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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 32585, 24 pages doi:10.1155/2007/32585 Research Article A Hardy Inequality with Remainder Terms in the Heisenberg Group and the Weighted Eigenvalue Problem Jingbo Dou, Pengcheng Niu, and Zixia Yuan Received 22 March 2007; Revised 26 May 2007; Accepted 20 October 2007 Recommended by L ´ aszl ´ oLosonczi Based on properties of vector fields, we prove Hardy inequalities with remainder terms in the Heisenberg group and a compact embedding in weighted Sobolev spaces. The best constants in Hardy inequalities are determined. Then we discuss the existence of solutions for the nonlinear eigenvalue problems in the Heisenberg group with weights for the p- sub-Laplacian. The asymptotic behaviour, simplicity, and isolation of the first eigenvalue are also considered. Copyright © 2007 Jingbo Dou et al. This is an open access article distributed under the Creative Commons Attribution License, w hich permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction Let L p,μ u =−Δ H,p u −μψ p |u| p−2 u d p ,0≤ μ ≤  Q − p p  p , (1.1) be the Hardy operator on the Heisenberg group. We consider the following weighted eigenvalue problem with a singular weight: L p,μ u = λf(ξ)|u| p−2 u,inΩ ⊂ H n , u = 0, on ∂Ω, (1.2) where 1 <p<Q = 2n +2, λ ∈ R, f (ξ) ∈ Ᏺ p :={f : Ω→R + | lim d(ξ)→0 (d p (ξ) f (ξ)/ (ψ p (ξ)) = 0, f (ξ) ∈ L ∞ loc (Ω \{0})}, Ω is a bounded domain in the Heisenberg group, and the definitions of d(ξ)andψ p (ξ); see below. We investigate the weak solution of (1.2)and the asymptotic behavior of the first eigenvalue for different singular weights as μ increases to ((Q − p)/p) p . Furthermore, we show that the first eigenvalue is simple and isolated, as 2 Journal of Inequalities and Applications well as the eigenfunctions corresponding to other eigenvalues change sign. Our proof is mainly based on a Hardy inequality with remainder terms. It is established by the vec- tor field method and an elementary integral inequality. In addition, we show that the constants appearing in Hardy inequality are the best. Then we conclude a compact em- bedding in the weighted Sobole v space. Themaindifficulty to study the properties of the first eigenvalue is the lack of regu- larit y of the weak solutions of the p-sub-Laplacian in the Heisenberg group. Let us note that the C α regularity for the weak solutions of the p-subelliptic operators formed by the vector field satisfy ing H ¨ ormander’s condition was given in [1] and the C 1,α regularity of the weak solutions of the p-sub-Laplacian Δ H,p in the Heisenberg group for p near 2 was proved in [2]. To obtain results here, we employ the Picone identity and Harnack inequality to avoid effectively the use of the regularity. The eigenvalue problems in the Euclidean space have been studied by many authors. We refer to [3–11]. These results depend usually on Hardy inequalities or improved Hardy inequalities (see [4, 12–14]). Let us recall some elementary facts on the Heisenberg group (e.g., see [15]). Let H n be a Heisenberg group endowed with the group law ξ ◦ ξ  =  x + x  , y + y  ,t + t  +2 n  i=1  x i y  i − x  i y i   , (1.3) where ξ = (z,t) = (x, y,t) = (x 1 ,x 2 , ,x n , y 1 , , y n ,t), z = (x, y), x ∈ R n , y ∈ R n , t ∈ R, n ≥ 1; ξ  = ( x  , y  ,t  ) ∈ R 2n+1 . This group multiplication endows H n with a structure of nilpotent Lie group. A family of dilations on H n is defined as δ τ (x, y,t) =  τx,τy,τ 2 t  , τ>0. (1.4) The homogeneous dimension with respect to dilations is Q = 2n +2. The left invariant vector fields on the Heisenberg group have the form X i = ∂ ∂x i +2y i ∂ ∂t , Y i = ∂ ∂y i − 2x i ∂ ∂t , i = 1,2, ,n. (1.5) We denote the horizontal gradient by ∇ H = (X 1 , ,X n ,Y 1 , ,Y n ), and write div H (v 1 ,v 2 , ,v 2n ) =  n i =1 (X i v i + Y i v n+i ). Hence, the sub-Laplacian Δ H and the p-sub- Laplacian Δ H,p are expressed by Δ H = n  i=1 X 2 i + Y 2 i =∇ H ·∇ H , Δ H,p u =∇ H    ∇ H u   p−2 ∇ H u  = div H    ∇ H u   p−2 ∇ H u  , p>1, (1.6) respectively . The distance function is d(ξ, ξ  ) =  (x − x  ) 2 +(y − y  ) 2  2 +  t − t  − 2(x·y  − x  ·y)  2  1/4 ,forξ, ξ  ∈ H n . (1.7) Jingbo Dou et al. 3 If ξ  = 0, we denote d(ξ) = d(ξ,0) =  | z| 4 + t 2  1/4 ,with|z|=  x 2 + y 2  1/2 . (1.8) Note that d(ξ) is usually called the homogeneous norm. For d = d(ξ), it is easy to calculate ∇ H d = 1 d 3  | z| 2 x + yt |z| 2 y − xt  ,   ∇ H d   p = | z| p d p = ψ p , Δ H,p d = ψ p Q − 1 d . (1.9) Denote by B H (R) ={ξ ∈ H n | d(ξ) <R} the ball of r adius R centered at the origin. Let Ω 1 = B H (R 2 )\ B H (R 1 )with0≤ R 1 <R 2 ≤∞ and u(ξ) = v(d(ξ)) ∈ C 2 (Ω 1 )bearadial function with respect to d(ξ). Then Δ H,p u = ψ p   v    p−2  (p − 1)v  + Q − 1 d v   . (1.10) Let us recall the change of polar coordinates (x, y,t) →(ρ,θ,θ 1 , ,θ 2n−1 )in[16]. If u(ξ) = ψ p (ξ)v(d(ξ)), then  Ω1 u(ξ)dξ = s H  R 2 R 1 ρ Q−1 v(ρ)dρ, (1.11) where s H = ω n  π 0 (sinθ) n−1+p/2 dθ, ω n is the 2n-Lebesgue measure of the unitary Euclidean sphere in R 2n . TheSobolevspacein H n is written by D 1,p (Ω)={u : Ω→R;u,|∇ H u|∈L p (Ω)}. D 1,p 0 (Ω) is the closure of C ∞ 0 (Ω) with respect to the norm u D 1,p 0 (Ω) = (  Ω |∇ H u| p dξ) 1/p . In the sequel, we denote by c, c 1 , C, and so forth some positive constants usually except special narr ating. This paper is organized as follows. In Section 2, we prove the Hardy inequality with remainder terms by the vector field method in the Heisenberg group. In Section 3,we discuss the optimality of the constants in the inequalities which is of its independent interest. In Section 4, we show some useful properties concerning the Hardy operator (1.1), and then check the existence of solutions of the eigenvalue problem (1.2)(1<p< Q) and the asymptotic behavior of the first eigenvalue as μ increases to ((Q − p)/p) p .In Section 5, we study the simplicity and isolation of the first eigenvalue. 2. The Hardy inequality with remainder terms D’Ambrosio in [17] has proved a Hardy inequality in the bounded domain Ω ⊂ H n :let p>1andp =Q.Foranyu ∈ D 1,p 0 (Ω,|z| p /d 2p ), it holds that C Q,p  Ω ψ p |u| p d p dξ ≤  Ω   ∇ H u   p dξ, (2.1) where C Q,p =|(Q − p)/p| p .Moreover,if0∈ Ω, then the constant C Q,p is best. In this section, we give the Hardy inequality with remainder terms on Ω, based on the careful 4 Journal of Inequalities and Applications choice of a suitable vector field and an elementary integral inequality. Note that we also require that 0 ∈ Ω. Theorem 2.1. Let u ∈ D 1,p 0 (Ω / {0}). Then (1) if p =Q and there exists a positive constant M 0 such that sup ξ∈Ω d(ξ)e 1/M 0 := R 0 < ∞, then for any R ≥ R 0 ,  Ω   ∇ H u   p dξ ≥     Q − p p     p  Ω ψ p |u| p d p dξ + p − 1 2p     Q − p p     p−2  Ω ψ p |u| p d p  ln  R d  −2 dξ; (2.2) moreover, if 2 ≤ p<Q, then choose sup ξ∈Ω d(ξ) = R 0 ; (2) if p = Q and there exits M 0 such that sup ξ∈Ω d(ξ)e 1/M 0 <R, then  Ω   ∇ H u   p dξ ≥  p − 1 p  p  Ω ψ p |u| p  d ln(R/d)  p dξ. (2.3) Before we prove the theorem, let us recall that Γ  d(ξ)  = ⎧ ⎨ ⎩ d(ξ) (p−Q)/(p−1) if p=Q, −lnd(ξ)ifp = Q (2.4) is the solution of Δ H,p at the origin, that is, Δ H,p Γ(d(ξ)) = 0onΩ \{0}. Equation (2.4) is useful in our proof. For convenience, write Ꮾ(s) =−1/ ln(s), s ∈ (0,1), and A = (Q − p)/p. Thus, for some positive constant M>0, 0 ≤ Ꮾ  d(ξ) R  ≤ M,sup ξ∈Ω d(ξ) <R, ξ ∈ Ω. (2.5) Furthermore, ∇ H Ꮾ γ  d R  = γ Ꮾ γ+1 (d/R)∇ H d d , dᏮ γ (ρ/R) dρ = γ Ꮾ γ+1 (ρ/R) ρ ∀γ ∈ R, (2.6)  b a Ꮾ γ+1 (s) s ds = 1 γ  Ꮾ γ (b) − Ꮾ γ (a)  . (2.7) Proof. Let T be a C 1 vector field on Ω and let it be specified later. For any u ∈ C ∞ 0 (Ω \{0}), we use H ¨ older’s inequality and Young’s inequality to get  Ω  div H T  | u| p dξ =−p  Ω  T,∇ H u  | u| p−2 udξ ≤ p   Ω   ∇ H u   p dξ  1/p   Ω |T| p/(p−1) |u| p dξ  (p−1)/p ≤  Ω   ∇ H u   p dξ +(p − 1)  Ω |T| p/(p−1) |u| p dξ. (2.8) Jingbo Dou et al. 5 Thus, the following elementary integral inequality:  Ω   ∇ H u   p dξ ≥  Ω  div H T − (p − 1)|T| p/(p−1)  | u| p dξ (2.9) holds. (1) Let a be a free parameter to be chosen later. Denote I 1 (Ꮾ) = 1+ p − 1 pA Ꮾ  d R  + aᏮ 2  d R  , I 2 (Ꮾ) = p − 1 pA Ꮾ 2  d R  +2aᏮ 3  d R  , (2.10) and pick T(d) = A|A| p−2 (|∇ H d| p−2 ∇ H d/d p−1 )I 1 . An immediate computation shows div H  A|A| p−2   ∇ H d   p−2 ∇ H d d p−1  = A|A| p−2 dΔ H,p d − (p − 1)   ∇ H d   p d p = A|A| p−2 (Q − 1 − p +1)   ∇ H d   p d p = p|A| p   ∇ H d   p d p . (2.11) By (2.6), div H T = p|A| p   ∇ H d   p d p I 1 + A|A| p−2   ∇ H d   p−2 ∇ H d d p−1 × ∇ H d d  p − 1 pA Ꮾ 2  d R  +2aᏮ 3  d R  = p|A| p   ∇ H d   p d p I 1 + A|A| p−2   ∇ H d   p d p I 2 , div H T − (p − 1)|T| p/(p−1) = p|A| p   ∇ H d   p d p I 1 + A|A| p−2   ∇ H d   p d p I 2 − (p − 1)|A| p   ∇ H d   p d p I p/(p−1) 1 =|A| p   ∇ H d   p d p  pI 1 + 1 A I 2 − (p − 1)I p/(p−1) 1  . (2.12) We claim div H T − (p − 1)|T| p/(p−1) ≥|A| p   ∇ H d   p d p  1+ p − 1 2pA 2 Ꮾ 2  d R  . (2.13) In fact, arguing as in the proof of [13, Theorem 4.1], we set f (s): = pI 1 (s)+ 1 A I 2 (s) − (p − 1)I p/(p−1) 1 (s) (2.14) 6 Journal of Inequalities and Applications and M = M(R):= sup ξ∈Ω Ꮾ(d(ξ)/R), and distinguish three cases (i) 1 <p<2 <Q, a> (2 − p)(p − 1) 6p 2 A 2 , (2.15) (ii) 2 ≤ p<Q, a = 0, (2.16) (iii) p>Q, a< (2 − p)(p − 1) 6p 2 A 2 < 0. (2.17) It yields that f (s) ≥ 1+ p − 1 2pA 2 s 2 ,0≤ s ≤ M, (2.18) (see [13]) and then fol lows (2.13). Hence (2.2)isproved. (2) If p = Q, then we choose the vector field T(d) = ((p − 1)/p) p−1 (|∇ H d| p−2 ∇ H d/ d p−1 )Ꮾ p−1 (d/R). It gives div H T =  p − 1 p  p−1   Q − 1 − (p − 1)    ∇ H d   p d p Ꮾ p−1  d R  +(p − 1)Ꮾ p  d R    ∇ H d   p d p  = p  p − 1 p  p Ꮾ p  d R    ∇ H d   p d p , (2.19) and hence div H T − (p − 1)|T| p/(p−1) =  p − 1 p  p Ꮾ p  d R    ∇ H d   p d p . (2.20) Combining (2.20)with(2.9)follows(2.3).  Remark 2.2. The domain Ω in (2.9) may be bounded or unbounded. In addition, if we select that T(d) = A|A| p−2 (|∇ H d| p−2 ∇ H d/d p−1 ), then div H T − (p − 1)|T| p/(p−1) = p|A| p   ∇ H d   p d p − (p − 1)|A| p   ∇ H d   p d p =|A| p   ∇ H d   p d p . (2.21) Hence, from (2.9)weconclude(2.1) on the bounded domain Ω and on H n (see [15]), respectively . We will prove in next section that the constants in (2.2)and(2.3)arebest. Now, we state the Poincar ´ einequalityprovedin[17]. Jingbo Dou et al. 7 Lemma 2.3. Let Ω be a subset of H n bounded in x 1 direction, that is, there ex ists R>0 such that 0 <r =|x 1 |≤R for ξ = (x 1 ,x 2 , ,x n , y 1 , , y n ,t) ∈ Ω. Then for any u ∈ D 1,p 0 (Ω), then c  Ω |u| p dξ ≤  Ω   ∇ H u   p dξ, (2.22) where c = ((p − 1)/pR) p . Using (2.9) by choosing T =−((p − 1)/p) p−1 (∇ H r/r p−1 ) immediately provides a dif- ferent proof to (2.22). In the following, we describe a compactness result by using (2.1)and(2.22). Theorem 2.4. Suppose p =Q and f (ξ) ∈ Ᏺ p . Then there exists a positive constant C f ,Q,p such that C f ,Q,p  Ω f (ξ)|u| p dξ ≤  Ω   ∇ H u   p dξ, (2.23) and the embedding D 1,p 0 (Ω)  L p (Ω, fdξ) is compact. Proof. Since f (ξ) ∈ Ᏺ p ,wehavethatforany > 0, there exist δ>0andC δ > 0suchthat sup B H (δ)⊆Ω d p ψ p f (ξ) ≤  , f (ξ)| Ω\B H (δ) ≤ C δ . (2.24) By (2.1)and(2.22), it follows  Ω f (ξ)|u| p dξ =  B H (δ) f |u| p dξ +  Ω\B H (δ) f |u| p dξ ≤   B H (δ) ψ p |u| p d p dξ + C δ  Ω\B H (δ) |u| p dξ ≤ C −1 f ,Q,p  Ω   ∇ H u   p dξ, (2.25) then (2.23)isobtained. Now, we prove the compactness. Let {u m }⊆D 1,p 0 (Ω) be a bounded sequence. By re- flexivity of the space D 1,p 0 (Ω) and the Sobolev embedding for vector fields (see [18]), it yields u m j  u weakly in D 1,p 0 (Ω), u m j −→ u strongly in L p (Ω) (2.26) for a subsequence {u m j } of {u m } as j→∞.WriteC δ =f  L ∞ (Ω\B H (δ)) .From(2.1),  Ω f   u m j − u   p dξ =  B H (δ) f   u m j − u   p dξ +  Ω\B H (δ) f   u m j − u   p dξ ≤   B H (δ) ψ p   u m j − u   p d p dξ + C δ  Ω\B H (δ)   u m j − u   p dξ ≤ C −1 Q,p  Ω   ∇ H  u m j − u    p dξ + C δ  Ω   u m j − u   p dξ. (2.27) 8 Journal of Inequalities and Applications Since {u m }⊆D 1,p 0 (Ω) is bounded, we have  Ω f   u m j − u   p dξ ≤  M + C δ  Ω   u m j − u   p dξ, (2.28) where M>0 is a constant depending on Q and p.By(2.26), lim j→∞  Ω f   u m j − u   p dξ ≤  M. (2.29) As  is arbitrary, lim j→∞  Ω f |u m j − u| p dξ = 0. Hence D 1,p 0 (Ω)  L p (Ω, fdξ)iscompact.  Remark 2.5. The class of the functions f (ξ) ∈ Ᏺ p has lower-order singularity than d −p (ξ) at the origin. The examples of such functions are (a) any bounded function, (b) in a small neighborhood of 0, f (ξ) = ψ p (ξ)/d β (ξ), 0 <β<p, (c) f (ξ) = ψ p (ξ)/d p (ξ)(ln(1/d(ξ))) 2 in a small neighborhood of 0. 3.Proofofbestconstantsin(2.2)and(2.3) In this section, we prove that the constants appearing in Theorem 2.1 are the best. To do this, we need two lemmas. First we introduce some notations. For some fixed small δ>0, let the test function ϕ(ξ) ∈ C ∞ 0 (Ω) satisfy 0 ≤ ϕ ≤ 1and ϕ(ξ) = ⎧ ⎪ ⎨ ⎪ ⎩ 1ifξ ∈ B H  0, δ 2  , 0ifξ ∈ Ω \ B H (0,δ), (3.1) with |∇ H ϕ| < 2|∇ H d|/d. Let  > 0 small enough, and define V  (ξ) = ϕ(ξ)  ,with  = d −A+ Ꮾ −κ  d R  , 1 p <κ< 2 p , J γ () =  Ω ϕ p (ξ)   ∇ H d   p d Q−p Ꮾ −γ  d R  dξ, γ ∈ R. (3.2) Lemma 3.1. For  > 0 small, it holds (i) c  −1−γ ≤ J γ () ≤ C −1−γ , γ>−1, (ii) J γ () = (p/(γ +1))J γ+1 ()+O  (1), γ>−1, (iii) J γ () = O  (1), γ<−1. Proof. By the change of p olar coordinates (1.11)and0 ≤ ϕ ≤ 1, we have J γ () ≤  B H (δ)   ∇ H d   p d Q−p Ꮾ −γ  d R  dξ = s H  ρ<δ ρ −Q+p Ꮾ −γ  ρ R  ρ Q−1 dρ = s H  ρ<δ ρ −1+p Ꮾ −γ  ρ R  dρ. (3.3) Jingbo Dou et al. 9 By (2.7)weknowthatforγ< −1 the right-hand side of (3.3) has a finite limit, hence (iii) follows from  → 0. To show (i), we set ρ = Rτ 1/ .Thus,dρ = (1/)Rτ 1/ − 1 dτ, Ꮾ −γ (τ 1/ ) =  −γ Ꮾ −γ (τ), and J γ () ≤ s H  ρ<δ ρ −1+p Ꮾ −γ  ρ R  dρ = s H  (δ/R)  0  Rτ 1/  −1+p Ꮾ −γ  Rτ 1/ R  1  Rτ 1/ − 1 dτ = s H R p  −1−γ  (δ/R)  0 τ p−1 Ꮾ −γ (τ)dτ. (3.4) It follows the right-hand side of (i). Using the fact that ϕ = 1inB H (δ/2), J γ () ≥  B H (δ/2)   ∇ H d   p d Q−p Ꮾ −γ  d R  dξ = s H R p  −1−γ  (δ/2R)  0 τ p−1 Ꮾ −γ (τ)dτ, (3.5) and the left-hand side of (i) is proved. Now we prove (ii). Let Ω η : ={ξ ∈ Ω | d(ξ) >η}, η>0, be small and note the boundary term −  d=η  ϕ p   ∇ H d   p−2 ∇ H d d Q−1−p  Ꮾ −γ−1  d R  ∇ H d·  ndS −→ 0asη −→ 0. (3.6) From (2.6),  Ω div H  ϕ p   ∇ H d   p−2 ∇ H d d Q−1−p  Ꮾ −γ−1  d R  dξ =−  Ω ϕ p   ∇ H d   p−2 d Q−1−p  ∇ H d,∇ H Ꮾ −γ−1  d R  dξ = (γ +1)  Ω ϕ p   ∇ H d   p d Q−p Ꮾ −γ  d R  dξ = (γ +1)J γ (). (3.7) On the other hand,  Ω div H  ϕ p   ∇ H d   p−2 ∇ H d d Q−1−p  Ꮾ −γ−1  d R  dξ = p  Ω ϕ p−1   ∇ H d   p−2  ∇ H d,∇ H ϕ  d Q−1−p Ꮾ −γ−1  d R  dξ +(1 − Q + p + Q − 1)  Ω ϕ p   ∇ H d   p d Q−p Ꮾ −γ−1  d R  dξ = p  Ω ϕ p−1   ∇ H d   p−2  ∇ H d,∇ H ϕ  d Q−1−p Ꮾ −γ−1  d R  dξ + pJ γ+1 (). (3.8) 10 Journal of Inequalities and Applications We claim that p  Ω ϕ p−1 (|∇ H d| p−2 ∇ H d,∇ H ϕ/d Q−1−p )Ꮾ −γ−1 (d/R)dξ = O  (1). In fact, by (3.1)and(1.11),  Ω ϕ p−1   ∇ H d   p−2  ∇ H d,∇ H ϕ  d Q−1−p Ꮾ −γ−1  d R  dξ ≤ 2  B H (δ)   ∇ H d   p d Q−p Ꮾ −γ−1  d R  dξ ≤ 2s H  B H (δ) ρ −Q+p Ꮾ −γ−1  ρ R  ρ Q−1 dρ = 2s H  B H (δ) ρ p − 1 Ꮾ −γ−1  ρ R  dρ. (3.9) Using the estimate (i) follows that  B H (δ) ρ p − 1 Ꮾ −γ−1 (ρ/R)dρ = O  (1). Combining (3.7) with (3.8)gives (γ +1)J γ () = pJ γ+1 ()+O  (1). (3.10) This allows us to conclude (ii).  We next estimate the quantity I[V  ] =  Ω   ∇ H V    p dξ −|A| p  Ω   ∇ H d   p   V    p d p dξ. (3.11) Lemma 3.2. As  → 0,itholds (i) I(V  ) ≤ (κ(p − 1)/2)|A| p−2 J pκ−2 ()+O  (1); (ii)  B H (δ) |∇ H V  | p dξ ≤|A| p J pκ ()+O  ( 1−pκ ). Proof. By the definition of V  (ξ), we see ∇ H V  (ξ) = ϕ(ξ)∇ H   +   ∇ H ϕ. Using the ele- mentary inequality |a + b| p ≤|a| p + c p  | a| p−1 |b| + |b| p  , a,b ∈ R 2n , p>1, (3.12) one has   ∇ H V    p ≤ ϕ p   ∇ H     p + c p    ∇ H ϕ     ϕ p−1   ∇ H     p−1 +   ∇ H ϕ   p       p  ≤ ϕ p   ∇ H     p + c p    2∇ H d   d   ϕ p−1   ∇ H     p−1 +    2∇ H d   d  p       p  . (3.13) [...]... 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It is established by the vec- tor field method and an elementary integral inequality. . Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 32585, 24 pages doi:10.1155/2007/32585 Research Article A Hardy Inequality with Remainder Terms. inequality. In addition, we show that the constants appearing in Hardy inequality are the best. Then we conclude a compact em- bedding in the weighted Sobole v space. Themaindifficulty to study the properties