# Proof for Mean Value Property using a specific limit

I am trying to prove the following:

Suppose $$u \in C^2(\Omega)$$. For some $$x \in \Omega$$ we have that

\begin{align} \Delta u(x) = \lim_{r \to 0} \frac{2n}{r^2} \left[ \frac{1}{w_n} \int_{\partial B_{1}} u(x + ry)\mathrm{d}S_y - u(x) \right]. \end{align}

If $$u$$ is harmonic, then $$u$$ has the Mean Value Property.

I believe that we basically need to see this limit in some smart way, but I still couldnt do that. Can someone give me a hint on this? But it must use the limit above.

Version 1 Maybe stepping back one step actually make things clearer. You want to proof that if $$\Delta u(x) \equiv 0$$, then for any $$x \in \Omega$$ and $$r > 0$$ we have $$u(x) = \frac{1}{|\partial B_r(x)|} \int_{\partial B_r(x)} u(y) dS(y).$$
Now, the right hand side turns with transformation of variables into $$\frac{1}{\omega_n r^{n-1}} \int_{\partial B_1} u(x+ry)dS(ry) = \frac{1}{\omega_n} \int_{\partial B_1} u(x+ry) dS(y).$$ Now lets differentiate the right hand side with respect to $$r$$ (where by dominated convergence we can differentiate under the integral): $$\frac{d}{dr}\;\frac{1}{\omega_n} \int_{\partial B_1} u(x+ry) dS(y) = \frac{1}{\omega_n} \int_{\partial B_1} \frac{d}{dr}\;u(x+ry) dS(y) = \frac{1}{\omega_n} \int_{\partial B_1} (\nabla u)(x+ry)\cdot y\, dS(y)$$ Now, changing the varibales back again, we get: $$\frac{1}{\omega_n} \int_{\partial B_1} (\nabla u)(x+ry)\cdot y\, dS(y) = \frac{1}{\omega_n r^{n-1}} \int_{\partial B_r(x)} \nabla u(x)\cdot \frac{(x-y)}{r}\, dS(y)$$ which with the divergence theorem and the fact that $$u$$ is harmonic turns into $$\frac{d}{dr}\;\frac{1}{\omega_n} \int_{\partial B_1} u(x+ry) dS(y) = \frac{1}{\omega_n r^{n-1}} \int_{\partial B_r(x)} \Delta u dy = 0.$$
From this we can infer that the mean value representation actually is constant for $$r$$. So, it is left to show that the right hand side in the very first equation tends to $$u(x)$$ when radius $$r$$ shrinks (from above) to zero. But this is also not to difficult to see.
I am aware that I didn't use your cited formula and to be honest I am a bit baffled by it. But isn't it just this argument: Suppose $$u$$ is harmonic then $$\Delta u(x) \equiv 0$$. Now, this means we have $$0 = \lim_{r\to 0} \frac{2n}{r^2}\left(\frac{1}{\omega_n} \int_{\partial B_1} u(x+ry)dS(y) - u(x)\right).$$ But for this to be zero we need $$\lim_{r\to 0} \frac{1}{r}\left(\frac{1}{\omega_n} \int_{\partial B_1} u(x+ry)dS(y) - u(x)\right) = 0$$ and $$\lim_{r\to 0} \left(\frac{1}{\omega_n} \int_{\partial B_1} u(x+ry)dS(y) - u(x)\right) = 0$$ to be zero and now that I typed it, I think that is actually the same argument I used :). Because that is the statement.