## 2 Preliminaries
Hence, by Proposition 1, we get  
\[\lim_{n\to\infty}\|S_{n}x_{n}-Sx_{n}\|=0. \tag{35}\]  
Thus, combining (31) with (35) 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,  
\[\lim_{n\to\infty}\|x_{n}-Sx_{n}\|=0. \tag{36}\]  
Define \(\overline{S}:=(2I-S)^{-1}\). \(\overline{S}:C\to C\) 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, \(\overline{S}\) is nonexpansive and the fixed point set \(\operatorname{Fix}(\overline{S})=\operatorname{Fix}(S)=\bigcap_{n=0}^{ \infty}\operatorname{Fix}(S_{n})\). Observe that  
\[ \begin{array}{ll}\|x_{n}-\overline{S}x_{n}\|&=\|\overline{S}^{-1}x_{n}- \overline{S}x_{n}\|\\ &\leq\|\overline{S}^{-1}x_{n}-x_{n}\|\\ &=\|x_{n}-Sx_{n}\|.\end{array} \]  
From (36), it follows that  
\[\lim_{n\to\infty}\|x_{n}-\overline{S}x_{n}\|=0. \tag{37}\]  
**Step 6. ** We aim to present  
\[\limsup_{n\to\infty}\langle(I-f)x^{*},x^{*}-x_{n}\rangle\leq 0, \tag{38}\]  
where \(\{x^{*}\}=\operatorname{VI}(\operatorname{VI}(\Omega,A-g),I-f)\). Indeed, we choose a subsequence \(\{x_{n_{i}}\}\) of \(\{x_{n}\}\) such that  
\[\lim_{i\to\infty}\langle(I-f)x^{*},x^{*}-x_{n_{i}}\rangle=\limsup_{n\to\infty }\langle(I-f)x^{*},x^{*}-x_{n}\rangle.\]  
We suppose a subsequence \(x_{n_{i}}\rightharpoonup\bar{x}\in C\). Observe that \(G\) and \(\overline{S}\) have the nonexpansivity and that \(T\) has the asymptotically nonexpansivity. Since \((I-G)x_{n}\to 0,\ (I-T)x_{n}\to 0\) and \((I-\overline{S})x_{n}\to 0\), 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, \(\bar{x}\in\Omega=\bigcap_{n=0}^{\infty}\operatorname{Fix}(S_{n})\cap \operatorname{GSVI}(C,B_{1},B_{2})\cap\operatorname{Fix}(T)\). We present that \(\bar{x}\in\operatorname{VI}(\Omega,A-g)\). As a fact, let \(y\in\Omega\) be a arbitrarily fixed point. Then, it follows from (6), (8), and the monotonicity of \(A-g\) that  
\[ \begin{array}{ll}&\|y_{n}-y\|^{2}\leq\|(z_{n}-y)-\alpha_{n}(Az_{n}-g(x_{n})) \|^{2}\\ &=\|z_{n}-y\|^{2}+2\alpha_{n}\langle Az_{n}-g(x_{n}),y-z_{n}\rangle+\alpha_{n} ^{2}\|Az_{n}-g(x_{n})\|^{2}\\ &\leq\|x_{n}-y\|^{2}+2\alpha_{n}\langle Az_{n}-g(z_{n}),y-z_{n}\rangle+2\alpha _{n}\|z_{n}-x_{n}\|\|y-z_{n}\|+\alpha_{n}^{2}\|Az_{n}-g(x_{n})\|^{2}\\ &\leq\|x_{n}-y\|^{2}+2\alpha_{n}\langle Ay-g(y),y-z_{n}\rangle+2\alpha_{n}\|z _{n}-x_{n}\|\|\langle y-z_{n}\|+\alpha_{n}^{2}\|Az_{n}-g(x_{n})\|^{2},\end{array} \]which implies that, for all \(n\geq 0\),  
\[ \begin{array}{ll}0\leq\frac{\|x_{n}-y\|^{2}-\|y_{n}-y\|^{2}}{a_{n}}+2\langle(A- g)y,y-z_{n}\rangle+2l\|z_{n}-x_{n}\|\|y-z_{n}\|+\alpha_{n}\|Az_{n}-g(x_{n})\|^{2}\\ \leq\frac{(\|x_{n}-y\|+\|y_{n}-y\|)\|x_{n}-y\|}{a_{n}}+2\langle(A-g)y,y-z_{n} \rangle+2l\|z_{n}-x_{n}\|\|y-z_{n}\|+\alpha_{n}\|Az_{n}-g(x_{n})\|^{2}.\end{array} \]  
From (29), it is easy to see that \(x_{n_{i}}\rightharpoonup\bar{x}\) leads to \(z_{n_{i}}\rightharpoonup\bar{x}\). Since \(\lim_{n\to\infty}\alpha_{n}=0\) and \(\|x_{n}-y_{n}\|=o(\alpha_{n})\) (due to the assumption), we have  
\[\begin{array}{ll}0&\leq\liminf_{n\to\infty}\{\frac{(\|x_{n}-y\|+\|y_{n}-y\| )\|x_{n}-y  
Therefore, applying Lemma 3 to relation (40), we conclude that \(x_{n}\to x^{*}\) as \(n\to\infty\). This completes the proof. These authors contributed equally to this work. This research was funded by the Natural Science Foundation of Shandong Province of China (ZR2017 LA001) and Youth Foundation of Linyi University (LYDX2016 BS023). The first author was partially supported by the Innovation Program of Shanghai Municipal Education Commission (15 ZZ068), Ph.D.