Lecture 25. The Sobolev inequality proof of the Myer’s diameter theorem

Posted on November 5, 2013 by Fabrice Baudoin

It is a well-known result that if \mathbb{M} is a complete n-dimensional Riemannian manifold with \mathbf{Ricci} \ge \rho, for some \rho > 0, then \mathbb{M} has to be compact with diameter less than \pi \sqrt{\frac{n-1}{\rho} }. The proof of this fact can be found in any graduate book about Riemannian geometry and classically relies on the study of Jacobi fields. We propose here an alternative proof of the diameter theorem that relies on the sharp Sobolev inequality proved in the previous Lecture. The beautiful argument goes back to Bakry and Ledoux. We only sketch the main arguments and refer the readers to the original article.

The theorem by Bakry and Ledoux is the following:

Theorem: Assume that for some p > 2, we have the inequality,
\| f \|_p^2 \le \| f \|_2^2 + A \int_\mathbb{M} \Gamma(f) d\mu, \quad f \in C_0^\infty(\mathbb{M}),
then \mathbb{M} is compact with diameter less than \pi \frac{ \sqrt{2pA}}{p-2}.

Combining this with the inequality
\frac{n \rho}{(n-1)(p-2)} \left( \left( \int_{\mathbb{M}} | f |^p d\mu\right)^{2/p}-\int_\mathbb{M} f^2 d\mu \right) \le \int_\mathbb{M} \Gamma(f) d\mu,
that was proved in the previous Lecture gives \mathbf{diam} (\mathbb{M}) \le \pi \sqrt{\frac{2p}{p-2} } \sqrt{\frac{n-1}{n\rho}}. When n=2 we conclude then by letting p \to \infty and when n > 2, we conclude by choosing p=\frac{2n}{n-2}.

By using a scaling argument it is easy to see that it is enough to prove that if for some n > 2,
\| f \|_{\frac{2n}{n-2}}^2 \le \| f \|_2^2 + \frac{4}{n(n-2)} \int_\mathbb{M} \Gamma(f) d\mu,
then \mathbf{diam} (\mathbb{M}) \le \pi .

The main idea is to apply the Sobolev inequality to the functions which are the extremals functions on the sphere. Such extremals are solutions of the fully non linear PDE
f^{(n+2)/(n-2)} -f =- \frac{4}{n(n-2)} Lf
and on the spheres the extremals are explicitly given by
f=(1+\lambda \sin d)^{1-n/2}
where -1 > \lambda > 1 and d is the distance to a fixed point. So, on our manifold \mathbb{M}, that satisfies the inequality
\| f \|_{\frac{2n}{n-2}}^2 \le \| f \|_2^2 + \frac{4}{n(n-2)} \int_\mathbb{M} \Gamma(f) d\mu,
we consider the functional
F(\lambda) =\int_\mathbb{M} ( 1+\lambda \sin (f) )^{2-n} d\mu, \quad -1 < \lambda <1,
where f is a function on \mathbb{M} that satisfies \| \Gamma(f) \|_\infty \le 1. The first step is to prove a differential inequality on F. For k > 0, we denote by D_k the differential operator on (-1,1) defined by
D_k=\frac{1}{k} \lambda \frac{\partial}{\partial \lambda} +I.

Lemma: Denoting G=D_{n-1} F, we have
(D_{n-2} G)^{(n-2)/n} +\frac{n-2}{n} (1-\lambda^2) D_{n-2}G \le \left( 1+\frac{n-2}{n}\right)G.

Proof: We denote \alpha=\frac{n-2}{n} and f_\lambda=(1+\lambda \sin f)^{1-n/2}, -1 < \lambda < 1. By the chain-rule and the hypothesis that \Gamma(f) \le 1, we get
\int_\mathbb{M} \Gamma(f_\lambda) d\mu \le \left( \frac{n}{2} -1 \right)^2 \int_\mathbb{M} (1+\lambda \sin f)^{-n} (1-\sin^2 f) d\mu.
From the Sobolev inequality applied to f_\lambda, we thus have,
\left( \int_\mathbb{M} ( 1+\lambda \sin (f) )^{-n} d\mu \right)^\alpha \le \int_\mathbb{M} ( 1+\lambda \sin (f) )^{2-n} d\mu +\alpha \lambda^2 \int_\mathbb{M} (1+\lambda \sin f)^{-n} (1-\sin^2 f) d\mu.
It is then an easy calculus exercise to deduce our claim \square

The next idea is then to use a comparison theorem to bound F in terms of solutions of the equation
(D_{n-2} H)^{(n-2)/n} +\frac{n-2}{n} (1-\lambda^2) D_{n-2}H \le \left( 1+\frac{n-2}{n}\right)H.
Actually, such solutions are given by
H_c(\lambda)=\frac{1}{1+\alpha} U_c(\lambda) ^{\frac{2\alpha}{1-\alpha}} +\frac{\alpha}{1+\alpha} (1-\lambda^2) U_c(\lambda)^{\frac{2}{1-\alpha}},
where c \in \mathbb{R}, \alpha=\frac{n-2}{n} and
U_c(\lambda)=\frac{ c\lambda +\sqrt{c^2 \lambda^2 +(1-\lambda^2) }}{1-\lambda^2}.
We have then the following comparison result:

Lemma: Let G be such that
(D_{n-2} G)^{(n-2)/n} +\frac{n-2}{n} (1-\lambda^2) D_{n-2}G \le \left( 1+\frac{n-2}{n}\right)G,
and assume that G(\lambda_0) < H_c (\lambda_0) for some \lambda_0 \in [0,1). Then for every \lambda_0 \le \lambda < 1,
G(\lambda) \le H_c(\lambda).

Using the previous lemma, we see (again we refer to the original article for the details) that \int_{\mathbb{M}} \sin f d\mu > 0 implies that \int_{\mathbb{M}} (1+\sin f )^{n-1}d\mu < +\infty and \int_{\mathbb{M}} \sin f d\mu < 0 implies that \int_{\mathbb{M}} (1-\sin f )^{n-1}d\mu < \infty . Iterating this result on the basis of the Sobolev inequality again, we actually have
\| (1 \pm \sin f)^{-1} \|_{\infty} < + \infty.
from which the conclusion easily follows.

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