Research Article | Open Access
Yunyan Yang, Xiaobao Zhu, "Trudinger-Moser Embedding on the Hyperbolic Space", Abstract and Applied Analysis, vol. 2014, Article ID 908216, 4 pages, 2014. https://doi.org/10.1155/2014/908216
Trudinger-Moser Embedding on the Hyperbolic Space
Let be the hyperbolic space of dimension . By our previous work (Theorem 2.3 of (Yang (2012))), for any , there exists a constant depending only on and such that where , is the measure of the unit sphere in , and . In this note we shall improve the above mentioned inequality. Particularly, we show that, for any and any , the above mentioned inequality holds with the definition of replaced by . We solve this problem by gluing local uniform estimates.
Let be a bounded smooth domain in . The classical Trudinger-Moser inequality [1–3] says for some constant depending only on , where is the usual Sobolev space and denotes the Lebesgue measure of . In the case where is an unbounded domain of , the above integral is infinite, but it was shown by Cao , Panda , and do Ó  that for any and any there holds Later Ruf , Li and Ruf , and Adimurthi and Yang  obtained (2) in the critical case .
The study of Trudinger-Moser inequalities on compact Riemannian manifolds can be traced back to Aubin , Cherrier [11, 12], and Fontana . A particular case is as follows. Let be an -dimensional compact Riemannian manifold without boundary. Then there holds
In view of (2), it is natural to consider extension of (3) on complete noncompact Riemannian manifolds. In  we obtained the following results. Let be a complete noncompact Riemannian manifold. If the Trudinger-Moser inequality holds on it, then there holds . If the Ricci curvature has lower bound, say , the injectivity radius has a positive lower bound then for any there exists a constant depending only on , , , and such that Since depends on , (4) is weaker than (2) when is replaced by . Moreover, the condition that has lower bound is not necessary for the validity of the Trudinger-Moser inequality.
In this note, we will continue to study (4) in whole by gluing local uniform estimates. Particularly, we have the following.
Theorem 1. Let be an -dimensional hyperbolic space, , where is the measure of the unit sphere in . Then for any , any , and any satisfying , there exists some constant depending only on and such that
2. Local Estimates
Lemma 2. For any , any , and any with , there exists some constant depending only on such that where denotes the geodesic ball of which is centered at with radius .
Proof. It is well known (see, e.g., , II.5, Theorem 1) that there exists a homomorphism such that , that in these coordinates the Riemannian metric can be represented by
where is the standard Euclidean metric on , and that
where denotes a ball centered at with radius . Moreover, the corresponding polar coordinates read
where is the standard metric on .
Denote ; then , , and . Calculating directly, we have Since , we have . Noting that , we have by (10) The standard Trudinger-Moser inequality (1) implies where is a constant depending only on . This together with (10) immediately leads to This is exactly (6) and thus ends the proof of the lemma.
Corollary 3. For any , any , and any with , there exists some constant depending only on such that
Proof. Since it follows from (13) that there exists some constant depending only on such that In particular, Here and in the sequel we often denote various constants by the same ; the reader can easily distinguish them from the context. Noting that for any , , we conclude Combining (16) and (19), we obtain (14).
3. Proof of Theorem 1
Proof of Theorem 1. Let be a positive real number which will be determined later. By (, Lemma 1.6) we can find a sequence of points such that , that for any , and that for any , belongs to at most balls , where depends only on . Let be the cut-off function satisfies the following conditions: (i) ; (ii) on and on ; (iii) . Let be fixed. For any satisfying we have . For any , using an elementary inequality , we find some constant depending only on and such that where in the last inequality we choose a sufficiently large to make sure . Let and . Noting that , we have by (21) and Lemma 2 where is a constant depending only on and . By the choice of and (22), we have for some constant depending only on and . For any , we can choose sufficiently small such that . This ends the proof of Theorem 1.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
This work is supported by the NSFC 11171347. The authors thank the referee for pointing out some grammar mistakes and the reference .
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