## 2 Preliminaries
** Let us prove \(\|x_{n}-S_{n}x_{n}\|\to 0,\ \|x_{n}-w_{n}\|\to 0\) and \(\|x_{n}-Tx_{n}\|\to 0\) as \(n\to\infty\). Indeed, combining (5) and (8), we obtain that  
\[\quad-\|u_{n}-v_{n}-(p-q)\|^{2}-\|v_{n}-z_{n}+(p-q)\|^{2}+2\mu_{1} \|B_{1}q-B_{1}v_{n}\|\|z_{n}-p\|\]1  
That is,  
\[\quad-\|u_{n}-v_{n}-(p-q)\|^{2}-\|v_{n}-z_{n}+(p-q)\|^{2}+2\mu_{1} \|B_{1}q-B_{1}v_{n}\|\|z_{n}-p\|\]2Observe \(\{S_{n}\}_{n=0}^{\infty}\) is \(\ell\)-uniformly Lipschitzian. We further get from (28) and (30) that  
\[ \begin{array}{ll}\|S_{n}x_{n}-x_{n}\|&\leq\|S_{n}x_{n}-S_{n}u_{n}\|+\|S_{n}u_{n }-u_{n}\|+\|u_{n}-x_{n}\|\\ &\leq\ell\|x_{n}-u_{n}\|+\|S_{n}u_{n}-u_{n}\|+\|u_{n}-x_{n}\|\\ &=(\ell+1)\|x_{n}-u_{n}\|+\|S_{n}u_{n}-u_{n}\|\to 0\quad(n\to\infty).\end{array} \]  
That is,  
\[\lim_{n\to\infty}\|x_{n}-S_{n}x_{n}\|=0. \tag{31}\]  
We note that \(\{\alpha_{n}+\beta_{n}\}\subset[c,d]\subset(0,1)\) for some \(c,d\in(0,1)\), and observe that  
\[ \begin{array}{ll}\|x_{n}-T^{n}y_{n}\|\leq\|x_{n}-x_{n+1}\|+\|x_{n+1}-T^{n}y_ {n}\|\\ \leq\|x_{n}-x_{n+1}\|+\alpha_{n}\|f(x_{n})-T^{n}y_{n}\|+\beta_{n}\|x_{n}-T^{n }y_{n}\|\\ \quad+(1-\alpha_{n}-\beta_{n})\|P_{\infty}\|a_{n}x_{n}+(I-\alpha_{n}\mu F)T^{ n}y_{n}\|-T^{n}y_{n}\|\\ \leq\|x_{n}-x_{n+1}\|+\alpha_{n}\|f(x_{n})-T^{n}y_{n}\|+\beta_{n}\|x_{n}-T^{n }y_{n}\|+a_{n}\|x_{n}-\mu F(T^{n}y_{n})\|.\end{array} \]  
Then,  
\[ \begin{array}{ll}\|x_{n}-T^{n}y_{n}\|&\leq\frac{1}{1-\beta_{n}}\{\|x_{n}-x_{ n+1}\|+\alpha_{n}(\|f(x_{n})-T^{n}y_{n}\|+\|x_{n}-\mu F(T^{n}y_{n})\|)\}\\ &\leq\frac{1}{1-d}\{\|x_{n}-x_{n+1}\|+\alpha_{n}(\|f(x_{n})-T^{n}y_{n}\|+\|x_{ n}-\mu F(T^{n}y_{n})\|)\}.\end{array} \]  
Hence, we get  
\[ \begin{array}{ll}\|y_{n}-T^{n}y_{n}\|&\leq\|y_{n}-x_{n}\|+\|x_{n}-T^{n}y_{n} \|\\ &\leq\|y_{n}-x_{n}\|+\frac{1}{1-d}\{\|x_{n}-x_{n+1}\|+\alpha_{n}(\|f(x_{n})-T^ {n}y_{n}\|+\|x_{n}-\mu F(T^{n}y_{n})\|)\}.\end{array} \]  
Consequently, from (15), (29) and \(\lim_{n\to\infty}\alpha_{n}=0\), we obtain that  
\[\lim_{n\to\infty}\|x_{n}-T^{n}y_{n}\|=0\quad\text{and}\quad\lim_{n\to\infty} \|y_{n}-T^{n}y_{n}\|=0. \tag{32}\]  
So it follows that  
\[ \begin{array}{ll}\|x_{n}-w_{n}\|&\leq\|x_{n}-\alpha_{n}x_{n}-(I-\alpha_{n} \mu F)T^{n}y_{n}\|\\ &\leq\|x_{n}-T^{n}y_{n}\|+\alpha_{n}\|x_{n}-\mu F(T^{n}y_{n})\|\to 0\quad(n\to\infty). \end{array} \]  
That is,  
\[\lim_{n\to\infty}\|x_{n}-w_{n}\|=0. \tag{33}\]  
We also note that  
\[ \begin{array}{ll}\|y_{n}-Ty_{n}\|&\leq\|y_{n}-T^{n}y_{n}\|+\|T^{n}y_{n}-T^{n +1}y_{n}\|+\|T^{n+1}y_{n}-Ty_{n}\|\\ &\leq\|y_{n}-T^{n}y_{n}\|+\|T^{n}y_{n}-T^{n+1}y_{n}\|+(1+\theta_{1})\|T^{n}y_{n }-y_{n}\|\\ &=\|T^{n}y_{n}-T^{n+1}y_{n}\|+(2+\theta_{1})\|T^{n}y_{n}-y_{n}\|.\end{array} \]  
By the condition (v) and (32), we get  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0.\]  
Further, noticing that  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|\leq\|y _{n}-Ty_{n}\|+(2+\theta_{1})\|x_{n}-y_{n}\|,\]  
we deduce from (29) that  
\[\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0. \tag{34}\]  
**Step 5. ** Set \(\overline{S}:=(2I-S)^{-1}\). We aim to prove \(\|x_{n}-\overline{S}x_{n}\|\to 0\) as \(n\to\infty\). We show that \(S:C\to C\) is pseudocontractive and \(\ell\)-Lipschitzian such that \(\lim_{n\to\infty}\|Sx_{n}-x_{n}\|=0\), where \(Sx=\lim_{n\to\infty}S_{n}x\,\forall x\in C\). Observe that for all \(x,y\in C\), \(\lim_{n\to\infty}\|S_{n}x-Sx\|=0\) and \(\lim_{n\to\infty}\|S_{n}y-Sy\|=0\). Since each \(S_{n}\) is a pseudocontractive operator, we get  
\[\langle Sx-Sy,x-y\rangle=\lim_{n\to\infty}\langle S_{n}x-S_{n}y,x-y\rangle\leq \|x-y\|^{2}.\]  
This presents that \(S\) is pseudocontractive. Note that \(\{S_{n}\}_{n=0}^{\infty}\) is \(\ell\)-uniformly Lipschitzian  
\[\|Sx-Sy\|=\lim_{n\to\infty}\|S_{n}x-S_{n}y\|\leq\ell\|x-y\|,\quad\forall x,y \in C.\]  
This means that \(S\) is \(\ell\)-Lipschitzian. Since the boundedness of \(\{x_{n}\}\) and putting \(D=\overline{\operatorname{conv}}\{x_{n}:n\geq 0\}\) (the closure of convex hull of the set \(\{x_{n}:n\geq 0\}\)), we have \(\sum_{n=1}^{\infty}\sup_{x\in D}\|S_{n}x-S_{n-1}x\|<\infty\).