# Systems of Variational Inequalities with Nonlinear Operators
## 3 Main Results
\]  
Since \(\lim_{u\to\infty}c_{n}=0\) and \(\lim_{u\to\infty}\alpha_{n}=0\), we conclude from (17) and (23) that  
\[\lim_{n\to\infty}g_{1}(\|u_{n}-z_{n}-(p-q)\|)=0\quad\text{and}\quad\lim_{n\to \infty}g_{2}(\|z_{n}-y_{n}+(p-q)\|)=0.\]Using the properties of \(g_{1}\) and \(g_{2}\), we obtain  
\[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]  
It follows that  
\[\|u_{n}-y_{n}\|\leq\|u_{n}-z_{n}-(p-q)\|+\|z_{n}-y_{n}+(p-q)\|\to 0\quad(n\to\infty).\]  
That is,  
\[\lim_{n\to\infty}\|u_{n}-Gu_{n}\|=\lim_{n\to\infty}\|u_{n}-y_{n}\|=0. \tag{28}\]  
Also, according to (8) we have  
\[\|u_{n}-p\|^{2}\leq\beta_{n}\langle x_{n}-p,j(u_{n}-p)\rangle+(1-\beta_{n})\|u _{n}-p\|^{2},\]  
which together with Lemma 5, yields  
\[2\|u_{n}-p\|^{2} \leq 2\langle x_{n}-p,j(u_{n}-p)\rangle\] \[\leq[\|x_{n}-p\|^{2}+\|u_{n}-p\|^{2}-g(\|x_{n}-u_{n}\|)].\]  
This immediately implies that  
\[\|u_{n}-p\|^{2}+g(\|x_{n}-u_{n}\|)\leq\|x_{n}-p\|^{2},\]  
which together with (22), yields  
\[\|x_{n+1}-p\|^{2} \leq\|u_{n}-p\|^{2}+c_{n}M_{2}+\alpha_{n}M_{3}\] \[\leq\|x_{n}-p\|^{2}-g(\|x_{n}-u_{n}\|)+c_{n}M_{2}+\alpha_{n}M_{3}.\]  
Hence we have  
\[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]0 \[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]1  
Since \(\lim_{n\to\infty}c_{n}=0\) and \(\lim_{n\to\infty}\alpha_{n}=0\), we obtain from (17) that \(\lim_{n\to\infty}g(\|x_{n}-u_{n}\|)=0\). Using the properties of \(g\), we have  
\[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]2  
Also, observe that  
\[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]3  
and  
\[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]4  
Then from (28) and (29) it follows that  
\[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]5  
**Step 4. ** We claim that \(\|Tx_{n}-x_{n}\|\to 0\) and \(\|S_{n}x_{n}-x_{n}\|\to 0\) as \(n\to\infty\). Indeed, combining (8) with (29), we obtain that  
\[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]6That is,  
\[\lim_{n\to\infty}\|u_{n}-z_{n}-(p-q)\|=0\quad\text{and}\quad\lim_{n\to\infty}\|z _{n}-y_{n}+(p-q)\|=0. \tag{27}\]7  
Since \(\{S_{n}\}_{n=0}^{\infty}\) is \(\ell\)-uniformly Lipschitzian on \(C\), we deduce from (29) and (31) that  
\[ \begin{array}{ll}\|S_{n}x_{n}-x_{n}\|&\leq\|S_{n}x_{n}-S_{n}u_{n}\|+\|S_{n}u_ {n}-u_{n}\|+\|u_{n}-x_{n}\|\\ &\leq(\ell+1)\|x_{n}-u_{n}\|+\|S_{n}u_{n}-u_{n}\|\to 0\quad(n\to\infty). \end{array} \]  
That is,  
\[\lim_{n\to\infty}\|x_{n}-S_{n}x_{n}\|=0. \tag{32}\]  
We note that  
\[ \begin{array}{ll}\|x_{n}-T^{n}y_{n}\|&\leq\|x_{n}-x_{n+1}\|+\|x_{n+1}-T^{n}y _{n}\|\\ &=\leq\|x_{n}-x_{n+1}\|+\alpha_{n}\|f(x_{n})-\theta FT^{n}y_{n}\|.\end{array} \]  
Then we have  
\[ \begin{array}{ll}\|y_{n}-T^{n}y_{n}\|&\leq\|y_{n}-x_{n}\|+\|x_{n}-T^{n}y_{n} \|\\ &\leq\|y_{n}-x_{n}\|+\|x_{n}-x_{n+1}\|+\alpha_{n}\|f(x_{n})-\theta FT^{n}y_{n} \|.\end{array} \]  
Consequently, from (17), (30) and \(\lim_{n\to\infty}\alpha_{n}=0\), it follows that  
\[\lim_{n\to\infty}\|y_{n}-T^{n}y_{n}\|=0. \tag{33}\]  
We also note that  
\[\|y_{n}-Ty_{n}\|\leq\|y_{n}-T^{n}y_{n}\|+\|T^{n}y_{n}-T^{n+1}y_{n}\|+\|T^{n+1} y_{n}-Ty_{n}\|.\]  
By the assumption \(\sum_{n=0}^{\infty}\|T^{n+1}y_{n}-T^{n}y_{n}\|<\infty\), (33) and the condition that \(T:C\to C\) is uniformly continuous, we get  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]  
In addition, noticing that  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]  
we deduce from (30), (34) and the uniform continuity of \(T\) that  
\[\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0. \tag{35}\]  
**Step 5.