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Let $U\subset\mathbb{R}^n$ be an open set, $x\in U$ and $u\in C^2(U)$ a harmonic function. I would like know what is the theorem that is used to conclude that $$\lim_{r\to0}\int_{\partial B(x,r)}u(y)\;dS(y)=u(x).$$

This equality was taken of page 26 of PDE Evans book. The author gives no explanation about it. Maybe it's quite obvious, but I need help to understand it.

Thanks.

EDITED: Sorry. The correct equality is $$\lim_{r\to0} \left(\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)\;dS(y)\right)=u(x),$$ where $\alpha(n)$ is the volume of unit ball in $\mathbb{R}^n$.

Pedro
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    Evans was using the notation that there is a bar in the middle the integral sign, which means the integral is divided by the measure the set that it is integrated on. – Shuhao Cao Sep 01 '13 at 17:09

4 Answers4

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The equality is false as stated. You have to normalize the integral so that it becomes an average over $\partial B(x,r)$. That is, replace by $$\lim_{r\to 0} \frac{1}{n\omega_n r^{n-1}} \int_{\partial B(x,r)} u(y) dS(y) = u(x),$$ where $w_n = | B(0,1)|$ is the volume of the unit ball in $\mathbb{R}^n$.

This equality follows easily from Taylor expanding $u$ around $x$: $$ u(y) = u(x) +\nabla u(x) \cdot (y-x) + O(|y-x|^2).$$ Then \begin{multline} \frac{1}{n\omega_n r^{n-1}} \int_{\partial B(x,r)} u(y) dS(y) = \frac{1}{n\omega_n r^{n-1}} \int_{\partial B(x,r)} u(x) dS(y) \\ + \frac{1}{n\omega_n r^{n-1}} \nabla u(x) \cdot \int_{\partial B(x,r)} (y-x) dS(y) + \frac{1}{n\omega_n r^{n-1}} \int_{\partial B(x,r)} O(r^2) dS(y) \\ = u(x) + 0 + o(r). \end{multline} Send $r \to 0$ and $o(r) \to 0$, and we're done.

Notice that this does not use the fact that $u$ is harmonic. You don't actually need that for this part.

Glitch
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$$ \lim_{r\rightarrow 0} \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(y)\,dS(y) = u(x)+\lim_{r\rightarrow 0} \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(y)-u(x)\,dS(y) $$ and $$ \frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)} u(y)-u(x)\,dS(y)\le \frac{M}{|\partial B(x,r)|}\int_{\partial B(x,r)} r\; dS(y) \rightarrow 0 \mbox{ as } r\rightarrow 0 $$

Pocho la pantera
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It is a consequence of Lebesgue Besicovitch Theorem ( see Evans Gariepy Book, Theorem 1.7.1, pag. 43) and Divergence Theorem. Try to prove it with this suggestion.

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No no no!! There is a much more elementary proof using only calculus!

$$\frac{1}{\omega _n} \int_{\partial B(x,r)}u(y)\;dS(y)=\int_{\partial B(0,1)}u(x+ry)\;dS(y)$$ (the jacobian takes care of the r powers)

Take derivative with respect to $r$. Since we have made the integration domain independent of $r$, this becomes doable. After the derivative apply Integration by parts (maybe also switch back to radius $r$ ball at $x$) to see that the value is in fact zero!

This shows that the quantity is constant. But for very small $r$, by continuity of $u$, all $u(y)$'s are close to $u(x)$, thus the integral (the average) is close to $u(x)$ arbitrarily as well.

Behnam Esmayli
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