## 3 Main Results
 \tag{48}\]  
Thus, in terms of (47) and (48), we get  
\[ \begin{array}{ll}\|x_{n+1}-x^{*}\|^{2}\\ \leq[1-\alpha_{n}(\tau-\delta)]\|y_{n}-x^{*}\|^{2}+\theta_{n}(2+\theta_{n})\| y_{n}-x^{*}\|^{2}+2\alpha_{n}\langle\mu A_{2}x^{*},x^{*}-q_{n}\rangle\\ \quad+2\beta_{n}\langle f(x^{*})-x^{*},x_{n+1}-x^{*}\rangle\\ \leq[1-\alpha_{n}(\tau-\delta)]\{\|x_{n}-x^{*}\|^{2}+(1-\sigma_{n})[2\delta_{n }\langle A_{1}x^{*},x^{*}-z_{n}\rangle+\delta_{n}^{2}\|A_{1}z_{n}\|^{2}]\}\\ \quad+\theta_{n}(2+\theta_{n})\|y_{n}-x^{*}\|^{2}+2\alpha_{n}\langle\mu A_{2}x ^{*},x^{*}-q_{n}\rangle+2\beta_{n}\|f(x^{*})-x^{*}\|\|x_{n+1}-x^{*}\|\\ \leq[1-\alpha_{n}(\tau-\delta)]\|x_{n}-x^{*}\|^{2}+(1-\alpha_{n}(\tau-\delta)) (1-\sigma_{n})2\delta_{n}\langle A_{1}x^{*},x^{*}-z_{n}\rangle+\alpha_{n}^{2} \|A_{1}z_{n}\|^{2}\\ \quad+\theta_{n}(2+\theta_{n})\|y_{n}-x^{*}\|^{2}+2\alpha_{n}\langle\mu A_{2}x ^{*},x^{*}-q_{n}\rangle+2\beta_{n}\|f(x^{*})-x^{*}\|\|x_{n+1}-x^{*}\|\\ =[1-\alpha_{n}(\tau-\delta)]\|x_{n}-x^{*}\|^{2}+\alpha_{n}(\tau-\delta)\{\frac {(1-\alpha_{n}(\tau-\delta))(1-\sigma_{n})2\delta_{n}}{a_{\tau}(\tau-\delta)} \langle A_{1}x^{*},x^{*}-z_{n}\rangle+\frac{a_{n}\|A_{1}z_{n}\|^{2}}{\tau- \delta}\\ \quad+\frac{\theta_{n}(2+\theta_{n})\|y_{n}-x^{*}\|^{2}}{\alpha_{n}(\tau- \delta)}+\frac{2}{\tau-\delta}\langle\mu A_{2}x^{*},x^{*}-q_{n}\rangle+\frac{2 \beta_{n}\|f(x^{*})-x^{*}\|\|x_{n+1}-x^{*}\|}{a_{\tau}(\tau-\delta)}\}.\end{array}  \tag{49}\]  
It can be readily seen that (45) guarantees that  
\[\limsup_{n\to\infty}\frac{(1-\alpha_{n}(\tau-\delta))(1-\sigma_{n})2\delta_{n }}{\alpha_{n}(\tau-\delta)}\langle A_{1}x^{*},x^{*}-z_{n}\rangle\leq 0\]  
and  
\[\limsup_{n\to\infty}\frac{2}{\tau-\delta}\langle\mu A_{2}x^{*},x^{*}-q_{n} \rangle\leq 0.\]  
In fact, from \(\limsup_{n\to\infty}\langle A_{1}x^{*},x^{*}-z_{n}\rangle\leq 0\), it follows that for any given \(\varepsilon>0\) there exists an integer \(n_{0}\geq 1\) such that \(\langle A_{1}x^{*},x^{*}-z_{n}\rangle\leq\varepsilon,\ \forall n\geq n_{0}\). Then, from \(\delta_{n}\leq\alpha_{n}\), we get  
\[ \begin{array}{ll}\frac{(1-\alpha_{n}(\tau-\delta))(1-\sigma_{n})2\delta_{n}} {\alpha_{n}(\tau-\delta)}\langle A_{1}x^{*},x^{*}-z_{n}\rangle&\leq\frac{(1- \alpha_{n}(\tau-\delta))(1-\sigma_{n})2\delta_{n}}{\alpha_{n}(\tau-\delta)} \varepsilon\\ &\leq\frac{2}{\tau-\delta}\varepsilon,\quad\forall n\geq n_{0},\end{array} \]  
which hence yields  
\[\limsup_{n\to\infty}\frac{(1-\alpha_{n}(\tau-\delta))(1-\sigma_{n})2\delta_{n} }{\alpha_{n}(\tau-\delta)}\langle A_{1}x^{*},x^{*}-z_{n}\rangle\leq\frac{2}{ \tau-\delta}\varepsilon.\]  
Letting \(\varepsilon\to 0\), we get \(\limsup_{n\to\infty}\frac{(1-\alpha_{n}(\tau-\delta))(1-\sigma_{n})2\delta_{n}}{ \alpha_{n}(\tau-\delta)}\langle A_{1}x^{*},x^{*}-z_{n}\rangle\leq 0\). Since \(\sum_{n=0}^{\infty}\alpha_{n}=\infty\), \(\lim_{n\to\infty}\frac{\theta_{n}}{\alpha_{n}}=0\) and \(\lim_{n\to\infty}\frac{\theta_{n}}{\alpha_{n}}=0\) (due to Conditions (i) and (ii)), we deduce that \(\sum_{n=0}^{\infty}\alpha_{n}(\tau-\delta)=\infty\) and  
\[ \begin{array}{ll}\limsup_{n\to\infty}\{\frac{(1-\alpha_{n}(\tau-\delta))(1- \sigma_{n})2\delta_{n}}{\alpha_{n}(\tau-\delta)}\langle A_{1}x^{*},x^{*}-z_{n} \rangle+\frac{\alpha_{n}\|A_{1}z_{n}\|^{2}}{\tau-\delta}\\ \qquad\qquad+\frac{\theta_{n}(2+\theta_{n})\|y_{n}-x^{*}\|^{2}}{\alpha_{n}(\tau- \delta)}+\frac{2}{\tau-\delta}\langle\mu A_{2}x^{*},x^{*}-q_{n}\rangle+\frac{2 \theta_{n}\|f(x^{*})-x^{*}\|\|x_{n+1}-x^{*}\|}{\alpha_{n}(\tau-\delta)}\}\leq 0.\end{array} \]  
We can apply Lemma 3 to the relation (49) and conclude that \(x_{n}\to x^{*}\) as \(n\to\infty\). This completes the proof. The following results can be obtained by Theorem 1 easily, and hence we omit the details. **Corollary 1**. : _We suppose \(C\) is a convex nonempty closed set of a real Hilbert space \(H\) and \(f:C\to C\) is a contraction with the parameter \(\delta\in[0,1)\).