Multivalued Common Fixed Points Theorem in Complex b-Metric Spaces

Abstract

In this paper, we establish a result for the existence of common fixed points for multi-valued mappings, satisfying some contractions for complex-valued b-metric spaces. Finally, we present an example to illustrate and support our results.

1. Introduction and Preliminaries

The notion of metric space was introduced in 1906 by M. Fréchet and developed shortly after by F. Hausdorff. In 1920, S. Banach [1] published a significant result in the field of fixed points, often referred to as the Banach fixed-point theorem. This theorem provides conditions under which a contraction mapping on a complete metric space has a unique fixed point. This result opened up new avenues for generalizing and expanding upon the theory of metric spaces. At the level of contractions, many researchers have explored various types of mappings that generalize the concept of a contraction.

In 1989, Bakhtin [2] generalized the notion of metric space to b-metric spaces and Czerwik [3,4] extended many proprieties of metric spaces to b-metric spaces. Subsequently, several authors published papers relating to the existence and uniqueness of fixed points for single-valued and multi-valued mappings defined on b-metric spaces and their applications, see for example [5,6,7,8,9,10,11,12,13,14].

In 2011, Azam et al. [15] introduced the notion of complex valued metric space and proved the existence of common fixed points of a pair of mappings satisfying contractive-type conditions and involving rational inequalities. The notion of complex-valued b-metric space was introduced by Rao et al. [16] in 2013. Many researchers have studied fixed point theorems in complex-valued metric and b-metric spaces [3,17,18,19,20,21,22]. In this paper, we establish a result for the existence of common fixed points for multi-valued mappings, satisfying some contractions involving rational inequalities for complex-valued b-metric spaces, we illustrate our results using an example.

We start the preliminaries with the definition of a partial order relation “$⪯$” for the complex numbers set $\mathbb{C}$, as follows:

$$\mathrm{for}\mathrm{all}{z}_1,z_2\in \mathbb{C}:z_1⪯z_2\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}Re\left(z_1\right)\le Re\left(z_2\right)\mathrm{and}Im\left(z_1\right)\le Im\left(z_2\right).$$

Clearly, $z_1⪯z_2$ implies that one of the following cases holds:

$$\begin{array}{cc}\hfill Re\left(z_1\right)& =Re\left(z_2\right)\mathrm{and}Im\left(z_1\right)=Im\left(z_2\right);\hfill \\ \hfill Re\left(z_1\right)& <Re\left(z_2\right)\mathrm{and}Im\left(z_1\right)=\mathrm{Im}\left(z_2\right);\hfill \\ \hfill Re\left(z_1\right)& =Re\left(z_2\right)\mathrm{and}Im\left(z_1\right)<\mathrm{Im}\left(z_2\right);\hfill \\ \hfill Re\left(z_1\right)& <Re\left(z_2\right)\mathrm{and}Im\left(z_1\right)<\mathrm{Im}\left(z_2\right).\hfill \end{array}$$

We write

$$z_1⋨z_2\mathrm{if}{z}_1⪯z_2\mathrm{and}{z}_1\ne z_2;$$

and

$$z_1\prec z_2\mathrm{if}Re\left(z_1\right)<Re\left(z_2\right)\mathrm{and}Im\left(z_1\right)<Im\left(z_2\right).$$

One can easily prove the following:

$$\begin{array}{cc}\hfill \mathrm{For}\mathrm{all}{z}_1,z_2& \in \mathbb{C}:0⪯z_1⋨z_2\mathrm{implies}\left|z_1\right|<\left|z_{{}_2}\right|;\hfill \\ \hfill a,b& \in \mathbb{R}\mathrm{and}a\le b\mathrm{implies}az⪯bz,\mathrm{for}\mathrm{all}0⪯z\in \mathbb{C};\hfill \\ \hfill a,b& \in \mathbb{R}\mathrm{and}0\le a\le b\mathrm{implies}a{z}_1⪯b{z}_2,\mathrm{for}\mathrm{all}0⪯z_1,z_2\in \mathbb{C}.\hfill \end{array}$$

Definition 1

([15]). Let X be a nonempty set. A function $d_c$: $X\times X$→$\mathbb{C}$ is called a complex-valued metric on X if, for all $x,y,z\in X$, the following conditions are satisfied:

(cm-1) 

$0⪯d_c(x,y)$ and $d_c(x,y)=0$ if and only if $x=y;$

(cm-2) 

$d_c(x,y)=d_c(y,x);$

(cm-3) 

$d_c(x,y)⪯d_c(x,z)+d_c(z,y).$

The pair $(X,d_c)$ is called a complex-valued metric space.

Example 1

([23]). Define $d_c:\mathbb{C}\times \mathbb{C}\to \mathbb{C}$ by

$$d_c(z_1,z_2)=\left|Re\left(z_1\right)-Re\left(z_2\right)|+i|Im\left(z_1\right)-Im\left(z_2\right)\right|;$$

then $(X,d_c)$ is a complex-valued metric space.

Definition 2

([16]). Let X be a nonempty set and $s⩾1$ be a given real number. A function $d_c:X\times X\to \mathbb{C}$ is called a complex-valued b-metric on X if, for all $x,y,z\in X$, the following conditions are satisfied:

(cbm-1) 

$0⪯d_c(x,y)$ and $d_c(x,y)=0$ if and only if $x=y;$

(cbm-2) 

$d_c(x,y)=d_c(y,x);$

(cbm-3) 

$d_c(x,y)⪯s[d_c(x,z)+d_c(z,y)].$

The pair $(X,d_c)$ is called a complex-valued b-metric space.

Example 2

([24]). Let $X=[0,1]$. Define the mapping $d_c:X\times X\to \mathbb{C}$ by

$$d_c(x,y)={\left|x-y\right|}^2+i{\left|x-y\right|}^2\mathit{for}\mathit{all}x,y\in X.$$

Then $(X,d_c)$ is a complex-valued b-metric space with $s=2$.

Remark 1.

Clearly, a complex-valued b-metric space with $s=1$ is a complex valued metric space; however, the converse is not true in general, see for example Example 2.1 in [25].

Definition 3

([16]). Let $(X,d_c)$ be a complex valued b-metric space.

(i) 

A point $x\in X$ is called interior point of a set A $\subseteq X$ whenever there exists $0\prec r\in \mathbb{C}$ such that $B(x,r)=\{y\in X:d_c(x,y)⪯r\}\subseteq A;$

(ii) 

A point $x\in X$ is called limit point of a set A whenever for every 0 $\prec r\in \mathbb{C}:$ $B(x,r) \cap$ $(A-\{x\left\}\right)\ne \varphi$;

(iii) 

A subset $A\subseteq X$ is called open whenever each element of A is an interior point of A;

(iv) 

A subset $A\subseteq X$ is called closed whenever each element of A belongs to A;

(v) 

A sub-basis for a Hausdorff topology τ on X is a family $F=\left\{B\right(x,r):x\in X$ and $0\prec r\}.$

Definition 4

([16]). Let $(X,d_c)$ be a complex-valued b-metric space and $\left\{x_n\right\}$ be a sequence in X and $x\in X$.

• If for every $c\in \mathbb{C}$, with $0\prec c$ there is $N\in \mathbb{N}$ such that for all $n>N$, $d_c(x_n,x)\prec c$, then $\left\{x_n\right\}$ is said to be convergent, $\left\{x_n\right\}$ converges to x and and x is the limit point of $\left\{x_n\right\}$, we denotes this by $\underset{n\to +\infty }{lim}{x}_n=x$ or $x_n\to x$ as n $\to +\infty ;$
• If for every $c\in \mathbb{C}$, with $0\prec c$, there is $N\in \mathbb{N}$ such that, for all $n>N$, $d_c(x_n,x_{n+m})⪯c$, where $m\in \mathbb{N},$ then $\left\{x_n\right\}$ is said to be Cauchy sequence;
• If every Cauchy sequence in X is convergent, then $(X,d_c)$ is said to be a complete complex-valued b-metric space.

Definition 5

([18]). Let $(X,d_c)$ be a complex-valued b-metric space and let $\left\{x_n\right\}$ be a sequence in X. Then

• $\left\{x_n\right\}$ converge to $x\in X$ if and only if $\underset{n\to +\infty }{lim}\left|d_c(x_n,x)\right|=0$;
• $\left\{x_n\right\}$ is a Cauchy sequence if and only if $\underset{n\to +\infty }{lim}\left|d_c(x_{n+m},x_n)\right|=0$, where $m\in \mathbb{N}$.

Example 3.

Let ${\displaystyle X=[0,1],0<\theta \le \frac{\pi }{2}}$ and $p>1$ be real numbers. Define the mapping $d_c:X\times X\to \mathbb{C}$ by

$$d_c(x,y)={|x-y|}^p{e}^{i\theta }={|x-y|}^p\left(cos\theta +isin\theta \right),\mathit{for}\mathit{all}x,y\in [0,1];$$

then, $(X,d_c)$ is a complete complex-valued b-metric space with $s=2^{p-1}$.

Indeed, clearly, the properties (cbm-1) and (cbm-2) of Definition 2 are satisfied by $d_c$. On the other hand, for all $x,y,z\in X$, we have

$$\begin{array}{cc}\hfill {|x-z+z-y|}^p& \le {\left(\right|x-z|+|z-y\left|\right)}^p\hfill \\ \hfill & \le 2^{p-1}{\left(\right|x-z|}^p{+|z-y|}^p).\hfill \end{array}$$

Since $0⪯e^{i\theta }$, then

$${|x-z+z-y|}^p{e}^{i\theta }⪯2^{p-1}{\left(\right|x-z|}^p{+|z-y|}^p)e^{i\theta },$$

which implies

$$\begin{array}{cc}\hfill d_c(x,z)& ={|x-z+z-y|}^p{e}^{i\theta }\hfill \\ \hfill & ⪯2^{p-1}{\left(\right|x-z|}^p{+|z-y|}^p)e^{i\theta }\hfill \\ \hfill & =2^{p-1}(d_c(x,z)+d_c(y,z)),\hfill \end{array}$$

then $d_c$ satisfies (cb-3) and $(X,d_c)$ is a complex-valued b-metric space with $s=2^{p-1}$.

Now, let us prove that $(X,d_c)$ is complete. If $\left\{x_n\right\}$ is a Cauchy sequence in $(X,d_c)$ then for $m\in \mathbb{N}$

$$\begin{array}{cc}\hfill 0& =\underset{n\to +\infty }{lim}\left|d_c(x_{m+n},x_n)\right|\hfill \\ \hfill & =\underset{n\to +\infty }{lim}\left|{\left(x_{m+n}-x_n\right)}^p{e}^{i\theta }\right|\hfill \\ \hfill & =\underset{n\to +\infty }{lim}{|x_{m+n}-x_n|}^p\hfill \end{array}$$

so, $\left\{x_n\right\}$ is a Cauchy sequence in $(X,d)$, with $d(x,y)$= ${|x-y|}^p.$ Since $(X,d)$ is a complete b-metric space (see [26]), there exists $x\in X$ such that

$$0=\underset{n\to +\infty }{lim}\left|d(x_n,x)\right|=\underset{n\to +\infty }{lim}\left|d_c(x_n,x)\right|,$$

then $\left\{x_n\right\}$ converges to x in $(X,d_c)$ and $(X,d_c)$ is a complete complex-valued b-metric space.

Let $(X,d_c)$ be a complex-valued b-metric space. For $A,B\subseteq X$, we denote by $CB\left(X\right)$ the set of all nonempty closed and bounded subsets of X.

For all $z_1\in \mathbb{C},$ we denote $\sigma \left(z_1\right)$ $=\{z_2\in \mathbb{C}$: $z_1⪯$ $z_2\}$ and

$$\sigma (a,B)={\displaystyle \bigcup }_{b\in B}\sigma \left(d_c(a,b)\right)={\displaystyle \bigcup }_{b\in B}\{z\in \mathbb{C}:d_c(a,b)⪯z\},\mathrm{for}\mathrm{all}a\in X\mathrm{and}B\in CB\left(X\right).$$

For all $A,B\in CB\left(X\right)$, we denote

$$\sigma (A,B)=\left({\displaystyle \bigcap }_{a\in A}\sigma (a,B)\right)\cap \left({\displaystyle \bigcap }_{b\in B}\sigma (b,A)\right).$$

Definition 6

([16]). Let $(X,d_c)$ be a complex-valued b-metric space and $T:X\to CB\left(X\right)$ be a multi-valued map. For all $x\in X$ and A∈$CB\left(X\right)$ we put $W_x\left(A\right)=\left\{d_c(x,a):a\in A\right\}.$ Thus, for all $x,y\in X:$ $W_x\left(Ty\right)$ $=\left\{d_c(x,u):u\in Ty\right\}$.

Definition 7

([16]). Let $(X,d_c)$ be a complex-valued b-metric space. A subset A of X is called bounded from below if there exists some z∈X such that $z⪯a$, for all $a\in A$.

Definition 8

([16]). Let $(X,d_c)$ be a complex-valued b-metric space. A multi-valued mapping $T:X\to 2^{\mathbb{C}}$ is called bounded from below if for each $x\in X$ there exists some $z_x\in \mathbb{C}$ such that $z_x⪯u$, for all $u\in Tx$.

Definition 9

([16]). Let $(X,d_c)$ be a complex-valued metric space. A multi-valued mapping $T:X\to CB\left(X\right)$ is said to have the lower bound property (l.b property) on $(X,d_c)$ if for any $x\in X$, the multi-valued mapping $F_x:X$ $\to \mathbb{C}$ defined by $F_x\left(y\right)=W_x\left(Ty\right)$ is bounded from below. That is, for $x,y\in X$, there exists an element $l_x\left(Ty\right)\in \mathbb{C}$ such that $l_x\left(Ty\right)⪯$u for all $u\in W_x\left(Ty\right)$, where $l_x\left(Ty\right)$ is called a lower bound of T associated with $(x,y)$.

Definition 10

([16]). Let $(X,d_c)$ be a complex-valued b-metric space. The multi-valued mapping $T:X\to CB\left(X\right)$ is said to have the greatest lower bound property (g.l.b property) on $(X,d_c)$ if a greatest lower bound of $W_x\left(Ty\right)$ exists in $\mathbb{C}$ for all $x,y\in X$. We denote $d_c(x,Ty)$ by the g.l.b of $W_x\left(Ty\right).$ That is, $d_c(x,Ty)$ $=inf\left\{d_c(x,u):u\in Ty\right\}$.

We finish this section with the notion of “max” for the partial order relation “$⪯$”:

Definition 11

([27]). The max function for the partial order relation “$⪯$” is defined by the following.

(i) 

$max\{z_1,z_2\}=z_2$ if and only if $z_1⪯z_2;$

(ii) 

If $z_1⪯$ $max\{z_2,z_3\}$ then $z_1⪯z_2$ or $z_1⪯z_3;$

(iii) 

$max\{z_1,z_2\}=z_2$ if and only if $z_1⪯z_2$ or $\left|z_1\right|\le \left|z_2\right|$.

2. Main Result

In this section, we prove a common fixed-point theorem for multi-valued mappings on complex-valued b-metric spaces. Our main result is stated as

Theorem 1.

Let $(X,d_c)$ be a complete complex-valued b-metric space and let $S,T:X\to CB\left(X\right)$ be multi-valued mappings with a g.l.b property such that

$$\alpha max\left(d_c(x,Ty),d_c(y,Sx),\frac{d_c(x,Ty)d_c(y,Sx)}{1+d_c(x,y)}\right)\in \sigma \left(Sx,Ty\right),\mathit{for}\mathit{all}x,y\in X;$$

(1)

where α is a real, such that ${\displaystyle 0<\alpha <\frac{1}{2s}.}$ Then, S and T have a common fixed point.

Proof. 

Let $x_0$ be an arbitrary point in X and $x_1\in S{x}_0$. From (1) (with $x=x_0$ and $y=x_1$), we have

$$\alpha d_c(x_0,T{x}_1)=\alpha max\left(d_c(x_0,T{x}_1),d_c(x_1,S{x}_0),\frac{d_c(x_0,T{x}_1)d_c(x_1,S{x}_0)}{1+d_c(x_0,x_1)}\right)\in \sigma (S{x}_0,T{x}_1);$$

then

$$d_c(x_0,T{x}_1)\in {\displaystyle \bigcap }_{x\in S{x}_0}\sigma (x,T{x}_1);$$

i.e.,

$$d_c(x_0,T{x}_1)\in \sigma (x,T{x}_1),\mathrm{for}\mathrm{all}x\in S{x}_0.$$

Since $x_1\in S{x}_0$, we have

$$d_c(x_0,T{x}_1)\in \sigma (x_1,T{x}_1)\underset{x\in T{x}_1}{={\displaystyle \bigcup }}\sigma \left(d_c(x_1,x)\right),$$

then, there exists $x_2\in T{x}_1$, such that

$$\alpha d_c(x_0,T{x}_1)\in \sigma \left(d_c(x_1,x_2)\right),$$

then

$$d_c(x_1,x_2)⪯\alpha d_c(x_0,T{x}_1).$$

By using the greatest lower bound property (g.l.b property) of T, we obtain

$$d_c(x_1,x_2)⪯d_c(x_0,T{x}_1)⪯\alpha d_c(x_0,x_2),$$

so,

$$\left|d_c(x_1,x_2)\right|\le \alpha \left|d_c(x_0,x_2)\right|.$$

Now, using the triangular inequality we obtain

$$\left|d_c(x_1,x_2)\right|\le \alpha s\left|d_c(x_0,x_1)\right|+\alpha s\left|d_c(x_1,x_2)\right|,$$

then

$${\displaystyle |d_c(x_1,x_2)|\le \frac{\alpha s}{1-\alpha s}\left|d_c(x_0,x_1)\right|.}$$

Inductively, we construct a sequence $\left\{x_n\right\}$ of elements in X, such that

$$\forall n\in \mathbb{N}:\left|d_c(x_n,x_{n+1})\right|\le {\varrho }^n\left|d_c(x_0,x_1)\right|,$$

with ${\displaystyle \varrho =\frac{\alpha s}{1-\alpha s}<1,x_{2n+1}\in S{x}_{2n}}$ and $x_{2n+2}\in T{x}_{2n+1}$.

For $m,n\in \mathbb{N}$, we have

$$\begin{array}{cc}\hfill \left|d_c(x_n,x_{n+m})\right|& \le s\left|d_c(x_n,x_{n+1})\right|+s\left|d_c(x_{n+1},x_{n+2})\right|+...+s\left|d_c(x_{n+m-1},x_{n+m})\right|\hfill \\ \hfill & \le s\left({\varrho }^n+{\varrho }^{n+1}+...+{\varrho }^{n+m}\right)\left|d_c(x_0,x_1)\right|\hfill \\ \hfill & \le s\frac{{\varrho }^n}{1-\varrho }\left|d_c(x_0,x_1)\right|.\hfill \end{array}$$

then

$$\underset{n\to +\infty }{lim}\left|d_c(x_n,x_{n+m})\right|=0.$$

So, $\left\{x_n\right\}$ is a Cauchy sequence in $(X,d_c)$ which is a complete complex-valued b-metric space, then, there exists $v\in X$ such that $x_n$ $\to v$ as n$\to +\infty$.

Let us now prove that $v\in Tv\cap Sv$. From (1) we have

$$\alpha max\left(d_c(x_{2n},Tv),d_c(v,S{x}_{2n}),\frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)}\right)\in \sigma \left(d_c(x_{2n},Tv)\right),$$

and then

$$\alpha max\left(d_c(x_{2n},Tv),d_c(v,S{x}_{2n}),\frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)}\right)\in {\displaystyle \bigcap }_{x\in S{x}_{2n}}\sigma \left(d_c(x,Tv)\right).$$

On the other hand

$$\alpha max\left(d_c(x_{2n},Tv),d_c(v,S{x}_{2n}),\frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)}\right)\in \sigma \left(d_c(x,Tv)\right),\forall x\in S{x}_{2n}.$$

Since $x_{2n+1}\in S{x}_{2n},$ we have

$$\begin{array}{cc}\hfill \alpha max\left(d_c(x_{2n},Tv),d_c(v,S{x}_{2n}),\frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)}\right)& \in \sigma \left(d_c(x_{2n+1},Tv)\right)\hfill \\ \hfill & ={\displaystyle \bigcup }_{u^{\prime }\in Tu}\sigma \left(d_c(x_{2n+1},u^{\prime })\right).\hfill \end{array}$$

Then, there exists $v_n\in Tv$, such that

$$\alpha max\left(d_c(x_{2n},Tv),d_c(v,S{x}_{2n}),\frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)}\right)\in \sigma \left(d_c(x_{2n+1},v_n)\right)$$

i.e.,

$$d_c(x_{2n+1},v_n)⪯\alpha max\left(d_c(x_{2n},Tv),d_c(v,S{x}_{2n}),\frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)}\right).$$

With the property $\left(ii\right)$ of Definition 11 it holds

$$d_c(x_{2n+1},v_n)⪯\alpha d_c(x_{2n},Tv)$$

or

$$d_c(x_{2n+1},v_n)⪯\alpha d_c(v,S{x}_{2n})$$

or

$$d_c(x_{2n+1},v_n)⪯\alpha \frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)},$$

which implies

$$\left|d_c(x_{2n+1},v_n)\right|⪯\alpha \left|d_c(x_{2n},Tv)\right|$$

or

$$\left|d_c(x_{2n+1},v_n)\right|⪯\alpha \left|d_c(v,S{x}_{2n})\right|$$

or

$$\left|d_c(x_{2n+1},v_n)\right|⪯\alpha \left|\frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)}\right|.$$

Now, we distinguish three cases:

Case 1: $\left|d_c(x_{2n+1},v_n)\right|⪯\alpha \left|d_c(x_{2n},Tv)\right|.$

By the greatest lower bound property of T, we have

$$\left|d_c(x_{2n+1},v_n)\right|\le \alpha \left|d_c(x_{2n},v_n)\right|,$$

with the triangular inequality

$$\left|d_c(x_{2n+1},v_n)\right|\le \alpha s(\left|d_c(x_{2n},x_{2n+1})\right|+\left|d_c(x_{2n+1},v_n)\right|),$$

which implies

$$\left|d_c(x_{2n+1},v_n)\right|\le \frac{s\alpha }{1-s\alpha }\left|d_c(x_{2n},x_{2n+1})\right|.$$

On the other hand, by the triangular inequality, we have

$$\left|d_c(v,v_n)\right|\le s(\left|d_c(v,x_{2n+1})\right|+\left|d_c(x_{2n+1},v_n)\right|)$$

then

$$\left|d_c(v,v_n)\right|\le s\left|d_c(v,x_{2n+1})\right|+\frac{s^2\alpha }{1-s\alpha }\left|d_c(x_{2n},x_{2n+1})\right|.$$

Letting $n\to +\infty$, we find

$$\underset{n\to +\infty }{lim}\left|d_c(v,v_n)\right|=0.$$

Case 2: $\left|d_c(x_{2n+1},v_n)\right|⪯\alpha \left|d_c(v,S{x}_{2n})\right|.$

With the greatest lower bound property of S

$$\left|d_c(x_{2n+1},v_n)\right|\le \alpha \left|d_c(v,x_{2n+1})\right|.$$

With the triangular inequality

$$\begin{array}{cc}\hfill \left|d_c(v,v_n)\right|& \le s(\left|d_c(v,x_{2n+1})\right|+\left|d_c(x_{2n+1},v_n)\right|)\hfill \\ \hfill & \le s\left|d_c(v,x_{2n+1})\right|+\alpha s\left|d_c(v,x_{2n+1})\right|.\hfill \end{array}$$

Letting $n\to +\infty$, we obtain

$$\underset{n\to +\infty }{lim}\left|d_c(v,v_n)\right|=0.$$

Case 3: ${\displaystyle \left|d_c(x_{2n+1},v_n)\right|⪯\alpha \left|\frac{d_c(x_{2n},Tv)d_c(v,S{x}_{2n})}{1+d_c(x_{2n},v)}\right|.}$

With the greatest lower bound property of S and T

$$\left|d_c(x_{2n+1},v_n)\right|⪯\alpha \left|\frac{d_c(x_{2n},v)d_c(v,x_{2n})}{1+d_c(x_{2n},v)}\right|.$$

With the triangular inequality

$$\begin{array}{cc}\hfill \left|d_c(v,v_n)\right|& \le s(\left|d_c(v,x_{2n+1})\right|+\left|d_c(x_{2n+1},v_n)\right|)\hfill \\ \hfill & \le s\left|d_c(v,x_{2n+1})\right|+s\alpha \left|\frac{d_c(x_{2n},v)d_c(v,x_{2n})}{1+d_c(x_{2n},v)}\right|.\hfill \end{array}$$

Letting $n\to +\infty$, we find

$$\underset{n\to +\infty }{lim}\left|d_c(v,v_n)\right|=0.$$

In these three cases, we have $\left|d_c(v,v_n)\right|\to 0$ as $n\to +\infty$ then $\left\{v_n\right\}$ converges to $v.$

Since $\left\{v_n\right\}\subseteq Tv$ and $Tv$ is closed, we obtain v ∈$Tv$.

By the same method we prove that $v\in Sv$.

So, v is a common fixed point for T and S. □

By setting $T=S$ in Theorem 1, we have the following corollary:

Corollary 1.

Let $(X,d_c)$ be a complete complex-valued b-metric space and let $S:X\to CB\left(X\right)$ be a multi-valued mapping with g.l.b property, such that

$$\alpha max\left(d_c(x,Sy),d_c(y,Sx),\frac{d_c(x,Sy)d_c(y,Sx)}{1+d_c(x,y)}\right)\in \sigma \left(Sx,Sy\right),\mathit{for}\mathit{all}x,y\in X,$$

where α is real, such that ${\displaystyle 0<\alpha <\frac{1}{2s}.}$ Then, S has a fixed point.

Finally, by setting $s=1$ in Theorem 1 and Corollary 1, we have the following corollaries,

Corollary 2.

Let $(X,d_c)$ be a complete complex-valued metric space and let $S,T:X\to CB\left(X\right)$ be multi-valued mappings with g.l.b property such that

$$\alpha max\left(d_c(x,Ty),d_c(y,Sx),\frac{d_c(x,Ty)d_c(y,Sx)}{1+d_c(x,y)}\right)\in \sigma \left(Sx,Ty\right),\mathit{for}\mathit{all}x,y\in X,$$

where α is a real such that ${\displaystyle 0<\alpha <\frac{1}{2}}$. Then S and T have a common fixed point.

Corollary 3.

Let $(X,d_c)$ be a complete complex-valued metric space and let $S:X\to CB\left(X\right)$ be multi-valued mapping with g.l.b property, such that

$$\alpha max\left(d_c(x,Sy),d_c(y,Sx),\frac{d_c(x,Sy)d_c(y,Sx)}{1+d_c(x,y)}\right)\in \sigma \left(Sx,Sy\right),\mathit{for}\mathit{all}x,y\in X,$$

where α is a real, such that ${\displaystyle 0<\alpha <\frac{1}{2}}$. Then S has a fixed point.

Now, we illustrate our main results using the following example:

Example 4.

Let $(X,d_c)$ be the complete complex-valued b-metric space given in Example 3 with $p=2$ and ${\displaystyle \theta =\frac{\pi }{2}}$ that is

$$X=[0,1]\mathit{and}{d}_c={|x-y|}^{2}i,\mathit{for}\mathit{all}x,y\in X.$$

We define the mappings $S,T:X\to CB\left(X\right)$ by

$$Sx=\{t\in X:0\le t\le \frac{x}{7}\}\mathit{and}Tx=\{t\in X:0\le t\le \frac{x}{6}\},\mathit{for}\mathit{all}x\in X.$$

For all $x,y\in X$, we have

$$\begin{array}{cc}\hfill W_x\left(Ty\right)& =\left\{d_c(x,t):t\in Ty\right\}\hfill \\ \hfill & =\left\{{|x-t|}^{2}i:0\le t\le \frac{y}{6}\right\}.\hfill \end{array}$$

For all ${|x-t|}^{2}i\in W_x\left(Ty\right),$ we have

$$0⪯{|x-t|}^{2}i\mathit{if}x\le \frac{y}{6}$$

and

$$|x-\frac{y}{6}{|}^{2}i⪯{|x-t|}^{2}i\mathit{if}x⩾\frac{y}{6},$$

then T has the greatest lower bound property and

$$\mathit{for}\mathit{all}x,y\in X:d(x,Ty)=\left\{\begin{array}{c}0\mathit{if}0\le x<\frac{y}{6}\\ \left|x-\frac{y}{6}\right|i\mathit{if}\frac{y}{6}\le x\le 1.\end{array}\right.$$

In a similar way we prove that S has the greatest lower bound property and

$$\mathit{for}\mathit{all}x,y\in X:d(y,Sx)=\left\{\begin{array}{c}0\mathit{if}0\le y<\frac{x}{7}\\ \left|y-\frac{x}{7}\right|i\mathit{if}\frac{x}{7}\le y\le 1.\end{array}\right.$$

If $x=y=0$, then the condition (1) is obviously satisfied. Without losing the generality, we can suppose $0<x<y\le 1$. Then

$$\begin{array}{cc}\hfill \sigma (m,Sx)& ={\displaystyle \bigcup }_{t\in Sx}\left\{z\in \mathbb{C}:d_c(t,m)⪯z\right\}\hfill \\ \hfill & ={\displaystyle \bigcup }_{0\le t\le \frac{x}{7}}\left\{{z\in \mathbb{C}:|m-t|}^{2}i⪯z\right\}\hfill \\ \hfill & ={\displaystyle \bigcup }_{0\le t\le \frac{x}{7}}\left\{a+ib\in \mathbb{C}:0\le {a\mathit{and}|m-t|}^2\le b\right\}\hfill \\ \hfill & =\left\{a+ib\in \mathbb{C}:0\le a\mathit{and}|m-\frac{x}{7}{|}^2\le b\right\},\hfill \end{array}$$

hence

$$\begin{array}{cc}\hfill {\displaystyle \bigcap }_{m\in Ty}\sigma (m,Sx)& ={\displaystyle \bigcap }_{m\in Ty}\left\{a+ib\in \mathbb{C}:0\le a\mathit{and}|m-\frac{x}{7}{|}^2\le b\right\}\hfill \\ \hfill & ={\displaystyle \bigcap }_{0\le m\le \frac{y}{6}}\left\{a+ib\in \mathbb{C}:0\le a\mathit{and}|m-\frac{x}{7}{|}^2\le b\right\}\hfill \\ \hfill & =\left\{a+ib\in \mathbb{C}:0\le a\mathit{and}{\left|\frac{y}{6}-\frac{x}{7}\right|}^2\le b\right\}\hfill \\ \hfill & =\left\{z\in \mathbb{C}:{\left|\frac{y}{6}-\frac{x}{7}\right|}^{2}i⪯z\right\}\hfill \\ \hfill & =\sigma \left({\left|\frac{y}{6}-\frac{x}{7}\right|}^{2}i\right).\hfill \end{array}$$

In a similar way

$$\begin{array}{cc}\hfill \sigma (\ell ,Ty)& ={\displaystyle \bigcup }_{t\in Ty}\left\{z\in \mathbb{C}:d_c(t,\ell )⪯z\right\}\hfill \\ \hfill & ={\displaystyle \bigcup }_{0\le t\le \frac{y}{6}}\left\{{z\in \mathbb{C}:|\ell -t|}^{2}i⪯z\right\}\hfill \\ \hfill & ={\displaystyle \bigcup }_{0\le t\le \frac{y}{6}}\left\{a+ib\in \mathbb{C}:0\le {a\mathit{and}|\ell -t|}^2\le b\right\}\hfill \\ \hfill & =\left\{a+ib\in \mathbb{C}:0\le a\mathit{and}|\ell -\frac{y}{6}{|}^2\le b\right\}\hfill \end{array}$$

then

$$\begin{array}{cc}\hfill {\displaystyle \bigcap }_{\ell \in Sx}\sigma (\ell ,Ty)& ={\displaystyle \bigcap }_{\ell \in Sx}\left\{a+ib\in \mathbb{C}:0\le a\mathit{and}|\ell -\frac{y}{6}{|}^2\le b\right\}\hfill \\ \hfill & ={\displaystyle \bigcap }_{0\le \ell \le \frac{x}{7}}\left\{a+ib\in \mathbb{C}:0\le a\mathit{and}|\ell -\frac{y}{6}{|}^2\le b\right\}\hfill \\ \hfill & =\left\{a+ib\in \mathbb{C}:0\le a\mathit{and}{\left|\frac{x}{7}-\frac{y}{6}\right|}^2\le b\right\}\hfill \\ \hfill & =\left\{z\in \mathbb{C}:{\left|\frac{x}{7}-\frac{y}{6}\right|}^{2}i⪯z\right\}\hfill \\ \hfill & =\sigma \left({\left|\frac{y}{6}-\frac{x}{7}\right|}^{2}i\right).\hfill \end{array}$$

In summary, for all, $0<x<y\le 1$, we have

$$\mathit{for}\mathit{all}x,y\in X:d(x,Ty)=\left\{\begin{array}{c}0\mathit{if}0\le x<\frac{y}{6}\\ \left|x-\frac{y}{6}\right|i\mathit{if}\frac{y}{6}\le x\le 1\end{array}\right.;$$

$$\mathit{for}\mathit{all}x,y\in X:d(y,Sx)=\left\{\begin{array}{c}0\mathit{if}0\le y<\frac{x}{7}\\ \left|y-\frac{x}{7}\right|i\mathit{if}\frac{x}{7}\le y\le 1\end{array}\right.$$

and

$$\sigma (Sx,Ty)=\sigma \left({\left|\frac{y}{6}-\frac{x}{7}\right|}^{2}i\right).$$

If $0\le x\le \frac{y}{6}$ then

$$d_c(x,Ty)=\frac{d_c(x,Ty)d_c(y,Sx)}{1+dc(x,y)}=0.$$

if $\frac{y}{6}<x\le 1$ then

$$d_c(x,Ty)={\left|x-\frac{y}{6}\right|}^{2}i⪯{\left|y-\frac{x}{7}\right|}^{2}i=d(y,Sx).$$

and

$$\begin{array}{cc}\hfill \frac{d_c(x,Ty)d_c(y,Sx)}{1+d_c(x,y)}& =\frac{-{\left|x-\frac{y}{6}\right|}^2{\left|y-\frac{x}{7}\right|}^2}{{1+|x-y|}^{2}i}\hfill \\ \hfill & ={\left|x-\frac{y}{6}\right|}^2{\left|y-\frac{x}{7}\right|}^2\frac{{-1+|x-y|}^{2}i}{{1+|x-y|}^4}\hfill \\ \hfill & ⪯{\left|y-\frac{x}{7}\right|}^{2}i=d(y,Sx).\hfill \end{array}$$

then for all $0<x<y\le 1$

$$max\left(d_c(x,Ty),d_c(y,Sx),\frac{d_c(x,Ty)d_c(y,Sx)}{1+d_c(x,y)}\right)={\left|y-\frac{x}{7}\right|}^{2}i=d(y,Sx).$$

For all $0<x<y\le 1,$ we have

$${\left|\frac{y}{6}-\frac{x}{7}\right|}^{2}i\le \frac{1}{5}{\left|y-\frac{x}{7}\right|}^{2}i,$$

then

$$\frac{1}{5}max\left(d_c(x,Ty),d_c(y,Sx),\frac{d_c(x,Ty)d_c(y,Sx)}{1+d_c(x,y)}\right)=\frac{1}{5}d(y,Sx)\in \sigma (Sx,Ty)$$

that is S and T satisfies (1) with ${\displaystyle \alpha =\frac{1}{5}<\frac{1}{4}=\frac{1}{2s}}$. Then, S and T have a common fixed point which is equal to 0.

3. Conclusions

This paper makes a significant contribution by establishing a result concerning the existence of common fixed points for multi-value maps that satisfy contraction conditions for complex- valued b-metric spaces. The study deepens the concept of fixed points in the context of multi-value mapping and introduces the framework of complex-value b-metric spaces. By proving that these multi-valued mappings satisfy contraction properties, the paper offers a solid basis for new research and applications in various areas, including functional analysis, fixed-point theory, and complex analysis. The results obtained in this manuscript help to advance our understanding of the theory of fixed points in the context of b-metric spaces and open up possibilities for exploring related concepts and properties in the future.