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% 16/11/2023 10AM

\subsubsection*{Facts}

\begin{enumerate}[(1)]
  \item $\normp\|f \conv \gt\|_1 \le \normp\|f\|_1$.
  \item $f \conv \gt$ is continuous.
  \item $f \conv \gt$ is bounded.
  \item $\ft{f \conv \gt}(w) = \ft{f}(u)
    \ft{\gt}(u) = \ft{f}(u)
    e^{-\frac{t|u|^2}{2}}$.
  \item $\ft{f \conv \gt}$ is bounded
    continuous.
  \item $\normp\|\ft{f \conv \gt}\|_1 \le c_t
    \normp\|\ft{f}\|_\infty \le c_t
    \normp\|f\|_1$.
  \item For $\mu$ a probability measure, and any
    $t > 0$, $\mu \conv \gt$ is a \gls{gconv}.
    Note that
    \[ \mu \conv \gt = \mu \conv (\gt[\frac{t}{2}]
    \conv \gt[\frac{t}{2}]) = \ub{(\mu \conv
    \gt[\frac{t}{2}])}_{\in \Lp[1]} \conv
    \gt[\frac{t}{2}] .\]
\end{enumerate}

\begin{flashcard}[FI-for-gconv-lemma]
\begin{lemma*}
  \gls{FI} holds for \glspl{gconv}.
\end{lemma*}

\begin{proof}
  \cloze{Let $f \in \Lp[1]$, $t > 0$. Then
  \begin{align*}
    (2\pi)^d f \conv \gt(x)
    &= (2\pi)^d \int f(x - y) \gt(y) \dd y \\
    &= \int f9x -y ) \cdot (2\pi)^d \gt(y) \dd y
    \\
    &= \int f(x - y) \int \ft{\gt}(u) e^{-i\langle
    u, y\rangle} \dd u \dd y
    &&\text{(\gls{FI})} \\
    &= \iint f(x - y) \ft{\gt}(u) e^{-i\langle u,
    y\rangle} \dd u \dd y \\
    &= \int \ft{\gt}(u) \left( \int f(x - y)
    e^{-\langle u, y \rangle} \dd y \right) \dd u
    &&\text{(\nameref{fubini_tonelli})} \\
    &= \int \ft{\gt}(u) \left( \int f(y)
    e^{-i\langle u, x - y\rangle} \dd y \right)
    \dd u \\
    &= \int \ft{\gt}(u) e^{-i\langle u, x\rangle}
    \left( \int f(y) e^{i\langle u, y\rangle} \dd
    y\right)\dd u \\
    &= \int \ft{\gt}(u) e^{-i\langle u, x\rangle}
    \ft{f}(u) \dd u \\
    &= \int \ft{\gt}(u) \ft{f}(u) e^{-\langle u,
    x\rangle} \dd u \\
    &= \ft{f \conv \gt}(u) e^{-i\langle u,
    x\rangle} \dd u \qedhere
  \end{align*}
  }
\end{proof}
\end{flashcard}

$f \conv \gt$ as $t \to 0$, $f \conv \gt$
``$\to$'' $f \conv \delta_0 = f$.

\begin{flashcard}[gaussian-converge-to-delta-lemma]
\begin{lemma*}
  Let $f \in \Lp(\RR^d)$, $1 \le p < \infty$. Then
  $\cloze{\normp\|f \conv \gt - f\|_p} \to 0$ as $t \to
  0$.
\end{lemma*}

\begin{proof}
  \cloze{Given $\eps > 0$, there exists $h \in
  C_c(\RR^d)$ (continuous functions with compact
  support), such that $\normp\|f - h\|_p \le
  \frac{\eps}{3}$. Then by linearity of $\conv$,
  \[ \normp\|f \conv \gt - h \conv \gt\|_p
  = \normp\|\ub{(f - h)}_{\in \Lp} \conv \gt\|_p
  \le \normp\|f - h\|_p \le \frac{\eps}{3} .\]
  So by \nameref{minkowski_ineq},
  \begin{align*}
    \normp\|f \conv \gt - f\|_p
    &\le \ub{\normp\|f \conv \gt - h \conv
    \gt\|_p}_{\le \frac{\eps}{3}}
    + \ub{\normp\|f - h\|_p}_{\le \frac{\eps}{3}}
    + \normp\|h \conv \gt - h\|_p \\
    &\le 2 \frac{\eps}{3} + \normp\|h \conv \gt -
    h\|_p
  \end{align*}
  So it is enough to prove that $\normp\|h \conv
  \gt - h\|_p \to 0$. So $h$ is bounded and $h$ is
  supported on $[-M, M]^d$ say, for some $M > 0$.
  Define
  \[ e(y) = \int |h(x - y) - h(x)|^p \dd x .\]
  Then as $y \to 0$, $|h(x - y) - h(x)|^p \to 0$
  as $h$ is continuous. Also,
  \[ |h(x - y) - h(x)|^p \le 2^p
  \normp\|h\|_\infty^p \indicator{|x| \le M + 1} \]
  for $|y| < 1$. Hence, by \nameref{dct}, $e(y)
  \to 0$ as $y \to 0$. Also,
  \begin{align*}
    \normp\|h \conv \gt - h\|_p^p
    &= \int \left| \int h(x - y) \gt(y) \dd y -
    \int h(x) \gt (y) \dd y \right|^p \dd x \\
    &= \int \left| \int (h(x - y) - h(x)) \gt(y)
    \dd y \right|^p \dd x \\
    &\le \iint |h(x - y) - h(x)|^p \gt(y) \dd y
    \dd x
    &&\text{($p \ge 1$, \nameref{jensens_ineq})}
    \\
    &= \int \left( \int |h(x - y) - h(x)|^p \dd x
    \right) \gt(y) \dd y
    &&\text{(\nameref{fubini_tonelli})} \\
    &= \int e(y) \gt(y) \dd y \\
    &= \int e(y) \frac{1}{t^{d / 2}} \gt[1] \left(
    \frac{y}{\sqrt{t}}\right) \dd y \\
    &= \int e(\sqrt{t} y) \gt[1](y) \dd y
    \\
    &\to 0
    &&\text{(\nameref{dct})}
  \end{align*}
  ($e$ is bounded so $e(\sqrt{t} y) \gt[1] \le
  C\gt[1]$).}
\end{proof}
\end{flashcard}

\begin{flashcard}[FI-formula-thm]
\begin{theorem*}[Fourier Inversion]
  Let $f \in \Lp[1](\RR^d)$ and $\ft{f} \in
  \Lp[1](\RR^d)$. Then
  \[ f(x) = \frac{1}{(2\pi)^d} \int e^{-i\langle
  u, x\rangle} \ft{f}(u) \dd u \]
  \gls{al_ev} in $\RR^d$.
\end{theorem*}

\begin{proof}
  \cloze{
  Consider $f \conv \gt$ and
  \[ f_t(x) = \frac{1}{(2\pi)^d} \int e^{-i\langle
  x, u\rangle} \ub{\ft{f}(u)
  \ub{e^{-\frac{t|u|^2}{2}}}_{\ft{\gt}(u)}}_{\ft{f
  \conv \gt}(u)} \dd u . \label{lec19_l144_eq} \tag{$*$} \]
  As \gls{FI} holds for $f \conv \gt$ ($f \conv
  \gt$ is a \gls{gconv}), we have $f
  \conv \gt = f_t$. So, $\normp\|f_t - f\|_1
  \stackrel{t \to 0}{\longrightarrow} 0$, i.e.
  there exists a subsequence $t_n \downarrow 0$
  such that $f_{t_n} \to f$ \gls{al_ev} (so $f_t
  \convP f$). But from \eqref{lec19_l144_eq}, as
  $t \to 0$, $e^{-i\langle x, u\rangle} \ft{f}(u)
  e^{-t \frac{|u|^2}{2}} \to e^{i\langle x,
  u\rangle} \ft{f}(u)$ and bounded by
  $|\ft{f}(u)|$ which is integrable. So by
  \nameref{dct},
  \[ f_t(x) \stackrel{t \to 0}{\longrightarrow}
  \frac{1}{(2\pi)^d} \int e^{-i\langle x,
  u\rangle} \ft{f}(u) \dd u \]
  \gls{al_ev}. Hence
  \[ f = \frac{1}{(2\pi)^d} \int e^{-i\langle x,
  u\rangle} \ft{f}(u) \dd u \]
  \gls{al_ev}.}
\end{proof}
\end{flashcard}

\begin{theorem*}[Plancherel]
  For $f, g \in \Lp[1] \cap \Lp[2](\RR^d)$,
  \[ \normp\|f\|_2 = \frac{1}{(2\pi)^{d/2}}
  \normp\|\ft{f}\|_2 \qquad \text{and} \qquad
  \langle f, g \rangle_2 =
  \frac{1}{(2\pi)^{\frac{d}{2}}} \langle \ft{f},
  \ft{g} \rangle .\]
  $f \mapsto \ft{f}$.
\end{theorem*}

\begin{remark*}
  Since $\Lp[1] \cap \Lp[2]$ is dense in $\Lp[2]$,
  the linear operator
  \begin{align*}
    F_0 : \Lp[1] \cap \Lp[2]
    &\to \Lp[2] \\
    F_0(f)
    &= (2\pi)^{-\frac{d}{2}} \ft{f}
  \end{align*}
  extends to an isometry $F : \Lp[2] \to \Lp[2]$,
  which is an isometry by Fourier inversion
  formula.
\end{remark*}