$\textbf{ Statement of Theorem:}$ If $u \in C^2(U)$ is harmonic, then $$u(x) = \frac{1}{m(\partial B(x,r))}\int_{\partial B(x,r)} u dS = \frac{1}{m(B(x,r))}\int_{B(x,r)} u dy$$ for each $B(x,r) \subset U$.
$\textbf{Proof:}$ Let $$ \phi(r) := \frac{1}{m(\partial B(x,r))} u(y) d S(y) = \frac{1}{m(B(0,1))} \int u(x+rz) dS(z)$$
Then, $$ \phi'(r) = \frac{1}{m(\partial B(0,1))} \int_{\partial B(0,1)} Du(x + rz) \cdot z dS(z)$$
Using Green's formula we compute, $$ \phi'(r) = \frac{1}{m(\partial B(x,r))} \int_{\partial B(x,r)} D u(y) \cdot \frac{y-x}{r} dS(y)$$
$$\frac{1}{m(\partial B(x,r))} \int_{ \partial B(x,r)} \frac{\partial u}{\partial \nu} dS(y) \hspace{10mm} (*)$$ where $\nu$ is the outward facing normal vector.
$$ = \frac{r}{n} \frac{1}{m(B(x,r))} \int_{B(x,r)} \Delta u(y) dy, \hspace{10mm} (**)$$
$\textbf{Question:}$ What happened between steps $(*)$ and $(**)$. I can see we used the surface area of the ball, however I am not sure how $\frac{\partial u}{\partial \nu}$ turned in to the Laplacian, $\Delta u$.
Best Answer
Divergence Theorem : $$ \int_U {\rm div}\ \nabla f =\int_{\partial U} \nabla f\cdot \nu $$ where $\nu$ is an unit outnormal to $\partial U$