## 3 Main Results
Suppose that \(\{\alpha_{n}\},\{\beta_{n}\},\{\gamma_{n}\},\{\sigma_{n}\}\subset(0,1]\) and \(\{\delta_{n}\}\subset(0,2\zeta]\) are the sequences such that_  
* \(\lim_{n\to\infty}\alpha_{n}=0\)_,_ \(\sum_{n=0}^{\infty}\alpha_{n}=\infty\) _and_ \(\sum_{n=0}^{\infty}|\alpha_{n+1}-\alpha_{n}|<\infty\)_._
* \(\lim_{n\to\infty}\frac{\theta_{n}}{\alpha_{n}}=0\)_,_ \(\lim_{n\to\infty}\frac{\theta_{n}}{\alpha_{n}}=0\)_,_ \(\sum_{n=0}^{\infty}|\beta_{n+1}-\beta_{n}|<\infty\) _and_ \(\sum_{n=0}^{\infty}|\delta_{n+1}-\delta_{n}|<\infty\)_._
* \(0<\liminf_{n\to\infty}\sigma_{n}\leq\limsup\sup_{n\to\infty}\sigma_{n}<1\) _and_ \(\sum_{n=0}^{\infty}|\sigma_{n+1}-\sigma_{n}|<\infty\)_._
* \(0<\liminf_{n\to\infty}\gamma_{n}\leq\limsup_{n\to\infty}\gamma_{n}<1\) _and_ \(\sum_{n=0}^{\infty}|\gamma_{n+1}-\gamma_{n}|<\infty\)_._
* \(\delta_{n}\leq\alpha_{n}\) _and_ \(\sum_{n=0}^{\infty}\|T^{n+1}y_{n}-T^{n}y_{n}\|<\infty\)_._ _Then, the sequence_ \(\{x_{n}\}_{n=0}^{\infty}\) _generated by Algorithm_ 1 _satisfies the following properties:_
* \(\{x_{n}\}_{n=0}^{\infty}\) _is bounded._
* \(\lim_{n\to\infty}\|x_{n}-y_{n}\|=0\)_,_ \(\lim_{n\to\infty}\|x_{n}-Gx_{n}\|=0\)_,_ \(\lim_{n\to\infty}\|x_{n}-Tx_{n}\|=0\) _and_ \(\lim_{n\to\infty}\|x_{n}-Sx_{n}\|=0\)
_(c)_ \(\{x_{n}\}_{n=0}^{\infty}\) _converges to the unique solution of Problem 1 if_ \(\frac{\|x_{n}-y_{n}\|}{\delta_{n}}\to 0\) _as_ \(n\to\infty\)_._  
Proof. : First, let us show that \(P_{\text{VI}(\Omega,A_{1})}(I-\mu A_{2})\) is a contractive mapping. Indeed, by Lemma 4, we have  
\[ \begin{array}{ll}(1-\tau)\|x-y\|&\geq\|(I-\mu A_{2})x-(I-\mu A_{2})y\|\\ &\geq\|P_{\text{VI}(\Omega,A_{1})}(I-\mu A_{2})x-P_{\text{VI}(\Omega,A_{1})}( I-\mu A_{2})y\|,\end{array} \]  
for any \(x,y\in C\), which implies that \(P_{\text{VI}(\Omega,A_{1})}(I-\mu A_{2})\) is a contraction mapping. [PERSON]'s Contraction Mapping Principle tell us that \(P_{\text{VI}(\Omega,A_{1})}(I-\mu A_{2})\) has a fixed point and further it is unique. For example, \(x^{*}\in C\), that is, \(x^{*}=P_{\text{VI}(\Omega,A_{1})}(I-\mu A_{2})x^{*}\). Hence, by Lemma 8, we get  
\[\{x^{*}\}=\text{Fix}(P_{\text{VI}(\Omega,A_{1})}(I-\mu A_{2}))=\text{VI}( \text{VI}(\Omega,A_{1}),A_{2}).\]  
That is, Problem 1 has a unique solution. Taking into account that  
\[0<\liminf_{n\to\infty}\gamma_{n}\leq\limsup_{n\to\infty}\gamma_{n}<1,\]  
we usually suppose \(\{\gamma_{n}\}\subset[a,b]\subset(0,1)\) for some \(a,b\in(0,1)\). Note that the mapping \(G:C\to C\) is defined as \(G:=P_{C}(I-\mu_{1}B_{1})P_{C}(I-\mu_{2}B_{2})\), where \(\mu_{1}\in(0,2\alpha)\) and \(\mu_{2}\in(0,2\beta)\). Thus, by Lemma 6, we know that \(G\) is nonexpansive. It is easy to see that there exists an element \(u_{n}\in C\) such that  
\[u_{n}=\gamma_{n}x_{n}+(1-\gamma_{n})S_{n}u_{n}. \tag{11}\]  
In fact, it is a unique element. Thus, we can consider the mapping  
\[F_{n}x=\gamma_{n}x_{n}+(1-\gamma_{n})S_{n}x,\quad\forall x\in C.\]  
Since \(S_{n}:C\to C\) is a continuous pseudocontraction mapping, we deduce that all \(x,y\in C\),  
\[\langle F_{n}x-F_{n}y,x-y\rangle=(1-\gamma_{n})\langle S_{n}x-S_{n}y,x-y \rangle\leq(1-\gamma_{n})\|x-y\|^{2}.\]  
In addition, from \(\{\gamma_{n}\}\subset[a,b]\subset(0,1)\) we get \(0<1-\gamma_{n}<1\) 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 there exists a unique element \(u_{n}\in C\), for each \(n\geq 0\), satisfying (11). Thus, it can be readily seen that the hybrid Mann viscosity implicit iterative scheme (10) can be rewritten as  
\[\left\{ \begin{array}{ll}&u_{n}=\gamma_{n}x_{n}+(1-\gamma_{n})S_{n}u_{n},\\ &z_{n}=Gu_{n},\\ &y_{n}=\sigma_{n}x_{n}+(1-\sigma_{n})P_{C}(I-\delta_{n}A_{1})z_{n},\\ &x_{n+1}=(1-\beta_{n})P_{C}(I-\alpha_{n}\mu A_{2})T^{n}y_{n}+\beta_{n}f(y_{n}),\quad\forall n\geq 0.\end{array} \right. \tag{12}\]  
Next, we divide the rest of the proof into several steps. **Step 1. ** We claim that \(\{x_{n}\},\{y_{n}\},\{z_{n}\},\{u_{n}\},\{v_{n}\},\{T^{n}y_{n}\}\) and \(\{A_{2}(T^{n}y_{n})\}\) are bounded.