## 3 Main Results
Now, if \(r_{k_{j}}=0\), then following the Lipschitz continuity of \(D\), (12)-(14) and (40), we have that  
\[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]1 \[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]2 \[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]3 \[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]4 \[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]5 \[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]6 \[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]7  
This implies that  
\[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]8  
By Lemma 6, we have that \(\lim_{j\to\infty}\lambda_{k_{j}}=\lambda\), and this implies that \(\frac{\lambda_{k_{j}}}{\lambda_{k_{j}+1}}=1\). Furthermore, by assumption \((A_{4})\), we have that \(\lim_{j\to\infty}t_{k_{j}}=0=\lim_{j\to\infty}s_{k_{j}}\) and \(0<s<s_{1}^{\prime}<1\). Due to (36) and (42), we have that  
\[1-\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}>0,\;\;\forall k\geq K_{2}.\]9  
Next, if \(r_{k_{j}}=0\), then due to (14), we know that \(\lim_{j\to\infty}\|w_{k_{j}}-v_{k_{j}}\|=0\) also holds. Now, by the boundedness of \(\{p_{k_{j}}\}\), then there exists a subsequence \(\{p_{k_{j}}\}\) of \(\{p_{k_{j}}\}\), which is weakly convergent to some \(q\in\mathcal{H}\); furthermore,  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}\in( 0,1),\;\forall k\geq K.\]0  
From (38), we have that \(w_{k_{j}}\rightharpoonup q\) as \(j\to\infty\). This implies from Lemma 9 and (43) that \(q\in(D+E)^{-1}(0)\). By (44) and the assumption that \(p^{\star}\) is the unique solution to problem (8), we have  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}\in( 0,1),\;\forall k\geq K.\]1The combination of (39) and (45) yields  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}\in( 0,1),\;\forall k\geq K.\]2  
Using Remark 1 and (46), we obtain  
\[c_{1}+t_{k}\in(0,2)\;\;\text{and}\;\;\frac{\lambda_{k}(s_{k}+s)}{\lambda_{k+1}}\in( 0,1),\;\forall k\geq K.\]3  
Thus, from **Claim 2**, assumption \((A_{4})\), (47), and Lemma 3, it follows that \(\|p_{k}-p^{*}\|=0\), as required. Next, we present the second method in Algorithm 2. ```
Initialization: Choose \(\phi>0\), \(\lambda_{1}>0\), \(\varrho\in\left(0,\frac{2\eta}{l_{1}^{2}}\right)\) and \(\{\theta_{k}\}\subset[a,b]\subset(0,1]\). Let \(p_{0}\), \(p_{1}\in\mathcal{H}\) and set \(k=1\). Iterative Steps: Calculate the next iteration point \(p_{k+1}\) as follows: Step 1: Choose \(\phi_{k}\), such that \(\phi_{k}\in[0,\tilde{\phi}_{k}]\), where

\[\tilde{\phi}_{k}= \begin{cases}\min\Bigl{\{}\frac{k-1}{k+\phi-1}, \frac{\epsilon_{k}}{\|p_{k}-p_{k-1}\|}\Bigr{\}},&\text{if }\ p_{k}\
eq p_{k-1},\\ \frac{k-1}{k+\phi-1},&\text{otherwise}.\end{cases} \] (48) Step 2: Set \[w_{k}=p_{k}+\phi_{k}(p_{k}-p_{k-1}),\] (49) and compute \[v_{k}=(I+\lambda_{k}E)^{-1}(I-\lambda_{k}D)w_{k},\] (50) Step 3: Compute \[z_{k}=(1-\theta_{k})w_{k}+\theta_{k}(v_{k}+\lambda_{k}(Dw_{k}-Dv_{k})).\] (51) Step 4: Compute \[p_{k+1}=(I-\alpha_{k}\varrho\mathcal{F})z_{k},\ \forall k\geq 1.\] (52) Update \[\lambda_{k+1}= \begin{cases}\min\Bigl{\{}\frac{(\alpha_{k}+\varsigma)\|w_{k}-p_ {k}\|}{\|Dw_{k}-Dv_{k}\|},\lambda_{k}+q_{k}\Bigr{\}},&\text{if }\ Dw_{k}\
eq Dv_{k},\\ \lambda_{k}+q_{k},&\text{otherwise}.\end{cases} \] (53) Put \(k:=k+1\) and return to Step 1. ```  
**Algorithm 2** A modified accelerated Tseng splitting method  
**Remark 3**. : _From (48), and Assumption 1\((A_{4})\), we observe that_  
\[\lim_{k\to\infty}\phi_{k}\|p_{k}-p_{k-1}\|=0\text{ and }\] \[\lim_{k\to\infty}\frac{\phi_{k}}{\alpha_{k}}\|p_{k}-p_{k-1}\|=0.\]  
**Lemma 11**. : _If assumption \((A_{5})\) is performed, then a positive integer \(K\) exists, such that_  
\[\frac{(s_{k}+s)^{2}\lambda_{k}^{2}}{\lambda_{k+1}^{2}}\in(0,1),\ \forall k\geq K.\]  
Proof. : The proof is similar to the proof of Lemma 7.