Im trying to solve the following problem.
Let $\Omega \in \mathbb{C}$ an open bounded set, let $f\colon \Omega \to \mathbb{C}$ be holomorphic, and supose there exists a constant $M \geq 0$ wich satisfies that if $(z_n)_{n\geq 1}$ is a convergent sequence of points in $\Omega$ such that if $\lim\limits_{n \to \infty} z_n \in \partial\Omega$ then $\limsup\limits_{n \to \infty} |f(z_n)| \leq M $.
Prove that for all $z \in \Omega$ we have $|f(z)| \leq M$.
My attempt: First supose $f$ is not constant because if it is constant i have nothing to prove, because the lim sup is equal to $f(z)$ in any point and it would be true. Then, as we said, supose $f$ is not constant, and supose the condition doesnt satisfies, that is there exists some $z$ such that $|f(z)| > M$, then the all the $z$ that do this cannot be a discrete set, because I can pick the maximum of the set and would get that the function is constant by the Maximum Modulus Principle. So the set $A =$ {$z$ such that $|f(z)| > M$} is infinite, now if $\overline A = \Omega$ i would reach a contradiction, because I can pick a sequence $(z_n) \subset A$, convergent to the border which would yield that $M < limsup |f(z_n)| \leq M$. The thing is if A is not dense in $\Omega$ then Im a bit lost.