Zygmund s Fourier restriction theorem and Bernstein s inequality

Transcription

1 Zygmund s Fourier restriction theorem and Bernstein s inequality Jordan Bell jordanbell@gmailcom Department of Mathematics, University of Toronto February 13, Zygmund s restriction theorem Write T d = R d /Z d Write λ d for the Haar measure on T d for which λ d T d = 1 For ξ Z d, we define e ξ : T d S 1 by e ξ x = e 2πiξ x, x T d For f L 1 T d, we define its Fourier transform ˆf : Z d C by ˆfξ = fe ξ dλ d = T d fxe 2πiξ x dx, T d ξ Z d For x R d, we write x = x 2 = x x2 d, x 1 = x x d, and x = max{ x j : 1 j d} For 1 p <, we write 1/p f p = fx p dx T d For 1 p q, f p f q Parseval s identity tells us that for f L 2 T d, ˆf = ˆfξ 2 l 2 ξ Z d 1/2 = f 2, and the Hausdorff-Young inequality tells us that for 1 p 2 and f L p T d, ˆf l q = ˆfξ q ξ Z d 1/q f p, 1

2 where 1 p + 1 q = 1; ˆf = sup l ξ Z d ˆfξ Zygmund s theorem is the following 1 Theorem 1 Zygmund s theorem For f L 4/3 T 2 and r > 0, ˆfξ 2 1/2 5 1/4 f 4/3 1 Proof Suppose that For ξ Z 2, we define Then We have S = ˆfξ 2 1/2 c ξ = ˆfξ S χ ζ =r c ξ 2 = S 2 = ˆfξ 2 = = > 0 ˆfξ 2 S 2 = 1 2 ˆfξ ˆfξ ˆfξc ξ S, hence, defining c : T 2 C by cx = ξ Z d c ξ e 2πiξ x = c ξ e 2πiξ x, x T 2, we have, applying Parseval s identity, S = ˆfξc ξ = fxcxdx T 2 For p = 4 3, let 1 p + 1 q = 1, ie q = 4 Hölder s inequality tells us fxcx dx f 4/3 c 4 T 2 1 Mark A Pinsky, Introduction to Fourier Analysis and Wavelets, p 236, Theorem

3 For ρ Z 2, we define γ ρ = c µ c ν µ ν=ρ Then define Γx = cx 2, which satisfies Γx = cxcx = c ξ c ζ e 2πiξ ζ x = γ ρ e 2πiρ x ξ Z 2 ζ Z 2 ρ Z 2 Parseval s identity tells us c 4 4 = Γ 2 2 = γ ρ 2 ρ Z 2 First, γ 0 = µ Z 2 c µ c µ = µ Z 2 c µ 2 = 1 Second, suppose that ρ Z 2, ρ = 2r If ρ/2 Z 2, then γ ρ = c ρ/2 c ρ/2, and if ρ/2 Z 2 then γ ρ = 0 It follows that γ ρ 2 = γ 2µ 2 = c µ 2 c µ 2 3 ρ =2r Third, suppose that ρ Z 2, 0 < ρ < 2r Then, for C ρ = {µ Z 2 : µ = r, µ ρ = ρ }, we have C ρ 2 If C ρ = 0 then γ ρ = 0 If C ρ = 1 and C ρ = {µ}, then γ ρ = c µ c µ ρ and so γ ρ 2 = c µ 2 c µ ρ 2 If C ρ = 2 and C ρ = {µ, m}, then γ ρ = c µ c µ ρ + c m c m ρ and so It follows that 0< ρ <2r Using 3 and then 2, γ ρ 2 4 0< ρ 2r γ ρ 2 2 c µ 2 c µ ρ c m 2 c m ρ γ ρ 2 4, ν =r,0< µ ν <2r, ν =r,0< µ ν <2r, ν =r,0< µ ν <2r, ν =r = 4 c µ 2 = 4 c µ 2 c ν 2 2 c µ 2 c ν 2 c µ 2 c ν 2 + c µ 2 c µ 2 c µ 2 c ν c µ 2 c µ 2 3

4 Fourth, if ρ Z 2, ρ > 2r then γ ρ = 0 Putting the above together, we have ρ Z 2 γ ρ = 5 Hence c 4 4 5, and therefore S = fxcxdx fxcx dx f 4/3 c 4 f 4/3 5 T T 1/4, 2 2 proving the claim 2 Tensor products of functions For f 1 : X 1 C and f 2 : X 2 C, we define f 1 f 2 : X 1 X 2 C by f 1 f 2 x 1, x 2 = f 1 x 1 f 2 x 2, x 1, x 2 X 1 X 2 For f 1 L 1 T d1 and f 2 L 1 T d2, it follows from Fubini s theorem that f 1 f 2 L 1 T d1+d2 For ξ 1 Z d1 and ξ 2 Z d2, Fubini s theorem gives us f 1 f 2 ξ 1, ξ 2 = f 1 f 2 x 1, x 2 e 2πiξ1,ξ2 x1,x2 dλ d1+d 2 x 1, x 2 T d 1 +d 2 = f 1 f 2 x 1, x 2 e 2πiξ1,ξ2 x1,x2 dλ d2 x 2 dλ d1 x 1 T d 1 T d 2 = f 1 x 1 e 2πiξ1 x1 f 2 x 2 e 2πiξ2 x2 dλ d2 x 2 dλ d1 x 1 T d 1 = ˆf 1 ξ 1 ˆf 2 ξ 2 = ˆf 1 ˆf 2 ξ 1, ξ 2, showing that the Fourier transform of a tensor product is the tensor product of the Fourier transforms T d 2 3 Approximate identities and Bernstein s inequality for T An approximate identity is a sequence k in L T d such that i sup k 1 <, ii for each, and iii for each 0 < δ < 1 2, lim n T d k xdλ d x = 1, δ x 1 δ k x dλ d x = 0 4

5 Suppose that k is an approximate identity It is a fact that if f CT d then k f f in CT d, if 1 p < and f L p T d then k f f in L p T d, and if µ is a complex Borel measure on T d then k µ weak-* converges to µ 2 The Riesz representation theorem tells us that the Banach space MT d = rcat d of complex Borel measures on T d, with the total variation norm, is the dual space of the Banach space CT d A trigonometric polynomial is a function P : T d C of the form P x = a ξ e 2πiξ x, x T d ξ Z d for which there is some 0 such that a ξ = 0 whenever ξ > We say that P has degree ; thus, if P is a trigonometric polynomial of degree then P is a trigonometric polynomial of degree M for each M For f L 1 T, we define S f CT by S fx = ˆfje 2πijx, x T j We define the Dirichlet kernel D : T C by D x = e 2πijx, x T, which satisfies, for f L 1 T, j D f = S f We define the Fejér kernel F CT by We can write the Fejér kernel as F x = 1 j + 1 j F = D n, n=0 e 2πijx = j Z χ [,] j 1 j e 2πijx, + 1 where χ A is the indicator function of the set A It is straightforward to prove that F is an approximate identity We define the d-dimensional Fejér kernel F,d CT d by F,d = F F }{{ } d 2 Camil Muscalu and Wilhelm Schlag, Classical and Multilinear Harmonic Analysis, volume I, p 10, Proposition 15 5

6 We can write F,d as F,d x = ξ 1 ξ 1 1 ξ d e 2πiξ x, x T d Using the fact that F is an approximate identity on T, one proves that F,d is an approximate identity on T d The following is Bernstein s inequality for T Theorem 2 Bernstein s inequality If P is a trigonometric polynomial of degree, then P 4π P Proof Define Q = e P F 1 e e P F 1 e The Fourier transform of the first term on the right-hand side is, for j Z, e P F 1 ê j = ê P j k F 1 j kê k k Z = ê P j F 1 j = P j F 1 j, and the Fourier transform of the second term is P j F 1 j + Therefore, for j Z, using P = χ [,] P, Qj = P j F 1 j F 1 j + = P j j χ [ +1, 1] j 1 χ [ +1, 1] 1 = P j χ [1,] j 1 + j + χ [,2 1] j 1 j χ [ 2+1, ] j 1 + j + χ [, 1] j 1 j + = P j χ [1,] j 1 + j χ [, 1] j 1 j + = P j j χ [1,]j + j χ [, 1]j = j P j j + On the other hand, P j = 2πij P j, 6

7 so that P = 2πiQ, ie P = 2πie P F 1 e e P F 1 e Then, by Young s inequality, P = 2π e P F 1 e e P F 1 e 2π e P F 1 e + 2π e P F 1 e = 2π e P F 1 + 2π e P F 1 2π e P F π e P F 1 1 = 4π P 7