The Fourier transform can be defined on $L^1(\mathbb{R}^n) \cap L^2(\mathbb{R}^n)$, and we can extend this to $X:=L^2(\mathbb{R}^n)$ by a density argument.
Now, by Plancherel we know that $\|\widehat{f}\|_{L^2(\mathbb{R}^n)} = \|f\|_{L^2(\mathbb{R}^n)}$, so the Fourier transform is an isometry on this space.
My question now is, what is a theorem that guarantees that the Fourier transform has a fixed point on $L^2$? I know the Gaussian is a fixed point, but I'm also interested in other integral transforms, but I just take the Fourier transform as an example.
The Banach Fixed Point Theorem does not work here since we don't have a contraction (operator norm $< 1$). Can we apply the Tychonoff fixed point theorem? Then we would need to show that there exists a non-empty compact convex set $C \subset X$ such that the Fourier transform restricted to $C$ is a mapping from $C$ to $C$. Is this possible?
If we have a fixed point, what would be a way to show it is unique? By linearity we obviously have infinitely many fixed points of we have at least two of them.
Best Answer
My Functional Analysis Fu has gotten bit weak lately, but I think the following should work:
The Schauder fixed point theorem says, that a continuous function on a compact convex set in a topological vector space has a fixed point. Because of isometry, the Fourier transform maps the unit ball in $L^2$ to itself. Owing to the Banach Alaoglu theorem, the unit ball in $L^2$ is compact with respect to the weak topology. The Fourier transform is continuous in the weak topology, because if $( f_n, \phi ) \to (f, \phi)$ for all $\phi \in L^2$, then $$ (\hat{f}_n, \phi) = (f_n, \hat{\phi}) \to (f, \hat{\phi}) = (\hat{f}, \phi). $$