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I have the following exercise in my complex analysis course:

Let $u(x,y)$ be harmonic function on a domain $D$ such that $u$ has a global maximum on $D$.
Prove that $u$ is constant on $D$.

My efforts so far:
If I knew that $D$ was simply connected, then I'd know how to prove it, but because it's not given, I started by using the fact that there is some point $(x_0,y_0)\in D$ such that for all $(x,y)\in D$, it holds $u(x,y)\leq u(x_0,y_0)$.
Now, since $D$ is open, there is an open disk around $(x_0,y_0)$ that is contained in $D$, and I managed to prove that $u(x_0,y_0)$ is constant within that disk. I'm missing the final part where I take a general $(x,y)\in D$ and show that $u(x,y)=u(x_0,y_0)$
I know that D is path connected, so there must be a path that connects $(x,y)$ to $(x_0,y_0)$ but I don't know how to continue from here (I thought about covering this path by open disks but I can't really write it down formally).

Any help would be much appreciated.

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$\{(x,y)\in D: u(x,y)=u(x_0,y_0)\}$ is a closed subset of $D$, by continuity of $u$. Use what you have already proved to show that it is open. Since $D$ is connected it follows that $u(x,y)=u(x_0,y_0)$ for all $(x,y) \in D$.

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  • $\begingroup$ How come the set $\{(x,y)\in D : u(x,y)=u(x_0,y_0)\}$ is closed (even though $u$ is continuous)? Let's say that we knew in advance that $u$ is constant on $D$, so it is continuous in particular. But, in this case, the above set is simply $D$, which is open (and not necessarily closed). On top of that, if this set is closed, then since it is both open and closed it must be all of $\mathbb{C}$. Am I getting it wrong? $\endgroup$
    – Vegetal605
    Jan 12, 2023 at 14:56
  • $\begingroup$ @Vegetal605 It is a closed subset of $D$ in the subspace topology of $D$. $\endgroup$ Jan 12, 2023 at 23:13

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