## 3 Main Results
Since \(p^{\star}\in(D+E)^{-1}(0)\), it follows that  
\[-\lambda_{k}Dp^{\star}\in\lambda_{k}Ep^{\star}.\]  
Now, due to the maximal monotonicity of \(E\), we have that  
\[\langle w_{k}-\lambda_{k}Dw_{k}-v_{k}+\lambda_{k}Dp^{\star},v_{k}-p^{\star} \rangle\geq 0.\]  
Thus,  
\[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]0  
From (13), it implies that  
\[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]1  
By the monotonicity of \(D\), it follows that  
\[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]2  
By (19), it follows that  
\[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]3 \[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]4  
Since \(z_{k}=w_{k}-m_{k}r_{k}\), we have that \(\|m_{k}\cdot r_{k}\|^{2}=\|z_{k}-w_{k}\|^{2}\). From (14), if \(r_{k}\
eq 0\), we have \(m_{k}\|r_{k}\|^{2}=(c_{1}+t_{k})\langle w_{k}-v_{k},r_{k}\rangle\). From Lemma 2 and (20), we obtain  
\[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]5 \[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]6 \[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]7 \[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]8 \[\frac{(s_{k}+s)\|w_{k}-v_{k}\|}{\|Dw_{k}-Dv_{k}\|}\geq\frac{(s_{k}+s)\|w_{k}-v _{k}\|}{L_{2}\|w_{k}-v_{k}\|}\geq\frac{\mu}{L_{2}}.\]9  
**Lemma 9**. : _Let \(\{w_{k}\}\) and \(\{v_{k}\}\) be sequences generated by Algorithm 1. Let \(\{w_{k_{j}}\}\) and \(\{v_{k_{j}}\}\) be subsequences of \(\{w_{k}\}\) and \(\{v_{k}\}\), respectively. If \(w_{k_{j}}\rightharpoonup x^{*}\in\mathcal{H}\) and \(\lim_{j\to\infty}\|w_{k_{j}}-v_{k_{j}}\|=0\), then \(x^{*}\in(D+E)^{-1}(0)\)._Proof. : The proof is similar to that of Lemma 7 in [5]. Thus, we omit it here. Lemma 10.: _Let \(\{p_{k}\}\) be the sequence generated by Algorithm 1. Then, \(\{p_{k}\}\) is bounded._  
Proof. : Let \(p^{\star}\in\Omega\). From (10) and Assumption 1\((A_{4})\), we have \(\phi_{k}\|p_{k}-p_{k-1}\|\leq\epsilon_{k}\), \(\forall k\in\mathbb{N}\), and this implies that  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}} \in(0,1),\;\forall k\geq K.\]0  
It implies from (22) that there exists \(K_{3}>0\), such that  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}} \in(0,1),\;\forall k\geq K.\]1  
Using (10) and (23), we have  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}} \in(0,1),\;\forall k\geq K.\]2 \[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}} \in(0,1),\;\forall k\geq K.\]3 \[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}} \in(0,1),\;\forall k\geq K.\]4 \[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}} \in(0,1),\;\forall k\geq K.\]5  
By Lemma 7, we know that a positive integer \(K\) exists, such that \(0<c_{1}+t_{k}<2\).