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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?

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How $u$ is important in your question? Are you just asking that $C^2$ convex function is subharmonic? –  Ilya May 25 '12 at 12:03
    
@Ilya, thank you. I meant $f(u)$. I want to show $-\Delta f(u) \leq 0$. –  Herband May 25 '12 at 12:07
    
I hope, you meant $-\Delta f\leq 0$ –  Ilya May 25 '12 at 12:10
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use Jensen's inequality to show averages over balls smaller than value at center, (or bigger, whichever is right). –  mike May 25 '12 at 12:12
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1 Answer

up vote 1 down vote accepted

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.

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just that simple! Thanks very much. –  Herband May 25 '12 at 12:22
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