Contents

1 What is Fourier Restriction Theory?

Main object: $f:{ℝ}^d\to ℂ$, $f(x)={\sum }_{\xi \in ℝ}{b}_{\xi }{e}^{2\pi ix\cdot \xi }$, $b_{\xi }\in ℂ$.

$x\in {ℝ}^d$ is a spatial variable, and $\xi \in {ℝ}^d$ is the frequency variable.

The frequencies (or Fourier transform) of $f$ is restricted to a set $R$ (where $R$ we will always be finite – so no need to worry about convergence issues).

Goal: Understand the behaviour of $f$ in terms of properties of $R$.

Guiding principle: if properties of an object avoid (linear) structure, then we expect some random or average behaviour.

The above examples avoid linear structure using some notion fo curvature. See Bourgain $\mathrm{\Lambda }(p)$ paper: $\to$ extra behaviour.

Square root cancellation: If we add $\pm 1$ randomly $N$ times, then we expect a quantity with size $N^{\frac{1}{2}}$.

Theorem 1.1 (Khinchin’s inequality). Assuming that:

• ${\{{𝜀}_n\}}_{n=1}^N$ be IID random variables with $\mathbb{P}({𝜀}_n=1)=\mathbb{P}({𝜀}_n=-1)=\frac{1}{2}$

• $1<p<\infty$

• $x_1,\dots ,x_N\in ℂ$

Then

Proof. Without loss of generality, $x_1,\dots ,x_n\in ℝ$. Without loss of generality, $\parallel x{\parallel }_2=1$.

$p=2$: want to show $𝔼\left({\left|{\sum }_n{𝜀}_n{x}_n\right|}^2\right)\sim 1$.

What about general exponents $p$?

The equality here is the Layer cake formula, which is true for any $p\in (0,\infty )$.

Let $\lambda >0$. Study the random variable $e^{\lambda {\sum }_n{𝜀}_n{x}_n}\in (0,\infty )$.

Fact: $\frac{1}{2}{e}^z+\frac{1}{2}{e}^{-z}\le e^{\frac{z^2}{2}}$ (to check, use the Taylor series). So we can get

$$\begin{array}{llllllll}\hfill \alpha \mathbb{P}(e^{\lambda \sum _n{𝜀}_n{x}_n}>\alpha )& \le 𝔼(e^{\lambda \sum _n{𝜀}_n{x}_n})& \hfill & (\text{Chebyshev’s inequality})& \hfill & & \hfill & \\ \hfill & \le \prod _n{e}^{{\lambda }^2|x_n{|}^2∕2}& \hfill & & \hfill & \\ \hfill & =e^{{\lambda }^2∕2}& \hfill & & \hfill & \end{array}$$

By symmetry,

Choose $\alpha =e^{{\lambda }^2}$:

Use in Layer cake:

Lower bound: use Hölder’s inequality. $X={\sum }_n{𝜀}_n{x}_n$.

$\frac{1}{p}+\frac{1}{q}=1$. □

Can you find a more intuitive proof? E-mail Dominique Maldague.

Useful for exercises!

Return to Fourier restriction context.

Both $f,g$ are $1$-periodic. So study them on $𝕋=[0,1]$. $f(0)=N$, $|f(x)|\sim N$ for $\in \left[0,c\frac{1}{N}\right]$. $g(0)iN$, $|g(x)|\sim N$ for $x\in \left[0,c\frac{1}{N^2}\right]$.

2 Exponential sums in $L^p$

Recall: studying $f(x)={\sum }_{\xi \in R}e(\xi \cdot x)$. When $R$ does not have (linear) structure, expect $|f(x)|\sim |R{|}^{\frac{1}{2}}$ (“sqrt cancellation”) in an appropriate sense (in $L^p$), more than when $R$ is structured.

Linear $R$ $f(x)={\sum }_{n=1}^{N}e(nx)$ vs convex $R$ $g(x)={\sum }_{n=1}^{N}e(n^{2}x)$.

Have

$L^2$ does not distinguish $f,g$. $L^{\infty }$ does not distinguish. However, $2<p<\infty$ does.

What about $1\le p\le 2$? We don’t usually study this range because the estimates tend to be trivial / not interesting.

Focus on $2<p<\infty$. Preduct size of

Square root cancellation lower bound:

Constant integral lower bound:

$\int |f{|}^p\lesssim N^{p∕2}+N^{p-1}$, $\int |g{|}^p\lesssim N^{p∕2}+N^{p-2}$.

Note that for the $f$ bound, $N^{p-1}$ is bigger than $N^{p-1}$ (as long as $p>2$), so the constant integral dominates!

For $g$, if $2\le p\le 4$, the square root cancellation dominates, but for $p>4$ the constant integral takes over.

Assuming:

Theorem 2.1. - $p>2$ - $b_n\in ℂ$ Then:

Note that this is sharp when $b_n\equiv 1$.

${\sum }_{n=1}^N{b}_{n}e(n^{2}x)$. Focus on $p=4$.

Number Theory lemma: If $m\in \mathbb{Z}$, then

For fixed $n_1,n_3$,

We will now use $\precsim$ to mean $\lesssim$ up to powers of $\log N$.

$2<p<4$: $\frac{1}{p}=\frac{𝜃}{2}+\frac{1-𝜃}{4}$, $𝜃\in [0,1]$, $h={\sum }_n{b}_{n}e(n^{2}x)$, $\parallel h{\parallel }_p\le \parallel h{\parallel }_2^{𝜃}\parallel h{\parallel }_4^{1-𝜃}\precsim \parallel b_n{\parallel }_2^{𝜃}\parallel b_n{\parallel }_2^{1-𝜃}\approx \parallel b_n{\parallel }_2$.

$p\ge 4$:

Assuming:

Theorem 2.2. - $2>p$ Then:

Sharp by $b_n\equiv 1$.

Positive take away: estimates are sharp, proofs are elementary. Easy to think of sharp examples.

Number Theory counting idea shows:

Sharp, $6,4=p_{\text{crit}}$.

Strichartz estimate for periodic Schrödinger equation, observed by Bourgain in 1990s.

Negatives: on ${𝕋}^3$ $\to$ $p_{\text{crit}}=\frac{10}{3}$, but this technique can only work on even integer values of $p$.

${𝕋}^1$, ${𝕋}^2$ only sharp Strichartz estimates per Schrödinger until 2015!

2015: Bourgain-Demeter proved $(l^2,L^p)$ sharp decoupling estimate. Gives sharp Strichartz estimate for Schrödinger in ${𝕋}^d$ for all $d$.

Proved earlier:

Conjecture:

$a_n\in [0,1]$, $a_{n+1}-a_n\sim \frac{1}{N}$, $a_{n+2}-a_{n+1}-(a_{n+1}-a_n)\sim \frac{1}{N^2}$.

Example: $\{a_n\}\sim {\left\{\frac{n^3}{N^3}\right\}}_{n=N}^{2N}$.

3 Introduction to Fourier Transform

Correction for lecture 2: Number Theory Lemma.

True statements:

See Terence Tao notes online.

Explaining $\#÷(n){\lesssim }_{𝜀}{n}^{𝜀}$: $\forall 𝜀>0$, $\exists c_{𝜀}\in (0,\infty )$ such that $\#÷(n)\le C_{𝜀}{n}^{𝜀}$ for all $n\ge 1$.

We will be using $\precsim$ to mean “$\le$ but up to sub-polynomial in $n$”.

Question:

Recall: $n\sim N$ means $N\le n\le 2N$.

$\{a_n\}\subseteq [0,10]$, $a_{n+1}-a_n\sim \frac{1}{N}$, $(a_{n+2}-a_{n+1})-(a_{n+1}-a_n)\sim \frac{1}{N^2}$.

Reasonable conjecture?

Yes, reasonable.

Khinchin’s inequality: May select $b_n\in \{\pm 1\}$ so that

Constant integral: $b_n\equiv 1$. $(1,1,1,\dots ,1)$.

Warning. Enemy scenario: $\{a_n\}={\left\{\sqrt{\frac{n}{N}}\right\}}_{n=N}^{2N}$ (technically $\left\{-\sqrt{\frac{3N-n}{N}}\right\}$ if we want to satisfy the conditions mentioned above).

$(b_n)=(1,0,\dots ,0,1,0,\dots ,0,1,\dots ,0)$. Length $N$ vector with $N^{\frac{1}{2}}$ many $1$s. Have:

We can calculate that the above expression is in fact $\ge N^{\frac{p}{2}-\frac{1}{2}}$ (which breaks the conjecture until $p>6$).

It turns out that this is (roughly speaking) the only problem.

Why do we care?

$b_n\equiv 1$, $p=4$. Then

Decoupling doesn’t know how to take advantage of $b_n\equiv 1$.

3.1 Fourier Transform on ${ℝ}^n$

$f:{ℝ}^n\to ℂ$, $f\in S({ℝ}^n)$ Schwartz function: $\parallel x^{\alpha }{\partial }^{\beta }f{\parallel }_{\infty }<\infty$ for all $\alpha ,\beta$.

Facts:

• If $f\in S$, then $\widehat{f}\in S$.

• Plancherel’s Theorem: $\int f(x)g(x)dx=\int \widehat{f}(\xi )ĝ(\xi )d\xi$.

• $f(x)=e^{-\pi x^2}$, $\widehat{f}(\xi )=e^{-\pi {\xi }^2}$

• $\lambda >1$

  

  On the left: mass of $f$ is smashed by a factor of $\lambda$. On the right: the mass of $\widehat{f}$ is stretched by a factor of $\lambda$, “$L^1$ normalized”. $\parallel \widehat{f_{\lambda }}{\parallel }_1$ is independent of $\lambda$.

  There is a general formula for $\widehat{f\circ A}$ where $A$ is an affine transformation.

• $\widehat{cf+g}=c\widehat{f}+\widehat{g}$.

• Translations are dual to modulations:

  $$\begin{array}{llll}\hfill f_{\tau }(x)& =f(x-\tau )& \hfill & \\ \hfill \widehat{f_{\tau }}(\xi )& =e^{-2\pi i\xi \cdot \tau }\widehat{f}(\xi )& \hfill & \\ \hfill f^{t}au(x)& =e^{2\pi i\tau \cdot x}f(x)& \hfill & \\ \hfill \widehat{f^{\tau }}(\xi )& =\widehat{f}(\xi -\tau )& \hfill & \end{array}$$

Basic question about $\widehat{f}$: $L^p$ to $L^q$ boundedness?

Plancherel’s:

$p=1$, $q=\infty$:

By interpolation (Marcinkiewicz): $\parallel \widehat{f}{\parallel }_q\le \parallel f{\parallel }_p$ for $1\le p\le 2$, $\frac{1}{p}+\frac{1}{q}=1$ (Hausdorff-Young inequality).

Are there any other $(p,q)$ for which

Attempt 1: Let $\phi \in C_c^{\infty }({ℝ}^n)$ (compactly supported smooth function on ${ℝ}^n$), with $\mathrm{supp}\phi \subseteq B_1(0)$.

Consider $f(x)={\sum }_{k=1}^N{𝜀}_k\phi (x-v_k)$.

Choose $v_k$ so that $\{B_1(v_k)\}$ are not overlapping. Then

$\parallel \widehat{f}{\parallel }_q\sim N^{\frac{1}{2}}\parallel \widehat{\phi }{\parallel }_q$.

4 Introduction to Fourier Restriction

Theorem (Hausdorff-Young inequality). Assuming that:

• $1\le p\le 2$

• $\frac{1}{p}+\frac{1}{q}=1$

Then

Are there any other $(p,q)$ for which

We saw that $1\le p\le 2$ was necessary (translations / modulations, Khinchin’s inequality).

Scaling: Plug in $f_{\lambda }(x)=f(\lambda x)$ which is $L^{\infty }$-normalised ($\parallel f_{\lambda }{\parallel }_{\infty }=\parallel f{\parallel }_{\infty }$). Then $\widehat{f_{\lambda }}(\xi )={\lambda }^{-n}\widehat{f}({\lambda }^{-1}\xi )$ which is $L^1$-normalised ($\parallel \widehat{f_{\lambda }}{\parallel }_1=\parallel \widehat{f}{\parallel }_1$).

Classical questions

What is $C_{p,q}$ the smallest constant such that $\parallel \widehat{f}{\parallel }_q\le C_{p,q}\parallel f{\parallel }_p$?

Which functions $f$ satisfy $\frac{\parallel \widehat{f}{\parallel }_q}{\parallel f{\parallel }_p}=C_{p,q}$?

2014:

$R_R(p\to q)$ means

In the second example, only this trivial statement is true (i.e. $R_R(p\to q)$ is false for all other values for $p,q$).

Let $R=S^{n-1}$ be the unit sphere in ${ℝ}^n$. Consider

Notation.

We may call $ˇ$ the “inverse Fourier transform”.

Let $\phi \in C_c^{\infty }({ℝ}^n)$, $ℝ$-valued, $\ge \frac{1}{2}$ on $B_1(0)$, with support in $B_2(0)$.

May also assume $ˇ\phi$ is $ℝ$-valued, $ˇ\phi \gtrsim 1$ on $B_c(0)$. $ˇ\phi$ bounded, $|ˇ\phi (x)|{\lesssim }_m{(|x{|}^2+1)}^{-m}\forall m$. $|ˇ\phi (x)|$ behaves like ${\chi }_{B_1(0)}$ in $L^p$.

Consider dilates $f_R(x)=ˇ\phi (R^{-1}x)e^{2\pi ix\cdot v_r}$, $R\gg 1$.

Frequency side $|\widehat{f_R}(\xi )|\sim R^n{\chi }_{B_{R^{-1}}(0)}(x)$ ($L^1$-norm).

height $1$, ${\int }_{{ℝ}^n}|f_R(x){|}^{p}dx\sim R^n$.

$R^{n-\frac{n-1}{q}}\lesssim R^{\frac{n}{p}}$, $n-\frac{n-1}{q}\le \frac{n}{p}$.

Consider: $g_R(x)=e^{ix\cdot w_r}ˇ\phi (R^{-1}{x}_1,\dots ,R^{-1}{x}_{n-1},R^{-2}{x}_n)$.

Frequency: $|\widehat{g_r}(\xi )|\approx |cylinder{|}^{-1}{\chi }_{cylinder}(\xi )$. $|cylinder{|}^{-1}|$ $\to$ ${(R-1\cdot R^{-1}{R}^{-2})}^{-1}=R^{n+1}$.

Spatial side: $|g_{R^{\prime }}(x)|\approx {\chi }_{cylinder}(x)$ ($L^{\infty }$-norm).

$\int |g_R(x){|}^p=R^{n-1+2}=R^{n+1}$.

$\to$ $R^{n+1-\frac{n-1}{q}}\lesssim R^{\frac{n+1}{q}}$, $⟹n+1-\frac{n-1}{q}\le \frac{n+1}{p}$. Implies B.

On Monday, we will build examples $h$ such that $\widehat{h}$ sees all of $S^{n-1}$.

$R=S^{n-1}\subseteq R^n$.

Consider the statement

Fix $\phi \in C_c^{\infty }({ℝ}^n)$, $\phi \sim {\chi }_{B_1(0)}$. For computing $L^p$ norms, $|\widehat{\phi }|\sim {\chi }_{B_c(0)}$.

Wave packet function with localised spatial and frequency behaviour.

Last time: $f_R(x)=e^{ix\cdot v_R}ˇ\phi (R^{-1}{x}_1,\dots ,R^{-1}{x}_{n-1},R^{-2}{x}_n)$.

Frequency: $|\widehat{f_R}(\xi )|$

${\int }_{S^{n-1}}|\widehat{f_R}{|}^{q}d\sigma$

Spatial: $|f_R(x)|$

${\int }_{{ℝ}^1}|f_R{|}^p$

Note: sphere near $n$ looks like $({\xi }^1,1-\frac{1}{2}|{\xi }^{\prime }{|}^2)$, $S^{n-1}\cap \mathrm{supp}\widehat{f_R}\sim \text{cap of radius }{R}^{-1}$.

Naive attempt: $g_R(x)=R^nˇ\phi (Rx)$

Frequency: $|\widehat{g_R}(\xi )|$

${\int }_{S^{n-1}}|\widehat{g_R}(\xi ){|}^{q}d\xi \sim 1$

Spatial: $|g_R(x)|$

$\int |g_R{|}^p\sim R^{-n}\cdot R^{np}$.

Deduce: $1\lesssim R^{-\frac{n}{p}+n}$, so

$|\widehat{g_R}|\sim 1$ on $S^{n-1}$ made $\parallel \widehat{g_R}{\parallel }_{L^q(S^{n-1})}$ easy to compute.

Could we improve things?

Could think about $R\sim 1$: then $\parallel \widehat{g_R}{\parallel }_{L^q(S^{n-1})}\sim 1$, $\parallel g_R{\parallel }_p\sim 1$. This is more efficient, but we can’t take a limit. So not so useful.

Build a function $H(x)$ which satisfies $|\widehat{H}(\xi )|\sim 1$ on $S^{n-1}$.

Let $\{𝜃\}$ be a maximal collection of $\sim R^{-1}$-spaced points on $S^{n-1}$ ($\#\{𝜃\}\sim R^{n-1}$).

For each $𝜃$, let $A_{𝜃}^{-1}:{ℝ}^n\to {ℝ}^n$ be an affine map which sends

$\widehat{H}(\xi )={\sum }_{𝜃}{\phi }_{𝜃}(\xi )\sim 1$ on $S^{n-1}$ (actually on $R^{-2}$-neighbourhood of $S^{n-1}$).

${\int }_{S^{n-1}}|\widehat{H}(\xi ){|}^{q}d\sigma \sim 1$.

$H(x)={\sum }_{𝜃}ˇ{\phi }_{𝜃}(x)$.

$|{\phi }_{𝜃}(x)|$ Frequency:

$|ˇ{\phi }_{𝜃}(x)|$ Spatial:

$\mathrm{ess}\mathrm{supp}H\supseteq {\bigcup }_{𝜃}\mathrm{ess}\mathrm{supp}ˇ{\phi }_{𝜃}$.

“bush of tubes”

$\sim R^{n-1}$ many $R\times \cdots \times R\times R^2$ tubes in $R^{-1}$-separated directions

Compute

(∗)

Consider overlap of $T_{𝜃}$ on $\lambda S^{n-1}$ ($\lambda \in (R,R^2)$).

Average overlap on $\lambda S^{n-1}$:

Not too hard to check that the number of active $T_{𝜃}$ on $\lambda S^{n-1}$ is $\sim {\lambda }^{-(n-1)}{R}^{2(n-1)}$.

Now calculate:

Two cases: either $R^{2n}$ dominates or the other term dominates. So $\boxed{p\le \frac{2n}{n+1}}$.

5 Equivalent Versions of Fourier Restriction

Searching for $(p,q)$ for which

• (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

Conjecture (Restriction Conjecture). $R_{S^{n-1}}(p\to q)$ if and only if $n+1-\frac{n-1}{q}\le \frac{n+1}{p}$ and $p<\frac{2n}{n+1}$.

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

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

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”:

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

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

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.

(∗)

using RPn−1∗(q′→ p′) Goal is to bound this last expression by

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

$s\ge 1$ (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: $x,f(x),Eg(x)$

Frequency $\xi$:

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

Aiming for

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

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

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

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

See Young’s convolution:

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

Example. $g={\chi }_B$ ($B$ is the unit ball in ${ℝ}^n$),

RHS is “average value of $f$ on $B(x)$”.

“$f\ast {\chi }_B$ is approximately constant on balls of radius $\sim 1$”.

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 $x\in B^{\prime }$,

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

Digesting $f\ge 0$.

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

Suppose this:

LHS has to be constant.

Lemma (Locally constant property). Assuming that:

• $f\in S({ℝ}^n)$

• $\mathrm{supp}\widehat{f}\subseteq B_1(0)$

Then for any unit ball $B^1\subseteq {ℝ}^n$ and any $x\in B^{\prime }$,

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

Proof of locally constant property. Let $f\in S({ℝ}^n)$, $\mathrm{supp}\widehat{f}\subseteq B_1(0)$. Let $\phi \in C_c^{\infty }({ℝ}^n)$ such that $\phi \equiv 1$ on $B_1(0)$, $\mathrm{supp}\phi \subseteq B_2(0)$.

By Fourier inversion:

Let $x_0,x\in B^1$ (unit ball in ${ℝ}^n$).

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

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

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

Repeat steps of proof:

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

6 Tube incidence implications of Fourier restriction

Last time: LOCALLY CONSTANT PROPERTY (think of it more as “heuristic”).

If $\mathrm{supp}\widehat{f}\subseteq B_1$ then

• (1) Imagine $|f|\sim \text{const}$ on unit balls.
• (2) $f=ˇ(\widehat{{\phi }_{B_1}})=f\ast ˇ{\phi }_{B_1}$.

What if $\mathrm{supp}\widehat{f}\subseteq B_{\lambda }(0)$ (so $|f|\sim \text{const}$ on ${\lambda }^{-1}$-balls)?

where ${\phi }_{B_{\lambda }}(\xi )={\phi }_{B_1}({\lambda }^{-1}\xi )$.

$\ast |ˇ{\phi }_{B_1}|$ is approximately averaging over a $1$-ball. $\ast |ˇ{\phi }_{B_{\lambda }}|$ is approximately averaging over ${\lambda }^{-1}$-balls.

What about $\mathrm{supp}\widehat{f}\subseteq B_{\lambda }(v)$?

Same thing happens, because $e^{ix\cdot \text{something}}f$ will have Fourier support in $B_{\lambda }(0)$, and taking absolute values means we don’t notice the $e^{ix\cdot \text{something}}$ (modulation).

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

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

Last lecture $\to$

Can choose $\widehat{{\phi }_{B_R}}$ such that

• $|\widehat{{\phi }_{B_R}}|\sim R^n{\chi }_{B_{R^{-1}}(0)}$.

• $\parallel \widehat{{\phi }_{B_R}}{\parallel }_1\sim 1$.

Case 1: $\frac{p^{\prime }}{q^{\prime }}\le 1$. LHS of $(\ast )$ (using Hölder) is

∼1 (Holder) ≲|IR−1|1−p′ q′ (∫ ℝn ∫ ℝn|f^|q′ (ξ − η)|φBR^|(η)dηdξ) p′ q′ ∼|IR−1|1−p′ q′ (∫ ℝn|f^|q′(ξ)dξ) p′ q′

Case 2: $\frac{p^{\prime }}{q^{\prime }}>1$. Use $P^1\subseteq {ℝ}^2$ for intuition.

Imagine a function $g$ which is approximately constant on each $R^{-1}$ cube $Q$. Think of $g$ as $g={\sum }_Q{g}_Q$.

Note

Therefore,

if $|{\xi }_2|\le c{R}^{-1}$. ($\le C$ for all ${\xi }_2\in I_{R^{-1}}$).

$|(P^{\prime }+(0,{\xi }_2))\cap Q|\sim R^{-1}$ for $|{\xi }_2|\le c{R}^{-1}$.

$\sim |I_{R^{-1}}{|}^{1-\frac{p^{\prime }}{q}+\frac{p}{q^{\prime }}}{C}^{\frac{p^{\prime }}{q^{\prime }}}=|I_{R^{-1}}{|}^{1-\frac{p^{\prime }}{q^{\prime }}}{(|I_{R^{-1}}|C)}^{\frac{p^{\prime }}{q^{\prime }}}{\left({\int }_{R^{-1}}{\int }_{|t|<1}g(t,t^2+{\xi }_2)dtd{\xi }_2\right)}^{\frac{p^{\prime }}{q^{\prime }}}$

Important: locally constant property means we didn’t need $\frac{p^{\prime }}{q^{\prime }}<1$, like before.

Make the intuition rigorous.

Consider the integral:

$$\begin{array}{llllll}\hfill {\int }_{|{\xi }^{\prime }|<1}{(|\widehat{f}|\ast |\widehat{{\phi }_{B_R}}|({\xi }^{\prime },|{\xi }^{\prime }{|}^2+{\xi }_n))}^{q^{\prime }}d{\xi }^{\prime }& ={\int }_{|{\xi }^{\prime }|<1}{\left({\int }_{{ℝ}^n}|\widehat{f}|(\eta )|\widehat{{\phi }_{B_R}}|(({\xi }^{\prime },|{\xi }^{\prime }{|}^2+{\xi }_n)-\eta )d\eta \right)}^{q^{\prime }}d{\xi }^{\prime }& \hfill & & \hfill & \\ \hfill & \le {\int }_{|{\xi }^{\prime }|<1}\left({\int }_{{ℝ}^n}|\widehat{f}{|}^{q^{\prime }}(\eta )|\widehat{{\phi }_{B_R}}|(({\xi }^{\prime },|{\xi }^{\prime }{|}^2+{\xi }_n)-\eta )d\eta \right)d{\xi }^{\prime }& \hfill & \text{Same pointwise Holder}& \hfill & & \hfill & \\ \hfill & \sim R^n{(R^{-1})}^{n-1}{\int }_{{ℝ}^n}|\widehat{f}{|}^{q^{\prime }}(\eta )d\eta & \hfill & & \hfill & \\ \hfill & \sim R^1{\int }_{{ℝ}^n}|\widehat{f}{|}^{q^{\prime }}& \hfill & & \hfill & \\ \hfill & \lesssim {(R^{-1})}^1\cdot R^{\frac{p^{\prime }}{q^{\prime }}}{\left(\int |\widehat{f}{|}^{q^{\prime }}\right)}^{\frac{p^{\prime }}{q^{\prime }}}& \hfill & & \hfill & \end{array}$$

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