Ambarzumian-type problems for discrete Schrödinger operators
Jerik Eakins, William Frendreiss, Burak Hatino?lu, Lucille Lamb, Sithija Manage, Alejandra Puente
aa r X i v : . [ m a t h . SP ] J a n AMBARZUMIAN-TYPE PROBLEMS FOR DISCRETESCHR ¨ODINGER OPERATORS
JERIK EAKINS, WILLIAM FRENDREISS, BURAK HAT˙INO ˘GLU, LUCILLE LAMB,SITHIJA MANAGE, AND ALEJANDRA PUENTE
Abstract.
We discuss the problem of unique determination of the finite free dis-crete Schr¨odinger operator from its spectrum, also known as Ambarzumian problem,with various boundary conditions, namely any real constant boundary condition atzero and Floquet boundary conditions of any angle. Then we prove the followingAmbarzumian-type mixed inverse spectral problem: Diagonal entries except the firstand second ones and a set of two consecutive eigenvalues uniquely determine the finitefree discrete Schr¨odinger operator. Introduction
The Jacobi matrix is a three-diagonal matrix defined as b a a b a . . . 00 a b . . . 00 . . . . . . . . . a n − . . . a n − b n where n ∈ N , a k > k ∈ { , , . . . , n − } and b k ∈ R for any k ∈ { , , . . . , n } .When a k = 1 for each k ∈ { , , . . . , n − } , this matrix defines the finite discreteSchr¨odinger operator.Direct spectral problems aim to get spectral information from the sequences { a k } n − k =1 and { b k } nk =1 . In inverse spectral problems one tries to recover these sequences fromspectral information such as the spectrum, the spectral measure or Weyl m -function.Early inverse spectral problems for finite Jacobi matrices appear as discrete analogsof inverse spectral problems for the Schr¨odinger (Sturm-Liouville) equations − u ′′ ( t ) + q ( t ) u ( t ) = zu ( t ) , on the interval [0 , π ] with the boundary conditions u (0) cos α − u ′ (0) sin α = 0 u ( π ) cos β + u ′ ( π ) sin β = 0 , where the potential function q ∈ L (0 , π ) is real-valued and α, β ∈ [0 , π ).The first inverse spectral result on Schr¨odinger operators is given by Ambarzumian[1]. He considered continuous potential with Neumann boundary conditions at both Key words and phrases. inverse spectral theory, discrete Schr¨odinger operators, Jacobi matrices,Ambarzumian-type problems. endpoints ( α = β = π/
2) and showed that q ≡ L -potential is uniquelyrecovered from two spectra, which share the same boundary condition at π ( β = β )and one of which is with Dirichlet boundary condition at 0 ( α = 0 , α ∈ (0 , π )).A few years later, Levinson [23] removed the Dirichlet boundary condition restrictionfrom Borg’s result. This famous theorem is also known as two-spectra theorem. ThenMarchenko [24] observed that the spectral measure (or Weyl-Titchmarsh m -function)uniquely recovers an L -potential. Another classical result is due to Hochstadt andLiebermann [21], which says that if the first half of an L -potential is known, onespectrum recovers the whole. One can find the statements of these classical theoremsand some other results from the inverse spectral theory of Schr¨odinger operators e.g.in [16] and references therein.Finite Jacobi matrix analogs of Borg’s and Hochstadt and Lieberman’s theorems wereconsidered by Hochstadt [18–20], where the potential q is replaced by the sequences { a k } n − k =1 and { b k } nk =1 . These classical theorems led to various other inverse spectral re-sults on finite Jacobi matrices (see [2, 5, 12, 15, 25, 31] and references therein) and othersettings such as semi-infinite, infinite, generalized Jacobi matrices and matrix-valuedJacobi operators (see e.g. [4, 6–11, 13, 14, 17, 26, 27, 29] and references therein). In gen-eral, these problems can be divided into two groups. In Borg-Marchenko-type spectralproblems, one tries to recover the sequences { a k } n − k =1 and { b k } nk =1 from the spectraldata. On the other hand, Hochstadt-Lieberman-type (or mixed) spectral problemsrecover the sequences { a k } n − k =1 and { b k } nk =1 using a mixture of partial information onthese sequences and the spectral data.Ambarzumian-type problems focus on inverse spectral problems for free discreteSchr¨odinger operators, i.e. a k = 1 and b k = 0 for every k , or similar cases when b k = 0for some k . In this paper, we first revisit the classical Ambarzumian problem for thefinite discrete Sch¨odinger operator in Theorem 3.3, which says that the spectrum of thefree operator uniquely determines the operator. Then we provide a counter-example,Example 3.6, which shows that knowledge of the spectrum of the free operator witha non-zero boundary condition is not sufficient for unique recovery. In Theorem 3.7,we observe that a non-zero boundary condition along with the corresponding spectrumof the free operator is needed for the uniqueness result. However, in Theorem 3.8 weprove that for the free operator with Floquet boundary conditions, the set of eigenvaluesincluding multiplicities is sufficient to get uniqueness up to transpose.We also answer the following mixed Ambarzumian-type inverse problem positivelyin Theorem 4.2. Inverse Spectral Problem.
Let us define the discrete Schr¨odinger matrix S n as a k = 1 for k ∈ { , . . . , n − } and b , b ∈ R , b k = 0 for k ∈ { , . . . , n } . Let us alsodenote the free discrete Schr¨odinger operator by F n , which is defined as a k = 1 for k ∈ { , . . . , n − } and b k = 0 for k ∈ { , . . . , n } . If S n and F n share two consecutiveeigenvalues, then do we get b = b = 0, i.e. S n = F n ?The paper is organized as follows. In Section 2 we recall necessary definitions andresults we use in our proofs. In Section 3 we consider the problem of unique determi-nation of the finite free discrete Schr¨odinger operator from its spectrum, with various MBARZUMIAN-TYPE PROBLEMS FOR DISCRETE SCHR ¨ODINGER OPERATORS 3 boundary conditions, namely any real constant boundary condition at zero, and Flo-quet boundary conditions of any angle. In Section 4 we prove the above mentionedAmbarzumian-type mixed inverse spectral problem.2.
Preliminaries
Let us start by fixing our notation. Let J n represent the finite Jacobi matrix of size n × n (2.1) J n := b a · · · a b a . . . ...0 a b . . . 0... . . . . . . . . . a n − · · · a n − b n , where a k > , b k ∈ R . Given J n , let us consider the Jacobi matrix where all a k ’s and b k ’s are the same as J n except b and b n are replaced by b + b and b n + B respectivelyfor b, B ∈ R , i.e.(2.2) J n + b ( δ , · ) δ + B ( δ n , · ) δ n . The Jacobi matrix (2.2) is given by the Jacobi difference expression a k − f k − + b k f k + a k f k +1 , k ∈ { , · · · , n } with the boundary conditions f = bf and f n +1 = Bf n . Let us note that we assume a = 1 and a n = 1.In order to get a unique Jacobi difference expression with boundary conditions fora given Jacobi matrix, we can see the first and the last diagonal entries of the matrix J n , defined in (2.1), as the boundary conditions at 0 and n + 1 respectively. Thereforelet J n ( b, B ) denote the Jacobi matrix J n satisfying b = b and b n = B .If we consider the Jacobi difference expression with the Floquet boundary conditions f = f n e πiθ and f n +1 = f e − πiθ , θ ∈ [0 , π )then we get the following matrix representation, which we denote by J n ( θ ).(2.3) J n ( θ ) := b a · · · e πiθ a b a . . . 00 a b . . . 0... . . . . . . . . . a n − e − πiθ · · · a n − b n . Let us denote the discrete Sch¨odinger matrix accordingly, i.e. S n ( b, B ) denotes thematrix J n such that b = b , b n = B and a k = 1 for each k ∈ { , , · · · , n − } . Similarly EAKINS, FRENDREISS, HAT˙INO ˘GLU, LAMB, MANAGE, AND PUENTE S n ( θ ) denotes the matrix J n ( θ ) such that a k = 1 for each k ∈ { , , · · · , n − } . Let usdenote the free discrete Sch¨odinger matrix of size n × n by F n : F n = · · ·
01 0 1 . . . ...0 1 0 . . . 0... . . . . . . . . . 10 · · · , so F n ( b, B ) and F n ( θ ) denote the free discrete Sch¨odinger matrices with boundaryconditions b at 0, B at n + 1 and Floquet boundary conditions for θ , respectively.Let us state some basic properties of the free discrete Sch¨odinger matrix. If λ , λ , ..., λ n denote the eigenvalues of F n , they have the following properties: • For all k , λ k ∈ [ − , • The free discrete Sch¨odinger matrix F n has n distinct eigenvalues, so we canreorder the eigenvalues such that λ < λ < · · · < λ n . • Let F n − be the ( n − × ( n −
1) submatrix of F n obtained by removing thelast row and the last column of F n . If µ , µ , · · · , µ n − denote the eigenvaluesof F n − taken in increasing order, then we have the interlacing property ofeigenvalues, i.e. λ < µ < λ < µ < ... < λ n − < µ n − < λ n The second and third properties are valid for any Jacobi matrix J n . These basicproperties can be found in [30], which provides an extensive study of Jacobi operators.The following results show smoothness of simple eigenvalues and corresponding eigen-vectors of a smooth matrix-valued function. We will use them in Section 4 in order toprove the mixed inverse spectral problem mentioned in Introduction. Theorem 2.1. ([22] , Theorem 9.7 ) Let A ( t ) be a differentiable square matrix-valuedfunction of the real variable t . Suppose that A (0) has an eigenvalue a of multiplicityone, in the sense that a is a simple root of the characteristic polynomial of A (0) . Thenfor t small enough, A ( t ) has an eigenvalue a ( t ) that depends differentiably on t, andwhich equals a o at zero, that is, a (0) = a . Theorem 2.2. ([22] , Theorem 9.8 ) Let A ( t ) be a differentiable matrix-valued functionof t , a ( t ) an eigenvalue of A ( t ) of multiplicity one. Then we can choose an eigenvector X ( t ) of A ( t ) pertaining to the eigenvalue a ( t ) to depend differentiably on t . Once we obtain smoothness of an eigenvalue and the corresponding eigenvector of asmooth self-adjoint matrix, the Hellmann-Feynman theorem relates the derivatives ofthe eigenvalue and the matrix with the corresponding eigenvector.
Theorem 2.3 (Hellmann-Feynman) . ([28] , Theorem 1.4.7 ) Let A ( t ) be self-adjointmatrix-valued, X ( t ) be vector-valued and λ ( t ) be real-valued functions. If A ( t ) X ( t ) = λ ( t ) X ( t ) and || X ( t ) || = 1 , then λ ′ ( t ) = h X ( t ) , A ′ ( t ) X ( t ) i . MBARZUMIAN-TYPE PROBLEMS FOR DISCRETE SCHR ¨ODINGER OPERATORS 5 Ambarzumian problem with various boundary conditions
In addition to the notations we introduced in the previous section, let p n ( x ) be thecharacteristic polynomial of F n with zeroes λ , . . . , λ n and let q n ( x ) be the characteristicpolynomial of S n with zeroes µ , . . . , µ n .Let us start by obtaining the first three leading coefficients of q n ( x ). This is awell-known result, but we give a proof in order make this section self-contained. Lemma 3.1.
The characteristic polynomial q n ( x ) of the discrete Schr¨odinger matrix S n has the form q n ( x ) = x n − n X i =1 b i ! x n − + X ≤ i 1, then the product will be zero. Thus we are looking for transpositions where | i − j | = 1. There are n − , , (2 , , . . . , ( n − , n − , ( n − , n ).The product is of the form n Y i =1 α i,σ ( i ) = ( − − x − b )( x − b ) · · · ( x − b n )( x − b i )( x − b j ) = x n − + r n − ( x ) , where r n − ( x ) is a polynomial of degree at most n − 3. Since the signature of atransposition is negative, we derive − x n − for each product. Summing over all n − P ≤ i The characteristic polynomial p n ( x ) of the free discrete Schr¨odingermatrix F n has the form p n ( x ) = x n − ( n − x n − + P n − ( x ) , where P n − ( x ) is a polynomial of degree at most n − .Proof. Simply set b i = 0 for each i ∈ { , · · · , n } and apply Lemma 3.1. (cid:3) Let us start by giving a proof of Ambarzumian problem with Dirichlet-Dirichletboundary conditions, i.e. for the matrix F n (0 , 0) = F n in our notation. EAKINS, FRENDREISS, HAT˙INO ˘GLU, LAMB, MANAGE, AND PUENTE Theorem 3.3. Suppose S n shares all of its eigenvalues with F n . Then S n = F n .Proof. In order for the two matrices to have all the same eigenvalues, they must haveequal characteristic polynomials. Comparing the results from Lemma 3.1 to Corollary3.2, we must have(3.2) n X i =1 b i = 0 and X ≤ i MBARZUMIAN-TYPE PROBLEMS FOR DISCRETE SCHR ¨ODINGER OPERATORS 7 Example 3.6. Let us define the discrete Schr¨odinger matrices A := and B := − / (1 + √ 5) 1 01 1 10 1 (1 + √ / .The matrices A and B have the same characteristic polynomial x − x − x + 2, sothey share the same spectrum.This example shows that Theorem 3.3 was a special case, so in order to get uniquenessof a rank-one perturbation of the free operator, we also need to know the non-zeroboundary condition along with the spectrum. Theorem 3.7. Suppose S n ( b, b n ) shares all of its eigenvalues with F n ( b, . Then S n ( b, b n ) = F n ( b, .Proof. Comparing coefficients of characteristic polynomials of S n ( b, b n ) and F n ( b, b + n X i =2 b i = b and b n X j =2 b j + X ≤ i 1, respectively: S n ( θ ) = b · · · e πiθ b b . . . 0... . . . . . . . . . 1 e − πiθ · · · b n , F n ( φ ) = · · · e πiφ e − πiφ · · · . The following theorem shows that with Floquet boundary conditions, the knowledgeof the spectrum of the free operator is sufficient for the uniqueness of the operator upto transpose. Theorem 3.8. Suppose that S n ( θ ) shares all of its eigenvalues with F n ( φ ) , includingmultiplicity, for ≤ θ, φ < . Then b = · · · = b n = 0 and θ = φ or θ = 1 − φ , i.e. S n ( θ ) = F n ( φ ) or S n ( θ ) = F T n ( φ ) EAKINS, FRENDREISS, HAT˙INO ˘GLU, LAMB, MANAGE, AND PUENTE Proof. Let us define D [ k, l ] as the following determinant of a ( l − k + 1) × ( l − k + 1)matrix for 1 ≤ k < l ≤ n : D [ k, l ] := (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x − b k − · · · − x − b k +1 . . . . . . ...0 . . . . . . . . . 0... . . . . . . x − b l − − · · · − x − b l (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( l − k +1) Let us consider the characteristic polynomial of S n ( θ ) by using cofactor expansion onthe first row: | x I n − S n ( θ ) | = ( x − b ) D [2 , n ] + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − − · · · − x − b − − x − b n − − − e − πiθ · · · − x − b n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( n − + ( − n +1 (cid:0) − e πiθ (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − x − b − · · · − x − b . . . . . .... . . . . . . . . . − · · · − x − b n − − e − πiθ · · · − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( n − Then by using cofactor expansions on the first row of the determinant in the secondterm and on the first column of the determinant in the third term we get | x I n − S n ( θ ) | = ( x − b ) D [2 , n ] + ( − D [3 , n ] + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − · · · x − b − − − − e − πiθ · · · − x − b n (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( n − + ( − n +1 (cid:0) − e πiθ (cid:1) ( − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − x − b − · · · − x − b . . . . . .... . . . . . . . . . − · · · . . . − x − b n − · · · · · · − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( n − + ( − n +1 (cid:0) − e πiθ (cid:1) ( − n (cid:0) − e − πiθ (cid:1) D [2 , n − MBARZUMIAN-TYPE PROBLEMS FOR DISCRETE SCHR ¨ODINGER OPERATORS 9 Now let’s use cofactor expansion on the first column of the determinant in the thirdterm. Also note that the determinant in the fourth term is the determinant of an uppertriangular matrix. Therefore, | x I n − S n ( θ ) | = ( x − b ) D [2 , n ] − D [3 , n ]+ ( − n − (cid:0) − e πiθ (cid:1) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − · · · x − b − − x − b . . . . . . 00 . . . . . . − − e − πiθ − x − b n − − (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( n − + ( − n +1 (cid:0) − e πiθ (cid:1) (cid:2) ( − − n − + ( − n (cid:0) − e − πiθ (cid:1) D [2 , n − (cid:3) Finally, noting again that the determinant in the third term is that of a lower triangularmatrix, we get(3.5) | x I n − S n ( θ ) | = ( x − b ) D [2 , n ] − D [3 , n ] + ( − n − (cid:0) − e πiθ (cid:1) ( − n − + ( − n (cid:0) − e − πiθ (cid:1) + ( − n +1 D [2 , n − x − b ) D [2 , n ] − D [3 , n ] − D [2 , n − − e πiθ − e − πiθ At this point note that D [ k, l ] is the characteristic polynomial of the following discreteSchr¨odinger matrix b k · · · b k +1 b l − · · · b l . Therefore using Lemma 3.1 and equation (3.5), we obtain(3.6) | x I n − S n ( θ ) | = x n − n X i =1 b i ! x n − + X ≤ i 3, which is independent of θ . Usingthe same steps for F n ( φ ), we obtain(3.7) | x I n − F n ( φ ) | = x n − ( n − x n − + g n − ( x ) − e πiφ − e − πiφ where g n − is a polynomial of degree at most n − 3, which is independent of φ .Comparing equations (3.6) and (3.7), like we did in the proof of Theorem 3.3, wecan conclude that the diagonal entries { b i } ni =1 of S n ( θ ) must be zero. Note that the expression consisting of the first three terms in the right end of (3.5),( x − b ) D [2 , n ] − D [3 , n ] − D [2 , n − 1] is independent of θ . In addition, we observed that b = · · · = b n = 0. Therefore using the equivalence of the characteristic polynomials of S n ( θ ) and F n ( φ ), we obtain e πiθ + e − πiθ = e πiφ + e − πiφ , which can be written using Euler’s identity as(3.8) 2 cos(2 πθ ) = 2 cos(2 πφ ) . Equation (3.8) is valid if and only if θ differs from φ or − φ by an integer. Since0 ≤ θ, φ < 1, the only possible values for θ are φ and 1 − φ . This completes theproof. (cid:3) An Ambarzumian-type mixed inverse spectral problem Let us introduce the following n × n discrete Schr¨odinger matrix for 1 ≤ m ≤ n : S n,m := b . . . 01 . . . 1 0 . . . 00 1 b m . . . . . . Let us also recall that F n denotes the free discrete Schr¨odinger matrix of size n × n .In this section our goal is to answer the following Ambarzumian-type mixed spectralproblem positively for the m = 2 case. Inverse Spectral Problem. If S n,m and F n share m consecutive eigenvalues, thendo we get b = · · · = b m = 0, i.e. S n,m = F n ?When m = 1, this problem becomes a special case of the following result of Gesztesyand Simon [15]. For a Jacobi matrix given as (2.1), let us consider the sequences { a k } and { b k } as a single sequence b , a , b , a , · · · , a n − , b n = c , c , · · · , c n − , i.e. c k − := b k and c k := a k . Theorem 4.1. ([15] , Theorem 4.2 ) Suppose that ≤ k ≤ n and c k +1 , · · · , c n − areknown, as well as k of the eigenvalues. Then c , · · · , c k are uniquely determined. By letting k = 1, we get the inverse spectral problem stated above for m = 1. Nowlet us prove the m = 2 case. Let λ < λ < · · · < λ n denote the eigenvalues of F n , andlet ˜ λ < ˜ λ < · · · < ˜ λ n denote the eigenvalues of S n, . Theorem 4.2. Let λ k = ˜ λ k and λ k +1 = ˜ λ k +1 for some k ∈ { , , . . . , n − } . Then b = 0 and b = 0 , i.e. S n, = F n .Proof. For simplicity, let us use S n instead of S n, . We start by proving the followingclaim. MBARZUMIAN-TYPE PROBLEMS FOR DISCRETE SCHR ¨ODINGER OPERATORS 11 Claim: If λ k = ˜ λ k and λ k +1 = ˜ λ k +1 , then either b = b = 0 or b = λ k + λ k +1 and b = 1 /λ k + 1 /λ k +1 .Let us consider the characteristic polynomial of S n using cofactor expansion on thelast row of λI − S n .det( λI − S n ) =( λ − b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) λ − b − . . . − λ − − λ . . . 0... . . . . . . . . . − . . . − λ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( n − + (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − − . . . − λ − − λ . . . 0... . . . . . . . . . − . . . − λ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( n − Using cofactor expansion on the first row for the first term and the first column forthe second term, we getdet( λI − S n ) = ( λ − b )( λ − b ) det( λI − F n − )+ ( λ − b ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) − − . . . λ − − λ . . . 0... . . . . . . . . . − . . . − λ (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( n − − det( λI − F n − )Finally, using cofactor expansion on the first column of the second term, we get(4.1) det( λI − S n ) = [( λ − b )( λ − b ) − 1] det( λI − F n − ) − ( λ − b ) det( λI − F n − ) . Since ˜ λ k = λ k and ˜ λ k +1 = λ k +1 , right hand side of (4.1) is zero when λ = λ k or λ = λ k +1 . Therefore for λ = λ k or λ = λ k +1 we get(4.2) ( λ − b )( λ − b ) − λ − b = det( λI − F n − )det( λI − F n − ) . Note that equation (4.2) is also valid for F n , i.e. when b = b = 0, and the righthand side of the equation does not depend on b or b and hence identical for S n and F n . Therefore the left hand side of (4.2) should also be identical for S n and F n , when˜ λ k = λ k and ˜ λ k +1 = λ k +1 . Hence,(4.3) ( λ − b )( λ − b ) − λ − b = ( λ − λ − − λ − λ = λ k or λ = λ k +1 . Therefore, λ ( λ − b )( λ − b ) − λ = ( λ − λ − b ) λ − ( b + b ) λ + b b λ − λ = λ − b λ − λ + b − b λ + b b λ − b = 0 for λ = λ or λ = λ . If b = 0, then b = 0 from the last equation above, so we canassume b = 0. Then λ − b λ + b /b = 0 for λ = λ k or λ = λ k +1 .Since x − b x + b /b is a monic polynomial with two distinct roots x = λ k and x = λ k +1 , we get x − b x + b /b = ( x − λ k )( x − λ k +1 )which implies x − b x + b /b = x − ( λ k + λ k +1 ) x + λ k λ k +1 Comparing coefficients we get our claim, since b = λ k + λ k +1 , and b /b = λ k λ k +1 implies b = b λ k λ k +1 = λ k + λ k +1 λ k λ k +1 = 1 λ k + 1 λ k +1 Now our goal is to get a contradiction for the second case of the claim, i.e. when b = λ k + λ k +1 and b = 1 /λ k + 1 /λ k +1 , so let us assume b = λ k + λ k +1 and b = 1 /λ k + 1 /λ k +1 . First let us show that b and b have the same sign. If n is even and k = n/ 2, then λ k = − λ k +1 . Hence b = b = 0. If n is odd and k = ( n − / k = ( n + 1) / 2, thenone of the eigenvalues λ k or λ k +1 is zero, so b is undefined. For all other values of k ,two consecutive eigenvalues λ k and λ k +1 and hence b and b have the same sign.Without loss of generality let us assume both λ k and λ k +1 are negative and b ≤ b .Let us define the matrices C n and M n ( t ) with the real parameter t as follows: C n := b . . . b . . . and M n ( t ) := − t . . . − t . . . . Note that kth eigenvalue of C n , denoted by λ k ( C n ), is greater than or equal to˜ λ k , since C n ≥ A n . Let us also note that M n ( − b ) = C n and M n (0) = F n . Letus denote the kth eigenvalue of M n ( t ) by λ k ( t ) and the corresponding eigenvector by X ( t ), normalized as || X ( t ) || = 1. Since M n ( t ) is a smooth function of t around 0, sameis true for λ k ( t ) and X ( t ) by Theorem 2.1 and Theorem 2.2. Let us recall that M n ( t )is self-adjoint, || X ( t ) || = 1 and M n ( t ) X ( t ) = λ k ( t ) X ( t ). Therefore by Theorem 2.3,the Hellmann-Feynman Theorem, we get(4.4) λ ′ k ( t ) = h X ( t ) , M ′ n ( t ) X ( t ) i = − X ( t ) − X ( t ) , where X ( t ) T = [ X ( t ) , X ( t ) , · · · , X n ( t )]. Since X ( t ) is a non-zero eigenvector of thetridiagonal matrix M n ( t ), at least one of X ( t ) and X ( t ) is non-zero. Therefore byequation (4.4), there exists an open interval I ⊂ R containing 0 such that λ ′ k ( t ) < t ∈ I , i.e. λ k ( t ) is decreasing on I . This implies existence of 0 < t < − b satisfying λ k < λ k ( t ) ≤ λ k ( − b ) = λ k ( C n ) ≤ ˜ λ k . MBARZUMIAN-TYPE PROBLEMS FOR DISCRETE SCHR ¨ODINGER OPERATORS 13 This contradicts with our assumption that λ k = ˜ λ k . Therefore only the first case ofthe claim is true, i.e. b = b = 0 and hence S n = F n . (cid:3) Acknowledgement The authors would like to thank Wencai Liu for introducing them this project andhis constant support. This work was partially supported by NSF DMS-2015683 andDMS-2000345. References [1] V. 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