Complex analysis problem, using limsup

Question

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.

Answer 1

You don't need that $A$ is dense, all you need is that there is at least one sequence converging to (a point of) $\partial\Omega$ with $\limsup \lvert f(z_n)\rvert > M$.

On the path you started on, you can modify $A$, choose an $\varepsilon > 0$ such that $\lvert f(z)\rvert > M+\varepsilon$ for some $z\in\Omega$, and use $\tilde{A} = \{z \in \Omega : \lvert f(z)\rvert > M + \varepsilon\}$. When only demanding $\lvert f(z)\rvert > M$, you cannot deduce that $\limsup \lvert f(z_n)\rvert > M$ when $z_n$ is a sequence in $A$ converging to a point of $\partial \Omega$, therefore we only consider points where the value of $\lvert f(z)\rvert$ is larger than something fixed that is larger than $M$.

Then the problem is to deduce that there is at least one sequence in $\tilde{A}$ converging to a point of $\partial \Omega$. Now, if there weren't such a sequence, then $\tilde{A}$ would have a positive distance from $\partial \Omega$, that means there is a $\delta > 0$ such that

$$\tilde{A}\subset \Omega_\delta = \{ z \in \Omega : D_\delta(z) \subset \Omega\}.$$

But then $f$ would be holomorphic on $\Omega_\delta$ and continuous on $\overline{\Omega}_\delta$, and

$$\sup_{z\in\partial \Omega_\delta} \lvert f(z)\rvert \leqslant M+\varepsilon < \sup_{z\in\tilde{A}} \lvert f(z)\rvert$$

contradicts the maximum modulus principle.

Answer 2

Take the closure of the curve union lim sup. Take max of closure.
By Max mod principle , the max is on the boundary.
Only lim sup could be on the boundary. QED