Zygmund’s Fourier restriction theorem and Bernstein’s inequality

1 Zygmund’s restriction theorem

Write ${𝕋}^d={ℝ}^d/{\mathbb{Z}}^d$. Write ${\lambda }_d$ for the Haar measure on ${𝕋}^d$ for which ${\lambda }_d({𝕋}^d)=1$. For $\xi \in {\mathbb{Z}}^d$, we define $e_{\xi }:{𝕋}^d\to S^1$ by

$$e_{\xi }(x)=e^{2\pi i\xi \cdot x},x\in {𝕋}^d.$$

For $f\in L^1({𝕋}^d)$, we define its Fourier transform $\widehat{f}:{\mathbb{Z}}^d\to ℂ$ by

$$\widehat{f}(\xi )={\int }_{{𝕋}^d}f\overline{e_{\xi }}𝑑{\lambda }_d={\int }_{{𝕋}^d}f(x)e^{-2\pi i\xi \cdot x}𝑑x,\xi \in {\mathbb{Z}}^d.$$

For $x\in {ℝ}^d$, we write $|x|={|x|}_2=\sqrt{x_1^2+\mathrm{\cdots }+x_d^2}$, ${|x|}_1=|x_1|+\mathrm{\cdots }+|x_d|$, and ${|x|}_{\mathrm{\infty }}=\max \{|x_j|:1\le j\le d\}$.

For , we write

$${\parallel f\parallel }_p={\left({\int }_{{𝕋}^d}{|f(x)|}^p𝑑x\right)}^{1/p}.$$

For $1\le p\le q\le \mathrm{\infty }$, ${\parallel f\parallel }_p\le {\parallel f\parallel }_q$.

Parseval’s identity tells us that for $f\in L^2({𝕋}^d)$,

$${\parallel \widehat{f}\parallel }_{{\mathrm{\ell }}^2}={\left(\sum _{\xi \in {\mathbb{Z}}^d}{|\widehat{f}(\xi )|}^2\right)}^{1/2}={\parallel f\parallel }_2,$$

and the Hausdorff-Young inequality tells us that for $1\le p\le 2$ and $f\in L^p({𝕋}^d)$,

$${\parallel \widehat{f}\parallel }_{{\mathrm{\ell }}^q}={\left(\sum _{\xi \in {\mathbb{Z}}^d}{|\widehat{f}(\xi )|}^q\right)}^{1/q}\le {\parallel f\parallel }_p,$$

where $\frac{1}{p}+\frac{1}{q}=1$; ${\parallel \widehat{f}\parallel }_{{\mathrm{\ell }}^{\mathrm{\infty }}}={sup}_{\xi \in {\mathbb{Z}}^d}|\widehat{f}(\xi )|$.

Zygmund’s theorem is the following.

Theorem 1 (Zygmund’s theorem).

For $f\mathrm{\in }{L}^{\mathrm{4}\mathrm{/}\mathrm{3}}\mathit{}\mathrm{(}{\mathrm{T}}^{\mathrm{2}}\mathrm{)}$ and $r\mathrm{>}\mathrm{0}$,

$${\left(\sum _{|\xi |=r}{|\widehat{f}(\xi )|}^2\right)}^{1/2}\le 5^{1/4}{\parallel f\parallel }_{4/3}.$$ (1)

Proof.

Suppose that

$$S={\left(\sum _{|\xi |=r}{|\widehat{f}(\xi )|}^2\right)}^{1/2}>0.$$

For $\xi \in {\mathbb{Z}}^2$, we define

$$c_{\xi }=\frac{\overline{\widehat{f}(\xi )}}{S}{\chi }_{|\zeta |=r}.$$

Then

$$\sum _{|\xi |=r}{|c_{\xi }|}^2=\sum _{|\xi |=r}\frac{{|\widehat{f}(\xi )|}^2}{{|S|}^2}=1.$$ (2)

We have

$S^2$ $={\displaystyle \sum _{|\xi |=r}}{|\widehat{f}(\xi )|}^2$

$={\displaystyle \sum _{|\xi |=r}}\widehat{f}(\xi )\overline{\widehat{f}(\xi )}$

$=\left({\displaystyle \sum _{|\xi |=r}}\widehat{f}(\xi )c_{\xi }\right)S,$

hence, defining $c:{𝕋}^2\to ℂ$ by

$$c(x)=\sum _{\xi \in {\mathbb{Z}}^d}{c}_{\xi }{e}^{2\pi i\xi \cdot x}=\sum _{|\xi |=r}{c}_{\xi }{e}^{2\pi i\xi \cdot x},x\in {𝕋}^2,$$

we have, applying Parseval’s identity,

$$S=\sum _{|\xi |=r}\widehat{f}(\xi )c_{\xi }={\int }_{{𝕋}^2}f(x)\overline{c(x)}𝑑x.$$

For $p=\frac{4}{3}$, let $\frac{1}{p}+\frac{1}{q}=1$, i.e. $q=4$. Hölder’s inequality tells us

$${\int }_{{𝕋}^2}|f(x)\overline{c(x)}|𝑑x\le {\parallel f\parallel }_{4/3}{\parallel c\parallel }_4.$$

For $\rho \in {\mathbb{Z}}^2$, we define

$${\gamma }_{\rho }=\sum _{\mu -\nu =\rho }{c}_{\mu }\overline{c_{\nu }}.$$

Then define $\mathrm{\Gamma }(x)={|c(x)|}^2$, which satisfies

$$\mathrm{\Gamma }(x)=c(x)\overline{c(x)}=\sum _{\xi \in {\mathbb{Z}}^2}\sum _{\zeta \in {\mathbb{Z}}^2}{c}_{\xi }\overline{c_{\zeta }}{e}^{2\pi i(\xi -\zeta )\cdot x}=\sum _{\rho \in {\mathbb{Z}}^2}{\gamma }_{\rho }{e}^{2\pi i\rho \cdot x}.$$

Parseval’s identity tells us

$${\parallel c\parallel }_4^4={\parallel \mathrm{\Gamma }\parallel }_2^2=\sum _{\rho \in {\mathbb{Z}}^2}{|{\gamma }_{\rho }|}^2.$$

First,

$${\gamma }_0=\sum _{\mu \in {\mathbb{Z}}^2}{c}_{\mu }\overline{c_{\mu }}=\sum _{\mu \in {\mathbb{Z}}^2}{|c_{\mu }|}^2=1.$$

Second, suppose that $\rho \in {\mathbb{Z}}^2,|\rho |=2r$. If $\rho /2\in {\mathbb{Z}}^2$, then ${\gamma }_{\rho }=c_{\rho /2}\overline{c_{-\rho /2}}$, and if $\rho /2\notin {\mathbb{Z}}^2$ then ${\gamma }_{\rho }=0$. It follows that

$$\sum _{|\rho |=2r}{|{\gamma }_{\rho }|}^2=\sum _{|\mu |=r}{|{\gamma }_{2\mu }|}^2=\sum _{|\mu |=r}{|c_{\mu }|}^2{|c_{-\mu }|}^2.$$ (3)

Third, suppose that . Then, for

$$C_{\rho }=\{\mu \in {\mathbb{Z}}^2:|\mu |=r,|\mu -\rho |=|\rho |\},$$

we have $|C_{\rho }|\le 2$. If $|C_{\rho }|=0$ then ${\gamma }_{\rho }=0$. If $|C_{\rho }|=1$ and $C_{\rho }=\{\mu \}$, then ${\gamma }_{\rho }=c_{\mu }\overline{c_{\mu -\rho }}$ and so ${|{\gamma }_{\rho }|}^2={|c_{\mu }|}^2{|c_{\mu -\rho }|}^2$. If $|C_{\rho }|=2$ and $C_{\rho }=\{\mu ,m\}$, then ${\gamma }_{\rho }=c_{\mu }\overline{c_{\mu -\rho }}+c_m\overline{c_{m-\rho }}$ and so

$${|{\gamma }_{\rho }|}^2\le 2{|c_{\mu }|}^2{|c_{\mu -\rho }|}^2+2{|c_m|}^2{|c_{m-\rho }|}^2.$$

It follows that

Using (3) and then (2),

$\le 4{\displaystyle \sum _{|\mu |=r,|\nu |=r}}{|c_{\mu }|}^2{|c_{\nu }|}^2$

$=4{\left({\displaystyle \sum _{|\mu |=r}}{|c_{\mu }|}^2\right)}^2$

$=4.$

Fourth, if $\rho \in {\mathbb{Z}}^2,|\rho |>2r$ then ${\gamma }_{\rho }=0$. Putting the above together, we have

$$\sum _{\rho \in {\mathbb{Z}}^2}{|{\gamma }_{\rho }|}^2\le 1+4=5.$$

Hence ${\parallel c\parallel }_4^4\le 5$, and therefore

$$|S|=\left|{\int }_{{𝕋}^2}f(x)\overline{c(x)}𝑑x\right|\le {\int }_{{𝕋}^2}|f(x)\overline{c(x)}|𝑑x\le {\parallel f\parallel }_{4/3}{\parallel c\parallel }_4\le {\parallel f\parallel }_{4/3}{5}^{1/4},$$

proving the claim. ∎

2 Tensor products of functions

For $f_1:X_1\to ℂ$ and $f_2:X_2\to ℂ$, we define $f_1\otimes f_2:X_1\times X_2\to ℂ$ by

$$f_1\otimes f_2(x_1,x_2)=f_1(x_1)f_2(x_2),(x_1,x_2)\in X_1\times X_2.$$

For $f_1\in L^1({𝕋}^{d_1})$ and $f_2\in L^1({𝕋}^{d_2})$, it follows from Fubini’s theorem that $f_1\otimes f_2\in L^1({𝕋}^{d_1+d_2})$.

For ${\xi }_1\in {\mathbb{Z}}^{d_1}$ and ${\xi }_2\in {\mathbb{Z}}^{d_2}$, Fubini’s theorem gives us

$\widehat{f_1\otimes f_2}({\xi }_1,{\xi }_2)$ $={\displaystyle {\int }_{{𝕋}^{d_1+d_2}}}{f}_1\otimes f_2(x_1,x_2)e^{-2\pi i({\xi }_1,{\xi }_2)\cdot (x_1,x_2)}𝑑{\lambda }_{d_1+d_2}(x_1,x_2)$

$={\displaystyle {\int }_{{𝕋}^{d_1}}}\left({\displaystyle {\int }_{{𝕋}^{d_2}}}{f}_1\otimes f_2(x_1,x_2)e^{-2\pi i({\xi }_1,{\xi }_2)\cdot (x_1,x_2)}𝑑{\lambda }_{d_2}(x_2)\right)𝑑{\lambda }_{d_1}(x_1)$

$={\displaystyle {\int }_{{𝕋}^{d_1}}}{f}_1(x_1)e^{-2\pi i{\xi }_1\cdot x_1}\left({\displaystyle {\int }_{{𝕋}^{d_2}}}{f}_2(x_2)e^{-2\pi i{\xi }_2\cdot x_2}𝑑{\lambda }_{d_2}(x_2)\right)𝑑{\lambda }_{d_1}(x_1)$

$=\widehat{f_1}({\xi }_1)\widehat{f_2}({\xi }_2)$

$=\widehat{f_1}\otimes \widehat{f_2}({\xi }_1,{\xi }_2),$

showing that the Fourier transform of a tensor product is the tensor product of the Fourier transforms.

3 Approximate identities and Bernstein’s inequality for 𝕋

An approximate identity is a sequence $k_N$ in $L^{\mathrm{\infty }}({𝕋}^d)$ such that (i) , (ii) for each $N$,

$${\int }_{{𝕋}^d}{k}_N(x)𝑑{\lambda }_d(x)=1,$$

and (iii) for each ,

$$\underset{n\to \mathrm{\infty }}{lim}{\int }_{\delta \le x\le 1-\delta }|k_N(x)|𝑑{\lambda }_d(x)=0.$$

Suppose that $k_N$ is an approximate identity. It is a fact that if $f\in C({𝕋}^d)$ then $k_N*f\to f$ in $C({𝕋}^d)$, if and $f\in L^p({𝕋}^d)$ then $k_N*f\to f$ in $L^p({𝕋}^d)$, and if $\mu$ is a complex Borel measure on ${𝕋}^d$ then $k_N*\mu$ weak-* converges to $\mu$. (The Riesz representation theorem tells us that the Banach space $ℳ({𝕋}^d)=rca({𝕋}^d)$ of complex Borel measures on ${𝕋}^d$, with the total variation norm, is the dual space of the Banach space $C({𝕋}^d)$.)

A trigonometric polynomial is a function $P:{𝕋}^d\to ℂ$ of the form

$$P(x)=\sum _{\xi \in {\mathbb{Z}}^d}{a}_{\xi }{e}^{2\pi i\xi \cdot x},x\in {𝕋}^d$$

for which there is some $N\ge 0$ such that $a_{\xi }=0$ whenever ${|\xi |}_{\mathrm{\infty }}>N$. We say that $P$ has degree $N$; thus, if $P$ is a trigonometric polynomial of degree $N$ then $P$ is a trigonometric polynomial of degree $M$ for each $M\ge N$.

For $f\in L^1(𝕋)$, we define $S_{N}f\in C(𝕋)$ by

$$(S_{N}f)(x)=\sum _{|j|\le N}\widehat{f}(j)e^{2\pi ijx},x\in 𝕋.$$

We define the Dirichlet kernel $D_N:𝕋\to ℂ$ by

$$D_N(x)=\sum _{|j|\le N}{e}^{2\pi ijx},x\in 𝕋,$$

which satisfies, for $f\in L^1(𝕋)$,

$$D_N*f=S_{N}f.$$

We define the Fejér kernel $F_N\in C(𝕋)$ by

$$F_N=\frac{1}{N+1}\sum _{n=0}^N{D}_n,$$

We can write the Fejér kernel as

$$F_N(x)=\sum _{|j|\le N}\left(1-\frac{|j|}{N+1}\right)e^{2\pi ijx}=\sum _{j\in \mathbb{Z}}{\chi }_{[-N,N]}(j)\left(1-\frac{|j|}{N+1}\right)e^{2\pi ijx},$$

where ${\chi }_A$ is the indicator function of the set $A$. It is straightforward to prove that $F_N$ is an approximate identity.

We define the $d$-dimensional Fejér kernel $F_{N,d}\in C({𝕋}^d)$ by

$$F_{N,d}=\underset{d}{\underbrace{F_N\otimes \mathrm{\cdots }\otimes F_N}}.$$

We can write $F_{N,d}$ as

$$F_{N,d}(x)=\sum _{{|\xi |}_{\mathrm{\infty }}\le N}\left(1-\frac{|{\xi }_1|}{N+1}\right)\mathrm{\cdots }\left(1-\frac{|{\xi }_d|}{N+1}\right)e^{2\pi i\xi \cdot x},x\in {𝕋}^d.$$

Using the fact that $F_N$ is an approximate identity on $𝕋$, one proves that $F_{N,d}$ is an approximate identity on ${𝕋}^d$.

The following is Bernstein’s inequality for $\mathrm{T}$.

Theorem 2 (Bernstein’s inequality).

If $P$ is a trigonometric polynomial of degree $N$, then

$${\parallel P^{\prime }\parallel }_{\mathrm{\infty }}\le 4\pi N{\parallel P\parallel }_{\mathrm{\infty }}.$$

Proof.

Define

$$Q=((e_{-N}P)*F_{N-1})e_N-((e_{N}P)*F_{N-1})e_{-N}.$$

The Fourier transform of the first term on the right-hand side is, for $j\in \mathbb{Z}$,

$(\widehat{e_{-N}P*F_{N-1}})*\widehat{e_N}(j)$ $={\displaystyle \sum _{k\in \mathbb{Z}}}\widehat{e_{-N}P}(j-k)\widehat{F_{N-1}}(j-k)\widehat{e_N}(k)$

$=\widehat{e_{-N}P}(j-N)\widehat{F_{N-1}}(j-N)$

$=\widehat{P}(j)\widehat{F_{N-1}}(j-N),$

and the Fourier transform of the second term is

$$\widehat{P}(j)\widehat{F_{N-1}}(j+N).$$

Therefore, for $j\in \mathbb{Z}$, using $\widehat{P}={\chi }_{[-N,N]}\widehat{P}$,

$\widehat{Q}(j)$ $=\widehat{P}(j)\left(\widehat{F_{N-1}}(j-N)-\widehat{F_{N-1}}(j+N)\right)$

$=\widehat{P}(j)\left({\chi }_{[-N+1,N-1]}(j-N)\left(1-{\displaystyle \frac{|j-N|}{N}}\right)-{\chi }_{[-N+1,N-1]}\left(1-{\displaystyle \frac{|j+N|}{N}}\right)\right)$

$=\widehat{P}(j)({\chi }_{[1,N]}(j)(1+{\displaystyle \frac{j-N}{N}})+{\chi }_{[N,2N-1]}(j)(1-{\displaystyle \frac{j-N}{N}})$

$-{\chi }_{[-2N+1,-N]}(j)(1+{\displaystyle \frac{j+N}{N}})-{\chi }_{[-N,-1]}(j)(1-{\displaystyle \frac{j+N}{N}}))$

$=\widehat{P}(j)\left({\chi }_{[1,N]}(j)\left(1+{\displaystyle \frac{j-N}{N}}\right)-{\chi }_{[-N,-1]}(j)\left(1-{\displaystyle \frac{j+N}{N}}\right)\right)$

$=\widehat{P}(j)\left({\displaystyle \frac{j}{N}}{\chi }_{[1,N]}(j)+{\displaystyle \frac{j}{N}}{\chi }_{[-N,-1]}(j)\right)$

$={\displaystyle \frac{j}{N}}\widehat{P}(j).$

On the other hand,

$$\widehat{P^{\prime }}(j)=2\pi ij\widehat{P}(j),$$

so that

$$P^{\prime }=2\pi iNQ,$$

i.e.

$$P^{\prime }=2\pi iN(((e_{-N}P)*F_{N-1})e_N-((e_{N}P)*F_{N-1})e_{-N}).$$

Then, by Young’s inequality,

${\parallel P^{\prime }\parallel }_{\mathrm{\infty }}$ $=2\pi N{\parallel ((e_{-N}P)*F_{N-1})e_N-((e_{N}P)*F_{N-1})e_{-N}\parallel }_{\mathrm{\infty }}$

$\le 2\pi N{\parallel ((e_{-N}P)*F_{N-1})e_N\parallel }_{\mathrm{\infty }}+2\pi N{\parallel ((e_{N}P)*F_{N-1})e_{-N}\parallel }_{\mathrm{\infty }}$

$=2\pi N{\parallel (e_{-N}P)*F_{N-1}\parallel }_{\mathrm{\infty }}+2\pi N{\parallel (e_{N}P)*F_{N-1}\parallel }_{\mathrm{\infty }}$

$\le 2\pi N{\parallel e_{-N}P\parallel }_{\mathrm{\infty }}{\parallel F_{N-1}\parallel }_1+2\pi N{\parallel e_{N}P\parallel }_{\mathrm{\infty }}{\parallel F_{N-1}\parallel }_1$

$=4\pi N{\parallel P\parallel }_{\mathrm{\infty }}.$

∎