Some fixed and coincidence point theorems for expansive maps in cone metric spaces

Abstract

In this article, we establish some common fixed and common coincidence point theorems for expansive type mappings in the setting of cone metric spaces. Our results extend some known results in metric spaces to cone metric spaces. Also, we introduce some examples the support the validity of our results.

Mathematics Subject Classification: 54H25; 47H10; 54E50.

1. Introduction

Huang and Zhang [1] introduced the notion of cone metric spaces as a generalization of metric spaces. They replacing the set of real numbers by an ordered Banach space. Huang and Zhang [1] presented the notion of convergence of sequences in cone metric spaces and proved some fixed point theorems. Then after, many authors established many fixed point theorems in cone metric spaces. For some fixed point theorems in cone metric spaces we refer the reader to [230].

In the present article, E stands for a real Banach space.

Definition 1.1. Let P be a subset of E with$\text{Int}\left(P\right)\ne 0̸$. Then P is called a cone if the following conditions are satisfied:

1. (1)

P is closed and P ≠ {θ}.

2. (2)

a, b R + , x, y P implies ax + by P.

3. (3)

x P ∩ -P implies x = θ.

For a cone P, define a partial ordering with respect to P by x y if and only if y - x P. We shall write x y to indicate that x y but xy, while x y will stand for y - x Int P. It can be easily shown that λInt(P) Int(P) for all positive scalar λ.

Definition 1.2. [1]Let X be a nonempty set. Suppose the mapping d : X × XE satisfies

1. (1)

θ d(x, y) for all x, y X and d(x, y) = θ if and only if x = y.

2. (2)

d(x, y) = d(y, x) for all x, y X.

3. (3)

d(x, y) d(x, z) + d(y, z) for all x, y X.

Then d is called a cone metric on X, and (X, d) is called a cone metric space.

Definition 1.3. [1]Let (X, d) be a cone metric space. Let (x n ) be a sequence in X and x X. If for every c E with θ c, there is an N N such that d(x n , x) c for all nN, then (x n ) is said to be convergent and (x n ) converges to x and x is the limit of (x n ). We denote this by limn→+∞x n = x or x n x as n → +∞. If for every c E with θ c there is an N N such that d(x n ,x m ) c for all n,mN, then (x n ) is called a Cauchy sequence in X. The space (X,d) is called a complete cone metric space if every Cauchy sequence is convergent.

The cone P in a real Banach space E is called normal if there is a number λ > 0 such that for all x,y E,

Iiker Ṣahin and Mustafa Telci [30] studied a theorem on common fixed points of expansive type mappings in cone metric spaces.

Definition 1.4. [30]Let E and F be real Banach spaces and P and Q be cones in E and F, respectively. Let (X, d) and (Y, ρ) be cone metric spaces, where d : X × XE and ρ : Y × YF. A function f : XY is said to be continuous at x0 X, if for every c F with 0 c, there exists b E with 0 b such that ρ(fx, fx0) c whenever x X and d(x, x0) b.

Lemma 1.1. [30]Let (X, d) and (Y, ρ) be cone metric spaces. A function f : XY is continuous at a point x0 X if and only if whenever a sequence (x n ) in X converges to x0, the sequence (fx n ) converges to fx0.

Theorem 1.1. [30]Let (X, d) be a complete cone metric space and P be a cone. Let f and g be surjective self-mappings of X satisfying the following inequalities

$\begin{array}{c}d\left(g\phantom{\rule{0.3em}{0ex}}fx,fx\right)\succcurlyeq \phantom{\rule{0.3em}{0ex}}ad\left(fx,x\right),\\ d\left(f\phantom{\rule{0.3em}{0ex}}gx,gx\right)\succcurlyeq \phantom{\rule{0.3em}{0ex}}bd\left(gx,x\right)\end{array}$

for all x X, where a,b > 1. If either f or g is continuous, then f and g have a common fixed point.

The aim of this article is to study new theorems of common fixed and coincidence point for expansive mappings in cone metric spaces under a set of conditions. Our results generalize several well known comparable results in the literature. Also, we introduced some examples to support the validity of our results.

2. Main results

Theorem 2.1. Let (X, d) be a cone metric space with a solid cone P. Let T, f : XX be mappings satisfying:

$d\left(Tx,Ty\right)\succcurlyeq ad\left(fx,fy\right)+bd\left(fx,Tx\right)+cd\left(fy,Ty\right)$
(1)

for all x, y X where a,b,c ≥ 0 with a + b +c > 1. Suppose the following hypotheses:

1. (1)

b < 1 or c < 1.

2. (2)

fX TX.

3. (3)

TX is a complete subspace of X.

Then T and f have a coincidence point.

Proof. Let x0 X. Since fX TX, we choose x1 X such that Tx1 = fx0. Again, we can choose x2 X such that Tx2 = fx1. Continuing in the same way, we construct a sequence (x n ) in X such that Txn+ 1= fx n for all n N {0}.

If fxm-1= fx m for some m N, then Tx m = fx m . Thus x m is a coincidence point of T and f.

Now, assume that xn-1x n for all n N.

Case (1): Suppose b < 1.

By (1), we have

$\begin{array}{ll}\hfill d\left(f{x}_{n-1},f{x}_{n}\right)& =d\left(T{x}_{n},T{x}_{n+1}\right)\phantom{\rule{2em}{0ex}}\\ \succcurlyeq ad\left(f{x}_{n},f{x}_{n+1}\right)+bd\left(f{x}_{n},T{x}_{n}\right)+cd\left(f{x}_{n+1},T{x}_{n+1}\right)\phantom{\rule{2em}{0ex}}\\ =ad\left(f{x}_{n},f{x}_{n+1}\right)+bd\left(f{x}_{n},f{x}_{n-1}\right)+cd\left(f{x}_{n+1},f{x}_{n}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Thus, we have

Hence

$d\left(f{x}_{n+1},f{x}_{n}\right)\preccurlyeq \frac{1-b}{a+c}d\left(f{x}_{n-1},f{x}_{n}\right).$
(2)

Case (2): Suppose c < 1.

By (1), we have

$\begin{array}{ll}\hfill d\left(f{x}_{n},f{x}_{n-1}\right)& =d\left(T{x}_{n+1},T{x}_{n}\right)\phantom{\rule{2em}{0ex}}\\ \succcurlyeq ad\left(f{x}_{n+1},f{x}_{n}\right)+bd\left(f{x}_{n+1},T{x}_{n+1}\right)+cd\left(f{x}_{n},T{x}_{n}\right)\phantom{\rule{2em}{0ex}}\\ =ad\left(f{x}_{n},f{x}_{n+1}\right)+bd\left(f{x}_{n+1},f{x}_{n}\right)+cd\left(f{x}_{n},f{x}_{n-1}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Thus, we have

$\left(1-c\right)d\left(f{x}_{n-1},f{x}_{n}\right)\succcurlyeq \left(a+b\right)d\left(f{x}_{n+1},f{x}_{n}\right).$

Hence

$d\left(f{x}_{n+1},f{x}_{n}\right)\preccurlyeq \frac{1-c}{a+b}d\left(f{x}_{n-1},f{x}_{n}\right).$
(3)

In Case (1), we let

$\lambda =\frac{1-b}{a+c}$

and in Case (2), we let

$\lambda =\frac{1-c}{a+b}.$

Thus in both cases, we have λ < 1 and

$d\left(f{x}_{n+1},f{x}_{n}\right)\preccurlyeq \lambda d\left(f{x}_{n-1},f{x}_{n}\right).$
(4)

By (4), we have

$\begin{array}{ll}\hfill d\left(f{x}_{n+1},f{x}_{n}\right)& \preccurlyeq \lambda d\left(f{x}_{n-1},f{x}_{n}\right)\phantom{\rule{2em}{0ex}}\\ \preccurlyeq {\lambda }^{2}d\left(f{x}_{n-2},f{x}_{n-1}\right)\phantom{\rule{2em}{0ex}}\\ ⋮\phantom{\rule{2em}{0ex}}\\ \preccurlyeq {\lambda }^{n}d\left(f{x}_{0},f{x}_{1}\right).\phantom{\rule{2em}{0ex}}\end{array}$

So for m > n, we have

$\begin{array}{ll}\hfill d\left(f{x}_{n},f{x}_{m}\right)& \preccurlyeq d\left(f{x}_{n},f{x}_{n+1}\right)+d\left(f{x}_{n+1},f{x}_{n+2}\right)+\cdots +d\left(f{x}_{m-1},f{x}_{m}\right)\phantom{\rule{2em}{0ex}}\\ \preccurlyeq \left({\lambda }^{n}+{\lambda }^{n+1}+\cdots +{\lambda }^{m-1}\right)d\left(f{x}_{0},f{x}_{1}\right)\phantom{\rule{2em}{0ex}}\\ \preccurlyeq {\lambda }^{n}\sum _{i=0}^{+}\infty {\lambda }^{i}d\left(f{x}_{0},f{x}_{1}\right)\phantom{\rule{2em}{0ex}}\\ =\frac{{\lambda }^{n}}{1-\lambda }d\left(f{x}_{0},f{x}_{1}\right).\phantom{\rule{2em}{0ex}}\end{array}$

Let θ c be given, choose δ > 0 such that c + Nδ(0) P, where

${N}_{\delta }\left(0\right)=\left\{y\in E:∥y∥<\delta \right\}.$

Also, choose a natural number N1 such that

$\frac{{\lambda }^{n}}{1-\lambda }d\left(f{x}_{0},f{x}_{1}\right)\in {N}_{\delta }\left(0\right),$

for mN1. Then

$\frac{{\lambda }^{n}}{1-\lambda }d\left(f{x}_{0},f{x}_{1}\right)\ll c,$

for all mN1. Thus,

$d\left(f{x}_{n},f{x}_{m}\right)\preccurlyeq \frac{{\lambda }^{n}}{1-\lambda }d\left(f{x}_{0},f{x}_{1}\right)\ll c,$

for all m > n. Therefore (Tx n ) is a cauchy sequence in (TX, d). Since (TX, d) is a complete cone metric space, there is u X such that (Tx n ) converges to Tu as n → +∞. Hence fx n converges to Tu as n → +∞. Since a + b + c > 1, we have a, b and c are not all 0. So we have the following cases.

Case 1: If a ≠ 0, then

$d\left(T{x}_{n},Tu\right)\succcurlyeq ad\left(f{x}_{n},fu\right)+bd\left(f{x}_{n},T{x}_{n}\right)+cd\left(fu,Tu\right)\succcurlyeq ad\left(f{x}_{n},fu\right).$

Hence

$d\left(f{x}_{n},fu\right)\preccurlyeq \frac{1}{a}d\left(T{x}_{n},Tu\right).$

Let θ c be given, choose δ > 0 such that c + Nδ(0) P, where

${N}_{\delta }\left(0\right)=\left\{y\in E:∥y∥<\delta \right\}.$

Since $\frac{1}{a}d\left(T{x}_{n},Tu\right)\to \theta$. We choose a natural number n0 N such that

$\frac{1}{a}d\left(T{x}_{n},Tu\right)\in {N}_{\delta }\left(0\right),$

for nn0. Then

$\frac{1}{a}d\left(T{x}_{n},Tu\right)\ll c,$

for all nn0. Thus,

$d\left(f{x}_{n},fu\right)\preccurlyeq \frac{1}{a}d\left(T{x}_{n},Tu\right)\ll c,$

for all nn0. Thus fx n fu as n → +∞. By uniqueness of limit, we have Tu = fu. Therefore T and f have a coincidence point.

Case 2: If b ≠ 0, then

$d\left(Tu,T{x}_{n}\right)\succcurlyeq ad\left(f{x}_{n},fu\right)+bd\left(fu,Tu\right)+cd\left(f{x}_{n},T{x}_{n}\right)\succcurlyeq bd\left(fu,Tu\right).$

Hence

$d\left(fu,Tu\right)\preccurlyeq \frac{1}{b}d\left(T{x}_{n},Tu\right).$

As similar proof of Case (1), we can show that fu = Tu. Thus f and T have a coincidence point.

Case 3: If c ≠ 0, then

$d\left(T{x}_{n},Tu\right)\succcurlyeq ad\left(f{x}_{n},fu\right)+bd\left(f{x}_{n},T{x}_{n}\right)+cd\left(Tu,fu\right)\succcurlyeq cd\left(fu,Tu\right).$

Hence

$d\left(fu,Tu\right)\preccurlyeq \frac{1}{c}d\left(T{x}_{n},Tu\right).$

As similar proof of Case (1), we can show that fu = Tu. Thus f and T have a coincidence point.

Corollary 2.1. Let (X,d) be a cone metric space with a solid cone P. Let T, f : XX be mappings satisfying:

$d\left(Tx,Ty\right)\succcurlyeq ad\left(fx,fy\right)+bd\left(fx,Tx\right)$

for all x, y X where a, b ≥ 0 with a + b > 1 and b < 1. Suppose the following hypotheses:

1. (1)

fX TX.

2. (2)

TX is a complete subspace of X.

Then T and f have a coincidence point.

Corollary 2.2. Let (X, d) be a complete cone metric space with a solid cone P. Let T, f :XX be mappings satisfying:

$d\left(Tx,Ty\right)\succcurlyeq ad\left(fx,fy\right)$

for all x, y X where a > 1. Suppose the following hypotheses:

1. (1)

fX TX.

2. (2)

TX is a complete subspace of X.

Then T and f have a coincidence point.

Corollary 2.3. Let (X, d) be a complete cone metric space with a solid cone P. Let T :XX be a surjective mapping satisfying:

$d\left(Tx,Ty\right)\succcurlyeq ad\left(x,y\right)+bd\left(x,Tx\right)+cd\left(y,Ty\right)$

for all x, y X where a,b,c ≥ 0 with a + b + c > 1. Suppose b < 1 or c < 1. Then T has a fixed point.

Proof. Follows from Theorem 2.1 by taking f = I, the identity map.

Corollary 2.4. Let (X, d) be a complete cone metric space with a solid cone P. Let T :XX be a surjective mapping satisfying:

$d\left(Tx,Ty\right)\succcurlyeq ad\left(x,y\right)$

for all x, y X where with a > 1. Then T has a fixed point.

Putting E = R, P = {x R : x ≥ 0} and d : X × XR in Corollaries 2.1 and 2.2, we have the following results:

Corollary 2.5. Let (X, d) be a complete metric space. Let T : XX be a surjective mapping satisfying:

$d\left(Tx,Ty\right)\ge ad\left(x,y\right)+bd\left(x,Tx\right)$

for all x, y X where a, b ≥ 0 with a + b > 1 and b < 1. Then T has a fixed point.

Corollary 2.6. Let (X, d) be a complete metric space. Let T : XX be a surjective mapping satisfying:

$d\left(Tx,Ty\right)\ge ad\left(x,y\right)+bd\left(y,Ty\right)$

for all x, y X where a, b ≥ 0 with a + b > 1 and b < 1. Then T has a fixed point.

Now, we present a fixed point theorem for two maps.

Theorem 2.2. Let T, S : XX be two surjective mappings of a complete cone metric space (X, d) with a solid cone P. Suppose that T and S satisfying the following inequalities

$d\left(T\left(Sx\right),Sx\right)+kd\left(T\left(Sx\right),x\right)\succcurlyeq ad\left(Sx,x\right)$
(5)

and

$d\left(S\left(Tx\right),Tx\right)+kd\left(S\left(Tx\right),x\right)\succcurlyeq bd\left(Tx,x\right)$
(6)

for all x X and some nonnegative real numbers a, b and k with a > 1 + 2k and b >1 + 2k. If T or S is continuous, then T and S have a common fixed point

Proof. Let x0 be an arbitrary point in X. Since T is surjective, there exists x1 X such that x0 = Tx1. Also, since S is surjective, there exists x2 X such that x2= Sx1. Continuing this process, we construct a sequence (x n ) in X such that x2n= Tx2n+1and x2n+1= Sx2n+2for all n N {0}. Now, for n N {0}, we have

$d\left(T\left(S{x}_{2n+2}\right),S{x}_{2n+2}\right)+kd\left(T\left(S{x}_{2n+2}\right),{x}_{2n+2}\right)\succcurlyeq ad\left(S{x}_{2n+2},{x}_{2n+2}\right).$

Thus, we have

$d\left({x}_{2n},{x}_{2n+1}\right)+kd\left({x}_{2n},{x}_{2n+2}\right)\succcurlyeq ad\left({x}_{2n+1},{x}_{2n+2}\right),$

which implies that

$d\left({x}_{2n},{x}_{2n+1}\right)+kd\left({x}_{2n},{x}_{2n+1}\right)+kd\left({x}_{2n+1},{x}_{2n+2}\right)\succcurlyeq ad\left({x}_{2n+1},{x}_{2n+2}\right).$

Hence

$d\left({x}_{2n+1},{x}_{2n+2}\right)\preccurlyeq \frac{1+k}{a-k}d\left({x}_{2n},{x}_{2n+1}\right)$
(7)

On other hand, we have

$d\left(S\left(T{x}_{2n+1}\right),T{x}_{2n+1}\right)+kd\left(S\left(T{x}_{2n+1}\right),{x}_{2n+1}\right)\succcurlyeq bd\left(T{x}_{2n+1},{x}_{2n+1}\right).$

Thus, we have

$d\left({x}_{2n-1},{x}_{2n}\right)+kd\left({x}_{2n-1},{x}_{2n+1}\right)\succcurlyeq bd\left({x}_{2n,}{x}_{2n+1}\right).$

Since d(x2n- 1,x2n) + d(x2n, x2n+1) d(x2n- 1,x2n+ 1), we have

$d\left({x}_{2n-1},{x}_{2n}\right)+kd\left({x}_{2n-1},{x}_{2n}\right)+kd\left({x}_{2n},{x}_{2n+1}\right)\succcurlyeq bd\left({x}_{2n},{x}_{2n+1}\right).$

Hence

$d\left({x}_{2n},{x}_{2n+1}\right)\preccurlyeq \frac{1+k}{b-k}d\left({x}_{2n-1},{x}_{2n}\right)$
(8)

Let

$\lambda =\text{max}\left\{\frac{1+k}{a-k},\frac{1+k}{b-k}\right\}.$

Then by combining (7) and (8), we have

$d\left({x}_{n},{x}_{n+1}\right)\le \lambda d\left({x}_{n-1},{x}_{n}\right)\phantom{\rule{1em}{0ex}}\forall n\in \mathbf{N}\cup \left\{0\right\}.$
(9)

Repeating (9) n-times, we get

$d\left({x}_{n},{x}_{n+1}\right)\preccurlyeq {\lambda }^{n}d\left({x}_{0},{x}_{1}\right).$

Thus, for m > n, we have

$\begin{array}{ll}\hfill d\left({x}_{n},{x}_{m}\right)& \preccurlyeq d\left({x}_{n},{x}_{n+1}\right)\cdots +d\left({x}_{m-1},{x}_{m}\right)\phantom{\rule{2em}{0ex}}\\ \preccurlyeq \left({\lambda }^{n}+\cdots +{\lambda }^{m-1}\right)d\left({x}_{0},{x}_{1}\right)\phantom{\rule{2em}{0ex}}\\ \preccurlyeq \frac{{\lambda }^{n}}{1-\lambda }d\left({x}_{0},{x}_{1}\right).\phantom{\rule{2em}{0ex}}\end{array}$

As similar arguments to proof of Theorem 2.1, we can show that (x n ) is a Cauchy sequence in the complete cone metric space (X, d). Then there exists v X such that x n v as n → +∞. Therefore x2n+ 1v and x2n+ 2v as n → +∞. Without loss of generality, we may assume that T is continuous, then Tx2n+ 1Tv as n → +∞. But Tx2n+ 1= x 2n v as n → +∞. Thus, we have Tv = v. Since S is surjective, there exists w X such that Sw = v. Now,

$d\left(T\left(Sw\right),Sw\right)+kd\left(T\left(Sw\right),w\right)\succcurlyeq ad\left(Sw,w\right),$

$d\left(v,w\right)\preccurlyeq \frac{k}{a}d\left(v,w\right).$

Since a > k, we conclude that d(v, w) = θ. So v = w. Hence Tv = Sv = v. Therefore v is a common fixed point of T and S.

By taking b = a in Theorem 2.2, we have the following result.

Corollary 2.7. Let T, S : XX be two surjective mappings of a complete cone metric space (X, d) with a solid cone P. Suppose that T and S satisfying the following inequalities

$d\left(T\left(Sx\right),Sx\right)+kd\left(T\left(Sx\right),x\right)\succcurlyeq ad\left(Sx,x\right)$
(10)

and

$d\left(S\left(Tx\right),Tx\right)+kd\left(S\left(Tx\right),x\right)\succcurlyeq ad\left(Tx,x\right)$
(11)

for all x X and some nonnegative real numbers a and k with a > 1 + 2k. If T or S is continuous, then T and f have a common fixed point

By taking S = T in Corollary 2.7, we have the following corollary.

Corollary 2.8. Let T : XX be a surjective mapping of a complete cone metric space (X, d) with a solid cone P. Suppose that T satisfying

$d\left(T\left(Tx\right),Tx\right)+kd\left(T\left(Tx\right),x\right)\succcurlyeq ad\left(Tx,x\right)$
(12)

for all x X and some nonnegative real number a and k with a > 1 + 2k. If T is continuous, then T has a fixed point.

Now, we present some examples to illustrate the useability of our results.

Example 2.1. (The case of normal cone) Let X = [0,+∞), E = R2. Let P = {(a, b) :a0,b ≥ 0} be the cone with d(x, y) = (|x - y|, |x - y|). Then (X, d) is a complete cone metric space. Define T : XX by Tx = 2x. Then T has a fixed point.

Proof. Note that

$d\left(T\left(Tx\right),Tx\right)+d\left(T\left(Tx\right),x\right)\ge 4d\left(Tx,x\right)$

for all x X. Thus T satisfies all the hypotheses of Corollary 2.8 and hence T has a fixed point. Here 0 is the fixed point of T.

Example 2.2. (The case of non-normal cone) Let X = [0, 1], $E={C}_{\mathbf{R}}^{1}\left(\left[0,1\right]\right)$. Let P = {ϕ E: ϕ(t) ≥ 0, t [0, 1]}. Define the mapping d : X × XE by

$d\left(x,y\right)\left(t\right):=\left|x-y\right|\varphi \left(t\right),$

where ϕ P is a fixed function, for example ϕ(t) = et. Define T, f : XX by$Tx=\frac{1}{4}x$and$fx=\frac{1}{16}x$. Then T and f have a coincidence point.

Proof. Note that

$\begin{array}{ll}\hfill d\left(Tx,Ty\right)\left(t\right)& =\left|\frac{1}{4}x-\frac{1}{4}y\right|{e}^{t}\phantom{\rule{2em}{0ex}}\\ =4\left|\frac{1}{16}x-\frac{1}{16}y\right|{e}^{t}\phantom{\rule{2em}{0ex}}\\ =4d\left(fx,fy\right)\left(t\right)\phantom{\rule{2em}{0ex}}\end{array}$

for all x, y X and t [0, 1]. Thus T and f satisfy all the hypotheses of Corollary 2.2 and hence T and f have a coincidence point. Here 0 is the coincidence point of T and f.

Remarks:

1. (1)

Theorem 4.1 of [29] is a special case of Theorem 2.2.

2. (2)

Corollary 4.1 of [29] is a special case of Corollary 2.8.

3. (3)

Theorem 4 of [31] is a special case of Corollary 2.8.

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Shatanawi, W., Awawdeh, F. Some fixed and coincidence point theorems for expansive maps in cone metric spaces. Fixed Point Theory Appl 2012, 19 (2012). https://doi.org/10.1186/1687-1812-2012-19