A truncated indefinite Stieltjes moment problem
A truncated indefinite Stieltjes moment problem in the class Nkκ of generalized Stieltjes functions is studied. The set of solutions of Stieltjes moment problem is described by Schur step-by-step algorithm, which is based on the expansion of the solutions in a generalized Stieltjes continued fractio...
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irk-123456789-1450852019-01-15T01:23:48Z A truncated indefinite Stieltjes moment problem Kovalyov, I.M. A truncated indefinite Stieltjes moment problem in the class Nkκ of generalized Stieltjes functions is studied. The set of solutions of Stieltjes moment problem is described by Schur step-by-step algorithm, which is based on the expansion of the solutions in a generalized Stieltjes continued fraction. The resolvent matrix is represented in terms of generali-zed Stieltjes polynomials. A factorization formula for the resolvent matrix is found. 2016 Article A truncated indefinite Stieltjes moment problem / I.M. Kovalyov // Український математичний вісник. — 2016. — Т. 13, № 4. — С. 473-498. — Бібліогр.: 24 назв. — англ. 1810-3200 2010 MSC. 30E05, 15B57, 46C20, 47A57 http://dspace.nbuv.gov.ua/handle/123456789/145085 en Український математичний вісник Інститут прикладної математики і механіки НАН України |
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A truncated indefinite Stieltjes moment problem in the class Nkκ of generalized Stieltjes functions is studied. The set of solutions of Stieltjes moment problem is described by Schur step-by-step algorithm, which is based on the expansion of the solutions in a generalized Stieltjes continued fraction. The resolvent matrix is represented in terms of generali-zed Stieltjes polynomials. A factorization formula for the resolvent matrix is found. |
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Kovalyov, I.M. |
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Kovalyov, I.M. A truncated indefinite Stieltjes moment problem Український математичний вісник |
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Kovalyov, I.M. |
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Kovalyov, I.M. |
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A truncated indefinite Stieltjes moment problem |
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A truncated indefinite Stieltjes moment problem |
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A truncated indefinite Stieltjes moment problem |
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A truncated indefinite Stieltjes moment problem |
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A truncated indefinite Stieltjes moment problem |
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truncated indefinite stieltjes moment problem |
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Інститут прикладної математики і механіки НАН України |
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A truncated indefinite Stieltjes moment problem / I.M. Kovalyov // Український математичний вісник. — 2016. — Т. 13, № 4. — С. 473-498. — Бібліогр.: 24 назв. — англ. |
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Український математичний вісник |
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AT kovalyovim atruncatedindefinitestieltjesmomentproblem AT kovalyovim truncatedindefinitestieltjesmomentproblem |
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2025-07-10T20:49:00Z |
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2025-07-10T20:49:00Z |
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1837294472590786560 |
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Український математичний вiсник
Том 13 (2016), № 4, 473 – 498
A truncated indefinite Stieltjes moment problem
Ivan Kovalyov
(Presented by M. M. Malamud)
Abstract. A truncated indefinite Stieltjes moment problem in the
class Nk
κ of generalized Stieltjes functions is studied. The set of solu-
tions of Stieltjes moment problem is described by Schur step-by-step
algorithm, which is based on the expansion of the solutions in a general-
ized Stieltjes continued fraction. The resolvent matrix is represented in
terms of generali-zed Stieltjes polynomials. A factorization formula for
the resolvent matrix is found.
2010 MSC. Primary 30E05; Secondary 15B57, 46C20, 47A57.
Key words and phrases. Stieltjes moment problem, Continued frac-
tion, Generalized Stieltjes fraction, Schur algorithm, Solution matrix.
1. Introduction
The classical Stieltjes moment problem was studied in [23]. It consists
in the following:
Given a sequence of real numbers {si}∞i=0, find a positive measure σ
with a support on R+, such that
si =
∫
R+
tidσ(t), i ∈ Z+ = N ∪ {0}. (1.1)
The problem (1.1) with a finite data set {si}2ni=0 is called the truncated
Stieltjes moment problem. The following inequalities
Sn+1 := (si+j)
n
i,j=0 ≥ 0, S+
n := (si+j+1)
n−1
i,j=0 ≥ 0 (1.2)
are necessary for solvability of the truncated Stieltjes moment problem.
If, additionally, the matrices Sn+1 and S+
n are nondegenerate, then the
inequalities
Sn+1 > 0 and S+
n > 0
Received 20.11.2016
The author thanks V. A. Derkach for many suggestions and observations.
ISSN 1810 – 3200. c⃝ Iнститут математики НАН України
474 A truncated indefinite Stieltjes moment problem
are also sufficient for solvability of the truncated moment problem (1.1)
with the data set {si}2ni=0 (see [16]). The degenerate case of the truncated
Stieltjes moment problem was studied in [3].
Recall that a function f holomorphic on C\R is said to belong to
the class N (see [1, Section 3.1]) [22, Appendix]), if Imf(z) ≥ 0 and
f(z) = f(z) for all z ∈ C+. Clearly, the Stieltjes transform of σ
f(z) =
∫
R+
dσ(t)
t− z
z ∈ C\R+ (1.3)
belongs to N. Moreover, f belongs to the Stieltjes class S consisting of
functions f ∈ N which admit holomorphic and nonnegative continuations
to R−. By M.G. Krein’s criterion [15]
f ∈ S⇐⇒ f ∈ N and zf ∈ N. (1.4)
By the Hamburger–Nevanlinna Theorem (see [1]) the truncated Stielt-
jes moment problem can be reformulated in terms of the Stieltjes trans-
form (1.3) of σ as the following interpolation problem at ∞: Find f ∈ S
such that
f(z) = −s0
z
− s1
z2
− · · · − s2n
z2n+1
+ o
(
1
z2n+1
)
, z→̂∞. (1.5)
The notation z→̂∞ means that z → ∞ nontangentially, that is inside
the sector ε < arg z < π − ε for some ε > 0.
A function f meromorphic on C\R with the set of holomorphy hf is
said to be in the generalized Nevanlinna class Nκ (κ ∈ N), if for every
set zi ∈ C+ ∩ hf (j = 1, . . . , n) the form
n∑
i,j=1
f(zi)− f(zj)
zi − zj
ξiξj
has at most κ and for some choice of zi (i = 1, . . . , n) it has exactly κ
negative squares. For f ∈ Nκ let us write κ−(f) = κ. In particular,
if κ = 0 then the class N0 coincides with the class N of Nevanlinna
functions.
A function f ∈ Nκ is said to belong to the class N+
κ (see [17, 18]) if
zf ∈ N and to the class Nk
κ (k ∈ N) if zf ∈ Nk
κ (see [5,6]). In particular,
if k = 0, then N0
κ := N+
κ , and if κ = 0, k ̸= 0 Nk
0 coincides with the
generalized Stieltjes class S+
κ introduced in [12,13].
In the present paper the following indefinite moment problem in the
classes Nk
κ is studied.
I. Kovalyov 475
Problem MPk
κ(s, ℓ). Given ℓ, κ, k ∈ Z+, and a sequence s = {si}ℓi=0 of
real numbers, describe the set Mk
κ(s) of functions f ∈ Nk
κ, which have
the following asymptotic expansion
f(z) = −s0
z1
− s1
z2
− · · · − sℓ
zℓ+1
+ o
(
1
zℓ+1
)
, z→̂∞. (1.6)
Indefinite moment problems in the classes Nκ were studied in [4, 5, 14,
19]. Indefinite moment problems in the classes N+
κ and Nk
κ were studied
in [19,20] and [7, 10], respectively.
This paper is a continuation of [10], where a Schur type algorithm for
the moment problem MPk
κ(s, ℓ) was elaborated. We restrict ourselves to
the case of a nondegenerate problem. Namely, if ℓ = 2n − 1 the even
moment problem MP k
κ (s, 2n− 1) is called nondegenerate if detSn ̸= 0.
Recall ([9]), that a number nj ∈ N is called a normal index of the
sequence s, if detSnj ̸= 0. The ordered set of all normal indices
n1 < n2 < · · · < nN
of the sequence s is denoted byN (s). With this notation the even moment
problem MP k
κ (s, 2n−1) is nondegenerate if n ∈ N (s). Let us set n := nN
and ℓ = 2nN − 1. In Theorem 4.2 we show that the nondegenerate even
moment problem MP k
κ (s, 2nN − 1) is solvable if and only if
κ+N := ν−(SnN ) ≤ κ and k+N := ν−(S
+
nN
) ≤ k,
where ν−(SnN ) denotes the number of negative eigenvalues of SnN with
account of multiplicities. Every solution f of the even moment problem
MP k
κ (s, 2nN − 1) admits the following representation
f(z) =
1
−zm1(z) +
1
l1(z) + · · ·+
1
−zmN (z) +
1
lN (z) + τ(z)
, (1.7)
where mi(z) and li(z) are some polynomials determined by the data
s = {si}2nN−1
i=0 , and τ ∈ N
k−k+N
κ−κ+
N
and τ(z)−1 = o(1), as z→̂∞.
Furthermore, the continued fraction (1.7) is associated with the fol-
lowing system of difference equations{
y2i+1 − y2i−1 = −zmi+1(z)y2i,
y2i+2 − y2i = li+1(z)y2i+1.
(1.8)
476 A truncated indefinite Stieltjes moment problem
see [24, Section 1]. The polynomials P+
i (z) and Q+
i (z), which satisfy the
system (1.8) and the following initial conditions
P+
−1(z) ≡ −1, P+
0 (z) ≡ 0; Q+
−1(z) ≡ 0, Q+
0 (z) ≡ 1
are called generalized Stieltjes polynomials.
In Theorem 5.5 it is shown that the formula (1.7) can be rewritten in
terms of the polynomials Q+
2N−1, Q
+
2N , P+
2N−1 and Q+
2N as follows
f(z) =
Q+
2N−1(z)τ(z) +Q+
2N (z)
P+
2N−1(z)τ(z) + P+
2N (z)
. (1.9)
The resolvent matrix of the even moment problem MP k
κ (s, 2nN − 1)
W2N (z) =
(
Q+
2N−1(z) Q+
2N (z)
P+
2N−1(z) P+
2N (z)
)
(1.10)
admits the following factorization
W2N (z) =M1(z)L1(z) . . .MN (z)LN (z), (1.11)
where the matrices Mj(z) and Lj(z) are defined by (4.12).
Analogous results for odd moment problem MP k
κ (s, 2nN − 2) are
presented in Theorem 4.1 and Theorem 5.2. Sequences s = {si}ℓi=0 which
satisfy the condition
detS+
nj
̸= 0 j = 1, . . . , N, (1.12)
are called regular, [10]. The moment problem MP k
κ (s, ℓ) in the class
of regular sequences s = {si}ℓi=0 was studied in [10]. As was shown
in [10] the polynomials lj(z) in this case are reducing to constants and
the resolvent matrices Lj(z) are changing accordingly.
2. Preliminaries
2.1. Generalized Nevanlinna and Stieltjes classes
Every real polynomial P (t) = pνt
ν+pν−1t
ν−1+ . . .+p1t+p0 of degree
ν belongs to a class Nκ, where the index κ = κ−(P ) can be evaluated by
(see [17, Lemma 3.5])
κ−(P ) =
{ [
ν+1
2
]
, if pν < 0; and ν is odd ;[
ν
2
]
, otherwise . (2.1)
Proposition 2.1. ([17]) Let f ∈ Nκ, f1 ∈ Nκ1, f2 ∈ Nκ2. Then
I. Kovalyov 477
(1) −f−1 ∈ Nκ;
(2) f1 + f2 ∈ Nκ′, where κ′ ≤ κ1 + κ2;
(3) If, in addition, f1(iy) = o(y) as y → ∞ and f2 is a polynomial,
then
f1 + f2 ∈ Nκ1+κ2 . (2.2)
(4) If a function f ∈ Nκ has an asymptotic expansion (1.6), then there
exists κ′ ≤ κ, such that {sj}ℓj=0 ∈ Hκ′,ℓ.
Proposition 2.2. ([10]) The following equivalences hold:
(1) f ∈ Nk
κ ⇐⇒ − 1
f ∈ N−k
κ ;
(2) f ∈ Nk
κ ⇐⇒ zf ∈ N−κ
k , in particular, f ∈ N+
κ ⇐⇒ zf ∈ S−
κ ;
(3) If a function f ∈ Nk
κ has an asymptotic expansion (1.6) then
{sj}ℓj=0 ∈ Hk′
κ′,ℓ with κ′ ≤ κ, k′ ≤ k. (2.3)
2.2. Normal indices
Recall that the set N (s) = {nj}Nj=1 of normal indices of the sequence
s = {sj}ℓj=0 is defined by
N (s) = {nj : Dnj ̸= 0, j = 1, 2, . . . , N}, Dnj := det(si+k)
nj−1
i,k=0. (2.4)
Let us set D+
n := det(si+j+1)
n−1
i,j=0. By the Sylvester identity (see [9,
Proposition 3.1] or [7, Lemma 5.1] for detail), the set N (s) is the union
of two not necessarily disjoint subsets
N (s) = {νj}N1
j=1 ∪ {µj}
N2
j=1, (2.5)
which are selected by
Dνj ̸= 0 and D+
νj−1 ̸= 0, for all j = 1, N1 (2.6)
and
Dµj ̸= 0 and D+
µj
̸= 0, for all j = 1, N2. (2.7)
Moreover, the normal indices νj and µj satisfy the following inequalities
0 < ν1 ≤ µ1 < ν2 ≤ µ2 < . . . (2.8)
478 A truncated indefinite Stieltjes moment problem
For every nj ∈ N (s) polynomials of the first and the second kind
Pnj (z) and Qnj (z) can be defined by standard formulas
Pnj (z) =
1
Dnj
det
s0 s1 · · · snj
· · · · · · · · · · · ·
snj−1 snj · · · s2nj−1
1 z · · · znj
,
Qnj (z) = St
(
Pnj (z)−Pnj (t)
z−t
)
,
(2.9)
where St is the linear functional on the set of polynomial of formal degree
ℓ, defined by
St(t
i) = si, i = 0, 1, . . . , ℓ.
Definition 2.3. The sequence s = {si}ℓi=0 is called regular (s ∈ Hk,reg
κ,ℓ ) if
and only if one of the following equivalent conditions holds
( [9, Lemma 3.1])
(1) Pnj (0) ̸= 0 for every j ≤ N ;
(2) D+
nj−1 ̸= 0 for every j ≤ N ;
(3) D+
nj
̸= 0 for every j ≤ N ;
(4) νj = µj for all j, such that νj , µj ∈ N (s).
2.3. Class Uκ(J) and linear fractional transformations
Let κ1 ∈ N and let J be the 2× 2 signature matrix
J =
(
0 −i
i 0
)
.
A 2×2 matrix valued function W (z) = (wi,j(z))
2
i,j=1 that is meromorphic
in C+ is said to belong to the class Uκ(J) of generalized J-inner matrix
valued functions if (see [2], [8]):
(i) the kernel
KW
ω (z) =
J −W (z)JW (ω)∗
−i(z − ω̄)
(2.10)
has κ negative squares in H+
W × H+
W and
(ii) J −W (µ)JW (µ)∗ = 0 for a.e. µ ∈ R,
I. Kovalyov 479
where H+
W denotes the domain of holomorphy of W in C+.
Consider the linear fractional transformation
TW [τ ] = (w11τ(z) + w12)(w21τ(z) + w22)
−1 (2.11)
associated with the matrix valued function W (z). The linear fractional
transformation associated with the product W1W2 of two matrix valued
function W1(z) and W2(z), coincides with the composition TW1 ◦ TW2 .
As is known, if W ∈ Uκ1(J) and τ ∈ Nκ2 then TW [τ ] ∈ Nκ′ , where
κ′ ≤ κ1 + κ2, cf. [17, Satz 4.1]
In the present paper two partial cases, in which the preceding inequal-
ity becomes equality, will be needed.
Lemma 2.4. ([10]) Let m(z) be a real polynomial κ1 = κ−(zm), k1 =
κ−(m), let M be a 2× 2 matrix valued function
M(z) =
(
1 0
−zm(z) 1
)
(2.12)
and let τ be a meromorphic function, such that τ(z)−1 = o(z) as z→̂∞.
Then M ∈ Uκ1(J) and the following equivalences hold:
τ ∈ Nκ2 ⇐⇒ TM [τ ] ∈ Nκ1+κ2 , (2.13)
τ ∈ Nk2
κ2
⇐⇒ TM [τ ] ∈ Nk1+k2
κ1+κ2
. (2.14)
Lemma 2.5. ([10]) Let l(z) be a real polynomial and indices κ1 = κ−(l),
k1 = κ−(zl(z)), let L be a 2× 2 matrix valued function
L(z) =
(
1 l(z)
0 1
)
(2.15)
and let τ be a meromorphic function, such that τ(z)−1 = o(1) as z→̂∞.
Then L ∈ Uk1(J) and the following equivalences hold:
τ ∈ Nκ2 ⇐⇒ TL[τ ] ∈ Nκ1+κ2 ,
τ ∈ Nk2
κ2
⇐⇒ TL[τ ] ∈ Nk1+k2
κ1+κ2
.
3. Basic moment problem in Nk
κ
In this section we expose some material from [10] concerning the basic
odd and even moment problems in generalized Stieltjes class Nk
κ and
describe their solutions.
480 A truncated indefinite Stieltjes moment problem
3.1. Basic odd moment problem MP k
κ (s, 2ν1 − 2)
An odd moment problem MP k
κ (s, 2n− 2) is called nondegenerate if
Dn ̸= 0 and D+
n−1 ̸= 0. (3.1)
If, in addition, n = ν1 ∈ N (s), then the nondegenerate moment problem
MP k
κ (s, 2ν1 − 2) is called basic. In this case
N (s) = {ν1} and s0 = . . . = sν1−2 = 0, sν1−1 ̸= 0. (3.2)
The basic moment problem MP k
κ (s, 2ν1 − 2) can be reformulated as
follows:
Given a sequence s = {si}2ν1−2
i=0 , such that (3.2) holds, or equivalently
N (s) = {ν1}. Find all functions f ∈ Nk
κ, which admit the asymptotic
expansion
f(z) = −sν1−1
zν1
− · · · − s2ν1−2
z2ν1−1
+ o
(
1
z2ν1−1
)
, z→̂∞. (3.3)
Let s = {si}2ν1−2
i=0 be a sequence of real numbers from H and let (3.2)
hold. Then s ∈ Hk1
κ1,2ν1−2, where κ1 and k1 are defined by
κ1 = ν−(Sν1) =
{ [
ν1+1
2
]
, if ν1 is odd and sν1−1 < 0;[
ν1
2
]
, otherwise. (3.4)
k1 = ν−(S
+
ν1−1) =
{
[ν12 ], if ν1 is even and sν1−1 < 0;
[ν1−1
2 ], otherwise. (3.5)
Let us define the polynomial m1, associated with the sequence
s= {si}2ν1−2
i=0 , by
m1(z) =
(−1)ν1+1
Dν1
∣∣∣∣∣∣∣∣∣
0 . . . 0 sν1−1 sν1
... . . . . . .
...
sν1−1 . . . . . . . . . s2ν1−2
1 z . . . zν1−2 zν1−1
∣∣∣∣∣∣∣∣∣ (Dν1 := detSν1).
(3.6)
Obviously, the leading coefficient of m1 is
(−1)ν1+1
D+
ν1−1
Dν1
=
1
sν1−1
(3.7)
and by Proposition 2.1, m1 ∈ Nκ1
k1
, i.e. the indices κ1 and k1 are con-
nected with m1 by
κ1 = κ−(zm1), k1 = κ−(m1). (3.8)
I. Kovalyov 481
Lemma 3.1. (cf. [4, 10]) Let a function f ∈ Nk
κ admit the asymp-
totic expansion (3.3) and let ν1 be the first normal index of the sequence
s = {si}2ν1−2
i=0 , let polynomial m1, indices κ1 and k1 be defined by (3.6)
and (3.8), respectively. Then f admits the following representation
f(z) = TM1 [τ ] =
τ(z)
−zm1(z)τ(z) + 1
, (3.9)
where
τ ∈ Nk−k1
κ−κ1
and τ−1 = o(z), z→̂∞. (3.10)
Furthermore, the matrix valued function
M1(z) =
(
1 0
−zm1(z) 1
)
(3.11)
belongs to the class Uκ1(J).
Conversely, if τ satisfies (3.10) and f is defined by (3.9), then f ∈
Nk
κ.
Proof. Assume that f ∈ Nk
κ and f admits the asymptotic expansion
(3.3). Then by [10, Lemma 3.1]
f(z) = − 1
zm1(z) + g(z)
, (3.12)
where the polynomial m1 is defined by (3.6), g ∈ Nκ−κ1 and g(z) = o(z)
as z→̂∞. On the other hand, we can rewrite (3.12) as follows
−1/f(z) = zm1(z) + g(z). (3.13)
Replacing g by −τ−1 in (3.13), we obtain τ ∈ Nκ−κ1 . Due to the as-
sumption zf ∈ Nk one gets − 1
zf ∈ Nk and hence the equality
−1/zf(z) = m1(z)− 1/zτ(z), (3.14)
Proposition 2.1 and (3.8) imply −(zτ(z))−1 ∈ Nk−k1 . Therefore, τ ∈
Nk−k1
κ−κ1
and τ−1 = o(z) as z→̂∞. Replacing g by −τ−1 in (3.12) one
obtains (3.9). Furthermore, by Lemma 2.4M1 ∈ Uκ1(J). This completes
the proof.
A sequence (c0, . . . , cn) of real numbers determines an upper triangu-
lar Toeplitz matrix T (c0, . . . , cn) of order (n + 1) × (n + 1) with entries
ti,j = cj−i for i ≤ j and ti,j = 0 for i > j:
T (c0, . . . , cn) =
c0 . . . cn
. . .
...
c0
. (3.15)
482 A truncated indefinite Stieltjes moment problem
Theorem 3.2. ([10]) Let ν1 be the first normal index of the sequence
s = {si}2ν1−2
i=0 , let m1, κ1 and k1 be defined by (3.6), (3.4) and by (3.5),
respectively, and let ℓ ≥ 2ν1 − 2. Then:
(1) The problem MP k
κ(s, ℓ) is solvable if and only if
κ1 ≤ κ and k1 ≤ k. (3.16)
(2) f ∈Mk
κ(s, 2ν1 − 2) if and only if f admits the representation
f = TM1 [τ ], (3.17)
where τ satisfies the following conditions
τ ∈ Nk−k1
κ−κ1
and
1
τ(z)
= o(z), z→̂∞. (3.18)
(3) If ℓ > 2ν1 − 2, then f ∈ Mk
κ(s, ℓ) if and only if f admits the
representation f = TM1 [τ ], where τ ∈ Nk−k1
κ−κ1
and τ admits the
following asymptotic expansion
−τ−1(z) = −s(1)−1 −
s
(1)
0
z
− · · · −
s
(1)
ℓ−2ν1
zℓ−2ν1+1
+ o
(
1
zℓ−2ν1+1
)
, z→̂∞,
(3.19)
where the sequence
{
s
(1)
i
}ℓ−2ν1
i=−1
is determined by the matrix equa-
tion
T (m
(1)
ν1−1, . . . ,m
(1)
0 ,−s(1)−1, . . . ,−s
(1)
ℓ−2ν1
)T (sν1−1, . . . , sℓ) = Iℓ−ν1+2.
(3.20)
Remark 3.3. On the other hand, the sequence {s(1)i }
n−2ν1
i=−1 can be found
by the following equivalent formulas (see [4, Proposition 2.1])
s
(1)
−1 =
(−1)ν1+1
sν1−1
D+
ν1
Dν1
, (3.21)
s
(1)
i =
(−1)i+ν1
si+ν1+2
ν1−1
∣∣∣∣∣∣∣∣∣∣∣∣∣
sν1 sν1−1 0 . . . 0
...
. . .
. . .
. . .
...
...
. . .
. . . 0
...
. . . sν1−1
s2ν1+i . . . . . . . . . sν1
∣∣∣∣∣∣∣∣∣∣∣∣∣
i = 0, n− 2ν1.
(3.22)
I. Kovalyov 483
3.2. Basic even moment problem MP k
κ (s, 2µ1 − 1)
An even moment problem MP k
κ (s, 2n− 1) is called nondegenerate, if
the following conditions hold
Dn ̸= 0 and D+
n ̸= 0. (3.23)
The nondegenerate even moment problem MP k
κ (s, 2n−1) is called basic,
if n is the smallest normal index of the sequence {si}2n−1
i=0 such that (3.23)
holds. In view of the classification of normal indices in (2.6) and (2.7), the
basic even moment problem coincides with the problem MP k
κ (s, 2µ1−1).
In this case
either N (s) = {ν1} or N (s) = {ν1, µ1} ,
regarding to the conditions
ν1 = µ1 or ν1 < µ1.
The basic even moment problem MP k
κ (s, 2µ1−1) can be reformulated
as follows:
Given a sequence s = {si}2µ1−1
i=0 ∈ H, where µ1 is the smallest index
n such that (3.23) holds, find all functions f ∈ Nk
κ, such that
f(z) = −sν1−1
zν1
− · · · − s2µ1−1
z2µ1
+ o
(
1
z2µ1
)
, z→̂∞.
Solution of the basic even moment problem will be splitted into two
steps. On the first step one applies Lemma 3.1 to construct a sequence
{s(1)i }
2(µ1−ν1)−1
i=−1 from the asymptotic expansion of the function −τ−1. If
f ∈Mk
κ(s, 2µ1−1) then by Theorem 3.2 f admits the representation (3.9)
which can be rewritten as
− 1
f(z)
= zm1(z)−
1
g1(z)
, (3.24)
where we use g1 instead of τ and−g−11 has the following asymptotic ex-
pansion
− 1
g1(z)
= −s(1)−1 −
s
(1)
0
z
− · · · −
s
(1)
2(µ1−ν1)−1
z2(µ1−ν1)
+ o
(
1
z2(µ1−ν1)
)
, z→̂∞,
(3.25)
with s
(1)
i defined by (3.20). By Lemma 2.5
κ− κ−(zm1) = κ−(g1) ≥ κ−(l1) + κ−(τ),
κ− κ−(m1) = κ−(zg1) ≥ κ−(zl1) + κ−(zτ).
(3.26)
484 A truncated indefinite Stieltjes moment problem
Therefore, f ∈ Nk
κ if and only if g1 ∈ N
k−κ−(m1)
κ−κ−(zm1)
and g1 is represented
as
g1(z) = TL1 [τ ] := l1(z) + τ(z), (3.27)
where τ ∈ Nk−κ−(m1)−κ−(zl1)
κ−κ−(zm1)−κ−(l1)
and l1(z) is calculated as follows:
(1) if ν1 = µ1, then
l1 =
1
s
(1)
−1
= (−1)ν1+1sν1−1
Dν1
D+
ν1
; (3.28)
(2) if ν1 < µ1, then
l1(z)=
1
s
(1)
µ1−ν1−1 det(S
(1)
µ1−ν1)
∣∣∣∣∣∣∣∣∣
s
(1)
0 . . . s
(1)
µ1−ν1−1 s
(1)
µ1−ν1
· · · · · · · · · · · ·
s
(1)
µ1−ν1−1 . . . s
(1)
2µ1−2ν1−2 s
(1)
2µ1−2ν1−1
1 . . . zµ1−ν1−1 zµ1−ν1
∣∣∣∣∣∣∣∣∣ ,
(3.29)
where the matrix S(1)µ1−ν1 is defined as in (1.2), i.e.
S(1)µ1−ν1 = (s
(1)
i+j−1)
µ1−ν1−1
i,j=0 .
Theorem 3.4. ([10]) Let s = {si}2µ1−1
i=0 be a sequence from Hk
κ, such
that N(s) = {ν1, µ1} (ν1 ≤ µ1), and let m1, l1 be defined by (3.6), (3.28)
and (3.29), respectively. Then:
(1) The problem MP k
κ (s, 2µ1 − 1) is solvable if and only if
κ+1 := ν−(Sµ1) ≤ κ and k+1 := ν−(S
+
µ1
) ≤ k. (3.30)
(2) f ∈Mk
κ(s, 2µ1 − 1) if and only if f admits the following represen-
tation
f = TM1L1 [τ ] =
1
−zm1(z) +
1
l1(z) + τ(z)
, (3.31)
where
τ ∈ N
k−k+1
κ−κ+
1
and τ(z) = o(1) as z→̂∞. (3.32)
The indices κ+1 and k+1 can be expressed in terms of m1 and l1 by
κ+1 = κ−(zm1) + κ−(l1), k+1 = κ−(m1) + κ−(zl1). (3.33)
I. Kovalyov 485
(3) If ℓ > 2µ1 − 1, then f ∈ Mk
κ(s, ℓ), if and only if f admits the
representation (3.31), where
τ ∈Mk−k+1
κ−κ1
(s(1), ℓ− 2µ1), (3.34)
κ+1 and k+1 are determined by (3.30) and the sequence {s(1)i }
ℓ−2µ1
i=−1
is determi-ned by the matrix equation
T (l1,−s(1)0 , . . . ,−s(1)ℓ−2µ1
)T (s
(1)
−1, . . . , s
(1)
ℓ−2µ1
) = Iℓ−2µ1+2, (3.35)
if µ1 = ν1, and if ν1 < µ1 by the following equation
T (l
(1)
µ1−ν1 , . . . , l
(1)
0 ,−s(1)0 , . . . ,−s(1)ℓ−2µ1
)T (s
(1)
µ1−ν1−1, . . . , s
(1)
ℓ−2ν1
)
= Iℓ−µ1−ν1+2.
(3.36)
Proof. (1)–(3) are implied by the above considerations, in particular,
(3.30) follows from (3.26) and (3.33) follows from (3.27) and Proposition
2.1.
Remark 3.5. The sequence {s(1)i }
ℓ−2µ1
i=0 can also be found by the follow-
ing formula (see [4, Proposition 2.1], [10, (3.38)])
s
(1)
i =
(−1)i+µ1−ν1
(s
(1)
µ1−ν1−1)
i+µ1−ν1+2
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
s
(1)
µ1−ν1 s
(1)
µ1−ν1−1 0 . . . 0
...
. . .
. . .
. . .
...
...
. . .
. . . 0
...
. . . s
(1)
µ1−ν1−1
s
(1)
2(µ1−ν1)+i
. . . . . . . . . s
(1)
µ1−ν1
∣∣∣∣∣∣∣∣∣∣∣∣∣∣
,
(3.37)
where i = 0, ℓ− 2µ1.
Remark 3.6. The resolvent matrix of the basic even moment problem
Mk
κ(s, 2µ1 − 1) takes the form
W2(z) =
(
1 l1(z)
−zm1(z) −zm1(z)l1(z) + 1
)
. (3.38)
Furthermore, W2(z) admits the following factorization
W2(z) =M1(z)L1(z), (3.39)
where the matrices M1(z) and L1(z) are defined by (2.12), (2.15) and
the corresponding linear fractional transform is defined by
TW2 [f1] =
f1(z) + l1(z)
−zm1(z)f1(z)− zm1(z)l1 + 1
. (3.40)
486 A truncated indefinite Stieltjes moment problem
Remark 3.7. If the sequence s = {si}2µ1−1
i=0 belongs to Hk,reg
κ,2µ1−1, then
l1(z) is a constant,
l1 =
1
s
(1)
−1
and L1 =
(
1 l1
0 1
)
.
In this case, the resolvent matrix W2(z) of the basic even moment
prob-lemMk
κ(s, 2µ1 − 1) admits the factorization
W2(z) =M1(z)L1.
and (3.38) takes the form
W2(z) =
(
1 l1
−zm1(z) −zm1(z)l1 + 1
)
. (3.41)
4. The Schur algorithm
In this section we study a step-by-step algorithm, which describes all
solutions of the general nondegenerate indefinite moment problem in the
class Nk
κ. This algorithm is based on the elementary steps introduced in
the previous section.
4.1. Odd moment problem
Let MP k
κ (s, 2νN − 2) be a nondegenerate odd moment problem, i.e.
DνN ̸= 0 and D+
νN−1 ̸= 0. (4.1)
Theorem 4.1. Let s = {si}2νN−2
i=0 ∈ Hk
κ,2νN−2, let N(s) = {νj}Nj=1 ∪
{µj}N−1
j=1 , and let mj(z) and lj(z) be defined by (4.10) and (4.11), respec-
tively. Then:
(1) A nondegenerate odd moment problem MP k
κ (s, 2νN −2) is solvable
if and only if
κN := ν−(SνN ) ≤ κ and kN := ν−(S
+
νN−1) ≤ k. (4.2)
(2) f ∈Mk
κ(s, 2νN − 2) if and only if f admits the following represen-
tation
f = TW2N−1
[τ ], (4.3)
where
W2N−1(z) :=M1(z)L1(z) . . . LN−1(z)MN (z) (4.4)
and τ(z) satisfies the conditions
τ ∈ Nk−kN
κ−κN
and
1
τ(z)
= o(z), z→̂∞. (4.5)
I. Kovalyov 487
(3) The representation (4.3) can be rewritten as a continued fraction
expansion
f(z) =
1
−zm1(z) +
1
l1(z) +
1
−zm2(z) + · · ·+
1
−zmN (z) + 1
τ(z)
.
(4.6)
(4) The indices κN and kN are related to mj and lj by
κN =
N∑
j=1
κ−(zmj)+
N−1∑
j=1
κ−(lj), kN =
N∑
j=1
κ−(mj)+
N−1∑
j=1
κ−(zlj).
Proof. Let f ∈ Nk
κ and f have the asymptotic
f(z) = −s0
z
− s1
z2
− · · · − s2νN−2
z2νN−1
+ o
(
1
z2νN−1
)
, z→̂∞.
Then by Theorem 3.4, the function f can be represented as follows
f(z) =
1
−zm1(z) +
1
l1(z) + f1(z)
,
where the polynomialsm1 and l1 are defined by (3.6) and (3.29), respecti-
vely, and
κ+1 = κ−(zm1) + κ−(l1) ≤ κ and k+1 = κ−(m1) + κ−(zl1) ≤ k. (4.7)
In this case f1 ∈ N
k−k+1
κ−κ+
1
and f1 has the following asymptotic expansion
f1(z) = −
s
(1)
0
z
− s
(1)
1
z2
− · · · −
s
(1)
2(νN−ν1)−2
z2(νN−ν1)−1
+ o
(
1
z2(νN−ν1)−1
)
, z→̂∞,
where the sequence s(1) = {s(1)i }
2(νN−ν1)−2
i=1 is found recursively by (3.20),
(3.35) and (3.36). Moreover, by [10, Lemma 2.5] the set of normal indices
of the sequence s(1) is
N(s(1)) = {nj − ν1}Nj=2.
488 A truncated indefinite Stieltjes moment problem
Continuing this process and applying Theorem 3.4 N − 1 times, one
const-ructs sequences of polynomials mj , lj and functions fj , gj , such
that
− 1
fj−1(z)
= zmj(z)−
1
gj(z)
, 1 ≤ j ≤ N,
gj(z) = lj(z) + fj(z), 1 ≤ j ≤ N − 1.
The indices κ+j and k+j are defined by
κ+j =
j∑
i=1
κ−(zmi) +
j∑
i=1
κ−(li) ≤ κ,
k+j =
N∑
i=1
κ−(mi) +
j∑
i=1
κ−(zli) ≤ k.
(4.8)
Hence
gj ∈ N
k−kj
κ−κj
and fj ∈ N
k−k+j
κ−κ+
j
, 1 ≤ j ≤ N − 1.
Moreover, gj and fj have the following induced asymptotic expan-
sions
gj(z) = −s(j)−1−
s
(j)
0
z
− s
(j)
1
z2
−· · ·−
s
(j)
2(νN−νj)−2
z2(νN−νj)−1
+o
(
1
z2(νN−νj)−1
)
, z→̂∞,
fj(z) = −
s
(j)
0
z
− s
(j)
1
z2
− · · · −
s
(j)
2(νN−µj)−2
z2(νN−µj)−1
+ o
(
1
z2(νN−µj)−1
)
, z→̂∞,
where the sequences {s(j)i }
2(νN−νi)−2
i=−1 and {s(j)i }
2(νN−µi)−2
i=0 are found from
the equalities
T (m
(j)
νj−1, . . .,m
(j)
0 ,−s(j)−1, . . . ,−s
(j)
ℓj−2νj
)T (s
(j)
νj−1, . . . , s
(j)
ℓj
)=Iℓj−ν1+2, ℓj=ℓ−2µj−1,
T (l
(j)
µj−νj
, . . . , l
(j)
0 ,−s(j)0 , . . . ,−s(j)ℓ−2µj
)T (s
(j)
µj−νj−1, . . . , s
(j)
ℓj−2νj
)=Iℓ−µj−νj+2.
Therefore, fj−1 takes the following representation in terms of fj :
fj−1(z) =
1
−zmj(z) +
1
lj(z) + fj(z)
(j = 1, . . . , N − 1), (4.9)
I. Kovalyov 489
Here the sequence s(j) = {s(j)i }
2(νN−µj)−2
i=0 is determined recursively by
(3.20) and (3.36) and polynomials mj and lj are defined by the formulas
mj(z) =
(−1)ν+1
detS
(j)
ν
∣∣∣∣∣∣∣∣∣∣
0 . . . 0 s
(j−1)
ν−1 s
(j−1)
ν
... . . . . . .
...
s
(j−1)
ν−1 . . . . . . . . . s
(j−1)
2ν−2
1 z . . . zν−2 zν−1
∣∣∣∣∣∣∣∣∣∣
, (4.10)
lj(z) =
1
s
(j)
−1
= (−1)ν+1s
(j)
ν−1
D
(j)
ν
D
(j)+
ν
, if νj = µj ;
1
s
(j)
µ−1 det(S
(j)
µ )
∣∣∣∣∣∣∣∣∣
s
(j)
0 . . . s
(j)
µ−1 s
(j)
µ
· · · · · · · · · · · ·
s
(j)
µ−1 . . . s
(j)
2µ−2 s
(j)
2µ−1
1 . . . zµ−1 zµ
∣∣∣∣∣∣∣∣∣ , if νj < µj .
,
(4.11)
where ν = νj − µj−1 and µ = µj − νj for all j = 1, . . . , N − 1.
Let the matrix functions Mj(z) and Lj(z) be defined by
Mj(z) =
(
1 0
−zmj(z) 1
)
and Lj(z) =
(
1 lj(z)
0 1
)
, (j = 1, . . . , N − 1).
(4.12)
Then it follows from (4.9) that
fj−1(z) = TMj(z)Lj(z)[fj(z)] (j = 1, . . . , N − 1). (4.13)
On the last step we get the function fN−1(z), which is a solution of the
basic moment problemMP k
κ (s
(N−1), 2(νN−µN−1)−2). By Theorem 3.2,
the function fN−1(z) can be represented as
fN−1(z) =
1
−zmN (z) +
1
fN (z)
= TMN (z)[fN (z)], (4.14)
where the polynomial mN (z) is defined by (4.10) and fN (z) is a function
from Nk−kN
κ−κN
, such that fN (z)(−1) = o(z) as z→̂∞ and
κN = κ+N−1 + κ−(zmN ) ≤ κ and kN = k+N−1 + κ−(mN ) ≤ k. (4.15)
Now (4.2) is implied by (4.8) and (4.15).
The converse statements of Theorem 4.1 are also implied by Theorem
3.2 and Theorem 3.4. Replacing fN (z) by τ(z), we get (2) and (3).
Combining (4.9), (4.13) and Lemmas 2.4–2.5, we obtain the statement
(4).
490 A truncated indefinite Stieltjes moment problem
4.2. Even moment problem
Let s = {si}2µN−1
i=0 ∈ Hk
κ,2µN−1, let the set of normal indices N (s) =
{νj}Nj=1 ∪ {µj}Nj=1 and let MP k
κ (s, 2µN − 1) be a nondegenerate even
moment problem, i.e.
DµN ̸= 0 and D+
µN
̸= 0. (4.16)
Theorem 4.2. Let s = {si}2µN−1
i=0 ∈ Hk
κ,2µN−1 and let N (s) = {νj}Nj=1 ∪
{µj}Nj=1.
(1) A nondegenerate odd moment problem MP k
κ (s, 2µN−1) is solvable,
if and only if
κ+N := ν−(SµN ) ≤ κ and k+N := ν−(S
+
µN
) ≤ k; (4.17)
(2) f ∈Mk
κ(s, 2µN − 1) if and only if f admits the representation
f = TW2N
[τ ], (4.18)
where
W2N (z) :=W2N−1(z)LN (z) =M1(z)L1(z) . . .MN (z)LN (z)
(4.19)
and τ(z) satisfies the following conditions
τ ∈ N
k−k+N
κ−κ+
N
and
1
τ(z)
= o(1), z→̂∞; (4.20)
(3) The representation (4.18) can be rewritten as the continued fraction
expansion
f(z) =
1
−zm1(z) +
1
l1(z) + · · ·+
1
−zmN (z) +
1
lN (z) + τ(z)
,
(4.21)
where mj(z) and lj(z) are defined by (3.6) and (4.11), respectively;
(4) The indices κ+N and k+N can be found by
κ+N =
N∑
j=1
κ−(zmj) +
N∑
j=1
κ−(lj), k+N =
N∑
j=1
k−(mj) +
N∑
j=1
κ−(zlj).
I. Kovalyov 491
Proof. Applying Theorem 3.4 N −1 times in the same way as in the odd
case one obtains the sequence of fj ∈ N
k−k+j
κ−κ+
j
and polynomials mj and lj
defined by (4.10) and (4.11), respectively, such that (4.8) and (4.9) hold.
On the last step we obtain the function fN−1(z), which is a solution of
the basic even moment problemMP
k−k+N−1
κ−κ+
N−1
(s(N−1), 2(µN − µN−1) − 1).
By Theorem 3.4, the function fN−1 can be represented as follows:
fN−1(z) =
1
−zmN (z) +
1
lN (z) + fN (z)
, (4.22)
the inequalities
κ+N = κ+N−1 + κ−(zmN ) + κ−1(lN ) ≤ κ,
k+N = k+N−1 + κ−(mN ) + κ−(zlN ) ≤ k (4.23)
hold and fN (z) is a function from N
k−k+N
κ−κ+
N
, such that fN (z) = o(1) as
z→̂∞.
Replacing fN by τ and combining the statements (4.9) and (4.22)
one obtains (2)–(4).
By (4.9) and (4.22) the inequality (4.17) is implied by (4.8), (4.23).
Conversely, if (4.17) holds, one can apply Theorem 3.2 N − 1 times and
then Theorem 3.4. By these theorems the function f determined by
(4.18) belongs to MP k
κ (s, 2µN − 1). This completes the proof.
5. Resolvent matrices in odd and even cases
5.1. Odd moment problem
In the present section resolvent matrices W2N−1 and W2N for odd
and even moment problem will be studied.
Recall some facts concerning continued fractions
Proposition 5.1. ([24, Chapter I]) Let a1, a2, . . . , an, ω ∈ C and let
fn =
1
a1 +
1
a2 + · · ·
1
an + ω
. (5.1)
Then fn can be represented as follows
An−1ω +An
Bn−1ω +Bn
, (5.2)
492 A truncated indefinite Stieltjes moment problem
where the quantities Ai, Bi (i ∈ N) are solutions of the following recur-
rence system
yi+1 − yi = ai+1yi−1, i = 0, n− 1, (5.3)
subject to the initial conditions
A−1 = 1, A0 = 0, B−1 = 0, B0 = 1. (5.4)
Continued fractions (4.6) and (4.21) have partial denominators of two
types
a2i−1 = −zmi(z) and a2i = li(z), i = 1, N. (5.5)
Therefore, it is reasonable to write (5.3) separately for odd and even
indices. The numerator and denominator of the n−th convergent of (5.1)
will be denoted by
Q+
i (z) = Ai and P+
i (z) = Bi. (5.6)
Then the equality (5.3) takes the form
y2i+1 − y2i−1 = −zmi+1(z)y2i,
y2i+2 − y2i = li+1(z)y2i+1.
(5.7)
By Proposition 5.1 P+
i (z) and Q+
i (z) are solutions of the system(5.7)
subject to the initial conditions
P+
−1(z) ≡ 0, P+
0 (z) ≡ 1, Q+
−1(z) ≡ 1, Q+
0 (z) ≡ 0.
(5.8)
Polynomials P+
i (z) and Q+
i (z) will be called generalized Stieltjes
polyno-mials of the first and the second kind, respectively. In the case
of a regular sequence {si}ℓi=0 ∈ H
k,reg
κ explicit formulas for P+
i (z) and
Q+
i (z) were found in [10]. In the definite case (i.e. s ∈ H0
0) see [22, v.4.2]
and [11], [12, (10.29)].
The results of Theorems 4.1 and 4.2 can be reformulated in terms of
generalized Stieltjes polynomials.
Theorem 5.2. Let s ∈ Hk
κ,2νN−2, let (4.2) hold and let polynomials
mj(z) (1 ≤ j ≤ N) and lj(z) (1 ≤ j ≤ N − 1) be defined by (4.10) and
(4.11), respectively. Let P+
i (z) and Q+
i (z) be generalized Stieltjes polyno-
mials of the first and the second kind, respectively. Then any solution of
the moment problemMP k
κ (s, 2νN−2) admits the following representation
f(z) =
Q+
2N−1(z)τ(z) +Q+
2N−2(z)
P+
2N−1(z)τ(z) + P+
2N−2(z)
, (5.9)
I. Kovalyov 493
where τ satisfies the conditions
τ(z) ∈ Nk−kN
κ−κN
and
1
τ(z)
= o(z), (z→̂∞). (5.10)
Furthermore, the resolvent matrix of the odd moment problem
MP k
κ (s,2νN− 2)
W2N−1(z) =
(
Q+
2N−1 Q+
2N−2
P+
2N−1 P+
2N−2
)
(5.11)
belongs to the UκN (J) and admits the following factorization
W2N−1(z) =M1(z)L1(z) . . . LN−1(z)MN (z), (5.12)
where the matrices Mj(z) and Lj(z) are defined by (4.12).
Proof. Assume f belongs to the Nevanlinna class Nk
κ and f has the
asymptotic expansion
f(z) = −s0
z
− · · · − s2νN−2
z2νN−1
+ o
(
1
z2νN−1
)
, z→̂∞.
Then, by Proposition 4.1, the function f takes the following form
f(z) =
1
−zm1(z) +
1
l1(z) + · · ·+
1
lN−1(z) +
1
−zmN (z) +
1
τ(z)
, (5.13)
where (5.10) holds and by Proposition 5.1, we can rewrite f as follows
f(z) =
Q+
2N−1(z)τ(z) +Q+
2N−2(z)
P+
2N−1(z)τ(z) + P+
2N−2(z)
, (5.14)
where the polynomials Q+
2N−2, Q
+
2N−1 and P+
2N−2, P
+
2N−1 are defined by
the recurrence relations (5.7)–(5.8).
Hence, the solution matrixW2N−1(z) is well defined by (5.11). Apply-
ing the induction, we show thatW2N−1(z) admits the factorization (5.12),
i.e.
(i) if i = 1, then W1(z) =M1(z) and
W1(z) =
(
Q+
1 (z) Q+
0 (z)
P+
1 (z) P+
0 (z)
)
; (5.15)
494 A truncated indefinite Stieltjes moment problem
(ii) if i = N − 1, then (4.12) and (5.12) hold (assumption of induc-
tion);
(iii) if i = N , then
W2N−1(z) =M1(z)L1(z) . . . LN−1(z)MN (z)=W2N−3(z)LN−1MN (z)
=
(
Q+
2N−3(z) Q+
2N−4(z)
P+
2N−3(z) P+
2N−4(z)
)(
1 lN−1(z)
0 1
)(
1 0
−zmN (z) 1
)
=
(
Q+
2N−3(z) lN−1(z)Q
+
2N−3(z) +Q+
2N−4(z)
P+
2N−3(z) lN−1(z)P
+
2N−3(z) + P+
2N−4(z)
)(
1 0
−zmN (z) 1
)
=
(
Q+
2N−3(z) Q+
2N−2(z)
P+
2N−3(z) P+
2N−2(z)
)(
1 0
−zmN (z) 1
)
=
(
Q+
2N−3(z)− zmN (z)Q+
2N−2(z) Q+
2N−2(z)
P+
2N−3(z)− zmN (z)P+
2N−2(z) P+
2N−2(z)
)
=
(
Q+
2N−1(z) Q+
2N−2(z)
P+
2N−1(z) P+
2N−2(z)
)
. (5.16)
So, (5.12) is proved.
By Lemmas 2.4 and 2.5 the matrices valued functionMi(z) and Li(z)
belong to the classes
Mi(z) ∈ Uκ−(zmi)(J) and Li(z) ∈ Uκ−(li)(J), i = 1, N.
As is known the product of mvf’s from the classes Uκ1(J) and Uκ2(J)
belongs to the class Uκ′(J), where κ′ ≤ κ1 + κ2.
Therefore
W2N−1(z) =M1(z)L1(z) . . . LN−1(z)MN (z) ∈ Uκ′(J),
where
κ′ ≤
N∑
j=1
κ−(zmj) +
N−1∑
j=1
κ−(lj) = κN . (5.17)
By [8, Lemma 3.4] the function f = TW2N−1
[1], corresponding to the
parameter τ(z) ≡ 1, belongs to the class Nκ′′ , with
κ′′ ≤ κ′. (5.18)
On the other hand, by Theorem 4.1 f = TW2N−1
[1] ∈ NκN , i.e.
κ′′ = κN . (5.19)
Comparing (5.17), (5.18) and (5.19) yields
κ′ = κ′′ = κN
and thus W2N−1 ∈ UκN (J). This completes the proof.
I. Kovalyov 495
Remark 5.3. In the case, where s ∈ H+
κ , deg(mi) ≤ 1 and li = const > 0
in (4.6), the moment problem MP+
κ (s, 2νN − 2) was studied in [21] and
these results are the special case of Theorem 5.2.
Remark 5.4. In the case. where f ∈ Nk,reg
κ , the odd moment problem
MP k
κ (s, 2nN −2) was studied in [10]. These results are contained in pre-
vious Theorem. Moreover, the polynomials lj(z) are non-zero constants
in (5.13), such that
lj(z) =
1
s
(j)
−1
. (5.20)
5.2. Even moment problem
Now we study the even moment problem MP k
κ (s, 2µN − 1). In this
case we also find all solutions of MP k
κ (s, 2µN − 1) by the following state-
ment
Theorem 5.5. Let s ∈ Hk
κ,2µN− 1, let (4.17) hold and let polynomials
mj(z) and lj(z) (1 ≤ j ≤ N) be defined by (4.10) and (4.11), respectively.
Let P+
i (z) and Q+
i (z) be generalized Stieltjes polynomials of the first and
the second kind, respectively. Then any solution of the moment problem
MP k
κ (s, 2µN − 1) admits the following representation
f(z) =
Q+
2N−1(z)τ(z) +Q+
2N (z)
P+
2N−1(z)τ(z) +Q+
2N (z)
, (5.21)
where τ satisfies the following conditions
τ(z) ∈ N
k−k+N
κ−κ+
N
and τ(z) = o(1), z→̂∞. (5.22)
Furthermore, the resolvent matrix of the even moment problem
MP k
κ (s, 2µN − 1)
W2N (z) =
(
Q+
2N−1(z) Q+
2N (z)
P+
2N−1(z) P+
2N (z)
)
(5.23)
belongs to the Uκ+
N
(J) and admits the following factorization
W2N (z) =M1(z)L1(z) . . .MN (z)LN (z), (5.24)
where the matrices Mj(z) and Lj(z) are defined by (4.12).
496 A truncated indefinite Stieltjes moment problem
Proof. Suppose f belongs to the Nevanlinna class Nk
κ and f has the
asymptotic expansion
f(z) = −s0
z
− · · · − s2µN−1
z2µN
+ o
(
1
z2µN
)
, z→̂∞.
By Proposition 4.2, the function f takes the form (4.21), where (5.22)
holds. By [24, chapter I], the function f can be rewritten in the form
(5.21), where the polynomials P+
i (z) and Q+
i (z) can be found as the
solutions of the recurrence relations (5.7)–(5.8).
Hence, the resolvent matrix of the even moment problem
MP k
κ (s, 2µN − 1) takes the form (5.23). By Theorem 5.2 (see (5.11)
and (5.12)), we obtain
M1(z)L1(z) . . . LN−1(z)MN (z)LN (z) =
(
Q+
2N−1 Q+
2N−2
P+
2N−1 P+
2N−2
)(
1 lN (z)
0 1
)
=
(
Q+
2N−1(z) lNQ
+
2N−1(z)+Q
+
2N−2(z)
P+
2N−1(z) lNP
+
2N−1(z)+P
+
2N−2(z)
)
=
(
Q+
2N−1(z) Q+
2N (z)
P+
2N−1(z) P+
2N (z)
)
=W2N (z).
By Lemmas 2.4 and 2.5 the mvf’s Mi(z) and Li(z) belong to the classes
Mi(z) ∈ Uκ−(zmi)(J) and Li(z) ∈ Uκ−(li)(J), i = 1, N.
As is known the product of mvf’s from the classes Uκ1(J) and Uκ2(J)
belongs to the class Uκ′(J), where κ′ ≤ κ1 + κ2. Hence
W2N (z) =M1(z)L1(z) . . . LN−1(z)MN (z)LN (z) ∈ Uκ′(J),
where
κ′ ≤
N∑
j=1
κ−(zmj) +
N∑
j=1
κ−(lj) = κ+N . (5.25)
By [8, Lemma 3.4] the function f = TW2N
[z], corresponding to the
parameter τ(z) = z, belongs to the class Nκ′′ , with
κ′′ ≤ κ′. (5.26)
On the other hand, by Theorem 4.2 f = TW2N
[z] ∈ Nκ+
N
, i.e.
κ′′ = κ+N . (5.27)
Comparing (5.25), (5.26) and (5.27) yields
κ′ = κ′′ = κ+N
and thus W2N ∈ Uκ+
N
(J). This completes the proof.
I. Kovalyov 497
Remark 5.6. The even moment problem MP k
κ (s, 2nN − 1) in the class
Nk,reg
κ was studied in [10] and the results in [10, Theorem 5.9] is the
special case of Theorem 5.5.
Acknowledgements
This work is supported by the grant of Volkswagen Foundation and by
Ministry of Education and Science of Ukraine (project number
0115U000556).
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Contact information
Ivan Kovalyov Dragomanov National
Pedagogical University,
Kiev, Ukraine
E-Mail: i.m.kovalyov@gmail.com
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