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    Hardy-Rellich Type Inequalities Associated with Dunkl Operators?

    2022-04-26 08:28:02LiTANGHaitingCHENShoufengSHENYongyangJIN

    Li TANG Haiting CHEN Shoufeng SHEN Yongyang JIN

    Abstract In this paper, the authors obtain the Dunkl analogy of classical Lp Hardy inequality for p>N+2γ with sharp constant ,where 2γ is the degree of weight function associated with Dunkl operators, and Lp Hardy inequalities with distant function in some G-invariant domains.Moreover they prove two Hardy-Rellich type inequalities for Dunkl operators.

    Keywords Hardy inequalities, Hardy-Rellich inequalities, Best constant, Dunkl operators

    1 Introduction

    The classical Hardy inequality

    If RNis replaced by a bounded convex domain ?, the following sharp inequality holds for 1

    where δ(x):= dist(x,??)(see [18]).Mazya proved in [19] that (1.2)can be characterized in terms of p-capacity.When ? is non-convex,the problem is more complicated.For domains such that ??δ is nonnegative in the distributional sense, some results were obtained by Barbatis,Filippas and Tertikas in[4].It is equivalent between non-negativity of ??δ in the distributional sense and the mean-convexity of the domain when the boundary is smooth enough, see [9–10,17, 20].Ancona [1] obtained some results in planar simply connected domains by using Koebe one-quarter theorem; some other Hardy inequalities for special domains see [8].

    The Rellich inequality

    is a generalization of Hardy inequality,which holds for u ∈C∞0(RN)and the constantis sharp when N ≥5.In [22], Tertikas and Zographopoulos obtained a Hardy-Rellich type inequality which reads as

    In the setting of Dunkl operators, the author in [23] proved a sharp analogical inequality of(1.1)for Dunkl operators

    and the following inequality for

    however the sharpness of the constant for p ≠ 2 in (1.6)is not known.They also obtained an analogical inequality of (1.3)for Dunkl Laplacian

    The plan of this paper is as follows: We introduce some definitions and basic facts of Dunkl operators in the second section.Then, in section three, we obtain some LpHardy inequalities associated with distant function for Dunkl operators by choosing specific vector fields,especially a Hardy inequality on a non-convex domain ? = B(0,R)c, which leads to classical LpHardy inequality associated with Dunkl operators for p > N +2γ.In the last section, we obtain two Hardy-Rellich type inequalities for Dunkl operators by the method of spherical h-harmonic decomposition.

    2 Preliminaries

    Dunkl theory is a generalization of Fourier analysis and special function theory about root system.It generalizes Bessel functions on flat symmetric spaces, also Macdonald polynomials on affine buildings.Moreover,Dunkl theory has extensive applications in algebra(double affine Hecke algebras), probability theory (Feller processes with jump)and mathematical physics(quantum many body problems, Calogero-Moser-Sutherland molds).

    In this section, we will introduce some fundamental concepts and notations of Dunkl operators, see also [6, 21] for more details.

    If a finite set R ?RN{0}such that R ∩αR={?α,α} and σα(R)=R for all α ∈R, then we call R a root system.Denote σαas the reflection on the hyperplane which is orthogonal to the root α, written as

    We write G as the group generated by all the reflections σαfor α ∈R, it is a finite group.Let k:R →[0,∞)be a G-invariant function, i.e., k(α)=k(vα)for all v ∈G and all α ∈R, simply written as kα=k(α).R can be denoted as R=R+∪(?R+), when α ∈R+, then ?α ∈?R+,and R+is called a positive subsystem.We fix a positive subsystem R+in a root system R.Without loss of generality, we assume that |α|2=2 for all α ∈R.

    Definition 2.1Fori=1,··· ,N,the Dunkl operators onC1(RN)is defined as follows

    By this definition,we can see that even if the decomposition of R is not unique,the different choices of positive subsystems make no difference in the definitions due to the G-invariance of k.Denote by ?k= (T1,··· ,TN)the Dunkl gradient,the Dunkl-Laplacian.Especially, for k = 0 we have ?0= ?and ?0= ?.The Dunkl-Laplacian can be written in terms of the usual gradient and Laplacian as follows:

    The weight function naturally associated to Dunkl operators is

    This is a homogeneous function of degree 2γ, where

    We will work in spaces Lp(μk),where dμk=ωkdx is the weighted measure.About this weighted measure we have the formula of integration by parts

    If at least one of the functions u, v is G-invariant, the following Leibniz rule

    Ti(uv)=uTiv+vTiu

    holds.In general, we have

    3 Lp Hardy Inequalities

    In this section we prove a general Hardy inequality with remainder terms for Dunkl operators in G-invariant domains,then we get the Dunkl analogy of Hardy inequality(1.1)for p>N+2γ.

    Firstly, we review some basic facts of distant function.

    Lemma 3.1(see [3])Let? ?RNbe an open set such that?? ≠ ?.The following propositions hold true.

    (i)The functionδ(x)is differentiable at a pointx ∈?if and only if there exists a uniquepointN(x)=y ∈??such thatδ(x)=|x ?y|.Ifδ(x)is differentiable, thenand|?δ|=1.

    (ii)DenoteΣ(?)as the set of points whereδ(x)is not differentiable.If?is bounded withC2,1boundary, then|Σ(?)|=0.

    (iii)Assume that?is convex.Then?δ ≤0in the sense of distributions, i.e.,

    For ? ?RN, if for all x ∈?, g ∈G, we have gx ∈?, then ? is called a G-invariant domain.

    Lemma 3.2If? ?RNis G-invariant,g ∈G,x ∈?Σ(?), then

    ProofFrom the proof of Theorem 5.2 in [23], the function δ(x)is G-invariant.For any x ∈?, we have y =N(x)∈??, δ(x)=|x ?y| and

    δ(gx)=δ(x)=|x ?y|=|g(x ?y)|=|gx ?gy|.

    Due to the uniqueness of N(x), we get that N(gx)=gy.Therefore

    Remark 3.1If F =h1x+h2?δ, where h1, h2are G-invariant functions, then by Lemma 3.2, we have that 〈α,F(σαx)〉=?〈α,F〉.

    Theorem 3.1Let? ?RNbe a G-invariant domain with|Σ(?)| = 0.Then for allu ∈C∞0(?), we have the inequality

    where

    ProofIf F satisfies that 〈α,F(σαx)〉=?〈α,F〉, then

    Let x=σαy.Then

    Because of 〈α,F(σαy)〉 = ?〈α,F(y)〉, 〈α,σαy〉 = ?〈α,y〉, dμk(σαy)= ωk(σαy)d(σαy)=ωk(y)|J|dy, where

    Straightforward calculation shows that J =?1.

    Thus dμk(σαy)=dμk(y), and we have

    Putting (3.5)into (3.3), we get

    we used H¨older inequality and Young inequality in the last inequality above.Then,

    The last inequality above is obtained by using H¨older inequality

    we thus complete the proof of Theorem 3.1.

    Remark 3.2If the root systemsatisfies span()?RN?1.Then the following inequality holds for any u ∈C∞0(RN?1×R+),

    Let SNdenote the symmetric group in N elements.A root system of SNis given by R={±(ei?ej),1 ≤i

    (span(R))⊥=e1+···+eN=:η,

    see [3] for more details.Let the domain ? = span(R)×η+, where η+is the positive direction of the straight line coinciding with η.Then ? is G-invariant, δ(x)=dist(x,spanR)and

    Fix R+= {ei?ej,1 ≤i < j ≤N}, then ??δ = 0 and 〈ρ,?δ〉 = 0, by Theorem 3.1, we have the following corollary.

    Corollary 3.1ForR={±(ei?ej),1 ≤i

    wherek =kα=kβ,?α,β ∈R,=(x1,··· ,xi?1,xj,xi+1,··· ,xj?1,xi,xj+1,··· ,xN).

    ProofIt is easy to prove v ?σα?v?1=σvαfor all v ∈G, as there is one conjugate class in R,so kα=kβfor all α,β ∈R,see also[6].Straightforward computation shows σei?ej(x)=.

    By (3.8)in the proof of Theorem 3.1, it is easy to see that the following theorem holds.

    Theorem 3.2If? ?RNsatisfies|Σ(?)| =0,〈ρ,?δ〉 ≥0.The following inequality holds for allu ∈C∞0(?),

    where?is a positive constant.

    Remark 3.3If a domain ? satisfies that |Σ(?)| = 0, 〈ρ,?δ〉 ≥0 and δ?kδ ≤θ < p ?1,where θ is a positive constant, i.e., then there is a positive constant C =C(θ,p)such that

    Corollary 3.2Suppose that? = B(0,r)c,p > N +2γ, the following inequality holds for allu ∈C∞0(?),

    ProofWhen ? =B(0,r)c, then |Σ(?)| =0, δ =|x|?r,and

    Let r tend to zero, the following sharp inequality follows from Corollary 3.2.

    Corollary 3.3Suppose thatp > N + 2γ.The following inequality holds for allu ∈(RN{0}),

    ProofThere only remains to prove the optimality of the constant.For any?>0 we choose

    We can write dμk= rN+2γ?1ωk(ξ)drdν(ξ), where ν is the surface measure on the sphere SN.Thus by directly computing, we have

    4 Hardy-Rellich Type Inequality

    Spherical h-harmonicsWe will introduce some concepts and fundermental facts for spherical h-harmonic theory, see [6] for more details.If a homogeneous polynomial p of degree n that satisfies

    ?kp=0,

    then we call it an h-harmonic polynomial of degree n.Spherical h-harmonics (or just hharmonics)of degree n are defined as the restrictions of h-harmonic polynomials of degree n to the unit sphere SN?1.Denote Pnthe space of h-harmonics of degree n.Denote d(n)the dimension of Pn, it is finite and given by following formula:

    Moreover,the space L2(SN?1,ωk(ξ)dξ)can be decomposed as the orthogonal direct sum of the spaces Pn, for n=0,1,2,···.

    where ?k,0is an analogue of the classical Laplace-Beltrami operator on the sphere, and it only acts on the ξ variable.Then the spherical h-harmonics Yniare eigenfunctions of ?k,0, and its eigenvalues are given by

    The h-harmonic expansion of a function u ∈L2(μk)can be expressed as

    where

    and ν is the surface measure on the sphere SN?1.

    Theorem 4.1Let≠2.Then we have the inequality

    where:=N +2γ, and the constantis sharp.

    ProofOur goal is to find best constant C satisfying

    Using spherical decomposition:

    By integration by parts, we have

    By using the following two weighted Hardy inequalities,

    we get

    Thus (4.1)holds.Finally, we show the optimality of.For any ?>0,

    Straightforward calculation shows

    Theorem 4.2AssumeN ≥5+2γ.Then, for anyu ∈(RN), we have the inequality

    where the constantis sharp.

    ProofBy integration by parts,

    where

    Then

    Therefore

    Let

    where

    From Parseval identity, we have

    Also

    Then we have

    By using spherical decomposition,

    By integration by parts, we obtain

    since λ0=0, one has B0=0.

    Using the following weighted Hardy inequality

    and denoting

    we have

    For n=0,

    here

    So

    When N ≥5+2γ, D1≥0,

    Dn≥D2≥0 (n=3,4,···), so (4.5)holds.

    Next we prove the optimality of the constantFor any ?>0, take

    By directly computing, it follows that

    AcknowledgementThe authors thank the anonymous reviewer’s suggestions for improving this manuscript.

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