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\section{Introduction to Fourier Transform}

Correction for lecture 2: Number Theory Lemma.

True statements:
\[
  \frac{1}{N} \sum_{1 \le n \le N} \# \div(n)
  \lesssim \log N
,\]
and $\# \div(n) \lesssim_\eps n^\eps$.

See Terence Tao notes online.

Explaining $\# \div(n) \lesssim_\eps n^\eps$:
$\forall \eps > 0$, $\exists c_\eps \in (0,
\infty)$ such that $\#\div(n) \le C_\eps n^\eps$
for all $n \ge 1$.

We will be using $\lessapprox$ to mean ``$\le$ but
up to sub-polynomial in $n$''.

\textbf{Question:}
\[
  \average_{[0, N^2]} \left| \sum_{n \sim N} b_n
  e(a_n x) \right|^p \dd x
  \lessapprox (1 + N^{\frac{p}{2} - 2})
  \|b_n\|_2^2
.\]
For example, $a_n = \frac{n^2}{N^2}$.

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.

\nameref{thm:khinchin}: May select $b_n \in \{\pm
1\}$ so that
\[
  \average_{[0, N^2]} \left| \sum_n b_n e(a_n x)
  \right|^p \dd x
  \sim \average_{[0, N^2]} \left| \sum_n |b_n|^2
  \right|^{\frac{p}{2}} \dd x
.\]

Constant integral: $b_n \equiv 1$.
$(1, 1, 1, \ldots, 1)$.
\[
  \average_{[0, N^2]} \left| \sum_n e(a_n x)
  \right|^p \dd x
  \gtrsim \frac{1}{N^2} \int_{[0, c]} N^p \dd x
  \sim N^{p - 2} = N^{\frac{p}{2} - 2} \|b_n\|_2^p
.\]

\begin{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).

  $(b_n) = (1, 0, \ldots, 0, 1, 0, \ldots, 0, 1,
  \ldots, 0)$. Length $N$ vector with $N^{\half}$
  many $1$s. Have:
  \[
    \average_{[0, N^2]}
    \left| \sum_{N \le m^2 \le 2N} e \left(
    \sqrt{\frac{m^2}{N}} x \right) \right|^p \dd x
    = \average_{[0, N^2]} \left| \sum_{m^2 \sim
    N} e \left( \frac{m}{\sqrt{N}} x \right)
    \right|^p \dd x
  \]
  \begin{center}
    \includegraphics[width=0.6\linewidth]{images/2d279d2f53204ef2.png}
  \end{center}
  We can calculate that the above expression is in
  fact $\ge N^{\frac{p}{2} - \half}$ (which breaks
  the conjecture until $p > 6$).
\end{warning*}

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

Why do we care?

$b_n \equiv 1$, $p = 4$. Then
\[
  \average_{[0, N^2]} \left| \sum_n e(a_n x)
  \right|^4 \dd x \lessapprox N^2 = |\{a_n\}|^2
.\]
\[
  \implies \#\{a_{n_1} + a_{n_2} = a_{n_3} +
  a_{n_4}\} \lesssim |\{a_n\}|^2
.\]
$\to$ Convex sequences have minimal additive
energy.

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

\subsection{Fourier Transform on $\Rbb^n$}

$f : \Rbb^n \to \Cbb$, $f \in
\mathcal{S}(\Rbb^n)$ Schwartz function:
$\|x^\alpha \partial^\beta f\|_\infty < \infty$
for all $\alpha, \beta$.
\[
  \hat{f}(\xi) = \int_{\Rbb^n} e^{-2\pi i x \cdot
  \xi} f(x) \dd x
.\]
$x$ is the spatial variable, and $\xi$ is the
frequency variable.

\textbf{Facts:}
\begin{itemize}
  \item If $f \in \mathcal{S}$, then $\hat{f} \in
    \mathcal{S}$.
  \item Plancherel's Theorem: $\int f(x) \ol{g(x)}
    \dd x = \int \hat{f}(\xi) \ol{\hat{g}(\xi)}
    \dd \xi$.
  \item $f(x) = e^{-\pi x^2}$, $\hat{f}(\xi) =
    e^{-\pi \xi^2}$
  \item $\lambda > 1$
    \begin{center}
      \includegraphics[width=0.6\linewidth]{images/610f9662fddc4068.png}
    \end{center}
    On the left: mass of $f$ is smashed by a
    factor of $\lambda$. On the right: the mass of
    $\hat{f}$ is stretched by a factor of
    $\lambda$, ``$L^1$ normalized''.
    $\|\hat{f_\lambda}\|_1$ is independent of
    $\lambda$.

    There is a general formula for $\widehat{f
    \circ A}$ where $A$ is an affine
    transformation.
  \item $\ft{cf + g} = c \ft{f} +
    \ft{g}$.
  \item Translations are dual to modulations:
    \begin{align*}
      f_\tau(x)
      &= f(x - \tau) \\
      \ft{f_\tau}(\xi)
      &= e^{-2\pi i \xi \cdot \tau} \ft{f}(\xi) \\
      f^tau(x)
      &= e^{2\pi i \tau \cdot x} f(x) \\
      \ft{f^\tau}(\xi)
      &= \ft{f}(\xi - \tau)
    \end{align*}
\end{itemize}

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

Plancherel's:
\[
  \|\ft{f}\|_2^2
  = \int \ft{f} \ol{\ft{f}}
  = \int f \ol{f} = \|f\|_2^2
\]
(isometry on $L^2$).

$p = 1$, $q = \infty$:
\[
  \left| \int e^{-2\pi i x \cdot \xi} f(x) \dd x \right|
  \le \int |f(x)| \dd x = \|f\|_1
\]
(contraction from $L^1$ to $L^\infty$).

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

Are there any other $(p, q)$ for which
\[
  \|\ft{f}\|_q
  \lesssim_{p, q}
  \|f\|_q
?\]

Attempt 1: Let $\varphi \in C_c^\infty(\Rbb^n)$
(compactly supported smooth function on $\Rbb^n$),
with $\supp \varphi \subseteq B_1(0)$.

Consider $f(x) = \sum_{k = 1}^N \eps_k \varphi(x - v_k)$.

Choose $v_k$ so that $\{B_1(v_k)\}$ are not
overlapping. Then
\[
  \|f\|_p^p = \int \left| \sum_k \eps_k \varphi(x - v_k)
  \right|^p
  = \sum_k \int |\varphi(x - v_k)|^p \dd x
  = N \|\varphi\|_p^p
.\]
Also
\[
  \ft{f}(\xi)
  = \sum_{k = 1}^N \eps_k e^{-2 \pi i \xi v_k}
  \ft{\varphi}(\xi)
  = \left( \sum_{k = 1}^N \eps_k e^{-2 \pi i \xi v_k}
  \right)
  \ft{\varphi}(\xi)
.\]
We will use \nameref{thm:khinchin}.

$\|\ft{f}\|_q \sim N^{\half} \|\ft{\varphi}\|_q$.
