Iterative approximation of fixed points of quasi-contraction mappings in cone metric spaces

Abstract

In this paper, we establish a full statement of Ćirić’s fixed point theorem in the setting of cone metric spaces. More exactly, we obtain a priori and a posteriori error estimates for approximating fixed points of quasi-contractions in a cone metric space. Our result complements recent results of Zhang (Comput. Math. Appl. 62:1627-1633, 2011), Ding et al. (J. Comput. Anal. Appl. 15:463-470, 2013) and others.

MSC: Primary 54H25; secondary 47H10, 46A19.

1 Introduction

In this paper, we study fixed points of quasi-contraction mappings in a cone metric space $(X,d)$ over a solid vector space $(Y,⪯)$. Cone metric spaces have a long history (see Collatz [1], Zabrejko [2], Janković et al. [3], Proinov [4] and references therein). A unified theory of cone metric spaces over a solid vector space was developed in a recent paper of Proinov [4]. Recall that an ordered vector space with convergence structure $(Y,⪯)$ is called:

• a solid vector space if it can be endowed with a strict vector ordering $(\prec )$;

• a normal vector space if the convergence of Y has the sandwich property.

Every metric space $(X,d)$ is a cone metric space over ℝ (with usual ordering and usual convergence). On the other hand, every cone metric space over a solid vector space is a metrizable topological space (see Proinov [4] and references therein). It is well known that a lot of fixed point results in cone metric setting can be directly obtained from their metric versions (see Du [5], Amini-Harandi and Fakhar [6], Feng and Mao [7], Kadelburg et al. [8], Asadi et al. [9], Proinov [4], and Ercan [10]).

For instance, for this purpose we can use the following theorem. This theorem shows that every cone metric is equivalent to a metric which preserves the completeness as well as some inequalities.

Theorem 1.1 ([[4], Theorem 9.3])

Let $(X,d)$ be a cone metric space over a solid vector space $(Y,⪯)$. Then there exists a metric ρ on X such that the following statements hold true.

• (i) The topology of $(X,d)$ coincides with the topology of $(X,\rho )$.

• (ii) $(X,d)$ is complete if and only if $(X,\rho )$ is complete.

• (iii) For $x,x_1,\dots ,x_n\in X$, $y,y_1,\dots ,y_n\in X$ and ${\lambda }_1,\dots ,{\lambda }_n\in \mathbb{R}$,

  $$d(x,y)⪯\sum _{i=1}^n{\lambda }_{i}d(x_i,y_i)\mathit{\text{implies}}\rho (x,y)\le \sum _{i=1}^n{\lambda }_i\rho (x_i,y_i).$$

In 1922, Banach [11] proved his famous fixed point theorem for contraction mappings. Banach’s contraction principle is one of the most useful theorems in the fixed point theory. It has two versions: a short version and a full version. In a metric space setting its full statement can be seen, for example, in the monograph of Berinde [[12], Theorem 2.1]. Recently, full statements of Banach’s fixed point theorem in a cone metric spaces over a solid vector space were given by Radenović and Kadelburg [[13], Theorem 3.3] and Proinov [[4], Theorem 11.1].

Definition 1.2 ([14])

Let $(X,d)$ be a metric space. A mapping $T:X\to X$ is called a quasi-contraction (with contraction constant λ) if there exists $\lambda \in [0,1)$ such that

$$d(Tx,Ty)\le \lambda max\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$$

(1)

for all $x,y\in X$.

There are a large number of generalizations of Banach’s contraction principle (see, for example, [14–18] and references therein). In 1974, Ćirić [14] introduced contraction mappings and proved the following well known generalization of Banach’s fixed point theorem.

Theorem 1.3 Let $(X,d)$ be a complete metric space and $T:X\to X$ be a quasi-contraction with contraction constant λ. Then the following statements hold true:

• (i) Existence and uniqueness. T has a unique fixed point ξ in X.

• (ii) Convergence of Picard iteration. For every starting point $x\in X$ the Picard iteration sequence $(T^{n}x)$ converges to ξ.

• (iii) A priori error estimate. For every point $x\in X$ the following a priori error estimate holds:

  $$d(T^{n}x,\xi )\le \frac{{\lambda }^n}{1-\lambda }d(x,Tx)\mathit{\text{for all }}n\ge 0.$$

  (2)

Following Zhang [19], in the next definition, we define a useful binary relation between an ordered vector space Y and the set of all subsets of Y. It plays a very important role in this paper as it is used to prove our main result.

Definition 1.4 ([19])

Let $(Y,⪯)$ be an ordered vector space, $x\in Y$ and $A\subset Y$. We say that $x⪯A$ if there exists at least one vector $y\in A$ such that $x⪯y$.

In 2009, Ilić and Rakočević [20] generalized the concept of quasi-contraction to cone metric space as follows: A selfmapping T of a cone metric space $(X,d)$ over an ordered vector space $(Y,⪯)$ is called a quasi-contraction on X if there exists $\lambda \in [0,1)$ such that

$$d(Tx,Ty)⪯\lambda \{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$$

(3)

for all $x,y\in X$. They proved the following result [[20], Theorem 2.1]: Let $(X,d)$ be a cone metric space over a normal solid Banach space $(Y,⪯)$; then every quasi-contraction T of the type (3) has a unique fixed point in X, and for all $x\in X$ the Picard iterative sequence $(T^{n}x)$ converges to this fixed point. Kadelburg et al. [[21], Theorem 2.2] improved this result by omitting the assumption of normality provided that $\lambda \in [0,1/2)$. Gajić and Rakočević [[22], Theorem 3] proved this result for any contraction constant $\lambda \in [0,1)$. Rezapour et al. [[23], Theorem 2.1] proved this result in the case when Y is a solid topological vector space and $\lambda \in [0,1)$. Furthermore, Kadelburg et al. [[8], Theorem 3.5(b)] proved that this result is equivalent to the short version of Ćirić’s fixed point theorem.

In 2011, Zhang [19] presented the following new definition for quasi-contractions in cone metric spaces.

Definition 1.5 ([19])

Let $(X,d)$ be a cone metric space over an ordered vector space $(Y,⪯)$. A mapping $T:X\to X$ is called a quasi-contraction (with contraction constant λ) if there exists $\lambda \in [0,1)$ such that

$$d(Tx,Ty)⪯\lambda co\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}$$

(4)

for all $x,y\in X$.

By applying Theorem 1.1 to the first two conclusions of Theorem 1.3, we obtain the following fixed point theorem in a cone metric setting.

Theorem 1.6 Let $(X,d)$ be a complete cone metric space over a solid vector space $(Y,⪯)$ and $T:X\to X$ be a quasi-contraction. Then the following statements hold true:

• (i) Existence and uniqueness. T has a unique fixed point ξ in X.

• (ii) Convergence of Picard iteration. For every starting point $x\in X$ the Picard iteration sequence $(T^{n}x)$ converges to ξ.

In 2011, Zhang [19] proved Theorem 1.6 in the case when Y is a normal solid Banach space. In 2013, Ding et al. [24] proved this theorem in the case when Y is a solid topological vector space.

In this paper, we establish a full statement of Ćirić’s fixed point theorem in the setting of cone metric spaces. Our result complements Theorem 1.6. Thus it extends and complements the corresponding results of Zhang [19], Ding et al. [24], and others.

For some recent results on the topic, we refer the reader to [25–38]. In the papers [37, 38] one can find some applications of cone metric spaces to iterative methods for finding all zeros of polynomial simultaneously.

2 Preliminaries

In this section, we introduce some basic definitions and theorems of cone metric spaces over a solid vector space.

Definition 2.1 ([4])

Let Y be a real vector space and S be the set of all infinite sequences in Y. A binary relation → between S and Y is called a convergence on Y if it satisfies the following axioms:

(C1) If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$.

(C2) If $x_n\to x$ and $\lambda \in \mathbb{R}$, then $\lambda x_n\to \lambda x$.

(C3) If ${\lambda }_n\to \lambda$ in ℝ and $x\in Y$, then ${\lambda }_{n}x\to \lambda x$.

A real vector space Y endowed with convergence is said to be a vector space with convergence. If $x_n\to x$, then $(x_n)$ is said to be a convergent sequence in Y, and the vector x is said to be a limit of $(x_n)$.

Definition 2.2 ([4])

Let $(Y,\to )$ be a vector space with convergence. An ordering ⪯ on Y is said to be a vector ordering if it is compatible with the algebraic and convergence structures on Y in the sense that the following are true:

(V1) If $x⪯y$, then $x+z⪯y+z$.

(V2) If $\lambda \ge 0$ and $x⪯y$, then $\lambda x⪯\lambda y$.

(V3) If $x_n\to x$, $y_n\to y$, $x_n⪯y_n$ for all n, then $x⪯y$.

A vector space with convergence Y endowed with vector ordering is called an ordered vector space with convergence.

Definition 2.3 ([4])

Let $(Y,⪯,\to )$ be an ordered vector space with convergence. A strict ordering ≺ on Y is said to be a strict vector ordering if it is compatible with the vector ordering, the algebraic structure and the convergence structure on Y in the sense that the following are true:

(S1) If $x\prec y$, then $x⪯y$.

(S2) If $x⪯y$ and $y\prec z$, then $x\prec z$.

(S3) If $x\prec y$, then $x+z\prec y+z$.

(S4) If $\lambda >0$ and $x\prec y$, then $\lambda x\prec \lambda y$.

(S5) If $x_n\to x$, $y_n\to y$ and $x\prec y$, then $x_n\prec y_n$ for all but finitely many n.

It turns out that an ordered vector space can be endowed with at most one strict vector ordering (see Proinov [[4], Theorem 5.1]).

Definition 2.4 (Solid vector space)

An ordered vector space with convergence endowed with a strict vector ordering is said to be a solid vector space.

Let us consider an important example of a solid vector space.

Example 2.5 Let $(Y,\tau )$ be a topological vector space and $K\subset Y$ be a cone with nonempty interior $K^{\circ }$. Define the vector ordering ⪯ on Y and the strict vector ordering ≺ on Y, respectively, by means of

$$\begin{array}{r}x⪯y\text{if and only if}y-x\in K,\\ x\prec y\text{if and only if}y-x\in K^{\circ }.\end{array}$$

Then Y is a solid vector space called a solid topological vector space.

Now let us recall the definition of a cone metric space known also as ‘K-metric spaces’ (see Zabrejko [2], Proinov [4] and references therein).

Definition 2.6 (Cone metric space)

Let X be a nonempty set, and let $(Y,⪯)$ be an ordered vector space with convergence. A vector-valued function $d:X\times X\to Y$ is said to be a cone metric on Y if the following conditions hold:

• (i) $d(x,y)⪰0$ for all $x,y\in X$ and $d(x,y)=0$ if and only if $x=y$;

• (ii) $d(x,y)=d(y,x)$ for all $x,y\in X$;

• (iii) $d(x,y)⪯d(x,z)+d(z,y)$ for all $x,y,z\in X$.

The pair $(X,d)$ is called a cone metric space over Y.

Let $(X,d)$ be a cone metric space over a solid vector space $(Y,⪯,\prec )$, $x_0\in X$ and $r\in Y$ with $r\succ 0$. Then the set $U(x_0,r)=\{x\in X:d(x,x_0)\prec r\}$ is called an open ball with center $x_0$ and radius r.

Every cone metric space X over a solid vector space Y is a Hausdorff topological space with topology generated by the basis of all open balls. Then a sequence $(x_n)$ of points in X converges to $x\in X$ if and only if for every vector $c\in Y$ with $c\succ 0$, $d(x_n,x)\prec c$ for all but finitely many n.

Recall also that a sequence $(x_n)$ in X is called a Cauchy sequence if for every $c\in Y$ with $c\succ 0$ there is $N\in \mathbb{N}$ such that $d(x_n,x_m)\prec c$ for all $n,m>N$. A cone metric space X is called complete if each Cauchy sequence in X is convergent.

In order to prove our main result we need the following two theorems.

Theorem 2.7 ([4])

Let $(X,d)$ be a complete cone metric space over a solid vector space $(Y,⪯)$. Suppose $(x_n)$ is a sequence in X satisfying

$$d(x_n,x_m)⪯b_n\mathit{\text{for all }}n,m\ge 0\mathit{\text{ with }}m\ge n,$$

where $(b_n)$ is a sequence in Y which converges to 0. Then $(x_n)$ converges to a point $\xi \in X$ with error estimate

$$d(x_n,\xi )⪯b_n\mathit{\text{for all }}n\ge 0.$$

Theorem 2.8 ([4])

Let $(X,d)$ be a cone metric space over a solid vector space $(Y,⪯)$ and $T:X\to X$. Suppose that for some $x\in X$, the Picard iteration $(T^{n}x)$ converges to a point $\xi \in X$. Suppose also that there exist nonnegative numbers α and β such that

$$d(\xi ,T\xi )⪯\alpha d(x,\xi )+\beta d(Tx,\xi )\mathit{\text{for each }}x\in X.$$

(5)

Then ξ is a fixed point of T.

3 Auxiliary results

Let A be a subset of a real vector space Y. Recall that the convex hull of A, denoted coA, is the smallest convex set including A. Suppose $x,x_1,\dots ,x_n\in Y$. It is well known that $x\in co\{x_1,\dots ,x_n\}$ if and only if there exist nonnegative numbers ${\alpha }_1,\dots ,{\alpha }_n$ such that ${\sum }_{i=1}^n{\alpha }_i=1$ and $x={\sum }_{i=1}^n{\alpha }_i{x}_i$.

Lemma 3.1 Let $(Y,⪯)$ be an ordered vector space. Suppose that x, y, $x_1,\dots ,x_n$, $y_1,\dots ,y_m$ are vectors in Y and λ is a real number. Then:

(P1) $x⪯co\{x_1,\dots ,x_n\}\Rightarrow x⪯co\{x_1,\dots ,x_n,y\}$;

(P2) $x⪯co\{x_1,\dots ,x_n\}$ and $x_i⪯y_i$ for all $i\Rightarrow x⪯co\{y_1,\dots ,y_n\}$;

(P3) $x⪯co\{x_1,\dots ,x_n,y\}$ and $y⪯co\{y_1,\dots ,y_m\}\Rightarrow x⪯co\{x_1,\dots ,x_n,y_1,\dots ,y_m\}$;

(P4) $x⪯co\{0,x_1,\dots ,x_n\}\iff x⪯co\{x_1,\dots ,x_n\}$ if $x_i⪰0$ for some i;

(P5) $x⪯co\{\lambda x,x_1,\dots ,x_n\}\iff x⪯co\{x_1,\dots ,x_n\}$ if $\lambda <1$ and $x_i⪰0$ for some i;

(P6) $x⪯co\{x_1,\dots ,x_n,y\}\iff x⪯co\{x_1,\dots ,x_n\}$ if $y=x_i$ for some i.

Proof We only prove the necessity of (P5) since the proofs of the other properties are similar. The inequality $x⪯co\{\lambda x,x_1,\dots ,x_n,\}$ implies that there exist nonnegative numbers $\alpha ,{\alpha }_1,\dots ,{\alpha }_n$ such that $\alpha +{\sum }_{i=1}^n{\alpha }_i=1$ and $x⪯\alpha \lambda x+{\sum }_{i=1}^n{\alpha }_i{x}_i$. From this inequality and $\alpha \lambda <1$, we deduce

$$x⪯\sum _{i=1}^n{\beta }_i{x}_i,$$

(6)

where ${\beta }_i={\alpha }_i/(1-\alpha \lambda )$. We have ${\sum }_{i=1}^n{\beta }_i=(1-\alpha )/(1-\alpha \gamma )<1$. By the assumptions, we have $x_i⪰0$ for some i. Without loss of generality we may assume that $x_1⪰0$. Define the nonnegative numbers ${\gamma }_1,\dots ,{\gamma }_n$ by ${\gamma }_1=1-{\sum }_{j=2}^n{\beta }_j$ and ${\gamma }_i={\beta }_i$ for $i\ge 2$. From (6) and ${\beta }_1\le {\gamma }_1$, we obtain

$$x⪯\sum _{i=1}^n{\gamma }_i{x}_i.$$

This implies $x⪯co\{x_1,\dots ,x_n\}$ since ${\sum }_{i=1}^n{\gamma }_i=1$. □

Remark 3.2 Note that Lemma 3.1 remains true if we omit the expression ‘co’ from its formulation.

The following lemma was given by Zhang [[19], Lemma 6] in a slightly different form. We give a simple proof of this lemma.

Lemma 3.3 Let $(X,d)$ be a cone metric space over an ordered vector space $(Y,⪯)$, $T:X\to X$ be a quasi-contraction with contraction constant $\lambda \in [0,1)$, and let $x\in X$. Then for every $m\in \mathbb{N}$, we have

$$d(T^{i}x,T^{m}x)⪯{\lambda }^{i}co\{d(x,Tx),\dots ,d(x,T^{m}x)\}\mathit{\text{for }}i=1,\dots ,m.$$

(7)

Proof We prove the statement by induction on m. It is obviously true for $m=1$. Assume that $n\in \mathbb{N}$ and assume that (7) is satisfied for any natural number $m\le n$. We have to prove that

$$d(T^{i}x,T^{n+1}x)⪯{\lambda }^{i}co\{d(x,Tx),\dots ,d(x,T^{n+1}x)\}\text{for }i=1,\dots ,n+1.$$

(8)

We divide the proof of (8) into three steps.

Step 1. We claim that for every natural number $i\le n$ the following inequality holds:

$$d(T^{i}x,T^{n+1}x)⪯co\{{\lambda }^{i}d(x,Tx),\dots ,{\lambda }^{i}d(x,T^{n}x),\lambda d(T^{n}x,T^{n+1}x),\lambda d(T^{i-1}x,T^{n+1}x)\}.$$

By the definition of the quasi-contraction mapping, we obtain

$$\begin{array}{r}d(T^{i}x,T^{n+1}x)\\ =d(T\left(T^{i-1}x\right),T\left(T^{n}x\right))\\ ⪯\lambda co\{d(T^{i-1}x,T^{n}x),d(T^{i-1}x,T^{i}x),d(T^{n}x,T^{n+1}x),d(T^{i-1}x,T^{n+1}x),d(T^{i}x,T^{n}x)\}.\end{array}$$

From the induction hypothesis and properties (P1) and (P2), we get the following three inequalities:

$$\begin{array}{c}d(T^{i-1}x,T^{n}x)⪯{\lambda }^{i-1}co\{d(x,Tx),\dots ,d(x,T^{n}x)\},\hfill \\ d(T^{i-1}x,T^{i}x)⪯{\lambda }^{i-1}co\{d(x,Tx),\dots ,d(x,T^{n}x)\},\hfill \\ d(T^{i}x,T^{n}x)⪯{\lambda }^{i-1}co\{d(x,Tx),\dots ,d(x,T^{n}x)\}.\hfill \end{array}$$

From the last four inequalities and properties (P3) and (P6), we obtain the desired inequality.

Step 2. We claim that for every natural number $i\le n$ the following inequality holds:

$$d(T^{i}x,T^{n+1}x)⪯co\{{\lambda }^{i}d(x,Tx),\dots ,{\lambda }^{i}d(x,T^{n+1}x),\lambda d(T^{n}x,T^{n+1}x)\}.$$

We prove this by finite induction on i. Setting $i=1$ in the claim of Step 1, we immediately arrive at the following inequality:

$$d(Tx,T^{n+1}x)⪯co\{\lambda d(x,Tx),\dots ,\lambda d(x,T^{n}x),\lambda d(T^{n}x,T^{n+1}x),\lambda d(x,T^{n+1}x)\},$$

which proves the claim of Step 2 for $i=1$. Assume that for some $i\le n$, the claim of Step 2 holds. Now we shall show that

$$d(T^{i+1}x,T^{n+1}x)⪯co\{{\lambda }^{i+1}d(x,Tx),\dots ,{\lambda }^{i+1}d(x,T^{n+1}x),\lambda d(T^{n}x,T^{n+1}x)\}.$$

(9)

It follows from Step 1 that

$$\begin{array}{r}d(T^{i+1}x,T^{n+1}x)\\ ⪯co\{{\lambda }^{i+1}d(x,Tx),\dots ,{\lambda }^{i+1}d(x,T^{n+1}x),\lambda d(T^{n}x,T^{n+1}x),\lambda d(T^{i}x,T^{n+1}x)\}.\end{array}$$

(10)

By the finite induction hypothesis and property (P2), we have

$$d(T^{i}x,T^{n+1}x)⪯co\{{\lambda }^{i}d(x,Tx),\dots ,{\lambda }^{i}d(x,T^{n+1}x),d(T^{n}x,T^{n+1}x)\}.$$

(11)

From (10), (11), and properties (P3) and (P6), we obtain (9).

Step 3. Now we shall prove (8). From the claim of Step 2 with $i=n$, we get

$$d(T^{n}x,T^{n+1}x)⪯co\{{\lambda }^{n}d(x,Tx),\dots ,{\lambda }^{n}d(x,T^{n+1}x),\lambda d(T^{n}x,T^{n+1}x)\}.$$

According to the property (P5), this inequality is equivalent to

$$d(T^{n}x,T^{n+1}x)⪯{\lambda }^{n}co\{d(x,Tx),\dots ,d(x,T^{n+1}x)\},$$

which by (P2) implies

$$d(T^{n}x,T^{n+1}x)⪯{\lambda }^{i-1}co\{d(x,Tx),\dots ,d(x,T^{n+1}x)\}.$$

(12)

Finally, by the claim of Step 2 and the inequality (12), taking into account the properties (P3) and (P6), we obtain (8). This completes the proof of the lemma. □

In the following lemma, we show that if T is a quasi-contraction of a cone metric space X, then for every starting point $x\in X$, the Picard iteration sequence $(T^{n}x)$ is bounded in the space X.

Lemma 3.4 Let $(X,d)$ be a cone metric space over an ordered vector space $(Y,⪯)$, $T:X\to X$ be a quasi-contraction with contraction constant $\lambda \in [0,1)$, and let $x\in X$. Then for every $m\in \mathbb{N}$, we have

$$d(x,T^{m}x)⪯\frac{1}{1-\lambda }d(x,Tx).$$

(13)

Proof We prove the statement by induction on m. If $m=1$, then inequality (13) holds since $0\le \lambda <1$. Assume that $n\in \mathbb{N}$ and assume that (7) is satisfied for any natural number $m\le n$. Then we have to prove that

$$d(x,T^{n+1}x)⪯\frac{1}{1-\lambda }d(x,Tx).$$

(14)

From the triangle inequality, we obtain

$$d(x,T^{n+1}x)⪯d(x,Tx)+d(Tx,T^{n+1}x).$$

(15)

By Lemma 3.3, we get

$$d(Tx,T^{n+1}x)⪯\lambda co\{d(x,Tx),\dots ,d(x,T^{n+1}x)\}.$$

(16)

By the induction hypothesis, we have that (13) holds for all $m\le n$. Then it follows from (16), (P2), and (P6) that

$$d(Tx,T^{n+1}x)⪯co\{\frac{\lambda }{1-\lambda }d(x,Tx),\lambda d(x,T^{n+1}x)\}.$$

This inequality implies that there exists $\alpha \in [0,1]$ such that

$$d(Tx,T^{n+1}x)⪯\alpha \frac{\lambda }{1-\lambda }d(x,Tx)+(1-\alpha )\lambda d(x,T^{n+1}x).$$

(17)

Combining (15) and (17), we get

$$d(x,T^{n+1}x)⪯d(x,Tx)+\alpha \frac{\lambda }{1-\lambda }d(x,Tx)+(1-\alpha )\lambda d(x,T^{n+1}x),$$

which is equivalent to the following inequality:

$$(1-\lambda +\alpha \lambda )d(x,T^{n+1}x)⪯\frac{1-\lambda +\alpha \lambda }{1-\lambda }d(x,Tx).$$

Multiplying both sides of this inequality by $1/(1-\lambda +\alpha \lambda )$, we obtain (14). This completes the proof of the lemma. □

Lemma 3.5 Let $(X,d)$ be a cone metric space over an ordered vector space $(Y,⪯)$, and let $T:X\to X$ be a quasi-contraction with contraction constant $\lambda \in [0,1)$. Then for all $x,y\in X$, we have

$$d(x,Tx)⪯\alpha d(x,y)+\beta d(x,Ty),$$

(18)

where $\alpha =\lambda /(1-\lambda )$ and $\beta =(1+\lambda )/(1-\lambda )$.

Proof Let $x,y\in X$ be fixed. First we shall prove that

$$d(Tx,Ty)⪯\lambda (d(x,y)+d(x,Tx)+d(x,Ty)).$$

(19)

It follows from Definition 1.5 that there exist five nonnegative numbers $\alpha ,\beta ,\gamma ,\mu ,\nu$ such that $\alpha +\beta +\gamma +\mu +\nu =1$ and

$$d(Tx,Ty)⪯\lambda (\alpha d(x,y)+\beta d(x,Tx)+\gamma d(y,Ty)+\mu d(x,Ty)+\nu d(y,Tx)).$$

From this and the inequalities

$$d(y,Ty)⪯d(x,y)+d(x,Ty)\text{and}d(y,Tx)⪯d(x,y)+d(x,Tx),$$

we obtain

$$d(Tx,Ty)⪯\lambda ((\alpha +\gamma +\nu )d(x,y)+(\beta +\nu )d(x,Tx)+(\gamma +\mu )d(x,Ty)).$$

This inequality yields (19) since $\alpha +\gamma +\nu \le 1$, $\beta +\nu \le 1$ and $\gamma +\mu \le 1$. Now we are ready to prove (18). From the triangle inequality, we get

$$d(x,Tx)⪯d(x,Ty)+d(Tx,Ty).$$

From this and (19), we obtain

$$d(x,Tx)⪯d(x,Ty)+\lambda (d(x,y)+d(x,Tx)+d(x,Ty)),$$

which can be presented in the following equivalent form:

$$(1-\lambda )d(x,Tx)⪯\lambda d(x,y)+(1+\lambda )d(x,Ty).$$

Multiplying both sides of this inequality by $1/(1-\lambda )$, we get (18). □

Lemma 3.6 Let $(X,d)$ be a cone metric space over an ordered vector space $(Y,⪯)$. Then every quasi-contraction $T:X\to X$ has at most one fixed point in X.

Proof Suppose that x and y are two fixed points of T. It follows from the inequality (4) and properties (P4) and (P6) that $d(x,y)⪯\lambda d(x,y)$ which implies $d(x,y)⪯0$. On the other hand, $d(x,y)⪰0$. Hence, $d(x,y)=0$, which yields $x=y$. □

4 Main result

Now we are ready to state the main result of this paper. Let $(X,d)$ be a complete cone metric space over an ordered vector space Y. Recall that for a point $x_0\in X$ and a vector $r\in Y$ with $r⪰0$, the set $\overline{U}(x_0,r)=\{x\in X:d(x,x_0)⪯r\}$ is called a closed ball with center $x_0$ and radius r.

Theorem 4.1 Let $(X,d)$ be a complete cone metric space over a solid vector space $(Y,⪯)$, and let $T:X\to X$ be a quasi-contraction with contraction constant $\lambda \in [0,1)$. Then the following statements hold true:

• (i) Existence, uniqueness and localization. T has a unique fixed point ξ which belongs to the closed ball $\overline{U}(x,r)$ with radius

  $$r=\frac{1}{1-\lambda }d(x,Tx),$$

where x is any point in X.

• (ii) Convergence of Picard iteration. Starting from any point $x\in X$ the Picard sequence $(T^{n}x)$ remains in the closed ball $\overline{U}(x,r)$ and converges to ξ.

• (iii) A priori error estimate. For every point $x\in X$ the following a priori estimate holds:

  $$d(T^{n}x,\xi )⪯\frac{{\lambda }^n}{1-\lambda }d(x,Tx)\mathit{\text{for all }}n\ge 0.$$

  (20)

• (iv) A posteriori error estimates. For every point $x\in X$ the following a posteriori estimate holds:

  $$d(T^{n}x,\xi )⪯\frac{1}{1-\lambda }d(T^{n}x,T^{n+1}x)\mathit{\text{for all }}n\ge 0,$$

  (21)
  $$d(T^{n}x,\xi )⪯\frac{\lambda }{1-\lambda }d(T^{n}x,T^{n-1}x)\mathit{\text{for all }}n\ge 1.$$

  (22)

Proof Let x be an arbitrary point in X. By Lemma 3.3, for all $m,n\in \mathbb{N}$ with $m\ge n$, we have

$$d(T^{n}x,T^{m}x)⪯{\lambda }^{n}co\{d(x,Tx),\dots ,d(x,T^{m}x)\}.$$

From this, Lemma 3.4, and properties (P2) and (P6), we deduce

$$d(T^{n}x,T^{m}x)⪯b_n,\text{where }{b}_n=\frac{{\lambda }^n}{1-\lambda }d(x,Tx).$$

(23)

Note that $(b_n)$ is a sequence in Y which converges to 0 since ${\lambda }^n\to 0$ in ℝ. Now applying Theorem 2.7 to the Picard sequence $(T^{n}x)$, we conclude that there exists a point $\xi \in X$ such that $(T^{n}x)$ converges to ξ and $d(T^{n}x,\xi )⪯b_n$ for every $n\ge 0$. The last inequality coincides with the estimate (20). Setting $n=0$ in (20), we get

$$d(x,\xi )⪯\frac{1}{1-\lambda }d(x,Tx)$$

(24)

which means that $\xi \in \overline{U}(x,r)$. The inequality (24) holds for every point $x\in X$. Applying (24) to the point $T^{n}x$, we obtain (21). Setting $n=1$ in (20), we get

$$d(Tx,\xi )⪯\frac{\lambda }{1-\lambda }d(x,Tx).$$

(25)

Applying (25) to the point $T^{n-1}x$, we get (22). Setting $n=0$ in (23), we obtain $d(x,T^{m}x)⪯b_0$ for every $m\ge 0$. Hence, the sequence $(T^{n}x)$ lies in the ball $\overline{U}(x,r)$ since $r=b_0$.

It follows from Lemma 3.5 that ξ satisfies condition (5). Hence, by Theorem 2.8, we conclude that ξ is a fixed point of T. The uniqueness of the fixed point follows from Lemma 3.6. □

Theorem 4.1 extends and complements the recent results of Ding et al. [[24], Theorem 3.1] and Zhang [[19], Theorem 3] as well as previous results due to Ilić and Rakočević [[20], Theorem 2.1], Kadelburg et al. [[21], Theorem 2.2], Rezapour et al. [[23], Theorem 2.1] and Kadelburg et al. [[8], Theorem 3.5(b)] who have studied quasi-contraction mappings of the type (3).

Theorem 4.1 also extends and complements the results of Abbas and Rhoades [[39], Corollary 2.3], Olaleru [[40], Theorem 2.1], Azam et al. [[41], Theorem 2.2], Song et al. [[42], Corollary 2.1]. These authors have studied the class of mappings satisfying a contractive condition of the type

$$d(Tx,Ty)⪯\alpha d(x,y)+\beta d(x,Tx)+\gamma d(y,Ty)+\mu d(x,Ty)+\nu d(y,Tx)$$

(26)

for all $x,y\in X$, where α, β, γ, μ and ν are five nonnegative constants such that $\alpha +\beta +\gamma +\mu +\nu <1$. In this case Theorem 4.1 holds with $\lambda =\alpha +\beta +\gamma +\mu +\nu$ since condition (26) implies condition (4) with this λ. Let us note that in this special case Theorem 4.1 holds even with $\lambda =(\alpha +\delta )/(1-\delta )$, where $\delta =(\beta +\gamma +\mu +\nu )/2$.

5 Examples

Zhang [[19], Example 1] gives an example showing that the set of all quasi-contractions of the type (3) is a proper subset of the set of all quasi-contractions defined by (4). In order to prove this, he considers a selfmapping of a cone metric space X over a normal solid vector space Y. Ding et al. [[24], Example 4.1] provide a similar example, but for the case of a non-normal solid vector space Y.

The aim of this section is to unify these two examples. Let ℬ denote the set of all quasi-contractions of the type (3), and let denote the set of all quasi-contractions defined by (4), that is,

$$\begin{array}{c}\mathcal{B}=\{T:X\to X\mid d(Tx,Ty)⪯\lambda \{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}\},\hfill \\ \mathcal{C}=\{T:X\to X\mid d(Tx,Ty)⪯\lambda co\{d(x,y),d(x,Tx),d(y,Ty),d(x,Ty),d(y,Tx)\}\}.\hfill \end{array}$$

Now we shall construct a family of examples which show that ℬ is a proper subset of . In particular, this family contains both the example of Zhang [19] and the example of Ding et al. [24].

Definition 5.1 Let $(Y,⪯)$ be an ordered vector space, and let $a,b,c⪰0$ be three vectors in Y. We say that the triple $(a,b,c)$ satisfies property (C) if the following two statements hold:

• $a⪯\lambda co\{b,c\}$ for some $\lambda \in [0,1)$, $b⪯a+c$ and $c⪯a+b$.

• $a⪯k\{b,c\}$ is wrong for every $k\in [0,1)$.

Proposition 5.2 Let $Y=\mathbb{R}$ be endowed with the usual ordering ≤. Then there are no triples $(a,b,c)$ in Y satisfying property (C).

Proof Assume that there is a triple $(a,b,c)$ in Y with property (C). Then $a⪯\lambda max\{b,c\}$ for some $\lambda \in [0,1)$. On the other hand, $a⪯kmax\{b,c\}$ is wrong for every $k\in [0,1)$. This is a contradiction which proves the proposition. □

Proposition 5.3 Let $Y={\mathbb{R}}^n$ ($n\ge 2$) be endowed with coordinate-wise ordering ⪯. Then in Y there exist infinitely many triples $(a,b,c)$ satisfying property (C).

Proof Choose three real numbers α, β and γ such that

$$\beta <\gamma <3\beta \text{and}max\{\beta ,\gamma -\beta \}\le \alpha <\frac{\beta +\gamma }{2}.$$

Then the vectors $a=(\alpha ,\dots ,\alpha )$, $b=(\beta ,\dots ,\beta ,\gamma )$ and $c=(\gamma ,\dots ,\gamma ,\beta )$ satisfy property (C) with $\lambda =\frac{2\alpha }{\beta +\gamma }$. □

Proposition 5.4 Let $Y=C^n[0,1]$ ($n\ge 2$) be endowed with point-wise ordering ⪯. Then in Y there exist infinitely many triples $(a,b,c)$ satisfying property (C).

Proof Choose three real numbers α, β and γ as in the proof of Proposition 5.3, then choose a real number δ such that

$$\gamma -\alpha \le \delta <\gamma \text{and}\delta \le \frac{\alpha +\gamma -\beta }{2}.$$

Then the functions $a(t)=\alpha$, $b(t)=\beta +\delta t$ and $c(t)=\gamma -\delta t$ satisfy property (C) with $\lambda =\frac{2\alpha }{\beta +\gamma }$. □

Example 5.5 Let $(Y,⪯)$ be an arbitrary solid vector space, and let $(a,b,c)$ be any triple in Y with property (C). Furthermore, let $X=\{x,y,z\}$, and let $d:X\times X\to Y$ be defined by

$$d(x,y)=a,d(x,z)=b,d(y,z)=c,$$

$d(u,v)=d(u,v)$ and $d(u,u)=0$ for $u,v\in X$. Then $(X,d)$ is a complete cone metric space over Y since the triple $(a,b,c)$ satisfies property (C). Consider now the mapping $T:X\to X$ defined by

$$Tx=x,Ty=x,Tz=y.$$

Using Lemma 3.1, it is easy to prove that $T\in \mathcal{C}$ if and only if $a⪯\lambda co\{b,c\}$ for some $\lambda \in [0,1)$. Analogously, taking into account Remark 3.2, one can easily prove that $T\in \mathcal{B}$ if and only if $a⪯k\{b,c\}$ for some $k\in [0,1)$. Now taking into account that the triple $(a,b,c)$ satisfies property (C), we conclude that $T\in \mathcal{C}$ and $T\notin \mathcal{B}$. Hence, ℬ is a proper subset of .