Laplacian of a function as the limit over the ball

Question

I've been trying to prove the following identity $$ \lim_{r \to 0}\frac{1}{r^2}\left(\frac{1}{|B_{r}(x)|}\int_{B_{r}(x)} u(y)dy - u(x)\right) = \frac{1}{2(N+2)} \Delta u(x), $$ where this is done in $\mathbb{R}^{N}$ with $u \in C^2(\Omega)$ for the bounded domain we work in.

I've seen the following post, where a similar integral is computer over the boundary instead. I tried to replace $$ \int_{B_{r}(x)} u(y)dy = \int_{0}^{r} \int_{\partial B_{r}(x)} u(\sigma)d\sigma dr $$ to then replicate the same Taylor approximation, but now I'm stuck with the constants not cancelling out properly. Is there perhaps an easier way, or am I missing a step?

Answer 1

The post you linked shows that $$ \frac 1 {\vert \partial B_r \vert}\int_{\partial B_r(x)} u(y) \, d\mathcal H^{n-1}_x= u(x)+ \frac 1 {2n} \Delta u(x) r^2 + o(r^2) \qquad \text{as } r\to 0^+.$$ Here I'm using little-o notation, that is, $f(r)=o(g(r))$ if $\lim_{r\to 0}f(r)/g(r)=0$. Since $\vert \partial B_r \vert = n \omega_n r^{n-1} $ where $\omega_n = \vert B_1 \vert$, we have that \begin{align*} \frac 1 {\vert B_r \vert}\int_{\partial B_r(x)} u(y) \, d x &= \frac 1 {\omega_n r^n} \int_0^r \int_{\partial B_\rho(x)} u(y) \, d \mathcal H^{n-1}_y \, d\rho \\ &= \frac 1 {\omega_n r^n} \int_0^r n \omega_n \rho^{n-1}\big ( u(x)+ \frac 1 {2n} \Delta u(x) \rho^2 + o(\rho^2)\big ) \, d\rho \\ &= \frac 1 {r^n}\bigg [ u(x) \int_0^r n\rho^{n-1} \, d \rho + \frac 12 \Delta u(x) \int_0^r \rho^{n+1} \, d\rho+\int_0^r o(r^{n+1}) \, d \rho\bigg]\\ &=\frac 1 {r^n}\bigg [ u(x)r^n + \frac 1{2(n+2)} \Delta u(x)r^{n+2} +o(r^{n+2})\bigg]\\ &= u(x) + \frac 1{2(n+2)} \Delta u(x)r^2 +o(r^2). \end{align*}