13/11/10 下午2:51Lecture 25. The Sobolev inequality proof of the Myer’s diameter theorem | Research and Lecture notes

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Lecture 25. The Sobolev inequality proof of the Myer’s diameter theoremPosted on November 5, 2013

It is a well-known result that if is a complete -dimensional Riemannian manifold with , for some , then has to be compact with diameter less than . 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 herean alternative proof of the diameter theorem that relies on the sharp Sobolev inequality proved in the previousLecture. The beautiful argument goes back to Bakry and Ledoux. We only sketch the main arguments and referthe readers to the original article.

The theorem by Bakry and Ledoux is the following:

Theorem: Assume that for some , we have the inequality,

then is compact with diameter less than .

Combining this with the inequality

that was proved in the previous Lecture gives . When we conclude then by letting

and when , we conclude by choosing .

By using a scaling argument it is easy to see that it is enough to prove that if for some ,

then .

The main idea is to apply the Sobolev inequality to the functions which are the extremals functions on thesphere. Such extremals are solutions of the fully non linear PDE

and on the spheres the extremals are explicitly given by

where and is the distance to a fixed point. So, on our manifold , that satisfies the inequality

we consider the functional

where is a function on that satisfies . The first step is to prove a differential inequality on . For , we denote by the differential operator on defined by

Lemma: Denoting , we have

Research and Lecture notesby Fabrice Baudoin

13/11/10 下午2:51Lecture 25. The Sobolev inequality proof of the Myer’s diameter theorem | Research and Lecture notes

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Proof: We denote and , . By the chain-rule and the hypothesis that ,we get

From the Sobolev inequality applied to , we thus have,

It is then an easy calculus exercise to deduce our claim

The next idea is then to use a comparison theorem to bound in terms of solutions of the equation

Actually, such solutions are given by

where , and

We have then the following comparison result:

Lemma: Let be such that

and assume that for some . Then for every ,

Using the previous lemma, we see (again we refer to the original article for the details) that impliesthat and implies that . Iterating this result on the basisof the Sobolev inequality again, we actually have

from which the conclusion easily follows.

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