showing a convex function s subharmonic

Question

Given a $C^2$ convex function $f$ and $u$ a harmonic function in an open subset of $\mathbb{R^2}$, how can I show that $f(u)$ is sub-harmonic?

Answer 1

You can also simply compute derivatives since everything is differentiable enough: $$ (f(u))_x = f'\cdot u_x\Rightarrow (f(u))_{xx} = f'' u_x^2+f'u_{xx} $$ and clearly, $(f(u))_{yy} = f'' u_y^2+f' u_{yy}$. As a result, $$ \Delta f(u) = f'' (u_x^2+u_y^2)\geq 0 $$ since $f$ is a convex $C^2$ function.