Let
\(z^{\dagger}=\operatorname{proj}_{\Gamma}(\gamma f+\mathcal{I}-\mathcal {B})z^{\dagger}\). Subsequently, we obtain
\(z^{\dagger}\in\mathcal{C}\cap \operatorname{Fix}(\mathcal{T})\) and
\(\mathcal{A}z^{\dagger}\in\mathcal{Q}\cap \operatorname{Fix}(\mathcal{S})\). Note that
\(\operatorname{proj}_{\mathcal{Q}}\) is firmly nonexpansive. From (
2.1), we deduce
$$ \bigl\Vert z_{n}-\mathcal{A}z^{\dagger}\bigr\Vert ^{2}=\bigl\Vert \operatorname{proj}_{\mathcal{Q}}\mathcal {A}x_{n}-\operatorname{proj}_{\mathcal{Q}}\mathcal{A}z^{\dagger} \bigr\Vert ^{2}\le\bigl\Vert \mathcal {A}x_{n}- \mathcal{A}z^{\dagger}\bigr\Vert ^{2}-\Vert \mathcal{A}x_{n}-z_{n}\Vert ^{2}. $$
(3.4)
Applying Proposition
3.4 and noting conditions (C2) and (C3), we have
$$\operatorname{Fix}\bigl(\mathcal{S}\bigl((1-\eta_{n})\mathcal{I}+ \eta_{n}\mathcal{S}\bigr)\bigr)=\operatorname{Fix}(\mathcal{S}) $$
and
$$\operatorname{Fix}\bigl(\mathcal{T}\bigl((1-\gamma_{n})\mathcal{I}+ \gamma_{n}\mathcal {T}\bigr)\bigr)=\operatorname{Fix}(\mathcal{T}) $$
for all
\(n\in\mathbb{N}\).
By condition (C2) and Proposition
3.5, we derive
$$\begin{aligned} \bigl\Vert v_{n}-\mathcal{A}z^{\dagger}\bigr\Vert =&\bigl\Vert \bigl[(1-\zeta_{n})\mathcal{I}+\zeta_{n}\mathcal {S} \bigl((1-\eta_{n})\mathcal{I}+\eta_{n}\mathcal{S}\bigr) \bigr]z_{n}-\mathcal{A}z^{\dagger}\bigr\Vert \\ =&\bigl\Vert \bigl[(1-\zeta_{n})\mathcal{I}+\zeta_{n} \mathcal{S}\bigl((1-\eta_{n})\mathcal {I}+\eta_{n}\mathcal{S} \bigr)\bigr]z_{n} \\ &{} -\bigl[(1-\zeta_{n})\mathcal{I}+\zeta_{n}\mathcal{S} \bigl((1-\eta_{n})\mathcal {I}+\eta_{n}\mathcal{S}\bigr)\bigr] \mathcal{A}z^{\dagger}\bigr\Vert \\ \le&\bigl\Vert z_{n}-\mathcal{A}z^{\dagger}\bigr\Vert . \end{aligned}$$
This together with (
3.4) implies that
$$ \bigl\Vert v_{n}-\mathcal{A}z^{\dagger}\bigr\Vert ^{2}\le\bigl\Vert \mathcal{A}x_{n}- \mathcal{A}z^{\dagger}\bigr\Vert ^{2}-\Vert \mathcal{A}x_{n}-z_{n}\Vert ^{2}. $$
(3.5)
By condition (C3) and Proposition
3.5, we derive
$$\begin{aligned} \bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert =&\bigl\Vert \bigl[(1-\beta_{n})\mathcal{I}+\beta_{n}\mathcal {T} \bigl((1-\gamma_{n})\mathcal{I}+\gamma_{n}\mathcal{T}\bigr) \bigr]u_{n}-z^{\dagger}\bigr\Vert \\ =&\bigl\Vert \bigl[(1-\beta_{n})\mathcal{I}+\beta_{n} \mathcal{T}\bigl((1-\gamma_{n})\mathcal {I}+\gamma_{n} \mathcal{T}\bigr)\bigr]u_{n} \\ &{} -\bigl[(1-\beta_{n})\mathcal{I}+\beta_{n}\mathcal{T} \bigl((1-\gamma_{n})\mathcal {I}+\gamma_{n}\mathcal{T}\bigr) \bigr]z^{\dagger}\bigr\Vert \\ \le&\bigl\Vert u_{n}-z^{\dagger}\bigr\Vert . \end{aligned}$$
(3.6)
Noting that
\(\operatorname{proj}_{\mathcal{C}}\) is nonexpansive, we obtain
$$ \bigl\Vert u_{n}-z^{\dagger}\bigr\Vert =\bigl\Vert \operatorname{proj}_{\mathcal{C}} y_{n}-\operatorname{proj}_{\mathcal{C}}z^{\dagger} \bigr\Vert \le\bigl\Vert y_{n}-z^{\dagger}\bigr\Vert . $$
(3.7)
From (
3.1), we get
$$\begin{aligned} \bigl\Vert y_{n}-z^{\dagger}\bigr\Vert =&\bigl\Vert \alpha_{n}\gamma f(x_{n})+(\mathcal{I}- \alpha_{n}\mathcal {B}) \bigl(x_{n}-\delta\mathcal{A}^{*}( \mathcal{A}x_{n}-v_{n})\bigr)-z^{\dagger}\bigr\Vert \\ =&\bigl\Vert \alpha_{n}\gamma\bigl(f(x_{n})-f \bigl(z^{\dagger}\bigr)\bigr)+\alpha_{n}\bigl(\gamma f \bigl(z^{\dagger }\bigr)-\mathcal{B}z^{\dagger}\bigr) \\ &{} +(\mathcal{I}-\alpha_{n}\mathcal{B}) \bigl(x_{n}-z^{\dagger}- \delta\mathcal {A}^{*}(\mathcal{A}x_{n}-v_{n})\bigr)\bigr\Vert \\ \le&\alpha_{n}\gamma\bigl\Vert f(x_{n})-f \bigl(z^{\dagger}\bigr)\bigr\Vert +\alpha_{n}\bigl\Vert \gamma f \bigl(z^{\dagger }\bigr)-\mathcal{B}z^{\dagger}\bigr\Vert \\ &{} +\Vert \mathcal{I}-\alpha_{n}\mathcal{B}\Vert \bigl\Vert x_{n}-z^{\dagger}-\delta\mathcal {A}^{*}(\mathcal{A}x_{n}-v_{n}) \bigr\Vert \\ \le&\alpha_{n}\gamma\rho\bigl\Vert x_{n}-z^{\dagger} \bigr\Vert +\alpha_{n}\bigl\Vert \gamma f\bigl(z^{\dagger } \bigr)-\mathcal{B}z^{\dagger}\bigr\Vert \\ &{} +(1-\alpha_{n}\sigma)\bigl\Vert x_{n}-z^{\dagger}+ \delta\mathcal {A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\Vert . \end{aligned}$$
(3.8)
Observe that
$$\begin{aligned} \bigl\langle x_{n}-z^{\dagger}, \mathcal{A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\rangle =&\bigl\langle \mathcal{A}\bigl(x_{n}-z^{\dagger}\bigr), v_{n}- \mathcal{A}x_{n}\bigr\rangle \\ =&\bigl\langle \mathcal{A}x_{n}-\mathcal{A}z^{\dagger}+v_{n}- \mathcal {A}x_{n}-(v_{n}-\mathcal{A}x_{n}), v_{n}-\mathcal{A}x_{n}\bigr\rangle \\ =&\bigl\langle v_{n}-\mathcal{A}z^{\dagger}, v_{n}- \mathcal{A}x_{n}\bigr\rangle -\| v_{n}-\mathcal{A}x_{n} \|^{2}. \end{aligned}$$
(3.9)
Using (
3.3), we obtain
$$ \bigl\langle v_{n}-\mathcal{A}z^{\dagger}, v_{n}-\mathcal{A}x_{n}\bigr\rangle =\frac{1}{2} \bigl( \bigl\Vert v_{n}-\mathcal{A}z^{\dagger}\bigr\Vert ^{2}+\Vert v_{n}-\mathcal{A}x_{n}\Vert ^{2}-\bigl\Vert \mathcal{A}x_{n}-\mathcal{A}z^{\dagger} \bigr\Vert ^{2} \bigr). $$
(3.10)
From (
3.5), (
3.9) and (
3.10), we get
$$\begin{aligned} \bigl\langle x_{n}-z^{\dagger}, \mathcal{A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\rangle =& \frac{1}{2} \bigl(\bigl\Vert v_{n}-\mathcal{A}z^{\dagger} \bigr\Vert ^{2}+\Vert v_{n}-\mathcal{A}x_{n} \Vert ^{2}-\bigl\Vert \mathcal{A}x_{n}- \mathcal{A}z^{\dagger}\bigr\Vert ^{2} \bigr) \\ &{} -\Vert v_{n}-\mathcal{A}x_{n}\Vert ^{2} \\ \le&\frac{1}{2} \bigl(\bigl\Vert \mathcal{A}x_{n}- \mathcal{A}z^{\dagger}\bigr\Vert ^{2}-\Vert z_{n}- \mathcal{A}x_{n}\Vert ^{2}+\Vert v_{n}- \mathcal{A}x_{n}\Vert ^{2} \\ &{} -\bigl\Vert \mathcal{A}x_{n}-\mathcal{A}z^{\dagger}\bigr\Vert ^{2} \bigr)-\Vert v_{n}-\mathcal {A}x_{n} \Vert ^{2} \\ =&-\frac{1}{2}\Vert z_{n}-\mathcal{A}x_{n}\Vert ^{2}-\frac{1}{2}\Vert v_{n}-\mathcal{A}x_{n} \Vert ^{2}. \end{aligned}$$
(3.11)
According to equality (
3.3), we get
$$\begin{aligned} \bigl\Vert x_{n}-z^{\dagger}+\delta\mathcal{A}^{*}(v_{n}- \mathcal{A}x_{n})\bigr\Vert ^{2} =&\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}+\delta^{2} \bigl\Vert \mathcal{A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\Vert ^{2} \\ &{} +2\delta\bigl\langle x_{n}-z^{\dagger}, \mathcal{A}^{*}(v_{n}- \mathcal {A}x_{n})\bigr\rangle . \end{aligned}$$
Combining the above equality and (
3.11), we deduce
$$\begin{aligned} \bigl\Vert x_{n}-z^{\dagger}+\delta \mathcal{A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\Vert ^{2} \le&\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}+\delta^{2}\Vert \mathcal{A}\Vert ^{2} \Vert v_{n}-\mathcal{A}x_{n}\Vert ^{2} \\ &{} -\delta \Vert z_{n}-\mathcal{A}x_{n}\Vert ^{2}-\delta \Vert v_{n}-\mathcal{A}x_{n}\Vert ^{2} \\ =&\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}+ \bigl(\delta^{2}\Vert \mathcal{A}\Vert ^{2}-\delta\bigr) \Vert v_{n}-\mathcal {A}x_{n}\Vert ^{2} \\ &{} -\delta \Vert z_{n}-\mathcal{A}x_{n}\Vert ^{2}. \end{aligned}$$
(3.12)
In view of condition (C4), we know that
\(\delta^{2}\|\mathcal{A}\| ^{2}-\delta<0\). From (
3.12), we have
$$ \bigl\Vert x_{n}-z^{\dagger}+\delta\mathcal{A}^{*}(v_{n}- \mathcal{A}x_{n})\bigr\Vert ^{2}\le\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}. $$
Therefore,
$$ \bigl\Vert x_{n}-z^{\dagger}+\delta \mathcal{A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\Vert \le\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert . $$
(3.13)
Substituting (
3.13) into (
3.8) we deduce
$$\begin{aligned} \begin{aligned}[b] \bigl\Vert y_{n}-z^{\dagger}\bigr\Vert &\le \alpha_{n}\gamma\rho\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert +\alpha_{n}\bigl\Vert \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}\bigr\Vert +(1-\alpha_{n}\sigma)\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert \\ &=\bigl[1-(\sigma-\gamma\rho)\alpha_{n}\bigr]\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert +\alpha_{n}\bigl\Vert \gamma f\bigl(z^{\dagger}\bigr)-\mathcal{B}z^{\dagger}\bigr\Vert . \end{aligned} \end{aligned}$$
(3.14)
From (
3.6), (
3.7) and (
3.14), we get
$$\begin{aligned} \bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert \le&\bigl[1-(\sigma- \gamma\rho)\alpha_{n}\bigr]\bigl\Vert x_{n}-z^{\dagger} \bigr\Vert +\alpha_{n}\bigl\Vert \gamma f\bigl(z^{\dagger} \bigr)-\mathcal{B}z^{\dagger}\bigr\Vert \\ =&\bigl[1-(\sigma-\gamma\rho)\alpha_{n}\bigr]\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert +(\sigma-\gamma\rho ) \alpha_{n}\frac{\Vert \gamma f(z^{\dagger})-\mathcal{B}z^{\dagger} \Vert }{\sigma-\gamma \rho}. \end{aligned}$$
By induction, we get
$$ \bigl\Vert x_{n}-z^{\dagger}\bigr\Vert \le\max\biggl\{ \bigl\| x_{0}-z^{\dagger}\bigr\| , \frac{\|\gamma f(z^{\dagger })-\mathcal{B}z^{\dagger}\|}{\sigma-\gamma\rho}\biggr\} . $$
Hence, the sequence
\(\{x_{n}\}\) is bounded.
Using the firm nonexpansiveness of
\(\operatorname{proj}_{\mathcal{C}}\), we have
$$\begin{aligned} \bigl\Vert u_{n}-z^{\dagger}\bigr\Vert ^{2} =&\bigl\Vert \operatorname{proj}_{\mathcal{C}}y_{n}-z^{\dagger} \bigr\Vert ^{2} \\ \le&\bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2}- \Vert \operatorname{proj}_{\mathcal{C}} y_{n}-y_{n}\Vert ^{2} \\ =&\bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2}- \Vert u_{n}-y_{n}\Vert ^{2}. \end{aligned}$$
(3.15)
From (
3.6), (
3.14) and (
3.15), we deduce
$$\begin{aligned} \bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert ^{2} \le& \bigl\Vert u_{n}-z^{\dagger}\bigr\Vert ^{2} \\ \le&\bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2}- \Vert u_{n}-y_{n}\Vert ^{2} \\ \le&\bigl[1-(\sigma-\gamma\rho)\alpha_{n}\bigr]\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}+\frac{\alpha _{n}}{\sigma-\gamma\rho} \bigl\Vert \gamma f\bigl(z^{\dagger}\bigr)-\mathcal{B}z^{\dagger}\bigr\Vert ^{2}-\Vert u_{n}-y_{n}\Vert ^{2}. \end{aligned}$$
It follows that
$$ \Vert u_{n}-y_{n}\Vert ^{2}\le \bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}-\bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert ^{2}+ \frac{\alpha _{n}}{\sigma-\gamma\rho}\bigl\Vert \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}\bigr\Vert ^{2}. $$
(3.16)
Next, we consider two possible cases: the sequence
\(\{\|x_{n}-z^{\dagger}\|\} \) is either monotone decreasing at infinity (Case 1) or not (Case 2).
Case 1.
There exists \(n_{0}\) such that the sequence \(\{\|x_{n}-z^{\dagger}\|\} _{n\ge n_{0}}\) is decreasing.
Case 2.
For any \(n_{0}\), there exists an integer \(m\ge n_{0}\) such that \(\| x_{m}-z^{\dagger}\|\le\|x_{m+1}-z^{\dagger}\|\).
In Case 1, we assume that there exists some integer
\(m>0\) such that
\(\{ \|x_{n}-z^{\dagger}\|\}\) is decreasing for all
\(n\ge m\). Then
\(\lim_{n\to\infty}\|x_{n}-z^{\dagger}\|\) exists. From (
3.16), we deduce
$$ \lim_{n\to\infty}\|u_{n}-y_{n}\|=0. $$
(3.17)
From (
3.8), we have
$$\begin{aligned} \bigl\Vert y_{n}-z^{\dagger}\bigr\Vert \le& \alpha_{n}\gamma\rho\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert +\alpha_{n}\bigl\Vert \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}\bigr\Vert \\ &{} +(1-\alpha_{n}\sigma)\bigl\Vert x_{n}-z^{\dagger}+ \delta\mathcal {A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\Vert \\ =&\alpha_{n}\sigma\frac{\gamma\rho \Vert x_{n}-z^{\dagger} \Vert +\Vert \gamma f(z^{\dagger })-\mathcal{B}z^{\dagger} \Vert }{\sigma} \\ &{} +(1-\alpha_{n}\sigma)\bigl\Vert x_{n}-z^{\dagger}+ \delta\mathcal {A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\Vert . \end{aligned}$$
(3.18)
Since
\(\{x_{n}\}\) is bounded, there exists a constant
M> such that
$$\sup_{n}\biggl\{ \frac{\gamma\rho\|x_{n}-z^{\dagger}\|+\|\gamma f(z^{\dagger })-\mathcal{B}z^{\dagger}\|}{\sigma}\biggr\} < M. $$
By (
3.18), we deduce
$$ \bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2}\le\alpha_{n}\sigma M^{2}+(1- \alpha_{n}\sigma)\bigl\Vert x_{n}-z^{\dagger }+\delta \mathcal{A}^{*}(v_{n}-\mathcal{A}x_{n})\bigr\Vert ^{2}. $$
(3.19)
Combining (
3.12) and (
3.19), we obtain
$$\begin{aligned} \bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert ^{2} \le& \bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2} \\ \le&(1-\sigma\alpha_{n})\bigl\Vert x_{n}-z^{\dagger} \bigr\Vert ^{2}+(1-\sigma\alpha_{n}) \bigl( \delta^{2}\Vert \mathcal{A}\Vert ^{2}-\delta\bigr)\Vert v_{n}-\mathcal{A}x_{n}\Vert ^{2} \\ &{} -(1-\sigma\alpha_{n})\delta \Vert z_{n}- \mathcal{A}x_{n}\Vert ^{2}+\alpha_{n}\sigma M^{2}. \end{aligned}$$
Hence,
$$\begin{aligned} 0 \le&(1-\sigma\alpha_{n}) \bigl(\delta-\delta^{2}\Vert \mathcal{A}\Vert ^{2}\bigr)\Vert v_{n}-\mathcal {A}x_{n}\Vert ^{2}+(1-\sigma\alpha_{n})\delta \Vert z_{n}-\mathcal{A}x_{n}\Vert ^{2} \\ \le&(1-\sigma\alpha_{n})\bigl\Vert x_{n}-z^{\dagger} \bigr\Vert ^{2}-\bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert ^{2}+\alpha _{n}\sigma M^{2}, \end{aligned}$$
which implies that
$$ \lim_{n\to\infty}\|v_{n}- \mathcal{A}x_{n}\|=\lim_{n\to\infty}\| z_{n}- \mathcal{A}x_{n}\|=0. $$
(3.20)
Therefore,
$$ \lim_{n\to\infty}\|v_{n}-z_{n}\|=0. $$
(3.21)
Note that
\(v_{n}-z_{n}=\zeta_{n}[\mathcal{S}((1-\eta_{n})\mathcal{I}+\eta _{n}\mathcal{S})z_{n}-z_{n}]\). Thus,
$$ \lim_{n\to\infty}\bigl\Vert z_{n}- \mathcal{S}\bigl((1-\eta_{n})\mathcal{I}+\eta_{n}\mathcal {S} \bigr)z_{n}\bigr\Vert =\lim_{n\to\infty}\bigl\Vert \mathcal{A}x_{n}-\mathcal{S}\bigl((1-\eta _{n})\mathcal{I}+ \eta_{n}\mathcal{S}\bigr)\mathcal{A}x_{n}\bigr\Vert =0. $$
(3.22)
Since
$$\begin{aligned} \Vert \mathcal{A}x_{n}-\mathcal{S}\mathcal{A}x_{n}\Vert \le&\bigl\Vert \mathcal {A}x_{n}-S\bigl((1-\eta_{n}) \mathcal{I}+\eta_{n}\mathcal{S}\bigr)\mathcal{A}x_{n}\bigr\Vert \\ &{} +\bigl\Vert \mathcal{S}\bigl((1-\eta_{n})\mathcal{I}+ \eta_{n}\mathcal{S}\bigr)\mathcal {A}x_{n}-\mathcal{S} \mathcal{A}x_{n}\bigr\Vert \\ \le&\bigl\Vert \mathcal{A}x_{n}-\mathcal{S}\bigl((1- \eta_{n})\mathcal{I}+\eta_{n}\mathcal {S}\bigr) \mathcal{A}x_{n}\bigr\Vert +\mathcal{L}_{1} \eta_{n}\Vert \mathcal{A}x_{n}-\mathcal {S} \mathcal{A}x_{n}\Vert , \end{aligned}$$
it follows that
$$ \Vert \mathcal{A}x_{n}-\mathcal{S}\mathcal{A}x_{n}\Vert \le\frac{1}{1-\mathcal {L}_{1}\eta_{n}}\bigl\Vert \mathcal{A}x_{n}-\mathcal{S}\bigl((1- \eta_{n})\mathcal{I}+\eta _{n}\mathcal{S}\bigr) \mathcal{A}x_{n}\bigr\Vert . $$
This together with (
3.22) implies that
$$ \lim_{n\to\infty}\|\mathcal{A}x_{n}- \mathcal{S}\mathcal{A}x_{n}\|=0. $$
(3.23)
According to (
3.1), we have
$$\begin{aligned} \Vert y_{n}-x_{n}\Vert =&\bigl\Vert \alpha_{n}\gamma f(x_{n})-\delta\mathcal{A}^{*}(\mathcal {A}x_{n}-v_{n})-\alpha_{n}\mathcal{B} \bigl(x_{n}-\delta\mathcal{A}^{*}(\mathcal {A}x_{n}-v_{n}) \bigr)\bigr\Vert \\ \le&\delta \Vert \mathcal{A}\Vert \Vert v_{n}- \mathcal{A}x_{n}\Vert +\alpha_{n}\bigl\Vert \gamma f(x_{n})-\mathcal{B}\bigl(x_{n}-\delta\mathcal{A}^{*}( \mathcal{A}x_{n}-v_{n})\bigr)\bigr\Vert . \end{aligned}$$
It follows from (
3.20) and (C1) that
$$ \lim_{n\to\infty}\|x_{n}-y_{n}\|=0. $$
(3.24)
From (
3.1) and (
3.2), we have
$$\begin{aligned} \bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert ^{2} =&\bigl\Vert (1-\beta_{n}) \bigl(u_{n}-z^{\dagger} \bigr)+\beta_{n}\bigl[\mathcal {T}\bigl((1-\gamma_{n})u_{n}+ \gamma_{n}\mathcal{T}u_{n}\bigr)-z^{\dagger}\bigr]\bigr\Vert ^{2} \\ =&(1-\beta_{n})\bigl\Vert u_{n}-z^{\dagger}\bigr\Vert ^{2}+\beta_{n}\bigl\Vert \mathcal{T}\bigl((1-\gamma _{n})u_{n}+\gamma_{n}\mathcal{T}u_{n} \bigr)-z^{\dagger}\bigr\Vert ^{2} \\ &{} -\beta_{n}(1-\beta_{n})\bigl\Vert \mathcal{T} \bigl((1-\gamma_{n})u_{n}+\gamma _{n} \mathcal{T}u_{n}\bigr)-u_{n}\bigr\Vert ^{2}. \end{aligned}$$
(3.25)
Applying Proposition
3.5, we get
$$\begin{aligned}& \bigl\Vert \mathcal{T}\bigl((1-\gamma_{n})u_{n}+ \gamma_{n} \mathcal{T}u_{n}\bigr)-z^{\dagger}\bigr\Vert ^{2} \\& \quad \le\bigl\Vert u_{n}-z^{\dagger}\bigr\Vert ^{2}+(1-\gamma_{n})\bigl\Vert u_{n}-\mathcal{T} \bigl((1-\gamma _{n})u_{n}+\gamma_{n} \mathcal{T}u_{n}\bigr)\bigr\Vert ^{2}. \end{aligned}$$
(3.26)
From (
3.19), (
3.25) and (
3.26), we deduce
$$\begin{aligned} \bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert ^{2} \le& \bigl\Vert u_{n}-z^{\dagger}\bigr\Vert ^{2}- \beta_{n}(\gamma_{n}-\beta_{n})\bigl\Vert u_{n}-\mathcal{T}\bigl((1-\gamma_{n})u_{n}+ \gamma_{n}\mathcal{T}u_{n}\bigr)\bigr\Vert ^{2} \\ \le&\alpha_{n}\sigma M^{2}+(1-\alpha_{n}\sigma) \bigl\Vert x_{n}-z^{\dagger}+\delta\mathcal {A}^{*}(v_{n}- \mathcal{A}x_{n})\bigr\Vert ^{2} \\ &{} -\beta_{n}(\gamma_{n}-\beta_{n})\bigl\Vert u_{n}-\mathcal{T}\bigl((1-\gamma_{n})u_{n}+\gamma _{n}\mathcal{T}u_{n}\bigr)\bigr\Vert ^{2} \\ \le&\alpha_{n}\sigma M^{2}+\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}-\beta_{n}( \gamma_{n}-\beta_{n})\bigl\Vert u_{n}-\mathcal{T} \bigl((1-\gamma_{n})u_{n}+\gamma_{n} \mathcal{T}u_{n}\bigr)\bigr\Vert ^{2}. \end{aligned}$$
It follows that
$$\begin{aligned}& \beta_{n}(\gamma_{n}-\beta_{n})\bigl\Vert u_{n}-\mathcal{T}\bigl((1-\gamma_{n})u_{n}+\gamma _{n}\mathcal{T}u_{n}\bigr)\bigr\Vert ^{2} \\& \quad \le\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}-\bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert ^{2}+\alpha_{n}\sigma M^{2}. \end{aligned}$$
Therefore,
$$ \lim_{n\to\infty}\bigl\Vert u_{n}- \mathcal{T}\bigl((1-\gamma_{n})u_{n}+\gamma_{n} \mathcal {T}u_{n}\bigr)\bigr\Vert =0. $$
(3.27)
Observe that
$$\begin{aligned} \Vert u_{n}-\mathcal{T}u_{n}\Vert \le&\bigl\Vert u_{n}-\mathcal{T}\bigl((1-\gamma_{n})u_{n}+\gamma _{n}\mathcal{T}u_{n}\bigr)\bigr\Vert +\bigl\Vert \mathcal{T}\bigl((1-\gamma_{n})u_{n}+\gamma_{n} \mathcal {T}u_{n}\bigr)-\mathcal{T}u_{n}\bigr\Vert \\ \le&\bigl\Vert u_{n}-\mathcal{T}\bigl((1-\gamma_{n})u_{n}+ \gamma_{n}\mathcal{T}u_{n}\bigr)\bigr\Vert +\mathcal {L}_{2}\gamma_{n}\Vert u_{n}- \mathcal{T}u_{n}\Vert . \end{aligned}$$
Thus,
$$ \|u_{n}-\mathcal{T}u_{n}\|\le\frac{1}{1-\mathcal{L}_{2}\gamma_{n}}\bigl\Vert u_{n}-\mathcal{T}\bigl((1-\gamma_{n})u_{n}+ \gamma_{n}\mathcal{T}u_{n}\bigr)\bigr\Vert . $$
This together with (
3.27) implies that
$$ \lim_{n\to\infty}\|u_{n}- \mathcal{T}u_{n}\|=0. $$
(3.28)
Next, we show that
$$\limsup_{n\to\infty}\bigl\langle \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}, y_{n}-z^{\dagger}\bigr\rangle \le0. $$
Choose a subsequence
\(\{y_{n_{i}}\}\) of
\(\{y_{n}\}\) such that
$$ \limsup_{n\to\infty}\bigl\langle \gamma f \bigl(z^{\dagger}\bigr)-\mathcal{B}z^{\dagger}, y_{n}-z^{\dagger} \bigr\rangle =\lim_{i\to\infty}\bigl\langle \gamma f\bigl(z^{\dagger } \bigr)-\mathcal{B}z^{\dagger}, y_{n_{i}}-z^{\dagger}\bigr\rangle . $$
(3.29)
Since the sequence
\(\{y_{n_{i}}\}\) is bounded, we can choose a subsequence
\(\{y_{n_{i_{j}}}\}\) of
\(\{y_{n_{i}}\}\) such that
\(y_{n_{i_{j}}}\rightharpoonup z\). For the sake of convenience, we assume (without loss of generality) that
\(y_{n_{i}}\rightharpoonup z\). Subsequently, we derive from the above conclusions that
$$ \textstyle\begin{cases} x_{n_{i}}\rightharpoonup z, \\ y_{n_{i}}\rightharpoonup z, \\ u_{n_{i}}\rightharpoonup z \end{cases} $$
(3.30)
and
$$ \textstyle\begin{cases} \mathcal{A}x_{n_{i}}\rightharpoonup\mathcal{A}z, \\ \mathcal{A}y_{n_{i}}\rightharpoonup\mathcal{A}z, \\ \mathcal{A}u_{n_{i}}\rightharpoonup\mathcal{A}z. \end{cases} $$
(3.31)
Note that
\(u_{n_{i}}=\operatorname{proj}_{\mathcal{C}}y_{n_{i}}\in\mathcal{C}\) and
\(z_{n_{i}}=\operatorname{proj}_{\mathcal{Q}}\mathcal{A}x_{n_{i}}\in\mathcal{Q}\). From (
3.30), we deduce
\(z\in\mathcal{C}\) and
\(\mathcal{A}z\in\mathcal {Q}\) by (
3.31). By the demiclosedness of
\(\mathcal{T}-\mathcal {I}\) and
\(\mathcal{S}-\mathcal{I}\), we deduce
\(z\in \operatorname{Fix}(\mathcal{T})\) (by (
3.28)) and
\(\mathcal{A}z\in \operatorname{Fix}(\mathcal{S})\) (by (
3.23)). To this end, we deduce
\(z\in\mathcal{C}\cap \operatorname{Fix}(\mathcal{T})\) and
\(\mathcal{A}z\in\mathcal{Q}\cap \operatorname{Fix}(\mathcal{S})\). That is to say,
\(z\in\Gamma\).
Therefore,
$$\begin{aligned} \limsup_{n\to\infty}\bigl\langle \gamma f \bigl(z^{\dagger}\bigr)-\mathcal{B}z^{\dagger}, y_{n}-z^{\dagger} \bigr\rangle =&\lim_{i\to\infty}\bigl\langle \gamma f \bigl(z^{\dagger }\bigr)-\mathcal{B}z^{\dagger}, y_{n_{i}}-z^{\dagger} \bigr\rangle \\ =&\lim_{i\to\infty}\bigl\langle \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}, z-z^{\dagger}\bigr\rangle \\ \le&0. \end{aligned}$$
(3.32)
From (
3.1), we have
$$\begin{aligned} \bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2} =&\bigl\Vert \alpha_{n}\gamma\bigl(f(x_{n})-f\bigl(z^{\dagger} \bigr)\bigr)+\alpha_{n}\bigl(\gamma f\bigl(z^{\dagger }\bigr)- \mathcal{B}z^{\dagger}\bigr) \\ &{} +(\mathcal{I}-\alpha_{n}\mathcal{B}) \bigl(x_{n}-z^{\dagger}- \delta\mathcal {A}^{*}(\mathcal{A}x_{n}-v_{n})\bigr)\bigr\Vert ^{2} \\ \le&\Vert \mathcal{I}-\alpha_{n}\mathcal{B}\Vert ^{2} \bigl\Vert x_{n}-z^{\dagger}-\delta\mathcal {A}^{*}( \mathcal{A}x_{n}-v_{n})\bigr\Vert ^{2} \\ &{} +2\alpha_{n}\gamma\bigl\langle f(x_{n})-f \bigl(z^{\dagger}\bigr), y_{n}-z^{\dagger}\bigr\rangle +2 \alpha_{n}\bigl\langle \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}, y_{n}-z^{\dagger }\bigr\rangle \\ \le&(1-\alpha_{n}\sigma)^{2}\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}+2\alpha_{n} \gamma\bigl\Vert f(x_{n})-f\bigl(z^{\dagger}\bigr)\bigr\Vert \bigl\Vert y_{n}-z^{\dagger}\bigr\Vert \\ &{} +2\alpha_{n}\bigl\langle \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}, y_{n}-z^{\dagger}\bigr\rangle \\ \le&(1-\alpha_{n}\sigma)^{2}\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}+\alpha_{n} \gamma\rho\bigl\Vert x_{n}-z^{\dagger}\bigr\Vert ^{2}+ \alpha_{n}\gamma\rho\bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2} \\ &{} +2\alpha_{n}\bigl\langle \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}, y_{n}-z^{\dagger}\bigr\rangle . \end{aligned}$$
It follows that
$$\begin{aligned} \bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2} \le& \biggl[1-\frac{2(\sigma-\gamma\rho)\alpha_{n}}{1-\gamma\rho\alpha _{n}} \biggr]\bigl\Vert x_{n}-z^{\dagger} \bigr\Vert ^{2}+\frac{\sigma^{2}\alpha_{n}^{2}}{1-\gamma\rho\alpha _{n}}\bigl\Vert x_{n}-z^{\dagger} \bigr\Vert ^{2} \\ &{} +\frac{2\alpha_{n}}{1-\gamma\rho\alpha_{n}}\bigl\langle \gamma f\bigl(z^{\dagger }\bigr)- \mathcal{B}z^{\dagger}, y_{n}-z^{\dagger}\bigr\rangle . \end{aligned}$$
Therefore,
$$\begin{aligned} \bigl\Vert x_{n+1}-z^{\dagger}\bigr\Vert ^{2} \le&\bigl\Vert y_{n}-z^{\dagger}\bigr\Vert ^{2} \\ \le& \biggl[1-\frac{2(\sigma-\gamma\rho)\alpha_{n}}{1-\gamma\rho\alpha _{n}} \biggr]\bigl\Vert x_{n}-z^{\dagger} \bigr\Vert ^{2}+\frac{\sigma^{2}\alpha_{n}^{2}}{1-\gamma\rho\alpha _{n}}\bigl\Vert x_{n}-z^{\dagger} \bigr\Vert ^{2} \\ &{}+\frac{2\alpha_{n}}{1-\gamma\rho\alpha_{n}}\bigl\langle \gamma f\bigl(z^{\dagger }\bigr)- \mathcal{B}z^{\dagger}, y_{n}-z^{\dagger}\bigr\rangle . \end{aligned}$$
(3.33)
Applying Lemma
2.6 and (
3.32) to (
3.33), we deduce
\(x_{n}\to z^{\dagger}\).
Case 2. Assume that there exists an integer
\(n_{0}\) such that
$$\bigl\Vert x_{n_{0}}-z^{\dagger}\bigr\Vert \le\bigl\Vert x_{n_{0}+1}-z^{\dagger}\bigr\Vert . $$
Set
\(\omega_{n}=\{\|x_{n}-z^{\dagger}\|\}\). Then we have
$$\omega_{n_{0}}\le\omega_{n_{0}+1}. $$
Define an integer sequence
\(\{\tau_{n}\}\) for all
\(n\ge n_{0}\) as follows:
$$\tau(n)=\max\{l\in\mathbb{N}| n_{0}\le l\le n, \omega_{l} \le\omega_{l+1}\}. $$
It is clear that
\(\tau(n)\) is a nondecreasing sequence satisfying
$$\lim_{n\to\infty}\tau(n)=\infty $$
and
$$\omega_{\tau(n)}\le\omega_{\tau(n)+1} $$
for all
\(n\ge n_{0}\).
By a similar argument as that of Case 1, we can obtain
$$\begin{aligned} \begin{aligned} &\lim_{n\to\infty} \Vert u_{\tau(n)}-y_{\tau(n)}\Vert = \lim_{n\to\infty} \Vert x_{\tau(n)}-y_{\tau(n)}\Vert =0, \\ &\lim_{n\to\infty} \Vert \mathcal{S}x_{\tau(n)}- \mathcal{A}x_{\tau(n)}\Vert =0 \end{aligned} \end{aligned}$$
and
$$ \lim_{n\to\infty} \Vert u_{\tau(n)}-\mathcal{T}u_{\tau(n)} \Vert =0. $$
This implies that
$$\omega_{w}(y_{\tau(n)})\subset\Gamma. $$
Thus, we obtain
$$ \limsup_{n\to\infty}\bigl\langle \gamma f \bigl(z^{\dagger}\bigr)-\mathcal{B}z^{\dagger },y_{\tau(n)}-z^{\dagger} \bigr\rangle \le0. $$
(3.34)
Since
\(\omega_{\tau(n)}\le\omega_{\tau(n)+1}\), we have from (
3.33) that
$$\begin{aligned} \omega_{\tau(n)}^{2} \le&\omega_{\tau(n)+1}^{2} \\ \le& \biggl[1-\frac{2(\sigma-\gamma\rho)\alpha_{\tau(n)}}{1-\gamma\rho \alpha_{\tau(n)}} \biggr]\omega_{\tau(n)}^{2}+ \frac{\sigma^{2}\alpha_{\tau (n)}^{2}}{1-\gamma\rho\alpha_{\tau(n)}}\omega_{\tau(n)}^{2} \\ &{}+\frac{2\alpha_{\tau(n)}}{1-\gamma\rho\alpha_{\tau(n)}}\bigl\langle \gamma f\bigl(z^{\dagger}\bigr)- \mathcal{B}z^{\dagger}, y_{\tau(n)}-z^{\dagger}\bigr\rangle . \end{aligned}$$
(3.35)
It follows that
$$ \omega_{\tau(n)}^{2}\le\frac{2}{2(\sigma-\gamma\rho)-\sigma^{2}\alpha_{\tau (n)}}\bigl\langle \gamma f\bigl(z^{\dagger}\bigr)-\mathcal{B}z^{\dagger}, y_{\tau(n)}-z^{\dagger }\bigr\rangle . $$
(3.36)
Combining (
3.34) and (
3.36), we have
$$\limsup_{n\to\infty}\omega_{\tau(n)}\le0, $$
and hence
$$ \lim_{n\to\infty}\omega_{\tau(n)}=0. $$
(3.37)
By (
3.35), we obtain
$$\limsup_{n\to\infty}\omega_{\tau(n)+1}^{2}\le\limsup _{n\to\infty}\omega _{\tau(n)}^{2}. $$
This together with (
3.37) implies that
$$\lim_{n\to\infty}\omega_{\tau(n)+1}=0. $$
Applying Lemma
2.7 we get
$$0\le\omega_{n}\le\max\{\omega_{\tau(n)}, \omega_{\tau(n)+1}\}. $$
Therefore,
\(\omega_{n}\to0\). That is,
\(x_{n}\to z^{\dagger}\). This completes the proof. □