The concept of cone b-Banach space and fixed point theorems

Definition 1.1

(see [39]) Let s ≥ 1 be a given real number. Suppose that X is a nonempty set. We call a mapping ρ : X × X → E a cone b-metric if it satisfies:

1. θ < ρ ( x , y ) with x ≠ y and ρ ( x , y ) = θ if and only if x = y ;

2. ρ ( x , y ) = ρ ( y , x ) ;

3. ρ ( x , y ) ≤ s [ ρ ( x , z ) + ρ ( z , y ) ] ,

Definition 1.2

(see [39]) If ( X , ρ ) is a CbMS, P ⊂ E is a solid cone, x ∈ X , and { x n } n ≥ 1 is a sequence in X . Then

1. { x n } is called a Cauchy sequence, if for every c ∈ E with θ ≪ c , there is an N such that for n , m > N , ρ ( x n , x m ) ≪ c ;

2. { x n } converges to x ∈ X , if for every c ∈ E with θ ≪ c , there is an N such that for n > N , ρ ( x n , x ) ≪ c for some fixed x ∈ X ;

3. ( X , ρ ) is complete if every Cauchy sequence in X is convergent in X .

Lemma 1.3

(see [39]) If ( X , ρ ) is a CbMS, then one has the following properties:

1. If x ≤ y and y ≪ z , then x ≪ z .

2. If x ≪ y and y ≪ z , then x ≪ z .

3. If θ ≤ x ≪ c for each c ∈ int P , then x = θ .

4. If c ∈ int P , θ ≤ x n and x n → θ , then there exists n 0 such that x n ≪ c for all n > n 0 .

5. Let θ ≪ c , if θ ≤ ρ ( x n , x ) ≤ y n and y n → θ , then eventually ρ ( x n , x ) ≪ c , where x ∈ X and { x n } n ≥ 1 is a sequence in X .

6. If θ ≤ x n ≤ y n and x n → x , y n → y , then x ≤ y , for each cone P .

7. If x ≤ λ x where x ∈ P and 0 ≤ λ < 1 , then x = θ .

Definition 1.4

(see [35]) If X is a vector space over R, cone P ⊂ E , and a mapping ‖ ⋅ ‖ E : X → E satisfies

1. ‖ x ‖ E ≥ θ for all x ∈ X and ‖ x ‖ E = θ if and only if x = θ ;

2. ‖ x + y ‖ E ≤ ‖ x ‖ E + ‖ y ‖ E for x , y ∈ X ;

3. ‖ k x ‖ E = ∣ k ∣ ‖ x ‖ E for k ∈ R ,

Definition 1.5

Let 1 ≤ s ≤ 2 be a given real number. Assume X is a vector space over R, cone P ⊂ E . If ‖ ⋅ ‖ P : X → E satisfies

1. ‖ x ‖ P ≥ θ for x ∈ X and ‖ x ‖ P = θ if and only if x = θ ;

2. ‖ x + y ‖ P ≤ s [ ‖ x ‖ P + ‖ y ‖ P ] for x , y ∈ X ;

3. ‖ k x ‖ P = ∣ k ∣ s ‖ x ‖ P for k ∈ R ,

Definition 1.6

( X , ‖ ⋅ ‖ P ) is a CbNS, P ⊂ E is a solid cone, x ∈ X , and { x n } n ≥ 1 is a sequence in X . Then

1. we say that { x n } n ≥ 1 converges to x if for every c ∈ E with θ ≪ c , there is a natural number N satisfying ‖ x n − x ‖ P ≪ c for each n ≥ N . We denote lim n → ∞ x n = x or x n → x ;

2. we say that { x n } n ≥ 1 is a Cauchy sequence if for every c ∈ E with θ ≪ c , there exists a natural number N satisfying ‖ x n − x m ‖ P ≪ c for all n , m ≥ N ;

3. we say that ( X , ‖ ⋅ ‖ P ) is complete if every Cauchy sequence is convergent.

Lemma 1.7

Suppose ( X , ‖ ⋅ ‖ P ) is a CbNS, P is a solid cone, x ∈ X , and { x n } n ≥ 1 is a sequence in X . Then the following statements hold:

1. ‖ x n − x ‖ P → θ ( n → ∞ ) if and only if { x n } converges to x .

2. ‖ x n − x m ‖ P → θ ( n , m → ∞ ) if and only if { x n } is a Cauchy sequence.

Proof

First, we prove (i). If ‖ x n − x ‖ P → θ ( n → ∞ ) , then for any c ∈ int P , θ ≤ ‖ x n − x ‖ P , and ‖ x n − x ‖ P → θ , by (p4), one can obtain that there exists n 0 such that for n > n 0 we have ‖ x n − x ‖ P ≪ c , then { x n } converges to x .

In turn, if { x n } converges to x , using (p3) we get that for any c ∈ E , θ ≪ c ( c ∈ int P ) , θ ≤ ‖ x n − x ‖ P ≪ c , then ‖ x n − x ‖ P = θ .

The proof of (ii) is similar to (i). On one hand, if ‖ x n − x m ‖ P → θ ( n , m → ∞ ) , then for any c ∈ int P , θ ≤ ‖ x n − x m ‖ P , and ‖ x n − x m ‖ P → θ , by (p4), one can obtain that there exists N such that for all n , m > N we have ‖ x n − x m ‖ P ≪ c , then { x n } is a Cauchy sequence.

On the other hand, if { x n } is a Cauchy sequence, then for any c ∈ E , θ ≪ c ( c ∈ int P ) , θ ≤ ‖ x n − x m ‖ P ≪ c , by (p3) we can obtain that ‖ x n − x m ‖ P = θ .□

Example 1.8

Let X = R 2 , E = R 2 , and P = { ( x , y ) ∈ E : x ≥ 0 , y ≥ 0 } , we define ‖ ( x , y ) ‖ P = ( ∣ x ∣ 2 , ∣ y ∣ 2 ) . Then ( X , ‖ ⋅ ‖ P ) is a cone b -Banach space.

Proof

First, by the definition of ‖ ( x , y ) ‖ P , we get

for all u = ( x , y ) ∈ X . It is obvious that ‖ ( x , y ) ‖ P = θ if and only if ( x , y ) = θ .

Next, according to

We see for any ( x 1 , y 1 ) , ( x 2 , y 2 ) ∈ X , it follows that

For any u = ( x , y ) ∈ X , k ∈ R , we have

Thus, from Definition 1.5, ( X , ‖ ⋅ ‖ P ) is a CbNS with the coefficient s = 2 > 1 .

Finally, we suppose { u n } ∈ X is a Cauchy sequence in X , where u n = ( x n , y n ) , u m = ( x m , y m ) . By Definition 1.6 and Lemma 1.3, we deduce that

Therefore,

which means that { x n } and { y n } are Cauchy sequences in R. Since R is complete, there are x , y ∈ R , such that x n → x ( n → ∞ ) and y n → y ( n → ∞ ) . Let u = ( x , y ) ∈ X , then

By Lemma 1.3 (p4), one can obtain that { u n } converges to u in X , that is, ( X , ‖ ⋅ ‖ P ) is a cone b-Banach space.□

Remark 1.9

The new definition of cone b-Banach space generalizes cone Banach space. Clearly, a cone b-Banach space with s = 1 is exactly a cone Banach space. The above example is also sufficient to support the cone b-Banach space in Definition 1.5.

Theorem 2.1

Let X be a cone b-Banach space with the coefficient 1 ≤ s ≤ 2 and the b-norm ‖ x ‖ P = ρ ( x , θ ) , D ⊂ X is closed and convex, S : D → D is a map, satisfying

where q ≥ 0 and ρ ( ⋅ , ⋅ ) is a cone b -metric, then S has at least one fixed point.

Proof

When q = 0 , the conclusion holds. So we only consider the case q > 0 . Let x = y in (2.1), we get that

Then ρ ( x , S x ) = θ , that is S x = x for all x ∈ D . Thus, every x in D is a fixed point of S . Next we assume that x ≠ y and S is not an identity mapping on D . First, we prove that condition (2.1) implies that q ≥ 1 . Otherwise, q < 1 . Suppose x is not a fixed point of S and let y = S x , then we get that

Thus, ρ ( S x , S 2 x ) < θ . This is not true. Thus, q ≥ 1 .

When q = 1 , substituting y = S x in (2.1) implies that

Thus, ρ ( S x , S 2 x ) = θ , that is S x = S 2 x , so S x is a fixed point of S .

When q > 1 , we introduce a sequence { x n } , defined by:

where x 0 ∈ D is arbitrary and a , b > 0 which satisfy 1 q ≤ b s ( a + b ) s < 2 q . Since D is a convex set, { x n } ⊂ D .

Notice that

which is equivalent to

Taking into account the equation (2.7) and the condition (2.1), we have

Therefore, ρ ( x n , x n + 1 ) ≤ k 1 ρ ( x n − 1 , x n ) , where k 1 = q b s ( a + b ) s − 1 . Since 1 q ≤ b s ( a + b ) s < 2 q , we know 0 ≤ k 1 < 1 . Then, we get

For any m ≥ 1 , p ≥ 1 , it follows that

Let θ ≪ c be given. Note that s p k 1 m + 1 s − k 1 ρ ( x 1 , x 0 ) + s p − 1 k 1 m ρ ( x 1 , x 0 ) → θ , as m → ∞ . According to Lemma 1.3 (p4), we find m 0 ∈ N , such that

for each m > m 0 . Therefore,

for all m > m 0 and any p . Thus by Lemma 1.3 (p1), we can get that Cauchy sequence { x n } converges to z ∈ D . Considering the following inequality

and from Lemma 1.3 (p5)

Combining (2.7) with (2.1), we substitute x = z and y = x n into (2.1) to get that

Thus when n → ∞ , from Lemma 1.3, we can get d ( z , S z ) ≤ θ , that is, S z = z .□

Corollary 2.2

Let X be a cone Banach space with the norm ‖ x ‖ E = ρ ( x , θ ) , D ⊂ X is closed and convex, S : D → D is a map, satisfying

where q ≥ 0 , then S has at least one fixed point.

Remark 2.3

If s = 1 in Theorem 2.1, we can obtain Corollary 2.2 which improves and extends Theorem 2.4 in [35]. We do not need the conditions: cone P is normal and 2 ≤ q ≤ 4 , and we only require q ≥ 0 . If 1 ≤ s ≤ 2 in Theorem 2.1, we extend this fixed point theorem to our newly defined cone b-Banach space. According to the proof of Theorem 2.1, we can easily get the following inference.

Corollary 2.4

Let X be a cone b-Banach space with the coefficient 1 ≤ s ≤ 2 and the b-norm ‖ x ‖ P = ρ ( x , θ ) , D ⊂ X is closed and convex, S : D → D is a map, satisfying

where q ≥ 0 and x ≠ y . Then (i) q ≥ 1 ; (ii) S has at least one fixed point.

Corollary 2.5

Let X be a cone Banach space with the norm ‖ x ‖ E = ρ ( x , θ ) , D ⊂ X is closed and convex, S : D → D is a map, satisfying

where q ≥ 0 and x ≠ y . Then (i) q ≥ 1 ; (ii) S has at least one fixed point.

Theorem 2.6

Let X be a cone b-Banach space with the coefficient 1 ≤ s ≤ 2 and the b-norm ‖ x ‖ P = ρ ( x , θ ) , D ⊂ X is closed and convex, S : D → D is a map, satisfying

where p ≥ 0 . Then S has at least one fixed point. Moreover, when 0 ≤ p < 2 , S has an unique fixed point.

Proof

If x = y in (2.12), we have 2 ρ ( x , S x ) ≤ p ρ ( x , x ) . One can obtain ρ ( x , S x ) = θ , then S x = x for all x ∈ D . Thus, every x in D is a fixed point of S . Next we consider the case x ≠ y . From the triangle inequality (iii) in Definition 1.1, we have

By (2.12),

Thus, letting q = s ( 2 + p ) implies that

Hence, by Theorem 2.1, S has at least one fixed point.

When 0 ≤ p < 2 , we show the uniqueness of fixed points. Assume that there are x , y ∈ D such that S x = x and S y = y . By (2.12), we have

that is, ρ ( x , y ) ≤ p 2 ρ ( x , y ) , since 0 ≤ p 2 < 1 , then by Lemma 1.3 (p7), we get ρ ( x , y ) = θ . It is clear that S has an unique fixed point.□

Corollary 2.7

Let X be a cone Banach space with the norm ‖ x ‖ E = ρ ( x , θ ) , D ⊂ X is closed and convex, S : D → D is a map, satisfying

where p ≥ 0 . Then S has at least one fixed point. Especially, when 0 ≤ p < 2 , S has an unique fixed point.

Remark 2.8

When s = 1 in Theorem 2.6, our result exists in cone Banach space, that is Corollary 2.7. Clearly, Corollary 2.7 amends and improves Theorem 2.5 in [35] and we particularly discuss the uniqueness of fixed points. When 1 < s ≤ 2 , the condition is in cone b-Banach space, we extend this fixed point theorems to our newly defined cone b-Banach space.

Theorem 2.9

Let X be a cone b-Banach space with the coefficient 1 ≤ s ≤ 2 and the b-norm ‖ x ‖ P = ρ ( x , θ ) , D ⊂ X is closed and convex, S : D → D is a map, satisfying

where r > 1 . Then S has at least one fixed point.

Proof

If x = y , by (2.17) we have ρ ( x , S x ) = θ , then S x = x for all x ∈ D . That is, S is an identity mapping on D . Next we consider the case x ≠ y . Introduce a sequence { x n } defined by (2.5) and the following equalities

hold. The s -triangle inequality implies

Then by (2.19) and (2.7) we have

Substituting x = x n − 1 and y = x n in (2.17) and considering (2.7) and (2.21), we can obtain

and thus ρ ( x n , x n + 1 ) ≤ k 2 ρ ( x n − 1 , x n ) , where k 2 = r s b s + s a s − s ( a + b ) s ( 1 + s ) ( a + b ) s . We can find a , b such that ( a + b ) s + a s b s < r < ( 1 + 2 s ) ( a + b ) s s b s + a b s . Then 0 < k 2 < 1 . Hence, Cauchy sequence { x n } converges to some z ∈ D . Since { S x n } also converges to z as in the proof of Theorem 2.1, the inequality (2.17) yields that ρ ( S z , z ) + ρ ( z , S z ) ≤ θ , that is, S z = z .□