: the set A = {k; k Õ 0} is contained in A, but T (A) = {k : k Õ 2} Ã A (use 2 = 2 1 À A, but 2 1 + 1 = 3 Ã A).
∑∑ Problem 7.4 Solution: (i) First of all we remark that Ti*1 (Ai ) is itself a -algebra, cf. Example 3.3(vii).
If C is a -algebra of subsets of X such that Ti : (X, C) ô (Xi , Ai ) becomes measurable,
we know from the very definition that T *1 (Ai ) œ C. From this, however, it is clear that T *1 (Ai ) is the minimal -algebra that renders T measurable.
(ii) From part (i) we know that (Ti , i À I) necessarily contains Ti*1 (Ai ) for every i À I. Since ⇠∑ ⇡ ∑ *1 *1 (A ) œ (T , i À I). On the T (A ) is, in general, not a -algebra, we have T i i i i i i i ∑ *1 *1 other hand, each Ti is, because of Ti (Ai ) œ i Ti (Ai ) œ (Ti , i À I) measurable w.r.t. ⇠∑ ⇡ *1 (A ) and this proves the claim. T i i i ∑∑ Problem 7.5 Solution: (i), (ii) 1T *1 (A® ) (x) = 1 € x À T *1 (A® ) € T (x) À A® € 1A® (T (x)) = 1 € (1A® ˝T )(x) = 1 Since an indicatior function can only assume the values 0 and 1, the claimed equality follows for the value 0 by negating the previously shown equivalence.
(iii) “Ÿ”: Assume that T is measurable. We have T *1 (A® ) À A ≈A® À A ® and since A is a -algebra, we conclude
(T ) = ({T *1 (A® ) A® À A ® }) œ (A) = A. “◊”: (T ) œ A implies, in particular, T *1 (A® ) À A ≈A® À A ® , i.e., T is measurable. (iii) Theorem 7.6 shows that image measures are measures. By the definition of T , we have
T *1 (E ® ) = E and ⌫˝T *1 (E ® ) < ÿ, resp., ⌫˝T *1 (E ® ) = 1 follows from the definition of image measures.
The image measure obtained from a -finite measure need not be -finite! Counterexemple: Let While 72
be the counting measure on Z2 and define T ((x, y)) = x.
is -finite, the image measure T ( ) isn’t.
Solution Manual. Chapter 1–13. Last update 23rd August 2017 ∑∑ Problem 7.6 Solution: We have T *1 (G) œ T *1 ( (G)) ´≠≠≠Ø≠≠≠¨
-Ÿ
(T *1 (G)) œ T *1 ( (G)).
is itself a -algebra
For the converse consider T : (X, (T *1 (G))) ô (Y , (G)). By the very choice of the -algebras and since T *1 (G) œ
(T *1 (G)) we find that T is (T *1 (G))_ (G) measurable—mind that we
only have to check measurability at a generator (here: G) in the image region. Thus, T *1 ( (G)) œ (T *1 (G)). Alternative: We have T *1 (G) œ T *1 ( (G)) ´≠≠≠Ø≠≠≠¨
-Ÿ
(T *1 (G)) œ T *1 ( (G)).
is itself a -algebra
For the converse, set ⌃ := {G À (G) : T *1 (G) À (T *1 (G))}. It is not hard to see that ⌃ is itself a -algebra and that G œ ⌃ œ (G). Thus, (G) = ⌃ and so T *1 ( (G)) œ (T *1 (G)).
∑∑ Problem 7.7 Solution: We have to show that f : (F , F) ô (X, (Ti , i À I)) measurable ◊Ÿ
≈ i À I : Ti ˝f : (F , F) ô (Xi , Ai ) measurable.
Now ≈ i À I : (Ti ˝f )*1 (Ai ) œ F ◊Ÿ ≈ i À I : f *1 Ti*1 (Ai ) œ F ⇠Õ ⇡ ◊Ÿ f *1 Ti*1 (Ai ) œ F iÀI 4 ⇠Õ ⇡5 ( ↵} we find some (symmetric) neighbourhood (a ‘ball’) of the type (x * h, x + h) œ {u > ↵}. What does this
mean? Obviously, that u(y) > ↵ for any y À (x * h, x + h) and, in other words, u(y) > ↵ whenever y is such that x * y < h. And this is the hint of how to use continuity: we use it in order to find the value of h.
u being continuous at x means that ≈✏ > 0
« >0
≈y : x * y <
: u(x) * u(y) < ✏.
Since u(x) > ↵ we know that for a sufficiently small ✏ we still have u(x) Œ ↵ + ✏. Take this ✏ and find the corresponding . Then
u(x) * u(y) Õ u(x) * u(y) < ✏
≈ x * y < 81
R.L. Schilling: Measures, Integrals & Martingales and since ↵ + ✏ Õ u(x) we get ↵ + ✏ * u(y) < ✏
≈ x * y <
i.e. u(y) > ↵ for y such that x * y < . This means, however, that h =
does the job. ∑∑
Problem 8.8 Solution: The minimum/maximum of two numbers a, b À R can be written in the form 1 a + b * a * b 2 1 max{a, b} = a + b + a * b 2 min{a, b} =
which shows that we can write min{x, 0} and max{x, 0} as a combination of continuous functions. As such they are again continuous, hence measurable. Thus, u+ = max{u, 0},
u* = * min{u, 0}
are compositions of measurable functions, hence measurable. ∑∑ Problem 8.9 Solution: (i) From the definition of the supremum we get sup fi (x) > i
◊Ÿ «i0 À I : fi0 (x) > ◊Ÿ «i0 À Ifi0 (x) > Õ ◊Ÿ x À {fi > }. i
(ii) Let x À {supi fi < }. Then we have fj (x) Õ supiÀI fi (x) < ∂ means x À {fj < } for all j À I and so x À jÀI {fj < }.
for all j À I; this
(Note: ‘–’ is, in general, wrong. To see this, use e.g. fi (x) := * 1i , i À N, and ∂ = 0. Then we have {supi fi < 0} = Á ë E = i {fi < 0}.) ∑ (iii) Let x À i {fi Œ }. Then there is some i0 À I such that x À {fi0 Œ }, hence sup f (x) Œ fi0 (x) Œ . iÀI
(iv) This follows from sup fi (x) Õ iÀI
◊Ÿ ≈i À I : fi (x) Õ ◊Ÿ ≈i À I : x À {fi Õ } à ◊Ÿ x À {fi Õ }. iÀI
82
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (v)–(viii) can be proved like parts (i)–(iv). ∑∑ Problem 8.10 Solution: The fj are step-functions where the bases of the steps are the sets Ajk and Aj . Since they are of the form, e.g. k2*j Õ u < (k + 1)2*j = k2*j Õ u „ u < (k + 1)2*j , it is clear that they are not only in A but in (u).
∑∑ Problem 8.11 Solution: Corollary 8.12 If u± are measurable, it is clear that u = u+ * u* is measurable since differences of measurable functions are measurable.
(For the converse we could use the previous Problem 8.10, but we give an alternative proof...) Conversely, let u be measurable. Then sn ~ u (this is short for: limnôÿ sn (x) = u(x) and this
+ is an increasing limit) for some sequence of simple functions sn . Now it is clear that s+ n ~u ,
+ + * * + and s+ n is simple, i.e. u is measurable. As u = u * u we conclude that u = u * u is again
measurable as difference of two measurable functions. (Notice that in no case ‘ÿ * ÿ’ can occur!)
Corollary 8.13 This is trivial if the difference u * v is defined. In this case it is measurable as difference of measurable functions, so
{u < v} = {0 < u * v} etc. is measurable. Let us be a bit more careful and consider the case where we could encounter expressions of the type ‘ÿ * ÿ’. Since sn ~ u for simple functions (they are always R-valued...) we get ( 0 n1 ˘ ˘ ˘ ˘ n ([(* b) ‚ (*1), a) ‰ [ a, b · 1)), 0 < a < 1 j2 h b Õ 0 or a > 1 n0, =l ˘ ˘ n2 12 ([0 ‚ a · 1, b · 1)), otherwise j h b Õ 0 or a > 1 n0, =l˘ ˘ n( b · 1)) * (0 · a · 1), otherwise j ∑∑ Problem 8.18 Solution:
87
R.L. Schilling: Measures, Integrals & Martingales • clear, since u(x * 2) is a combination of the measurable shift ⌧2 and the measurable function u. • this is trivial since u ;ô eu is a continuous function, as such it is measurable and combinations of measurable functions are again measurable.
• this is trivial since u ;ô sin(u + 8) is a continuous function, as such it is measurable and combinations of measurable functions are again measurable.
• iterate Problem 8.13 • obviously, sgn x = (*1) 1(*ÿ,0) (x) + 0 1{0} (x) + 1 1(0,ÿ) (x), i.e. a measurable function. Using the first example, we see now that sgn u(x * 7) is a combination of three measurable functions.
∑∑ Problem 8.19 Solution: Betrachte zum Beispiel T : [0, 1) ô [0, 1) mit T (x) = R mit wn (x) =
(*1)n 1
[1_2,1) (x).
x 2
und wn : [0, 1) ô ∑∑
Problem 8.20 Solution: Let A œ R be such that A à B. Then it is clear that u(x) = 1A (x)*1Ac (x) is NOT measurable (take, e.g. A = {f = 1} which should be measurable for measurable functions), but clearly, f (x) = 1 and as constant function this IS measurable.
∑∑ Problem 8.21 Solution: We want to show that the sets {u Õ ↵} are Borel sets. We will even show that they are intervals, hence Borel sets. Imagine the graph of an increasing function and the line y = ↵ cutting through. Essentially we have three scenarios: the cut happens at a point where (a) u is continuous and strictly increasing or (b) u is flat or (c) u jumps—i.e. has a gap; these three cases
are shown in the following pictures: From the three pictures it is clear that we get in any case an 6
6
↵ -
b a
c
interval for the sub-level sets {u Õ } where
6
↵ b
-
a
is some level (in the pic’s
b
a
↵ -
= ↵ or = ), you can
read off the intervals on the abscissa where the dotted lines cross the abscissa.
Now let’s look at the additional conditions: First the intuition: From the first picture, the continuous and strictly increasing case, it is clear that we can produce any interval (*ÿ, b] to (*ÿ, a] by looking at {u Õ } to {u Õ ↵} my moving up the -line to level ↵. The point is here that we get all intervals, so we get a generator of the Borel sets, so we should get all Borel sets. 88
Solution Manual. Chapter 1–13. Last update 23rd August 2017 The second picture is bad: the level set {u Õ } is (*ÿ, b] and all level sets below will only come up to the point (*ÿ, c], so there is no chance to get any set contained in (c, b), i.e. we cannot get all Borel sets.
The third picture is good again, because the vertical jump does not hurt. The only ‘problem’ is
whether {u Õ } is (*ÿ, b] or (*ÿ, b) which essentially depends on the property of the graph whether u(b) =
or not, but this is not so relevant here, we just must make sure that we can get
more or less all intervals. The reason, really, is that jumps as we described them here can only happen countably often, so this problem occurs only countably often, and we can overcome it therefore.
So the point is: we must disallow flat bits, i.e. (u) is the Borel -algebra if, and only, if u is strictly
increasing, i.e. if, and only if, u is injective. (Note that this would have been clear already from Problem 8.15, but our approach here is much more intuitive.)
∑∑ Problem 8.22 Solution: For every n À N the function n …
gn (x) :=
i=1
2*i 1Gi (x),
x À X,
is A_B(R)-measurable. Therefore, g = limnôÿ gn is A_B(R)-measurable (pointwise limit of measurable functions), and so (g) œ A. For the inclusion A œ (g) we define ⌃ := {A À A : A À (g)}. ⌃ ist a -Algebra: (⌃1 ) X À ⌃ since X À A and X À (g). (⌃2 ) For A À ⌃ we have A À Ac À ⌃. (⌃3 ) For (An )nÀN œ ⌃ we see
(g); since (g) is a -algebra, we see that Ac À
∑
nÀN An
À (g), thus
∑
n An
(g); hence,
À ⌃.
Since Gi = {g = 2*i } À (g) we see that G œ ⌃. Consequently, A = (G) œ (g). ∑∑ Problem 8.23 Solution: Without loss of generality, assume that u is right-continuous (left-continuity works analogously). Approximate u with simple functions: 2
un (x) :=
2n … i=1
u(xni+1 )1[xn ,xn ) (x) i
i+1
where xni := *n + ni . The functions un are obviously Borel measurabl. We claim: u(x) = lim un (x). nôÿ
89
R.L. Schilling: Measures, Integrals & Martingales Indeed: For each x À R there is some N À N such that x À [*N, N]. By definition, we find for all n Œ N,
0 un (x) = u
(
‚nx„+1 n
‚nx„ + 1 n
1
is the smallest number of the form kn , k À Z, which exceeds x.) Because of the right-
continuity of u we get un (x) ô u(x) as n ô ÿ. Therefore, u is Borel-measurable (pointwise limit of measurable functions).
∑∑ Problem 8.24 Solution: Every linear map on a finite-dimensional vector space is continuous, hence Borel measurable.
Note that f : R ô R2 , f (x) := (x, 0)Ò , is continuous, hence Borel measurable. This map is, however, not measurable with respect to the completed Borel -algebras:
To see this, let A œ R, A à B(R), be a subset of a Lebesgue null set. For A ù {0} we see that A ù {0} À B(R2 ); this follows from A ù {0} œ N := R ù {0} and Problem 4.15, Problem 6.7). On the other hand,
f : (R, B(R)) ô (R2 , B(R2 )) is not measurable.
f *1 (A
2 (N)
= 0 (cf.
ù {0}) = A à B(R) œ B(R), i.e.
∑∑ Problem 8.25 Solution: Without loss of generality we consider the right-continuous situation. The left-continuous counterpart is very similar.
• Fix ! À ⌦. Note that it is enough to show that t ;ô ⇠(t, !)1[a,b] (t) =: ⇠ a,b (t, !) is measurable for all a < b.
Indeed: Because of ⇠(t, !) = lim ⇠ *R,R (t, !) Rôÿ
the map t ;ô ⇠(t, !) is measurable (pointwise limit of measurable functions, cf. Corollary 8.10.
In order to keep notation simple, we assume that a = 0 and b = 1; the general case is similar. Define
… ⇠ i+1 ⇡ ⇠ 2n , ! 14
2n *1
⇠n (t, !) := For any t À [0, 1] we have
‚2n t„+1 2n
0 ⇠n (t, !) = ⇠
i=0
i i+1 , ·1 2n 2n
1 (t).
ö t, and because of right-continuity,
1 t À [0, 1] ‚2n t„ + 1 , ! ,,,,,,,,,ô , ⇠(t, !) = ⇠ 0,1 (t, !). nôÿ 2n
For t à [0, 1] we have ⇠n (t, !) = 0 = ⇠ 0,1 (t, !) and, thus, ⇠ 0,1 (t, !) = lim ⇠n (t, !) nôÿ
90
≈t À R, ! À ⌦.
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Consequently, it is enough to show (by Corollary 8.10) that each t ;ô ⇠n (t, !) is measurable. For ↵ À R we get
1 i i+1 {t : ⇠n (t, !) Õ ↵} = , À B(R) 2n 2n iÀI ´≠≠≠≠Ø≠≠≠≠¨ Õ
4
ÀB(R)
where
$ ⇠ ⇡ % i+1 I := i À {0, … , 2n * 1}; ⇠ , ! Õ ↵ . 2n This proves that t ;ô ⇠n (t, !) is measurable.
• Since t ;ô ⇠(t, !) is right-continuous, we have
sup ⇠(t, !) = sup ⇠(t, !). tÀR
tÀQ
(?)
Indeed: The estimate ‘Œ’ is clear, i.e. we only have to show ‘Õ’. Using the definition of the supremum, there is for each ✏ > 0 some s À R such that
⇠(s, !) Œ sup ⇠(t, !) * ✏. tÀR
Because of right-continuity we find some r À Q, r > s, such that ⇠(r, !) * ⇠(s, !) Õ ✏. Therefore,
sup ⇠(t, !) Œ ⇠(r, !) Œ ⇠(s, !) * ✏ Œ sup ⇠(t, !) * 2✏. tÀQ
Since ✏ > 0 is arbitrary, the claim follows.
tÀR
From (?) we get that the map ! ;ô suptÀR ⇠(t, !) is measurable (as supremum of countably many measurable functions, cf. Corollary 8.10).
∑∑ Problem 8.26 Solution: ‘◊’: Assume that there are A_B(R)-measurable functions f , g : X ô R satisfying f Õ
Õ g and {f ë g} = 0. For any x À R we get
{ Õ x} = { Õ x, f = g} ‰ { Õ x, f ë g}
= {g Õ x, f = g} ‰ { Õ x, f ë g} . ´≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠¨ ´≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠¨ =:A
=:N
Since f and g are measurable, we see that A À A. For N we only get N œ {f ë g}, i.e. N is a subset of a -null set. By the definition of A (see Problem 4.15) we find { Õ x} À A. ‘Ÿ’: Assume, first, that
is a simple function, i.e. (x) =
N … i=1
ci 1Ai (x),
x À X,
with ci À R, Ai À A (i = 1, … , n). From the definition of A we get that the Ai are of the form Ai = Bi + Ni 91
R.L. Schilling: Measures, Integrals & Martingales with Bi À A and Ni being a subset of a -null set Mi À A. Define f (x) :=
n … i=1
ci 1Bi (x),
g(x) :=
n … i=1
ci 1Bi ‰Mi (x),
x À X.
These are clearly A_B(R)-measurable functions and f Õ Õ g. Moreover, H n I n Õ … (f ë g) Õ Mi Õ (Mi ) = 0. i=1
i=1
This proves that the claim holds for simple functions. Let
be any A_B(R)-measurable function. Using Corollary 8.9, we get a sequence (
A_B(R)-measurable simple functions such that
n )nÀN
ô (x) for all x À X. By the first part of
n (x)
this proof, there are A_B(R)-messbare Funktionen fn , gn , n À N, such that mit fn Õ and (fn ë gn ) = 0. Set
f (x) := lim inf fn (x), nôÿ
g(x) := lim inf gn (x), nôÿ
(f ë g) Õ
H
Õ
nÀN
I
{fn ë gn }
Õ
… nÀN
n
Õ gn
x À X.
The functions f and g are again A_B(R)-measurable (Corollary 8.10) and we have f Õ Moreover,
of
Õ g.
(fn ë gn ) = 0. ∑∑
92
9 Integration of positive functions. Solutions to Problems 9.1–9.14 Problem 9.1 Solution: We know that for any two simple functions f , g À E+ we have I (f + g) = I (f ) + I (g) (= additivity), and this is easily extended to finitely many, say, m different positive simple functions. Observe now that each ⇠n 1An is a positive simple function, hence H I
m … n=1
I ⇠n 1 An
=
m … n=1
I
m m ⇠ ⇡ … ⇠ ⇡ … ⇠n 1An = ⇠n I 1An = ⇠n n=1
n=1
An .
Put in other words: we have used the linearity of I . ∑∑ Problem 9.2 Solution: We use indicator functions. Note that any fixed x can be contained in k À {0, 1, … , N} of the sets An . Then x is contained in A1 ‰ 5 ‰ AN as well as in An ‰ Ak where n < k; as usual: … n
1An
m n
= 0 if m < n. This gives
k 2
of the pairs
0 1 … k =kÕ1+ = 1A1 ‰5‰AN + 1An 1Ak 2 nuŒk} = (u * 1)1{uŒ1} Œ u1{uŒ1} * 1{uŒ0} . k=1
ÿ … i=1
1{uŒi} Õ u1{uŒ1} Õ u Õ 1 +
ÿ …
1{uŒi} =
i=1
ÿ …
1{uŒi} .
i=0
Second proof: For x À X, there is some k À N0 such that k Õ u(x) < k + 1. Therefore, x À {u Œ i} ≈i À {0, … , k} and x à {u Œ i} ≈i Œ k + 1. 103
R.L. Schilling: Measures, Integrals & Martingales Thus,
…
1{uŒi} (x) = k + 1.
iÀN0
Since k Õ u(x) Õ k + 1 we get …
H 1{uŒi} (x) = k + 1 Œ u(x) Œ k = (k + 1) * 1 =
iÀN0
…
I 1{uŒi} (x)
* 1.
iÀN0
As 1 = 1{uŒ0} (u Œ 0, by assumption) we get the claimed estimates. Integrating these inequalities we get ÿ …
{u Œ i} Õ
i=1
u d Õ
ÿ …
{u Πi},
i=0
≥ and (ii) follows. If u À L1 ( ), then we get iŒ1 (u Œ 1) < ÿ. On the other hand, if u is ≥ measurable, and i (u Œ i) < ÿ, then we get î u d < ÿ, i.e. u À L1 ( ) and (i) follows. was only used for î 1 d < ÿ or {u Œ 0} < ÿ – which is only
The finiteness of the measure
needed for the second estimate in (ii). Hence, the lower estimate in (ii) holds for all measures!
∑∑ Problem 10.7 Solution: One possibility to solve the problem is to follow the hint. We provide an alternative (and shorter) solution.
(i) Observe that uj * v Œ 0 is a sequence of positive and integrable functions. Applying Fatou’s
lemma (in the usual form) yields (observing the rules for lim inf, lim sup from Appendix A, compare also Problem 9.10):
lim inf uj d * j
vd =
lim inf (uj * v) d j
Õ lim inf
(uj * v) d
= lim inf
uj d *
j j
vd
and the claim follows upon subtraction of the finite (!) number î v d . (ii) Very similar to (i) by applying Fatou’s lemma to the positive, integrable functions w*uj Œ 0:
wd *
lim sup uj d = j
lim inf (w * uj ) d j
Õ lim inf j
=
(w * uj ) d
w d * lim sup
Now subtract the finite number î w d on both sides. 104
j
uj d
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (iii) We had the counterexample, in principle, already in Problem 9.10. Nevertheless... Consider Lebesgue measure on R. Put fj (x) = *1[*2j,*j] (x) and gj (x) = 1[j,2j] (x). Then lim inf fj (x) = 0 and lim sup gj (x) = 0 for every x and neither admits an integrable minorant resp. majorant.
∑∑ Problem 10.8 Solution: For u = 1B and v = 1C we have, because of independence,
uv dP = P (A „ B) = P (A)P (B) =
For positive, simple functions u =
≥ j
uv dP =
… j,k
=
… j,k
=
… j,k
0 =
k
↵j
k P (Aj
↵j
k P (Aj )P (Bk )
…
v dP .
we find
1Aj 1Bk dP „ Bk )
10 ↵j P (Aj )
u dP
u dP
k k 1 Ck
↵j
j
=
≥
↵j 1Bj and v =
… k
1 k P (Bk )
v dP .
For measurable u À M+ (B) and v À M+ (C) we use approximating simple functions uk À E + (B), uk ~ u, and vk À E + (C), vk ~ v. Then, by Beppo Levi,
uv dP = lim k
uk vk dP = lim k
=
uk dP lim j
u dP
vj dP
v dP .
Integrable independent functions: If u À L1 (B) and v À L1 (C), the above calculation when applied to u, v shows that u v is integrable since
uv dP Õ
u dP
v dP < ÿ.
Considering positive and negative parts finally also gives
uv dP =
u dP
v dP .
Counterexample: Just take u = v which are integrable but not square integrable, e.g. u(x) = v(x) = x*1_2 . Then î(0,1) x*1_2 dx < ÿ but î(0,1) x*1 dx = ÿ, compare also Problem 10.2.
∑∑ Problem 10.8 Solution:
105
R.L. Schilling: Measures, Integrals & Martingales (i) Since the map g : C ô R2 is continuous, we have g *1 (B(R2 )) œ B(C). On the other hand, for z À C and ✏ > 0 we have B✏ (z) = g *1 (Bg(z) (✏)) À g *1 (B(R2 )); thus, (OC ) œ g *1 (B(R2 )) (Note that the -algebra (OC ) is generated by the open balls B✏ (z), z À C, ✏ > 0, cf. the proof of Problem 3.12.)
(ii) Part (i) shows that a map h : E ô C is A_C-measurable if, and only if, g˝h : E ô R2 is A_B(R2 )-measurable.
Indeed: The map h : (E, A) ô (C, C) is, by definition, measureable if h*1 (A) À A for all A À C. Since C = g *1 (B(R2 )), this is the same as h*1 (g *1 (B)) = (g˝h)*1 (B) À A
for all B À B(R2 ), hence it is the same as the measurability of g˝h.
"Ÿ": Assume that h : E ô C is A_C-measurable. Then we have that H I Re h (g˝h) = Im h is A_B(R2 )-measurable. Since the projections ⇡j : R2 Œ (x1 , x2 ) ;ô xj À R are
Borel measurable (due to continuity!), we get that Re h = ⇡1 (g˝h) and Im h = ⇡2 (g˝h) are measurable (composition of measurable functions).
"◊": Assume that Re h and Im h are A_B(R)-measurable. Then the map (g˝h) = (Re h, Im h) is A_B(R2 )-measurable. With the above arguments we conclude that h : (E, A) ô (C, C) is measurable.
(iii) We show first additivity: let g, h À L1C ( ). From Re(g +h) Õ Re g+ Re h À L1 ( ),
Im(g +h) Õ Im(g)+ Im(h) À L1 ( )
we conclude that g + h À L1 ( ). Since Re(g + h) = Re(g) + Re(h) and Im(g + h) = Im(g) + Im(h), we get from the definition of the integral
(g + h) d =
Re(g + h) d + i
=
(Re(g) + Re(h)) d + i(Im(g) + Im(h)) d
=
0
= =
Im(g + h) d
Re(h) d + i Im(g) d + i Im(h) d 1 0 1 Re(g) d + i Im(g) d + Re(h) d + i Im(h) d Re(g) d +
gd +
hd .
Note that we have used the R-linearity of the integral for real-valued functions. The homogeneity of the complex integral is shown in a very similar way.
(iv) Since Re h and Im h are real, we get î Re h d À R and î Im h d À R. Therefore, 0 1 0 1 Re hd = Re Re h d + i Im h d 106
Solution Manual. Chapter 1–13. Last update 23rd August 2017 = Similarly, we see
0
Im
Re h d .
1 hd
0 = Im =
Re h d + i
1 Im h d
Im h d .
(v) We follow the hint: as î h d À C we can pick some ✓ À (*⇡, ⇡] such that ei✓ î h d Œ 0. Thus, (iii) and (iv) entail
Û Û hd Û Û
Û Û = ei✓ hd Û Û 0 = Re e
i✓
1 hd
=
Re(ei✓ h) d
Õ
ei✓ h d
=
h d .
(vi) We know from (ii) that h : (E, A) ô (C, C) is measurable if, and only if, Re h and Im h are A_B(R2 )-measurable. If Re h and Im h are -integrable, then so is h =
˘ (Re h)2 + (Im h)2 Õ Re h + Im h.
If h À L1R ( ), then we conclude from Re h Õ h and Im h Õ h, that Re h and Im h are -integrable.
∑∑
107
11 Null sets and the ‘almost everywhere’. Solutions to Problems 11.1–11.12 Problem 11.1 Solution: True, we can change an integrable function on a null set, even by setting it to the value +ÿ or *ÿ on the null set. This is just the assertion of Theorem 11.2 and its Corollaries 11.3, 11.4.
∑∑ Problem 11.2 Solution: We have seen that a single point is a Lebesgue null set: {x} À B(R) for all x À R and ({x}) = 0, see e.g. Problems 4.13 and 6.7. If N is countable, we know that Ω N = {xj : j À N} = jÀN {xj } and by the -additivity of measures H (N) =
”
jÀN
I {xj }
=
…
{xj } =
jÀN
…
0 = 0.
jÀN
The Cantor set C from Problem 7.12 is, as we have seen, uncountable but has measure (C) = 0. This means that there are uncountable sets with measure zero. In R2 and for two-dimensional Lebesgue measure
2
the situation is even easier: every line L in
the plane has zero Lebesgue measure and L contains certainly uncountably many points. That 2 (L)
= 0 is seen from the fact that L differs from the ordinate {(x, y) À R2 : x = 0} only
2 ({x
= 0}) = 0 as seen in Problem 6.7.
by a rigid motion T which leaves Lebesgue measure invariant (see Chapter 4, Theorem 4.7) and
∑∑ Problem 11.3 Solution: (i) Since {u > c} œ {u Œ c} and, therefore, ({u > c}) Õ
({u Πc}), this follows
immediately from Proposition 11.5. Alternatively, one could also mimic the proof of this Proposition or use part (iii) of the present problem with (t) = t, t Π0.
(ii) This will follow from (iii) with (t) = tp , t Œ 0, since ({u > c}) Õ
({u Πc}) as
{u > c} œ {u Œ c}. (iii) We have, since
is increasing, ({u Πc}) = ({ (u) Π(c)}) =
1{x:
(u(x))Π(c)} (x)
(dx)
109
R.L. Schilling: Measures, Integrals & Martingales =
Õ
Õ
=
(u(x)) 1 (u(x)) {x: (u(x))Π(u(x)) 1{x: (u(x))Π(c) (u(x)) (dx) (c)
1 (c)
(c)} (x)
(dx)
(c)} (x)
(dx)
(u(x)) (dx)
(iv) Let us set b = ↵ î u d . Then we follow the argument of (iii), where we use that u and b are strictly positive.
({u Πb}) =
1{x:u(x)Œb} (x) (dx)
u(x) 1 (x) (dx) u(x) {x:u(x)Œb} u(x) Õ 1 (x) (dx) b {x:u(x)Œb} u Õ d b 1 = ud b =
and substituting ↵ î u d for b shows the inequality. (v) Using the fact that
is decreasing we get {u < c} = { (u) >
(c)}—mind the change
of the inequality sign—and going through the proof of part (iii) again we use there that increases only in the first step in a similar role as we used the decrease of that the argument of (iii) is valid after this step and we get, altogether, ({u < c}) = ({ (u) > =
=
Õ
Õ
=
1{x:
(c)})
(u(x))> (c)} (x)
(dx)
(u(x)) 1 (u(x)) {x: (u(x))> (u(x)) 1{x: (u(x))> (c) (u(x)) (dx) (c)
1 (c)
(c)} (x)
(dx)
(c)} (x)
(dx)
(u(x)) (dx)
(vi) This follows immediately from (ii) by taking Then
˘ = P, c = ↵ V⇠, u = ⇠ * E⇠ and p = 2.
1 ⇠ * E⇠2 dP ˘ 2 (↵ V⇠) 1 1 = 2 V⇠ = 2 . ↵ V⇠ ↵
P(⇠ * E⇠ Œ ↵E⇠) Õ
110
here! This means
Solution Manual. Chapter 1–13. Last update 23rd August 2017 ∑∑ Problem 11.4 Solution: We mimic the proof of Corollary 11.6. Set N = {u = ÿ} = {up = ÿ}. ∂ Then N = kÀN {up Œ k} and using Markov’s inequality (MI) and the ‘continuity’ of measures, Proposition 4.3(vii), we find
H
(N) =
Ã
I {up Πk}
4.3(vii)
=
kÀN
lim ({up Πk})
kôÿ
MI
Õ lim
kôÿ
1 up d = 0. k ´≠≠Ø≠≠¨ ↵} À A < . Using that u Õ u< Õ v we find that {u > ↵} œ {u< > ↵} œ {v > ↵} but {v > ↵}, {u > ↵} À A and {u > ↵} ‰ {v > ↵} œ {u ë v} is a -null set. Because of Problem 4.15 we conclude that {u< > ↵} À A < .
∑∑ Problem 11.6 Solution: Throughout the solution the letters A, B are reserved for sets from A. (i)
a) Let A œ E œ B. Then (A) Õ < (E)
Õ
(B) and going to the supAœE and inf EœB proves
< (E).
b) By the definition of
<
and
<
we find some A œ E such that
< (E)
* (A) Õ ✏.
Since Ac – E c we can enlarge A, if needed, and achieve
<
(E c ) * (Ac ) Õ ✏.
Thus, (X) *
< (E)
*
< (E)
Õ
<
(E c )
* (A) +
<
(E c ) * (Ac )
Õ 2✏, and the claim follows as ✏ ô 0. c) Let A – E and B – F be arbitrary majorizing A-sets. Then A ‰ B – E ‰ F and <
(E ‰ F ) Õ (A ‰ B) Õ (A) + (B).
Now we pass on the right-hand side, separately, to the inf A–E and inf B–F , and obtain <
(E ‰ F ) Õ
<
(E) +
<
(F ).
d) Let A œ E and B œ F be arbitrary minorizing A-sets. Then A ⁄ B œ E ⁄ F and < (E
⁄ F ) Œ (A ⁄ B) = (A) + (B).
Now we pass on the right-hand side, separately, to the supAœE and supBœF , where we stipulate that A „ B = Á, and obtain < (E
⁄ F) Œ
< (E)
+
< (F ).
113
R.L. Schilling: Measures, Integrals & Martingales (ii) By the definition of the infimum/supremum we find sets An œ E œ An such that
< (A)
<
* (An ) +
1 (A) * (An ) Õ . n
Without loss of generality we can assume that the An increase and that the An decrease. Now ∑ ∂ A< := n An , A< := n An are A-sets with A< œ A œ A< . Now, (An ) ö (A< ) as well as (An ) ô (An ) ô
< (E)
< (E)
which proves (A< ) =
which proves (A< ) =
< (E).
< (E).
Analogously, (An ) ~ (A< ) as well as
(iii) In view of Problem 4.15 and (i), (ii), it is clear that EœX:
< (E)
=
<
(E) =
E œ X : « A, B À A, A œ E œ B, (B ‰ A) = 0 but the latter is the completed -algebra A < . That <
and
<
coincide on
A< .
0 everywhere, and using Beppo Levi’s theorem
fd = =
ÿ … j=1 ÿ
=
ÿ … j=1
…
1 1 j 2 ( (Aj ) + 1) Aj
1 d
1 1 d 2j ( (Aj ) + 1) Aj (Aj )
… j=1 ÿ
Õ
0
2j ( (Aj ) + 1) 2*j = 1.
j=1
Thus, set P (A) := îA f d . We know from Problem 9.7 that P is indeed a measure. If N À N , then, by Theorem 11.2,
P (N) = so that N œ NP .
N
fd
11.2
= 0
Conversely, if M À MP , we see that M
fd =0
but since f > 0 everywhere, it follows from Theorem 11.2 that 1M f = 0 -a.e., i.e. (M) = 0. Thus, NP œ N .
Remark. We will see later (cf. Chapter 20 or Chapter 25, Radon–Nikod˝m theorem) that N = NP if and only if P = f
(i.e., if P has a density w.r.t. ) such that f > 0.
∑∑ Problem 11.10 Solution: Well, the hint given in the text should be good enough. ∑∑ Problem 11.11 Solution: Observe that C
ud =
C
wd
◊Ÿ
C
(u+ + w* ) d =
C
(u* + w+ ) d
holds for all C À C. The right-hand side can be read as the equality of two measures A ;ô
îA (u+ + w* ) d , A ;ô îA (u* + w+ ) d , A À A which coincide on a generator C which satisfies the conditions of the uniqueness theorem of measures (Theorem 5.7). This shows that A
ud =
A
wd
≈A À A.
115
R.L. Schilling: Measures, Integrals & Martingales Now the direction ‘Ÿ’ follows from Corollary 11.7 where B = A. The converse implication ‘◊’ follows directly from Corollary 11.6 applied to u1C and w1C . ∑∑ Problem 11.12 Solution: (i) “œ”: Let x À Cf , i.e. f (x) = limnôÿ fn (x) exists; in particular, (fn (x))nÀN is Cauchy: for all k À N there is some l À N such that
fn (x) * fm (x) Õ
1 k
Šm, n Πl.
∂ ∑ ∂ 1 This shows that x À kÀN lÀN ÿ n,m=l {fn (x) * fm (x) Õ k }. ∂ ∑ ∂ 1 “–”: Assume that kÀN lÀN ÿ n,m=l {fn (x) * fm (x) Õ k }. This means that for every k À N there is some l À N with
fn (x) * fm (x) Õ
1 k
Šm, n Πl.
This shows that (fn (x))nÀN is a Cauchy sequence in R. The claim follows since R is complete.
(ii) From the definition of limits we get (as in part (i)) Cf = Observe that Akn
~
ÿ Ã Õ Ã kÀN lÀN m=l ÿ ÿ Õ Ã l=1 m=l
{fm (x) * f (x) Õ k1 };
{fm (x) * f (x) Õ k1 } – Cf
as n ô ÿ. Using the continuity of measures, we get Hÿ ÿ I % ÕÃ$ 1 k (An ) ~ fm (x) * f (x) Õ k = (X). l=1 m=l
(Note: if A œ B is measurable and (A) =
(X), then we have (B) =
(X).) In
particular we can pick n = n(k, ✏) in such a way that (Akn ) Œ (X) * ✏2*k . Therefore, (X ‰ Akn(k,✏) ) = (X) * (Akn(k,✏) ) Õ ✏2*k . (iii) Fix ✏ > 0, pick n = n(k, ✏) as in part (ii), and define A✏ :=
à kÀN
Akn(k,✏) À A.
Using the sub-additivity of we get H I ⇡ Õ⇠ … … k (X ‰ A✏ ) = X ‰ An(k,✏) Õ (X ‰ Akn(k,✏) ) Õ ✏2*k Õ ✏. kÀN
116
kÀN
kÀN
Solution Manual. Chapter 1–13. Last update 23rd August 2017 It remains to show that fn converges uniformly to f on the set A✏ . By definition, A✏ =
ÿ Ã n(k,✏) Õ Ã kÀN l=1 m=l
{f * fm Õ k1 },
i.e. for all x À A✏ and k À N there is some l(x) Õ n(k, ✏) such that f (x) * fm (x) Õ Since l(x) Õ n(k, ✏) we get, in particular, f (x) * fm (x) Õ
1 k
1 k
Šm Πl(x).
≈x À A✏ , m Œ n(k, ✏).
Since k À N is arbitrary, the uniform convergence A✏ follows.
(iv) Consider one-dimensional Lebesgue measure, set f (x) := x and fn (x) := x1[*n,n] .
Then we have fn (x) ~ f (x) for every x, but the set {fn * f > ✏} = [*n, n]c has infinite measure for any ✏ > 0.
∑∑
117
12 Convergence theorems and their applications. Solutions to Problems 12.1–12.37 Problem 12.1 Solution: We start with the simple remark that a * bp Õ (a + b)p Õ (max{a, b} + max{a, b})p = 2p max{a, b}p = 2p max{ap , bp } Õ 2p (ap + bp ). Because of this we find that uj * up Õ 2p g p and the right-hand side is an integrable dominating function.
Proof alternative 1: Apply Theorem 12.2 on dominated convergence to the sequence up
of integrable functions. Note that
j (x)
and independent of j. Thus, jôÿ
lim
ô 0 and that 0 Õ jôÿ
uj * up d = lim
j
j
d =
=
Õ
where
jôÿ
lim
j
0d
= 0.
=
2p g p
j
:= uj *
is integrable
d
Proof alternative 2: Mimic the proof of Theorem 12.2 on dominated convergence. To do so we remark that the sequence of functions 0Õ Since the limit limj
j
get
j
:= 2p g p * uj * up ,,,,,,,,,ô , 2p g p jôÿ
exists, it coincides with lim inf j
2p g p d =
lim inf
j,
and so we can use Fatou’s Lemma to
jôÿ
j
d
Õ lim inf
j
d
= lim inf
2p g p * uj * up d
jôÿ jôÿ
119
R.L. Schilling: Measures, Integrals & Martingales 0
=
2 g d + lim inf
=
2p g p d * lim sup
*
p p
jôÿ
jôÿ
1 uj * u d p
uj * up d
where we use that lim inf j (*↵j ) = * lim supj ↵j . This shows that lim supj î uj * up d hence
0 Õ lim inf jôÿ
uj * up d Õ lim sup jôÿ
= 0,
uj * up d Õ 0
showing that lower and upper limit coincide and equal to 0, hence limj î uj * up d = 0. ∑∑ Problem 12.2 Solution: Assume that, as in the statement of Theorem 12.2, uj ô u and that uj Õ f À L1 ( ). In particular,
*f Õ uj and uj Õ f
(j À N) is an integrable minorant resp. majorant. Thus, using Problem 10.7 at < below,
ud = <
lim inf uj d jôÿ
Õ lim inf jôÿ
uj d
uj d
Õ lim sup <
Õ This proves î u d = limj î uj d .
jôÿ
lim sup uj d jôÿ
=
ud .
Addition: since 0 Õ u * uj Õ limj uj + uj Õ 2f À L1 ( ), the sequence u * uj has an integrable majorant and using Problem 10.7 we get 0 Õ lim sup jôÿ
uj * u d Õ
lim sup uj * u d = jôÿ
0d = 0
and also (i) of Theorem 12.2 follows... ∑∑ Problem 12.3 Solution: By assumption we have 0 Õ fk * gk ,,,,,,,,,ô , f * g, kôÿ
0 Õ Gk * fk ,,,,,,,,,ô , G * f. kôÿ
Using Fatou’s Lemma we find
120
(f * g) d =
lim(fk * gk ) d
=
lim inf (fk * gk ) d
k
k
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Õ lim inf
(fk * gk ) d
= lim inf
fk d *
k k
and
(G * f ) d =
lim(Gk * fk ) d
=
lim inf (Gk * fk ) d
k
k
Õ lim inf k
=
(Gk * fk ) d
G d * lim sup k
Adding resp. subtracting î g d resp. î G d therefore yields lim sup k
and the claim follows.
fk d Õ
gd ,
f d Õ lim inf k
fk d .
fk d
∑∑ Problem 12.4 Solution: Using Beppo Levi’s theorem in the form of Corollary 9.9 we find
ÿ … j=1
uj d =
which means that the positive function
≥ÿ
ÿ … j=1
(*)
uj d < ÿ,
j=1 uj is finite almost everywhere, i.e. the series
converges (absolutely) almost everywhere.
≥ÿ
j=1 uj
Moreover, N …
j=1
uj d =
and, using the triangle inequality both quantities N Û … Û uj d Û Û j=n
can be estimated by
N … j=n
N … j=1
(**)
uj d
N Û Û… Û and Û uj d Û Û Û Û j=n
Û Û Û Û
uj d ,,,,,,,,,,,,,,ô , 0 n,Nôÿ
because of (*). This shows that both sides in (**) are Cauchy sequences, i.e. they are convergent. ∑∑
Problem 12.5 Solution: Since L1 ( ) Œ uj ö 0 we find by monotone convergence, Theorem 12.1, that î uj d ö 0. Therefore, =
ÿ … j=1
(*1) uj and S = j
ÿ … j=1
(*1)j
uj d
converge
121
R.L. Schilling: Measures, Integrals & Martingales (conditionally, in general). Moreover, for every N À N,
N … j=1
(*1) uj d = j
N … j=1
(*1)j uj d ,,,,,,,,,,,ô , S. Nôÿ
All that remains is to show that the right-hand side converges to î ≥N j j=1 (*1) uj we have
d . Observe that for SN :=
S2N Õ S2N+2 Õ … Õ S
and we find, as Sj À L1 ( ), by monotone convergence that Nôÿ
lim
S2N d =
d . ∑∑
Problem 12.6 Solution: Consider uj (x) := j 1(0,1_j) (x), j À N. It is clear that uj is measurable and Lebesgue integrable with integral
uj d = j
1 =1 j
≈ j À N.
Thus, limj î uj d = 1. On the other hand, the pointwise limit is u(x) := lim uj (x) í 0 j
so that 0 = î u d = î limj uj d ë 1.
The example does not contradict dominated convergence as there is no uniform dominating integrable function.
Alternative: a similar situation can be found for vk (x) := k1 1[0,k] (x) and the pointwise limit v í 0. Note that in this case the limit is even uniform and still limk î vk d = 1 ë 0 = î v d . Again
there is no contradiction to dominated convergence as there does not exist a uniform dominating integrable function.
∑∑
Problem 12.7 Solution: Using the majorant (e*rx Õ 1 À L1 ( ), r, x Œ 0) we find with dominated convergence
rôÿ [0,ÿ)
lim
e*rx (dx) =
lim e*rx (dx) = 1 (dx) = {0}. [0,ÿ) rôÿ [0,ÿ) {0} ∑∑
Problem 12.8 Solution:
(i) Let ✏ > 0. As u À L1 ( ), monotone convergence shows that Rôÿ B (0)c R
lim
122
u d = 0.
Solution Manual. Chapter 1–13. Last update 23rd August 2017 In particular, we can pick an R > 0 such that BR (0)c
u d Õ ✏.
Since K is compact (in fact: bounded), there is some r = r(R) > 0, such that x + K œ BR (0)c for all x satisfying x Œ r. Thus, we have x+K
u d Õ
BR (0)c
u d Õ ✏
≈ x À Rn , x Œ r.
(ii) Fix ✏ > 0. By assumption, u is uniformly continuous. Therefore, there is some that
u(y) * u(x) Õ ✏
> 0 such
≈ x À Rn , y À x + K := x + B (0) = B (x).
Hence, 1 u(x)p d (y) (K + x) K+x 1 Õ u(y) * u(x) +u(y) (K) K+x ´≠≠≠≠Ø≠≠≠≠¨
u(x)p =
(a + b)p Õ (2 max{a, b})p Õ 2p (ap + bp ), 0
a, b Π0
(?)
1 u(y) d (y)
✏ d (y) + K+x K+x (K + x) C Õ C✏ p + u(y) d (y). (K) (K) K+x ´≠≠Ø≠≠¨
C u(x) Õ (K) p
d (y).
Õ✏
Using the elementary inequality
we get for C = 2p
p
p
1
Part (i) now implies
✏ô0
lim sup u(x)p Õ C✏ p ,,,,,,,ô , 0 xôÿ
and this is the same as to say limxôÿ u(x) = 0. ∑∑ Problem 12.9 Solution: (i) Fix ✏ > 0, R > 0 and consider B := {u Õ R}. By definition, supxÀB u(x) < ÿ. On the other hand, dominated convergence and Corollary 11.6 show that Rôÿ u>R
lim
u(x) dx =
u=ÿ
u(x) dx = 0.
In particular, we can choose R so large, that îB u(x) dx < ✏. Using Markov’s inequality (Proposition 11.5) yields
(B) = {u Œ R} Õ
1 u(x) dx < ÿ. R
123
R.L. Schilling: Measures, Integrals & Martingales (ii) Fix ✏ > 0 and let B À B(Rn ) be as in (i). Further, let A À B(Rn ) with (A) < ✏. Then we have
A
u d =
A„B
u d +
A„Bc
u d
(A „ B) + u d ´≠Ø≠¨ Bc
Õ sup u(x) xÀB
Õ (A)
Õ sup u(x) ✏ + ✏. xÀB
(Observe that supxÀB u(x) < ÿ.) This proves (A)ô0 A
lim
u d = 0. ∑∑
Problem 12.10 Solution:
(i) From un À L1 ( ) and Òun * uÒÿ Õ 1 (for all sufficiently large n) we infer
u d Õ
un * u d +
un d Õ Òun * uÒÿ (X) +
i.e. u À L1 ( ). A very similar argument gives Û Û u d * ud Û n Û
Û Û Û = Û (u * u) d Û Û n Û Û
un d < ÿ,
Û ÛÕ Û un * u d Õ Òun * uÒÿ (X). Û
Since (X) < ÿ, uniform convergence Òun * uÒÿ ô 0 implies that Û lim ÛÛ un d * ud nôÿ Û (ii) False. Counterexample: (R, B(R), uniformly, un À
L1 ( 1 ),
but
nôÿ
lim
1 ) and u
n (x)
:=
un d = 1 ë 0 =
Û Û = 0. Û Û 1 1 (x), x 2n [*n,n]
À R. Clearly, un ô 0
ud . ∑∑
Problem 12.11 Solution: Without loss of generality we assume that u is increasing. Because of the monotonicity of u, we find for every sequence (an )nÀN œ (0, 1) such that an ö 0, that u(an ) ô u(0+) := inf u(t). t>0
If an := tn , t À (0, 1), we get u(tn ) ö 0 and by monotone convergence nôÿ 0
lim
1
nÀN 0
u(tn ) dt = inf
1
u(tn ) dt =
0
1
inf u(tn ) dt =
nÀN
0
1
u(0+) dt = u(0+). ∑∑
124
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 12.12 Solution: Set un (t) := tn u(t), t À (0, 1). Since tn Õ 1 for t À (0, 1), we have un (t) = tn f (t) Õ f (t) À L1 (0, 1).
Since tn ,,,,,,,,,ô , 0 for all t À (0, 1) and f (t) < ÿ a.e. (Corollary 11.6), we have un (t) ô 0 a.e. nôÿ
An application of dominated convergence (Theorem 12.2 and Remark 12.3) yields nôÿ 0
lim
1
nôÿ 0
tn u(t) dt = lim
1
un (t) dt =
0
1
lim u (t) nôÿ n
dt = 0.
´≠Ø≠¨ 0
∑∑ Problem 12.13 Solution: From the geometric series we know that This implies that for all t > 0 et
1 1*x
=
≥
n nŒ0 x
for x À [0, 1).
… … 1 1 1 = t = e*t (e*t )n = e*nt *t *1 e 1*e nŒ0 nŒ1 ≥k
(observe that e*t < 1 for t > 0!). Set uk (t) := sin(t)
*nt , n=1 e
then we get the estimate
Û… Û k … … Û k *nt Û sin t Û Û uk (t) Õ sin t Û e Û = sin t e*nt Õ sin t e*nt = t e *1 Û n=1 Û n=1 nŒ1 Û Û
( 0. Using the elementary inequalities et * 1 Πt (t Π0) and et * 1 Πet_2 (t Π1) we see
uk (t) Õ 1[0,1] (t) + e*t_2 1(1,ÿ) (t) =: w(t).
Let us now show that w À L1 (0, ÿ). This can be done with Beppo Levi’s theorem: 0
ÿ
w(t) dt =
0
1
w(t) dt + 1 ´Ø¨ 1
nÀN 1
= 1 + sup
n
ÿ
w(t) dt ´Ø¨ e*t_2
⌅ ⇧n e*t_2 dt = 1 + sup * 2e*t_2 t=1 < ÿ. nÀN
We use here that every Riemann-integrable function f : [a, b] ô C, *ÿ < a < b < ÿ,
is Lebesgue integrable and that Riemann and Lebesgue intgrals coincide (in this case, see Theorem 12.8). By dominated convergence, 0
ÿ
k ÿ ÿ … sin(t) dt = lim u (t) dt = lim sin(t)e*nt dt. k 0 kôÿ 0 kôÿ et * 1 n=1
With Im ei t = sin t we get 0
ÿ
sin(t)e
*nt
0 dt = Im
0
ÿ
t(i*n)
e
1 dt ,
(cf. Problem 10.9). Again by dominated convergence, 0 1 ÿ R sin(t)e*nt dt = Im lim et(i*n) dt 0 Rôÿ 1_R 125
R.L. Schilling: Measures, Integrals & Martingales H = Im = Im
4 lim
⇠
Rôÿ
⇡
et(i*n) i*n
5R
I
t=1_R
1 1 = 2 . n*i n +1
∑∑ Problem 12.14 Solution: We know that the exponential function is given by ezx = uk (x) := u(x) By the triangle inequality,
k … (zx)n
n!
n=0
≥ nŒ0
(zx)n . n!
Thus,
,,,,,,,,,ô , u(x)ezx . kôÿ
k … … zxn Û (zx)n Û Û Û uk (x) Õ u(x) = u(x)ezx . Û n! Û Õ u(x) n! Û n=0 Û nŒ0
As x ;ô e x u(x) is integrable for fixed
= ±z, we get
uk (x) Õ u(x)e*zx 1(*ÿ,0) (x) + u(x)ezx 1[0,ÿ) (x) À L1 (R).
An application of dominated convergence and the linearity of the integral give
u(x)ezx dx =
lim uk (x) dx
kôÿ
kôÿ
= lim
uk (x) dx
k … 1 = lim (zx)n u(x) dx kôÿ n! n=0
=
ÿ n … z n=0
n!
xn u(x) dx. ∑∑
Problem 12.15 Solution: We get ÛÛîA u d ÛÛ Õ îA u d straight from the triangle inequality. Therefore, it is enough to prove the second estimate. Fix ✏ > 0.
Solution 1: The Sombrero lemma ensures that there is a sequence (un )nÀN œ E(A) with un Õ u and limnôÿ un = u (Corollary 8.9). From dominated convergence we get î un * u d ,,,,,,,,,ô , 0;
in particular, we can choose n À N such that î un * u d Õ ✏. Since each un is bounded (b/o the nôÿ
definition of a simple function) we get
for any A À A with (A) < A
u d Õ
A
(A) < ✏
un d Õ Òun Òÿ
:= ✏_Òun Òÿ . Using the triangle inequality we get
un * u d +
for any A À A with (A) < . 126
A
A
un d Õ
un * u d +
A
un d Õ 2✏
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Solution 2: Obviously, A
u d =
A„{uŒR}
u d +
A„{u 0 with
For the second expression in (?) we have A„{u 0. Since (x) * t Õ (x) + t Õ (x) + R À L1 ([0, 1], dx) and since t ;ô (x)*t is continuous, the continuity lemma, Theorem 12.4, shows that the mapping
(*R, R) Œ t ;ô f (t) =
[0,1]
(x) * t dx
is continuous. Since R > 0 is arbitrary, the claim follows. Alternative solution: Using the lower triangle inequality we get that f (t) * f (s) Õ
Û Û s * t dx = s * t, Û (x) * t * (x) * sÛ dx Õ Û [0,1] Û [0,1]
i.e. f is Lipschitz continuous. (ii) ‘◊’: Let t À R and assume that { = t} = 0. For h À R we define f (t + h) * f (t) = h
Õt*h
(x) * (t + h) * (x) * t dx h 133
R.L. Schilling: Measures, Integrals & Martingales (x) * (t + h) * (x) * t dx t*h< t}. hô0
By our assumptions, {t * h <
< t + h} ,,,,,,,,ô , { hô0
convergence we arrive at I2 (h) = (notice that
t*h<
t} + { < t}. hô0 h lim
‘Ÿ’: We use the notation introduced in the direction ‘◊’. If f is differentiable at t À R, we find as in the first part of the proof that
lim I2 (h) = f ® (t) * lim I1 (h) * lim I3 (h)
hô0
hô0
hô0
exists. We split I2 once again: I2 (h) =
(x) * (t + h) * (x) * t dx h 0. Then we get
Û sin2 (x) Û Û *tx Û )t u(t, x) = Û (*x)e Û Û x2 Û Û Û *tx Õ 1[0,1] (x) + xe 1[1,ÿ) (x) À L1 ([0, ÿ)) Û sin2 (x) Û Û 2 *tx Û )t2 u(t, x) = Û (*x) e Û Û x2 Û Û Û 2 *tx Õ 1[0,1] (x) + x e 1[1,ÿ) (x) À L1 ([0, ÿ)). Now the differentiability lemma shows that f has two derivatives which are given by f ® (t) = * f ®® (t) =
0
0
ÿ
ÿ
sin2 (x) *tx e dx, x
sin2 (x)e*tx dx.
(ii) In order to calculate f ®® we use that Riemann and Lebesgue integrals auszurechnen coincide if a function is Riemann integrable (Theorem 12.8). Using sin2 (x) = 12 (1 * cos(2x)) =
1 2
Re(1 * ei 2x ) we get
0 ÿ 1 1 i 2x *tx f (t) = Re (1 * e )e dx , 0 2 ®®
(cf. Problem 10.9). Using dominated convergence, we see 0
ÿ
Rôÿ 0
(1 * ei 2x )e*tx dx = lim
R
(1 * ei 2x )e*tx dx.
Since x ;ô (1 * ei 2x )e*tx is Riemann integrable, we can integrate ‘as usual’: 0
ÿ
(1 * ei 2x )e*tx dx = lim
Rôÿ
⌧
1 *tx e *t
R x=0
* lim
Rôÿ
⌧
1 x(2i*t) e 2i * t
R x=0
=
1 1 * . t t * 2i 135
R.L. Schilling: Measures, Integrals & Martingales Thus,
0 1 ⇠ ⇡ 1 1 1 1 1 t 2 f (t) = Re * = * 2 = 2 . 2 t t * 2i 2 t t +4 t(t + 4) ®®
The limits limtôÿ f (t) and limtôÿ f ® (t) follow again with dominated convergence (the necessary majorants are those from part (i)): 0 2 1 ÿ sin (x) *tx lim f (t) = lim e dx = 0, tôÿ 0 tôÿ x2 0 2 1 ÿ sin (x) *tx ® lim f (t) = * lim e dx = 0. tôÿ 0 tôÿ x
(iii) We begin with a closed expression for f ® : from the fundamental theorem of (Riemann) integration we know
f ® (R) * f ® (t) = Letting R ô ÿ we get using (ii) Rôÿ t
f ® (t) = * lim
R
t
R
f ®® (s) ds.
f ®® (s) ds
⌧ 1 1 lim log s * log(s2 + 4) 2 Rôÿ 2 ⇠ ⇡ 1 1 = log t * log(t2 + 4) 2 2 1 t = log ˘ . 2 t2 + 4 =*
Finally,
Rôÿ t
f (t) = * lim
R
f ® (s) ds = *
1 2 t
ÿ
log ˘
R s=t
s s2 + 4
ds.
(In this part we have again used the fact that the Lebesgue integral extends the Riemann integral.)
∑∑ Problem 12.26 Solution: We follow the hint: since e*tx Œ 0 we can use Beppo Levi to get 0
ÿ
nÀN 0
e*xt dx = sup
n
nôÿ 0
e*xt dx = lim
n
e*xt dx.
Moreover, x ;ô e*tx is continuous, hence measurable and Riemann-integrable on compact intervals, and we may (Theorem 12.8) use the Riemann integral to evaluate things. 4 *tx 5n n nôÿ 1 e *xt e dx = ,,,,,,,,,ô , . 0 *t x=0 t Thus, e*tx À L1 (0, ÿ) and î0 e*xt dx = ÿ
orem 12.5. For u(t, x) :=
e*tx
we have
1 . t
Now we use the differentiability lemma, The-
)t u(t, x) = xe*tx Õ xe*ax À L1 (0, ÿ)
136
≈ t À (a, ÿ), a > 0,
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (cf. Example 12.14). Therefore (use the differentiability lemma) d dt 0
ÿ
e*tx dx =
0
ÿ
(*x)e*tx dx
≈ t À (a, ÿ).
Since a > 0 is arbitrary, we get differentiability on (0, ÿ). Iterating this argument, we inver that we can swap derivatives of any order with the integral. Morover, 0 ÿ 1 ⇠ ⇡ dn dn 1 *xt e dx = dtn dtn t 0 ÿ 0 1 (*1)n n! Ÿ (*x)n e*xt dx = n+1 . 0 t If t = 1, the claim follows. ∑∑ Problem 12.27 Solution: Fix throughout (a, b) œ (0, ÿ) and take x À (a, b). Let us remark that, just as in Problem 12.17, we prove that (0,1)
t* dt < ÿ
≈ 1.
(i) That the integrand function x ;ô (t, x) is continuous on (a, b) is clear. It is therefore enough to find an integrable dominating function. We have e*t tx*1 Õ ta*1
≈ t À (0, 1),
x À (a, b)
Št Π1,
x À (a, b)
which is clearly integrable on (0, 1) and e*t tx*1 Õ Ma,b t*2
where we use that t⇢ e*t , ⇢ > 0, is continuous and limtôÿ t⇢ e*t = 0 to find Ma,b . This
function is integrable over [1, ÿ). Both estimates together give the wanted integrable dom-
inating function. The Continuity lemma (Theorem 12.4) applies. The well-definedness of (x) comes for free as a by-product of the existence of the dominating function.
(ii) Induction Hypothesis: lem.
(m)
exists and is of the form as claimed in the statement of the prob-
Induction Start m = 1: We have to show that (x) is differentiable. We want to use the
Differentiability lemma, Theorem 12.5. For this we remark first of all, that the integrand function x ;ô (t, x) is differentiable on (a, b) and that
)x (t, x) = )x e*t tx*1 = e*t tx*1 log t. We have now to find a uniform (for x À (a, b)) integrable dominating function for )x (t, x). Since log t Õ t for all t > 0 (the logarithm is a concave function!), Û *t x*1 Û log tÛ = e*t tx*1 log t Ûe t Û Û 137
R.L. Schilling: Measures, Integrals & Martingales Õ e*t tx Õ e*t tb Õ Cb t*2
Št Π1,
x À (a, b)
(use for the last step the argument used in part (i) of this problem). Moreover, Û *t x*1 Û log tÛ Õ ta*1 log t Ûe t Û Û 1 = ta*1 log Õ Ca t*1_2 t
≈ t À (0, 1),
x À (a, b)
where we use the fact that limtô0 t⇢ log 1t = 0 which is easily seen by the substitution t = e*u and u ô ÿ and the continuity of the function t⇢ log 1t .
Both estimates together furnish an integrable dominating function, so the Differentiability lemma applies and shows that ®
(x) =
(0,ÿ)
)x (t, x) dt =
Induction Step m ‰ m + 1: Set tiability Lemma to find that
(m+1) (t, x)
(m) (x).
= )x
(m) (t, x)
(0,ÿ)
e*t tx*1 log t dt =
(1)
(x).
= e*t tx*1 (log t)m . We want to apply the Differen-
With very much the same arguments as in the induction start we
(m) (t, x)
exists (obvious) and satisfies the following bounds
Û *t x*1 Û Ûe t (log t)m+1 Û = e*t tx*1 (log t)m+1 Û Û Õ e*t tx+m Õ e*t tb+m Õ Cb,m t*2
≈ t Œ 1, x À (a, b)
Û *t x*1 Û Ûe t (log t)m+1 Û Õ ta*1 log tm Û Û ⇠ ⇡ 1 m+1 = ta*1 log t Õ Ca,m t*1_2
≈ t À (0, 1), x À (a, b)
and the Differentiability lemma applies completing the induction step. (iii) Using a combination of Beppo Levi (indicated by ‘B-L’), Riemann=Lebesgue (if the Riemann integral over an interval exists) and integration by parts (for the Riemann integral, indicated by ‘I-by-P’) techniques we get
nôÿ (1_n,n)
x (x) = lim
n
e*t )t tx dt 1_n ⌅ ⇧n = lim e*t tx t=1_n * lim (R) = lim (R)
B-L
e*t xtx*1 dt
nôÿ
nôÿ
= lim (R) nôÿ
nôÿ
n
1_n
nôÿ (1_n,n)
= lim
138
e*t t(x+1)*1 dt
e*t t(x+1)*1 dt
n
1_n
)t e*t tx dt
I-by-P
Solution Manual. Chapter 1–13. Last update 23rd August 2017 =
(0,ÿ)
B-L
e*t t(x+1)*1 dt
= (x + 1). ∑∑ Problem 12.28 Solution: (i) The function x ;ô x ln x is bounded and continuous in [0, 1], hence Riemann integrable.
Since in this case Riemann and Lebesgue integrals coincide, we may use Riemann’s integral and the usual rules for integration. Thus, changing variables according to x = e*t , dx = *e*t dt and then s = (k + 1)t, ds = (k + 1) ds we find, 0
1
(x ln x)k dx =
0
ÿ⌅
⇧k e*t (*t) e*t dt
= (*1)
k
= (*1)k
0 0 ⇠
ÿ
tk e*t(k+1) dt
ÿ⇠
s k+1 ⇡k+1
⇡k
e*s ÿ
ds k+1
1 s(k+1)*1 e*s ds 0 k+1 ⇠ ⇡k+1 1 = (*1)k (k + 1). k+1 = (*1)k
(ii) Following the hint we write *x
x
=e
*x ln x
=
ÿ …
(*1)k
k=0
(x ln x)k . k!
Since for x À (0, 1) the terms under the sum are all positive, we can use Beppo Levi’s theorem and the formula (k + 1) = k! to get (0,1)
ÿ …
1 (x ln x)k dx (0,1) k! k=0 ÿ ⇠ ⇡k+1 … 1 1 = (*1)k (*1)k (k + 1) k! k+1 k=0 ÿ ⇠ ⇡k+1 … 1 = k+1 k=0 ÿ ⇠ ⇡ … 1 n = . n n=1
x*x dx =
(*1)k
∑∑ Problem 12.29 Solution: Fix (a, b) œ (0, 1) and let always u À (a, b). We have for x Œ 0 and L À N0 Û eux Û Û xL f (u, x) = xL ÛÛ x Û Ûe + 1Û eux = xL x e +1 139
R.L. Schilling: Measures, Integrals & Martingales Õ xL
eux ex
= xL e(u*1)x Õ 1[0,1] (x) + Ma,b 1(1,ÿ) (x) x*2 where we use that u * 1 < 0, the continuity and boundedness of x⇢ e*ax for x À [1, ÿ) and ⇢ Œ 0. If x Õ 0 we get
Û eux Û Û xL f (u, x) = xL ÛÛ x Û Ûe + 1Û = xL e*ux Õ 1[*1,0] (x) + Na,b 1(*ÿ,1) (x) x*2 .
Both inequalities give dominating functions which are integrable; therefore, the integral îR xL f (u, x) dx exists.
To see m-fold differentiability, we use the Differentiability lemma (Theorem 12.5) m-times. Formally, we have to use induction. Let us only make the induction step (the start is very similar!). For this, observe that
)um (xn f (u, x)) = )um
xn eux xn+m eux = ex + 1 ex + 1
but, as we have seen in the first step with L = n + m, this is uniformly bounded by an integrable function. Therefore, the Differentiability lemma applies and shows that )um
R
xn f (u, x) dx =
R
xn )um f (u, x) dx =
R
xn+m f (u, x) dx. ∑∑
Problem 12.30 Solution: Because of the binomial formual we have (1 + x2 )n Π1 + nx2 ; this yields, in particular,
Û 1 + nx2 Û Û Û Û (1 + x2 )n Û Õ 1. Û Û
Since
1 + nx2 =0 nôÿ (1 + x2 )n lim
≈ x À (0, 1)
(exponential growth is always stronger than polynomial growth!) we can use dominated convergence and find
nôÿ 0
lim
1
1 + nx2 dx = 0. (1 + x2 )n ∑∑
Problem 12.31 Solution: (i) We begin by showing that f is well defined, i.e. the integral expression makes sense. Recall the following estimates
arctan(y) Õ y,
140
arctan(y) Õ
⇡ , 2
y À R,
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (the first inequality follows from the mean value theorem, the second from the definition of arctan.) Moreover,
1 1 11 2 sinh x = (ex * e*x ) Œ (ex * 1) Œ x 2 2 22 ⇠ ⇡ For u(t, x) := arctan sinht x we see
≈x Œ 1.
Û t Û ⇡ Û 1[1,ÿ) (x) 1(0,1) (x) + ÛÛ Û 2 Û sinh x Û ⇡ 1 1 Õ 1[0,1] (x) + 1 (x) À L1 ((0, ÿ)). 2 4 x2 [1,ÿ)
u(t, x) Õ
This proves that the integral f (t) = î(0,ÿ) u(t, x) dx exists. In order to check differentiability of f , we have to find (Theorem 12.5) a majorizing function for the derivative of the integrand. Fix R > 0 and let t À (R*1 , R). By the chain rule ) u(t, x) = )t
1+
1 R*2 +sinh x
t sinh x
1 ⇡2 sinh x
1
= Since x ;ô
1
⇠
t2 sinh x
+ sinh x
.
is continuous, there is a constant C1 > 0 such that sup
xÀ[0,1]
R*2
1 Õ C1 . + sinh x
Using 0 Õ sinh x Õ 1 for x À [0, 1], we get )t u(t, x) Õ
1 R*2 sinh x
+ sinh x
Õ
R*2
1 Õ C1 + sinh x
≈ x À [0, 1].
Similarly we get for x > 1 )t u(t, x) Œ
1 1 2 1 =2 x = x À L1 ((1, ÿ)). sinh x e * e*x e 1 * e*2x ´≠Ø≠¨ ÕC2 0 is arbitrary, f is differentiable on (0, ÿ). That limtö0 f ® (t) does not exist, follows directly from the closed expresson for f ® in part (ii).
141
R.L. Schilling: Measures, Integrals & Martingales (ii) Note that f (0) = 0. In order to find an expression for f ® , we perform the following substitution: u = cosh x and we get, observing that cosh2 x * sinh2 x = 1: f ® (t) = = (Observe: x ;ô
1
t2 +sinh x sinh x
(1,ÿ)
t2
˘ u2 *1
1 1 du ˘ ˘ + u2 * 1 u2 * 1
1 du. (1,ÿ) t2 * 1 + u2
is continuous, hence Riemann-integrable. Since we have estab-
lished in part (i) the existence of the Lebesgue integral, we can use Riemann integrals (b/o Theorem 12.8).) There are two cases:
• t > 1: We have t2 * 1 > 0 and so f ® (t) =
˘
• t < 1: Then C :=
1 t2 * 1 (1,ÿ)
0 1+
1 ˘
u
12 du
t2 *1
L H IMÿ ˘ 1 u = 2 t2 * 1 arctan ˘ t *1 t2 * 1 H H II u=1 1 ⇡ 1 =˘ * arctan ˘ 2 2 2 t *1 t *1 ⇠˘ ⇡ 1 =˘ arctan t2 * 1 . 2 t *1 1 * t2 makes sense and we get u2 + t2 * 1 = u2 * C 2 = (u + C)(u * c).
Moreover, by partial fractions, u2
and so (1,ÿ)
1 1 1 1 1 = * 2 2C u + C 2C u * C *C
1 u2 * C 2 du = du (1,ÿ) u2 + t 2 * 1 0 R 1 R 1 1 1 = lim du * du 1 u * C 2C Rôÿ 1 u + C ⇠ ⇠ ⇡ ⇠ ⇡⇡ 1+C R+C 1 = lim ln + ln 2C Rôÿ 1*C R*C ⇠ ⇡ 1 1+C = ln 2C 1 * CH I ˘ 1 + 1 * t2 1 = ˘ ln . ˘ 2 1 * t2 1 * 1 * t2
The first part of our argument shows, in particular, 1
ÿ
f ® (t) dt = ÿ.
Since f (t) = f (1) + î1 f ® (s) ds, t Œ 1, we get limtôÿ f (t) = ÿ. t
142
Solution Manual. Chapter 1–13. Last update 23rd August 2017 ∑∑ Problem 12.32 Solution: (i) Since
Û d m *tX Û Û m *tX Û m Û Û Û dtm e Û = ÛÛX e ÛÛ Õ X Û Û
m applications of the differentiability lemma, Theorem 12.5, show that that
(m) (0+) X
= (*1)m
(m) (0+) X
exists and
X m dP.
(ii) Using the exponential series we find that *tX
e
ÿ … (*1)k tk (*1)k tk * X = Xk k! k! k=0 k=m+1 m …
k
=t
m+1
ÿ …
X m+1+j
j=0
(*1)m+1+j tj . (m + 1 + j)!
Since the left-hand side has a finite P-integral, so has the right, i.e. and we see that
ÿ ⇠…
X m+1+j
j=0
(*1)m+1+j tj ⇡ dP (m + 1 + j)!
converges
m ⇠ … (*1)k tk ⇡ e*tX * Xk dP = o(tm ) k! k=0
as t ô 0.
(iii) We show, by induction in m, that … (*u)k Û um Û *u m*1 Ûe * ÛÕ Û k! ÛÛ m! Û k=0
Šu Π0.
(*)
Because of the elementary inequality e*u * 1 Õ u
≈u Œ 0
the start of the induction m = 1 is clear. For the induction step m ô m + 1 we note that m m*1 Û *u … (*u)k ÛÛ ÛÛ u ⇠ *y … (*y)k ⇡ ÛÛ Ûe * = e * dyÛ Û Û Û k! k! Û Û Û Û 0 k=0 k=0 m*1 … (*y)k Û Ûe*y * Û dy Û Û k! Û Û k=0
0
uÛ
Õ
0
u
=
um+1 , (m + 1)!
Õ (*)
ym dy m!
and the claim follows. 143
R.L. Schilling: Measures, Integrals & Martingales Setting x = tX in (*), we find by integration that 0 ±
e
*tX
*
m*1
…
î X k dP
(*1) t
k k
k!
k=0
1 Õ
tm î X m dP m!
.
(iv) If t is in the radius of convergence of the power series, we know that lim
tm î X m dP
môÿ
which, when combined with (iii), proves that X (t) = lim
môÿ
m!
m*1
…
(*1)k tk
k=0
=0 î X k dP k!
. ∑∑
Problem 12.33 Solution: (i) Wrong, u is NOT continuous on the irrational numbers. To see this, just take a sequence of rationals qj À Q „ [0, 1] approximating p À [0, 1] ‰ Q. Then
lim u(qj ) = 1 ë 0 = u(p) = u(lim qj ). j
j
(ii) True. Mind that v is not continuous at 0, but {n*1 , n À N} ‰ {0} is still countable. (iii) True. The points where u and v are not 0 (that is: where they are 1) are countable sets, hence measurable and also Lebesgue null sets. This shows that u, v are measurable and almost everywhere 0, hence î u d = 0 = î v d .
(iv) True. Since Q „ [0, 1] as well as [0, 1] ‰ Q are dense subsets of [0, 1], ALL lower resp. upper Darboux sums are always
S⇡ [u] í 0 resp. S ⇡ [u] í 1
(for any finite partition ⇡ of [0, 1]). Thus upper and lower integrals of u have the value 0 resp. 1 and it follows that u cannot be Riemann integrable.
∑∑ Problem 12.34 Solution: Note that every function which has finitely many discontinuities is Riemann integrable. Thus, if {qj }jÀN is an enumeration of Q, the functions uj (x) := 1{q1 ,q2 ,…,qj } (x) are Riemann integrable (with Riemann integral 0) while their increasing limit uÿ = 1Q is not Riemann integrable.
∑∑ Problem 12.35 Solution: Of course we have to assume that u is Borel measurable! By assumption we know that uj := u1[0,j] is (properly) Riemann integrable, hence Lebesgue integrable and [0,j] 144
ud =
[0,j]
uj d = (R)
0
j
u(x) dx ,,,,,,,,,ô , jôÿ
0
ÿ
u(x) dx.
Solution Manual. Chapter 1–13. Last update 23rd August 2017 The last limit exists because of improper Riemann integrability. Moreover, this limit is an increasing limit, i.e. a ‘sup’. Since 0 Õ uj ~ u we can invoke Beppo Levi’s theorem and get
u d = sup j
uj d =
0
ÿ
u(x) dx < ÿ
proving Lebesgue integrability. ∑∑ Problem 12.36 Solution: Observe that x2 = k⇡ ◊Ÿ x =
˘ k⇡, x Œ 0, k À N0 . Thus, Since
sin x2 is continuous, it is on every bounded interval Riemann integrable. By a change of variables, y = x2 , we get
˘ b
˘a
sin(x2 ) dx =
a
b
b sin y dy sin y ˘ = ˘ dy 2 y a 2 y
which means that for a = ak = k⇡ and b = bk = (k + 1)⇡ = ak+1 the values î˘ sin(x2 ) dx ak ⌅˘ ˘ ⇧ are a decreasing sequence with limit 0. Since on ak , ak+1 the function sin x2 has only one ˘ ak+1
sign (and alternates its sign from interval to interval), we can use Leibniz’ convergence criterion to see that the series
… k
˘ ak+1
˘ak
(*)
sin(x2 ) dx
converges, hence the improper integral exists. The function cos x2 can be treated similarly. Alternatively, we remark that sin x2 = cos(x2 * ⇡_2). The functions are not Lebesgue integrable. Either we show that the series (*) does not converge absolutely, or we argue as follows:
sin x2 = cos(x2 *⇡_2) shows that î sin x2 dx and î cos x2 dx either both converge or diverge.
If they would converge (this is equivalent to Lebesgue integrability...) we would find because of sin2 + cos2 í 1 and sin , cos Õ 1, ÿ=
0
ÿ
1 dx = = Õ
0 0 0
ÿ⌅
⇧ (sin x2 )2 + (cos x2 )2 dx
ÿ ÿ
(sin x2 )2 dx + sin x2 dx +
0 0
ÿ ÿ
(cos x2 )2 dx cos x2 dx < ÿ,
which is a contradiction. ∑∑ Problem 12.37 Solution: Let r < s and, without loss of generality, a Õ b. A change of variables yields
r
s
s s f (bx) * f (ax) f (bx) f (ax) dx = dx * dx r r x x x
145
R.L. Schilling: Measures, Integrals & Martingales = =
bs
br
bs
as
as f (y) f (y) dy * dy ar y y br f (y) f (y) dy * dy y y ar
Using the mean value theorem for integrals, I.12, we get r
s
bs br f (bx) * f (ax) 1 1 dx = f (⇠s ) dy * f (⇠r ) dy as y ar y x
= f (⇠s ) ln ab * f (⇠r ) ln ab . Since ⇠s À (as, bs) and ⇠r À (ar, br), we find that ⇠s ,,,,,,,,,ô , ÿ and ⇠r ,,,,,,,ô , 0 which means that sôÿ
r
s
rô0
⌅ ⇧ sôÿ f (bx) * f (ax) dx = f (⇠s ) * f (⇠r ) ln ab ,,,,,,,,,,,ô , (M * m) ln ab . rô0 x ∑∑
146
13 The function spaces Lp. Solutions to Problems 13.1–13.26 Problem 13.1 Solution: (i) We use Hölder’s inequality for r, s À (1, ÿ) and ÒuÒqq =
uq d = Õ
0
Now let us choose r and s. We take r= hence
u d u d
0
1q_p 1_q
0 =
u d
p
u d
= 1 to get
11_r
qr
and
ÒuÒq =
1 t
11_r 0
qr
p 1 q > 1 -Ÿ = q r p
p
+
uq 1 d
0
=
1 s
s
11_s
1 d
( (X))1_s .
q 1 1 =1* =1* , s r p
( (X))(1*q_p)(1_q)
1q_p 1_q
( (X))1_q*1_p
= ÒuÒp ( (X))1_q*1_p .
(ii) If u À Lp we know that u is measurable and ÒuÒp < ÿ. The inequality in (i) then shows that ÒuÒq Õ const ÒuÒp < ÿ,
hence u À Lq . This gives Lp œ Lq . The inclusion Lq œ L1 follows by taking p ‰ q, q ‰ 1.
Let (un )nÀN œ Lp be a Cauchy sequence, i.e. limm,nôÿ Òun * um Òp = 0. Since by the inequality in (i) also
lim Òun * um Òq Õ (X)1_q*1_p lim Òun * um Òp = 0 Lq
m,nôÿ
we get that (un )nÀN œ
is also a Cauchy sequence in Lq . m,nôÿ
(iii) No, the assertion breaks down completely if the measure example:
has infinite mass. Here is an
= Lebesgue measure on (1, ÿ). Then the function f (x) =
over [1, ÿ), but
f 2 (x)
=
1 x2
is. In other words: f Ã
L1 (1, ÿ)
1 x
is not integrable
but f À L2 (1, ÿ), hence
L2 (1, ÿ) — L1 (1, ÿ). (Playing around with different exponents shows that the assertion also fails for other p, q Œ 1....).
147
R.L. Schilling: Measures, Integrals & Martingales ∑∑ Problem 13.2 Solution: This is going to be a bit messy and rather than showing the ‘streamlined’ solution we indicate how one could find out the numbers oneself. Now let (0, 1) and let ↵, inequality
be conjugate indices:
ur d =
0
Õ 0 =
1 ↵
+
= 1 where ↵,
1
ÒuÒr Õ
À (1, ÿ). Then by the Hölder
ur ur(1* ) d
11 0 u
r ↵
u
r ↵
↵
d
1r
r ↵
d
0
1
ur
↵
d
= ÒuÒr ↵ ÒuÒ1* r(1*
)
r ↵
0
u
Taking rth roots on both sides yields 0
be some number in
11
r(1* )
u
d
r(1* )
ur(1*
)
1 r(1* ) d
1 d
r(1* )
.
(1* ) r(1* )
.
This leads to the following system of equations: p = r ↵,
q = r(1 * ) ,
1=
1 1 + ↵
=
q*p . r*p
with unknown quantities ↵, , . Solving it yields =
1 r 1 p
*
1 q
*
1 q
,
↵=
q*p q*r
∑∑ Problem 13.3 Solution:
(i) If u, v À Lp ( ), then u + v and ↵u are again in Lp ( ); this follows from the homogeneity
of the integral and Minkowski’s inequality (Corollary 13.4. Using the Cauchy–Schwarz inequality, the product uv is in Lp ( ), if u, v À L2p ( ). More generally: if there are conjugate numbers ↵, then uv À Lp ( ).
À [1, ÿ] (i.e. ↵ *1 +
*1
= 1), such that u À L↵p and v À L p ,
(ii) Consider the measure space ((0, 1), B(0, 1), ) and set u(x) := v(x) := x*1_3 . This gives
i.e. u, v À
L2 ( 0
0
1
u(x)2 dx =
0
1
⌅ ⇧1 x*2_3 dx = 3 x1_3 x=0 = 3 < ÿ,
). On the other hand, u v à L2 ( ) as 1
u(x)v(x)2 dx =
0
1
⌅ ⇧1 x*4_3 dx = lim *3x*1_3 x=r = ÿ. rô0
This proves that L2 ( ) is not an algebra. Define õ u := u2 and võ := v2 , we get a similar counterexample which works in L1 ( ). 148
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (iii) From Minkowski’s inequality we get ÒuÒp = Ò(u * v) + vÒp Õ Òu * vÒp + ÒvÒp -Ÿ ÒuÒp * ÒvÒp Õ Òu * vÒp . If we change the rôles of u and v, we obtain ÒvÒp * ÒuÒp Õ Òv * uÒp = Òu * vÒp and, therefore, Û Û ÛÒuÒp * ÒvÒp Û = max{ÒuÒp * ÒvÒp , ÒvÒp * ÒuÒp } Õ Òu * vÒp . Û Û ∑∑ Problem 13.4 Solution: (i) We consider the three cases separately. (a) Every map u : (⌦, {Á, ⌦}) ô (R, {Á, R}) is measurable. Indeed: u is measurable if, and only if, u*1 (A) À {Á, ⌦} for all A À A = {Á, R}. Since u*1 (Á) = Á
u*1 (R) = ⌦
this is indeed true for any map u. (b) Every measurable map u : (⌦, {Á, ⌦}) ô (R, B(R)) is constant Indeed: Suppose, u is not constant, i.e. there are !1 , !2 À ⌦ and x, y À R, x ë y, such that u(!1 ) = x,
u(!2 ) = y. Then u*1 ({x}) Ã {Á, ⌦} as !1 À u*1 ({x}) (and so u*1 ({x}) ë Á) and !2 Ã u*1 ({x}) (and so u*1 ({x}) ë ⌦).
(c) Every measurable map u : (⌦, {Á, ⌦}) ô (R, P(R)) is clearly {Á, ⌦}_B(R)-measurable. From (b) we know that such functions are constant. On the other hand, constant maps are measurable for any -algebra. Therefore, every {Á, ⌦}_P(R)-measurable map is constant.
(ii) We determine first the (B)-measurable maps. We claim: every (B)_B(R)-measurable map is of the form
u(!) = c1 1B (!) + c2 1Bc (!),
! À ⌦,
(?)
for c1 , c2 À R. Indeed: If u is given by (?), then h n⌦, n nB, u*1 (A) = l nB c , n nÁ, j
c1 , c2 À A, c1 À A, c2 Ã A, c1 Ã A, c2 À A, c1 , c2 Ã A
149
R.L. Schilling: Measures, Integrals & Martingales for any A À B(R). Therefore, u is (B)_B(R)-measurable. Conversely, assume that the function u is (B)_B(R)-measurable. Choose any !1 À B, !2 À B c and define c1 = u(!1 ), c2 = u(!2 ). If u were not of the form (?), then there would be some ! À ⌦
such that u(!) Ã {c1 , c2 }. In this case A := {u(!)} satisfies u*1 (A) Ã {Á, ⌦, B, B c }, contradicting the measurability of u. By definition, L (⌦, (B), ) =
<
p
u : (⌦, (B)) ô (R, B(R)) messbar :
= u d < ÿ p
.
We have already shown that the (B)-measurable maps are given by (?). Because of the linearity of the integral we see that
up d = c1 p (B) + c2 p (B c ).
Consequently, u À Lp (⌦, (B), ) if, and only if, • c1 = 0 or (B) < ÿ • c2 = 0 or (B c ) < ÿ.
In particular, every map of the form (?) is in Lp (⌦, (B), ) if
is a finite measure. ∑∑
Problem 13.5 Solution: Proof by induction in N. Start N = 2: this is just Hölder’s inequality. Hypothesis: the generalized Hölder inequality holds for some N Œ 2. Step N ‰ N + 1:. Let u1 , … , uN , w be N + 1 functions and let p1 , … , pN , q > 1 be such that
p*1 + p*1 + … + p*1 + q *1 = 1. Set p*1 := p*1 + p*1 + … + p*1 . Then, by the ordinary Hölder N N 1 2 1 2 inequality,
0 u1 u2 … uN w d Õ 0 =
u1 u2 … uN d p
u1
p
u2
p
11_p
… uN d p
ÒuÒq 11_p
ÒuÒq
Now use the induction hypothesis which allows us to apply the generalized Hölder inequality for ≥ *1 N (!) factors j := pj _p, and thus N j=1 j = p_p = 1, to the first factor to get
0 u1 u2 … uN w d =
u1
p
u2
p
… uN d p
11_p
ÒuÒq
Õ ÒuÒp1 ÒuÒp2 … ÒuÒpN ÒuÒq . ∑∑
150
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 13.6 Solution: Draw a picture similar to the one used in the proof of Lemma 13.1 (note that the increasing function need not be convex or concave....). Without loss of generality we can assume that A, B > 0 are such that (A) Œ B which is equivalent to A Œ inverses. Thus,
AB = Using the fact that
0
B
(⌘) d⌘ +
increases, we get that
0
( (B)) = B -Ÿ
(B)
(⇠) d⇠ +
(C) ΠB
A
(B)
≈C Œ
(B) since
and
are
B d⇠.
(B)
and we conclude that AB = Õ = =
0
B
0
B
0
B
(⌘) d⌘ + (⌘) d⌘ + (⌘) d⌘ +
0
(B)
0
(B)
0
A
(⇠) d⇠ +
(⇠) d⇠ +
A (B) A
B d⇠
(B)
(⇠) d⇠
(⇠) d⇠
(B) + (A). ∑∑
Problem 13.7 Solution: Let us show first of all that Lp -limkôÿ uk = u. This follows immediately ≥ from limkôÿ Òu * uk Òp = 0 since the series ÿ k=1 Òu * uk Òp converges. Therefore, we can find a subsequence (uk(j) )jÀN such that lim u (x) jôÿ k(j)
= u(x)
almost everywhere.
Now we want to show that u is the a.e. limit of the original sequence. For this we mimic the trick from the Riesz–Fischer theorem 13.7 and show that the series
ÿ K … … (uj+1 * uj ) = lim (uj+1 * uj ) = lim uK j=0
Kôÿ
j=0
Kôÿ
(again we agree on u0 := 0 for notational convenience) makes sense. So let us employ Lemma 13.6 used in the proof of the Riesz–Fischer theorem to get
Ù… Ù Ù… Ù Ùÿ Ù Ùÿ Ù Ù (uj+1 * uj )Ù Õ Ù uj+1 * uj Ù Ù Ù Ù Ù Ù j=0 Ù Ù Ù Ù Ùp Ù j=0 Ùp ÿ … Õ Òuj+1 * uj Òp j=0
Õ
ÿ … j=0
Òuj+1 * uÒp + Òu * uj Òp
1. This shows that the Lp -limit of this subsequence—let us call it w if it exists at all—cannot be (not even a.e.) u = 0.
On the other hand, we know that a sub-subsequence (õ uk(j) )j of (uk(j) )j converges pointwise almost everywhere to the Lp -limit:
lim õ uk(j) (x) = w(x). j
Since the full sequence limn un (x) = u(x) = 0 has a limit, this shows that the sub-sub-sequence limit w(x) = 0 almost everywhere—a contradiction. Thus, w does not exist in the first place.
∑∑ Problem 13.9 Solution: Using Minkowski’s and Hölder’s inequalities we find for all ✏ > 0 Òuk vk * uvÒ1 = Òuk vk * uk v + uk v * uvÒ Õ Òuk (vk * v)Ò + Ò(uk * u)vÒ Õ Òuk Òp Òvk * vÒq + Òuk * uÒp ÒvÒq Õ (M + ÒvÒq )✏ for all n Œ N✏ . We use here that the sequence (Òuk Òp )kÀN is bounded. Indeed, by Minkowski’s inequality
Òuk Òp = Òuk * uÒp + ÒuÒp Õ ✏ + ÒuÒp =: M. ∑∑
152
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 13.10 Solution: We use the simple identity Òun * um Ò22 =
(un * um )2 d
=
(u2n * 2un um + um ) d
= Òun Ò22 + Òum Ò22 * 2
(*)
un um d .
Case 1: un ô u in L2 . This means that (un )nÀN is an L2 Cauchy sequence, i.e. that limm,nôÿ Òun * um Ò22 = 0. On the other hand, we get from the lower triangle inequality for norms lim ÛÒun Ò2 * ÒuÒ2 ÛÛ Õ lim Òun * uÒ2 = 0
nôÿ Û
nôÿ
so that also limnôÿ Òun Ò22 = limmôÿ Òum Ò22 = ÒuÒ22 . Using (*) we find 2
un um d = Òun Ò22 + Òum Ò22 * Òun * um Ò22 ,,,,,,,,,,,,,ô , ÒuÒ22 + ÒuÒ22 * 0 n,môÿ
= 2ÒuÒ22 . Case 2: Assume that limn,môÿ î un um d = c for some number c À R. By the very definition of this double limit, i.e.
≈✏ > 0
« N✏ À N
:
Û Û Û un um d * c Û < ✏ Û Û Û Û
≈n, m Œ N✏ ,
we see that limnôÿ î un un d = c = limmôÿ î um um d hold (with the same c!). Therefore, again by (*), we get
Òun * um Ò22 = Òun Ò22 + Òum Ò22 * 2
un um d
,,,,,,,,,,,,,ô , c + c * 2c = 0, n,môÿ
i.e. (un )nÀN is a Cauchy sequence in L2 and has, by the completeness of this space, a limit. ∑∑ Problem 13.11 Solution: Use the exponential series to conclude from the positivity of h and u(x) that
exp(hu) = Integrating this gives
j=0
j!
Œ
hN N u . N!
hN uN d Õ exp(hu) d < ÿ N!
and we find that u À LN . Since we have LN œ Lp .
ÿ … h j uj
is a finite measure we know from Problem 13.1 that for N > p ∑∑
153
R.L. Schilling: Measures, Integrals & Martingales Problem 13.12 Solution: (i) We have to show that un (x)p := np↵ (x + n)*p has finite integral—measurability is clear since un is continuous. Since np↵ is a constant, we have only to show that (x + n)*p is in L1 . Set
:= p > 1. Then we get from a Beppo Levi and a domination argument (0,ÿ)
(x + n)*
(dx) Õ Õ
(0,ÿ) (0,1)
(x + 1)*
(dx)
1 (dx) +
(1,ÿ)
kôÿ (1,k)
Õ 1 + lim
x*
(x + 1)*
(dx)
(dx).
Now using that Riemann=Lebesgue on intervals where the Riemann integral exists, we get kôÿ (1,k)
lim
x*
k
kôÿ 1
(dx) = lim
x* dx
⌅ ⇧k = lim (1 * )*1 x1* 1 kôÿ
= (1 * )*1 lim k1* * 1 kôÿ
= ( * 1)
*1
< ÿ
which shows that the integral is finite.
(ii) We have to show that vn (x)q := nq e*qnx is in L1 —again measurability is inferred from continuity. Since nq is a constant, it is enough to show that e*qnx is integrable. Set Since
lim ( x)2 e*
and since e* that
xôÿ
x
x
=0
and
e*
x
Õ1
= qn.
Šx Π0,
is continuous on [0, ÿ), we conclude that there are constants C, C( ) such
* x
e
< = C Õ min 1, ( x)2 $ % 1 Õ C( ) min 1, 2 x ⇠ ⇡ 1 = C( ) 1(0,1) (x) + 1[1,ÿ) 2 x
but the latter is an integrable function on (0, ÿ). ∑∑ Problem 13.13 Solution: Without loss of generality we may assume that ↵ Õ between the case x À (0, 1) and x À [1, ÿ). If x Õ 1, then
1_2 1 1 1 Œ ↵ Œ ↵ = ↵ ↵ ↵ x x +x x + x↵ x +x
≈ x Õ 1;
this shows that (x↵ + x )*1 is in L1 ((0, 1), dx) if, and only if, ↵ < 1. 154
. We distinguish
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Similarly, if x Œ 1, then 1_2 1 1 1 Œ ↵ Œ = x x +x x +x x +x
this shows that (x↵ + x )*1 is in L1 ((1, ÿ), dx) if, and only if,
Thus, (x↵ + x )*1 is in L1 (R, dx) if, and only if, both ↵ < 1 and
≈x Œ 1 > 1. > 1. ∑∑
Problem 13.14 Solution: If we use X = {1, 2, … , n}, x(j) = xj , 0
n … j=1
xj
p
11_p
= ÒxÒp(
= ✏1 + 5 + ✏n we have
)
and it is clear that this is a norm for p Π1 and, in view of Problem 13.19 it is not a norm for
p < 1 since the triangle (Minkowski) inequality fails. (This could also be shown by a direct counterexample.
∑∑ Problem 13.15 Solution: Without loss of generality we can restrict ourselves to positive functions—
else we would consider positive and negative parts. Separability can obviously considered separately!
Assume that L1+ is separable and choose u À Lp+ . Then up À L1 and, because of separability, there is a sequence (fn )n œ D1 œ L1 such that
in L1
in L1
nôÿ
nôÿ
fn ,,,,,,,,,ô , up -Ÿ upn ,,,,,,,,,ô , up if we set un := fn
1_p
À Lp . In particular, un(k) (x) ô u(x) almost everywhere for some subsequence
and Òun(k) Òp ,,,,,,,,,ô , ÒuÒp . Thus, Riesz’s theorem 13.10 applies and proves that kôÿ
in Lp
Lp Œ un(k) ,,,,,,,,,ô , u. kôÿ
Obviously the separating set Dp is essentially the same as D1 , and we are done. The converse is similar (note that we did not make any assumptions on p Œ 1 or p < 1—this is immaterial in the above argument).
∑∑ Problem 13.16 Solution: We have seen in the lecture that, whenever limnôÿ Òu * un Òp = 0, there is a subsequence un(k) such that limkôÿ un(k) (x) = u(x) almost everywhere. Since, by assumption,
limjôÿ uj (x) = w(x) a.e., we have also that limjôÿ un(j) (x) = w(x) a.e., hence u(x) = w(x) almost everywhere.
∑∑
155
R.L. Schilling: Measures, Integrals & Martingales Problem 13.17 Solution:
We remark that y ;ô log y is concave. Therefore, we can use Jensen’s
inequality for concave functions to get for the probability measure _ (X) = (X)*1 1X 0 1 d d (log u) Õ log u (X) (X) H I î ud = log (X) 0 1 1 = log , (X) and the claim follows.
∑∑ Problem 13.18 Solution: As a matter of fact, (0,1)
u(s) ds
(0,1)
log u(t) dt Õ
(0,1)
u(x) log u(x) dx.
We begin by proving the hint. log x Œ 0 ◊Ÿ x Œ 1. So, ⇡ log y Õ y log y ◊Ÿ 1 Õ y ⇠ ⇡ ≈y Õ 1 : log y Õ y log y ◊Ÿ 1 Œ y .
≈y Œ 1 : and
⇠
Assume now that î(0,1) u(x) dx = 1. Substituting in the above inequality y = u(x) and integrating over (0, 1) yields
(0,1)
log u(x) dx Õ
(0,1)
log
(0,1)
u(x) log u(x) dx.
Now assume that ↵ = î(0,1) u(x) dx. Then î(0,1) u(x)_↵ dx = 1 and the above inequality gives u(x) u(x) u(x) dx Õ log dx (0,1) ↵ ↵ ↵
which is equivalent to (0,1)
log u(x) dx * log ↵ =
(0,1)
log u(x) dx *
(0,1)
log ↵ dx
u(x) dx (0,1) ↵ u(x) u(x) Õ log dx (0,1) ↵ ↵ u(x) 1 = u(x) log dx ↵ (0,1) ↵ =
156
log
=
1 1 u(x) log u(x) dx * u(x) log ↵ dx ↵ (0,1) ↵ (0,1)
=
1 1 u(x) log u(x) dx * u(x) dx log ↵ ↵ (0,1) ↵ (0,1)
Solution Manual. Chapter 1–13. Last update 23rd August 2017 1 u(x) log u(x) dx * log ↵. ↵ (0,1)
=
The claim now follows by adding log ↵ on both sides and then multiplying by ↵ = î(0,1) u(x) dx. ∑∑ Problem 13.19 Solution: (i) Let p À (0, 1) and pick the conjugate index q := p_(p*1) < 0. Moreover, s := 1_p À (1, ÿ) and the conjugate index t,
1 s
1 t
+
= 1, is given by 1 p
s t= = s*1
1 p
=
*1
1 À (1, ÿ). 1*p
Thus, using the normal Hölder inequality for s, t we get
up d =
0
Õ 0 =
wp d wp
up
up w p
s
11_s 0 d
1p 0 uw d
w
11_t
w*pt d
p_(p*1)
11*p d
.
Taking pth roots on either side yields 0
p
11_p
0 Õ
u d
0 =
10
uw d 10
uw d
w
p_(p*1)
1(1*p)_p d
1*1_q
q
w d
and the claim follows. (ii) This ‘reversed’ Minkowski inequality follows from the ‘reversed’ Hölder inequality in exactly the same way as Minkowski’s inequality follows from Hölder’s inequality, cf. Corollary 13.4. To wit:
(u + v)p d =
(u + v) (u + v)p*1 d
=
u (u + v)p*1 d +
(i)
v (u + v)p*1 d
Ù Ù Ù Ù Œ ÒuÒp Ù(u + v)p*1 Ù + ÒvÒp Ù(u + v)p*1 Ù . Ù Ùq Ù Ùq
Dividing both sides by Òu + vp*1 Òq proves our claim since Ù Ù Ù(u + v)p*1 Ù = Ù Ùq
0
(p*1)q
(u + v)
11_q d
0 =
(u + v) d
Problem 13.20 Solution: By assumption, u Õ ÒuÒÿ Õ C < ÿ and u ì 0.
p
11*1_p . ∑∑
157
R.L. Schilling: Measures, Integrals & Martingales (i) We have Mn = Note that Mn > 0.
un d Õ C n
d = C n (X) À (0, ÿ).
(ii) By the Cauchy–Schwarz-Inequality, Mn = =
un d
0
u
n+1 2
u
n*1 2
u d ˘ = Mn+1 Mn*1 . n+1
Õ
d 11_2 0
u
n*1
11_2 d
(iii) The upper estimate follows from Mn+1 =
un+1 d Õ
un ÒuÒÿ d = ÒuÒÿ Mn .
Set P := _ (X); the lower estimate is equivalent to 0 0 ◊Ÿ 0 ◊Ÿ
u
n
d (X)
u dP
n
n
u dP
11_n
î un
Õ
11+1_n Õ 1(n+1)_n
î un+1
Õ
d (X) d (X)
un+1 dP
un+1 dP
and the last inequality follows easily from Jensen’s inequality since P is a probability measure:
0
n
u dP
1(n+1)_n
un
n+1 n
dP =
un+1 dP .
(iv) Following the hint we get ÒuÒn Œ
⇠
u > ÒuÒÿ * ✏
⇡1_n
i.e. lim inf ÒuÒn Œ ÒuÒÿ . nôÿ
Combining this with the estimate from (iii), we get ÒuÒÿ Õ lim inf (X)*1_n ÒuÒn (iii)
nôÿ
Mn+1 nôÿ Mn Mn+1 Õ lim sup Mn nôÿ Õ lim inf
Õ ÒuÒÿ .
158
nôÿ
ÒuÒÿ * ✏ ,,,,,,,,,ô , ÒuÒÿ , ✏ô0
Solution Manual. Chapter 1–13. Last update 23rd August 2017 ∑∑ Problem 13.21 Solution: The hint says it all.... Maybe, you have a look at the specimen solution of Problem 13.20, too.
∑∑ Problem 13.22 Solution: We begin with two observations • If r Õ s Õ q, then ÒuÒr Õ ÒuÒs . This follows from Jensen’s inequality (Theorem 13.13) and
the fact that V (x) := xs_r , x À R, is convex (cf. also Problem 13.1). In particular, ÒuÒr < ÿ for all r À (0, q).
• We have
log u d Õ log ÒuÒp
(?)
≈ p À (0, q).
This follows again from Jensen’s inequality applied to the convex function V (x) := * log x: 0 1 p * log u d Õ * log(up ) d * p log u d ; therefore,
0 1 1 p log ÒuÒp = log u d Œ log u d . p
Because of (?) it is enough to show that limpô0 ÒuÒp Õ exp(î ln u d ). (Note: by the monotonicity of ÒuÒp as p ö 0 we know that the limit limpô0 ÒuÒp exists.) Note that ap * 1 , p>0 p
log a = inf (Hint: show by differentiation that p ;ô
ap *1 p
Use l’Hospital’s rule to show that limpô0
log u d
(??)
a > 0.
is increasing. = log a.) From monotone convergence (mc) we get
ap *1 p mc
up * 1 d p>0 p î up d * 1
= inf
Π=
p *1
ÒuÒpp p
(??)
Œ log ÒuÒp
for all p > 0. Letting p ô 0 finishes the proof. ∑∑ Problem 13.22 Solution: (A ) > 0 and
Case 1: ÒuÒLÿ < ÿ. For A := {u Œ ÒuÒÿ * }, 0
ÒuÒp Œ
A
11 (ÒuÒÿ * ) d p
p
> 0, we gave
1
= (ÒuÒÿ * ) (A ) p .
159
R.L. Schilling: Measures, Integrals & Martingales Therefore,
0 lim inf ÒuÒp Œ lim inf pôÿ
Since
(ÒuÒÿ * ) (A )
nôÿ
1 p
1 = ÒuÒÿ * .
> 0 is arbitrary, this shows that lim inf pôÿ ÒuÒp Œ ÒuÒÿ .
On the other hand, we have for p > q
u(x)p d =
q u(x)p*q u(x)q d Õ ÒuÒp*q ÿ ÒuÒq .
Taking pth roots on both sides of the inequality, we get 0 q1 p*q p lim sup ÒuÒp Õ lim sup ÒuÒÿ ÒuÒqp = ÒuÒÿ . pôÿ
pôÿ
This finishes the proof for all ÒuÒLÿ < ÿ. Case 2: ÒuÒLÿ = ÿ. The estimate
lim sup ÒuÒp Õ ÒuÒÿ pôÿ
is trivially true. The converse inequality follows like this: Define AR := {u Œ R}, R > 0. We have (Ar ) > 0 (otherwise ÒuÒLÿ < ÿ!) and, as in the first part of the proof, we find 0 ÒuÒp Œ
A R
11 p
p
R d
1
= R (AR ) p .
Thus, lim inf pôÿ ÒuÒp Œ R and since R > 0 is arbitrary, the claim follows: lim inf ÒuÒp Œ ÿ = ÒuÒÿ . pôÿ
∑∑ Problem 13.23 Solution: Without loss of generality we may assume that f Œ 0. We use the following standard representation of f , see (8.7):
f=
N …
j 1Aj
j=0
with 0 =
0
A1 ⁄ 5 ⁄ A N .
<
1
< … <
N
< ÿ and mutually disjoint sets Aj . Clearly, {f ë 0} =
Assume first that f À E „ Lp ( ). Then ÿ>
f d = p
N … j=1
thus ({f ë 0}) < ÿ.
p j
(Aj ) Œ
N … j=1
p 1
(Aj ) =
p 1
({f ë 0});
Conversely, if ({f ë 0}) < ÿ, we get
160
f d = p
N … j=1
p j
(Aj ) Õ
N … j=1
p N
(Aj ) =
p N
({f ë 0}) < ÿ.
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Since this integrability criterion does not depend on p Œ 1, it is clear that E + „Lp ( ) = E + „L1 ( ), and the rest follows since E = E + * E + .
∑∑ Problem 13.24 Solution: (i) ◊Ÿ (ii) and (iii) ◊Ÿ (iv), since ⇤ is concave if, and only if, V = *⇤ is convex. Moreover, (iii) generalizes (i) and (iv) gives (ii). It is, therefore, enough to verify (iii). Since u is integrable and takes values in (a, b), we get a=
a (dx) <
u(x) (dx) <
b (dx) = b.
This shows that the l.h.S. of the Jensen inequality is well-defined. The rest of the proof is similar to the one of Theorem 13.13: take some affine-linear l(x) = ↵x + x À (a, b) – and notice that 0 1 l ud =↵ ud +
=
Õ V (x) – here we only consider
(↵u + ) d Õ
V (u) d .
Now go to the sup over all affine-linear l below V and the claim follows. ∑∑ Problem 13.25 Solution: (i) Note that ⇤(x) = x1_q is concave—e.g. differentiate twice and show that it is negative—and using Jensen’s inequality for positive f , g Œ 0 yields
fg d = Õ
0
Õ
gf *p_q 1{f ë0} f p d H
î g q f *p 1{f ë0} f p d
I1_q
î fp d 11*1_q 0 11_q p q f d g d fp d
where we use 1{f ë0} Õ 1 in the last step.
Note that f g À L1 follows from the fact that g q f *p 1{f ë0} f p = g q À L1 .
(ii) The function ⇤(x) = (x1_p + 1)p has second derivative ⇤®® (x) =
1*p 1 + x*1_p x*1*1_p Õ 0 p
showing that ⇤ is concave. Using Jensen’s inequality gives for f , g Œ 0 ⇠g ⇡p (f + g)p 1{f ë0} d = 1{f ë0} + 1 f p 1{f ë0} d f L0 Mp 11_p p1 î g d {f ë0} Õ fp d +1 {f ë0} î{f ë0} f p d 40 11_p 0 11_p 5p p p = g d + f d . {f ë0} {f ë0}
161
R.L. Schilling: Measures, Integrals & Martingales Adding on both sides î{f =0} (f + g)p d
= î{f =0} g p d yields, because of the elementary
inequality Ap + B p Õ (A + B)p , A, B Œ 0, p Œ 1,
(f + g)p d 40
Õ 40 Õ
{f ë0}
p
11_p
p
g d 11_p
g d
0 +
0 +
f d {f ë0} 11_p 5p p f d . p
11_p 5p
4 +
{f =0}
p
5p_p
g d
∑∑ Problem 13.26 Solution: Using Hölder’s inequality we get f * ap Õ (f + a)p = (1 f + 1 a)p Õ 2p*1 (f p + ap ). Since (X) < ÿ, this shows that both sides of the asserted integral inequality are finite. Without loss of generality we may assume that a > 0, otherwise we would consider *f instead of f.
Without loss of generality we may assume that m = î f d
= 0, otherwise we would consider
f * î f d instead of f . Observe that
{0 0 is arbitrary and ⌧ =
M « j=1
Òfj Ò⌧
for all M = 1, 2, … , N * 1
Mt . (M*1)t+1
The induction step uses Young’s inequality: Òf1 ? f2 ? 5 ? fN Òr Õ Òf1 Òp Òf2 ? 5 ? fN Òq where p =
Nr (N*1)r+1
and q is given by 1 1 1 (N * 1)r + 1 1 1 1 1 +1= + = + =1+ * + r q q Nr q q N Nr
so that
q= Using the induction hypothesis we now get
Nr . N +r*1
Òf1 ? 5 ? fN Òr Õ Òf1 Òp Òf2 ? 5 ? fN Òq Õ Òf1 Òp
Òf2 Òs 5 ÒfN Òs
where s is, because of the induction assumption, given by s= =
(N * 1)q (N * 2)q + 1 Nr (N * 1) N+r*1 Nr (N * 2) N+r*1 +1
(N * 1)Nr (N * 2)Nr + N + r * 1 (N * 1)Nr = 2 N r * 2Nr + r + (N * 1) (N * 1)Nr = (N * 1)2 r + (N * 1) Nr = =p (N * 1)r + 1 =
and we are done. ∑∑ Problem 15.16 Solution: Note that v(x) = (i) u ? v(x) =
194
0
2⇡
d (1 dx
* cos x)1[0,2⇡) (x) = 1(0,2⇡) (x) sin x. Thus,
1R (x * y) sin y dy =
0
2⇡
sin y dy = 0
≈ x.
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (ii) Since all functions u, v, w,
are continuous, we can use the usual rules for the (Riemann)
integral and get, using integration by parts and the fundamental theorem of integral calculus, x
v ? w(x) =
=
*
=
d (x * y) dy
=
(x * y) (y) dy
=
d dx
(x * y) d dy
*ÿ
(x * y) y
*ÿ
(t) dt dx y
*ÿ
(t) dt dx
(t) dt dx
? (x).
If x À (0, 4⇡), then x * y À (0, 2⇡) for some suitable y = y= and even for all y from an interval (y0 * ✏, y0 + ✏) œ (0, 2⇡). Since follows.
(iii) Obviously,
is positive with support [0, 2⇡], the positivity
(i)
(u ? v) ? w = 0 ? w = 0 while u ? (v ? w)(x) =
1R (x * y)v ? w(y) dy
=
v ? w(y) dy
=
? (y) dy
> 0. Note that w is not an (pth power, p < ÿ) integrable function so that we cannot use Fubini’s theorem to prove associativity of the convolution.
∑∑
195
16 Integrals of images and Jacobi’s transformation rule. Solutions to Problems 16.1–16.12 Problem 16.1 Solution: Since F and Fi are F -sets, we get Õ
F =
kÀN
Fi =
Ck ,
Õ kÀN
Cki
for closed sets Ck resp. Cki . Since complements of closed sets are open, we find, using the rules for (countable) unions and intersections that (i)
n à i=1
(ii)
Õ iÀN
Fi =
Fi =
n Ã Õ i=1 kÀN
ÕÕ iÀN kÀN
Moreover,
à iÀN
(iii)
F =
Õ kÀN
(iv)
n Õ Ã
=
Cki
kÀN i=1
Cki .
´Ø¨ closed set
Cki =
Õ
(i,k)ÀNùN
Cki .
´≠≠≠≠Ø≠≠≠≠¨
countable union!
Fic
à à ⌅ ⇧c = Cki = iÀN kÀN
4 Ck -Ÿ F = c
à (i,k)ÀNùN
⌅ i ⇧c Ck .
´≠≠≠≠≠≠Ø≠≠≠≠≠≠¨
5c
Õ kÀN
Ck
countable intersection!
=
Ã
Ckc . kÀN ´Ø¨
Set c1 := C and Ci = Á, i Œ 2. Then C =
open
Õ
iÀN
Ci is an F -set. ∑∑
Problem 16.2 Solution: Write
=
n
and B = B(Rn ). Fix B À B. According to Lemma 16.12
there are sets F À F and G À G such that
F œ B œ G and
(F ) = (B) = (G).
Since for closed sets Cj and open sets Uj we have F = and suitable M = M✏ À N, N = N✏ À N that
∑
Cj and G =
∂
Uj we get for some ✏ > 0
C1 ‰ 5 ‰ CN œ B œ U1 „ 5 „ UM 197
R.L. Schilling: Measures, Integrals & Martingales and Û Û Û (U1 „ 5 „ UM ) * (B)Û Õ ✏, Û Û Û Û Û (B) * (C1 ‰ 5 ‰ CN )Û Õ ✏. Û Û
(*) (**)
Since finite unions of closed sets are closed and finite intersections of open sets are open, (*) proves outer regularity while (**) proves inner regularity (w.r.t. close sets).
To see inner regularity with compact sets, we note that the closed set C ® := C1 ‰ 5 ‰ CN is approximated by the following compact sets
Kl := Bl (0) „ C ® ~ C ® as l ô ÿ and, because of the continuity of measures, we get for suitably large L = L✏ À N that Û Û Û (KL ) * (C1 ‰ 5 ‰ CN )Û Õ ✏ Û Û which can be combined with (**) to give Û Û Û (KL ) * (B)Û Õ 2✏. Û Û This shows inner regularity for the compact sets. ∑∑ Problem 16.3 Solution: Notation (for brevity): Write
=
n,
Ñ =
n,
B = B(Rn ) and B < =
B < (Rn ). By definition, B < = B ‰ N < where N < is a subset of a B-measurable null set N. (We indicate B < -sets by an asterisk, C (with and without ornaments and indices C ® ...) is always a closed set and U etc. is always an open set.
Solution 1: Following the hint we get (with the notation of Problem 11.6) (B) = Ñ (B < ) = = = Õ
<
(B < ) inf
BŒA–B <
(by 11.6)
(A)
inf
inf
(U )
(by 16.2)
inf
inf
(U )
(as B 0 = 35 , 45 . This does not contradict Lemma 16.6 since G is not a semi-ring. 200
Solution Manual. Chapter 1–13. Last update 23rd August 2017 ∑∑ Problem 16.6 Solution: We want to show that a)
n (x
b)
n (t
c)
+ B) =
n (B),
(Theorem 5.8(i));
B À B(Rn ), x À Rn
(Problem 5.9);
B) = tn n (B), B À B(Rn ), t Œ 0
A( n ) = det A*1
n,
A À Rnùn , det A ë 0
(Theorem 7.10).
From Theorem 16.4 we know that for any C 1 diffeomorphism n
( (B)) =
holds. Thus a), b), c) follow upon setting a) b) c)
B
the formula n
det D d
(y) = x + y -Ÿ D í 1 -Ÿ det D í 1;
(y) = t y -Ÿ D í t id -Ÿ det D í tn ;
(y) = A*1 y -Ÿ D (y) í A*1 -Ÿ det D í det A*1 . ∑∑
Problem 16.7 Solution: (i) The map
: R Œ x ;ô (x, f (x)) is obviously bijective and differentiable with deriv-
ative D (x) = (1, f ® (x)) so that D (x)2 = 1 + (f ® (x))2 . The inverse of by
is given
: (x, f (x)) ;ô x which is clearly differentiable. ˘ (ii) Since D (x) = 1 + (f ® (x))2 is positive and measurable, it is a density function *1
and
is a measure, cf. Lemma 10.8, while
:= D (x)
=
measure in the sense of Definition 7.7.
( ) is an image
(iii) This is Theorem 15.1 and/or Problem 15.1. (iv) The normal is, by definition, orthogonal to the gradient: D (x) = (1, f ® (x)); obviously n(x) = 1 and
H
*f ® (x) 1 ˘
n(x) D (x) = Further,
I H
I
1 f ® (x)
= 0.
1 + (f ® (x))2 ®
` x * ˘ rf (x) 1+[f ® (x)]2 õ (x, r) = r rf (x) + ˘ r ® p 1+[f
so that
0 D õ (x, r) =
) õ (x, r) )(x, r)
a s, s (x)]2 q
1 ®
`1 * r ) ˘ f (x) )x 1+[f ® (x)]2 =r ® r * ˘ f (x) ® 2 p 1+[f (x)]
) ˘ f ® (x) + r )x ˘
1
1
a
1+[f ® (x)]2 s
1+[f ® (x)]2
s q 201
R.L. Schilling: Measures, Integrals & Martingales For brevity we write f , f ® , f ®® instead of f (x), f ® (x), f ®® (x). Now ˘ ® ®® f ®® 1 + [f ® ]2 * f ® ˘ f f ® 2 ® f (x) 1+[f ] ) = ˘ ® 2 )x 1 + [f ® (x)]2 1 + [f ] and
* ˘f
® f ®®
) 1 = . ˘ )x 1 + [f ® (x)]2 1 + [f ® ]2 1+[f ® ]2
Thus, det D õ (x, r) becomes 0
1
1* ˘ 1 + [f ® ]2 f®
˘ r f ®® 1 + [f ® ]2 *
r [f ® ]2 f ®® ˘ 1+[f ® ]2
1
1 + [f ® ]2 0
®
r f ® f ®® ˘ 1+[f ® ]2
1
+˘ f * 1 + [f ® ]2 1 + [f ® ]2 =˘
1
r f ®® *
r [f ® ]2 f ®® 1+[f ® ]2
* 1 + [f ® ]2 1 + [f ® ]2 1 + [f ® ]2 r f ®® =˘ * ® 2 1 + [f ® ]2 1 + [f ] ˘ r f ®® = 1 + [f ® ]2 * 1 + [f ® ]2
[f ® ]2
r [f ® ]2 f ®® 1+[f ® ]2
+˘ * ® 2 1 + [f ® ]2 1 + [f ]
If x is from a compact set, say [c, d], we can, because of the continuity of f , f ® and f ®® , achieve that for sufficiently small values of r < ✏ we get that det D õ > 0, i.e. õ is a local C 1 -diffeomorphism.
(v) The set is a ‘tubular’ neighbourhood of radius r around the graph f for x À [c, d]. Measurability follows, since õ is a diffeomorphism, from the fact that the set C(r) is the image of the cartesian product of measurable sets.
(vi) Because of part (iv) we have, for fixed x and sufficiently small values of r, that the determinant is positive so that
1 Û Û Ûdet D õ (x, s)Û 1 (ds) Û 2r (*r,r) Û ®® ˘ Û Û 1 Û 1 + (f ® (x))2 * s f (x) Û 1 (ds) = lim Û ® 2 rö0 2r (*r,r) Û 1 + (f (x)) ÛÛ 0˘ 1 s f ®® (x) 1 1 ® 2 = lim 1 + (f (x)) * (ds) rö0 2r (*r,r) 1 + (f ® (x))2 ˘ 1 = lim 1 + (f ® (x))2 1 (ds) rö0 2r (*r,r)
lim rö0
s f ®® (x) 1 1 (ds) rö0 2r (*r,r) 1 + (f ® (x))2 ˘ f ®® (x) 1 = 1 + (f ® (x))2 * lim s ® 2 1 + (f (x)) rö0 2r (*r,r) ˘ = 1 + (f ® (x))2 * lim
202
1
(ds)
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Û Û = Ûdet D õ (x, 0)Û. Û Û (vii) We have 1 1 (x, y) 2 (dx, dy) 2r R2 C(r) 1 = 1 õ *1 (x, y) 2 (dx, dy) 2r R2 ( (C)ù(*r,r)) 1 Û Û = 1 *1 (C)ù(*r,r) (z, s) Ûdet D õ (z, s)Û 2 (dz, ds) Û Û 2r R2 4 5 1 Û Û 1 õ = 1 *1 (C) (z) Ûdet D (z, s)Û (ds) 1 (dz). Û R 2r (*r,r) Û ´≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠≠¨ ,,,,rö0 ,,ô , det D õ (z,0) Since
*1 (C)
(Thm 16.4) (Tonelli)
is a bounded subset of R, we can use the result of part (vii) and domin-
ated convergence and the proof is finished. (viii) This follows from (i)–(iii) and the fact that
Û Û ˘ Ûdet D õ (x, 0)Û = 1 + (f ® (x))2 Û Û and the geometrical meaning of the weighted area tubular neighbourhood of the graph.
1 2r
2 (C(r))—recall
that C(r) was a ∑∑
Problem 16.8 Solution: (i) det D (x) is positive and measurable, hence a density and, by Lemma 10.8, det D d
is a measure. Therefore,
Definition 7.7.
(det D
d)
is an image measure in the sense of
Using the rules for densities and integrals w.r.t. image measures we get (cf. e.g. Theorem 15.1 and/or Problem 15.1) M
ud
M
=
M
ud
(ii) This is the formula from part (i) with (iii) The equality
ud
n
=
d
det D
=
*1 (M)
det D d
u˝
d
.
= ✓r ; observe that ✓r (Rn ) = Rn .
(0,ÿ) {ÒxÒ=1}
1
u(r x) rn*1 (dx)
(dr)
is just Theorem 16.22. The equality (0,ÿ) {ÒxÒ=r} =
u(x) (dx)
(0,ÿ) {ÒxÒ=1}
1
(dr)
u(r x) rn*1 (dx)
1
(dr)
follows from part (ii). 203
R.L. Schilling: Measures, Integrals & Martingales ∑∑ Problem 16.9 Solution: We have 1 2
=
(0,ÿ)
y*1_2 e*y (dy).
Using the change of variables y = (x) = x2 , we get D (x) = 2x and 1 2
=2
(0,ÿ)
e*x
2
(dx) = 2
(*ÿ,ÿ)
e*x
2
16.16
(dx) =
˘
⇡. ∑∑
Problem 16.10 Solution: Write
=(
1,
`) 1 r )r D (r, ✓, !) = r ) 2 )r r) 3 p )r
2, ) 1 )✓ ) 2 )✓ ) 3 )✓
3 ).
Then
) 1 a )! ) 2s )! s ) 3s )! q
`cos ✓ cos ! *r sin ✓ cos ! *r cos ✓ sin !a r s = r sin ✓ cos ! r cos ✓ cos ! *r sin ✓ sin ! s r s 0 r cos ! q p sin ! Developing according to the bottom row we calculate for the determinant detD (r, ✓, !) = sin ! det
H I *r sin ✓ cos ! *r cos ✓ sin !
r cos ✓ cos ! *r sin ✓ sin ! H I cos ✓ cos ! *r sin ✓ cos ! + r cos ! det sin ✓ cos ! r cos ✓ cos ! ⇠ ⇡ = sin ! r2 sin2 ✓ cos ! sin ! + r2 cos2 ✓ cos ! sin ! ⇠ ⇡ + r cos ! r cos2 ✓ cos2 ! + r sin2 ✓ cos2 ! = r2 sin2 ! cos ! + r2 cos ! cos2 ! = r2 cos ! where we use repeatedly the elementary relation sin2
+ cos2
= 1.
Thus, Ã
u(x, y, z) d
3
(x, y, z)
R3
=
Ã
u˝ (r, ✓, !) det D (r, ✓, !) d
3
(r, ✓, !)
*1 (R3 )
=
ÿ
2⇡
⇡_2
0 0 *⇡_2
U (r cos ✓ cos !, r sin ✓ cos !, r sin !) r2 cos ! dr d✓ d!. ∑∑
204
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 16.11 Solution: (i) We change in (x) = variables according to u2 = t, and get (x) = 2
ÿ
0
0
ÿ
e*t tx*1 dt
2
e*u u2x*1 du.
Using Tonelli’s theorem we find 0 ÿ 10 ÿ 1 *u2 2x*1 *v2 2y*1 (x) (y) = 4 e u du e v dv 0 0 =4
(0,ÿ)2
e*u
2 *v2
u2x*1 v2y*1 d(u, v).
(ii) We have to show that B(x, y) (x + y) = (x) (y). Using polar coordinates in (i) we see (x) (y) = 4 =4
ÿ
r=0 0 ÿ r=0
2⇡
2
e*r r2x+2y*1 (cos )2x*1 (sin )2y*1 d dr =0 1H *r2 2x+2y*1
e
r
dr
⇡_2
=0
2x*1
(cos )
(sin )
I
2y*1
d
(?)
.
Setting s := r2 we see ÿ
r=0
2
e*r r2x+2y*1 dr =
ÿ
1 1 e*s s(x+y)*1 ds = (x + y). 2 s=0 2
Change variables in the second integral of (?) according to t = cos2 cos2
= 1. This yields
⇡_2 =0
(cos )2x*1 (sin )2y*1 d =
and use sin2
+
1
1 1 t2x*1 (1 * t)2y*1 dt = B(x, y). 2 0 2 ∑∑
Problem 16.12 Solution: We introduce planar polar coordinates as in Example 16.15: (x, y) = (r cos ✓, r sin ✓),
r > 0, ✓ À [0, 2⇡).
Thus, À
xm yn d
2
(x, y)
ÒxÒ2 +ÒyÒ2 0
for any choice of signs, thus À
xm yn d
2
(x, y) = 4
ÒxÒ2 +ÒyÒ2 0
Introducing planar polar coordinates yields, as seen above, for even m and n, 4
À
xm yn d
2
(x, y) =
ÒxÒ2 +ÒyÒ2 0, y>0
4 m + n + 2 0
⇡_2
cosm ✓ sinn ✓ d✓
1
m*1 n*1 4 (1 * t2 ) 2 (t2 ) 2 t dt m + n + 2 0 ˘ ˘ where we use the substitution t = sin ✓ and cos ✓ = 1 * sin2 ✓ = 1 * t2 . A further substitution
=
s = t2 yields
=
1
m*1 n*1 2 (1 * s) 2 s 2 ds m + n + 2 0
1
m+1 n+1 2 (1 * s) 2 *1 s 2 *1 ds m+n+2 0 2 = B m+1 , n+1 2 2 m+n+2
=
which is Euler’s Beta function. There is a well-known relation between the Euler Beta- and Gamma functions:
B(x, y) =
206
(x) (y) (x + y)
(*)
Solution Manual. Chapter 1–13. Last update 23rd August 2017 so that, finally,
À
ÒxÒ2 +ÒyÒ2 0. It is enough to show that there is some h À C
such that Òf * hÒp Õ ✏. Since D is dense in Lp ( ), there exists some g À D satisfying Òf * gÒp Õ ✏_2. On the other hand, as C is dense in D, there is some h À D such that Òg * hÒp Õ ✏_2. Now the triangle inequality gives
Òf * hÒp Õ Òf * gÒp + Òg * hÒp Õ
✏ ✏ + . 2 2 ∑∑
Problem 17.2 Solution: (i) Continuity follows from the continuity of the function x ;ô d(x, A), cf. (17.1). Clearly,
0 Õ uk Õ 1 and uk K = 1 and uUkc = 0. Since UK ö K, we get uk ö 1K . Since U k is closed and bounded, it is clear that U k is compact, i.e. supp uk is compact.
(ii) This follows from (i) and monotone convergence.
(iii) We have (K) = ⌫(K) for all compact sets K œ Rn and the compact sets generate
the Borel -algebra. In particular, this holds for [*k, k]n ~ Rn , so that the conditions for the uniqueness theorem for measures (Theorem 5.7) are satisfied. We conclude that = ⌫.
(iv) Since each x has a compact neighbourhood, we can choose k so large that B1_k (x) ∑ becomes compact. In particular, K œ xÀK B1_k(x) (x) is an open cover. We can choose
each k(x) so large, that B1_k(x) (x) has a compact closure. Since K is compact, we find ∑ finitely many xi such that K œ i B1_k(xi ) (xi ) = Uk where k := maxi ki . In particular, ∑ U k œ i B1_(xi ) (xi ) is compact. This produces a sequence of Uk ö K. The rest follows almost literally as in the previous steps.
∑∑ Problem 17.3 Solution:
(i) We have to show that Ò⌧h f Òpp = Òf Òp for all p À Lp (dx). This is an immediate consequence of the invariance of Lebesgue measure under translations: Ò⌧h f Òpp =
R
f (x * h)p dx =
R
f (y)p dy = Òf Òpp .
209
R.L. Schilling: Measures, Integrals & Martingales (ii) We show the assertion first for f À Cc (R). If f À Cc (R), then K := supp f is compact. Pick R > 0 in such a way that K + B1 (0) œ BR (0). Since limhô0 f (x * h) = f (x) and f (x * h) * f (x) Õ 2Òf Òÿ 1B
R (0)
(x) À Lp (dx)
for any h < 1, we can use dominated convergence to get Ò⌧h f * f Òpp =
Now take f À
Lp (dx).
f (x * h) * f (x)p dx ,,,,,,,,ô , 0.
Since Cc (R) is dense in
Lp (dx),
hô0
cf. Theorem 17.8, there is a
sequence (fn )nÀN œ Cc (R) such that Òfn * f Òp ô 0. From part (i) we get Ò⌧h f * f Òp Õ Ò⌧h (f * fn )Òp +Ò⌧h fn * fn Òp + Òfn * f Òp ´≠≠≠≠≠≠Ø≠≠≠≠≠≠¨ ÕÒfn *f Òp
,,,,,,,,ô , 2Òfn * f Òp ,,,,,,,,,ô , 0. nôÿ
hô0
This finishes the proof of the first assertion. The second claim follows in a similar way. Consider first f À Cc (R) and K := supp f . Since K is compact, there is some R > 0 with (h + K) „ K = Á1 for all h > R. If h > R, then
f (x * h) * f (x)p = f (x * h)p 1K (x + h) + f (x)p 1K (x) and so Ò⌧h f * f Òpp = =
K+h
K
f (x * h)p dx +
f (y)p dy +
K
K
f (x)p dx
f (x)p dx
= 2Òf Òpp . This proves the assertion for f À Cc (R), and the general case follows via density as in the first part of (ii).
∑∑ Problem 17.4 Solution: (i) Continuity is an immediate consequence of the dominated convergence theorem: assume
that (xn )nÀN is a sequence converging to x À R. Since 1[xn *h,xn +h] ô 1[x*h,x+h] a.e. and f À L1 (dx), we see that Mh f (xn ) ô Mh f (x) as n ô ÿ. Contractivity of Mh follows from
Mh f (x) dx = Õ
1 2h
Û x+h Û Û Û f (t) dtÛ dx Û Û x*h Û Û Û
h
1 f (x + t) dx dt Õ Òf Ò1 2h *h ´≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠¨ î f (y) dy=Òf Ò1
(use Tonelli’s theorem to interchange the order of integrations).
1
We use the notation h + K := {h + x; x À K}.
210
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (ii) Assume first that f À Cc (R). Because of the continuity of the function f we find Mh f (x) * f (x) Õ
h
1 f (x + t) * f (x) dx Õ sup f (x + t) * f (x) ,,,,,,,,ô , 0 hô0 2h *h tÀ[*h,h]
for all x À R. Since the support of f , K := supp f , is compact, there is some R > 0 such that K + B1 (0) ” BR (0). For h < 1 we get Mh f (x) = 0 = f (x) if x à BR (0). Since Mh f (x) Õ f (x) for x À R, we get
Mh f (x) * f (x) = Mh f (x) * f (x)1B
R (0)
(x) Õ 2Òf Òÿ 1B
R (0)
(x) À L1 (dx).
An application of the dominated convergence theorem reveals ÒMh f * f Ò1 =
hô0
Mh f (x) * f (x) dx ,,,,,,,,ô , 0,
i.e. the claim is true for any f À Cc (R). Now we take a general f À L1 (dx). Because of Theorem 17.8 there is a sequence (fn )nÀN œ Cc (R) such that Òfn * f Ò1 ô 0. Therefore, ÒMh f * f Ò1 Õ ÒMh (f * fn )Ò1 +ÒMh fn * fn Ò1 + Òfn * f Ò1 ´≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠¨ =Òfn *f Ò1
,,,,,,,,ô , 2Òfn * f Ò1 ,,,,,,,,,ô , 0. nôÿ
hô0
∑∑ Problem 17.5 Solution:
(i) Let A À B(X) such that f := 1A À Lp ( ). Clearly, (A) < ÿ and because of the outer regularity of
there is an open set U œ X such that A œ U and (U ) < ÿ.
Literally as in the proof of Lemma 17.3 we can construct some with Òf *
✏ Òp
✏
À CLip (X) „ Lp ( )
Õ ✏ (just replace in the proof Cb (X) with CLip (X)).
(ii) If f À Lp ( ), then the Sombrero lemma shows that there is a sequence of simple functions (fn )nÀN satisfying 0 Õ fn Õ f , fn ~ f . Using the monotone convergence theorem, we see î (f * fn )p d
ö 0; in particular, there is some n À N such that
Òfn * f Òp Õ ✏. Using linearity and the result of part (i), we get some such that Òfn *
✏ Òp
Õ ✏. Therefore,
Òf *
✏ Òp
Õ Òf * fn Òp + Òfn *
✏ Òp
✏
À CLip (X)
Õ 2✏.
(iii) We use the decomposition f = f + * f * . Since f + , f * À Lp ( ), part (ii) furnishes functions ,
Consequently,
À CLip (X) „ Lp ( ) such that Òf + * Òp Õ ✏ and Òf * *
Òp Õ ✏.
Òf * ( * )Òp Õ Òf + * Òp + Òf * * Òp Õ 2✏. ∑∑
211
R.L. Schilling: Measures, Integrals & Martingales Problem 17.6 Solution: A set U œ X is said to be relatively compact if it closure U is compact. (i) Let (xn )nÀN be a countable dense subset of X. By assumption, each xn has a relatively
compact open neighbourhood: xn À Vn and V n is compact. Since B1_k (xn ) œ Vn for sufficiently large values of k Œ k0 (xn ), we see that the balls B1_k (xn ), k Œ k0 (xn ), are also relatively compact. Thus,
{B1_k (xn ) : n À N, k Œ k0 (xn )} =: {Un ; n À N} is a sequence of relatively compact, open sets. For any open set U œ X we find U=
Õ nÀN Un œU
Un .
(The inclusion ‘–’ is obvious. In order to see ‘œ’ we observe that for any x À U there is some r > 0 with Br (x) œ U . Since (xn )nÀN is dense, we may choose n À N and k Œ k0 (xn ) such that B1_k (xn ) œ Br (x) œ U .)
(ii) The sets Kn := U 1 ‰ 5 ‰ U n are compact and increase towards X.
(iii) Assume that U œ X is an open set such that (U ) < ÿ and let (Un )nÀN be the sequence
from part (i). Because of (i), there is a subsequence (Un(k) )kÀN œ (Un )nÀN such that ∑ ∑ U = k Un(k) . Set Wn := nk=1 Un(k) and observe that Wn À D. Since Wn ~ U , Beppo Levi’s theorem shows that
This tells us that 1U À D.
(iv) First we show that
Ò1Wn * 1U Òp ,,,,,,,,,ô , 0. nôÿ
is outer regular. Set Gn :=
n Õ k=1
Uk .
Obviously, the Gn are open sets, Gn ~ X and (Gn ) < ÿ – here we use that the Uk are relatively compact and that
is finite on compact sets. This means that the assumptions
of Theorem H.3 are satisfied, and we see that
is outer regular.
Let B À B(X), (B) < ÿ and fix ✏ > 0. Since
is outer regular, there is a sequence
of open sets (Un )nÀN such that Un – B and (Un ) < ÿ. By monotone convergence, Ò1Un * 1B Òp ô 0 as n ô ÿ. Pick n À N such that Ò1Un * 1B Òp Õ ✏. Because of (iii), there is some D À D with Ò1Un * 1D Òp Õ ✏. Consequently,
Ò1B * 1D Òp Õ ÒIB * 1Un Òp + Ò1Un * 1D Òp Õ 2✏. (v) By definition, D œ Lp ( ), i.e. it is enough to show that for every f À Lp ( ) and
✏ > 0 there is some D À D such that Òf * 1D Òp Õ ✏. Using the Sombrero lemma
(Corollary 8.9) and the dominated convergence theorem we can construct a sequence 212
Solution Manual. Chapter 1–13. Last update 23rd August 2017 of simple functions (fn )nÀN œ Lp ( ) such that Òf * fn Òp ô 0. If n is sufficiently large, we have Òf * fn Òp Õ ✏. Since fn is of the form fn (x) =
N … j=1
cj 1Bj (x)
where cj À R, Bj À B(X), j = 1, … , N, we can use part (iv) to get D À D with Òfn * 1D Òp Õ ✏. With the triangle inequality we see that Òf * 1D Òp Õ 2✏. The separability of Lp ( ) now follows from the fact that D is a countable set.
∑∑ Problem 17.7 Solution:
(i) Assume first that A is an open set. Without loss of generality A ë Á. Fix ✏ > 0. Since $ % x À A : d(x, Ac ) < 1n ö Á as n ô ÿ the continuity of measures furnishes some N À N such that $ % d( , Ac ) < 1n < ✏ ≈n Œ N. Define
n (x)
Since 0 Õ
n
:= min{nd(x, Ac ), 1}. Clearly, Õ 1A À
therefore, {1A ë ,,,,,,ô , n Òp ,,,nôÿ
functions
n
Lp
n}
we even have
n À Cb (X) and Ò ✏ Òÿ p L ( ). Moreover,
À $ % {1A ë n } œ d( , Ac ) < 1n ; n
Õ 1 = Ò1A Òÿ .
Õ ✏ for all n Œ N. Using dominated convergence gives Ò1A *
0. If n Œ N is large enough, we get Ò1A * satisfy all requirements of the theorem.
n Òp
Õ ✏. For such n, the
In order to show the claim for any Borel set A À B(X), we proceed as in the proof of Lemma 17.3: let U œ X, (U ) < ÿ, and define D := {A À B(U ) : ≈✏ > 0
«
✏
À Cb (X)„Lp ( ) satisfying the assertion for f = 1A }.
As in the proof of Lemma 17.3 we see that D is a Dynkin system. By construction, the open sets are contained in D, and so B(U ) œ D.
If A À B(X) is an arbitrary Borel set with 1A À Lp ( ), we have (A) < ÿ. Since
is outer regular, there exists an open set U œ X such that A œ U and (U ) < ÿ. Since A À B(U ) œ D, the claim follows.
(ii) Let f À Lp ( ), 0 Õ f Õ 1, and fix ✏ > 0. Without loss of generality we may assume
that Òf Òÿ = 1, otherwise we would use f _Òf Òÿ . The (proof of the) Sombrero lemma (Theorem 8.8) shows that n2n *1
… k 1$ k Õf < k+1 % + n1{f >n} fn := n 2 2n 2n k=0
0Õf Õ1
=
2n *1
… k 1$ k Õf < k+1 % , n 2 2n 2n k=0
n À N,
213
R.L. Schilling: Measures, Integrals & Martingales monotonically converges to f . With f0 := 0 we get n … … … 1 (fj * fj*1 ) = (fj * fj*1 ) = nôÿ 2j j=1 jŒ1 jŒ1
f = lim (fn * f0 ) = lim nôÿ
for
j
j
:= 2j (fj * fj*1 ). We claim that
$ % k ≈x À fj*1 = j*1 .. 2 $ % $ k k Indeed: By definition, fj attains on fj*1 = 2j*1 = 2j*1 Õf <
(??)
j (x) À {0, 1}
ues
2k 2j
and
2k+1 . 2j
In the first case, we have
j
= 0, in the latter
j
k+1 2j*1
%
only the val-
= 1. Thus,
happens if, and only if, $ % $ % 2k + 1 2k + 1 2k + 2 x À fj = = Õf < . 2j 2j 2j Therefore, we can write Aj := {
j
=1
= 1} in the following form
2n*1 *1 $
Õ
Aj = Since
j
j (x)
k=0
% 2k + 1 2k + 2 Õ f < . 2j 2j
= 1Aj , we get f=
… 1 1 . 2j Aj jŒ1
Observe that 1Aj Õ 2j f À Lp ( ). Because of part (i), there is for every j Œ 1 a function j,✏
À Cb (X) „ Lp ( ) such that Ò
j,✏
*
The function •
✏
✏
j Òp
:=
✏ , { 2j
≥
j,✏
jŒ1 2j
j,✏
ë
j}
Õ
and
✏ 2j
Ò
j,✏ Òÿ
ÕÒ
j Òÿ
Õ 1.
enjoys all required properties:
is continuous (since it is the uniform limit of continuous functions): Ù Ù Ù Ù Ù Ù
• Ò
Õ
✏
≥
*
Ù
n …
j,✏ Ù Ù j Ù 2 Ù j=1 Ùÿ
≥
ÿ … 1 Õ Ò 2j j=n+1
ÿ … 1 ,,,,,,,,,ô , 0. j,✏ Òÿ Õ 2j nôÿ j=n+1
= 1 = Òf Òÿ . ≥ • Ò ✏ * f Òp Õ jŒ1 21j Ò j,✏ * j Òp Õ ✏ jŒ1 21j Õ ✏. In particular, ≥ ≥ • { ✏ ë f } Õ jŒ1 { j,✏ ë j } Õ jŒ1 ✏2*j = ✏. ✏ Òÿ
Õ
Ò
jŒ1
≥
j,✏ Òÿ 2j
Õ
1 jŒ1 2j
✏
À Lp ( ).
(iii) Observe, first of all, that the theorem holds for all g À Lp ( ) with 0 Õ g Õ ÒgÒÿ < ÿ;
for this, apply part (ii) to g_ÒgÒÿ . Without loss of generality we may assume for such g that ✏ Œ 0; otherwise we would consider õ✏ := ✏ ‚ 0. Let f À Lp ( ) and Òf Òÿ < ÿ. We write f = f + * f * and, because of the preceding remark, there are functions Ò
214
✏ Òÿ
✏,
Õ Òf + Òÿ ,
✏
À Cb (X) „ Lp ( ), f+ ë
✏
Õ✏
✏
and
Π0,
✏
Π0, such that
Òf + *
✏ Òp
Õ✏
Solution Manual. Chapter 1–13. Last update 23rd August 2017 and Ò For
✏
:=
*
✏
✏
À Cb (X) „ Lp ( ) we find
✏
{
ë f} Õ {
✏
as well as Ò
✏ Òÿ
(this step requires that
✏
Õ✏
and
ë f +} + {
✏
Òf * *
✏ Òp
Õ ✏.
ë f * } Õ 2✏
Õ max{Òf + Òÿ , Òf * Òÿ } = Òf Òÿ
Π0 and
✏
Òf * Consequently,
f* ë
Õ Òf * Òÿ ,
✏ Òÿ
✏
Π0). The triangle inequality yields
Õ Òf + *
✏ Òp
✏ Òp
+ Òf * *
✏ Òp
Õ 2✏.
satisfies the conditions of the theorem for f .
✏
(iv) Fix f À Lp ( ) and ✏ > 0. Using the Markov inequality we get {f Œ R} Õ
1 f p d . Rp
In particular, we can pick a sufficiently large R > 0 such that {f Œ R} Õ ✏. Using monotone convergence, we see
{f >R}
f p d < ✏
if R > 0 is large. Setting fR := (*R) ‚ f · R, we can use (iii) to construct a function ✏
À Cb (X) „ Lp ( ) with Ò
✏ Òÿ
Obviously, Ò Ò
✏
Õ ÒfR Òÿ ,
✏ Òÿ
fR ë
Õ Òf Òÿ . Moreover,
* f Òpp
=
Õ
✏
✏
* f p d +
{f ÕR}
✏
✏ Rp
and
* f p d +
{f >R} „{ ✏ =fR }
´≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠¨ =:I1
ÕÒ
✏
*
ÒfR *
✏ Òp
Õ ✏.
* f p d
✏
{f >R} „{ ✏ ëfR }
´≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠¨ =:I2
+ I1 + I2 .
fR Òpp
Let us estimate I1 and I2 separately. Since fR {f >R} = R, we get I1 =
{f >R}„{
Õ
{f >R}„{
✏ =fR }
(f * R)p d +
fp d + {f R}
{f R}„{ {
✏ ëfR }
p ✏ Òÿ
Õ C(p)Rp
✏ + C(p)✏. Rp
✏
Ò Since ✏ > 0 is arbitrary, Ò
{f ë
This shows that
✏
✏
p
d + C(p)
ë fR } + C(p)
Õ C(p)Ò
Therefore,
✏
{f >R}„{
{f >R}
✏ ëfR }
f p d
f p d
* f Òpp Õ ✏ p + ✏ + 2C(p)✏.
* f Òp is as small as we want it to be. Finally,
✏
✏}
Õ {fR ë
✏}
+ {f Œ R} Õ 2✏.
enjoys all required properties.
Remark: (]) follows from Hölder’s inequality I1 H n I1 Û… Û H… n p q … Û n Û p q Û xj yj Û Õ x y j j Û Û Û j=1 Û j=1 j=1 Û Û
for x, y À Rn and conjugate indices p, q Œ 1. If we take, in particular, d = 2, x = (a, b), y = (1, 1), then
1
1
a 1 + b 1 Õ (ap + bp ) p 2 q . Raising both sides to the pth power proves the estimate. ∑∑
Problem 17.8 Solution: We see immediately that îa p(x)f (x) dx = 0 for all polynomials p. Fix b
g À C[a, b] and ✏ > 0. By Weierstraß’ theorem, there is some polynomial p such that Òg*pÒÿ Õ ✏. Therefore,
b Û b Û Û b Û Û Û g(x)f (x) dxÛ = ÛÛ (g(x) * p(x))f (x) dx + p(x)f (x) dx ÛÛ Û Û a Û Û a a Û Û Û ´≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠¨ =0
Õ
b
p(x) * g(x) f (x) dx a ´≠≠≠≠≠Ø≠≠≠≠≠¨ Õ✏
Õ✏ From this we conclude that a 216
b
a
b
f (x) dx.
g(x)f (x) dx = 0
≈g À C[a, b].
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Define measures
±
by
± (dx)
:= 1[a,b] (x)1{±f >0} (x) dx. Then î g d
+
= î gd
g À C[a, b]. According to Theorem 17.12, C[a, b] is a determining set, and so only possible if
+
= 0, hence f = 0 Lebesgue a.e.
=
* *.
for all
This is ∑∑
Problem 17.9 Solution: (i) First of all, we note that it is enough to know that the polynomials are uniformly dense
in the set C[*1, 1]. This follows immediately from the observation that any function in C[0, 1] can be mapped onto C[a, b] using the affine transform a + t(b * a), t À [0, 1] – and vice versa. Fix u À C[*1, 1] and define a sequence of polynomials (pn )nÀN by 1 cn
pn (x) :=
0
x2 *1 16
1n
x À R,
,
where cn := î*4 (x2 _16*1)n dx. Since u À C[*1, 1], there is some õ u À C(R) such that 4
õ u(x) = 0 for x > 2 and õ u(x) = u(x) for x À [*1, 1]. Define põn (x) := pn (x)1[*4,4] (x) and
un (x) := õ u ? põn (x) = We find un (x) = since
õ u(x * y)õ pn (y) dy,
x À R.
õ u(x * y)pn (y) dy ≈x À [*2, 2],
x Õ 2 -Ÿ õ u(x * y) = 0
≈y > 2.
Using the fact that un (x) =
õ u(y)pn (x * y) dy,
x À [*2, 2]
we see that un [*2,2] is a polynomial. Let us show that un ô õ u converges uniformly – and since õ u[*1,1] = u, the claim follows. Using that põn Œ 0 and î põn dx = 1 we get Û Û Ûu (x) * õ u(x)ÛÛ = ÛÛ (õ u(x * y) * õ u(x))õ pn (y) dyÛÛ Û n Û Û Õ
⌧ * R1 , R1
+
R‰
⌧
õ u(x * y) * õ u(x)õ pn (y) dy
* R1 , R1
õ u(x * y) * õ u(x)õ pn (y) dy
=: I1 (x) + I2 (x) for all R > 0. Let us bound I1 and I2 separately. Since õ u(x) = 0 for x > 2, the function õ u is uniformly continuous and we get I1 (x) Õ
sup
⌧ yÀ * R1 , R1
õ u(x * y) * õ u(x)
⌧ * R1 , R1
põn (y) dy
217
R.L. Schilling: Measures, Integrals & Martingales Õ
sup
⌧ yÀ * R1 , R1
õ u(x * y) * õ u(x)
,,,,,,,,,,ô , 0 Rôÿ
uniformly for all x. Because of the boundedness of õ u we see that I2 (x) Õ 2Òõ uÒÿ
R‰
⌧ * R1 , R1
põn (y) dy.
Since põn (y) ö 0 for all y ë 0, we can use the monotone convergence theorem to conclude that I2 ,,,,,,,,,ô , 0 uniformly in x. This proves the claim. nôÿ
(ii) Fix u À Cc [0, ÿ). Since u has compact support, u(x) = 0 for large x; in particular, u˝(* log)(x) = 0 if x is small. Therefore,
h nu˝(* log)(x), x À (0, 1] l x = 0, n0, j defines a continuous function on [0, 1]. According to (i), there is a sequence of polynomials (pn )nÀN with pn ô u˝(* log) uniformly.
(iii) For p(x) := xn we obviously have p(e*t ) = e*nt = ✏n (t) and, by assumption,
p(e*t ) (dt) =
✏n (t) (dt) =
✏n (t) ⌫(dt) =
p(e*t ) ⌫(dt).
(?)
Using the linearity of the integral, this equality extends to arbitrary polynomials p.
Assume that u À Cc [0, ÿ) and (pn )nÀN as in (ii). Since pn converges uniformly to u˝(* log), we can interchange integration and limit to get
ud =
(u˝(* log))(e*t ) (dt)
nôÿ
= lim (?)
pn (e*t ) (dt)
nôÿ
= lim
pn (e*t ) ⌫(dt)
=
(u˝(* log))(e*t ) ⌫(dt)
=
u d⌫. ∑∑
218
18 Hausdorff measure. Solutions to Problems 18.1–18.7 Problem 18.1 Solution: This is clear from the monotonicity of the infimum and the fact that there are more P- -covers than C- -covers, i.e. we have H
,P (A)
ÕH
,C (A).
∑∑ Problem 18.2 Solution: measures
From the proof of Corollary 18.10 we know, using the monotonicity of
H (A) = H (G)
lim H (U k )
=
kôÿ
Uk – A
Œ
inf H (U ) : U – A, U open
ΠH (A).
U –A
When using the monotonicity we must make sure that H (U k ) < ÿ – this we can enforce by U k ‰ U k „ U (where U is the open set with finite Hausdorff measure).
For counting measure this is clearly violated: Any open set U – A := {a} has infinitely many points! Nevertheless A is itself a G -set.
∑∑ Problem 18.3 Solution: By Corollary 18.10 there are open sets Ui such that H := H (H ‰ B) = 0 or H (H) = H (B). Now we can write each Ui as an F -set: Ui =
Õ Br (x)œUi ,xÀUi
∂
i Ui
– B and
Br_2 (x)
is indeed a countable union of closed sets, since Ui œ X contains a countable dense subset. So we have
Ui =
Õ k
Fik
for closed sets Fik .
Without loss of generality we may assume that the sets Fik increase in k, otherwise we would
consider Fi1 ‰ 5 ‰ Fik . By the continuity of measure (here we require the measurability of B!) we have
lim H (B „ Fik ) = H (B „ Ui ) = H (B).
kôÿ
219
R.L. Schilling: Measures, Integrals & Martingales In particular, for every ✏ > 0 there is some k(i) with
H (B ‰ Fik(i) ) Õ ✏_2i ,
Consider the closed set F =
∂
i Fik(i)
and observe that
H (F ) Œ H (F „ B) Œ H (B) *
Since F œ
∂
i Ui ,
i À N.
… i
we get H (F ‰ B) Õ H
0
H (B ‰ Fik(i) ) Œ H (B) *
à i
1 Ui ‰ B
= H (H ‰ B) = 0.
By Corollary 18.10, the set F ‰ B is contained in a G -set G = such that H (G) = 0 = H (F ‰ B). Thus,
F ‰G =F „
Õ i
Vic =
… ✏ = H (B) * ✏. 2i i
Õ i
∂
i Vi
(where the Vi are open sets)
F „ Vic ´Ø¨ closed
is an F -set inside B – we have F ‰ G œ F ‰ (F ‰ B) œ B – and
H (F ‰ G) Œ H (F ) * H (G) Œ H (B) * ✏.
Now consider ✏ =
1 n
and take unions of the thus obtained F -sets. But, clearly, countable unions
of F -sets are still F .
∑∑ Problem 18.4 Solution: Fix A œ Rn . We have to show that for any Q œ Rn the equality #Q = #(Q „ A) + #(Q ‰ A) holds. We distinguish between two cases. Case 1: #Q = ÿ. Then at least one of the terms #(Q „ A), #(Q ‰ A) on the right-hand side must be infinite, so the equality is clear.
Case 2: #Q < ÿ. Then both sets (Q „ A), (Q ‰ A) are finite and, as such, they are metrically separated. Therefore we can use the fact that H (A) = #(A) is a metric outer measure (Theorem 18.5) 0
to get equality.
∑∑ Problem 18.5 Solution:
Use Lemma 18.17 to see 0 Õ dimH B Õ dimH Rn as B œ Rn . From
Example 18.18 we know that dimH Rn = n.
If B contains an open set U (or a set of non-zero Lebesgue measure), we see H n (B) ΠH n (U ) > 0;
intersect with a large open ball K to make sure that H n (B „ K) < ÿ and U „ K œ B „ K. This shows n = dimH (B „ K) Õ dimH (B) Õ n.
∑∑
220
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 18.6 Solution: By self-similarity, we see for the Sierpinski triangle of generation i, S i*1
and its follow-up stage S i = S1i ‰ S2i ‰ S3i that the Ski ’s are scaled versions of S with a factor 12 . So,
H s (S i*1 ) = H s (S1i ) + H s (S2i ) + H s (S3i ) = 3 2*s H s (S i*1 )
and dividing by H s (S i*1 ) and solving the equality 1 = 3 2*s ◊Ÿ 2s = 3 ◊Ÿ s = log 3_ log 2 Koch’s snowflake S has in each subsequent generation stage 4 new parts, each scaled by 1_3, so H s (S) = H s (S1 ) + H s (S2 ) + H s (S3 ) + H s (S4 ) = 4 3*s H s (S)
and dividing by H s (S) and solving the equality 1 = 4 3*s ◊Ÿ 3s = 4 ◊Ÿ s = log 4_ log 3. ∑∑ Problem 18.7 Solution: Let (Si )iÀN be an ✏-cover of A. Then we have ÿ … i=1
ÿ … (diam Ui ) (diam Ui ) = (diam Ui ) (diam Ui ) i=1
Õ
ÿ …
sup
i=1 xÕ✏
(x) (diam Ui ) (x) ÿ
(x) … = sup (diam Ui ). (x) i=1 xÕ✏ Taking the inf over all admissible ✏-covers shows H ✏ (A) Õ sup xÕ✏
(x) (x) H ✏ (A) Õ sup H (A). (x) (x) xÕ✏
Letting ✏ ô 0 yields
H (A) = lim H ✏ (A) Õ lim sup ✏ô0
✏ô0 xÕ✏
(x) (x) H (A) = lim sup H (A) = 0. (x) (x) xô0 ∑∑
221
19 The Fourier transform. Solutions to Problems 19.1–19.9 Problem 19.1 Solution: (a) By definition, 1 1¶ 1 (x)e*i x⇠ dx [*1,1] (⇠) = 2⇡ [*1,1] 4 *i x⇠ 51 1 e = * 2⇡ i ⇠ x=*1 1 1 i⇠ = e * e*i ⇠ 2⇡ i ⇠ 1 sin ⇠ = ⇡ ⇠
for ⇠ ë 0. Here we use that sin ⇠ = Im ei⇠ =
1 i⇠ (e 2i
* e*i ⇠ ). For ⇠ = 0 we have
1 1 1¶ 1 (x) dx = . [*1,1] (0) = 2⇡ [*1,1] ⇡ (Note that expect.)
sin ⇠ ⇠
ô 1 as ⇠ ô 0, i.e. the Fourier transform is continuous at ⇠ = 0 – as one would
(b) The convolution theorem, Theorem 19.11, shows that f£ < g = (2⇡)fÇ g. Ç Because of part (a) we get
0 F(1[*1,1] < 1[*1,1] )(⇠) = (2⇡)
1 sin ⇠ ⇡ ⇠
12
=
2
2 sin ⇠ . ⇡ ⇠2
(c) We get from the definition that ÿ
1 e*x e*i x⇠ dx 2⇡ 0 ÿ 1 = e*x(1+i ⇠) dx 2⇡ 0 1 1 ⌅ *x(1+i ⇠) ⇧ÿ =* e x=0 2⇡ 1 + i ⇠ 1 1 = . 2⇡ 1 + i ⇠
F(e*( ) 1[0,ÿ) ( ))(⇠) =
(d) Obviously, we have
e*i x⇠ e*x =
(*ÿ,0)
e*i x⇠ ex dx +
(0,ÿ)
e*i x⇠ e*x dx
223
R.L. Schilling: Measures, Integrals & Martingales =
(0,ÿ)
ei y⇠ e*y dy +
(0,ÿ)
e*i x⇠ e*x dx.
Thus, F(e* )(⇠) = F(e* 1[0,ÿ) )(*⇠) + F(e* 1[0,ÿ) )(⇠) 0 1 (c) 1 1 1 = + 2⇡ 1 * i ⇠ 1 + i ⇠ 1 1 = . ⇡ 1 + ⇠2 (e) From (d) and F˝Fu(x) = (2⇡)*1 u(*x) (cf. Corollary 19.24) we find 0 F
1 1 + x2
1
(d) 1 1 (⇠) = ⇡ F˝F(e* )(⇠) = e**⇠ = e*⇠ . 2 2
(f) Note that [*1,1]
(1 * x)e*i x⇠ dx =
[*1,1]
e*i x⇠ dx +
[*1,0]
xe*i x⇠ dx *
[0,1]
xe*i x⇠ dx
=
[*1,1]
e*i x⇠ dx +
[0,1]
(*y)ei y⇠ dy *
[0,1]
xe*i x⇠
=
[*1,1]
e*i x⇠ dx *
x (ei x⇠ + e*i x⇠ ) dx. [0,1] ´≠≠≠≠≠Ø≠≠≠≠≠¨ 2 cos(x⇠)
The first expression is as in part (a). For the second integral we use integration by parts: 0
1
4
sin(x⇠) x cos(x⇠) dx = x ⇠
51
1
1 sin(x⇠) dx ⇠ 0 x=0 4 51 sin(⇠) 1 cos(x⇠) = * ⇠ ⇠ ⇠ x=0 sin(⇠) cos(⇠) 1 = * + 2. ⇠ ⇠2 ⇠
Thus, 1 sin ⇠ 1 F(1[*1,1] (1 * ))(⇠) = * ⇡ ⇠ ⇡
0
*
sin ⇠ cos ⇠ 1 * 2 + 2 ⇠ ⇠ ⇠
1 =
1 1 * cos ⇠ . ⇡ ⇠2
(g) By definition, H F
ÿ k … t *t e k! k=0
I k
(⇠) =
ÿ k ÿ … 1 t *t 1 … tk *t *i k⇠ e*i x⇠ e k (dx) = e e . 2⇡ k! 2⇡ k! k=0 k=0
Since e*i k⇠ = (e*i ⇠ )k , we conclude that Hÿ I ÿ … tk 1 … (te*i ⇠ )k *t 1 *t te*i ⇠ 1 t(e*i ⇠ *1) *t F e k (⇠) = e = e e = e . k! 2⇡ k! 2⇡ 2⇡ k=0 k=0
224
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (h) The same calculation as in (g) yields H k 0 1 I n 0 1 … n … 1 n k n*k k n*k *i x⇠ F p q e p q k (⇠) = k (dx) k 2⇡ k n=0 k=0 n 0 1 1 … n k n*k *i ⇠k = p q e 2⇡ k=0 k n 0 1 1 … n = (pe*i ⇠ )k q n*k 2⇡ k=0 k =
1 pe*i⇠ + q 2⇡
n
.
In the final step we use the binomial theorem.
∑∑ Problem 19.2 Solution: Observe that for complex numbers u, v À C u + v2 = (u + v)(u + v) = (u + v)(uÑ + v) Ñ = uuÑ + uvÑ + vuÑ + vvÑ = u2 + 2 Re uvÑ + v2 and so, setting v ‰ *v u * v2 = u2 * 2 Re uvÑ + v2 and so, setting v ‰ iv u + iv2 = u2 + 2 Im uvÑ * v2 and so, setting v ‰ *iv u * iv2 = u2 * 2 Im uvÑ * v2 And this gives u + v2 * u * v2 + iu + iv2 * iu * iv2 = 4 Re uvÑ + 4i Im uvÑ = 4uv. Ñ Thus, we have the following ‘polarization’ formula 4 5 1 2 2 2 2 uvÑ dx = u + v dx * u * v dx + i u + iv dx * i u * iv dx 4 ⇧ 1⌅ = Òu + vÒ22 * Òu * vÒ22 + iÒu + ivÒ22 * iÒu * ivÒ22 4 and now the claim follows directly from the statement of Plancherel’s theorem.
Alternative solution: Mimic the proof of Theorem 19.20: We have u, v, ö u, vö À L2 ( n ) (as a result of Theorem 19.20), and so u vÑ and ö u vö are integrable. Therefore,
ö u(⇠)ö v(⇠) d⇠ = (2⇡)*n
ö u(⇠)ú v(⇠) d⇠
225
R.L. Schilling: Measures, Integrals & Martingales 19.12
= (2⇡)*n
19.9
= (2⇡)*n
⌅ ⇧ u(x)F ú v (x) dx
u(x)v(x) dx. ∑∑
Problem 19.3 Solution: Assume that õ = . We have (⇠)
e*ix⇠ (dx)
e*ix⇠ õ(dx)
=
e*i(*x)⇠ (dx)
=
e*ix⇠ (dx)
=
e*ix⇠ (dx)
= õ=
=
15.1
= Therefore,
is real-valued. On the other hand, the above calculation shows that (⇠) =
This means that transform.
(⇠).
e*ix⇠ õ(dx).
entails F = F õ, and so
=
= õ because of the injectivity of the Fourier ∑∑
Problem 19.4 Solution: From linear algebra we know that a symmetric positive definite matrix has
a unique symmetric positive square root, i.e. there is some B À Rnùn which is symmetric and ˘ positive definite such that B 2 = A. Since det(B 2 ) = (det B)2 , we see that det B = det A > 0. Now we change coordinates according to y := Bx
e*iÍx,⇠Î e*Íx,AxÎ dx =
e*iÍx,⇠Î e*ÍBx,BxÎ dx
*1 2 1 e*iÍB y,⇠Î e*y dy det B *1 2 1 =˘ e*iÍy,B ⇠Î e*y dy. det A
=
If we set g1_2 (x) :=
1 ⇡ n_2
exp *x2 ,
cf. Example 19.2(iii), then the calculation from above gives ⇡ n_2 F(e*Í ,A Î )(⇠) = ˘ F(g1_2 ) B *1 ⇠ . det A 226
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Example 19.2(iii) now shows *Í ,A Î
F(e
0 1 B *1 ⇠2 ⇡ n_2 1 )(⇠) = ˘ exp * . n 4 det A (2⇡)
Finally, since B *1 = (B *1 )Ò , B *1 ⇠2 = ÍB *1 ⇠, B *1 ⇠Î = Í⇠, (B *1 B *1 )⇠Î, ´≠Ø≠¨ A*1
we infer that F(e
*Í ,A Î
0 1 Í⇠, A*1 ⇠Î 1 )(⇠) = ˘ exp * . n_2 n_2 4 det A 2 (2⇡) 1
1
∑∑ 2 _2t
Problem 19.5 Solution: gt (x) = (2⇡t)*1_2 e*x
and göt (⇠) = (2⇡)*1 e*t⇠
2 _2
orem (Theorem 19.20, plus polarization) or by Problem 19.2 we see that
2 _2
ö u(⇠)e*t⇠
d⇠ = (2⇡)
. By Plancherel’s the-
ö u(⇠)ö gt (⇠) d⇠
=
u(x)gt (x) dx
=
u(x)(2⇡t)*1_2 e*x
= (2⇡)*1_2
2 _2t
u(ty)e*y
2 _2
dx dy
Õ cÒuÒÿ . (In fact, c = 1, see Example 14.11). Now let t ~ 0 using monotone convergence and use that, by assumption, ö u Œ 0.
The same argument holds for L2 -functions since gt À L2 . ∑∑ Problem 19.6 Solution: We follow the hint and find using Fubini’s theorem 2
1_R ⇠ ⇡n 1_R R … (1 * eiÍx,⇠Î ) (dx) d⇠1 … d⇠d 2 *1_R *1_R Rn 1_R ⇠ ⇡n 1_R R =2 … (1 * eiÍx,⇠Î ) d⇠1 … d⇠d (dx) Rn 2 *1_R *1_R 1_R
1_R
R R … (1 * eiÍx,⇠Î ) d⇠1 … d⇠d (dx) Rn 2 *1_R 2 *1_R H I 1_R 1_R R R iÍx,⇠Î =2 1* … e d⇠1 … d⇠d (dx) Rn 2 *1_R 2 *1_R H I n 1_R « R =2 1* eixn ⇠n d⇠n (dx) Rn 2 *1_R =2
n=1
227
R.L. Schilling: Measures, Integrals & Martingales H
4 5⇠ =1_R I n « R eixn ⇠n n =2 1* (dx) Rn 2 ixn ⇠n =*1_R n=1 H I n « eixn _R * e*ixn _R =2 1* (dx) Rn 2ixn _R n=1 H I n « sin(xn _R) =2 1* (dx) Rn xn _R n=1 H I n « sin(xn _R) Œ2 1* (dx). Rn ‰[*2R,2R]n xn _R n=1 In the last step we use that the integrand is positive since sin y_y Õ 1. Observe that x À Rn ‰ [*2R, 2R]n ◊Ÿ «n = 1, … , n : xn > 2R and so
n « sin(xn _R) n=1
hence 2
xn _R
Õ
1 2
1_R ⇠ ⇡n 1_R R … (1 * eiÍx,⇠Î ) (dx) d⇠1 … d⇠d 2 *1_R *1_R Rn H I n « sin(xn _R) Œ2 1* (dx) Rn ‰[*2R,2R]n xn _R n=1 ⇠ ⇡ 1 Œ2 1* (dx) Rn ‰[*2R,2R]n 2
Œ
Rn ‰[*2R,2R]n
(dx).
Remark. A similar inequality exists for the Fourier transform (instead of the inverse Fourier transform). This has the form
(Rn ‰ [*2R, 2R]n ) Õ 2(⇡R)n
[*1_R,1_R]n
ö(0) * Re ö(⇠) d⇠. ∑∑
Problem 19.7 Solution: (i) Let ⇠1 , … , ⇠n À Rn and we get
n … i,k=1
1, … , n
(⇠j * ⇠k ) j Ñ k =
À C. From the definition of the Fourier transform
n 1 … (2⇡)n j,k=1
j k
Ñ
e*i x(⇠j *⇠k ) d (x)
n 1 … Ñ e*i x⇠j e*i x⇠k d (x) (2⇡)n j,k=1 j k H n IH n I … … 1 *i x⇠j *i x⇠k d (x) = je ke (2⇡)n j=1 k=1
=
228
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Û… Û n 1 Û = (2⇡)n ÛÛ j=1 Û
je
Û2 Û Û d (x) Œ 0. Û Û
*i x⇠j Û
Note that this already implies that (*⇠) = (⇠). The argument is as follows: If we ≥ have for a matrix (ajk ) that jk ajk j Ñ j Œ 0, then 0Õ
… jk
ajk
Ñ =
j k
… jk
ajk
Ñ =
j k
… jk
ajk Ñ j
k
=
… kj
akj Ñ k
j
which means that ajk = akj . Apply this to the matrix ajk = (⇠j * ⇠k ) with m = 2 and ⇠1 = ⇠ and ⇠2 = 0 to infer that (⇠) = (*⇠).
(ii) We want to use the differentiability lemma for parameter-dependent integrals. For this we define
u(⇠, x) := Since
1 *i x⇠ e . (2⇡)n
is a finite measure and u(x, ⇠) Õ (2⇡)*n , we find u(⇠, ) À L1 ( ). Moreover, )⇠j u(⇠, x) = (2⇡)*d xj Õ (2⇡)*d x Õ (2⇡)*d 1[*1,1] (x) + xm 1R‰[*1,1] (x) =: w(x) À L1 ( )
is an integrable majorant. With Theorem 12.5 we find )⇠j (⇠) = )⇠j
u(⇠, x) (dx) =
1 (*ixj )e*i x⇠ (dx). (2⇡)n
Iterating this argument, we see that ) ↵ exists for any ↵ À Nn0 such that ↵ Õ m.
(iii) We follow the hint and consider first the case d = 1 and n = 1. We can rewrite the expression (2h) * 2 (0) + (*2h) using Fourier transforms:
1 (e*i 2hx * 2 + ei 2hx ) (dx) 2⇡ 1 = (cos(2hx) * 1) (dx). ⇡
(2h) * 2 (0) + (*2h) =
L’Hospital’s theorem applies and gives 1 * cos(2y) yô0 1 ,,,,,,,,ô , . 2 4y2 Now we can use Fatou’s lemma
x2
1 * cos(2hx) 1 (dx) = x2 lim (dx) hô0 2 4(hx)2 1 Õ lim inf 2 (1 * cos(2hx)) (dx) hô0 4h 1 = *⇡ lim inf 2 (2h) * 2 (0) + (*2h)) hô0 4h = *⇡
®®
(0) < ÿ.
229
R.L. Schilling: Measures, Integrals & Martingales If n Π1, we use induction. Assume that proved for n*1. Since
À C 2n (R) Ÿ
À C 2n (R) and that the assertion has been
À C 2(n*1) , we see by the induction assumption
that î x2(n*1) d (x) < ÿ. Thus, ⌫(dx) := x2(n*1) (dx) is a measure and 1 x2(n*1) e*i x⇠ d (x) 2⇡ 1 1 d 2(n*1) = e*i x⇠ d (x). 2⇡ (*i)2(n*1) d⇠ 2(n*1)
⌫ö(⇠) =
Consequently, we see that ⌫Ç À C 2 (R). The first part of the proof (n = 1) gives
x2n d (x) =
x2 d⌫(x) < ÿ.
If d Œ 1, then we set ⇡j (x) := xj , x À Rn , j À {1, … , d}. Apply the case d = 1 to the measures ⇡j ( ).
(iv) Assume that z À Cn . If K := supp is compact, then we get, because of the continuity of e*i zx , that M := supxÀK e*i zx < ÿ. From we conclude that
ud =
i.e.
supp
ud
for any u Π0
e*i zx d (x) Õ M (Rn ) < ÿ, 1 e*i zx d (x) (2⇡)n
(z) = is well-defined. Setting
n 1 … (*i zx)k un (x) := , (2⇡)n k=0 k!
x À Rn ,
we get n 1 … zxk 1 zx 1 un (x) Õ Õ e Õ sup ezx < ÿ. (2⇡)n k=0 k! (2⇡)n (2⇡)n xÀK
Since
is a finite measure, we can use the dominated convergence theorem to get (z) =
lim u (x) nôÿ n
nôÿ
= lim
(dx)
un (x) d (x)
ÿ 1 … 1 = (zx)k d (x). (2⇡)n k=0 k!
This proves that
is analytic. ∑∑
230
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 19.8 Solution: Note that ei x_n ,,,,,,,,,ô , 1 for all x À R. On the other hand, we gather from nôÿ
îB eix_n dx = 0 that 1B ei _n À L1 (dx). As ei x_n = 1, we get convergence
nôÿ B
0 = lim
ei x_n dx =
lim ei x_n dx = B nôÿ ´≠≠Ø≠≠¨
1 (B)
1
< ÿ. By dominated
(B).
1
Alternative solution: Set f (x) := 1B (x); by assumption, fÇ(1_n) = 0. Since the Fourier transform is continous, cf. 19.3, we get
⇠ ⇡ 1 fÇ(0) = lim fÇ = 0. nôÿ n
On the other hand, fÇ(0) = (2⇡)*1 1 (B).
∑∑ Problem 19.9 Solution: (i) ◊: Since (R ‰
2⇡ Z) ⇠
= 0 we find =
… jÀZ
pj
2⇡ Z ⇠
with pj := ( 2⇡ j). From the definition of the Fourier transform we get ⇠ 1 e*i x⌘ (dx) 2⇡ 4 0 1 5 1 … 2⇡ = p exp *i j ⌘ 2⇡ jÀZ j ⇠
Ç (⌘) =
for all ⌘ À R. Setting ⌘ = ⇠, we see Ç (⇠) =
1 … p exp(*i2⇡j) 2⇡ jÀZ j ´≠≠≠Ø≠≠≠¨ 1
1 … = p exp(*i 0) = Ç (0). 2⇡ jÀZ j Ÿ: From Ç (⇠) = Ç (0) we conclude 2⇡( Ç (0) * Ç (⇠)) =
In particular, î (1 * e*i x⇠ ) (dx) À R, i.e.
(1 * e*i x⇠ ) (dx) = Re
(1 * e*i x⇠ ) (dx) = 0.
(1 * e*i x⇠ ) (dx) =
Since 1 * cos(x⇠) Œ 0, this implies
(1 * cos(x⇠)) (dx) = 0.
{x À R; 1 * cos(x⇠) > 0} = 0. Consequently, 0=
{x À R; cos(x⇠) ë 1} =
0
1 2⇡ R‰ Z . ⇠
231
R.L. Schilling: Measures, Integrals & Martingales (ii) Because of ö(⇠1 ) = ö(0) there is some z1 À R such that ö(⇠1 ) = ö(0)ei z1 ⇠1 . Therefore,
1 e*i ⇠1 (x+z1 ) (dx) = Ç (0). 2⇡
Observe that the left-hand side is just the Fourier transform of the measure ⌫(B) := (B * z1 ), B À B(R), and so
⌫(⇠ Ç 1 ) = Ç (0) = ⌫(0). Ç From part (i) we get that ⌫(R ‰
2⇡ Z) ⇠1
<
= 0. This is the same as 0
2⇡ R ‰ z1 + Z ⇠1
1= = 0.
Using the same argument we find some z2 À R, such that < 0 1= 2⇡ R ‰ z2 + Z = 0. ⇠2 Setting
0 A :=
1 0 1 2⇡ 2⇡ z1 + Z „ z2 + Z ⇠1 ⇠2
we see that (R ‰ A) = 0. Let us show that A contains at most one element: Assume, on the contrary, that there are two distinct points in A, then there are n, n® À Z and m, m® À Z such that
2⇡ n = z2 + ⇠1 2⇡ z1 + m = z2 + ⇠1 z1 +
2⇡ ® n, ⇠2 2⇡ ® m. ⇠2
Subtracting these identities, we get 2⇡ (n * m) = ⇠1 ⇠ Ÿ 2 = ⇠1 This is clearly contradicting the assumption
2⇡ ® (n * m® ) ⇠2 n® * m® À Q. n*m ⇠1 ⇠2
à Q. ∑∑
232
20 The Radon–Nikodým theorem. Solutions to Problems 20.1–20.9 Problem 20.1 Solution: The assumption ⌫ Õ
immediately implies ⌫ ~ . Indeed,
(N) = 0 -Ÿ 0 Õ ⌫(N) Õ (N) = 0 -Ÿ ⌫(N) = 0. Using the Radon–Nikod˝m theorem we conclude that there exists a measurable function f À M+ (A) such that ⌫ = f
. Assume that f > 1 on a set of positive -measure. Without loss
of generality we may assume that the set has finite measure, otherwise we would consider the intersection Ak „ {f > 1} with some exhausting sequence Ak ~ X and (Ak ) < ÿ. Then, for sufficiently small ✏ > 0 we know that ({f Œ 1 + ✏}) > 0 and so ⌫({f Œ 1 + ✏}) =
{f Œ1+✏}
Œ (1 + ✏)
fd
{f Œ1+✏}
d
Œ (1 + ✏) ({f Œ 1 + ✏}) Œ ({f Œ 1 + ✏}) which is impossible. ∑∑ Problem 20.2 Solution: Because of our assumption both know
⌫=f
and
~ ⌫ and ⌫ ~
which means that we
= g⌫
for positive measurable functions f , g which are a.e. unique. Moreover, ⌫=f
= f g⌫
so that f g is almost everywhere equal to 1 and the claim follows. Because of Problem 20.4 (which is just Corollary 25.6) it is clear that f , g < ÿ a.e. and, by the same argument, f , g > 0 a.e.
Note that we do not have to specify w.r.t. which measure we understand the ‘a.e.’ since their null sets coincide anyway.
∑∑
233
R.L. Schilling: Measures, Integrals & Martingales Problem 20.3 Solution: Take Lebesgue measure x + ÿ 1[0,1]c (x). Then f
:=
is certainly not -finite.
1
on (R, B(R)) and the function f (x) := ∑∑
Problem 20.4 Solution: See the proof of Corollary 25.6. ∑∑ Problem 20.5 Solution: See the proof of Theorem 25.9. ∑∑ Problem 20.6 Solution: (i) If F is AC, continuity is trivial, just take N = 2 in the very definition of AC functions.
To see that F is also BV, we take ✏ = 1 and choose > 0 such that for any subcollection a Õ ≥ ≥ x1 < y1 < 5 < xN < yN Õ b with n (yn * xn ) < we have n F (yn ) * F (xn ) < 1. Let M = [(b*a)_ ]+1 and ai = a+i(b*a)_M for i = 0, 1, … , M. Clearly, ai *ai*1 = (b*a)_M < and, in particular, V (f , [ai*1 , ai ]) < 1 for all i = 0, 1, … M. Thus, V (f ; [a, b]) Õ
M … i=1
V (f , [ai*1 , ai ]) < M.
(ii) Following the hint, we see that f is increasing. Define g := F * f . We have to show that g is increasing. Let x < y. Obviously,
V (f ; [a, x]) + F (y) * F (x) Õ V (f ; [a, x]) + F (y) * F (x) Õ V (f ; [a, y]) (since the points x < y can be added to extend any partition of [a, x] to give a partition of [a, y]). This gives g(x) Õ g(y).
(iii) Fix ✏ > 0 and pick x1 < y1 < x2 < y2 < 5 < xN < yN with F (yn ) * F (xn ) Õ =
[xn ,yn )
≥N
n=1 yn
* xn < . Then
f (t) (dt)
[xn ,yn )„{f ÕR}
f (t) (dt) +
[xn ,yn )„{f >R}
f (t) (dt)
Õ Ryn * xn + ({f > R} „ [xn , yn )) Õ Ryn * xn +
1 f d . R [xn ,yn )
Summing over n = 1, … , N gives N … n=1
F (yn ) * F (xn ) Õ R
N … n=1
yn * xn +
Now we choose first R = 12 ✏ î f d and then 234
N
1… 1 f d . Õ R + f d . R n=1 [xn ,yn ) R = 12 ✏_R.
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (iv) Write F = f1 * f2 with fi increasing (see part (ii)). From (ii) we know that we can pick
f1 (x) = V (F , [a, x]). Since F is absolutely continuous, so is f1 , hence f2 . This follows from the observation that
V (F , [a, y]) * V (F , [a, x]) = V (F , [x, y]) Since the fi are continuous, the set-functions
i [a, x)
≈ x < y.
:= fi (x)*fi (a) are pre-measures and extend
to measures on the Borel -algebra – see also Problem 6.1.
Now let N be a Lebesgue null-set. For every > 0 we can cover N by finitely many intervals ≥ [xi , yi ] such that i (yi * xi ) < . Without loss of generality we can make the intervals nonoverlapping and their length is still < . Since the fi are AC, we find for every ✏ some that
… n
In particular,
such
fi (yn ) * fi (xn ) < ✏.
i (N)
Õ
… n
i ([xn , yn ])
which shows that the Lebesgue null-set is also a follows from the Radon–Nikod˝m theorem.
i -null
< ✏,
set, i.e.
i
~
and therefore the claim ∑∑
Problem 20.7 Solution: This problem is somewhat ill-posed. We should first embed it into a suitable context, say, on the measurable space (R, B(R)). Denote by measure. Then
= 1[0,2]
=
1
one-dimensional Lebesgue
and ⌫ = 1[1,3]
and from this it is clear that ⌫ = 1[1,2] ⌫ + 1(2,3] ⌫ = 1[1,2] + 1(2,3] and from this we read off that 1[1,2] ⌫ ~ while 1(2,3] ⌫Ú . It is interesting to note how ‘big’ the null-set of ambiguity for the Lebesgue decomposition is—it
is actually R ‰ [0, 3] a, from a Lebesgue (i.e. ) point of view, huge and infinite set, but from a -⌫-perspective a negligible, namely null, set.
∑∑ Problem 20.8 Solution: Since we deal with a bounded measure we can use F (x) := (*ÿ, x) rather
than the more cumbersome definition for F employed in Problem 6.1 (which is good for locally finite measures!).
235
R.L. Schilling: Measures, Integrals & Martingales With respect to one-dimensional Lebesgue measure we can decompose 20.4 into
= Now define that
˝
2
:=
˝
˝
and F2 :=
is a finite measure (since
+
Ú
where
˝ (*ÿ, x). ˝
F (yj ) * F (xj ) = =
˝
Ú
~ ,
Ú .
We have to prove property (2). For this we observe
and that, therefore,
Õ
Thus, for every R > 0
˝
according to Theorem
=f
˝
with a function f À L1 ( ).
(xj , yj )
(xj ,yj )
f (t) (dt)
ÕR
(xj ,yj )
(dt) +
ÕR
(xj ,yj )
(dt) +
{f Œ R} „ (xj , yj ) 1 fd R (xj ,yj )
where we use the Markov inequality, cf. Proposition 11.5, in the last step. Summing over j = 1, 2, … , N gives
since
Ω
j (xj , yj )
N … j=1
F2 (yj ) * F2 (xj ) Õ R
+
1 fd R
œ R. Now we choose for given ✏ > 0 first R :=
î fd
to confirm that
✏ N … j=1
this settles b). Now consider the measure
Ú.
and then
:=
✏ ✏2 = R î fd
F2 (yj ) * F2 (xj ) Õ 2✏
Its distribution function F Ú (x) :=
Ú (*ÿ, x)
is increasing, left-
continuous but not necessarily continuous. Such a function has, by Lemma 14.14 at most countably many discontinuities (jumps), which we denote by J . Thus, we can write Ú
=
1
+
3
with the jump (or saltus) F (y) := F (y+) * F (y*) if y À J . 1
:=
… yÀJ
1
F (y)
y,
and
is clearly a measure (the sum being countable) with
3
1
:=
Õ
Ú
Ú
*
1;
and so is, therefore,
defining difference is always positive). The corresponding distribution functions are F1 (x) :=
… yÀJ ,y 0. Because of (i) there is some g À Lq ( ), ÒgÒq Õ 1 such that
f g d Œ Òf Òp * ✏.
Since D is dense, there is some h À D with Òg * hÒq Õ ✏. The Hölder inequality now shows
fh d =
f (h * g) d +
fg d
fg d
Œ *Òf Òp Òh * gÒq + Œ *Òf Òp ✏ +
fg d
Œ *Òf Òp ✏ + Òf Òp * ✏ = Òf Òp (1 * ✏) * ✏. Letting ✏ ô 0 proves the claim. 241
R.L. Schilling: Measures, Integrals & Martingales (iii) If f g À L1 ( ) for all g À Lq ( ), then If (g) := î f g d is a positive linear functional on Lq ( ). From Theorem 21.5 we know that there exists a unique fõ À Lq ( ) such that If (g) =
fõg d
≈g À Lq ( ).
Therefore, f = fõ À Lq ( ). ∑∑ Problem 21.2 Solution: (i) We use a classical diagonal argument (as in the proof of Theorem 21.18). Let (gn )nÀN denote an enumeration of Dq . Hölder’s inequality (13.5) tells us Û Û u g d Û n i Û
Û Û Õ Òu Ò Òg Ò Õ n p i q Û Û
0
1
sup Òun Òp Ògi Òq
nÀN
for all i, n À N. If i = 1, the sequence (î un g1 d )nÀN is bounded. Therefore, the Bolzano–Weierstraß theorem shows the existence of a subsequence (u1n )nÀN such that the limit
nôÿ
lim
u1n g1 d
i exists. We pick recursively subsequences (ui+1 n )nÀN œ (un )nÀN such that the limits nôÿ
lim
ui+1 n gi+1 d
exist. Because of the recursive thinning, we see that nôÿ
lim
uin gk d
exists for all k = 1, 2, … , i. Thus, for the diagonal sequence vn := unn the limits limnôÿ î vn gi d exist for each i À N.
(ii) Let g À Lq ( ) and (un(i) )iÀN be the diagonal sequence constructed in (i). Since R is complete, it is enough to show that î un(i) g d By assumption, Dq
is dense in Lq (
is a Cauchy sequence. Fix ✏ > 0.
), i.e. there exists some h À Dq such that Òg *hÒq Õ iÀN
✏. Part (i) shows that we can take N À N with
Û Û u hd * u hd Û n(i) n(k) Û
Û ÛÕ✏ Û Û
≈i, k Œ N.
Hölder’s inequality and the triangle inequality show Û Û Û un(i) g d * un(k) g d ÛÛ Û Û Û Û = ÛÛ (un(i) * un(k) )(g * h) d Û Û Õ ÛÛ (un(i) * un(k) )(g * h) d Û
Û (un(i) * un(k) )h d ÛÛ Û Û Û Û Û + Û (u * u )h d Û n(k) Û Û n(i) Û Û Û Û ´≠≠≠≠≠≠≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠≠≠≠≠≠≠¨ +
Õ✏ b/o (?)
242
(?)
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Õ Òun(i) * un(k) Òp Òg * hÒq + ✏ Õ (Òun(i) Òp + Òun(k) Òp )Òg * hÒq + ✏ Õ 2 sup Òun Òp Òg * hÒq + ✏ nÀN 0 1 Õ 2 sup Òun Òp + 1 ✏ nÀN
for any i, k Œ N. This proves that î un(i) g d
iÀN
is Cauchy.
(iii) Without loss of generality we may assume that the limits iôÿ
I(g) := lim
iôÿ
and J (g) := lim
u+ gd , n(i)
u* gd n(i)
exist for all g À Lq ( ). Indeed: From (i),(ii) we see that there is a subsequence such that I(g) exists for all g À Lq ( ). Thinning out this subsequence once again, we see
that J (g) exists for all g À Lq ( ). Since I and J are positive linear functionals on Lq ( ), Theorem 21.5 proves that there are unique functions v, w À Lq ( ), v, w Œ 0 representing these functionals: I(g) = Therefore, iôÿ
lim
and
vg d
iôÿ
un(i) g d = lim =
J (g) =
wg d .
iôÿ
u+ g d * lim n(i)
u* gd n(i)
(v * w)g d .
The claim follows if we use u := v * w À Lq ( ). ∑∑ Problem 21.3 Solution: (i) By Problem 19.7(i) or 21.4(a), ök is positive semidefinite, i.e. for any choice of m À N, 1, … , m
À C and ⇠1 , … , ⇠m À Rn we have m … i,k=1
Ñ Œ 0.
ök (⇠i * ⇠k )
i k
(⇠i * ⇠k )
i k
Since limiôÿ öi (⇠) = (⇠), we see m … i,k=1
Ñ Œ 0.
Since öi (*⇠) = öi (⇠), this also holds for the limit (*⇠) = lim öi (*⇠) = lim öi (⇠) = (⇠) This shows that
iôÿ
iôÿ
≈⇠ À Rn .
is positive semidefinite. If m = 1 resp. m = 2, we see that the matrices H I ⇠ ⇡ (0) (*⇠) and (0) (⇠) (0)
243
R.L. Schilling: Measures, Integrals & Martingales are positve hermitian for all ⇠ À Rn . Since determinants of positive hermitian matrices are positive, we find (0) Œ 0 and
0 Õ (0)2 * (⇠) (*⇠) = (0)2 * (⇠) (⇠) = (0)2 * (⇠)2 . (ii) First of all we show that the limit exists. Pick u À Ccÿ (Rn ). Because of Theorem 19.23, F *1 u À S(Rn ) and we can use Plancherel’s theorem (Theorem 19.12), to get
ud
i
=
F (F *1 u) d
=
i
F *1 u(⇠)öi (⇠) d⇠.
Since öi (⇠) Õ öi (0) ô (0) is uniformly bounded, we can use dominated convergence and find that
iôÿ
⇤(u) := lim
ud
i
=
F *1 u(⇠) (⇠) d⇠
is well-defined. The linearity of ⇤ follows from the linearity of the integral Moreover, if u Œ 0, then
iôÿ
⇤u = lim
ud
i
Π0.
(iii) The continuity of ⇤ follows from ⇤u Õ lim sup iôÿ
u d
i
Õ ÒuÒÿ lim sup iôÿ
n n i (R ) = (2⇡) (0)ÒuÒÿ . ´Ø¨
(2⇡)n öi (0)
Since Ccÿ (Rn ) is uniformly dense in Cc (Rn ), (see Problem 15.13, the proof resembles the argument of Theorem 15.11), we can extend ⇤ to a positive linear functional on Cc (Rn ): For u À Cc (Rn ) we take (ui )iÀN œ Ccÿ (Rn ), such that Òui * uÒÿ ô 0. Since ⇤(ui ) * ⇤(uk ) = ⇤(ui * uk ) Õ (2⇡)n (0)Òui * uk Òÿ , we conclude that (⇤ui )iÀN is a Cauchy sequence in R. Therefore, the limit ⇤u := limiôÿ ⇤ui exists and defines a positive linear functional on Cc (Rn ). By Riesz’s rep-
resentation theorem, Theorem 21.8, there exists a unique regular measure representing the functional ⇤
⇤u = (iv) Let ✏ > 0. Since
ud
≈u À Cc (Rn ).
is continuous at ⇠ = 0, there is some (⇠) * (0) < ✏
> 0 such that
≈⇠ Õ .
Because of Lévy’s truncation inequality, Problem 19.6, n i (R
‰ [*R, R]n ) Õ 2(R⇡)n
[*1_R,1_R]n
(öi (0) * Re öi (⇠)) d⇠
(note that úi (⇠) = (2⇡)n öi (*⇠)). With the dominated convergence theorem we get lim sup iôÿ
244
n i (R
‰ [*R, R]n ) Õ 2(R⇡)n
[*1_R,1_R]n
( (0) * Re (⇠)) d⇠
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Õ 2(2⇡)n ✏ for R Œ 1 . In particular we find for i Œ n0 (✏), get
(v) Let (
n i (R
n i (R
‰ [*R, R]n ) Õ 3(2⇡)n ✏. In order to
‰ [*R, R]n ) Õ 3(2⇡)n ✏ for i = 1, … , n0 (✏), we can increase R, if needed.
k )kÀN
œ Cc (Rn ) be a sequence of functions such that 0 Õ
(use, e.g. Urysohn functions, cf. page 239, or construct the we have
k
k
Õ 1 and
k
~ 1Rn
directly). Because of (iii)
d = ⇤( k ) Õ (2⇡)n (0). k The monotone convergence theorem shows that is a finite measure: kÀN
(Rn ) = sup Moreover, M := supiÀN show that
i
n i (R )
Õ (2⇡)n (0).
kd
< ÿ since
n i (R )
converges weakly to . First of all,
ud
,,,,,,ô , i ,,,iôÿ
ud
= (2⇡)n öi (0) ô (0). It remains to (?)
≈u À Cc (Rn ).
Let u À Cc (Rn ). Since Ccÿ (Rn ) is dense in Cc (Rn ), there is a sequence (fk )kÀN œ Ccÿ (Rn ) such that Òfk * uÒÿ ô 0. Thus,
Û Û Û ud i * u d ÛÛ Û Û Û Û Û Û Û Û Û Õ ÛÛ (u * fk ) d i ÛÛ + ÛÛ fk d i * fk d ÛÛ + ÛÛ (fk * u) d ÛÛ Û Û Û Û Û Û Û Û Õ Òu * fk Òÿ i (Rn ) + ÛÛ fk d i * f d Û + Òfk * uÒÿ (Rn ) k ÛÛ Û Û Û Õ Òu * fk Òÿ (M + (Rn )) + ÛÛ fk d i * fk d ÛÛ Û Û (ii)
,,,,,,,,,ô , Òu * fk Òÿ (M + (Rn )) ,,,,,,,,,ô , 0. iôÿ
Assume that f À Cb
kôÿ
(Rn ).
with K := [*R, R]n
For ✏ > 0, Party (iv) shows that there is some R > 0 such that c i (Kn )
=
n i (R
‰ K) Õ ✏.
Without loss of generality we may assume that (Rn ‰ K) Õ ✏. Pick 0Õ
Õ 1 and K = 1. Then
Since f
Û Û fd Û Û Û Õ ÛÛ f Û Û Õ ÛÛ f Û Û Õ ÛÛ f Û
i
*
À Cc (Rn ),
Û f d ÛÛ Û
Û Û Û f d ÛÛ + ÛÛ (1 * )f d i + (1 * )f d ÛÛ Û Û 0 1Û Û d i* f d ÛÛ + Òf Òÿ 1 cd i+ 1 cd K K Û Û d i* f d ÛÛ + 2Òf Òÿ ✏. Û n À Cc (R ), the first term on the right vanishes as i ô ÿ, cf. (iii). So, d
i
*
Û lim sup ÛÛ f d iôÿ Û
i
*
Û f d ÛÛ Õ 2Òf Òÿ ✏ ,,,,,,,ô , 0. ✏ô0 Û
245
R.L. Schilling: Measures, Integrals & Martingales (vi) Let ( ⇠À
k )kÀN be a Rn , we get
weakly convergent sequence of finite measures. Define f (x) := e*ix ⇠ ,
ök (⇠) =
1 e*ix ⇠ d (2⇡)n
, k (x) ,,,,,,,,,ô kôÿ
1 e*ix ⇠ (dx) = ö(⇠), (2⇡)n
i.e. the Fourier transforms converge pointwise. From part (iv) we know that the sequence (
k )kÀN
is tight. For ✏ > 0 there is some R > 0 such that
n k (R
‰ K) Õ ✏ for K :=
[*R, R]n . Without loss of generality we can enlarge R to make sure that (Rn ‰ K) Õ ✏,
too. Because of the (uniform) continuity of the function R Œ r ;ô eir on compact sets, there is some
> 0 such that
ei(⇠*⌘) x * 1 Õ ✏
≈⇠ * ⌘ < , x À K.
If k À N, ⇠, ⌘ À Rn with ⇠ * ⌘ < , then we see 1 1 ei⇠ x * ei⌘ x k (dx) = ei(⇠*⌘) x * 1 k (dx) (2⇡)n (2⇡)n 1 1 = ei(⇠*⌘) x * 1 k (dx) + ei(⇠*⌘) x * 1 (dx) n (2⇡) K ´≠≠≠≠≠Ø≠≠≠≠≠¨ (2⇡)n K c ´≠≠≠≠≠Ø≠≠≠≠≠¨ k
ök (⇠) * ök (⌘) Õ
Õ Õ
k
(2⇡)n
Õ✏
✏+
2 (2⇡)n
Õ2
i (K
c
)
1 (M + 2)✏ (2⇡)n
where M := supkÀN (ök )kÀN .
(Rn )
n k (R )
< ÿ. This proves the equicontinuity of the sequence
(vii) Let ⇠ À Rn and ✏ > 0. Use equicontinuity of the sequence (ök )kÀN to pick some Since ö is continuous, we can ensure that
is such that
ö(⇠) * ö(⌘) Õ ✏
> 0.
≈⇠ * ⌘ Õ .
This entails for all ⌘ À Rn satisfying ⌘ * ⇠ Õ : ök (⌘) * ö(⌘) Õ ök (⌘) * ök (⇠) +ök (⇠) * ö(⇠) + ö(⇠) * ö(⌘) ´≠≠≠≠≠Ø≠≠≠≠≠¨ ´≠≠≠≠≠≠≠Ø≠≠≠≠≠≠≠¨ Õ✏
Õ✏
-Ÿ
sup ök (⌘) * ö(⌘) Õ 2✏ + ök (⇠) * ö(⇠) ,,,,,,,,,ô , 2✏ ,,,,,,,ô , 0.
⌘ÀB (⇠)
kôÿ
✏ô0
Here we use that ök converges pointwise to ö, cf. (vi). The calculation shows that ök con-
verges locally uniformly to ö. Since locally uniform convergence is the same as uniform convergence on compact sets, we are done.
∑∑ Problem 21.4 Solution:
246
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (i) Since
is a finite measure, the continuity of ö follows directly from the continuity
lemma, Theorem 12.4 (cf. also 19.3). In order to show positive definiteness, pick m À N, ⇠1 , … , ⇠m À Rn and m … j,k=1
(⇠j * ⇠k )
1, … , m
À C. We get
m 1 … (2⇡)n j,k=1
Ñ j k =
j k
Ñ
e*ix (⇠j *⇠k ) (dx)
m 1 … Ñ = e*ix ⇠j e*ix ⇠k (dx) (2⇡)n j,k=1 j k Hm IH m I … … 1 *ix ⇠j *ix ⇠k = (dx) je ke (2⇡)n j=1
Û… Ûm 1 Û = (2⇡)n ÛÛ j=1 Û
k=1
Û2
je
Û Û Û Û
*ix ⇠j Û
(dx) Π0.
(ii) For m = 1 and ⇠ = 0 the definition of positive definiteness implies that the matrix ( (0)) is positive definite; in particular, (0) Œ 0. ≥ If we have for a matrix (aik ) that ik aik i Ñ j Œ 0, then 0Õ
… ik
aik
Ñ =
i k
… ik
aik
Ñ =
i k
… ik
aik Ñ i
k
=
… ki
aki Ñ k
i
which means that aik = aki . Apply this to the matrix aik = (⇠i * ⇠k ) with m = 2 and ⇠1 = ⇠ and ⇠2 = 0 to infer that (⇠) = (*⇠). Moreover, the matrix H I (0) (*⇠) (⇠)
(0)
is positive semidefinite; in particular its determinant is positive: 0 Õ (0)2 * (*⇠) (⇠). Since (*⇠) = (⇠), we get the inequality as claimed. (iii) Because of (⇠) Õ (0) we see that ⇠ ⇡ Û Û ix ⇠ *2✏⇠2 Û (⇠ * ⌘) e e eix ⌘ e*2✏⌘2 d⇠ d⌘ ÛÛ ÛÀ Û Û ⇠ ⇡ *2✏⇠2 *2✏⌘2 Õ (0) e e d⇠ d⌘ < ÿ, À i.e. ⌫✏ is well-defined. Let us show that ⌫✏ Œ 0. For this we cover Rn with countably
many disjoint cubes (Iik )iÀN with side-length 1_k and we pick any ⇠ik À Iik . Using the dominated convergence theorem and the positive definiteness of the function ⌫✏ (x) = lim
kôÿ
= lim
kôÿ
… m,jÀN
…
m,jÀN
I k I k m
we get
⇠ ⇡⇠ ⇡ k k 2 k k 2 (⇠mk * ⇠jk ) eix ⇠j e*2✏⇠j eix ⇠m e*2✏⇠m d⇠ d⌘
j
⇠ ⇡⇠ ⇡ k k 2 k k 2 (⇠jk * ⇠mk ) k*n eix ⇠j e*2✏⇠j k*n eix ⇠j e*2✏⇠j
247
R.L. Schilling: Measures, Integrals & Martingales Œ 0. Because of the parallelogram identity 2⇠2 + 2⌘2 = ⇠ * ⌘2 + ⇠ + ⌘2 we obtain ⌫✏ (x) =
À
=
À
⇠
⇡
eix ⇠ eix ⌘ e*2✏⌘2 *2✏⇠2
d⇠ d⌘
⇠ ⇡ 2 2 eix (⇠*⌘) e*⇠*⌘ *⇠+⌘ d⇠ d⌘.
Changing variables according to H I t s
H I ⇠*⌘
:=
⇠+⌘
=
H IH I idn * idn ⇠ idn
idn
⌘
=: A
H I ⇠ ⌘
leads to 2 2 1 (t)eix t e*✏(t +s ) dt ds det A À 2 1 = (t)e*✏t eix t dt c 1 = (t)eix t dt. c ✏
⌫✏ (x) =
(iv) Define
(?)
0 1 x2 1 gt (x) := exp * . 2t (2⇡t)n_2
Applying Theorem 19.12 for the finite measure (dx) := e*tx dx yields 2
t
2
(?)
⌫✏ (x)e* 2 x dx =
for all t > 0 (observe: Therefore, F *1 (e
* 2t 2
thus get
✏
1 F *1 ( c
✏ )(x)e
* 2t x2
dx =
1 c
✏ (⇠)F
2
(0) and î gt (x) dx = 1 we
t
2
⌫✏ (x)e* 2 x dx =
(2⇡)n c
✏ (⇠)gt (⇠) d⇠
Õ
(2⇡)n (0). c
Fatou’s lemma (Theorem 9.11) finally shows ⌫✏ (x) dx =
1
2
lim ⌫✏ (x)e* 2k x dx
kôÿ
Õ lim inf
Õ
(0).
kôÿ (2⇡)n
c
1
2
⌫✏ (x)e* 2k x dx
Since ⌫✏ Œ 0, see (iii), this means that ⌫✏ À L1 (Rn ). 248
t
(e* 2 )(⇠) d⇠
À L1 (Rn )). Example 19.2(iii) shows F (gt )(x) = (2⇡)*n exp(*tx2 _2).
)(⇠) = (2⇡)n gt (⇠). Since (⇠) Õ
*1
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (v) Parts (iii) and (iv) show that ö✏ = ✏
✏
for the finite measure
✏ (dx)
:= c⌫✏ (x) dx. Since
ô , Lévy’s continuity theorem (Problem 21.3) shows that there exists a measure
which is the weak limit of the family
✏
as ✏ ô 0 and ö = .
∑∑ Problem 21.5 Solution: (i) Since uniform convergence preserves continuity, we see that every u À Cc (X) is continuous. By construction, the set {u Œ ✏} is compact since there is some u✏ À Cc (X)
such that Òu * u✏ Òÿ < ✏. This means that u vanishes at infinity. In particular Cc (X) œ Cÿ (X).
Conversely, if u À Cÿ (X) and ✏ > 0, there is some compact set K✏ such that u Õ ✏ outside of K✏ . Now we use Urysohn’s lemma and construct a function that 1K✏ Õ
✏
Õ 1. Then we get u✏ :=
✏u
À Cc (X) as well as
u * u✏ = (1 *
✏ )u
✏
À Cc (X) such
Õ✏
uniformly for all x. (ii) It is obvious that Cÿ (X) is a vector space and that Ò÷Òÿ is a norm in this space. The completeness follows from part (i) since Cÿ (X) = Cc (X) = Cc (X).
(iii) Let u À Cÿ (X) and ✏ > 0. Urysohn’s lemma shows that there is a 0Õ
Õ 1, such that u Õ ✏ on the set { < 1} = { =
1}c .
Therefore,
Û Û Û ud n * u d ÛÛ Û Û Û Û Û Û Õ ÛÛ u d n * u d ÛÛ + ÛÛ u(1 * ) d n * u(1 * ) d Û Û Û Û Û ⌅ ⇧ Õ ÛÛ u d n * u d ÛÛ + ✏ n (X) + (X) Û Û 21.16 Û Û Õ ÛÛ u d n * u d ÛÛ + 2✏ sup m (X). Û Û mÀN
À Cc (X),
Û Û Û Û
Since u À Cc (X), we find as n ô ÿ Û lim sup ÛÛ u d nôÿ Û
n
*
Û u d ÛÛ Õ 2✏ sup Û mÀN
, m (X) ,,,,,,,ô ✏ô0
0. ∑∑
Problem 21.6 Solution:
(i) First we consider u À Ccÿ (Rn ). According to Theorem 19.23, F *1 u À S(Rn ), and Plancherel’s theorem (Theorem 19.12) gives
ud
i
=
F (F *1 u) d
i
=
F *1 u(⇠)öi (⇠) d⇠.
249
R.L. Schilling: Measures, Integrals & Martingales Since öi (⇠) Õ öi (0) ô (0) is uniformly bounded, we can use the dominated convergence theorem to see
iôÿ
⇤(u) := lim
ud
i
=
F *1 u(⇠) (⇠) d⇠
i.e. ⇤(u) is well-defined. Moreover, n i (R )
i.e. M := supi
n i (R )
= (2⇡)n öi (0) ,,,,,,,,,ô , (2⇡)n (0), iôÿ
< ÿ. Assume now that u À Cc (X). Since Ccÿ (Rn ) is dense in
Cc (Rn ) (with respect to uniform convergence, cf. Problem 15.13), there is a sequence (uk )kÀN œ Ccÿ (Rn ) such that Òuk * uÒÿ ô 0. Thus, Û Û ud Û Û
i
Û Û Û
Û Õ ÛÛ (u * uk ) d Û
*
ud
jÛ
Û Û Û Û Û + Û (u * u ) d Û + Û u d * u d k jÛ k i Û Û k Û Û Û Û Û Û Õ Òu * uk Òÿ i (Rn ) + j (Rn ) + ÛÛ uk d i * uk d j ÛÛ Û Û Û Û Õ 2Òu * uk Òÿ M + ÛÛ uk d i * u d Û k j ÛÛ Û iÛ
Û Û Û
jÛ
,,,,,,,,,,,ô , 2Òu * uk Òÿ M ,,,,,,,,,ô , 0. i,jôÿ
This shows that î u d limiôÿ î u d
i
kôÿ
i iÀN
is a Cauchy sequence in R. Thus, the limit ⇤(u) :=
exists. Since convergent sequences are bounded, we see Û sup ÛÛ u d iÀN Û
Û Û < ÿ. Û
iÛ
Since u À Cc (Rn ) -Ÿ u À Cc (Rn ), we get nÀN
sup
u d
i
0. Since
is continuous at ⇠ = 0, there is some (⇠) * (0) < ✏
> 0 such that
≈⇠ Õ .
From Lévy’s truncation inequality, Problem 19.6, we get n i (R
‰ [*R, R]n ) Õ 2(R⇡)n
[*1_R,1_R]n
(öi (0) * Re öi (⇠)) d⇠
(observe, that úi (⇠) = (2⇡)n öi (*⇠)). Now we can use dominated convergence to get lim sup
n i (R
iôÿ
‰ [*R, R]n ) Õ 2(R⇡)n
[*1_R,1_R]n
( (0) * Re (⇠)) d⇠
Õ 2(2⇡)n ✏ for all R Œ 1 . In particular, we find order to ensure
n i (R
need be.
‰ [*R, R]n ) Õ
n n n i (R ‰ [*R, R] ) Õ 3(2⇡) ✏ for i Œ n0 (✏). In 3(2⇡)n ✏ for i = 1, … , n0 (✏), we can enlarge R, if
∑∑ Problem 21.7 Solution: Since
B
ud
n
=
B„supp u
ud
n
we can assume, without loss of generality, that B is contained in a compact set. Denote by K := B
the closure of B and by U := B ˝ the open interior of B. Moreover, we can assume that u Œ 0 – otherwise we consider u± separately.
According to Urysohn’s lemma (Lemma B.2 or (21.6) & (21.7)), there are sequences (wk )kÀN œ Cc (X), (vk )kÀN œ Cc (X), 0 Õ vk Õ 1, 0 Õ wk Õ 1, with wk ~ 1U and vk ö 1K . By assumption v , n ,,,,ô
and so
B
ud
n
Õ
Beppo Levi’s theorem implies lim sup nôÿ
Similarly, we get from B
ud
n
Œ
K
ud
n
Õ
B
ud
n
Õ inf
U
ud
n
kÀN
Œ
,,,,,,ô , n ,,,nôÿ
u vk d .
u vk d =
K
ud .
u wk d .
u vk d
u wk d
,,,,,,ô , n ,,,nôÿ
251
R.L. Schilling: Measures, Integrals & Martingales and the monotone convergence theorem lim inf nôÿ
B
ud
n
kÀN
Πsup
u wk d =
U
ud .
Finally, since (K ‰ U ) = ()B) = 0, we see that lim sup nôÿ
B
ud
n
Õ
K
ud =
U
u d Õ lim inf nôÿ
B
ud
n.
∑∑
252
22 Uniform integrability and Vitali’s convergence theorem. Solutions to Problems 22.1–22.17 Problem 22.1 Solution: First, observe that lim uj (x) = 0 ◊Ÿ lim uj (x) = 0. j
j
Thus, x À {lim uj = 0} ◊Ÿ ≈ ✏ > 0
« N✏ À N
≈ j Œ N✏ : uj (x) Õ ✏
◊Ÿ ≈ ✏ > 0
« N✏ À N
: sup uj (x) Õ ✏
◊Ÿ ≈ ✏ > 0
« N✏ À N
: x À { sup uj Õ ✏}
j
◊Ÿ ≈ ✏ > 0 : x À
Õ
à Õ
jŒN✏
{sup uj Õ ✏}
NÀN jŒN
◊Ÿ ≈ k À N : x À ◊Ÿ x À
jŒN✏
Õ
{sup uj Õ 1_k}
NÀN jŒN
{sup uj Õ 1_k}.
kÀN NÀN jŒN
Equivalently, {lim uj = 0}c = j
Õ Ã
{sup uj > 1_k}.
kÀN NÀN jŒN
By assumption and the continuity of measures, 0 1 ⇠ ⇡ Ã {sup uj > 1_k} = lim {sup uj > 1_k} = 0 N
NÀN jŒN
jŒN
and, since countable unions of null sets are again null sets, we conclude that {lim uj = 0} has full measure. j
∑∑ Problem 22.2 Solution: Note that xÀ
sup uj > ✏ jŒk
◊Ÿ sup uj (x) > ✏ jŒk
253
R.L. Schilling: Measures, Integrals & Martingales ◊Ÿ « j Œ k : uj (x) > ✏ Õ ◊Ÿ x À {uj > ✏} jŒk
and since
Õ ÃÕ def {uj > ✏} ö {uj > ✏} = lim sup{uj > ✏} jŒk
jôÿ
kÀN jŒk
we can use the continuity of measures to get 0 1 0 1 ⇠ ⇡ Õ ÃÕ lim sup uj > ✏ = lim {uj > ✏} = {uj > ✏} . k
k
jŒk
jŒk
kÀN jŒk
This, and the result of Problem 22.1 show that either of the following two equivalent conditions ⇠ ⇡ lim sup uj Œ ✏ = 0 ≈ ✏ > 0; kôÿ
⇠
jŒk
⇡ lim sup{uj Œ ✏} = 0 jôÿ
≈ ✏ > 0;
ensure the almost everywhere convergence of limj uj (x) = 0. ∑∑ Problem 22.3 Solution: • Assume first that uj ô u in -measure, that is, ≈ ✏ > 0,
≈ A À A, (A) < ÿ : lim j
{uj * u > ✏} „ A = 0.
Since uj * uk Õ uj * u + u * uk ≈j, k À N we see that {uj * uk > 2✏} œ {uj * u > ✏} ‰ {u * uk > ✏} (since, otherwise uj * uk Õ ✏ + ✏ = 2✏). Thus, we get for every measurable set A with finite -measure that
{uj * uk > 2✏} „ A ⌅ ⇧ Õ ({uj * u > ✏} „ A) ‰ ({uk * u > ✏} „ A) ⌅ ⇧ ⌅ ⇧ Õ {uj * u > ✏} „ A + {uk * u > ✏} „ A and each of these terms tend to infinity as j, k ô ÿ. • Assume now that uj * uk ô 0 in -measure as j, k ô ÿ. Let (Al )l be an exhausting sequence such that Al ~ X and (Aj ) < ÿ.
The problem is to identify the limiting function. Fix l. By assumption, we can choose Nj À N, j À N, such that ≈ m, n Œ Nj : 254
{um * un > 2*j } „ Al < 2*j .
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (Note that Nj may depend on l, but we suppress this dependency as l is fixed.) By enlarging Nj , if needed, we can always assume that
N1 < N2 < 5 < Nj < Nj+1 ô ÿ. Consequently, there is an exceptional set Ej œ Al with (Ej „ Al ) < 2*j such that uNj+1 (x) * uNj (x) Õ 2*j and, if Ei< :=
∑
jŒi Ej
≈ x À Al ‰ Ej
we have (Ei „ Al ) Õ 2 2*i as well as
uNj+1 (x) * uNj (x) Õ 2*j
Šj Πi,
≈ x À Al ‰ Ei< .
This means that … j
(uNj+1 * uNj ) converges uniformly for x À Al ‰ Ei<
so that limj uNj exists uniformly on Al ‰ Ei< for all i. Since (Ei< „ Al ) < 2 2*i we conclude that
lim uNj 1Al = u(l) 1Al exists almost everywhere j
for some u(l) . Since, however, a.e. limits are unique (up to a null set, that is) we know that
u(l) = u(m) a.e. on Al „Am so that there is a (up to null sets) unique limit function u satisfying lim uNj = u exists a.e., hence in measure by Lemma 22.4. j
(*)
Thus, we have found a candidate for the limit of our Cauchy sequence. In fact, since uk * u Õ uk * uNj + uNj * u we have ({uk * u > ✏} „ Al ) Õ ({uk * uNj > ✏} „ Al ) + ({uNj * u > ✏} „ Al ) and the first expression on the right-hand side tends to zero (as k, N(j) ô ÿ) because of the assumption, while the second term tends to zero (as N(j) ô ÿ) because of (*))
∑∑ Problem 22.4 Solution:
(i) This sequence converges in measure to f í 0 since for ✏ À (0, 1) (fn,j > ✏) = [(j * 1)_n, j_n] = 1n ,,,,,,,,,ô , 0. nôÿ
This means, however, that potential a.e. and Lp -limits must be f í 0, too. Since for every x
lim inf fn,j (x) = 0 < ÿ = lim sup fn,j 255
R.L. Schilling: Measures, Integrals & Martingales the sequence cannot converge at any point.
Also the Lp -limit (if p Π1) does not exist, since
fn,j p d = np [(j * 1)_n, j_n] = np*1 .
(ii) As in (i) we see that gn ,,,,,ô , g í 0. Similarly,
gn p d = np (0, 1_n) = np*1
so that the Lp -limit does not exist. The pointwise limit, however, exists since lim n1(0,n) (x) = 0.
nôÿ
for every x À (0, 1). (iii) The shape of gn is that of a triangle with base [0, 1_n]. Thus, for every ✏ > 0, (hn > ✏) Õ [0, 1_n] =
1 n
which shows that hn ,,,,,ô , h í 0. This must be, if the respective limits exist, also the limiting function for a.e. and Lp -convergence. Since
hn p d = apn
1 2
[0, 1_n] =
apn 2n
we have Lp -convergence if, and only if, the sequence apn _n tends to zero as n ô ÿ.
We have, however, always a.e. convergence since the support of the function hn is [0, 1_n] and this shrinks to {0} which is a null set. Thus, lim an (1 * nx)+ = 0 n
except, possibly, at x = 0. ∑∑ Problem 22.5 Solution: We claim that (i) (ii) (iii) (iv) Note that
auj + bwj ô au + bw; max(uj , wj ) ô max(u, w); min(uj , wj ) ô min(u, w); uj ô u. auj + bwj * au * bw Õ auj * u + bwj * w
so that {auj + bwj * au * bw > 2✏} œ {uj * u > ✏_a} ‰ {wj * w > ✏_b}. 256
Solution Manual. Chapter 1–13. Last update 23rd August 2017 This proves the first limit. Since, by the lower triangle inequality, uj * u Õ uj * u we get {uj * u > ✏} œ {uj * u > ✏} and uj ô u follows. Finally, since
⇠ ⇡ 1 uj + wj + uj * wj 2 we get max uj , wj ô max u, w by using rules (i) and (iv) several times. The minimum is treated max uj , wj =
similarly.
∑∑ Problem 22.6 Solution: The hint is somewhat misleading since this construction is not always possible (or sensible). Just imagine R with the counting measure. Then X
f
would be all of R...
What I had in mind when giving this hint was a construction along the following lines: Consider Lebesgue measure
in R and define f := 1F + ÿ1F c where F = [*1, 1] (or any other
set of finite Lebesgue measure). Then
:= f
sequence un ,,,,ô , u converging in -measure. Then
is a not -finite measure. Moreover, Take any
({un * u > ✏} „ A) = ({un * u > ✏} „ A) since all sets A with (A) < ÿ are contained in F and (F ) = (F ) < ÿ. Thus, un ,,,,,ô , u.
However, changing u arbitrarily on 1F c also yields a limit point in -measure since, as mentioned above, all sets of finite -measure are within F .
This pathology cannot happen in a -finite measure space, cf. Lemma 22.6. ∑∑ Problem 22.7 Solution: (i) Fix ✏ > 0. Then
u * uj d =
A
u * uj d
=
A„{u*uj Õ✏}
u * uj d +
Õ
A„{u*uj Õ✏}
✏d +
Õ ✏ (A) + 2C
A„{u*uj >✏}
A„{u*uj >✏}
u * uj d
(u + uj ) d
A „ {u * uj > ✏}
,,,,,,,,,ô , ✏ (A) jôÿ
,,,,,,,ô , 0. ✏ô0
257
R.L. Schilling: Measures, Integrals & Martingales (ii) Note that uj converges almost everywhere and in -measure to u í 0. However,
uj d = [j, j + 1] = 1 ë 0
so that the limit—if it exists—cannot be u í 0. Since this is, however, the canonical candidate, we conclude that there is no L1 convergence.
(iii) The limit depends on the set A which is fixed. This means that we are, essentially, dealing with a finite measure space.
∑∑ Problem 22.8 Solution: A pseudo-metric is symmetric (d2 ) and satisfies the triangle inequality (d3 ). (i) First we note that ⇢ (⇠, ⌘) À [0, 1] is well-defined. That it is symmetric (d2 ) is obvi-
ous. For the triangle inequality we observe that for three random variables ⇠, ⌘, ⇣ and numbers ✏, > 0 we have
⇠ * ⇣ Õ ⇠ * ⌘ + ⌘ * ⇣ implying that {⇠ * ⇣ > ✏ + } œ {⇠ * ⌘ > ✏} ‰ {⌘ * ⇣ > } so that P(⇠ * ⇣ > ✏ + ) Õ P(⇠ * ⌘ > ✏) + P(⌘ * ⇣ > ). If ✏ > ⇢P (⇠, ⌘) and
> ⇢P (⌘, ⇣) we find
P(⇠ * ⇣ > ✏ + ) Õ P(⇠ * ⌘ > ✏) + P(⌘ * ⇣ > ) Õ ✏ + which means that ⇢P (⇠, ⇣) Õ ✏ + . Passing to the infimum of all possible - and ✏-values we get ⇢P (⇠, ⇣) Õ ⇢P (⇠, ⌘) + ⇢P (⌘, ⇣). (ii) Assume first that ⇢P (⇠j , ⇠) ,,,,,,,,,ô , 0. Then jôÿ
⇢P (⇠j , ⇠) ,,,,,,,,,ô , 0 ◊Ÿ « (✏j )j œ R+ : P(⇠ * ⇠j > ✏j ) Õ ✏j jôÿ
-Ÿ ≈ ✏ > ✏j : P(⇠ * ⇠j > ✏) Õ ✏j . Thus, for given ✏ > 0 we pick N = N(✏) such that ✏ > ✏j for all j Œ N (possible as ✏j ô 0). Then we find
≈ ✏ > 0 « N(✏) À N ≈ j Œ N(✏) : P(⇠ * ⇠j > ✏) Õ ✏j ; this means, however, that P(⇠ * ⇠j > ✏) ,,,,,,,,,ô , 0 for any choice of ✏ > 0. jôÿ
258
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Conversely, assume that ⇠j ,,,,,ô , 0. Then P
≈ ✏ > 0 : lim P(⇠ * ⇠j > ✏) = 0 j
◊Ÿ ≈ ✏, > 0 « N(✏, ) ≈ j Œ N(✏, ) : P(⇠ * ⇠j > ✏) < -Ÿ ≈ ✏ > 0 « N(✏) ≈ j Œ N(✏) : P(⇠ * ⇠j > ✏) < ✏ -Ÿ ≈ ✏ > 0 « N(✏) ≈ j Œ N(✏) : ⇢P (⇠, ⇠j ) Õ ✏ -Ÿ lim ⇢P (⇠, ⇠j ) = 0. j
(iii) We have (ii)
P
⇢(⇠j , ⇠k ) ,,,,,,,,,,,,,,ô , 0 ◊Ÿ ⇠j * ⇠k ,,,,,,,,,,,,,,ô , 0 j,kôÿ
P22.3
j,kôÿ P
◊Ÿ « ⇠ : ⇠k ,,,,,,,,,ô , ⇠ (ii)
kôÿ
◊Ÿ « ⇠ : ⇢(⇠, ⇠k ) ,,,,,,,,,,,,ô , 0 kôÿ
(iv) Note that for x, y > 0
and
x+y y y x x = + Õ + 1+x+y 1+x+y 1+x+y 1+x 1+y h nx + y = (x · 1) + (y · 1) (x + y) · 1 = l n1 Õ (x · 1) + (y · 1) j
if x + y Õ 1; if x + y Œ 1.
This means that both gP and dP satisfy the triangle inequality, that is (d3 ). Symmetry, i.e. (d2 ), is obvious.
Moreover, since for all x Œ 0 x x Õx·1Õ2 1+x 1+x
(consider the cases x Õ 1 and x Œ 1 separately), we have gP (⇠, ⌘) Õ dP (⇠, ⌘) Õ 2gP (⇠, ⌘) which shows that gP and dP have the same Cauchy sequences. Moreover, for all ✏ Õ 1, P(⇠ * ⌘ > ✏) = P(⇠ * ⌘ · 1 > ✏) 1 ⇠ * ⌘ · 1 dP ✏ 1 = dP (⇠, ⌘) ✏ Õ
so that (because of (iii)) any dP Cauchy sequence is a ⇢P Cauchy sequence. And since for all ✏ Õ 1 also
dP (⇠, ⌘) =
⇠*⌘>✏
⇠ * ⌘ · 1 dP +
⇠*⌘Õ✏
⇠ * ⌘ · 1 dP
259
R.L. Schilling: Measures, Integrals & Martingales Õ
⇠*⌘>✏
1 dP +
⇠*⌘Õ✏
✏ dP
Õ P(⇠ * ⌘ > ✏) + ✏, all ⇢P Cauchy sequences are dP Cauchy sequences, too. ∑∑ Problem 22.9 Solution: (i) Fix ✏ > 0. We have
un * u · 1A d =
{un *uÕ✏}
un * u · 1A d +
{un *u>✏}
un * u · 1A d
Õ ✏ (A) + ({un * u > ✏} „ A). Letting first n ô ÿ and then ✏ ô 0 yields lim sup n
un * u · 1A d Õ ✏ (A) ,,,,,,,ô , 0. ✏ô0
(ii) WLOG we show that (un )n contains an a.e. convergent subsequence. Let (Ak )k be as in
the hint and fix i. By (i) we know that u * un · 1Ai ô 0 in L1 . By Corollary 13.8 we (i) see that there is a subsequence u(i) n such that u * un · 1Ai ô 0 almost everywhere.
Now take repeatedly subsequences as i ‰ i + 1 ‰ i + 2 ‰ … etc. and then take the diagonal sequence. This will furnish a subsequence (u®®n )n œ (un )n which converges a.e. ∑ to u on i Ai = X.
(iii) We are now in the setting of Corollary 13.8: un , u Õ w for some w À Lp ( ) and un ,,ô ,
u. Thus, every subsequence (u®n )n œ (un )n converges in measure to the same limit u and a.e.
by (ii) there is some (u®®n )n œ (u®n )n such that u®®n ,,,,,ô , u. Now we can use the dominated convergence theorem (Theorem 12.2 or Theorem 13.9) to show that limn Òu®®n *uÒp = 0. Assume now that un does not converge to u in Lp . This means that lim supn Òun * uÒp >
0, i.e. there is some subsequence such that lim inf n Òu®n * uÒp > 0. On the other hand, there is some (u®®n )n œ (u®n )n such that
0 = lim Òu®®n * uÒp Œ lim inf Òu®n * uÒp > 0 n
n
and this is a contradiction. ∑∑ Problem 22.10 Solution: Note that the sets Aj are of finite -measure. Observe that the functions fj := u1Aj
• converge in -measure to f í 0: ({fj > ✏} „ Aj ) Õ (Aj ) ,,,,,,,,,ô , 0. jôÿ
260
Solution Manual. Chapter 1–13. Last update 23rd August 2017 • are uniformly integrable: sup j
{fj >u}
fj d = 0
since fj = u1Aj Õ u and u is integrable.
Therefore, Vitali’s Theorem shows that fj ô 0 in L1 so that î fj d = îA u d ô 0. j
∑∑ Problem 22.11 Solution: (i) Trivial. More interesting is the assertion that A sequence (xn )n œ R converges to 0 if, and only if, every subsequence (xnk )k contains some sub-subsequence (õ xnk )k which converges to 0.
Necessity is again trivial. Sufficiency: assume that (xn )n does not converge to 0. Then
the sequence (min{xn , 1})n is bounded and still does not converge to 0. Since this sequence is bounded, it contains a convergent subsequence (xnk )k with some limit ↵ ë 0. But then (xnk )k cannot contain a sub-subsequence (õ xnk )k which is a null sequence.
(ii) If un ,,,,,ô , u, then every subsequence unk ,,,,,ô , u. Thus, using the argument from the proof of Problem 22.3 we can extract a sub-subsequence (õ unk )k œ (unk )k such that lim õ unk (x)1A (x))u(x)1A (x) almost everywhere.
(*)
k
Note that (unless we are in a -finite measure space) the exceptional set may depend on the testing set A.
Conversely, assume that every subsequence (unk )k œ (un )n has a sub-subsequence (õ unk )k satisfying (*). Because of Lemma 22.4 we have lim k
{õ unk * u > ✏} „ A = 0.
Assume now that un does not converge in -measure on A to u. Then xn := ({un * u > ✏} „ A) ô ° 0. Since the whole sequence (xn )n is bounded (by (A)) there exists some subsequence (xnk )k given by (unk )k such that
xnk = ({unk * u > ✏} „ A) ô ↵ ë 0.
This contradicts, however, the fact that xnk has itself a subsequence converging to zero.
(iii) Fix some set A of finite -measure. All conclusions below take place relative to resp. on this set only.
If un ,,,,,ô , u we have for every subsequence (unk )k a sub-subsequence (õ unk )k with õ unk ô u a.e. Since
is continuous, we get
˝õ unk ô
˝u a.e.
This means, however, that every subsequence ( ˝unk )k of ( ˝un )n has a sub-subsequence ( ˝õ unk )k which converges a.e. to
˝u. Thus, part (ii) says that
˝un ,,,,,ô ,
˝u.
261
R.L. Schilling: Measures, Integrals & Martingales ∑∑
Problem 22.12 Solution: functions f✏ , g✏ À
L1+
Since F and G are uniformly integrable, we find for any given ✏ > 0
such that
sup
f ÀF
{f >f✏ }
gÀG {g>g✏ }
f d Õ ✏ and sup
We will use this notation throughout.
g d Õ ✏.
(i) Since f := f1 + 5 + fn À L1+ we find that {fj >f }
fj d =
Á
fj d = 0
uniformly for all 1 Õ j Õ n. This proves uniform integrability. (ii) Instead of {f1 , … , fN } (which is uniformly integrable because of (i)) we show that F ‰ G is uniformly integrable.
Set h✏ := f✏ + g✏ . Then h✏ À L1+ and {w Œ f✏ + g✏ } œ {w Œ f✏ } „ {w Œ g✏ } which means that we have h w d Õ ✏ nî w d Õ l {w>f✏ } {w>h✏ } nî{w>g } w d Õ ✏ ✏ j
if w À F
if w À G.
Since this is uniform for all w À F ‰ G, the claim follows.
(iii) Set h✏ := f✏ + g✏ À L1+ . Since f + g Õ f + g we have {f + g > h✏ } œ {f > h✏ } ‰ {g > h✏ } ⌅ ⇧ = {f > h✏ } „ {g > h✏ } ⌅ ⇧ ‰ {f > h✏ } „ {g Õ h✏ } ⌅ ⇧ ‰ {f Õ h✏ } „ {g > h✏ } which implies that
f + g d
{f +g>h✏ }
Õ
f + g d +
{f >h✏ } „{g>h✏ }
=
{f >h✏ }
262
f d +
{g>h✏ }
f d +
{f >h✏ } „{gÕh✏ }
g d +
f ‚ g d
{f Õh✏ } „{g>h✏ }
g d +
{f >h✏ } „{g>h✏ }
f d +
f ‚ g d +
{f >h✏ } „{gÕh✏ }
{f >h✏ } „{g>h✏ }
Õ
{f >h✏ }
g d
{f Õh✏ } „{g>h✏ }
f d +
{g>h✏ }
g d
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Õ
f d +
{f >f✏ }
g d +
{g>g✏ }
f d +
{f >f✏ }
g d
{g>g✏ }
Õ 4✏
uniformly for all f À F and g À G.
(iv) This follows from (iii) if we set •
tF ‰ F ,
•
tf✏ ‰ f✏ ,
• •
(1 * t)F ‰ G, (1 * t)f✏ ‰ g✏ ,
and observe that the calculation is uniform for all t À [0, 1].
(v) Without loss of generality we can assume that F is convex, i.e. coincides with its convex hull.
Let u be an element of the L1 -closure of (the convex hull of) F . Then there is a sequence (fj )j œ F : lim Òu * fj Ò1 = 0. j
We have, because of u Õ u * fj + fj , {u > f✏ } œ {u * fj > f✏ } ‰ {fj > f✏ } ⌅ ⇧ = {u * fj > f✏ } „ {fj > f✏ } ⌅ ⇧ ‰ {u * fj > f✏ } „ {fj Õ f✏ } ⌅ ⇧ ‰ {u * fj Õ f✏ } „ {fj > f✏ } that
u d
{u>f✏ }
Õ
u d +
u * fj d +
{u*fj >f✏ } „{fj >f✏ }
Õ
u d +
{u*fj >f✏ } „{fj Õf✏ }
{u*fj >f✏ } „{fj >f✏ }
+
u d
{u*fj Õf✏ } „{fj >f✏ }
fj d
{u*fj >f✏ } „{fj >f✏ }
u * fj ‚ fj d +
{u*fj >f✏ } „{fj Õf✏ }
Õ Òu * fj Ò1 +
u * fj ‚ fj d
{u*fj Õf✏ } „{fj >f✏ }
fj d + Òu * fj Ò1 +
{fj >f✏ }
fj d
{fj >f✏ }
Õ 2 Òu * fj Ò1 + 2✏ ,,,,,,,,,ô , 2✏. jôÿ
Since this holds uniformly for all such u, we are done. 263
R.L. Schilling: Measures, Integrals & Martingales ∑∑ Problem 22.13 Solution: By assumption,
≈ ✏ > 0 « w✏ À L1+ : sup
f ÀF
{f >w✏ }
f d Õ ✏.
Now observe that sup f d {sup1ÕjÕk fj >w✏ } 1ÕjÕk j Õ Õ Õ
k … l=1 k
…
l=1 k
…
{sup1ÕjÕk fj >w✏ }„{fl =sup1ÕjÕk fj } {fl >w✏ }
fl d
fl d
✏
l=1
= k ✏. Therefore,
and we get
sup fj d
1ÕjÕk
Õ
sup f d + sup f d {sup1ÕjÕk fj Õw✏ } 1ÕjÕk j {sup1ÕjÕk fj >w✏ } 1ÕjÕk j
Õ
w✏ d + k✏ 1 kôÿ k lim
1 w✏ d + ✏ = ✏ kôÿ k
sup fj d Õ lim
1ÕjÕk
which proves our claim as ✏ > 0 was arbitrary.
∑∑
Problem 22.14 Solution: Since the function u í R, R > 0, is integrable w.r.t. the probability measure P, we get
{uj >R}
uj dP Õ = Õ
{uj >R} 1
Rp*1 1 Rp*1
uj
uj p*1 Rp*1
{uj >R}
dP
uj p dP
uj p dP
1 sup uk p dP Rp*1 k 1 = p*1 sup Òuk Òpp R k
Õ
264
Solution Manual. Chapter 1–13. Last update 23rd August 2017 which converges to zero as R ô ÿ. This proves uniform integrability. Counterexample: Vitali’s theorem implies that a counterexample should satisfy P
uj does not converge in L1 .
Òuj Ò1 = 1,
uj ,,,,,,,,,ô , u, jôÿ
Consider, for example, the probability space ((0, 1), B(0, 1), dx) and the sequence uj := j 1(0,1_j) .
Then uj ô 0 pointwise (everywhere!), hence in measure. This is also the expected L1 limit, if it exists. Moreover,
Òuj Ò1 =
uj dx = 1
which means that uj cannot converge in L1 to the expected limit u í 0, i.e. it does not converge in L1 .
Vitali’s theorem shows now that (uj )j cannot be uniformly integrable. We can verify this fact also directly: for R > 0 and all j > R we get {uj >R} which proves sup j
uj dx =
{uj >R}
uj dx = 1
uj dx = 1
≈R > 0
and (uj )j cannot be uniformly integrable (in view of the equivalent characterizations of uniform integrability on finite measure spaces, cf. Theorem 22.9)
∑∑ Problem 22.15 Solution: We have ÿ …
j (j < f Õ j + 1) =
j=k
ÿ … j=k ÿ
Õ
… j=k
=
{jk}
f d Ù sup
ÿ …
f ÀF j=k
{f >k}
f d
j (j < f Õ j + 1).
This proves the claim (since we are in a finite measure space where u í k is an integrable function!) Problem 22.16 Solution: Fix ✏ > 0. By assumption there is some w = w✏ À L1+ such that sup i
{fi >w}
∑∑
fi d Õ ✏.
Since ui Õ fi we infer that {ui > w} œ {fi > w}, and so {ui >w}
ui d Õ
{fi >w}
fi d Õ ✏ uniformly for all i À I. ∑∑
Problem 22.17 Solution: Let g À L1+ ( ). Then 0Õ
(u * g · u) d =
{uŒg}
(u * g) d Õ
{uŒg}
u d .
This implies that uniform integrability of the family F implies that the condition of Problem 22.17 holds. On the other hand,
{uŒg}
u d =
{uŒg}
(2u * u) d
Õ
{uŒg}
(2u * g) d
Õ
{2uŒg}
=2
(2u * g) d
{uΠ1 g}
u * 12 g d
{uΠ1 g}
u *
2
=2
2
⌅1 ⇧ g · u d 2
and since g À L1 if, and only if, g À L1 , we see that the condition given in Problem 22.17 entails uniform integrability.
1 2
In finite measure spaces this conditions is simpler: constants are integrable functions in finite measure spaces; thus we can replace the condition given in Problem 22.17 by lim sup
Rôÿ uÀF
(u * R · u) d = 0. ∑∑
266
23 Martingales. Solutions to Problems 23.1–23.16 Problem 23.1 Solution: Since A0 = {Á, X} an A0 -measurable function u must satisfy {u = s} = Á or = X, i.e. all A0 -measurable functions are constants.
So if (uj )jÀN0 is a martingale, u0 is a constant and we can calculate its value because of the martingale property:
X
u0 d =
X
u1 d
-Ÿ u0 = (X)*1
Conversely, since A0 = {Á, X} and since Á
u0 d =
Á
X
u1 d .
(*)
u1 d
always holds, it is clear that the calculation and choice in (*) is necessary and sufficient for the claim.
∑∑ Problem 23.2 Solution: We consider only the martingale case, the other two cases are similar. (a) Since Bj œ Aj we get A -Ÿ
uj d =
B
A
uj d =
showing that (uj , Bj )j is a martingale.
uj+1 d
B
uj+1 d
≈ A À Aj ≈ B À Bj
(b) It is clear that the above implication cannot hold if we enlarge Aj to become Cj . Just consider
the following ‘extreme’ case (to get a counterexample): Cj = A for all j. Any martingale (uj , C)j must satisfy,
A
uj d =
A
uj+1 d
≈ A À A.
Considering the sets A := {uj < uj+1 } À A and A® := {uj > uj+1 } À A we conclude that 0=
{uj >uj+1 }
(uj * uj+1 ) d
-Ÿ
({uj > uj+1 }) = 0
and, similarly ({uj < uj+1 }) = 0 so that uj = uj+1 almost everywhere and for all j. This
means that, if we start with a non-constant martingale (uj , Aj )j , then this can never be a martingale w.r.t. the filtration (Cj )j .
267
R.L. Schilling: Measures, Integrals & Martingales ∑∑ Problem 23.3 Solution: For the notation etc. we refer to Problem 4.15. Since the completion A j is given by
A j = (Aj , N),
N := M œ X : « N À A, N – M,
(N) = 0
we find that for all A l} = {⌧k*1 Õ l} À Al
while, by definition {
k
= N} = Á À AN . ∑∑
Problem 24.2 Solution: Theorem 24.7 becomes for supermartingales: Let (ul )lÀ*N be a backwards supermartingale and assume that A*ÿ is -finite. Then limjôÿ u*j = u*ÿ À (*ÿ, ÿ] exists a.e.
Moreover, L1 -limjôÿ u*j = u*ÿ if, and only if, supj î u*j d < ÿ; in this case (ul , Al )lÀ*N is a supermartingale and u*ÿ is finitely-valued.
Using this theorem the claim follows immediately from the supermartingale property: *ÿ <
A
u*1 d Õ
A
u*j d Õ
and, in particular, for A = X À A*ÿ .
A
u*ÿ d < ÿ
≈ j À N, A À A*ÿ
∑∑ Problem 24.3 Solution: Corollary 24.3 shows pointwise a.e. convergence. Using Fatou’s lemma we get
jôÿ
0 = lim
uj d = lim inf jôÿ
uj d
275
R.L. Schilling: Measures, Integrals & Martingales Œ
lim inf uj d
=
uÿ d Œ 0
jôÿ
so that uÿ = 0 a.e.
Moreover, since î uj d ,,,,,,,,,ô , 0 = î uÿ d , Theorem 24.6 shows that uj ô uÿ in L1 -sense. jôÿ
∑∑
Problem 24.4 Solution: From L1 -limjôÿ uj = f we conclude that supj î uj d < ÿ and we get that limjôÿ uj exists a.e. Since L1 -convergence also implies a.e. convergence of a subsequence, the limiting functions must be the same.
∑∑ Problem 24.5 Solution:
The quickest solution uses the famous Chung-Fuchs result that a simple
random walk (this is just Sj := ⇠1 + 5 + ⇠j with ⇠k iid Bernoulli p = q = 12 ) does not converge and that *ÿ = lim inf j Sj < lim supj Sj = ÿ a.e. Knowing this we are led to 1 P (uj converges) = P (⇠0 + 1 = 0) = . 2 It remains to show that uj is a martingale. For A À (⇠1 , … , ⇠j ) we get A
uj+1 dP = = = =
A A A A
(⇠0 + 1)(⇠1 + 5 + ⇠j + ⇠j+1 ) dP (⇠0 + 1)(⇠1 + 5 + ⇠j ) dP + uj dP +
A
(⇠0 + 1) dP
⌦
A
(⇠0 + 1)⇠j+1 dP
⇠j+1 dP
uj dP
where the last step follows because of independence. If you do not know the Chung-Fuchs result, you could argue as follows: assume that for some finite random variable S the limit Sj (!) ô S(!) takes place on a set A œ ⌦. Since the ⇠j ’s are iid, we have
⇠2 + ⇠3 + 5 ô S and ⇠1 + ⇠2 + 5 ô S which means that S and S + ⇠1 have the same probability distribution. But this entails that S is necessarily ±ÿ, i.e., Sj cannot have a finite limit.
∑∑ Problem 24.6 Solution: (i) Cf. the construction in Scholium 23.4. 276
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (ii) Note that n2 * (n * 1)2 * 1 = 2n * 2 is even. The function f : R2n*2 ô R, f (x1 , … , x2n*2 ) = x1 + 5 + xn2 *(n*1)2 is clearly Borel measurable, i.e. the function
f (⇠(n*1)2 +2 , … , ⇠n2 ) = ⇠(n*1)2 +2 + 5 + ⇠n2 is An -measurable and so is the set An .
Moreover, x À An if, and only if, exactly half of ⇠(n*1)2 +2 , … , ⇠n2 are +1 and the other half is *1. Thus,
0 10 1n*1 0 1n*1 0 10 12n*2 2n * 2 1 1 2n * 2 1 (An ) = = n*1 2 2 n*1 2
Using Stirling’s formula, we get 0 1 (2k)! 1 2k = k! k! 22k k ˘
2⇡2k(2k)2k ek ek ˘ ˘ 22k 2⇡k 2⇡kkk kk e2k 1 = ˘ ,,,,,,,,,ô , 0. k⇡ kôÿ Ì
Setting k = n * 1 this shows both
…
lim (An ) = 0 and n
(An ) Ì
n
… 1 ˘ = ÿ. n n
Finally, lim supn 1An = 1lim supn An = 1 a.e. while, by Fatou’s lemma 0Õ
lim inf 1An d Õ lim inf n
n
1An d = lim inf (An ) = 0, n
i.e., lim inf n 1An = 0 a.e. This means that 1An does not have a limit as n ô ÿ.
(iii) For A À An we have because of independence A = = = (iv) We have
Mn+1 d
A A A
Mn (1 + ⇠n2 +1 ) d + Mn d
[0,1]
A
1An ⇠n2 +1 d
(1 + ⇠n2 +1 ) d +
A
1An d
[0,1]
⇠n2 +1 d
Mn d .
{Mn+1 ë 0}
= {Mn+1 ë 0, ⇠n2 +1 = *1} ‰ {Mn+1 ë 0, ⇠n2 +1 = +1} œ An ‰ {Mn ë 0, ⇠n2 +1 = +1}.
277
R.L. Schilling: Measures, Integrals & Martingales (v) By definition, Mn+1 * Mn = Mn ⇠n2 +1 + 1An ⇠n2 +1 = (Mn + 1An )⇠n2 +1 so that Mn+1 * Mn = Mn + 1An ⇠n2 +1 = Mn + 1An . This shows that for x À {limn Mn exists} the limit limn 1An (x) exists. But, because of (ii), the latter is a null set, so that the pointwise limit of Mn cannot exist. On the other hand, using the inequality (iv), shows (Mn+1 ë 0) Õ
and iterating this gives
(Mn+k ë 0) Õ Õ
1 2k 1 2k
1 2
(Mn ë 0) + (An )
(Mn ë 0) + (An ) + 5 (An+k*1 ) + (An ) + 5 (An+k*1 ).
Letting first n ô ÿ and then k ô ÿ yields
lim sup (Mj ë 0) = 0 j
so that limj (Mj = 0) = 0. ∑∑ Problem 24.7 Solution: Note that for A À {1}, {2}, … , {n}, {n + 1, n + 2, …} we have A
A
⇠n+1 dP =
(n + 2)1[n+2,ÿ)„N dP
h n0 =l (n + 2)1[n+2,ÿ)„N dP n j [n+1,ÿ)„N
if A is a singleton else
and in the second case we have [n+1,ÿ)„N
(n + 2)1[n+2,ÿ)„N dP =
[n+2,ÿ)„N
= (n + 2)
(n + 2) dP
ÿ …
P ({j})
j=n+2 ÿ 0
= (n + 2)
…
j=n+2
= 1, The same calculation shows A 278
⇠n dP =
A
(n + 1)1[n+1,ÿ)„N dP
1 1 * j j+1
1
Solution Manual. Chapter 1–13. Last update 23rd August 2017 h if A is a singleton n0 =l (n + 1)1[n+1,ÿ)„N dP = 1 else n j [n+1,ÿ)„N so that
A
A
⇠n+1 dP =
⇠n dP
for all A from a generator of the -algebra which contains an exhausting sequence. This shows, by Remark 23.2(i) that (⇠n )n is indeed a martingale.
The second calculation from above also shows that î ⇠n dP = 1 while and
sup ⇠n = ÿ n
lim ⇠n = 0 n
are obvious. ∑∑ Problem 24.8 Solution: (i) Using Problem 23.6 we get
(uj * uj*1 )2 d =
u2j d * 2
uj uj*1 d +
=
u2j d * 2
u2j*1 d +
=
u2j d *
which means that
u2N d =
and the claim follows.
N … j=1
u2j*1 d
u2j*1 d
u2j*1 d
(uj * uj*1 )2 d
(ii) Because of Example 23.3(vi), p = 2, we conclude that (u2j )j is a submartingale which,
due to L2 -boundedness, satisfies the assumptions of Theorem 24.2 on submartingale
convergence. This means that limj u2j = u2 exists a.e. This is, alas, not good enough to get uj ô u a.e., it only shows that uj ô u a.e.
The following trick helps: let (Ak )k œ A0 be an exhausting sequence with Ak ~ X
and (Ak ) < ÿ. Then (1Ak uj )j is an L1 -bounded martingale: indeed, if A À An then A „ Ak À An and it is clear that A
1 A k un d =
A„Ak
un d =
A„Ak
un+1 d =
A
1Ak un+1 d
while, by the Cauchy–Schwarz inequality,
1Ak un d Õ
˘
v (Ak )
sup n
u2n d Õ ck .
279
R.L. Schilling: Measures, Integrals & Martingales Thus, we can use Theorem 24.2 and conclude that 1Ak un ,,,,,,,,,ô , 1 Ak u nôÿ
almost everywhere with, because of almost-everywhere-uniqueness of the limits on each of the sets Ak , a single function u. This shows un ô u a.e.
(iii) Following the hint and using the arguments of part (i) we find
(uj+k * uj )2 d =
(u2j+k * u2j ) d
j+k …
=
l=j+1 j+k …
=
l=j+1
(u2l * u2l*1 ) d
(ul * ul*1 )2 d .
Now we use Fatou’s lemma and the result of part (ii) to get
lim inf (u * uj )2 d Õ lim inf j
j
(u * uj )2 d
(u * uj )2 d
Õ lim sup j
Õ lim sup j
since, by L2 -boundedness,
k=1 î
≥ÿ
ÿ … l=j+1
(ul * ul*1 )2 d
=0 (uk * uk*1 )2 d < ÿ.
(iv) Since (⇠) < ÿ, constants are integrable and we find using the Cauchy–Schwarz and Markov inequalities
uk >R
uk d Õ
˘
1 Õ R Õ
v (uk > R) v
u2k d
v
u2k d u2k d
1 sup u2 d R k k
from which we get uniform integrability; the claim follows now from parts (i)–(iii) and Theorem 24.6.
∑∑ Problem 24.9 Solution:
(i) Note that î ✏j dP = 0 and î ✏j2 dP = 1. Moreover, ⇠n := w.r.t. the filtration An := (✏1 , … , ✏n ) and
280
⇠n2 dP
=
n … j=1
y2j .
≥n
j=1 ✏j yj
is a martingale
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 24.8 now shows that
≥ÿ
2 j=1 yj
< ÿ means that the martingale (⇠n )n is L2 -
bounded, i.e. ⇠n converges a.e. The converse follows from part (iii).
(ii) This follows with the same arguments as in part (i) with An = (⇠1 , … , ⇠n ).
(iii) We show that Sn2 * An is a martingale. Now for A À An A
Mn+1 dP = = = =
A A A A
2 (Sn+1 * An+1 ) dP 2 (Sn2 + 2⇠n+1 Sn + ⇠n+1 * An *
(Sn2 * An ) dP + A
Mn dP +
But, because of independence, A =
2 (2⇠n+1 Sn + ⇠n+1 *
A
2⇠n+1 dP
= 0 + P (A)
⌦
2 n+1
A
2 ) dP n+1
2 (2⇠n+1 Sn + ⇠n+1 *
2 (2⇠n+1 Sn + ⇠n+1 *
2 ) dP n+1
2 ) dP n+1
2 ) dP n+1
Sn dP + P (A)
* P (A)
2 ⇠n+1 dP * P (A)
2 n+1
2 n+1
= 0. and the claim is established. Now define ⌧ := ⌧ := inf {j : Mj > }. By optional sampling, (Mn·⌧ )n is again a martingale and we have Mn·⌧ = Mn 1{nt,sôt f (s) exists and is finite for every t and that {f ë F } is at most countable (since it is contained in the set of discontinuities of f ), hence a Lebesgue null set.
If f is right-continuous, (a, b] := f (b) * f (a) extends uniquely to a measure on the Borel-sets and this measure is locally finite and -finite. If we apply Theorem 25.9 to write
=
˝
+
and we get a.e.
Ú
with
˝
~
and
ÚÚ
. By Corollary 25.22 D
Ú
and
= 0 a.e. and D
= ˝
1
we can
exists a.e.
˝ (x * r, x + r) (x * r, x + r) = lim +0 rô0 rô0 2r 2r and we can set f ® (x) = D (x) which is a.e. defined. Where it is not defined, we put it equal to 0.
D (x) = lim
Now we get
f (b) * f (a) = (a, b] Π(a, b) =
(a,b)
d
Œ
(a,b)
d
=
(a,b)
D (x) (dx)
=
(a,b)
f ® (x) (dx).
˝
The above estimates show that we get equality if f is continuous and also absolutely continuous w.r.t. Lebesgue measure.
∑∑ Problem 25.15 Solution: Without loss of generality we may assume that fj (a) = 0, otherwise we would consider the (still increasing) functions x ;ô fj (x) * fj (a) resp. their sum x ;ô s(x) * s(a). The derivatives are not influenced by this operation. As indicated in the hint call sn (x) := f1 (x) + 5 + fn (x) the nth partial sum. Clearly, s, sn are increasing
sn (x + h) * sn (x) sn+1 (x + h) * sn+1 (x) s(x + h) * s(x) Õ Õ . h h h 291
R.L. Schilling: Measures, Integrals & Martingales and possess, because of Problem 25.14, almost everywhere positive derivatives: s®n (x) Õ s®n+1 (x) Õ 5 s® (x),
≈x à E
Note that the exceptional null-sets depend originally on the function sn etc. but we can consider
their (countable!!) union and get thus a universal exceptional null set E. This shows that the formally differentiated series
ÿ … j=1
fj® (x)
converges for all x à E.
Since the sequence of partial sums is increasing, it will be enough to check that s® (x) * s®n (x) ,,,,,,,,,ô , 0 k
kôÿ
≈x à E.
Since, by assumption the sequence sk (x) ô s(x) we can choose a subsequence nk in such a way that
s(b) * snk (b) < 2*k
≈ k À N.
Since 0 Õ s(x) * snk (x) Õ s(b) * snk (b) the series
ÿ … k=1
(s(x) * snk (x)) Õ
ÿ …
2*k < ÿ
≈ x À [a, b].
k=1
By the first part of the present proof, we can differentiate this series term-by-term and get that ÿ … (s® (x) * s®n (x)) converges ≈ x À (a, b) ‰ E k
k=1
and, in particular, s® (x) * s®n (x) ,,,,,,,,,ô , 0 for all x À (a, b) ‰ E which was to be proved. k
kôÿ
∑∑
292
26 Abstract Hilbert space. Solutions to Problems 26.1–26.19 Problem 26.1 Solution: If we set
=
1
+5+
n,
X = {1, 2, … , n}, A = P(X) or
X = N, A = P(X), respectively, we can deduce 26.5(i) and (ii) from 26.5(iii).
=
≥
jÀN j ,
Let us, therefore, only verify (iii). Without loss of generality (see the complexification of a real inner product space in Problem 26.3) we can consider the real case where L2 = L2R . • L2 is a vector space — this was done in Remark 13.5. • Íu, vÎ is finite on L2 ù L2 — this is the Cauchy–Schwarz inequality 13.3. • Íu, vÎ is bilinear — this is due to the linearity of the integral. • Íu, vÎ is symmetric — this is obvious. • Ív, vÎ is definite, and ÒuÒ2 is a Norm — cf. Remark 13.5. ∑∑ Problem 26.2 Solution: (i) We prove it for the complex case—the real case is simpler. Observe that Íu ± w, u ± wÎ = Íu, uÎ ± Íu, wÎ ± Íw, uÎ + Íw, wÎ = Íu, uÎ ± Íu, wÎ ± Íu, wÎ + Íw, wÎ = Íu, uÎ ± 2 ReÍu, wÎ + Íw, wÎ. Thus, Íu + w, u + wÎ + Íu * w, u * wÎ = 2Íu, uÎ + 2Íw, wÎ. Since ÒvÒ2 = Ív, vÎ we are done. (ii) (SP1 ): Obviously, 0 < (u, u) =
1 4
Ò2vÒ2 = ÒvÒ2 -Ÿ v ë 0.
(SP1 ): is clear. (iii) Using at the point (*) below the parallelogram identity, we have 4(u + v, w) = 2(u + v, 2w) =
1 2
Òu + v + 2wÒ2 * Òu + v * 2wÒ2 293
R.L. Schilling: Measures, Integrals & Martingales 1 2 < 1 =2
=
Ò(u + w) + (v + w)Ò2 * Ò(u * w) + (v * w)Ò2
⌧ 2 Òu + wÒ2 + Òv + wÒ2 * Òu * wÒ2 * Òv * wÒ2
= 4(u, w) + 4(v, w) and the claim follows.
(iv) We show (qv, w) = q(v, w) for all q À Q. If q = n À N0 , we iterate (iii) n times and have
(nv, w) = n(v, w)
≈ n À N0
(*)
(the case n = 0 is obvious). By the same argument, we get for m À N (v, w) = m m1 v, w = m
which means that
1 m
v, w =
1 m
(v, w)
Combining (*) and (**) then yields ( mn v, w) =
1 m
v, w
≈ m À N. n m
(**)
(v, w). Thus,
(pu + qv, w) = p(u, w) + q(v, w)
≈ p, q À Q.
(v) By the lower triangle inequality for norms we get for any s, t À R Û Û ÛÒtv ± wÒ * Òsv ± wÒÛ Õ Ò(tv ± w) * (sv ± w)Ò Û Û = Ò(t * s)vÒ = t * s ÒvÒ. This means that the maps t ;ô tv ± w are continuous and so is t ;ô (tv, w) as the sum of two continuous maps. If t À R is arbitrary, we pick a sequence (qj )jÀN œ Q such that limj qj = t. Then
(tv, w) = lim(qj v, w) = lim qj (qv, w) = t(v, w) j
so that
j
(su + tv, w) = (su, w) + (tv, w) = s(u, w) + t(v, w). ∑∑ Problem 26.3 Solution: This is actually a problem on complexification of inner product spaces... . Since v and iw are vectors in V ‚ iV and since ÒvÒ = Ò ± ivÒ, we get (v, iw)R = = =
1 4 1 4 1 4
Òv + iwÒ2 * Òv * iwÒ2 Òi(w * iv)Ò2 * Ò(*i)(w + iv)Ò2 Òw * ivÒ2 * Òw + ivÒ2
= (w, *iv)R = *(w, iv)R . 294
(*)
Solution Manual. Chapter 1–13. Last update 23rd August 2017 In particular, (v, iv) = *(v, iv) -Ÿ (v, iv) = 0
≈v,
and we get (v, v)C = (v, v)R > 0 -Ÿ v = 0. Moreover, using (*) we see that (v, w)C = (v, w)R + i(v, iw)R <
= (w, v)R * i(w, iv)R = (w, v)R + iÑ (w, iv)R = (w, v)R + i(w, iv)R = (w, v)C . Finally, for real ↵,
À R the linearity property of the real scalar product shows that
(↵u + v, w)C = ↵(u, w)R + (v, w)R + i↵(u, iw)R + i (v, iw)R = ↵(u, w)C + (v, w)C . Therefore to get the general case where ↵,
À C we only have to consider the purely imaginary
case:
<
(iv, w)C = (iv, w)R + i(iv, iw)R = *(v, iw)R * i(v, *w)R = *(v, iw)R + i(v, w)R = i i(v, iw)R + (v, w)R = i(v, w)C , where we use twice the identity (*). This shows complex linearity in the first coordinate, while skew-linearity follows from the conjugation rule (v, w)C = (w, v)C .
∑∑ Problem 26.4 Solution: The parallelogram law (stated for L1 ) would say: 0
0
1
u + w dx
12
0 +
0
1
u * w dx
12
0 =2
0
1
12 0 u dx + 2
0
1
w dx
12 .
If u ± w, u, w have always only ONE sign (i.e. +ve or *ve), we could leave the modulus signs ÷
away, and the equality would be correct! To show that there is no equality, we should therefore choose functions where we have some sign change. We try: u(x) = 1_2,
w(x) = x
(note: u * w does change its sign!) and get 0
1
u + w dx =
0
1
( 12 + x) dx = [ 12 (x + x2 )]10 = 1
295
R.L. Schilling: Measures, Integrals & Martingales 0
1
u * w dx =
0
1_2
( 12 * x) dx + 1_2
= [ 12 (x * x2 )]0 = 0 0 This shows that
1
1
u dx =
w dx =
1 4
*
0 0
12 + ( 14 )2 =
1
1
1 8
*
1 8
1 2
dx =
+
1 4
1
1_2
+ [ 12 (x2 * x)]11_2 =
1 4
1 2
x dx = [ 12 x2 ]10 =
17 16
(x * 12 ) dx
1 2
ë 1 = 2( 12 )2 + 2( 12 )2 .
We conclude, in particular, that L1 cannot be a Hilbert space (since in any Hilbert space the Parallelogram law is true....).
∑∑ Problem 26.5 Solution:
(i) If k = 0 we have ✓ = 1 and everything is obvious. If k ë 0, we use the summation formula for the geometric progression to get S :=
n
n
1 … jk 1 … k ✓ = ✓ n j=1 n j=1
j
=
✓ 1 * (✓ k )n n 1 * ✓k
but (✓ k )n = exp(2⇡ ni k n) = exp(2⇡ik) = 1. Thus S = 0 and the claim follows.
(ii) Note that ✓ j = ✓ *j so that
Òv + ✓ j wÒ2 = Ív + ✓ j w, v + ✓ j wÎ = Ív, vÎ + Ív, ✓ j wÎ + Í✓ j w, vÎ + Í✓ j w, ✓ j wÎ = Ív, vÎ + ✓ *j Ív, wÎ + ✓ j Íw, vÎ + ✓ j ✓ *j Íw, wÎ = Ív, vÎ + ✓ *j Ív, wÎ + ✓ j Íw, vÎ + Íw, wÎ. Therefore, n
1… j ✓ Òv + ✓ j wÒ2 n j=1 n
n
n
n
1… j 1… 1 … 2j 1… j = ✓ Ív, vÎ + Ív, wÎ + ✓ Íw, vÎ + ✓ Íw, wÎ n j=1 n j=1 n j=1 n j=1 = 0 + Ív, wÎ + 0 + 0 where we use the result from part (i) of the exercise. (iii) Since the function
;ô ei Òv + ei wÒ2 is bounded and continuous, the integral exists as
a (proper) Riemann integral, and we can use any Riemann sum to approximate the integral,
296
Solution Manual. Chapter 1–13. Last update 23rd August 2017 see 12.6–12.12 in Chapter 12 or Corollary I.6 and Theorem I.8 of Appendix I. Before we do that, we change variables according to
= ( + ⇡)_2⇡ so that d
= d _2⇡ and
1 Ù Ù2 Ù Ù2 ei Ùv + ei wÙ d = * e2⇡i Ùv * e2⇡i wÙ d . Ù Ù Ù Ù (0,1] 2⇡ (*⇡,⇡] Now using equidistant Riemann sums with step 1_n and nodes ✓nj = e2⇡i yields, because of part (ii) of the problem, *
Ù Ù2 e2⇡i Ùv * e2⇡i wÙ d Ù Ù (0,1]
1 n
j
, j = 1, 2, … , n
n
1… j ✓n Òv * ✓nj wÒ2 nôÿ n j=1
= * lim
= * lim Ív, *wÎ nôÿ
= Ív, wÎ. ∑∑ Problem 26.6 Solution: We assume that V is a C-inner product space. Then, Òv + wÒ2 = Ív + w, v + wÎ = Ív, vÎ + Ív, wÎ + Íw, vÎ + Íw, wÎ = ÒvÒ2 + Ív, wÎ + Ív, wÎ + ÒwÒ2 = ÒvÒ2 + 2 ReÍv, wÎ + ÒwÒ2 . Thus
Òv + wÒ2 = ÒvÒ2 + ÒwÒ2 ◊Ÿ ReÍv, wÎ = 0 ◊Ÿ vÚw. ∑∑
Problem 26.7 Solution: Let (hk )k œ H such that limk Òhk * hÒ = 0. By the triangle inequality Òhk * hl Ò Õ Òhk * hÒ + Òh * hl Ò ,,,,,,,,,,,,,,ô , 0. ´≠Ø≠¨ ´≠≠Ø≠≠¨ k,lôÿ ô0
ô0
Problem 26.8 Solution: Let g, gõ À H. By the Cauchy–Schwarz inequality 26.3
∑∑
ÛÍg, hÎ * Íõ g , hÎÛÛ Õ ÛÛÍg * gõ, hÎÛÛ Õ ÒhÒ Òõ g * gÒ Û which proves continuity. Incidentally, this calculation shows also that, since g ;ô Íg, hÎ is linear, it would have been enough to check continuity at the point g = 0 (think about it!).
∑∑ Problem 26.9 Solution: Definiteness (N1 ) and positive homogeneity (N2 ) are obvious. The triangle inequality reads in this context (g, g ® , h, h® À H):
Ò(g, h) + (g ® , h® )Ò Õ Ò(g, h)Ò + Ò(g ® , h® )Ò ◊Ÿ 297
R.L. Schilling: Measures, Integrals & Martingales Òg + g ® Òp + Òh + h® Òp Since
1_p
Õ ÒgÒp + ÒhÒp 1_p
Òg + g ® Òp + Òh + h® Òp
Õ
⌅
ÒgÒÒg ® Ò
1_p
⇧p
+ Òg ® Òp + Òh® Òp
⌅ ⇧p + ÒhÒ + Òh® Ò
1_p
.
1_p
we can use the Minkowski inequality for sequences resp. in R2 —which reads for numbers a, A, b, B Œ 0 (a + b)p + (A + B)p
1_p
Õ a p + Ap
1_p
+ bp + B p
1_p
—and the claim follows. Since R2 is only with the Euclidean norm a Hilbert space—the parallelogram identity fails for the norms (xp + yp )1_p —this shows that also in the case at hand only p = 2 will be a Hilbert space norm.
∑∑
Problem 26.10 Solution: For the scalar product we have for all g, g ® , h, h® À H such that Òg * g ® Ò2 + Òh * h® Ò2 < 1
⌅ ⇧1_2 Û Û ÛÍg * g ® , h * h® ÎÛ Õ Òg * g ® Ò Òh * h® Ò Õ Òg * g ® Ò2 + Òh * h® Ò2 Û Û where we use the elementary inequality ˘ 1 ab Õ (a2 + b2 ) Õ a2 + b2 Õ a2 + b2 . 2 ´≠≠≠≠Ø≠≠≠≠¨ if a2 +b2 Õ1
⌅ ⇧1_2 Since (g, h) ;ô ÒgÒ2 + ÒhÒ2 is a norm on H ù H we are done. Essentially the same calculation applies to (t, h) ;ô t h. ∑∑
Problem 26.11 Solution: Assume that H has a countable maximal ONS, say (ej )j . Then, by definition, every vector h À H can be approximated by a sequence made up of finite linear combinations of the (ej )j :
hk :=
n(k) … j=1
↵j ej
(note that ↵j = 0 is perfectly possible!). In view of problem 26.10 we can even assume that the ↵j are rational numbers. This shows that the set D := is a countable dense subset of H.
n $… j=1
↵j ej : ↵j À Q, n À N
%
Conversely, if D œ H is a countable dense subset, we can use the Gram-Schmidt procedure and obtain from D an ONS. Then Theorem 26.24 proves the claim.
∑∑
298
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 26.12 Solution: Let us, first of all, show that for a closed subspace C œ H we have C = (C Ú )Ú .
Because of Lemma 26.12 we know that C œ (C Ú )Ú and that C Ú is itself a closed linear subspace of H. Thus,
C ‚ C Ú = H = C Ú ‚ (C Ú )Ú .
Thus C cannot be a proper subspace of (C Ú )Ú and therefore C = (C Ú )Ú . Applying this to the obviously closed subspace C := K w = span(w) we conclude that span(w) = span(w)ÚÚ .
By assumption, Mw = {w}Ú and MwÚ = {w}ÚÚ and we have w À {w}ÚÚ . The last expression is a (closed) subspace, so
w À {w}ÚÚ -Ÿ span(w) œ {w}ÚÚ also. Further {w} œ span(w) -Ÿ {w}Ú – span(w)Ú -Ÿ {w}ÚÚ œ span(w)ÚÚ = span(w) and we conclude that {w}ÚÚ = span(w) which is either {0} or a one-dimensional subspace. ∑∑ Problem 26.13 Solution: (i) By Pythagoras’ Theorem 26.19 Òej * ek Ò2 = Òej Ò2 + Òek Ò2 = 2
≈ j ë k.
This shows that no subsequence (ej )jÀ= can ever be a Cauchy sequence, i.e. it cannot converge.
If h À H we get from Bessel’s inequality 26.19 that the series … j
Íej , hÎ2 Õ ÒhÒ2
is finite, i.e. converges. Thus the sequence with elements Íej , hÎ must converge to 0 as j ô ÿ.
(ii) Parseval’s equality 26.19 shows that ÒhÒ2 =
ÿ … j=1
Íej , hÎ2 =
ÿ … j=1
cj 2 Õ
ÿ … 1 1.
j=1
≥
Since the remaining tail of the series
a2j = ÿ we can construct recursively a
j>j1
strictly increasing sequence (jk )kÀN0 œ N, j0 := 1, such that … jÀJk
(ii) Define the numbers
k
where Jk := (jk , jk+1 ] „ N.
a2j > 1
as, say, k
1 := t . ≥ 2 k a jÀJk j
Then … j
b2j = =
…… k jÀJk
… k
=
Moreover, since
t≥ we get
j
=
k jÀJk k
k
…1 k
a2j
Π1,
2 k aj
…
…1
(iii) We want to show (note that we renamed
≈
a2j
jÀJk
a2j
…
k
for any sequence ↵ = (↵j )j we have:
≥
……
k
Œ
jÀJk
2 jÀJk aj
k
=
jÀJk
a2j
… 1 < ÿ. 2 k k
jÀJk
aj bj =
…
≥ k2
≥
…
2 k
… k
=
2 2 k aj
jÀJk
a2j
≥
jÀJk
t≥
a2j
jÀJk
a2j
= ÿ.
:= a and ↵ := b for notational reasons) that
À l 2 : Í↵, Î < ÿ -Ÿ ↵ À l 2 . 303
R.L. Schilling: Measures, Integrals & Martingales Assume, to the contrary, that ↵ Ã l 2 . Then
≥ j
↵j2 = ÿ and, by part (i), we can find
a sequence of jk with the properties described in (i). Because of part (ii) there is a sequence
= ( j )j À l 2 such that the scalar product Í↵, Î = ÿ. This contradicts our
assumption, i.e. ↵ should have been in l 2 in the first place.
(iv) Since, by Theorem 26.24 every separable Hilbert space has a basis (ej )jÀN œ H, we can identify h À H with the sequence of ‘coordinates’ (Íh, ej Î)jÀN and it is clear that (iii) implies (iv).
∑∑ Problem 26.17 Solution: (i) Since P 2 = P is obvious by the uniqueness of the minimizing element, this part follows already from Remark 26.15.
(ii) Note that for u, v À H we have
≈ h À H : Íu, hÎ = Ív, hÎ -Ÿ u = v.
Indeed, consider h := u * v. Then Íu, hÎ = Ív, hÎ -Ÿ 0 = Íu * v, hÎ = Íu * v, u * vÎ = u * v2 so that u = v. Linearity of P : Let ↵,
À K and f , g, h À H. Then
ÍP (↵f + g), hÎ = Í↵f + g, P hÎ = ↵Íf , P hÎ + Íg, P hÎ = ↵ÍP f , hÎ + ÍP g, hÎ = Í↵P f + P g, hÎ and we conclude that P (↵f + g) = ↵P f + P g. Continuity of P : We have for all h À H
ÒP hÒ2 = ÍP h, P hÎ = ÍP 2 h, hÎ = ÍP h, hÎ Õ ÒP hÒ ÒhÒ and dividing by ÒP hÒ shows that P is continuous, even a contraction.
Closedness of P (H): Note that f À P (H) if, and only if, f = P h for some h À H. Since P 2 = P we get
f = P h ◊Ÿ f * P h = 0 ◊Ÿ f * P 2 h = 0 ◊Ÿ f * P f = 0 ◊Ÿ f À (id *P )*1 ({0})
304
Solution Manual. Chapter 1–13. Last update 23rd August 2017 and since P is continuous and {0} is a closed set, (id *P )*1 ({0}) is closed and the above line shows P (H) = (id *P )*1 ({0}) is closed.
Projection: In view of Corollary 26.14 we have to show that P h * h is for any h À H orthogonal to f À P (H). But
ÍP h * h, f Î = ÍP h, f Î * Íh, f Î = Íh, P f Î * Íh, f Î = Íh, f Î * Íh, f Î = 0. (iii) Since, by assumption, ÒP hÒ Õ ÒhÒ, P is continuous and closedness follows just as in (ii). It is, therefore, enough to show that P is an orthogonal projection. We will show that N := {h À H : P h = 0} satisfies N Ú = P (H).
For this we observe that if h À H, P (P h * h) = P 2 h * P h = P h * P h = 0 so that P h * h À N. In particular
h À N Ú -Ÿ y = P h * h À N
-Ÿ P h = h + y with hÚy.
(*)
Thus, ÒhÒ2 + ÒyÒ2 = ÒP hÒ2 Õ ÒhÒ2 -Ÿ ÒyÒ2 -Ÿ y = 0. We conclude that h À N Ú -Ÿ P h * h = 0 -Ÿ P h = h -Ÿ h À P (H) and we have shown that N Ú œ P (H).
To see the converse direction we pick h À P (H) and find P h = h. Since H = N ‚ N Ú we have h = x + xÚ with x À N and xÚ À N Ú . Thus,
((2v)2 }
u2n d Õ
A n
u2n d
Õ
A n
u2 d
n
0
0
,,,,,,,,,ô , 0 nôÿ
by dominated convergence since An0 ô Á and u2 À L1 ( ).
Using the convergence theorem for UI (sub)martingales, Theorem 24.6, we conclude that u2j converges pointwise and in L1 -sense to some u2ÿ À L1 (Aÿ ) and that (u2j )jÀN‰{ÿ}
is again a submartingale. By Riesz’s convergence theorem 13.10 we conclude that uj ô uÿ in L2 -norm.
Remark: We can also identify uÿ with v: since E Aj v = uj = E Aj uÿ it follows that for k = 1, 2, … , j and all j
0 = ÍE Aj v * E Aj uÿ , 1Ak Î = Ív * uÿ , E Aj 1Ak Î = Ív * uÿ , 1Ak Î 309
R.L. Schilling: Measures, Integrals & Martingales i.e. v = uÿ on all sets of the „-stable generator of Aÿ which can easily be extended to contain an exhausting sequence A1 ⁄ 5 ⁄ An of sets of finite -measure.
(vi) The above considerations show that the functions T U n … D := ↵0 1An + ↵j 1Aj : n À N, ↵j À R 0
j=1
(if (An0 ) = ÿ, then ↵0 = 0) are dense in L2 (Aÿ ). It is easy to see that T
E :=
q0 1An + 0
n … j=1
U
qj 1Aj : n À N, ↵j À Q
(if (An0 ) = ÿ, then q0 = 0) is countable and dense in D so that the claim follows. ∑∑
310
27 Conditional expectations. Solutions to Problems 27.1–27.19 Problem 27.1 Solution: In Theorem 27.4(vii) we have seen that EH EG u = EH u. Since, by 27.4(i) and 27.1 EH u À L2 (H) œ L2 (G) we have, because of 27.4 EG EH u = EH u. ∑∑
Problem 27.2 Solution: Note that by the Markov inequality {u > 1} Õ î u2 d < ÿ, i.e. u1{u>1} is an integrable function (use Cauchy-Schwarz). We have 1 {u > 1} =
{u>1}
1d
(*)
<
{u>1}
assumption
Õ
ud
{u > 1}.
In the step marked (*) we really (!) need that {u > 1} > 0 — otherwise we could not get a strict inequality. Thus, {u > 1} < {u > 1} which is a contradiciton. Therefore, {u > 1} = 0 and we have u Õ 1 a.e.
If you are unhappy with strict inequalities, you can extend the argument as follows: By assumption ∑ {u > 1} > 0. Since {u > 1} = nŒ1 {u Œ 1+1_n}, there is some N such that {u Œ 1+1_n} > 0 for all n Œ N — use a continuity of measure argument. Now we get for all n Œ N
1d =
1 uŒ1+ n
uŒ1+
< 1+
Observe that {u>1}
1d =
ÿ … n=N+1 ÿ
Õ
…
n=N+1
=
Õ
1 n
uŒ1+
1 uŒ1+ n uŒ1+
1 n
1 n
1+1_nÕu 1}. Alternative: From î{u>1} u d Õ (u > 1) we get {u>1}
(u * 1) d Õ 0.
Observe that (u * 1)1{u>1} Π0 implies {u>1}
(u * 1) d Π0.
Therefore, î{u>1} (u * 1) d = 0 and we see that (u * 1)1{u>1} = 0 a.e., hence 1{u>1} = 0 a.e. ∑∑ Problem 27.3 Solution: Note that, since EG is (currently...) only defined for L2 -functions the problem implicitly requires that f À L2 (A, ). (A look at the next section reveals that this is not really necessary...). Below we will write Í÷, ÷ÎL2 ( is meant.
)
resp. Í÷, ÷ÎL2 (⌫) to indicate which scalar product
We begin with a general consideration: Let u, w be functions such that u2 , v2 À L2 ( ). Then we have u w Õ 12 (u2 + w2 ) À L2 ( ) and, using again the elementary inequality xy Õ
x2 y2 + 2 2
t t for x = u_ EG (u2 ) and y = w_ EG (w2 ) we conclude that on Gn := {EG (u2 ) > {EG (w2 ) > 1n }
u w 1Gn Õ t t EG (w2 ) EG (w2 )
L
M u2 w2 + 1Gn . 2 EG (u2 ) 2 EG (w2 )
Taking conditional expectations on both sides yields, since Gn À G: EG u w 1Gn Õ 1Gn . t t G 2 G 2 E (w ) E (w ) Multiplying through with the denominator of the lhS and letting n ô ÿ gives Û G Û ÛE (uw)Û1G< Õ EG uw 1G< Õ Û Û
t
EG (u2 )
on the set G< := Gu „ Gw := {EG u2 > 0} „ {EG w2 > 0}. 312
t
EG (w2 )
1 } n
„
Solution Manual. Chapter 1–13. Last update 23rd August 2017 (i) Set G< := {EG f > 0} and Gn := {EG f > 1n }. Clearly, using the Markov inequality, (Gn ) Õ n2
(EG f )2 d Õ n
f2 d < ÿ
so that by monotone convergence we find for all G À G „ G< ⌫(G) = Íf , 1G ÎL2 (
)
= supÍf , 1G„Gn ÎL2 ( n
)
= supÍf , EG 1G„Gn ÎL2 (
)
= supÍEG f , 1G„Gn ÎL2 (
)
n n
= ÍEG f , 1G ÎL2 ( which means that ⌫G„G< = EG f
)
G„G< .
(ii) We define for bounded u À L2 (⌫) P u :=
EG (f u) EG f
1G < .
Let us show that P À L2 (⌫). Set G˘f u := {EG f u2 > 0}. Then, for bounded u À L2 (⌫)
Ù E (f u) Ù2 Ù Ù 1 < ˘ ˘ G „G „G Ù EG f fu fÙ 2 Ù ÙL (⌫) ⌅ G ⇧2 E (f u) d⌫ = G< „G˘ „G˘ ⌅EG f ⇧2 fu f ⌅ G ⇧2 E (f u) = fd G< „G˘ „G˘ ⌅EG f ⇧2 fu f ⌅ G ⇧2 E (f u) = EG f d G< „G˘ „G˘ ⌅EG f ⇧2 fu f ⌅ G ⇧2 E (f u) = d G< „G˘ „G˘ EG f fu f ⌅ G ⌅˘ ˘ ⇧⇧2 E f ( f u) = d G< „G˘ „G˘ EG f fu f ⌅ ⇧ E G f E G f u2 Õ d G< „G˘ „G˘ EG f fu f G
=
G< „G˘
fu
= sup n
„G˘f
⌅ ⇧ EG f u2 d
1Gn „G˘
fu
„G˘f E
G
⌅
⇧ f u2 d
313
R.L. Schilling: Measures, Integrals & Martingales = sup
EG 1Gn „G˘
= sup
1Gn „G˘
n n
= Õ
fu
fu
1G< „G˘
fu
„G˘f
„G˘f
„G˘f
f u2 d
f u2 d
f u2 d
Ù˘ Ù2 f u2 d = Ù f uÙ 2 Ù ÙL (
)
= ÒuÒ2L2 (⌫) < ÿ.
Still for bounded u À L2 (⌫), Gn „{f 0} ∑∑ Problem 27.4 Solution: Since G = {G1 , … , Gn } such that the Gj ’s form a mutually disjoint partition of the whole space X, we have
2
L (G) =
T n … j=1
U ↵j 1Gj : ↵j À R
.
It is, therefore, enough to determine the values of the ↵j . Using the symmetry and idempotency of the conditional expectation we get for k À {1, 2, … , n}
ÍEG u, 1Gk Î = Íu, EG 1Gk Î = Íu, 1Gk Î =
Gk
ud .
On the other hand, using that EG u À L2 (G) we find ÍEG u, 1Gk Î =
n (…
and we conclude that ↵k =
j=1
n ) … ↵j 1Gj , 1Gk = ↵j Í1Gj , 1Gk Î = ↵k (Gk ) j=1
(dx) 1 ud = u(x) . Gk (Gk ) Gk (Gk ) ∑∑
315
R.L. Schilling: Measures, Integrals & Martingales Problem 27.5 Solution: We follow the hint. Let u À Lp ( ) and define un = [(*n) ‚ u · n]1{uŒ1_n} . Clearly, un is bounded, and by the Markov inequality (11.4) {u Œ 1_n} = {up Œ 1_np } Õ np
up d < ÿ.
Therefore, un À Lr ( ) for all r Œ 1:
un r d =
{un Œ1_n}
(u · n)r Õ nr {un Œ 1_n} Õ nr+p
up d < ÿ.
Since un ô u a.e., dominated convergence (use the majorant up ) shows that un ô u in Lp . Thus, we see as in the remark before Theorem 27.5 that (T un )nÀN is a Cauchy sequence in Lp ( ), i.e. the
limit Lp -limn T un exists. If (wn )n is a further approximating sequence such that wn ô u in Lp ( ), we get
ÒT un * T wn Òp = ÒT (un * wn )Òp Õ cÒun * wn Òp Õ cÒun * uÒp + cÒu * wn Òp ,,,,,,,,,ô , 0 nôÿ
which shows that limn T un = limn T wn , i.e. Tõ u := limn T un (as an Lp -limit) is well-defined since it is independent of the approximating sequence. Linearity is clear from the linearity of the limit.
Assume now that 0 Õ un ~ u where un À Lp ( ) „ L2 ( ). By the first part, Tõ u = limn T un in Lp , so there is a subsequence such that Tõ u = limk T un a.e. Because of monotonicity we have k
T unk Õ T un
≈ n Œ n(k) -Ÿ 0 Õ Tõ u * T un Õ Tõ u * T unk .
So, 0 Õ lim sup(Tõ u * T un ) Õ Tõ u * T unk ,,,,,,,,,ô , 0, nôÿ
kôÿ
which shows that limn (Tõ u * T un ) = 0. ∑∑ Problem 27.6 Solution: Let Gu := {EG up > 0}, Gw := {EG wq > 0} and G := Gu „ Gw . Following the hint we get u
w
up uq 1 Õ 1 + ⌅ ⇧1_p ⌅ ⇧1_q G p EG (up ) G q EG (wq ) 1G EG (up ) EG (wq ) Since 1G is bounded and G-measurable, we can apply EG on both sides of the above inequality and get
EG (uw) EG (up ) EG (uq ) 1 Õ 1 + ⌅ ⇧1_p ⌅ ⇧1_q G p EG (up ) G q EG (wq ) 1G = 1G EG (up ) EG (wq ) or ⌅ ⇧1_p ⌅ G ⇧1_q EG (uw)1G Õ EG (up ) E (wq ) 1G ⌅ G ⇧ ⌅ ⇧ 1_p 1_q Õ E (up ) EG (wq ) .
316
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Denote by Gn an exhaustion of X such that Gn À G, Gn ~ X and (Gn ) < ÿ. Then G c
up d = sup n
u
Gc „Gn
up d
u
= supÍ1Gc „Gn , up Î u
n
= supÍEG 1Gc „Gn , up Î u
n
= supÍ1Gc „Gn , EG (up )Î u
n
=0 which means that 1Gu u = u almost everywhere. Thus, EG (uw)1G = EG (uw1G ) = EG (u1Gu w1Gw ) = EG (uw) and the inequality follows since
Û G Û ÛE (uw)Û Õ EG (uw). Û Û ∑∑
Problem 27.7 Solution: In this problem it is helpful to keep the distinction between EG defined on L2 (A) and the extension E G defined on LG (A).
Since A is -finite we can find an exhausting sequence of sets An ~ X with (An ) < ÿ. Setting for u, w À LG (A) with uE G w À L1 (A) un := (*n) ‚ u · n 1An and wn := (*n) ‚ w · n 1An
we have found approximating sequences such that un , wn À L1 (A) „ Lÿ (A) and, in particular, À L2 (A).
(iii): For u, w Œ 0 we find by monotone convergence, using the properties listed in Theorem 27.4: ÍE G u, wÎ = limÍEG un , wÎ n
= lim limÍEG un , wm Î n
m
= lim limÍun , EG wm Î n
m
= limÍun , E G wÎ n
= Íu, E G wÎ. In the general case we write ÍE G u, wÎ = ÍE G u+ , w+ Î * ÍE G u* , w+ Î * ÍE G u+ , w* Î + ÍE G u* , w* Î and consider each term separately. The equality ÍE G u, wÎ = ÍE G u, E G wÎ follows similarly. (iv): we have u = w -Ÿ uj = wj
≈ j -Ÿ EG uj = EG wj
≈j 317
R.L. Schilling: Measures, Integrals & Martingales and we get E G u = lim EG uj = lim EG wj = w. j
j
(ix): we have 0 Õ u Õ 1 -Ÿ 0 Õ un Õ 1
≈n
-Ÿ 0 Õ EG un Õ 1
≈n
-Ÿ 0 Õ E G u = lim EG un Õ 1. n
(x): u Õ w -Ÿ 0 Õ w * u -Ÿ 0 Õ E G (u * w) = E G u * E G w. (xi):
Û Û ±u Õ u -Ÿ ±E G u Õ E G u -Ÿ ÛE G uÛ Õ E G u. Û Û ∑∑
Problem 27.8 Solution: (Mind the typo in the hint: EG = EG should read EG = E G .) Assume first
that G is -finite and denote by Gk À G, Gk ~ X and (Gk ) < ÿ an exhausting sequence. Then 1Gk À L2 (G), 1Gk ~ 1 and
E G 1 = sup EG 1Gk = sup 1Gk = 1. k
k
Conversely, let E G 1 = 1. Because of Lemma 27.7 there is a sequence (uk )k œ L2 (A) with uk ~ 1. By the very definition of E G we have
E G 1 = sup EG uk = 1, k
i.e. there is a sequence gk := EG uk À L2 (G) such that gk ~ 1. Set Gk := {gk > 1 * 1_k} and observe that Gk ~ X as well as
(Gk ) Õ = Õ
1 (1 * 1 (1 * 1 (1 *
1 2 ) k 1 2 ) k 1 2 ) k
gk2 d
ÒEG uk Ò2L2 Òuk Ò2L2
< ÿ. This shows that G is -finite. If G is not -finite, e.g. if G = {Á, G, Gc , X} where (G) < ÿ and (Gc ) = ÿ we find that L2 (G) = {c1G : c À R} 318
Solution Manual. Chapter 1–13. Last update 23rd August 2017 which means that E G 1 = 1G since for every A œ Gc , A À A and (A) < ÿ we find E G 1A⁄G = E G (1A + 1G ) = E G 1A + E G 1G = E G 1A + 1G Since this must be an element of L2 (G), we have necessarily E G 1A = c1G or Íc1G , 1G Î = ÍE G 1A , 1G Î = Í1A , E G 1G Î = Í1A , 1G Î = (A „ G) = 0, hence c = 0 or E G 1A = 0. This shows that E G 1 = 1G Õ 1 is best possible. ∑∑ Problem 27.9 Solution: For this problem it is helpful to distinguish between EG (defined on L2 ) and the extension E G .
Without loss of generality we may assume that g Œ 0—otherwise we would consider positive and negative parts separately. Since g À Lp (G) we have that {g > 1_j} Õ j p
gp d < ÿ
which means that the sequence gj := (j · g)1{g>1_j} À L2 (G). Obviously, gj ~ g pointwise as well as in Lp -sense. Using the results from Theorem 27.4 we get
EG gj = gj -Ÿ E G = sup EG gj = sup gj = g. j
j
∑∑ Problem 27.10 Solution: For this problem it is helpful to distinguish between EG (defined on L2 ) and the extension E G .
For u À Lp (A) we get E H E G u = E H u because of Theorem 27.11(vi) while the other equality E G E H u = E H u follows from Problem 27.9.
If u À M + (A) (mind the misprint in the problem!) we get a sequence uj ~ u of functions uj À
L2+ (A). From Theorem 27.4 we know that EG uj À L2 (G) increases and, by definition, it increases towards E G u. Thus,
E H E G uj = E H uj ~ E H u while EH EG uj ~ E H sup EG uj = E H E G u. The other equality is similar.
j
∑∑
319
R.L. Schilling: Measures, Integrals & Martingales Problem 27.11 Solution: We know that Lp (An ) =
T n … j=1
U cj 1[j*1,j) : cj À R
since c0 1[n,ÿ) À Lp if, and only if, c0 = 0. Thus, E An u is of the form E u(x) = An
n … j=1
cj 1[j*1,j) (x)
and integrating over [k * 1, k) yields [k*1,k)
E An u(x) dx = ck .
Since [k*1,k)
E An u(x) dx = ÍE An u, 1[k*1,k) Î = Íu, E An 1[k*1,k) Î = Íu, 1[k*1,k) Î =
we get E u(x) = An
n … j=1
[j*1,j)
[k*1,k)
u(x) dx
u(t) dt1[j*1,j) (x). ∑∑
Problem 27.12 Solution: For this problem it is helpful to distinguish between EG (defined on L2 ) and the extension E G .
If (X) = ÿ and if G = {Á, X}, then L1 (G) = {0} which means that E G u = 0 for any u À L1 (A). Thus for integrable functions u > 0 and G not -finite we can only have ‘Õ’.
If G is -finite and if Gj ~ X, Gj À G, (Gj ) < ÿ is an exhausting sequence, we find for any u À L1+ (A)
E G u d = sup j
Gj
EG u d
= supÍE G u, 1Gj Î j
= supÍu, E G 1Gj Î j
= supÍu, 1Gj Î j
= Íu, 1Î =
320
ud .
Solution Manual. Chapter 1–13. Last update 23rd August 2017 If G is not -finite and if u Œ 0, we perform a similar calculation with an exhausting sequence
Aj À A, Aj ~ X, (Aj ) < ÿ (it is implicit that A is -finite as otherwise the conditional expectation would not be defined!):
E G u d = sup j
Aj
EG u d
= supÍE G u, 1Aj Î j
= supÍu, E G 1Aj Î j
Õ Íu, 1Î =
ud . ∑∑
Problem 27.13 Solution: Proof of Corollary 27.14: Since lim inf uj = sup inf uj jôÿ
we get
k jŒk
EG inf uj Õ EG um
≈m Œ k
jŒk
thus
EG inf uj Õ inf EG um Õ sup inf EG um = lim inf EG um . jŒk
mŒk
môÿ
k mŒk
Since on the other hand the sequence inf jŒk uj increases, as k ô ÿ, towards supk inf jŒk uj we can use the conditional Beppo Levi theorem 27.13 on the left-hand side and find
EG lim inf uj = EG sup inf uj = sup EG inf uj Õ lim inf EG um . jôÿ
k jŒk
The Corollary is proved.
jŒk
k
môÿ
Proof of Corollary 27.15: Since uj Õ w we conclude that u = limj uj Õ w and that 2w * u * uj Œ 0. Applying the conditional Fatou lemma 27.14 we find EG (2w) = EG lim inf 2w * u * uj j
Õ lim inf EG 2w * u * uj j
= EG (2w) * lim sup EG (u * uj ) j
which shows that
lim sup EG (u * uj ) = 0 -Ÿ lim EG (u * uj ) = 0. Since, however, the claim follows.
j
j
Û G Û Û Û , 0 ÛE uj * EG uÛ = ÛEG (uj * u)Û Õ EG uj * u ,,,,,,,,,ô Û Û Û Û jôÿ ∑∑
321
R.L. Schilling: Measures, Integrals & Martingales Problem 27.14 Solution:
(i) -Ÿ (ii): Let A À Aÿ be such that (A) < ÿ. Then, by Hölder’s
inequality with 1_p + 1_q = 1,
Û Û u d * ud Û j A Û A
Û 1_q ÛÕ ,,,,,,ô , 0. Û uj * u d Õ Òuj * uÒp (A) ,,,jôÿ Û A
Thus, if uÿ := EAÿ u, we find by the martingale property for all k > j and A À Aj such that (A) < ÿ
A
uj d =
A
kôÿ A
uk d = lim
uk d =
A
ud =
A
uÿ d ,
and since we are in a -finite setting, we can apply Theorem 27.12(i) and find that uj = EAj uÿ . (ii) -Ÿ (iii): Assume first that uÿ À L1 „ Lp . Then uj = EAj uÿ À L1 „ Lp and Theorem 27.19(i) shows that uj ,,,,,,,,,ô , uÿ both in L1 and a.e. In particular, we get jôÿ
Íuÿ * uj , Î Õ Òuÿ * uj Ò1 Ò Òÿ ô 0
≈
À Lÿ .
In the general case where uÿ À Lp (Aÿ ) we find for every ✏ > 0 an element u✏ÿ À L1 (Aÿ ) „ Lp (Aÿ ) such that
Òuÿ * u✏ÿ Òp Õ ✏ (indeed, since we are working in a -finite filtered measure space, there is an exhaustion Ak ~ X
such that Ak À Aÿ and for large enough k = k✏ the function u✏ÿ := uÿ 1Ak will to the job). Similarly, we can approximate any fixed
À Lq by
✏
À Lq „ L1 such that Ò *
✏Ò
q
Õ ✏.
Now we set u✏j := EAj u✏ÿ and observe that Òuj * u✏j Òp = ÒEAj uÿ * EAj u✏ÿ Òp Õ Òuÿ * u✏ÿ Òp Õ ✏. Thus, for any
À Lq ,
Íuj * uÿ , Î = Íuj * u✏j * uÿ + u✏ÿ , Î + Íu✏j * u✏ÿ , Î = Íuj * u✏j * uÿ + u✏ÿ , Î + Íu✏j * u✏ÿ ,
*
✏
Òq + Íu✏j * u✏ÿ ,
✏
Î + Íu✏j * u✏ÿ ,
✏
Î
Õ Òuj * u✏j Òp + Òuÿ * u✏ÿ Òp Ò Òq + Òu✏j * u✏ÿ Òp Ò *
✏
Î Õ const.✏
Õ 2✏ Ò Òq + ✏ Òu✏j * u✏ÿ Òp + Íu✏j * u✏ÿ , ✏ Î ´≠≠≠Ø≠≠≠¨ ´≠≠≠≠≠Ø≠≠≠≠≠¨ Õ2Òu✏ÿ Òp Õ2(✏+Òuÿ Òp ) ,,,,jôÿ ,,,,,ô , 0 for sufficiently large j’s, and the claim follows. (iii) -Ÿ (ii): Let un(j) be a subsequence converging weakly to some u À Lp , i.e., limÍun(k) * u, Î = 0 k
322
≈
À Lq .
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Then, in particular, limÍun(k) * u, EAn Î = 0
≈
k
or
limÍEAn un(k) * EAn u, Î = 0
À Lq , n À N ≈
k
À Lq , n À N.
Since uj is a martingale, we find that EAn un(k) if n < n(k), i.e., Íun * EAn u, Î = 0
≈
À Lq , n À N.
and we conclude that un = EAn u. Because of the tower property we can always replace u by uÿ := EAÿ u:
un = EAn u = EAn EAÿ u = EAn uÿ
and the claim follows.
(ii) -Ÿ (i): We show that we can take u = uÿ . First, if uÿ À L1 „ Lÿ we find by the closability of martingales, Theorem 27.19(i), that
lim Òuj * uÒ1 = 0. Moreover, using that a *
j
br
Õ (a + b)r Õ 2r (ar + br ), we find
Òuj * uÒpp =
uj * up d
=
uj * u uj * up*1 d
p*1 Õ 2p*1 (Òuj Òp*1 ÿ + ÒuÒÿ )
uj * u d
Õ 2p ÒuÒp*1 Òuj * uÒ1 ÿ ,,,,,,,,,ô , 0 where we use that
jôÿ
A Òuj Òÿ = ÒEA j uÒÿ Õ Ej ÒuÒÿ Õ ÒuÒÿ .
Now for the general case where uÿ À Lp . Since we are in a -finite setting, we can set u✏ :=
(u 1Aj ) · j, j = j(✏) sufficiently large and Aj ô X an exhausting sequence of sets from Aÿ , and can guarantee that
Òu * u✏ Òp Õ ✏.
At the same time, we get for u✏j := EAj u✏ À L1 „ Lÿ that Òuj * u✏j Òp = ÒEAj u * EAj u✏ Òp Õ Òu * u✏ Òp Õ ✏. Thus, by the consideration for the special case where u✏ À L1 „ Lÿ , Òuj * uÒp Õ Òuj * u✏j Òp + Òu✏j * u✏ Òp + Òu✏ * uÒp Õ ✏ + Òu✏j * u✏ Òp + ✏ ,,,,,,,,,ô , 2✏ ,,,,,,,ô , 0. jôÿ
✏ô0
∑∑
323
R.L. Schilling: Measures, Integrals & Martingales Problem 27.15 Solution: Obviously, mk = mk*1 + (uk * E Ak*1 uk ). Since m1 = u1 À L1 A1 , this shows, by induction, that mk À L1 (Ak ). Applying E Ak*1 to both sides of the displayed equality yields
E Ak*1 mk = E Ak*1 mk*1 + E Ak*1 (uk * E Ak*1 uk ) = mk*1 + E Ak*1 uk * E Ak*1 uk = mk*1 which shows that mk is indeed a martingale. ∑∑ Problem 27.16 Solution: Problem 27.15 shows that sk is a martingale, so that s2k is a sub-martingale (use Jensen’s inequality for conditional expectations). Now … … s2k d = u2k d + 2 uu d j k j and if j < k
j 0) Õ P (Mk > 0) = P (Xj > 0 P (M
≈j = 1, 2, … , k) = 2*k ,,,,,,,,,ô , 0 kôÿ
which yields a contradiction. ∑∑ Problem 27.19 Solution: (Compare this problem with Problem 22.16.) Recall that in finite measure spaces uniform integrability follows from (and is actually equivalent to) lim sup
Rôÿ n
{un >R}
un d = 0;
this is true since in a finite measure space the constant function w í R is integrable. Observe now that {un >R}
un d Õ = = Õ Õ
This shows that
{un >R} {un >R}
E An f d fd
{un >R}„{f ÕR_2} {un >R}„{f ÕR_2}
fd +
{un >R}„{f >R_2}
fd
1 u d + fd {un >R}„{f >R_2} 2 n
1 u d + fd {un >R} 2 n {f >R_2}
Rôÿ 1 un d Õ f d ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,ô , 0. {f >R_2} uniformly for all n 2 {un >R}
∑∑
325
28 Orthonormal systems and their convergence behaviour. Solutions to Problems 28.1–28.11 Problem 28.1 Solution: Since Jk(↵, ) is a polynomial of degree k, it is enough to show that Jk(↵, ) is orthogonal in L2 (I, ⇢(x) dx) to any polynomial p(x) of degree j < k. We write ) k for u(x) = (x * 1)k+↵ (x + 1)k+ . Then we get by repeatedly integrating by parts 1
*1
dk dxk
and
Jk(↵, ) (x)p(x)(x * 1)↵ (x + 1) dx
(*1)k 1 = p(x)) k u(x) dx k k! 2 *1 ⌧ = p(x) ) k*1 u(x) * ) 1 p(x) ) k*2 u(x) + 5 + (*1)k*1 ) k*1 p(x) u(x) + (*1)k
1
*1
1 *1
u(x)) k p(x) dx.
Obviously, ) l u(*1) = ) l u(1) = 0 for all 0 Õ l Õ k * 1 and ) k p í 0 since p is a polynomial of degree j < k.
∑∑ Problem 28.2 Solution: It is pretty obvious how to go about this problem. The calculations themselves are quite tedious and therefore omitted.
∑∑ Problem 28.3 Solution: Theorem 28.6: The polynomials are dense in C[a, b] with respect to uniform convergence. Proof 1: mimic the proof of 28.6 with the obvious changes; Proof 2: Let f À C[a, b]. Then fõ(y) := f (a + (b * a)y), y À [0, 1] satisfies fõ À C[0, 1] and, because of Theorem 28.6, there is a sequence of polynomials põn such that lim sup fõ(y) * põn (y) = 0.
nôÿ yÀ[0,1]
Define pn (x) := põn
x*a b*a
, x À [a, b]. Clearly pn is a polynomial and we have sup pn (x) * f (x) = sup õ pn (y) * fõ(y).
xÀ[a,b]
yÀ[0,1]
327
R.L. Schilling: Measures, Integrals & Martingales Corollary 28.8: The monomials are complete in L1 ([a, b], dt). Proof 1: mimic the proof of 28.8 with the obvious changes; Proof 2: assume that for all j À N0 we have a
Since 0
1
b
u(x)xj dx = 0.
u((b * a)t + a)t dx = j
=
a
b
j … k=0
4
x*a u(x) b*a
ck
a
b
5j dx
u(x)xk dx
=0 we get from Corollary 28.8 that u((b * a)t + a) = 0
Lebesgue almost everywhere on [0, 1]
and since the map [0, 1] Œ t ;ô x = (b * a)t + a À [a, b] is continuous, bijective and with a continuous inverse, we also get
u(x) = 0
Lebesgue almost everywhere on [a, b]. ∑∑
Problem 28.4 Solution: Observe that ⇠ ⇡ ⌧ ⇠ ⇡ Re ei(x*y) * ei(x+y) = Re eix e*iy * eiy ⌧ = Re * 2ieix sin y = 2 sin x sin y, and that
⇠ ⇡ ⌧ ⇠ ⇡ Re ei(x+y) + ei(x*y) = Re eix eiy + e*iy ⌧ = Re 2eix cos y = 2 cos x cos y.
Moreover, we see that for N À N0 h eiNx Û⇡ Û = 0, if N ë 0; n eiNx dx = l iN Û*⇡ *⇡ n2⇡, if N = 0. j ⇡
Thus, if k ë l ⇡
*⇡ 328
0 2 cos kx cos lx dx = Re
⇡
*⇡
e
i(k+l)x
dx +
⇡
*⇡
e
i(k+l)x
1 dx = 0
Solution Manual. Chapter 1–13. Last update 23rd August 2017 and if k = l Œ 1 ⇡
*⇡
0 2 cos kx cos kx dx = Re
and if k = l = 0,
⇡
*⇡
⇡
*⇡
2ikx
dx +
e
2 cos kx cos kx dx =
⇡
*⇡
1 1 dx = 2⇡
2 dx = 4⇡.
The proof for the pure sines integral is similar while for the mixed sine-cosine integrals the integrand
x ;ô cos kx sin lx is always an odd function, the integral over the symmetric (w.r.t. the origin) interval (*⇡, ⇡) is always zero.
∑∑ Problem 28.5 Solution: (i) We have
⇠ ix ⇡ e + e*ix k 2k cosk (x) = 2k 2 ⇠ ix ⇡ e + e*ix k = 2 k 0 1 … k = eijx e*i(k*j)x j j=0 k 0 1 … k i(2j*k)x = e j j=0
Adding the first and last terms, second and penultimate terms, term no. j and k * j, etc. under the sum gives, since the binomial coefficients satisfy – if k = 2n is even
k j
=
k k*j
,
0 1 n*1 0 1 … 2n 2n i(2j*2n)x i(2n*2j)x 2 cos (x) = (e +e )+ j n j=0 0 1 n 0 1 … 2n 2n = 2 cos(2j * 2n) + j n j=0 2n
2n
– if k = 2n * 1 is odd 2
2n*1
cos
2n*1
1 n*1 0 … 2n * 1 (x) = (ei(2j*2n+1)x + ei(2n*2j*1)x ) j j=0 1 n*1 0 … 2n * 1 = 2 cos(2n * 2j * 1)x. j j=0
In a similar way we compute sink x: 2k sink (x) = 2k
⇠
eix * e*ix 2i
⇡k
329
R.L. Schilling: Measures, Integrals & Martingales ⇡ eix * e*ix k 2 k 0 1 … k = i*k (*1)k*j eijx e*i(k*j)x j j=0 k 0 1 … k *k =i (*1)k*j ei(2j*k)x . j j=0 = i*k
⇠
Adding the first and last terms, second and penultimate terms, term no. j and k * j, etc. under the sum gives, since the binomial coefficients satisfy – if k = 2n is even
k j
=
k k*j
,
22n sin2n (x) 0 1 n*1 0 1 … 2n 2n 2n*j i(2j*2n)x j i(2n*2j) = (*1) (*1) e + (*1) e + j n j=0 n
0 1 n*1 0 1 … 2n 2n n*j i(2j*2n)x i(2n*2j) = (*1) e +e + j n j=0 =
0 1 n*1 0 1 … 2n 2n (*1)n*j 2 cos(2n * 2j)x + j n j=0
– if k = 2n * 1 is odd 22n*1 sin2n*1 (x) 1 n*1 0 … 2n * 1 n = i(*1) (*1)2n*1*j ei(2j*2n+1)x + (*1)*j ei(2n*2j*1) j j=0 =i =i =
1 n*1 0 … 2n * 1 (*1)n*j * ei(2j*2n+1)x + ei(2n*2j*1) j j=0 1 n*1 0 … 2n * 1 (*1)n*j 2i sin(2n * 2j + 1)x j j=0
1 n*1 0 … 2n * 1 (*1)n*j*1 2 sin(2n * 2j + 1)x. j j=0
(ii) We have cos kx + i sin kx = eikx = eix
k
= cos x + i sin x
and we find, using the binomial formula, k 0 1 … k cos kx + i sin kx = cosj x ik*j sink*j x j j=0
and the claim follows by separating real and imaginary parts. (iii) Since a trigonometric polynomial is of the form Tn (x) = a0 +
330
n … k=1
ak cos kx + bk sin kx
k
Solution Manual. Chapter 1–13. Last update 23rd August 2017 it is a matter of double summation and part (ii) to see that Tn (x) can be written like Un (x). Conversely, part (i) enables us to rewrite any expression of the form Un (x) as Tn (x).
∑∑ Problem 28.6 Solution: By definition, DN (x) =
N
1 … + cos jx. 2 j=1
Multiplying both sides by sin x2 and using the formula cos ax sin bx =
⇠ (a + b)x (a * b)x ⇡ 1 sin * sin 2 2 2
where j = (a + b)_2 and 1_2 = (a * b)_2, i.e. a = (2j + 1)_2 and b = (2j * 1)_2 we arrive at DN (x) sin
x 2
=
1 2
sin
x 2
+
1 2
N ⇠ … j=1
⇡ (2j*1)x sin (2j+1)x * sin = sin (2N+1)x . 2 2 2 ∑∑
Problem 28.7 Solution: We have sin x =
2 4 * ⇡ ⇡
0
1 cos 2x cos 4x cos 6x + + +5 . 1 3 3 5 5 7
Indeed, let us calculate the Fourier coefficients 28.8. First, bk =
⇡
1 sin x sin kx dx = 0, ⇡ *⇡
k À N,
since the integrand is an odd function. So no sines appear in the Fourier series expansion. Further, using the symmetry properties of the sine function
⇡
1 sin x dx 2⇡ *⇡ ⇡ 1 = sin x dx ⇡ 0 1 Û⇡ = (* cos x)Û Û0 ⇡ 2 = ⇡
a0 _2 =
and using the elementary formula 2 sin a cos b = sin(a * b) + sin(a + b) we get ⇡
1 sin x cos jx dx ⇡ *⇡ ⇡ 2 = sin x cos jx dx ⇡ 0 ⇡ 2 1 = sin((j + 1)x) * sin((j * 1)x) dx ⇡ 0 2 4 5⇡ 1 cos((j * 1)x) cos((j + 1)x) = * ⇡ j*1 j+1 0
aj =
331
R.L. Schilling: Measures, Integrals & Martingales 4 5 1 cos((j * 1)⇡) cos((j + 1)⇡) 1 1 = * * + . ⇡ j*1 j+1 j*1 j+1
If j is odd, we get aj = 0 and if j is even, we have 4 5 1 *1 *1 1 1 4 1 aj = * * + =* . ⇡ j*1 j+1 j*1 j+1 ⇡ (j * 1)(j + 1) This shows that we have only evenly indexed cosines in the Fourier series.
∑∑ Problem 28.8 Solution: This is not as trivial as it looks in the first place! Since u is itself a Haar function, we have
sN (u, x) = u(x)
≈N À N
(it is actually the first Haar function) so that sN converges in any Lp -norm, 1 Õ p < ÿ to u.
The same applies to the right tail of the Haar wavelet expansion. The left tail, however, converges
only for 1 < p < ÿ in Lp . The reason is the calculation of Step 5 in the proof of Theorem 28.20 which goes in the case p = 1: EA*M u = 2*M
[*2M ,0)
u(x) dx1[*2M ,0) + 2*M
[0,2M )
u(x) dx1[0,2M )
= 2*M 1[0,2M ) , but this is not L1 -convergent to 0 as it would be required. For p > 1 all is fine, though.... ∑∑ Problem 28.9 Solution: Assume that u is uniformly continuous (Cc and Cÿ -functions are!). Since H
sn (u; x) = EAn u(x) is the projection onto the sets in AnH , see e.g. Step 2 in the proof of Theorem 28.17, we have sn (u; x) =
1 u(y) dx1I (x) (I) I
where I is an dyadic interval from the generator of AnH as in Step 2 of the proof of Theorem 28.17. Thus, if x is from I we get
Û 1 Û sn (u; x) * u(x) = ÛÛ (u(y) * u(x)) dxÛÛ Û (I) I Û 1 Õ u(y) * u(x) dx (I) I 1 Õ ✏ dx (I) I =✏
if (I) < is some
for small enough
> 0. This follows from uniform continuity: for given ✏ > 0 there
> 0 such that for x, y À I (this entails x * y Õ !) we have u(x) * u(y) Õ ✏.
The above calculation holds uniformly for all x and we are done.
∑∑
332
Solution Manual. Chapter 1–13. Last update 23rd August 2017 Problem 28.10 Solution: The calculation for the right tail is more or less the same as in Problem
28.9. Only the left tail differs. Here we argue as in Step 5 of the proof of Theorem 28.20: if u À Cc (R) we can assume that supp u œ [*R, R] and we see EA*M u(x) = 2*M
[*R,0]
u(x) dx1[*2M ,0) + 2*M
[0,R]
u(x) dx1[0,2M )
Õ 2*M R ÒuÒÿ 1[*2M ,0) + 2*M R ÒuÒÿ 1[0,2M ) = 2*M R ÒuÒÿ 1[*2M ,2M ) Môÿ
Õ 2*M R ÒuÒÿ ,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,ô , 0 uniformly for all x
If u À Cÿ we can use the fact that Cc is dense in Cÿ , i.e. we can find for every ✏ > 0 functions v = v✏ À Cc and w = w✏ À Cÿ such that
u=v+w
and
ÒwÒÿ Õ ✏.
Then Û A*M Û Û Û Û Û u(x)Û Õ ÛEA*M v(x)Û + ÛEA*M w(x)Û ÛE Û Û Û Û Û Û Û A*M Û A*M Õ ÛE v(x)Û + E ÒwÒÿ Û Û Û Û Õ ÛEA*M v(x)Û + ✏ Û Û and, by the first calculation for Cc -functions, the right-hand side converges, since v À Cc , to 0 + ✏ uniformly for all x, and letting ✏ ô 0 we conclude the proof.
∑∑ Problem 28.11 Solution:
See the picture at the end of this solution. Since the function u(x) :=
1[0,1_3) (x) is piecewise constant, and since for each Haar function î
k,j
dx = 0 unless j = k = 1,
we see that only a single Haar function contributes to the value of sN (u; 13 ), namely where
1 3
The idea of the proof is now pretty clear: take values N where x =
N,k ,
supp
n,j .
i.e. where i.e. where
1 N,k ( 3 ) = 1 M,l ( 3 ) =
1 and values M such that x =
1 3
1 3
is in the left ‘half’ of
is in the opposite, negative ‘half’ of
À
M,l ,
*1. Of course, k, l depend on x, N and M respectively. One should expect
that the partial sums for these different positions lead to different limits, hence different upper and lower limits.
The problem is to pick N’s and M’s. We begin with the simple observation that the dyadic (i.e. base-2) representation of 1_3 is the periodic, infinite dyadic fraction
and that the finite fractions
ÿ … 1 1 = 0.01010101 5 = 2k 3 2 k=1 n … 1 dn := 0. 0101 5 01 = ´≠≠≠Ø≠≠≠¨ k=1 22k 2n
333
R.L. Schilling: Measures, Integrals & Martingales approximate 1_3 from the left in such a way that ÿ ÿ … … 1 1 1 1 1 * dn = < = 2n+2 2k l 3 2 2 2 1* k=n+1 l=2n+2
1 2
=
1
22n+1
Now consider those Haar functions whose support consists of intervals of the length 2*2n , i.e. the
supp
2n,j ’s 2n,j
and agree that j = j(1_3, n) is the one value where
1 3
À supp
= [dn , dn + 1_22n ] and we get for the Haar-Fourier partial sum s2n (u, 13 ) *
1_3
1 = 3 dn = 22n = 4n = 4n
dn
…
1 2k 2 k=n+1 ÿ … 1 k 4 k=n+1
= 4n 4*n*1 1 = . 3
1
1*
1 4
The shift by *1_3 comes from the starting ‘atypical’ Haar function 1 . 3
0,0 since Íu,
Using the next smaller Haar functions with support of length 2*2n*1 , i.e. the with j as above dn ) and that
1 2n+1,2j*1 ( 3 )
By construction
1 2n,j ( 3 )
2n dx
1 * 3 ÿ
2n,j .
0,0 Î
= î0
2n+1,k ’s,
1_3
dx =
we see that
= *1 (since twice as many Haar functions appear in the run-up to
1 s2n+1 (u, 13 ) * 3 4 dn +1_22n+2 5 1_3 1 n+1 n+1 = 2 dx * 2 dx 2n+1,2j*1 ( 3 ) dn dn +1_22n+2 4 5 1 1 1 = dn + 2n+2 * dn * + dn + 2n+2 2n+1 (*2n+1 ) 3 2 2 4 5 1 2 = dn * + 2n+2 (*22n+2 ) 3 2 ⇠ ⇡ 1 = 4 22n * dn * 2 3 1 =4 *2 (using the result above) 3 2 =* 3 This shows that
2 1 > * = s2n+1 (u, 13 ) 3 3 and the claim follows since because of the above inequality, s2n (u; 13 ) =
1 2 lim inf sN (u; 13 ) Õ * Õ Õ lim sup sN (u; 13 ). N 3 3 N 334
Solution Manual. Chapter 1–13. Last update 23rd August 2017
∑∑
335
R.L. Schilling: Measures, Integrals & Martingales
6
2
2n+1 2
and
2n,j
2n+1,2j*1
2n
22
dn+1 dn
1 3
-
dn +
1 22n+1
Picture is not to scale! 2 336
dn +
1 22n
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