When reading some papers on PDEs, the following shows up several times:
For a $C^{\infty}$ function $u$, $\frac{\int_{\mathbb{S}_r(x)}u-u(x)}{r^2}$ converges to $1/2n\Delta u(x)$ uniformly on compact sets as $r\to 0$.
Here $\mathbb{S}_r(x)$ is the sphere of radius $r$ centred at $x$, and $n$ is the dimension of the Euclidean space.
I am wondering how we can prove this. I think we might use Taylor's approximation, but ended up with something of the form \begin{equation} \langle D^2u(x)(y-x),(y-x)\rangle/\lvert y-x\rvert^2,\end{equation} and we need to show this goes to $1/n\Delta u(x)$. So the difference is that $\Delta u$ is a diagonal matrix while the Hessian is not necessarily so.
I guess in the end we might be able to show the significant part is the diagonal, but can someone give a hint on how to do this?
Thanks!
Hint: Let $$A(f,x,r)=\frac{1}{|B(x,t)|}\int_{B(x,r)}f,\ \mbox{and}\ B(f,x,r)=\frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}f,$$
where $|\cdot|$ is Lebesgue measure. Green identity implies that $$\int_{B(x,r)}\Delta f=\int_{\partial B(x,r)}\frac{\partial f}{\partial \eta}=r^{N-1}\frac{d}{dr}\int_{\partial B(0,1)}f(x+r\sigma)d\sigma,$$
whence $$r^{N} A(\Delta f,x,r)=Nr^{N-1}\frac{d}{dr}B(f,x,r)\tag{1}.$$
Can you conclude from here?