n 1 f = f m, m=0 n 1 k=0 Note that if n = 2, we essentially get example (i) (but for complex functions).

Transcription

1 Chapter 4 Fourier Analysis Intuition This discussion is borrowed from [T. Tao, Fourier Transform]. Fourier transform/series can be viewed as a way to decompose a function from some given space V into a superposition of symmetric functions (whatever that means). To define the symmetry, we will have a certain group G action of some maps V V and a function f from V will be viewed as symmetric if action of any map from G on f preserves f up to a multiplicative factor (character) of modulus 1. Some examples of this: Examples 4.1. (i) (even odd decomposition) V is the space of real-valued functions f : R R with the group G of maps f(x) f(x) and f(x) f( x) (isomorphic to Z 2 ). Symmetric functions are then of two types: even or odd. The Fourier decomposition of a function f : R R then is f = f e + f o, where f e = f(x)+f( x) 2 (even part), f o = f(x) f( x) 2 (odd part). (ii) (decomposition with respect to rotations by roots of unity) Fix n N. Let V be the space of complexvalued functions f : C C with the group G of maps f(z) f(e 2πij/n z) for j = 0,..., n 1 (isomorphic to Z n ). Symmetric functions are then of n types: each type is completely determined by action of the generating element of G: f(e 2πi/n z) = e 2πim/n f(z) for some m = 0, 1,..., n 1 (action of the other elements are then uniquely determined). The Fourier decomposition of a function f : C C then is where f = n 1 m=0 f m, n 1 f m (z) := 1 n f(e 2πik/n z)e 2πimk/n. Note that if n = 2, we essentially get example (i) (but for complex functions). k=0 (iii) (Fourier series) Now let V consists of (smooth) functions f : D C on the unit circle D := {z C : z = 1}. Let G consists of (rotations) maps f(z) f(e iφ z) for some choice of e iφ D (so that G is D viewed as a group with the usual multiplication operation). Observe that a function f n (z) := c n z n (with some n Z) satisfies f(e iφ z) = e inθ f(z), and it can be shown that these are the only examples of symmetric functions. Then the Fourier decomposition takes any smooth f and decomposes it into f(z) = n= f(n)z n, 39

2 where f(n) := 2π 0 f(e iθ inθ dθ )e 2π. Note that this expression can be viewed as n limit of example (ii). Note that we start with G = D and end up with another group Z which indexes the symmetry types. This is known as the Pontryagin dual group of D. This example can be generalized to Fourier series for functions f : D n C. (iv) (Fourier transform) Finally, let us consider V to consist of (smooth) functions f : R C, and let the group G consist of (translations) maps f(x) f(x + t) for some choice of t R (so that G is R viewed as a group with the usual addition operation). Now we have more than countably many symmetric functions: indeed, note that for an arbitrary choice of ξ R, function f ξ (x) = c ξ e 2πixξ satisfies the symmetry f(x+t) = e 2πitξ f(x), and it can be shown that these are the only examples of symmetric functions (see [F, 8.19]). Then the Fourier decomposition takes any (smooth) f and decomposes it into f(z) = f(ξ)e 2πixξ dξ, where f(ξ) := R R f(x)e 2πixξ dx. Note that the superposition of symmetric components is interpreted now through an integral rather than through a sum. Note that here G = R, whose Pontryagin dual is again R. This example can be generalized to Fourier series for functions f : R n C Preliminaries: some notation Let us set some basic notation: Dot product and norm: for x, y R n, let x y = n j=1 x jy j and x = x x. Partial derivatives: j mf := m f x. m j For a multi-index (or, an ordered n-tuple of nonnegative integers) α = (α 1,..., α n ), we define α = α α n, α := α αn n α! := α j!, x α := j=1 n j=1 ( is confusing but it should be clear from the context if the argument is an index or a vector). C k denotes the space of all functions whose all derivatives of total order k are continuous. C denotes the space of all infinitely differentiable functions. Cc denotes C functions with compact support. x αj j n, 40

3 Throughout this chapter, whenever we write L p (X) (for various choices of the space X), we always assume the Lebesgue measure on X. The Lebesgue integration will be denoted with dx (where x will typically be in R n ). Let us denote f(x) := f( x). Given y R n, let us denote τ y (f)(x) := f(x y) Preliminaries: the n-torus T n and functions on T n Define the torus T as [0, 1] with endpoints identified (i.e., R/Z). Let us regard functions on T as 1-periodic functions on the real line: f : R C with f(x) = f(x + 1). It is sometimes useful to identify T with the unit circle D via the map from [0, 1] to {z C : z = 1} given by x e 2πix. This sets up the correspondence g(x) = f(e 2πix ) between functions f(z) : D C and functions g(x) : T C. Similarly, we will view T n (the n-torus ) as the cube [0, 1] n with opposite sides identified, i.e., T n = R n /Z n. Consequently, functions on T n are functions f : R n C that are 1-periodic in each coordinate: f(x+m) = f(x) for all x R n and all m Z n. Analogously, it is sometimes useful to identify T n with D n C n via the map from [0, 1] n to D n given by (x 1,..., x n ) (e 2πix1,..., e 2πixn ) Preliminaries: convolutions Definition 4.2. (i) For measurable functions f, g : R n C we define their convolution (f g)(x) := f(x y)g(y) dy R n for all x where this integral exists. (ii) Convolution for functions f, g : T n C are defined analogously, but now the integral is over T n and we treat x y as the subtraction in R n /Z n (i.e., modulo 1 in each coordinate). Remark 4.3. It is elementary to see that f g = g f. Theorem 4.4 (Young s Inequality for convolutions). Suppose 1 p, q, r and Suppose f L p and g L q. Then f g L r, and 1 p + 1 q = r. Remarks 4.5. (a) This works both in R n and in T n. (b) The most common special cases of Young s inequality are: f g r f p g q. (4.0.1) f g f p g q if 1 p + 1 q = 1, (4.0.2) and f g p f p g 1 if 1 p. (4.0.3) 41

4 Proof. This can be proven via a careful application of generalized Hölder s inequality. Alternatively: Firstly, for p = q = 1 (so that r = 1), the inequality follows from the triangle inequality and linear change of variables: f g 1 = f(x y)g(y) dy dx f(x y) g(y) dy dx = f 1 g 1. (4.0.4) Secondly, for r = and any conjugate p, q (i.e., 1 p + 1 q = 1, 1 p ), we have: f g = f(x y)g(y) dy f p g q (4.0.5) by the usual Hölder s inequality. Now for any p, q, r with 1 p + 1 q = r, let Now let us interpolate these two results using two applications of the Riesz Thorin theorem 3.35: first we interpolate between L 1 L 1 and L L to obtain the L p L p result, and then we interpolate between L 1 L p and L p/(p 1) L to get the general L q L r result. Here are the details: First, the convolution operator T (f) := f g is linear and satisfies T (f) 1 g 1 f 1 when viewed as in L 1 L 1 ; and T (f) g 1 f when viewed as in L L. Then the Riesz Thorin theorem says that T (f) s g 1 f s for any s [1, ] viewed as in L s L s. Now, as we just proved, the convolution operator S(g) := f g is linear and satisfies T (g) p f p g 1 when viewed as in L 1 L p ; also T (g) f p g p/(p 1) when viewed as in L p/(p 1) L. Then the Riesz Thorin theorem says that T (g) r f p g s viewed as in L s L r, where 1 s = 1 t t 1 + p/(p 1), 1 r = 1 t p + t. Getting rid of t, we get 1 s + 1 p = 1 r + 1, which is what we need. Proposition 4.6. (i) If 1 p + 1 q = 1, 1 < p < and f Lp (T n ), g L q (T n ), then f g C(T n ). (ii) If 1 p + 1 q = 1, 1 < p < and f Lp (R n ), g L q (R n ), then f g C 0 (R n ). Remark 4.7. If p = 1 or p = continuity in fact still holds [F241]. Proof. (i) and (ii): If f, g C c, then the statement is clear. Now for any f L p, g L q, choose f n, g n C c such that f n f in L p and g n g in L q (see Theorem 3.24). By the Young s inequality, f g f n g n f (g g n ) + (f f n ) g n f p g g n q + f f n p g n q 0. So f g is a uniform limit of continuous functions with compact support, so f g C(T n ) or f g C 0 (T n ), respectively Preliminaries: approximate identity Definition 4.8. An approximate identity on T n is a sequence of functions {k j (x)} j=1 on Tn that satisfy (i) k j (x) 0; (ii) k T n j (x) dx = 1 for all j; (iii) For any ε > 0, lim k j (x) dx = 0. j x >ε 42

5 Remarks 4.9. (a) x > ε is meant in the periodic sense, e.g., in T 1 it means that x (ε, 1 ε). (b) One can relax the requirement k j 0 but then instead needs to require k j 1 C for some choice of C > 0. (c) Similarly, one defines an approximate identity on R n : just replace each T n with R n. (d) For an example of an approximate identity, we can set k j (x) = j n φ(x/j) for any φ L + L 1 (T n ) (or on R n ). Proposition Let {k j } j=1 be an approximate identity on Tn or on R n. Then: (i) If f C(T n ) then k j f f 0. The same statement holds for R n with the only change that in f needs to be bounded and uniformly continuous on R n. (ii) If f L p (T n ) for 1 p <, then The same statement holds for f L p (R n ). k j f f p 0. Proof. Let us prove only the T 1 case cases T n and R n follow along the same lines. (i) By periodicity and Definition 4.8(ii), we have (k j f)(x) f(x) = 1 Now divide the integral into regions y (ε, 1 ε) and y ε: k j f f 0 k j (y)(f(x y) f(x)) dy. sup f(x y) f(x) + 2 f k j (y) dy, y ε,x [0,1] y >ε so we get lim sup k j f f sup y ε,x [0,1] f(x y) f(x). But the latter expression goes to 0 as ε 0 since any continuous function on [0, 1] is uniformly continuous. (ii) The statement for f continuous with compact support follow from (i) since k j f f p k j f f on T n (for R n one should make use of compact support and property (iii) of Definition 4.8). Now for arbitrary f L p, find g C c such that f g p < ε. Then using Young s inequality (4.0.3), we get: lim sup j k j f f p lim sup ( k j (f g) p + k j g g p + f g p ) j Then we take ε 0. lim sup k j 1 f g p f g p < 2ε. j 4.1 Fourier Analysis on T n Fourier series on L 2 (T n ): definition Recall that if H is a Hilbert space and {φ j } j=1 is an orthonormal basis, then for any x H, we have x = x, φ j φ j, j=1 43

6 where the infinite sum is meant as the converging in the norm of H (that is, x n j=1 x, φ j φ j 0 as n ). Coefficients x, φ j above are called the (abstract) Fourier coefficients. Now we consider L 2 (T n ) with inner product f, g := f(x)g(x) dx. Note that the sequence {e 2πim x } T n m Z n is orthonormal (basic application of the Fubini theorem): { e 2πim x e 2πik x 1, if m = k, dx = [0,1] n 0, if m k. In fact, {e 2πim x } m Z n is an orthonormal basis of L 2 (T n ) (it is proven below in Theorem 4.26). Definition For f L 2 (T n ): (i) Define its Fourier coefficients to be f(m) := f(x)e 2πim x dx, T n m Z n. (4.1.1) (ii) Define its Fourier series to be m Z m f(m)e 2πim x. (4.1.2) As a corollary of the general Hilbert space theory, we immediately get: Corollary For any f, g L 2 (T n ): (i) f(t) is the L 2 (T n )-limit of the sequence of its partial Fourier sums S N (f) := f(m)e m N 2πim t : S N (f) f 2 0. (ii) (Plancherel s identity) f 2 2 = m Z n f(m) 2 (iii) (Parseval s relation) T n f(t)g(t) dt = m Z n f(m)ĝ(m). (iv) The map f { f(m)} m Z n is an isometry from L 2 (T n ) onto l 2 (Z n ). Proof. (i) and (ii) are standard facts about orthonormal bases. (iii) follows from (ii) and the polarization identity ( f, g = 1 4 x + y 2 x y 2 + i x iy 2 i x + iy 2). (iv) is then almost proved except for surjectivity: given {a m } m Z n we can form partial Fourier sums f M (t) = m M a me 2πim t and show that it s Cauchy in L 2 (T n ), therefore converging to some f L 2 (T n ). Then Fourier s coefficients of f are a m s and we are done. 44

7 4.1.2 Fourier series on L 1 (T n ): definition and basic properties Note that the definition (4.1.1) and (4.1.2) work under the mere assumption that f L 1 (T n ). However the convergence of (4.1.2) is then no longer clear. Let us state some of the properties of the Fourier series on L 1. Recall the notation f(x) := f( x) and τ y (f)(x) := f(x y). Proposition Let f, g be in L 1 (T n ). Then for all m, k Z n, λ C, y T n : (i) f + g(m) = f(m) + ĝ(m); (ii) λf(m) = λ f(m); (iii) f(m) = f( m); (iv) f(m) = f( m); (v) τ y (f)(m) = f(m)e 2πim y ; (vi) ( e 2πik x f(x) ) (m) = f(m k); (vii) sup m Z n f(m) f 1, in particular, if f L 1 (T n ) then f l (Z n ); (viii) f g(m) = f(m)ĝ(m); (ix) α f(m) = (2πim) α f(m) for f C α. Proof. To prove (viii), we use Tonelli+Fubini theorems to interchange the integrals: f g(m) = f(x y)g(y)e 2πim (x y) e 2πim y dy dx = f(m)ĝ(m). T n T n Fourier series on L 1 (T n ): various types of pointwise convergence Intuition. Recall that an infinite series j=1 a j is called summable if its partial sums s n := n j=1 a j has a 1 n finite limit, and Cesàro summable if the arithmetic mean of its partial sum has a finite limit: n j=1 s j. It is elementary to show that if a series is summable then it is Cesàro summable (with the same limit), but not necessarily vice versa. We want to discuss the issue of pointwise convergence of Fourier series, and Cesàro summation is natural here. Definition Let f L 1 (T n ) and N N. (i) The expression S N (f)(x) := m Z n m j N is called the square partial sum of the Fourier series of f. ˆf(m)e 2πim x (4.1.3) 45

8 (ii) Let S k1,...,k n (f)(x) := m Z ˆf(m)e n (the rectangular partial sum of Fourier series ). The m j k j expression 1 C N (f)(x) := (N + 1) n S m1,...,m n (f)(x) m Z n 0 m j N is called the square Cesàro mean of the Fourier series of f. Remarks (i) One can also form circular (spherical) partial sums and circular (spherical) Cesàro means: now one would sum over m Z n, m j 2 M instead of m Z n, m j M. (ii) Yet another common way to sum the Fourier series is via the Abel Poisson summation (convolution with a Poisson kernel). There re also notions of Riesz and Bochner Riesz summations Fourier series on L 1 (T n ): Dirichlet and Fejér kernels Intuition. One of the themes in the discussion that follows will be that partial Fourier sums S M (f) do not behave as well as their Cesàro means C M (f) do. The main explanation for this is that the Fejér kernels form an approximate identity which we explore in this section. Definition Let N N. (i) On T 1 : the Dirichlet kernel on T 1 is defined to be D N (x) = (ii) On T 1 : the Fejér kernel on T 1 is defined to be N m= N F N (x) = 1 N + 1 e 2πimx. N D j (x) (iii) On T n : the square Dirichlet kernel on T n is defined to be D(n, N)(x) = m Z n m j N (iv) On T n : the square Fejér kernel on T n is defined to be F (n, N)(x) = 1 (N + 1) n j=0 e 2πim x = D N (x 1 )... D N (x n ). N... k 1=0 = F N (x 1 ) F N (x n ) N D k1 (x 1 ) D kn (x n ) k n=0 Proposition (i) On T 1 : for x [0, 1]. D N (x) = sin((2n + 1)πx) sin(πx) 46

9 (ii) On T 1 : F N (x) = N j= N ( 1 j ) e 2πijx = 1 N + 1 N + 1 Remark Combining (ii) with Definition 4.16(iv), we get F (n, N)(x) = ( 1 m ) 1 N + 1 m Z n m j N ( ) 2 sin(π(n + 1)x). sin(πx) ( 1 m ) n e 2πim x. (4.1.4) N + 1 Proof. (i) follows from from the formula for the geometric series. (ii) The first expression is immediate from the definition. The second uses (i) and a bit more of trigonometry. Proposition Let f L 1 (T n ) and M N. (i) Square partial sum of the Fourier series of f is a convolution with the square Dirichlet kernel D(n, N): S N (f)(x) = (f D(n, N))(x). (ii) Square Cesàro mean of the Fourier series of f is a convolution with the square Fejér kernel F (n, N): C N (f)(x) = (f F (n, N))(x) = ( ˆf(m) 1 m ) ( 1 1 m ) n e 2πim x. (4.1.5) N + 1 N + 1 m Z n m j N Proof. (i) follows directly from the definitions of convolution and of D N. Note that by Proposition 4.13(viii) and Definition 4.16 we can immediately see that the Fourier series of left-hand side and right-hand side coincide, but we do not know the uniqueness result yet (see Proposition 4.27 below). (ii) follows from in the exact same way by using also (4.1.4). Proposition Fejér kernel {F (n, N)} N=1 forms an approximate identity on Tn. Remark Dirichlet kernel, on the other hand, has D N 1 log n. Proof. For T 1 this is an easy check using Proposition 4.17(ii). For T n this follows if one uses the fact that F (n, N)(x) is the product of F (1, N)(x j ) Fourier series on L 1 (T n ): uniform and L p convergence of C N (f) Theorem on T n ). (i) (Fejér s Theorem) For any f C(T n ), C N (f) f in the norm (i.e., uniformly (ii) For any f L p (T n ) (1 p < ), C N (f) f in the L p norm. Remarks (a) Note that continuity is necessary for (i) to hold as uniform limit of continuous functions is continuous. (b) Part (i) for the usual partial sums S N (f) fails: there exist continuous functions whose Fourier series do not converge at infinitely many points (however, see Theorem 4.39 below). So for the uniform convergence of S N (f) one needs some stronger condition than mere continuity e.g., differentiability or Hölder continuity suffices for T 1, see Theorem (c) Part (ii) for the usual partial sums S N (f) works only for 1 < p <, see Theorem

10 Proof. Both (i) and (ii) follow immediately from the fact that Fejér kernel forms an approximate identity and combining it with Proposition Fourier series on L 1 (T n ): some corollaries of Fejér Theorem 4.24 (Weierstrass approximation theorem on the circle/torus). The set of trigonometric polynomials (i.e., functions T n C of the form c m e 2πim x m Z n m N with N N and c j C) is -dense in C(T n ). Remarks (a) If we chose different scaling, then this theorem in dimension 1 says that polynomials in ±e iθ (sometimes called Laurent polynomials) are -dense in C( D). (b) If one takes real parts, then one obtains yet another restatement of the same theorem: function of the type N j=1 a j cos(jθ) + b j sin(jθ) are -dense in the space of 2π-periodic functions on R. (c) There is also the Weierstrass approximation theorem on the real line which states that polynomials in x are -dense in C([a, b]) for any bounded interval [a, b] R. (d) All of these Weierstrass theorems are in fact special cases of the much more general Stone Weierstrass theorem which states that if X is compact Hausdorff, then any sub-algebra of C R (X) (index R here means real-valued) that contains the function f(x) 1 and separates points is necessarily -dense in C R (X). Proof. Given any f C(T n ), we know (Theorem 4.22(i)) that sequence of Cesàro means converges to f uniformly. But each Cesàro mean is a trigonometric polynomial, so we are done. Now we are able to prove the following theorem which we stated without a proof before: Theorem (i) For any 1 p <, the set of trigonometric polynomials is p -dense in L p (T n ). (ii) Moreover, {e 2πim x } m Z n is an orthonormal basis of L 2 (T n ). Proof. For (i), just use Theorem 4.22(ii) as in the previous proof. (ii) then follows as orthonormality (follows from direct computation along with Tonelli+Fubini theorems) and density is enough. Proposition 4.27 (Uniqueness of Fourier coefficients). Let f, g L 1 (T n ) satisfy f(m) = ĝ(m) for all m Z n. Then f = g a.e. Remark For f, g L 2 (T n ) this follows from the previous theorem. However for L 1 we need a proof. Proof. By linearity we can assume g = 0, so that ˆf(m) = ĝ(m) = 0 for all m Z n. From the definition of the Cesàro means (Definition 4.14(ii)), C N (f)(x) = 0 for all N. But we know that f C N (f) 1 0 by Theorem 4.22, which implies f 1 = 0, i.e., f = 0 a.e. Theorem 4.29 (Fourier inversion). Suppose f L 1 (T n ) and ˆf l 1 (Z n ) (that is, f(m) < ). Then S N (f) converges uniformly (and absolutely) to a continuous function that is equal to f a.e. Remarks (a) While the conclusion of the theorem is very nice, it is very unsatisfying as there are conditions on both f and f. (b) Compare this Theorem with Theorem 4.58 for Fourier inversion on R n. 48

11 Proof. Uniform convergence of S N (f) is clear from the l 1 condition and elementary convergence test. The limit function g must be continuous. It is easy to see that both g and f have the same Fourier coefficients. By Proposition 4.27, we get f = g a.e. Theorem 4.31 (Riemann Lebesgue lemma). Let f L 1 (T n ). Then f(m) 0 as m. Remarks (a) The converse to this is not true: not every sequence satisfying a m 0 is the Fourier sequence of some f L 1. (b) On the other hand, the following can be proved (see [G, Thm 3.2.2]): given any sequence d m 0, there exists f L 1 (T n ) whose Fourier coefficients have slower rate of decay than d m, i.e., f(m) d m for all m Z n. Proof. Given any ε > 0, find a trigonometric polynomial P such that f P 1 < ε (use Theorem 4.26(i)). Then for any large enough m Z n we have P (m) = 0, which implies f(m) = f(m) P (m) = (f(x) P (x))e 2πim x dx f P 1 ε. T n This proves that f(m) 0 as m Fourier series on L 1 (T): uniform and L p convergence of S N (f) Theorem Suppose f : T 1 C is Hölder continuous of order α (0, 1], which means that there is some C > 0 such that for all x, y T 1 f(x) f(y) C x y α. Then S N (f) f in the norm (i.e., uniformly on T 1 ). Remarks (a) If α = 1, such functions are also called Lipschitz continuous. (b) E.g., functions in C 1 (T) are Hölder continuous with α = 1. (c) Note that if we assume Hölder continuity at one point x 0 only, the the proof shows that S N (f)(x 0 ) goes f(x 0 ). f(x) f(x 0) x x 0 dx < (d) Instead of the Hölder continuity, one could instead impose a weaker Dini condition T for the convergence at a point x 0, or a uniform Dini-type condition for the uniform convergence. The proof goes in the exact same manner as below. (e) There is an analogue of this in T n due to Tonelli see [G, Sect ]. Proof. Let us prove pointwise convergence only, and leave uniform estimates as an exercise. So let us fix x T. Note that the Dirichlet kernel (see Propositions 4.19(i) and 4.17(i)), satisfies T D N(y) dy = 1. So we can write S N (f)(x) f(x) = D N (y)(f(x y) f(x)) dy = sin((2n + 1)πy)h x (y) dy, (4.1.6) where h x (y) := 1 f(x y) f(x) sin(πy) T T C y α sin(πy) C y 1+α. Note that h x L 1 (T). Using sin((2n + 1)πy) = 2i (e(2n+1)πiy + e (2N+1)πiy ), we can rewrite (4.1.6) as ( 1 2i (e πiy h x (y)) ( N) (e πiy h x (y)) (N) ) which goes to 0 by the Riemann Lebesgue lemma (Theorem 4.31). 49

12 Theorem If 1 < p < and f L p (T n ), then S N (f) f in the p -norm. Remarks (a) This fails for circular partial Fourier sums if n 2 and p 2. (b) This fails for p = 1 even in the one-dimensional case n = 1! Proof. For p = 2 we already discussed it. For p 2, see, e.g., [G, Sect ] (requires a lot more machinery than we currently have) Fourier series on L 1 (T 1 ): pointwise convergence of S N (f) and C N (f) We saw that C N (f) f if f C(T) and S N (f) f if f C(T) and f is Hölder continuous. We also saw that Hölder or Dini conditions at a point x 0 (both of which are stronger than continuity at x 0 ) suffice for the convergence of C N (f)(x 0 ) to f(x 0 ). In this section we show that a jump discontinuity (under some extra mild conditions) usually do not spoil the picture but the limit of S N (x) or C N (x) will no longer be f(x). Theorem as N. (i) Suppose f L 1 (T) has one sided limits f(x 0 ) and f(x 0 +). Then C N (f)(x 0 ) 1 2 (f(x 0+) + f(x 0 )) (ii) Suppose f : T C is a function of bounded variation (=difference of two bounded monotonically non-decreasing functions), and f has one sided limits f(x 0 ) and f(x 0 +). Then as N. S N (f)(x 0 ) 1 2 (f(x 0+) + f(x 0 )) Proof. (i) See [G, Sect ] for all the details. By translation, we may assume x 0 = 0. We write C N (f)(0) 1 2 (f(0+) + f(0 )) = 1/2 1/2 F N (0 x)f(x) dx 1 (f(0+) + f(0 )) 2 = 1/2 0 (f(x) f(0+))f N (x) dx + 0 1/2 (f(x) f(0 ))F N (x) dx. We divide 1/2 into δ /2, where δ > 0 is small enough to make f(δ) f(0+) < ε. Then δ can be bounded by ε δ δ 0 0 F N < ε, while the second integral can be bounded by 1/2 ( f(x) + f(0+) ) dx sup δ [δ,1/2] F N (x) and then showing that sup [δ,1/2] F N (x) 0 for any δ > 0. (ii) See [F, Thm 8.43] for all the details. We just sketch some ideas similar to (i). First we may assume that x 0 = 0 and f is real-valued, non-decreasing, and right-continuous. Given ε > 0, we write S N (f)(0) 1 1/2 2 (f(0+) + f(0 )) = (f(x) f(0+))d N (x) dx /2 (f(x) f(0 ))D N (x) dx. The difference from the proof of Theorem 4.33 is that at t = 0 we no longer have a bound on f(x) f(0) Cx α. However we still have f(x) f(0+) 0. The difference with (i) is that D N doesn t behave as well as F N. Just like in (i), we split 1/2 0 into δ 0 + 1/2 δ, where δ > 0 is small enough to make f(δ) f(0+) < ε. Then δ 0 can be shown to be comparable with ε δ η D N (x) dx for some η [0, δ] (important that f is monotone here! otherwise oscillations in D N could add up and one would get a contribution of order D N dx which is large); and the second integral is small by the Riemann Lebesgue lemma as in the proof of Theorem

13 Note that alongside we proved the following Corollary Suppose f n f in p -norm (for any 1 p ). Then there exists an subsequence n k N such that f nk (x) f(x) for almost all x Fourier series on L 1 (T 1 ): a.e. pointwise convergence of S N (f) and C N (f) If we only require convergence a.e., then the conditions can be significantly relaxed: Theorem (i) If f L 1 (T) then C N (f)(x) f(x) for a.e. x T. (ii) (Carleson s theorem) If f L p (T) (for some 1 < p < ), then S N (f)(x) f(x) for a.e. x T. Remarks (a) Both parts work for T n without a change. (b) Part (ii) fails for the case p = 1 (so, again, C N behaves better than S N!). (c) There are examples of (continuous!) functions where S N (f) f fails on an arbitrary set of Lebesgue measure 0. Proof. We will omit the proofs. (i) is proved in [G, Thm 3.3.3], (ii) in [G, Thm ] Fourier series on L p (T n ): summability of coefficients and Hausdorff Young Recall that we showed: f L 2 (T n ) iff f l 2 (Z n ); f L 1 (T n ) implies f l (Z n ) (in fact, more is true: f(m) 0 by the Riemann Lebesgue lemma). Theorem 4.41 (Hausdorff Young Inequality). Suppose that 1 p 2, and let 1 p + 1 q = 1. If f Lp (T n ), then f l q (Z n ). Moreover, f q f p. Remark The optimal constants are in fact f q p 1/2p q 1/2q f p. Proof. Use Riesz Thorin Theorem 3.35 to interpolate between the above two L 2 l 2 and L 1 l results Fourier series on L 1 (T n ): decay of the Fourier coefficients Intuition. By the Riemann Lebesgue lemma, f(m) 0 as m for any f L 1 (T n ). In this section we discuss the speed of decay of f. This is closely connected with the smoothness properties of f: the smoother f is, the faster is the decay of f at infinity. Note that the speed of decay of f is also related to the summability properties ( f l q as in the previous section). Theorem Let s N. 51

14 (i) Suppose α f exist and α f L 1 for α s. Then ( ) s f(m) n sup α =s α f(m) 2π m s, (4.1.7) in particular f(m) goes to 0 as m faster than 1 m s. (ii) Suppose that for some C > 0 and all m Z n. Then α f exist for all α s 1. f(m) 1 C 1 + m s+n (4.1.8) Proof. (i) Given m Z n, choose j such that m j = max 1 k n m k. Note that m j 1 n m (here m = m m). Then we use integration by parts with respect to xj (see property (ix) in Proposition 4.13) s times to get: f(m) = f(x)e 2πix m dx = ( 1)s (2πim j ) s ( j s f)(x)e 2πix m dx. (4.1.9) This proves (4.1.7). The statement about the limit follows from α f L 1 and the Riemann Lebesgue lemma. (ii) Since 1 R n 1+ x dx <, we have that (4.1.8) implies f l 1, so Theorem 4.29 tells us that n+1 f(m)e 2πix m = f uniformly. We can apply α (with α s 1) term-wise since the resulting equality α f = m Z n f(m)(2πim) α e 2πix m is still uniformly convergent by (4.1.8). Note that with α = s the uniform convergence of the last series is no longer clear. 4.2 Fourier Analysis on R n Schwartz space Intuition. Instead of defining the Fourier transform for L 1 (R n ) functions, let us introduce the class of Schwartz functions (smooth functions that decay fast an infinity) and work with Fourier transform for only those. Because of the decay, technicalities will be very mild, and then afterwards we will be able to extend the Fourier transform and its properties to the general setting. Definition (i) A function f : R n C is called a Schwartz function (we write f S(R n )) if f C (R n ) and for any pair of multi-indices α and β, f α,β := sup x R n x α β f(x) <. (4.2.1) (ii) Let f n, f S(R n ). We say that the sequence f n converges to f in S(R n ) iff for all α, β: as n. f n f α,β 0 52

15 Remarks (a) C c (R n ) S(R n ). (b) Any function of the form x κ e x 2 (where κ is some multi-index) is in S(R n ). (c) We defined a Schwartz function via sup x R n x α β f(x) < for all α, β. Equivalently, one can use sup x R n β (x α f(x)) < for all α, β. Yet another way to characterize Schwartz functions is by saying that for all N N and all α there is C α,n > 0 such that ( α f)(x) C α,n 1 (1 + x ) N. Definition Let X be a Hausdorff topological vector space. Let { j } j=1 be a family of semi-norms on X which generate the topology on X. We say that X is a Fréchet space if X is complete with respect to this family of semi-norms: that is, if x j X is a Cauchy sequence (for every j: x k x m j 0 as k, m ), then x j converges to some element x in X (that is, for every j: x n x j 0 as n ). Remark One can define a complete metric that induces the same topology on X. d(x, y) = k=0 2 j x y j 1 + x y j Proposition S(R n ) with the family of semi-norms { α,β } forms a Fréchet space. Proof. To prove completeness, let f j f k α,β 0 as j, k. Then for any β, { β f k } k=1 is Cauchy in, so β f k g β as k uniformly, for some continuous g β. Let e j be the j-th standard vector in R n. Then by the Fundamental Theorem of Calculus, we have f k (x + te j ) f k (x) = Taking k (using uniform convergence), we get g 0 (x + te j ) g 0 (x) = t 0 t 0 j f k (x + se j ) ds. g ej (x + se j ) ds. Differentiating this, we get j g 0 = g ej. By a similar argument and an induction, we get β g 0 = g β for all β. Now note that {x α f k } k=1 is also uniformly convergent, and it is easy to see that the limit therefore must be x α g 0. Similarly for {x α β f k } k=1, which shows that g 0 S(R n ) and f k g 0 in S(R n ). Proposition (i) If f, g S(R n ) then fg and f g are in S(R n ). Moreover, α (f g) = ( α f) g = g ( α g) for any multi-index α. (ii) For any 1 p <, S(R n ) is p -dense in L p (R n ). Proof. Left as an exercise. 53

16 4.2.2 Fourier transform on S(R n ): definition and basic properties Definition Let f S(R n ). Define the Fourier transform of f to be f(ξ) = f(x)e 2πix ξ dx. (4.2.2) R n Recall the notation f(x) := f( x) and τ y (f)(x) := f(x y). Proposition Let f, g be in S(R n ). Then for all y R n, λ C: (i) f + g = f + ĝ; (ii) λf = λ f; (iii) f(m) = f; (iv) f = f; (v) τ y (f)(ξ) = f(ξ)e 2πiξ y ; (vi) ( e 2πiy x f(x) ) (ξ) = τ y ( f)(ξ); (vii) f f 1 ; (viii) f g = fĝ; (ix) α f(ξ) = (2πiξ) α f(ξ); (x) ( α f)(ξ) = (( 2πix) α f(x)) (ξ); (xi) f S(R n ) implies f S(R n ). Example If f = e πa x 2 with a > 0 (note that f S(R n )), then its Fourier transform is a n/2 e π ξ 2 /a. To prove this for n = 1, we use properties (x) and then (ix) (see above) to see that d dξ f(ξ) = ( 2πixf(x)) (ξ) = ( 2πixe πax2 ) (xi) = i a (f ) (ξ) = 2π a ξ f(ξ). This is an ODE which integrates to e πξ2 /a f(ξ) = c. Taking ξ = 0 gives c = f(0) = e πax2 dx = 1 a. The case for general n follows from Fubini Fourier transform on S(R n ): the inverse Fourier transform and Fourier inversion Definition For a Schwartz function f S(R n ), define its inverse Fourier transform to be f (x) := f(ξ)e 2πix ξ dξ = f(x). R n Remark It is clear that all the properties of Fourier transform from the previous section, hold for the inverse transform with the straightforward modifications. Theorem Let f, g S(R n ). 54

17 (i) f(x)ĝ(x) dx = f(x)g(x) dx. R n R n (ii) (Fourier Inversion) ( f) = f = (f ) ; (iii) (Parseval s relation) f, g = f, ĝ ; (iv) (Plancherel s identity) f 2 = f 2 = f 2 ; Proof. (i) follows from Fubini theorem: f(x)ĝ(x) dx = f(x)g(y)e R n R R 2πiy x dy dx = f(y)g(y) dy. n n (ii) Let ε > 0, y R n, and g(x) = e 2πiy x e πε2 x 2. By combining Example 4.52 and Proposition 4.51(vi), we have ĝ(ξ) = 1 2 ε n e π ξ y /ε 2. The important observation is that {ĝ(ξ)} ε 0 is an approximate identity. By (i) we get f(x) 1 2 R ε n e π x y /ε 2 dx = n R n f(x)e 2πiy x e πε 2 x 2 dx. (4.2.3) Now take ε 0. The left-hand side of (4.2.3) goes to f(y) since f is continuous and {ĝ(ξ)} ε 0 is an approximate identity (see Proposition 4.10). The right-hand side of (4.2.3) converges to ( f) (y) by the Lebesgue Dominated Convergence, so we are done. (iii) is obtained by plugging in g = ( h) into (i). (iv) follows by choosing f = g in (iii) Fourier transform on L 1 (R n ): definition and basic properties Note that Definition 4.50 of the Fourier transform works for any f L 1 (R n ). Proposition Let f, g L 1 (R n ). Then properties (i) (viii) of Proposition 4.51 hold. Theorem 4.57 (Riemann Lebesgue lemma). Let f L 1 (R n ). Then f C 0 (R n ). Proof. Continuity (in fact, uniform continuity) follows from lim f(ξ + h) f(ξ) = lim f(x)e 2πix ξ (e 2πix h 1) dx = 0 h 0 h 0 R n where we use the Dominated Convergence Theorem the integrand is bounded by 2 f(x) L 1. Convergence to 0 can be proved by first proving this for simple functions (Fourier transform of χ [a,b] (x) is 1 2πiξ (e 2πiξa e 2πiξb ) which goes to 0), and then approximating f L 1 by a simple function φ with φ f 1 ε, and then f(ξ) φ(ξ) f φ 1 ε. 55

18 4.2.5 Fourier transform on L 1 (R n ): the Fourier Inversion formula Similarly, the inverse Fourier transform (Definition 4.53) work for any f L 1 (R n ). However the Fourier Inversion formula does not necessarily holds for all functions in L 1, compare with Theorem Theorem Suppose f L 1 (T n ) and f L 1 (T n ). Then ( f) = (f ) is a continuous function which is equal to f a.e. Proof. The proof follows exactly same lines as the proof of Theorem 4.55(i) (ii). The main difference is that f L 1, so that we do not get pointwise convergence of the left-hand side of (4.2.3) to f but only L 1 -convergence. The right-hand side of (4.2.3) still converges pointwise to ( f), so we are done Fourier transform on L 2 (R n ) Unlike the case of n-torus, L 2 (R n ) is not a subspace of L 1 (R n ), and the definition of the Fourier transform does not work for any f L 2 (R n ). However the Fourier transform is well-defined on L 1 (R n ) L 2 (R n ) which is dense in L 2 (R n ) (e.g., because it contains the class of simple functions which we know are 2 -dense in L 2 ). Moreover, on L 1 L 2 the Fourier transform is an isometry (see Theorem 4.55(iv) these properties can be extended to f L 1 L 2 ). Definition For f L 2 (R n ), choose any f N L 1 (R n ) L 2 (R n ) such that f N f 2 0. Then the L 2 (R n )-limit of f N is defined to be the Fourier transform of f, denoted by F(f). Remarks (a) If f L 1 (R n ) L 2 (R n ) then f C 0 (R n ) and F(f) L 2 (R n ). It can be shown that these two functions are equal a.e.. Therefore it is not much abuse of notation to use again f for F(f). (b) In particular, we can take f N (x) := f(x)χ x N L 1 L 2 in the definition to get that F(f)(ξ) is the L 2 -limit of (f(x)χ x N ). F(f)(ξ) = L 2 - lim f(x)e 2πix ξ dx. N x N The existence of L 2 -limit implies that we also have a pointwise (a.e.) convergence along a subsequence. Definition For f L 2 (R n ) we define the inverse Fourier transform of f to be the L 2 (R n ) function given by F 1 (f)(x) := F(f)( x). Remarks (a) Again, if f L 1 L 2, it can be shown that F 1 (f) = f a.e., so we will use either notation interchangeably. (b) Because we have (the Fourier inversion) F F 1 = F 1 F on a dense subset S(R n ) of L 2, we get that it holds for all f L 2 (R n ). (c) Part (b) says that if f L 2, then the inverse Fourier transform ( f) of f converges to f in the L 2 -sense. Just like in the Remark 4.60(b), we can take an approximating sequence to be f(x)χ max xj N. In other words, if ( ) s N (f) := f(x)χmax xj N = f(x)e 2πix ξ dx. (4.2.4) max x j N (this is the analogue of the square partial Fourier sum S N (f) that we had on the n-torus), then (b) says that f s N 2 0 (compare with Corollary 4.12(i)). Of course there is no significance in the square max x j N here and circles or any other expanding sequence of domains would also work (just like on the n-torus for L 2 -convergence). 56

19 4.2.7 Fourier transform on L 1 (R n ): various types of convergence Intuition. Recall (Sections ) that on T n we had different ways to sum the Fourier series (pointwise, L p, Cesàro, Abel (see HW), etc) which allowed us to recover f (at least a.e.) from the Fourier series. Similar theory can be developed for the Fourier transform. As a prototype, let us discuss the analogue of Theorem 4.22 for the uniform and L p convergence of the Cesàro means of Fourier series. Given f L 1 (R n ) and its Fourier transform f, define ( ) s N (f) := f(x)χmax xj N = max x j N f(x)e 2πix ξ dx, (4.2.5) and n σ N (f) := f(x) j=1 ( 1 x ) j N χ max xj N = n max x j N j=1 ( 1 x ) j f(x)e 2πix ξ dx. (4.2.6) N These should be viewed as the analogues of S N (f) and C N (f) from the n-torus: cf. (4.1.3) and (4.1.5), respectively. To make the analogy even more transparent, let us note the following: Lemma Let f L 1 (R n ). Then (i) s N (f) = f d N, where d N (x) := n j=1 sin(2πnx j ) πx j = ( χ max xj N(x) ). (ii) σ N (f) = f k N, where k N (x) := n j=1 sin 2 (πnx j ) π 2 Nx 2 j n = j=1 ( 1 x ) j N χ max xj N(x) Remarks (a) More generally, both (i) and (ii) are consequences of the following fact: assume that f, h L 1 (R n ) and that ĥ L1 (R n ), then we have that (f h)(x) = f(ξ) ĥ(ξ)e 2πix ξ dξ. R n (b) d N and k N are the real-line analogues of the Dirichlet and Fejér kernels D N and K N that we had on the n-torus. (c) Compare with Proposition 4.19 as well as Proposition Proof. Let us prove Remark (a). By the Property (viii) of Proposition 4.51 we have f h = f ĥ = fh. Since H L 1 and f C 0, we have that fh L 1, so we can apply the Inversion Formula (Theorem 4.58) to get f h = ( fh) which is what we wanted to show. Proposition {k N (x)} N=1 forms an approximate identity on Rn.. Remark Just like before, the Dirichlet kernels {d N } N=1 satisfy d N dx = 1 for all N, but fail d N 0 or d N dx < C. do not form an approximate identity: they Proof. Just a calculation. 57

20 Corollary (i) For any f L p (R n ) (1 p < ), σ N (f) f in the p -norm. (ii) For any f L 1 (R n ) C(R n ), σ N (f) f uniformly on compacts of R n. Remark This is the analogue of Theorem Proof. Immediate from Propositions 4.65 and Fourier transform on L p (R n ): summability of f and Hausdorff Young Intuition. This is the exact analogue of Section Recall that we showed: f L 2 (R n ) iff f L 2 (R n ); f L 1 (T n ) implies f C 0 (R n ), in particular f L (R n ). Theorem 4.69 (Hausdorff Young Inequality). Suppose that 1 p 2, and let 1 p + 1 q = 1. If f Lp (R n ), then f L q (R n ). Moreover, f q f p. Proof. Use Riesz Thorin Theorem 3.35 to interpolate between the above two L 2 L 2 and L 1 L results Fourier series on L 1 (R n ): decay of f at Intuition. The analogue of the results from can also be proved in the same way: the smoother f is, the faster is the decay of f at infinity. We omit the details The Poisson Summation Formula Intuition. Given a function f L 1 (R n ) C(R n ), let us form its periodization F (x) = m Z n f(x + m). (4.2.7) F can now be viewed as a periodic T n C function (in fact, F L 1 (T n ) since f L 1 (R n )). It is not hard to see that the Fourier coefficients F (m) of F coincide with the Fourier transform f(ξ) evaluated at ξ = m. Assuming the Fourier series of F converge to F, we therefore should expect F (x) = f(m)e 2πim x, in particular F (0) = f(m) which is the Poisson Summation Formula, see (4.2.10). Theorem 4.70 (Poisson Summation Formula). Suppose f L 1 (R n ) C(R n ) and f L 1 (R n ) satisfy for some C > 0, δ > 0. Then for all x R n we have m Z n f(x + m) = f(x) + f(x) C(1 + x ) n δ (4.2.8) m Z n f(m)e 2πim x, (4.2.9) 58

21 and in particular f(m) = f(m). (4.2.10) m Z n m Z n Remark Note that (4.2.8) automatically implies f L 1 (R n ) Proof. Define F as in (4.2.7). Applying Theorem 2.26 (together with f L 1 (R n )) we obtain that (4.2.7) converges a.e. and F L 1 (T n ). In fact, it converges uniformly on T n from the condition f(x) C(1 + x ) n δ ; in particular F is continuous. Applying Theorem 2.26 again we get F (m) = f(m) for each m Z n. Since F (m) = f(m) 1 C <, (1 + m ) n+δ m Z n m Z n m Z n we get (see Theorem 4.29) that the Fourier series of F converge uniformly to F. This is precisely (4.2.9). 59