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Spectral theory of certain unbounded Jacobi matrices
Jaouad Sahbani
Institut de Mathématiques de Jussieu Université Paris 7-Denis Diderot, case 7012, 2, place Jussieu, 75251 Paris Cedex 05, France Received 13 July 2007
Available online 23 December 2007 Submitted by R. Curto
Abstract
Using the conjugate operator method of Mourre we study the spectral theory of a class of unbounded Jacobi matrices. We especially focus on the case where the off-diagonal entriesan=nα(1+o(1))and diagonal onesbn=λnα(1+o(1))withα >0, λ∈R.
©2007 Elsevier Inc. All rights reserved.
Keywords:Essential spectrum; Absolutely continuous spectrum; Mourre estimate
1. Introduction
Let H=l2(N)be the Hilbert space of square summable sequences (ψn)n1 endowed with the scalar product φ, ψ =n1φnψn. Consider the formal difference expressionτ given by
(τ ψ )n=an−1ψn−1+bnψn+anψn+1, for alln1, (1.1) withan>0,bn∈Randψ0=0. LetHminbe the restriction ofτ to the subspacel02of sequences with only finitely
many non-zero coordinates. It is easy to verify thatHmin∗ =HmaxwhereHmaxis the restriction ofτ toD(Hmax)= {ψ∈H|τ ψ∈H}. LetH:=Hmin=Hmax∗ be the closure ofHminthat will be called here the Jacobi operator realized by the sequencesanandbn. It is known thatHminis essentially self-adjoint onl02if and only ifHmaxis symmetric. In such a case,H=Hmaxand we will say thatH is essentially self-adjoint onl02. Otherwise,H has uncountably many self-adjoint extensions and each one of them has a purely discrete spectrum, see [3, p. 528].
Our aim in this paper is to study the spectral properties of the Jacobi operatorH mainly in the case where the matrix elements are of the form
an=nα+ηn, (1.2)
bn=λ
nα+(n−1)α+βn, (1.3)
whereα >0,λ∈Randηn/nα andβn/nα are small in certain sense. This subject is extensively studied and there
is a large literature devoted to it, see [6–13,15–21,24,25] and references therein. One meets their various methods E-mail address:[email protected].
0022-247X/$ – see front matter ©2007 Elsevier Inc. All rights reserved. doi:10.1016/j.jmaa.2007.12.044
of investigation using orthogonal polynomials theory, positive commutator methods, or Gilbert–Pearson theory of subordinacy.
For example, [6,8–11,15,21] are based on the positive commutator method of Putnam and Kato. The latter states that if there is a bounded self-adjoint operatorAsuch that the commutator[H, iA]is positive thenH has a purely absolutely continuous spectrum. This approach allowed the resolution of great spectral problems, see [22]. However, the boundedness ofAis rather annoying since it excludes natural candidates to this role. Moreover, the global character of the required commutator positivity makes it very unstable under perturbations, even compact. Mourre’s method [2] resolves these problems, since the conjugate operator Ahas not to be bounded anymore and[H, iA]has only to be positive locally and up to a compact operator. In particular, such a positivity is stable under a large class of relatively compact perturbation. Furthermore, it is not a one-dimensional approach as the Gilbert–Pearson method is. Surprisingly, it has never been applied to Jacobi matrices with unbounded entries before [24,25]. There we only consider the cases whereλ=0 and 0< α <1, andλ= ±2 withα=1. Even in these cases the results of the present paper improve significantly those of [24,25]. This is due, in part, to the sharpness of our criteria of relative boundedness and compactness with respect toHestablished in Sections 3–4.
We start our study by decomposing the operatorHas the sumH=H0+V of the unperturbed model corresponding to the case whereηn=βn=0 and the perturbationV. Then by an elementary analysis one gets:
Theorem 1.1.
(a) If|λ|>1thenH0is essentially self-adjoint and has no essential spectrum for allα >0. (b) If|λ|<1thenαcomes in the game as follows:
(b1) ifα >1thenH0is not essentially self-adjoint onl20;
(b2) if0< α1thenH0is essentially self-adjoint onl02and has a purely absolutely continuous spectrum filling the whole real axisR.
(c) Ifλ= ±1thenH0is essentially self-adjoint onl02regardless ofα, but: (c1) ifα >2and then the essential spectrum ofH0is empty;
(c2) if0< α <2then the spectrum ofH0is absolutely continuous and fills the semi-axis[0,±∞)(i.e.(−∞,0] ifλ= −1and[0,+∞)ifλ=1);
(c3) ifα=2then the spectrum ofH0is purely absolutely continuous and fills the semi-axis[1/4,±∞).
Remarks.
(i) The point (a) is a particular case of Theorem 4.1 of [17].
(ii) The assertion (b1) extends Theorem 1.5 of page 507 of [3] which only covers the case whereλ=0.
(iii) The assertion (b2) can be deduced from [15,17]. See however the remark of Section 2.1 below where a simplified proof of this assertion is sketched.
(iv) Finally, the point (c1) is similar to Theorem 8 of [13] by Hinton and Lewis. Here we give a quite elementary proof of this assertion based on a simple computation of the resolvent which turns out to be a compact operator inH.
In the cases where H0 has a purely absolutely continuous spectrum, since the constancy of the multiplicity of the spectrum of H0, there exists, in abstract level, a locally conjugate operatorAtoH0in the Mourre’s sense, see Proposition 7.2.17 of [2]. It is natural to construct such an operatorAwhich is a key step of this study. As one may guess, the operatorAdepends on the parametersλ andα, see (7.1), (9.1) and (10.1). For example, the operatorA defined by (7.1) does not work for the borderline case|λ|<1 andα=1 since it becomes bounded in this case and no Mourre estimate can be expected with it, see (5.2). Notice that Pedersen tried in [21] to employ Mourre’s theory to a similar model by choosingAof the same form such thatB= [H, iA]and[B, iA]are bounded. But thenAitself should be necessarily bounded and so he finally used Putnam–Kato method. With our operator defined by (9.1) the commutator [H, iA]is equal to a positive constant plus an unbounded component which turns out to be relatively compact with respect toH0(thanks to our criterion ofH0-compactness).
Note that if α >1 and|λ|<1 then we also have a Mourre estimate between any self-adjoint extensionH0s of H0andAdefined by (7.1) which is clearly bounded. In particular, by (5.2) below we re-obtain the emptiness of the
essential spectrum ofH0s. This idea is extended to a wider class of Jacobi matrices, see Theorem 7.3. This shows that Mourre’s theory can be used also to show the absence of the continuous spectrum.
Combining Theorem 1.1 with our criteria of relative boundedness and compactness we answer the question of self-adjointness ofH and compute its essential spectrum for a large class of perturbation coefficientsηnandβn. For
a sequencer=(rn)nwe define∂rn=rn+1−rnand∂k+1rn=∂(∂kr)n. Put
Δn:=
ηn
nα and β
n:=βn−λ(ηn+ηn−1). (1.4)
Theorem 1.2.Iflim supn→∞|Δn|<1andsupn(|nα∂Δn| + |βn|) <∞thenHis essentially self-adjoint onl02if and only if(λ, α) /∈ ]−1,1[ × ]1,∞[.
In the sequel we focus on the case where(λ, α) /∈ ]−1,1[ × ]1,∞[. Concerning the essential spectrumσess(H )of Hwe have:
Theorem 1.3.Assume thatlimn→∞(|Δn| + |nα∂Δn| + |βn|)=0. Thenσess(H )=σess(H0). In particular, we have: (i) if|λ|<1and0< α1thenσess(H )=R;
(ii) ifλ= ±1and0< α <2thenσess(H )= [0,±∞); (iii) and ifλ= ±1andα=2thenσess(H )= [1/4,±∞).
Remark.If 0< α1 then conditions of Theorem 1.3 are equivalent to lim n→∞ ηn nα + |∂ηn| +βn =0. (1.5)
That∂ηntends to zero at infinity means thatηndoes not oscillate strongly. In contrast, ifηnis 2-periodic andη1=0 andη2=1 thenHhas a spectral gap, see Theorem 1.1 of [7]. This illustrates the sharpness of our conditions.
Theorem 1.4.Let|λ|<1and0< α <1. Suppose that(1.5)holds true and lim n→∞n 1−α∂β n+n 1−2α∂η n+n1−α∂2ηn=0. (1.6) Then the setσp(H )of the eigenvalues ofH has no finite accumulation point.
Remarks.
(i) Similar results are obtained for the borderline case where|λ|<1 andα=1 in Section 9 andλ= ±1 in Sections 10 and 11.
(ii) In Theorem 6.1 of [16] Janas and Naboko constructed an operatorH withbn=0,ηn=O(n−(1−α))with 1/2<
α <1 and such thatσp(H ) = ∅. Combining this with Theorem 1.4 it is reasonable to conjecture that:there exists
ηn such thatlimn→∞(|ηn/nα| + |∂ηn|)=0, limn→∞n1−α∂ηn= ∞ arbitrarily slowly and the corresponding Jacobi operatorH (withbn=0for simplicity)has a dense point spectrum inR.
Theorem 1.5.In addition to the hypotheses of Theorem1.4we assume that sup
n1
n2−3α∂2η
n+n2(1−α)∂3ηn+n2(1−α)∂2βn<∞. (1.7) Then the singular continuous spectrumσsc(H )ofH is empty.
LetB(H)be the Banach algebra of bounded operators inHand consider the positive operatorN acting inHby (N ψ )n=nψn.
Theorem 1.6.Letσ >1−2α and suppose the assumptions of Theorem1.5hold. Then, in the sense of the norm topology ofB(H), the limits
R(λ±i0):= lim
μ→0+N
exist locally uniformly onR\σp(H ), and the maps
R\σp(H )λ −→ R(λ±i0)∈B(H)
are locally Hölder continuous of orderθ=min(1−σα −12,12). Moreover, for allϕ∈C0∞(R\σp(H ))we have
N−σe−iH tϕ(H )N−σC1+ |t|−θ.
The fact that σ is not greater than 1/2 comes from the fact thatH has no spectral gap, see [23] for a precise discussion.
The paper is organized as follows:
• An elementary analysis of the unperturbed modelH0is given in Section 2.
• In Sections 3–4 we study the relative boundedness and compactness with respect to a given Jacobi matrix. • Section 5 contains a brief review on what we need from Mourre’s theory.
• In Section 6 we show a useful property for the resolvent ofH.
• In Section 7 we establish a Mourre estimate forH0when|λ|<1 andα <1, which is extended toHin Section 8. • The borderline case when|λ|<1 andα=1 is considered in Section 9.
• In Sections 10 and 11 we analyze the borderline casesλ= ±1.
• In Appendix A we consider some special cases where one has simple commutation rules and so Mourre’s method gives more than what we obtain in the general case and in a quite elegant manner.
2. Elementary analysis ofH0
As we already explained in the introduction some assertions of Theorem 1.1 follow from different known results, see [6,10,15,17]. This section can be considered partially as a review of these results.
In this section let H be the Jacobi operator realized by an>0 and bn ∈R (arbitrary). First, the point (a) of
Theorem 1.1 follows form Theorem 4.1 of [17] stating:if lim
n→∞an= +∞ and lim infn→∞
b2n a2
n+an2−1 >2,
thenH is essentially self-adjoint onl02and has no essential spectrum. 2.1. Proof of the point (b) of Theorem 1.1
Clearly, the points (b1) of Theorem 1.1 follows immediately from the following:
Proposition 2.1. Assume thatlimn→∞an/an+1=1,limn→∞bn/an=2λ and the sequences∂(an/an+1), ∂(bn/an) are both summable. If|λ|<1andn1a1
n<∞thenHis not essentially self-adjoint onl
2 0.
Proof. It suffices to show that any generalized solution of the difference equationanψn+1+bnψn+an−1ψn−1=0 is square summable (limit circle case). It is enough to show thatun:=√anψnis bounded. But then by substituting in
the last equation we get thatunsatisfies the equation
hnun+1+vnun+hn−1un−1= −2λun (2.1)
wherehn= √
an/an+1andvn=(bn/an)−2λ. By hypothesis we have
hn>0, lim n→∞hn=1 and n1 |hn+1−hn|<∞, (2.2) lim n→∞vn=0 and n1 |vn+1−vn|<∞. (2.3)
But it is known that under these conditions, if |λ|<1 then any solution of Eq. (2.1) is bounded, see for example Theorem 3.19 of [26]. 2
Remark. Of course, if in Proposition 2.1 we assume that n1an1 = ∞ thenH is essentially self-adjoint onl02 (Carleman condition). Moreover, ifn1|∂an1|<∞thenH has a purely absolutely continuous spectrum filling the whole real axisRand so the assertion (b2) of Theorem 1.1 follows. Indeed, as above, one can show that for allE∈R and any solution ofτ ψ=Eψwe have thatun:=√anψnis bounded. To prove a similar assertion in [17] the authors
estimate a product of large number of transfer matrices which is rather more complicated. The proof can be now completed exactly as in [17].
2.2. Proof of the point (c1) of Theorem 1.1
This is clearly a particular case of the following more general theorem, see [13] for a different proof of this theorem:
Theorem 2.1.Assume thatlimn→∞n2/an=0andβn−=bn+(an+an−1)(respectivelyβn+=bn−(an+an−1))is bounded. Then the Jacobi operatorHis essentially self-adjoint onl02andσess(H )= ∅.
Proof. (i) By Theorem 1.4 in page 505 and the Corollary in page 506 of [3] the operatorH is self-adjoint and semi-bounded. Moreover, using the unitary operator given by(U ψ )n=(−1)nψn, it is enough to study the minus case.
(ii) Assume thatβn−=0 and denote the corresponding operatorH0. We will prove thatH0is invertible and its inverse is compact inH. Recall that, according to Hardy’s inequality, the Carleman operator defined by
(T1f )n= 1 n j=n j=1 fj
is bounded inH. Thus the operator defined by (T2f )n= 1 √a n j=n j=1 fj
is compact inHsincen/√antends to zero at infinity. Then its adjointT2∗ and the productT2∗T2are both compact. ButT2∗T2is given explicitly by
T2∗T2fn= k=∞ k=n 1 ak j=k j=1 fj for allf ∈H.
Thus, it is easy to verify that for anyf∈Hwe have an T2∗T2f n+1−(an−1+an) T2∗T2f n+an−1 T2∗T2f n−1=fn, (2.4) which shows thatT2∗T2f∈D(H0)and thatH0T2∗T2f = −f. Similarly, one may verify easily thatT2∗T2H0f = −f for eachf ∈D(H0). HenceH0is invertible andH0−1= −T2∗T2which is compact. Henceσess(H0)= ∅.
(iii) Since the sequenceβnis bounded, the multiplication operator(Rψ )n=βnψnis bounded. ButH=H0+R
andH0has a compact resolvent, soHhas a compact resolvent as well. 2
Theorem 2.2. Let H0± be the Jacobi operators realized by the sequences an >0 and bn±= ±(an +an−1). If {n2/an}n1is bounded then0∈/σ (H0±).
Proof. By repeating the same argument above for H0− we get thatH0− is invertible and its inverse equals−T2∗T2 which is only bounded here. 2
2.3. Proof of the points (c2) and (c3) of Theorem 1.1
(1) Again it suffices to study the caseλ=1. To do so we recall a known trick from [9,10]. Lethn>0 such that the
operatorLdefined by
is essentially self-adjoint onl02. One can show thatL2leaves invariant the spanH0of{e2k−1}k1and its restriction to H0is unitarily equivalent to the Jacobi operatorLdefined inHby
(Lψ )k=akψk+1+bkψk+ak−1ψk−1
whereak =h2k−1h2kandbk=h22k−1+h22k−2. In particular,σ∗(L)=σ∗(L2)whereσ∗(·)is any spectral component
of the underlying operators.
(2)Proof of the statement(c2)of Theorem1.1. LetLbe the operator defined by (2.5) withhk=(k+12)α/2. By the
remark of Section 2.1 and since 0< α <2, the spectrum ofLis purely absolutely continuous and fills the whole real axisR. In particular,σ (L)=σ (L2)= [0,∞). But, 2−αL−H0is a compact operator inHsince
2−αak=kα− α 32k α−2+Okα−4, (2.6) 2−αbk=kα+(k−1)α−3α(α−1) 16 k α−2+Okα−3, (2.7) and 0< α <2. In particular, σess(H0)=σess 2−αL= [0,∞).
Since H00 (cf. Proposition 11.1 below) we getσ (H0)=σess(H0)= [0,∞). In fact, according to Corollary 4.1 below we also have that the absolutely continuous components ofH0and 2−αLare unitarily equivalent. In particular, the absolutely continuous spectrum ofH0isσac(H0)= [0,∞).
(3)Proof of the statement(c3)of Theorem1.1. LetLbe the operator defined by (2.5) withhk=k+1/2. According
to the remark of Section 2.1 the spectrum ofL is purely absolutely continuous and fills the whole real axisR. In particular,σ (L)=σ (L2)= [0,∞). On the other hand,
ak=h2k−1h2k=4k2−1/4,
bk=h22k−1+h22k−2=4k2+(k−1)2−3 2.
Consequently, L=4H0−14J1−1 where J1 is the operatorJλ withλ=1 of Corollary 4.3 below which isH
0-compact. Thus σess(H0)= [14,∞). Moreover, according to Theorem 11.1, the operatorH0 has a purely absolutely continuous spectrum.
3. Relative boundedness and proof of Theorem 1.2
In this sectionH denotes the Jacobi operator realized by the sequencesan=an(0)+ηnandbn=b(n0)+βnwhere
an(0), b(n0), ηn andβn are arbitrary real sequences such thatan>0 andan(0)>0. We shall denote byH0the operator corresponding to the case whereηn=βn=0.
Lemma 3.1.Assume that we have
lim sup n→∞ ηn an(0) <1, (3.1) sup n1 ηn an(0−)1 an(0) −ηn−1 +ηn b(n0) an(0) −βn <∞. (3.2)
ThenD(H0)=D(H ), andH is essentially self-adjoint onl02if and only ifH0is essentially self-adjoint onl02.
Proof. One can assume, without loss of generality, that supn|ηn/a(n0)|<1 for alln. Otherwise we have only to change
a finite number of matrix elements ofH to be in this situation which is a compact perturbation. Then the operators D, K defined inHby
(Dψ )n= ηn a(n0) ψn, (3.3) (Kψ )n= ηn a(n0−)1 a(n0) −ηn−1 ψn−1+ ηn b(n0) an(0) −βn ψn (3.4)
are bounded inHandD<1. Moreover, for allψ∈l02we have (DH0ψ )n=
(H−H0)ψn+(Kψ )n.
It follows that, for allψ∈l20we have
(H−H0)ψD · H0ψ + K · ψ. (3.5) SinceD<1 and according to Theorem X.13 of [22] the assertions of the lemma follows immediately. The proof is complete. 2
Remark.LetV be the difference operator given by (V ψ )n=ηn−1ψn−1+βnψn+ηnψn+1
withD(V ):= {u∈H|V ψ∈H}. Clearly,V coincides withH−H0onl20. Moreover, if instead of (3.1) and (3.2) we only have sup n1 ηn an(0) +ηn a(n0−)1 a(n0) −ηn−1 +ηn b(n0) a(n0) −βn <∞ (3.6)
then, by the same argument as in the last proof,V isH0-bounded.
Proof of Theorem 1.2. Consider the particular case wherea(n0)=nα andbn(0)=λ(nα+(n−1)α). Then (3.1) and
(3.2) are equivalent to the hypotheses of Theorem 1.2. Hence a direct application of Lemma 3.1 shows that under such hypotheses D(H )=D(H0)andH is essentially self-adjoint if and only ifH0 is. The proof is complete by Theorem 1.1. 2
4. Relative compactness and proof of Theorem 1.3
We keep the notation of the last section and assume thatH0is essentially self-adjoint onl02. If the hypothesis of Lemma 3.1 holds then V isH0-bounded withH0-bound less than 1. The latter is 0 if limn→∞ηn/an=0. At first
sight, one can think that this condition may ensure the relativeH0-compactness ofV. This is not true as it has been already mentioned in the remark following Theorem 1.3. In fact, we have
Lemma 4.1.The operatorV isH0-compact provided that lim n→∞ ηn an(0) +ηn a(n0−)1 a(n0) −ηn−1 +ηn b(n0) an(0) −βn =0. (4.1)
In particular, the operatorV E0(Δ)is compact inHfor all compact Borel setsΔ(E0denotes the spectral measure ofH0).
Proof. It suffices to show thatV isH0-compact. According to (4.1), the operatorsDandKdefined by (3.3) and (3.4) are compact inH. Moreover, we have seen that (the requirement of Lemma 3.1 being satisfied)D(H0)is included in D(V )and
V =DH0−K onD(H0).
Letz∈ρ(H0). By multiplying this identity by(H0−z)−1we get
But then the two terms of the right-hand side of the last identity are compact. Hence V (H0−z)−1 is a compact operator inH, that isV isH0-compact. 2
One can deduce from the last proof and the formula (4.2) thatV (H0−z)−1is a trace class operator as soon as ∞ n=1 ηn a(n0) +ηn an(0−)1 an(0) −ηn−1 +ηn bn(0) an(0) −βn <∞. (4.3)
In particular, by Kuroda–Birman theorem, see [22], one can deduce
Corollary 4.1.If (4.3)holds true then the absolutely continuous components ofH andH0are unitarily equivalent. By Weyl’s criterion we deduce immediately from Lemma 4.1 the following:
Corollary 4.2.Under the assumption of Lemma4.1the sumH=H0+V defined onD(H )=D(H0)is self-adjoint andσess(H0+V )=σess(H0).
Corollary 4.3.Assume thatlimn→∞an= +∞,limn→∞an+1/an=1andlimn→∞bn/an=2λ. LetJλbe the Jacobi operator defined inHby
(Jλψ )n=ψn−1+2λψn+ψn+1.
ThenJλP (Jλ)isH-compact, for any polynomial functionP.
Proof of Theorem 1.3. A straightforward application of Lemma 4.1 combined with Theorem 1.1. The proof is com-plete. 2
5. The conjugate operator theory
The following brief review on the conjugate operator theory is based on [2,23]. Let H,A be two self-adjoint operators in a Hilbert spaceH. For a complex numberzin the resolvent setρ(H )ofH we setR(z)=(H−z)−1.
Definition 5.1.Letkbe a positive integer and 0< σ <1 a real number. (1) We say that a bounded operatorT inHis of classCk(A)if the map
t−→T (t )=e−iAtT eiAt∈B(H)
is strongly of classCk on R. If thekth derivative is Hölder continuous of order σ we say that H is of class
Ck+σ(A).
(2) We say thatH is of class Ck(A), respectively Ck+σ(A), ifR(z)is of classCk(A), respectivelyCk+σ(A), for some (and so for all)z∈ρ(H ).
Let us equipD(H )with the graph normfH= f + Hf. One has (see Theorem 6.2.10 of [2])
Proposition 5.1.The operatorH is of classC1(A)if and only if the following two conditions hold:
(i) for somez∈ρ(H )the setD(z)= {ψ∈D(A)|R(z)ψ andR(z)ψ∈D(A)}is dense inD(A)equipped with the graph norm · A, and
(ii) there is a constantc >0such that for allψ∈D(A)∩D(H ): ψ,[H, iA]ψ=2H ψ, iAψcψ2H.
From now on, we assume thatH is at least of classC1(A). Then[H, iA]defines a bounded operator fromD(H ) into its adjointD(H )∗that we still denote by the same symbol[H, iA].
LetEbe the spectral measure ofH. We define the open setμ˜A(H )of real numbersλsuch that for some constant a >0, a neighborhoodΔofλand a compact operatorKinHthe following Mourre estimate holds
E(Δ)[H, iA]E(Δ)aE(Δ)+K. (5.1)
Remarks.
(1) It is important to stress here that (5.1) holds for any realλbut witha∈Rinstead ofa >0. One may therefore define the function
ρ(λ):=supa∈R|(5.1) holds.
With this notation one has:λ∈ ˜μA(H )if and only ifρ(λ) >0.
(2) One can prove that ifλ∈σess(H )andAis bounded then ρ(λ)=0, and therefore no Mourre estimate can be expected, see Chapter 7 of [2]. In other words, we have
A∈B(H) ⇒ σess(H )∩ ˜μA(H )= ∅. (5.2) The first main consequence of the Mourre estimate and the Virial lemma is:
Theorem 5.1.The setμ˜A(H )contains at most a discrete set of eigenvalues ofHand all these eigenvalues are finitely degenerate.
PutμA(H ):= ˜μA(H )\σp(H ), whereσp(H )denotes the set of eigenvalues ofH, andA =(1+A2)1/2. One
has, see [23],
Theorem 5.2.Assume thatHis of classCs+21(A)for some1/2< s <1. ThenHhas no singular continuous spectrum inμA(H ). Moreover, the limits
Rs(λ±i0):= lim μ→0+A
−sR(λ+iμ)A−s
exist locally uniformly onμA(H )for the norm topology ofB(H)and they are locally Hölder continuous of order s−12as function ofλ∈μA(H ). In particular, for everyϕ∈C0∞(μA(H ))we have
A−se−iH tϕ(H )A−sCt−(s−12).
6. A useful property of the resolvent ofH
Our goal in this section is to show the following lemma which will be used in the proof of theC1(A)-regularity in the next sections.
Theorem 6.1.LetH be the Jacobi operator realized byan>0andbn∈Rsuch thatsupn1an/n <∞. Then for any
p >0there existsC >0such that for anyz∈ρ(H )withd=distance(z, σ (H )) > Cwe have sup
n1
npen, (H−z)−1ek<∞, for allk1. (6.1) In particular, for anyσ >0 the estimate (6.1) withp=σ+2 shows that(H−z)−1l02and(H−z)−1l02are included inD(Nσ)for somez∈ρ(H ).
Proof. By hypothesis Carleman condition is clearly satisfied and so H is essentially self-adjoint on l20. We will establish (6.1) which is a Combes–Thomas-like estimate by adapting the argument given in Appendix II of [1].
Let us fixnand define the operator of multiplicationLnby
(Lnψ )j=
jpψj ifjn,
which is bounded and invertible in H. Moreover, it is obvious that Ln and L−n1 leave invariant D(H ), and
induce two bounded operators in D(H ) endowed with · H that we denote by the same symbols. Then
Ln(H −z)L−n1:D(H )→H is invertible and (Ln(H −z)Ln−1)−1=Ln(H −z)−1L−n1. Moreover, a direct
com-putation shows that onD(H )we have the following identity:
Ln(H−z)L−n1=H+Kn−z (6.2)
whereKnis defined inHby(Knψ )j=kj,j+1ψj+1+kj,j−1ψj−1with the only non-zero matrix coefficients ofKn
are kj,j+1=aj 1+1 j −p −1 if 1jn−1, kj,j−1=aj−1 1−1 j −p −1 if 1jn.
Since supn1an/n <∞, there is a constantC >0 independent ofnsuch thatKn< C. But then by (6.2) and the
resolvent equation we get Ln(H−z)L−n1 −1 = p0 (H−z)−1−Kn(H−z)−1 p
where the right-hand side converges in norm as soon asKn(H−z)−1<1. The latter is ensured by choosingzsuch
thatd=distance(z, σ (H )) > C, which can be done independently of bothnandk. In this case, one has Ln(H−z)L−n1−1 1
d−C. Now observe that for everynkwe have
npen, (H−z)−1ek=kpen, Ln(H−z)−1L−n1ek
=kpen,Ln(H−z)L−n1
−1 ek. Thus for everynk(kbeing fixed) we have
npen, (H−z)−1ekkpLn(H−z)L−n1
−1
kp 1
z−C ≡C , which is the desired property, since the constantCis independent ofn. 2
7. Mourre estimate forH0with 0< α <1 and|λ|<1
In this section H0 denotes the Jacobi operator realized by the sequences an=nα andbn=λ(an+an−1)with 0< α <1 and |λ|<1. According to (b2) of Theorem 1.1, H0 is essentially self-adjoint on l02 and has a purely absolutely continuous spectrum filling the real axis. Our goal now is to establish a Mourre estimate forH0. LetAbe the operator defined inHby
(Aψ )n=i(αn−1ψn−1−αnψn+1) and αn=n1−α. (7.1)
Since 1−α <1, the operatorAis clearly essentially self-adjoint onl20. This operator has been used for example by Dombrowski and Dombrowski and Pedersen mainly forλ=0. SinceAis unbounded Putnam–Kato approach does not applied and so they only proved the absence of the point spectrum by a kind of Virial theorem. Here we prove that Mourre’s theory applies very easily:
Theorem 7.1.The operatorH0is of classC1(A)andμ˜A(H0)=R.
Proof. (i) In the sequel for anyz∈C\σ (H0)we denote by R0(z)the resolvent ofH0. By Theorem 6.1, for any ψ∈l02,R0(z)ψ andR0(z)ψ belong toD(N2)⊂D(A), for somez∈ρ(H0). Thusl02is contained in the subspace D(z)= {u∈D(A)|R0(z)uandR0(z)u∈D(A)}. In particular,D(z)is dense inD(A)which is exactly the point (i) of Proposition 5.1. On the other hand, a direct computation shows that forψ∈l02, we have
ψ,[H0, iA]ψ =(H0ψ, iAψ )+(iAψ, H0ψ ) = n ψn(cn−2ψn−2+dn−1ψn−1+2ψn+dnψn+1+cnψn+2) where cn=an+1αn−anαn+1, dn=λαn(an+1−an−1).
Using the obvious fact(1+1n)x=1+xn+x(x2n−21)+O(1/n3),we get cn= −1+2α+εn withεn=O 1/n2, dn=2λα+κn withκn=O 1/n2.
Thus there exists a constantC >0 such that for anyψ∈l20we have ψ,[H0, iA]ψCψ2.
Hence, by Proposition 5.1,H0is of classC1(A). Moreover, the form[H0, iA]defines a unique bounded operator in Hdenoted by the same symbol and
[H0, iA] =B0+K1, (7.2)
(B0ψ )n=(2α−1)ψn−2+2λαψn−1+2ψn+2λαψn+1+(2α−1)ψn+2, (7.3)
(K1ψ )n=εn−2ψn−2+κn−1ψn−1+κnψn+1+εnψn+2. (7.4)
(ii) It remains to prove the Mourre estimate. Sinceεnandκn tend to zero at infinity, the operatorK1is compact. Moreover, we easily verify that
B0=4
1−λ2(1−α)+(2α−1)Jλ2+2λ(2−3α)Jλ (7.5)
wherea=4(1−λ2)(1−α) >0, sinceα <1 and|λ|<1; andJλis the operator of Corollary 4.3. The latter ensures
that(2α−1)Jλ2+2λ(2−3α)JλisH0-compact. So for any bounded real intervalΔone has
E(Δ)[H0, iA]E(Δ)=E(Δ)B0E(Δ)+E(Δ)K1E(Δ)=aE(Δ)+K for some compact operatorKand thereforeμ˜A(H )=R. 2
Remark.
(i) From (7.5) we see how the Mourre estimate breaks down ifα=1. This is not surprising since in this case the operatorAbecomes bounded and therefore it could not play the role of a conjugate operator of any self-adjoint operatorH with non-trivial essential spectrum.
(ii) Assume thatα >1 and letH0s be a self-adjoint extension ofH0. Then by replacingAby−Aand by the same argument (7.5) implies that for any bounded real intervalΔone has
E(Δ)H0s,−iAE(Δ)=aE(Δ)+K,
witha=4(1−λ2)(α−1) >0. This proves thatμ˜A(H0s)=R. Since the operatorAin this case is bounded, by (5.2) the operatorH0s has no essential spectrum. This is a new proof of the emptiness of the essential spectrum of any self-adjoint extension ofH0ifα >1 without using the orthogonal polynomial theory. See Theorem 7.3 for a generalization of this idea.
Theorem 7.2.The operatorH0is of classC2(A).
Proof. Letz∈C\σ (H0). According to Theorem 7.1 the operatorR0(z)is of classC1(A)and
R0(z), iA
It suffices to show that each term of the right-hand side is of classC1(A). But the matrix elementsεn andκn ofK1 areO(1/n2)and so the range ofK1is included inD(A). HenceAK1is clearly bounded and soK1is of classC1(A). ConcerningB0, according to (7.5) and the boundedness ofJλ, it suffices to show thatJ˜λ:= [Jλ, iA]defines a bounded
operator inH, which is true since
(J˜λψ )n=(αn−αn+1)ψn+2+2(αn−αn−1)ψn+(αn−2−αn−1)ψn−2 andαn−αn+2=O(n−α). The proof is complete. 2
Remark. It should be clear that, by a direct application of Mourre’s theory recalled in Section 5, we get various information on the boundary values of the resolvent ofH0but we will not state them separately.
Notice that our argument works for a class of more general coefficients than those of power-like. Indeed, letH be the Jacobi operator realized byan>0 andbn=λ(an+an−1). Assume that limn→∞an= +∞and
an+1 an =
1+xn and lim
n→∞nxn=α. (7.7)
In particular, α0 andan/an+1tends to 1. Therefore, if|λ|>1 thenH is essentially self-adjoint operator and its spectrum is purely discrete. In the opposite case we have the following:
Theorem 7.3.Assume that(7.7)holds true and consider the operatorAdefined by(7.1)withαn=n/an. If|λ|<1 then we have:
(i) if 0α <1 thenH is essentially self-adjoint onl02 andμ˜A(H )=R(and so the point spectrum ofH has no accumulation point);
(ii) ifα >1then any self-adjoint realization ofHhas no essential spectrum.
Proof. (i) Ifα <1 thenan1 = ∞and thereforeH is essentially self-adjoint onl02. Moreoveran/n→0 and so
Lemma 6.1 applies. Then the point (i) of Proposition 5.1 can be verified in this case as we have done forH0above. Furthermore, evaluating the matrix elements of commutatorB= [H, iA], exactly as in the proof of Theorem 7.1 and using (7.7) we get cn=n an+1 an − (n+1) an an+1 = − 1+2α+εn withεn→0, dn=λn an+1 an − an−1 an =2λα+κn withκn→0.
The proof now can be finished as forH0of Theorem 7.1.
(ii) Ifα >1 thenan1 <∞and{n/an}n1is bounded. In particular, the operatorAis bounded, while we do not know whetherH is essentially self-adjoint or not (except forλ=0). LetHs be a self-adjoint realization ofHinH. By repeating the same calculation we getμ˜A(Hs)=R. The proof is finished by (5.2). 2
Example.It is clear that ifan=nα(lnn)β withα0 then (7.7) holds.
8. Proofs of Theorems 1.4–1.6
Throughout this section we suppose that 0< α <1 and (1.5) holds true. ThenV isH0-compact,H=H0+V is essentially self-adjoint onl02andσess(H )=σess(H0)=R. We start by isolating a sufficient condition that guarantees theH0-compactness of[V , iA]whereAis defined by (7.1).
Proposition 8.1.Under the hypotheses of Theorem1.4, the commutatorW= [V , iA] =V (iA)−(iA)V defined on l20coincides with anH-compact operator denoted also byW= [V , iA].
Proof. (i) Let us definetn:=αn∂ηn,εn:= −(∂α)nηn, n:=(∂α)n−1ηn andκn:=λ(∂α)n−1(∂η)n−1+αn(∂β)n.
(W ψ )n=γn−2ψn−2+rn−1ψn−1+2dnψn+rnψn+1+γnψn+2, γn:=αnηn+1−αn+1ηn=tn+εn,
rn:=αn(∂β)n=λ(tn+tn−1)+κn,
dn:=αnηn−αn−1ηn−1=tn−1+n. (8.1)
Consider the operators defined inHby
(T ψ )n=tn−1ψn−1+λ(tn+tn−1)ψn+tnψn+1, (8.2) (Kψ )n=εn−2ψn−2+κn−1ψn−1+2nψn+κnψn+1+εnψn+2. (8.3) Hypotheses of Theorem 1.4 implies that
lim n→∞ tn nα + |∂tn| + |εn| + |n| + |κn| =0. (8.4)
Hence the operatorKis compact inH, while by Lemma 4.1 the operatorT isH0-compact. SinceH=H0+V and V isH0-compact,T isH-compact. On the other hand, letSbe the shift operator defined by(Sψ )n=ψn+1. Then one
can verify easily that onl02we have
W=T S+(T S)∗+K. (8.5)
The proof is clearly complete. 2
Remark.If we only assume that the sequences inside the limit in (8.4) are bounded thenW will beH-bounded.
Corollary 8.1.Under the hypotheses of Theorem1.4the operatorHis of classC1(A)andμ˜A(H )=R. In particular, the operatorHhas at most a discrete set of eigenvalues inR(which is the claim of Theorem1.4).
Proof. The point (i) of Proposition 5.1 can be proved exactly as we have done forH0. Moreover, according to Theo-rem 7.1 and Proposition 8.1, there is a constantC >0 such that for anyψ∈l20we have
ψ,[H, iA]ψ=ψ,[H0, iA]ψ
+ψ,[V , iA]ψCψ2H.
ThenH is of classC1(A). In particular, this quadratic form has a unique extension to D(H ) and the associated operator will be denoted by the same symbol[H, iA]. Combining (7.2) and (7.5) with Proposition 8.1 we obtain
[H, iA] = [H0, iA] +W=4
1−λ2(1−α)+(2α−1)Jλ2+2λ(2−3α)Jλ+K1+W.
ButJλis bounded andH0-compact, and soJλ andJλ2areH-compact. Moreover,W isH-compact as well asK1is compact. Thus for any bounded real intervalΔone has
E(Δ)[H, iA]E(Δ)=41−λ2(1−α)E(Δ)+K for some compact operatorK. This shows thatμ˜A(H )=R. 2
Remark.With the notationΔn=ηn/nα we have
an+1 an = 1+xn withnxn=α+n∂Δn 1+O(Δn) .
Under the hypothesis (1.6) the last expression need not to have a limit (it is unbounded in general). Comparing this with Theorem 7.3 we see that the best choice for the conjugate operator A for H is with αn=n1−α rather than
αn=n/an. It is the latter choice that we used in [24]. On the other hand, theorems of [15,17] required that after some
Proposition 8.2.Under the hypothesis of Theorem1.5the operatorHis of classC2(A).
Proof. Letz∈C\σ (H )andR=R(z). It suffices to show that[R, iA]is of classC1(A). But, by Corollary 8.1, we have
[R, iA] = −R[H, iA]R= −R[H0, iA]R−R[V , iA]R.
SinceRand[H0, iA]are bounded inHand of classC1(A), the first termR[H0, iA]Ris clearly of classC1(A). It is enough to prove thatR[V , iA]Ris of classC1(A). But by (8.5) one has
R[V , iA]R=RT SR+R(T S)∗R+RKR.
ConcerningRKRit suffices to show that the commutator[K, iA]is bounded inH. A direct calculation shows that the matrix elements of [K, iA]are bounded which implies that third term is of classC1(A). Moreover,RT SRand R(T S)∗R are of the same class. So it is enough to studyRT SR. For let us remark first thatT has the same form asV. So by applying the remark following Proposition 9.1 toT (replaceηnbytnandβnbyλ(tn+tn−1)), we see that [T , iA]isH-bounded as soon as (1.7) holds true. On the other hand, we have
[RT SR, iA] =R[T , iA]SR+RT[S, iA]R+ [R, iA]T SR+RT S[R, iA].
The first term is bounded inH. SinceRT is bounded andRis of classC1(A), the two last terms are clearly bounded inH. Moreover, it is easy to verify that all the matrix elements of[S, iA]tends to zero, which finishes the proof. 2
Proof of Theorems 1.5 and 1.6. Combining Corollary 8.1, Proposition 8.2 and Theorem 5.2 we get thatH has no singular continuous spectrum which is Theorem 1.5. Moreover, letσ > (1−α)/2 or equivalentlyσ =r(1−α)for somer >1/2. By Theorem 2.2 the limits
lim
μ→0+A
−r(H−x∓iμ)−1A−r
exist locally uniformly onR\σp(H )for the norm topology ofB(H)and these boundary values are Hölder continuous
of order min(r −12,1/2)as function of the spectral parameterx∈R\σp(H ). But obviously one hasN−σAr =
N−r(1−α)Ar∈B(H). This enables us to get that the limits R0,s(x∓i0):= lim
μ→0+N
−σ(H−x∓iμ)−1N−σ
exist locally uniformly onR\σp(H )for the norm topology ofB(H)and these boundary values are Hölder continuous
of order min(r−12,1/2)as function ofx∈R\σp(H ). The proof of the remaining assertion is similar and will be
omitted. 2
9. The borderline cases|λ|<1 andα=1
This section is entirely devoted to the borderline cases where|λ|<1 andα=1 in (1.2) and (1.3). According to Theorem 1.1, the unperturbed modelH0has a purely absolutely continuous spectrum that fills the whole real axis. Our goal is to exhibit a conjugate operator forH0and to study the preservation of the absolute continuity of the spectrum ofH0after perturbation byηnandβn.
LetAbe the operator defined inHby
(Aψ )n=i(αn−1ψn−1−αnψn+1) and αn=lnn (9.1)
which is essentially self-adjoint onl02. We have
Theorem 9.1.The operatorH0is of classC2(A)andAis locally conjugate toH0onR, i.e.μ˜A(H0)=R.
Proof. The point (i) of Proposition 5.1 follows from Theorem 6.1 exactly as we have done forα <1 in Section 7. Moreover, the commutatorB0= [H0, iA]is given, in the form sense onl20, by
(B0ψ )=cn−2ψn−2+dn−1ψn−1+knψn+dnψn+1+cnψn+2, cn=an+1αn−anαn+1=lnn−1+εn,
dn=λαn(an+1−an−1)=2λlnn,
kn=2(anαn−an−1αn−1)=2 lnn+2+κn
whereεnandκnbehaves likeO(1/n)at infinity. Here-above we have used the obvious fact thatαn+1−αn=1/n+
O(1/n2). Lettn=lnn−1 and consider the operatorsT andKdefined by
(T ψ )n=tn−1ψn−1+2λtnψn+tn+1ψn+1, (9.2)
(K1ψ )n=εn−2ψn−2+nψn+εnψn+2. (9.3) WhileKis clearly a compact operator inH, the operatorT isH0-compact, by Lemma 4.1. Moreover, we have in the form sense onl02,
B0=4
1−λ2+T S+(T S)∗+2λJλ+K (9.4)
whereJλis the operator of Corollary 4.3, which isH0-compact. Thus there exists a constantC >0 such that for any
ψ∈l20we have
ψ,[H0, iA]ψCψ2H.
ThenH0is of classC1(A)and the last quadratic form has a unique extension toD(H0)and the associated operator will be denoted by the same symbolB0= [H0, iA]. Moreover, for any bounded real intervalΔthe productsE(Δ)T E(Δ), E(Δ)JλE(Δ)andE(Δ)KE(Δ)are all compact operators inH. Thus
E(Δ)[H0, iA]E(Δ)=4
1−λ2E(Δ)+K for some compact operatorKand thereforeμ˜A(H )=R.
(iii) It remains to check thatH0is of classC2(A). Letz∈C\σ (H0)andR0=(H0−z)−1. SinceH0is of class C1(A),R0is of classC1(A)and
[R0, iA] = −R0 4−λ2+K+λJλ R0−R0 T S+(T S)∗R0.
The first term can be proved to be of class C1(A) exactly as we have done in Section 7. Clearly R0T SR0 and R0(T S)∗R0 are similar and it suffices to deal withR0T SR0. SinceR0 is of classC1(A)andR0T S is bounded, it suffices to prove that the[T SR0, iA]is bounded inH. We have
[T SR0, iA] =R0[T , iA]S+R0T[S, iA] + [R0, iA]T S=R0[T , iA]S+R0T[S, iA] +R0B0R0T S.
The last term is bounded as a product of the bounded operatorsR0B0andR0T S. The middle term is bounded since [S, iA]is, by simple calculation. Concerning the first term,T has the same form asH0so the same calculation shows that the matrix elements of[T , iA]are bounded (in fact tend to zero). The proof is complete. 2
Recall that, by Theorem 1.3, if lim
n→∞
|ηn/n| + |∂ηn| +βn=0
thenσess(H )=σess(H0)=R. Concerning the point spectrum we have
Theorem 9.2.In addition to the hypotheses of Theorem1.3we suppose that lim n→∞lnn∂β n+∂ 2η n=0.
ThenHis of classC1(A)andμA(H )=R. In particular, the possible eigenvalues ofHhave no accumulation point.
Again by mimicking the proof of Proposition 8.2 withαn=lnn, we get
Theorem 9.3.In addition to the hypotheses of the last theorem assume that sup
n1
(lnn)2∂3ηn+∂2βn<∞. ThenH has no singular continuous spectrum.
Similar statement to Theorem 1.6 holds as well.
10. The borderline casesλ= ±1: Birth and death Jacobi matrices
In this section we focus on the Jacobi operators realized by the sequences given by (1.2) and (1.3) withλ= ±1 that we denote by H±. Notice thatH− is the infinitesimal matrix of certain birth and death processes see [14]. In Theorems 1.1 and 1.3 we describe the essential spectrum of those operators for everyαand for a quite large class of perturbations byηnandβn. Our aim now is to exhibit a conjugate operator forH0±and then forH±.
Consider the operatorAdefined inHby
(iAψ )n=αn−1ψn−1−αnψn+1 withαn=n. (10.1)
It is clearly essentially self-adjoint in H. If α=1 then we proved in [25] that H0+ is A-homogeneous, that is [H0+, iA] =2H0+. In particular, H0+ has purely absolutely continuous spectrum filling the semi-axis [0,∞). See Appendix A for further results. More generally, we have
Theorem 10.1.Assume that0< α <1. Then the operatorH0± is of classC2−α(A)and Ais locally conjugate to H0± onR\ {0}, i.e.μ˜A(H0±)=R\ {0}. In particular, H0± has no singular continuous spectrum and the possible eigenvalues ofH0±can only accumulate to0.
Proof. (i) We will show the theorem forH0+ realized by the sequencesan=nα andbn=an+an−1. ForH0− the proof is similar. The point (i) of Proposition 5.1 follows from Theorem 6.1. Moreover, the commutatorB0= [H0+, iA] is given, in the form sense onl02, by
(B0ψ )n=cn−2ψn−2+dn−1ψn−1+knψn+dnψn+1+cnψn+2, cn=αnan+1−αn+1an=(α−1)nα+κn,
dn=αn(bn+1−bn)=2αnα+n,
kn=2(αnan−αn−1an−1)=2(1+α)nα+εn
whereκn, nandεnbehave likeO(nα−1)near to infinity. Lettn=nα/2. By Lemma 4.1 the operatorT defined by
(T ψn)=tn−1ψn−1+(tn+tn−1)ψn+tnψn+1 (10.2) isH0+-compact. Moreover, direct computation shows that
H0+, iA=(4−2α)H0++(α−1)T2+K, (10.3) (Kψn)=κn−2ψn−2+n−1ψn−1+εnψn+nψn+1+κnψn+2. (10.4)
Sinceα <1, all the matrix elementsεn,nandκnofKtend to zero at infinity and therefore the operatorKis compact
inH. Then for allψ∈l02:
ψ,H0+, iAψ=(4−2α)ψ, H0+ψ+(1−α)T ψ2+ ψ, Kψ. SinceT isH0+-compact there exists a constantC >0 such that for allψ∈l02
ψ,H0+, iAψCψ2
This shows thatH0+is of classC1(A).
(ii) LetΔ=(x, y)be a real interval such thatx >0. Then
E(Δ)H0+, iAE(Δ)=2(2−α)E(Δ)H0+E(Δ)+E(Δ)KE(Δ)aE(Δ)+E(Δ)KE(Δ) witha=2(2−α)x. This is the desired Mourre estimate.
(iii) It remains to show thatH0+is of classC2−α(A). SinceH0+is of classC1(A), it suffices to show that[R(z), iA] is of classC1−α(A), for somez∈C\σ (H ). But we have
R(z), iA= −R(z)H0+, iAR(z) and one can verify that
H0+, iA=(α−1)SH0++H0+S∗+2H0++K
whereKhas similar form toKabove with matrix coefficients behaving asO(nα−1)and soKis of classC1−α(A). One can also prove thatSis of classC1(A)and so the contribution ofR(z)((α−1)(SH0++(H0+S)∗)+2H0+)R(z) is of classC1(A). The proof is finished. 2
Remarks.
(i) According to Theorem 1.1, if α >2 then the essential spectrum ofH0±is empty. So no interest to try to get a Mourre estimate in this case.
(ii) Concerning the remaining case where 1< α <2 the last proof works at formal level. More precisely, by devel-oping further (at order 2) the matrix elements of the commutator[H0+, iA]we get
H0+, iA=(4−2α)H0++(α−1)T2+α(1−α)T+K
whereKis a compact operator andThas the same form ofT but by replacingtnbytn =n1−α which isH0+ -compact by Lemma 4.1. However, we are not able to show theC1(A)-regularity since we do not have a needed property of the resolvent ofH0+similar to Theorem 6.1.
(iii) The case whereα=2 is postponed to the next section.
Let us now study the perturbed Hamiltonian. Assume that the sequences Δn:=ηn/nα, nα(∂Δ)n and βn :=βn±(ηn+ηn−1)
tend to zero. ThenV isH0∓-compact andσess(H±)=σess(H0±)= [0,∞).
Theorem 10.2.Letα∈(0,1]. Assume that lim n→∞n(∂β ) n+n1−α(∂η)n+n ∂2η n=0.
ThenμA(H±)=R\0. In particular, the only possible accumulation point of the eigenvalues ofH±is0.
Proof. By repeating the proof of Proposition 8.1 and its Corollary 8.1 but withαn=nandλ= ±1. 2
Concerning the singular continuous spectrum and by similar argument to above, we can prove the following
Theorem 10.3.Under the hypothesis of Theorem 10.2we assume that, for someδ >0, the sequencesn2(∂2β)n,
n2−α(∂2η)n,n2(∂3η)nare bounded. Then the operatorH±has no singular continuous spectrum.
Remark.Notice that in the caseα=1 our present results improve those of [25]. This is due to our relative bounded-ness/compactness criterion.
11. Birth and death Jacobi matrices withα=2
For this model we start by the following lemma borrowed from [8]:
Proposition 11.1.LetH±be the self-adjoint Jacobi operator realized by an arbitraryan>0andbn= ±(an+an−1). Then for anyψ∈D(H±)
ψ, H±ψ= ±
n=∞
n=1
an|ψn+1±ψn|2.
In particular,σ (H±)⊂ [0,±∞)and0is not an eigenvalue ofH±.
Recall that hereafterH0±are the unperturbed models withα=2 andλ= ±1.
Corollary 11.1.The operatorAdefined by(10.1)isH0±-bounded.
Proof. Letψ∈D(H0+)and write (Aψ )n 2 =n(ψn+1+ψn)−ψn−(n−1)(ψn+ψn−1) 2 3n2|ψn+1+ψn|2+(n−1)2|ψn+ψn−1|2+ |ψn+1|2 . Then, by Proposition 11.1,{(Aψ )n}nis square summable and
Aψ2C ψ2+ n1 an|ψn+1+ψn|2 Cψ2+ψ, H0+ψCH0+ψ2+ ψ2
which is the desired property. The same can be done for the minus case. 2 In particular, the Putnam–Kato theory applies:
Theorem 11.1.Letα=2. Then the commutator[H0±, iA] =C±2 whereC±is Jacobi matrix realized byan=nand
bn= ±(2n−1). In particular,Hhas a purely absolutely continuous spectrum on[1/4,∞).
Proof. (i) The calculation that we have done in the last section shows that the commutator is given in the form sense onl02, by[H0+, iA] =C+2. Since 0 is not an eigenvalue ofC+the theorem is proved. The minus case is similar. 2
Remark. It is of interest to construct a conjugate operator in the Mourre’s sense for the model considered in this section.
Appendix A. Some special cases
LetH0, Abe a pair of self-adjoint operators in a Hilbert spaceHsuch thatH0is of classC1(A)and satisfying [H0, iA] =aH0+b with(a, b)∈R2\
(0,0).
In this case the conjugate operator theory is not only very efficient but also quite elegant. One may see [4] where the caseb=0 is developed in details. Their argument extends without real efforts to the general case(a, b) =(0,0). See also [5] for optimal propagation properties in this case. In this appendix we give three examples from the class of Jacobi matrices considered in this paper satisfying the last commutation rule.
The first example is the Jacobi operator related to the family of Hermite polynomials. It is the operatorH0realized byan=
√
n/2,bn=0. In this case, the spectral measure and its Stieltjes and Fourier transforms are well known. By
repeating the argument of Section 7 withα=1/2, λ=0, it turns out thatH0is of classC∞(A)and that
[H0, iA] =1. (A.1)
The second example is related to Laguerre polynomials,that is the operatorH0realized byan=n, bn=2n−1.
Here again the spectral measure is explicitly known. On the other hand, our method of Section 10 withα=1,λ=2 andαn=nshows thatH0is of classC∞(A)and that
[H0, iA] =H0. (A.2)
More generally, letH0be the Jacobi operatorH0realized byan=
αn2+βn+γandb
n=δn. Here-aboveα0,
β, γ∈Rsuch thatαn2+βn+γ→ +∞asn→ +∞. Ifδ2>4α0 then the spectrum ofH0is purely discrete, see also [12,19] where forα >0 this spectrum has been computed and proved to be independent of the parameterγ. In contrast, ifδ2<4αorδ=α=0 then, by Proposition 2.1,H0has a purely absolutely continuous spectrum filling the whole real axis.
Our third example is the borderline case whereδ2=4α. In [17] the authors proved thatH0has a purely absolutely continuous spectrum provided thatγ =β4α2 −16α. We proved in [24] that this condition is not necessary. Indeed, by repeating the argument of Section 10 withan=αn=
αn2+βn+γ we get thatH
0is of classC1(A)and
[H0, iA] =δH0+2(β−α). (A.3)
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