This question arose when I tried to discuss the spectrum of a special $L^{2}(\mathbb{R})$-operator. Let $g(x) = e^{-x}$. For which $\lambda \in \mathbb{C}$ is there a function $f \in L^{2}(\mathbb{R})$ satisfying
$$g(x)f(x-2) = \lambda f(x)$$
almost everywhere?
If $\lambda=0$ then clearly $f$ must be zero, so let us assume $\lambda\neq 0$. Then given any function $f_0:[0,2)\to \mathbb{C}$, there is a unique function $f:\mathbb{R}\to\mathbb{C}$ extending $f_0$ and satisfying the functional equation. Let us take $f_0$ to be the constant function $1$, and let $f$ be the unique extension satisfying the functional equation.
We can rewrite the functional equation as $f(x+2)=\lambda^{-1}e^{-x-2}f(x)$. In particular, when $x>0$, by induction we find that $\log |f(x)|=\Theta(-x^2)$ (for $x\in[2n,2n+2)$, $\log |f(x)|$ is roughly $-2-4-6-\dots-2n\approx -n^2$ plus terms that are linear in $n$). In particular, this means $f$ is square-integrable as $x\to +\infty$.
On the other hand, for $x<0$, we use the functional equation in the form $f(x-2)=\lambda e^xf(x)$. Again by induction, we find that $\log|f(x)|=\Theta(-x^2)$ (for $x\in [-2n,-2n+2)$ we again get a sum like $-2-4-6-\dots-2n\approx -n^2$ plus linear terms). So $f$ is also square-integrable as $x\to-\infty$.
Thus we conclude that $f\in L^2(\mathbb{R})$. So a nonzero $f\in L^2(\mathbb{R})$ satisfying the functional equation exists for all $\lambda\neq 0$.