eigen functions of finite filters

Question

In the context of DSP (Digital Signal Processing) denote $$H=\{h|h:\mathbb Z \rightarrow \mathbb C\}$$ as the set of filters. one can prove that the functions $$\omega\in\mathbb R:e^{i\omega x}\in H$$ are eigen functions of every filter (that keeps the convolution sum finite): $$\forall \omega \in \mathbb R\space \forall h\in H \space \exists a\in\mathbb C: e^{i\omega x}*h=a\cdot e^{i\omega x}$$ where $*$ is the discrete convolution operator: $$x,y\in H\qquad(x*y)|_k=\sum_{j=-\infty}^{\infty}{x(j)y(k-j)}$$ My question is, are $e^{i\omega x}$ the only eigen functions? I managed to convince myself that the claim holds when testing against every filter in $H$:
Suppose $y$ is such an eigen function, since $e^{i\omega x} \in H$ and convolution is commutative: $$b\cdot y=y*e^{i\omega x}=a\cdot e^{i\omega x}$$ But in the real world a large part of the interesting filters are finite, i.e $$h:\exists k_0\in \mathbb N: |k|>k_0\rightarrow h(k)=0$$ Does the claim hold if the candidate function is restricted to be eigen function of such filters? Since $e^{i\omega x}$ is not a "finite" filter, I cannot use the trick I used for proving the previous claim.

Answer 1

Suppose $x$ is such an eigen function, and take for example $h=(1,1)\in H$
$$(x*h)|_k = x(k)h(k-k)+x(k-1)h(k-(k-1))=x(k)+x(k-1)$$ Suppose that for some $\alpha \in \mathbb C$: $x*h=\alpha\cdot x$, then $(x*h)|_k = \alpha\cdot x(k)$. by substitution $$x(k)+x(k-1)=\alpha\cdot x(k)$$ $$x(k-1)=(\alpha -1)x(k)$$ $k$ is arbitrary: $$\forall k\in \mathbb Z\> \forall j \in \mathbb N:x(k+j)=(\alpha -1)^j x(k)$$ We got 4 options for $x$:
1. if $\alpha -1 = 1$ then $x$ is constant
2. if $\alpha -1 \neq 1$ and $|\alpha -1|=1$ then $x$ is of the form $e^{i(\omega x+\phi)}$
3. if $|\alpha -1|>1$ then $x$ is not periodical and $x(n)\rightarrow\infty$
4. if $|\alpha -1|<1$ then $x$ is not periodicaland $x(n)\rightarrow0$
So the only viable periodical option is 2.