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This question arose when I investigated into the proof of the mean-value property of a harmonic function. Let $u$ be a harmonic function of class $C^2$ defined on an open set in $\mathbb{R}^n$. The mean-value property asserts that $$u(x)=\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} u\mathrm{d}S,$$ where $\alpha(n)$ denotes the volume of a unit ball in $\mathbb{R}^n$ and $B(x,r)$ is a ball of center $x$ and radius $r$ that is contained in $U$. To prove, the author invoked the fact that $$\lim_{t\to 0^+}\frac{1}{n\alpha(n)t^{n-1}}\int_{\partial B(x,t)} u\mathrm{d}S=u(x).$$ Intuitively, this is absolutely right. But how do I prove it rigorously? Thanks.

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    $\begingroup$ Noticed that $\frac{1}{n\alpha(n) r^{n-1}}\int_{\partial B(x,r)} dS=1$ and u is continuous at point x, then you can use $\epsilon-\delta$ definition to finish the proof. $\endgroup$ – ling Oct 11 '19 at 2:48
  • $\begingroup$ Thank you, but I still can't get it. $\endgroup$ – Steve Oct 11 '19 at 2:57
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Let $\epsilon > 0$. Then there exists $\delta > 0$ such that $|u(x)-u(y)|<\epsilon$ for all $y\in B(x,\delta)$. So, for $r<\delta$, \begin{align} \left|u(x) - \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)} u\,\mathrm{d}S\right| &= \left|\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}(u(x)-u)\,\mathrm{d}S\right|\\ &\le \frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}|u(x)-u|\,\mathrm{d}S\\ &< \epsilon. \end{align}

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