It is well-known that a continuous function on satisfies the mean value property over spheres if and only if it is harmonic i.e. . 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 of , each geodesic sphere near 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 on an -dimensional Riemannian manifold , it holds that near , its average value over a geodesic sphere satisfies

Here denotes the -dimensional area of . Actually, the term is explicit and has been given in Theorem 1. In particular, if is harmonic,

When is Einstein, this can be improved to

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 on an Einstein manifold, it holds that near ,

His approach involves directly differentiating with respect to , which would yield an integral involving the mean curvature of . He then examined the expansion of to obtain the derivative estimate , thus integrating with respect to would lead to his result. His method does not seem to be adaptable to the case where . Moreover, the order seems to be not optimal.

On the other hand, Gray and Willmore [GW] Theorem 4.5 improved Friedman’s result when 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 in the normal coordinates. They were able to express their difference in terms of the curvature of 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 is not intrinsic to the geometry of (e.g. it is not true that when , , even at the point ), 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 is small enough, 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 be an -dimensional Riemannian manifold and be a sufficiently smooth function which is well defined at least near . From now on we will fix such and . Let be the geodesic sphere of radius around and denotes its dimensional area element. For convenience, we sometimes allow and still define . Here be the standard dimensional unit sphere, whose area element will be denoted by , and define . Clearly we have and for , we have , i.e. is even. We define to be the open geodesic ball of radius centered at .

We sometimes denote by . We use the following notations for the curvatures on : denotes the Riemannian curvature tensor ( in index form) and denotes the Ricci curvature. We write e.g. . We use the convention that is the sectional curvature. The scalar curvature is denoted by . Moreover, unless otherwise stated, all the curvature terms are computed at the point . Denote the -times covariant derivative of by , thus is the Hessian of . We will not distinguish the gradient vector and the covariant derivative of , e.g. .

Finally, by we mean a function on for some such that , where depends on and but is independent of . This constant may vary from line to line.

Our main theorem is

Theorem 1On a smooth -dimensional Riemannian manifold , for any smooth function defined near , we have

Corollary 2If is a Riemannian manifold such that all harmonic functions satisfies the property that at all point ,

Then is Einstein.

There is also a version of the theorem for the average of over geodesic balls:

Theorem 3On a smooth -dimensional Riemannian manifold , for any smooth function defined near , we have

where is given in (3).

Our method can be illustrated in the proof of the following lemma. Firstly, by an analytic Riemannian manifold , 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 4Assume is an analytic Riemannian manifold. Suppose is analytic, then is an analytic function on an interval containing , where is defined to be .

**Proof**: By pulling back using the geodesic spherical coordinates, we simply regard . i.e. we write as by abuse of notations. Simply write the -th covariant derivative of at as . Then by analyticity, we have

which converges locally uniformly, and in particular uniformly in for for some .

Therefore for each , we can integrate this series and get

where we simply write by and . We claim that this converges locally uniformly in .

It is not hard to see that (cf. Equation (5)). So it suffices to show that

converges locally uniformly. But this follows immediately as converges locally uniformly, and in particular uniformly for when . From the above, we also know that and is also analytic in . Therefore is also analytic.

By the above lemma, since is even, its Taylor series about 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 by substituting the Taylor’s expansion of in the averaged integral. So instead of differentiating the averaged value with respect to , we will substitute both the Taylor expansion of and the expansion of the area element into the integral . 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 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 is an odd function on , in the sense that , then

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

Lemma 5(cf. [G] Equation 11

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

Lemma 6

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

Lemma 7

**Proof**: Define the vector field on . Then

So the divergence theorem gives

By symmetry and using spherical coordinates, it is easy to see that . From this the result follows.

Lemma 8We have the following equations:

and

**Proof**: We just prove the first equation. We omit in the following computation. By Lemma 7,

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

We remark that there is a more systematic way of integrating polynomials of the type 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 :

Lemma 9

We can now prove Theorem 1.

**Proof**: By Taylor’s theorem and (5),

We do not give the term in the expansion of here, but remark that it is of degree in (which is naturally the same as the power of by Taylor’s theorem. cf. [G] Equation 10. By the observation that the integral of an odd function on vanishes, we can simplify the above as

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

Combining (8) and Lemma 9, we obtain

In particular, if is Einstein and is harmonic, since is constant, we have

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

* On the other hand, integrating (7) gives
*

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

Remark 1Using our method, it is clear that we can go on to obtain the order 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 10Let be a Riemannian manifold.

- Suppose the Ricci curvature , and , then
Here is the area of the geodesic sphere of radius in the space form of curvature . This is equivalent to

where denoted the -dimensional volume of the geodesic ball of radius in the space form of curvature .

- Suppose , and , then
This is equivalent to

**Proof**: Suppose , then by Laplacian comparison theorem,

Here and . So

which implies

As , we have

Integrating with respect to , we also have

The proof for the subharmonic case is similar except that we require the sectional curvature to be in order to use the Hessian comparison theorem.

Nice post

Very nice post. Thanks!