2002
DOI: 10.1016/s1631-073x(02)02529-3
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A sharp Sobolev inequality on Riemannian manifolds

Abstract: Let (M, g) be a smooth compact Riemannian manifold without boundary of dimension n ≥ 6. We prove thatThe inequality is sharp in the sense that on any (M, g), K can not be replaced by any smaller number and Rg can not be replaced by any continuous function which is smaller than Rg at some point. If (M, g) is not locally conformally flat, the exponent 2n/(n + 2) can not be replaced by any smaller number. If (M, g) is locally conformally flat, a stronger inequality, with 2n/(n + 2) replaced by 1, holds in all dim… Show more

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Cited by 13 publications
(15 citation statements)
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“…Comparing with some existing results for higher dimensional Yamabe type problem (see, for example, [18] and [20]), our proof seems quite standard. We start with a local sharp inequality (caution: this local inequality does not rely on Moser's inequality), which allows us to obtain a rough sharp inequality (or, sometimes, we refer it as an ǫ-level inequality) in Proposition 2.1.…”
Section: 3supporting
confidence: 73%
“…Comparing with some existing results for higher dimensional Yamabe type problem (see, for example, [18] and [20]), our proof seems quite standard. We start with a local sharp inequality (caution: this local inequality does not rely on Moser's inequality), which allows us to obtain a rough sharp inequality (or, sometimes, we refer it as an ǫ-level inequality) in Proposition 2.1.…”
Section: 3supporting
confidence: 73%
“…For manifolds with boundary, the same type of result (see Li and Zhu [14]) is true. More precisely, if (V , g) is a compact Riemannian n-manifold with smooth boundary and n ≥ 3, then, for all u ∈ C ∞ (V ),…”
Section: Introductionmentioning
confidence: 55%
“…However, several difficulties appear because we have to study functions on M while we have informations only on ∂M . The specific problems of trace inequalities can also be seen in the paper of Li and Zhu [12] who obtained the same result for trace Sobolev inequality. Their proof is really different from the one of Hebey and Vaugon [8] who solved the problem for standard Sobolev inequality (without trace).…”
Section: Introductionmentioning
confidence: 63%
“…One easily gets that, in addition, limL k = limλ 2 k = 1. Coming back to (12), it follows that, if (c α ) α is such that lim Aα cα = 0,…”
Section: Emmanuel Humbert Nodeamentioning
confidence: 99%