On manifolds with non-negative Ricci curvature and Sobolev inequalities
On manifolds with non-negative Ricci curvature and Sobolev inequalities
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DOI：
10.4310/cag.1999.v7.n2.a7
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发表时间：
1999
影响因子：
0.7
通讯作者：
M. Ledoux
中科院分区：
文献类型：
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作者：
M. Ledoux
- 设M是具有非负Ricci曲率的完备n维Riemann流形，其中Sobolev不等式（？|F| dv）1/p ≤ C（λ）|f| qdv）1/q，f ∈ C ∞ 0（M），1 ≤ q1.此外，对于q> 1，（1）中的等式由函数（λ + λ）得到，|X| q/（q − 1））1 −（n/q），λ> 0，其中|X|是IR中向量x的欧几里得长度。我们实际上在这里感兴趣的是那些流形M的几何，对于这些流形M，Sobolev不等式（1）之一满足IR的最佳常数C = K（n，q）。定理设M是具有非负Ricci曲率的完备n维黎曼流形。如果索伯列夫不等式（1）中的一个满足C = K（n，q），则M与IR等距。q = 1（p = n/（n − 1））的特殊情况当然是众所周知的。在这种情况下，索伯列夫不等式等价于等周不等式（voln（Ω））（n − 1）/n ≤ K（n，1）voln − 1（n Ω），其中n Ω是M中光滑有界开集Ω的边界。如果我们让V（x0，s）= V（s）是测地线球B（x0，s）= B（s）的体积，中心为x0，半径为s，在M中，我们有d ds voln（B（s））= voln − 1（B（s））。因此，在等周不等式中设Ω = B（s），我们得到V（s）（n − 1）/n ≤ K（n，1）V ′（s）对所有s。积分得到V（s）≥（nK（n，1））− nsn，并且由于K（n，1）= n − 1 ω n，对于每个s，（2）V（s）≥ V0（s）其中V0（s）= ω ns是IR中半径为s的欧几里德球的体积。例如[Ch]）V（s）≤ V0（s），对于每个s，由（2）和等式的情况，M与IR等距。因此，定理的主要兴趣在于q> 1的情况。通常，经典值q = 2（和p = 2n/（n − 2））特别令人感兴趣（见下文）。应该注意的是，已知的结果已经意味着M的标量曲率在这种情况下为零（参见。[He]4.10）。定理的证明。它的灵感来自于最近的工作[B-L]中发展的技术，其中得到了满足Sobolev不等式的紧致黎曼流形的直径的一个尖锐的界，扩展了经典的Myers定理。因此，我们假设Sobolev不等式（1）满足C = K（n，q），其中q> 1。首先回想一下，IR中这个不等式的极值函数是函数（λ +| X| q）1 −（n/q），λ> 0，其中q ′ = q/（q − 1）。设x0是M中的一个不动点，θ> 1。设f = θ − 1d（·，x0）其中d是M上的距离函数。其思想是将索伯列夫不等式（1）应用于（λ + f ′）1 −（n/q），其中C = K（n，q），对于每个λ> 0，推导出一个微分不等式，其解可以与极值欧几里德情况进行比较。设对于每个λ> 0，F（λ）= 1 n − 1 <$1（λ + fq）n − 1 dv。首先注意F是定义良好的，并且在λ中连续可微。事实上，根据Fubini定理，对于每个λ> 0，
— Let M be a complete n-dimensional Riemanian manifold with non-negative Ricci curvature in which one of the Sobolev inequalities (∫ |f |dv )1/p ≤ C(∫ |∇f |qdv)1/q, f ∈ C∞ 0 (M), 1 ≤ q 1. Moreover, for q > 1, the equality in (1) is attained by the functions (λ + |x|q/(q−1))1−(n/q), λ > 0, where |x| is the Euclidean length of the vector x in IR. We are actually interested here in the geometry of those manifolds M for which one of the Sobolev inequalities (1) is satisfied with the best constant C = K(n, q) of IR. The result of this note is the following theorem. Theorem. Let M be a complete n-dimensional Riemannian manifold with nonnegative Ricci curvature. If one of the Sobolev inequalities (1) is satisfied with C = K(n, q), then M is isometric to IR. The particular case q = 1 (p = n/(n − 1)) is of course well-known. In this case indeed, the Sobolev inequality is equivalent to the isoperimetric inequality ( voln(Ω) )(n−1)/n ≤ K(n, 1)voln−1(∂Ω) where ∂Ω is the boundary of a smooth bounded open set Ω in M . If we let V (x0, s) = V (s) be the volume of the geodesic ball B(x0, s) = B(s) with center x0 and radius s in M , we have d ds voln ( B(s) ) = voln−1 ( ∂B(s) ) . Hence, setting Ω = B(s) in the isoperimetric inequality, we get V (s)(n−1)/n ≤ K(n, 1)V ′(s) for all s. Integrating yields V (s) ≥ (nK(n, 1))−nsn, and since K(n, 1) = n−1ω n , for every s, (2) V (s) ≥ V0(s) where V0(s) = ωns is the volume of the Euclidean ball of radius s in IR. IfM has nonnegative Ricci curvature, by Bishop’s comparison theorem (cf. e.g. [Ch]) V (s) ≤ V0(s) for every s, and by (2) and the case of equality,M is isometric to IR. The main interest of the Theorem therefore lies in the case q > 1. As usual, the classical value q = 2 (and p = 2n/(n − 2)) is of particular interest (see below). It should be noticed that known results already imply that the scalar curvature of M is zero in this case (cf. [He], Prop. 4.10). Proof of the Theorem. It is inspired by the technique developed in the recent work [B-L] where a sharp bound on the diameter of a compact Riemannian manifold satisfying a Sobolev inequality is obtained, extending the classical Myers theorem. We thus assume that the Sobolev inequality (1) is satisfied with C = K(n, q) for some q > 1. Recall first that the extremal functions of this inequality in IR are the functions (λ + |x|q)1−(n/q), λ > 0, where q′ = q/(q − 1). Let now x0 be a fixed point in M and let θ > 1. Set f = θ−1d(·, x0) where d is the distance function on M . The idea is then to apply the Sobolev inequality (1), with C = K(n, q), to (λ+ f ′ )1−(n/q), for every λ > 0 to deduce a differential inequality whose solutions may be compared to the extremal Euclidean case. Set, for every λ > 0, F (λ) = 1 n− 1 ∫ 1 (λ+ fq)n−1 dv. Note first that F is well defined and continuously differentiable in λ. Indeed, by Fubini’s theorem, for every λ > 0,