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    Extremal Functions for an Improved Trudinger-Moser Inequality Involving Lp-Norm in Rn

    2023-04-16 18:51:16YANGLiuandLIXiaomeng

    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).

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