Prove $\max_{\bar \Omega} |\nabla u|^2=\max_{\partial \Omega} |\nabla u|^2$.

Let $\Omega \in \mathbb R^n$, $n>1$, be a bounded domain with smooth boundary. Let $u \in C^1 (\bar \Omega)$ be harmonic in $\Omega$.

(a) Prove $\max_{\bar \Omega} |\nabla u|^2=\max_{\partial \Omega} |\nabla u|^2$.

(b) Let $u$ satisfies $u \in C^2 (\mathbb R^n)$, $\Delta u =0$ on $\mathbb R^n$. Show that if $u \in L^2 (\mathbb R^n)$, then $u$ is identically equal to zero.

Thoughts: For part (a), I think I need to apply the maximum principle but based on the maximum principle, do I need to show that $\nabla u$ is harmonic and is in $C^2$?

Further thoughts: I can't comment because of low points. For John Ma, we know the definition of $C^1$ function, then we can just required $u$ to satisfy $\Delta u =0$.

Final thoughts: Just checked my previous homework and I think I need to show $|\nabla u |^2$ is subharmonic, then apply the maximum principle to subharmonic functions.

Any hints would be appreciated.

I think I need to show $\mid \nabla u \mid ^2$ is subharmonic, then apply the maximum principle to subharmonic functions.

• @JohnMa I don't see how it contradicts to the condition. Like I said (in the edit), $u$ is a $C^1$ function and at the same time $\Delta u=0$. – Chris Gartland Oct 23 '15 at 8:37

$\nabla u$ isn't harmonic in general. $\nabla u$ is a gradient vector, $|\nabla u|^2=\sum_i (\partial_i u)^2$ thus, $$\Delta|\nabla u|^2=\sum_i \Delta (\partial_i u)^2= \sum_i \left(2 (\partial_i u)\Delta (\partial_i u)+|\nabla\partial_i u|^2 \right)= \sum_{i ,j} \partial_i\partial_j u\ne0$$ The last equality is due to the fact that $u$ is given harmonic and $\Delta (\partial_i u) =\partial_i( \Delta u)$.