Fourier Series. ,..., e ixn ). Conversely, each 2π-periodic function φ : R n C induces a unique φ : T n C for which φ(e ix 1

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1 Fourier Series Let {e j : 1 j n} be the standard basis in R n. We say f : R n C is π-periodic in each variable if f(x + πe j ) = f(x) x R n, 1 j n. We can identify π-periodic functions with functions on a torus. Let S 1 = {e iθ : θ R} C, and T n = S 1 S 1 C n. To each function φ : T n C we can identify a π-periodic function φ : R n C by φ(x 1,...,x n ) = φ(e ix 1,..., e ixn ). Conversely, each π-periodic function φ : R n C induces a unique φ : T n C for which φ(e ix 1,...,e ixn ) = φ(x 1,...,x n ). If φ : R n C is π-periodic, φ is uniquely determined by its values φ(x) for x [ π, π) n or for x [0, π) n. Let ν n = (π) n λ n, where λ n is n-dimensional Lebesgue measure. Then ν n induces a measure ν n on T n for which φd νn = φ dν n. T n [0,π] n From here on, we blur the distinction between φ and φ and between ν n and ν n, and we will abuse these notations. Note: ν n (T n ) = ν n ([0, π] n ) = 1. Let L p (T n ) denote L p ([0, π] n ) with measure ν n (1 p ). L (T n ) is a Hilbert space with inner product (ψ, φ) = ψφ dνn = ψφ dνn. T n [0,π] n Theorem. {e ix ξ : ξ Z n } is an orthonormal system in L (T n ). Proof. { 1, ξ = η (e ix η, e ix ξ ) = e ix (ξ η) dν n = [0,π] 0, ξ η. n Definition. A trigonometric polynomial is a finite linear combination of {e ix ξ : ξ Z n }. Note: since {e ix ξ, e ix ξ } and {cos(x ξ), sin(x ξ)} span the same two-dimensional space, we could use sines and cosines as our basic functions. Definition. C(T n ) is the space of all continuous π-periodic functions φ : R n C. Note that C(T n ) C([0, π] n ). We will use the uniform norm φ u = sup x φ(x) on C(T n ). C k (T n ) (for k 0, k Z) is the space of all C k π-periodic functions φ : R n C. Again C k (T n ) C k ([0, π] n ). We will use the norm φ C k = α k α φ u. 93

2 94 Fourier Series Fourier Coefficients For f L 1 (T n ), define f(ξ) = e ix ξ f(x)dν n (x) T n for ξ Z n. Then f(ξ) f 1. We regard f as a map f : Z n C; then f l (Z n ) and f f 1. Since ν(t n ) = 1, we have L (T n ) L 1 (T n ) and f 1 f. For f L (T n ), f can be expressed as an inner product: f(ξ) = (e ix ξ, f). The Fourier series of f is the formal series ξ Z n f(ξ)e ix ξ. We will study in what sense this series converges to f. A key role is played by what is called a summability kernel; this is a sequence Q k with properties (1), (), (3), (4) below. Define Q k (x) = a 1 k ( n j=1 1 (1 + cosx j) ) k where Lemma. a k = T n ( n j=1 k 1 (1 + cosx j)) dν n (x). (1) Q k is a trigonometric polynomial. () Q k (x) 0 (3) Q T n k (x)dν n (x) = 1. (4) For 0 < δ < π, set η k (δ) = max{q k (x) : x [ π, π] n \( δ, δ) n }. Then lim k η k (δ) = 0. Proof. The first three properties are clear. To prove (4), we first show that the sequence a k satisfies a k ((k + 1)) n. In fact, since 1 + cosx is non-negative on [0, π] and concave on [0, π ], we have 1 π ( 1 π (1 + cosx)) k 1 π ( dx = 1 π π (1 + cosx)) k dx 0 1 π/ ( 1 x k dx π π) Now if x [ π, π] n \( δ, δ) n, then 0 = (k + 1) 1 (1 k 1 ) ((k + 1)) 1. Q k (x) a 1 k ( 1 (1 + cosδ)) k ((k + 1)) n ( 1 (1 + cosδ)) k 0

3 Fourier Series 95 as k. Theorem. Given f C(T n ), let p k (x) = (f Q k )(x) = f(x y)q k (y)dν n (y). T n Then p k is a trigonometric polynomial, and p k f u 0 as k. Proof. Since Q k (ξ) = 0 for ξ sufficiently large, p k (x) = f(y)q k (x y)dν n (y) = f(y) T n T n ξ Q k (ξ)e i(x y) ξ dν n (y) = ξ f(ξ) Q k (ξ)e ix ξ is a trigonometric polynomial. Given ǫ > 0, the uniform continuity of f implies that there is δ > 0 such that x 1 x < δ f(x 1 ) f(x ) < ǫ. Then p k (x) f(x) = (f(x y) f(x))q k (y)dν n (y) T n T n f(x y) f(x) Q k (y)dν n (y) = I 1 + I, where I 1 is the integral over y ( δ, δ) n and I is the integral over y [ π, π] n \( δ, δ) n. Now I 1 ǫ Q k (y) dν n (y) ǫ, and ( δ,δ) n I f u η k (δ) dν n f u η k (δ) < ǫ [ π,π] n \( δ,δ) n for k sufficiently large, so the result follows. Corollary. Trigonometric polynomials are dense in C(T n ). Remark. The proof of the Theorem above used only the properties (1) (4) of the Q k. Therefore the same result holds for any summability kernel. Another example for n = 1 is the Féjer kernel, which is given by ( 1 F k (x) = (k + 1) 1sin (k + 1)x) sin ( 1 x) = k ξ= k ( 1 ξ ) e ixξ. k + 1 If we define S k (f) = k ξ= k f(ξ)e ix ξ and σ k (f) = (k + 1) 1 (S 0 (f) + + S k (f)), then σ k (f) = f F k. It follows that for f C(T), σ k (f) f uniformly. This is the classical result that the Fourier series of f C(T) is Cesàro summable to f.

4 96 Fourier Series The partial sums S k (f) themselves are obtained by convolving f with the Dirichlet kernel : S k (f) = f D k, where k D k (x) = e ixξ = sin (( ) ) k + 1 x sin ( 1 x). ξ= k The Dirichlet kernel however, is not a summability kernel: D k is not nonnegative (not horrible), and more crucially it does not satisfy condition (4) of a summability kernel. This is the reason that pointwise convergence of Fourier series is a delicate matter. Corollary. Trigonometric polynomials are dense in L (T n ). Proof. Given f L (T n ) and ǫ > 0, there exists g C(T n ) such that f g < 1 ǫ and there exists a trigonometric polynomial p such that p g u < 1 ǫ, so since ν n(t n ) = 1, f p f g + g p f g + g p u < ǫ. Corollary. {e ix ξ : ξ Z n } is a complete orthonormal system in L (T n ). Hence if f L (T n ), the Fourier series of f (any arrangement) converges to f in L. Also, the map F : L (T n ) l (Z n ) given by f f is a Hilbert space isomorphism and f l (Z n ) = f L (T n ). Theorem. If f L 1 (T n ), then p k f in L 1 (T n ), where p k = f Q k and Q k is as above. Proof. The proof is similar to the proof of the Theorem above, except we use continuity of translation in L 1 instead of uniform continuity. Given ǫ > 0, choose δ f(x α) f(x) 1 < ǫ whenever α < δ. By Fubini, [ ] p k (x) f(x) dν n (x) Q k (y) f(x y) f(x) dν n (x) dν n (y) = I 1 + I, T n T n T n and Q k (y) ǫ dν n (y) = ǫ I 1 I f 1 η k (δ) 0 as k. Corollary. (Uniqueness Theorem). If f L 1 (T n ) and ( ξ Z n ) f(ξ) = 0, then f = 0 a.e. Thus if f, g L 1 (T n ) and f ĝ, then f = g a.e. Proof. If f 0, then p k (x) = ξ f(ξ) Q k (ξ)e ix ξ = 0, and p k f in L 1. Theorem. (Riemann-Lebesgue Lemma). If f L 1 (T n ), then f(ξ) 0 as ξ. Proof. This follows from the analogous result for the Fourier transform, which will be proved later. The statement for the Fourier transform is that if f L 1 (R n ), then e ix ξ f(x) dx R n 0 as ξ, ξ R n. The Fourier series version stated here follows by applying the R n version to f(x)χ [ π,π] n(x) L 1 (R n ) and restricting to ξ Z n.

5 Fourier Series 97 Absolutely Convergent Fourier Series Theorem. Suppose f L 1 (T n ) and f l 1 (Z n ). Then the Fourier series of f converges absolutely and uniformly to a g C(T n ), and g = f a.e. Proof. Let g(x) = ξ Z n f(ξ)e ix ξ. Since f l 1 (Z n ), this series converges uniformly and absolutely, and g C(T n ). By the Dominated Convergence Theorem, ĝ(ξ) = T n e ix ξ ( η Z n f(η)e ix η ) dν n (x) = η Z n f(η) T n e ix ξ e ix η dν n (x) = f(ξ), so g = f a.e. Decay of Fourier Coefficients Smoothness of f Lemma. Suppose α(ξ) l 1 (Z n ) and iξ j α(ξ) l 1 (Z n ). Let f = ξ α(ξ)eix ξ and g = ξ iξ jα(ξ)e ix ξ. Then f, g C(T n ), xj f exists everywhere, and xj f = g. Proof. The two series of continuous functions converge absolutely and uniformly to f and g, respectively. Since xj (α(ξ)e ix ξ ) = iξ j α(ξ)e ix ξ, the result follows from a standard theorem in analysis. Theorem. Suppose f L 1 (T n ) and (1 + ξ m ) f(ξ) l 1 (Z n ) for some integer m 0. Then the Fourier series of f converges absolutely and uniformly to a g C m (T n ), and f = g a.e. Proof. By the theorem above, we only have to show that g C m (T n ). For each ν with ν m, (iξ) ν f(ξ) l 1 (Z n ), so ξ (iξ)ν f(ξ)e ix ξ converges absolutely and uniformly to some g ν C(T n ). By the Lemma and induction, g ν = ν g. Theorem. Suppose f C m (T n ). (a) For ν m, ν x f(ξ) = (iξ)ν f(ξ). (b) (1 + ξ m ) f(ξ) l (Z n ) and (1 + ξ m ) f(ξ) l (Z n ) c n,m f C m (T n ) for some constant c n,m depending only on n, m. (c) f(ξ) c n,m (1 + ξ ) m f C m (T n ) for ξ Z n. Proof.

6 98 Fourier Series (a) follows by integration by parts. Set x = (x,...,x n ), so x = (x 1, x ). Then f x 1 (ξ) = = Iterate for higher derivatives. (b) Part (a) gives for 1 j n: T n 1 T n 1 = (iξ 1 ) f(ξ). [ ] ix ξ f e (x) dν 1 (x 1 ) dν n 1 (x ) T x [ 1 ] (iξ 1 ) e ix ξ f(x) dν 1 (x 1 ) dν n 1 (x ) T ξ m j f l (Z n ) = m x j f l (Z n ) = m x j f L (T n ) m x j f u. Adding gives and (b) follows. (1 + ξ 1 m + + ξ n m ) f(ξ) l (Z n ) f C m (T n ), (c) is immediate from (b) and the fact that on functions on Z n. In comparing the last two theorems, we see that in order to conclude that a given f L 1 (T n ) is C m, it suffices to know that (1+ ξ ) m f l 1, and in the other direction, if f is C m, then (1 + ξ ) m f l. Thus the space of Fourier coefficients of C m functions lies between (1 + ξ ) m l 1 and (1 + ξ ) m l. So faster decay of f corresponds to more smoothness of f.