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Suppose we are given a convolution operator $$ \mathcal{K}[f\,](t):=\int K(t-s)f(s)ds $$ acting on $f\in H_1$ where $H_1$ is a vector space with orthonormal basis $\{\phi_n(t)\}_{n=0}^{N-1}$. If needed, $N=\infty$. Suppose $K$ having infinite support.

I wonder for which sets of functions $\{\phi_n(t)\}$ it results $\mathcal{K}[f\,](t)\in H_1$. In other words, for which $H_1$ the operator $\mathcal{K}$ is invariant. In other words, when $\mathcal{K}[f\,](t)$ can be expressed in terms of (i.e. as a linear combination of) $\{\phi_n(t)\}$. Therefore my question is not strictly about eigenfunctions of the convolution operator.

I figure out the particular solution $$ \phi_n(t)=\frac{1}{\sqrt{W}}e^{-2\pi i t n/W} 1_{[-W/2,W/2]}(t), \quad n\in\mathbb{Z},$$ where $1_A(x)$ is the indicator function of the set $A$, $1_{A}(x)=1$ if $x\in A$ and $0$ otherwise.

In this case $\{\phi_n(t)\}$ has compact support. I wonder if there are other solutions having compact support and if there are solutions having unbounded support.

Feel free to add as many hypotheses as you want if they allow to solve the problem.


Update: I made a step forward.

Denote Fourier transforms of $K(t)$ and $f(t)$ with $\hat{K}(\xi)$ and $\hat{f}(\xi)$, respectively, assuming that they exist. I am trying to find a set of functions $\{\hat{\phi}_n(t)\}$ spanning an invariant subspace of, say, $L^2([a,b])$. Suppose that it exists. If there are no constraints on $\hat{K}(\xi)$, then I claim that $$\mathrm{span}\{\hat{\phi}_n(t)\} = L^2(I), $$ where $I = \bigcup_n \mathrm{supp}(\hat{K}(\xi)\hat{\phi}_n(\xi))$. In fact, consider $\hat{\phi}_0(t)$: there exists $\{c_{n\ell}\}$ such that $$ \hat{K}(\xi)\hat{\phi}_n(\xi) = \sum_\ell c_{n\ell} \hat{\phi}_\ell(\xi) $$ where $c_{n\ell}$ are given by the projection of the LHS on the basis supposed orthonormal w.l.o.g. . But the LHS is arbitrary where it is non-null, since $\hat{K}(\xi)$ is arbitrary, therefore the basis has to represent an "arbitrary" function for $\xi\in \mathrm{supp}(\hat{K}(\xi)\hat{\phi}_n(\xi))$. By extending this argument to remaining basis vectors, the claim follows.

If this argument is correct, the problem is closed for arbitrary $\hat{K}$, that is, trying to find a basis that is "not a function" of $\hat{K}$.

The problem remains for a given (fixed) $\hat{K}$.

In fact, following the suggestion of Davide, we can build the following sequence iterating the operator $\mathcal{K}$ acting on a generic subspace $H$: $$ H \to \mathcal{K}H \to \mathcal{K}^2 H \to \cdots \to \mathcal{K}^n H \to \cdots $$ where it is denoted with $\mathcal{K}H$ the range of $\mathcal{K}$ acting on $H$. In the finite dimensional case, where $H\subseteq V$ and $\mathrm{dim}V<+\infty$, the problem of finding a subspace $H$ such that there exists $i\geq 0$ such that $\mathcal{K}^i H = \mathcal{K}^{i+1}H$, is solved by the Fitting's lemma (used in the Fitting decomposition). I don't know if there are extensions to the infinite-dimensional case.

Any feedback or opinion or suggestion is very welcome.

Update 2 I think this is strictly linked with the "invariant subspace problem" that is, in general, an open problem, but with many particular cases already solved. Perhaps the problem with a convolutional operator one of these.

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Have you tried the case $K(t)=e^{-t^2/2}$? –  Davide Giraudo Aug 4 '12 at 10:14
Thanks for your reply! Actually I am looking at $\phi_n(t)$ or $f(t)$ but having fixed $K(t)$. Just think about $K(t)$ as a generic function (of course with some hypotheses of regularity and so on) and try to find those $\{\phi_n(t)\}$ that are solutions irrespective of $K(t)$. –  Gweedo Aug 4 '12 at 10:25
I guess by infinite support you mean unbounded support. $H_1$ has to contain all the iterate of the convolution. –  Davide Giraudo Aug 4 '12 at 11:24
Thank you Davide, your suggestion is very useful. Now the problem can be stated in general as follows: Given a convolution operator $\mathcal{K}\colon V\to V$, find its $\mathcal{K}$-invariant subspaces. Basically I think of $V$ as $L^2(\Omega)$ with $\Omega$ either compact or unbounded. P.S. I really appreciate if you correct my terminology. –  Gweedo Aug 4 '12 at 13:36

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