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, $$\lim_{t \to 0} \avint_{\partial B(x,t)} u(y) \, dS(y)=u(x)?$$


1 Answer 1



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

  • $\begingroup$ 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. $\endgroup$
    – Cookie
    Dec 22, 2014 at 21:05
  • $\begingroup$ @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$. $\endgroup$
    – Martin R
    Dec 22, 2014 at 21:08
  • $\begingroup$ @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. $\endgroup$
    – Martin R
    Dec 22, 2014 at 21:13
  • $\begingroup$ 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. $\endgroup$
    – Cookie
    Dec 22, 2014 at 21:14
  • $\begingroup$ Do you know lebesgue points? If yes, this is just the result of Lebesgue points. $\endgroup$
    – spatially
    Dec 23, 2014 at 2:49

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