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If $u$ is harmonic , is it necessarily convex as well ? What my main interest is to show that $|$u$|^p $ is subharmonic .

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    Compute $\Delta |u|^p$ directly to see it doesn't change sign, which depends on $p$.2012-04-23

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On $\mathbb{R}^2$ $x\,y$ and $x^2-y^2$ are examples of harmonic functions that are neither concave nor convex.

If $u$ is harmonic on an open set $\Omega\subset\mathbb{R}^n$, then it verifies the mean value property: $ u(x)=\frac{1}{|B_R(x)|}\int_{B_R(x)}u(y)\,dy,\quad B_R(x)\subset\Omega $ where $B_R(x)$ is the ball of radius $R>0$ centred at $x$ and $|B_R(x)|$ its measure. You can show that if $p>1$ then $v=|u|^p$ is subharmonic using Jensen's or Hölder's inequality to show that it satisfies the inequality $ v(x)\le\frac{1}{B_R(x)}\int_{|B_R(x)|}v(y)\,dy,\quad B_R(x)\subset\Omega. $

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    Because it is a local condition. If $u\in C^2(\Omega)$ then $u$ is subharmonic if $-\Delta u\le0$, which is a local condition. For continuous (or more generally lower semicontinuous) functions, one possible definition is that it satisfies a mean value inequality for all sufficiently small balls, which is again local.2012-04-24
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Suppose $u$ is harmonic, i.e. $\Delta u=0$. Then if $u$ would be necessarily convex it is easy to see that $-u$ is also harmonic, and $u$ would be necessarily concave also.

Therefore if $u$ harmonic implies $u$ convex it follows that every harmonic function is both convex and concave (i.e. the images of $u$ are all in the same hyperplane), which is not true for every harmonic function.

So $u$ is not necessarily convex.

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    @ Beni , i didn't understand, can you please tell me why if $u$ is subharmonic locally that $\Delta u\ge 0$ should be valid in every ball.Thanks2012-04-23