On the structure of Hausdorff moment sequences of complex matrices

TL;DR

This paper investigates the intrinsic structure of truncated matricial Hausdorff moment sequences, revealing their bounded variation within specific intervals and identifying extremal solutions using algebraic methods and matrix parallel sums.

Contribution

It characterizes the structure of matricial Hausdorff moment sequences and introduces extremal solutions based on endpoint conditions, employing algebraic techniques and matrix parallel sums.

Findings

Each moment sequence varies within a closed bounded interval.

Endpoints of the interval correspond to distinguished solutions.

Parallel sum of matrices is key in the proofs.

Abstract

The paper treats several aspects of the truncated matricial $[\alpha,\beta]$-Hausdorff type moment problems. It is shown that each $[\alpha,\beta]$-Hausdorff moment sequence has a particular intrinsic structure. More precisely, each element of this sequence varies within a closed bounded matricial interval. The case that the corresponding moment coincides with one of the endpoints of the interval plays a particular important role. This leads to distinguished molecular solutions of the truncated matricial $[\alpha,\beta]$-Hausdorff moment problem, which satisfy some extremality properties. The proofs are mainly of algebraic character. The use of the parallel sum of matrices is an essential tool in the proofs.

Equations

$${\int }_{\Omega }\Phi \mathrm{d}\mu {\Psi }^{\ast }:={\int }_{\Omega }\Phi {\mu }_{\tau }^{\prime }{\Psi }^{\ast }\mathrm{d}\tau$$

$${\int }_{\Omega }f\mathrm{d}\mu :={\int }_{\Omega }(f{I}_q)\mathrm{d}\mu I_q^{\ast }$$

$$s_j^{[\sigma ]}:={\int }_{\Omega }{x}^j\sigma (\mathrm{d}x)$$

$${\mathcal{M}}_{\ge }^q([\alpha ,\beta ])={\mathcal{M}}_{\ge ,\infty }^q([\alpha ,\beta ]).$$

$${\sigma }^{\vee }(B):=\sigma \left(R_{\Omega }^{-1}(B)\right).$$

$$[-\alpha s_{j+k}+s_{j+k+1}{]}_{j,k=0}^{n-1}$$

$$[s_{j+k}{]}_{j,k=0}^n$$

$$H_n=\left[\begin{array}{cc}{H}_{n-1}& y_{n,2n-1}\\ z_{n,2n-1}& s_{2n}\end{array}\right].$$

$$J_{q,n}:=\mathrm{diag}{\left((-1{)}^j{I}_q\right)}_{j=0}^n.$$

$$(\text{fourIdx}\alpha s{)}_j$$

$$(\text{fourIdx}\alpha \beta s{)}_j$$

$$\mathcal{P}\left[[\alpha ,\beta ];(s_j{)}_{j=0}^m\right]:=\left\{{\int }_{[\alpha ,\beta ]}{x}^{m+1}\sigma (\mathrm{d}x):\sigma \in {\mathcal{M}}_{\ge }^q\left[[\alpha ,\beta ];(s_j{)}_{j=0}^m,=\right]\right\}.$$

$$(\beta -\alpha )s_j$$

$$(\text{fourIdx}\alpha \beta s{)}_j$$

$$(\text{fourIdx}\alpha \beta H{)}_n$$

$$-(-\beta )H_n^{⟨r⟩}+K_n^{⟨r⟩}$$

$$\begin{gathered} -(-\beta )(-\alpha )H_{n-1}^{⟨r⟩}+\left[(-\beta )+(-\alpha )\right]K_{n-1}^{⟨r⟩}-G_{n-1}^{⟨r⟩} \\ =J_{p,n-1}^{\ast }(\text{fourIdx}\alpha \beta H{)}_{n-1}{J}_{q,n-1}. \end{gathered}$$

$$\begin{gathered} -(-\beta )(-\alpha )H_{n-1}^{⟨r⟩}+\left[(-\beta )+(-\alpha )\right]K_{n-1}^{⟨r⟩}-G_{n-1}^{⟨r⟩} \\ =-\beta \alpha J_{p,n-1}^{\ast }{H}_{n-1}{J}_{q,n-1}+(\beta +\alpha )J_{p,n-1}^{\ast }{K}_{n-1}{J}_{q,n-1}-J_{p,n-1}^{\ast }{G}_{n-1}{J}_{q,n-1} \\ =J_{p,n-1}^{\ast }(\text{fourIdx}\alpha \beta H{)}_{n-1}{J}_{q,n-1}.\text{qed} \end{gathered}$$

$${\mathcal{H}}_{q,2n}^{\ge ,\mathrm{e}}=\left\{(s_j{)}_{j=0}^{2n}\in {\mathcal{H}}_{q,2n}^{\ge }:\mathcal{R}(y_{n+1,2n})\subseteq \mathcal{R}(H_{n-1})\right\}.$$

$${\mathcal{H}}_{q,2n}^{\ge ,\mathrm{e}}=\left\{(s_j{)}_{j=0}^{2n}\in {\mathcal{H}}_{q,2n}^{\ge }:\mathcal{N}(L_{n-1})\subseteq \mathcal{N}(L_n)\right\}.$$

$$\left\{s_{2n+2}\in {\mathbb{C}}^{q\times q}:(s_j{)}_{j=0}^{2n+2}\in {\mathcal{H}}_{q,2n+2}^{\ge }\right\}=\left\{X\in {\mathbb{C}}_{\mathrm{H}}^{q\times q}:X\ge {\Theta }_{n+1}\right\}.$$

$$\left\{s_{2n+2}\in {\mathbb{C}}^{q\times q}:(s_j{)}_{j=0}^{2n+2}\in {\mathcal{H}}_{q,2n+2}^{>}\right\}=\left\{X\in {\mathbb{C}}_{\mathrm{H}}^{q\times q}:X>{\Theta }_{n+1}\right\}.$$

$$A\overline{\pm }B:=A(A+B{)}^{†}B.$$

$$\mathcal{R}(y_{n+1,2n+1})$$

$$(\text{fourIdx}\alpha \beta H{)}_{n-1}=(\beta -\alpha ){\Delta }_{p,n}^{\ast }\left[(\text{fourIdx}\alpha H{)}_n\overline{\pm }(\text{fourIdx}\beta H{)}_n\right]{\Delta }_{q,n}.$$

$$(\text{fourIdx}\alpha H{)}_n$$

$$H_n$$

$$(\text{fourIdx}\alpha H{)}_n$$

$$=-\alpha \beta H_n{H}_n^{†}{H}_n+\alpha H_n{H}_n^{†}{K}_n+\beta K_n{H}_n^{†}{H}_n-K_n{H}_n^{†}{K}_n$$

$$=-\alpha \beta H_n+(\alpha +\beta )K_n-K_n{H}_n^{†}{K}_n.$$

$${\Delta }_{p,n}^{\ast }{K}_n{H}_n^{†}{K}_n{\Delta }_{q,n}={\nabla }_{p,n}^{\ast }{H}_n{H}_n^{†}{H}_n{\nabla }_{q,n}={\nabla }_{p,n}^{\ast }{H}_n{\nabla }_{q,n}=G_{n-1}.$$

Full text

Mathematics Subject Classification (2010):

44A60, 47A57.

Keywords:

truncated matricial Hausdorff moment problem, canonical molecular solutions, matricial intervals associated with matricial Hausdorff moment sequences, parallel sum of matrices.

1 Introduction

The starting point of this paper was a question connected to matricial versions of the truncated power moment problem on a compact interval $[\alpha,\beta]$ of the real axis. In joint work with A. E. Choque Rivero and Yu. M. Dyukarev (see [8, 9]), the first and second authors could extend the characterizations of solvability of this moment problem, which were given in the scalar case by M. G. Krein [26] (see also Krein/Nudelman [27, Ch. III]) to the matrix case. In the case that $q\in\mathbb{N}$, $n\in\mathbb{N}$, and $(s_{j})_{j=0}^{2n-1}$ is a sequence of complex ${q\times q}$ matrices for which the moment problem in question is solvable, in their joint paper [13] with Yu. M. Dyukarev the authors constructed a concrete molecular solution (,i. e., a discrete non-negative Hermitian ${q\times q}$ measure concentrated on finitely many points of the interval $[\alpha,\beta]$) of the moment problem. The motivation for this paper was to find an explicit molecular solution for the case of a given sequence $(s_{j})_{j=0}^{2n}$ of prescribed moments. A closer look at our method used in [13] shows that the realization of our aim can be reached by a thorough study of the structure of finite $[\alpha,\beta]$-Hausdorff non-negative definite sequences of complex ${q\times q}$ matrices (see Definition 4.2). The key information comes from Theorem 4.10, which says that if $m\in\mathbb{N}$ and if $(s_{j})_{j=0}^{m}$ is an $[\alpha,\beta]$-Hausdorff non-negative definite sequence, then we can always find a complex ${q\times q}$ matrix $s_{m+1}$ such that the sequence $(s_{j})_{j=0}^{m+1}$ is $[\alpha,\beta]$-Hausdorff non-negative definite. We are even able to describe all complex ${q\times q}$ matrices $s_{m+1}$ which can be chosen to realize this aim. More precisely, the set of all these matrices $s_{m+1}$ turns out to be a closed matricial interval of complex ${q\times q}$ matrices. As well the left and right endpoint as the midpoint of this interval play (as clearly expected) a distinguished role (see Section 10). The main part of this paper is concerned with the investigation of the structure of $[\alpha,\beta]$-Hausdorff non-negative definite sequences of complex ${q\times q}$ matrices and the study of the above mentioned extension problem for such sequences. These results lead us to interesting new insights concerning a whole family of particular molecular solutions of the matrix version of the truncated $[\alpha,\beta]$-Hausdorff moment problem. In particular, we guess that the choice of the endpoints of the interval exactly leads to those extremal molecular solutions which were studied by M. G. Krein [26] (see also Krein/Nudelman [27, Ch. III, §5]). M. G. Krein found them via the lower and upper principal representation of the given moment sequence (see Section 12).

A more careful view shows that the situation is in some sense similar as in the case of non-negative definite sequences from $\mathbb{C}^{q\times q}$ (see [16]) or ${p\times q}$ Schur sequences (see [17]). If $n\in\mathbb{N}$ and if $(C_{j})_{j=0}^{n}$ is a sequence of one of the just mentioned types, then, for each $m\in\{1,2,\dotsc,n\}$, the matrix $C_{m}$ belongs to a matrix ball the parameters of which depend on the sequence $(C_{j})_{j=0}^{m-1}$. Having in mind the geometry of a matrix ball, we see that there are two types of distinguished points, namely the center of the matrix ball at the one hand and its boundary points on the other hand. The ${q\times q}$ non-negative definite sequences or ${p\times q}$ Schur sequences which are starting from some index consist only of the centers of the matrix balls in question occupy an extremal position in the set of all sequences of the considered types. Similar things can be said about those sequences which contain an element of the boundary of the relevant matrix ball. A similar situation will be met for $[\alpha,\beta]$-Hausdorff non-negative definite sequences. This will be discussed in detail in Section 10.

The study of moment spaces was initiated by C. Carathéodory [4, 5] in the context of the trigonometric moment problem. The approach of Carathéodory was based on the theory of convexity. O. Toeplitz [34] observed that the results of Carathéodory can be reformulated in terms of non-negative Hermitian Toeplitz matrices. This view was then generally accepted and is also the basis for the approach in the matrix case in [16, 17].

In the study of the moment space connected with the $[\alpha,\beta]$-Hausdorff moment problem, the theory of convexity played an important role from the very beginning. These developments were strongly influenced by M. G. Krein’s landmark paper [26], which essentially determined the further direction of research reflected in the monographs Karlin/Shapley [23], Karlin/Studden [24], and Krein/Nudelman [27]. It should be mentioned that Skibinsky [31, 32] considered probability measures on $[0,1]$ and observed that the $(n+1)$-th moment of such measures can vary within a closed bounded interval of the real axis. The work of Skibinsky was also an important source of inspiration for the theory of canonical moments, which was worked out by Dette/Studden [10]

2 On the matrix version of the truncated $[\alpha,\beta]$-Hausdorff moment problem

In order to formulate the moment problem, we are going to study, we first give some notation. Let $\mathbb{C}$, $\mathbb{R}$, $\mathbb{N}_{0}$, and $\mathbb{N}$ be the set of all complex numbers, the set of all real numbers, the set of all non-negative integers, and the set of all positive integers, resp. For every choice of $\kappa,\tau\in\mathbb{R}\cup\{-\infty,\infty\}$, let $\mathbb{Z}_{\kappa,\tau}$ be the set of all integers $\ell$ for which $\kappa\leq\ell\leq\tau$ holds. Throughout this paper, let $p$ and $q$ be positive integers. If $\mathcal{X}$ is a non-empty set, then $\mathcal{X}^{p\times q}$ stands for the set of all ${p\times q}$ matrices each entry of which belongs to $\mathcal{X}$, and $\mathcal{X}^{p}$ is short for $\mathcal{X}^{p\times 1}$. We will write $\mathbb{C}_{\mathrm{H}}^{q\times q}$, $\mathbb{C}_{\geq}^{q\times q}$, and $\mathbb{C}_{>}^{q\times q}$ for the set of all Hermitian complex ${q\times q}$ matrices, the set of all non-negative Hermitian complex ${q\times q}$ matrices, and the set of all positive Hermitian complex ${q\times q}$ matrices, resp. If $A$ and $B$ are complex ${q\times q}$ matrices, then $A\geq B$ or $B\leq A$ (resp. $A>B$ or $B<A$) means that $A$ and $B$ are Hermitian and $A-B$ is non-negative Hermitian (resp. positive Hermitian).

Let $(\Omega,\mathfrak{A})$ be a measurable space. Each countably additive mapping whose domain is ${\mathfrak{A}}$ and whose values belong to $\mathbb{C}_{\geq}^{q\times q}$ is called a non-negative Hermitian ${q\times q}$ measure on $(\Omega,{\mathfrak{A}})$. For the integration theory with respect to non-negative Hermitian measures, we refer to Kats [25] and Rosenberg [30]. If $\mu=[\mu_{jk}]_{j,k=1}^{q}$ is a non-negative Hermitian measure on $(\Omega,\mathfrak{A})$, then each entry function $\mu_{jk}$ is a complex measure on $(\Omega,\mathfrak{A})$. In particular, $\mu_{11},\mu_{22},\dotsc,\mu_{qq}$ are finite non-negative real-valued measures. For each $H\in\mathbb{C}_{\geq}^{q\times q}$, the inequality $H\leq(\operatorname{tr}H)I_{q}$ holds true. Hence, each entry function $\mu_{jk}$ is absolutely continuous with respect to the so-called trace measure $\tau\vcentcolon=\sum_{j=1}^{q}\mu_{jj}$ of $\mu$, i. e., for each $M\in\mathfrak{A}$ which satisfies $\tau(M)=0$, it follows $\mu(M)=0_{q\times q}$. The Radon-Nikodym derivatives $\mathrm{d}\mu_{jk}/\mathrm{d}\tau$ are thus well defined up to sets of zero $\tau$-measure. Obviously, the matrix-valued function $\mu_{\tau}^{\prime}\vcentcolon=[\mathrm{d}\mu_{jk}/\mathrm{d}\tau]_{j,k=1}^{q}$ is $\mathfrak{A}$-measurable and integrable with respect to $\tau$. The matrix-valued function $\mu_{\tau}^{\prime}$ is said to be the trace derivative of $\mu$. If $\nu$ is a non-negative real-valued measure on $\mathfrak{A}$, then let the class of all $\mathfrak{A}$-measurable ${p\times q}$ matrix-valued functions $\Phi=[\phi_{jk}]_{\begin{subarray}{c}j=1,\dotsc,p\\ k=1,\dotsc,q\end{subarray}}$ on $\Omega$ such that each $\phi_{jk}$ is integrable with respect to $\nu$ be denoted by $[\mathcal{L}^{1}(\Omega,\mathfrak{A},\nu;\mathbb{C})]^{p\times q}$. An ordered pair $(\Phi,\Psi)$ consisting of an $\mathfrak{A}$-measurable ${p\times q}$ matrix-valued function $\Phi$ on $\Omega$ and an $\mathfrak{A}$-measurable ${r\times q}$ matrix-valued function $\Psi$ on $\Omega$ is said to be integrable with respect to a non-negative Hermitian measure $\mu$ on $(\Omega,\mathfrak{A})$ if $\Phi\mu_{\tau}^{\prime}\Psi^{*}$ belongs to $[\mathcal{L}^{1}(\Omega,\mathfrak{A},\tau;\mathbb{C})]^{p\times r}$, where $\tau$ is the trace measure of $\mu$. In this case, the integral of $(\Phi,\Psi)$ with respect to $\mu$ is defined by

[TABLE]

and for any $M\in\mathfrak{A}$, the pair $(1_{M}\Phi,1_{M}\Psi)$ is integrable with respect to $\mu$, where $1_{M}$ is the indicator function of the set $M$. Then the integral of $(\Phi,\Psi)$ with respect to $\mu$ over $M$ is defined by $\int_{M}\Phi\mathrm{d}\mu\Psi^{*}\vcentcolon=\int_{\Omega}(1_{M}\Phi)\mu_{\tau}^{\prime}(1_{M}\Psi)^{*}\mathrm{d}\tau$. An $\mathfrak{A}$-measurable complex-valued function $f$ on $\Omega$ is said to be integrable with respect to a non-negative Hermitian measure $\mu$ on $(\Omega,\mathfrak{A})$ if the pair $(fI_{q},I_{q})$ is integrable with respect to $\mu$. In this case the integral of $f$ with respect to $\mu$ is defined by

[TABLE]

and for any $M\in\mathfrak{A}$, the function $1_{M}f$ is integrable with respect to $\mu$. Then the integral of $f$ with respect to $\mu$ over $M$ is defined by $\int_{M}f\mathrm{d}\mu\vcentcolon=\int_{M}(fI_{q})\mathrm{d}\mu I_{q}^{*}$. We denote by $\mathcal{L}^{1}(\Omega,\mathfrak{A},\mu;\mathbb{C})$ the set of all $\mathfrak{A}$-measurable complex-valued functions $f$ on $\Omega$ which are integrable with respect to a non-negative Hermitian measure $\mu$ on $(\Omega,\mathfrak{A})$.

Let ${\mathfrak{B}_{\mathbb{R}}}$ (resp. ${\mathfrak{B}}_{\mathbb{C}}$) be the $\sigma$-algebra of all Borel subsets of $\mathbb{R}$ (resp. $\mathbb{C}$). For all $\Omega\in{\mathfrak{B}_{\mathbb{R}}}\setminus\{\emptyset\}$, let ${\mathfrak{B}_{\Omega}}$ be the $\sigma$-algebra of all Borel subsets of $\Omega$ and let ${\mathcal{M}_{\geq}^{q}(\Omega)}$ be the set of all non-negative Hermitian ${q\times q}$ measures on $(\Omega,{\mathfrak{B}_{\Omega}})$. For all $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, let ${\mathcal{M}_{\geq,\kappa}^{q}(\Omega)}$ be the set of all $\sigma\in{\mathcal{M}_{\geq}^{q}(\Omega)}$ such that for each $j\in\mathbb{Z}_{0,\kappa}$ the function $f_{j}\colon\Omega\to\mathbb{C}$ defined by $f_{j}(x)\vcentcolon=x^{j}$ belongs to $\mathcal{L}^{1}(\Omega,{\mathfrak{B}_{\Omega}},\sigma;\mathbb{C})$. If $\sigma\in\mathcal{M}^{q}_{\geq,\kappa}(\Omega)$, then let

[TABLE]

for all $j\in\mathbb{Z}_{0,\kappa}$. Obviously, ${\mathcal{M}_{\geq}^{q}(\Omega)}\subseteq\mathcal{M}^{q}_{\geq,k}(\Omega)\subseteq\mathcal{M}^{q}_{\geq,k+1}(\Omega)\subseteq\mathcal{M}^{q}_{\geq,\infty}(\Omega)$ holds true for every choice of $\Omega\in{\mathfrak{B}}_{\mathbb{R}}\setminus\{\emptyset\}$ and $k\in\mathbb{N}_{0}$. Furthermore, if $\Omega$ is a non-empty bounded Borel subset of $\mathbb{R}$, then one can easily see that ${\mathcal{M}_{\geq}^{q}(\Omega)}=\mathcal{M}^{q}_{\geq,\infty}(\Omega)$. In particular, for all $\alpha\in\mathbb{R}$ and $\beta\in(\alpha,\infty)$ we have

[TABLE]

Let $\mathcal{M}^{q,\mathrm{mol}}_{\geq}(\Omega)$ be the set of all $\sigma\in{\mathcal{M}_{\geq}^{q}(\Omega)}$ for which there exists a finite subset $B$ of $\Omega$ satisfying $\sigma(\Omega\setminus B)=0_{q\times q}$. The elements of $\mathcal{M}^{q,\mathrm{mol}}_{\geq}(\Omega)$ are called molecular. Obviously, $\mathcal{M}^{q,\mathrm{mol}}_{\geq}(\Omega)$ is the set of all $\sigma\in{\mathcal{M}_{\geq}^{q}(\Omega)}$ for which there exist an $m\in\mathbb{N}$ and sequences $(\xi_{\ell})_{\ell=0}^{m}$ and $(M_{\ell})_{\ell=0}^{m}$ from $\Omega$ and $\mathbb{C}_{\geq}^{q\times q}$ such that $\sigma=\sum_{\ell=1}^{m}\delta_{\xi_{\ell}}M_{\ell}$, where $\delta_{\xi_{\ell}}$ is the Dirac measure on ${\mathfrak{B}_{\Omega}}$ with unit mass at $\xi_{\ell}$, $\ell\in\mathbb{Z}_{1,m}$. In particular, we have $\mathcal{M}^{q,\mathrm{mol}}_{\geq}(\Omega)\subseteq\mathcal{M}^{q}_{\geq,\infty}(\Omega)$.

The following matricial moment problem lies in the background of our considerations:

M[$\Omega$; $(s_{j})_{j=0}^{\kappa}$, $=$]

Let $\Omega\in{\mathfrak{B}_{\mathbb{R}}}\setminus\{\emptyset\}$ and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{q\times q}$. Parametrize the set ${\mathcal{M}_{\geq}^{q}[{\Omega};(s_{j})_{j=0}^{\kappa},=]}$ of all $\sigma\in{\mathcal{M}_{\geq,\kappa}^{q}(\Omega)}$ such that $s_{j}^{[\sigma]}=s_{j}$ for all $j\in\mathbb{Z}_{0,\kappa}$.

In this paper, we often use the procedure of reflecting measures on the real axis. For this reason, we introduce some terminology. Let $R\colon\mathbb{R}\to\mathbb{R}$ be defined by $x\mapsto-x$. Then $R$ is a continuous involution of $\mathbb{R}$ and consequently ${\mathfrak{B}_{\mathbb{R}}}$-${\mathfrak{B}_{\mathbb{R}}}$-measurable. In view of $\Omega\in{\mathfrak{B}_{\mathbb{R}}}\setminus\{\emptyset\}$, the set $\Omega^{\vee}\vcentcolon=R^{-1}(\Omega)$ belongs to ${\mathfrak{B}_{\mathbb{R}}}\setminus\{\emptyset\}$ and the mapping $R_{\Omega}\vcentcolon=\operatorname{Rstr}_{\Omega}R$ is ${\mathfrak{B}_{\Omega}}$-${\mathfrak{B}_{\Omega^{\vee}}}$-measurable, whereas the mapping $R_{\Omega^{\vee}}\vcentcolon=\operatorname{Rstr}_{\Omega^{\vee}}R$ is ${\mathfrak{B}_{\Omega^{\vee}}}$-${\mathfrak{B}_{\Omega}}$-measurable. Moreover $R_{\Omega}\circ R_{\Omega^{\vee}}=\operatorname{Id}_{\Omega}$ and $R_{\Omega^{\vee}}\circ R_{\Omega}=\operatorname{Id}_{\Omega^{\vee}}$. For each $\sigma\in{\mathcal{M}_{\geq}^{q}(\Omega)}$, we denote by $\sigma^{\vee}$ the image measure of $\sigma$ with respect to $R_{\Omega}$, i. e., for $B\in{\mathfrak{B}_{\Omega^{\vee}}}$, we have

[TABLE]

By construction then $\sigma^{\vee}\in{\mathcal{M}_{\geq}^{q}(\Omega^{\vee})}$. If $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$ and if $\sigma\in\mathcal{M}^{q}_{\geq,\kappa}(\Omega)$ then it is easily checked that $\sigma^{\vee}\in{\mathcal{M}_{\geq,\kappa}^{q}(\Omega^{\vee})}$ and that $s_{j}^{[\sigma^{\vee}]}=(-1)^{j}s_{j}^{[\sigma]}$ for all $j\in\mathbb{Z}_{0,\kappa}$.

Using the preceding considerations we get the following result:

Remark 2.1*.*

Let $\Omega\in{\mathfrak{B}_{\mathbb{R}}}\setminus\{\emptyset\}$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{q\times q}$. Let $\Omega^{\vee}\vcentcolon=\{x\in\mathbb{R}\colon-x\in\Omega\}$ and let $r_{j}\vcentcolon=(-1)^{j}s_{j}$ for all $j\in\mathbb{Z}_{0,\kappa}$. Then $\Omega^{\vee}\in{\mathfrak{B}_{\mathbb{R}}}\setminus\{\emptyset\}$ and, using the notation introduced in (2.3), we get ${\mathcal{M}_{\geq}^{q}[{\Omega^{\vee}};(r_{j})_{j=0}^{\kappa},=]}=\{\sigma^{\vee}\colon\sigma\in{\mathcal{M}_{\geq}^{q}[{\Omega};(s_{j})_{j=0}^{\kappa},=]}\}$.

The discussions of this paper are mostly concentrated on the case that the set $\Omega$ is a bounded and closed interval of the real axis. Such moment problems are called to be of Hausdorff-type. The following special case of Remark 2.1 is of particular interest for us.

Remark 2.2*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{q\times q}$. Let $r_{j}\vcentcolon=(-1)^{j}s_{j}$ for all $j\in\mathbb{Z}_{0,\kappa}$. Then ${\mathcal{M}_{\geq}^{q}[{[-\beta,-\alpha]};(r_{j})_{j=0}^{\kappa},=]}=\{\sigma^{\vee}\colon\sigma\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\kappa},=]}\}$.

The solvability of the truncated matricial $[\alpha,\beta]$-Hausdorff-type moment problem can be characterized as follows:

Theorem 2.3** ([8, Theorem 1.2]).**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}$, and let $(s_{j})_{j=0}^{2n-1}$ be a sequence from $\mathbb{C}^{q\times q}$. Then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{2n-1},=]}\neq\emptyset$ if and only if the block Hankel matrices

[TABLE]

are both non-negative Hermitian.

Theorem 2.4** ([9, Theorem 1.3]).**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}$, and let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{2n},=]}\neq\emptyset$ if and only if the block Hankel matrices

[TABLE]

are both non-negative Hermitian.

In the scalar case $q=1$, Theorems 2.3 and 2.4 are due to M. G. Krein [26] (see also [27, Ch. III]). The approach of M. G. Krein is essentially based on the use of the theory of convexity along the lines applied by C. Carathéodory [4, 5] in the context of the trigonometric moment problem. The proofs of Theorems 2.3 and 2.4 given in [8, 9], resp., are rather complicated and not constructive. They are based on Potapov’s method of fundamental matrix inequalities and an extensive explicit solving procedure of the corresponding system of these inequalities in the non-degenerate case. Indeed, if the block Hankel matrices in (2.4) and (2.5) are both positive Hermitian, then [8, Corollary 6.13] and [9, Corollary 6.15] show that Problem M[$[\alpha,\beta]$; $(s_{j})_{j=0}^{2n-1}$, $=$] and Problem M[$[\alpha,\beta]$; $(s_{j})_{j=0}^{2n}$, $=$] have solutions. In the proofs of [8, Theorem 1.2] and [9, Theorem 1.3] a perturbation argument for the construction of a sequence of corresponding approximating non-degenerate problems is used and then a matricial version of the Helly-Prokhorov Theorem (see [18, Satz 9] or [12, Lemma 2.2.1]) is applied.

3 On Hankel non-negative definite sequences

In this section, we summarize essential properties on two classes of sequences of complex ${q\times q}$ matrices which are a main tool for the investigations of this paper. This material is mainly taken from [14, 19].

Definition 3.1**.**

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Let $H_{n}\vcentcolon=[s_{j+k}]_{j,k=0}^{n}$. Then $(s_{j})_{j=0}^{2n}$ is called Hankel non-negative definite (resp. Hankel positive definite) if $H_{n}\in\mathbb{C}_{\geq}^{(n+1)q\times(n+1)q}$ (resp. $\mathbb{C}_{>}^{(n+1)q\times(n+1)q}$). We denote by $\mathcal{H}^{\geq}_{q,2n}$ (resp. $\mathcal{H}^{>}_{q,2n}$) the set of all Hankel non-negative definite (resp. Hankel positive definite) sequences $(s_{j})_{j=0}^{2n}$ from $\mathbb{C}^{q\times q}$.

Definition 3.2**.**

1. (a)

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n}$ is called Hankel non-negative definite extendable (resp. Hankel positive definite extendable) if there exist matrices $s_{2n+1}$ and $s_{2n+2}$ from $\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{2n+2}\in\mathcal{H}^{\geq}_{q,2n+2}$ (resp. $(s_{j})_{j=0}^{2n+2}\in\mathcal{H}^{>}_{q,2n+2}$). 2. (b)

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n+1}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n+1}$ is called Hankel non-negative definite extendable (resp. Hankel positive definite extendable) if there exists a matrix $s_{2n+2}$ from $\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{2n+2}\in\mathcal{H}^{\geq}_{q,2n+2}$ (resp. $(s_{j})_{j=0}^{2n+2}\in\mathcal{H}^{>}_{q,2n+2}$).

If $m\in\mathbb{N}_{0}$, then the notation $\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$ (resp. $\mathcal{H}^{>,\mathrm{e}}_{q,m}$) stands for the set of all sequences $(s_{j})_{j=0}^{m}$ from $\mathbb{C}^{q\times q}$ which are Hankel non-negative definite extendable (resp. Hankel positive definite extendable).

The importance of the class $\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$ in the context of moment problems is caused by the following observation:

Theorem 3.3** (see [14, Theorem 4.17]).**

Let $m\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then ${\mathcal{M}_{\geq}^{q}[{\mathbb{R}};(s_{j})_{j=0}^{m},=]}\neq\emptyset$ if and only if $(s_{j})_{j=0}^{m}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$.

Now we are going to indicate some essential features of the structure of Hankel non-negative definite sequences. First we introduce some matrices which occupy a key role in the sequel. For each matrix $A\in\mathbb{C}^{p\times q}$, we denote by $A^{\dagger}$ its Moore-Penrose inverse. This means that $A^{\dagger}$ is the unique matrix $X$ from $\mathbb{C}^{q\times p}$ which satisfies the four equations $AXA=A$, $XAX=X$, $(AX)^{*}=AX$, and $(XA)^{*}=XA$. For every choice of $n\in\mathbb{N}$ and $A_{1},A_{2},\dotsc,A_{n}\in\mathbb{C}^{p\times q}$, let $\operatorname{col}(A_{j})_{j=1}^{n}\vcentcolon=[A_{1}^{*},A_{2}^{*},\dotsc,A_{n}^{*}]^{*}$ and let $\operatorname{row}(A_{j})_{j=1}^{n}\vcentcolon=[A_{1},A_{2},\dotsc,A_{n}]$. The null matrix which belongs to $\mathbb{C}^{p\times q}$ is denoted by $0_{p\times q}$, whereas $I_{p}$ is the identity matrix belonging to $\mathbb{C}^{p\times p}$.

Notation 3.4**.**

Let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$ and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{p\times q}$.

1. (a)

Let $H_{n}\vcentcolon=[s_{j+k}]_{j,k=0}^{n}$ for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$, let $K_{n}\vcentcolon=[s_{j+k+1}]_{j,k=0}^{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$, and let $G_{n}\vcentcolon=[s_{j+k+2}]_{j,k=0}^{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+2\leq\kappa$. 2. (b)

Let $y_{\ell,m}\vcentcolon=\operatorname{col}(s_{j})_{j=\ell}^{m}$ and $z_{\ell,m}\vcentcolon=\operatorname{row}(s_{j})_{j=\ell}^{m}$ for all $\ell,m\in\mathbb{N}_{0}$ with $\ell\leq m\leq\kappa$. 3. (c)

Let $\Theta_{0}\vcentcolon=0_{p\times q}$ and, for all $n\in\mathbb{N}$ with $2n-1\leq\kappa$, let $\Theta_{n}\vcentcolon=z_{n,2n-1}H_{n-1}^{\dagger}y_{n,2n-1}$. Furthermore, for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$, let $L_{n}\vcentcolon=s_{2n}-\Theta_{n}$. 4. (d)

Let $M_{0}\vcentcolon=0_{p\times q}$, and, for all $n\in\mathbb{N}$ with $2n\leq\kappa$, let $M_{n}\vcentcolon=z_{n,2n-1}H_{n-1}^{\dagger}y_{n+1,2n}$.

If we build the matrices introduced in Notation 3.4 from an other given sequence, e. g. $(t_{j})_{j=0}^{\kappa}$, then this is indicated by a superscript $\langle t\rangle$, i. e. $H_{n}^{\langle t\rangle}\vcentcolon=[t_{j+k}]_{j,k=0}^{n}$, etc.

Remark 3.5*.*

Let $n\in\mathbb{N}$ and let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Then the block Hankel matrix $H_{n}$ admits the block partition

[TABLE]

Let $\mathcal{R}(A)\vcentcolon=\{Ax\colon x\in\mathbb{C}^{q}\}$ and $\mathcal{N}(A)\vcentcolon=\{x\in\mathbb{C}^{q}\colon Ax=0_{p\times 1}\}$ be the column space and the null space of a matrix $A\in\mathbb{C}^{p\times q}$.

Remark 3.6*.*

Let $n\in\mathbb{N}$ and let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Then:

1. (a)

$(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$ if and only if $(s_{j})_{j=0}^{2n-2}\in\mathcal{H}^{\geq}_{q,2n-2}$, $L_{n}\in\mathbb{C}_{\geq}^{q\times q}$, $\mathcal{R}(y_{n,2n-1})\subseteq H_{n-1}$, and $s_{2n+1}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$. 2. (b)

$(s_{j})_{j=0}^{2n}\in\mathcal{H}^{>}_{q,2n}$ if and only if $(s_{j})_{j=0}^{2n-2}\in\mathcal{H}^{>}_{q,2n-2}$, $L_{n}\in\mathbb{C}_{>}^{q\times q}$, and $s_{2n+1}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$.

Proof.

Taking into account (3.1), this is a consequence of Lemma A.12. ∎

A sequence $(s_{j})_{j=0}^{\infty}$ from $\mathbb{C}^{q\times q}$ is called Hankel non-negative definite (resp. Hankel positive definite) if $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$ (resp. $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{>}_{q,2n}$) for all $n\in\mathbb{N}_{0}$. We denote by $\mathcal{H}^{\geq}_{q,\infty}$ (resp. $\mathcal{H}^{>}_{q,\infty}$) the set of all Hankel non-negative definite (resp. Hankel positive definite) sequences $(s_{j})_{j=0}^{\infty}$ from $\mathbb{C}^{q\times q}$.

Remark 3.7*.*

Let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$. Then:

1. (a)

If $(s_{j})_{j=0}^{2\kappa}\in\mathcal{H}^{\geq}_{q,2\kappa}$, then $s_{j}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ for all $j\in\mathbb{Z}_{0,2\kappa}$ and $s_{2k}\in\mathbb{C}_{\geq}^{q\times q}$ for all $k\in\mathbb{Z}_{0,\kappa}$. 2. (b)

If $(s_{j})_{j=0}^{2\kappa}\in\mathcal{H}^{>}_{q,2\kappa}$, then $s_{j}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ for all $j\in\mathbb{Z}_{0,2\kappa}$ and $s_{2k}\in\mathbb{C}_{>}^{q\times q}$ for all $k\in\mathbb{Z}_{0,\kappa}$.

Lemma 3.8**.**

Let $n\in\mathbb{N}$ and let $(s_{j})_{j=0}^{2n-1}$ be a sequence from $\mathbb{C}_{\mathrm{H}}^{q\times q}$. Then:

1. (a)

If $(s_{j})_{j=0}^{2n-2}\in\mathcal{H}^{\geq}_{q,2n-2}$, then $\Theta_{n}\in\mathbb{C}_{\geq}^{q\times q}$. 2. (b)

If $(s_{j})_{j=0}^{2n-2}\in\mathcal{H}^{>}_{q,2n-2}$ and $n\geq 2$, then $\Theta_{n}\in\mathbb{C}_{>}^{q\times q}$.

Proof.

(a) This follows immediately from Remarks A.10 and A.2.

(b) Now suppose that $(s_{j})_{j=0}^{2n-2}\in\mathcal{H}^{>}_{q,2n-2}$ and $n\geq 2$. Then $H_{n-1}$ is positive Hermitian. In particular, $H_{n-1}^{-1}$ is positive Hermitian. From Remark 3.7(b) we obtain $s_{2n-2}\in\mathbb{C}_{>}^{q\times q}$. In particular, $\det s_{2n-2}\neq 0$. Because of $n\geq 2$, the matrix $s_{2n-2}$ is a block in $y_{n,2n-1}$. Thus, $\operatorname{rank}y_{n,2n-1}=q$. Hence, $\Theta_{n}\in\mathbb{C}_{>}^{q\times q}$ follows from Remark A.2. ∎

Remark 3.9*.*

Let $n\in\mathbb{N}_{0}$. Then $\mathcal{H}^{\geq,\mathrm{e}}_{q,2n}\subseteq\mathcal{H}^{\geq}_{q,2n}$ and, in the case $n\in\mathbb{N}$, furthermore $\mathcal{H}^{\geq,\mathrm{e}}_{q,2n}\neq\mathcal{H}^{\geq}_{q,2n}$.

For all $n\in\mathbb{N}_{0}$ let

[TABLE]

Remark 3.10*.*

Let $n\in\mathbb{N}_{0}$. Then $J_{q,n}^{*}=J_{q,n}$ and $J_{q,n}^{2}=I_{(n+1)q}$. In particular, $J_{q,n}$ is unitary.

Lemma 3.11**.**

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Let the sequence $(r_{j})_{j=0}^{2n}$ be given by $r_{j}\vcentcolon=(-1)^{j}s_{j}$. Then:

1. (a)

Let $H_{n}^{\langle s\rangle}\vcentcolon=[s_{j+k}]_{j,k=0}^{n}$ and $H_{n}^{\langle r\rangle}\vcentcolon=[r_{j+k}]_{j,k=0}^{n}$. Then $H_{n}^{\langle r\rangle}=J_{q,n}^{*}H_{n}^{\langle s\rangle}J_{q,n}$. 2. (b)

$(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$* if and only if $(r_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$.* 3. (c)

$(s_{j})_{j=0}^{2n}\in\mathcal{H}^{>}_{q,2n}$* if and only if $(r_{j})_{j=0}^{2n}\in\mathcal{H}^{>}_{q,2n}$.*

Proof.

Part (a) follows by direct computation. Parts (b) and (c) follow from (a). ∎

Lemma 3.12**.**

Let $m\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Let the sequence $(r_{j})_{j=0}^{m}$ be given by $r_{j}\vcentcolon=(-1)^{j}s_{j}$. Then $(s_{j})_{j=0}^{m}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$ if and only if $(r_{j})_{j=0}^{m}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$.

Proof.

Use Lemma 3.11(b). ∎

Now we turn our attention to a subclass of Hankel non-negative definite sequences, which plays a central role in the sequel. If $n\in\mathbb{N}_{0}$ and if $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$, then $(s_{j})_{j=0}^{2n}$ is called Hankel completely degenerate if $L_{n}=0_{q\times q}$, where $L_{n}$ is given in Notation 3.4(c). We denote by $\mathcal{H}^{\geq,\mathrm{cd}}_{q,2n}$ the set of all sequences $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$ which are Hankel completely degenerate. If $n\in\mathbb{N}_{0}$, then a sequence $(s_{j})_{j=0}^{\infty}\in\mathcal{H}^{\geq}_{q,\infty}$ is said to be Hankel completely degenerate of order $n$ if $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq,\mathrm{cd}}_{q,2n}$. A sequence $(s_{j})_{j=0}^{\infty}\in\mathcal{H}^{\geq}_{q,\infty}$ is called Hankel completely degenerate if there exists an $n\in\mathbb{N}_{0}$ such that $(s_{j})_{j=0}^{\infty}$ is Hankel completely degenerate of order $n$.

4 On $[\alpha,\beta]$-Hausdorff non-negative definite sequences from $\mathbb{C}^{q\times q}$

Against to the background of Theorems 2.3 and 2.4 now we are going to introduce one of the central notions of this paper. Before doing this, we introduce some notation.

Notation 4.1**.**

Let $\alpha,\beta\in\mathbb{C}$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{q\times q}$. Then let the sequences $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{\kappa-1}$ and $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{\kappa-1}$ be given by

[TABLE]

Furthermore, if $\kappa\geq 2$, then let $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{\kappa-2}$ be defined by

[TABLE]

When using the sequences from Notation 4.1, we write $(\fourIdx{}{\alpha}{}{}{H})_{n}$, $(\fourIdx{}{}{}{\beta}{H})_{n}$, $(\fourIdx{}{\alpha}{}{\beta}{H})_{n}$, etc.

Definition 4.2**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $n\in\mathbb{N}_{0}$.

1. (a)

Let $(s_{j})_{j=0}^{2n+1}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n+1}$ is called $[\alpha,\beta]$-Hausdorff non-negative definite (resp. $[\alpha,\beta]$-Hausdorff positive definite) if both sequences $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}$ and $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}$ are Hankel non-negative definite (resp. Hankel positive definite). 2. (b)

Let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n}$ is called $[\alpha,\beta]$-Hausdorff non-negative definite (resp. $[\alpha,\beta]$-Hausdorff positive definite) if $(s_{j})_{j=0}^{2n}$ and, in the case $n\geq 1$, moreover $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2(n-1)}$ is Hankel non-negative definite (resp. Hankel positive definite).

If $m\in\mathbb{N}_{0}$, then the symbol $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ (resp. $\mathcal{F}^{>}_{q,m,\alpha,\beta}$) stands for the set of all $[\alpha,\beta]$-Hausdorff non-negative definite (resp. $[\alpha,\beta]$-Hausdorff positive definite) sequences $(s_{j})_{j=0}^{m}$ from $\mathbb{C}^{q\times q}$.

In view of Definition 4.2, now Theorems 2.3 and 2.4 can be summarized and reformulated as follows:

Theorem 4.3**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\neq\emptyset$ if and only if $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$.

Definition 4.4**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{m}$ is called $[\alpha,\beta]$-Hausdorff non-negative definite extendable (resp. $[\alpha,\beta]$-Hausdorff positive definite extendable) if there exists a matrix $s_{m+1}\in\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{m+1}$ is $[\alpha,\beta]$-Hausdorff non-negative definite (resp. $[\alpha,\beta]$-Hausdorff positive definite). We denote by $\mathcal{F}^{\geq,\mathrm{e}}_{q,m,\alpha,\beta}$ (resp. $\mathcal{F}^{>,\mathrm{e}}_{q,m,\alpha,\beta}$) the set of all $[\alpha,\beta]$-Hausdorff non-negative definite extendable (resp. $[\alpha,\beta]$-Hausdorff positive definite extendable) sequences $(s_{j})_{j=0}^{m}$ from $\mathbb{C}^{q\times q}$.

Using Theorem 4.3, we derive now several algebraic results on the matrix sequences introduced in Definitions 4.2 and 4.4, resp.

Proposition 4.5**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then $(s_{j})_{j=0}^{\ell}\in\mathcal{F}^{\geq}_{q,\ell,\alpha,\beta}$ for all $\ell\in\mathbb{Z}_{0,m}$.

Proof.

In view of Theorem 4.3 we have ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\neq\emptyset$. Let $\ell\in\mathbb{Z}_{0,m}$. Then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\subseteq{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\ell},=]}$. Hence ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\ell},=]}\neq\emptyset$. Thus, again applying Theorem 4.3, we get $(s_{j})_{j=0}^{\ell}\in\mathcal{F}^{\geq}_{q,\ell,\alpha,\beta}$. ∎

Corollary 4.6**.**

Let $\alpha\in\mathbb{R}$, $\beta\in(\alpha,\infty)$, and $m\in\mathbb{N}_{0}$. Then $\mathcal{F}^{\geq,\mathrm{e}}_{q,m,\alpha,\beta}\subseteq\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$.

Proof.

Use Proposition 4.5. ∎

Proposition 4.5 leads us to the following notions:

Definition 4.7**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $(s_{j})_{j=0}^{\infty}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{\infty}$ is called $[\alpha,\beta]$-Hausdorff non-negative definite (resp. $[\alpha,\beta]$-Hausdorff positive definite) if for all $m\in\mathbb{N}_{0}$ the sequence $(s_{j})_{j=0}^{m}$ is $[\alpha,\beta]$-Hausdorff non-negative definite (resp. $[\alpha,\beta]$-Hausdorff positive definite). The notation $\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ (resp. $\mathcal{F}^{>}_{q,\infty,\alpha,\beta}$) stands for set of all $[\alpha,\beta]$-Hausdorff non-negative definite (resp. $[\alpha,\beta]$-Hausdorff positive definite) sequences $(s_{j})_{j=0}^{\infty}$ from $\mathbb{C}^{q\times q}$.

Lemma 4.8**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $\sigma\in{\mathcal{M}_{\geq}^{q}([\alpha,\beta])}$. Then $\sigma\in\mathcal{M}^{q}_{\geq,\infty}([\alpha,\beta])$ and the sequence $(s_{j}^{[\sigma]})_{j=0}^{\infty}$ given via (2.1) belongs to $\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$.

Proof.

By the choice of $\sigma$ we have $\sigma\in\bigcap_{m=0}^{\infty}\mathcal{M}^{q}_{\geq,m}([\alpha,\beta])$. Thus, Theorem 4.3 implies $(s_{j}^{[\sigma]})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ for each $m\in\mathbb{N}_{0}$. Hence $(s_{j}^{[\sigma]})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$. ∎

Proposition 4.9**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then there is a sequence $(s_{j})_{j=m+1}^{\infty}$ from $\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$.

Proof.

In view of Theorem 4.3 we have ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\neq\emptyset$. Let $\sigma\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}$. In view of Lemma 4.8, we have then $\sigma\in\mathcal{M}^{q}_{\geq,\infty}([\alpha,\beta])$, $(s_{j}^{[\sigma]})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$, and $(s_{j}^{[\sigma]})_{j=0}^{m}=(s_{j})_{j=0}^{m}$. ∎

Theorem 4.10**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $m\in\mathbb{N}_{0}$. Then $\mathcal{F}^{\geq,\mathrm{e}}_{q,m,\alpha,\beta}=\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$.

Proof.

Combine Corollary 4.6 and Propositions 4.9 and 4.5. ∎

Comparing Theorem 4.10 with Remark 3.9, we see that both statements are completely different. Against to the background of Theorems 4.3 and 3.3 we can now immediately see the reason for this phenomenon, namely, in view of (2.2), we have ${\mathcal{M}_{\geq}^{q}([\alpha,\beta])}=\mathcal{M}^{q}_{\geq,\infty}([\alpha,\beta])$, whereas on the other hand it can be easily checked that, for each $k\in\mathbb{N}$, the proper inclusion $\mathcal{M}^{q}_{\geq,k+1}(\mathbb{R})\subset\mathcal{M}^{q}_{\geq,k}(\mathbb{R})$ is satisfied.

In view of Lemma 4.8, we will consider the following problem:

P[$[\alpha,\beta]$; $(s_{j})_{j=0}^{m}$, $=$]

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Describe the set

[TABLE]

Proposition 4.11**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $\mathcal{P}[[\alpha,\beta];(s_{j})_{j=0}^{m}]\neq\emptyset$ if and only if $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. In this case, $\mathcal{P}[[\alpha,\beta];(s_{j})_{j=0}^{m}]=\{s_{m+1}\in\mathbb{C}^{q\times q}\colon(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}\}$.

Proof.

Let $\mathcal{P}\vcentcolon=\mathcal{P}[[\alpha,\beta];(s_{j})_{j=0}^{m}]$. First Assume $\mathcal{P}\neq\emptyset$. Let $s\in\mathcal{P}$. Then there is a $\sigma\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}$ such that $\int_{[\alpha,\beta]}x^{m+1}\sigma(\mathrm{d}x)=s$. In particular, ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\neq\emptyset$. Thus, $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ according to Theorem 4.3. Furthermore, setting $s_{m+1}\vcentcolon=s$, we have $\sigma\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m+1},=]}$. Again, Theorem 4.3 yields $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}$.

Now suppose that $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\neq\emptyset$ by virtue of Theorem 4.3. Let $\sigma\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}$. In view of Lemma 4.8, then $\int_{[\alpha,\beta]}x^{m+1}\sigma(\mathrm{d}x)\in\mathcal{P}$ follows. In particular, $\mathcal{P}\neq\emptyset$. Finally, we consider an arbitrary $s_{m+1}\in\mathbb{C}^{q\times q}$ with $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}$. Using Theorem 4.3, we get ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m+1},=]}\neq\emptyset$. Let $\sigma\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m+1},=]}$. Then $\sigma\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}$ and $s_{m+1}=\int_{[\alpha,\beta]}x^{m+1}\sigma(\mathrm{d}x)$. Consequently, $s_{m+1}\in\mathcal{P}$. ∎

Proposition 4.11 leads us to the problem of describing the set $\{s_{m+1}\in\mathbb{C}^{q\times q}\colon(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}\}$.

Remark 4.12*.*

Let $\alpha,\beta\in\mathbb{C}$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{p\times q}$. Then

[TABLE]

for all $j\in\mathbb{Z}_{0,\kappa-1}$. Furthermore, if $\kappa\geq 2$, then

[TABLE]

and $(\beta-\alpha)s_{j+2}=\beta^{2}(\fourIdx{}{\alpha}{}{}{s})_{j}+\alpha^{2}(\fourIdx{}{}{}{\beta}{s})_{j}-(\beta-\alpha)(\fourIdx{}{\alpha}{}{\beta}{s})_{j}$ for all $j\in\mathbb{Z}_{0,\kappa-2}$.

Remark 4.13*.*

Let $\alpha,\beta\in\mathbb{C}$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{p\times q}$. Then $(\fourIdx{}{\alpha}{}{}{H})_{n}=-\alpha H_{n}+K_{n}$ and $(\fourIdx{}{}{}{\beta}{H})_{n}=\beta H_{n}-K_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$. Furthermore, if $\kappa\geq 2$, then $(\fourIdx{}{\alpha}{}{\beta}{H})_{n}=-\alpha\beta H_{n}+(\alpha+\beta)K_{n}-G_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+2\leq\kappa$.

Using Remark 4.13, we obtain:

Remark 4.14*.*

Let $\alpha,\beta\in\mathbb{C}$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{p\times q}$. Then $(\beta-\alpha)H_{n}=(\fourIdx{}{\alpha}{}{}{H})_{n}+(\fourIdx{}{}{}{\beta}{H})_{n}$ and $(\beta-\alpha)K_{n}=\beta(\fourIdx{}{\alpha}{}{}{H})_{n}+\alpha(\fourIdx{}{}{}{\beta}{H})_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$. Furthermore, if $\kappa\geq 2$, then

[TABLE]

and $(\beta-\alpha)G_{n}=\beta^{2}(\fourIdx{}{\alpha}{}{}{H})_{n}+\alpha^{2}(\fourIdx{}{}{}{\beta}{H})_{n}-(\beta-\alpha)(\fourIdx{}{\alpha}{}{\beta}{H})_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+2\leq\kappa$.

Lemma 4.15**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{2n+1}$ be a sequence from $\mathbb{C}^{q\times q}$. If $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{>}_{q,2n+1,\alpha,\beta}$), then $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$ (resp. $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{>}_{q,2n}$).

Proof.

By virtue of Remark 4.14, we have $H_{n}=(\beta-\alpha)^{-1}[(\fourIdx{}{\alpha}{}{}{H})_{n}+(\fourIdx{}{}{}{\beta}{H})_{n}]$. In view of Definition 4.2(a) and Definition 3.1, the assertions follow. ∎

Lemma 4.16**.**

Let $\alpha\in\mathbb{R}$ and $\beta\in(\alpha,\infty)$. Then $\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}\subseteq\mathcal{H}^{\geq}_{q,\infty}$ and $\mathcal{F}^{>}_{q,\infty,\alpha,\beta}\subseteq\mathcal{H}^{>}_{q,\infty}$.

Proof.

This follows in view of Definition 4.7 from Lemma 4.15. ∎

Lemma 4.17**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then $s_{j}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ for all $j\in\mathbb{Z}_{0,m}$ and $s_{2k}\in\mathbb{C}_{\geq}^{q\times q}$ for all $k\in\mathbb{N}_{0}$ with $2k\leq m$.

Proof.

In the case $m=2n$ with some $n\in\mathbb{N}_{0}$, the assertion follows from Definition 4.2 and Remark 3.7(a).

Now suppose that $m=2n+1$ for some $n\in\mathbb{N}_{0}$. By virtue of Lemma 4.15, then $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$. According to Remark 3.7(a), we obtain $s_{j}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ for all $j\in\mathbb{Z}_{0,2n}$ and $s_{2k}\in\mathbb{C}_{\geq}^{q\times q}$ for all $k\in\mathbb{Z}_{0,n}$. In view of Definition 4.2(a), furthermore $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$. Because of Remark 3.7(a), in particular $(\fourIdx{}{\alpha}{}{}{s})_{2n}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$. Since the matrix $s_{2n+1}$ admits the representation $s_{2n+1}=(\fourIdx{}{\alpha}{}{}{s})_{2n}+\alpha s_{2n}$, we see that $s_{2n+1}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ holds true as well. ∎

Lemma 4.18**.**

Let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{p\times q}$, and let the sequence $(r_{j})_{j=0}^{\kappa}$ be given by $r_{j}\vcentcolon=(-1)^{j}s_{j}$. Then $H_{n}^{\langle r\rangle}=J_{p,n}H_{n}J_{q,n}$ for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$ and $K_{n}^{\langle r\rangle}=-J_{p,n}K_{n}J_{q,n}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$ and $G_{n}^{\langle r\rangle}=J_{p,n}G_{n}J_{q,n}$ for all $n\in\mathbb{N}_{0}$ with $2n+2\leq\kappa$.

Proof.

Apply Lemma 3.11(a). ∎

Lemma 4.19**.**

Let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{p\times q}$, and let the sequence $(r_{j})_{j=0}^{\kappa}$ be given by $r_{j}\vcentcolon=(-1)^{j}s_{j}$. Then:

1. (a)

$\Theta_{n}^{\langle r\rangle}=\Theta_{n}$* for all $n\in\mathbb{N}_{0}$ with $2n-1\leq\kappa$.* 2. (b)

$M_{n}^{\langle r\rangle}=-M_{n}$* and $L_{n}^{\langle r\rangle}=L_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$.*

Proof.

In view of Notation 3.4, we have $\Theta_{0}^{\langle r\rangle}=0_{p\times q}=\Theta_{0}$ and $M_{0}^{\langle r\rangle}=0_{p\times q}=-M_{0}$.

Suppose that $n\in\mathbb{N}$ with $2n-1\leq\kappa$. We have $z_{n,2n-1}^{\langle r\rangle}=(-1)^{n}z_{n,2n-1}J_{q,n-1}$ and $y_{n,2n-1}^{\langle r\rangle}=(-1)^{n}J_{p,n-1}y_{n,2n-1}$. Lemma 4.18 and Remarks 3.10 and A.9 yield $(H_{n-1}^{\langle r\rangle})^{\dagger}=(J_{p,n-1}H_{n-1}J_{q,n-1})^{\dagger}=J_{q,n-1}(H_{n-1})^{\dagger}J_{p,n-1}$. In view of Remark 3.10, we obtain $\Theta_{n}^{\langle r\rangle}=(-1)^{2n}z_{n,2n-1}H_{n-1}^{\dagger}y_{n,2n-1}=\Theta_{n}$.

Now suppose that $n\in\mathbb{N}$ with $2n\leq\kappa$. We have then furthermore $y_{n+1,2n}^{\langle r\rangle}=(-1)^{n+1}J_{p,n-1}y_{n+1,2n}$. Using Remark 3.10, we obtain $M_{n}^{\langle r\rangle}=(-1)^{2n+1}z_{n,2n-1}H_{n-1}^{\dagger}y_{n+1,2n}=-M_{n}$. The rest follows from (a). ∎

Lemma 4.20**.**

Let $\alpha,\beta\in\mathbb{C}$, let $m\in\mathbb{N}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{p\times q}$. Let the sequence $(r_{j})_{j=0}^{m}$ be given by $r_{j}\vcentcolon=(-1)^{j}s_{j}$.

1. (a)

Let $m=2n+1$ with some $n\in\mathbb{N}_{0}$. Then

[TABLE] 2. (b)

Let $m=2n$ with some $n\in\mathbb{N}$. Then

[TABLE]

Proof.

(a) According to Lemma 4.18 and Remark 3.10, we have $H_{n}^{\langle r\rangle}=J_{p,n}^{*}H_{n}J_{q,n}$ and $K_{n}^{\langle r\rangle}=-J_{p,n}^{*}K_{n}J_{q,n}$. Hence, $-(-\beta)H_{n}^{\langle r\rangle}+K_{n}^{\langle r\rangle}=\beta J_{p,n}^{*}H_{n}J_{q,n}-J_{p,n}^{*}K_{n}J_{q,n}=J_{p,n}^{*}(\fourIdx{}{}{}{\beta}{H})_{n}J_{q,n}$ and, analogously, $(-\alpha)H_{n}^{\langle r\rangle}-K_{n}^{\langle r\rangle}=J_{p,n}^{*}(\fourIdx{}{\alpha}{}{}{H})_{n}J_{q,n}$.

(b) According to Lemma 4.18 and Remark 3.10, we have $H_{n-1}^{\langle r\rangle}=J_{p,n-1}^{*}H_{n-1}J_{q,n-1}$, $K_{n-1}^{\langle r\rangle}=-J_{p,n-1}^{*}K_{n-1}J_{q,n-1}$, and $G_{n-1}^{\langle r\rangle}=J_{p,n-1}^{*}G_{n-1}J_{q,n-1}$. Hence,

[TABLE]

The following technical result is useful for our further purposes.

Lemma 4.21**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Let the sequence $(r_{j})_{j=0}^{m}$ be given by $r_{j}\vcentcolon=(-1)^{j}s_{j}$. Then:

1. (a)

$(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$* if and only if $(r_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,-\beta,-\alpha}$.* 2. (b)

$(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$* if and only if $(r_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,-\beta,-\alpha}$.* 3. (c)

$(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq,\mathrm{e}}_{q,m,\alpha,\beta}$* if and only if $(r_{j})_{j=0}^{m}\in\mathcal{F}^{\geq,\mathrm{e}}_{q,m,-\beta,-\alpha}$.* 4. (d)

$(s_{j})_{j=0}^{m}\in\mathcal{F}^{>,\mathrm{e}}_{q,m,\alpha,\beta}$* if and only if $(r_{j})_{j=0}^{m}\in\mathcal{F}^{>,\mathrm{e}}_{q,m,-\beta,-\alpha}$.*

Proof.

(a) The case $m=0$ is trivial. Assume that $m=2n+1$ with some $n\in\mathbb{N}_{0}$. If $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$, then $\{(\fourIdx{}{\alpha}{}{}{H})_{n},(\fourIdx{}{}{}{\beta}{H})_{n}\}\subseteq\mathbb{C}_{\geq}^{(n+1)q\times(n+1)q}$. Hence, $\{-(-\beta)H_{n}^{\langle r\rangle}+K_{n}^{\langle r\rangle},(-\alpha)H_{n}^{\langle r\rangle}-K_{n}^{\langle r\rangle}\}\subseteq\mathbb{C}_{\geq}^{(n+1)q\times(n+1)q}$ follows from Lemma 4.20(a). Thus, $(r_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,-\beta,-\alpha}$. If $(r_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,-\beta,-\alpha}$, then, in view of $(s_{j})_{j=0}^{2n+1}=((-1)^{j}r_{j})_{j=0}^{2n+1}$, we get from the already proved implication that $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$. Assume that $m=2n$ with some $n\in\mathbb{N}$. If $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$, then the matrices $H_{n}$ and $(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}$ are non-negative Hermitian. From $H_{n}^{\langle r\rangle}=J_{q,n}^{*}H_{n}J_{q,n}$ and Lemma 4.20(b) we see that the matrices $H_{n}^{\langle r\rangle}$ and $-(-\beta)(-\alpha)H_{n-1}^{\langle r\rangle}+\left[(-\beta)+(-\alpha)\right]K_{n-1}^{\langle r\rangle}-G_{n-1}^{\langle r\rangle}$ are non-negative Hermitian. Hence, $(r_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,-\beta,-\alpha}$. If $(r_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,-\beta,-\alpha}$, then, by $(s_{j})_{j=0}^{2n}=((-1)^{j}r_{j})_{j=0}^{2n}$, we get from the already proved implication that $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$.

(b) This can be proved analogously to (a) using $\det J_{q,n}\neq 0$.

(c)–(d) The assertions of (c) and (d) follow immediately from (a) and (b), resp. ∎

5 On the canonical CD-measure associated with a sequence belonging to $\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$

In our joint paper [13] with Yu. M. Dyukarev, we constructed explicitly a molecular solution of Problem M[$[\alpha,\beta]$; $(s_{j})_{j=0}^{2n-1}$, $=$]. Now we are looking for an appropriate modification of our method used in [13], which allows us to handle Problem M[$[\alpha,\beta]$; $(s_{j})_{j=0}^{2n}$, $=$] in a similar way as in [13]. First we sketch some essential features of our construction of a distinguished molecular solution of M[$[\alpha,\beta]$; $(s_{j})_{j=0}^{2n-1}$, $=$] in [13].

Theorem 5.1**.**

Let $n\in\mathbb{N}$ and let $(s_{j})_{j=0}^{2n-1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$. Then:

1. (a)

There is a unique sequence $(s_{k})_{k=2n}^{\infty}$ from $\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{\infty}$ is a Hankel completely degenerate Hankel non-negative definite sequence of order $n$. 2. (b)

The set ${\mathcal{M}_{\geq}^{q}[{\mathbb{R}};(s_{j})_{j=0}^{\infty},=]}$ contains exactly one element $\sigma_{n}$. 3. (c)

The measure $\sigma_{n}$ is molecular.

Proof.

From [14, Proposition 2.38] we get (a). In view of (a) the assertions of (b) and (c) follow from parts (a) and (b) of Proposition 4.9 in [14], resp. ∎

If $n\in\mathbb{N}$ and if $(s_{j})_{j=0}^{2n-1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$, then the measure $\sigma_{n}$ given via Theorem 5.1(b) is called the Hankel completely degenerate non-negative Hermitian measure (short CD-measure) associated with $(s_{j})_{j=0}^{2n-1}$. (For additional information, we refer to [14, Chs. 4 and 5].) In [13], we found a molecular solution for Problem M[$[\alpha,\beta]$; $(s_{j})_{j=0}^{2n-1}$, $=$] in the case of a sequence $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. This will be explained now in more detail.

Theorem 5.2**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}$, and let $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Then $(s_{j})_{j=0}^{2n-1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$. Moreover, the CD-measure $\sigma_{n}$ associated with $(s_{j})_{j=0}^{2n-1}$ fulfills $\sigma_{n}(\mathbb{R}\setminus[\alpha,\beta])=0_{q\times q}$ and the measure $\underline{\sigma}_{2n-1}\vcentcolon=\operatorname{Rstr}_{\mathfrak{B}_{[\alpha,\beta]}}\sigma_{n}$ is molecular and satisfies $\underline{\sigma}_{2n-1}\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{2n-1},=]}$.

Proof.

Because of [13, Lemma 2.21], the sequence $(s_{j})_{j=0}^{2n-1}$ belongs to $\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$. In view of $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$, we see from formulas (1.2) and (1.3) in [13] that in [13, Theorem 1.1] we can particularly choose $\Omega=[\alpha,\beta]$. Then from [13, Proof of Theorem 1.1, p. 921] we see that $\sigma_{n}(\mathbb{R}\setminus[\alpha,\beta])=0_{q\times q}$. From Theorem 5.1(c) we see that $\sigma_{n}$ is molecular. In view of $\sigma_{n}(\mathbb{R}\setminus[\alpha,\beta])=0_{q\times q}$ and Theorem 5.1(b), we obtain the rest. ∎

Proposition 5.3**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq,\mathrm{e}}_{q,2n,\alpha,\beta}$. Further, let $s_{2n+1}\in\mathbb{C}^{q\times q}$ be such that $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$. Then $\underline{\sigma}_{2n+1}$ given in Theorem 5.2 belongs to ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{2n},=]}\cap\mathcal{M}^{q,\mathrm{mol}}_{\geq}([\alpha,\beta])$.

Proof.

Use Theorem 5.2. ∎

Corollary 5.4**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq,\mathrm{e}}_{q,2n,\alpha,\beta}$. Then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{2n},=]}\cap\mathcal{M}^{q,\mathrm{mol}}_{\geq}([\alpha,\beta])\neq\emptyset$.

Proof.

In view of Definition 4.4, we have $\{s_{2n+1}\in\mathbb{C}^{q\times q}\colon(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}\}\neq\emptyset$. Now the assertion follows from Proposition 5.3. ∎

In the situation of Proposition 5.3, we see that each matrix $s_{2n+1}\in\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$ generates a molecular solution of Problem M[$[\alpha,\beta]$; $(s_{j})_{j=0}^{2n}$, $=$]. In order to get more information about this family of molecular measures, we are led to the problem of describing the set $\{s_{2n+1}\in\mathbb{C}^{q\times q}\colon(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}\}$. This problem is of purely algebraic nature as well as our method to handle it. We will carefully study the intrinsic structure of finite $[\alpha,\beta]$-Hausdorff non-negative definite sequences. More precisely, if $m\in\mathbb{N}$ and if $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$, then we will show in Section 10 that, for each $k\in\mathbb{Z}_{1,m}$, the matrix $s_{k}$ belongs to some matrix interval $[\mathfrak{a}_{k-1},\mathfrak{b}_{k-1}]$, where $\mathfrak{a}_{k-1}$ and $\mathfrak{b}_{k-1}$ will be explicitly computed in terms of the sequence $(s_{j})_{j=0}^{k-1}$. Particular attention is drawn to the case that $s_{k}$ coincides with one of the interval ends $\mathfrak{a}_{k-1}$ or $\mathfrak{b}_{k-1}$.

6 On Hankel non-negative definite extendable sequences

The class of $[\alpha,\beta]$-Hausdorff non-negative definite sequences stands in the center of this paper. If we look back at Definition 4.2, where we have introduced it, then we see immediately that such a sequence is determined by the interplay of two Hankel non-negative definite sequences. Hence, our further considerations will be mainly based on the theory of Hankel non-negative definite sequences as well as the theory of Hankel non-negative definite extendable sequences. For this reason, in this section we sketch some features of this theory, which turn out to be important for our further considerations.

Lemma 6.1** ([22, Lemma 3.1(a)–(d)]).**

Let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$ and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,\kappa}$. Then:

1. (a)

$s_{j}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$* for all $j\in\mathbb{Z}_{0,\kappa}$ and $s_{2k}\in\mathbb{C}_{\geq}^{q\times q}$ for all $k\in\mathbb{N}_{0}$ with $2k\leq\kappa$.* 2. (b)

$\bigcup_{j=2k}^{\kappa}\mathcal{R}(s_{j})\subseteq\mathcal{R}(s_{2k})$* and $\mathcal{N}(s_{2k})\subseteq\bigcap_{j=2k}^{\kappa}\mathcal{N}(s_{j})$ for all $k\in\mathbb{N}_{0}$ with $2k\leq\kappa$.*

Lemma 6.2** (cf. [14, Lemma 2.7]).**

Let $n\in\mathbb{N}$. Then

[TABLE]

Remark 6.3*.*

For each $n\in\mathbb{N}_{0}$ we see from [14, Proposition 2.24] that $\mathcal{H}^{>,\mathrm{e}}_{q,2n}=\mathcal{H}^{>}_{q,2n}$.

Proposition 6.4** (cf. [14, Proposition 2.13]).**

Let $n\in\mathbb{N}$. Then

[TABLE]

Proposition 6.4 indicates an essential difference between the set $\mathcal{H}^{\geq,\mathrm{e}}_{q,2n}$ and the set of finite non-negative definite sequences from $\mathbb{C}^{q\times q}$. Proposition 6.4 tells us that there is an inclusion of the null spaces of the consecutive associated Schur complements introduced in Notation 3.4(c), whereas [16, Lemma 6] shows that the corresponding consecutive Schur complements associated with a (Toeplitz-)non-negative definite sequence from $\mathbb{C}^{q\times q}$ are even decreasing with respect to the Löwener semi-ordering in the set $\mathbb{C}_{\mathrm{H}}^{q\times q}$. Clearly, this implies the inclusion for the corresponding null spaces.

Lemma 6.5** (cf. [14, Lemma 2.16]).**

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n+1}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n+1}$ if and only if $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$, $s_{2n+1}^{*}=s_{2n+1}$, and $\mathcal{R}(y_{n+1,2n+1})\subseteq\mathcal{R}(H_{n})$.

Remark 6.6*.*

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n+1}$ be a sequence from $\mathbb{C}^{q\times q}$. Then from [14, Proposition 2.24] we see that $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{>,\mathrm{e}}_{q,2n+1}$ if and only if $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{>}_{q,2n}$ and $s_{2n+1}^{*}=s_{2n+1}$.

Proposition 6.7** (cf. [14, Proposition 2.22(a)]).**

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n+1}$. Then

[TABLE]

Proposition 6.8**.**

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{>,\mathrm{e}}_{q,2n+1}$. Then

[TABLE]

Proof.

This follows from Remark 6.6 and [14, Proposition 2.24]. ∎

We denote by $\mathcal{D}_{p\times q,\kappa}$ the set of all sequences $(s_{j})_{j=0}^{\kappa}$ from $\mathbb{C}^{p\times q}$ which fulfill $\bigcup_{j=0}^{\kappa}\mathcal{R}(s_{j})\subseteq\mathcal{R}(s_{0})$ and $\mathcal{N}(s_{0})\subseteq\bigcap_{j=0}^{\kappa}\mathcal{N}(s_{j})$.

Proposition 6.9** ([22, Proposition 4.24]).**

Let $m\in\mathbb{N}_{0}$. Then $\mathcal{H}^{\geq,\mathrm{e}}_{q,m}\subseteq\mathcal{D}_{q\times q,m}$.

7 Algebraic approach to Proposition 4.5 on the sections of $[\alpha,\beta]$-Hausdorff non-negative definite sequences

A main theme of our following considerations is to gain more insights into the intrinsic structure of $[\alpha,\beta]$-Hausdorff non-negative definite sequences. Such sequences are characterized by a particular interplay between some Hankel non-negative definite sequences. More precisely, we have to organize the interplay between several non-negative Hermitian block Hankel matrices. The difficulty of the situation results from the fact that the block Hankel matrix of step $n$ is not a principle submatrix of the block Hankel matrix of stage $n+1$. It will turn out (see Lemma 7.2) that the parallel sum of matrices (see Appendix B) will be the essential tool to overcome this difficulty.

The construction of a purely algebraic approach to Proposition 4.5 stands in the center of this section. Our proof of Proposition 4.5 above is based on Theorem 4.3, which gives a characterization of the solvability of the truncated $[\alpha,\beta]$-Hausdorff moment problem. Since Proposition 4.5 is a statement of purely algebraic character, for our further purposes it is advantageous to have an independent algebraic proof as well.

The key instrument of our purely algebraic approach to Theorem 4.10 is to find convenient expressions for the relevant block Hankel matrices in terms of parallel sums of matrices and applying Lemma B.2. For every choice of complex ${p\times q}$ matrices $A$ and $B$, let $A\mathbin{\overline{\pm}}B$ be the parallel sum of $A$ and $B$, i. e., let

[TABLE]

Remark 7.1*.*

Let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$ and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{p\times q}$. For each $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$, then the block Hankel matrix $H_{n}$ admits the block representations $H_{n}=\operatorname{row}(y_{k,k+n})_{k=0}^{n}$ and $H_{n}=\operatorname{col}(z_{j,j+n})_{j=0}^{n}$.

For all $n\in\mathbb{N}$, let \Delta_{q,n}\vcentcolon=\bigl{[}\begin{smallmatrix}I_{nq}\\ 0_{q\times nq}\end{smallmatrix}\bigr{]} and \nabla_{q,n}\vcentcolon=\bigl{[}\begin{smallmatrix}0_{q\times nq}\\ I_{nq}\end{smallmatrix}\bigr{]}.

In [33, proof of Lemma 2.5], the following result (formulated for the square case $p=q$) was already proved. Now we give an alternative proof.

Lemma 7.2**.**

Let $n\in\mathbb{N}$ and let $(s_{j})_{j=0}^{2n+1}$ be a sequence from $\mathbb{C}^{p\times q}$ such that

[TABLE]

For every choice of complex numbers $\alpha$ and $\beta$ with $\alpha\neq\beta$, then

[TABLE]

Proof.

Using Remark 4.14, we see that

[TABLE]

Setting $Y\vcentcolon=\operatorname{row}(y_{k,k+n})_{k=1}^{n}$ and $Z\vcentcolon=\operatorname{col}(z_{j,j+n})_{j=1}^{n}$, from Remark 7.1 we see that $H_{n}$ and $K_{n}$ can be represented via

[TABLE]

Thus, (7.1) implies $\mathcal{R}(K_{n})\subseteq\mathcal{R}(H_{n})$ and $\mathcal{N}(H_{n})\subseteq\mathcal{N}(K_{n})$. Consequently, Remarks A.5 and A.6 provide us $H_{n}H_{n}^{\dagger}K_{n}=K_{n}$ and $K_{n}H_{n}^{\dagger}H_{n}=K_{n}$. Hence, in view of Remark 4.13, we get then

[TABLE]

Obviously, $\Delta_{p,n}^{*}K_{n}=Z=\nabla_{p,n}^{*}H_{n}$ and $K_{n}\Delta_{q,n}=Y=H_{n}\nabla_{q,n}$ and, consequently,

[TABLE]

Furthermore, Remark 3.5 yields $\Delta_{p,n}^{*}H_{n}\Delta_{q,n}=H_{n-1}$ and $\Delta_{p,n}^{*}K_{n}\Delta_{q,n}=K_{n-1}$. Using the last two equations, (7.6), (7.3), and (7.2), we get then

[TABLE]

Remark 7.3*.*

Let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$ and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{p\times q}$. For each $n\in\mathbb{N}$ with $2n\leq\kappa$, then the block Hankel matrix $H_{n}$ admits the block representations

[TABLE]

Lemma 7.4**.**

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n+2}$ be a sequence from $\mathbb{C}^{p\times q}$. For every choice of $\alpha$ and $\beta$ in $\mathbb{C}$, then

[TABLE]

Proof.

Using Remark 7.3, we conclude $\Delta_{p,n+1}^{*}H_{n+1}\Delta_{q,n+1}=H_{n}$, $\Delta_{p,n+1}^{*}H_{n+1}\nabla_{q,n+1}=K_{n}$, $\nabla_{p,n+1}^{*}H_{n+1}\Delta_{q,n+1}=K_{n}$, and $\nabla_{p,n+1}^{*}H_{n+1}\nabla_{q,n+1}=G_{n}$. For each $\alpha\in\mathbb{C}$ and each $\beta\in\mathbb{C}$, this implies (7.7) and (7.8). ∎

Lemma 7.5**.**

Let $\alpha\in\mathbb{R}$, $\beta\in(\alpha,\infty)$, and $n\in\mathbb{N}_{0}$. Then $\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}\subseteq\mathcal{H}^{\geq,\mathrm{e}}_{q,2n+1}$.

Proof.

This can be seen from [13, Lemma 2.21]. ∎

Remark 7.6*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $n\in\mathbb{N}_{0}$. Then, we see from Lemma 7.5 and Proposition 6.9 that $\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}\subseteq\mathcal{D}_{q\times q,2n+1}$.

Proposition 7.7**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then:

1. (a)

If $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$, then $(s_{j})_{j=0}^{\ell}\in\mathcal{F}^{\geq}_{q,\ell,\alpha,\beta}$ for each $\ell\in\mathbb{Z}_{0,m}$. 2. (b)

If $(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$, then $(s_{j})_{j=0}^{\ell}\in\mathcal{F}^{>}_{q,\ell,\alpha,\beta}$ for each $\ell\in\mathbb{Z}_{0,m}$.

Proof.

The case $m=0$ is trivial. Assume that $m\geq 1$ and $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$). It is sufficient to prove that $(s_{j})_{j=0}^{m-1}$ belongs to $\mathcal{F}^{\geq}_{q,m-1,\alpha,\beta}$ (resp. $\mathcal{F}^{>}_{q,m-1,\alpha,\beta}$).

First we suppose that there is an integer $n\in\mathbb{N}_{0}$ such that $m=2n+1$. Then Lemma 4.15 yields that the matrix $H_{n}$ is non-negative Hermitian (resp. positive Hermitian). If $n=0$, then $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{>}_{q,2n,\alpha,\beta}$) immediately follows from Definition 4.2. Now we assume that $n\geq 1$. In view of Lemma 4.17, we have $s_{j}^{*}=s_{j}$ for all $j\in\mathbb{Z}_{0,2n+1}$. The combination of Lemmata 7.5 and 6.5 provides us $\mathcal{R}(y_{n+1,2n+1})\subseteq\mathcal{R}(H_{n})$ and $\mathcal{N}(H_{n})=\mathcal{N}(H_{n}^{*})=\mathcal{R}(H_{n})^{\bot}\subseteq\mathcal{R}(y_{n+1,2n+1})^{\bot}=\mathcal{N}(y_{n+1,2n+1}^{*})=\mathcal{N}(z_{n+1,2n+1})$. From Lemma 7.2 we get then

[TABLE]

Because of $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{>}_{q,2n+1,\alpha,\beta}$), Definition 4.2(a) shows that the matrices $(\fourIdx{}{\alpha}{}{}{H})_{n}$ and $(\fourIdx{}{}{}{\beta}{H})_{n}$ are both non-negative Hermitian (resp. positive Hermitian). Thus, Lemma B.2 (resp. Remark B.1) implies that the matrix $(\fourIdx{}{\alpha}{}{}{H})_{n}\mathbin{\overline{\pm}}(\fourIdx{}{}{}{\beta}{H})_{n}$ is non-negative Hermitian (resp. positive Hermitian) as well. Because of $\beta-\alpha>0$ and $\operatorname{rank}\Delta_{q,n}=nq$, then (7.9) und Remark A.2 yield that the matrix $(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}$ is non-negative Hermitian (resp. positive Hermitian). Consequently, in view of Definition 4.2(b), we see that $(s_{j})_{j=0}^{2n}$ belongs to $\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$ (resp. $\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$).

It remains to consider the case that $m=2n$ holds true with some $n\in\mathbb{N}$. By virtue of $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{>}_{q,2n,\alpha,\beta}$) and Definition 4.2(b), the matrix $(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}$ is non-negative Hermitian (resp. positive Hermitian). Lemma 7.4 and $\beta>\alpha$ yield $(\fourIdx{}{\alpha}{}{}{H})_{n-1}\geq(\beta-\alpha)^{-1}(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}$ and $(\fourIdx{}{}{}{\beta}{H})_{n-1}\geq(\beta-\alpha)^{-1}(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}$. Hence, the matrices $(\fourIdx{}{\alpha}{}{}{H})_{n-1}$ and $(\fourIdx{}{}{}{\beta}{H})_{n-1}$ are both non-negative Hermitian (resp. positive Hermitian). Thus, Definition 4.2(a) implies $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{>}_{q,2n-1,\alpha,\beta}$). ∎

Corollary 7.8**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{q\times q}$. Then:

1. (a)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$, then $L_{n}\in\mathbb{C}_{\geq}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$, furthermore $\{(\fourIdx{}{\alpha}{}{}{L})_{n},(\fourIdx{}{}{}{\beta}{L})_{n}\}\subseteq\mathbb{C}_{\geq}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$, and $(\fourIdx{}{\alpha}{}{\beta}{L})_{n}\in\mathbb{C}_{\geq}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n+2\leq\kappa$. 2. (b)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{>}_{q,\kappa,\alpha,\beta}$, then $L_{n}\in\mathbb{C}_{>}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$, furthermore $\{(\fourIdx{}{\alpha}{}{}{L})_{n},(\fourIdx{}{}{}{\beta}{L})_{n}\}\subseteq\mathbb{C}_{>}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$, and $(\fourIdx{}{\alpha}{}{\beta}{L})_{n}\in\mathbb{C}_{>}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n+2\leq\kappa$.

Proof.

Use Proposition 7.7, Definitions 4.2 and 4.7, and Remark 3.6. ∎

Corollary 7.9**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $m\in\mathbb{N}_{0}$. Then $\mathcal{F}^{\geq,\mathrm{e}}_{q,m,\alpha,\beta}\subseteq\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ and $\mathcal{F}^{>,\mathrm{e}}_{q,m,\alpha,\beta}\subseteq\mathcal{F}^{>}_{q,m,\alpha,\beta}$.

Proof.

Use Proposition 7.7. ∎

Proposition 7.10**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $m\in\mathbb{N}_{0}$. Then $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}\subseteq\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$ and $\mathcal{F}^{>}_{q,m,\alpha,\beta}\subseteq\mathcal{H}^{>,\mathrm{e}}_{q,m}$.

Proof.

We first show $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}\subseteq\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$. Because of $\mathcal{H}^{\geq,\mathrm{e}}_{q,0}=\mathcal{H}^{\geq}_{q,0}$, the case $m=0$ is trivial. If $m=2n+1$ for some $n\in\mathbb{N}_{0}$, then Lemma 7.5 yields the assertion.

Now let $m=2n$ with some $n\in\mathbb{N}$. We consider an arbitrary sequence $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$. Then Definition 4.2(b) and Remark 7.3 provide us $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$ and $\{(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1},G_{n-1}\}\subseteq\mathbb{C}_{\geq}^{nq\times nq}$. Thus, Remark A.1 yields $\mathcal{R}(G_{n-1}+(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1})=\mathcal{R}(G_{n-1})+\mathcal{R}((\fourIdx{}{\alpha}{}{\beta}{H})_{n-1})$. According to Proposition 7.7(a), the sequence $(s_{j})_{j=0}^{2n-1}$ belongs to $\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Consequently, Definition 4.2(a) yields $\{(\fourIdx{}{\alpha}{}{}{H})_{n-1},(\fourIdx{}{}{}{\beta}{H})_{n-1}\}\subseteq\mathbb{C}_{\geq}^{nq\times nq}$. Hence, using Remark A.1 again, we conclude $\mathcal{R}((\fourIdx{}{\alpha}{}{}{H})_{n-1}+(\fourIdx{}{}{}{\beta}{H})_{n-1})=\mathcal{R}((\fourIdx{}{\alpha}{}{}{H})_{n-1})+\mathcal{R}((\fourIdx{}{}{}{\beta}{H})_{n-1})$. Since Remark 4.14 shows that $H_{n-1}=\frac{1}{\beta-\alpha}[(\fourIdx{}{\alpha}{}{}{H})_{n-1}+(\fourIdx{}{}{}{\beta}{H})_{n-1}]$ and $G_{n-1}+(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}=\frac{1}{\beta-\alpha}[\beta^{2}(\fourIdx{}{\alpha}{}{}{H})_{n-1}+\alpha^{2}(\fourIdx{}{}{}{\beta}{H})_{n-1}]$, we get then

[TABLE]

Taking into account that $y_{n+1,2n}$ is the last ${nq\times q}$ block column of $G_{n-1}$, we infer $\mathcal{R}(y_{n+1,2n})\subseteq\mathcal{R}(H_{n-1})$. Thus, Lemma 6.2 yields $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n}$.

Now we show $\mathcal{F}^{>}_{q,m,\alpha,\beta}\subseteq\mathcal{H}^{>,\mathrm{e}}_{q,m}$. If $m=2n$ with some $n\in\mathbb{N}_{0}$, this follows from Remark 6.3 and Definition 4.2(b). If $m=2n+1$ with some $n\in\mathbb{N}_{0}$, this follows from Remark 6.6 in view Lemmata 4.15 and 4.17. ∎

8 On Stieltjes non-negative definite sequences

In this section, we discuss two dual classes of finite sequences of complex ${q\times q}$ matrices which turn out to be intimately connected with the class $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. The first and second types of this classes will be associated with the left endpoint $\alpha$ and the right endpoint $\beta$ of the interval $[\alpha,\beta]$, resp. These classes were already studied by the authors in earlier work (see [14, 20, 21]). First we consider the class of sequences connected with the left endpoint $\alpha$ of $[\alpha,\beta]$.

Definition 8.1**.**

Let $\alpha\in\mathbb{R}$ and let $n\in\mathbb{N}_{0}$.

1. (a)

Let $(s_{j})_{j=0}^{2n+1}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n+1}$ is called $[\alpha,\infty)$-Stieltjes non-negative definite (resp. $[\alpha,\infty)$-Stieltjes positive definite) if both sequences $(s_{j})_{j=0}^{2n}$ and $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}$ are Hankel non-negative definite (resp. Hankel positive definite). 2. (b)

Let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n}$ is called $[\alpha,\infty)$-Stieltjes non-negative definite (resp. $[\alpha,\infty)$-Stieltjes positive definite) if $(s_{j})_{j=0}^{2n}$ and, in the case $n\geq 1$, also $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2(n-1)}$ is Hankel non-negative definite (resp. Hankel positive definite).

If $m\in\mathbb{N}_{0}$, then, $\mathcal{K}^{\geq}_{q,m,\alpha}$ (resp. $\mathcal{K}^{>}_{q,m,\alpha}$) stands for the set of all $[\alpha,\infty)$-Stieltjes non-negative definite (resp. $[\alpha,\infty)$-Stieltjes positive definite) sequences $(s_{j})_{j=0}^{m}$ from $\mathbb{C}^{q\times q}$.

Let $\alpha\in\mathbb{R}$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{m}$ is called $[\alpha,\infty)$-Stieltjes non-negative definite extendable (resp. $[\alpha,\infty)$-Stieltjes positive definite extendable) if there exists a matrix $s_{m+1}\in\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{m+1}\in\mathcal{K}^{\geq}_{q,m+1,\alpha}$ (resp. $(s_{j})_{j=0}^{m+1}\in\mathcal{K}^{>}_{q,m+1,\alpha}$). We denote by $\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}$ (resp. $\mathcal{K}^{>,\mathrm{e}}_{q,m,\alpha}$) the set of all $[\alpha,\infty)$-Stieltjes non-negative definite extendable (resp. $[\alpha,\infty)$-Stieltjes positive definite extendable) sequences $(s_{j})_{j=0}^{m}$ from $\mathbb{C}^{q\times q}$. The importance of the class $\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}$ in the context of moment problems is caused by the following observation:

Theorem 8.2** ([13, Theorem 1.3]).**

Let $\alpha\in\mathbb{R}$, $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\infty)};(s_{j})_{j=0}^{m},=]}\neq\emptyset$ if and only if $(s_{j})_{j=0}^{m}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}$.

Lemma 8.3** ([20, Lemma 2.9(a)–(b)]).**

Let $\alpha\in\mathbb{R}$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{K}^{\geq}_{q,m,\alpha}$. Then $s_{j}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ for all $j\in\mathbb{Z}_{0,m}$ and $s_{2k}\in\mathbb{C}_{\geq}^{q\times q}$ for all $k\in\mathbb{N}_{0}$ with $2k\leq m$. Furthermore, $(\fourIdx{}{\alpha}{}{}{s})_{j}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ for all $j\in\mathbb{Z}_{0,m-1}$ and $(\fourIdx{}{\alpha}{}{}{s})_{2k}\in\mathbb{C}_{\geq}^{q\times q}$ for all $k\in\mathbb{N}_{0}$ with $2k+1\leq m$.

Remark 8.4*.*

Let $\alpha\in\mathbb{R}$ and let $m\in\mathbb{N}_{0}$. Then $\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}\subseteq\mathcal{K}^{\geq}_{q,m,\alpha}$ and, in the case $m\in\mathbb{N}$, furthermore $\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}\neq\mathcal{K}^{\geq}_{q,m,\alpha}$.

Remark 8.5*.*

Let $\alpha\in\mathbb{R}$, let $m\in\mathbb{N}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{K}^{\geq}_{q,m,\alpha}$. Then $(s_{j})_{j=0}^{\ell}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,\ell,\alpha}$ for all $\ell\in\mathbb{Z}_{0,m-1}$.

Remark 8.6*.*

Let $\alpha\in\mathbb{R}$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}$ (resp. $\mathcal{K}^{>}_{q,m,\alpha}$). Then $(s_{j})_{j=0}^{\ell}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,\ell,\alpha}$ (resp. $\mathcal{K}^{>}_{q,\ell,\alpha}$) for all $\ell\in\mathbb{Z}_{0,\kappa}$.

Let $\alpha\in\mathbb{R}$. A sequence $(s_{j})_{j=0}^{\infty}$ from $\mathbb{C}^{q\times q}$ is called $[\alpha,\infty)$-Stieltjes non-negative definite (resp. $[\alpha,\infty)$-Stieltjes positive definite) if $(s_{j})_{j=0}^{m}\in\mathcal{K}^{\geq}_{q,m,\alpha}$ (resp. $(s_{j})_{j=0}^{m}\in\mathcal{K}^{>}_{q,m,\alpha}$) for all $m\in\mathbb{N}_{0}$. We denote by $\mathcal{K}^{\geq}_{q,\infty,\alpha}$ (resp. $\mathcal{K}^{>}_{q,\infty,\alpha}$) the set of all $[\alpha,\infty)$-Stieltjes non-negative definite (resp. $[\alpha,\infty)$-Stieltjes positive definite) sequences $(s_{j})_{j=0}^{\infty}$ from $\mathbb{C}^{q\times q}$. Furthermore, let $\mathcal{K}^{\geq,\mathrm{e}}_{q,\infty,\alpha}\vcentcolon=\mathcal{K}^{\geq}_{q,\infty,\alpha}$ and let $\mathcal{K}^{>,\mathrm{e}}_{q,\infty,\alpha}\vcentcolon=\mathcal{K}^{>}_{q,\infty,\alpha}$.

Lemma 8.7**.**

Let $\alpha\in\mathbb{R}$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{K}^{\geq}_{q,\kappa,\alpha}$. Then $L_{n}\in\mathbb{C}_{\geq}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$, and $(\fourIdx{}{\alpha}{}{}{L})_{n}\in\mathbb{C}_{\geq}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$.

Proof.

Combine Definition 8.1, Remarks 8.5 and 8.4, and Remark 3.6(a). ∎

Proposition 8.8** ([20, Proposition 2.19]).**

Let $\alpha\in\mathbb{R}$. Then:

1. (a)

Let $n\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{2n+1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n+1,\alpha}$, then

[TABLE]

and

[TABLE] 2. (b)

Let $n\in\mathbb{N}$ and let $(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}$, then

[TABLE]

and

[TABLE]

Now we consider the constructions connected with the right endpoint $\beta$ of the interval $[\alpha,\beta]$.

Definition 8.9**.**

Let $\beta\in\mathbb{R}$ and let $n\in\mathbb{N}_{0}$.

1. (a)

Let $(s_{j})_{j=0}^{2n+1}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n+1}$ is called $(-\infty,\beta]$-Stieltjes non-negative definite (resp. $(-\infty,\beta]$-Stieltjes positive definite) if both sequences $(s_{j})_{j=0}^{2n}$ and $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}$ are Hankel non-negative definite (resp. Hankel positive definite). 2. (b)

Let $(s_{j})_{j=0}^{2n}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{2n}$ is called $(-\infty,\beta]$-Stieltjes non-negative definite (resp. $(-\infty,\beta]$-Stieltjes positive definite) if $(s_{j})_{j=0}^{2n}$ and, in the case $n\geq 1$, also $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2(n-1)}$ is Hankel non-negative definite (resp. Hankel positive definite).

If $m\in\mathbb{N}_{0}$, then $\mathcal{L}^{\geq}_{q,m,\beta}$ (resp. $\mathcal{L}^{>}_{q,m,\beta}$) stands for the set of all $(-\infty,\beta]$-Stieltjes non-negative definite (resp. $(-\infty,\beta]$-Stieltjes positive definite) sequences $(s_{j})_{j=0}^{m}$ from $\mathbb{C}^{q\times q}$.

Let $\beta\in\mathbb{R}$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then $(s_{j})_{j=0}^{m}$ is called $(-\infty,\beta]$-Stieltjes non-negative definite extendable (resp. $(-\infty,\beta]$-Stieltjes positive definite extendable) if there exists a matrix $s_{m+1}\in\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{m+1}\in\mathcal{L}^{\geq}_{q,m+1,\beta}$ (resp. $(s_{j})_{j=0}^{m+1}\in\mathcal{L}^{>}_{q,m+1,\beta}$). We denote by $\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$ (resp. $\mathcal{L}^{>,\mathrm{e}}_{q,m,\beta}$) the set of all $(-\infty,\beta]$-Stieltjes non-negative definite extendable (resp. $(-\infty,\beta]$-Stieltjes positive definite extendable) sequences $(s_{j})_{j=0}^{m}$ from $\mathbb{C}^{q\times q}$.

A sequence $(s_{j})_{j=0}^{\infty}$ from $\mathbb{C}^{q\times q}$ is called $(-\infty,\beta]$-Stieltjes non-negative definite (resp. $(-\infty,\beta]$-Stieltjes positive definite) if $(s_{j})_{j=0}^{m}\in\mathcal{L}^{\geq}_{q,m,\beta}$ (resp. $(s_{j})_{j=0}^{m}\in\mathcal{L}^{>}_{q,m,\beta}$) for all $m\in\mathbb{N}_{0}$. We denote by $\mathcal{L}^{\geq}_{q,\infty,\beta}$ (resp. $\mathcal{L}^{>}_{q,\infty,\beta}$) the set of all $(-\infty,\beta]$-Stieltjes non-negative definite (resp. $(-\infty,\beta]$-Stieltjes positive definite) sequences $(s_{j})_{j=0}^{\infty}$ from $\mathbb{C}^{q\times q}$. Furthermore, let $\mathcal{L}^{\geq,\mathrm{e}}_{q,\infty,\beta}\vcentcolon=\mathcal{L}^{\geq}_{q,\infty,\beta}$ and let $\mathcal{L}^{>,\mathrm{e}}_{q,\infty,\beta}\vcentcolon=\mathcal{L}^{>}_{q,\infty,\beta}$.

Lemma 8.10**.**

Let $\beta\in\mathbb{R}$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Let the sequence $(r_{j})_{j=0}^{m}$ be given by $r_{j}\vcentcolon=(-1)^{j}s_{j}$. Then:

1. (a)

$(s_{j})_{j=0}^{m}\in\mathcal{L}^{\geq}_{q,m,\beta}$* if and only if $(r_{j})_{j=0}^{m}\in\mathcal{K}^{\geq}_{q,m,-\beta}$.* 2. (b)

$(s_{j})_{j=0}^{m}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$* if and only if $(r_{j})_{j=0}^{m}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,m,-\beta}$.* 3. (c)

$(s_{j})_{j=0}^{m}\in\mathcal{L}^{>}_{q,m,\beta}$* if and only if $(r_{j})_{j=0}^{m}\in\mathcal{K}^{>}_{q,m,-\beta}$.* 4. (d)

$(s_{j})_{j=0}^{m}\in\mathcal{L}^{>,\mathrm{e}}_{q,m,\beta}$* if and only if $(r_{j})_{j=0}^{m}\in\mathcal{K}^{>,\mathrm{e}}_{q,m,-\beta}$.*

Proof.

This can be verified analogously to the proof of Lemma 4.21. ∎

The importance of the class $\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$ in the context of moment problems is caused by the following observation:

Theorem 8.11**.**

Let $\beta\in\mathbb{R}$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then ${\mathcal{M}_{\geq}^{q}[{(-\infty,\beta]};(s_{j})_{j=0}^{m},=]}\neq\emptyset$ if and only if $(s_{j})_{j=0}^{m}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$.

Proof.

Combine Lemma 8.10(b), Theorem 8.2, and Remark 2.1. ∎

Remark 8.12*.*

Let $\beta\in\mathbb{R}$ and let $m\in\mathbb{N}_{0}$. Then $\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}\subseteq\mathcal{L}^{\geq}_{q,m,\beta}$ and, in the case $m\in\mathbb{N}$, furthermore $\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}\neq\mathcal{L}^{\geq}_{q,m,\beta}$.

Remark 8.13*.*

Let $\beta\in\mathbb{R}$, let $m\in\mathbb{N}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{L}^{\geq}_{q,m,\beta}$. Then $(s_{j})_{j=0}^{\ell}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,\ell,\beta}$ for all $\ell\in\mathbb{Z}_{0,m-1}$.

Remark 8.14*.*

Let $\beta\in\mathbb{R}$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$ (resp. $\mathcal{L}^{>}_{q,m,\beta}$). Then $(s_{j})_{j=0}^{\ell}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,\ell,\beta}$ (resp. $\mathcal{L}^{>}_{q,\ell,\beta}$) for all $\ell\in\mathbb{Z}_{0,m}$.

Remark 8.15*.*

Let $\beta\in\mathbb{R}$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{L}^{\geq}_{q,\kappa,\beta}$. In view of Definition 8.9, Remarks 8.13 and 8.12, and Remark 3.6(a), then $L_{n}\in\mathbb{C}_{\geq}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$, and $(\fourIdx{}{}{}{\beta}{L})_{n}\in\mathbb{C}_{\geq}^{q\times q}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$.

Remark 8.16*.*

For each $n\in\mathbb{N}$, it is readily checked that

[TABLE]

Proposition 8.17**.**

Let $\alpha,\beta\in\mathbb{R}$ and let $n\in\mathbb{N}$. Then:

1. (a)

$\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-1,\alpha}=\{(s_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq}_{q,2n-1,\alpha}\colon\mathcal{N}(L_{n-1})\subseteq\mathcal{N}((\fourIdx{}{\alpha}{}{}{L})_{n-1})\}$. 2. (b)

$\mathcal{L}^{\geq,\mathrm{e}}_{q,2n-1,\beta}=\{(s_{j})_{j=0}^{2n-1}\in\mathcal{L}^{\geq}_{q,2n-1,\beta}\colon\mathcal{N}(L_{n-1})\subseteq\mathcal{N}((\fourIdx{}{}{}{\beta}{L})_{n-1})\}$. 3. (c)

$\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}=\{(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq}_{q,2n,\alpha}\colon\mathcal{N}((\fourIdx{}{\alpha}{}{}{L})_{n-1})\subseteq\mathcal{N}(L_{n})\}$. 4. (d)

$\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}=\{(s_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq}_{q,2n,\beta}\colon\mathcal{N}((\fourIdx{}{}{}{\beta}{L})_{n-1})\subseteq\mathcal{N}(L_{n})\}$.

Proof.

Parts (a) and (c) are proved in [13, Lemmata 4.15 and 4.16].

(b) Let $r_{j}\vcentcolon=(-1)^{j}s_{j}$ for all $j\in\mathbb{Z}_{0,2n-1}$. Then $s_{j}\vcentcolon=(-1)^{j}r_{j}$ for each $j\in\mathbb{Z}_{0,2n-1}$. First we assume $(s_{j})_{j=0}^{2n-1}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n-1,\beta}$. Lemma 8.10(b) provides us then $(r_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-1,-\beta}$. Thus, part (a) shows $(r_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq}_{q,2n-1,-\beta}$ and

[TABLE]

In particular, Lemma 8.10(a) implies

[TABLE]

Taking into account that

[TABLE]

holds true for each $j\in\mathbb{Z}_{0,2n-1}$, Lemma 4.19(b) implies $(_{-\beta}L^{\langle r\rangle})_{n-1}=(\fourIdx{}{}{}{\beta}{L})_{n-1}$. Since Lemma 4.19(b) also yields $L_{n-1}^{\langle r\rangle}=L_{n-1}$, from (8.1) we get then

[TABLE]

Conversely, now suppose that (8.2) and (8.4) are valid. Because of (8.2) and Lemma 8.10(a), then $(r_{j})_{j=0}^{2n-1}$ belongs to $\mathcal{K}^{\geq}_{q,2n-1,-\beta}$. From Lemma 4.19(b) we get again $L_{n-1}^{\langle r\rangle}=L_{n-1}$ and, in view of (8.3), the equation $(\fourIdx{}{-\beta}{}{}{L^{\langle r\rangle}})_{n-1}=(\fourIdx{}{}{}{\beta}{L})_{n-1}$ as well. Thus, (8.4) implies (8.1). Consequently, part (a) yields $(r_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-1,-\beta}$. Hence, Lemma 8.10(b) shows that $(s_{j})_{j=0}^{2n-1}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n-1,\beta}$.

(d) Analogously, as part (b) can be proved by virtue of part (a), part (d) can be checked using part (c). ∎

If $\mathcal{M}$ is a non-empty subset of $\mathbb{C}^{q}$, then let $\mathcal{M}^{\bot}$ be the set of all $x\in\mathbb{C}^{q}$ which fulfill $\langle x,y\rangle_{\mathrm{E}}=0$ for each $y\in\mathcal{M}$, where $\langle\cdot,\cdot\rangle_{\mathrm{E}}$ is the Euclidean inner product in $\mathbb{C}^{q}$.

Proposition 8.18**.**

Let $\alpha,\beta\in\mathbb{R}$ and let $n\in\mathbb{N}$. Then:

1. (a)

$\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-1,\alpha}=\{(s_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq}_{q,2n-1,\alpha}\colon\mathcal{R}((\fourIdx{}{\alpha}{}{}{L})_{n-1})\subseteq\mathcal{R}(L_{n-1})\}$. 2. (b)

$\mathcal{L}^{\geq,\mathrm{e}}_{q,2n-1,\beta}=\{(s_{j})_{j=0}^{2n-1}\in\mathcal{L}^{\geq}_{q,2n-1,\beta}\colon\mathcal{R}((\fourIdx{}{}{}{\beta}{L})_{n-1})\subseteq\mathcal{R}(L_{n-1})\}$. 3. (c)

$\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}=\{(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq}_{q,2n,\alpha}\colon\mathcal{R}(L_{n})\subseteq\mathcal{R}((\fourIdx{}{\alpha}{}{}{L})_{n-1})\}$. 4. (d)

$\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}=\{(s_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq}_{q,2n,\beta}\colon\mathcal{R}(L_{n})\subseteq\mathcal{R}((\fourIdx{}{}{}{\beta}{L})_{n-1})\}$.

Proof.

In view of the Definition of the set $\mathcal{K}^{\geq}_{q,2n-1,\alpha}$ and Lemma 8.7, we see that, for each $(s_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq}_{q,2n-1,\alpha}$, the matrices $L_{n-1}$ and $(\fourIdx{}{\alpha}{}{}{L})_{n-1}$ are both non-negative Hermitian. Thus, $\mathcal{N}(L_{n-1})=\mathcal{N}(L_{n-1}^{*})=\mathcal{R}(L_{n-1})^{\bot}$ and $\mathcal{N}((\fourIdx{}{\alpha}{}{}{L})_{n-1})=\mathcal{N}((\fourIdx{}{\alpha}{}{}{L})_{n-1}^{*})=\mathcal{R}((\fourIdx{}{\alpha}{}{}{L})_{n-1})^{\bot}$. Consequently, the application of Proposition 8.17 completes the proof of (a). Parts (b), (c), and (d) can be checked analogously. ∎

Now we are going to study several aspects of the interplay between those four Hankel non-negative definite sequences which determine the sections of an $[\alpha,\beta]$-Hausdorff non-negative definite sequence. In particular, we derive formulas connecting the matrices which were introduced in parts (c) and (d) of Notation 3.4 for each of the sequences $(s_{j})_{j=0}^{\kappa}$, $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{\kappa-1}$, $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{\kappa-1}$, and $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{\kappa-2}$. These formulas play an important role in the proof of Theorem 10.14, which is of central importance for our further considerations.

Lemma 8.19**.**

Let $\alpha,\beta\in\mathbb{R}$ and let $n\in\mathbb{N}$. Then:

1. (a)

If $(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq}_{q,2n,\alpha}$, then

[TABLE] 2. (b)

If $(s_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq}_{q,2n,\beta}$, then

[TABLE]

Proof.

A proof of part (a) is given in [20, Proposition 6.4]. To prove part (b), we consider an arbitrary sequence $(s_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq}_{q,2n,\beta}$. Setting $r_{j}\vcentcolon=(-1)^{j}s_{j}$ for all $j\in\mathbb{Z}_{0,2n}$, we see from Lemma 8.10(a) that $(r_{j})_{j=0}^{2n}$ belongs to $\mathcal{K}^{\geq}_{q,2n,-\beta}$. Because of part (a), we have then

[TABLE]

Obviously, $r_{2n-1}=-s_{2n-1}$ and $r_{2n}=s_{2n}$. Lemma 4.19 yields $L_{\ell}^{\langle r\rangle}=L_{\ell}$ for $\ell\in\{n-1,n\}$ and $M_{n-1}^{\langle r\rangle}=-M_{n-1}$. Taking into account that (8.3) holds true for all $j\in\mathbb{Z}_{0,2n-1}$, from Lemma 4.19 we also get $(\fourIdx{}{-\beta}{}{}{M^{\langle r\rangle}})_{n-1}=-(\fourIdx{}{}{}{\beta}{M})_{n-1}$ and $(\fourIdx{}{-\beta}{}{}{L^{\langle r\rangle}})_{n-1}=(\fourIdx{}{}{}{\beta}{L})_{n-1}$. Thus, in view of (8.5), the proof is complete. ∎

Lemma 8.20**.**

Let $\alpha,\beta\in\mathbb{R}$ and let $n\in\mathbb{N}$. Then:

1. (a)

If $(s_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-1,\alpha}$, then

[TABLE] 2. (b)

If $(s_{j})_{j=0}^{2n-1}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n-1,\beta}$, then

[TABLE]

Proof.

First we suppose that $(s_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-1,\alpha}$. Then there is an $s_{2n}\in\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq}_{q,2n,\alpha}$. Using Lemma 8.19(a), the equation $\Theta_{n}=s_{2n}-L_{n}$, and the equation $(\fourIdx{}{\alpha}{}{}{s})_{2n-1}=-\alpha s_{2n-1}+s_{2n}$, we get (8.6). Part (a) is proved. In view of Lemma 8.19(b), part (b) can be checked analogously. ∎

Lemma 8.21**.**

Let $\alpha,\beta\in\mathbb{R}$ and let $n\in\mathbb{N}$. Then:

1. (a)

If $(s_{j})_{j=0}^{2n+1}\in\mathcal{K}^{\geq}_{q,2n+1,\alpha}$, then

[TABLE] 2. (b)

If $(s_{j})_{j=0}^{2n+1}\in\mathcal{L}^{\geq}_{q,2n+1,\beta}$, then

[TABLE]

Proof.

Part (a) is proved in  [20, Proposition 6.5]. In order to check part (b), we consider an arbitrary sequence $(s_{j})_{j=0}^{2n+1}\in\mathcal{L}^{\geq}_{q,2n+1,\beta}$. Let $r_{j}\vcentcolon=(-1)^{j}s_{j}$ for each $j\in\mathbb{Z}_{0,2n+1}$. According to Lemma 8.10(a), the sequence $(r_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{K}^{\geq}_{q,2n+1,-\beta}$. Thus, part (a) provides us

[TABLE]

Obviously, $r_{2n+1}=-s_{2n+1}$ and, by (8.3), furthermore $(\fourIdx{}{-\beta}{}{}{r})_{2n-1}=-(\fourIdx{}{}{}{\beta}{s})_{2n-1}$. Lemma 4.19 yields $M_{n}^{\langle r\rangle}=-M_{n}$ and $L_{n}^{\langle r\rangle}=L_{n}$. Since (8.3) holds true for all $j\in\mathbb{Z}_{0,2n}$, Lemma 4.19 also shows that $(\fourIdx{}{-\beta}{}{}{L^{\langle r\rangle}})_{\ell}=(\fourIdx{}{}{}{\beta}{L})_{\ell}$ for $\ell\in\{n-1,n\}$ and that $(\fourIdx{}{-\beta}{}{}{M^{\langle r\rangle}})_{n-1}=-(\fourIdx{}{}{}{\beta}{M})_{n-1}$. Thus, (8.7) follows from (8.8). ∎

Lemma 8.22**.**

Let $\alpha,\beta\in\mathbb{R}$ and let $n\in\mathbb{N}$. Then:

1. (a)

If $(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}$, then

[TABLE] 2. (b)

If $(s_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}$, then

[TABLE]

Proof.

Choose a matrix $s_{2n+1}\in\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{K}^{\geq}_{q,2n+1,\alpha}$ (resp. $\mathcal{L}^{\geq}_{q,2n+1,\beta}$) and apply Lemma 8.21. ∎

Lemma 8.23**.**

Let $\alpha,\beta\in\mathbb{R}$ and let $\kappa\in\mathbb{N}\cup\{\infty\}$. Then:

1. (a)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{K}^{\geq}_{q,\kappa,\alpha}$, then $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{m}\in\mathcal{K}^{\geq}_{q,m,\alpha}$ for all $m\in\mathbb{Z}_{0,\kappa-1}$. 2. (b)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{L}^{\geq}_{q,\kappa,\beta}$, then $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{m}\in\mathcal{L}^{\geq}_{q,m,\beta}$ for all $m\in\mathbb{Z}_{0,\kappa-1}$.

Proof.

(a) In view of Remarks 8.5 and 8.4, part (a) is proved in [21, Proposition 5.1].

(b) Use Lemma 8.10 and part (a). ∎

The following result gives us a first impression of the interplay between $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ and the classes studied in this section.

Proposition 8.24**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $n\in\mathbb{N}_{0}$. Then $\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}=\mathcal{K}^{\geq}_{q,2n+1,\alpha}\cap\mathcal{L}^{\geq}_{q,2n+1,\beta}$.

Proof.

Let $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$. Then Definition 4.2(a) shows that $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}\}\subseteq\mathcal{H}^{\geq}_{q,2n}$. Moreover, from Lemma 4.15 we know that $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$. Hence, $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{K}^{\geq}_{q,2n+1,\alpha}\cap\mathcal{L}^{\geq}_{q,2n+1,\beta}$.

Conversely, now assume that $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{K}^{\geq}_{q,2n+1,\alpha}\cap\mathcal{L}^{\geq}_{q,2n+1,\beta}$. Then the definitions of the sets $\mathcal{K}^{\geq}_{q,2n+1,\alpha}$ and $\mathcal{L}^{\geq}_{q,2n+1,\beta}$ in combination with Definition 4.2(a) yield immediately $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$. ∎

9 On the inclusion $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}\subseteq\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$

In this section we investigate interrelations between the set $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ on the one hand and the intersection $\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$ on the other hand.

Proposition 9.1**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then:

1. (a)

$\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{\kappa-1},((\fourIdx{}{}{}{\beta}{s})_{j})_{j-0}^{\kappa-1}\}\subseteq\mathcal{F}^{\geq}_{q,\kappa-1,\alpha,\beta}$. 2. (b)

If $\kappa\geq 2$, then $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{\kappa-2}\in\mathcal{F}^{\geq}_{q,\kappa-2,\alpha,\beta}$.

Proof.

(a) First we consider the case that $\kappa=2n+1$ with some $n\in\mathbb{N}_{0}$. Then Definition 4.2(a) yields $\{(\fourIdx{}{\alpha}{}{}{H})_{n},(\fourIdx{}{}{}{\beta}{H})_{n}\}\subseteq\mathbb{C}_{\geq}^{(n+1)q\times(n+1)q}$. If $n=0$, then (a) is proved. Assume $n\geq 1$. Then Proposition 7.7(a) provides us $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$ and, consequently, $(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}\in\mathbb{C}_{\geq}^{nq\times nq}$. From Remark 4.14 we have $(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}=(_{\alpha}(H_{\beta}))_{n-1}=((_{\alpha}H)_{\beta})_{n-1}$. Thus, $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq}_{q,2n,\alpha}$ and $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq}_{q,2n,\beta}$. Applying Lemma 8.23(a) to $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}$, from Remark 4.12 we get $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq}_{q,2n-1,\alpha}$. In particular, $-\alpha(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}+(\fourIdx{}{\alpha}{}{\beta}{K})_{n-1}\in\mathbb{C}_{\geq}^{nq\times nq}$. Taking into account that Remark 4.12 shows that $-\alpha(\fourIdx{}{\alpha}{}{\beta}{s})_{j}+(\fourIdx{}{\alpha}{}{\beta}{s})_{j+1}=-\alpha\beta(\fourIdx{}{\alpha}{}{}{s})_{j}+(\alpha+\beta)(\fourIdx{}{\alpha}{}{}{s})_{j+1}-(\fourIdx{}{\alpha}{}{}{s})_{j+2}$ holds true for all $j\in\mathbb{Z}_{0,2n-2}$, then $-\alpha\beta(\fourIdx{}{\alpha}{}{}{H})_{n-1}+(\alpha+\beta)(\fourIdx{}{\alpha}{}{}{K})_{n-1}-(\fourIdx{}{\alpha}{}{}{G})_{n-1}\in\mathbb{C}_{\geq}^{nq\times nq}$ and $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$ follow. Applying Lemma 8.23(b) to the sequence $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}$, we obtain from Remark 4.12 that $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2n-1}$ belongs to $\mathcal{L}^{\geq}_{q,2n-1,\beta}$. In particular, $\beta(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}-(\fourIdx{}{\alpha}{}{\beta}{K})_{n-1}\in\mathbb{C}_{\geq}^{nq\times nq}$. Since Remark 4.12 yields $\beta(\fourIdx{}{\alpha}{}{\beta}{s})_{j}-(\fourIdx{}{\alpha}{}{\beta}{s})_{j+1}=-\alpha\beta(\fourIdx{}{}{}{\beta}{s})_{j}+(\alpha+\beta)(\fourIdx{}{}{}{\beta}{s})_{j+1}-(\fourIdx{}{}{}{\beta}{s})_{j+2}$ for all $j\in\mathbb{Z}_{0,2n-2}$, we conclude then $-\alpha\beta(\fourIdx{}{}{}{\beta}{H})_{n-1}+(\alpha+\beta)(\fourIdx{}{}{}{\beta}{K})_{n-1}-(\fourIdx{}{}{}{\beta}{G})_{n-1}\in\mathbb{C}_{\geq}^{nq\times nq}$ and $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$ and, because of Remark 4.12, furthermore $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$.

Now we consider the case that $\kappa=2n$ with some $n\in\mathbb{N}$. Then the matrices $H_{n}$ and $(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}$ are both non-negative Hermitian. Proposition 7.7(a) provides us $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Hence, $\{(\fourIdx{}{\alpha}{}{}{H})_{n-1}(\fourIdx{}{}{}{\beta}{H})_{n-1}\}\in\mathbb{C}_{\geq}^{nq\times nq}$. Thus, $(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq}_{q,2n,\alpha}\cap\mathcal{L}^{\geq}_{q,2n,\beta}$. Applying Lemma 8.23(a) to $(s_{j})_{j=0}^{2n}$, we infer $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq}_{q,2n-1,\alpha}$. Thus, $-\alpha(\fourIdx{}{\alpha}{}{}{H})_{n-1}+(\fourIdx{}{\alpha}{}{}{K})_{n-1}\in\mathbb{C}_{\geq}^{nq\times nq}$. Since Remark 4.14 shows that $\beta(\fourIdx{}{\alpha}{}{}{H})_{n-1}-(\fourIdx{}{\alpha}{}{}{K})_{n-1}=(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}$ is valid, from Definition 4.2(a) we get $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Lemma 8.23(b) provides us $(\fourIdx{}{}{}{\beta}{s})_{j=0}^{2n-1}\in\mathcal{L}^{\geq}_{q,2n-1,\beta}$. Consequently, the matrix $\beta(\fourIdx{}{}{}{\beta}{H})_{n-1}-(\fourIdx{}{}{}{\beta}{K})_{n-1}$ is non-negative Hermitian. Since Remark 4.14 shows that $-\alpha(\fourIdx{}{}{}{\beta}{H})_{n-1}+(\fourIdx{}{}{}{\beta}{K})_{n-1}=(\fourIdx{}{\alpha}{}{\beta}{H})_{n-1}$ holds true, then Definition 4.2(a) shows that $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$.

(b) In view of Remark 4.12, this follows by applying part (a) to one of the sequences $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{\kappa-1}$ or $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{\kappa-1}$ which, according to part (a), belong to $\mathcal{F}^{\geq}_{q,\kappa-1,\alpha,\beta}$. ∎

Obviously, $\mathcal{F}^{\geq}_{q,0,\alpha,\beta}=\mathcal{K}^{\geq,\mathrm{e}}_{q,0,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,0,\beta}$ and $\mathcal{F}^{\geq}_{q,0,\alpha,\beta}=\mathcal{K}^{\geq}_{q,0,\alpha}\cap\mathcal{L}^{\geq}_{q,0,\beta}$ for all $\alpha,\beta\in\mathbb{R}$ with $\alpha<\beta$.

Proposition 9.2**.**

Let $\alpha\in\mathbb{R}$ and let $\beta\in(\alpha,\infty)$. Then $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}\subseteq\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$ for all $m\in\mathbb{N}_{0}$.

Proof.

The case $m=0$ is trivial. Let $m\in\mathbb{N}$ and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$.

We consider now the case that $m=2n+1$ with some $n\in\mathbb{N}_{0}$. From Lemma 7.5 we know that $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n+1}$. Furthermore, Definition 4.2(a) yields $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}\}\subseteq\mathcal{H}^{\geq}_{q,2n}$. Thus, Remark 8.16 shows that $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{K}^{\geq,\mathrm{e}}_{q,2n+1,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,2n+1,\beta}$.

It remains to consider the case that $m=2n$ with some $n\in\mathbb{N}$. Since $(s_{j})_{j=0}^{2n}$ belongs to $\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$, from Definition 4.2(b) we have $(s_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$. Proposition 9.1(a) yields $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-1},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\}\subseteq\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Consequently, Lemma 7.5 provides us $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-1},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\}\subseteq\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$. Hence, Remark 8.16 yields that $(s_{j})_{j=0}^{2n}$ belongs to $\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}$. ∎

Alternative proof of Proposition 9.2.

Let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. In view of Theorem 4.3, then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\neq\emptyset$. Defining $\sigma_{[\alpha]}\colon{\mathfrak{B}_{{[\alpha,\infty)}}}\to\mathbb{C}^{q\times q}$ by $\sigma_{[\alpha]}(B)\vcentcolon=\sigma(B\cap[\alpha,\beta])$ and $\sigma^{[\beta]}\colon{\mathfrak{B}_{(-\infty,\beta]}}\to\mathbb{C}^{q\times q}$ by $\sigma^{[\beta]}(C)\vcentcolon=\sigma(C\cap[\alpha,\beta])$, we immediately see $\sigma_{[\alpha]}\in{\mathcal{M}_{\geq}^{q}[{[\alpha,\infty)};(s_{j})_{j=0}^{m},=]}$ and $\sigma^{[\beta]}\in{\mathcal{M}_{\geq}^{q}[{(-\infty,\beta]};(s_{j})_{j=0}^{m},=]}$, resp. Thus, Theorems 8.2 and 8.11 provide $(s_{j})_{j=0}^{m}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,m,\alpha}$ and $(s_{j})_{j=0}^{m}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,m,\beta}$. ∎

Proposition 9.3**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $n\in\mathbb{N}_{0}$. Then $\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}=\mathcal{K}^{\geq,\mathrm{e}}_{q,2n+1,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,2n+1,\beta}$.

Proof.

In view of Definition 4.2(a), Remark 8.16 provides us $\mathcal{K}^{\geq,\mathrm{e}}_{q,2n+1,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,2n+1,\beta}\subseteq\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$. Thus, the application of Proposition 9.2 completes the proof. ∎

In view of Proposition 9.2, the following example shows that $\mathcal{F}^{\geq}_{1,2,\alpha,\beta}$ is a proper subset of $\mathcal{K}^{\geq,\mathrm{e}}_{1,2,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{1,2,\beta}$ if $\alpha$ and $\beta$ are real numbers such that $\alpha<\beta$.

Example 1*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let

[TABLE]

Then H_{1}=\bigl{[}\begin{smallmatrix}2&0\\ \alpha+\beta&\sqrt{3}(\beta-\alpha)\end{smallmatrix}\bigr{]}\bigl{[}\begin{smallmatrix}2&0\\ \alpha+\beta&\sqrt{3}(\beta-\alpha)\end{smallmatrix}\bigr{]}^{*} is positive Hermitian and $L_{1}=3(\beta-\alpha)$, $(\fourIdx{}{\alpha}{}{}{H})_{0}=(\fourIdx{}{\alpha}{}{}{L})_{0}=2(\beta-\alpha)$, and $(\fourIdx{}{}{}{\beta}{H})_{0}=(\fourIdx{}{}{}{\beta}{L})_{0}=2(\beta-\alpha)$ are positive. Thus, parts (c) and (d) of Proposition 8.17 show that $(s_{j})_{j=0}^{2}$ belongs to $\mathcal{K}^{\geq,\mathrm{e}}_{1,2,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{1,2,\beta}$. However, taking into account that $(\fourIdx{}{\alpha}{}{\beta}{H})_{0}=-2(\beta-\alpha)^{2}$ is negative, we see that $(s_{j})_{j=0}^{2}\notin\mathcal{F}^{\geq}_{1,2,\alpha,\beta}$.

10 On the structure of $[\alpha,\beta]$-Hausdorff non-negative definite sequences

In this section, we discuss essential aspects of the structure of the elements of the set $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. In particular, we will see that each element of such a sequence varies within a closed matricial subinterval of $\mathbb{C}_{\mathrm{H}}^{q\times q}$ the endpoints of which are completely determined by the preceding elements of the sequence. First we introduce two types of matricial intervals:

Let $A,B\in\mathbb{C}_{\mathrm{H}}^{q\times q}$. Then we set

[TABLE]

The matrix $A$ (resp. $B$) is called the matricial left (resp. right) endpoint of $[A,B]$ or $(A,B)$. The matrix $C\vcentcolon=\frac{1}{2}(A+B)$ (resp. $D\vcentcolon=B-A$) is called the matricial midpoint (resp. length) of $[A,B]$ or $(A,B)$.

The following result indicates that closed and open matricial intervals are intimately related to closed and open matricial matrix balls.

Lemma 10.1**.**

Let $A,B\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ and let $D\vcentcolon=B-A$. Then:

1. (a)

$[A,B]\neq\emptyset$* if and only if $D\in\mathbb{C}_{\geq}^{q\times q}$. In this case,*

[TABLE] 2. (b)

$(A,B)\neq\emptyset$* if and only if $D\in\mathbb{C}_{>}^{q\times q}$. In this case,*

[TABLE]

Proof.

(a) If $[A,B]\neq\emptyset$, then there exists a matrix $X\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ with $A\leq X\leq B$. In particular, $A\leq B$, i. e., $D\in\mathbb{C}_{\geq}^{q\times q}$. Conversely, assume $D\in\mathbb{C}_{\geq}^{q\times q}$. Obviously, $[0_{q\times q},I_{q}]\neq\emptyset$. Let $K\in[0_{q\times q},I_{q}]$. Then $X\vcentcolon=A+\sqrt{D}K\sqrt{D}$ is Hermitian. By virtue of Remark A.2, furthermore $X-A=\sqrt{D}K\sqrt{D}\geq 0_{q\times q}$ and $B-X=D-\sqrt{D}K\sqrt{D}=\sqrt{D}(I_{q}-K)\sqrt{D}\geq 0_{q\times q}$. Hence, $X\in[A,B]$. In particular, $[A,B]\neq\emptyset$. Now let $X\in[A,B]$. In view of Remark A.8, then $K\vcentcolon=\sqrt{D}^{\dagger}(X-A)\sqrt{D}^{\dagger}$ is Hermitian and we have

[TABLE]

Hence, $\mathcal{R}(X-A)\subseteq\mathcal{R}(D)$ and $\mathcal{N}(D)\subseteq\mathcal{N}(X-A)$ by virtue of Remark A.11. According to Remark A.10, thus $\mathcal{R}(X-A)\subseteq\mathcal{R}(\sqrt{D})$ and $\mathcal{N}(\sqrt{D})\subseteq\mathcal{N}(X-A)$. Using Remarks A.5 and A.6, we obtain then $\sqrt{D}K\sqrt{D}=X-A$. Hence, $X=A+\sqrt{D}K\sqrt{D}$. With Remark A.2, we conclude from (10.1) that $0_{q\times q}\leq K\leq\sqrt{D}^{\dagger}D\sqrt{D}^{\dagger}$. Using Remarks A.8 and A.3 and Proposition A.4, we have

[TABLE]

Thus, $0_{q\times q}\leq K\leq I_{q}$, i. e., $K\in[0_{q\times q},I_{q}]$.

(b) Since every positive Hermitian matrix is invertible, this can be shown more easy in a similar manner. ∎

From Lemma 10.1 we conclude the following result:

Remark 10.2*.*

Let $A,B\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ and let $C\vcentcolon=\frac{1}{2}(A+B)$ and $D\vcentcolon=B-A$. Then:

1. (a)

If $[A,B]\neq\emptyset$, then $\{A,C,B\}\subseteq\{A+\eta D\colon\eta\in[0,1]\}\subseteq[A,B]$. 2. (b)

If $(A,B)\neq\emptyset$, then $C\in\{A+\eta D\colon\eta\in(0,1)\}\subseteq(A,B)$.

Now we are going to introduce those sequences of matrices which will be needed to describe that sequences of matricial intervals which turn out to be associated with an $[\alpha,\beta]$-Hausdorff non-negative definite sequence:

Definition 10.3**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. For all $k\in\mathbb{N}_{0}$ with $2k\leq\kappa$, let

[TABLE]

For all $k\in\mathbb{N}_{0}$ with $2k+1\leq\kappa$, let

[TABLE]

Then the sequences $(\mathfrak{a}_{j})_{j=0}^{\kappa}$ and $(\mathfrak{b}_{j})_{j=0}^{\kappa}$ are called the sequence of left matricial interval endpoints associated with $(s_{j})_{j=0}^{\kappa}$ and $[\alpha,\beta]$ and the sequence of right matricial interval endpoints associated with $(s_{j})_{j=0}^{\kappa}$ and $[\alpha,\beta]$, resp.

Remark 10.4*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then $\mathfrak{a}_{0}=\alpha s_{0}$ and $\mathfrak{b}_{0}=\beta s_{0}$. If $\kappa\geq 1$, further $\mathfrak{a}_{1}=s_{1}s_{0}^{\dagger}s_{1}$ and $\mathfrak{b}_{1}=-\alpha\beta s_{0}+(\alpha+\beta)s_{1}$.

Remark 10.5*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Because of Definition 10.3, Notation 3.4, and Remarks 4.17 and A.7, then $\{\mathfrak{a}_{j},\mathfrak{b}_{j}\}\subseteq\mathbb{C}_{\mathrm{H}}^{q\times q}$ for all $j\in\mathbb{Z}_{0,\kappa}$.

Definition 10.6**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then the sequences $(A_{j})_{j=0}^{\kappa}$ and $(B_{j})_{j=1}^{\kappa}$ given with $\mathfrak{a}_{-1}\vcentcolon=0_{q\times q}$ by

[TABLE]

are called the sequence of lower Schur complements associated with $(s_{j})_{j=0}^{\kappa}$ and $[\alpha,\beta]$ and the sequence of upper Schur complements associated with $(s_{j})_{j=0}^{\kappa}$ and $[\alpha,\beta]$, resp.

Taking into account that Proposition 10.15(a) will show $\mathfrak{b}_{j-1}-\mathfrak{a}_{j-1}\in\mathbb{C}_{\geq}^{q\times q}$, the matrices $A_{j}$ and $B_{j}$ introduced in Definition 10.6 determine the position of the matrix $s_{j}$ in the interval $[\mathfrak{a}_{j-1},\mathfrak{b}_{j-1}]$.

Remark 10.7*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then $A_{0}=s_{0}$. If $\kappa\geq 1$, then furthermore $A_{1}=(\fourIdx{}{\alpha}{}{}{s})_{0}$ and $B_{1}=(\fourIdx{}{}{}{\beta}{s})_{0}$. If $\kappa\geq 2$, then moreover $B_{2}=(\fourIdx{}{\alpha}{}{\beta}{s})_{0}$.

Remark 10.8*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then $A_{2n}=L_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n\leq\kappa$ and $A_{2n+1}=(\fourIdx{}{\alpha}{}{}{L})_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$. Furthermore, $B_{2n+1}=(\fourIdx{}{}{}{\beta}{L})_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+1\leq\kappa$ and $B_{2n+2}=(\fourIdx{}{\alpha}{}{\beta}{L})_{n}$ for all $n\in\mathbb{N}_{0}$ with $2n+2\leq\kappa$.

From Remark 10.8 and Notation 3.4(c) we see now that the matrices introduced in Definition 10.6 are indeed Schur complements in block Hankel matrices which are responsible for the property of belonging to the set $\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. For this reason, we had chosen the terminology introduced in Definition 10.6. In the sequel, we investigate the interplay between the ranges of two consecutive elements of the sequence of matrices introduced in Definition 10.6.

Lemma 10.9**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{q\times q}$.

1. (a)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$, then $A_{j}\in\mathbb{C}_{\geq}^{q\times q}$ for all $j\in\mathbb{Z}_{0,\kappa}$ and $B_{j}\in\mathbb{C}_{\geq}^{q\times q}$ for all $j\in\mathbb{Z}_{1,\kappa}$. 2. (b)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{>}_{q,\kappa,\alpha,\beta}$, then $A_{j}\in\mathbb{C}_{>}^{q\times q}$ for all $j\in\mathbb{Z}_{0,\kappa}$ and $B_{j}\in\mathbb{C}_{>}^{q\times q}$ for all $j\in\mathbb{Z}_{1,\kappa}$.

Proof.

Use Remark 10.8 and Corollary 7.8. ∎

In the sequel, we need detailed information on the interplay between the ranges of consecutive elements of the sequences of lower and upper Schur complements associated with an $[\alpha,\beta]$-Hausdorff non-negative definite sequence. The next result gives us a first impression on this theme.

Lemma 10.10**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then $\mathcal{R}(A_{j})+\mathcal{R}(B_{j})\subseteq\mathcal{R}(A_{j-1})\cap\mathcal{R}(B_{j-1})$ for each $j\in\mathbb{Z}_{2,\kappa}$.

Proof.

Suppose $\kappa\geq 2$. We consider an arbitrary $n\in\mathbb{N}$ with $2n\leq\kappa$. In view of Proposition 7.7(a), we have $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$. Thus, Proposition 9.2 yields $(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}$. Hence, using Proposition 8.18, we get

[TABLE]

Proposition 9.1(a) provide us $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-1},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\}\subseteq\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Hence, from Proposition 9.2 we conclude $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-1,\alpha}$ and $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-1}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n-1,\beta}$. Using Remark 4.12 and Proposition 8.18, we infer then

[TABLE]

From (10.2), (10.3), and Remark 10.8 we get $\mathcal{R}(A_{2n})+\mathcal{R}(B_{2n})\subseteq\mathcal{R}(A_{2n-1})\cap\mathcal{R}(B_{2n-1})$.

Now we consider an arbitrary $n\in\mathbb{N}$ with $2n+1\leq\kappa$. Then Proposition 7.7(a) shows that $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$. According to Proposition 9.2, this implies $(s_{j})_{j=0}^{2n+1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n+1,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,2n+1,\beta}$. Thus, from Proposition 8.18 we obtain

[TABLE]

Proposition 9.1(a) yields $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}\}\subseteq\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$. Consequently, Proposition 9.2 provides us $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n-1,\beta}$ and $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}$. Taking into account Remark 4.12, then Proposition 8.18 shows that $\mathcal{R}((\fourIdx{}{\alpha}{}{}{L})_{n})\cup\mathcal{R}((\fourIdx{}{}{}{\beta}{L})_{n})\subseteq\mathcal{R}((\fourIdx{}{\alpha}{}{\beta}{L})_{n-1})$. Therefore, in view of (10.4) and Remark 10.8, we get then the inclusion $\mathcal{R}(A_{2n+1})+\mathcal{R}(B_{2n+1})\subseteq\mathcal{R}(A_{2n})\cap\mathcal{R}(B_{2n})$. ∎

Corollary 10.19 will show, that in the situation of Lemma 10.10 we even have the equality $\mathcal{R}(A_{j})+\mathcal{R}(B_{j})=\mathcal{R}(A_{j-1})\cap\mathcal{R}(B_{j-1})$ for all $j\in\mathbb{Z}_{2,\kappa}$.

Definition 10.11**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then the sequences $(\mathfrak{c}_{j})_{j=0}^{\kappa}$ and $(\mathfrak{d}_{j})_{j=0}^{\kappa}$ given by

[TABLE]

are called the sequence of matricial interval midpoints associated with $(s_{j})_{j=0}^{\kappa}$ and $[\alpha,\beta]$ and the sequence of matricial interval lengths associated with $(s_{j})_{j=0}^{\kappa}$ and $[\alpha,\beta]$, resp.

Remark 10.12*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. If $m\in\mathbb{Z}_{0,\kappa}$, then it is easily checked that $(\mathfrak{d}_{j})_{j=0}^{m}$ is the sequence of matricial interval lengths associated with $(s_{j})_{j=0}^{m}$ and $[\alpha,\beta]$. Obviously, $\mathfrak{d}_{0}=(\beta-\alpha)s_{0}$. If $\kappa\geq 1$, then $\mathfrak{d}_{1}=-\alpha\beta s_{0}+(\alpha+\beta)s_{1}-s_{1}s_{0}^{\dagger}s_{1}$ and Definitions 10.11 and 10.6 also show that $\mathfrak{d}_{j}=A_{j+1}+B_{j+1}$ for all $j\in\mathbb{Z}_{0,\kappa-1}$.

Remark 10.13*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Because of Definition 10.12 and Remark 10.5, then $\{\mathfrak{c}_{j},\mathfrak{d}_{j}\}\subseteq\mathbb{C}_{\mathrm{H}}^{q\times q}$ for all $j\in\mathbb{Z}_{0,\kappa}$.

The next result is of fundamental importance for our further considerations. Again we make essential use of the parallel sum of matrices:

Theorem 10.14**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then $\mathfrak{d}_{0}=(\beta-\alpha)A_{0}$. If $\kappa\geq 1$, then $\mathfrak{d}_{k}=(\beta-\alpha)(A_{k}\mathbin{\overline{\pm}}B_{k})$ and $\mathfrak{d}_{k}=(\beta-\alpha)(B_{k}\mathbin{\overline{\pm}}A_{k})$ for all $k\in\mathbb{Z}_{1,\kappa}$.

Proof.

In view of Notation 3.4(c) and Definition 10.6, we have $s_{0}=L_{0}=A_{0}$ and, consequently, $\mathfrak{d}_{0}=\mathfrak{b}_{0}-\mathfrak{a}_{0}=(\beta-\alpha)s_{0}$. Assume $\kappa\geq 1$. According to Lemma 10.9(a), we know that $\{A_{k},B_{k}\}\subseteq\mathbb{C}_{\geq}^{q\times q}$ for each $k\in\mathbb{Z}_{1,\kappa}$. Thus, Lemmata B.2 and B.3 provide us $A_{k}\mathbin{\overline{\pm}}B_{k}=B_{k}\mathbin{\overline{\pm}}A_{k}$ for each $k\in\mathbb{Z}_{1,\kappa}$. From Remark 10.12 we see that

[TABLE]

is valid for each $k\in\mathbb{Z}_{1,\kappa}$. Using Proposition 7.7(a) and Remark 7.6, we conclude that $(s_{j})_{j=0}^{1}$ belongs to $\mathcal{D}_{q\times q,1}$. Hence, Remarks A.5 and A.6 yield $s_{0}s_{0}^{\dagger}s_{1}=s_{1}$ and $s_{1}s_{0}^{\dagger}s_{0}=s_{1}$. Taking into account (10.5) and Remarks 10.8 and 10.12, we obtain then

[TABLE]

In the cases $\kappa=0$ and $\kappa=1$, the proof is complete. Now suppose $\kappa\geq 2$. We assume that $m\in\mathbb{Z}_{2,\kappa}$ and that

[TABLE]

Since the matrices $A_{m-1}$ and $B_{m-1}$ are both non-negative Hermitian, from Lemma B.2 and Proposition B.5 we get

[TABLE]

where $\mathcal{R}_{m-1}\vcentcolon=\mathcal{R}(A_{m-1})\cap\mathcal{R}(B_{m-1})$. By virtue of Lemma 10.10, we infer $\mathcal{R}(A_{m})+\mathcal{R}(B_{m})\subseteq\mathcal{R}_{m-1}$ and, consequently, $P_{\mathcal{R}_{m-1}}A_{m}=A_{m}$ and $P_{\mathcal{R}_{m-1}}B_{m}=B_{m}$. Since the matrices $P_{\mathcal{R}_{m-1}}$ and $A_{m}$ are Hermitian, then $A_{m}P_{\mathcal{R}_{m-1}}=A_{m}$ also holds true. Using (10.5), (10.6), and (10.7), it follows

[TABLE]

First we consider now the case that $m=2n$ with some positive integer $n$. Then from (10.8) and Remark 10.8 we obtain

[TABLE]

According to Proposition 7.7(a), the sequence $(s_{j})_{j=0}^{2n}$ belongs to $\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$. Thus, Proposition 9.2 yields $(s_{j})_{j=0}^{2n}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}$. Hence, Lemma 8.22 shows that (8.9) and (8.10) are true. This implies

[TABLE]

Since $(s_{j})_{j=0}^{2n}$ belongs to $\mathcal{K}^{\geq,\mathrm{e}}_{q,2n,\alpha}\cap\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}$, we get from Proposition 8.17 that $\mathcal{N}((\fourIdx{}{\alpha}{}{}{L})_{n-1})\cup\mathcal{N}((\fourIdx{}{}{}{\beta}{L})_{n-1})\subseteq\mathcal{N}(L_{n})$. Consequently, Remark A.6 shows that

[TABLE]

Remark 4.12 and Notation 3.4 yield

[TABLE]

Using (10.11) for $n=1$, from (10.12) and (10.13) we obtain then $L_{1}(\fourIdx{}{\alpha}{}{}{L})_{0}^{\dagger}(\fourIdx{}{\alpha}{}{\beta}{L})_{0}=L_{1}(\fourIdx{}{\alpha}{}{}{L})_{0}^{\dagger}[(\fourIdx{}{\alpha}{}{}{M})_{0}-(\fourIdx{}{\alpha}{}{}{s})_{1}]+\beta L_{1}$ and $L_{1}(\fourIdx{}{}{}{\beta}{L})_{0}^{\dagger}(\fourIdx{}{\alpha}{}{\beta}{L})_{0}=L_{1}(\fourIdx{}{}{}{\beta}{L})_{0}^{\dagger}[(\fourIdx{}{}{}{\beta}{s})_{1}-(\fourIdx{}{}{}{\beta}{M})_{0}]-\alpha L_{1}$. This implies

[TABLE]

Comparing (10.9) and (10.10) for $n=1$ with (10.14), we conclude $\mathfrak{d}_{2}=(\beta-\alpha)(A_{2}\mathbin{\overline{\pm}}B_{2})$.

Now we suppose that $n\geq 2$. From Proposition 7.7(a) we know that $(s_{j})_{j=0}^{2n-1}$ belongs to $\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Hence, Proposition 9.1(a) implies $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-2},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-2}\}\subseteq\mathcal{F}^{\geq}_{q,2n-2,\alpha,\beta}$. Consequently, because of Proposition 9.2, we have $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-2}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n-2,\beta}$ and $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-2}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-2,\alpha}$. In view Remark 4.12, the application of Lemma 8.22(b) to the sequences $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-2}$ and of Lemma 8.22(a) to the sequences $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-2}$ provides us

[TABLE]

and, analogously,

[TABLE]

By virtue of Remark 4.12, we get

[TABLE]

and, analogously,

[TABLE]

Using (10.11), we conclude then

[TABLE]

and

[TABLE]

This implies

[TABLE]

Comparing the foregoing equation with (10.9) and (10.10), we see that the equation $\mathfrak{d}_{2n}=(\beta-\alpha)(A_{2n}\mathbin{\overline{\pm}}B_{2n})$ holds true.

It remains to consider the case that $m=2n+1$ with some $n\in\mathbb{N}$. Because of Remark 10.8 and (10.8), we have

[TABLE]

From Proposition 7.7(a) we know that $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$. Proposition 9.1(a) shows that $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$. Consequently, Proposition 9.2 yields $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}$. Using Remark 4.12 and applying Lemma 8.22(b) to the sequence $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}$, we get

[TABLE]

Since $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$, we see from Proposition 9.2 that $(s_{j})_{j=0}^{2n+1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n+1,\alpha}$ is valid. Thus, Lemma 8.20(a) provides us

[TABLE]

In view of Definitions 10.11 and 10.3, (10.16), and (10.17), we conclude

[TABLE]

According to Proposition 7.7(a), the sequence $(s_{j})_{j=0}^{2n}$ belongs to $\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$. Consequently, Proposition 9.2 shows that $(s_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}$. Therefore, because of Lemma 8.22(b), equation (8.10) holds true. By virtue of Notation 3.4(c) and (8.10), we have then

[TABLE]

Since $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{K}^{\geq,\mathrm{e}}_{q,2n+1,\alpha}$, we see from Proposition 8.17(a) that $\mathcal{N}(L_{n})\subseteq\mathcal{N}((\fourIdx{}{\alpha}{}{}{L})_{n})$. Hence, Remark A.6 shows that $(\fourIdx{}{\alpha}{}{}{L})_{n}L_{n}^{\dagger}L_{n}=(\fourIdx{}{\alpha}{}{}{L})_{n}$. Thus, because of (10.19), we obtain

[TABLE]

Taking $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$ and Proposition 9.1(a) into account, we get $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. By virtue of Proposition 9.2, this implies $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\in\mathcal{K}^{\geq,\mathrm{e}}_{q,2n-1,\alpha}$. Applying Lemma 8.20(a) to the sequence $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}$ and using Remark 4.12, we infer

[TABLE]

In view of Notation 3.4(c) and Remark 4.12, then

[TABLE]

follows. Since $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n}\in\mathcal{L}^{\geq,\mathrm{e}}_{q,2n,\beta}$ is true, Proposition 8.17(d) yields $\mathcal{N}(((_{\alpha}L)_{\beta})_{n-1})\subseteq\mathcal{N}((\fourIdx{}{\alpha}{}{}{L})_{n})$. Since Remark 4.12 shows that $((_{\alpha}L)_{\beta})_{n-1}=(\fourIdx{}{\alpha}{}{\beta}{L})_{n-1}$, then Remark A.6 yields $(\fourIdx{}{\alpha}{}{}{L})_{n}(\fourIdx{}{\alpha}{}{\beta}{L})_{n-1}^{\dagger}(\fourIdx{}{\alpha}{}{\beta}{L})_{n-1}=(\fourIdx{}{\alpha}{}{}{L})_{n}$. Consequently, from (10.21) we obtain

[TABLE]

Combining (10.20) and (10.22), we conclude

[TABLE]

Comparing (10.23) with (10.18) and (10.15), we get $\mathfrak{d}_{2n+1}=(\beta-\alpha)(A_{2n+1}\mathbin{\overline{\pm}}B_{2n+1})$. ∎

The following result contains an important property of the interval lengths introduced in Definition 10.11:

Proposition 10.15**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{q\times q}$. Then:

1. (a)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$, then $\mathfrak{d}_{j}\in\mathbb{C}_{\geq}^{q\times q}$ for each $j\in\mathbb{Z}_{0,\kappa}$. 2. (b)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{>}_{q,\kappa,\alpha,\beta}$, then $\mathfrak{d}_{j}\in\mathbb{C}_{>}^{q\times q}$ for each $j\in\mathbb{Z}_{0,\kappa}$.

Proof.

Use Theorem 10.14, Lemmata 10.9 and B.2, and Remark B.1. ∎

Observe that Proposition 10.15 is the reason that $(\mathfrak{d}_{j})_{j=0}^{\kappa}$ is said to be the sequence of matricial interval lengths associated with $(s_{j})_{j=0}^{\kappa}$ and $[\alpha,\beta]$.

Proposition 10.16**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m+1}$ be a sequence from $\mathbb{C}^{q\times q}$. Then:

1. (a)

If $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}$, then $s_{m+1}\in[\mathfrak{a}_{m},\mathfrak{b}_{m}]$. 2. (b)

If $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{>}_{q,m+1,\alpha,\beta}$, then $s_{m+1}\in(\mathfrak{a}_{m},\mathfrak{b}_{m})$.

Proof.

Suppose $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. According to Remark 10.5, we have $\mathfrak{a}_{m}^{*}=\mathfrak{a}_{m}$ and $\mathfrak{b}_{m}^{*}=\mathfrak{b}_{m}$. Proposition 7.10 provides us $(s_{j})_{j=0}^{m}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$. From $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}$ and Lemma 4.17 we know that $s_{m+1}^{*}=s_{m+1}$. Since $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}$ and Lemma 10.9(a) show that $\{A_{m+1},B_{m+1}\}\subseteq\mathbb{C}_{\geq}^{q\times q}$ holds true, in view of Definition 10.6, then (a) follows. If $(s_{j})_{j=0}^{m}$ even belongs to $\mathcal{F}^{>}_{q,m,\alpha,\beta}$, Lemma 10.9(b) shows that $A_{m+1}$ and $B_{m+1}$ are positive Hermitian. In view of Definition 10.6, then (b) follows. ∎

Corollary 10.17**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}$ be a sequence from $\mathbb{C}^{q\times q}$. Then:

1. (a)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$,then,

[TABLE]

for all $k\in\mathbb{N}$ with $2k\leq\kappa$, and

[TABLE]

for all $k\in\mathbb{N}$ with $2k+1\leq\kappa$. 2. (b)

If $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{>}_{q,\kappa,\alpha,\beta}$,then,

[TABLE]

for all $k\in\mathbb{Z}_{2,\infty}$ with $2k\leq\kappa$, and

[TABLE]

for all $k\in\mathbb{Z}_{2,\infty}$ with $2k+1\leq\kappa$.

Proof.

Suppose $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{>}_{q,\kappa,\alpha,\beta}$). In view of Lemma 4.17 and Remarks 10.5 and 10.13, all the matrices involved are Hermitian. We consider a number $k\in\mathbb{N}$ (resp. $k\in\mathbb{Z}_{2,\infty}$).

First assume that $2k\leq\kappa$. Because of Proposition 10.15, then $\mathfrak{d}_{2k}$ is non-negative Hermitian (resp. positive Hermitian). Since $\mathfrak{c}_{2k}-\mathfrak{a}_{2k}=\frac{1}{2}\mathfrak{d}_{2k}$ and $\mathfrak{b}_{2k}-\mathfrak{c}_{2k}=\frac{1}{2}\mathfrak{d}_{2k}$, by virtue of Definition 10.11, the two inner inequalities in (10.24) (resp. (10.26)) follow. According to Proposition 7.7, we have $(s_{j})_{j=0}^{2k-1}\in\mathcal{F}^{\geq}_{q,2k-1,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2k-1}\in\mathcal{F}^{>}_{q,2k-1,\alpha,\beta}$). In view of Definition 4.2(a), hence $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2k-2},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2k-2}\}\subseteq\mathcal{H}^{\geq}_{q,2k-2}$ (resp. $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2k-2},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2k-2}\}\subseteq\mathcal{H}^{>}_{q,2k-2}$). Since the matrices $(\fourIdx{}{\alpha}{}{}{s})_{2k-1}$ and $(\fourIdx{}{}{}{\beta}{s})_{2k-1}$ are, according to Lemma 4.17 and Notation 4.1, both Hermitian, we see from Lemma 3.8 that $(\fourIdx{}{\alpha}{}{}{\Theta})_{k}$ and $(\fourIdx{}{}{}{\beta}{\Theta})_{k}$ are both non-negative Hermitian (resp. positive Hermitian). Consequently, the two outer inequalities in (10.24) (resp. (10.26)) can be seen from Definition 10.3.

Now assume that $2k+1\leq\kappa$. Because of Proposition 10.15, then $\mathfrak{d}_{2k+1}$ is non-negative Hermitian (resp. positive Hermitian). Since $\mathfrak{c}_{2k+1}-\mathfrak{a}_{2k+1}=\frac{1}{2}\mathfrak{d}_{2k+1}$ and $\mathfrak{b}_{2k+1}-\mathfrak{c}_{2k+1}=\frac{1}{2}\mathfrak{d}_{2k+1}$ by virtue of Definition 10.11, the two inner inequalities in (10.25) (resp. (10.27)) follow. According to Proposition 7.7, we have $(s_{j})_{j=0}^{2k}\in\mathcal{F}^{\geq}_{q,2k,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2k}\in\mathcal{F}^{>}_{q,2k,\alpha,\beta}$). In view of Definition 4.2(b), hence $(s_{j})_{j=0}^{2k}\in\mathcal{H}^{\geq}_{q,2k}$ and $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2k-2}\in\mathcal{H}^{\geq}_{q,2k-2}$ (resp. $(s_{j})_{j=0}^{2k}\in\mathcal{H}^{>}_{q,2k}$ and $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2k-2}\in\mathcal{H}^{>}_{q,2k-2}$). Since the matrices $s_{2k+1}$ and $(\fourIdx{}{\alpha}{}{\beta}{s})_{2k-1}$ are, according to Lemma 4.17 and Notation 4.1, both Hermitian, we see from Lemma 3.8 that $\Theta_{k+1}$ and $(\fourIdx{}{\alpha}{}{\beta}{\Theta})_{k}$ are both non-negative Hermitian (resp. positive Hermitian). Consequently, the two outer inequalities in (10.25) (resp. (10.27)) can be seen from Definition 10.3. ∎

Proposition 10.18**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}_{0}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then $\mathcal{R}(\mathfrak{d}_{0})=\mathcal{R}(A_{0})$ and $\mathcal{N}(\mathfrak{d}_{0})=\mathcal{N}(A_{0})$. If $\kappa\geq 1$, then

[TABLE]

hold true for all $j\in\mathbb{Z}_{1,\kappa}$ and

[TABLE]

are fulfilled for all $j\in\mathbb{Z}_{0,\kappa-1}$.

Proof.

Because of $\alpha<\beta$ and Theorem 10.14, we have $\mathcal{R}(\mathfrak{d}_{0})=\mathcal{R}(A_{0})$ and $\mathcal{N}(\mathfrak{d}_{0})=\mathcal{N}(A_{0})$. Assume now $\kappa\geq 1$. Then Lemma 10.9(a) provides us $\{A_{j},B_{j}\}\subseteq\mathbb{C}_{\geq}^{q\times q}$ for all $j\in\mathbb{Z}_{1,\kappa}$. Using Theorem 10.14, Lemmata B.2 and B.4, and $\alpha<\beta$, we get (10.28) for all $j\in\mathbb{Z}_{1,\kappa}$. By virtue of Remarks 10.12 and A.1, we also obtain (10.29) for each $j\in\mathbb{Z}_{0,\kappa-1}$. ∎

The following result improves Lemma 10.10.

Corollary 10.19**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then $\mathcal{R}(A_{j})+\mathcal{R}(B_{j})=\mathcal{R}(A_{j-1})\cap\mathcal{R}(B_{j-1})$ and $\mathcal{N}(A_{j-1})+\mathcal{N}(B_{j-1})=\mathcal{N}(A_{j})\cap\mathcal{N}(B_{j})$ for all $j\in\mathbb{Z}_{2,\kappa}$

Proof.

This is a direct consequence of Proposition 10.18. ∎

Corollary 10.20**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Then $\mathcal{R}(\mathfrak{d}_{j})\subseteq\mathcal{R}(\mathfrak{d}_{j-1})$, $\mathcal{N}(\mathfrak{d}_{j-1})\subseteq\mathcal{N}(\mathfrak{d}_{j})$, $\mathcal{R}(A_{j})\subseteq\mathcal{R}(A_{j-1})$, and $\mathcal{N}(A_{j-1})\subseteq\mathcal{N}(A_{j})$ for all $j\in\mathbb{Z}_{1,\kappa}$. Furthermore, if $\kappa\geq 2$, then $\mathcal{R}(B_{j})\subseteq\mathcal{R}(B_{j-1})$ and $\mathcal{N}(B_{j-1})\subseteq\mathcal{N}(B_{j})$ for each $j\in\mathbb{Z}_{2,\kappa}$.

Proof.

Proposition 10.18 provides us $\mathcal{R}(\mathfrak{d}_{j})=\mathcal{R}(A_{j})\cap\mathcal{R}(B_{j})\subseteq\mathcal{R}(A_{j})+\mathcal{R}(B_{j})=\mathcal{R}(\mathfrak{d}_{j-1})$ for all $j\in\mathbb{Z}_{1,\kappa}$ and $\mathcal{R}(A_{j})\subseteq\mathcal{R}(A_{j})+\mathcal{R}(B_{j})=\mathcal{R}(\mathfrak{d}_{j-1})=\mathcal{R}(A_{j-1})\cap\mathcal{R}(B_{j-1})\subseteq\mathcal{R}(A_{j-1})$ as well as $\mathcal{R}(B_{j})\subseteq\mathcal{R}(\mathfrak{d}_{j-1})\subseteq\mathcal{R}(B_{j-1})$ for all $j\in\mathbb{Z}_{2,\kappa}$. Proposition 10.18, Remark 10.12, and Definition 10.6 also yield $\mathcal{R}(A_{1})\subseteq\mathcal{R}(A_{1})+\mathcal{R}(B_{1})=\mathcal{R}(\mathfrak{d}_{0})=\mathcal{R}(s_{0})=\mathcal{R}(A_{0})$. The asserted inclusions for the null spaces follow analogously. ∎

Corollary 10.21**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. For all $j\in\mathbb{Z}_{1,\kappa}$, then

[TABLE]

Proof.

In view of (B.1), this is a consequence of Theorem 10.14 and Remark 10.12. ∎

Now we are going to investigate the sequence of matricial interval lengths $(\mathfrak{d}_{j})_{j=0}^{\kappa}$ associated with a sequence $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. Using Theorem 10.14 and the arithmetics of the parallel sum of matrices, we derive a formula connecting two consecutive interval lengths $\mathfrak{d}_{j+1}$ and $\mathfrak{d}_{j}$:

Proposition 10.22**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, and let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$. For all $j\in\mathbb{Z}_{0,\kappa-1}$, then

[TABLE]

Proof.

We consider an arbitrary $j\in\mathbb{Z}_{0,\kappa-1}$. According to Lemma 10.9(a), the matrices $A_{j+1}$ and $B_{j+1}$ are both non-negative Hermitian. In view of Lemmata B.2 and B.6, then $(A_{j+1}+B_{j+1})-4(A_{j+1}\mathbin{\overline{\pm}}B_{j+1})=(A_{j+1}-B_{j+1})(A_{j+1}+B_{j+1})^{\dagger}(A_{j+1}-B_{j+1})$. Using Remark 10.12 and Theorem 10.14, we get $(\beta-\alpha)\mathfrak{d}_{j}-4\mathfrak{d}_{j+1}=(\beta-\alpha)(A_{j+1}-B_{j+1})\mathfrak{d}_{j}^{\dagger}(A_{j+1}-B_{j+1})$. Since $2(s_{j+1}-\mathfrak{c}_{j})=A_{j+1}-B_{j+1}$ holds true, we obtain $4\mathfrak{d}_{j+1}=(\beta-\alpha)\mathfrak{d}_{j}-4(\beta-\alpha)(s_{j+1}-\mathfrak{c}_{j})\mathfrak{d}_{j}^{\dagger}(s_{j+1}-\mathfrak{c}_{j})$. ∎

Proposition 10.22 leads us now to an important monotonicity property for the sequence of interval lengths. Moreover, the extremal role of the midpoint of the interval in this context will be clear.

Proposition 10.23**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $\kappa\in\mathbb{N}\cup\{\infty\}$, let $(s_{j})_{j=0}^{\kappa}\in\mathcal{F}^{\geq}_{q,\kappa,\alpha,\beta}$, and let $m\in\mathbb{Z}_{1,\kappa}$. Then $\mathfrak{d}_{m}\leq\frac{\beta-\alpha}{4}\mathfrak{d}_{m-1}$. Furthermore, $\mathfrak{d}_{m}=\frac{\beta-\alpha}{4}\mathfrak{d}_{m-1}$ if and only if $s_{m}=\mathfrak{c}_{m-1}$.

Proof.

The matrices $\mathfrak{d}_{m-1}$ and $\mathfrak{d}_{m}$ are both non-negative Hermitian according to Proposition 10.15. Let $D\vcentcolon=\frac{\beta-\alpha}{4}\mathfrak{d}_{m-1}-\mathfrak{d}_{m}$. Because of Proposition 10.22, we have $D=(\beta-\alpha)(s_{m}-\mathfrak{c}_{m-1})\mathfrak{d}_{m-1}^{\dagger}(s_{m}-\mathfrak{c}_{m-1})$. Since the matrices $s_{m}$ and $\mathfrak{c}_{m-1}$ are both Hermitian by virtue of Lemma 4.17 and Remark 10.13, hence

[TABLE]

According to Remark A.10, the matrix $\mathfrak{d}_{m-1}^{\dagger}$ is non-negative Hermitian. In view of $\beta>\alpha$, then $D\in\mathbb{C}_{\geq}^{q\times q}$ follows.

Now assume $D=0_{q\times q}$. Because of $\beta>\alpha$ and $\mathfrak{d}_{m-1}^{\dagger}\in\mathbb{C}_{\geq}^{q\times q}$, we obtain from (10.30) then $\sqrt{\mathfrak{d}_{m-1}^{\dagger}}(s_{m}-\mathfrak{c}_{m-1})=0_{q\times q}$. In particular, $\mathfrak{d}_{m-1}\mathfrak{d}_{m-1}^{\dagger}(s_{m}-\mathfrak{c}_{m-1})=0_{q\times q}$. In view of Definitions 10.11 and 10.6, we get $s_{m}-\mathfrak{c}_{m-1}=\frac{1}{2}(A_{m}-B_{m})$. Hence, $\mathcal{R}(s_{m}-\mathfrak{c}_{m-1})\subseteq\mathcal{R}(A_{m})+\mathcal{R}(B_{m})$. Since $\mathcal{R}(A_{m})+\mathcal{R}(B_{m})=\mathcal{R}(\mathfrak{d}_{m-1})$ by virtue of Proposition 10.18, we have $\mathcal{R}(s_{m}-\mathfrak{c}_{m-1})\subseteq\mathcal{R}(\mathfrak{d}_{m-1})$. Using Remark A.5, we obtain $s_{m}-\mathfrak{c}_{m-1}=\mathfrak{d}_{m-1}\mathfrak{d}_{m-1}^{\dagger}(s_{m}-\mathfrak{c}_{m-1})$ and hence $s_{m}-\mathfrak{c}_{m-1}=0_{q\times q}$.

Conversely, if $s_{m}=\mathfrak{c}_{m-1}$, then $D=0_{q\times q}$ follows from (10.30). ∎

We turn our the attention to an interesting subclass of $[\alpha,\beta]$-Hausdorff non-negative definite sequences:

Definition 10.24**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then $(s_{j})_{j=0}^{m}$ is called $[\alpha,\beta]$-Hausdorff completely degenerate if $\mathfrak{d}_{m}=0_{q\times q}$, where $\mathfrak{d}_{m}$ is given in Definition 10.11. We denote by $\mathcal{F}^{\geq,\mathrm{cd}}_{q,m,\alpha,\beta}$ the set of all sequences $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ which are $[\alpha,\beta]$-Hausdorff completely degenerate.

Definition 10.25**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$.

1. (a)

Let $m\in\mathbb{N}_{0}$. Then $(s_{j})_{j=0}^{\infty}$ is called $[\alpha,\beta]$-Hausdorff completely degenerate of order $m$ if $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m,\alpha,\beta}$. 2. (b)

The sequence $(s_{j})_{j=0}^{\infty}$ is called $[\alpha,\beta]$-Hausdorff completely degenerate if there exists an $m\in\mathbb{N}_{0}$ such that $(s_{j})_{j=0}^{\infty}$ is Hankel completely degenerate of order $m$.

We denote by $\mathcal{F}^{\geq,\mathrm{cd}}_{q,\infty,\alpha,\beta}$ the set of all sequences $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ which are $[\alpha,\beta]$-Hausdorff completely degenerate.

Lemma 10.26**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then $(s_{j})_{j=0}^{m}$ is $[\alpha,\beta]$-Hausdorff completely degenerate if and only if $\mathcal{R}(A_{m})\cap\mathcal{R}(B_{m})=\{0_{q\times 1}\}$.

Proof.

Use Proposition 10.18. ∎

Lemma 10.27**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ be $[\alpha,\beta]$-Hausdorff completely degenerate of order $m$. Then:

1. (a)

If $m=2n$ with some $n\in\mathbb{N}_{0}$, then $(s_{j})_{j=0}^{\infty}$ is Hankel completely degenerate of order $n+1$ and the sequences $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{\infty}$, $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{\infty}$, and $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{\infty}$ are Hankel completely degenerate of order $n$. 2. (b)

If $m=2n+1$ with some $n\in\mathbb{N}_{0}$, then $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{\infty}$ is Hankel completely degenerate of order $n$ and the sequences $(s_{j})_{j=0}^{\infty}$, $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{\infty}$, and $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{\infty}$ are Hankel completely degenerate of order $n+1$.

Proof.

First observe that the sequences $(s_{j})_{j=0}^{\infty}$, $((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{\infty}$, $((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{\infty}$, and $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{\infty}$ belong to $\mathcal{H}^{\geq}_{q,\infty}$ by virtue of Lemma 4.16 and Proposition 9.1. In view of Definitions 10.25 and 10.24, we have $\mathfrak{d}_{m}=0_{q\times q}$. According to Proposition 10.18, we get $\mathcal{R}(A_{m+1})+\mathcal{R}(B_{m+1})=\{0_{q\times 1}\}$. In particular, $\mathcal{R}(A_{m+1})=\{0_{q\times 1}\}$ and $\mathcal{R}(B_{m+1})=\{0_{q\times 1}\}$. Using Corollary 10.20, this implies $\mathcal{R}(A_{m+2})=\{0_{q\times 1}\}$ and $\mathcal{R}(B_{m+2})=\{0_{q\times 1}\}$. Thus, each of the matrices $A_{m+1}$, $B_{m+1}$, $A_{m+2}$, and $B_{m+2}$ is the zero matrix $0_{q\times q}$.

(a) In this situation each of the matrices $A_{2n+1}$, $B_{2n+1}$, $A_{2n+2}$, and $B_{2n+2}$ is the zero matrix $0_{q\times q}$. In view of Remark 10.8, hence $(\fourIdx{}{\alpha}{}{}{L})_{n}=0_{q\times q}$, $(\fourIdx{}{}{}{\beta}{L})_{n}=0_{q\times q}$, $L_{n+1}=0_{q\times q}$, and $(\fourIdx{}{\alpha}{}{\beta}{L})_{n}=0_{q\times q}$.

(b) In this situation each of the matrices $A_{2n+2}$, $B_{2n+2}$, $A_{2n+3}$, and $B_{2n+3}$ is the zero matrix $0_{q\times q}$. In view of Remark 10.8, hence $L_{n+1}=0_{q\times q}$, $(\fourIdx{}{\alpha}{}{\beta}{L})_{n}=0_{q\times q}$, $(\fourIdx{}{\alpha}{}{}{L})_{n+1}=0_{q\times q}$, and $(\fourIdx{}{}{}{\beta}{L})_{n+1}=0_{q\times q}$. ∎

Lemma 10.28**.**

Let $\alpha\in\mathbb{R}$ and let $\beta\in(\alpha,\infty)$. Then $\mathcal{F}^{\geq,\mathrm{cd}}_{q,\infty,\alpha,\beta}\subseteq\mathcal{H}^{\geq,\mathrm{cd}}_{q,\infty}$.

Proof.

This is a direct consequence of Lemma 10.27. ∎

Lemma 10.29**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$, and let $m\in\mathbb{N}_{0}$. If $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff completely degenerate of order $m$, then $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff completely degenerate of order $\ell$ for all $\ell\in\mathbb{Z}_{m,\infty}$.

Proof.

This can be seen from Corollary 10.20 ∎

Proposition 10.30**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ be $[\alpha,\beta]$-Hausdorff completely degenerate of order $m$. Then $s_{j}=\mathfrak{c}_{j-1}=\mathfrak{a}_{j-1}=\mathfrak{b}_{j-1}$ for all $j\in\mathbb{Z}_{m+1,\infty}$.

Proof.

Let $\ell\in\mathbb{Z}_{m,\infty}$. In view of Lemma 10.29 and Definitions 10.25 and 10.24, then $\mathfrak{d}_{\ell}=0_{q\times q}$. Because of Definition 10.11, hence $\mathfrak{a}_{\ell}=\mathfrak{b}_{\ell}$ and thus $\mathfrak{c}_{\ell}=\mathfrak{a}_{\ell}$. In particular, $[\mathfrak{a}_{\ell},\mathfrak{b}_{\ell}]=\{\mathfrak{c}_{\ell}\}$. According to Definition 4.7, furthermore $(s_{j})_{j=0}^{\ell+1}\in\mathcal{F}^{\geq}_{q,\ell+1,\alpha,\beta}$. Proposition 10.16(a) yields then $s_{\ell+1}\in[\mathfrak{a}_{\ell},\mathfrak{b}_{\ell}]$. Since $[\mathfrak{a}_{\ell},\mathfrak{b}_{\ell}]=\{\mathfrak{c}_{\ell}\}$, this implies $s_{\ell+1}=\mathfrak{c}_{\ell}$. ∎

Proposition 10.31**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m,\alpha,\beta}$. Then there exists a unique sequence $(s_{j})_{j=m+1}^{\infty}$ from $\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$.

Proof.

Combine Propositions 4.9 and 10.30. ∎

Proposition 10.31 leads us to the following notion:

Definition 10.32**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m,\alpha,\beta}$. Let $(s_{j})_{j=m+1}^{\infty}$ be the unique sequence from $\mathbb{C}^{q\times q}$ such that $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$. Then $(s_{j})_{j=0}^{\infty}$ is called the $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{m}$.

The considerations of this section lead us to a further interesting subclass of $\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$:

Definition 10.33**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$. If $m\in\mathbb{N}$ is such that $s_{j}=\mathfrak{c}_{j-1}$ for all $j\in\mathbb{Z}_{m,\infty}$, where $\mathfrak{c}_{j-1}$ is given by Definition 10.11, then $(s_{j})_{j=0}^{\infty}$ is called $[\alpha,\beta]$-Hausdorff central of order $m$. If there exists an $\ell\in\mathbb{N}$ such that $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff central of order $\ell$, then $(s_{j})_{j=0}^{\infty}$ is simply called $[\alpha,\beta]$-Hausdorff central.

Remark 10.34*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$, and let $m\in\mathbb{N}$. If $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff central of order $m$, then $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff central of order $\ell$ for all $\ell\in\mathbb{Z}_{m,\infty}$.

Proposition 10.35**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}$, and let $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$. Then $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff central of order $m$ if and only if $\mathfrak{d}_{\ell}=\frac{\beta-\alpha}{4}\mathfrak{d}_{\ell-1}$ for all $\ell\in\mathbb{Z}_{m,\infty}$.

Proof.

This is a direct consequence of Proposition 10.23. ∎

A closer look at Propositions 10.23 and 10.35 shows that the role of the central sequences in the context of $[\alpha,\beta]$-Hausdorff non-negative definite sequences is comparable with the role of the central sequences in the context of non-negative definite sequences from $\mathbb{C}^{q\times q}$ or of the central sequences of ${p\times q}$ Schur sequences. On the one side we see that starting from some index the interval lengths are maximal with respect to the Löwner semi-ordering in the set $\mathbb{C}_{\mathrm{H}}^{q\times q}$. On the other side [16, Lemma 6] and [17, Lemma 7] indicate that the semi-radii of the corresponding matrix balls are maximal with respect to the Löwner semi-ordering in the set $\mathbb{C}_{\mathrm{H}}^{q\times q}$ as well for central non-negative definite sequences from $\mathbb{C}^{q\times q}$ as for central ${p\times q}$ Schur sequences, resp.

In view of Definition 10.33, we obtain from Proposition 10.30 the following result:

Remark 10.36*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ be $[\alpha,\beta]$-Hausdorff completely degenerate of order $m$. Then $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff central of order $m+1$.

Lemma 10.37**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m,\alpha,\beta}$. Then the $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{m}$ is $[\alpha,\beta]$-Hausdorff completely degenerate of order $m$ and $[\alpha,\beta]$-Hausdorff central of order $m+1$.

Proof.

The $[\alpha,\beta]$-Hausdorff completely degenerate sequence $(s_{j})_{j=0}^{\infty}$ associated to $(s_{j})_{j=0}^{m}$ belongs to $\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ by definition and fulfills $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m,\alpha,\beta}$. In view of Definition 10.25(a), hence $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff completely degenerate of order $m$. According to Remark 10.36, then $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff central of order $m+1$. ∎

From Lemmata 10.37 and 10.27 we obtain the following result:

Remark 10.38*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m,\alpha,\beta}$. Then the $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{m}$ is Hankel completely degenerate of order $n+1$, where $n$ is the unique number from $\mathbb{N}_{0}$ with $m=2n$ or $m=2n+1$.

Section 10 has some points of touch with the paper Dette/Studden [11], where the moment space of a matrix measure $\sigma\in{\mathcal{M}_{\geq}^{q}([0,1])}$ is studied and where it is assumed that the corresponding block Hankel matrices are non-singular. The strategy used in [11] is mainly based on using convexity techniques and constructing canonical representations for the points of the moment space as this was done in the classical scalar case by M. G. Krein [26] (see also Krein/Nudelman [27, Ch. 3]). An important tool used in [11] is given by a matricial generalization of the theory of classical canonical moments (see the monograph Dette/Studden [10]).

11 On the problem of $[\alpha,\beta]$-Hausdorff non-negative definite extension

Let $m\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then against to the background of Proposition 10.16 we will be able to describe the set $\{s_{m+1}\in\mathbb{C}^{q\times q}\colon(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}\}$.

Lemma 11.1**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then:

1. (a)

If $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$, then $[\mathfrak{a}_{m},\mathfrak{b}_{m}]\neq\emptyset$ and $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}$ for all $s_{m+1}\in[\mathfrak{a}_{m},\mathfrak{b}_{m}]$. 2. (b)

If $(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$, then $(\mathfrak{a}_{m},\mathfrak{b}_{m})\neq\emptyset$ and $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{>}_{q,m+1,\alpha,\beta}$ for all $s_{m+1}\in(\mathfrak{a}_{m},\mathfrak{b}_{m})$.

Proof.

Assume $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$). From Proposition 10.15 we know that $\mathfrak{d}_{m}$ it non-negative Hermitian (resp. positive Hermitian). In view of Definition 10.11, Lemma 10.1 yields then $[\mathfrak{a}_{m},\mathfrak{b}_{m}]\neq\emptyset$ (resp. $(\mathfrak{a}_{m},\mathfrak{b}_{m})\neq\emptyset$). Proposition 7.10 provides us $(s_{j})_{j=0}^{m}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,m}$ (resp. $(s_{j})_{j=0}^{m}\in\mathcal{H}^{>,\mathrm{e}}_{q,m}$). We consider an arbitrary $s_{m+1}\in[\mathfrak{a}_{m},\mathfrak{b}_{m}]$ (resp. $s_{m+1}\in(\mathfrak{a}_{m},\mathfrak{b}_{m})$). Then $s_{m+1}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ and $\mathfrak{a}_{m}\leq s_{m+1}\leq\mathfrak{b}_{m}$ (resp. $\mathfrak{a}_{m}<s_{m+1}<\mathfrak{b}_{m}$).

First we discuss the case $m=0$. In view of Remark 10.4 and Notation 4.1, then $(\fourIdx{}{\alpha}{}{}{H})_{0}$ and $(\fourIdx{}{}{}{\beta}{H})_{0}$ are both non-negative Hermitian (resp. positive Hermitian). Thus, $(s_{j})_{j=0}^{1}\in\mathcal{F}^{\geq}_{q,1,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{1}\in\mathcal{F}^{>}_{q,1,\alpha,\beta}$) follows from Definition 4.2(a).

Now let $m=1$. Then Remark 10.4 and Notation 4.1 yield $\Theta_{1}\leq s_{2}$ and $(\fourIdx{}{\alpha}{}{\beta}{H})_{0}\in\mathbb{C}_{\geq}^{q\times q}$ (resp. $\Theta_{1}<s_{2}$ and $(\fourIdx{}{\alpha}{}{\beta}{H})_{0}\in\mathbb{C}_{>}^{q\times q}$). Since $(s_{j})_{j=0}^{1}$ belongs to $\mathcal{H}^{\geq,\mathrm{e}}_{q,1}$ (resp. $\mathcal{H}^{>,\mathrm{e}}_{q,1}$), then we see from Propositions 6.7 and 6.8 that $(s_{j})_{j=0}^{2}$ belongs to $\mathcal{H}^{\geq}_{q,2}$, (resp. $\mathcal{H}^{>}_{q,2}$) i. e., the matrix $H_{1}$ is non-negative Hermitian (resp. positive Hermitian). Thus, Definition 4.2(b) shows that $(s_{j})_{j=0}^{2}\in\mathcal{F}^{\geq}_{q,2,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2}\in\mathcal{F}^{>}_{q,2,\alpha,\beta}$).

Now we consider the case that $m=2n$ is valid where $n$ is some positive integer. According to $\mathfrak{a}_{2n}\leq s_{2n+1}\leq\mathfrak{b}_{2n}$ (resp. $\mathfrak{a}_{2n}<s_{2n+1}<\mathfrak{b}_{2n}$), Definition 10.3 and Notation 4.1, then

[TABLE]

Because of Proposition 9.1(a) and Proposition 7.10, we have $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-1},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\}\subseteq\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}\subseteq\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$. For $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{>}_{q,2n,\alpha,\beta}$, we obtain $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{>}_{q,2n-1,\alpha,\beta}$ by virtue of Proposition 7.7(b) and $\{(\fourIdx{}{\alpha}{}{}{s})_{2n-1},(\fourIdx{}{}{}{\beta}{s})_{2n-1}\}\subseteq\mathbb{C}_{\mathrm{H}}^{q\times q}$ according to Lemma 4.17. Because of Definition 4.2(a) and Remark 6.6, thus $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n-1},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n-1}\}\subseteq\mathcal{H}^{>,\mathrm{e}}_{q,2n-1}$ if $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{>}_{q,2n,\alpha,\beta}$. Using (11.1) and Propositions 6.7 and 6.8, we conclude $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}\}\subseteq\mathcal{H}^{\geq}_{q,2n}$ (resp. $\{((\fourIdx{}{\alpha}{}{}{s})_{j})_{j=0}^{2n},((\fourIdx{}{}{}{\beta}{s})_{j})_{j=0}^{2n}\}\subseteq\mathcal{H}^{>}_{q,2n}$). Hence, Definition 4.2(a) implies $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{>}_{q,2n+1,\alpha,\beta}$).

Finally, we consider now the case that there is a positive integer $n$ such that $m=2n+1$. From $\mathfrak{a}_{2n+1}\leq s_{2n+2}\leq\mathfrak{b}_{2n+1}$ (resp. $\mathfrak{a}_{2n+1}<s_{2n+2}<\mathfrak{b}_{2n+1}$), Definition 10.3 and Notation 4.1 we get then

[TABLE]

From $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{>}_{q,2n+1,\alpha,\beta}$) and Proposition 7.10 we get $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n+1}$ (resp. $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{>,\mathrm{e}}_{q,2n+1}$). In view of Proposition 9.1(b) and Proposition 7.10, we have $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}\subseteq\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$. For $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{>}_{q,2n+1,\alpha,\beta}$, we obtain $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{>}_{q,2n,\alpha,\beta}$ by virtue of Proposition 7.7(b) and $(\fourIdx{}{\alpha}{}{\beta}{s})_{2n-1}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ according to Lemma 4.17 and Notation 4.1. Because of Definition 4.2(b) and Remark 6.6, thus $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2n-1}\in\mathcal{H}^{>,\mathrm{e}}_{q,2n-1}$ if $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{>}_{q,2n+1,\alpha,\beta}$. Taking into account (11.2) and Propositions 6.7 and 6.8, we infer then $(s_{j})_{j=0}^{2n+2}\in\mathcal{H}^{\geq}_{q,2n+2}$ and $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2n}\in\mathcal{H}^{\geq}_{q,2n}$ (resp. $(s_{j})_{j=0}^{2n+2}\in\mathcal{H}^{>}_{q,2n+2}$ and $((\fourIdx{}{\alpha}{}{\beta}{s})_{j})_{j=0}^{2n}\in\mathcal{H}^{>}_{q,2n}$). Because of Definition 4.2(b), we conclude $(s_{j})_{j=0}^{2n+2}\in\mathcal{F}^{\geq}_{q,2n+2,\alpha,\beta}$ (resp. $(s_{j})_{j=0}^{2n+2}\in\mathcal{F}^{>}_{q,2n+2,\alpha,\beta}$). ∎

Theorem 11.2**.**

Let $m\in\mathbb{N}_{0}$ and let $(s_{j})_{j=0}^{m}$ be a sequence from $\mathbb{C}^{q\times q}$. Then:

1. (a)

If $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$, then $[\mathfrak{a}_{m},\mathfrak{b}_{m}]\neq\emptyset$ and

[TABLE] 2. (b)

If $(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$, then $(\mathfrak{a}_{m},\mathfrak{b}_{m})\neq\emptyset$ and

[TABLE]

Proof.

Combine Lemma 11.1 and Proposition 10.16. ∎

A closer look at the proof of Theorem 11.2 shows that we now immediately obtain purely algebraic proofs of some statements formulated in Section 4, which were proved there using results on moment problems. This concerns Theorem 4.3, Proposition 4.5, Corollary 4.6, Proposition 4.9, and Theorem 4.10.

Theorem 11.3**.**

Let $\alpha\in\mathbb{R}$, $\beta\in(\alpha,\infty)$, and $m\in\mathbb{N}_{0}$. Then $\mathcal{F}^{>,\mathrm{e}}_{q,m,\alpha,\beta}=\mathcal{F}^{>}_{q,m,\alpha,\beta}$.

Proof.

Combine Corollary 7.9 and Theorem 11.2(b). ∎

Proposition 11.4**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$, and let $s_{m+1}\in\{\mathfrak{a}_{m},\mathfrak{b}_{m}\}$. Then $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m+1,\alpha,\beta}$.

Proof.

From Theorem 11.2(a) and Remark 10.2(a) we obtain $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq}_{q,m+1,\alpha,\beta}$. In view of Definition 10.6, we have furthermore $A_{m+1}=0_{q\times q}$ or $B_{m+1}=0_{q\times q}$. In particular, $\mathcal{R}(A_{m+1})\cap\mathcal{R}(B_{m+1})=\{0_{q\times 1}\}$. Thus, Lemma 10.26 yields $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m+1,\alpha,\beta}$. ∎

In view of Propositions 10.31, 11.4, and 10.30, the following notions seem to be natural:

Definition 11.5**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Let the sequence $(s_{j})_{j=m+1}^{\infty}$ be recursively defined by $s_{j}\vcentcolon=\mathfrak{a}_{j-1}$ (resp. $s_{j}\vcentcolon=\mathfrak{b}_{j-1}$). Then $(s_{j})_{j=0}^{\infty}$ is called the lower (resp. upper) $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{m}$.

Proposition 11.6**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then the lower (resp. upper) $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{m}$ belongs to $\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ and is $[\alpha,\beta]$-Hausdorff completely degenerate of order $m+1$ as well as $[\alpha,\beta]$-Hausdorff central of order $m+2$.

Proof.

Denote by $(s_{j})_{j=0}^{\infty}$ the lower (resp. upper) $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{m}$. In view of Definition 4.7 and Remark 10.2(a), we conclude $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ by successive application of Theorem 11.2(a). According to Definition 11.5 and Proposition 11.4, furthermore $(s_{j})_{j=0}^{m+1}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,m+1,\alpha,\beta}$. In view of Definition 10.25(a), thus $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff completely degenerate of order $m+1$. By virtue of Remark 10.36, then $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff central of order $m+2$. ∎

From Proposition 11.6 and Lemma 10.27 we obtain the following result:

Remark 11.7*.*

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then the lower as well as the upper $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{m}$ is Hankel completely degenerate of order $n+1$, where $n$ is the unique number from $\mathbb{N}_{0}$ with $m=2n-1$ or $m=2n$.

Lemma 11.8**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}$, and let $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Then the lower $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n-1}$ is Hankel completely degenerate of order $n$.

Proof.

Denote by $(s_{j})_{j=0}^{\infty}$ the lower $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n-1}$. According to Proposition 11.6, then $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$. In particular, $(s_{j})_{j=0}^{\infty}\in\mathcal{H}^{\geq}_{q,\infty}$ by virtue of Lemma 4.16. In view of Definition 11.5, we have furthermore $s_{2n}=\mathfrak{a}_{2n-1}$. Thus, $s_{2n}=\Theta_{n}$ according to Definition 10.3. From Notation 3.4(c) we see then $L_{n}=0_{q\times q}$. ∎

Definition 11.9**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Let the sequence $(s_{j})_{j=m+1}^{\infty}$ be recursively defined by $s_{j}\vcentcolon=\mathfrak{c}_{j-1}$, where $\mathfrak{c}_{j-1}$ is given by Definition 10.11. Then $(s_{j})_{j=0}^{\infty}$ is called the $[\alpha,\beta]$-Hausdorff central sequence associated with $(s_{j})_{j=0}^{m}$.

Proposition 11.10**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then the $[\alpha,\beta]$-Hausdorff central sequence associated with $(s_{j})_{j=0}^{m}$ is $[\alpha,\beta]$-Hausdorff non-negative definite and $[\alpha,\beta]$-Hausdorff central of order $m+1$.

Proof.

Denote by $(s_{j})_{j=0}^{\infty}$ the $[\alpha,\beta]$-Hausdorff central sequence associated with $(s_{j})_{j=0}^{m}$. In view of Definition 4.7 and Remark 10.2(a), we conclude $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$ by successive application of Theorem 11.2(a). By virtue of Definitions 11.9 and 10.33, then $(s_{j})_{j=0}^{\infty}$ is $[\alpha,\beta]$-Hausdorff central of order $m+1$. ∎

Proposition 11.11**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$. Then the $[\alpha,\beta]$-Hausdorff central sequence associated with $(s_{j})_{j=0}^{m}$ is $[\alpha,\beta]$-Hausdorff positive definite.

Proof.

In view of Definition 4.7 and Remark 10.2(b), we conclude by successive application of Theorem 11.2(b), that the $[\alpha,\beta]$-Hausdorff central sequence associated with $(s_{j})_{j=0}^{m}$ belongs to $\mathcal{F}^{>}_{q,\infty,\alpha,\beta}$. ∎

Proposition 11.12**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$. Then there exists a sequence $(s_{j})_{j=m+1}^{\infty}$ of complex ${q\times q}$ matrices such that $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{>}_{q,\infty,\alpha,\beta}$.

Proof.

Use Proposition 11.11. ∎

12 Applications to the moment problem M[$[\alpha,\beta]$; $(s_{j})_{j=0}^{m}$, $=$]

In this section, we discuss first applications of the preceding investigations on the structure of matricial $[\alpha,\beta]$-Hausdorff non-negative definite sequences. The following statement concretizes Proposition 4.11.

Theorem 12.1**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then $\mathcal{P}[[\alpha,\beta];(s_{j})_{j=0}^{m}]$ given by (4.1) admits the representation $\mathcal{P}[[\alpha,\beta];(s_{j})_{j=0}^{m}]=[\mathfrak{a}_{m},\mathfrak{b}_{m}]$ where $\mathfrak{a}_{m}$ and $\mathfrak{b}_{m}$ are given in Definition 10.3.

Proof.

Combine Proposition 4.11 and Theorem 11.2(a). ∎

Theorem 12.2**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, and let $(s_{j})_{j=0}^{\infty}$ be a sequence from $\mathbb{C}^{q\times q}$. Then ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}\neq\emptyset$ if and only if $(s_{j})_{j=0}^{\infty}\in\mathcal{F}^{\geq}_{q,\infty,\alpha,\beta}$. In this case, the set ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}$ contains exactly one element.

Proof.

In view of a matricial version of the Helly-Prohorov Theorem (see, e. g. [18, Satz 9, Bemerkung 2]), it is readily checked that ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}\neq\emptyset$ if and only if ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\neq\emptyset$ for all $m\in\mathbb{N}_{0}$. (The essential idea of this argumentation is originated in [1, proof of Theorem 2.1.1]). In view of Definition 4.7, then the asserted equivalence follows using Theorem 4.3. Since the interval $[\alpha,\beta]$ is bounded, it is a well-known fact that the set ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}$ contains at most one element. This completes the proof. ∎

Now we turn our attention to the special molecular solutions of truncated matricial $[\alpha,\beta]$-Hausdorff moment problems, which were constructed in Section 5 (see Theorem 5.2 and Proposition 5.3).

Proposition 12.3**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then:

1. (a)

The lower (resp. upper) $[\alpha,\beta]$-Hausdorff completely degenerate sequence $(\underline{s}_{j})_{j=0}^{\infty}$ (resp. $(\overline{s}_{j})_{j=0}^{\infty}$) associated to $(s_{j})_{j=0}^{m}$ belongs to $\mathcal{F}^{\geq,\mathrm{cd}}_{q,\infty,\alpha,\beta}$. 2. (b)

The set ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(\underline{s}_{j})_{j=0}^{\infty},=]}$ contains exactly one element $\underline{\sigma}_{m}$ and the set ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(\overline{s}_{j})_{j=0}^{\infty},=]}$ contains exactly one element $\overline{\sigma}_{m}$.

Proof.

(a) Use Proposition 11.6.

(b) In view of (a), this follows from Theorem 12.2. ∎

Definition 12.4**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then the non-negative Hermitian ${q\times q}$ measure $\underline{\sigma}_{m}$ (resp. $\overline{\sigma}_{m}$) is called the lower (resp. upper) CD-measure associated with $(s_{j})_{j=0}^{m}$ and $[\alpha,\beta]$.

Proposition 12.5**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}$, and let $(s_{j})_{j=0}^{2n-1}\in\mathcal{F}^{\geq}_{q,2n-1,\alpha,\beta}$. Then:

1. (a)

Denote by $\underline{\sigma}_{2n-1}$ the lower CD-measure associated with $(s_{j})_{j=0}^{2n-1}$ and $[\alpha,\beta]$. Then $(s_{j})_{j=0}^{2n-1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$ and $\underline{\sigma}_{2n-1}$ is the restriction onto ${\mathfrak{B}_{[\alpha,\beta]}}$ of the CD-measure $\sigma_{n}$ associated with $(s_{j})_{j=0}^{2n-1}$. 2. (b)

Denote by $\overline{\sigma}_{2n-1}$ the upper CD-measure associated with $(s_{j})_{j=0}^{2n-1}$ and $[\alpha,\beta]$. Let $s_{2n}\vcentcolon=\mathfrak{b}_{2n-1}$ and let $s_{2n+1}\vcentcolon=\mathfrak{b}_{2n}$. Then $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n+1}$ and $\overline{\sigma}_{2n-1}$ is the restriction onto ${\mathfrak{B}_{[\alpha,\beta]}}$ of the CD-measure $\sigma_{n+1}$ associated with $(s_{j})_{j=0}^{2n+1}$.

Proof.

(a) In view of Lemma 7.5, we have $(s_{j})_{j=0}^{2n-1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n-1}$. Denote by $(s_{j})_{j=0}^{\infty}$ the lower $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n-1}$. According to Lemma 11.8, then $(s_{j})_{j=0}^{\infty}$ is Hankel completely degenerate of order $n$. By virtue of Theorem 5.1, thus $(s_{j})_{j=2n}^{\infty}$ coincides with the unique sequence from Theorem 5.1(a) and therefore ${\mathcal{M}_{\geq}^{q}[{\mathbb{R}};(s_{j})_{j=0}^{\infty},=]}=\{\sigma_{n}\}$. By definition, $\underline{\sigma}_{2n-1}$ belongs to ${\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}$. Then, $\mu\colon{\mathfrak{B}_{\mathbb{R}}}\to\mathbb{C}_{\geq}^{q\times q}$ defined by $\mu(B)\vcentcolon=\underline{\sigma}_{2n-1}(B\cap[\alpha,\beta])$ belongs to ${\mathcal{M}_{\geq}^{q}[{\mathbb{R}};(s_{j})_{j=0}^{\infty},=]}$. Consequently, $\mu=\sigma_{n}$. Since the restriction of $\mu$ onto ${\mathfrak{B}_{[\alpha,\beta]}}$ is $\underline{\sigma}_{2n-1}$, the proof of part (a) is complete.

(b) Applying Proposition 11.4 twice, we see that $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{F}^{\geq,\mathrm{cd}}_{q,2n+1,\alpha,\beta}$. Denote by $(s_{j})_{j=0}^{\infty}$ the upper $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n+1}$. In view of Definition 11.5, the sequence $(s_{j})_{j=0}^{\infty}$ coincides with the upper $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n-1}$. Hence, $\{\overline{\sigma}_{2n-1}\}={\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}$ by definition. In view of $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,2n+1,\alpha,\beta}$, Proposition 10.30 yields $s_{j}=\mathfrak{a}_{j-1}$ for all $j\in\mathbb{Z}_{2n+2,\infty}$. According to Definition 11.5, then $(s_{j})_{j=0}^{\infty}$ is the lower $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n+1}$. By definition, then $\{\underline{\sigma}_{2n+1}\}={\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}$. Hence, $\overline{\sigma}_{2n-1}=\underline{\sigma}_{2n+1}$. The application of part (a) to the sequence $(s_{j})_{j=0}^{2n+1}$ completes the proof. ∎

Lemma 12.6**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $n\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{2n}\in\mathcal{F}^{\geq}_{q,2n,\alpha,\beta}$. Then:

1. (a)

Denote by $\underline{\sigma}_{2n}$ the lower CD-measure associated with $(s_{j})_{j=0}^{2n}$ and $[\alpha,\beta]$ and let $s_{2n+1}\vcentcolon=\mathfrak{a}_{2n}$. Then $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n+1}$ and $\underline{\sigma}_{2n}$ is the restriction onto ${\mathfrak{B}_{[\alpha,\beta]}}$ of the CD-measure $\sigma_{n+1}$ associated with $(s_{j})_{j=0}^{2n+1}$. 2. (b)

Denote by $\overline{\sigma}_{2n}$ the upper CD-measure associated with $(s_{j})_{j=0}^{2n}$ and $[\alpha,\beta]$ and let $s_{2n+1}\vcentcolon=\mathfrak{b}_{2n}$. Then $(s_{j})_{j=0}^{2n+1}\in\mathcal{H}^{\geq,\mathrm{e}}_{q,2n+1}$ and $\overline{\sigma}_{2n}$ is the restriction onto ${\mathfrak{B}_{[\alpha,\beta]}}$ of the CD-measure $\sigma_{n+1}$ associated with $(s_{j})_{j=0}^{2n+1}$.

Proof.

(a) From Proposition 11.4 we see that $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{F}^{\geq}_{q,2n+1,\alpha,\beta}$. In view of Definition 11.5, the lower $[\alpha,\beta]$-Hausdorff completely degenerate sequence $(s_{j})_{j=0}^{\infty}$ associated with $(s_{j})_{j=0}^{2n}$ coincides with the lower $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n+1}$. Hence, $\underline{\sigma}_{2n}$ is the lower CD-measure $\underline{\sigma}_{2n+1}$ associated with $(s_{j})_{j=0}^{2n+1}$ and $[\alpha,\beta]$. The application of Proposition 12.5(a) to the sequence $(s_{j})_{j=0}^{2n+1}$ completes the proof of (a).

(b) From Proposition 11.4 we see that $(s_{j})_{j=0}^{2n+1}$ belongs to $\mathcal{F}^{\geq,\mathrm{cd}}_{q,2n+1,\alpha,\beta}$. Denote by $(s_{j})_{j=0}^{\infty}$ the upper $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n+1}$. In view of Definition 11.5, then $(s_{j})_{j=0}^{\infty}$ is the upper $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n}$. Hence, $\{\overline{\sigma}_{2n}\}={\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}$ by definition. In view of $(s_{j})_{j=0}^{2n+1}\in\mathcal{F}^{\geq,\mathrm{cd}}_{q,2n+1,\alpha,\beta}$, Proposition 10.30 yields $s_{j}=\mathfrak{a}_{j-1}$ for all $j\in\mathbb{Z}_{2n+2,\infty}$. According to Definition 11.5, then $(s_{j})_{j=0}^{\infty}$ is the lower $[\alpha,\beta]$-Hausdorff completely degenerate sequence associated with $(s_{j})_{j=0}^{2n+1}$. By definition, then $\{\underline{\sigma}_{2n+1}\}={\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{\infty},=]}$. Hence, $\overline{\sigma}_{2n}=\underline{\sigma}_{2n+1}$. The application of Proposition 12.5(a) to the sequence $(s_{j})_{j=0}^{2n+1}$ completes the proof of part (b). ∎

Corollary 12.7**.**

Let $\alpha\in\mathbb{R}$, let $\beta\in(\alpha,\infty)$, let $m\in\mathbb{N}_{0}$, and let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{q,m,\alpha,\beta}$. Then $\{\underline{\sigma}_{m},\overline{\sigma}_{m}\}\subseteq{\mathcal{M}_{\geq}^{q}[{[\alpha,\beta]};(s_{j})_{j=0}^{m},=]}\cap\mathcal{M}^{q,\mathrm{mol}}_{\geq}([\alpha,\beta])$

Proof.

Combine Propositions 12.5 and 12.6 with Theorem 5.1(c). ∎

Now we want to add some comments concerning the case $q=1$. Let $(s_{j})_{j=0}^{m}\in\mathcal{F}^{\geq}_{1,m,\alpha,\beta}$. Then it can be expected that the lower (resp. upper) CD-measure $\underline{\sigma}_{m}$ (resp. $\overline{\sigma}_{m}$) associated with $(s_{j})_{j=0}^{m}$ and $[\alpha,\beta]$ coincides with the lower (resp. upper) principal solution constructed by M. G. Krein in [26] in [27, Ch. III]. Furthermore, the other molecular solutions obtained in Proposition 5.3 should be canonical solutions in the terminology of M. G. Krein. The case of a sequence $(s_{j})_{j=0}^{m}\in\mathcal{F}^{>}_{q,m,\alpha,\beta}$ deserves particular attention. In this case, the theory of orthogonal matrix polynomials can be effectively applied and promises to produce a whole collection of useful explicit formulas. This will be confirmed by the strategy used by M. G. Krein [26] (see also [27, Ch. III]). It should be mentioned that first steps in this direction are already contained in the papers [6, 7] by A. E. Choque Rivero, where the resolvent matrices for the non-degenerate truncated $[\alpha,\beta]$-Hausdorff moment problem were expressed in terms of orthogonal matrix polynomials and moreover multiplicative decompositions of these resolvent matrices were derived. We will handle this theme in separate work.

Appendix A Some facts from matrix theory

Remark A.1*.*

Let $n\in\mathbb{N}$ and let $(A_{j})_{j=0}^{n}$ be a sequence of non-negative Hermitian complex ${q\times q}$ matrices. Then $\mathcal{R}(\sum_{j=1}^{n}A_{j}))=\sum_{j=1}^{n}\mathcal{R}(A_{j})$ and $\mathcal{N}(\sum_{j=1}^{n}A_{j})=\bigcap_{j=1}^{n}\mathcal{N}(A_{j})$.

Remark A.2*.*

If $A\in\mathbb{C}_{\geq}^{q\times q}$ and if $X\in\mathbb{C}^{q\times p}$, then $X^{*}AX\in\mathbb{C}_{\geq}^{p\times p}$. If $A\in\mathbb{C}_{>}^{q\times q}$ and if $X\in\mathbb{C}^{q\times p}$ with $\operatorname{rank}X=p$, then $X^{*}AX\in\mathbb{C}_{>}^{p\times p}$.

Denote by $P_{\mathcal{U}}$ the matrix associated with the orthogonal projection in the Euclidean space $\mathbb{C}^{q}$ onto a linear subspace $\mathcal{U}$.

Remark A.3*.*

Let $\mathcal{U}$ be a linear subspace of $\mathbb{C}^{q}$, then $P_{\mathcal{U}}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ and $0_{q\times q}\leq P_{\mathcal{U}}\leq I_{q}$.

For the convenience of the reader, we state some well-known and some special results on Moore-Penrose inverses of matrices (see e. g., Rao/Mitra [29] or [12]Section 1). If $A\in\mathbb{C}^{p\times q}$, then (by definition) the Moore-Penrose inverse $A^{\dagger}$ of $A$ is the unique matrix $A^{\dagger}\in\mathbb{C}^{q\times p}$ which satisfies the four equations

[TABLE]

Proposition A.4** (see, e. g. [12]Theorem 1.1.1).**

If $A\in\mathbb{C}^{p\times q}$ then a matrix $X\in\mathbb{C}^{q\times p}$ is the Moore-Penrose inverse of $A$ if and only if $AX=P_{\mathcal{R}(A)}$ and $XA=P_{\mathcal{R}(X)}$.

Remark A.5*.*

Let $A\in\mathbb{C}^{p\times q}$ and $B\in\mathbb{C}^{{p\times r}}$. Then $\mathcal{R}(B)\subseteq\mathcal{R}(A)$ if and only if $AA^{\dagger}B=B$.

Remark A.6*.*

Let $A\in\mathbb{C}^{p\times q}$ and $B\in\mathbb{C}^{{n\times q}}$. Then $\mathcal{N}(A)\subseteq\mathcal{N}(B)$ if and only if $BA^{\dagger}A=B$.

Remark A.7*.*

$(A^{*})^{\dagger}=(A^{\dagger})^{*}$ for each $A\in\mathbb{C}^{p\times q}$.

Remark A.8*.*

Let $A\in\mathbb{C}_{\mathrm{H}}^{q\times q}$. Then $A^{\dagger}\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ and $A^{\dagger}A=AA^{\dagger}$.

Remark A.9*.*

Let $A\in\mathbb{C}^{p\times q}$. If $U\in\mathbb{C}^{{n\times p}}$ fulfills $U^{*}U=I_{p}$ and if $V\in\mathbb{C}^{{q\times r}}$ is such that $VV^{*}=I_{q}$, then $(UAV)^{\dagger}=V^{*}A^{\dagger}U^{*}$.

Remark A.10*.*

Let $A\in\mathbb{C}_{\geq}^{q\times q}$. Then $A^{\dagger}\in\mathbb{C}_{\geq}^{q\times q}$ and $\sqrt{A^{\dagger}}=\sqrt{A}^{\dagger}$. Furthermore, $\mathcal{R}(\sqrt{A})=\mathcal{R}(A)$ and $\mathcal{N}(\sqrt{A})=\mathcal{N}(A)$.

Remark A.11*.*

Let $A,B\in\mathbb{C}_{\mathrm{H}}^{q\times q}$ with $0_{q\times q}\leq A\leq B$. Then $\mathcal{R}(A)\subseteq\mathcal{R}(B)$ and $\mathcal{N}(B)\subseteq\mathcal{N}(A)$.

Lemma A.12** ([2, 15]).**

Let M=\bigl{[}\begin{smallmatrix}A&B\\ C&D\end{smallmatrix}\bigr{]} be the block representation of a complex ${(p+q)\times(p+q)}$ matrix $M$ with ${p\times p}$ block $A$. Then:

1. (a)

$M$* is non-negative Hermitian if and only if $A$ and $D-CA^{\dagger}B$ are both non-negative Hermitian, $\mathcal{R}(B)\subseteq\mathcal{R}(A)$, and $C=B^{*}$.* 2. (b)

$M$* is positive Hermitian if and only if $A$ and $D-CA^{\dagger}B$ are both positive Hermitian and $C=B^{*}$.*

Appendix B Parallel sum of matrices

For every choice of complex ${p\times q}$ matrices $A$ and $B$, the parallel sum $A\mathbin{\overline{\pm}}B$ of $A$ and $B$ is defined by

[TABLE]

Furthermore, let $\mathcal{P}_{{p\times q}}$ be the set of all pairs $(A,B)\in\mathbb{C}^{p\times q}\times\mathbb{C}^{p\times q}$ such that $\mathcal{R}(A)\subseteq\mathcal{R}(A+B)$ and $\mathcal{N}(A+B)\subseteq\mathcal{N}(A)$ hold true.

Let $A$ and $B$ be non-singular matrices from $\mathbb{C}^{q\times q}$ such that $\det(A+B)\neq 0$ and $\det(A^{-1}+B^{-1})\neq 0$. Then $(A,B)\in\mathcal{P}_{{q\times q}}$ and $A\mathbin{\overline{\pm}}B=(A^{-1}+B^{-1})^{-1}$.

Remark B.1*.*

If $A,B\in\mathbb{C}_{>}^{q\times q}$, then $(A,B)\in\mathcal{P}_{{q\times q}}$ and $A\mathbin{\overline{\pm}}B\in\mathbb{C}_{>}^{q\times q}$.

Lemma B.2** ([3, Lemma 4]).**

If $A,B\in\mathbb{C}_{\geq}^{q\times q}$, then $(A,B)\in\mathcal{P}_{{q\times q}}$ and $A\mathbin{\overline{\pm}}B\in\mathbb{C}_{\geq}^{q\times q}$.

Lemma B.3** ([28, Theorem 2.2(a)]).**

If $(A,B)\in\mathcal{P}_{{p\times q}}$, then $(B,A)\in\mathcal{P}_{{p\times q}}$ and $A\mathbin{\overline{\pm}}B=B\mathbin{\overline{\pm}}A$.

Lemma B.4** ([28, Theorem 2.2(f)]).**

If $(A,B)\in\mathcal{P}_{{p\times q}}$, then $\mathcal{R}(A\mathbin{\overline{\pm}}B)=\mathcal{R}(A)\cap\mathcal{R}(B)$ and $\mathcal{N}(A\mathbin{\overline{\pm}}B)=\mathcal{N}(A)+\mathcal{N}(B)$.

Proposition B.5** ([28, Theorem 2.2(g)]).**

If $(A,B)\in\mathcal{P}_{{q\times q}}$, then $(A\mathbin{\overline{\pm}}B)^{\dagger}=P_{\mathcal{R}(A^{*})\cap\mathcal{R}(B^{*})}(A^{\dagger}+B^{\dagger})P_{\mathcal{R}(A)\cap\mathcal{R}(B)}$.

Lemma B.6**.**

If $(A,B)\in\mathcal{P}_{{p\times q}}$, then $(A+B)-4(A\mathbin{\overline{\pm}}B)=(A-B)(A+B)^{\dagger}(A-B)$.

Proof.

Let $(A,B)\in\mathcal{P}_{{p\times q}}$ and let $C\vcentcolon=A+B$. In view of Lemma B.3, we get

[TABLE]

Hence,

[TABLE]