Extraction new results of common fixed point theorems for $({T}, {\alpha }_{{s}}, {F})$-contraction of six mappings in a tripled b-metric space with an application of integral equations

Abstract

The aim of this work is to usher in tripled b-metric spaces, triple weakly $\alpha _{s}$-admissible, triangular partially triple weakly $\alpha _{s}$-admissible and their properties for the first time. Also, we prove some theorems about coincidence and common fixed point for six self-mappings. On the other hand, we present a new model, talk over an application of our results to establish the existence of common solution of the system of Volterra-type integral equations in a triple b-metric space. Also, we give some example to illustrate our theorems in the section of main results. Finally, we show an application of primary results.

1 Introduction and preliminaries

The Banach contraction principle plays a central part in metric fixed point theory, and a great number of researchers revealed many fruitful generalizations of this resolution in diverse ways. In 1989, Bakhtin investigated the concept of b-metric space [1]. However, Czerwik initiated the study of fixed point of self-mappings in a b-metric space and proved an analogue of Banach’s fixed point theorem [2]. Since then, numerous research articles have been published comprising fixed point theorems for several classes of single-valued and multi-valued operators in b-metric spaces (for example, consider [3–6]). In 2012, the concept of F-contraction, which is one of these generalizations, was introduced by Wardowski [7]. He presented that every F-contraction defined in a complete metric space has a unique fixed point. Subsequently, the subject of F-contraction proved to be a milestone in the fixed point theory, and numerous research papers on F-contraction have been published (for instance, see [4, 8–19]). In the same year, Samet et al. investigated the idea of ($\alpha , \psi$)-contractive and α-admissible mappings and established some significant fixed point solutions for such a variety of functions defined on a complete metric space (for more details, see [20]). Some authors such as Salimi, Latif, Hussain et al. improved the concept of α-admissibility and proved some important (common) fixed point theorems as well (for more information, see [21–24].

Recently, Cosentino and Vetro established a fixed point result for Hardy–Rogers-type F-contraction [25]. Also, Minak, Helvaci, and Altun presented a fixed point result for Ćirić-type generalized F-contraction [26]. In 2018, Nazam, Muhammad, and Postolache investigated some common fixed point results for four self-mappings satisfying such kind of contractions on the $\alpha _{s}$-complete b-metric space and applied their conclusion to infer several new and old results, based on the idea of Ćirić-type and Hardy–Rogers-type ($\alpha _{s}, F$)-contractions [27].

In this study, motivated by [27] and among these achievements, we are working to stretch out the Ćirić-type and Hardy–Rogers-type ($\alpha _{s}, F$)-contractions based on six self-mappings defined on a b-metric space. Also, some common fixed point results for six self-mappings satisfying such kind of contractions are shown in the ($T, \alpha _{s}, F$)-complete tripled b-metric space. Consequently, we discuss an application of the main result to show the existence of common solution of the system of Volterra-type integral equations.

Let X be a nonempty set, $\mathbb{R}^{+} =(0, \infty )$, $\mathbb{R}_{0}^{+} = [ 0, \infty )$, and $s >1$ be a real constant. Suppose that $d_{b}$ maps $X\times X\times X$ into $\mathbb{R}_{0}^{+}$ somehow that for all x, y, z, and $a_{i}$ with $i\in \{1,2,3,4\}$ belong to X satisfying the following conditions [9]:

• $d_{b} ( x, y, z )=0$ if and only if $x = y = z$.

• $d_{b} ( x, y, z )>0$ if and only if $x\neq y$ or $x\neq z$ or $y\neq z$.

• $d_{b} ( x, y, z )= d_{b} ( x, z, y )= d_{b} ( z, y, x )= d_{b} ( y, x, z )= d_{b} ( z, x, y )= d_{b} ( y, z, x )$.

• $d_{b} ( x, x, y )= d_{b} ( x, y, y )$.

• $d_{b} ( x, x, y ) \leq d_{b} ( x, y, z )$, $d_{b} ( x, x, z ) \leq d_{b} ( x, y, z )$, $d_{b} ( y, y, z ) \leq d_{b} ( x, y, z )$.

• $d_{b} ( x, y, z ) \leq s [ d_{b} ( x, a_{1}, a_{2} ) + d_{b} ( y, a_{3}, a_{4} ) + d_{b} ( z, a_{2}, a_{3} )]$.

We say that ($X, d_{b}, s$) is a tripled b-metric space.

Example 1.1

Let $X= \mathbb{R}_{0}^{+}$. We define $d_{b}: X^{3} \rightarrow \mathbb{R}_{0}^{+}$ as follows:

$$ d_{b} ( x, y, z )=\max \bigl\{ \vert x-y \vert ^{2}, \vert x-z \vert ^{2}, \vert y-z \vert ^{2} \bigr\} . $$

Then ($X, d_{b}, s$) is a tripled b-metric space with $s =2$.

We bring back into reader’s mind some definitions and properties of b-metric.

Definition 1.2

(see [2])

Let A be a nonempty set, and let $s>1$ be a real number. A mapping $d^{*}: A^{2} \rightarrow \mathbb{R}_{0}^{+}$ is said to be a b-metric if, for all a, b, and $c\in A$, we have:

• $a = b$ if and only if $d^{*} ( a, b )=0$;

• $d^{*} ( a, b )= d^{*} ( b, a )$;

• $d^{*} ( a, b ) \leq s [ d^{*} ( a, c ) + d^{*} ( c, b )]$.

In this case, the triple ($A, d^{*}, s$) is called a b-metric space (with coefficient s).

Remark 1.3

Definition 1.2 allows us to remark that b-metric space is effectually more general than metric space as a b-metric is a metric when $s=1$. It is worth to mention that the b-metric structure produces some differences to the classical case of metric spaces: the b-metric on a nonempty set M need not be continuous, open balls in such spaces need not be open sets, and so on. The following example describes the significance of a b-metric.

For the notions like convergence, completeness, Cauchy sequence in the setting of b-metric spaces, the reader is referred to Aghajani et al. [28], Czerwik [2], Amini-Harandi [29], Huang et al. [3], Khamsi and Hussain [5]. In line with Wardowski [7], Cosentino et al. [30] investigated a nonlinear function $F: \mathbb{R}^{+} \rightarrow \mathbb{R}$ complying with the following axioms:

• F is strictly increasing;

• $\lim_{r\rightarrow \infty } r_{n} =0$ if and only if $\lim_{n\rightarrow \infty } F ( r_{n} )= -\infty$;

• $\lim_{r\rightarrow \infty } r_{n} =0$ there exists $a\in (0,1)$ such that $\lim_{r_{n} \rightarrow 0^{+}} ( r_{n} )^{a} F ( r_{n} )=0$;

• $\tau +F ( s r_{n} ) \leq F ( r_{n- 1} )$ implies $\tau +F ( s^{n} r_{n} ) \leq F ( s^{n- 1} r_{n- 1} )$ for each $n\in \mathbb{N}$ and some $\tau >0$

for all sequence $\{ r_{n} \}$ of positive numbers. We denote the set of all functions satisfying the conditions ($F_{1}$), ($F_{2}$), ($F_{3}$), and ($F_{4}$) by $\mathcal{F}_{s}$.

Example 1.4

(see [30])

Let $F: \mathbb{R}^{+} \rightarrow \mathbb{R}$ be defined by $F(r)=\ln r$ or $F(r)=r+\ln r$. Then F satisfies in the conditions.

Theorem 1.5

(see [31])

Let ($X,d$) be a complete metric space and $T:X\rightarrow X$ be a bijective ($\xi ,\alpha ,\eta$)-expansive mapping of type B satisfying the following conditions:

• $T^{- 1}$ is α-admissible with respect to η;

• There exists $x_{0} \in X$ such that $\alpha ( x_{0}, T^{- 1} x_{0} ) \geq \eta ( x_{0}, T^{- 1} x_{0} )$;

• T is continuous.

Then T has a fixed point.

Definition 1.6

(see [32])

Let ($X, p_{b}$) be a partial b-metric space with the coefficient $s\geq 1$. A mapping $T:X\rightarrow X$ is said to be a generalized $\alpha -\eta -\psi$-Geraghty contractive type mapping if there exist $\psi \in \Psi$, $\alpha ,\eta :X\times X\rightarrow [ 0,\infty )$, and $\beta \in \mathcal{F}$ such that

$$ \alpha ( x, y ) \geq \eta ( x, y ) \quad \text{implies}\quad \psi \bigl( s p_{b} ( Tx, Ty )\bigr) \leq \beta \bigl( \psi \bigl( M_{s}^{T} ( x, y )\bigr)\bigr) \psi \bigl( M_{s}^{T} ( x, y )\bigr) $$

(1.1)

for all $x, y\in X$, where

$$ M_{s}^{T} ( x, y )=\max \biggl\{ p_{b} ( x, y ), p_{b} ( x, Tx ), p_{b} ( y, Ty ), \frac{p_{b} ( x, Ty ) + p_{b} ( y, Tx )}{2 s} \biggr\} . $$

Theorem 1.7

(see [32])

Let ($X, p_{b}$) be a $p_{b}$-complete partial b-metric space with the coefficient $s\geq 1$. Let $T:X\rightarrow X$ be a generalized $\alpha -\eta -\psi$-Geraghty contractive type mapping. Suppose that the following conditions hold:

• T is a triangular α-orbital admissible mapping with respect to η;

• There exists $x_{1} \in X$ such that $\alpha ( x_{1}, T x_{1} ) \geq \eta ( x_{1}, T x_{1} )$;

• $\{ x_{n} \}$ is α-regular with respect to η.

Then T has a fixed point.

Example 1.8

(see [32])

Let $X=[0,\infty )$ and with the partial b-metric $p_{b}:X\times X\rightarrow [0,\infty )$ defined by $p_{b} (x,y)=\max \{x,y \}^{2}$ for all $x,y\in X$. Obviously, ($X, p_{b}$) is a partial b-metric space with $s=2$. Define the mapping $T:X\rightarrow X$ given by

$$ Tx = \textstyle\begin{cases} \frac{x}{9} & \text{if } x\in [0,1];\\ \ln x+ 3 & \text{if } x\in (1, \infty ). \end{cases} $$

Define $\psi :[0, \infty ) \rightarrow [0, \infty )$ and $\beta :[0, \infty ) \rightarrow [0,1)$ by $\psi ( t )= t$ and

$$ \beta ( t )= \textstyle\begin{cases} \frac{e^{-t}}{1 +t} & \text{if } x\in (0, \infty );\\ \frac{1}{2} & \text{if } t =0. \end{cases} $$

Let $\alpha , \eta : X\times X\rightarrow [0, \infty )$ be defined by

$$ \alpha ( x, y )= \textstyle\begin{cases} 6 & \text{if } x\in [0,1];\\ 0 & \text{if } x\in (1, \infty ), \end{cases} $$

and

$$ \eta ( x, y )= \textstyle\begin{cases} 2 & \text{if } x\in [0,1];\\ 1 & \text{if } x\in (1, \infty ). \end{cases} $$

Let $\alpha ( x, Tx ) \geq \eta ( x, Tx )$. Thus $x, Tx\in [0,1]$ and so $T^{2} x = T ( Tx ) \in [0,1]$, which implies that $\alpha ( Tx, T^{2} x ) \geq \eta ( Tx, T^{2} x )$, that is, T is α-orbital admissible with respect to η. Now, let $\alpha ( x, y ) \geq \eta ( x, y )$ and $\alpha ( y, Ty ) \geq \eta ( y, Ty )$, we get that $x, y, Ty\in [0,1]$ and so $\alpha ( x, Ty ) \geq \eta ( x, Ty )$. Therefore T is triangular α-orbital admissible with respect to η. Let $\{ x_{n} \}$ be a sequence such that $\{ x_{n} \}$ is $p_{b}$-convergent to z and $\alpha ( x_{n}, x_{n+ 1} ) \geq \eta ( x_{n}, x_{n+ 1} )$ for all $n\in \mathbb{N}$. Then $\{ x_{n} \} \subseteq [0,1]$ for any $n\in \mathbb{N}$ and so $z\in [ 0,1 ]$, from which we have $\alpha ( x_{n}, z ) \geq \eta ( x_{n}, z )$. That is, $\{ x_{n} \}$ is α-regular with respect to η. The condition (ii) of Theorem 1.7 is satisfied with $x_{1} =1 \in X$ since ($\alpha (1, T 1)=2 \geq 2= \eta (1, T 1)$. We next prove that T is a generalized α-η-ψ-Geraghty contraction type mapping. Let $x, y\in X$ with $\alpha ( x, y ) \geq \eta ( x, y )$. Thus $x, y\in [0,1]$. Without loss of generality, we may assume that $0 \leq y\leq x\leq 1$. Therefore

$$ p_{b} ( Tx, Ty )= \biggl[ \max \biggl\{ \frac{x}{9}, \frac{y}{9} \biggr\} \biggr]^{2} = \frac{x^{2}}{81} $$

and

$$ M_{s}^{T} ( x, y )=\max \biggl\{ x^{2}, x^{2}, y^{2}, \frac{x^{2} + [ \max \{ y, \frac{x}{9} \} ]^{2}}{4} \biggr\} = x^{2}. $$

Since $\frac{2}{81} \leq \frac{1}{2 e} \leq \frac{e^{- x^{2}}}{1 + x^{2}}$, we obtain that

$$\begin{aligned} \psi \bigl( s p_{b} ( Tx, Ty )\bigr)&= \psi \biggl(2 \frac{x^{2}}{81} \biggr)= \frac{2 x^{2}}{81} \leq \frac{e^{- x^{2}}}{1 + x^{2}} \cdot x^{2} \\ &\leq \beta \bigl( \psi \bigl( x^{2} \bigr)\bigr) \psi \bigl( x^{2} \bigr) \\ &\leq \beta \bigl( \psi \bigl( M_{s}^{T} ( x, y )\bigr) \bigr) \psi \bigl( M_{s}^{T} ( x, y )\bigr). \end{aligned}$$

Thus T is a generalized α-η-ψ-Geraghty contraction type mapping. Hence all the assumptions in Theorem 1.7 are satisfied and thus T has a fixed point which is $x =0$.

Definition 1.9

(see [27])

Let ($M, d^{*},s$) be a b-metric space, $S:M\rightarrow M$ and $\alpha _{s}:M\times M\rightarrow \mathbb{R}_{0}^{+}$ be two mappings. The mapping S is said to be $\alpha _{s}$-admissible if

$$ \alpha _{s} ( r_{1}, r_{2} ) \geq s^{2} \Rightarrow \alpha _{s} \bigl( S ( r_{1} ), S ( r_{2} )\bigr) \geq s^{2} \quad \text{for all } r_{1}, r_{2} \in M. $$

Theorem 1.10

(see [27])

Let M be a nonempty set and $\alpha _{s}$ be as defined in Definition 1.9. Let f, g, S, T be $\alpha _{s} -b$-continuous self-mappings defined on an $\alpha _{s}$-complete b-metric space ($M, d^{*},s$) such that $f(M)\subseteq T(M)$, $g(M)\subseteq S(M)$. Suppose that, for all $( r_{1}, r_{2} )\in \gamma _{f,g, \alpha _{s}}$, there exist $F\in \mathcal{F}_{s}$ and $\tau >0$ such that

$$ \tau +F \bigl( s d^{*} \bigl( f ( r_{1} ), g ( r_{2} )\bigr)\bigr) \leq F \bigl( \mathcal{M}_{1} ( r_{1}, r_{2} )\bigr). $$

(1.2)

Assume that the pairs ($f, S$), ($g, T$) are $\alpha _{s}$-compatible and the pairs ($f, g$) and ($g, f$) are triangular partially weakly $\alpha _{s}$-admissible with respect to T and S, respectively. Then the pairs ($f, S$), ($g, T$) have the coincidence point (say) v in M. Moreover, if $\alpha _{s} ( Sv, Tv ) \geq s^{2}$, then v is a common fixed point of f, g, S, T.

Remark 1.11

(see [27])

If we suppose that $\alpha _{s} (v,w)\geq s^{2}$ for each pair of common fixed point of f, g, S, T, then v is unique. Indeed, if w is another fixed point of f, g, S, T and assuming on the contrary that $d^{*} (fv,gw)>0$, then from (1.2) we have

$$ F \bigl( s d^{*} ( v, w )\bigr)= F \bigl( s d^{*} \bigl( S ( v ), T ( w )\bigr)\bigr) \leq F \bigl( \mathcal{M}_{1} ( v, w )\bigr) -\tau , $$

(1.3)

where

$$ \begin{aligned} \mathcal{M}_{1} ( v, w )&=\max \biggl\{ d^{*} \bigl( S ( v ), T ( w )\bigr), d^{*} \bigl( f ( v ), S ( v )\bigr),\\ &\quad d^{*} \bigl( g ( w ), T ( w )\bigr) \frac{d^{*} ( S ( v ), g ( w )) + d^{*} ( f ( v ), T ( w ))}{2 s} \biggr\} . \end{aligned} $$

Thus, by (1.3), we have

$$ F \bigl( s d^{*} ( v, w )\bigr)< F \bigl( d^{*} ( v, w ) \bigr), $$

which is a contradiction. Hence, $v = w$ and v is a unique common fixed point of self-mappings f, g, S, T.

Theorem 1.12

(see [27])

Let f, g, S, T be self-mappings defined on an $\alpha _{s}$-regular and $\alpha _{s}$-complete metric space ($M, d^{*},s$) such that $f(M)\subseteq T(M)$, $g(M)\subseteq S(M)$, and $T(M)$ and $S(M)$ are closed subsets of M. Suppose that, for all $( r_{1}, r_{2} )\in \gamma _{f,g, \alpha _{s}}$, there exist $F\in \mathcal{F}_{s}$ and $\tau >0$ such that

$$ \tau +F ( s d^{*} \bigl( f ( r_{1} ), g ( r_{2} ) \bigr) \leq F \bigl( \mathcal{M}_{1} ( r_{1}, r_{2} )\bigr). $$

(1.4)

Assume that the pairs ($f, S$), ($g, T$) are weakly compatible and the pairs ($f, g$) and ($g, f$) are triangular partially weakly $\alpha _{s}$-admissible with respect to T and S, respectively. Then the pairs ($f, S$), ($g, T$) have the coincidence point v in M. Moreover, if $\alpha _{s} ( Sv, Tv ) \geq s^{2}$, then v is a coincidence point of f, g, S, T.

Theorem 1.13

(see [27])

Let f, g, S, T be $\alpha _{s}$-continuous self-mappings defined on an $\alpha _{s}$-complete b-metric space ($M, d^{*},s$) such that $f(M)\subseteq T(M)$, $g(M)\subseteq S(M)$. Suppose that, for all $( r_{1}, r_{2} )\in \gamma _{f,g, \alpha _{s}}$, there exist $F\in \mathcal{F}_{s}$ and $\tau >0$ such that

$$ \tau +F ( s d^{*} \bigl( f ( r_{1} ), g ( r_{2} ) \bigr) \leq F \bigl( \mathcal{M}_{i} ( r_{1}, r_{2} )\bigr) $$

(1.5)

holds for one of $i =2,3,4,5,6$, where

$$\begin{aligned} \mathcal{M}_{2} ( r_{1}, r_{2} )&= a_{1} d^{*} \bigl( S ( r_{1} ), T ( r_{2} )\bigr) + a_{2} d^{*} \bigl( f ( r_{1} ), S ( r_{1} )\bigr) + a_{3} d^{*} \bigl( g ( r_{2} ), T ( r_{2} )\bigr) \\ &\quad {}+ a_{4} \bigl[ d^{*} \bigl( S ( r_{1} ), g ( r_{2} )\bigr) + d^{*} \bigl( f ( r_{1} ), T ( r_{2} )\bigr)\bigr] \end{aligned}$$

with $a_{i} \geq 0$, $i =1,2,3,4$, such that $a_{1} + a_{2} + a_{3} + 2 s a_{4} =1$;

$$ \mathcal{M}_{3} ( r_{1}, r_{2} )= a_{1} d^{*} \bigl( S ( r_{1} ), T ( r_{2} )\bigr) + a_{2} d^{*} \bigl( f ( r_{1} ), S ( r_{1} )\bigr) + a_{3} d^{*} \bigl( g ( r_{2} ), T ( r_{2} )\bigr), $$

with $a_{1} + a_{2} + a_{3} =1$;

$$\begin{aligned}& \mathcal{M}_{4} ( r_{1}, r_{2} )= k \max \bigl\{ d^{*} \bigl( f ( r_{1} ), S ( r_{1} )\bigr), d^{*} \bigl( g ( r_{2} ), T ( r_{2} )\bigr)\bigr\} \quad \textit{with } k\in [0,1); \\& \begin{aligned} \mathcal{M}_{5} ( r_{1}, r_{2} )&= a_{1} ( r_{1}, r_{2} ) d^{*} \bigl( S ( r_{1} ), T ( r_{2} )\bigr) + a_{2} ( r_{1}, r_{2} ) d^{*} \bigl( f ( r_{1} ), S ( r_{1} )\bigr)\\ &\quad {} + a_{3} ( r_{1}, r_{2} ) d^{*} \bigl( g ( r_{2} ), T ( r_{2} )\bigr) \\ &\quad {}+ a_{4} ( r_{1}, r_{2} )\bigl[ d^{*} \bigl( S ( r_{1} ), g ( r_{2} )\bigr)) + d^{*} \bigl( f ( r_{1} ), T ( r_{2} )\bigr)\bigr] \end{aligned} \end{aligned}$$

with $a_{i} ( r_{1}, r_{2} )$, $i =1,2,3,4$ are nonnegative functions such that

$$\begin{aligned}& \sup_{r_{1}, r_{2} \in M} \bigl\{ a_{1} ( r_{1}, r_{2} ) + a_{2} ( r_{1}, r_{2} ) + a_{3} ( r_{1}, r_{2} ) + 2 s a_{4} ( r_{1}, r_{2} )\bigr\} =1; \\& \begin{aligned} \mathcal{M}_{6} ( r_{1}, r_{2} )&= a_{1} d^{*} \bigl( S ( r_{1} ), T ( r_{2} )\bigr) + \frac{a_{2} + a_{3}}{2} \bigl[ d^{*} \bigl( f ( r_{1} ), S ( r_{1} )\bigr) + d^{*} \bigl( g ( r_{2} ), T ( r_{2} )\bigr)\bigr] \\ &\quad {}+ \frac{a_{4} + a_{5}}{2 s} \bigl[ d^{*} \bigl( S ( r_{1} ), g ( r_{2} )\bigr) + d^{*} \bigl( f ( r_{1} ), T ( r_{2} )\bigr)\bigr] \end{aligned} \end{aligned}$$

with $a_{1} + a_{2} + a_{3} + a_{4} + a_{5} =1$.

Assume that the pairs ($f, S$), ($g, T$) are $\alpha _{s}$-compatible and the pairs ($f, g$) and ($g, f$) are triangular partially weakly $\alpha _{s}$-admissible pairs of mappings with respect to T and S, respectively. Then the pairs ($f, S$), ($g, T$) have the coincidence point v in M. Moreover, if $\alpha _{s} ( Sv, Tv ) \geq s^{2}$, then v is a common point of f, g, S, T.

2 Main results

In this section, first we introduce some definitions in a tripled b-metric space ($X, d_{b}$) and present several examples.

Definition 2.1

Let ($X, d_{b},s$) be a tripled b-metric space, $T:X\rightarrow X$ and $\alpha _{s}: X^{3} \rightarrow \mathbb{R}_{0}^{+}$ be two mappings. The mapping T is said to be $\alpha _{s}$-admissible if $\alpha _{s} ( x,y,z ) \geq s^{2}$, then $\alpha _{s} (Tx,Ty,Tz)\geq s^{2}$ for all $x,y,z\in X$.

Definition 2.2

Let ($X, d_{b},s$) be a tripled b-metric space, $T:X\rightarrow X$ and $\alpha _{s}: X^{3} \rightarrow \mathbb{R}_{0}^{+}$ be two mappings. The mapping T is said to be triangular $\alpha _{s}$-admissible if

• $\alpha _{s} ( x, y, z ) \geq s^{2}$ implies that $\alpha _{s} ( Tx, Ty, Tz ) \geq s^{2}$ for all $x, y, z\in X$;

• $\alpha _{s} ( x, y, z ) \geq s^{2}$ and $\alpha _{s} ( y, z, w ) \geq s^{2}$ imply $\alpha _{s} ( x, z, w ) \geq s^{2}$for all $x, y, z, w\in X$.

Definition 2.3

Let ($X, d_{b},s$) be a tripled b-metric space, $f,g,h:X\rightarrow X$ and $\alpha _{s}: X^{3} \rightarrow \mathbb{R}_{0}^{+}$ be four mappings. The tripled ($f,g,h$) is said to be

• triple weakly $\alpha _{s}$-admissible if $\alpha _{s} ( f ( x ), gf ( x ), hgf ( x )) \geq s^{2}$, $\alpha _{s} ( g ( x ), hg ( x ), fhg ( x )) \geq s^{2}$, and $\alpha _{s} ( h ( x ), fh ( x ), gfh ( x )) \geq s^{2}$ for all $x\in X$;

• partially weakly $\alpha _{s}$-admissible if $\alpha _{s} ( f ( x ), gf ( x ), hgf ( x )) \geq s^{2}$ for all $x\in X$.

Definition 2.4

Let ($X, d_{b},s$) be a tripled b-metric space and $f,g,h,\phi :X\rightarrow X$ be four mappings such that $f(X)\cup g(X)\cup h(X)\subseteq \phi (X)$. The triple of mappings ($f,g,h$) is said to be

• triple weakly $\alpha _{s}$-admissible with respect to ϕ if and only if $\alpha _{s} ( f ( x ), g ( y ), h ( z )) \geq s^{2}$ for all $x\in X$, for all $y\in \phi ^{- 1} gf ( x )$, for all $z\in \phi ^{- 1} hgf ( x )$ and $\alpha _{s} ( h ( x ), g ( y ), f ( z )) \geq s^{2}$ for all $x\in X$, for all $y\in \phi ^{- 1} gh ( x )$, for all $z\in \phi ^{- 1} fgh ( x )$ and $\alpha _{s} ( g ( x ), f ( y ), h ( z )) \geq s^{2}$ for all $x\in X$, for all $y\in \phi ^{- 1} fg ( x )$, for all $z\in \phi ^{- 1} hfg ( x )$;

• partially triple weakly $\alpha _{s}$-admissible with respect to ϕ if and only if

  $$ \alpha _{s} \bigl( f ( x ), g ( y ), h ( z )\bigr) \geq s^{2} $$

  for all $x\in X$, $y\in \phi ^{- 1} gf ( x )$, and $z\in \phi ^{- 1} hgf ( x )$.

Definition 2.5

Let ($X, d_{b},s$) be a tripled b-metric space and $f,g,h,\phi :X\rightarrow X$ be four mappings such that $f(X)\cup g(X)\cup h(X)\subseteq \phi (X)$. The triple of mappings ($f,g,h$) is said to be triangular triple weakly $\alpha _{s}$-admissible with respect to ϕ if

• $\alpha _{s} ( h ( x ), g ( y ), f ( z )) \geq s^{2}$ for all $x\in X$, for all $y\in \phi ^{- 1} gf ( x )$, $z\in \phi ^{- 1} hgf ( x )$, and

  $$ \alpha _{s} \bigl( h ( x ), g ( y ), f ( z )\bigr) \geq s^{2} $$

  for all $x\in X$, for all $y\in \phi ^{- 1} gh ( x )$, for all $z\in \phi ^{- 1} fgh ( x )$, and $\alpha _{s} ( g ( x ), f ( y ), h ( z )) \geq s^{2}$ for all $x\in X$, for all $y\in \phi ^{- 1} fg ( x )$, for all $z\in \phi ^{- 1} hfg ( x )$;

• $\alpha _{s} ( x, y, z ) \geq s^{2}$ and $\alpha _{s} ( y, z, w ) \geq s^{2}$ imply $\alpha _{s} ( x, z, w ) \geq s^{2}$ for all $x, y, z, w\in X$.

Example 2.6

Let $X= \mathbb{R}_{0}^{+}$ and

$$ d_{b} ( x, y, z )=\max \bigl\{ \vert x-y \vert ^{2}, \vert x-z \vert ^{2}, \vert y-z \vert ^{2} \bigr\} $$

for all $x, y, z\in X$. Then ($X, d_{b}, s$) is a tripled b-metric with $s =2$. We define $f ( x )= x$, $g ( x )= x^{\frac{1}{2}}$, $h ( x )= x^{\frac{1}{4}}$, and $S ( x )= x^{4}$ if $x\in [0,1)$ and $f ( x )= g ( x )= h ( x )= S ( x )=1$, whenever $x\in [1, \infty )$ and $\alpha _{s}: X^{3} \rightarrow \mathbb{R}_{0}^{+}$ as follows:

$$ \alpha ( x, y, z )= \textstyle\begin{cases} \max \{ 4 +y-x,4 +z-x,4 +z-x \} , & x, y, z\in [0,1),\\ 0, & \text{otherwise}. \end{cases} $$

Then, for all $x\in [0,1)$, $y\in S^{- 1} ( g ( f ( x )))$, $z\in S^{- 1} ( h ( g ( x )))$, we have $y = x^{\frac{1}{8}}$, $z = x^{\frac{1}{32}}$,

$$ \alpha _{s} \bigl( x, g \bigl( x^{\frac{1}{8}} \bigr), h \bigl( x^{\frac{1}{32}} \bigr) \bigr) = \alpha _{s} \bigl( x, x^{\frac{1}{16}}, x^{\frac{1}{32 \times 4}} \bigr) \geq s^{2}. $$

Thus the triple of mappings ($f, g, h$) is triangular weakly $\alpha _{s}$-admissible with respect to S. Indeed, if $\alpha _{s} ( x, y, z ) \geq s^{2}$ and $\alpha _{s} ( y, z, w ) \geq s^{2}$, then $\alpha _{s} ( x, z, w ) \geq s^{2}$. Since $y-x\geq 0$ or $z-x\geq 0$ or $z-y\geq 0$ and $z-y\geq 0$ or $w-z\geq 0$ or $w-y\geq 0$. Thus $w-x\geq 0$ or $w-z\geq 0$ or $z-x\geq 0$.

Definition 2.7

Let $f,g,h,\phi :X\rightarrow X$ be four self-mappings defined on a tripled b-metric space such that $f(X)\cup g(X)\cup h(X)\subseteq \phi (X)$. The triple of mappings ($f,g,h$) is said to be triangular triple partially weakly $\alpha _{s}$-admissiblewith respect to ϕ if

• $\alpha _{s} ( f ( x ), g ( y ), h ( z ) ) \geq s^{2}$ for all $x\in X$, $y\in \phi ^{- 1} ( g ( f ( x )))$, $z\in \phi ^{- 1} ( hg ( f ( x )))$,

• $\alpha _{s} ( x, y, z ) \geq s^{2}$, $\alpha _{s} ( y, z, w ) \geq s^{2}$ imply $\alpha _{s} ( x, z, w ) \geq s^{2}$ for all $x, y, z\in X$.

Definition 2.8

Let ($X, d_{b},s$) be a tripled b-metric space. The tripled b-metric space X is said to be $\alpha _{s}$-complete if and only if every Cauchy sequence $\{ x_{n} \}$ in X such that $\alpha _{s} ( x_{n}, x_{n+1}, x_{n+2} ) \geq s^{2}$ for all $n \in \mathbb{N}$ converges in X. That is,

$$ \lim_{n\rightarrow \infty } d_{b} ( x_{n}, x, x )= \lim _{n\rightarrow \infty } d_{b} ( x_{n}, x_{n}, x )=0. $$

If X is a complete tripled metric space, then X is also an $\alpha _{s}$-complete tripled metric space, but the converse is not true. The following example explains this fact.

Example 2.9

Let $X= \mathbb{R}^{+}$ and $d_{b}: X^{3} \rightarrow \mathbb{R}_{0}^{+}$ be the tripled b-metric. Define $\alpha _{s}: X^{3} \rightarrow \mathbb{R}_{0}^{+}$,

$$ \alpha ( x, y, z )= \textstyle\begin{cases} 4\max \{ e^{| x-y | }, e^{| y-z | }, e^{| x-z | } \} , & x, y, z\in [0, \frac{5}{2} ),\\ 0, & \text{otherwise}. \end{cases} $$

It is easy to see that ($X, d_{b}, S$) in not a complete tripled b-metric space, but ($X, d_{b}, s$) is an $\alpha _{s}$-complete tripled b-metric.

Definition 2.10

Let ($X, d_{b},s$) be a tripled b-metric space. We say that the self-mapping T is an $\alpha _{s}$-continuous mapping on ($X, d_{b},s$) if, for given $x\in X$ and sequence $\{ x_{n} \}$,

$$ \lim_{n\rightarrow \infty } d_{b} ( x_{n}, x, x )= \lim _{n\rightarrow \infty } d_{b} ( x_{n}, x_{n}, x )=0, $$

and $\alpha ( x_{n}, x_{n+ 1}, x_{n+ 2} ) \geq s^{2}$ for all $n\in \mathbb{N}$ implies

$$ \lim_{n\rightarrow \infty } d_{b} ( T x_{n}, Tx, Tx )= \lim_{n\rightarrow \infty } d_{b} ( T x_{n}, T x_{n}, Tx )=0. $$

Example 2.11

Let $X= \mathbb{R}_{0}^{+}$ and $d_{b}: X^{3} \rightarrow \mathbb{R}_{0}^{+}$ for all $x,y,z\in X$, define by $d_{b} (x,y,z)=\max \{ | x-y | ^{2},| x-z | ^{2},| y-z | ^{2} \}$ and

$$\begin{aligned}& T ( x )= \textstyle\begin{cases} \sin \pi x, & x\in [0,1],\\ \cos \pi x+ 2, & x\in (1, \infty ), \end{cases}\displaystyle \\& \alpha _{s} ( x, y, z )= \textstyle\begin{cases} x^{2} + y^{2} + 4, & x, y, z\in [0,1],\\ 0, & \text{otherwise}. \end{cases}\displaystyle \end{aligned}$$

Then T is not continuous on X; however, T is $\alpha _{s}$-continuous.

Definition 2.12

Let ($X, d_{b},s$) be a tripled b-metric space. The pairs of self-mappings ($f,g$), ($g,h$), and ($f,h$) are said to be $\alpha _{s}$-compatible if

$$\begin{aligned}& \lim_{n\rightarrow \infty } d_{b} \bigl( gh ( x_{n} ), hg ( x_{n} ), g ( x_{n} ) \bigr) =0, \\& \lim_{n\rightarrow \infty } d_{b} \bigl( fg ( x_{n} ), gf ( x_{n} ), f ( x_{n} ) \bigr) =0, \\& \lim_{n\rightarrow \infty } d_{b} \bigl( hf ( x_{n} ), fh ( x_{n} ), h ( x_{n} ) \bigr) =0, \end{aligned}$$

or $\lim_{n\rightarrow \infty } d_{b} ( gh ( x_{n} ), hg ( x_{n} ), h ( x_{n} ) ) =0$ or $\lim_{n\rightarrow \infty } d_{b} ( fg ( x_{n} ), gf ( x_{n} ), g ( x_{n} ) ) =0$ or

$$ \lim_{n\rightarrow \infty } d_{b} \bigl( hf ( x_{n} ), fh ( x_{n} ), f ( x_{n} ) \bigr) =0, $$

whenever $\{ x_{n} \}$ is a sequence in X such that $\alpha ( x_{n}, x_{n+ 1}, x_{n+ 1} ) \geq s^{2}$, and

$$ \lim_{n\rightarrow \infty } f ( x_{n} )= \lim_{n\rightarrow \infty } g ( x_{n} )= \lim_{n\rightarrow \infty } h ( x_{n} )= t $$

for some $t\in X$.

Example 2.13

Let $X=[1,\infty )$ and $d_{b}:X\times X\times X\rightarrow \mathbb{R}_{0}^{+}$ be defined by

$$ d_{b} ( x, y, z )=\max \bigl\{ \vert x-y \vert ^{2}, \vert x-z \vert ^{2}, \vert y-z \vert ^{2} \bigr\} $$

for all $x, y, z\in X$, then ($X, d_{b}, s =2$) is a tripled b-metric space. Define $f ( x )=4$, $g ( x )=16 - 3 x$ if $x\in [1,4]$ and $f ( x )=8$ and $g ( x )=9$ whenever $x\in (4, \infty )$ and

$$ \alpha ( x, y, z )= \textstyle\begin{cases} 6, & x, y, z\in [1,4],\\ 0, & \text{otherwise}. \end{cases} $$

Let us consider $\{ x_{n} \}$ to be a sequence such that $\alpha ( x_{n}, x_{n+ 1}, x_{n+ 2} ) \geq s^{2}$, and let

$$ \lim_{n\rightarrow \infty } f ( x_{n} )= \lim_{n\rightarrow \infty } g ( x ), $$

then $x_{n} =4$. It is clear that $\lim_{n\rightarrow \infty } f ( x_{n} )= \lim_{n\rightarrow \infty } g ( x )=4$. We obtain that

$$\begin{aligned} \lim_{n\rightarrow \infty } d_{b} \bigl( fg ( x_{n} ), gf ( x_{n} ), f ( x_{n} )\bigr)&= \lim_{n\rightarrow \infty } d_{b} \bigl( fg ( x_{n} ), gf ( x_{n} ), g ( x_{n} )\bigr) \\ &= d_{b} (4,4,4)=0. \end{aligned}$$

Hence ($f, g$) is an $\alpha _{s}$-compatible pair. Now, if we consider $x_{n} =4 - \frac{1}{n}$, then

$$ \lim_{n\rightarrow \infty } f ( x_{n} )= \lim_{n\rightarrow \infty } g ( x_{n} )=4. $$

But $\lim_{n\rightarrow \infty } gf ( x_{n} )=4$,

$$ \lim_{n\rightarrow \infty } fg ( x_{n} )= \lim_{n\rightarrow \infty } f \biggl(16 - 3\biggl(4 - \frac{1}{n} \biggr)\biggr)= \lim _{n\rightarrow \infty } f \biggl(4 + \frac{3}{n} \biggr)=8, $$

and $\lim_{n\rightarrow \infty } d_{b} ( fg ( x_{n} ), gf ( x_{n} ), f x_{n} ) \neq 0$. Consequently, ($f, g$) is not compatible.

Definition 2.14

Let $f,g$, and T be self-mappings defined on a nonempty set X. If $f(x)=g(x)=T(x)$ for some $x\in X$, then x is called a coincidence point of $f,g$, and T. Three self-mappings $f,g$, and T defined on X are said to be weakly compatible if $\{f,g\}$, $\{g,T\}$, and $\{f,T\}$ commute at their coincidence points.

Definition 2.15

Let ($X, d_{b},s$) be a tripled b-metric space. The space ($X, d_{b},s$) is said to be $\alpha _{s}$-regular if, for any sequence $\{ x_{n} \}$ in X, the following condition holds: if $x_{n} \rightarrow x$ as $n\rightarrow \infty$ and $\alpha _{s} ( x_{n}, x_{n+1}, x_{n+2} )\geq s^{2}$ for all $n \in \mathbb{N}$, then $\alpha _{s} ( x_{n},x,x)\geq s^{2}$ and $\alpha _{s} ( x_{n}, s_{n},x)\geq s^{2}$ for all $n \in \mathbb{N}$.

Now, we are ready to prove our results.

Lemma 2.16

Let ($X, d_{b},s$) be a tripled b-metric space. If there exist three sequence $\{ x_{n} \}$, $\{ y_{n} \}$, and $\{ z_{n} \}$ such that $\lim_{n\rightarrow \infty } d_{b} ( x_{n}, y_{n}, z_{n} )=0$ and $\lim_{n\rightarrow \infty } x_{n} = \lim_{n\rightarrow \infty } y_{n} =t$ for some $t\in X$, then $\lim_{n\rightarrow \infty } z_{n} =t$.

Proof

By the triangle inequality, we have

$$ d_{b} ( z_{n}, t, t ) \leq s \bigl[ d_{b} ( z_{n}, x_{n}, y_{n} ) + d_{b} ( t, t, t ) + d_{b} ( t, y_{n}, t ) \bigr]. $$

By taking limit as $n\rightarrow \infty$, the result follows. □

Definition 2.17

Let ($X, d_{b},s$) be a tripled b-metric space, $f,g,h, S_{1}, S_{2}, S_{3}:X\rightarrow X$ be self-mappings, and $\alpha _{s}$ be as defined in Definition 2.1. We define the set $\lambda _{f,g,h, \alpha _{s}}$ by

$$ \begin{aligned}[b] \lambda _{f, g, h, \alpha _{s}} & = \bigl\{ ( \alpha , \beta , \gamma ) \in X^{3}: \alpha _{s} \bigl( S_{1} ( \alpha ), S_{2} ( \beta ), S_{3} ( \gamma )\bigr) \geq s^{2}, \\ & \quad \text{and } d_{b} \bigl( f ( \alpha ), g ( \beta ), h ( \gamma )\bigr)>0 \bigr\} . \end{aligned} $$

(2.1)

Let

$$ \begin{aligned}[b] & M ( \alpha , \beta , \gamma ) \\ &\quad =\max \biggl\{ d_{b} \bigl( S_{1} ( \alpha ), S_{2} ( \beta ), S_{3} ( \gamma ) \bigr), d_{b} \bigl( f ( \alpha ), S_{2} ( \alpha ), S_{3} ( \alpha ) \bigr), \\ &\qquad d_{b} \bigl( g ( \beta ), S_{1} ( \beta ), S_{3} ( \beta ) \bigr), d_{b} \bigl( h ( \gamma ), S_{1} ( \gamma ), S_{2} ( \gamma ) \bigr), \\ &\qquad \frac{d_{b} ( S_{1} ( \alpha ), g ( \beta ), h ( \gamma ) ) + d_{b} ( f ( \alpha ), S_{2} ( \beta ), h ( \gamma ) ) + d_{b} ( S_{3} ( \gamma ), g ( \beta ), f ( \alpha ) )}{3 s} \biggr\} . \end{aligned} $$

(2.2)

The following theorem is one of our main results.

Theorem 2.18

Let X be a nonempty set and $\alpha _{s}$ be as defined in Definition 2.1. Let f, g, h, $S_{1}$, $S_{2}$, $S_{3}$ be $\alpha _{s} - b$-continuous self-mappings defined an $\alpha _{s}$-complete tripled b-metric space ($X, d_{b},s$) such that $f(X)\subseteq S_{1} (X)$, $g(X)\subseteq S_{2} (X)$, and $h(X)\subseteq S_{3} (X)$. Suppose that, for all $(x,y,z)\in \lambda _{f,g,h, \alpha _{s}}$, there exist $F\in \mathcal{F}_{s}$ and $r>0$ such that

$$ r+F \bigl( s d_{b} \bigl( f ( x ), g ( y ), h ( z )\bigr)\bigr) \leq F \bigl( M ( x, y, z )\bigr). $$

(2.3)

Assume that the pairs ($f, S_{1}$), ($g, S_{2}$), and ($h, S_{3}$) are $\alpha _{s}$-compatible and the triples ($f, g, h$), $( g, f, h )$, and ($h, g, f$) are triangular partially weakly $\alpha _{s}$-admissible with respect to $S_{1}$, $S_{2}$, and $S_{3}$, respectively. Then the pairs ($f, S_{1}$), $( g, S_{2} )$, and ($h, S_{3}$) have the coincidence fixed point say v in X. Moreover, if $\alpha _{s} ( S_{1} ( v ), S_{2} ( v ), S_{3} ( v )) \geq s^{2}$, then v is a common fixed point of f, g, h, $S_{1}$, $S_{2}$, $S_{3}$.

Proof

Let $x_{0} \in X$ be an arbitrary point. As $f ( X ) \subseteq S_{1} ( X )$, there exists $x_{1} \in X$ such that $f ( x_{0} )= S_{1} ( x_{1} )$. Since $g ( x_{1} ) \in S_{2} ( X )$, we can choose $x_{2} \in X$ such that $g ( x_{1} )= S_{2} ( x_{2} )$. Since $h ( x_{2} ) \in S_{3} ( X )$, there exists $x_{3} \in X$ such that $h ( x_{2} )= S_{3} ( x_{3} )$. In general, $x_{2 n}$, $x_{2 n+ 1}$, and $x_{2 n+ 2}$ are chosen in X such that $f ( x_{2 n} )= S_{1} ( x_{2 n+ 1} )$, $g ( x_{2 n+ 1} )= S_{2} ( x_{2 n+ 2} )$, and $h ( x_{2 n+ 2} )= S_{3} ( x_{2 n+ 3} )$. Define a sequence $\{ J_{n} \} \in X$ such that, for all $n\in \mathbb{N}$, $J_{2 n+ 1} = f ( x_{2 n} )= S_{1} ( x_{2 n+ 1} )$, $J_{2 n+ 2} = g ( x_{2 n+ 1} )= S_{2} ( x_{2 n+ 2} )$, and $J_{2 n+ 2} = h ( x_{2 n+ 2} )= S_{3} ( x_{2 n+ 3} )$. As $x_{1} \in S_{1}^{- 1} ( f ( x_{0} ))$, $x_{2} \in S_{2}^{- 1} ( g ( x_{1} ))$, $x_{3} \in S_{3}^{- 1} ( h ( x_{2} ))$, and ($f, g, h$), ($h, g, f$), and ($g, f, h$) are triangular partially weakly $\alpha _{s}$-admissible triples of mappings with respect to $S_{1}$, $S_{2}$, and $S_{3}$, respectively, we have

$$\begin{aligned}& \alpha _{s} \bigl( f ( x_{0} ), g ( x_{1} ), h ( x_{2} ) \bigr) = \alpha _{s} \bigl( S_{1} ( x_{1} ), S_{2} ( x_{2} ), S_{3} ( x_{3} ) \bigr) \geq s^{2}, \\& \alpha _{s} \bigl( h ( x_{2} ), g ( x_{1} ), f ( x_{0} ) \bigr) = \alpha _{s} \bigl( S_{3} ( x_{3} ), S_{2} ( x_{2} ), S_{1} ( x_{1} ) \bigr) \geq s^{2}, \end{aligned}$$

and

$$ \alpha _{s} \bigl( g ( x_{1} ), f ( x_{0} ), h ( x_{2} ) \bigr) = \alpha _{s} \bigl( S_{2} ( x_{2} ), S_{1} ( x_{1} ), S_{3} ( x_{3} ) \bigr) \geq s^{2}. $$

Continuing this way, we obtain

$$\begin{aligned}& \alpha _{s} \bigl( S_{1} ( x_{2 n+ 1} ), S_{2} ( x_{2 n+ 2} ), S_{3} ( x_{2 n+ 3} ) \bigr) \geq s^{2}, \\& \alpha _{s} \bigl( S_{3} ( x_{2 n+ 3} ), S_{2} ( x_{2 n+ 2} ), S_{1} ( x_{2 n+ 1} ) \bigr) \geq s^{2}, \end{aligned}$$

and $\alpha _{s} ( S_{2} ( x_{2 n+ 2} ), S_{1} ( x_{2 n+ 1} ), S_{3} ( x_{2 n+ 3} ) ) \geq s^{2}$. Thus, we have

$$\begin{aligned}& \alpha _{s} ( J_{2 n+ 1}, J_{2 n+ 2}, J_{2 n+ 3} ) \geq s^{2}, \\& \alpha _{s} ( J_{2 n+ 3}, J_{2 n+ 2}, J_{2 n+ 1} ) \geq s^{2}, \end{aligned}$$

and $\alpha _{s} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n+ 3} ) \geq s^{2}$ for all $n\in \mathbb{N}$. At present, we prove that

$$ \lim_{l\rightarrow \infty } d_{b} ( J_{l}, J_{l+ 1}, J_{l+ 2} ) =0. $$

Set $d_{l} = d_{b} ( J_{l}, J_{l+ 1}, J_{l+ 2} )$. Suppose that $d_{l_{0}} =0$ for some $l_{0}$. Then $J_{l_{0}} = J_{l_{0} + 1}$. If $l_{0} =2 n$, then $J_{2 n} = J_{2 n+ 1}$ gives $J_{2 n+ 1} = J_{2 n+ 2}$. Indeed, by contractive condition (2.3), we get

$$\begin{aligned} F \bigl( s d_{b} ( J_{2 n+ 1}, J_{2 n+ 2}, J_{2 n+ 3} ) \bigr) &= F \bigl( s d_{b} \bigl( f ( x_{2 n} ), g ( x_{2 n+ 1} ), h ( x_{2 n+ 2} ) \bigr) \bigr) \\ &\leq F \bigl( M ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) \bigr) -r \end{aligned}$$

for all $n\in \mathbb{N} \cup \{0\}$, where

$$\begin{aligned} M ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) &=\max \biggl\{ d_{b} \bigl( S_{1} ( x_{2 n} ), S_{2} ( x_{2 n+ 1} ), S_{3} ( x_{2 n+ 2} ) \bigr), \\ &\quad d_{b} \bigl( f ( x_{2 n} ), S_{2} ( x_{2 n} ), S_{3} ( x_{2 n} ) \bigr), \\ &\quad d_{b} \bigl( g ( x_{2 n+ 1} ), S_{1} ( x_{2 n+ 1} ), S_{3} ( x_{2 n+ 1} ) \bigr), \\ &\quad d_{b} \bigl( h ( x_{2 n+ 2} ), S_{1} ( x_{2 n+ 2} ), S_{2} ( x_{2 n+ 2} ) \bigr), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} \bigl( S_{1} ( x_{2 n} ), g ( x_{2 n+ 1} ), h ( x_{2 n+ 2} ) \bigr) \\ &\quad {}+ d_{b} \bigl( f ( x_{2 n} ), S_{2} ( x_{2 n+ 1} ), h ( x_{2 n+ 2} ) \bigr) \\ &\quad {}+ d_{b} \bigl( S_{3} ( x_{2 n+ 2} ), g ( x_{2 n+ 1} ), f ( x_{2 n} ) \bigr) \bigr]\biggr\} \\ &=\max \biggl\{ d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ), d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ), \\ &\quad d_{b} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n} ), d_{b} ( J_{2 n+ 2}, J_{2 n+ 2}, J_{2 n+ 2} ), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} ( J_{2 n}, J_{2 n+ 2}, J_{2 n+ 2} ) + d_{b} ( J_{2 n+ 1}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ &\quad {}+ d_{b} ( J_{2 n+ 1}, J_{2 n+ 2}, J_{2 n+ 1} ) \bigr]\biggr\} . \end{aligned}$$

So

$$ \begin{aligned} M ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) & = \max \biggl\{ d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 1} ), d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ), \\ &\quad d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} ( J_{2 n}, J_{2 n+ 2}, J_{2 n+ 2} ) + d_{b} ( J_{2 n+ 1}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ &\quad{}+ d_{b} ( J_{2 n+ 1}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr]\biggr\} \\ & \leq \max \bigl\{ d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ), d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ), \\ &\quad d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ), d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr\} \\ & =\max \bigl\{ d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ), d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr\} . \end{aligned} $$

Since $d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) =0$, therefore $M ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) = d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} )$. Then

$$ F \bigl( s d_{b} ( J_{2 n+ 1}, J_{2 n+ 2}, J_{2 n+ 3} ) \bigr) = F \bigl( d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr) -r. $$

By ($F_{1}$), we have

$$ s d_{b} ( J_{2 n+ 1}, J_{2 n+ 2}, J_{2 n+ 3} ) \leq d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) -r. $$

Let $l =2 n$, then we have $s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \leq d_{b} ( J_{2 n- 2}, J_{2 n- 1}, J_{2 n} ) -r$. Thus, for all n,

$$ d_{b} ( J_{n}, J_{n+ 1}, J_{n+ 2} ) \leq \frac{1}{s} d_{b} ( J_{n- 1}, J_{n}, J_{n+ 1} ). $$

That is, a sequence $\{ d_{b} ( J_{n}, J_{n+ 1}, J_{n+ 2} ) \}$ is nonincreasing and $d_{b} ( J_{n}, J_{n+ 1}, J_{n+ 2} ) \rightarrow 0$ as $n\rightarrow \infty$. Hence $\lim_{l\rightarrow \infty } d_{b} ( J_{l}, J_{l+ 1}, J_{l+ 2} ) =0$ holds true. Now, suppose that $d_{l} = d_{b} ( J_{l}, J_{l+ 1}, J_{l+ 2} ) >0$ for each $l\in \mathbb{N}$. We claim that $\lim_{n\rightarrow \infty } d_{b} ( J_{n}, J_{n+ 1}, J_{n+ 2} ) = -\infty$. Let $l =2 n$. As

$$\alpha _{s} \bigl( S_{1} ( x_{2 n} ), S_{2} ( x_{2 n+ 1} ), S_{3} ( x_{2 n+ 2} ) \bigr) \geq s^{2}, $$

$d_{b} ( f ( x_{2 n} ), g ( x_{2 n} ), h ( x_{2 n+ 1} ) ) >0$, so $( x_{2 n- 1}, x_{2 n}, x_{2 n+ 1} ) \in \lambda _{f, g, h, \alpha _{s}}$, by (2.3), we obtain

$$ F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) \leq F \bigl( d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr) -r $$

(2.4)

for all $n\in \mathbb{N}$. Similarly, for $\mathcal{l} =2 n- 1$,

$$ F \bigl( s d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr) \leq F \bigl( d_{b} ( J_{2 n- 2}, J_{2 n- 1}, J_{2 n} ) \bigr) -r $$

(2.5)

for all $n\in \mathbb{N}$. Hence, by (2.4) and (2.5), we have

$$ F \bigl( s d_{b} ( J_{n}, J_{n+ 1}, J_{n+ 2} ) \bigr) \leq F \bigl( d_{b} ( J_{n- 1}, J_{n}, J_{n+ 1} ) \bigr) -r $$

(2.6)

for all $n\in \mathbb{N}$. Let $a_{n} = d_{b} ( J_{n}, J_{n+ 1}, J_{n+ 2} )$ for each $n\in \mathbb{N}$. By (2.6) and property ($F_{4}$), we have $r+F ( s^{n} a_{n} ) \leq F ( s^{n- 1} a_{n- 1} )$ for all $n\in \mathbb{N}$. Continuing this process, we obtain

$$ F \bigl( s^{n} a_{n} \bigr) \leq F ( a_{n} ) -nr $$

(2.7)

for all $n\in \mathbb{N}$. On taking limit $n\rightarrow \infty$ in (2.7), we have $\lim_{n\rightarrow \infty } F ( s^{n} a_{n} )= -\infty$. By property ($F_{2}$), we get $\lim_{n\rightarrow \infty } s^{n} a_{n} =0$ and ($F_{2}$) implies that there exists $k\in (0,1)$ such that $\lim_{n\rightarrow \infty } ( s^{n} a_{n} )^{k} F ( s^{n} a_{n} )=0$. By (2.7), for all $n\in \mathbb{N}$, we obtain

$$ \bigl( s^{n} a_{n} \bigr)^{k} F \bigl( s^{n} a_{n} \bigr) - \bigl( s^{n} a_{n} \bigr)^{k} F ( a_{0} ) \leq - \bigl( s^{n} a_{n} \bigr)^{k} nr\leq 0. $$

(2.8)

On taking limit $n\rightarrow \infty$ in (2.8), we have $\lim_{n\rightarrow \infty } n ( s^{n} a_{n} )^{k} =0$. This implies there exists $n_{1} \in \mathbb{N}$ such that $n ( s^{n} a_{n} )^{k} \leq 1$ for all $n\geq n_{1}$, or $s^{n} a_{n} \leq \frac{1}{n^{\frac{1}{k}}}$ for all $n\geq n_{1}$. To prove $\{ J_{n} \}$ is a Cauchy sequence, by the triangular inequality, we have

$$\begin{aligned} d_{b} ( x_{n}, x_{m}, x_{m} ) &\leq s \bigl[ d_{b} ( x_{n}, x_{n+ 1}, x_{n+ 2} ) + d_{b} ( x_{m}, x_{m}, x_{m} ), d_{b} ( x_{m}, x_{m+ 2}, x_{m} ) \bigr] \\ &= s d_{b} ( x_{n}, x_{n+ 1}, x_{n+ 2} ) +s d_{b} ( x_{n+ 2} ), x_{m}, x_{m} ) \\ &\leq s d_{b} ( x_{n}, x_{n+ 1}, x_{n+ 2} ) + s^{2} \bigl[ d_{b} ( x_{n+ 2}, x_{n+ 3}, x_{n+ 4} ) \\ &\quad {} + d_{b} ( x_{m}, x_{m}, x_{m} ) + d_{b} ( x_{m}, x_{n+ 3}, x_{n+ 1} ) \bigr] \\ &= s d_{b} ( x_{n}, x_{n+ 1}, x_{n+ 2} ) + s^{2} d_{b} ( x_{n+ 2}, x_{n+ 3}, x_{n+ 4} ) + s^{2} d_{b} ( x_{n+ 3}, x_{m}, x_{m} ) \\ &\leq s d_{b} ( x_{n}, x_{n+ 1}, x_{n+ 2} ) + s^{2} d_{b} ( x_{n+ 2}, x_{n+ 3}, x_{n+ 4} ) \\ &\quad{}+ s^{3} d_{b} ( x_{n+ 3}, x_{n+ 4}, x_{n+ 5} ) + s^{3} d_{b} ( x_{n+ 4}, x_{m}, x_{m} ). \end{aligned}$$

Take $m = n+p$, ($n, p\in \mathbb{N}$), then we have

$$\begin{aligned} d_{b} ( x_{n}, x_{m}, x_{m} ) &\leq s d_{b} ( x_{n}, x_{n+ 1}, x_{n+ 2} ) + s^{2} d_{b} ( x_{n+ 2}, x_{n+ 3}, x_{n+ 4} ) \\ &\quad {}+ s^{3} d_{b} ( x_{n+ 3}, x_{n+ 4}, x_{n+ 5} ) +\cdots + s^{n- 1} d_{b} ( x_{n+p- 1}, x_{n+p}, x_{n+p} ) \\ &\leq \frac{s}{s^{n} n^{\frac{1}{k}}} + \frac{s^{2}}{s^{n+ 2} ( n+ 2 )^{\frac{1}{k}}} + \frac{s^{3}}{s^{n+ 3} ( n+ 3 )^{\frac{1}{k}}} \\ &\quad {}+\cdots + \frac{s^{p- 1}}{s^{n+P- 1} ( n+p- 1 )^{\frac{1}{k}}} \\ &= \frac{s^{1 -n}}{n^{\frac{1}{k}}} + \frac{s^{-n}}{( n+ 2 )^{\frac{1}{k}}} + \frac{s^{-n}}{( n+ 3 )^{\frac{1}{k}}} +\cdots + \frac{s^{-n}}{( n+p- 1 )^{\frac{1}{k}}} \\ &= \frac{s^{1 -n}}{n^{\frac{1}{k}}} + s^{-n} \sum_{i =2}^{p- 1} \frac{1}{( n+i )^{\frac{1}{k}}}. \end{aligned}$$

Since $\sum_{i =2}^{p- 1} \frac{1}{( n+i )^{\frac{1}{k}}}$ is convergent and $s^{-n} \rightarrow 0$ as $n\rightarrow \infty$, thus we conclude that

$$ \lim_{n, m\rightarrow \infty } d_{b} ( x_{n}, x_{m}, x_{m} ) =0. $$

This implies that $\{ J_{n} \}$ is a Cauchy sequence in the $\alpha _{s}$-complete tripled b-metric space X and

$$ \alpha _{s} ( J_{n}, J_{n+ 1}, J_{n+ 2} ) \geq s^{2}, $$

there exists $v\in X$ such that

$$ \lim_{n\rightarrow \infty } d_{b} ( J_{2 n+ 1}, v, v ) = \lim _{n\rightarrow \infty } d_{b} ( f x_{2 n}, v, v ) = \lim _{n\rightarrow \infty } d_{b} \bigl( S_{1} ( x_{2 n+ 1} ), v, v \bigr) =0. $$

Consequently, $f ( x_{2 n} ) \rightarrow v$ and $S_{1} ( x_{2 n+ 1} ) \rightarrow v$ as $n\rightarrow \infty$. So

$$ \lim_{n\rightarrow \infty } d_{b} ( J_{2 n+ 1}, v, v ) = \lim _{n\rightarrow \infty } d_{b} ( g x_{2 n}, v, v ) = \lim _{n\rightarrow \infty } d_{b} \bigl( S_{2} ( x_{2 n+ 1} ), v, v \bigr) =0. $$

Thus $g ( x_{2 n} ) \rightarrow v$ and $S_{2} ( x_{2 n+ 1} ) \rightarrow v$ as $n\rightarrow \infty$. Again, we have

$$ \lim_{n\rightarrow \infty } d_{b} ( J_{2 n}, v, v ) = \lim _{n\rightarrow \infty } d_{b} ( h x_{2 n}, v, v ) = \lim _{n\rightarrow \infty } d_{b} \bigl( S_{3} ( x_{2 n+ 1} ), v, v \bigr) =0. $$

Hence $h ( x_{2 n} ) \rightarrow v$ and $S_{3} ( x_{2 n+ 1} ) \rightarrow v$ as $n\rightarrow \infty$. Now, since ($f, S_{1}$) is an $\alpha _{s}$-compatible pair and

$$ \alpha _{s} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \geq s^{2}. $$

Therefore, we have $\lim_{n\rightarrow \infty } d_{b} ( f S_{1} ( x_{2 n} ), S_{1} f ( x_{2 n} ), x_{2 n} ) =0$ and ($g, S_{2}$) is an $\alpha _{s}$-compatible pair and

$$ \alpha _{s} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \geq s^{2}. $$

We have $\lim_{n\rightarrow \infty } d_{b} ( g S_{2} ( x_{2 n} ), S_{2} g ( x_{2 n} ), x_{2 n} ) =0$ and ($h, S_{3}$) is an $\alpha _{s}$-compatible pair, we get

$$ \lim_{n\rightarrow \infty } d_{b} \bigl( h S_{3} ( x_{2 n} ), S_{3} h ( x_{2 n} ), x_{2 n} \bigr) =0. $$

Since $\lim_{n\rightarrow \infty } d_{b} ( f ( x_{2 n} ), v, v ) =0$, $\lim_{n\rightarrow \infty } d_{b} ( S_{1} ( x_{2 n} ), v, v ) =0$, and f, $S_{1}$ is $\alpha _{s}$-continuous. Thus $\lim_{n\rightarrow \infty } d_{b} ( S_{1} f ( x_{2 n} ), S_{1} v, S_{1} v ) =0$, $\lim_{n\rightarrow \infty } d_{b} ( f S_{1} ( x_{2 n} ), fv, fv ) =0$, and

$$ \lim_{n\rightarrow \infty } d_{b} \bigl( g ( x_{2 n} ), v, v \bigr) =0, $$

so g, $S_{2}$ is $\alpha _{s}$-continuous, we have $\lim_{n\rightarrow \infty } d_{b} ( S_{2} g ( x_{2 n} ), S_{2} v, S_{2} v ) =0$ and

$$ \lim_{n\rightarrow \infty } d_{b} \bigl( g S_{2} ( x_{2 n} ), gv, gv \bigr) =0. $$

Again in this way, $\lim_{n\rightarrow \infty } d_{b} ( S_{3} h ( x_{2 n} ), S_{3} v, S_{3} v ) =0$ and $\lim_{n\rightarrow \infty } d_{b} ( h S_{3} g ( x_{2 n} ), hv, hv ) =0$. By the triangle inequality, we have

$$ \begin{aligned}[b] d_{b} \bigl( fv, S_{1} v, S_{1} ( x_{2 n} ) \bigr) & \leq s \bigl[ d_{b} \bigl( fv, fv, f S_{1} ( x_{2 n} ) \bigr) + d_{b} \bigl( S_{1} v, S_{1} f ( x_{2 n} ), S_{1} v \bigr) \\ &\quad {} + d_{b} \bigl( S_{1} x_{2 n}, f S_{1} x_{2 n}, S_{1} f ( x_{2 n} ) \bigr) \bigr]. \end{aligned} $$

(2.9)

Applying limit as $n\rightarrow \infty$, we obtain $d_{b} ( fv, S_{1} v, v ) \leq 0$, which yields that $fv = S_{1} v = v$. Thus v is a coincidence and common fixed point of f, $S_{1}$. Arguing in a similar manner, we can prove that $gv = S_{2} v = v$ and $hv = S_{1} v = v$. Thus $fv = gv = hv = S_{1} v = S_{2} v = S_{3} v = v$ and v is a common fixed point of f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$. □

Remark 2.19

If we suppose that $\alpha _{s} ( v,w,w ) \geq s^{2}$ for each pair of common fixed points of f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$, then v is unique. Indeed, if w is another fixed point of f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$ and assuming on contrary $d_{b} ( fv,gw,hw ) >0$, then from (2.3) we have

$$ F \bigl( d_{b} ( v, w, w ) \bigr) = F \bigl( s d_{b} \bigl( S_{1} ( v ), S_{2} ( w ), S_{3} ( w ) \bigr) \bigr) \leq F \bigl( M ( v, w, w ) \bigr) -r, $$

(2.10)

where

$$ \begin{aligned} M ( v, w, w ) & =\max \biggl\{ d_{b} \bigl( S_{1} ( v ), S_{2} ( w ), S_{3} ( w ) \bigr), d_{b} \bigl( f ( v ), S_{2} ( v ), S_{3} ( v ) \bigr), \\ &\quad d_{b} \bigl( g ( w ), S_{1} ( w ), S_{3} ( w ) \bigr), d_{b} \bigl( h ( w ), S_{1} ( w ), S_{2} ( w ) \bigr), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} \bigl( S_{1} ( v ), g ( w ), h ( w ) \bigr) \\ &\quad {} + d_{b} \bigl( f ( v ), S_{2} ( w ), h ( w ) \bigr) + d_{b} \bigl( S_{3} ( w ), g ( w ), f ( v ) \bigr) \bigr]\biggr\} \\ & =\max \biggl\{ d_{b} ( v, w, w ), d_{b} ( v, v, v ), d_{b} ( w, w, w ), d_{b} ( w, w, w ), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} ( v, w, w ), d_{b} ( v, w, w ) + d_{b} ( w, w, v ) \bigr]\biggr\} . \end{aligned} $$

Thus, by (2.10), we have $F ( s d_{b} ( v, w, w ) ) \leq F ( d_{b} ( v, w, w ) ) -r < F ( d_{b} ( v, w, w ) )$, which is a contradiction. Hence $v = w$ and v is a unique common fixed point of self-mappings f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$.

The following example elucidates Theorem 2.18.

Example 2.20

Let $X= \mathbb{R}_{0}^{+}$ and $d_{b}:X\times X\times X\rightarrow \mathbb{R}_{0}^{+}$ be defined by

$$ d_{b} ( x, y, z )=\max \bigl\{ \vert x-y \vert ^{2}, \vert x-z \vert ^{2}, \vert y-z \vert ^{2} \bigr\} $$

for all $x, y, z\in X$. Define $\alpha _{s}: X\times X\times X\rightarrow \mathbb{R}_{0}^{+}$ by

$$ \alpha _{s} ( x, y, z )= \textstyle\begin{cases} 4\max \{ e^{x-y}, e^{x-z}, e^{y-z} \} , & x\geq y\geq z,\\ 4\max \{ e^{y-x}, e^{z-x}, e^{z-y} \} , & x\leq y\leq z . \end{cases} $$

So ($S, d_{b}, s$) is an $\alpha _{s}$-complete tripled b-metric with $s =2$. Define the mappings f, g, h, $S_{1}$, $S_{2}$, and $S_{3}: X\rightarrow X$ for all $x\in X$ by

$$ \begin{gathered} f ( x ) =\ln \biggl( 1 + \frac{x}{5} \biggr), \\ g ( x ) =\ln \biggl( 1 + \frac{x}{6} \biggr), \\ h ( x ) =\ln \biggl( 1 + \frac{x}{7} \biggr), \end{gathered} $$

$S_{1} ( x )= e^{6 x} - 1$, $S_{2} ( x )= e^{7 x} - 1$, and $S_{3} ( x )= e^{8 x} - 1$. Clearly, f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$ are $\alpha _{s}$-continuous self-mappings complying with $f ( X )= g ( X )= h ( X )= S_{1} ( X )= S_{2} ( X )= S_{3} ( X )$. We note that the pair ($f, S_{1}$) is $\alpha _{s}$-compatible. Indeed, let $\{ x_{n} \}$ be a sequence in X satisfying $\alpha _{s} ( x_{n}, x_{n+ 1}, x_{n+ 2} ) \geq s^{2}$ and

$$ \lim_{n\rightarrow \infty } f ( x_{n} )= \lim_{n\rightarrow \infty } \ln \biggl( 1 + \frac{x_{n}}{5} \biggr) = \lim_{n\rightarrow \infty } S_{1} ( x_{n} )= t $$

for some $t\in X$. Then $\lim_{n\rightarrow \infty } | f ( x_{n} ) -t | ^{2} = \lim_{n\rightarrow \infty } | S_{1} ( x_{n} ) -t | ^{2} =0$, equivalently

$$ \lim_{n\rightarrow \infty } \biggl\vert \ln \biggl( 1 + \frac{x_{n}}{5} \biggr) -t \biggr\vert ^{2} = \lim_{n\rightarrow \infty } \bigl\vert e^{6 x_{n}} - 1 -t \bigr\vert ^{2} =0 $$

implies

$$ \lim_{n\rightarrow \infty } \bigl\vert x_{n} - \bigl(5 e^{t} - 5\bigr) \bigr\vert ^{2} = \lim _{n\rightarrow \infty } \biggl\vert x_{n} - \frac{\ln ( t+ 1)}{6} \biggr\vert ^{2} =0. $$

Uniqueness of limit gives that $5 e^{t} - 5= \frac{\ln ( t+ 1)}{6}$, thus $t =0$ is only possible solution. Due to $alph a_{s}$-continuity of f and $S_{1}$, for $t =0 \in X$, we have

$$\begin{aligned}& \lim_{n\rightarrow \infty } d_{b} \bigl( f S_{1} ( x_{n} ), S_{1} f ( x_{n} ), f ( x_{n} ) \bigr)\\& \quad =\max \Bigl\{ \lim_{n\rightarrow \infty } \bigl\vert f S_{1} ( x_{n} ) - S_{1} f ( x_{n} ) \bigr\vert ^{2}, \\& \qquad \lim_{n\rightarrow \infty } \bigl\vert S_{1} f ( x_{n} ) -f ( x_{n} ) \bigr\vert ^{2}, \lim_{n\rightarrow \infty } \bigl\vert f S_{1} ( x_{n} ) -f ( x_{n} ) \bigr\vert ^{2} \Bigr\} \\& \quad =\max \bigl\{ \bigl\vert f ( t ) - S_{1} ( t ) \bigr\vert ^{2}, \bigl\vert S_{1} ( t ) -t \bigr\vert ^{2}, \bigl\vert f ( t ) -t \bigr\vert ^{2} \bigr\} \\& \quad =0. \end{aligned}$$

Similarly, the pair ($g, S_{2}$) and ($h, S_{3}$) is $\alpha _{s}$-compatible. To prove that ($f, g, h$) is a partially weakly $\alpha _{s}$-admissible triple of mappings with respect to $S_{!}$, let $x\in X$ and $y\in S_{1}^{- 1} ( g ( f ( x )))$, that is, $S_{1} ( y )= g ( f ( x ))$ and

$$ e^{6 y} - 1= g \biggl( \ln \biggl( 1 + \frac{x}{5} \biggr) \biggr) =\ln \biggl( 1 + \frac{\ln ( 1 + \frac{x}{5} )}{6} \biggr). $$

Thus $y = \frac{1}{6} \ln ( 1 + \ln ( 1 + \frac{\ln ( 1 + \frac{x}{5} )}{6} ) )$. We have

$$ f ( x )=\ln \biggl( 1 + \frac{x}{5} \biggr) \geq g ( y )=\ln \biggl( 1 + \frac{y}{6} \biggr) =\ln \biggl( 1 + \frac{1}{36} \ln \biggl( 1 + \ln \biggl( 1 + \frac{\ln ( 1 + \frac{x}{5} )}{6} \biggr) \biggr) \biggr). $$

We have $z\in S_{1}^{- 1} ( hg ( f ( x )) )$, that is, $S_{1} ( z )= hg ( f ( x ))$, $S_{1} ( z )= h ( S_{1} ( y ))$, $e^{z} - 1=\ln ( 1 + \frac{S_{1} ( y )}{7} )$,

$$ e^{6 z} - 1=\ln \biggl( 1 + \frac{1}{7} \ln \biggl( 1 + \frac{\ln ( 1 + \frac{x}{5} )}{6} \biggr) \biggr), $$

and

$$ z = \frac{1}{6} \ln \biggl( 1 + \ln \biggl( 1 + \frac{1}{7} \ln \biggl( \frac{\ln ( 1 + \frac{x}{5} )}{5} \biggr) \biggr) \biggr). $$

We conclude that

$$\begin{aligned} g ( y )&=\ln \biggl( 1 + \frac{y}{6} \biggr) =\ln \biggl( 1 + \frac{1}{42} \ln \biggl( 1 + \ln \biggl( 1 + \frac{\ln ( 1 + \frac{x}{5} )}{6} \biggr) \biggr) \biggr) \\ &\geq h ( z )=\ln \biggl( 1 + \frac{z}{7} \biggr) \\ &=\ln \biggl( 1 + \frac{1}{42} \ln \biggl( 1 + \ln \biggl( 1 + \frac{1}{7} \ln \biggl( 1 + \frac{\ln ( 1 + \frac{x}{5} )}{6} \biggr) \biggr) \biggr) \biggr). \end{aligned}$$

Thus $\alpha _{s} ( f ( x ), g ( y ), h ( z ) ) =4\max \{ e^{x-y}, e^{x-z}, e^{y-z} \} \geq s^{2}$. In this process, we can prove that ($g, f, h$) is a partially weakly $\alpha _{s}$-admissible triple of mappings with respect to $S_{2}$ and ($h, g, f$) is a partially weakly $\alpha _{s}$-admissible triple of mappings with respect $S_{1}$. Now, for each $x, y, z\in X$, consider

$$\begin{aligned}& d_{b} \bigl( f ( x ), g ( y ), h ( z ) \bigr) =\max \bigl\{ \bigl\vert f ( x ) -g ( y ) \bigr\vert ^{2}, \bigl\vert g ( y ) -h ( z ) \bigr\vert ^{2}, \bigl\vert f ( x ) -h ( z ) \bigr\vert ^{2} \bigr\} , \\& \begin{aligned} \bigl\vert f ( x ) -g ( y ) \bigr\vert ^{2} &= \biggl\vert \ln \biggl( 1 + \frac{x}{5} \biggr) - \ln \biggl( 1 + \frac{y}{6} \biggr) \biggr\vert ^{2} \\ &\leq \biggl( \frac{x}{5} - \frac{y}{6} \biggr)^{2} \\ &= \frac{1}{900} ( 6 x- 5 y )^{2} \\ &\leq \frac{1}{900} \bigl( e^{6 x} - e^{5 y} \bigr)^{2}, \end{aligned} \\& \begin{aligned} \bigl\vert g ( y ) -h ( z ) \bigr\vert ^{2} &= \biggl\vert \ln \biggl( 1 + \frac{y}{6} \biggr) - \ln \biggl( 1 + \frac{z}{7} \biggr) \biggr\vert ^{2} \\ &\leq \biggl( \frac{y}{6} - \frac{z}{7} \biggr)^{2} \\ &= \frac{1}{1764} ( 7 y- 6 z )^{2} \\ &\leq \frac{1}{1764} \bigl( e^{7 y} - e^{6 z} \bigr)^{2}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \bigl\vert f ( x ) -h ( z ) \bigr\vert ^{2} &= \biggl\vert \ln \biggl( 1 + \frac{x}{5} \biggr) - \ln \biggl( 1 + \frac{z}{7} \biggr) \biggr\vert ^{2} \\ &\leq \biggl( \frac{x}{5} - \frac{z}{7} \biggr)^{2} \\ &= \frac{1}{1225} ( 7 x- 5 z )^{2} \\ &\leq \frac{1}{1225} \bigl( e^{7 x} - e^{5 z} \bigr)^{2}. \end{aligned}$$

Thus

$$\begin{aligned} d_{b} \bigl( f ( x ), g ( y ), h ( z ) \bigr) &\leq \frac{1}{900} \max \bigl\{ \bigl( e^{6 x} - e^{5 y} \bigr)^{2}, \bigl( e^{7 y} - e^{6 z} \bigr)^{2}, \bigl( e^{7 x} - e^{5 z} \bigr)^{2} \bigr\} \\ &= \frac{1}{900} d_{b} \bigl( S_{1} ( x ), S_{2} ( y ), S_{3} ( z ) \bigr). \end{aligned}$$

Define the function $F: \mathbb{R}^{+} \rightarrow \mathbb{R}$ by $F ( x )=\ln x$ for all $x\in \mathbb{R}^{+}$. Hence, for all $x, y, z\in X$ such that $d_{b} ( f ( x ), g ( y ), h ( z ) ) >0$, $r =\ln (900)$, we obtain

$$ r+F \bigl( d_{b} \bigl( f ( x ), g ( y ), h ( z ) \bigr) \bigr) \leq F \bigl( M ( x, y, z ) \bigr). $$

Thus the contractive condition (2.3) is satisfied for all $x, y, z\in X$. Hence, all the hypotheses of Theorem 2.18 are satisfied. Note that f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$ have a unique common fixed point $x =0$.

We have obtained some results from Theorem 2.18, which we express in order.

Corollary 2.21

Let X be a nonempty set and $\alpha _{s}:X\times X\times X\rightarrow \mathbb{R}_{0}^{+}$ be a function. Let ($X, d_{b},s$) be an $\alpha _{s}$-complete tripled metric space and f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$ be $\alpha _{s}$-continuous self-mappings on ($X, d_{b},s$) such that for all $(x,y,z)\in \lambda _{f,g,h, \alpha _{s}}$ the inequality

$$ s d_{b} \bigl( f ( x ), g ( y ), h ( z ) \bigr) \leq kM ( x, y, z ) $$

(2.11)

holds. Assume that the pairs ($f, S_{1}$), $( g, S_{2} )$, and ($h, S_{3}$) are $\alpha _{s}$-compatible and the triples of mappings ($f, g, h$), ($g, f, h$), and ($h, g, f$) are triangular partially weakly $\alpha _{s}$-admissible with respect to $S_{1}$, $S_{2}$, and $S_{3}$, respectively. Then the pairs ($f, S_{1}$), ($g, S_{2}$), and ($h, S_{3}$) have the coincidence point v in X. Moreover, if $\alpha _{s} ( S_{1} v, S_{2} v, S_{3} v ) \geq s^{2}$, then v is a common fixed point of f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$.

Proof

For all $( x, y, z ) \in \lambda _{f, g, h, \alpha _{s}}$, we have $s d_{b} ( f ( x ), g ( y ), h ( z ) ) \leq kM ( x, y, z )$. It follows that $r+ \ln ( d_{b} ( f ( x ), g ( y ), h ( z ) ) ) \leq \ln ( M ( x, y, z ) )$, where $r =\ln ( \frac{s}{k} )>0$. Then the contraction condition (2.11) reduces to (2.3) with $F ( x )=\ln x$, and the application of Theorem 2.18 ensures the existence of a fixed point. □

If we set $S = S_{1} = S_{2} = S_{3}$ in Theorem 2.18, we obtain the following corollaries.

Corollary 2.22

Let $f,g,h$, and S be self-mappings defined on an $\alpha _{s}$-complete tripled metric space ($X, d_{b},s$) such that $f(X)\cup g(X)\cup h(X)\subseteq S(X)$ with $\alpha _{s}$-continuous. Suppose that, for all $x,y,z\in X$ with $\alpha _{s} (Tx,Ty,Tz)\geq s^{2}$, there exist $F\in \mathcal{F}_{s}$ and $r>0$ such that $d_{b} ( f(x),g(y),h(z) ) >0$, then

$$ r+F \bigl( s d_{b} \bigl( f ( x ), g ( y ), h ( z ) \bigr) \bigr) \leq F \bigl( M ( x, y, z ) \bigr), $$

where

$$\begin{aligned} M ( x, y, z ) &=\max \biggl\{ d_{b} \bigl( S ( x ), S ( y ), S ( z ) \bigr), d_{b} \bigl( f ( x ), S ( x ), S ( x ) \bigr), \\ &\quad d_{b} \bigl( g ( y ), S ( y ), S ( y ) \bigr), d_{b} \bigl( h ( z ), S ( z ), S ( z ) \bigr), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} \bigl( S ( x ), g ( y ), h ( z ) \bigr) + d_{b} \bigl( f ( x ), S ( y ), h ( z ) \bigr) \\ &\quad {}+ d_{b} \bigl( S ( z ), g ( y ), f ( x ) \bigr) \bigr]\biggr\} . \end{aligned}$$

Assume that either the pair ($f, S$) is $\alpha _{s}$-compatible and f is $\alpha _{s}$-continuous or ($g, S$) is $\alpha _{s}$-compatible and g is $\alpha _{s}$-continuous, or ($h, S$) is $\alpha _{s}$-compatible and h is $\alpha _{s}$-continuous. Then the pairs ($f, S$), $( g, S )$, and ($h, S$) have the coincidence point v in X provided that the triple of mappings ($f, g, h$) is triangular weakly $\alpha _{s}$-admissible with respect to S. Moreover, if $\alpha _{s} ( Sv, Sv, Sv ) \geq s^{2}$, then v is a common fixed point of f, g, h, and S.

If we set $S_{1} = S_{2} = S_{3}$ and $f = g = h$ in Theorem 2.18, we obtain the following corollary.

Corollary 2.23

Let f and S be $\alpha _{s}$-continuous self-mappings defined on an $\alpha _{s}$-complete tripled metric space ($X, d_{b},s$) such that $f(X)\subseteq S(X)$. Suppose that, for all $x,y,z\in X$ with $\alpha _{s} (Sx,Sy,Sz)\geq s^{2}$, there exist $F\in \mathcal{F}_{s}$ and $r>0$ such that $d_{b} ( f(x),f(y),f(z) ) >0$, then

$$ r+F \bigl( s d_{b} \bigl( f ( x ), f ( y ), f ( z ) \bigr) \bigr) \leq F \bigl( M ( x, y, z ) \bigr), $$

where

$$\begin{aligned} M ( x, y, z ) &=\max \biggl\{ d_{b} \bigl( S ( x ), S ( y ), S ( z ) \bigr), d_{b} \bigl( f ( x ), S ( x ), S ( x ) \bigr), \\ &\quad d_{b} \bigl( f ( y ), S ( y ), S ( y ) \bigr), d_{b} \bigl( f ( z ), S ( z ), S ( z ) \bigr), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} \bigl( S ( x ), f ( y ), f ( z ) \bigr) + d_{b} \bigl( f ( x ), S ( y ), f ( z ) \bigr) \\ &\quad {}+ d_{b} \bigl( S ( z ), f ( y ), f ( x ) \bigr) \bigr]\biggr\} . \end{aligned}$$

Assume that the pair ($f, S$) is $\alpha _{s}$-compatible. Then the mappings f and S have the coincidence fixed point in X provided that fg is a triangular weakly $\alpha _{s}$-admissible mapping with respect to S. Moreover, if $\alpha _{s} ( Sv, Sv, Sv ) \geq s^{2}$, then f, S has a common point v.

Corollary 2.24

Let $f,g,h$, and S be self-mappings defined on an $\alpha _{s}$-regular and $\alpha _{s}$-complete tripled metric space ($X, d_{b},s$) such that $f(X),g(X),h(X)\subseteq S(X)$, and $S(X)$ is a closed subset of X. Suppose that, for all $x,y,z\in X$ with $\alpha _{s} (Sx,Sy,Sz)\geq s^{2}$, there exist $F\in \mathcal{F}_{s}$, and $r>0$ such that $d_{b} ( f(x),g(y),h(z) ) >0$, then $r+F ( s d_{b} ( f(x),g(y),h(z) ) ) \leq F ( M(x,y,z) )$, where

$$\begin{aligned} M ( x, y, z ) &=\max \biggl\{ d_{b} \bigl( S ( x ), S ( y ), S ( z ) \bigr), d_{b} \bigl( f ( x ), S ( x ), S ( x ) \bigr), \\ &\quad d_{b} \bigl( g ( y ), S ( y ), S ( y ) \bigr), d_{b} \bigl( h ( z ), S ( z ), S ( z ) \bigr), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} \bigl( S ( x ), g ( y ), h ( z ) \bigr) + d_{b} \bigl( f ( x ), S ( y ), h ( z ) \bigr) \\ &\quad {}+ d_{b} \bigl( S ( z ), g ( y ), f ( x ) \bigr) \bigr]\biggr\} . \end{aligned}$$

Assume that the pairs ($f, S$), ($g, S$), and ($h, S$) are weakly compatible and the triple of mappings ($f, g, h$) is triangular weakly $\alpha _{s}$-admissible with respect to S. Then the pairs ($f, S$), ($g, S$), and ($h, S$) have the coincidence point v in X. Moreover, if $\alpha _{s} ( Sv, Sv, Sv ) \geq s^{2}$, then v is a coincidence point of f, g, h, and S.

Corollary 2.25

Let f and S be self-mappings defined on an $\alpha _{s}$-regular and $\alpha _{s}$-complete tripled metric space ($X, d_{b},s$) such that $f(X)\subseteq S(X)$, and $S(X)$ is a closed subset of X. Suppose that, for all $x,y,z\in X$ with $\alpha _{s} (Sx,Sy,Sz)\geq s^{2}$, there exist $F\in \mathcal{F}_{s}$ and $r>0$ such that $d_{b} ( f(x),f(y),f(z) ) >0$, then $r+F ( s d_{b} ( f(x),f(y),f(z) ) ) \leq F ( M(x,y,z) )$, where

$$\begin{aligned} M ( x, y, z ) &=\max \biggl\{ d_{b} \bigl( S ( x ), S ( y ), S ( z ) \bigr), d_{b} \bigl( f ( x ), S ( x ), S ( x ) \bigr), \\ &\quad d_{b} \bigl( f ( y ), S ( y ), S ( y ) \bigr), d_{b} \bigl( f ( z ), S ( z ), S ( z ) \bigr), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} \bigl( S ( x ), f ( y ), f ( z ) \bigr) + d_{b} \bigl( f ( x ), S ( y ), f ( z ) \bigr) \\ &\quad {}+ d_{b} \bigl( S ( z ), f ( y ), f ( x ) \bigr) \bigr]\biggr\} . \end{aligned}$$

Assume that the pair ($f, S$) is weakly compatible and f is a triangular weakly $\alpha _{s}$-admissible mapping with respect to S. Then the pair ($f, S$) has the coincidence point v in X.

Corollary 2.26

Let $f,g$, and h be self-mappings defined on a complete tripled metric space ($X, d_{b},s$). Suppose that, for all $x,y,z\in X$ with $\alpha _{s} (x,y,z)\geq s^{2}$, there exist $F\in \mathcal{F}_{s}$ and $r>0$ such that $d_{b} ( f(x),g(y),h(z) ) >0$, then $r+F ( s d_{b} ( f(x),g(y),h(z) ) ) \leq F ( M(x,y,z) )$, where

$$\begin{aligned} M ( x, y, z ) &=\max \biggl\{ d_{b} ( x, y, z ), d_{b} \bigl( f ( x ), x, x \bigr), \\ &\quad d_{b} \bigl( g ( y ), y, y \bigr), d_{b} \bigl( h ( z ), z, z \bigr), \\ &\quad \frac{1}{3 s} \bigl[ d_{b} \bigl( x, g ( y ), h ( z ) \bigr) + d_{b} \bigl( f ( x ), y, h ( z ) \bigr) \\ &\quad {}+ d_{b} \bigl( z, g ( y ), f ( x ) \bigr) \bigr]\biggr\} . \end{aligned}$$

Assume that the triple of mappings ($f, g, h$) is triangular weakly $\alpha _{s}$-admissible. Then f, g, and h have a common fixed point v in X provided that either f or g or h is $\alpha _{s}$-continuous, or X is $\alpha _{s}$-regular.

Theorem 2.27

Let f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$ be $\alpha _{s}$-continuous self-mappings defined on an $\alpha _{s}$-complete tripled b-metric space ($X, d_{b},s$) such that $f(X)\subseteq S_{1} (X)$, $g(X)\subseteq S_{2} (X)$, and $h(X)\subseteq S_{3} (X)$. Suppose that, for all $( x,y,z ) \in \lambda _{f,g,h, \alpha _{s}}$, there exist $F\in \mathcal{F}_{s}$ and $r>0$ such that

$$ r+F \bigl( s d_{b} \bigl( f ( x ), g ( y ), h ( z )\bigr)\bigr) \leq F \bigl( M_{i} ( x, y, z )\bigr) $$

(2.12)

holds for one of $i =1,2,3,4,5$, where

$$ \begin{aligned} M_{1} ( x, y, z ) & = a_{1} d_{b} \bigl( S_{1} ( x ), S_{2} ( y ), S_{3} ( z ) \bigr) + a_{2} d_{b} \bigl( f ( x ), S_{2} ( x ), S_{3} ( x ) \bigr) \\ &\quad {} + a_{3} d_{b} \bigl( g ( y ), S_{1} ( y ), S_{3} ( y ) \bigr) + a_{4} d_{b} \bigl( h ( z ), S_{1} ( z ), S_{2} ( z ) \bigr) \\ &\quad {} + a_{5} \bigl[ d_{b} \bigl( S_{1} ( x ), g ( y ), h ( z ) \bigr) + d_{b} \bigl( f ( x ), S_{2} ( y ), h ( z ) \bigr) \\ &\quad {} + d_{b} \bigl( S_{3} ( z ), g ( y ), f ( x ) \bigr) \bigr] \end{aligned} $$

with $a_{i} \geq 0$, $i =1,2,3,4,5$, such that $a_{1} + a_{2} + a_{3} + 3 a_{5} = s$,

$$ \begin{aligned} M_{2} ( x, y, z ) & = a_{1} d_{b} \bigl( S_{1} ( x ), S_{2} ( y ), S_{3} ( z ) \bigr) + a_{2} d_{b} \bigl( f ( x ), S_{2} ( x ), S_{3} ( x ) \bigr) \\ &\quad {} + a_{3} d_{b} \bigl( g ( y ), S_{1} ( y ), S_{3} ( y ) \bigr) + a_{4} d_{b} \bigl( h ( z ), S_{1} ( z ), S_{2} ( z ) \bigr) \end{aligned} $$

with $a_{1} + a_{2} + a_{3} = s$,

$$ \begin{aligned} M_{3} ( x, y, z ) & = k \max \bigl\{ d_{b} \bigl( f ( x ), S_{2} ( x ), S_{3} ( x ) \bigr), d_{b} \bigl( g ( y ), S_{1} ( y ), S_{3} ( y ) \bigr), \\ &\quad d_{b} \bigl( h ( z ), S_{1} ( z ), S_{2} ( z ) \bigr) \bigr\} \end{aligned} $$

with $k\in [0,1)$,

$$ \begin{aligned} M_{4} ( x, y, z ) & = a_{1} ( x, y, z ) d_{b} \bigl( S_{1} ( x ), S_{2} ( y ), S_{3} ( z ) \bigr) \\ &\quad {} + a_{2} ( x, y, z ) d_{b} \bigl( f ( x ), S_{2} ( x ), S_{3} ( x ) \bigr) \\ &\quad {} + a_{3} ( x, y, z ) d_{b} \bigl( g ( y ), S_{1} ( y ), S_{3} ( y ) \bigr) \\ &\quad {} + a_{4} d_{b} \bigl( h ( z ), S_{1} ( z ), S_{2} ( z ) \bigr) \\ &\quad{}+ a_{5} ( x, y, z )\bigl[ d_{b} \bigl( S_{1} ( x ), g ( y ), h ( z ) \bigr) \\ &\quad {} + d_{b} \bigl( f ( x ), S_{2} ( y ), h ( z ) \bigr) \\ &\quad {} + d_{b} \bigl( S_{3} ( z ), g ( y ), f ( x ) \bigr) \bigr] \end{aligned} $$

with $a_{i} ( x, y, z )$, $i =1,2,3,4,5$, are nonnegative functions such that

$$ \sup_{x, y, z\in X} \bigl[ a_{1} ( x, y, z ) + a_{2} ( x, y, z ) + a_{3} ( x, y, z ) + 3 a_{5} ( x, y, z ) \bigr] = s. $$

Suppose that the pairs ($f, S_{1}$), ($g, S_{2}$), and ($h, S_{3}$) are $\alpha _{s}$-compatible and the triples of mappings ($f, g, h$), $( g, f, h )$, and ($h, g, f$) are triangular partially triple weakly $\alpha _{s}$-admissible with respect to $S_{1}$, $S_{2}$, and $S_{3}$, respectively. Then the pairs ($f, S_{1}$), $( g, S_{2} )$, and ($h, S_{3}$) have the coincidence point v in X. Moreover, if $\alpha _{s} ( S_{1} ( v ), S_{2} ( v ), S_{3} ( v )) \geq s^{2}$, then v is a common fixed point of f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$.

Proof

In line with the beginning part of Theorem 2.18, for all $( x, y, z ) \in \lambda _{f, g, h, \alpha _{s}}$ for some $F\in \mathcal{F}_{s}$ and $r >0$, from contractive condition (2.12) we get

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) & = F \bigl( s d_{b} \bigl( f ( x_{2 n} ), g ( x_{2 n+ 1} ), h ( x_{2 n+ 2} ) \bigr) \bigr) \\ & \leq F \bigl( M_{1} ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) \bigr) -r \end{aligned} $$

(2.13)

for all $n\in \mathbb{N}$, where

$$ \begin{aligned} M_{1} ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) & = a_{1} d_{b} \bigl( S_{1} ( x_{2 n} ), S_{2} ( x_{2 n+ 1} ), S_{3} ( x_{2 n+ 2} ) \bigr) \\ &\quad {} + a_{2} d_{b} \bigl( f ( x_{2 n} ), S_{2} ( x_{2 n} ), S_{3} ( x_{2 n} ) \bigr) \\ &\quad {} + a_{3} d_{b} \bigl( g ( x_{2 n+ 1} ), S_{1} ( x_{2 n+ 1} ), S_{3} ( x_{2 n+ 1} ) \bigr) \\ &\quad {} + a_{4} d_{b} \bigl( h ( x_{2 n+ 2} ), S_{1} ( x_{2 n+ 2} ), S_{2} ( x_{2 n+ 2} ) \bigr) \\ &\quad {} + a_{5} \bigl[ d_{b} \bigl( S_{1} ( x_{2 n} ), g ( x_{2 n+ 1} ), h ( x_{2 n+ 2} ) \bigr) \\ &\quad {} + d_{b} \bigl( f ( x_{2 n} ), S_{2} ( x_{2 n+ 1} ), h ( x_{2 n+ 2} ) \bigr) \\ &\quad {} + d_{b} \bigl( S_{3} ( x_{2 n+ 2} ), g ( x_{2 n+ 1} ), f ( x_{2 n} ) \bigr) \bigr] \\ & = a_{1} d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 1} ) + a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ) \\ &\quad {} + a_{3} d_{b} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n} ) + a_{4} d_{b} ( J_{2 n+ 2}, J_{2 n+ 2}, J_{2 n+ 2} ) \\ &\quad {} + a_{5} \bigl[ d_{b} ( J_{2 n}, J_{2 n+ 2}, J_{2 n+ 2} ) + d_{b} ( J_{2 n+ 1}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ &\quad {} + d_{b} ( J_{2 n+ 1}, J_{2 n+ 2}, J_{2 n+ 1} ) \bigr] \\ & \leq a_{1} d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 1} ) + a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ) \\ &\quad {} + a_{3} d_{b} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n} ) \\ &\quad {} + a_{5} \bigl[3 d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr] \\ & =( a_{1} + a_{3} + 3 a_{5} ) d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ &\quad {} + a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ). \end{aligned} $$

Now from (2.13) we have

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) & = F \bigl( ( a_{1} + a_{3} + 3 a_{5} ) d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ &\quad {} + a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ) \bigr) -r. \end{aligned} $$

(2.14)

Since F is strictly increasing, (2.14) implies

$$ \begin{aligned} s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) & \leq ( a_{1} + a_{3} + 3 a_{5} ) d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ &\quad {} + a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ). \end{aligned} $$

So

$$ \begin{aligned} ( s- a_{1} - a_{3} - 3 a_{5} ) d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) & \leq a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ). \end{aligned} $$

Hence

$$ \begin{aligned} d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) & \leq \frac{a_{2}}{s- a_{1} - a_{3} - 3 a_{5}} d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ). \end{aligned} $$

Since $a_{1} + a_{2} + a_{3} + 3 a_{5} = s$, therefore $d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \leq d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} )$. Thus from (2.14) we obtain

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) & \leq F \bigl( d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr) -r \end{aligned} $$

(2.15)

for all $n\in \mathbb{N}$. Similarly,

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr) & \leq F \bigl( d_{b} ( J_{2 n- 2}, J_{2 n- 1}, J_{2 n} ) \bigr) -r \end{aligned} $$

(2.16)

for all $n\in \mathbb{N}$. Hence, from (2.15) and (2.16), we have

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{n}, J_{n+ 1}, J_{n+ 2} ) \bigr) & = F \bigl( d_{b} ( J_{n- 1}, J_{n}, J_{n+ 1} ) \bigr) -r. \end{aligned} $$

(2.17)

Inequality (2.17) leads to remark that $\{ x_{n} \}$ is a Cauchy sequence, and the remaining part of the proof can easily be followed from the finishing part of the proof of Theorem 2.18. For $M_{2} ( x, y, z )$, in line with the beginning part of the proof of Theorem 2.18, for all $( x, y, z ) \in \lambda _{f, g, h, \alpha _{s}}$, for some $F\in \mathcal{F}_{s}$, and $r >0$, from contractive condition (2.11), we get

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) & = F \bigl( s d_{b} \bigl( f ( x_{2 n} ), g ( x_{2 n+ 1} ), h ( x_{2 n+ 2} ) \bigr) \bigr) \\ & \leq F \bigl( M_{2} ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) \bigr) -r \end{aligned} $$

(2.18)

for all $n\in \mathbb{N} \cup \{0\}$, where

$$ \begin{aligned} M_{2} ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) & = a_{1} d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 1} ) + a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ) \\ &\quad{}+ a_{3} d_{b} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n} ) \\ &\quad{}+ a_{4} d_{b} ( J_{2 n+ 2}, J_{2 n+ 2}, J_{2 n+ 2} ) \\ & \leq a_{1} d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) + a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ) \\ &\quad{}+ a_{3} d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ & =( a_{1} + a_{3} ) d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ &\quad{}+ a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ). \end{aligned} $$

From (2.18), we have

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) & \leq F \bigl( ( a_{1} + a_{3} ) d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) \\ &\quad{}+ a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ) -r. \end{aligned} $$

(2.19)

Since F is strictly increasing, (2.19) implies

$$ \begin{aligned} s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) & \leq ( a_{1} + a_{2} ) d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \\ &\quad{}+ a_{2} d_{b} ( J_{2 n+ 1}, J_{2 n}, J_{2 n- 1} ), \end{aligned} $$

so $( s- a_{1} - a_{3} ) d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \leq a_{2} d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} )$. Hence

$$ \begin{aligned} d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) & \leq \frac{a_{2}}{s- a_{1} - a_{3}} d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ). \end{aligned} $$

Thus, from (2.19), we obtain

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) & \leq F \bigl( d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr) -r \end{aligned} $$

(2.20)

for all $n\in \mathbb{N}$. Similarly,

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr) & \leq F \bigl( d_{b} ( J_{2 n- 2}, J_{2 n- 1}, J_{2 n} ) \bigr) -r \end{aligned} $$

(2.21)

for all $n\in \mathbb{N}$. Hence, from (2.20) and (2.21), we have

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{n}, J_{n+ 1}, J_{n+ 2} ) \bigr) & \leq F \bigl( d_{b} ( J_{n- 1}, J_{n}, J_{n+ 1} ) \bigr) -r. \end{aligned} $$

(2.22)

Inequality (2.22) leads to remark that $\{ J_{n} \}$ is a Cauchy sequence, and the remaining part of the proof can easily be followed from the finishing part of the proof of Theorem 2.18. For $M_{3} ( x, y, z )$, in line with the beginning part of the proof of Theorem 2.18, for all $( x, y, z ) \in \lambda _{f, g, h, \alpha _{s}}$, for some $F\in \mathcal{F}_{s}$, and $r >0$, from contractive condition (2.12), we get

$$ \begin{aligned}[b] F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) & = F \bigl( s d_{b} \bigl( f ( x_{2 n} ), g ( x_{2 n+ 1} ), h ( x_{2 n+ 2} ) \bigr) \bigr) \\ & \leq F \bigl( M_{3} ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) \bigr) -r \end{aligned} $$

(2.23)

for all $n\in \mathbb{N} \cup \{0\}$, where

$$ \begin{aligned}[b] M_{3} ( x_{2 n}, x_{2 n+ 1}, x_{2 n+ 2} ) & = k \max \bigl\{ d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ), d_{b} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n} ),0\bigr\} \\ & = k \max \bigl\{ d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ), d_{b} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n} ) \bigr\} . \end{aligned} $$

If

$$ \max \bigl\{ d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ), d_{b} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n} ) \bigr\} = d_{b} ( J_{2 n+ 2}, J_{2 n+ 1}, J_{2 n} ), $$

then from (2.23) we have $F ( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) ) \leq F ( k d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) ) -r$. Since F is strictly increasing, we have $s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) < k d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} )$. It is a contradiction. Thus we have

$$ \begin{aligned} F \bigl( s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \bigr) & \leq F \bigl( k d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ) \bigr) -r, \end{aligned} $$

and $s d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) \leq k d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} )$. So

$$ \begin{aligned} d_{b} ( J_{2 n}, J_{2 n+ 1}, J_{2 n+ 2} ) & \leq \frac{k}{s} d_{b} ( J_{2 n- 1}, J_{2 n}, J_{2 n+ 1} ). \end{aligned} $$

The emaining part of the proof can easily be followed from the proof of Theorem 2.18. Similar arguments hold from $M_{4} ( x, y, z )$. □

Theorem 2.28

Let f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$ be self-mappings defined on a complete tripled b-metric space ($X, d_{b},s$) such that $f(X)\subseteq S_{1} (X)$, $g(X)\subseteq S_{2} (X)$, and $h(X)\subseteq S_{3} (X)$. If there exist $F\in \mathcal{F}_{s}$ and $r>0$ such that $d_{b} ( f(x),g(y),h(z) ) >0$, then

$$ r+F ( s d_{b} \bigl( f ( x ), g ( y ), h ( z ) \bigr) \leq F \bigl( M ( x, y, z )\bigr) $$

for all $x, y, z\in X$. Then f, g, h, $S_{1}$, $S_{2}$, and $S_{3}$ have a unique common fixed point in X provided that $S_{1}$, $S_{2}$, and $S_{3}$ are continuous and pairs ($f, S_{1}$), ($g, S_{2}$), and ($h, S_{3}$) are compatible.

Proof

The arguments follow the same lines as in the proof of Theorem 2.18. □

3 Application to a system of integral equations

Let $X = C ([0,1], \mathbb{R})$ be the space of all continuous real-valued functions defined on [$0,1$]. Let $d_{b}: X\times X\times X\rightarrow \mathbb{R}_{0}^{+}$ be defined

$$ d_{b} ( u, v, w ) =\max \Bigl\{ \sup_{t\in [0,1]} \bigl\vert u ( t ) -v ( t ) \bigr\vert ^{2}, \sup_{t\in [0,1]} \bigl\vert u ( t ) -w ( t ) \bigr\vert ^{2}, \sup _{t\in [0,1]} \bigl\vert v ( t ) -w ( t ) \bigr\vert ^{2} \Bigr\} $$

for all $u, v, w\in C ( [ 0,1 ], \mathbb{R} )$, and define $\alpha _{s}: X\times X\times X\rightarrow \mathbb{R}_{0}^{+}$ by $\alpha _{s} ( u, v, w )= s^{2}$ for all $u, v, w\in X$. Obviously, ($X, d_{b}, s$) is an $\alpha _{s}$-complete tripled b-metric space. We will apply Theorem 2.18 to show the existence of a common solution of the system of Volterra-type integral equations given by

$$ \begin{gathered} u ( t ) = p ( t ) + \int _{0}^{t} K \bigl( t, r, S_{1} \bigl( u ( t )\bigr)\bigr) dr, \\ v ( t ) = p ( t ) + \int _{0}^{t} J \bigl( t, r, S_{2} \bigl( v ( t )\bigr)\bigr) dr, \\ w ( t ) = p ( t ) + \int _{0}^{t} I \bigl( t, r, S_{3} \bigl( w ( t )\bigr)\bigr) dr \end{gathered} $$

(3.1)

for all $t\in [0,1]$, where $p:[0,1] \rightarrow \mathbb{R}$ is a continuous function and $K, J, I:[0,1] \times [0,1] \times X\rightarrow \mathbb{R}$ are lower semi-continuous operators. Now, we prove the following theorem to ensure the existence of solution for the system of integral equations.

Theorem 3.1

Let $X=C([0,1], \mathbb{R)}$ and define the mappings $f,g,h:X\rightarrow X$ by

$$ \begin{gathered} f \bigl( u ( t )\bigr) = p ( t ) + \int _{0}^{t} K \bigl( t, r, S_{1} \bigl( u ( t )\bigr)\bigr) dr, \\ g \bigl( v ( t )\bigr) = p ( t ) + \int _{0}^{t} J \bigl( t, r, S_{2} \bigl( v ( t )\bigr)\bigr) dr, \\ h \bigl( w ( t )\bigr) = p ( t ) + \int _{0}^{t} I \bigl( t, r, S_{3} \bigl( w ( t )\bigr)\bigr) dr \end{gathered} $$

for all $t\in [0,1]$. Assume that the following conditions are satisfied.

• There exists a continuous function $\phi _{i}: X\rightarrow \mathbb{R}_{0}^{+}$, $i =1,2,3$, such that

  $$ \begin{gathered} \bigl\vert K ( t, r, S_{1} ) -J ( t, r, S_{2} ) \bigr\vert \leq \phi _{1} ( r ) \bigl\vert S_{1} \bigl( u ( t )\bigr) - S_{2} \bigl( v ( t )\bigr) \bigr\vert , \\ \bigl\vert K ( t, r, S_{1} ) -I ( t, r, S_{3} ) \bigr\vert \leq \phi _{2} ( r ) \bigl\vert S_{1} \bigl( u ( t )\bigr) - S_{3} \bigl( w ( t )\bigr) \bigr\vert , \\ \bigl\vert J ( t, r, S_{2} ) -I ( t, r, S_{3} ) \bigr\vert \leq \phi _{3} ( r ) \bigl\vert S_{2} \bigl( v ( t ) \bigr) - S_{3} \bigl( w ( t )\bigr) \bigr\vert \end{gathered} $$

  for each $t, r\in [0,1]$ and $S_{1}$, $S_{2}$, and $S_{3} \in X$;

• There exists $\tau >0$ such that

  $$ \int _{0}^{t} \phi _{1} ( r ) dr, \int _{0}^{t} \phi _{2} ( r ) dr, \int _{0}^{t} \phi _{3} ( r ) dr\leq \sqrt{\frac{e^{-\tau }}{s}}. $$

Then the system of integral Eqs. (3.1) has a solution.

Proof

By assumptions (i) and (ii), we have

$$ \begin{aligned} d_{b} \bigl( f \bigl( u ( t )\bigr), g \bigl( v ( t ) \bigr), h \bigl( w ( t )\bigr) \bigr) & =\max \Bigl\{ \sup_{t\in [0,1]} \bigl\vert f \bigl( u ( t )\bigr) -g \bigl( v ( t )\bigr) \bigr\vert ^{2}, \\ &\quad \sup_{t\in [0,1]} \bigl\vert g \bigl( v ( t )\bigr) -h \bigl( w ( t )\bigr) \bigr\vert ^{2}, \\ &\quad \sup_{t\in [0,1]} \bigl\vert f \bigl( u ( t )\bigr) -h \bigl( w ( t )\bigr) \bigr\vert ^{2} \Bigr\} , \end{aligned} $$

where

$$\begin{aligned}& \begin{aligned} \sup_{t\in [0,1]} \bigl\vert f \bigl( u ( t ) \bigr) -g \bigl( v ( t )\bigr) \bigr\vert ^{2}& = \biggl( \sup _{t\in [0,1]} \int _{0}^{t} \bigl\vert K \bigl( t, r, S_{1} \bigl( u ( t ) \bigr) \bigr) - J \bigl( t, r, S_{2} \bigl( v ( t )\bigr)\bigr) \bigr\vert dr \biggr)^{2} \\ &\leq \biggl( \sup_{t\in [0,1]} \int _{0}^{t} \phi _{1} ( r ) \bigl\vert S_{1} \bigl( u ( t )\bigr) - S_{2} \bigl( v ( t ) \bigr) \bigr\vert dr \biggr)^{2} \\ &\leq \biggl( \sqrt{\sup_{t\in [0,1]} \bigl\vert S_{1} \bigl( u ( t )\bigr) - S_{2} \bigl( v ( t )\bigr) \bigr\vert ^{2}} \int _{0}^{t} \phi _{1} ( r ) dr \biggr)^{2} \\ &= \sup_{t\in [0,1]} \bigl\vert S_{1} \bigl( u ( t ) \bigr) - S_{2} \bigl( v ( t )\bigr) \bigr\vert ^{2} \biggl( \int _{0}^{t} \phi _{1} ( r ) dr \biggr)^{2}, \end{aligned} \\& \sup_{t\in [0,1]} \bigl\vert g \bigl( v ( t )\bigr) -h \bigl( w ( t )\bigr) \bigr\vert ^{2} \leq \sup_{t\in [0,1]} \bigl\vert S_{2} \bigl( v ( t )\bigr) - S_{3} \bigl( w ( t ) \bigr) \bigr\vert ^{2} \biggl( \int _{0}^{t} \phi _{2} ( r ) dr \biggr)^{2}, \\& \sup_{t\in [0,1]} \bigl\vert f \bigl( u ( t )\bigr) -h \bigl( w ( t )\bigr) \bigr\vert ^{2} \leq \sup_{t\in [0,1]} \bigl\vert S_{1} \bigl( u ( t )\bigr) - S_{3} \bigl( w ( t ) \bigr) \bigr\vert ^{2} \biggl( \int _{0}^{t} \phi _{3} ( r ) dr \biggr)^{2}. \end{aligned}$$

Consequently, we have

$$ \begin{aligned} d_{b} \bigl( f \bigl( u ( t )\bigr), g \bigl( v ( t ) \bigr), h \bigl( w ( t )\bigr) \bigr) &= \frac{e^{-\tau }}{s} \max \Bigl\{ \sup _{t\in [0,1]} \bigl\vert S_{1} \bigl( u ( t )\bigr) - S_{2} \bigl( v ( t )\bigr) \bigr\vert ^{2}, \\ &\quad \sup_{t\in [0,1]} \bigl\vert S_{2} \bigl( v ( t ) \bigr) - S_{3} \bigl( w ( t )\bigr) \bigr\vert ^{2}, \sup _{t\in [0,1]} \bigl\vert S_{1} \bigl( u ( t )\bigr) - S_{3} \bigl( w ( t )\bigr) \bigr\vert ^{2} \Bigr\} \\ & = \frac{e^{-\tau }}{s} d_{b} \bigl( S_{1} \bigl( u ( t ) \bigr), S_{2} \bigl( v ( t )\bigr), S_{3} \bigl( w ( t )\bigr) \bigr) \\ & \leq \frac{e^{-\tau }}{s} M \bigl( u ( t ), v ( t ), Sw ( t ) \bigr). \end{aligned} $$

Thus, we obtain

$$ s d_{b} \bigl( f \bigl( u ( t )\bigr), g \bigl( v ( t )\bigr), h \bigl( w ( t )\bigr) \bigr) \leq e^{-\tau } M \bigl( u ( t ), v ( t ), w ( t ) \bigr), $$

which implies that

$$ \tau + \ln \bigl( s d_{b} \bigl( f \bigl( u ( t )\bigr), g \bigl( v ( t )\bigr), h \bigl( w ( t )\bigr) \bigr) \bigr) \leq \ln \bigl( M \bigl( u ( t ), v ( t ), w ( t ) \bigr) \bigr). $$

For $F ( r )=\ln r$, all the hypotheses of Theorem 2.28 are satisfied. Hence the system of integral equations has a unique common solution. □