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\section{Exponential sums in $L^p$}

Recall: studying $f(x) = \sum_{\xi \in
\mathcal{R}} e(\xi \cdot x)$. When $\mathcal{R}$
does not have (linear) structure, expect $|f(x)|
\sim |\mathcal{R}|^{\half}$ (``sqrt
cancellation'') in an appropriate sense (in
$L^p$), \emph{more} than when $\mathcal{R}$ is
structured.

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

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

\begin{center}
  \includegraphics[width=0.6\linewidth]{images/e8c277e0ed1c415d.png}
\end{center}
$L^2$ 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:
\[
  N = \int_{[0, 1]} |f|^2
  \stackrel{\text{H\"older's}}{\le} \left(
  \int_{[0, 1]} |f|^{2 \cdot \frac{p}{2}}
  \right)^{\frac{2}{p}} (1)^{\square}
\]
$\implies N^{p / 2} \le \int_{[0, 1]} |f|^p$.

Constant integral lower bound:
\begin{align*}
  \int_{[0, 1]} |f|^p
  &\ge \int_{[0, cN^{-1}]} |f|^p
  \gtrsim N^p N^{-1} \\
  \int_{[0, 1]} |g|^p
  &\ge \int_{[0, cN^{-2}]} |g|^p
  \gtrsim N^p N^{-2}
\end{align*}
$\int |f|^p \lesssim N^{p / 2} + N^{p - 1}$, $\int
|g|^p \lesssim N^{p / 2} + N^{p - 2}$.

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.

\begin{fcthm}
  Assuming:
  - $p > 2$
  - $b_n \in \Cbb$
  Then:
  \[
    \int \left| \sum_n b_n e(nx) \right|^p
    \le N^{\frac{p}{2} - 1} \|b_n\|_2^p
  .\]
\end{fcthm}

\begin{proof}
  Consider: $\sum_{n = 1}^N b_n e(nx)$, $b_n \in
  \Cbb$.
  \begin{align*}
    \int_{[0, 1]} \left| \sum_n b_n e(nx) \right|^p
    \dd x
    &\le \left\| \sum_n b_n e(nx) \right\|_\infty^{p
    - 2}
    \int_{[0, 1]} \left| \sum_n b_n e(nx) \right|^2
    \dd x \\
    &\le (N^{\half} \|b_n\|_2)^{p - 2} \|b_n\|_2^2
    &&\text{CS} \\
    &= N^{\frac{p}{2} - 1} \|b_n\|_2^p \qedhere
  \end{align*}
\end{proof}

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

$\sum_{n = 1}^{N} b_n e(n^2 x)$. Focus on $p = 4$.
\[
  \int_{[0, 1]} \left| \sum_n b_n e(n^2 x)
  \right|^4 \dd x
  = \sum_{n_1} \sum_{n_2} \sum_{n_3} \sum_{n_4}
  b_{n_1} b_{n_2} \ol{b_{n_3}} \ol{b_{n_4}}
  \int_{[0, 1]} e((n_1^2 + n_2^2 - n_3^2 - n_4^2)
  x) \dd x
.\]
The integral vanishes unless $n_1^2 - n_3^2 =
n_4^2 - n_2^2$.

Number Theory lemma: If $m \in \Zbb$, then
\[
  \text{\# divisors of $m$} \lesssim \log m
.\]
Follows from unique prime factorisation.

\begin{warning*}
  The above lemma is false.

  See correction later.
\end{warning*}

For fixed $n_1, n_3$,
\[
  \#\ub{\{(n_2, n_4) : n_1^2 - n_3^2 = (n_4 + n_2)(n_4
  - n_2)\}}_{\eqdef \mathcal{S}_{n_1, n_3}} \lesssim \log N
.\]
Hence
\begin{align*}
  \int_{[0, 1]} \left| \sum_n b_n e(n^2 x)
  \right|^4 \dd x
  &\le \sum_{n_1} \sum_{n_3} |b_{n_1} b_{n_3}|
  \left| \sum_{(n_2, n_4) \in \mathcal{S}_{n_1,
  n_3}} b_{n_2} \ol{b_{n_4}} \right| \\
  &\lesssim (\log N) \|b_n\|_2^4
\end{align*}
We will now use $\lessapprox$ to mean $\lesssim$
up to powers of $\log N$.

$2 < p < 4$: $\frac{1}{p} = \frac{\theta}{2} +
\frac{1 - \theta}{4}$, $\theta \in [0, 1]$, $h =
\sum_n b_n e(n^2 x)$, $\|h\|_p \le \|h\|_2^\theta
\|h\|_4^{1 - \theta} \lessapprox \|b_n\|_2^\theta
\|b_n\|_2^{1 - \theta} \approx \|b_n\|_2$.

$p \ge 4$:
\[
  \int_{[0, 1]} |h|^p
  \lessapprox \|h\|_\infty^{p - 4} \|b_n\|_2^4
  \stackrel{CS}{\le} (N^{\half} \|b_n\|_2)^{p - 4}
  \|b_n\|_2^4
  \approx N^{\frac{p}{2} - 2} \|b_n\|_2^p
.\]

\begin{fcthm}
  Assuming:
  - $2 > p$
  Then:
  \[
    \int \left| \sum_n b_n e(n^2 x) \right|^p \dd
    x
    \lessapprox (1 = N^{\frac{p}{2} - 2})
    \|b_n\|_2^p
  .\]
\end{fcthm}

\noproof

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{align*}
  \int_{[0, 1]^2} |u(x, t)|^6 \dd x \dd t
  &\lessapprox \|b_n\|_2^6
  &
  u(x, t)
  &= \sum_{\substack{|n| \sim N \\ n \in \Zbb}}
  b_n e(nx + n^2 t) \\
  \int_{[0, 1]^3} |u(x, t)|^4 \dd x \dd t
  \lessapprox \|b_n\|_2^4
  &
  u(x, t)
  &= \sum_{\substack{|n| \sim N \\ n \in \Zbb^2}}
  b_n e(n \cdot x + |n|^2 t)
\end{align*}
Sharp, $6, 4 = p_{\text{crit}}$.

Strichartz estimate for periodic Schr\"odinger
equation, observed by Bourgain in 1990s.

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

$\Tbb^1$, $\Tbb^2$ only sharp Strichartz estimates
per Schr\"odinger until 2015!

2015: Bourgain-Demeter proved $(l^2, L^p)$ sharp
decoupling estimate. Gives sharp Strichartz
estimate for Schr\"odinger in $\Tbb^d$ for all
$d$.

Proved earlier:
\[
  N^{-2} \int_{[0, N^2]} \left| \sum_{n \sim N}
  e \left( \frac{n^2}{N^2} x \right) \right|^p \dd
  x
  \le (1 + N^{\frac{p}{2} - 2}) \|b_n\|_2^p
.\]
(where $n \sim N$ means $N \le n \le 2N$).

Conjecture:
\[
  \average_{[0, N^2]} \left| \sum_{n \in N} b_n
  e(a_n x) \right|^p \dd x
  \lesssim (1 + N^{\frac{p}{2} - 2}) \|b_n\|_2^p
\]

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

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