ZHU Maochunand ZHENG Yifeng

School of Mathematical Sciences,Institute of Applied System Analysis,Jiangsu University,Zhenjiang 212013,China.

Abstract. In this paper,we are concerned with a sharp fractional Trudinger-Moser type inequality in bounded intervals of R under the Lorentz-Sobolev norms constraint.For any ,we obtainand βq is optimal in the sense thatfor any β>βq.Furthermore,when q is even,we obtainfor any function h:[0,∞)→[0,∞) with limt→∞h(t)=∞.As for the key tools of proof,we use Green functions for fractional Laplace operators and the rearrangement of a convolution to the rearrangement of the convoluted functions.

Key Words:Trudinger-Moser inequality;Lorentz-Sobolev space;bounded intervals.

1 Introduction

Let ? ?Rn(n≥2) denote a bounded domain in Rn.Then the Sobolev embedding theorem assures thatW1,n0 (?)?Lq(?) for anyq∈[1,+∞),but easy examples show thatW1,n0 (?)/?L∞(?).One knows from the works by Yudovich [1],Pohozaev [2]and Trudinger [3] thatembeds into the Orlicz spaceL?(?),with N-functionThis embedding was made more precise by J.Moser[4],and obtained the following inequality(now is calledTrudinger-Moser inequality)

whereωn?1denotes the area of the unit sphere in Rn.This result has led to many related results:extension of Trudinger-Moser inequality to unbounded domains,see[5–9],an analog of (1.1) for higher order derivatives is established by Adams [10],extension of Trudinger-Moser inequality in Lorentz spaces,see[11–13].For other important extensions can see[14–18],and the reference therein.

Recently,S.Lula,A.Maalaoui and L.Martinazzi[19]established the Trudinger-Moser type inequality in one dimension,which can be sated as the following inequality:

is optimal.

In this work,we are interested in the Trudinger-Moser type inequality in one dimension space under the Lorentz-Sobolev norms constraint.In order to state the main results of the paper we first focus our attention on the fractional Sobolev spaces and introduce some relevant function spaces.

Let us consider the space

wheres∈(0,1).For functions?∈Ls(Rn),we define the fractional Laplacian(??)s/2?as follows.

where F is the Fourier transform.Foru∈Ls(Rn),the fractional Laplacian (??)s/2uis defined as an element ofS′,the tempered distributions on Rn,by the relation

where S denotes the Schwartz space of rapidly decreasing smooth functions.

Also note that it could happen that supp((??)s/2u)/?? even if supp(u)?? for some open set ? in Rn.By using the above notion,we define the Bessel potential space

and its subspace

for a set ? ?Rnis a bounded interval (possibly unbounded),s≥0 andp∈(1,∞).Both spaces are endowed with the norm

On the other hand,we define the Sobolev-Slobodeckij spaceWs,p(R)as follows

From[20],we know thatHs,2(Rn)=Ws,2(Rn),however in general,Hs,p(Rn)/=Ws,p(Rn)forp/=2.

In the following,we consider the simplest one-dimensional case,that is,n=1,andp=2.In this case,the Bessel potential spaceH1/2,2(R) coincides with the Sobolev-Slobodeckij spaceW1/2,2(R),and both seminorms are related as

Now we focus on Lorentz space,let?be a measurable function in an open bounded setI?R.Denoting by|I|the Lebesgue measure of a measurable setI?R,let

be the distribution function of?.The decreasing rearrangement??(x)of?is defined as the distribution function ofμ?,that is

and the spherically symmetric rearrangement?#(x)of?is defined by

whereI#is the open ball with center in the origin which satisfies

Since??is nonincreasing,the maximal function???of the rearrangement of??,defined by

is also nonincreasing and??≤???.For more information of the rearrangement we refer the reader to[21].

The Lorentz spacesL(p,q),with 1

It is well known that whenq>pthe quantity‖?‖p,qis not a norm.But the quantity

is a norm for anypandq.It is easy to prove by using the Hardy’s inequalities that these two quantities are equivalent in the sense that

We state thatL(p,p)=Lp,L(p,∞)=Mp(the Marcinkiewicz space),and for 1

Now,we state the main results of this paper.

Theorem 1.1.Let I be a bounded subset ofR,and let u∈such that:

where‖·‖2,q is Lorentz norms.Then for any,we have:

Moreover,for any β>βq,we have

Furthermore,when q is even we have

for any function h:[0,∞)→[0,∞)withlimt→∞h(t)=∞.

2 The proof of Theorem 1.1

In this section,we will prove Theorem 1.1.To do so,we will apply some ideas used in [10] and [19].First,we will writeuin terms of a Green representation formula and state a lemma which was originally proved in [10].This is a key tool for proving the theorem.

Lemma 2.1.For any p∈(1+∞),setand

Let I be a bounded subset ofRand u∈(I),,then we have

Proof.From Lemma A.1 in the appendix,we get

whereG1/4is the Green’s function of the interval defined in Lemma A.2 in the appendix,Therefore,we have

In particular,ifg(x)=I1/2(x)=|x|?1/2,then by directly computing,we get

Thus by O’Neil’s lemma[22,Lemma 1.5],for anyt>0,we can obtain

This completes the proof.

Lemma 2.2([10,Lemma 1]).Let a(s,t)be a nonnegative measurable function in(?∞,+∞)×[0,+∞)such that for some p∈[1,∞],

Suppose that for φ(s)≥0

holds.Then a constant c0=c0(p,λ)exists such that

Now we can give the proof for Theorem 1.1.

Proof of Theorem1.1.Taking into account the result of Lemma 2.1,where as usual,we have putBy the change of variablet=|I|e?s,then we have

It is clear that(2.3)is satisfied.On the other hand,for any 1

and then(2.4)holds withλ=(4/q′)1/q′.

Finally we observe that

This means that also(2.5)is satisfied.Then Lemma 2.2 implies:

Taking into account(2.7),we get

Now,taking into account thatu?(s)≤u??(s),we get

It remains to show(1.3) and (1.4).By a simple scaling argument it suffices to prove(1.3)and(1.4)for a given interval,sayI=(?1,1).The proof is based on the construction of suitable test functions and it is split into two steps,which was first used in[19].

Step I.Definition of the test functions.We fixτ≥1 and set

Now we check thatf(y)∈L(2,q).By directly calculating,we have

Now letu=uτ∈solve

in the sense of Theorem 4 in the appendix.As in[19],we obtainu∈.

Step II.Conclusion.Recalling that(??)1/4u=finI,from Lemma A.2,we have forx∈I

whereG1/4(x,y)andH1/4(x,y)are as in Lemma A.2.

We now want a lower bound foruin the interval[?r,r].We fix 0

SinceH1/4is bounded on[?r,r]×[?1/2,1/2],we have

asτ→∞.We now setwτ:=,where

That isF(τ)～τ(τ→∞).

Whenβ>βq,we rewriteβasβ=βq(1+ε),for someε>0.Then we get

Consequently,we obtain

In particular,whenqis even,we putq=2n<∞.Therefore,we obtain

wheneverhsatisfies limt→∞h(t)=∞.

Acknowledgments

The first author was partially supported by Natural Science Foundation of China(120-71185,11971202,12061010),Natural Science Foundation of Jiangsu Province(BK20160483),Outstanding Young foundation of Jiangsu Province (BK20200042) and Jiangsu University Foundation Grant(16JDG043).

Appendix

Theorem A.1.Given s∈(0,1),f∈L2(I)and g:R→Rsuch that

there exists a unique function u∈+g solving the problem

Moreover such u satisfies(??)su=in I in the sense of distributions,i.e.

Lemma A.1([19,Lemma 2.1]).For s∈(0,1)the fundamental solution of(??)s/2onRis

i.e.,Fs(x)=in the sense of tempered distributions.

Lemma A.2([19,Lemma 2.2]).Fix s∈(0,1),for any x∈I=(?1,1),let gx∈C∞(R)be any function with gx(y)=Fs(x?y)for y∈Ic.Then there exists a unique solution Hs(x,·)∈+gx to

and the function Gs(x,y):=Fs(x?y)?Hs(x,y)is the Green function of(??)s/2on I,i.e.for x∈I it satisfies

moreover

For any u∈we have

where the right-hand side is well defined for a.e.x∈I thanks to(A.4)and Fubini’s theorem.