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Volume 2007, Article ID 32585, 24 pagesdoi:10.1155/2007/32585 Research Article A Hardy Inequality with Remainder Terms in the Heisenberg Group and the Weighted Eigenvalue Problem Jingbo

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 ´o Losonczi

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 bestconstants in Hardy inequalities are determined Then we discuss the existence of solutionsfor the nonlinear eigenvalue problems in the Heisenberg group with weights for the p-

sub-Laplacian The asymptotic behaviour, simplicity, and isolation of the first eigenvalueare also considered

Copyright © 2007 Jingbo Dou et al This is an open access article distributed under theCreative Commons Attribution License, which permits unrestricted use, distribution,and reproduction in any medium, provided the original work is properly cited

to ((Q − p)/ p) p Furthermore, we show that the first eigenvalue is simple and isolated, as

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well as the eigenfunctions corresponding to other eigenvalues change sign Our proof ismainly 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 theconstants appearing in Hardy inequality are the best Then we conclude a compact em-bedding in the weighted Sobolev space

The main difficulty to study the properties of the first eigenvalue is the lack of larity of the weak solutions of thep-sub-Laplacian in the Heisenberg group Let us note

regu-that theC α regularity for the weak solutions of the p-subelliptic operators formed by

the vector field satisfying H¨ormander’s condition was given in [1] and theC1,αregularity

of the weak solutions of thep-sub-LaplacianΔH,pin the Heisenberg group for p near 2

was proved in [2] To obtain results here, we employ the Picone identity and Harnackinequality 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 Hardyinequalities (see [4,12–14])

Let us recall some elementary facts on the Heisenberg group (e.g., see [15]) LetHnbe

a Heisenberg group endowed with the group law

The homogeneous dimension with respect to dilations isQ =2n + 2 The left invariant

vector fields on the Heisenberg group have the form

i =1(X i v i+Y i v n+i) Hence, the sub-LaplacianΔH and the

p-sub-LaplacianΔH,pare expressed by

Ifξ  =0, we denote

d(ξ) = d(ξ, 0) =| z |4+t2  1/4

, with| z | =x2+y2  1/2

Note thatd(ξ) is usually called the homogeneous norm.

Ford = d(ξ), it is easy to calculate

Denote byBH(R) = { ξ ∈ H n | d(ξ) < R }the ball of radiusR centered at the origin Let

Ω1= BH(R2)\ BH(R1) with 0≤ R1< R2≤ ∞andu(ξ) = v(d(ξ)) ∈ C2(Ω1) be a radialfunction with respect tod(ξ) Then

The Sobolev space inHnis written byD1,p(Ω)={ u :Ω→R;u, |∇ Hu |∈ L p(Ω)}.D1,0p(Ω)

is the closure ofC ∞0(Ω) with respect to the norm u  D1,p

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 domainΩ⊂ H n: let

p > 1 and p = Q For any u ∈ D1,0p(Ω,| z | p /d2p), it holds that

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choice of a suitable vector field and an elementary integral inequality Note that we alsorequire that 0∈ Ω.

Theorem 2.1 Let u ∈ D01,p(Ω / {0} ) Then

(1) if p = Q and there exists a positive constant M0such that sup ξ ∈Ωd(ξ)e1/M0:= R0< ∞,

then for any R ≥ R0,

moreover, if 2 ≤ p < Q, then choose sup ξ ∈Ωd(ξ) = R0;

(2) if p = Q and there exits M0such that sup ξ ∈Ωd(ξ)e1/M0< R, then

is the solution ofΔH,pat the origin, that is,ΔH,p Γ(d(ξ)) =0 onΩ\ {0} Equation (2.4)

is useful in our proof For convenience, writeᏮ(s) = −1/ ln(s), s ∈(0, 1), andA =(Q −

p)/ p Thus, for some positive constant M > 0,

Proof Let T be a C1vector field onΩ 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

Thus, the following elementary integral inequality:



Ω H u p dξ ≥

 Ω

divHT−(p −1)|T| p/(p −1) | u | p dξ (2.9)holds

(1) Leta be a free parameter to be chosen later Denote

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andM = M(R) : =supξ ∈ΩᏮ(d(ξ)/R), and distinguish three cases

(see [13]) and then follows (2.13) Hence (2.2) is proved

(2) If p = Q, then we choose the vector field T(d) =((p −1)/ p) p −1(|∇ H d | p −2∇ H d/

H d p

d p ,

(2.19)and hence

Remark 2.2 The domainΩ in (2.9) may be bounded or unbounded In addition, if weselect that T(d) = A | A | p −2(|∇ H d | p −2∇ H d/d p −1), then

We will prove in next section that the constants in (2.2) and (2.3) are best

Now, we state the Poincar´e inequality proved in [17]

Lemma 2.3 Let Ω be a subset ofHn bounded in x1direction, that is, there exists R > 0 such that 0 < r = | x1| ≤ R for ξ =(x1,x2, , x n,y1, , y n,t) ∈ Ω Then for any u ∈ D01,p(Ω),

dif-In the following, we describe a compactness result by using (2.1) and (2.22)

Theorem 2.4 Suppose p = Q and f (ξ) ∈Ᏺp Then there exists a positive constant Cf ,Q,p such that

and the embedding D01,p(Ω)  L p(Ω, f dξ) is compact.

Proof Since f (ξ) ∈Ᏺp, we have that for any > 0, there exist δ > 0 and Cδ > 0 such that

Now, we prove the compactness Let{ u m } ⊆ D1,0p(Ω) be a bounded sequence By flexivity of the spaceD1,0p(Ω) and the Sobolev embedding for vector fields (see [18]), ityields

re-um j  u weakly in D1,p

0 (Ω),

u m j −→ u strongly in L p(Ω) (2.26)for a subsequence{ um j }of{ um }as j →∞ WriteCδ = f L ∞( Ω\ B H(δ)) From (2.1),

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Since{ u m } ⊆ D01,p(Ω) is bounded, we have

Remark 2.5 The class of the functions f (ξ) ∈Ᏺphas lower-order singularity thand − 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) = ψp(ξ)/d p(ξ)(ln (1/d(ξ)))2in a small neighborhood of 0

3 Proof of best constants in ( 2.2 ) and ( 2.3 )

In this section, we prove that the constants appearing inTheorem 2.1are the best To dothis, we need two lemmas First we introduce some notations

For some fixed smallδ > 0, let the test function ϕ(ξ) ∈ C ∞0(Ω) satisfy 0≤ ϕ ≤1 and

By (2.7) we know that forγ < −1 the right-hand side of (3.3) has a finite limit, hence (iii)follows from→0.

To show (i), we setρ = Rτ1/  Thus,dρ =(1/ )Rτ1/ −1dτ,Ꮾ− γ(τ1/ )=  − γᏮ− γ(τ), and



dξ = s H R p   −1− γ(δ/2R) 

0 τ p −1Ꮾ− γ(τ)dτ, (3.5)

and the left-hand side of (i) is proved

Now we prove (ii) LetΩη:= { ξ ∈Ω| d(ξ) > η },η > 0, be small and note the boundary



∇ H d · ndS −→0 asη −→0. (3.6)From (2.6),

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(γ + 1)Jγ()= p  Jγ+1() +O (1). (3.10)

We next estimate the quantity

Since∇ H   = − d − A+ −1Ꮾ− κ(d/R)(A − +κ Ꮾ(d/R)) ∇ Hd, it follows

By (i) ofLemma 3.1, it derivesΠ1,Π2= O (1), as→0

From (3.14), (3.15) and the definition ofJγ(), it clearly shows

plicity, denoteζ =  − κ Ꮾ(d/R) Since ζ is small compared to A, we use Taylor’s expansion

to yield

| A − ζ | p − | A | p ≤ − pA | A | p −2ζ + p(p −1)

2 | A | p −2ζ2+C | ζ |3. (3.18)Thus, we can estimateΠ3by

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We will show that

We are now ready to give the proof of the best constants inTheorem 2.1

Theorem 3.3 Let 0 ∈ Ω be a bounded domain inHn and p = Q Suppose that for some constants B > 0, D ≥ 0, and ι > 0, the following inequality holds for any u(ξ) ∈ C ∞0(Ω\ {0} ):

(i)B ≤ | A | p ;

(ii) if B = | A | p and D > 0, then ι ≥ 2;

(iii) if B = | A | p and ι = 2, then D ≤((p −1)/2p) | A | p −2.

Proof Choose u(ξ) = V (ξ).

(i) By (ii) ofLemma 3.2, we have

(ii) SetB = | A | pand assume by contradiction thatι < 2 Since pκ − ι > −1, using (i)

ofLemma 3.2and (i) ofLemma 3.1leads to

which is a contradiction Henceι ≥2

(iii) IfB = | A | pandι =2, then by (i) ofLemma 3.2,

Theorem 3.4 Set 0 ∈ Ω and p = Q Suppose that there exist some constants D ≥ 0 and

ι > 0 such that the following inequality holds for all u(ξ) ∈ C0∞(Ω\ {0} ):

Proof The proof is essentially similar to one ofTheorem 3.3 Let the test functionϕ be

as before (see (3.1)) For > 0, κ > (p −1)/ p, define V  = ϕ with  = d (−ln(d/R)) κ

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4 The weighted eigenvalue problem

This section is devoted to the problem (1.2) by using the Hardy inequality with remainderterms

We begin with some properties concerning the Hardy operator (1.1)

Lemma 4.1 Suppose that u(ξ) ∈ D1,0p(Ω) and p= Q Then

(1)L p,μ is a positive operator if μ ≤ C Q,p ; in particular, if μ = C Q,p , then v(ξ) =

d(p − Q)/ p(ξ) is a solution of L p,μ u = 0;

(2)Lp,μ is unbounded from below if μ > CQ,p

Proof (1) It is obvious from (2.1) thatLp,μis a positive operator

We now suppose thatμ = CQ,pand verify thatv = d(p − Q)/ psatisfiesLp,μu =0 For thepurpose, setv  = d(p − Q)/ p+  ∈ D01,p(Ω) and A=(Q − p)/ p Since

(2) By the density argument, we selectφ(ξ) ∈ C ∞0(Ω), φ L p =1, such thatC Q,p =

In order to prove the main result (Theorem 4.6below) we need the following twopreliminary lemmas

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Lemma 4.2 Let { gm } ⊂ L p(Ω)(1≤ p < ∞ ) be such that as m →∞ ,

Lemma 4.3 Suppose that { um } ⊂ D1,0p(Ω)(1≤ p < ∞ ) satisfies

(Ω)(p = p/(p − 1)), gm is bounded in M( Ω) (the space of

Radon measures), that is,

for all ϕ ∈ Ᏸ(Ω) with supp(ϕ) ⊂ K, where CK is a constant which depends on the compact set K Then there exists a subsequence { u m j } of { u m } such that

u m j −→ u strongly in D1,0q(Ω),∀ q < p. (4.10)Its proof is similar to one of [20, Theorem 2.1]

Lemma 4.4 Let p = Q and

Proof (i) If f (ξ) ∈Ip, then

(ii) We write f (ξ) = ψph(ξ)/d p(ξ)(ln 1/d(ξ))2, whereh(ξ) →∞asd(ξ) →0 Then, forthe sufficiently small > 0, we select u(ξ) = V (ξ) = ϕ(ξ)d − A+ (ξ)Ꮾ− κ(d(ξ)/R) and get

from (i) ofLemma 3.2,

ψp

d p | u | p −2uϕ dξ = λ



Ωf (ξ) | u | p −2uϕ dξ (4.17)for any ϕ ∈ C ∞0(Ω) In this case, we call that u is the eigenfunction of problem (1.2)associated to the eigenvalueλ.

Theorem 4.6 Suppose that 1 < p < Q, 0 ≤ μ < ((Q − p)/ p) p , and f (ξ) ∈Ᏺp The problem (1.2) admits a positive weak solution u ∈ D1,0p(Ω), corresponding to the first eigenvalue λ=

λ1

μ(f ) > 0 Moreover, as μ increases to ((Q − p)/ p) p , λ1

μ(f ) → λ( f ) ≥ 0 for f (ξ) ∈Ᏺp and the limit λ( f ) > 0 for f (ξ) ∈Ip If f (ξ) ∈Ip and ψ −1

p (ξ)d p(ξ) f (ξ)(ln 1/d(ξ))2→∞ , as

d(ξ) → 0, then the limit λ( f ) =0.

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Jμ(um)→ λ1

μ andJ μ (um)→0 strongly inD −01,p (Ω), when the component of J

μ(um) is stricted toᏹ The coercivity of J μimplies that{ u m }is bounded and then there exists asubsequence, still denoted by{ um }, such that

asm →∞ ByTheorem 2.4 inSection 2 we know thatD01,p(Ω) is compactly embedded

inL p(Ω, f dξ), and it follows that ᏹ is weakly closed and hence u ∈ ᏹ Moreover, u m

satisfies

−ΔH,p u m − ψ p μ

d p u m p −2u m = λ m u m p −2u m f + f m, inᏰ(Ω), (4.21)wherefm →0 strongly inD −1,p 

by applyingLemma 4.2toumand∇ H um, whereo(1) →0 asm →∞ ThusCQ,p > μ,  um −

u  L p p(Ω,ψp d − p)→0, and∇ H(u m − u)  L p p(Ω)→0 asm →∞ It shows thatJ μ(u) = λ1

μandλ = λ1

μ.SinceJ μ(| u |)= J μ(u), we can take u > 0 inΩ ByLemma 4.3,u is a distribution solution

of (1.2) and sinceu ∈ D1,0p(Ω), it is a weak solution to eigenvalue problem (1.2) sponding toλ = λ1

corre-μ Moreover, if f (ξ) ∈Ip, then byLemma 4.4,

asμ increases to ((Q − p)/ p) p When f (ξ) ∈Ip, usingLemma 4.4again, it follows that(4.12) is not true and henceλ( f ) =0 This completes the proof 

Remark 4.7 The set ᏹ is a C1manifold inD1,0p(Ω) By Ljusternik-Schnirelman criticalpoint theory onC1manifold, there exists a sequence{ λ m }of eigenvalues of (1.2), that

is, writingΓm = { A ⊂Ᏻ| A is symmetric, compact, and γ(A) ≥ m }, whereγ(A) is the

Krasnoselski’s genus ofA (see [19]), then for any integerm > 0,

is an eigenvalue of (1.2) Moreover, limm →∞ λ m →∞

5 Simplicity and isolation for the first eigenvalue

This section is to consider the simplicity and isolation for the first eigenvalue We alwaysassume that f satisfies the conditions inTheorem 4.6 From the previous results we knowclearly that the first eigenvalue is

In what follows we need the Picone identity proved in [15]

Proposition 5.1 (Picone identity) For di fferentiable functions u ≥0,v > 0 onΩ⊂ H n , with Ω a bounded or unbounded domain inHn , then

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UsingProposition 5.1and (5.4),

The right-hand side of (5.5) tends to zero whenε →0 It follows thatL(u, v) =0 and by

Proposition 5.1there exists a constantc such that u = cv.

(ii) Letu > 0 and v > 0 be eigenfunctions corresponding to λ1

μandλ, respectively

Lemma 5.3 If u ∈ D1,0p(Ω) is a nonnegative weak solution of (1.2), then either u(ξ) ≡ 0 or

u(ξ) > 0 for all ξ ∈ Ω.

Proof For any R > r, BH(0,R) ⊃ BH(0,r), let u ∈ D01,p(Ω) be a nonnegative weak solution

of (1.2) In virtue of Harnack’s inequality (see [1]), there exists a constantCR > 0 such that

Theorem 5.4 Every eigenfunction u1corresponding to λ1

μ does not change sign in Ω: either

u1> 0 or u1< 0.

Proof From the proof of existence of the first eigenvalue we see that there exists a positive

eigenfunction, that is, ifv is an eigenfunction, then u1= | v |is a solution of the tion problem and also an eigenfunction Thus, fromLemma 5.3it follows that| v | > 0 and

Lemma 5.5 For u ∈ C(Ω\ {0})∩ D01,p(Ω), let ᏺ be a component of{ ξ ∈Ω| u(ξ) > 0 } Then u |ᏺ∈ D1,0p(ᏺ).

Proof Let um ∈ C(Ω\ {0})∩ D01,p(Ω) be such that u m → u in D1,0p(Ω) Therefore, u+

with|∇ H ϕ R | ≤ C |∇ H d | /d(ξ), for some positive constant C Now, consider the sequence

ωm(ξ) = ϕR(ξ)vm(ξ) |ᏺ SinceϕR(ξ)vm(ξ) ∈ C( Ω), we claim that ω m ∈ C( ᏺ) and ω m =0

on the boundary∂ ᏺ In fact, if ξ ∈ ∂ ᏺ and ξ =0, thenϕ R =0, and soω m =0 Ifξ ∈

∂ᏺ∩ Ω and ξ =0, thenu =0 (sinceu is continuous except at {0}), and hencevm =0 If

ξ ∈ ∂ Ω, then u m =0 and sov m =0 Therefore,ω m =0 on∂ ᏺ, and ω m ∈ D01,p(ᏺ) Noting

Theorem 5.6 The eigenvalue λ1

μ is isolated in the spectrum, that is, there exists δ > 0 such that there is no other eigenvalues of (1.2) in the interval (λ1

μ,λ1

μ+δ) Moreover, if v is an eigenfunction corresponding to the eigenvalue λ = λ1

μ and ᏺ is a nodal domain of v, then

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Proof Let u1 be the eigenfunction corresponding to the eigenvalueλ1

μ Let { λm }be asequence of eigenvalues such thatλm > λ1

0<



Ω H u m p dξ − μ

 Ω

ψ p

d p u m p dξ = λ m



Ωf (ξ) u m p dξ = λ m, (5.14)

it follows thatum is bounded ByLemma 4.3, there exists a subsequence (still denoted

by{ um }) of { um } such thatum  u weakly in D1,p

0 (Ω), um → u strongly in L p(Ω) and

∇ H u m →∇ H u a.e in Ω Letting m →∞in (5.13) yields

Lp,μu = λ1

μ f (ξ) | u | p −2u. (5.15)Therefore,u = ± u1 Using (iii) ofTheorem 5.2we see thatum changes sign For conve-nience, we assume thatu =+u1 Then

ψp

d p u − m p dξ = λm



Ωf (ξ) u − m p dξ. (5.17)Using the Hardy inequality and Sobolev inequality yields

Next, we prove (5.12) Assumev > 0 in ᏺ (the case v < 0 being treated similarly) In

view ofLemma 5.5, we havev |ᏺ∈ D1,0p(ᏺ) Define the function

Thus, the following elementary integral inequality:

...

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choice of a suitable vector field and an elementary integral... parameter to be chosen later Denote

6 Journal of Inequalities and Applications

and< i>M