This post’ll be pretty quick—the Plancherel theorem, a basic result on Fourier transforms, is a quick corollary of what I’ve already talked about.

We have shown that the Fourier transform is an isomorphism of ${\mathcal{S}}$ onto itself; the inverse is given by the inverse Fourier transform. The next step is to extend this to an isometry of ${L^2}$ onto itself. Since ${\mathcal{S}}$ is dense in ${L^2}$, it will be sufficient to show that

$\displaystyle ||f||_2 = ||\hat{f}||_2$

for ${f \in \mathcal{S}}$. We will do this by proving the identity

$\displaystyle (\hat{f},g) = (f, \tilde{g})$

for ${f,g \in L^2}$. This is a simple computation:

$\displaystyle (\hat{f},g) = \iint f(y) e^{-2 \pi i x.y} \overline{ g(x)} dy dx = (f, \tilde{g}).$

In other words, the Fourier transform and its inverse are adjoints. If we take ${g = \hat{f}}$ and use the inversion formula, it becomes clear that the Fourier transform preserves the ${L^2}$-norm, whence follows

Theorem 1 (Plancherel) The Fourier transform extends to an isometry of ${L^2}$ onto itself.

Incidentally, for a ${L^2}$ function ${f}$, it is not necessarily true that

$\displaystyle \hat{f}(x) = \int f(y) e^{-2 \pi i x.y} dy$

because that integral need not exist. It is, however, true that the integral will exist almost everywhere in a “principal value” sense, which we do not need to bother with here.

In a sense, this is a continuous analog of the Parseval theorem, which states that the Fourier coefficient map from ${L^2 \rightarrow l^2}$ (for ${l^2}$ the space of double-sided, square-summable sequences) is an isometry.