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

$${\int }_{[0,1]}|f(x){|}^{2}dx=N={\int }_{[0,1]}|g(x){|}^{2}dx.$$

Have $|f(x)|\sim N$ on $[0,c{N}^{-1}]$, and $|g(x)|\sim N$ on $[0,c{N}^{-2}]$.

$L^2$ 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

$${\int }_{[0,1]}|f{|}^p.$$

Square root cancellation lower bound:

$$N={\int }_{[0,1]}|f{|}^2\stackrel{\text{Hölder’s}}{\le }{\left({\int }_{[0,1]}|f{|}^{2\cdot \frac{p}{2}}\right)}^{\frac{2}{p}}{(1)}^{\square }$$

$⟹N^{p∕2}\le {\int }_{[0,1]}|f{|}^p$.

Constant integral lower bound:

$$\begin{array}{llll}\hfill {\int }_{[0,1]}|f{|}^p& \ge {\int }_{[0,c{N}^{-1}]}|f{|}^p\gtrsim N^p{N}^{-1}& \hfill & \\ \hfill {\int }_{[0,1]}|g{|}^p& \ge {\int }_{[0,c{N}^{-2}]}|g{|}^p\gtrsim N^p{N}^{-2}& \hfill & \end{array}$$

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

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.

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

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

$${\int }_{[0,1]}{\left|\sum _n{b}_{n}e(n^{2}x)\right|}^{4}dx=\sum _{n_1}\sum _{n_2}\sum _{n_3}\sum _{n_4}{b}_{n_1}{b}_{n_2}\overline{b_{n_3}}\overline{b_{n_4}}{\int }_{[0,1]}e((n_1^2+n_2^2-n_3^2-n_4^2)x)dx.$$

The integral vanishes unless $n_1^2-n_3^2=n_4^2-n_2^2$.

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

$$\text{\# divisors of }m\lesssim \log m.$$

Follows from unique prime factorisation.

For fixed $n_1,n_3$,

Hence

$$\begin{array}{llll}\hfill {\int }_{[0,1]}{\left|\sum _n{b}_{n}e(n^{2}x)\right|}^{4}dx& \le \sum _{n_1}\sum _{n_3}|b_{n_1}{b}_{n_3}|\left|\sum _{(n_2,n_4)\in S_{n_1,n_3}}{b}_{n_2}\overline{b_{n_4}}\right|& \hfill & \\ \hfill & \lesssim (\log N)\parallel b_n{\parallel }_2^4& \hfill & \end{array}$$

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

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

$p\ge 4$:

$${\int }_{[0,1]}|h{|}^p\precsim \parallel h{\parallel }_{\infty }^{p-4}\parallel b_n{\parallel }_2^4\stackrel{CS}{\le }{(N^{\frac{1}{2}}\parallel b_n{\parallel }_2)}^{p-4}\parallel b_n{\parallel }_2^4\approx N^{\frac{p}{2}-2}\parallel b_n{\parallel }_2^p.$$

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:

$$\begin{array}{llllllll}\hfill {\int }_{{[0,1]}^2}|u(x,t){|}^{6}dxdt& \precsim \parallel b_n{\parallel }_2^6& \hfill u(x,t)& =\sum _{\begin{array}{c}|n|\sim N\\ n\in \mathbb{Z}\end{array}}{b}_{n}e(nx+n^{2}t)& \hfill & & \hfill & \\ \hfill {\int }_{{[0,1]}^3}|u(x,t){|}^{4}dxdt\precsim \parallel b_n{\parallel }_2^4& u(x,t)& \hfill =\sum _{\begin{array}{c}|n|\sim N\\ n\in {\mathbb{Z}}^2\end{array}}{b}_{n}e(n\cdot x+|n{|}^{2}t)& & \hfill & & \hfill \end{array}$$

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:

$$N^{-2}{\int }_{[0,N^2]}{\left|\sum _{n\sim N}e\left(\frac{n^2}{N^2}x\right)\right|}^{p}dx\le (1+N^{\frac{p}{2}-2})\parallel b_n{\parallel }_2^p.$$

(where $n\sim N$ means $N\le n\le 2N$).

Conjecture:

$$-{\int }_{[0,N^2]}{\left|\sum _{n\in N}{b}_{n}e(a_{n}x)\right|}^{p}dx\lesssim (1+N^{\frac{p}{2}-2})\parallel b_n{\parallel }_2^p$$

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

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