Inverse of Infinite Hankel Moment Matrices

Inverse of Infinite Hankel Moment Matrices††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

Christian BERG † and Ryszard SZWARC ‡

C. Berg and R. Szwarc

† Department of Mathematical Sciences, University of Copenhagen,
† Universitetsparken 5, DK-2100 Copenhagen, Denmark<EMAIL_ADDRESS>http://www.math.ku.dk/~berg/

‡ Institute of Mathematics, University of Wroclaw,
‡ pl. Grunwaldzki 2/4, 50-384 Wrocław, Poland<EMAIL_ADDRESS>http://www.math.uni.wroc.pl/~szwarc/

Received January 19, 2018, in final form October 02, 2018; Published online
October 06, 2018

Let $(s_{n})_{n\geq 0}$ denote an indeterminate Hamburger moment sequence and
let $\mathcal{H}=\\{s_{m+n}\\}$ be the corresponding positive definite Hankel
matrix. We consider the question if there exists an infinite symmetric matrix
$\mathcal{A}=\\{a_{j,k}\\}$, which is an inverse of $\mathcal{H}$ in the sense
that the matrix product $\mathcal{A}\mathcal{H}$ is defined by absolutely
convergent series and $\mathcal{A}\mathcal{H}$ equals the identity matrix
$\mathcal{I}$, a property called (aci). A candidate for $\mathcal{A}$ is the
coefficient matrix of the reproducing kernel of the moment problem, considered
as an entire function of two complex variables. We say that the moment problem
has property (aci), if (aci) holds for this matrix $\mathcal{A}$. We show that
this is true for many classical indeterminate moment problems but not for the
symmetrized version of a cubic birth-and-death process studied by Valent and
co-authors. We consider mainly symmetric indeterminate moment problems and
give a number of sufficient conditions for (aci) to hold in terms of the
recurrence coefficients for the orthonormal polynomials. A sufficient
condition is a rapid increase of the recurrence coefficients in the sense that
the quotient between consecutive terms is uniformly bounded by a constant
strictly smaller than one. We also give a simple example, where (aci) holds,
but an inverse matrix of $\mathcal{H}$ is highly non-unique.

indeterminate moment problems; Jacobi matrices; Hankel matrices; orthogonal
polynomials 42C05; 44A60; 47B36; 33D45; 60J80

##  1 Introduction and summary of results

Consider a Hamburger moment sequence $(s_{n})$ given as

$\displaystyle s_{n}=\int_{-\infty}^{\infty}x^{n}{\rm d}\mu(x),\qquad n\geq
0,$

where $\mu$ is a probability measure with infinite support and moments of any
order. The corresponding orthonormal polynomials $P_{n}$, obtained from the
sequence of monomials $x^{n}$ by the Gram–Schmidt procedure, satisfy $P_{0}=1$
and

$\displaystyle
xP_{n}(x)=b_{n}P_{n+1}(x)+a_{n}P_{n}(x)+b_{n-1}P_{n-1}(x),\qquad n\geq 0,$
(1.1)

where $b_{n}>0,$ $a_{n}\in\mathbb{R}$ with the convention $b_{-1}=1$,
$P_{-1}=0$. For $n\geq 0$ we have $\deg P_{n}=n$ and the leading coefficient
of $P_{n}$ is equal to

$\displaystyle{1\over b_{0}b_{1}\cdots b_{n-1}}.$ (1.2)

We are interested in the so-called indeterminate case, when the moments
$s_{n}$ do not determine the measure $\mu$ uniquely. Such measures must have
unbounded support. In particular the numbers $s_{2n}$ must grow faster than
any exponential.

The infinite Hankel matrix $\mathcal{H}=\\{h_{m,n}\\}$

$\displaystyle
h_{m,n}=s_{m+n}=\big{\langle}x^{m},x^{n}\big{\rangle}_{L^{2}(\mu)},\qquad
m,n\geq 0$

is positive definite in the sense that

$\displaystyle(\mathcal{H}v,v)=\sum_{m,n=0}^{\infty}s_{m+n}v_{n}\overline{v}_{m}>0,\qquad
v\in\mathcal{F}_{c}(\mathbb{N}_{0}),\qquad v\neq 0,$ (1.3)

where $\mathcal{F}_{c}(\mathbb{N}_{0})$ denotes the set of complex sequences
$v=(v_{n})_{n\geq 0}$, with only finitely many non-zero entries.

Berg, Chen and Ismail [3] showed that the quadratic form associated with the
matrix $\mathcal{H}$ is bounded below, i.e., for a suitable constant $C>0$

$\displaystyle(\mathcal{H}v,v)\geq C\|v\|^{2},\qquad
v\in\mathcal{F}_{c}(\mathbb{N}_{0}),$

if and only if the moment problem associated with $(s_{n})$ is indeterminate.

Hamburger obtained a related characterization of indeterminacy in terms of the
quadratic forms associated with the matrices $\\{s_{j+k}\\}$ and
$\\{s_{j+k+2}\\}$, see [23, p. 70]. For a comparison between Hamburger’s
result and the result of Berg–Chen–Ismail, see [3].

Yafaev [25] contains a characterization of closability of the Hankel form
(1.3) leading to the determinate case.

The smallest eigenvalue $\lambda_{N}$ of the truncated matrix
$\mathcal{H}_{N}=\\{s_{m+n}\\}_{0\leq m,n\leq N}$ has been studied in a number
of papers starting with Szegő and Widom–Wilf, see [7] and references therein.

The inverse of $\mathcal{H}_{N}$ can simply be described as the matrix

$\displaystyle\mathcal{A}^{(N)}=\\{a_{j,k}(N)\\}_{0\leq j,k\leq N},$

where

$\displaystyle
K_{N}(z,w)=\sum_{n=0}^{N}P_{n}(z)P_{n}(w)=\sum_{j,k=0}^{N}a_{j,k}(N)z^{j}w^{k}.$
(1.4)

A proof and historical details can be found in [7, Theorem 1.1].

In the indeterminate case it is possible to let $N\to\infty$ in (1.4) and
obtain the infinite matrix $\mathcal{A}=\\{a_{j,k}\\}$ with
$a_{j,k}=\lim\limits_{N\to\infty}a_{j,k}(N)$ as coefficient matrix of the
entire function of two complex variables

$\displaystyle
K(z,w)=\sum_{n=0}^{\infty}P_{n}(z)P_{n}(w)=\sum_{j,k=0}^{\infty}a_{j,k}z^{j}w^{k},$

called the reproducing kernel of the indeterminate moment problem.

The coefficients of $\mathcal{H}$ can grow fast, but on the other hand the
coefficients of $\mathcal{A}$ decay fast, so it is natural to examine if it is
possible to let $N$ tend to infinity in the equations

$\displaystyle\sum_{k=0}^{N}a_{i,k}(N)s_{k+j}=\delta_{i,j},\qquad 0\leq
i,j\leq N,$

and to obtain the infinite matrix equation

$\displaystyle\mathcal{A}\mathcal{H}=\mathcal{H}\mathcal{A}=\mathcal{I},$
(1.5)

i.e.,

$\displaystyle\sum_{k=0}^{\infty}a_{i,k}s_{k+j}=\delta_{i,j},\qquad i,j\geq
0,$ (1.6)

meaning that all the above series are convergent with the given sum. Notice
that since both matrices $\mathcal{H}$ and $\mathcal{A}$ are symmetric, it is
enough to prove $\mathcal{A}\mathcal{H}=\mathcal{I}$ in (1.5).

Since absolute convergence is often more easy to establish than convergence,
we investigate the question if the series in (1.6) are absolutely convergent
for all $i$, $j$.

For the Stieltjes–Wigert moment sequence (sometimes called log-normal moments)

$\displaystyle s_{n}=q^{-\tfrac{1}{2}(n+1)^{2}},\qquad 0<q<1,$

it was shown that (1.5) holds and all the series in (1.6) are absolutely
convergent, see [7, Section 5]. After this was done the same results were
established but not published for several other known indeterminate moment
problems, and we began to search for general theorems about (1.5). In this
paper we present the general results obtained so far. At present we have no
applications of equation (1.5).

If $\mathcal{H}$ is replaced by an infinite positive definite Toeplitz matrix
$M$, the question of existence of a classical inverse of $M$ in the sense of
(1.5) has been studied in [15].

The sequence $c_{k}:=\sqrt{a_{k,k}}$ plays a crucial role in our
investigations. It decays quickly to zero in the sense that
$\lim\limits_{k\to\infty}k\root k \of{c_{k}}=0$, see (2.13).

Concerning the matrix $\mathcal{A}$, it was shown in [7, Section 4] that it is
of trace class, and that

$\displaystyle\operatorname{tr}\mathcal{A}=\frac{1}{2\pi}\int_{0}^{2\pi}\sum_{k=0}^{\infty}\big{|}P_{k}\big{(}{\rm
e}^{{\rm i}t}\big{)}\big{|}^{2}{\rm d}t.$

In the indeterminate case the infinite Hankel matrix $\mathcal{H}$ does not
correspond to an operator on $\ell^{2}$ defined on the subspace spanned by the
standard orthonormal basis $(\delta_{n})_{n\geq 0}$ in $\ell^{2}$, where
$\delta_{n}=(\delta_{n,m})_{m\geq 0}$.

In fact, by a theorem of Carleman we necessarily have
$\sum\limits_{n=0}^{\infty}s_{2n}^{-1/(2n)}<\infty$, hence $s_{2n}\geq 1$ for
$n$ sufficiently large, and therefore

$\displaystyle\sum_{m=0}^{\infty}s_{n+m}^{2}=\infty\qquad\mbox{for all}\ n.$

In connection with the study of (1.5) we introduce the following:

######  Definition 1.1.

An indeterminate moment problem has property

  1. (c)

if all the series in (1.6) are convergent,

  2. (ac)

if all the series in (1.6) are absolutely convergent,

  3. (aci)

if it has property (ac) and (1.5) holds,

  4. (cs)

if $\sum\limits_{n=0}^{\infty}c_{2n}s_{2n}<\infty$,

  5. (cs*)

if $\sum\limits_{l=0}^{\infty}c_{l}|s_{l+m}|<\infty$ for all $m\geq 0$.

######  Theorem 1.2.

For an indeterminate moment problem the following holds:

  1. (i)

(cs*) $\implies$ (aci) $\implies$ (ac) $\implies$ (c),

  2. (ii)

(cs*) $\implies$ (cs).

Concerning (i), the claim (cs*) $\implies$ (aci) is proved in Lemma 2.1, the
rest is obvious. To see (ii) we put $m=0$ in (cs*) and get

$\displaystyle\sum_{n=0}^{\infty}c_{2n}s_{2n}\leq\sum_{l=0}^{\infty}c_{l}|s_{l}|<\infty.$

We shall mainly focus on the case of symmetric moment problems, where the
central recurrence coefficients $a_{n}=0$ for all $n$, and the moment problem
is characterized by the sequence $(b_{n})$.

######  Remark 1.3.

We are not able to decide if some of these implications can be reversed. Note
however, that for Stieltjes problems and for symmetric moment problems the
properties (ac) and (aci) are equivalent, cf. Propositions 2.4 and 2.5.

We next define some classes of sequences $(b_{n})$ which will be considered in
connection with the properties of Definition 1.1.

######  Definition 1.4.

A sequence $(b_{n})$ of positive numbers is called eventually log-convex if
there exists $n_{0}\in\mathbb{N}$ such that

$\displaystyle b_{n}^{2}\leq b_{n-1}b_{n+1},\qquad n\geq n_{0},$

and it is called eventually log-concave if the reverse inequality holds. For
$n_{0}=1$ we drop “eventually”.

The following lemma is given in [8, Lemma 4.1].

######  Lemma 1.5.

Let $(b_{n})$ be an unbounded positive sequence which is either eventually
log-convex or eventually log-concave. Then $(b_{n})$ is eventually strictly
increasing to infinity.

The following concept is crucial for several results.

######  Definition 1.6.

A sequence $(b_{n})$ of positive numbers is called eventually $q$-increasing,
if there exist a number $0<q<1$ and $n_{0}\in\mathbb{N}$ such that

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq q,\qquad n\geq n_{0}.$ (1.7)

For $n_{0}=1$ we drop “eventually”.

Note that if $(b_{n})$ is unbounded and eventually log-convex, then it is
eventually $q$-increasing with $q=b_{n_{0}-1}/b_{n_{0}}<1$, where $n_{0}$ is
so large that $(b_{n})$ is log-convex and strictly increasing for $n\geq
n_{0}$.

We recall that an indeterminate moment problem has an order $\rho$ satisfying
$0\leq\rho\leq 1$. This is the common order of the four functions of the
Nevanlinna matrix, see [5].

######  Proposition 1.7.

  1. $(i)$

The symmetric moment problem for an eventually $q$-increasing sequence
$(b_{n})$ is indeterminate and of order $0$.

  2. $(ii)$

The symmetric moment problem for a positive sequence $(b_{n})$ satisfying

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq{\rm e}^{-\beta/n},\qquad n\geq
n_{0},\qquad\beta>1,$ (1.8)

is indeterminate of order $\leq 1/\beta$.

###### Proof.

Let $(Q_{n})$ denote the sequence of polynomials of the second kind. By [8,
Theorem 1.2], it suffices to prove that
$\big{(}P_{n}^{2}(0)\big{)},\big{(}Q_{n}^{2}(0)\big{)}\in\ell^{\alpha}$ for
all $0<\alpha\leq 1$ in the first case and for all $\alpha>1/\beta$ in the
second case.

By the symmetry assumption we know that $P_{2n+1}(0)=Q_{2n}(0)=0$ and by [8,
Remark 4.5] we have

$\displaystyle P_{2n}(0)=(-1)^{n}\frac{b_{0}b_{2}\cdots
b_{2n-2}}{b_{1}b_{3}\cdots b_{2n-1}},\qquad
Q_{2n+1}(0)=(-1)^{n}\frac{b_{1}b_{3}\cdots b_{2n-1}}{b_{0}b_{2}\cdots
b_{2n}}.$ (1.9)

For $n>n_{0}$ such that (1.7) holds, we have

$\displaystyle P_{2n}^{2}(0)\leq\left(\frac{b_{0}b_{2}\cdots
b_{2n_{0}-2}}{b_{1}b_{3}\cdots b_{2n_{0}-1}}\right)^{2}q^{2(n-n_{0})},\qquad
Q_{2n+1}^{2}(0)\leq\left(\frac{b_{1}b_{3}\cdots b_{2n_{0}-1}}{b_{0}b_{2}\cdots
b_{2n_{0}}}\right)^{2}q^{2(n-n_{0})},$

and the first assertion follows.

For $n>n_{0}$ such that (1.8) holds, we have

$\displaystyle P_{2n}^{2}(0)\leq\left(\frac{b_{0}b_{2}\cdots
b_{2n_{0}-2}}{b_{1}b_{3}\cdots
b_{2n_{0}-1}}\right)^{2}\exp\left(-2\beta\sum_{j=n_{0}}^{n-1}\frac{1}{2j+1}\right),$

and a similar inequality for $Q_{2n+1}^{2}(0)$. Using the inequality

$\displaystyle\sum_{j=n_{0}}^{n-1}\frac{1}{2j+1}\geq\int_{n_{0}}^{n}\frac{{\rm
d}x}{2x+1}=\frac{1}{2}\log\frac{2n+1}{2n_{0}+1},$

we get

$\displaystyle P_{2n}^{2}(0)\leq\frac{C}{(2n+1)^{\beta}}$

for a suitable constant $C$, and the second assertion follows. ∎

The following examples have $q$-increasing coefficients $b_{n}$.

######  Example 1.8.

The discrete $q$-Hermite polynomials of type II are given by

$\displaystyle
x\tilde{h}_{n}(x;q)=\tilde{h}_{n+1}(x;q)+q^{-2n+1}\big{(}1-q^{n}\big{)}\tilde{h}_{n-1}(x;q),$

cf. [20], where $0<q<1$. This is the monic version and the corresponding
coefficients for the orthonormal version are

$\displaystyle b_{n}=q^{-n-1/2}\sqrt{1-q^{n+1}}.$

It follows that

$\displaystyle\frac{b_{n-1}}{b_{n}}=q\sqrt{\frac{1-q^{n}}{1-q^{n+1}}}<q,$

so $(b_{n})$ is $q$-increasing and log-concave.

It is known that $s_{2n}=\big{(}q;q^{2}\big{)}_{n}q^{-n^{2}}$, where the
shifted factorials are defined in (3.10).

In [14] Christiansen and Ismail studied symmetric Al-Salam–Chihara polynomials
of type II given by the recursion

$\displaystyle
2xp^{n}Q_{n}(x;\beta)=\big{(}1-p^{n+1}\big{)}Q_{n+1}(x;\beta)+\big{(}\beta+p^{n-1}\big{)}Q_{n-1}(x;\beta),$

where $0<p<1$, $\beta\geq 0$. The recurrence coefficients for the orthonormal
version are

$\displaystyle
b_{n}=\tfrac{1}{2}p^{-n-1/2}\big{[}\big{(}1-p^{n+1}\big{)}\big{(}\beta+p^{n}\big{)}\big{]}^{1/2}.$

It follows that

$\displaystyle\frac{b_{n-1}}{b_{n}}=p\left[\frac{1-p^{n}}{1-p^{n+1}}\frac{\beta+p^{n-1}}{\beta+p^{n}}\right]^{1/2}\leq
p\left[\frac{\beta+1}{\beta+p}\right]^{1/2}\leq\sqrt{p},$

so $(b_{n})$ is $\sqrt{p}$-increasing.

One can prove that $(b_{n})$ is eventually log-convex when $0<\beta<1/p$ and
eventually log-concave when $\beta\geq 1/p$ or $\beta=0$. The case $\beta=0$
yields the $p^{-1}$-Hermite polynomials studied by Ismail and Masson in [19].

We shall now formulate our main results for symmetric moment problems given in
terms of $a_{n}=0$ and $b_{n}>0$ from (1.1). Our results are based on a study
of two quantities $U_{n}$, $V_{n}$ defined in Section 2, see (2.11) and
(2.12). As far as we know these quantities have not been investigated
previously. They are treated in Sections 3 and 4, and to avoid duplication we
shall not mention these technical results here.

We need the following decreasing functions

$\displaystyle u(\alpha):=\max_{2^{-1/\alpha}\leq x\leq
1}(x+1)\int_{x}^{1}\log\left(\frac{y^{\alpha}}{1-y^{\alpha}}\right)\frac{{\rm
d}y}{(1+y)^{2}},\qquad\alpha>0,$ (1.10) $\displaystyle
v(\alpha):=\int_{1}^{\infty}\log\big{(}\big{[}1-x^{-\alpha}\big{]}^{-1}\big{)}{\rm
d}x,\qquad\alpha>1,$ (1.11) $\displaystyle
k(\alpha):=4^{-\alpha}\exp(2u(\alpha)+v(\alpha)),\qquad\alpha>1,$ (1.12)

which are motivated by Theorems 3.9, 4.10, and 5.2 respectively. The limiting
values of these functions at the end-points of their interval of definition
are also given there. Further properties of the functions $u$, $v$ are given
in Appendix A.

######  Theorem 1.9.

Consider a symmetric moment problem given in terms of an eventually
$q$-increasing sequence $(b_{n})$, cf. Definition 1.6. We then have

  1. $(i)$

$(U_{n})$, $(V_{n})$, $\big{(}c_{n}\sqrt{s_{2n}}\big{)}$ are bounded
sequences.

  2. $(ii)$

Property (cs*) holds.

This result follows from Theorems 3.7 and 4.5, Corollary 4.6 and Theorem 5.1.

We next relax the condition of $(b_{n})$ being eventually $q$-increasing.

######  Theorem 1.10.

Consider a symmetric moment problem given in terms of $(b_{n})$ satisfying

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq{\rm e}^{-f(n)},\qquad n\geq n_{0},$

where $f(n)>0$ for $n\geq n_{0}$ and
$\alpha:=\liminf\limits_{n\to\infty}nf(n)\in[0,\infty]$.

We then have

  1. $(i)$

$\limsup\limits_{n\to\infty}\root n \of{U_{n}}\leq{\rm e}^{2u(\alpha)}$,
$0\leq\alpha\leq\infty$,

  2. $(ii)$

$\limsup\limits_{n\to\infty}\root n \of{V_{n}}\leq{\rm e}^{v(\alpha)/2}$,
$1\leq\alpha\leq\infty$,

  3. $(iii)$

Property (cs*) holds if $k(\alpha)<1$, which in turn holds for $\alpha\geq
1.68746$, cf. Remark 5.3.

This result follows from Theorems 3.10, 4.11, and 5.2.

The last theorem applied to the symmetric indeterminate moment problem with
$b_{n}=(n+1)^{c}$, $c>1$, yields that (cs*) holds for $c\geq 1.68746$. This is
not optimal, because in Section 7 we prove by completely different methods
that (cs*) actually holds for $c>3/2$ and that property (cs) (and hence (cs*))
does not hold for $1<c<3/2$. The behaviour for the border case $c=3/2$ is not
known.

A main difficulty is that almost nothing is known about the orthonormal
polynomials and related quantities in case $b_{n}=(n+1)^{c}$, see the preamble
to Section 7. We have relied on the symmetrized version of a birth-and-death
process with cubic rates studied by Valent in [24]. In Section 6 we find an
estimate of the moments and explicit expressions for the entries of the matrix
$\mathcal{A}$ in that case.

A symmetric moment problem given by an eventually log-convex or eventually
log-concave sequence $(b_{n})$ is indeterminate if and only if the Carleman
condition $\sum 1/b_{n}<\infty$ holds. In the indeterminate case the order
$\rho$ of the moment problem is equal to the exponent of convergence
$\mathcal{E}(b_{n})$ of $(b_{n})$ defined as

$\displaystyle\mathcal{E}(b_{n})=\inf\left\\{\alpha>0\colon\sum_{n=0}^{\infty}\frac{1}{b_{n}^{\alpha}}<\infty\right\\}.$

See Theorems 1.4 and 4.11 in [8]. The log-concave case builds on technique
introduced by Berezanskiĭ, see [1, p. 26].

In Appendix A we take up certain questions related to the above, but which are
not needed in the main results. We prove that $\liminf_{n}U_{n}^{1/n}\geq 4$
for all bounded sequences $(b_{n})$. In Remark A.4 we notice that although
$\mathcal{A}$ can be an inverse of $\mathcal{H}$, it is not uniquely
determined by being so. We also give some properties of the functions $u$,
$v$.

##  2 Preliminaries

In this section we shall summarize some notation and results from our papers
[7, 8, 9], which will be needed in the following. We use the following
notation for the orthonormal polynomials

$\displaystyle
P_{n}(x)=b_{n,n}x^{n}+b_{n-1,n}x^{n-1}+\cdots+b_{1,n}x+b_{0,n},$
$\displaystyle
x^{n}=c_{n,n}P_{n}(x)+c_{n-1,n}P_{n-1}(x)+\cdots+c_{1,n}P_{1}(x)+c_{0,n}P_{0}(x).$
(2.1)

By (1.2) we get

$\displaystyle b_{n,n}={1\over b_{0}b_{1}\cdots b_{n-1}},\qquad
c_{n,n}=b_{0}b_{1}\cdots b_{n-1}.$ (2.2)

The matrices $\mathcal{B}=\\{b_{i,j}\\}$ and $\mathcal{C}=\\{c_{i,j}\\}$ with
the assumption

$\displaystyle b_{i,j}=c_{i,j}=0\qquad\text{for}\quad i>j$

are upper-triangular. Since $\mathcal{B}$ and $\mathcal{C}$ are transition
matrices between two sequences of linearly independent systems of functions,
we have

$\displaystyle\mathcal{B}\mathcal{C}=\mathcal{C}\mathcal{B}=\mathcal{I}.$

Observe that (for say $m\leq n$)

$\displaystyle
s_{m+n}=h_{m,n}=\big{\langle}x^{m},x^{n}\big{\rangle}_{L^{2}(\mu)}=c_{0,m}c_{0,n}+c_{1,m}c_{1,n}+\cdots+c_{m,m}c_{m,n}=\big{(}\mathcal{C}^{t}\mathcal{C}\big{)}_{m,n},$

so $\mathcal{H}=\mathcal{C}^{t}\mathcal{C}$.

Hence we have

$\displaystyle\mathcal{I}=\mathcal{B}\mathcal{C}=\mathcal{B}\big{(}\mathcal{B}^{t}\mathcal{C}^{t}\mathcal{C}\big{)}=\mathcal{B}\big{(}\mathcal{B}^{t}\mathcal{H}\big{)}.$
(2.3)

The multiplication is well defined as $\mathcal{B}$, $\mathcal{C}$ are upper-
triangular matrices and $\mathcal{B}^{t}$, $\mathcal{C}^{t}$ are lower-
triangular, so we are always dealing with finite sums in the matrix products.

Let us now assume that the moment problem is indeterminate. We recall from [7,
Proposition 4.2] that $\mathcal{B}$ is a Hilbert–Schmidt matrix and that

$\displaystyle\mathcal{A}=\mathcal{B}\mathcal{B}^{t},$ (2.4)

This means

$\displaystyle a_{k,l}=\sum^{\infty}_{n=\max(k,l)}b_{k,n}b_{l,n},$ (2.5)

hence

$\displaystyle
a_{k,l}^{2}\leq\left(\sum_{n=k}^{\infty}b_{k,n}^{2}\right)\left(\sum_{n=l}^{\infty}b_{l,n}^{2}\right).$

Let us recall some of the arguments leading to these results. Because of
indeterminacy it is known that the series

$\displaystyle\sum_{n=0}^{\infty}|P_{n}(z)|^{2}$

is convergent uniformly for $z$ in bounded subsets of $\mathbb{C}$. Moreover,
the sum has subexponential growth by a result of M. Riesz, cf. [1, p. 56],
i.e., it is dominated by a multiple of ${\rm e}^{\varepsilon|z|}$ for any
$\varepsilon>0$. Fixing $r>0$, we have

$\displaystyle P_{n}\big{(}r{\rm e}^{{\rm
i}t}\big{)}=\sum_{k=0}^{n}b_{k,n}r^{k}{\rm e}^{{\rm i}tk}.$

By Parseval’s identity we obtain

$\displaystyle{1\over 2\pi}\int_{0}^{2\pi}\big{|}P_{n}\big{(}r{\rm e}^{{\rm
i}t}\big{)}\big{|}^{2}{\rm d}t=\sum_{k=0}^{n}b_{k,n}^{2}r^{2k}.$

Next

$\displaystyle{1\over
2\pi}\int_{0}^{2\pi}\sum_{n=0}^{\infty}\big{|}P_{n}\big{(}r{\rm e}^{{\rm
i}t}\big{)}\big{|}^{2}{\rm
d}t=\sum_{n=0}^{\infty}\sum_{k=0}^{n}b_{k,n}^{2}r^{2k}=\sum_{k=0}^{\infty}r^{2k}\sum_{n=k}^{\infty}b_{k,n}^{2}.$

Therefore, denoting

$\displaystyle
c_{k}=\left(\sum_{n=k}^{\infty}b_{k,n}^{2}\right)^{1/2}=\sqrt{a_{k,k}},$ (2.6)

we have

$\displaystyle\sum_{k=0}^{\infty}c_{k}^{2}r^{2k}={1\over
2\pi}\int_{0}^{2\pi}\sum_{n=0}^{\infty}\big{|}P_{n}\big{(}r{\rm e}^{{\rm
i}t}\big{)}\big{|}^{2}{\rm d}t.$

(For $r=1$ this proves that $\mathcal{B}$ is Hilbert–Schmidt, and in
particular the series in (2.5) is absolutely convergent.)

######  Lemma 2.1.

Consider an indeterminate moment problem. If

$\displaystyle\sum_{n,l=0}^{\infty}|b_{k,n}b_{l,n}s_{l+m}|<\infty\qquad\mbox{for
all}\quad k,m\geq 0,$ (2.7)

then property (aci) from Definition 1.1 holds.

For (2.7) to hold it suffices that

$\displaystyle\sum_{l=0}^{\infty}c_{l}|s_{l+m}|<\infty\qquad\mbox{for
all}\quad m\geq 0.$ (2.8)

In particular (cs*) implies (aci).

###### Proof.

We first notice that (2.7) implies (ac) because of (2.5). We next note that
(2.7) implies

$\displaystyle\mathcal{B}\big{(}\mathcal{B}^{t}\mathcal{H}\big{)}=\big{(}\mathcal{B}\mathcal{B}^{t}\big{)}\mathcal{H},$

so by (2.3) and (2.4) we get

$\displaystyle\mathcal{I}=\mathcal{B}\big{(}\mathcal{B}^{t}\mathcal{H}\big{)}=\big{(}\mathcal{B}\mathcal{B}^{t}\big{)}\mathcal{H}=\mathcal{A}\mathcal{H}.$

Suppose next that (2.8) holds. By the Cauchy–Schwarz inequality

$\displaystyle\sum_{n,l=0}^{\infty}|b_{k,n}b_{l,n}s_{l+m}|\leq\sum_{l=0}^{\infty}c_{k}c_{l}|s_{l+m}|<\infty,$

where we have used (2.6). ∎

######  Remark 2.2.

It is tempting to use $|s_{l+m}|\leq\sqrt{s_{2l}}\sqrt{s_{2m}}$ to assert that
the conditions of Lemma 2.1 are satisfied if $\sum c_{l}\sqrt{s_{2l}}<\infty$.
This is not possible because $c_{l}\sqrt{s_{2l}}\geq 1$ by Lemma 2.6.

We shall mainly be concerned about (1.5) in the symmetric case, characterized
by $a_{n}\equiv 0$, but since there is a one-to-one correspondence between
Stieltjes moment problems and symmetric moment problems, we shall first
compare (1.5) in the two cases.

Suppose

$\displaystyle t_{n}=\int_{0}^{\infty}t^{n}{\rm d}\nu(t),\qquad n\geq 0$

is a Stieltjes moment sequence of a probability measure $\nu$ on $[0,\infty[$
with infinite support. The corresponding orthonormal polynomials $(P_{n})$
satisfy (1.1), where now $a_{n}>0$. Let $(R_{n})$ denote the orthonormal
polynomials with respect to the measure $t{\rm d}\nu(t)$. Then

$\displaystyle S_{2n}(x):=P_{n}\big{(}x^{2}\big{)},\qquad
S_{2n+1}(x):=xR_{n}\big{(}x^{2}\big{)}$

are the orthonormal polynomials with respect to the unique symmetric
probability measure $\mu$ on $\mathbb{R}$ such that for all non-negative Borel
functions $f\colon[0,\infty[\to\mathbb{R}$

$\displaystyle\int_{-\infty}^{\infty}f\big{(}x^{2}\big{)}{\rm
d}\mu(x)=\int_{0}^{\infty}f(t){\rm d}\nu(t).$

The moments $s_{n}$ of $\mu$ are $s_{2n}=t_{n}$, $s_{2n+1}=0$.

The three terms recurrence relation for $(S_{n})$ has the form

$\displaystyle xS_{n}(x)={}^{s}b_{n}S_{n+1}(x)+{}^{s}b_{n-1}S_{n-1}(x)$ (2.9)

with

$\displaystyle a_{0}=({}^{s}b_{0})^{2},\qquad
b_{n}=({}^{s}b_{2n})({}^{s}b_{2n+1}),\qquad n\geq 0,\qquad
a_{n}=({}^{s}b_{2n})^{2}+({}^{s}b_{2n-1})^{2},\qquad n\geq 1.$

It is known that $\mu$ is indeterminate if and only if $\nu$ is Stieltjes
indeterminate, i.e., there are measures on $[0,\infty[$ different from $\nu$
with the same moments as $\nu$, cf. [13, p. 333]. In the affirmative case we
denote the reproducing kernels

$\displaystyle
K(z,w)=\sum_{n=0}^{\infty}P_{n}(z)P_{n}(w)=\sum_{j,k=0}^{\infty}a_{j,k}z^{j}w^{k},$
$\displaystyle
K^{\prime}(z,w)=\sum_{n=0}^{\infty}R_{n}(z)R_{n}(w)=\sum_{j,k=0}^{\infty}a^{\prime}_{j,k}z^{j}w^{k}.$

(Notice that also $t{\rm d}\nu(t)$ is Stieltjes indeterminate.)

The reproducing kernel for the symmetric measure $\mu$ is given as

$\displaystyle
K_{s}(z,w)=\sum_{n=0}^{\infty}S_{n}(z)S_{n}(w)=\sum_{n=0}^{\infty}P_{n}\big{(}z^{2}\big{)}P_{n}\big{(}w^{2}\big{)}+zw\sum_{n=0}^{\infty}R_{n}\big{(}z^{2}\big{)}R_{n}\big{(}w^{2}\big{)}$
$\displaystyle\hphantom{K_{s}(z,w)}{}=\sum_{j,k=0}^{\infty}a_{j,k}z^{2j}w^{2k}+\sum_{j,k=0}^{\infty}a^{\prime}_{j,k}z^{2j+1}w^{2k+1}=\sum_{j,k=0}^{\infty}a^{(s)}_{j,k}z^{j}w^{k},$
(2.10)

where

$\displaystyle a_{j,k}^{(s)}=\begin{cases}a_{j/2,k/2},&j,\ k\ \text{even},\\\
0,&j+k\ \text{odd},\\\ a^{\prime}_{(j-1)/2,(k-1)/2},&j,\ k\
\text{odd}.\end{cases}$

We then easily get the following about property (aci) from Definition 1.1:

######  Proposition 2.3.

The symmetric moment problem with moments $(s_{n})$ has property (aci) if and
only if the two Stieltjes problems with moments $(t_{n})$ and $(t_{n+1})$ have
property (aci).

For a Stieltjes problem we notice that the coefficients $b_{k,n}$ of the
orthonormal polynomials $P_{n}$ have the sign pattern
$b_{k,n}=(-1)^{n-k}|b_{k,n}|$ because the zeros of $P_{n}$ are all positive.
This implies that

$\displaystyle
a_{k,l}=\sum_{n=0}^{\infty}b_{k,n}b_{l,n}=(-1)^{k+l}\sum_{n=0}^{\infty}|b_{k,n}||b_{l,n}|,$

hence

$\displaystyle\sum_{l=0}^{\infty}|a_{k,l}s_{l+m}|=\sum_{l,n=0}^{\infty}|b_{k,n}||b_{l,n}|s_{l+m}.$

By Lemma 2.1 this shows

######  Proposition 2.4.

For an indeterminate Stieltjes problem the properties (ac) and (aci) from
Definition 1.1 are equivalent.

Combining this with Proposition 2.3 we immediately get:

######  Proposition 2.5.

For a symmetric indeterminate moment problem the properties (ac) and (aci)
from Definition 1.1 are equivalent.

In order to verify convergence of the series in (2.8), we shall estimate the
sequences $(c_{n})$, defined in (2.6), and $(s_{2n})$ in terms of the
quantities

$\displaystyle U_{n}:=\frac{s_{2n}}{b_{0}^{2}b_{1}^{2}\cdots
b_{n-1}^{2}},\qquad n\geq 1,\qquad U_{0}:=s_{0}=1$ (2.11)

and

$\displaystyle V_{n}:=b_{0}b_{1}\cdots b_{n-1}c_{n},\qquad n\geq 1,\qquad
V_{0}:=c_{0}>1.$ (2.12)

This will be done in the next two sections.

The quantity $U_{n}$ is defined for any moment problem, while $V_{n}$ only
makes sense for indeterminate problems.

It is easy to see that both quantities are scale invariant in the sense that
if $(a_{n})$, $(b_{n})$ from the recurrence relation (1.1) are replaced by a
constant multiple $(\lambda a_{n})$, $(\lambda b_{n})$ for some $\lambda>0$,
then $U_{n}$, $V_{n}$ remain unchanged.

In the following we shall often compare quantities like
$s_{n},U_{n},c_{n},\ldots$ depending on the recurrence coefficients $(a_{n})$,
$(b_{n})$ with those corresponding to other recurrence coefficients, say
$(\tilde{a}_{n})$, $\big{(}\tilde{b}_{n}\big{)}$. We then denote the
corresponding quantities $\tilde{s}_{n},\tilde{U}_{n},\tilde{c}_{n},\ldots$.

######  Lemma 2.6.

  1. $(i)$

For any moment problem we have $U_{n}\geq 1$.

  2. $(ii)$

For any indeterminate moment problem we have $V_{n}\geq 1$ and
$V_{n}\sqrt{U_{n}}=c_{n}\sqrt{s_{2n}}\geq 1$.

###### Proof.

By (2.1) we have

$\displaystyle s_{2n}=\langle
x^{n},x^{n}\rangle=c_{0,n}^{2}+c_{1,n}^{2}+\dots+c_{n,n}^{2}\geq
c_{n,n}^{2}=b_{0}^{2}b_{1}^{2}\cdots b_{n-1}^{2},$

hence by (2.6) and (2.2)

$\displaystyle c_{n}=\left(\sum_{j=n}^{\infty}b_{n,j}^{2}\right)^{1/2}\geq
b_{n,n}={1\over b_{0}b_{1}\cdots b_{n-1}}\geq 1/\sqrt{s_{2n}}.$ ∎

In [8, Section 3] we considered the power series

$\displaystyle\Phi(z)=\sum_{n=0}^{\infty}c_{n}z^{n},$

and proved that $\Phi$ is an entire function of minimal exponential type. In
terms of $(c_{n})$ this can be expressed

$\displaystyle\lim_{k\to\infty}kc_{k}^{1/k}=0.$ (2.13)

It was also proved there that $\Phi$ has the same order $\rho$ and type $\tau$
as the moment problem, which by definition is the common order and type of the
entire functions in the Nevanlinna matrix, see [5, 6]. From the general
formula for type of an entire function in terms of its power series
coefficients, see [21], we then get the following result.

######  Theorem 2.7.

Consider an indeterminate moment problem of order $0<\rho<1$ and type
$0\leq\tau\leq\infty$. Then

$\displaystyle\limsup_{k\to\infty}k^{1/\rho}c_{k}^{1/k}=({\rm
e}\rho\tau)^{1/\rho}.$ (2.14)

##  3 Estimates of the even moments via $\boldsymbol{U_{n}}$

In all of this section we assume that the moment problem is symmetric or
equivalently that $a_{n}\equiv 0$, so the recurrence relation (1.1) takes the
simplified form

$\displaystyle xP_{n}=b_{n}P_{n+1}+b_{n-1}P_{n-1}.$ (3.1)

######  Proposition 3.1.

Let $(b_{n})$ and $\big{(}\tilde{b}_{n}\big{)}$ denote two sequences of
positive numbers satisfying $b_{n}\leq\tilde{b}_{n}$, $n\geq 0$. Then the
corresponding even moments satisfy $s_{2n}\leq\tilde{s}_{2n}$, $n\geq 0$.

###### Proof.

Let $J$ denote the Jacobi matrix associated to the moment problem, i.e.,

$\displaystyle J=\begin{pmatrix}0&b_{0}&0&0&\cdots\\\
b_{0}&0&b_{1}&0&\cdots\\\ 0&b_{1}&0&b_{2}&\cdots\\\ 0&0&b_{2}&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix}.$ (3.2)

It is well-known that $s_{n}=(J^{n}\delta_{0},\delta_{0})$, and the conclusion
follows. ∎

In the symmetric case $P_{k}$ involves only monomials with the same parity as
$k$. Thus we have

$\displaystyle{x^{n}\over b_{0}b_{1}\cdots
b_{n-1}}=\sum_{k=0}^{[n/2]}u_{n,k}P_{n-2k}=u_{n,0}P_{n}+u_{n,1}P_{n-2}+\cdots,$
(3.3)

hence

$\displaystyle U_{n}:={s_{2n}\over b_{0}^{2}b_{1}^{2}\cdots
b_{n-1}^{2}}=\sum_{k=0}^{[n/2]}u_{n,k}^{2}.$ (3.4)

(Compared with (2.1) we have $u_{n,k}=c_{n-2k,n}/(b_{0}\cdots b_{n-1})$, while
$c_{n-2k-1,n}=0$.) By (1.2) we have $u_{n,0}=1$.

By the recurrence relation (3.1) the coefficients $u_{n,k}$ satisfy

$\displaystyle u_{n+1,k}={b_{n-2k}\over b_{n}}u_{n,k}+{b_{n+1-2k}\over
b_{n}}u_{n,k-1},\qquad 1\leq k\leq[n/2].$ (3.5)

In fact, multiplying (3.3) by $x/b_{n}$ and using the recurrence relation
(3.1), we get

$\displaystyle\frac{x^{n+1}}{b_{0}b_{1}\cdots
b_{n}}=\sum_{k=0}^{[n/2]}\frac{u_{n,k}}{b_{n}}\left(b_{n-2k}P_{n-2k+1}+b_{n-2k-1}P_{n-2k-1}\right),$

and (3.5) follows. We may extend the definition by setting $u_{n,k}=0$ for
$k>[n/2]$ or $k<0$ and $b_{n}=0$ for $n<0$. In this way it is not difficult to
see that

$\displaystyle u_{n+1,k}={b_{n-2k}\over b_{n}}u_{n,k}+{b_{n+1-2k}\over
b_{n}}u_{n,k-1},\qquad n,k\geq 0,$ (3.6)

and it follows by induction that all coefficients $u_{n,k}\geq 0$.

By (3.4) and (3.5) this immediately implies the following.

######  Proposition 3.2.

Assume $(b_{n})$ and $\big{(}\tilde{b}_{n}\big{)}$ are positive sequences
satisfying

$\displaystyle{b_{n-1}\over
b_{n}}\leq{\tilde{b}_{n-1}\over\tilde{b}_{n}},\qquad n\geq 1,$

and let $s_{2n}$ and $\tilde{s}_{2n}$ denote the corresponding even moments.
Then

$\displaystyle U_{n}:={s_{2n}\over b_{0}^{2}b_{1}^{2}\cdots
b_{n-1}^{2}}\leq\tilde{U}_{n}:={\tilde{s}_{2n}\over\tilde{b}_{0}^{2}\tilde{b}_{1}^{2}\cdots\tilde{b}_{n-1}^{2}}.$

For $n_{0}\in\mathbb{N}$ the shifted recurrence sequence
$b_{n}^{(n_{0})}:=b_{n+n_{0}}$, $n\geq 0$ leads to moments $s_{n}^{(n_{0})}$
and

$\displaystyle
U_{n}^{(n_{0})}:=\frac{s_{2n}^{(n_{0})}}{\big{(}b_{0}^{(n_{0})}\cdots
b_{n-1}^{(n_{0})}\big{)}^{2}}=\frac{s_{2n}^{(n_{0})}}{b_{n_{0}}^{2}\cdots
b_{n_{0}+n-1}^{2}}.$

######  Proposition 3.3.

The expressions above satisfy

$\displaystyle s_{2n+2}\geq b_{0}^{2}s_{2n}^{(1)},\qquad U_{n+1}\geq
U_{n}^{(1)}.$ (3.7)

If $b_{n}\leq b_{n}^{(n_{0})}$ for $n\geq 0$ and $\lim b_{n}^{1/n}=1$, then

$\displaystyle\limsup_{n\to\infty}U_{n}^{1/n}=\limsup_{n\to\infty}\big{(}U_{n}^{(n_{0})}\big{)}^{1/n}.$
(3.8)

The latter holds for $b_{n}=\lambda(n+1)^{c}$, $\lambda,c>0$.

###### Proof.

Define

$\displaystyle J_{(1)}=\begin{pmatrix}0&0&0&0&\cdots\\\ 0&0&b_{1}&0&\cdots\\\
0&b_{1}&0&b_{2}&\cdots\\\ 0&0&b_{2}&0&\cdots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{pmatrix}=\begin{pmatrix}0&0\\\
0&J^{(1)}\end{pmatrix},$

where $J^{(1)}$ is the Jacobi matrix (3.2) corresponding to $b_{n}^{(1)}$ and
the matrix to the right is a block matrix. The upper left $0$ is a scalar and
the upper right $0$ is an infinite row vector of zeros. In block matrix
notation we then have

$\displaystyle\big{(}J_{(1)}\big{)}^{n}=\begin{pmatrix}0&0\\\
0&\big{(}J^{(1)}\big{)}^{n}\end{pmatrix},$

hence

$\displaystyle
s_{2n}^{(1)}=\big{(}(J_{(1)})^{2n}\delta_{1},\delta_{1}\big{)}.$

This gives

$\displaystyle
s_{2n+2}=\big{(}J^{2n+2}\delta_{0},\delta_{0}\big{)}=\big{(}J^{2n}J\delta_{0},J\delta_{0}\big{)}=b_{0}^{2}\big{(}J^{2n}\delta_{1},\delta_{1}\big{)}\geq
b_{0}^{2}\big{(}(J_{(1)})^{2n}\delta_{1},\delta_{1}\big{)}=b_{0}^{2}s_{2n}^{(1)},$

hence $U_{n+1}\geq U_{n}^{(1)}$.

If $b_{n}\leq b_{n}^{(n_{0})}$ we have $s_{2n}\leq s_{2n}^{(n_{0})}$ by
Proposition 3.1, and therefore for $n>n_{0}$

$\displaystyle U_{n}\leq\frac{b_{n_{0}}^{2}b_{n_{0}+1}^{2}\cdots
b_{n_{0}+n-1}^{2}}{b_{0}^{2}b_{1}^{2}\cdots
b_{n-1}^{2}}U_{n}^{(n_{0})}\leq\frac{b_{n}^{2}b_{n+1}^{2}\cdots
b_{n+n_{0}-1}^{2}}{b_{0}^{2}b_{1}^{2}\cdots b_{n_{0}-1}^{2}}U_{n+n_{0}}.$

If $\lim b_{n}^{1/n}=1$ we get (3.8). ∎

######  Lemma 3.4.

Assume $(b_{n})$ satisfies

$\displaystyle\sup_{i,j\geq 0}\frac{b_{i}}{b_{i+j}}=M<\infty.$

Then

$\displaystyle U_{n+1}\leq 4M^{2}U_{n},\qquad s_{2n+2}\leq
4M^{2}b_{n}^{2}s_{2n}.$ (3.9)

###### Proof.

By the recurrence equation (3.5) we get

$\displaystyle u_{n+1,k}\leq M(u_{n,k}+u_{n,k-1}),$

hence

$\displaystyle u_{n+1,k}^{2}\leq
2M^{2}\big{(}u_{n,k}^{2}+u_{n,k-1}^{2}\big{)}.$

This gives

$\displaystyle U_{n+1}=\sum_{k=0}^{[(n+1)/2]}u_{n+1,k}^{2}\leq 4M^{2}U_{n},$

hence $s_{2n+2}\leq 4M^{2}b_{n}^{2}s_{2n}$. ∎

In the following we shall use the notation from [16] about $q$-shifted
factorials. For $z,q\in\mathbb{C}$ we write

$\displaystyle(z;q)_{n}=\prod_{j=0}^{n-1}\big{(}1-zq^{j}\big{)},\qquad
n=1,2,\ldots,\qquad(z;q)_{0}=1.$ (3.10)

When $|q|<1$ we define the absolutely convergent infinite product

$\displaystyle(z;q)_{\infty}=\prod_{j=0}^{\infty}\big{(}1-zq^{j}\big{)}.$
(3.11)

######  Lemma 3.5.

Assume $(b_{n})$ is log-concave and strictly increasing. Define
$q_{n}:=b_{n-1}/b_{n}$, $n\geq 1$. Then $0<q_{n}\leq q_{n+1}<1$ and

$\displaystyle u_{n,k}\leq
q_{n}^{k^{2}}\prod_{j=1}^{k}\big{(}1-q_{n}^{2j}\big{)}^{-1}=\frac{q_{n}^{k^{2}}}{\big{(}q_{n}^{2};q_{n}^{2}\big{)}_{k}},\qquad
n\geq 1,\qquad k\geq 0.$ (3.12)

###### Proof.

The inequality trivially holds when $k=0$ independent of $n$, because both
sides are 1. Since $u_{1,k}=0$ for $k\geq 1$ the inequality holds in
particular for $n=1$ and all $k$. Notice that by assumption

$\displaystyle\frac{b_{n-j}}{b_{n}}=q_{n-j+1}q_{n-j+2}\cdots q_{n}\leq
q_{n}^{j}.$

Let us define

$\displaystyle\tau_{n,k}:=\frac{q_{n}^{k^{2}}}{\big{(}q_{n}^{2};q_{n}^{2}\big{)}_{k}}.$

Since $q_{n}\leq q_{n+1}$ we have $\tau_{n,k}\leq\tau_{n+1,k}$. Suppose now
that (3.12) holds for some $n\geq 1$ and all $k$. For $k\geq 1$ we get by
(3.5) and the induction hypothesis

$\displaystyle u_{n+1,k}\leq q_{n}^{2k}u_{n,k}+q_{n}^{2k-1}u_{n,k-1}\leq
q_{n}^{2k}\tau_{n,k}+q_{n}^{2k-1}\tau_{n,k-1}$
$\displaystyle\hphantom{u_{n+1,k}}{}\leq
q_{n+1}^{2k}\tau_{n+1,k}+q_{n+1}^{2k-1}\tau_{n+1,k-1}=\tau_{n+1,k}.$ ∎

######  Theorem 3.6.

Assume $(b_{n})$ is $q$-increasing, cf. Definition 1.6, and let $U_{n}$ be
given by (2.11). Then

$\displaystyle
U_{n}\leq\big{(}q^{2};q^{2}\big{)}_{\infty}^{-2}\sum_{k=0}^{\infty}q^{2k^{2}},\qquad
n\geq 0.$

This holds in particular if $(b_{n})$ is log-convex and strictly increasing
with $q=b_{0}/b_{1}<1$.

###### Proof.

For $l\geq 1$ we have

$\displaystyle\frac{b_{n}}{b_{n+l}}=\frac{b_{n}}{b_{n+1}}\frac{b_{n+1}}{b_{n+2}}\cdots\frac{b_{n+l-1}}{b_{n+l}}\leq
q^{l},$ (3.13)

so from equation (3.5) we get

$\displaystyle u_{n+1,k}\leq q^{2k}u_{n,k}+q^{2k-1}u_{n,k-1}.$ (3.14)

We claim that

$\displaystyle u_{n,k}\leq
q^{k^{2}}\prod_{j=1}^{k}\big{(}1-q^{2j}\big{)}^{-1}=\frac{q^{k^{2}}}{\big{(}q^{2};q^{2}\big{)}_{k}},$
(3.15)

which is certainly true for $n=0,k\geq 0$ since $u_{0,0}=1$ and $u_{0,k}=0$
for $k\geq 1$. It is also true when $k=0$ independent of $n$ since
$u_{n,0}=1$.

Assume that (3.15) holds for some $n$ and all $k$. From (3.14) we get for
$k\geq 1$

$\displaystyle
u_{n+1,k}\leq\frac{q^{k^{2}+2k}}{\big{(}q^{2};q^{2}\big{)}_{k}}+\frac{q^{(k-1)^{2}+2k-1}}{\big{(}q^{2};q^{2}\big{)}_{k-1}}=\frac{q^{k^{2}}}{\big{(}q^{2};q^{2}\big{)}_{k}},$

which finishes the proof of (3.15). In particular, using the notation from
(3.11)

$\displaystyle
u_{n,k}\leq\frac{q^{k^{2}}}{\big{(}q^{2};q^{2}\big{)}_{\infty}},$

hence by (3.4)

$\displaystyle
U_{n}\leq\big{(}q^{2};q^{2}\big{)}_{\infty}^{-2}\sum_{k=0}^{\infty}q^{2k^{2}}<\infty.$
∎

Theorem 3.6 can be slightly extended.

######  Theorem 3.7.

Assume $(b_{n})$ is eventually $q$-increasing, cf. Definition 1.6. Then the
sequence $(U_{n})$ from (2.11) is bounded.

###### Proof.

By (1.7) we see that $(b_{n})$ is strictly increasing for $n\geq n_{0}$ and
tends to infinity. For $k\geq 1$ we then have

$\displaystyle\max\\{b_{n}\colon 0\leq
n<n_{0}+k\\}=\max\left(\max\\{b_{n}\colon 0\leq
n<n_{0}\\},b_{n_{0}+k-1}\right)$

and the latter is $<b_{n_{0}+k}$ for $k$ sufficiently large.

By increasing $n_{0}$ if necessary we may therefore assume that

$\displaystyle b_{n_{0}}>\max\\{b_{n}\colon 0\leq n<n_{0}\\}.$

Defining

$\displaystyle q_{1}=\max_{0\leq
n<n_{0}}\left(\frac{b_{n}}{b_{n_{0}}}\right)^{1/(n_{0}-n)},\qquad
p=\max(q,q_{1}),$

we have $0<p<1$.

Let

$\displaystyle\tilde{b}_{n}:=\begin{cases}b_{n_{0}}p^{n_{0}-n},&0\leq
n<n_{0},\\\ b_{n},&n_{0}\leq n.\end{cases}$

Then $\big{(}\tilde{b}_{n}\big{)}$ is $p$-increasing and
$b_{n}\leq\tilde{b}_{n}$. By Proposition 3.1 we get $s_{2n}\leq\tilde{s}_{2n}$
and for $n>n_{0}$ we find

$\displaystyle U_{n}=\frac{s_{2n}}{b_{0}^{2}b_{1}^{2}\cdots
b_{n-1}^{2}}\leq\frac{\tilde{b}_{0}^{2}\cdots\tilde{b}_{n_{0}-1}^{2}}{b_{0}^{2}\cdots
b_{n_{0}-1}^{2}}\tilde{U}_{n}.$

This shows the boundedness of $(U_{n})$ since $(\tilde{U}_{n})$ is bounded by
Theorem 3.6. ∎

######  Proposition 3.8.

Assume $(b_{n})$ is strictly increasing and log-concave. Then $u_{n,k}$ from
(3.3) satisfy

$\displaystyle u_{n,0}=1,\qquad u_{n,k}\leq\prod_{j=1}^{k}{b_{n-2k+j-1}\over
b_{n-j}-b_{n-2k+j-2}},\qquad 1\leq k\leq{n\over 2}.$ (3.16)

###### Proof.

We will use induction with respect to $n$. Equation (3.16) is certainly true
for $n=1$. Assume the conclusion is satisfied for $n\geq 1$. For $1\leq j\leq
k$, $k\geq 2$ we use log-concavity to obtain

$\displaystyle\frac{b_{n-2k+j-1}}{b_{n-j}-b_{n-2k+j-2}}\leq\frac{b_{n-2k+j}}{b_{n+1-j}-b_{n-2k+j-1}},$

hence by the induction hypothesis

$\displaystyle u_{n+1,k}={b_{n-2k}\over b_{n}}u_{n,k}+{b_{n-2k+1}\over
b_{n}}u_{n,k-1}$ $\displaystyle\hphantom{u_{n+1,k}}{}\leq{b_{n-2k}\over
b_{n}}\prod_{j=1}^{k}{b_{n-2k+j-1}\over b_{n-j}-b_{n-2k+j-2}}+{b_{n-2k+1}\over
b_{n}}\prod_{j=1}^{k-1}{b_{n-2k+j+1}\over b_{n-j}-b_{n-2k+j}}$
$\displaystyle\hphantom{u_{n+1,k}}{}\leq{b_{n-2k}\over
b_{n}}\prod_{j=1}^{k}{b_{n-2k+j}\over b_{n+1-j}-b_{n-2k+j-1}}+{b_{n-2k+1}\over
b_{n}}\prod_{j=2}^{k}{b_{n-2k+j}\over b_{n+1-j}-b_{n-2k+j-1}}$
$\displaystyle\hphantom{u_{n+1,k}}{}=\prod_{j=1}^{k}{b_{n-2k+j}\over
b_{n+1-j}-b_{n-2k+j-1}}\left[{b_{n-2k}\over b_{n}}+{b_{n-2k+1}\over
b_{n}}{b_{n}-b_{n-2k}\over b_{n-2k+1}}\right]$
$\displaystyle\hphantom{u_{n+1,k}}{}=\prod_{j=1}^{k}{b_{n-2k+j}\over
b_{n+1-j}-b_{n-2k+j-1}}.$

For $k=1$ we similarly have

$\displaystyle u_{n+1,1}={b_{n-2}\over b_{n}}u_{n,1}+{b_{n-1}\over
b_{n}}\leq{b_{n-2}\over b_{n}}{b_{n-2}\over b_{n-1}-b_{n-3}}+{b_{n-1}\over
b_{n}}$ $\displaystyle\hphantom{u_{n+1,1}}{}\leq{b_{n-2}\over
b_{n}}{b_{n-1}\over b_{n}-b_{n-2}}+{b_{n-1}\over b_{n}}={b_{n-1}\over
b_{n}-b_{n-2}}.$ ∎

######  Theorem 3.9.

Assume $(b_{n})$ satisfies

$\displaystyle{b_{n-1}\over b_{n}}\leq{\rm e}^{-\beta/n},\qquad n\geq
1,\qquad\beta>0.$ (3.17)

The quantity $U_{n}$ from (2.11) satisfies

$\displaystyle\limsup_{n\to\infty}U_{n}^{1/n}\leq{\rm e}^{2u(\beta)},$ (3.18)

where

$\displaystyle u(\beta):=\max_{2^{-1/\beta}\leq x\leq
1}(x+1)\int_{x}^{1}\log\left(\frac{y^{\beta}}{1-y^{\beta}}\right)\frac{{\rm
d}y}{(1+y)^{2}}>0.$ (3.19)

The function $u(\beta)$ is decreasing with $\lim\limits_{\beta\to
0}u(\beta)=\infty$, $\lim\limits_{\beta\to\infty}u(\beta)=0$.

###### Proof.

By (3.17) we get

$\displaystyle b_{n}\geq b_{0}{\rm e}^{\beta(1+1/2+\dots+1/n)},$

showing that $(b_{n})$ is strictly increasing to infinity.

Defining

$\displaystyle\tilde{b}_{0}:=b_{0},\qquad\tilde{b}_{n}:=b_{0}{\rm
e}^{\beta(1+1/2+\dots+1/n)},\qquad n\geq 1,$

then $\big{(}\tilde{b}_{n}\big{)}$ is strictly increasing and log-concave.

For $\widehat{b}_{n}=(n+1)^{\beta}$ we then have

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq\frac{\tilde{b}_{n-1}}{\tilde{b}_{n}}\leq\frac{\widehat{b}_{n-1}}{\widehat{b}_{n}},$
(3.20)

so by Proposition 3.2

$\displaystyle U_{n}\leq\tilde{U}_{n}\leq\widehat{U}_{n}.$

We claim that

$\displaystyle\prod_{j=1}^{k}{\tilde{b}_{n-2k+j-1}\over\tilde{b}_{n-j}-\tilde{b}_{n-2k+j-2}}\leq\prod_{j=1}^{k}{\widehat{b}_{n-2k+j-1}\over\widehat{b}_{n-j}-\widehat{b}_{n-2k+j-2}}=:\sigma_{n,k},$
(3.21)

which follows from the second inequality in (3.20) because of

$\displaystyle{\tilde{b}_{n-2k+j-1}\over\tilde{b}_{n-j}-\tilde{b}_{n-2k+j-2}}=\frac{\tilde{b}_{n-2k+j-1}/\tilde{b}_{n-j}}{1-\tilde{b}_{n-2k+j-2}/\tilde{b}_{n-j}}$

and a similar expression for $\widehat{b}_{n}$. Let $k_{n}$ be the smallest
index so that

$\displaystyle\sigma_{n,k_{n}}=\max\\{\sigma_{n,k}\colon 1\leq k\leq n/2\\}.$

We have $n=k_{n}(\gamma_{n}+2),$ where $\gamma_{n}\geq 0$. By Proposition 3.8
and the inequality (3.21) we have $\tilde{u}_{n,k}\leq\sigma_{n,k}$, hence

$\displaystyle\limsup_{n}U_{n}^{1/n}\leq\limsup_{n}\tilde{U}_{n}^{1/n}\leq\limsup_{n}\left(\sum_{k=0}^{[n/2]}\sigma_{n,k}^{2}\right)^{1/n}\leq\limsup_{n}\big{(}1+[n/2]\sigma_{n,k_{n}}^{2}\big{)}^{1/n}$
$\displaystyle\hphantom{\limsup_{n}U_{n}^{1/n}}{}\leq\limsup_{n}(1+[n/2])^{1/n}\sigma_{n,k_{n}}^{2/n}\leq\limsup_{n}\sigma_{n,k_{n}}^{2/n},$
(3.22)

where we have used that $\sigma_{n,0}=1$ and

$\displaystyle\sigma_{n,k_{n}}\geq\sigma_{n,1}=\frac{(n-1)^{\beta}}{n^{\beta}-(n-2)^{\beta}}\to\infty$

for $n\to\infty$. By the right-hand side of (3.21) we get

$\displaystyle\limsup_{n}\sigma_{n,k_{n}}^{1/n}=\limsup_{n}\left(\prod_{j=0}^{k_{n}-1}{\widehat{b}_{n-2k_{n}+j}\over\widehat{b}_{n-j-1}-\widehat{b}_{n-2k_{n}+j-1}}\right)^{1/n}.$

Furthermore

$\displaystyle\log\left(\prod_{j=0}^{k_{n}-1}{\widehat{b}_{n-2k_{n}+j}\over\widehat{b}_{n-j-1}-\widehat{b}_{n-2k_{n}+j-1}}\right)^{1/n}=\frac{1}{(\gamma_{n}+2)k_{n}}\sum_{j=0}^{k_{n}-1}E_{n}(\beta,j),$
(3.23)

where

$\displaystyle
E_{n}(\beta,j)=\beta\log(\gamma_{n}+{j/k_{n}}+1/k_{n})-\log\big{[}(\gamma_{n}+2-{j/k_{n}})^{\beta}-(\gamma_{n}+{j/k_{n}})^{\beta}\big{]}.$
(3.24)

Let $\gamma$ be an accumulation point of $(\gamma_{n})$.

We consider first the case, where $\gamma=\infty$ is the only accumulation
point, so $\gamma_{n}\to\infty$. We have

$\displaystyle\log(\gamma_{n}+{j/k_{n}}+1/k_{n})\leq\log(\gamma_{n}+1).$
(3.25)

The inequality $(0<y<x)$

$\displaystyle{x^{\beta}-y^{\beta}\over x-y}\geq\begin{cases}\beta
y^{\beta-1},&\beta\geq 1,\\\ \beta x^{\beta-1},&0<\beta<1,\end{cases}$

implies

$\displaystyle(\gamma_{n}+2-{j/k_{n}})^{\beta}-(\gamma_{n}+{j/k_{n}})^{\beta}\geq\begin{cases}2\beta(1-j/k_{n})(\gamma_{n}+{j/k_{n}})^{\beta-1},&\beta\geq
1,\\\
2\beta(1-j/k_{n})(\gamma_{n}+2-{j/k_{n}})^{\beta-1},&0<\beta<1\end{cases}$
$\displaystyle\hphantom{(\gamma_{n}+2-{j/k_{n}})^{\beta}-(\gamma_{n}+{j/k_{n}})^{\beta}}{}\geq\begin{cases}2\beta(1-j/k_{n})\gamma_{n}^{\beta-1},&\beta\geq
1,\\\ 2\beta(1-j/k_{n})(\gamma_{n}+2)^{\beta-1},&0<\beta<1.\end{cases}$ (3.26)

We have

$\displaystyle-\frac{1}{k_{n}}\sum_{j=0}^{k_{n}-1}\log(1-j/k_{n})\leq-\int_{0}^{1}\log
x{\rm d}x=1.$ (3.27)

By (3.25), (3.26) and (3.27) we get that the right-hand side of (3.23) is
bounded above by $C\log(\gamma_{n}+2)/(\gamma_{n}+2)$ for suitable $C>0$,
hence tends to 0. In this case (3.22) gives

$\displaystyle\limsup_{n}U_{n}^{1/n}\leq 1,$

and (3.18) holds.

Assume now that $\gamma<\infty$. Then, in view of (3.23) and (3.24) the
quantity

$\displaystyle
I={1\over\gamma+2}\int_{0}^{1}\big{\\{}\beta\log(\gamma+x)-\log\big{[}(\gamma+2-x)^{\beta}-(\gamma+x)^{\beta}\big{]}\big{\\}}{\rm
d}x$

is an accumulation point of (3.23). The substitution

$\displaystyle y={\gamma+x\over\gamma+2-x}$

leads to

$\displaystyle
I={2\gamma+2\over\gamma+2}\int_{\gamma\over\gamma+2}^{1}\log\left({y^{\beta}\over
1-y^{\beta}}\right){{\rm d}y\over(1+y)^{2}}.$

Set $\sigma=\gamma/(\gamma+2)$. Then

$\displaystyle I=(\sigma+1)\int_{\sigma}^{1}\log\left({y^{\beta}\over
1-y^{\beta}}\right){{\rm d}y\over(1+y)^{2}},$

hence

$\displaystyle I\leq\max_{0\leq x\leq
1}\left\\{(x+1)\int_{x}^{1}\log\left({y^{\beta}\over 1-y^{\beta}}\right){{\rm
d}y\over(1+y)^{2}}\right\\}.$

Clearly the maximum is attained for $x\geq 2^{-1/\beta}$, as in this case the
integrated function is nonnegative. Now (3.18) follows.

The last assertion is easy to prove. ∎

######  Theorem 3.10.

Assume $(b_{n})$ satisfies

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq{\rm e}^{-f(n)},\qquad n\geq n_{0}\geq
1,$ (3.28)

where $f(n)>0$ for $n\geq n_{0}$ and

$\displaystyle\alpha:=\liminf_{n\to\infty}nf(n)>0.$

The quantity $U_{n}$ from (2.11) satisfies

$\displaystyle\limsup_{n}U_{n}^{1/n}\leq{\rm e}^{2u(\alpha)},$ (3.29)

where $u(\alpha)$ is defined in (3.19) and $u(\infty):=0$.

###### Proof.

Let $0<\beta<\alpha$ be arbitrary. There exists $N\geq n_{0}$ such that

$\displaystyle nf(n)\geq\beta,\qquad n\geq N.$ (3.30)

Combining (3.28) and (3.30), we see that $(b_{n})$ is eventually strictly
increasing to infinity. By replacing $N$ by a larger integer if necessary, we
may assume that

$\displaystyle\max\\{b_{n}\colon 0\leq n\leq N-1\\}<b_{N},$

see the proof of Theorem 3.7. Defining

$\displaystyle\tilde{b}_{n}:=\begin{cases}b_{n},&0\leq n\leq N,\\\
\displaystyle
b_{N}\exp\left[\beta\left(\frac{1}{N+1}+\frac{1}{N+2}+\dots+\frac{1}{n}\right)\right],&n>N,\end{cases}$

we have

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq\frac{\tilde{b}_{n-1}}{\tilde{b}_{n}},\qquad
n\geq 1,$

hence $U_{n}\leq\tilde{U}_{n}$ by Proposition 3.2.

We see that $\tilde{b}_{n}\leq\tilde{b}_{n}^{(N)}:=\tilde{b}_{n+N}$ and
$\tilde{b}_{n}^{1/n}\to 1$, so by Proposition 3.3 we get

$\displaystyle\limsup_{n}U_{n}^{1/n}\leq\limsup_{n}\tilde{U}_{n}^{1/n}=\limsup_{n}\big{(}\tilde{U}_{n}^{(N)}\big{)}^{1/n}.$
(3.31)

Define now

$\displaystyle\widehat{b}_{0}=b_{N}\exp\left[-\beta\left(1+\frac{1}{2}+\dots+\frac{1}{N}\right)\right],$
$\displaystyle\widehat{b}_{n}=\widehat{b}_{0}\exp\left[\beta\left(1+\frac{1}{2}+\ldots+\frac{1}{n}\right)\right],\qquad
n\geq 1.$

Then $\big{(}\widehat{b}_{n}\big{)}$ is strictly increasing and satisfies
$\widehat{b}_{n-1}/\widehat{b}_{n}=\exp(-\beta/n)$, $n\geq 1$. Furthermore,
the conditions of Proposition 3.3 are satisfied so

$\displaystyle\limsup_{n}\widehat{U}_{n}^{1/n}=\limsup_{n}\big{(}\widehat{U}_{n}^{(N)}\big{)}^{1/n}.$
(3.32)

Notice that $\widehat{b}_{n}^{(N)}=\tilde{b}_{n}^{(N)}$, so by (3.31), (3.32)
and Theorem 3.9

$\displaystyle\limsup_{n}U_{n}^{1/n}\leq\limsup_{n}\widehat{U}_{n}^{1/n}\leq{\rm
e}^{2u(\beta)}.$

Since $\beta<\alpha$ is arbitrary, we finally get (3.29) by continuity. ∎

######  Remark 3.11.

The assumptions of Theorem 3.10 are satisfied for the following sequences,
where it is assumed that $n$ is so large that the expressions are defined and
positive:

$\displaystyle b_{n}={\rm e}^{n^{\gamma}},$ $\displaystyle 0<\gamma<1,$
$\displaystyle f(n)=\gamma n^{\gamma-1},$ $\displaystyle\alpha=\infty,$
$\displaystyle b_{n}={\rm e}^{(\log n)^{\gamma}},$ $\displaystyle\gamma>1,$
$\displaystyle f(n)={\gamma\over n}\log^{\gamma-1}n,$
$\displaystyle\alpha=\infty,$ $\displaystyle b_{n}={\rm e}^{\log n(\log\log
n)^{\gamma}},\ $ $\displaystyle\gamma>1,$ $\displaystyle f(n)={\log\log
n+\gamma\over n}(\log\log n)^{\gamma-1},$ $\displaystyle\alpha=\infty,$
$\displaystyle b_{n}=n^{\gamma},$ $\displaystyle\gamma>0,$ $\displaystyle
f(n)={\gamma\over n},$ $\displaystyle\alpha=\gamma.$

######  Remark 3.12.

Assume $(b_{n})$ is increasing. Then by (3.6) we get

$\displaystyle u_{n+1,k}\leq u_{n,k}+u_{n,k-1},\qquad k\geq 0.$

This implies, by an induction argument, that

$\displaystyle u_{n,k}\leq{n\choose k}.$

Thus by (3.4) we get

$\displaystyle U_{n}\leq\sum_{k=0}^{[n/2]}{n\choose k}^{2}\leq{2n\choose
n}\leq 4^{n},$ (3.33)

hence

$\displaystyle\limsup_{n}U_{n}^{1/n}\leq 4.$

This estimate is weaker than the one obtained in Theorem 3.10 for
$\alpha>0.30873$. In fact, $\exp(2u(\alpha))\leq 4$ for $u(\alpha)\leq\log 2$
and $u(0.30872)>\log 2$, $u(0.30783)<\log 2$.

However, for bounded sequences $(b_{n})$ the estimate (3.33) cannot be
improved, see Appendix A.

##  4 Estimates of $\boldsymbol{c_{n}}$ via $\boldsymbol{V_{n}}$

In all of this section we consider symmetric moment problems given by a
positive sequence $(b_{n})$, and it is tacitly assumed that the moment problem
is indeterminate in order to be able to define $c_{n}$ from (2.6), hence $\sum
1/b_{n}<\infty$ by a classical result due to Carleman, cf. [1, p. 24].

For $k,l\geq 0$ define

$\displaystyle v_{k,l}:=(-1)^{l}b_{0}b_{1}\cdots b_{k-1}b_{k,k+2l},$ (4.1)

where $b_{k,k+2l}$ is the coefficient to $x^{k}$ in $P_{k+2l}$, so
$(-1)^{l}b_{k,k+2l}>0$ and hence $v_{k,l}>0$.

Observe that $v_{k,0}=1$ and since $b_{k,j}=0$ if $j<k$ it makes sense to
define $v_{k,l}=0$ if $l<0$.

From (1.9) we know

$\displaystyle
P_{2l}(0)=b_{0,2l}=(-1)^{l}v_{0,l}=(-1)^{l}\prod_{j=1}^{l}\frac{b_{2j-2}}{b_{2j-1}}.$
(4.2)

Notice that by (2.6) and (2.12)

$\displaystyle V_{n}=b_{0}b_{1}\cdots
b_{n-1}c_{n}=\left(\sum_{l=0}^{\infty}v_{n,l}^{2}\right)^{1/2}.$ (4.3)

######  Lemma 4.1.

The coefficients $v_{k,l}$ satisfy

$\displaystyle v_{k+1,l}={b_{k}\over b_{k+2l}}v_{k,l}+{b_{k+2l-1}\over
b_{k+2l}}v_{k+1,l-1},\qquad k,l\geq 0.$ (4.4)

###### Proof.

We have

$\displaystyle xP_{k+2l}=b_{k+2l}P_{k+2l+1}+b_{k+2l-1}P_{k+2l-1}.$

By equating the coefficients of $x^{k+1}$ we obtain

$\displaystyle b_{k,k+2l}=b_{k+2l}b_{k+1,k+2l+1}+b_{k+2l-1}b_{k+1,k+2l-1}.$

Multiplying the last identity by $(-1)^{l}b_{0}b_{1}\cdots b_{k-1}$ leads to

$\displaystyle v_{k,l}={b_{k+2l}\over b_{k}}v_{k+1,l}-{b_{k+2l-1}\over
b_{k}}v_{k+1,l-1}.$

This gives (4.4). ∎

The following result follows easily from (4.2) and (4.4).

######  Proposition 4.2.

Assume $(b_{n})$ and $\big{(}\tilde{b}_{n}\big{)}$ are positive sequences
satisfying

$\displaystyle{b_{n-1}\over
b_{n}}\leq{\tilde{b}_{n-1}\over\tilde{b}_{n}},\qquad n\geq 1.$

Then $v_{k,l}\leq\tilde{v}_{k,l}$ and $V_{k}\leq\tilde{V}_{k}$.

For $n_{0}\in\mathbb{N}$ the shifted sequence $b_{n}^{(n_{0})}:=b_{n+n_{0}}$
also leads to an indeterminate moment problem, cf. [1, p. 28], and we shall
consider the associated quantities $v_{k,l}^{(n_{0})}$, $V_{k}^{(n_{0})}$.

######  Proposition 4.3.

The expressions above satisfy

$\displaystyle v_{k,l}^{(1)}\leq v_{k+1,l},\qquad V_{k}^{(1)}\leq
V_{k+1},\qquad k,l\geq 0,$ (4.5) $\displaystyle v_{k+1,l}\leq
Lv_{k,l}^{(1)},\qquad V_{k+1}\leq LV_{k}^{(1)},\qquad k,l\geq 0,$ (4.6)

where $L:=\sum\limits_{n=0}^{\infty}P_{n}^{2}(0)$. For $n_{0}\in\mathbb{N}$ we
similarly have

$\displaystyle v_{k,l}^{(n_{0})}\leq v_{k+n_{0},l}\leq
L(n_{0})v_{k,l}^{(n_{0})},\qquad V_{k}^{(n_{0})}\leq V_{k+n_{0}}\leq
L(n_{0})V_{k}^{(n_{0})},$ (4.7)

for a suitable constant $L(n_{0})>0$. Furthermore,

$\displaystyle\limsup_{k\to\infty}V_{k}^{1/k}=\limsup_{k\to\infty}\big{(}V_{k}^{(n_{0})}\big{)}^{1/k}.$
(4.8)

###### Proof.

From (4.4) we get

$\displaystyle
v_{k+1,l}^{(1)}=\frac{b_{k+1}}{b_{k+2l+1}}v_{k,l}^{(1)}+\frac{b_{k+2l}}{b_{k+2l+1}}v_{k+1,l-1}^{(1)},\qquad
k,l\geq 0,$ (4.9) $\displaystyle
v_{k+2,l}=\frac{b_{k+1}}{b_{k+2l+1}}v_{k+1,l}+\frac{b_{k+2l}}{b_{k+2l+1}}v_{k+2,l-1},\qquad
k,l\geq 0.$ (4.10)

On the other hand

$\displaystyle
v_{1,l}=\frac{b_{0}}{b_{2l}}v_{0,l}+\frac{b_{2l-1}}{b_{2l}}v_{1,l-1}\geq\frac{b_{2l-1}}{b_{2l}}v_{1,l-1}.$
(4.11)

Iterating this we get using $v_{1,0}=1$

$\displaystyle
v_{1,l}\geq\prod_{j=1}^{l}\frac{b_{2j-1}}{b_{2j}}=v_{0,l}^{(1)},$ (4.12)

where the last equality follows from (4.2). This proves the left inequality in
(4.5) for $k=0$ and all $l$, and it is clear for $l=0$ and all $k$. In
particular it is true for $k+l\leq 1$. Assume that it holds for $k+l\leq n$
for some $n\geq 1$. To prove it for $k+l=n+1$, it suffices to prove
$v_{k+1,l}^{(1)}\leq v_{k+2,l}$ when $l\geq 1$, but this follows immediately
by (4.9) and (4.10) and the induction hypothesis. The right inequality in
(4.5) is an immediate consequence of the left inequality.

Defining $t_{l}=v_{1,l}/v_{0,l}^{(1)}$, $l\geq 0$ we get by (4.12)

$\displaystyle t_{l}=v_{1,l}\prod_{j=1}^{l}\frac{b_{2j}}{b_{2j-1}},\qquad
l\geq 1,$

so by (4.11) and (4.2)

$\displaystyle
t_{l}=\left(\frac{b_{0}}{b_{2l}}v_{0,l}+\frac{b_{2l-1}}{b_{2l}}v_{1,l-1}\right)\prod_{j=1}^{l}\frac{b_{2j}}{b_{2j-1}}=\frac{b_{0}}{b_{2l}}\prod_{j=1}^{l}\frac{b_{2j-2}b_{2j}}{b_{2j-1}^{2}}+t_{l-1}.$

We then get for $l\geq 1$

$\displaystyle
t_{l}=t_{0}+\sum_{m=1}^{l}\frac{b_{0}}{b_{2m}}\prod_{j=1}^{m}\frac{b_{2j-2}b_{2j}}{b_{2j-1}^{2}}=1+\sum_{m=1}^{l}P_{2m}^{2}(0)\leq\sum_{n=0}^{\infty}P_{n}^{2}(0)=L,$

hence $v_{1,l}\leq Lv_{0,l}^{(1)}$ for all $l$. By induction as above we then
get the left side of (4.6), and the right side follows.

The inequalities (4.7) easily follow by iteration with a suitable constant
$L(n_{0})>0$ independent of $k$, and (4.8) follows. ∎

######  Theorem 4.4.

Assume $(b_{n})$ is $q$-increasing, cf. Definition 1.6. The quantity $V_{n}$
from (2.12) satisfies

$\displaystyle
V_{n}\leq\frac{1}{\big{(}q^{2};q^{2}\big{)}_{\infty}\sqrt{1-q^{2}}}.$ (4.13)

This holds in particular if $(b_{n})$ is log-convex and strictly increasing
with $q=b_{0}/b_{1}<1$.

###### Proof.

Note that the moment problem is indeterminate by Proposition 1.7. From (4.4)
and (3.13) we obtain

$\displaystyle v_{k+1,l}\leq q^{2l}v_{k,l}+qv_{k+1,l-1}.$ (4.14)

We claim that

$\displaystyle v_{k,l}\leq
q^{l}\prod_{j=1}^{l}\big{(}1-q^{2j}\big{)}^{-1}=\frac{q^{l}}{\big{(}q^{2};q^{2}\big{)}_{l}},\qquad
k,l\geq 0,$ (4.15)

which is certainly true for $l=0$ since $v_{k,0}=1$ and for $k=0$ since by
(4.2)

$\displaystyle v_{0,l}\leq
q^{l}\leq\frac{q^{l}}{\big{(}q^{2};q^{2}\big{)}_{l}}.$

We shall proceed by induction in $k+l$, and as already observed (4.15) holds
when $k+l\leq 1$.

Suppose (4.15) holds for $k+l\leq n$ for some $n\geq 1$ and we shall prove it
for $k+l=n+1$. From (4.14) and the induction hypothesis we get for $l\geq 1$

$\displaystyle v_{k+1,l}\leq
q^{2l}v_{k,l}+q\,v_{k+1,l-1}\leq\frac{q^{3l}}{\big{(}q^{2};q^{2}\big{)}_{l}}+\frac{q^{l}}{\big{(}q^{2};q^{2}\big{)}_{l-1}}=\frac{q^{l}}{\big{(}q^{2};q^{2}\big{)}_{l}}.$

From (4.15) we get

$\displaystyle v_{k,l}\leq\frac{q^{l}}{\big{(}q^{2};q^{2}\big{)}_{\infty}}$

and finally

$\displaystyle
V_{n}^{2}=\sum_{l=0}^{\infty}v_{n,l}^{2}\leq\big{(}q^{2};q^{2}\big{)}_{\infty}^{-2}\sum_{l=0}^{\infty}q^{2l},$

leading to (4.13). ∎

The boundedness of $(V_{n})$ also holds in a slightly more general case.

######  Theorem 4.5.

Assume $(b_{n})$ is eventually $q$-increasing, cf. Definition 1.6. The
sequence $(V_{n})$ from (2.12) is bounded.

###### Proof.

Note that the moment problem is indeterminate by Proposition 1.7. We assume
that (1.7) holds. Then the sequence $\big{(}b_{n}^{(n_{0})}\big{)}$ is
$q$-increasing, so $\big{(}V_{n}^{(n_{0})}\big{)}$ is bounded by Theorem 4.4.
The boundedness of $(V_{n})$ follows from (4.7). ∎

Combining Theorem 3.7 and Theorem 4.5 we get:

######  Corollary 4.6.

Assume $(b_{n})$ is eventually $q$-increasing. Then
$c_{n}\sqrt{s_{2n}}=V_{n}\sqrt{U_{n}}$ is a bounded sequence.

We now turn to the log-concave case.

######  Proposition 4.7.

Assume $(b_{n})$ is log-concave and strictly increasing. Then

$\displaystyle V_{k}\leq V_{k+1},\qquad c_{k}\leq b_{k}c_{k+1},\qquad k\geq
0.$

###### Proof.

We start by showing that

$\displaystyle v_{0,1}<v_{1,1}.$ (4.16)

Indeed, as

$\displaystyle P_{2}(x)={1\over b_{0}b_{1}}x^{2}-{b_{0}\over b_{1}},\qquad
P_{3}(x)={1\over b_{0}b_{1}b_{2}}x^{3}-{b_{0}^{2}+b_{1}^{2}\over
b_{0}b_{1}b_{2}}x,$

we have

$\displaystyle v_{0,1}={b_{0}\over b_{1}},\qquad
v_{1,1}={b_{0}^{2}+b_{1}^{2}\over b_{1}b_{2}},$

and using $b_{0}/b_{1}\leq b_{1}/b_{2}$ we get (4.16).

We are going to show that

$\displaystyle v_{k,l}\leq v_{k+1,l},$ (4.17)

by induction with respect to $k+l$. It is certainly true for $l=0$ and for
$k=0$, $l=1$ by (4.16), and it therefore holds for $k+l\leq 1$.

We first show (4.17) for $k=0$, i.e.,

$\displaystyle v_{0,l}\leq v_{1,l},\qquad l\geq 0,$

and we use induction in $l$. Assuming it holds for $l-1\geq 0$ we get by (4.4)

$\displaystyle
v_{1,l}=\frac{b_{0}}{b_{2l}}v_{0,l}+\frac{b_{2l-1}}{b_{2l}}v_{1,l-1}\geq\frac{b_{0}}{b_{2l}}v_{0,l}+\frac{b_{2l-1}}{b_{2l}}v_{0,l-1}=v_{0,l}\left(\frac{b_{0}}{b_{2l}}+\frac{b_{2l-1}}{b_{2l}}\frac{b_{2l-1}}{b_{2l-2}}\right)>v_{0,l},$

where we have used (4.2) and $b_{2l-1}^{2}\geq b_{2l}b_{2l-2}$.

Assume now that (4.17) holds for $k+l\leq n$ for some $n\geq 1$. By log-
concavity

$\displaystyle\frac{b_{n}}{b_{n+l}}\leq\frac{b_{n+1}}{b_{n+l+1}},\qquad
n,l\geq 0,$

and therefore for $l\geq 1$

$\displaystyle
v_{k+1,l}=\frac{b_{k}}{b_{k+2l}}v_{k,l}+\frac{b_{k+2l-1}}{b_{k+2l}}v_{k+1,l-1}\leq\frac{b_{k+1}}{b_{k+2l+1}}v_{k+1,l}+\frac{b_{k+2l}}{b_{k+2l+1}}v_{k+2,l-1}=v_{k+2,l}.$

Finally

$\displaystyle
V_{k}=\left(\sum_{l=0}^{\infty}v_{k,l}^{2}\right)^{1/2}\leq\left(\sum_{l=0}^{\infty}v_{k+1,l}^{2}\right)^{1/2}=V_{k+1},$

hence $c_{k}\leq b_{k}c_{k+1}$. ∎

######  Theorem 4.8.

Assume $(b_{n})$ is log-concave and strictly increasing. Set $b_{-1}=0$. The
quantities $v_{k,l}$ from (4.1) satisfy

$\displaystyle v_{k,l}\leq\prod_{j=1}^{l}{b_{k+2j-2}\over
b_{k+2j-1}-b_{k-1}},\qquad k,l\geq 0.$

###### Proof.

Define

$\displaystyle w_{k,l}:=\prod_{j=1}^{l}{b_{k+2j-2}\over
b_{k+2j-1}-b_{k-1}},\qquad k,l\geq 0,$

where as usual empty products are equal to 1. We are going to show that

$\displaystyle v_{k,l}\leq w_{k,l}.$ (4.18)

Clearly $v_{k,0}=w_{k,0}=1$. Moreover, by (4.2)

$\displaystyle v_{0,l}=w_{0,l}=\prod_{j=1}^{l}{b_{2j-2}\over b_{2j-1}},\qquad
l\geq 1.$

Therefore (4.18) is valid for $l=0$ and all $k$ and for $k=0$ and all $l$.
Observe that by log-concavity we have

$\displaystyle w_{k,l}\leq w_{k+1,l}.$ (4.19)

We also have

$\displaystyle{w_{k+1,l-1}\over w_{k+1,l}}={b_{k+2l}-b_{k}\over b_{k+2l-1}}.$
(4.20)

Assume (4.18) is valid for $k+l\leq n,n\geq 1$ and let us prove it for
$k+l=n+1$. We may assume $k,l\geq 1$ and write $k=k^{\prime}+1$. By the
induction hypothesis, in view of (4.4), (4.19) and (4.20), we get

$\displaystyle v_{k,l}=v_{k^{\prime}+1,l}\leq{b_{k^{\prime}}\over
b_{k^{\prime}+2l}}w_{k^{\prime},l}+{b_{k^{\prime}+2l-1}\over
b_{k^{\prime}+2l}}w_{k^{\prime}+1,l-1}$
$\displaystyle\hphantom{v_{k,l}}{}\leq{b_{k^{\prime}}\over
b_{k^{\prime}+2l}}w_{k^{\prime}+1,l}+{b_{k^{\prime}+2l-1}\over
b_{k^{\prime}+2l}}w_{k^{\prime}+1,l-1}=w_{k^{\prime}+1,l}=w_{k,l}.$ ∎

The previous theorem can be extended to the eventually log-concave case.

######  Theorem 4.9.

Assume $(b_{n})$ is log-concave and strictly increasing for $n\geq n_{0}$.
Then there exists $L(n_{0})\geq 1$ such that the quantities $v_{k,l}$ from
(4.1) satisfy

$\displaystyle v_{k,l}\leq
L(n_{0})\prod_{j=1}^{l}\frac{b_{k+2j-2}}{b_{k+2j-1}-b_{k-1}},\qquad k\geq
n_{0},\qquad l\geq 0.$

###### Proof.

The sequence $\big{(}b_{n}^{(n_{0})}\big{)}$ is log-concave and strictly
increasing, so by Theorem 4.8

$\displaystyle v_{k,l}^{(n_{0})}\leq\prod_{j=1}^{l}{b_{k+n_{0}+2j-2}\over
b_{k+n_{0}+2j-1}-b_{k+n_{0}-1}},\qquad k,l\geq 0,$

and since $v_{k+n_{0},l}\leq L(n_{0})v_{k,l}^{(n_{0})}$ by (4.7), the
assertion follows. ∎

######  Theorem 4.10.

Assume $(b_{n})$ satisfies

$\displaystyle{b_{n-1}\over b_{n}}\leq{\rm e}^{-\beta/n},\qquad n\geq
1,\qquad\beta>1.$

The quantities $V_{n}$ from (2.12) satisfy

$\displaystyle\limsup_{n}V_{n}^{1/n}\leq{\rm e}^{v(\beta)/2},$ (4.21)

where

$\displaystyle
v(\beta):=\int_{1}^{\infty}\log\big{(}\big{[}1-x^{-\beta}\big{]}^{-1}\big{)}{\rm
d}x.$ (4.22)

The function $v(\beta)$ is positive and decreasing with $\lim\limits_{\beta\to
1}v(\beta)=\infty$ and $\lim\limits_{\beta\to\infty}v(\beta)=0$.

###### Proof.

Note that the moment problem is indeterminate by Proposition 1.7. Defining
$\big{(}\tilde{b}_{n}\big{)}$ by

$\displaystyle\tilde{b}_{0}=b_{0},\qquad\tilde{b}_{n}=b_{0}{\rm
e}^{\beta(1+1/2+\dots+1/n)},\qquad n\geq 1,$

then $\tilde{b}_{n-1}/\tilde{b}_{n}={\rm e}^{-\beta/n}$, so
$\big{(}\tilde{b}_{n}\big{)}$ is strictly increasing, log-concave and

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq\frac{\tilde{b}_{n-1}}{\tilde{b}_{n}},$

hence $V_{n}\leq\tilde{V}_{n}$ by Proposition 4.2.

By Theorem 4.8

$\displaystyle\tilde{v}_{k,l}\leq\prod_{j=1}^{l}{\tilde{b}_{k+2j-2}\over\tilde{b}_{k+2j-1}-\tilde{b}_{k-1}}=\prod_{j=1}^{l}\frac{\tilde{b}_{k+2j-2}/\tilde{b}_{k+2j-1}}{1-\tilde{b}_{k-1}/\tilde{b}_{k+2j-1}}$
$\displaystyle\hphantom{\tilde{v}_{k,l}}{}=\prod_{j=1}^{l}{{\rm
e}^{-{\beta\over k+2j-1}}\over 1-{\rm
e}^{-\beta\sum\limits_{i=k}^{k+2j-1}{1\over i}}}={\rm
e}^{-\beta\sum\limits_{j=1}^{l}{1\over k+2j-1}}\prod_{j=1}^{l}\bigg{(}1-{\rm
e}^{-\beta\sum\limits_{i=k}^{k+2j-1}{1\over i}}\bigg{)}^{-1}.$

We need the estimates

$\displaystyle\sum_{i=k}^{k+2j-1}{1\over i}\geq\int_{k}^{k+2j}{1\over x}{\rm
d}x=\log(k+2j)-\log k$ (4.23)

and

$\displaystyle\sum_{j=1}^{l}{1\over k+2j-1}={1\over 2}\sum_{j=1}^{l}{1\over
j+{k-1\over 2}}\geq{1\over 2}\int_{k+1\over 2}^{l+{k+1\over 2}}{1\over x}{\rm
d}x={1\over 2}[\log(k+2l+1)-\log(k+1)].$

Thus, for $k,l\geq 1$,

$\displaystyle\tilde{v}_{k,l}\leq\left({k+1\over
k+2l+1}\right)^{\beta/2}\prod_{j=1}^{l}\left[1-\left({k\over
k+2j}\right)^{\beta}\right]^{-1}$
$\displaystyle\hphantom{\tilde{v}_{k,l}}{}\leq 2^{\beta/2}\left({k\over
k+2l}\right)^{\beta/2}\prod_{j=1}^{l}\left[1-\left({k\over
k+2j}\right)^{\beta}\right]^{-1}.$

For $k=2u,u\in\mathbb{N}$ we then get

$\displaystyle\tilde{v}_{2u,l}\leq\zeta_{u,l}:=2^{\beta/2}\left({u\over
u+l}\right)^{\beta/2}\prod_{j=1}^{l}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-1}$ (4.24)

and for $k=2u+1$, $u\in\mathbb{N}$

$\displaystyle\tilde{v}_{2u+1,l}=\left(\frac{u+1}{u+l+1}\right)^{\beta/2}\prod_{j=1}^{l}\left[1-\left(\frac{2u+1}{2u+2j+1}\right)^{\beta}\right]^{-1}<\zeta_{u+1,l}.$
(4.25)

Next, for any $N\in\mathbb{N}$

$\displaystyle\sum_{l=1}^{\infty}\zeta_{u,l}^{2}\leq
2^{\beta}\sum_{l=1}^{\infty}\left({u\over
u+l}\right)^{\beta}\prod_{j=1}^{l}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2}$
$\displaystyle\hphantom{\sum_{l=1}^{\infty}\zeta_{u,l}^{2}}{}=2^{\beta}\sum_{l=1}^{N}\left({u\over
u+l}\right)^{\beta}\prod_{j=1}^{l}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2}$
$\displaystyle\hphantom{\sum_{l=1}^{\infty}\zeta_{u,l}^{2}\leq}{}+2^{\beta}\prod_{j=1}^{N}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2}\sum_{l=N+1}^{\infty}\left({u\over
u+l}\right)^{\beta}\prod_{j=N+1}^{l}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2}$
$\displaystyle\hphantom{\sum_{l=1}^{\infty}\zeta_{u,l}^{2}}{}\leq
2^{\beta}(N+S_{N})\prod_{j=1}^{N}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2},$

where

$\displaystyle S_{N}:=\sum_{l=N+1}^{\infty}\left({u\over
u+l}\right)^{\beta}\prod_{j=N+1}^{l}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2}.$

Let now $N=un_{u}$, where $n_{u}\in\mathbb{N}$ will be chosen later and write

$\displaystyle
S_{N}=\sum_{n=n_{u}+1}^{\infty}\sum_{l=(n-1)u+1}^{nu}\left({u\over
u+l}\right)^{\beta}\prod_{j=N+1}^{l}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2}$ $\displaystyle\hphantom{S_{N}}{}\leq
u\sum_{n=n_{u}+1}^{\infty}\frac{1}{n^{\beta}}\prod_{j=N+1}^{nu}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2}$
$\displaystyle\hphantom{S_{N}}{}=u\sum_{n=n_{u}+1}^{\infty}\frac{1}{n^{\beta}}\prod_{m=n_{u}+1}^{n}\;\prod_{j=(m-1)u+1}^{mu}\left[1-\left({u\over
u+j}\right)^{\beta}\right]^{-2}$ $\displaystyle\hphantom{S_{N}}{}\leq
u\sum_{n=n_{u}+1}^{\infty}\frac{1}{n^{\beta}}\prod_{m=n_{u}+1}^{n}\left[1-\frac{1}{m^{\beta}}\right]^{-2u}<u\sum_{n=n_{u}+1}^{\infty}\frac{1}{n^{\beta}}\prod_{m=n_{u}+1}^{\infty}\left[1-\frac{1}{m^{\beta}}\right]^{-2u}.$

We use the estimate

$\displaystyle\sum_{n=n_{u}+1}^{\infty}\frac{1}{n^{\beta}}<\int_{n_{u}}^{\infty}\frac{{\rm
d}x}{x^{\beta}}=\frac{1}{(\beta-1)n_{u}^{\beta-1}}\leq\frac{1}{\beta-1}$

to get

$\displaystyle
S_{N}<\frac{u}{\beta-1}\prod_{m=n_{u}+1}^{\infty}\left[1-\frac{1}{m^{\beta}}\right]^{-2u},$

hence

$\displaystyle\sum_{l=1}^{\infty}\zeta_{u,l}^{2}\leq
2^{\beta}u\prod_{j=1}^{un_{u}}\left[1-\left(\frac{u}{u+j}\right)^{\beta}\right]^{-2}\left(n_{u}+\frac{1}{\beta-1}\prod_{m=n_{u}+1}^{\infty}\left[1-\frac{1}{m^{\beta}}\right]^{-2u}\right).$

For $n_{u}=u$ we get

$\displaystyle\sum_{l=1}^{\infty}\zeta_{u,l}^{2}\leq\frac{2^{\beta}\beta}{\beta-1}u^{2}\prod_{j=1}^{u^{2}}\left[1-\left(\frac{u}{u+j}\right)^{\beta}\right]^{-2}\prod_{m=u+1}^{\infty}\left[1-\frac{1}{m^{\beta}}\right]^{-2u},$

hence

$\displaystyle\limsup_{u\to\infty}\left(\sum_{l=1}^{\infty}\zeta_{u,l}^{2}\right)^{1/u}\leq\limsup_{u\to\infty}\prod_{j=1}^{u^{2}}\left[1-\left(\frac{u}{u+j}\right)^{\beta}\right]^{-2/u}$

because

$\displaystyle\lim_{u\to\infty}\prod_{m=u+1}^{\infty}\left[1-\frac{1}{m^{\beta}}\right]^{-2}=1.$

Furthermore, for $u\to\infty$ we have

$\displaystyle-\frac{2}{u}\sum_{j=1}^{u^{2}}\log\big{[}1-(1+j/u)^{-\beta}\big{]}\to-2\int_{1}^{\infty}\log\big{[}1-x^{-\beta}\big{]}{\rm
d}x,$

hence

$\displaystyle\limsup_{u\to\infty}\left(\sum_{l=1}^{\infty}\zeta_{u,l}^{2}\right)^{1/u}\leq{\rm
e}^{2v(\beta)}.$ (4.26)

From (4.3), (4.24) and (4.25) we get

$\displaystyle
V_{2u}^{1/(2u)}\leq\tilde{V}_{2u}^{1/(2u)}\leq\left(1+\sum_{l=1}^{\infty}\zeta_{u,l}^{2}\right)^{1/(4u)},$
$\displaystyle
V_{2u+1}^{1/(2u+1)}\leq\tilde{V}_{2u+1}^{1/(2u+1)}\leq\left(1+\sum_{l=1}^{\infty}\zeta_{u+1,l}^{2}\right)^{1/(4u+2)}.$

We next use that

$\displaystyle\zeta_{u,1}=\left(\frac{2u}{u+1}\right)^{\beta/2}\left[1-\left(\frac{u}{u+1}\right)^{\beta}\right]^{-1}\to\infty\qquad\text{for}\quad
u\to\infty,$

so (4.26) implies that (4.21) holds. The last assertion is easy to prove. ∎

######  Theorem 4.11.

Assume $(b_{n})$ satisfies

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq{\rm e}^{-f(n)},\qquad n\geq n_{0}\geq
1,$

where $f(n)>0$ for $n\geq n_{0}$ and

$\displaystyle\liminf_{n}nf(n)=\alpha>1.$

The symmetric moment problem given by $(b_{n})$ is indeterminate of order
$\leq 1/\alpha$ and the quantity $V_{n}$ from (2.12) satisfies

$\displaystyle\limsup_{n}V_{n}^{1/n}\leq{\rm e}^{v(\alpha)/2},$ (4.27)

where $v(\alpha)$ is defined in (4.22) and $v(\infty):=0$.

###### Proof.

Let $1<\beta<\alpha$ be arbitrary. There exists $N\geq n_{0}$ such that

$\displaystyle nf(n)\geq\beta,\qquad n\geq N,$

and it follows by Proposition 1.7 that the moment problem is indeterminate of
order $\leq 1/\beta$, hence of order $\leq 1/\alpha$ since $\beta$ can be
arbitrarily close to $\alpha$. Defining

$\displaystyle\tilde{b}_{n}=\begin{cases}b_{n},&0\leq n\leq N,\\\
\displaystyle
b_{N}\exp\left[\beta\left(\frac{1}{N+1}+\frac{1}{N+2}+\dots+\frac{1}{n}\right)\right],&n>N,\end{cases}$

we have

$\displaystyle\frac{b_{n-1}}{b_{n}}\leq\frac{\tilde{b}_{n-1}}{\tilde{b}_{n}},\qquad
n\geq 1,$

hence $V_{n}\leq\tilde{V}_{n}$ by Proposition 4.2.

By (4.8) we get

$\displaystyle\limsup_{n}V_{n}^{1/n}\leq\limsup_{n}\tilde{V}_{n}^{1/n}=\limsup_{n}\big{(}\tilde{V}_{n}^{(N)}\big{)}^{1/n}.$
(4.28)

Define now

$\displaystyle\widehat{b}_{0}=b_{N}\exp\left[-\beta\left(1+\frac{1}{2}+\dots+\frac{1}{N}\right)\right],$
$\displaystyle\widehat{b}_{n}=\widehat{b}_{0}\exp\left[\beta\left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)\right],\qquad
n\geq 1.$

Then $(\widehat{b}_{n})$ satisfies
$\widehat{b}_{n-1}/\widehat{b}_{n}=\exp(-\beta/n),\;n\geq 1$, so by (4.8) and
Theorem 4.10 we get

$\displaystyle\limsup_{n}\big{(}\widehat{V}_{n}^{(N)}\big{)}^{1/n}=\limsup_{n}\widehat{V}_{n}^{1/n}\leq{\rm
e}^{v(\beta)/2}.$ (4.29)

Using $\widehat{b}_{n}^{(N)}=\tilde{b}_{n}^{(N)}$, we see by (4.28) and (4.29)
that

$\displaystyle\limsup_{n}V_{n}^{1/n}\leq{\rm e}^{v(\beta)/2}.$

Since $1<\beta<\alpha$ is arbitrary, we finally get (4.27) by continuity. ∎

##  5 Existence of the matrix product $\boldsymbol{\mathcal{A}\mathcal{H}}$

In all of this section we consider a symmetric indeterminate moment problem
given by a positive sequence $(b_{n})$ necessarily satisfying $\sum
1/b_{n}<\infty$, cf. the beginning of Section 4. We shall give sufficient
conditions ensuring property (cs*).

Since the odd moments vanish, (2.8) takes the form

$\displaystyle\sum_{n\geq j/2}^{\infty}c_{2n-j}s_{2n}<\infty\qquad\text{for
all}\quad j\geq 0.$ (5.1)

Using the expressions (2.11) and (2.12) for $U_{n}$ and $V_{n}$ we find for
$n\geq j+1$

$\displaystyle c_{2n-j}s_{2n}=V_{2n-j}U_{n}\frac{b_{0}^{2}b_{1}^{2}\cdots
b_{n-1}^{2}}{b_{0}b_{1}\cdots b_{2n-j-1}}=V_{2n-j}U_{n}\frac{b_{0}b_{1}\cdots
b_{n-1}}{b_{n}b_{n+1}\cdots b_{2n-j-1}}.$ (5.2)

######  Theorem 5.1.

Assume $(b_{n})$ is eventually $q$-increasing, cf. Definition 1.6. For $j\geq
0$ we have

$\displaystyle\lim_{n\to\infty}(c_{2n-j}s_{2n})^{1/n}=0.$

In particular the series (5.1) is convergent, so the moment problem has
property (cs*) of Definition 1.1.

###### Proof.

By Theorem 3.7 and Theorem 4.5 $(U_{n})$, $(V_{n})$ are bounded. By increasing
$n_{0}$ from (1.7) if necessary, we may assume that $b_{n}\geq 1$ for $n\geq
n_{0}$ and that $n_{0}>j$. For suitable $C>0$ and $n>n_{0}$ we find

$\displaystyle c_{2n-j}s_{2n}\leq C{b_{0}b_{1}\cdots b_{n-1}\over
b_{n}b_{n+1}\cdots b_{2n-j-1}}=Cb_{0}b_{1}\cdots
b_{n_{0}-1}{b_{n_{0}}b_{n_{0}+1}\cdots b_{n-1}\over b_{n}b_{n+1}\cdots
b_{2n-j-1}}$ $\displaystyle\hphantom{c_{2n-j}s_{2n}}{}\leq Cb_{0}b_{1}\cdots
b_{n_{0}-1}{b_{n_{0}}b_{n_{0}+1}\cdots b_{n-1}\over b_{n}b_{n+1}\cdots
b_{2n-n_{0}-1}}\leq Cb_{0}b_{1}\cdots b_{n_{0}-1}q^{(n-n_{0})^{2}},$

and therefore

$\displaystyle\lim_{n\to\infty}(c_{2n-j}s_{2n})^{1/n}=0.$ ∎

Now we turn to a case which includes eventually log-concave sequences
$(b_{n})$ of sufficient growth.

######  Theorem 5.2.

Assume $(b_{n})$ satisfies

$\displaystyle{b_{n-1}\over b_{n}}\leq{\rm e}^{-f(n)},\qquad n\geq n_{0}\geq
1,$

where $f(n)>0$ for $n\geq n_{0}$ and

$\displaystyle\alpha=\liminf_{n}nf(n)>1.$

For $j\geq 0$ we then have

$\displaystyle\limsup_{n\to\infty}(c_{2n-j}s_{2n})^{1/n}\leq
k(\alpha):=\begin{cases}4^{-\alpha}\exp(2u(\alpha)+v(\alpha)),&1<\alpha<\infty,\\\
0,&\alpha=\infty,\end{cases}$

where $u(\alpha)$, $v(\alpha)$ are defined in (3.19), (4.22). In particular
the series (5.1) is convergent when $k(\alpha)<1$, and in this case the moment
problem has property (cs*) of Definition 1.1.

###### Proof.

Let $j\geq 0$ be given and assume $1<\beta<\alpha$. By assumptions the
sequence $(b_{n})$ is strictly increasing for $n\geq n_{0}$ and tends to
infinity. Thus, there exists $n_{1}\geq\max(n_{0},j+1)$ such that

$\displaystyle b_{n}\geq 1,\qquad f(n)\geq{\beta\over n},\qquad n\geq n_{1}$
(5.3)

and for $n>n_{1}$ we then have

$\displaystyle{b_{0}b_{1}\cdots b_{n-1}\over b_{n}b_{n+1}\cdots
b_{2n-j-1}}\leq b_{0}b_{1}\cdots b_{n_{1}-1}{b_{n_{1}}b_{n_{1}+1}\cdots
b_{n-1}\over b_{n}b_{n+1}\cdots b_{2n-n_{1}-1}}.$

By assumptions and by (5.3) we get for $k\geq 1$

$\displaystyle{b_{n_{1}+k-1}\over b_{n+k-1}}\leq{\rm
e}^{-\sum\limits_{i=n_{1}+k}^{n+k-1}f(i)}\leq{\rm
e}^{-\beta\sum\limits_{i=n_{1}+k}^{n+k-1}{1\over i}}\leq{\rm
e}^{-\beta[\log(n+k)-\log(n_{1}+k)]},$

and we have used (4.23) to get the last inequality.

Therefore we have

$\displaystyle{b_{n_{1}}b_{n_{1}+1}\cdots b_{n-1}\over b_{n}b_{n+1}\cdots
b_{2n-n_{1}-1}}\leq{\rm
e}^{-\beta\sum\limits_{k=1}^{n-n_{1}}[\log(n+k)-\log(n_{1}+k)]}.$

As the function $x\longmapsto\log(n+x)-\log(n_{1}+x)$ is decreasing for $x>0,$
we get

$\displaystyle\sum_{k=1}^{n-n_{1}}[\log(n+k)-\log(n_{1}+k)]\geq\int_{1}^{n-n_{1}+1}[\log(n+x)-\log(n_{1}+x)]{\rm
d}x$
$\displaystyle\hphantom{\sum_{k=1}^{n-n_{1}}[\log(n+k)-\log(n_{1}+k)]}{}=\int_{n+1}^{2n-n_{1}+1}\log
x{\rm d}x-\int_{n_{1}+1}^{n+1}\log x{\rm d}x=2n\log 2+{\rm o}(n).$

Summarizing we obtain

$\displaystyle\limsup_{n\to\infty}\left({b_{0}b_{1}\cdots b_{n-1}\over
b_{n}b_{n+1}\cdots b_{2n-j-1}}\right)^{1/n}\leq 4^{-\beta}.$

Since $1<\beta<\alpha$ is arbitrary we get

$\displaystyle\limsup_{n\to\infty}\left({b_{0}b_{1}\cdots b_{n-1}\over
b_{n}b_{n+1}\cdots b_{2n-j-1}}\right)^{1/n}\leq 4^{-\alpha}.$ (5.4)

Finally, by juxtaposing (5.2), (5.4), (3.29) and (4.27), we get the
conclusion. ∎

######  Remark 5.3.

The function (1.12), i.e.,

$\displaystyle k(\alpha)=4^{-\alpha}\exp(2u(\alpha)+v(\alpha))$

is decreasing for $\alpha>1$. Using the software Maple we get

$\displaystyle k(1.68745)\approx 1.00001,\qquad k(1.68746)\approx 0,99997.$

Thus the series (5.1) is convergent for $\alpha\geq 1.68746$.

Examples with $\alpha=\infty$ are given in Remark 3.11. For $b_{n}=(n+1)^{c}$
we have $\alpha=c$, so property (cs*) is satisfied for $c\geq 1.68746$.

In Section 7 we will analyze the case $b_{n}=(n+1)^{c}$ by different methods
and show that(cs*) actually holds for $c>3/2$, but (cs) does not hold for
$1<c<3/2$.

######  Remark 5.4.

Chen–Ismail [12] considered the symmetrized version of the birth-and-death
process with quartic rates studied in [10]. In [12] we have

$\displaystyle b_{n}=2(n+1)\sqrt{(2n+1)(2n+3)},\qquad n\geq 0,$

so Theorem 5.2 holds with

$\displaystyle
f(n)=\log(1+1/n)+\frac{1}{2}\log\left(\frac{2n+3}{2n-1}\right),\qquad n\geq
1,$

and therefore $\alpha:=\lim\limits_{n\to\infty}nf(n)=2$. This shows that (cs*)
holds in this case.

More generally, consider the symmetrized version of the birth-and-death
process with polynomial rates of degree $p\geq 3$, where $(b_{n})$ is given in
[9, Proposition 3.2]. Take the special case $d_{j}=e_{j}-p/2$, $j=1,\ldots,p$
so that $(b_{n})$ is eventually log-concave by [9, Proposition 6.2] and hence
eventually strictly increasing by Lemma 1.5. In this case it is easy to prove
that Theorem 5.2 yields $\alpha=p/2$, and therefore (cs*) holds when $p\geq
4$.

In the next section we shall look at the cubic case $p=3$.

##  6 The symmetrized cubic case

In a series of papers Valent and coauthors have studied birth-and-death
processes with polynomial rates leading to indeterminate Stieltjes moment
problems, if the polynomial degree is at least three, see [24] and references
therein. In [24] Valent formulated conjectures about order and type of these
Stieltjes problems. The conjecture about order was settled in Romanov [22].
Another proof was given in [9] together with expressions for the type using
multi-zeta values. These formulas have lead to a proof of the Valent
conjecture about type in Bochkov [11].

Valent’s conjectures were based on concrete calculation of order and type in a
quartic case [10] and a cubic case [24]. In the latter Valent considered the
Stieltjes moment problem associated with the birth-and-death process with the
cubic rates

$\displaystyle\lambda_{n}=(3n+1)(3n+2)^{2},\qquad\mu_{n}=(3n)^{2}(3n+1),$
(6.1)

leading to the Stieltjes moment problem with recurrence coefficients from
(1.1) given by

$\displaystyle a_{n}=\lambda_{n}+\mu_{n},\qquad
b_{n}=\sqrt{\lambda_{n}\mu_{n+1}},\qquad n\geq 0.$

The purpose of this section is to estimate the moments in this cubic case.
Using these estimates it was possible to prove that property (cs) does not
hold for the corresponding symmetric moment problem. This was a surprise to us
because (cs) turned out to be true in all examples, we had been able to
calculate so far.

In order to distinguish the quantities associated with this special case from
the general quantities, we shall equip them with a star $*$.

The recurrence coefficients from (3.1) are

$\displaystyle b_{2n}^{*}=\sqrt{\lambda_{n}}=\sqrt{3n+1}(3n+2),\qquad
b_{2n-1}^{*}=\sqrt{\mu_{n}}=3n\sqrt{3n+1},$ (6.2)

cf. [9, p. 343].

######  Theorem 6.1.

For the symmetric moment problem given by $(b_{n}^{*})$ and corresponding to
the Stieltjes problem with rates (6.1) we have

$\displaystyle\sum_{n=0}^{\infty}c_{2n}^{*}s_{2n}^{*}=\infty,$

so property (cs) does not hold. The series

$\displaystyle\sum_{l=0}^{\infty}a_{2k,2l}^{*}s_{2l+2m}^{*}$

is not absolutely convergent for $k\geq 0$, $m\geq 1$, so property (ac) does
not hold.

The proof will be given after Proposition 6.5.

###  6.1 Estimate of the moments $\boldsymbol{s_{2n}^{*}}$

In accordance with [24] define

$\displaystyle\theta:=\int_{0}^{1}\frac{{\rm
d}u}{(1-u^{3})^{2/3}}=\frac{1}{3}B(1/3,1/3)=\frac{\Gamma^{3}(1/3)}{2\pi\sqrt{3}},\qquad
a:=\theta^{3},$ (6.3)

where $B$ is the beta-function. Note that

$\displaystyle\theta\approx 1.76663,\qquad a\approx
5.51370,\qquad\Gamma(1/3)\approx 2.67893.$

The Stieltjes problem given by (6.1) is indeterminate and Valent found the $B$
and $D$ functions from the Nevanlinna matrix. To give his results let

$\displaystyle\varphi_{0}(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(3n)!}z^{n},\qquad\varphi_{2}(z)=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(3n+2)!}z^{n}.$

They are entire functions of order 1/3 and type 1.

Valent’s result is that

$\displaystyle D(z)=cz\varphi_{2}(az),\qquad
B(z)-\frac{1}{t_{0}}D(z)=-\varphi_{0}(az),$ (6.4)

where

$\displaystyle c=\frac{\sqrt{3}}{(2\pi)^{3}}\Gamma(1/3)^{6}\approx 2.58105,$

and $t_{0}$ is a certain negative constant defined as

$\displaystyle t_{0}=\lim_{x\to-\infty}D(x)/B(x),$

cf. [10, Lemma 2.2.1], where it was called $\alpha$. Valent deduced from (6.4)
that the order and type of the Stieltjes problem are $1/3$ and $\theta$.

Associated with the constant $t+i\gamma$, where $t\in\mathbb{R}$, $\gamma>0$,
there is a solution to the moment problem with the density on $\mathbb{R}$,
cf. [4] and [10, equation (2.17)],

$\displaystyle\nu_{t+{\rm
i}\gamma}(x)=\frac{\gamma}{\pi}\frac{1}{(tB(x)-D(x))^{2}+\gamma^{2}B^{2}(x)}.$

We now use

$\displaystyle
t=\frac{t_{0}}{1+t_{0}^{2}},\qquad\gamma=\frac{t_{0}^{2}}{1+t_{0}^{2}}$

leading to

$\displaystyle\nu_{t+{\rm
i}\gamma}(x)=\frac{1}{\pi}\frac{1}{D^{2}(x)+(B(x)-D(x)/t_{0})^{2}},$

hence to a density for the Stieltjes problem with moments

$\displaystyle
s_{n}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x^{n}}{\varphi_{0}^{2}(ax)+c^{2}x^{2}\varphi_{2}^{2}(ax)}{\rm
d}x,\qquad n\geq 0.$

Let $B_{s}$, $D_{s}$ denote the functions from the Nevanlinna matrix in the
symmetrized case. Then we know (cf., e.g., [2, p. 48])

$\displaystyle
B_{s}(z)=B\big{(}z^{2}\big{)}-\frac{1}{t_{0}}D\big{(}z^{2}\big{)}=-\varphi_{0}\big{(}az^{2}\big{)},\qquad
D_{s}(z)=(1/z)D\big{(}z^{2}\big{)}=cz\varphi_{2}\big{(}az^{2}\big{)}.$

This gives the symmetric density

$\displaystyle\big{(}\pi\big{(}B_{s}^{2}(x)+D_{s}^{2}(x)\big{)}\big{)}^{-1}=\frac{1}{\pi}\big{(}\varphi_{0}^{2}\big{(}ax^{2}\big{)}+c^{2}x^{2}\varphi_{2}^{2}\big{(}ax^{2}\big{)}\big{)}^{-1}$

for the symmetric problem with even moments $s_{2n}^{*}$ and odd moments 0,
i.e.,

$\displaystyle
s_{2n}^{*}=\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{x^{2n}}{\varphi_{0}^{2}\big{(}ax^{2}\big{)}+c^{2}x^{2}\varphi_{2}^{2}\big{(}ax^{2}\big{)}}{\rm
d}x=\frac{a^{-n-1/2}}{\pi}\int_{0}^{\infty}\frac{t^{n-1/2}}{\varphi_{0}^{2}(t)+(c^{2}/a)t\varphi_{2}^{2}(t)}{\rm
d}t.$ (6.5)

Using (6.5) we shall prove the following estimates of $s_{2n}^{*}$.

######  Proposition 6.2.

There exist constants $K_{i}>0$, $i=1,2$, such that

$\displaystyle
K_{1}a^{-n}\Gamma(3n+3/2)<s_{2n}^{*}<K_{2}a^{-n}n\Gamma(3n+3/2),\qquad n\geq
1,$ (6.6)

where $a$ is given by (6.3). In particular

$\displaystyle\lim_{n\to\infty}\frac{\root n
\of{s_{2n}^{*}}}{n^{3}}=\frac{27}{a{\rm e}^{3}}=\left(\frac{3}{{\rm
e}\theta}\right)^{3}$ (6.7)

and

$\displaystyle\lim_{n\to\infty}\root n
\of{U_{n}^{*}}=\left(\frac{2}{\theta}\right)^{3}.$ (6.8)

###### Proof.

In [17] some interesting formulas are given about the trigonometric functions
of order 3. These functions are defined as

$\displaystyle\sigma_{l}(u)=\sum_{n=0}^{\infty}(-1)^{n}\frac{u^{3n+l}}{(3n+l)!},\qquad
l=0,1,2.$

With ${\rm j}={\rm e}^{{\rm i}\pi/3}$ we have

$\displaystyle\sigma_{0}(u)=\frac{1}{3}\big{(}{\rm e}^{-u}+{\rm e}^{{\rm
j}u}+{\rm e}^{\overline{{\rm j}}u}\big{)}=\frac{1}{3}\left({\rm
e}^{-u}+2\cos\left(\frac{\sqrt{3}}{2}u\right){\rm
e}^{u/2}\right)=\varphi_{0}\big{(}u^{3}\big{)},$
$\displaystyle\sigma_{2}(u)=\frac{1}{3}\big{(}{\rm e}^{-u}-{\rm j}{\rm
e}^{ju}-\overline{{\rm j}}{\rm e}^{\overline{{\rm
j}}u}\big{)}=\frac{1}{3}\left({\rm
e}^{-u}-2\cos\left(\frac{\sqrt{3}}{2}u+\pi/3\right){\rm
e}^{u/2}\right)=u^{2}\varphi_{2}\big{(}u^{3}\big{)},$

and there is a similar expression for $\sigma_{1}(u)$, which we do not need.

Substituting $t=u^{3}$ in (6.5) we get

$\displaystyle
s_{2n}^{*}=\frac{3a^{-n-1/2}}{\pi}\int_{0}^{\infty}\frac{u^{3n+1/2}}{\sigma_{0}^{2}(u)+\big{(}c^{2}/(au)\big{)}\sigma_{2}^{2}(u)}{\rm
d}u$

or

$\displaystyle
s_{2n}^{*}=\frac{27a^{-n-1/2}}{\pi}\int_{0}^{\infty}\frac{u^{3n+1/2}{\rm
e}^{-u}}{N(u)}{\rm d}u$

with

$\displaystyle N(u)=\left({\rm
e}^{-3u/2}+2\cos\left(\frac{\sqrt{3}}{2}u\right)\right)^{2}+\frac{c^{2}}{au}\left({\rm
e}^{-3u/2}-2\cos\left(\frac{\sqrt{3}}{2}u+\pi/3\right)\right)^{2}.$

One can prove that $N(u)\leq 9$ for $u\geq 0$. (The constant comes from the
software Maple. That $N(u)$ is bounded for $u\geq 0$ is however elementary,
since the first term is bounded by 9, and using that

$\displaystyle{\rm e}^{-3u/2}-2\cos\left(\frac{\sqrt{3}}{2}u+\pi/3\right)$

vanishes for $u=0$, also the second term is bounded.)

Furthermore,

$\displaystyle N(u)\geq k_{1}:=\big{(}{\rm
e}^{-3/2}+2\cos\big{(}\sqrt{3}/2\big{)}\big{)}^{2}\approx 2.30690,\qquad 0\leq
u\leq 1,$

and using that $c^{2}/a\approx 1.20823>1$, we get for $u\geq 1$:

$\displaystyle uN(u)>\left({\rm
e}^{-3u/2}+2\cos\left(\frac{\sqrt{3}}{2}u\right)\right)^{2}+\left({\rm
e}^{-3u/2}-2\cos\left(\frac{\sqrt{3}}{2}u+\pi/3\right)\right)^{2}$
$\displaystyle\hphantom{uN(u)}{}=2{\rm
e}^{-3u}+4\cos^{2}\left(\frac{\sqrt{3}}{2}u\right)+4\cos^{2}\left(\frac{\sqrt{3}}{2}u+\pi/3\right)$
$\displaystyle\hphantom{uN(u)>}{}+4{\rm
e}^{-3u/2}\left(\cos\left(\frac{\sqrt{3}}{2}u\right)-\cos\left(\frac{\sqrt{3}}{2}u+\pi/3\right)\right)$
$\displaystyle\hphantom{uN(u)}{}\geq f(u):=2{\rm e}^{-3u}-8{\rm
e}^{-3u/2}+2\geq f(1),$

where we have used that $\cos^{2}(x)+\cos^{2}(x+\pi/3)\geq 1/2$ and that
$f(u)$ is increasing (for $u\geq 0$). This shows that

$\displaystyle uN(u)\geq k_{2}:=f(1)\approx 0.31453,\qquad u\geq 1.$

We then get

$\displaystyle\frac{3a^{-n-1/2}}{\pi}\Gamma(3n+3/2)<s_{2n}^{*}<\frac{27a^{-n-1/2}}{\pi}\left(\int_{0}^{1}\frac{u^{3n+1/2}{\rm
e}^{-u}}{k_{1}}{\rm d}u+\int_{1}^{\infty}\frac{u^{3n+3/2}{\rm
e}^{-u}}{k_{2}}{\rm d}u\right)$
$\displaystyle\hphantom{\frac{3a^{-n-1/2}}{\pi}\Gamma(3n+3/2)<s_{2n}^{*}}{}<\frac{27a^{-n-1/2}}{\pi}\left(\frac{1}{k_{1}}+\frac{1}{k_{2}}\Gamma(3n+5/2)\right),$

which shows (6.6).

Note that by Stirling’s formula

$\displaystyle\Gamma(3n+\delta)\sim 3^{\delta-1/2}\sqrt{2\pi}\big{(}27/{\rm
e}^{3}\big{)}^{n}n^{3n+\delta-1/2},$ (6.9)

so (6.6) leads to (6.7).

From (6.2) we get

$\displaystyle b_{0}^{*}b_{1}^{*}\cdots b_{2n}^{*}=\Gamma(3n+2),\qquad
b_{0}^{*}b_{1}^{*}\cdots b_{2n-1}^{*}=\sqrt{3n+1}\Gamma(3n+1),$ (6.10)

which together with (6.7) gives (6.8). ∎

###  6.2 The matrix $\boldsymbol{\mathcal{A}}$ in the symmetrized cubic case

Let $(S_{n})$ denote the orthonormal polynomials in the symmetrized cubic
case. They are given by (2.9), with ${}^{s}b_{n}=b_{n}^{*}$ from (6.2).

It is known that, cf. (2.10),

$\displaystyle
K_{s}(z,w)=\sum_{n=0}^{\infty}S_{n}(z)S_{n}(w)=\frac{D_{s}(z)B_{s}(w)-B_{s}(z)D_{s}(w)}{w-z}=\sum_{k,l=0}^{\infty}a_{k,l}^{(s)}z^{k}w^{l},$

where the third expression can be found in [2, equation (1.8)].

We shall find explicit formulas for $a_{k,l}^{(s)}$ based on the following
lemma.

######  Lemma 6.3.

Let $f$, $g$ be entire functions with power series

$\displaystyle f(z)=\sum_{n=0}^{\infty}\alpha_{n}z^{n},\qquad
g(z)=\sum_{n=0}^{\infty}\beta_{n}z^{n}.$

Then the entire function of two variables

$\displaystyle L(z,w)=\frac{f(z)g(w)-f(w)g(z)}{z-w}$

has the power series

$\displaystyle L(z,w)=\sum_{k,l=0}^{\infty}\gamma_{k,l}z^{k}w^{l},$

where

$\displaystyle\gamma_{k,l}=\sum_{j=0}^{l}(\alpha_{j+k+1}\beta_{l-j}-\beta_{j+k+1}\alpha_{l-j}).$
(6.11)

######  Remark 6.4.

By construction $L(z,w)=L(w,z)$ and therefore $\gamma_{k,l}=\gamma_{l,k}$, but
this is not obvious from (6.11).

###### Proof.

We have

$\displaystyle L(z,w)=\frac{f(z)-f(w)}{z-w}g(w)-f(w)\frac{g(z)-g(w)}{z-w}$

and

$\displaystyle\frac{f(z)-f(w)}{z-w}=\sum_{n=1}^{\infty}\alpha_{n}\frac{z^{n}-w^{n}}{z-w}=\sum_{n=1}^{\infty}\alpha_{n}\sum_{j=0}^{n-1}z^{n-1-j}w^{j}$
$\displaystyle\hphantom{\frac{f(z)-f(w)}{z-w}}{}=\sum_{j=0}^{\infty}w^{j}\sum_{n=j+1}^{\infty}\alpha_{n}z^{n-1-j}=\sum_{j,k=0}^{\infty}\alpha_{j+k+1}z^{k}w^{j}.$

Therefore,

$\displaystyle\frac{f(z)-f(w)}{z-w}g(w)=\sum_{m=0}^{\infty}\beta_{m}w^{m}\sum_{j,k=0}^{\infty}\alpha_{j+k+1}z^{k}w^{j}=\sum_{k,l=0}^{\infty}z^{k}w^{l}\left(\sum_{j=0}^{l}\alpha_{j+k+1}\beta_{l-j}\right),$

and by symmetry we get (6.11). ∎

Write

$\displaystyle
f(z)=B_{s}(z)=\sum_{n=0}^{\infty}\alpha_{n}z^{n},\qquad\alpha_{2n}=\frac{-(-a)^{n}}{(3n)!},\qquad\alpha_{2n+1}=0,$
$\displaystyle
g(z)=D_{s}(z)=\sum_{n=0}^{\infty}\beta_{n}z^{n},\qquad\beta_{2n}=0,\qquad\beta_{2n+1}=\frac{c(-a)^{n}}{(3n+2)!}.$

Then $K_{s}(z,w)=L(z,w)$ from Lemma 6.3, and we get

$\displaystyle
a_{k,l}^{(s)}=a_{k,l}^{*}=\sum_{j=0}^{l}(\alpha_{j+k+1}\beta_{l-j}-\beta_{j+k+1}\alpha_{l-j}).$

In particular $a_{0,0}^{*}=-\alpha_{0}\beta_{1}=c/2$ and for $k\geq 1$

$\displaystyle
a_{k,k}^{*}=\sum_{j=0}^{k}(\alpha_{j+k+1}\beta_{k-j}-\beta_{j+k+1}\alpha_{k-j}).$

Let us first focus on $k$ even ($\geq 2$):

$\displaystyle
a_{2k,2k}^{*}=\sum_{j=0}^{2k}(\alpha_{j+2k+1}\beta_{2k-j}-\beta_{j+2k+1}\alpha_{2k-j})=\sum_{j=0}^{k-1}\alpha_{2j+2k+2}\beta_{2k-2j-1}-\sum_{j=0}^{k}\beta_{2j+2k+1}\alpha_{2k-2j}$
$\displaystyle\hphantom{a_{2k,2k}^{*}}{}=c\sum_{j=0}^{k-1}\frac{-(-a)^{j+k+1}}{(3j+3k+3)!}\frac{(-a)^{k-j-1}}{(3k-3j-1)!}-c\sum_{j=0}^{k}\frac{(-a)^{j+k}}{(3k+3j+2)!}\frac{-(-a)^{k-j}}{(3k-3j)!}$
$\displaystyle\hphantom{a_{2k,2k}^{*}}{}=ca^{2k}\left(\frac{1}{(6k+2)!}+\sum_{j=0}^{k-1}\frac{6j+3}{(3k+3j+3)!(3k-3j)!}\right),$

hence

$\displaystyle
a_{2k,2k}^{*}=\frac{ca^{2k}}{(6k+3)!}\sum_{j=0}^{k}(6j+3)\binom{6k+3}{3k-3j}.$
(6.12)

For $k$ odd we similarly get

$\displaystyle
a_{2k-1,2k-1}^{*}=\frac{6ca^{2k-1}k}{(6k+1)!}\sum_{j=0}^{k-1}(6j+3)\binom{6k+1}{3k-3j-1}.$

Using the following estimates

$\displaystyle
3\binom{6k+3}{3k}<\sum_{j=0}^{k}(6j+3)\binom{6k+3}{3k-3j}<\binom{6k+3}{3k}\sum_{j=0}^{k}(6j+3)=3(k+1)^{2}\binom{6k+3}{3k},$

we get for suitable $C_{i}>0$, $i=1,2$,

$\displaystyle C_{1}\big{(}a{\rm
e}^{3}/27\big{)}^{k}k^{-3k-2}<c_{2k}^{*}=\sqrt{a_{2k,2k}^{*}}<C_{2}\big{(}a{\rm
e}^{3}/27\big{)}^{k}k^{-3k}.$ (6.13)

From (6.13) and similar inequalities for
$c_{2k-1}^{*}=\sqrt{a_{2k-1,2k-1}^{*}}$ together with (6.10), we get

######  Proposition 6.5.

$\displaystyle\lim_{n\to\infty}n^{3/2}\root n \of{c_{n}^{*}}=\left(\frac{2{\rm
e}\theta}{3}\right)^{3/2}.$ (6.14)

and

$\displaystyle\lim_{n\to\infty}\root{n} \of{V_{n}^{*}}=\theta^{3/2}.$ (6.15)

Equation (6.14) is a sharpening of equation (2.14), by replacing $\limsup$ by
a limit. Note that $\rho=2/3$ is the order and $\tau=\theta$ is the type of
the symmetric moment problem. Equation (6.15) is a sharpening of (4.27)
applied to $b_{n}^{*}$ given by (6.2).

###### Proof of Theorem 6.1.

Putting (6.6) and (6.13) together and using (6.9), we get

$\displaystyle L_{1}n^{-1}<s_{2n}^{*}c_{2n}^{*}<L_{2}n^{2}$

for suitable $L_{i}>0$, $i=1,2$.

This shows divergence of $\sum s_{2n}^{*}c_{2n}^{*}$, which is the first part
of Theorem 6.1.

We shall next show that most of the series involved in defining
$\mathcal{A}\mathcal{H}$ are not absolutely convergent. We know that
$a_{2k,2l+1}^{*}=0$ and equation (6.12) can easily be generalized to

$\displaystyle
a_{2k,2l}^{*}=\frac{c(-a)^{k+l}}{(3k+3l+3)!}\sum_{j=0}^{l}\binom{3k+3l+3}{3l-3j}(3k-3l+6j+3).$

By symmetry we can assume that $k\geq l$, so all terms in the sum are
positive. We then get

$\displaystyle(3(k-l)+3)\binom{3k+3l+3}{3l}<\sum_{j=0}^{l}(3(k-l)+6j+3)\binom{3k+3l+3}{3l-3j}$
$\displaystyle\hphantom{(3(k-l)+3)\binom{3k+3l+3}{3l}}{}<3(k+1)(l+1)\binom{3k+3l+3}{3l},$

because

$\displaystyle\binom{3k+3l+3}{3l-3j}\leq\binom{3k+3l+3}{3l},\qquad 0\leq j\leq
l\leq k,$

hence

$\displaystyle
ca^{k+l}\frac{3(k-l)+3}{(3k+3)!(3l)!}<|a_{2k,2l}^{*}|<ca^{k+l}\frac{3(k+1)(l+1)}{(3k+3)!(3l)!}.$

If we look at $\sum_{k}|a_{2k,2l}^{*}|s_{2k+2m}^{*}$, for each fixed $l$, $m$,
it is enough to examine convergence of the series for $k>l$, i.e., of the
terms

$\displaystyle\frac{a^{k}k}{(3k+2)!}s_{2k+2m}^{*}.$

Using (6.6), we get that for $k$ sufficiently large

$\displaystyle
M_{1}k^{3m-3/2}<\frac{a^{k}k}{(3k+2)!}s_{2k+2m}^{*}<M_{2}k^{3m-1/2}$

for constants $M_{1}$, $M_{2}$ depending on $l$, $m$ but independent of $k$.
This shows that the series $\sum_{k}|a_{2k,2l}^{*}|s_{2k+2m}^{*}$ diverges for
$m\geq 1$, but we do not get information about convergence/divergence when
$m=0$.

We conclude that most of the series involved in the calculation of
$\mathcal{A}\mathcal{H}$ are not absolutely convergent. ∎

##  7 The symmetric moment problem with $\boldsymbol{b_{n}=c^{c}(n+1)^{c}}$,
$\boldsymbol{c>0}$

The sequence $b_{n}=b_{n}(c)=c^{c}(n+1)^{c}$ is log-concave for $c>0$. By the
Carleman condition the moment problem is determinate for $0<c\leq 1$, and for
$c>1$ it is indeterminate of order $\rho=1/c$, cf. [8, Theorem 4.11]. The
factor $c^{c}$ does not influence the order of the moment problem, but it is
convenient when $c=3/2$ for comparison with the symmetrized cubic case.

The corresponding orthonormal polynomials $P_{n}(c;x)$ do not seem to be known
except in the cases $c=1/2$ and $c=1$, where they are scaled versions of the
Hermite polynomials and Meixner–Pollaczek polynomials. In a precise manner

$\displaystyle
P_{n}(1/2;x)=(2^{n}n!)^{-1/2}H_{n}\big{(}x/\sqrt{2}\big{)},\qquad
P_{n}(1;x)=P_{n}^{(1/2)}(x/2;\pi/2),$

where $H_{n}$ are the Hermite polynomials and $P_{n}^{(\lambda)}(x;\varphi)$
the Meixner–Pollaczek polynomials in the notation of [20].

######  Definition 7.1.

The type of the symmetric indeterminate moment problem with
$b_{n}=c^{c}(n+1)^{c}$, $c>1$ is denoted by $\tau_{c}$.

By [9, Proposition 2.3] we know that $b_{n}=(n+1)^{c}$ leads to a moment
problem with order $1/c$ and type

$\displaystyle\frac{\tau_{c}}{(c^{-c})^{1/c}}=c\tau_{c}.$

For real $c>1$ and $n\geq 1$ define the multi-zeta value111Observe that the
inequalities between the indices $k_{j}$ are alternating between $\leq$ and
$<$: $k_{2j-1}\leq k_{2j}<k_{2j+1}\leq k_{2j+2}$.

$\displaystyle\gamma_{n}(c)=\sum_{1\leq k_{1}\leq k_{2}<\cdots<k_{2n-1}\leq
k_{2n}}\left(k_{1}k_{2}\cdots k_{2n-1}k_{2n}\right)^{-c}.$

Next define the function

$\displaystyle G_{c}(z)=\sum_{n=1}^{\infty}\gamma_{n}(c)z^{n}.$

By [9, Remark 1.6] it is known that the order of $G_{c}$ is $1/(2c)$ and by
[9, Theorem 1.11] we get $c\tau_{c}=T_{c}/2$, where $T_{c}$ is the type of
$G_{c}$ given by

$\displaystyle
T_{c}=\frac{2c}{e}\limsup_{n\to\infty}\big{(}n(\gamma_{n}(c))^{1/(2cn)}\big{)},$

based on [21]. (Unfortunately there is a misprint in equation (18) in [9, p.
339].) By Bochkov’s proof in [11] of the Valent conjecture about type follows
that

$\displaystyle
T_{c}=B\left(\frac{1}{2c},1-\frac{1}{c}\right),\qquad\tau_{c}=\int_{0}^{1}\frac{{\rm
d}t}{\big{(}1-t^{2c}\big{)}^{1/c}}.$

We shall first compare quantities defined in terms of
$b_{n}=b_{n}(3/2)=(3/2)^{3/2}(n+1)^{3/2}$ with quantities in Section 6 for the
symmetrized cubic case, where $b_{n}=b_{n}^{*}$ is defined in (6.2).

######  Lemma 7.2.

Let $s_{2n},c_{n}$ denote the quantities corresponding to $b_{n}=b_{n}(3/2)$.
Then

$\displaystyle\lim_{n\to\infty}\frac{\root n
\of{s_{2n}}}{n^{3}}=\left(\frac{3}{{\rm e}\theta}\right)^{3},$ (7.1)

$($i.e., the same value as in (6.7)$)$ and

$\displaystyle\limsup_{n\to\infty}n^{3/2}\root n
\of{c_{n}}=\limsup_{n\to\infty}(2n)^{3/2}\root 2n
\of{c_{2n}}=\left(\frac{2{\rm e}\theta}{3}\right)^{3/2},$ (7.2)

$($i.e., the same value as in (6.14)$)$. In particular

$\displaystyle\limsup_{n\to\infty}\root{n} \of{c_{2n}s_{2n}}=1.$ (7.3)

###### Proof.

First note that

$\displaystyle b_{n}(3/2)\leq b_{n}^{*}\leq b_{n+1}(3/2),$ (7.4)

i.e.,

$\displaystyle(3/2)^{3/2}(2n+1)^{3/2}\leq\sqrt{3n+1}(3n+2)\leq(3/2)^{3/2}(2n+2)^{3/2},$
(7.5) $\displaystyle(3/2)^{3/2}(2n)^{3/2}\leq
3n\sqrt{3n+1}\leq(3/2)^{3/2}(2n+1)^{3/2}.$ (7.6)

The two inequalities to the right are obvious, and so is the inequality to the
left in (7.6). The inequality to the left in (7.5) is equivalent to

$\displaystyle(n+1/2)^{3}\leq(n+1/3)(n+2/3)^{2}.$

Setting $x=n+1/2$ the inequality is reduced to

$\displaystyle
x^{3}\leq(x-1/6)(x+1/6)^{2}=x^{3}+\frac{1}{36}\big{(}\big{(}2x^{2}-x\big{)}+\big{(}4x^{2}-1/6\big{)}\big{)},$

which is valid for $x\geq 1/2$.

By (7.4) we get from Proposition 3.1, (3.7) and (3.9)

$\displaystyle s_{2n}\leq s_{2n}^{*}\leq
s_{2n}^{(1)}\leq\left(\frac{2}{3}\right)^{3}s_{2n+2}\leq 4(n+1)^{3}s_{2n}.$

It follows that

$\displaystyle\frac{\root n \of{s_{2n}}}{n^{3}}\leq\frac{\root n
\of{s_{2n}^{*}}}{n^{3}}\leq\big{(}4(n+1)^{3}\big{)}^{1/n}\frac{\root n
\of{s_{2n}}}{n^{3}}$

and (7.1) follows from (6.7).

The function $\sum\limits_{n=0}^{\infty}c_{n}z^{n}$ has the same order and
type as the moment problem for $(b_{n}(3/2))$ by Theorem 3.1 in [8], so by
(2.14)

$\displaystyle\limsup_{n\to\infty}n^{3/2}\root n \of{c_{n}}=\left(\frac{2{\rm
e}\tau_{3/2}}{3}\right)^{3/2}.$

However, by Remark 1.10 in [9] we have $\tau_{3/2}=\theta$, where $\theta$ is
the type of the cubic Stieltjes case of Section 6, and the type is unchanged
when we consider the symmetrized version, cf. Proposition 3.1 in [9].

Therefore

$\displaystyle\limsup_{n\to\infty}n^{3/2}\root n \of{c_{n}}=\left(\frac{2{\rm
e}\theta}{3}\right)^{3/2}=\limsup_{n\to\infty}n^{3/2}\root n \of{c_{n}^{*}},$

where the last equality sign comes from (6.14).

By Proposition 4.7 we have $c_{k}\leq b_{k}(3/2)c_{k+1}$, and from this it is
easy to see that

$\displaystyle\limsup_{n\to\infty}(2n)^{3/2}\root 2n
\of{c_{2n}}=\limsup_{n\to\infty}(2n+1)^{3/2}\root 2n+1 \of{c_{2n+1}},$

and (7.2) follows. ∎

######  Theorem 7.3.

Define $b_{n}(c)=c^{c}(n+1)^{c}$, $c>1$ and let $s_{2n}(c)$, $c_{n}(c)$ denote
the corresponding quantities. Then the series $\sum c_{2n}(c)s_{2n}(c)$ is
convergent for $c>3/2$ and divergent for $1<c<3/2$, i.e., property (cs) holds
for $c>3/2$, but not for $1<c<3/2$.

###### Proof.

From (5.2) we get

$\displaystyle c_{2n}(c)s_{2n}(c)\binom{2n}{n}^{c}=V_{2n}(c)U_{n}(c),$

hence

$\displaystyle\frac{c_{2n}(c)s_{2n}(c)}{c_{2n}(3/2)s_{2n}(3/2)}=\frac{V_{2n}(c)U_{n}(c)}{V_{2n}(3/2)U_{n}(3/2)}\binom{2n}{n}^{3/2-c}.$

By Propositions 3.2 and 4.2 $U_{n}(c)$ and $V_{n}(c)$ are decreasing in $c$,
hence

$\displaystyle\frac{c_{2n}(c)s_{2n}(c)}{c_{2n}(3/2)s_{2n}(3/2)}\
\begin{cases}\leq\binom{2n}{n}^{3/2-c}&\text{for}\ 3/2<c,\\\
\geq\binom{2n}{n}^{3/2-c}&\text{for}\ 1<c<3/2.\end{cases}$

By Stirling’s formula

$\displaystyle\lim_{n\to\infty}\binom{2n}{n}^{1/n}=4,$

so by (7.3) we get

$\displaystyle\limsup_{n\to\infty}\root{n} \of{c_{2n}(c)s_{2n}(c)}\
\begin{cases}\leq 4^{3/2-c}<1&\text{for}\ 3/2<c,\\\ \geq
4^{3/2-c}>1&\text{for}\ 1<c<3/2.\end{cases}$ ∎

######  Theorem 7.4.

The symmetric moment problem with $b_{n}(c)=c^{c}(n+1)^{c}$ has property (cs*)
of Definition 1.1 when $c>3/2$.

###### Proof.

By Proposition 4.7 we have $c_{k}(c)\leq b_{k}(c)c_{k+1}(c)$, hence for $j\geq
0$

$\displaystyle c_{2n-j}(c)\leq
c^{cj}\big{(}(2n-j+1)(2n-j+2)\cdots(2n)\big{)}^{c}c_{2n}(c),$

and it follows that

$\displaystyle\limsup_{n\to\infty}\root{n} \of{c_{2n-j}(c)s_{2n}(c)}\leq
4^{3/2-c}<1.$

This shows that (5.1) holds. ∎

######  Remark 7.5.

We do not know if (ac) or (cs) holds in case $c=3/2$. We also do not know if
(aci) holds when $1<c<3/2$. In order to answer these questions, it seems to be
necessary to have information about the orthogonal polynomials $P_{n}(3/2;x)$.

##  Appendix A Appendix

###  A.1 A lower bound of $\boldsymbol{(U_{n})}$ when $\boldsymbol{(b_{n})}$
is bounded

######  Proposition A.1.

Let $(b_{n})$ be a bounded sequence. Then

$\displaystyle\liminf_{n\to\infty}U_{n}^{1/n}\geq 4.$

If $\lim b_{n}=b\geq 0$, then $\liminf$ is a limit.

###### Proof.

Let $J$ denote the Jacobi matrix associated with the sequence $(b_{n}),$
acting on $\ell^{2}(\mathbb{N}_{0})$ by

$\displaystyle J\delta_{n}=b_{n}\delta_{n+1}+b_{n-1}\delta_{n-1},\qquad n\geq
0,$

where $\delta_{n}$, $n\in\mathbb{N}_{0}$ is the standard orthonormal basis in
$\ell^{2}(\mathbb{N}_{0})$ with the convention $b_{-1}=0,$ $\delta_{-1}=0$.
The operator $J$ is bounded on $\ell^{2}(\mathbb{N}_{0})$ and its operator
norm is given by

$\displaystyle\|J\|^{2}=\lim_{n\to\infty}s_{2n}^{1/n}.$

Let $S$ denote the unilateral shift defined by $S\delta_{n}=\delta_{n+1}$.
Define

$\displaystyle J_{N}={1\over N}\sum_{j=0}^{N-1}(S^{*})^{j}JS^{j}.$

Then

$\displaystyle\|J_{N}\|\leq\|J\|.$

The operator $J_{N}$ is corresponds to the Jacobi matrix with coefficients

$\displaystyle b_{n}^{[N]}={1\over N}\sum_{j=0}^{N-1}b_{n+j}.$

For any fixed $n$ we have

$\displaystyle\lim_{N\to\infty}\big{[}b_{n+1}^{[N]}-b_{n}^{[N]}\big{]}=\lim_{N\to\infty}{b_{n+N}-b_{n}\over
N}=0,$

hence

$\displaystyle\lim_{N\to\infty}\big{[}b_{n}^{[N]}-b_{0}^{[N]}\big{]}=0.$ (A.1)

We will consider two cases.

Case (a). The sequence $b_{0}^{[N]}$, $N\geq 1$ does not converge to $0$. Let
$b$ denote the greatest accumulation point of $b_{0}^{[N]}$, $N\geq 1$. Then
there exists a sequence $N_{1}<N_{2}<\cdots$ from $\mathbb{N}$ so that

$\displaystyle\lim_{{k\to\infty}}b_{0}^{[N_{k}]}=b.$

By (A.1) for any fixed $n$ we have

$\displaystyle\lim_{{k\to\infty}}b_{n}^{[N_{k}]}=b.$

The latter implies that the sequence of the operators $J_{N_{k}}$ converges
$*$-weakly to the operator $\widetilde{J}$ with constant Jacobi coefficients
$b$. It is well known that

$\displaystyle\big{\|}\widetilde{J}\big{\|}=2b.$

Thus, by Fatou’s lemma, we get

$\displaystyle
2b=\big{\|}\widetilde{J}\big{\|}\leq\liminf_{k\to\infty}\|J_{N_{k}}\|\leq\|J\|.$
(A.2)

On the other hand we have

$\displaystyle(b_{0}b_{1}\cdots b_{N-1})^{1/N}\leq{1\over
N}\sum_{j=0}^{N-1}b_{j}=b_{0}^{[N]}.$ (A.3)

Hence

$\displaystyle\limsup_{N}(b_{0}b_{1}\cdots b_{N-1})^{1/N}\leq b.$ (A.4)

Therefore by (A.2) and (A.4) we obtain

$\displaystyle\liminf_{N\to\infty}{{s_{2N}}^{1/N}\over\big{(}b_{0}^{2}b_{1}^{2}\cdots
b_{N-1}^{2}\big{)}^{1/N}}=\liminf_{N\to\infty}{\|J\|^{2}\over\big{(}b_{0}^{2}b_{1}^{2}\cdots
b_{N-1}^{2}\big{)}^{1/N}}$
$\displaystyle\hphantom{\liminf_{N\to\infty}{{s_{2N}}^{1/N}\over\big{(}b_{0}^{2}b_{1}^{2}\cdots
b_{N-1}^{2}\big{)}^{1/N}}}{}={\|J\|^{2}\over\limsup\limits_{N\to\infty}\big{(}b_{0}^{2}b_{1}^{2}\cdots
b_{N-1}^{2}\big{)}^{1/N}}\geq{4b^{2}\over b^{2}}=4.$

Case (b). The sequence $b_{0}^{[N]}$, $N\geq 1$ converges to $0$. Then by
(A.3) we get

$\displaystyle\lim_{N\to\infty}(b_{0}b_{1}\cdots b_{N-1})^{1/N}=0.$

Next

$\displaystyle\lim_{N\to\infty}{s_{2N}^{1/N}\over\big{(}b_{0}^{2}b_{1}^{2}\cdots
b_{N-1}^{2}\big{)}^{1/N}}=\lim_{N\to\infty}{\|J\|^{2}\over\big{(}b_{0}^{2}b^{2}_{1}\cdots
b^{2}_{N-1}\big{)}^{1/N}}=\infty.$

The last part of the result follows by inspection of the proof given for the
first part. ∎

######  Remark A.2.

The proof of Proposition A.1 shows that if $\lim b_{n}=1/2$ and if the moment
representing measure $\mu$ is supported by $[-1,1]$, then $\|J\|=1$ and $\lim
U_{n}^{1/n}=4$. Examples of this are given by the Gegenbauer weights

$\displaystyle
w_{\lambda}(x)=\frac{(1-x^{2})^{\lambda-1/2}}{B(1/2,\lambda+1/2)},\qquad-1<x<1,\qquad\lambda>-1/2.$

It is known that

$\displaystyle s_{2n}=\frac{(1/2)_{n}}{(\lambda+1)_{n}},\qquad
b_{n}=\frac{1}{2}\sqrt{\frac{(n+1)(n+2\lambda)}{(n+\lambda)(n+\lambda+1)}},$

where the expression for $b_{n}$ shall be interpreted for $\lambda=0$ as
$b_{0}=1/\sqrt{2}$, $b_{n}=1/2$, $n\geq 1$, cf. [20], so it is easy to see
that $\lim U_{n}^{1/n}=4$ independent of $\lambda$.

###  A.2 Non-unicity of the inverse of $\boldsymbol{\mathcal{H}}$

A positive definite Hankel matrix $\mathcal{H}$ acts injectively on
$\mathcal{F}_{c}(\mathbb{N}_{0})$ in the sense that $\mathcal{H}v=0$ for
$v\in\mathcal{F}_{c}(\mathbb{N}_{0})$ implies $v=0$.

Nevertheless, it is possible that there exist sequences $v\neq 0$ with
$\mathcal{H}v=0$, and such that $\mathcal{H}v$ is absolutely convergent in the
sense that $\sum_{n}|s_{m+n}v_{n}|<\infty$ for all $m\geq 0$.

We show this for the following variant of the log-normal moment sequence. Let
$0<q<1$ and consider

$\displaystyle s_{n}=q^{-n^{2}/2}=\int_{0}^{\infty}x^{n}w_{q}(x)/x{\rm
d}x,\qquad n\geq 0,$ (A.5)

where

$\displaystyle w_{q}(x)=\frac{1}{\sqrt{2\pi\log(1/q)}}\exp\left(-\frac{(\log
x)^{2}}{2\log(1/q)}\right).$

Note that $w_{q}(x/q)=\sqrt{q}w_{q}(x)/x$, so from [7, equation (42)], we get
that the orthonormal polynomials for (A.5) are

$\displaystyle
P_{n}(x)=(-1)^{n}\frac{q^{n/2}}{\sqrt{(q;q)_{n}}}\sum_{k=0}^{n}\binom{n}{k}_{q}(-1)^{k}q^{k^{2}-k/2}x^{k},$
(A.6)

with

$\displaystyle\binom{n}{k}_{q}=\frac{(q;q)_{n}}{(q;q)_{k}(q;q)_{n-k}}.$

######  Theorem A.3.

Let $\mathcal{H}$ be the Hankel matrix for the indeterminate Stieltjes problem
with moments (A.5). The set $\mathcal{V}$ of real sequences $v$ such that
$\mathcal{H}v=0$ and $\mathcal{H}v$ is absolutely convergent, is an infinite
dimensional vector space. For any $k\geq 1$ and
$(a_{0},a_{1},\ldots,a_{k-1})\in\mathbb{R}^{k}$ there exists $v\in\mathcal{V}$
such that $v_{j}=a_{j}$, $j=0,1,\ldots,k-1$.

###### Proof.

Let $v=(v_{n})_{n\geq 0}$ be defined by

$\displaystyle v_{n}=(-1)^{n}\frac{q^{n^{2}-n/2}}{(q;q)_{n}},\qquad n\geq 0.$
(A.7)

It is well-known that

$\displaystyle(z;q)_{\infty}=\sum_{n=0}^{\infty}(-1)^{n}\frac{q^{\binom{n}{2}}}{(q;q)_{n}}z^{n},\qquad
z\in\mathbb{C},$

see [16]. For $m\geq 0$ we therefore get

$\displaystyle\sum_{n=0}^{\infty}s_{m+n}v_{n}=q^{-m^{2}/2}\sum_{n=0}^{\infty}(-1)^{n}\frac{q^{\binom{n}{2}-mn}}{(q;q)_{n}}=q^{-m^{2}/2}\big{(}q^{-m};q\big{)}_{\infty}=0,$

and

$\displaystyle\sum_{n=0}^{\infty}s_{m+n}|v_{n}|=q^{-m^{2}/2}\sum_{n=0}^{\infty}\frac{q^{\binom{n}{2}-mn}}{(q;q)_{n}}=q^{-m^{2}/2}\big{(}{-}q^{-m};q\big{)}_{\infty}<\infty,$

showing that $v\in\mathcal{V}$ and $v_{0}=1$.

We next construct $v\in\mathcal{V}$ with $k$ initial values and this implies
that $\dim\mathcal{V}=\infty$.

For $(c_{0},\ldots,c_{k-1})\in\mathbb{R}^{k}$ to be chosen later, we define
the entire function

$\displaystyle
g_{k}(z)=\left(\sum_{j=0}^{k-1}c_{j}z^{j}\right)(z;q)_{\infty}=\sum_{n=0}^{\infty}b_{n}z^{n},$

where

$\displaystyle
b_{n}=\sum_{j=0}^{n\wedge(k-1)}c_{j}(-1)^{n-j}\frac{q^{\binom{n-j}{2}}}{(q;q)_{n-j}},\qquad
n\geq 0.$

Defining $v_{n}=q^{n^{2}/2}b_{n}$, $n\geq 0$, we get for $m\geq 0$

$\displaystyle\sum_{n=0}^{\infty}s_{m+n}v_{n}=q^{-m^{2}/2}\sum_{n=0}^{\infty}b_{n}q^{-mn}=q^{-m^{2}/2}g_{k}\big{(}q^{-m}\big{)}=0.$

To see the absolute convergence, we write

$\displaystyle\sum_{n=k-1}^{\infty}s_{m+n}|v_{n}|\leq
q^{-m^{2}/2}\sum_{n=k-1}^{\infty}q^{-mn}\sum_{j=0}^{k-1}|c_{j}|\frac{q^{\binom{n-j}{2}}}{(q;q)_{n-j}}\leq
q^{-m^{2}/2}\sum_{j=0}^{k-1}|c_{j}|\sum_{n=j}^{\infty}q^{-mn}\frac{q^{\binom{n-j}{2}}}{(q;q)_{n-j}}$
$\displaystyle\hphantom{\sum_{n=k-1}^{\infty}s_{m+n}|v_{n}|}{}=q^{-m^{2}/2}\sum_{j=0}^{k-1}|c_{j}|q^{-mj}\sum_{n=0}^{\infty}q^{-mn}\frac{q^{\binom{n}{2}}}{(q;q)_{n}}<\infty.$

For $i=0,1,\ldots,k-1$ we shall solve the linear equations

$\displaystyle
v_{i}=q^{i^{2}/2}\sum_{j=0}^{i}c_{j}(-1)^{i-j}\frac{q^{\binom{i-j}{2}}}{(q;q)_{i-j}}=a_{i},$

which have a unique solution $(c_{0},\ldots,c_{k-1})$. ∎

By (A.6) we see that the matrix $\mathcal{A}=\\{a_{j,k}\\}$ is given for
$j\geq k$ as

$\displaystyle
a_{j,k}=(-1)^{j+k}q^{j^{2}+k^{2}-(j+k)/2}\sum_{n=j}^{\infty}\frac{\binom{n}{j}_{q}\binom{n}{k}_{q}}{(q;q)_{n}}q^{n}$
$\displaystyle\hphantom{a_{j,k}}{}=(-1)^{j+k}\frac{q^{j^{2}+k^{2}-(j+k)/2}}{(q;q)_{j}(q;q)_{k}}\sum_{n=j}^{\infty}\frac{(q;q)_{n}}{(q;q)_{n-j}(q;q)_{n-k}}q^{n}$
$\displaystyle\hphantom{a_{j,k}}{}=(-1)^{j+k}\frac{q^{j^{2}+k^{2}+(j-k)/2}}{(q;q)_{j-k}(q;q)_{k}}\sum_{p=0}^{\infty}\frac{(q^{j+1};q)_{p}}{(q;q)_{p}(q^{j-k+1};q)_{p}}q^{p},$

hence

$\displaystyle|a_{j,k}|\leq\frac{q^{j^{2}+k^{2}}}{(q;q)_{j}(q;q)_{k}(q;q)_{\infty}}\sum_{p=0}^{\infty}\frac{q^{p}}{(q;q)_{p}}.$

The last sum equals $(q;q)_{\infty}^{-1}$, and by symmetry we get

$\displaystyle|a_{j,k}|\leq\frac{q^{j^{2}+k^{2}}}{(q;q)_{j}(q;q)_{k}(q;q)_{\infty}^{2}},\qquad
j,k\geq 0.$

This shows that $\mathcal{A}\mathcal{H}$ is absolutely convergent, and it is
easy to verify that property (cs*) holds.

For $v\in\mathcal{V}\setminus\\{0\\}$ we have

$\displaystyle v=(\mathcal{A}\mathcal{H})v\neq\mathcal{A}(\mathcal{H}v)=0.$

######  Remark A.4.

There are symmetric matrices $\widetilde{\mathcal{A}}$ different from
$\mathcal{A}$ which satisfy $\widetilde{\mathcal{A}}\mathcal{H}=\mathcal{I}$.

We simply construct a non-zero symmetric matrix $\mathcal{M}$ such that
$\mathcal{H}\mathcal{M}=0$ and such that the product is absolutely convergent,
and we then define $\widetilde{\mathcal{A}}=\mathcal{A}+\mathcal{M}$.

As the first column of $\mathcal{M}$ we can take $v$ defined by (A.7). As the
second column we can choose $v^{(1)}\in\mathcal{V}$ with the initial condition
$v^{(1)}_{0}=v_{1}$. As the third column we choose $v^{(2)}\in\mathcal{V}$,
now with two initial conditions in order to make $\mathcal{M}$ symmetric and
so on.

###  A.3 Behaviour of the functions $\boldsymbol{u}$, $\boldsymbol{v}$ at
infinity

######  Proposition A.5.

The function $u$ given in (1.10) satisfies

$\displaystyle\lim_{\alpha\to\infty}\alpha u(\alpha)=\frac{\pi^{2}}{24},$
(A.8)

and the function $v$ given by (1.11) satisfies

$\displaystyle\lim_{\alpha\to\infty}\alpha v(\alpha)=\frac{\pi^{2}}{6}.$ (A.9)

###### Proof.

From (1.10) we get

$\displaystyle u(\alpha)=\max_{2^{-1}\leq x\leq
1}\big{(}x^{1/\alpha}+1\big{)}\int_{x^{1/\alpha}}^{1}\log\left(\frac{y^{\alpha}}{1-y^{\alpha}}\right)\frac{{\rm
d}y}{(1+y^{2})}.$

After the substitution $t=y^{\alpha}$ in the integral we find for fixed
$\alpha>0$

$\displaystyle\alpha u(\alpha)=\max_{2^{-1}\leq x\leq
1}\big{(}x^{1/\alpha}+1\big{)}\int_{x}^{1}\log\left(\frac{t}{1-t}\right)\frac{t^{1/\alpha}}{(1+t^{1/\alpha})^{2}}\frac{{\rm
d}t}{t}$ $\displaystyle\hphantom{\alpha
u(\alpha)}{}\leq\frac{1}{2}\int_{1/2}^{1}\log\left(\frac{t}{1-t}\right)\frac{{\rm
d}t}{t}=\frac{\pi^{2}}{24}.$

For the evaluation of the last integral we use that

$\displaystyle\frac{1}{2}\log^{2}(t)+\sum_{k=1}^{\infty}\frac{t^{k}}{k^{2}}$

is a primitive of the integrand, and we next use formulas (0.233.3) and
(0.241.2) from [18].

On the other hand for any $x\in\big{[}2^{-1},1\big{]}$ we have

$\displaystyle\alpha
u(\alpha)\geq\big{(}x^{1/\alpha}+1\big{)}\int_{x}^{1}\log\left(\frac{t}{1-t}\right)\frac{t^{1/\alpha}}{(1+t^{1/\alpha})^{2}}\frac{{\rm
d}t}{t}$ $\displaystyle\hphantom{\alpha
u(\alpha)}{}\geq\big{(}2^{-1/\alpha}+1\big{)}\int_{x}^{1}\log\left(\frac{t}{1-t}\right)\frac{t^{1/\alpha}}{(1+t^{1/\alpha})^{2}}\frac{{\rm
d}t}{t},$

and hence

$\displaystyle\alpha
u(\alpha)\geq\big{(}2^{-1/\alpha}+1\big{)}\int_{1/2}^{1}\log\left(\frac{t}{1-t}\right)\frac{t^{1/\alpha}}{(1+t^{1/\alpha})^{2}}\frac{{\rm
d}t}{t}.$

This gives

$\displaystyle\liminf_{\alpha\to\infty}\alpha
u(\alpha)\geq\frac{1}{2}\int_{1/2}^{1}\log\left(\frac{t}{1-t}\right)\frac{{\rm
d}t}{t},$

so (A.8) follows.

The substitution $t=x^{\alpha}$ in the expression (1.11) for $v$ yields

$\displaystyle\alpha
v(\alpha)=\int_{1}^{\infty}\log\big{(}\big{[}1-t^{-1}\big{]}^{-1}\big{)}t^{1/\alpha}\frac{{\rm
d}t}{t},$

and hence

$\displaystyle\lim_{\alpha\to\infty}\alpha
v(\alpha)=\int_{1}^{\infty}\log\big{(}\big{[}1-t^{-1}\big{]}^{-1}\big{)}\frac{{\rm
d}t}{t}=-\int_{0}^{1}\log(1-t)\frac{{\rm d}t}{t}=\frac{\pi^{2}}{6},$

which shows (A.9). ∎

### Acknowledgements

We are grateful to all the referees. Their remarks improved the exposition
substantially.

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