Perturbation determinants and discrete spectra of semi-infinite non-self-adjoint Jacobi operators
aa r X i v : . [ m a t h . SP ] J a n PERTURBATION DETERMINANTS AND DISCRETESPECTRA OF SEMI-INFINITE NON-SELF-ADJOINTJACOBI OPERATORS
L. GOLINSKII
Abstract.
We study the trace class perturbations of the half-line, dis-crete Laplacian and obtain a new bound for the perturbation determi-nant of the corresponding non-self-adjoint Jacobi operator. Based onthis bound, we obtain the Lieb–Thirring inequalities for such operators.The spectral enclosure for the discrete spectrum and embedded eigen-values are also discussed.
In memory of Sergey Naboko (1950–2020)
Introduction
In the last two decades there was a splash of activity around the spectraltheory of non-self-adjoint perturbations of some classical operators of mathe-matical physics, such as the Laplace and Dirac operators on the whole space,their fractional powers, and others. Recently, there has been some interestin studying certain discrete models of the above problem. In particular, thestructure of the spectrum for compact, non-self-adjoint perturbations of thefree Jacobi and the discrete Dirac operators has attracted much attentionlately. Actually the problem concerns the discrete component of the spec-trum and the rate of its accumulation to the essential spectrum. Such typeof results under the various assumptions on the perturbations are unitedunder a common name
Lieb–Thirring inequalities . For a nice account of theexisting results on the problem for non-self-adjoint, two-sided Jacobi oper-ators, the reader may consult two recent surveys [14] and [9, Section 5.13]and references therein.The spectral theory of semi-infinite, self-adjoint
Jacobi matrices is quitepopular owing to their tight relation to the theory of orthogonal polynomialson the real line [18]. In contrast, there are only a few papers where semi-infinite, non-self-adjoint Jacobi matrices are examined [17, 1, 2, 7, 12, 13].The main object under consideration is a semi-infinite Jacobi matrix(0.1) J ( { a j } , { b j } , { c j } ) j ∈ N = b c a b c a b c . . . . . . . . . , Date : January 15, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Non-self-adjoint Jacobi matrices; discrete spectrum; perturba-tion determinant; Jost solutions. with uniformly bounded complex entries, and a j c j = 0, j ∈ N := { , , . . . } .The spectral theory of the underlying non-self-adjoint Jacobi operator in-cludes, among others, the structure of their spectra. We denote by J thesemi-infinite discrete Laplacian, i.e., J = J ( { } , { } , { } ). If J − J is acompact operator, that is,lim k →∞ a k − k →∞ c k − k →∞ b k = 0 , the geometric image of the spectrum is plainly evident σ ( J ) = σ ess ( J ) ∪ σ d ( J ) = [ − , ∪ σ d ( J ) , the discrete component σ d ( J ) is an at most countable set of points in C \ [ − ,
2] with the only possible limit points on [ − , traceclass perturbations of the discrete Laplacian J (0.2) J − J ∈ S ⇔ ∞ X k =1 ( | − a k | + | b k | + | − c k | ) < ∞ . Now the discrete spectrum is the set of isolated eigenvalues of finite algebraicmultiplicity.Our strategy is similar to that in [7] and [11]. The key issue is the boundfor the perturbation determinant (0.3) L ( λ, J ) := det( I + ( J − J )( J − λ ) − )introduced by M.G. Krein [10] in the late 1950s. The main feature of thisanalytic function on the resolvent set ρ ( J ) = C \ [ − ,
2] is that the set of itszeros agrees with the discrete spectrum of the perturbed operator J , andmoreover, the multiplicity of each zero equals the algebraic multiplicity ofthe corresponding eigenvalue. So the original problem of the spectral theorycan be restated as the classical problem of zero distributions of analyticfunctions, which goes back to Jensen and Blaschke.The argument is pursued in two steps. The first one results in the boundfor the perturbation determinant, typical for the functions of non-radialgrowth. The classes of such analytic (and subharmonic) functions in theunit disk were introduced and studied in [4, 8] (for some advances see [5]).The Blaschke-type conditions for the zero sets (Riesz measures) were provedtherein, with an important amplification in [14, Theorem 4], better adaptedfor applications. The second step is just the latter result applied to thebound mentioned above.Such two-step algorithm is applied, by and large, in [14] to the two-sidedJacobi operators. In our approach to the problem the argument in the firststep is totally different. The point is that for semi-infinite Jacobi operatorsneither the Fourier transform machinery, nor a simple matrix representationfor the resolvent of the free Jacobi operator are at our disposal. Instead, wedeal with the associated three-term recurrence relation(0.4) u k − + b k u k + a k c k u k +1 = λ ( z ) u k , k ∈ N , λ ( z ) = z + 1 z , and its solutions. Here λ ( · ) is the Zhukovsky function which maps theunit disk onto the resolvent set ρ ( J ) = C \ [ − , ACOBI MATRICES 3 u = ( u k ) k ≥ is the eigenvector of J with the eigenvalue λ if and only if (0 , u )is the ℓ -solution of (0.4). The solution u + = ( u + k ) k ≥ of (0.4) is called the Jost solution if(0.5) lim k →∞ z − k u + k ( z ) = 1 , z ∈ D := D \{ } . We study the Jost solutions by reducing the difference equation (0.4) to aVolterra-type discrete integral equation, the standard idea in analysis, see,e.g., [19, Section 7.5], [7, 11]. The bounds for the Jost solutions stem fromthe successive approximations method. The crucial point in such an ap-proach is that the perturbation determinant agrees with the first coordinate u +0 of the Jost solution, known as the Jost function , see the non-self-adjointversion of [16, Theorem 2.16] (the calculation there has nothing to do withself-adjointness).For two-sided Jacobi operators the Lieb–Thirring inequality is due toHansmann–Katriel [14, Theorem 1]. It states that for each ε ∈ (0 ,
1) thereis a constant C ( ε ) > X λ ∈ σ d ( J ) dist( λ, [ − , ε | λ − | + ε ≤ C ( ε ) k J − J k . The result is known to be sharp [3] in the sense that (0.6) is false for ε = 0.As it turns out, the small parameter in the numerator of the left side (0.6)can be dropped, the result obtained recently in [11].The Lieb–Thirring inequality for semi-infinite Jacobi operators is exactlythe same. Theorem 0.1.
Let J − J ∈ S . Then for each ε ∈ (0 , there is a constant C ( ε ) > so that (0.7) X λ ∈ σ d ( J ) dist( λ, [ − , | λ − | − ε ≤ C ( ε )∆ , ∆ := ∞ X n =1 ( | b n | + | − a n c n | ) . If J is the discrete Schr¨odinger operator, that is, a n = c n ≡
1, then(0.8) X λ ∈ σ d ( J ) dist( λ, [ − , | λ − | − ε ≤ C ( ε ) k J − J k . Remark 0.2.
The value ∆ in (0.7) in place of k J − J k looks quite natural,at least for small perturbations. Indeed, given a Jacobi matrix J , considera class S ( J ) of Jacobi matrices S ( J ) = { b J := T − J T, T = diag( t j ) j ∈ N is a diagonal isomorphism of ℓ ( N ) } , b J = J (cid:0) { a j r j } , { b j } , { c j r − j } (cid:1) , r n = t n t n +1 , n ∈ N . Clearly, σ d ( b J ) = σ d ( J ) since b J is similar to J . Hence the left side of (0.7)is constant within the class S ( J ) and so is ∆, in contrast to k J − J k . Forthe class S ( J ) both sides of (0.7) vanish, whereas k J − J k , J ∈ S ( J ),can be arbitrarily large.Next, | − a n c n | ≤ | − a n | + | − c n | + | − a n || − c n | , and so(0.9) ∆ ≤ k J − J k + k J − J k . L. GOLINSKII
We see that for small perturbations the value ∆ has at least the same orderas k J − J k .As we mentioned earlier, the result of Hansmann–Katriel (0.6) is sharp.To prove that, the authors of [3] introduce and study a special class of two-sided Jacobi operators with rectangular (step) potentials. In Section 3 wedeal with an obvious counterpart of this class in semi-infinite setting. Forsuch operators some quantitative information about the discrete spectrum,which provides sharpness of Theorem 0.1, can be gathered. The argumentin [3] relies heavily on a simple matrix representation for the resolvent ofthe whole-line free Jacobi matrix and the theory of Kac–Murdock–Szeg˝omatrices, neither of which is available in the semi-infinite case. Instead, weanalyze directly the recurrence relations for eigenvectors and apply Rouch´e’sTheorem to the roots of the special algebraic equations.1. Jost solutions and discrete Volterra equations
We derive the bounds for the Jost solution u + by reducing the differenceequation (0.4) to the Volterra-type discrete integral equation. The unity ofthe first coefficient (0.4) appears to be crucial.Define a non-symmetric Green kernel for k, m ∈ Z (as a two-sided Toeplitzmatrix) by(1.1) G ( k, m ; z ) := ( z m − k − z k − m z − z − , m ≥ k, , m ≤ k, z ∈ D . The basic properties of this kernel can be verified directly: G ( k, m − z ) + G ( k, m + 1; z ) − (cid:16) z + 1 z (cid:17) G ( k, m ; z ) = δ k,m ,G ( k − , m ; z ) + G ( k + 1 , m ; z ) − (cid:16) z + 1 z (cid:17) G ( k, m ; z ) = δ k,m . (1.2)We make use of these properties for k, m ∈ N := { , , . . . } . The kernel(1.3) T ( k, m ; z ) := − b m G ( k, m ; z ) + (1 − a m − c m − ) G ( k, m − z ) , is a key player of the game. Note that the values of b , a − , c − are imma-terial. Theorem 1.1.
The Jost solution u + = ( u + k ) k ≥ of the difference equation (0.4) satisfies the discrete Volterra equation (1.4) u + k ( z ) = z k + ∞ X m = k +1 T ( k, m ; z ) u + m ( z ) , k ∈ N , z ∈ D . Conversely, each solution u = ( u k ) k ≥ of (1.4) solves (0.4) .Proof. We multiply the first relation (1.2) for G by u + m , relation (0.4) for u + by G ( k, m ), and subtract the later from the former h G ( k, m + 1) u + m − G ( k, m ) u + m − i + h − b m G ( k, m ) + G ( k, m − i u + m − a m c m G ( k, m ) u + m +1 = δ k,m u + m . ACOBI MATRICES 5
Next, we sum up over m from k + 1 to N , taking into account that G ( k, k + 1) = 1, G ( k, k ) = 0 G ( k, N + 1) u + N + N X m = k +1 h − b m G ( k, m ) + G ( k, m − i u + m − N X m = k a m c m G ( k, m ) u + m +1 = u + k , or u + k = G ( k, N + 1) u + N − a N c N G ( k, N ) u + N +1 + N X m = k +1 T ( k, m ) u + m . The latter equality holds for arbitrary solutions of (0.4). If u + is the Jostsolution, then, by (1.1),lim N →∞ G ( k, N + 1) u + N − a N c N G ( k, N ) u + N +1 = z k , and (1.4) follows.To prove the converse statement, let u = ( u k ) k ≥ be any solution of (1.4).Then u k − + u k +1 = (cid:16) z + 1 z (cid:17) z k + T ( k − , k ) u k + T ( k − , k + 1) u k +1 + ∞ X m = k +2 h T ( k − , m ) + T ( k + 1 , m ) i u m . But T ( k − , k ) u k = − b k u k ,T ( k − , k + 1) u k +1 = h − b k +1 G ( k − , k + 1) + (1 − a k c k ) G ( k − , k ) i u k +1 = − (cid:16) z + 1 z (cid:17) b k +1 u k +1 + (1 − a k c k ) u k +1 = (cid:16) z + 1 z (cid:17) T ( k, k + 1) u k +1 + (1 − a k c k ) u k +1 ,T ( k − , m ) + T ( k + 1 , m ) = (cid:16) z + 1 z (cid:17) T ( k, k + 1) . Finally, u k − + u k +1 = − b k u k + (1 − a k c k ) u k +1 + (cid:16) z + 1 z (cid:17) (cid:16) z k + ∞ X m = k +1 T ( k, m ) u m (cid:17) , which is (0.4). The proof is complete. (cid:3) The further study of the Volterra equation (1.4) relies upon the modifiedkernel(1.5) e T ( k, m ; z ) := T ( k, m ; z ) z m − k . L. GOLINSKII
It is easy to verify that e T ( k, m ; · ) are polynomials of z . Indeed, z m − k G ( k, m ; z ) = z m − k z m − k − z k − m z − z − = z z m − k ) − z − , m ≥ k ; z m − k G ( k, m − z ) = z m − k z m − k − − z k − m +1 z − z − = z z m − k − − z − , m ≥ k + 1 , as claimed. The bounds for the kernel e T follow from the latter relations (cid:12)(cid:12) z m − k G ( k, m ; z ) (cid:12)(cid:12) ≤ | z | · min n ( m − k ) + , | − z | o , (cid:12)(cid:12) z m − k G ( k, m − z ) (cid:12)(cid:12) ≤ | z | · min n ( m − k − + , | − z | o , and so (cid:12)(cid:12) e T ( k, m ; z ) (cid:12)(cid:12) ≤ δ m | z | min n ( m − k ) + , | − z | o , z ∈ D ,δ m := | b m | + | − a m − c m − | , a = c = 1 . (1.6)In particular,(1.7) (cid:12)(cid:12) e T ( k, m ; z ) (cid:12)(cid:12) ≤ | ω ( z ) | δ m , ω ( z ) := 2 z − z , z ∈ D := D \{± } . Theorem 1.2. ( i ) . Assume that (1.8) ∆ = ∞ X j =1 δ j < ∞ . Then the equation (1.4) has a unique solution u + = ( u + k ) k ≥ so that u + k areanalytic in D , continuous in D , and for k ∈ N and z ∈ D (1.9) | u + k ( z ) − z k | ≤ | z | k | ω ( z ) | s ( k ) exp n | ω ( z ) | s ( k ) o , s ( k ) := ∞ X j = k +1 δ j . ( ii ) . Assume that (1.10) ∆ := ∞ X j =1 jδ j < ∞ . Then (1.4) has a unique solution u + = ( u + k ) k ≥ so that u + k are analytic in D , continuous in D , and for k ∈ N and z ∈ D (1.11) | u + k ( z ) − z k | ≤ | z | k +1 s ( k ) exp (cid:8) s ( k ) (cid:9) , s ( k ) := ∞ X j = k +1 jδ j . Proof.
It is advisable to introduce new variables in (1.4) f j ( z ) := u + j ( z ) z − j − , j ∈ N , ACOBI MATRICES 7 so the equation (1.4) turns into f k ( z ) = g k ( z ) + ∞ X m = k +1 e T ( k, m ; z ) f m ( z ) ,g k ( z ) := ∞ X m = k +1 e T ( k, m ; z ) , k ∈ N , z ∈ D . (1.12)It is clear from (1.7) and the assumption (1.8) that the latter series convergesabsolutely and uniformly on compact subsets of D and so represents ananalytic function in D . Moreover, | g k ( z ) | ≤ | ω ( z ) | s ( k ) , k ∈ N , z ∈ D . We are going to solve (1.12) by using the successive approximationsmethod. Let f k, ( z ) := g k ( z ) , f k, ( z ) := ∞ X m = k +1 e T ( k, m ; z ) f m, ( z ) . As above, the latter series converges absolutely and uniformly on compactsubsets of D and | f k, ( z ) | ≤ (cid:0) | ω ( z ) | s ( k ) (cid:1) . We define consecutively f k,j +1 ( z ) := ∞ X m = k +1 e T ( k, m ; z ) f m,j ( z ) , and show by induction that(1.13) | f k,p ( z ) | ≤ (cid:0) | ω ( z ) | s ( k ) (cid:1) p ( p − , p ∈ N . Indeed, the bound is true for p = 1 ,
2, assume that it holds for p = 1 , , . . . , j and k ∈ N . We have | f k,j +1 ( z ) | ≤ | ω ( z ) | ∞ X m = k +1 δ m | f m,j ( z ) | ≤ | ω ( z ) | j +1 ( j − ∞ X m = k +1 δ m s j ( m ) . The elementary inequality ( a + b ) j +1 − a j +1 ≥ ( j + 1) ba j , a, b ≥ a = s ( p ), b = δ p , a + b = s ( p − ∞ X m = k +1 δ m s j ( m ) ≤ j + 1 ∞ X m = k +1 (cid:0) s j +10 ( p − − s j +10 ( p ) (cid:1) = s j +10 ( k ) j + 1 , and so | f k,j +1 ( z ) | ≤ | ω ( z ) | j +1 s j +10 ( k )( j − j + 1) ≤ (cid:0) | ω ( z ) | s ( k ) (cid:1) j +1 j ! , as claimed.Hence, the series f k := ∞ X j =1 f k,j ( z ) L. GOLINSKII converges absolutely and uniformly on compact subsets of D and representsan analytic in D function, continuous in D . It satisfies (1.12) f k ( z ) − g k ( z ) = f k ( z ) − f k, ( z ) = ∞ X j =1 f k,j +1 ( z ) = ∞ X j =1 ∞ X m = k +1 e T ( k, m ; z ) f m,j ( z )= ∞ X m = k +1 e T ( k, m ; z ) f m ( z ) . This solution admits the bound, see (1.13), | f k ( z ) | ≤ ∞ X j =1 | f k,j ( z ) | ≤ | ω ( z ) | s ( k ) ∞ X p =0 (cid:0) | ω ( z ) | s ( p ) (cid:1) p p ! = | ω ( z ) | s ( k ) e | o ( z ) | s ( k ) , which is (1.9).As far as uniqueness goes, suppose that there are two solutions of (1.4) u + = ( u + k ) k ≥ , v + = ( v + k ) k ≥ . Assume further that z = 0. By (1.7), we have(1.14) | u + k ( z ) − v + k ( z ) | ≤ | ω ( z | ) X m ≥ k +1 δ m | u + m ( z ) − v + m ( z ) | =: q k ( z ) . Clearly, q k ց k → ∞ .We fix z and assume first that q p > p ∈ N . By (1.14), q p − − q p q p = δ p | u + p ( z ) − v + p ( z ) || ω ( z ) | q p ≤ | ω ( z ) | δ p , q p ≤ q N N Y j = p +1 (cid:0) | ω ( z ) | δ j (cid:1) . The latter product converges as N → ∞ , so q p = 0 in contradiction withour assumption.Next, let l ∈ N exist so that q l = 0. Then, in view of monotonicity, q l +1 = q l +2 = . . . = 0, and, in the opposite way, successively, | u + l − v + l | = 0 ⇒ q l − = 0 ⇒ | u + l − − v + l − | = 0 ⇒ q l − = 0 . . . ⇒ | u +0 − v +0 | = 0 , so the uniqueness follows.(ii). The proof goes along the same line of reasoning with the auxiliarybounds for z ∈ D (cid:12)(cid:12)(cid:12) e T ( k, m ; z ) (cid:12)(cid:12)(cid:12) ≤ | z | mδ m , | g k ( z ) | ≤ | z | s ( k ) , | f k,p ( z ) | ≤ (cid:0) | z | s ( k ) (cid:1) p ( p − . The proof is complete. (cid:3) Perturbation determinant and the Lieb–Thirring inequality
Under our main assumption (0.2), the perturbation determinant L ( λ, J )(0.3) is a well-defined analytic function in D . Corollary 2.1.
Let J − J ∈ S . Then the bound holds (2.1) | L ( λ ( z ) , J ) − | ≤ | ω ( z ) | ∆ exp n | ω ( z ) | ∆ o , z ∈ D . Under assumption (1.10) , (2.2) | L ( λ ( z ) , J ) − | ≤ | z | ∆ e ∆ , z ∈ D . ACOBI MATRICES 9
Proof.
By the non-self-adjoint version of [16, Section 2] (the calculation thereis algebraic and so immediately extends to the non-self-djoint case), the Jostsolution u + of (0.4) equals u + k ( z ) = z k L ( λ ( z ) , J ( k ) ) , k ∈ N , where J ( k ) , the k -stripped Jacobi matrix, is obtained from J by droppingthe first k rows and columns. So, (2.1) and (2.2) are (1.9) and (1.11),respectively, with k = 0. (cid:3) We are now ready for
Proof of Theorem 0.1 . The following bound for L is immediate from (2.1)and an elementary inequality 1 + xe x ≤ e x , x ≥ | L ( z ) | ≤ | z | ∆ | − z | , L (0) = 1 , z ∈ D . The rest is standard nowadays. According to [14, Theorem 4], for each ε ∈ (0 ,
1) there is a constant C ( ε ) > Z ( L ) X ζ ∈ Z ( L ) (1 − | ζ | ) | ζ − | ε | ζ | ε ≤ C ( ε )∆ , (each zero is taken with its multiplicity). The latter inequality turns into(0.7) when we go over to the Zhukovsky images, taking into account thedistortion for the Zhukovsky function [14, Lemma 7].For the discrete Schr¨odinger operators ( a j = c j ≡ ∞ X j = −∞ | b j | = k J − J k , and (0.8) follows. The proof is complete.It might be worth comparing the key inequality (2.3) and (0.9) withlog | L ( z ) | ≤ C abs | − z | ( k J − J k + k J − J k ) , z ∈ D , the result obtained in [16, Theorem 2.8] in the self-adjoint case.There is yet another consequence of Theorem 1.2, (i), which concernsembedded eigenvalues of Jacobi operator J . The result is likely to be known(cf. [15] for two-sided discrete Schr¨odinger operators), so we briefly outlineits proof. Corollary 2.2.
Let J − J ∈ S . Then the operator J has no embeddedeigenvalues, i.e., eigenvalues on ( − , .Proof. Assume on the contrary, that λ = 2 cos θ , 0 < θ < π , is the eigen-value of J . In this case λ is also the eigenvalue for the modified Jacobioperator b J := J ( { } , { b j } , { a j c j } ) . Let h = ( h k ) k ≥ be the correspondingeigenvector, so we have an ℓ -solution h ′ = (0 , h ) of (0.4) with z = e iθ .On the other hand, the Jost solution is known to be continuous in D , sothe second solution u + = ( u + k ) k ≥ of (0.4) comes up. It is clear from (1.9)that(2.4) | u + k ( e iθ ) | = 1 + o (1) , k → ∞ . The Wronskian of these two solutions W ( u + , h ′ ) = ∞ Y j = k +1 ( a j c j ) − ( u + k h k +1 − u + k +1 h k )at the point e iθ is k -independent, and as h ∈ ℓ , and u + is bounded, we seethat W ( u + , h ′ ) ≡
0. Hence, u + and h are to be linearly dependent, thatcontradicts (2.4) and h ∈ ℓ . (cid:3) Remark 2.3.
As a byproduct, the bound (2.1) for the perturbation de-terminant provides some information on location of the discrete spectrum(spectral enclosure). Indeed, let κ be a unique positive root of the equation xe x = 1 , κ ≈ . . Then L = 0 in D as long as | ω ( z ) | ∆ < κ, | z || − z | < κ , or in terms of the Zhukovsky images(2.5) σ d ( J ) ⊂ ( λ ∈ C \ [ − ,
2] : | λ − | ≤ (cid:18) κ (cid:19) ) . So, the discrete spectrum lies in a certain Cassini oval.The spectral enclosures are normally derived from the Birman–Schwingerprinciple. Precisely, λ ( z ) ∈ σ d ( J ) ⇒ k K ( z ) k ≤ ,K is the Birman–Schwinger operator. In our case one has(2.6) σ d ( J ) ⊂ (cid:8) λ ∈ C \ [ − ,
2] : | λ − | ≤ k J − J k (cid:9) . For two-sided discrete Schr¨odinger operators the sharp oval which con-tains the discrete spectrum is known [15](2.7) σ d ( J ) ⊂ (cid:8) λ ∈ C \ [ − ,
2] : | λ − | ≤ k J − J k (cid:9) . For semi-infinite discrete Schr¨odinger operators one has(2.8) σ d ( J ) ⊂ n λ ∈ C \ [ − ,
2] : | λ − | ≤ κ k J − J k o . As another curious consequence of (2.2), we note that under the assump-tion (1.10), the discrete spectrum is empty as long as(2.9) ∞ X j =1 j ( | b j | + | − a j c j | ) < κ. This effect is inherent to semi-infinite Jacobi matrices.
ACOBI MATRICES 11 Semi-infinite discrete Schr¨odinger operators withrectangular potentials
Given n ∈ N and h >
0, we study a semi-infinite discrete Schr¨odingeroperator with a pure imaginary step potential(3.1) J n,h := b b b
1. . . . . . . . . , b j = b j ( h ) = (cid:26) ih, j = 1 , . . . , n ;0 , j ≥ n + 1.It is clear that J n,h = J + ih P n , P n is the orthogonal projection onto thelinear span of the first n basis vectors { e , . . . , e n } . The following result isa slight refinement of [3, Lemma 4]. Proposition 3.1.
The discrete spectrum of J n,h is contained in the rectangle (3.2) σ d ( J n,h ) ⊂ { λ ∈ C : | Re λ | < , < Im λ < h } . Proof. If J n,h ϕ = λϕ , k ϕ k = 1, ϕ = { ϕ k } k ≥ , then λ = h J n,h ϕ, ϕ i = h J ϕ, ϕ i + ih hP n ϕ, ϕ i . With ϕ = 0 we have | Re λ | = (cid:12)(cid:12) h J ϕ, ϕ i (cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)X j ≥ ( ϕ j − + ϕ j +1 ) ϕ j (cid:12)(cid:12)(cid:12) = 2 (cid:12)(cid:12)(cid:12)X j ≥ Re ( ϕ j − ϕ j ) (cid:12)(cid:12)(cid:12) ≤ X j ≥ | ϕ j − ϕ j | < X j ≥ | ϕ j | = 2 , (the latter inequality is strict since 2 | ϕ j − ϕ j | = | ϕ j − | + | ϕ j | would imply | ϕ j − | = | ϕ j | for all j that is impossible).Next, obviously 0 ≤ Im λ = h hP n ϕ, ϕ i ≤ h . Moreover,Im λ = 0 ⇒ hP n ϕ, ϕ i = 0 ⇒ P n ϕ = 0 , and so J ϕ = λϕ that is also impossible. Similarly,Im λ = h ⇒ hP n ϕ, ϕ i = kP n ϕ k = 1 ⇒ P n ϕ = ϕ, and again, the conclusion J ϕ = ( λ − ih ) ϕ is false. (cid:3) Let J n,h g = λg = (cid:16) z + 1 z (cid:17) g, z ∈ D ; g = { g j } j ≥ ∈ ℓ ( N ) , g = 0 . The system of recurrence relations for the coordinates looks as follows ihg + g = λg ,g j − + ihg j + g j +1 = λg j , j = 2 , . . . , n,g j − + g j +1 = λg j , j = n + 1 , . . . . (3.3)We begin with the characteristic equation for the second series of relations(3.4) t − ( λ − ih ) t + 1 = ( t − t )( t − t ) = 0 , t j = t j ( z, h ) , t = t − . Furthermore, t + t = t + t − = λ − ih , and since Im λ < h due toProposition 3.1, we have 0 < | t | < < | t | . Put ζ := t , so(3.5) ζ + 1 ζ = λ − ih = z + 1 z − ih, ζ, z ∈ D . Yet another relation between z and ζ comes from the adjustment in (3.3).Indeed, g j = aζ j + bζ − j , j = 1 , . . . , n + 1 , | a | + | b | > g j = g n z j − n , j = n, n + 1 , . . . , since g ∈ ℓ . For j = 2 we have g = aζ + bζ − . On the other hand, thefirst relation in (3.3) gives g = ( λ − ih ) g = ( ζ + ζ − )( aζ + bζ − ) = aζ + bζ − + a + b ⇒ a + b = 0 . For j = n + 1 g n +1 = aζ n +1 + bζ − n − = ( aζ n + bζ − n ) z ⇒ aζ n ( z − ζ ) + bζ − n ( z − ζ − ) = 0 . The system of two homogeneous, linear equations has a nontrivial solutionif and only if its determinant vanishes(3.6) (cid:12)(cid:12)(cid:12)(cid:12) ζ n ( z − ζ ) ζ − n ( z − ζ − ) (cid:12)(cid:12)(cid:12)(cid:12) = ζ − n ( z − ζ − ) − ζ n ( z − ζ ) = 0 . Hence, if λ ∈ σ d ( J n,h ) then (3.5) – (3.6) hold.Conversely, let a pair z, ζ ∈ D satisfy (3.5) – (3.6), so ζ is the root ofthe characteristic equation (3.4) in D . It is a matter of direct calculation tocheck that g = ( g j ) j ≥ with g j := ζ j − ζ − j , j = 1 , , . . . , n + 1; g j = g n z j − n , j = n + 1 , . . . is the eigenvector of J n,h with the eigenvalue λ .Let us express z from (3.6)(3.7) z = ζ n +1 − ζ − n − ζ n − ζ − n = ζ n +2 − ζ n +1 − ζ ∈ D and plug this expression into (3.5). We come to the main algebraic equationassociated with J n,h ζ + 1 ζ + ih = ζ n +2 − ζ n +1 − ζ + ζ n +1 − ζζ n +2 − , or(3.8) P n ( ζ ) := ζ n − ( ζ − − ih (1 − ζ n )(1 − ζ n +2 ) = 0 . Conversely, let ζ be a root of (3.8) in D so that (3.7) holds. Then (3.6) istrue, and (3.5) is a simple consequence of the equality( ζ n +2 − + ( ζ n +1 − ζ ) − ( ζ + ζ − )( ζ n +2 − ζ n +1 − ζ ) = ζ n ( ζ − . Thereby, we come to the following result.
Proposition 3.2.
Let z ∈ D . The number λ = z + z − ∈ σ d ( J n,h ) if andonly if there is a root ζ ∈ D of the polynomial P n (3.8) so that (3.7) holds. Our next goal is to show that the number of such roots is large enough(proportional to n ).From now on we put h := n − α , 0 < α < J n,h = J n,α .The parameter n is assumed to be large.Take 0 < c < α < c < r j = r j ( n ) := 1 − c j log n n + 1 , j = 1 , . ACOBI MATRICES 13
Clearly,(3.10) r n +1 j = n − c j (cid:18) O (cid:18) log nn (cid:19)(cid:19) , j = 1 , . Proposition 3.3.
The circular segment S = S ( n, a ) := n w = re iϕ : r < r < r , a π < ϕ < (cid:16) − a (cid:17) π o , < a < , contains at least N = (1 / − a ) n + O (1) simple roots of P n (3.8) .Proof. Write P n = P n, − P n, with P n, ( w ) = 2 w n +1 + in − α ,P n, ( w ) = w n +1 ( w + w − ) + in − α ( w n + w n +2 − w n +2 ) . We wish to compare the roots of P n with the roots of a simple binomial P n, . The latter are known explicitly P n, ( w ) = 0 ⇔ w = w n,k = γ n exp n iπ n + 1) (4 k − o , k = 1 , . . . , n + 1 ,γ n = | w n,k | = (cid:16) n − α (cid:17) / n +1 = 1 − α log n n + 1 + O (cid:16) n (cid:17) . Define the values ϕ k = ϕ k ( n ) by(3.11) ϕ k := π (4 k + 1)2(2 n + 1) ⇒ ϕ k − < arg w n,k < ϕ k , k = 1 , . . . , n + 1 , and the segments S k = S k ( n ) by S k := (cid:8) w = re iϕ : r < r < r , ϕ k − < ϕ < ϕ k (cid:9) . Since r ( n ) < γ n < r ( n ) for large n , each segment S k contains exactly oneroot w n,k of P n, . As it turns out, P n, dominates P n, on the boundary ∂S k for certain values of k , specified below.We have P n, ( re iϕ ) = 2 r n +1 cos(2 n + 1) ϕ + i (2 r n +1 sin(2 n + 1) ϕ + n − α ) , and so(3.12) | P n, ( re iϕ ) | ≥ | r n +1 sin(2 n + 1) ϕ + n − α | . For the second polynomials P n, ( re iϕ ) = r n +1 e i (2 n +1) ϕ (cid:0) r e iϕ + r − e − iϕ + in − α ( re iϕ + r − e − iϕ − r n +1 e i (2 n +1) ϕ ) (cid:1) , and since r e iϕ + r − e − iϕ = ( r − e iϕ + 2 cos 2 ϕ + ( r − − e − iϕ = 2 cos 2 ϕ + O (cid:18) log nn (cid:19) , r ≤ r ≤ r , (3.13)we come to the bound(3.14) | P n, ( re iϕ ) | ≤ r n +1 (cid:0) | cos 2 ϕ | + O ( n − α ) (cid:1) , r ≤ r ≤ r , with O uniform in r and ϕ . Let first ϕ = ϕ k , r ≤ r ≤ r . By (3.11), sin(2 n + 1) ϕ k = 1 for all k , so(3.12) , (3.14) imply | P n, ( re iϕ k ) | ≥ r n +1 + n − α , | P n, ( re iϕ k ) | ≤ r n +1 | cos 2 ϕ k | + o ( n − α ) , (see (3.10) for the last term). Hence, | P n, ( re iϕ k ) | > | P n, ( re iϕ k ) | for all k ,that is, the domination holds on the radial sides of S k .Next, let r = r , ϕ k − ≤ ϕ ≤ ϕ k , be the exterior arc of S k . By (3.10), | P n, ( re iϕ ) | = | r n +1 e i (2 n +1) ϕ + in − α | = r n +1 (cid:16) O (cid:0) n − ( α − c ) (cid:1)(cid:17) , | P n, ( re iϕ k ) | ≤ r n +1 (cid:0) | cos 2 ϕ k | + O ( n − α (cid:1) ) . We choose k in such a way that the value | cos 2 ϕ k | is separated from 1. Anelementary calculation shows that for k ∈ Λ = Λ n,a := [ a ′ n , a ′′ n ] ,a ′ n = a n + 14 + 1 , a ′′ n = (cid:16) − a (cid:17) n + 14 − , (3.15)the inequality holds(3.16) aπ < ϕ k − < ϕ k < (cid:16) − a (cid:17) π. So, again | P n, ( r e iϕ ) | > | P n, ( r e iϕ ) | , ϕ k − ≤ ϕ ≤ ϕ k , k ∈ Λ . Finally, along the interior arc of S k , that is, r = r , ϕ k − ≤ ϕ ≤ ϕ k , wehave r n +11 = O ( n − c ) = o ( n − α ), and so | P n, ( r e iϕ ) | ≥ n − α , | P n, ( r e iϕ ) | = O ( n − c ) , as needed.By Rouch´e’s Theorem, each region S k with k ∈ Λ contains one simpleroot of P n . The number of such regions is N = (1 / − a ) n + O (1), and, by(3.16), all of them lie in S . The proof is complete. (cid:3) In the rest of the paper we focus on the roots of P n just found. Denotethem by ζ k = ζ k ( n ) = ρ k e iθ k , ρ k = ρ k ( n ), θ k = θ k ( n ). The equality (3.8)with ζ = ζ k reads − in α ζ n +1 k (cid:0) ζ k − ζ − k (cid:1) = (1 − ζ nk )(1 − ζ n +2 k ) . As in (3.13) above, (cid:0) ζ k − ζ − k (cid:1) = − θ k (cid:18) O (cid:16) log nn (cid:17)(cid:19) . Since ρ n +1 k = O (cid:0) n − c (cid:1) , we end up with(3.17) 4 i sin θ k n α ρ n +1 k e i (2 n +1) θ k = 1 + O (cid:0) n − c (cid:1) . Take the absolute value and the real part in (3.17):4 sin θ k n α ρ n +1 k = 1 + O (cid:0) n − c (cid:1) , − θ k n α ρ n +1 k sin(2 n + 1) θ k = 1 + O (cid:0) n − c (cid:1) , (3.18) ACOBI MATRICES 15 with O uniform in k ∈ Λ. By plugging the first equality in (3.18) into thesecond one, we have(3.19) sin(2 n + 1) θ k = − O (cid:0) n − c (cid:1) . Regarding θ k themselves, it is clear from the construction of S k that(3.20) θ k = π n + 1) (4 k + 1) + O (cid:16) n (cid:17) , k ∈ Λ . For the range of arguments in S we have(3.21) 0 < a π ≤ θ k ≤ (cid:16) − a (cid:17) π < , so it follows from (3.18) that(3.22) ρ k = 1 − α log n n + 1 + O (cid:16) n (cid:17) , with O uniform in k ∈ Λ (cf. with γ n = | w n,k | ).Here is the main result of the section. Theorem 3.4.
For the discrete Schr¨odinger operators J n,α the followingrelation holds (3.23) lim n →∞ k J n,α − J k X λ ∈ σ d ( J n,α ) dist( λ, [ − , | λ − | / = + ∞ . Proof.
In view of Proposition 3.2, to make sure that the roots { ζ k } k ∈ Λ gen-erate eigenvalues of J n,α , we have to check the relation (3.7). Indeed, let z k = z k ( n ) := ζ n +2 k − ζ n +1 k − ζ k . Then 1 − | z k | = ( ρ k − − ρ n +2 k ) − ρ n +2 k sin(2 n + 1) θ k sin θ k | ζ n +1 k − ζ k | . By (3.22), the first term in the numerator is( ρ k − − ρ n +2 k ) = O (cid:16) log nn (cid:17) . As for the second one, the first equality in (3.18) and (3.21) imply ρ n +1 k = n − α θ k (1 + O ( n − c )) > n − α , and in view of (3.19) and (3.21), the second term is − n + 1) θ k sin θ k > c > . Hence, 1 − | z k | > n , as claimed.Therefore, the numbers λ k = λ k ( n ) := ζ k + ζ − k + in − α = ρ k e iθ k + ρ − k e − iθ k + in − α are distinct eigenvalues of J n,α for all large enough n and k ∈ Λ (3.15). Wehave, as in (3.13) above,(3.24) λ k = 2 cos θ k + in − α + O (cid:16) log nn (cid:17) , and so, by Proposition 3.1,(3.25) dist ( λ k , [ − , λ n,k = n − α (1 + o (1)) . As we restrict ourselves with the eigenvalues λ k , k ∈ Λ, X λ ∈ σ d ( J n,α ) dist( λ, [ − , | λ − | / ≥ X k ∈ Λ dist( λ k , [ − , | λ k − | / =: I n,a . Next, by the location of the discrete spectrum and (3.24), we see that | λ k + 2 | / ≤ (4 + n − α ) / < √ , | λ k − | / ≤ θ k n − α/ (1 + o (1)) ≤ θ k + n − α/ (1 + o (1)) , (3.26)and so (3.25) and (3.26) imply (see also (3.20)) I n,a ≥ n − α (1 + o (1)) √ X k ∈ Λ h π n + 1) (4 k + 1) + n − α/ (1 + o (1)) i − ≥ n − α (1 + o (1)) π √ X k ∈ Λ k + u n , u n := n − α/ π (1 + o (1)) + 1 . An elementary bound m X m k + u n > Z m m dxx + u n = log m + u n m + u n with m := [ a ′ n ] + 1, m := [ a ′′ n ] (see (3.15)) gives I n,a ≥ n − α (1 + o (1)) π √ − a + 4 n − α/ (1 + o (1)) a + 2 n − α/ (1 + o (1)) . Hence, taking into account, that k J n,α − J k = n X j =1 | b j ( n ) | = n − α , we end up with the relationlim inf n →∞ k J n,α − J k X λ ∈ σ d ( J n,α ) dist( λ, [ − , | λ − | / ≥ C log 1 − aa , and it remains only to tend a →
0. The proof is complete. (cid:3)
Remark 3.5.
The Jost solution u + = ( u + j ) j ≥ for Jacobi operators withthe step potential can be found explicitly. Indeed, the recurrence relation u + k − ( z ) + ihu + k ( z ) + u + k +1 ( z ) = (cid:16) z + 1 z (cid:17) u + k ( z ) , k = 1 , , . . . , n ; u + k ( z ) = z k , k = n, n + 1 . . . can be resolved, and we come to the following expression for the Jost function u +0 (which is the same as the perturbation determinant) L ( z, J n,h ) = z n U n (cid:16) z + z − − ih (cid:17) − z n +1 U n − (cid:16) z + z − − ih (cid:17) ,U k is the Chebyshev polynomial of the second kind. It might be a challengingproblem analyzing the roots of the polynomial on the right side directly. ACOBI MATRICES 17
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