5 Equivalent Versions of Fourier Restriction

Searching for $(p,q)$ for which

$$\parallel \widehat{f}{\parallel }_{L^q(S^{n-1})}\le C_{p,q,n}\parallel f{\parallel }_{L^p({ℝ}^n)}.$$

• (1)

  

  $R={[0,1]}^{n-1}\times \{0\}$

  

• (2)

  

  $\to {\int }_{B_{R^2}}{\left|{\sum }_{𝜃}{\chi }_{T_{𝜃}}\right|}^{\frac{p}{2}}$. “continuum incidence geometry problem”.

  

  integral equals $={\sum }_{B_R\subseteq B_{R^2}}{\int }_{B_R}{\left|{\sum }_{𝜃}{\chi }_{T_{𝜃}}\right|}^{\frac{p}{2}}$.

  “fractal geometry”.

  

Reminder: $R_{S^{n-1}}(p\to q)$ means

$$\parallel \widehat{f}{\parallel }_{L^q(S^{n-1})}\le C_{p,q,n}\parallel f{\parallel }_{L^q({ℝ}^n)}\forall f\in S({ℝ}^n).$$

First proved for $S^1\subseteq {ℝ}^2$ by Fefferman (1970) and Zygmund (1974).

Special things happen in ${ℝ}^2$, classical harmonic analysis techniques apply.

Same conjecture for $R_{P^{n-1}}(p\to q)$, where

$$P^{n-1}=\{(\xi ,|\xi {|}^2)\in {ℝ}^n:|\xi |<1\}.$$

$\parallel \widehat{f}{\parallel }_{L^q(P^{n-1})}^q={\int }_{|\xi |<1}|\widehat{f}(\xi ,|\xi {|}^2){|}^{q}d\xi$.

For $n\ge 3$, open and active!

Restriction theory can be used to deduce continuum incidence geometry estimates.

Surprisingly, we can go the other way too (very recent progress, whereas the above direction has been well-known since at least the $90$s).

Equivalent formulations of $R_{P^{n-1}}(p\to q)$

Dual version is called “Fourier extension”:

$$\parallel \widehat{f}{\parallel }_{L^q(P^{n-1})}=\underset{\begin{array}{c}g\in L^{q^{\prime }}(P^{n-1})\\ \parallel g{\parallel }_{L^{q^{\prime }}(P^{n-1})}=1\end{array}}{\sup }\left|{\int }_{\begin{array}{c}|\xi |<1\\ \xi \in {ℝ}^{n-1}\end{array}}\widehat{f}(\xi ,|\xi {|}^2)g(\xi )d\xi \right|.$$

($\frac{1}{q}+\frac{1}{q^{\prime }}=1$). The integral equals:

$$\begin{array}{llll}\hfill {\int }_{|\xi |<1}{\int }_{{ℝ}^n}{e}^{-2\pi ix\cdot (\xi ,|\xi {|}^2)}f(x)dxg(\xi )d\xi & ={\int }_{{ℝ}^n}f(x)\underset{Eg(x)}{\underbrace{{\int }_{|\xi |<1}{e}^{-2\pi ix\cdot (\xi ,|\xi {|}^2)}g(\xi )d\xi }}dx& \hfill & \\ \hfill \parallel Eg{\parallel }_{L^{p^{\prime }}({ℝ}^n)}& \le C_{p,q,n}\parallel g{\parallel }_{L^{q^{\prime }(P^{n-1})}}\forall S(P^{n-1})& \hfill & \end{array}$$

Call the last inequality $R_{P^{n-1}}^{\ast }(q^{\prime }\to p^{\prime })$.

Local, dual version: allows us to work with functions, F.T.

For any $R\ge 1$, any $B_R\subseteq {ℝ}^n$, we have

$${\left({\int }_{{ℝ}^n}|f(x){|}^{p^{\prime }}\right)}^{\frac{1}{p^{\prime }}}\lesssim R^{\frac{1}{q}}{\left(\int |\widehat{f}(\xi ){|}^{q^{\prime }}d\xi \right)}^{\frac{1}{q^{\prime }}}$$

for all $f\in S({ℝ}^n)$ with $\mathrm{supp}\widehat{f}\subseteq N_{R^{-1}}(P^{n-1})$.

Call this $R_{P^{n-1}}^{\ast ,\text{loc}}(q^{\prime }\to p^{\prime })$.

$R_{P^{n-1}}^{\ast }(q^{\prime }\to p^{\prime })⟹R_{p^{n-1}}^{\ast ,\text{loc}}(q^{\prime }\to p^{\prime })$. First thing bounds $Eg$, while secound thing bounds $f$ when we have $\mathrm{supp}\widehat{f}\subseteq N_{R^{-1}}(P^{n-1})$.

Let $f\in S({ℝ}^n)$, $\mathrm{supp}\widehat{f}\subseteq N_{R^{-1}}(P^{n-1})$. Try to express $f$ in trems of ext. op.

$$\begin{array}{llllll}\hfill f(x)& \stackrel{\text{Fourier inversion}}{=}{\int }_{{ℝ}^n}{e}^{2\pi ix\cdot \xi }\widehat{f}(\xi )d\xi & \hfill & & \hfill & \\ \hfill & ={\int }_{{ℝ}^{n-1}}{\int }_{ℝ}{e}^{2\pi ix\cdot ({\xi }^{\prime },{\xi }_n)}\widehat{f}({\xi }^{\prime },{\xi }_n)d{\xi }_{n}d{\xi }^{\prime }& \hfill & & \hfill & \\ \hfill & ={\int }_{ℝ}{e}^{2\pi x_n{\xi }_n}{\int }_{{ℝ}^{n-1}}{e}^{2\pi ix\cdot ({\xi }^{\prime },|{\xi }^{\prime }{|}^2)}\underset{=:g_{{\xi }_n}({\xi }^{\prime })}{\underbrace{\widehat{f}({\xi }^{\prime },|{\xi }^{\prime }{|}^2+{\xi }_n)}}d{\xi }^{\prime }d{\xi }_n& \hfill & {\xi }_n\mapsto {\xi }_n+|{\xi }^{\prime }{|}^2& \hfill & & \hfill & \\ \hfill & ={\int }_{[-R^{-1},R^{-1}]}{e}^{2\pi x_n{\xi }_n}\underset{E{g}_{{\xi }_n}(x)}{\underbrace{{\int }_{|{\xi }^{\prime }|<1}{e}^{2\pi ix\cdot ({\xi }^{\prime },|{\xi }^{\prime }{|}^2)}\underset{=:g_{{\xi }_n}({\xi }^{\prime })}{\underbrace{\widehat{f}({\xi }^{\prime },|{\xi }^{\prime }{|}^2+{\xi }_n)}}d{\xi }^{\prime }}}d{\xi }_n& \hfill & & \hfill & \\ \hfill {\int }_{{ℝ}^n}|f(x){|}^{p^{\prime }}dx& ={\int }_{{ℝ}^n}{\left|{\int }_{I_{R^{-}1}}{e}^{2\pi i{x}_n{\xi }_n}E{g}_{{\xi }_n}(x)d{\xi }_n\right|}^{p^{\prime }}dx& \hfill & R^{\ast }\text{ bounds }\int |Eg{|}^{p^{\prime }}& \hfill & & \hfill & \\ \hfill & \stackrel{\text{Hölders’s}}{\le }{\int }_{{ℝ}^n}|I_{R^{-}1}{|}^{p^{\prime }-1}{\int }_{I_{R^{-1}}}|E{g}_{{\xi }_n(x)}{|}^{p^{\prime }}d{\xi }_{n}dx& \hfill & & \hfill & \\ \hfill & \lesssim \underset{(\ast )}{\underbrace{{(R^{-1})}^{p^{\prime }-1}{\int }_{I_{R^{-1}}}{\left({\int }_{|{\xi }^{\prime }|<1}|g_{{\xi }_n}({\xi }^{\prime }){|}^{q^{\prime }}d{\xi }^{\prime }\right)}^{\frac{p^{\prime }}{q^{\prime }}}d{\xi }_n}}& \hfill & \text{using }{R}_{P^{n-1}}^{\ast }(q^{\prime }\to p^{\prime })& \hfill & & \hfill & \end{array}$$ Goal is to bound this last expression by

$$R^{-\frac{p^{\prime }}{q}}\underset{\parallel \widehat{f}{\parallel }_{L^{q^{\prime }}(N_{R^{-1}}(P^{n-1}))}}{\underbrace{{\int }_{I_{R^{-1}}}{\int }_{|{\xi }^{\prime }|<1}|g_{{\xi }_n}({\xi }^{\prime }){|}^{q^{\prime }}d{\xi }^{\prime }d{\xi }_n}}$$

Lucy case: $\frac{p^{\prime }}{q^{\prime }}<1$, i.e. $1<\frac{q^{\prime }}{p^{\prime }}$. Then

$$(\ast )\le R^{-p^{\prime }+1}{\int }_{I_{R^{-1}}}$$

$${\int }_{A}h\stackrel{\text{Hölder’s}}{\le }|A{|}^{1-\frac{1}{s}}{\left({\int }_A{h}^s\right)}^{\frac{1}{s}},$$

$s\ge 1$ (actually $\approx$ since $h$ approximately constant on $A$)

$R_{P^{n-1}}^{\ast }(q^{\prime }\to p^{\prime })⟹R_{P^{n-1}}^{\ast ,\text{loc}}(q^{\prime }\to p^{\prime })$ continued.

Spatial: $x,f(x),Eg(x)$

$$\boxed{{\int }_{{ℝ}^n}|f(x){|}^{p^{\prime }}dx}$$

$${\int }_{{ℝ}^n}|Eg(x){|}^{p^{\prime }}dx$$

Frequency $\xi$:

$g({\xi }^{\prime },|{\xi }^{\prime }{|}^2)=g({\xi }^{\prime })$, $\mathrm{supp}\widehat{f}\subseteq N_{R^{-1}}(P^{n-1})$,

$${\int }_{N_{R^{-1}}(P^{n-1})}|\widehat{f}(\xi ){|}^{q^{\prime }}d\xi$$

$${\int }_{|{\xi }^{\prime }|<1}|g({\xi }^{\prime }){|}^{q^{\prime }}d{\xi }^{\prime }.$$

$${\int }_{B_R}|f(x){|}^{p^{\prime }}dx\le \cdots \stackrel{(\ast )}{\le }{R}^{-(p^{\prime }-1)}{\int }_{I_{R^{-1}}}{\left({\int }_{|{\xi }^{\prime }|<1}|g_{{\xi }_n}({\xi }^{\prime }){|}^{q^{\prime }}d{\xi }^{\prime }\right)}^{\frac{p^{\prime }}{q^{\prime }}}d{\xi }_n.$$

Aiming for

$$(\ast )\le R^{-\frac{p^{\prime }}{q}}{\left({\int }_{I_{R^{-1}}}{\int }_{|{\xi }^{\prime }|<1}|g_{{\xi }_n}({\xi }^{\prime }){|}^{q^{\prime }}d{\xi }^{\prime }\right)}^{\frac{p^{\prime }}{q^{\prime }}}.$$

Lucky case: $\frac{p^{\prime }}{q^{\prime }}\le 1$.

$$(\ast )\stackrel{\text{Hölder’s inequality in }{\xi }_n}{\le }{R}^{-(p^{\prime }-1)}|I_{R^{-1}}{|}^{1-\frac{p^{\prime }}{q^{\prime }}}{\left({\int }_{I_{R^{-1}}}{\int }_{|{\xi }^{\prime }|<1}|g_{{\xi }_n}({\xi }^{\prime }){|}^{q^{\prime }}d{\xi }^{\prime }d{\xi }_n\right)}^{\frac{p^{\prime }}{q^{\prime }}}$$

$R^{-(p^{\prime }-1)}{(R^{-1})}^{1-\frac{p^{\prime }}{q^{\prime }}}=R^{-\frac{p^{\prime }}{q}}$ ($1=\frac{1}{q}+\frac{1}{q^{\prime }}$)

Unlucky case: $\frac{p^{\prime }}{q^{\prime }}>1$. Hölder’s inequality goes in the wrong direction:

$${\int }_{A}h\le |A{|}^{1-\frac{1}{s}}{\left({\int }_A{h}^s\right)}^{\frac{1}{s}}$$

for all $s\ge 1$, $h\ge 0$.

This is equality if $h$ is constant on $A$.

PAUSE THIS.

Useful Harmonic Analysis tool

The locally constant property.

Convolution: Let $f,g\in S({ℝ}^n)$. Define $f\ast g\in S({ℝ}^n)$ by

$$f\ast g(x)={\int }_{{ℝ}^n}f(x-y)g(y)dy={\int }_{{ℝ}^n}f(Y)g(x-y)dy.$$

See Young’s convolution:

$$\parallel f\ast g{\parallel }_{L^r}\le \parallel f{\parallel }_p\parallel g{\parallel }_q$$

when $1+\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$.

Support property: $\mathrm{supp}f=A$, $\mathrm{supp}g=B$, $\mathrm{supp}f\ast g\subseteq A+B=\{a+b:a\in A,b\in B\}$.

Convolution and Fourier Transform: $\widehat{fg}=\widehat{f}\ast \widehat{g}$.

Locally constant property: Let $f\in S({ℝ}^n)$, $\mathrm{supp}\widehat{f}\subseteq B_1(0)$. Then for any unit ball $B^{\prime }\subseteq {ℝ}^n$ and any $x\in B^{\prime }$,

$$\parallel f{\parallel }_{L^{\infty }(B^{\prime })}{\lesssim }_m{\int }_{{ℝ}^n}|f|(x-y)\frac{1}{{(1+|y{|}^2)}^m}dy.$$

$\frac{1}{{(1+|y{|}^2)}^m}$ is $\sim 1$ on $|y|\lesssim 1$.

Digesting $f\ge 0$.

For any unit interval $I^{\prime }\subseteq {ℝ}^n$, $x\in I^{\prime }$,

$$\parallel f{\parallel }_{L^{\infty }(I^{\prime })}\le {\int }_{\left(-\frac{1}{2},\frac{1}{2}\right)}f(x-y)dy.$$

Suppose this:

LHS has to be constant.

$$\parallel f{\parallel }_{L^{\infty }(I^{\prime })}\lesssim {\int }_{\left(-\frac{1}{2},\frac{1}{2}\right)}f(x-y)dy.$$

$$\parallel f{\parallel }_{L^{\infty }(I^{\prime })}\stackrel{(\ast )}{\lesssim }{\int }_{\left(-\frac{1}{2},\frac{1}{2}\right)}f(x-y)dy$$

The proof of this fact is more important than the statement – we will be using the strategy in future.

Returning to $R_{P^{n-1}}^{\ast }(q^{\prime }\to p^{\prime })⟹R_{P^{n-1}}^{\ast ,\text{loc}}(q^{\prime }\to p^{\prime })$

$${\int }_{B_R}|f(x){|}^{p^{\prime }}dx\lesssim {\int }_{B_R}|f{\phi }_{B_R}{|}^{p^{\prime }}dx$$

where ${\phi }_{B_R}\in S({ℝ}^n)$ satisfies ${\phi }_{B_R}\sim 1$ on $B_R$, $\mathrm{supp}\widehat{{\phi }_{B_R}}\subseteq B_{R^{-1}}(0)$.

$$\widehat{f{\phi }_{B_R}}=\underset{R^{-1}\text{ neighbourhood of }{P}^{n-1}}{\underbrace{\widehat{f}}}\ast \underset{R^{-1}\text{ ball}}{\underbrace{\widehat{{\phi }_{B_R}}}}(\xi )$$

supported in $2R^{-1}$-neighbourhood of $P^{n-1}$.

Repeat steps of proof:

$$\begin{array}{llll}\hfill {\int }_{B_R}|f(x){|}^{p^{\prime }}dx& \lesssim R^{-(p^{\prime }-1)}{\int }_{I_{R^{-1}}}{\left({\int }_{|{\xi }^{\prime }|<1}|h_{{\xi }_n}({\xi }^{\prime }){|}^{q^{\prime }}d{\xi }^{\prime }\right)}^{\frac{p^{\prime }}{q^{\prime }}}d{\xi }_n& \hfill & \end{array}$$

$$h({\xi }_n)=\boxed{\widehat{f}\ast \widehat{{\phi }_{B_R}}({\xi }^{\prime },|{\xi }^{\prime }{|}^2+{\xi }_n)}$$