We start this section with the notion of asymptotic center - essentially due to Edelstein [21] - of a sequence which is not only useful in proving a fixed point result but also plays a key role to define the concept of △-convergence in hyperbolic spaces. In 1976, Lim [22] introduced the concept of △-convergence in the general setting of metric spaces. In 2008, Kirk and Panyanak [23] further analyzed this concept in geodesic spaces. They showed that many Banach space results involving weak convergence have a precise analog version of △-convergence in geodesic spaces.
Let \(\{x_{n}\}\) be a bounded sequence in a hyperbolic space X. For \(x\in X\), define a continuous functional \(r(\cdot,\{x_{n}\}):X\rightarrow {}[ 0,\infty)\) by
$$ r \bigl(x,\{x_{n}\} \bigr)=\limsup_{n\rightarrow\infty}d(x,x_{n}). $$
The asymptotic radius and asymptotic center of the bounded sequence \(\{x_{n}\}\) with respect to a subset K of X is defined and denoted thus:
$$ r_{K} \bigl(\{x_{n}\} \bigr)=\inf \bigl\{ r \bigl(x, \{x_{n}\} \bigr):x\in K \bigr\} $$
and
$$ A_{K} \bigl(\{x_{n}\} \bigr)= \bigl\{ x\in K:r \bigl(x, \{x_{n}\} \bigr)\leq r \bigl(y,\{x_{n}\} \bigr) \text{ for all }y\in K \bigr\} , $$
respectively.
Recall that a sequence \(\{x_{n}\}\) in X is said to △-converge to \(x\in X\) if x is the unique asymptotic center of \(\{u_{n}\}\) for every subsequence \(\{u_{n}\}\) of \(\{x_{n}\}\). In this case, we write \(\triangle\mbox{-}\!\lim_{n}x_{n}=x\) and call x the △-limit of \(\{x_{n}\}\).
A mapping \(T:K\rightarrow K\) is semi-compact if every bounded sequence \(\{x_{n}\}\subset K\) satisfying \(\lim_{n\rightarrow \infty}d(x_{n},Tx_{n})= 0\), has a convergent subsequence.
We now list some useful lemmas as well as establish some auxiliary results required in the sequel.
Lemma 2.1
([24])
Let
\((X,d,W)\)
be a complete uniformly convex hyperbolic space with monotone modulus of uniform convexity
η. Then every bounded sequence
\(\{x_{n}\}\)
in
X
has a unique asymptotic center with respect to any nonempty closed convex subset
K
of
X.
Proposition 2.2
([25])
Let
\((X,d,W)\)
be a complete uniformly convex hyperbolic space with monotone modulus of uniform convexity
η. The intersection of any decreasing sequence of nonempty bounded closed convex subsets of
X
is nonempty.
Lemma 2.3
([26])
Let
\(\{ a_{n} \} \), \(\{ b_{n} \}\), and
\(\{ c_{n} \} \)
be sequences of nonnegative real numbers such that
\(\sum_{n=1}^{\infty }b_{n}<\infty\)
and
\(\sum_{n=1}^{\infty}c_{n}<\infty\). If
\(a_{n+1} \leq(1+b_{n})a_{n}+c_{n}\), \(n\geq1\), then
\(\lim_{n\rightarrow\infty}a_{n}\)
exists.
Lemma 2.4
([27])
Let
\((X,d,W)\)
be a uniformly convex hyperbolic space with monotone modulus of uniform convexity
η. Let
\(x\in X\)
and
\(\{\alpha_{n}\}\)
be a sequence in
\([a,b]\)
for some
\(a,b\in(0,1)\). If
\(\{x_{n}\}\)
and
\(\{y_{n}\}\)
are sequences in
X
such that
\(\limsup_{n\rightarrow\infty}d(x_{n},x)\leq c\), \(\limsup_{n\rightarrow\infty}d(y_{n},x)\leq c\)
and
\(\lim_{n\rightarrow\infty}d(W(x_{n},y_{n},\alpha _{n}),x)=c\)
for some
\(c\geq0\), then
\(\lim_{n\rightarrow\infty }d(x_{n},y_{n})=0\).
Lemma 2.5
([27])
Let
K
be a nonempty, closed, and convex subset of a uniformly convex hyperbolic space
X
and
\(\{x_{n}\}\)
a bounded sequence in
K
such that
\(A_{K}(\{x_{n}\})=\{y\}\)
and
\(r_{K}(\{x_{n}\})=\rho\). If
\(\{y_{m}\}\)
is another sequence in
K
such that
\(\lim_{m\rightarrow\infty}r(y_{m},\{x_{n}\})=\rho\), then
\(\lim_{m\rightarrow\infty}y_{m}=y\).
Lemma 2.6
Let
K
be a nonempty, closed, and convex subset of a hyperbolic space
X
and let
\(\{ T_{i} \} _{i=1}^{m}:K\rightarrow K\)
be a finite family of total asymptotically nonexpansive mappings with sequences
\(\{k_{in}\}\)
and
\(\{\varphi_{in}\}\), \(n\geq1\), \(i=1,2,\ldots,m\)
such that
\(F=\bigcap_{i=1}^{m}F ( T_{i} ) \neq\phi\). For
\(i=1,2,\ldots,m\)
if the following conditions are satisfied:
-
(C1)
\(\sum_{n=1}^{\infty}k_{in}<\infty\)
and
\(\sum_{n=1}^{\infty}\varphi_{in}<\infty\);
-
(C2)
there exist constants
\(M_{i}, M_{i}^{\ast}>0\)
such that
\(\xi_{i} ( \lambda_{i} ) \leq M_{i}^{\ast}\lambda _{i}\)
for all
\(\lambda_{i}\geq M_{i}\), \(i=1,2,3,\ldots,m\),
then the sequence
\(\{ x_{n} \} \)
given by (1.1) is bounded and
\(\lim_{n\rightarrow\infty}d ( x_{n},p ) \)
exists for each
\(p\in F\).
Proof
Let \(p\in F\) and \(m=1\) in (1.1), we have
$$\begin{aligned} d ( x_{n+1},p ) & =d \bigl( W \bigl( x_{n},T_{1}^{n}x_{n}, \alpha _{n} \bigr) ,p \bigr) \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n}d \bigl( T_{1}^{n}x_{n},p \bigr) \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n} \bigl\{ d ( x_{n},p ) +k_{1n }\xi_{1} \bigl( d ( x_{n},p ) \bigr) +\varphi_{1n} \bigr\} \\ & \leq d(x_{n},p)+bk_{1n}\xi_{1} \bigl( d ( x_{n},p ) \bigr) +b\varphi_{1n}, \end{aligned}$$
(2.1)
where \(\alpha_{n}\leq b\) for some \(b>0\).
Since \(\xi_{1}\) is an increasing function, \(\xi ( d ( x_{n},p ) ) \leq\xi ( M_{1} ) \) for \(d ( x_{n},p ) \leq M_{1}\). Moreover, \(\xi ( d ( x_{n},p ) ) \leq d ( x_{n},p ) M_{1}^{\ast}\) for \(d ( x_{n},p ) \geq M_{1}\) (by (C2)). In either case, we have
$$ \xi \bigl( d ( x_{n},p ) \bigr) \leq\xi ( M_{1} ) +d ( x_{n},p ) M_{1}^{\ast}, $$
(2.2)
where \(M_{1}, M_{1}^{\ast}>0\).
Utilizing (2.2) in (2.1), we have
$$ d ( x_{n+1},p ) \leq d ( x_{n},p ) +bk_{1n}\xi _{1} ( M_{1} ) +bk_{1n}M_{1}^{\ast}d ( x_{n},p ) +b\varphi_{1n}. $$
On setting \(a_{1}=\max\{b,bM_{1}^{\ast},b\xi_{1} ( M_{1} ) \}>0\), the above inequality becomes
$$\begin{aligned} d ( x_{n+1},p ) \leq&d ( x_{n},p ) +a_{1}k_{1n}+a_{1}k_{1n}d ( x_{n},p ) +a_{1}\varphi_{1n} \\ =&(1+a_{1}k_{1n})d ( x_{n},p ) +a_{1}k_{1n}+a_{1}\varphi_{1n}. \end{aligned}$$
If \(m=2\) in (1.1), we have the following iteration scheme:
$$\begin{aligned}& x_{1}\in K, \\& x_{n+1}=W \bigl( x_{n},T_{1}^{n}y_{1n}, \alpha_{n} \bigr) , \\& y_{1n}=W \bigl( x_{n},T_{2}^{n}x_{n}, \alpha_{n} \bigr) . \end{aligned}$$
In view of the above iteration scheme, we have the following estimates:
$$\begin{aligned} d ( x_{n+1},p ) & =d \bigl( W \bigl( x_{n},T_{1}^{n}y_{1n}, \alpha _{n} \bigr) ,p \bigr) \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n}d \bigl( T_{1}^{n}y_{1n},p \bigr) \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n} \bigl\{ d ( y_{1n},p ) +k_{1n }\xi_{1} \bigl( d ( y_{1n},p ) \bigr) +\varphi_{1n} \bigr\} \end{aligned}$$
(2.3)
and
$$\begin{aligned} d ( y_{1n},p ) & =d \bigl( W \bigl( x_{n},T_{2}^{n}x_{n}, \alpha _{n} \bigr) ,p \bigr) \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n}d \bigl( T_{2}^{n}x_{n},p \bigr) \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n} \bigl\{ d ( x_{n},p ) +k_{2n }\xi_{2} \bigl( d ( x_{n},p ) \bigr) +\varphi_{2n} \bigr\} . \end{aligned}$$
(2.4)
Again, we follow the method of proof as discussed for \(m=1\).
In the light of (C2) and the increasing functions \(\xi_{i}\) (\(i=1,2\)), there exist \(M_{i}, M_{i}^{\ast}>0\) (\(i=1,2\)) such that
$$ \xi_{1} \bigl( d ( y_{1n},p ) \bigr) \leq\xi_{1} ( M_{1} ) +d ( y_{1n},p ) M_{1}^{\ast} $$
(2.5)
and
$$ \xi_{2} \bigl( d ( x_{n},p ) \bigr) \leq\xi_{2} ( M_{2} ) +d ( x_{n},p ) M_{2}^{\ast}. $$
(2.6)
Substituting (2.5) in (2.3) and (2.6) in (2.4), we get
$$\begin{aligned} d ( x_{n+1},p ) \leq& ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha_{n} \bigl\{ d ( y_{1n},p ) +k_{1n }\xi _{1} \bigl( d ( y_{1n},p ) \bigr) + \varphi_{1n} \bigr\} \\ \leq&(1-\alpha_{n})d(x_{n},p)+\alpha_{n}d(y_{1n},p)+ \alpha _{n}k_{1n}\xi _{1}(M_{1}) \\ &{} +\alpha_{n}k_{1n}d(y_{1n},p)M_{1}^{\ast}+ \alpha_{n}\varphi_{n} \\ =&(1-\alpha_{n})d(x_{n},p)+\alpha_{n} \bigl(1+k_{1n}M_{1}^{\ast } \bigr)d(y_{1n},p)+ \alpha_{n}k_{1n}\xi_{1}(M_{1})+ \alpha_{n}\varphi_{n} \end{aligned}$$
(2.7)
and
$$\begin{aligned} d ( y_{1n},p ) & =d \bigl( W \bigl( x_{n},T_{2}^{n}x_{n}, \alpha _{n} \bigr) ,p \bigr) \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n}d \bigl( T_{2}^{n}x_{n},p \bigr) \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n} \bigl\{ d ( x_{n},p ) +k_{2n } \bigl( \xi_{2} \bigl( d ( x_{n},p ) \bigr) \bigr) +\varphi_{2n} \bigr\} \\ & \leq ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha _{n} \bigl\{ d ( x_{n},p ) +k_{2n } \bigl( \xi_{2} ( M_{2} ) +d ( x_{n},p ) M_{2}^{\ast} \bigr) + \varphi _{2n} \bigr\} \\ & =d ( x_{n},p ) +\alpha_{n}k_{2n } \xi_{2} ( M_{2} ) +\alpha_{n}k_{2n}d ( x_{n},p ) M_{2}^{\ast}+\alpha _{n}\varphi _{2n}. \end{aligned}$$
(2.8)
Putting (2.8) in (2.7), we have
$$\begin{aligned} \begin{aligned} d ( x_{n+1},p ) \leq{}& ( 1-\alpha_{n} ) d ( x_{n},p ) +\alpha_{n} \bigl( 1+k_{1n}M_{1}^{\ast} \bigr) \bigl[d ( x_{n},p ) +\alpha_{n}k_{2n } \xi_{2} ( M_{2} ) \\ &{}+\alpha_{n}k_{2n}d ( x_{n},p ) M_{2}^{\ast}+\alpha _{n}\varphi _{2n} \bigr]+\alpha_{n}k_{1n }\xi_{1} ( M_{1} ) +\alpha _{n}\varphi_{1n} \\ ={}& \bigl( 1+\alpha_{n}k_{1n}M_{1}^{\ast}+ \alpha _{n}^{2}k_{2n}M_{2}^{\ast }+ \alpha_{n}^{2}k_{1n}k_{2n}M_{1}^{\ast}M_{2}^{\ast} \bigr) d ( x_{n},p ) \\ &{}+\alpha_{n} \bigl( 1+k_{1n}M_{1}^{\ast} \bigr) \bigl[\alpha_{n}k_{2n }\xi _{2} ( M_{2} ) +\alpha_{n}\varphi_{2n} \bigr] \\ &{}+\alpha_{n}k_{1n }\xi_{1} ( M_{1} ) +\alpha_{n}\varphi_{1n} \\ \leq{}& \bigl( 1+\alpha_{n}k_{1n}M_{1}^{\ast}+ \alpha _{n}k_{2n}M_{2}^{\ast }+ \alpha_{n}k_{1n}k_{2n}M_{1}^{\ast}M_{2}^{\ast} \bigr) d ( x_{n},p ) \\ &{}+\alpha_{n}k_{2n }\xi_{2} ( M_{2} ) + \alpha _{n}k_{1n}k_{2n }M_{1}^{\ast} \xi_{2} ( M_{2} ) \\ &{}+\alpha_{n}\varphi_{2n}+\alpha_{n}k_{1n} \varphi_{2n}M_{1}^{\ast }+\alpha_{n}k_{1n } \xi_{1} ( M_{1} ) +\alpha_{n} \varphi_{1n}. \end{aligned} \end{aligned}$$
Again \(\alpha_{n}\leq b\) for all \(n\geq1\). Further we can bound the convergent sequences \(\{k_{1n}\}\), \(\{k_{2n}\}\), and \(\{\varphi_{2n}\}\) as \(k_{1n}\leq c_{1}\), \(k_{2n}\leq c_{2}\), and \(\varphi_{2n}\leq d_{2}\) for all \(n\geq1\).
Utilizing these bounds, the above estimate becomes
$$\begin{aligned} d ( x_{n+1},p ) \leq& \bigl[ 1+bk_{1n}M_{1}^{\ast}+ \bigl( bM_{2}^{\ast}+bc_{1}M_{1}^{\ast}M_{2}^{\ast} \bigr) k_{2n } \bigr] d ( x_{n},p ) \\ &{}+b\xi_{2} ( M_{2} ) k_{2n }+bc_{2}M_{1}^{\ast} \xi _{2} ( M_{2} ) k_{1n} \\ &{}+b\varphi_{2n}+bc_{1}M_{1}^{\ast} \varphi_{2n}+bk_{1n }\xi _{1} ( M_{1} ) +b\varphi_{1n} \\ =& \bigl[ 1+bM_{1}^{\ast}k_{1n}+b \bigl( M_{2}^{\ast}+c_{1}M_{1}^{\ast }M_{2}^{\ast} \bigr) k_{2n} \bigr] d ( x_{n},p ) \\ &{}+b \bigl( c_{2}M_{1}^{\ast}\xi_{2} ( M_{2} ) +\xi_{1} ( M_{1} ) \bigr) k_{1n}+b\xi_{2} ( M_{2} ) k_{2n } \\ &{}+b\varphi_{1n}+b \bigl( 1+c_{1}M_{1}^{\ast} \bigr) \varphi_{2n}. \end{aligned}$$
That is,
$$ d ( x_{n+1},p ) \leq \bigl( 1+ ( k_{1n }+k_{2n } ) a_{2} \bigr) d ( x_{n},p ) + ( k_{2n }+k_{1n }+ \varphi _{2n}+\varphi_{1n} ) a_{2}, $$
where
$$\begin{aligned} a_{2}&=\max \bigl\{ b,bM_{1}^{\ast},b ( 1+c_{1} ) M_{1}^{\ast},b \bigl( M_{2}^{\ast}+c_{1}M_{1}^{\ast}M_{2}^{\ast } \bigr) ,b \bigl( c_{2}M_{1}^{\ast}\xi_{2} ( M_{2} ) +\xi_{1} ( M_{1} ) \bigr) ,b \xi_{2} ( M_{2} ) \bigr\} \\ &>0. \end{aligned}$$
That is,
$$ d ( x_{n+1},p ) \leq \Biggl( 1+a_{2}\sum _{j=1}^{2}k_{jn } \Biggr) d ( x_{n},p ) +a_{2}\sum_{j=1}^{2} ( k_{jn }+\varphi_{jn} ) . $$
Continuing in a similar fashion for any \(m\geq1\), we have the following compact form of the above estimate:
$$ d ( x_{n+1},p ) \leq \Biggl( 1+a_{m}\sum _{j=1}^{m}k_{jn } \Biggr) d ( x_{n},p ) +a_{m}\sum_{j=1}^{m} ( k_{jn }+\varphi_{jn} ) $$
for some constant \(a_{m}>0\).
Appealing to Lemma 2.3, the above inequality implies that \(\lim_{n\rightarrow\infty}d ( x_{n},p ) \) exists and hence the sequence \(\{ x_{n} \} \) is bounded. □
Lemma 2.7
Let
K
be a nonempty, closed, and convex subset of a uniformly convex hyperbolic space
X
with monotone modulus of uniform convexity
η
and let
\(\{ T_{i} \} _{i=1}^{m}:K\rightarrow K\)
be a finite family of uniformly continuous total asymptotically nonexpansive mappings with sequences
\(\{k_{in}\}\)
and
\(\{\varphi_{in}\}\), \(n\geq1\), \(i=1,2,\ldots,m\)
such that
\(F:=\bigcap_{i=1}^{m}F ( T_{i} ) \neq\phi\). For
\(i=1,2,\ldots,m\), if the following conditions are satisfied:
-
(C1)
\(\sum_{n=1}^{\infty}k_{in}<\infty\)
and
\(\sum_{n=1}^{\infty}\varphi_{in}<\infty\);
-
(C2)
there exist constants
\(M_{i}, M_{i}^{\ast}>0\)
such that
\(\xi_{i} ( \lambda_{i} ) \leq M_{i}^{\ast}\lambda _{i}\)
for all
\(\lambda_{i}\geq M_{i}\),
then for the sequence
\(\{ x_{n} \} \)
given by (1.1), we have
\(\lim_{n\rightarrow\infty}d ( x_{n},T_{i}x_{n} ) =0\), \(i=1,2,\ldots,m\).
Proof
It follows from Lemma 2.6 that \(\lim_{n\rightarrow \infty}d ( x_{n},p ) \) exists. Without loss of generality, we can assume that \(\lim_{n\rightarrow\infty}d ( x_{n},p ) =r>0\). We first distinguish two cases to show that \(\lim_{n\rightarrow \infty}d ( x_{n},T_{i}^{n}x_{n} ) =0\), \(i=1,2,\ldots ,m\).
Case 1. For \(m=1\), we proceed as follows:
$$ \lim_{n\rightarrow\infty}d \bigl( W \bigl( x_{n},T_{1}^{n}x_{n}, \alpha_{n} \bigr) ,p \bigr) =\lim_{n\rightarrow \infty}d ( x_{n+1},p ) =r. $$
(2.9)
It follows from the total asymptotically nonexpansiveness of \(T_{1}\), that is,
$$\begin{aligned} d \bigl( T_{1}^{n}x_{n},p \bigr) =&d \bigl( T_{1}^{n}x_{n},T_{1}^{n}p \bigr) \leq d ( x_{n},p ) +k_{1n }\xi_{1} \bigl(d ( x_{n},p ) \bigr)+\varphi_{1n} \\ \leq&d ( x_{n},p ) +k_{1n }M_{1}^{\ast}d ( x_{n},p ) +k_{1n}\xi_{1} ( M_{1} ) + \varphi_{1n} \\ =& \bigl(1+k_{1n }M_{1}^{\ast} \bigr)d ( x_{n},p ) +k_{1n}\xi_{1} ( M_{1} ) + \varphi_{1n}. \end{aligned}$$
Taking lim sup on both sides of the above estimate and utilizing the fact that \(k_{1n}\rightarrow0\) and \(\varphi_{1n}\rightarrow0\) as \(n\rightarrow\infty\), we get
$$ \limsup_{n\rightarrow\infty}d \bigl( T_{1}^{n}x_{n},p \bigr) \leq r. $$
Moreover, \(\limsup_{n\rightarrow\infty}d ( x_{n},p ) =r\). Hence, the conclusion, i.e., \(\lim_{n\rightarrow\infty }d ( T_{1}^{n}x_{n},x_{n} ) =0\) follows from Lemma 2.4.
Case 2. For \(m=2\), the iteration (1.1) reduces to
$$\begin{aligned}& x_{n+1} = W \bigl( x_{n},T_{1}^{n}y_{1n}, \alpha_{n} \bigr) , \\& y_{1n} = W \bigl( x_{n},T_{2}^{n}x_{n}, \alpha_{n} \bigr) . \end{aligned}$$
As calculated in Lemma 2.6, we have
$$\begin{aligned} d ( y_{1n},p ) =&d \bigl( W \bigl( x_{n},T_{2}^{n}x_{n}, \alpha _{n} \bigr) ,p \bigr) \\ \leq& \bigl(1+k_{2n }M_{2}^{\ast} \bigr)d ( x_{n},p ) +\alpha _{n}k_{2n}\xi _{2} ( M_{2} ) +\alpha_{n}\varphi_{2n}. \end{aligned}$$
Taking lim sup on both sides in the above estimate, we have
$$ \limsup_{n\rightarrow\infty}d ( y_{1n},p ) \leq r. $$
(2.10)
Now observe that
$$ d \bigl( T_{1}^{n}y_{1n},p \bigr) \leq \bigl(1+k_{1n }M_{1}^{\ast} \bigr)d ( y_{1n},p ) +k_{1n}\xi_{1} ( M_{1} ) + \varphi_{1n}. $$
Taking lim sup on both sides of the above estimate and utilizing (2.10), we get
$$ \limsup_{n\rightarrow\infty}d \bigl( T_{1}^{n}y_{1n},p \bigr) \leq r. $$
(2.11)
Since \(\limsup_{n\rightarrow\infty}d ( x_{n},p ) \leq r\), it follows from (2.11) and Lemma 2.4 that
$$ \lim_{n\rightarrow\infty}d \bigl( x_{n},T_{1}^{n}y_{1n} \bigr) =0. $$
(2.12)
Observe that
$$\begin{aligned} d ( x_{n},p ) \leq&d \bigl( x_{n},T_{1}^{n}y_{1n} \bigr) +d \bigl( T_{1}^{n}y_{1n},p \bigr) \\ \leq&d \bigl( x_{n},T_{1}^{n}y_{1n} \bigr) + \bigl(1+k_{1n }M_{1}^{\ast } \bigr)d ( y_{1n},p ) +k_{1n}\xi_{1} ( M_{1} ) + \varphi_{1n}. \end{aligned}$$
Hence, we deduce from the above estimate that
$$ r\leq\liminf_{n\rightarrow\infty}d ( y_{1n},p ) . $$
(2.13)
The estimates (2.10) and (2.13) collectively imply that
$$ \lim_{n\rightarrow\infty}d ( y_{1n},p ) =d \bigl( W \bigl( x_{n},T_{2}^{n}x_{n},\alpha_{n} \bigr) ,p \bigr) =r. $$
(2.14)
Utilizing the total asymptotically nonexpansiveness of \(T_{2}\) and the fact that \(k_{2n}\rightarrow0\), \(\varphi_{2n}\rightarrow0\) as \(n\rightarrow \infty\), we have
$$ \limsup_{n\rightarrow\infty}d \bigl( T_{2}^{n}x_{n},p \bigr) \leq r. $$
(2.15)
Now (2.14), (2.15), and Lemma 2.4, imply that
$$ \lim_{n\rightarrow\infty}d \bigl( T_{2}^{n}x_{n},x_{n} \bigr) =0. $$
(2.16)
With the help of \(d ( y_{1n},x_{n} ) \leq\alpha_{n}d ( T_{2}^{n}x_{n},x_{n} ) \leq bd ( T_{2}^{n}x_{n},x_{n} ) \) and (2.16), we obtain
$$ \lim_{n\rightarrow\infty}d ( y_{1n},x_{n} ) =0. $$
(2.17)
It follows from (2.12) and (2.17) that
$$ \lim_{n\rightarrow\infty}d \bigl( y_{1n},T_{1}^{n}y_{1n} \bigr) =0. $$
(2.18)
Moreover
$$ d \bigl( T_{1}^{n}x_{n},x_{n} \bigr) \leq d \bigl( T_{1}^{n}x_{n},T_{1}^{n}y_{1n} \bigr) +d \bigl( T_{1}^{n}y_{1n},x_{n} \bigr) . $$
Since \(T_{1}\) is uniformly continuous, letting \(n\rightarrow \infty\) in the above estimate and utilizing (2.12) and (2.17), we get
$$ \lim_{n\rightarrow\infty}d \bigl( T_{1}^{n}x_{n},x_{n} \bigr) =0. $$
Hence
$$ \lim_{n\rightarrow\infty}d \bigl( T_{i}^{n}x_{n},x_{n} \bigr) =0\quad \text{for }i=1,2. $$
Continuing in a similar way, we get
$$ \lim_{n\rightarrow\infty}d \bigl( T_{i}^{n}x_{n},x_{n} \bigr) =0\quad \text{for }i=1,2,\ldots,m. $$
Now, utilizing the uniform continuity of \(T_{i}\), the following estimate:
$$ d ( x_{n},T_{i}x_{n} ) \leq d \bigl( x_{n},T_{i}^{n}x_{n} \bigr) +d \bigl( T_{i}^{n}x_{n},T_{i}x_{n} \bigr) $$
implies that
$$ \lim_{n\rightarrow\infty}d ( T_{i}x_{n},x_{n} ) =0\quad \text{for }i=1,2,\ldots,m. $$
This completes the proof. □