Let $D$ be some interval in $\mathbb{R}$ and fix a subset $A \subset D$.
Let $v$ be a function that is periodic with fundamental period $|D|$.
We want to find pairs $(g,\lambda)$, where $\operatorname{supp} g\subset A$, such that
$$ (v*g)\mathbb{I}_A = \lambda g,$$
where $\mathbb{I}_A$ is the indicator function of $A$ and $*$ denotes convolution. Thus, we are looking for functions that have an eigenfunction property but only restricted to the support $A$.
Now, since a convolution is involved, one might want to apply the Fourier transform. Appyling the Fourier transform $\mathcal{F}$ to the left hand side yields $$\mathcal{F}((v*g) \mathbb{I}_A) = \mathcal{F}(v*g)*\mathcal{F}(\mathbb{I}_A) = (\mathcal{F}(v)\mathcal{F}(g))* \mathcal{F}(\mathbb{I}_A).$$
Applying the transform to the rhs of our original equation and setting both terms equal would read
$$(\mathcal{F}(v)\mathcal{F}(g))* \mathcal{F}(\mathbb{I}_A) = \lambda \mathcal{F}(g).$$
One could also use the fact that $g = g\mathbb{I}_A$ and write
$$(\mathcal{F}(v)\mathcal{F}(g))* \mathcal{F}(\mathbb{I}_A) = \lambda \mathcal{F}(g)*\mathcal{F}(\mathbb{I}_A).$$
Now unfortunately one cannot cancel the convolution on both sides - in any case, this would then yield that $g$ would have to be an eigenfunction of the convolution and thus could not have support different from $D$.
I would appreciate any ideas on how to tackle this problem, be it with or without using Fourier transforms.
Edit
I tried to 'guess' the solution in an simple case, but also this seemed harder than I thought:
Consider $D=[0,2\pi]$, $A = [a,b]$ an interval and $v = \cos(x)$.
Then one might wonder if a solution of the form $g(x) = \mathbb{I}_{[a,b]}(x)\cos(nx)$ is possible (analogously to the unconstrained case)
We have $$ v*g(x) = \int_a^b \cos(x-y) \cos(ny) dy = \\ \frac{1}{2(n+1)}\sin\left((n+1)y -x\right)\bigg \vert_a^b + \frac{1}{2(n-1)}\sin\left((n-1)y +x\right)\bigg \vert_a^b.$$
It does not seem very obvious if there is a choice for $a,b$ such that this evaluates to $\lambda \cos(nx)$ for some $\lambda$.