1
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Apr 3 '18 at 21:25
  • $\begingroup$ @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? $\endgroup$ – Error 404 Apr 4 '18 at 9:01
  • 1
    $\begingroup$ 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 $\Omega$. $\endgroup$ – Daniel Fischer Apr 4 '18 at 9:14
  • $\begingroup$ @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? $\endgroup$ – Error 404 Apr 4 '18 at 9:27
  • 1
    $\begingroup$ 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.) $\endgroup$ – Daniel Fischer Apr 4 '18 at 9:30
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.