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\begin{flashcard}[fourier-inversion-defn]
\begin{definition*}
  \glsnoundefn{FI}{Fourier Inversion}{N/A}
  \cloze{For $f \in \Lp[1](\RR^d)$, with $\ft{f} \in
  \Lp[1](\RR^d)$, we say that the} \emph{Fourier
  Inversion} \cloze{holds for $f$ if
  \[ f(x) = \frac{1}{(2\pi)^d} \int_{\RR^d}
  \ft{f}(u) e^{-i\langle u, x\rangle} \dd u \]
  \gls{al_ev} in $\RR^d$.}
\end{definition*}
\end{flashcard}

\begin{remark*}
  \phantom{}
  \begin{enumerate}[(1)]
    \item The $\RHS$ is continuous by
      \nameref{dct}, so for $f$ continuous, the
      equality is everywhere.
    \item $f \mapsto \ft{f}$, $\Lp[1] \to C_0$
      is $1-1$. (for $f, g
      \in \Lp[1]$ with $\ft{f} = \ft{g}$, then
      $f - g \in \Lp[1]$ and $\ft{f - g} =
      \ft{f} - \ft{g} = 0$. So by \gls{FI}, $f -
      g = 0$ \gls{al_ev}).
  \end{enumerate}
\end{remark*}
\vspace{-1em}

A key concept in Fourier analysis is
\emph{convolution}.

\begin{flashcard}[convolution-defn]
\begin{definition*}[Convolution]
  \glssymboldefn{fourier_convolution}{$*$}{$*$}
  \glsnoundefn{fconv}{convolution}{convolutions}
  \cloze{For $f \in \Lp(\RR^d)$, $1 \le p < \infty$, and
  $\nu$ a probability measure,}
  \[ f * \nu(x) = \cloze{\begin{cases}
    \int_{\RR^d} f(x - y) \nu (\dd y) & \text{if
    the integral exists} \\
    0 & \text{otherwise}
  \end{cases} \qquad x \in \RR^d} \]
\end{definition*}
\end{flashcard}
\vspace{-1em}

\begin{align*}
  \int |f \conv \nu(x)|^p \dd x
  &\le \int \left( \int |f(x - y)| \nu(\dd y)
  \right)^p \dd x \\
  &\le \iint |f(x - y)|^p \nu(\dd y) \dd x
  &&\text{(as $p \ge 1$, use
  \nameref{jensens_ineq})} \\
  &= \int \left( \int |f(x - y)^p| \dd x
  \right)\nu(\dd y)
  &&\text{(\nameref{fubini_tonelli})} \\
  &= \int \left( \int |f(x)|^p \dd x \right)
  \nu(\dd y)
  &&\text{($\lambda$ is translation invariant)} \\
  &= \normp\|f\|_p^p \\
  &< \infty
\end{align*}
Hence $f \conv \nu$ is defined \gls{al_ev}, and
$\normp\|f \conv \nu\|_p \le \|f\|_p < \infty$.
When $\nu$ has pdf $g \in \Lp[1]$,
\[ f \conv \nu(x) = \int f(x - y) g(y) \dd y = f
\conv g(x) .\]
For 2 probability measures $\mu, \nu$ on $\RR^d$,
the \gls{fconv} $\mu \conv \nu$ is a new
probability measure defined as
\[ \mu \conv \nu(A) = \iint \indicator{A}(x +
y) \mu(\dd x) \nu(\dd y) = \mu \pmeas \nu(x + y
\in A) = \PP(X + Y \in A) .\]
where $X, Y$ are independent, $X \sim \mu$, $Y
\sim \nu$. In other words, $(X + Y) \sim \mu \conv
\nu$.

If $\mu$ has pdf $f \in \Lp[1]$, then
\begin{align*}
  \mu \conv \nu(A)
  &= \int \left( \int \indicator{A}(x + y) f(x)
  \dd x \right)\nu(\dd y) \\
  &= \int \left( \int \indicator{A}(x) f(x - y)
  \dd x \right)\nu(\dd y)
  &&\text{(translation invariance)} \\
  &= \int \indicator{A}(x) \left( \int \ob{f(x - y)
  \nu(\dd y)}^{f \conv \nu(x)} \right) \dd x
  &&\text{(\nameref{fubini_tonelli})} \\
  &= \int \indicator{A} f \conv \nu(x) \dd x
\end{align*}
So, $\mu \conv \nu$ has the pdf $f \conv \nu$.

Easy to check:
\begin{enumerate}[(1)]
  \item $\ft{f \conv \nu}(u) = \ft{f}(u)
    \ft{\nu} (u)$ for all $f \in \Lp[1]$, $\nu$
    a probability measure.
  \item $\ft{\mu \conv \nu}(u) = \ft{u}(u)
    \cdot \ft{\nu}(u)$ for all $\mu, \nu$
    probability measures. $X, Y$ independent, $X
    \sim \mu$, $Y \sim \nu$, then $X + Y \sim \nu
    \conv \mu$, then
    \[ \ft{\mu \conv \nu} = \EE e^{i\langle u, X
    + Y\rangle} \stackrel{\text{ind}}{=} \EE
    e^{i\langle u, X\rangle} \cdot \EE e^{i\langle
    u, Y \rangle} = \ft{\mu}(u) \ft{\nu}(u) .\]
\end{enumerate}

\subsubsection*{Fourier transform of Gaussians}

If $\phi_z$ is the \gls{char_func} of $Z \sim
\normaldist(0, 1)$, i.e.
\[ \phi_Z(u) = \EE e^{iuZ} = \int
\frac{1}{\sqrt{2\pi}} e^{-\frac{z^2}{2}} e^{iuz}
\dd z \]
then by a previous theorem, $\phi_Z$ is
differentiable and can be differentiated under the
integral sign, i.e.
\begin{align*}
  \frac{\dd}{\dd u} \phi_Z(u)
  &= \frac{1}{\sqrt{2\pi}} \int \frac{\dd}{\dd u}
  (e^{-\frac{z^2}{2}} e^{iuz}) \dd z \\
  &= \frac{1}{\sqrt{2\pi}} \int iz e^{iuz}
  e^{-\frac{z^2}{2}} \dd z \\
  &= \frac{i}{\sqrt{2\pi}} \int e^{iuz}
  (ze^{-\frac{z^2}{2}}) \dd z \\
  &= \frac{i}{\sqrt{2\pi}} \int iu e^{iuz}
  e^{-\frac{z^2}{2}} \dd z
  &&\text{(integration by parts)} \\
  &= -u \int e^{iuz}
  \frac{e^{-\frac{z^2}{2}}}{\sqrt{2\pi}} \dd z \\
  &= -u\phi_Z(u) \label{lec18_l136_eq} \tag{$*$}
\end{align*}
Hence,
\[ \frac{\dd}{\dd u}(e^{\frac{u^2}{2}} \phi_Z(u))
= e^{\frac{u^2}{2}} \phi_Z'(u) + \phi_Z(u) u
e^{\frac{u^2}{2}}
\stackrel{\text{\eqref{lec18_l136_eq}}}{=} 0 .\]
i.e. $e^{\frac{u^2}{2}} \phi_Z(u) = \phi_Z(0) =
1$, so $\phi_Z(u) = e^{-\frac{u^2}{2}}$.

Consider for $t \in (0, \infty)$, the centered
Gaussian random variable on $\RR^d$ which has pdf
$g_t$ (with respect to $\lambda^d$),
\[ g_t(x) = \frac{1}{(2\pi t)^{d/2}}
e^{-\frac{\|x\|^2}{2t}}, \qquad \|x\|^2 = \sum_{i
= 1}^d x_i^2 .\]
So, if $(Z_1, \ldots, Z_d)$ are IID
$\normaldist(0, 1)$, then $\sqrt{t}Z$ has density
$g$. So,
\begin{align*}
  \ft{g_t}(y)
  &= \EE(e^{i\langle u, \sqrt{t} z \rangle}) \\
  &= \EE \left( e^{i\sum_{i = 1}^d u_i \sqrt{t}
  z_i} \right) \\
  &= \EE \left( \prod_{i = 1}^d e^{iu_i \sqrt{t}
  z_i} \right) \\
  &= \prod_{i = 1}^d \EE (e^{iu_i \sqrt{t} z_i})
  &&\text{($z_i$ independent)}
  \\
  &= \prod_{i = 1}^d \phi_Z(\sqrt{t}u_i) \\
  &= \prod_{i = 1}^d e^{-t \frac{u_i^2}{2}} \\
  &= e^{-\frac{t\|u\|^2}{2}}
\end{align*}
Hence,
\[ \ft{g_t}(y) = e^{-t\|u\|^2/2} =
\frac{(2\pi)^{d/2}}{t^{d/2}} = \left(
\frac{t}{2\pi} \right)^{d/2} e^{-t\|u\|^2/2} =
\frac{(2\pi)^{d/2}}{t^{d/2}} g_{1/t}(u) .\]
So,
\[ \ft{\ft{g_t}}(u) =
\frac{(2\pi)^{d/2}}{t^{d/2}} \ft{g_{1/t}}(u)
= (2\pi)^d g_t(u) .\]
Then
\[ g_t(x) = g_t(-x) = (2\pi)^{-d}
\ft{\ft{g_t}}(-x) = (2\pi)^{-d} \int
\ft{g_t}(u) e^{-i\langle u, x\rangle} \dd u .\]
Thus the \gls{FI} holds for $g_t$.

\begin{flashcard}[gaussian-conv-defn]
\begin{definition*}[Gaussian convolution]
  \glssymboldefn{gt}{$g_t$}{$g_t$}
  \glsnoundefn{gconv}{Gaussian
  convolution}{Gaussian convolutions}
  \cloze{
  For $f \in \Lp[1](\RR^d)$, the \emph{Gaussian
  convolution} of $f$ is $f \conv g_t$ where
  \[ g_t(u) = \frac{1}{(2\pi)^{\frac{d}{2}}}
  e^{-\frac{|u|^2}{2t}} .\]}
\end{definition*}
\end{flashcard}
% $f \conv g_t$ where $f \in \Lp[1]$.
% \[ f \conv g_t(x) = \int f(x - y) g_t(y) \dd y
% \qquad \forall x \in \RR^d, f \in \Lp[1], t > 0 .\]
% Hence, $f \conv g_t \in \Lp[1]$ and $\normp\|f
% \conv g_t\|_1 \le \normp\|f\|_1 < \infty$. $f
% \conv g_t$ is continuous since
% \[ f \conv g_t(x) = \int f(y) g_t(x - y) \dd y \]
% (by translation invariance of $\lambda$), hence
% continuout by \nameref{dct}, since $g_t$ is
% continuous.