If
\(\kappa_{2n^{*}}=\kappa_{2n^{*}+1}\) for some
\(n^{*}\in \mathbb{N} _{0}\), then
\(\kappa_{2n^{*}}\) is a common fixed point of
Φ and
Ψ. If
\(\kappa_{2n}\neq \kappa_{2n+1}\), for all
\(n\in \mathbb{N} _{0}\). Then
\(d(\varPhi \kappa_{2n} ,\varPsi \kappa_{2n+1}) > 0\), for all
\(n\in \mathbb{N}_{0}\) and using assumption
\((C_{2})\), we obtain
$$ \begin{aligned} &(\kappa_{1}, \kappa_{2})= ( \varPhi \kappa_{0},\varPsi \kappa_{1})\in \mathcal{R}, \\ &(\kappa_{2}, \kappa_{3})= (\varPsi \kappa_{1}, \varPhi \kappa_{2})\in \mathcal{R}, \\ &(\kappa_{3}, \kappa_{4})= (\varPhi \kappa_{2}, \varPsi \kappa_{3})\in \mathcal{R}, \\ &(\kappa_{4}, \kappa_{5})= (\varPsi \kappa_{3}, \varPhi \kappa_{4})\in \mathcal{R}, \\ \vdots \end{aligned} $$
In general,
$$ \begin{aligned} &(\kappa_{2n}, \kappa_{2n+1})=( \varPsi \kappa_{2n-1},\varPhi \kappa_{2n}) \in \mathcal{R}. \end{aligned} $$
Thus
\((\kappa_{2n}, \kappa_{2n+1})\in \mathcal{X}\), for all
\(n\in \mathbb{N}_{0}\). Now, taking in (
3.1)
\(\kappa_{1}= \kappa_{2n}\) and
\(\kappa_{2}=\kappa_{2n-1}\), we have
$$ \begin{aligned} \mathsf{F}\bigl(d(\kappa_{2n}, \kappa_{2n+1})\bigr) &=\mathsf{F}\bigl(d(\kappa_{2n+1}, \kappa_{2n})\bigr) \\ &=\mathsf{F}\bigl(d(\varPhi \kappa_{2n},\varPsi \kappa_{2n-1}) \bigr) \\ &\leq \mathsf{F} \biggl( d(\kappa_{2n},\kappa_{2n-1})+ \frac{d(\kappa _{2n-1},\varPhi \kappa_{2n})d(\kappa_{2n},\varPsi \kappa_{2n-1})}{1 + d( \kappa_{2n},\kappa_{2n-1})} \biggr) -\xi \\ &=\mathsf{F} \biggl( d(\kappa_{2n},\kappa_{2n-1})+ \frac{d(\kappa_{2n-1}, \kappa_{2n+1})d(\kappa_{2n},\kappa_{2n})}{1 + d(\kappa_{2n},\kappa _{2n-1})} \biggr) -\xi \\ &=\mathsf{F} \bigl( d(\kappa_{2n},\kappa_{2n-1})\bigr) ) -\xi, \end{aligned} $$
for all
\(n\in \mathbb{N}\). Similarly, setting
\(\kappa_{1}=\kappa_{2n}\) and
\(\kappa_{2}=\kappa_{2n+1}\) in (
3.1), we can write
$$\begin{aligned} \mathsf{F}\bigl(d(\kappa_{2n+1}, \kappa_{2n+2})\bigr) =&\mathsf{F}\bigl(d(\varPhi \kappa _{2n}, \varPsi \kappa_{2n+1})\bigr) \\ \leq& \mathsf{F} \biggl( d(\kappa_{2n},\kappa_{2n+1})+ \frac{d(\kappa _{2n+1},\varPhi \kappa_{2n})d(\kappa_{2n},\varPsi \kappa_{2n+1})}{1 + d( \kappa_{2n},\kappa_{2n+1})} \biggr) -\xi \\ =&\mathsf{F} \biggl( d(\kappa_{2n},\kappa_{2n+1})+ \frac{d(\kappa_{2n+1}, \kappa_{2n+1})d(\kappa_{2n},\kappa_{2n+2})}{1 + d(\kappa_{2n}, \kappa_{2n+1})} \biggr) -\xi \\ =&\mathsf{F} \bigl( d(\kappa_{2n},\kappa_{2n+1})\bigr) ) -\xi. \end{aligned}$$
In general,
$$ \begin{aligned} \mathsf{F}\bigl(d(\kappa_{n}, \kappa_{n+1})\bigr) &\leq \mathsf{F} \bigl( d(\kappa _{n-1}, \kappa_{n})\bigr) ) -\xi, \end{aligned} $$
(3.3)
where
\(n\in \mathbb{N}\). Now, using inequality (
3.3), we can write
$$\begin{aligned} \mathsf{F}\bigl(d(\kappa_{n},\kappa_{n+1})\bigr) \leq & \mathsf{F}\bigl(d(\kappa_{n-1}, \kappa_{n})\bigr)-\xi \\ \leq & \mathsf{F}\bigl(d(\kappa_{n-2},\kappa_{n-1})\bigr)-2 \xi \\ \leq & \mathsf{F}\bigl(d(\kappa_{n-3},\kappa_{n-2})\bigr)-3 \xi \\ \leq & \mathsf{F}\bigl(d(\kappa_{n-4},\kappa_{n-3})\bigr)-4 \xi \\ \vdots& \\ \leq & \mathsf{F}\bigl(d(\kappa_{0},\kappa_{1})\bigr)-n \xi, \end{aligned}$$
that is,
$$ \begin{aligned} \mathsf{F}\bigl(d(\kappa_{n}, \kappa_{n+1})\bigr)\leq & \mathsf{F}\bigl(d(\kappa_{0}, \kappa_{1})\bigr)-n\xi, \end{aligned} $$
(3.4)
where
\(n\in \mathbb{N}\). Thus
\(\underset{n\rightarrow \infty }{\lim } \mathsf{F}(d(\kappa_{n},\kappa_{n+1}))=-\infty \), by condition (2) of Definition
2.1, we get
$$ \underset{n\rightarrow \infty }{\lim }d(\kappa_{n}, \kappa_{n+1})=0 \quad \text{or}\quad \underset{n\rightarrow \infty }{ \lim }d(\kappa_{n},\kappa_{n+1})=0^{+}. $$
(3.5)
From condition
\((3)\) of Definition
2.1, we can find
\(\varepsilon \in \mathopen{]}0,1[\) such that
$$ \underset{n\rightarrow \infty }{\lim } \bigl( d( \kappa_{n},\kappa_{n+1}) \bigr) ^{\varepsilon }\mathsf{F} \bigl( d(\kappa_{n},\kappa_{n+1}) \bigr) =0. $$
(3.6)
Using (
3.4), we have
$$ \begin{aligned} \bigl( d(\kappa_{n}, \kappa_{n+1}) \bigr) ^{\varepsilon } \bigl( \mathsf{F}\bigl(d( \kappa_{n},\kappa_{n+1})\bigr)-\mathsf{F}\bigl(d( \kappa_{0},\kappa _{1})\bigr) \bigr)\leq & - \bigl( d( \kappa_{n},\kappa_{n+1}) \bigr) ^{\varepsilon }n\xi \leq 0. \end{aligned} $$
(3.7)
Taking the limit
\(n\rightarrow \infty \) in (
3.7), and using (
3.5) and (
3.6), we get
$$ \begin{aligned} \underset{n\rightarrow \infty }{\lim }n \bigl( d(\kappa_{n},\kappa_{n+1}) \bigr) ^{\varepsilon }=0. \end{aligned} $$
(3.8)
Hence there exists
\(n_{0}\in \mathbb{N}\) such that, for all
\(n\geq n_{0}\),
\(n ( d(\kappa_{n},\kappa_{n+1}) ) ^{\varepsilon }\leq 1\). Consequently, we have
$$ \begin{aligned} d(\kappa_{n}, \kappa_{n+1})\leq \frac{1}{n^{\frac{1}{\varepsilon }}}, \quad \forall n\geq n_{0}. \end{aligned} $$
(3.9)
Now, we show that
\(\{\kappa_{n}\}\) is a Cauchy sequence. For this purpose, using (
3.9) and the triangular inequality, for all
\(m>n\geq n_{1}\), we have
$$ \begin{aligned} d(\kappa_{n},\kappa_{m}) &\leq d( \kappa_{n},\kappa_{n+1})+d(\kappa_{n+1}, \kappa_{n+2})+d(\kappa_{n+2},\kappa_{n+3})+\cdots+d( \kappa_{m-1},\kappa _{m}) \\ &\leq \frac{1}{n^{\frac{1}{\varepsilon }}}+\frac{1}{(n+1)^{\frac{1}{ \varepsilon }}}+\frac{1}{(n+2)^{\frac{1}{\varepsilon }}}+\cdots+ \frac{1}{(m-1)^{\frac{1}{ \varepsilon }}} \\ &= \sum^{m-1}_{i=n}\frac{1}{i^{\frac{1}{\varepsilon }}}. \end{aligned} $$
Since
\(d(\kappa_{n},\kappa_{m})\leq \sum^{m-1}_{i=n} \frac{1}{i^{\frac{1}{\varepsilon }}}<\infty \), the sequence
\(\{\kappa_{n}\}\) is Cauchy in
M. Due to completeness of
M, one can find
\(t\in M\) such that
\(\kappa_{n}\rightarrow t\) as
\(n \rightarrow \infty \).