## 3 Main Results
Indeed, take an element \(p\in\Omega=\bigcap_{n=0}^{\infty}\text{Fix}(S_{n})\cap\text{GSVI}(C,B_{1},B_{2 })\cap\text{Fix}(T)\) arbitrarily. Then, we have \(S_{n}p=p\), \(Gp=p\) and \(Tp=p\). Since each \(S_{n}:C\to C\) is a pseudocontraction mapping, it follows that  
\[ \begin{array}{ll}\|u_{n}-p\|^{2}&=\gamma_{n}\langle x_{n}-p,u_{n}-p\rangle+( 1-\gamma_{n})\langle S_{n}u_{n}-p,u_{n}-p\rangle\\ &\leq\gamma_{n}\|x_{n}-p\|\|u_{n}-p\|+(1-\gamma_{n})\|u_{n}-p\|^{2},\end{array} \]  
which hence yields  
\[\|u_{n}-p\|\leq\|x_{n}-p\|,\quad\forall n\geq 0. \tag{13}\]  
This shows that \(\{v_{n}\}\) is bounded. **Step 2. ** We claim that \(\|x_{n+1}-x_{n}\|\to 0\) and \(\|y_{n+1}-y_{n}\|\to 0\) as \(n\to\infty\). Indeed, we set \(p_{n}=P_{\mathbb{C}}(I-\delta_{n}A_{1})z_{n}\) and \(q_{n}=P_{\mathbb{C}}(I-\alpha_{n}\mu A_{2})T^{n}y_{n}\). Then, from (12), we have  
\[\left\{ \begin{array}{ll}&u_{n}=\gamma_{n}x_{n}+(1-\gamma_{n})S_{n}u_{n},\\ &y_{n}=\sigma_{n}x_{n}+(1-\sigma_{n})p_{n},\\ &x_{n+1}=\beta_{n}f(y_{n})+(1-\beta_{n})q_{n}.\end{array} \right.\]  
Simple calculations show that  
\[\left\{ \begin{array}{ll}&u_{n}-u_{n-1}=\gamma_{n}(x_{n}-x_{n-1})+(\gamma_{n} -\gamma_{n-1})(x_{n-1}-S_{n-1}u_{n-1})+(1-\gamma_{n})(S_{n}u_{n}-S_{n-1}u_{n- 1}),\\ &y_{n}-y_{n-1}=\sigma_{n}(x_{n}-x_{n-1})+(\sigma_{n}-\sigma_{n-1})(x_{n-1}-p_{ n-1})+(1-\sigma_{n})(p_{n}-p_{n-1}),\\ &x_{n+1}-x_{n}=\beta_{n}(f(y_{n})-f(y_{n-1}))+(\beta_{n}-\beta_{n-1})(f(y_{n-1 })-q_{n-1})+(1-\beta_{n})(q_{n}-q_{n-1}).\end{array} \right. \tag{16}\]  
It follows that  
\[ \begin{array}{ll}&\|u_{n}-u_{n-1}\|^{2}\\ &=\gamma_{n}\langle x_{n}-x_{n-1},u_{n}-u_{n-1}\rangle+(1-\gamma_{n})\langle S _{n}u_{n}-S_{n-1}u_{n-1},u_{n}-u_{n-1}\rangle\\ &\quad+(\gamma_{n}-\gamma_{n-1})\langle x_{n-1}-S_{n-1}u_{n-1},u_{n}-u_{n-1}\rangle \\ &=\gamma_{n}\langle x_{n}-x_{n-1},u_{n}-u_{n-1}\rangle+(1-\gamma_{n})[\langle S _{n}u_{n}-S_{n-1}u_{n},u_{n}-u_{n-1}\rangle\\ &\quad+\langle S_{n-1}u_{n}-S_{n-1}u_{n-1},u_{n}-u_{n-1}\rangle]+(\gamma_{n}- \gamma_{n-1})\langle x_{n-1}-S_{n-1}u_{n-1},u_{n}-u_{n-1}\rangle\\ &\leq\gamma_{n}\|x_{n-1}-x_{n}\|\|u_{n}-u_{n-1}\|+(1-\gamma_{n})[\|S_{n}u_{n}- S_{n-1}u_{n}\|\|u_{n}-u_{n-1}\|\\ &\quad+\|u_{n-1}-u_{n}\|^{2}]+|\gamma_{n}-\gamma_{n-1}|\|x_{n-1}-S_{n-1}u_{n-1} |\|u_{n}-u_{n-1}\|,\end{array} \]  
which hence yields  
\[ \begin{array}{ll}\|u_{n}-u_{n-1}\|&\leq\gamma_{n}\|x_{n-1}-x_{n}\|+(1-\gamma _{n})[\|S_{n}u_{n}-S_{n-1}u_{n}\|\\ &\quad+\|u_{n-1}-u_{n}\|]+|\gamma_{n}-\gamma_{n-1}|\|x_{n-1}-S_{n-1}u_{n-1}\|. \end{array} \]  
This immediately leads to  
\[ \begin{array}{ll}\|u_{n}-u_{n-1}\|&\leq\|x_{n-1}-x_{n}\|+\frac{1-\gamma_{n} }{\gamma_{n}}\|S_{n}u_{n}-S_{n-1}u_{n}\|+|\gamma_{n}-\gamma_{n-1}\frac{\|x_{n- 1}-S_{n-1}u_{n-1}\|}{\gamma_{n}}\\ &\leq\|x_{n}-x_{n-1}\|+\frac{1}{a}\|S_{n}u_{n}-S_{n-1}u_{n}\|+|\gamma_{n}- \gamma_{n-1}|\frac{\|x_{n-1}-S_{n-1}u_{n-1}\|}{a}.\end{array}  \tag{17}\]  
Putting \(D=\{u_{n}:n\geq 0\}\), we know that \(D\) is a bounded subset of \(C\). Then, by the assumption, we get \(\sum_{n=1}^{\infty}\sup_{x\in D}\|S_{n}x-S_{n-1}x\|<\infty\). Noticing \(\|S_{n}u_{n}-S_{n-1}u_{n}\|\leq\sup_{x\in D}\|S_{n}x-S_{n-1}x\|\ \forall n\geq 1\), we have  
\[\sum_{n=1}^{\infty}\|S_{n}u_{n}-S_{n-1}u_{n}\|<\infty. \tag{18}\]  
In addition, from \(p_{n}=P_{\mathbb{C}}(I-\delta_{n}A_{1})z_{n}\) and \(\{\delta_{n}\}\subset(0,2\zeta]\), we observe that  
\[ \begin{array}{ll}\|p_{n}-p_{n-1}\|&\leq\|(I-\delta_{n}A_{1})z_{n}-(I-\delta_ {n-1}A_{1})z_{n-1}\|\\ &=\|(I-\delta_{n}A_{1})z_{n}-(I-\delta_{n}A_{1})z_{n-1}-(\delta_{n}-\delta_{n- 1})A_{1}z_{n-1}\|\\ &\leq\|(I-\delta_{n}A_{1})z_{n}-(I-\delta_{n}A_{1})z_{n-1}\|+|\delta_{n}-\delta _{n-1}|\|A_{1}z_{n-1}\|\\ &\leq\|z_{n}-z_{n-1}\|+|\delta_{n}-\delta_{n-1}|\|A_{1}z_{n-1}\|\\ &\leq\|u_{n}-u_{n-1}\|+M_{0}|\delta_{n}-\delta_{n-1}|,\end{array}  \tag{19}\]where \(\sup_{n\geq 0}\|A_{1}z_{n}\|\leq M_{0}\) for some \(M_{0}>0\). Thus, from (16), (17) and (19), we get  
\[