Differentiation and function spaces

Transcription

1 209: Honors Analysis in R n Differentiation and function spaces 1 Banach Space, Hilbert Space A vector (linear) space E over a field K is a set endowed with two operations + : E E E and : K E E that satisfy the familiar axioms [(1) : f + g = g + f], [(2) : 0, f + 0 = f], [(3) : (f + g) + h = f + (g + h)], [(4) : f, f + ( f) = 0], [(5) : 1 f = f], [(6) : λ, µ K, (λµ) f = λ (µ f)], [(6) : λ (f + g) = λ f + λ g]. Definition 1 Let E be a real or complex vector space. A function : E R + is a norm on E if f = 0 f = 0 λf = λ f, λ K (K = R or K = C f + g f + g. A vector space endowed with a norm is called a normed vector space. The norm defines a topology. An set U is open if for any f U there exists r > 0 so that B(f, r) U where B(f, r) = {g f g < r}. The norm defines a distance d(f, g) = f g. Definition 2 We say that E is a Banach space if it is a normed vector space that is complete, i.e. every Cauchy sequence converges. Basic examples of Banach spaces are: R n, C(K), L p (dµ), for µ a complete measure on a measurable space (X, Σ), with special case l p (Z), 1 p. For K R n a compact set with norm f C = sup x K f(x). C(K) = {f : K R f continuous} Exercise 1 C(K) is a real Banach space. Tha analogous space of complex valued functions is a complex Banach space. L p (dµ) is a real or complex Banach space. 1

2 Exercise 2 Let T : X Y be a linear operator between two normed linear spaces. Linear means that T (f + g) = T f + T g and T (λf) = λt f. The following are equivalent: 1. T is continuous. 2. T is continuous at C 0 such that T (x) C x ho lds x X. Exercise 3 If T : X Y is linear and continuous then T x sup x =0 x = sup T x {x; x 1} x = holds. The common value is denoted T. sup T x {x; x =1} Exercise 4 If T : X Y is linear and continuous and if S : Y Z is linear and continuous, then ST : X Z defined by ST (x) = S(T (x)) is linear and continuous. Exercise 5 Let E = R n, F = R m, e j = (δ jk ) j,k=1,...n and f α = (δ αβ ) α,β=1,...,m the canonical bases. Let T : E F be a linear operator. Prove that it is continuous. Associate to T the matrix (t αi ) α=1,...m,i=1,...n defined by m β=1 t βif β = T (e i ). Prove that the matrix associated to the composition of two operators T S is the product of the corresponding matrices. Proposition 1 Let L(E, F ) = {T : E F T linear and continuous}. Then T T is a norm on L(E, F ) and L(E, F ) is a Banach space. Theorem 1 (Hahn-Banach, real). Let T : S R be a linear map, defined on a linear subspace of the real vector space X. Assume that there exists a real valued function p : X R + such that p(x + y) p(x) + p(y) and p(tx) = tp(x) for all x, y in X and all t 0. Assume that T (x) p(x) holds for all x S. Then, there exists a linear functional F : X R that satisfies F (x) = T (x) for all x S and F (x) p(x) for all x X. Theorem 2 (Hahn-Banach, complex). Let X be a complex vector space, S a linear subspace, p a nonnegative real valued function that satsifies p(x + y) p(x)+p(y) and p(zx) = z p(x) for all x, y in X and all z C. Let T : S C be a linear functional satisfying T (x) p(x) for all x S. Then there exists a linear functional F defined on X such that F (x) = T (x) for allx S and F (x) p(x) for all x X 2

3 Proofs: Please see Kolmogorov and Fomin, Chapter 4, section 14. Definition 3 A scalar product on a real (complex) vector space H is a bilinear (sesquilinear) map H H C, denoted (f, g), that is, a map with the properties (f, g) = (g, f), (λf 1 + µf 2, g) = λ(f 1, g) + µ(f 2, g), (f, f) 0, (f, f) = 0 f = 0. λ µ C The scalar product (hermitian product, dot product) defines a norm The Schwarz inequality holds f = (f, f). (f, g) f g. Definition 4 We say that H is a Hilbert space if it is a real or complex vector space endowed with a scalar product such that H is complete, i.e. any Cauchy sequence converges. A basic example of a Hilbert space is L 2 (dµ) = {f f 2 dµ < } where µ is a complete measure on a measurable space (X, Σ). The scalar product is (f, g) = f(x)g(x)dµ In any Hilbert space we have the parallelogram identity X f + g 2 + f g 2 = 2( f 2 + g 2 ). A set C H is convex if, whenever f, g C then (1 t)f + tg C holds for all t [0, 1]. Proposition 2 Let C H be a closed convex subset of a Hilbert space. Then the minimum norm is attained: c C such that c = inf f C f 3

4 Proof. Let d = inf f C f. Let f n C be such that lim n f n = d. By the parallelogram identity [ f n f n+p 2 1 ( = 4 fn 2 + f n+p 2) f n + f n+p ] Suppose that ɛ > 0 is given, and we choose N such that, for n N we have f n 2 d 2 + ɛ4 4. Then, because of C s convexity, we have 1 2 (f n + f n+p ) C, hence, by the definition of d, 1 2 (f n + f n+p ) 2 d 2. It follows then that f n f n+p ɛ holds for all n N, p 0. Thus, the sequence f n is Cauchy. Because H is Hilbert, the sequence converges, f n c, and because C is closed and f n C, the limit c C. Because the norm is continuous we obtain c = d.. If S H is any set, we define S = {f H (f, s) = 0, s S} Exercise 6 S is a closed, linear subspace of H. (That means that S is a closed set and that it is a vectorial subspace of H). Proposition 3 Let M H be a closed linear sunbspace. Then for every h H there exists a unique m M such that h m M. The identity h 2 = m 2 + h m 2 holds. The map P : H M that assigns P h = m is linear, continuous and onto M. P is idempotent, i.e. composition with itself does not change it, P 2 = P. P is selfadjoint, i.e. (P f, g) = (f, P g) holds. The map P is called the orthogonal projector onto M. The proof follows from the previous proposition by observing that h + M is closed and convex. Then, let q x + M be of minimum norm. We can write q = x m with m M (because M is a linear space). Assume by contradiction that there exists n M such that (q, n) 0. Without loss of generality we may assume that (q, n) > 0. (Indeed, this is achieved by setting n = zn with z = (q, n).) Then a simple calculation shows that q 2 > q tn 2 can be achieved by taking 0 < t small enough. This contradicts the minimality. 4

5 Theorem 3 (Riesz) Let L : H C be a continuous linear map from the Hilbert space H. Then there exists a unique f H such that Lh = (h, f) holds for all h H. Proof. Consider kerl = {h Lh = 0} and assume it is not the entire H. (If it is then f = 0 will do the job). Then let g (kerl). The existence of g 0 is guaranteed by the orthogonal projector theorem, by taking h / kerl and setting g = h P h. Then, for any element f H we note that h = L(g)f (Lf)g kerl, and thus (h, g) = 0 holds. This implies that Lf = (f, g) holds for all f with g = [ g 2 L(g)]g.. A family {e a } a A of elements in a Hilbert space is said to be orthonormal if (e a, e b ) = δ ab where δ ab = 1 if a = b and δ ab = 0 if a b. We define for any f H, f(a) = (f, e a ). Let F A be a finite set. Consider M to be the linear subspace generated by {e a } a F. Then the orthogonal projector onto M is given by P f = a F f(a)e a. Note that one can think of P f as a solution to a variational problem: Find the element m of M that best approximates f, i.e. that minimizes the distance f m. An orthonormal set {e a } A is said to be complete if it is maximal (under inclusion). Theorem 4 TFAE 1. {e a } A is maximal orthonormal 2. {e a } A is orthonormal and the linear space it generates is dense. 3. f 2 = a A f(a) 2 The sum in 3 is the integral with respect to counting measure H 0. i.e. the supremum of all finite sums. Complete orthonormal sets are called orthonormal bases. The cardinality of the base is an invariant of the Hilbert space. Hilbert spaces with countable bases are called separable Hilbert spaces. 2 Differentiation Definition 5 Let U E be an open set in the Banach space E and let f : U F where F is another Banach space. We say that f is Fréchet 5

6 differentiable at x 0 U if there exists a bounded linear operator A : E F such that, for every ɛ > 0 there exists δ > 0 such that f(x) f(x 0 ) A(x x 0 ) ɛ x x 0 holds for all x U with x x 0 δ. The map A is called the derivative of f at at x 0. When E = R n and F = R m then this is the usual definition of derivative. We will use the notation The chain rule: A = (Df)(x 0 ) Proposition 4 Let E, F, G be Banach spaces, U open in E, x 0 U, V open in F, f(x 0 ) V, and f : U E F, g : V G. Assume that f is Fréchet differentiable at x 0, g is continuous on V and and Fréchet differentiable at f(x 0 ). Then h = g f defined on U g 1 (V ) is is differentiable at x 0 and Dh(x 0 ) = (Dg(f(x 0 )))(Df)(x 0 ) Definition 6 The function f : U F is Gateaux differentiable at x 0 U if, for any v E, with v = 1, the map defined on a neighborhood of 0 in R is differentiable at t = 0, i.e. t f(x 0 + tv) f(x 0 + tv) f(x 0 ) (D v f)(x 0 ) = lim t 0, t 0 t exists. The derivative is called the directional derivative of f in the direction v at x 0. When E = R n, v = e i = (δ ij ) j=1,...n then the directional derivative is called the partial derivative. We will use the notation f(x + te i ) f(x) ( i f)(x) = lim t 0, t 0 t Proposition 5 If f : U F is Fréchet differentiable at x 0 Gateaux differentiable and then it is (D v f)(x 0 ) = Df(x 0 )v 6

7 3 Function Spaces A multiindex α = (α 1,..., α n ) is an element of Z n with nonnegative entries. Its length is α = α α n. The Euclidean norm of an element x R n is denoted the same way, x = x x 2 n. An open ball in R n of radius r is B(x, r) = {y; x y < r}. The factorial of a multiindex is α! = i α i!. The derivatives are denoted α : α f = α f α 1 x 1... x αn n The vector f = ( 1 f,..., n f) is the gradient, and i f = f xi. If U is an open subset of R n and m is nonnegative, we say that s function f : U R is in C m (U) if f is m times differentiable and α f are continuous for α m. When m = 0 we sometimes write C(U). The support of a continuous function is the complement of the largest open set where the function is identically zero. The subset of C m (U) formed with compactly supported functions is denoted C0 m (U). If a function f has continuous derivatives of any order we say that f is infinitely differentiable, f C (U). If the function has compact support, we write f C0 (U). If 1 p < we define ( ) 1 f p = f p p dx for any measurable function in U. If the integral is finite we say that f L p (U). Here dx = λ is Lebesgue measure. For p = we define U f = inf{c 0 ; f c λ a.e. in U} Sometimes we write L p short for L p (U). The L p spaces can be defined starting from any complete measure µ. Theorem 5 Let 1 p <. Let f be a measurable function in U. Assume that there exists a constant C such that fφdx C φ p holds for all φ C 0 (U). Then f L p (U) and U f p C. 7

8 Remark 1 Note carefully that p = is not allowed in the theorem. The theorem together with Hölder s inequality imply f p = sup φfdx. φ p 1 where the supremum is taken over all C 0 (U) functions with L p norm less or equal to 1. The range of p is (1, ]. Definition 7 If f L p (R n ) and φ C0 (R n ) we define the convolution φ f by (φ f)(x) = φ(y)f(x y)dy Proposition 6 For φ, ψ, χ C 0 (R n ): (i) φ ψ = ψ φ (ii) (φ ψ) χ = φ (ψ χ) (iii) α (φ ψ) = ( α φ) ψ = φ ( α ψ) (iv.) supp(φ ψ) suppφ + suppψ. Here suppφ means the support of the function φ and A + B = {x; x = a + b, a A, b B}. Proposition 7 (Young) Let f L p, 1 p and φ C 0 (R n ). Then holds U φ f L p φ 1 f p Proposition 8 Let 1 p <. Then C 0 (U) is dense in L p (U), that is for any f L p (U) there exists a sequence of functions φ j C 0 (U) that converges in L p to f. Remark 2 Note that p = is not allowed. Note also that C 0 (U) L p (U). Exercise 7 Give an example of a function f L (R) such that there is no sequence of compactly supported continuous functions that converges to it in L. (Hint: what can you say about the uniform limit of a sequence of continuous functions?) 8

9 Definition 8 Let h R n. We denote the translate of f by h by τ h (f) τ h (f)(x) = f(x h). We denote the finite difference by δ h δ h (f) = f(x h) f(x) Exercise 8 Let f L p (R n ), 1 p <. Let ɛ > 0. Then there exists δ > 0 so that δ h f p ɛ holds for all h R n with h δ. (Hint: approximate first f in L p by a continuous function with compact support. Then use the uniform continuity of that function.) Exercise 9 Let f L p and g L p that f g is uniformly continuous. with p, p conjugate exponents. Prove Proposition 9 Let φ L 1 (R n ) and assume that φdx = 1 Consider for ɛ > 0 ( x ) φ ɛ (x) = ɛ n φ ɛ Let f L p, 1 p <. Then lim f (φ ɛ f) p = 0 ɛ 0 Idea of proof. Because φdx = 1 we have φ ɛ f f = φ(z)δ ɛz (f)dz Then φ ɛ f f p φ(z) δ ɛz f p dz holds. The proof is completed by using Lebesgue dominated convergence. 9

10 Definition 9 Take φ C0 (R n ), φ 0, φdx = 1, suppφ {x; x < 1}. The family ( x ) φ ɛ = ɛ n φ ɛ is called a standard mollifier. Proposition 10 Assume that φ C 0 (R n ), f L p (R n ), 1 p <. Then φ f C (R n ) and α (φ f) = ( α φ) f Proposition 11 C 0 (U) is dense in L p (U), 1 p <. Idea of proof. By Proposition 13 it is enough to approximate continuous functions with compact support. Convolution with a standard mollifier will do the trick. Proposition 12 Let K U be a compact included in an open set in R n. Then there exists ψ C 0 (U) such that 0 ψ 1 and ψ(x) = 1 holds for all x K. Idea of proof. Take ɛ > 0 small enough so that dist(k, U) > 3ɛ. (Any point in K has a ball centered at x and of radius 3ɛ included in U). K ɛ = {x : k K, k x ɛ}. Take φ ɛ a standard mollifier and put ψ = φ ɛ 1 Kɛ. Note that if x K and z suppφ ɛ then x z K ɛ and therefore ψ(x) = φɛ (z)dz = 1. Proposition 13 (Partition of unity). Let U = a A U a be a union of open sets in R n. There exists a countable family ψ j C 0 (R n ), j N, such that the following properties hold. (i) j N, a A, supp ψ j U a. (ii) K U, K compact, {j; suppψ j K } is finite. (iii) j N ψ j(x) = 1. 10

11 Idea of proof. First construct a family B (1) j B (1) j B (2) j of open balls such that U = j B (1) j and j N, a A, B (2) j U a. (B is the closure of the ball B.) This is done, for instance, by choosing as the family B (2) j the collection of all open balls with rational centers (all coordinates) and rational radii, so that their closure is contained in some U a, and for B (1) j the family of balls with same centers as the balls in B (2) j but half the radius. If x U, a A, r > 0, B(x, 3r) U a. Then y Q n, q Q so that x B(y, q) and 0 < q < r. Then B(y, 2q) U a which puts B(y, q) in the list B (1) j. Thus B (1) j U. By the previous proposition there exists φ j C0 (B (2) j ) so that φ j 1 on B (1) j. Define now ψ 1 = φ 1 and ψ j+1 = φ j j+1 i=1 (1 φ i) for j 1. Note that supp ψ j B (2) j and also, by induction, ψ ψ j = 1 j i=1 (1 φ i). If K U ( a notation for K compact, included in U), then finitely many B (1) j cover K. Let J be the largest index of in the cover,.i.e K J j=1b (1) j. If j J + 1 then ψ j (x) = 0 at x K and ψ 1 (x) + + ψ J (x) = 1. 4 Fourier series Definition 10 Let f : [0, 2π] C, f L 1 ([0, 2π]). For j Z we define the jth Fourier coefficient of f by ˆf(j) = 1 2π 2π 0 f(θ)e ijθ dθ. Definition 11 If f, g L 2 ([0, 2π]) we define their scalar product and norm as f, g = 1 2π f(θ)g(θ)dθ, f 2 L 2π 2 (T) = f, f. 0 Let n 0 and set n (P n f) (θ) = ˆf(j)e ijθ. j= n Note that P n f is a trigonometric polynomial. Fourier coefficients we see that (P n f) (θ) = 1 2π 2π 0 11 D n (θ t)f(t)dt Using the definition of the

12 with D n (θ) the Dirichlet kernel n D n (θ) = e ijθ = sin ( ) n θ. sin θ 2 j= n Exercise 10 Verify the last equality above using trigonometric identities. Theorem 6 Let f L 2 ([0, 2π]). Then (i) P n f L 2 (T) f L 2 (T), (ii) lim n f P n f L 2 (T) = 0, (iii) f 2 L 2 (T) = j Z f(j) 2. Consider now the average of n + 1 operators P j, and observe that where F n is the Fejér kernel F n (θ) = S n = 1 n + 1 (P P n ) (S n f) (θ) = 1 2π F n (θ t)f(t)dt 2π 0 n j= n ( 1 j ) e ijθ = 1 n + 1 n + 1 ( sin n+1 sin θ 2 Exercise 11 Verify that the average 1 n+1 n j=0 P j is given by the integral above and verify the identities defining the Fejér kernel. The following are properties of the Fejér kernel: 2 θ ) 2. (i) F n 0, F n is continuous, 2π periodic, 1 2π (ii) F 2π 0 n (θ)dθ = 1, (iii) δ > 0, C δ > 0, so that sup θ [δ,2π δ] F n (θ) C δ n+1. Theorem 7 Let f be a continuous, complex valued function so that f(0) = f(2π). Then S n f converge uniformly to f, that is lim sup n θ [0,2π] f(θ) (S n f)(θ) = 0. 12

13 Remark 3 This theorem implies that any continuous, periodic function can be approximated uniformly by trigonometric polynomials. Remark 4 The same statement is not true for P n. Functional analysis arguments show that there exist many continuous functions f for which the uniform convergence of P n f to f fails. In fact, even pointwise convergence fails: one can specify any point θ 0, and find (many) functions such that, for each such f, the sequence of numbers P n f(θ 0 ) is not even bounded, as n. Idea of proof. First, in view of periodicity, we have (S n f)(θ) = 1 2π F n (θ t)f(t)dt = 1 2π F n (t)f(θ t)dt. 2π 0 2π 0 Then we write (S n f f) (θ) = 1 2π F n (t) (δ t f(θ)) dt 2π 0 with (δ t f)(θ) = f(θ t) f(θ), the finite difference. This is possible because of property (ii) of F n. Now pick ɛ > 0. Because f is (uniformly) continuous and 2π-periodic, there exists δ > 0 so that, if t J δ = [0, δ] [2π δ, 2π] then δ t f(θ) ɛ 2 holds for all θ [0, 2π]. Because F n is nonnegative then (S n f f) (θ) 1 F n (t) δ t (f)(θ) dt + 1 2π δ F n (t) δ t (f)(θ) dt 2π J δ 2π δ holds, and by the argument above and normalization (ii) of F n we see that 1 F n (t) δ t (f)(θ) dt ɛ 2π J δ 2 is true for all θ [0, 2π]. On the other hand, after δ has been fixed, using property (iii) of F n we may choose n large enough so that 1 2π δ F n (t) δ t (f)(θ) dt 2C δm 2π δ n + 1 ɛ 2 holds for all θ. (Above, M = sup t [0,2π] f(t) ). 13

14 Definition 12 Let f : U R n C be a continuous bounded function defined in the open set U. Let 0 < α < 1. We say that is Hölder continuous of order α, f C 0, α (U) if f(x) f(y) [f] 0,α = sup <. x y x y α The same definition is used if U is any set. When the function f is periodic in each direction with period 2π we write f C 0,α (T n ). The quantity is the Hölder norm. f 0,α = f + [f] 0,α Proposition 14 Let f : [0, 2π] C. Then (i) If f, df,... dk 1 f, are continuous and periodic, dθ dθ k 1 and if dk f L 1 ([0, 2π]), then dθ f(j) C j k holds. k (ii) If f C 0,α ([0, 2π])is periodic, then f(j) C j α holds. Proposition 15 (Bernstein) Let α > 1. There exists a constant C 2 α > 0 so that, for any f C 0,α (T), f(j) C α f 0,α holds. j Z Idea of proof. Let m be a nonnegative integer, and consider the annulus A m = {j Z; 2 m j < 2 m+1 }. Take h = 2π 3 2 m and note that, for any j A m we have jh [ 2π consequently e ijh , 4π 3 ], and Therefore j A m f(j) j A m e ijh 1 2 f(j) 2 14

15 But (e ijh 1) f(j) = δ h f(j) and consequently, from Parseval, f(j) δ hf 2 L 2 (T). j A m Now, because f C 0,α (T) we have Consequently, δ h f 2 L 2 (T) C h 2α f 2 0,α j A m f(j) 2 C f 2 0,α2 2αm holds (in many proofs from now on C might change harmlessly from line to line).using Schwartz: f(j) C #{j A m }2 αm f 0,α = C f 0,α 2 ( 1 2 α)m. j A m Because α > 1 the right hand side is summable in m, and that implies that 2 the left hand side is summable. Definition 13 A sequence of numbers {λ n } is said to be lacunary if there exists a constant q > 1 such that λ n+1 > qλ n. A trigonometric series is lacunary if all the frequencies appearing in it are of the form ±λ n, with {λ n } lacunary. Lemma 1 Let f L 1 (T) and assume that f(j) = 0 for all j such that 1 n j 2N. Assume also that f(t) = O(t) ast t 0. Then, f(n) 2π 4 N 1 sup t 1 f(t) + N 2 f L 1 (T) t N 1 4 Proof.. We use the condition f(j) = 0 as follows. If g N is a trigonometric polynomial of degree 2N that satisfies ĝ N (0) = 1, then f(n) = 1 2π 2π 0 e int f(t)g N (t)dt. 15

16 We choose g N We note that = F N 2 L 2 (T) F 2 N (t) where F N is the Fejér kernel of order N. 0 F N (t) π 2 (n + 1)t 2 follows from the formula for the kernel, and also F N 2 L 2 (T) = ( 1 j ) 2 N N so we deduce j N 0 g N (t) 2π 4 N 3 t 4. Now we break the integral into three pieces: f(n) I 1 + I 2 + I 3. The second piece is Finally, the third piece is I 1 = 1 2π t N 1 f(t) g N (t)dt N 1 sup t N 1 t 1 f(t) 1 2π = N 1 sup t N 1 t 1 f(t) 2π 0 g N (t)dt I 2 = 1 2π N 1 t N 1 f(t) g N(t)dt 4 π 3 N 3 sup t N 4 1 t 1 f(t) N 1 4 t 3 dt N 1 π 3 N 1 sup t N 1 t 1 f(t) 4 I 3 = 1 2π and this concludes the proof. A corollary is t N 1 4 f(t) g N(t)dt π 3 N 2 f L 1 (T) Corollary 1 Let {λ n } be a a lacunary sequence and assume that f a n cos(λ n t) is in L 1 (T). Assume that f is differentiable at one point. Then a n = o(λ n 1 ). 16

17 Proof. WLOG we assume that f is differentiable at t = 0. Replacing f by f f(0) cos(t) f (0) sin(t), we may assume that f(t) = o(t) as t 0. The lacunarity condition implies that there exists a positive constant c such that for all integers j such that 1 λ n j cλ n we have f(j) = 0. We may apply then the lemma with λ n in place of n in the lemma, and 2N = cλ n. We obtain that a n = 2 f(λ n ) = o(λ 1 n ). Corollary 2 The Weierstrass function 2 n cos(2 n t) is continuous and nowhere differentiable. Exercise 12 Let f a n cos(λ n t) is Hölder continuous with exponent 0 < α < 1 at t = t 0 and assume that {λ n } is lacunary. Deduce that a n = o(λ α n ). Prove that this implies that f is everywhere Hölder continuous with exponent α. 5 Fourier Transform Definition 14 Let f L 1 (R n ) and ξ R n. The Fourier transform of f at ξ is F(f)(ξ) = ˆf(ξ) = e iξ x f(x)dx. Here ξ x = n j=1 ξ jx j is scalar product in R n. The function f might as well be taken complex valued and f L 1 means that f is integrable, or, equivalently, both the real part Re(f) and the imaginary part Im(f) are in L 1. Proposition 16 Let φ, ψ C 0 (R n ). (i) F(φ ψ) = ˆφ ˆψ (ii) F(τ h φ)(ξ) = e ih ξ ˆφ(ξ) (iii) F(e ih x φ) = τ h ˆφ (iv) F( j φ)(ξ) = iξ j ˆφ (v) (vi) F(ix j φ) = j ( ˆφ) φ ˆψdx = ˆφψdx 17

18 Proposition 17 Let f L 1 (R n ). Then (i) ˆf L (R n ) and ˆf f 1 (ii) lim ξ ˆf(ξ) = 0 (iii) ˆf is uniformly continuous. Point (ii) is the Riemann-Lebesgue lemma. Idea of proof. For (iii) let ɛ > 0. Take φ C0 (R n ) so that f φ 1 ɛ. 3 Then use ˆf(ξ) ˆφ(ξ) ɛ together with the fact that 3 ˆφ(ξ + h) ˆφ(ξ) = (e ih x 1 ) e iξ x φ(x)dx which implies that sup ˆφ(ξ + h) ˆφ(ξ) sup e ih x 1 φ 1. ξ x supp φ Choosing h δ with δ small enough we have sup ˆφ(ξ + h) ˆφ(ξ) ɛ ξ 3. For (ii) Take ɛ > 0 use the same φ. Note that, in view of the previous proposition ˆφ(ξ) 1 ξ φ 1 Thus, for ξ large enough ˆφ(ξ) ɛ 2. Exercise 13 Let Φ(x) = e ɛ x 2. Then φ = ˆΦ is φ(ξ) = ( π ) n 2 e ξ 2 4ɛ ɛ and φ, Φ 0, φ, Φ L 1 L and Φ(0) = 1, φdξ = (2π) n. Lemma 2 Assume φ 0, φ = ˆΦ, Φ(0) = 1 φ, Φ L 1 L and (2π) n φdξ = 1. 18

19 Then, for any f L 1 e ix ξ ˆf(ξ)Φ(ξ)dξ = f(y)φ(y x)dy. Moreover, the function f ɛ (x) = (2π) n e ix ξ ˆf(ξ)Φ(ɛξ)dξ converges in L 1 to f. Idea of proof. After verifying the identity and rescaling we note that f ɛ (x) = f ψ ɛ where ψ ɛ (x) = (2πɛ) n φ ( ) x ɛ. Because ψ has integral equal to one and is positive, the convergence follows from Proposition 14. Exercise 14 Let f j be a sequence of L 1 functions that converges to f in L 1. Then there exists a subsequence of f j that converges almost everywhere to f. Theorem 8 (Fourier Inversion). Assume that f L 1 (R n ) and ˆf L 1 (R n ). Then f(x) = (2π) n e ix ξ ˆf(ξ)dξ holds dx-a.e. Idea of proof. Pick Φ, φ like in the lemma. By the Lemma, f ɛ converges to f in L 1, and hence, by the exercise, a subsequence converges to f(x) almost everywhere. On the other hand, by Lebesgue dominated convergence f ɛ (x) converges (for each fixed x) to (2π) n e ix ξ ˆf(ξ)dξ. Theorem 9 Let f L 1 L 2. Then f(x) 2 dx = (2π) n ˆf(ξ) 2 dξ. Idea of proof. WLOG f is smooth, compactly supported. Consider f(x) = f( x) where z is the complex conjugate. Then let g = f f. By construction, ĝ(ξ) = ˆf(ξ) 2 0 and g(0) = f(x) 2 dx. We may apply the Fourier inversion formula. 19