Two Sandwich Theorems for Linear Operators and the Moment Problem

Octav Olteanu

Abstract

We give a direct proof for theorem 2 [13] (which is equivalent to theorem 1 [15]). Then we apply theorems 1 [15] and 4 [15] to some concrete spaces of sequences or functions which have a Schauder basis. The polynomials x_{j}(t) = t^{j}, j ∈ N considered in the classical moment problem, are replaced by the elements of the Schauder basis.

Mathematics Subject Classification: 46A22, 47A57. Key words: extension of linear operators, moment problems.

1 Introduction

Before stating the abstract moment problem, we recall some definitions. A subset X_{+} of a real vector space X is said to be a convex cone if X_{+} + X_{+} ⊂ X_{+} and αX_{+} ⊂ X_{+} for any α ∈ [0, ∞). An ordered vector space is a vector space X endowed with an order relation defined by a convex cone X_{+} ⊂ X in the following way: x_{1} ≤ x_{2} iff x_{2} −x_{1} ∈ X_{+}. X_{+} is said to be the positive cone of X. A vector lattice is an ordered vector space Y such that for any y_{1}, y_{2} ∈ Y, there exists the supremum sup{y_{1}, y_{2}} =: y_{1} ∨ y_{2}. An order complete vector lattice is a vector lattice in which any family {y_{j} : j ∈ J} ⊂ Y bounded from above has a supremum sup{y_{j} : j ∈ J} =: ∨_{j∈J} y_{j} ∈ Y. For a deep study of ordered vector spaces see [6] or [20].

The abstract moment problem may be stated in the following way. One give two ordered vector spaces X, Y and two families of elements {x_{j} : j ∈ J} ⊂ X, {y_{j} : j ∈ J} ⊂ Y. One also gives a convex operator p : X → Y. The problem is to find necessary and sufficient conditions on y_{j} (which are called moments since they generalize the classical moments), for the existence of a linear operator f ∈ L(X, Y), with the following properties:

(1.1)  f(x_{j}) = y_{j} ∀j ∈ J (the moment conditions),

Balkan Journal of Geometry and Its Applications, Vol.1, No.2, 1996, pp. 75-85 \begin{matrix}c \\ ◯ \end{matrix}Balkan Society of Geometers, Geometry Balkan Press

(1.2)  f(X_{+}) ⊂ Y_{+} (positivity),

(1.3)  f(x) ≤ p(x) ∀x ∈ X (the generalization of the continuity).

In the classical moment problem we have Y = R, X is a space of functions on an interval I ⊂ R containing the polynomials x_{j}(t) = t^{j}, j ∈ N, t ∈ I and p : X → R is a seminorm on X. X_{+} is usually a convex cone such that p is monotone (0 ≤ x_{1} ≤ x_{2} ⇒ p(x_{1}) ≤ p(x_{2})). In the moment problem we are interested in the existence, unicity and construction of the solution f ∈ L(X, Y). The main purpose of the present work is to characterize the existence of f. If we note X_{0} := Sp{x_{j} : j ∈ J} and if we suppose that

f_{0} : X_{0} → Y, f_{0} (∑_{j∈F} λ_{j}x_{j}) := ∑_{j∈F} λ_{j}y_{j}

(F ⊂ J being a finite subset), is well defined, then the problem of the existence of the solution is in fact the problem of extending f_{0} to a linear operator f ∈ L(X, Y) which has the properties (1.2) and (1.3). When Y is an order complete vector lattice, the abstract moment problem is solved by theorem 1′, which is equivalent to theorem 1, both of them being stated and proved below (here we give a direct proof for theorem 1). The Hahn-Banach theorem is a particular case of theorem 1. It may be obtained taking in theorem 1 X_{+} := {0}, when the order relation on X is the equality relation. The theorem of H. Bauer (see [3] or [20]) may be easily obtained from theorem 1. The corollary 2 [9, p.336] may be also got using theorem 1′. In [16] we applied theorem 1′ to some calssical spaces X of functions (we considered X = C^{1}([0, b]), C^{2}([0, b]), C^{1}([0, b_{1}] × [0, b_{2}]), L^{1}([0, b_{1}] × [0, b_{2}]), BV ([a, b])). On the other hand, in [17] we proved theorem 2 of the present work and we applied it to some spaces of functions and measures. In section 3 of the present work we prove some applications of the two general theorems of section 2.

2 General theorems

Theorem 1. Let X be an ordered vector space, let Y be an order complete vector lattice and let p : X → Y be a convex operator. Let X_{0} ⊂ X be a vector subspace and let f_{0} ∈ L(X_{0}, Y) be a linear operator. The following statements are equivalent:

(a) there exists a linear and positive extension f ∈ L(X, Y) of f_{0} such that f(x) ≤ p(x) ∀x ∈ X;

(b) f_{0}(x′) ≤ p(x) ∀(x′, x) ∈ X_{0} × X with x′ ≤ x.

This theorem was published in [13], without proof. An indirect proof was published in [14], where we deduced it from a more general result. Here we give a direct proof.

Proof of theorem 1.

(a) ⇒ (b) is almost obvious (f_{0}(x′) = f(x′) ≤ f(x) ≤ p(x) ∀(x′, x) ∈ X_{0} × X with x′ ≤ x, since f(x′) ≤ f(x) by the positivity and linearity of f).

(b) ⇒ (a) Let S := {(S, f_{s}) : X_{0} ⊂ S ⊂ X, S is a vector subspace of X, f_{s} ∈ L(S, Y), f_{s}/X_{0} = f_{0}, f_{s}(x) ≥ 0 ∀x ∈ S ∩ X_{+} and (x′, x) ∈ S × X, x′ ≤ x, imply f_{s}(x′) ≤ p(x)}.

We consider the following natural order relation on S : (S_{1}, f_{s_{1}}) < (S_{2}, f_{s_{2}}) iff S_{1} ⊂ S_{2} and f_{s_{2}}/S_{1} = f_{s_{1}}. It is easy to see that S is inductively ordered with respect to this order relation. Let (M, f_{M}) ∈ S be a maximal element of S, which exists by Zorn’s lemma. To finish the proof, it is sufficient to prove that M = X. Supposing the contrary, let x̃ ∈ X \ M. We construct an element (M̃, fM̃) ∈ S, where M̃ := M ⊕ Sp{x̃}, fM̃ : M̃ → Y being a linear extension of f_{M}. This will contradict the maximality of (M, f_{M}) in S. We have to choose ỹ ∈ Y such that defining fM̃ : M̃ → Y by fM̃(m + λx̃) := f_{M}(m) + λỹ, to have (M̃, fM̃) ∈ S. So, we must show that

(2.1)  m_{1} + αx̃ ∈ X_{+}, m_{1} ∈ M, α ∈ R, imply f_{M}(m_{1}) + αỹ ≥ 0 in Y,

(2.2) m_{2} + βx̃ ≤ x, m_{2} ∈ M, β ∈ R, x ∈ X, imply f_{M}(m_{2}) + βỹ ≤ p(x) in Y.

For α = 0, (2.1) is true since (M, f_{M}) ∈ S. For β = 0, (2.2) is accomplished by the same reason. For α ≠ 0 (2.1) is equivalent to (2.1.1) and (2.1.2) taken together, where:

(2.1.1)  \begin{matrix}m_{1} + λ_{1}x̃ ∈ X_{+}, m_{1} ∈ M, λ_{1} > 0 ⇒ \\ ⇒ f_{M}(m_{1}) + λ_{1}ỹ ≥ 0, i.e. ỹ ≥ −f_{M}(m_{1})/λ_{1}, \end{matrix}

(2.1.2)  \begin{matrix}m̃_{1} + μ_{1}x̃ ∈ X_{+}, m̃_{1} ∈ M, μ_{1} < 0 ⇒ \\ f_{M}(m̃_{1}) + μ_{1}ỹ ≥ 0, i.e. ỹ ≤ −f_{M}(m̃_{1})/μ_{1}. \end{matrix}

Hence (2.1) is equivalent to (2.1’), where:

(2.1′)  \begin{matrix}y_{1} := −f_{M}(m_{1})/λ_{1} ≤ ỹ ≤ −f_{M}(m̃_{1})/μ_{1} =: ỹ_{1}, \\ λ_{1} > 0, μ_{1} < 0, m_{1} + λ_{1}x̃ ∈ X_{+}, m̃_{1} + μ_{1}x̃ ∈ X_{+}. \end{matrix}

Similarly, (2.2) is equivalent to (2.2’):

(2.2′) ỹ_{2} := (1/μ_{2})[p(x̃′) − f_{M}(m̃_{2})] ≤ ỹ ≤ (1/λ_{2})[p(x′) − f_{M}(m_{2})] =: y_{2},

where

λ_{2} > 0, μ_{2} < 0, m_{2} + λ_{2}x̃ ≤ x′, m̃_{2} + μ_{2}x̃ ≤ x̃′.

To find an ỹ ∈ Y which fulfills (2.1′) and (2.2′), we must prove the following four inequalities:

(2.3)  y_{1} ≤ ỹ_{1}, y_{1} ≤ y_{2}, ỹ_{2} ≤ ỹ_{1}, ỹ_{2} ≤ y_{2}.

Supposing that (2.3) are proved, we may choose ỹ such that

y_{1} ∨ ỹ_{2} ≤ ỹ ≤ ỹ_{1} ∧ y_{2}.

The proof of the inequalities (2.3) is not difficult. Here we prove that y_{1} ≤ y_{2}. Let λ_{1} > 0, λ_{2} > 0, m_{1}, m_{2} ∈ M such that m_{1}+λ_{1}x̃ ≥ 0 and m_{2}+λ_{2}x̃ ≤ x′ ∈ X. Then we get:

−(1/λ_{1})m_{1} ≤ x̃ ≤ (1/λ_{2})(x′ − m_{2}),

which imply

(2.4)  λ_{2}[−(1/λ_{1})m_{1} + (1/λ_{2})m_{2}] ≤ x′.

On the other hand, (M, f_{M}) ∈ S and (2.4) imply:

λ_{2}[−(1/λ_{1})f_{M}(m_{1}) + (1/λ_{2})f_{M}(m_{2})] ≤ p(x′),

which may be rewritten as follows:

−f_{M}(m_{1})λ_{1} ≤ (1/λ_{2})[p(x′) − f_{M}(m_{2})],

i.e.

y_{1} ≤ y_{2}.

Theorem 1′. (Theorem 1 [15]). Let X, Y, p be as in theorem 1 stated above. Let {x_{j} : j ∈ J} ⊂ X, {y_{j} : j ∈ J} ⊂ Y. The following statements are equivalent:

(a) there exists f ∈ L(X, Y) such that f(x) ≥ 0 ∀x ∈ X_{+}, f(x_{j}) = y_{j} ∀j ∈ J and f(x) ≤ p(x) ∀x ∈ X;

(b) for any finite subset F ⊂ J and any {λ_{j} : j ∈ F } ⊂ R, the relation ∑_{j∈F} λ_{j}x_{j} ≤ x in X implies ∑_{j∈F} λ_{j}y_{j} ≤ p(x) in Y.

Theorem 1′ is a rewritting of theorem 1 (we take in theorem 1 X_{0} := Sp{x_{j} : j ∈ J}, etc). Theorem 2. (Theorem 4 [15] and 2.1. [17]). Let X, Y, {x_{j} : j ∈ J}, {y_{j} : j ∈ J} be as in theorem 1′ and let f_{1}, f_{2} ∈ L(X, Y). Let us consider the following statements:

(a) there exists f ∈ L(X, Y) such that f(x_{j}) = y_{j}, ∀j ∈ J and f_{1}(z) ≤ f(z) ≤ f_{2}(z) ∀z ∈ X_{+};

(b) for any finite subset F ⊂ J and any {λ_{j} : J ∈ F } ⊂ R, we have:

(2.5)  \begin{matrix}∑_{j∈F} λ_{j}x_{j} = z_{2} − z_{1} with \\ z_{1}, z_{2} ∈ X_{+} ⇒ ∑_{j∈F} λ_{j}y_{j} ≤ f_{2}(z_{2}) − f_{1}(z_{1}); \end{matrix}

If X is a vector lattice, we also consider the statement (b′):

(b′) f_{1}(z) ≤ f_{2}(z) ∀z ∈ X_{+} and for any finite subset F ⊂ J and any {λ_{j} : j ∈ F } ⊂ R, we have

(2.6)  ∑_{j∈F} λ_{j}y_{j} ≤ f_{2} ((∑_{j∈F} λ_{j}x_{j})^{+}) − f_{1} ((∑_{j∈F} λ_{j}x_{j})^{−});

(for each x ∈ X, we note x^{+} := x ∨ 0, x^{−} := (−x) ∨ 0 and we have x = x^{+} − x^{−} ∀x ∈ X);

(c) if x_{j} ∈ X_{+} ∀j ∈ J, then f_{1}(x_{j}) ≤ y_{j} ≤ f_{2}(x_{j}) ∀j ∈ J.

Then (b) ⇔ (a) ⇒ (c) and, if X is a vector lattice, we have (b′) ⇔ (b) ⇔ (a) ⇒ (c).

3 Applications

Theorem 3. Let X be a real separable Hilbert space and let {x_{j} : j ∈ N} a fixed orthonormal basis in X. Let X_{+} := {x ∈ X : < x, x_{j} >≥ 0 ∀j ∈ N} and let {y_{j} : j ∈ N} ⊂ X_{+} such that if we note ρ_{m} := ∑_{j=0}^{∞} < y_{j}, x_{m} > we must have ∑_{m=0}^{∞} ρ_{m}^{2} < ∞. Then there exists f ∈ L(X, X), f(X_{+}) ⊂ X_{+}, f(x_{j}) = y_{j} ∀j ∈ N and f(x) ≤∥ x ∥ ỹ, where

ỹ:= ∑_{m=0}^{∞} ρ_{m} x_{m}, x ∈ X.

Proof. We shall apply theorem 1′, (b) ⇒ (a). Let n ∈ N, let {λ_{0}, ..., λ_{n}} ⊂ R and let x ∈ X such that

∑_{j=0}^{n} λ_{j}x_{j} ≤ x = ∑_{m=0}^{∞} < x, x_{m} > x_{m}

By the definition of X_{+}, this implies

(3.1)  λ_{j} ≤< x, x_{j} > ∀j ∈ {0, 1, ..., n}.

On the other hand, y_{j} ∈ X_{+} ∀j ∈ N, is equivalent to

(3.2)  < y_{j}, x_{m} >≥ 0 ∀(j, m) ∈ N^{2}.

So, from (3.1) and (3.2) we deduce

∑_{j=0}^{n} λ_{j}y_{j} = ∑_{j=0}^{n} λ_{j} ( ∑_{m=0}^{∞} < y_{j}, x_{m} > x_{m}) = ∑_{m=0}^{∞} (∑_{j=0}^{n} λ_{j} < y_{j}, x_{m} >) x_{m} ≤

≤ ∑_{m=0}^{∞} (∑_{j=0}^{n} < x, x_{j} >< y_{j}, x_{m} >) x_{m} ≤ ∥x∥ [ ∑_{m=0}^{∞} (∑_{j=0}^{n} < y_{j}, x_{m} >) x_{m}] ≤

∥x∥ ( ∑_{m=0}^{∞} ρ_{m}x_{m}) = ∥x∥ỹ =: p(x),

where ∥x∥ =< x, x >^{1/2}. By theorem 1′, (b) ⇒ (a), the conclusion follows. The theorem is proved. We go on by two applications of theorem 2. We recall the following notations:

l^{1} := {(α_{0}, ..., α_{n}, ...) ∈ R^{N} : ∑_{j=0}^{∞} | α_{j} |< ∞},

l^{∞} := {(β_{0}, ..., β_{n}, ...) ∈ R^{N} : sup_{j∈N} | β_{j} |< ∞},

We consider the operator f_{2} : l^{1} → l^{∞} defined by:

f_{2}((α_{0}, ..., α_{n}, ...)) = (β_{0}, ..., β_{n}, ...),

where

β_{n} := ∑_{k=0}^{n} α_{k}.

In l^{1} and l^{∞} we consider the convex cone of sequences which have all their components positive. It is clear that l^{1} ⊂ l^{∞} and ∀x ∈ l_{+}^{1}, we have x ≤ f_{2}(x). It is also well known that l^{∞} is an order complete vector lattice. So, we may apply theorem 2, (b′) ⇒ (a), for X := l^{1}, Y := l^{∞}, f_{1}(x) = x ∀x ∈ X and f_{2} defined as above. We get the following result. Theorem 4. Let X := l^{1}, Y := l^{∞}, let x_{j} ∈ X, x_{j} := (0, ..., 0, 1, 0, ..., 0, ...), j ∈ N, {y_{j} : j ∈ N} ⊂ X ⊂ Y, y_{j} = ∑_{m=0}^{∞} α_{m}^{(j)}x_{m}. The following statements are equivalent:

(a) there exists f ∈ L(X, Y), f(x_{j}) = y_{j} ∀j ∈ N, x ≤ f(x) ≤ f_{2}(x) ∀x ∈ X_{+};

(b) for any n ∈ N and any {λ_{0}, ..., λ_{n}} ⊂ R, we have:

(b_{1})  ∑_{j=0}^{n} λ_{j} α_{m}^{(j)} ≤ λ_{0}^{+} + ... + λ_{m−1}^{+} + λ_{m}, if 0 ≤ m ≤ n,

and

(b_{2})  ∑_{j=0}^{n} λ_{j} α_{m}^{(j)} ≤ λ_{0}^{+} + ... + λ_{p}^{+} + ... + λ_{n}^{+}, if m ≥ n + 1, m ∈ N

(c) the following three conditions are fulfilled:

(c_{1})  α_{m}^{(j)} = 0 ∀j ∈ N, ∀m ∈ N such that 0 ≤ m ≤ j − 1,

(c_{2})  α_{j}^{(j)} = 1 ∀j ∈ N,

(c_{3}).  α_{m}^{(j)} ∈ [0, 1] ∀j ∈ N, ∀m ≥ j + 1, m ∈ N

Proof. To prove (a) ⇔ (b), we use (a) ⇔ (b′) of theorem 2. We have only to show that the relations (b_{1}) and (b_{2}) (together) are equivalent to (2.6). We have:

∑_{j=0}^{n} λ_{j} y_{j} = ∑_{j=0}^{n} λ_{j} ( ∑_{m=0}^{∞} α_{m}^{(j)}x_{m}) = ∑_{m=0}^{∞} (∑_{j=0}^{n} λ_{j}α_{m}^{(j)}) x_{m}

and so, (2.6) may be written as follows:

∑_{m=0}^{∞} (∑_{j=0}^{n} λ_{j}α_{m}^{(j)}) x_{m} ≤ f_{2}((λ_{0}^{+}, ..., λ_{k}^{+}, ..., λ_{n}^{+}, 0, 0, ...)−(λ_{0}^{−}, ..., λ_{k}^{−}, ..., λ_{n}^{−}, 0, 0, ...) =

= (λ_{0}^{+}, λ_{0}^{+} + λ_{1}^{+}, ..., λ_{0}^{+} + ... + λ_{n}^{+}, λ_{0}^{+} + ... + λ_{n}^{+}, 0, ...) − (λ_{0}^{−}, λ_{1}^{−}, ..., λ_{n}^{−}, 0, ...) =

= (λ_{0}, λ_{0}^{+}+λ_{1}, ..., λ_{0}^{+}+...+λ_{n−1}^{+}+λ_{n}, λ_{0}^{+}+...+λ_{n−1}^{+}+λ_{n}^{+}, λ_{0}^{+}+...+λ_{n−1}^{+}+λ_{n}^{+}, ...).

By the definition of Y_{+}, (2.6) is equivalent to the statement (b) of theorem 4.

(a) ⇒ (c) is almost obvious. Indeed, since x_{j} ∈ X_{+}, we have:

x_{j} = (0, ..., 0, 1, 0, ...) ≤ y_{j} = (α_{0}^{(j)}, ..., α_{j}^{(j)}, ..., α_{m}^{(j)}, ...) ≤ T_{2}(x_{j}) = (0, ..., 0, 1, 1, 1, ...)

which imply (c).

(c) ⇒ (b) To prove (c) ⇒ (b_{1}), let n ∈ N and m ∈ N, m ≤ n.

From (c_{1}), (c_{2}) and (c_{3}) we deduce:

∑_{j=0}^{n} λ_{j}α_{m}^{(j)} = ∑_{j=0}^{m−1} λ_{j}α_{m}^{(j)} + λ_{m}α_{m}^{(m)} + ∑_{j=m+1}^{n} λ_{j}α_{m}^{(j)} =

∑_{j=0}^{m−1} λ_{j}α_{m}^{(j)} + λ_{m} ≤ ∑_{j=0}^{m−1} λ_{j}^{+} + λ_{m} = λ_{0}^{+} + ... + λ_{m−1}^{+} + λ_{m}.

This proves (b_{1}). To finish the proof, we have to show that (c) ⇒ (b_{2}).

Let m, n ∈ N, m ≥ n + 1. If j ∈ N, j ≤ n ≤ m − 1, then m ≥ j + 1 and, by (c_{3}), α_{m}^{(j)} ∈ [0, 1]. This implies:

∑_{j=0}^{n} λ_{j} α_{m}^{(j)} ≤ ∑_{j=0}^{n} λ_{j}^{+},

i.e. (b_{2}). The theorem is proved.

We go on by an application of theorem 2 to a space of analitic functions.

Let ρ > 0. We denote by A_{ρ} the set of all complex functions, defined on the open disk | z |< ρ of the complex plane, which can be represented as the sum of an absolutely convergent series

x(z) = ∑_{j=0}^{∞} α_{j}z^{j},

the coefficients α_{j} being real numbers. Then X = A_{ρ} is a real vector space which can be ordered by the convex cone

(3.3)  X_{+} := {x ∈ A_{ρ} : x(z) = ∑_{j=0}^{∞} α_{j} z^{j}, α_{j} ≥ 0 ∀j ∈ N} .

It is easy to see that X is an order complete vector lattice. Theorem 5. Let X = A_{ρ} and let X_{+} be the cone defined by (3.3). Let us denote x_{j}(z) := z^{j}, j ∈ N, | z |< ρ and let us consider the function g ∈ X, g(z) = 1 + z. On the other hand, let {y_{j} : j ∈ N} ⊂ X be a sequence in X, y_{j}(z) = ∑_{m=0}^{∞} α_{m}^{(j)}z^{m}, j ∈ N, | z |< ρ. Let us consider the following statements:

(a) there exists f ∈ L(X, X) such that f(x_{j}) = y_{j} ∀j ∈ N, x ≤ f(x) ≤ xg ∀x ∈ X_{+} ;

(b) for any n ∈ N and any {λ_{0}, λ_{1}, ..., λ_{n}} ⊂ R, we have:

(b_{1})  ∑_{j=0}^{n} λ_{j}α_{0}^{(j)} ≤ λ_{0},

(b_{2})  ∑_{j=0}^{n} λ_{j}α_{m}^{(j)} ≤ λ_{m−1}^{+} + λ_{m}, ∀m ∈ {1, 2, ..., n},

(b_{3})  ∑_{j=0}^{n} λ_{j}α_{n+1}^{(j)} ≤ λ_{n}^{+},

(b_{4})  ∑_{j=0}^{n} λ_{j}α_{m}^{(j)} ≤ 0 ∀m ∈ {n + 2, n + 3, ...};

(c) the α_{m}^{(j)} fulfill the conditions:

(c_{1})  α_{m}^{(j)} = 0 j ∈ N, ∀m ∈ N\{j, j + 1},

(c_{2})  α_{j}^{(j)} = 1 ∀j ∈ N,

(c_{3})  0 ≤ α_{j+1}^{(j)} ≤ 1 ∀j ∈ N.

Then we have (a) ⇔ (b) ⇔ (c). Proof. For (a) ⇔ (b) we apply theorem 2, (a)⇔ (b′), for Y = X, f_{1}(x) = x, f_{2}(x) = xg, ∀x ∈ X. We check that the assertion (b) of theorem 5 is equivalent to the assertion (b′) of theorem 2. We remark that for any x = ∑_{m=0}^{∞} α_{m}x_{m} ∈ X_{+}, the relation f_{1}(x) := x ≤ f_{2}(x) := xg is true. Indeed, we have:

(xg)(z) = x(z)g(z) = ( ∑_{m=0}^{∞} α_{m}z^{m}) (1 + z) = ∑_{0}^{∞} α_{m}z^{m}+

+ ∑_{0}^{∞} α_{m}z^{m+1} = ∑_{m=0}^{∞} α_{m}z^{m}+

+ ∑_{m=1}^{∞} α_{m−1}z^{m} = α_{0} + ∑_{m=1}^{∞}(α_{m} + α_{m−1})z^{m}.

So, we have got:

xg = α_{0}x_{0} + ∑_{m=1}^{∞}(α_{m} + α_{m−1})x_{m} ≥ α_{0}x_{0} + ∑_{m=1}^{∞} α_{m}x_{m}

since α_{m−1} ≥ 0 ∀m ∈ {1, 2, ...} by the definition of X_{+} ∋ x. So, we have only to verify the equivalence (2.6) ⇔ (b) of theorem 5. Let us write (2.6) in our particular case. Let n ∈ N, {λ_{0}, λ_{1}, ..., λ_{n}} ⊂ R. Then (2.6) may be written in the following way:

∑_{j=0}^{n} λ_{j}y_{j} = ∑_{j=0}^{n} λ_{j} (∑_{m=0}^{∞} α_{m}^{(j)}x_{m}) = ∑_{m=0}^{∞} (∑_{j=0}^{n} λ_{j}α_{m}^{(j)}) x_{m} ≤
≤ f_{2}(((∑_{j=0}^{n} λ_{j}x_{j})^{+}) − f_{1} ((∑_{j=0}^{n} λ_{j}x_{j})^{−}) =
= f_{2} (∑_{j=0}^{n} λ_{j}^{+}x_{j}) − f_{1} (∑_{j=0}^{n} λ_{j}^{−}x_{j}) =
= (∑_{m=0}^{n} λ_{m}^{+}x_{m}) (1 + x_{1}) − ∑_{m=0}^{n} λ_{m}^{−}x_{m} =
= ∑_{m=0}^{n} λ_{m}^{+}x_{m} + ∑_{m=0}^{n} λ_{m}^{+}x_{m+1} − ∑_{m=0}^{n} λ_{m}^{−}x_{m} =
= ∑_{m=0}^{n} λ_{m}x_{m} + ∑_{m=1}^{n+1} λ_{m−1}^{+}x_{m} =
= ∑_{m=1}^{n} (λ_{m−1}^{+} + λ_{m}) x_{m} + λ_{0}x_{0} + λ_{n}^{+}x_{n+1}.

By the definition of X_{+}, this is equivalent to (b) of theorem 5.

(a) ⇒ (c) Since x_{j} ∈ X_{+}, we have from (a):

x_{j} = f_{1}(x_{j}) ≤ ∑_{m=0}^{∞} α_{m}^{(j)}x_{m} = y_{j} = f(x_{j}) ≤ f_{2}(x_{j}) = x_{j}(1 + x_{1}) = x_{j} + x_{j+1}.

By the definition of X_{+}, (c) follows.

(c) ⇒ (b)

(b_{1})  ∑_{j=0}^{n} λ_{j}α_{0}^{(j)} = λ_{0}α_{0}^{(0)} + ∑_{j=1}^{n} λ_{j}α_{0}^{(j)} = λ_{0},

by (c_{1}) and (c_{2}).

(b_{2})  m ∈ {1, 2, ..., n} ⇒ ∑_{j=0}^{n} λ_{j}α_{m}^{(j)} = λ_{m}α_{m}^{(m)} + λ_{m−1}α_{m}^{(m−1)} =

= λ_{m} + λ_{m−1}α_{m}^{(m−1)} ≤ λ_{m} + λ_{m−1}^{+}

by (c_{1}), (c_{2}) and (c_{3}).

(b_{3})  ∑_{j=0}^{n} λ_{j}α_{n+1}^{(j)} = λ_{n}α_{n+1}^{(n)} ≤ λ_{n}^{+}, by (c_{1}) and (c_{3}).

(b_{4})  ∑_{j=0}^{n} λ_{j}α_{m}^{(j)} = 0

if m ≥ n + 2, by (c_{1}). The theorem is proved.

References

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[2] N.I.Ahiezer, The classical moment problem and some related questions in analysis, Oliver and Boyd, Edinburgh and London, 1965.

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[14] O.Olteanu, Théorèmes de prolongement d’opérateurs linéaires, Rev. Roumaine Math. Pures et Appl. XXVIII, 10(1983), 953-983.

[15] O.Olteanu, Application de théorèmes de prolongement d’opérateurs linéaires au problème des moments et à une généralisation d’un théorème de Mazur-Orlicz, C.R.Acad. Sci. Paris 313 (1991), 739-742.

[16] O.Olteanu, Applications of the solution of the abstract moment problem to the classical moment problem, St. Cerc. Mat. (Mathematical Reports), to appear.

[17] O.Olteanu, Applications of a general sandwich theorem for operators to the moment problem, Rev. Roum. Math. Pures et Appl., 41, 7-8(1996), 513-521.

[18] G.Păltineanu, Elements of approximation theory of continuous functions (Theorems of Stone-Weierstrass type), Romanian Academy Publishing House, Bucharest, 1982 (in Romanian).

[19] W.Rudin, Real and complex analysis, New York, Mc.Graw-Hill, 1966.

[20] H.H.Schaefer, Topological vector spaces, Macmillian Company, New York, London, 1966.

Politehnica University of Bucharest Department of Mathematics I Splaiul Independenţei 313 77206 Bucharest, ROMANIA