# Systems of Variational Inequalities with Nonlinear Operators
## 3 Main Results
It is easy to see that for each \(n\geq 0\) there exists a unique element \(u_{n}\in C\) such that  
\[u_{n}=\beta_{n}(x_{n}-S_{n}u_{n})+S_{n}u_{n}. \tag{8}\]  
As a matter of fact, consider the mapping  
\[F_{n}x=\beta_{n}(x_{n}-S_{n}x)+S_{n}x,\quad\forall x\in C.\]  
Since \(S_{n}:C\to C\) is a continuous pseudocontraction mapping, we deduce that  
\[(1-\beta_{n})\|x-y\|^{2}\geq(1-\beta_{n})\langle S_{n}x-S_{n}y,j(x-y)\rangle= \langle F_{n}x-F_{n}y,j(x-y)\rangle,\quad\forall x,y\in C.\]  
Also, from \(\{\beta_{n}\}\subset[a,b]\subset(0,1)\) we get \(0<1-\beta_{n}<1\) for all \(n\geq 0\). Thus, \(F_{n}\) is a continuous and strong pseudocontraction mapping of \(C\) into itself. By Proposition 2, we know that for each \(n\geq 0\) there exists a unique element \(u_{n}\in C\), satisfying (8). Therefore, it can be readily seen that the hybrid viscosity implicit iterative scheme (7) can be rewritten as  
\[\left\{ \begin{array}{rl}&u_{n}=\beta_{n}(x_{n}-S_{n}u_{n})+S_{n}u_{n},\\ &y_{n}=Gu_{n},x_{n+1}=\Pi_{\rm C}[(T^{n}y_{n}-\alpha_{n}\theta FT^{n}y_{n})+ \alpha_{n}f(x_{n})],\quad\forall n\geq 0,\end{array} \right. \tag{9}\]  
Next, we divide the rest of the proof into several steps. **Step 1. ** We claim that \(\{x_{n}\}\), \(\{y_{n}\}\), \(\{z_{n}\}\), \(\{u_{n}\}\), \(\{f(x_{n})\}\), \(\{T^{n}y_{n}\}\) and \(\{F(T^{n}y_{n})\}\) are bounded vector sequences. Indeed, take an element \(p\in\Omega=\bigcap_{n=0}^{\infty}\operatorname{Fix}(S_{n})\cap\operatorname{ GSVI}(C,B_{1},B_{2})\cap\operatorname{Fix}(T)\) arbitrarily. Thenwe have \(S_{n}p=p\), \(Gp=p\) and \(Tp=p\). Since each \(S_{n}:C\to C\) is a pseudocontraction mapping, it follows that  
\[\|p-u_{n}\|^{2} =\langle\beta_{n}(p-x_{n})+(1-\beta_{n})(p-S_{n}u_{n}),j(p-u_{n})\rangle\] \[=(1-\beta_{n})\langle p-S_{n}u_{n},j(p-u_{n})\rangle+\beta_{n} \langle p-x_{n},j(p-u_{n})\rangle\] \[\leq(1-\beta_{n})\|p-u_{n}\|^{2}+\beta_{n}\|x_{n}-p\|\|p-u_{n}\|,\]  
which hence yields  
\[\|p-u_{n}\|\leq\|p-x_{n}\|,\quad\forall n\geq 0. \tag{10}\]  
Then we observe  
\[\|y_{n}-p\|=\|Gu_{n}-p\|\leq\|u_{n}-p\|\leq\|x_{n}-p\|. \tag{11}\]  
Combining (9) and (11), we have  
\[\|x_{n+1}-p\| \leq\|\alpha_{n}(f(x_{n})-\theta F(p))+(I-\alpha_{n}\theta F)T^{n} y_{n}-(I-\alpha_{n}\theta F)p\|\] \[\leq\alpha_{n}\gamma\|x_{n}-p\|+\alpha_{n}\|f(p)-\theta F(p)\|+( 1-\alpha_{n}\bar{\gamma}\theta)\|T^{n}y_{n}-p\|\] \[\leq\alpha_{n}\gamma\|x_{n}-p\|+\alpha_{n}\|f(p)-\theta F(p)\|+( 1-\alpha_{n}\bar{\gamma}\theta)(\|y_{n}-p\|+c_{n})\] \[\leq c_{n}+\alpha_{n}\|f(p)-\theta F(p)\|+\alpha_{n}\gamma\|x_{n} -p\|+(1-\alpha_{n}\bar{\gamma}\theta)\|x_{n}-p\|\] \[\leq c_{n}+\max\{\|x_{n}-p\|,\frac{\|f(p)-\theta F(p)\|}{\bar{ \gamma}\theta-\gamma}\}.\]  
By induction, we get  
\[=(1-\beta_{n})\langle p-S_{n}u_{n},j(p-u_{n})\rangle+\beta_{n} \langle p-x_{n},j(p-u_{n})\rangle\]0  
It immediately follows that \(\{x_{n}\}\) is bounded, and so are the sequences \(\{y_{n}\},\{u_{n}\},\{f(x_{n})\},\{T^{n}y_{n}\}\) and \(\{F(T^{n}y_{n})\}\) (due to (10), (11) and the Lipschitz continuity of \(f\)). Taking into account that \(\{S_{n}\}\) is \(\ell\)-uniformly Lipschitzian on \(C\), we know that  
\[=(1-\beta_{n})\langle p-S_{n}u_{n},j(p-u_{n})\rangle+\beta_{n} \langle p-x_{n},j(p-u_{n})\rangle\]1  
which implies that \(\{S_{n}u_{n}\}\) is bounded. In addition, from Lemma 3 and \(p\in\Omega\subset\mathrm{GSVI}(C,B_{1},B_{2})\), it also follows that \((p,q)\) is a solution of GSVI (3) where \(q=\Pi_{C}(p-\mu_{2}B_{2}p)\). Please note that \(z_{n}=\Pi_{C}(I-\mu_{2}B_{2})u_{n}\) for all \(n\geq 0\). Then by Lemma 4 we get  
\[=(1-\beta_{n})\langle p-S_{n}u_{n},j(p-u_{n})\rangle+\beta_{n} \langle p-x_{n},j(p-u_{n})\rangle\]2 \[=(1-\beta_{n})\langle p-S_{n}u_{n},j(p-u_{n})\rangle+\beta_{n} \langle p-x_{n},j(p-u_{n})\rangle\]3 \[=(1-\beta_{n})\langle p-S_{n}u_{n},j(p-u_{n})\rangle+\beta_{n} \langle p-x_{n},j(p-u_{n})\rangle\]4  
This shows that \(\{z_{n}\}\) 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, from (9) we have  
\[