Showing $f$ holomorph $\Rightarrow$ $\nabla^2(|f|^2)\ge0$

Question

I want to show $f$ holomorph $\Rightarrow$ $\nabla^2(|f|^2)\ge0$

I already showed that for $f(x+iy)=u(x,y)+iv(x,y)$ that $\nabla^2v=\nabla^2u=0$ (therefore harmonic).

And if I'm correct $\nabla^2(|f^2|)$ should be $\Leftrightarrow \nabla^2(u^2+v^2)=\nabla^2(u^2)+\nabla^2(v^2)$ but I can't get any further.

Thanks for your help!

Answer 1

This is very easy. $|f|^2=u^2+v^2$; just differentiate! Writing $N=|f|^2$: $$N_x=2uu_x+2vv_x,$$so $$N_{xx}=2u_x^2+2uu_{xx}+2v_x^2+2vv_{xx}.$$Similarly for $N_{yy}$. Now add; since $u_{xx}+u_{yy}=0$ and similarly for $v$ you get $$\nabla^2N=2||\nabla u||^2+2||\nabla v||^2.$$