The following is an old qualifying exam question in Complex Analysis.
Let $G$ be the open strip $\{z \in \Bbb C | 1 < \Im(z) < 2\}$. Prove that there exists a function $f$ that is analytic on $G$ which does not extend analytically beyond any boundary point of $G$
I would need to show that for any $z \in \partial G$ and any $\epsilon > 0$, there is no function $g$ analytic on $B(z;\epsilon)$ such that $g(z) = f(z)$ on $G \cap B(z,\epsilon)$. My first thought is to turn this into a problem on the unit disk by applying the necessary transformation (I believe it's $\phi(z) = \frac{1 - ie^{\pi z}}{1 + ie^{\pi z}}$). But I don't know of any functions on the unit disk which cannot be analytically extended on any boundary point of the disk; at least, I wouldn't know how to prove the existence of one. Any hints?