Fourier Inversion Theorem - Proof of The Inversion Theorem

Proof of The Inversion Theorem

First we will consider Fourier transforms of functions in the Schwartz space; these are smooth functions such that, for any multi-indices and ,

These functions are clearly seen to be absolutely integrable, and the Fourier transform of a Schwarz function is also a Schwartz function. An example is the Gaussian function, which we will actually use in proving the inversion formula. We will use the convention that, and the claim is that for a Schwartz function ,

To do this, we will need a few facts.

1. For and Schwartz functions, Fubini's theorem implies that .
2. If and, then .
3. If and, then
4. Define ; then
5. Set . Then with denoting convolution, is an approximation to the identity:, where the convergence is uniform on bounded sets for bounded and uniformly continuous and the convergence is in the p-norm for .

We can now prove the inversion formula. First, note that by the dominated convergence theorem

Define . Applying the second and then third fact from above, With as before, we can push the Fourier transform onto in the last integral to get

the convolution of ƒ with an approximate identity. Hence by the last fact

This establishes that the Fourier transform is an invertible map of the Schwartz space to itself. In particular, it is an isometry in the norm, and Schwartz functions are dense in . The Fourier transform and its inverse then extend to unitary operators on all of for which, with the identity map.

While the integral defining the Fourier transform or its inverse may not make sense for general functions, one can always integrate over a symmetric rectangle and take the limits as its length tends to infinity. In other words, taking an increasing sequence of relatively compact sets growing to, and taking the limit of, where denotes the indicator function of a set. Since is compactly supported, the integral defining its Fourier transform exists. But clearly in, hence as well.