If $u$ is harmonic, is it necessarily convex?

Question

If $u$ is harmonic on an arbitrary open ball, is it necessarily convex as well?

My main interest is to show that $|u|^p$ is subharmonic.

Answer 1

On $\mathbb{R}^2$ $x\,y$ and $x^2-y^2$ are examples of harmonic functions that are neither concave nor convex.

If $u$ is harmonic on an open set $\Omega\subset\mathbb{R}^n$, then it verifies the mean value property: $$ u(x)=\frac{1}{|B_R(x)|}\int_{B_R(x)}u(y)\,dy,\quad B_R(x)\subset\Omega $$ where $B_R(x)$ is the ball of radius $R>0$ centred at $x$ and $|B_R(x)|$ its measure. You can show that if $p>1$ then $v=|u|^p$ is subharmonic using Jensen's or Hölder's inequality to show that it satisfies the inequality $$ v(x)\le\frac{1}{B_R(x)}\int_{|B_R(x)|}v(y)\,dy,\quad B_R(x)\subset\Omega. $$

Answer 2

Suppose $u$ is harmonic, i.e. $\Delta u=0$. Then if $u$ would be necessarily convex it is easy to see that $-u$ is also harmonic, and $u$ would be necessarily concave also.

Therefore if $u$ harmonic implies $u$ convex it follows that every harmonic function is both convex and concave (i.e. the images of $u$ are all in the same hyperplane), which is not true for every harmonic function.

So $u$ is not necessarily convex.