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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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Harmonic on what? On $\Bbb{R}^2$, on an open ball? – Beni Bogosel Apr 23 '12 at 15:40
@ Beni , on a arbitrary open ball . – Theorem Apr 23 '12 at 15:42
Compute $\Delta |u|^p$ directly to see it doesn't change sign, which depends on $p$. – Andrew Apr 23 '12 at 19:12
up vote 4 down vote accepted

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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Thank you . I would like to know why subharmonic functions are called "sub" harmonic . what is the idea behind it ? – Theorem Apr 23 '12 at 21:15
Because in some sense they are below harmonic functions. They satisfy the mean value property with $\le$ instead of equality. If they are $C^2$, they satisfy $-\Delta u\le0$. – Julián Aguirre Apr 23 '12 at 21:22
My notes say that if a function is subharmonic locally some open ball in a region $R$ then its subharmonic in the whole $R$. What is the reason behind it ? – Theorem Apr 23 '12 at 21:29
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. – Julián Aguirre Apr 24 '12 at 9:29
Thank you Professor. – Theorem Apr 24 '12 at 9:50

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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Does it make a difference if $u$ is subharmonic ? Because my notes say that $|u|^p$ is subharmonic without any explaination , i am not able to understand how ! – Theorem Apr 23 '12 at 15:51
I'll think about it. Probably it is possible to prove that $u$ harmonic implies $|u|^p$ subharmonic. What can you say about $p$? $p>1,2$? – Beni Bogosel Apr 23 '12 at 16:00
$p>=1$ and any idea why subharmonic functions are calls "subharmonic". – Theorem Apr 23 '12 at 16:04
Here is the definition for subharmonic functions: The idea is that a subharmonic function on a ball, for example, is below any harmonic function which takes the same values on the boundary of the ball. If $u$ is $C^2$ then it is subharmonic if and only if $\Delta u \geq 0$. – Beni Bogosel Apr 23 '12 at 16:09
@ Beni can you explain me why if $u$ is subharmonic locally i.e in some ball of radius $r$ in region $R$ it should be subharmonic in the whole region $R$? any idea ?? – Theorem Apr 23 '12 at 16:23

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