Absence of embedded eigenvalues for Hamiltonian with crossed magnetic and electric fields
aa r X i v : . [ m a t h . SP ] D ec ABSENCE OF EMBEDDED EIGENVALUES FOR HAMILTONIAN WITHCROSSED MAGNETIC AND ELECTRIC FIELDS
MOUEZ DIMASSI, MASAKI KAWAMOTO AND VESSELIN PETKOV
Abstract.
In the presence of the homogeneous electric field E and the homogeneous perpendicularmagnetic field B , the classical trajectory of a quantum particle on R moves with drift velocity α which is perpendicular to the electric and magnetic fields. For such Hamiltonians the absence ofthe embedded eigenvalues of perturbed Hamiltonian has been conjectured. In this paper one provesthis conjecture for the perturbations V ( x, y ) which have sufficiently small support in direction ofdrift velocity. Introduction
We consider the quantum dynamics on the plane R in the presence of a homogeneous constantelectric field which lies on this plane and a constant magnetic field which is perpendiculars to thisplane. Therefore the quantum system can be described by the following magnetic Stark Hamiltonianacting on L ( R ) H LS := 12 m (cid:18) D X + B Y (cid:19) + 12 m (cid:18) D Y − B X (cid:19) − qE · X + V, where D X = − i∂ X , D Y = − i∂ Y , X = ( X, Y ) ∈ R , m > q = 0 are the position, the mass andthe charge of a quantum particle and E = E = ( E , E ) = (0 , B = (0 , , B ), B = 0 stand for theelectric field and the magnetic field, respectively. Next V : R → R is the multiplication operator by V ( X ). We assume that V ( X ) is bounded and decays as | X | → ∞ . Under some decaying conditionsfor the potential V , in [5] and [1] it was established that σ ( H LS − V ) = σ ac ( H LS − V ) = R , σ ess ( H LS ) = R . Here σ ( L ) , σ ac ( L ) , σ ess ( L ) , σ pp ( L ) denote the spectrum, the absolutely continuous spectrum, theessential spectrum and the point spectrum, respectively, of the operator L . In the physical literatureit was conjectured that σ pp ( H LS ) = ∅ . This property has been proved in the following cases:(I) | qE | − qE · ∇ V > X ∈ R (see [4]),(II) | qE | is sufficiently large [1] or sufficiently small [4].Moreover, it was shown in [5] that(III) There exists R > σ pp ( H LS ) ∩ (cid:16) ( −∞ , − R ] ∪ [ R , ∞ ) (cid:17) = ∅ and, moreover, thereexist at most a finite number of eigenvalues with finite multiplicities.In particular, (II) implies that if eigenvalues exist, then | qE | is not small as well as not large. Thiscondition seems very strange and it is natural to show that for any | qE | , H LS has no eigenvalues. Key words and phrases.
Crossed magnetic and electric fields, Embedded eigenvalues.
The absence of point spectrum of H LS is an open and challenging problem. There are two majordifficulties in the investigation of this problem. If we consider the operator H = H LS − V , first H has double characteristics and second the electric and magnetic fields are not decreasing as | X | → ∞ .Consequently, even for H it is quite difficult to obtain weighted estimates for the resolvent k h X i − s ( H − λ − ± iǫ ) − h X i − s k , s > / , | λ | ≫ ǫ = 0 . In the literature there are a lot of works treating weighted resolventestimates for the perturbations of the Laplacian. Recently the proof of suitable Carleman estimatesled to several important results. We cite only some recent works [16], [17], [18], [8], where thereader may find other references. However, in these papers some decay of the potentials is assumedand this plays a crucial role in the analysis. Studying H , we cannot treat H as a perturbationof − ∆ since electromagnetic potentials do not decrease in X but increases quadratically. Usually,a Hamiltonian with quadratic potential may have only bound states. Nevertheless, the presenceof electric potential qE · X implies that H has only a continuous spectrum. In this direction theHamiltonian H is an exceptional model in quantum physics. In the case with a potential V , it isnatural to consider H as an unperturbed operator and to obtain resolvent estimates for H .We examine the situation when the support of V ( X, Y ) in direction of drift velocity α = ( E /B, − E /B )is sufficiently small. Passing to new coordinates ( x, y ), this means that the support of V ( x, y ) hassmall support with respect to y (see Assumption 1.1 below). We do not impose conditions on | qE | .Concerning the velocity α , notice that according to Proposition 4.4 in Adachi-Kawamoto [1], wehave the estimate s − lim t →∞ χ ( q B α · X ≤ c t ) ϕ ( H LS ) e − itH LS = 0 , where χ is the characteristic function such that χ ( s ≤ a ) = 1 if s ≤ a and χ ( s ≤ a ) = 0 if s > a , ϕ ∈ C ∞ ( R ) and c > α . By using this proposition,Kawamoto [12] characterized the space L pp ( H LS ) of all eigenstates of H LS , as follows: ψ ∈ L pp ( H LS ) ⇔ lim R →∞ sup t ∈ R (cid:13)(cid:13) χ ( R ≤ | α · X | ) e − itH LS ψ (cid:13)(cid:13) L ( R ) = 0 . Hence the norms of the eigenfunctions over the region | α · X | ≥ R goes to 0 as R → ∞ , it isexpected that the behavior of the potential in direction perpendicular to α must be negligible forthe existence of eigenvalues. It is easy to see that | α | − ( α · X ) + | qE | − ( qE · X ) = | X | , α · qE = 0 , (see § α is perpendicular to qE .In the following up to the end of the paper for simplicity we assume m = 1 / , B = 1 , q = 1 . Introduce the change of variables | E | x = − E · X , | α | y = − α · X , (1.1)hence x = − E X + E Y | E | , y = − E X + E Y | E | , X = − E x + E y | E | , Y = − E x − E y | E | and ∂ X = − E | E | ∂ x − E | E | ∂ y , ∂ Y = − E | E | ∂ x + E | E | ∂ y . MBEDDED EIGENVALUES 3
By using these variables, the Hamiltonian H LS is reduced to H LS = (cid:18) D x + 12 y (cid:19) + (cid:18) D y − x (cid:19) + | E | x + V ( x, y )and with the unitary transform e ixy/ , we have e − ixy/ H LS e ixy/ = ( D x + y ) + D y + | E | x + V ( x, y ) . The potential V changes but we will denote again the new potential by V ( x, y ) . Throughout thispaper we assume | E | = 1 and consider the reduced Hamiltonians H := H + V,H := ( D x + y ) + D y + x, acting on D ( H ) = D ( H ) ⊂ L ( R ). In the exposition we will use the notation h r i = (1+ r ) / , r =( x + y ) / and similar notation for h x i , h y i . The purpose of this paper is the study two problems: (A)
Estimates for the resolvent h r i − δ ( H − λ − iν ) − h r i − δ for | λ | ≫ , ν > δ > . (B) Absence of eigenvalues of the operator H .The problem (A) is examined in Section 3 and we prove the estimate k h r i − δ ( H − λ − iν ) − h r i − δ k L → L ≤ C δ ν − | λ | − δ/ , (1.2)where 0 < δ ≤ , < ν ≤ | λ | ≫ . For δ = 2 we obtains the optimal decay O ( | λ | − / ).The proof of Proposition 3.1 is based on a representation of operator e itH established in [1]. Itseems that this is the first result where we have an estimate of the resolvent of H as | λ | → ∞ . For operators with magnetic and electric potentials having some decay a similar result with bound O ( | λ | − / ) and constant C > ν > H LS , it is an open problem to improve (1.2) with a constant independent of ν > (B) is studied under the assumption Assumption 1.1.
We have V ( x, y ) , ∂ x V ( x, y ) ∈ C ( R ) and there exists η > such that supp( V ) ⊂ (cid:8) ( x, y ) ∈ R : − η ≤ y ≤ η (cid:9) . (1.3) Moreover, the potential V ( x, y ) satisfies the estimates sup x,y ∈ R h x i s | V ( x, y ) | ≤ A , sup x,y ∈ R h x i | V x ( x, y ) | ≤ A , (1.4) with constants s > / , A k > , k = 0 , . Remark 1.2.
Under Assumption . , it is easy to prove that the operators V ( H + i ) − , ( H − i ) − ( ∂ x V ) ( H + i ) − are compact ones. M. DIMASSI, M. KAWAMOTO AND V. PETKOV
Our goal is to prove the absence of embedded eigenvalues of H , when V satisfies (1.3) with small η . To examine the non-existence of eigenvalues of H , first we prove in Proposition 4.2 that without any assumption on the support of V there exist R > , R > independent of the support of V such that σ pp ( H ) ∩ (cid:16) ( −∞ , − R ) ∪ ( R , ∞ ) (cid:17) = ∅ . This statement is more precise that the result in [5], where the dependence of the support of V wasimplicit. Moreover, in contrast to [5], the constants R , R are explicitly given and we have R = C + k V k L ∞ , R = C k h x i h y i V x k L ∞ , where C > , C > η and V . We see that the eigenvalues-free region depends only of the amplitudes A , A . The argument in Section 4 is based on Lemmas 2.2, 2.3 and2.7. and we show that with a suitable weight ϕ ( x ) > k p ϕ ( x )( D x + y )( H − λ − i ) − k L → L ≤ C h λ i / , k p ϕ ( x ) D y ( H − λ − i ) − k L → L ≤ C h λ i / , with constant C > λ . In the literature such type of estimates with λ = 0have been used without a weight p ϕ ( x ). However we show in Appendix A that the operators( D x + y )( H − i ) − , D y ( H − i ) − are unbounded (see Remark 2.4). We expect that the properties ofthese operators as well as Appendix A will be useful for further analysis.Obviously, if V satisfies Assumption 1.1 with s ≥ / , then V satisfies (1.4) with s < /
4. Next inthe exposition without loss of generality we assume that 1 / < s < / . Fixing R = max { R , R } >
0, we establish a Mourre type estimate for the operator H . More precisely, setting γ = 2 s − < / , there exists a constant C R,γ independent of η such thatsup λ ∈ [ − R,R ] ,ν> (cid:13)(cid:13)(cid:13)(cid:13) | y | − γ F (cid:16) yη (cid:17) h x i − / − γ/ ( H − λ ∓ iν ) − h x i − / − γ/ F (cid:16) yη (cid:17) | y | − γ (cid:13)(cid:13)(cid:13)(cid:13) L → L ≤ C R,γ , (1.5)where F ( t ) ∈ C ∞ ( R : [0 , F ( t ) = 1 for | t | ≤ , F ( t ) = 0 for | t | ≥ H has no eigenvalues in R , however the above estimatefor the resolvent of H is non-trivial. Since H has only continuous spectrum, the starting point isthe Mourre estimate (5.1). Since γ < / , the weight | y | − γ is integrable around 0 and this plays anessential role.Our main result is the following Theorem 1.3.
Let V satisfy the Assumption . , γ = 2 s − and let C R,γ > be the constant in (1 . . Assume that η γ C R,γ A = c R,γ,η < . (1.6) Then the operator H has no embedded eigenvalues. The condition (1.6) does not imply the smallness of the potential, it is not related to q | E | as wellas to the cases (I) and (II). In fact, given a potential satisfying Assumption 1.1, one can choosethe constant η small enough in function of A and C R,γ , to obtain that there are no embeddedeigenvalues of H .Our result may be generalized to cover the case when the support of V is included in a strip { ( x, y ) ∈ R : | y − β | ≤ η } with fixed β >
0. Also, we can consider potentials having some
MBEDDED EIGENVALUES 5 singularities for | x | + | y | ≤ K . These generalizations need some technical modifications, but theidea of the proof is the same. For simplicity of the exposition we are not going to treat them.Considering the case η →
0, one can choose A large enough and such a case is closely related tothe one where the potential is a delta function. When the potential V is a delta function, there areinteresting results due to Hauge-van Leeuwen [10], Gyger-Martin [9] concerning the non-existenceof embedded eigenvalues. Finally, notice that the absence of embedded eigenvalues is important forthe analysis of the resonances widths (see [6], [7]) .Setting s = 1 / γ/
2, the proof of Theorem 1.3 is based on the equality | y | − γ F (cid:16) yη (cid:17) h x i − s ( H − λ − iν ) − | D x + i | − = | y | − γ F (cid:16) yη (cid:17) h x i − s ( H − λ − iν ) − | D x + i | − −| y | − γ F (cid:16) yη (cid:17) h x i − s ( H − λ − iν ) − V ( H − λ − iν ) − | D x + i | − , ν > , < γ < / , where λ ∈ R and F (cid:16) yη (cid:17) is a cut-off function equal to 1 for | y | ≤ η . By the condition of V , we have V = | y | − γ F ( y/η ) · | y | γ V and for η γ C R,γ k h x i γ V k L ∞ = c R,γ,η < λ ∈ [ − R,R ] ,ν> (cid:13)(cid:13)(cid:13) | y | − γ F (cid:16) yη (cid:17) h x i − s ( H − λ − iν ) − | D x + i | − (cid:13)(cid:13)(cid:13) L → L ≤ B R,γ − c R,γ,η . If ψ ∈ L ( R ) is an eigenfunction of H with eigenvalues λ ∈ [ − R, R ], we show in the AppendixB that D x ψ ∈ L and one obtains easily a contradiction with the above estimate. Following thisapproach, one needs to establish uniform estimate (1.5). To cover more general cases of potentials,it is necessary to obtain estimate similar to (1.5) with more general weights and this is an interestingopen problem.The plan of the paper is as follows. In Section 2 we prove some preliminary results includingLemma 2.3, Lemma 2.7 and Proposition 2.9 which are used in the next sections. In Section 3 weexamine the estimates of the resolvent of H . The absence of large eigenvalues of H is studied inSection 4. Mourre type estimates are proved in Section 5 and Theorem 1.3 is established in Section6. In the Appendix A and B we prove some technical results. Finally, notice that we use theAssumption 1.1 for the support of V ( x, y ) with respect to y only in Section 6. The results in othersections concerning H hold without any restriction on the support of V .2. Preliminaries
In this section we prove some lemmas which are necessary for the exposition. Throughout thissection to the end of this paper, k·k L ( R ) and k·k B ( L ( R )) are denoted as k · k and ( · , · ) denotesthe inner product on L ( R ). Also we denote by D ( A ) the domain of the operator A . We write r := p x + y and h·i = (1 + · ) / . Lemma 2.1 (Interpolation Theorem) . Let A and B be positive selfadjoint operators on L ( R ) andlet T be a bounded operator on L ( R ) . Assume that with constants α , β , α , β ≥ and C , C > we have (cid:13)(cid:13)(cid:13) A α T B β (cid:13)(cid:13)(cid:13) ≤ C , (cid:13)(cid:13)(cid:13) A α T B β (cid:13)(cid:13)(cid:13) ≤ C . M. DIMASSI, M. KAWAMOTO AND V. PETKOV
Then for all < θ ′ < , setting α θ ′ = α (1 − θ ′ ) + α θ ′ and β θ ′ = β (1 − θ ′ ) + β θ ′ , one has (cid:13)(cid:13)(cid:13) A α θ ′ T B β θ ′ (cid:13)(cid:13)(cid:13) ≤ C − θ ′ C θ ′ . Proof.
We can find the proof of this lemma for example in § f ( z ) be an analytic function on Ω := { z = x + iy : 0 < x < , −∞ < y < ∞} which is boundedon Ω := { z = x + iy : 0 ≤ x ≤ , −∞ < y < ∞} . Then if one hassup −∞ 1, we get | f ( θ ′ ) | ≤ C − θ ′ C θ ′ k u kk v k . Since C , C are independent of I , by taking I → R , one completes the proof of Lemma 2.1. (cid:3) Introduce a positive function ρ ( x ) ∈ C ∞ ( R ) such that for some fixed a > ρ ( x ) = − x , x ≤ − a , ρ ( x ) = 2 x, x > a. Lemma 2.2. For f ∈ D ( H ) we have Z Z R ρ ( x ) (cid:16) | ( D x + y ) f | + | D y f | (cid:17) dxdy ≤ C ( k Hf k + k f k ) . (2.1) Proof. Consider for f ∈ C ∞ ( R ) the integral0 ≤ Z Z R | ( H − ρ ) f | dxdy = Z Z R (cid:16) | H f | + ( ρ − xρ + ρ ′′ ) | f | (cid:17) dxdy − Z Z R ρ (cid:16) | ( D x + y ) f | + | D y f | (cid:17) dxdy. Here we have used that by integration by parts one obtains − Re Z Z R h (( D x + y ) f ) ρ ¯ f + ρf ( D x + y ) f i dxdy = Z Z R (cid:16) − ρ | D x + y ) f | + ρ ′′ | f | (cid:17) dxdy and similarly one transforms the integral with D y f. Clearly with a constant C > , one has ρ − xρ + ρ ′′ < C , hence 2 Z Z R ρ ( x ) (cid:16) | ( D x + y ) f | + | D y f | (cid:17) dxdy ≤ C ( k H f k + k f k ) ≤ C ( k Hf k + k f k ) . MBEDDED EIGENVALUES 7 Since H is a closed operator, for every f ∈ D ( H ) there exists a sequence of functions f n ∈ C ∞ ( R )such that f n → f, Hf n → Hf in L . Taking the limit n → ∞ , we obtain the result. (cid:3) The above Lemma is analogous to Lemma 1 in [15] for Stark Hamiltonian.Consider a function 0 < ϕ ( x ) ≤ A defined by ϕ ( x ) = tan − ( x − a + π/ , x ≥ a,ϕ ( x ) , − a < x < a − x , x ≤ − a, where a ≤ ϕ ( x ) ≤ , | x | ≤ a is a smooth function so that ϕ ( x ) ∈ C ( R ) and ϕ ′ ( x ) > x ∈ R . Lemma 2.3. We have the estimates (cid:13)(cid:13)(cid:13) ϕ / ( x )( D x + y )( H − λ − i ) − (cid:13)(cid:13)(cid:13) L → L ≤ C h λ i / , (2.2) (cid:13)(cid:13)(cid:13) ϕ / ( x ) D y ( H − λ − i ) − (cid:13)(cid:13)(cid:13) L → L ≤ C h λ i / (2.3) with a constant C = C a > independent of λ. Proof. By using the resolvent equality( H − λ − i ) − = ( H − λ − i ) − − ( H − λ − i ) − V ( H − λ − i ) − , it is sufficient to prove the estimates with H replaced by H . We apply the unitary operator e iλD x giving a shift x → x + λ, and obtain e iλD x ϕ / ( x )( D x + y ) e − iλD x e iλD x ( H − λ − i ) − e − iλD x = ϕ / ( x + λ ) ρ − / ( x ) (cid:16) ρ / ( x )( D x + y )( H − i ) − (cid:17) . On the other hand, (cid:13)(cid:13)(cid:13) ρ / ( x )( D x + y )( H − i ) − (cid:13)(cid:13)(cid:13) L → L ≤ C. In fact, we apply (2.1) replacing H by H and choose f = ( H − i ) − g . This yields k ρ / ( x )( D x + y )( H − i ) − g k ≤ C (cid:16) k H ( H − i ) − g k + k ( H − i ) − g k (cid:17) ≤ C k g k . It remains to prove the estimate ϕ ( x + λ ) ρ − ( x ) ≤ C (1 + | λ | ) (2.4)with C = C ( a ) > λ. For x ≥ − a the function ρ − ( x ) is bounded by a constant B a depending on a and ϕ ( x + λ ) ≤ A , hence we have (2.4) . We are going to study the case x < − a. We have three subcases: (i) | λ | ≤ a, x < − a , (ii) | λ | > a, − | λ | ≤ x < − a , (iii) | λ | > a, x < − | λ | . Clearly, in the subcase (i) one has x + λ ≤ − a + | λ | ≤ − a and ϕ ( x + λ ) ρ − ( x ) = | x || x + λ | ≤ | λ | a . In the subcase (ii) we have ρ − ( x ) = | x | ≤ | λ | . In the subcase (iii) we have x + λ < −| λ | ≤ − a and ϕ ( x + λ ) ρ − ( x ) = | x || x + λ | ≤ | λ | a . Thus we obtain (2.4). For D y we apply the same argument. (cid:3) Remark 2.4. The presence of the factor ϕ / ( x ) in the estimates (2 . , (2 . is important forthe boundedness of these operators. In fact, the domain D ( H ) is not included in the domains D ( D x + y ) , D ( D y ) and both operators ( D x + y )( H − i ) − , D y ( H − i ) − are unbounded. We provethis property in Appendix A. Remark 2.5. It is clear that the estimates (2 . , (2 . hold with ϕ ( x ) replaced by h x i − . M. DIMASSI, M. KAWAMOTO AND V. PETKOV Corollary 2.6. For every < γ ≤ we have the estimate (cid:13)(cid:13)(cid:13) h x i − γ/ h D y i γ ( H − λ − i ) − (cid:13)(cid:13)(cid:13) L → L ≤ C γ h λ i γ/ (2.5) Proof. Writing h D y i = h D y i D y + i ( D y + i ) , we deduce that (2.5) holds with γ = 1 . Next we apply theinterpolation Lemma 2.1 between k h x i h D y i h x i ( H − λ − i ) − k ≤ (cid:13)(cid:13)(cid:13) h x i − / h D y i h x i − / ( H − λ − i ) − (cid:13)(cid:13)(cid:13) ≤ C h λ i / . (cid:3) Notice that we have ϕ ′ ( x ) = (cid:16) x − a + π/ (cid:17) − , x ≥ a,ϕ ′ ( x ) , − a ≤ x < a,x − , x < − a, which implies ( ϕ ′ ( x )) / ≤ C p ϕ ( x ) , hence the estimates (2.2), (2.3) hold with ϕ ( x ) replaced by ϕ ′ ( x ) . For the eigenfunctions of H we need a more precise result. Lemma 2.7. Let ψ and λ be an eigenfunction and eigenvalue of H . Moreover, suppose that k ψ k = 1 . Then we have the estimates (cid:13)(cid:13)(cid:13)p ϕ ′ ( x )( D x + y ) ψ (cid:13)(cid:13)(cid:13) ≤ C h λ i / , (2.6) (cid:13)(cid:13)(cid:13)p ϕ ( x ) p ϕ ′ ( y ) D y ψ (cid:13)(cid:13)(cid:13) ≤ C h λ i / (2.7) with C = C a > independent of ψ and λ and the support of V .Proof. By a direct calculus we obtain a representation for the commutator0 = (cid:16) i [ H, ϕ ( x )( D x + y ) + ( D x + y ) ϕ ( x )] ψ, ψ (cid:17) = (cid:16)(cid:0) D x + y ) ϕ ′ ( x )( D x + y ) + 4 ϕ ( x ) D y − ϕ ( x )(1 + V x ) + 2 ϕ ′′′ ( x ) (cid:1) ψ, ψ (cid:17) . Hence we have (cid:13)(cid:13)(cid:13)p ϕ ′ ( x )( D x + y ) ψ (cid:13)(cid:13)(cid:13) ≤ C + k ϕ ( x ) D y ψ k , where the constant C > ϕ ( x ) and k V x k L ∞ . Applying Lemma 2.3, one deduces k ϕ ( x ) D y ψ k ≤ C √ A k ϕ / ( x ) D y ( H − λ − i ) − ψ k ≤ C h λ i / and (cid:13)(cid:13)(cid:13)p ϕ ′ ( x )( D x + y ) ψ (cid:13)(cid:13)(cid:13) ≤ C h λ i / . MBEDDED EIGENVALUES 9 Now we pass to the analysis of the estimate containing D y . In a similar way one has0 = (cid:16) i [ H, D y ϕ ( x ) ϕ ( y ) + ϕ ( y ) ϕ ( x ) D y ] ψ, ψ (cid:17) = (cid:16) (4 D y ϕ ( x ) ϕ ′ ( y ) D y − ϕ ( x ) ϕ ′′′ ( y )) ψ, ψ (cid:17) + 4Re (cid:16) ( D x + y ) ϕ ′ ( x ) ϕ ( y ) D y ψ, ψ (cid:17) − D y ϕ ( y ) ϕ ( x ) ψ, V ψ ) − ϕ ( x ) ϕ ( y ) D y ψ, V ψ ) ≥ (cid:13)(cid:13)(cid:13)p ϕ ( x ) ϕ ′ ( y ) D y ψ (cid:13)(cid:13)(cid:13) − A (cid:13)(cid:13)(cid:13)p ϕ ′ ( x )( D x + y ) ψ (cid:13)(cid:13)(cid:13) (cid:13)(cid:13)(cid:13)p ϕ ′ ( x ) D y ψ (cid:13)(cid:13)(cid:13) − C (cid:16) (cid:13)(cid:13)(cid:13)p ϕ ( x ) D y ψ (cid:13)(cid:13)(cid:13)(cid:17) . Applying the estimates (cid:13)(cid:13)(cid:13)p ϕ ′ ( x )( D x + y ) ψ (cid:13)(cid:13)(cid:13) ≤ C a h λ i / , (cid:13)(cid:13)(cid:13)p ϕ ′ ( x ) D y ψ (cid:13)(cid:13)(cid:13) ≤ C a k h x i − D y ψ k ≤ C a k p ϕ ( x ) D y ψ k ≤ C a h λ i / , we obtain the result. (cid:3) It is obvious that the estimates (2.6) and (2.7) hold with p ϕ ′ ( x ) and p ϕ ′ ( y ) replaced by h x i − and h y i − , respectively. Remark 2.8. Notice that by Lemma . , we obtain h r i − / ( D x + y ) ϕ ∈ L , h r i − / D y ϕ ∈ L , ϕ ∈ D ( H ) . Let H ev be the space generated by the eigenfunctions of H . By the closed graph theorem the operators h r i − / ( D x + y ) and h r i − / D y are bounded as operators from H ev to L ( R ) . Therefore for everyeigenfunction ψ of H we have the estimate k h r i − / ( D x + y ) ψ k + k h r i − / D y ψ k ≤ B k ψ k , (2.8) where B > is independent of ψ . However, the constant B in general could depend on the supportof V . Let F ( t ) ∈ C ∞ ( R ) be a function such that 0 ≤ F ( t ) ≤ , F ( t ) = 1 for | t | ≤ , F ( t ) = 0 for | t | ≥ . For c > F c ( t ) = F ( tc ) . Proposition 2.9. Let < γ < / , β ∈ R and < η < . Then the operator | y − β | − γ F η ( y − β ) h x i − γ/ h H i − γ is bounded and its bound is independent of η and β .Proof. First consider the case β = 0. Let f ( t ) ∈ C ∞ ( R ) be a function such that f ( t ) = 1 for | t | ≤ . Then f ( t ) F η ( t ) = F η ( t ) and for any φ ∈ C ∞ ( R ), it is enough to prove that A := (cid:13)(cid:13)(cid:13) | y | − γ f ( y ) h x i − γ/ h H i − γ φ (cid:13)(cid:13)(cid:13) ≤ C γ k φ k with a constant C γ dependent of γ but independent of η . By simple calculation one has A ≤ (cid:13)(cid:13)(cid:13) | y | − γ f ( y ) h x i − γ/ h H i − γ φ (cid:13)(cid:13)(cid:13) ≤ (cid:13)(cid:13) | y | − γ h D y i − γ (cid:13)(cid:13) L ( R y ) → L ( R y ) (cid:13)(cid:13)(cid:13) h D y i γ f ( y ) h x i − γ/ h H i − γ φ (cid:13)(cid:13)(cid:13) . Here we apply the fractional Sobolev inequality (see, e.g., Stein-Weiss [14] and Yafaev [19]) (cid:13)(cid:13) | y | − γ u (cid:13)(cid:13) L ( R y ) ≤ C γ (cid:13)(cid:13)(cid:13) ( D y ) γ/ u (cid:13)(cid:13)(cid:13) L ( R y ) for u ∈ D (cid:0) ( D y ) γ/ (cid:1) ⊂ L ( R ). Then A can be estimated by A ≤ C γ k h D y i γ f ( y ) h x i − γ/ h H i − γ φ k . On the other hand, the norms of the operators (cid:13)(cid:13)(cid:13) h D y i h x i · f ( y ) · h H i − (cid:13)(cid:13)(cid:13) L → L , (cid:13)(cid:13)(cid:13) h D y i h x i − / · f ( y ) · h H i − (cid:13)(cid:13)(cid:13) L → L ≤ (cid:13)(cid:13)(cid:13) h x i − / h D y i f ( y )( H − i ) − · ( H − i ) h H i − (cid:13)(cid:13)(cid:13) L → L are bounded. In fact, we write ( H − i ) − = ( H − i ) − − ( H − i ) − V ( H − i ) − and one appliesCorollary 2.6 to estimate h x i − / D y ( H − i ) − . Since h x i − / h D y i h x i − / is selfadjoint, by usingthe interpolation Lemma 2.1, we conclude that (cid:13)(cid:13)(cid:13) h D y i γ h x i − γ/ · f ( y ) · h H i − γ (cid:13)(cid:13)(cid:13) L → L is bounded. Therefore A ≤ C γ k φ k and we obtain the estimate.Now consider the case when β = 0. We have (cid:13)(cid:13)(cid:13) | y − β | − γ f ( y − β ) h x i − γ/ h H i − γ (cid:13)(cid:13)(cid:13) L → L ≤ (cid:13)(cid:13) | y − β | − γ h D y i − γ (cid:13)(cid:13) B ( L ( R y )) (cid:13)(cid:13)(cid:13) h D y i γ f ( y − β ) h x i − γ/ h H i − γ (cid:13)(cid:13)(cid:13) L → L ≤ (cid:13)(cid:13)(cid:13) e − iβD y | y | − γ h D y i − γ e iβD y (cid:13)(cid:13)(cid:13) B ( L ( R y )) (cid:13)(cid:13)(cid:13) h D y i γ f ( y − β ) h x i − γ/ h H i − γ (cid:13)(cid:13)(cid:13) L → L ≤ C, noting that sup y | f ′ ( y − β ) | is independent of β. (cid:3) Estimates of the resolvent of H In the section we establish a decay estimate for k f ( H − λ − iν ) − g k L → L with ν > | λ | → ∞ . In [7] the case when f, g ∈ C ∞ ( R ) has been studied, while in [1] thesituation with f, g ∈ L p ( R ) , p > Proposition 3.1. Consider the operator M δ ( λ, ν ) := h r i − δ ( H − λ − iν ) − h r i − δ , < δ ≤ , where λ ∈ R , < ν ≤ . Then for < θ ≤ / and | λ | ≥ , there exists a constant C = C ( θ ) > such that k M δ ( λ, ν ) k L → L ≤ Cν − (cid:16) | λ | − θ + (1 + ν ) | λ | − + (1 + ν − ) | λ | θ − (cid:17) δ/ . (3.1) Proof. We consider only the case λ > 0, since for λ < ω n := n t ∈ [0 , ∞ ) : | t − nπ | ≤ λ − θ o , n ∈ N ∪ { } , MBEDDED EIGENVALUES 11 and Ω = S ∞ n =0 ω n . By the integral formula for the resolvent, we have M ( λ, ν ) = i h r i − Z ∞ e − it ( H − λ − iν ) h r i − dt = K + K with K := i h r i − Z Ω e − it ( H − λ − iν ) h r i − dt and K := i h r i − Z [0 , ∞ ) \ Ω e − it ( H − λ − iν ) h r i − dt = i h r i − Z [0 , ∞ ) \ Ω iλ (cid:18) ddt e itλ (cid:19) e − it ( H − iν ) h r i − dt = 1 λ ∞ X n =0 h h r i − e − it ( H − λ − iν ) h r i − i (cid:12)(cid:12) t =( n +1) π − λ − θ t = nπ + λ − θ + iλ Z [0 , ∞ ) \ Ω h r i − e − it ( H − λ − iν ) ( H − iν ) h r i − dt. For φ ∈ L one gets k K φ k ≤ ∞ X n =0 (cid:13)(cid:13)(cid:13)(cid:13)Z ω n h r i − e − it ( H − λ − iν ) h r i − φdt (cid:13)(cid:13)(cid:13)(cid:13) L ( R ) ≤ ∞ X n =0 Z ω n (cid:13)(cid:13)(cid:13) h r i − e − it ( H − λ − iν ) h r i − φ (cid:13)(cid:13)(cid:13) L ( R ) dt ≤ ∞ X n =0 Z nπ + λ − θ nπ − λ − θ (cid:13)(cid:13)(cid:13) h r i − (cid:13)(cid:13)(cid:13) L ∞ k φ k L ( R ) e − tν dt = k h r i − k L ∞ k φ k L ( R ) ν (cid:16) e νλ − θ − e − νλ − θ (cid:17) ∞ X n =0 e − nνπ ≤ C e νλ − θ − e − νπ λ − θ k φ k L ( R ) ≤ C ν − λ − θ k φ k L ( R ) . (3.2)Here for 0 < ν ≤ − e − νπ ≥ πν e π . Next1 λ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ∞ X n =0 h h r i − e − it ( H − λ − iν ) h r i − i (cid:12)(cid:12) t =( n +1) π − λ − θ t = nπ + λ − θ φ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ λ − k h r i − k L ∞ k φ k L ( R ) ∞ X n =0 (cid:16) e − ν (( n +1) π − λ − θ ) + e − ν ( nπ + λ − θ ) (cid:17) ≤ Cλ − e νλ − θ k φ k L ∞ X n =0 e − νnπ ≤ C e νλ − θ − e − νπ λ − kk φ k L ≤ Cν − λ − kk φ k L and 1 λ (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)Z [0 , ∞ ) \ Ω h r i − e − it ( H − λ − iν ) ν h r i − dt (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ≤ νλ k h r i − k Z ∞ e − νt dt = λ − k h r i − k . Therefore, k K k L → L ≤ Cν − λ − k h r i − k + λ − k K k L → L (3.3)with K = Z [0 , ∞ ) \ Ω h r i − H e − it ( H − λ − iν ) h r i − dt. Now we estimate k K k L → L . Let x = ( x, y ). By an application of the formula (4.6) in [1], wehave the following representation of the operator e − itH (For the operator H LS in [1] one choosesthe constants q = B = 1, m = 1 / ω = 2, E = − , E = 0, ν = 0, ˜ ν = ω = 2, θ = π and E = 1) (cid:0) e − itH φ (cid:1) ( x ) = ( e − ixy/ e − itH LS e ixy/ φ )( x )= 14 πi sin( t ) Z R e − ia ( t ) e − ixy/ e ib ( t ) · x e − ic ( t ) · A ( x ) e − i w · A ( x − c ( t )) e i (cot t )( x − c ( t ) − w ) / e iw w / φ ( w ) d w =: 14 πi sin( t ) Z R K ( t, x , w ) φ ( w ) d w with w = ( w , w ) ∈ R , A ( x ) = ( − y/ , x/ , a ( t ) = Z t (cid:0) b ( s ) + 2 b ( s ) · A ( c ( s )) (cid:1) ds and b ( t ) = ( b ( t ) , b ( t )) , c ( t ) = ( c ( t ) , c ( t )) with b ( t ) = − (sin(2 t )) / , b ( t ) = (1 − cos(2 t )) / ,c ( t ) = cos(2 t ) , c ( t ) = t − sin(2 t ) . Simple calculation shows that y∂ x K ( t, x , w ) = y (cid:18) − i y ib ( t ) − i c ( t )2 − i w i cot t x − c ( t ) − w ) (cid:19) K ( t, x , w ) ,∂ x K ( t, x , w ) = (cid:18) − i y ib ( t ) − i c ( t )2 − i w i cot t x − c ( t ) − w ) (cid:19) + i cot t ! K ( t, x , w )and ∂ y K ( t, x , w )= (cid:18) − i x ib ( t ) + i c ( t )2 + i w i cot t y − c ( t ) − w ) (cid:19) + i cot t ! K ( t, x , w ) . Thus we deduce h r i − H K ( t, x , w ) h w i − = h r i − (cid:0) D x + 2 yD x + y + D y + x (cid:1) K ( t, x , w ) h w i − = X k =1 Q ,k ( t, x ) K ( t, x , w ) Q ,k ( t, w ) , where for k = 1 , ..., Q ,k ( t, x ) = q ,k ( t ) m ,k ( x ) , Q ,k ( t, w ) = q ,k ( t ) m ,k ( w )and | q ,k ( t ) q ,k ( t ) | ≤ C (cid:16) | cot t | (1 + t ) + | cot t | (1 + t + t ) (cid:17) , MBEDDED EIGENVALUES 13 k m ,k ( x ) k L ∞ x ≤ C, k m ,k ( w ) k L ∞ w ≤ C. Hence we have a smoothing effect k K φ k L ( R ) = C (cid:13)(cid:13)(cid:13)(cid:13)(cid:13)Z [0 , ∞ ) \ Ω (sin t ) − e − it ( − λ − iν ) Z R X k =1 Q ,k ( t, x ) K ( t, x , w ) Q ,k ( t, w ) φ ( w ) d w dt (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ( R ) ≤ C Z [0 , ∞ ) \ Ω (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) X k =1 Q ,k ( t, x ) (cid:16) e − it ( H − λ − iν ) Q ,k ( t, · ) φ ( · ) (cid:17) ( x ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) L ( R x ) dt ≤ C Z [0 , ∞ ) \ Ω 7 X k =1 kQ ,k ( t, x ) k L ∞ x (cid:13)(cid:13)(cid:13)(cid:16) e − it ( H − λ − iν ) Q ,k ( t, · ) φ ( · ) (cid:17) ( x ) (cid:13)(cid:13)(cid:13) L ( R x ) dt ≤ C Z [0 , ∞ ) \ Ω 7 X k =1 kQ ,k ( t, · ) k L ∞ kQ ,k ( t, · ) k L ∞ k φ k L ( R ) e − νt dt ≤ C X n ∈ N Z ( n +1) π − λ − θ nπ + λ − θ e − νt (cid:16) λ θ (1 + t ) + (sin t ) − (1 + t + t ) (cid:17) k φ k L ( R ) dt ≤ Cν − λ θ (1 + ν − + ν − ) k φ k L ( R ) . (3.4)Here we used an integration by parts for the term involving (sin t ) − combined with the fact thatfor t ∈ ( nπ + λ − θ , ( n + 1) π − λ − θ ) and λ − θ ≤ π one has a lower bound | sin t | = | sin( t − nπ ) | ≥ | sin( λ − θ ) | ≥ λ − θ . Taking together (3.2), (3.3) and (3.4), we get k M ( λ, ν ) k L → L ≤ C ( θ ) ν − (cid:16) λ − θ + (1 + ν ) λ − + (1 + ν − ) λ θ − (cid:17) . Clearly k M ( λ, ν ) k L → L ≤ ν − and by Lemma 2.1 with A = B = h r i − , T = ( H − λ − iν ) − , α = β = 2, α = β = 0 and θ ′ = 1 − δ/ 2, one deduces k M δ ( λ, ν ) k L → L ≤ ( C ( θ )) δ/ ν − (cid:16) λ − θ + (1 + ν ) λ − + (1 + ν − ) λ θ − (cid:17) δ/ . (cid:3) Absence of large embedded eigenvalues In this section we study the relation σ pp ( H ) ∩ (cid:16) ( −∞ , − R ) ∪ ( R , ∞ ) (cid:17) = ∅ . and we work without any assumption on the support of V . The absence of large eigenvalues hasbeen established by Dimassi-Petkov [5]. However, the fact that R , R > V has not been proven in [5]. Here we establish this result and, moreover, we obtainbounds for R , R . In Appendix B we prove the following Proposition 4.1. Assume that we have kh x i / h y i V k L ∞ ≤ A , h x i / V → , h x i / V x → , h y i V x → x + y ) → ∞ . (4.1) Let ψ be an eigenfunction of H with eigenvalues λ . Then D x ψ ∈ L ( R ) . Proposition 4.2. Assume that V satisfies the conditions (4 . and sup ( x,y ) ∈ R | h x i h y i V x ( x, y ) | ≤ A . Then there exist constants R > , R > independent of η such that σ pp ( H ) ∩ (cid:16) ( −∞ , − R ) ∪ ( R , ∞ ) (cid:17) = ∅ . Moreover, we have R ≤ ( C a A ) , R ≤ C a + A , (4.2) where C a > is a constant depending on the choice of the function ϕ ( x ) in Section and a > . Notice that in the case when V satisfies Assumption 1.1 the conditions of Proposition 4.2 arefulfilled. Proof. Let ψ and λ be an eigenfunction and an eigenvalue of H , respectively. Let k ψ k = 1 and let | λ | ≥ 1. The operator D x is a conjugated operator for H in the sense of [13] and D x satisfies theconditions (a)-(e) in [13] (see for more details Section 3 in [4]). In particular, the condition (c) in[13] means that for Ψ ∈ D ( H ) ∩ D ( D x ) the symmetric form(Ψ , i [ H, D x ]Ψ) = i ( H Φ , D x Ψ) − i ( D x Ψ , H Ψ)is bounded from below and closable and we can define the self-adjoint operator [ H, D x ] o associatedto its closure ([13]). According to Proposition 4.1, we have ψ ∈ D ( H ) ∩ D ( D x ). Thus i [ H, D x ] o ψ is well defined and 0 = ( ψ, i [ H, D x ] o ψ ) = ((1 + V x ) ψ, ψ ) . Consequently,0 = (cid:12)(cid:12)(cid:12) ( i [ H, D x ] o ψ, ψ ) (cid:12)(cid:12)(cid:12) ≥ − (cid:12)(cid:12)(cid:12) (( ∂ x V ) ψ, ψ ) (cid:12)(cid:12)(cid:12) ≥ − kh x i h y i ∂ x V k L ∞ (cid:13)(cid:13)(cid:13) h x i − h y i − ψ (cid:13)(cid:13)(cid:13) . (4.3)Let ϕ ( x ) be the function introduced in Section 2. Obviously, with a constant c a > | ϕ ′′ ( x )( ϕ ′ ( x )) − | ≤ c a , ∀ x ∈ R . (4.4)Recall that from Lemma 2.6 we have k p ϕ ′ ( x )( D x + y ) ψ k ≤ C | λ | / , k p ϕ ′ ( x ) ϕ ′ ( y ) D y ψ k ≤ C | λ | / . (4.5)We need the following Lemma 4.3. We have the equality Γ( ψ ) := k p ϕ ′ ( x ) ϕ ′ ( y )( D x + y ) ψ k + k p ϕ ′ ( x ) ϕ ′ ( y ) D y ψ k − Im (cid:16) ϕ ′′ ( x )( ϕ ′ ( x )) − p ϕ ′ ( x ) ϕ ′ ( y )( D x + y ) ψ, p ϕ ′ ( x ) ϕ ′ ( y ) ψ (cid:17) − Im (cid:16) ( ϕ ′′ ( y )( ϕ ′ ( y )) − p ϕ ′ ( x ) ϕ ′ ( y ) D y ψ, p ϕ ′ ( x ) ϕ ′ ( y ) ψ (cid:17) + (cid:16)p ϕ ′ ( x ) ϕ ′ ( y )( x + V ) ψ, p ϕ ′ ( x ) ϕ ′ ( y ) ψ (cid:17) (4.6)= λ ( ψ, ϕ ′ ( x ) ϕ ′ ( y ) ψ ) . Notice that by (4.5) all scalar products in (4.6) are well defined. MBEDDED EIGENVALUES 15 Proof. Choose a sequence of functions f n ∈ C ∞ ( R ) such that f n → ψ, Hf n → Hψ in L . Clearly,(( H − λ ) f n , ϕ ′ ( x ) ϕ ′ ( y ) f n ) → (( H − λ ) ψ, ϕ ′ ( x ) ϕ ( y ) ψ ) = 0 . By integration by parts, we will show thatΓ( f n ) = ( Hf n , ϕ ′ ( x ) ϕ ′ ( y ) f n ) (4.7)which yields (( H − λ ) f n , ϕ ′ ( x ) ϕ ′ ( y ) f n ) = Γ( f n ) − λ ( f n , ϕ ′ ( x ) ϕ ′ ( y ) f n ) . To do this, we transform the term(( D x + y ) f n + D y f n , ϕ ′ ( x ) ϕ ′ ( y ) f n ) . First consider ( D y f n , ϕ ′ ( x ) ϕ ′ ( y ) f n ) = ( ϕ ′ ( y ) D y p ϕ ′ ( x ) D y f n , p ϕ ′ ( x ) f n )= ( p ϕ ′ ( x ) ϕ ′ ( y ) D y f n , p ϕ ′ ( x ) ϕ ′ ( y ) D y f n )+ i ( ϕ ′′ ( y )( ϕ ′ ( y )) − p ϕ ′ ( x ) ϕ ′ ( y ) D y f n , p ϕ ′ ( x ) ϕ ′ ( y ) f n ) . Second, by the same argument we get(( D x + y ) f n , ϕ ′ ( x ) ϕ ′ ( y ) f n ) = (cid:16) ϕ ′ ( x )( D x + y ) p ϕ ′ ( y )( D x + y ) f n , p ϕ ′ ( y ) f n (cid:17) = ( p ϕ ′ ( x ) ϕ ′ ( y )( D x + y ) f n , p ϕ ′ ( x ) ϕ ′ ( y )( D x + y ) f n )+ i ( ϕ ′′ ( x )( ϕ ′ ( x )) − p ϕ ′ ( x ) ϕ ′ ( y )( D x + y ) f n , p ϕ ′ ( x ) ϕ ′ ( y ) f n ) . Thus we obtain (4.7). We take the limit n → ∞ and deduce Γ( f n ) → Γ( ψ ) . Indeed, by Lemma 2.2we have in L the convergence p ϕ ′ ( x )( D x + y ) f n → p ϕ ′ ( x )( D x + y ) ψ, p ϕ ′ ( x ) D y f n → p ϕ ′ ( x ) D y ψ and the function ϕ ′ ( x ) x is bounded for all x ∈ R . (cid:3) Applying (4.4) and (4.5), one has (cid:12)(cid:12)(cid:12)(cid:16) ϕ ′′ ( y )( ϕ ′ ( y )) − p ϕ ′ ( x ) ϕ ′ ( y ) D y ψ, p ϕ ′ ( x ) ϕ ′ ( y ) ψ (cid:17)(cid:12)(cid:12)(cid:12) ≤ C k p ϕ ( x ) ϕ ′ ( y ) D y ψ k ≤ C | λ | / , (cid:12)(cid:12)(cid:12) ( ϕ ′′ ( x )( ϕ ′ ( x )) − p ϕ ′ ( x ) ϕ ′ ( y )( D x + y ) ψ, p ϕ ′ ( x ) ϕ ′ ( y ) ψ ) (cid:12)(cid:12)(cid:12) ≤ C k p ϕ ( x ) ϕ ′ ( y )( D x + y ) ψ k ≤ C | λ | / . Consequently, from (4.6) one deduces k p ϕ ′ ( x ) ϕ ′ ( y ) ψ k ≤ C ( | λ | / + 1) | λ | − ≤ C | λ | − / , | λ | ≥ , hence kh x i − h y i − ψ k ≤ C k p ϕ ′ ( x ) ϕ ′ ( y ) ψ k ≤ C | λ | − / . Going back to (4.3), we deduce that for | λ | ≥ (2 C A ) we have no eigenvalues of H .For λ ≤ k p ϕ ′ ( x ) ϕ ′ ( y )( D x + y ) ψ k = B , k p ϕ ′ ( x ) ϕ ′ ( y ) D y ψ k = B , k p ϕ ′ ( x ) ϕ ′ ( y ) ψ k = D. Since − λ k p ϕ ′ ( x ) ϕ ′ ( y ) ψ k ≥ , the equality (4.6) implies B + B − C B D − C B D + (cid:16)p ϕ ′ ( x ) ϕ ′ ( y )( x + V ) ψ, p ϕ ′ ( x ) ϕ ′ ( y ) ψ (cid:17) ≤ −| λ | D with constants C > , C > λ . Therefore, (cid:16) B − C D (cid:17) + (cid:16) B − C D (cid:17) − (cid:16) C C (cid:17) D ≤ ( C + A − | λ | ) D with a constant C > ϕ ( x ) and independent of λ. Consequently, one deduces | λ | D ≤ (cid:16) C C (cid:17) D + ( C + A ) D = ( C + A ) D If | λ | > C + A , we have k ϕ ′ ( x ) ϕ ′ ( y ) ψ k = 0 , hence ψ = 0 . (cid:3) Mourre type estimate for the operator H In this section we fix R ≥ max { R , R } , where R k , k = 1 , , are given by Proposition 4.2. Thefollowing result follows from [13]. Proposition 5.1. There exists a constant C R > such that sup λ ∈ [ − R,R ] ,ν> (cid:13)(cid:13) | D x + β + i | − ( H − λ ∓ iν ) − | D x + β + i | − (cid:13)(cid:13) L → L ≤ C R . (5.1)We have i [ D x + β, H ] = 1 . As it was mentioned in the previous section, the conjugate operator D x + β satisfies the conditions(a)-(e) in [13] and the principal theorem in [13] implies the estimate (5.1). Proposition 5.2. Let < γ < / , s = 1 / γ/ , β ∈ R and λ ∈ [ − R, R ] . Then we have theestimate sup λ ∈ [ − R,R ] ,ν> (cid:13)(cid:13) | y − β | − γ F η ( y − β ) h x i − s ( H − λ ∓ iν ) − s h x i − s F η ( y − β ) | y − β | − γ (cid:13)(cid:13) L → L ≤ C R,γ (5.2) with a constant C R,γ > independent of η and β .Proof. For simplicity we treat the case β = 0. Define J H := | D x + i | − ( H − λ ∓ iν ) − | D x + i | − . We write | y | − γ F η ( y ) h x i − s ( H − λ ∓ iν ) − h x i − s F η ( y ) | y | − γ = | y | − γ F η ( y ) h x i − s F R ( H )( H − λ ∓ iν ) − F R ( H ) h x i − s F η ( y ) | y | − γ + | y | − γ F η ( y ) h x i − s (1 − F R ( H ))( H − λ ∓ iν ) − h x i − s F η ( y ) | y | − γ = I J H I ∗ + I with I := | y | − γ F η ( y ) h x i − s F R ( H ) | D x + i | and I := | y | − γ F η ( y ) h x i − s (1 − F R ( H ))( H − λ ∓ iν ) − h x i − s F η ( y ) | y | − γ . MBEDDED EIGENVALUES 17 First, we show that k I k L → L ≤ C ,R,γ . To do this, one considers the product I = I , I , with I , := | y | − γ F η ( y ) h x i − s F R ( H )( D x + i ) ,I , := ( D x + i ) − | D x + i | . Clearly, I , is a bounded operator.Next we write I , = | y | − γ F η ( y ) h x i − s ( D x + i ) F R ( H )+ | y | − γ F η ( y ) h x i − s h H i − γ h H i γ [ F R ( H ) , D x ] = J + J . The term J can be estimated by (cid:13)(cid:13) | y | − γ F ( y ) h x i − s ( D x + y + i ) F R ( H ) (cid:13)(cid:13) + (cid:13)(cid:13) | y | − γ F ( y ) h x i − s F R ( H ) (cid:13)(cid:13) ≤ C (cid:13)(cid:13) h D y i γ F ( y ) h x i − s ( D x + y + i ) F R ( H ) (cid:13)(cid:13) + C ≤ C (cid:13)(cid:13) h D y i γ F ( y ) h x i − s ( D x + y + i )( H + i ) − (cid:13)(cid:13) + C . To handle the operator on the right hand side, write h D y i γ F ( y ) h x i − s ( D x + y + i )( H + i ) − = h D y i γ F ( y ) h x i − γ/ ( H + i ) − h x i − / ( D x + y + i )( H + i ) − (5.3)+ h D y i γ F ( y ) h x i − γ/ [ h x i − / ( D x + y + i ) , ( H + i ) − ]( H + i ) − . According to (2.2), (2.3) and Corollary 2.6, the first term in right hans side of (5.3) is bounded.For the second term one has h D y i γ F ( y ) h x i − γ/ [ h x i − / ( D x + y + i ) , ( H + i ) − ]( H + i ) − = h D y i γ F ( y ) h x i − γ/ ( H + i ) − [ H , h x i − / ( D x + y + i )]( H + i ) − . Clearly, [ H , h x i − / ( D x + y + i )]( H + i ) − = [( D x + y ) + D y + x, h x i − / ( D x + y ) + i h x i − / ]( H + i ) − = ix h x i − / ( D x + y ) ( H + i ) − / B , where B is a bounded operator. It remains to show that the operator B = h x i − ( D x + y ) ( H − i ) − is bounded. Set Q = ( D x + y ) + D y . Then( D x + y ) ( H − i ) − = ( D x + y ) ( Q − i ) − + ( D x + y ) ( Q − i ) − x ( H − i ) − . The pseudoffiferential operator ( D x + y ) ( Q − i ) − has symbol in S ( R x,y,ξ,η ) ), hence it is bounded(see [4]). Consequently, the operator h x i − ( D x + y ) ( Q − i ) − x is also bounded since by composition of pseudodifferential operators its principal symbol is in S ( R x,y,ξ,η ) ) . This implies that B is bounded. To prove the boundedness of J , let ˜ g ( z ) ∈ C ∞ ( C ) be an almost analytic continuation of g ( s ) = F R ( s ) such that ¯ ∂ z ˜ g ( z ) = O ( | Im z | N ) , ∀ N ∈ N . Consider the representation F R ( H ) = 1 π Z ¯ ∂ z ˜ g ( z )( H − z ) − L ( dz ) , where L ( dz ) is the Lebesgue measure on C . Therefore i [ F R ( H ) , D x ] = iπ Z ¯ ∂ z ˜ g ( z )[( H − z ) − , D x ] L ( dz )= iπ Z ¯ ∂ z ˜ g ( z )( H − z ) − [ H , D x ]( H − z ) − L ( dz )= − π Z ¯ ∂ z ˜ g ( z )( H − z ) − L ( dz ) . On the other hand, the operator | y | − γ F η ( y ) h x i − s h H i − γ = h x i − / | y | − γ F η ( y ) h x i − γ/ h H i − γ is bounded applying Proposition 2.9 with H replaced by H , while1 π Z ¯ ∂ z ˜ g ( z ) h H i γ ( H − z ) − L ( dz )is trivially bounded. Combining the above estimates, one concludes that k I k L → L ≤ C ,R,γ . Concerning I , notice that for | λ | ≤ R by the spectral Theorem the operator h H i γ (1 − ( F R ( H )) )( H − λ ∓ iν ) − h H i γ is bounded. Next one obtains the estimate k I k L → L ≤ C (cid:13)(cid:13)(cid:13) | y | − γ F η ( y ) h x i − γ/ h H i − γ (cid:13)(cid:13)(cid:13) L → L ≤ C ,R,γ by applying once more Proposition 2.9. The case β = 0 can be treated by a similar argument. (cid:3) In the next section we need a modification of Proposition 5.2 when we have a product with aright factor | D x + i | − . Proposition 5.3. Let < γ < / , s = 1 / γ/ , β ∈ R . Then we have sup λ ∈ [ − R,R ] ,ν> (cid:13)(cid:13) | y − β | − γ F η ( y − β ) h x i − s ( H − λ ∓ iν ) − | D x + i | − (cid:13)(cid:13) L → L ≤ B R,γ (5.4) with constant B R,γ > independent of η and β .Proof. We use the notations of the proof of Proposition 5.2. For β = 0 one has | y | − γ F η ( y ) h x i − s ( H − λ ∓ iν ) − | D x + i | − = I J H + J , where I and J H are the same as in the proof of Proposition 5.2 and J = | y | − γ F η ( y ) h x i − s (1 − F R ( H ))( H − λ ∓ iν ) − | D x + i | − . Notice that the operator J can be bounded by C ,R,γ by a calculation similar to that used for I in the proof of Proposition 5.2 and we leave the details to the reader. The case β = 0 is treated bya similar argument. (cid:3) MBEDDED EIGENVALUES 19 Absence of embedded eigenvalues for potentials with small support In this section we prove Theorem 1.3. Proof. Concerning H and 0 < γ < / , s = 1 / γ/ , we have the estimates (5.2) with β = 0 and(5.4). For the operator H = H + V with supp V ⊂ { ( x, y ) : | y | ≤ η } write( H − λ − iν ) − = ( H − λ − iν ) − h − V ( H − λ − iν ) − i (6.1)which yields | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − | D x + i | − = | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − | D x + i | − − h | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − h x i − s | y | − γ F η ( y ) i(cid:16) h x i s | y | γ V (cid:17) × h | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − | D x + i | − i . Therefore, (cid:16) I + h | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − h x i − s | y | − γ F η ( y ) i(cid:16) h x i γ | y | γ V (cid:17)(cid:17) × | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − | D x + i | − = | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − | D x + i | − . Clearly, (cid:13)(cid:13)(cid:13) h x i γ | y | γ V (cid:13)(cid:13)(cid:13) L ( R ) → L ( R ) ≤ η γ k h x i γ V k L ∞ ( R ) . Consequently, assuming η γ C R,γ k h x i γ V k L ∞ ( R ) = c R,γ,η < 1, we deduce that the operator inthe brackets (cid:16) ... (cid:17) is invertible andsup | λ |≤ R,ν> (cid:13)(cid:13)(cid:13) | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − | D x + i | − (cid:13)(cid:13)(cid:13) L → L ≤ B R,γ − c R,γ,η . This estimate implies that H has no eigenvalues in [ − R, R ] . In fact, let ψ be an eigenfunction of H with eigenvalue λ ∈ [ − R, R ] . By Proposition 4.1 we know that D x ψ ∈ L ( R ) , hence | D x + i | ψ = | D x + i | ( D x + i ) − ( D x + i ) ψ ∈ L ( R ) . Then we conclude that | y | − γ F η ( y ) h x i − s ( H − λ − iν ) − | D x + i | − | D x + i | ψ = | y | − γ F η ( y ) h x i − s iν − ψ. If F η ( y ) ψ ( x, y ) = 0, then V ( x, y ) ψ ( x, y ) = 0 and ψ will be an eigenfunction of H which is impossi-ble. Thus | y | − γ F η ( y ) h x i − s ψ = 0 and as ν ց L ( R ) norm of the function | y | − γ F η ( y ) h x i − s ν − ψ is not bounded. We obtain a contradiction and the proof is complete. (cid:3) Appendix A We prove in this Appendix the following Lemma A.1. The operators x ( H − i ) − , ( D x + y ) k ( H − i ) − , ( D y ) k ( H − i ) − , k = 1 , are unbounded from L ( R ) into L ( R ) . Proof. Set U = e iD x D y . We have U − ( D x + y ) U = y, U − xU = x − D y . Combining this with the fact that U commutes with D y , we get U − ( D x + y ) k U U − (cid:0) ( D x + y ) + D y + x − i (cid:1) − U = y k (cid:0) y + D y + x − D y − i (cid:1) = y k (cid:18) y + ( D y − 12 ) + x − − i (cid:19) − . Hence, applying the unitary transformation e iy/ , one deduces that ( D x + y ) k ( H + i ) − is unitarilyequivalent to L k = y k (cid:18) y + D y + x − − i (cid:19) − =: y k ( B − i ) − . Next, we prove that L is unbounded from L ( R ) into L ( R ) . Let ϕ ∈ C ∞ (]1 , R ) be a functionsuch that R ϕ ( x ) dx = 1, and let ψ n ( y ) be the normalized eigenfunction of the harmonic oscillatorcorresponding to λ n = 2 n + 1, that is( D y + y ) ψ n ( y ) = (2 n + 1) ψ n ( y ) , k ψ n k = 1 . (A.1)Set Ψ n ( x, y ) = ψ n ( y ) ϕ ( x + 2 n + 1). Clearly, L Ψ n ( x, y ) = y (cid:18) n + 1 + x − − i (cid:19) − Ψ n ( x, y ) , k Ψ n k = 1 . Therefore, k L Ψ n k = Z R y ψ n ( y ) dy Z R ϕ ( x + 2 n + 1)( x + 2 n + ) + 1 dx. On the support of ϕ ( x + 2 n + 1) we have ≤ x + 2 n + ≤ , hence1( x + 2 n + ) + 1 ≥ . This yields k L Ψ n k ≥ Z R y ψ n ( y ) dy Z R ϕ ( x + 2 n + 1) dx = 1665 Z R y ψ n ( y ) dy. (A.2)By using the Fourier transform F y → η with respect to y , one obtains F ( D y + y ) F − = D η + η and k ψ n ( y ) k = k ˆ ψ n ( η ) k . Thus we deduce that ˆ ψ n ( η ) is also a solution of (A.1) and ˆ ψ n ( η ) = ψ ( η ). Therefore k ψ ′ n ( y ) k = k ηψ n ( η ) k = k yψ n ( y ) k . Combining this with the obvious equality2 n + 1 = h ( D y + y ) ψ n , ψ n i = k ψ ′ n k + k yψ n k , we deduce that k yψ n k = n +12 . Consequently, (A.2) yields k L Ψ n k ≥ n + 1)130 . Letting n → ∞ , we conclude that L is unbounded from L into L . On the other hand, from k L u k = |h L u, ( B − i ) − u i| ≤ k L u kk u k , MBEDDED EIGENVALUES 21 we deduce that L is also unbounded. This shows that ( D x + y ) k ( H − i ) − , k = 1 , x ( H − i ) − and ( D y ) k ( H − i ) − are unbounded. (cid:3) Appendix B In this Appendix we establish Proposition 4.1. Proof of Proposition . ψ be a normalized by k ψ k = 1 . Suppose that D x ψ L ( R ) (B.1)and for ǫ > 0, introduce the function f ǫ ( x ) = ln (cid:16) h x i ǫ h x i (cid:17) . The operators F ǫ = e f ǫ ( D x ) = h D x i ǫ h D x i and its inverse F − ǫ = e − f ǫ ( D x ) = ǫ h D x ih D x i are bounded. Therefore, F ǫ ψ ∈ L ( R ). The condition(B.1) implies lim ǫ ց k F ǫ ψ k = ∞ . Let F x = F x → ξ denotes the Fourier transform with respect to x .The dominated convergence theorem yieldslim ǫ ց Z Z R e f ǫ ( D x ) ψ ( x, y ) F − x g ( ξ, y ) dxdy = lim ǫ ց Z Z R e f ǫ ( ξ ) ( F x ψ )( ξ, y ) g ( ξ, y ) dξdy = Z Z R h ξ i ( F x ψ )( ξ, y ) g ( ξ, y ) dξdy, for all g ( ξ, y ) ∈ F x ( C ∞ ( R )). This implieslim ǫ ց Z Z R e f ǫ ( D x ) ψ ( x, y ) h ( x, y ) dxdy k F ǫ ψ k = 0 , ∀ h ∈ C ∞ ( R ) . Consequently, the normalized function ϕ ǫ := F ǫ ψ k F ǫ ψ k converges weakly to zero.By using F − ǫ xF ǫ = x + i ( ∂ x f ǫ )( D x ), and taking into account that F ǫ commutes with the operator H − x − V , we get Hϕ ǫ = F ǫ (cid:0) H + i ( ∂ x f ǫ )( D x ) − V + F − ǫ V F ǫ (cid:1) ψ k F ǫ ψ k = ( λ + i ( ∂ x f ǫ )( D x ) + V − F ǫ V F − ǫ ) ϕ ǫ . (B.2)Notice that the operators ǫ h D x i and ǫ h D x i ǫ h D x i are bounded from L into L uniformly with respectto ǫ ∈ [0 , V, ∂ x V ∈ L ∞ ( R ), the operator h D x i V h D x i − is bounded. Hence, F ǫ V F − ǫ = h D x i ǫ h D x i V ǫ h D x ih D x i = 11 + ǫ h D x i (cid:0) h D x i V h D x i − (cid:1) + ǫ h D x i ǫ h D x i V, (B.3)is uniformly bounded for ǫ ∈ [0 , K ǫ := i ( ∂ x f ǫ )( D x ) + V − F ǫ V F − ǫ . Let G ( x, y ) be a continuous function going to zero as ( x + y ) → ∞ . It is well known that φ h D x i − s ( H + i ) − , is a compact operator for every φ ∈ C ∞ ( R ) and all s ≥ G h D x i − s ( H + i ) − is also compact. We claim that G h D x i − s ϕ ǫ converges strongly to zero as ǫ ց . (B.4)To prove this, we use (B.2). Write G h D x i − s ϕ ǫ = G h D x i − s ( H + i ) − ( H + i ) ϕ ǫ = G h D x i − s ( H + i ) − ( λ + i + K ǫ ) ϕ ǫ . Since ( λ + i + K ǫ ) is bounded uniformly for ǫ ∈ [0 , ϕ ǫ converges weakly to zero, it followsfrom the compactness of G h D x i − s ( H + i ) − that the right hand side of the above equality convergesstrongly to zero.For t > , let χ t ( x ) be an odd smooth function satisfying χ t ( x ) = x, ≤ x ≤ t, t, x ≥ t, (B.5) χ ( k ) t ( x ) = O ( t − k +1 ) , k ≥ , and χ ′ t ( x ) ≥ 0. Clearly, i [ x, − χ t ( D x )] = χ ′ t ( D x ) and i lim t →∞ i ([ x, − χ t ( D x )] ϕ ǫ , ϕ ǫ ) = ( ϕ ǫ , ϕ ǫ ) . (B.6)Next, we claim that for every fixed ǫ > t →∞ i ([ V, − χ t ( D x )] ϕ ǫ , ϕ ǫ ) = − t →∞ Im( ϕ ǫ , χ t ( D x ) V ϕ ǫ ) = ( V x ϕ ǫ , ϕ ǫ ) . (B.7)First, it follows from (B.2) that h ǫ := ( H − i ) ϕ ǫ is uniformly bounded in L with respect to ǫ ∈ [0 , D x V ( H − i ) − h ǫ = V x ( H − i ) − h ǫ + V ( D x + y )( H − i ) − h ǫ − yV ( H − i ) − h ǫ ∈ L . Combining this with the fact that | χ t ( ξ ) − ξ | ≤ C | ξ | (uniformly for t ≥ | ( χ t ( ξ ) − ξ ) H ǫ ( ξ ) | ≤ C | ξ | | H ǫ ( ξ ) | ∈ L , where H ǫ ( ξ ) = F x → ξ (cid:16) V ( H − i ) − h ǫ (cid:17) ( ξ ) . Hence, the dominated convergence theorem yieldslim t → + ∞ χ t ( D x ) V ( H − i ) − h ǫ = V x ( H − i ) − h ǫ = V x ϕ ǫ , in L , and the proof of the claim is complete. Taking together (B.6), (B.7) and the equality [ H, χ t ( D x )] =[ x + V, χ t ( D x )], we obtain lim t → + ∞ i ([ H, − χ t ( D x )] ϕ ǫ , ϕ ǫ ) = ((1 + V x ) ϕ ǫ , ϕ ǫ ) . Now applying (B.4) with G = ∂ x V and s = 0, we deduce that((1 + ∂ x V ) ϕ ǫ , ϕ ǫ ) ≥ 12 (B.8)for ǫ small enough. To complete the proof, we will show that the left hand side of (B.8) is less than for ǫ small enough. This leads to a contradiction.Equation (B.2) implies i ([ H, − χ t ( D x )] ϕ ǫ , ϕ ǫ ) = i ( χ t ( D x ) Hϕ ǫ , ϕ ǫ ) − i ( χ t ( D x ) ϕ ǫ , Hϕ ǫ ) (B.9)= i ( χ t ( D x )( λ + K ǫ ) ϕ ǫ , ϕ ǫ ) − i ( χ t ( D x ) ϕ ǫ , ( λ + K ǫ ) ϕ ǫ )= − χ t ( D x ) K ǫ ϕ ǫ , ϕ ǫ ) . On the other hand, the inequality χ t ( x ) ( ∂ x f ǫ ) ( x ) = xχ t ( x ) h x i (1 + ǫ h x i ) ≥ , yields χ t ( D x ) ( ∂ x f ǫ ) ( D x ) ≥ , MBEDDED EIGENVALUES 23 in the sense of self-adjoint operators. Consequently, − m ( χ t ( D x ) K ǫ ϕ ǫ , ϕ ǫ )= − m (cid:0) iχ t ( D x )( ∂ x f ǫ )( D x ) + V − F ǫ V F − ǫ ϕ ǫ , ϕ ǫ (cid:1) ≤ m (cid:0) χ t ( D x ) (cid:0) F ǫ V F − ǫ − V (cid:1) ϕ ǫ , ϕ ǫ (cid:1) . (B.10)From (B.3), we have F ǫ V F − ǫ − V = 11 + ǫ h D x i (cid:0) h D x i V h D x i − (cid:1) − 11 + ǫ h D x i V = 11 + ǫ h D x i [ h D x i , V ] h D x i − . Thereforelim t →∞ (cid:16) χ t ( D x ) (cid:0) F ǫ V F − ǫ − V (cid:1) ϕ ǫ , ϕ ǫ (cid:17) = (cid:16) D x (cid:0) F ǫ V F − ǫ − V (cid:1) ϕ ǫ , ϕ ǫ (cid:17) = (cid:16) D x ǫ h D x i [ h D x i , V ] h D x i − ϕ ǫ , ϕ ǫ (cid:17) (B.11)is bounded uniformly for ǫ ∈ [0 , t → ∞ , we deduce from (B.7), (B.10) and (B.11)((1 + ∂ x V ) ϕ ǫ , ϕ ǫ ) ≤ (cid:16) D x ǫ h D x i [ h D x i , V ] h D x i − ϕ ǫ , ϕ ǫ (cid:17) . (B.12)To complete the proof of Proposition 4.1, we apply the following Lemma B.1. We have lim ε → (cid:16) 11 + ε h D x i D x [ h D x i , V ] h D x i − ϕ ε , ϕ ε (cid:17) = 0 . Proof. Write D x [ h D x i , V ] h D x i − = (cid:16) h D x i D x V − D x V h D x i (cid:17) h D x i − = h h D x i ( V x + V D x ) − ( V x + V D x ) h D x i i h D x i − = [ h D x i , V x ] h D x i − + (cid:16) h D x i V − V h D x i (cid:17) D x h D x i − = [ h D x i , V x ] h D x i − + [ h D x i , V ] D x h D x i − and set L := 11 + ε h D x i [ h D x i , V x ] h D x i − , L := 11 + ε h D x i (cid:16) [ h D x i , V ] D x h D x i − (cid:17) . Therefore, D x h D x i − ϕ ε = D x h D x i − ( H − i ) − ( H − i ) ϕ ε = ( H − i ) − D x h D x i − h ε + ( H − i ) − [ D x h D x i − , x ]( H − i ) − h ε . Clearly, the operator [ D x h D x i − , x ] = h D x i − (1 − D x h D x i ) is bounded and this implies that D x h D x i − ϕ ε = ( H − i ) − ˜ h ε (B.13)with ˜ h ε bounded in L uniformly with respect to ε. Recall that the operator h x i − / ( D x + y )( H − i ) − is bounded by Lemma 2.3. By using this, one deduces that the operator( H + i ) − V D x ( H − i ) − = ( H + i ) − V h x i / h x i − / ( D x + y )( H − i ) − − ( H + i ) − V y ( H − i ) − 14 M. DIMASSI, M. KAWAMOTO AND V. PETKOV is compact since V h x i / → , V y → x + y ) → ∞ by conditions (4.1). To handle the operator h D x i , we exploit the following representation V h D x i ( H − i ) − = V ( D x + i ) h D x i ( D x + i ) − ( H − i ) − = V ( D x + i )( H − i ) − h D x i ( D x + i ) − + V ( D x + i )( H − i ) − [ h D x i ( D x + i ) − , x ]( H − i ) − . Obviously, the commutator [ h D x i ( D x + i ) − , x ] is a bounded operator and ( D x + i ) = ( D x + y ) − ( y − i ). So as above we obtain that ( H + i ) − V h D x i ( H − i ) − is compact. In the same way weshow that the operators( H + i ) − V x D x ( H − i ) − , ( H + i ) − V x h D x i ( H − i ) − are compact because V x h x i / → , V x y → x + y ) → ∞ according to conditions (4.1).To deal with the operator L , write (cid:16) 11 + ε h D x i [ h D x i , V x ] h D x i − ϕ ε , ϕ ε (cid:17) = (cid:16) 11 + ε h D x i [ h D x i , V x ] h D x i − ϕ ε , ( H − i ) − h ε (cid:17) = (cid:16) 11 + ε h D x i ( H + i ) − [ h D x i , V x ] h D x i − ϕ ε , h ε (cid:17) − (cid:16) ( H + i ) − εD x h D x i (1 + ε h D x i ) ( H + i ) − [ h D x i , V x ] h D x i − ϕ ε , h ε (cid:17) . We have ( H + i ) − [ h D x i , V x ] h D x i − = ( H + i ) − ( h D x i V x h D x i − − V x ) . The analysis of the termwith V x is easy since ( H + i ) − V x is compact. For the other term we get h D x i − ϕ ε = h D x i − ( H − i ) − h ε = ( H − i ) − h D x i − h ε − ( H − i ) − εD x h D x i (1 + ε h D x i ) ( H − i ) − h ε and notice that ( H + i ) − h D x i V x ( H − i ) − is compact.Passing to the analysis of the operator L , we have [ h D x i , V ] = h D x i V − V h D x i . For h D x i V D x h D x i − we repeat the above argument by using (B.13) and the fact that ( H + i ) − h D x i V ( H − i ) − is com-pact since its adjoint ( H + i ) − V h D x i ( H − i ) − is compact. On the other hand, applying (B.13)once more, we have V h D x i D x h D x i − ϕ ε = V h D x i ( H − i ) − ˜ h ε . The operator V h D x i ( H − i ) − has been treated above and the proof is complete. (cid:3) Acknowledgment. Thanks are due to the referees for their critical comments and useful sug-gestions leading to an improvement of the previous version of the paper. References [1] T. Adachi and M. 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