Some Periodic Point and Fixed Point Results in Multiplicative Metric Spaces

Abstract

We investigate the periodic points and common fixed point of generalized contraction mappings self-mappings in the setup of multiplicative metric spaces. We also study the well-posedness for the obtained results. The common fixed point results of mappings involved in the cyclic representation are also obtained. Moreover, some applications to obtain the common solution of integral equations are presented.

1 Introduction

Fixed point theory is a very effective and powerful tool for solving various kind of   mathematical problems. The study of fixed points of mappings has several applications in the solution of optimization problems, differential equations and integral equations (see for example, [6, 7, 18, 19, 21, 25]). Theorems dealing with fixed point of certain mappings inspired and motivated the investigations of many other important kinds of points like periodic points, intersection points, sectional points, etc.

It is an obvious fact that if S is a self-map which has a fixed point x,  that is, $Sx=x,$ then x is also a fixed point of $S^{n}$ for every natural number n,  that is, $S^{n}x=x.$ However, fixed points of $S^{n}$ for a natural number $n>1$ do not need to be fixed points for S as they can be periodic points of the period larger than 1. For example, consider, $X=\mathbb {R},$ and S defined as $S\left( x\right) =a-x$ for some $a\in \mathbb {R}\backslash \{0\}.$ Then S has a unique fixed point at $x=\dfrac{a}{2},$ but every even iterate of S is the identity map, which has every point of $\mathbb {R}$ as a fixed point. On the other hand, if $X=[0,\pi ]$ and $S\left( x\right) =\cos x,$ then every iterate of S has the same fixed point as S (see [4, 11, 12, 14, 20, 22, 26, 28]). If a self-map satisfies $F(S)=F(S^{n})$ for each $n\in \mathbb {N},$ where F(S) denotes a set of all fixed point of S,  then it is said that S has property P. Also if two self-maps S and T satisfy $F(S)\bigcap F(T)=F(S^{n})\bigcap F \left( T^{n}\right)$ for each $n\in \mathbb {N},$ then it is said that pair (S, T) has property Q.

Jeong and Rhoades [20] showed that maps satisfying many contractive conditions have property P. Abbas and Rhoades in [5] studied the same problem in cone metric spaces (see also, [22, 26]), and in [25] considered the mappings satisfying a contractive condition of integral type for which fixed point and periodic point coincide. Chaipunya, Cho and Kumam [11] studied the property P and periodic points of order $\infty$. Chen, Karapınar and Rakočević [12] considered mappings satisfying a contractive condition in the setting of generalized quasi metric spaces. It could be also interesting to mention, that in a totally different context of interplay of dynamical systems and $C^*$-algebras, Silvestrov and Tomiyama [27] obtained several general equivalent conditions for the coincidence of the sets of recurrent and periodic points of homeomorphism dynamical systems of topological spaces, and discussed some examples and classes of homeomorphism dynamical systems satisfying property of coincidence of the sets of periodic and fixed points (property P).

Recently, Özavşar and Çevikel [23] proved an analogous of Banach contraction principle in the framework of multiplicative metric spaces. They also studied some topological properties of the relevant multiplicative metric space. Bashirov, Kurpınar and Ozyapıcı [9] studied the concept of multiplicative calculus and proved a fundamental theorem of multiplicative calculus. They also illustrated the usefulness of multiplicative calculus with some interesting applications. Multiplicative calculus provides natural and straightforward way to compute the derivative of product and quotient of two functions [10]. Florack and van Assen [13] gave applications of multiplicative calculus in biomedical image analysis. He, Song and Chen [16] studied common fixed points for weak commutative mappings on a multiplicative metric space (see also, [1, 3]). Recently, Yamaod and Sintunavarat [29] obtained some fixed point results for generalized contraction mappings with cyclic $(\alpha ,\beta )$-admissible mapping in multiplicative metric spaces.

We study the common fixed point problems that satisfy property Q of mappings in the framework of multiplicative metric spaces. We also show the well-posedness of these results. We also study the sufficient conditions for the existence of common fixed points of pair of power contractive type mappings involved in cyclic representation of a non-empty subset of a multiplicative metric space. Some applications of obtained results are also shown.

By $\mathbb {R},$ $\mathbb {R}_{>0}$, $\mathbb {R}_{>0}^n$ and $\mathbb {N}$ we denote the set of all real numbers, the set of all positive real numbers, the set of all n-tuples of positive real numbers and the set of all natural numbers, respectively.

The following definitions and results will be needed in the sequel [9, 23].

Definition 1

(multiplicative metric space) Let X be a non-empty set. A mapping $d:X\times X\rightarrow \mathbb {R}_{>0}$ is said to be a multiplicative metric on X if for any $x,y,z\in X,$ the following conditions hold:

• (i) $d(x,y)\ge 1$ and $d(x,y)=1$ if and only if $x=y$;

• (ii) $d(x,y)=d(y,x);$

• (iii) $d(x,y)\le d(x,z)\cdot d(z,y).$

The pair (X, d) is called a multiplicative metric space. 

Definition 2

([23]) A sequence $\{x_{n}\}$ in a multiplicative metric space (X, d) is multiplicative convergent to x in X if and only if $d(x_{n},x)\rightarrow 1$ as $n\rightarrow \infty$.

Definition 3

Let $(X,d_{X})$ and $(Y,d_{Y})$ be two multiplicative metric spaces, and $x_{0}$ an arbitrary but fixed element of X. A mapping $S:X\rightarrow Y$ is said to be multiplicative continuous at $x_{0}$ if and only if $x_{n}\rightarrow x_{0}$ in $(X,d_{X})$ implies that $S(x_{n})\rightarrow S(x_{0})$ in $(Y,d_{Y})$, that is, for any $\varepsilon >1$, there exists $\delta >1$, which depends on $x_{0}$ and $\varepsilon$, such that $d_{Y}(Sx,Sx_{0})<\varepsilon$ for all those x in X for which $d_{X}(x,x_{0})<\delta$.

Definition 4

([23]) A sequence $\{x_{n}\}$ in a multiplicative metric space (X, d) is said to be multiplicative Cauchy sequence if for any $\varepsilon >1$, there exists $n_{0}\in \mathbb {N}$ such that $d(x_{n},x_{m})<\varepsilon$ for all $m,n\ge n_{0}$.

A multiplicative metric space (X, d) is said to be complete if every multiplicative Cauchy sequence $\{x_{n}\}$ in X is multiplicative convergent in X.

Theorem 1

([23]) A sequence $\{x_{n}\}$ in a multiplicative metric space (X, d) is multiplicative Cauchy if and only if $d(x_{n},x_{m})\rightarrow 1$ as $n,m\rightarrow \infty$.

The multiplicative absolute-value function $\left| \cdot \right| _* :\mathbb {R}\rightarrow \mathbb {R}_{>0}$ is defined as

$$ \left| \alpha \right| _* =\left\{ \begin{array}{cl} \underset{}{\alpha } &{} \text {if }\alpha \ge 1, \\ \dfrac{1}{\underset{}{\alpha }} &{} \text {if }\alpha \in (0,1), \\ \underset{}{1} &{} \text {if }\alpha =0, \\ -\dfrac{1}{\underset{}{\alpha }} &{} \text {if }\alpha \in (-1,0), \\ -\alpha &{} \text {if }\alpha \le -1. \end{array} \right. $$

For arbitrary $x,y\in \mathbb {R}_{>0}$, the multiplicative absolute value function $\left| \cdot \right| _* :\mathbb {R}_{>0}\rightarrow \mathbb {R}_{>0}$ has the following basic properties: $\ \left| x\right| _* \ge 1$, $x\le \left| x\right| _*$, $\ x\le \dfrac{1}{\left| x\right| _* }$ if $x\le 0$ and $\dfrac{1}{\left| x\right| _* }\le x$ if $x>0$, as well as $\left| x\cdot y\right| _* \le \left| x\right| _* \left| y\right| _*$.

Example 1

Let $X=C^{*}[a,b]$ be the collection of all real-valued multiplicative continuous functions over $[a,b]\subset \mathbb {R}_{>0}$ with the multiplicative metric d defined for arbitrary $S,\ T\in X$ by $d(S,T)=\sup \limits _{x\in [a,b]}\left| \frac{S(x)}{T(x)} \right| _*,$ where $\left| \cdot \right| _* :\mathbb {R}_{>0}\rightarrow \mathbb {R}_{>0}$ is the multiplicative absolute value function. Then $(C^{*}[a,b],d)$ is complete multiplicative metric space.

The notion of well-posedness of fixed point problem has evoked much interest to several mathematicians. Recently, Karapinar [21] studied well-posed problem for a cyclic weak contraction mapping on a complete metric space (see also, [2, 24]). We define well-posedness of common fixed point problem in multiplicative metric space.

Definition 5

Let (X, d) be a multiplicative metric space and S and T be two self-maps on X. A common fixed point problem of S and T is called well-posed on X if S and T have at most one common fixed point (say u) and for any sequence $\{x_{n}\}$ in X such that $\lim \limits _{n\rightarrow \infty }d(Sx_{n},x_{n})=1$ or $\lim \limits _{n\rightarrow \infty }d(Tx_{n},x_{n})=1\$implies that $\lim \limits _{n\rightarrow \infty }d(x_{n},u)=1.$

2 Periodic Point Results

In this section, we obtain several periodic point and common fixed point results of self-maps satisfying certain power contractive conditions in the framework of multiplicative metric space. We start with the following result.

Theorem 2

Let $\left( X,d\right)$ be a complete multiplicative metric space and $S,T:X\rightarrow X$. Suppose that there exists an upper semi-continuous and nondecreasing function $\phi : [1,\infty )\rightarrow [1,\infty )$, obeying $\phi (t)<t$ for all $t>1$, such that for all $x,y\in X$,

$$\begin{aligned} d(Sx,Ty)^{\delta }\le \phi (d(x,y)^{\alpha }\cdot d(x,Sx)^{\beta }\cdot d(y,Ty)^{\gamma }), \end{aligned}$$

(17.1)

where $\alpha ,\beta ,\gamma \ge 0$ with $\delta =\alpha +\beta +\gamma \in \left( 0,\infty \right)$. Then $F(S)\bigcap F\left( T\right)$ is singleton and pair $\left( S,T\right)$ has property Q.

Proof

$\square$First, we show that $F(S)\bigcap F\left( T\right) \ne \phi .$ We divide the proof in four facts.

Fact 1. If S or T has a fixed point u in X, then u is a common fixed point of S and T.

Indeed, let u be a fixed point of S. Assume that $d(u,Tu)>1.$ From (17.1) with $x=y=u$, we have

$$\begin{aligned}&d(u,Tu)^{\delta } =d(Su,Tu)^{\delta } \le \phi (d(u,u)^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(u,Tu)^{\gamma }) \\&\quad =\phi (d(u,u)^{\alpha +\beta }\cdot d(u,Tu)^{\gamma }) \le \phi (d(u,Tu)^{\gamma }) \le \phi \left( d(u,Tu)^{\delta }\right) <d(u,Tu)^{\delta }, \end{aligned}$$

a contradiction. Hence $d(u,Tu)=1$ and so $u=Tu.$ Therefore u is a common fixed point of S and T. Similarly, if u is a fixed point of T, then it is also fixed point of S.

Fact 2. The sequence $\{x_{n}\}$ constructed by S and T satisfies $\lim \limits _{n\rightarrow \infty }d(x_{n},x_{n+1})=1$ in a multiplicative metric space.

Indeed, let $x_{0}$ be an arbitrary point of X. If $Sx_{0}=x_{0}$, then the proof is finished, so we assume that $Sx_{0}\ne x_{0}$. Define a sequence $\{x_{n}\}$ in X as $Sx_{2n}=x_{2n+1}$ and $Tx_{2n+1}=x_{2n+2}$ for $n\in \mathbb {N}.$

We may assume that $d(x_{2n},x_{2n+1})>1,$ for all $n\in \mathbb {N}.$ If not, then $x_{2k}=x_{2k+1}$ for some k,  so $Sx_{2k}=x_{2k+1}=x_{2k},$ and thus $x_{2k}$ is a fixed point of S. Hence $x_{2k}$ is also a fixed point of T by Fact 1. Now, by taking $d(x_{2n},x_{2n+1})>1$ for all $n\in \mathbb {N},$ from (17.1), we consider

$$\begin{aligned} d(x_{2n+1},x_{2n+2})^{\delta }= & {} d(Sx_{2n},Tx_{2n+1})^{\delta } \\\le & {} \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},Sx_{2n})^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\= & {} \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},x_{2n+1})^{\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }) \\= & {} \phi (d(x_{2n},x_{2n+1})^{\alpha +\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }) \\< & {} d(x_{2n},x_{2n+1})^{\alpha +\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }, \end{aligned}$$

which implies that $d(x_{2n+1},x_{2n+2})^{\alpha +\beta }<d(x_{2n},x_{2n+1})^{\alpha +\beta }.$ If $\alpha +\beta =0,$ then a contradiction arises. Thus $\alpha +\beta >0$, and we have $d(x_{2n+1},x_{2n+2})<d(x_{2n},x_{2n+1}),$ for all $n\in \mathbb {N}.$ Again from (17.1), we have

$$\begin{aligned} d(x_{2n+2},x_{2n+3})^{\delta }= & {} d(Tx_{2n+1},Sx_{2n+2})^{\delta } =d(Sx_{2n+2},Tx_{2n+1})^{\delta } \\\le & {} \phi (d(x_{2n+2},x_{2n+1})^{\alpha }\cdot d(x_{2n+2},Sx_{2n+2})^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\= & {} \phi (d(x_{2n+1},x_{2n+2})^{\alpha }\cdot d(x_{2n+2},x_{2n+3})^{\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }) \\= & {} \phi (d(x_{2n+1},x_{2n+2})^{\alpha +\gamma }\cdot d(x_{2n+2},x_{2n+3})^{\beta }) \\< & {} d(x_{2n+1},x_{2n+2})^{\alpha +\gamma }\cdot d(x_{2n+2},x_{2n+3})^{\beta }, \end{aligned}$$

which implies that $d(x_{2n+2},x_{2n+3})^{\alpha +\gamma }< d(x_{2n+1},x_{2n+2})^{\alpha +\gamma }.$ If $\alpha +\gamma =0,$ then a contradiction arises. Hence $\alpha +\gamma >0$ and $d(x_{2n+2},x_{2n+3})<d(x_{2n+1},x_{2n+2}),$ for all $n\in \mathbb {N}$. Consequently, for all $n \in \mathbb {N}$,

$$\begin{aligned} d(x_{n},x_{n+1})<d(x_{n-1},x_{n}). \end{aligned}$$

(17.2)

Therefore, the decreasing sequence of positive real numbers $\{d(x_{n},x_{n+1})\}$ converges to some $c\ge 1.$ If we assume that $c>1,$ then from (17.2) we deduce that

$$\begin{aligned} c^{\delta }= & {} \lim _{n\rightarrow \infty } d(x_{2n+1},x_{2n+2})^{\delta }=\lim _{n\rightarrow \infty } d(Sx_{2n},Tx_{2n+1})^{\delta } \\\le & {} \limsup \limits _{n\rightarrow \infty } \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},Sx_{2n})^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\= & {} \limsup \limits _{n\rightarrow \infty } \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},x_{2n+1})^{\beta }\cdot d(x_{2n+1},x_{2n+2}))^{\gamma } \\\le & {} \phi (c^{\alpha +\beta +\gamma })=\phi (c^{\delta }) <c^{\delta }, \end{aligned}$$

a contradiction, so $c=1,$ that is, $\lim \limits _{n\rightarrow \infty }d(x_{n},x_{n+1})^{\delta }=1$ and so $\lim \limits _{n\rightarrow \infty }d(x_{n},x_{n+1})=1.$

Fact 3. The sequence $\{x_{n}\}$ constructed in Fact 2 is a multiplicative Cauchy in (X, d).

We have $d(x_{n},x_{n+1})^{\delta } \le \phi (d(x_{n-1},x_{n})^{\delta })\le \ldots \le \phi ^{n}(d(x_{0},x_{1})^{\delta }).$ For $m,n\in \mathbb {N}$ with $m>n$,

$$\begin{aligned} d(x_{n},x_{m})^{\delta }\le & {} d(x_{n},x_{n+1})^{\delta }\cdot d(x_{n+1},x_{n+2})^{\delta }\cdot \ldots \cdot d(x_{m-1},x_{m})^{\delta } \\\le & {} \phi ^{n}(d(x_{0},x_{1})^{\delta })\cdot \phi ^{n+1}(d(x_{0},x_{1})^{\delta })\cdot \ldots \cdot \phi ^{m-1}(d(x_{0},x_{1})^{\delta }), \end{aligned}$$

which implies that $d(x_{n},x_{m})^{\delta }$ converges to 1 as $n,m\rightarrow \infty$. Thus, $\lim \limits _{n,m\rightarrow \infty } d(x_{n},x_{m})=1,$ that is, $\{x_{n}\}$ is a multiplicative Cauchy sequence in (X, d).

Fact 4. $F\left( S\right) \bigcap F\left( T\right) \ne \emptyset$.

Indeed, since (X, d) is complete multiplicative space, there exists u in X such that $\lim \limits _{n\rightarrow \infty }d(u,x_{n})=1.$ Assume on contrary that, $d(Su,u)>1$, then from (17.1), we have

$$\begin{aligned} \begin{array}{lll} d(Su,x_{2n+2})^{\delta } &{}=&{}d(Su,Tx_{2n+1})^{\delta } \\ &{}\le &{}\phi (d(u,x_{2n+1})^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\ &{}=&{}\phi (d(u,x_{2n+1})^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }), \end{array} \end{aligned}$$

(17.3)

we deduce, by taking upper limit as $n\rightarrow \infty$ into account (17.3), that

$$\begin{aligned} d(Su,u)^{\delta }\le & {} \limsup \limits _{n\rightarrow \infty } \phi (d(u,x_{2n+1})^{\alpha } \cdot d(u,Su)^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\\le & {} \phi (d(u,u)^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(u,u)^{\gamma }) \le \phi (d(u,Su)^{\beta }) <d(u,Su)^{\delta }, \end{aligned}$$

a contradiction. Hence $u=Su$ and thus u is the common fixed point of S and T by Fact 1.

Now, let us show that $F\left( S\right) \bigcap F\left( T\right)$ is singleton set. Assume on contrary that $Su=Tu=u$ and $Sv=Tv=v$ but $u\ne v$. Then

$$\begin{aligned} d(u,v)^{\delta }= & {} d(Su,Tv)^{\delta } \le \phi (d(u,v)^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(v,Tv)^{\gamma }) \\= & {} \phi (d(u,v)^{\alpha }\cdot d(u,u)^{\beta }\cdot d(v,v))^{\gamma } \le \phi (d(u,v)^{\delta }), \end{aligned}$$

a contradiction because $d(u,v)^{\delta }>1.$ Hence $u=v$ and $F\left( S\right) \bigcap F\left( T\right) =\{u\}.$

Let $u\in F(S^{n})\bigcap F(T^{n})$ be arbitrary for $n>1$, since the statement for $n=1$ is trivial. Now,

$$\begin{aligned} d(u,Tu)^{\delta }= & {} d(S(S^{n-1}u),T(T^{n}u))^{\delta } \\\le & {} \phi (d(S^{n-1}u,T^{n}u)^{\alpha }\cdot d(S^{n-1}u,S^{n}u)^{\beta }\cdot d(T^{n}u,T^{n+1}u))^{\gamma } \\= & {} \phi (d(S^{n-1}u,T^{n}u)^{\alpha }\cdot d(S^{n-1}u,T^{n}u)^{\beta }\cdot d(u,Tu)^{\gamma }) \\< & {} d(S^{n-1}u,T^{n}u)^{\alpha +\beta }\cdot d(u,Tu)^{\gamma }, \end{aligned}$$

which implies that $d(u,Tu)<d(S^{n-1}u,T^{n}u)^{\lambda },$ where $\lambda =\dfrac{\alpha +\beta }{\delta -\gamma }=1.$ Thus

$$\begin{aligned} d(u,Tu)= & {} d(S^{n}u,T^{n+1}u)<d(S^{n-1}u,T^{n}u) \\< & {} d(S^{n-2}u,T^{n-1}u)<\ldots <d(u,Tu), \end{aligned}$$

a contradiction. Hence $d(u,Tu)=1,$ that is, $u=Tu.$ Similarly, it can be shown that $Su=u.$ Thus $u\in F(S)\bigcap F(T),$ and hence S and T have property Q.

Theorem 3

Let $\left( X,d\right)$ be a complete multiplicative metric space and $S,T:X\rightarrow X$. If there exists an upper semi-continuous nondecreasing function $\phi :[1,\infty )\rightarrow [1,\infty )$, obeying $\phi (t)<t$ for all $t>1$, such that for all $x,y\in X$,

$$ d(Sx,Ty)^{\delta }\le \phi (d(x,y)^{\alpha }\cdot d(x,Sx)^{\beta }\cdot d(y,Ty)^{\gamma }), $$

where $\alpha ,\beta ,\gamma >0$ with $\delta =\alpha +\beta +\gamma \in \left( 0,\infty \right)$, then the common fixed point problem of S and T is well-posed on X.

Proof

$\square$Due to Theorem 2, for any $x_{0}\in X,$ $u\in X$ is the unique common fixed point of S and T. Let $\{x_{n}\}$ be a sequence in X such that $d(Sx_{n},x_{n})\rightarrow 1$ as $n\rightarrow \infty .$ Then,

$$\begin{aligned} d(x_{n},u)^{\delta }\le & {} d(x_{n},Sx_{n})^{\delta }\cdot d(Sx_{n},Tu)^{\delta } \\\le & {} d(x_{n},Sx_{n})^{\delta }\cdot \phi (d(x_{n},u)^{\alpha }\cdot d(x_{n},Sx_{n})^{\beta }\cdot d(u,Tu)^{\gamma }) \\= & {} d(x_{n},Sx_{n})^{\delta }\cdot \phi (d(x_{n},u)^{\alpha }\cdot d(x_{n},Sx_{n})^{\beta }) \le d(x_{n},Sx_{n})^{\delta +\beta }\cdot d(x_{n},u)^{\alpha }, \end{aligned}$$

which implies that $d(x_{n},u)^{\beta +\gamma }\le d(x_{n},Sx_{n})^{\delta +\beta },$ that is, $d(x_{n},u)\le [d(x_{n},Sx_{n})]^{\frac{\delta +\beta }{\beta +\gamma }}.$ Taking limit as $n\rightarrow \infty$ implies $d(x_{n},u)\rightarrow 1.$

Corollary 1

Let $\left( X,d\right)$ be a complete multiplicative metric space and $S,T:X\rightarrow X$. If there exists an upper semi-continuous nondecreasing function $\phi :[1,\infty )\rightarrow [1,\infty )$, with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for each $t>1$, such that for all $x,y\in X$,

$$ d(S^{s}x,T^{t}y)^{\delta }\le \phi (d(x,y)^{\alpha }\cdot d(x,S^{s}x)^{\beta }\cdot d(y,T^{t}y))^{\gamma }, $$

where $\alpha ,\beta ,\gamma \ge 0$ with $\delta =\alpha +\beta +\gamma \in \left( 0,\infty \right)$ and $s,t\in \mathbb {N}$, then $F(S)\bigcap F\left( T\right)$ is singleton and pair $\left( S,T\right)$ has property Q.

Proof

$\square$It follows from Theorem 2, that $S^{s}$ and $T^{t}$ have a unique common fixed point w. Now $S(w)=S(S^{s}(w))=S^{s+1}(w)=S^{s}(S(w))$ and $T(w)=T(T^{t}(w))=T^{t+1}(w)=T^{t}(T(w))$ implies that Sw and Tw are also fixed points for $S^{s}\$and $T^{t}$. Since the common fixed point of $S^{s}\$and $T^{t}$ is unique, we deduce that $w=Sw=Tw.$ It is obvious that every fixed point of S is a fixed point of T and conversely.

If we take $\phi (t)=t^{k}$ for $k\in [0,1)$ in Theorem 2, we have the following Corollary.

Corollary 2

Let $\left( X,d\right)$ be a complete multiplicative metric space and $S,T:X\rightarrow X$. Suppose that there exists $k\in [0,1)$ such that for all $x,y\in X$,

$$ d(Sx,Ty)^{\delta }\le (d(x,y)^{\alpha }\cdot d(x,Sx)^{\beta }\cdot d(y,Ty)^{\gamma })^{k}, $$

where $\alpha ,\beta ,\gamma \ge 0$ with $\delta =\alpha +\beta +\gamma \in \left( 0,\infty \right)$. Then $F(S)\bigcap F\left( T\right)$ is singleton and pair $\left( S,T\right)$ has property Q.

Corollary 3

Let $\left( X,d\right)$ be a complete multiplicative metric space, $S,T:X\rightarrow X$, and suppose that there exists some upper semi-continuous nondecreasing function $\phi :[1,\infty )\rightarrow [1,\infty )$, with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for each $t>1$, such that one of the following conditions is satisfied for all $x,y\in X$,

• (i) $d(Sx,Ty)\le \phi (d(x,y)),$

• (ii) $d(Sx,Ty)\le \phi (d(x,Sx)),$

• (iii) $d(Sx,Ty)\le \phi (d(y,Ty)).$

Then $F(S)\bigcap F\left( T\right)$ is singleton and pair $\left( S,T\right)$ has property Q.

Proof

$\square$Taking (i) $\alpha =1$ and $\beta =\gamma =0;$ (ii) $\beta =1,$ $\alpha =\gamma =0;$ (iii) $\gamma =1,$ $\alpha =\beta =0$ in Theorem 2, respectively, then the conclusion of Corollary 3 can be obtained from Theorem 2 immediately.

Corollary 4

Let $\left( X,d\right)$ be a complete multiplicative metric space and $S,T:X\rightarrow X$. If there exists an upper semi-continuous nondecreasing function $\phi :[1,\infty )\rightarrow [1,\infty )$, with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for each $t>1$, such that one of the following condition is satisfied for all $x,y\in X$,

• (i) $d(Sx,Ty)^{2}\le \phi (d(x,y)\cdot d(x,Sx)),$

• (ii) $d(Sx,Ty)^{2}\le \phi (d(x,y)\cdot d(y,Ty)),$

• (iii) $d(Sx,Ty)^{2}\le \phi (d(x,Sx)\cdot d(y,Ty)),$

then $F(S)\bigcap F\left( T\right)$ is singleton and pair $\left( S,T\right)$ has property Q.

Proof

$\square$Taking (i) $\alpha =\beta =1$ and $\gamma =0;$ (ii) $\alpha =\gamma =1,$ $\beta =0;$ (iii) $\beta =\gamma =1,$ $\alpha =0$ in Theorem 2, respectively, the conclusion of Corollary 4 can be obtained from Theorem 2 immediately.

Corollary 5

Let $\left( X,d\right)$ be a complete multiplicative metric space and $S,T:X\rightarrow X$. Suppose that there exists $\phi :[1,\infty )\rightarrow [1,\infty )$ an upper semi-continuous nondecreasing function $\phi :[1,\infty )\rightarrow [1,\infty )$, with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for each $t>1$, such that for all $x,y\in X$, $d(Sx,Ty)^{3}\le \phi (d(x,y)\cdot d(x,Sx)\cdot d(y,Ty)).$ Then $F(S)\bigcap F\left( T\right)$ is singleton and pair $\left( S,T\right)$ has property Q.

Proof

$\square$Taking $\alpha =\beta =\gamma =1$ in Theorem 2, then the conclusion of Corollary 5 can be obtained from Theorem 2 immediately.

Corollary 6

Let $\left( X,d\right)$ be a complete multiplicative metric space and $T:X\rightarrow X$. If there exists an upper semi-continuous nondecreasing function $\phi :[1,\infty )\rightarrow [1,\infty )$, with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for each $t>1$, such that for all $x,y\in X$,

$$ d(Tx,T^{2}y)^{\delta }\le \phi (d(x,y)^{\alpha }\cdot d(x,Tx)^{\beta }\cdot d(y,T^{2}y)^{\gamma }), $$

where $\alpha ,\beta ,\gamma \ge 0$ with $\delta =\alpha +\beta +\gamma \in \left( 0,\infty \right)$, then $F(T)\bigcap F\left( T^{2}\right)$ is singleton and pair $\left( T,T^{2}\right)$ has property Q.

Proof

$\square$Take $S=T^{2}$ in (17.1), then Corollary 6 follows from Theorem 2.

3 Cyclic Contractions

Now we obtain common fixed point result for self-maps satisfying cyclic contraction defined on a multiplicative metric space.

Definition 6

Let $\{X_{i}: i=1,2,\ldots ,p\}$ be a finite collection of non-empty subsets of a set X,  where p is some positive integer and $S,T:X\rightarrow X$. The set X is said to have a cyclic representation with respect to the collection $\{X_{i}: i=1,2,\ldots ,p\}$ and a pair (S, T) if

• (1) $X=\bigcup \limits _{i=1}^{p}X_{i},$

• (2) $S(X_{1})\subseteq X_{2},$ $T(X_{2})\subseteq X_{3},$ $\ldots ,$ $S(X_{p-1})\subseteq X_{p},$ $T(X_{p})\subseteq X_{1}.$

Theorem 4

Let $\left( X,d\right)$ be a multiplicative metric space, $A_{1},A_{2},\ldots ,A_{p}$ non-empty closed subsets of X and $Y=\bigcup \limits _{i=1}^{p}A_{i}$. Suppose that $S,T:Y\rightarrow Y$ are such that

• (i) Y has a cyclic representation with respect to pair (S, T) and to the collection $\{A_{i}: i=1,2,\ldots ,p\};$

• (ii) there exists an upper semi-continuous nondecreasing function $\phi :[1,\infty )\rightarrow [1,\infty )$, with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for each $t>1$, such that for any $\left( x,y\right) \in A_{i}\times A_{i+1}$, $i=1,2,\ldots ,p,$

  $$\begin{aligned} d(Sx,Ty)^{\delta }\le \phi (d(x,y)^{\alpha }\cdot d(x,Sx)^{\beta }\cdot d(y,Ty)^{\gamma }) \end{aligned}$$

  (17.4)

  holds with $A_{p+1}=A_{1}$ where $\alpha ,\beta ,\gamma \ge 0$ with $\delta =\alpha +\beta +\gamma \in \left( 0,\infty \right)$.

Then $F(S)\bigcap F\left( T\right)$ is singleton with $F(S)\bigcap F\left( T\right) \subseteq \bigcap \limits _{i=1}^{p}A_{i}$.

Proof

$\square$To show that $F(S)\bigcap F\left( T\right)$ is singleton, we divide the proof in four facts.

Fact 1. The sequence $\{x_{n}\}$ constructed by S and T satisfies $\lim \limits _{n\rightarrow \infty }d(x_{n},x_{n+1})=1$ in a multiplicative metric space.

Let $x_{0}$ be a given point in $\bigcup _{i=1}^{p}A_{i}.$ Choose point $x_{1}$ in $A_{i_{0}+1}$ and point $x_{2}$ in $A_{i_{0}+2}$ such that $S(x_{0})=x_{1}$ and $T(x_{1})=x_{2}.$ This can be done because $S(A_{i_{0}})\subseteq A_{i_{0}+1}$ and $T(A_{i_{0}+1}) \subseteq A_{i_{0}+2}$. Continuing this process, for $n>0,$ there exists $i_{n}\in \{1,2,\ldots ,p\}$ such that having chosen $x_{2n}$ in $A_{2i_{n}},$ we obtain $x_{2n+1}$ in $A_{2i_{n}+1}$ and $x_{2n+2}$ in $A_{2i_{n}+2}$ such that $S(x_{2n})=x_{2n+1}$ and $T(x_{2n+1})=x_{2n+2}$. If for some $n_{0}\ge 0,$ we have $x_{2n_{0}}=x_{2n_{0}+1},$ then $x_{2n_{0}}=S\left( x_{2n_{0}}\right)$ implies that $x_{n_{0}}$ is the fixed point of S. And from (17.4),

$$\begin{aligned} d(x_{2n+1},x_{2n+2})^{\delta }= & {} d(Sx_{2n},Tx_{2n+1})^{\delta } \\\le & {} \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},Sx_{2n})^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\= & {} \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},x_{2n+1})^{\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }) \\= & {} \phi (d(x_{2n+1},x_{2n+2})^{\gamma }) \le d(x_{2n+1},x_{2n+2})^{\gamma }, \end{aligned}$$

which implies that $x_{2n+1}=x_{2n+2}.$ Thus $x_{2n_{0}}=S\left( x_{2n_{0}}\right) =T\left( x_{2n_{0}}\right)$ implies that $x_{n_{0}}$ is the common fixed point of S and T.

Now, by taking $x_{2n}\ne x_{2n+1}$ for all $n\in \mathbb {N},$ from (17.4), we consider

$$\begin{aligned} d(x_{2n+1},x_{2n+2})^{\delta }= & {} d(Sx_{2n},Tx_{2n+1})^{\delta } \\\le & {} \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},Sx_{2n})^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\= & {} \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},x_{2n+1})^{\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }) \\= & {} \phi (d(x_{2n},x_{2n+1})^{\alpha +\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }) \\< & {} d(x_{2n},x_{2n+1})^{\alpha +\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }, \end{aligned}$$

which implies that $d(x_{2n+1},x_{2n+2})^{\alpha +\beta }<d(x_{2n},x_{2n+1})^{\alpha +\beta }.$ If $\alpha +\beta =0,$ then a contradiction arises. Thus $\alpha +\beta >0$, and for all $n\in \mathbb {N},$ $d(x_{2n+1},x_{2n+2})<d(x_{2n},x_{2n+1}).$ Again from (17.4), we have

$$\begin{aligned} d(x_{2n+2},x_{2n+3})^{\delta }= & {} d(Tx_{2n+1},Sx_{2n+2})^{\delta } =d(Sx_{2n+2},Tx_{2n+1})^{\delta } \\\le & {} \phi (d(x_{2n+2},x_{2n+1})^{\alpha }\cdot d(x_{2n+2},Sx_{2n+2})^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\= & {} \phi (d(x_{2n+1},x_{2n+2})^{\alpha }\cdot d(x_{2n+2},x_{2n+3})^{\beta }\cdot d(x_{2n+1},x_{2n+2})^{\gamma }) \\= & {} \phi (d(x_{2n+1},x_{2n+2})^{\alpha +\gamma }\cdot d(x_{2n+2},x_{2n+3})^{\beta }) \\< & {} d(x_{2n+1},x_{2n+2})^{\alpha +\gamma }\cdot d(x_{2n+2},x_{2n+3})^{\beta }, \end{aligned}$$

which implies that $d(x_{2n+2},x_{2n+3})^{\alpha +\gamma }<d(x_{2n+1},x_{2n+2})^{\alpha +\gamma }.$ If $\alpha +\gamma =0,$ then a contradiction arises. Hence $\alpha +\gamma >0$, and for all $n\in \mathbb {N},$ $d(x_{2n+2},x_{2n+3})<d(x_{2n+1},x_{2n+2}),$ Consequently, for all $n\in \mathbb {N},$

$$\begin{aligned} d(x_{n},x_{n+1})<d(x_{n-1},x_{n}). \end{aligned}$$

(17.5)

Therefore, the decreasing sequence of positive real numbers $\{d(x_{n},x_{n+1})\}$ converges to some $c\ge 1.$ If we assume that $c>1,$ then from (17.5) we deduce that

$$\begin{aligned} c^{\delta }= & {} \lim _{n\rightarrow \infty } d(x_{2n+1},x_{2n+2})^{\delta }=\lim _{n\rightarrow \infty } d(Sx_{2n},Tx_{2n+1})^{\delta } \\\le & {} \limsup \limits _{n\rightarrow \infty } \phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},Sx_{2n})^{\beta }\cdot d(x_{2n+1},Tx_{2n+1})^{\gamma }) \\= & {} \limsup \limits _{n\rightarrow \infty }\phi (d(x_{2n},x_{2n+1})^{\alpha }\cdot d(x_{2n},x_{2n+1})^{\beta }\cdot d(x_{2n+1},x_{2n+2}))^{\gamma } \\\le & {} \phi (c^{\alpha +\beta +\gamma })=\phi (c^{\delta }) <c^{\delta }, \end{aligned}$$

a contradiction, so $c=1,$ that is, $\lim \limits _{n\rightarrow \infty }d(x_{n},x_{n+1})^{\delta }=1$, and so $\lim \limits _{n\rightarrow \infty }d(x_{n},x_{n+1})=1.$

Fact 2. The sequence $\{x_{n}\}$ constructed in Fact 1 is a multiplicative Cauchy in (X, d).

Indeed, $d(x_{n},x_{n+1})^{\delta } \le \phi (d(x_{n-1},x_{n}))^{\delta } \le \ldots \le \phi ^{n}(d(x_{0},x_{1})^{\delta }).$ Now, for $m,n\in \mathbb {N}$ such that $m>n$,

$$\begin{aligned} d(x_{n},x_{m})^{\delta }\le & {} d(x_{n},x_{n+1})^{\delta }\cdot d(x_{n+1},x_{n+2})^{\delta }\cdot \ldots \cdot d(x_{m-1},x_{m})^{\delta } \\\le & {} \phi ^{n}(d(x_{0},x_{1})^{\delta })\cdot \phi ^{n+1}(d(x_{0},x_{1}))^{\delta }\cdot \ldots \cdot \phi ^{m-1}(d(x_{0},x_{1})^{\delta }), \end{aligned}$$

which implies that $d(x_{n},x_{m})^{\delta }$ converges to 1 as $n,m\rightarrow \infty$. Thus, $\lim \limits _{n,m\rightarrow \infty }d(x_{n},x_{m})=1,$ that is, $\{x_{n}\}$ is a multiplicative Cauchy sequence in (X, d).

Fact 3. $F\left( S\right) \bigcap F\left( T\right) \ne \emptyset$.

Indeed, since (X, d) is complete multiplicative space, there exists u in X such that $\lim \limits _{n\rightarrow \infty }d(u,x_{n})=1.$

Now we show that $u\in \bigcap \limits _{i=1}^{p}A_{i}.$ From condition 4, and $x_{0}\in A_{i_{0}}$ for some $i_{0}\in \{1,2,\ldots ,p\}$, we can choose a subsequence $\{x_{n_{k}}\}$ in $A_{i_{0}}$ out of the sequence $\{x_{n}\}.$ Obviously, $\{x_{n_{k}}\} \subseteq S(A_{i_{0}})\subseteq A_{i_{0}+1}$. As $A_{i_{0}+1}\$is closed, so $u\in A_{i_{0}+1}$. Similarly, we can choose a subsequence $\{x_{n_{k}+1}\}$ in $A_{i_{0}+1}$ out of the sequence $\{x_{n}\}.$ Obviously, $\{x_{n_{k}+1}\} \subseteq T(A_{i_{0}+1})\subseteq A_{i_{0}+2}$. As $A_{i_{0}+2}\$is closed, so $u\in A_{i_{0}+2}$. Continuing this way, we obtain that $u\in \bigcap \limits _{i=1}^{p}A_{i}\$and hence $\bigcap _{i=1}^{p}A_{i}\ne \emptyset$.

Now we show that $S(z)=z$. Since $u\in \bigcap \limits _{i=1}^{p}A_{i},$ there exists some i in $\{1,2,\ldots ,p\}$ such that $u\in A_{i}$. Choose a subsequence $\{x_{2n_{k}+1}\}$ of $\{x_{n}\}$ with $x_{2n_{k}+1}\in A_{i+1}$. Assume on contrary that, $d(Su,u)>1$, then from (17.4), we have

$$\begin{aligned} d(Su,x_{2n_{k}+2})^{\delta }= & {} d(Su,Tx_{2n_{k}+1})^{\delta } \nonumber \\\le & {} \phi (d(u,x_{2n_{k}+1})^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(x_{2n_{k}+1},Tx_{2n_{k}+1})^{\gamma }) \nonumber \\= & {} \phi (d(u,x_{2n_{k}+1})^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(x_{2n_{k}+1},x_{2n_{k}+2})^{\gamma }), \end{aligned}$$

(17.6)

we deduce, by taking upper limit as $k\rightarrow \infty$ into account (17.6), that

$$\begin{aligned} d(Su,u)^{\delta }\le & {} \limsup \limits _{n\rightarrow \infty } \phi (d(u,x_{2n_{k}+1})^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(x_{2n_{k}+1},Tx_{2n_{k}+1})^{\gamma }) \\\le & {} \phi (d(u,u)^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(u,u)^{\gamma }) \le \phi (d(u,Su)^{\beta })<d(u,Su)^{\delta }, \end{aligned}$$

a contradiction. Hence $u=Su.$ Similarly, we have $u=Tu$ and thus u is the common fixed point of S and T.

Fact 4. The set $F\left( S\right) \bigcap F\left( T\right)$ is a singleton set.

Assume on contrary that $Su=Tu=u$ and $Sv=Tv=v$ but $u\ne v$. Then

$$\begin{aligned} d(u,v)^{\delta }= & {} d(Su,Tv)^{\delta } \le \phi (d(u,v)^{\alpha }\cdot d(u,Su)^{\beta }\cdot d(v,Tv)^{\gamma }) \\= & {} \phi (d(u,v)^{\alpha }\cdot d(u,u)^{\beta }\cdot d(v,v)^{\gamma }) \le \phi (d(u,v)^{\delta }), \end{aligned}$$

a contradiction because $d(u,v)^{\delta }>1.$ Hence $u=v$ and $F\left( S\right) \bigcap F\left( T\right) =\{u\}$.

Example 2

Let $X=\mathbb {R}$, and d a multiplicative metric on X defined by $d(x,y)=a^{\left| x-y\right| },$ where $a>1$ is a real number. For some $c>1$ , set $A_{1}=[-c,0]$, $A_{2}=[0,c]$ and $A_{3}=A_{1}$. Define $S,T:\bigcup \limits _{i=1}^{2}A_{i}\rightarrow \bigcup \limits _{i=1}^{2}A_{i}$ by $S\left( x\right) =-\dfrac{k_{1}}{c}x$ and $T\left( x\right) =-\dfrac{ k_{2}}{c}x,$ where $0<k_{1}\le k_{2}\le \frac{1}{2}c.$ Note that $S(A_{1})=[0,k_{1}] \subseteq [0,c]=A_{2}$ and $T(A_{2})=[-k_{2},0]\subseteq [-c,0]=A_{1}.$ $Y=A_{1}\bigcup A_{2}$ has a cyclic representation with respect to pair (S, T).

Define $\phi :[1,\infty )\rightarrow [1,\infty )$ by $\phi (t){=}\left\{ \begin{array}{l} t^{4/5},\text { \ \ if }t\in [1,a^{\delta c}), \\ a^{\delta c},\text { \ if }a^{\delta c}\le t, \end{array} \right.$ where $\delta \in \left( 0,\infty \right) .$ Clearly $\phi$ is upper semi-continuous and nondecreasing with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for all $t>1$.

We show that condition 4 is satisfied for $\alpha =\dfrac{\delta }{2},$ $\beta =\gamma =\dfrac{\delta }{4}$.

Now, for $x\in A_{1},$ $y\in A_{2},$

$$\begin{aligned} d(Sx,Ty)^{\delta }= & {} d(-\frac{k_{1}x}{c},-\frac{k_{2}y}{c})^{\delta } =a^{\frac{\delta }{c}(k_{2}y-k_{1}x)} \\\le & {} a^{\frac{4}{5}[\frac{\delta }{2}(y-x)+\frac{\delta }{4}(1+\frac{k_{1} }{c})\left| x\right| +\frac{\delta }{4}(1+\frac{k_{2}}{c})y]} \\= & {} \phi (a^{\alpha (y-x)+\beta (1+\frac{k_{1}}{c})\left| x\right| +\gamma (1+\frac{k_{2}}{c})y}) \\= & {} \phi (d(x,y)^{\alpha }\cdot d(x,Sx)^{\beta }\cdot d(y,Ty)^{\gamma }). \end{aligned}$$

When $x\in A_{2},$ $y\in A_{1},$

$$\begin{aligned} d(Sx,Ty)^{\delta }= & {} d(-\frac{k_{1}x}{c},-\frac{k_{2}y}{c})^{\delta } =a^{\frac{\delta }{c}(k_{1}x-k_{2}y)} \\\le & {} a^{\frac{4}{5}[\frac{\delta }{2}(x-y)+\frac{\delta }{4}(1+\frac{k_{1} }{c})x+\frac{\delta }{4}(1+\frac{k_{2}}{c})\left| y\right| ]} \\= & {} \phi (a^{\alpha (x-y)+\beta (1+\frac{k_{1}}{c})x+\gamma (1+\frac{k_{2}}{c} )\left| y\right| }) \\= & {} \phi (d(x,y)^{\alpha }\cdot d(x,Sx)^{\beta }\cdot d(y,Ty)^{\gamma }). \end{aligned}$$

Thus, S and T satisfy all the conditions of Theorem 4. Moreover, S and T have at most one common fixed point.

4 Applications

Let $\Omega =[0,1]$ be a bounded set in $\mathbb {R}$, $L^{2}(\Omega ),$ the set of comparable functions on $\Omega$ whose square is integrable on $\Omega .$ Consider the integral equations

$$\begin{aligned} \begin{array}{c} x(t)=\int \limits _{\Omega }q_{1}(t,s,x(s))ds+k(t), \\ y(t)=\int \limits _{\Omega }q_{2}(t,s,y(s))ds+k(t), \end{array} \end{aligned}$$

(17.7)

where $q_{1},q_{2}:\Omega \times \Omega \times \mathbb {R}\rightarrow \mathbb {R}$ and $k:\Omega \rightarrow \mathbb {R}$ be given continuous mappings. Altun and Simsek [8] obtained the common solution of integral equations (17.7) as an application in ordered Banach spaces. We shall study sufficient condition for existence of common solution of integral equations in framework of multiplicative metric spaces. Define $d:X\times X\rightarrow [1,\infty )$ by $d(x,y)=e^{\ \underset{t\in \Omega }{\sup }\left| x(t)-y(t)\right| }.$ Then $\left( X,d\right)$ is a complete multiplicative metric space. Suppose that there exists an upper semi-continuous nondecreasing function $\phi :[1,\infty )\rightarrow [1,\infty )$, with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for each $t>1$, such that

$$ \underset{t\in \Omega }{\sup }\int \limits _{\Omega }\left| q_{1}(t,s,u(s))-q_{2}(t,s,v(s))\right| ds\le \ln \left( \phi \left( e^{\ \delta \underset{t\in \Omega }{\sup }\left| u(t)-v(t)\right| }\right) ^{1/\delta }\right) $$

for each $s\in \Omega ,\$where $\delta \in \left( 0,\infty \right) .$

Then the integral equations (17.7) have a common solution in $L^{2}(\Omega )$.

Proof

$\square$Let $(Sx)(t)=\int \limits _{\Omega }q_{1}(t,s,x(s))ds+k(t)$ and $(Tx)(t)=\int \limits _{\Omega }q_{2}(t,s,x(s))ds+k(t).$ For all $x,y\in X$,

$$\begin{aligned} d(Sx,Ty)^{\delta }= & {} e^{\delta \underset{t\in \Omega }{ \sup } \left| (Sx)(t)-(Ty)(t)\right| } =e^{\delta \underset{t\in \Omega }{\sup }\mid \int \limits _{\Omega }q_{1}(t,s,x(s))ds-\int \limits _{\Omega }q_{2}(t,s,y(s))ds\mid } \\\le & {} e^{\delta \underset{t\in \Omega }{\sup }\int \limits _{\Omega }\left| q_{1}(t,s,x(s))-q_{2}(t,s,y(s))\right| ds} \le \phi \left( e^{\delta \underset{t\in \Omega }{\sup }\left| x(t)-y(t)\right| }\right) \\= & {} \phi (d(x,y)^{\alpha }\cdot d(x,Sx)^{\beta }\cdot d(y,Ty)^{\gamma }). \end{aligned}$$

Thus (17.1) is satisfied for $\delta =\alpha +\beta +\gamma \in \left( 0,\infty \right)$ with $\beta =\gamma =0$. Now Theorem 2 yields the common solutions of integral equations (17.7) in $L^{2}(\Omega ).$

Now, consider another integral equation

$$\begin{aligned} p(t,x(t))=\int \limits _{\Omega }q(t,s,x(s))ds, \end{aligned}$$

(17.8)

where $p:\Omega \times \mathbb {R}\rightarrow \mathbb {R}$ and $q:\Omega \times \Omega \times \mathbb {R}\rightarrow \mathbb {R}$ be two mappings. Hussain, Khan and Agarwal [17] obtained the solution of implicit integral equation (17.8) as an application of Ky Fan type fixed point theorem in ordered Banach spaces (see also, [15]). We shall study sufficient condition for existence of solution of integral equation in framework of multiplicative metric spaces.

We assume that there exists a function $G:\Omega \times \mathbb {R}\rightarrow \mathbb {R}_{>0}$ such that

• (i) $p(s,v(t))\ge \int \limits _{\Omega }q(t,s,u(s))ds\ge G(s,v(t))$ for each $s,t\in \Omega$.

• (ii) $\underset{t\in \Omega }{\sup }[p(s,v(t))-G(s,v(t))]\le \ln \left( \phi (e^{\ \delta \underset{t\in \Omega }{\sup }\left| p(s,v(t))-v(t)\right| })\right) ^{1/\delta },$ for each $s\in \Omega ,$ where $\phi :[1,\infty )\rightarrow [1,\infty )$ is an upper semi-continuous and nondecreasing function with $\sum \limits _{n=1}^{\infty }\phi ^{n}(t)$ convergent for each $t>1$ and $\delta \in \left( 0,\infty \right) .$

Then integral equation (17.8) has a solution in $L^{2}(\Omega )$.

Proof

$\square$Define $(Sx)(t)=p(t,x(t))$ and $(Tx)(t)=\int \limits _{\Omega }q(t,s,x(s))ds.$ Now

$$\begin{aligned} d(Sx,Ty)^{\delta }= & {} e^{\ \delta \underset{t\in \Omega }{\sup } \left| (Sx)(t)-(Ty)(t)\right| } =e^{\ \delta \underset{t\in \Omega }{\sup }\mid p(t,x(t))-\int \limits _{\Omega }q(t,s,y(t))dt\mid } \\\le & {} e^{\ \delta \underset{t\in \Omega }{\sup }\left| p(t,x(t))-T(t,x(t))\right| } \le \phi (e^{\ \beta \underset{t\in \Omega }{\sup }\left| p(t,x(t))-x(t)\right| }) =\phi \left( d(x,Sx)^{\beta }\right) . \end{aligned}$$

Thus, for all $x,y\in X$, $d(Sx,Ty)^{\delta }\le \phi (d(x,y)^{\alpha }\cdot d(x,Sx)^{\beta }\cdot d(y,Ty)^{\gamma }),$ where $\delta =\beta \in \left( 0,\infty \right)$. Now Theorem 2 yields the solution of integral equation (17.8) in $L^{2}(\Omega ).$