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\section{What is Fourier Restriction Theory?}

Main object: $f : \Rbb^d \to \Cbb$, $f(x) =
\sum_{\xi \in \Rbb} b_\xi e^{2\pi i x \cdot \xi}$,
$b_\xi \in \Cbb$.

\begin{notation*}
  We will write $e(x \cdot \xi)i e^{2\pi i x \cdot
  \xi}$.
\end{notation*}

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

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

Goal: Understand the behaviour of $f$ in terms of
properties of $\mathcal{R}$.

\begin{example*}
  \phantom{}
  \begin{enumerate}[(i)]
    \item Schr\"odinger equation:
      \[
        u(x, t) = \sum_{n = 1}^{N} b_n e(nx + n^2 t)
      .\]
      Easy: $(2\pi i \partial_+ - \Delta) u = 0$, with
      initial data $u(x, 0) = \sum_{n - 1}^{N} b_n
      e(nx)$.

      $(x, t) = (x_1, x_2)$. Then since $e(nx + n^2
      t)i e((n, n^2) \cdot (x, t))$, we might consider
      $\mathcal{R} = \{(n, n^2) : n = 1, \ldots, N\}$.
    \item Dirichlet polynomials:
      \[
        D(t) = \sum_{n = N}^{2N} b_n e^{i(\log n)t}
      .\]
      with $b_n \equiv 1$, partial sums of Riemann
      Zeta function. We might consider $\mathcal{R} =
      \left\{\frac{1}{2\pi}\log n\right\}_{n = N}^{2N}$.
  \end{enumerate}
  Both avoid linear structure.

  $\{\log n\}$ is a concave set (getting closer
  and closer together).

  $\{(n, n^2)\}$ lie on a parabola.
\end{example*}

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
$\Lambda(p)$ paper: $\to$ extra behaviour.

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

\begin{fcthm}[Khinchin's inequality]
  \label{thm:khinchin}
  Assuming:
  - $\{\eps_n\}_{n = 1}^N$
  be IID random variables with $\Pbb(\eps_n = 1) =
  \Pbb(\eps_n = -1) = \half$
  - $1 < p < \infty$
  - $x_1, \ldots, x_N \in \Cbb$
  Then:
  \[
    \left( \Ebb \left| \sum_{n = 1}^{N} \eps_n x_n
    \right|^p \right)^{\frac{1}{p}} \sim_p
    \left( \sum_{n = 1}^{N} |x_n|^2 \right)^{\half}
    = \|x_n\|_2
  .\]
\end{fcthm}

\begin{notation*}
  $\sim_p$ means $\sim$ but the constant may
  depend on $p$.
\end{notation*}

\begin{proof}
  Without loss of generality, $x_1, \ldots, x_n
  \in \Rbb$. Without loss of generality, $\|x\|_2
  = 1$.

  $p = 2$: want to show $\Ebb \left( \left| \sum_n
  \eps_n x_n \right|^2 \right) \sim 1$.
  \[
    \Ebb \left( \sum_n \eps_n x_n \ol{\sum_m
    \eps_m x_m} \right)
    = \sum_{n, m} \Ebb(\eps_n \eps_m x_n \ol{x_m})
    = \sum_n |x_n|^2 + \sum_n \sum_{m \neq n} x_n
    \ol{x_m} \ub{\Ebb \eps_n}_{= 0} \Ebb \eps_m
  .\]
  What about general exponents $p$?
  \[
    \Ebb \left( \left| \sum_n \eps_n x_n \right|^p \right)
    = \int_0^\infty \Pbb \left( \left| \sum_n \eps_n
    x_n \right|^p > \alpha\right) \dd \alpha
  .\]
  The equality here is the Layer cake formula,
  which is true for any $p \in (0, \infty)$.

  Let $\lambda > 0$. Study the random variable
  $e^{\lambda \sum_n \eps_n x_n} \in (0, \infty)$.
  \[
    \Ebb \left( e^{\lambda \sum_n \eps_n x_n} \right)
    = \Ebb \left( \prod_n e^{\lambda \eps_n x_n} \right)
    = \prod_n \Ebb e^{\lambda \eps_n x_n}
    = \prod_n \left(\half e^{\lambda x_n} + \half
    e^{-\lambda x_n}\right)
  .\]
  Fact: $\half e^z + \half e^{-z} \le
  e^{\frac{z^2}{2}}$ (to check, use the Taylor
  series).
  So we can get
  \begin{align*}
    \alpha \Pbb (e^{\lambda \sum_n \eps_n x_n} >
    \alpha)
    &\le \Ebb(e^{\lambda \sum_n \eps_n x_n})
    &&(\text{Chebyshev's inequality})\\
    &\le \prod_n e^{\lambda^2 |x_n|^2 / 2} \\
    &= e^{\lambda^2 / 2}
  \end{align*}
  By symmetry,
  \[
    \alpha \Pbb(e^{\left| \lambda \sum_n \eps_n
    x_n \right|} > \alpha) \lesssim e^{\lambda^2 /
    2}
  .\]
  Choose $\alpha = e^{\lambda^2}$:
  \[
    \Pbb\left(\left| \sum_n \eps_n x_n \right| >
    \lambda \right)
    = \Pbb\left(\left| \sum_n \eps_n x_n \right|^p >
    \lambda^p \right)
    \lesssim e^{-\lambda^2 / 2}
  .\]
  Use in Layer cake:
  \[
    \Ebb \left( \left| \sum_n \eps_n x_n \right|^p
    \right) \lesssim \int_0^\infty
    e^{-\alpha^{2 / p} / 2} \dd \alpha
    \sim_p 1
  .\]
  Lower bound: use H\"older's inequality. $X =
  \sum_n \eps_n x_n$.
  \[
    \ub{\Ebb(X \ol{X})}_{= 1} \le \ub{(\Ebb(|X|^p))^{1
    / p}}_{\lesssim_p 1} (\Ebb |X|^q)^{1 / q}
  .\]
  $\frac{1}{p} + \frac{1}{q} = 1$.
\end{proof}

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

\begin{corollary*}
  $\Ebb \left( \int \left| \sum_{n = 1}^N \eps_n
  f_n(x) \right|^p \dd x \right) \sim_p \int
  \left| \sum_{n = 1}^{N} |f_n(x)|^2 \right|^{p /
  2} \dd x$.
\end{corollary*}

Useful for exercises!

Return to Fourier restriction context.
\begin{align*}
  f(x)
  &= \sum_{n = 1}^N e(nx)
  &&
  \mathcal{R}
  &= \{1, \ldots, N\} \\
  g(x)
  &= \sum_{n = 1}^{N} e(n^2 x)
  &&
  \mathcal{R}
  &= \{1^2, 2^2, \ldots, N^2\}
\end{align*}
Both $f, g$ are $1$-periodic. So study them on
$\Tbb = [0, 1]$. $f(0) = N$, $|f(x)| \sim N$ for
$ \in \left[ 0, c \frac{1}{N} \right]$.
$g(0)i N$, $|g(x)| \sim N$ for $x \in \left[ 0, c
\frac{1}{N^2} \right]$.
\[
  \int_{[0, 1]} |f(x)|^2 \dd x
  = \sum_{n, m} \int_{[0, 1]} e((n - mx) \dd x = N)
.\]
\[
  \int_{[0, 1]} |g(x)|^2
  = \sum_{n, m} \int_{[0, 1]} e((n^2 - m^2) x) \dd
  x = N
.\]
\begin{center}
  \includegraphics[width=0.6\linewidth]{images/cab612ba286f4e61.png}
\end{center}
\[
  \int_{[0, 1]} |f(x)|^p \dd x
  \ge N^{p / 2} + N^{p - 1}
\]
\[
  \int_{[0, 1]} |f(x)|^p \dd x \ge N{p / 2} + N^{p
  - 2}
.\]
For the first one, $N^{p - 1}$ (organised
behaviour) dominates as soon as $p > 2$, and for
the second one, $N^{p / 2}$ dominates for $2 \le p
\le 4$ (``square root cancellation behaviour lasts
for longer'').