# Zygmund’s Fourier restriction theorem and Bernstein’s inequality

Jordan Bell
February 13, 2015

## 1 Zygmund’s restriction theorem

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

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

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

 $\hat{f}(\xi)=\int_{\mathbb{T}^{d}}f\overline{e_{\xi}}d\lambda_{d}=\int_{% \mathbb{T}^{d}}f(x)e^{-2\pi i\xi\cdot x}dx,\qquad\xi\in\mathbb{Z}^{d}.$

For $x\in\mathbb{R}^{d}$, we write $|x|=|x|_{2}=\sqrt{x_{1}^{2}+\cdots+x_{d}^{2}}$, $|x|_{1}=|x_{1}|+\cdots+|x_{d}|$, and $|x|_{\infty}=\max\{|x_{j}|:1\leq j\leq d\}$.

For $1\leq p<\infty$, we write

 $\left\|f\right\|_{p}=\left(\int_{\mathbb{T}^{d}}|f(x)|^{p}dx\right)^{1/p}.$

For $1\leq p\leq q\leq\infty$, $\left\|f\right\|_{p}\leq\left\|f\right\|_{q}$.

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

 $\left\|\hat{f}\right\|_{\ell^{2}}=\left(\sum_{\xi\in\mathbb{Z}^{d}}|\hat{f}(% \xi)|^{2}\right)^{1/2}=\left\|f\right\|_{2},$

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

 $\left\|\hat{f}\right\|_{\ell^{q}}=\left(\sum_{\xi\in\mathbb{Z}^{d}}|\hat{f}(% \xi)|^{q}\right)^{1/q}\leq\left\|f\right\|_{p},$

where $\frac{1}{p}+\frac{1}{q}=1$; $\left\|\hat{f}\right\|_{\ell^{\infty}}=\sup_{\xi\in\mathbb{Z}^{d}}|\hat{f}(\xi)|$.

Zygmund’s theorem is the following.11 1 Mark A. Pinsky, Introduction to Fourier Analysis and Wavelets, p. 236, Theorem 4.3.11.

###### Theorem 1 (Zygmund’s theorem).

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

 $\left(\sum_{|\xi|=r}|\hat{f}(\xi)|^{2}\right)^{1/2}\leq 5^{1/4}\left\|f\right% \|_{4/3}.$ (1)
###### Proof.

Suppose that

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

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

 $c_{\xi}=\frac{\overline{\hat{f}(\xi)}}{S}\chi_{|\zeta|=r}.$

Then

 $\sum_{|\xi|=r}|c_{\xi}|^{2}=\sum_{|\xi|=r}\frac{|\hat{f}(\xi)|^{2}}{|S|^{2}}=1.$ (2)

We have

 $\displaystyle S^{2}$ $\displaystyle=\sum_{|\xi|=r}|\hat{f}(\xi)|^{2}$ $\displaystyle=\sum_{|\xi|=r}\hat{f}(\xi)\overline{\hat{f}(\xi)}$ $\displaystyle=\left(\sum_{|\xi|=r}\hat{f}(\xi)c_{\xi}\right)S,$

hence, defining $c:\mathbb{T}^{2}\to\mathbb{C}$ by

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

we have, applying Parseval’s identity,

 $S=\sum_{|\xi|=r}\hat{f}(\xi)c_{\xi}=\int_{\mathbb{T}^{2}}f(x)\overline{c(x)}dx.$

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

 $\int_{\mathbb{T}^{2}}|f(x)\overline{c(x)}|dx\leq\left\|f\right\|_{4/3}\left\|c% \right\|_{4}.$

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

 $\gamma_{\rho}=\sum_{\mu-\nu=\rho}c_{\mu}\overline{c_{\nu}}.$

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

 $\Gamma(x)=c(x)\overline{c(x)}=\sum_{\xi\in\mathbb{Z}^{2}}\sum_{\zeta\in\mathbb% {Z}^{2}}c_{\xi}\overline{c_{\zeta}}e^{2\pi i(\xi-\zeta)\cdot x}=\sum_{\rho\in% \mathbb{Z}^{2}}\gamma_{\rho}e^{2\pi i\rho\cdot x}.$

Parseval’s identity tells us

 $\left\|c\right\|_{4}^{4}=\left\|\Gamma\right\|_{2}^{2}=\sum_{\rho\in\mathbb{Z}% ^{2}}|\gamma_{\rho}|^{2}.$

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 $\rho/2\not\in\mathbb{Z}^{2}$ 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 $\rho\in\mathbb{Z}^{2},0<|\rho|<2r$. Then, for

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

we have $|C_{\rho}|\leq 2$. 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

 $|\gamma_{\rho}|^{2}\leq 2|c_{\mu}|^{2}|c_{\mu-\rho}|^{2}+2|c_{m}|^{2}|c_{m-% \rho}|^{2}.$

It follows that

 $\sum_{0<|\rho|<2r}|\gamma_{\rho}|^{2}\leq 4\sum_{|\mu|=r,|\nu|=r,0<|\mu-\nu|<2% r}|c_{\mu}|^{2}|c_{\nu}|^{2}.$

Using (3) and then (2),

 $\displaystyle\sum_{0<|\rho|\leq 2r}|\gamma_{\rho}|^{2}$ $\displaystyle\leq 4\sum_{|\mu|=r,|\nu|=r,0<|\mu-\nu|<2r}|c_{\mu}|^{2}|c_{\nu}|% ^{2}+\sum_{|\mu|=r}|c_{\mu}|^{2}|c_{-\mu}|^{2}$ $\displaystyle\leq 4\sum_{|\mu|=r,|\nu|=r,0<|\mu-\nu|<2r}|c_{\mu}|^{2}|c_{\nu}|% ^{2}+4\sum_{|\mu|=r}|c_{\mu}|^{2}|c_{-\mu}|^{2}$ $\displaystyle\leq 4\sum_{|\mu|=r,|\nu|=r}|c_{\mu}|^{2}|c_{\nu}|^{2}$ $\displaystyle=4\left(\sum_{|\mu|=r}|c_{\mu}|^{2}\right)^{2}$ $\displaystyle=4.$

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

 $\sum_{\rho\in\mathbb{Z}^{2}}|\gamma_{\rho}|^{2}\leq 1+4=5.$

Hence $\left\|c\right\|_{4}^{4}\leq 5$, and therefore

 $|S|=\left|\int_{\mathbb{T}^{2}}f(x)\overline{c(x)}dx\right|\leq\int_{\mathbb{T% }^{2}}|f(x)\overline{c(x)}|dx\leq\left\|f\right\|_{4/3}\left\|c\right\|_{4}% \leq\left\|f\right\|_{4/3}5^{1/4},$

proving the claim. ∎

## 2 Tensor products of functions

For $f_{1}:X_{1}\to\mathbb{C}$ and $f_{2}:X_{2}\to\mathbb{C}$, we define $f_{1}\otimes f_{2}:X_{1}\times X_{2}\to\mathbb{C}$ by

 $f_{1}\otimes f_{2}(x_{1},x_{2})=f_{1}(x_{1})f_{2}(x_{2}),\qquad(x_{1},x_{2})% \in X_{1}\times X_{2}.$

For $f_{1}\in L^{1}(\mathbb{T}^{d_{1}})$ and $f_{2}\in L^{1}(\mathbb{T}^{d_{2}})$, it follows from Fubini’s theorem that $f_{1}\otimes f_{2}\in L^{1}(\mathbb{T}^{d_{1}+d_{2}})$.

For $\xi_{1}\in\mathbb{Z}^{d_{1}}$ and $\xi_{2}\in\mathbb{Z}^{d_{2}}$, Fubini’s theorem gives us

 $\displaystyle\widehat{f_{1}\otimes f_{2}}(\xi_{1},\xi_{2})$ $\displaystyle=\int_{\mathbb{T}^{d_{1}+d_{2}}}f_{1}\otimes f_{2}(x_{1},x_{2})e^% {-2\pi i(\xi_{1},\xi_{2})\cdot(x_{1},x_{2})}d\lambda_{d_{1}+d_{2}}(x_{1},x_{2})$ $\displaystyle=\int_{\mathbb{T}^{d_{1}}}\left(\int_{\mathbb{T}^{d_{2}}}f_{1}% \otimes f_{2}(x_{1},x_{2})e^{-2\pi i(\xi_{1},\xi_{2})\cdot(x_{1},x_{2})}d% \lambda_{d_{2}}(x_{2})\right)d\lambda_{d_{1}}(x_{1})$ $\displaystyle=\int_{\mathbb{T}^{d_{1}}}f_{1}(x_{1})e^{-2\pi i\xi_{1}\cdot x_{1% }}\left(\int_{\mathbb{T}^{d_{2}}}f_{2}(x_{2})e^{-2\pi i\xi_{2}\cdot x_{2}}d% \lambda_{d_{2}}(x_{2})\right)d\lambda_{d_{1}}(x_{1})$ $\displaystyle=\hat{f_{1}}(\xi_{1})\hat{f_{2}}(\xi_{2})$ $\displaystyle=\hat{f_{1}}\otimes\hat{f_{2}}(\xi_{1},\xi_{2}),$

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 $L^{\infty}(\mathbb{T}^{d})$ such that (i) $\sup_{N}\left\|k_{N}\right\|_{1}<\infty$, (ii) for each $N$,

 $\int_{\mathbb{T}^{d}}k_{N}(x)d\lambda_{d}(x)=1,$

and (iii) for each $0<\delta<\frac{1}{2}$,

 $\lim_{n\to\infty}\int_{\delta\leq x\leq 1-\delta}|k_{N}(x)|d\lambda_{d}(x)=0.$

Suppose that $k_{N}$ is an approximate identity. It is a fact that if $f\in C(\mathbb{T}^{d})$ then $k_{N}*f\to f$ in $C(\mathbb{T}^{d})$, if $1\leq p<\infty$ and $f\in L^{p}(\mathbb{T}^{d})$ then $k_{N}*f\to f$ in $L^{p}(\mathbb{T}^{d})$, and if $\mu$ is a complex Borel measure on $\mathbb{T}^{d}$ then $k_{N}*\mu$ weak-* converges to $\mu$.22 2 Camil Muscalu and Wilhelm Schlag, Classical and Multilinear Harmonic Analysis, volume I, p. 10, Proposition 1.5. (The Riesz representation theorem tells us that the Banach space $\mathcal{M}(\mathbb{T}^{d})=rca(\mathbb{T}^{d})$ of complex Borel measures on $\mathbb{T}^{d}$, with the total variation norm, is the dual space of the Banach space $C(\mathbb{T}^{d})$.)

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

 $P(x)=\sum_{\xi\in\mathbb{Z}^{d}}a_{\xi}e^{2\pi i\xi\cdot x},\qquad x\in\mathbb% {T}^{d}$

for which there is some $N\geq 0$ such that $a_{\xi}=0$ whenever $|\xi|_{\infty}>N$. 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 $M\geq N$.

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

 $(S_{N}f)(x)=\sum_{|j|\leq N}\hat{f}(j)e^{2\pi ijx},\qquad x\in\mathbb{T}.$

We define the Dirichlet kernel $D_{N}:\mathbb{T}\to\mathbb{C}$ by

 $D_{N}(x)=\sum_{|j|\leq N}e^{2\pi ijx},\qquad x\in\mathbb{T},$

which satisfies, for $f\in L^{1}(\mathbb{T})$,

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

We define the Fejér kernel $F_{N}\in C(\mathbb{T})$ by

 $F_{N}=\frac{1}{N+1}\sum_{n=0}^{N}D_{n},$

We can write the Fejér kernel as

 $F_{N}(x)=\sum_{|j|\leq N}\left(1-\frac{|j|}{N+1}\right)e^{2\pi ijx}=\sum_{j\in% \mathbb{Z}}\chi_{[-N,N]}(j)\left(1-\frac{|j|}{N+1}\right)e^{2\pi ijx},$

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 $F_{N,d}\in C(\mathbb{T}^{d})$ by

 $F_{N,d}=\underbrace{F_{N}\otimes\cdots\otimes F_{N}}_{d}.$

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

 $F_{N,d}(x)=\sum_{|\xi|_{\infty}\leq N}\left(1-\frac{|\xi_{1}|}{N+1}\right)% \cdots\left(1-\frac{|\xi_{d}|}{N+1}\right)e^{2\pi i\xi\cdot x},\qquad x\in% \mathbb{T}^{d}.$

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

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

###### Theorem 2 (Bernstein’s inequality).

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

 $\left\|P^{\prime}\right\|_{\infty}\leq 4\pi N\left\|P\right\|_{\infty}.$
###### 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 $j\in\mathbb{Z}$,

 $\displaystyle(\widehat{e_{-N}P*F_{N-1}})*\widehat{e_{N}}(j)$ $\displaystyle=\sum_{k\in\mathbb{Z}}\widehat{e_{-N}P}(j-k)\widehat{F_{N-1}}(j-k% )\widehat{e_{N}}(k)$ $\displaystyle=\widehat{e_{-N}P}(j-N)\widehat{F_{N-1}}(j-N)$ $\displaystyle=\widehat{P}(j)\widehat{F_{N-1}}(j-N),$

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 $\widehat{P}=\chi_{[-N,N]}\widehat{P}$,

 $\displaystyle\widehat{Q}(j)$ $\displaystyle=\widehat{P}(j)\left(\widehat{F_{N-1}}(j-N)-\widehat{F_{N-1}}(j+N% )\right)$ $\displaystyle=\widehat{P}(j)\left(\chi_{[-N+1,N-1]}(j-N)\left(1-\frac{|j-N|}{N% }\right)-\chi_{[-N+1,N-1]}\left(1-\frac{|j+N|}{N}\right)\right)$ $\displaystyle=\widehat{P}(j)\bigg{(}\chi_{[1,N]}(j)\left(1+\frac{j-N}{N}\right% )+\chi_{[N,2N-1]}(j)\left(1-\frac{j-N}{N}\right)$ $\displaystyle-\chi_{[-2N+1,-N]}(j)\left(1+\frac{j+N}{N}\right)-\chi_{[-N,-1]}(% j)\left(1-\frac{j+N}{N}\right)\bigg{)}$ $\displaystyle=\widehat{P}(j)\left(\chi_{[1,N]}(j)\left(1+\frac{j-N}{N}\right)-% \chi_{[-N,-1]}(j)\left(1-\frac{j+N}{N}\right)\right)$ $\displaystyle=\widehat{P}(j)\left(\frac{j}{N}\chi_{[1,N]}(j)+\frac{j}{N}\chi_{% [-N,-1]}(j)\right)$ $\displaystyle=\frac{j}{N}\widehat{P}(j).$

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,

 $\displaystyle\left\|P^{\prime}\right\|_{\infty}$ $\displaystyle=2\pi N\left\|((e_{-N}P)*F_{N-1})e_{N}-((e_{N}P)*F_{N-1})e_{-N}% \right\|_{\infty}$ $\displaystyle\leq 2\pi N\left\|((e_{-N}P)*F_{N-1})e_{N}\right\|_{\infty}+2\pi N% \left\|((e_{N}P)*F_{N-1})e_{-N}\right\|_{\infty}$ $\displaystyle=2\pi N\left\|(e_{-N}P)*F_{N-1}\right\|_{\infty}+2\pi N\left\|(e_% {N}P)*F_{N-1}\right\|_{\infty}$ $\displaystyle\leq 2\pi N\left\|e_{-N}P\right\|_{\infty}\left\|F_{N-1}\right\|_% {1}+2\pi N\left\|e_{N}P\right\|_{\infty}\left\|F_{N-1}\right\|_{1}$ $\displaystyle=4\pi N\left\|P\right\|_{\infty}.$