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Fourier Series: Convergence and Summability

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Fourier Series: Convergence and Summability CHAPTER 1 Fourier Series: Convergence and Summability Let T = R/Z be the one-dimensional torus (circle). We consider various function spaces on it, namely C(T),Cα(T), and Lp(T). The space of complex Borel measures on T will be denoted by M(T). Any µ ∈ M(T) has associated with it a Fourier series X∞ µ ∼ µˆ(n)e(nx) n=−∞ where we let e(x) = e2πix and Z 1 µˆ(n) = e(−nx) dµ(x) . 0 We first consider the classical question of convergence of Fourier series. The partial sums of f ∈ L1(T) are defined as XN ˆ SN f(x) = f(n)e(nx) n=−N XN Z 1 = e(−ny)f(y) dy e(nx) n=−N 0 Z 1 XN = e(n(x − y))f(y) dy 0 n=−N Z 1 = DN (x − y)f(y) dy 0 PN where DN (x) = n=−N e(nx) is the Dirichlet kernel. Exercise 1. Check that sin((2N + 1)πx) D (x) = , N sin(πx) and draw the graph of DN . One can also write S f(x) = (D ∗ f)(x) R N N where f ∗ g(x) := T f(x − y)g(y) dy is the convolution of f and g. You should think of f ∗ g as an average of translates of f. Exercise 2. Prove the following properties of the convolution: 1 p 1 a) kf ∗ gkp ≤ kfkpkgk1 for all 1 ≤ p ≤ ∞, f ∈ L , g ∈ L . This is called Young’s inequality. You should pay careful attention to the fact that the integral defining f ∗ g is not necessarily absolutely convergent for every x. b) More generally, if f ∈ C(T), µ ∈ M(T) then f ∗ µ is well defined. Show that, for 1 ≤ p ≤ ∞, kf ∗ µkp ≤ kfkpkµk which allows you to extend f ∗ µ to f ∈ L1. p p0 1 1 c) If f ∈ L (T) and g ∈ L (T) where 1 < p < ∞, and p + p0 = 1 then f ∗ g, originally defined only almost every where, extends to a continuous function on T and kf ∗ gk∞ ≤ kfkpkgkp0 . Is this still true of p = 1 or p = ∞? d) For f, g ∈ L1(T) show that for all n ∈ Z f[∗ g(n) = fˆ(n)ˆg(n) . It is typically difficult to understand convergence of SN f. This can be seen as an instance of the fact that the Dirichlet kernel is not an approximate identity, see below. One (standard) positive result is the following theorem. α Theorem 1. If f ∈ C (T), 0 < α ≤ 1, then kSN f − fk∞ −→ 0 as N −→ ∞. Proof. One has, with δ > 0 to be determined, Z 1 SN f(x) − f(x) = (f(x − y) − f(x))DN (y) dy Z0 = (f(x − y) − f(x))DN (y) dy (1) |yZ|≤δ + (f(x − y) − f(x))DN (y) dy . |y|>δ There is the obvious bound µ ¶ 1 |D (y)| ≤ C min N, . N |y| Here and in what follows, C will denote a numerical constant that can change from line to line. The first integral in (1) can be estimated as follows Z Z 1 α−1 α |f(x) − f(x − y)| dy ≤ [f]α |y| dy ≤ C[f]αδ (2) |y|≤δ |y| |y|≤δ where we have set |f(x) − f(x − y)| [f]α = sup α . x,y |y| 2 To bound the second term in (1) one needs to exploit the oscillation of DN (y). In fact, Z B := (f(x − y) − f(x))DN (y) dy = |y|>δ Z f(x − y) − f(x) = sin((2N + 1)πy) dy sin(πy) |yZ|>δ 1 = − hx(y) sin((2N + 1)π(y + )) dy |y|>δ 2N + 1 f(x−y)−f(x) where hx(y) := sin(πy) . Therefore, Z 2B = hx(y) sin((2N + 1)πy) dy |y|>δ Z µ ¶ 1 − hx y − sin((2N + 1)πy) dy |y− 1 |>δ 2N + 1 Z 2N+1 1 = (h (y) − h (y − )), sin((2N + 1)πy) dy x x 2N + 1 |yZ|>δ 1 + hx(y − ) sin((2N + 1)πy) dy [−δ,−δ+ 1 ] 2N + 1 Z 2N+1 1 − hx(y − ) sin((2N + 1)πy) dy . 1 2N + 1 [δ,δ+ 2N+1 ] These integrals are estimated by putting absolute values inside. To do so we use the bounds kfk |h (y)| < C ∞ , x δ µ ¶ |τ|α[f] kfk |h (y) − h (y + τ)| < C α + ∞ |τ| x x δ δ2 if |y| > δ > 2τ. In view of the preceding one checks µ ¶ N −α[f] N −1kfk |B| ≤ C α + ∞ , (3) δ δ2 1 −α/2 provided δ > N . Choosing δ = N one concludes from (1), (2), and (3) that ³ ´ −α2/2 −α/2 −1+α |(SN f)(x) − f(x)| ≤ C N + N + N (kfk∞ + [f]α) , which proves the theorem. ¤ Remark. We shall see later that the theorem fails for continuous functions i.e., α = 0. 3 Better convergence properties are achieved by means of Cesaro means, i.e., 1 NX−1 σ f := S f . N N n n=0 1 PN−1 Setting KN = N n=0 Dn, which is called the Fejer kernel, one therefore has σN f = KN ∗ f. ³ ´2 1 sin(Nπx) Exercise 3. Check that KN (x) = N sin(πx) . b It is important to realize that KN looks like a triangle, i.e., for all n ∈ Z µ ¶ |n| + Kb (n) = 1 − . N N The importance of KN with respect to convergence properties lies with the fact that the Fejer kernels form an approximate identity (abbreviated a.i.). ∞ ∞ Definition 1. {ΦN }N=1 ⊂ L (T) are an approximate identity provided R A1) 1 Φ (x) dx = 1 for all N 0 NR A2) sup 1 |Φ (x)| dx < ∞ N 0 N R A3) for all δ > 0 one has |x|>δ |ΦN (x)|dx −→ 0 as N −→ ∞. ∞ Lemma 1. The Fejer kernels {KN }N=1 form an a.i. R Proof. We clearly have 1 K (x) dx = 1 (why?) and K (x) ≥ 0 so that A1) and 0 N ¡ N ¢ C 2 1 A2) hold. A3) follows from the bound |KN (x)| ≤ N min N , x2 . ¤ ∞ Lemma 2. For any a.i. {ΦN }N=1 one has a) If f ∈ C(T), then kΦN ∗ f − fk∞ −→ 0 as N −→ ∞ p b) If f ∈ L (T) where 1 ≤ p < ∞, then kΦN ∗ f − fkp −→ 0 as N −→ ∞. Proof. a) Since T is compact, f is uniformly continuous. Given ² > 0, let δ > 0 be such that sup sup |f(x − y) − f(x)| < ² . x |y|<δ Then, by A1)–A3), ¯ Z ¯ ¯ ¯ |(ΦN ∗ f)(x) − f(x)| = ¯ (f(x − y) − f(x))ΦN (y) dy¯ TZ Z ≤ sup sup |f(x − y) − f(x)| kΦN (t)|dt + |ΦN (y)|2kfk∞ dy x∈T |y|<δ T |y|≥δ <C² provided N is large. 4 b) Let g ∈ C(T) with kf − gkp < ². Then kΦN ∗ f − fkp ≤ kΦN ∗ (f − g)kp + kf − gkp + kΦN ∗ g − gkp µ ¶ ≤ sup kΦN k1 + 1 kf − gkp + kΦN ∗ g − gk∞ N where we have used Young’s inequality (Exercise 2 part a)) to obtain the first term on the right-hand side. Using A2), the assumption on g, as well as part a) finishes the proof. ¤ Corollary 1. a) Trigonmetric polynomials are dense in C(T), Lp(T), 1 ≤ p < ∞. 2 b) For any f ∈ L (T) X 2 ˆ 2 kfk2 = |f(n)| n∈Z 2 c) R{e(nx)}n∈Z form a complete orthonormal basis in L (T). 2 d) P ˆ for all f, g ∈ L (T) Tf(x)¯g(x) dx= n∈Z f(n)gˆ(n) Proof. ∞ a) By Lemma 1, {KN }N=1 form an a.i. and Lemma 2 applies. Since σN f = KN ∗ f is a trigonometric polynomial, we are done. b), c), d) are well-known to be equivalent by basic Hilbert space theory. The point to make here is of course that Z 1 en(x)em(x) dx = δ0(n − m) 0 (where δ0(j) = 1 if j = 0 and δ0(j) = 0 otherwise). Generally speaking one thus has Bessel’s inequality X ˆ 2 2 |f(n)| ≤ kfk2 2 and equality is equivalent to span {en} being dense in L (T). That, however, is guaran- teed by part a). ¤ Remark. Parts b), c), d) go under the name Plancherel, Riesz-Fischer, and Parse- val. Corollary 2. (uniqueness theorem): If f ∈ L1(T) and fˆ(n) = 0 for all n ∈ Z, then f = 0. Proof. σN f = 0 for all N by assumption and kσN f − fk1 −→ 0. ¤ Corollary 3. (Riemann-Lebesgue): If f ∈ L1(T), then fˆ(n) −→ 0 as n −→ ∞ Proof. Given ² > 0, let N be so large that kσN f − fk1 < ². ˆ d ˆ Then |f(n)| = |σN f(n) − f(n)| ≤ kσN f − fk1 < ² for |n| > N. ¤ 5 Try to do problem 41 from the appendix. p We now return to the issue of convergence of the partial sums SN f in L (T) or C(T) (observe that it makes no sense to ask about uniform convergence of SN f for general f ∈ L∞(T) because uniform limits of continuous functions are continuous). Lemma 3. The following statements are equivalent: For any 1 ≤ p ≤ ∞ a) For every f ∈ Lp(T) (or f ∈ C(T) if p = ∞) kSN f − fkp −→ 0 as N −→ ∞. b) supN kSN kp→p < ∞ Proof. The implication b) =⇒ a) follows from the fact that trigonometric polynomi- als are dense. The implication a) =⇒ b) can be deduced immediately from the uniform boundedness principle of functional analysis.
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