# New Results on Markov Moment Problem

Authors: Octav Olteanu
Journal: International Journal of Analysis
 (2013)

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## Abstract

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.


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## Body

## 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 [1] for details), finds the repartition of the positive mass on the semiaxis[0,∞),  if the moments of arbitrary ordersk(k=0,1,2,…)   are given. Precisely, in the Stieltjes moment problem, a sequence of real numbers(sk)k≥0is given, and one looks for a nondecreasing real functionσ(t) (t≥0)     that verifies the moment conditions:
(1)∫0∞tkdσ=sk(k=0,1,2,…).
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   Rn,  n≥2, 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 [15]). It says that iffis a linear positive operator on a subspaceSof the ordered vector spaceX,the target space being an order-complete vector latticeY,    and for eachx∈X, there iss∈S,x≤s,thenfhas a linear positive extensionF:X→Y. 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 A with some properties with respect to the subspace S (e.g.,  S∩ ri(A)≠Φ), where ri(A) is the relative interior of A. 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 [16], without loosing convexity, but strongly generalizing the classical result. The first detailed proof was published in [17]. A proof in terms of the moment problem was published in [12]. Parts of these generalizations of the Hahn-Banach principle are involved in the present work too. Throughout this first part,  X will be a real vector space,  Y an order-complete vector lattice, A,B⊂X a convex subsets, q:A→Y a concave operator, p:B→Y  a convex operator,   S⊂X  a vector subspace,  f:S→Y a linear operator.Theorem 1.
Assume that
f
∣
S
∩
A
≥
q
∣
S
∩
A
,
f
∣
S
∩
B
≤
p
∣
S
∩
B.
The following assertions are equivalent:
(a)
there is a linear extensionF:X→Y of the operator f such that
(2)F∣A≥q,F∣B≤p;
(b)
there arep1:A→Y convex and  q1:B→Y concave operators such that for all
(3)(ρ,t,λ′,a1,a′,b1,b′,v)∈[0,1]2×(0,∞)×A2×B2×S,
One has:
(4)
(
1
-
t
)
a
1
-
t
b
1
=
v
+
λ
′
[
(
1
-
ρ
)
a
′
-
ρ
b
′
]
⇒
(
1
-
t
)
p
1
(
a
1
)
-
t
q
1
(
b
1
)
≥
f
(
v
)
+
λ
′
[
(
1
-
ρ
)
q
(
a
′
)
-
ρ
p
(
b
′
)
]
.The following result related to the theorem of  H. Bauer (see [15, Section 5.4]) is deduced.Theorem 2.
Let
X
be a preordered vector space with its positive cone
X
+
,
p
:
X
→
Y
a convex operator,
S
⊂
X
a vector subspace, and
f
:
S
→
Y
a linear positive operator. The following assertions are equivalent:
(a)
there is a linear positive extensionF:X→Y  of  f such that
(5)F(x)≤p(x)∀x∈X;
(b)
f
(
x
′
)
≤
p
(
x
) for all  (x′,x)∈S×X  such that  x′≤x.Now, we can deduce the main results on the abstract moment problem [11].Theorem 3.
Let
X
,
Y, p:X→Ybe as in Theorem2,{xj}j∈J⊂X,  {yj}j∈J⊂Ygiven families. The following statements are equivalent:(a)
there is a linear positive operatorF:X→Y such that
(6)F(xj)=yj∀j∈J,F(x)≤p(x)∀x∈X;(b)
for any finite subsetJ0⊂J  and any  {λj}j∈J0⊂R, one has
(7)∑j∈J0λjxj≤x⇒∑j∈J0λjyj≤p(x).
A clearer sandwich-moment problem variant is the following one.Theorem 4.
Let
X
,
Y
,
{
x
j
}
j
∈
J
,
{
y
j
}
j
∈
J
be as in Theorem3 and
F
1,
F2∈L(X,Y)two linear operators. The following statements are equivalent:(a)
there is a linear operatorF∈L(X,Y) such that
(8)F1(x)≤F(x)≤F2(x)∀x∈X+,F(xj)=yj∀j∈J;(b)
for any finite subsetJ0⊂J and any {λj}j∈J0⊂R,  one has:
(9)(∑j∈J0λjxj=φ2-φ1,φ1,φ2∈X+)⇒∑j∈J0λjyj≤F2(φ2)-F1(φ1).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 (Section4).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 [26], 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

The results of this section are using those published in [9, 10]. We recall the following recent approximation lemmas concerning the approximation on closed unbounded finite-dimensional subsets.Lemma 5 (see [9] and [10, Lemma 1.3(d)]).
Ifx∈C0([0,∞)×[0,∞))  is a nonnegative continuous function with compact support, then there exists a sequence (pm)m of positive polynomials on  [0,∞)×[0,∞), such that
(10)pm(t)>x(t)∀t≥0,∀m∈Z+,pm→x
uniformly on compact subsets of  [0,∞)×[0,∞).The idea of the proof is to add the∞ point and to apply the Stone-Weierstrass theorem to the subalgebra generated by the functions  exp(-mt1-nt2),   m,n∈Z+. Then, one uses for each such exp-function suitable majorizing or minorizing partial polynomial sums, as well as the relation
(11)exp(s)-(1+s1!+s22!+⋯+smm!)=exp(s)m!∫0sexp(-t)·tmdt,m∈Z+.
Note that [10, Lemma 1.4] asserts the density of positive polynomials in (Lν1(A))+, for any closed subset A of a finite-dimensional space, ν, being a positive regular Borel M-determinate measure. Here is the exact formulation of this result.Lemma 6 (see [10, Lemma 1.4]).
LetA⊂Rn be an arbitrary closed subset and ν an M-determinate positive regular Borel measure onA,with finite moments of all natural orders. Then, for anyx∈(C0(A))+such thatx≤p∈Pfor some polynomialp,there exists a sequence(pm)m,pm∈P+,pm≥x,pm→xinLν1(A).In particular, one has:(12)lim∫Apm(t)dν=∫Ax(t)dν,P+is dense in(Lν1(A))+,andPis dense inLν1(A),Recall thatP is the subspace of all polynomial functions on A.Lemma 7.
For any simple function of the form(13)s=∑j=1NαjχIj,αj≥0j=1,…,N,Ij⊂[0,∞)j=1,…,N,
being intervals, there exists a sequence of polynomial
(14)(pm)m,pm→sinLν1([0,∞)),pm≥s∀m,
where ν is an M-determinate positive regular Borel measure with finite moments of all natural orders. If s is extended to an even function in R, then the polynomials pm,m∈Z+ are restrictions to the positive semiaxes of even polynomials in R.Note that in the preceding Lemmas5–7, the sequence (pm)m 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(15)p2(t)=p2,12(t2)+t2p2,22(t2),t∈R,p2,j∈R[t],j=1,2.
We start by proving some results on the one-dimensional moment problem.Theorem 9 (see [10, Theorem 2.1]).
Letν be a positive determinate Borel regular measure on R with finite moments m2k=∫Rt2kdν(t2) of all orders, k∈Z+ such that ∫Rx1(t)dν=0 for any odd function x1∈Lν1(R).  Let (y2k)k∈Z+ be a given sequence of real numbers.
The following statements are equivalent:
(a)
there exists a Borel functionh such that
(16)∫Rt2kh(t2)dν(t2)=y2k,∫Rt2k+1h(t2)dν=0,∀k∈Z+,0≤h(t2)≤1dν-a.e.inR;
(b)
for any finite subsetJ0⊂Z+ and any {λj;j∈J0}⊂R, one has
(17)0≤∑i,j∈J0λiλjy2i+2j≤∑i,j∈J0λiλj∫Rt2i+2jdν,0≤∑i,j∈J0λiλjy2i+2j+2≤∑i,j∈J0λiλj∫Rt2i+2j+2dν.Proof.
The hypothesis  (a) implies that (b) is obvious because of the qualities ofh.
(b)⇒(a). Since ν vanishes on the subspace of odd functions, the same property is valid for h(t2)dν. The problem can be reduced to a corresponding one on the positive semiaxis u=t2≥0. Using the form of the positive polynomials in u∈R+ [1], the hypothesis (b) yields(18)f(∑j∈J0λjx~j):=∑j∈J0λjy2j2,x~j(u)=uj,j∈Z+⇒0≤f(p)≤∫0∞p(u)dν∀p∈P+.
Since P is a majorizing subspace in
(19)X:={x∈Lν1([0,∞));∃p∈P,|x(u)|≤p(u),∀u≥0},
there exists a positive linear extension F∈X* of f. By Haviland-Riesz representation theorem, there exists a representing positive regular Borel measure  μ  on  [0,∞)  such that
(20)μ(p)=F(p)=f(p)≤ν(p),p∈P+.
Applying Haviland-Riesz theorem once more, we infer the existence of a positive regular Borel measure λ on  [0,∞), such that
(21)λ(p)=(ν-μ)(p)⇒ν(p)=λ(p)+μ(p),p∈P.
Since  ν is M-determinate, application of Lemma 7 and the last relation lead to
(22)ν(B)=λ(B)+μ(B),∀B⊂[0,∞),
Borel subset. In particular, μ is absolutely continuous with respect to ν, hence there exists h1∈Lν1([0,∞)) such that dμ=h1dν on simple functions. From [28, Theorem 1.40], we infer that  0≤h1≤1ν-a.e. Since ν was a positive regular Borel measure, we can find a Borel function h such that  h=h1ν-a.e. The previous relations lead to
(23)∫0∞ujh(u)dν=∫0∞ujdμ=F(x~j)=y2j2⇒∫-∞∞t2jh(t2)dν(t2)=2y2j2=y2j
This concludes the proof.Corollary 10.
Letα>0 and {y2k}k∈Z+⊂R. The following statements are equivalent:(a)
there exists a Borel functionh, such that
(24)∫Rt2kh(t2)dt=y2k,∫Rt2k+1h(t2)dt=0,∀k∈Z+,0≤h(t2)≤12παexp(-t22α2)a.e.inR;(b)
for any finite subsetJ0⊂Z+ and any {λj;j∈J0}⊂R,  one has(25)0≤∑i,j∈Jλiλjy2i+2j≤∑i,j∈Jλiλjα2(i+j)[2(i+j)-1]!!,0≤∑i,j∈Jλiλjy2i+2j+2≤∑i,j∈Jλiλjα2(i+j+1)[2(i+j+1)-1]!!.Proof.
One applies Theorem9 to  dν=(1/(2πα))exp(-t2/2α2)dt, as a measure on R.
Due to Carleman criterion,dν is M-determinate. A straightforward computation shows that
(26)∫Rt2kdν=α2k(2k-1)!!,k≥1,k∈Z+.
The conclusion follows.Using a quite different proof with respect to that of Theorem9, we proved the following “abstract operatorial version.” Let ν be a positive measure as in Theorem 9, vanishing on the subspace of odd functions from Lν1(R). Let Y be a Banach lattice, which is order complete. Let xj(t)=tj,   t∈R,  t2=u∈[0,∞),  x~j(u)=uj. Then, the following theorem holds.Theorem 11 (see [10, Theorem 2.1]).
Let
F
2
:
L
ν
1
(
R
)
→
Y
be a linear continuous positive operator vanishing on the subspace of odd functions and
{
y
2
k
}
k
⊂
Y.
The following statements are equivalent:
(a)
there exists a linear operatorF:Lν1(R)→Y such that
(27)F(x2k)=y2k,F(x2k+1)=0,k∈Z+;0≤F(x)≤F2(x)∀x∈(Lν1(R))+⇒∥F∥≤∥F2∥;
(b)
for any finite subsetJ0⊂Z+ and any {λj;j∈J0}⊂R, one has
(28)0≤∑i,j∈J0λiλjy2i+2j≤∑i,j∈J0λiλjF2(x2i+2j),0≤∑i,j∈J0λiλjy2i+2j+2≤∑i,j∈J0λiλjF2(x2i+2j+2).In the sequel,Y will be the order-complete vector lattice, which is also a commutative Banach algebra of selfadjoint operators [6, 14]. Let H be a Hilbert space A a self-adjoint operator from H into H,
(29)Y1=Y1(A)={T∈Α(H);AT=TA},Y={T∈Y1;TU=UT,∀U∈Y1}.
Let X be the space of all continuous functions  x applying R into R, such that |x| is majorized by a polynomial on R. Denote by X2 the subspace of even functions, and S(A) will be the spectrum of A.Theorem 12.
Let{B2k}k∈Z+⊂Y=Y(A). The following statements are equivalent:(a)
there existsF∈L(X,Y)such that(30)F(x2k)=B2k,F(x2k+1)=0∀k∈Z+,0≤F(φ2·χS(A))≤∫S(A)φ2(t)dEA,φ∈X2,+;(b)
for any finite subsetJ0⊂Z+and any{λj;j∈J0}⊂R,
one has(31)0≤∑i,j∈J0λiλjB2i+2j≤∑i,j∈J0λiλjA2i+2j,0≤∑i,j∈J0λiλjB2i+2j+2≤∑i,j∈J0λiλjA2i+2j+2.Proof.
(b)⇒(a). Let
(32)f:P∩X2→Y,f(∑j∈J0λjx2j)=∑j∈J0λjB2j.
Then, the assumptions of (b) become
(33)0≤f(p2)≤∫S(A)p2(t)dEA,∀p2∈P2,+.
Since  P2  is a majorizing subspace in  X2, f has a positive linear extension
(34)F2∈L+(X2,Y).
Define(35)F:X→Y,F(x)=F(x1+x2)=F2(x2),x2(t)=12(x(t)+x(-t)),t∈R.
Then, F is a linear positive extension of f to the space X. The first two properties of the solution F hold. Using Urysohn’s lemma, for any  φ2∈(X2)+, there is a sequence
(36){φ~2,l}l⊂(Cc(R))+∩X2,φ~2,l↓φ2·χS(A).
Applications of the one-dimensional variant of Lemma 5 lead to the existence of a sequence of polynomial functions (pm)m such that
(37)pm∣S(A)→φ2χS(A)uniformly onS(A).
These arguments and the positivity of F yield
(38)0≤F(φ2·χS(A))≤infF(pm)=inff(pm)≤inf∫S(A)pm(t)dEA=∫S(A)φ2(t)dEA.
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(39)X={x:R2→R;xcontinuous,∃p∈Ps.t.|x(t1,t2)|≤p(t1,t2),∀(t1,t2)∈R2},X2={x∈X;x(t1,t2)=x(±t1,±t2),(t1,t2)∈R2}.
If H is a Hilbert space and A1,A2∈Α(H) are two bounded positive commuting selfadjoint operators, one defines
(40)Y1=Y1(A1,A2)={T∈Α(H);TAj=AjT,j=1,2},Y={U∈Y1;TU=UT,∀T∈Y1}.
Then, Y endowed with the usual order relation on selfadjoint operators is an order-complete vector lattice and a commutative Banach algebra.Theorem 13.
Let{B(2m,2n)}(m,n)∈Z+2⊂Y(A1,A2)  be a given sequence. The following statements are equivalent:(a)
there existsF∈L(X,Y)such that(41)F(x(2m,2n))=B(2m,2n),(m,n)∈Z+2,F(x(k,l))=0,∀(k,l)∈(Z+)2∖(2Z+)2,0≤F(x2)≤∫S(A1)×S(A2)x2(t1,t2)dEA1dEA2∀x2∈(X2)+,|F(φ)|≤∫S(A1)×S(A2)|φ(t1,t2)|dEA1dEA2∀φ∈X2;(b)
for allJ1,J2⊂Z+finite subsets, for all{aj;j∈J1},
and for all{bm;m∈J2}⊂R,
one has(42)0≤∑i,j∈J1,m,n∈J2‍aiajbmbnB(2i+2j,2m+2n)≤∑i,j∈J1,m,n∈J2‍aiajbmbnA12i+2jA22m+2n0≤∑i,j∈J1,m,n∈J2‍aiajbmbnB(2i+2j+2,2m+2n)≤∑i,j∈J1,m,n∈J2‍aiajbmbnA12i+2j+2A22m+2n0≤∑i,j∈J1,m,n∈J2‍aiajbmbnB(2i+2j,2m+2n+2)≤∑i,j∈J1,m,n∈J2‍aiajbmbnA12i+2jA22m+2n+20≤∑i,j∈J1,m,n∈J2‍aiajbmbnB(2i+2j+2,2m+2n+2)≤∑i,j∈J1,m,n∈J2‍aiajbmbnA12i+2j+2A22m+2n+2.Proof.
Since (a)⇒(b) is obvious, we have to prove the implication (b)⇒(a). By Stone-Weierstrass theorem, we know that the vector subspace S generated by the functions
(43)f(t1,t2)=f1(t1)·f2(t2)=(f1⊗f2)(t1,t2),fj∈C(S(Aj)),j=1,2,
is a dense subalgebra of C(S(A1)×S(A2)). Hence, for any x∈(C(S(A1)×S(A2)))+, there exists a sequence (fm)m of elements from S, which converges uniformly to x on S(A1)×S(A2), such that fm>x≥0 in S(A1)×S(A2). For any function from C(S(A1)×S(A2)), we consider its extension to [0,∞)×[0,∞) which vanishes outside S(A1)×S(A2), and then we extend again this function to R2, such that x~(t1,t2)=x~(±t1,±t2),  for all (t1,t2)∈R2. If we write
(44)fm(t1,t2)=fm,1(t1)fm,2(t2),(t1,t2)∈R2,
where fm,j are extended to R+ such that to vanish outside  S(Aj),j=1,2, and to R  such that to be even functions, we infer that fm,1,fm,2 are even bounded measurable nonnegative functions on R 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 fm,j  on  S(Aj),j=1,2,  by even positive polynomials on  R. Using the representation of these polynomials given in Remark 8, (b) is rewritten as
(45)f((∑ajt12j)(∑bmt22m)):=∑ajbmB(2j,2m)⇒0≤f(p1p2)≤∫S(A1)×S(A2)p1(t1)·p2(t2)dEA1dEA2,pj∈(R[tj])+,j=1,2.
In particular, f is a linear positive operator on the subspace generated by the products of even polynomials in each variable, which is a majorizing subspace in X2. This yields the existence of a positive extension F of f to the whole space X2. The positivity of F and the uniform convergence of (pm)m to x2≥0 on S(A1)×S(A2) lead to
(46)F(x2)=limF(pm)=limf(pm)≤lim∫S(A1)×S(A2)pm,1(t1)pm,2(t2)dEA1dEA2=∫S(A1)×S(A2)x2(t1,t2)dEA1dEA2.
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, X will be the space of all absolutely convergent power series:(47)φ(z)=∑j=0∞αjzj,|z|<b,X+={φ∈X;αj∈R+∀j∈Z+},X=X+-X+.
One assumes that the elements of the space X are continuous in the closed disk {|z|≤b}. Let H be a Hilbert space, U0 a selfadjoint operator acting on H, and Y=Y(U0) the order-complete vector lattice and commutative Banach algebra of selfadjoint operators defined in Section 2 (see also [6, 14]). We denote that  φj(z)=zj,  |z|≤b,   j∈Z+.Theorem 14.
LetX, H, U0, Y,  φj be as previously mentioned. Assume that b>1 and let A∈Y,  ∥A∥<b. Let  ε>0  and {Bj}j∈Z+⊂Y.  The following statements are equivalent:(a)
there exists a linear operatorFapplyingXintoYsuch that(48)F(φj)=Bj∀j∈Z+,φ(A)-ε·φ(I)≤F(φ)≤φ(A)+ε·φ(I)∀φ∈X+,∥F(φ)∥≤∥φ∥+ε·|φ(1)|,φ∈X;(b)
one has(49)Aj-ε·I≤Bj≤Aj+ε·I,j∈Z+.Proof.
(a)⇒(b) is obvious, since from (a) we derive
(50)Bj=F(φj)∈[φj(A)-ε·φj(I),φj(A)+ε·φj(I)]=[Aj-ε·I,Aj+ε·I],j∈Z+.
(b)⇒(a). One applies Theorem 4 to  xj=φj∈X,j∈Z+. We have
(51)∑j∈Fλjφj=ψ2-ψ1=∑j∈Z+ajφj-∑j∈Z+bjφj,aj,bj≥0⇒-(aj+bj)≤-bj≤λj≤aj≤aj+bj⇒|λj|≤aj+bj,j∈F⇒λjBj≤λjAj+ε·λjI,∀j∈F+,λjBj≤λjAj-ε·λjI,∀j∈F-,F+={j∈F;λj≥0},F-={j∈F;λj<0}.
The preceding relations yield(52)∑j∈FλjBj≤∑j∈F+λjAj+ε·(∑j∈F+λj)·I+∑j∈F-λjAj-ε·(∑j∈F-λj)·I=∑j∈FλjAj+ε·(∑j∈F|λj|)·I=ψ2(A)-ψ1(A)+ε·(∑j∈F|λj|)·I≤ψ2(A)-ψ1(A)+ε·(∑j∈Z+|λj|)·I≤ψ2(A)-ψ1(A)+ε·(∑j∈Z+(aj+bj))·I=ψ2(A)-ψ1(A)+ε·∑j∈Z+(aj+bj)·I=ψ2(A)+ε·(∑j∈Z+aj)·I-[ψ1(A)-ε·(∑j∈Z+bj)·I]=F2(ψ2)-F1(ψ1).
Theorem 4 implies the existence of a linear operator
(53)F∈L(X,Y),F1≤F≤F2onX+,F(φj)=Bj∀j∈Z+.
One obtains(54)|F(φ)|≤|F(φ+)|+|F(φ-)|≤φ+(A)+ε·φ+(I)+φ-(A)+ε·φ-(I)=|φ(A)|+ε|φ(I)|≤∥φ∥·I+ε·|φ(1)|·I⇒∥F(φ)∥≤∥φ∥+ε·|φ(1)|,∀φ∈X.
This concludes the proof.In the next theorems, the notion of a resolvent appears naturally during the proofs.Theorem 15.
Let
b
>
1
,
X
,
Y
,
{
φ
j
}
j
be as previously mentioned. Let
A
∈
Y
,
S
(
A
)
⊂
(
0
,
b
)
,
{
B
j
}
j
∈
Z
+
⊂
Y
,
ε
>
0. 						Assume that(55)0≤Bj≤Aj+ε·I∀j∈Z+.Then, there exists a linear positive operatorF∈L+(X,Y),
such that(56)F(φj)=Bj∀j∈Z+,|F(φ)|≤∥φ∥·[(I-b-1A)-1+εbb-1I]⇒∥F∥≤bb-∥A∥+εbb-1.Proof.
The following implications hold(57)∑j∈Fλjφj≤φ=∑j∈Z+ajφj⇒λj≤aj≤|aj|≤∥φ∥∞bj,j∈F⇒∑j∈FλjBj≤∑j∈F+λj(Aj+ε·I)≤∥φ∥·[∑j∈Z+b-jAj+ε·(∑j∈Z+b-j)·I]=∥φ∥·[(I-b-1A)-1+ε·bb-1I]=p(φ),φ∈X.
Note that in the preceding relations the Cauchy inequalities for the function φ were used. Application of Theorem 3 leads to the existence of a linear positive operator F from X into Y, satisfying the moment conditions, such that F≤p on X. Now, the conclusion follows easily, by using the monotony of the norm on Y:
(58)∥F∥≤∥(I-b-1A)-1∥+ε·bb-1≤1+∥A∥b+∥A∥2b2+⋯+ε·bb-1=bb-∥A∥+ε·bb-1.
The proof is complete.Similar results are valid for the multidimensional moment problem. LetXρ- be the space of analytic functions φ in the polydisc  Tρ-=∏k=1n{|zk|≤ρk},  ρ-=(ρ1,…,ρn).  Whence,
(59)φ(z1,…,zn)=∑j∈Z+ncjzj=∑j∈Z+ncjz1j1⋯znjn,j=(j1,…,jn),|zj|≤ρj,j=1,…,n,φj(z1,…,zn)=z1j1⋯znjn.
The order relation on X is defined by the positive cone
(60)X+={φ∈X;Recj≥0,Imcj≥0}.
Consider an arbitrary real or complex Hilbert space H,A1,…,An∈A(H) commuting selfadjoint operators (AiAj=AjAi,i,j=1,…,n). We introduce the usual spaces:
(61)Y1={U∈A(H);UAj=AjU,j=1,…,n},Y={U∈Y1;UV=VU,∀V∈Y1},Y+={U∈Y;〈U(h),h〉≥0,∀h∈H}.Theorem 16 (see [8, Theorem 3.1]).
With the previous notations, assume that(62)ρj>∥Aj∥,j=1,…,n,
and let {Bj;j∈Z+n}⊂Y. Consider the following statements:(a)
there exists anR-linear positive continuous operator F:Xρ-→Y such that
(63)F(xj)=Bj,j∈Z+n,F(φ)≤∥φ∥∞∏k=1nρk(ρkI-Ak)-1,∥F∥≤∏k=1nρkρk-∥Ak∥;(b)
one has(64)0≤Bj≤A1j1⋯Anjn,∀j∈Z+n;(c)
one has(65)0≤Bj≤∏k=1nρkjk+1(ρkI-Ak)-1,∥Bj∥≤∏k=1nρkjk+1ρk-∥Ak∥∀j=(j1,…,jn)∈Z+n.Then,(b)⇒(a)⇒(c).Proof.
(b)⇒(a). Let F⊂Z+n be a finite subset and let {λj}j∈F⊂R be such that
(66)∑j∈Fλjφj≤φ=∑j∈Z+ncjφj.
Then, the order relation on X and Cauchy’s inequalities yield
(67)λj≤Recj≤|cj|≤∥φ∥∞ρ1j1⋯ρnjn,j=(j1,…,jn).
Using the hypothesis (b) and the previous relations, one obtains
(68)∑j∈FλjBj≤∑j∈F+λjBj≤∑j∈F+λjA1j1⋯Anjn≤∥φ∥∞(∑j∈Z+n(A1ρ1)j1⋯(Anρn)jn)=∥φ∥∞(∑j1∈Z+(A1ρ1)j1)⋯(∑jn∈Z+(Anρn)jn)=∥φ∥∞(I-A1ρ1)-1⋯(I-Anρn)-1=∥φ∥∞∏k=1nρk(ρkI-Ak)-1=p(φ),φ∈X.
Application of Theorem 3 leads to the existence of an R-linear positive operator F applying X into Y such that
(69)F(φj)=Bj,j∈Z+n,F(φ)≤p(φ)=p(-φ)∀φ∈X⇒|F(φ)|≤p(φ),φ∈X.
Using the monotony of the norm on Y, we deduce
(70)∥F(φ)∥≤∥φ∥∞∏k=1∞∥(I-Akρk)-1∥≤∥φ∥∞×[∑j1∈Z+(∥A1∥ρ1)j1]⋯[∑jn∈Z+(∥An∥ρn)jn]=∥φ∥∞∏k=1∞ρkρk-∥Ak∥,∀φ∈X.
Thus, the implication (b)⇒(a) is proved.
(a)⇒(c) The hypothesis on positivity of the solution F and of functions φj yield
(71)Bj=F(φj)≤∥φj∥∞∏k=1nρk(ρkI-Ak)-1=∏k=1nρkjk+1(ρkI-Ak)-1⇒∥Bj∥=∥F(φj)∥≤∥φj∥∏k=1nρkρk-∥Ak∥=∏k=1nρkjk+1ρk-∥Ak∥,j=(j1,…,jn)∈Z+n.
The proof is complete.Applying Theorem16 to the real Hilbert space H=Ldt2([0,π/2]), n=2,  Ak,  k=1,2 being the multiplication operator with the functions h1(t)=sint,  h2(t)=cost, one obtains the following.Corollary 17.
Let{Bj;j∈Z+2}⊂Y=Y(A1,A2),   ρ>1, ρ1=ρ2=ρ.
Consider the following statements:
(a)
there existsF∈L+(X,Y) such that
(72)F(φj)=Bj,j∈Z+2,∫0π/2(F(φ)(h))(t)·h(t)dt≤∥φ∥ρ2∫0π/2h2(t)(ρ-sint)(ρ-cost)dt,∀h∈Ldt2([0,π2]);
(b)
one has(73)0≤∫0π/2Bj(h)(t)·h(t)dt≤∫0π/2sinj1t·cosj2t·h2(t)dt,∀h∈H,j=(j1,j2)∈Z+2;
(c)
one has(74)0≤∫0π/2B(j1,j2)(h)(t)·h(t)dt≤ρj1+j2+2∫0π/2h2(t)(ρ-sint)(ρ-cost)dt,∀h∈H.Then,(b)⇒(a)⇒(c).For some related results see [8].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(un)n=0∞ in an ordered vector space Y is said to be nonnegative with respect to the interval I⊂R if for any n∈Z+, we have
(75)λ0+λ1t+⋯+λntn≥0∀t∈I⇒λ0u0+λ1u1+⋯+λnun≥0inY.
It is clear that for compact intervals and for Y=R, this condition is necessary and sufficient for the existence of a unique positive solution of the moment problem associated to the moments yj=uj∈R,  j∈Z+. The continuity with respect to the supnorm is also obvious. But no information concerning dominating L1 norm for the solution holds. Such information would be useful for integral representation of the extension of the solution to the space L1(I) (see the next results).Theorem 19.
Letb∈(0,∞). Consider the following statements:(a)
there exists a uniqueh∈L∞([0,b])such that(76)0≤h(t)≤1a.e.,∫0btjh(t)dt=yj,∀j∈Z+;(b)
the sequence(77)(1,y0,2y1,…,nyn-1,…)is  nonnegative  with  respect  to[0,b].Then, (b)⇒(a).Proof.
We have(78)∑j=0nλjτj=φ2(τ)-φ1(τ),φ1,φ2∈X+,X=L1([0,b])⇒λ0t+λ1t22+⋯+λntn+1n+1≤∫0bφ2(τ)dτ=F2(φ2)-F1(φ1),(F1≡0),∀t∈[0,b]⇒∫0bφdτ2-λ0t-λ1t22-⋯-λntn+1n+1≥0,∀t∈[0,b]⇒∫0bφ2(τ)dτ-λ0y0-λ12y12-⋯-λn(n+1)ynn+1≥0⇒∑j=0nλjyj≤F2(φ2)-F1(φ1).
Application of Theorem 4, (b)⇒(a), leads to the existence of a linear functional F on X=L1([0,b]) such that
(79)F(xj)=yj,j∈Z+,0=F1(φ)≤F(φ)≤F2(φ)=∫0bφdt,∀φ∈X+.
Using the representation of a linear positive functional on L1, there exists
(80)h∈L∞([0,b]),0≤h,F(φ)=∫0bφ·hdt,∀φ∈L1([0,b]).
From the last equality written for φ=χB, B⊂[0,b] being a Borel subset, we infer (via measure theory) that
(81)h≤1a.e.,∫0btjh(t)dt=F(xj)=yj,∀j∈Z+.
The proof is complete.Lemma 20.
Leth:[a,b]→R be a positive function in L1([a,b]) and p:[a,b]→R a continuous convex function on [a,b]. Then, one has
(82)∫abp(t)h(t)dt≥(∫abh(t)dt)·p(∫abth(t)dt∫abh(t)dt).Corollary 21.
Let(yj)j∈Z+ be a sequence of real numbers such that the sequence
(83)(1,y0,2y1,…,nyn-1,(n+1)yn,…)
is nonnegative with respect to [0,b] and y0≠0. Then,
(84)y01-jy1j≤yj≤bj+1j+1,j∈Z+,y1≤b·y0.Proof.
Applying Theorem19 and Lemma 20 to p(t)=tj,  t∈[0,b],  j∈Z+, one obtains
(85)yj=∫0btjh(t)dt≤∫0btjdt=bj+1j+1,yj≥(∫0bth(t)dt)j(∫0bh(t)dt)j-1=y01-j·y1j,j∈Z+,y1=∫0bt·h(t)dt≤b∫0bh(t)dt=b·y0.
The proof is finished.


## 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 [20], also using some results from [27]. Let us denote that
(86)ψj(t)=j·tj-1,t∈[0,b],j∈NTheorem 22.
For a given family of numbers(mj)j=1n,  consider the following statements: (a), (b), and (c):(a)
there existsh∈L∞([0,b])such that(87)0≤h(t)≤1a.e.,mj=j∫0btj-1h(t)dt,j=1,2,…,n;(b)
for any family of scalars(λj)j=0n,
one has:(88)∑j=1nλjmj≤∑j=1nλjbj;(c)
there exists a Borel subsetEsuch that(89)∫Bj·tj-1dt=mj,j=1,…,n.Then,(b)⇒(a)⇔(c)Proof.
Let the point(b) be accomplished and assume that
(90)∑j=1nλj·j·tj-1=φ2-φ1,φj∈(L1([0,b]))+.
Then the integration on [0,b] yields
(91)∑j=1nλjbj≤∫0bφ2(t)dt=F2(φ2)⇒∑j=1nλjmj≤∫0bφ2(t)dt.
Application of Theorem 4 to F1≡0,F2 defined previously, leads to the existence of a linear positive form F on L1([0,b]) such that(92)F(ψj)=mj,j=1,…,n,F(ψ)≤∫0bψ·dt,ψ∈(L1([0,b]))+⇒|F(φ)|≤F(φ+)+F(φ-)≤∫0b|φ(t)|·dt,φ∈L1([0,b]).
If characteristic functions of Borel subsets stand forψ, the conclusion (a) follows by measure theory [28]. The implication (a)⇒(c) is a consequence of equality (15.14) in [27] (see also [27, Exercise 2.57]). The set of values for the control function u, namely,  [-1,1], is replaced by [0,1], the set of values for h, that stands for the control function u. The extreme points of the positive part of the unit ball of L∞([0,b]) are the characteristic functions of measurable sets. The converse is obvious.Corollary 23.
Under the equivalent conditions (a), (c) of Theorem22, there are sequences
(93)y1,n<x1,n<y2,n<x2,n<⋯<yl,n<xl,n<⋯,n∈N
such that the following relations hold:
(94)mk=infn∈N(∑j=1∞(xj,nk-yj,nk)),k=1,…,n.Proof.
One uses the fact that any Borel subset is a joint of aGδ set and a set of measure zero [28].Remark 24.
In order to approximate the numbersxj,nk,yj,nk, one can make use of Fourier approximate expansion of h with respect to the orthonormal sequence attached to the functions ktk-1 via Gram-Schmidt algorithm, also using the values of the moments mk. Thus, one obtains a smooth approximation h~ of h, and the intervals of ends yl,n,xl,n are connected components of the open sets of the form
(95){jl2p(n)<h~(t)<jl+12p(n)}.Remark 25.
A similar result to that of Theorem22 in several dimensions holds, with the same proof. We state it for the two-dimensional case.Theorem 26.
Let(m(j1,j2))1≤j1≤n1,1≤j2≤n2 be a given family of real numbers and consider the functions
(96)ψ(j1,j2)(t1,t2)=j1j2t1j1-1t2j2-1,1≤jp≤np,p=1,2,(t1,t2)∈K2=[0,b1]×[0,b2].
Consider the following assertions: (a), (b), (c). Then  (b)⇒(a)⇔(c)(a)
there exists a Borel functionh,0≤h(t1,t2)≤1  a.e. such that
(97)∬K2(ψ(j1,j2)·h)(t1,t2)dt1dt2=m(j1,j2),1≤j1≤n1,1≤j2≤n2;(b)
for any family of scalars{λ(j1,j2)}1≤j1≤n11≤j2≤n2,  one has
(98)∑1≤j1≤n11≤j2≤n2λ(j1,j2)m(j1,j2)≤∑1≤j1≤n11≤j2≤n2λ(j1,j2)b1j1b2j2;(c)
there exists a Borel subsetB2⊂K2 such that
(99)∬B2ψ(j1,j2)(t1,t2)dt1dt2=m(j1,j2),1≤jp≤np,p=1,2.Corollary 27.
If one of the conditions (a), (c) from Theorem26  is accomplished, there are sequences
(100)y1,n<x1,n<⋯<yl,n<xl,n<⋯,v1,n<u1,n<⋯<vl,n<ul,n<⋯,n∈N,
such that
(101)m(k1,k2)=infn∈N{∑l∈N(xl,nk1-yl,nk1)·(ul,nk2-vl,nk2)}.Proof.
The Borel subsetB2 is the joint of a Gδ set and a null set. For an open subset Dn⊃B2, we consider its decomposition into cells [28], used in the construction of Lebesgue measure. If we determine smooth approximations h~ of h by means of Gram-Schmidt algorithm for the functions ψ(j1,j2) and the given moments m(j1,j2) via Fourier expansion, then the sequences from the present corollary can be determined by means of cell decomposition of the open subsets
(102){(t1,t2);m(l)2p(n)<h~(t1,t2)<m(l)+12p(n)}.
Of course, this way one obtains approximations of these numbers.The previous statements solve the truncated moment problems and sketch an algorithm for determining the numbersyl,n,xl,n,vl,n,ul,n. In the last twenty years the truncated moment problem was studied intensively. In [26], the author proves that solving the truncated (not necessarily complex) moment problem for any natural n yields to the solution of the full moment problem. The idea is of passing through the limit, which is allowed by a weakly compact  argument. The following theorem proposes a similar construction, by using Krein-Milman theorem. We state it firstly in the one-dimensional case, although the several dimensional case one proves similarly.Theorem 28.
With the notations from Theorem22, let (mk)k≥1 be a sequence of real numbers. Consider the following assertions:(a)
there exists a Borel functionh such that
(103)0≤h(t)≤1a.e.,mk=k∫abtk-1h(t)dt,k∈N∖{0};(b)
for any natural numbern≥1, and any ε>0, there exists nonnegative scalars βj,j=1,…,n, and sequences:
(104)yj,1<xj,1<⋯<yj,l<xj,l<⋯,j=1,…,nsuch as(105)1-ε≤∑j=1nβj≤1,mk=limn(∑j=1nβj(∑l(xj,lk-yj,lk)));(c)
for anyn∈N∖{0},  there exists a Borel subset Bn such that
(106)mk=k∫Bntk-1dt,k=1,…,n;(d)
for any naturaln≥1 and any {λ1,…,λn}⊂R, the following relation holds true:
(107)∑k=1nλkmk≤∑k=1nλk(bk-ak).Then  (d)⇒(a)⇔(b)⇔(c).Proof.
(a)⇒(b). One applies Krein-Milman Theorem for the weakly compact subset formed by intersecting the unit ball of L∞([a,b]) with the positive cone of the same space. Then, we must have
(108)h=limn(∑j=1nαjχBj)=limn(∑j=1nβjχDj),αj≥0,∑j=1nαj=1,βj=m(Bj)m(Dj)·αj∈[(1-ε)αj,αj],Bj⊂Dj,j=1,…,n,Dj
which are suitably chosen open subsets, where the limit is in the weak topology on L∞, with respect to the dual pair (L1,L∞);   m is the Lebesgue measure. This leads to
(109)mk=k∫abtk-1h(t)dt=limn(∑j=1nβj∫Djktk-1dt).
Since each open subset Dj has an at most countable decomposition,
(110)Dj=⋃l(yj,l,xj,l),
the conclusion (b) follows. For the converse implication, observe that each step function,
(111)hn:=∑j=1nβjχDj,Dj=⋃l(yj,l,xj,l),j=1,…,n′βj≥0,1-ε≤∑j=1nβj≤1,
is an element of the positive part of the unit ball in L∞, and this subset is weakly compact. Let h be the (w)-limit of a subsequence of the sequence (hn)n. Then, we have
(112)mk=limn∫abhmn(t)k·tk-1dt=k∫abh(t)·tk-1dt,
by Lebesgue dominated convergence theorem. Hence, (b)⇒(a) is proved. The implication (a)⇒(c) follows from (a)⇒(c) of Theorem 22, since a solution of the full moment problem is a solution of all truncated moment problems. The proof of  (c)⇒(a) is similar to that of (b)⇒(a). It remains to prove that (d)⇒(a). This is a consequence of the implication (b)⇒(a) of Theorem 4. If J0⊂N∖{0} is a finite subset and {λj;j∈J0}⊂R, then the following implications hold true:
(113)∑j∈J0λj·j·tj-1=φ2-φ1,φp∈L+1⇒∑j∈J0λj·(bj-aj)≤∫abφ2dt=F2(φ2)-F1(φ1),F1≡0⇒∑j∈J0λj·mj≤∫abφ2dt=F2(φ2).
Application of Theorem4 leads to the existence of a linear functional F on L1([a,b]), verifying
(114)F(ktk-1)=mk,k≥1,k∈N,0≤F(ψ)≤∫abψdt,∀ψ∈L+1([a,b]).
Now, the conclusion follows by measure theory [28].Remark 29.
For the full moment problem, the following algorithm holds in determiningyj,l,xj,l.
Step  1. Approximating the function h.
Let(en)n≥1 be a Hilbert base constructed by the aid of Gram-Schmidt procedure, applied to the system of linearly independent functions
(115)φn(t)=n·tn-1,n∈N∖{0}.
Then, for each fixed  n≥1, one has
(116)en=∑j=1naj(n)φj⇒〈h,en〉=∑j=1naj(n)〈h,φj〉=∑j=1naj(n)mj,
where the coefficients aj(n) are known from the Gram-Schmidt procedure. Hence, we can determine each Fourier coefficient of h that is, we can approximate h in L2-norm by a sequence of polynomial functions hn,  n≥1. Then, there exists a subsequence [28] hkn→h pointwise almost everywhere in [a,b].
Step  2. For each n∈N∖{0}, the subsets,
(117){t;ml2p<hkn(t)<ml+12p},p,ml∈N,
are open subsets. Their connected components have end points approximations of the unknowns yj,l, xj,l. From [26], using a weakly compact standard argument, we can obtain h as the limit of a subsequence of (χBn)n,  where Bn are as at point (c) of Theorem 28. Considering a suitable open set Dkn⊃Bkn, from (a) one obtains:(118)mk≈∫Dknk·tk-1dt≈∑l(xn,lk-yn,lk),Dkn=⋃l]yn,l,xn,l[⋃l.Note that all the results of this section can be adapted to the multidimensional moment problem, with similar proofs.


## 5. Distanced Convex Subsets: Extension of  Linear Operators and the Moment Problem

The case when the subsetA from Theorem 1 is distanced with respect to the subspace S, while the restriction involving B is missing, has a nice geometric meaning and leads to interesting results concerning the extension of linear functionals and operators (see [5]).IfV is a convex neighborhood of the origin in a locally convex space, we denote by pV the gauge attached to  V.Theorem 30.
LetX be a locally convex space,  Y an order-complete vector lattice with strong order unit u0, and S⊂X a vector subspace. Let A⊂X be a convex subset with the following qualities:(i)
there exists a neighborhoodV of the origin such that  (S+V)∩A=Φ (A and S are distanced);(ii)
A is bounded.
Then, for any equicontinuous family of linear operators
{
f
j
}
j
∈
J

⊂

L
(
S
,
Y
)
and for any
y
~

∈

Y
+

∖

{
0
},
there exists an equicontinuous family{Fj}j∈J⊂L(X,Y)such that(119)Fj∣S=fj,Fj∣A≥y~,∀j∈J.Moreover, ifVis a neighborhood of the origin such that(120)fj(V∩S)⊂[-u0,u0],(S+V)∩A=Φ,0<α∈Rs.t.pV|A≤α,α1>0s.t.y~≤α1u0,then the following relations hold:(121)Fj(x)≤(1+α+α1)pV(x)·u0,x∈X,j∈J.The last relation of Theorem30 gives a relationship between an upper bound and the lower bound of Fj,j∈J. The next result is a consequence of the preceding one, in terms of the moment problem. The topology on Y is the norm topology for which the unit closed ball is the interval  [-u0,u0].Theorem 31.
Let(122)0<b∈R,X=C([0,b]),xj(t)=tj,j∈Z+,j≥1,t∈[0,b],{φk}k∈Z+⊂X,∥φk∥≤1,φ0≡1,φk(0)=1,k∈Z+.
LetY be an order-complete vector lattice with strong order unit u0and (yn)n≥1 a sequence in Y such that (u0,y1,y2,…) is positive with respect to [0,b]. Then, for any α1∈R+, there exists F∈L(X,Y) such that
(123)F(xj)=yj,j∈Z+,j≥1,F(φk)≥α1u0,k∈Z+,F(x)≤(2+α1)·∥x∥·u0,x∈X.
Moreover, if α1≥1 and Y is endowed with a linear topology such that its positive cone is closed and normal, then F is continuous and positive.Proof.
Let(124)S=Sp{xj;j∈Z+,j≥1},A=co{φk;k∈Z+}.
For all p∈S,a∈A, we have
(125)∥p-a∥≥|p(0)-a(0)|=1,
so that d(S,A)≥1 holds. This leads to (S+B(0,1))∩A=Φ, so we can take in Theorem 30   V=B(0,1),pV=∥·∥∞. From the hypothesis on ∥φk∥,  k∈Z+, we infer that(126)∥φk∥≤1∀k∈Z+⇒∥·∥|A≤1⇒α:=1,
where α is the constant from Theorem 30. We define that
(127)y~:=α1·u0,f(∑j=1nλjxj):=∑j=1nλjyj.
Let∑j=1nλjxj∈S∩B(0,1). Using the hypothesis on the sequence which is positive with respect to [0,b],  this yields
(128)sup{|∑j=1nλjtj|;t∈[0,b]}<1⇒∑j=1nλjtj+1>0,1-∑j=1nλjtj>0⇒u0+∑j=1nλjyj≥0,u0-∑j=1nλjyj≥0⇒f(∑j=1nλjxj)=∑j=1nλjyj∈[-u0,u0]⇒f(S∩B(0,1))⊂[-u0,u0].
The conclusion is thatf verifies the conditions in the statement of Theorem 30. Application of the latter theorem leads to the existence of an extension F of f such that
(129)F|A≥α1y0,F(x)≤(2+α1)·∥x∥·u0,x∈X.
Whence, the first part of the theorem is proved. It remains to prove the continuity and positivity of the extensionF, under the hypothesis mentioned in the statement. Replacing x by -x, it is easy to see that
(130)|F(x)|≤(2+α1)·∥x∥·u0,x∈C([0,b]).
From these relations, also using that the positive cone Y+ is normal, the continuity of F follows. In order to prove that F is positive, it is sufficient to show that F(p)≥0 for all positive polynomials p. Hypothesis yields
(131)p(t)=λ0+∑j=1nλjtj≥0∀t∈[0,b]⇒λ0u0+∑j=1nλjyj≥0⇒F(p)=λ0F(φ0)+∑j=1nλjyj≥λ0(α1·u0-u0)=λ0(α1-1)·u0≥0.
The proof is complete.A consequence of Theorem31 is the following one.Corollary 32.
LetH be a Hilbert space, A∈A(H) a positive selfadjoint operator, and Y=Y(A) the commutative algebra of selfadjoint operators defined in Section 2. Let b∈R+ be such that the spectrum S(A)⊂[0,b] and α1≥1. Then, there exists a nondecreasing mapping σ:[0,b]→Y such that
(132)∫0btjdσ(t)=Aj,j∈Z+,j≥1,∫0be-ktdσ(t)≥α1I,k∈Z+,∫0bx(t)dσ(t)≤(2+α1)·∥x∥·I,x∈C([0,b]).
In particular, for such a mapping σ, the following relation holds:
(133)α1+1≤∥σ(b)-σ(0)∥+∥exp(-kA)∥,k∈Z+.Proof.
We apply Theorem31 to φk(t)=e-kt, k∈Z+,   t∈[0,b],  u0=I. The sequence (I,A,A2,…,An,…) is positive with respect to [0,b] because of the positivity of the spectral measure attached to A. Due to the hypothesis of Theorem 31 being accomplished, there exists a positive continuous solution F satisfying the relations in the statement of the latter theorem. This solution applies norm-bounded subsets into order-bounded subsets of Y(A). Application of [14, page 272], yields the existence of a “representing” nondecreasing mapping σ, such that the vector measure dσ represents F. Next, we prove the last relation of the statement:
(134)α1I≤∫0be-ktdσ=σ(b)-σ(0)+∑m=1∞(-k)mm!Am=σ(b)-σ(0)-I+exp(-kA).
Further computation yields
(135)(α1+1)I≤σ(b)-σ(0)+exp(-kA)⇒α1+1≤∥σ(b)-σ(0)∥+∥exp(-kA)∥,k∈Z+.
The proof is complete.Application of Theorem31 to positive sequences of real numbers with respect to [0,b], the functions φk,  k∈Z+ being as in Corollary 32, yields the following consequence, which we state without proof.Corollary 33 (see [5, Corollary 3.3]).
Let(1,y1,y2,…) be a sequence of real numbers, which is positive with respect to [0,b], b>0 being arbitrary. Then, the following inequalities hold:
(136)y1≤3(j-1)/j·yj1/j,j∈Z+,j≥1.
In particular, one has: y1≤3liminfyj1/j.Next, we consider the analogue of Theorem31, for a space X of analytic functions. Let b>1 and X be the space of all functions φ which can be represented as an absolutely convergent series,
(137)φ(z)=∑j=0∞λjzj,|z|<b,λj∈R,φ
being continuous in the closed disc |z|≤b. Let φj(z)=zj,   j∈Z+,  Y=L∞(Ω),(Ω,μ) being a measurable space, and u0(ω)=1,for all  ω∈Ω.Theorem 34.
Let{ψk}k∈Z+⊂X be a sequence such that
(138)∥ψk∥∞≤M,ψk(0)=1,∀k∈Z+.
Assume that b>1 and let (yj)j≥1 be a sequence in Y such that
(139)∥yj∥≤b-1,j≥1,j∈Z.
Then, for any y~∈Y+∖{0}, there exists F∈L(X,Y) such that
(140)F(φj)=yj,j≥1,F(ψk)≥y~,k∈Z+,F(x)≤(1+M+∥y~∥∞)·∥x∥∞·u0,x∈X.Proof.
IfS=Sp{φj;j≥1},A=co{ψk;k∈Z+}, we have
(141)d(S,A)≥1,(S+B(0,1))∩A=Φ,V=B(0,1),pV=∥·∥∞.
Let  φ∈B(0,1)∩S,   φ=∑j=1nλjφj, ∥φ∥<1. Using the Cauchy inequalities, we derive
(142)|λj|≤∥φ∥bj<1bj,j∈{1,2,…,n}.
Define f(∑j=1nλjφj)=∑j=1nλjyj, n∈Z, n≥1,  λj∈R. Using also the hypothesis on ∥yj∥,j≥1, one obtains
(143)|∑j=1nλjyj|≤∑j=1n|λj|·|yj|≤(∑j=1n(1bj)·∥yj∥∞)·u0≤(∑j=1∞(1bj))·(b-1)u0=u0.
Whence,  φ=∑j=1nλjφj∈S∩B(0,1)⇒f(φ)∈[-u0,u0]. Since
(144)∥·∥|A≤M,y~≤∥y~∥∞·u0,
one takes in Theorem 30  α=M,  α1=∥y~∥∞. Now, the conclusion follows via Theorem 30, for which all requirements of the hypothesis are accomplished.


## 6. Geometric Aspects Related to the Moment Problem

In the first part of this section, we give sufficient conditions for the existence of a solution, in a very general setting. The method works in an arbitrary measurable space, which may not involve polynomials. In the particular case of subsets ofRn, the form of positive polynomials is not necessarily used.Theorem 35.
Let(T,ν) be a measurable space, where ν is a positive σ-finite measure on T.  Let X=Lν1(T), {xj}j∈J⊂X, {yj}j∈J⊂R. Consider the following statements:(a)
there existsh∈Lν∞(T) such that
(145)∫Txj(t)·h(t)dν=yj,j∈J,-1≤h(t)≤1ν-a.e.;(b)
for any finite subsetJ0⊂J and any {λj;j∈J0}⊂R,  one has
(146)∑i,j∈J0λiλjyiyj≤∑i,j∈J0λiλj∫Txi(t)dν·∫Txj(t)dν;(c)
for any finite subsetJ0⊂J and any  {λj;j∈J0}⊂R, one has
(147)∑i,j∈J0λiλjyiyj≤∑i,j∈J0|λi|·|λj|·∫T|xi(t)|dν·∫T|xj(t)|dν.
Then, (b)⇒(a)⇒(c).Proof.
(b)⇒(a). The following implications hold:(148)∑j∈J0λjxj=φ2-φ1,φk∈X+,k=1,2⇒-∫Tφ1dν-∫Tφ2dν≤∫Tφ2dν-∫Tφ1dν=∑j∈J0λj∫Txjdν≤∫Tφ2dν+∫Tφ1dν=∫Tφ2dν-(-∫Tφ1dν)=F2(φ2)-F1(φ1)⇒|∑j∈J0λj∫Txjdν|≤F2(φ2)-F1(φ1).
On the other hand, the condition (b) can be rewritten as
(149)(∑j∈J0λjyj)2≤(∑j∈J0λj∫Txjdν)2⟺|∑j∈J0λjyj|≤|∑j∈J0λj∫Txjdν|.
From the preceding relation, we conclude
(150)∑j∈J0λjyj≤|∑j∈J0λjyj|≤|∑j∈J0λj∫Txj(t)dν|≤F2(φ2)-F1(φ1).
Whence, the conditions from the statement of Theorem 4 are accomplished, so that there exists a linear form  F on X such that
(151)-∫Tφdν=F1(φ)≤F(φ)≤F2(φ)=∫Tφdν⇒|F(φ)|≤∫Tφdν∀φ∈X+⇒|F(φ)|≤|F(φ+)|+|F(φ-)|≤∫T(φ++φ-)dν=∫T|φ|dν=∥φ∥1⇒∥F∥≤1.
Following the representation of continuous linear functionals on L1 spaces, there exist
(152)h∈Lν∞(T),F(φ)=∫Tφ(t)h(t)dν∀φ∈Lν1(T),∥h∥∞=∥F∥≤1⇒-1≤h(t)≤1ν-a.e.
We also infer that
(153)yj=F(xj)=∫Txj(t)h(t)dν,∀j∈J.
Thus, the proof of the implication (b)⇒(a) is finished. The implication (a)⇒(c) is almost obvious:
(154)|∑j∈J0λjyj|=|∑j∈J0λj∫Txj·h·dν|≤∑j∈J0|λj|·∫T|xj(t)|·|h(t)|·dν≤∑j∈J0|λj|·∫T|xj(t)|·dν.
Taking the squares of the two members, the inequality is preserved. The proof is complete.Corollary 36.
Let(155)T={(t1,t2)∈R2;0≤t1<∞,0≤t2≤exp(-t1)},{y(j1,j2)}(j1,j2)∈Z+2⊂R.Consider the following statements:(a)
there exists a Lebesgue measurable functionh on T such that
(156)∫Tt1j1t2j2h(t1,t2)dt1dt2=y(j1,j2),∀(j1,j2)∈Z+2,-1≤h(t1,t2)≤1a.e.;(b)
for any finite subsetJ0⊂Z+2 and any {λ(j1,j2);(j1,j2)∈J0}⊂R, one has
(157)∑(i1,i2),(j1,j2)∈J0‍λ(i1,i2)λ(j1,j2)y(i1,i2)y(j1,j2)≤∑(i1,i2),(j1,j2)∈J0‍λ(i1,i2)λ(j1,j2)i1!(i2+1)i1+2·j1!(j2+1)j1+2.
Then, (b)⇒(a).Note thatT is a closed unbounded and non-semi-algebraic subset.Details can be found in [8]. In [8, Theorem 2.5], the construction of the solution of a moment problem on the ellipse is given.



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