22 Approximation Theorems and Convolutions

22.1 Density Theorems

In this section, (X, ,µ) will be a measure space will be a subalgebra of . M A M Notation 22.1 Suppose (X, ,µ) is a measure space and is a sub- M A ⊂ M algebra of . Let S( ) denote those simple functions φ : X C such that 1 M A → φ− ( λ ) for all λ C and let Sf ( ,µ) denote those φ S( ) such that µ(φ =0){ } <∈ A. ∈ A ∈ A 6 ∞ p p Remark 22.2. For φ Sf ( ,µ) and p [1, ), φ = z=0 z 1 φ=z and hence ∈ A ∈ ∞ | | 6 | | { } P φ p dµ = z pµ(φ = z) < (22.1) | | | | ∞ z=0 Z X6 p p so that Sf ( ,µ) L (µ). Conversely if φ S( ) L (µ), then from Eq. (22.1) it follows thatA µ (⊂φ = z) < for all z =0∈andA therefore∩ µ (φ =0)< . Hence we have shown, for any 1 ∞p< , 6 6 ∞ ≤ ∞ p Sf ( ,µ)=S( ) L (µ). A A ∩ Lemma 22.3 (Simple Functions are Dense). The simple functions, p Sf ( ,µ), form a dense subspace of L (µ) for all 1 p< . M ≤ ∞

Proof. Let φn ∞ be the simple functions in the approximation Theo- { }n=1 rem 18.42. Since φn f for all n, φn Sf ( ,µ) and | | ≤ | | ∈ M p p p p 1 f φn ( f + φn ) 2 f L (µ) . | − | ≤ | | | | ≤ | | ∈ Therefore, by the dominated convergence theorem,

p p lim f φn dµ = lim f φn dµ =0. n | − | n | − | →∞ Z Z →∞ 420 22 Approximation Theorems and Convolutions

The goal of this section is to ﬁnd a number of other dense subspaces of Lp (µ) for p [1, ). The next theorem is the key result of this section. ∈ ∞ Theorem 22.4 (Density Theorem). Let p [1, ), (X, ,µ) be a mea- ∈ ∞ M sure space and M be an algebra of bounded F —valued(F = R or F = C) measurable functions such that

1. M Lp (µ, F) and σ (M)= . ⊂ M 2. There exists ψk M such that ψk 1 boundedly. ∈ → 3. If F = C we further assume that M is closed under complex conjugation. p Then to every function f L (µ, F) , there exists φn M such that ∈ p ∈ limn f φn Lp(µ) =0, i.e. M is dense in L (µ, F) . →∞ k − k Proof. Fix k N for the moment and let denote those bounded — ∈ H M measurable functions, f : X F, for which there exists φn ∞ M such → { }n=1 ⊂ that limn ψkf φn Lp(µ) =0. Aroutinecheckshows is a subspace →∞ k − k H of ∞ ( , F) such that 1 ,M and is closed under complex M ∈ H ⊂ H H conjugation if F = C. Moreover, is closed under bounded convergence. H To see this suppose fn and fn f boundedly. Then, by the dominated ∈ H → 1 convergence theorem, limn ψk (f fn) Lp(µ) =0. (Take the dominating →∞pk − k function to be g =[2C ψk ] where C is a constant bounding all of the | | 1 fn n∞=1 .) We may now choose φn M such that φn ψkfn Lp(µ) n {|then|} ∈ k − k ≤

lim sup ψkf φn Lp(µ) lim sup ψk (f fn) Lp(µ) n k − k ≤ n k − k →∞ →∞ + lim sup ψkfn φn Lp(µ) =0 (22.2) n k − k →∞ which implies f . An application of Dynkin’s Multiplicative System The- ∈ H orem 18.51 if F = R or Theorem 18.52 if F = C now shows contains all bounded measurable functions on X. H Let f Lp (µ) be given. The dominated convergence theorem implies ∈ limk ψk1 f k f f Lp(µ) =0. (Take the dominating function to be →∞ p {| |≤ } − g =[2C f ] where C is a bound on all of the ψk .) Using this and what we |° | ° | | have just° proved, there exists° φk M such that ∈ 1 ψk1 f k f φk p . {| |≤ } − L (µ) ≤ k ° ° The same line of reasoning° used in Eq. (22.2)° now implies limk f φk Lp(µ) = 0. →∞ k − k 1 It is at this point that the proof would break down if p = . ∞ 22.1 Density Theorems 421

Definition 22.5. Let (X, τ) be a topological space and µ be a measure on X = σ (τ) . A locally integrable function is a Borel measurable function B f : X C such that K f dµ < for all compact subsets K X. We will →1 | | ∞ ⊂ write Lloc(µ) for the space of locally integrable functions. More generally we p R say f L (µ) iff 1K f p < forallcompactsubsetsK X. ∈ loc k kL (µ) ∞ ⊂ Definition 22.6. Let (X, τ) be a topological space. A K-finite measure on X is Borel measure µ such that µ (K) < for all compact subsets K X. ∞ ⊂ Lebesgue measure on R is an example of a K-finite measure while counting measure on R is not a K-finite measure.

Example 22.7. Suppose that µ is a K-ﬁnite measure on Rd . An application of p d B Theorem 22.4 shows Cc (R, C) is dense in L (R , Rd ,µ; C). To apply Theorem d B d 22.4, let M := Cc R , C and ψk (x):=ψ (x/k) where ψ Cc R , C with ψ (x)=1in a neighborhood of 0. The proof is completed by∈ showing σ (M)= d ¡ ¢ ¡ ¢ σ Cc R , C = Rd , which follows directly from Lemma 18.57. B d We may also give a more down to earth proof as follows. Let x0 R ,R> ¡ ¡ ¢¢ c 1/n ∈ 0,A:= B (x0,R) and fn (x):=d (x) . Then fn M and fn 1 A ∈ → B(x0,R) as n which shows 1B(x0,R) is σ (M)-measurable, i.e. B (x0,R) σ (M) . →∞ d ∈ Since x0 R and R>0 were arbitrary, σ (M)= d . ∈ BR More generally we have the following result.

Theorem 22.8. Let (X, τ) be a second countable locally compact Hausdorﬀ space and µ : X [0, ] be a K-ﬁnite measure. Then Cc(X) (the space of continuous functionsB → with∞ compact support) is dense in Lp(µ) for all p [1, ). (See also Proposition 25.23 below.) ∈ ∞

Proof. Let M := Cc(X) and use Item 3. of Lemma 18.57 to ﬁnd functions ψk M such that ψk 1 to boundedly as k . The result now follows from∈ an application of→ Theorem 22.4 along with→∞ the aid of item 4. of Lemma 18.57.

Exercise 22.1. Show that BC (R, C) is not dense in L∞(R, R,m; C). Hence the hypothesis that p< in Theorem 22.4 can not be removed.B ∞ n Corollary 22.9. Suppose X R is an open set, X is the Borel σ — algebra ⊂ B p on X and µ be a K-ﬁnite measure on (X, X ) . Then Cc(X) is dense in L (µ) for all p [1, ). B ∈ ∞ Corollary 22.10. Suppose that X is a compact subset of Rn and µ is a ﬁnite p measure on (X, X ), then polynomials are dense in L (X, µ) for all 1 p< . B ≤ ∞ Proof. Consider X to be a metric space with usual metric induced from Rn. Then X is a locally compact separable metric space and therefore 422 22 Approximation Theorems and Convolutions

p Cc(X, C)=C(X, C) is dense in L (µ) for all p [1, ). Since, by the dominated convergence theorem, uniform convergence∈ implies∞ Lp(µ) — convergence, it follows from the Weierstrass approximation theorem (see Theorem 8.34 and Corollary 8.36 or Theorem 12.31 and Corollary 12.32) that polynomials are also dense in Lp(µ).

Lemma 22.11. Let (X, τ) be a second countable locally compact Hausdorﬀ 1 space and µ : X [0, ] be a K-ﬁnite measure on X. If h Lloc(µ) is a function such thatB → ∞ ∈

fhdµ =0for all f Cc(X) (22.3) ∈ ZX then h(x)=0for µ —a.e.x. (See also Corollary 25.26 below.)

Proof. Let dν(x)= h(x) dx, then ν is a K-ﬁnite measure on X and hence 1 | | Cc(X) is dense in L (ν) by Theorem 22.8. Notice that

f sgn(h)dν = fhdµ =0for all f Cc(X). (22.4) · ∈ ZX ZX

Let Kk k∞=1 be a sequence of compact sets such that Kk X as in Lemma { } 1 ↑ 11.23. Then 1Kk sgn(h) L (ν) and therefore there exists fm Cc(X) such ∈ 1 ∈ that fm 1K sgn(h) in L (ν). So by Eq. (22.4), → k

ν(Kk)= 1K dν = lim fmsgn(h)dν =0. k m ZX →∞ ZX

Since Kk X as k , 0=ν(X)= X h dµ, i.e. h(x)=0for µ —a.e.x. As an↑ application→∞ of Lemma 22.11 and| Example| 12.34, we will show that the Laplace transform is injective. R

Theorem 22.12 (Injectivity of the Laplace Transform). For f L1([0, ),dx), the Laplace transform of f is deﬁned by ∈ ∞

∞ λx f(λ):= e− f(x)dx for all λ>0. L Z0 If f(λ):=0then f(x)=0for m -a.e. x. L Proof. Suppose that f L1([0, ),dx) such that f(λ) 0. Let g ∈ ∞ L ≡ ∈ C0([0, ), R) and ε>0 be given. By Example 12.34 we may choose aλ λ>0 ∞ { } such that #( λ>0:aλ =0 ) < and { 6 } ∞ λx g(x) aλe− <εfor all x 0. | − | ≥ λ>0 X Then 22.1 Density Theorems 423

∞ ∞ λx g(x)f(x)dx = g(x) aλe− f(x)dx − ¯Z0 ¯ ¯Z0 Ã λ>0 ! ¯ ¯ ¯ ¯ X ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ∞ λx ¯ ¯ g(x) aλe− f(x) dx ¯ ε f 1. ≤ 0 ¯ − ¯ | | ≤ k k Z ¯ λ>0 ¯ ¯ X ¯ ¯ ¯ Since ε>0 is arbitrary, it follows¯ that ∞ g(x¯)f(x)dx =0for all g 0 ∈ C0([0, ), R). The proof is ﬁnished by an application of Lemma 22.11. Here∞ is another variant of Theorem 22.8.R

Theorem 22.13. Let (X, d) be a metric space, τd be the topology on X generated by d and X = σ(τd) be the Borel σ — algebra. Suppose µ : X [0, ] B B → ∞ is a measure which is σ — ﬁnite on τd and let BCf (X) denote the bounded continuous functions on X such that µ(f =0)< . Then BCf (X) is a dense subspace of Lp(µ) for any p [1, ). 6 ∞ ∈ ∞

Proof. Let Xk τd be open sets such that Xk X and µ(Xk) < and let ∈ ↑ ∞ ψk(x)=min(1,k dXc (x)) = φk(dXc (x)), · k k see Figure 22.1 below. It is easily veriﬁed that M := BCf (X) is an algebra,

y 1

0.75

0.5

0.25

0 0 0.5 1 1.5 2

Fig. 22.1. The plot of φn for n =1, 2, and 4. Notice that φn 1(0, ). → ∞

ψk M for all k and ψk 1 boundedly as k . Given V τ and ∈ → →∞ ∈ k, n N,let ∈ c fk,n (x):=min(1,n d(V Xk) (x)). · ∩ Then fk,n =0 = V Xk so fk,n BCf (X). Moreover { 6 } ∩ ∈

lim lim fk,n = lim 1V Xk =1V k n k ∩ →∞ →∞ →∞ which shows V σ (M) and hence σ (M)= X . The proof is now completed by an application∈ of Theorem 22.4. B 424 22 Approximation Theorems and Convolutions

Exercise 22.2. (BRUCE: Should drop this exercise.) Suppose that (X, d) is a metric space, µ is a measure on X := σ(τd) which is finite on bounded B measurable subsets of X. Show BCb(X, R), defined in Eq. (19.26), is dense in p L (µ) . Hints: let ψk be as defined in Eq. (19.27) which incidentally may be used to show σ (BCb(X, R)) = σ (BC(X, R)) . Then use the argument in the proof of Corollary 18.55 to show σ (BC(X, R)) = X . B Theorem 22.14. Suppose p [1, ), is an algebra such that σ( )= ∈ ∞ A ⊂ M p A and µ is σ — finite on . Then Sf ( ,µ) is dense in L (µ). (See also Remark 25.7M below.) A A

Proof. Let M := Sf ( ,µ). By assumption there exits Xk such that A ∈ A µ(Xk) < and Xk X as k . If A , then Xk A and ∞ ↑ →∞ ∈ A ∩ ∈ A µ (Xk A) < so that 1Xk A M. Therefore 1A = limk 1Xk A is σ (M) — measurable∩ ∞ for every A ∩ ∈. So we have shown that→∞ σ∩(M) and therefore = σ ( ) ∈ Aσ (M) , i.e. σ (M)=A ⊂. The theorem⊂ M M A ⊂ ⊂ M M now follows from Theorem 22.4 after observing ψk := 1Xk M and ψk 1 boundedly. ∈ → Theorem 22.15 (Separability of Lp — Spaces). Suppose, p [1, ), is a countable algebra such that σ( )= and µ is σ — ﬁnite∈ on∞. ThenA ⊂ LMp(µ) is separable and A M A

D = aj1Aj : aj Q + iQ,Aj with µ(Aj) < { ∈ ∈ A ∞} is a countable denseX subset.

Proof. It is left to reader to check D is dense in Sf ( ,µ) relative to the p A L (µ) — norm. The proof is then complete since Sf ( ,µ) is a dense subspace of Lp (µ) by Theorem 22.14. A n Example 22.16. The collection of functions of the form φ = k=1 ck1(ak,bk] p p with ak,bk Q and ak

Corollary 22.17 (Riemann Lebesgue Lemma). Suppose that f L1(R,m), then ∈ lim f(x)eiλxdm(x)=0. λ →±∞ ZR 22.1 Density Theorems 425

n Proof. By Example 22.16, given ε>0 there exists φ = k=1 ck1(ak,bk] with ak,bk R such that ∈ P f φ dm < ε. | − | ZR Notice that

n iλx iλx φ(x)e dm(x)= ck1(ak,bk](x)e dm(x) R R k=1 Z Z X n bk n iλx 1 iλx bk = ck e dm(x)= ckλ− e |ak k=1 Zak k=1 X n X 1 iλbk iλak = λ− ck e e 0 as λ . − → | | →∞ k=1 X ¡ ¢ Combining these two equations with

f(x)eiλxdm(x) (f(x) φ(x)) eiλxdm(x) + φ(x)eiλxdm(x) ≤ − ¯ZR ¯ ¯ZR ¯ ¯ZR ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ f φ dm + φ(x)eiλx¯dm(¯x) ¯ ¯ ¯ ≤ ¯ | − | ¯ ¯ ¯ ZR ¯ZR ¯ ¯ ¯ ε + φ(x)eiλx¯dm(x) ¯ ≤ ¯ ¯ ¯ZR ¯ ¯ ¯ ¯ ¯ we learn that ¯ ¯

lim sup f(x)eiλxdm(x) ε + lim sup φ(x)eiλxdm(x) = ε. λ ≤ λ | |→∞ ¯ZR ¯ | |→∞ ¯ZR ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ Since ε>0 is¯ arbitrary, this completes¯ the proof¯ of the Riemann¯ Lebesgue lemma.

Corollary 22.18. Suppose is an algebra such that σ( )= and µ is σ — ﬁnite on . Then forA every⊂ M B such that µ(B) 0 there exists D A such that µ(B D) ∈<ε.M(See also Remark 25.7∞ below.) ∈ A 4 n Proof. By Theorem 22.14, there exists a collection, Ai , of pairwise { }i=1 disjoint subsets of and λi R such that X 1B f dµ < ε where f = n A n∈ | − | λi1A . Let A0 := X Ai then i=1 i \ ∪i=1 ∈ A R P 426 22 Approximation Theorems and Convolutions

n 1B f dµ = 1B f dµ | − | | − | ZX i=0 ZAi Xn = µ (A0 B)+ 1B λi dµ + 1B λi dµ ∩ Ai B | − | Ai B | − | i=1 "Z ∩ Z \ # Xn = µ (A0 B)+ [ 1 λi µ (B Ai)+ λi µ (Ai B)] (22.5) ∩ | − | ∩ | | \ i=1 Xn µ (A0 B)+ min µ (B Ai) ,µ(Ai B) (22.6) ≥ ∩ { ∩ \ } i=1 X where the last equality is a consequence of the fact that 1 λi + 1 λi . Let ≤ | | | − | 0 if µ (B Ai) <µ(Ai B) αi = ∩ \ 1 if µ (B Ai) µ (Ai B) ½ ∩ ≥ \ n and g = i=1 αi1Ai =1D where

P D := Ai : i>0 & αi =1 . ∪ { } ∈ A

Equation (22.5) with λi replaced by αi and f by g implies

n 1B 1D dµ = µ (A0 B)+ min µ (B Ai) ,µ(Ai B) . | − | ∩ { ∩ \ } X i=1 Z X

The latter expression, by Eq. (22.6), is bounded by X 1B f dµ < ε and therefore, | − | R µ(B D)= 1B 1D dµ < ε. 4 | − | ZX

Remark 22.19. We have to assume that µ(B) < as the following example ∞ shows. Let X = R, = ,µ= m, be the algebra generated by half open M B A intervals of the form (a, b], and B = n∞=1(2n, 2n +1]. It is easily checked that for every D , that m(B∆D)= ∪. ∈ A ∞

22.2 Convolution and Young’s Inequalities

Throughout this section we will be solely concerned with d — dimensional Lebesgue measure, m, and we will simply write Lp for Lp Rd,m .

Definition 22.20 (Convolution). Let f,g : Rd C be¡ measurable¢ functions. We define → f g(x)= f(x y)g(y)dy (22.7) ∗ d − ZR 22.2 Convolution and Young’s Inequalities 427 whenever the integral is defined, i.e. either f (x ) g ( ) L1(Rd,m) or − · · ∈ f (x ) g ( ) 0. Notice that the condition that f (x ) g ( ) L1(Rd,m) is equivalent− · · to≥ writing f g (x) < . Byconvention,iftheintegralinEq.− · · ∈ (22.7) is not defined, let| f| ∗ |g(|x):=0∞. ∗ d Notation 22.21 Given a multi-index α Z+, let α = α1 + + αd, ∈ | | ··· d ∂ α d ∂ αj xα := xαj , and ∂α = := . j x ∂x ∂x j=1 j=1 j Y µ ¶ Y µ ¶ d d d For z R and f : R C, let τzf : R C be defined by τzf(x)=f(x z). ∈ → → − Remark 22.22 (The Significance of Convolution). 1. Suppose that f,g L1 (m) are positive functions and let µ be the measure 2 ∈ on Rd defined by ¡ ¢ dµ (x, y):=f (x) g (y) dm (x) dm (y) . Then if h : R [0, ] is a measurable function we have → ∞ h (x + y) dµ (x, y)= h (x + y) f (x) g (y) dm (x) dm (y) d 2 d 2 Z(R ) Z(R ) = h (x) f (x y) g (y) dm (x) dm (y) d 2 − Z(R ) = h (x) f g (x) dm (x) . d ∗ ZR In other words, this shows the measure (f g) m isthesameasS µ where S (x, y):=x+y. In probability lingo, the distribution∗ of a sum of∗ two “in- dependent” (i.e. product measure) random variables is the the convolution of the individual distributions. α 2. Suppose that L = α k aα∂ is a constant coefficient differential operator and suppose that| we|≤ can solve (uniquely) the equation Lu = g in the P form u(x)=Kg(x):= k(x, y)g(y)dy d ZR where k(x, y) is an “integral kernel.” (This is a natural sort of assumption since, in view of the fundamental theorem of calculus, integration is the d inverse operation to differentiation.) Since τzL = Lτz for all z R , (this is another way to characterize constant coefficient differential∈ operators) 1 and L− = K we should have τzK = Kτz. Writing out this equation then says

k(x z,y)g(y)dy =(Kg)(x z)=τzKg(x)=(Kτzg)(x) d − − ZR = k(x, y)g(y z)dy = k(x, y + z)g(y)dy. d − d ZR ZR 428 22 Approximation Theorems and Convolutions

Since g is arbitrary we conclude that k(x z,y)=k(x, y + z). Taking y =0then gives − k(x, z)=k(x z,0) =: ρ(x z). − − We thus find that Kg = ρ g. Hence we expect the convolution operation to appear naturally when∗ solving constant coefficient partial differential equations. More about this point later. Proposition 22.23. Suppose p [1, ],f L1 and g Lp, then f g(x) exists for almost every x, f g ∈Lp and∞ ∈ ∈ ∗ ∗ ∈ f g f g . k ∗ kp ≤ k k1 k kp Proof. This follows directly from Minkowski’s inequality for integrals, Theorem 21.27. p p Proposition 22.24. Suppose that p [1, ), then τz : L L is an iso- p ∈ ∞d p → metric isomorphism and for f L ,z R τzf L is continuous. ∈ ∈ → ∈ p p Proof. The assertion that τz : L L is an isometric isomorphism follows from translation invariance of Lebesgue→ measure and the fact that τ z τz = id. For the continuity assertion, observe that − ◦ τzf τyf = τ y (τzf τyf) = τz yf f k − kp k − − kp k − − kp p d from which it follows that it is enough to show τzf f in L as z 0 R . d → → ∈ When f Cc(R ),τzf f uniformly and since the K := z 1supp(τzf) is ∈ → ∪| |≤ compact, it follows by the dominated convergence theorem that τzf f in p d p d → L as z 0 R . For general g L and f Cc(R ), → ∈ ∈ ∈ τzg g τzg τzf + τzf f + f g k − kp ≤ k − kp k − kp k − kp = τzf f +2 f g k − kp k − kp and thus

lim sup τzg g p lim sup τzf f p +2 f g p =2 f g p . z 0 k − k ≤ z 0 k − k k − k k − k → → d p Because Cc(R ) is dense in L , the term f g p may be made as small as we please. k − k

Exercise 22.4. Compute the operator norm, τz I L(Lp(m)) , of τz I and d p k − k − use this to show z R τz L (L (m)) is not continuous. ∈ → ∈ Deﬁnition 22.25. Suppose that (X, τ) is a topological space and µ is a measure on X = σ(τ). For a measurable function f : X C we deﬁne the essentialB support of f by →

suppµ(f)= x X : µ( y V : f(y) =0 ) > 0 neighborhoods V of x . { ∈ { ∈ 6 }} ∀ (22.8)} Equivalently, x/supp (f) iﬀ thereexistsanopenneighborhoodV of x such ∈ µ that 1V f =0a.e. 22.2 Convolution and Young’s Inequalities 429

It is not hard to show that if supp(µ)=X (see Deﬁnition 21.41) and f C(X) then supp (f) = supp(f):= f =0 , see Exercise 22.7. ∈ µ { 6 } Lemma 22.26. Suppose (X, τ) is second countable and f : X C is a mea- → surable function and µ is a measure on X . Then X := U supp (f) may B \ µ be described as the largest open set W such that f1W (x)=0for µ —a.e.x. Equivalently put, C := suppµ(f) is the smallest closed subset of X such that f = f1C a.e.

Proof. To verify that the two descriptions of suppµ(f) are equivalent, suppose supp (f) is deﬁned as in Eq. (22.8) and W := X supp (f). Then µ \ µ W = x X : τ V x such that µ( y V : f(y) =0 )=0 { ∈ ∃ 3 3 { ∈ 6 }} } = V o X : µ (f1V =0)=0 ∪ { ⊂ 6 } = V o X : f1V =0for µ —a.e. . ∪ { ⊂ }

So to ﬁnish the argument it suﬃces to show µ (f1W =0)=0. To to this let be a countable base for τ and set 6 U

f := V : f1V =0a.e. . U { ∈ U }

Then it is easily seen that W = f and since f is countable ∪U U

µ (f1W =0) µ (f1V =0)=0. 6 ≤ 6 V f X∈U

Lemma 22.27. Suppose f,g,h : Rd C aremeasurablefunctionsandas- → sume that x is a point in Rd such that f g (x) < and f ( g h )(x) < , then | |∗| | ∞ | |∗ | | ∗ | | ∞ 1. f g(x)=g f(x) 2. f ∗ (g h)(x)=(∗ f g) h(x) ∗ ∗ d ∗ ∗ 3. If z R and τz( f g )(x)= f g (x z) < , then ∈ | | ∗ | | | | ∗ | | − ∞

τz(f g)(x)=τzf g(x)=f τzg(x) ∗ ∗ ∗ 4. If x/supp (f) + supp (g) then f g(x)=0and in particular, ∈ m m ∗ supp (f g) supp (f) + supp (g) m ∗ ⊂ m m

where in deﬁning suppm(f g) we will use the convention that “f g(x) = 0”when f g (x)= . ∗ ∗ 6 | | ∗ | | ∞ Proof. For item 1.,

f g (x)= f (x y) g (y)dy = f (y) g (y x)dy = g f (x) | | ∗ | | d | | − | | d | | | | − | | ∗ | | ZR ZR 430 22 Approximation Theorems and Convolutions where in the second equality we made use of the fact that Lebesgue measure invariant under the transformation y x y. Similar computations prove all of the remaining assertions of the→ﬁrst− three items of the lemma. Item 4. Since f g(x)=f˜ g˜(x) if f = f˜ and g =˜g a.e. we may, by replacing f f ∗ g∗ g f by 1suppm(f) and by 1suppm(g) if necessary, assume that =0 supp (f) and g =0 supp (g). So if x/(supp (f)+supp{ (g6 )) then} ⊂ m { 6 } ⊂ m ∈ m m x/( f =0 + g =0 ) and for all y Rd, either x y/ f =0 or y/ g∈=0{ .6 That} is{ to6 say} either x y ∈ f =0 or y− ∈g =0{ 6 and} hence∈ {f(x6 y})g(y)=0for all y and therefore− ∈f {g(x)=0} . This∈ shows{ that} f g =0 −d ∗ ∗ on R suppm(f)+suppm(g) and therefore \ ³ ´ d d R suppm(f) + suppm(g) R suppm(f g), \ ⊂ \ ∗ ³ ´ i.e. supp (f g) supp (f) + supp (g). m ∗ ⊂ m m Remark 22.28. Let A, B be closed sets of Rd, it is not necessarily true that A + B is still closed. For example, take

A = (x, y):x>0 and y 1/x and B = (x, y):x<0 and y 1/ x , { ≥ } { ≥ | |} then every point of A + B has a positive y - component and hence is not zero. On the other hand, for x>0 we have (x, 1/x)+( x, 1/x)=(0, 2/x) A + B for all x and hence 0 A + B showing A + B is− not closed. Nevertheless∈ if one of the sets A or B∈is compact, then A + B is closed again. Indeed, if A is d compact and xn = an + bn A + B and xn x R , then by passing to a ∈ → ∈ subsequence if necessary we may assume limn an = a A exists. In this case →∞ ∈ lim bn =lim(xn an)=x a B n n →∞ →∞ − − ∈ exists as well, showing x = a + b A + B. ∈ Proposition 22.29. Suppose that p, q [1, ] and p and q are conjugate p q ∈ ∞ d exponents, f L and g L , then f g BC(R ), f g f p g q ∈ ∈ d ∗ ∈ k ∗ k∞ ≤ k k k k and if p, q (1, ) then f g C0(R ). ∈ ∞ ∗ ∈ Proof. The existence of f g(x) and the estimate f g (x) f g for ∗ | ∗ | ≤ k kp k kq all x Rd is a simple consequence of Holders inequality and the translation in- ∈ variance of Lebesgue measure. In particular this shows f g f p g q . By relabeling p and q if necessary we may assume thatkp ∗ [1k∞, ≤).kSincek k k ∈ ∞

τz (f g) f g = τzf g f g k ∗ − ∗ ku k ∗ − ∗ ku τzf f g 0 as z 0 ≤ k − kp k kq → → it follows that f g is uniformly continuous. Finally if p, q (1, ), we learn ∗ ∈ ∞ d from Lemma 22.27 and what we have just proved that fm gm Cc(R ) ∗ ∈ where fm = f1 f m and gm = g1 g m. Moreover, | |≤ | |≤ 22.2 Convolution and Young’s Inequalities 431

f g fm gm f g fm g + fm g fm gm k ∗ − ∗ k∞ ≤ k ∗ − ∗ k∞ k ∗ − ∗ k∞ f fm g + fm g gm ≤ k − kp k kq k kp k − kq f fm g + f g gm 0 as m ≤ k − kp k kq k kp k − kq → →∞ d showing, with the aid of Proposition 12.23, f g C0(R ). ∗ ∈ Theorem 22.30 (Young’s Inequality). Let p, q, r [1, ] satisfy ∈ ∞ 1 1 1 + =1+ . (22.9) p q r

If f Lp and g Lq then f g (x) < for m —a.e.x and ∈ ∈ | | ∗ | | ∞ f g f g . (22.10) k ∗ kr ≤ k kp k kq In particular L1 is closed under convolution. (The space (L1, ) is an example of a “Banach algebra” without unit.) ∗

Remark 22.31. Before going to the formal proof, let us ﬁrst understand Eq. (22.9) by the following scaling argument. For λ>0, let fλ(x):=f(λx), then after a few simple change of variables we ﬁnd

d/p d fλ = λ− f and (f g)λ = λ fλ gλ. k kp k k ∗ ∗ Therefore if Eq. (22.10) holds for some p, q, r [1, ], we would also have ∈ ∞ d/r d/r (d+d/r d/p d/q) f g = λ (f g) λ λ fλ gλ = λ − − f g k ∗ kr k ∗ λkr ≤ k kp k kq k kp k kq for all λ>0. This is only possible if Eq. (22.9) holds.

Proof. By the usual sorts of arguments, we may assume f and g are 1 1 1 positive functions. Let α, β [0, 1] and p1,p2 (0, ] satisfy p1− +p2− +r− = 1. Then by Hölder’s inequality,∈ Corollary 21.3,∈ ∞

(1 α) (1 β) α β f g(x)= f(x y) − g(y) − f(x y) g(y) dy ∗ d − − ZR h i 1/r 1/p1 (1 α)r (1 β)r αp1 f(x y) − g(y) − dy f(x y) dy ≤ d − d − × µZR ¶ µZR ¶ 1/p2 g(y)βp2 dy × d µZR ¶ 1/r = f(x y)(1 α)rg(y)(1 β)rdy f α g β . − − αp1 βp2 d − k k k k µZR ¶ Taking the rth power of this equation and integrating on x gives 432 22 Approximation Theorems and Convolutions

f g r f(x y)(1 α)rg(y)(1 β)rdy dx f α g β r − − αp1 βp2 k ∗ k ≤ d d − ·k k k k ZR µZR ¶ (1 α)r (1 β)r αr βr = f − g − f g . (22.11) (1 α)r (1 β)r αp1 βp2 k k − k k − k k k k

Let us now suppose, (1 α)r = αp1 and (1 β)r = βp2, in which case Eq. (22.11) becomes, − − f g r f r g r k ∗ kr ≤ k kαp1 k kβp2 which is Eq. (22.10) with

p := (1 α)r = αp1 and q := (1 β)r = βp2. (22.12) − − So to finish the proof, it suffices to show p and q are arbitrary indices in 1 1 1 [1, ] satisfying p− +q− =1+r− . If α, β, p1,p2 satisfy the relations above, then∞ r r α = and β = r + p1 r + p2 and 1 1 1 1 1 r + p 1 r + p + = + = 1 + 2 p q αp1 αp2 p1 r p2 r 1 1 2 1 = + + =1+ . p1 p2 r r Conversely, if p, q, r satisfy Eq. (22.9), then let α and β satisfy p =(1 α)r and q =(1 β)r, i.e. − − r p p r q q α := − =1 1 and β = − =1 1. r − r ≤ r − r ≤ Using Eq. (22.9) we may also express α and β as 1 1 α = p(1 ) 0 and β = q(1 ) 0 − q ≥ − p ≥ and in particular we have shown α, β [0, 1]. If we now define p1 := p/α ∈ ∈ (0, ] and p2 := q/β (0, ], then ∞ ∈ ∞ 1 1 1 1 1 1 + + = β + α + p1 p2 r q p r 1 1 1 =(1 )+(1 )+ − q − p r 1 1 =2 1+ + =1 − r r µ ¶ as desired. Theorem 22.32 (Approximate δ — functions). Let p [1, ],φ 1 d d ∈ ∞ ∈ L ( ),a:= d f(x)dx, and for t>0 let φt(x)=t φ(x/t). Then R R − R 22.2 Convolution and Young’s Inequalities 433

p p 1. If f L with p< then φt f af in L as t 0. ∈ d ∞ ∗ → ↓ 2. If f BC(R ) and f is uniformly continuous then φt f af 0 as t ∈0. k ∗ − k∞ → ↓ d 3. If f L∞ and f is continuous on U o R then φt f af uniformly on compact∈ subsets of U as t 0. ⊂ ∗ → ↓ Proof. Making the change of variables y = tz implies

φt f(x)= f(x y)φt(y)dy = f(x tz)φ(z)dz ∗ d − d − ZR ZR so that

φt f(x) af(x)= [f(x tz) f(x)] φ(z)dz ∗ − d − − ZR = [τtzf(x) f(x)] φ(z)dz. (22.13) d − ZR Hence by Minkowski’s inequality for integrals (Theorem 21.27), Proposition 22.24 and the dominated convergence theorem,

φt f af p τtzf f p φ(z) dz 0 as t 0. k ∗ − k ≤ d k − k | | → ↓ ZR Item 2. is proved similarly. Indeed, form Eq. (22.13)

φt f af τtzf f φ(z) dz k ∗ − k ≤ d k − k | | ∞ ZR ∞ which again tends to zero by the dominated convergence theorem because limt 0 τtzf f =0uniformly in z by the uniform continuity of f. ↓ k − k∞ d Item 3. Let BR = B(0,R) be a large ball in R and K @@ U, then

sup φt f(x) af(x) x K | ∗ − | ∈ [f(x tz) f(x)] φ(z)dz + [f(x tz) f(x)] φ(z)dz ≤ B − − Bc − − ¯Z R ¯ ¯Z R ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ φ(z) dz sup f(x ¯tz)¯ f(x) +2 f φ(z) dz¯ ¯ c ¯ ≤ B | | · x K,z BR | − − | k k∞ B | | Z R ∈ ∈ Z R

lim sup sup φt f(x) af(x) 2 f φ(z) dz 0 as R . t 0 x K | ∗ − | ≤ k k∞ z >R | | → →∞ ↓ ∈ Z| |

See Theorem 8.15 if Folland for a statement about almost everywhere convergence. 434 22 Approximation Theorems and Convolutions

Exercise 22.5. Let e 1/t if t>0 f(t)= − 0 if t 0. ½ ≤ Show f C∞(R, [0, 1]). ∈ d Lemma 22.33. There exists φ Cc∞(R , [0, )) such that φ(0) > 0, supp(φ) B¯(0, 1) and φ(x)dx ∈=1. ∞ ⊂ Rd Proof. Deﬁne h(t)=R f(1 t)f(t +1) where f is as in Exercise 22.5. − 2 Then h Cc∞(R, [0, 1]), supp(h) [ 1, 1] and h(0) = e− > 0. Deﬁne c = ∈ ⊂ 2− h( x 2)dx. Then φ(x)=c 1h( x ) is the desired function. Rd − The| | reader asked to prove the following| | proposition in Exercise 22.9 below. R 1 d 1 d Proposition 22.34. Suppose that f Lloc(R ,m) and φ Cc (R ), then 1 d ∈ ∈ d f φ C (R ) and ∂i(f φ)=f ∂iφ. Moreover if φ Cc∞(R ) then ∗ ∈ d ∗ ∗ ∈ f φ C∞(R ). ∗ ∈ d Corollary 22.35 (C∞ — Uryhson’s Lemma). Given K @@ U o R , there d ⊂ exists f Cc∞(R , [0, 1]) such that supp(f) U and f =1on K. ∈ ⊂ d Proof. Let φ be as in Lemma 22.33, φt(x)=t− φ(x/t) be as in Theorem 22.32, d be the standard metric on Rd and ε = d(K, Uc). Since K is compact c d and U is closed, ε>0. Let Vδ = x R : d(x, K) <δ and f = φε/3 1Vε/3 , then ∈ ∗ supp(f) supp(©φ )+V V¯ ª U. ⊂ ε/3 ε/3 ⊂ 2ε/3 ⊂ Since V¯ is closed and bounded, f C∞(U) and for x K, 2ε/3 ∈ c ∈

f(x)= 1d(y,K)<ε/3 φε/3(x y)dy = φε/3(x y)dy =1. d · − d − ZR ZR The proof will be ﬁnished after the reader (easily) veriﬁes 0 f 1. Here is an application of this corollary whose proof is left≤ to≤ the reader, Exercise 22.10.

Lemma 22.36 (Integration by Parts). Suppose f and g are measur- d able functions on R such that t f(x1,...,xi 1,t,xi+1,...,xd) and t → − → g(x1,...,xi 1,t,xi+1,...,xd) are continuously diﬀerentiable functions on R − d ∂f for each ﬁxed x =(x ,...,xd) . Moreover assume f g, g and 1 R ∂xi ∂g ∈ · · f are in L1(Rd,m). Then · ∂xi ∂f ∂g gdm = f dm. d ∂x · − d · ∂x ZR i ZR i With this result we may give another proof of the Riemann Lebesgue Lemma. 22.2 Convolution and Young’s Inequalities 435

Lemma 22.37 (Riemann Lebesgue Lemma). For f L1(Rd,m) let ∈ d/2 iξ x fˆ(ξ):=(2π)− f(x)e− · dm(x) d ZR ˆ d ˆ d/2 be the Fourier transform of f. Then f C0(R ) and f (2π)− f . ∈ ≤ k k1 d/2 ° °ˆ∞ (The choice of the normalization factor, (2π)− , in° °f is for later conve- nience.) ° °

Proof. The fact that fˆ is continuous is a simple application of the dominated convergence theorem. Moreover,

ˆ d/2 f(ξ) f(x) dm(x) (2π)− f 1 ≤ d | | ≤ k k ¯ ¯ ZR ¯ ¯ so it only remains¯ to¯ see that fˆ(ξ) 0 as ξ . First suppose that d d ∂2 → | | →∞ d f Cc∞(R ) and let ∆ = j=1 ∂x2 be the Laplacian on R . Notice that ∈ j ∂ iξ x iξ x iξ x 2 iξ x e− · = iξje− · and P∆e− · = ξ e− · . Using Lemma 22.36 re- ∂xj − − | | peatedly,

k iξ x k iξ x 2k iξ x ∆ f(x)e− · dm(x)= f(x)∆xe− · dm(x)= ξ f(x)e− · dm(x) d d − | | d ZR ZR ZR = (2π)d/2 ξ 2k fˆ(ξ) − | | for any k N. Hence ∈ d/2 2k k (2π) fˆ(ξ) ξ − ∆ f 0 ≤ | | 1 → ¯ ¯ ˆ d ¯ ¯ ° °1 d as ξ and f C0(R )¯. Suppose¯ that°f L° (m) and fk Cc∞(R ) is | | →∞ ∈ ∈ ˆ ∈ ˆ a sequence such that limk f fk 1 =0, then limk f fk =0. →∞ k − k →∞ − ˆ d ° °∞ Hence f C0(R ) by an application of Proposition 12.23. ° ° ∈ ° ° Corollary 22.38. Let X Rd be an open set and µ be a Radon measure on ⊂ X . B p 1. Then Cc∞(X) is dense in L (µ) for all 1 p< . 2. If h L1 (µ) satisﬁes ≤ ∞ ∈ loc

fhdµ =0for all f C∞(X) (22.14) ∈ c ZX then h(x)=0for µ —a.e.x.

Proof. Let f Cc(X),φbe as in Lemma 22.33, φt be as in Theorem ∈ 22.32 and set ψt := φt (f1X ) . Then by Proposition 22.34 ψt C∞(X) and ∗ ∈ by Lemma 22.27 there exists a compact set K X such that supp(ψt) K ⊂ ⊂ for all t suﬃciently small. By Theorem 22.32, ψt f uniformly on X as t 0 → ↓ 436 22 Approximation Theorems and Convolutions

1. The dominated convergence theorem (with dominating function being p f 1K ), shows ψt f in L (µ) as t 0. This proves Item 1., since k k∞ → ↓ p Theorem 22.8 guarantees that Cc(X) is dense in L (µ). 2. Keeping the same notation as above, the dominated convergence theorem (with dominating function being f h 1K ) implies k k∞ | |

0 = lim ψthdµ = lim ψthdµ = fhdµ. t 0 t 0 ↓ ZX ZX ↓ ZX The proof is now ﬁnished by an application of Lemma 22.11.

22.2.1 Smooth Partitions of Unity

We have the following smooth variants of Proposition 12.16, Theorem 12.18 and Corollary 12.20. The proofs of these results are the same as their continuous counterparts. One simply uses the smooth version of Urysohn’s Lemma of Corollary 22.35 in place of Lemma 12.8. Proposition 22.39 (Smooth Partitions of Unity for Compacts). Sup- d n pose that X is an open subset of R ,K X is a compact set and = Uj ⊂ U { }j=1 is an open cover of K. Then there exists a smooth (i.e. hj C∞(X, [0, 1])) n ∈ partition of unity hj of K such that hj Uj for all j =1, 2,...,n. { }j=1 ≺ Theorem 22.40 (Locally Compact Partitions of Unity). Suppose that X is an open subset of Rd and is an open cover of X. Then there exists a UN smooth partition of unity of hi (N = is allowed here) subordinate to { }i=1 ∞ the cover such that supp(hi) is compact for all i. U Corollary 22.41. Suppose that X is an open subset of Rd and = U Uα α A τ is an open cover of X. Then there exists a smooth partition { } ∈ ⊂ of unity of hα α A subordinate to the cover such that supp(hα) Uα for { } ∈ U ⊂ all α A. Moreover if U¯α is compact for each α A we may choose hα so ∈ ∈ that hα Uα. ≺

22.3 Exercises

Exercise 22.6. Let (X, τ) be a topological space, µ ameasureon X = B σ(τ) and f : X C be a measurable function. Letting ν be the measure, → dν = f dµ, show supp(ν) = suppµ(f), where supp(ν) is deﬁnedinDeﬁnition 21.41).| |

Exercise 22.7. Let (X, τ) be a topological space, µ ameasureon X = σ(τ) B such that supp(µ)=X (see Deﬁnition 21.41). Show suppµ(f) = supp(f)= f =0 for all f C(X). { 6 } ∈ 22.3 Exercises 437

Exercise 22.8. Prove the following strong version of item 3. of Proposition 10.52, namely to every pair of points, x0,x1, in a connected open subset V d of R there exists σ C∞(R,V) such that σ(0) = x0 and σ(1) = x1. Hint: ∈ First choose a continuous path γ :[0, 1] V such that γ (t)=x0 for t near 0 → and γ (t)=x1 for t near 1 and then use a convolution argument to smooth γ. Exercise 22.9. Prove Proposition 22.34 by appealing to Corollary 19.43.

d 1 Exercise 22.10 (Integration by Parts). Suppose that (x, y) R R − d 1 ∈ × → f(x, y) C and (x, y) R R − g(x, y) C are measurable functions ∈ ∈ × → ∈ such that for each ﬁxed y Rd,x f(x, y) and x g(x, y) are continuously ∈ → → diﬀerentiable. Also assume f g, ∂xf g and f ∂xg areintegrablerelativeto d · 1 · · d Lebesgue measure on R R − , where ∂xf(x, y):= f(x + t, y) t=0. Show × dt |

∂xf(x, y) g(x, y)dxdy = f(x, y) ∂xg(x, y)dxdy. (22.15) d 1 d 1 R R − · − R R − · Z × Z × (Note: this result and Fubini’s theorem proves Lemma 22.36.) Hints: Let ψ Cc∞(R) be a function which is 1 in a neighborhood of ∈ 0 R and set ψε(x)=ψ(εx). First verify Eq. (22.15) with f(x, y) replaced ∈ by ψε(x)f(x, y) by doing the x —integralﬁrst. Then use the dominated convergence theorem to prove Eq. (22.15) by passing to the limit, ε 0. ↓ iλ x Exercise 22.11. Let µ be a ﬁnite measure on d , then D := span e · : BR { λ Rd is a dense subspace of Lp(µ) for all 1 p< . Hints: By Theorem ∈ } d p ≤ ∞ d 22.8, Cc(R ) is a dense subspace of L (µ). For f Cc(R ) and N N, let ∈ ∈

fN (x):= f(x +2πNn). n Zd X∈ d Show fN BC(R ) and x fN (Nx) is 2π — periodic, so by Exercise 12.13, ∈ → x fN (Nx) can be approximated uniformly by trigonometric polynomials. → ¯ Lp(µ) p Use this fact to conclude that fN D . After this show fN f in L (µ). ∈ → Exercise 22.12. Suppose that µ and ν are two ﬁnite measures on Rd such that iλ x iλ x e · dµ(x)= e · dν(x) (22.16) d d ZR ZR for all λ Rd. Show µ = ν. Hint:∈Perhaps the easiest way to do this is to use Exercise 22.11 with the measure µ being replaced by µ + ν. Alternatively, use the method of proof of Exercise 22.11 to show Eq. (22.16) implies fdµ(x)= fdν(x) for all Rd Rd d f Cc(R ) and then apply Corollary 18.58. ∈ R R

Exercise 22.13. Again let µ be a ﬁnite measure on Rd . Further assume that M x B d CM := d e dµ(x) < for all M (0, ). Let ( ) be the space of R | | R ∞ α ∈ ∞ d P p polynomials, ρ(x)= α N ραx with ρα C, on R . (Notice that ρ(x) R | |≤ ∈ | | ≤ P 438 22 Approximation Theorems and Convolutions

M x d p Ce | | for some constant C = C(ρ, p, M), so that (R ) L (µ) for all P ⊂ 1 p< .) Show (Rd) is dense in Lp(µ) for all 1 p< . Here is a possible≤ outline.∞ P ≤ ∞ d n Outline: Fix a λ R and let fn(x)=(λ x) /n! for all n N. ∈ · ∈ α Mt α α 1. Use calculus to verify supt 0 t e− =(α/M) e− for all α 0 where ≥ ≥ (0/M )0 := 1. Usethisestimatealongwiththeidentity

pn pn pn pn M x pn M x λ x λ x = x e− | | λ e | | | · | ≤ | | | | | | | | ³ ´ to ﬁnd an estimate on fn . k kp 2. Use your estimate on fn to show ∞ fn < and conclude k kp n=0 k kp ∞ PN iλ ( ) n lim e · · i fn =0. N ° − ° →∞ ° n=0 °p ° X ° ° ° 3. Now ﬁnish by appealing to° Exercise 22.11. °

Exercise 22.14. Again let µ be a ﬁnite measure on Rd but now assume there exists an ε>0 such that C := eε x dµ(x) < B. Also let q>1 and Rd | | q α ∞ d h L (µ) be a function such that d h(x)x dµ(x)=0for all α N0. (As ∈ RR ∈ mentioned in Exercise 22.14, ( d) Lp(µ) for all 1 p< , so x h(x)xα R R is in L1(µ).) Show h(x)=0forP µ—a.e.⊂ x using the following≤ ∞ outline.→ d n Outline: Fix a λ R , let fn(x)=(λ x) /n! for all n N, and let p = q/(q 1) be the conjugate∈ exponent to q.· ∈ − α εt α α 1. Use calculus to verify supt 0 t e− =(α/ε) e− for all α 0 where ≥ ≥ (0/ε)0 := 1. Use this estimate along with the identity

pn pn pn pn ε x pn ε x λ x λ x = x e− | | λ e | | | · | ≤ | | | | | | | | ³ ´ to ﬁnd an estimate on fn . k kp 2. Use your estimate on fn p to show there exists δ>0 such that k k iλ x n∞=0 fn p < when λ δ and conclude for λ δ that e · = p k k n ∞ | | ≤ | | ≤ L (µ)- ∞ i f (x). Conclude from this that P n=0 n

P iλ x h(x)e · dµ(x)=0when λ δ. d | | ≤ ZR d itλ x 3. Let λ R ( λ not necessarily small) and set g(t):= d e · h(x)dµ(x) ∈ | | R for t R. Show g C∞(R) and ∈ ∈ R

(n) n itλ x g (t)= (iλ x) e · h(x)dµ(x) for all n N. d · ∈ ZR 22.3 Exercises 439

4. Let T =sup τ 0:g 0 . By Step 2., T δ. If T< , then { ≥ |[0,τ] ≡ } ≥ ∞

(n) n iT λ x 0=g (T )= (iλ x) e · h(x)dµ(x) for all n N. d · ∈ ZR iT λ x Use Step 3. with h replaced by e · h(x) to conclude

i(T +t)λ x g(T + t)= e · h(x)dµ(x)=0for all t δ/ λ . d ≤ | | ZR This violates the deﬁnition of T and therefore T = andinparticular we may take T =1to learn ∞

iλ x d h(x)e · dµ(x)=0for all λ R . d ∈ ZR 5. Use Exercise 22.11 to conclude that

h(x)g(x)dµ(x)=0 d ZR for all g Lp(µ). Now choose g judiciously to ﬁnish the proof. ∈