A well-known feature of harmonic functions on (domains of) $\mathbb{R}^n$ is the mean-value property: that is, if $\Delta u = 0$, then $$ u(x_0) = \frac{1}{\text{Vol}(\partial B_r(x_0))}\int_{\partial B_r(x_0)}{u\,dS} = \frac{1}{\text{Vol}(B_r(x_0))}\int_{B_r(x_0)}{u\,dV}. $$ Is the same true on manifolds in general? That is, given a manifold $(M,g)$ and a smooth function $u:M\rightarrow\mathbb{R}$ satisfying $$\Delta_gu = 0\quad\text{where}\quad\Delta_g = \frac{1}{\sqrt{\det g}}\frac{\partial}{\partial x^i}g^{ij}\sqrt{\det g}\frac{\partial}{\partial x^j},$$ is it true that $$ u(x_0) = \frac{1}{\text{Vol}(\partial B_r(x_0))}\int_{\partial B_r(x_0)}{u\,dS}\quad\text{or}\quad u(x_0) = \frac{1}{\text{Vol}(B_r(x_0))}\int_{B_r(x_0)}{u\,dV}?$$ Here, $B_r(x_0)$ is the geodesic ball of radius $r$ around $x_0$. If the equalities do not hold, how would the average value (either over a sphere or over a ball) change with $r$? For example, on a surface the Gaussian curvature appears in the higher-order terms in the Taylor expansion of the length/area of a circle/ball of radius $r$--does a similar phenomenon occur for the mean value?

  • $\begingroup$ I've worked this out before for the case of submanifolds of $\mathbb{S}^2\subseteq\mathbb{R}^3$. You can obtain the following mean value property: Let $\Omega\subseteq\mathbb{S}^n$ be a smooth domain. Then $u$ is harmonic with respect to the Laplace-Beltrami operator on $\mathbb{S}^2$ (or on $\Omega$) if and only if for every $x_0\in\Omega$ and $r>0$ such that $B_r(x_0)\Subset\Omega$ (the geodesic ball is compactly contained in $\Omega$) then $u(x_0)=\frac{1}{2\pi\sqrt{r(1-r)}}\int_{\partial B_r(x_0)}u(y)~d\sigma(y)$. $\endgroup$ Commented Aug 15, 2017 at 6:37
  • $\begingroup$ I derived this formula by using an explicit representation of the Laplace-Beltrami operator in spherical coordinates. Then I fully developed the theory for the solvability of the Poisson equation on these regions; in particular, I came up with representation formulas for solutions of the Poisson equation in $\Omega$, which allowed me to derive this mean value property for sub-2-manifolds of $\mathbb{S}^2$. $\endgroup$ Commented Aug 15, 2017 at 6:41
  • $\begingroup$ I think that this mean value property will also hold for any open submanifold of $\mathbb{S}^n$, and I wouldn't be surprised if it also holds for open submanifolds of $\mathbb{H}^n$ (the hyperbolic space with constant negative curvature). On the other hand, I'm not sure if any this will hold in the more general setting where the curvature of $M$ is not constant... I'll think about this some more, and will write more here if I come up with anything. $\endgroup$ Commented Aug 15, 2017 at 6:46

1 Answer 1


After doing a quick Google search I came across this blog post: https://cuhkmath.wordpress.com/2015/08/14/mean-value-theorems-for-harmonic-functions-on-riemannian-manifolds/

The main points of the article are as follows:

  • The mean value property of harmonic functions holds on an arbitrary manifold $M$ only when for every point $p\in M$ every geodesic sphere near $p$ has constant mean curvature. Such manifolds are called harmonic manifolds. This coincides with the intuition I had in my comments above.

  • For any smooth function $u$ on an $n$-manifold $(M,g)$ it holds that $$\frac{1}{\mathrm{vol}(\partial B_r(p)}\int_{\partial B_r(p)}u(y)~d\sigma(y)=u(p)+\frac{\Delta u(p)}{2n}r^2 + \mathscr{O}(r^4).$$ The $\mathscr{O}(r^4)$ term is given by $B(n)r^4 + \mathscr{O}(r^6)$, where $$B(n):=\frac{1}{24(n+2)}\left(3\Delta^2u - 2\langle\nabla^2 u,\rho\rangle-3\langle\nabla u,\nabla\tau\rangle + \frac{4}{n}\tau\Delta u\right).$$ In the above $\rho(x,y)=\mathrm{tr}\left(R(\cdot,x,y,\cdot)\right)$, $R$ is the Riemann curvature tensor, and $\tau=\mathrm{tr}(\rho)$.

  • We have a similar mean value property for geodesic balls. For any smooth function $u$ it holds that $$\frac{1}{\mathrm{vol}(B_r(p))}\int_{B_r(p)}u(y)~d\mu(y)=u(p) + \frac{\Delta u(p)}{2(n+2)}r^2 + B(n+2)r^4 + \mathscr{O}(r^6).$$

  • You have similar sub-mean value properties for subharmonic functions given bounds on the Ricci and Riemann curvatures.

The blog provides plenty of references and proofs (they seem to be correct, but I haven't had the time to work through them yet).

  • 1
    $\begingroup$ Awesome, this is what I was looking for. Thanks! $\endgroup$
    – Joey Zou
    Commented Aug 17, 2017 at 5:21
  • $\begingroup$ Is the computation to see that a sphere $\mathbb S$ (in any dimension) is harmonic manifold easy? I can't seem to do it. Could someone point me in the right direction? I would boil down to showing each geodesic sphere on a sphere has constant mean curvature. $\endgroup$
    – MEG
    Commented Sep 21, 2021 at 19:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .