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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..

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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 $U=\{y:\epsilon<|y-x|<r\}$. Then the boundary integral is $$\int_{\partial U}\left(v\frac{\partial u}{\partial n}-u\frac{\partial v}{\partial n}\right) =-(n-2)\,r^{1-n}\int_{|y-x|=r} u(y) -\int_{|y-x|=\epsilon} \left( v\frac{\partial u}{\partial n} - (n-2)\,\epsilon^{1-n}u(y) \right) \\ \to (n-2)\left\{|\mathbb S^{n-1}| u(x)-r^{1-n}\int_{|y-x|=r} u(y)\right\} \qquad \text{as }\epsilon\to 0$$ This is encouraging. By the identity, the above integral is equal to $$\int_U (v\Delta u-u\Delta v)=\int_U v\;\Delta u \approx \Delta u(x)\, \int_U v$$ where $\approx $ is justified by the continuity of $\Delta u$. As $\epsilon\to 0$, the last integral converges to $$\int_{0}^r (t-r^{2-n}t^{n-1})\,dt = \frac{r^2}{2}-\frac{r^2}{n}=\frac{n-2}{2n}r^2.$$ Hurray! $\Box$

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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Using symmetry and Landau $o$-notation keeps it from being too boring (I hope) :-) – robjohn Sep 13 '12 at 14:30

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 $$ \lim_{r\to0}\frac{2n}{r^2}\left(u(x)-\frac1{|\partial B_r(x)|}\int_{\partial B_r(x)}u(y)\,\mathrm{d}\sigma_y\right)=-\Delta u(x)\tag{3} $$

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