Strong maxima and minima

Question

I'm stuck with this problem, in particular at b):

Let $u:D \subseteq \mathbb{R}^2 \to \mathbb{R} \in C^2$ a harmonic function. $u$ has a local maximum at point $\vec{p} \in D$. Then:

(a) Show that, $u$ is a constant function in some disk with center $\vec{p}$

(b) Suppose, that $D$ is arc-connected and $\vec{p}$ is a global maximum. Then, $u$ is constant at $D$

So far, I know that if $\nabla^2 u = 0 $, then at point $\vec{p}$ $$ u(\vec{p})=\frac{1}{\pi R^2}\iint_{B_R} u \, ds$$ Where, $B_R=B(\vec{p},R)\subseteq \mathbb{R}^2$. So my idea is. Let, $\vec{q} \in B_R$, where $B_R$ is the open ball where $u(\vec{p})$ is local maximum, and take the difference: $$0 \leq u(\vec{p})-u(\vec{q}) \in C^2$$ So, if I prove $$ \frac{1}{\pi R^2}\iint_{B_R} u(\vec{q})-u(\vec{p}) \, ds = 0$$ Then, for any $\vec{q} \in B_R$ $$u(\vec{q})-u(\vec{p})=0$$ Therefore $u$ is constant at $B_R$

Any hint is appreciate

Answer 1

Suppose $\vec q$ is a point in $B(\vec p ,R)$ with $u(\vec q)<u(\vec p)$. Then you can find an $r$ with $B_{r}=B(\vec q,r)\subset B(\vec p,R)$ where for every $\vec s\in B(\vec q, r)$, $u(\vec s)<u(\vec p)$. Now $$u(\vec p)=\frac {1}{\pi R^2} \iint_{B_R}u\, ds =\frac {1}{\pi R^2} \iint_{B_R\setminus B_r}u\, ds + \frac {1}{\pi R^2} \iint_{B_r}u\, ds<\frac {1}{\pi R^2} \iint_{B_R\setminus B_r}u\, ds + \frac {u(\vec p)}{\pi R^2}\pi r^2<u(\vec p) $$, which is impossible. So there cannot be such a $\vec q$ in $B_R$ and $\vec p$ is a local maximum, so $u$ must be constant on $B_R$.

Answer 2

When two discs overlap, apply the Poisson formula to each one to infer that the harmonic functions are real parts of holomorphic functions in each disc, which holomorphic functions are uniquely determined up to an imaginary constant. (Rudin Real and Complex Analysis 11.13) Holomorphic functions which coincide on an uncountable set must be equal, a simple consequence of Rudin 10.18. The intersection of the discs is uncountable. Thus, if the harmonic function is constant on one disc, it is constant on the second overlapping disc too.

Consider a path from $p$ to another point $p'\in D$, which path exists because of the assumption that $D$ is path-connected. Cover the path with overlapping discs contained in $D$ and centered on path points, with the first disc (centered on $p$) being one where the harmonic function is constant. By compactness there is a finite subset of overlapping discs which covers the path. The harmonic function is constant on the first disc. As a result of overlap, the harmonic function will be constant on all the other discs, and thus it equals that constant at $p'$.

Part (a) can be proved easily with the Poisson formula too, see Rudin 11.15.