## 3 Main Results
By (5) and (28), we obtain that  
\[0\leq\|S_{n}u_{n}-u_{n}\|=\frac{\gamma_{n}}{1-\gamma_{n}}\|x_{n}-u_{n}\|\leq \frac{b}{1-b}\|x_{n}-u_{n}\|\to 0\quad(n\to\infty). \tag{30}\]  
Since \(\{S_{n}\}_{n=0}^{\infty}:C\to C\) is \(\ell\)-uniformly Lipschitzian, we deduce from (28) and (30) that  
\[ \begin{array}{ll}\|S_{n}x_{n}-x_{n}\|&\leq\|S_{n}x_{n}-S_{n}u_{n}\|+\|S_{n} u_{n}-u_{n}\|+\|u_{n}-x_{n}\|\\ &\leq(\ell+1)\|x_{n}-u_{n}\|+\|S_{n}u_{n}-u_{n}\|\to 0\quad(n\to\infty).\end{array}  \tag{31}\]  
Since \(1\geq\limsup_{n\to\infty}\beta_{n}\) and \(\liminf_{n\to\infty}\beta_{n}\geq 0\), One supposes, without loss of generality, that \(\{\beta_{n}\}\subset[c,d]\subset(0,1)\) for some \(c,d\in(0,1)\). From \(x_{n+1}=\beta_{n}x_{n}+(1-\beta_{n})\Pi_{C}(I-\alpha_{n}F)T^{n}y_{n}\), we obtain  
\[\|x_{n}-T^{n}y_{n}\|\leq\|x_{n}-x_{n+1}\|+\|x_{n+1}-T^{n}y_{n}\|\leq\|x_{n}-x_ {n+1}\|+\beta_{n}\|x_{n}-T^{n}y_{n}\|+(1-\beta_{n})\alpha_{n}\|F(T^{n}y_{n})\|.\]  
Then we have \((1-d)\|x_{n}-T^{n}y_{n}\|\leq\|x_{n}-x_{n+1}\|+(1-d)\alpha_{n}\|F(T^{n}y_{n})\|\). Hence we get  
\[\|y_{n}-T^{n}y_{n}\|\leq\|y_{n}-x_{n}\|+\|x_{n}-T^{n}y_{n}\|\leq\|y_{n}-x_{n} \|+\frac{1}{1-d}\|x_{n}-x_{n+1}\|+\alpha_{n}\|F(T^{n}y_{n})\|.\]  
Consequently, from (14), (29) and \(\lim_{n\to\infty}\alpha_{n}=0\), it follows that  
\[\lim_{n\to\infty}\|x_{n}-T^{n}y_{n}\|=0\quad\text{and}\quad\lim_{n\to\infty} \|y_{n}-T^{n}y_{n}\|=0. \tag{32}\]  
We also note that \(\|y_{n}-Ty_{n}\|\leq\|y_{n}-T^{n}y_{n}\|+\|T^{n}y_{n}-T^{n+1}y_{n}\|+\|T^{n+1} y_{n}-Ty_{n}\|\). By the assumption \(\lim_{n\to\infty}\|T^{n+1}y_{n}-T^{n}y_{n}\|=0\), (32) and the condition that \(T:C\to C\) is uniformly continuous, we get  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{33}\]  
In addition, noticing that \(\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|Ty_{n}-Tx_{n}\|+\|y_{n}-Ty_{n}\|\), we deduce from (29), (33) and the uniform continuity of \(T\) that  
\[\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0. \tag{34}\]  
**Step 5. ** We claim that \(\|x_{n}-\overline{S}x_{n}\|\to 0\) as \(n\to\infty\) where \(\overline{S}:=(I+I-S)^{-1}\). Indeed, we first 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 pseudocontractive, we get \(\langle Sx-Sy,j(x-y)\rangle=\lim_{n\to\infty}\langle S_{n}x-S_{n}y,j(x-y) \rangle\leq\|x-y\|^{2}\). Thus, \(S\) is pseudocontractive. Also, since \(\{S_{n}\}_{n=0}^{\infty}\) is \(\ell\)-uniformly Lipschitzian on \(C\), we get \(\|Sx-Sy\|=\lim_{n\to\infty}\|S_{n}x-S_{n}y\|\leq\ell\|x-y\|\ \forall x,y\in C\). This means that \(S\) is \(\ell\)-Lipschitzian. Noticing 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 \(\sum_{n=1}^{\infty}\sup_{x\in D}\|S_{n}x-S_{n-1}x\|<\infty\). Therefore, by Proposition 1 we get \(\lim_{n\to\infty}\sup_{x\in D}\|S_{n}x-Sx\|=0\), which immediately yields  
\[\lim_{n\to\infty}\|S_{n}x_{n}-Sx_{n}\|=0. \tag{35}\]  
Thus, combining (31) and (35) implies  
\[0\leq\|x_{n}-Sx_{n}\|\leq\|x_{n}-S_{n}x_{n}\|+\|S_{n}x_{n}-Sx_{n}\|\to 0\quad(n\to\infty). \tag{36}\]Next, putting \(\overline{S}:=(2I-S)^{-1}\), one sees that \(\overline{S}\) is nonexpansive, and \(\operatorname{Fix}(\overline{S})=\operatorname{Fix}(S)=\bigcap_{n=0}^{\infty} \operatorname{Fix}(S_{n})\) (see [16]). From (36) it follows that  
\[ \begin{array}{ll}\|x_{n}-\overline{S}x_{n}\|&=\|\overline{S}\overline{S}^{-1 }x_{n}-\overline{S}x_{n}\|\leq\|\overline{S}^{-1}x_{n}-x_{n}\|\\ &=\|(2I-S)x_{n}-x_{n}\|=\|x_{n}-Sx_{n}\|\to 0\quad(n\to\infty).\end{array}  \tag{37}\]  
**Step 6. ** One proves that  
\[\limsup_{n\to\infty}\langle F(x^{*}),j(x^{*}-w_{n})\rangle\leq 0, \tag{38}\]  
where \(x^{*}=\Pi_{\Omega}(I-\rho F)(x^{*})\) with \(\rho\in(0,\|F\|^{-1})\). There is a sequence \(\{w_{n_{i}}\}\) such that \(\limsup_{n\to\infty}\langle F(x^{*}),j(x^{*}-w_{n})\rangle=\lim_{i\to\infty} \langle F(x^{*}),j(x^{*}-w_{n_{i}})\rangle\). Now we show that \(\Pi_{\Omega}(I-\rho F)\) is a contraction.