2
$\begingroup$

I know the following classical maximum principle for harmonic functions:

If $\Omega \subset \mathbb{C}$ is open and connected and $u \in C^2(\Omega)$ is harmonic, then

$u$ has maximum (or minimum) in $\Omega$ $\implies$ $u$ constant.

How can I prove that the theorem is true if the hypothesis is that $u$ has a local maximum (or minimum).

$\endgroup$
  • 1
    $\begingroup$ Harmonic functions are real-analytic. $\endgroup$ – Daniel Fischer Jan 6 '17 at 14:35
  • $\begingroup$ @DanielFischer What does this imply here? $\endgroup$ – user393664 Jan 6 '17 at 14:40
  • $\begingroup$ An identity theorem. Then use what the principle you know implies in case of a local extremum. $\endgroup$ – Daniel Fischer Jan 6 '17 at 14:41
  • $\begingroup$ @DanielFischer The issue is still not clear to me. Could you add an answer to this question? $\endgroup$ – user393664 Jan 17 '17 at 15:17
2
$\begingroup$

If $u$ has a local maximum at $x_0 \in \Omega$, that means that there is an open neighbourhood $U\subset \Omega$ of $x_0$ such that $u\lvert_U$ has a global maximum at $x_0$. Since clearly $u\lvert_U$ is harmonic on $U$, by the form of the maximum principle that you already know, it follows that $u\lvert_U$ is constant.

The two harmonic functions $u$ and $v \colon z \mapsto u(x_0)$ hence coincide on a nonempty open set (namely on $U$), and since $\Omega$ is connected, and harmonic functions are real-analytic, the following identity theorem (the case $n = 2$ with the usual identification $\mathbb{C}\cong \mathbb{R}^2$) implies that $u \equiv v$ on $\Omega$, i.e. $u$ is constant.

Identity theorem: Let $W\subset \mathbb{R}^n$ a connected open set, and $f,g \colon W \to \mathbb{R}$ real-analytic. If there is a nonempty open set $V\subset W$ such that $f\lvert_V \equiv g\lvert_V$, then $f \equiv g$ on all of $W$.

Proof: It suffices to consider $g \equiv 0$, since $f\equiv g \iff f-g \equiv 0$. Consider the set

$$Z = \{ x \in W : \text{there is a neighbourhood } U \text{ of } x \text{ such that } f\lvert_U \equiv 0\}.$$

By its definition, $Z$ is an open subset of $W$ (if $x\in W$ and $U$ is a neighbourhood of $x$ such that $f\lvert_U \equiv 0$, then $y\in Z$ for all $y$ in the interior of $U$). Since $f$ is real analytic, we can also describe $Z$ as

$$Z = \{ x \in W : D^{\alpha}f(x) = 0 \text{ for all } \alpha \in \mathbb{N}^n\},$$

for the power series representation of $f$ about $x$,

$$f(y) = \sum_{\alpha \in \mathbb{N}^n} \frac{D^{\alpha}f(x)}{\alpha!}(y-x)^{\alpha},$$

vanishes in a neighbourhood of $x$ if and only if all coefficients vanish. By continuity of $D^{\alpha}f$, the set

$$Z_{\alpha} = (D^{\alpha}f)^{-1}(0)$$

is closed (in $W$) for every $\alpha \in \mathbb{N}^n$, and hence so is the intersection

$$Z = \bigcap_{\alpha \in \mathbb{N}^n} Z_{\alpha}.$$

Thus $Z$ is an open and closed subset of $W$, and since $W$ is connected, we have either $Z = W$ or $Z = \varnothing$. But $Z \neq \varnothing$ was part of the hypotheses of the theorem, so $Z = W$ and indeed $f \equiv 0$ on $W$.

$\endgroup$
0
$\begingroup$

Assume that you can reach the local maximal $M$ in $x_0$, then by the same argument as the maximal situation, you can find a neighbor $U$ of $x_0$, s.t. $u=M$ in $U$, so $W=\{x:u(x)=M\}$ is open. Moreover $\{x:u(x)\neq M\}$ is open. By connectedness of $X$, we much have $W=X.$ Maybe the following is better:

enter image description here

$\endgroup$
  • 1
    $\begingroup$ We know $u(x_0) = M$ is a local max and therefore $u=M$ in a neighborhood of $x_0.$ How does that imply $\{x\in \Omega: u(x) = M\}$ is open? $\endgroup$ – zhw. Jan 6 '17 at 20:36

Your Answer

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