Showing that a harmonic function maps open sets to open sets.

Question

I am trying to show that a harmonic function maps open sets to open sets. I have written down a proof based on the hint provided by Theo Bendit here :

Proof : Let $u : \Omega \to \Bbb R$ be a non-constant harmonic function, where $\Omega$ is an open subset of $\Bbb C$. Let $V$ be any open subset of $\Omega$. Then $V=\bigcup_{i \in I}D_i$ where $D_i'$s are open balls and $I$ is an indexing set.

Thus $u(V)=u(\bigcup_{i \in I}D_i)=\bigcup_{i \in I}u(D_i).$

Since each $D_i$ is simply connected, there is a holomorphic function $f_i$ on each $D_i$ such that $u=\text {Re}f_i$ on $D_i$. But $\text {Re}f_i=p\circ(f_i)$ where $p$ is a projection map as described in the other answer of the linked question.

$\therefore u(V)=\bigcup_{i \in I}p(f_i(D_i)).$ By open mapping theorem, $f_i(D_i)$ is an open set in $\Bbb C$. This together with $p$ being an open map implies that $u(V)$ is open in $\Bbb R$.

• Are there any errors in my proof?
• Also I am curious to know whether there alternate ways to do this.

Thanks!

Answer 1

As per inputs by Daniel Fischer, here are the two approaches to this result :

Approach 1 : Let $u : \Omega \to \Bbb R$ be a non-constant harmonic function on open set $\Omega$ in $\Bbb C$ which is non-constant on every component of $\Omega$. Let $V$ be an open subset of $\Omega$. Since $V$ is open, $V=\bigcup_j \{D_j\}$ where each $D_j$ is an open disc.

Suppose $u$ is constant on some $D_k$ then by identity theorem, $u$ is constant on the connected component of $\Omega$ which contains $D_k$. This is a contradiction. Thus $u$ is non-constant on each $D_j$. As each $D_j$ is simply connected, there is a holomorphic function $f_j$ such that $\text {Re}f_j=u$ on each $D_j$. But $\text {Re}f_j=p \circ f_j$ where $p$ is a projection map on first component. $\therefore u(V)=\bigcup_{i \in I}p(f_i(D_i)).$ This together with both $p$ and $f_i$ being open maps implies that $u(V)$ is open in $\Bbb C$.

Approach 2 (By Maximum/Minimum principle): Let $u,\Omega,V$ be as before in approach 1. Let $\{S_i\}$ be the set of connected components of $\Omega$ which intersect with $V$. Let $V_i=V \cap S_i$ Then $V=\bigcup_i V_i \Rightarrow u(V)=\bigcup_i u(V_i).$

Since each $V_i$ is connected, $u(V_i)$ is also connected in $\Bbb R$. Thus each $u(V_i)$ is an interval in $\Bbb R.$ Observe that $u$ is a non-constant harmonic function on open set $V_i$ (similar argument as $D_i$), therefore $u$ does not have extremums on each $V_i$ by maximum/minimum principles. Thus each $u(V_i)$ is an open interval in $\Bbb R$ $\Rightarrow u(V)$ is an open set in $\Bbb R$ since arbitrary union of open sets is open.