I'm trying to prove that if $\Omega$ is an open subset of $\mathbb{R}^n$ and $u$ a $C^2$ function then $\lim_{r\to 0}\frac{2n}{r^2}\left(u(x)-\frac{1}{|\partial B_r(x)|}\int_{\partial B_r(x)}u(y)d\sigma_y\right)=-\Delta u(x).$ I tried to use the representation formula on balls, but nothing came out of it..
Limit involving the laplacian
2 Answers
One can get this by writing out Taylor's expansion, but I'll try Green's second identity just to see what happens. Assume $n\ge 3$. Let $v(y)=|y-x|^{2-n}-r^{2-n}$ (noting that this is a fundamental solution of $\Delta$) and $U=\{y:\epsilon<|y-x|
Same idea works in two dimensions, but with $v(y)=\log \frac{|y-x|}{r}$. Again, this is not the quickest proof, but I did not want to type out the boring Taylor's approximation argument.
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0Using symmetry and Landau $o$-notation keeps it from being too boring (I hope) :-) – 2012-09-13
Taking the mean value of the Taylor expansion over $\partial B_r(x)$ is a simple way to show this. $ \begin{align} &\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)}(u(y)-u(x))\,\mathrm{d}\sigma_y\\ &=\color{#C00000}{\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)}\left(\sum_{k=1}^nu_k(x)(y_k-x_k)+\sum_{j< k}u_{jk}(x)(y_j-x_j)(y_k-x_k)\right)\,\mathrm{d}\sigma_y}\\ &+\color{#00A000}{\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)}\frac12\sum_{k=1}^nu_{kk}(x)(y_k-x_k)^2\,\mathrm{d}\sigma_y}\\ &+\color{#0000FF}{\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)}o(|y-x|^2)\,\mathrm{d}\sigma_y}\tag{1} \end{align} $ By cancellation due to the symmetry of $\partial B_r(x)$, the red integral is $0$. It is simple to see that the blue integral is $o(r^2)$. All that is left is the green integral.
By symmetry, the mean value of $(y_k-x_k)^2$ over $\partial B_r(x)$ is independent of $k$, and the mean value of $\sum\limits_{k=1}^n(y_k-x_k)^2$ over $\partial B_r(x)$ is $r^2$. Thus, the mean value of $(y_k-x_k)^2$ over $\partial B_r(x)$ is $r^2/n$.
Thus, $ \begin{align} &\lim_{r\to0}\frac{2n}{r^2}\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)}(u(y)-u(x))\,\mathrm{d}\sigma_y\\ &=\lim_{r\to0}\frac{2n}{r^2}\color{#C00000}{0}\\ &+\lim_{r\to0}\frac{2n}{r^2}\color{#00A000}{\frac{r^2}{n}\frac12 \sum_{k=1}^nu_{kk}}\\ &+\lim_{r\to0}\frac{2n}{r^2}\color{#0000FF}{o(r^2)}\\ &=\sum_{k=1}^nu_{kk}\tag{2} \end{align} $ Rearranging $(2)$ yields $ \lim_{r\to0}\frac{2n}{r^2}\left(u(x)-\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)}u(y)\,\mathrm{d}\sigma_y\right)=-\Delta u(x)\tag{3} $