Applications and theorem on common fixed point in complex valued b-metric space

1.   Introduction and preliminaries

Banach contraction fundamental was an authority and a reference for many researchers through the last decades in the field of nonlinear analysis, it was used to establish the existence of a unique solution for a nonlinear integral equation ^[4]. In 1989, Bakthtin ^[3] initiated the motif of b-metric space after that Czerwik in ^[7,8] defined it such as current structure which is considere generalization of metric spaces. The complex valued b-metric spaces concept was introduced in 2013 by Rao et al. ^[13], which was more general than the well-known complex valued metric spaces that were introduced in 2011 by Azam et al. ^[2] which proved some common fixed point theorems for mapping satisfying rational inequalities which are not worthwhile in cone metric spaces ^[1,10,11,16]. Sundry authors have studied and proved the fixed point results for mappings with satisfying different type contraction conditions in the framework of complex valued metric (b-metric) spaces(see ^{[5,6,9,13,17]}).

The main purpose of this paper is to present common fixed point results of four self-mappings to satisfy a rational inequality on complex valued b-metric spaces. and we establish the existence and the uniqueness of a common solution for the system of Urysohn integral equations. Also we prove the existence and the uniqueness of solution for linear system in complete complex valued b-metric space. In ^[2] the authors introduced the notion of complex-valued metric space and obtained a common fixed-point theorems of contraction type mappings using the partial inequality in a complex-valued metric space.

To do so, let us recall a natural relation $\leq$ on $\mathbb{C}$, the set of complex numbers as follows: let $z_{1}, z_2$ in $\mathbb{C}$

In ^[2], the authors defined a partial order relation $z_1 \precsim z_2$ on $\mathbb{C}$ as follows:

As a result, one can infer that $z_1 \precsim z_2$ if one of the following conditions is satisfied:

(ⅰ) $\operatorname{Re}(z_1) = \operatorname{Re}(z_2)$, $\operatorname{Im}(z_1) < \operatorname{Im}(z_2)$,

(ⅱ) $\operatorname{Re}(z_1) < \operatorname{Re}(z_2)$, $\operatorname{Im}(z_1) = \operatorname{Im}(z_2)$,

(ⅲ) $\operatorname{Re}(z_1) < \operatorname{Re}(z_2)$, $\operatorname{Im}(z_1) < \operatorname{Im}(z_2)$,

(ⅳ) $\operatorname{Re}(z_1) = \operatorname{Re}(z_2)$, $\operatorname{Im}(z_1) = \operatorname{Im}(z_2)$.

In (ⅰ), (ⅱ) and (ⅲ) we have $\lvert z_1\rvert$ $<$ $\lvert z_2\rvert$. In (ⅳ) we have $\lvert z_1\rvert$ $=$ $\lvert z_2\rvert$, so that, $\lvert z_1\rvert$ $\leq$ $\lvert z_2\rvert$ In particular, $z_1 \precsim z_2$ if $z_1\neq z_2$ and one of (ⅰ), (ⅱ) and (ⅲ) is satisfied. In this case $\lvert z_1\rvert$ $<$ $\lvert z_2\rvert$. We will write $z_1\prec z_2$ if only (ⅲ) is satisfied. Further,

In ^[2], the authors defined the complex-valued metric space $(X, d)$ in the following way:

Definition 1.1. Let $X$ be a non-empty set. A mapping $d:\: X\times X \rightarrow \mathbb{C}$ is called a complex valued metric on $X$ if the following conditions are satisfied:

$\bf{(a)}$ $0\precsim d(x, y)$ for all $x, y \in X$ and $d(x, y) = 0$ $\Leftrightarrow$ $x = y$,

$\bf{(b)}$ $d(x, y) = d(y, x)$, for all $x, y\in X$,

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

Then $d$ is called a complex valued metric in $X$, and $(X, d)$ is a complex valued metric space.

Example 1. Let $X = \mathbb{C}$ define the mapping $d:X \times X \longrightarrow \mathbb{C}$ by:

Then $(X, d)$ is a complex valued metric space.

Definition 1.2. ^[13] Let $X$ be a nonempty set and let $s\geq 1$ be given real number. A mapping $d:\: X\times X \rightarrow \mathbb{C}$ is called a complex valued b-metric on $X$ if the following conditions are satisfied:

$\bf{(a)}$ $0\precsim d(x, y)$ for all $x, y \in X$ and $d(x, y) = 0$ $\Leftrightarrow$ $x = y$,

$\bf{(b)}$ $d(x, y) = d(y, x)$, for all $x, y\in X$,

$\bf{(c)}$ $d(x, y) \precsim s[d(x, z)+d(z, y)]$ for all $x, y, z\in X$.

Then $(X, d)$ is a complex valued b-metric space.

Example 2. ^[13] Let $X = \mathbb{C}$ define the mapping $d:X \times X \longrightarrow \mathbb{C}$ by:

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

Definition 1.3. ^[13] Suppose that $(X, d)$ is a complex valued b-metric space and $\lbrace z_n \rbrace$ is a sequence in $X$ and $z\in X$ then

$\bf{(i)}$ We say that a sequence $\lbrace z_n \rbrace$ converges to an element $z_0\in X$ if for every $0\prec c\in \mathbb{C}$, there exists an integer $N$ such that $d(z_n, z_0) \prec c$ for all $n\geq N.$ In this case, we write $z_n \longrightarrow z_{0}$.

$\bf{(ii)}$ We say that $\lbrace z_n \rbrace$ is a Cauchy sequence if for every $0\prec c\in \mathbb{C}$, there exists an integer $N$ such that $d(z_n, z_m) \prec c$ for all $n, m\geq N$.

$\bf{(iii)}$ We say that $(X, d)$ is complete, if every Cauchy sequence in $X$ converges to a point in $X$.

Definition 1.4. Let $S$ and $T$ be self mappings of a nonemplty set $X$. If $w = Sz = Tz$ for some $z$ in $X$, then $z$ is called a coincidence points of $S$ and $T$ and $w$ is called a point of coincidence of $S$ and $T$.

Definition 1.5. ^[15] Let $S$ and $T$ be a self-mappings of a complex valued metric space $\left(X, d \right)$. The mappings $S$ and $T$ are said to be compatible if:

whenever $\left\{ z_{n}\right\}$ is a sequence in $X$ such that:

Definition 1.6. ^[12] Let $S$ and $T$ be self mappings of a nonemplty set $X$. $S$ and $T$ are said to be weakly compatible if they commute at their coincidence points, i.e, $Sz = Tz$ for some $z$ in $X$ implies that $STz = TSz$.

Definition 1.7. A matrix norm induced by vectors norms is given by:

Example 3. Let $X = \mathbb{C}^{n}$. A vector norm in a complex valued b-metric given by:

where $z, w\in X$ such that $z = (z_1, \ldots, z_{n})^{t}$ and $w = (w_1, \ldots, w_{n})^{t},$ then $(X, d_2)$ is a complex valued b-metric space.

Definition 1.8. ^[14] defined the $\max$ function for the partial order relation by:

$\textbf{(i)}$ $\max \lbrace z_1, z_2 \rbrace = z_2 \Leftrightarrow z_1\precsim z_2$,

$\textbf{(ii)}$ $z_1\precsim \max \lbrace z_1, z_3\rbrace \Rightarrow z_1\precsim z_2,$ or $z_1\precsim z_3$,

$\textbf{(iii)}$ $\max \lbrace z_1, z_2\rbrace = z_2 \Leftrightarrow z_1\precsim z_2$ or $\lvert z_1\rvert$ $\leq$ $\lvert z_2\rvert$.

Using definition $(8)$ we have the following Lemma.

Lemma 1. ^[14] Let $z_1, z_2, z_{3, \cdots }\in\mathbb{C}$ and the partial order relation $\precsim$ is defined on $\mathbb{C}$, the following statements are achieve:

(i) if $z_1 \precsim \max \lbrace z_2, z_3 \rbrace$ then $z_1 \precsim z_2$ if $z_3 \precsim z_2$

(ii) if $z_1 \precsim \max \lbrace z_2, z_3, z_4 \rbrace$ then $z_1\precsim z_2$ if $\max \lbrace z_3, z_4 \rbrace \precsim z_2$,

(iii) if $z_1\precsim \max \left\{z_2, z_3, z_4, z_5\right\}$ then $z_1\precsim z_2$ if $\max\left\{ z_3, z_4, z_5\right\} \precsim z_2,$ $and\ so\ on$.

2.   Main results

In this section, we prove common fixed point theorem for four mappings in a complete complex valued b-metric spaces using rational type contraction condition and we give some examples. Our first new result is the following:

Theorem 2.1. Let $(X, d)$ be a complete complex valued b-metric space and $S, T, P, Q:X\rightarrow X$ be a self mappings satisfying the conditions:

$C_1$ $S(X)\subset Q(X)$ and $T(X)\subset P(X)$,

$C_2$ $d(Sz, Tw)\precsim \frac{\lambda}{s^{2}} R(z, w),$ if $s\geq1$ and $\lambda\in (0, 1)$ for all $z, w\in X$ where

$C_3$ the pair $(S, P)$ is compatible and the pair $(T, Q)$ is weakly compatible,

$C_4$ either $P$ or $S$ is continuous.

Then $S, T, P$ and $Q$ have a unique common fixed point in $X$.

Proof. Let $z_0\in X$ be arbitrary. From the condition $C_1$, there exist $z_{1}, z_{2}$ such that $w_{0} = Q z_{1} = Sz_{0}$ and $w_{1} = P z_{2} = Tz_{1}$. We can construct successively the sequences $\left\lbrace w_{n} \right\rbrace$ and $\left\lbrace z_{n} \right\rbrace$ in $X$ as follows:

Using $(2.1)$ in $C_2$ we get:

where

we have:

and we have

which is implies

from (2.2) and (2.3) we get:

with

If

then,

which is a contradiction, since $\lambda\in(0, 1), s\geq 1$. We conclude that $d(w_{2n}, w_{2n+1})\precsim \frac{\lambda}{s^{2}} d(w_{2n-1}, w_{2n})$.

Similarly we get $d(w_{2n+1}, w_{2n+2})\precsim \frac{\lambda}{s^{2}} d(w_{2n}, w_{2n+1}).$

It follows that

which implies

for $m < n$ we have:

Therefore,

hence,

Thus, $\lbrace w_{n} \rbrace$ is a Cauchy sequence in $X$. Since $X$ is complete, so there exists some $u \in X$ such that $w_{n}\rightarrow u$ as $n\rightarrow\infty$. For its sub-sequences we also have $Q z_{2n+1}\rightarrow u, Sz_{2n}\rightarrow u, Pz_{2n+1}\rightarrow u$ and $Tz_{2n}\rightarrow u$

from $C_4$ if $P$ is continuous

as $P$ is continuous, then $PPz_{2n}\rightarrow Pu$ and $PSz_{2n}\rightarrow Pu$, as $n\rightarrow\infty$. Also, since the pair $(S, P)$ is compatible, this implies that $SPz_{2n}\rightarrow Pu$. Indeed,

So,

We prove $Pu = u$. On the contrary we suppose that $Pu\neq u$

using $C_2$ with $z = Pz_{2n}, w = z_{2n+1}$, we get:

where

let $n\rightarrow\infty$ we get:

Further,

So, $(1-\frac{\lambda}{s^{2}}) \vert d(Pu, u) \vert \leq 0,$ which is a contradiction that is $\vert d(Pu, u) \vert = 0$ then $Pu = u$.

We prove $Su = u$. On the contrary we suppose that $Su\neq u$

Using $C_2$ with $z = u, w = z_{2n+1}$, we get: $d(Su, Tz_{2n+1})\precsim \frac{\lambda}{s^{2}} R(u, z_{2n+1})$ where

let $n\rightarrow\infty$ we get:

Then, $d(Su, u) \precsim \frac{\lambda}{s^{2}} d(Su, u)$, further, $\vert d(Su, u) \vert \leq \frac{\lambda}{s^{2}} \vert d(Su, u) \vert,$ which is a contradiction that is $\vert d(Su, u) \vert = 0$ then, $Su = u$. We prove $Qu = Tu$, as $S(X)\subset Q(X)$, so there exists $v\in X$ such that $u = Su = Qv$. First, we shall show that $Qv = Tv$ for this we get:

where,

then,

Then, $d(Qv, Tv)\precsim \frac{\lambda}{s^{2}} d(Qv, Tv)$, further, $\vert d(Qv, Tv) \vert \leq \frac{\lambda}{s^{2}} \vert d(Qv, Tv) \vert$, which is a contradiction that is $\vert d(Qv, Tv) \vert = 0$, then, $Qv = Tv = u$. As the pair $(T, Q)$ is weakly compatible, so we have $TQv = QTv$, therefore $Qu = Tu$.

We prove $u = Tu$, On the contrary we suppose that $Tu\neq u$,

where,

then,

Then, $d(u, u)\precsim \frac{\lambda}{s^{2}} d(u, u),$ further, $\vert d(u, Tu) \vert \leq \frac{\lambda}{s^{2}} \vert d(u, Tu) \vert,$ which is a contradiction that is $\vert d(u, Tu) \vert = 0$ then $u = Tu$.

Now we prove that $Qu = u$, On the contrary we suppose that $Qu\neq u,$, we have:

from $C_2$ we get:

where,

Further, $\vert d(u, Qu) \vert \leq \frac{\lambda}{s^{2}} \vert d(u, Qu)\vert,$ which is contradiction that is $\vert d(u, Qu) \vert = 0$ then $u = Qu$.

On conclude $Su = Tu = Pu = Qu = u$ when $P$ is continuous, we get the same results when $S$ is continuous.

Now we prove the uniqueness, Let $u^{*}$ be another common fixed point of $S, T, P$ and $Q$, then

Putting $z = u, w = u^{*}$ in $C_2$, we get: $d(u, u^{*}) = d(Su, Tu^{*})\precsim \frac{\lambda}{s^{2}} R(u, u^{*})$, where,

Further, $\vert d(u, u^{*}) \vert \leq \frac{\lambda}{s^{2}} \vert d(u, u^{*}) \vert,$ which is a contradiction that is $\vert d(u, u^{*}) \vert = 0$, which implies that $u = u^{*}$. Thus $u$ is the unique common fixed point of $S, T, P$ and $Q$ in $X$.

Corollary 1. Let $(X, d)$ be a complete complex valued b-metric space, if we put $S = T$ and $P = Q = I$ with exceeding the $\max$ of the rest of terms, we confirm the inequality of contraction of $T$ in the complete complex valued b-metric space. So we get: $d(Tz, Tw)\precsim \frac{\lambda}{s^{2}}d(z, w)$, where, $\lambda\in(0, 1), s\geq 1$ for all $z, w\in X.$ Then, $T$ have unique fixed point in $X$.

Example 4. Let $X = [0.1]$, for all $z, w\in X$. Define $d:\: X\times X \rightarrow \mathbb{C}$ a complex valued b-metric with $s = 2$ by:

Now define the mappings $S, T, P, Q:X\rightarrow X$ by:

.

Thus all the conditions of Theorem 2.1 are satisfied where $\lambda = \frac{1}{64}$ and $s = 2$. Then legibly $'0'$ is the unique common fixed point of the mappings $S, T, P$ and $Q$.

Example 5. Let $X = B(0, r), r > 1$, for all $z, w\in X$. Define $d:\: X\times X \rightarrow \mathbb{C}$ by:

a complete complex valued b-metric where $\Gamma$ is a closed path in $X$ containing a zero. We prove that $d$ is a complex b-metric with $s = 2$

Now we define the mappings $S, T, P, Q:X \rightarrow X$ by:

Using the Cauchy formula when the mappings $S, T, P$ and $Q$ are analytics we get:

Further,

Thus all the conditions of Theorem 2.1 are satisfied then the mappings $S, T, P$ and $Q$ have a unique common fixed point in $X$.

3.   Application to integral equations

Our first new results in this section is the following:

Theorem 3.1. Let $X = C ([a, b], \mathbb{R}^n), a > 0$ and $d:X\times X\rightarrow \mathbb{C}$ is defined as follows:

Consider the Urysohn integral equations

where $u\in[a, b]\subset\mathbb{R} \: and \: z, g, h\in X$.

Assume that $K_{1}, K_{2}:[a, b]\times[a, b]\times \mathbb{R}^{n}\rightarrow \mathbb{R}^{n}$ such that $F_{z}, G_{z} \in X$ for each $z\in X$, where

If there exist $s\geq 1, \:\lambda\in(0, 1)$ such that the inequality:

where,

and

holds for all $z, w\in X$. then, the system of Urysohn integral equations has a unique common solution in $X$.

Proof. Define $S, T:X\rightarrow X$ by:

Then,

From assumption $3.1$, for each $u\in[a, b]$ we have:

which implies that

Therefore,

Thus all the conditions of Theorem $2.1$ with $P = Q = I_{X}$ are satisfied. Therefore, the system of Urysohn integral equations has a unique common solution in $X$.

4.   Application to linear system

In this section we give an application using the Corollary 1 in $(X = \mathbb{C}^{n}, d_2)$ the complete complex valued b-metric space where,

Theorem 4.1. Let $(X = \mathbb{C}^{n}, d_2)$ a complex valued b-metric space where $z = (z_1, \ldots, z_{n})^{t}\in X$ and $w = (w_1, \ldots, w_{n})^{t}\in X,$ if $\beta < \frac{1}{n}$ where,

then, the following linear system of $n$ equations and $n$ unknowns $AZ = B$ has a unique solution.

Where $z = (z_1, \ldots, z_{n})^{t}\in X$ and $a_{ij} \in \mathbb{R}$ where $1\leq i, j \leq n$ and $b_1, b_2, b_{n} \in \mathbb{C}$

Proof. Define $T:X\rightarrow X$ by $Tz = (A+I)Z-B.$ for proving that linear system $AZ = B$ have a unique solution, its enough to prove that $T$ is a contraction.

Since

where,

Then,

So, we get finally that:

$d_2(Tz, Tw)\precsim n \beta d_2(z, w)$ or $\beta = \max \left\{ | a_{ij} |, | a_{ij} + 1 | \; \; \; \; \forall 1\leq i, j \leq n\right\}.$

We conclude that $T$ is contraction mapping. by applying Corollary 1, the linear system has a unique solution.

Acknowledgments

The authors are thankful to the learned referees for their valuable comments which helped in bringing this paper to its present form. Khaled Berrah acknowledges the financial support of Larbi Tebessi University, Tebessa, Algeria.

Conflict of interest

The authors declare that there is no conflicts of interest in this paper.