Applied Mathematical Sciences, Vol. 7, 2013, no. 134, 6659 - 6668 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2013.310544 Construction of a Cone by Using Weak*- Total Families Nashat Faried Mathematics Department, Faculty of Science Ain Shams university, Cairo, Egypt Hossam A Ghany Mathematics Department, Faculty of Science Taif University, Taif, Saudi Arabia Mathematics Department, Faculty of Industrial Education Helwan university, Cairo, Egypt Copyright © 2013 Nashat Faried and Hossam A. Ghany. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract This paper is devoted to construct a cone by using the so called weak*-total families in different spaces ( Banach space, Hilbert space, metric space,...etc.). We were advised some forms for weak*-total subsets to construct an exact forms for cones like ( Normal cones, Allows plastering cones, Solid cones,...etc.). Keywords: cone; weak*-total subsets 1. Introduction In the study of various problems in general system theory, mechanics , operator theory ... etc, an important rule is played by monotonicity of the corresponding " inputs" and "outputs", and by the positiveness or monotonicity of various linear and nonlinear mappings. The mathematical models of such objects lead in many cases to 6660 Nashat Faried and Hossam A. Ghany operator equations in spaces, which partially ordered by some cone. The study of ordered linear spaces was initiated in the last 1950's by Riesz. Many authors like M. G. Krein, who worked on the geometry of cone in Banach space and has given proofs of main theorems for cone construction in his famous paper [ 1], there are many relations between cones, the class of positive definite functions, the class of negative definite functions, the class of completely monotone functions and the class of completely alternating functions [2-5]. In this paper, we will discuss the relations between the construction of cones and the families of weak*-total subset. This paper is devoted to give an exact method to construct cones using a weak*-total families. Then we will search some types of weak*-total families that grantees the implementation of definite cones such as solid cone or allows plastering cone or normal cone...etc. 2. Weak*-total subset Definition 2.1 : (Total subset) [7] A subset S of a topological vector space E is said to be total in E if each continuous linear from on E satisfying f (x) = 0 for all x belonging to S necessarily satisfies f = 0 . We suggest to call this notion of totality "weak totality" and parallel we give the following definition of weak*-totality. Definition 2.2: (weak*-total subset) We will call a subset F of the dual space E * of a topological vector space E weak *-total if for each element χ belonging to E and satisfying f (χ ) = 0 for all f in F necessarily satisfies χ = 0 . Using a weak*-total subset F of the dual space E * of a Banach space E we define a partial ordering on E associated with F as follows x ≤ F y if and only if f (x) ≤ f ( y) for all f ∈ F Remark. This partial ordering is well defined. 1- Reflexivity of the relation " ≤ F " is trivial. 2- The relation " ≤ F " Is anti-symmetric (we remark that, the weak*- totality guarantees the anti symmetricity of the above defined partial ordering). In fact, let x ≤ F y and y ≤ F x Construction of a cone by using weak*- total families 6661 Hence, f (x) ≤ f ( y) and f (x) ≤ f ( y) for all f ∈ F So, for all f ∈ F f (x) = f (y) hence , for all f ∈ F f (x − y) = 0 Since the family F is weak*-total then, x − y = 0 Hence x = y . 3- The transitivity of the relation " ≤ F " . In fact, x ≤ F y and y ≤ F z Implies that f (x) ≤ f ( y) and f (y) ≤ f (z) for all f ∈ F Hence, for all f ∈ F f (x) ≤ f (z) So x ≤ F z . Example 2.1 ∞ The family F = {en : n ∈ N} is a weak*- total family in the space l of all bounded sequences which is the dual space of l1 of all absolutely summable sequences . In fact , ∞ < x , en >= xn = 0 for all en ∈l implies that x = 0. we remarks that " The family F = {en : n ∈ N} is weak*- total in l∞ , but F is not abases for l∞ (l∞ being not separable i.e., l∞ has no Schauder bases)". The partial ordering associated to F in this case is defined as follow x ≤ F y if and only if xi ≤ yi for all i ∈ N This is the well known ordering in l1. Remark. On a Banach space E with a cone K a partial ordering is introduced in the following manner : x ≤ F y if and only if y − x ∈ K. Definition 2.3: Using a weak*- total subset F we can define a corresponding cone K F as follow: K F = {z ∈ X : f (z) ≥ 0 for all f ∈ F} We will call K F the cone generated by the weak*-total subset F . 6662 Nashat Faried and Hossam A. Ghany Theorem 2.1. Let F be a bounded, closed and convex subset of a normed space E , which does not contain the zero of E . Then the set K (F ) = {x ∈ E : x = t z where z ∈ F and t ≥ 0} is a cone . ∞ Proof. First we will prove that K(F) is closed . Actually , let (un )n=1 ∈ K(F) and || un − v ||→ 0 (without loss of generality let v ≠0 ) . Every element un has the representation un = tn zn where zn ∈ F and tn ≥ 0 for all n∈ N . Since F is bounded , then there exist m, M ≥ 0 such that m ≤ || zn || ≤ M for all n∈ N ∞ since (un )n=1 is convergent then it is bounded , so we have || un ||= tn || zn || ≤ C for some C > 0 ∞ ∞ since (un )n=1 is bounded , so (tn )n=1 is bounded, hence there exists a convergent subsequence (t ) ⊆ (t ) and t > 0such that: n j n 0 t ⎯⎯→j→⎯∞ t n j 0 We contract a subsequence (z )∞ of the sequence (z )∞ such that : n j j =1 n n=1 v tn tn v || z − ||=|| z − j z + j z − || n j t n j t n j t n j t 0 0 0 0 1 ≤ {|| z || || t − t || + || t z − v} n j 0 n j n j n j t0 Hence, v || z − || ⎯⎯→n→⎯∞ 0 n j t0 Consequently, v z = z = 0 lim n j n→∞ t0 Since z ∈ F for all n ∈N , then n j j v = t0 z0 , t0 > 0 and z0 ∈ F Hence v ∈ K(F) , so K(F) is closed . Secondly, we will prove that K(F) is a convex set . Actually , let u,v ∈ K (F ) and α , β ≥ 0 , hence there exists t1,t2 ≥ 0 and z1, z2 ∈ F such that : u = t1z1 and v = t2z2 Construction of a cone by using weak*- total families 6663 then α t1 β t2 α u + β v =(α t1 + β t2 ){ z1 + z2} α t1 + β t2 α t1 + β t2 =(α t1 + β t2 ){λ z1 + (1− λ)z2} α t where λ = 1 , so α u + β v∈ K(F) i.e., K(F) is a convex set. α t1 + β t2 Finally, we will prove that K(F) I− K(F) = 0 . Let ω, −ω∈K(F), hence there exists z0 ∈F such that : −t0z0 ∈K(F), for somet0 > 0. Consequently −t0z0 = t1z1 ∈K(F), for somet1 > 0 and z1 ∈F. Therefore t0 t1 θ = z0 + z1 t0 +t1 t0 +t1 which gives a contradiction. Definition 2.3 : Let E be a topological vector space, a cone K ⊆ E allows the plastering K1 ( K1 is another cone in E ), if every point x0 ∈ K lies in the cone K1together with a spherical neighborhood {x :|| x − x0 ||≤ b || x0 ||} , where b > 0 does not depend on x0 ; the cone K1 is called the plastering of K . Theorem 2.2 : Let E be a normed space . A necessary and sufficient condition for a uniformly positive linear functional f on a cone K ⊆ E is that the cone K allows plastering . Proof . Let f be a uniformly positive linear functional on K , hence f (x) ≥ a || x || for all x ∈ K Let N = {x ∈ E; f (x) = 1} It is clear that M = K I N is a closed convex set . Denote by F the set F = {x ∈ N; || x − x* ||≤ 2e} where x* is a fixed element of M and e is a positive number satisfies the inequality : || x − x1 ||≤ e 6664 Nashat Faried and Hossam A. Ghany where x∈ N and x1 ∈ M . Construct a cone K1 such that K1 = K (F ) . We will show that the cone K1 is the plastering of K . Indeed, let x0 ∈ K and e a2 || h ||≤ || x ||. To prove x + h ∈ K it is suffices to show that 2 || f || +ea || f || 0 0 1 x + h x || 0 − 0 ||≤ e f (x0 + h) f (x0 ) x + h x x Since 0 , 0 ∈ N and 0 ∈ M . Hence f (x0 + h) f (x0 ) f (x0 ) e a2 || f (x )h − x f (h) ||≤ 2 || f |||| x || || x || 0 0 0 2 || f || +ea || f || 0 But ef (x0 ){ f (x0 ) + f (h)} ≥ ea || x0 ||{a || x0 || + f (h)} ≥ ea || x0 ||{a || x0 || − || f |||| h ||} e a2 ≥ ea || x ||2 {a − } 0 2 + ea 2e a2 ≥ || x ||2 (2 + ea) || f || 0 Therefore, x + h x || 0 − 0 ||≤ e f (x0 + h) f (x0 ) x0 + h x0 + h So, ∈ F , this implies x0 + h = f (x0 + h) ∈ K(F) , f (x0 + h) f (x0 + h) hence x0 ∈ K is an interior point of K1.