On the Moment Problem and Related Problems

Abstract

Firstly, we recall the classical moment problem and some basic results related to it. By its formulation, this is an inverse problem: being given a sequence ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ of real numbers and a closed subset $F\subseteq {ℝ}^n$, $n\in \left\{1,2,\dots \right\},$ find a positive regular Borel measure $\mu$ on $F$ such that ${{\displaystyle \int }}_F^{}{t}^{j}d\mu =y_j, j\in {\mathbb{N}}^n.$ This is the full moment problem. The existence, uniqueness, and construction of the unknown solution $\mu$ are the focus of attention. The numbers $y_j, j\in {\mathbb{N}}^n$ are called the moments of the measure $\mu .$ When a sandwich condition on the solution is required, we have a Markov moment problem. Secondly, we study the existence and uniqueness of the solutions to some full Markov moment problems. If the moments $y_j$are self-adjoint operators, we have an operator-valued moment problem. Related results are the subject of attention. The truncated moment problem is also discussed, constituting the third aim of this work.

1. Introduction

The moment problem is recalled in the Abstract. Originally, it was formulated by T. Stieltjes in 1894–1895 (see [1]): find the repartition of the positive mass on the non-negative semi-axis, if the moments of arbitrary orders $k$ ($k=0,1,2,\dots$) are given. Specifically, in the Stieltjes moment problem, a sequence of real numbers ${\left(y_k\right)}_{k\ge 0}$ is given, and one looks for a nondecreasing real function $\sigma \left(t\right) \left(t\ge 0\right)$, which verifies the moment conditions:${{\displaystyle \int }}_0^{\infty }{t}^{k}d\sigma =y_k, \left(k=0,1,2,\dots \right).$If such a function $\sigma$ does exist, the sequence ${\left(y_k\right)}_{k\ge 0}$ is called a Stieltjes moment sequence. A Hamburger moment sequence is a sequence ${\left(y_k\right)}_{k\ge 0}$ for which there exists a positive regular Borel measure $\mu$ on $ℝ$, such that ${{\displaystyle \int }}_{ℝ}^{}{t}^{k}d\mu =y_k, k=0,1,\dots$.

Going back to the problem formulated by T. Stieltjes, this is a one-dimensional moment problem, on an unbounded interval. Specifically, it is an interpolation problem with the constraint on the positivity of the measure $d\sigma .$The numbers $y_k, k\in \mathbb{N}=\left\{0,1,2,\dots \right\}$ are called the moments of the measure $d\sigma .$ The existence, uniqueness, and construction of the solution $\sigma$are studied. The moment problem is an inverse problem: we are looking for an unknown measure, starting from its given moments. The direct problem might be presented as follows: being given the measure $d\sigma ,$ find its moments ${{\displaystyle \int }}_0^{\infty }{t}^{k}d\sigma .k=0,1,2,\dots .$ For general knowledge in analysis and functional analysis related to the present paper, see [1,2,3,4,5,6,7,8,9,10]. Various aspects of the moment problem are discussed in [11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26]. For other applications, results on Markov moment problem and connections with other fields of analysis see [27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42]. The moment problem has applications in physics, probability theory, and statistics, as discussed in the Introduction of [2]. We will use the notations:

$$\mathbb{N}=\left\{0,1,2,\dots \right\}, {ℝ}_{+}=\left[0,\infty \right[.$$

All of the measures appearing in this paper are positive regular Borel measures on a closed subset $F$ of ${ℝ}^{n },$ with finite moments of all orders. To simplify the writing, we will name a measure with these properties “measure on $F$”. Other notations will be specified in the following text when they are used. Going back to the $n$ dimensional moment problem formulated in the Abstract, if

$${\phi }_j\left(t\right)=t^j=t_1^{j_1}\cdots t_n^{j_n}, j=\left(j_1,\dots ,j_n\right)\in {\mathbb{N}}^n, t=\left(t_1,\dots ,t_n\right)\in F, n\in \mathbb{N}, n\ge 1,$$

and a sequence ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ of real numbers is given, then one studies the existence, uniqueness, and construction of a linear positive form $T$ defined on a function space $X_1$ containing polynomials and continuous compactly supported real functions, such that the moment conditions

$$T\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^n$$

(1)

are satisfied (for details and motivation of using this subspace, see Section 3.2, Theorem 28). Usually, if $F\subseteq {ℝ}^n$ is a closed unbounded subset, and $\nu$ is a positive regular measure on $F,$ with finite moments of all orders, we put $X=L_{\nu }^p\left(F\right),$ where $1\le p<\infty$and $X_1=\{f\in X;\exists p\in 𝒫,|f|\le p\}$. Going back to the moment conditions in Equation (1), due to Haviland’s theorem [11], for the existence of such a positive linear functional $T,$ it is sufficient (and necessary) that the linear form

$$T_0:𝒫\to ℝ, T_0\left({\displaystyle \sum }_{j\in J_0}{\alpha }_j{\phi }_j\right):={\displaystyle \sum }_{j\in J_0}{\alpha }_j{y}_j,$$

(2)

defined on the subspace $𝒫$ of polynomial functions, satisfies the condition

$$p\in 𝒫, p\left(t\right)\ge 0 \forall t\in F\Rightarrow T_0\left(p\right)\ge 0.$$

(3)

If the implication (3) holds true, Haviland’s theorem ensures the existence of a positive regular Borel measure $\mu$ on $F$, such that

$$y_j=T_0\left({\phi }_j\right)={{\displaystyle \int }}_F^{}{t}^{j}d\mu , j\in {\mathbb{N}}^n.$$

(4)

If Equation (4) holds, we say that ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ is a moment sequence (or a sequence of moments) on $F.$

In the case of a Markov moment problem, more powerful extension results for linear operators can be applied. Aside from positivity, the extension $T$ of $T_0$ is dominated by a given continuous sublinear operator $P$ (or by a continuous convex operator). Usually, this leads not only to the continuity of $T,$ but also to evaluating (or even determining exactly) its norm, in terms of the norm of the sublinear operator $P.$ Extension results on linear operators, constrained by sandwich conditions, have been recently applied in [27] in order to characterize the monotone increasing convex operators defined on convex cones, in terms of their subgradients. In [28], some density results in a different framework with respect to that discussed in the present Section 3.2 are highlighted.

A serious problem related to the multidimensional moment problem arises from the fact that, unlike the one-dimensional case, the form of non-negative polynomials on ${ℝ}^n, n\ge 2,$ in terms of sums of squares, is not known. There exist non-negative polynomials on ${ℝ}^2$, which are not expressible as sums of squares (see [2,12,29]). The article [22] gives the form of non-negative polynomials on a strip (in ${ℝ}^2$) in terms of sums of squares (see Section 3 below). The case of compact subsets $K$ of ${ℝ}^n, n\ge 2$ was studied in detail, and in the case of semi-algebraic compact subsets the form of positive polynomials defined on such a compact was determined, in terms of sums of squares and of the polynomials defining the semi-algebraic set (Positivstellensatz; see [16]). This study was continued and generalized in [2,17,18,19]. Before these results, in [15], the moment problem for a compact set with a non-empty interior in ${ℝ}^n$ was solved. A related result on the decomposition of positive polynomials on a compact subset with a non-empty interior was deduced. The main difference between the one-dimensional and the multidimensional cases of the moment problem, in terms of a sequence of numbers ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$, can be formulated in the following way: for any $n\in \left\{1,2\dots \right\},$ any moment sequence ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ is positive definite; that is:

$${\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{y}_{i+j}\ge 0$$

for any finite subset $J_0\subset {\mathbb{N}}^n$ and any $\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ.$ For $n=1,$ the converse is true, since any non-negative polynomial on $ℝ$ is a sum of (two) squares of polynomials, and a square of a polynomial has the form ${\left({\displaystyle \sum }_{j\in J_0}{\lambda }_j{\phi }_j\right)}^2={\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{\phi }_{i+j}$; then, one applies Haviland’s theorem. For $n\ge 2$, there exist positive definite sequences that are not moment sequences (see Remark 2 for more details). Section 3.1 briefly reviews such results, in the end adding sufficient checkable criteria for the determinacy of measures on $ℝ$ and on ${ℝ}_{+}$.

A major part of the present work focuses on Markov moment problems on ${ℝ}^n$ and on ${ℝ}_{+}^n$. A first main general result is Lemma 18, which claims that for any moment-determinate measure $\nu$ on an unbounded closed subset $F$of ${ℝ}^n,$ and any non-negative continuous compactly supported function $\psi :F\to ℝ,$ there exists a sequence of polynomials ${\left(p_m\right)}_{m\in \mathbb{N}}, p_m\ge \psi$ for all $m, p_m\to \psi$ in the norm of the space $L_{\nu }^1\left(F\right)$; a related main result is Theorem 19. Moreover, if $\nu ={\nu }_1\times \cdots \times {\nu }_n$, where each ${\nu }_j$, $j=1,\dots ,n$ is a moment-determinate measure on $ℝ$ (respectively on ${ℝ}_{+}$), with finite moments of all orders, we prove the approximation of a non-negative continuous compactly supported function $\psi$ by sums of special non-negative polynomials of the form

$$(p_1\otimes \cdots \otimes p_n)\left(t_1,\dots ,t_n\right)=p_1\left(t_1\right)\cdots p_n\left(t_n\right),$$

(5)

where each $p_j, j=1,\dots ,n$ is a non-negative polynomial on the whole real axes (or on ${ℝ}_{+}).$ In particular, each $p_j, j=1,\dots ,n$ has a known analytic expression in terms of sums of squares: $p\left(t\right)\ge 0 \forall t\in ℝ\iff p\left(t\right)=p_1^2\left(t\right)+p_2^2\left(t\right), p\left(t\right)\ge 0 \forall t\in {ℝ}_{+}\iff p\left(t\right)=p_1^2\left(t\right)+t{p}_2^2\left(t\right), t\in {ℝ}_{+}$ for some $p_1,p_2\in ℝ\left[t\right].$ Thus the determinacy of such products of measures follows from the determinacy of each factor. Hence, $\psi$ is approximated by sums of polynomials whose form in terms of sums of squares is known. The approximation holds in $L_{\nu }^1\left({ℝ}^n\right)$, (respectively in $L_{\nu }^1\left({ℝ}_{+}^n\right)).$ In particular, the convex cone ${𝒫}_{++}$ of all sums of non-negative polynomials (Equation (5)) is dense in ${\left(L_{\nu }^1\left({ℝ}^n\right)\right)}_{+}$ (respectively in ${\left(L_{\nu }^1\left({ℝ}_{+}^n\right)\right)}_{+}$), and $𝒫$ is dense in $L_{\nu }^1\left({ℝ}^n\right)$ (respectively in $L_{\nu }^1\left({ℝ}_{+}^n\right))$(see Lemmas 4 and 5). The sums of polynomials given by Equation (5) dominate the function $\psi .$ We obtain characterization of the existence and uniqueness of a linear operator solution $T$verifying the interpolation moment conditions of Equation (1), and the constraints $T_1\le T\le T_2$ on the positive cone of the domain space (Theorem 19). Here, $T_1,T_2$ are given bounded linear operators. Such characterizations are partially formulated in terms of quadratic forms (with vector coefficients). The paper [29] contains the first example of a non-negative polynomial on ${ℝ}^2$ that is not a sum of squares. The papers [30,31] emphasize optimization problems related to the moment problem, while in [32,33,34,35,36,37,38,39,40,41,42] various aspects of the full and of the truncated Markov moment problem are examined. All of the main results of the present work concerning the Markov moment problem are valid for the vector-valued problem. The case of the corresponding scalar-valued problem follows as a consequence. Although the full moment problem is discussed in detail in Section 3 (because it could have a unique solution, whose existence is characterized in terms of quadratic forms (see [36,37,38]), the truncated moment problem is also one of the aims of the same section. Major results on this subject have been published in [24,25,30,31,32,33,37,38,39]. In the end, as an application of polynomial approximation in another direction, the positivity of a bounded linear operator is also characterized in terms of quadratic forms (see [36], Theorem 7, and Theorem 18 proved in Section 3.2 of the present work). From this point of view, the difficulty arising from the fact that non-negative polynomials on ${ℝ}^n, n\ge 2$ are not sums of squares is solved. Another application of the solution to an abstract moment problem (Theorem 2, stated below) to an unexpected sandwich result is briefly reviewed with the aid of the referenced citations (see [40,41], Theorems 2 and 4).

The rest of the paper is organized as follows: Section 2 is devoted to some basic general methods of extension for linear positive operators and abstract moment problems that are applied throughout this work. In Section 3, earlier and recent main results on the moment problem and on a related problem are stated; most of these results refer to the Markov moment problem, and are accompanied by their proofs. In Section 4, we briefly discuss some of the results mentioned above.

2. Methods

In the following section, we describe the methods that are more or less related to the results stated below. A motivation of proving extension results for positive linear operators was to solve the existence problem of solutions for moment problems. Theorems 2 and 3 stated below are motivated by solving Markov moment problems. All of the next three theorems work not only for extension of linear functionals, but also for a large class of linear operators. The polynomial approximation results have been motivated by the Markov moment problem, but they also led to characterization of the positivity of some bounded linear operators only in terms of quadratic forms, even in the case of spaces of functions of several variables (see Theorem 18 of Section 3.2). Although Theorems 3 and 4 are equivalent (where ${\left\{x_j\right\}}_{j\in J}$ stands for a Hamel base of $X),$ we must recall that Hahn–Banach theorems have many more applications in various fields of research that have no connection with the moment problem. To name two of them, a first application of the geometric form of the Hahn–Banach theorem is the Krein–Milman theorem (and its corresponding finite-dimensional result, Carathéodory’s theorem). One of the directions of application of the Krein–Milman theorem is the representation theory. For example, any “abstract” function satisfying some conditions has an integral representation involving concrete functions, since it is a limit of convex combinations of such functions, which are extreme points of a certain convex compact subset of a function space. Another application of the Krein–Milman theorem is to the notion of a barycenter of a probability measure on a convex compact subset of a (Hausdorff) locally convex space. The related notion in physics is that of the center of mass of a finite system of points $\left(x_k,m_k\right), k=1,\dots ,N,$ where $x_k$ indicates the position and $m_k$ the mass (see [8], p. 75). For an infinite system of points, one takes the limit as $N\to \infty .$ Finally, we recall that the Hahn–Banach theorem is applied in optimization theory, for both cases: one objective function (one criterion) and the corresponding vector-valued objective function (optimization following several different criteria: Pareto optimization). Pareto optimization has applications in economics, finance, etc. On the other hand, polynomial approximation in $L^1$ spaces—mentioned at point (3) below—is far from being the only efficient type of approximation. As is well known, Fourier approximation holds in $L^2$spaces and in any Hilbert space (see [9], and eventually additional information on special orthogonal functions).

(1)

Extension of linear operators, with one condition (Theorem 1) or two constraints (sandwich condition, in Theorems 2 and 3) on the linear extension. Here is one of the old results on this subject, with many applications to the scalar and the vector-valued moment problems. Let $X_1$ be an ordered vector space whose positive cone $X_{1,+}$ generates $X_1$ ($X_1=X_{1,+}-X_{1,+}).$ Recall that in such an ordered vector space $X_1,$ a vector subspace $S$ is called a majorizing subspace if for any $x\in X_1$ there exists $s\in S$ such that $x\le s.$ The following Kantorovich theorem on the extension of positive linear operators holds true:

Theorem 1.

(see [10], Theorem 1.2.1). Let$X_1$ be an ordered vector space whose positive cone generates $X_1$, $X_0\subset X_1$ —a majorizing vector subspace, $Y$ an order complete vector space, and $T_0:X_0\to Y$ a positive linear operator. Then, $T_0$ admits a positive linear extension $T:X_1\to Y$.

The next two results refer to the Markov moment problem.

Theorem 2.

(see [35], Theorem 4). Let $X$ be an ordered vector space, $Y$ an order complete vector lattice,${\left\{x_j\right\}}_{j\in J}$, ${\left\{y_j\right\}}_{j\in J}$ families of elements in $X$ and $Y,$ respectively, and $T_1,T_2\in L\left(X,Y\right)$ two linear operators. The following statements are equivalent:

(1)

There is a linear operator$T\in L\left(X,Y\right),$such that

$$T_1\left(x\right)\le T\left(x\right)\le T_2\left(x\right), x\in X_{+}, T\left(x_j\right)=y_j, j\in J;$$

(2)

For any finite subset$J_0\subset J,$and any${\left\{{\alpha }_j\right\}}_{j\in J_0}\subset ℝ$, the following implication holds true:

$$\left({\displaystyle \sum }_{j\in J_0}{\alpha }_j{x}_j={\psi }_2-{\psi }_1, {\psi }_1,{\psi }_2\in X_{+}\right)\Rightarrow {\displaystyle \sum }_{j\in J_0}{\alpha }_j{y}_j\le T_2\left({\psi }_2\right)-T_1\left({\psi }_1\right).$$

If$X$is a vector lattice, then assertions (a) and (b) are equivalent to (c), where (c)$T_1\left(w\right)\le T_2\left(w\right)$for all$w\in X_{+}$and for any finite subset$J_0\subset J$and$\forall \left\{{\alpha }_j;j\in J_0\right\}\subset ℝ,$we have

$${\displaystyle \sum }_{j\in J_0}{\alpha }_j{y}_j\le T_2\left({\left({\displaystyle \sum }_{j\in J_0}{\alpha }_j{x}_j\right)}^{+}\right)-T_1\left({\left({\displaystyle \sum }_{j\in J_0}{\alpha }_j{x}_j\right)}^{-}\right).$$

Remark 1.

It is worth noting that Theorem 2 leads to a sandwich-type theorem on the existence of an affine functional$h, f\le h\le g$on$S$, where$S$is a finite-simplicial (convex) subset of a vector space, while$f$and$–g$are convex functionals such that$f\le g$on$S.$The novelty is that a finite-simplicial set might be unbounded in any locally convex topology on the domain space containing$S$(see [40,41]). A topological version has been published in [41], Theorem 4.

Theorem 3.

(see [35], Theorem 1 and [41], Theorem 5). Let $X$ be a preordered vector space, $Y$ an order complete vector lattice, $P:X\to Y$ a convex operator, and ${\left\{x_j\right\}}_{j\in J}\subset X, {\left\{y_j\right\}}_{j\in J}\subset Y$ given families. The following statements are equivalent:

(a)

There exists a linear positive operator$T:X\to Y$, such that

$$T\left(x_j\right)=y_j, j\in J, T\left(x\right)\le P\left(x\right), x\in X;$$

(b)

For any finite subset $J_0\subseteq J, and any \left\{{\lambda }_j;j\in J_0\right\}\subseteq ℝ,$we have

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{x}_j\le x\in X\Rightarrow {\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j\le P\left(x\right).$$

It is worth noting that both Theorems 2 and 3 can be obtained as consequences of a much more general result on the extension of linear operators, preserving a sandwich condition (see the corresponding references in [35]). For instance, Theorem 3 can clearly be reformulated as the next result (see the original source cited in [35], first published in 1978, and recalled in [39] as Theorem 1).

Theorem 4.

Let $X$be an ordered vector space,$Y$an order complete vector space,$M\subset X$a vector subspace,$T_1:M\to Y$a linear operator, and$P:X\to Y$a convex operator. The following statements are equivalent:

(a)

There exists a positive linear extension$T:X\to Y$of$T_1$, such that$T\le P$on$X;$

(b)

We have$T_1\left(u\right)\le P\left(x\right)$for all$\left(u,x\right)\in M\times X$, such that$u\le x.$

It is clear that by applying Theorem 4 to the very particular case when $X_{+}=\left\{\mathbf{0}\right\}$, one obtains the Hahn–Banach theorem. Another version of Theorems 3 and 4 is the next one, recently applied in [27] to find a characterization of isotone (monotone increasing) convex operators defined on a convex con, in terms of its subgradients. This result admits a direct sharp proof (see [41], Theorem 5), without using the more general theorems mentioned as references in [35] or Theorem 1 of [39].

Theorem 5.

Let$X$be an ordered vector space,$Y$an order complete vector space,$M\subset X$a vector subspace,$T_1:M\to Y$a linear operator, and$P:X_{+}\to Y$a convex operator. The following statements are equivalent:

(a)

There exists a positive linear extension$T:X\to Y$of$T_1$, such that$T{|}_{X_{+}}\le P;$

(b)

We have$T_1\left(h\right)\le P\left(x\right)$for all$\left(h,x\right)\in M\times X_{+}$, such that$h\le x.$

In fact, the simple direct proof of Theorem 5—recently published in [41] Theorem 5, works with insignificant modifications for Theorem 4 as well.

(2)

We recall basic statements of [2,15,16], referring to the classical moment problem. These methods can be applied to the Markov moment problem as well (see [35], Theorem 2 and [2], Corollary 12.29). Sufficient criteria for determinacy and indeterminacy are also under attention (see [2,20]);

(3)

We review our earlier results on polynomial approximation on unbounded subsets, and its applications to the vector-valued Markov moment problem, recently published, completed, and generalized in [38] (see also [36]). To this aim, we essentially use the notion of a moment-determinate ($M-$ determinate) measure (see [2,13,20]);

(4)

Two results on truncated scalar-valued Markov moment problem [37] are briefly discussed.

3. Results

3.1. The Moment Problem on Compact Subsets of ${ℝ}^{\mathrm{n}}$

The next results are chronologically ordered following their date of publication. We start by recalling some main theorems of [15], concerning the moment problem on a compact subset $K$ with a non-empty interior in ${ℝ}^n.$ Observe that this condition is quite natural. The author aimed to include all convex compact subsets. If a convex subset of ${ℝ}^n$ has an empty interior, it is easy to see that it is contained in a linear variety of dimension at most $n-1.$ If we consider the linear $X_K$ generated by such a compact $K,$ then the convex compact $K$ has a non-empty interior in $X_K.$ Therefore, for convex compact finite-dimensional subsets $K,$ we may always assume that the interior of $K$ is non-empty. Going back to the case of an arbitrary compact subset $K$ of ${ℝ}^n,$ with a non-empty interior in ${ℝ}^n,$ the open set $D={ℝ}^n\backslash K$ can be written as a union of sets of the form

$${\mathit{D}}_{\mathit{i}}={\mathit{p}}_{\mathit{i}}^{-\mathbf{1}}\left(\left]-\infty ,\mathbf{0}\right[\right), \mathit{i}\in \mathit{I},$$

where $p_i$ are polynomials with real coefficients, of degrees smaller than or equal to two. Consequently,

$$\mathit{K}={ℝ}^{\mathit{n}}\backslash {{{\displaystyle \cup }}^{ }}_{\mathit{i}\in \mathit{I}}{\mathit{D}}_{\mathit{i}}={{{\displaystyle \cap }}^{ }}_{\mathit{i}\in \mathit{I}}{\mathit{p}}_{\mathit{i}}^{-1}\left(\left[0,\infty \right[\right)={{{\displaystyle \cap }}^{ }}_{\mathit{i}\in \mathit{I}}{\mathit{p}}_{\mathit{i}}^{-1}\left({ℝ}_{+}\right).$$

One denotes by $E\left(K\right)$ the vector space generated by the polynomials of degree at most one and the polynomials $p_i, i\in I.$ The space$E\left(K\right)$ is clearly a vector subspace of the space ${ℝ}_2\left[t_1,\dots ,t_n\right]$ of all polynomials of degrees at most two. We denote by $E_{+}\left(K\right)$ the convex cone of all polynomials in $E\left(K\right)$, which takes non-negative values at all points of $K$, and one denotes by $G\left(K\right)$ the set of those $p\in E_{+}\left(K\right)$ that generates an extreme ray of $E_{+}\left(K\right).$ An important subset is

$${\mathit{G}}_{\mathbf{1}}\left(\mathit{K}\right)=\left\{\mathit{p}\in \mathit{G}\left(\mathit{K}\right);{\Vert \mathit{p}\Vert }_{\mathit{K}}=\mathbf{1}\right\}.$$

If $K$ is convex, then we can take as $E\left(K\right)$ the space ${ℝ}_1\left[t_1,\dots ,t_n\right]$ of all polynomials of degree at most one. With $n=1$ and $K=\left[0,1\right]$, it is easy to see that $G_1\left(K\right)$ consists in only two elements: the polynomials $t$ and $1-t.$ From this, we infer that the set of polynomials $t^k{\left(1-t\right)}^m, k,m\in \mathbb{N},$ which appear naturally in the classical Hausdorff moment problem, should be replaced, in the general case, by the set $\Delta \left(K\right)$ of polynomials, which are finite products of elements of $G_1\left(K\right).$ Since $\Delta \left(K\right)$ has only a multiplicative structure (the sum of two elements of this set is not, in general, an element of $\Delta \left(K\right)$), one introduces the convex cone $\mathcal{C}$ of all linear combinations with non-negative coefficients of elements of $\Delta \left(K\right).$ The interesting step is that of introducing the set $\pi \left(K\right)$ of all linear forms $T$ on $𝒫=ℝ[t_1\dots t_n]$, such that $T(\P )=1,T\left(p\right)\ge 0$ for all $p\in \Delta \left(K\right).$ Then, the article goes on with three key lemmas, followed by several basic theorems partially deduced from the lemmas that precede them. In what follows, we review three of these main theorems.

Theorem 6.

(See [15], Theorem 1). For each $\mathit{T}\in \mathit{\pi }\left(\mathit{K}\right),$ there exists a unique probability measure $\mathit{\mu }$ on $\mathit{K}$, such that $\mathit{T}\left(\mathit{p}\right)={{\displaystyle \int }}_{\mathit{K}}^{}\mathit{p}\mathit{d}\mathit{\mu }$ for any polynomial $\mathit{p}\in ℝ\left[{\mathit{t}}_1,\dots ,{\mathit{t}}_{\mathit{n}}\right].$

Theorem 6 leads to the next result, which gives a necessary and sufficient condition for the existence of a solution to the moment problem. This condition is formulated only in terms of the moments $y_j, j\in {\mathbb{N}}^n$ and the special polynomials that are elements of $\Delta \left(K\right);$ therefore, we say that the next theorem solves the moment problem. From the point of view of the next theorem, we can say that it represents the multidimensional case of the Hausdorff moment problem (the moment problem on $K=\left[0,1\right]$). On the other hand, the next theorem works for non-convex compact subsets as well. The set $G_1\left(K\right)$ is much more difficult to determine in this general setting; consequently, the proofs are more difficult.

Theorem 7.

(See [15], Theorem 2). Let $\mathit{K}$ be a compact subset of ${ℝ}^{\mathit{n}},$ with a non-empty interior. A necessary and sufficient condition for a sequence ${\left({\mathit{y}}_{\mathit{j}}\right)}_{\mathit{j}\in {\mathbb{N}}^{\mathit{n}}}$ being a moment sequence on $\mathit{K}$ is that the linear form $\mathit{T}$ defined on $ℝ\left[{\mathit{t}}_{\mathbf{1}},\dots ,{\mathit{t}}_{\mathit{n}}\right]$ by $\mathit{T}\left({\mathit{t}}^{\mathit{j}}\right)={\mathit{y}}_{\mathit{j}}, \mathit{j}\in {\mathbb{N}}^{\mathit{n}}$ satisfies the condition $\mathit{T}\left(\mathit{p}\right)\ge \mathbf{0}$ for each polynomial $\mathit{p}\in \Delta \left(\mathit{K}\right).$

The next result gives the expression of any polynomial $p$ that is positive at each point of $K$, by means of some polynomials that are elements of $\Delta \left(K\right).$ Since any such polynomial $p$ is a linear combination with positive coefficients of elements of $\Delta \left(K\right),$ the next result is called the decomposition theorem.

Theorem 8.

(See [15], Theorem 4). Each polynomial that has positive values at all points of a compact subset $\mathit{K}$ with a non-empty interior in ${ℝ}^{\mathit{n}}$ is a linear combination with positive coefficients of elements of $\Delta \left(\mathit{K}\right).$

Next, we consider the moment problem on compact semi-algebraic subsets of ${ℝ}^n.$ If $y={\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ $n\ge 2$is a sequence of real numbers, one denotes by $T_y$ the linear functional defined on $ℝ\left[t_1,\dots ,t_n\right]$ by

$$T_y\left({\displaystyle \sum }_{j\in J_0}{\alpha }_j{t}^j\right)={\displaystyle \sum }_{j\in J_0}{\alpha }_j{y}_j,$$

where $J_0\subset {\mathbb{N}}^n$ is a finite subset and ${\alpha }_j$ are arbitrary real coefficients. If $\left\{f_1,\dots ,f_k\right\}$ is a finite subset of $ℝ\left[t_1,\dots ,t_n\right]$, then the closed subset given by

$$K =\left\{t\in {ℝ}^n;{ f}_1\left(t\right)\ge 0,\dots ,{ f}_k\left(t\right)\ge 0\right\}$$

(6)

is called a semi-algebraic set.

Theorem 9.

(see [2,16]). If $K$ is a compact semi-algebraic set, as defined above, then there is a positive Borel measure $\mu$ supported on $K$, such that

$${{\displaystyle \int }}_K^{}{t}^{j}d\mu =y_j, \forall j\in {\mathbb{N}}^n,$$

if, and only if:

$$T_y\left(f_1^{e_1}\cdots f_k^{e_k}{p}^2\right)\ge 0, \forall p\in ℝ\left[t_1,\dots ,t_n\right],\forall e_1,\dots ,e_k\in \left\{0,1\right\}.$$

Corollary 1.

(see [16]). With the above notations, if $p\in ℝ\left[t_1,\dots ,t_n\right]$ is such that $p\left(t\right)>0$ for all $t$ in the semi-algebraic compact $K$ defined by Equation (6), then $p$ is a finite sum of special polynomials of the form

$$f_1^{e_1}\cdots f_k^{e_k}{q}^2\ge 0,$$

for some $q\in ℝ\left[t_1,\dots ,t_n\right]$ and $e_1,\dots ,e_k\in \left\{0,1\right\}.$

Corollary 1 is named Schmüdgen’s Positivstellensatz. There also exists Putinar’s Positivstellensatz. These are representations of positive polynomials on basic closed semi-algebraic sets, in terms of sums of squares and polynomials defining the semi-algebraic set under attention.

Remark 2.

Let$F\subseteq {ℝ}^n$be an arbitrary closed subset, and$\mu$a positive regular Borel measure on$F$with finite moments$y_j={{\displaystyle \int }}_F^{}{t}^{j}d\mu$of all orders$j\in {\mathbb{N}}^n$. This yields

$${\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{y}_{i+j}={\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{{\displaystyle \int }}_F^{}{t}^{i+j}d\mu ={{\displaystyle \int }}_F^{}{\left({\displaystyle \sum }_{j\in J_0}{\lambda }_j{t}^j\right)}^{2}d\mu \ge 0,$$

for any finite subset$J_0\subset {\mathbb{N}}^n$and any$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ.$In other words, any moment sequence${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$is positive definite. If$n=1,$the converse is true, since any non-negative polynomial on$ℝ$is a sum of squares. Then, one applies Haviland’s theorem. However, for$n\ge 2$there exist positive definite sequences${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$, which are not moment sequences (see [4]).

In what follows, we review some known aspects of the problem of determinacy of a measure, in the one-dimensional case. A Hamburger moment sequence is determinate if it has a unique representing measure, while a Stieltjes moment sequence is called determinate if it has only one representing measure supported on $\left[0,\infty \right[.$ The Carleman theorem (the next result) contains a powerful sufficient condition for determinacy.

Theorem 10.

(Carleman’s condition; see [2], Theorem 4.3). Suppose that $\mathit{y}={\left(y_n\right)}_{n\in \mathbb{N}}$ is a positive semi-definite sequence (${{\displaystyle \sum }}_{i,j=0}^n{y}_{i+j}{\lambda }_i{\lambda }_j\ge 0$ for all $n\in \mathbb{N}$ and arbitrary ${\lambda }_j\in ℝ, j=0,\dots ,n).$

(i)

If$\mathit{y}$satisfies Carleman’s condition

$${\displaystyle \sum }_{n=1}^{\infty }{y}_{2n}^{-\frac{1}{2n}}=+\infty ,$$

then$\mathit{y}$is a determinate Hamburger moment sequence.

(ii)

If, in addition${\left(y_{n+1}\right)}_{n\in \mathbb{N}}$is positive semidefinite, and

$${\displaystyle \sum }_{n=1}^{\infty }{y}_n^{-\frac{1}{2n}}=+\infty ,$$

then$\mathit{y}$is a determinate Stieltjes moment sequence.

The following theorem of Krein consists of a sufficient condition for indeterminacy (for measures given by densities).

Theorem 11.

(See [2], Theorem 4.14). Let $f$ be a non-negative Borel function on $ℝ.$ Suppose that the measure $\mu$ defined by $d\mu =f\left(t\right)dt$ is a Radon measure on $ℝ,$ and has finite moments $y_n:={{\displaystyle \int }}_{ℝ}^{}{t}^{n}d\mu$ for all $n\in \mathbb{N}.$ If

$${{\displaystyle \int }}_{ℝ}^{}\frac{ln\left(f\left(x\right)\right)}{1+x^2}dx>-\infty ,$$

then the moment sequence $y={\left(y_n\right)}_{n\in \mathbb{N}.}$ is $M-$indeterminate.

Next, we present new checkable sufficient conditions on distributions of random variables that imply Carleman’s condition, ensuring determinacy. Consider two random variables: $V ~ \mathsf{\Psi },$ $V$with values in $ℝ, \mathrm{and} W ~ \mathsf{\Phi }, W$ with values in ${ℝ}_{+}.$ Assume that both $\mathsf{\Psi }$ and $\mathsf{\Phi }$ have continuous derivatives, and let $f={\mathsf{\Psi }}^{\prime }$ and $g={\mathsf{\Phi }}^{\prime }$, respectively, be the corresponding densities. All moments of $\mathsf{\Psi }$, $\mathsf{\Phi }$ are assumed to be finite. The symbol $\nearrow$ used below has the usual meaning of “monotone increasing”.

Theorem 12.

(See [20], Theorem 1; Hamburger case). Assume that the density $f$ of $\Psi$ is symmetrical on $ℝ$, and continuous and strictly positive outside an interval $\left(-t_0,t_0\right), t_0>1,$ such that the following conditions hold:

$$\begin{gathered} {{\displaystyle \int }}_{\left|t\right|\ge t_0}^{}\frac{-\ln f\left(t\right)}{t^2 \ln \left(\left|t\right|\right)}dt=\infty , \\ \frac{-\ln f\left(t\right)}{\ln t}\nearrow \infty \mathrm{as} t_0\le t\to \infty . \end{gathered}$$

then $V~ \Psi$ satisfies Carleman’s condition and, hence, is $M-$determinate.

Theorem 13.

(See [20], Theorem 2; Stieltjes case). Assume that the density $g$ of $\mathsf{\Phi }$ is continuous and strictly positive on $\left[a,\infty \right)$ for some $a>1$, such that the following conditions hold:

$$\begin{gathered} {{\displaystyle \int }}_a^{\infty }\frac{-\ln g\left(t^2\right)}{t^2\ln t}dt=\infty , \\ \frac{-\ln g\left(t\right)}{\ln t}\nearrow \infty \mathrm{as} a\le t\to \infty . \end{gathered}$$

Under these conditions, $W~ \mathsf{\Phi }$ satisfies Carleman’s condition and, hence, is $M-$determinate.

Example 1.

The distribution function$\Phi$with a density of g$\left(u\right)=exp\left(-u\right), u\in {ℝ}_{+},$satisfies the conditions of Theorem 13; hence, it is$M-$determinate.

Remark 3.

The problem of determinacy of measures on${ℝ}^n, {ℝ}_{+}^n, n\ge 2,$is much more difficult than that for the case$n=1.$In the next subsection, we partially solve this problem (see Lemmas 3 and 4, Theorems 15 and 19, as well as their corollaries, stated or proved in Section 3.2).

3.2. Polynomial Approximation on Unbounded Subsets, Markov Moment Problem, and Related Problems

The present section is based on results published in [36,38], also using the expression of non-negative polynomials on a strip [22]. We start by recalling a polynomial approximation result on ${ℝ}_{+}$, and some of its applications. Its proof follows via the Stone–Weierstrass theorem.

Lemma 1.

(See [38], Lemma 1). Let $\psi :{ℝ}_{+}=\left[0,\infty \right[\to {ℝ}_{+}$ be a continuous function, such that $\underset{t\to \infty }{\mathit{lim}}\psi \left(t\right)$ exists in ${ℝ}_{+}.$ Then, there is a decreasing sequence ${\left(h_l\right)}_l$ in $Span\left\{e_k;k\in \mathbb{N}\right\}$, where the functions $e_k;k\in \mathbb{N}$ are defined as follows:

$$e_k\left(t\right)=exp\left(-kt\right), k\in \mathbb{N}, t\in \left[0,\infty \right),$$

such that $h_l\left(t\right)\ge \psi \left(t\right)$, $t\ge 0, l\in \mathbb{N}=\left\{0,1,2,\dots \right\}$, $lim{h}_l=\psi$ uniformly on $\left[0,\infty \right[$ . There exists a sequence of polynomial functions ${\left({\tilde{p}}_l\right)}_{l\in \mathbb{N}}$, ${\tilde{p}}_l\ge h_l\ge \psi , lim {\tilde{p}}_l=\psi$, uniformly on compact subsets of $\left[0,\infty \right).$ In particular, such polynomial approximation holds for non-negative continuous compactly supported functions $\psi :{ℝ}_{+}\to {ℝ}_{+}$.

Proof. 

The idea is to consider the sub-algebra $\widehat{S}=Span\left\{{\widehat{e}}_k;k\ge 0\right\}$ of $C\left(\left[0,\infty \right]\right)$, where $\left[0,\infty \right]$ is the Alexandroff extension of $\left[0,\infty \right[,$ and ${\widehat{e}}_k$ is the continuous extension of $e_k$ to $\left[0,\infty \right], {\widehat{e}}_k(\infty )=0, k\ge 1, {\widehat{e}}_0(\infty )=1.$ This sub-algebra clearly separates the points of $\left[0,\infty \right],$ and contains the constant functions. According to the Stone–Weierstrass theorem,$\widehat{S}$ is dense in $C\left(\left[0,\infty \right]\right).$ It follows that any continuous function $\psi :{ℝ}_{+}\to {ℝ}_{+}$ with the property that the limit $\underset{t\to \infty }{\lim }\psi \left(t\right)$ exists in ${ℝ}_{+}$ can be uniformly approximated on ${ℝ}_{+}$ by elements from $Span\left\{e_k;k\ge 0\right\}.$ As is well known, when the convergence is uniform, the approximating sequence ${\left(h_l\right)}_l$for $\psi$ can be chosen such that $h_l\left(t\right)\ge \psi \left(t\right)\ge 0$ for all $t\in \left[0,\infty \right[.$ Assume the following: $h_l={{\displaystyle \sum }}_{k=0}^{m_l}{\alpha }_{l,k}{e}_k, {\alpha }_{l,k}\in ℝ,l=0,1,2,\dots$ If ${\alpha }_{l,k}\ge 0,$ we obtain ${\alpha }_{l,k}{e}_k\le {\alpha }_{l,k}{p}_{l,k}$, where $p_{l,k}$ is a majorizing partial sum of the power series of $e_k, e_k=\underset{l\to \infty }{\lim }{p}_{l,k}$, the convergence being uniform on any compact subset of ${ℝ}_{+}.$ If ${\alpha }_{l,k}<0,$ we deduce ${\alpha }_{l,k}{e}_k<{\alpha }_{l,k}{q}_{l,k}$, where $q_{l,k}$ is a minorizing partial sum of the power series of $e_k=\underset{l\to \infty }{\lim }{p}_{l,k},$ and the convergence is uniform on compact subsets of the non-negative semi-axes. Summing as $k=0,1,\dots ,m_l,$ one obtains a polynomial ${\tilde{p}}_l\ge h_l\ge \psi \ge \mathbf{0}$ on ${ℝ}_{+}.$ Since the sum defining ${\tilde{p}}_l$ has a finite number of terms of such partial sums, we conclude ${\tilde{p}}_l\to \psi$ uniformly on compact subsets of ${ℝ}_{+},$ as $l\to \infty .$ This ends the proof.  □

In applications, the preceding lemma could be useful in order to prove a similar type of result for continuous functions defined only on a compact subset $K\subset {ℝ}_{+},$ taking values in ${ℝ}_{+}.$ For such a function as $\phi :K\to {ℝ}_{+},$ one denotes by ${\phi }_0:{ℝ}_{+}\to {ℝ}_{+}$ the extension of $\phi$, which satisfies ${\phi }_0\left(t\right)=0$ for all $t\in {ℝ}_{+}\backslash K.$ From Lemma 13 we infer the next result.

Lemma 2.

(See [38], Lemma 2). If $K\subset {ℝ}_{+}$ is a compact subset, and $\phi :K\to {ℝ}_{+}$ a continuous function, then there exists a sequence ${\left({\tilde{p}}_l\right)}_{l\in \mathbb{N}}$ of polynomial functions, such that ${\tilde{p}}_l\ge {\phi }_0$ on ${ℝ}_{+}, {\tilde{p}}_l{|}_K\to \phi , l\to \infty$, uniformly on $K.$

Proof. 

The idea is to reduce the proof to that of the preceding Lemma 1. Specifically, we can easily construct a continuous extension $\psi :{ℝ}_{+}\to {ℝ}_{+}$ of $\phi ,$ with compact support $supp\left(\psi \right), \psi \ge {\phi }_0.$ Assuming this is done, if ${\left({\tilde{p}}_l\right)}_{l\in \mathbb{N}}$ are as in Lemma 1, since ${\tilde{p}}_l\to \psi$ uniformly on the compact $K$ and $\psi \left(t\right)=\phi \left(t\right)$for all $t$ in $K,$ it results in the following first conclusion:

$$\underset{t\in K}{\mathit{sup}}\left| {\tilde{p}}_l\left(t\right)-\phi \left(t\right)\right|=\underset{t\in K}{\mathit{sup}}\left| {\tilde{p}}_l\left(t\right)-\psi \left(t\right)\right|\to 0, l\to \infty .$$

Moreover, according to Lemma 1, we have ${\tilde{p}}_l\ge \psi \ge {\phi }_0\ge \mathbf{0}$ on ${ℝ}_{+}.$This will end the proof. To construct $\psi$, let $a=infK, b=supK, 0\le a\le b<\infty .$ It is clear that ${\phi }_0$ might have discontinuities at the ends of the intervals representing connected components of $\left[0,\infty \right[\backslash K.$ If $\phi \left(b\right)=0,$ then ${\phi }_0$ is continuous at $b$ and on the entire interval [b,$\infty [.$ If $\phi \left(b\right)>0$, for an arbitrary $\epsilon >0,$ define $\psi$ on the interval $\left[b,b+\epsilon \right]$ as the affine function whose graph is the line segment joining the points

$$\left(b,\phi \left(b\right)\right)\mathrm{and} \left(b+\epsilon ,0\right), \psi \left(t\right)=0 \mathrm{for} \mathrm{all} t>b+\epsilon , \psi \left(t\right)=\phi \left(t\right) \mathrm{for} \mathrm{all} t \mathrm{in}K.$$

It remains to define $\psi$ on each bounded connected component of $\left[0,\infty \right)\backslash K.$ Let ]$t_1,t_2[$ be such an interval, $t_1,t_2\in K, t_1<t_2$and $0<\epsilon <\left(t_2-t_1\right)/2.$ On the interval $\left[t_2-\epsilon ,t_2\right]$, we define $\psi$ as the affine function whose graph is the line segment of ends $\left(t_2-\epsilon ,0\right), \left(t_2,\phi \left(t_2\right)\right)$. Similarly, on the interval $\left[t_1,t_1+\epsilon \right]$, we consider the line segment joining the points $\left(t_1,\phi \left(t_1\right)\right), \left(t_1+\epsilon ,0\right)$. The definition at points $t_1,t_2$ is in accordance with the previous condition and $\psi \left(t\right)=\phi \left(t\right)$for all $t\in K.$ On the interval $\left]t_1+\epsilon , t_2-\epsilon \right[,$ $\psi \equiv \mathbf{0}.$ Finally, if $a>0$ and $\phi \left(a\right)>0,$ taking $0<\epsilon <a,$ we define $\psi$ on the interval $\left[a-\epsilon ,a\right]$ as being the function whose graph is the line segment joining the points $\left(a-\epsilon ,0\right), \left(a,\phi \left(a\right)\right), \psi \equiv \mathbf{0}$ on $\left[0,a-\epsilon \right].$ If $a=0,$ we have $\psi \left(a\right)=\phi \left(a\right),$since $a\in K$ and the interval $\left]0,a\right[$ is empty. If $a>0$ and $\phi \left(a\right)=0,$ we define $\psi \equiv \mathbf{0}$ on $\left[0,a\right].$ Thus, $\psi$ is defined, non-negative, and continuous on [0,$\infty [, \psi {|}_K\equiv \phi ,$ while $supp\left(\psi \right)$ is compact, contained in $\left[0,b+\epsilon \right].$ The proof is complete.  □

From Lemma 2 and Theorem 1 (where $X_1$ stands for $C\left(K\right)$ and $X_0$ stands for the subspace $𝒫$ of all polynomial functions), the next corollary follows easily. We recall a well-known important example of an order complete Banach lattice $Y$ of self-adjoint operators acting on a complex or real Hilbert space $H.$ Let $\mathcal{A}=\mathcal{A}\left(H\right)$ be the ordered vector space of all of the self-adjoint operators acting on $H,$ and let $A\in \mathcal{A}.$ The natural order relation on $\mathcal{A}$ is $A\le B$ if, and only if:

$$\langle Ah,h\rangle \le \langle Bh,h\rangle \mathrm{for} \mathrm{all} h \mathrm{in} H.$$

One can prove that $\mathcal{A}$ with this ordering is not a lattice. Therefore, it is interesting to fix $A\in \mathcal{A}$ and define the following:

$$Y_1\left(A\right)=\left\{V\in \mathcal{A};AV=VA\right\}, Y\left(A\right) =\left\{V\in Y_1\left(A\right);WV=VW, \forall W\in Y_1\left(A\right)\right\}.$$

(7)

Then, $Y\left(A\right)$ is an order complete Banach lattice (and a commutative real algebra), as discussed in [5]. If $V\in \mathcal{A},$ we denote by $\sigma \left(V\right)$ the spectrum of $V$, and by$d{E}_V$the spectral measure attached to $V.$

Corollary 2.

With the above notations, assume that$A$is a positive self-adjoint operator acting on$H$,$Y=Y\left(A\right)$is the space defined by Equation (7), and${\left(B_j\right)}_{j\in \mathbb{N}}$is a sequence of operators in$Y\left(A\right).$The following statements are equivalent:

(a)

There exists a unique positive linear operator$T:C\left(\sigma \left(A\right)\right)\to Y$, such that

$$T\left({\phi }_j\right)=B_j, j\in \mathbb{N}, T\left(\phi \right)\le {{\displaystyle \int }}_{\sigma \left(A\right)}^{}\phi \left(t\right)d{E}_A, \phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+},T\le 1.$$

(b)

For any polynomial${{\displaystyle \sum }}_{j=0}^m{\alpha }_j{\phi }_j\ge \mathbf{0}$ on$\sigma \left(A\right),$the result is${{\displaystyle \sum }}_{j=0}^m{\alpha }_j{B}_j\ge \mathbf{0};$if$J_0\subset \mathbb{N}$is an arbitrary finite subset, and$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$then the following inequalities hold:

$${\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{B}_{i+j+l}\le {\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{A}^{i+j+l}, l\in \left\{0,1\right\}.$$

Proof. 

$\left(b\right)\Rightarrow \left(a\right).$We define $T_0:𝒫\to Y\left(A\right)$ by $T_0\left({\displaystyle \sum }_{j\in J_0}{\alpha }_j{\phi }_j\right):={\displaystyle \sum }_{j\in J_0}{\alpha }_j{B}_j$, where $J_0\subset \mathbb{N}$ is an arbitrary finite subset, ${\alpha }_j\in ℝ, j\in J_0.$ Then, $T_0$ is linear and, according to the first condition (b), $T_0\left(p\right)\ge \mathbf{0}$ for all polynomials $p$ with $p\left(t\right)\ge 0 \forall t\in \sigma \left(A\right).$ On the other hand, for each $g\in C\left(\sigma \left(A\right)\right)$, there exists a constant function $c$, such that $g\left(t\right)\le c$ for all $t\in \sigma \left(A\right).$ According to Theorem 1, $T_0$ has a linear positive extension $T:C\left(\sigma \left(A\right)\right)\to Y$. Next, we prove that $T$ is continuous (and its norm can be determined). This can be shown for any positive linear operator, in a more general framework. Namely, any positive linear operator acting between two ordered Banach spaces is continuous (see [6] and/or [27]). Here, we are interested only in our problem, when the norm of the involved positive linear operator can be determined. Indeed, for an arbitrary $g\in C\left(\sigma \left(A\right)\right),$ we can write:

$$\pm T\left(g\right)=T\left(\pm g\right)\le T\left(\left|g\right|\right)\le T\left(\Vert g\Vert \mathbf{1}\right)=\Vert g\Vert T\left(\mathbf{1}\right)\Rightarrow \left|T\left(g\right)\right|\le \Vert g\Vert T\left(\mathbf{1}\right)$$

Since the norm on $Y\left(A\right)$ is solid $\left(\left|y_1\right|\le \left|y_2\right|\Rightarrow \Vert y_1\Vert \le \Vert y_2\Vert \right)$, the preceding inequality leads to:

$$\Vert T\left(g\right)\Vert \le \Vert g\Vert \Vert T\left(\mathbf{1}\right)\Vert , g\in C\left(\sigma \left(A\right)\right)\Rightarrow \Vert T\Vert \le \Vert T\Vert \left(\mathbf{1}\right)=\Vert T_0\left({\phi }_0\right)\Vert =\Vert B_0\Vert .$$

On the other hand, clearly $\Vert T\left(\mathbf{1}\right)\Vert \le \Vert T\Vert$; hence, $\Vert T\Vert =\Vert T\left(\mathbf{1}\right)\Vert =\Vert B_0\Vert .$ To finish the proof of the basic implication $\left(b\right)\Rightarrow \left(a\right),$ we only have to show that $T\left(\phi \right)\le {{\displaystyle \int }}_{\sigma \left(A\right)}^{}\phi \left(t\right)d{E}_A, \phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+},\Vert T\Vert \le 1.$

Let $\phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+}.$ According to Lemma 2, where $K$ stands for $\sigma \left(A\right),$ there exists a sequence ${\left({\tilde{p}}_l\right)}_{l\in \mathbb{N}}$ of polynomial functions such that ${\tilde{p}}_l\ge {\phi }_0\ge \mathbf{0}$ on ${ℝ}_{+}, {\tilde{p}}_l{|}_{\sigma \left(A\right)}\to \phi , l\to \infty$, uniformly on$\sigma \left(A\right).$ The continuity of $T$ and of $T_2:C\left(\sigma \left(A\right)\right)\to Y\left(A\right), T_2\left(g\right)={{\displaystyle \int }}_{\sigma \left(A\right)}^{}g{d}_{E_A},$ also using the last property on the sequence ${\left(B_j\right)}_{j\in \mathbb{N}}$ stated at (b), lead to:

$$T\left(\phi \right)=\underset{l}{\lim }T\left({\tilde{p}}_{\iota }\right)\ge \underset{l}{\lim }{T}_2{\tilde{p}}_l=T_2\left(\phi \right)={{\displaystyle \int }}_{\sigma \left(A\right)}^{}\phi d_{E_A}$$

In particular, for $\phi =\mathbf{1}$, we have: $T\left(\mathbf{1}\right)\le {{\displaystyle \int }}_{\sigma \left(A\right)}^{}\mathbf{1}{d}_{E_A}=I\Rightarrow \Vert T\Vert =\Vert T\left(\mathbf{1}\right)\Vert \le \Vert I\Vert =1.$ Here, $I:H\to H$is the identity operator; this proves $\left(b\right)\Rightarrow \left(a\right).$ It is worth noting that we have used the expression of non-negative polynomials on ${ℝ}_{+}:p\left(t\right)\ge 0 \forall t\in {ℝ}_{+}\iff p\left(t\right)=p_1^2\left(t\right)+t{p}_2^2\left(t\right), t\in {ℝ}_{+}$ for some $p_1,p_2\in ℝ\left[t\right].$ The implication $\left(a\right)\Rightarrow \left(b\right)$ is obvious.  □

Next, we generalize the sandwich condition $0\le T\le T_2$ on $X_{+}$, appearing in Corollary 2 (where $X=C\left(\sigma \left(A\right)\right)),$ to the condition $T_1\le T\le T_2$ on $X_{+},$ where $T_1$, $T_2$ are two given linear operators on $X, \mathbf{0}\le T_1\le T_2$ on $X_{+}.$ We start with a general result. Let $K\subset {ℝ}_{+}$ be an arbitrary compact subset. We denote by $X=C\left(K\right)$ the Banach lattice of all real-valued continuous functions on $K$, and let $Y$ be an arbitrary order complete Banach lattice. One denotes by ${\phi }_j$ the monomials ${\phi }_j\left(t\right)=t^j, t\in {ℝ}_{+}, j\in \mathbb{N}.$

Theorem 14.

(See [38], Theorem 3). Let $T_1,T_2$ be two linear operators from $X$ to $Y,$ such that $\mathbf{0}\le T_1\le T_2$ on the positive cone of $X,$ while ${\left(y_n\right)}_{n\ge 0}$ is a given sequence of elements in $Y.$ The following statements are equivalent:

(a)

There exists a unique (bounded) linear operator$T:X\to Y$, such that

$$T\left({\phi }_j\right)=y_j, j\in \mathbb{N}, T_1\le T\le T_2 \mathrm{on} \mathrm{the} \mathrm{positive} \mathrm{cone} \mathrm{of} X, \Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert ;$$

(b)

For any polynomial${{\displaystyle \sum }}_{j=0}^m{\alpha }_j{\phi }_j\ge \mathbf{0}$ on$K,$ we have${{\displaystyle \sum }}_{j=0}^m{\alpha }_j{T}_1\left({\phi }_j\right)\le {{\displaystyle \sum }}_{j=0}^m{\alpha }_j{y}_j ;$if$J_0\subset \mathbb{N}$is a finite subset, and$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$then the following conditions are satisfied:

$${\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{y}_{i+j+l}\le {\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{T}_2\left({\phi }_{i+j+l}\right), l\in \left\{0,1\right\};$$

(c)

$T_1\le T_2$on$X_{+}$, and for any polynomial${\displaystyle \sum }_{j\in J_0}{\lambda }_j{\phi }_j$, the following inequality holds:

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j\le T_2\left({\left({\displaystyle \sum }_{j\in J_0}{\lambda }_j{\phi }_j\right)}^{+}\right)-T_1\left({\left({\displaystyle \sum }_{j\in J_0}{\lambda }_j{\phi }_j\right)}^{-}\right).$$

Proof. 

According to the notations and assertions of (a), the implication $\left(a\right)\Rightarrow \left(b\right)$ is clear. To prove the converse implication, we observe that the first assertion of (b) says that in defining the following:

$$T_0\left({{\displaystyle \sum }}_{j=0}^m{\alpha }_j{\phi }_j\right)={{\displaystyle \sum }}_{j=0}^m{\alpha }_j{y}_j, m\in \mathbb{N},{\alpha }_j\in ℝ,$$

we obtain a linear operator defined on the subspace of polynomial functions, which verifies the moment conditions:

$$T_0\left({\phi }_j\right)=y_j,j\in \mathbb{N},$$

($T_0\ge T_1$ holds on the convex cone ${𝒫}_{+}$ of all polynomial functions which are non-negative on $K).$ On the other hand, any element from $X=C\left(K\right)$ is dominated by a constant function, so that the subspace $𝒫$ of polynomial functions defined on ${ℝ}_{+}$ verifies the hypothesis of Theorem 1, where $X_1$ stands for $X,$ and $X_0$ stands for $𝒫$. According to Theorem 1, the linear operator $T_0-T_1:𝒫\to Y$ which is positive on ${𝒫}_{+}=𝒫\cap X_{+}$, admits a positive linear extension $U:X\to Y.$ We define $T=T_1+U\ge T_1$ on $X_{+}.$ In addition, $T\in L_{+}\left(X,Y\right)$ verifies the following:

$$T\left({\phi }_j\right)=T_1\left({\phi }_j\right)+U\left({\phi }_j\right)=T_1\left({\phi }_j\right)+T_0\left({\phi }_j\right)-T_1\left({\phi }_j\right)=T_0\left({\phi }_j\right)=y_j,j\in \mathbb{N}.$$

In other words, $T:X\to Y$ is a linear extension of $T_0:𝒫\to Y$, which dominates $T_1$ on $X_{+}$. Next, we prove that $T\le T_2$ on $X_{+}.$ To this end, observe that according to the second assertion of (b), we already know that $T\le T_2$ on special polynomial functions, which are non-negative on the entirety of the semi-axes ${ℝ}_{+}.$ Indeed, any non-negative polynomial $p=p\left(t\right)$ on ${ℝ}_{+}$ has the explicit form $p\left(t\right)=q^2\left(t\right)+t{r}^2\left(t\right)$ for some $q,r\in ℝ\left[t\right].$ On the other hand, since

$$T\ge T_1\ge \mathbf{0},$$

the linear operator $T$ is positive and, hence, is also continuous; $T_2$ is continuous as well, thanks to its positivity. We now apply Lemma 2 for an arbitrary $\phi \in X_{+}.$ Using the notations of Lemma 2, and the above-discussed assertions, we infer the following:

$$\mathbf{0}\le T_1\left(\phi \right)\le T\left(\phi \right)=\underset{l\to \infty }{\lim }T\left( {\tilde{p}}_l\right)\le \underset{l\to \infty }{\lim }{T}_2\left( {\tilde{p}}_l\right)=T_2\left(\phi \right), \phi \in X_{+}.$$

It remains to prove the last relation of (a). If $\psi$ is an arbitrary function in $X$, then the preceding inequality yields

$$T\left(\psi \right)\le T\left(\left|\psi \right|\right)\le T_2\left(\left|\psi \right|\right)$$

and, similarly, $-T\left(\psi \right)=T\left(-\psi \right)\le T_2\left(\left|\psi \right|\right).$ These inequalities yield $\left|T\left(\psi \right)\right|\le T_2\left(\left|\psi \right|\right)$ and, since $Y$ is a Banach lattice, the conclusion is $\Vert T\left(\psi \right)\Vert \le \Vert T_2\left(\left|\psi \right|\right)\Vert \le \Vert T_2\Vert \Vert \left|\psi \right|\Vert =\Vert T_2\Vert \Vert \psi \Vert , \psi$in $X.$ Thus, $\Vert T\Vert \le \Vert T_2\Vert .$Similarly, $\Vert T_1\Vert \le \Vert T\Vert .$The equivalence $\left(a\right)\iff \left(c\right)$ follows directly from Theorem 2. This completes the proof.  □

Corollary 3.

With the notation of Corollary 2, assume that the spectrum$\sigma \left(A\right)$is contained in the interval$\left[0,1\right].$The following statements are equivalent:

(a)

There exists a unique positive linear operator$T:C\left(\sigma \left(A\right)\right)\to Y$, such that

$$\begin{gathered} T\left({\phi }_j\right)=B_j, j\in \mathbb{N}, {{\displaystyle \int }}_{\sigma \left(A\right)}^{}t\phi \left(t\right)d{E}_A\le T\left(\phi \right)\le {{\displaystyle \int }}_{\sigma \left(A\right)}^{}\phi \left(t\right)d{E}_A, \\ \forall \phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+}, \Vert A\Vert \le \Vert T\Vert \le 1; \end{gathered}$$

(b)

For any polynomial${{\displaystyle \sum }}_{j=0}^m{\alpha }_j{t}^j\ge 0 \forall t\in \sigma \left(A\right)$ $,$the result is${{\displaystyle \sum }}_{j=0}^m{\alpha }_j{B}_j\ge {{\displaystyle \sum }}_{j=0}^m{\alpha }_j{A}^{j+1};$if$J_0\subset \mathbb{N}$is a finite subset, and$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$then the following inequalities hold:

$${\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{B}_{i+j+l}\le {\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{A}^{i+j+l}, l\in \left\{0,1\right\}.$$

Proof. 

One applies Theorem 14, where $K$ stands for $\sigma \left(A\right),$

$$T_1\left(\phi \right)={{\displaystyle \int }}_{\sigma \left(A\right)}^{}t\phi \left(t\right)d{E}_A, T_2\left(\phi \right)={{\displaystyle \int }}_{\sigma \left(A\right)}^{}\phi \left(t\right)d{E}_A, \phi \in C\left(\sigma \left(A\right)\right).$$

We observe that $T_1\left(\phi \right)={{\displaystyle \int }}_{\sigma \left(A\right)}^{}td{E}_A\Rightarrow {{\displaystyle \int }}_{\sigma \left(A\right)}^{}\phi \left(t\right)d{E}_A=A{T}_2\left(\phi \right)\in \left[\mathbf{0},T_2\left(\phi \right)\right],$ for all $\phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+},$ since $0\le t\le 1$ for all $t\in \sigma \left(A\right), \phi \left(t\right)\ge 0, t\in \sigma \left(A\right),$ and the algebra $Y\left(A\right)$ is commutative. Moreover, as in the proof of Corollary 2,

$$\Vert T_1\Vert =\Vert T_1\left(\mathbf{1}\right)\Vert =\Vert {{\displaystyle \int }}_{\sigma \left(A\right)}^{}td{E}_A\Vert ·\Vert I\Vert =\Vert A\Vert .$$

On the other hand, the first condition of (b) says that

$${{\displaystyle \sum }}_{j=0}^m{\alpha }_j{B}_j\ge {{\displaystyle \sum }}_{j=0}^m{\alpha }_j{T}_1\left({\phi }_j\right).$$

The second condition of (b) is exactly the second condition of (b) written in Theorem 14, where $y_n$ stands for $B_n, n\in \mathbb{N}.$ The conclusion follows via Theorem 14.  □

Next, we review the key polynomial approximation result of this subsection, which works on arbitrary closed subsets $F\subseteq {ℝ}^n.$

Lemma 3.

(See [36], Lemma 3). Let $F\subseteq {ℝ}^{n }$ be an unbounded closed subset, and $\nu$ an M-determinate measure on $F$ (with finite moments of all natural orders). Then, for any $x\in C_0\left(F\right), x\left(t\right)\ge 0, \forall t\in F,$ there exists a sequence ${\left(p_m\right)}_m,p_m\ge x, m\in \mathbb{N}, p_m\to x$ in $L_{\nu }^1\left(F\right)$. In particular, we have

$$\underset{m}{\lim }{{\displaystyle \int }}_F^{}{p}_m\left(t\right)d\nu ={{\displaystyle \int }}_F^{}x\left(t\right)d\nu ,$$

${𝒫}_{+}$ is dense in ${\left(L_{\nu }^1\left(F\right)\right)}_{+}$, and ${𝒫}_{+}$ is dense in $L_{\nu }^1\left(F\right).$

Proof. 

To prove the assertions of the statement, it is sufficient to show that for any $x\in {\left(C_0\left(F\right)\right)}_{+}$, we have

$$Q_1\left(x\right):=inf\left\{{{\displaystyle \int }}_F^{}p\left(t\right)d\nu ;p\ge x, p\in 𝒫\right\}={{\displaystyle \int }}_F^{}x\left(t\right)d\nu .$$

Obviously, one has

$$Q_1\left(x\right)\ge {{\displaystyle \int }}_F^{}x\left(t\right)d\nu$$

(8)

To prove the converse, we define the linear form

$$T_0:X_0:=𝒫\oplus Sp\left\{x\right\}\to ℝ, F_0\left(p+\alpha x\right):={{\displaystyle \int }}_F^{}p\left(t\right)d\nu +\alpha Q_1\left(x\right), p\in 𝒫, \alpha \in ℝ.$$

Next, we show that $F_0$ is positive on $X_0$. In fact, for $\alpha <0$, one has (from the definition of $Q_1$, which is a sublinear functional on $X_1$):

$$p+\alpha x\ge 0\Rightarrow p\ge -\alpha x\Rightarrow \left(-\alpha \right)Q_1\left(x\right)=Q_1\left(-\alpha x\right)\le \underset{F}{\overset{}{{\displaystyle \int }}}p\left(t\right)d\nu \Rightarrow T_0\left(p+\alpha x\right)\ge 0.$$

If $a\ge 0$, we infer that:

$$\begin{gathered} 0=Q_1\left(0\right)=Q_1\left(\alpha x-\alpha x\right)\le \alpha Q_1\left(x\right)+Q_1\left(-\alpha x\right)\Rightarrow \\ \underset{F}{\overset{}{{\displaystyle \int }}}p\left(t\right)d\nu \ge Q_1\left(-\alpha x\right)\ge -\alpha Q_1\left(x\right)\Rightarrow T_0\left(p+\alpha x\right)\ge 0 \end{gathered}$$

where, in both possible cases, we have $x_0\in {\left(X_0\right)}_{+}\Rightarrow T_0\left(x_0\right)\ge 0.$ Since $X_0$ contains the space of the polynomials’ functions, which is a majorizing subspace of $X_1$, there exists a linear positive extension $T:X\to ℝ$ of $T_0,$ which is continuous on $C_0\left(F\right)$ with respect to the sup-norm. Therefore, $T$ has a representation by means of a positive Borel regular measure $\mu$ on $F$, such that

$$T\left(x\right)={{\displaystyle \int }}_F^{}x\left(t\right)d\mu , x\in C_0\left(F\right).$$

Let $p\in {𝒫}_{+}$ be a non-negative polynomial function. There is a nondecreasing sequence ${\left(x_m\right)}_m$ of continuous non-negative function with compact support, such that $x_m\nearrow p$ pointwise on $F$. Positivity of $T$ and Lebesgue’s dominated convergence theorem for $\mu$ Yield

$$\underset{F}{\overset{}{{\displaystyle \int }}}p\left(t\right)d\nu =T\left(p\right)\ge \mathit{sup}T\left(x_m\right)=\mathit{sup}\underset{F}{\overset{}{{\displaystyle \int }}}{x}_m\left(t\right)d\mu =\underset{F}{\overset{}{{\displaystyle \int }}}p\left(t\right)d\mu ,p\in {𝒫}_{+}$$

Thanks to Haviland’s theorem, there exists a positive Borel regular measure $\lambda$ on $F$, such that

$$\lambda \left(p\right)=\nu \left(p\right)-\mu \left(p\right)\iff \nu \left(p\right)=\lambda \left(p\right)+\mu \left(p\right), p\in 𝒫.$$

Since $\nu$ is assumed to be M-determinate, it follows that

$$\nu \left(B\right)=\lambda \left(B\right)+\mu \left(B\right)$$

for any Borel subset $B$ of $F$. From this last assertion, approximating each $x\in {\left(L_{\nu }^1(F\right)}_{+}$, by a nondecreasing sequence of non-negative simple functions, and also using Lebesgue’s convergence theorem, one obtains firstly for positive functions, then for arbitrary $\nu$-integrable functions, $\phi :$

$${{\displaystyle \int }}_F^{}\phi d\nu ={{\displaystyle \int }}_F^{}\phi d\lambda +{{\displaystyle \int }}_F^{}\phi d\mu , \phi \in L_{\nu }^1\left(F\right).$$

In particular, we must have

$${{\displaystyle \int }}_F^{}xd\nu \ge {{\displaystyle \int }}_F^{}xd\mu =T\left(x\right)=T_0\left(x\right) =Q_1\left(x\right)$$

(9)

Then, Equations (8) and (9) conclude the proof.  □

Remark 4.

We recall that the preceding Lemma 3 is no more valid when we replace$L_{\nu }^1\left(F\right)$with the Hilbert space$L_{\nu }^2\left(F\right), F={ℝ}^n, n\ge 2$(see [13], Theorem 4.4, where the authors construct such a measure$\nu$).

Theorem 15.

(See [36], Theorem 2). Let $F$ be a closed unbounded subset of ${ℝ}^n, Y$ an order complete Banach lattice, ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ a given sequence in $Y,$ and $\nu$ an $M-$determinate measure on $F.$ Let $T_2\in B\left(L_{\nu }^1\left(F\right),Y\right)$ be a linear positive (bounded) operator from $L_{\nu }^1\left(F\right)$ to $Y.$ The following statements are equivalent:

(a)

There exists a unique linear operator$T\in B\left( L_{\nu }^1\left(F\right),Y\right)$, such that$T\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^n,$ $F$is between$\mathbf{0}$and$T_2$on the positive cone of$L_{\nu }^1\left(F\right),$and$\Vert T\Vert \le \Vert T_2\Vert ;$

(b)

For any finite subset$J_0\subset {\mathbb{N}}^n,$and any${\left\{a_j\right\}}_{j\in J_0}\subset ℝ,$we have

$${{\displaystyle \sum }}_{j\in J_0}{a}_j{\phi }_j\ge 0 on F \Rightarrow 0\le {{\displaystyle \sum }}_{j\in J_0}{a}_j{y}_j\le {{\displaystyle \sum }}_{j\in J_0}{a}_j{T}_2\left({\phi }_j\right).$$

Proof. 

Observe that the assertion (b) says that

$$0\le T_0\left(p\right)\le T_2\left(p\right), p\in {𝒫}_{+},$$

(10)

where $T_0:𝒫\to Y$ is the unique linear operator that verifies the interpolation conditions of Equation (1). Thus, $\left(a\right)\Rightarrow \left(b\right)$ is obvious. To prove the converse, consider the vector subspace $X_1\subset L_{\nu }^1\left(F\right)$ of all functions $\phi \in L_{\nu }^1\left(F\right)$, verifying

$$\left|\phi \left(t\right)\right|\le p\left(t\right) \forall t\in F$$

for some polynomial $p.$ Clearly, $X_1$ contains the subspace of polynomials as well as the subspace of continuous compactly supported real-valued functions. On the other hand, the subspace of polynomials is a majorizing subspace in $X_1$, and according to the first inequality of Equation (10), $T_0$ is positive as a linear operator on $𝒫.$ Application of Theorem 1 yields the existence of a positive linear extension $T:X_1\to Y$ of $T_0.$ Let $x$ be a non-negative continuous compactly supported function on $F$, and ${\left(p_m\right)}_m$ a sequence of polynomials with the properties specified in Lemma 3. According to the second inequality of Equation (10), we have

$$T\left(p_m\right)=T_0\left(p_m\right)\le T_2(p_m) \mathrm{for} \mathrm{all} m\in \mathbb{N}.$$

Our next goal is to prove that

$$T\left(x\right)\le T_2\left(x\right)$$

(11)

Assuming the contrary, we should have $T_2\left(x\right)-T\left(x\right)\notin Y_{+}$. Since $Y_{+}$ is closed, a Hahn–Banach separation theorem leads to the existence of a positive linear form $y^{\star }$ in the dual $Y^{\star }$ of $Y,$ verifying

$$y^{\star }\left(T_2\left(x\right)-T\left(x\right)\right)<0.$$

(12)

The positive linear form $y^{\star }\circ T$ has a representing positive regular Borel measure $\mu$, for which Fatou’s lemma can be applied; we infer that

$$y^{\star }\left(T\left(x\right)\right)\le \mathrm{limin}{f}_m{y}^{\star }\left(T\left(p_m\right)\right)\le \mathrm{limin}{f}_m{y}^{\star }\left(T_2\left(p_m\right)\right)=y^{\star }\left(T_2\left(\underset{m}{\lim }{p}_m\right)\right)=y^{\star }\left(T_2\left(x\right)\right).$$

(13)

Equations (12) and (13) yield

$$y^{\star }\left(T_2\left(x\right)\right)<y^{\star }\left(T\left(x\right)\right)\le y^{\star }\left(T_2\left(x\right)\right),$$

implying the contradiction $y^{\star }\left(T_2\left(x\right)\right)<y^{\star }\left(T_2\left(x\right)\right).$ Hence, the assumption $T_2\left(x\right)-T\left(x\right)\notin Y_{+}$ was false, such that we must have $T_2\left(x\right)-T\left(x\right)\in Y_{+},$ i.e., Equation (11) is proven. Now, let $\phi \in C_0\left(F\right)$ be arbitrary. According to the preceding considerations, we obtain

$$\left|T\left(\phi \right)\right|=\left|T\left({\phi }^{+}\right)-T\left({\phi }^{-}\right)\right|\le T\left({\phi }^{+}\right)+T\left({\phi }^{-}\right)\le T_2\left({\phi }^{+}\right)+T_2\left({\phi }^{-}\right)=T_2\left(\left|\phi \right|\right).$$

Since the norm on $Y$ is solid ($\left|y_1\right|\le \left|y_2\right|\Rightarrow \Vert y_1\Vert \le \Vert y_2\Vert ),$ we infer that

$$\Vert T\left(\phi \right)\Vert \le \Vert T_2\left(\left|\phi \right|\right)\Vert \le \Vert T_2\Vert {\Vert \left|\phi \right|\Vert }_1=\Vert T_2\Vert {\Vert \phi \Vert }_1, \forall \phi \in C_0\left(F\right).$$

Using the fact that $C_0\left(F\right)$ is dense in $L_{\nu }^1\left(F\right)$ (see [9]), the last evaluation leads to the existence of a linear extension $\tilde{T}:L_{\nu }^1\left(F\right)\to Y$of $T,$ such that

$$\Vert \tilde{T}\left(\phi \right)\Vert \le \Vert T_2\Vert {\Vert \phi \Vert }_1, {\Vert \phi \Vert }_1=\underset{A}{\overset{}{{\displaystyle \int }}}\left|\phi \left(t\right)\right|d\nu , \phi \in L_{\nu }^1\left(F\right).$$

It follows that $\Vert \tilde{T}\Vert \le \Vert T_2\Vert$, and the positivity of $\tilde{T}$ is a consequence of the positivity of $T,$ via continuity of the extension $\tilde{T}$ and the density of ${\left(C_0\left(F\right)\right)}_{+}$ in ${\left(L_{\nu }^1\left(F\right)\right)}_{+}.$ We also note that

$$\tilde{T}\left({\phi }_j\right)=T_0\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^n.$$

This concludes the proof.  □

We continue by recalling a major result on the form of non-negative polynomials in a strip [22], which leads to a simple solution for the related Markov moment problem.

Theorem 16.

(See [22]). Supposing that $p\left(t_1,t_2\right)\in ℝ\left[t_1,t_2\right]$ is non-negative on the strip $F=\left[0,1\right]\times ℝ$, then $p\left(t_1,t_2\right)$ is expressible as

$$p\left(t_1,t_2\right)=\sigma \left(t_1,t_2\right)+\tau \left(t_1,t_2\right)t_1\left(1-t_1\right),$$

where $\sigma \left(t_1,t_2\right),\tau \left(t_1,t_2\right)$ are sums of squares in $ℝ\left[t_1,t_2\right].$

Let $F=\left[0,1\right]\times ℝ,$ $\nu$ a $M-$ determinate measure on$F,$ and $X=L_{\nu }^1\left(F\right),{\phi }_j\left(t_1,t_2\right):=t_1^{j_1}{t}_2^{j_2}, j=\left(j_1,j_2\right)\in {\mathbb{N}}^2, \left(t_1,t_2\right)\in F.$ Let $Y$ be an order complete Banach lattice, and ${\left(y_j\right)}_{j\in {\mathbb{N}}^2}$ a sequence of given elements in $Y.$ The next result follows as a consequence of Theorems 16 and 15.

Theorem 17.

Let$T_2\in B_{+}\left(X,Y\right)$be a linear (bounded) positive operator from$X$to$Y.$The following statements are equivalent:

(a)

There exists a unique (bounded) linear operator$T:X\to Y$, such that$T\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^2,$where$T$is between zero and$T_2$on the positive cone of$X, \Vert T\Vert \le \Vert T_2\Vert ;$

(b)

For any finite subset$J_0$⊂${\mathbb{N}}^2,$and any$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$we have

$$\begin{gathered} 0\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{y}_{i+j}\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{T}_2\left({\phi }_{i+j}\right); \\ 0\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j\left(y_{i_1+j_1+1, i_2+j_2}-y_{i_1+j_1+2, i_2+j_2}\right)\le \\ {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j\left(T_2\left({\phi }_{i_1+j_1+1, i_2+j_2}-{\phi }_{i_1+j_1+2, i_2+j_2}\right)\right), i=\left(i_1,i_2\right), j=\left(j_1,j_2\right)\in J_0. \end{gathered}$$

Lemma 4.

(See [36], Lemma 4 and the references of [36]). Let $\nu ={\nu }_1\times \cdots \times {\nu }_n$ be a product of n$M-$ determinate measures on ${ℝ}_{+}=\left[0,\infty \right[.$ Then we can approximate any non-negative continuous compactly supported function in $X=L_{\nu }^1\left({ℝ}_{+}^n\right)$ with sums of products

$$p_1\otimes \cdots \otimes p_n,$$

$p_j$ positive polynomial on the real non-negative semi-axis, in variable $t_j\in {ℝ}_{+}, j=1,\dots ,n,$ where

$$(p_1\otimes \cdots \otimes p_n)\left(t_1,\dots ,t_n\right)=p_1\left(t_1\right)\cdots p_n\left(t_n\right).$$

Proof. 

Let $f\in {\left(C_0\left({ℝ}_{+}^n\right)\right)}_{+}$, $K_i=p{r}_i\left(supp\left(f\right)\right), a_i=inf{K}_i$, $b_i=sup{K}_i, i=1,\dots ,n, K=\left[a_1,b_1\right]\times \cdots \times \left[a_n,b_n\right]$.

The restriction of $f$ to the parallelepiped $K$ can be approximated uniformly on $K$ by Bernstein polynomials $B_m$ in $n$ variables. Any such polynomial $B_m$ is a sum of products of the form $q_{m,1}\otimes \cdots \otimes q_{m,n}$, where each $q_{m,i}$ is a polynomial non-negative on $\left[a_i,b_i\right], i=1,\dots ,n,m\in \mathbb{N}.$ Thus, $B_m$ can be written as

$$B_m={\displaystyle \sum }_{\begin{array}{c}{k}_i=0,\dots ,m,\\ i=1,\dots ,n\end{array}}{q}_{m,k_1}\otimes \cdots \otimes q_{m,k_n},$$

where $q_{m,k_i}$ is a non-negative polynomial on $\left[a_i,b_i\right], i=1,\dots ,n,m\in \mathbb{N}.$ Based on the Weierstrass–Bernstein uniform approximation theorem, we have:

$$\Vert f-B_m\Vert {}_{\infty }:=\underset{t\in K}{sup}\left|f\left(t\right)-B_m\left(t\right)\right|\to 0, m\to \infty .$$

By abuse of notation, we write $q_{m,i}=q_{m,k_i}.$ We need a similar approximation, with sums of tensor products of non-negative polynomials $p_i$, $p_i\left(t_i\right)\ge 0$ for all $t_i\in {ℝ}_{+}$, $i=1,\dots ,n$ in the space $L_{\nu }^1\left({ℝ}_{+}^n\right).$ To this aim, the idea is to use Lemma 18 for $n=1, F={ℝ}_{+}$, followed by Fubini’s theorem. Specifically, we define $q_{0,m,i}=q_{m,i}·{\chi }_{\left[a_i,b_i\right]}, i=1,\dots ,n$and $f_i\left(t\right)=q_{m,i}\left(t\right),$ $t\in \left[a_i,b_i\right], f_i\left(t\right)=0$ for $t$ outside an interval $\left[a_i-\epsilon ,b_i+\epsilon \right]$ with small $\epsilon >0,$ the graph of $f_i$ on $\left[b_i,b_i+\epsilon \right]$ being the line segment joining the points $(b_i,q_i\left(b_i)\right)$ and $\left(b_i+\epsilon ,0\right).$ We proceed similarly on an interval $\left[a_i-\epsilon ,a_i\right].$ Clearly, for $\epsilon >0$ small enough, $f_i$ approximates $q_{0, m,i}$ in $L_{{\nu }_i}^1\left({ℝ}_{+}\right)$, as accurately as we wish. On the other hand, $f_i$ is non-negative, compactly supported, and continuous on ${ℝ}_{+},$ such that Lemma 18 ensures the existence of an approximating polynomial $p_i$, with respect to the norm of $L_{d{\nu }_i}^1\left( {ℝ}_{+}\right)$, $p_i\left(t\right)\ge 0$for all $t\in {ℝ}_{+}, i=1,\dots ,n.$ According to Fubini’s theorem, the preceding reasoning yields the following: $p_1\otimes \cdots \otimes p_n$approximates $f_1\otimes \cdots \otimes f_n$, and $f_1\otimes \cdots \otimes f_n$ approximates $q_{0,m,1}\otimes \cdots \otimes q_{0,m,n}=q_{0,m,k_1}\otimes \cdots \otimes q_{0,m,k_n,}$ The approximations holds for finite sums of these products in $L_{\nu }^1\left({ℝ}_{+}^n\right)$. Moreover, finite sums of the functions $q_{0,m,1,}\otimes \cdots \otimes q_{0,m,n}$approximate $f$ uniformly on $K,$ since their restrictions to $K$ define the restriction to $K$ of the approximating Bernstein polynomials ${\left(B_m\right)}_{m\in \mathbb{N}}$ associated with $f.$ Since $f$ and $q_{0,m,1,}\otimes \cdots \otimes q_{0,m,n}$ vanish outside $K,$ we infer that

$$\begin{gathered} {\Vert f-{\displaystyle \sum }_{\begin{array}{c}{k}_i=0,\dots ,m,\\ i=1,\dots ,n\end{array}}^{}{q}_{0,m,k_1,}\otimes \cdots \otimes q_{0,m,k_n}\Vert }_1={{\displaystyle \int }}_K^{}\left|f-{\displaystyle \sum }_{\begin{array}{c}{k}_i=0,\dots ,m,\\ i=1,\dots ,n\end{array}}^{}{q}_{m,,k_1}\otimes \cdots \otimes q_{m,k_n}\right|d\nu \le \\ \underset{t\in K}{\sup }\left|f\left(t\right)-B_m\left(t\right)\right|\Rightarrow \nu \left(K\right)\to 0, m\to \infty . \end{gathered}$$

The conclusion is that $f$ can be approximated in ${\Vert \Vert }_{1,{ℝ}_{+}^n}$ by sums of products $p_1\otimes \cdots \otimes p_n,$ where $p_i$ is non-negative on ${ℝ}_{+}$ for all $i=1,\dots ,n.$ This ends the proof.  □

Note that, in the proof of Lemma 4, the previous approximation Lemma 3 was applied only for $n=1.$ By means of the same proof as that of the preceding result, we have:

Lemma 5.

Let$\nu ={\nu }_1\times \cdots \times {\nu }_n$be a product of n$M-$determinate measures on$ℝ.$Then, we can approximate any non-negative continuous compactly supported function in$X=L_{\nu }^1\left({ℝ}^n\right)$with sums of products

$$p_1\otimes \cdots \Rightarrow p_n,$$

$p_j$non-negative polynomial on the entire real line,$j=1,\dots ,n$.

Corollary 4.

(See [36], Theorem 5). Let $X$ be as in Lemma 4, ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ a sequence in $Y,$ where $Y$ is an order complete Banach lattice; and let $T_2\in B_{+}\left(X,Y\right)$ be a positive bounded linear operator. The following statements are equivalent:

(a)

There exists a unique (bounded) linear operator$T\in B\left(X,Y\right),$such that$T\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^n;$ $T$is between zero and$T_2$on the positive cone of$X, \Vert T\Vert \le \Vert T_2\Vert ;$

(b)

For any finite subset$J_0\subset {\mathbb{N}}^n,$and any$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$we have

$${{\displaystyle \sum }}_{j\in J_0}{\lambda }_j{\phi }_j\left(t\right)\ge 0 \forall t\in {ℝ}_{+}^n\Rightarrow {{\displaystyle \sum }}_{j\in j_0}^{}{\lambda }_j{y}_j\in Y_{+}.$$

For any finite subsets$J_k\subset \mathbb{N}, k=1,\dots ,n,$and any${\left\{{\lambda }_{j_k}\right\}}_{j_k\in J_k}\subset ℝ, k=1,\dots ,n,$the following relations hold:

$$\begin{gathered} {\displaystyle \sum }_{i_1,j_1\in J_1}\left(\cdots \left({\displaystyle \sum }_{i_n,j_n\in J_n}{\lambda }_{i_1}{\lambda }_{j_1}\dots {\lambda }_{i_n}{\lambda }_{j_n}{y}_{i_1+j_1+l_1,\dots ,i_n+j_n+l_n}\right)\cdots \right)\le \\ {\displaystyle \sum }_{i_1,j_1\in J_1}\left(\cdots \left({\displaystyle \sum }_{i_n,j_n\in J_n}{\lambda }_{i_1}{\lambda }_{j_1}\dots {\lambda }_{i_n}{\lambda }_{j_n}{T}_2({\phi }_{i_1+j_1+l_1,\dots ,i_n+j_n+l_n}\right)\cdots \right), \left(l_1,\dots ,l_n\right)\in {\left\{0,1\right\}}^n. \end{gathered}$$

Proof. 

One repeats the proof of Theorem 19, where the convergent sequence of non-negative polynomials for the continuous non-negative compactly supported function $x$ can be chosen in terms of sums of tensor products of non-negative polynomials ${ℝ}_{+}$. Such convergent sequences do exist, as shown by the preceding Lemma 22. The last inequality of (b) in the present corollary says that on each term $p$ of such a sequence, the inequality of Equation (10) holds true, where ${𝒫}_{+}$ must be replaced by the convex cone ${𝒫}_{++}$ of all special non-negative polynomials generated by the tensor products emphasized in Lemma 22. The motivation of the condition $\left(l_1,\dots ,l_n\right)\in {\left\{0,1\right\}}^n$ of (b) in the present statement comes from the fact that $p_j\left(t\right)\ge 0$ for all $t\in {ℝ}_{+}\iff p_j\left(t\right)=q_j^2\left(t\right)+t{r}_j^2\left(t\right), t\in {ℝ}_{+}$, for some $q_j,r_j\in ℝ\left[t\right].$ This ends the proof.  □

Corollary 5.

Let$\nu$be an$M-$determinate measure on${ℝ}_{+}$, with finite moments of all natural orders,$X=L_{\nu }^1\left({ℝ}_{+}\right).$Let$Y$be an order complete Banach lattice, and$T_2\in B_{+}\left(X,Y\right), {\left(y_j\right)}_{j\in \mathbb{N}}$a sequence in$Y.$The following statements are equivalent:

(a)

There exists a unique (bounded) linear operator$T\in B\left(X,Y\right)$that satisfies the conditions

$$T\left({\phi }_j\right)=y_j, j\in \mathbb{N}, 0\le T\le T_2 on X_{+}, \Vert T\Vert \le \Vert T_2\Vert ;$$

(b)

For any finite subset$J_0\subset \mathbb{N}$, and any$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$the following inequalities hold true:

$$0\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{y}_{i+j+k}\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{T}_2\left({\phi }_{i+j+k}\right), k\in \left\{0,1\right\}.$$

Similarly to Corollary 4, also using Lemma 5 and the fact that any non-negative polynomial on the real axes is the sum of squares of (two) polynomials, if we denote ${\phi }_j\left(t\right):=t_1^{j_1}\cdots t_n^{j_n}$, $j=\left(j_1,\dots ,j_n\right)\in {\mathbb{N}}^n, t=\left(t_1,\dots ,t_n\right)\in {ℝ}^n,$then the following result holds true:

Corollary 6.

(See [36], Theorem 6). Let $X=L_{\nu }^1\left({ℝ}^n\right)$, where $\nu ={\nu }_1\times \cdots \times {\nu }_n$ is a product of $n M-$ determinate measures on $ℝ$ ; let $Y$ be an order complete Banach lattice, and ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ a sequence in $Y, T_2\in B_{+}\left(X,Y\right)$. The following statements are equivalent:

(a)

There exists a unique (bounded) linear operator$T\in B\left(X,Y\right)$, such that$T\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^n, T$is between zero and$T_2$on the positive cone of$X, \Vert T\Vert \le \Vert T_2\Vert ;$

(b)

For any finite subset$J_0\subset {\mathbb{N}}^n,$and any$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$we have

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{\phi }_j\left(t\right)\ge 0 \forall t\in {ℝ}^n\Rightarrow {\displaystyle \sum }_{j\in j_0}^{}{\lambda }_j{y}_j\in Y_{+}.$$

For any finite subsets$J_k\subset \mathbb{N}, k=1,\dots ,n,$and any${\left\{{\lambda }_{j_k}\right\}}_{j_k\in J_k}\subset ℝ, k=1,\dots ,n,$the following relations hold:

$$\begin{gathered} {\displaystyle \sum }_{i_1,j_1\in J_1}\left(\dots \left({\displaystyle \sum }_{i_n,j_n\in J_n}{\lambda }_{i_1}{\lambda }_{j_1}\dots {\lambda }_{i_n}{\lambda }_{j_n}{y}_{i_1+j_1,\dots ,,i_n+j_n}\right)\cdots \right)\le \\ {\displaystyle \sum }_{i_1,j_1\in J_1}\left(\dots \left({\displaystyle \sum }_{i_n,j_n\in J_n}{\lambda }_{i_1}{\lambda }_{j_1}\dots {\lambda }_{i_n}{\lambda }_{j_n}{T}_2({\phi }_{i_1+j_1,\dots ,,i_n+j_n}\right)\cdots \right). \end{gathered}$$

Corollary 7.

Let us consider the hypothesis and notations from Corollary 5, where we replace${ℝ}_{+}$with$ℝ.$The following statements are equivalent:

(a)

There exists a unique (bounded) linear operator$T\in B\left(X,Y\right)$, which verifies

$$T\left({\phi }_j\right)=y_j, j\in \mathbb{N}, 0\le T\le T_2 on X_{+}, \Vert T\Vert \le \Vert T_2\Vert ;$$

(b)

For any finite subset$J_0\subset \mathbb{N},$and any$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$the following inequalities hold true:

$$0\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{y}_{i+j}\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{T}_2\left({\phi }_{i+j}\right).$$

Theorem 18.

(See [36], Theorem 7). Let $X$ be as in Corollary 6, and $Y$ be a Banach lattice. Assume that $T$ is a linear bounded operator from $X$ to $Y.$ The following statements are equivalent:

(a)

$T$is positive on the positive cone of$X;$

(b)

For any finite subsets$J_k\subset \mathbb{N}, k=1,\dots ,n,$and any${\left\{{\lambda }_{j_k}\right\}}_{j_k\in J_k}\subset ℝ, k=1,\dots ,n,$the following relations hold:

$$0\le {\displaystyle \sum }_{i_1,j_1\in J_1}\left(\cdots \left({\displaystyle \sum }_{i_n.j_n\in J_n}{\lambda }_{i_1}{\lambda }_{j_1}\cdots {\lambda }_{i_n}{\lambda }_{j_n}T\left(x_{i_1+j_1},\dots {,}_{i_n+j_n}\right)\right)\cdots \right).$$

Proof. 

Note that (b) says that $T$ is positive on the convex cone generated by special positive polynomials $p_1\otimes \cdots \otimes p_n,$ each factor of any term in the sum being non-negative on the whole real axis. Consequently, (a)⇒(b) is clear. In order to prove the converse, observe that any non-negative element of $X$ can be approximated by non-negative continuous compactly supported functions. Such functions can be approximated by the sums of tensor products of positive polynomials in each separate variable, the latter being sums of squares. The conclusion is that any non-negative function from $X$ can be approximated in $X=L_{\nu }^1\left({ℝ}^n\right)$ by the sums of tensor products of squares of polynomials in each separate variable. We know that on such special polynomials, $T$ admits values in $Y_{+}$, according to the condition (b). Now, the desired conclusion is a consequence of the continuity of $T,$ also using the fact that the positive cone of $Y$ is closed. This concludes the proof.  □

In what follows, we review some of the results of [38]. If $F\subseteq {ℝ}^n$ is an arbitrary closed unbounded subset, then we denote by ${𝒫}_{+}$ the convex cone of all polynomial functions (with real coefficients), taking non-negative values at any point of $F.$ ${𝒫}_{++}$ will be a sub-cone of ${𝒫}_{+}$, generated by special non-negative polynomials expressible in terms of sums of squares.

Theorem 19.

(See [38], Theorem 4). Let $F\subseteq {ℝ}^n$ be a closed unbounded subset; $\nu$ a moment-determinate measure on $F,$ having finite moments of all orders; and $X=L_{\nu }^1\left(F\right), {\phi }_j\left(t\right)=t^j,t\in F, j\in {\mathbb{N}}^n.$ Let $Y$ be an order complete Banach lattice, ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ a given sequence of elements in $Y, T_1$ and $T_2$ two bounded linear operators from $X$ to $Y.$ Assume that there exists a sub-cone ${𝒫}_{++}\subseteq {𝒫}_{+}$ such that each $f\in {\left(C_0\left(F\right)\right)}_{+}$ can be approximated in $X$ by a sequence ${\left(p_l\right)}_l,p_l\in {𝒫}_{++},p_l\ge f$ for all $l$. The following statements are equivalent:

(a)

There exists a unique (bounded) linear operator$T:X\to Y, T\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^n,0\le T_1\le T\le T_2$on$X_{+}, \Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert ;$

(b)

For any finite subset$J_0\subset {\mathbb{N}}^n,$and any$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$the following implications hold true:

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{\phi }_j\in {𝒫}_{+}\Rightarrow {\displaystyle \sum }_{j\in J_0}{\lambda }_j{T}_1\left({\phi }_j\right)\le {\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j,$$

(14)

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{\phi }_j\in {𝒫}_{++}\Rightarrow {\displaystyle \sum }_{j\in J_0}{\lambda }_j{T}_1\left({\phi }_j\right)\ge 0, {\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j \le {\displaystyle \sum }_{j\in J_0}{\lambda }_j{T}_2\left({\phi }_j\right).$$

(15)

Proof. 

We start by observing that the first condition of Equation (15) implies the positivity of the bounded linear operator $T_1,$ via its continuity. Indeed, if $f\in {\left(C_0\left(F\right)\right)}_{+} , p_l\in {𝒫}_{++}, p_l\ge f$ for all $l, p_l\to f$ in $L_{\nu }^1\left(F\right)$ then, according to the first condition of Equation (15), $T_1\left( p_l\right)\ge 0$ for all $l\in \mathbb{N}$, and the continuity of $T_1$ yields the following:

$$T_1\left(f\right)=\underset{l}{\lim }{T}_1\left(p_l\right)\ge 0.$$

Since ${\left(C_0\left(F\right)\right)}_{+}$ is dense in $X_{+}$ as explained by measure theory, the continuity of $T_1$ implies $T_1\ge 0$ on $X_{+}.$ Thus, $T_1$ is a positive linear operator. Next, we define $T_0:𝒫\to Y, T_0\left({{\displaystyle \sum }}_{j\in J_0}{\lambda }_j{\phi }_j\right)={{\displaystyle \sum }}_{j\in J_0}{\lambda }_j{y}_j$, where the sums are finite and the coefficients ${\lambda }_j$ are arbitrary real numbers. Equation (14) says that $T_0-T_1\ge 0$ on ${𝒫}_{+}.$ If we consider the vector subspace $X_1$ of $X$ formed by all functions $\psi \in X$ having the modulus $\left|\psi \right|$ dominated by a polynomial $p\in {𝒫}_{+}$ on the entire set $F,$ then $𝒫$ is a majorizing subspace of $X_1$, and $T_0-T_1$ is a positive linear operator on $𝒫.$ The application of Theorem 1 leads to the existence of a positive linear extension $U:X_1\to Y$ of $T_0-T_1.$ Clearly, $X_1$ contains $C_0\left(F\right)+𝒫.$ Indeed, since $\phi \in C_0\left(S\right)\Rightarrow \left|\phi \right|\in {\left(C_0\left(F\right)\right)}_{+}\Rightarrow \left|\phi \right|\le b1\in 𝒫$ (according to Weierstrass’ theorem), we infer that $\phi \in X_1$; here, $b<\infty$ is a real number. Hence, $C_0\left(F\right)\subset X_1.$ Now, if $p\in 𝒫,$ we observe the following:

$$1+p^2-2\left|p\right|={(1-|p\left|\right)}^2\ge 0$$

which can be written as follows:

$$\left|p\right|\le \frac{1+p^2}{2}\in 𝒫.$$

According to the definition of $X_1,$ it results that $𝒫\subset X_1.$ Consequently, $C_0\left(F\right)+𝒫\subset X_1.$ Going back to the positive linear extension $U:X_1\to Y$ of $T_0-T_1,$ we conclude that ${\widehat{T}}_0=U+T_1:X_1\to Y$ is an extension of $T_0, {\widehat{T}}_0\ge T_1$ on ${\left(X_1\right)}_{+}$, and ${\widehat{T}}_0\left(p\right)=T_0\left(p\right)\le T_2\left(p\right)$ for all $p\in {𝒫}_{++}$, according to the last requirement of Equation (15). A first conclusion is as follows:

$$T_1\left(p\right)\le {\widehat{T}}_0\left(p\right)\le T_2\left(p\right) \mathrm{for} \mathrm{all} p\in {𝒫}_{++}, {\widehat{T}}_0\left(\psi \right)\ge T_1\left(\psi \right)\ge 0, \psi \in {\left(X_1\right)}_{+}.$$

(16)

Our next goal is to prove the continuity of ${\widehat{T}}_0$ on $C_0\left(S\right).$ Let ${\left(f_l\right)}_{l\ge 0}$ be a sequence of non-negative continuous compactly supported functions, such that $f_l\to 0$ in $X_1,$ and take a sequence of polynomials $p_l\ge f_l\ge 0,p_l\in {𝒫}_{++}$ for all $l,$ such that the following convergence result holds: $\Vert p_l-f_l\Vert {}_1\to 0, l\to \infty .$Then, apply the following:

$$\Vert p_l\Vert {}_1\le \Vert p_l-f_l\Vert {}_1+\Vert f_l\Vert {}_1\to 0, l\to \infty .$$

Now, Equation (16) and the continuity of $T_1,$ $T_2$ yield the following:

$$0\leftarrow T_1\left(p_l\right)\le {\widehat{T}}_0\left(p_l\right)\le T_2\left(p_l\right)\to 0.$$

Hence, ${\widehat{T}}_0\left(p_l\right)\to 0;$ this results in the following:

$$0\le T_1\left(f_l\right)\le {\widehat{T}}_0\left(f_l\right)\le {\widehat{T}}_0\left(p_l\right)\to 0.$$

Hence, ${\widehat{T}}_0\left(f_l\right)\to 0.$ If ${\left(g_n\right)}_{n\ge 0}$ is an arbitrary sequence of compactly supported and continuous functions, such that $g_n\to 0$ in $X_1$, then $g_n^{+}\to 0,g_n^{-}\to 0.$ According to what we already have proven, we can write ${\widehat{T}}_0\left(g_n^{+}\right)\to 0$ and ${\widehat{T}}_0\left(g_n^{-}\right)\to 0,$ which further yield ${\widehat{T}}_0\left(g_n\right)\to 0.$ This proves the continuity of ${\widehat{T}}_0$ on $C_0\left(F\right),$ and the subspace $C_0\left(F\right)$ is dense in $X.$ Hence, there exists a unique continuous linear extension $T\in B\left(X,Y\right)$ of ${\widehat{T}}_0.$ This results in $0\le T_1\le T\le T_2$ on $X_{+}, \Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert , T\left({\phi }_j\right)=T_0\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^n.$ Indeed, $T_1, T,T_2$ are linear and continuous, and ${𝒫}_{++}$ is dense in ${\left(C_0\left(F\right)\right)}_{+};$ hence, it is dense in $X_{+}$ as well. For an arbitrary $\phi \in X,$ the following inequalities hold true, via the preceding remarks:

$$\begin{gathered} \pm T\left(\phi \right)=T\left(\pm \phi \right)\le T\left(\left|\phi \right|\right)\le T_2\left(\left|\phi \right|\right)\Rightarrow \\ \left|T\left(\phi \right)\right|\le T_2\left(\left|\phi \right|\right)\Rightarrow \Vert T\left(\phi \right)\Vert \le \Vert T_2\left(\left|\phi \right|\right)\Vert \le \Vert T_2\Vert \phi \Vert . \end{gathered}$$

It follows that $\Vert T\Vert \le \Vert T_2\Vert$ and, similarly, $\Vert T_1\Vert \le \Vert T\Vert .$ The uniqueness of the solution $T$ follows according to the density of polynomials in $X,$ via the continuity of the linear operator $T$ and application of Lemma 18. This ends the proof.  □

Corollary 8.

(See [38], Corollary 3). Let $\nu ={\nu }_1\times \cdots \times {\nu }_n, n\ge 2,{\nu }_j$ being an $M-$ determinate (moment-determinate) measure on $ℝ, j=1,\dots ,n, X=L_{\nu }^1\left({ℝ}^n\right), {\phi }_j\left(t\right)=t^j,t\in {ℝ}^n, j\in {\mathbb{N}}^n.$ Additionally, assume that ${\nu }_j$ has finite moments of all orders, $j=1,\dots ,n.$ Let $Y$ be an order complete Banach lattice, ${\left(y_j\right)}_{j\in {\mathbb{N}}^n}$ a given sequence of elements in $Y, and T_1$ and $T_2$ two bounded linear operators from $X$ to $Y.$ The following statements are equivalent:

(a)

There exists a unique (bounded) linear operator$T:X\to Y, T\left({\phi }_j\right)=y_j, j\in {\mathbb{N}}^n,0\le T_1\le T\le T_2$on$X_{+}, \Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert ;$

(b)

For any finite subset$J_0\subset {\mathbb{N}}^n,$and any$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$the following implication holds true:

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{\phi }_j\in {𝒫}_{+}\Rightarrow {\displaystyle \sum }_{j\in J_0}{\lambda }_j{T}_1\left({\phi }_j\right)\le {\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j.$$

For any finite subsets$J_k\subset \mathbb{N}, k=1,\dots ,n,$and any${\left\{{\lambda }_{j_k}\right\}}_{j_k\in J_k}\subset ℝ,$the following inequalities hold true:

$$\begin{gathered} 0\le {\displaystyle \sum }_{i_1,j_1\in J_1}\left(\cdots \left({\displaystyle \sum }_{i_n.j_n\in J_n}{\lambda }_{i_1}{\lambda }_{j_1}\cdots {\lambda }_{i_n}{\lambda }_{j_n}{T}_1\left({\phi }_{i_1+j_1,\dots ,i_n+j_n}\right)\right)\cdots \right); \\ {\displaystyle \sum }_{i_1,j_1\in J_1}\left(\cdots \left({\displaystyle \sum }_{i_n.j_n\in J_n}{\lambda }_{i_1}{\lambda }_{j_1}\cdots {\lambda }_{i_n}{\lambda }_{j_n}{y}_{i_1+j_1,\dots ,i_n+j_n}\right)\cdots \right)\le \\ {\displaystyle \sum }_{i_1,j_1\in J_1}\left(\cdots \left({\displaystyle \sum }_{i_n.j_n\in J_n}{\lambda }_{i_1}{\lambda }_{j_1}\cdots {\lambda }_{i_n}{\lambda }_{j_n}{T}_2\left({\phi }_{i_1+j_1,\dots ,i_n+j_n}\right)\right)\cdots \right). \end{gathered}$$

Proof. 

One applies Theorem 19 and Lemma 5.  □

Corollary 9.

(See [38], Corollary 2). Let $X= L_{\nu }^1(ℝ),$ where $\nu$ is a moment-determinate measure on $ℝ.$ Assume that $Y$ is an arbitrary order complete Banach lattice, and ${\left(y_n\right)}_{n\ge 0}$ is a given sequence with its terms in $Y.$ Let $T_1,T_2$ be two linear operators from $X$ to $Y$, such that $0\le T_1\le T_2$ on $X_{+}.$ The following statements are equivalent:

(a)

There exists a unique bounded linear operator$T$from$X$to$Y, T_1\le T\le T_2$on$X_{+},$ $\Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert$, such that$T\left({\phi }_n\right)=y_n$for all$n\in \mathbb{N};$

(b)

If$J_0\subset \mathbb{N}$is a finite subset, and$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$then

$${\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{T}_1\left({\phi }_{i+j}\right)\le {\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{y}_{i+j}\le {\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{T}_2\left({\phi }_{i+j}\right).$$

For $Y=ℝ,$ based on the measure theory arguments discussed in [9], Corollary 30 can be written as follows:

Corollary 10

. Let $\nu$ be a moment-determinate measure on $ℝ.$ Assume that $h_1,h_2$ are two functions in $L_{\nu }^{\infty }(ℝ)$, such that $0\le h_1\le h_2$ almost everywhere. Let ${\left(y_n\right)}_{n\ge 0}$ be a given sequence of real numbers. The following statements are equivalent:

(a)

There exists$h\in L_{\nu }^{\infty }(ℝ),$such that$h_1\le h\le h_2 \nu -$almost everywhere,${{\displaystyle \int }}_{ℝ}^{}{t}^{j}h\left(t\right)d\nu =y_j$for all$j\in \mathbb{N};$

(b)

If$J_0\subset \mathbb{N}$is a finite subset, and$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$then

$${\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j\underset{ℝ}{\overset{}{{\displaystyle \int }}}{t}^{i+j}{h}_1\left(t\right)d\nu \le {\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{y}_{i+j}\le {{\displaystyle \sum }}_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{{\displaystyle \int }}_{ℝ}^{}{t}^{i+j}{h}_2\left(t\right)d\nu .$$

Similarly to Corollary 10, replacing $ℝ$ with ${ℝ}_{+}$ we can derive the following:

Corollary 11.

Let$X=L_{\nu }^1\left({ℝ}_{+}\right),$where$\nu$is a moment-determinate measure on${ℝ}_{+}.$Assume that$Y$is an arbitrary order complete Banach lattice, and${\left(y_n\right)}_{n\ge 0}$is a given sequence with its terms in$Y.$Let$T_1,T_2$be two linear operators from$X$to$Y$, such that$0\le T_1\le T_2$on$X_{+}.$As usual, we denote${\phi }_j\left(t\right)=t^j, j\in \mathbb{N}, t\in {ℝ}_{+}.$The following statements are equivalent:

(a)

There exists a unique bounded linear operator$T$from$X$to$Y, T_1\le T\le T_2$on$X_{+},$ $\Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert$, such that$T\left({\phi }_n\right)=y_n$for all$n\in \mathbb{N};$

(b)

If$J_0\subset \mathbb{N}$is a finite subset, and$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$then

$${\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{T}_1\left({\phi }_{i+j+k}\right)\le {\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{y}_{i+j+k}\le {\displaystyle \sum }_{i,j\in J_0}^{}{\lambda }_i{\lambda }_j{T}_2\left({\phi }_{i+j+k}\right), k\in \left\{0,1\right\}.$$

In the scalar-valued case, we derive the following consequences:

Corollary 12.

Let$\nu$be a moment-determinate measure on${ℝ}_{+}.$Assume that$h_1,h_2$are two functions in$L_{\nu }^{\infty }\left({ℝ}_{+}\right)$, such that$0\le h_1\le h_2$almost everywhere. Let${\left(y_n\right)}_{n\ge 0}$be a given sequence of real numbers. The following statements are equivalent:

(a)

There exists$h\in L_{\nu }^{\infty }\left({ℝ}_{+}\right),$such that$h_1\le h\le h_2 \nu -$almost everywhere,${{\displaystyle \int }}_{{ℝ}_{+}}^{}{t}^{j}h\left(t\right)d\nu =y_j$for all$j\in \mathbb{N};$

(b)

If$J_0\subset \mathbb{N}$is a finite subset, and$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$then:

$$\begin{gathered} {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j\underset{{ℝ}_{+}}{\overset{}{{\displaystyle \int }}}{t}^{i+j+k}{h}_1\left(t\right)d\nu \le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{y}_{i+j+k}\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j\underset{{ℝ}_{+}}{\overset{}{{\displaystyle \int }}}{t}^{i+j+k}{h}_2\left(t\right)d\nu , \\ k\in \left\{0,1\right\}. \end{gathered}$$

Corollary 13.

Assume that$h_1$is a function in$L_{exp\left(-t\right)dt}^{\infty }\left({ℝ}_{+}\right)$, such that$\mathbf{0}\le h_1\le \mathbf{1}$almost everywhere. Let${\left(y_n\right)}_{n\ge 0}$be a given sequence of real numbers. The following statements are equivalent:

(a)

There exists$h\in L_{exp\left(-t\right)dt}^{\infty }\left({ℝ}_{+}\right),$such that$h_1\le h\le \mathbf{1}$almost everywhere,${{\displaystyle \int }}_{{ℝ}_{+}}^{}{t}^{j}h\left(t\right)e^{-t}dt=y_j$for all$j\in \mathbb{N};$

(b)

If$J_0\subset \mathbb{N}$is a finite subset, and$\left\{{\lambda }_j;j\in J_0\right\}\subset ℝ,$then

$$\begin{gathered} {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j\underset{{ℝ}_{+}}{\overset{}{{\displaystyle \int }}}{t}^{i+j+k}{h}_1\left(t\right)e^{-t}dt\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j{y}_{i+j+k}\le {\displaystyle \sum }_{i,j\in J_0}{\lambda }_i{\lambda }_j\left(i+j+k\right)!, \\ k\in \left\{0,1\right\}. \end{gathered}$$

3.3. On the Truncated Moment Problem

The truncated moment problem is important in mathematics because it involves only a finite number of moments (of limited order), which are assumed to be known (or given, or measurable); therefore, it can be related to optimization problems [24,30,31], as well as to constructive methods for finding solutions [32,33]. The results that follow have been published in [37]. Previous results on this subject, or related problems, have been published in [24,25,30,31,32,33,39], and many other articles. We start by studying the existence of some scalar-valued truncated (reduced) Markov moment problems on a closed-bounded or unbounded subset $F$ of ${ℝ}^n$, where $n\ge 1$ is an integer. We denote by ${ℝ}_d\left[t_1,\dots t_n\right]$the real vector subspace of all polynomial functions $𝒫$ of $n$ real variables, with real coefficients, generated by $t^k=t_1^{k_1}\cdots t_n^{k_n},$ $k_i\in \left\{0,1,\dots ,d\right\}, i=1,\dots ,n$, where $d\ge 1$ is a fixed integer. The dimension of this subspace is clearly equal to $N={\left(d+1\right)}^n.$ Given a finite set ${\left\{m_k\right\}}_{\begin{array}{c}0\le k_i\le d,\\ i=1,\dots ,n\end{array}}$of real numbers, and a positive Borel measure μ on $F$, with finite absolute moments of all orders smaller than or equal to $d$ (i.e., ${{\displaystyle \int }}_F^{}{\left|t\right|}^{k}d\mu <\infty$ for all $k=\left(k_1,\dots ,k_n\right)\in {\mathbb{N}}^n$ with $k_i$ ≤ d, $i=1,\dots ,n$), the existence and, eventually, construction or approximation of a Lebesgue measurable real non-negative function $h\in L_{\mu }^q\left(F\right)$, satisfying the moment conditions

$${{\displaystyle \int }}_F^{}{t}^{k}h\left(t\right)d\mu =m_k, k_i \text{≤}d i=1,\dots ,n,$$

and

$$0\le h\left(t\right) \mathrm{for} \mathrm{almost} \mathrm{all} t\in F,{\Vert h\Vert }_q\le 1$$

are under attention, where $q$ is the conjugate of a given number $p\in \left[1,\infty \right[.$Here, ${\Vert ·\Vert }_q$ is the usual norm on $L_{\mu }^q\left(F\right).$ The solution $h$ appears as the representing function from $L_{\mu }^q\left(F\right)$ for a positive linear functional $T$ on $L_{\mu }^p\left(F\right)$ $, p\in \left[1,\infty \right), 1/p+1/q=1,\Vert T\Vert \le 1.$We start with the case when $p=1.$ The next result follows from Theorem 3, where $X=L_{\mu }^1\left(F\right),$ $Y$ stands for $ℝ$, and $y_j=m_j,j=\left(j_1,\dots ,j_n\right), j_l\le d, l=1,\dots ,n, P\left(g\right)={\Vert g\Vert }_1, g\in L_{\mu }^1\left(F\right), x_j\left(t\right)=t^j, t\in F, j\in {\left\{0,1,\dots ,d\right\}}^n.$ Standard measure theory results are also applied.

Theorem 20.

(See [37]). Assume that all absolute moments ${{\displaystyle \int }}_F^{}{\left|t\right|}^{k}d\mu <\infty$ for $k=\left(k_1,\dots ,k_n\right)$ with $k_i$ ≤ d, $i=1,\dots ,n,$ are finite. Let ${\left\{m_k\right\}}_{\begin{array}{c}{k}_i\le d,\\ i=1,\dots ,n\end{array}}$ be a given (finite) sequence of real numbers. The following statements are equivalent:

(a)

There exists$h\in L_{\mu }^{\infty }\left(F\right)$, such that$0\le h\left(t\right)\le 1$almost everywhere,

$${{\displaystyle \int }}_F^{}{t}^{k}h\left(t\right)d\mu =m_k,k_i \le d i=1,\dots ,n;$$

(b)

For any family of scalar${\left\{a_k\right\}}_{\begin{array}{c}{k}_i\le d,\\ i=1,\dots ,n\end{array}}$the inequality${{\displaystyle \sum }}_{\begin{array}{c}{k}_i\le d,\\ i=1,\dots ,n\end{array}}^{}{a}_k{\phi }_k\le g\in L_{\mu }^1\left(F\right)$implies

$${{\displaystyle \sum }}_{\begin{array}{c}{k}_i\le d,\\ i=1,\dots ,n\end{array}}^{}{a}_k{m}_k\le {{\displaystyle \int }}_F^{}\left|g\left(t\right)\right|d\mu ={\Vert g\Vert }_1.$$

Proof. 

Since $\left(a\right)\Rightarrow \left(b\right)$ is almost obvious, we only have to prove the converse implication. Application of Theorem 3 to the particular case when $X=L_{\mu }^1\left(F\right), Y=ℝ, P\left(g\right)={\Vert g\Vert }_1$ for all $g$ in $L_{\mu }^1\left(F\right)$ leads to the existence of a positive linear functional $T:L_{\mu }^1\left(F\right)\to ℝ,$ such that

$$T\left({\phi }_k\right)=y_k, k_i\le d, i=1,\dots ,n, T\left(g\right)\le {\Vert g\Vert }_1, g\in L_{\mu }^1\left(F\right).$$

Writing the last inequality where we replace $g$ with $-g,$ we get $\Vert T\Vert \le 1.$ In particular, $T$ is continuous, of norm smaller than or equal to one. As is well known (see [9], Theorem 6.16), the vector topological dual of $L_{\mu }^1\left(F\right)$ is isometrically isomorphic to $L_{\mu }^{\infty }\left(F\right).$ Therefore, there exists a unique $h\in L_{\mu }^{\infty }\left(F\right)$, with ${\Vert h\Vert }_{\infty }=\Vert T\Vert ,$ such that $T\left(g\right)={{\displaystyle \int }}_F^{}g\left(t\right)h\left(t\right)d\mu$ for all $g\in L_{\mu }^1\left(F\right).$ Next, we use the positivity of $T:g\in {\left(L_{\mu }^1\left(F\right)\right)}_{+}\Rightarrow 0\le T\left(g\right)={{\displaystyle \int }}_F^{}g\left(t\right)h\left(t\right)d\mu .$ Writing this for $g={\chi }_E$, where $E\subseteq F$ is an arbitrary Borel subset, with $\mu \left(E\right)>0$, we obtain ${{\displaystyle \int }}_E^{}h\left(t\right)d\mu /\mu \left(E\right)\ge 0.$ According to [9], Theorem 1.40, $h\ge \mathbf{0}$ in $L_{\mu }^{\infty }\left(F\right).$ Since ${\Vert h\Vert }_{\infty }=\Vert T\Vert \le 1$, the desired conclusion $\mathbf{0}\le h\le \mathbf{1}$ in $L_{\mu }^{\infty }\left(F\right)$ follows. This ends the proof.  □

Using a similar proof to that of Theorem 20, where $\infty$ is replaced by $q\in \left]1,\infty \right[,$ (since $1$is replaced by $p\in \left]1,\infty \right[),$the following result also holds:

Theorem 21.

(see [37]). Let $p\in \left]1,\infty \right[, X=L_{\mu }^p\left(F\right), {\left\{m_k\right\}}_{\begin{array}{c}{k}_i\le d,\\ i=1,\dots ,n\end{array}}$ a given (finite) sequence of real numbers, and let $q$ be the conjugate of $p.$ The following statements are equivalent:

(a)

There exists$h\in L_{\mu }^q\left(F\right), such that 0\le h\left(t\right)$almost everywhere,

$${\Vert h\Vert }_q\le 1, {{\displaystyle \int }}_{\mathrm{F}}^{}{t}^{k}h\left(t\right)d\mu =m_k,k_i\le d ,i=1,\dots ,n;$$

(b)

For any family of scalars${\left\{a_k\right\}}_{\begin{array}{c}{k}_i\le d,\\ i=1,\dots ,n\end{array}}$the inequality

$${{\displaystyle \sum }}_{\begin{array}{c}{k}_i\le d,\\ i=1,\dots ,n\end{array}}^{}{a}_k{\phi }_k\le g\in L_{\mu }^p\left(F\right) \mathit{implies} {{\displaystyle \sum }}_{\begin{array}{c}{k}_i\le d,\\ i=1,\dots ,n\end{array}}^{}{a}_k{m}_k\le {\Vert g\Vert }_p.$$

More information on solutions to truncated moment problems and related problems can be found in [2,24,25,30,31,32,33,37,38,39]. Solutions to full moment problems as limits of weakly convergent sequences of solutions of truncated moment problems have been published in [25,39]. Finding step function solutions to truncated moment problems is discussed in [32,33]. Such a problem requires solving a system of nonlinear equations and matrix theory.

4. Discussion

Referring to the methods used throughout this paper, our extension results for linear operators, constrained by sandwich conditions, have found recent applications, such as those proven in [27]. Theorem 2 found unexpected applications as noted in Remark 1 (Section 2). On the other hand, Theorem 3 can be applied to the existence of solutions of full or truncated Markov moment problems. This paper unifies different methods used by many other authors, in order to prove the existence and uniqueness of the solution to the Markov moment problem, by means of polynomial approximation on unbounded subsets. An example is the statement and the proof of the key Lemma 3, followed by its applications. The statements of the results and related comments could be of interest to beginners, while reading the detailed proofs of the main lemmas and theorems addresses any mathematician interested in functional analysis. On the other hand, further relationships with other fields of research are revealed by means of the references.