# Fixed point Fourier transform (and similar transforms)

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 $\|\widehat{f}\|_2 = \|f\|_2$, 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.

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mathoverflow.net/questions/12045/… – anonymous Sep 12 '10 at 13:52
There they directly compute it but I just want to know if we can use one of the fixed point theorems. – Jonas Teuwen Sep 12 '10 at 14:29
I don't think you can get the results you want with fixed point theorems. – John D. Cook Sep 12 '10 at 22:29

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 $$(\hat{f}_n, \phi) = (f_n, \hat{\phi}) \to (f, \hat{\phi}) = (\hat{f}, \phi).$$
True, but if say the continuous function is $f\mapsto -f$ then the only fixed point is zero, which I don't think Jonas is looking for. – Robin Chapman Sep 13 '10 at 6:53