This question came up on a recent qualifying exam and I am unsure how to solve it and unsure if the question is stated correctly.
Here is the question:
Suppose $U$ is an open bounded connected subset of the complex plane, and $f$ is a non-constant holomorphic function on $U$. Suppose there exists some $M > 0$ such that $M$ is equal to the supremum of the limit of all convergent sequences of the form $\{ |f(z_{n})|\}_{n=1}^{\infty}$ where $z_{n} \in U$ for all $n \geq 1$ and $z_{n} \rightarrow z_{0} \in \partial U$. Prove that the $|f(z)| < M$ for all $z \in U$.
My interpretation:
I don't think the problem was stated correctly. As a counter-example, take $U$ as the annulus $\{z : 0 < |z| < 1\}$ and $f(z) := 1/z$. Then the conditions of the problem are satisfied with $M := 1$, since for any $z_{0}$ with $|z_{0}| = 1$, $|f(z)| \rightarrow 1$ as $z \rightarrow z_{0}$. But if $z_{0} = 0$, any sequence of the form $\{|f(z_{n})|\}_{n=1}^{\infty}$ where $z_{n} \in U$ and $z_{n} \rightarrow 0$, does not converge.
I think the problem should be stated with the assumption that all sequences of the form $\{|f(z_{n})|\}$, where $z_{n} \rightarrow z_{0} \in \partial U$, are convergent, or at least bounded. Yet even with this additional assumption, I am not sure how to proceed. My initial thought was to try and extend $f$ to a function that is continuous on $\partial U$ and then apply the Maximum Modulus Theorem. However, I don't think it is possible to extend $f$ this way in general (take $\log z$ for example), right?
Am I interpreting the problem correctly? Is the statement true with the additional assumption mentioned above?
The statement is true if you allow $|f(z_n)|$ to converge in $[0,\infty]$. Otherwise your counterexample is correct.
To prove this, let $U(\epsilon) := \{z\in U: d(z,\partial U)>\epsilon\}$. Assume there is $z\in U$ with $|f(z)|\geq M$. Then there is $\epsilon_0>0$ such that $z\in U(\epsilon)$ for all $\epsilon<\epsilon_0$. But now we have $\partial U(\epsilon)\subset U$ for all $\epsilon<\epsilon_0$, letting us construct (by maximum modulus principle) a sequence $(z_n)_{n\in\mathbb N}$ with $|f(z_n)|> M$ and $d(z_n,\partial U)\to 0$. We can assume $|f(z_n)|$ to be increasing, hence convergent in $(M,\infty]$ (again by maximum modulus principle). By boundedness of $U$, $(z_n)_{n\in\mathbb N}$ has a convergent subsequence which converges to $z_0\in\partial U$ contradicting our assumption.