2000
DOI: 10.1002/1097-0312(200102)54:2<229::aid-cpa4>3.0.co;2-i
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On the Caffarelli-Kohn-Nirenberg inequalities: Sharp constants, existence (and nonexistence), and symmetry of extremal functions

Abstract: Consider the following inequalities due to Caffarelli, Kohn, and Nirenberg [6]:. We shall answer some fundamental questions concerning these inequalities such as the best embedding constants, the existence and nonexistence of extremal functions, and their qualitative properties. While the case a ≥ 0 has been studied extensively and a complete solution is known, little has been known for the case a < 0. Our results for the case a < 0 reveal some new phenomena which are in striking contrast with those for the ca… Show more

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Cited by 467 publications
(498 citation statements)
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“…In the following, let [4], the solutions of (1.2) are in one-to-one correspondence to solutions of the following type of problem:…”
Section: Introductionmentioning
confidence: 99%
“…In the following, let [4], the solutions of (1.2) are in one-to-one correspondence to solutions of the following type of problem:…”
Section: Introductionmentioning
confidence: 99%
“…The idea is to use a conformal transformation to convert the problem to an equivalent one defined in a domain on a cyliner C ¼ R Â S NÀ1 : This idea has been used in [8] to study the symmetry property of extremal functions for the Caffarelli-Kohn-Nirenberg inequalities (1). More precisely, to a function uAC N 0 ðO\f0gÞ we associate vAC N 0 ð * OÞ by the transformation…”
Section: Hardy Inequalities With Remainder Termsmentioning
confidence: 99%
“…In [8], it was proved that when O ¼ R N ; the above transformation defines a Hilbert space isomorphism between D 1;2 a ðR N Þ and H 1 ðCÞ whose norm is given by jjvjj 2…”
Section: Hardy Inequalities With Remainder Termsmentioning
confidence: 99%
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“…Also, if 0 < s < 2, the constant μ 0,s (R n ) is again explicit, and the extremals are also known (see Ghoussoub and Yuan [33], Lieb [44], Catrina and Wang [15]). More precisely, μ 0,s (R n ) = (n − 2)(n − s) ω n−1 2 − s · 2 ( n−s 2−s ) ( 2n−2s 2−s )…”
mentioning
confidence: 99%