A note on fixed point results in complex-valued metric spaces

Abstract

In this paper, we prove that the fixed point results in the context of complex-valued metric spaces can be obtained as a consequence of corresponding existing results in the literature in the setting of associative metric spaces. In particular, we deduce that any complex metric space is a special case of cone metric spaces with a normal cone.

1 Introduction and preliminaries

The notion of complex-valued metric spaces was introduced by Azam et al. [1], as a generalization of metric spaces to investigate the existence and uniqueness of fixed point results for mappings satisfying a rational inequalities. Following this paper, a number of authors have reported several fixed point results for various mapping satisfying a rational inequalities in the context of complex-valued metric spaces; see e.g. [1–3] and the related references therein.

The aim of this short note is to emphasize that the complex-valued metric space is an example of the cone metric space that was introduced in [4–6] under the name K-metric and K-normed spaces and re-introduced by Huang and Zhang [7]. It is well known that if the cone is normal then the corresponding cone metric associates a metric. There are some other approaches to induce a metric from cone metric; see e.g. [8–16]. As a consequence of these observations, we notice that fixed point results in the context of complete complex-valued metric spaces can be deduced the corresponding fixed point results on (associative) complete metric space. Based on the discussion above, for our purpose, we first prove the existence of common fixed point theorems for multi-valued mapping in the context of complete metric space. Then we derive the main results of the recent paper of Ahmad et al. [2] as corollaries of our results.

For the sake of completeness we recollect some basic definitions and fundamentals results on the topic in the literature. We mainly follow the notions and notations of Azam et al. in [1].

Let ℂ be the set of complex numbers and $z_{1},z_{2}\in \mathbb{C}$. Define a partial order ≾ on ℂ as follows:

$$z_{1} \precsim z_{2} \quad\mbox{if and only if} \quad \operatorname{Re}(z_{1}) \leq \operatorname{Re}(z_{2}), \operatorname{Im}(z_{1}) \leq \operatorname{Im}(z_{2}). $$

It follows that

$$z_{1} \precsim z_{2} $$

if one of the following conditions is satisfied:

(h₁):

$\operatorname{Re}(z_{1})=\operatorname{Re}(z_{2})$; $\operatorname{Im}(z_{1})<\operatorname{Im}(z_{2})$,

(h₂):

$\operatorname{Re}(z_{1})<\operatorname{Re}(z_{2})$; $\operatorname{Im}(z_{1})=\operatorname{Im}(z_{2})$,

(h₃):

$\operatorname{Re}(z_{1})<\operatorname{Re}(z_{2})$; $\operatorname{Im}(z_{1})<\operatorname{Im}(z_{2})$,

(h₄):

$\operatorname{Re}(z_{1})=\operatorname{Re}(z_{2})$; $\operatorname{Im}(z_{1})=\operatorname{Im}(z_{2})$.

In particular, we shall write $z_{1} \precnsim z_{2}$ if $z_{1} \neq z_{2}$ and one of (h₁), (h₂), and (h₃) is satisfied. Further we write $z_{1} \prec z_{2}$ if only (h₃) is satisfied. Note that

$$0\precsim z_{1} \precnsim z_{2} \quad \Longrightarrow\quad |z_{1}|<|z_{2}|, $$

where $|\cdot|$ represents the modulus or magnitude of z, and

$$z_{1}\precsim z_{2},\qquad z_{2} \prec z_{3} \quad \Longrightarrow\quad z_{1} \prec z_{3}. $$

Definition 1

[1]

Let X be a nonempty set. A function $d:X \times X\to \mathbb{C}$ is called a complex-valued metric on X, if it satisfies the following conditions:

(b₁):

$0 \precsim d(x,y)$ for all $x,y\in X$ and $d(x,y) = 0$, if and only if $x = y$,

(b₂):

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

(b₃):

$d(x,y) \precsim d(x,z) + d(y,z)$, for all $x,y,z\in X$.

Furthermore, the pair $(X,d)$ is called a complex-valued metric space.

Let $\{x_{n}\}$ be a sequence in X and $x\in X$. If for every $c\in \mathbb{C}$, with $0\prec c$ there is $n_{0}\in \mathbb{N}$ such that for all $n > n_{0}$, $d(x_{n},x) \prec c$, then $\{x_{n}\}$ is said to be convergent, $\{x_{n}\}$ converges to x and x is the limit point of $\{ x_{n}\}$. We denote this by $\lim_{n}x_{n} = x$, or $x_{n}\to x$, as $n\to\infty$. If for every $c\in \mathbb{C}$ with $0\prec c$ there is $n_{0}\in \mathbb{N}$ such that for all $n > n_{0}$, $d(x_{n}, x_{n+m})\prec c$, then $\{x_{n}\}$ is called a Cauchy sequence in $(X, d)$. If every Cauchy sequence is convergent in $(X, d)$, then $(X, d)$ is called a complete complex-valued metric space.

Lemma 2

[1, Lemma 2, Azam et al.]

Let $(X, d)$ be a complex-valued metric space and let $\{x_{n}\}$ be a sequence in X. Then $\{x_{n}\}$ converges to x if and only if $|d(x_{n},x)|\to0$ as $n\to\infty$.

Lemma 3

[1, Lemma 3, Azam et al.]

Let $(X, d)$ be a complex-valued metric space and let $\{x_{n}\}$ be a sequence in X. Then $\{x_{n}\}$ is a Cauchy sequence if and only if $|d(x_{n},x_{n+m})|\to0$ as $n\to\infty$.

Let E be a real Banach space. A subset P of E is called a cone, if the followings hold:

(a₁):

P is closed, nonempty, and $P \neq\{0\}$,

(a₂):

$a, b \in \mathbb{R}$, $a,b\geq0$, and $x, y \in P$ imply that $ax+by\in P$,

(a₃):

$x \in P$ and $-x \in P$ imply that $x = 0$.

Given a cone $P\subset E$, we define a partial ordering ≤ with respect to P by $x \leq y$, if $y - x \in P$. We write $x < y$ to indicate that $x \leq y$ but $x\neq y$, while $x \ll y$ stands for $y - x \in \operatorname{int} P$, where intP denotes the interior of P.

The cone P is called normal, if there exist a number $K\geq1$ such that $0 \leq x \leq y$ implies $\|x\| \leq K\|y\|$, for all $x, y \in E$. The least positive number satisfying this, called the normal constant [7, 17].

In this paper, E denotes a real Banach space, P denotes a cone in E with $\operatorname{int} P\neq\emptyset$, and ≤ denotes a partial ordering with respect to P. For more details on the cone metric, we refer e.g. to [7, 17, 18].

Definition 4

[7]

Let X be a nonempty set. A function $d:X \times X\to E$ is called a cone metric on X, if it satisfies the following conditions:

(b₁):

$d(x,y) \geq0$ for all $x,y\in X$ and $d(x,y) = 0$, if and only if $x = y$,

(b₂):

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

(b₃):

$d(x,y) \leq d(x,z) + d(y,z)$, for all $x,y,z\in X$.

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

The following definitions and lemmas have been taken from [7, 18].

Definition 5

Let $(X,d)$ be a cone metric space and $\{x_{n}\}_{n\in \mathbb{N}}$ be a sequence in X and $x\in X$. If for all $c \in E$ with $0\ll c$, there is $n_{0} \in \mathbb{N}$ such that for all $n > n_{0}$, $d(x_{n}, x_{0})\ll c$, then $\{x_{n}\}_{n\in \mathbb{N}}$ is said to be convergent and $\{x_{n}\}_{n\in \mathbb{N}}$ converges to x and x is the limit of $\{x_{n}\}_{n\in \mathbb{N}}$.

Definition 6

Let $(X,d)$ be a cone metric space and $\{x_{n}\}_{n\in \mathbb{N}}$ be a sequence in X. If for all $c \in E$ with $0\ll c$, there is $n_{0} \in \mathbb{N}$ such that for all $m,n > n_{0}$, $d(x_{n},x_{m})\ll c$, then $\{x_{n}\}_{n\in \mathbb{N}}$ is called a Cauchy sequence in X.

Definition 7

Let $(X,d)$ be a cone metric space. If every Cauchy sequence is convergent in X, then X is called a complete cone metric space.

Definition 8

Let $(X,d)$ be a cone metric space. A self-map T on X is said to be continuous, if $\lim_{n\rightarrow\infty}x_{n}=x$ implies $\lim_{n\rightarrow\infty}T(x_{n})=T(x)$ for all sequences $\{x_{n}\}_{n\in \mathbb{N}}$ in X.

Lemma 9

Let $(X,d)$ be a normal cone metric space and P be a cone. Let $\{x_{n}\}_{n\in \mathbb{N}}$ be a sequence in X. Then $\{x_{n}\}_{n\in \mathbb{N}}$ converges to x, if and only if

$$ \lim_{n\to \infty}d(x_{n},x)=0. $$

(1.1)

Lemma 10

Let $(X,d)$ be a normal cone metric space and $\{x_{n}\}_{n\in \mathbb{N}}$ be a sequence in X. If $\{x_{n}\}_{n\in \mathbb{N}}$ is convergent, then it is a Cauchy sequence.

Lemma 11

Let $(X,d)$ be a cone metric space and P be a cone in E. Let $\{x_{n}\}_{n\in \mathbb{N}}$ be a sequence in X. Then $\{x_{n}\}_{n\in \mathbb{N}}$ is a Cauchy sequence, if and only if $\lim_{m,n\rightarrow\infty }d(x_{m},x_{n}) = 0$.

2 Main result

In this section, we represent our main results. First of all, we represent some simple observations. Let $(X,d_{\mathbb{C}})$ be a complex-valued metric space. Now, we define the following set:

$$\mathcal{P}_{\mathbb{C}}=\{x+iy:x\geq0,y\geq0\}. $$

It is apparent that $\mathcal{P}_{\mathbb{C}}\subset \mathbb{C}$. Note that $(\mathbb{C},|\cdot|)$ is a real Banach space.

Lemma 12

$\mathcal{P}_{\mathbb{C}}$ is a normal cone in a real Banach space $(\mathbb{C},|\cdot|)$.

Proof

Precisely, $\mathcal{P}_{\mathbb{C}}$ is nonempty, closed and $\mathcal {P}_{\mathbb{C}}\neq(\mathbf{0}_{\mathbb{C}})$. Also for all $\alpha,\beta\in \mathbb{R}^{+}$ and $p,q\in\mathcal{P}_{\mathbb{C}}$ we have $\alpha p+\beta q\in\mathcal{P}_{\mathbb{C}}$ and $\mathcal {P}_{\mathbb{C}}\cap(-\mathcal{P}_{\mathbb{C}})=(\mathbf{0}_{\mathbb{C}})$. Also the normality of $\mathcal{P}_{\mathbb{C}}$ is apparent. □

Lemma 13

Any complex-valued metric space $(X,d_{\mathbb{C}})$ is a cone metric space.

Proof

For all $p,q\in \mathbb{C}$ define

$$p\subseteqq q \quad\mbox{if and only if}\quad q-p\in\mathcal {P}_{\mathbb{C}}. $$

⫅ defines a partial ordered on ℂ and one can easily verify that $(X,d_{\mathbb{C}})$ is a cone metric space with respect to ⫅. □

Lemma 14

The partial ordered ⫅ defined in Lemma  13 is equivalent to ≲.

Proof

Assume $p=p_{1}+ip_{2}$ and $q=q_{1}+iq_{2}$. $p \subseteqq q$, if and only if $q-p\in\mathcal{P}_{\mathbb{C}}$, if and only if $q_{1}-p_{1}\geq0$, $q_{2}-p_{2}\geq0$. In other words, $\operatorname{Re}(p)\leq \operatorname{Re}(q)$, $\operatorname{Im}(p)\leq \operatorname{Im}(q)$, if and only if $p\lesssim q$. □

Lemma 15

A sequence $\{x_{n}\}$ in $(X,d_{\mathbb{C}})$ is convergent according to the concept of complex-valued metric space if and only if $\{x_{n}\}$ is convergent according to the concept of a cone metric space.

Proof

Let $\{x_{n}\}$ be sequence in X. Sequence $\{x_{n}\}$ converges to $x\in X$ according to the concept of complex-valued metric space if and only if $|d(x_{n},x)|\to0$ as $n\to\infty$ if and only if $\{x_{n}\}$ converges to x according to the concept of a cone metric space by considering ℂ as the Banach space endowed with the cone $\mathcal{P}_{\mathbb{C}}$ (see Lemma 9). □

Finally, we recall some fundamental definition for multi-valued mappings and related metric spaces. Let $(X,d)$ be a metric space. Let $\mathcal{P}(X)= \{ Y\mid Y\subset X \}$ and $P(X):= \{ Y\in\mathcal{P} ( X ) \mid Y\neq\emptyset \}$. Let us define the gap functional $D:P(X)\times P(X)\rightarrow \mathbb{R}_{+}\cup\{+\infty\}$, as

$$ D(A,B)=\operatorname{inf} \bigl\{ d(a,b) \mid a\in A, b\in B \bigr\} . $$

In particular, if $x_{0}\in X$, then $D ( x_{0},B ) :=D ( \{ x_{0} \} ,B )$.

We denote by $\mathcal{C}(X)$ the family of all nonempty closed subsets of X and $\mathcal{CB}(X)$ the family of all nonempty closed and bounded subsets of X. A function $\mathcal{H}:\mathcal{CB}(X)\times \mathcal{CB}(X)\rightarrow{}[0,\infty)$ defined by

$$\mathcal{H}(A,B)= \max \Bigl\{ \sup_{x\in B}D(x,A)\textit{,}\sup _{x\in A}D(x,B) \Bigr\} $$

is said to be the Hausdorff metric on $\mathcal{CB}(X)$ induced by the metric d on X where $D(x,A)=\inf\{d(x,y):y\in A\}$ for each $A\in \mathcal{CB}(X)$. A point v in X is a fixed point of a map T if $v=Tv$ (when $T:X\rightarrow X$ is a single-valued map) or $v\in Tv$ (when $T:X\rightarrow\mathcal{P}(X)$ is a multi-valued map). We say that T has an endpoint if there exists $v\in X$ such that $Tv=\{v\}$. The set of fixed points of T is denoted by $\mathcal{F}(T)$ and the set of common fixed points of two multi-valued mappings T, S is denoted by $\mathcal{F}(T,S)$.

Definition 16

For two multi-valued mappings $T,S:X\to\mathcal{CB}(X)$, we say that T, S satisfy the common approximate endpoint property if there exists a sequence $\{x_{n}\}\subset X$ such that

$$\lim_{n\to\infty}\mathcal{H} \bigl(\{x_{n} \},Tx_{n} \bigr)=\lim_{n\to\infty }\mathcal{H} \bigl( \{x_{n}\},Sx_{n} \bigr)=0. $$

Definition 17

For two mappings $T,S:X\to X$, we say that T, S have a common approximate fixed point if there exists a sequence $\{x_{n}\}\subset X$ such that

$$\lim_{n\to\infty}d(x_{n},Tx_{n})=\lim _{n\to\infty}d(x_{n},Sx_{n})=0. $$

Throughout the paper, we assume that $\{a,b,c,d,e\} \subset[0,1)$.

The following is the fundamental theorem of this paper.

Theorem 18

Let $(X,d)$ be a complete metric space and let $T,S:X\to\mathcal {CB}(X)$ be two multi-valued functions such that for each $x,y\in X$,

$$\begin{aligned} \mathcal{H}(Tx,Sy) \leq& ad(x,y)+bD(x,Sx)D(y,Ty) \\ &{}+c\sqrt{D(y,Sx)D(x,Ty)}+dD(x,Sx)D(x,Ty) \\ &{}+eD(y,Sx)D(y,Ty), \end{aligned}$$

where $a+b+c+2d+2e<1$. Then T, S have a unique endpoint, if and only if they satisfy the common approximate endpoint property.

Proof

If T, S have a unique endpoint then they precisely satisfy the common approximate endpoint property. Conversely, suppose that T, S satisfy the common approximate endpoint property, then there exists a sequence $\{x_{n}\}$ such that $\lim_{n\to\infty}\mathcal{H}(\{x_{n}\},Tx_{n})=\lim_{n\to\infty }\mathcal{H}(\{x_{n}\},Sx_{n})=0$. We claim that $\{x_{n}\}$ is a Cauchy sequence. For convenience suppose that $\alpha_{n}=\mathcal{H}(\{ x_{n}\},Tx_{n})$ and $\beta_{n}=\mathcal{H}(\{x_{n}\},Sx_{n})$; we have

$$\begin{aligned} d(x_{n},x_{m}) \leq&\alpha_{n}+ \beta_{m}+\mathcal{H}(Tx_{n},Sx_{m}) \\ \leq& \alpha_{n}+\beta_{m}+ad(x_{n},x_{m})+bD(x_{n},Sx_{n})D(x_{m},Tx_{m}) \\ &{}+c\sqrt{D(x_{m},Sx_{n})D(x_{n},Tx_{m})}+dD(x_{n},Sx_{n})D(x_{n},Tx_{m}) \\ &{}+eD(x_{m},Sx_{n})D(x_{m},Tx_{m}) \\ \leq&\alpha_{n}+\beta_{m}+ad(x_{n},x_{m})+b \beta_{n}\alpha_{m}+c\sqrt { \bigl(d(x_{n},x_{m})+ \beta_{n} \bigr) \bigl(d(x_{n},x_{m})+ \alpha_{m} \bigr)} \\ &{}+d \bigl(\beta_{n} \bigl(d(x_{n},x_{m})+ \alpha_{m} \bigr) \bigr)+e \bigl(\alpha_{m} \bigl(d(x_{n},x_{m})+ \beta_{n} \bigr) \bigr) \\ \leq&\alpha_{n}+\beta_{m}+ad(x_{n},x_{m})+c \frac{(d(x_{n},x_{m})+\beta _{n})+(d(x_{n},x_{m})+\alpha_{m})}{2} \\ &{}+d\beta_{n}d(x_{n},x_{m})+d\beta_{n} \alpha_{m}+e\alpha_{m}d(x_{n},x_{m})+e \alpha _{m}\beta_{n}. \end{aligned}$$

It means that

$$d(x_{n},x_{m}) (1-a-c-d\beta_{n}-e \alpha_{m})\leq\alpha_{n}+\beta _{m}+(d+b) \beta_{n}\alpha_{m}+e\alpha_{m}\beta_{n}+ \frac{c}{2}(\beta _{n}+\alpha_{m}). $$

In other words,

$$d(x_{n},x_{m})\leq\frac{\alpha_{n}+\beta_{m}+(d+b)\beta_{n}\alpha_{m}+e\alpha _{m}\beta_{n}+\frac{c}{2}(\beta_{n}+\alpha_{m})}{1-a-c-d\beta_{n}-e\alpha_{m}}. $$

Since $a+c<1$ we have $d(x_{n},x_{m})\to0$ as ${n,m\to\infty}$. Thus, $\{x_{n}\}$ is a Cauchy sequence. Since $(X,d)$ is complete metric space, it converges to some $z\in X$. Also we have

$$\begin{aligned} \mathcal{H}(z,Tz) \leq&d(z,x_{n})+\beta_{n}+ \mathcal{H}(Tz,Sx_{n}) \\ \leq&d(z,x_{n})+\beta_{n}+ad(x_{n},z)+bD(z,Sz)D(x_{n},Tx_{n}) \\ &{}+c\sqrt{D(x_{n},Sz)D(z,Tx_{n})}+dD(z,Sz)D(z,Tx_{n}) +eD(x_{n},Sz)D(x_{n},Tx_{n}) \\ \leq&d(z,x_{n})+\beta_{n}+ad(x_{n},z)+b \alpha_{n}D(z,Sz) \\ &{}+c\sqrt{D(x_{n},Sz) \bigl(d(x_{n},z)+ \alpha_{n} \bigr)}+dD(z,Sz) \bigl(d(z,x_{n})+ \alpha_{n} \bigr) \\ &{}+eD(x_{n},Sz)\alpha_{n}. \end{aligned}$$

(2.1)

By taking the limit on both sides of (2.1) we have $\mathcal {H}(\{z\},Tz)=0$ and so $Tz=\{z\}$. By a similar argument we deduce that $Sz=\{z\}$. □

Theorem 19

Let $(X,d)$ be a complete metric space and let $T,S:X\to\mathcal {CB}(X)$ be two multi-valued functions such that for each $x,y\in X$,

$$\begin{aligned} \mathcal{H}(Tx,Sy) \leq& ad(x,y)+b\frac{D(x,Sx)D(y,Ty)}{1+d(x,y)} \\ &{}+c\frac{\sqrt{D(y,Sx)D(x,Ty)}}{1+d(x,y)}+d\frac {D(x,Sx)D(x,Ty)}{1+d(x,y)} +e\frac{D(y,Sx)D(y,Ty)}{1+d(x,y)}, \end{aligned}$$

where $a+b+c+2d+2e<1$. Then T, S have a unique endpoint, if and only if they satisfy the common approximate endpoint property.

Proof

By Theorem 18 since

$$\begin{aligned} \mathcal{H}(Tx,Sy) \leq& ad(x,y)+b\frac{D(x,Sx)D(y,Ty)}{1+d(x,y)} \\ &{}+c\frac{\sqrt{D(y,Sx)D(x,Ty)}}{1+d(x,y)}+d\frac {D(x,Sx)D(x,Ty)}{1+d(x,y)} \\ &{}+e\frac{D(y,Sx)D(y,Ty)}{1+d(x,y)} \\ \leq&ad(x,y)+bD(x,Sx)D(y,Ty) \\ &{}+c\sqrt{D(y,Sx)D(x,Ty)}+dD(x,Sx)D(x,Ty) \\ &{}+eD(y,Sx)D(y,Ty), \end{aligned}$$

we conclude the desired result. □

Theorem 20

Let $(X,d)$ be a complete metric space and let $f,g:X\to X$ be two multi-valued functions such that for each $x,y\in X$,

$$\begin{aligned} d(fy,gx) \leq& ad(x,y)+bd(x,gx)d(y,fy) \\ &{}+cd(y,gx)d(x,fy)+dd(x,gx)d(x,fy) \\ &{}+ed(y,gx)d(y,fy), \end{aligned}$$

where $a+b+c+2d+2e<1$. Then f, g have a common fixed point.

Proof

Let $x_{0}\in X$ be arbitrary and let $x_{2n-1}=fx_{2n-2}$ and $x_{2n}=gx_{2n-1}$. We have

$$\begin{aligned} d(x_{2n+1},x_{2n}) =&d(fx_{2n},gx_{2n-1}) \\ \leq&ad(x_{2n},x_{2n-1})+bd(x_{2n-1},gx_{2n-1})d(x_{2n},fx_{2n}) \\ &{}+cd(x_{2n},gx_{2n-1})d(x_{2n-1},fx_{2n})+dd(x_{2n-1},gx_{2n-1}) d(x_{2n-1},fx_{2n}) \\ &{}+ed(x_{2n},gx_{2n-1})d(x_{2n},fx_{2n}) \\ =&ad(x_{2n},x_{2n-1})+bd(x_{2n-1},x_{2n})d(x_{2n},x_{2n+1}) \\ &{}+cd(x_{2n},x_{2n})d(x_{2n-1},x_{2n+1})+dd(x_{2n-1},x_{2n})d(x_{2n-1},x_{2n+1}) \\ &{}+ed(x_{2n},x_{2n})d(x_{2n},x_{2n+1}) \\ =&ad(x_{2n},x_{2n-1})+bd(x_{2n-1},x_{2n})d(x_{2n},x_{2n+1}) \\ &{}+dd(x_{2n-1},x_{2n})d(x_{2n-1},x_{2n+1}) \\ \leq&ad(x_{2n},x_{2n-1})+bd(x_{2n},x_{2n+1}) \\ &{}+dd(x_{2n-1},x_{2n}) \bigl(d(x_{2n-1},x_{2n})+d(x_{2n},x_{2n+1}) \bigr) \\ \leq&ad(x_{2n},x_{2n-1})+bd(x_{2n},x_{2n+1})+d \bigl(d(x_{2n-1},x_{2n})+d(x_{2n},x_{2n+1}) \bigr). \end{aligned}$$

It means that

$$d(x_{2n+1},x_{2n})\leq\frac{a+d}{1-b-d}d(x_{2n-1},x_{2n}). $$

Also

$$\begin{aligned} d(x_{2n-1},x_{2n}) =&d(fx_{2n-2},gx_{2n-1}) \\ \leq&ad(x_{2n-2},x_{2n-1})+bd(x_{2n-1},gx_{2n-1})d(x_{2n-2},fx_{2n-2}) \\ &{}+cd(x_{2n-2},gx_{2n-1})d(x_{2n-1},fx_{2n-2})+dd(x_{2n-1},gx_{2n-1})d(x_{2n-1},fx_{2n-2}) \\ &{}+ed(x_{2n-2},gx_{2n-1})d(x_{2n-2},fx_{2n-2}) \\ =&ad(x_{2n-2},x_{2n-1})+bd(x_{2n-1},x_{2n})d(x_{2n-2},x_{2n-1}) \\ &{}+cd(x_{2n-2},x_{2n})d(x_{2n-1},x_{2n-1})+dd(x_{2n-1},x_{2n})d(x_{2n-1},x_{2n-1}) \\ &{}+ed(x_{2n-2},x_{2n})d(x_{2n-2},x_{2n-1}) \\ =&ad(x_{2n-2},x_{2n-1})+bd(x_{2n-1},x_{2n})d(x_{2n-2},x_{2n-1}) \\ &{}+ed(x_{2n-2},x_{2n})d(x_{2n-2},x_{2n-1}) \\ \leq&ad(x_{2n-2},x_{2n-1})+bd(x_{2n-1},x_{2n}) \\ &{}+e \bigl(d(x_{2n-2},x_{2n-1})+d(x_{2n-1},x_{2n}) \bigr)d(x_{2n-2},x_{2n-1}) \\ \leq&ad(x_{2n-2},x_{2n-1})+bd(x_{2n-1},x_{2n})+e \bigl(d(x_{2n-2},x_{2n-1})+d(x_{2n-1},x_{2n}) \bigr). \end{aligned}$$

It means that

$$d(x_{2n-1},x_{2n})\leq\frac{a+e}{1-b-e}d(x_{2n-2},x_{2n-1}). $$

Now taking $\lambda=\max \{\frac{a+d}{1-b-d},\frac {a+e}{1-a-e} \}<1$ we conclude that for each $n\in \mathbb{N}$

$$d(x_{n+1},x_{n})\leq\lambda d(x_{n},x_{n-1}). $$

By a standard technique, one can show that $\{x_{n}\}$ is Cauchy. Since $(X,d)$ is a complete metric space, there exists $u\in X$ such that $x_{n} \to u$.

We claim that $fu=u$. By the triangle inequality, we have

$$\begin{aligned} d(u,fu) \leq& d(u,gx_{2n-1})+d(gx_{2n-1},fu) \\ \leq& d(u,gx_{2n-1})+ ad(x_{2n-1},u)+bd(x_{2n-1},gx_{2n-1})d(u,fu) \\ &{}+cd(u,gx_{2n-1})d(x_{2n-1},fu)+dd(x_{2n-1},gx_{2n-1})d(x_{2n-1},fu) \\ &{}+ed(u,gx_{2n-1})d(u,fu). \end{aligned}$$

Letting $n\to\infty$ in the inequality above, we get

$$d(u,fu)\leq0 \mbox{ and hence we find } fu=u. $$

Analogously, we derive that $gu=u$. Hence, we conclude that u is the common fixed point of f and g. □

Corollary 21

Let $(X,d)$ be a complete metric space and let $f,g:X\to X$ be two multi-valued functions such that for each $x,y\in X$,

$$\begin{aligned} d(fy,gx) \leq& ad(x,y)+bd(x,gx)d(y,fy) \\ &{}+dd(x,gx)d(x,fy)+ed(y,gx)d(y,fy), \end{aligned}$$

where $a+b +2d+2e<1$. Then f, g have a unique common fixed point.

Proof

By following the lines in the proof of Theorem 20 we get that f and g have a common fixed point, say $u \in X$. We shall show that u is the unique common fixed point of f and g. Suppose, on the contrary, that u and v are distinct common fixed points of f and g. Hence, we have

$$\begin{aligned} d(u,v) =&d(fu,gv)\leq ad(v,u)+bd(v,gv)d(u,fu) \\ &{}+ dd(v,gv)d(v,fu)+ ed(u,gv)d(u,fu) \\ \leq& ad(v,u)+bd(v, v)d(u, u) \\ &{}+dd(v, v)d(v, u)+ ed(u, v)d(u, u), \end{aligned}$$

which implies that

$$d(u,v) \leq ad(v,u). $$

Since $a<1$, $d(u,v)=0$, which is a contradiction. Hence, u is the unique common fixed point of f and g. □

Theorem 22

Let $(X,d)$ be a complex-valued metric space and let $f,g:X\to X$ be two functions and a, b, c, d, e be such that $a+b+c+2d+2e<1$. Let

$$\begin{aligned} d(fy,gx) \precsim& ad(x,y)+bd(x,gx)d(y,fy) \\ &{}+cd(y,gx)d(x,fy)+dd(x,gx)d(x,fy) \\ &{}+ed(y,gx)d(y,fy) \end{aligned}$$

for all $x,y\in X$. Then f, g have a unique common fixed point.

Proof

Taking $\rho(x,y)=|d(x,y)|$, $(X,\rho)$ is a complete metric space and applying Theorem 20 we conclude that f and g have a common fixed point. As in the proof of Corollary 21, uniqueness of the common fixed point of f and g can be derived easily by reductio ad absurdum. □

The following results, the main results of Ahmad et al. [2], can be considered as a consequence of Theorem 20.

Theorem 23

Let $(X,d)$ be a complex-valued metric space and let $f,g:X\to X$ be two functions and a, b, c, d, e be such that $a+b+c+2d+2e<1$. Let

$$\begin{aligned} d(fy,gx) \precsim & ad(x,y)+b\frac{d(x,gx)d(y,fy)}{1+d(x,y)} \\ &{}+c\frac{d(y,gx)d(x,fy)}{1+d(x,y)}+d\frac{d(x,gx)d(x,fy)}{1+d(x,y)} \\ &{}+e\frac{d(y,gx)d(y,fy)}{1+d(x,y)} \end{aligned}$$

for all $x,y\in X$. Then f, g have a unique common fixed point.

Proof

We have $a+b+c+d+e< a+b+c+2d+2e<1$ and

$$\begin{aligned} d(fy,gx) \precsim& ad(x,y)+b\frac{d(x,gx)d(y,fy)}{1+d(x,y)} \\ &{}+c\frac{d(y,gx)d(x,fy)}{1+d(x,y)}+d\frac{d(x,gx)d(x,fy)}{1+d(x,y)} \\ &{}+e\frac{d(y,gx)d(y,fy)}{1+d(x,y)} \\ \precsim&ad(x,y)+bd(x,gx)d(y,fy) \\ &{}+cd(y,gx)d(x,fy)+dd(x,gx)d(x,fy) \\ &{}+ed(y,gx)d(y,fy). \end{aligned}$$

By Theorem 20 we conclude that f and g have a common fixed point. Uniqueness can be derived easily verbatim as in the proof of Corollary 21. □

Remark 24

By Theorem 20, one can derive the other results in [2] but we prefer not to list these here.

In what follows we state a theorem that is just a variation of Theorem 20.

Corollary 25

Let $(X,d)$ be a complete metric space and let $f,g:X\to X$ be two multi-valued functions such that for each $x,y\in X$,

$$\begin{aligned} d(fy,gx) \leq& ad(x,y)+bd(x,gx)d(y,fy) \\ &{}+c\sqrt{d(y,gx)d(x,fy)}+dd(x,gx)d(x,fy) \\ &{}+ed(y,gx)d(y,fy), \end{aligned}$$

where $a+b+c+2d+2e<1$. Then f, g have a unique common fixed point.

Proof

By following the lines in the proof of Theorem 20, one can easily observe that

$$\lim_{n\to\infty}d(x_{n},fx_{n})=\lim _{n\to\infty}d(x_{n},gx_{n})=0. $$

Therefore, f, g have a common approximate fixed point. Thus taking $Tx=\{fx\}$ and $Sx=\{gx\}$ in Theorem 18 we conclude that T, S satisfy the common approximate endpoint property and so f, g have a unique common fixed point. □