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Advances in Nonlinear Analysis

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Weighted Caffarelli–Kohn–Nirenberg type inequalities related to Grushin type operators

Manli Song
  • School of Natural and Applied Sciences, Northwestern Polytechnical University, Xi’an, Shaanxi 710129, P. R. China
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/ Wenjuan Li
  • Corresponding author
  • School of Natural and Applied Sciences, Northwestern Polytechnical University, Xi’an, Shaanxi 710129, P. R. China
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Published Online: 2016-12-20 | DOI: https://doi.org/10.1515/anona-2016-0146

Abstract

We consider the Grushin type operator on xd×yk of the form

Gμ=i=1dxi2+(i=1dxi2)2μj=1kyj2

and derive weighted Hardy–Sobolev type inequalities and weighted Caffarelli–Kohn–Nirenberg type inequalities related to Gμ.

Keywords: Grushin type operator; weighted Hardy–Sobolev inequality; weighted Caffarelli–Kohn–Nirenberg type inequality

MSC 2010: 26D10; 35H20

1 Introduction

Hardy–Sobolev inequalities and Caffarelli–Kohn–Nirenberg inequalities on the Euclidean space play an important role in mathematics and applied fields. They are very useful tools to study various interesting problems in partial differential equations, such as eigenvalue problems, existence problems of equations with singular weights, regularity problems, etc.

The initial work of first order interpolation inequalities with weights (Caffarelli–Kohn–Nirenberg inequalities) was given by Caffarelli, Kohn and Nirenberg [2]. The result is stated as follows.

Theorem 1.

Let p, q, r, α, β, γ, σ and a satisfy

{p,q1,r>0,0a1,1p+αn,1q+βn,1r+γn>0,(1.1)

where γ=aσ+(1-a)β. Then there exists a positive constant C such that for all uC0(Rn) the inequality

|x|γuLrC|x|α|u|Lpa|x|βuLq1-a(1.2)

holds if and only if the following relations hold:

1r+γn=a(1p+α-1n)+(1-a)(1q+βn)(this is dimensional balance),(1.3)

and

0α-σif a>0α-σ1if a>0 and 1p+α-1n=1r+γn.

Furthermore, on any compact set in the parameter space in which (1.1), (1.3) and 0α-σ1 hold, the constant C is bounded.

When a=1, we see that (1.2) are reduced to Hardy–Sobolev inequalities, i.e., Hardy–Sobolev inequalities are special cases of Caffarelli–Kohn–Nirenberg inequalities. Later, Lin [13] generalized (1.2) to cases including derivatives of any order. Badiale and Tarantello [1] derived a class of more general Hardy–Sobolev inequalities with singular weights depending only on partial variables. Recently, Hardy–Sobolev type inequalities have been extended to noncommutative field vectors. In the Heisenberg group setting, we refer the readers to [3, 7, 8, 9, 10], etc.

In this paper, we shall prove weighted Hardy–Sobolev type inequalities and weighted Caffarelli–Kohn–Nirenberg type inequalities related to the Grushin type operator

Gμ=Δx+|x|2μΔy,

where xd and yk. Let Q=d+(1+μ)k be the homogeneous dimension, let

ρ=ρ(x,y)=(|x|2+2μ+(1+μ)2|y|2)12+2μ

be the distance function from the origin to (x,y) on xd×yk and let μ=(x,|x|μy) be the gradient operator. We have Gμ=μμ.

For the Grushin type operator, D’Ambrosio [4] proved some Hardy type inequalities and gave sharp estimates in some cases. Here, we recall a result from [4].

Theorem 2.

Let p>1 and αR satisfy 1p+αQ>1Q. Then there exists a positive constant C=(pQ-p+αp)p such that for any uDα1,p(Rd+k) we have

d+kραp|x|μpρμp|u|pρp𝑑x𝑑yCd+kραp|μu|p𝑑x𝑑y,(1.4)

where Dα1,p(Rd+k) denotes the closure of C0(Rd+k) with respect to the norm

uα1,p=(d+k|ραμu|p𝑑x𝑑y)1p.

Niu and Dou [14] established Hardy–Sobolev inequalities related to Gμ. Zhang et al. [15] obtained a class of weighted Hardy–Sobolev inequalities and a class of weighted Caffarelli–Kohn–Nirenberg inequalities in the special case p=2. In the sequel, let p*(s,p,Q)=p(Q-s)Q-p for any 1<p<Q. The weighted Hardy–Sobolev inequalities are listed as follows.

Theorem 3 (see [15]).

If 0s2<Q, α>2-Q2, there exists a positive constant C=C(s,α,μ,Q) such that for any uDα1,2(Rd+k) we have

d+k|x|μsρμs|ραu|2*(s,2,Q)ρs𝑑x𝑑yC(d+k|ραμu|2𝑑x𝑑y)Q-sQ-2.

The weighted Caffarelli–Kohn–Nirenberg inequalities related to Gμ for the case p=2 are given by the following theorem.

Theorem 4 (see [15]).

Let q, r, α, β, γ, σ and a satisfy

{q1,r>0,0a1,d+μ(α-γ)r>0,d+μ(α-β)q>0,γr+Q>0,βq+Q>0,2α+Q>0,

where γ=aσ+(1-a)β. Then there exists a positive constant C such that for all uC0(Rd+k) the inequality

(d+k(|x|μρμ)(α-γ)rργr|u|r𝑑x𝑑y)1rC(d+kρ2α|μu|2𝑑x𝑑y)a2(d+k(|x|μρμ)(α-β)qρβq|u|q𝑑x𝑑y)1-aq

holds if and only if the following relations hold:

1r+γQ=a(12+α-1Q)+(1-a)(1q+βQ)(this is dimensional balance),

and

0α-σif a>0α-σ1if a>0 and 12+α-1Q=1r+γQ.

In this paper, we are going to establish the following weighted Caffarelli–Kohn–Nirenberg inequalities related to Gμ for 1<p<Q.

Theorem 5.

Let p, q, r, α, β, γ, σ and a satisfy

{1<p<Q,q1,r>0,0a1,d+μ(α-γ)r>0,d+μ(α-β)q>0,αp+Q>0,βq+Q>0,γr+Q>0,

where

γ=aσ+(1-a)β.(1.5)

Then there exists a positive constant C such that for all uC0(Rd+k) the inequality

(d+k(|x|μρμ)(α-γ)rργr|u|r𝑑x𝑑y)1rC(d+kραp|μu|p𝑑x𝑑y)ap(d+k(|x|μρμ)(α-β)qρβq|u|q𝑑x𝑑y)1-aq(1.6)

holds if and only if the following relations hold:

1r+γQ=a(1p+α-1Q)+(1-a)(1q+βQ)(this is dimensional balance),(1.7)

and

0α-σif a>0(1.8)α-σ1if a>0 and 1p+α-1Q=1r+γQ.(1.9)

To prove Theorem 5 by employing the idea in [7, 15], we first need to obtain a class of weighted Hardy–Sobolev type inequalities for 1<p<Q.

Theorem 6.

If 1<p<Q, 0sp and α>p-Qp, there exists a positive constant C=C(s,p,α,μ,Q) such that for any uDα1,p(Rd+k) we have

d+k|x|μsρμs|ραu|p*(s,p,Q)ρs𝑑x𝑑yC(d+k|ραμu|p𝑑x𝑑y)Q-sQ-p.(1.10)

Remark 7.

When a=1, the conditions of Theorem 5 imply

0α-σ=α-γ1,1r+σQ=1p+α-1Q,

and then prp*=QpQ-p. Therefore, there exists t[0,1] satisfying

r=tp+(1-t)p*=p(Q-tp)Q-pand(α-σ)r=tp.

Replacing r and α-σ into (1.6), we easily see that (1.6) is reduced to (1.10), which is exactly a weighted Hardy–Sobolev type inequality.

Remark 8.

Since the methods based on radial symmetry in [2] are no longer suitable for Gμ, Zhang, Han and Dou [15] adopted a different idea. They first proved weighted Hardy–Sobolev type inequalities related to Gμ for the case p=2 and then derived the associated weighted Caffarelli–Kohn–Nirenberg inequalities for the special case. Inspired by their work, we extended their results to all cases 1<p<Q. However, it is still open for the cases pQ.

This paper is organized as follows: Section 2 introduces some definitions and basic facts related to Gμ. In Section 3, we establish a Sobolev–Stein embedding theorem and Hardy–Sobolev type inequalities related to Gμ. Furthermore, we prove Theorem 6. Section 4 is devoted to the proof of Theorem 5.

2 Preliminary

We shall introduce some notions and basic facts about the Grushin type operators. Let μ be a positive real number and (x,y)xd×yk=d+k with d,k1. We denote by |x| (resp. |y|) the Euclidean norm in d (resp. k), i.e., |x|2=i=1dxi2 (resp. |y|2=j=1kyj2). The symbol x (resp. y) and Δx (resp. Δy) stand respectively for the usual gradient operator and the Laplace operator on d (resp. k).

The Grushin type vector fields are defined by

Xi=xi,Yj=|x|μyj,i=1,2,,d,j=1,2,,k,

and the corresponding gradient operator and divergent operator are denoted respectively by

μ=(X1,X2,,Xd,Y1,Y2,,Yk)=(x,|x|μy),divμ(u1,u2,,ud+k)=i=1dXiui+j=1kYjuj+d.

Denote the Grushin type operator by

Gμ=i=1dXi2+j=1kYj2=Δx+|x|2μΔy=μμ.

A family of dilations {δλ:λ>0} on d+k is defined by

δλ(x,y)=(λx,λ1+μy),(x,y)d+k,

and Q=d+(1+μ)k is the corresponding homogeneous dimension. It is easy to see that the vector fields Xi and Yj are homogeneous of degree one with respect to the dilation, i.e., Xi(δλ)=λδλ(Xi), Yj(δλ)=λδλ(Yj), and hence μ(δλ)=λδλ(μ) and Gμ(δλ)=λ2δλ(Gμ).

The distance function from the origin to (x,y) on d+k is defined by

ρ=ρ(x,y)=((i=1dxi2)1+μ+(1+μ)2j=1kyj2)12+2μ=(|x|2+2μ+(1+μ)2|y|2)12+2μ.

It is not difficult to check that ρ is homogeneous of degree one with respect to δλ and

|μρ|=|x|μρμ,Gμρ=(Q-1)|x|2μρ2μ+1.(2.1)

Furthermore, Γ=Cμρ2-Q is the fundamental solution at the origin of Gμ (see [5]).

Denote the open ball of radius R centered at the origin by

BR={(x,y)d+k:ρ(x,y)<R}.

Recalling the explicit polar transform defined by D’Ambrosio [3], we have

dxdy=ρQ-1dρdσ,

where dσ=(11+μ)k|sinθ|d2-1|cosθ|k-1dθdωddωk, and ωd and ωk denote the usual surface measures on d and k, respectively. In addition, the criteria for the integrability of |x|pρq was given as follows:

  • (i)

    If p+d>0 and p+q+Q>0, then

    B2|x|pρq𝑑x𝑑y<+.

  • (ii)

    If p+d>0 and p+q+Q<0, then

    d+kB1|x|pρq𝑑x𝑑y<+.

3 Proof of Theorem 6

Firstly, we need to prove the Sobolev–Stein embedding result related to Grushin type operators.

Theorem 1.

If 1<p<Q, there exists a positive constant C=C(p,μ,Q) such that for any uD01,p(Rd+k) we have

(d+k|u|p*𝑑x𝑑y)1p*C(d+k|μu|p𝑑x𝑑y)1p.

Next, we shall prove the associated Hardy–Sobolev type inequalities.

Theorem 2.

If 1<p<Q, 0sp, there exists a positive constant C=C(s,p,μ,Q) such that for any uD01,p(Rd+k) we have

d+k|x|μsρμs|u|p*(s,p,Q)ρs𝑑x𝑑yC(d+k|μu|p𝑑x𝑑y)Q-sQ-p.

In order to prove Theorem 1, we consider the fractional integral operator

Iνf(x,y)=d+kρ((x,y)-(x,y))ν-Qf(x,y)𝑑x𝑑y,0<ν<Q.

The Hardy–Littlewood–Sobolev theorem for Iν holds (see [6]).

Theorem 3.

Let 0<ν<Q and 1p<Qν. Then we have the following:

  • (i)

    If 1<p<Q , then the condition 1p-1q=νQ is necessary and sufficient for the boundedness of Iν from Lp(d+k) to Lq(d+k).

  • (ii)

    If p=1 , then the condition 1-1q=νQ is necessary and sufficient for the boundedness of Iν from L1(d+k) to Lq,(d+k) , where Lq, denotes the weak Lq space.

Proof of Theorem 1.

For any uC0(d+k), using the integral representation formula for the fundamental solution of Gμ, we have

u(x,y)=d+kΓ((x,y)-(x,y))Gμu(x,y)𝑑x𝑑y.(3.1)

Noting Gμ=μμ and μ*=-μ, and integrating by parts on the right-hand side of (3.1), we obtain

u(x,y)=d+k(μΓ)((x,y)-(x,y))μu(x,y)𝑑x𝑑y.

Since

|μΓ|=Cμ|μ(ρ2-Q)|=Cμ(Q-2)ρ1-Q|μρ|Cμ(Q-2)ρ1-Q,

we obtain

|u(x,y)|Cμ(Q-2)d+kρ((x,y)-(x,y))1-Q|μu(x,y)|𝑑x𝑑y=Cμ(Q-2)I1(|μu|).

Now, the application of Theorem 3 yields

(d+k|u|q𝑑x𝑑y)1qCμ(Q-2)(d+k(I1(|μu|))q𝑑x𝑑y)1qC(d+k|μu|p𝑑x𝑑y)1p,

where

1p-1q=1Q(q=QpQ-p=p*)

and C is a suitable positive constant depending only on p, μ and Q. ∎

Proof of Theorem 2.

Recall p*(s,p,Q)=p(Q-s)Q-p, where 1<p<Q and 0sp. If s=0, then we have p*(p,p,Q)=p and Theorem 2 is reduced to Theorem 1. If s=p, then p*(0,p,Q)=p* and Theorem 2 is reduced to Theorem 2 in the case α=0. Therefore, it suffices to deal with the case 0<s<p.

Denoting p*(s,p,Q)=(1-sp)p*+s, by the Hölder inequality, (1.4) in the case α=0 and Theorem 1, we have

d+k|x|μsρμs|u|p*(s,p,Q)ρs𝑑x𝑑y(d+k|x|μpρμp|u|pρp𝑑x𝑑y)sp(d+k|u|p*𝑑x𝑑y)1-sp((pQ-p)pd+k|μu|p𝑑x𝑑y)sp(C(p,μ,Q)p*(d+k|μu|p𝑑x𝑑y)p*p)1-sp=C(d+k|μu|p𝑑x𝑑y)Q-sQ-p,

where

C=(pQ-p)sC(p,μ,Q)(1-sp)p*.

To prove Theorem 6, we shall introduce two results.

Lemma 4 (see [11]).

Let p1. For all ξ1,ξ2Rn the following inequalities hold:

  • (i)

    If p2 , then

    |ξ1+ξ2|p-|ξ1|p-p|ξ1|p-2ξ1,ξ2C(p)|ξ2|p,|ξ2|p-|ξ1|p-p|ξ1|p-2ξ1,ξ2-ξ1C(p)|ξ2-ξ1|p(|ξ1|+|ξ2|)2-p.

  • (ii)

    If p>2 , then

    |ξ1+ξ2|p-|ξ1|p-p|ξ1|p-2ξ1,ξ2p(p-1)2(|ξ1|+|ξ2|)p-2|ξ2|2,|ξ2|p-|ξ1|p-p|ξ1|p-2ξ1,ξ2-ξ1C(p)|ξ2-ξ1|p,

    where , represents the common inner product on the Euclidean space n.

Lemma 5 (see [12]).

For any ξ,ηRn and λ>0, we have

|ξ|λ+1+λ|η|λ+1-(λ+1)|η|λ-1ξ,η0,

and the equality holds if and only if ξ=η.

Proof of Theorem 6.

The condition α>p-Qp implies

αp*(s,p,Q)-s+Q>0,pα+Q>0,

which ensures that the left and right integral of (1.10) are well defined on C0(d+k). For any uDα1,p(d+k), taking w=ραu, by (2.1), we have

|μw|p=|ραμu+αρα-1uμρ|p(ρα|μu|+|α||ρα-1u||μρ|)p=2p(ρα|μu|+|α||ρα-1u||μρ|2)p2p(ραp|μu|p2+|α|p|ρα-1u|p|μρ|p2)2p-1(ραp|μu|p+|α|pραp|x|μpρμp|u|pρp).

It follows from (1.4) that

d+k|μw|p𝑑x𝑑y2p-1d+k(ραp|μu|p+|α|pραp|x|μpρμp|u|pρp)𝑑x𝑑y2p-1(|α|p(pQ-p+αp)p+1)d+kραp|μu|p𝑑x𝑑y,

which implies wD01,p(d+k). In addition, a straightforward computation gives

d+k|x|μsρμs|ραu|p*(s,p,Q)ρs𝑑x𝑑y=d+k|x|μsρμs|w|p*(s,p,Q)ρs𝑑x𝑑y,d+k|ραμu|p𝑑x𝑑y=d+k|μw-αρ-1wμρ|p𝑑x𝑑y.

Therefore, (1.10) is equivalent to the following inequality:

d+k|x|μsρμs|w|p*(s,p,Q)ρs𝑑x𝑑yC(d+k|μw-αρ-1wμρ|p𝑑x𝑑y)Q-sQ-p.

By Theorem 2, it suffices to prove

d+k|μw-αρ-1wμρ|p𝑑x𝑑yCd+k|μw|p𝑑x𝑑y(3.2)

for some suitable constant C>0.

According to Lemma 4, we will investigate (3.2) in the case 1<p2 and p>2, respectively.

Case 1: 1<p2. Taking ξ1=αρ-1wμρ and ξ2=μw-αρ-1wμρ in the first case of Lemma 4, we obtain

C(p)d+k|μw-αρ-1wμρ|p𝑑x𝑑yd+k{|μw|p-|α|p|w|pρp|μρ|p-pα|α|p-2w|w|p-2ρp-1|μρ|p-2μρ,μw-αρ-1wμρ}𝑑x𝑑y=d+k{|μw|p+(p-1)|α|p|w|pρp|μρ|p-pα|α|p-2w|w|p-2ρp-1|μρ|p-2μρ,μw}𝑑x𝑑y=d+k{|μw|p+(p-1)|α|p|x|μpρμp|w|pρp-α|α|p-2ρ1-p|μρ|p-2μρ,μ|w|p}dxdy.(3.3)

Integrating by parts, we have

d+kρ1-p|μρ|p-2μρ,μ|w|pdxdy=-d+k|w|pdivμ(ρ1-p|μρ|p-2μρ)dxdy.(3.4)

Since

μ(|x|(p-2)μ)μρ=(p-2)μ|x|pμρ2μ+1,

by (2.1), a straightforward computation implies

divμ(ρ1-p|μρ|p-2μρ)=divμ(|x|(p-2)μρ1-p+(2-p)μμρ)=ρ1-p+(2-p)μμ(|x|(p-2)μ)μρ+(1-p+(2-p)μ)|x|(p-2)μρ(2-p)μ-p|μρ|2+|x|(p-2)μρ1-p+(2-p)μGμρ=(Q-p)|x|pμρ(p+1)μ.(3.5)

Putting (3.4) and (3.5) into (3.3), we have

C(p)d+k|μw-αρ-1wμρ|p𝑑x𝑑yd+k|μw|p𝑑x𝑑y+α|α|p-2(Q-p+(p-1)α)d+k|x|μpρμp|w|pρp𝑑x𝑑y.(3.6)

Note that the condition α>p-Qp implies that Q-p+(p-1)α>0.

If α0, it follows from (1.4) and (3.6) that

C(p)d+k|μw-αρ-1wμρ|p𝑑x𝑑yC1d+k|μw|p𝑑x𝑑y,(3.7)

where

C1=1+α|α|p-2(Q-p+(p-1)α)(pQ-p)p=(pQ-p)p{(Q-pp)p+(p-1)|α|p+(Q-p)α|α|p-2}.

Taking ξ=p-Qp, η=α and λ=p-1>0 in Lemma 5, we obtain C1>0.

If α>0, then (3.7) holds naturally with C1=1.

In conclusion, we prove (3.2) with C=C(p)-1C1>0.

Case 2: p>2. A direct calculation gives

2p-2p(p-1)2(|αρ-1wμρ|+|μw|)p-2|μw-αρ-1wμρ|2p(p-1)2(2|αρ-1wμρ|+|μw|)p-2|μw-αρ-1wμρ|2p(p-1)2(|αρ-1wμρ|+|μw-αρ-1wμρ|)p-2|μw-αρ-1wμρ|2.

As in Case 1, applying the estimate in the second case of Lemma 4 to the above inequality, we deduce

2p-2p(p-1)2(|αρ-1wμρ|+|μw|)p-2|μw-αρ-1wμρ|2|μw|p-|αρ-1wμρ|p-p|αρ-1wμρ|p-2αρ-1wμρ,μw-αρ-1wμρ=|μw|p+(p-1)|α|p|x|μpρμp|w|pρp-α|α|p-2ρ1-p|μρ|p-2μρ,μ|w|p.

Therefore, argued as in Case 1,

2p-2p(p-1)2d+k(|αρ-1wμρ|+|μw|)p-2|μw-αρ-1wμρ|2𝑑x𝑑yd+k{|μw|p+(p-1)|α|p|x|μpρμp|w|pρp-α|α|p-2ρ1-p|μρ|p-2μρ,μ|w|p}dxdy=d+k|μw|p𝑑x𝑑y+α|α|p-2(Q-p+(p-1)α)d+k|x|μpρμp|w|pρp𝑑x𝑑yC1d+k|μw|p𝑑x𝑑y(3.8)

holds for a suitable C1>0.

In addition, exploiting the Hölder inequality and the Minkowski inequality, we deduce

d+k(|αρ-1wμρ|+|μw|)p-2|μw-αρ-1wμρ|2𝑑x𝑑y(d+k(|αρ-1wμρ|+|μw|)p𝑑x𝑑y)p-2p(d+k|μw-αρ-1wμρ|p𝑑x𝑑y)2p{(d+k|αρ-1wμρ|p𝑑x𝑑y)1p+(d+k|μw|p𝑑x𝑑y)1p}p-2(d+k|μw-αρ-1wμρ|p𝑑x𝑑y)2p.(3.9)

We conclude from (1.4)

d+k|αρ-1wμρ|p𝑑x𝑑y=|α|pd+k|x|μpρμp|w|pρp𝑑x𝑑y|α|p(pQ-p)pd+k|μw|p𝑑x𝑑y.(3.10)

Hence, by (3.9) and (3.10),

d+k(|αρ-1wμρ|+|μw|)p-2|μw-αρ-1wμρ|2𝑑x𝑑y(|α|pQ-p+1)p-2(d+k|μw|p𝑑x𝑑y)p-2p(d+k|μw-αρ-1wμρ|p𝑑x𝑑y)2p.(3.11)

Combining (3.8) and (3.11), we deduce

d+k|μw-αρ-1wμρ|p𝑑x𝑑yCd+k|μw|p𝑑x𝑑y,

where

C=2p(3-p)2[p(p-1)]-p2(|α|pQ-p+1)p(2-p)2C1p2>0.

In conclusion, (3.2) is proved. ∎

4 Proof of Theorem 5

4.1 Necessity

Necessity of (1.7). Let 0uC0(d+k) satisfy (1.6). Then uλ=uδλ(λ>0) also satisfies (1.6). A direct computation shows

d+k(|x|μρμ)(α-γ)rργr|uλ|r𝑑x𝑑y=λ-γr-Qd+k(|x|μρμ)(α-γ)rργr|u|r𝑑x𝑑y,d+kραp|μuλ|p𝑑x𝑑y=λ-(α-1)p-Qd+kραp|μu|p𝑑x𝑑y,d+k(|x|μρμ)(α-β)qρβq|uλ|q𝑑x𝑑y=λ-βq-Qd+k(|x|μρμ)(α-β)qρβq|u|q𝑑x𝑑y.

Applying (1.6) to uλ, we deduce

λ-γ-Qrλa(-(α-1)-Qp)+(1-a)(-β-Qq),

which is true for any λ>0, so the powers of λ on the two sides must be equal, i.e.,

-γ-Qr=a[-(α-1)-Qp]+(1-a)(-β-Qq),

which is exactly (1.7).

Necessity of (1.8). Let 0uC0(B1) satisfy (1.6). Take (x0,y0)d+k, x00 and for sufficiently large λ>0 define

uλ(x,y)=u((x,y)-δλ(x0,y0))=u(x-λx0,y-λ1+μy0).

Since

d+k(|x|μρμ)(α-γ)rργr|uλ|r𝑑x𝑑y=B1(|x+λx0|ρ(x+λx0,y+λ1+μy0))μ(α-γ)rρ(x+λx0,y+λ1+μy0)γr|u|r𝑑x𝑑y=λγrB1(|λ-1x+x0|ρ(λ-1x+x0,λ-1y+y0))μ(α-γ)rρ(λ-1x+x0,λ-1y+y0)γr|u|r𝑑x𝑑yC1λγr,

and

d+k(|x|μρμ)(α-β)qρβq|uλ|q𝑑x𝑑yC3λβq,

and further

d+kραp|μuλ|p𝑑x𝑑y=d+kρ(x+λx0,y+λ1+μy0)αp|μu|p𝑑x𝑑y=λαpd+kρ(λ-1x+x0,λ-1y+y0)αp|μu|p𝑑x𝑑yC2λαp,

applying (1.6) to uλ, we have

C11rλγC2apC31-aqλaα+(1-a)β.

This yields γ=aσ+(1-a)βaα+(1-a)β, namely (1.8).

Necessity of (1.9). We conclude from (1.7) that

1p+α-1Q=1q+βQ=1r+γQ.

Choose the function

uε={0for ρ1,ρ-γ-Qrlog1ρfor ερ1,ε-γ-Qrlog1εfor ρε.

In polar coordinates, this leads to

d+k(|x|μρμ)(α-γ)rργr|uε|r𝑑x𝑑y=C(0εργr+Q-1(ε-γ-Qrlog1ε)r𝑑ρ+ε1ρ-1(log1ρ)r𝑑ρ)=C((log1ε)rγr+Q+(log1ε)r+1r+1)C1(log1ε)r+1,

and

d+k(|x|μρμ)(α-β)qρβq|uε|q𝑑x𝑑yC2(log1ε)q+1,

and further

d+kραp|μuε|p𝑑x𝑑y2p-1ερ1ρ-Q((Q-pp)plogp1ρ+1)|μρ|p𝑑x𝑑yC3(log1ε)p+1.

Applying (1.6) to uε, we have

C11r(log1ε)1+1rC2apC31-aq(log1ε)a(1+1p)+(1-a)(1+1q),

which implies that

1+1ra(1+1p)+(1-a)(1+1q).

This immediately leads to

1rap+1-aq.(4.1)

The combination of (1.5), (1.7) and (4.1) yields (1.9).

4.2 Sufficiency

If a=0, inequality (1.6) obviously holds true. If a=1, the proof is complete by Remark 7. In the sequel, we deal only with the case 0<a<1.

Case (I): 0<a<1, 0α-σ1. In this case, we have p(1p+α-σ-1Q)-1p*. Analogous to the argument in Remark 7, there exists t[0,1] satisfying

(1p+α-σ-1Q)-1=p(Q-tp)Q-pand(α-σ)(1p+α-σ-1Q)-1=tp.

Applying (1.10) with 0s=tpp, we obtain

d+k(|x|μρμ)(α-σ)(1p+α-σ-1Q)-1|ραu|(1p+α-σ-1Q)-1ρ(α-σ)(1p+α-σ-1Q)-1𝑑x𝑑yC(d+k|ραμu|p𝑑x𝑑y)Q-tpQ-p.(4.2)

From (1.5) and (1.7) we have

1r=a(1p+α-σ-1Q)+1-aq<1,(r>1),(|x|μρμ)α-γργ|u|={(|x|μρμ)a(α-σ)ρaσ|u|a}{(|x|μρμ)(1-a)(α-β)ρ(1-a)β|u|1-a}.

By the Hölder inequality and (4.2), there follows

(d+k(|x|μρμ)(α-γ)rργr|u|r𝑑x𝑑y)1r(d+k(|x|μρμ)(α-σ)(1p+α-σ-1Q)-1|ραu|(1p+α-σ-1Q)-1ρ(α-σ)(1p+α-σ-1Q)-1𝑑x𝑑y)a(1p+α-σ-1Q)(d+k(|x|μρμ)(α-β)qρβq|u|q𝑑x𝑑y)1-aqC(d+kραp|μu|p𝑑x𝑑y)ap(d+k(|x|μρμ)(α-β)qρβq|u|q𝑑x𝑑y)1-aq.

Case (II): 0<a<1, α-σ>1. Putting

Ap=d+k|ραLu|p𝑑x𝑑y,Bq=d+k(|x|μρμ)(β-α)qρβq|u|q𝑑x𝑑y,

we see that (1.6) can be written as

(d+k(|x|μρμ)(α-γ)rργr|u|r𝑑x𝑑y)1rCAaB1-a.

Rescaling u such that AaB1-a=1, we aim to prove

(d+k(|x|μρμ)(α-γ)rργr|u|r𝑑x𝑑y)1rC.(4.3)

Note that in Case (I), inequality (1.6) has been proved for α-σ=0 and α-σ=1. Therefore,

(d+k(|x|μρμ)(α-δ)sρδs|u|s𝑑x𝑑y)1sC,(d+k(|x|μρμ)(α-ε)tρεt|u|t𝑑x𝑑y)1tC,

provided that δ, s, ε and t satisfy

δ=bα+(1-b)β,1s=bp+1-bq-bQ,ε=c(α-1)+(1-c)β,1t=cp+1-cq

for some choices of b and c, 0b,c1 and

d+μ(α-δ)s>0,δs+Q>0,d+μ(α-ε)t>0,εt+Q>0.

One computes that

1t+εQ=c(1p+α-1Q)+(1-c)(1q+βQ),1r+γQ=a(1p+α-1Q)+(1-a)(1q+βQ),1s+δQ=b(1p+α-1Q)+(1-b)(1q+βQ).

Since 0<a<1, α-σ>1, this ensures 1p+α-1Q1r+γQ from (1.9) and then 1p+α-1Q1q+βQ.

Case (II.1): 1p+α-1Q<1q+βQ. Taking b<a<c, we have

1t+εQ<1r+γQ<1s+δQ.(4.4)

A direct computation shows

1r-1s=(a-b)(1p-1q-1Q)+a(α-σ)Q,1r-1t=(a-c)(1p-1q)+a(α-σ-1)Q,

and

(1r+μ(α-γ)d)-(1t+μ(α-ε)d)=(a-c)(1p-1q+μd(β+1-α))+a(α-σ-1)(1Q+μd),(1r+μ(α-γ)d)-(1s+μ(α-σ)d)=(a-b)(1p-1q-1Q+μd(β-α))+a(α-σ)(1Q+μd).

The conditions 0<a<1 and α-σ>1 imply

0<a(α-σ-1)Q<a(α-σ)Q,0<a(α-σ-1)(1Q+μd)<a(α-σ)(1Q+μd),

and for sufficiently small |b-a| and |a-c|,

1r>1s,1r+μ(α-γ)d>1s+μ(α-σ)d,1r>1t,1r+μ(α-γ)d>1t+μ(α-ε)d.

Combining (4.4) and (4.5), we have

d+μ(ε-γ)trt-r>0,(γ-ε)trt-r+Q>0,d+μ(δ-γ)srs-r>0,(γ-δ)srs-r+Q<0.

Choose a fixed C0(d+k) function Φ(x,y) (0Φ1) such that

Φ(x,y)={1if ρ(x,y)<1,0if ρ(x,y)>2.

We shall investigate the left-hand side of (4.3) by splitting it into two parts. One obtains by the Hölder inequality that

(d+k(|x|μρμ)(α-γ)rργrΦ|u|r𝑑x𝑑y)1r(d+k(|x|μρμ)(α-ε)tρεt|u|t𝑑x𝑑y)1t(d+k(|x|μρμ)(ε-γ)rtt-rρ(γ-ε)rtt-rΦtt-r𝑑x𝑑y)1r-1tC(B2|x|μ(ε-γ)rtt-rρ(μ+1)(γ-ε)rtt-r𝑑x𝑑y)1r-1t(4.7)

and

(d+k(|x|μρμ)(α-γ)rργr(1-Φ)|u|r𝑑x𝑑y)1r(d+k(|x|μρμ)(α-δ)sρδs|u|s𝑑x𝑑y)1s(d+k(|x|μρμ)(δ-γ)rtt-rρ(γ-δ)sts-r(1-Φ)ss-r𝑑x𝑑y)1r-1sC(d+kB1|x|μ(δ-γ)rss-rρ(μ+1)(γ-δ)rss-r𝑑x𝑑y)1r-1s.(4.8)

Moreover, (4.6) ensures that the integrals on the right-hand side in (4.7) and (4.8) are bounded, which easily leads to (4.3).

Case (II.2): 1p+α-1Q>1q+βQ. Take c<a<b such that |c-a| and |a-b| are sufficiently small. Now inequalities (4.4)–(4.8) still hold true, and then the desired result (4.3) is derived.

Acknowledgements

The authors deeply thank Professor Pengcheng Niu for his suggestions and encouragement on the paper.

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About the article

Received: 2016-06-27

Revised: 2016-09-14

Accepted: 2016-10-02

Published Online: 2016-12-20


Funding Source: National Natural Science Foundation of China

Award identifier / Grant number: 11371036

Award identifier / Grant number: 11601427

The authors were supported by the National Natural Science Foundation of China (grant no. 11371036 and grant no. 11601427) and the Fundamental Research Funds for the Central Universities (grant no. 3102015ZY068).


Citation Information: Advances in Nonlinear Analysis, Volume 8, Issue 1, Pages 130–143, ISSN (Online) 2191-950X, ISSN (Print) 2191-9496, DOI: https://doi.org/10.1515/anona-2016-0146.

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© 2019 Walter de Gruyter GmbH, Berlin/Boston. This work is licensed under the Creative Commons Attribution 4.0 Public License. BY 4.0

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