## 3 Main Results
\begin{array}{ll}\|x_{n}-q_{n}\|&\leq\|x_{n}-(I-\alpha_{n}\mu A_{2})T^{n}y_ {n}\|\\ &\leq\|x_{n}-T^{n}y_{n}\|+\alpha_{n}\|\mu A_{2}(T^{n}y_{n})\|\to 0\quad(n\to\infty). \end{array} \]  
That is,  
\[\lim_{n\to\infty}\|x_{n}-q_{n}\|=0. \tag{40}\]  
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 Condition (v) and (39), we get  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0.\]  
In addition, noticing that  
\[ \begin{array}{ll}\|x_{n}-Tx_{n}\|&\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty _{n}-Tx_{n}\|\\ &\leq\|y_{n}-Ty_{n}\|+(2+\theta_{1})\|x_{n}-y_{n}\|,\end{array} \]  
we deduce from (36) that  
\[\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0. \tag{41}\]  
**Step 5. ** We claim that \(\|x_{n}-\overline{S}x_{n}\|\to 0\) as \(n\to\infty\) where \(\overline{S}:=(2I-S)^{-1}\). Indeed, first, let us 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 pseudocontractive, 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 means that \(S\) is pseudocontractive. Noting that \(\{S_{n}\}_{n=0}^{\infty}\) is \(\ell\)-uniformly Lipschitzian on \(C\), we have  
\[\|Sx-Sy\|=\lim_{n\to\infty}\|S_{n}x-S_{n}y\|\leq\ell\|x-y\|,\quad\forall x,y\in C.\]  
This means that \(S\) is \(\ell\)-Lipschitzian. Taking into account the boundedness of \(\{x_{n}\}\) and putting \(D=\overline{\operatorname{conv}}\{x_{n}:n\geq 0\}\) (the closure of convex hull of the set \(\{x_{n}:n\geq 0\}\)), by Assumption (C6) we have \(\sum_{n=1}^{\infty}\sup_{x\in D}\|S_{n}x-S_{n-1}x\|<\infty\). Hence, by Proposition 1, we get  
\[\lim_{n\to\infty}\sup_{x\in D}\|S_{n}x-Sx\|=0,\]  
which immediately yields  
\[\lim_{n\to\infty}\|S_{n}x_{n}-Sx_{n}\|=0. \tag{42}\]  
Thus, combining (38) with (42), we have  
\[\|x_{n}-Sx_{n}\|\leq\|x_{n}-S_{n}x_{n}\|+\|S_{n}x_{n}-Sx_{n}\|\to 0\quad(n\to\infty).\]  
That is,  
\[\lim_{n\to\infty}\|x_{n}-Sx_{n}\|=0. \tag{43}\]  
Now, let us show that. if we define \(\overline{S}:=(2I-S)^{-1}\), then \(\overline{S}:C\to C\) is nonexpansive, \(\operatorname{Fix}(\overline{S})=\operatorname{Fix}(S)=\bigcap_{n=0}^{ \infty}\operatorname{Fix}(S_{n})\) and \(\lim_{n\to\infty}\|x_{n}-\overline{S}x_{n}\|=0\). Indeed, put \(\overline{S}:=(2I-S)^{-1}\), where \(I\) is the identity mapping of \(H\). Then, it is known that \(\overline{S}\) is nonexpansive and the fixed point set \(\operatorname{Fix}(\overline{S})=\operatorname{Fix}(S)=\bigcap_{n=0}^{ \infty}\operatorname{Fix}(S_{n})\). From (43), it follows that  
\[ \begin{array}{ll}\|x_{n}-\overline{S}x_{n}\|&=\|\overline{S}^{-1}x_{n}- \overline{S}x_{n}\|\\ &\leq\|\overline{S}^{-1}x_{n}-x_{n}\|\\ &=\|(2I-S)x_{n}-x_{n}\|=\|x_{n}-Sx_{n}\|\to 0\quad(n\to\infty).\end{array} \]  
That is,  
\[\lim_{n\to\infty}\|x_{n}-\overline{S}x_{n}\|=0. \tag{44}\]  
**Step 6. ** We claim that  
\[\limsup_{n\to\infty}\langle A_{2}x^{*},x^{*}-q_{n}\rangle\leq 0\quad\text{and} \quad\limsup_{n\to\infty}\langle A_{1}x^{*},x^{*}-z_{n}\rangle\leq 0, \tag{45}\]  
where \(\{x^{*}\}=\operatorname{VI}(\operatorname{VI}(\Omega,A_{1}),A_{2})\). Indeed, we fix sequence \(\{q_{n_{i}}\}\) of \(\{q_{n}\}\) such that  
\[\limsup_{n\to\infty}\langle A_{2}x^{*},x^{*}-q_{n}\rangle=\lim_{i\to\infty} \langle A_{2}x^{*},x^{*}-q_{n_{i}}\rangle.\]  
Since \(\{q_{n}\}\) is a bounded sequence in \(C\), we may assume, without loss of generality, that \(q_{n_{i}}\rightharpoonup\bar{x}\in C\). Since \(\lim_{n\to\infty}\|x_{n}-q_{n}\|=0\) (due to (40)), it follows from \(q_{n_{i}}\rightharpoonup\bar{x}\) that \(x_{n_{i}}\rightharpoonup\bar{x}\).