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From PDE Evans, 2nd edition, pages 25-26.

THEOREM 2 (Mean Value Formulas for Laplace's equation). If $u \in C^2(U)$ is harmonic, then $$u(x)=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{\partial B(x,r)}u \, dS=\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits} \avint_{B(x,r)} u \, dy$$ for each ball $B(x,r)\subset U$.

Proof. 1. Set $$\phi(r) := \avint_{\partial B(x,r)} u(y) \, dS(y) = \avint_{\partial B(0,1)} u(x+rz) \, dS(z).$$ Then $$\phi'(r)=\avint_{\partial B(0,1)} Du(x+rz) \cdot z \, dS(z)$$ and consequently, using Green's formulas from §C.2, we compute \begin{align}\phi'(r) &= \avint_{\partial B(x,r)} Du(y) \cdot \frac{u-x}r dS(y) \\ &= \avint_{\partial B(x,r)} \frac{\partial u}{\partial \nu} dS(y) \\ &= \frac rn \avint_{B(x,r)} \Delta u(y) \, dy = 0.\end{align}

Hence $\phi$ is constant, and so $$\phi(r)=\lim_{t \to 0} \phi(t) = \lim_{t \to 0} \avint_{\partial B(x,t)} u(y) \, dS(y) = u(x).$$

The proof continues on but I'm stopping here for this question I need to ask:

Can anyone explain the last line? I understand that $\phi$ is constant since we established that $\phi'(r)=0$ for all harmonic functions $u$. Hence, this is why we can say $\phi(r)=\lim_{t\to 0} \phi(t)$. But how can we establish the last equality, that is, $$\lim_{t \to 0} \avint_{\partial B(x,t)} u(y) \, dS(y)=u(x)?$$

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Cookie
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1 Answers1

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$\def\avint{\mathop{\,\rlap{-}\!\!\int}\nolimits}$

$$ \avint_{\partial B(x,t)} \, dS(y) = 1 $$ and therefore

$$ \bigl| u(x) - \avint_{\partial B(x,t)} u(y) \, dS(y)\bigr| = \bigl| \avint_{\partial B(x,t)} (u(x) - u(y)) \, dS(y) \bigr| \le \avint_{\partial B(x,t)} |u(x) - u(y)| \, dS(y) \\ \le \max_{y \in \partial B(x,t)} \{ |u(x) - u(y)| \} \quad . $$

This converges to zero for $t \to 0$ because $u$ is continuous at $x$.

Martin R
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  • Now, I get that $u$ is continuous, but if we are dealing with $x,y$, does $u$ also need to be uniformly continuous? I still can't understand the justification in the last line of your answer. – Cookie Dec 22 '14 at 21:05
  • @dragon: $u$ is uniformly continuous on each compact set that is not needed here. For fixed $x$ and $\varepsilon > 0$ there is a $\delta > 0$ such that $|u(x)−u(y)| < \varepsilon $ for all $y$ with $|y - x| < \delta$. If follows that the last expression is less than $\varepsilon$ for $ 0 < t < \delta$. – Martin R Dec 22 '14 at 21:08
  • @dragon: I do not know what your $\alpha(n)$ is. But anyway, I am sure that the definition is equivalent to $\int f , dS $ divided by $\int 1 , dS$. The average of a constant function is always defined to be that constant. – Martin R Dec 22 '14 at 21:13
  • Maybe that definition is equivalent to what you stated. Evans seems to present complicated notation all the time in the book; I too have no idea what $\alpha(n)$ means, but I think it probably has something to do with volume of a ball in higher dimensions. – Cookie Dec 22 '14 at 21:14
  • Do you know lebesgue points? If yes, this is just the result of Lebesgue points. – spatially Dec 23 '14 at 02:49
  • Lebesgue points? Not familiar with that term, though I have taken a course in introduction to measure theory. I looked the term up on Wikipedia, and they had this: $$\lim_{r\rightarrow 0^+}\frac{1}{|B(x,r)|}\int_{B(x,r)} !|f(y)-f(x)|,\mathrm{d}y=0.$$ The result by the author is probably derived algebraicallly from this as you already said. – Cookie Dec 23 '14 at 05:58
  • @ Martin: I do not understand why is the surface area of $|u(x)-u(y)|\le\max_{y\in\partial B(x,t)}{|u(x)-u(y)|}$. Is it because the average should roughly less than the maximum of the function? Could you please explain further? – math101 Jun 02 '15 at 13:42
  • @math101: Yes. Formally: $\def\avint{\mathop{,\rlap{-}!!\int}\nolimits}$ $\avint_{\partial B(x,t)} |u(x) - u(y)| , dS(y) \le \avint_{\partial B(x,t)} \max_{z \in \partial B(x,t)} { |u(x) - u(z)| } , dS(y) = \max_{z \in \partial B(x,t)} { |u(x) - u(z)| } \avint_{\partial B(x,t)} 1 , dS(y) = \max_{z \in \partial B(x,t)} { |u(x) - u(z)| }$ – Martin R Jun 02 '15 at 17:16
  • In Evans, $\alpha(n) =$ volume of unit ball $B(0,1)$ in $\mathbb R^n$ and $n\alpha(n) = $ surface area of the unit sphere $\partial B(0,1)$ in $\mathbb R^n$. – Axion004 Aug 03 '22 at 22:40