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

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!

• As said in the comments under the other question, you need an additional assumption. That can be connectedness of $\Omega$, or the assumption that $u$ isn't constant on any component of $\Omega$. If there is a component of $\Omega$ on which $u$ is constant, then $u$ isn't open. When that assumption is added, the question remains why the $f_i$ are all non-constant. Assuming $\Omega$ connected, why can't $u$ be constant on some of the $D_i$? An alternative way is using the maximum/minimum principle, by which $u$ can't have a local extremum. – Daniel Fischer Apr 3 '18 at 21:25
• @DanielFischer Oh Maximum/Minimum principle seems a good approach here. For that we only need the additional assumption that $u$ isn't constant on any component of $\Omega$. Am I Right? – Error 404 Apr 4 '18 at 9:01
• But it does matter whether $\Omega$ is connected or not. If e.g. $\Omega = \mathbb{C}\setminus \mathbb{R}$ and $u$ is $1$ on the upper half-plane and $u(z) = \operatorname{Re} z$ on the lower half-plane, then $u$ is non-constant, but it's not an open mapping. A harmonic $u \colon \Omega \to \mathbb{R}$ is an open mapping if and only if $u$ is not constant on any component of $\Omega$. – Daniel Fischer Apr 4 '18 at 9:14
• @DanielFischer In your response to my proof, assuming $u$ isn't constant on any component of $\Omega$, if $u$ happens to be constant on a $D_j$, then $u$ has to be constant on that component (In which $D_j$ lies) due to uniqueness theorem of Harmonic functions. Am I right? – Error 404 Apr 4 '18 at 9:27
• Yes. If $u$ is constant on some nonempty open set $U$ (like a $D_i$), then by the identity theorem it's constant on all components of $\Omega$ that intersect $U$. (If $U$ is connected, that's a single component of course.) – Daniel Fischer Apr 4 '18 at 9:30

## 1 Answer

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.