## 3 Main Results
Note that \(G\) and \(\overline{S}\) are nonexpansive and that \(T\) is asymptotically nonexpansive. Since \((I-G)x_{n}\to 0\), \((I-T)x_{n}\to 0\) and \((I-\overline{S})x_{n}\to 0\) (due to (36), (41) and (44)), by Lemma 7 we have that \(\bar{x}\in\operatorname{Fix}(G)=\operatorname{GSVI}(C,B_{1},B_{2})\), \(\bar{x}\in\operatorname{Fix}(T)\) and \(\bar{x}\in\operatorname{Fix}(\overline{S})=\bigcap_{n=0}^{\infty} \operatorname{Fix}(S_{n})\). Then, \(\bar{x}\in\Omega=\operatorname{Fix}(G)\). Hence, by Proposition 1, we get  
\[\lim_{n\to\infty}\|x_{n}-Sx_{n}\|\leq\|x_{n}-Sx_{n}\|+\|S_{n}x_{n}-Sx_{n}\|\to 0 \quad(n\to\infty).\]  
That is,  
\[\lim_{n\to\infty}\|x_{n}-Sx_{  
\(\bigcap_{n=0}^{\infty}\operatorname{Fix}(S_{n})\cap\operatorname{GSVI}(C,B_{1},B_{2 })\cap\operatorname{Fix}(T)\). We claim that \(\tilde{x}\in\operatorname{VI}(\Omega,A_{1})\). In fact, let \(y\in\Omega\) be fixed arbitrarily. Then, it follows from (12), (14) and the \(\zeta\)-inverse-strong monotonicity of \(A_{1}\) that  
\[ \begin{array}{ll}\|y_{n}-y\|^{2}&\leq\sigma_{n}\|x_{n}-y\|^{2}+(1-\sigma_{n}) \|P_{\mathbb{C}}(z_{n}-\delta_{n}A_{1}z_{n})-P_{\mathbb{C}}y\|^{2}\\ &\leq\sigma_{n}\|x_{n}-y\|^{2}+(1-\sigma_{n})\|(z_{n}-y)-\delta_{n}A_{1}z_{n}\| ^{2}\\ &=\sigma_{n}\|x_{n}-y\|^{2}+(1-\sigma_{n})\{\|z_{n}-y\|^{2}+2\delta_{n}\langle A _{1}z_{n},y-z_{n}\rangle+\delta_{n}^{2}\|A_{1}z_{n}\|^{2}\}\\ &\leq\|x_{n}-y\|^{2}+(1-\sigma_{n})[2\delta_{n}\langle A_{1}y,y-z_{n}\rangle+ \delta_{n}^{2}\|A_{1}z_{n}\|^{2}],\end{array} \]  
which together with \(\{\sigma_{n}\}\subset[c,d]\), implies that for all \(n\geq 0\),  
\[ \begin{array}{ll}0&\leq\frac{1}{\delta_{n}(1-\sigma_{n})}(\|x_{n}-y\|^{2}-\| y_{n}-y\|^{2})+2\langle A_{1}y,y-z_{n}\rangle+\frac{\delta_{n}}{1-\sigma_{n}}\|A _{1}z_{n}\|^{2}\\ &\leq(\|x_{n}-y\|+\|y_{n}-y\|)\frac{\|x_{n}-y_{n}\|}{\delta_{n}(1-d)}+2\langle A _{1}y,y-z_{n}\rangle+\frac{\delta_{n}}{1-d}\|A_{1}z_{n}\|^{2}.\end{array} \]  
From (36) it is easy to see that \(x_{n_{i}}\rightharpoonup\tilde{x}\) leads to \(z_{n_{i}}\rightharpoonup\tilde{x}\). Since \(\lim_{n\to\infty}\delta_{n}=0\) and \(\|x_{n}-y_{n}\|=o(\delta_{n})\) (due to the assumption), we have  
\[\begin{array}{ll}0&\leq\liminf_{n\to\infty}\{(\|x_{n}-y\|+\|y_{n}-y\|)\frac{ \|x_{n}-y_{n}\|}{\delta_{n}(1-d)}+2\langle A_{1}y,y-z_{n}\rangle+\frac{\delta _{n}}{1-d}\|A_{1}z_{n}\|^{2}\}\\ &=\liminf_{n\to\infty}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! \!\!From (12) and the \(\zeta\)-inverse-strong monotonicity of \(A_{1}\), it follows that  
\[ \begin{array}{ll}\|y_{n}-x^{*}\|^{2}&\leq\sigma_{n}\|x_{n}-x^{*}\|^{2}+(1- \sigma_{n})\|P_{\zeta}(z_{n}-\delta_{n}A_{1}z_{n})-x^{*}\|^{2}\\ &\leq\sigma_{n}\|x_{n}-x^{*}\|^{2}+(1-\sigma_{n})\|(z_{n}-x^{*})-\delta_{n}A_{1 }z_{n}\|^{2}\\ &=\sigma_{n}\|x_{n}-x^{*}\|^{2}+(1-\sigma_{n})\|(\|z_{n}-x^{*}\|^{2}+2\delta_{n }\langle A_{1}z_{n},x^{*}-z_{n}\rangle+\delta_{n}^{2}\|A_{1}z_{n}\|^{2}]\\ &\leq\sigma_{n}\|x_{n}-x^{*}\|^{2}+(1-\sigma_{n})\|(\|x_{n}-x^{*}\|^{2}+2\delta _{n}\langle A_{1}x^{*},x^{*}-z_{n}\rangle+\delta_{n}^{2}\|A_{1}z_{n}\|^{2}]\\ &=\|x_{n}-x^{*}\|^{2}+(1-\sigma_{n})[2\delta_{n}\langle A_{1}x^{*},x^{*}-z_{ n}\rangle+\delta_{n}^{2}\|A_{1}z_{n}\|^{2}].\end{array}