LECTURE 4 Fourier transform 4.1. Schwartz functions Recall that L1(Rn) denotes the Banach space of functions f : Rn ! C that are absolutely integrable, i.e., jfj is Lebesgue integrable over Rn: The norm on this space is given by Z kfk1 = jf(x)j dx: Rn Given ξ 2 Rn and x 2 Rn; we put ξx := ξ1x1 + ··· + ξnxn: For each ξ 2 Rn; the exponential function eiξ : x 7! eiξx; Rn ! C; has absolute value 1 everywhere. Thus, if f 2 L1(Rn) then e−iξf 2 L1(Rn) for all ξ 2 Rn: Definition 4.1.1. For a function f 2 L1(Rn) we define its Fourier transform f^ = Ff : Rn ! C by Z (4.1) Ff(ξ) = f(x)e−iξx dx: Rn n We will use the notation Cb(R ) for the Banach space of bounded continuous functions Rn ! C equipped with the sup-norm. Lemma 4.1.2. The Fourier transform maps L1(R) continuous linearly to the n Banach space Cb(R ): Proof Let f be any function in L1(Rn): The functions fe−iξ are all dominated −iξ n by jfj in the sense that jfe j ≤ jfj (almost) everywhere. Let ξ0 2 R ; then it follows by Lebesgue's dominated convergence theorem that Ff(ξ) !Ff(ξ0) if ξ ! ξ0: This implies that Ff is continuous. It follows that F defines a linear map from L1(R) to C(Rn): It remains to be shown that F maps L1(R) continuously into Cb(R): 63 64 BAN-CRAINIC, ANALYSIS ON MANIFOLDS For this we note that for f 2 L1(Rn) and ξ 2 Rn; Z Z −iξx −iξx jFf(ξ)j = j f(x) e dx j ≤ jf(x) e j dx = kfk1: Rn Rn 1 n Thus, sup jFfj ≤ kfk1: It follows that F is a linear map L (R) ! Cb(R ) which is bounded for the Banach topologies, hence continuous. n n Remark 4.1.3. We denote by C0(R ) the subspace of Cb(R ) consisting of functions f that vanish at infinity. By this we mean that for any ϵ > 0 there exists a compact set K ⊂ Rn such that jfj < ϵ on the complement Rn n K: It n n is well known that C0(R ) is a closed subspace of Cb(R ); thus a Banach space of its own right. The well known Riemann-Lebesgue lemma asserts that, actually, F maps 1 n n L (R ) into C0(R ): The above amounts to the traditional way of introducing the Fourier trans- form. Unfortunately, the source space L1(Rn) is very different from the target n 1 n space Cb(R ): We shall now introduce a subspace of L (R ) which has the ad- vantage that it is preserved under the Fourier transform: the so-called Schwartz space. Definition 4.1.4. A smooth function f : Rn ! C is called rapidly decreasing, or Schwartz, if for all α; β 2 Nn; (4.2) sup jxβ@αf(x)j < 1: x2Rn The linear space of these functions is denoted by S(Rn): Exercise 4.1.5. Show that the function 2 f(x) = e−∥xk belongs to S(x): Condition (4.2) for all α; β is readily seen to be equivalent to the following condition, for all N 2 N; k 2 N : N α νN;k(f) := max sup (1 + kxk) j@ f(x)j < 1: jα|≤k x2Rn We leave it to the reader to check that ν = νN;k defines a norm, hence in particular a seminorm, on S(Rn): We equip S(Rn) with the locally convex topology generated by the set of norms νN;k; for N; k 2 N: The Schwartz space behaves well with respect to the operators (multiplica- tion by) xα and @β: Exercise 4.1.6. Let α; β be multi-indices. Show that xα : f 7! xαf and @β : f 7! @βf define continuous linear endomorphisms of S(Rn): Exercise 4.1.7. (a) Show that S(Rn) ⊂ L1(Rn); with continuous inclusion map. LECTURE 4. FOURIER TRANSFORM 65 (b) Show that 1 Rn ⊂ S Rn ⊂ 1 Rn Cc ( ) ( ) C ( ); with continuous inclusion maps. Lemma 4.1.8. The space S(Rn) is a Fr´echetspace. Proof As the given collection of seminorms is countable it suffices to show completeness, i.e., every Cauchy sequence in S(Rn) should be convergent. Let n (fn) be a Cauchy sequence in S(R ): Then by continuity of the second inclusion in Exercise 4.1.7 (b), the sequence is Cauchy in C1(Rn): By completeness of the latter space, the sequence fn converges to f, locally uniformly, in all derivatives. n n We will show that f 2 S(R ) and fn ! f in S(R ): First, since (fn) is Cauchy, n it is bounded in S(R ): Let N; k 2 N; then there exists a constant CN;k > 0 such n α α that νN;k(fn) ≤ CN;k; for all n 2 N: Let x 2 R ; then from @ fn(x) ! @ f(x) it follows that N α N α (1 + kxk) @ fn(x) ! (1 + kxk) @ f(x); as n ! 1: N α In view of the estimates νN;k(fn) ≤ CN;k; it follows that j(1 + kxk) @ f(x)j ≤ CN;k; for all α with jαj ≤ k: This being true for arbitrary x; we conclude that νN;k(f) ≤ CN;k: Hence f belongs to the Schwartz space. n Finally, we turn to the convergence of the sequence fn in S(R ): Let N; k 2 N: Let ϵ > 0: Then there exists a constant M such that n; m > M ) νN;k(fn − fm) ≤ ϵ/2: Let jαj ≤ k and fix x 2 Rn: Then it follows that ϵ (1 + kxk)N j@αf (x) − @αf (x)j ≤ n m 2 α α As @ fn ! @ f locally uniformly, hence in particular pointwise, we may pass to the limit for m ! 1 and obtain the above estimate with fm replaced by f; n for all x 2 R : It follows that νN;k(fn − f) < ϵ for all n ≥ M: Another important property of the Schwartz space is the following. 1 Rn S Rn Lemma 4.1.9. The space Cc ( ) is dense in ( ): 2 1 Rn ≤ ≤ Proof Fix a function ' Cc ( ) such that 0 ' 1 and ' = 1 on the closed unit ball in Rn: For k 2 N we put k k j α j ' Ck := max sup @ '(x) : jα|≤k x2Rn 2 Z 2 1 Rn For j + define the function 'j Cc ( ) by 'j(x) = '(x=j): 2 S Rn 2 1 Rn 2 Z Let now f ( ): Then 'jf Cc ( ) for all j +: We will complete the n proof by showing that 'jf ! f in S(R ) as j ! 1: Fix N; k 2 N: Our goal is to find an estimate for νN;k('jf −f); independent β of f: To this end, we first note that for every multi-index β we have @ 'j(x) = (1=j)jβj@β'(x=j): It follows that j β j ≤ 1k k 2 Z j j ≤ sup @ 'j ' Ck ; (j +; 0 < β k): Rn j 66 BAN-CRAINIC, ANALYSIS ON MANIFOLDS Let jαj ≤ k: Then by application of Leibniz' rule we obtain, for all x 2 Rn; that X ( ) α α 1 α α−β j@ (' f − f)(x)j ≤ j(' (x) − 1) @ f(x)j + k'k k j@ f(x)j: j j j C β 0=6 β≤α The first term on the right-hand side is zero for kxk ≤ j: For kxk ≥ j it can be estimated as follows: α −1 α j('j(x) − 1)@ f(x)j ≤ (1 + sup j'j)(1 + j) (1 + kxk)j@ f(x)j ≤ 2j−1(1 + kxk)j@αf(x)j: We derive that there exists a constant Ck > 0; only depending on k; such that for every N 2 N; C ν (' f − f) ≤ k ν (f): N;k j j N+1;k n It follows that 'jf ! f in S(R ): The following lemma is a first confirmation of our claim that the Schwartz space provides a suitable domain for the Fourier transform. Lemma 4.1.10. The Fourier transform is a continuous linear map S(Rn) ! S(Rn): Moreover, for each f 2 S(Rn) and all α 2 Nn; the following hold. (a) F(@αf) = (iξ)αFf; α α (b) F(x f) = (i@ξ) Ff: Proof Let f 2 S(Rn) and let 1 ≤ j ≤ n: Then it follows by differentiation under the integral sign that Z Z @ −iξx −iξx f(x) e dx = f(x)(−ixj)e dx: @ξj Rn Rn The interchange of integration and differentiation is justified by the observa- tion that the integrand on right-hand side is continuous and dominated by −n−1 the integrable function (1 + kxk) νn+1;0(f) (check this). It follows that F(−xjf) = @jFf: By repeated application of this formula, we see that Ff is a smooth function and that (b) holds. Since the inclusion map S(Rn) ! L1(Rn) 1 n n and the Fourier transform L (R ) ! Cb(R ) are continuous, it follows that F n n α is continuous from S(R ) to Cb(R ): As multiplication by x is a continuous endomorphism of the Schwartz space, it follows by application of (b) that F is a continuous linear map S(Rn) ! C1(Rn): Let f 2 C1(Rn) and 1 ≤ j ≤ n: Then by partial integration it follows that c Z Z −iξx −iξx @jf(x)e dx = (iξj) f(x)e dx Rn Rn so that F(@jf) = (iξj)F(f)(ξ): By repeated application of this formula, it 2 1 Rn 1 Rn S Rn follows that (a) holds for all f Cc ( ): By density of Cc ( ) in ( ) combined with continuity of the endomorphism @α 2 End(S) and continuity of F as a map S(Rn) ! C(Rn) it now follows that (a) holds for all f 2 S(Rn): It remains to establish the continuity of F as an endomorphism of S(Rn): For this it suffices to show that ξα@βF is continuous linear as a map S(Rn) ! n α β α β Cb(R ): This follows from ξ @ F = F ◦ (−i@) (−ix) (by (a), (b)) and the fact that (−i@)α ◦ (−ix)β is a continuous linear endomorphism of S(Rn): LECTURE 4.