In this section, we state and prove our main result.
Theorem 17
Let
\(( X,d ) \)
be a complete BMS, \(T:X\longrightarrow X\)
be a given map and let
\(\alpha:X\times X\longrightarrow[ 0,\infty ) \)
be a mapping. Suppose that the following conditions hold:
-
(1)
there exist
\(\theta\in\Theta\)
and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X, \quad d ( Tx,Ty ) \neq0\quad\Longrightarrow\quad\alpha ( x,y ) \cdot \theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( R(x,y) \bigr) \bigr] ^{k}, $$
where
$$ R(x,y)=\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)d(y,Ty)}{1+d(x,y)}\biggr\} , $$
-
(2)
there exists
\(x_{1}\in X\)
such that
\(\alpha ( x_{1},Tx_{1} ) \geq1\)
and
\(\alpha ( x_{1},T^{2}x_{1} ) \geq1\),
-
(3)
T
is a triangular
α-orbital admissible mapping,
-
(4)
T
is continuous.
Then
T
has a fixed point
\(x_{\ast}\in X\)
and
\(\{ T^{n}x_{1} \} \)
converges to
\(x_{\ast}\).
Proof
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 \(n\geq1\). Now from Lemma 16, we get
$$ \alpha \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \geq1\quad\mbox{for all } n\geq1, $$
(2.1)
also
$$ \alpha \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \geq1\quad\mbox{for all } n\geq1. $$
(2.2)
From condition (1) and (2.1), for every \(n\geq1\), we write
$$\begin{aligned} &\theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \\ &\quad\leq\alpha \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \cdot\theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) \\ &\quad\leq \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n}x_{1} ) ,d ( T^{n-1}x_{1},TT^{n-1}x_{1} ) , \\ d ( T^{n}x_{1},TT^{n}x_{1} ) ,\frac{d ( T^{n-1}x_{1},TT^{n-1}x_{1} ) d ( T^{n}x_{1},TT^{n}x_{1} ) }{1+d ( T^{n-1}x_{1},T^{n}x_{1} ) }\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad= \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n}x_{1} ) ,d ( T^{n}x_{1},T^{n+1}x_{1} ) , \\ \frac{d ( T^{n-1}x_{1},T^{n}x_{1} ) d ( T^{n}x_{1},T^{n+1}x_{1} ) }{1+d ( T^{n-1}x_{1},T^{n}x_{1} ) }\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad= \bigl[ \theta \bigl( \max \bigl\{ d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) ,d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr\} \bigr) \bigr] ^{k}. \end{aligned}$$
(2.3)
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 (2.3) 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 (2.3), 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
$$ 1\leq\theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) \leq \bigl[ \theta \bigl( d(x_{1},Tx_{1}) \bigr) \bigr] ^{k^{n}} \quad\mbox{for all } n\geq1. $$
(2.4)
Letting \(n\longrightarrow\infty\), we obtain
$$ \lim_{n\longrightarrow\infty}\theta \bigl( d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr) =1, $$
(2.5)
which together with (Θ2) gives as
$$ \lim_{n\longrightarrow\infty}d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) =0. $$
From condition (Θ3), there exist \(r\in ( 0,1 ) \) and \(\ell \in \left ( 0,\infty \right ] \) 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 (2.4), we get
$$ n \bigl[ d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \bigr] ^{r}\leq An \bigl( \bigl[ \theta \bigl( d(x_{1},Tx_{1}) \bigr) \bigr] ^{k^{n}}-1 \bigr) \quad\mbox{for all }n\geq n_{0}. $$
(2.6)
Letting \(n\longrightarrow\infty\) in the inequality (2.6), 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
$$ d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) \leq\frac{1}{n^{\frac {1}{r}}}\quad\mbox{for all }n\geq n_{1}. $$
(2.7)
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 \(n,m\geq1\) such that \(n\neq m \). Using condition (1) and (2.2), we get
$$\begin{aligned} &\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) \\ &\quad\leq\alpha \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \cdot\theta \bigl( d \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \bigr) \\ &\quad\leq \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n+1}x_{1} ) ,d ( T^{n-1}x_{1},TT^{n-1}x_{1} ) , \\ d ( T^{n+1}x_{1},TT^{n+1}x_{1} ) ,\frac{d ( T^{n-1}x_{1},TT^{n-1}x_{1} ) d ( T^{n+1}x_{1},TT^{n+1}x_{1} ) }{1+d ( T^{n-1}x_{1},T^{n+1}x_{1} ) }\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad=\left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n+1}x_{1} ) ,d ( T^{n-1}x_{1},T^{n}x_{1} ) , \\ d ( T^{n+1}x_{1},T^{n+2}x_{1} ) ,\frac{d ( T^{n-1}x_{1},T^{n}x_{1} ) d ( T^{n+1}x_{1},T^{n+2}x_{1} ) }{1+d ( T^{n-1}x_{1},T^{n+1}x_{1} ) }\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad=\left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n-1}x_{1},T^{n+1}x_{1} ) ,d ( T^{n-1}x_{1},T^{n}x_{1} ) , \\ d ( T^{n+1}x_{1},T^{n+2}x_{1} )\end{array}\displaystyle \right \} \right ) \right ] ^{k}. \end{aligned}$$
(2.8)
Since θ is non-decreasing, we obtain from (2.8)
$$ \theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) \leq \left [ \max \left \{ \textstyle\begin{array}{@{}c@{}} \theta ( d ( T^{n-1}x_{1},T^{n+1}x_{1} ) ) ,\theta ( d ( T^{n-1}x_{1},T^{n}x_{1} ) ) , \\ \theta ( d ( T^{n+1}x_{1},T^{n+2}x_{1} ) )\end{array}\displaystyle \right \} \right ] ^{k}. $$
(2.9)
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 (2.9)
$$\begin{aligned} 1 \leq&\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) \leq \bigl[ \max \bigl\{ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n}x_{1} \bigr) \bigr) ,\theta \bigl( d \bigl( T^{n+1}x_{1},T^{n+2}x_{1} \bigr) \bigr) \bigr\} \bigr] ^{k} \end{aligned}$$
for all \(n\geq N\). Letting \(n\longrightarrow\infty\) in the above inequality and using (2.5), 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 (2.9)
$$\begin{aligned} 1 \leq&\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) \leq \bigl[ \theta \bigl( d \bigl( T^{n-1}x_{1},T^{n+1}x_{1} \bigr) \bigr) \bigr] ^{k} \\ \leq& \bigl[ \theta \bigl( d \bigl( T^{n-2}x_{1},T^{n}x_{1} \bigr) \bigr) \bigr] ^{k^{2}}\leq\cdots\leq \bigl[ \theta \bigl( d \bigl( x_{1},T^{2}x_{1} \bigr) \bigr) \bigr] ^{k^{n}} \end{aligned}$$
for n large. Letting \(n\longrightarrow\infty\) in the above inequality, we obtain
$$ \lim_{n\longrightarrow\infty}\theta \bigl( d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \bigr) =1. $$
(2.10)
Then in all cases, (2.10) holds. Using (2.10) 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
$$ d \bigl( T^{n}x_{1},T^{n+2}x_{1} \bigr) \leq\frac{1}{n^{\frac {1}{r}}}\quad\mbox{for all }n\geq n_{2}. $$
(2.11)
Let \(h=\max \{ n_{0},n_{1} \} \). we consider two cases.
Case 1: If \(m>2\) is odd, then writing \(m=2L+1\), \(L\geq1\), using (2.7), for all \(n\geq h\), we obtain
$$\begin{aligned} d \bigl( T^{n}x_{1},T^{n+m}x_{1} \bigr) \leq&d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) +d \bigl( T^{n+1}x_{1},T^{n+2}x_{1} \bigr)+\cdots \\ &{}+d \bigl( T^{n+2L}x_{1},T^{n+2L+1}x_{1} \bigr) \\ \leq&\frac{1}{n^{\frac{1}{r}}}+\frac{1}{ ( n+1 ) ^{\frac {1}{r}}}+\cdots+\frac{1}{ ( n+2L ) ^{\frac{1}{r}}} \\ \leq&\sum_{i=n}^{\infty}\frac{1}{i^{\frac{1}{r}}}. \end{aligned}$$
Case 2: If \(m>2\) is even, then writing \(m=2L\), \(L\geq2\), using (2.7) and (2.11), 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 is \(x_{\ast}\in X\) such that \(T^{n}x_{1}\longrightarrow x_{\ast }\) as \(n\longrightarrow\infty\). Now, since T is continuous we have
$$ x_{\ast}=\lim_{n\longrightarrow\infty}T^{n+1}x_{1}= \lim_{n\longrightarrow\infty}T \bigl( T^{n}x_{1} \bigr) =T \Bigl( \lim_{n\longrightarrow\infty}T^{n}x_{1} \Bigr) =Tx_{\ast}. $$
We obtain \(x_{\ast}=Tx_{\ast}\), which is 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 (2.1), 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). □
Since a metric space is a Branciari metric space, we can obtain the following result from Theorem 17.
Corollary 18
Let
\(( X,d ) \)
be a complete metric space, \(T:X\longrightarrow X\)
be a given map and let
\(\alpha:X\times X\longrightarrow [ 0,\infty ) \)
be a mapping. Suppose that the following conditions hold:
-
(1)
there exist
\(\theta\in\Theta\)
and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X,\quad d ( Tx,Ty ) \neq0\quad\Longrightarrow\quad\alpha ( x,y ) \cdot\theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( R(x,y) \bigr) \bigr] ^{k}, $$
where
$$ R(x,y)=\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)d(y,Ty)}{1+d(x,y)}\biggr\} , $$
-
(2)
there exists
\(x_{1}\in X\)
such that
\(\alpha ( x_{1},Tx_{1} ) \geq1\)
and
\(\alpha ( x_{1},T^{2}x_{1} ) \geq1\),
-
(3)
T
is a triangular
α-orbital admissible mapping,
-
(4)
T
is continuous.
Then
T
has a fixed point
\(x_{\ast}\in X\)
and
\(\{ T^{n}x_{1} \} \)
converges to
\(x_{\ast}\).
In the next theorem we omit the continuity hypothesis of T.
Theorem 19
Let
\(( X,d ) \)
be a complete BMS, \(T:X\longrightarrow X\)
be a given map and let
\(\alpha:X\times X\longrightarrow [ 0,\infty ) \)
be a mapping. Suppose that the following conditions hold:
-
(1)
there exist
\(\theta\in\Theta\)
and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X,\quad d ( Tx,Ty ) \neq0\quad \Longrightarrow\quad\alpha ( x,y ) \cdot \theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( R(x,y) \bigr) \bigr] ^{k}, $$
where
$$ R(x,y)=\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)d(y,Ty)}{1+d(x,y)}\biggr\} , $$
-
(2)
there exists
\(x_{1}\in X\)
such that
\(\alpha ( x_{1},Tx_{1} ) \geq1\)
and
\(\alpha ( x_{1},T^{2}x_{1} ) \geq1\),
-
(3)
T
is a triangular
α-orbital admissible mapping,
-
(4)
if
\(\{ T^{n}x_{1} \} \)
is a sequence in
X
such that
\(\alpha ( T^{n}x_{1},T^{n+1}x_{1} ) \geq1\)
for all
n
and
\(x_{n}\longrightarrow x\in X\)
as
\(n\longrightarrow\infty\), then there exists a subsequence
\(\{ T^{{n(k)}}x_{1} \} \)
of
\(\{ T^{n}x_{1} \} \)
such that
\(\alpha ( T^{n ( k ) }x_{1},x ) \geq1\)
for all
k,
-
(5)
θ
is continuous.
Then
T
has a fixed point
\(x_{\ast}\in X\)
and
\(\{ T^{n}x_{1} \} \)
converges to
\(x_{\ast}\).
Proof
Let \(x_{1}\in X\) be such that \(\alpha ( x_{1},Tx_{1} ) \geq 1\) and \(\alpha ( x_{1},T^{2}x_{1} ) \geq1\). Following the proof of Theorem 17, we see that the sequence \(\{ T^{n}x_{1} \} \) defined by \(x_{n}=Tx_{n-1}=T^{n}x_{1}\) for all \(n\geq1\) converges to \(x_{\ast}\in X\). From condition (4), we see that there exists a subsequence \(\{ T^{n ( k ) }x_{1} \} \) of \(\{ T^{n}x_{1} \} \) such that \(\alpha ( T^{n ( k ) }x_{1},x_{\ast} ) \geq1\) for all k. We can suppose \(T^{n ( k ) +1}x_{1}\neq Tx_{\ast}\), then, from condition (1), we have
$$\begin{aligned} &\theta \bigl( d \bigl( T^{n ( k ) +1}x_{1},Tx_{\ast} \bigr) \bigr) \\ &\quad=\theta \bigl( d \bigl( T \bigl( T^{n ( k ) }x_{1} \bigr) ,Tx_{\ast} \bigr) \bigr) \\ &\quad\leq\alpha \bigl( T^{n ( k ) }x_{1},x_{\ast} \bigr) \cdot\theta \bigl( d \bigl( T \bigl( T^{n ( k ) }x_{1} \bigr) ,Tx_{\ast } \bigr) \bigr) \\ &\quad\leq \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n ( k ) }x_{1},x_{\ast} ) ,d ( T^{n ( k ) }x_{1},T ( T^{n ( k ) }x_{1} ) ) , \\ d ( x_{\ast},Tx_{\ast} ) ,\frac{d(T^{n ( k ) }x_{1},T(T^{n ( k ) }x_{1}))d ( x_{\ast},Tx_{\ast} ) }{1+(dT^{n ( k ) }x_{1},x_{\ast})}\end{array}\displaystyle \right \} \right ) \right ] ^{k} \\ &\quad= \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( T^{n ( k ) }x_{1},x_{\ast} ) ,d ( T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1} ) , \\ d ( x_{\ast},Tx_{\ast} ) ,\frac{d(T^{n ( k ) }x_{1},T^{n ( k ) +1}x_{1}))d ( x_{\ast},Tx_{\ast} ) }{1+(dT^{n ( k ) }x_{1},x_{\ast})}\end{array}\displaystyle \right \} \right ) \right ] ^{k}. \end{aligned}$$
(2.12)
Now, we suppose that \(d ( x_{\ast},Tx_{\ast} ) >0\). Taking the limit as \(k\longrightarrow\infty\) in (2.12), and by using the continuity of θ, and Lemma 14, we obtain
$$ \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) \leq \bigl[ \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) \bigr] ^{k}< \theta \bigl( d ( x_{\ast},Tx_{\ast} ) \bigr) , $$
which is a contradiction. Thus we have \(x_{\ast}=Tx_{\ast}\), 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 (2.1), 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). □
Example 20
Let \(X= [ -2,-1 ] \cup \{ 0 \} \cup [ 1,2 ] \). Define \(d:X\times X\longrightarrow [ 0,\infty ) \) as follows:
$$\begin{aligned}& d ( x,x ) =0,\quad \mbox{for all }x\in X, \\& d ( 1,2 ) =d ( 2,1 ) =3, \\& d ( 1,-1 ) =d ( -1,1 ) =d ( -1,2 ) =d ( 2,-1 ) =1, \\& d ( x,y ) =\vert x-y\vert ,\quad\mbox{otherwise}. \end{aligned}$$
It is clear that \(( X,d ) \) is a complete BMS, but it is not metric space because d does not satisfy triangle inequality on X. Indeed,
$$ 3=d ( 1,2 ) >d ( 1,-1 ) +d ( -1,2 ) =1+1=2. $$
Let \(T:X\longrightarrow X\) be the mapping defined by
$$ Tx=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} -x & \mbox{if }x\in [ -2,-1 ) \cup ( 1,2 ] , \\ 0 & \mbox{if }x\in \{ -1,0,1 \} .\end{array}\displaystyle \right . $$
Let \(\alpha:X\times X\longrightarrow [ 0,\infty ) \) be given by
$$ \alpha ( x,y ) =\left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1, & \mbox{if }xy\geq0, \\ 0, & \mbox{otherwise}. \end{array}\displaystyle \right . $$
Also define \(\theta: ( 0,\infty ) \longrightarrow ( 1,\infty ) \) by
$$ \theta ( t ) =e^{\sqrt{te^{t}}}. $$
Obviously, T is triangular α-orbital admissible mapping. Also the hypotheses of Theorem 19 are satisfied by T and, hence, T has a fixed point.
Corollary 21
Let
\(( X,d ) \)
be a complete metric space, \(T:X\longrightarrow X\)
be a given map and let
\(\alpha:X\times X\longrightarrow [ 0,\infty ) \)
be a mapping. Suppose that the following conditions hold:
-
(1)
there exist
\(\theta\in\Theta\)
and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X,\quad d ( Tx,Ty ) \neq0\quad\Longrightarrow\quad\alpha ( x,y ) \cdot \theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( R(x,y) \bigr) \bigr] ^{k}, $$
where
$$ R(x,y)=\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)d(y,Ty)}{1+d(x,y)}\biggr\} , $$
-
(2)
there exists
\(x_{1}\in X\)
such that
\(\alpha ( x_{1},Tx_{1} ) \geq1\)
and
\(\alpha ( x_{1},T^{2}x_{1} ) \geq1\),
-
(3)
T
is a triangular
α-orbital admissible mapping,
-
(4)
if
\(\{ T^{n}x_{1} \} \)
is a sequence in
X
such that
\(\alpha ( T^{n}x_{1},T^{n+1}x_{1} ) \geq1\)
for all
n
and
\(x_{n}\longrightarrow x\in X\)
as
\(n\longrightarrow\infty\), then there exists a subsequence
\(\{ T^{{n(k)}}x_{1} \} \)
of
\(\{ T^{n}x_{1} \} \)
such that
\(\alpha ( T^{n ( k ) }x_{1},x ) \geq1\)
for all
k,
-
(5)
θ
is continuous.
Then
T
has a fixed point
\(x_{\ast}\in X\)
and
\(\{ T^{n}x_{1} \} \)
converges to
\(x_{\ast}\).
To ensure the uniqueness of the fixed point, we shall consider the following hypothesis.
-
(H)
for all \(x\neq y\in X\), there exists \(v\in X\) such that \(\alpha ( x,v ) \geq1\), \(\alpha ( y,v ) \geq1\), and \(\alpha ( v,Tv ) \geq1\).
Theorem 22
Adding condition (H) to the hypothesis of Theorem
17
or Corollary
18 (respectively, Theorem
19
or Corollary
21) the uniqueness of the fixed point is obtained.
Proof
Suppose that \(x_{\ast}\) and \(y_{\ast}\) are two fixed points of T such that \(x_{\ast}\neq y_{\ast}\). Then by (H), there exists \(v\in X\) such that
$$ \alpha ( x_{\ast},v ) \geq1,\qquad\alpha ( y_{\ast},v ) \geq1 \quad\mbox{and}\quad\alpha ( v,Tv ) \geq1. $$
Since T is a triangular α-orbital admissible mapping, we see that
$$ \alpha \bigl( x_{\ast},T^{n}v \bigr) \geq1,\qquad\alpha \bigl( y_{\ast },T^{n}v \bigr) \geq1\quad\mbox{for all }n\geq1. $$
By Theorem 17 (respectively, Theorem 19) we deduce that the sequence \(\{ T^{n}v \} \) converges to a fixed point \(z_{\ast}\) of T. We can suppose that \(x_{\ast}\neq T^{n+1}v\) for all \(n\geq1\), then from condition (1), we have
$$\begin{aligned}[b] \theta \bigl( d \bigl( x_{\ast},T^{n+1}v \bigr) \bigr) &= \theta \bigl( d \bigl( Tx_{\ast},T^{n+1}v \bigr) \bigr) \leq\alpha \bigl( x_{\ast},T^{n}v \bigr) \cdot\theta \bigl( d \bigl( Tx_{\ast },T^{n+1}v \bigr) \bigr) \\ &\leq \left [ \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( x_{\ast},T^{n}v ) ,d ( x_{\ast},Tx_{\ast} ) , \\ d ( T^{n}v,T^{n+1}v ) ,\frac{d(x_{\ast},Tx_{\ast})d ( T^{n}v,T^{n+1}v ) }{1+(x_{\ast},T^{n}v)}\end{array}\displaystyle \right \} \right ) \right ] ^{k}. \end{aligned} $$
This implies
$$ \theta \bigl( d \bigl( x_{\ast},T^{n+1}v \bigr) \bigr) < \theta \left ( \max \left \{ \textstyle\begin{array}{@{}c@{}} d ( x_{\ast},T^{n}v ) ,d ( x_{\ast},Tx_{\ast} ) , \\ d ( T^{n}v,T^{n+1}v ) ,\frac{d(x_{\ast},Tx_{\ast})d ( T^{n}v,T^{n+1}v ) }{1+(x_{\ast},T^{n}v)}\end{array}\displaystyle \right \} \right ) . $$
Letting \(n\longrightarrow\infty\) in the above equality, if \(x_{\ast }\neq z_{\ast}\), then we get
$$ d ( x_{\ast},z_{\ast} ) < d ( x_{\ast},z_{\ast} ) , $$
which is a contradiction. Therefore, \(x_{\ast}=z_{\ast}\). Similarly, we get \(y_{\ast}=z_{\ast}\). Hence, \(x_{\ast}=y_{\ast}\), which is a contradiction. □
Corollary 23
Let
\(( X,d ) \)
be a complete BMS and
\(T:X\longrightarrow X\)
be a given mapping. Suppose that there exist
\(\theta\in\Theta\)
and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X,\quad d ( Tx,Ty ) \neq0\quad\Longrightarrow\quad\theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( R ( x,y ) \bigr) \bigr] ^{k}, $$
where
$$ R ( x,y ) =\max \biggl\{ d ( x,y ) ,d ( x,Tx ) ,d ( y,Ty ) ,\frac{d(x,Tx)d(y,Ty)}{1+d(x,y)} \biggr\} . $$
Then
T
has a unique fixed point.
Proof
Setting \(\alpha ( x,y ) =1\) for all \(x,y\in X\) in Theorem 22, we get this result. □
Corollary 24
[18]
Let
\(( X,d ) \)
be a complete BMS and
\(T:X\longrightarrow X\)
be a given mapping. Suppose that there exist
\(\theta \in\Theta\)
that is continuous and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X, \quad d ( Tx,Ty ) \neq0\quad\Longrightarrow\quad\theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( M ( x,y ) \bigr) \bigr] ^{k}, $$
where
$$ M ( x,y ) =\max \bigl\{ d ( x,y ) ,d ( x,Tx ) ,d ( y,Ty ) \bigr\} . $$
Then
T
has a unique fixed point.
Corollary 25
[1]
Let
\(( X,d ) \)
be a complete BMS and
\(T:X\longrightarrow X\)
be a given mapping. Suppose that there exist
\(\theta \in\Theta\)
and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X,\quad d ( Tx,Ty ) \neq0\quad\Longrightarrow\quad\theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( d ( x,y ) \bigr) \bigr] ^{k}. $$
Then
T
has a unique fixed point.
Example 26
Let \(X= \{ 0,1,2 \} \) endow with the metric d given by \(d ( x,y ) =\vert x-y\vert \) for all \(x,y\in X\). It is easy to show that \(( X,d ) \) is a complete metric space. Let \(T:X\longrightarrow X\) be the mapping defined by
$$ T ( 0 ) =0,\qquad T ( 1 ) =2,\qquad T ( 2 ) =1, $$
and \(\alpha:X\times X\longrightarrow [ 0,\infty ) \) be given by
$$ \alpha ( x,y ) =\left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1 & \mbox{if } ( x,y ) \in \left \{ \textstyle\begin{array}{@{}c@{}} ( 0,1 ) , ( 0,2 ) , ( 1,1 ) , ( 2,2 ) , \\ ( 1,2 ) , ( 2,1 )\end{array}\displaystyle \right \} , \\ 0 & \mbox{otherwise}. \end{array}\displaystyle \right . $$
Also define \(\theta: ( 0,\infty ) \longrightarrow ( 1,\infty ) \) by
$$ \theta ( t ) =e^{\sqrt{t}}. $$
It is not difficult to show that T is triangular α-orbital admissible mapping. Also the hypotheses of Theorem 22 are satisfied by T and hence, T has a fixed point. But the result of Jleli et al. (Corollary 25) cannot be applied to T. Indeed for \(x=1\), \(y=0\), we have
$$\begin{aligned} \theta \bigl( d \bigl( T ( 1 ) ,T ( 0 ) \bigr) \bigr) =&\theta \bigl( d ( 2,0 ) \bigr) =e^{\sqrt{2}} \\ \nleq& [ e ] ^{k}= \bigl[ \theta \bigl( d ( 1,0 ) \bigr) \bigr] ^{k},\quad \mbox{for all }k\in ( 0,1 ) . \end{aligned}$$
Now we will use the concept of an α-orbital attractive mapping introduced in [2].
Theorem 27
Let
\(( X,d ) \)
be a complete BMS, \(T:X\longrightarrow X\)
be a given map and let
\(\alpha:X\times X\longrightarrow [ 0,\infty ) \)
be a mapping. Suppose that the following conditions hold:
-
(1)
there exist
\(\theta\in\Theta\)
and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X,\quad d ( Tx,Ty ) \neq0\quad\Longrightarrow\quad\alpha ( x,y ) \cdot\theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( R(x,y) \bigr) \bigr] ^{k}, $$
where
$$ R(x,y)=\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)d(y,Ty)}{1+d(x,y)}\biggr\} , $$
-
(2)
there exists
\(x_{1}\in X\)
such that
\(\alpha ( x_{1},Tx_{1} ) \geq1\)
and
\(\alpha ( x_{1},T^{2}x_{1} ) \geq1\),
-
(3)
T
is an
α-orbital admissible mapping,
-
(4)
T
is an
α-orbital attractive mapping,
-
(5)
θ
is continuous.
Then
T
has a unique fixed point
\(x_{\ast}\in X\)
and
\(\{ T^{n}x_{1} \} \)
converges to
\(x_{\ast}\).
Proof
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 1: If \(m>2\) is odd, then writing \(m=2L+1\), \(L\geq1\), using (2.19), for all \(n\geq h\), we obtain
$$\begin{aligned} d \bigl( T^{n}x_{1},T^{n+m}x_{1} \bigr) \leq&d \bigl( T^{n}x_{1},T^{n+1}x_{1} \bigr) +d \bigl( T^{n+1}x_{1},T^{n+2}x_{1} \bigr)+\cdots \\ &{}+d \bigl( T^{n+2L}x_{1},T^{n+2L+1}x_{1} \bigr) \\ \leq&\frac{1}{n^{\frac{1}{r}}}+\frac{1}{ ( n+1 ) ^{\frac {1}{r}}}+\cdots+\frac{1}{ ( n+2L ) ^{\frac{1}{r}}} \\ \leq&\sum_{i=n}^{\infty}\frac{1}{i^{\frac{1}{r}}}. \end{aligned}$$
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. □
Corollary 28
Let
\(( X,d ) \)
be a complete metric space, \(T:X\longrightarrow X\)
be a given map, and let
\(\alpha:X\times X\longrightarrow [ 0,\infty ) \)
be a mapping. Suppose that the following conditions hold:
-
(1)
there exist
\(\theta\in\Theta\)
and
\(k\in ( 0,1 ) \)
such that
$$ x,y\in X,\quad d ( Tx,Ty ) \neq0 \quad\Longrightarrow\quad\alpha ( x,y ) \cdot \theta \bigl( d ( Tx,Ty ) \bigr) \leq \bigl[ \theta \bigl( R(x,y) \bigr) \bigr] ^{k}, $$
where
$$ R(x,y)=\max\biggl\{ d(x,y),d(x,Tx),d(y,Ty),\frac{d(x,Tx)d(y,Ty)}{1+d(x,y)}\biggr\} , $$
-
(2)
there exists
\(x_{1}\in X\)
such that
\(\alpha ( x_{1},Tx_{1} ) \geq1\)
and
\(\alpha ( x_{1},T^{2}x_{1} ) \geq1\),
-
(3)
T
is an
α-orbital admissible mapping,
-
(4)
T
is an
α-orbital attractive mapping.
Then
T
has a unique fixed point
\(x_{\ast}\in X\)
and
\(\{ T^{n}x_{1} \} \)
converges to
\(x_{\ast}\).
Example 29
Let \(X= \{ 0,6,7,8 \} \) endow with the metric d given by \(d ( x,y ) =\vert x-y\vert \) for all \(x,y\in X\). It is easy to show that \(( X,d ) \) is a complete metric space. Let \(T:X\longrightarrow X\) be the mapping defined by
$$ T ( 0 ) =T ( 6 ) =7\quad\mbox{and}\quad T ( 7 ) =T ( 8 ) =8, $$
and \(\alpha:X\times X\longrightarrow [ 0,\infty ) \) be given by
$$ \alpha ( x,y ) =\left \{ \textstyle\begin{array}{@{}l@{\quad}l} 0 & \mbox{if } ( x,y ) \in \{ ( 6,7 ) , ( 7,6 ) \} , \\ 1 & \mbox{otherwise}. \end{array}\displaystyle \right . $$
Also define \(\theta: ( 0,\infty ) \longrightarrow ( 1,\infty ) \) by
$$ \theta ( t ) =e^{t\sqrt{t}}. $$
It is easy to show that T is an α-orbital admissible and α-orbital attractive mapping. Also the hypotheses of Theorem 27 (Corollary 28) are satisfied by T, and hence T has a fixed point. But the result of Jleli et al. (Corollary 25) cannot be applied to T. Indeed for \(x=6\), \(y=7\), we have
$$\begin{aligned} \theta \bigl( d \bigl( T ( 6 ) ,T ( 7 ) \bigr) \bigr) =&\theta \bigl( d ( 7,8 ) \bigr) =e \\ \nleq& [ e ] ^{k}= \bigl[ \theta \bigl( d ( 6,7 ) \bigr) \bigr] ^{k},\quad \mbox{for all }k\in ( 0,1 ) . \end{aligned}$$