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world:inverse_fourier


2017/10/07 23:39 ·


# The inversion formula for the Fourier transform

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 $f(x)=\lim_{y \to x}f(y)$.

$\bproof$ Recall that for any complex number $\lambda \in \bbC$, and for all real number $\sigma \neq 0$, $$\label{eq:caract} \int \frac{\rme^{-x^2/(2\sigma^2)}}{\sqrt{2\pi \sigma^2}} \rme^{\lambda x} \rmd x=\rme^{\lambda^2 \sigma^2/2}$$ 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. $\eproof$