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I am trying to prove that if $h:\mathbb{R}^3\rightarrow \mathbb{R}$ is a smooth function which is zero outside some ball centered at the origin , $B_R(0)$, then

$h(0)=\frac{1}{4\pi}\int_{\mathbb{R}^3}\frac{1}{|x|}\triangle h(x)dx$

Now my idea was to see what happens to $\lim_{\epsilon\rightarrow 0}\frac{1}{4\pi}\int_{B_R(0)-B_{\epsilon(0)}}\frac{1}{|x|}h(x)dx$, and using green's theorem. I was able to get that in the end this will equal $\lim_{\epsilon\rightarrow 0}\frac{1}{4\pi}\int_{B_{\epsilon}(0)}\frac{\partial \frac{1}{|x|}}{\partial n}h(x)dx$. Now I would like to justify why this last integral converges to $h(0)$ but I am not sure if there is a simple way to do it, without using some change of coordinates and taylor expand $h(x)$ at the origin. Any help is appreciated. Thanks in advance.

Someone
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1 Answers1

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On the boundary of $\mathbb R^3 - B_\epsilon(0)$, we have : $$\frac{\partial}{\partial n} =- \frac{x}{\epsilon}\cdot\nabla$$ Therefore, $$\frac{\partial}{\partial n}\frac{1}{|x|} = -\frac{x}{\epsilon}\cdot \nabla\frac{1}{\sqrt{x\cdot x}} = \frac{x}{\epsilon}\cdot \frac{x}{(x\cdot x)^{3/2}} = \epsilon^{-2}$$

Therefore, we have : $$\frac1{4\pi}\int_{B_\epsilon(0)}h(x)\frac{\partial}{\partial n}\frac{1}{|x|} \text d x = \frac{1}{\text{vol}(B_\epsilon(0))}\int_{B_\epsilon(0)}h(x)\text d x$$

This is the mean value of $h$ on the sphere of radius $\epsilon$ and, by continuity, it converges to $h(0)$ as $\epsilon \to 0$

SolubleFish
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  • The last limit is also a consequence of Lebesgue's differentiation theorem which requires $h$ to be only $L^1$ near 0. Nice answer though! – Nicolas Jun 11 '21 at 08:32
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    Thanks for pointing this out ! In the context of the OP, $h$ is a compactly supported smooth function, so I had no remorse using a result with stronger hypotheses than what was actually needed. – SolubleFish Jun 11 '21 at 08:35