Consider the two disjoint semicircles $A = \{-e^{i \phi} | -\pi/2 \leq \phi \leq \pi / 2\}$ and $B = \{1 + e^{i \phi} | -\pi/2 \leq \phi \leq \pi / 2\}$ in the complex plane:
Let $f$ be defined and analytic on $\mathbb{C} \setminus A \setminus B$.
Can we construct $f$ so that $g(z) = f(z) - f(z + 1)$ fulfills $g(z) = 0$ for $|z| < 1$, but $g$ does not vanish identically? That is, $f$ is "periodic", but only inside a subset of the plane.
The reason I believe this is possible is because $g$ is only defined on $\mathbb{C} \setminus A \setminus B \setminus (A - 1) \setminus (B - 1)$, which is not connected:
Since $g$ is not defined for $|z| = 1$, the identity theorem does not say anything about its behavior for $|z| > 1$.
If we can find a single solution for $f$, we can also generate new ones of the form $a(f(z))$. However, actually constructing such a function has been challenging to me.
My attempt is to take an arbitrary entire function $e$ (possibly $e(z) = 0$) and then write $$f(z) = e(z) + \int_{L} \mathrm{d}t \frac{u(t)}{z-t}$$ where $L = A + B$ and $u$ is an unknown weight function. From $f(z) = f(z + 1)$ we can then derive a singular integral equation for $u$, for which there seem to be existence theorems. However, this seems very circuitous since $u$ itself is not all that interesting to me.
Is there a simpler formulation that lets us construct $f$ directly? Obviously, any literature on similar problems would be appreciated, since I have a whole host of possible follow-up questions.
This question relates to a previous one. After reorganizing my thoughts I hopefully have extracted the actual mathematical content in a more understandable way.