Cone Normed Linear Spaces

Let V be a real vector space and ( • ) be a norm defined on V . Our aim is to study cone normed linear spaces. First we go through the definition of Huang and Zhang [1]; Let E be a Banach space and C be a non empty subset of E. C is called a cone if and only if (i) C is closed. (ii) a, b ∈ R, a, b ≥ 0, x, y ∈ C ⇒ ax + by ∈ C (iii) x ∈ C orx ∈ C for every x ∈ E. Now we define the order relation ′ ≤ ′ on E. For any given cone C ⊆ E, we define linearly ordered set C with respect to ′ ≤ ′ by x ≤ y if and only if yx ∈ C. x < y will indicate x ≤ y and x = y; while x ≪ y will stands for yx ∈ intC, intC denotes the interior of C. The cone C is called normal cone if there is a number K > 0 such that for all x, y ∈ E, 0 ≤ x ≤ y ⇔ x ≤ K y .

The least positive number satisfying the above inequality is called the normal constant of C.

Unless otherwise stated throughout this paper we shall denote θ as the null element of E.

Definition 2.1 Let V be a vector space over the field R. The mapping • c : V -→ E is said to be a cone norm if it satisfies the following conditions:

Proof. Let {x n } n converges to x. For every real p > 0, choose ǫ ∈ E with ǫ ≫ 0 such that K ǫ < p. Then there exist a positive integer n 0 such that Lemma 2.9 Let x, y, ǫ 1 , ǫ 2 ∈ E such that x ≪ ǫ 1 and y ≪ ǫ 2 then

Hence by corollary 2.8 we have

Theorem 2.10 In a cone normed linear space (V, • c ), if x n -→ x and y n -→ y then x n + y n -→ x + y Proof. By lemma 2.9 the theorem directly follows.

Theorem 2.11 In a cone normed linear space (V, • c ), if x n -→ x and real λ n -→ λ then λ n x n -→ λx.

Theorem 2.12 In a cone normed linear space (V, • c ), if {x n } n and {y n } n are cauchy sequences then {x n + y n } n is a cauchy sequence.

Proof. By lemma 2.9 the theorem directly follows.

Theorem 2.13 In a cone normed linear space (V, • c ), if {x n } n and {λ n } n ∈ R are cauchy sequences then {λ n x n } n is a cauchy sequence.

Definition 2.14 Let (U, • c ) and (V, • c ) be two cone normed linear spaces and f : U -→ V be a function, then f is said to be cone continuous at a point

Lemma 2.15 Let V be a cone normed linear space and x, y ∈ V then x cy c ≤ xy c and y c -

Hence cone norm function is cone continuous. Definition 2.18 A cone normed linear space (V, • c ) is said to be cone complete if every cauchy sequence in V converges to a point of V .

Theorem 2.19 Let (V, • c ) be a cone normed linear space such that every cauchy sequence in V has a convergent subsequence then V is cone complete.

Proof. Let {x n } n be a cauchy sequence in V and {x n k } k be a convergent subsequence of {x n } n . Since {x n } n is a cauchy sequence then for any ǫ 1 ∈ E with ǫ 1 ≫ θ there exists a positive integer n 1 such that

Let {x n k } k converges to x. then for any given ǫ 2 ∈ E with ǫ 2 ≫ θ there exists a positive integer n 2 such that

by lemma 2.9). Hence the proof.

Theorem 2.20 Let (V, • c ) be a cone normed linear space and C be a normal cone with normal constant K. Then every subsequence of a convergent sequence is convergent to the same limit.

Proof. Let {x n } n be a convergent sequence in V and converges to the point

Since x n converges to x, then for this ǫ ∈ E ∃ a positive integer n 0 such that

If possible let {x n k } k converges to y also. Then ∃ a positive integer n 1 such that

Let n 2 = max{n 0 , n 1 }. Now

This implies that xy c = 0. Hence the proof.

3 Finite dimensional cone normed linear spaces Lemma 3.1 Let {x 1 , x 2 , …, x n } be a linearly independent subset of a cone normed linear space (V, • c ). C be a normal cone with normal constant K, then there exist an element c ∈ intC such that for every set of real scalars α 1 , α 2 , …, α n we have

then each α i is zero and hence (1) is true. So we now assume that α > 0. Then (1) becomes

It is sufficient to prove that there exists an element c ∈ intC such that (2) is true for any set of scalars β 1 , β 2 , …, β n with n i=1 |β i | = 1 If possible let this is not true. Then there exists a sequence {y m } m ∈ V where

where n i=1 |β i | = 1 by theorem 2.10 and theorem 2.11. This implies that not all β i can be zero. Since {x 1 , x 2 , …, x n } is linearly independent. Therefore y = θ V . Now we show that y n,m → y implies y n,m c → y c For every real ǫ > 0, choose c ∈ E with c ≫ θ and K 2 c < ǫ. Since y n,m → y as m → ∞,then for this element c we can find a positive integer n 0 such that y n,m -

As {y n,m } m is a subsequence of {y m } m and y m c → θ as m → ∞ Therefore y n,m c → θ as m → ∞ and so y c = θ which gives y = θ V . This contradiction proves the lemma. Theorem 3.2 Every finite dimensional cone normed linear space with normal constant K is cone complete.

Proof:Let (V, • c ) be a cone normed linear space and C be a normal cone with normal constant K. Let {x n } be an arbitrary cauchy sequence in V .

We should show that {x n } converges to some element x ∈ V . Suppose that the dimension of V is m and let {e 1 , e 2 , …, e m } be a basis of V . Then each {

…(Full text truncated)…

This content is AI-processed based on ArXiv data.