Section 4.3 The mean value property
A very useful property of harmonic functions is the mean value principle, which states that the value of a harmonic function at a point is equal to its average value over spheres or balls centred at that point.
Compared with several years ago, I have tried to de-emphasise the mean value property in favour of more maximum principle arguments.
Notation 4.9. Averages over balls and spheres.
We denote the volume of a ball \(B \subset \R^N\) by \(\abs B\) and the surface area by \(\abs{\partial B}\text{.}\) Obviously these depend only on the radius of the ball and the dimension \(N\text{.}\) The corresponding averages of a function \(f\) are given by
\begin{equation*}
\frac 1{\abs{B}} \int_B f\, dx
\qquad \text{and} \qquad
\frac 1{\abs{\partial B}}\int_{\partial B} f\, dS\text{.}
\end{equation*}
Another common notation (which does not typeset well in the html output) is to drop the factors of \(1/\abs B\) or \(1/\abs{\partial B}\) and instead put a dash through the integral sign.
Theorem 4.10. Mean value property.
If \(u \in C^2(\Omega)\) is harmonic then
\begin{align}
u(x)
\amp = \frac 1{\abs{\partial B_r}}\int_{\partial B_r(x)} u\, dS \tag{4.11}\\
\amp = \frac 1{\abs{B_r}} \int_{B_r(x)} u(y)\, dy \tag{4.12}
\end{align}
for each closed ball \(\overline{B_r(x)} \subseteq \Omega\text{.}\)
Proof.
The idea of the proof is to fix \(x\text{,}\) consider the average over \(\partial B_r(x)\) as a function of \(r\text{,}\) and show that that the derivative of this function is zero.
Changing variables to switch to an \(r\)-independent domain of integration, we find that
\begin{align*}
\phi (r) \amp:= \frac 1{\abs{\partial B_r}} \int_{\partial B_r(x)}u(y)\, dS(y)\\
\amp= \frac 1{\abs{\partial B_1}}\int_{\partial B_1(0)}u(x+rz)\, dS(z).
\end{align*}
Note the Jacobian of the transformation cancels thanks since we are working with averages. Differentiating inside the integral gives
\begin{align*}
\phi'(r) \amp= \frac 1{\abs{\partial B_1}} \int_{\partial B_1(0)}\nabla u(x+rz)\cdot z\, dS(z)\\
\amp= \frac 1{\abs{\partial B_r}} \int_{\partial B_r(x)}\nabla u(y)\cdot \Big( \frac{y-x}{r} \Big)\, dS(y)\\
\amp= \frac{1}{\abs{\partial B_r}} \int_{B_r(x)} \nabla \cdot (\nabla u(y))\, dy,
\end{align*}
where in the last step we have used the Divergence theorem for balls. Thus
\begin{align*}
\phi'(r)
\amp= \frac{1}{\abs{\partial B_r}}\int_{B_r(x)} \Delta u\, dy
= 0
\end{align*}
since \(u\) is harmonic.
Therefore \(\phi\) is constant, and so to prove (4.11) it suffices to show
\begin{equation*}
\phi (0) = \lim_{t \to 0} \phi(t)
= \lim_{t \to 0} \frac 1{\abs{\partial B_t}} \int_{\partial B_t(x)}u\, dS
= u(x),
\end{equation*}
which follows from the continuity of \(u\text{.}\) Indeed, as \(t \to 0\) we have
\begin{align*}
\left|\frac 1{\abs{\partial B_t}} \int_{\partial B_t(x)}u\, dS -u(x)\right|
\amp\le \frac 1{\abs{\partial B_t}}\int_{\partial B_t(x)}\abs{u(y)-u(x)}\, dS(y) \\
\amp\le \sup_{y\in \overline{B_t(x)}} \abs{u(y)-u(x)} \to 0.
\end{align*}
To prove (4.12), we integrate using Theorem 2.51:
\begin{align*}
\int_{B_r(x)}u\, dy \amp= \int_0^r \bigg( \int_{\partial B_\rho(x)}u\, dS \bigg)d\rho\\
\amp= u(x) \int_0^r \bigg( \int_{\partial B_\rho(x)}dS \bigg)d\rho \\
\amp= u(x) \int_{B_r(x)}\, dy,
\end{align*}
where in the second step we have used (4.11). Re-arranging gives the result.
The mean value property (4.11) in fact characterises harmonic functions in the following sense. Note that here we only require \(u\) to be continuous!
Theorem 4.11. Converse to mean value property.
Let \(u \in C^0(\Omega)\text{,}\) and suppose that for each \(x \in \Omega\) and \(\varepsilon \gt 0\text{,}\) there exists \(0 \lt r \lt \varepsilon\) such that
\begin{equation*}
u(x)= \frac 1{\abs{\partial B_r}}\int_{\partial B_r(x)} u\, dS\text{.}
\end{equation*}
Then \(u \in C^2(\Omega)\) and is harmonic.
Proof.
Consider a ball \(\overline{B_R(x_0)} \subset \Omega\text{,}\) and let \(\bar u\) be the solution to the Dirichlet problem
\begin{equation*}
\left\{
\begin{alignedat}{2}
\Delta \bar u \amp= 0 \amp\quad\amp \ina B_R(x_0) \\
\bar u \amp= u \amp\quad\amp \ona \partial B_R(x_0)
\end{alignedat}
\right.
\end{equation*}
guaranteed by Theorem 4.8. We claim that \(u = \bar u\) on \(B_R(x_0)\text{,}\) which will complete the proof.
Suppose for the sake of contradiction that \(\bar u-u\) is positive at some point in \(\overline{B_R(x_0)}\text{,}\) and let \(E\) be the subset of \(\overline{B_R(x_0)}\) where \(\bar u-u\) achieves its maximum \(M \gt
0\text{.}\) Since \(E\) is compact, using Theorem 2.40 one can show that there exists a point \(x \in E\) which is furthest from \(x_0\text{.}\) Certainly \(x \notin \partial B_R(x_0)\text{,}\) since \(\bar u-u=0\) there. By assumption, we can therefore find \(r \gt 0\) such that \(B_r(x) \subset B_R(x_0)\) and
\begin{equation*}
u(x)= \frac 1{\abs{\partial B_r}}\int_{\partial B_r(x)} u\, dS\text{.}
\end{equation*}
Since \(\bar u\) is harmonic, applying Theorem 4.10 we conclude that
\begin{equation}
M = (\bar u-u)(x)= \frac 1{\abs{\partial B_r}}\int_{\partial B_r(x)} (\bar u-u)\, dS\text{.}\tag{4.13}
\end{equation}
But this is a contradiction: By our choice of \(x\) we have \((\bar
u-u)(y) \le (\bar u-u)(x)\) for all \(y \in \partial B_r(x)\text{.}\) Moreover the inequality is strict on a nonempty subset of \(\partial B_r(x)\text{,}\) namely the points in \(\partial B_r(x)\) which are still further from \(x_0\) than \(x\) is. Thus \(\bar u-u \le 0\) on \(B_R(x_0)\text{.}\) A similar argument shows that \(\bar u-u \ge 0\) on \(B_R(x_0)\text{.}\)
Exercises Exercises
1. (PS7) Reflecting harmonic functions.
Let \(B=B_1(0)\) and let \(B^+ = B \cap \{x_N \gt 0\}\) be its upper half. Suppose that \(u \in C^2(B^+) \cap C^0(\overline{B^+})\) is harmonic and that \(u=0\) on \(\partial B^+ \cap \{x_N=0\}\text{.}\) Define the function \(v\) on \(B\) by
\begin{align*}
v(x) =
\begin{cases}
u(x) \amp x_N \ge 0, \\
-u(x_1,\ldots,x_{N-1},-x_N) \amp x_N \lt 0.
\end{cases}
\end{align*}
Hint.
By Theorem 4.11, it is enough to show that, for every \(x \in
\Omega\text{,}\) \(u\) satisfies the mean-value property on \(B_r(x)\) for \(r\gt0\) sufficiently small. If \(x_N \gt 0\) or \(x_N \lt 0\text{,}\) pick \(r\) small enough that you can apply the mean-value property for \(u\) (or its reflection). If \(x_N = 0\text{,}\) use symmetry to argue that the average has to be \(0=u(x)\text{.}\)
Solution.
Let \(x \in B\text{.}\) If \(x_N \gt 0\text{,}\) then we can pick \(\varepsilon \gt 0\) small enough that \(B_\varepsilon(x) \subseteq B^+\text{.}\) Since \(v|_{B^+} = u\) is harmonic, we have
\begin{equation*}
v(x)
= \frac 1{\abs{\partial B_r}}\int_{\partial B_r(x)} v\, dS
\end{equation*}
for all \(0 \lt r \lt \varepsilon\) by the mean value property.
If \(x_N \lt 0\text{,}\) then instead we pick \(\varepsilon \gt 0\) small enough that \(B_\varepsilon(x) \subseteq B^-\text{,}\) the bottom half of the ball. One can check that
\begin{gather*}
v|_{B^-}(x) = -u(x_1,\ldots,x_{N-1},-x_n)
\end{gather*}
is harmonic on this set in several ways. We could simply calculate \(\Delta v\) using the chain rule (or recall Exercise 2.6.2), or we could change variables inside the integral to show that \(v|_{B^-}\) satisfies the mean value property.
It remains to consider the case where \(x_N = 0\text{.}\) Pick \(\varepsilon \gt 0\) small enough that \(B_\varepsilon(x) \subseteq B\text{.}\) Using a change of variables we find that, for any \(0 \lt r \lt \varepsilon\text{,}\)
\begin{align*}
\int_{\partial B_r(x)} v\, dS
\amp=
\int_{\partial B_r(x) \cap \{x_N \gt 0\}} v\, dS
+\int_{\partial B_r(x) \cap \{x_N \lt 0\}} v\, dS\\
\amp=
\int_{\partial B_r(x) \cap \{x_N \gt 0\}} u\, dS
-\int_{\partial B_r(x) \cap \{x_N \gt 0\}} u\, dS\\
\amp= 0 = u(x) = v(x).
\end{align*}
We can now can now conclude by applying Theorem 4.11.
Comment.
Note that the “for each \(x \in \Omega\)” in Theorem 4.11 is doing a lot of work. If we only know
\begin{gather*}
w(x)= \frac 1{\abs{\partial B_r}}\int_{\partial B_r(x)} w\, dS
\end{gather*}
for balls centred at a single \(x \in \Omega\) (and all sufficiently small \(r \gt 0\)), then we can conclude nothing about the function \(w\) whatsoever. On past exams students have sometimes gotten confused about this, especially when \(\Omega\) is itself a ball (or related to a ball).
