To What Extent is the Fourier Inversion Theorem Due to the Self-Adjointedness of the Laplacian

Question

I've tried looking this up (I looked at various spectral theorems) but couldn't find anything that talks about the connection between Fourier transforms and the eigenfunctions of the Laplacian (we may stick to $1$-D if it doesn't make a difference). Of course this stuff is used all the time in quantum mechanics, and I've seen the proof of Fourier inversion for $L^2$ in analysis, but nothing about spectral theory is mentioned. So can we say that Fourier inversion holds on $L^2$ due to the fact that the Laplacian is self-adjoint (well, I guess it's not everywhere defined, and maybe isn't quite self-adjoint, but maybe if we use a Sobolev space)?

Answer 1

Consider $L^{2}(\mathbb{R})$. The Fourier transform and its inverse implement the Spectral Theorem for the selfadjoint operator $Af = \frac{1}{i}\frac{d}{dx}f$ on the domain $\mathcal{D}(A)$ consisting of absolutely continuous $f \in L^{2}(\mathbb{R})$ for which $f'\in L^{2}(\mathbb{R})$. The spectral measure $E$ is $$ E[a,b]f = \frac{1}{2\pi}\int_{a}^{b}e^{isx}\int_{-\infty}^{\infty}f(t)e^{-ist}\,dt\,ds = (\chi_{[a,b]}f^{\wedge})^{\vee}. $$ For a general Borel subset $S$ of $\mathbb{R}$, the spectral measure is $E(S)f = (\chi_{S}f^{\wedge})^{\vee}$. The one-dimensional Laplacian is the square of $A$: $$ -\frac{d^{2}}{dx^{2}}f = A^{2}f = \int_{-\infty}^{\infty}t^{2}dE(t)f=(t^{2}f^{\wedge})^{\vee}. $$ It is not terribly difficult to use the Spectral Theorem to derive these facts, to derive the Fourier transform $\vee$ and its inverse $\wedge$, and to show that $\wedge$ and $\vee$ are isometric inverses. One can show that $f \in \mathcal{D}(A)$ iff $sf^{\wedge}(s)\in L^{2}(\mathbb{R})$. That is, $\int_{-\infty}^{\infty}s^{2}|f^{\wedge}(s)|^{2}\,ds < \infty$ iff $f \in L^{2}$ is absolutely continuous with $f' \in L^{2}$. I assume that's basically what you had in mind?