# 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 $$\|\widehat{f}\|_{L^2(\mathbb{R}^n)} = \|f\|_{L^2(\mathbb{R}^n)}$$, 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.

• 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

## 1 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 $$(\hat{f}_n, \phi) = (f_n, \hat{\phi}) \to (f, \hat{\phi}) = (\hat{f}, \phi).$$

• 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
• @Robin Chapman: Could you elaborate on that remark? What function do you mean? – Jonas Teuwen Sep 13 '10 at 17:24
• @Jonas: Robin is quite rightly pointing out, that my approach just shows that there is a fixed point in the unit ball of L^2. However, since 0 is in the unit ball and it is trivially a fixed point of the Fourier transform, this does not tell us anything new. – Michael Ulm Sep 14 '10 at 5:02
• Ah, right. Okay, then this is not the answer I'm looking for, sorry ;-). Maybe John D. Cook has a point that it might not work with fixed-point theorems? – Jonas Teuwen Sep 14 '10 at 8:52
• Since nobody else answers I will accept this as answer. The used technique can be of some value anyway. – Jonas Teuwen Oct 11 '10 at 18:57