Here is the proof
I don't really understand about the part beginning using Green's formula. How can $Du(y)$ become $du/dv$. Is it using the directional derivative formula? Also, how can you get/pull out the $r/n$? It seems to me he is using divergent theorem here.
Best Answer
A key point here, which might be causing some confusion, is that the dash through the integral sign indicates that we are computing a mean value. It's different than just a regular integral sign.
$B(x,r)$ is the ball of radius $r$ centered at $x$. If $y \in \partial B(x,r)$, then $n = \frac{y - x}{r}$ is the outward pointing unit normal vector to $\partial B(x,r)$ at $y$. The directional derivative of $u$ at $y$ in the direction $n$ is equal to $\nabla u(y) \cdot n$.
From the divergence theorem, \begin{align} \int_{\partial B(x,r)} \nabla u(y) \cdot n \, dS(y) &= \int_{B(x,r)} \text{div } \nabla u(y) \, dy \\ &= \int_{B(x,r)} \Delta u(y) \, dy. \end{align}
Let $c$ be the measure of the unit ball in $\mathbb R^n$. The measure of $B(x,r)$ is $c r^n$. The "surface area" of $\partial B(x,r)$ is $c n r^{n-1}$. (For example, the volume of a ball in $R^3$ is $(4/3) \pi r^3$, and the surface area of a sphere is $4 \pi r^2$.)
From the above equation, it follows that \begin{equation} \frac{1}{cn r^{n-1}}\int_{\partial B(x,r)} \nabla u(y) \cdot n \, dS(y) = \frac{r}{n} \frac{1}{cr^n} \int_{B(x,r)} \Delta u(y) \, dy. \end{equation} This is the equation at issue in your question. You can see now where the factor of $\frac{r}{n}$ comes from.