Let
\(x_{1}\in X\) be such that
\(\alpha ( x_{1},Tx_{1} ) \geq 1\) and
\(\alpha ( x_{1,}T^{2}x_{1} ) \geq1\). We define the iterative sequence
\(\{ x_{n} \} \) in
X by the rule
\(x_{n}=Tx_{n-1}=T^{n}x_{1}\) for all
\(n\geq1\). Obviously, if there exists
\(n_{0}\geq1\) for which
\(T^{n_{0}}x_{1}=T^{n_{0}+1}x_{1}\) then
\(T^{n_{0}}x_{1} \) shall be a fixed point of
T. Thus, we suppose that
\(T^{n}x_{1}\neq T^{n+1}x_{1}\) for every
\(n\geq1\). Since
T is
α -orbital admissible, we have
$$ \alpha ( x_{1},Tx_{1} ) \geq1\quad\mbox{implies}\quad \alpha \bigl( Tx_{1},T^{2}x_{1} \bigr) \geq1 $$
and
$$ \alpha \bigl( x_{1},T^{2}x_{1} \bigr) \geq1\quad \mbox{implies}\quad\alpha \bigl( Tx_{1},T^{3}x_{1} \bigr) \geq1. $$
By continuing this process, we get
$$ \alpha \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \geq1 \quad\mbox{for all }n\geq1 $$
(2.13)
and
$$ \alpha \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \geq1 \quad\mbox{for all }n\geq1. $$
(2.14)
From condition (1) and (
2.13), then for every
\(n\geq1\), we write
$$\begin{aligned} &\theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \\ &\quad\leq\alpha \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \cdot\theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) \\ &\quad\leq \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n}x_{1} ) ,d ( T^{n-1}x_{1},TT^{n-1}x_{1} ) , \\ d ( T^{n}x_{1},TT^{n}x_{1} ) ,\frac{d ( T^{n-1}x_{1},TT^{n-1}x_{1} ) d ( T^{n}x_{1},TT^{n}x_{1} ) }{1+d ( T^{n-1}x_{1},T^{n}x_{1} ) }\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad= \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n}x_{1} ) ,d ( T^{n}x_{1},T^{n+1}x_{1} ) , \\ \frac{d ( T^{n-1}x_{1},T^{n}x_{1} ) d ( T^{n}x_{1},T^{n+1}x_{1} ) }{1+d ( T^{n-1}x_{1},T^{n}x_{1} ) }\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad= \bigl[ \theta \bigl( \max \bigl\{ d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) ,d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr\} \bigr) \bigr] ^{k}. \end{aligned}$$
(2.15)
If there exists
\(n\geq1\) such that
\(\max \{ d ( T^{n-1}x_{1},T^{n}x_{1} ) ,d ( T^{n}x_{1},T^{n+1}x_{1} ) \} =d ( T^{n}x_{1},T^{n+1}x_{1} ) \), then inequality (
2.15) turns into
$$ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \leq \bigl[ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \bigr] ^{k}, $$
this implies
$$ \ln \bigl[ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \bigr] \leq k\ln \bigl[ \theta \bigl( d \bigl(T^{n}x_{1},T^{n+1}x_{1}\bigr) \bigr) \bigr] , $$
which is a contradiction with
\(k\in ( 0,1 ) \). Therefore
\(\max \{ d ( T^{n-1}x_{1},T^{n}x_{1} ) ,d ( T^{n}x_{1},T^{n+1}x_{1} ) \} =d ( T^{n-1}x_{1},T^{n}x_{1} ) \) for all
\(n\geq1\). Thus, from (
2.15), we have
$$ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \leq \bigl[ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) \bigr] ^{k}\quad\mbox{for all }n\geq1. $$
This implies
$$\begin{aligned} \theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \leq & \bigl[ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) \bigr] ^{k} \\ \leq& \bigl[ \theta \bigl( d \bigl( T^{n-2}x_{1},T^{n-1}x_{1} \bigr) \bigr) \bigr] ^{k^{2}}\leq\cdots\leq \bigl[ \theta \bigl( d(x_{1},Tx_{1}) \bigr) \bigr] ^{k^{n}}. \end{aligned}$$
Thus we have
$$ 1\leq\theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \leq \bigl[ \theta \bigl( d(x_{1},Tx_{1}) \bigr) \bigr] ^{k^{n}} \quad\mbox{for all } n\geq1. $$
(2.16)
Letting
\(n\longrightarrow\infty\), we obtain
$$ \lim_{n\longrightarrow\infty}\theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) =1, $$
(2.17)
which together with (Θ2) gives
$$ \lim_{n\longrightarrow\infty}d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) =0. $$
From condition (Θ3), there exist
\(r\in ( 0,1 ) \) and
\(\ell \in( 0,\infty ] \) such that
$$ \lim_{n\longrightarrow\infty}\frac{\theta ( d ( T^{n}x_{1},T^{n+1}x_{1} ) ) -1}{ [ d ( T^{n}x_{1},T^{n+1}x_{1} ) ] ^{r}}=\ell. $$
Suppose that
\(\ell<\infty\). In this case, let
\(B=\frac{\ell}{2}>0\). From the definition of the limit, there exists
\(n_{0}\geq1\) such that
$$ \biggl\vert \frac{\theta ( d ( T^{n}x_{1},T^{n+1}x_{1} ) ) -1}{ [ d ( T^{n}x_{1},T^{n+1}x_{1} ) ] ^{r}}-\ell \biggr\vert \leq B \quad\mbox{for all }n \geq n_{0}. $$
This implies
$$ \frac{\theta ( d ( T^{n}x_{1},T^{n+1}x_{1} ) ) -1}{ [ d ( T^{n}x_{1},T^{n+1}x_{1} ) ] ^{r}}\geq\ell-B=B \quad\mbox{for all }n\geq n_{0}. $$
Then
$$ n \bigl[ d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr] ^{r}\leq An \bigl[ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) -1 \bigr] \quad\mbox{for all }n\geq n_{0}, $$
where
\(A=\frac{1}{B}\). Suppose now that
\(\ell=\infty\). Let
\(B>0\) be an arbitrary positive number. From the definition of the limit, there exists
\(n_{0}\geq1\) such that
$$ \frac{\theta ( d ( T^{n}x_{1},T^{n+1}x_{1} ) ) -1}{ [ d ( T^{n}x_{1},T^{n+1}x_{1} ) ] ^{r}}\geq B \quad\mbox{for all } n\geq n_{0}. $$
This implies
$$ n \bigl[ d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr] ^{r}\leq An \bigl[ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) -1 \bigr] \quad\mbox{for all }n\geq n_{0}, $$
where
\(A=\frac{1}{B}\). Thus, in all cases, there exist
\(A>0\) and
\(n_{0}\geq 1 \) such that
$$ n \bigl[ d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr] ^{r}\leq An \bigl[ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) -1 \bigr] \quad\mbox{for all }n\geq n_{0}. $$
By using (
2.16), we get
$$ n \bigl[ d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr] ^{r}\leq An \bigl( \bigl[ \theta \bigl( d(x_{1},Tx_{1}) \bigr) \bigr] ^{k^{n}}-1 \bigr) \quad\mbox{for all }n\geq n_{0}. $$
(2.18)
Letting
\(n\longrightarrow\infty\) in the inequality (
2.18), we obtain
$$ \lim_{n\longrightarrow\infty}n \bigl[ d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr] ^{r}=0. $$
Thus, there exists
\(n_{1}\in \mathbb{N} \) such that
$$ d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \leq\frac{1}{n^{\frac {1}{r}}}\quad\mbox{for all }n\geq n_{1}. $$
(2.19)
Now, we will prove that
T has a periodic point. Suppose that it is not the case, then
\(T^{n}x_{1}\neq T^{m}x_{1}\) for all
\(m,n\geq1\) such that
\(n\neq m \). Using condition (1) and (
2.14), we get
$$\begin{aligned} &\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) \\ &\quad\leq\alpha \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \cdot\theta \bigl( d \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \bigr) \\ &\quad\leq \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{c} d ( T^{n-1}x_{1},T^{n+1}x_{1} ) ,d ( T^{n-1}x_{1},TT^{n-1}x_{1} ) , \\ d ( T^{n+1}x_{1},TT^{n+1}x_{1} ) ,\frac{d ( T^{n-1}x_{1},TT^{n-1}x_{1} ) d ( T^{n+1}x_{1},TT^{n+1}x_{1} ) }{1+d ( T^{n-1}x_{1},T^{n+1}x_{1} ) }\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad= \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n+1}x_{1} ) ,d ( T^{n-1}x_{1},T^{n}x_{1} ) , \\ d ( T^{n+1}x_{1},T^{n+2}x_{1} ) ,\frac{d ( T^{n-1}x_{1},T^{n}x_{1} ) d ( T^{n+1}x_{1},T^{n+2}x_{1} ) }{1+d ( T^{n-1}x_{1},T^{n+1}x_{1} ) }\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad= \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n+1}x_{1} ) ,d ( T^{n-1}x_{1},T^{n}x_{1} ) , \\ d ( T^{n+1}x_{1},T^{n+2}x_{1} )\end{array}\displaystyle \right \} \right ) \right ] ^{k}. \end{aligned}$$
(2.20)
Since
θ is non-decreasing, we obtain from (
2.20)
$$ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) \leq \left [ \max \left \{ \textstyle\begin{array}{@{}c@{}} \theta ( d ( T^{n-1}x_{1},T^{n+1}x_{1} ) ) ,\theta ( d ( T^{n-1}x_{1},T^{n}x_{1} ) ) , \\ \theta ( d ( T^{n+1}x_{1},T^{n+2}x_{1} ) )\end{array}\displaystyle \right \} \right ] ^{k}. $$
(2.21)
Let
I be the set of
\(n\in \mathbb{N} \) such that
$$\begin{aligned} u_{n} =&\max \bigl\{ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \bigr) ,\theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) ,\theta \bigl( d \bigl( T^{n+1}x_{1},T^{n+2}x_{1} \bigr) \bigr) \bigr\} \\ =&\theta \bigl( d \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \bigr) . \end{aligned}$$
If
\(\vert I\vert <\infty\) then there is
\(N\geq1\) such that, for all
\(n\geq N\),
$$\begin{aligned} &\max \bigl\{ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \bigr) ,\theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) ,\theta \bigl( d \bigl( T^{n+1}x_{1},T^{n+2}x_{1} \bigr) \bigr) \bigr\} \\ &\quad=\max \bigl\{ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) ,\theta \bigl( d \bigl( T^{n+1}x_{1},T^{n+2}x_{1} \bigr) \bigr) \bigr\} . \end{aligned}$$
In this case, we get from (
2.21)
$$\begin{aligned} 1 \leq&\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) \leq \bigl[ \max \bigl\{ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) ,\theta \bigl( d \bigl( T^{n+1}x_{1},T^{n+2}x_{1} \bigr) \bigr) \bigr\} \bigr] ^{k} \end{aligned}$$
for all
\(n\geq N\). Letting
\(n\longrightarrow\infty\) in the above inequality and using (
2.17), we obtain
$$ \lim_{n\longrightarrow\infty}\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) =1. $$
If
\(\vert I\vert =\infty\), we can find a subsequence of
\(\{ u_{n} \} \), then we denote also by
\(\{ u_{n} \} \), such that
$$ u_{n}=\theta \bigl( d \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \bigr) \quad\mbox{for }n\mbox{ large enough}. $$
In this case, we obtain from (
2.21)
$$\begin{aligned}[b] 1 &\leq\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) \leq \bigl[ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \bigr) \bigr] ^{k} \\ &\leq \bigl[ \theta \bigl( d \bigl( T^{n-2}x_{1},T^{n}x_{1} \bigr) \bigr) \bigr] ^{k^{2}}\leq\cdots\leq \bigl[ \theta \bigl( d \bigl( x_{1},T^{2}x_{1} \bigr) \bigr) \bigr] ^{k^{n}} \end{aligned} $$
for
n large. Letting
\(n\longrightarrow\infty\) in the above inequality, we obtain
$$ \lim_{n\longrightarrow\infty}\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) =1. $$
(2.22)
Then in all cases, (
2.22) holds. Using (
2.22) and (Θ2), we have
$$ \lim_{n\longrightarrow\infty}\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) =0. $$
Similarly from (Θ3) there exists
\(n_{2}\geq1\) such that
$$ d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \leq\frac{1}{n^{\frac {1}{r}}}\quad\mbox{for all }n\geq n_{2}. $$
(2.23)
Let
\(h=\max \{ n_{0},n_{1} \} \). We consider two cases.
Case 2: If
\(m>2\) is even, then writing
\(m=2L\),
\(L\geq2\), using (
2.19) and (
2.23), for all
\(n\geq h\), we have
$$\begin{aligned} d \bigl( T^{n}x_{1},T^{n+m}x_{1} \bigr) \leq&d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) +d \bigl( T^{n+2}x_{1},T^{n+3}x_{1} \bigr) +\cdots \\ &{}+d \bigl( T^{n+2L-1}x_{1},T^{n+2L}x_{1} \bigr) \\ \leq&\frac{1}{n^{\frac{1}{r}}}+\frac{1}{ ( n+2 ) ^{\frac {1}{r}}}+\cdots+\frac{1}{ ( n+2L-1 ) ^{\frac{1}{r}}} \\ \leq&\sum_{i=n}^{\infty}\frac{1}{i^{\frac{1}{r}}}. \end{aligned}$$
Thus, combining all cases, we have
$$ d \bigl( T^{n}x_{1},T^{n+m}x_{1} \bigr) \leq\sum_{i=n}^{\infty}\frac {1}{i^{\frac{1}{r}}}\quad \mbox{for all }n\geq h, m\geq1. $$
Since the series
\(\sum_{i=n}^{\infty}\frac{1}{i^{\frac{1}{r}}}\) is convergent (since
\(\frac{1}{r}>1\)), we deduce that
\(\{ T^{n}x_{1} \} \) is a Cauchy sequence. From the completeness of
X, there
\(x_{\ast}\in X\) such that
\(T^{n}x_{1}\longrightarrow x_{\ast}\) as
\(n\longrightarrow\infty\). Now, we prove that
\(x_{\ast}=Tx_{\ast}\). Since
T is
α-orbital attractive, we have for all
\(n\geq1\)
$$ \alpha \bigl( T^{n}x_{1},x_{\ast} \bigr) \geq1\quad \mbox{or}\quad\alpha \bigl( x_{\ast},T^{n+1}x_{1} \bigr) \geq1. $$
Hence there exists a subsequence
\(\{ T^{n ( k ) }x_{1} \} \) of
\(\{ T^{n}x_{1} \} \) such that
$$ \alpha \bigl( T^{n ( k ) }x_{1},x_{\ast} \bigr) \geq1 \quad\mbox{or}\quad \alpha \bigl( x_{\ast},T^{n ( k ) }x_{1} \bigr) \geq1 \quad\mbox{for all }k\geq1. $$
In the first case, without restriction of the generality, we can suppose that
\(T^{n ( k ) }x_{1}\neq x_{\ast}\) for all
k. Using condition (1), we have
$$\begin{aligned} \theta \bigl( d \bigl( T^{n ( k ) +1}x_{1},Tx_{\ast} \bigr) \bigr) =&\theta \bigl( d \bigl( TT^{n ( k ) }x_{1},Tx_{\ast} \bigr) \bigr) \\ \leq&\alpha \bigl( T^{n ( k ) }x_{1},x_{\ast} \bigr) \cdot\theta \bigl( d \bigl( TT^{n ( k ) }x_{1},Tx_{\ast} \bigr) \bigr) \\ \leq& \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n(k)}x_{1},x_{\ast} ) ,d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) , \\ d ( x_{\ast},Tx_{\ast} ) ,\frac{d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) d(x_{\ast},Tx_{\ast})}{1+d(T^{n ( k ) }x_{1},x_{\ast})}\end{array}\displaystyle \right \} \right ) \right ] ^{k}. \end{aligned}$$
This implies
$$ \theta \bigl( d \bigl( T^{n ( k ) +1}x_{1},Tx_{\ast} \bigr) \bigr) \leq \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n(k)}x_{1},x_{\ast} ) ,d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) , \\ d ( x_{\ast},Tx_{\ast} ) ,\frac{d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) d(x_{\ast},Tx_{\ast})}{1+d(T^{n ( k ) }x_{1},x_{\ast})}\end{array}\displaystyle \right \} \right ) \right ] ^{k}. $$
Letting
\(k\longrightarrow\infty\) in the above equality, using the continuity of
θ and Lemma
14, we get
$$ \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) \leq \bigl[ \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) \bigr] ^{k}< \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) , $$
which is a contradiction. The second case is similar. Therefore,
\(x_{\ast }=Tx_{\ast}\), which is also a contradiction with the assumption that
T does not have a periodic point. Thus
T has a periodic point, say
\(x_{\ast }\) of period
q. Suppose that the set of fixed points of
T is empty. Then we have
$$ q>1\quad\mbox{and}\quad d ( x_{\ast},Tx_{\ast} ) >0. $$
By using condition (1) and (
2.13), we get
$$\begin{aligned} \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) =&\theta \bigl( d \bigl( T^{q}x_{\ast},T^{q+1}x_{\ast} \bigr) \bigr) \leq\alpha \bigl( T^{q-1}x_{\ast},T^{q}x_{\ast} \bigr) \cdot\theta \bigl( d \bigl( T^{q}x_{\ast },T^{q+1}x_{\ast} \bigr) \bigr) \\ \leq& \bigl[ \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) \bigr] ^{k^{q}}< \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) , \end{aligned}$$
which is a contradiction. Thus the set of fixed points of
T is non-empty (that is,
T has at least one fixed point).
If
\(y_{\ast}\) is another fixed point of
T such that
\(x_{\ast}\neq y_{\ast}\), since
T is
α-orbital attractive, we deduce that
$$ \alpha \bigl( T^{n}x_{1},y_{\ast} \bigr) \geq1 \quad\mbox{or}\quad\alpha \bigl( y_{\ast},T^{n+1}x_{1} \bigr) \geq1. $$
Hence there exists a subsequence
\(\{ T^{n ( k ) }x_{1} \} \) of
\(\{ T^{n}x_{1} \} \) such that
$$ \alpha \bigl( T^{n ( k ) }x_{1},y_{\ast} \bigr) \geq1 \quad\mbox{or}\quad \alpha \bigl( y_{\ast},T^{n ( k ) }x_{1} \bigr) \geq1 \quad\mbox{for all }k\geq1. $$
In the first case, we have
$$\begin{aligned} \theta \bigl( d \bigl( T^{n ( k ) +1}x_{1},y_{\ast} \bigr) \bigr) =&\theta \bigl( d \bigl( T^{n ( k ) +1}x_{1},Ty_{\ast} \bigr) \bigr) =\theta \bigl( d \bigl( TT^{n ( k ) }x_{1},Ty_{\ast } \bigr) \bigr) \\ \leq&\alpha \bigl( T^{n ( k ) }x_{1},y_{\ast} \bigr) \cdot \theta \bigl( d \bigl( TT^{n ( k ) }x_{1},Ty_{\ast} \bigr) \bigr) \\ \leq& \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{c} d ( T^{n(k)}x_{1},y_{\ast} ) ,d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) , \\ d ( y_{\ast},Ty_{\ast} ) ,\frac{d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) d(y_{\ast},Ty_{\ast})}{1+d(T^{n ( k ) }x_{1},y_{\ast})}\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ =& \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n(k)}x_{1},y_{\ast} ) ,d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) , \\ d ( y_{\ast},Ty_{\ast} ) ,\frac{d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) d(y_{\ast},Ty_{\ast})}{1+d(T^{n ( k ) }x_{1},y_{\ast})}\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ < &\theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n(k)}x_{1},y_{\ast} ) ,d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) , \\ d ( y_{\ast},Ty_{\ast} ) ,\frac{d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) d(y_{\ast},Ty_{\ast})}{1+d(T^{n ( k ) }x_{1},y_{\ast})}\end{array}\displaystyle \right \} \right ) . \end{aligned}$$
This implies
$$ \theta \bigl( d \bigl( T^{n ( k ) +1}x_{1},y_{\ast} \bigr) \bigr) < \theta \left ( \max \left \{ \textstyle\begin{array}{c} d ( T^{n(k)}x_{1},y_{\ast} ) ,d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) , \\ d ( y_{\ast},Ty_{\ast} ) ,\frac{d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) d(y_{\ast},Ty_{\ast})}{1+d(T^{n ( k ) }x_{1},y_{\ast})}\end{array}\displaystyle \right \} \right ). $$
Letting
\(k\longrightarrow\infty\) in the above equality, we get
$$ \theta \bigl( d ( x_{\ast},y_{\ast} ) \bigr) < \theta \bigl( d ( x_{\ast},y_{\ast} ) \bigr) . $$
This is a contradiction. The second case is similar. □