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. 2017 Jun 2;2017(1):127. doi: 10.1186/s13660-017-1398-8

Singular Moser-Trudinger inequality with the exact growth condition on hyperbolic space

Zhao Liu 1, Lu Chen 2,
PMCID: PMC5488064  PMID: 28680230

Abstract

In this paper, we are concerned with a singular version of the Moser-Trudinger inequality with the exact growth condition in the n-dimension hyperbolic space Hn. Our result is a natural extension of the work of Lu and Tang in (J. Geom. Anal. 26:837-857, 2016).

Keywords: exact growth, hyperbolic space, singular Moser-Trudinger inequality

Introduction

Let W01,p(Ω) denote the usual Sobolev space, i.e., the completion of C0(Ω) under the Sobolev norm

uW01,p(Ω)=(Ω|u|p+|u|pdx)1p.

The classical Sobolev embedding theorem states W01,n(Ω)Lq(Ω) for all 1q< but W01,n(Ω)L(Ω). One can check this by choosing the function u(x)=ln(ln4R|xx0|) for some R>0, where x0 is a fixed point in Ω. It is natural to ask whether there exists an optimal embedding in this limiting case. Yudovic [2], Pohozaev [3] and Trudinger [4] showed that, for some α>0, W01,n(Ω) is continuously embedded into an Orlicz space with the Young function ϕα(t)=exp(α|t|nn11). Moser in [5] sharpened the exponent α and obtained the following result.

Theorem A

[5]

Let Ω be a bounded domain i n Rn (n2). Then there exists a positive constant Cn and a sharp constant αn=nωn11n1 such that

1|Ω|Ωexp(α|u|nn1)dxCn

for any ααn and uC0(Ω) with Ω|u|ndx1, where ωn1 is the area of the surface of the unit ball.

The above inequality is often referred as the Moser-Trudinger inequality. Carleson and Chang [6] employed a symmetrization and rearrangement arguments to obtain the extremals of the Moser-Trudinger inequality when Ω is a disk in R2. Later, the results of Carleson and Chang were extended by Flucher [7] to arbitrary domains in R2, and by Lin [8] for arbitrary domains in Rn. The existence of extremals of the Moser-Trudinger inequality was also extended to compact Riemannian manifold cases by Li in [9, 10]. Cohn and Lu [11] were concerned with the sharp constants for the Moser-Trudinger inequality on a bounded domain on Heisenberg group Hn. The singular version of the Moser-Trudinger inequality on the Heisenberg group was also proved by Lam, Lu and Tang in [12]. For more results as regards the Moser-Trudinger inequalities and applications in partial differential equations, please see [1319] and the references therein.

A natural idea is to consider the Moser-Trudinger inequality in the whole space Rn. Adachi and Tanaka in [20] proved the following nice result.

Theorem B

[20]

For 0<α<αn, there exists a positive constant Cn such that

supuW1,n(Rn),Rn|u|ndx1RnΦ(α|u(x)|nn1)dxCnRn|u(x)|ndx,

where Φ(t):=eti=0n2tii!. Moreover, the constant αn is sharp in the sense that if ααn, the supremum will become infinite.

They proved the sharpness of the exponent α by modifying a sequence of test functions introduced by Moser. In order to obtain the Moser-Trudinger inequality in the critical case α=αn, Ruf [21] (in the dimension n=2) and Li and Ruf [22] (in the dimension n3) replaced the Dirichlet norm with the standard Sobolev norm, i.e.

uW01,n(Rn)=(Rn|u|n+|u|ndx)1n,

and obtained the Moser-Trudinger inequality in the whole space Rn in the case of α=αn. Masmoudi and Sani in their elegant papers [23, 24] kept the two conditions α=αn and Rn|u|ndx1. They proved the following.

Theorem C

[24]

For n2, there exists a positive constant Cn  such that

RnΦ(αn|u|nn1)(1+|u|)nn1dxCnRn|u(x)|ndx,uW1,n(Rn) with Rn|u|ndx1.

Moreover, this inequality fails if the power nn1 in the denominator is replaced by any p<nn1.

The above result was also extended by Chen and Liu [25] to singular version through the change of variables developed by Dong and Lu in [17].

This paper is concerned with a singular version of Moser-Trudinger inequality with the exact growth condition on hyperbolic space Hn. The hyperbolic space Hn (n2) is a complete and simply connected Riemannian manifold and has constant curvature equal to −1. There exist many types of models for hyperbolic space Hn. However, the most important models are the half-space model, the ball model, and the hyperboloid (or Lorentz) model. Throughout this paper, we are concerned about the ball model because we can use symmetry and rearrangement argument in this setting.

We define Bn={xRn:|x|<1} as a pen unit ball in Rn equipped with the Riemannian metric gij=(11|x|2)2δij, which is referred to as the ball model of the hyperbolic space Hn. A direct computation shows that the volume element of hyperbolic space Hn is given by dV=(21|x|2)ndx and d(0,x)=ln1+|x|1|x|, where dx denotes the Lebesgue measure in Rn and d(0,x) denotes the hyperbolic distance between the origin and x. It is well known that the hyperbolic gradient g is defined as the g=(1|x|22)2, where ∇ denotes the general gradient in Rn.

Let Ω be a domain with finite measure on hyperbolic space Hn. Denote fLn(Ω)=(Ω|f|ndV)1n. A straightforward calculation yields

gfLn(Ω)=(Ωgf,gfgn/2dV)1n=(Ω|f|ndV)1n

and

gfLn(Hn)=(Hngf,gfgn/2dV)1n=(Bn|f|ndV)1n.

We also define W1,n(Hn) as the completion of C0(Hn) with the norm

uW1,n(Hn)=(Hn|gu|n+|u|ndV)1n.

The Moser-Trudinger inequality on hyperbolic space was first established by Mancini and Sandeep [26], they proved the Moser-Trudiner inequality on conformal discs. Lu and Tang in [27] considered the subcritical Moser-Trudinger inequality on high dimensional hyperbolic space. They proved the following result.

Theorem D

For any uW1,n(Hn) satisfying Hn|gu|ndV1, there exists a positive constant Cn such that

HnΦ(α(1βn)|u|nn1)(d(0,x))βdVCnHn|u(x)|n(d(0,x))βdV

for any α<αn. Furthermore the constant αn is sharp in the sense that the inequality does not hold if we replace the constant α with any ααn.

Recently, Lu and Tang in [1] considered the sharp Moser-Trudinger inequality with the exact growth condition on hyperbolic space, they proved the following results.

Theorem E

For any uW1,n(Hn) satisfying Hn|gu|ndV1, there exists a positive constant Cn such that

HnΦ(αn|u|nn1)(1+u)nn1dVCnHn|u(x)|ndV. 1

Furthermore, the power nn1 is sharp in the sense if the power nn1 is replaced by any p<nn1, the (1) become infinite.

Motivated by the above results, we consider the singular version of the Moser-Trudinger inequality with the exact growth condition on hyperbolic space. We state our results as follows.

Theorem 1

For any radially decreasing function uW1,n(Hn) satisfying Hn|gu|ndV1, there exists a positive constant Cn independent of u such that

HnΦ(αn(1βn)|u|nn1)(1+u)nn1(d(0,x))βdVCnHn|u(x)|n(d(0,x))βdV. 2

We verify that the power nn1 is optimal.

Theorem 2

If the power nn1 is replaced by any p<nn1, there exists a sequence of functions {uk} such that Hn|guk|ndV1, but

HnΦ(αn(1βn)|uk|nn1)(1+uk)p(d(0,x))βdV(Hn|uk(x)|n(d(0,x))βdV)1.

This paper is organized as follows. In Section 2, we give some important lemmas which will play key roles in the proof of Theorem 1. In Section 3, we establish a singular version of Moser-Trudinger inequality with the exact growth condition on hyperbolic space (Theorem 1). In Section 4, we give the proof of the sharpness of the singular Moser-Trudinger inequality with the exact growth condition in Theorem 1.

Some important lemmas

In this section, we give some key lemmas which play an important role in the proof of Theorem 1.

Lemma 3

see [1, 23]

Given any sequence a={ak}k1, let a1=k=0+|ak|, an=(k=0+|ak|n)1/n, a(e)=(k=0+|ak|nek(1βn))1/n and μ(h)=inf{a(e):a1=h,an1}. Then, for any h>1, we have

μn(h)e(1βn)hnn1hnn1.

With the help of Lemma 3, one can obtain the following lemma.

Lemma 4

There exists a constant C such that for, any nonnegative decreasing function u with u(R)>K1n and ωn1R|u|ntn1dtK for some R,K>0,

eαn(1βn)K1n1unn1(R)unn1(R)RnβCRuntnβ1dtKnn1.

Proof

By scaling, it suffices to show that, for any nonnegative decreasing function u satisfying u(1)>1 and ωn11|u|ntn1dt1,

eαn(1βn)unn1(1)unn1(1)1untnα1dt.

Let hk=αnn1nu(ek/n), ak=hkhk+1 and a={ak}. Then ak0 and

k0|ak|=h0=αnn1nu(1).

Since

hkhk+1=αnn1n(u(ekn)u(ek+1n))=αnn1nek+1neknu(t)dt=αnn1n(eknek+1n|u(t)|ntn1dt)1n(eknek+1n1tdt)n1n=(ωn11|u(t)|ntn1dt)1n,

we have

an=(k0|ak|n)1/n=(k0|hkhk+1|n)1/n1.

Moreover,

1untnβ1dt=k0eknek+1nuntnβ1dt=k0(u(ek+1n))neknek+1ntnβ1dtk0(u(ek+1n))ne(1βn)(k+1)=k1(u(ekn))ne(1βn)kk1hkne(1βn)kk1akne(1βn)k.

Therefore,

a(e)=a0n+k1akne(1βn)kh0n+k1akne(1βn)kh0n+1untnβ1dt. 3

Next, we start to estimate h0. Set 1<r<e1/4n, then

h0αnn1nu(r)=αnn1n1r|u(t)|dt=αnn1n(1r|u(t)|ntn1dt)1n(1r1tdt)n1n=4n1n(ωn11r|u(t)|ntn1dt)1nh02

and

1untnβ1dt1e1/4nuntnβ1dth0n. 4

By (3) and (4), we derive that

a(e)1untnβ1dt.

Then we apply Lemma 3 to conclude that

1untnβ1dte(1βn)h0nn1h0nn1=eαn(1βn)unn1(1)αnunn1(1).

This completes the proof of Lemma 4. □

Singular Moser-Trudinger inequality with the exact growth condition

In this section, we shall establish a singular version of Moser-Trudinger inequality with the exact growth condition on hyperbolic space. Namely, we will give the proof of Theorem 1. By the density, we can assume that u(x) is compactly supported in Hn. We use the idea of Moser [5]. Set d(0,x)=t and u(x)=v(d(0,x))=v(t), then

HnΦ(αn(1βn)|u|nn1)(1+u)nn1(d(0,x))βdV=ωn10Φ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dt,Hn|gu(x)|ndV=ωn10|v(t)|n(sinht)n1dt,Hn|u(x)|n(d(0,x))βdV=ωn10|v(t)|n(sinht)n1tβdt.

Thus, it suffices to show that there exists a positive constant Cn such that

0Φ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dtCn0|v(t)|n(sinht)n1tβdt

for any v(t) satisfying v(t)0,v(t)0,v(t0)=0 for some t0R and

ωn10|v(t)|n(sinht)n1dt1.

Set R=sup{tR:v(t)1}. We can use v(t)(0,1) for t(R,) to obtain

Φ(αn(1βn)|v|nn1)Cn|v(t)|n

for any tR. It follows that

RΦ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dtRΦ(αn(1βn)|v|nn1)tβ(sinht)n1dtCnR|v(t)|ntβ(sinht)n1dt. 5

Next, we focus on the integral over (0,R]. Set 0<ε0<1 and let R1(u)>0 such that

ωn10R1|v(t)|n(sinht)n1dtρε0

and

ωn1R1|v(t)|n(sinht)n1dtρ(1ε0),

where 0<ρ1.

In order to estimate the integral over (0,R], we need to consider two cases: R1R and R1R.

First, we consider the case that R1R. For 0<tR, we can write

v(t)=v(R)+Rtv(s)dsv(R)+(tR|v(s)|n(sinhs)n1ds)1n(tR1sinhsds)n1n1+(ρε0wn1)1n(ln(eR1eR+1et+1et1))n1n.

For any ε>0, one can apply the following well-known inequality:

1+sn1n((1+ε)s+Cε)n1n 6

to obtain

|v(t)|nn1(1+ε)(ρε0wn1)1n1(ln(eR1eR+1et+1et1))+Cε.

Pick ε sufficiently small such that (1+ε)n1ε0<1, then

|v(t)|nn1(ρ(1+ε)n1ε0wn1)1n1(ln(eR1eR+1et+1et1))+Cε0.

Denote c0=(nβ)(ρ(1+ε)n1ε0)1n1. It follows that 0<c0<nβ and

0RΦ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dt0RΦ(αn(1βn)|v|nn1)tβ(sinht)n1dt0RΦ(αn(1βn)((ρ(1+ε)n1ε0wn1)1n1ln(eR1eR+1et+1et1)+Cε0))×tβ(sinht)n1dtexp(αn(1βn)Cε0)(eR1eR+1)c00R(et+1)c0+n1(et1)c0n+1tβ(2et)n1dt.

For R0, it is easy to check that

exp(αn(1βn)Cε0)(eR1eR+1)c00R(et+1)c0+n1(et1)c0n+1tβ(2et)n1dtRnβ.

For R+, one can calculate

exp(αn(1βn)Cε0)(eR1eR+1)c00R(et+1)c0+n1(et1)c0n+1tβ(2et)n1dt0R(et+1)c0+n1(et1)c0n+1tβ(2et)n1dt.

On the other hand,

0R|v(t)|n(sinht)n1tβdt0R(sinht)n1tβdt.

Therefore, for R0,

0R(sinht)n1tβdtRnβ.

Thus,

limR00R|v(t)|n(sinht)n1tβdt0RΦ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dtlimR00R(sinht)n1tβdtexp(αn(1βn)Cε0)(eR1eR+1)c00R(et+1)c0+n1(et1)c0n+1tβ(2et)n1dtC0 7

and

limR+0R|v(t)|n(sinht)n1tβdt0RΦ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dtlimR+0R(sinht)n1tβdtexp(αn(1βn)Cε0)(eR1eR+1)c00R(et+1)c0+n1(et1)c0n+1tβ(2et)n1dtlimR+(sinhR)n1Rβexp(αn(1βn)Cε0)Rβ(eR)2n2(2eR)1n=C1. 8

We can combine (7) and (8) to derive that

0RΦ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dtCn0R|v(t)|n(sinht)n1tβdt. 9

Then by (5) and (9), we obtain the desired inequality of Theorem 1 for R1R.

Now, we consider the case R1<R. First, we consider the integral over (R1,R). By ωn1R1|v(t)|n(sinht)n1dtρε0, we have

v(t)=v(R)+Rtv(s)dsv(R)+(R1|v(s)|n(sinhs)n1ds)1n(tR1sinhsds)n1n1+(ρε0wn1)1n(ln(eR1eR+1et+1et1))n1n,

where R1<t<R. Set ε=(1+ε0)1n11. It is easy to check that

|v(t)|nn1(ρ(1ε02)wn1)1n1(ln(eR1eR+1et+1et1))+Cε0

by (6). Denote c1=(nβ)(ρ(1ε02))1n1, then 0<c1<nβ and

R1RΦ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dtR1RΦ(αn(1βn)|v|nn1)tβ(sinht)n1dt0RΦ(αn(1βn)(ρ(1ε02)wn1)1n1(ln(eR1eR+1et+1et1))+Cε0)tβ(sinht)n1dtexp(αn(1βn)Cε0)(eR1eR+1)c10R(et+1)c1+n1(et1)c1n+1tβ(2et)n1dt.

Using the same calculation as we did in the case R1R, we can derive

R1RΦ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dtCn0R|v(t)|n(sinht)n1tβdt. 10

Now, we only need to consider the integral on [0,R1). Set w(t)=v(t)v(R1), then

ωn10R1|w(t)|n(sinht)n1dtρε0.

One can employ the equality (6) to derive that

|v(t)|nn1=(1+ε)w(t)nn1+Cεv(R1)nn1.

Then we obtain

0R1Φ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dt0R1Φ(αn(1βn)|v|nn1)(v(R1))nn1tβ(sinht)n1dtexp(αn(1βn)Cεv(R1)nn1)(v(R1))nn1×0R1exp(αn(1βn)(1+ε)wnn1)tβ(sinht)n1dt. 11

Set ε=ε011n1, then

Cε=(11(1+ε)n1)11n=(1ε0)11n.

Since

ωn1R1|v(t)|n(sinht)n1dtρ(1ε0)1ε0.

We can apply Lemma 4 to obtain

exp(αn(1βn)Cεv(R1)nn1)(v(R1))nn1Cε0R1|v|ntn1dt(1ε0)nn1(R1)nβ. 12

Let Ω={x:d(0,x)<R1} and v1(x)=(1+ε)n1nw(d(0,x)) in Ω, then

Hn|gv1|ndV=ωn10R1(1+ε)n1|w(t)|n(sinht)n1dtρ1.

We can apply the singular Moser-Trudinger inequality on a bounded domain to obtain

Ωexp(αn(1βn)|v1|nn1)(d(0,x))βdVCΩ(d(0,x))βdV. 13

That is,

0R1exp(αn(1βn)(1+ε)|w|nn1)tβ(sinht)n1dtC0R1tβ(sinht)n1dt.

Since sinhtt is monotone increasing on (0,), by (11), (12) and (13) we derive

0R1Φ(αn(1βn)|v|nn1)(1+v)nn1tβ(sinht)n1dtCε0R1|v|ntn1βdt(R1)nβ0R1tβ(sinht)n1dtCε0R1|v|n(sinht)n1(tsinht)n1tβdt(R1)nβ0R1tn1β(sinhtt)n1dtCε0R1|v|n(sinht)n1tβdt. 14

Combining (10) with (14), we obtain the desired inequality of Theorem 1 for R1R. This accomplishes the proof of Theorem 1.

Sharpness

In this section, we show that the desired inequality in Theorem 1 does not hold if the power nn1 is replaced by any p<nn1.

We choose {uk}k=1 as follows:

uk(x)=ωn11nCk{kn1n,if 0d(0,x)ek,|lnd(0,x)|k1n,if ek<d(0,x)1,0,if d(0,x)>1,

where Ck=(k1ek1tn(sinht)n1dt)1n. It is easy to check that Ck1 and Cknn1kk0 as k. Let vk(d(0,x))=uk(x), then

vk(t)=ωn11nCk{kn1n,if 0tek,|lnt|k1n,if ek<t1,0,if t>1.

By calculation, we derive that

Hn|guk|ndV=ωn10|vk(t)|n(sinht)n1dt=(Ck)nkek1tn(sinht)n1dt=1

and

Hn|uk(x)|n(d(0,x))βdV=ωn10|vk(t)|n(sinht)n1tβdt=(Ck)n0ekkn1(sinht)n1tβdt+(Ck)nek1k1|lnt|n(sinht)n1tβdt=O(kn1e(nβ)k)+O(1k)=O(1k). 15

It follows that

HnΦ(αn(1βn)|uk|nn1)(1+uk)p(d(0,x))βdV=ωn10Φ(αn(1βn)|vk|nn1)(1+vk)ptβ(sinht)n1dtΦ(αn(1βn)|ωn11nCkkn1n|nn1)(1+|ωn11nCkkn1n|)p0ek(sinht)n1tβdtΦ((nβ)kCknn1)(1+|ωn11nCkkn1n|)pe(nβ)kkpn1n 16

as k.

When p<nn1, we can apply (15) and (16) to obtain

HnΦ(αn(1βn)|uk|nn1)(1+uk)p(d(0,x))βdV(Hn|uk(x)|n(d(0,x))αdV)1k1pn1n. 17

Thus, we accomplish the proof of Theorem 2.

Conclusions

In this paper, we prove a singular version of Moser-Trudinger inequality with the exact growth condition in the n-dimension hyperbolic space Hn. It is well known that the Moser-Trudinger inequality plays an important role in nonlinear analysis and can be applied to study the ground state solutions of N-Laplacian equation with critical exponential growth. Our results represent very good progress on modern analysis and geometric inequalities.

Acknowledgements

The authors would like to thank the anonymous referee for his/her valuable comments. The work is partly supported by the Scientific Research Fund of Jiangxi Provincial Education Department (No. GJJ160797).

Footnotes

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment. All authors read and approved the final manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

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