composition of convex function with harmonic function. Consider $u:\Omega \to \mathbb R$ be harmonic and $f$ be convex function. How do i prove that $f\circ u$ is subharmonic?
It seems straight forward : $\Delta (f\circ u (x)) \ge f(\Delta u(x))=0 $.
Is this all to this problem? Is there a better way?
Also $|u|^p$ is subharmonic for $p\ge1$ , this seems also very obvious because the map $t\to |t|^p$ is convex mapping. 
Any comments, improvements or answers are welcome. 

Basically you do what Davide Giraudo suggested. We use the definition as given on the EOM

An upper semicontinuous function $v:\Omega \to\mathbb{R}\cup \{-\infty\}$ is called subharmonic if for every $x_0 \in \Omega$ and for every $r > 0$ such that $\overline{B_r(x_0)} \subset \Omega$, that 
  $$ v(x_0) \leq \frac{1}{|\partial B_r(x_0)|} \int_{\partial B_r(x_0)} v(y) \mathrm{d}y $$

Now, using that $f$ is convex, Jensen's inequality takes the following form:
$$ f\left( \frac{1}{|\partial B_r(x_0)|} \int_{\partial B_r(x_0)} v(y)\mathrm{d}y \right) \leq \frac{1}{|\partial B_r(x_0)|}\int_{\partial B_r(x_0)} f\circ v(y) \mathrm{d}y $$
If $u$ is harmonic, we have that
$$ \frac{1}{|\partial B_r(x_0)|} \int_{\partial B_r(x_0)} u(y)\mathrm{d}y = u(x_0) $$
so combining the two facts you have that
$$ f\circ u(x_0) \leq \frac{1}{|\partial B_r(x_0)|}\int_{\partial B_r(x_0)} f\circ u(y) \mathrm{d}y $$
which is precisely the definition for $f\circ u$ to be subharmonic. 
