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Article

On the Truncated Multidimensional Moment Problems in Cn

by
Sergey Zagorodnyuk
School of Mathematics and Computer Sciences, V. N. Karazin Kharkiv National University, Svobody Sq., 4, 61022 Kharkiv, Ukraine
Axioms 2022, 11(1), 20; https://doi.org/10.3390/axioms11010020
Submission received: 26 November 2021 / Revised: 30 December 2021 / Accepted: 4 January 2022 / Published: 5 January 2022
(This article belongs to the Special Issue Orthogonal Polynomials, Special Functions and Applications)

Abstract

:
We consider the problem of finding a (non-negative) measure μ on B ( C n ) such that C n z k d μ ( z ) = s k , k K . Here, K is an arbitrary finite subset of Z + n , which contains ( 0 , , 0 ) , and s k are prescribed complex numbers (we use the usual notations for multi-indices). There are two possible interpretations of this problem. Firstly, one may consider this problem as an extension of the truncated multidimensional moment problem on R n , where the support of the measure μ is allowed to lie in C n . Secondly, the moment problem is a particular case of the truncated moment problem in C n , with special truncations. We give simple conditions for the solvability of the above moment problem. As a corollary, we have an integral representation with a non-negative measure for linear functionals on some linear subspaces of polynomials.

1. Introduction

Throughout the whole paper, n means a fixed positive integer. Let us introduce some notations. As usual, we denote by R , C , N , Z , Z + the sets of real numbers, complex numbers, positive integers, integers and non-negative integers, respectively. By Z + n we mean Z + × × Z + , and R n = R × × R , C n = C × × C , where the Cartesian products are taken with n copies. Let k = ( k 1 , , k n ) Z + n , z = ( z 1 , , z n ) C n . Then, z k means the monomial z 1 k 1 z n k n , and | k | = k 1 + + k n . By B ( M ) , we denote the set of all Borel subsets of a set M C n .
Let K be an arbitrary finite subset of Z + n , which contains 0 : = ( 0 , , 0 ) . Let S = ( s k ) k K be an arbitrary set of complex numbers. We shall consider the problem of finding a (non-negative) measure μ on B ( C n ) such that
C n z k d μ ( z ) = s k , k K .
There are two possible interpretations of this problem. Firstly, one may consider this problem as an extension of the truncated multidimensional moment problem on R n , where the support of the measure μ is allowed to lie in C n . A similar situation is known in the cases of the classical Stieltjes and Hamburger moment problems, where the support of the measure lies in [ 0 , + ) and in R , respectively. Secondly, and more directly, the moment problem (1) is a particular case of the truncated moment problem in C n (see ([1], Chapter 7), [2,3]), with special truncations. These truncations do not include conjugate terms.
It is well-known that the multidimensional moment problems are much more complicated than their one-dimensional prototypes [1,4,5,6,7,8]. An operator-theoretical interpretation of the full multidimensional moment problem was given by Fuglede in [9]. Important ideas in the operator approach to moment problems go back to the works of Naimark in 1940–1943, and then, they were developed by many authors, see historical notes in [10]. In [11], we presented the operator approach to the truncated multidimensional moment problem in R n . Other approaches to truncated moment problems can be found in [1,2,3,6,12,13]. A detailed exposition of the theory of (full and truncated) multidimensional moment problems is given in a recent Schmüdgen’s book [8]. Recent results can be also found in [14,15].
In the case of the moment problem (1), we shall need a modification of the operator approach, since we have no positive definite kernels here. However, this problem can be passed and we shall come to some commuting bounded operators. At first, we shall provide an auxiliary commuting extension for this tuple inside the original space. After this, we shall not use extensions of operators, but dilations. We shall apply the dilation theory for commuting contractions to obtain the required measure. Consequently and surprisingly, we have very simple conditions for the solvability of the moment problem (1) (Theorem 1). As a corollary, we have an integral representation with a non-negative measure for linear functionals L on some linear subspaces of polynomials (Corollary 1).
Notations. Besides the given notations above, we shall use the following conventions. If H is a Hilbert space, then ( · , · ) H and · H mean the scalar product and the norm in H, respectively. Indices may be omitted in obvious cases. For a linear operator A in H, we denote by D ( A ) its domain, by R ( A ) its range, and A * means the adjoint operator if it exists. If A is invertible, then A 1 means its inverse. A ¯ means the closure of the operator, if the operator is closable. If A is bounded, then A denotes its norm. For a set M H , we denote by M ¯ the closure of M in the norm of H. By Lin M , we mean the set of all linear combinations of elements from M, and span M : = Lin M ¯ . By E H , we denote the identity operator in H, i.e., E H x = x , x H . In obvious cases, we may omit the index H. If H 1 is a subspace of H, then P H 1 = P H 1 H denotes the orthogonal projection of H onto H 1 .

2. Truncated Moment Problems on C n

A solution to the moment problem (1) is given by the following theorem.
Theorem 1.
Let the moment problem (1) with some prescribed S = ( s k ) k K be given. The moment problem (1) has a solution if and only if one of the following conditions holds:
(a) 
s ( 0 , , 0 ) > 0 ;
(b) 
s k = 0 , k K .
If one of conditions ( a ) , ( b ) is satisfied, then there exists a solution μ with a compact support.
Proof. 
The necessity part of the theorem is obvious. Let moment problem (1) be given and one of conditions ( a ) , ( b ) holds. If ( b ) holds, then μ 0 is a solution of the moment problem. Suppose in what follows that s ( 0 , , 0 ) > 0 . Observe that we can include the set K into the following set:
K d : = { k = ( k 1 , , k n ) Z + n : k j d , j = 1 , 2 , , n } ,
for some large d 1 . Namely, d may be chosen greater than the maximum value of all possible indices k j in K . We now set s k : = 0 , for k K d \ K . Consider another moment problem of type (1), having a new set of indices K ˜ = K d . We are going to construct a solution to this moment problem, which, of course, will be a solution to the original problem.
Consider the usual Hilbert space l 2 of square summable complex sequences c = ( c 0 , c 1 , c 2 , ) , c l 2 2 = j = 0 | c j | 2 . We intend to construct a sequence { x k } k K ˜ of elements of l 2 , such that
( x k , x 0 ) l 2 = s k , k K ˜ .
The elements of the finite set K ˜ can be indexed by a single index, i.e., we assume
K ˜ = k 0 , k 1 , , k ρ ,
with ρ + 1 = | K ˜ | , and k 0 = ( 0 , , 0 ) . Denote a : = s ( 0 , , 0 ) ( > 0 ) . Set
x 0 : = a e 0 , x k j : = e j + s k j a e 0 , j = 1 , 2 , , ρ .
Here, e j means the vector c = ( c 0 , c 1 , c 2 , ) from l 2 , with c j = 1 , and 0’s in other places. Observe that for this choice of elements x k , conditions (2) hold true. Moreover, it is important for our future purposes that these elements x k are linearly independent.
Consider a finite-dimensional Hilbert space H : = Lin { x k } k K ˜ . Set
K d ; l : = { k = ( k 1 , , k n ) K d : k l d 1 } , l = 1 , 2 , , n .
Consider the following operator W j on Z + n :
W j ( k 1 , , k j 1 , k j , k j + 1 , , k n ) = ( k 1 , , k j 1 , k j + 1 , k j + 1 , , k n ) ,
for j = 1 , , n . Thus, the operator W j increases the j-th coordinate. We introduce the following operators M j , j = 1 , , n , in H:
M j k K d ; j α k x k = k K d ; j α k x W j k , α k C ,
with D ( M j ) = Lin { x k } k K d ; j . Since elements x k are linearly independent, we conclude that M j are well-defined operators. Operators M j can be extended to a commuting tuple of bounded operators on H. In fact, consider the following operators A j M j , j = 1 , , n :
A j k K d α k x k = k K d ; j α k x W j k , α k C .
Operators A j are well-defined linear operators on the whole H. It can be directly verified that they commute pairwisely. Notice that
A 1 k 1 A 2 k 2 A n k n x 0 = x ( k 1 , k 2 , , k n ) , k = ( k 1 , , k n ) K d .
Relation (8) can be verified using the induction argument. Since H is finite-dimensional, then
A j R , j = 1 , 2 , , n ;
for some R > 0 . Set
B j : = 1 C A j , j = 1 , , n ,
where C is an arbitrary number greater than n R . Then,
j = 1 n B j 2 < 1 .
In this case, there exists a commuting unitary dilation U = ( U 1 , , U n ) of ( B 1 , , B n ) , in a Hilbert space H ˜ H , see Proposition 9.2 in [16] (p. 37). Namely, we have:
P H H ˜ U 1 k 1 U 2 k 2 U n k n H = B 1 k 1 B 2 k 2 B n k n , k 1 , , k n Z + .
Moreover, we can choose U to be minimal, that is, the subspaces U 1 k 1 U n k n H will span the space H ˜ (see Theorem 9.1 in [16] (p. 36)):
H ˜ = span U 1 k 1 U n k n H , k 1 , , k n Z .
Then, the Hilbert space H ˜ will be separable. By (2), (8), (9) and (11), we may write for an arbitrary k = ( k 1 , , k n ) K ˜ :
s k = ( x k , x 0 ) l 2 = ( A 1 k 1 A 2 k 2 A n k n x 0 , x 0 ) l 2 = C | k | ( B 1 k 1 B 2 k 2 B n k n x 0 , x 0 ) l 2 =
= C | k | ( U 1 k 1 U 2 k 2 U n k n x 0 , x 0 ) l 2 = ( ( C U 1 ) k 1 ( C U 2 ) k 2 ( C U n ) k n x 0 , x 0 ) l 2 =
= ( N 1 k 1 N 2 k 2 N n k n x 0 , x 0 ) l 2 ,
where N j : = C U j , j = 1 , , n . Applying the spectral theorem for commuting bounded normal operators N j (or, equivalently, to their commuting real and imaginary parts), we obtain that
N j = C n z j d F ( z 1 , , z n ) , j = 1 , , n ,
where F ( z 1 , , z n ) is some spectral measure on B ( C n ) . Then,
s k = C n z 1 k 1 z n k n d ( F ( z 1 , , z n ) x 0 , x 0 ) l 2 , k = ( k 1 , , k n ) K ˜ .
This means that μ = ( F ( z 1 , , z n ) x 0 , x 0 ) l 2 is a solution of the moment problem. Since N j were bounded, μ has compact support. □
Corollary 1.
Let K be an arbitrary finite subset of Z + n , which contains 0 . Let L be a complex-valued linear functional on
M = M ( K ) : = Lin { z 1 k 1 z n k n } k = ( k 1 , , k n ) K ,
such that L ( 1 ) > 0 . Then, L has the following integral representation:
L ( p ) = C n p ( z 1 , , z n ) d μ , p M ,
where μ is a (non-negative) measure μ on B ( C n ) , having compact support.
Proof. 
It follows directly from Theorem 1. □

3. Conclusions

Let K be an arbitrary finite subset of Z + n , and S = ( s k ) k K an arbitrary set of real numbers. Recall that the truncated multidimensional moment problem consists of finding a (non-negative) measure μ on B ( R n ) such that [17]
t k d μ ( t ) = s k , k K .
Assume additionally that K contains 0 : = ( 0 , , 0 ) . Theorem 1 shows that if we extend the admissible support for the sought-for measure to lie in C n , then this helps essentially. Such a “complexification” of the problem leads to simple conditions of the solvability.
Corollary 1 can be compared with a well-known theorem of Boas, which gives a representation for functionals (see [18] (p. 74)). In our case, we have a non-negative measure in the representation. This provides a Hilbert space structure and corresponding tools for further investigations.
It is of interest to consider similar problems with infinite truncations and full moment problems. This will be studied elsewhere.

Funding

This research received no external funding.

Conflicts of Interest

The authors declare no conflict of interest.

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Zagorodnyuk, S. On the Truncated Multidimensional Moment Problems in Cn. Axioms 2022, 11, 20. https://doi.org/10.3390/axioms11010020

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Zagorodnyuk S. On the Truncated Multidimensional Moment Problems in Cn. Axioms. 2022; 11(1):20. https://doi.org/10.3390/axioms11010020

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Zagorodnyuk, Sergey. 2022. "On the Truncated Multidimensional Moment Problems in Cn" Axioms 11, no. 1: 20. https://doi.org/10.3390/axioms11010020

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