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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.

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If you already used Jensens's inequality, it's straight forward –  ulead86 Jul 19 '12 at 12:03
4  
If $f$ is not differentiable, $f\circ u$ maybe be not differentiable. An alternative way is to check that $$f\circ u(z)\leq \frac 1{2\pi r}\int_0^{2\pi}f\circ u(z+re^{it})dt,$$ using Jensen's inequality and one of your previous threads. –  Davide Giraudo Jul 19 '12 at 12:05
    
@DavideGiraudo : But will it not still be a problem when i move derivative inside ? I am a bit confused here but eager to know . –  Theorem Jul 19 '12 at 12:17
2  
Subharmonic functions do not need to be differentiable. The usual definition just requires its graph to lie below any harmonic function which is larger on the boundary. –  Willie Wong Jul 19 '12 at 12:35
    
@WillieWong I should have guessed it was this definition which was used. –  Davide Giraudo Jul 19 '12 at 12:55

1 Answer 1

up vote 5 down vote accepted

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.

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Thank you sir . –  Theorem Jul 19 '12 at 15:27

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