Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
add comment

2 Answers 2

up vote 3 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 verifeis 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 $$ v(x)\le\frac{1}{B_R(x)}\int_{|B_R(x)|}v(y)\,dy,\quad B_R(x)\subset\Omega. $$

share|improve this answer
    
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
1  
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
1  
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
add comment

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.

share|improve this answer
    
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
1  
Here is the definition for subharmonic functions: en.wikipedia.org/wiki/Subharmonic_function 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
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.