YANG Liu and LI Xiaomeng

1 College of Education,Huaibei Institute of Technology,Huaibei 235000,China.

2 School of Mathematics and Big Data,Chaohu University,Hefei 230000,China.

3 School of Mathematical Science,Huaibei Normal University,Huaibei 235000,China.

Abstract. Let W1,n(Rn)be the standard Sobolev space.For any τ＞0 and p＞n＞2,we denoteDefine a norm in W1,n(Rn)bywhere 0 ≤α＜λn,p.Using a rearrangement argument and blow-up analysis,we will provecan be attained by some function u0∈W1,n(Rn)∩C1(Rn) with ‖u0‖n,p=1,here αn=and ωn-1 is the measure of the unit sphere in Rn.

Key Words: Trudinger-Moser inequality;extremal function;blow-up analysis.

1 Introduction

Letn ≥2,and denotewhereωn-1is the area of the unit sphere in Rn.The famous Trudinger-Moser inequality[1-5]states that,for a bounded domain Ω?Rnand 0＜γ≤αn,

Ifγ＞αn,the integrals in(1.1)are still finite,but the supremum is infinity.

One of the interesting questions about Trudinger-Moser inequalities is whether extremal function exists or not.The first result in this direction was obtained by Carleson-Chang[6]in the case that Ω is a unit disk in Rn,then by Struwe[7]when Ω is a close to the ball in the sense of measure,by Flucher [8] for any bounded smooth domain in R2,and by Lin[9]to an arbitrary domain in Rn.

The Trudinger-Moser inequality(1.1)was extended by Cao[10],Panda[11],do ó[12],Ruf[13],and Li-Ruf[14]to the entire Euclidean space Rn(n≥2).Precisely,for anyγ≤αn,

Adimurthi-Yang[15]generalized(1.2)to a singular version.That is,for allτ＞0,n≥2,0＜β＜1 and 0＜η ≤1-β,one has

Obviously,for allτ∈(0,+∞),the normsare equivalent to the standard norms onW1,n(Rn).Then Li-Yang[16]obtained the existence of extremal functions for(1.3)using blow-up analysis.Later,(1.3)was extended by Li[17]to the following modified form.Letp＞n≥2 and

For 0＜β＜1 and 0≤α＜λn,p,the supremum

can be attained.Here and in the sequel

Clearly,(1.3)is a special case of(1.5).

In recent work,Li[18]proved that forp＞2 and 0≤α＜λ2,p,the supremum

can be achieved by some functionu0∈W1,2(R2)with‖u0‖2,p=1.

In[19],do ó and Souza proved that for 0≤?＜1,

wheremoreover,the extremal function for(1.8)exists.For results related to Trudinger-Moser inequality we refer to[20-24]and references therein.

Inspired by [17-19],we shall establish in this note the following extension of the Trudinger-Moser inequality(1.7)in high dimension.For simplicity we define a function Φ:N×RR by

Now we state our main result as follows:

Theorem 1.1.Let p＞n be a real number,λn,p and‖u‖n,p be defined as in(1.4),(1.6)respectively.For any fixed α,0≤α＜λn,p,there exists some u0∈W1,n(Rn)∩C1(Rn)with‖u0‖n,p=1such that

We prove Theorem 1.1 via the method of blow-up analysis.This method originally introduced by Ding-Jost-Li-Wang [25] and Li [26].Then,it has been successfully applied in the proof of Trudinger-Moser inequalities (see [27-32]).We have divided the proof into the following parts.In Section 2,for 0＜?＜αn,we prove that the subcritical Trudinger-Moser functionalΦ(n,(αn-?)|u|n/(n-1))dxhas a maximizer,denoted byu?.In Section 3,we perform the blow-up procedure.In Section 4,applying the result of Carleson-Chang[6],we derive an upper bound ofΦ(n,(αn-?)|u?|n/(n-1))dx.In Section 5,we prove the existence result Theorem 1.1 by constructing a test function sequence.

Throughout this note,various constants are often denoted by the sameC.‖·‖pdenotes theLp-norm with respect to the Lebesgue measure.Bris the ball of radiusrcentered at 0.

Before starting the next section,we quote some results for our use later.

Lemma 1.1(Radial Lemma).For any x∈Rn{0},if u*∈Ln(Rn)is a nonnegative decreasing radially and symmetric function,then one has

Lemma 1.2.Let R＞0be fixed.Suppose that u∈W1,n(BR)is a weak solution of

then we have

?If u≥0and f∈Lp(BR)for some p＞1,then there exists some constant C=C(n,R,p)such thatsupBR/2u≤C(infBR/2u+‖f‖Lp(BR)).

?If ‖u‖L∞(BR)≤L and ‖f‖Lp(BR)≤M for some p ＞1,then there exist two constants C=C(n,R,p,L,M)and0＜θ ≤1such that u∈and

?If‖u‖L∞(BR)≤L and‖f‖L∞(BR)≤M,then there exist two constants C=C(n,R,L,M)and0＜θ ≤1such that u∈C1,θ

Lemma 1.1 was due to Berestycki and Lions [33].The first two estimates in Lemma 1.2 were proved by Serrin[34],while the third estimate was obtained by Tolksdorf[35].

2 The subcritical case

For any 0＜?＜αn,we prove the existence of maximizer for the subcritical functional

For simplicity,writingαn,?=αn-?,Δnu=div(|?u|n-2?u),

Lemma 2.1.Let p＞n and0≤α＜λn,p be fixed.Then for any0＜?＜αn,there exists some nonnegative decreasing radially and symmetric function u?∈W1,n(Rn)∩C1(Rn)satisfying‖u?‖n,p=1and

Moreover,the Euler-Lagrange equation of u? is

Proof.By the Schwarz rearrangement(see[36]),we have

Since‖uk‖n,p ≤1 and 0≤α＜λn,p,we have

Therefore,ukis bounded inW1,n(Rn).Up to a subsequence,as∞,

The factuk ?u?weakly inW1,n(Rn)leads to

Combining now(2.5)and(2.6),we get

For anyu∈H,we have from(2.4)that

Then,it follows from Lemma 1.1 that

For allυ＞0,there exists a sufficiently larger0＞0 such that

Note that,

The mean value theorem and the factuk u?strongly infor anyq ＞1 implies that

Sinceυ＞0 is arbitrary,we have by(2.7)and(2.8)

We employ (2.3) and (2.9),and so identity (2.1) is verified.Hereu?is a maximizer forFα,p,?.

Next we proveu?0 and‖u?‖n,p=1.Suppose not,ifu?=0,clearly this leads to a contradiction asdx=0.Let‖u?‖n,p＜1.Then it follows that

This is obviously impossible.

A straightforward computation shows thatu?satisfies the Euler-Lagrange equation(2.2).Applying Lemma 1.2 to(2.2),we haveu?∈C1(Rn).

Lemma 2.2.Let λ? be as in(2.2),it holds that

Proof.Clearly,we have

On the other hand,

which implies that

Taking the supremum overu∈W1,n(Rn)with‖u‖n,p ≤1,we obtain

Note that for anyt≥0,

One has from(2.10)and(2.11)that

Thus we obtain the desired result.

3 Blow-up analysis

Denotec?=u?(0)=maxRn u?(x).Since‖u?‖n,p=1 and 0≤α＜λn,p,one can find some functionu0∈W1,n(Rn)such thatu??u0weakly inW1,n(Rn),u?u0strongly inLrloc(Rn)for allr＞1,andu? u0a.e.in Rn.

We may first assumec?is bounded,and have the following:

Lemma 3.1.If c? is bounded,then Fα,p is attained.

Proof.For anyR＞0,there holds

In addition,applying Lemma 1.1,we have

which together with(3.1)gives that

Applying Lemma 1.2 to (2.2),we conclude thatu? u0in(Rn).Therefore,u0is a desire extremal function and Theorem 1.1 holds.

Next,we assumec?+∞as0.We have the following:

Lemma 3.2.There holds u0≡0,and up to a subsequence|?u?|ndx ?δ0,where δ0denotes the Dirac measure centered at0∈Rn.

Proof.We frist prove|?u?|ndx?δ0.Suppose not,there exists0 such that

fort≥0 andq≥1.Then we immediately get

here we use Lemma 1.1.As a result,

Choosingυ＞0 sufficiently small andsufficiently close to 1,such that

By classical Trudinger-Moser inequality(1.1),we conclude that

Next we proveu0≡0.In view of‖u?‖n,p=1 and|?u?|ndx?δ0,we get‖u?‖n=o?(1),‖u?‖p=o?(1).Then

which impliesu0≡0.

Let

Then we have the following:

Lemma 3.3.For any κ＜αn/n,there holds

Proof.By definition ofr?,one has for anyR＞0

We now estimateI1andI2respectively.Note that

Then we have

Therefore,we obtain

hereo?(R) denotes that=0 for a fixedR ＞0.On the other hand,in view ofwe estimate

wheres1＞1,1/s1+1/s2=1 and 1＜s2＜αn/nκ.As in Lemma 3.2,we can see that

Recall thatu?0 in(Rn)forp＜∞,andc?+∞as0,we have

The desire result follows from(3.4)and(3.5).

Define two blow-up functions

Lemma 3.4.Let vn,? and wn,? be defined as in(3.6)and(3.7).Then vn,?1in(Rn),wn,? wn in(Rn),where

Moreover,

Proof.One has

By Lemma 3.3 and(3.6),we have

Then applying Lemma 1.2 to(3.9),we obtain

Noticing thatvnis a solution of the equation-Δnvn=0 in the distributional sense.The Liouville theorem implies thatvn ≡1 on Rn.

From the result in Lemma 1.2,we know that supBR wn,?(x)≤C(R)forR＞0.Applying Lemma 1.2 to(3.10),we have

Similar as in[16],we get

Consequently,wnsatisfies

with

Using the classification result for the quasi-linear Liouville equation of Esposito[38],we get

We can also refer to[14]for this kind of solution.Integration by parts,it then follows that

We next consider the convergence ofu?away from the concentration point 0.Similar to[26],defineu?,β=min{βc?,u?}for 0＜β＜1.Then we have the following:

Lemma 3.5.For any0＜β＜1,it holds that

Proof.Testing(2.2)with(u?-βc?)+,for any fixedR＞0,we obtain

here we use the estimates as below:

and

Letting+∞,we have

thanks to(3.8).Similarly,testing(2.2)withu?,β,we get

Note that,

This along with(3.11)and(3.12)gives the desire result.

Lemma 3.6.Under the assumption c?+∞0,we have

and for any θ＜n/(n-1),there holds+∞0.

Proof.Notice that,

for allt≥0.Then we have

and therefore,

On the other hand,

We claim that,

In fact,applying the mean value theorem to function Φ(n,t)and then using(3.13)again,we obtain

From the H¨older inequality and(3.2),one has

where 1/k1+1/k2=1 andk1＜1/β.In view of the definition ofu?,β,we obtain that

With the help of Trudinger-Moser inequality (1.2),and the fact‖u?,β‖p=o?(1),(3.16)follows.Due to(3.15)and(3.16),then letting1,we conclude

Combining(3.14)and(3.17),we obtain the desired result.

This is impossible since0.We finish the proof of the lemma.

Lemma 3.7.For any ζ(x)∈,there holds

Proof.We write

Let 0＜β＜1 be fixed,we divide Rninto three parts

We estimate the integrals ofζ(x)h?(x)over the right three domains of(3.18)respectively.Notice first that

Also,it follows from Lemma 3.3 that BRr? ?{u? ＞βc?}for sufficiently small? ＞0.Thus we have from(3.8)

In addition,we obtain

Consequently,letting0 and+∞,the desired result will now follow from the above estimates.

To proceed,we state the result as below,which can be proved by the similar idea in[13,Lemma 7].We omit the details.

Lemma 3.8.For any1＜q＜n,we have

where G is a Green’s function and satisfies

in a distributional sense.

Moreover,Gtakes the form

whereAis a constant,g(x)=O(|x|nlogn-1|x|)as0 andg∈C1(Rn).

4 An upper bound

In this section,we will derive an upper bound fordx.

By Lemma 3.8,we compute,for any fixedδ＞0,

Also,

and

Set

Using the result of Carleson-Chang[6],we obtain

From (4.1),we see thatτ? ≤1 for?andδ ＞0 sufficiently small.By Lemma 3.4,we getu?=c?+o?(1) on BRr?for a fixedR ＞0.This together with Lemma 3.8 leads to that on BRr? ?Bδ,

As a result,we obtain

Letting0,now(4.2)and(4.3)imply that

for anyR＞0.On the other hand,

and therefore,

Then we conclude from Lemma 3.6

5 Test function computation

Proof of Theorem1.1.To finish the prove of Theorem 1.1,we will construct a family of test functionφ?(x)∈W1,n(Rn)satisfying‖φ?‖n,p=1 and

for?＞0 sufficiently small.The contradiction between(4.4)and(5.1)tells us thatc?must be bounded.Then applying Lemma 1.2 to(2.2),we get the desired extremal function.For this purpose,define

Recall(3.19)andR=(-log?)2,we get

Integration by parts along with Lemma 3.8,we calculate

Also,a direct calculation shows that

It is easy to check that

Combining the above estimates(5.2)-(5.5)yields

Setting‖φ?‖n,p=1,we have

Then we conclude

Plugging(5.6)and(5.7)into the following estimate,we have

Making a change of variablest=we have

Moreover,on RnBR?,we have the estimate

Therefore,we conclude for?＞0 sufficiently small

Acknowledgement

This work is supported by National Science Foundation of China(Grant No.12201234),Natural Science Foundation of Anhui Province of China(Grant No.2008085MA07)and the Natural Science Foundation of the Education Department of Anhui Province(Grant No.KJ2020A1198).