The Inverse Spectral Problem for Jacobi-Type Pencils
In this paper we study the inverse spectral problem for Jacobi-type pencils. By a Jacobi-type pencil we mean the following pencil J₅−λJ₃, where J₃ is a Jacobi matrix and J₅ is a semi-infinite real symmetric five-diagonal matrix with positive numbers on the second subdiagonal. In the case of a specia...
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| description | In this paper we study the inverse spectral problem for Jacobi-type pencils. By a Jacobi-type pencil we mean the following pencil J₅−λJ₃, where J₃ is a Jacobi matrix and J₅ is a semi-infinite real symmetric five-diagonal matrix with positive numbers on the second subdiagonal. In the case of a special perturbation of orthogonal polynomials on a finite interval the corresponding spectral function takes an explicit form.
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 13 (2017), 085, 16 pages
The Inverse Spectral Problem for Jacobi-Type Pencils
Sergey M. ZAGORODNYUK
School of Mathematics and Computer Sciences, V.N. Karazin Kharkiv National University,
Svobody Square 4, Kharkiv 61022, Ukraine
E-mail: Sergey.M.Zagorodnyuk@gmail.com
Received June 10, 2017, in final form October 24, 2017; Published online October 28, 2017
https://doi.org/10.3842/SIGMA.2017.085
Abstract. In this paper we study the inverse spectral problem for Jacobi-type pencils. By
a Jacobi-type pencil we mean the following pencil J5 − λJ3, where J3 is a Jacobi matrix
and J5 is a semi-infinite real symmetric five-diagonal matrix with positive numbers on the
second subdiagonal. In the case of a special perturbation of orthogonal polynomials on
a finite interval the corresponding spectral function takes an explicit form.
Key words: operator pencil; recurrence relation; orthogonal polynomials; spectral function
2010 Mathematics Subject Classification: 42C05; 47B36
1 Introduction
The theory of operator pencils in Banach spaces has many applications and it is actively develo-
ping nowadays, see [7, 8, 10]. By operator pencils or operator polynomials one means polynomials
of complex variable λ whose coefficients are linear bounded operators acting in a Banach space X
L(λ) =
m∑
j=0
λjAj ,
where Aj : X → X, j = 0, . . . ,m. As it was noted in Rodman’s book [10], there exist three
broad topics in this theory which are the main foci of interest:
(a) linearization, i.e., reduction to a linear polynomial;
(b) various types of factorizations;
(c) the problems of multiple completness of eigenvectors and generalized eigenvectors.
If X is a Hilbert space and all operators Aj are self-adjoint, the pencil L(λ) is said to be self-
adjoint. In the case m = 1 (m = 2) the pencil is called linear (respectively quadratic). For
the recent progress on quadratic operator pencils and their applications we refer to the book of
Möller and Pivovarchik [8].
Linear self-adjoint operator pencils has no spectral theory which could be compatible with
the theory of single or commuting self-adjoint operators. The generalized eigenvalue problem
(A0 + λA1)x = 0, x ∈ X,
is essentially more generic than the usual eigenvalue problem for a self-adjoint operator. We
can explain this difference by the following astonishing result from the theory of linear matrix
pencils.
This paper is a contribution to the Special Issue on Orthogonal Polynomials, Special Functions and Applica-
tions (OPSFA14). The full collection is available at https://www.emis.de/journals/SIGMA/OPSFA2017.html
mailto:Sergey.M.Zagorodnyuk@gmail.com
https://doi.org/10.3842/SIGMA.2017.085
https://www.emis.de/journals/SIGMA/OPSFA2017.html
2 S.M. Zagorodnyuk
Theorem 1.1 ([9, Theorem 15.2.1, p. 341]). Any real square matrix B can be written as
B = AM−1 or B = M−1A where A and M are suitable symmetric matrices.
Nevertheless, the general spectral theory for linear operator pencils has old and new contri-
butions. In particular, it has applications to quotients of bounded operators and to the study
of electron waveguides in graphene, see [5] and references therein.
Recently, Ben Amara, Vladimirov and Shkalikov investigated the following linear pencil of
differential operators [2]
(py′′)′′ − λ(−y′′ + cry) = 0. (1.1)
The initial conditions are y(0) = y′(0) = y(1) = y′(1) = 0, or y(0) = y′(0) = y′(1) = (py′′)′(1) +
λαy(1) = 0. Here p, r ∈ C[0, 1] are uniformly positive, while the parameters c and α are real.
Equation (1.1) has several physical applications, including a motion of a partially fixed bar with
additional constraints in the elasticity theory [2].
It turns out that equation (1.1) is related to certain generalized orthogonal polynomials, just
as the Sturm–Liouville differential operator is related to the orthogonal polynomials on the real
line (OPRL).
The theory of orthogonal polynomials has numerous old and new contributions and applica-
tions, see [4, 6, 11, 12, 13, 14]. This theory is closely related to spectral problems for Jacobi
matrices. The direct and inverse spectral problems for Jacobi matrices (with matrix elements)
were described, e.g., in [1, 3]. For various other types of spectral problems we refer to historical
notes in [15]. Recall the following basic definition from [16].
Definition 1.2. A set Θ = (J3, J5, α, β), where α > 0, β ∈ R, J3 is a Jacobi matrix and J5
is a semi-infinite real symmetric five-diagonal matrix with positive numbers on the second
subdiagonal, is said to be a Jacobi-type pencil (of matrices).
From this definition we see that matrices J3 and J5 have the following form
J3 =
b0 a0 0 0 0 · · ·
a0 b1 a1 0 0 · · ·
0 a1 b2 a2 0 · · ·
...
...
...
. . .
, ak > 0, bk ∈ R, k ∈ Z+, (1.2)
J5 =
α0 β0 γ0 0 0 0 · · ·
β0 α1 β1 γ1 0 0 · · ·
γ0 β1 α2 β2 γ2 0 · · ·
0 γ1 β2 α3 β3 γ3 · · ·
...
...
...
...
. . .
, αn, βn ∈ R, γn > 0, n ∈ Z+. (1.3)
With a Jacobi-type pencil of matrices Θ one associates a system of polynomials {pn(λ)}∞n=0,
such that
p0(λ) = 1, p1(λ) = αλ+ β,
and
(J5 − λJ3)~p(λ) = 0, (1.4)
where ~p(λ) = (p0(λ), p1(λ), p2(λ), . . . )T. Here the superscript T means the transposition. Poly-
nomials {pn(λ)}∞n=0 are said to be associated to the Jacobi-type pencil of matrices Θ.
The Inverse Spectral Problem for Jacobi-Type Pencils 3
Observe that for each system of orthonormal polynomials on the real line with p0 = 1 one may
choose J3 to be the corresponding Jacobi matrix (which elements are the recurrence coefficients),
J5 = J2
3 , and α, β being the coefficients of p1 (p1(λ) = αλ+ β).
One can rewrite relation (1.4) in the scalar form
γn−2pn−2(λ) + (βn−1 − λan−1)pn−1(λ) + (αn − λbn)pn(λ)
+ (βn − λan)pn+1(λ) + γnpn+2(λ) = 0, n ∈ Z+, (1.5)
where p−2(λ) = p−1(λ) = 0, γ−2 = γ−1 = a−1 = β−1 = 0.
Let us return to the differential equation (1.1) and explain how it is related to polynomials
from (1.5). The idea is the same as in the classical case. We consider a partition of the inter-
val [0, 1]
0 = x0 < x1 < x2 < · · · < xN−1 < xN = 1,
with a uniform step h. Set yj := y(xj), pj := p(xj), rj := r(xj), 0 ≤ j ≤ N . Then we replace the
derivatives in (1.1) by the corresponding discretizations y′(xj) ≈ yj+1−yj
h , y′′(xj) ≈ yj+1−2yj+yj−1
h2
.
We obtain the following equation
pj−1yj−2 − 2(pj + pj−1)yj−1 + (pj+1) + 4pj + pj−1)yj − 2(pj+1 + pj)yj+1
+ pj+1yj+2 − λh2
(
yj−1 −
(
2− h2crj
)
yj + yj+1
)
= 0.
Denote
αj = pj+1 + 4pj + pj−1, βj = −2(pj+1 + pj), γj = pj+1,
aj = h2, bj =
(
−2− h2crj
)
h2.
Then we get
γj−2yj−2 + βj−1yj−1 + αjyj + βjyj+1 + γjyj+2 + λ
(
aj−1yj−1 + bjyj + ajyj+1
)
= 0.
If we set λ̃ = −λ, we obtain a relation of the form (1.5). Thus, relation (1.5) forms a discrete
grid model for the equation (1.1). The boundary conditions y(0) = 0, y′(0) = 0 correspond to
the convention y−2 = y−1 = 0. Instead of [0, 1] we can consider [0,+∞). The conditions on the
right end of the bar we replace by conditions on the left end which are controlled by α and β.
In this paper we shall introduce and study the direct and inverse spectral problems for
Jacobi type pencils. We shall also consider a special perturbation of the class of orthonormal
polynomials on a finite real interval. In that case it is possible to obtain a more convenient
integral representation for the corresponding spectral function.
Notations. As usual, we denote by R, C, N, Z, Z+, the sets of real numbers, complex
numbers, positive integers, integers and non-negative integers, respectively. By P we denote the
set of all polynomials with complex coefficients.
By l2 we denote the usual Hilbert space of all complex sequences c = (cn)∞n=0 = (c0, c1, c2, . . . )
T
with the finite norm ‖c‖l2 =
√
∞∑
n=0
|cn|2. Here T means the transposition. The scalar product
of two sequences c = (cn)∞n=0, d = (dn)∞n=0 ∈ l2 is given by (c, d)l2 =
∞∑
n=0
cndn. We denote
~em = (δn,m)∞n=0 ∈ l2, m ∈ Z+. By l2,fin we denote the set of all finite vectors from l2, i.e.,
vectors with all, but finite number, components being zeros. By diag(c1, c2, c3, . . . ) we mean
a semi-infinite matrix with cj ∈ C in j-th column and j-th row and zeros outside the main
diagonal.
4 S.M. Zagorodnyuk
By B(R) we denote the set of all Borel subsets of R. If σ is a (non-negative) bounded measure
on B(R) then by L2
σ we denote a Hilbert space of all (classes of equivalences of) complex-valued
functions f on R with a finite norm ‖f‖L2
σ
=
√∫
R |f(x)|2dσ. The scalar product of two functions
f, g ∈ L2
σ is given by (f, g)L2
σ
=
∫
R f(x)g(x)dσ. By [f ] we denote the class of equivalence in L2
σ
which contains the representative f . By P we denote a set of all (classes of equivalence which
contain) polynomials in L2
σ. As usual, we sometimes use the representatives instead of their
classes in formulas. Let B be an arbitrary linear operator in L2
σ with the domain P. Let
f(λ) ∈ P be nonzero and of degree d ∈ Z+, f(λ) =
d∑
k=0
dkλ
k, dk ∈ C. We set
f(B) =
d∑
k=0
dkB
k, B0 := E|P .
If f ≡ 0, then f(B) := 0|P .
If H is a Hilbert space then (·, ·)H and ‖ · ‖H mean the scalar product and the norm in H,
respectively. Indices may be omitted in obvious cases. For a linear operator A in H, we denote
by D(A) its domain, by R(A) its range, by KerA its null subspace (kernel), and A∗ means the
adjoint operator if it exists. If A is invertible then A−1 means its inverse. A means the closure
of the operator, if the operator is closable. If A is bounded then ‖A‖ denotes its norm. For
a set M ⊆ H we denote by M the closure of M in the norm of H. By LinM we mean the set
of all linear combinations of elements of M , and spanM := LinM . By E = EH (0 = 0H) we
denote the identity operator in H, i.e., EHx = x, x ∈ H (respectively the null operator in H,
i.e., 0Hx = 0, x ∈ H). If H1 is a subspace of H, then PH1 = PHH1
is an operator of the orthogonal
projection on H1 in H.
2 Preliminaries
In this section, for the convenience of the reader, we recall basic definitions and results from [16].
Then we state the direct and inverse spectral problems for the Jacobi-type pencils. The direct
problem will be solved immediately, while the inverse problem will be considered in the next
section. Set
un := J3~en = an−1~en−1 + bn~en + an~en+1,
wn := J5~en = γn−2~en−2 + βn−1~en−1 + αn~en + βn~en+1 + γn~en+2, n ∈ Z+.
Here and in what follows by ~ek with negative k we mean (vector) zero. The following operator
Af =
ζ
α
(~e1 − β~e0) +
∞∑
n=0
ξnwn,
f = ζ~e0 +
∞∑
n=0
ξnun ∈ l2,fin, ζ, ξn ∈ C, (2.1)
with D(A) = l2,fin is said to be the associated operator for the Jacobi-type pencil Θ. Notice that
in the sums in (2.1) only finite number of ξn are nonzero. We shall always assume this in the
case of elements from the linear span. In particular, we have
AJ3~en = J5~en, n ∈ Z+,
and therefore
AJ3 = J5.
The Inverse Spectral Problem for Jacobi-Type Pencils 5
As usual, the matrices J3 and J5 define linear operators with the domain l2,fin which we denote
by the same letters.
For an arbitrary non-zero polynomial f(λ) ∈ P of degree d ∈ Z+, f(λ) =
d∑
k=0
dkλ
k, dk ∈ C,
we set f(A) =
d∑
k=0
dkA
k. Here A0 := E|l2,fin . For f(λ) ≡ 0, we set f(A) = 0|l2,fin . The following
relations hold [16]
~en = pn(A)~e0, n ∈ Z+, (2.2)(
pn(A)~e0, pm(A)~e0
)
l2
= δn,m, n,m ∈ Z+. (2.3)
Denote by {rn(λ)}∞n=0, r0(λ) = 1, the system of polynomials satisfying
J3~r(λ) = λ~r(λ), ~r(λ) = (r0(λ), r1(λ), r2(λ), . . . )T. (2.4)
These polynomials are orthonormal on the real line with respect to a (possibly non-unique)
non-negative finite measure σ on B(R) (Favard’s theorem). Consider the following operator
U
∞∑
n=0
ξn~en =
[ ∞∑
n=0
ξnrn(x)
]
, ξn ∈ C, (2.5)
which maps l2,fin onto P. Here by P we denote a set of all (classes of equivalence which contain)
polynomials in L2
σ. Denote
A = Aσ = UAU−1. (2.6)
The operator A = Aσ is said to be the model representation in L2
σ of the associated operator A.
Theorem 2.1 ([16]). Let Θ = (J3, J5, α, β) be a Jacobi-type pencil. Let {rn(λ)}∞n=0, r0(λ) = 1,
be a system of polynomials satisfying (2.4) and σ be their (arbitrary) orthogonality measure
on B(R). The associated polynomials {pn(λ)}∞n=0 satisfy the following relations∫
R
pn(A)(1)pm(A)(1)dσ = δn,m, n,m ∈ Z+, (2.7)
where A is the model representation in L2
σ of the associated operator A.
We have now recalled all basic notations and results which we shall need in the present paper.
We introduce the following definition.
Definition 2.2. Let Θ = (J3, J5, α, β) be a Jacobi-type pencil and {pn(λ)}∞n=0 be the associated
polynomials to Θ. A sesquilinear functional S(u, v), u, v ∈ P, satisfying the following relation
S(pn, pm) = δn,m, n,m ∈ Z+,
is said to be the spectral function of the Jacobi-type pencil Θ.
In a usual manner, we may introduce the corresponding direct and inverse spectral problems.
The direct spectral problem for a Jacobi-type pencil Θ consists in searching for answers on the
following questions:
1) Does the spectral function exist?
2) If the spectral function exists, is it unique?
3) If the spectral function exists, how to find it (or them)?
6 S.M. Zagorodnyuk
By relation (2.3) we see that the spectral function always exists. By the linearity it follows
that the spectral function is unique. It has the following representation
S(u, v) =
(
u(A)~e0, v(A)~e0
)
l2
, u, v ∈ P, (2.8)
where A is the associated operator for Θ. By (2.8) we obtain that
S(u, v) = S(v, u), u, v ∈ P, (2.9)
and
S(u, u) ≥ 0, u ∈ P. (2.10)
Moreover, the following integral representation holds
S(u, v) =
∫
R
u(Aσ)(1)v(Aσ)(1)dσ, u, v ∈ P. (2.11)
Thus, the direct spectral problem for Θ is solved in full.
The inverse spectral problem for a Jacobi-type pencil Θ consists in searching for answers on
the following questions:
(a) Is it possible to reconstruct the Jacobi-type pencil Θ = (J3, J5, α, β) using its spectral
function? If it is possible, what is the procedure of the reconstruction?
(b) What are necessary and sufficient conditions for a sesquilinear functional σ(u, v), u, v ∈ P,
to be the spectral function of a Jacobi-type pencil Θ = (J3, J5, α, β)?
Questions (a) and (b) will be investigated in the next section.
3 The inverse spectral problem for a Jacobi-type pencil
An answer to the question (b) of the inverse spectral problem is given by the following theorem.
Theorem 3.1. A sesquilinear functional S(u, v), u, v ∈ P, satisfying relations (2.9), (2.10),
is the spectral function of a Jacobi-type pencil if and only if it admits the following integral
representation
S(u, v) =
∫
R
u(A)(1)v(A)(1)dσ, u, v ∈ P, (3.1)
where σ is a non-negative measure on B(R) with all finite power moments,∫
R
dσ = 1,
∫
R
|g(x)|2dσ > 0,
for any non-zero complex polynomial g, and A is a linear operator in L2
σ with the following
properties:
(i) D(A) = P;
(ii) for each k ∈ Z+ it holds
Axk = ξk,k+1x
k+1 +
k∑
j=0
ξk,jx
j , (3.2)
where ξk,k+1 > 0, ξk,j ∈ R, 0 ≤ j ≤ k;
The Inverse Spectral Problem for Jacobi-Type Pencils 7
(iii) the operator AΛ0 is symmetric, here by Λ0 we denote the operator of the multiplication by
an independent variable in L2
σ restricted to P.
Proof. Necessity. Let Θ = (J3, J5, α, β) be a Jacobi-type pencil and A be the associated
operator of Θ. Define polynomials {rn(λ)}, the measure σ, the operators U and A = Aσ as in
the introduction, see (2.4), (2.5), (2.6). Let S(u, v) be the spectral function of the pencil Θ. It
has an integral representation (2.11). It remains to check that σ and A possess all the required
properties in the statement of the theorem.
Since σ is generated by the Jacobi matrix J3, it has all required properties as in the statement
of the theorem. The operator Aσ is linear and defined on P.
Lemma 3.2. The associated operator A of a Jacobi-type pencil Θ has the following properties:
1) for each n ∈ Z+ it holds
~en = dn,nA
n~e0 +
n−1∑
j=0
dn,jA
j~e0, (3.3)
where dn,n > 0, dn,j ∈ R, 0 ≤ j ≤ n− 1;
2) for each n ∈ Z+ it holds
An~e0 = cn,n~en +
n−1∑
j=0
cn,j~ej , (3.4)
where cn,n > 0, cn,j ∈ R, 0 ≤ j ≤ n− 1;
3) for each n ∈ Z+ it holds
A~en = fn,n+1~en+1 +
n∑
j=0
fn,j~ej , (3.5)
where fn,n+1 > 0, fn,j ∈ R, 0 ≤ j ≤ n.
Proof of Lemma 3.2. Let pn(λ) =
n∑
j=0
dn,jλ
j , dn,n > 0, dn,j ∈ R, be the associated polyno-
mials of the pencil. By (2.2) we conclude that relation (3.3) holds.
Let us check relation (3.4) by the induction argument. For the cases n = 0, 1 relation (3.4)
obviously holds, see the definition of A. Assume that (3.4) holds for n = 1, 2, . . . , r, r ∈ N.
By (3.3) with n = r + 1 we may write
Ar+1~e0 =
1
dr+1,r+1
~er+1 −
1
dr+1,r+1
r∑
j=0
dr+1,jA
j~e0
=
1
dr+1,r+1
~er+1 +
r∑
j=0
j∑
m=0
(−1)dr+1,jcj,m
dr+1,r+1
~em.
Finally, in order to prove relation (3.5) we apply the operator A to the both sides of relation (3.3)
and use (3.4)
A~en = dn,nA
n+1~e0 +
n−1∑
j=0
dn,jA
j+1~e0
= dn,ncn+1,n+1~en+1 +
n∑
j=0
dn,ncn+1,j~ej +
n−1∑
j=0
j+1∑
k=0
dn,jcj+1,k~ek, n ∈ Z+.
Lemma 3.2 is proved. �
8 S.M. Zagorodnyuk
Return to the proof of Theorem 3.1. Choose an arbitrary k ∈ Z+. Observe that
xk =
k∑
j=0
gk,jrj(x), gk,j ∈ R, gk,k > 0.
By (3.5) we may write
Aσ
[
xk
]
=
k∑
j=0
gk,jUA~ej =
k∑
j=0
gk,j
(
fj,j+1[rj+1] +
j∑
l=0
fj,l[rl]
)
= gk,kfk,k+1[rk+1] +
k−1∑
j=0
gk,jfj,j+1rj+1 +
k∑
j=0
gk,j
j∑
l=0
fj,lrl
=
[
gk,kfk,k+1 ˜µk+1,k+1x
k+1 + gk,kfk,k+1
k∑
m=0
µ̃k+1,mx
m
+
k−1∑
j=0
gk,jfj,j+1rj+1 +
k∑
j=0
gk,j
j∑
l=0
fj,lrl
, (3.6)
where we set rl(λ) =
l∑
j=0
µ̃l,jλ
j , µ̃l,l > 0, µl,j ∈ R. Since we get a real polynomial of degree k+ 1
on the right-hand side of (3.6) and it has a positive leading coefficient gk,kfk,k+1 ˜µk+1,k+1, then
relation (3.2) is proved.
It remains to verify (iii). For arbitrary n,m ∈ Z+ we may write
(AσΛ0[rn(x)], [rm(x)]) =
(
UAU−1[an−1rn−1(x) + bnrn(x) + anrn+1(x)], [rm(x)]
)
= (AJ3~en, ~em) = (J5~en, ~em) = (~en, J5~em)
= (~en, A(am−1~em−1 + bm~em + am~em+1))
= ([rn(x)],AσΛ0[rm(x)]). (3.7)
By the linearity we conclude that AσΛ0 is symmetric.
Sufficiency. Suppose that a sesquilinear functional S(u, v), u, v ∈ P, satisfying relations (2.9),
(2.10), is given. Assume that it has the integral representation (3.1) where σ is a non-negative
probability measure on B(R) with all finite moments,
∫
R |g(x)|2dσ > 0, for any non-zero complex
polynomial g, and A is a linear operator in L2
σ with properties (i)–(iii). By condition (ii) and
the induction argument it can be directly verified that for each n ∈ Z+ it holds
An[1] = an,n[xn] +
n−1∑
j=0
an,j
[
xj
]
,
where an,n > 0, an,j ∈ R, 0 ≤ j ≤ n− 1.
Suppose that S(u, u) = 0 for a complex polynomial u. Then
0 = ‖u(A)(1)‖2L2
σ
.
Assume that u is nonzero, u(λ) =
n∑
k=0
dkλ
k, dn 6= 0, dk ∈ C, n ∈ Z+. Observe that
u(A)(1) =
(
n∑
k=0
dkAk
)
[1] = dnAn[1] +
n−1∑
k=0
dkAk[1]
The Inverse Spectral Problem for Jacobi-Type Pencils 9
= dn
an,n[xn]+
n−1∑
j=0
an,jx
j
+
n−1∑
k=0
dk
k∑
j=0
ak,j
[
xj
]
=
dnan,nxn + dn
n−1∑
j=0
an,jx
j +
n−1∑
k=0
k∑
j=0
dkak,jx
j
=: [r(x)].
We have a nonzero polynomial r (of degree n) with ‖r‖L2
σ
= 0. This contradicts to our assump-
tions on the measure σ. Consequently, u ≡ 0.
The functional S defines an inner product on P. Thus, the complex vector space P becomes
a space H with a scalar product. It is a normed space with the norm ‖p‖ =
√
S(p, p). We shall
not need its completion. Set
p0(λ) =
1
‖1‖H
= 1, p1(λ) =
λ− S(x, 1)1
‖λ− S(x, 1)1‖H
.
Let us check that p1(λ) is well-defined. Denote by {sk}∞k=0 the power moments of σ
sk =
∫
R
xkdσ, k ∈ Z+.
Let
∆n := det(sk+l)
n
k,l=0, n ∈ Z+, ∆−1 := 1,
be the corresponding Hankel determinants. Observe that
S(x, 1) =
∫
A(1)dσ =
∫
(ξ0,1λ+ ξ0,0)dσ = ξ0,1s1 + ξ0,0 ∈ R,
S(x, x) =
∫
A(1)A(1)dσ =
∫
(ξ0,1λ+ ξ0,0)
2dσ = ξ20,1s2 + 2ξ0,1ξ0,0s1 + ξ20,0.
We may write
‖λ− S(x, 1)1‖2H = S(λ− S(x, 1)1, λ− S(x, 1)1) = S(λ, λ)− (S(λ, 1))2
= ξ20,1
(
s2 − s21
)
= ξ20,1∆1 > 0,
where we have used our assumptions on the measure σ. Then
p1(λ) =
λ− ξ0,1s1 − ξ0,0
ξ0,1
√
∆1
.
Set
α =
1
ξ0,1
√
∆1
, β = −ξ0,1s1 + ξ0,0
ξ0,1
√
∆1
. (3.8)
Let {rn(λ)}∞n=0 be orthonormal polynomials with respect to the measure σ (having positive
leading coefficients). Denote by J3 the corresponding Jacobi matrix (formed by the recurrence
coefficients of rn). Denote
J5 = (gm,n)∞m,n=0, gm,n := (AΛ0[rn(λ)], [rm(λ)])L2
σ
.
By condition (iii) of the theorem we conclude that J5 is a symmetric semi-infinite matrix. Let
rn(λ) =
n∑
k=0
ηn,kλ
k, ηn,k ∈ R, ηn,n > 0, n ∈ Z+.
10 S.M. Zagorodnyuk
For an arbitrary n ∈ Z+ by condition (ii) we may write
AΛ0[rn(λ)] =
n∑
k=0
ηn,kA
[
λk+1
]
=
[
ηn,nξn+1,n+2λ
n+2 + dn+1(λ)
]
, (3.9)
where dn+1(λ) is a zero polynomial or a polynomial with real coefficients, deg p ≤ n+ 1. Then
gm,n = 0, m, n ∈ Z+, m > n+ 2,
and
gn+2,n = ηn,nξn+1,n+2
([
λn+2
]
, [rn+2(λ)]
)
=
ηn,nξn+1,n+2
ηn+2,n+2
> 0, n ∈ Z+.
By (3.9) we also see that gm,n are real numbers. We conclude that J5 is real five-diagonal and
it has positive numbers on the second sub-diagonal.
Consider a Jacobi-type pencil Θ̃ = (J3, J5, α, β). Let {pn(λ)}∞n=0 be the associated poly-
nomials to Θ̃. For the pencil Θ̃ we define the standard objects from the introduction. The
polynomials rn(λ) (see (2.4)) coincide with rn(λ). We may also choose the given σ as the
orthogonality measure (the orthogonality measure can be non-unique). The operators A, U
and Aσ we define in the standard way, see (2.1), (2.5), (2.6). We only can not use the brief
notation A for Aσ, since A already denotes the operator in the integral representation of S.
For the pencil Θ̃ we may apply our arguments in the proved Necessity. By relation (3.7) we
may write
(AσΛ0[rn], [rm])L2
σ
= (J5~en, ~em)l2 = gm,n = (AΛ0[rn], [rm])L2
σ
, n,m ∈ Z+.
Here the last equality follows by the definition of the matrix J5. Therefore
Aσ[λp(λ)] = A[λp(λ)], ∀ p ∈ P. (3.10)
Notice that
Aσ[1] = UA~e0 =
1
α
U(~e1 − β~e0) =
[
1
α
(r1(λ)− β)
]
. (3.11)
The orthonormal polynomial r1(λ) has the following form
r1(λ) =
1√
∆1
(λ− s1). (3.12)
By (3.11), (3.12) and (3.8) we conclude that
Aσ[1] = [ξ0,1λ+ ξ0,0].
On the other hand, by property (ii) we have A[1] = [ξ0,1λ+ ξ0,0]. Therefore
Aσ[1] = A[1]. (3.13)
Relations (3.10) and (3.13) show that Aσ = A. Comparing relations (2.11) and (3.1) we see
that the spectral function of Θ̃ coincides with S. �
Theorem 3.1 provides characteristic properties for the model representation A of the asso-
ciated operator of a Jacobi-type pencil. It is seen that these properties are close to the properties
of the multiplication operator Λ0. Of course, Λ0 itself satisfies properties (i)–(iii).
The Inverse Spectral Problem for Jacobi-Type Pencils 11
Corollary 3.3. Let σ be a non-negative measure on B(R) with all finite power moments,∫
R dσ = 1,
∫
R |g(x)|2dσ > 0, for any non-zero complex polynomial g. A linear operator A
in L2
σ is a model representation in L2
σ of the associated operator of a Jacobi-type pencil if and
only if properties (i)–(iii) of Theorem 3.1 hold.
Proof. It follows directly from our constructions in the proof of Theorem 3.1. �
Let Θ = (J3, J5, α, β) be a Jacobi-type pencil and A be a model representation in L2
σ of the
associated operator of Θ. By the latter corollary we conclude that AΛ0 is symmetric
(AΛ0[u(λ)], [v(λ)])L2
σ
= ([u(λ)],AΛ0[v(λ)])L2
σ
, u, v ∈ P. (3.14)
Suppose that the orthogonality measure σ is supported inside a finite real segment [a, b], 0 <
a < b < +∞, i.e., σ(R\[a, b]) = 0. In this case the operator Λ of the multiplication by an
independent variable has a bounded inverse on the whole L2
σ. By (3.14) we may write(
Λ−1A[λu(λ)], [λv(λ)]
)
L2
σ
=
(
Λ−1[λu(λ)],A[λv(λ)]
)
L2
σ
, u, v ∈ P.
Denote P0 = ΛP and A0 = A|P0 . Then(
Λ−1A0f, g
)
L2
σ
=
(
Λ−1f,A0g
)
L2
σ
, f, g ∈ P0.
Thus, in this case A0 is symmetric with respect to the form (Λ−1·, ·)L2
σ
. Analogous arguments
were used in the theory of operator pencils, see [7, Chapter IV, p. 163].
Example 3.4. Let an = 1, bn = c, c > 2, αn ∈ R, βn = 0, γn = 1, n ∈ Z+. Define J3 and J5
by (1.2), (1.3) with the above parameters. Consider a Jacobi-type pencil Θ = (J3, J5, α, β), with
arbitrary α > 0 and β ∈ R. In this case
rn(x) = Un
(
x− c
2
)
, n ∈ Z+,
where Un(t) = sin((n+1) arccos t)√
1−t2 is Chebyshev’s polynomial of the second kind. The orthonormality
relations for rn(x) have the following form∫ c+2
c−2
rn(λ)rm(λ)
√
1−
(
λ− c
2
)2 1
π
dλ = δn,m, n,m ∈ Z+.
By the recurrence relation (1.5) we calculate
p2(λ) = αλ2 + (c+ β)λ− α0.
Since (c + β)2 + 4αα0 can be made zero or negative by a proper choice of α0, then we have
a non-classical case. Thus, this case is worthy of an additional investigation which will be done
elsewhere.
We can state the following moment problem for Jacobi-type pencils: find a non-negative
measure σ on B(R) with all finite power moments,
∫
R dσ = 1,
∫
R |g(x)|2dσ > 0, for any non-
zero complex polynomial g, and a linear operatorA in L2
σ with properties (i)–(iii) of Theorem 3.1
such that∫
R
Am(1)An(1)dσ = sm,n, m, n ∈ Z+,
where {sm,n}∞m,n=0 is a prescribed set of real numbers (called moments).
As usual, there appear three important questions: 1) the solvability of the moment problem;
2) the uniqueness of a solution (the determinateness); 3) a description of all solutions. This
moment problem will be studied elsewhere.
12 S.M. Zagorodnyuk
4 A special perturbation of orthogonal polynomials
on a finite interval
If we look at the orthogonality relations (2.7) or at the representation of the spectral func-
tion (2.11), we can see that we need to calculate a polynomial of the operator A. However,
we do not have at hand any functional calculus for the operator A. It remains to calculate
the powers of A recurrently. It would be helpful to omit this procedure and to have a more
transparent relation. It is possible to do this in the following special case.
Consider a Jacobi-type pencil Θ = (J3, J5, α, β), with J3, J5 defined by (1.2), (1.3), satisfying
the following conditions
σ(R\[−c, c]) = 0, 0 < c < 1, (4.1)
J5 = aJ2
3 + bJ3 + ddiag(1, 0, 0, 0, . . . ), a > 0, b, d ∈ R,
α =
1
aa0
, β = − b0
a0
− b
aa0
. (4.2)
Here σ, as before (see the introduction), is the orthogonality measure for polynomials {rn(x)}∞n=0
(related to J3).
Introduce other related objects from the introduction: the space L2
σ, the operators A, U and
A = UAU−1. Observe that
UJ3U
−1[p(x)] = [xp(x)], p ∈ P.
For an arbitrary n ∈ Z+ we may write
A[xrn(x)] = UAU−1UJ3U
−1[rn(x)] = UAJ3~en = UJ5~en
= U
(
aJ2
3 + bJ3 + ddiag(1, 0, 0, 0, . . . )
)
~en
= aUJ3U
−1UJ3~en + bUJ3~en + dU diag(1, 0, 0, 0, . . . )~en
=
[
(ax+ b)xrn(x) + d(rn, r0)L2
σ
]
.
By the linearity we get
A[xp(x)] =
[
(ax+ b)xp(x) + d(p, 1)L2
σ
]
, p ∈ P.
Moreover, we have
A[1] = UAU−1U~e0 = UA~e0 = U
1
α
(~e1 − β~e0) =
[
1
α
(r1(x)− β)
]
.
Choose an arbitrary polynomial q(x) ∈ P. We may write q(x) = x q(x)−q(0)x + q(0). Then
A[q(x)] = A
[
x
q(x)− q(0)
x
]
+A[q(0)]
=
[(
ax2 + bx
)q(x)− q(0)
x
+ d
(
q(x)− q(0)
x
, 1
)
L2
σ
+
q(0)
α
(r1(x)− β)
]
=
[
(ax+ b)q(x) + d
(
q(x)− q(0)
x
, 1
)
L2
σ
]
.
Denote
sk :=
∫
R
xkdσ, k ∈ Z+, s−1 := 0.
The Inverse Spectral Problem for Jacobi-Type Pencils 13
Observe that
A
[ ∞∑
k=0
ckx
k
]
=
∞∑
k=0
ckA
[
xk
]
=
∞∑
k=0
ck
[
axk+1 + bxk + dsk−1
]
, ck ∈ C,
where all but finite number of ck are zeros (i.e., A acts on polynomials).
Observe that the operatorA can be unbounded: the corresponding example will be considered
below. Thus, there appears a question: how to simplify the calculation of u(A) for a complex
polynomial u in the integral representation (2.11)? Is there any (at least) polynomial calculus
for A? The answer is affirmative. Consider the following transformation from L2
σ to l2
G
[ ∞∑
k=0
ckx
k
]
=
∞∑
k=0
ck~ek, ck ∈ C.
Here all but finite number of ck are zeros. This will be assumed in what follows when dealing
with operators on polynomials. Since σ satisfies conditions for the integral representation (3.1)
(see the Necessity of the proof of Theorem 3.1), then
∫
R |g(x)|2dσ > 0, for any non-zero complex
polynomial g. This shows that two different polynomials can not belong to the same class of the
equivalence in L2
σ. Thus, the operator G is well-defined. It is a linear operator with D(G) = P
and R(G) = l2,fin. Moreover, G is invertible, since the polynomial is uniquely determined by its
coefficients. Set
A = GAG−1.
Then A is a linear operator in l2 with D(A) = l2,fin. Consider the following shift operator on
the whole l2
Sx =
∞∑
k=0
ck~ek+1, x ∈ l2, x = (ck)
∞
k=0, ck ∈ C.
Notice that S is linear and ‖Sx‖ = ‖x‖, ∀x ∈ l2. We may write
A
( ∞∑
k=0
ck~ek
)
= (aS + bE)
∞∑
k=0
ck~ek +
(
d
∞∑
k=0
cksk−1
)
~e0.
Here all but finite number of ck ∈ C are zeros. This will be assumed in the sequel for operators
on l2,fin. By condition (4.1) we see that
|sn| =
∣∣∣∣∫
R
xnχ[−c,c](x)dσ
∣∣∣∣ ≤ ∫
R
∣∣xnχ[−c,c](x)
∣∣dσ ≤ cn, n ∈ Z+,
where χ[−c,c](x) is the characteristic function of the segment [−c, c]. Therefore ~s :=
∞∑
k=1
sk−1~ek
belongs to l2. Consequently, we may write
Aw = (aS + bE)w + d(w,~s)l2~e0, w ∈ l2,fin.
Set
Âw = (aS + bE)w + d(w,~s)l2~e0, w ∈ l2.
The linear operator  is an extension of A. Observe that the operator  is bounded and∥∥Â∥∥ ≤ a+ |b|+ |d|‖~s‖l2 .
14 S.M. Zagorodnyuk
We can apply Riesz’s calculus for Â. For an arbitrary polynomial u ∈ P we may write
G
(
u
(
A
)
[1]
)
= u
(
Â
)
~e0 = − 1
2πi
∫
γ
u(z)Rz
(
Â
)
dz · ~e0 = − 1
2πi
∫
γ
u(z)
(
Rz
(
Â
)
~e0
)
dz, (4.3)
where γ is a circle centered at zero with a radius ρ bigger then a+ |b|+ |d|‖~s‖l2 , and Rz
(
Â
)
=(
Â− zE
)−1
. Here the last integral converges in the norm of l2.
Let us calculate Rz(Â)~e0 =: ~f , z ∈ γ. We may write
~e0 =
(
Â− zE
)
~f = (aS + bE)~f + d
(
~f,~s
)
l2
~e0 − z ~f.
Then
1
a
(
1− d
(
~f,~s
)
l2
)
~e0 =
(
S −
(
z − b
a
)
E
)
~f.
Since z ∈ γ, then
∣∣ z−b
a
∣∣ > 1 and
~f =
1− d(~f,~s)l2
a
(
S −
(
z − b
a
)
E
)−1
~e0. (4.4)
Denote ~u =
∞∑
k=0
uk~ek :=
(
S −
(
z−b
a
)
E
)−1
~e0. Then
~e0 =
(
S −
(
z − b
a
)
E
)
~u = S~u+
(
b− z
a
)
~u.
For the components of ~u we obtain the following equations(
b− z
a
)
u0 = 1, un−1 +
(
b− z
a
)
un = 0, n ∈ N.
Then
~u =
∞∑
k=0
(−1)
(
a
z − b
)k+1
~ek.
By (4.4) we may write
~f = τ
∞∑
k=0
(−1)
(
a
z − b
)k+1
~ek, τ ∈ C,
where
τ =
1− d
(
~f,~s
)
l2
a
.
Then
aτ = 1− d
(
τ
∞∑
k=0
(−1)
(
a
z − b
)k+1
~ek, ~s
)
l2
,
and
τ =
1
a+ d
∞∑
k=0
(−1)
(
a
z−b
)k+1
sk−1
.
The Inverse Spectral Problem for Jacobi-Type Pencils 15
Denote
s(z) :=
∞∑
k=0
(−1)
(
a
z − b
)k+1
sk−1, z ∈ γ,
~v(z) :=
∞∑
k=0
(−1)
(
a
z − b
)k+1
~ek, z ∈ γ.
Then
Rz
(
Â
)
~e0 = ~f =
1
a+ ds(z)
~v(z), z ∈ γ.
Thus, we obtained a transparent expression for Rz
(
Â
)
~e0. It can be used in relation (4.3) and it
gives a transparent expression for u(A)[1].
Example 4.1. Let Θ be a Jacobi-type pencil with α = β =
√
2, ak =
√
2, bk = 2, k ∈ Z+,
αn = βn = 0, γn = 1, n ∈ Z+, and J3, J5 have form (1.2), (1.3). This Jacobi type pencil was
considered in [16] and explicit formulas for the associated polynomials pn(λ) were obtained.
Let κ be an arbitrary positive number bigger than 2 + 2
√
2. Consider a Jacobi-type pencil
Θ̂ =
(
Ĵ3, Ĵ5, α̂, β̂
)
, where
Ĵ3 =
1
κ
J3, Ĵ5 =
1
κ
J5, α̂ = α, β̂ = β.
Notice that the associated polynomials and the associated operators for Θ and Θ̂ are the same.
It was shown in [16] that the associated operator for Θ is unbounded. Thus, the associated
operator for the pencil Θ̂ is unbounded, as well. The pencil Θ̂ satisfies conditions (4.1), (4.2)
with
c =
2 + 2
√
2
κ
, a =
1
2
κ, b = −2, d =
1
κ
.
Acknowledgements
The author is grateful to referees for their valuable comments and suggestions which led to an
essential improvement of the paper.
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16 S.M. Zagorodnyuk
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https://arxiv.org/abs/1508.01794
1 Introduction
2 Preliminaries
3 The inverse spectral problem for a Jacobi-type pencil
4 A special perturbation of orthogonal polynomials on a finite interval
References
|
| id | nasplib_isofts_kiev_ua-123456789-149264 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1815-0659 |
| language | English |
| last_indexed | 2025-12-07T18:43:15Z |
| publishDate | 2017 |
| publisher | Інститут математики НАН України |
| record_format | dspace |
| spelling | Zagorodnyuk, S.M. 2019-02-19T19:31:38Z 2019-02-19T19:31:38Z 2017 The Inverse Spectral Problem for Jacobi-Type Pencils / S.M. Zagorodnyuk // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 16 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 42C05; 47B36 DOI:10.3842/SIGMA.2017.085 https://nasplib.isofts.kiev.ua/handle/123456789/149264 In this paper we study the inverse spectral problem for Jacobi-type pencils. By a Jacobi-type pencil we mean the following pencil J₅−λJ₃, where J₃ is a Jacobi matrix and J₅ is a semi-infinite real symmetric five-diagonal matrix with positive numbers on the second subdiagonal. In the case of a special perturbation of orthogonal polynomials on a finite interval the corresponding spectral function takes an explicit form. This paper is a contribution to the Special Issue on Orthogonal Polynomials, Special Functions and Applications (OPSFA14). The full collection is available at https://www.emis.de/journals/SIGMA/OPSFA2017.html.
 The author is grateful to referees for their valuable comments and suggestions which led to an
 essential improvement of the paper. en Інститут математики НАН України Symmetry, Integrability and Geometry: Methods and Applications The Inverse Spectral Problem for Jacobi-Type Pencils Article published earlier |
| spellingShingle | The Inverse Spectral Problem for Jacobi-Type Pencils Zagorodnyuk, S.M. |
| title | The Inverse Spectral Problem for Jacobi-Type Pencils |
| title_full | The Inverse Spectral Problem for Jacobi-Type Pencils |
| title_fullStr | The Inverse Spectral Problem for Jacobi-Type Pencils |
| title_full_unstemmed | The Inverse Spectral Problem for Jacobi-Type Pencils |
| title_short | The Inverse Spectral Problem for Jacobi-Type Pencils |
| title_sort | inverse spectral problem for jacobi-type pencils |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/149264 |
| work_keys_str_mv | AT zagorodnyuksm theinversespectralproblemforjacobitypepencils AT zagorodnyuksm inversespectralproblemforjacobitypepencils |