Approximation Theorems and Convolutions

Transcription

1 22 Approximation Theorems and Convolutions 22.1 Density Theorems In this section, (, M,µ) will be a measure space A will be a subalgebra of M. Notation 22.1 Suppose (, M,µ) is a measure space and A M is a subalgebra of M. Let S(A) denote those simple functions φ : C such that φ 1 ({λ}) A for all λ C and let S f (A,µ) denote those φ S(A) such that µ(φ 6= 0)<. emark For φ S f (A,µ) and p [1, ), φ p = P z6=0 z p 1 {φ=z} and hence φ p dµ = z p µ(φ = z) < (22.1) z6=0 so that S f (A,µ) L p (µ). Conversely if φ S(A) L p (µ), then from Eq. (22.1) it follows that µ (φ = z) < for all z 6= 0and therefore µ (φ 6= 0)<. Hence we have shown, for any 1 p<, S f (A,µ)=S(A) L p (µ). Lemma 22.3 (Simple Functions are Dense). The simple functions, S f (M,µ), form a dense subspace of L p (µ) for all 1 p<. Proof. Let {φ n } n=1 be the simple functions in the approximation Theorem Since φ n f for all n, φ n S f (M,µ) and f φ n p ( f + φ n ) p 2 p f p L 1 (µ). Therefore, by the dominated convergence theorem, f φ n p dµ = lim n lim f φ n p dµ =0. n

2 Approximation Theorems and Convolutions The goal of this section is to find a number of other dense subspaces of L p (µ) for p [1, ). The next theorem is the key result of this section. Theorem 22.4 (Density Theorem). Let p [1, ), (, M,µ) be a measure space and M be an algebra of bounded F valued (F = or F = C) measurable functions such that 1. M L p (µ, F) and σ (M) =M. 2. There exists ψ k M such that ψ k 1 boundedly. 3. If F = C we further assume that M is closed under complex conjugation. Then to every function f L p (µ, F), there exists φ n lim n kf φ n k Lp (µ) =0, i.e. M is dense in Lp (µ, F). M such that Proof. Fix k N for the moment and let H denote those bounded M measurable functions, f : F, for which there exists {φ n } n=1 M such that lim n kψ k f φ n k L p (µ) =0. AroutinecheckshowsH is a subspace of (M, F) such that 1 H, M H and H is closed under complex conjugation if F = C. Moreover, H is closed under bounded convergence. To see this suppose f n H and f n f boundedly. Then, by the dominated convergence theorem, lim n kψ k (f f n )k L p (µ) =0.1 (Take the dominating function to be g = [2C ψ k ] p where C is a constant bounding all of the { f n } n=1.) We may now choose φ n M such that kφ n ψ k f n k L p (µ) 1 n then lim sup kψ k f φ n k L n p (µ) lim sup kψ k (f f n )k L n p (µ) + lim sup kψ k f n φ n k L n p (µ) =0 (22.2) which implies f H. An application of Dynkin s Multiplicative System Theorem if F = or Theorem if F = C now shows H contains all bounded measurable functions on. Let f L p (µ) be given. The dominated convergence theorem implies lim ψk k 1 { f k} f f L = 0. (Take the dominating function to be p (µ) g =[2C f ] p where C is a bound on all of the ψ k.) Using this and what we have just proved, there exists φ k M such that ψk 1 { f k} f φ L k 1 p (µ) k. The same line of reasoning used in Eq. (22.2) now implies lim k kf φ k k Lp (µ) = 0. 1 It is at this point that the proof would break down if p =.

3 22.1 Density Theorems 421 Definition Let (, τ) be a topological space and µ be a measure on B = σ (τ). A locally integrable function is a Borel measurable function f : C such that f dµ < for all compact subsets K. We will K write L 1 loc (µ) for the space of locally integrable functions. More generally we say f L p loc (µ) iff k1 Kfk L p (µ) < forallcompactsubsetsk. Definition Let (, τ) be a topological space. A K-finite measure on is Borel measure µ such that µ (K) < for all compact subsets K. Lebesgue measure on is an example of a K-finite measure while counting measure on is not a K-finite measure. Example Suppose that µ is a K-finite measure on B. An application of Theorem 22.4 shows C c (, C) is dense in L p (, B,µ; C). To apply Theorem 22.4, let M := C c, C and ψ k (x) :=ψ (x/k) where ψ C c, C with ψ (x) =1in a neighborhood of 0. The proof is completed by showing σ (M) = σ C c, C = B, which follows directly from Lemma We may also give a more down to earth proof as follows. Let x 0,> 0, A:= B (x 0,) c and f n (x) :=d 1/n A (x). Then f n M and f n 1 B(x0,) as n which shows 1 B(x0,) is σ (M)-measurable, i.e. B (x 0,) σ (M). Since x 0 and >0 were arbitrary, σ (M) =B. More generally we have the following result. Theorem Let (, τ) be a second countable locally compact Hausdorff space and µ : B [0, ] be a K-finite measure. Then C c () (the space of continuous functions with compact support) is dense in L p (µ) for all p [1, ). (See also Proposition below.) Proof. Let M := C c () and use Item 3. of Lemma to find functions ψ k M such that ψ k 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 Exercise Show that BC (, C) is not dense in L (, B,m; C). Hence the hypothesis that p< in Theorem 22.4 can not be removed. Corollary Suppose n is an open set, B is the Borel σ algebra on and µ be a K-finite measure on (, B ). Then C c () is dense in L p (µ) for all p [1, ). Corollary Suppose that is a compact subset of n and µ is a finite measure on (, B ), then polynomials are dense in L p (, µ) for all 1 p<. Proof. Consider to be a metric space with usual metric induced from n. Then is a locally compact separable metric space and therefore

4 Approximation Theorems and Convolutions C c (, C) =C(, C) is dense in L p (µ) for all p [1, ). Since, by the dominated convergence theorem, uniform convergence implies L p (µ) convergence, it follows from the Weierstrass approximation theorem (see Theorem 8.34 and Corollary 8.36 or Theorem and Corollary 12.32) that polynomials are also dense in L p (µ). Lemma Let (, τ) be a second countable locally compact Hausdorff space and µ : B [0, ] be a K-finite measure on. If h L 1 loc (µ) is a function such that fhdµ =0for all f C c () (22.3) then h(x) =0for µ a.e.x. (See also Corollary below.) Proof. Let dν(x) = h(x) dx, then ν is a K-finite measure on and hence C c () is dense in L 1 (ν) by Theorem Notice that f sgn(h)dν = fhdµ =0for all f C c (). (22.4) Let {K k } k=1 be a sequence of compact sets such that K k as in Lemma Then 1 Kk sgn(h) L 1 (ν) and therefore there exists f m C c () such that f m 1 Kk sgn(h) in L 1 (ν). So by Eq. (22.4), ν(k k )= 1 Kk dν = lim f m sgn(h)dν =0. m Since K k as k, 0=ν() = h dµ, i.e. h(x) =0for µ a.e.x. As an application of Lemma and Example 12.34, we will show that the Laplace transform is injective. Theorem (Injectivity of the Laplace Transform). For f L 1 ([0, ),dx), the Laplace transform of f is defined by Lf(λ) := If Lf(λ) :=0then f(x) =0for m -a.e. x. 0 e λx f(x)dx for all λ>0. Proof. Suppose that f L 1 ([0, ),dx) such that Lf(λ) 0. Let g C 0 ([0, ), ) and ε>0 be given. By Example we may choose {a λ } λ>0 such that #({λ >0:a λ 6=0}) < and g(x) λ>0 a λ e λx <εfor all x 0. Then

5 0 g(x)f(x)dx = Density Theorems 423 Ã g(x)! a λ e λx f(x)dx λ>0 g(x) a λ e λx f(x) dx εkfk 1. λ>0 Since ε > 0 is arbitrary, it follows that 0 g(x)f(x)dx = 0 for all g C 0 ([0, ), ). The proof is finished by an application of Lemma Here is another variant of Theorem Theorem Let (, d) be a metric space, τ d be the topology on generated by d and B = σ(τ d ) be the Borel σ algebra. Suppose µ : B [0, ] is a measure which is σ finite on τ d and let BC f () denote the bounded continuous functions on such that µ(f 6= 0)<. Then BC f () is a dense subspace of L p (µ) for any p [1, ). Proof. Let k τ d be open sets such that k and µ( k ) < and let ψ k (x) =min(1,k d c k (x)) = φ k (d c k (x)), see Figure 22.1 below. It is easily verified that M := BC f () is an algebra, y x Fig The plot of φ n for n =1, 2, and 4. Notice that φ n 1 (0, ). ψ k M for all k and ψ k 1 boundedly as k. Given V τ and k, n N,let f k,n (x) :=min(1,n d (V k ) c(x)). Then {f k,n 6=0} = V k so f k,n BC f (). Moreover lim lim f k,n = lim 1 V k n k =1 V k which shows V σ (M) and hence σ (M) =B. The proof is now completed by an application of Theorem 22.4.

6 Approximation Theorems and Convolutions Exercise (BUCE: Should drop this exercise.) Suppose that (, d) is a metric space, µ is a measure on B := σ(τ d ) which is finite on bounded measurable subsets of. Show BC b (, ), defined in Eq. (19.26), is dense in L p (µ). Hints: let ψ k be as defined in Eq. (19.27) which incidentally may be used to show σ (BC b (, )) = σ (BC(, )). Then use the argument in the proof of Corollary to show σ (BC(, )) = B. Theorem Suppose p [1, ), A M is an algebra such that σ(a) = M and µ is σ finite on A. Then S f (A,µ) is dense in L p (µ). (See also emark 25.7 below.) Proof. Let M := S f (A,µ). By assumption there exits k A such that µ( k ) < and k as k. If A A, then k A A and µ ( k A) < so that 1 k A M. Therefore 1 A = lim k 1 k A is σ (M) measurable for every A A. So we have shown that A σ (M) M and therefore M = σ (A) σ (M) M, i.e. σ (M) =M. The theorem now follows from Theorem 22.4 after observing ψ k := 1 k M and ψ k 1 boundedly. Theorem (Separability of L p Spaces). Suppose, p [1, ), A M is a countable algebra such that σ(a) =M and µ is σ finite on A. Then L p (µ) is separable and D = { a j 1 Aj : a j Q + iq, A j A with µ(a j ) < } is a countable dense subset. Proof. It is left to reader to check D is dense in S f (A,µ) relative to the L p (µ) norm. The proof is then complete since S f (A,µ) is a dense subspace of L p (µ) by Theorem Example The collection of functions of the form φ = P n k=1 c k1 (ak,b k ] with a k,b k Q and a k <b k are dense in L p (, B,m; C) and L p (, B,m; C) is separable for any p [1, ). To prove this simply apply Theorem with A being the algebra on generated by the half open intervals (a, b] with a<band a, b Q {± }, i.e. A consists of sets of the form `n k=1 (a k,b k ], where a k,b k Q {± }. Exercise Show L ([0, 1], B,m; C) is not separable. Hint: Suppose Γ is a dense subset of L ([0, 1], B,m; C) and for λ (0, 1), let f λ (x) := 1 [0,λ] (x). For each λ (0, 1), choose g λ Γ such that kf λ g λ k < 1/2 and then show the map λ (0, 1) g λ Γ is injective. Use this to conclude that Γ must be uncountable. Corollary (iemann Lebesgue Lemma). Suppose that f L 1 (,m), then f(x)e iλx dm(x) =0. lim λ ±

7 22.1 Density Theorems 425 Proof. By Example 22.16, given ε>0 there exists φ = P n k=1 c k1 (ak,b k ] with a k,b k such that f φ dm < ε. Notice that φ(x)e iλx dm(x) = = n c k 1 (ak,b k ](x)e iλx dm(x) k=1 n bk c k k=1 a k e iλx dm(x) = n k=1 c k λ 1 e iλx b k ak n = λ 1 c k e iλb k e iλa k 0 as λ. k=1 Combining these two equations with f(x)e iλx dm(x) (f(x) φ(x)) e iλx dm(x) + φ(x)e iλx dm(x) f φ dm + φ(x)e iλx dm(x) ε + φ(x)e iλx dm(x) we learn that lim sup λ f(x)e iλx dm(x) ε + lim sup λ φ(x)e iλx dm(x) = ε. Since ε>0 is arbitrary, this completes the proof of the iemann Lebesgue lemma. Corollary Suppose A M is an algebra such that σ(a) =M and µ is σ finite on A. Then for every B M such that µ(b) < and ε>0 there exists D A such that µ(b4d) <ε.(see also emark 25.7 below.) Proof. By Theorem 22.14, there exists a collection, {A i } n i=1, of pairwise disjoint subsets of A and λ i such that 1 B f dµ < ε where f = P n i=1 λ i1 Ai. Let A 0 := \ n i=1 A i A then

8 Approximation Theorems and Convolutions 1 B f dµ = = µ (A 0 B)+ = µ (A 0 B)+ µ (A 0 B)+ n i=0 1 B f dµ A i " n # 1 B λ i dµ + 1 B λ i dµ A i B A i\b n [ 1 λ i µ (B A i )+ λ i µ (A i \ B)] (22.5) i=1 i=1 n min {µ (B A i ),µ(a i \ B)} (22.6) i=1 where the last equality is a consequence of the fact that 1 λ i + 1 λ i. Let ½ 0 if µ (B Ai ) <µ(a α i = i \ B) 1 if µ (B A i ) µ (A i \ B) and g = P n i=1 α i1 Ai =1 D where D := {A i : i>0 & α i =1} A. Equation (22.5) with λ i replaced by α i and f by g implies n 1 B 1 D dµ = µ (A 0 B)+ min {µ (B A i ),µ(a i \ B)}. i=1 The latter expression, by Eq. (22.6), is bounded by 1 B f dµ < ε and therefore, µ(b4d) = 1 B 1 D dµ < ε. emark We have to assume that µ(b) < as the following example shows. Let =, M = B, µ= m, A be the algebra generated by half open intervals of the form (a, b], and B = n=1(2n, 2n +1]. It is easily checked that for every D A, that m(b D) = Convolution and Young s Inequalities Throughout this section we will be solely concerned with d dimensional Lebesgue measure, m, and we will simply write L p for L p,m. Definition (Convolution). Let f,g : C be measurable functions. We define f g(x) = f(x y)g(y)dy (22.7)

9 22.2 Convolution and Young s Inequalities 427 whenever the integral is defined, i.e. either f (x ) g ( ) L 1 (,m) or f (x ) g ( ) 0. Notice that the condition that f (x ) g ( ) L 1 (,m) is equivalent to writing f g (x) <. Byconvention,iftheintegralinEq. (22.7) is not defined, let f g(x) :=0. Notation Given a multi-index α d +, let α = α α d, x α := dy j=1 µ α x α j j, and α x = := x dy µ j=1 x j αj. For z and f : C, let τ z f : C be defined by τ z f(x) =f(x z). emark (The Significance of Convolution). 1. Suppose that f,g L 1 (m) are positive functions and let µ be the measure on 2 defined by dµ (x, y) :=f (x) g (y) dm (x) dm (y). Then if h : [0, ] is a measurable function we have h (x + y) dµ (x, y) = h (x + y) f (x) g (y) dm (x) dm (y) ( ) 2 ( ) 2 = h (x) f (x y) g (y) dm (x) dm (y) ( ) 2 = h (x) f g (x) dm (x). 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 independent (i.e. product measure) random variables is the the convolution of the individual distributions. 2. Suppose that L = P α k a α α is a constant coefficient differential operator and suppose that we can solve (uniquely) the equation Lu = g in the form u(x) =Kg(x) := k(x, y)g(y)dy 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 inverse operation to differentiation.) Since τ z L = Lτ z for all z, (this is another way to characterize constant coefficient differential operators) and L 1 = K we should have τ z K = Kτ z. Writing out this equation then says k(x z, y)g(y)dy =(Kg)(x z) =τ z Kg(x) =(Kτ z g)(x) = k(x, y)g(y z)dy = k(x, y + z)g(y)dy.

10 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 Suppose p [1, ], f L 1 and g L p, then f g(x) exists for almost every x, f g L p and kf gk p kfk 1 kgk p. Proof. This follows directly from Minkowski s inequality for integrals, Theorem Proposition Suppose that p [1, ), then τ z : L p L p is an isometric isomorphism and for f L p,z τ z f L p is continuous. Proof. The assertion that τ z : L p L p is an isometric isomorphism follows from translation invariance of Lebesgue measure and the fact that τ z τ z = id. For the continuity assertion, observe that kτ z f τ y fk p = kτ y (τ z f τ y f)k p = kτ z y f fk p from which it follows that it is enough to show τ z f f in L p as z 0. When f C c ( ),τ z f f uniformly and since the K := z 1 supp(τ z f) is compact, it follows by the dominated convergence theorem that τ z f f in L p as z 0. For general g L p and f C c ( ), kτ z g gk p kτ z g τ z fk p + kτ z f fk p + kf gk p and thus = kτ z f fk p +2kf gk p lim sup kτ z g gk p lim sup kτ z f fk p +2kf gk p =2kf gk p. z 0 z 0 Because C c ( ) is dense in L p, the term kf gk p may be made as small as we please. Exercise Compute the operator norm, kτ z Ik L(L p (m)), of τ z I and use this to show z τ z L (L p (m)) is not continuous. Definition Suppose that (, τ) is a topological space and µ is a measure on B = σ(τ). For a measurable function f : C we define the essential support of f by supp µ (f) ={x : µ({y V : f(y) 6= 0}}) > 0 neighborhoods V of x}. (22.8) Equivalently, x/ supp µ (f) iff thereexistsanopenneighborhoodv of x such that 1 V f =0a.e.

11 22.2 Convolution and Young s Inequalities 429 It is not hard to show that if supp(µ) = (see Definition 21.41) and f C() then supp µ (f) = supp(f) :={f 6= 0}, see Exercise Lemma Suppose (, τ) is second countable and f : C is a measurable function and µ is a measure on B. Then := U \ supp µ (f) may be described as the largest open set W such that f1 W (x) =0for µ a.e.x. Equivalently put, C := supp µ (f) is the smallest closed subset of such that f = f1 C a.e. Proof. To verify that the two descriptions of supp µ (f) are equivalent, suppose supp µ (f) is defined as in Eq. (22.8) and W := \ supp µ (f). Then W = {x : τ 3 V 3 x such that µ({y V : f(y) 6= 0}}) =0} = {V o : µ (f1 V 6=0)=0} = {V o : f1 V =0for µ a.e.}. So to finish the argument it suffices to show µ (f1 W 6=0)=0. To to this let U be a countable base for τ and set U f := {V U : f1 V =0a.e.}. Then it is easily seen that W = U f and since U f is countable µ (f1 W 6=0) V U f µ (f1 V 6=0)=0. Lemma Suppose f,g,h : C aremeasurablefunctionsandassume that x is a point in 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) 3. If z and τ z ( f g )(x) = f g (x z) <, then τ z (f g)(x) =τ z f g(x) =f τ z g(x) 4. If x/ supp m (f) + supp m (g) then f g(x) =0and in particular, supp m (f g) supp m (f) + supp m (g) where in defining supp m (f g) we will use the convention that f g(x) 6= 0 when f g (x) =. Proof. For item 1., f g (x) = f (x y) g (y)dy = f (y) g (y x)dy = g f (x)

12 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 by f1 suppm (f) and g by g1 suppm (g) if necessary, assume that {f 6= 0} supp m (f) and {g 6= 0} supp m (g). So if x / (supp m (f)+supp m (g)) then x / ({f 6= 0} + {g 6= 0}) and for all y, either x y / {f 6= 0} or y / {g 6= 0}. That is to say either x y {f =0} or y {g =0} and hence f(x y)g(y) ³ =0for all y and therefore f g(x) =0. This shows that f g =0 on \ supp m (f)+supp m (g) and therefore ³ \ supp m (f) + supp m (g) \ supp m (f g), i.e. supp m (f g) supp m (f) + supp m (g). emark Let A, B be closed sets of, 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 compact and x n = a n + b n A + B and x n x, then by passing to a subsequence if necessary we may assume lim n a n = a A exists. In this case lim b n = lim (x n a n )=x a B n n exists as well, showing x = a + b A + B. Proposition Suppose that p, q [1, ] and p and q are conjugate exponents, f L p and g L q, then f g BC( ), kf gk kfk p kgk q and if p, q (1, ) then f g C 0 ( ). Proof. The existence of f g(x) and the estimate f g (x) kfk p kgk q for all x is a simple consequence of Holders inequality and the translation invariance of Lebesgue measure. In particular this shows kf gk kfk p kgk q. By relabeling p and q if necessary we may assume that p [1, ). Since kτ z (f g) f gk u = kτ z f g f gk u kτ z f fk p kgk q 0 as z 0 it follows that f g is uniformly continuous. Finally if p, q (1, ), we learn from Lemma and what we have just proved that f m g m C c ( ) where f m = f1 f m and g m = g1 g m. Moreover,

13 22.2 Convolution and Young s Inequalities 431 kf g f m g m k kf g f m gk + kf m g f m g m k kf f m k p kgk q + kf m k p kg g m k q kf f m k p kgk q + kfk p kg g m k q 0 as m showing, with the aid of Proposition 12.23, f g C 0 ( ). Theorem (Young s Inequality). Let p, q, r [1, ] satisfy 1 p + 1 q =1+1 r. (22.9) If f L p and g L q then f g (x) < for m a.e.x and kf gk r kfk p kgk q. (22.10) In particular L 1 is closed under convolution. (The space (L 1, ) is an example of a Banach algebra without unit.) emark Before going to the formal proof, let us first understand Eq. (22.9) by the following scaling argument. For λ>0, let f λ (x) :=f(λx), then after a few simple change of variables we find kf λ k p = λ d/p kfk and (f g) λ = λ d f λ g λ. Therefore if Eq. (22.10) holds for some p, q, r [1, ], we would also have kf gk r = λ d/r k(f g) λ k r λ d/r λ kf λ k p kg λ k q = λ (d+d/r d/p d/q) kfk p kgk q 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 positive functions. Let α, β [0, 1] and p 1,p 2 (0, ] satisfy p 1 1 +p 1 2 +r 1 = 1. Then by Hölder s inequality, Corollary 21.3, h f g(x) = f(x y) (1 α) g(y) (1 β)i f(x y) α g(y) β dy µ 1/r µ 1/p1 f(x y) (1 α)r g(y) (1 β)r dy f(x y) αp1 dy µ d 1/p2 g(y) βp2 dy µ d 1/r = f(x y) (1 α)r g(y) (1 β)r dy kfk α αp 1 kgk β βp 2. Taking the r th power of this equation and integrating on x gives

14 Approximation Theorems and Convolutions µ kf gk r r f(x y) (1 α)r g(y) (1 β)r dy dx kfk α αp 1 kgk β βp 2 = kfk (1 α)r (1 α)r kgk(1 β)r (1 β)r kfkαr αp 1 kgk βr βp 2. (22.11) Let us now suppose, (1 α)r = αp 1 and (1 β)r = βp 2, in which case Eq. (22.11) becomes, kf gk r r kfkr αp 1 kgk r βp 2 which is Eq. (22.10) with p := (1 α)r = αp 1 and q := (1 β)r = βp 2. (22.12) So to finish the proof, it suffices to show p and q are arbitrary indices in [1, ] satisfying p 1 +q 1 =1+r 1. If α, β, p 1,p 2 satisfy the relations above, then α = r and β = r r + p 1 r + p 2 and 1 p + 1 q = = 1 r + p r + p 2 αp 1 αp 2 p 1 r p 2 r = 1 p p r =1+1 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. α := r p r =1 p r 1 and β = r q r Using Eq. (22.9) we may also express α and β as =1 q r 1. α = p(1 1 q ) 0 and β = q(1 1 p ) 0 and in particular we have shown α, β [0, 1]. If we now define p 1 := p/α (0, ] and p 2 := q/β (0, ], then as desired. 1 p p r = β 1 q + α1 p + 1 r =(1 1 q )+(1 1 p )+1 r µ = r r =1 Theorem (Approximate δ functions). Let p [1, ], φ L 1 ( ),a:= f(x)dx, and for t>0 let φ t (x) =t d φ(x/t). Then

15 22.2 Convolution and Young s Inequalities If f L p with p< then φ t f af in L p as t If f BC( ) and f is uniformly continuous then kφ t f afk 0 as t If f L and f is continuous on U o then φ t f af uniformly on compact subsets of U as t 0. Proof. Making the change of variables y = tz implies φ t f(x) = f(x y)φ t (y)dy = f(x tz)φ(z)dz so that φ t f(x) af(x) = [f(x tz) f(x)] φ(z)dz d = [τ tz f(x) f(x)] φ(z)dz. (22.13) Hence by Minkowski s inequality for integrals (Theorem 21.27), Proposition and the dominated convergence theorem, kφ t f afk p kτ tz f fk p φ(z) dz 0 as t 0. Item 2. is proved similarly. Indeed, form Eq. (22.13) kφ t f afk kτ tz f fk φ(z) dz which again tends to zero by the dominated convergence theorem because lim t 0 kτ tz f fk =0uniformly in z by the uniform continuity of f. Item 3. Let B = B(0,) be a large ball in and U, then sup φ t f(x) af(x) x K [f(x tz) f(x)] φ(z)dz + [f(x tz) f(x)] φ(z)dz B B c φ(z) dz sup f(x tz) f(x) +2kfk φ(z) dz B x K,z B B c kφk 1 sup f(x tz) f(x) +2kfk φ(z) dz x K,z B z > so that using the uniform continuity of f on compact subsets of U, lim sup t 0 sup φ t f(x) af(x) 2 kfk x K z > φ(z) dz 0 as. See Theorem 8.15 if Folland for a statement about almost everywhere convergence.

16 Approximation Theorems and Convolutions Exercise Let Show f C (, [0, 1]). f(t) = ½ e 1/t if t>0 0 if t 0. Lemma There exists φ C c (, [0, )) such that φ(0) > 0, supp(φ) B(0, 1) and φ(x)dx =1. Proof. Define h(t) =f(1 t)f(t +1) where f is as in Exercise Then h C c (, [0, 1]), supp(h) [ 1, 1] and h(0) = e 2 > 0. Define c = h( x 2 )dx. Then φ(x) =c 1 h( x 2 ) is the desired function. The reader asked to prove the following proposition in Exercise 22.9 below. Proposition Suppose that f L 1 loc (d,m) and φ Cc 1 ( ), then f φ C 1 ( ) and i (f φ) = f i φ. Moreover if φ Cc ( ) then f φ C ( ). Corollary (C Uryhson s Lemma). Given U o, there exists f C c (, [0, 1]) such that supp(f) U and f =1on K. Proof. Let φ be as in Lemma 22.33, φ t (x) =t d φ(x/t) be as in Theorem 22.32, d be the standard metric on and ε = d(k, U c ). Since K is compact and U c is closed, ε>0. Let V δ = x : d(x, K) <δ ª and f = φ ε/3 1 Vε/3, then supp(f) supp(φ ε/3 )+V ε/3 V 2ε/3 U. Since V 2ε/3 is closed and bounded, f Cc (U) and for x K, f(x) = 1 d(y,k)<ε/3 φ ε/3 (x y)dy = φ ε/3 (x y)dy =1. 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 Lemma (Integration by Parts). Suppose f and g are measurable functions on such that t f(x 1,...,x i 1,t,x i+1,...,x d ) and t g(x 1,...,x i 1,t,x i+1,...,x d ) are continuously differentiable functions on for each fixed x = (x 1,...,x d ). Moreover assume f g, f g x i are in L 1 (,m). Then f gdm = f g dm. x i x d i f x i g and With this result we may give another proof of the iemann Lebesgue Lemma.

17 22.2 Convolution and Young s Inequalities 435 Lemma (iemann Lebesgue Lemma). For f L 1 (,m) let ˆf(ξ) :=(2π) d/2 f(x)e iξ x dm(x) d be the Fourier transform of f. Then ˆf C 0 ( ) and ˆf (2π) d/2 kfk 1. (The choice of the normalization factor, (2π) d/2, in ˆf is for later convenience.) Proof. The fact that ˆf is continuous is a simple application of the dominated convergence theorem. Moreover, ˆf(ξ) f(x) dm(x) (2π) d/2 kfk 1 so it only remains to see that ˆf(ξ) 0 as ξ. First suppose that f Cc ( ) and let = P d j=1 2 be the Laplacian on. Notice that x 2 j x j e iξ x = iξ j e iξ x and e iξ x = ξ 2 e iξ x. Using Lemma repeatedly, k f(x)e iξ x dm(x) = f(x) k xe iξ x dm(x) = ξ 2k f(x)e iξ x dm(x) = (2π) d/2 ξ 2k ˆf(ξ) for any k N. Hence (2π) d/2 ˆf(ξ) ξ 2k k f 1 0 as ξ and ˆf C 0 ( ). Suppose that f L 1 (m) and f k Cc ( ) is a sequence such that lim k kf f k k 1 =0, then lim k ˆf ˆfk =0. Hence ˆf C 0 ( ) by an application of Proposition Corollary Let be an open set and µ be a adon measure on B. 1. Then Cc () is dense in L p (µ) for all 1 p<. 2. If h L 1 loc (µ) satisfies fhdµ =0for all f Cc () (22.14) then h(x) =0for µ a.e.x. Proof. Let f C c (), φbe as in Lemma 22.33, φ t be as in Theorem and set ψ t := φ t (f1 ). Then by Proposition ψ t C () and by Lemma there exists a compact set K such that supp(ψ t ) K for all t sufficiently small. By Theorem 22.32, ψ t f uniformly on as t 0

18 Approximation Theorems and Convolutions 1. The dominated convergence theorem (with dominating function being kfk 1 K ), shows ψ t f in L p (µ) as t 0. This proves Item 1., since Theorem 22.8 guarantees that C c () is dense in L p (µ). 2. Keeping the same notation as above, the dominated convergence theorem (with dominating function being kfk h 1 K ) implies 0 = lim t 0 ψ t hdµ = lim ψ t hdµ = t 0 fhdµ. The proof is now finished by an application of Lemma Smooth Partitions of Unity We have the following smooth variants of Proposition 12.16, Theorem and Corollary The proofs of these results are the same as their continuous counterparts. One simply uses the smooth version of Urysohn s Lemma of Corollary in place of Lemma Proposition (Smooth Partitions of Unity for Compacts). Suppose that is an open subset of,k is a compact set and U = {U j } n j=1 is an open cover of K. Then there exists a smooth (i.e. h j C (, [0, 1])) partition of unity {h j } n j=1 of K such that h j U j for all j =1, 2,...,n. Theorem (Locally Compact Partitions of Unity). Suppose that is an open subset of and U is an open cover of. Then there exists a smooth partition of unity of {h i } N i=1 (N = is allowed here) subordinate to the cover U such that supp(h i ) is compact for all i. Corollary Suppose that is an open subset of and U = {U α } α A τ is an open cover of. Then there exists a smooth partition of unity of {h α } α A subordinate to the cover U such that supp(h α ) U α for all α A. Moreover if Ūα is compact for each α A we may choose h α so that h α U α Exercises Exercise Let (, τ) be a topological space, µ ameasureonb = σ(τ) and f : C be a measurable function. Letting ν be the measure, dν = f dµ, show supp(ν) = supp µ (f), where supp(ν) is definedindefinition 21.41). Exercise Let (, τ) be a topological space, µ ameasureonb = σ(τ) such that supp(µ) = (see Definition 21.41). Show supp µ (f) = supp(f) = {f 6= 0} for all f C().

19 22.3 Exercises 437 Exercise Prove the following strong version of item 3. of Proposition 10.52, namely to every pair of points, x 0, x 1, in a connected open subset V of there exists σ C (,V) such that σ(0) = x 0 and σ(1) = x 1. Hint: First choose a continuous path γ :[0, 1] V such that γ (t) =x 0 for t near 0 and γ (t) =x 1 for t near 1 and then use a convolution argument to smooth γ. Exercise Prove Proposition by appealing to Corollary Exercise (Integration by Parts). Suppose that (x, y) 1 f(x, y) C and (x, y) 1 g(x, y) C are measurable functions such that for each fixed y,x f(x, y) and x g(x, y) are continuously differentiable. Also assume f g, x f g and f x g areintegrablerelativeto Lebesgue measure on 1, where x f(x, y) := d dt f(x + t, y) t=0. Show x f(x, y) g(x, y)dxdy = f(x, y) x g(x, y)dxdy. (22.15) 1 1 (Note: this result and Fubini s theorem proves Lemma ) Hints: Let ψ Cc () be a function which is 1 in a neighborhood of 0 and set ψ ε (x) =ψ(εx). First verify Eq. (22.15) with f(x, y) replaced by ψ ε (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. Exercise Let µ be a finite measure on B, then D := span{e iλ x : λ } is a dense subspace of L p (µ) for all 1 p<. Hints: By Theorem 22.8, C c ( ) is a dense subspace of L p (µ). For f C c ( ) and N N, let f N (x) := n d f(x +2πNn). Show f N BC( ) and x f N (Nx) is 2π periodic, so by Exercise 12.13, x f N (Nx) can be approximated uniformly by trigonometric polynomials. Use this fact to conclude that f N D Lp (µ). After this show f N f in L p (µ). Exercise Suppose that µ and ν are two finite measures on such that e iλ x dµ(x) = e iλ x dν(x) (22.16) for all λ. Show µ = ν. Hint: Perhaps the easiest way to do this is to use Exercise with the measure µ being replaced by µ + ν. Alternatively, use the method of proof of Exercise to show Eq. (22.16) implies fdµ(x) = fdν(x) for all f C c ( ) and then apply Corollary Exercise Again let µ be a finite measure on B. Further assume that C M := e M x dµ(x) < for all M (0, ). Let P( ) be the space of polynomials, ρ(x) = P α N ρ αx α with ρ α C, on. (Notice that ρ(x) p

20 Approximation Theorems and Convolutions Ce M x for some constant C = C(ρ, p, M), so that P( ) L p (µ) for all 1 p<.) Show P( ) is dense in L p (µ) for all 1 p<. Here is a possible outline. Outline: Fix a λ and let f n (x) =(λ x) n /n! for all n N. 1. Use calculus to verify sup t 0 t α e Mt =(α/m) α e α for all α 0 where (0/M ) 0 := 1. Usethisestimatealongwiththeidentity λ x pn λ pn x pn = ³ x M x pn e λ pn e M x to find an estimate on kf n k p. 2. Use your estimate on kf n k p to show P n=0 kf nk p < and conclude N lim eiλ ( ) i n f N n =0. p n=0 3. Now finish by appealing to Exercise Exercise Again let µ be a finite measure on B but now assume there exists an ε>0 such that C := e ε x dµ(x) <. Also let q>1and h L q (µ) be a function such that h(x)x α dµ(x) =0for all α N d 0. (As d mentioned in Exercise 22.14, P( ) L p (µ) for all 1 p<, so x h(x)x α is in L 1 (µ).) Show h(x) =0for µ a.e.x using the following outline. Outline: Fix a λ, let f n (x) =(λ x) n /n! for all n N, and let p = q/(q 1) be the conjugate exponent to q. 1. Use calculus to verify sup t 0 t α e εt =(α/ε) α e α for all α 0 where (0/ε) 0 := 1. Use this estimate along with the identity λ x pn λ pn x pn = ³ x ε x pn e λ pn e ε x to find an estimate on kf n k p. 2. Use your estimate on kf n k p to show there exists δ > 0 such that P n=0 kf nk p < when λ δ and conclude for λ δ that e iλ x = L p (µ)- P n=0 in f n (x). Conclude from this that h(x)e iλ x dµ(x) =0when λ δ. 3. Let λ ( λ not necessarily small) and set g(t) := e itλ x h(x)dµ(x) for t. Show g C () and g (n) (t) = (iλ x) n e itλ x h(x)dµ(x) for all n N.

21 22.3 Exercises Let T =sup{τ 0:g [0,τ] 0}. By Step 2., T δ. If T<, then 0=g (n) (T )= (iλ x) n e it λ x h(x)dµ(x) for all n N. Use Step 3. with h replaced by e it λ x h(x) to conclude g(t + t) = e i(t +t)λ x h(x)dµ(x) =0for all t δ/ λ. This violates the definition of T and therefore T = andinparticular we may take T =1to learn h(x)e iλ x dµ(x) =0for all λ. 5. Use Exercise to conclude that h(x)g(x)dµ(x) =0 for all g L p (µ). Now choose g judiciously to finish the proof.