Alternative proof of the converse of the mean value theorem. I have proven the converse of the MVT before, but I have not seen it done like this and I do not understand it.
Consider the limit:
$ \lim_{r \to 0} \frac{2n}{r^2}[\frac{1}{n \alpha (n)} \int_{\delta B(0,1)} u(x+r\omega)dS(\omega)-u(x)]$
In the above, $\alpha(n)$ represents the volume of the unit ball.
It says that you can use a Taylor expansion with remainder, and note that
$\int_{\delta B(0,1)}x_i=\int_{\delta B(0,1)}x_jx_k $ for $j\neq k$,
then it can be shown that this limit is equal to $ \Delta u(x)$, and then we conclude that the limit is $0$. I cannot begin to see how to reduce this limit to the laplacian of $u$. I am not very familiar with PDEs, but I am trying to learn. Can someone demonstrate how to manipulate this limit?
 A: 
Proposition: Suppose that $x_0\in \mathbb R^n$ and $u \in C^2(B_{2r}(x_0))$ for some $r>0$. Then $$\lim_{\rho \to 0^+} \frac1{\rho^2}\bigg \{  \frac1{\mathcal H^{n-1}(\partial B_\rho(x_0))}\int_{\partial B_\rho(x_0)} u(x) \, d\mathcal H^{n-1}_x -u(x_0) \bigg\}=\frac1{2n} \Delta u(x_0). $$
Proof . By Taylor's Theorem, $$u(x) = u(x_0) +Du(x_0) \cdot x + \frac12 x \cdot (D^2u(x_0)x) + o(\vert x \vert^2) \tag{$\ast$}$$ as $\vert x\vert \to 0$. Since $ x \mapsto D_iu(x_0) x_i$ is odd across the $x_i$-plane it follows that $$ \int_{\partial B_\rho(x_0)} D_iu(x_0) x_i \, d\mathcal H^{n-1}_x =0$$ for all $i=1,\dots n$. Similarly, $x \mapsto D_{ij}u(x_0)x_ix_j$ for $i\neq j$ is odd across the $x_i$-plane so $$\int_{\partial B_\rho(x_0)} D_{ij}u(x_0) x_ix_j \, d\mathcal H^{n-1}_x =0 $$ for all $i,j=1,\dots, n$, $i\neq j$. Integrating $(\ast)$ over $\partial B_r(x_0)$ gives \begin{align*}\frac1{\mathcal H^{n-1}(\partial B_\rho(x_0))}\int_{\partial B_\rho(x_0)} u(x) \, d\mathcal H^{n-1}_x &=u(x_0)+\frac12 \frac1{\mathcal H^{n-1}(\partial B_\rho(x_0))}\sum_{i=1}^n\int_{\partial B_\rho(x_0)} D_{ii}u(x_0)x_i^2 \, d\mathcal H^{n-1}_x +o (\rho^2). \\
&=u(x_0)+\frac12 \frac1{\mathcal H^{n-1}(\partial B_\rho(x_0))}\sum_{i=1}^nD_{ii}u(x_0)\int_{\partial B_\rho(x_0)} x_i^2 \, d\mathcal H^{n-1}_x +o (\rho^2).  \end{align*} Then, by symmetry, $$\int_{\partial B_\rho(x_0)} x_i^2 \, d\mathcal H^{n-1}_x = \frac1n\int_{\partial B_\rho(x_0)} \vert x \vert^2 \, d\mathcal H^{n-1}_x=\frac{\rho^2\mathcal H^{n-1}(\partial B_\rho(x_0))}{n}. $$ Thus, \begin{align*}
\frac1{\mathcal H^{n-1}(\partial B_\rho(x_0))}\int_{\partial B_\rho(x_0)} u(x) \, d\mathcal H^{n-1}_x &=u(x_0)+\frac{\rho^2}{2n}\sum_{i=1}^nD_{ii}u(x_0) +o (\rho^2)\\
& =u(x_0)+\frac{\rho^2}{2n}\Delta u(x_0) +o (\rho^2)
\end{align*} which gives the result. $\square$

The converse to the mean value properties say if $$ u(x_0)= \frac1{\mathcal H^{n-1}(\partial B_\rho(x_0))}\int_{\partial B_\rho(x_0)} u(x) \, d\mathcal H^{n-1}_x $$ for all $B_r(x_0) \subset \subset \Omega$ then $\Delta u=0$ in $\Omega$. This directly follows from the Proposition since $$\frac1{\mathcal H^{n-1}(\partial B_\rho(x_0))}\int_{\partial B_\rho(x_0)} u(x) \, d\mathcal H^{n-1}_x  -u(x_0)=0 $$for all $B_r(x_0) \subset \subset \Omega$.
