3.1. Markov Moment Problems on Compact Subsets
The following results are reviewed in the chronological order of the dates of their publication. A few completions are added, stated, and eventually proved as corollaries. We start with a short presentation of some of the results published firstly in [
14]. Let
be a compact subset of
with a non-empty interior. The open set
can be written as a union of sets of the form
, hence
where
are polynomials of degrees smaller or equal to two. One denotes by
the vector space generated by the polynomials of degree at most one and the polynomials
The space
is clearly a vector subspace of the space
of all polynomials of degrees at most two. We denote by
the convex cone of all polynomials in
, which takes non-negative values at all points of
, and one denotes by
the set of those
that generates an extreme ray of
An important subset is
If is convex, then we can take as the space of all polynomials of degree at most one. With and , it is easy to see that consists in only two elements: the polynomials and From this, we infer that the set of polynomials which appear naturally in the classical Hausdorff moment problem, should be replaced, in the general case, by the set of polynomials, which are finite products of elements of Since has only a multiplicative structure (the sum of two elements of this set is not, in general, an element of ), one introduces the convex cone of all linear combinations with nonnegative coefficients of elements of The next result 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 and the special polynomials that are elements of Therefore, we say that the next theorem solves the moment problem. From the point of view of the next theorem, it represents the multidimensional case of the Hausdorff moment problem (the moment problem on ). On the other hand, the next theorem works for non-convex compact subsets having non-empty interiors as well. We say that a sequence of real numbers is a moment sequence on the closed subset if there exists a positive regular Borel measure on such that
Theorem 1. (See [
14]).
Let be a compact subset of with a non-empty interior. A necessary and sufficient condition for a sequence being a moment sequence on is that the linear form defined on by
satisfies the condition for all polynomial .
The next result gives the expression of any polynomial that is positive at each point of , by means of some polynomials that are elements of Since any such polynomial is a linear combination with nonnegative coefficients of elements of (see below), the next result is called the decomposition theorem.
Theorem 2. (See [
14]).
Each polynomial that has positive values at all points of a compact subset with a non-empty interior in is a linear combination with nonnegative coefficients of elements of Here is a vector valued version of a solution of a Markov moment problem similar to the solution for the Hausdorff moment problem. In this case, we denote
Next, we pass to the vector valued Markov moment problem for compact subsets with nonempty interior.
Theorem 3. Letbe an arbitrary order complete Banach lattice,a sequence intwo linear operators fromintosuch thatonThe following statements are equivalent:
- (a)
There exists a unique bounded linear operatorapplyingintosuch that - (b)
The following inequalities hold:whereis defined by (3).
Proof. Sine all the polynomials in
are elements of
the implication (a)
(b) is obvious. Here
is the unique linear operator from
into
satisfying the moment conditions
for all
To prove the implication (b)
(a), we need Theorem 2. Let us denote by
the set of all polynomials
which take only positive values at each point of
If
then clearly
for all
Conversely, if
then
where
On the other hand, according to Theorem 2, any polynomial
from
is a linear combination with nonnegative coefficients of polynomials from
Thus, inequalities (8) lead to
Passing to the limit as
we find that:
Hence
is a positive linear operator from
into
According to Kantorovich extension theorem [
12], there exists a positive linear extension
of
, because
is a majorizing subspace in
The operators
are continuous, since they are positive, acting between Banach lattices ([
6] or [
29]). If
there exists a sequence
for all
in
Using (9) and the continuity of the involved operators, this leads to:
Thus,
on To prove the last inequalities in the statement, observe that
The inequality
follows by means of the same type of reasons. Finally, the operator
clearly verifies the interpolation moment conditions stated at point (a), since
This concludes the proof. □
Now the Hausdorff Markov vector valued moment problem follows consequently.
Theorem 4. Letbe an arbitrary order complete Banach lattice,a sequence intwo linear operators fromintosuch thaton. The following statements are equivalent:
- (a)
There exists a unique bounded linear operatorapplyingintosuch that - (b)
For any
the following inequalities hold:
Proof. We start by observing that
for all
hence
for all
. Thus, point (b) says that
This assertion is equivalent to that form (b) of Theorem 3, since we have
The conclusion follows. □
The next result is an application of theorem 4 to the case when the order complete Banach lattice is a space of self-adjoint operators acting on an arbitrary real or complex Hilbert space
We recall that a linear operator
(where
is a vector subspace of
), is called self-adjoint if it is bounded (continuous) and symmetric, that is
for all
According to this definition, any self-adjoint operator is symmetric. When
the converse is true, since any symmetric operator from
into itself is bounded, according to the closed graph theorem. A self-adjoint operator
acting on
is called positive if
for all
The real vector space of all self-adjoint operators acting on
will be denoted by
If
, the following equalities for the operatorial norm of
holds:
For any
, we define
In [
5], it was proved that
is an order complete Banach lattice (which is also a commutative real Banach algebra). In the case when
there exists a natural isomorphism between the ordered Banach space
and the real vector space of all symmetric
matrices with real entries. If
are two such matrices, then, according to the above definitions,
if and only if
is positive semi-definite. For an arbitrary Hilbert space
with
the ordered Banach space
is not a lattice. If
is the spectrum of
and
we denote by
the element obtained by means of functional calculus. Namely,
where
is the spectral measure attached to
Corollary 1. Letbe a Hilbert space,with the spectrumletbe a positive integer andbe defined by (10). Letbe a sequence of elements inThe following statements are equivalent:
- (a)
There exists a unique bounded linear operatorapplyingintosuch that - (b)
For anythe following inequalities hold:
In the sequel, moment problems on semi-algebraic compact subsets are reviewed. If
is a sequence of real numbers, one denotes by
the linear functional defined on
by
where
is a finite subset and
are arbitrary real coefficients. If
is a finite subset of
, then the closed subset given by
is called a semi-algebraic set.
Theorem 5. (See [
4,
15]).
If is a compact semi-algebraic set, as defined above, then there is a positive Borel measure supported on , such thatif, and only if: Corollary 2. (See [
15]).
With the above notations, if is such that for all in the semi-algebraic compact defined by Equation (11), then is a finite sum of special polynomials of the formfor some and 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. For other results on this subject, as well as for Markov moment problems not reviewed in the present work see [
3,
4,
17,
18,
19,
20,
33]. In [
19], moment problems on unbounded subsets are also discussed.
Lemma 1 (See [
36])
. Let be a continuous function, such that exists in Then, there is a decreasing sequence in , where the functions are defined as follows:such that uniformly on . There exists a sequence of polynomial functions ,
, uniformly on compact subsets of . In particular, such polynomial approximation holds for any continuous compactly supported function .
If is an arbitrary compact subset, then for any function , one denotes by the extension of , which satisfies for all .
Lemma 2 (See [
36] and the references therein)
. If is a compact subset, and a continuous function, then there exists a sequence of polynomial functions, such that on , uniformly on Corollary 3. Letbe a compact subset,a continuous function, a Banach lattice. Assume thatare positive linear operators fromintoThe following statements are equivalent:
- (a)
The following sandwich conditions hold:on
- (b)
For any finite subset, and anythe following inequalities hold:
Proof. The implication
is obvious since any polynomial
is nonnegative on
. For the converse, we know form Lemma 2 that there exists a sequence
of nonnegative polynomials on
, uniformly on
. Condition (b) implies
for all
because each nonnegative polynomial on
has the analytic form given in (12). Now the first part of assertion (a) follows from the continuity of the three involved positive operators
. Namely, we have:
The inequalities have been established in the proof of Theorem 3. This concludes the proof. □
Corollary 4. Letbe a Hilbert space, anda self-adjoint operator acting onwith the spectrumLetbe the Banach lattice defined by (10) and let denote. The following statements are equivalent.
- (a)
- (b)
For any finite subset, and anythe following inequalities hold:
Proof. One also uses the elementary inequalities
It follows that hence since the norm is monotone increasing on the positive cone of This concludes the proof. □
Example 1. We consider the real Hilbert spacethe operator defined by. Thenis a positive (linear) symmetric operator, We go on with a general Hahn-Banach type result, namely Mazur-Orlicz theorem, followed by an application.
Theorem 6. (See [
10,
16,
36]).
Let be a preordered vector space, an order complete vector space, a sublinear operator, and given families. The following statements are equivalent:- (a)
There exists a linear positive operator, such that - (b)
For any finite subset, we have
Let be a Hilbert space, a positive self-adjoint operator acting on the spectrum of the spectral measure attached to Let be the order complete Banach lattice defined by (10), and a sequence of elements in As usual, we denote
Theorem 7. (See [
10] and the references therein).
The following statements are equivalent.- (a)
There exists a positive linear operatorfrominto, such that - (b)
for all
Proof. Since the self-adjoint operator
is positive, its spectrum
is contained in
The implication
is almost obvious. Namely, if (a) holds, then:
To prove
we use the corresponding implication of Theorem 6, where
stands for
stands for
stands for
stands for
and
defines a sublinear continuous operator, which is symmetric, that is
. Since
from (b) of Theorem 6 are non-negative, and
from the assumption (b), for any finite subset
we infer that
On the other hand,
in
means
for all
. This yield:
According to Theorem 6, there exists a positive linear operator
from int
with
Consider the condition
, which is equivalent to
It results:
This implies
and, since
is a Banach lattice, the conclusion is
Thus, . This concludes the proof. □
3.2. Markov Moment Problems on Unbounded Subsets
As it is well known, unlike the case the form of nonnegative polynomials on and on in terms of sums of squares is not known. This is the motivation for the next polynomial approximation results, which solve this problem for the Markov moment problem in several dimensions. The general idea is to approximate an arbitrary continuous compactly supported nonnegative function on unbounded closed subset of by nonnegative polynomials on If or , the approximating polynomials are sums of polynomials defined by (1), hence they are particular cases of sums of squares of polynomials in real variables. The approximation holds in spaces, where is a moment determinate measure. From now on, by a determinate measure we mean a positive regular Borel moment determinate measure, with finite moments of all orders.
Lemma 3 (See [
34])
. Let be an unbounded closed subset, and a determinate measure on (with finite moments of all natural orders). Then, for any there exists a sequence in . We have: is dense in , and is dense in Proof. To prove the assertions of the statement, it is sufficient to show that for any
, we have
To prove the converse, we define the linear form
Next, we show that
is positive on
. In fact, for
, one has (from the definition of
, which is a sublinear functional on
The following implications hold.
If
, we infer that:
where, in both possible cases, we have
. Since
contains the space of the polynomials’ functions, which is a majorizing subspace of
, there exists a linear positive extension
of
which is continuous on
with respect to the sup-norm. Therefore,
has a representation by means of a positive Borel regular measure
on
, such that
Let
be a non-negative polynomial function. There is a nondecreasing sequence
of continuous non-negative function with compact support, such that
pointwise on
. Positivity of
and Lebesgue’s dominated convergence theorem for
yield
Thanks to Haviland’s theorem, there exists a positive Borel regular measure
on
, such that
Since
is assumed to be
M-determinate, it follows that
for any Borel subset
of
. From this last assertion, approximating each
, by a nondecreasing sequence of non-negative simple functions, and using Lebesgue’s convergence theorem, one obtains firstly for positive functions, then for arbitrary
-integrable functions,
In particular, we must have
The conclusion follows. □
Remark 1. We recall that the preceding Lemma 3 is no more valid when we replacewith the Hilbert space(see [23], where the authors construct such a measure). Lemma 4 (See [
37])
. Let be a product of n determinate measures on . Then we can approximate any nonnegative continuous compactly supported function with sums of products nonnegative polynomial on the entire real line, , and any such sum of special polynomials dominates on Lemma 5 (See [
37])
. Let be a product of n determinate measures on Then, we can approximate any nonnegative continuous compactly supported function with sums of products nonnegative polynomial on the entire nonnegative semi axes, , and any such sum of special polynomials dominates on Example 2. (i) The measureis moment determinate onIndeed, forthe measureis moment determinate onaccording to [24]. Forthe product of determinate measures onis a determinate measureonsince the polynomials are dense inaccording to Lemma 4 and measure theory results. (ii) For anythe measureis moment determinate on, because of similar reasons to those mentioned at point (i), accompanied by Lemma 5. Application of Lemma 3 and other appropriate results lead to the following theorem, which holds on arbitrary unbounded (or bounded) closed subsets.
Theorem 8. Letbe a closed unbounded subset ofan order complete Banach lattice,a given sequence ina determinate measure onLettwo linear bounded operators fromintoThe following statements are equivalent:
- (a)
there exists a unique bounded linear operatorsuch thatis betweenandon the positive cone of
- (b)
For any finite subsetand anywe have
In the cases when the analytic form of nonnegative polynomials in terms of sums of squares is known, Theorem 8 solves the existence and uniqueness of the solution for the full Markov moment problem in terms of quadratic forms. Here are a few such examples, formulated as consequences of Theorem 8. The simplest cases are and , regarded as closed subsets of
Corollary 5. Letwhereis a determinate measure onAssume thatis an arbitrary order complete Banach lattice, andis a given sequence with its terms inLetbe two linear operators fromto, such thatonThe following statements are equivalent:
- (a)
There exists a unique bounded linear operatorfromtoon, such thatfor all
- (b)
Ifis a finite subset, andthen
Corollary 6. Letwhereis a moment-determinate measure onAssume thatis an arbitrary order complete Banach lattice, andis a given sequence with its terms inLetbe two linear operators fromto, such thaton. As usual, we denote. The following statements are equivalent:
- (a)
There exists a unique bounded linear operatorfromtoon, such thatfor all
- (b)
Ifis a finite subset, andthen
Example 3. Letanda sequence of real numbers. The following statements are equivalent:
- (a)
There existsfor almost all, such that - (b)
Ifis a finite subset, and, then
To obtain the equivalence from Example 3, one applies Corollary 6 to
also using measure theory arguments [
9] and properties of the Gamma function.
For the expression of nonnegative polynomials on or on in terms of sums of squares is not known. A way of avoiding this difficulty is to apply approximation Lemmas 4 and 5, supplied by the following general theorem.
Theorem 9 (See [
34])
. Let be a closed unbounded subset; a moment-determinate measure on having finite moments of all orders; and Let be an order complete Banach lattice, a given sequence of elements in and two bounded linear operators from to Assume that there exists a sub-cone such that each can be approximated in by a sequence for all . The following statements are equivalent:- (a)
There exists a unique (bounded) linear operatoron
- (b)
For any finite subsetand anythe following implications hold true:
When the convex cone is the set of all finite sums of polynomials defined by (1), where each is nonnegative on the entire real axis. Thus for some From Theorem 9 and Lemma 4 we obtain:
Theorem 10. Letbeing a determinate measure onLetbe an order complete Banach lattice,a given sequence of elements inandtwo bounded linear operators mappingintoThe following statements are equivalent:
- (a)
There exists a unique (bounded) linear operator - (b)
For any finite subsetand any, the following implication holds true:For any finite subsetsand any, the following inequalities hold true:
If then will be the convex cone generated by the polynomials defined by (1), where each is nonnegative on hence for some .