Spectral theorem for Fourier transform

Question

The Fourier transform restricts to an isometry $F$ of the subspace of real even functions in $L^2(\mathbb{R}^n)$. Also $F$ is self adjoint since $$\int F(u)v = \int u F(v)$$ for real even functions. If this is correct, how does the spectral theorem reads when applied to $F$? Namely, what are the spectrum (I guess $-1$ and $1$) and the projection-valued measure? Can we describe eigenspaces explicitly if they exist?

Answer 1

The complex Fourier transform $\mathscr{F}$ satisfies $\mathscr{F}^4=I$. There are eigenvalues $1,i,-1,-i$, and the corresponding normalized eigenfunctions are the Hermite functions: $$ h_n(x)=\frac{(-1)^n e^{x^2/2}}{(2\pi)^{1/4}2^{n/2-1/4}\sqrt{n!}}\frac{d^n}{dx^{n}}e^{-x^2}. $$ These are polynomial functions times $e^{-x^2/2}$; the polynomials are the Hermite polynomials. These functions satisfy $$ \widehat{h_n}(s)=(-i)^nh_n(s). $$ And $\{ h_n \}$ is an orthonormal basis of $L^2(\mathbb{R})$. The spectral projection associated with $\mathscr{F}$ has 4 atoms, one at $1,i,-1,-i$ and the corresponding projections $E\{1\}$, $E\{i\}$, $E\{-1\}$, $E\{-i\}$ are the orthogonal projections on the the corresponding $h_n$. That is, $$ E\{1\}f = \sum_{k=0}^{\infty}\langle f,h_{4k}\rangle h_{4k}, \mbox{ etc..} $$ You may directly expression $E$ using the Lagrange interpolating polynomials. Knowing that $$ (\mathscr{F}-I)(\mathscr{F}-iI)(\mathscr{F}+I)(\mathscr{F}+iI)=0, $$ it follows that the spectral projection $E\{1\}$ is $$ E\{1\}= \frac{1}{(1-i)(1+1)(1-i)}(\mathscr{F}-iI)(\mathscr{F}+I)(\mathscr{F}+iI) $$ You can see that $(\mathscr{F}-I)E\{1\}=0$. This is $p(\mathscr{F})$ where $p(1)=1,p(i)=p(-1)=p(-i)=0$.

The real even functions can be studied from this, too.