In this section, we will present some fixed point theorems for contractive mappings in the setting of cone *b*-metric spaces. Furthermore, we will give examples to support our main results.

**Theorem 2.1** *Let* (X,d) *be a complete cone* *b*-*metric space with the coefficient* s\ge 1. *Suppose the mapping* T:X\to X *satisfies the contractive condition*

d(Tx,Ty)\le \lambda d(x,y),\phantom{\rule{1em}{0ex}}\mathit{\text{for}}x,y\in X,

*where* \lambda \in [0,1) *is a constant*. *Then* *T* *has a unique fixed point in* *X*. *Furthermore*, *the iterative sequence* \{{T}^{n}x\} *converges to the fixed point*.

*Proof* Choose {x}_{0}\in X. We construct the iterative sequence \{{x}_{n}\}, where {x}_{n}=T{x}_{n-1}, n\ge 1, *i.e.*, {x}_{n+1}=T{x}_{n}={T}^{n+1}{x}_{0}. We have

d({x}_{n+1},{x}_{n})=d(T{x}_{n},T{x}_{n-1})\le \lambda d({x}_{n},{x}_{n-1})\le \cdots \le {\lambda}^{n}d({x}_{1},{x}_{0}).

For any m\ge 1, p\ge 1, it follows that

\begin{array}{rcl}d({x}_{m+p},{x}_{m})& \le & s[d({x}_{m+p},{x}_{m+p-1})+d({x}_{m+p-1},{x}_{m})]\\ =& sd({x}_{m+p},{x}_{m+p-1})+sd({x}_{m+p-1},{x}_{m})\\ \le & sd({x}_{m+p},{x}_{m+p-1})+{s}^{2}[d({x}_{m+p-1},{x}_{m+p-2})+d({x}_{m+p-2},{x}_{m})]\\ =& sd({x}_{m+p},{x}_{m+p-1})+{s}^{2}d({x}_{m+p-1},{x}_{m+p-2})+{s}^{2}d({x}_{m+p-2},{x}_{m})\\ \le & sd({x}_{m+p},{x}_{m+p-1})+{s}^{2}d({x}_{m+p-1},{x}_{m+p-2})+{s}^{3}d({x}_{m+p-2},{x}_{m+p-3})+\cdots \\ +{s}^{p-1}d({x}_{m+2},{x}_{m+1})+{s}^{p-1}d({x}_{m+1},{x}_{m})\\ \le & s{\lambda}^{m+p-1}d({x}_{1},{x}_{0})+{s}^{2}{\lambda}^{m+p-2}d({x}_{1},{x}_{0})+{s}^{3}{\lambda}^{m+p-3}d({x}_{1},{x}_{0})+\cdots \\ +{s}^{p-1}{\lambda}^{m+1}d({x}_{1},{x}_{0})+{s}^{p-1}{\lambda}^{m}d({x}_{1},{x}_{0})\\ =& (s{\lambda}^{m+p-1}+{s}^{2}{\lambda}^{m+p-2}+{s}^{3}{\lambda}^{m+p-3}+\cdots +{s}^{p-1}{\lambda}^{m+1})d({x}_{1},{x}_{0})\\ +{s}^{p-1}{\lambda}^{m}d({x}_{1},{x}_{0})\\ =& \frac{s{\lambda}^{m+p}[{(s{\lambda}^{-1})}^{p-1}-1]}{s-\lambda}d({x}_{1},{x}_{0})+{s}^{p-1}{\lambda}^{m}d({x}_{1},{x}_{0})\\ \le & \frac{{s}^{p}{\lambda}^{m+1}}{s-\lambda}d({x}_{1},{x}_{0})+{s}^{p-1}{\lambda}^{m}d({x}_{1},{x}_{0}).\end{array}

Let \theta \ll c be given. Notice that \frac{{s}^{p}{\lambda}^{m+1}}{s-\lambda}d({x}_{1},{x}_{0})+{s}^{p-1}{\lambda}^{m}d({x}_{1},{x}_{0})\to \theta as m\to \mathrm{\infty} for any *k*. Making full use of Lemma 1.8, we find {m}_{0}\in \mathbb{N} such that

\frac{{s}^{p}{\lambda}^{m+1}}{s-\lambda}d({x}_{1},{x}_{0})+{s}^{p-1}{\lambda}^{m}d({x}_{1},{x}_{0})\ll c,

for each m>{m}_{0}. Thus,

d({x}_{m+p},{x}_{m})\le \frac{{s}^{p}{\lambda}^{m+1}}{s-\lambda}d({x}_{1},{x}_{0})+{s}^{p-1}{\lambda}^{m}d({x}_{1},{x}_{0})\ll c

for all m>{m}_{0} and any *p*. So, by Lemma 1.9, \{{x}_{n}\} is a Cauchy sequence in (X,d). Since (X,d) is a complete cone *b*-metric space, there exists {x}^{\ast}\in X such that {x}_{n}\to {x}^{\ast}. Take {n}_{0}\in \mathbb{N} such that d({x}_{n},{x}^{\ast})\ll \frac{c}{s(\lambda +1)} for all n>{n}_{0}. Hence,

d(T{x}^{\ast},{x}^{\ast})\le s[d(T{x}^{\ast},T{x}_{n})+d(T{x}_{n},{x}^{\ast})]\le s[\lambda d({x}^{\ast},{x}_{n})+d({x}_{n+1},{x}^{\ast})]\ll c,

for each n>{n}_{0}. Then, by Lemma 1.10, we deduce that d(T{x}^{\ast},{x}^{\ast})=\theta, *i.e.*, T{x}^{\ast}={x}^{\ast}. That is, {x}^{\ast} is a fixed point of *T*.

Now we show that the fixed point is unique. If there is another fixed point {y}^{\ast}, by the given condition,

d({x}^{\ast},{y}^{\ast})=d(T{x}^{\ast},T{y}^{\ast})\le \lambda d({x}^{\ast},{y}^{\ast}).

By Lemma 1.11, {x}^{\ast}={y}^{\ast}. The proof is completed. □

**Example 2.2** Let X=[0,1], E={\mathbb{R}}^{2} and p>1 be a constant. Take P=\{(x,y)\in E:x,y\ge 0\}. We define d:X\times X\to E as

d(x,y)=({|x-y|}^{p},{|x-y|}^{p})\phantom{\rule{1em}{0ex}}\text{for all}x,y\in X.

Then (X,d) is a complete cone *b*-metric space. Let us define T:X\to X as

Tx=\frac{1}{2}x-\frac{1}{4}{x}^{2}\phantom{\rule{1em}{0ex}}\text{for all}x\in X.

Therefore,

\begin{array}{rcl}d(Tx,Ty)& =& ({|Tx-Ty|}^{p},{|Tx-Ty|}^{p})\\ =& ({|\frac{1}{2}(x-y)-\frac{1}{4}(x-y)(x+y)|}^{p},{|\frac{1}{2}(x-y)-\frac{1}{4}(x-y)(x+y)|}^{p})\\ =& ({|x-y|}^{p}\cdot {|\frac{1}{2}-\frac{1}{4}(x+y)|}^{p},{|x-y|}^{p}\cdot {|\frac{1}{2}-\frac{1}{4}(x+y)|}^{p})\\ \le & \frac{1}{{2}^{p}}({|x-y|}^{p},{|x-y|}^{p})\\ =& \frac{1}{{2}^{p}}d(x,y).\end{array}

Here 0\in X is the unique fixed point of *T*.

**Theorem 2.3** *Let* (X,d) *be a complete cone* *b*-*metric space with the coefficient* s\ge 1. *Suppose the mapping* T:X\to X *satisfies the contractive condition*

d(Tx,Ty)\le {\lambda}_{1}d(x,Tx)+{\lambda}_{2}d(y,Ty)+{\lambda}_{3}d(x,Ty)+{\lambda}_{4}d(y,Tx),\phantom{\rule{1em}{0ex}}\mathit{\text{for}}x,y\in X,

*where the constant* {\lambda}_{i}\in [0,1) *and* {\lambda}_{1}+{\lambda}_{2}+s({\lambda}_{3}+{\lambda}_{4})<min\{1,\frac{2}{s}\}, i=1,2,3,4. *Then* *T* *has a unique fixed point in* *X*. *Moreover*, *the iterative sequence* \{{T}^{n}x\} *converges to the fixed point*.

*Proof* Fix {x}_{0}\in X and set {x}_{1}=T{x}_{0} and {x}_{n+1}=T{x}_{n}={T}^{n+1}{x}_{0}. Firstly, we see

\begin{array}{rcl}d({x}_{n+1},{x}_{n})& =& d(T{x}_{n},T{x}_{n-1})\\ \le & {\lambda}_{1}d({x}_{n},T{x}_{n})+{\lambda}_{2}d({x}_{n-1},T{x}_{n-1})+{\lambda}_{3}d({x}_{n},T{x}_{n-1})+{\lambda}_{4}d({x}_{n-1},T{x}_{n})\\ \le & {\lambda}_{1}d({x}_{n},{x}_{n+1})+{\lambda}_{2}d({x}_{n-1},{x}_{n})+s{\lambda}_{4}[d({x}_{n-1},{x}_{n})+d({x}_{n},{x}_{n+1})]\\ =& ({\lambda}_{1}+s{\lambda}_{4})d({x}_{n},{x}_{n+1})+({\lambda}_{2}+s{\lambda}_{4})d({x}_{n},{x}_{n-1}).\end{array}

It follows that

(1-{\lambda}_{1}-s{\lambda}_{4})d({x}_{n+1},{x}_{n})\le ({\lambda}_{2}+s{\lambda}_{4})d({x}_{n},{x}_{n-1}).

(2.1)

Secondly,

\begin{array}{rcl}d({x}_{n+1},{x}_{n})& =& d(T{x}_{n},T{x}_{n-1})=d(T{x}_{n-1},T{x}_{n})\\ \le & {\lambda}_{1}d({x}_{n-1},T{x}_{n-1})+{\lambda}_{2}d({x}_{n},T{x}_{n})+{\lambda}_{3}d({x}_{n-1},T{x}_{n})+{\lambda}_{4}d({x}_{n},T{x}_{n-1})\\ \le & {\lambda}_{1}d({x}_{n-1},{x}_{n})+{\lambda}_{2}d({x}_{n},{x}_{n+1})+s{\lambda}_{3}[d({x}_{n-1},{x}_{n})+d({x}_{n},{x}_{n+1})]\\ =& ({\lambda}_{2}+s{\lambda}_{3})d({x}_{n},{x}_{n+1})+({\lambda}_{1}+s{\lambda}_{3})d({x}_{n},{x}_{n-1}).\end{array}

This establishes that

(1-{\lambda}_{2}-s{\lambda}_{3})d({x}_{n+1},{x}_{n})\le ({\lambda}_{1}+s{\lambda}_{3})d({x}_{n},{x}_{n-1}).

(2.2)

Adding up (2.1) and (2.2) yields

d({x}_{n+1},{x}_{n})\le \frac{{\lambda}_{1}+{\lambda}_{2}+s({\lambda}_{3}+{\lambda}_{4})}{2-{\lambda}_{1}-{\lambda}_{2}-s({\lambda}_{3}+{\lambda}_{4})}d({x}_{n},{x}_{n-1}).

Put \lambda =\frac{{\lambda}_{1}+{\lambda}_{2}+s({\lambda}_{3}+{\lambda}_{4})}{2-{\lambda}_{1}-{\lambda}_{2}-s({\lambda}_{3}+{\lambda}_{4})}, it is easy to see that 0\le \lambda <1. Thus,

d({x}_{n+1},{x}_{n})\le \lambda d({x}_{n},{x}_{n-1})\le \cdots \le {\lambda}^{n}d({x}_{1},{x}_{0}).

Following an argument similar to that given in Theorem 2.1, there exists {x}^{\ast}\in X such that {x}_{n}\to {x}^{\ast}. Let c\gg \theta be arbitrary. Since {x}_{n}\to {x}^{\ast}, there exists *N* such that

d({x}_{n},{x}^{\ast})\ll \frac{2-s{\lambda}_{1}-s{\lambda}_{2}-{s}^{2}{\lambda}_{3}-{s}^{2}{\lambda}_{4}}{2{s}^{2}+2s}c\phantom{\rule{1em}{0ex}}\text{for all}nN.

Next we claim that {x}^{\ast} is a fixed point of *T*. Actually, on the one hand,

\begin{array}{rcl}d(T{x}^{\ast},{x}^{\ast})& \le & s[d(T{x}^{\ast},T{x}_{n})+d(T{x}_{n},{x}^{\ast})]=sd(T{x}^{\ast},T{x}_{n})+sd({x}_{n+1},{x}^{\ast})\\ \le & s[{\lambda}_{1}d({x}^{\ast},T{x}^{\ast})+{\lambda}_{2}d({x}_{n},T{x}_{n})+{\lambda}_{3}d({x}^{\ast},T{x}_{n})+{\lambda}_{4}d({x}_{n},T{x}^{\ast})]+sd({x}_{n+1},{x}^{\ast})\\ =& s[{\lambda}_{1}d({x}^{\ast},T{x}^{\ast})+{\lambda}_{2}d({x}_{n},{x}_{n+1})+{\lambda}_{3}d({x}^{\ast},{x}_{n+1})+{\lambda}_{4}d({x}_{n},T{x}^{\ast})]+sd({x}_{n+1},{x}^{\ast})\\ \le & s{\lambda}_{1}d({x}^{\ast},T{x}^{\ast})+{s}^{2}{\lambda}_{2}d({x}_{n},{x}^{\ast})+{s}^{2}{\lambda}_{2}d({x}^{\ast},{x}_{n+1})+s{\lambda}_{3}d({x}^{\ast},{x}_{n+1})\\ +{s}^{2}{\lambda}_{4}d({x}_{n},{x}^{\ast})+{s}^{2}{\lambda}_{4}d({x}^{\ast},T{x}^{\ast})+sd({x}_{n+1},{x}^{\ast})\\ =& (s{\lambda}_{1}+{s}^{2}{\lambda}_{4})d({x}^{\ast},T{x}^{\ast})+({s}^{2}{\lambda}_{2}+{s}^{2}{\lambda}_{4})d({x}_{n},{x}^{\ast})\\ +({s}^{2}{\lambda}_{2}+s{\lambda}_{3}+s)d({x}^{\ast},{x}_{n+1}),\end{array}

which implies that

(1-s{\lambda}_{1}-{s}^{2}{\lambda}_{4})d({x}^{\ast},T{x}^{\ast})\le ({s}^{2}{\lambda}_{2}+{s}^{2}{\lambda}_{4})d({x}_{n},{x}^{\ast})+({s}^{2}{\lambda}_{2}+s{\lambda}_{3}+s)d({x}^{\ast},{x}_{n+1}).

(2.3)

On the other hand,

\begin{array}{rcl}d({x}^{\ast},T{x}^{\ast})& \le & s[d({x}^{\ast},T{x}_{n})+d(T{x}_{n},T{x}^{\ast})]=sd({x}^{\ast},{x}_{n+1})+sd(T{x}_{n},T{x}^{\ast})\\ \le & sd({x}^{\ast},{x}_{n+1})+s[{\lambda}_{1}d({x}_{n},T{x}_{n})+{\lambda}_{2}d({x}^{\ast},T{x}^{\ast})+{\lambda}_{3}d({x}_{n},T{x}^{\ast})+{\lambda}_{4}d({x}^{\ast},T{x}_{n})]\\ =& sd({x}^{\ast},{x}_{n+1})+s[{\lambda}_{1}d({x}_{n},{x}_{n+1})+{\lambda}_{2}d({x}^{\ast},T{x}^{\ast})+{\lambda}_{3}d({x}_{n},T{x}^{\ast})+{\lambda}_{4}d({x}^{\ast},{x}_{n+1})]\\ \le & sd({x}^{\ast},{x}_{n+1})+{s}^{2}{\lambda}_{1}d({x}_{n},{x}^{\ast})+{s}^{2}{\lambda}_{1}d({x}^{\ast},{x}_{n+1})+s{\lambda}_{2}d({x}^{\ast},T{x}^{\ast})\\ +{s}^{2}{\lambda}_{3}d({x}_{n},{x}^{\ast})+{s}^{2}{\lambda}_{3}d({x}^{\ast},T{x}^{\ast})+s{\lambda}_{4}d({x}^{\ast},{x}_{n+1})\\ =& (s{\lambda}_{2}+{s}^{2}{\lambda}_{3})d({x}^{\ast},T{x}^{\ast})+({s}^{2}{\lambda}_{1}+{s}^{2}{\lambda}_{3})d({x}_{n},{x}^{\ast})\\ +({s}^{2}{\lambda}_{1}+s{\lambda}_{4}+s)d({x}^{\ast},{x}_{n+1}),\end{array}

which means that

(1-s{\lambda}_{2}-{s}^{2}{\lambda}_{3})d({x}^{\ast},T{x}^{\ast})\le ({s}^{2}{\lambda}_{1}+{s}^{2}{\lambda}_{3})d({x}_{n},{x}^{\ast})+({s}^{2}{\lambda}_{1}+s{\lambda}_{4}+s)d({x}^{\ast},{x}_{n+1}).

(2.4)

Combining (2.3) and (2.4) yields

\begin{array}{c}(2-s{\lambda}_{1}-s{\lambda}_{2}-{s}^{2}{\lambda}_{3}-{s}^{2}{\lambda}_{4})d({x}^{\ast},T{x}^{\ast})\hfill \\ \phantom{\rule{1em}{0ex}}\le {s}^{2}({\lambda}_{1}+{\lambda}_{2}+{\lambda}_{3}+{\lambda}_{4})d({x}_{n},{x}^{\ast})+({s}^{2}{\lambda}_{1}+{s}^{2}{\lambda}_{2}+s{\lambda}_{3}+s{\lambda}_{4}+2s)d({x}^{\ast},{x}_{n+1})\hfill \\ \phantom{\rule{1em}{0ex}}\le {s}^{2}d({x}_{n},{x}^{\ast})+({s}^{2}+2s)d({x}^{\ast},{x}_{n+1}).\hfill \end{array}

Simple calculations ensure that

d({x}^{\ast},T{x}^{\ast})\le \frac{{s}^{2}d({x}_{n},{x}^{\ast})+({s}^{2}+2s)d({x}^{\ast},{x}_{n+1})}{2-s{\lambda}_{1}-s{\lambda}_{2}-{s}^{2}{\lambda}_{3}-{s}^{2}{\lambda}_{4}}\ll c.

It is easy to see from Lemma 1.10 that d({x}^{\ast},T{x}^{\ast})=\theta, *i.e.*, {x}^{\ast} is a fixed point of *T*. Finally, we show the uniqueness of the fixed point. Indeed, if there is another fixed point {y}^{\ast}, then

\begin{array}{rcl}d({x}^{\ast},{y}^{\ast})& =& d(T{x}^{\ast},T{y}^{\ast})\\ \le & {\lambda}_{1}d({x}^{\ast},T{x}^{\ast})+{\lambda}_{2}d({y}^{\ast},T{y}^{\ast})+{\lambda}_{3}d({x}^{\ast},T{y}^{\ast})+{\lambda}_{4}d({y}^{\ast},T{x}^{\ast})\\ \le & s{\lambda}_{3}[d({x}^{\ast},{y}^{\ast})+d({y}^{\ast},T{y}^{\ast})]+s{\lambda}_{4}[d({y}^{\ast},{x}^{\ast})+d({x}^{\ast},T{x}^{\ast})]\\ =& s({\lambda}_{3}+{\lambda}_{4})d({x}^{\ast},{y}^{\ast}).\end{array}

Owing to 0\le s({\lambda}_{3}+{\lambda}_{4})<1, we deduce from Lemma 1.11 that {x}^{\ast}={y}^{\ast}. Therefore, we complete the proof. □

**Remark 2.4** Theorem 2.1 extends the famous Banach contraction principle to that in the setting of cone *b*-metric spaces.

**Remark 2.5** Any fixed point theorem in the setting of a metric space, a *b*-metric space or a cone metric space cannot cope with Example 2.2. So, Example 2.2 shows that the fixed point theory of cone *b*-metric spaces offers independently a strong tool for studying the positive fixed points of some nonlinear operators and the positive solutions of some operator equations.

**Remark 2.6** The main results are some valuable additions to the available references regarding cone *b*-metric spaces since we have known few fixed point theorems of contractive mappings in the setting of cone *b*-metric spaces.