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if $\log u$ and $\log v$ are both subharmonic function on an open set $U$, then $\log(u+v)$ is also subharmonic on $U$.

How can I prove this?

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Are you assuming that $u,v \neq 0$ and $u,v$ twice differentiable? – Willie Wong Mar 4 '11 at 17:03
yes i assume that – user7428 Mar 4 '11 at 23:11

If $w$ is strictly positive and $C^2$ then a direct calculation shows that $\log w$ is subharmonic if and only if $w\Delta w+|\nabla w|^2 \ge 0$. Taking $w=u+v$ we see that $\log(u+v)$ is subharmonic if and only if $$ (u\Delta u+|\nabla u|^2)+(v\Delta v+|\nabla v|^2) +\Delta(uv) \ge 0. $$ If $\log u$ and $\log v$ are subharmonic then the first two terms on the left are non-negative. And so is the third because $$ \Delta(uv) = uv\cdot\left[\Delta(\log u+\log v) +|\nabla(\log uv|^2\right], $$ which is clearly non-negative.

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I'm afraid this answer is wrong. There should be a minus sign in your criterium ("minus the norm of the gradient"), which complicates things considerably. Additionally, the first manipulation is incorrect. This is not the way to prove this. – A. Ramos Oct 16 '12 at 16:57

The assumption about the regularity is not necessary here.

Indeed, we know that we can find a sequence of smooth functions $u_n$ and $v_n$ decreasing to $u$ and $v$ respectively and such that $\log u_n, \log v_n$ are subharmonic (just use a convolution with a radial smoothing kernel).

We may then use the argument of JohnD which guarantees that $\log(u_n+v_n)$ is subharmonic. But a limit of a decreasing sequence of subharmonic functions is still subharmonic, so that $\log(u+v)$ is subharmonic.

Moreover, a direct proof (with no assumption on regularity) may be obtained once we know some basic stuff about sh functions. Namely, $f(x,y)=\log (e^x+e^y)$ defines a a convex function, non-decreasing in each variable, therefore $f(\log u, \log v)=\log(u+v)$ is subharmonic.

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