# Systems of Variational Inequalities with Nonlinear Operators
## 3 Main Results
Since \((I-G)x_{n}\to 0\) and \((I-\overline{S})x_{n}\to 0\) (due to (30) and (37)), by Lemma 6 we have that \(\bar{x}\in\operatorname{Fix}(G)=\operatorname{GSVI}(C,B_{1},B_{2})\) and \(\bar{x}\in\operatorname{Fix}(\overline{S})=\operatorname{Fix}(S)=\bigcap_{n=0}^ {\infty}\operatorname{Fix}(S_{n})\). From (35), we have that \(\lim_{i\to\infty}\|x_{n_{i}}-Tx_{n_{i}}\|=0\) for the subsequence \(\{x_{n_{i}}\}\) of \(\{x_{n}\}\). It follows that \(\bar{x}\in\operatorname{Fix}(T)\). Then, \(\bar{x}\in\Omega=\bigcap_{n=0}^{\infty}\operatorname{Fix}(S_{n})\cap \operatorname{Fix}(S_{n})\). Hence, \(\bar{x}\in\Omega=\bigcap_{n=0}^{\infty}\operatorname{Fix}(S_{n})\cap \operatorname{Fix}(S_{n})\). \(\mathrm{GSVI}(C,B_{1},B_{2})\cap\mathrm{Fix}(T)\). Since \(E\) admits a weakly sequentially continuous duality mapping \(j(\cdot)\) and \(x_{n_{i}}\rightharpoonup\tilde{x}\), we obtain  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]0 \[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]1  
which implies that (39) holds. Noticing that \(j(\cdot)\) is also norm-to-norm uniformly continuous on bounded subsets of \(E\), we obtain from (17) that  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]2  
**Step 7. ** We claim that \(x_{n}\to x^{*}\) as \(n\to\infty\). Indeed, it follows from \(x_{n+1}=\Pi_{C}v_{n}\) and Proposition 3 (iii) that  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]3  
which leads to  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]4  
Then, from (11) we have  
\[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]5 \[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]6 \[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]7 \[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]8 \[\lim_{n\to\infty}\|y_{n}-Ty_{n}\|=0. \tag{34}\]9 \[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]0  
where \(M_{4}=\sup_{n\geq 0}\|x_{n+1}-x^{*}\|\). This immediately implies that  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]1  
Applying Lemma 7 to (41), we infer that \(x_{n}\to x^{*}\) as \(n\to\infty\). The proof is completed. It is remarkable that according to the proof of Theorem 1, we know that \(\{y_{n}\}\) is bounded. We now give two examples to illustrate partial conditions of Theorem 1 to be satisfied. **Example 1**. : _Let \(T:C\to C\) be a contraction mapping with a constant \(\beta\in(0,1)\). We take \(S_{n}:=T^{n}\) and obtain_  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]2  
_for any bounded subset \(D\) of \(C\). Therefore, it follows that_  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]3  
In particular, whenever \(D\) is a bounded sequence \(\{x_{n}\}_{n=0}^{\infty}\) in \(C\), we have  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]4Since \(T\) is a contraction mapping, Banach's Contraction Mapping Principle guarantees that \(T\) has a unique fixed point. Say \(p\in C\). We define \(Sx:=p\) for all \(x\in C\). It is easy to see that \(Sx=\lim_{n\to\infty}S_{n}x\) for all \(x\in C\) and \(\operatorname{Fix}(S)=\bigcap_{n=0}^{\infty}\operatorname{Fix}(S_{n})\). **Example 2**. : _Let \(E=\mathbf{R}\) and \(C=[-1,1]\), and let \(T:C\to C\) be an identity mapping, i.e., \(Tx=x\) for all \(x\in C\). Moreover, we define_  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]5  
Then we obtain  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]6 \[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]7  
and  
\[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]8 \[\|x_{n}-Tx_{n}\|\leq\|x_{n}-y_{n}\|+\|y_{n}-Ty_{n}\|+\|Ty_{n}-Tx_{n}\|,\]9  
Therefore, it follows that for any bounded subset \(D\) of \(C\),  
\[\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0. \tag{35}\]0  
In addition, whenever \(\{x_{n}\}_{n=0}^{\infty}\) is a bounded sequence in \(C\), it is clear that  
\[\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0. \tag{35}\]1  
Also, we define \(Sx:=0\) for all \(x\in C\).