## 3 Results
\begin{split}&\|(I-P_{Q})A\bar{x}\|^{p}\\ &=\langle I_{E_{2}}^{p}(A\bar{x}-P_{Q}A\bar{x}),A\bar{x}-P_{Q}A\bar{ x}\rangle\\ &=\langle I_{E_{2}}^{p}(A\bar{x}-P_{Q}A\bar{x}),A\bar{x}-Au_{n_{i} }\rangle+\langle I_{E_{2}}^{p}(A\bar{x}-P_{Q}A\bar{x}),Au_{n_{i}}-P_{Q}Au_{n_{i }}\rangle\\ &\quad+\langle I_{E_{2}}^{p}(A\bar{x}-P_{Q}A\bar{x}),P_{Q}Au_{n_{ i}}-P_{Q}A\bar{x}\rangle\\ &\leq\langle I_{E_{2}}^{p}(A\bar{x}-P_{Q}A\bar{x}),A\bar{x}-Au_{n _{i}}\rangle+\langle I_{E_{2}}^{p}(A\bar{x}-P_{Q}A\bar{x}),Au_{n_{i}}-P_{Q}Au_{n _{i}}\rangle.\end{split}  \tag{38}\]  
By the continuity of \(A\) and \(\|x_{n}-u_{n}\|\to 0\) as \(n\to\infty\), we have \(Au_{n_{i}}\rightharpoonup A\bar{x}\) as \(n\to\infty\). Thus, letting \(i\to\infty\), we have  
\[\|A\bar{x}-P_{Q}A\bar{x}\|=0.\]  
Hence \(A\bar{x}=P_{Q}A\bar{x}\), that is, \(A\bar{x}\in Q\). Therefore, we have that \(\bar{x}\in F(T)\cap\mathcal{F}\). Next, we prove that \(\{x_{n}\}\) converges strongly to \(\Pi_{P(T)\cap\mathcal{F}}u\). Now, we have  
\[ \begin{split}& d_{p}(y_{n},x_{n})\\ &=d_{p}(J_{E_{1}^{+}}^{q}[\alpha_{n}]_{E_{1}}^{p}(u)+(1-\alpha_{n} )(\beta_{n}J_{E_{1}}^{p}(u)+\gamma_{n}]_{E_{1}}^{p}(x_{n})+\delta_{n}J_{E_{1}} ^{p}(Tx_{n}))],x_{n})\\ &\leq\alpha_{n}d_{p}(u,x_{n})+(1-\alpha_{n})\beta_{n}d_{p}(u,x_{n })\\ &\quad+(1-\alpha_{n})\gamma_{n}d_{p}(x_{n},x_{n})+(1-\alpha)\delta _{n}d_{p}(Tx_{n},x_{n})\\ &\to 0\end{split} \]  
as \(n\to\infty\). Thus we have  
\[\|y_{n}-x_{n}\|\to 0\]  
as \(n\to\infty\). Let \(\widehat{x}=\Pi_{P(T)\cap\mathcal{F}}u\). From (24), we have  
\[d_{p}(x_{n+1},\widehat{x})\leq(1-\alpha_{n})d_{p}(x_{n},\widehat{x})+\alpha_{ n}\langle I_{E_{1}}^{p}(u)-J_{E_{1}}^{p}(\widehat{x}),y_{n}-\widehat{x}\rangle.\]  
Then there exists a subsequence \(\{x_{n_{j}}\}\) of \(\{x_{n}\}\) such that  
\[\limsup_{n\to\infty}\langle J_{E_{1}}^{p}(u)-J_{E_{1}}^{p}(\widehat{x}),x_{n}- \widehat{x}\rangle=\lim_{j\to\infty}\langle J_{E_{1}}^{p}(u)-J_{E_{1}}^{p}( \widehat{x}),x_{n_{j}}-\widehat{x}\rangle\]  
and \(x_{n_{j}}\to\bar{x}\). Thus, from (15), it follows that  
\[ \begin{split}\limsup_{n\to\infty}\langle J_{E_{1}}^{p}(u)-J_{E_{1 }}^{p}(\widehat{x}),x_{n}-\widehat{x}\rangle&=\lim_{j\to\infty} \langle J_{E_{1}}^{p}(u)-J_{E_{1}}^{p}(\widehat{x}),x_{n_{j}}-\widehat{x}\rangle \\ &=\langle I_{E_{1}}^{p}(u)-J_{E_{1}}^{p}(\widehat{x}),\bar{x}- \widehat{x}\rangle\\ &\leq 0.\end{split} \]  
Since \(\|x_{n}-y_{n}\|\to 0\) as \(n\to\infty\), we have  
\[\limsup_{n\to\infty}\langle J_{E_{1}}^{p}(u)-J_{E_{1}}^{p}(\widehat{x}),y_{n} -\widehat{x}\rangle\leq 0.\]  
Hence, by Lemma 2, we conclude that \(d_{p}(x_{n},\widehat{x})\to 0\) as \(n\to\infty\). Therefore, \(x_{n}\to\widehat{x}\) as \(n\to\infty\) and, since \(\|x_{n}-u_{n}\|\to 0\) as \(n\to\infty\), we have \(u_{n}\to\widehat{x}\) as \(n\to\infty\). **Case II. ** Suppose that there exists a subsequence \(\{n_{j}\}\) of \(\{n\}\) such that  
\[d_{p}(x_{n_{i}},\widehat{x})<d_{p}(x_{n_{i}+1},\widehat{x})\]for all \(j\in\mathbb{N}\). Then, by Lemma 1, there exists a nondecreasing sequence \(\{m_{k}\}\subset\mathbb{N}\) with \(m_{k}\to\infty\) as \(n\to\infty\) such that  
\[d_{p}(x_{m_{k}},\widehat{x})\leq d_{p}(x_{m_{k}+1},\widehat{x}),\quad d_{p}(x_{k },\widehat{x})\leq d_{p}(x_{m_{k}+1},\widehat{x})\]  
for all \(k\in\mathbb{N}\). Thus it follows from (27) and the same methods in the proof of Case I that  
\[\|x_{m_{k}+1}-x_{m_{k}}\|\to 0,\quad\|Tx_{m_{k}}-x_{m_{k}}\|\to 0\]  
as \(k\to\infty\). Therefore, we have  
\[\limsup\limits_{k\to\infty}\langle J_{E_{1}}^{p}(u)-J_{E_{1}}^{p}(\widehat{x} ),y_{m_{k}+1}-\widehat{x}\rangle=\limsup\limits_{k\to\infty}\langle J_{E_{1}}^ {p}(u)-J_{E_{1}}^{p}(\widehat{x}),y_{m_{k}}-\widehat{x}\rangle\leq 0. \tag{39}\]  
From (24), also, we have  
\[d_{p}(x_{m_{k}+1},\widehat{x})\leq(1-\alpha_{m_{k}})d_{p}(x_{m_{k}},\widehat{ x})+\alpha_{m_{k}}\langle J_{E_{1}}^{p}(u)-J_{E_{1}}^{p}(\widehat{x}),y_{m_{k}}- \widehat{x}\rangle. \tag{40}\]  
Since \(d_{p}(x_{m_{k}},\widehat{x})\leq d_{p}(x_{m_{k}+1},\widehat{x})\), it follows from (40) that  
\[