I learned the following theorem in Folland's Introduction to Partial Differential Equations(p.69 Chapter 2):

Suppose $u$ is harmonic on an open set $\Omega\subset{\mathbb R}^n$. If $x\in\Omega$ and $r>0$ is small enough so that $\overline{B_r(x)}\subset\Omega$, then $$u(x)=\frac{1}{r^{n-1}\omega_n}\int_{S_r(x)}u(y)d\sigma(y)=\frac{1}{\omega_n}\int_{S_1(0)}u(x+ry)d\sigma(y),$$ where $$\omega_n=\frac{2\pi^{n/2}}{\Gamma(n/2)}.$$

I found that I could not immediately reconstruct a proof for the theorem. A key point is that one needs to use the Green's identity, which is a basic property of harmonic functions. But I don't see any "clue" that how people actually come up with this theorem and such proof. (Maybe this is the common problem, at least for me, for most of the textbooks.) A curious search in Google returns nothing satisfactory to me. Since this is a basic property of harmonic functions, I am wondering that if one needs to know this history of harmonic functions in order to know this theorem well.

Here is my question:

  • Can any one here come up with a motivation of this theorem in PDE?

My second question may be more vague:

  • How can I approach the proof of this theorem more "naturally" instead of just remembering bunch of facts? (In the language of Polya, any heuristics here?)
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    $\begingroup$ Do you know Cauchy's integral theorem and about holomorphic functions? $\endgroup$ – t.b. Jul 16 '11 at 17:37
  • $\begingroup$ The motivation comes from the theory of complex analysis... a holomorphic function satisfies the property that its value at a point is given by the average of a contour around it. $\endgroup$ – mathmath8128 Jul 16 '11 at 18:10
  • $\begingroup$ Do you denote by $\omega_n$ the surface of the unit sphere in $\mathbb R^n$? I'd expect $\sigma_{n-1}$ for this. That's the number you need in the denominators. $\endgroup$ – Hendrik Vogt Jul 16 '11 at 18:14
  • $\begingroup$ See Tristan Needham's Visual Complex Analysis, Chapter 2, Section VIII, for an intuitive and elementary proof of the analogous result in complex function theory. $\endgroup$ – Viktor Blasjo Jul 16 '11 at 18:41
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    $\begingroup$ On a simply connected plane domain a real-valued harmonic function is the same thing as the real part of a holomorphic function. One way to prove the Cauchy integral theorem (and the Cauchy integral formula) is via Green's theorem. Read the wiki articles I gave to you closely. $\endgroup$ – t.b. Jul 17 '11 at 2:02

Here's a slightly-less-than-rigorous heuristic of an infinitesimal version of the mean value theorem, which provides a sort of motivation for the macroscopic version. (You can probably make this precise and intuitive using non-standard analysis, or you can just make it precise using $\epsilon-\delta$. Here I only give the intuition.)

Suppose we don't know anything about Harmonicity of a function, and we want to think about mean values. In particular, give a function $u$ and a point $x$, we want to think about how large $u(x)$ is compared with the average of its infinitesimal neighbours. So you consider

$$ \int_{B_{\delta}(x)} u(y) - u(x) dy $$

where $B_\delta$ is the ball of radius $\delta$. Of course, if $u$ is continuous, then as $\delta\to 0$ the above expression vanishes. So we need to renormalise by dividing by an appropriate factor of $\delta$. But forget that for the time being. Now, we can assume, by translation, that $x = 0$. And we assume $u$ is sufficiently smooth that we can Taylor expand $u(y)$

$$ u(y) = u(0) + \sum_{i = 1}^d y_i\partial_iu + \frac12 \sum_{i,j = 1}^d y_iy_j\partial^2_{ij} u + \ldots $$

Now, the $u(0) - u(0)$ term cancels out. The first order terms vanishes, because $\partial_iu(0)$ is just some constant, and you are integrating $y_i$, which is an odd function, over a symmetric domain. You also see that by the same token, the integral of $y_iy_j\partial^2_{ij}u(0)$ in $B_\delta$ is zero, if $i\neq j$. So you are left with that the lowest order term

$$ \int_{B_\delta}u(y) - u(0) dy \sim \int_{B_\delta} \sum_{i = 1}^d y_i^2 \partial_i^2 u(0) dy = \triangle u(0) \cdot \int_{B_\delta} y_1^2 dy$$

where, by spherical symmetry, the integral over the ball of $y_i^2$ is some fixed constant independent of $i$. So you have that the Laplacian of a function measures the infinitesimal deviation of a function from its mean.

Once you have the infinitesimal version, the macroscopic version should be something that suggests itself as possibly being true.

  • $\begingroup$ ‍@Willie: Your argument is very nice for showing the converse: If $u$ is not harmonic, then it doesn't satisfy the mean value property. However, I do not see how it can be used to prove what the OP wants. It does give a nice intuition, yes, but I fail to see how to make it rigorous. $\endgroup$ – Hendrik Vogt Jul 17 '11 at 14:06
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    $\begingroup$ @Hendrik: I was answering the part of the question about the "motivation" of the theorem. Given that harmonic functions are precisely those that satisfy (or even defined by) an infinitesimal version of the mean value property, it is reasonable to ask whether they also satisfy the same property in the large. I edited the post slightly to clarify. $\endgroup$ – Willie Wong Jul 17 '11 at 14:33
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    $\begingroup$ The motivation for the proof via Green's identity, however, likely follows from physical considerations drawing from electrostatics; though I am not familiar enough with the history to say that that is what Gauss had in mind. $\endgroup$ – Willie Wong Jul 17 '11 at 14:48
  • $\begingroup$ @Willie: +1. Hmm, here comes the non-standard analysis. I learned the topic from Terence Tao's blog post. I cannot fully understand this topic though, nice to know that it can be used here. $\endgroup$ – user9464 Jul 17 '11 at 16:42
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    $\begingroup$ @Willie: That's what I mean, it gives a nice intuition, and thus a good motivation. But in the parentheses you write "You can probably make this precise ...". I thought you mean that one can probably make a rigorous proof out of that, and if that was possible, I would be very much interested. (Incidentally, I don't think that non-standard analysis could help here any more than $\varepsilon$-$\delta$-arguments. Ping, @Jack.) $\endgroup$ – Hendrik Vogt Jul 17 '11 at 17:27

I can tell you my favourite proof of the mean value property, which I find more intuitive than the one via Green's identity. To make the proof rigorous, you need to know about integration over the manifold $O(n)$ of all orthogonal matrices, but you can just depict what I do as averaging over all orthogonal matrices. Let's take $x=0$ for simplicity. Heuristically, you'd expect that $$ \frac{1}{r^{n-1}\sigma_{n-1}}\int_{S_r(0)}u(y)\,d\sigma(y) = {\rlap{\;\bar{}}\int_{O(n)}} u(Az)\,d\sigma(A) =: f(z) $$ for $z\in\mathbb R^n$ with $|z|=r$. On the left hand side you take the average of $u(y)$ over all $y$ with $|y|=r$, on the right hand side it's the average of $u(Az)$ over all the orthogonal matrices $A$, which should be and is indeed the same.

Now you can differentiate the right hand side under the integral sign: take the Laplacian with respect to $z$. Then, due to $\Delta u=0$, you see that $f$ is harmonic. Moreover, $f$ is radially symmetric, i.e., $f(z)$ depends on $|z|$ only. Finally, use the fact that a radially symmetric harmonic function defined on all of $\mathbb R^n$ is constant. This yields $f(z) = f(0) = u(0)$, and we're done.

  • $\begingroup$ +1. This is a really nice proof, and a very nice way of thinking about it! (P.S. I think you mean "a radially symmetric harmonic function" defined on all of $\mathbb{R}^n$ is constant...) $\endgroup$ – Akhil Mathew Jul 17 '11 at 1:05
  • $\begingroup$ @Akhil: Glad that you like it. And thanks a lot for pointing out that omission! $\endgroup$ – Hendrik Vogt Jul 17 '11 at 14:11
  • $\begingroup$ @Hendrik: +1. I don't know much about integration over the manifold though, it's nice to know that we can think it in this way. Any references for your proof? $\endgroup$ – user9464 Jul 17 '11 at 16:25
  • $\begingroup$ @Jack: As I wrote, for the intuition you don't really need to know the integration stuff; just imagine that the matrices $A$ rotate the function $u$ around the origin, and one averages over all the possible rotations of $u$. And sorry, no, I don't have a reference, I came up with this myself (some years ago, actually). $\endgroup$ – Hendrik Vogt Jul 17 '11 at 17:30

The following proof is (almost verbatim) from the freely available Harmonic Function Theory [pdf] by Axler, Bourdon, and Ramey, which is also a nice book I'd recommend for studying harmonic functions.

Let $n>2$. Define $\Omega = \{ \epsilon<\|x\|<1, x\in\mathbb{R}^n \}$ and $v(x)=\|x\|^{2-n}$, and let $u(x)$ be a harmonic function. Denote the unit sphere by $S$. Use Green's second identity on $u$ and $v$ to get

$$ 0 = (2-n) \int_S u\, ds - (2-n)\epsilon^{1-n} \int_{\epsilon S} u\, ds - \int_S \frac{\partial u}{\partial n} ds -\epsilon^{2-n} \int_{\epsilon S} \frac{\partial u}{\partial n} ds.$$

We can use the fact that $\oint \partial u / \partial n ds = 0 $ (again by Green's) to take out the latter two integrals. Then we have, after normalization,

$$ \frac{1}{\omega_n} \int_S u\, ds = \frac{1}{\epsilon^{n-1} \omega_n} \int_{\epsilon S} u\, ds = u(x) \text{ as } \epsilon\to 0. $$

This reasoning generalizes from the unit sphere to any sphere by appropriately scaling or translating $u$. For the planar case with $n=2$, define $v(x) = \ln \| x\|$ and use the same reasoning.


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