Inverse of Fourier transform

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) $

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

or

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

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

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

so

$$f(x)=(2\pi)^{-n}\int_{R^n}\^{f}(\xi)e^{i<x,\xi>}d\xi$$
Average: 5 (2 votes)

Back to top