# Mean Value Property Implies "Volume" Mean Value Property

• Mar 16th 2011, 07:43 PM
mathematicalbagpiper
Mean Value Property Implies "Volume" Mean Value Property
Suppose that $u$ has the mean value property in a domain $\Omega\subset\mathbb{R}^3$ i.e. over any sphere $S(r_0,\delta)$ with interior $B(r_0,\delta)$

$u(r_0)=\frac{1}{4\pi\delta^2}\int_{S(r_0,\delta)}u (r)d\sigma$

For every $\delta>0$ such that $\bar{B}(r_0,\delta)\subset\Omega$. Show that $u$ also has the "volume" mean value property:

$u(r_0)=\frac{1}{\frac{4}{3}\pi\delta^3}\int_{B(r_0 ,\delta)}u(r)dv$

For every ball $B(r_0,\delta)$ such that $\bar{B}(r_0,\delta)\subset\Omega$.

Somehow I think I have to use the fact that $u$ has the mean value property, but I'm not seeing how to do that. I tried comparing the the values in front of the integrands and that was inconclusive (depends on the choice of delta whether one is less than or greater than the other), so I don't know where else to go....
• Mar 16th 2011, 08:28 PM
Jose27
If you assume f is continous this might help you get an idea how to prove it (in case you know a harmonic function satisfies both of the above). This can be done even in the case f is only locally integrable. However I don't know if the result would be valid with no regularity on f.
• Mar 16th 2011, 08:39 PM
Tinyboss
Seems to me that knowing the integral on every sphere centered at r0 should allow you to compute the integral over a solid ball centered there, without further reference to the behavior of u.
• Mar 16th 2011, 09:13 PM
Jose27
Quote:

Originally Posted by Tinyboss
Seems to me that knowing the integral on every sphere centered at r0 should allow you to compute the integral over a solid ball centered there, without further reference to the behavior of u.

Well, surely one must ask some kind of integrability condition...