© Wuhan University 2023

0 Introduction

Fixed point theory provides essential tools for solving the existence of unique solutions to many problems in mathematics and applications. The Banach contraction mapping principle, one of the fundamental and most widely applied fixed point theorems, has been generalized. These expansions generally proceed along two lines: one is extending the domain of mappings, and another is considering a more general contractive condition on mappings; the fixed points or common fixed points of mappings satisfying certain contraction conditions on a specific space have received much research^[1-6]. Bashirov et al introduced the notion of multiplicative metric spaces and studied some fundamental theorems of multiplicative calculus${}^{\left[\mathrm{7}\right]}$. $\ddot{\mathrm{O}}\mathrm{z}\mathrm{a}\mathrm{v}\mathrm{s}\mathrm{a}\mathrm{r}$ and Cevikel studied some topological properties of multiplicative metric spaces and proved an analogous result to the Banach contraction principle in multiplicative metric spaces${}^{\left[\mathrm{8}\right]}$. Since then, some fixed-point and common fixed-point results have been obtained in multiplicative metric spaces^[9-13].

This paper shows some common fixed-point results for two mappings satisfying specific multiplicative contraction conditions with exponents of fraction expression in multiplicative metric spaces.

1 Preliminaries

Definition 1   ${}^{\left[\mathrm{7}\right]}$ Let $X$ be a nonempty set. If mapping $d:X\times X\to {\mathbb{R}}^{+}$ satisfies the following conditions:

(1) $d\left(x,y\right)\ge \mathrm{1}$ for all $x,y\in X$, and $d\left(x,y\right)=\mathrm{1}$ if and only if $x=y$;

(2) $d\left(x,y\right)=d\left(y,x\right)$ for all $x,y\in X$;

(3) $d\left(x,z\right)\le d\left(x,y\right)d\left(y,z\right)$ for all $x,y,z\in X$(multiplicative triangle inequality).

We say $d$ is a multiplicative metric on $X$, and $\left(X,d\right)$ a multiplicative metric space.

Definition 2   ${}^{\left[\mathrm{8}\right]}$ Let $\left(X,d\right)$ be a multiplicative metric space, $\left\{x_n\right\}\subset X,x\in X$, $\left\{x_n\right\}$ is said to be multiplicative convergent to $x$, if for arbitrary $\epsilon >\mathrm{1}$, there exists a natural number $N$ such that $d\left(x_n,x\right)<\epsilon$ for all $n>N$, denoted by $x_n\to x\left(n\to \mathrm{\infty }\right)$.

Definition 3   ${}^{\left[\mathrm{8}\right]}$ Let $\left(X,d\right)$ be a multiplicative metric space, $\left\{x_n\right\}\subset X$, $\left\{x_n\right\}$ is called a multiplicative Cauchy sequence, if for arbitrary $\epsilon >\mathrm{1}$, there exists a natural number $N$ such that $d\left(x_n,x_m\right)<\epsilon$ for all $n,m>N$.

We say that $\left(X,d\right)$ is complete if every multiplicative Cauchy sequence in $\left(X,d\right)$ is multiplicative convergent to $x\in X$.

Proposition 1   ${}^{\left[\mathrm{8}\right]}$ The uniqueness of the limit holds for a convergent sequence in a multiplicative metric space.

Definition 4   ${}^{\left[\mathrm{7}\right]}$ Multiplicative absolute value function $|\cdot {|}^{\mathrm{*}}:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ is defined as:

$|x{|}^{\mathrm{*}}=\{\begin{array}{c}x,x\ge \mathrm{1}\\ \frac{\mathrm{1}}{x},x<\mathrm{1}\end{array}$

Remark 1   Multiplicative absolute value function $|\cdot {|}^{\mathrm{*}}:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ satisfies :

1) $|x{|}^{\mathrm{*}}\ge \mathrm{1}$; 2) $|\frac{x}{x_{\mathrm{0}}}{|}^{\mathrm{*}}<a\iff \frac{x_{\mathrm{0}}}{a}<x<ax\begin{array}{cc}& ,x_{\mathrm{0}},x\in {\mathbb{R}}^{+}\end{array}$.

Proposition 2   ${}^{\left[\mathrm{8}\right]}$ $\left({\mathbb{R}}^{+},|\cdot {|}^{\mathrm{*}}\right)$ is a complete multiplicative metric space.

Proposition 3   ${}^{\left[\mathrm{8}\right]}$ Let $\left(X,d\right)$ be a multiplicative metric space, $\left\{x_n\right\}\subset X,x\in X$, $\left\{x_n\right\}$ multiplicative converges to $x$ if and only if $d\left(x_n,x\right)\to \mathrm{1}\left(n\to \mathrm{\infty }\right)$.

$\left\{x_n\right\}$ is a multiplicative Cauchy sequence if and only if$d\left(x_n,x_m\right)\to \mathrm{1}\left(n,m\to \mathrm{\infty }\right)$.

Definition 5   ${}^{\left[\mathrm{7}\right]}$ The multiplicative derivative of a function $f:\mathbb{R}\to \mathbb{R}$ is defined by $\underset{h\to \mathrm{0}}{\mathrm{l}\mathrm{i}\mathrm{m}}{\left(\frac{f\left(x+h\right)}{f\left(x\right)}\right)}^{\frac{\mathrm{1}}{h}}$. Denoted it by $f^{\mathrm{*}}\left(x\right)$.

If $f\left(x\right)$ is a positive function and its derivative at $x$ exists, then $f^{\mathrm{*}}\left(x\right)={\mathrm{e}}^{{\left(\mathrm{l}\mathrm{n}f\left(x\right)\right)}^{\mathrm{\text{'}}}}$.

Definition 6   ${}^{\left[\mathrm{7}\right]}$ Let $f:\left[a,b\right]\to \mathbb{R}$ be a positive bounded function, $P=\left\{x_{\mathrm{0}},x_{\mathrm{1}},\cdots ,x_n\right\}$ be a partition of $\left[a,b\right]$, and ${\xi }_i\in \left[x_{i-\mathrm{1}},x_i\right]$. The function $f$ is said to be integral in the multiplicative sense if there exists a number $P$ having the properties: for every $\epsilon >\mathrm{0},$ there exists a partition $P_{\epsilon }$ of $\left[a,b\right]$ such that $|\stackrel{n}{\prod _{i=\mathrm{1}}}f{\left({\xi }_i\right)}^{\left(x_i-x_{i-\mathrm{1}}\right)}-P|<\epsilon$ for every refinement $P$ of $P_{\epsilon }$ independently on selection of the numbers ${\xi }_i\in \left[x_{i-\mathrm{1}},x_i\right]$$\left(i=\mathrm{1,2},\cdots ,n\right)$. $P$ is called the multiplicative integral of $f$ on $\left[a,b\right]$, we denote it with ${\int }_a^{b}f{\left(x\right)}^{\mathrm{d}x}$.

It is easily seen that if $f$ is positive and Riemann integral on $\left[a,b\right]$, then ${\int }_a^{b}f{\left(x\right)}^{\mathrm{d}x}={\mathrm{e}}^{{\int }_a^b\mathrm{l}\mathrm{n}f\left(x\right)\mathrm{d}x}$.

Let $X$ be a nonempty set, recall that mappings $f,g:X\to X$ are weakly compatible if, for every $x\in X$, $fgx=ggx$ holds whenever $fx=gx$. If $f$ and $g$ are weakly compatible and have an unique point of coincidence $w=fx=gx$, then $w$ is the unique common fixed point of $f$ and $g$.

2 Common Fixed Points

Theorem 1   Let $\left(X,d\right)$ be a multiplicative metric space, mappings $f,g:X\to X$ satisfy: for all $x,y\in X$,

$d\left(fx,fy\right)\le d{\left(gx,gy\right)}^{{\lambda }_{\mathrm{1}}}d{\left(fx,gx\right)}^{{\lambda }_{\mathrm{2}}}d{\left(fy,gy\right)}^{{\lambda }_{\mathrm{3}}}d{\left(fx,gy\right)}^{{\lambda }_{\mathrm{4}}}d{\left(fy,gx\right)}^{{\lambda }_{\mathrm{5}}}$(1)

holds, where ${\lambda }_i\in \left(\mathrm{0,1}\right)\left(i=\mathrm{1,2},\mathrm{3,4},\mathrm{5}\right)$, ${\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}<\mathrm{1},{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{5}}<\mathrm{1},$${\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{3}}+\mathrm{2}{\lambda }_{\mathrm{4}}<\mathrm{1}$.

If $g\left(X\right)$ is a complete subspace of $X$, $f\left(X\right)\subset g\left(X\right)$ and $f$ and $g$ are weakly compatible, then $f$ and $g$ have an unique common fixed point.

Proof   Let $x_{\mathrm{0}}$ is an arbitrary point in $X$, since $f\left(X\right)\subset g\left(X\right)$, there exists $x_{\mathrm{1}},x_{\mathrm{2}}\in X$ such that $f{x}_{\mathrm{0}}=g{x}_{\mathrm{1}},f{x}_{\mathrm{1}}=g{x}_{\mathrm{2}}$. Continuing this process, we can obtain a sequence $\left\{x_n\right\}\subset X$ such that $f{x}_n=g{x}_{n+\mathrm{1}}\left(n=\mathrm{0,1},\mathrm{2},\cdots \right)$. From (1), we have

$\begin{array}{l}d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)=d\left(f{x}_n,f{x}_{n-\mathrm{1}}\right)\le d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{1}}}d{\left(f{x}_n,g{x}_n\right)}^{{\lambda }_{\mathrm{2}}}d{\left(f{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{3}}}d{\left(f{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{4}}}d{\left(f{x}_{n-\mathrm{1}},g{x}_n\right)}^{{\lambda }_{\mathrm{5}}}\\ =d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{1}}}d{\left(g{x}_{n+\mathrm{1}},g{x}_n\right)}^{{\lambda }_{\mathrm{2}}}d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{3}}}d{\left(g{x}_{n+\mathrm{1}},g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{4}}}d{\left(g{x}_n,g{x}_n\right)}^{{\lambda }_{\mathrm{5}}}\\ \le d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{3}}}d{\left(g{x}_{n+\mathrm{1}},g{x}_n\right)}^{{\lambda }_{\mathrm{2}}}d{\left(g{x}_{n+\mathrm{1}},g{x}_n\right)}^{{\lambda }_{\mathrm{4}}}d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{4}}}\\ =d{\left(g{x}_{n+\mathrm{1}},g{x}_n\right)}^{{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{4}}}d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}\end{array}$

which implies that $d{\left(g{x}_{n+\mathrm{1}},g{x}_n\right)}^{\mathrm{1}-{\lambda }_{\mathrm{2}}-{\lambda }_{\mathrm{4}}}\le d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}$, and so $d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)\le d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\frac{{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}{\mathrm{1}-{\lambda }_{\mathrm{2}}-{\lambda }_{\mathrm{4}}}}$.

Let $q=\frac{{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}{\mathrm{1}-{\lambda }_{\mathrm{2}}-{\lambda }_{\mathrm{4}}}$, then

$d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)\le d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^q\le d{\left(g{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{2}}\right)}^{q^{\mathrm{2}}}\le \cdots \le d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{q^n}$(2)

for all $n=\mathrm{1,2},\cdots$. Since ${\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{3}}+\mathrm{2}{\lambda }_{\mathrm{4}}<\mathrm{1}$, we get $q<\mathrm{1}$.

For arbitrary natural numbers $n,m,n>m$, from the multiplicative inequality and (2), we have

$\begin{array}{l}d\left(g{x}_n,g{x}_m\right)\le d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)d\left(g{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{2}}\right)\cdots d\left(g{x}_{m+\mathrm{1}},g{x}_m\right)\le d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{q^{n-\mathrm{1}}}d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{q^{n-\mathrm{2}}}\cdots d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{q^m}\\ =d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{q^{n-\mathrm{1}}+q^{n-\mathrm{2}}+\cdots +q^m}=d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{\frac{q^m\left(\mathrm{1}-q^{n-m-\mathrm{1}}\right)}{\mathrm{1}-q}}<d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{\frac{q^m}{\mathrm{1}-q}}\end{array}$

For every $\epsilon >\mathrm{1}$, by taking $N=\left[\frac{\mathrm{l}\mathrm{n}\frac{\left(\mathrm{1}-q\right)\mathrm{l}\mathrm{n}\epsilon }{\mathrm{l}\mathrm{n}\left(d\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)\right)}}{\mathrm{l}\mathrm{n}q}\right]$, we have${\left|\frac{d\left(g{x}_n,g{x}_m\right)}{\mathrm{1}}\right|}^{\mathrm{*}}<\epsilon$ for $n>N$, that is, $\left\{g{x}_n\right\}$ is a multiplicative Cauchy sequence in $g\left(X\right)$.

Since $g\left(X\right)$ is a complete, there exists $p\in X$ such that $g{x}_n\to gp$. From (1) and the multiplicative inequality,

$\begin{array}{l}d\left(g{x}_n,fp\right)=d\left(f{x}_{n-\mathrm{1}},fp\right)\le d{\left(g{x}_{n-\mathrm{1}},gp\right)}^{{\lambda }_{\mathrm{1}}}d{\left(f{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{2}}}d{\left(fp,gp\right)}^{{\lambda }_{\mathrm{3}}}d{\left(f{x}_{n-\mathrm{1}},gp\right)}^{{\lambda }_{\mathrm{4}}}d{\left(fp,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{5}}}\\ \le d{\left(g{x}_{n-\mathrm{1}},gp\right)}^{{\lambda }_{\mathrm{1}}}d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{2}}}d{\left(fp,g{x}_n\right)}^{{\lambda }_{\mathrm{3}}}d{\left(g{x}_n,gp\right)}^{{\lambda }_{\mathrm{3}}}d{\left(g{x}_n,gp\right)}^{{\lambda }_{\mathrm{4}}}d{\left(fp,g{x}_n\right)}^{{\lambda }_{\mathrm{5}}}d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{5}}}\end{array}$

We get that $d{\left(g{x}_n,fp\right)}^{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}\le d{\left(g{x}_{n-\mathrm{1}},gp\right)}^{{\lambda }_{\mathrm{1}}}d{\left(g{x}_n,gp\right)}^{{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{5}}}$, that is

$d\left(g{x}_n,fp\right)\le d{\left(g{x}_{n-\mathrm{1}},gp\right)}^{\frac{{\lambda }_{\mathrm{1}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}d{\left(g{x}_n,gp\right)}^{\frac{{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\frac{{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{5}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}$

Since $g{x}_n\to gp$, for every $\epsilon >\mathrm{1}$, there exists a natural number $N$ such that

${\left|\frac{d{\left(g{x}_{n-\mathrm{1}},gp\right)}^{\frac{{\lambda }_{\mathrm{1}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}}{\mathrm{1}}\right|}^{\mathrm{*}}<\sqrt[\mathrm{3}]{\epsilon }{\begin{array}{cc},& \left|\frac{d{\left(g{x}_n,gp\right)}^{\frac{{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}}{\mathrm{1}}\right|\end{array}}^{\mathrm{*}}<\sqrt[\mathrm{3}]{\epsilon }{\begin{array}{cc},& \left|\frac{d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\frac{{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{5}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}}{\mathrm{1}}\right|\end{array}}^{\mathrm{*}}<\sqrt[\mathrm{3}]{\epsilon }$

for $n>N$. Therefore, we obtain that

${\left|\frac{d\left(g{x}_n,fp\right)}{\mathrm{1}}\right|}^{\mathrm{*}}=d\left(g{x}_n,fp\right)\le d{\left(g{x}_{n-\mathrm{1}},gp\right)}^{\frac{{\lambda }_{\mathrm{1}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}d{\left(g{x}_n,gp\right)}^{\frac{{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\frac{{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{5}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}$

$={\left|\frac{d{\left(g{x}_{n-\mathrm{1}},gp\right)}^{\frac{{\lambda }_{\mathrm{1}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}}{\mathrm{1}}\right|}^{\mathrm{*}}{\left|\frac{d{\left(g{x}_n,gp\right)}^{\frac{{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}}{\mathrm{1}}\right|}^{\mathrm{*}}{\left|\frac{d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\frac{{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{5}}}{\mathrm{1}-{\lambda }_{\mathrm{3}}-{\lambda }_{\mathrm{5}}}}}{\mathrm{1}}\right|}^{\mathrm{*}}<\sqrt[\mathrm{3}]{\epsilon }\cdot \sqrt[\mathrm{3}]{\epsilon }\cdot \sqrt[\mathrm{3}]{\epsilon }=\epsilon$

for $n>N$, which implies that $g{x}_n\to fp$. From the uniqueness of the limit, $fp=gp$.

If there exists another point $p^{\mathrm{*}}\in X$ such that $f{p}^{\mathrm{*}}=g{p}^{\mathrm{*}}$. From (1), we get

$\begin{array}{l}d\left(gp,g{p}^{\mathrm{*}}\right)=d\left(fp,f{p}^{\mathrm{*}}\right)\le d{\left(gp,g{p}^{\mathrm{*}}\right)}^{{\lambda }_{\mathrm{1}}}d{\left(fp,gp\right)}^{{\lambda }_{\mathrm{2}}}d{\left(f{p}^{\mathrm{*}},g{p}^{\mathrm{*}}\right)}^{{\lambda }_{\mathrm{3}}}d{\left(fp,g{p}^{\mathrm{*}}\right)}^{{\lambda }_{\mathrm{4}}}d{\left(f{p}^{\mathrm{*}},gp\right)}^{{\lambda }_{\mathrm{5}}}\\ =d{\left(gp,g{p}^{\mathrm{*}}\right)}^{{\lambda }_{\mathrm{1}}}d{\left(gp,g{p}^{\mathrm{*}}\right)}^{{\lambda }_{\mathrm{4}}}d{\left(g{p}^{\mathrm{*}},gp\right)}^{{\lambda }_{\mathrm{5}}}=d{\left(gp,g{p}^{\mathrm{*}}\right)}^{{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}}.\end{array}$

Since ${\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}<\mathrm{1}$, the above inequality implies that $d\left(gp,g{p}^{\mathrm{*}}\right)=\mathrm{1}$, so $gp=g{p}^{\mathrm{*}}$. Since $f$ and $g$ are weakly compatible, $fp\left(=gp\right)$ is a unique common fixed point of $f$ and $g$.

Corollary 1   Let $\left(X,d\right)$ be a multiplicative metric space, mappings $f,g:X\to X$ satisfy the followings: for arbitrary $x,y\in X$, inequality (1) holds, where ${\lambda }_i\left(i=\mathrm{1,2},\mathrm{3,4},\mathrm{5}\right)$ are non-negative real numbers and ${\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}\le \mathrm{1},$${\lambda }_{\mathrm{3}}<{\lambda }_{\mathrm{2}},$${\lambda }_{\mathrm{4}}<{\lambda }_{\mathrm{5}}$. If $g\left(X\right)$ is a complete subspace of $X$, $f\left(X\right)\subset g\left(X\right)$ and $f$ and $g$ are weakly compatible. Then $f$ and $g$ have a unique common fixed point.

Proof   Since ${\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}\le \mathrm{1},{\lambda }_{\mathrm{3}}<{\lambda }_{\mathrm{2}},$${\lambda }_{\mathrm{4}}<{\lambda }_{\mathrm{5}}$ , we get

${\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}<{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}\le \mathrm{1},{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{5}}<{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}\le \mathrm{1},{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{3}}+\mathrm{2}{\lambda }_{\mathrm{4}}<{\lambda }_{\mathrm{1}}+{\lambda }_{\mathrm{2}}+{\lambda }_{\mathrm{3}}+{\lambda }_{\mathrm{4}}+{\lambda }_{\mathrm{5}}\le \mathrm{1}$

The conditions in Theorem 1 are satisfied. The conclusion is true from Theorem 1.

Corollary 2   Let $\left(X,d\right)$ be a multiplicative metric space, mappings $f,g:X\to X$ satisfy the following: for arbitrary $x,y\in X$,

$d\left(fx,fy\right)\le d{\left(gx,gy\right)}^{\alpha }{\left(d\left(fx,gx\right)d\left(fy,gy\right)\right)}^{\beta }{\left(d\left(fx,gy\right)d\left(fy,gx\right)\right)}^{\gamma }$

holds, where $\alpha ,\beta ,\gamma$ are non-negative real numbers and $\alpha +\mathrm{2}\beta +\mathrm{2}\gamma <\mathrm{1}$. If $g\left(X\right)$ is a complete subspace of $X$, $f\left(X\right)\subset g\left(X\right)$ and $f$ and $g$ are weakly compatible. Then $f$ and $g$ have a unique common fixed point.

Proof   Let ${\lambda }_{\mathrm{1}}=\alpha ,{\lambda }_{\mathrm{2}}={\lambda }_{\mathrm{3}}=\beta ,{\lambda }_{\mathrm{4}}={\lambda }_{\mathrm{5}}=\gamma$, it is easy to see that the conclusion is true from Theorem 1.

Corollary 3   ${}^{\left[\mathrm{6}\right]}$ Let $\left(X,d\right)$ be a multiplicative metric space, mappings $f:X\to X$ satisfy: for arbitrary $x,y\in X$,

$d\left(fx,fy\right)\le d{\left(x,y\right)}^{\lambda }$

holds, where $\mathrm{0}<\lambda <\mathrm{1}$. Then $f$ has a unique fixed point.

Proof   By taking $g=I_X$ and ${\lambda }_{\mathrm{1}}=\lambda ,{\lambda }_{\mathrm{2}}={\lambda }_{\mathrm{3}}={\lambda }_{\mathrm{4}}={\lambda }_{\mathrm{5}}=\mathrm{0}$ in Theorem 1, we see that $f$ has a unique fixed point.

Theorem 2   Let $\left(X,d\right)$ be a multiplicative metric space, mappings $f,g:X\to X$ satisfying followings: for arbitrary $x,y\in X$,

$d\left(fx,fy\right)\le \mathrm{m}\mathrm{a}\mathrm{x}\left\{d{\left(gx,gy\right)}^{\alpha },{\left(d\left(fx,gx\right)d\left(fy,gy\right)\right)}^{\frac{\beta }{d\left(gx,gy\right)}},{\left(d\left(fy,gx\right)d\left(fx,gy\right)\right)}^{\frac{\gamma d\left(fx,gy\right)}{d\left(fx,fy\right)d\left(gx,gy\right)d\left(fy,gy\right)}}\right\}$(3)

holds, where $\alpha ,\beta ,\gamma$ are non-negative real numbers and,$\mathrm{2}\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}<\mathrm{1}$. If $g\left(X\right)$ is a complete subspace of $X$, $f\left(X\right)\subset g\left(X\right)$ , $f$ and $g$ are weakly compatible, then $f$ and $g$ have a unique common fixed point.

Proof   As we do in the proof of Theorem 1, we can obtain a sequence $\left\{x_n\right\}\subset X$ such that $f{x}_n=g{x}_{n+\mathrm{1}}\left(n=\mathrm{0,1},\mathrm{2},\cdots \right)$. From (3), we have

$\begin{array}{l}d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)=d\left(f{x}_n,f{x}_{n-\mathrm{1}}\right)\\ \le \mathrm{m}\mathrm{a}\mathrm{x}\left\{d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\alpha },{\left(d\left(f{x}_n,g{x}_n\right)d\left(f{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{1}}\right)\right)}^{\frac{\beta }{d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}},{\left(d\left(f{x}_{n-\mathrm{1}},g{x}_n\right)d\left(f{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\frac{\gamma d\left(f{x}_n,g{x}_{n-\mathrm{1}}\right)}{d\left(f{x}_n,f{x}_{n-\mathrm{1}}\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)d\left(f{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{1}}\right)}}\right\}.\end{array}$

Since

$\begin{array}{l}{\left(d\left(f{x}_n,g{x}_n\right)d\left(f{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{1}}\right)\right)}^{\frac{\beta }{d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}}={\left(d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\frac{\beta }{d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}}\\ \le {\left(d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\beta }\le {\left(d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}},\end{array}$

$\begin{array}{l}{\left(d\left(f{x}_{n-\mathrm{1}},g{x}_n\right)d\left(f{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\frac{\gamma d\left(f{x}_n,g{x}_{n-\mathrm{1}}\right)}{d\left(f{x}_n,f{x}_{n-\mathrm{1}}\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)d\left(f{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{1}}\right)}}\le {\left(d\left(g{x}_n,g{x}_n\right)d\left(g{x}_{n+\mathrm{1}},g{x}_{n-\mathrm{1}}\right)\right)}^{\frac{\gamma d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}{d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}}\\ ={\left(d\left(g{x}_n,g{x}_n\right)d\left(g{x}_{n+\mathrm{1}},g{x}_{n-\mathrm{1}}\right)\right)}^{\frac{\gamma }{d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}}\le {\left(d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}},\end{array}$

and

$d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\alpha }\le {\left(d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\alpha }\le {\left(d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}}$

We get

$d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)\le {\left(d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)\right)}^{\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}}$

and so

$d{\left(g{x}_{n+\mathrm{1}},g{x}_n\right)}^{\mathrm{1}-\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}}\le d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}}$

Let $\lambda =\frac{\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}}{\mathrm{1}-\mathrm{m}\mathrm{a}\mathrm{x}\left\{\alpha ,\beta ,\gamma \right\}}$, then $\mathrm{0}<\lambda <\mathrm{1}$ and

$d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)\le d{\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)}^{\lambda }$(4)

From (4), we get

$d\left(g{x}_{n+\mathrm{1}},g{x}_n\right)\le d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{{\lambda }^n}$(5)

For natural numbers $n,m,n>m$, from the multiplicative triangle inequality and (5), we get

$\begin{array}{l}d\left(g{x}_n,g{x}_m\right)\le d\left(g{x}_n,g{x}_{n-\mathrm{1}}\right)d\left(g{x}_{n-\mathrm{1}},g{x}_{n-\mathrm{2}}\right)\cdots d\left(g{x}_{m+\mathrm{1}},g{x}_m\right)\\ \le d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{{}^{{\lambda }^{n-\mathrm{1}}+{\lambda }^{n-\mathrm{2}}+\cdots +{\lambda }^m}}\le d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{\frac{{\lambda }^m\left(\mathrm{1}-{\lambda }^{n-m-\mathrm{1}}\right)}{\mathrm{1}-\lambda }}\le d{\left(g{x}_{\mathrm{1}},g{x}_{\mathrm{0}}\right)}^{\frac{{\lambda }^m}{\mathrm{1}-\lambda }}.\end{array}$