5.2 Complex Borel Measures on R

Home , Borel measure, Bounded variation, Complex measure, Regular measure, Signed measure, Total variation

MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Contents

1 Measure Theory 3 1.1 σ-algebras ...... 3 1.2 Measures ...... 4 1.3 Construction of non-trivial measures ...... 5 1.4 Lebesgue measure ...... 10 1.5 Measurable functions ...... 14

2 Integration 17 2.1 Integration of simple non-negative functions ...... 17 2.2 Integration of non-negative functions ...... 17 2.3 Integration of real and complex valued functions ...... 19 2.4 Lp-spaces ...... 20 2.5 Relation to Riemann integration ...... 22 2.6 Modes of convergence ...... 23 2.7 Product measures ...... 25 2.8 Polar coordinates ...... 28 2.9 The transformation formula ...... 31

3 Signed and Complex Measures 35 3.1 Signed measures ...... 35 3.2 The Radon-Nikodym theorem ...... 37 3.3 Complex measures ...... 40

4 More on Lp Spaces 43 4.1 Bounded linear maps and dual spaces ...... 43 4.2 The dual of Lp ...... 45 4.3 The Hardy-Littlewood maximal functions ...... 47

5 Diﬀerentiability 51 5.1 Lebesgue points ...... 51 5.2 Complex Borel measures on R ...... 54 5.3 The fundamental theorem of calculus ...... 58

6 Functional Analysis 61 6.1 Locally compact Hausdorﬀ spaces ...... 61 6.2 Weak topologies ...... 62 6.3 Some theorems in functional analysis ...... 65 6.4 Hilbert spaces ...... 67

1 CONTENTS MATH 245A (17F)

7 Fourier Analysis 73 7.1 Trigonometric series ...... 73 7.2 Fourier series ...... 74 7.3 The Dirichlet kernel ...... 75 7.4 Continuous functions and pointwise convergence properties ...... 77 7.5 Convolutions ...... 78 7.6 Convolutions and diﬀerentiation ...... 78 7.7 Translation operators ...... 79 7.8 The Fourier transform ...... 79

8 The Theory of Distributions 83 8.1 Sobolev functions and weak derivatives ...... 83 8.2 Distributions ...... 84 8.3 Localization and support of distributions ...... 85 8.4 Fourier transforms of distributions ...... 85

2 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

1 MEASURE THEORY

1.1 σ-ALGEBRAS

Deﬁnition 1.1.1 (σ-algebra). Let X be a set and A be a family of subsets of X. We say that A is an algebra on X if

(i) ∅ ∈ A; (ii) if A ∈ A, then X\A ∈ A; (iii) if A, B ∈ A, then A ∪ B ∈ A.

We say that A is a σ-algebra on X if closure under ﬁnite unions extends to countable unions. Proposition 1.1.2. Let A be an algebra on X. If A, B ∈ A, then

1. A ∩ B ∈ A; 2. A\B ∈ A.

Moreover, if A is a σ-algebra and An ∈ A, then ∞ \ An ∈ A. n=1 Example 1.1.3. 1. For any set X, A = P(X) is a σ-algebra. 2. For any set X, A = {A ⊂ X | A is countable or X\A is countable} is a σ-algebra. T 3. If {Ai | i ∈ I} is a family of σ-algebras on X, then so is A = i Ai. Proposition 1.1.4. Let F ⊂ P(X) be a family of subsets of X. Then there is a unique σ-algebra A such that F ⊂ A ⊂ A0, where A0 is any σ-algebra on X with F ⊂ A0.

Proof. Let F be the family of σ-algebras A0 containing F. The appropriate A is \ A = A0. A0∈F

Deﬁnition 1.1.5 (σ-algebra generated by a family of sets). The σ-algebra A is the σ-algebra generated by F, denoted by σ(F). Deﬁnition 1.1.6 (Borel σ-algebra). Let (X, O) be a topological space. The Borel σ-algebra on X is BX = σ(O), the σ-algebra generated by the open sets in X. Example 1.1.7. 1. Every open set is Borel, and by complementation, every closed set is Borel.

2. In R, the set Q is Borel, as it is a countable union of points, which are closed. n 3.A rectangle in R is a set of the form [a1, b1] × · · · × [an, bn] with ai ≤ bi for all i. If R is the n family of all rectangles, then σ(R) = BR .

3 1.2 Measures MATH 245A (17F)

1.2 MEASURES

Deﬁnition 1.2.1 (Measure). A pair (X, A) of a set X and a σ-algebra A on X is a measurable space.A (positive) measure µ on a measurable space (X, A) is a function µ : A → [0, ∞] such that

(i) µ(∅) = 0;

(ii) if An ∈ A (n ∈ N) are pairwise disjoint, then

∞ ! ∞ [ X µ An = µ(An). n=1 n=1

The triple (X, A, µ) is a measure space. Example 1.2.2. Let X be a set and A = P(X).

1. For a given a ∈ X, the Dirac measure at a is ( 0 a 6∈ M, δa(M) = 1 a ∈ M.

2. The counting measure is ( |M| M is ﬁnite, µ(M) = ∞ otherwise.

3. If X is a topological space, then a Borel measure is a measure deﬁned on (X, BX ). S Notation. 1. Write An % A if A1 ⊂ A2 ⊂ · · · and A = n An. T 2. Write An & A if A1 ⊃ A2 ⊃ · · · and A = n An. Theorem 1.2.3. Let (X, A, µ) be a measure space. Then

1. (monotonicity) if A, B ∈ A and A ⊂ B, then µ(A) ≤ µ(B);

2. (countable subadditivity) if An ∈ A, then

∞ ! ∞ [ X µ An ≤ µ(An); n=1 n=1

3. (continuity from below) if An ∈ A and An % A,

µ(A) = lim µ(An); n→∞

4. (continuity from above) if An ∈ A, An & A, and µ(A1) < ∞, then

µ(A) = lim µ(An). n→∞

4 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Proof. 1. Since A ⊂ B, we have a decomposition B = A ∪ (B\A) into disjoint subsets. Then µ(B) = µ(A) + µ(B\A) ≥ µ(A).

2. Deﬁne -n−1 [ Bn = An An ⊂ An. k=1 S S By construction, the Bn’s are pairwise disjoint and n Bn = n An. Then ∞ ! ∞ ! ∞ ∞ [ [ X X µ An = µ Bn = µ(Bn) ≤ µ(An). n=1 n=1 n=1 n=1

3. This time, deﬁne Bn = An\An−1 (suppose A0 = ∅). We get

∞ ! ∞ ! ∞ N [ [ X X µ An = µ Bn = µ(Bn) = lim µ(Bn) = lim µ(AN ). N→∞ N→∞ n=1 n=1 n=1 n=1

4. Let Bn = A1\An. Then Bn % A1\A in A and µ(A1) = µ(An)+µ(Bn), so applying continuity from below gives µ(A1) = µ(A1\A) + µ(A) = lim µ(Bn) + µ(A). n→∞

Since µ(A1) is ﬁnite, the result follows from writing

µ(A1) = lim (µ(A1) − µ(An)) + µ(A) = µ(A1) + µ(A) − lim µ(An). n→∞ n→∞

Remark 1.2.4. Continuity from above may fail if we allow µ(A1) = ∞. For example, on R with T the counting measure, let An = [n, ∞). Then n An = ∅, but µ(An) = ∞ for each n. Deﬁnition 1.2.5 (Null set). Let (X, A, µ) be a measure space. A null set is a measurable set N ⊂ X with µ(N) = 0. Deﬁnition 1.2.6 (“Almost everywhere”). Let P (x) be a statement about points x in a measure space (X, A, µ). We say that P (x) is true µ-almost everywhere (or for µ-almost every x ∈ X), abbreviated µ-a.e., if there exists a null set N for which P (x) is true for all x ∈ X\N.

1.3 CONSTRUCTION OF NON-TRIVIAL MEASURES

Let (X, A, µ) be a measure space. Deﬁnition 1.3.1 (Complete measure space). We say that (X, A, µ) is complete if every subset of a null set is a null set. (In particular, every subset of a null set is measurable.) Theorem 1.3.2. Let A = {A ∪ B | A ∈ A and there exists a null set N such that B ⊂ N}. Then A is a σ-algebra on X with A ⊂ A. Moreover, µ can be uniquely extended to a measure µ on A which is complete.

5 1.3 Construction of non-trivial measures MATH 245A (17F)

Proof. It is clear that A ⊂ A. To see that A is a σ-algebra, we check the axioms.

(i) Since A ⊂ A, we have ∅ ∈ A and X ∈ A.

(ii) Let A ∪ B ∈ A with B ⊂ N for some null set N. By replacing A with A\N and B with B ∪ (A ∩ N) if necessary, we can suppose that A ∩ N = ∅. Then

X\(A ∪ B) = (X\A) ∩ (X\B) = [(X\A) ∩ (X\N)] ∪ [N\B] ∈ A.

(iii) Let An ∪ Bn ∈ A with Bn ⊂ Nn for null sets Nn, and let

∞ ∞ ∞ [ [ [ A = An,B = Bn,N = Nn. n=1 n=1 n=1 P Then A, N ∈ A and µ(N) ≤ n µ(Nn) = 0, so N is a null set with B ⊂ N, and

∞ [ (An ∪ Bn) = A ∪ B ∈ A. n=1

For existence of an extension µ, we attempt to define µ(A ∪ B) = µ(A). To show existence, define µ by µ(A ∪ B) = µ(A). To see that it is well-defined, let A ∪ B = A0 ∪ B0 with A, A0 ∈ A and B,B0 subsets of null sets. The union of two null sets is a null set, so we can take B,B0 ⊂ N for some null set N. Then

µ(A ∪ B) = µ(A) = µ(A ∪ N) = µ(A0 ∪ N) = µ(A0) = µ(A0 ∪ B0).

The rest of the proof is omitted. If µ exists and A ∪ B ∈ A with B ⊂ N for some µ-null set N, we must have

µ(A) = µ(A) ≤ µ(A ∪ B) ≤ µ(A ∪ N) ≤ µ(A) + µ(N) = µ(A) + µ(N) = µ(A), hence µ(A ∪ B) = µ(A) is ﬁxed.

Deﬁnition 1.3.3 (Outer measure). An outer measure on a set X is a function µ∗ : P(X) → [0, ∞] such that

(i) µ∗(∅) = 0;

(ii) µ∗(A) ≤ µ∗(B) whenever A ⊂ B;

∗ S P ∗ (iii) µ ( n An) ≤ n µ (An). Lemma 1.3.4. Let A be an algebra on X.

1. Suppose that whenever An ∈ A with An % A, we have A ∈ A. Then A is a σ-algebra. S 2. Suppose that whenever An ∈ A are pairwise disjoint, we have A = n An ∈ A. Then A is a σ-algebra.

6 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Theorem 1.3.5 (Carath´eodory). Let µ∗ be an outer measure on X and A be the family of sets A ⊂ X such that µ∗(T ) = µ∗(T ∩ A) + µ∗(T ∩ Ac)

∗ for all T ⊂ X. Then A is a σ-algebra on X and µ = µ |A is a complete measure on (X, A).

Proof. To show that A is a σ-algebra, it is enough to show that A is an algebra which is closed under countable unions of pairwise disjoint sets. It is clear that ∅ ∈ A. Since the condition for A ∈ A is symmetric in A and Ac, it follows that Ac ∈ A. To get ﬁnite unions, let A, B ∈ A. Since

A ∪ B = (A ∩ Bc) t (A ∩ B) t (B ∩ Ac), we have

µ∗(T ) = µ∗(T ∩ A) + µ∗(T ∩ Ac) = µ∗(T ∩ A ∩ B) + µ∗(T ∩ A ∩ Bc) + µ∗(T ∩ Ac ∩ B) + µ∗(T ∩ Ac ∩ Bc) ≥ µ∗(T ∩ (A ∪ B)) + µ∗(T ∩ Ac ∩ Bc) = µ∗(T ∩ (A ∪ B)) + µ∗(T ∩ (A ∪ B)c)) ≥ µ∗(T ), hence equality holds everywhere and A ∪ B ∈ A. Sn Let An ∈ A be pairwise disjoint and Bn = i=1 Ai. Then for any test set T ,

n ∗ X ∗ µ (T ∩ Bn) = µ (T ∩ Ai), i=1 so n ∗ ∗ ∗ c X ∗ ∗ c µ (T ) = µ (T ∩ Bn) + µ (T ∩ Bn) = µ (T ∩ Ai) + µ (T ∩ Bn). i=1 S S Letting n → ∞ and writing B = n Bn = n An, we have by countable subadditivity

µ∗(T ) ≥ µ∗(T ∩ B) + µ∗(T ∩ Bc) ≥ µ∗(T ), so we have the required equality for B ∈ A. Using the test condition for T = B as above, we conclude

∞ ∗ X ∗ ∗¨¨ µ (B) = µ (An) +¨µ (∅). n=1 This means that µ∗ is countably additive on A, so µ is a measure. To show that µ∗ is complete, we must show that if µ∗(A) = 0, then A ∈ A. For any test set T ,

µ∗(T ) ≤ µ∗(T ∩ A) + µ∗(T ∩ Ac) = µ∗(T ∩ Ac) ≤ µ∗(T ), so equality holds everywhere and we have the required equality for A ∈ A.

7 1.3 Construction of non-trivial measures MATH 245A (17F)

Deﬁnition 1.3.6 (µ∗-measurable). If µ∗ is an outer measure on X, then a subset A ⊂ X is µ∗-measurable if it lies in the family A from the theorem. Deﬁnition 1.3.7 (Premeasure). Let A be an algebra on X.A premeasure ν : A → [0, ∞] is a function such that

(i) ν(∅) = 0, S (ii) if An ∈ A are pairwise disjoint with n An ∈ A, then

∞ ! ∞ [ X ν An = ν(An). n=1 n=1 Lemma 1.3.8. Let A be an algebra on X and ν : A → [0, ∞] be a premeasure. Deﬁne

( ∞ ∞ ) ∗ X [ µ (M) = inf ν(An) | An ∈ A and M ⊂ An n=1 n=1 for M ⊂ X. Then µ∗ is an outer measure extending ν. Moreover, each M ∈ A is µ∗-measurable.

Proof. First we show that µ∗ is an outer measure. That µ∗(∅) = 0 and that µ∗ is monotonic are ∗ clear, so it remains to show countable subadditivity. Let Mn ⊂ X. If some µ (Mn) = ∞, then the ∗ result is clear, so suppose µ (Mn) < ∞. Let > 0. For each n, there exists a countable cover of Mn by sets An,k ∈ A such that

∞ X µ∗(M ) ≤ ν(A ) ≤ µ∗(M ) + . n n,k n 2n n=1 Then ∞ [ [ M = Mn ⊂ An,k n=1 n,k∈N is a countable cover of M by elements of A, so

∞ ∞ ∗ X X ∗ µ (M) ≤ ν(An,k) ≤ µ (Mn) + . n,k=1 n=1 Since > 0 is arbitrary, we have countable subadditivity. To show that µ∗ and ν agree on A, suppose M ∈ A. Certainly we know that µ∗(M) ≤ ν(M), as M itself covers M. In the other direction, let An ∈ A cover M and let

-n−1 ! [ Bn = M ∩ An Ai . i=1 S Then the Bn are pairwise disjoint elements of A with n Bn = M, so ∞ ∞ X X ν(M) = ν(Bn) ≤ ν(An). n=1 n=1

8 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

∗ Taking the inﬁmum over all countable covers An of M, we have ν(M) ≤ µ (M). Finally, we show that each M ∈ A is µ∗-measurable. Let T ⊂ X and > 0 be arbitrary. If µ∗(T ) = ∞, then µ∗(T ) = ∞ ≥ µ∗(T ∩ M) + µ∗(T ∩ M c) ≥ µ∗(T ), ∗ so we have equality everywhere. If µ (T ) < ∞, then pick a cover An ∈ A of T such that

∞ ∗ X ∗ µ (T ) ≤ ν(An) ≤ µ (T ) + . n=1

c Since ν(M ∩ An) + ν(M ∩ An) = ν(An), we have

∞ ∗ X µ (T ) + ≥ ν(An) n=1 ∞ ∞ X X c = ν(M ∩ An) + ν(M ∩ An) n=1 n=1 = µ∗(M ∩ T ) + µ∗(M c ∩ T ) ≥ µ∗(T ).

Since is arbitrary, we have the required (Carathéodory) criterion. Theorem 1.3.9 (Carathéodory extension theorem). Let A be an algebra on X, ν : A → [0, ∞] be a premeasure, and M = σ(A). Then there exists a measure µ : M → [0, ∞] extending ν. Moreover, if ν is σ-finite, then the extension µ on M is unique.

Proof. Using Theorem 1.3.5 and Lemma 1.3.8, ν deﬁnes an outer measure µ∗ on X extending ν. If B is the family of µ∗-measurable sets, then B is a σ-algebra containing A, so M = σ(A) ⊂ B. The ∗ ∗ function µ = µ |B is then a measure extending ν through µ .

For uniqueness, letµ ˜ be another measure on M that extends ν. Let An ∈ A cover M. Then ∞ ∞ X X µ˜(M) ≤ µ˜(An) = ν(An). n=1 n=1 Taking the infimum over all countable covers of M by sets in A, we getµ ˜(M) ≤ µ(M). To get the other direction, we first show that µ(M) ≤ µ˜(M) for all M ∈ M with µ(M) < ∞. Let > 0. Then we can find sets An ∈ A covering M such that ∞ X ∗ ν(An) ≤ µ (M) + = µ(M) + . n=1 Sn−1 S S Define Bn = An\ i=1 Ai. These sets are pairwise disjoint in A and A = n An = n Bn, so ∞ ∞ ∞ X X X µ(A) = µ(Bn) = ν(Bn) ≤ ν(An) ≤ µ(M) + . n=1 n=1 n=1 Since µ(M) < ∞, it follows that

µ˜(A\M) ≤ µ(A\M) = µ(A) − µ(M) < .

9 1.4 Lebesgue measure MATH 245A (17F)

Hence n ! n ! [ [ µ(M) ≤ µ(A) = lim µ Ai = lim µ˜ Ai =µ ˜(A) < µ˜(M) + . n→∞ n→∞ i=1 i=1 As is arbitrary, we have µ(M) ≤ µ˜(M) when µ(M) < ∞. Having shown that µ(M) =µ ˜(M) whenever M has finite µ-measure, we show that this is true in general. Since ν is σ-finite, we can find pairwise disjoint sets Fn ∈ A with ν(Fn) = µ(Fn) < ∞ and S X = n Fn. For each M ∈ M, we have

∞ ∞ X X µ(M) = µ(M ∩ Fn) = µ˜(M ∩ Fn) =µ ˜(M). n=1 n=1

1.4 LEBESGUE MEASURE

Deﬁnition 1.4.1 (h-interval). An h-interval (half-open interval) is a set I ⊂ R of the form   (a, b] a, b ∈ R, a < b   (−∞, b] b ∈  R I = (a, ∞) a ∈ R   R   ∅.

The length of an h-interval I is l(I) = b − a for h-intervals of the ﬁrst form, ∞ for the second, third, and fourth forms, and 0 for the empty set. Lemma 1.4.2. Let C be a family of subsets of X such that

(i) ∅ ∈ C (ii) A ∩ B ∈ C whenever A, B ∈ C (iii) if A ∩ C, then Ac is a ﬁnite union of elements of C which is pairwise disjoint.

If A is the family of all subsets of X that can be represented as a ﬁnite union of pairwise disjoint sets in C, then A is an algebra on X.

Corollary 1.4.3. Let A be the family of all subsets of R that can be written as a ﬁnite union of pairwise disjoint h-intervals. Then A is an algebra on R (the algebra generated by h-intervals). Deﬁnition 1.4.4 (h-rectangle). An h-rectangle R ⊂ Rn is a set of the form

R = I1 × · · · × In where each Ik is an h-interval.

Proposition 1.4.5. The family A of all subsets of Rn that can be written as a ﬁnite union of pairwise disjoint h-rectangles is an algebra on Rn.

10 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Proof. We use Lemma 1.4.2 for the family R of h-rectangles in Rn. (i) We have ∅ = ∅ × · · · × ∅ ∈ R.

(ii) If R = I1 × · · · × In and S = J1 × · · · × Jn are in R, then

R ∩ S = (I1 ∩ J1) × · · · × (In ∩ Jn) ∈ R,

as intersections of h-intervals are h-intervals.

c (iii) If R = I1 × In ∈ R, then Ii = R\Ii is a disjoint union of at most two h-intervals. This implies

n ( c c n G Y Ij j = il for some 1 ≤ l ≤ k R = R \R = Ij otherwise 1≤i1<···

is a large but ﬁnite disjoint union of h-rectangles.

n Deﬁnition 1.4.6 (Content). If R = I1 × · · · × In is an h-rectangle in R , then the content of R is

|R| = l(I1) ····· l(In) ∈ [0, ∞], with the convention that 0 · ∞ = 0.

Lemma 1.4.7 (Basic lemma). Let {Ri}i∈I and {Sj}j∈J be two countable families of h-rectangles n S S in R . Suppose Rk ∩ Rl = ∅ for distinct k, l ∈ I and i Ri ⊂ j Sj. Then X X |Ri| ≤ |Sj|. i∈I j∈J

Proof. See Homework 3 Problem 1. Definition 1.4.8 (Premeasure on h-rectangles). Let A be the algebra generated by the h-rectangles n in R . To define a premeasure ν : A → [0, ∞], let M ∈ A have the form M = R1 t · · · t Rk for Pk h-rectangles Ri. Then set ν(M) = i=1|Ri|. Proposition 1.4.9. ν is a well-defined premeasure on A.

Proof. To see that ν is well-deﬁned, suppose M = R1 t · · · t Rk = S1 t · · · t Sl. The basic lemma Pk Pl then applies in both directions, so i=1|Ri| = j=1|Sj|. To see that ν is a premeasure, ﬁrst we have ν(∅) = 0. Then, given pairwise disjoint An ∈ A, S S with An = Rn,1 t · · · t Rn,kn . If A = n An = n,k Rn,k, then applying the basic lemma in both directions again, we have

∞ k ∞ X X Xn X ν(A) = |Rn,k| = |Rn,k| = ν(An). n,k n=1 k=1 n=1

11 1.4 Lebesgue measure MATH 245A (17F)

Deﬁnition 1.4.10 (Lebesgue outer measure). The Lebesgue outer measure is the outer measure induced by ν through Lemma 1.3.8, i.e.

( ∞ ) ∗ X [ λn(M) = inf ν(Ai) | Ai ∈ A and M ⊂ Ai . i=1 i

Deﬁnition 1.4.11 (Lebesgue measurable subset). A subset of Rn which is λ∗-measurable is (Lebesgue) measurable.

∗ Deﬁnition 1.4.12 (Lebesgue measure). The Lebesgue measure is the measure λn induced by λn through Theorem 1.3.5. Proposition 1.4.13. 1. Every h-rectangle R is measurable with λ(R) = |R|. 2. Every Borel set is measurable. 3. Lebesgue measure is complete.

Deﬁnition 1.4.14 (Rectangle). A rectangle is a compact set of the form R = [a1, b1]×· · ·×[an, bn].

Proposition 1.4.15. 1. If R = [a1, b1] × · · · × [an, bn] is a rectangle, then

λ(R) = (b1 − a1) ··· (bn − an).

2. If M ⊂ Rn, then ( ∞ ) ∗ X [ λ (M) = inf λ(Ri) | Ri rectangles with M ⊂ Ri . i=1 i

Theorem 1.4.16. Let M ⊂ Rn. The following are equivalent: (1) M is a null set, i.e. M is measurable and λ(M) = 0; P (2) for each > 0, there is a countable cover of M by rectangles Rk such that k λ(Rk) < ; (3) there exists a Borel set B ⊂ Rn with M ⊂ B and λ(B) = 0.

Proof. See Homework 3 Problem 2.

Corollary 1.4.17. A set M ⊂ Rn is measurable if and only if there exists a Borel set B and a null set N such that M = B ∪ N.

Proof. ( =⇒ ) Omitted.

∗ ( ⇐= ) If λ(M) = λ (M) < ∞, then for each k ∈ N, there exist rectangles Rk,i covering M such that ∞ ! ∞ [ X 1 λ(M) ≤ λ R ≤ λ(R ) ≤ λ(M) + . k,i k,i k i=1 i=1 Let [ \ Ak = Rk,i and A = Ak. i k

12 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

This is Borel with M ⊂ A, so by construction, λ(A) = λ(M). Thus λ(A\M) = 0, so there exists a Borel set C with A\M ⊂ C and λ(C) = 0. We then have

M = (A\C) ∪ (M ∩ C)

with B = A\C Borel and N = M ∩ C null. n For the general case, let Rk = [−k, k] . If Mk = M ∩ Rk, then each Mk has ﬁnite measure, S so we can ﬁnd Borel sets Bk and null sets Nk such that Mk = Bk ∪ Nk. Then if B = Bk S S k and N = k Nk, we have that B is Borel, N is null, and M = k Mk = B ∪ N.

Theorem 1.4.18. The Lebesgue measure on Rn is the unique measure λ such that

(i) λ is deﬁned on the σ-algebra of all Lebesgue measurable subsets;

(ii) λ is translation invariant, i.e. λ(M) = λ(t + M) for all t ∈ Rn and M ⊂ Rn measurable;

(iii) λ([0, 1]n) = 1.

Proof. See Homework 4 Problem 1.

Deﬁnition 1.4.19 (Regular measure). Let X be a topological space, A be a σ-algebra containing BX , and µ : A → [0, ∞] be a measure. We say that

1. µ is inner regular if for all A ∈ A,

µ(A) = sup{µ(K) | K ⊂ AcompactinX};

2. µ is outer regular if for all A ∈ A,

µ(A) = inf{µ(U) | U ⊃ AopeninX};

3. µ is regular if it is inner and outer regular.

Proposition 1.4.20. Lebesgue measure is regular.

Proof. See Homework 2 Problem 3.

Proposition 1.4.21. If T : Rn → Rn is a linear map and M ⊂ Rn is measurable, then T (M) is measurable and λ(T (M)) = |det T | · λ(M). In particular, orthogonal transformations preserve Lebesgue measure.

Proof. See Homework 4 Problems 3 and 4.

13 1.5 Measurable functions MATH 245A (17F)

Example 1.4.22 (Non-measurable subset of R). Deﬁne an equivalence relation on [0, 1] by

x ∼ y ⇐⇒ x − y ∈ Q. Invoking the axiom of choice, we can pick an element from each equivalence class to get a set E ⊂ [0, 1]. Suppose that E were measurable. Enumerate the rationals qi ∈ [−1, 1] and consider the translates qi + E ⊂ [−1, 2]. These are disjoint by construction, so

∞ ! ∞ [ X λ (qi + E) = λ(qi + E) = ∞ · λ(E). i=1 i=1 S Since all of these translates lie in [−1, 2], λ ( i(qi + E)) ≤ 3. However, given any x ∈ [0, 1], there must be some y ∈ [0, 1] with x ∼ y, so there is a rational q ∈ [−1, 1] such that x = y + q. This S means that λ ( i(qi + E)) ≥ 1. There is no possible value of λ(E) which makes ∞ · λ(E) lie in this range, so E cannot be measurable.

This can be slightly modiﬁed to prove the following result.

Proposition 1.4.23. Every measurable set M ⊂ R with λ(M) > 0 has a non-measurable subset. Example 1.4.24 (Lebesgue measurable set that is not Borel). Let c : [0, 1] → [0, 1] be the Cantor function and f : [0, 1] → [0, 2] be given by f(x) = c(x) + x. Then if C ⊂ [0, 1] is the Cantor set, we have λ(f([0, 1]\C)) = λ([0, 1]\C) = 1, from which it follows that µ(f(C)) = 1. Hence there is a non-measurable set N ⊂ f(C), which in particular is not Borel. Since f is strictly increasing, it maps Borel sets to Borel sets (see Homework 1 Problem 4). Thus f −1(N) is not Borel, but it is a subset of C, which has measure zero, so N is Lebesgue measurable (with measure zero).

1.5 MEASURABLE FUNCTIONS

Deﬁnition 1.5.1 (Measurable function). Let (X, A) and (Y, B) be measurable spaces. A map f : X → Y is called (A, B)-measurable if f −1(B) ∈ A for all B ∈ B.

If A and B are understood, we will simply say that f is measurable.

Proposition 1.5.2. If B = σ(S), then f is measurable if and only if f −1(S) ∈ A for all S ∈ S.

Corollary 1.5.3. Let X and Y be topological spaces with their Borel σ-algebras. If f is continuous, then f is (Borel) measurable.

Proposition 1.5.4. If f : X → Y and g : Y → Z are measurable, then g ◦ f is measurable.

In the case that X = Rn, there are two natural σ-algebras to put on Rn, namely the Borel σ-algebra B and the σ-algebra L of Lebesgue measurable sets.

Deﬁnition 1.5.5 (Measurable functions from Rn). A function f : Rn → Y is Lebesgue measurable, n or simply measurable, if f :(R , L) → (Y, BY ) is (L, BY )-measurable. We say f is Borel measurable n if f :(R , B) → (Y, BY ) is (B, BY )-measurable. Remark 1.5.6. 1. If f is Borel measurable, then f is (Lebesgue) measurable.

14 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

2. According to this convention, a function f : R → R is measurable if it is (L, B)-measurable, so the domain and codomain have diﬀerent σ-algebras. In particular, if f, g : R → R are measurable, then g ◦ f need not be measurable. On the other hand, if f is measurable and g is Borel measurable, then g ◦ f is measurable.

Lemma 1.5.7. Let (X, A) be a measurable space. Then

1. f : X → R is measurable if and only if f −1((a, ∞]) ∈ A for each a ∈ R.

2. if u, v : X → R are measurable, Z is a topological space, and F : R2 → Z is continuous, then F ◦ (u, v): X → Z is measurable.

Corollary 1.5.8. 1. If f, g : X → C are measurable, then so are f, f + g, and f · g.

2. If f : X → C is measurable and z is a constant, then u = Re f, v = Im f, |f|, 1/f, and zf are measurable.

Proposition 1.5.9. Let fn : X → R be measurable functions. Then sup fn, inf fn, lim sup fn, and lim inf fn are measurable.

Corollary 1.5.10. 1. If f and g are measurable, then f ∨ g = max(f, g) and f ∧ g = min(f, g) are measurable.

2. If fn : X → R are measurable and fn → f pointwise, then f is measurable.

Deﬁnition 1.5.11 (Positive and negative parts). Given f : X → R, the positive part f + and negative part f − are f + = max(f, 0), f − = − min(f, 0), so that f = f + − f −.

Proposition 1.5.12. If f : X → R are measurable, then f + and f − are measurable.

Deﬁnition 1.5.13 (Characteristic function). If A ⊂ X, then χA : X → R deﬁned by χ(x ∈ A) = 1 and χ(x 6∈ A) = 0 is the characteristic function (or indicator function) of A.

Proposition 1.5.14. If A ⊂ X is measurable, then χA : X → R is measurable.

Deﬁnition 1.5.15 (Simple function). A simple function is a function f : X → C of the form

n X f = αiχAi , i=1 where αi ∈ C and Ai is measurable for each i.

Notation. Write fn % f if f1 ≤ f2 ≤ · · · and fn → f pointwise.

Theorem 1.5.16. Let (X, A) be a measurable space and f : X → [0, ∞] be a measurable function. Then there exist simple functions sn : X → [0, ∞) such that sn % f and sn → f pointwise.

15 1.5 Measurable functions MATH 245A (17F)

Proof. Deﬁne k k + 1 E = f −1 , ,F = f −1((2n, +∞]), n,k 2n 2n n and set 22n−1 X k s = · χ + 2n · χ . n 2n En,k Fn k=0

16 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

2 INTEGRATION

2.1 INTEGRATION OF SIMPLE NON-NEGATIVE FUNCTIONS

Note that every simple function has ﬁnite image set. Conversely, every measurable function which has a ﬁnite image set is a simple function.

Deﬁnition 2.1.1 (Standard representation). If s : X → C is a simple function with image set s(X) = {α1, . . . , αn}, then the standard representation of s is

n X s = αkχAk k=1

−1 for Ak = s (αk). Notation. Write S+ be the space of all simple functions with values in [0, ∞).

Deﬁnition 2.1.2 (Integral of a non-negative simple function). Given a simple function s ∈ S+ in standard representation, the Lebesgue integral of s with respect to µ is

n Z X s dµ = αkµ(Ak). k=1 P Proposition 2.1.3. 1. If s = βkχB as a ﬁnite sum which is not necessarily the standard R P k k representation, then s = k βkµ(Bk). 2. If s ∈ S+ and c ≥ 0, then R cs = c R s.

3. If s, t ∈ S+, then R (s + t) = R s + R t.

4. If s, t ∈ S+ with s ≤ t, then R s ≤ R t.

Notation. If A ∈ A and s ∈ S+, then write Z Z s dµ = s · χA dµ. A

Proposition 2.1.4. Let s ∈ S+ and deﬁne for A ∈ A Z µ(A) = s dµ. A Then µ is a measure on (X, A).

2.2 INTEGRATION OF NON-NEGATIVE FUNCTIONS

Notation. Let (X, A, µ) be a measure space. Write L+(µ) for the set of non-negative measurable functions on X.

17 2.2 Integration of non-negative functions MATH 245A (17F)

Deﬁnition 2.2.1 (Integral of a non-negative function). For f ∈ L+, the Lebesgue integral of f with respect to µ is Z Z f dµ = sup s dµ | s ∈ S+ with s ≤ f .

Proposition 2.2.2. 1. This is consistent with the previous deﬁnition for simple functions. 2. If f, g ∈ L+ and f ≤ g, then R f dµ ≤ R g dµ.

+ Theorem 2.2.3 (Lebesgue monotone convergence theorem). If fn, f ∈ L and fn % f, then Z Z f dµ = lim fn dµ. n→∞ R Proof. Since { fn dµ} is an increasing sequence of extended real numbers which is bounded above by R f dµ, its limit exists and is at most R f dµ. To show the reverse inequality, let α ∈ (0, 1) and φ be a simple function with 0 ≤ φ ≤ f. Deﬁne An = {x ∈ X | fn(x) ≥ αφ(x)}. By construction, R R R An % X and fn dµ ≥ fn dµ ≥ α φ dµ. Taking the limit as n → ∞, we have An An Z Z Z lim fn dµ ≥ lim α φ dµ = α φ dµ. n→∞ n→∞ An Taking the supremum over all φ and letting α → 1−, we have Z Z f dµ ≤ lim fn dµ, n→∞ as required. Proposition 2.2.4. 1. If f ∈ L+ and α ≥ 0, then R αf dµ = α R f dµ. 2. If f, g ∈ L+, then R (f + g) dµ = R f dµ + R g dµ.

+ R P P R 3. If fn ∈ L , then ( n fn) dµ = n fn dµ. 4. If f ∈ L+, then R f = 0 if and only if f = 0 µ-a.e.

+ R R Corollary 2.2.5. 1. If fn, f ∈ L and fn % f µ-a.e., then fn dµ → f dµ. 2. If f, g ∈ L+ and f = g µ-a.e., then R f dµ = R g dµ.

Lemma 2.2.6 (Fatou). Let fn : X → [0, ∞] be measurable functions. Then Z Z lim inf fn dµ ≤ lim inf fn dµ. n→∞ n→∞

Proof. Let gk = inf fn. n≥k

Then gk ≤ fm for each m ≥ k, hence Z Z gk dµ ≤ inf fm dµ. m≥k

18 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Applying Theorem 2.2.3 to the sequence {gk}, Z Z Z Z lim inf fn dµ = lim gk dµ ≤ lim inf fm dµ = lim inf fn dµ. n→∞ k→∞ k→∞ m≥k n→∞

2.3 INTEGRATION OF REAL AND COMPLEX VALUED FUNCTIONS

Deﬁnition 2.3.1 (Integrable function to R). A function f : X → R is integrable (with respect to µ) if f is measurable and R |f| dµ < ∞.

Notation. The space of real-valued measurable functions is L1 (µ), or L1 if µ is understood. R R Proposition 2.3.2. If f ∈ L1 , then f +, f − ∈ L+ and R f + dµ, R f − dµ < ∞. R Deﬁnition 2.3.3 (Integral of a real-valued integrable function). For f ∈ L1 , the integral of f is R Z Z Z f dµ = f + dµ − f − dµ.

Proposition 2.3.4. A function f : X → is in L1 if and only if R f + dµ and R f − dµ are ﬁnite. R R Deﬁnition 2.3.5 (Integrable function to C). A function f : X → C is integrable if f is measurable and R |f| dµ < ∞. The space of complex-valued integrable functions is L1.

Notation. The space of complex-valued measurable functions is L1(µ), or L1 if µ is understood.

Proposition 2.3.6. If f ∈ L1, then Re f, Im f ∈ L1 . R Deﬁnition 2.3.7 (Integral of a complex-valued integrable function / integral over a subset). For f ∈ L1, the integral of f is Z Z Z f dµ = Re f dµ + i Im f dµ.

If A ∈ A and f ∈ L1, deﬁne Z Z f dµ = χAf dµ. A

Theorem 2.3.8. L1 is a C-vector space and the integral is a linear functional on L1. Proposition 2.3.9. If f ∈ L1, then |R f| dµ ≤ R |f| dµ. R R Proof. There exists α ∈ C with |α| = 1 such that | f dµ| = α f dµ. Then Z Z Z Z Z

f dµ = α f dµ = αf dµ = Re αf dµ = Re(αf) dµ

Z Z Z ≤ |Re(αf)| dµ ≤ |αf| dµ = |f| dµ.

19 2.4 Lp-spaces MATH 245A (17F)

1 Theorem 2.3.10 (Lebesgue dominated convergence theorem). Let fn be a sequence in L and 1 f : X → C. Suppose that fn → f pointwise and there exists a non-negative g ∈ L such that 1 R |fn| ≤ g for all n. Then f ∈ L and |fn − f| dµ → 0.

Proof. By taking real and imaginary parts, it suﬃces to assume that fn and f are real-valued. Both g + fn and g − fn are non-negative integrable functions, so by Lemma 2.2.6, Z Z Z Z Z Z g dµ + f dµ = lim inf(g + fn) dµ ≤ lim inf (g + fn) dµ = g dµ + lim inf fn dµ n→∞ n→∞ n→∞ Z Z Z Z Z Z g dµ − f dµ = lim inf(g − fn) dµ ≤ lim inf (g − fn) dµ = g dµ − lim sup fn dµ. n→∞ n→∞ n→∞ The result follows.

Corollary 2.3.11. Let fn, f : X → C be measurable. Suppose fn(x) → f(x) for µ-a.e. x ∈ X and R there exists a non-negative measurable function g : X → [0, ∞] with g < ∞ and |fn| ≤ g µ-a.e. R for each n. Then the functions fn, f are all integrable and |fn − f| dµ → 0 as n → ∞. R R R Proposition 2.3.12. If |fn − f| dµ → 0 as n → ∞, then fn dµ = f dµ.

2.4 Lp-SPACES

Deﬁnition 2.4.1 (Lp-spaces). Let 1 ≤ p < ∞. We deﬁne Lp(µ) (or Lp when µ is understood) as R p the space of all integrable functions f : X → C for which |f| dµ < ∞. Notation. If f : X → C is measurable, then write Z 1/p p kfkp = |f| dµ .

Proposition 2.4.2. kfkp = 0 if and only if f = 0 µ-a.e. Lemma 2.4.3. If a, b ≥ 0 and 0 < λ < 1, then

aλb1−λ ≤ λa + (1 − λ)b, with equality if and only if a = b.

Proof. Since log : (0, ∞) → R is strictly concave, Jensen’s inequality gives λ log a + (1 − λ) log b ≤ log(λa + (1 − λ)b).

Deﬁnition 2.4.4 (Conjugate exponent). If 1 < p < ∞, then the unique q for which 1 < q < ∞ and 1/p + 1/q = 1 is the conjugate exponent of p. Theorem 2.4.5 (H¨older’sinequality). Let 1 < p < ∞ and let q be the conjugate exponent of p. If f, g : X → C are measurable, then kfgk1 ≤ kfkpkgkq, with equality if and only if α|f|p = β|g|q µ-a.e. for some constants α, β which are not both zero.

20 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Proof. If one of kfkp or kgkq is 0 or ∞, then the result is clear. Furthermore, scaling f and g do not change the validity of the inequality, so it suﬃces to consider kfkp = kgkq = 1. By Lemma 2.4.3 with a = |f(x)|p, b = |g(x)|q, and λ = 1/p, we have

1 1 |f(x)g(x)| ≤ |f(x)|p + |g(x)|q. (∗) p q

Integrating both sides,

1 Z 1 Z 1 1 kfgk ≤ |f|p dµ + |g|q dµ = + = 1 = kfk kgk . 1 p q p q p q

Equality holds if and only if it holds µ-a.e. in (∗), which happens when |f|p = |g|q µ-a.e.

Theorem 2.4.6 (Minkowski’s inequality). If 1 ≤ p < ∞ and f, g ∈ Lp, then

kf + gkp ≤ kfkp + kgkp.

Proof. If f + g = 0 µ-a.e. or p = 1, then the result is clear. Otherwise,

|f + g|p ≤ (|f| + |g|)|f + g|p−1, so applying Theorem 2.4.5 twice, Z p p−1 p−1 |f + g| dµ ≤ kfkpk|f + g| kq + kgkpk|f + g| kq Z 1/q p = (kfkp + kgkp) |f + g| dµ

(1−q−1)−1/p kf + gkp ≤ (kfkp + kgkp) = kfkp + kgkp.

p Corollary 2.4.7. Let 1 ≤ p < ∞. Then L is a C-vector space on which k · kp is a seminorm.

Notation. For measurable functions f, g : X → C, write f ∼ g if f = g µ-a.e. This is an equivalence relation on the set of measurable functions and on each Lp.

Deﬁnition 2.4.8 (Lp-spaces). Let 1 ≤ p < ∞. Deﬁne Lp(µ) = Lp(µ)/ ∼.

p Theorem 2.4.9. L is a C-vector space on which k · kp is a norm. Deﬁnition 2.4.10 (Banach space). A normed (real or complex) vector space V is a Banach space if it is complete with respect to the metric induced by the norm.

Lemma 2.4.11. A normed vector space is a Banach space if and only if every absolutely convergent series converges.

Theorem 2.4.12. Lp is a Banach space for 1 ≤ p < ∞.

21 2.5 Relation to Riemann integration MATH 245A (17F)

p P Proof. Let fn ∈ L be such that n kfnkp = S < ∞. Deﬁne n n ∞ X X X Fn = fk,Gn = |fk|,G = |fk|. k=1 k=1 k=1

For each n, we have |Gn| ≤ B, so by Theorem 2.2.3, Z Z p p p G dµ = lim Gn dµ ≤ B . n→∞

p P This tells us that G ∈ L , so in particular G(x) < ∞ µ-a.e. If F = k fk (deﬁned µ-a.e.), we have |F | ≤ G, so F ∈ Lp. Furthermore,

n X p F − fk ≤ (2G) , k=1 so by Theorem 2.3.10, p p n Z n X X F − fk = F − fk dµ → 0, k=1 p k=1 i.e. the series for F converges in the Lp norm. P Proposition 2.4.13. For 1 ≤ p < ∞, the set of simple functions s = j ajχEj with µ(Ej) < ∞ for all j is dense in Lp.

Deﬁnition 2.4.14 (Essential supremum / L∞). Let f : X → C be a measurable function. The essential supremum of f is

kfk∞ = ess supx∈X |f(x)| = {λ ≥ 0 | µ ({x ∈ X | |f(x)| > a}) = 0} . The space of measurable functions with ﬁnite essential supremum is L∞, and L∞ = L∞/ ∼. Remark 2.4.15. We may regard ∞ as the conjugate exponent for 1 and vice versa. Theorem 2.4.16. 1. H¨older’sinequality extends to {p, q} = {1, ∞}.

∞ 2. L is a C-vector space on which k · k∞ is a norm. 3. L∞ is a Banach space. 4. Simple functions are dense in L∞.

2.5 RELATION TO RIEMANN INTEGRATION

Notation. Given a function f :[a, b] → R and a partition P of [a, b], write UP f and Lpf for the upper and lower sums of f on the partition.

Theorem 2.5.1. If f :[a, b] → R is Riemann integrable, then f is Lebesgue integrable and Z b Z f(x) dx = f dλ. a [a,b]

22 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Proof. For a partition P given by a = t0 < t1 < ··· < tn = b, let

Mi = sup f(x) and mi = inf f(x), x∈[t ,t ] x∈[ti−1,ti] i−1 i then deﬁne simple functions

n n X X ΦP = Miχ(tj−1,tj ] and φP = miχ(tj−1,tj ] i=1 i=1 R R so that UP f = [a,b] ΦP dλ and LP f = [a,b] φP dλ. By assumption, there are partitions Pk, each one R b a reﬁnement of the previous, such that UPk f and LPk f converge to a f(x) dx. Suppose ΦPk → Φ and φPk → φ; note that these limits are monotone. Then we have φPk ≤ φ ≤ f ≤ Φ ≤ ΦPk ≤ ΦP1 for each k, so by the dominated convergence theorem,

Z Z Z b Z Z Z b

φ dλ = lim φPk dλ = f(x) dx, Φ dλ = lim ΦPk dλ = f(x) dx. [a,b] k→∞ [a,b] a [a,b] k→∞ [a,b] a R Thus [a,b](Φ − φ) dλ = 0 with Φ − φ ≥ 0, so Φ = φ almost everywhere. Since φ ≤ f ≤ Φ, in fact φ = f almost everywhere. As φ is the limit of simple functions, φ is measurable, so f is also measurable. Furthermore,

Z b Z Z Z Z b f(x) dx = φ dλ ≤ f dλ = Φ dλ = f(x) dx, a [a,b] [a,b] [a,b] a

R R b so [a,b] f dλ = a f(x) dx.

Theorem 2.5.2. If f :[a, b] → R is bounded, then f is Riemann integrable if and only if the set of discontinuities of f has measure zero.

Proof. To be written. (Folland ex 2.23)

2.6 MODES OF CONVERGENCE

Deﬁnition 2.6.1 (Convergence in measure). Let fn, f : X → C be measurable functions. We say that fn → f in measure if for each > 0,

lim µ ({x ∈ X | |fn(x) − f(x)| > }) = 0. n→∞

Example 2.6.2. Consider three modes of convergence for measurable functions fn, f : X → C:

1. fn → f pointwise µ-a.e.;

p 2. fn → f in L for some 1 ≤ p < ∞;

3. fn → f in measure.

23 2.6 Modes of convergence MATH 245A (17F)

Given 1 ≤ p < ∞, the example on [0, 1] given by

1/p fn = n χ[0,1/n] and f = 0 shows that 1 =6 ⇒ 2 and 3 =6 ⇒ 2. The typewriter sequence on [0, 1] given by

fn = χ[j/2k,(j+1)/2k] and f = 0, where n = 2k + j with 0 ≤ j < 2k, shows that 2 =6 ⇒ 1 and 3 =6 ⇒ 1. The example on R given by

fn = χ[n,n+1] and f = 0 shows that 1 =6 ⇒ 3. The only remaining implication, 2 =⇒ 3, turns out to be correct. The implications 3 =⇒ 1 and 1 =⇒ 3 can also be made to work with slight modiﬁcations.

p Proposition 2.6.3. If fn → f in L for some 1 ≤ p < ∞, then fn → f in measure.

Proof. Let > 0 and consider En = {x ∈ X | |fn(x) − f(x)| ≥ }. We have Z Z |fn − f| dµ ≥ |fn − f| dµ ≥ · µ(En), En

−1 R so µ(En) ≤ |fn − f| dµ → 0 as n → ∞.

Proposition 2.6.4. If fn → f in measure, then there exists a subsequence (fnj )j such that fnj → f pointwise µ-a.e.

Proof. If fn → f in measure, then fn is also Cauchy in measure. Choose a subsequence gj = fnj such that for −j Ej = {x ∈ X | |gj(x) − gj+1(x)| ≥ 2 }, −k S k−l we have µ(Ek) ≤ 2 . Setting Fk = j≥k Ej, then µ(Fk) ≤ 2 and

i−1 X 1−j |gj(x) − gi(x)| ≤ |gl+1(x) − gl(x)| ≤ 2 l=j T for i ≥ j ≥ k and x 6∈ Fk. This means that {gj(x)} is Cauchy if x 6∈ Fk, for any k. Let F = k Fk. Then µ(F ) = 0 and f(x) = lim gj(x) is deﬁned µ-a.e. (Deﬁne f to be zero elsewhere.) We have that f is measurable and gj → f µ-a.e.

Theorem 2.6.5 (Egorov). If fn → f pointwise µ-a.e. and µ(X) < ∞, then for every > 0, there exists E ⊂ X such that µ(E) < and fn → f uniformly on X\E.

Proof. Without loss of generality, suppose fn → f everywhere. Given n and k, let

[ −1 En(k) = {x ∈ X | |fm(x) − f(x)| ≥ k }. m≥n T For fixed k, we have a decreasing sequence in n with n En(k) = ∅, so since the measure is finite, −k we have µ(En(k)) → µ(∅) = 0. Given > 0 and k, choose nk so that µ(Enk (k)) < · 2 and define S c E = k Enk (k). Then µ(E) < and fn → f uniformly on E by construction.

24 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Corollary 2.6.6. If fn → f pointwise µ-a.e. and µ(X) < ∞, then fn → f in measure. Deﬁnition 2.6.7 (Radon measure). A Borel measure µ on a topological space X is a Radon measure if it is ﬁnite on all compact sets, outer regular on all Borel sets, and inner regular on all open sets. Lemma 2.6.8 (Urysohn). Let X be a normal topological space and let A, B ⊂ X be disjoint non-empty closed subsets of X. Then there exists a continuous function f : X → [0, 1] such that f(A) = {0} and f(B) = {1}.

Proof. Let Q ⊂ [0, 1] consist of all dyadic rationals. We construct a collection of open sets Uq indexed by Q with Uq ⊂ Ur whenever q ≤ r. Set U1 = X. Since X is normal, there are open sets U and V separating A and B. Set U0 = U. Since X is normal, there is a set U1/2 with U0 ⊂ U1/2 ⊂ U1/2 ⊂ X\B. Repeating this, we inductively deﬁne all of the sets Uq for q ∈ Q. The continuous function we wish to deﬁne is then

f(x) = inf{q ∈ Q | x ∈ Uq}.

Theorem 2.6.9 (Lusin). Let µ be a Radon measure on a locally compact Hausdorﬀ space X. Suppose f : X → C is measurable and A ⊂ X is a Borel set of ﬁnite measure. Then for each > 0, there exists a closed E ⊂ A such that µ(E) < and f is continuous on A\E.

Proof. To be written.

p Corollary 2.6.10. The space Cc(X) of continuous functions with compact support is dense in L for 1 ≤ p < ∞.

Proof. To be written.

∞ Remark 2.6.11. In the case p = ∞, we still have Cc(X) ⊂ L , but Cc(X) need not be dense in ∞ L . For example, in R with Lebesgue measure, the ball of radius 1/2 around f = χ[0,1] contains no continuous function with compact support.

2.7 PRODUCT MEASURES

Deﬁnition 2.7.1 (Product σ-algebra). Let (X, A) and (Y, B) be measurable spaces. The product σ-algebra on X × Y is A ⊗ B = σ{A × B | A ∈ A and B ∈ B}. Deﬁnition 2.7.2 (Sections). Let (X, A) and (Y, B) be measurable spaces and E ⊂ X × Y . For y x ∈ X and y ∈ Y , the x-section Ex and the y-section E are

Ex = {y ∈ Y | (x, y) ∈ E},Ey = {x ∈ X | (x, y) ∈ E}.

y Given f : X ×Y → C, for x ∈ X and y ∈ Y , the x-section fx : Y → C and the y-section f : X → C are deﬁned by y fx(y) = f (x) = f(x, y).

25 2.7 Product measures MATH 245A (17F)

y Proposition 2.7.3. 1. If E ∈ A ⊗ B, then Ex ∈ B and E ∈ A for all x ∈ X and y ∈ Y . y 2. If f : X × Y → C is measurable, then fx and f are measurable. Lemma 2.7.4 (Monotone class theorem). Let (X, A) be a measurable space and F be a family of functions f : X → C such that

(i) there exists a π-system P ⊂ A with X ∈ P, σ(P) = A, and χA ∈ F for each A ∈ P; (ii) F is closed under linear combinations; (iii) F is closed under monotone limits.

Then F contains all measurable functions X → R. Proof. See Homework 6 Problem 1. Lemma 2.7.5. Let (X, A, µ) and (Y, B, ν) be σ-ﬁnite measure spaces. If E ∈ A ⊗ B, then the y functions x 7→ ν(Ex) and y 7→ µ(E ) are measurable and Z Z y ν(Ex) dµ = µ(E ) dν.

Proof. Suppose µ and ν are finite, and let F be the set of all E ∈ A ⊗ B for which this claim holds. y For E = A × B, we have ν(Ex) = χA(x)ν(B) and µ(E ) = µ(A)χB(y), so Z Z Z Z y ν(Ex) dµ = ν(B) χA(x) dµ = ν(B)µ(A), µ(E ) dν = µ(A) χB(y) dν = µ(A)ν(B), showing that E ∈ F. By additivity of the integral, finite disjoint unions of rectangles are in F, Thus it remains to show closure under monotone limits. Let En % E with En ∈ F. Then fn(y) = y y µ((En) ) and gn(x) = ν((En)x) are measurable with fn % f : y 7→ µ(E ) and gn % g : x 7→ ν(Ex). Hence f and g are measurable, and by the monotone convergence theorem (2.2.3), Z Z Z Z y y µ(E ) dν = lim µ((En) ) dν = lim ν((En)x) dµ = ν(Ex) dµ. n→∞ n→∞ Thus E ∈ F, as required to complete the proof in the finite measure case. Applying the monotone convergence theorem again gives the result for general σ-finite measures. Theorem 2.7.6 (Tonelli). Let (X, A, µ) and (Y, B, ν) be σ-finite measure spaces.

1. There exists a unique measure ω = µ × ν on (X × Y, A ⊗ B) such that ω(A × B) = µ(A)ν(B) whenever A ∈ A and B ∈ B. 2. If f : X × Y → [0, ∞] is measurable, then the functions Z Z y G(x) = fx dν and H(y) = f dµ

are also non-negative measurable, and Z Z Z f dω = G dµ = H dν.

26 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Proof. 1. See Homework 8 Problems 1 and 2. 2. The case where f is a characteristic function is Lemma 2.7.5, and we get non-negative simple functions by linearity of the integral. For a general measurable f, take simple functions sn % f and deﬁne Gn,Hn on the sn as above. By the monotone convergence theorem (2.2.3), Gn % G and Hn % H, so G and H are measurable. Applying it again, Z Z Z Z G dµ = lim Gn dµ = lim sn dω = f dω, n→∞ n→∞

and similarly for R H dν.

Theorem 2.7.7 (Fubini). Let (X, A, µ) and (Y, B, ν) be σ-ﬁnite measure spaces and ω = µ × ν.

1 1 y 1 1. If f ∈ L (ω), then fx ∈ L (ν) and f ∈ L (µ) for µ-a.e. x ∈ X and ν-a.e. y ∈ Y . 2. The functions Z Z y G(x) = fx dν and H(y) = f dµ

are deﬁned for µ-a.e. x ∈ X and ν-a.e. y ∈ Y , and by making arbitrary (re-)deﬁnitions on a null set, G ∈ L1(µ) and H ∈ L1(ν). 3. We have Z Z Z f dω = G dµ = H dν.

Proof. If f is non-negative integrable, then by Tonelli’s theorem (2.7.6), it follows that G < ∞ 1 y 1 µ-a.e. and H < ∞ ν-a.e., so fx ∈ L (ν) and f ∈ L (µ) for almost every x ∈ X and y ∈ Y . Fubini’s theorem then pops out by applying Tonelli’s theorem to the positive and negative parts of the real and imaginary parts of f.

Lemma 2.7.8. If (Z, A, µ) is the completion of (Z, A, µ) and f : Z → C is µ-measurable, then there exists a µ-measurable function g : Z → C such that f = g µ-a.e. y Lemma 2.7.9. If h = 0 ω-a.e. on X × Y , then hx = 0 ν-a.e. for µ-a.e. x ∈ X and h = 0 µ-a.e. for ν-a.e. y ∈ Y . Theorem 2.7.10 (Fubini-Tonelli for complete measures). Let (X, A, µ) and (Y, B, ν) be σ-ﬁnite complete measure spaces and (X × Y, C, ω) be the completion of (X × Y, A ⊗ B, µ × ν). Suppose that f : X × Y → C is a measurable function.

+ y 1. If f is in L , then fx and f are measurable for µ-a.e. x ∈ X and ν-a.e. y ∈ Y . Moreover, R R y the functions x 7→ fx dν and y 7→ f dµ are measurable.

1 y 2. If f is in L (ω), then fx and f are integrable for µ-a.e. x ∈ X and ν-a.e. y ∈ Y . Morevoer, R R y the functions x 7→ fx dν and y 7→ f dµ are integrable.

In both cases, Z ZZ ZZ f dω = f(x, y) dµ(x) dν = f(x, y) dν(y) dµ(x).

27 2.8 Polar coordinates MATH 245A (17F)

n Example 2.7.11. Let Bn be the Borel σ-algebra on R and Ln be the σ-algebra of Lebesgue n measurable subsets of R . Deﬁne a Borel measure βn on Bn by restricting the Lebesgue measure λn to Borel sets. Since Bm+n = Bm ⊗ Bn (see Homework 5 Problem 1), uniqueness of product measure implies βm+n = βm × βn. In the case of Lebesgue measure, it turns out that λm+n 6= λm × λn. However, it is the case that λm+n is the completion of λm × λn, so Fubini-Tonelli can be used for Lebesgue integrals.

Proposition 2.7.12. Let f : R → [0, ∞) be a measurable function. Then

2 M = {(x, y) | 0 ≤ y ≤ f(x)} ⊂ R is measurable, and Z λ2(M) = f dλ1.

Proof. To be written.

2.8 POLAR COORDINATES

Deﬁnition 2.8.1 (Restriction of a measure). Let (X, A, µ) be a measure space and M ⊂ A be measurable. The restriction of µ to M is the measure space (M, AM , µ|M ), where AM = {A ∈ A | A ⊂ M} and µ|M (A) = µ(A) for A ∈ AM .

Proposition 2.8.2. Let (X, Aµ) be a measure space, (M, AM , µ|M ) be the restriction of µ to M ∈ A, and f : M → C be a function. Deﬁne f˜ : X → C by ( 0 x 6∈ M, f˜ = f(x) x ∈ M.

Then

1. f is measurable (integrable) if and only if f˜ is measurable (integrable);

2. if f is integrable, then Z Z ˜ f dµ|M = f dµ. M X Deﬁnition 2.8.3 (Pushforward measure). Let (X, A, µ) be a measure space, (Y, B) be a measurable space, and T :(X, A) → (Y, B) be measurable. The pushforward measure of µ by T is the measure deﬁned on (Y, B) by −1 T∗µ(B) = µ(T (B)). Proposition 2.8.4. If g : Y → [0, ∞] is measurable, then g ◦ T : X → [0, ∞] is measurable and Z Z g d(T∗µ) = g ◦ T dµ.

Deﬁnition 2.8.5 (Polar coordinates). The polar coordinates for a point p ∈ Rn\{0} are the pair (r, ξ), with r ∈ (0, ∞) and ξ ∈ Sn−1 ⊂ Rn, such that p = rξ.

28 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Representation in polar coordinates induces a homeomorphism Φ : Rn\{0} → (0, ∞) × Sn−1. Let n−1 m = Φ∗(βn) be the pushforward measure of βn by Φ. That is, given A ⊂ (0, ∞) × S Borel, we have −1 −1 m(A) = βn(Φ (A)) = λn(Φ (A)). Let ρ be the Borel measure on (0, ∞) deﬁned as Z Z ρ(E) = rn−1 dr = xn−1 dx. E E Theorem 2.8.6. There exists a unique Borel measure σ on Sn−1 such that m = ρ × σ. If f is a n 1 Borel measurable function on R with values in [0, ∞] or f ∈ L (βn), then Z Z ∞ Z n−1 f dλn = f(rξ)r dσ(xi) dr. 0 Sn−1 Proof. To show existence of σ, if E ⊂ Sn−1 is Borel, then let E˜ = {rξ | 0 < r < 1 and ξ ∈ E} be the open cone with base E. This is Borel, and we can deﬁne a Borel measure

σ(E) = nλn(E˜). We claim that m = ρ×σ. By uniqueness of the product measure, it suﬃces to show that m(A×B) = ρ(A)σ(B) whenever A ⊂ (0, ∞) and B ⊂ Sn−1 are Borel. We have

−1 m(A × B) = λn(Φ (A × B)) = λn ({rξ | r ∈ A and ξ ∈ B}) . Fixing B ⊂ Sn−1 Borel, it suﬃces to do the proof on the π-system of intervals (0, α) with α > 0. We have

n λn ({rξ | r ∈ (0, α) and ξ ∈ B}) = λn(α · B˜) = alpha λn(B˜) an Z a = σ(B) = rn−1 dr · σ(B) = ρ((0, α))σ(B). n 0 For uniqueness, suppose m(A × B) = ρ(A)σ(B) = ρ(A)˜σ(B). Picking A = (0, 1), we have σ(B) = σ˜(B) for all Borel sets B ⊂ Sn−1. n To show that the integral is correct, it is enough to consider f = χM for M ⊂ R Borel and then apply the monotone class theorem. Then N = M\{0}. By Fubini, Z Z χM dλn = χN dβn n n R R \{0} Z −1 n = (χN ◦ Φ ) ◦ Φ d (βn|R \{0}) n R \{0} Z = χN (rξ) dm(r, ξ) (0,∞)×Sn−1 Z ∞ Z n−1 = χM (rξ)r dσ(ξ) dr. 0 Sn−1

29 2.8 Polar coordinates MATH 245A (17F)

Proposition 2.8.7. 1. σ is invariant under rotations.

2. If n = 2 and E ⊂ S1 is an arc of angle α, then σ(E) = α (arc length measure).

Lemma 2.8.8. Z −|x|2 n/2 In = e dλn(x) = π n R √ Proof sketch. Use Fubini to show that In+1 = In · I1. That I1 = π is a well-known argument via computing I2 with polar coordinates.

Deﬁnition 2.8.9 (Γ function). The Γ function on positive reals is

Z ∞ Γ(z > 0) = tz−1e−t dt. 0

This admits an analytic continuation to a meromorphic function on the complex plane.

Proposition 2.8.10. 1. Γ(z + 1) = zΓz;

2. Γ(z)Γ(1 − z) = π/ sin(πz);

3. Γ(n) = (n − 1)! for n ∈ N; √ 4. Γ(1/2) = π.

Proposition 2.8.11. For n ∈ N, 2πn/2 σ(Sn−1) = Γ(n/2)

If Bn = {x ∈ Rn | |x| < 1}, then πn/2 λ ( n) = . n R Γ(n/2 + 1)

Proof. We compute Z n/2 −|x|2 In = π = e dλn(x) n R ∞ Z Z 2 = e−r rn−1 dσ(ξ) dr 0 Sn−1 ∞ Z 2 = σ(Sn−1) e−r rn−1 dr 0 Z ∞ 1 = σ(Sn−1) e−xxn/2−1 dx, 0 2

30 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis which gives the ﬁrst formula. For the second formula,

n n λn(B ) = λn(B \{0}) n−1 = λn(S˜ ) 1 = σ(Sn−1) n 1 2πn/2 = n Γ(n/2) πn/2 = . Γ(n/2 − 1)

Example 2.8.12. For the ﬁrst few n,

4π π2 λ ( 1) = 2, λ ( 3) = , λ ( 4) = . 1 B 3 B 3 4 B 2 For odd dimensions, it is helpful to use the functional equation to reduce the Γ functions to the case Γ(1/2).

2.9 THE TRANSFORMATION FORMULA

Definition 2.9.1 (C1-diffeomorphism). Let U, V ⊂ Rn be open. A C1-diffeomorphism T : U → V is a bijection such that T and T −1 are both C1-smooth.

n m 1 Lemma 2.9.2. Let U ⊂ R and T : U → V ⊂ R be C -smooth with T (x) = (y1(x), . . . , ym(x)). 1 Then each component function yi is C -smooth, and the derivative of T is the Jacobian matrix   ∂y1 (x) ··· ∂y1 (x) ∂x1 ∂xn  . . .  DT (x) =  . .. .  .   ∂ym (x) ··· ∂ym (x) ∂x1 ∂xn Deﬁnition 2.9.3 (Jacobian). The Jacobian (determinant) is the determinant of the Jacobian matrix, JT (x) = det(DT (x)). Lemma 2.9.4 (Chain rule). Let T : U → V and S : V → W be C1-smooth. Then

D(S ◦ T )(x) = DS(T (x)) ◦ DT (x).

Corollary 2.9.5. If T : U → V is a C1-diﬀeomorphism, then DT −1(T (x)) = (DT (x))−1.

Lemma 2.9.6 (Whitney decomposition). Let Ω (Rn be non-empty and open. Then there is a S countable collection of cubes Qk such that Ω = k Qk and the Qk have disjoint interiors. Moreover, we can ensure that diam(Qk) ≤ dist(Qk, ∂Ω) ≤ 4 diam(Qk).

31 2.9 The transformation formula MATH 245A (17F)

Proof. To be written.

Theorem 2.9.7 (Transformation formula). Let U, V ⊂ Rn be open and T : U → V be a C1- diﬀeomorphism. If f is a (Lebesgue) measurable function on V taking values in [0, ∞], or if f is an integrable function on V taking values in C, then f ◦ T is measurable / integrable, Z Z f dλn = (f ◦ T )|JT | dλn. V U

Proof (outline). Since T is a homeomorphism, E ⊂ U is Borel if and only if T (E) ⊂ V is Borel. Furthermore, since T is a C1-diﬀeomorphism, N ⊂ U is a null set if and only if T (N) ⊂ V is a null set. Thus M ⊂ U is measurable if and only if T (M) is measurable, and f is a measurable function on V if and only if f ◦ T is a measurable function on U. Since |Jt| is continuous, (f ◦ T )|JT | is measurable. (These statements all work with measurable replaced by integrable.) First we show that if Q ⊂ U is a cube, then Z λn(T (Q)) ≤ |JT | dλn. Q

For this, decompose Q into small cubes Q1,...,QN on which T behaves like an aﬃne map, i.e. for x ∈ Qi, we have

T (x) = T (xi) + Ai(x) + o(|x − xi|),Ai(x) = T (xi) + DT (xi)(x − xi). where xi is the center of the cube Qi. Since the cubes overlap on sets of measure zero, which are preserved by C1-diﬀeomorphisms, we can say

N N X X λn(T (Q)) = λn(T (Qi)) . λn(Ai(Qi)) i=1 i=1 N N X X Z = |det(DT (xi))| · λn(Qi) = |JT (xi)| dλn i=1 i=1 Qi N X Z Z ≈ |JT | dλn = |JT | dλn. i=1 Qi Q Next we show that Z Z f dλn ≤ (f ◦ T ) · |JT | dλn V U for all measurable f ≥ 0 on V . This is true if f is the characteristic function of T (Q) for a cube Q ⊂ U. Since every open set in Rn can be decomposed into cubes with non-overlapping interiors (Whitney cube decomposition), the claim also holds for f = χW whenever W ⊂ V is open. By outer regularity and a limiting argument, the claim then holds for f = χT (M) whenever M ⊂ U is measurable. Then it holds for simple functions, and ﬁnally by monotone convergence it follows for arbitrary non-negative measurable functions. This claim implies the result with an inequality −1 in one direction, with g = (f ◦ T )|JT | ≥ 0. For the other direction, we apply the claim for T and g = (f ◦ T )|JT |.

32 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Corollary 2.9.8. If E ⊂ U is measurable, then T (E) ⊂ V is measurable and Z λn(T (E)) = |JT | dλn. E

MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

3 SIGNED AND COMPLEX MEASURES

3.1 SIGNED MEASURES

Deﬁnition 3.1.1 (Signed measure). Let (X, A) be a measurable space. A signed measure on (X, A) is a function µ : A → R such that

(i) µ(∅) = 0; (ii) µ takes at most one of the values +∞ or −∞;

(iii) if An ∈ A are pairwise disjoint, then

∞ ! ∞ [ X µ An = µ(An); n=1 n=1

in particular, the sum always converges in R. Example 3.1.2. 1. Every measure is a signed measure. For emphasis, we may refer to measures in the previous sense as positive measures.

2. Let ν be a positive measure on (X, A). We say that a measurable function f : X → R is ν-integrable in the extended sense if Z Z f + dν < ∞ or f − dν < ∞.

If f is ν-integrable in the extended sense, then Z Z Z µ(A) = f dν = f + dν − f − dν A A A is a signed measure. Definition 3.1.3 (Mutually singular measures). Let (X, A) be a measurable space and µ, ν are positive measures. We say that µ and ν are mutually singular, written µ ⊥ ν, if there exist disjoint E,F ∈ A such that E ∪ F = X and µ(E) = ν(F ) = 0. Definition 3.1.4 (Totally positive set). Let (X, A) be a measurable space and µ be a signed measure. A set P ∈ A is (totally) positive if µ(A) ≥ 0 for all A ∈ A with A ⊂ P . One can similarly define (totally) negative sets and total null sets. Lemma 3.1.5. Let µ be a signed measure on (X, A).

1. If An % A with An,A ∈ A, then µ(An) → µ(A).

2. If An & A with An,A ∈ A and µ(A1) ∈ R, then µ(An) → µ(A). 3. Measurable subsets and countable unions of positive sets are positive. 4. If µ < +∞ and µ(A) 6= −∞, then A contains a positive subset P with µ(P ) ≥ µ(A).

35 3.1 Signed measures MATH 245A (17F)

Proof. 1. Omitted. 2. Omitted. 3. Omitted.

4. For each > 0, we claim that there exists A ⊂ A in A with µ(A) ≥ µ(A) such that if B ∈ A and B ⊂ A, then µ(B) ≥ −. If A = A satisfies the required properties, then we are done. Otherwise, we can find B1 ⊂ A in A such that µ(B1) < −, and then redefine A = A\B1. If this satisfies the required properties, then we are done, otherwise we take B2 ⊂ A = A\B1 in A such that µ(B2) < −, and then set A = A\(B1 ∪ B2). Continuing in this manner, we either succeed in finding the required A or we find pairwise disjoint Bk ⊂ A S in A with µ(Bk) < −. Then if B = k Bk, we have µ(B) = −∞ by countable additivity, and µ(A\B) 6= +∞ by assumption, so

µ(A) = µ(B) + µ(A\B) = −∞,

a contradiction.

Using the claim, we construct measurable sets A ⊃ A1 ⊃ A2 ⊃ · · · such that µ(An) ≥ µ(An−1) for each n and µ(B) ≥ −1/n for any measurable B ⊂ An. (Since µ(An) 6= ±∞ for each n, as µ is never +∞ and µ(An) ≥ µ(A) > −∞, the claim applies to each An to give us the required T An+1.) We claim that P = n An ∈ A is positive with µ(P ) ≥ µ(A). For the second part, we have An & P and µ(A1) 6= ±∞, so by continuity from above,

µ(P ) = lim µ(An) ≥ µ(A), n→∞

as µ(An) ≥ µ(A) for each n. To see that P is positive, if B ⊂ P , then B ⊂ An for each n, so µ(B) ≥ −1/n for each n, hence µ(B) ≥ 0.

Theorem 3.1.6 (Hahn decomposition). Let µ be a signed measure on (X, A). Then there exist a positive set P and a negative set N such that P ∩ N = ∅ and P ∪ N = X. Moreover, if P 0,N 0 is another such pair, then the symmetric diﬀerences P 4P 0 and N4N 0 are total null.

Proof. Without loss of generality, suppose µ does not take the value +∞, and let

s = sup{µ(A) | A ∈ A} ∈ [0, +∞].

Let An ∈ A be measurable sets with µ(An) → s. By the lemma, for each n, there is a positive set S Pn ⊂ An with µ(Pn) ≥ µ(An). Then P = n Pn is positive and µ(P ) ≥ µ(Pn) ≥ µ(An) for each n, so we have s ≥ µ(P ) ≥ lim µ(An) = s. n→∞ Thus µ(P ) = s < +∞. The set N = X\P ∈ A is negative, as otherwise there exists E ⊂ N with µ(E) > 0, and then µ(P ∪ E) = µ(P ) + µ(E) > s, a contradiction. Hence we have existence for the required decomposition. For uniqueness, P 0\P ⊂ P 0 ∩ N is positive and negative, hence total null. Similarly P \P 0 is total null, so P 4P 0 is total null. The same proof works for N4N 0.

36 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Theorem 3.1.7 (Jordan decomposition). Let µ be a signed measure on (X, A). Then there exist unique positive measures µ+ and µ− such that µ = µ+ − µ− and µ+ ⊥ µ−.

Proof. Let X = P ∪ N be a Hahn decomposition for µ and deﬁne

µ+(A) = µ(A ∩ P ) and µ−(A) = −µ(A ∩ N).

It is clear that this satisﬁes the required conditions, so we have proved existence. For uniqueness, suppose that in addition to µ = µ+ −µ− as above, we have µ = ν+ −ν− for positive measures ν+ ⊥ ν−. Let E,F be disjoint with E ∪ F = X and ν+(E) = ν−(F ) = 0. For A ∈ A with A ⊂ E, we have µ(A) = µ(A ∩ E) + µ(A ∩ F ) = −ν−(A ∩ E), so E is a negative set for µ. Similarly, F is a positive set for µ, so X = E∪F is a Hahn decomposition for µ. By the uniqueness of Hahn decomposition, P 4F and N4E are total null sets for µ. Hence for A ∈ A, ν+(A) = ν+(A ∩ E) + ν+(A ∩ F ) = µ(A ∩ F ) = µ(A ∩ P ) = µ+(A), so ν+ = µ+, and similarly ν− = µ−. Deﬁnition 3.1.8 (Variations). The positive measures µ+ and µ− are the positive and negative variations of µ. The total variation of µ is the positive measure |µ| = µ+ + µ−. Remark 3.1.9. For a given A ∈ A, we have |µ(A)| ≤ |µ|(A), but equality might not hold.

3.2 THE RADON-NIKODYM THEOREM

Definition 3.2.1 (Finite signed measure). A signed measure µ is said to be finite if |µ| is finite as a positive measure. Similarly, µ is σ-finite if |µ| is σ-finite.

Notation. Let (X, A) be a measurable space, ν be a positive measure on X, and f : X → R be ν-integrable in the extended sense. We write dµ = f dν to mean that µ is the signed measure R deﬁned by µ(A) = A f dν. Proposition 3.2.2. If dµ = f dν, then {f = 0} is a total null set for µ.

Proof. The sets P = {f ≥ 0} and N = {f < 0} give a Hahn decomposition for µ, as do P 0 = {f > 0} and N 0 = {f ≤ 0}. Hence P 4P 0 = N4N 0 = {f = 0} is total null. Proposition 3.2.3. If dµ = f dν, then d|µ| = |f| dν.

Proof. The sets P = {f ≥ 0} and N = {f < 0} give a Hahn decomposition for µ, so the Jordan decomposition µ = µ+ − ν− satisﬁes dµ+ = f + dν and dµ− = f − dν. Then

d|µ| = dµ+ + dν− = f + dν + f − dν = |f| dν.

Deﬁnition 3.2.4 (Absolute continuity). Let (X, A) be a measurable space, µ be a positive measure, and ν be a signed measure. We say that ν is absolutely continuous with respect to µ, written ν µ, if ν(A) = 0 for A ∈ A whenever µ(A) = 0.

37 3.2 The Radon-Nikodym theorem MATH 245A (17F)

Proposition 3.2.5. 1. If ν is ﬁnite, then ν µ if and only if for all > 0, there exists δ > 0 such that µ(A) < δ implies |ν(A)| < for all A ∈ A.

2. ν µ if and only if ν+ µ and ν− µ, or equivalently if and only if |ν| µ.

3. If ν is a signed measure, µ is a positive measure, ν µ, and ν ⊥ µ, then ν = 0.

Lemma 3.2.6. Let ν, µ be positive measures on (X, A). Then either ν ⊥ µ or there exist > 0 and E ∈ A such that µ(E) > 0 and ν ≥ µ on E.

S Proof. For each n, let X = Pn ∪ Nn be a Hahn decomposition of ν − (1/n)µ. Deﬁne P = n Pn and N = X\P = ∩n(X\Pn). Then N ⊂ Nn and Nn is totally negative for ν − (1/n)µ, so 1 1 0 ≥ ν(N) − µ(N) =⇒ 0 ≤ ν(N) ≤ µ(N), n n for each n. This shows that ν(N) = 0. If µ(P ) = 0, then ν ⊥ µ. Otherwise, µ(P ) > 0, so µ(Pn) > 0 for some n. Since Pn is totally positive for ν − (1/n)µ, we can take E = Pn and = 1/n.

Theorem 3.2.7 (Lebesgue-Radon-Nikodym). Let (X, A) be a measurable space, µ be a σ-finite positive measure, and ν be a σ-finite signed measure. Then there are unique σ-finite signed measures λ, ρ with ν = λ + ρ such that λ ⊥ µ and ρ µ. Moreover, there exists a function f : X → R which is µ-integrable in the extended sense such that dρ = f dµ, and f is unique up to equality µ-a.e.

Proof. First suppose that µ and ν are ﬁnite positive measures, and let Z F = f : X → [0, ∞] measurable | f dµ ≤ ν(E) for all E ∈ A . E We will construct f ∈ F which is maximal in the sense that if g ∈ F, then µ({g ≥ f}) = 0. First, we note that F 6= ∅ as 0 ∈ F, so Z a = sup f dµ | f ∈ F > −∞.

Since each of these integrals is bounded above by ν(X) < +∞, we have that a is ﬁnite. Pick fn ∈ F R such that fn dµ → a as n → ∞.

Deﬁne the functions gn = max(f1, . . . , fn). We claim that gn ∈ F for each n. To see this, it is enough to show that if h1, h2 ∈ F, then h = max(h1, h2) ∈ F. Let A = {h1 > h2} ∈ A. For E ∈ A, Z Z Z h dµ = h1 dµ + h2 dµ ≤ ν(E ∩ A) + ν(E ∩ (X\A)) = ν(E), E E∩A E∩(X\A) so h ∈ F, as required.

By construction, gn %, so we can deﬁne f to be the pointwise limit. Then f is measurable, and to see that f ∈ F, we have by the monotone convergence theorem that Z Z Z f dµ = lim gn dµ = lim gn dµ ≤ ν(E) E E n→∞ n→∞ E

38 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis for all E ∈ A. In the case E = X, we obtain R f dµ = a < +∞, so f takes the value +∞ only on a set of measure zero. Redefining f by setting f(x) = 0 wherever it was +∞ before, we can suppose f takes values in [0, ∞). Consider the finite measure ρ defined by Z ρ(A) = f dµ A for A ∈ A, so dρ = f dµ. Since f ∈ F, the difference λ = ν − ρ is a finite positive measure. By construction, ρ µ. We claim that λ ⊥ µ. Otherwise, by the lemma, there exist E ∈ A and > 0 such that µ(E) > 0 and λ ≥ µ on E. Then

ν(A) ≥ µ(A) + ρ(A) for all A ∈ A with A ⊂ E, so Z Z ν(A) ≥ χE dµ + f dµ. A A Hence f + χ ∈ F, but E Z (f + χE) dµ = a + µ(E) > a, X contradicting the definition of a as the supremum of all such integrals. Thus we have shown existence in the case that µ and ν are finite. If µ and ν are σ-finite positive measures, then we can decompose X into countably many pairwise disjoint measurable pieces Xn on each of which both µ and ν are finite positive measures. Let ν = λn + ρn be a decomposition on Xn with λn ⊥ µ and ρn µ on Xn, then define

∞ ∞ X X λ(A) = λn(A ∩ Xn) and ρ(A) = ρn(A ∩ Xn) n=1 n=1 for A ∈ A. The decomposition ν = λ + ρ has the required properties. Finally, in the general case where ν need not be a positive measure, let ν = ν+ − ν− be the Jordan decomposition of ν and write ν± = λ± − ρ±. Then for λ = λ+ − λ− and ρ = ρ+ − ρ−, the required decomposition is ν = λ + ρ.

For uniqueness, first suppose µ and ν are finite. Given two decompositions ν = λ1 + ρ1 = λ2 + ρ2, it must be that λ1, λ2, ρ1, ρ2 are all finite, so then λ = λ1 − λ2 is mutually singular with µ, but also λ = ρ2 − ρ1 µ, so λ = 0, which implies λ1 = λ2 and ρ1 = ρ2. Then d(ρ1 − ρ2) = (f1 − f2) dµ = 0, so f1 − f2 is zero µ-a.e. In the general case of σ-finite measures, we use the same idea of partitioning the space into subsets on which the measures are finite.

Deﬁnition 3.2.8 (Lebesgue decomposition). A decomposition ν = λ + ρ as in the theorem is a Lebesgue decomposition for ν.

Definition 3.2.9 (Radon-Nikodym derivative). Let µ be a σ-finite positive measure and ν be a σ-finite signed measure with ν µ. The function f for which dν = f dµ (unique up to modification on a set of measure zero) is the Radon-Nikodym derivative of ν with respect to µ.

39 3.3 Complex measures MATH 245A (17F)

3.3 COMPLEX MEASURES

Deﬁnition 3.3.1 (Complex measure). Let (X, A) be a measurable space. A complex measure ν is a function ν : A → C such that

(i) ν(∅) = 0; S (ii) if An ∈ A are pairwise disjoint and A = n An, then

∞ X ν(A) = ν(An). n=1

If ν is a complex measure, then ν = νr + iνi for signed measures νr, νi defined by νr(A) = Re ν(A) and νi = Im ν(A). Since ν only takes finite values, both νr and νi take values in R, hence are finite signed measures. Theorem 3.3.2 (Lebesgue-Radon-Nikodym for complex measures). Let (X, A) be a measurable space, µ be a σ-finite positive measure, and ν be a complex measure. Then ν has a unique Lebesgue decomposition ν = λ + ρ with λ ⊥ µ and ρ µ. Moreover, there is a Radon-Nikodym derivative f ∈ L1(µ) for which dρ = f dµ, and f is unique µ-a.e.

Proof. This follows from applying Lebesgue-Radon-Nikodym to νr and νi. Deﬁnition 3.3.3 (Total variation for complex measures). Let ν be a complex measure on (X, A). The total variation of ν is

( ∞ ∞ ) X [ |ν|(A) = sup |ν(An)| | An ∈ A pairwise disjoint and An = A . n=1 n=1 Remark 3.3.4. For signed measures, this agrees with our earlier deﬁnition of the total variation (Homework B1 Problem 2). However, for complex measures, we need not have |ν|= 6 |νr| + |νi|. Proposition 3.3.5. If ν is a complex measure on (X, A), then |ν| is a ﬁnite positive measure.

Proof. The finite positive measure µ = |νr| + |νi| satisfies ν µ, so there is a Radon-Nikodym derivative f ∈ L1(µ) such that dν = f dµ, i.e. Z ν(A) = f dµ. A We claim that Z |ν|(A) = |f| d mu A for A ∈ A, i.e. d|ν| = |f| dµ, from which it follows that |ν| is a finite positive measure. S First, let A ∈ A and A = n An be a partition of A into countably many measurable sets. Then ∞ ∞ Z ∞ Z Z ∞ ! Z X X X X |ν(An)| = f dµ ≤ |f| dµ = |f|χAn dµ = |f| dµ,

n=1 n=1 An n=1 An n=1 A

40 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis where in the second to last equality, we use the monotone convergence theorem. To obtain the reverse inequality, let A ∈ A and > 0. Since simple functions are dense in L1(µ), PN there exists s = k=1 ckχBk in standard representation such that Z |f − s| dµ < .

Let N [ B = Bk,A0 = A ∩ (X\B),Ai = A ∩ Bi (i = 1,...,N). k=1

Then A0,...,AN partition A into pairwise disjoint measurable sets, and

N N Z Z N Z Z X X X |ν(Ai)| = (f − s) dµ + s dµ ≥ s dµ − |f − s| dµ

i=0 i=0 Ai Ai i=0 Ai N X Z Z ≥ |s| dµ − ≥ |f| dµ − 2, i=0 Ai A where in the second to last inequality, we use the fact that s is constant on each Ai. Since is arbitrary, it follows that Z |ν|(A) ≥ |f| dµ. A

MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

4 MORE ON LP SPACES

4.1 BOUNDED LINEAR MAPS AND DUAL SPACES

Throughout, let F = R or F = C. Deﬁnition 4.1.1 (Bounded linear map). Let X and Y be normed vector spaces over F and let T : X → Y be a linear map. We say that T is bounded if there exists a constant C ≥ 0 such that kT (x)k ≤ Ckxk for all x ∈ X. Theorem 4.1.2. Let X and Y be normed vector spaces and T : X → Y be a linear map. Then the following are equivalent:

1. T is continuous; 2. T is continuous at 0; 3. T is bounded.

Proof. 1 =⇒ 2. Obvious. 2 =⇒ 3. Since T is linear, T (0) = 0, so if T is continuous at 0, then there exists δ > 0 such that kxk ≤ δ implies kT (x)k ≤ 1. If u ∈ X is an arbitrary non-zero vector, then deﬁne x = (δ/kuk)u; we have kxk = δ, so kT (x)k ≤ 1 =⇒ kT (u)k ≤ (1/δ)kuk. 3 =⇒ 1. If T is bounded, then there exists C ≥ 0 such that kT (x)k ≤ Ckxk for all x ∈ X. By linearity, kT (x) − T (y)k = kT (x − y)k ≤ Ckx − yk, so T is Lipschitz, hence continuous.

Example 4.1.3. 1. If X and Y are finite-dimensional normed vector spaces, then any linear map T : X → Y is bounded. 2. On the space of smooth functions on [0, 1] with the supremum norm, the differentiation operator is unbounded. Definition 4.1.4 (Operator norm). Let T : X → Y be a bounded linear map. The operator norm of T is kT k = inf{C | kT (x)k ≤ Ckxk for all x ∈ X}. Proposition 4.1.5. kT (x)k kT k = sup | x ∈ X\{0} = sup{kT (x)k | x ∈ X and kxk ≤ 1}. kxk

Proposition 4.1.6. For vector spaces X and Y over F, the set L(X,Y ) of bounded linear operators X → Y is a vector space over F, and the operator norm is a norm on L(X,Y ). Proposition 4.1.7. If T ∈ L(X,Y ),S ∈ L(Y,Z), then S ◦ T ∈ L(X,Z) and kS ◦ T k ≤ kSk · kT k. Proposition 4.1.8. Let X be a normed vector space and let Y be a Banach space. Then L(X,Y ) is a Banach space.

43 4.1 Bounded linear maps and dual spaces MATH 245A (17F)

Proof. Let (Tn) be a Cauchy sequence in L(X,Y ), so for any > 0, there exists N such that for all n, m ≥ N, kTn − Tmk < . Then for each x ∈ X, the sequence (Tn(x)) is Cauchy in Y , so it has a limit. Hence deﬁne a function T : X → Y pointwise by

T (x) = lim Tn(x). n→∞

To see that Tn → T in operator norm, ﬁx > 0 and let x ∈ X with kxk ≤ 1. Then there exists N such that for all n ≥ N, we have

k(Tn − T )(x)k ≤ k(Tn − TN )(x)k + k(TN − T )(x)k ≤ kTn − TN k + k(TN − T )(x)k < 2, hence kTn − T k ≤ 2.

Deﬁnition 4.1.9 (Dual space). Let X be a normed vector space over F. The space L(X, F) of bounded linear functionals is the dual space of X, denoted X∗.

By construction, X∗ is a Banach space.

Theorem 4.1.10 (Hahn-Banach). Let X be a normed vector space over F and let M ⊂ X be a linear subspace. Let f : M → F be a bounded linear functional. Then there exists a bounded linear functional F : X → F such that F |M = f and kF k = kfk.

Proof. First suppose F = R. If kfk = 0, then f = 0 on M, so we can take F = 0 on X. Otherwise, without loss of generality, kfk = 1. Let Z be the set of pairs (U, g), where U ⊂ X is a subspace with M ⊂ U and g : U → R is a bounded linear functional with g|M = f and kgk = kfk = 1; then Z is non-empty, as (M, f) ∈ Z. A partial order can be deﬁned on Z by saying that (U, g) ≤ (U 0, g0) 0 0 if U ⊂ U and g |U = g. With this partial order, every chain in Z has an upper bound, so by Zorn’s lemma, Z has a maximal element (V,F ). We claim that V = X.

Suppose otherwise, and let x0 ∈ X\V , without loss of generality with kx0k = 1. Take W to be the span of V and x0. We wish to extend F to some G : W → R by taking G(x0) = α for some α such that kGk = 1 and extending linearly. To see that such an α exists, note that we already have kF k = 1 on V , so we only require that |G(v + tx0)| ≤ kv + tx0k, or equivalently,

−kv + tx0k ≤ F (v) + tα ≤ kv + tx0k.

We already have this inequality for t = 0 independent of α, and for t 6= 0, rearranging tells us that we must satisfy

−ku + x0k − F (u) ≤ α ≤ ku + x0k − F (u)(u = v/t) for all u ∈ V . It suﬃces to show that −ku1 + x0k − F (u1) ≤ ku2 + x0k − F (u2) for all u1, u2 ∈ V . Indeed, we have

F (u2) − F (u1) = F (u2 − u1) ≤ ku2 − u1k = k(u2 + x0) − (u1 + x0)k ≤ ku2 + x0k + ku1 + x0k, so the required α exists. We thus have a contradiction to the maximality of (V,F ). If F = C, then we can also regard X as a real normed space. Consider R-linear functionals X → R, i.e. real-valued functionals F with F (αx) = αF (x) only required for α ∈ R. Given a (C-linear)

44 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

functional F : X → C, it is immediate that u = Re F is an R-linear functional. Conversely, if we have an R-linear functional u : X → R, we obtain the corresponding C-linear functional F (x) = u(x) − iu(ix).

Looking at operator norms, for all x ∈ X,

|u(x)| = |Re F (x)| ≤ |F (x)| ≤ kF kkxk so kuk ≤ kF k. Conversely, for any x ∈ X, pick α ∈ C with |α| = 1 so that αF (x) ∈ R. Then |F (x)| = |αF (x)| = |F (αx)| = |u(αx)| ≤ kukkαxk = kukkxk.

We have thus showed that kuk = kF k. Returning to the theorem, let f : M → C be a C-linear functional and v = Re f : M → R. Then kfk = kvk, and by Hahn-Banach for R-linear functionals, we can extend v to an R-linear function u : X → R with kuk = kvk = kfk. Deﬁne F : X → C from u as above. Then F is a C-linear functional with kF k = kuk = kfk, and F extends f. Corollary 4.1.11. For each x ∈ X,

kxk = sup{|f(x)| | f ∈ X∗ with kfk ≤ 1}, and this supremum is attained as a maximum.

Proof. The result is clear for x = 0. Otherwise, if kfk ≤ 1, then |f(x)| ≤ kfkkxk ≤ kxk, so

kxk ≥ sup{|f(x)| | f ∈ X∗ with kfk ≤ 1}.

For the other direction, let M = Fx and consider the bounded linear functional g : M → F given by g(λx) = λkxk. This has kgk = 1, and by Hahn-Banach, we can extend g to a functional f ∈ X∗ with kfk = 1. Then |f(x)| = |g(x)| = kxk. Remark 4.1.12. In general, kfk = sup{|f(x)| | x ∈ X with kxk ≤ 1}, but the supremum is not generally attained as a maximum if X is inﬁnite-dimensional.

Definition 4.1.13 (Reflexive space). The bilinear form X ×X∗ → F given by hx, fi = f(x) defines an injection ev : X → X∗∗ by x 7→ hx, i. We say that X is reflexive if ev is an isomorphism.

Since the dual of any normed space is a Banach space, all reflexive spaces are Banach. Example 4.1.14. Every finite-dimensional normed vector space is reflexive.

4.2 THE DUAL OF LP

Let (X, A, µ) be a measure space, 1 ≤ p ≤ ∞, and q be the conjugate exponent of p. For g ∈ Lq(µ), deﬁne Z p Φg : L (µ) → C, Φg(f) = fg dµ.

p By H¨older’sinequality, Φg is a bounded linear functional on L (µ).

45 4.2 The dual of Lp MATH 245A (17F)

Theorem 4.2.1. Let 1 ≤ p < ∞ and q be the conjugate exponent of p, and suppose that (X, A, µ) is σ-ﬁnite. Then for each bounded linear functional Φ: Lp(µ) → C, there exists g ∈ Lq(µ) such that Φ = Φg. Moreover, g is unique up to a set of measure zero and kΦk = kgkq.

Proof (outline). Let 1 ≤ p < ∞, so 1 < q ≤ ∞. p ∗ p For existence, let Φ ∈ (L ) be arbitrary. If µ is a ﬁnite measure, then χA ∈ L for all A ∈ A, so we can deﬁne a complex measure ν(A) = Φ(χA). Indeed, ν(∅) = 0, and to see that ν is countable S additive, let An ∈ A be pairwise disjoint with A = n An and Bn = A1 ∪ · · · ∪ An. We use continuity of Φ to get

ν(A) = Φ(χA) = lim Φ(χB ) n→∞ n

= lim Φ(χA + ··· + χA ) n→∞ 1 n

= lim ν(A1) + ··· + ν(An) n→∞ ∞ X = ν(An). n=1

We have ν µ, as if µ(A) = 0 for some A ∈ A, then χA = 0 almost everywhere, i.e. χA = 0 in p 1 L , so ν(A) = Φ(χA) = 0. By Radon-Nikodym, there exists g ∈ L such that Φ(χA) = ν(A) = R R A g dµ = χAg dµ for all A ∈ A. By linearity, this extends to all simple functions, i.e. Z Φ(s) = sg dµ for all simple functions s ∈ Lp. Now let h ∈ L∞ ⊂ Lp. Since simple functions are dense in L∞, ∞ there is a sequence sn → h in L . Hence Z Z p p p p ksn − hkp = |sn − h| dµ ≤ ksn − hk∞ dµ = µ(X)ksn − hk∞ → 0,

p ∞ so sn → h in L . Hence for h ∈ L . Z Z Φ(n) = lim Φ(sn) = lim sng dµ = hg dmu, n→∞ n→∞ where to see that the last step is valid, we have

Z Z Z

lim sup sng dµ − hg dµ ≤ lim sup |sn − h||g| dµ n→∞ n→∞ Z = lim sup ksn − hk∞ |g| dµ = 0. n→∞

We now claim that g ∈ Lq. If 1 < p < ∞, then 1 < q < ∞. There exists a measurable function α on X such that |α| = 1 and g = α|g| (exercise). Let En = {|g| ≤ n} ∈ A and

46 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

q−1 ∞ p fn = α|g| χEn ∈ L ⊂ L . We can now apply Φ to get Z Z q q−1 |g| dµ = α|g| g dµ En En Z

= fng dµ

= |Φ(fn)| ≤ kΦkkfnkp Z 1/p = kΦk |g|(q−1)p En Z 1/p = kΦk |g|q , En so

kgχEn kq ≤ kΦk.

Taking the limit as n → ∞, we get kgkq ≤ kΦk < ∞. Since Φg = Φ for simple functions, which p are dense in L , continuity gives us Φg = Φ everywhere. For the norm we have kgkq ≤ kΦk = kΦgk ≤ kgkq. In the case p = 1 and q = ∞, one can show that kgk∞ ≤ kΦk, and then the argument proceeds in the same way. If µ is σ-ﬁnite, then there exist measurable sets En → x of ﬁnite measure. Use a limiting argument and monotone convergence. q R R p For uniqueness, suppose that Φg = Φg˜ for g, g˜ ∈ L . Then fg dµ = fg˜ dµ for all f ∈ L , R p R so equivalently, f · (g − g˜) dµ for all f ∈ L . Then 0 = χE(g − g˜) dµ whenever E ∈ A and µ(E) < ∞, so g − g˜ = 0 almost everywhere.

q p ∗ p ∗ Remark 4.2.2. 1. The map T : L (µ) → L (µ) given by T (g) = Φg ∈ L (µ) is a well-deﬁned surjective linear isometry which identiﬁes Lp(µ)∗ with Lq(µ) for 1 ≤ p < ∞. 2. On the other hand, L∞(µ)∗ need not be isomorphic with L1(µ). For example, (l∞)∗ 6= l1, where lp is the Lp norm on N with the counting measure. Theorem 4.2.3 (Basic 5B-covering lemma). Let (X, d) be a metric space and let B be a collection of open balls in X with uniformly bounded radii. Then there exists a collection B˜ of pairwise disjoint balls with [ [ B ⊂ 5B. B∈B B∈B˜

Proof. See Homework B4 Problem 3.

4.3 THE HARDY-LITTLEWOOD MAXIMAL FUNCTIONS

Deﬁnition 4.3.1 (Locally integrable function). A measurable function f : Rn → C is locally R n R n integrable if B|f| dλn < ∞ for all balls B ⊂ R , or equivalently, K |f| dλn < ∞ for all K ⊂ R 1 n compact. The space of locally integrable functions is denoted Lloc(R ). 1 Notation. For f ∈ Lloc, write Z 1 Z − |f| dλn = |f| dλn. B λn(B) B

47 4.3 The Hardy-Littlewood maximal functions MATH 245A (17F)

1 If f ∈ Lloc, we deﬁne the centered Hardy-Littlewood maximal function by Z (Hf)(x) = sup |f| dλn r>0 B(x,r)

There is also an uncentered Hardy-Littlewood maximal function given by Z (Mf)(x) = sup |f|, B3x B the supremum ranging over open balls.

1 n n Lemma 4.3.2. 1. If f ∈ Lloc, then Hf : R → [0, ∞] and Mf : R → [0, ∞] are measurable and Hf ≤ Mf.

2. The operators M and H are sublinear, i.e. M(f +g) ≤ M(f)+M(g) and M(αf) = |α|M(f), and similarly for H.

Proof. 1. It is clear that Hf ≤ Mf from deﬁnition. Measurability of Hf is in Folland, so only measurability of Mf is left. It suﬃces to show that {x ∈ Rn | (Mf)(x) > α} is open for each R α ∈ R. If x ∈ {Mf > α}, then there exists an open ball B with x ∈ B and B|f| > α. Then (Mf)(y) > α for all y ∈ B, and so B ⊂ {Mf > α}.

1 Remark 4.3.3. 1. For f ∈ Lloc, we have |f| ≤ Hf ≤ Mf almost everywhere, but Mf is typically not much larger than |f|. For f ∈ Lp with 1 < p ≤ ∞, we get

kMfkp ≤ C(n, p)kfkp.

2. For f ∈ Lp with 1 ≤ p ≤ ∞, we get

Z |f|p C λn{|f| > α} ≤ p ≤ p |f|>α α α

for α > 0 and C a constant independent of α.

3. In general, Mf 6∈ L1 if f ∈ L1, but Mf still satisﬁes a “weak-type L1 estimate”.

Theorem 4.3.4 (Weak-type L1-estimate for the Hardy-Littlewood maximal function). There exists a constant C = C(n) = 5n such that

C λ {Mf > α} ≤ kfk n α for α > 0 and f ∈ L1(Rn).

Proof. Let f ∈ L1 and α > 0. Let A = {x ∈ Rn | (Mf)(x) > α} and pick x ∈ A. Then we can ﬁnd n R a ball Bx ⊂ with x ∈ Bx and (Mf)(x) ≥ |f| dλn > α. R Bx

48 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

R Let B = {Bx | x ∈ A} be the collection of these balls. If B = B(a, r) ∈ B, then α < B|f| means that Z n 1 1 λn(B) = cnr < |f| ≤ kfk1 < ∞, α B α so the balls of B have uniformly bounded radii. Applying the 5B-covering lemma, there is a subfamily B˜ ⊂ B of pairwise disjoint balls with [ [ A ⊂ B ⊂ 5B. B∈B B∈B˜ ˜ ˜ Note that B is countable, say B = {Bi | i ∈ I ⊂ N}. Then ! [ λn(A) ≤ λn 5Bi i∈I X ≤ λn(5Bi) i∈I n X = 5 λn(Bi) i∈I 5n X Z ≤ |f| α i∈I Bi 5n Z = |f| S α Bi 5n ≤ kfk . α 1

Lemma 4.3.5. Let g : Rn → C and 1 ≤ p ≤ ∞. Then Z Z ∞ p p−1 |g| dλn = p α λn{|g| > α} dα. 0 Theorem 4.3.6 (Lp-boundedness of the maximal function). Let n ∈ N and 1 < p ≤ ∞. Then 1/p p·5n there exists a constant C = C(p, n) = 2 p−1 if p < ∞ (for p = ∞, we take C = 1) such that p n kMfkp ≤ Ckfkp for all f ∈ L (R ). Proof. The L∞ case is trivial. The Lp-boundedness of M is derived from an interpolation technique (see Marcinkiewicz interpolation theorem). Our basic idea is to split the function f ∈ Lp into a “small part” g and a “large part” h, then use diﬀerent estimates for g and h. We have Z Z ∞ p p−1 kMfkp = |Mf|p = p α λn{|g| > α} dα. 0

To show that kMfkp < ∞, we need good estimates for λ(α) = λn{Mf > α}. Fix α > 0. Deﬁne

g = f ◦ χ{|f|≤α/2}, h = f · χ{|f|>α/2}.

49 4.3 The Hardy-Littlewood maximal functions MATH 245A (17F)

Then f = g + h and kgk∞ ≤ α/2. We have

Z Z |f|p−1 khk1 = ≤ |f| p−1 {|f|>α/2} {|f|>α/2} (α/2) 2 p−1 Z ≤ |f|p < ∞, α so h ∈ L1. Hence α Mf ≤ Mg + Mh ≤ + Mh. 2 This implies that

λ(α) = λn{Mf > α} ≤ λn{Mh > α/2} 5n ≤ khk α/2 1 2 · 5n Z = |f|. α {|f|>α/2}

Substituting,

Z ∞ n Z ! p p−1 2 · 5 kMfkp ≤ p α |f| dα 0 α {|f|>α/2} Z ∞ Z n p−2 = 2p · 5 α χ{|f|(x)>α/2}|f| dα dλn n 0 R Z Z ∞ n p−2 = 2p · 5 |f| α χ{|f|(x)>α/2} dα dλn(x) n R 0 Z Z 2|f|(x) ! n p−2 = 2p · 5 |f|(x) α dα dλn(x) n R 0 n Z 2p · 5 p−1 p−1 = |f|(x) · 2 |f(x)| dλn(x) p − 1 n R p n Z p n 2 p · 5 p 2 p · 5 p = |f| = kfkp. p − 1 n p − 1 R

50 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

5 DIFFERENTIABILITY

5.1 LEBESGUE POINTS

Deﬁnition 5.1.1 (Lebesgue point). Let f : Rn → C be measurable. We say that x ∈ Rn is a Lebesgue point of f if Z lim |f(y) − f(x)| dλn(y) = 0. r→0+ B(x,r) Remark 5.1.2. If f is continuous, then every point is a Lebesgue point of f.

Example 5.1.3. Let f = χ[0,1]. Then the Lebesgue points of f are the points in R\{0, 1}. 1 n Theorem 5.1.4 (Lebesgue diﬀerentiation theorem). Let f ∈ Lloc(R ). Then almost every point is a Lebesgue point of f.

1 n 1 n Proof. Without loss of generality, f ∈ L (R ), as the Lebesgue points of f and f ·χB(0,n) ∈ Lloc(R ) are the same on B(0, n) for each n. For r > 0, deﬁne Z (Trf)(x) = |f − f(x)| dλn B(x,r) and T f(x) = lim sup Trf(x) ∈ [0, ∞]. r→0+ We must show that T f(x) = 0 for almost every x ∈ Rn. n 1 n n Let k ∈ N be arbitrary. Since Cc(R ) is dense in L (R ), we can ﬁnd gk ∈ Cc(R ) such that kf − gkk1 < 1/k. Since gk is continuous, T g ≡ 0. Let hk = f − gk. Then f = gk + hk with khkk1 < 1/k. We have Trf ≤ Trgk + Trhk, and so T f ≤ T hk. This gives Z Trhk(x) = |hk − hk(x)| dλn B(x,r) Z ≤ |hk| dλn + |hk(x)| B(x,r)

≤ (Mhk)(x) + |hk|(x).

Letting r → 0+, we get T hk(x) ≤ (Mhk)(x) + |hk|(x) for all x ∈ Rn. Let α > 0 be arbitrary and

A(k, α) = {Mhk ≥ α/2} ∪ {|hk| > α/2}, so then {T f < α} ⊂ A(k, α) for all k. By our weak-type L1-estimate for the maximal function and |hk|, we have 5n 1 1 2(5n + 1) λ (A(k, α)) ≤ kh k + kh k = → 0 n α/2 k 1 α/2 k 1 k α

51 5.1 Lebesgue points MATH 245A (17F) as k → ∞. Hence ∞ ! \ λn A(k, α) = 0, k=1 so we conclude that {T f > α} is Lebesgue measurable with measure zero. Since this is true for all α = 1/L with L ∈ N, we conclude that {T f > 0} has measure 0.

1 n Corollary 5.1.5. If f ∈ Lloc(R ), then Z lim f dλn = f(x) r→0+ B(x,r) for almost every x.

Proof. If x ∈ Rn is a Lebesgue point, then

Z Z

lim sup − f(x) ≤ lim sup |f − f(x)| = 0. r→0+ B(x,r) r→0+ B(x,r)

n n A family {Er}r>0 of measurable sets in R shrinks nicely to x ∈ R if

(i) Er ⊂ B(x, r) for r > 0;

(ii) there exists α > 0 such that λn(Er) ≥ αλn(B(x, r)).

Remark 5.1.6. The sets Er need not contain x.

1 n n Theorem 5.1.7. Let f ∈ Lloc(R ) and let x ∈ R be a Lebesgue point of f. Suppose {Er}r>0 is a family of sets shrinking nicely to x. Then

Z lim |f(y) − f(x)| dλn(y) = 0. r→0+ Er

Proof.

Z 1 Z |f(y) − f(x)| dλn(y) = |f − f(x)| Er λn(Er) Er 1 Z ≤ |f − f(x)| λn(Er) B(x,r) 1 Z ≤ |f − f(x)| → 0. α B(x,r)

52 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Let E ⊂ Rn be measurable. The (metric) density of E at x ∈ Rn is

λn(E ∩ B(x, r)) DE(x) = lim r→0+ λn(B(x, r)) if the limit exists. A point with DE(x) = 1 is called a Lebesgue density point of E.

n n Theorem 5.1.8. Let E ⊂ R be measurable. Then DE(x) exists for almost every x ∈ R and n DE(x) = 0 for almost every x ∈ R \E, while DE(x) = 1 for almost every x ∈ E.

1 Proof. Let f = χE ∈ Lloc. For each Lebesgue point x of f, Z λn(E ∩ B(x, r)) χE = → χE(x) ∈ {0, 1}. Bx,r λn(B(x, r)) as r → 0.

n Theorem 5.1.9. Let ν be a complex Borel measure on R and dν = dρ + f dλn be its Lebesgue n decomposition with respect to λn. Suppose for each x ∈ R we have a family {Er(x)}r>0 of Borel sets shrinking nicely to x. Then ν(E (x)) f(x) = lim r r→0+ λn(Er(x)) for λn-a.e. x.

Proof. We start by writing Z ν(Er(x)) ρ(Er(x)) = + f dλn. λn(Er(x)) λn(Er(x)) Er (x) For the second term, at each Lebesgue point x, Z lim f dλn = f(x). r→0+ Er (x) For the ﬁrst term,

ρ(Er(x)) |ρ|(Er(x)) ≤ λn(Er(x)) λn(Er(x)) 1 |ρ|(B(x, r)) ≤ · . α(x) λn(B(x, r))

Let µ = |ρ|, a ﬁnite positive Borel measure on Rn with µ ⊥ λ. Let A be a Borel set such that µ(A) = 0 and λ(Rn\A) = 0. For each k ∈ N, consider µ(B(x, r)) 1 Fk = x ∈ A | lim sup > . r→0+ λn(B(x, r)) k

Since µ is outer regular, for each > 0, there is an open set U ⊃ A such that µ(U\A) < . Then µ(U) = µ(U\A) + µ(A) < . For each x ∈ Fk, there exists a ball Bx ⊂ U of small radius (wlog

53 5.2 Complex Borel measures on R MATH 245A (17F)

bounded by 1) such that µ(Bx)/λn(Bx) > 1/k. By the 5B-covering lemma, we can ﬁnd a disjoint subfamily {Bm | m ∈ N} such that

∞ [ [ Fk ⊂ V = Bk ⊂ 5Bm.

x∈Fk m=1

Then

∞ X λn(Fk) ≤ λn(V) ≤ λn(5Bm) m=1 ∞ n X = 5 λn(Bm) m=1 ∞ ∞ ! n X n [ ≤ 5 k µ(Bm) = 5 kµ Bm n=1 m=1 n n ≤ 5 kµ(U) ≤ 5 k.

Letting → 0 and keeping k ﬁxed, we get λn(Fk) = 0. Hence

∞ ! n [ λn (R \A) ∪ Fk = 0, k=1 which implies the result.

5.2 COMPLEX BOREL MEASURES ON R

We wish to describe all complex (Borel) measures on R. The basic idea is that given a complex measure µ on R, we can deﬁne Fµ(x) = µ((−∞, x]) ∈ C. These functions will be a special type of function, namely a function of “bounded variation”.

Deﬁnition 5.2.1 (Functions of bounded variation). A function F : R → C has bounded variation (is a BV-function) if there exists a constant M > 0 such that

n X |F (xk) − F (xk−1)| ≤ M k=1 whenever n ∈ N and x0 < x1 < ··· < xn.

The set of all BV-functions on R is denoted by BV (R), or simply BV when understood. If F ∈ BV and v ∈ R, we deﬁne

( n ) X TF (x) = sup |F (xk) − F (xk−1)| | x0 < x1 < ··· < xn ≤ x , k=1 the total variation of F up to x.

54 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Theorem 5.2.2. 1. F ∈ BV if and only if Re F ∈ BV and Im F ∈ BV .

2. F : R → R is a BV-function if and only if F is the diﬀerence of two bounded increasing functions. 3. If F ∈ BV , then

F (x+) = lim F (y) and F (x−) = lim F (y) y→x+ y→x−

exist for all x ∈ R. Furthermore, F (+∞) and F (−∞) also exist. 4. If F ∈ BV , then F has at most countably many discontinuities.

Proof. 1. trivial

2. Suppose F = G − H for bounded increasing functions G, H. Let x0 < x1 < ··· < xn. Then

= G(xn) − G(x0) + H(xn) − H(x0)

≤ 2MG + 2MH ,

where |G| ≤ 2MG and |H| ≤ 2MH .

Conversely, let F : R → R be a BV-function and consider TF : R → R. If x ≤ y, then

T (x) + |F (y) − F (x)| ≤ TF (y),

so TF (x) − F (x) ≤ TF (y) − F (y),TF (x) + F (x) ≤ TF (y) + F (y).

Thus TF −F and TF +F are both bounded increasing functions, and F is half their diﬀerence. 3. This is true for all bounded increasing functions, hence for all BV-functions. 4. This is true for all bounded increasing functions (standard exercise), hence for all BV- functions.

Deﬁnition 5.2.3. A function F ∈ BV is called a normalized BV-function if F is right-continuous, i.e. F (x+) = F (x) for all x ∈ R, and F (−∞) = 0. The set of normalized BV-functions is denoted NBV .

Lemma 5.2.4. If F ∈ NBV , then TF ∈ NBV .

55 5.2 Complex Borel measures on R MATH 245A (17F)

Proof. Since TF is a bounded increasing function, TF ∈ BV . Let > 0 and b ∈ R be arbitrary. We can ﬁnd x0 < ··· < xn ≤ b such that

n X TF (b) − ≤ |F (xi) − F (xi−1)| ≤ TF (b). i=1 Then k X TF (xk) − ≤ |F (xi) − F (xi−1)| ≤ TF (xi) i=1 for all k ≤ n. To show that TF (−∞) = 0, we pick b = 0. Then for k = 0, we have

TF (x0) − ≤ 0, so 0 ≤ TF (−∞) ≤ TF (x0) ≤ . It remains to show that TF is right continuous. + To show that TF (x ) = TF (x) at x ∈ R, let b = x + 1. Throwing the point x into the sequence, we only increase the sum, but keep it below TF (b), so without loss of generality, x is one of the points + x0, . . . , xn. Since F (x ) = F (x), we can ﬁnd h ∈ (0, 1) so that |F (x + h) − F (x)| < . Without loss of generality, we can assume x + h is in the sequence and is the next point after x. If xk = x, then

k X TF (x) − ≤ |F (xi) − F (xi−1)| ≤ TF (x), i=1 so

k X TF (x) − |F (xi) − F (xi−1)| ≤ . i=1 Similarly,

k+1 X TF (x + h) − |F (xi) − F (xi−1)| ≤ . i=1

The two sums diﬀer precisely by |F (x + h) − F (x)| < , so |TF (x + h) − TF (x)| < 3.

Theorem 5.2.5 (Complex measures on R). F ∈ NBV if and only if there exists a complex Borel measure µ on R such that F = Fµ. Moreover, for a given F ∈ NBV , the corresponding complex measure µ is unique.

Proof. First, we show that if µ is a complex Borel measure, then Fµ ∈ NBV . Since Fαµ+βν = αFµ + βFν , the map µ 7→ Fµ is linear, so by splitting into real and imaginary parts, then applying Jordan decomposition, it suﬃces to suppose that µ is a ﬁnite positive Borel measure. To see that F ∈ BV , let x0 < ··· < xn be arbitrary. Then

F (xk) − F (xk−1) = µ(−∞, xk] − µ(−∞, xk−1] = µ(xk−1, xk] ≥ 0.

Hence n X |F (xk) − F (xk−1)| = µ(x0, xn] ≤ µ(R) < ∞. k=1

56 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

For right continuity, let x ∈ R be arbitrary and xn % x. Then F (xn) = µ(−∞, xn] → µ(−∞, x] = F (x) since the intervals are decreasing and have uniformly bounded measure. The proof that F (−∞) = 0 proceeds similarly.

Next, we show that if F ∈ NBV , then there exists a complex Borel measure µ for which Fµ = F . Without loss of generality, assume that F is real-valued. Since TF ∈ NBV by the lemma, we can write F = (TF +F )−TF as a difference of increasing functions, so without loss of generality, assume that F is increasing. We then attempt to define µ by µ(a, b] = F (b) − F (a). More specifically, we define a premeasure on the algebra generated by the h-intervals with this formula and use the Carathéodory extension theorem to get a measure. The proof that this has the required properties is in Homework B6. Finally, uniqueness of µ is in the same homework.

Proposition 5.2.6. Fµ is continuous if and only if µ has no atoms, i.e. points x for which µ({x}) 6= 0. Deﬁnition 5.2.7 (Absolutely continuous (AC) function). A function F : R → C is absolutely continuous if for all > 0, there exists δ > 0 so that if (a1, b1) ..., (an, bn) ⊂ R are disjoint intervals, then n n X X bi − ai < δ =⇒ |F (bi) − F (ai)| < . i=1 i=1

Denote by AC or AC(R) the set of all AC-functions on R.

Proposition 5.2.8. Let µ be a complex Borel measure on R and F = Fµ. Then µ λ1 if and only if F is AC.

1 Proof. ( =⇒ ) If µ λ1, then there exists f ∈ L (λ1) such that dµ = f dλ1. Then d|µ| = |f| dλ1, hence |µ| λ1, so for each > 0, there exists δ > 0 such that λ1(A) < δ implies |µ|(A) < for P all Borel sets A ⊂ R. Suppose (a1, b1),..., (an, bn) are disjoint intervals. If (bi − ai) < δ, S i then for A = i(ai, bi), we have λ1(A) < δ, so n n n X X X |F (bi) − F (ai)| = |µ(ai, bi]| ≤ |µ|(ai, bi) = |µ|(A) < . i=1 i=1 i=1

Note that we used absolute continuity to conclude that |µ|({bi}) = 0.

( ⇐= ) That µ λ1 is equivalent to saying that if E ⊂ R is Borel and λ1(E) = 0, then µ(E) = 0. Let E ⊂ R with λ1(E) = 0 and > 0 be arbitrary. Let δ > 0 be as in the definition of absolute continuity of F . Since |µ| and λ1 are outer regular, we can find open sets Vn containing E such that |µ|(Vn) → |µ|(E) as n → ∞ and an open set U ⊃ E with λ1(U) < δ. Define Un = Vn ∩U. Then Un is open, Vn ⊃ Un ⊃ E, and |µ|(Un) → |µ|(E). Furthermore, λ1(Un) < δ for each n ∈ N. Then

|µ(Un) − µ(E)| = |µ(Un\E)| ≤ |µ|(Un\E) = |µ|(Un) − |µ|(E) → 0, S so µ(Un) → µ(E). Since Un is open, we can write Un = i(ai, bi) with the (ai, bi) pairwise disjoint. Moreover, ∞ X (bi − ai) = λ1(Un) < δ. i=1

57 5.3 The fundamental theorem of calculus MATH 245A (17F)

By absolute continuity of F , for each N,

N N N N !

X X X [ > |F (bi) − F (ai)| = |µ(ai, bi]| = |µ(ai, bi)| ≥ µ (ai, bi) → |µ(Un)|. i=1 i=1 i=1 i=1 Here we have used absolute continuity of F to get that F is continuous, so µ has no atoms. Hence |µ(Un)| ≤ for all n, so |µ(E)| ≤ .

Remark 5.2.9. AC and BV are not the same class of functions. For example, F (x) = sin x is in AC, since it is Lipschitz, but F is not in BV .

Lemma 5.2.10. Let [a, b] ⊂ R be a ﬁnite interval. Then AC[a, b] ⊂ BV [a, b]. Proof. Let F ∈ AC[a, b] and pick δ > 0 for = 1 as in the deﬁnition. To show that F ∈ BV [a, b], let a ≤ x0 < ··· < xn ≤ b. By adding points spaced δ/2 apart, we may assume without loss of generality that xk+1 − xk < δ for all k and that the endpoints are in the collection. Greedily group the points starting from the left so that each group has size less than δ, so the total number of groups is at most N = b2(b − a)/δc + 1. Then

n−1 N X X X |F (xk+1) − F (xk)| = |F (xk+1) − F (x)| ≤ N. k=0 i=1 group of points Hence F ∈ BV [a, b].

5.3 THE FUNDAMENTAL THEOREM OF CALCULUS

Theorem 5.3.1. Let F : R → R be increasing. Then F 0(x) exists almost everywhere. Proof. Without loss of generality, suppose F is bounded by fixing a finite interval [−n, n] and redefining F outside [−n, n] by making F constant on each outward ray. We can recover the result for a general F by letting n → ∞, noting that a countable union of sets of measure zero is still of measure zero. Define G(x) = F (x+) for x ∈ R. Then G is increasing, bounded, and right continuous. Moreover, the set of x ∈ R where H(x) = G(x) − F (x) 6= 0 is countable. There exists a finite Borel measure µ on R such that µ(a, b] = G(b) − G(a). Then ( µ(x, x + h] h > 0, G(x + h) − G(x) = −µ(x + h, x] h < 0.

The sets (x, x + r] and (x − r, x] shrink nicely to x as r → 0. It follows that if dµ = f dλ1 + dν, then G(x + h) − G(x) µ(x, x + h] lim = = lim = f(x) h→0+ h h→0+ λ1(x, x + h] for almost every x and G(x + h) − G(x) µ(x − h, x] lim = = lim = f(x) h→0− h h→0+ λ1(x − h, x]

58 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis for almost every x. Hence G0(x) exists (and is f(x)) for almost every x. We claim that H0(x) exists and H0(x) = 0 for almost every x ∈ R. From this, it will follow that F 0(x) = G0(x) − H0(x) exists 0 for almost every x and F (x) = f(x) for such x. Let E = {xn} be an enumeration of the points P where H(x) 6= 0. Deﬁne a measure ρ = n H(xn) · δxn . This is a ﬁnite Borel measure which is singular with respect to Lebesgue measure. Then

H(x + h) − H(x) H(x + h) + H(x) ≤ h |h| 2ρ[x − |h|, x + |h|] ≤ |h| ρ[x − |h|, x + |h|] = 2 · → 0 λ1[x − |h|, x + |h|] as h → 0 for almost every x by Lebesgue diﬀerentiation. This completes the proof.

Corollary 5.3.2. Let F ∈ NBV . Then F 0(x) exists for almost every x ∈ R. Moreover, if µ is the unique complex Borel measure such that Fµ = F and dµ = f dλ1 + dν is the Lebesgue 0 0 1 decomposition of µ, then F (x) = f(x) for almost every x ∈ R. In particular, F ∈ L (λ1). Finally, if F ∈ AC ∩ NBV , then Z x Z 0 0 F (x) = F (t) dt = F dλ1 −∞ for each x ∈ R.

Proof. Everything but the last part follows from the theorem if F is increasing. In the general case, we split F into its real and imaginary parts and write each as diﬀerences of increasing functions in 0 NBV . If F ∈ AC ∩ NBV , then µ λ1, so dµ = f dλ1 = F dλ1. Hence for each x ∈ R, we have Z Z 0 F (x) = Fµ(x) = µ((−∞, x]) = f dλ1 = F dλ1.

Theorem 5.3.3 (Fundamental theorem of calculus for Lebesgue integrals). Let [a, b] ⊂ R be a ﬁnite interval and F :[a, b] → C be a function. The following are equivalent: (1) F ∈ AC[a, b]; (2) F is diﬀerentiable almost everywhere on [a, b] with F 0 ∈ L1[a, b], and Z x F (x) − F (a) = F 0(t) dt a for each x ∈ [a, b]; (3) there exists f ∈ L1[a, b] such that Z x F (x) − F (a) = f(t) dt a for x ∈ [a, b].

59 5.3 The fundamental theorem of calculus MATH 245A (17F)

Proof. By subtracting a constant, we may assume that F (a) = 0. We may extend F to R by setting F (x) = 0 for x < a and F (x) = F (b) for x > b.

(1) =⇒ (2) Since F ∈ AC[a, b], we have F ∈ BV [a, b], and so for the extended function F ∈ AC ∩ BV , we have F ∈ NBV ∩ AC. Then the corollary applies. (2) =⇒ (3) We take f = F 0.

(3) =⇒ (1) We extend f to be 0 outside [a, b]. Then f ∈ L1(R), so we can deﬁne a complex Borel measure dµ = f dλ1 on R with µ λ1. Then F = Fµ.

60 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

6 FUNCTIONAL ANALYSIS

6.1 LOCALLY COMPACT HAUSDORFF SPACES

Let X be a locally compact Hausdorﬀ space.

Proposition 6.1.1. 1. If K ⊂ U ⊂ X with K compact and U open, then there exists an open set V with V compact and K ⊂ V ⊂ V ⊂ U.

2. (Urysohn) If K ⊂ U ⊂ X with K compact and U open, then there exists a continuous function f : X → [0, 1] such that f = 1 on K and supp f ⊂ U is compact. We may write K ≺ f ≺ U.

3. (Tietze extension) If K ⊂ X is compact and f : K → C is continuous, then there exists F : X → C continuous with supp F compact and F |K = f.

4. (Partitions of unity) Let K ⊂ X be compact and U = {U1,...,Un} be an open cover of K. Then there exist continuous functions h1, . . . , hn : X → [0, 1] such that hi ≺ Ui, 0 ≤ h1 + ··· + hn ≤ 1, and h1 + ··· + hn = 1 on K. The functions h1, . . . , hn form a partition of unity on K subordinate to U.

Proof of 4. By a covering argument, we can find compact sets F1,...,Fn such that Fi ⊂ Ui for i = 1, . . . , n and K ⊂ F1 ∪ · · · ∪ Fn. By Urysohn, we may pick continuous functions gi such that Fi ≺ gi ≺ Ui. Then U = {g1 + ··· + gn > 0} ⊃ K is open, so by Urysohn again, we can find f with K ≺ f ≺ U. If g0 = 1 − f, then g0|K = 0 and g0|X\U = 1. Then g = g0 + ··· + gn > 0 on X, so we can define hi = gi/g for i = 1, . . . , n.

The set Cc(X) of continuous functions f : X → C with compact support is a vector space over C with uniform norm (or supremum norm)

kfk∞ = sup|f(x)| < ∞. x∈X

In general, Cc(X) is not a Banach space.

Definition 6.1.2 (Vanishing at infinity). A continuous function f : X → C vanishes at infinity if for all > 0, there exists a compact set K such that |f(x)| < for x ∈ X\K.

The set of continuous functions vanishing at inﬁnity is denoted C0(X).

Theorem 6.1.3. (C0(X), k k∞) is a Banach space, and Cc(X) is dense in C0(X).

Remark 6.1.4. The Alexandroff one-point compactification of X is the space Xb = X ∪ {∞}, with topology defined by saying that U ⊂ Xb is open if one of the following holds:

1. U ⊂ X and U is open in X;

2. ∞ ∈ U and Xb\U ⊂ X is compact.

If f ∈ C0(X), then f uniquely extends to Xb by setting f(∞) = 0, and conversely, if f : Xb → C is continuous with f(∞) = 0, then f restricts to a function in C0(X).

61 6.2 Weak topologies MATH 245A (17F)

Deﬁnition 6.1.5 (Complex Radon measure). A complex Radon measure µ is a complex Borel measure whose total variation measure |µ| is a (positive) Radon measure. Theorem 6.1.6 (Homework B2 Problem 1). Let M(X) denote the set of complex Radon measures on X. For µ ∈ M(X), deﬁne kµk = |µ|(X). Then M(X) is a vector space and k k is a norm.

Deﬁnition 6.1.7 (Positive linear functional). A linear functional I : Cc(X) → C is positive if I(f) ≥ 0 ≥ 0 whenever f ≥ 0. Theorem 6.1.8 (Riesz-Markov-Kakutani representation theorem). 1. There is an isometric iso- ∗ morphism M(X) → C0(X) given by Z µ 7→ Iµ : C0(X) → C,Iµ(f) = f dµ.

2. If I : Cc(X) → C is a positive linear functional (not necessarily bounded), then there exists a R unique positive Radon measure µ on X such that I(f) = f dµ for all f ∈ Cc(X).

Proof of uniqueness for 2. For uniqueness, by outer regularity, it suﬃces to show that µ(U) is determined by I for all open sets U ⊂ X. Let K ⊂ U be compact. Then there exists f ∈ Cc(X) with K ≺ f ≺ U by Urysohn, so χK ≤ f ≤ χU . This implies that Z µ(K) ≤ f dµ = I(f) ≤ µ(U).

Taking the supremum over all compact sets K, the inner regularity of U tells us that µ(U) = sup I(f). f≺U For existence, the idea is to deﬁne a set function on open sets in this way, then use this on open covers to construct an outer measure.

If X is compact, so C0(X) = C(X). This gives us the following corollary. Corollary 6.1.9. If X is compact, then C(X)∗ =∼ M(X). Remark 6.1.10. If µ is a positive Radon measure, then I : f 7→ R f dµ is a positive linear functional on Cc(X).

6.2 WEAK TOPOLOGIES

Let X be a vector space over F = R or F = C. Deﬁnition 6.2.1 (Seminorm). A seminorm on X is a map p : X → [0, ∞) such that for all x, y ∈ X and α ∈ F, we have p(αx) = |α|p(x) and p(x + y) ≤ p(x) + p(y). Example 6.2.2. If X is a Banach space and f ∈ X∗, then p : x 7→ |f(x)| is a seminorm on X.

Proposition 6.2.3. Let X be a vector over F and {pα}α∈A be a family of seminorms on X. Call a set U ⊂ X open if for all x ∈ U, there exist > 0 and a ﬁnite set F ⊂ A such that

NF,(x) = {y ∈ X | pα(y − x) < for all α ∈ F } ⊂ U. Then the open sets form a topology T on X.

62 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Proof. It is certainly the case that ∅ and X are open. Unions of open sets are clearly open. Let U, V ∈ T , with associated F,G and , δ. We can then take F ∪ G and min(, δ) to get that U ∩ V ∈ T .

Remark 6.2.4. Each set NF,(x) is an open set. Deﬁnition 6.2.5 (Weak and weak-∗ topologies). Let X be a Banach space. The weak topology on ∗ X is the topology induced by the family of seminorms {pf | f ∈ X } given by pf (x) = |f(x)|. The ∗ weak-∗ topology on X is the topology induced by the family of seminorms {qx | x ∈ X} given by qx(f) = |f(x)|. Remark 6.2.6. Since X∗ is a Banach space, it also has a weak topology. If X is reﬂexive, then the weak and weak-∗ topologies on X∗ coincide. Lemma 6.2.7. Let X be a Banach space. Every open set in the weak topology on X is also open in the norm topology. Similarly, every open subset of X∗ in the weak-∗ topology is also open in the norm topology.

Proof. Let U ⊂ X be weakly open and x ∈ U be arbitrary. Then there exist > 0 and f1, . . . , fn ∈ X∗ for which NF,(x) = {y ∈ X | |fi(y − x)| < } ⊂ U. Let δ = . min(kf1k,..., kfnk) + 1

Then B(x, δ) ⊂ NF,(x) ⊂ U, as required. The proof for the weak-∗ topology is similar. Remark 6.2.8. The weak topology is more coarse than the norm topology, but is still Hausdorff. However, it is not metrizable if X has infinite dimension. Similar statements hold for the weak-∗ topology. Definition 6.2.9 (Directed set). A directed set is a set A with a partial order ≤ such that for all α, β ∈ A, there exists γ ∈ A such that α, β ∈ γ.

Example 6.2.10. 1. With the usual order, N is a directed set. 2. The set of ﬁnite subsets F ⊂ [0, 1] is a directed set with inclusion.

3. The family Ux of all open neighborhoods of x ∈ X in some topological space is a directed set with reverse inclusion.

Deﬁnition 6.2.11 (Net). A net in a topological space X is a family {xα}α∈A of points xα ∈ X indexed by a directed set A.

Deﬁnition 6.2.12 (Convergence of nets). If X is a topological space, {xα}α∈A is a net, and x ∈ X, we say that {xα} converges to x, written xα → x, if for every neighborhood U ⊂ X, there exists α0 ∈ A such that xα ∈ U for all α ≥ α0. Example 6.2.13. Let f be Riemann integrable on [0, 1] and let A be the set of all ﬁnite partitions F of [0, 1]. Then the lower sums {LF f}F ∈A form a convergent net in R which converges to the integral of f.

63 6.2 Weak topologies MATH 245A (17F)

Proposition 6.2.14. 1. If U ⊂ X, then x ∈ U if and only if there is a net in U which converges to x. 2. A map f : X → Y of topological spaces is continuous if and only if for any convergent net in X with limit x, the image net in Y converges to f(x). Remark 6.2.15. The topology on a space is uniquely determined by the convergent nets and their limits.

Notation. Let X be a Banach space. If a net {xα}α ∈ A converges with respect to the weak w topology to x, we write xα → x or xα * x.

w ∗ Proposition 6.2.16. xα → x if and only if f(xα) → f(x) for all f ∈ X .

w ∗ Proof. ( =⇒ ) Suppose xα → x and let f ∈ X and > 0 be arbitrary. Then there exists α0 ∈ A such that

xα ∈ N{pf },(x) = {y ∈ X | |f(y − x)| < }

for all α ≥ α0. Then |f(xα − x)| = |f(xα) − f(x)| < for all α ≥ α0, so f(xα) → f(x).

∗ ( ⇐= ) Suppose f(xα) → f(x) for all f ∈ X . It suﬃces to check the convergence condition for

any neighborhood of the form NF,(x) in the weak topology. If F = {pf1 , . . . , pfn }, then fi(xα) → fi(x), so there exist αi ∈ A such that for each i, we have |fi(xα − x)| < for all α ≥ αi. Since A is a directed set, there is an α0 with α0 ≥ αi for all i = 1, . . . , n. We then w have |fi(xα − x)| < for all i and α ≥ α0, so xα → x.

Q Example 6.2.17. Let Xi for i ∈ I be topological spaces and = i Xi. If πj : X → Xj is the projection, then one can show that for a net {xα} in X, we have xα → x if and only if pj(xα) → pj(x) for all j. Example 6.2.18. Let (Z, A, µ) be a σ-ﬁnite measure space and 1 ≤ p < ∞. Then X = Lp(µ) is w R R a Banach space. If {fn} is a sequence in X, then fn → f if and only if fng dµ = fg dµ for all g ∈ Lq(µ). Example 6.2.19. Let Z be a locally compact Hausdorﬀ space and Y = M(Z) be the set of all ∗ complex Radon measures on Z. If X = C0(Z), then by the Riesz representation theorem, X = Y . w∗ R R Then µn → µ ∈ M(Z) if and only if f dµn → f dµ for all f ∈ C0(Z). Instead of weak-∗ convergence, this convergence of measures is sometimes called vague convergence.

Pn w∗ Example 6.2.20. Let Z = [0, 1] and µn = (1/n) k=1 δk/n. Then µn → λ1, Lebesgue measure on [0, 1]. Theorem 6.2.21 (Banach-Alaoglu). Let X be a Banach space and B = {f ∈ X∗ | kfk ≤ 1} be the closed unit ball in X∗. Then B is compact with respect to the weak-∗ topology. Q Proof. For each x ∈ X, let Dx = {z ∈ C | |z| ≤ kxk} and P = x Dx with the product topology. By Tychonoﬀ’s theorem, P is compact. Deﬁne a map I : B → P by setting I(f) = (f(x))x∈X . Then I is a homeomorphism onto its image if B has the weak-∗ topology and I(B) has the subspace topology from P , and I(B) ⊂ P is closed, hence compact. To see that I is a homeomorphism onto

64 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis its image, I : B → I(B) is certainly a bijection, so we must check continuity of I and I−1. It w∗ w∗ suﬃces to show that fα → f in B if and only if I(fα) → I(f) in I(B). Indeed, fα → f if and only if fα(x) → f(α) for all x ∈ X, which happens if and only if I(fα) → I(f) in I(B) (by the characterization of convergence of nets in the product topology). To see that I(B) ⊂ P is closed, consider a net in I(B), which has the form {I(fα)}α∈A, where {fα}α∈A is a net in B. Suppose I(fα) → z = (zx)x∈X . We must show that there exists f ∈ B such that z = I(f). Deﬁne f(x) = zx for x ∈ X. Then f is a linear functional on X, as if a, b ∈ F and x, y are arbitrary, then

fα(ax + by) = pax+by(I(fα)) → pax+by(z) = zax+by = f(ax + by), and by linearity and a similar computation,

fα(ax + by) = afα(x) + bfα(y) → azx + bzy = af(x) + bf(y).

Since z ∈ P , we have zx = f(x) ∈ Dx, so |f(x)| = |zx| ≤ kxk for all x ∈ X, so kfk ≤ 1.

Corollary 6.2.22. Let Z be a compact metric space and {µn} be a sequence of Borel probability measures on Z. Then there is a subsequence µnk and a Borel probability measure µ such that w∗ µnk → µ.

Proof (outline). We regard µn as complex measures with kµnk = µn(Z) = 1. By Banach-Alaoglu, there is a subsequence µnk → µ for some µ ∈ M(Z). (Here, we use the fact that C(Z) is separable and that the weak-∗ topology on the unit ball in C(Z)∗ is metrizable.) One can show by integrating suitable non-negative functions that µ is a positive measure, and µ(X) = 1 is seen by integrating the function 1.

6.3 SOME THEOREMS IN FUNCTIONAL ANALYSIS

Theorem 6.3.1 (Baire’s category theorem). Let X be a complete metric space. T 1. If Gn for n ∈ N are open and dense in X, then n Gn is dense in X. S 2. If A = n An is a countable union of closed subsets An with empty interior, then A has empty interior.

Proof. The two statements are equivalent, so we will show the first one. Let D = bigcapnGn and let B0 = B(x, ) for x ∈ X and > 0 be arbitrary. We inductively construct closed balls n Bn = B(xn, rn) such that 0 < rn ≤ /2 and Bn+1 ⊂ Bn ∩G1 ∩· · ·∩Gn+1 ⊂ B0. Since G1 is dense, G1 ∩ B0 6= ∅, so we can pick x1 ∈ G1 ∩ Int B0 and r1 ≤ /2 so that B1 = B(xn, rn) ⊂ G1 ∩ Int B0. If Bn has been chosen then Gn+1 ∩ Int Bn 6= ∅, so we can pick Bn+1 in accordance with the given specifications. From the nesting of these closed balls, it follows that {xn} is a Cauchy sequence, so it has a limit x since X is complete, and x ∈ Bn ⊂ Gn for all n. Remark 6.3.2. Such sets A are often called sets of the first category or meagre.

Example 6.3.3. Rn cannot be represented as a countable union of codimension 1 aﬃne subspaces.

Theorem 6.3.4 (Banach-Steinhaus). Let X be a Banach space, Y be a normed space, and {Ti}i∈I be a family of bounded linear operators Ti : X → Y . Then if supi kTi(x)k < ∞ for each x, we have supi kTik < ∞.

65 6.3 Some theorems in functional analysis MATH 245A (17F)

Proof. For n ∈ N, deﬁne \ −1 An = x ∈ X | sup kTi(x)k ≤ n = Ti (B(0, n)). i∈I i∈I S Then n An = X, so by Baire’s category theorem, one of the sets An has non-empty interior. Suppose that Ak has non-empty interior. Then there exists x0 ∈ X and r > 0 such that B(x0, r) ⊂ Ak. If x ∈ B(0, r) is arbitrary, then x0 + x ∈ B(x0, x), so

kTi(x)k = kTi(x + x0) − Ti(x0)k

≤ kTi(x + x0)k + kTi(x0)k ≤ 2k.

By homogeneity, 2k kT (x)k ≤ kxk i r for x ∈ X and i ∈ I, so kTik ≤ 2k/r for all i ∈ I.

Corollary 6.3.5. Let X be a Banach space, Y be a normed space, and Tn : X → Y be bounded linear operators for n ∈ N. If {Tn(x)} converges pointwise for each x ∈ X, then supn kTnk < ∞. Corollary 6.3.6. Let X be a Banach space. Then every weakly convergent sequence in X is bounded ∗ in norm. Similarly, every sequence fn in X converging in the weak-∗ topology is uniformly bounded in operator norm.

∗ Proof. Each x ∈ X can be considered a bounded linear functional Tx on X with kTxk = kxk. w ∗ If xn → x, then f(xn) → f(x) for all f ∈ X , so Txn (f) → Tx(f). By Banach-Steinhaus, ∗ supn kTxn k = supn kxnk < ∞. A similar argument applies in X , but more directly since we already have bounded linear functionals. Theorem 6.3.7 (Open mapping theorem). Let X and Y be Banach spaces and T : X → Y be a bounded linear operator. If T is surjective, then T is an open mapping.

Proof. We first claim that there exists C0 > 0 such that for all y ∈ Y and > 0, there exists S x ∈ X with ky − T (x)k < and kxk ≤ C0kyk. Since T is surjective, Y = k T (Bk), where Bk = B(0, k) ⊂ X. By Baire’s category theorem, there exists k for which T (Bk) has non-empty interior. Then there exists y0 ∈ Y and r > 0 such that B(y0, r) ⊂ T (Bk). Now let y ∈ Y be 0 arbitrary with y 6= 0. Then y0, y1 = y0 + (r/kyk)y ∈ T (Bk), so there exist sequences {zn} and {zn} 0 0 0 0 in Bk such that T (zn) → −y0 and T (zn) → y1. If xn = zn − zn, then T (xn) → (r/kyk)y. Then 0 xn = (kyk/r)xn ∈ B2kkyk/r, which proves the claim. Our second claim is that there is another constant C1 > 0 such that for all y ∈ Y , there exists x ∈ X with y = T x and kxk ≤ C1kyk. Let y0 = y ∈ Y be arbitrary, and without loss of generality suppose kyk = 1. By the first claim, there exists x0 ∈ X such that ky0 − T x0k ≤ 1/2 and kx0k ≤ C0. Let n+1 n y1 = y0 − T x0. By induction, we can pick xn ∈ X with kyn − T (xn)k ≤ 1/2 and kxnk ≤ C0/2 , and then set yn+1 = yn − T (xn). Define sn = x0 + x1 + ··· + xn. Then {sn} is a Cauchy sequence, so sn → x for some x ∈ X. By construction, we find that kxk ≤ 2C0 when kyk = 1 and

y − T (x) = lim y − T (sn) = lim (y0 − T (x0) − · · · − T (xn)) = lim yn+1 = 0. n→∞ n→∞ n→∞

66 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Finally, we prove that T is an open mapping. Let U ⊂ X be open and y0 ∈ T (U). There exists x0 ∈ U such that y0 = T (x0). Since U is open, there exists δ > 0 such that BX (x0, δ) = x0 + BX (0, δ) ⊂ U. By the second claim, BY (0, δ/C1) ⊂ T (BX (0, δ)), so if r = δ/C1, then

BY (y0, r) = y0 + B(0, r) = T (x0 + BX (0, δ)) = T (BX (x0, δ)).

Corollary 6.3.8. If T is bijective, then T −1 : Y → X is a bounded linear operator.

Proof. Since T is linear, T −1 is linear. To see that T −1 is bounded, it suﬃces to show that T −1 is continuous. The preimage of an open set U ⊂ X is T (U) ⊂ Y , which is open.

Theorem 6.3.9 (Closed graph theorem). Let X and Y be Banach spaces, and let T : X → Y be a linear map. Then T is continuous if and only if graph T ⊂ X × Y is closed.

Proof. ( =⇒ ) Let (zn) be a sequence in graph T converging to z = (x, y) ∈ X × Y . If zn = (xn,T (xn)), then xn → x and T (xn) → y. Since T is continuous, T (x) = y, so z = (x, T (x)) ∈ graph T .

( ⇐= ) Suppose graph T is closed. Then it is a Banach space with norm induced by the norm on X × Y . Let π1 and π2 be the projection maps; these are bounded linear operators, and R = π1|graph T is a bijection onto X. By the open mapping theorem, the inverse is a bounded linear operator, hence T = π2 ◦ S is a bounded linear operator, so continuous.

6.4 HILBERT SPACES

Let X be a vector space over F = R or F = C with an inner product. We say that X is a Hilbert space if X is complete with the metric induced by the inner product.

Example 6.4.1. 1. Rn and Cn with the usual inner products. 2. If f, g ∈ L2(µ), we can deﬁne Z hf, gi = fg dµ.

3. If X has the counting measure µ on its power set, then we write l2(X) for L2(µ).

Proposition 6.4.2. Let X be a Hilbert space with inner product h , i.

1. If (xn, yn) → (x, y), then hxn, yni → hx, yi. 2. kx + yk2 + kx − yk2 = 2kxk2 + 2kyk2.

We say that x, y ∈ X are orthogonal, written x ⊥ y, if hx, yi = 0.

2 2 Proposition 6.4.3. If x1, . . . , xn ∈ X and xi ⊥ xj for i 6= j, then kx1 + ··· + xnk = kx1k + ··· + 2 kxnk .

67 6.4 Hilbert spaces MATH 245A (17F)

Deﬁnition 6.4.4 (Convex set). A set K ⊂ X in a vector space X is convex if for all x, y ∈ K and λ ∈ [0, 1], we have (1 − λ)x + λy ∈ K. Theorem 6.4.5. Let X be a Hilbert space, K ⊂ X be closed and convex, and z ∈ X. Then there exists a unique x ∈ K such that kx − zk = inf{ky − zk | y ∈ K}.

Proof. For existence, suppose without loss of generality that z = 0. Let {xn} be an inﬁmizing sequence in K, meaning that kxnk → d = inf{kyk | y ∈ K}. Then

2 2 xn + xm 2 2 kxn − xmk + 4 = 2kxnk + 2kxmk , 2 so 2! 2 2 2 xn + xm 2 2 2 lim sup kxn − xmk = lim sup 2kxnk + 2kxmk − 4 ≤ 2d + 2d − 4d = 0, n,m→∞ n,m→∞ 2 where we have used the fact that (xn +xm)/2 ∈ K in the last non-trivial step. Thus (xn) is Cauchy, so xn → x with kxk = d. For uniqueness, suppose x, x0 ∈ K and kxk = kx0k = d. We use the same trick to get that kx − x0k2 ≤ 0, so x = x0.

Let M ⊂ X be a linear subspace. The orthogonal complement to M is {x ∈ X | x ⊥ y for all y ∈ M}. Theorem 6.4.6. Let X be a Hilbert space and M ⊂ X be a closed subspace. Then X = M ⊕ M ⊥.

Proof. Let x ∈ X be arbitrary. Since M is closed and convex, there exists y ∈ M such that kx − yk = inf{kx − y0k | y0 ∈ M}. We claim that x − y ∈ M ⊥. For all t ∈ F and m ∈ M, we have y + tm ∈ M, so

kx − yk2 ≤ kx − (y + tm)k2 = kx − yk2 + 2 Re thx − y, mi + |t|2kuk2.

Since this is true for all t ∈ F, we must have hx − y, mi. Theorem 6.4.7 (Riesz representation theorem). Let X be a Hilbert space and f ∈ X∗. Then there ∗ exists a unique y ∈ X such that f = fy, where fy(x) = hx, yi. The map y ∈ X 7→ fy ∈ X is a conjugate linear isometric isomorphism between Banach spaces.

Proof. Let f ∈ X∗ and M = ker f. Then M is a closed linear subspace of X, so X = M ⊕ M ⊥. If M ⊥ = 0, then X = M and f = 0, so we can pick y = 0. Otherwise, pick z ∈ M ⊥ non-zero. Then z 6∈ M, so f(z) 6= 0. We pick y = αz for some suitable α ∈ F. Given x ∈ X, consider u = f(x)z − f(z)x. Then f(u) = f(x)f(z) − f(z)f(x) = 0, so u ∈ M. Hence u ⊥ z, so

0 = hf(x)z − f(z)x, zi = f(x)kzk2 − f(z)hx, zi.

Hence f(x) = f(z)/kzk2hx, zi, so we take α = f(z)/kzk2.

A set {xα}α∈A in a Hilbert space is an orthonormal system if hxα, xβi = 0 if α 6= β and 1 if α = β.

68 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Lemma 6.4.8. Let {xα}α∈A be an orthonormal system in a Hilbert space X. Then the following are equivalent:

(1) {xα} is a maximal orthonormal system;

(2) U = span{xα} is dense in X;

(3) if x ∈ X and hx, xαi = 0 for all α ∈ A, then x = 0.

Proof. (1) =⇒ (2) Suppose {xα}α∈A is maximal and let M = U. Then M is a linear subspace of ⊥ ⊥ X, so X = M ⊕ M . If M 6= 0, then {xα} is not maximal, as we could adjoin a non-zero element of M ⊥. Hence M ⊥ = 0 and M = X.

(2) =⇒ (3) Suppose x ∈ X and hx, xαi = 0 for all α. Then there exist zn ∈ U such that zn → x. Since x ⊥ U, we have 2 kxk = hx, xi = lim hx, zni = 0 n→∞

(3) =⇒ (1) If not, then a vector that we could adjoin to {xα} would break (3).

Deﬁnition 6.4.9 (Complete ONS). An orthonormal system satisfying any of these conditions is a complete orthonormal system or a Hilbert basis.

Lemma 6.4.10. Let X be an inﬁnite-dimensional separable Hilbert space. Then X has a countable (necessarily inﬁnite) Hilbert basis {x1, x2,...}.

Proof. Let {zn} be a countable dense subset of X and {yn} be a maximal linearly independent subset. Then U = span{yn} is dense in X. Running Gram-Schmidt (valid for a countable inﬁnite set of vectors as well), we produce an orthonormal system {xn} with span dense in X. This is the required Hilbert basis.

Theorem 6.4.11 (Parseval identities). Let {xn} be a Hilbert basis of a separable Hilbert space X of inﬁnite dimension. Then

P 1. x = nhx, xnixn; P 2. hx, yi = nhx, xnihy, xni;

2 P 2 3. kxk = n|hx, xni .

Proof. We ﬁrst show that ∞ X 2 2 |hx, xni| ≤ kxk . (∗) n=1

69 6.4 Hilbert spaces MATH 245A (17F)

(This is Bessel’s inequality, which holds for arbitrary orthonormal systems.) Write αn = hx, xni. Then 2 N X 0 ≤ x − αnxn n=1 * N N + X X = x − αnxn, x − αnxn n=1 n=1 * N + N 2

2 X X = kxk − 2 Re x, αnxn + αnxn n=1 n=1 N ! N 2 X 2 X 2 = kxk − 2 Re |αn| + |αn| n=1 n=1 N 2 X 2 = kxk − |αn| n=1

P 2 2 Hence n|αn| converges and is bounded above by kxk . P Let the partial sums of n αnxn be sN . We claim that (sN ) is a Cauchy sequence. We have, for M ≤ N,

N 2 2 X lim sup N,M → ∞ksN − sM k = lim sup αnx)n N,M→0 n=M+1 N X 2 = lim sup |αn| = 0. N,M→0 n=M+1 P Since (sN ) is a Cauchy sequence, it converges. let s = n αnxn. Then for all N,

hx − s, xN i = αN − hs, xN i N X = αN − lim h αnxn, xni N→∞ n=1

= αN − αN = 0, so by completeness, x − s = 0. This proves 1. The Parseval identities (2 and 3) follow directly.

Theorem 6.4.12 (Weierstrass approximation theorem). Let [a, b] ⊂ R. Then for each f ∈ C[a, b] and > 0, there exists a polynomial P such that kf − P k∞ ≤ .

Proof. Without loss of generality [a, b] ⊂ (0, 1). Pick f ∈ C[a, b] and extend f to R with supp f ⊂ R [0, 1]. Deﬁne Pn(x) = (Kn ∗ f)(x) = Kn(x − u)f(u) du for ( 0 |u| > 1, Kn(u) = 2 n cn(1 − u ) |u| ≤ 1.

70 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

The constant cn is chosen so that Kn integrates to 1. Then |Pn(x) − f(x)| is uniformly small on [a, b] for n large, since f is continuous. Theorem 6.4.13 (Stone-Weierstrass theorem). Let X be a compact Hausdorﬀ space and A ⊂ C(X, R) be a subalgebra of C(X, R) which is closed with respect to the uniform norm. If A separates points in X and for all x ∈ X, there exists f ∈ A such that f(x) 6= 0, then A = C(X, R).

Proof. We claim that if f ∈ A, then |f| ∈ A. Without loss of generality, |f| ≤ 1. It follows from the Weierstrass approximation theorem that there exist polynomials Pn with Pn(0) = 0 such that Pn(x) → |x| uniformly on [−1, 1]. Since A is a subalgebra of C(X, R) and |f| ≤ 1, we have Pn(f) → |f| uniformly on X. If f, g ∈ A, then f ∨ g = max(f, g) and f ∧ g = min(f, g) are in A (so A is a lattice of functions), as 1 1 1 1 f ∨ g = (f + g) + |f − g|, f ∧ g = (f + g) − |f − g|. 2 2 2 2 It then follows, from the above and from A separating points with non-zero function values at both points, that for all x, y ∈ X with x 6= y and for all α, β ∈ R, there exists f ∈ A such that f(x) = α and f(y) = β.

Now let f ∈ C(X, R) and > 0 be arbitrary. For all x, y ∈ X, we can ﬁnd gxy ∈ A such that gxy(x) = f(x) and gxy(y) = f(y). Let Uxy = {z ∈ x | f(z) − < gxy(z)} and Vxy = {z ∈ X | S gxy(z) < f(z) +}. Then Uxy and Vxy are open, and x, y ∈ Uxy,Vxy. Fix y ∈ X. Then X = x Uxy is an open cover, so there is a ﬁnite subcover X = Ux1y ∪ · · · ∪ Uxny. Let gy = gx1y ∨ · · · ∨ gxny ∈ A.

Then gy > f − by construction. Moreover, for z ∈ Vy = Vx1y ∩· · ·∩Vxny, we have gxi (z) < f(z)+, so gy(z) < f(z) + for z ∈ Vy. By compactness again, we can write X = Vy1 ∪ · · · ∪ Vym and take g = gy1 ∧ · · · ∧ gym . This g is in A, and kg − fk∞ < . Pick = 1/n and get gn ∈ A with kf − gnk < 1/n. Then gn → f in C(X, R), and by closure of A, we have f ∈ A.

Remark 6.4.14. If A is not required to be closed, then we conclude that A is dense in C(X, R). Corollary 6.4.15. Let K ⊂ Rn be compact. The set of all polynomial functions on Rn is dense in C(K, R).

Proof. It is clear that the polynomial functions form a subalgebra. The coordinate functions are sufficient to separate points, and the constant function 1 is sufficient for non-vanishing. Hence Stone-Weierstrass applies. Theorem 6.4.16 (Complex Stone-Weierstrass). Let X be a compact Hausdorff space and A be a complex subalgebra of C(X). If, in addition to the previous conditions, A is closed under complex conjugation, then A = C(X).

Proof. If f ∈ A, then Re f and Im f are in A. Then A ∩ C(X, R) satisﬁes the conditions of Stone-Weierstrass, so C(X, R) ⊂ A. Since A is a complex subalgebra, A = C(X).

MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

7 FOURIER ANALYSIS

7.1 TRIGONOMETRIC SERIES

Let T be the unit circle. A function f : T → C is equivalently a function f : R → C which is 2π-periodic, or also equivalently, a function f :[−π, π) → C. For 1 ≤ p < ∞, we can regard Lp(T) as the set of (equivalence classes) of 2π-periodic Lebesgue- measurable functions f : R → C with

Z π 1/p 1 p kfkp = |f(t)| dt < ∞. 2π −π

We have f ∼ g if and only if f = g almost everywhere. Equivalently we look at functions f : T → C with the arc length measure on T. This is obtained by pushing Lebesgue measure on [−π, π) forward via t 7→ eit. The space L∞(T) consists of 2π-periodic Lebesgue measurable functions f : R → C with

kfk∞ = ess sup{|f(t)| | t ∈ R = inf{m | λ1{|f| > m} = 0} < ∞.

Let C(T) be the space of continuous functions F : T → C (or space of 2π-periodic continuous functions f : R → C) equipped with the norm kfk∞ = sup{f(t) | t ∈ R}. Each of these spaces is a Banach space. Moreover, we have the inclusions

∞ p q 1 C(T) ⊂ L (T) ⊂ L (T) ⊂ L (T) ⊂ L (T) whenever ∞ ≥ p ≥ q ≥ 1, by H¨older’sinequality.

Deﬁnition 7.1.1 (Trigonometric polynomial). A trigonometric polynomial is a function f ∈ C(T) of the form N X f(t) = a0 + (an cos nt + bn sin nt), n=1 where a0, an, bn ∈ C.

Equivalently, we may write N X int f(t) = cne . n=−N

Note that L2(T) is a Hilbert space with 1 Z π hf, gi = f(t)g(t) dt, 2π −π and C(T) ⊂ L2(T).

Theorem 7.1.2. The trigonometric polynomials are dense in C(T).

73 7.2 Fourier series MATH 245A (17F)

Proof. It is convenient to write P as the space of functions of the form

N X n z ∈ T 7−→ cnz . n=−N

That these are dense in C(T) follows from complex Stone-Weierstrass. Corollary 7.1.3. (a) Trigonometric polynomials are dense in L2(T). int 2 (b) Deﬁne un(t) = e for n ∈ Z and t ∈ R. Then {un} is a Hilbert space basis for L (T).

7.2 FOURIER SERIES

Deﬁnition 7.2.1 (Fourier series). Let f ∈ L1(T) ⊃ L2(T). The Fourier coeﬃcients of f are Z π Z π 1 1 −int fb(n) = f(t)un(t) dt = f(t)e dt. 2π −π 2π −π

2 If f ∈ L (T), then fb(n) = hf, uni. The Fourier series of f is n X f ∼ fb(n)eint. n=−∞ (Here ∼ is used to say that f has Fourier series given by the right hand side.) When f ∈ L2(T), we have actual equality.

Denote the partial sums by N X int (sN f)(t) = fb(n)e . n=−N

2 2 Theorem 7.2.2. (a) If f ∈ L (T), then sN f → f in L .

(b) The map f 7→ fb is an isomorphism between the Hilbert spaces L2(T) and `2(Z).

int Proof. (a) Since un(t) = e is a Hilbert space basis,

N ∞ X X sN f = hf, uniun → hf, uniun = f. n=−N n=−∞

(b) If f ∈ L2(T), then ∞ ∞ X 2 X 2 2 |fb(n)| = |hf, uni| = kfk2 < ∞, n=−∞ n=−∞

so T : f 7→ fb is well-deﬁned as a map L2(T) → `2(Z). Clearly T is linear, and the norm 2 P is preserved by T , so T is injective. For surjectivity, let (an) ∈ ` (Z). Then N anun = f 2 converges in L (T) and fb(n) = an by construction.

74 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Theorem 7.2.3 (Riemann-Lebesgue lemma). (a) Let f ∈ L1(T). Then |fb(n)| → 0 as |n| → ∞.

1 (b) If f, g ∈ L (T) and fb = gb, then f = g almost everywhere.

Proof. (a) Let f ∈ L1(T) and > 0. Since trigonometric polynomials are dense in L1(T), there PN int is a trigonometric polynomial −N ane such that kf − P k1 < . For n > N, we have Pb(n) = 0. Thus

(b) Let f 7→ fb for f ∈ L1(T). This is linear, so it is enough to show that if fb = 0, then f = 0 almost everywhere. In this case, fb is in `2(Z), so f ∈ L2(T) is zero almost everywhere.

Remark 7.2.4. Let c0 = c0(Z) be the set of sequences (an) with an → 0 as |n| → ∞. This is 1 a Banach space with the supremum norm. Deﬁne T : L (T) → c0 by f 7→ fb. By the Riemann- Lebesgue lemma, T is an injective linear map. It is bounded, as

Z π 1 −int |fb(n)| = f(t)e dt ≤ kfk1. 2π −π

One can show that T is not surjective.

7.3 THE DIRICHLET KERNEL

Write N Z π N ! X int 1 X in(t−x) sN f(t) = fb(n)e = f(x) e dx, 2π n=−N −π n=−N which is of the form f ∗ DN for

N X sin((N + 1/2)t) D (t) = eint = . N sin(t/2) n=−N

Lemma 7.3.1. kDnk1 → ∞ as n → ∞.

75 7.3 The Dirichlet kernel MATH 245A (17F)

Proof.

1 Corollary 7.3.2. The map T : L (T) → c0 given by f 7→ fb is not surjective.

Proof. Suppose otherwise. Then T is a bounded linear isomorphism, so by the open mapping −1 theorem, the inverse T is bounded, so there exists C such that kfk1 ≤ Ckfbk∞. Applying this to the Dirichlet kernel DN , we have kDN k1 ≤ CkDbN k∞ = C, a contradiction.

1 1 Theorem 7.3.3. There exists f ∈ L (T) such that sN f 6→ f in L (T).

Proof. Suppose otherwise. The bounded linear operators sN then have pointwise boundedness, so they are uniformly bounded in operator norm. Now deﬁne

1 1 sin((n + 1)t/2)2 K = (D + ··· + D ) = ≥ 0. n n + 1 0 n n + 1 sin(t/2)

Then kKnk1 = 1, and

n ! X n + 1 − |k| s (K ) = s u N n N n + 1 k k=−n N N X n + 1 − |k| X = u → u = D n + 1 k k N k=−N k=−N

1 in L (T) as n → ∞. Therefore, ksN (Kn)k1 → kDN k1 as n → ∞, so ksN kop ≥ kDN k1 → ∞ as N → ∞, a contradiction.

Theorem 7.3.4 (Riesz; 1920’s). For 1 < p < ∞, there is a constant Cp such that ksN kLp→Lp ≤ Cp.

76 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

7.4 CONTINUOUS FUNCTIONS AND POINTWISE CONVERGENCE PROPERTIES

If f ∈ C(T), it need not be the case that sN f → f in C(T), i.e. uniformly. We can instead ask the question of pointwise convergence, but again, the answer is that this is not guaranteed.

Theorem 7.4.1 (Homework C2 Problem 3). There are functions f ∈ C(T) such that sN (f)(0) 6→ f(0) as N → ∞.

The next weakest statement we may consider is convergence almost everywhere. This is indeed the p case, and in fact, for each f ∈ L (T), where 1 < p < ∞, we have sN (f) → f almost everywhere. However, it remains false for p = 1 and p = ∞. If f ∈ L1(T), then 1 Z π (sN f)(t) = DN (t − x)f(x) dx 2π −π 1 Z t+π = f(t − y)DN (y) dy 2π t−π 1 Z π = f(t − x)DN (x) dx 2π −π 1 Z π 1 Z 0 = f(t − x)DN (x) dx + f(t − x)DN (x) dx 2π 0 2π −π 1 Z π = [f(t − x) + f(t + x)]DN (x) dx. 2π 0

Let S ∈ C and consider (sN f)(t) − S. Then 1 Z π (sN f)(t) − S = [f(t − x) + f(t + x) − 2S]DN (x) dx. 2π 0 2 1 1 π Lemma 7.4.2. − ≤ for −π/2 ≤ x ≤ π/2. sin x x 24 Lemma 7.4.3. If h ∈ L1(T), then Z 1 h(x) sin N + x dx → 0 2 as N → ∞.

Theorem 7.4.4. Let f ∈ L1(T) and suppose that Z π |f(t + x) + f(t − x) − 2S| dx < ∞. 0 x

Then (sN f)(t) → S.

Proof. We consider some functions in L1(T) deﬁned initially on [−π, π) which are then extended by periodicity. Deﬁne

1 a(x) = (f(t + x) + f(t − x) − 2S) · χ(0,π)(x) ∈ L (T)

77 7.5 Convolutions MATH 245A (17F) and

2a(x) 1 1 2 1 g(x) = ∈ L (T), h(x) = − a(x) ∈ L (T). x sin(x/2) x Then Z π

2π|sN f(t) − S| = [f(t + x) + f(t − x) − 2S]DN (x) dx 0 Z π 1 2 2 = a(x) − + sin((N + 1/2)x) dx 0 sin(x/2) x x Z π Z π

≤ h(x) sin((N + 1/2)x) dx + g(x) sin((N + 1/2)x) dx 0 0 → 0.

Definition 7.4.5 (Höldercontinuity). A function f ∈ C(T) is Höldercontinuous if there exists a Hölderexponent α ∈ (0, 1] and C ≥ 0 such that for all t ∈ R and |x|π, we have |f(t+x)−f(t)| ≤ C|x|.

Corollary 7.4.6. Suppose f ∈ C(T) is H¨oldercontinuous. Then (sN f)(t) → f(t) for all t ∈ R.

7.5 CONVOLUTIONS

Deﬁnition 7.5.1 (Convolution). Let f, g be measurable functions on Rn. The convolution f ∗ g is deﬁned by Z (f ∗ g)(x) = f(x − y)g(y) dy, where the integration is respect to Lebesgue measure.

Theorem 7.5.2. Let f, g ∈ L1(Rn). Then (x, y) 7→ F (x, y) = f(x−y)g(y) is in L1(R2n). Moreover, y 7→ f(x − y)g(y) ∈ L1(Rn) for almost every x ∈ Rn, and f ∗ g ∈ L1(Rn) is in L1(Rn) with kf ∗ gk1 ≤ kfk1kgk1.

Proof. There exist Borel functions f˜ ∼ f andg ˜ ∼ g, and then (x, y) 7→ F˜(x, y) = f˜(x − y)˜g(y) is Borel measurable. Since F ∼ F˜, it follows that F is Lebesgue measurable. The rest F ∈ L1(R2n) can be seen by Fubini-Tonelli.

Theorem 7.5.3. L1(Rn) is a commutative C-algebra with addition and convolution.

Remark 7.5.4. One can deﬁne (f∗g)(x) for almost every x ∈ Rn when f ∈ L1(Rn) and g ∈ Lp(Rn), p n where 1 ≤ p ≤ ∞. Then f ∗ g ∈ L (R ) with kf ∗ gkp ≤ kfk1kgkp. This result is (a special case of) Young’s inequality, which is proved using Minkowski’s inequality for integrals.

7.6 CONVOLUTIONS AND DIFFERENTIATION

n Let f : R → C and write x1, . . . , xn for the standard coordinates. The partial derivatives will be written ∂kf. For higher order partial derivatives, it is convenient to introduce multi-index notation: let α = (α1, . . . , αn) be an ordered n-tuple of non-negative integers and |α| = α1 + ··· + αn be the

78 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

α order of α. Then ∂ f denotes taking αk partial derivatives in xk for each k, so the total number of derivatives is |α|. If f ∈ CN (Rn) for N ≥ |α|, this is well-deﬁned. α α1 αn Write α! = α1! ··· αn! and x = x1 ··· xn

Theorem 7.6.1 (Taylor). If f ∈ Ck(Rn) and a, x ∈ Rn, then

X ∂αf(a) f(x) = (x − a)α + R(x) α! |α|≤K where R(x) = o(kx − akK ) as x → a.

Theorem 7.6.2 (Leibniz rule). If f, g are smooth, then

X α! ∂α(fg) = ∂βf∂γ g. β!γ! β+γ=α

1 n k n k n α α Theorem 7.6.3. If f ∈ Lloc(R ) and g ∈ Cc (R ), then f ∗ g ∈ C (R ) and ∂ (f ∗ g) = f ∗ ∂ g whenever |α| ≤ k.

A similar statement is true if f ∈ L1(Rn) and g ∈ Ck(Rn) with ∂αg bounded for each α with |α| ≤ k.

7.7 TRANSLATION OPERATORS

n n Let f : R → R and y ∈ R . Deﬁne τyf(x) = f(x − y).

n n n Proposition 7.7.1. 1. If f ∈ Cc(R ) and y ∈ R , then τyf ∈ Cc(R ) and kτyf − fk∞ → 0 as y → 0.

p n n p n 2. If f ∈ L (R ) and y ∈ R for 1 ≤ p < ∞, then τyf ∈ L (R ) and kτyf − fkp → 0 as y → 0.

n 1 n Theorem 7.7.2. Let φ be a molliﬁer on R and let f ∈ Lloc(R ). Then φt ∗ f is smooth for all t → 0 and for almost every x, we have (φt ∗ f)(x) → f(x) as t → 0. p n p n p If f ∈ L (R ) for 1 ≤ p < ∞, then φt ∗ f ∈ L (R ) and φt ∗ f → f in L as t → 0.

∞ n p n Corollary 7.7.3. Cc (R ) is dense in L (R ).

7.8 THE FOURIER TRANSFORM

For a T -periodic function, we can construct

∞ X 2πint/T f(t) ∼ ane , n=−∞ where Z T/2 1 −2πint/T an = f(t)e dt. T −T/2

79 7.8 The Fourier transform MATH 245A (17F)

As T → ∞, taking ξ = 2πn/T , the Fourier coeﬃcient an goes to

1 Z ∞ a(ξ) = f(t)e−iξt dt. 2π −∞

The function a is deﬁned whenever f ∈ L1(R). We then have Z ∞ f(t) ∼ a(ξ)eiξt dξ. −∞

Remark√ 7.8.1. To hide the factor of 1/2π, one can introduce a “normalized Lebesgue measure” dm = 1/ 2π dλ. Then Z Z fb(ξ) f(t) ∼ fb(ξ)eiξt dm(ξ) = √ eiξt dξ, 2π and √ 1 Z Z fb(ξ) = 2πa(ξ) = √ f(t)e−itξ dt = f(t)e−iξt dm(t). 2π

Deﬁnition 7.8.2 (Fourier transform). Let f ∈ L1(Rn). Its Fourier transform fb : Rn → C is given by Z −iξ·t fb(ξ) = f(t)e dmn(t), where 1 dm (t) = dλ (t). n (2π)n/2 n

Remark 7.8.3. One can see that kfbk∞ ≤ kfk1.

iξt In terms of the phase factors eξ(t) = e , we have Z fb(ξ) = fe−ξ dmn.

If α is a multi-index, then α α ∂ eξ = (iξ) eξ.

For our purposes, we will slightly redeﬁne convolutions by Z (f ∗ g)(x) = f(x − y)g(y) dm(y).

Theorem 7.8.4. Let f, g ∈ L1(Rn) and u ∈ Rn.

1. F(τuf) = e−ufb.

2. F(euf) = τufb.

3. F(f ∗ g) = fbgb.

80 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

4. If T is an orthogonal linear transformation, then F(f ◦ T ) = fb◦ T .

n 5. If a > 0 and g(t) = f(t/a), then gb(ξ) = a fb(aξ). Let S denote the set of Schwartz functions. Proposition 7.8.5. 1. If f, g ∈ S, then f + g, fg, f ∗ g ∈ S.

2. If f ∈ S and p is a polynomial, then fp ∈ S. 3. If f ∈ S and α is any multi-index, then ∂αf ∈ S.

α Theorem 7.8.6. If f ∈ S, then fb ∈ S. Moreover, if α is any multi-index and Pα(x) = x , then

α |α| α |α| α ∂ fb(ξ) = (−i) Pdαf(ξ) and ∂df(ξ) = i ξ fb(ξ).

Proof. To be written.

2 Lemma 7.8.7. φ(x) = e−|x| /2 is a Schwartz function with φb = φ.

Proof. The general case can be deduced from the case of dimension one by Fubini’s theorem. For n = 1, we have φ0(x) = −xφ(x), so φ0(x) + xφ(x) = 0. Applying the Fourier transform, iξφb(ξ) + iφb0(ξ) = 0, so φb and φ satisfy the same diﬀerential equation. By diﬀerentiating the quotient, it follows that φ/φb is constant. By checking x = 0 and ξ = 0, the constant is 1.

1 R R Lemma 7.8.8. If f, g ∈ L , then fgb = fgb. Proof. To be written. (Fubini) Theorem 7.8.9 (Fourier inversion). 1. If f ∈ S, then Z f(x) = fb(ξ)eix·ξ dm(ξ)(∗)

for all x ∈ Rn. 2. If f ∈ L1 and fb ∈ L1, then (∗) holds for almost every x.

Proof. To be written.

Theorem 7.8.10 (Plancherel). There is a unique linear isometry F : L2 → L2 such that F(f) = fb almost everywhere whenever f ∈ L1 ∩ L2.

Proof. To be written.

It is conventional to write fb instead of F(f) for f ∈ L2, even if f 6∈ L1.

MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

8 THE THEORY OF DISTRIBUTIONS

n ∞ For our purposes, let Ω be a region of R , i.e. an open connected subset. Functions in Cc (Ω) will be referred to as test functions.

8.1 SOBOLEV FUNCTIONS AND WEAK DERIVATIVES

Lemma 8.1.1. (a) Let f, g ∈ L1(Ω). Then f = g almost everywhere if and only if R fφ = R gφ for all test functions φ.

1 R R (b) If f ∈ C (Ω), then (∂kf)φ = − f(∂kφ) for all test functions φ and 1 ≤ k ≤ n.

1 1 Deﬁnition 8.1.2 (Sobolev function). Let f ∈ L (Ω) and 1 ≤ k ≤ n. We say that gk ∈ L (Ω) is R R loc loc a weak partial derivative for f if gkφ = − f(∂kφ) for all test functions φ. If the weak partial derivatives exist for all k, then f is a Sobolev function. 1,1 The space of Sobolev functions is denoted Wloc (Ω).

Proposition 8.1.3. 1. If the weak partial derivative gk exists, then it is unique (as an element 1 of Lloc).

1 1 2. If f ∈ C (Ω), then ∂kf ∈ Lloc is the weak k-th partial derivative of f.

In particular, the second result justiﬁes the use of the notation ∂kf for the weak partial derivative, even if the classical partial derivative does not exist.

Example 8.1.4. The function f with f(x) = 0 for x ≤ 0 and f(x) = x for x ≥ 0 is a Sobolev function with weak derivative equal to 0 for x < 0 and 1 for x > 0. However, this weak derivative is not a Sobolev function.

Deﬁnition 8.1.5 (Sobolev space). Let 1 ≤ p ≤ ∞. The Sobolev space W 1,p(Ω) is the set of all Sobolev functions which are in Lp and whose weak derivatives are also in Lp. This is a normed vector space with Sobolev norm

Z n !1/p p X p kfkW 1,p = |f| + |∂kf| (p < ∞) k=1

kfkW 1,∞ = max{kfk∞, k∂1fk∞,..., k∂nfk∞}.

Theorem 8.1.6. W 1,p(Ω) is a Banach space.

Higher-order weak partial derivatives are deﬁned in a similar way, with Z Z |α| α gαφ = (−1) f∂ φ

∞ for all test functions φ ∈ Cc (Ω). As for ﬁrst-order weak partial derivatives, gα is unique if it exists and agrees with the classical partial derivative if f ∈ C|α|(Ω). We may then deﬁne the Sobolev

83 8.2 Distributions MATH 245A (17F) spaces W k,p(Ω) analogously, with Sobolev norms

 1/p Z X α p kfkW k,p =  |∂ f|  (p < ∞) |α|≤k α kfkW k,∞ = max k∂ fk∞. |α|≤k

Proposition 8.1.7. C∞(Ω) is dense in W k,p(Ω).

k,p ∞ k,p Notation. 1. W0 (Ω) is the closure of Cc (Ω) in W (Ω).

k,p α p 2. Wloc (Ω) is the space of locally integrable functions with ∂ f ∈ Lloc(Ω) for all |α| ≤ k. 3. Hk is the Hilbert space W k,2(Rn).

8.2 DISTRIBUTIONS

∞ Proposition 8.2.1. There is a topology on Cc (Ω) with the property that a sequence φk converges to φ if

(i) there exists L ⊂ Ω compact such that supp(φk) ⊂ L for all k;

α α (ii) ∂ φk → ∂ φ uniformly on Ω for all α.

∞ D Let D(Ω) be the space Cc (Ω) with this topology, and write φk → φ. Deﬁnition 8.2.2 (Distribution). A distribution (or generalized function) is a continuous linear functional on D(Ω). The vector space of distributions is denoted D0(Ω). Notation. For a distribution T ∈ D0(Ω) and test function φ ∈ D(Ω), we write hT, φi for T (φ).

1 Example 8.2.3. 1. If f ∈ Lloc(Ω), then f deﬁnes a distribution Tf via Z hTf , φi = f(x)φ(x) dλ(x).

Following this example, one may write hT, φi = R T (x)φ(x) dλ(x) for any distribution T , even if it is not represented by an actual function. R 2. Every positive Radon measure µ deﬁnes a distribution Tµ via hTµ, φi = φ dµ.

A special case of this is the Dirac delta distribution δp with hδp, φi = φ(p). Remark 8.2.4. 1. Convergence of distributions corresponds to weak-∗ convergence in Banach spaces, although D0(Ω) is not itself a Banach space. 2. There is no well-defined multiplication of distributions. Definition 8.2.5 (Distributional derivative). Let T be a distribution and α be a multi-index. The distributional derivative ∂αT is defined by

h∂αT, φi = (−1)|α|hT, ∂αφi.

84 MATH 245A (17F) (L) M. Bonk / (TA) A. Wu Real Analysis

Example 8.2.6. The Heaviside step function H with H(x) = 0 for x < 0 and H(x) = 1 for x > 0 has distributional derivative the Dirac delta at 0.

Proposition 8.2.7. The distributional derivative of a Sobolev function coincides with the weak derivative.

Lemma 8.2.8. Let T : D(Ω) → C be a linear map. The following are equivalent:

(1) T is a distribution;

D (2) if φk → 0, then T (φk) → 0;

(3) for all compact L ⊂ Ω, there exist N = N(L) and C = C(L) > 0 such that |T (φ)| ≤ CkφkN,∞ for all φ ∈ D(Ω) with supp φ ⊂ L.

8.3 LOCALIZATION AND SUPPORT OF DISTRIBUTIONS

Deﬁnition 8.3.1 (Restriction). Let U ⊂ Ω be open and T ∈ D0(Ω). The restriction of T to U is deﬁned by hT |U , φi = hT, φi for all φ ∈ D(U) ⊂ D(Ω).

0 We say that T vanishes on U ⊂ Ω open if T |U = 0 in D (U). Equivalently, hT, φi = 0 for all φ ∈ D(U), or hT, φi = 0 for all φ ∈ D(Ω) with supp(φ) ⊂ U.

Theorem 8.3.2. Let T be a distribution and let W be the union of all open sets in Ω on which T vanishes. Then T vanishes on W .

Proof. To be written. (partitions of unity)

Deﬁnition 8.3.3 (Support). The support of T is supp(T ) = Ω\W .

Theorem 8.3.4. 1. Every distribution T can be written as an inﬁnite sum of a locally ﬁnite family of distributions with compact support.

P α 2. Every distribution T with compact support can be written in the form T = |α|≤N ∂ fα for some N and locally integrable fα. 3. Every distribution T with support {p} for some p ∈ Ω can be written in the form T = P α |α|≤N cα∂ δp for some N and complex constants cα.

8.4 FOURIER TRANSFORMS OF DISTRIBUTIONS

n R R Suppose Ω = R . We would like to use the result fgb = fgb to analogously deﬁne the Fourier transform of a distribution T by hTb , φi = hT, φbi. However, the Fourier transform of a test function ∞ n ∞ φ ∈ Cc (R ) need not be in Cc . Therefore, we enlarge the space of allowed test functions to the Schwartz space and restrict the space of distributions accordingly.

Deﬁnition 8.4.1 (Schwartz seminorm). Fix N and a multi-index α. The Schwartz seminorm is deﬁned by N α n kφkN,α = supx∈R (1 + |x| )|∂ φ(x)|.

85 8.4 Fourier transforms of distributions MATH 245A (17F)

n Proposition 8.4.2. There is a topology on S(R ) with the property that φk → φ if kφk −φkN,α → 0 α α as k → ∞ for all N and α, or equivalently, if ∂ φk → ∂ φ uniformly for all α and, for any multi- β α n indices α, β, there exists C = C(α, β) > 0 such that |x ∂ φk(x)| ≤ C for all x ∈ R and k.

n S From now on, S(R ) refers to the space of Schwartz functions with this topology. We write φk → φ. Proposition 8.4.3. The inclusion map D(Rn) ,→ S(Rn) is continuous with their respective topologies, and the image is dense in S(Rn).