Some results in cone metric spaces with applications in homotopy theory

Definition 2.1

A binary relation ℛ over a set X ≠ ϕ defines a partial order if it has following axioms:

1. ℛ is reflexive;

2. ℛ is antisymmetric;

3. ℛ is transitive.

A set having partial order ℛ is known as a partially ordered set denoted by ( X , ℛ ) .

Let ( ℰ , ∥ ⋅ ∥ ) be a real Banach space. A subset C of ℰ is called a cone if and only if

1. C ≠ ϕ , closed and C ≠ { 0 } ;

2. for all a , b ∈ ℝ with a, b ≥ 0 and σ , ξ ∈ C , we have a σ + b ξ ∈ C ;

3. C ∩ ( − C ) = { 0 } .

Given C ⊂ ℰ , the partial order ≼ with respect to C is defined as follows:

We shall write σ ≺ ξ to indicate that σ ≼ ξ but σ ≠ ξ, while σ ≪ ξ stands for ξ − σ ∈ C ∘ (interior of C ).

The cone C is said to be a normal cone if there exists a positive constant K, such that

Throughout this article, we let X = ( X , ℛ ) and ≼ being a partial order with respect to cone C defined in ℰ . If X ⊆ ℰ , then ℛ and ≼ are identical, otherwise are different.

Definition 2.2

[7] A mapping d : X × X ↦ ℰ is said to be a cone metric if for all σ, ξ, ν, ∈ X the following conditions are satisfied:

1. 0 ≼ d ( σ , ξ ) and d(σ, ξ) = 0 if and only if σ = ξ;

2. d ( σ , ξ ) = d ( ξ , σ ) ;

3. d ( σ , ξ ) ≼ d ( σ , ν ) + d ( ν , σ ) .

The cone metric space is denoted by (X,d).

Definition 2.3

[7] Let ℰ be a real Banach space, (X, d) be a cone metric space, and c ∈ ℰ with 0 ≪ c. A sequence {σ _n } is called a Cauchy sequence if there exists a natural number N ∈ ℕ , such that d(σ _n , σ _m ) ≪ c for all n, m > N. The sequence {σ _n } is said to be convergent if there exists an N ∈ ℕ , such that d(σ _n , σ) ≪ c for all n ≥ N and σ ∈ X.

Example

Let ≼ be the partial order with respect to cone C , as defined in Section 2. Let ( ℰ , ∥ ⋅ ∥ ) be a real Banach space and A : ℰ 6 → ℰ is defined as

Then, the operator A ∈ G :

( A 1 ) . Let υ 5 ≼ γ 5 and υ 6 ≼ γ 6 , then γ 5 − υ 5 ∈ C and γ 6 − υ 6 ∈ C . Now, we show that A ( υ 1 , υ 2 , υ 3 , υ 4 , υ 5 , υ 6 ) − A ( υ 1 , υ 2 , υ 3 , υ 4 , γ 5 , γ 6 ) ∈ C . Consider,

Thus, A ( υ 1 , υ 2 , υ 3 , υ 4 , γ 5 , γ 6 ) ≼ A ( υ 1 , υ 2 , υ 3 , υ 4 , υ 5 , υ 6 ) .

( A 2 ) . Let υ , γ ∈ ℰ be such that 0 ≼ υ , 0 ≼ γ . If A ( γ , υ , υ , γ , γ + υ , 0 ) ≼ 0 , then we have

For if γ = 0, then α υ ∈ C . Thus, there exists T : ℰ → ℰ defined by T ( υ ) = η υ ( 0 ≤ η < 1 2 and η = α is a scalar), such that γ ≼ T ( υ ) . Now, if γ ≠ 0, then − γ + α max { υ , υ , γ , γ + υ , 0 } ∈ C implies α ( γ + υ ) − γ ∈ C , which then gives α υ − (1 − α ) γ ∈ C . Thus, (1 − α ) γ ≼ α υ , so there exists T : ℰ → ℰ defined by T ( υ ) = η υ ( η = α 1 − α is a scalar), such that γ ≼ T ( υ ) .

( A 3 ) . Let υ ∈ ℰ be such that ∥;υ∥; > 0 and consider 0 ≼ A ( υ , 0 , 0 , υ , υ , 0 ) , then υ − α max { 0 , 0 , υ , υ , 0 } ∈ C . This implies α υ ≼ υ , which holds whenever ∥;υ∥; > 0.

Remark 3.1

If T ∈ B ( ℰ , ℰ ) , the Neumann series I + T + T ² +…+ T ⁿ +… converges if ∥;T∥;₁ < 1 and diverges otherwise. Also if ∥;T∥;₁ < 1, then there exists λ > 0, such that ∥;T∥;₁ < λ < 1 and ∥;T ⁿ ∥;₁ ≤ λ ⁿ < 1.

Theorem 4.1

Let (X, d) be a complete cone metric space and C ⊂ ℰ be a cone. Let f: X → X. If there exist T ∈ B ( ℰ , ℰ ) with ∥;T∥;₁ < 1, identity operators I : ℰ → ℰ , and A ∈ G , such that for all comparable elements σ, κ ∈ X

and

1. there exists σ ∈ X, such that σ 0 ℛ f ( σ 0 ) ;

2. for all σ, κ ∈ X, σ ℛ κ implies f ( σ ) ℛ f ( κ ) ;

3. for a sequence {σ _n } with σ _n → x* whose all sequential terms are comparable, we have σ n ℛ x ⁎ for all n ∈ ℕ .

Then, x* = f(x*).

Proof

Let σ ₀ ∈ X be such that σ 0 ℛ f ( σ 0 ) . We construct a sequence {σ _n } by f(σ _n−1) = σ _n . Then, σ 0 ℛ σ 1 . By (4.1), for σ = σ ₀, we have

that is,

By (d3), we have

and so we rewrite (4.2) employing condition ( A 1 ) as follows:

and thus by ( A 2 ) , there exists T ∈ B ( ℰ , ℰ ) with ∥;T∥;₁ < 1, such that

Since σ 0 ℛ σ 1 , assumption (2) implies σ 1 = f ( σ 0 ) ℛ f ( σ 1 ) = σ 2 and by (4.1), for σ = σ ₁, we have

that is,

By (d3), we have

and ( A 1 ) implies

By ( A 2 ) , there exists T ∈ B ( ℰ , ℰ ) with ∥;T∥;₁ < 1, such that

By continuing this pattern, we can construct a sequence {σ _n }, such that

σ n ℛ σ n + 1 with σ _n+1 = f(σ _n ) and

implies

For m , n ∈ ℕ with m > n, consider

Since ∥;T∥;₁ < 1, so T ⁿ → 0 as n → ∞. Thus, lim n → ∞ d ( σ n , σ m ) = 0 , which implies that {σ _n } is a Cauchy sequence in X. Since (X, d) is a complete cone metric space, there exists x* ∈ X, such that σ _n → x* as n → ∞. Thus, there exists a natural number N ₂, such that

We claim that

We assume against our claim that

and

By (d3) and (4.1), we have

Thus,

which is an absurdity. Hence, for each n ≥ 1, we have

and

Assume that ∥;d(x*,f(x*)∥; > 0. By assumption (3), we have σ n ≼ x ⁎ for all n ∈ ℕ , and then by (4.1), we get

or

Letting n → ∞, we have

This is a contradiction to ( A 3 ) . Thus, ∥;d(x*,f(x*)∥; = 0. Hence, d(x*,f(x*)) = 0. It follows from (d1) that x* = f(x*).□

Theorem 4.2

Let (X,d) be a complete cone metric space and f be a self-mapping on X. If for all comparable elements σ, κ ∈ X, there exist T ∈ B ( ℰ , ℰ ) with ∥;T ₁∥; < 1, identity operator I : ℰ → ℰ , and A ∈ G , such that

and

1. there exists σ ₀ ∈ X, such that f ( σ 0 ) ℛ σ 0 ;

2. for any σ, κ ∈ X, σ ℛ κ implies f ( κ ) ℛ f ( σ ) ;

3. for a sequence {σ _n } with σ ₀ → x* whose all sequential terms are comparable, we have σ n ℛ x ⁎ for all n ∈ ℕ .

Then, f has a fixed point in X.

Proof

Let σ ₀ be an initial point in X. Define the sequence {σ _n } by σ _n = f(x _n−1) for all n. By assumption (1), we have σ 1 = f ( σ 0 ) ℛ σ 0 , and then, assumption (2) implies f ( σ 0 ) ℛ f ( σ 1 ) , i.e., σ 1 ℛ σ 2 . By (4.3), we have

or

By (d3), we have

and then using A 1 , we obtain

By ( A 2 ) , there exists T ∈ B ( ℰ , ℰ ) with ∥;T∥;₁ < 1, such that

Since σ 1 ℛ σ 2 , by assumption (2), we get σ 3 ℛ σ 2 , and thus by (4.3)

By (d3), ( A 1 ) , and ( A 2 ) , we get

By continuing the pattern, we construct a sequence {σ _n }, such that

Hence, by the same reasoning as in the proof of Theorem 4.1, we have x* = f(x*).□

Theorem 4.3

Let (X,d) be a complete cone metric space and f be a monotone self-mapping on X. If for all comparable elements σ, κ ∈ X, there exist T ∈ B ( ℰ , ℰ ) with ∥;T∥;₁ < 1, identity operator I : ℰ → ℰ , and A ∈ G , such that

and

1. there exists σ ₀ ∈ X, such that σ 0 ℛ f ( σ 0 ) or f ( σ 0 ) ℛ σ 0 ;

2. for a sequence {σ _n } with σ _n → x* whose all sequential terms are comparable, we have σ n ℛ x ⁎ for all n ∈ ℕ .

Then, f has a fixed point in X.

Proof

Let σ ₀ be any point in X. Define the sequence {σ _n } by σ _n = f(σ _n−1) for all n ∈ ℕ . By assumption (1), we have σ 0 ℛ f ( σ 0 ) = σ 1 . Also σ 2 ℛ σ 1 since f is monotone (either order preserving or order reversing). By (4.4), we have

that is,

By (d3) and ( A 1 ) ,

By ( A 2 ) , there exists T ∈ B ( ℰ , ℰ ) with ∥;T∥;₁ < 1, such that

Since σ 2 ℛ σ 1 , by assumption (2), we get σ 3 ℛ σ 2 (since f is monotone), and thus, by (4.3)

By (d3), ( A 1 ) , and ( A 2 ) , we get

By following the same pattern, we construct a sequence {σ _n }, such that

Rest of the proof is similar to the proof of Theorem 4.1.□

Remark 4.1

(1). Fixed point in Theorems 4.1–4.3 can be proved to be unique if additionally we assume that for every pair of elements σ, κ ∈ X, there exists either an upper bound or lower bound of σ, κ.

(2). If cone is taken as normal in the above theorems, then we can replace A 3 by

A ( σ , ξ , ν , α , α + ξ , α + ν ) ≼ 0 for all, σ ≪ c, ξ ≪ c, ν ≪ c and α ≪ c.

The following example illustrates the main theorem.

Example

Let ℰ = ( ℝ 3 , ∥ ⋅ ∥ ) with ∥ σ ∥ = max { ( | σ 1 | , | σ 2 | , | σ 3 | ) } , then ( ℰ , ∥; . ∥; ) is a real Banach space. Define C = { ( σ , ξ , ν ) ∈ ℝ 3 : σ , ξ , ν ≥ 0} , then, it is a cone in ℰ . Define the cone metric d by d ( σ , ξ ) = { | σ 1 − ξ 1 | , | σ 2 − ξ 2 | , | σ 3 − ξ 3 | } , where σ = ( σ 1 , σ 2 , σ 3 ) and ξ = ( ξ 1 , ξ 2 , ξ 3 ) . Let X = { (0,0,0), ( 0,0, 1 4 ) , ( 0, 1 4 ,0 ) } ⊂ ℰ and define f by

then the mapping f is monotone with respect to the partial order ≼ . Let

Define T ( σ ) = σ 3 and

Then, ∥ T ∥ = 1 3 < 1 , and hence, T ∈ B ( ℰ , ℰ ) . Also, T ( C ) ⊂ C . Now, if σ = (0, 0, 0) and κ = ( 0 , 0 , 1 4 ) , then σ ≼ κ . Also