De Branges Type Lemma and Approximation in Weighted Spaces

The purpose of this paper is to give some generalizations of de Branges Lemma for weighted spaces to obtain different approximation theorems in weighted spaces for algebras, vector subspaces or convex cones. We recall that the (original) de Branges Lemma (Proc Am Math Soc 10(5):822–824, 1959) was demonstrated for continuous scalar function on a compact space while, the weighted spaces are classes of continuous scalar functions on a locally compact space (e.g. the space of function with compact support, the space of bounded functions, the space of functions vanishing at infinity, the space of functions rapidly decreasing at infinity).


Introduction
Using two fundamental tools in functional analysis: Hahn-Banach and Krein-Milman theorems, in 1959, Louis de Branges [2] give a nice proof of Stone-Weierstrass theorem on algebras of real continuous functions on a compact Hausdorff space.
In this paper, we present, in a natural way, the duality theory for general weighted spaces, i.e. a class of scalar continuous functions on a locally compact space (Lemma 2.1, Theorem 2.1).
Also we extend de Branges Lemma in this new setting: Lemma 3.2 for linear subspaces and Lemma 5.1 for convex cones of a weighted space.
We mention that a characterization of the dual of a weighted space was obtained by Summers in [10], in the particular case V ≤ C + (X), i.e. for any v ∈ V there is w ∈ C + (X) such that v ≤ w and a version of de Remark 1.1. a) If for any point x ∈ Xthere exists v x ∈ V such that v x (x) > 0, then the weighted space CV 0 (X) is a locally convex Hausdorff space. b) If for any point x ∈ X there are v x ∈ V and r ∈ R, r > 0 such that the set is a neighborhood of x in X, then the locally convex spaces (CV 0 (X), ω V ) is a complete space. Further, we mention some particular weighted spaces: a) If V = {1} then CV 0 (X) = C 0 (X)-the space of continuous functions vanishing at infinity and the weighted topology ω V coincide with the uniform convergence topology. b) If V = C + 0 (X) then CV 0 (X) = C b (X)-the space of continuous bounded functions on X and the weighted topology ω V coincide with the strict topology β. c) Let X = R n , and let P n be the set of all polynomials defined on R n with values in K.
If V = {|p|; ∀p ∈ P n }, then CV 0 (R n ) coincide with the space of functions rapidly decreasing at infinity.

Duality for Weighted Spaces
In this part, X will be a Hausdorff locally compact space, K(X) will be the space of all continuous real or complex functions on X with compact support.
Obviously K(X) ⊂ CV 0 (X) and it is well known that the space K(X) is dense in CV 0 (X), with respect to the weighted topology ω V , i.e. K(X) = CV 0 (X).
Hence, any continuous functional on the vector space K(X) endowed with the trace of ω V on it (denoted also by ω V ) may be uniquely extended to a continuous linear functional on the locally convex space (CV 0 (X), ω V ) and, therefore, the locally convex spaces (K(X), ω V ) and (CV 0 (X), ω V ) have the same dual.
Further, we represent any element θ of this dual under the form: where μ is the corresponding scalar measure defined on the set B(X) of all Borel subsets of the locally compact space X.
To this purpose, for any compact subset Kof X, let us denote by K(X, K) the set of all continuous scalar functions on X which vanishes outside K. Obviously, we have On the vector space K(X), there are three remarkable topologies: the topology of uniform convergence denoted by τ u , the weighted topology ω V and the 120 Page 4 of 19 I. Bucur and G. Paltineanu MJOM inductive topology, denoted by τ ind , i.e. the finest locally convex topology on K(X) making continuous the injection maps: for all compact subsets K of X,where each subspace K(X, K) is endowed with the uniform topology. What is important here is the fact that giving an arbitrary locally convex space E and a linear map then T is continuous iff the restriction of T to any K(X, K) is continuous. It is no difficult to show that we have and, therefore, Any element θ ∈ (K(X), τ ind ) * is usually called a Radon integral on X. This means that θ is a scalar continuous map whose restriction to any K(X, K) is continuous with respect to the uniform topology, i.e. for each compact subset K of X there exists a number a K > 0 such that The linear, scalar map θ defined on K(X) belongs to (K(X), τ u ) * iff there exists a number a > 0 such that We recall that for any Radon integral θ on X there exists a smallest positive Radon integral, denoted by |θ|, such that In fact, for any f ∈ K + (X), we have Also, any real, positive (on K(X)) functional is a Radon integral on X and for any real, Radon integral θ on K(X), there exist two positive Radon integral θ + and θ − such that As usually, a countable additive positive map μ defined on B(X) is called a Radon measure on X if for any Borel set A ∈ B(X), we have μ(A) = sup{μ(K); K compact, K ⊂ A} and μ(K) < ∞ for any compact subset K of X.
A countable additive scalar map on B(X) is called a Radon measure on X if the positive measure |μ| is a Radon measure.
If μ is a positive Radon measure on X, then for any lower semicontinuous positive function ϕ on X, we have Obviously any function f ∈ K(X) is integrable with respect to μ and for any compact K ⊂ X and any f ∈ K(X, K), we have Let us denote by M(X) the set of all Radon measure on X. From the above considerations for any μ ∈ M + (X), the map is an element of the dual (K(X), τ ind ) * . Conversely, for any θ ∈ (K(X), τ ind ) * + (θ ≥ 0 on K(X)) there exists and it is uniquely determined a positive Radon measure μ θ , such that In the above correspondence θ → μ θ between (K(X), τ ind ) * + and M + (X), the elements θ ∈ (K(X), τ u ) * + are those positive Radon integrals for which the associated measures μ θ are finite on X, i.e. μ θ (X) < ∞.
If θ is a Radon real integral on X then |θ|, θ + , θ − are positive Radon integrals on X and we extend the above correspondence, associating with θ the real sign measure μ θ := μ θ+ − μ θ− . We have also μ |θ| = |μ θ | = μ θ+ + μ θ− . A similar extension may be done if K(X) is the set of all continuous complex functions with compact support in X.
In the sequel, we shall freely use the term Radon measure instead of Radon integral or conversely.
If μ is a Radon measure on X and f : X → R is μ-integrable then the map fμ given by is also a Radon measure on X and we have |fμ| = |f ||μ|. We remark also that if μ is a Radon measure such that 1 v d|μ| < ∞ for some weight v ∈ V then CV 0 (X) ⊂ L 1 (μ).
The following results concerning weighted spaces are known, in the particular case V ≤ C + (X) (see Summers [10]). We present them in a more general frame using different proof ideas.

Lemma 2.1.
Let v be an upper semicontinuous non-negative real function on X, let p v be the seminorm on K(X) associated with v, i.e.
, ∀x ∈ X , We have a) The dual of the locally convex space Proof. Indeed, if we consider a linear map θ : K(X) to K, which is continuous with respect to the seminorm p v : K(X) → K, then we have Since for any compact set K the lower semicontinuous function 1 v : X → (0, ∞] has a strictly positive infimum α K on K, we deduce that for any function ϕ ∈ K(X) which vanishes outside K, we have Hence, θ is a Radon measure on X. From the definition of the positive measure |θ|, we have, for any The compactness of the set B 0 v : with respect to the weak topology on M(X) given by the duality (f, μ) → μ(f ) defined on K(X) × M(X) follows now from Alaoglu's theorem applied to the locally convex space K(X) endowed with the seminorm p v .
Theorem 2.1. Let V be a Nachbin family on X and let CV 0 (X) be the weighted space associated with the family V , endowed with weighted topology ω V . Then the dual CV 0 (X) * of the locally convex space (CV 0 (X), ω V ) is identical with the dual of the space K(X) endowed with the induced ω V -topology. More precisely θ ∈ CV 0 (X) * iff there exist a Radon measure μ on X and v ∈ V such that In the other words, follows that any element θ ∈ CV 0 (X) * is totally determined by its restriction to K(X). On the other hand, there exists a weight v ∈ V and α ∈ R + such that Hence, the restriction of θ to K(X) satisfies the same inequality and therefore, using Lemma 2.1, there exists a Radon measure μ on X such that Since any function g ∈ CV 0 (X) is dominated at infinity by 1 v and 1 v is strictly positive on X, we deduce that there exists β ∈ R + such that |g| ≤ β · 1 v , onX and, therefore, g ∈ L 1 (|μ|). Moreover, there exists a sequence (f n ) n ∈ K(X) such that and so the proof is finished.

Lemma De Branges for Weighted Spaces
In this section, for any Radon measure μ on X, we shall use the notation σ(μ) for the support of μ-the smallest closed subset F of X for which |μ|(X\F ) = 0.
If the sequence (f n ) n is pointwisely convergent on F to a function g, then g ∈ L 1 (L) and lim n→∞ L(f n ) = gdL.
Particularly, we have and, therefore, the sequence (f in ) n is pointwisely convergent to g on the set and |L|(X\F ) = 0, |L|([v = 0]) = 0 we deduce that the function g belongs to L 1 (|L|) and we have The assertion b) has a similar proof using Lebesgue domination theorem.
In the sequel, for any v ∈ V , respectively, any μ ∈ [CV 0 (X)] * , we shall use the notations: respectively, σ(μ) the support of μ. Also, for any linear subspace W ⊂ CV 0 (X), we denote by W 0 its polar set, i.e.
and for any convex set S ⊂ [CV 0 (X)] * , we denote by Ext(S) the set of all extreme points of S.
Corollary 3.1. Let W ⊂ CV 0 (X) be a linear subspace and let (P i ) i∈I be a partition of X such that for any v ∈ V and any μ ∈ Ext The following result is a generalization of de Branges lemma. In the particular case V ≤ C + (X), this result was obtained by Prolla [5].
for some v ∈ V and let f be a real valued continuous and bounded function on σ(μ) such that μ(f · w) = 0, ∀w ∈ W. Then f is constant on σ(μ).
Proof. Let n, m ∈ N * be sufficiently large such that Obviously the function g = 1 n · 1 − 1 m · f has the same properties like f but 0 < g < 1 on σ(μ).
We denote also by g the positive Borel extension of g on X such that g = 0 on X\σ(μ). We consider the Radon measures μ 1 , μ 2 on X given by Using Lemma 2.1 and Theorem 2.1, we have for any Radon measure λ on X: and similarly μ 2 v = 1. On the other hand, we have μ 1 (w) = μ 2 (w) = 0 for all w ∈ W and, therefore, Hence, g α · μ = μ and so g = α on σ(μ). Therefore, the function f = m · (1 − n · g) is constant on σ(μ) and so the proof is finished.

Some Approximation Theorems in Weighted Spaces
We denote by S the family of all subsets of X which are antisymmetric with respect to A. Obviously, S = φ because for any x ∈ X, the set {x} ∈ S.
Remark 4.1. The family S has the following properties: (ii) The closure S of any S ∈ S belongs to S. (iii) Any element x ∈ X belongs to a maximal (with respect to the inclusion order relation) element of S denoted by S x . (iv) For any x, y ∈ X, we have one or other of the relations:

Theorem 4.1. Let A be a nonempty subset of C(X, C) such that any element a ∈ A is a bounded function on the set
Proof. First, we show that for any v ∈ V and any extreme element μ ∈ and, therefore, any function a ∈ A is bounded on σ(μ). We have μ ∈ W 0 and μ(a · w) = 0 for any a ∈ A and any w ∈ W. Using Lemma 3.2, we deduce that any element a ∈ A which is real on σ(μ) is constant on σ(μ). Therefore, σ(μ) ∈ S and so there exists x μ ∈ X such that σ(μ) ⊂ S xµ .

Remark 4.2. If A ⊂ C(X, C) is a self-adjoint algebra then any antisymmetric subset with respect to A is a set of constancy for A. Particularly for any
x ∈ X, we have Indeed, any element a ∈ A is of the form a = a + i · a , where a , a are real functions on X.
Since a = a+a 2 ∈ A, a = a−a 2·i ∈ A, we deduce that a and a are constant on any antisymmetric set with respect to A and, therefore, a is constant on any such set.
From previous remark, and Theorem 4.1, it follows: Nachbin [4] Let A be a subalgebra of C b (X), self-adjoint in the complex case, and let W ⊂ CV 0 (X) be a linear subspace such that A · W ⊂ W.
Then W is localizable with respect to A, i.e.
Corollary 4.1. Let A be a subalgebra of C b (X) containing the constant function 1, separating the points of X, and self-adjoint in the complex case. If W ⊂ CV 0 (X) is a linear subspace such that A · W ⊂ W and for any x ∈ X there exists a w ∈ W such that w(x) = 0, then W is dense in CV 0 (X), i.e. W = CV 0 (X).

Proof. Since A separates the points of X it follows that [x] A = {x}, ∀x ∈ X.
Let f ∈ CV 0 (X) be arbitrary. If f (x) = 0, then obviously If f (x) = 0, then, there exists w ∈ W such that w(x) = 0. Further, we have Therefore, we have Now, from Theorem 4.2, we deduce W = CV 0 (X).

Definition 4.2.
Let M ⊂ C(X, R) and W ⊂ CV 0 (X) be two nonempty subsets. A subset S of X will be called antialgebraic with respect to the pair (M, W) if any element m ∈ M such that m · w|S ∈ W|S, ∀w ∈ W is constant on S.
If we denote by T the family of all antialgebraic subsets of X with respect to the pair (M, W) then for any x ∈ X the singleton {x} belongs to T. The set T endowed with the inclusion order relation has similar properties as family S in the Remark 4.1. For any x ∈ X we denote by T x the maximal (M, W)antialgebraic subset containing x. We have X = x∈X T x and {T x } x∈X is a partition of X.
where for any x ∈ X, T ix denotes the maximal element from T i containing the point x.
Proof. Applying Corollary 3.1, it will be sufficient to show that for any v ∈ V and any μ ∈ Ext(B 0 v ∩ W 0 ) the set σ(μ) is included in T x0 for some x 0 ∈ X or equivalently to show that σ(μ) is an antialgebraic set with respect to the pair (M, W). Let m ∈ M be such that m · w|σ(μ) ∈ W|σ(μ) for all w ∈ W. Using Lemma 3.2, we deduce that m is constant on σ(μ). Hence, σ(μ) is an antialgebraic set with respect to the pair (M, W) and the proof is finished.
In the particular case V ≤ C + (X), this result was obtained in [8].
Proof. The assertion follows from Theorem 4.3 since any function of M is V -bounded.
where for any x ∈ X we have denoted by   then we have Proof. Since M · W ⊂ W and M separates the points of X it follows that On the other hand, if we denote W(X) = {w(x); w ∈ W}, then we have We finish the proof applying Corollary 4.2.
Let S ⊂ X be a closed set and let I S = {f ∈ CV 0 (X); f |S = 0}. Obviously, I S is an order ideal of CV 0 (X). We remark also that I S is closed with respect to the weighted topology. Indeed, let x 0 ∈ S be arbitrary and let v 0 ∈ V be a weight with the property v 0 (x 0 ) > 0. If g ∈ I S , then for any ε > 0 and any v ∈ V , there exists f ∈ I S such that In the particular case x = x 0 and v = v 0 , it results As ε > 0 is arbitrary, we deduce that g(x 0 ) = 0, so g ∈ I S . In [3], Lemma 3.8, C.Portenier states that any closed order ideal of CV 0 (X, R) has the preceding form. Using Corollary 4.2, we give a very simple proof of Portenier's result.
is a bijection between I and F just a decreasing one: This allows us to generalize some results involving different type of closed subset of X(antisymmetric, interpolating, antialgebraic sets) to the abstract case of closed order ideals in a locally convex lattices.

Stone-Weierstrass Theorem for Convex Cones in Weighted Spaces
In this section, we consider a convex cone C ⊂ CV 0 (X, R) and we denote by C 0 its polar set, i.e.
For any weight v ∈ V , we denote by Ext{B 0 v ∩ C 0 } the set of all extreme points of the compact convex setB 0 v ∩ C 0 of the dual [CV 0 (X, R)] * of the locally convex space (CV 0 (X, R), ω V ). We remember that the closure of any convex cone in an arbitrary locally convex space coincides with its bipolar with respect to the natural duality. In our case, The following result is a generalization of de Branges Lemma for a convex cone.
denotes the support of the Radon measure μ, then any function we have ϕ = 0 or ϕ = 1 on σ(μ). We suppose now that |μ|(ϕ) = 0 and |μ|(1 − ϕ) = 0 and we consider the measures μ 1 , μ 2 given by Further, we have Since ϕ · h|σ(μ) ∈ C|σ(μ) and (1 − ϕ) · h|σ(μ) ∈ C|σ(μ), ∀h ∈ C and using Lemma 3.1, having in mind that μ ∈ C 0 , we deduce Theorem 5.1. A function f ∈ CV 0 (X) belongs to the closure C of the convex cone C in the locally convex space (CV 0 (X, R, ω V ) if and only if for any weight v ∈ V and any μ ∈ Ext{B Since the following map: is a continuous affine function if we endow the set B 0 v ∩ C 0 with the trace of the weak topology on [CV 0 (X, R)] * and the maximum of this map is realized The following statement is a procedure to describe the closure of a convex cone in some circumstances.
Corollary 5.1. Let C ⊂ CV 0 (X, R) be a convex cone and let (P α ) α∈I be a partition of X such that for any v ∈ V , and any μ ∈ Ext{B 0 v ∩ C 0 }, there exists α ∈ I such that the support of μ,σ(μ) ⊂ P α . Then we have Further, we state such kind of such circumstances. A nonempty subset M of C(X, [0, 1]) is said to be a set with complement if for any ϕ ∈ M, we have 1 − ϕ ∈ M.
The following definition is analogous with Definition 4.6. of [9] Definition 5.1. A subset S ⊂ X is called antisymmetric with respect to the pair (M, C) if any function ϕ ∈ M with the properties: is a constant function on S.
Further, we denote by B the family of all subsets of X antisymmetric with respect to the pair (M, C). The following assertions are almost obvious.
For any x ∈ Xm, we denote B x = ∪{B; B ∈ B, x ∈ B}. We have The family (B x ) x∈X is a partition of X and for any B ∈ B, there exists , ∀f ∈ C, we deduce from Lemma 5.1, that ϕ is a constant function on σ(μ).
Theorem 5.2. Let X be a locally compact Hausdorff space, V be a Nachbin family of weights on X, C ⊂ CV 0 (X, R) be a convex cone and M ⊂ C(X, [0, 1]) be a nonempty subset with complement. Then where (B x ) x∈X is the family of all maximal subsets of X antisymmetric with respect to the pair (M, C).
Proof. The assertion follows from the above remark and from the Corollary 5.1. First, this result was obtained in the case V ≤ C + (X) in [7]. See also [6] for compact spaces.
Proof. Since M is a set with complement and M · C ⊂ C we deduce that all functions from M are constant on any antisymmetric set with respect to the pair (M, C) and for any x ∈ X the set [x] M is antisymmetric set with respect to the pair (M, C). Hence, the set [x] M = B x , ∀x ∈ X.
The assertion follows now from Theorem 5.2.
Then we have Example 5.1. Let X = R and let ϕ : R → (0, 1) be the strictly continuous homeomorphism given by ϕ(x) = 2 π · arctg(e x ). On the space C b (R) of all real bounded and continuous functions on R, we consider the strict topology β given by the family of seminorms: where g runs the set C 0 + (R) of all positive, continuous functions on R vanishing at infinity. In fact β = ω V , where the Nachbin family of weights V is just C 0 + (R). We know that CV 0 (R) = C b (R). a ij · x i · y j , a ij ≥ 0, n ∈ N ⎫ ⎬ ⎭ .
Obviously, C separates the points of Xsince ϕ ∈ C. Using the notations from Corollary 5.3, we have X 0 = φ, X + = R and, therefore, C = C b + (R), i.e. for any f ∈ C b + (R) and any g ∈ C 0 + (R), there exists a sequence (P k ) k of polynomials P k (x, y) = n k i,j=1 a ij k · x i · y j , a ij k ≥ 0 such that the sequence (g · P k (ϕ, 1 − ϕ)) k converges uniformly to g · f on R.
Example 5.2. Let X = (0, 1)and let V = C 0 + (X) be the set of all positive, continuous functions v on X vanishing at infinity, i.e. lim x→0 v(x) = 0 = lim x→1 v(x). On the space C b (X)of all real continuous functions on X we consider the strict topology β given by the family of seminorms: In fact, β = ω V and as it is known we have CV 0 (X) = C b (X). Let us consider the convex cone C in C + b (X) of all functions of the form: x ∈ X → c(x) = n i,j=1 where n runs the set N * , a i,j ∈ R, a i,j ≥ 0. The required conditions of Corollary 5.3 are satisfied if we take M = {x, 1 − x}. Hence, C = C b + (X), i.e. any continuous, positive and bounded function on X may be approximated with functions of the form n i,j=1 a ij · x i · (1 − x) j , a i,j ≥ 0. Example 5.3. On the space X = (0, 2), we consider V = C + 0 (X), C b (X), p v with v ∈ V , as in the previous example and we consider also the convex cone C of all functions of the form: x ∈ X → c(x) = n i,j,k=1 a ijk · x i · (1 − x) 2j−1 · (2 − x) k , n ∈ N * , a i,j,k ≥ 0. (1) We take M = 1 2 · x, 1 − 1 2 · x and we have, using the notations of Corollary 5.3, X + = (0, 1], X − = [1, 2) and, therefore, any continuous, real, bounded function on (0, 2) such that f (x) ≥ 0, ∀x ∈ (0, 1), f(1) = 0, f(x) ≤ 0, ∀x ∈ (1, 2) may be approximated with respect to the strict topology on the open interval (0, 2) by functions of the form (1).