Mean value properties for harmonic functions on Riemannian manifolds

It is well-known that a continuous function ${u}$ on ${\mathbb R^n}$ satisfies the mean value property over spheres if and only if it is harmonic i.e. ${\Delta u=0}$. More generally, this also holds on a Riemannian space form. However, this property is generally not true on a general Riemannian manifold. In fact, it is not hard to show that the mean-value property is true only when for each point ${p}$ of ${M}$, each geodesic sphere near ${p}$ has constant mean curvature ([Fr], [W]). Such a manifold is called a harmonic manifold.

This motivates us to investigate into how well the mean-value type property holds on a general Riemannian manifold, and how it may improve on an Einstein manifold (which is not necessarily harmonic).

In this note, we prove that (Theorem 1) for any smooth function ${u}$ on an ${n}$-dimensional Riemannian manifold ${(M,g)}$, it holds that near ${p}$, its average value over a geodesic sphere ${S_r(p)}$ satisfies

$\displaystyle \frac{1}{|S_r(p)|}\int_{S_r(p)}u dS_r= u(p)+ \frac{\Delta u(p)}{2n}r^2 +O(r^4). \ \ \ \ \ (1)$

Here ${|S_r(p)|}$ denotes the ${(n-1)}$-dimensional area of ${S_r(p)}$. Actually, the ${O(r^4)}$ term is explicit and has been given in Theorem 1. In particular, if ${u}$ is harmonic,

$\displaystyle \frac{1}{|S_r(p)|}\int_{S_r(p)}u dS_r= u(p)+O(r^4).$

When ${(M,g)}$ is Einstein, this can be improved to

$\displaystyle \frac{1}{|S_r(p)|}\int_{S_r(p)}u dS_r= u(p)+O(r^6). \ \ \ \ \ (2)$

These kinds of results have been previously obtained by Friedman [Fr], and also Gray-Willmore [GW]. Friedman showed that for any nonzero (and so locally either positive or negative) harmonic function ${u}$ on an Einstein manifold, it holds that near ${p}$,

$\displaystyle \frac{1}{|S_r(p)|}\int _{S_r(p)}u dS_r = u(p)+O(r^3).$

His approach involves directly differentiating ${\overline u(r)=\frac{1}{|S_r(p)|}\int _{S_r(p)}u dS_r }$ with respect to ${r}$, which would yield an integral involving the mean curvature ${H(S_r)}$ of ${S_r}$. He then examined the expansion of ${H(S_r)}$ to obtain the derivative estimate ${\frac{\overline u'(r)}{\overline u(r)}=O(r^2)}$, thus integrating with respect to ${r}$ would lead to his result. His method does not seem to be adaptable to the case where ${u(p)=0}$. Moreover, the order ${O(r^3)}$ seems to be not optimal.

On the other hand, Gray and Willmore [GW] Theorem 4.5 improved Friedman’s result when ${(M,g)}$ is an analytic Riemannian manifold: they showed that in this case, (1) and (2) are true. However, their approach is quite complicated and not very illuminating. Indeed, they proved the result by comparing the Riemannian Laplacian with the so called Euclidean Laplacian, which is just ${\sum_{i=1}^n \frac{\partial ^2}{\partial {x^i}^2}}$ in the normal coordinates. They were able to express their difference in terms of the curvature of ${M}$ and using various commutative formulas for the covariant derivatives, they can transfer the Euclidean mean-value property to the Riemannian setting, with some additional curvature terms. Unfortunately, the Euclidean Laplacian ${\widetilde \Delta}$ is not intrinsic to the geometry of ${M}$ (e.g. it is not true that when ${k\ge 2}$, ${(\widetilde \Delta)^k = \Delta^k}$, even at the point ${p}$), and the details are indeed quite hard to be checked completely.

In this note, we provide a somewhat more transparent proof of these kinds of results. Our main idea is that when ${r}$ is small enough, ${S_r}$ is sufficiently close to a Euclidean sphere, to the extent which enables us to use the symmetry of some spherical integrals to obtain a lot of cancelations and improve the error estimate of Friedman.

1. Main results

We explain our notations. Let ${(M,g)}$ be an ${n}$-dimensional Riemannian manifold ${p\in M}$ and ${u}$ be a sufficiently smooth function which is well defined at least near ${p}$. From now on we will fix such ${p}$ and ${u}$. Let ${S_r(p)}$ be the geodesic sphere of radius ${r}$ around ${p}$ and ${dS_r}$ denotes its ${n-1}$ dimensional area element. For convenience, we sometimes allow ${r\le0}$ and still define ${S_r=\{\exp_p (r v): v\in \mathbb S^{n-1}\}}$. Here ${\mathbb S^{n-1}}$ be the standard ${n-1}$ dimensional unit sphere, whose area element will be denoted by ${d\mathbb S^{n-1}}$, and define ${\alpha_n=\int_{\mathbb S^{n-1}}d\mathbb S^{n-1}=\mathrm{Area}(\mathbb S^{n-1})}$. Clearly we have ${S_{-r}= S_r}$ and for ${\overline u(r)=\frac{1}{|S_r|}\int_{S_r} udS_r}$, we have ${\overline u(-r)=\overline u(r)}$, i.e. ${\overline u}$ is even. We define ${B_r(p)}$ to be the open geodesic ball of radius ${r}$ centered at ${p}$.

We sometimes denote ${g}$ by ${\langle \cdot, \cdot\rangle}$. We use the following notations for the curvatures on ${M}$: ${\textrm{Rm}}$ denotes the Riemannian curvature tensor (${R_{xyzw}}$ in index form) and ${\rho(x, y)=\mathrm{tr }\;\mathrm{Rm}(\cdot, x, y, \cdot)}$ denotes the Ricci curvature. We write e.g. ${\nabla^2 _{xy} \rho_{zw}=(\nabla _x\nabla _y\rho)(z, w)}$. We use the convention that ${R_{1221}}$ is the sectional curvature. The scalar curvature is denoted by ${\tau=\textrm{tr}(\rho)}$. Moreover, unless otherwise stated, all the curvature terms are computed at the point ${p}$. Denote the ${k}$-times covariant derivative of ${u}$ by ${\nabla ^ku}$, thus ${\nabla ^2 u(x, y)=(\nabla _x \nabla u)(y)}$ is the Hessian of ${u}$. We will not distinguish the gradient vector and the covariant derivative of ${u}$, e.g. ${\nabla u(v)=\langle \nabla u, v\rangle}$.

Finally, by ${O(r^k)}$ we mean a function ${h(x)}$ on ${B_{r_0}(p)}$ for some ${r_0=r_0(p)>0}$ such that ${|h(x)|\le Cr^k}$, where ${C}$ depends on ${u}$ and ${B_{r_0}(p)}$ but is independent of ${x}$. This constant may vary from line to line.

Our main theorem is

 Theorem 1 On a smooth ${n}$-dimensional Riemannian manifold ${(M, g)}$, for any smooth function ${u}$ defined near ${p}$, we have $\displaystyle \frac{1}{|S_r(p)|}\int_{S_r(p)}udS_r= u(p)+ \frac{\Delta u(p)}{2n}r^2 + B(n)r^4+O(r^6)$ where $\displaystyle B(n) =\frac{1}{24n(n+2)}\left(3 \Delta^2 u - 2 \langle \nabla ^2 u, \rho\rangle -3\langle \nabla u, \nabla \tau\rangle+\frac{4}{n}\tau \Delta u \right)(p). \ \ \ \ \ (3)$ In particular, if ${(M,g)}$ is Einstein and ${u}$ is harmonic, we have $\displaystyle \frac{1}{|S_r|}\int_{S_r(p)}u dS_r= u(p)+O(r^6).$
 Corollary 2 If ${(M,g)}$ is a Riemannian manifold such that all harmonic functions ${u}$ satisfies the property that at all point ${p}$, $\displaystyle \frac{1}{|S_r|}\int_{S_r(p)}u dS_r= u(p)+O(r^6).$ Then ${(M,g)}$ is Einstein.

There is also a version of the theorem for the average of ${u}$ over geodesic balls:

 Theorem 3 On a smooth ${n}$-dimensional Riemannian manifold ${(M, g)}$, for any smooth function ${u}$ defined near ${p}$, we have $\displaystyle \frac{1}{|B_r(p)|}\int_{B_r(p)}udS_r= u(p)+ \frac{\Delta u}{2(n+2)}r^2 + B(n+2)r^4+O(r^6)$ where ${B(n)}$ is given in (3).

Our method can be illustrated in the proof of the following lemma. Firstly, by an analytic Riemannian manifold ${(M,g)}$, we mean that the underlying smooth manifold is analytic and the Riemannian metric is also an analytic tensor field. The following lemma is expected:

 Lemma 4 Assume ${(M,g)}$ is an analytic Riemannian manifold. Suppose ${ u}$ is analytic, then ${\overline u(r)}$ is an analytic function on an interval containing ${0}$, where ${\overline u(0)}$ is defined to be ${u(p)}$.

Proof: By pulling back using the geodesic spherical coordinates, we simply regard ${u=u(r, v)}$. i.e. we write ${u(\exp_p(rv))}$ as ${u(r, v)}$ by abuse of notations. Simply write the ${k}$-th covariant derivative of ${u}$ at ${p}$ as ${\nabla ^k u}$. Then by analyticity, we have

$\displaystyle u(r,v)= \sum_{k=0}^\infty \frac{1}{k!}\nabla ^ku ( v, \cdots, v)r^k$

which converges locally uniformly, and in particular uniformly in ${v\in \mathbb S^{n-1}}$ for ${r\le r_0}$ for some ${r_0>0}$.

Therefore for each ${r}$, we can integrate this series and get

$\displaystyle \int_{S_r}u dS_r= \sum_{k=0}^\infty \frac{r^k}{k!}\int_{v\in \mathbb S^{n-1}} \nabla^k u (v, \cdots, v)dS_r(v) \ \ \ \ \ (4)$

where we simply write ${\Phi_r^{*}dS_r}$ by ${dS_r(v)}$ and ${\Phi_r(v)= \exp_p(rv)}$. We claim that this converges locally uniformly in ${r}$.

It is not hard to see that ${dS_r(v)= r^{n-1}(1+O(r^2 ))d\mathbb S^{n-1}(v)}$ (cf. Equation (5)). So it suffices to show that

$\displaystyle r^{n-1}\sum_{k=0}^\infty \frac{r^{k}}{k!}\int_{\mathbb S^{n-1}} \nabla^k u (v, \cdots, v)d\mathbb S^{n-1}(v)$

converges locally uniformly. But this follows immediately as ${u(r,v)= \sum_{k=0}^\infty \frac{1}{k!}\nabla^ku ( v, \cdots, v)r^k}$ converges locally uniformly, and in particular uniformly for ${v\in \mathbb S^{n-1}}$ when ${r\le r_0}$. From the above, we also know that ${\int_{S_r} udS_r=O(r^{n-1})}$ and ${|S_r|= \alpha_nr^{n-1}(1+O(r^2))}$ is also analytic in ${r}$. Therefore ${\overline u(r)}$ is also analytic. $\Box$

By the above lemma, since ${\overline u}$ is even, its Taylor series about ${0}$ has only even terms. We aim to find the coefficients of this expansion. Strictly speaking, we do not need the above lemma to prove our result. We require analyticity just to avoid writing down the error term in the smooth version of Taylor’s theorem. However, from the above proof, we see that basically we can estimate ${\overline u}$ by substituting the Taylor’s expansion of ${u}$ in the averaged integral. So instead of differentiating the averaged value with respect to ${r}$, we will substitute both the Taylor expansion of ${u}$ and the expansion of the area element into the integral ${\int_{S_r}u dS_r}$. Thanks to the symmetry of the Euclidean sphere, there is a large amount of cancelation in the integral and we obtain a clean argument of our result and at the same time improve the estimate of Friedman.

Of course, if the mean value property holds then ${\overline u(r)=u(p)}$ has only one term in its expansion. In the general case, to compute the coefficients, we will frequently use the standard but useful result in calculus that if ${f(v)}$ is an odd function on ${\mathbb S^{n-1}}$, in the sense that ${f(-v)=-f(v)}$, then

$\displaystyle \int_{\mathbb S^{n-1}}f(v)d\mathbb S^{n-1}=0.$

We now provide various formulas to compute the expansion of the average integral. We have the following expansion of the area element of ${S_r}$:

 Lemma 5 (cf. [G] Equation 11 $\displaystyle dS_r(v)= r^{n-1}\left(1-\frac{r^2}{2\cdot 3} \rho_ {vv}-\frac{r^3}{2\cdot 3!}\nabla _v\rho_{vv} +\frac{r^4}{4!}\left(-\frac{3}{5}\nabla _{vv}^2\rho_{vv}+\frac{1}{3}\rho_{vv}^2-\frac{2}{15}R_{vavb}^2\right)+O(r^5)\right)d\mathbb S^{n-1}. \ \ \ \ \ (5)$

In order to compute the coefficients of the expansion, we need several lemmas about the integral of polynomial functions on the standard sphere. First of all, it is an exercise in calculus to see that

$\displaystyle \int_{\mathbb S^{n-1}}v^i v^j d\mathbb S^{n-1}(v)= \frac{\alpha_n}{n}\delta_{ij}. \ \ \ \ \ (6)$

From this we have

 Lemma 6 $\displaystyle \int _{\mathbb S^{n-1}}\nabla _p^2 u (v,v)d\mathbb S^{n-1}= \frac{\alpha_n}{n}\Delta u(p).$

The analogue of (6) for polynomials of higher degree is slightly more complicated:

 Lemma 7 $\displaystyle \int_{\mathbb S^{n-1}}v^i v^j v^k v^l d\mathbb S^{n-1}(v)= \frac{\alpha_n}{n(n+2)}(\delta_{li}\delta_{jk}+\delta_{lj}\delta_{ik}+\delta_{lk}\delta_{ij}).$

Proof: Define the vector field ${F(x)= x^i x^j x^k e_l}$ on ${\mathbb R^3}$. Then

$\displaystyle \begin{array}{rl} \int_{\mathbb S^{n-1}} v^i v^j v^k v^l d\mathbb S^{n-1}(v) =\int_{\mathbb S^{n-1}} \langle F(v), v \rangle d\mathbb S^{n-1}(v) \end{array}$

So the divergence theorem gives

$\displaystyle \begin{array}{rl} \int_{\mathbb S^{n-1}} v^i v^j v^k v^l d\mathbb S^{n-1}(v) =&\int_{\mathbb B^n}\mathrm{div}(F)dV(x)\\ =&\int_{\mathbb B^n}(\delta_{li} x^j x^k+ \delta_{lj}x^i x^k+\delta_{lk}x^ix^j)dV\\ =&\delta_{li} \int_{\mathbb B^n}x^j x^k dV+ \delta_{lj}\int_{\mathbb B^n}x^i x^kdV+\delta_{lk}\int_{\mathbb B^n}x^ix^jdV. \end{array}$

By symmetry and using spherical coordinates, it is easy to see that ${\int_{\mathbb B^n} x^i x^jdV= \begin{cases} \frac{\alpha_n}{n(n+2)}\quad &\textrm{if }i=j\\ 0 \quad &\textrm{if }i\ne j \end{cases} }$. From this the result follows.

$\Box$

 Lemma 8 We have the following equations: $\displaystyle \int_{\mathbb S^{n-1}} \nabla ^4 _p u(v,v,v,v)d\mathbb S^{n-1}= \frac{\alpha_n}{n(n+2)}\left(3 \Delta ^2 u+\langle \nabla \tau, \nabla u\rangle +2\langle \rho, \nabla ^2u\rangle\right)(p),$ $\displaystyle \int_{\mathbb S^{n-1}}\nabla ^2_pu(v,v)\rho_{vv}d\mathbb S^{n-1}(v)=\frac{\alpha_n}{n(n+2)}\left(2\langle \rho, \nabla ^2u\rangle+\tau \Delta u\right)(p),$ and $\displaystyle \int_{\mathbb S^{n-1}} \nabla _pu (v)\nabla _v\rho_{vv}d\mathbb S^{n-1}=\frac{ 2\alpha_n }{n(n+2)}\langle \nabla \tau, \nabla u\rangle(p).$

Proof: We just prove the first equation. We omit ${p}$ in the following computation. By Lemma 7,

$\displaystyle \begin{array}{rl} \frac{n(n+2)}{\alpha_n}\int_{\mathbb S^{n-1}} \nabla ^4 u(v,v,v,v)d\mathbb S^{n-1}(v) =& \Delta \Delta u+\sum_{i, j=1}^n (\nabla ^4_{ijji} u+\nabla ^4_{jiji}u)\\ =& \Delta ^2u+2\sum_{i, j=1}^n \nabla ^4 _{ijij}u\\ =& \Delta ^2 u+2\sum_{i, j=1}^n (\nabla ^4_{iijj}u-\nabla _i(R_{jijl}\nabla _lu))\\ =& \Delta ^2u+2\sum_{i, j=1}^n (\nabla ^4_{iijj}u+\nabla _i(\rho_{il}\nabla _lu))\\ =&3 \Delta ^2u+\langle \nabla \tau, \nabla u\rangle +2\langle \rho, \nabla ^2u\rangle. \end{array}$

Here we use the Ricci identity on the third line and the second Bianchi identity on the last line. In our notations, ${R_{xyxy}}$ is the Riemannian curvature tensor and . The other two equations are similarly proved.

$\Box$

We remark that there is a more systematic way of integrating polynomials of the type ${\int_{\mathbb S^{n-1}}v^iv^j \cdots d\mathbb S^{n-1}(v)}$ using the gamma function (cf. e.g. [F] ), justifying the appearance of the gamma function in [GW] . Thus it is possible to give a more unified statement concerning the above lemmas. We will not do it here.

By [G] Theorem 3.1 or by directly integrating (5), we also have the Taylor expansion for ${S_r}$:

 Lemma 9 $\displaystyle |S_r|= \alpha_n r^{n-1}\left(1-\frac{r^2}{6n}\tau(p) +\frac{r^4}{360n(n+2)}\left(-3|Rm|^2+8 |\rho|^2 +5 \tau ^2 -18 \Delta \tau\right)(p)+O(r^6) \right). \ \ \ \ \ (7)$

We can now prove Theorem 1.
Proof: By Taylor’s theorem and (5),

$\displaystyle \begin{array}{rcl} &&\int_{S_r}u(x) dS_r(x) -u(p)|S_r|\\ &=&\int _{S_r}\left(u(x)-u(p)\right)dS_r(x)\\ &=&\int _{\mathbb S^{n-1}}\left(r\nabla _pu(v)+\frac{r^2}{2!}\nabla _p^2u(v,v)+\frac{r^3}{3!}\nabla_p ^3 u(v,v,v)+\frac{r^4}{4!}\nabla_p ^4 u(v,v,v,v)+O(r^5) \right)dS_r(v)\\ &=&r^{n-1}\int _{\mathbb S^{n-1}}\left(r\nabla _pu(v)+\frac{r^2}{2!}\nabla _p^2u(v,v)+\frac{r^3}{3!}\nabla_p ^3 u(v,v,v)+\frac{r^4}{4!}\nabla_p ^4 u(v,v,v,v)+O(r^5)\right)\\ &&\left(1-\frac{r^2}{2\cdot 3}\rho_{vv}-\frac{r^3}{2\cdot 3!}\nabla _v\rho_{vv}+O(r^4)\right)d\mathbb S^{n-1}(v) \end{array}$

We do not give the ${O(r^4)}$ term in the expansion of ${dS_r(v)}$ here, but remark that it is of degree ${4}$ in ${v}$ (which is naturally the same as the power of ${r}$ by Taylor’s theorem. cf. [G] Equation 10. By the observation that the integral of an odd function on ${\mathbb S^{n-1}}$ vanishes, we can simplify the above as

$\displaystyle \begin{array}{rcl} &&\int_{S_r}u(x) dS_r(x) -u(p)|S_r|\\ &=&r^{n-1}\left[\frac{r^2}{2}\int \nabla _p^2 u(v,v)d\mathbb S^{n-1}(v)\right.\\ &&\left.+r^4 \int _{\mathbb S^{n-1}}\left(-\frac{1}{2\cdot 3!} \nabla _pu(v)\nabla _v\rho_{vv} -\frac{1}{2! 6}\nabla ^2_p u(v,v)\rho_{vv}+\frac{1}{4!}\nabla ^4_{p}u(v,v,v,v)\right) d\mathbb S^{n-1}(v)\right.\\ &&\left.+O(r^6)\right]. \end{array}$

The coefficients can be computed by using Lemma 6 and Lemma 8, from which we obtain

$\displaystyle \begin{array} {rcl} &&\int_{S_r(p)} u dS_r - u(p)|S_r(p)|\\ &=&\alpha_nr^{n-1}\left[ r^2 \frac{\Delta u(p)}{2n} +r^4\left(-\frac{1}{8n(n+2)}\langle \nabla \tau, \nabla u\rangle -\frac{1}{12n(n+2)}\langle \rho, \nabla ^2u\rangle\right.\right.\\ &&\left.\left.+\frac{1}{8n(n+2)}\Delta^2 u-\frac{1}{12n(n+2)} \tau \Delta u \right)(p)+O(r^6)\right]. \end{array} \ \ \ \ \ (8)$

Combining (8) and Lemma 9, we obtain

$\displaystyle \frac{1}{|S_r|}\int_{S_r}udS_r= u(p)+ \frac{\Delta u(p)}{2n}r^2 + B(n)r^4+O(r^6), \ \ \ \ \ (9)$

where

$\displaystyle B(n) =\frac{1}{24n(n+2)}\left(3 \Delta^2 u - 2 \langle \nabla ^2 u, \rho\rangle -3\langle \nabla u, \nabla \tau\rangle+\frac{4}{n}\tau \Delta u. \right)(p).$

In particular, if ${(M,g)}$ is Einstein ${(n\ge 3)}$ and ${u}$ is harmonic, since ${R}$ is constant, we have

$\displaystyle \frac{1}{|S_r|}\int_{S_r(p)}u dS_r= u(p)+O(r^6).$

$\Box$

Proof: [Proof of Theorem 3.] By integrating (8) and using the Fubini’s theorem,

$\displaystyle \begin{array}{rcl} &&\int_{B_r(p)}u dV - u(p) |B_r(p)|\\ &=&\int_0^r\alpha_nt^{n-1}\left[t^2 \frac{\Delta u(p)}{2n}+t^4\left(-\frac{1}{8n(n+2)}\langle \nabla \tau, \nabla u\rangle -\frac{1}{12n(n+2)}\langle \rho, \nabla ^2u\rangle\right.\right.\\ &&\left.\left.+\frac{1}{8n(n+2)}\Delta^2 u-\frac{1}{12n(n+2)} \tau \Delta u\right)(p)+O(t^6)\right]dt\\ &=&\frac{\alpha_n r^n}{n}\left(\frac{\Delta u(p)}{2(n+2)}r ^2 +t^4\left(-\frac{1}{8(n+2)(n+4)}\langle \nabla \tau, \nabla u\rangle -\frac{1}{12(n+2)(n+4)}\langle \rho, \nabla ^2u\rangle\right.\right.\\ &&\left.\left.+\frac{1}{8(n+2)(n+4)}\Delta^2 u-\frac{1}{12(n+2)(n+4)} \tau \Delta u\right)(p)+O(t^6)\right]dt\\ \end{array} \ \ \ \ \ (10)$

On the other hand, integrating (7) gives

$\displaystyle |B_r(p)|=\frac{\alpha_nr^n}{n}\left(1-\frac{\tau(p)}{6(n+2)}r^2+O(r^4)\right).$

Divide (10) by this expression, we can get the result.

$\Box$

 Remark 1 Using our method, it is clear that we can go on to obtain the order ${O(r^6)}$ term in the expansion in Theorem 1 and 3. However, it has a long (consists of six lines!) expression which does not seem particularly interesting to us. In any case, it has already been computed in [GW] (Equation 4.11) using a rather long computation.

2. Some related results

Although not proved using the method in Section 1, we state some other results related to sub- or super- mean-value property on a Riemannian manifold, which we cannot find a reference in the literature.

 Theorem 10 Let ${(M,g)}$ be a Riemannian manifold. Suppose the Ricci curvature ${\rho\ge (n-1)k}$, ${\Delta u\le 0}$ and ${u\ge 0}$, then $\displaystyle u(p)\ge \frac{1}{|\mathbb S_k(r)|} \int_{S_r(p)} u dS.$ Here ${|\mathbb S_k(r)|}$ is the area of the geodesic sphere of radius ${r}$ in the space form of curvature ${k}$. This is equivalent to $\displaystyle u(p)\ge \frac{1}{|\mathbb B_k(r)|} \int_{B_r(p)} udV$ where ${|\mathbb B_k(r)|}$ denoted the ${n}$-dimensional volume of the geodesic ball of radius ${r}$ in the space form of curvature ${k}$. Suppose ${\mathrm{Rm}\le k}$, ${\Delta u\ge 0}$ and ${u\ge 0}$, then $\displaystyle u(p)\le \frac{1}{|\mathbb S_k(r)|} \int_{S_r(p)} u dS.$ This is equivalent to $\displaystyle u(p)\le \frac{1}{|\mathbb B_k(r)|} \int_{B_r(p)} udV.$

Proof: Suppose ${\rho\ge (n-1)k}$, then by Laplacian comparison theorem,

$\displaystyle \begin{array}{rl} 0\ge\int_{B_r(p)} \Delta u dV =&\int_{S_r}\langle \nabla u, \nu \rangle dS\\ =&\int_{v\in \mathbb S^{n-1}} \frac{\partial }{\partial r}u (\exp_{p}rv) d S(v)\\ =&\frac{d}{dr}\left(\int _{S_r}u dS\right)- \int_{S_r}u HdS\\ \ge&\frac{d}{dr}\left(\int _{S_r}u dS\right)- \int_{S_r}u H_k(r) dS\\ =&\frac{d}{dr}\left(\int _{S_r}u dS\right)- (n-1)\frac{c_k(r)}{s_k(r)}\int_{S_r}udS. \end{array}$

Here ${s_k(r)= \begin{cases} r\quad &\textrm{if }k=0\\ \sin (\sqrt k r)&\textrm{if }k>0\\ \sinh (\sqrt {-k} r)&\textrm{if }k<0 \end{cases} }$ and ${c_k(r)= \begin{cases} 1\quad &\textrm{if }k=0\\ \cos (\sqrt k r)&\textrm{if }k>0\\ \cosh (\sqrt {-k} r)&\textrm{if }k<0 \end{cases} }$. So

$\displaystyle \begin{array}{rl} 0\ge \alpha_n s_k(r)^{n-1} \frac{d}{dr} \left(\int_{S_r} udS\right) -(n-1)w_{n}s_k(r)^{n-2}c_k(r) \int_{S_r} u dS \end{array}$

which implies

$\displaystyle \begin{array}{rl} 0\ge \frac{d}{dr} \left(\frac{\int_{S_r}u dS}{\alpha_n s_k(r)^{n-1}}\right) =\frac{d}{dr}\left(\frac{1}{|S_k(r)|}\int _{S_r(p)} u dS\right). \end{array}$

As ${\lim_{r\rightarrow 0}\frac{1}{|S_k(r)|}\int_{S_r(p)}u dS= u(p)}$, we have

$\displaystyle u (p)\ge \frac{1}{|S_k(r)|} \int_{S_r(p)} udS.$

Integrating with respect to ${r}$, we also have

$\displaystyle u(p)\ge \frac{1}{|B_k(r)|} \int_{B_r(p)} udV.$

The proof for the subharmonic case is similar except that we require the sectional curvature to be ${\le k}$ in order to use the Hessian comparison theorem.

$\Box$