From condition (iii), there exists
\(x_{0} \in\mathcal{J_{T}}\) such that
$$Ax_{0} \preceq_{1} Bx_{0} \quad\text{and}\quad \alpha(x_{0},Tx_{0})\geq1. $$
Define the sequence
\(\{x_{n}\}\) by
\(x_{n}=Tx_{n-1}\), for all
\(n\in\mathbb {N}\). Applying Proposition
1.22,
\(\{x_{n}\}\) is a comparable sequence. If
\(x_{n_{0}}=x_{n_{0}+1}\) for some
\(n_{0}\in\mathbb{N}\), then
\(Tx_{n_{0}}=x_{n_{0}+1}=x_{n_{0}}\), and hence the proof is completed. Now, let
\(x_{n}\neq x_{n+1}\),
\(n=0,1,2,\ldots \) . Since
\(Ax_{0}\preceq_{1} Bx_{0}\) and
T is
\((A,B,C,D,\preceq_{1},\preceq_{2})\)-stable, we have
$$Ax_{0} \preceq_{1} Bx_{0} \quad \Longrightarrow\quad CTx_{0} \preceq_{2} DTx_{0}, $$
that is,
\(Cx_{1} \preceq_{2} Dx_{1}\). Hence
$$Ax_{0} \preceq_{1} Bx_{0}\quad \text{and}\quad Cx_{1} \preceq_{2} Dx_{1}. $$
Continuing this process, by induction, for all
\(n\in\mathbb{N}\) we get
$$ Ax_{2n}\preceq_{1} Bx_{2n}\quad \text{and}\quad Cx_{2n+1}\preceq_{2} Dx_{2n+1}. $$
(2.1)
Also, applying Lemma
1.7 for all
\(m, n \in\mathbb{N}\) with
\(n < m\), we have
$$ \alpha(x_{n},x_{m})\geq1. $$
(2.2)
Since
\(\{x_{n}\}\) is comparable, applying (
2.1), (
2.2) and (vi), by symmetry, for
\(n=1,2,\ldots\) , we have
$$\begin{aligned} \phi\bigl(d(x_{n},x_{n+1})\bigr) \leq& \alpha(x_{n-1},x_{n})\phi \bigl(d(x_{n},x_{n+1}) \bigr) \\ =&\alpha(x_{n-1},x_{n})\phi\bigl(d(Tx_{n-1},Tx_{n}) \bigr) \\ \leq&h(x_{n-1},x_{n})\phi\bigl(M_{a}(x_{n-1},x_{n}) \bigr) \\ < & \phi\bigl(M_{a}(x_{n-1},x_{n})\bigr). \end{aligned}$$
(2.3)
Also, we have
$$\begin{aligned} M_{a}(x_{n-1},x_{n}) =&\max \biggl\{ d(x_{n-1},x_{n}),d(x_{n-1},Tx_{n-1}),d(x_{n},Tx_{n}), \frac {d(x_{n-1},Tx_{n})+d(x_{n},Tx_{n-1})}{2} \biggr\} \\ =&\max \biggl\{ d(x_{n-1},x_{n}),d(x_{n-1},x_{n}),d(x_{n},x_{n+1}), \frac {d(x_{n-1},x_{n+1})+d(x_{n},x_{n})}{2} \biggr\} \\ =&\max \biggl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}), \frac {d(x_{n-1},x_{n+1})}{2} \biggr\} \\ \leq& \max \biggl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}), \frac {d(x_{n-1},x_{n})+d(x_{n},x_{n+1})}{2} \biggr\} \\ =& \max\bigl\{ d(x_{n-1},x_{n}),d(x_{n},x_{n+1}) \bigr\} . \end{aligned}$$
If
\(M_{a}(x_{n-1},x_{n})=d(x_{n},x_{n+1})\), applying (
2.3), we deduce that
$$\begin{aligned} \phi\bigl(d(x_{n},x_{n+1})\bigr) < & \phi \bigl(M_{a}(x_{n-1},x_{n})\bigr) \\ =&\phi\bigl(d(x_{n},x_{n+1})\bigr), \end{aligned}$$
which is a contradiction. Thus, we conclude that
$$ M_{a}(x_{n-1},x_{n})=d(x_{n-1},x_{n}), \quad\forall n\in\mathbb{N}. $$
(2.4)
Now, from (
2.3) and (
2.4), we get
$$\phi\bigl(d(x_{n},x_{n+1})\bigr)< \phi\bigl(d(x_{n-1},x_{n}) \bigr), \quad \forall n\in \mathbb{N}. $$
The monotony of
ϕ implies that
$$d(x_{n},x_{n+1})< d(x_{n-1},x_{n}), \quad \forall n\in\mathbb{N}. $$
We deduce that the sequence
\(\{d(x_{n}, x_{n+1})\}\) is nonnegative and decreasing. Consequently, there exists
\(r \geq0\) such that
\(\lim_{n\rightarrow\infty }d(x_{n},x_{n+1})=r\). We prove that
\(r = 0\). In the contrary case, suppose that
\(r > 0\). Then from (
2.3) and (
2.4), we have
$$0< \frac{\phi(d(x_{n},x_{n+1}))}{\phi(d(x_{n-1},x_{n}))}\leq h(x_{n-1},x_{n}), $$
which implies that
\(\lim_{n\rightarrow\infty}h(x_{n-1},x_{n})=1\). Since
\(h \in\mathcal{H}\),
$$\lim_{n\rightarrow\infty}d(x_{n-1},x_{n})=0. $$
This implies that
\(r=0\), which is a contradiction. Therefore
$$\lim_{n\rightarrow\infty}d(x_{n},x_{n+1})=0. $$
Now, we shall prove that
\(\{x_{n}\}\) is a Cauchy sequence in comparable complete metric space
\((X,\preceq,d)\). Suppose, on the contrary, that
\(\{x_{n}\}\) is not a Cauchy sequence. Thus, there exists
\(\epsilon>0\) such that, for all
\(k\in\mathbb{N}\), there exist
\(n_{k}> m_{k}>k\) such that
$$d(x_{m_{k}},x_{n_{k}})\geq\epsilon. $$
Also, choosing
\(m_{k}\) as small as possible, it may be assumed that
$$d(x_{m_{k}},x_{n_{k}-1})< \epsilon. $$
Hence for each
\(k\in\mathbb{N}\), we have
$$\begin{aligned} \epsilon \leq& d(x_{m_{k}},x_{n_{k}})\leq d(x_{m_{k}},x_{n_{k}-1})+d(x_{n_{k}-1},x_{n_{k}}) \\ \leq& \epsilon+d(x_{n_{k}-1},x_{n_{k}}). \end{aligned}$$
Letting
\(k\rightarrow\infty\) in the above inequality, we get
$$\lim_{n\rightarrow\infty} d(x_{n_{k}},x_{m_{k}})=\epsilon. $$
The triangle inequality implies that
$$ \lim_{n\rightarrow\infty} d(x_{n_{k}+1},x_{m_{k}})= \epsilon,\qquad \lim_{n\rightarrow\infty} d(x_{n_{k}},x_{m_{k}-1})= \epsilon,\qquad \lim_{n\rightarrow\infty} d(x_{n_{k}+1},x_{m_{k}+1})= \epsilon. $$
(2.5)
We see that, for all
\(k\in\mathbb{N}\), there exists
\(i_{k}\in\{0,1\}\) such that
$$n_{k} - m_{k} + i_{k} \equiv1(2). $$
Now, applying (vi), for
\(k \in\mathbb{N}\), we conclude that
$$ \begin{aligned}[b] \phi\bigl(d(x_{n_{k}+1},x_{m_{k}-i_{k}+1})\bigr)&\leq \alpha(x_{n_{k}},x_{m_{k}-i_{k}})\phi \bigl(d(x_{n_{k}+1},x_{m_{k}-i_{k}+1}) \bigr) \\ &=\alpha(x_{n_{k}},x_{m_{k}-i_{k}})\phi\bigl(d(Tx_{n_{k}},Tx_{m_{k}-i_{k}}) \bigr) \\ & \leq h(x_{n_{k}},x_{m_{k}-i_{k}})\phi\bigl(M_{a}(x_{n_{k}},x_{m_{k}-i_{k}}) \bigr). \end{aligned} $$
(2.6)
Also, for any
\(k\in\mathbb{N}\), we have
$$\begin{aligned} M_{a}(x_{n_{k}},x_{m_{k}-i_{k}}) = & \max \biggl\{ d(x_{n_{k}},x_{m_{k}-i_{k}}),d(x_{n_{k}},Tx_{n_{k}}),d(x_{m_{k}-i_{k}},Tx_{m_{k}-i_{k}}),\\ & \frac {d(x_{n_{k}},Tx_{m_{k}-i_{k}})+d(x_{m_{k}-i_{k}},Tx_{n_{k}})}{2} \biggr\} \\ = & \max \biggl\{ d(x_{k},x_{m_{k}-i_{k}}),d(x_{n_{k}},x_{n_{k}+1}),d(x_{m_{k}-i_{k}},x_{m_{k}-i_{k}+1}),\\ & \frac {d(x_{n_{k}},x_{m_{k}-i_{k}+1})+d(x_{m_{k}-i_{k}},x_{n_{k}+1})}{2} \biggr\} \\ \leq& \max \biggl\{ d(x_{n_{k}},x_{m_{k}-i_{k}}),d(x_{n_{k}},x_{n_{k}+1}),d(x_{m_{k}-i_{k}},x_{m_{k}-i_{k}+1}),\\ & \frac {d(x_{n_{k}},x_{m_{k}-i_{k}})+d(x_{m_{k}-i_{k}},x_{m_{k}-i_{k}+1})}{2} + \frac{d(x_{m_{k}-i_{k}},x_{n_{k}})+d(x_{n_{k}},x_{n_{k}+1})}{2} \biggr\} . \end{aligned}$$
Since
\(\lim_{k\rightarrow\infty}d(x_{n_{k}},x_{n_{k}+1})=0\),
$$ \lim_{k\rightarrow\infty}M_{a}(x_{n_{k}},x_{m_{k}-i_{k}})= \lim_{k\rightarrow \infty}d(x_{n_{k}},x_{m_{k}-i_{k}}). $$
(2.7)
Combining (
2.6) and (
2.7) with the continuity of
ϕ, we get
$$\lim_{k\rightarrow\infty}\phi\bigl(d(x_{n_{k}+1},x_{m_{k}-i_{k}+1})\bigr) \leq\lim_{k\rightarrow\infty}h(x_{n_{k}},x_{m_{k}-i_{k}})\lim _{k\rightarrow\infty }\phi\bigl(d(x_{n_{k}},x_{m_{k}-i_{k}})\bigr). $$
Applying (
2.5), we deduce that
$$\lim_{k\rightarrow\infty}h(x_{n_{k}},x_{m_{k}-i_{k}})=1. $$
Since
\(h\in\mathcal{H}(X)\),
$$\lim_{k\rightarrow\infty}d(x_{n_{k}},x_{m_{k}-i_{k}})=0, $$
which is a contradiction. Thus,
\(\{x_{n}\}\) is Cauchy comparable and so there exists
\(x^{*}\in X\) such that
\(\lim_{n\rightarrow\infty} x_{n}=x^{*}\). Since
T is a comparable continuous function,
$$\lim_{n\rightarrow\infty} x_{n+1}=\lim_{n\rightarrow\infty}Tx_{n}=Tx^{*}. $$
Therefore
$$ Tx^{*}=x^{*}. $$
(2.8)
A and
B are comparable continuous and
\(\{x_{2n}\}\) is a comparable sequence, therefore
$$\lim_{n\rightarrow\infty}d\bigl(Ax_{2n},Ax^{*}\bigr)=\lim _{n\rightarrow\infty }d(Bx_{2n},Bx_{2n})=0. $$
Since ⪯
1 is
d-regular, (
2.1) implies that
$$ Ax^{*}\preceq_{1} Bx^{*}. $$
(2.9)
Since
T is
\((A,B,C,D,\preceq_{1},\preceq_{2})\)-stable, applying (
2.9), we have
$$CTx^{*}\preceq_{2} DTx^{*}. $$
This implies that
$$ Cx^{*}\preceq_{2} Dx^{*}. $$
(2.10)
Applying (
2.8), (
2.9) and (
2.10), we deduce that
\(x^{*}\) is a solution of (
2.1). □