Moment Problems and Integral Equations

Abstract

The first part of this work provides explicit solutions for two integral equations; both are solved by means of Fourier transform. In the second part of this paper, sufficient conditions for the existence and uniqueness of the solutions satisfying sandwich constraints for two types of full moment problems are provided. The only given data are the moments of all positive integer orders of the solution and two other linear, not necessarily positive, constraints on it. Under natural assumptions, all the linear solutions are continuous. With their value in the subspace of polynomials being given by the moment conditions, the uniqueness follows. When the involved linear solutions and constraints are positive, the sufficient conditions mentioned above are also necessary. This is achieved in the third part of the paper. All these conditions are written in terms of quadratic expressions.

1. Introduction

Generally, solving functional equations cannot be achieved using direct integration operations or by studying the derivative of the involved functions, even in the case of smooth functions. On the other hand, under a natural hypothesis, a first-order differential equation with initial conditions is equivalent to the corresponding integral equation. The latter equation cannot be solved explicitly. Successive approximations with error control can be obtained by applying the contraction principle or other approximation methods. An efficient method for solving such difficulties and other problems involves using the properties of the Fourier transform (see [1,2]). These two books provide other aspects and applications of the Fourier transform. To name only one of them, we invoke the Uncertainty Principle and the importance of the function $e^{-x^2/2}, x\in ℝ:=\left(-\infty ,+\infty \right).$ The connection of this function with Cauchy–Schwarz inequality is also outlined in [1]. Namely, the Gauss-type distribution is the only one for which equality occurs in the inequality claimed by the Uncertainty Principle, as well as in the corresponding Cauchy–Schwarz inequality. The fact that the Gauss function defined above is the sole fixed function for the Fourier transform is also used in Example 2 below. For the second part of this work and related general knowledge, refer to the monographs and books [1,2,3,4,5,6,7,8]. Very recent results on criterions of indeterminacy have been published in [9]. We recall that a positive regular Borel measure, $\mu ,$ having finite moments of all orders, is called determinate (or (M)-determinate, or moment-determinate), on the closed subset $F\subseteq {ℝ}^n$ if it is uniquely determinate by its moments, defined by

$${\int }_F{t}^{j}d\mu ,\hspace{1em}j=\left(j_1,\dots ,j_n\right)\in {\mathbb{N}}^n,$$

(1)

$$t^j:=t_1^{j_1}\cdots t_n^{j_n}, t=\left(t_1,\dots ,t_n\right)\in F,n\in \mathbb{N}:=\left\{0,1,2,\dots \right\}, n\ge 1.$$

(2)

In other words, the measure $\mu$ is (M)-determinate, if for any other measure $\nu$ with

$${\int }_F{t}^{j}d\mu ={\int }_F{t}^{j}d\nu , j\in {\mathbb{N}}^n,$$

we have $\mu =\nu$ as measures. A measure which is not (M)-determinate on $F$ is called indeterminate on $F.$ In ref. [10], other recent results, involving fractional moments and related approximation methods have been emphasized. Earlier and recent results, as well as various aspects of the moment problem, have been discussed in [5,8,9,10,11,12,13,14,15,16,17,18].

Significant results on the moment problem have been recently published in [19,20,21,22]. Article [23] provides interesting applications of Hahn–Banach-type results in Banach lattice frameworks to a direction without any connection with the moment problem. On the other hand, the existence of a solution for the classical moment problem is an extension problem of a linear functional or operator, from the subspace of polynomials to a larger function space, which is usually a Banach lattice. The codomain space is assumed to be an order-complete Banach lattice, so that Hahn–Banach-type extension results work in this setting. Besides the moment interpolations (see Equation (7) below), one or two constraints on the solution are required (see Equation (8) below). Earlier results using extensions of linear functionals or geometric forms of the Hahn–Banach theorem, applied to the moment problem or to solve other problems, have been published in [4,5,6,8,14,24,25]. From all these references, we directly or partially use ideas or results. In the present article, extensions of linear functionals and operators, polynomial approximation on unbounded intervals [24,25], and general results of analysis and functional analysis are applied. With respect to previously published articles on this subject, currently, our linear constraints on the solution are not positive on the positive cone of the domain space (see Theorems 1–3 and Corollaries 1 and 2). Consequently, the solution is not positive. However, the continuity of the constraints leads to the continuity of the solution, and the sufficient conditions (b) below are expressed in terms of quadratic expressions. We recall that any positive linear operator acting on ordered Banach spaces (in particular, on Banach lattices) is continuous (see [23]). At the end of this manuscript, the case of positive linear constraints is under attention. In this case, our sufficient conditions mentioned above are also necessary. All these conditions are written in terms of quadratic expressions involving the coefficients of the corresponding polynomials, the given moments, and the constraints, then passing the limit. Since the values of the solution on polynomials is given by the interpolation moment conditions, the uniqueness of the solution of the full moment problem follows by the density of polynomials in the considered domain function space. The rest of this paper is organized as follows: Section 2 refers to the methods applied to solve the problems of Section 3. Section 3 contains the results and their proofs. Section 4 (Discussion) comments on some of the results. Section 5 concludes the paper. Here, possible directions for future work are also mentioned.

2. Methods

Here are the methods applied in this work:

• Using properties of Fourier transform [1,2] in solving integral equations.
• Using the expression of nonnegative polynomials on $\left[0,+\infty \right)$ in terms of sums of squares of polynomials [8].
• Applying notions and results on determinacy of measures [19] to prove approximation on unbounded closed subsets $F\subseteq ℝ,$ in $L_{\mu }^1\left(F\right)$spaces, of functions from ${(L_{\mu }^1\left(F\right))}_{+}$ by nonnegative polynomials on $F$. Here, $\mu$ is a positive regular Borel moment determinate measure on $F,$ with finite moments of all orders. Applying uniform polynomial approximation on compact subsets, $S\subset {ℝ}_{+},$ of any function $f\in {\left(C\left(S\right)\right)}_{+}$ by restrictions to $S$ of nonnegative polynomials on the entire nonnegative semiaxes (see [24,25]).
• Applying a Hahn–Banach-type theorem [24,25] in solving full moment problems with two linear constraints on the solution (see also [5] for the reduced, scalar valued Markov moment problem). In the present work, the moments of all positive integer orders are prescribed, and operator valued moment problems are solved. Therefore, we use a much stronger version of a Hahn–Banach-type result, recently recalled in [25], which is valid for infinitely many (countable or uncountable) interpolation moment conditions and for the order-complete vector valued version of the problem (see also the references from [25] related to the original works on this subject). Although the constraints are not positive on the positive cone of the domain space, they are assumed to be continuous. From the proofs, the linear solution is continuous as well. Hence, it is the unique solution via the density of polynomials in the domain space. Our solutions hold true for vector valued and operator valued moment problems as well.
• Using the order-complete Banach lattice of self-adjoint operators studied in [6], which is also a commutative Banach algebra over the real field, as codomain of our solution for a moment problem.
• Using functional calculus for self-adjoint operators [4,6,7], we solve our operator valued moment problem as demonstrated in Theorems 3 and 6 below.
• Using general properties of Banach lattices and specific properties of the concrete Banach lattices appearing in Section 3 [6,7,23,24,25], we prove sufficient conditions for the existence and uniqueness of the solution of the full moment problem, with two constraints. In case of positive linear constraints, these conditions are also necessary.

3. Results

3.1. Solving Integral Equations by Means of Fourier Transform

This first part of the Results section is based on properties of the Fourier transform published in [1,2].

Example 1. 

The solution  $f$ of the integral equation:

$$\underset{ℝ}{\int }{\left(f\left(x\right)\right)}^3·\frac{x\left(1+x^2\right)}{arctan\left(x\right)}·cos\left(\omega x\right)dx=\frac{sin\left(\omega \right)}{\omega }, \omega \in ℝ,$$

(3)

is

$$f\left(x\right)=={\left(\frac{1}{2}·{\chi }_{\left[-1,1\right]}\left(x\right).·\frac{arctan\left(x\right)}{x·\left(1+x^2\right)}\right)}^{1/3},x\in ℝ, {\Vert f\Vert }_{L^{\infty }(ℝ)}=f\left(0\right)=\frac{1}{\sqrt[3]{2}}$$

Proof. 

Using Euler’s formula, $e^{-i\omega x}=cos\left(\omega x\right)+isin\left(\omega x\right)$, the fact that $f$ is an even function and the equality $ℱ\left({\chi }_{\left[-1,1\right]}\left(x\right)\right)\left(\omega \right)=\sqrt{\frac{2}{\pi }}·\frac{sin\left(\omega \right)}{\omega },$ Equation (3) can be written as

$$ℱ\left(\sqrt{2\pi }·{\left(f\left(x\right)\right)}^3·\frac{x}{arctan\left(x\right)}·\left(1+x^2\right)\right)\left(\omega \right)=\sqrt{\frac{\pi }{2}}ℱ\left({\chi }_{\left[-1,1\right]}\left(x\right)\right)\left(\omega \right)=ℱ\left(\sqrt{\frac{\pi }{2}}·{\chi }_{\left[-1,1\right]}\left(x\right)\right)\left(\omega \right).$$

Since the Fourier operator $ℱ$ is invertible, the last equality is equivalent to

$$\sqrt{2\pi }·{\left(f\left(x\right)\right)}^3·\frac{x}{\arctan \left(x\right)}·\left(1+x^2\right)=\sqrt{\frac{\pi }{2}}·{\chi }_{\left[-1,1\right]}\left(x\right).$$

Hence, the solution $f$ can be written as

$$f\left(x\right)={\left(\frac{1}{2}·{\chi }_{\left[-1,1\right]}\left(x\right).·\frac{arctan\left(x\right)}{x·\left(1+x^2\right)}\right)}^{1/3}.$$

To prove the last equality from the statement, regarding ${\Vert f\Vert }_{L^{\infty }(ℝ)},$ we recall that ${\chi }_{\left[-1,1\right]}\left(x\right)$ is null for $x\in ℝ\setminus \left[-1,1\right]$ and ${\chi }_{\left[-1,1\right]}\left(x\right)=1$ for all $x\in \left[-1,1\right].$ Thus ${\chi }_{\left[-1,1\right]}$ is an even function, with ${\Vert {\chi }_{\left[-1,1\right]}\Vert }_{{L^{\infty }}_{(ℝ)}}=1$. On the other hand, $\frac{1}{1+x^2}\le 1$ for all $x\in ℝ,$ equality occurs if and only if $x=0,$ and this function is also even. One can prove that the same remark holds for the even function $0<arctg\left(x\right)/x<1$ for all $x>0,$ extended by continuity with the value $1$ at $x=0.$ Hence, the maximum point of $f$ is $0$ and $f\left(0\right)=\frac{1}{\sqrt[3]{2}}.$ □

Example 2. 

The solution  $g$ of the integral equation:

$$\frac{1}{2\pi }\underset{ℝ}{\int }\left(\underset{ℝ}{\int }g\left(x-t\right)e^{-\left|t\right|}dt\right)e^{-i\omega x}dx=e^{-{\omega }^2/2},$$

(4)

is

$$g\left(x\right)=\sqrt{\pi /2}·\left(2-x^2\right)·e^{-x^2/2}, x\in ℝ,$$

$${\Vert g\Vert }_{L^{\infty }(ℝ)}=g\left(0\right)=\sqrt{2\pi }.$$

Proof. 

Equation (4) can be written as

$$\frac{1}{\sqrt{2\pi }}ℱ\left(\left(g\left(x\right)\ast e^{-\left|x\right|}\right)\right)\left(\omega \right)=e^{-{\omega }^2/2}, \omega \in ℝ.$$

(5)

Now, using a well-known property [1,2] of the Fourier transform with respect to the convolution operation, the left-hand side of Equation (5) becomes

$$\begin{array}{c}\frac{1}{\sqrt{2\pi }}·\sqrt{2\pi } ·ℱ\left(g\left(x\right)\right)\left(\omega \right)·ℱ\left(e^{-\left|x\right|}\right)\left(\omega \right)=\\ =ℱ\left(g\left(x\right)\right)\left(\omega \right)·\sqrt{\frac{2}{\pi }}·\frac{1}{1+{\omega }^2} .\end{array}$$

Inserting this into Equation (5), we find

$$\sqrt{\frac{2}{\pi }}·\frac{1}{1+{\omega }^2} ·ℱ\left(g\left(x\right)\right)\left(\omega \right)=e^{-{\omega }^2/2},$$

$$ℱ\left(g\left(x\right)\right)\left(\omega \right)=\sqrt{\frac{\pi }{2}}·\left(1+{\omega }^2\right)e^{-{\omega }^2/2}.$$

Thus, also using the Fourier inversion formula, we find the explicit integral formula for $f=f\left(x\right):$

$$\begin{array}{c}g\left(x\right)=\sqrt{\frac{\pi }{2}}·{ℱ}^{-1}\left(\left(1+{\omega }^2\right)e^{-{\omega }^2/2}\right)\left(x\right)=\\ \sqrt{\frac{\pi }{2}}·e^{-x^2/2}+\sqrt{\frac{\pi }{2}}·{ℱ}^{-1}\left({\omega }^2{e}^{-{\omega }^2/2}\right)\left(x\right)=\\ \sqrt{\frac{\pi }{2}}·e^{-x^2/2}+\sqrt{\frac{\pi }{2}}·\frac{1}{\sqrt{2\pi }}·\underset{ℝ}{\int }{\omega }^2{e}^{-{\omega }^2/2}{e}^{i\omega x}d\omega =\\ \sqrt{\frac{\pi }{2}}·e^{-x^2/2}+\frac{1}{2}·\underset{ℝ}{\int }{\omega }^2{e}^{-{\omega }^2/2}cos\left(\omega x\right)d\omega =\sqrt{\frac{\pi }{2}}·e^{-x^2/2}+\frac{1}{2}J\left(x\right). \end{array}$$

(6)

Hence, the problem is reduced to the calculation of the integral:

$$J\left(x\right):=\underset{ℝ}{\int }{\omega }^2{e}^{-{\omega }^2/2}cos\left(\omega x\right)d\omega .$$

This can be achieved through integration by parts, following the ideas from [1,2]:

$$\begin{array}{c}J\left(x\right)=-\underset{ℝ}{\int }\omega ·\left(-\omega e^{-{\omega }^2/2}\right)\cos \left(\omega x\right)d\omega =\\ -\omega ·cos\left(\omega x\right)·e^{-{\omega }^2/2}{|}_{-\infty }^{+\infty }+\underset{ℝ}{\int }{e}^{-{\omega }^2/2}\frac{d}{d\omega }\left(\omega ·cos\left(\omega x\right)\right)d\omega =\\ \underset{ℝ}{\int }{e}^{-{\omega }^2/2}\cos \left(\omega x\right)d\omega -x\underset{ℝ}{\int }\omega e^{-{\omega }^2/2}sin\left(\omega x\right)d\omega =\\ \sqrt{2\pi } ·e^{-x^2/2}+x\left(e^{-{\omega }^2/2}sin\left(\omega x\right){|}_{-\infty }^{+\infty }-x\underset{ℝ}{\int }{e}^{-{\omega }^2/2}cos\left(\omega x\right)d\omega \right)=\sqrt{2\pi } ·\left(1-x^2\right)e^{-x^2/2}.\end{array}$$

Substituting this into (6), we find the explicit expression for g$\left(x\right)$ in terms of the elementary analytic functions $e^{-x^2/2}$ and $x^2,$ as follows:

$$g\left(x\right)=e^{-x^2/2}\left(\sqrt{\frac{\pi }{2} }+\frac{1}{2}\sqrt{2\pi }·\left(1-x^2\right)\right)=\sqrt{\pi /2}·\left(2-x^2\right)·e^{-x^2/2}, x\in ℝ.$$

$${\Vert g\Vert }_{L^{\infty }(ℝ)}=g\left(0\right)=\sqrt{2\pi }.$$

□

3.2. On the Moment Problem on Subsets of $\left[0,+\infty \right):$ Sufficient Conditions

We start this subsection with a full Stieltjes moment problem in the $L_{\nu }^1\left(\left[0,+\infty \right)\right)$ space, with $\nu$ being a determinate measure on $\left[0,+\infty \right)$ (see [19] for determinacy and [20,21] for results on the reduced moment problem on $\left[0,+\infty \right)$ and related maximum entropy problems). In the next theorem, lower and upper dominated conditions on the linear operator solution are required. All the vector spaces involved in what follows are considered over the real field. Unlike previously published results, in the next theorem, none of the operators $T_1,T_2$ are assumed to be positive. In other words, the result holds for arbitrary bounded linear operators.

Theorem 1. 

Let  $\nu$ be a determinate positive regular Borel measure on $\left[0,+\infty \right),$ with finite moments of all orders; let $X$ be the Banach lattice $L_{\nu }^1\left(\left[0,+\infty \right)\right),$ $p_j\left(t\right)=t^j, t\in \left[0,+\infty \right), j\in \mathbb{N};$ let $Y$ be an order-complete Banach lattice. Let ${\left(y_j\right)}_{j\in \mathbb{N}}$ be a sequence of elements in $Y.$ Let $T_1,T_2$ be two bounded linear operators from $X$ into $Y, T_1\le T_2$ on the positive cone $X_{+}.$ Let us consider the following two statements:

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

  $$T\left(p_j\right)=y_j, j\in \mathbb{N},$$

  (7)

  $$T_1\left(\phi \right)\le T\left(\phi \right)\le T_2\left(\phi \right), \phi \in {\left(L_{\nu }^1\left(\left[0,+\infty \right)\right)\right)}_{+}.$$

  (8)
• (b) For $\epsilon >0$, any finite subsets $J_0,J_1=J_1\left(\epsilon \right),J_2=J_2\left(\epsilon \right)$ of $\mathbb{N}$ and any families of scalars ${\left\{{\lambda }_j\right\}}_{j\in J_0}, {\left\{{\alpha }_k\right\}}_{k\in J_2\left(\epsilon \right)}, {\left\{{\beta }_m\right\}}_{m\in J_1\left(\epsilon \right)},$ the following implication holds: if $q,r\in \left\{0,1\right\},$

  $${\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}-{\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j \searrow 0, \epsilon \searrow 0,$$

  (9)

  then

  $$\underset{\epsilon \searrow 0}{\lim }\left({\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{T}_2(p_{k+l+q}) -{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{T}_1\left(p_{m+n+r}\right)\right)\ge {\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j$$

  (10)

  Then,$\left(b\right)\Rightarrow \left(a\right)$ holds.

Proof. 

To prove that implication (b) implies (a), we use the corresponding implication of Theorem 2.38 from [25], accompanied by a polynomial approximation result of elements from ${\left(L_{\nu }^1\left(\left[0,+\infty \right)\right)\right)}_{+}$ by nonnegative polynomials, in the norm of the space $L_{\nu }^1\left(\left[0,+\infty \right)\right)$ (see [25], Lemma 4.11, p. 383). Let

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j={\phi }_2-{\phi }_1,$$

with

$${\phi }_1,{\phi }_2\in {\left(L_{\nu }^1\left(\left[0,+\infty \right)\right)\right)}_{+}.$$

According to the polynomial approximation result by nonnegative polynomials for the elements ${\phi }_1,{\phi }_2,$ for $\epsilon >0,$ there exist polynomials $u_{1,\epsilon },u_{2,\epsilon },$ with $u_{i,\epsilon }\left(t\right)\ge {\phi }_i\left(t\right)\ge 0$ for almost all $t\in \left[0,+\infty \right),$ such that

$${\Vert u_{i,\epsilon }-{\phi }_i\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}\le \epsilon \searrow 0, i=1,2.$$

In other words, we can write ${\phi }_i=\underset{\epsilon \searrow 0}{\lim }{u}_{i,\epsilon }, i=1,2.$We recall the explicit expression of polynomials taking nonnegative values at all points of the interval $\left[0,+\infty \right).$ As is well known (see [8]), any such polynomial function is a sum of squares of polynomials plus $p_1\left(t\right)=t$ multiplied by another sum of squares of polynomials. Hence, for $u_{1,\epsilon },u_{2,\epsilon },$ we can write

$$u_{2,\epsilon }:={\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}, u_{1,\epsilon }:={\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r},\hspace{1em}q,r\in \left\{0,1\right\}.$$

(11)

On the other hand, from (9) and (10) written for ${\lambda }_j=0, j\in J_0$, we derive that

$$\begin{array}{c}\Vert u_{1,\epsilon }-u_{2,\epsilon }-\left({\phi }_1-{\phi }_2\right){\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}\le \\ <\Vert u_{1,\epsilon }-{\phi }_1{\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}+\Vert u_{2,\epsilon }-{\phi }_2{\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}\le 2\epsilon \searrow 0.\\ \Vert {\displaystyle \sum _{j\in J_0}}{\lambda }_j{p}_j-\left(u_{2,\epsilon }-u_{1,\epsilon }\right){\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}\le \Vert {\displaystyle \sum _{j\in J_0}}{\lambda }_j{p}_j-\left({\phi }_2-{\phi }_1\right){\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}+\\ \Vert {\phi }_2-{\phi }_1-\left(u_{2,\epsilon }-u_{1,\epsilon }\right){\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}=\Vert u_{1,\epsilon }-u_{2,ϵ}-\left({\phi }_1-{\phi }_2\right){{\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}}_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}\le 2\epsilon \searrow 0.\end{array}$$

(12)

for some nonnegative polynomials $u_{1,\epsilon },u_{2,\epsilon }$ depending on the functions ${\phi }_1,{\phi }_2,$and on arbitrary small $\epsilon >0.$

In our hypothesis (b), $\epsilon \searrow 0;$ then, Equation (10) claims that for any $u_2,u_1$ as mentioned above and written in (11), we have

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j\le \underset{\epsilon \searrow 0}{\lim }\left(T_2\left(u_{2,\epsilon }\right)-T_1\left(u_{1,\epsilon }\right)\right).$$

(13)

Now, the continuity of the bounded linear operators $T_1,T_2$ converges

$$\Vert u_{i,\epsilon }-{\phi }_i{\Vert }_{L_{\nu }^1\left(\left[0,+\infty \right)\right)}\to 0, \epsilon \searrow 0, i=1,2,$$

as well as the fact that the positive cone of the Banach lattice $Y$ is topologically closed, passing to the limit as $\epsilon ·0.$ Using (13) and $\epsilon ·0,$ this yields

$${\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j\le T_2\left({\phi }_2\right)-T_1\left({\phi }_1\right).$$

Since all of this reasoning holds for arbitrary ${\phi }_1,{\phi }_2\in {\left(L_{\nu }^1\left(\left[0,+\infty \right)\right)\right)}_{+}$ with ${\sum }_{j\in J_0}{\lambda }_j{p}_j={\phi }_2-{\phi }_1,$ the conclusion (a) concerning the existence of the linear solution $T$ satisfying the properties (7) and (8) follows Theorem 2.38 from [25]. It remains to prove the continuity of the solution $T.$ Since $T$ is linear, it is sufficient (and necessary) to prove its continuity at zero. If $g_n\to 0$ in the Banach lattice $L_{\nu }^1\left(\left[0,+\infty \right)\right),$ then $g_n^{+}\to 0$ in the same space and $g_n^{+}\in {\left(L_{\nu }^1\left(\left[0,+\infty \right)\right)\right)}_{+}$for all $n.$ According to the continuity of $T_1, T_2,$ also using (8), since the positive cone of the Banach lattice $Y$ is closed and normal (see [7]), this implies $\underset{n}{\lim }T\left(g_n^{+}\right)=0.$ Similarly, $\underset{n}{\lim }T\left(g_n^{-}\right)=0.$ Hence,

$$\underset{n}{\lim }T\left(g_n\right)=\underset{n}{\lim }T\left(g_n^{+}\right)-\underset{n}{\lim }T\left(g_n^{-}\right)=0.$$

Now the uniqueness of the continuous solution $T$ follows from the density of polynomials in $L_{\nu }^1\left(\left[0,+\infty \right)\right).$ □

Corollary 1. 

Let  $\nu , X, p_j$ be as in Theorem 3,  ${\left(y_j\right)}_{j\in \mathbb{N}}$ a sequence of real numbers, and  $g_1,g_2$ be two given functions from $L_{\nu }^{\infty }\left(\left[0,+\infty \right)\right), g_1\le g_2$ almost everywhere in  $\left[0,+\infty \right).$ Let us consider the following two statements.

• (a) There exists a unique function  $g\in L_{\nu }^{\infty }\left(\left[0,+\infty \right)\right)$ such that

  $${\int }_0^{+\infty }{t}^{j}g\left(t\right)d\nu =y_j, j\in \mathbb{N}, g_1\left(t\right)\le g\left(t\right)\le g_2\left(t\right),$$

  for almost all  $t\in \left[0,+\infty \right).$
• (b) For any finite subsets  $J_0,J_1,J_2$ of $\mathbb{N}$ and any families of scalars ${\left\{{\lambda }_j\right\}}_{j\in J_0}, {\left\{{\alpha }_k\right\}}_{k\in J_2}, {\left\{{\beta }_m\right\}}_{m\in J_1},$ the following implication holds: if

  $${\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}-{\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j \searrow 0, \epsilon \searrow 0, q,r\in \left\{0,1\right\},$$

  then

  $$\underset{\epsilon \searrow 0}{\lim }\left({\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{\int }_0^{+\infty }{t}^{k+l+q}{g}_2\left(t\right)d\nu -{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{\int }_0^{+\infty }{t}^{m+n+r}{g}_1\left(t\right)d\nu \right)\ge {\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j.$$

  Then,  $\left(b\right)\Rightarrow \left(a\right)$ holds.

Proof. 

One applies Theorem 3, where $Y$ stands for $ℝ.$ The conclusion follows via measure theory results [1,3]. □

We can prove the following result in the same manner as that of Theorem 1, using another polynomial approximation result, namely Lemma 2 from [24].

Theorem 2. 

Let  $S\subset \left[0,+\infty \right)$ be a nonempty compact subset and  $C\left(S\right)$ be the Banach lattice of all real valued continuous functions on  $S.$ Let  $p_j,Y,y_j,T_1,T_2$ be as in the Theorem 1 proved above, where the space  $X=L_{\nu }^1\left(\left[0,+\infty \right)\right)$ is replaced by $C\left(S\right).$ Let us consider the following two statements:

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

  $$T\left(p_j\right)=y_j, j\in \mathbb{N},$$

  $$T_1\left(\phi \right)\le T\left(\phi \right)\le T_2\left(\phi \right), \phi \in {\left(C\left(S\right)\right)}_{+}.$$
• (b) For any finite subsets  $J_0,J_1,J_2$ of $\mathbb{N}$ and any families of scalars ${\left\{{\lambda }_j\right\}}_{j\in J_0}, {\left\{{\alpha }_k\right\}}_{k\in J_2(\left(\epsilon \right)}, {\left\{{\beta }_m\right\}}_{m\in J_1\left(\epsilon \right)},$ the following implication holds: if

  $${\displaystyle \sum }_{k,l\in J_2(\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}-{\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j\searrow 0, \epsilon \searrow 0, q,r\in \left\{0,1\right\},$$

  then

  $${\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j\le \underset{\epsilon \searrow 0}{\lim }\left({\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{T}_2(p_{k+l+q})-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{T}_1\left(p_{m+n+r}\right) \right).$$

  Then,$\left(b\right)\Rightarrow \left(a\right)$ holds.

Proof. 

The proof follows the ideas of Theorem 1. Here, the main difference is that the polynomial approximation provided by Lemma 2 from [24] is applied. Now, any function $\phi \in C\left(S\right)$ with

$$\phi \left(t\right)\ge 0\mathrm{for}\mathrm{all}t\in S,$$

can be approximated by a sequence of polynomials $r_m\ge {\phi }_0$ on $\left[0,+\infty \right),$ where ${\phi }_0$ is the extension of $\phi$ defined by ${\phi }_0\left(t\right):=0$ for all $t\in \left[0,+\infty \right)\setminus S.$ The convergence holds uniformly on $S;$ hence, it holds in $C\left(S\right).$ Then, $r_m\left(t\right)\ge {\phi }_0\left(t\right)\ge 0$ for all $t\ge 0,$ so that each polynomial $r_m$ can be expressed as $r_m\left(t\right)={\sum }_{k,l\in J_1}{\alpha }_k{\alpha }_l{p}_{k+l+q}$, $q\in \left\{0,1\right\},$ for finite subset $J_1\subset \mathbb{N}$ and real coefficients ${\alpha }_l.$

Let us notice that ${\phi }_0$ can be discontinuous at some points in $\left[0,+\infty \right).$ The conclusion follows in the same manner as that of Theorem 1. □

In the next result, $Y=Y\left(A\right)$ will be the order-complete Banach lattice of self-adjoint operators acting on a Hilbert space $H$, associated with a self-adjoint operator $A$ acting on $H$ (see [6], pp. 303–305). This Banach lattice is also a commutative Banach algebra. The norm on $Y\left(A\right)$ is the operatorial norm, and the order relation is the usual one: $U\le W$ in $Y\left(A\right)$ if and only if $\left(Uh;h\right)\le \left(Wh;h\right)$ for all $h\in H.$ We use the following type of self-adjoint operators. Namely, we call self-adjoint operator $A$ on a real or complex Hilbert space ($H)$ a bounded linear operator applying $H$ into $H,$ which satisfies the following symmetry condition: $\left(Ax;y\right)=\left(x;Ay\right)$ for all $x,y\in H.$ According to the definition of the adjoint $A^{*}$ of a bounded linear operator $\left(A\right)$ on $H$ [4], the above written condition says that $A^{*}=A.$ Thus, a self-adjoint operator equals its adjoint. This is the motivation of the terminology self-adjoint for such operators. If $H$ is finite dimensional, a self-adjoint operator is called a Hermitian operator.

Corollary 2. 

Let  $H$ be the real Hilbert space and  $A$ a positive self-adjoint operator from  $H$ into $H,$  with the spectrum  $\sigma \left(A\right).$  Assume that  $T_1,T_2$  are two bounded linear operators from  $X:=C\left(\sigma \left(A\right)\right)$  into  $Y\left(A\right).$

Let  ${\left(E_j\right)}_{j\in \mathbb{N}}$  be given a sequence of operators from  $Y\left(A\right).$  Let us consider the following two statements:

• (a) There exists a unique bounded linear operator  $T\in B\left(X,Y\left(A\right)\right)$  with

  $$T\left(p_j\right)=E_j, j\in \mathbb{N},$$

  $$T_1\left(\phi \right)\le T\left(\phi \right)\le T_2\left(\phi \right), \phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+}.$$
• (b) For any finite subsets  $J_0,J_1,J_2$ of $\mathbb{N}$ and any families of scalars ${\left\{{\lambda }_j\right\}}_{j\in J_0}, {\left\{{\alpha }_k\right\}}_{k\in J_2\left(\epsilon \right)}, {\left\{{\beta }_m\right\}}_{m\in J_1\left(\epsilon \right)},$ the following implication holds. If

  $${\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}-{\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j\searrow 0, \epsilon \searrow 0, q,r\in \left\{0,1\right\},$$

  then:

  $${\displaystyle \sum }_{j\in J_0}{\lambda }_j{E}_j\le \underset{\epsilon \searrow 0}{\lim }\left({\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{T}_2(p_{k+l+q})-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{T}_1\left(p_{m+n+r}\right),\right).$$

  Then,$\left(b\right)\Rightarrow \left(a\right)$ holds.

Proof. 

Since the self-adjoint operator $A$ is positive, its spectrum is contained in the interval $\left[0,+\infty \right).$ The conclusion follows via Theorem 4, where $S$ stands for $\sigma \left(A\right).$ □

Theorem 3. 

Let $H$ be the Hilbert space $L_{dt}^2\left(\left[0,1\right]\right)$ and $A$ a self-adjoint operator acting on $H,$ such that the spectrum $\sigma \left(A\right)$ is the entire interval $\left[0,1\right]$. With the notations and in the framework of Corollary 2, let us consider the following two statements.

• (a) There exists a unique bounded linear operator  $T\in B\left(X,Y\left(A\right)\right)$with

  $$T\left(p_j\right)=E_j, j\in \mathbb{N},$$

  $$\left(\frac{1}{2}I-A\right)\phi \left(A\right)\le T\left(\phi \right)\le \left(e^{-A}-\frac{1}{2}I\right)\phi \left(A\right), \phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+}.$$
• (b) For any finite subsets  $J_0,J_1,J_2$ of $\mathbb{N}$ and any families of scalars ${\left\{{\lambda }_j\right\}}_{j\in J_0}, {\left\{{\alpha }_k\right\}}_{k\in J_2}, {\left\{{\beta }_m\right\}}_{m\in J_1},$ the following implication holds: if

  $${\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}-{\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j\searrow 0, \epsilon \searrow 0, q,r\in \left\{0,1\right\},$$

  then

  $${\displaystyle \sum }_{j\in J_0}{\lambda }_j{E}_j\le \underset{\epsilon \searrow 0}{\lim }\left(\left(e^{-A}-\frac{1}{2}I\right)·\left({\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{A}^{k+l+q}\right)-\left(\frac{1}{2}I-A\right)·\left({\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{A}^{m+n+r}\right)\right).$$

  Then,$\left(b\right)\Rightarrow \left(a\right)$ holds.

Proof. 

We apply Corollary 2 to the bounded linear operators $T_1,T_2\in ℬ\left(C\left(\sigma \left(A\right)\right),Y_A\right),$ defined below, as follows:

$$g\left(t\right):=e^{-t}-\frac{1}{2}>h\left(t\right):=\frac{1}{2}-t, t\ge 0.$$

$$T_2\left(\phi \right)\left(t\right):=g\left(t\right)\phi \left(t\right)=(M_g\phi )\left(t\right)=\left(e^{-t}-\frac{1}{2}\right)\phi \left(t\right),$$

$$T_1\left(\phi \right)\left(t\right):=h\left(t\right)\phi \left(t\right)=\left(M_h\phi \right)\left(t\right)=\left(\frac{1}{2}-t\right)\phi \left(t\right), t\in \sigma \left(A\right), \phi \in C\left(\sigma \left(A\right)\right).$$

$$T_1\left(\phi \right)\left(A\right):=\left(\frac{1}{2}I-A\right)\phi \left(A\right), T_2\left(\phi \right)\left(A\right):=\left(e^{-A}-\frac{1}{2}I\right)\phi \left(A\right), \phi \in C\left(\sigma \left(A\right)\right).$$

Unlike our previously published results (see [25], Theorems 2.38, 2.39), now, the above defined operators are not positive since the subsets of the interval $\left[0,1\right]$ where the function $g$ takes positive and negative values, respectively, are open and nonempty. Consequently, the Lebesgue measure of these open subsets is positive. The same remark holds for $h.$ Now, the conclusion follows from Corollary 2 via functional calculus for self-adjoint operators and continuous functions on their spectrums. □

3.3. On the Moment Problem on Subsets of $\left[0,+\infty \right):$ Necessary and Sufficient Conditions

In this section, we study the case when the given linear operators $T_1, T_2$ are positive. Clearly, the solution $T$ will be positive as well. According to a more general result of [23], all these three positive linear operators will be continuous since they are acting on Banach lattices. Thus, their positivity allows the sufficient conditions from Section 3.2 to be proved as necessary conditions. We recall that a linear operator on ordered vector spaces is called positive if it is positive on the positive cone of the domain space. We also recall that every positive linear operator on ordered Banach spaces is continuous. An ordered Banach space is a Banach space $X$ endowed with a linear order relation, such that the positive cone $X_{+}$ is proper $(X_{+}\cap \left(-X_{+}\right)=\left\{0\right\})$, generating $\left(X=X_{+}-X_{+}\right);$ $X_{+}$ is topologically closed, and $0\le x\le y$ in $X$ implies $\Vert x\Vert \le \Vert y\Vert .$ A Banach lattice is a vector lattice $X$ which is also a Banach space, such that $\left|x\right|\le \left|y\right|$ in $X$ implies $\Vert x\Vert \le \Vert y\Vert .$ We recall that in a vector lattice $X$,

$$\left|x\right|:=sup\left\{x,-x\right\}=x\vee \left(-x\right), x\in X.$$

From these definitions, it is easy to see that that every Banach lattice is an ordered Banach space (see [6,7]).

Theorem 4. 

Let  $\nu ,X,p_j,y_j$ be as in Theorem 1,  $Y$ be an order-complete Banach lattice, and $T_1,T_2$ be two linear operators from $X$ into $Y.$ Assume that  $T_1$  is positive on the positive cone of  $X=L_{\nu }^1\left(\left[0,+\infty \right)\right).$  The following two statements are equivalent:

• (a) There exists a unique positive (bounded) linear operator  $T\in B\left(X,Y\right)$  with

  $$T\left(p_j\right)=y_j, j\in \mathbb{N},$$

  $$T_1\left(\phi \right)\le T\left(\phi \right)\le T_2\left(\phi \right), \phi \in {\left(L_{\nu }^1\left(\left[0,+\infty \right)\right)\right)}_{+}, \Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert .$$
• (b) For  $\epsilon >0$, any finite subsets  $J_0,J_1=J_1\left(\epsilon \right),J_2=J_2\left(\epsilon \right)$  of  $\mathbb{N}$  and any families of scalars  ${\left\{{\lambda }_j\right\}}_{j\in J_0}, {\left\{{\alpha }_k\right\}}_{k\in J_2\left(\epsilon \right)}, {\left\{{\beta }_m\right\}}_{m\in J_1\left(\epsilon \right)},$  the following implication holds: if  $q,r\in \left\{0,1\right\},$

  $${\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}-{\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j \searrow 0, \epsilon \searrow 0,$$

   then

  $$\underset{\epsilon \searrow 0}{\lim }\left({\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{T}_2(p_{k+l+q}) -{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{T}_1\left(p_{m+n+r}\right)\right)\ge {\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j.$$

Proof. 

Implication (b) implies that (a) has been proven in Theorem 1$.$ According to the proof of Theorem 1 and the positivity hypothesis on $T_1,$ we derive $T\ge T_1\ge 0$ on $X_{+}.$ This results in $T$ being a positive linear operator; hence, it is continuous. For the converse, observe that all the involved linear operators $T_1,T_2,T$are positive (and hence are continuous). Also, we notice that the arrow $\searrow$ from (b) proves that for any $J_1,J_2,$ the polynomial ${\sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}$ dominates the polynomial ${\sum }_{j\in J_0}{\lambda }_j{p}_j.$ Now, using (a) and the positivity of $T$, which implies its continuity, we infer that

$$\begin{array}{c}{\displaystyle \sum _{j\in J_0}}{\lambda }_j{y}_j=T\left({\displaystyle \sum _{j\in J_0}}{\lambda }_j{p}_j\right)\le \underset{\epsilon \searrow 0}{\lim }T\left(u_{2,\epsilon }\right)-\underset{\epsilon \searrow 0}{\lim }T\left(u_{1,\epsilon }\right)\le \\ \underset{\epsilon \searrow 0}{\lim }{T}_2\left(u_{2,\epsilon }\right)-\underset{\epsilon \searrow 0}{\lim } T_1\left(u_{1,\epsilon }\right)=\underset{\epsilon \searrow 0}{\lim }\left({\displaystyle \sum _{k,l\in J_2\left(\epsilon \right)}}{\alpha }_k{\alpha }_l{T}_2(p_{k+l+q})-{\displaystyle \sum _{m,n\in J_1\left(\epsilon \right)}}{\beta }_m{\beta }_n{T}_1\left(p_{m+n+r}\right)\right).\end{array}$$

Thus, (a) implies that (b) is proven. Next, we prove that the solution $T$ satisfies the following inequalities:

$$\Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert .$$

The linearity and positivity of $T$ and $T_1\le T\le T_2$ on $X_{+}$ lead to

$$T\left(\pm x\right)\le T\left(\left|x\right|\right)\le T_2\left(\left|x\right|\right)\mathrm{for}\mathrm{all}x\in X.$$

Hence, $\left|T\left(x\right)\right|\le T_2\left(\left|x\right|\right), x\in X.$ Since $Y,X$ are a Banach lattices, this implies

$$\Vert T\left(x\right)\Vert \le \Vert T_2\left(\left|x\right|\right)\Vert \le \Vert T_2\Vert ·\Vert \left|x\right|\Vert =\Vert T_2\Vert ·\Vert x\Vert$$

If $\Vert x\Vert \le 1,$ one obtains $\Vert T\left(x\right)\Vert \le \Vert T_2\Vert ;$ hence, $\Vert T\Vert \le \Vert T_2\Vert .$ The inequality $\Vert T_1\Vert \le \Vert T\Vert$ follows the same proof. □

Theorem 5. 

Let $S, X:=C\left(S\right), p_j, Y,y_j$ be as in Theorem 2. Assume that  $T_1,T_2$ are linear operators applying $C\left(S\right)$ into $Y,$ such that $T_1$ is positive on the positive cone ${(C\left(S\right))}_{+}.$ The following two statements are equivalent:

• (a) There exists a unique positive (hence bounded) linear operator$T\in B\left(X,Y\right)$with

  $$T\left(p_j\right)=y_j, j\in \mathbb{N},$$

  $$T_1\left(\phi \right)\le T\left(\phi \right)\le T_2\left(\phi \right), \phi \in {\left(C\left(S\right)\right)}_{+}, \Vert T_1\Vert \le \Vert T\Vert \le \Vert T_2\Vert .$$
• (b) For any finite subsets  $J_0,J_1,J_2$ of $\mathbb{N}$ and any families of scalars ${\left\{{\lambda }_j\right\}}_{j\in J_0}, {\left\{{\alpha }_k\right\}}_{k\in J_2(\left(\epsilon \right)}, {\left\{{\beta }_m\right\}}_{m\in J_1\left(\epsilon \right)},$ the following implication holds. If

  $${\displaystyle \sum }_{k,l\in J_2(\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}-{\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j\searrow 0, \epsilon \searrow 0, q,r\in \left\{0,1\right\},$$

  then

  $${\displaystyle \sum }_{j\in J_0}{\lambda }_j{y}_j\le \underset{\epsilon \searrow 0}{\lim }\left({\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{T}_2(p_{k+l+q})-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{T}_1\left(p_{m+n+r}\right) \right).$$

Proof. 

Implication (b) implies that (a) has been proven in Theorem 2. The implication in the reversed sense follows in the same manner as (a) implying (b) of Theorem 4. □

Theorem 6. 

Let  $H$ be the Hilbert space  $L_{dt}^2\left(\left[0,1\right]\right)$  and  $A$  a positive self-adjoint operator acting on  $H, {\left(E_j\right)}_{j\in \mathbb{N}},$  a sequence in  $Y\left(A\right).$  With the notations and in the setting of Corollary 2, assume that  $T_1,T_2·C\left(\sigma \left(A\right)\right)·Y\left(A\right)$  are defined by

$$T_1\left(\phi \right)\left(t\right):=\left(A+\frac{1}{2}I\right)\phi \left(A\right), T_2\left(\phi \right)\left(t\right):=\left(e^A-\frac{1}{2}I\right)\phi \left(A\right), \phi \in C\left(\sigma \left(A\right)\right).$$

The following two statements are equivalent:

• (a) There exists a unique bounded linear operator  $T\in B\left(X,Y\left(A\right)\right)$with

  $$T\left(p_j\right)=E_j, j\in \mathbb{N},$$

  $$\left(\frac{1}{2}I+A\right)\phi \left(A\right)\le T\left(\phi \right)\le \left(e^A-\frac{1}{2}I\right)\phi \left(A\right), \phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+},$$

  $$\Vert \frac{1}{2}I+A\Vert \le \Vert T\Vert \le \Vert e^A-\frac{1}{2}I\Vert .$$
• (b) For any finite subsets  $J_0,J_1,J_2$ of $\mathbb{N}$ and any families of scalars ${\left\{{\lambda }_j\right\}}_{j\in J_0}, {\left\{{\alpha }_k\right\}}_{k\in J_2}, {\left\{{\beta }_m\right\}}_{m\in J_1},$ the following implication holds: if

  $${\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{p}_{k+l+q}-{\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{p}_{m+n+r}-{\displaystyle \sum }_{j\in J_0}{\lambda }_j{p}_j\searrow 0, \epsilon \searrow 0, q,r\in \left\{0,1\right\},$$

  then

  $${\displaystyle \sum }_{j\in J_0}{\lambda }_j{E}_j\le \underset{\epsilon \searrow 0}{\lim }\left(\left(e^A-\frac{1}{2}I\right)·\left({\displaystyle \sum }_{k,l\in J_2\left(\epsilon \right)}{\alpha }_k{\alpha }_l{A}^{k+l+q}\right)-\left(\frac{1}{2}I+A\right)·\left({\displaystyle \sum }_{m,n\in J_1\left(\epsilon \right)}{\beta }_m{\beta }_n{A}^{m+n+r}\right)\right).$$

Proof. 

The first two properties from (a) on the solution $T$ follow from (b) in the same manner as the proof of Theorem 3. To obtain the last assertion of (a) on the estimation of the norm of $T,$ we use the fact that $Y\left(A\right)$ is a Banach lattice so that $U,V\in Y\left(A\right), 0\le U\le V$ implies $\Vert U\Vert \le \Vert V\Vert .$ With the notations from above, the norms $\Vert T_1\left(\phi \right)\Vert ,\Vert T_2\left(\phi \right)\Vert ,$ $\phi \in {\left(C\left(\sigma \left(A\right)\right)\right)}_{+}$ and then $\Vert T_1\Vert ,\Vert T_2\Vert$ can be determined via functional calculus for continuous real valued functions on the spectrum of the self-adjoint operator $A$ [4]. □

4. Discussion

We have proven theorems referring to integral equations and to the moment problem, and two of their consequences. The moment problem is an inverse problem because we find information on the solution (such as existence, values of polynomials, uniqueness, and sandwich properties), although the solution cannot be expressed in terms of elementary functions. To solve such problems, we use other authors and our previous results on this subject. In the present work, sufficient conditions for the existence and uniqueness of the solutions are expressed in terms of quadratic forms. Usually, proving sufficiency is the difficult part in case of our equivalent conditions. The cases of spaces $L_{\nu }^1\left(\left[0,+\infty \right)\right)$, with $\nu$ an (M)-determinate measure on $\left[0,+\infty \right),$ as well as that of the space of all real valued continuous functions on a compact subset of $\left[0,+\infty \right),$ are both under attentions. The results of the approximation of nonnegative functions from the domain space by nonnegative polynomials on $\left[0,+\infty \right)$ applied in these two cases are different. Their aim is to use the expression of such a polynomial in terms of sums of squares. In case of positive linear constraints, the solution is also positive. The sufficient conditions mentioned above are also necessary conditions. The norm of the solution can be estimated in terms of the norms of the given constraints. All these results are expressed in terms of the given moments and linear constraints. As a direction for future work, one can prove a version of Theorem 1 for functions of several variables with the space $L_{\nu }^1\left({\left[0,+\infty \right)}^n\right), n\in \mathbb{N}, n\ge 2, \nu ={\nu }_1\times \cdots \times {\nu }_n, {\nu }_j$ being a determinate measure in $\left[0,+\infty \right),$for any $j\in \left\{1,\dots ,n\right\}.$ Similar problems can be solved in spaces $L_{\nu }^1\left({ℝ}^n\right), n\ge 2.$ The connection of the present study with the notions of symmetry is pointed out by means of the results involving self-adjoint operators. Such results are stated and proved in Corollary 2 and Theorems 3 and 6. The paragraph preceding Corollary 2 reviews and discusses the corresponding terminology. On the other hand, the second problem appearing in the title, that of solving integral equations, is explicitly solved: the solutions are expressed in terms of elementary functions. This cannot be so easily achieved without using Fourier transform and its properties. Of course, there are other types of problems which can also be solved using the methods provided by the Fourier transform.

5. Conclusions

Referring to the moment problem, the main results of this article are contained in Section 3.2. These results work for arbitrary bounded (not necessarily positive) linear operators $T_1, T_2$ defining the constraints. They establish sufficient conditions for the existence and uniqueness of the bounded linear solution $T$ of the full moment problem, satisfying the sandwich constraints on the positive cone of the domain space. For the case of positive linear operators appearing in Section 3.3, they prove the main implication showing the sufficiency of the conditions stated in point (b) of the corresponding statements. This is true, since every positive linear operator acting between Banach lattices is continuous, when applying the results from Section 3.2. It is possible that other Hahn–Banach-type results can be applied, which are specific to the extension of positive linear operators. The converse implication (a) implies (b) and follows without using any extension-type result for linear operators. This is the easier part of the proof. However, this is also important since it ensures the characterization of the existence and uniqueness of the positive (bounded) linear solution of the full moment problem under attention. Regarding the integral equations of Section 3.1, it seems that applying properties of Fourier transform is an appropriate way of solving them. In the moment problems, as well as in the integral equations, the solutions are defined implicitly. Here is a possible relationship between Examples 1 and 2 and the rest of the results, referring to the moment problem: In both cases, the solutions are defined implicitly. This is a similarity property. Unlike the case of Examples 1 and 2, where the solutions can be expressed in terms of elementary functions, in the cases of the moment problems of Section 3.2 and Section 3.3, we have information on the properties of the solutions (mentioned in points (a) of the corresponding theorems), but we cannot express the solutions in terms of elementary functions. The moment problem is an inverse problem. Returning to Examples 1 and 2, both the solutions are derived with the aid of the inverse Fourier transform, an integral transform whose kernel is known. As directions for related future work, one can state similar results to those from Theorems 3 and 6, respectively, when the Hilbert space $H={ℝ}^n, n\in \mathbb{N}, n\ge 2.$ Such results involve commutative vector subspaces of the space $\mathcal{S}ym\left(n,ℝ\right)$ of all symmetric $n\times n$ matrices with real entries. These subspaces are also order-complete vector lattices with respect to the usual order relation on $\mathcal{S}ym\left(n,ℝ\right).$ The case of infinite dimensional Hilbert space, $H$, could be completed by passing from self-adjoint operators to similar results for normal operators (bounded linear operators $U\in B\left(H\right),$with $U{U}^{*}=U^{*}U$). Thus, applications of the general results of the present work to concrete spaces and operators could be continued and completed.