Inverse of Fourier transform

Submitted by Structure on Sat, 05/15/2010 - 06:57.

Let $\^{f}(\xi)=\int_{R^n}f(x)e^{-i<x,\xi>}dx$ be the Fourier transform of a function f in $\mathcal{S}(R^n)$ . We shall prove that

$f(x)=(2\pi)^{-n}\int_{R^n}\^{f}(\xi)e^{i<x,\xi>}d\xi$

Let g be another function in $\mathcal{S}(R^n)$ . We have

$\int_{R^n}\^{f}(\xi)g(\xi)e^{i<x,\xi>}d\xi=\int_{R^n}\int_{R^n}f(y)e^{-i<y,\xi>}dy\;g(\xi)e^{i<x,\xi>}d\xi=$

$\int_{R^n}f(y)\int_{R^n}g(\xi)e^{-i<y-x,\xi>}d\xi\;dy=<br /> \int_{R^n}f(y)\^{g}(y-x)dy$

From this relation for $g_{\epsilon}(\xi)=g(\epsilon\xi)$

$\int_{R^n}\^{f}(\xi)g_{\epsilon}(\xi)e^{i<x,\xi>}d\xi=\int_{R^n}f(y)\^{g}_{\epsilon}(y-x)dy$

$\int_{R^n}\^{f}(\xi)g(\epsilon\xi)e^{i<x,\xi>}d\xi=\int_{R^n}f(y)\^{g}(\epsilon^{-1}(y-x))\epsilon^{-n}dy=\int_{R^n}f(x+\epsilon z)\^g(z)dz$

Applying Lebesgue dominanted convergence theorem for $\epsilon \to 0$ we get

$g(0)\int_{R^n}\^{f}(\xi)e^{i<x,\xi>}d\xi=f(x)\int_{R^n}\^g(z)dz$

Taking $g(x)=e^{-\frac{||x||^2}{2}}$ we have g(0)=1 and $\^g(z)=(2\pi)^{\frac{n}{2}}e^{-\frac{||z||^2}{2}}$ and as

$\int_{R^n}e^{-\frac{||z||^2}{2}}dz=(2\pi)^{\frac{n}{2}}$

we finally get

$\int_{R^n}\^{f}(\xi)e^{i<x,\xi>}d\xi=(2\pi)^nf(x)$

$f(x)=(2\pi)^{-n}\int_{R^n}\^{f}(\xi)e^{i<x,\xi>}d\xi$