$Prove that $\lim_{\epsilon \rightarrow 0} \int_{\partial B(x,\epsilon)} \frac{\partial \Phi}{\partial \nu}(y)f(x-y)dy=f(x)$$

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$Prove that $\lim_{\epsilon \rightarrow 0} \int_{\partial B(x,\epsilon)} \frac{\partial \Phi}{\partial \nu}(y)f(x-y)dy=f(x)$$

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$$\lim_{\epsilon \rightarrow 0} \int_{\partial B(x,\epsilon)} \frac{\partial \Phi}{\partial \nu}(y)f(x-y)dy=f(x)$$

where $\phi$ is the fundamental solution of the laplacian in dimesnion $n\geq 3$ and $f$ is a continuous function.

Partial Differential Equations Parabolic PDEs Limits

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Note that for the fundamental solution of the Laplacian $\Phi$ we have
\[\nabla \Phi=\frac{-1}{n \alpha (n)}\frac{y}{|y|^n},\]
where $\alpha(n)$ is the volume of the unit ball in dimesnion $n$. Also
\[\nu=\frac{y}{|y|}=\frac{y}{\epsilon}.\]
So
\[\int_{\partial B(x,\epsilon)}\frac{\partial \Phi}{\partial \nu}(y)f(x-y)dS(y)=\frac{-1}{n \alpha (n)} \int_{\partial B(x,\epsilon)}\frac{y}{|y|^n}\cdot \frac{y}{|y|}f(x-y)dS(y)\]
\[=\frac{-1}{n \alpha (n)} \int_{\partial B(x,\epsilon)}\frac{|y|^2}{|y|^n}f(x-y)dS(y)=\frac{-1}{n \alpha (n)\epsilon^{n-1}} \int_{\partial B(x,\epsilon)}f(x-y)dS(y) \rightarrow -f(x)\]
as $\epsilon \rightarrow 0$, as the above is just the negative of the average integral of $f(x-y)$ over $\partial B(x, \epsilon)$.

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$Prove that $\lim_{\epsilon \rightarrow 0} \int_{\partial B(x,\epsilon)} \frac{\partial \Phi}{\partial \nu}(y)f(x-y)dy=f(x)$$

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