Let $f \in \mathcal{O}(D)$, $\ D$ is a bounded region in $\mathbb{C}$, be such that $$\limsup _{D \ni z \to z_0} |f(z)| < \infty$$ for any $z_0 \in \partial D$. Prove that $f$ is bounded.
So we would need to prove that $\lim_{|z| \to \infty}|f(z)| = 0$ or find $M$ such that $|f| \le M$.
I thought I could find a $g \in \mathcal{O}(D)$ such that $|f(z)| \le |g(z)|$ and then prove that $g$ is constant.
But I really don't see that and I'm stuck.
Could you help me locate the main theorem which would be helpful here?
Assume that $f$ is not bounded in $D$. Then there exists a sequence $(z_n)$ in $D$ such that $|f(z_n)| \to \infty$. $D$ is bounded implies that $\overline D$ is compact, and therefore $(z_n)$ has a subsequence $(z_{n_k})$ such that $z_{n_k} \to a \in \overline D$.
$a \in \partial D$ is not possible because of the assumption $\limsup _{D \ni z \to a} |f(z)| < \infty$.
$a \in D$ is also not possible because then $|f(z_{n_k})| \to |f(a) | < \infty$ in contradiction to $|f(z_n)| \to \infty$.
Note that only the continuity of $f$ is used here, not the holomorphy.