# Harmonic function squared and mean value

1. Nov 7, 2013

### arestes

1. The problem statement, all variables and given/known data
Let u be a harmonic function in the open disk K centered at the origin with radius a. and
$∫_K[u(x,y)]^2 dxdy = M < ∞$. Prove that
$|u(x,y)| \le \frac{1}{a-\sqrt{x^2+y^2}}\left( \frac{M}{\pi}\right)^{1/2}$ for all (x,y) in K.

2. Relevant equations
Mean value property for harmonic functions.

3. The attempt at a solution
I first thought this was easy and directly applied the mean property for harmonic functions, but of course, the square of a harmonic function is not harmonic (unless it's a constant).

I can see this problems begs for the mean value using a ball with radius $a-\sqrt{x^2+y^2}$ which would be entirely inside the disk K but I can't get around the fact that u squared is not harmonic. Help please

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Last edited: Nov 7, 2013
2. Nov 8, 2013

### brmath

Are you integrating over the whole disk K (i.e. a double integral) or as a line integral around the boundary of K.

3. Nov 8, 2013

### arestes

I integrated over the area of the smaller circle inside with radius $r=a-\sqrt{x^2+y^2}$. This is exactly what I did (I know it's wrong):

$u^2(x,y)=\frac{\int_{B_r(x,y)} u^2(w, z)dwdz}{\pi r^2}\leq \frac{\int_K u^2(w, z)dwdz}{\pi r^2}=\frac{M}{\pi r^2}\\ |u(x,y)|\leq \sqrt{\frac{M}{\pi r^2}}$
But of course, this would work if u^2 were harmonic, which is seldom the case.

Any ideas? I was thinking maybe applying the maximum principle and somehow get bounds...

Last edited: Nov 8, 2013
4. Nov 9, 2013