how to prove that this weak solution is subharmonic?

Question

My question is about this article http://hal.inria.fr/docs/00/12/87/60/PDF/fbpLaplacian.pdf.

My question is :

Consider a smooth, bounded and convex domain $K$ in $R^n$ such that $K\subset \{ x_1 = 0 \}$ and $\Omega $ a bounded convex domain such that $\partial \Omega \supset K$.

Let $u \in H^1(\Omega)$ a weak solution of the problem : $$ \left\{ \begin{array}{ccccccc} \Delta u = 0, \ in \ \Omega \\ u = 1 \ in \ K \\ u = 0 \ on \ \partial \Omega - K \\ \end{array} \right. $$

In the page 4 the authors say : $|\nabla u|²$ is subharmonic (i believe that is in the weak sense). I dont know to how to show this .. Someone can give me a help to prove (or say to me a book with the proof)?

Thanks in advance

Answer 1

Combine the following facts:

1. If $u$ is harmonic, then the partial derivatives of $u$ are harmonic.
2. If $u$ is harmonic and $\phi:\mathbb R\to\mathbb R$ is convex, then $\phi\circ u$ is subharmonic.
3. The function $\phi(t)=t^2$ is convex.
4. $|\nabla u|^2$ is the sum of squares of partial derivatives of $u$.

Any book that deals with subharmonic functions should have some version of 2 in it. If $\phi$ is smooth (as it is here), you can do direct computation, but it's not much fun. It's better to prove the sub-mean-value property of $\phi\circ u$ using Jensen's inequality.