Review Article | Open Access
New Results on Markov Moment Problem
The present work deals with the existence of the solutions of some Markov moment problems. Necessary conditions, as well as necessary and sufficient conditions, are discussed. One recalls the background containing applications of extension results of linear operators with two constraints to the moment problem and approximation by polynomials on unbounded closed finite-dimensional subsets. Two domain spaces are considered: spaces of absolute integrable functions and spaces of analytic functions. Operator valued moment problems are solved in the latter case. In this paper, there is a section that contains new results, making the connection to some other topics: bang-bang principle, truncated moment problem, weak compactness, and convergence. Finally, a general independent statement with respect to polynomials is discussed.
1. Introduction and Known Results
We recall that the classical formulation of the moment problem, under the terms of T. Stieltjes, given in 1894-1895 (see  for details), finds the repartition of the positive mass on the semiaxis ), if the moments of arbitrary orders are given. Precisely, in the Stieltjes moment problem, a sequence of real numbers is given, and one looks for a nondecreasing real function () that verifies the moment conditions: This is a one-dimensional moment problem, on an unbounded interval. The existence, the uniqueness, and the construction of the solution are considered. The present paper concerns firstly the existence problem. If the interval is replaced by a subset of , , we have a multidimensional moment problem. If the sequence of the real numbers moments becomes a sequence of operators, we have an operator-valued moment problem. Most of the problems appearing in applications require not only positive, but also an upper constraint on the solution. This is the Markov moment problem.
Applying extension Hahn-Banach type results in existence questions concerning the moment problem is a well-known technique [1–13]. One of the most useful results is the lemma of the majorizing subspace (see [14, Section 5.1.2] for the proof of the lattice version of this lemma; see also ). It says that if is a linear positive operator on a subspace of the ordered vector space , the target space being an order-complete vector lattice , and for each , there is , , then has a linear positive extension . Another geometric remark is that in the real case, the sublinear functional from Hahn-Banach theorem can be replaced by a convex one. The theorem remains valid when the convex dominating functional is defined on a convex subset with some properties with respect to the subspace (e.g., ri), where ri is the relative interior of . The problem is to find necessary and sufficient conditions for the existence of a solution, preserving sandwich conditions. Here we recall an answer published in 1978 , without loosing convexity, but strongly generalizing the classical result. The first detailed proof was published in . A proof in terms of the moment problem was published in . Parts of these generalizations of the Hahn-Banach principle are involved in the present work too. Throughout this first part, will be a real vector space, an order-complete vector lattice, a convex subsets, a concave operator, a convex operator, a vector subspace, a linear operator.
Theorem 1. Assume that , .
The following assertions are equivalent:(a)there is a linear extension of the operator such that (b)there are convex and concave operators such that for all
The following result related to the theorem of H. Bauer (see [15, Section 5.4]) is deduced.
Theorem 2. Let be a preordered vector space with its positive cone a convex operator, a vector subspace, and a linear positive operator. The following assertions are equivalent:(a)there is a linear positive extension of such that (b) for all such that .
Now, we can deduce the main results on the abstract moment problem .
Theorem 3. Let , , be as in Theorem 2, , given families. The following statements are equivalent: (a)there is a linear positive operator such that
(b)for any finite subset and any , one has
A clearer sandwich-moment problem variant is the following one.
Theorem 4. Let , , be as in Theorem 3 and , two linear operators. The following statements are equivalent:(a)there is a linear operator such that (b)for any finite subset and any , one has:
Some of the results of this work are applications of the theorems stated previously. Most of the results were published in the last ten years, a few of them being new (Section 4).
For uniqueness of the solution, see [1, 3, 4, 18–22]. Several other interesting related results are contained in [20, 23] (application of fixed-point principle, iterative methods, and complex functions methods) [8, 24], (construction of a solution) [13, 25]. For stable algorithms related to the finite Markov moment problem, see [20, 24]. Similar results are deduced in Section 4 of the present work, by using other methods: bang-bang principle, weak compactness, weak approximation , characteristic functions as extreme “points,” Gram-Schmidt procedure, and cell-decomposition. These methods seem to work in arbitrary several dimensions.
2. Approximation on Unbounded Subsets and Moment Problems in Spaces of “Even” Functions
Lemma 5 (see  and [10, Lemma 1.3(d)]). If is a nonnegative continuous function with compact support, then there exists a sequence of positive polynomials on , such that uniformly on compact subsets of .
The idea of the proof is to add the point and to apply the Stone-Weierstrass theorem to the subalgebra generated by the functions , . Then, one uses for each such -function suitable majorizing or minorizing partial polynomial sums, as well as the relation Note that [10, Lemma 1.4] asserts the density of positive polynomials in , for any closed subset of a finite-dimensional space, , being a positive regular Borel M-determinate measure. Here is the exact formulation of this result.
Lemma 6 (see [10, Lemma 1.4]). Let be an arbitrary closed subset and an M-determinate positive regular Borel measure on , with finite moments of all natural orders. Then, for any such that for some polynomial , there exists a sequence ,, in . In particular, one has: is dense in , and is dense in ,
Recall that is the subspace of all polynomial functions on .
Lemma 7. For any simple function of the form being intervals, there exists a sequence of polynomial where is an M-determinate positive regular Borel measure with finite moments of all natural orders. If is extended to an even function in , then the polynomials are restrictions to the positive semiaxes of even polynomials in .
Note that in the preceding Lemmas 5–7, the sequence is not monotone, but its elements are positive on their domain. The proof of Lemma 6 uses Hahn-Banach theorem, while that of Lemma 5 is more “constructive,” the latter being a statement independent with respect to the notion of a measure.
Remark 8. The general form of positive even polynomials on the real axes is We start by proving some results on the one-dimensional moment problem.
Theorem 9 (see [10, Theorem 2.1]). Let be a positive determinate Borel regular measure on with finite moments of all orders, such that for any odd function . Let be a given sequence of real numbers.
The following statements are equivalent:(a)there exists a Borel function such that (b)for any finite subset and any , one has
Proof. The hypothesis (a) implies that (b) is obvious because of the qualities of .
(b)(a). Since vanishes on the subspace of odd functions, the same property is valid for . The problem can be reduced to a corresponding one on the positive semiaxis . Using the form of the positive polynomials in , the hypothesis (b) yields Since is a majorizing subspace in there exists a positive linear extension of . By Haviland-Riesz representation theorem, there exists a representing positive regular Borel measure on such that Applying Haviland-Riesz theorem once more, we infer the existence of a positive regular Borel measure on , such that Since is M-determinate, application of Lemma 7 and the last relation lead to Borel subset. In particular, is absolutely continuous with respect to , hence there exists such that on simple functions. From [28, Theorem 1.40], we infer that -a.e. Since was a positive regular Borel measure, we can find a Borel function such that -a.e. The previous relations lead to This concludes the proof.
Corollary 10. Let and . The following statements are equivalent:(a)there exists a Borel function , such that (b)for any finite subset and any , one has
Proof. One applies Theorem 9 to , as a measure on .
Due to Carleman criterion, is M-determinate. A straightforward computation shows that The conclusion follows.
Using a quite different proof with respect to that of Theorem 9, we proved the following “abstract operatorial version.” Let be a positive measure as in Theorem 9, vanishing on the subspace of odd functions from . Let be a Banach lattice, which is order complete. Let , , , . Then, the following theorem holds.
Theorem 11 (see [10, Theorem 2.1]). Let be a linear continuous positive operator vanishing on the subspace of odd functions and .
The following statements are equivalent:(a)there exists a linear operator such that (b)for any finite subset and any , one has
In the sequel, will be the order-complete vector lattice, which is also a commutative Banach algebra of selfadjoint operators [6, 14]. Let be a Hilbert space a self-adjoint operator from into , Let be the space of all continuous functions applying into , such that is majorized by a polynomial on . Denote by the subspace of even functions, and will be the spectrum of .
Theorem 12. Let . The following statements are equivalent:(a)there exists such that(b)for any finite subset and any , one has
Proof. (b)(a). Let
Then, the assumptions of (b) become
Since is a majorizing subspace in , has a positive linear extension
Define Then, is a linear positive extension of to the space . The first two properties of the solution hold. Using Urysohn’s lemma, for any , there is a sequence Applications of the one-dimensional variant of Lemma 5 lead to the existence of a sequence of polynomial functions such that These arguments and the positivity of yield The proof is finished.
The last result of this section concerns the multidimensional moment problems with solutions vanishing on some subspaces. We prove the result only for the bidimensional case. Let If is a Hilbert space and are two bounded positive commuting selfadjoint operators, one defines Then, endowed with the usual order relation on selfadjoint operators is an order-complete vector lattice and a commutative Banach algebra.
Theorem 13. Let be a given sequence. The following statements are equivalent:(a)there exists such that(b)for all finite subsets, for all , and for all , one has
Proof. Since (a)(b) is obvious, we have to prove the implication (b)(a). By Stone-Weierstrass theorem, we know that the vector subspace generated by the functions is a dense subalgebra of . Hence, for any , there exists a sequence of elements from , which converges uniformly to on , such that in . For any function from , we consider its extension to which vanishes outside , and then we extend again this function to , such that , for all . If we write where are extended to such that to vanish outside , and to such that to be even functions, we infer that are even bounded measurable nonnegative functions on with compact support. An application of Luzin’s theorem and of the one-dimensional version of Lemma 5 [10, lemma 1.3(b)] leads to uniform approximation of on , by even positive polynomials on . Using the representation of these polynomials given in Remark 8, (b) is rewritten as In particular, is a linear positive operator on the subspace generated by the products of even polynomials in each variable, which is a majorizing subspace in . This yields the existence of a positive extension of to the whole space . The positivity of and the uniform convergence of to on lead to The last relation (a) follows too. This concludes the proof.
3. Applying the Solutions of the Abstract Moment Problems to Some Classical Moment Problems
The results of this section follow the papers in [7, 8]. In the next theorem, will be the space of all absolutely convergent power series: One assumes that the elements of the space are continuous in the closed disk . Let be a Hilbert space, a selfadjoint operator acting on , and the order-complete vector lattice and commutative Banach algebra of selfadjoint operators defined in Section 2 (see also [6, 14]). We denote that , , .
Theorem 14. Let , , , , be as previously mentioned. Assume that and let , . Let and . The following statements are equivalent:(a)there exists a linear operator applying into such that(b)one has
Proof. (a)(b) is obvious, since from we derive
(b)(a). One applies Theorem 4 to . We have
The preceding relations yield Theorem 4 implies the existence of a linear operator
One obtains This concludes the proof.
In the next theorems, the notion of a resolvent appears naturally during the proofs.
Theorem 15. Let be as previously mentioned. Let , , , . Assume thatThen, there exists a linear positive operator , such that
Proof. The following implications hold Note that in the preceding relations the Cauchy inequalities for the function were used. Application of Theorem 3 leads to the existence of a linear positive operator from into , satisfying the moment conditions, such that on . Now, the conclusion follows easily, by using the monotony of the norm on : The proof is complete.
Similar results are valid for the multidimensional moment problem. Let be the space of analytic functions in the polydisc , . Whence, The order relation on is defined by the positive cone Consider an arbitrary real or complex Hilbert space commuting selfadjoint operators . We introduce the usual spaces:
Theorem 16 (see [8, Theorem 3.1]). With the previous notations, assume that and let . Consider the following statements:(a)there exists an -linear positive continuous operator such that (b)one has (c)one has Then, (b)(a)(c).
Proof. (b)(a). Let be a finite subset and let be such that
Then, the order relation on and Cauchy’s inequalities yield
Using the hypothesis (b) and the previous relations, one obtains
Application of Theorem 3 leads to the existence of an -linear positive operator applying into such that
Using the monotony of the norm on , we deduce
Thus, the implication (b)(a) is proved.
(a)(c) The hypothesis on positivity of the solution and of functions yield The proof is complete.
Applying Theorem 16 to the real Hilbert space , , , being the multiplication operator with the functions , , one obtains the following.
Corollary 17. Let , , .
Consider the following statements: (a)there exists such that (b)one has (c)one has Then,(b)(a)(c).
For some related results see .
Our next goal is to consider some Markov moment problems in terms of nonnegative sequences with respect to an interval. We recall this notion, that is well known.
Definition 18. A sequence in an ordered vector space is said to be nonnegative with respect to the interval if for any , we have It is clear that for compact intervals and for , this condition is necessary and sufficient for the existence of a unique positive solution of the moment problem associated to the moments , . The continuity with respect to the supnorm is also obvious. But no information concerning dominating norm for the solution holds. Such information would be useful for integral representation of the extension of the solution to the space (see the next results).
Theorem 19. Let . Consider the following statements:(a)there exists a unique such that(b)the sequenceThen, (b)(a).
Proof. We have Application of Theorem 4, (b)(a), leads to the existence of a linear functional on such that Using the representation of a linear positive functional on , there exists From the last equality written for , being a Borel subset, we infer (via measure theory) that The proof is complete.
Lemma 20. Let be a positive function in and a continuous convex function on . Then, one has
Corollary 21. Let be a sequence of real numbers such that the sequence is nonnegative with respect to and . Then,
4. Extreme Points and the Markov Moment Problem
The aim of this section is to apply some ideas from [20, 26, 27] in solving truncated moment problems and their connection to the full moment problem. The results of this section are new. The following theorem concerns the finite Markov moment problem considered in , also using some results from . Let us denote that
Theorem 22. For a given family of numbers , consider the following statements: (a), (b), and (c):(a)there exists such that(b)for any family of scalars , one has:(c)there exists a Borel subset such that Then,(b)(a)(c)
Proof. Let the point be accomplished and assume that
Then the integration on yields
Application of Theorem 4 to defined previously, leads to the existence of a linear positive form on such that
If characteristic functions of Borel subsets stand for , the conclusion (a) follows by measure theory . The implication (a)(c) is a consequence of equality (15.14) in  (see also [27, Exercise 2.57]). The set of values for the control function , namely, , is replaced by , the set of values for , that stands for the control function . The extreme points of the positive part of the unit ball of are the characteristic functions of measurable sets. The converse is obvious.
Corollary 23. Under the equivalent conditions (a), (c) of Theorem 22, there are sequences such that the following relations hold:
Proof. One uses the fact that any Borel subset is a joint of a set and a set of measure zero .
Remark 24. In order to approximate the numbers , one can make use of Fourier approximate expansion of with respect to the orthonormal sequence attached to the functions via Gram-Schmidt algorithm, also using the values of the moments . Thus, one obtains a smooth approximation of , and the intervals of ends are connected components of the open sets of the form
Remark 25. A similar result to that of Theorem 22 in several dimensions holds, with the same proof. We state it for the two-dimensional case.
Theorem 26. Let be a given family of real numbers and consider the functions Consider the following assertions: (a), (b), (c). Then (b)(a)(c)(a)there exists a Borel function a.e. such that