# The average of a function over a sphere tends to its value at the center, as the radius goes to zero

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.

• 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. – ling Oct 11 '19 at 2:48
• Thank you, but I still can't get it. – Steve Oct 11 '19 at 2:57

## 1 Answer

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}