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{{page>:defs}}

$$
\newcommand{\bfourier}[1]{\bar{\mathcal F}_{#1}}
\newcommand{\lone}{\mathsf{L}_1}
\newcommand{\bbC}{\mathbb C}
$$

====== The inversion formula for the Fourier transform ======

<WRAP center round box 80%>
__**Proposition**__
 Let $f$ be a function on $\lone(\rset)$. Assume that its Fourier transform $\tilde f$ is also in $\lone(\rset)$. Then, $f(x)=\bfourier{\tilde{f}}(x)$ for every 
 $x\in \rset$ such that $f(x)=\lim_{y \to x}f(y)$.
</WRAP>
$\bproof$
Recall that for any complex number $\lambda \in \bbC$, and for all real number
$\sigma \neq 0$,
\begin{equation} \label{eq:caract}
\int \frac{\rme^{-x^2/(2\sigma^2)}}{\sqrt{2\pi \sigma^2}}
\rme^{\lambda x}  \rmd x=\rme^{\lambda^2 \sigma^2/2}
\end{equation}
Set
$$
A_\sigma(x)=\int \tilde f(\nu) \rme^{2 i \pi \nu x}e^{-\nu^2/(2 \sigma^2)}\rmd \nu
$$
As $\tilde f \in \lone(\rset)$, the dominated convergence theorem shows that $\lim_{\sigma\to \infty}A_\sigma(x)=\bfourier{\tilde f}(x)$. Then,
\begin{align*}
A_\sigma(x)&\stackrel{(a)}{=}\int  f(y) \lr{\int  \rme^{-2 i \pi \nu (y-x)} e^{-\nu^2/(2\sigma^2)}\rmd \nu} \rmd y\\
&\stackrel{(b)}{=}\int f(y) \rme^{-2 \pi^2 (y-x)^2 \sigma^2}\sqrt{2\pi \sigma^2} \rmd
y\\
&\stackrel{(c)}{=}\int f(x+s/\sigma) \rme^{-2 \pi^2 s^2}\sqrt{2\pi} \rmd
s
\end{align*}
where $(a)$ is obtained by replacing $\tilde f$ by its expression in terms of $f$ and by using Fubini, $(b)$ comes from \eqref{eq:caract} applied to $\lambda=-2 i \pi(y-x)$. And $(c)$ follows from the change of variable $y=x+s/\sigma$.
Set $\phi(s)=\rme^{-2 \pi^2 s^2}\sqrt{2\pi}$ and by applying \eqref{eq:caract} to $\sigma^2=1/(4\pi^2)$ et $\lambda=0$, we deduce $\int
\phi(s) \rmd s=1$, (we can shorten the proof if we note that $\phi$ is the density of a normal centered distribution with variance $1/(4 \pi^2)$. Posons $g_x(u)=f(x+u)-f(x)$. Since $x$ is a continuity point of  $f$, we have $\lim_{ u \to 0} g_x(u)=0$. Then, 
\begin{align*}
  |A_\sigma(x)-f(x)|&=\left|\int g_x(s/\sigma) \phi(s) \rmd s\right|  \leq
  \sup_{|s| < \sqrt{\sigma}} |g_x(s/\sigma)|+\int_{|s| \geq
    \sqrt{\sigma}} g_x(s/\sigma) \phi(s) \rmd s\\
  & \leq \sup_{|u| < 1/\sqrt{\sigma}} |g_x(u)|+\int_{|u| \geq1/
    \sqrt{\sigma}}  |g_x(u)| \phi(\sigma u) \sigma\rmd u\\
   & \leq \sup_{|u| < 1/\sqrt{\sigma}} |g_x(u)|+\underbrace{\int |g_x(u)| \rmd
    u}_{<\infty\, (\mbox{car }f \in \lone(\rset))}\times \lr{ \rme^{-2\pi^2
      \sigma} \sqrt{2\pi} \sigma}
\end{align*}
The proof is finished by letting  $\sigma$ goes to infinity.  
$\eproof$
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