For any
\(x,y\in C\), we see that
$$\begin{aligned}& \bigl\Vert (I-s_{n}A)x-(I-s_{n}A)y\bigr\Vert ^{2} \\& \quad =\|x-y\|^{2}-2s_{n}\langle x-y, Ax-Ay \rangle+s_{n}^{2}\| Ax-Ay\|^{2} \\& \quad \leq\|x-y\|^{2}-s_{n}(2\alpha-s_{n}) \|Ax-Ay\|^{2}. \end{aligned}$$
By using condition (e), we see that
\(\|(I-s_{n}A)x-(I-s_{n}A)y\|\leq\|x-y\|\). This proves that
\(I-s_{n}A\) is nonexpansive. Put
\(S_{n}=\delta_{n}I+(1-\delta_{n})S\). It follows from Lemma
1.4 that
\(S_{n}\) is nonexpansive and
\(F(S_{n})=F(S)\). Let
\(p\in\Omega\) be fixed arbitrarily. Hence, we have
$$\begin{aligned} \Vert x_{n+1}-p\Vert &\leq\alpha_{n}\bigl\Vert f(x_{n})-p\bigr\Vert +\beta_{n}\Vert x_{n}-p \Vert +\gamma_{n}\Vert S_{n}y_{n}-p\Vert \\ &\leq\alpha_{n}\beta \Vert x_{n}-p\Vert + \alpha_{n}\bigl\Vert f(p)-p\bigr\Vert +\beta_{n}\Vert x_{n}-p\Vert +\gamma _{n}\Vert y_{n}-p\Vert \\ &\leq\bigl(1-\alpha_{n}(1-\beta)\bigr)\Vert x_{n}-p\Vert +\alpha_{n}\bigl\Vert f(p)-p\bigr\Vert +\Vert e_{n} \Vert \\ &\leq\max\biggl\{ \Vert x_{n}-p\Vert ,\frac{\Vert f(p)-p\Vert }{1-\beta}\biggr\} + \Vert e_{n}\Vert . \end{aligned}$$
It follows that
\(\|x_{n}-p\|\leq\max\{\|x_{1}-p\|,\frac{\|f(p)-p\|}{1-\beta}\}+\sum_{n=1}^{\infty}\|e_{n}\|\). This shows that
\(\{x_{n}\}\) is bounded, so are
\(\{y_{n}\}\) and
\(\{z_{n}\}\). Let
\(\lambda_{n}=\frac{x_{n+1}-\beta_{n}x_{n}}{1-\beta_{n}}\). It follows that
$$\begin{aligned} \lambda_{n+1}-\lambda_{n} =&\frac{\alpha_{n+1} f(x_{n+1})+\gamma_{n+1}S_{n+1}y_{n+1}}{1-\beta_{n+1}} - \frac{\alpha_{n} f(x_{n})+\gamma_{n}S_{n}y_{n}}{1-\beta_{n}} \\ =&\frac{\alpha_{n+1} (f(x_{n+1})-S_{n+1}y_{n+1})+(1-\beta _{n+1})S_{n+1}y_{n}}{1-\beta_{n+1}} \\ &{} -\frac{\alpha_{n} (f(x_{n})-S_{n}y_{n})+(1-\beta_{n})S_{n}y_{n}}{1-\beta_{n}} \\ =&\frac{\alpha_{n+1} (f(x_{n+1})-S_{n+1}y_{n+1})}{1-\beta_{n+1}}-\frac {\alpha_{n} (f(x_{n})-S_{n}y_{n})}{1-\beta_{n}}+S_{n+1}y_{n+1}-S_{n}y_{n}. \end{aligned}$$
Hence, we have
$$\begin{aligned} \|\lambda_{n+1}-\lambda_{n}\| \leq&\frac{\alpha_{n+1}\|f(x_{n+1})-S_{n+1}y_{n+1}\|}{1-\beta_{n+1}} + \frac{\alpha_{n} \|f(x_{n})-S_{n}y_{n}\|}{1-\beta_{n}} \\ &{} +\|S_{n+1}y_{n+1}-S_{n}y_{n}\|. \end{aligned}$$
(2.1)
Since
\(z_{n}=T_{r_{n}}x_{n}\), we find that
\(F(z_{n},z)+\frac{1}{r_{n}}\langle z-z_{n},z_{n}-x_{n}\rangle\geq0\),
\(\forall z\in C \) and
\(F(z_{n+1},z)+\frac{1}{r_{n+1}}\langle z-z_{n+1},z_{n+1}-x_{n+1}\rangle\geq0\),
\(\forall z\in C\). It follows that
\(F(z_{n},z_{n+1})+\frac{1}{r_{n}}\langle z_{n+1}-z_{n},z_{n}-x_{n}\rangle\geq0 \) and
\(F(z_{n+1},z_{n})+\frac{1}{r_{n+1}}\langle z_{n}-z_{n+1},z_{n+1}-x_{n+1}\rangle\geq0\),
\(\forall z\in C\). By using condition (A2), we find that
\(\langle z_{n+1}-z_{n},\frac{z_{n}-x_{n}}{r_{n}}-\frac {z_{n+1}-x_{n+1}}{r_{n+1}}\rangle\geq0\). Hence, we have
$$\biggl\langle z_{n+1}-z_{n}, z_{n+1}-x_{n} -\frac{r_{n}}{r_{n+1}}(z_{n+1}-x_{n+1})\biggr\rangle \geq \|z_{n+1}-z_{n}\|^{2}. $$
This implies that
\(\langle z_{n+1}-z_{n}, x_{n+1}-x_{n} +(1-\frac {r_{n}}{r_{n+1}})(z_{n+1}-x_{n+1})\rangle\geq\|z_{n+1}-z_{n}\|^{2}\). Hence, we have
$$\|z_{n+1}-z_{n}\| \leq\|x_{n+1}-x_{n}\|+ \frac{|r_{n+1}-r_{n}|}{r_{n+1}}\| T_{r_{n+1}}x_{n}-x_{n+1}\|. $$
It follows that
$$\begin{aligned} \Vert y_{n+1}-y_{n}\Vert \leq&\bigl\Vert P_{C} (z_{n+1}-s_{n+1}Az_{n+1}+e_{n+1})-P_{C} (z_{n}-s_{n+1}Az_{n}+e_{n+1})\bigr\Vert \\ &{} +\bigl\Vert P_{C} (z_{n}-s_{n+1}Az_{n}+e_{n+1})-P_{C} (z_{n}-s_{n}Az_{n}+e_{n})\bigr\Vert \\ \leq&\bigl\Vert (I-s_{n+1}A)z_{n+1}-(I-s_{n+1}A)z_{n} \bigr\Vert \\ &{} +\bigl\Vert (z_{n}-s_{n+1}Az_{n}+e_{n+1})- (z_{n}-s_{n}Az_{n}+e_{n})\bigr\Vert \\ \leq&\Vert z_{n+1}-z_{n}\Vert +|s_{n+1}-s_{n}| \Vert Az_{n}\Vert +\Vert e_{n+1}\Vert +\Vert e_{n}\Vert \\ \leq&\Vert x_{n+1}-x_{n}\Vert +\frac{|r_{n+1}-r_{n}|}{r_{n+1}} \Vert T_{r_{n+1}}x_{n}-x_{n}\Vert \\ &{} +|s_{n+1}-s_{n}|\Vert Az_{n}\Vert + \Vert e_{n+1}\Vert +\Vert e_{n}\Vert . \end{aligned}$$
This implies that
$$\begin{aligned}& \|S_{n+1}y_{n+1}-S_{n}y_{n}\| \\& \quad \leq\|S_{n+1}y_{n+1}-S_{n+1}y_{n}\|+ \|S_{n+1}y_{n}-S_{n}y_{n}\| \\& \quad \leq\|y_{n+1}-y_{n}\|+\|S_{n+1}y_{n}-S_{n}y_{n} \| \\& \quad \leq \|x_{n+1}-x_{n}\|+\frac{|r_{n+1}-r_{n}|}{r_{n+1}}\| T_{r_{n+1}}x_{n}-x_{n}\| \\& \qquad {} +|s_{n+1}-s_{n}|\|Az_{n}\|+ \|e_{n+1}\|+\|e_{n}\|+|\delta_{n+1}-\delta _{n}|\|Sy_{n}-y_{n}\|. \end{aligned}$$
(2.2)
Substituting (
2.2) into (
2.1), we find that
$$\begin{aligned}& \|\lambda_{n+1}-\lambda_{n}\|-\|x_{n+1}-x_{n} \| \\& \quad \leq\frac{\alpha_{n+1}\|f(x_{n+1})-S_{n+1}y_{n+1}\|}{1-\beta_{n+1}} +\frac{\alpha_{n} \|f(x_{n})-S_{n}y_{n}\|}{1-\beta_{n}} \\& \qquad {} +\frac{|r_{n+1}-r_{n}|}{r_{n+1}}\|T_{r_{n+1}}x_{n}-x_{n} \| \\& \qquad {} +|s_{n+1}-s_{n}|\|Az_{n}\|+ \|e_{n+1}\|+\|e_{n}\|+|\delta_{n+1}-\delta _{n}|\|Sy_{n}-y_{n}\|. \end{aligned}$$
It follows from conditions (a)-(e) that
$$\limsup_{n\rightarrow\infty}\bigl(\Vert \lambda_{n+1}- \lambda_{n}\Vert -\|x_{n+1}-x_{n}\| \bigr)\leq0. $$
By using Lemma
1.5, we see that
\(\lim_{n\rightarrow\infty}\|\lambda_{n}-x_{n}\|=0\), which in turn implies that
$$ \lim_{n\rightarrow\infty}\|x_{n+1}-x_{n} \|=0. $$
(2.3)
Since
\(T_{r_{n}}\) is firmly nonexpansive, we find that
$$\begin{aligned} \|z_{n}-p\|^{2} &=\|T_{r_{n}}x_{n}-T_{r_{n}}p \|^{2} \\ &\leq\langle x_{n}-p,z_{n}-p\rangle \\ &= \frac{1}{2} \bigl(\|x_{n}-p\|^{2}+ \|z_{n}-p\|^{2}-\|x_{n}-z_{n} \|^{2} \bigr). \end{aligned}$$
That is,
$$\|z_{n}-p\|^{2} \leq\|x_{n}-p\|^{2}- \|x_{n}-z_{n}\|^{2}. $$
It follows that
$$\begin{aligned} \Vert x_{n+1}-p\Vert ^{2} &\leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\beta_{n}\Vert x_{n}-p\Vert ^{2}+\gamma_{n}\Vert S_{n}y_{n}-p\Vert ^{2} \\ &\leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\beta_{n}\Vert x_{n}-p\Vert ^{2}+\gamma_{n}\Vert y_{n}-p\Vert ^{2} \\ &\leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\beta_{n}\Vert x_{n}-p\Vert ^{2}+\gamma_{n}\Vert z_{n}-p\Vert ^{2}+f_{n} \\ &\leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\Vert x_{n}-p\Vert ^{2}- \gamma_{n}\Vert x_{n}-z_{n}\Vert ^{2}+f_{n}, \end{aligned}$$
where
\(f_{n}=\|e_{n}\|(\|e_{n}\|+2\|z_{n}-p\|)\). This further implies that
$$\begin{aligned} \gamma_{n}\|x_{n}-z_{n}\|^{2} &\leq \alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+ \|x_{n}-p\|^{2}-\| x_{n+1}-p\|^{2}+f_{n} \\ &\leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\bigl(\Vert x_{n}-p\Vert +\| x_{n+1}-p\| \bigr)\|x_{n}-x_{n+1}\|+f_{n}. \end{aligned}$$
By using conditions (a) and (b), we find from (
2.3) that
$$ \lim_{n\rightarrow\infty}\|z_{n}-x_{n}\|=0. $$
(2.4)
Since
A is
α-inverse-strongly monotone, we find that
$$\begin{aligned} \Vert y_{n}-p\Vert ^{2} &\leq\bigl\Vert (z_{n}-s_{n}Az_{n})-(p-s_{n}Ap)+e_{n} \bigr\Vert ^{2} \\ &\leq\bigl\Vert (z_{n}-p)-s_{n}(Az_{n}-Ap)\bigr\Vert ^{2}+f_{n} \\ &\leq \Vert x_{n}-p\Vert ^{2}-s_{n}(2 \alpha_{n}-s_{n})\Vert Az_{n}-Ap\Vert ^{2}+f_{n}. \end{aligned}$$
It follows that
$$\begin{aligned} \Vert x_{n+1}-p\Vert ^{2} &\leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\beta_{n}\Vert x_{n}-p\Vert ^{2}+\gamma_{n}\Vert S_{n}y_{n}-p\Vert ^{2} \\ &\leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\beta_{n}\Vert x_{n}-p\Vert ^{2}+\gamma_{n}\Vert y_{n}-p\Vert ^{2} \\ &\leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\Vert x_{n}-p\Vert ^{2}-s_{n}(2 \alpha-s_{n})\gamma_{n}\Vert Az_{n}-Ap\Vert ^{2}+f_{n}. \end{aligned}$$
This yields that
$$s_{n}(2\alpha-s_{n})\gamma_{n} \|Az_{n}-Ap\|^{2} \leq\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\|x_{n}-p \|^{2}-\| x_{n+1}-p\|^{2}+f_{n}. $$
By using (
2.3), we find from conditions (a), (c) and (d) that
$$ \lim_{n\rightarrow\infty}\|Az_{n}-Ap\|=0. $$
(2.5)
Since
\(P_{C}\) is firmly nonexpansive, we find that
$$\begin{aligned} \Vert y_{n}-p\Vert ^{2} \leq&\bigl\langle (I-s_{n}A)z_{n}+e_{n}-(I-s_{n}A)p, y_{n}-p\bigr\rangle \\ =&\frac{1}{2}\bigl\{ \bigl\Vert (I-s_{n}A)z_{n}+e_{n}-(I-s_{n}A)p \bigr\Vert ^{2}+\Vert y_{n}-p\Vert ^{2} \\ &{} -\bigl\Vert (I-s_{n}A)z_{n}+e_{n}-(I-s_{n}A)p-(y_{n}-p) \bigr\Vert ^{2}\bigr\} \\ \leq&\frac{1}{2}\bigl\{ \Vert z_{n}-p\Vert ^{2}+f_{n}+\Vert y_{n}-p\Vert ^{2}- \bigl\Vert z_{n}-y_{n}- \bigl(s_{n}(Az_{n}-Ap)-e_{n} \bigr)\bigr\Vert ^{2}\bigr\} \\ =&\frac{1}{2}\bigl\{ \Vert z_{n}-p\Vert ^{2}+f_{n}+\Vert y_{n}-p\Vert ^{2}- \Vert z_{n}-y_{n}\Vert ^{2} \\ &{} +2\bigl\langle z_{n}-y_{n},s_{n}(Az_{n}-Ap)-e_{n} \bigr\rangle -\bigl\Vert s_{n}(Az_{n}-Ap)-e_{n} \bigr\Vert ^{2}\bigr\} \\ \leq&\frac{1}{2}\bigl\{ \Vert x_{n}-p\Vert ^{2}+f_{n}+\Vert y_{n}-p\Vert ^{2}- \Vert z_{n}-y_{n}\Vert ^{2} \\ &{} +2\Vert z_{n}-y_{n}\Vert \bigl\Vert s_{n}(Az_{n}-Ap)-e_{n}\bigr\Vert -\bigl\Vert s_{n}(Az_{n}-Ap)-e_{n}\bigr\Vert ^{2} \bigr\} , \end{aligned}$$
which yields that
$$\begin{aligned} \|y_{n}-p\|^{2} \leq&\|x_{n}-p \|^{2}+f_{n}-\|z_{n}-y_{n} \|^{2}+2s_{n}\|z_{n}-y_{n}\| \|Az_{n}-Ap\| \\ &{} +2\|z_{n}-y_{n}\|\|e_{n}\|. \end{aligned}$$
It follows that
$$\begin{aligned} \Vert x_{n+1}-p\Vert ^{2} \leq&\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\beta_{n} \Vert x_{n}-p\Vert ^{2}+\gamma_{n}\Vert S_{n}y_{n}-p\Vert ^{2} \\ \leq&\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\beta_{n}\Vert x_{n}-p\Vert ^{2}+\gamma_{n}\Vert y_{n}-p\Vert ^{2} \\ \leq&\alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+\Vert x_{n}-p\Vert ^{2}+f_{n}- \gamma_{n}\Vert z_{n}-y_{n}\Vert ^{2} \\ &{} +2s_{n}\gamma_{n}\Vert z_{n}-y_{n} \Vert \Vert Az_{n}-Ap\Vert +2\Vert z_{n}-y_{n} \Vert \Vert e_{n}\Vert . \end{aligned}$$
This in turn leads to
$$\begin{aligned} \gamma_{n}\|z_{n}-y_{n}\|^{2} \leq& \alpha_{n} \bigl\Vert f(x_{n})-p\bigr\Vert ^{2}+ \|x_{n}-p\|^{2}-\| x_{n+1}-p\|^{2}+f_{n} \\ &{} +2s_{n}\gamma_{n}\|z_{n}-y_{n}\| \|Az_{n}-Ap\|+2\|z_{n}-y_{n}\|\|e_{n}\|. \end{aligned}$$
By use of (
2.3) and (
2.5), we find from restrictions (a), (b), (c) and (e) that
$$ \lim_{n\rightarrow\infty}\|z_{n}-y_{n}\|=0. $$
(2.6)
On the other hand, we have
$$\gamma_{n}\|S_{n}y_{n}-x_{n}\|\leq \|x_{n+1}-x_{n}\|+\alpha_{n}\bigl\Vert x_{n}-f(x_{n})\bigr\Vert . $$
By using conditions (a) and (b), we find from (
2.3) that
$$ \lim_{n\rightarrow\infty}\|x_{n}-S_{n}y_{n} \|=0. $$
(2.7)
Since
\(S_{n}\) is nonexpansive, we find that
$$\begin{aligned} \|S_{n}x_{n}-x_{n}\|&\leq \|S_{n}x_{n}-S_{n}y_{n} \|+\|S_{n}y_{n}-x_{n}\| \\ &\leq\|x_{n}-z_{n}\|+\|z_{n}-y_{n}\|+ \|S_{n}y_{n}-x_{n}\|. \end{aligned}$$
It follows from (
2.4), (
2.6) and (
2.7) that
$$ \lim_{n\rightarrow\infty}\|x_{n}-S_{n}x_{n} \|=0. $$
(2.8)
Notice that
$$\begin{aligned} \|Sx_{n}-x_{n}\| &\leq\bigl\Vert Sx_{n}- \bigl( \delta_{n}x_{n}+(1-\delta_{n})Sx_{n} \bigr)\bigr\Vert +\|S_{n}x_{n}-x_{n}\| \\ &\leq\delta_{n}\|Sx_{n}-x_{n}\|+ \|S_{n}x_{n}-x_{n}\|. \end{aligned}$$
By using condition (c), we find that
$$ \lim_{n\rightarrow\infty}\|x_{n}-Sx_{n} \|=0. $$
(2.9)
Now, we are in a position to show
\(\limsup_{n\rightarrow\infty}\langle f(q)-q,x_{n}-q\rangle\leq0\), where
\(q=P_{\Omega}f(q)\). To show it, we can choose a subsequence
\(\{x_{n_{i}}\}\) of
\(\{x_{n}\}\) such that
$$\limsup_{n\rightarrow\infty}\bigl\langle f(q)-q,x_{n}-q\bigr\rangle =\lim_{i\rightarrow\infty}\bigl\langle f(q)-q,x_{n_{i}}-q\bigr\rangle . $$
Since
\(\{x_{n_{i}}\}\) is bounded, we can choose a subsequence
\(\{x_{n_{i_{j}}}\}\) of
\(\{x_{n_{i}}\}\) which converges weakly to some point
\(\bar{x}\). We may assume, without loss of generality, that
\(\{x_{n_{i}}\}\) converges weakly to
\(\bar{x}\).
Next, we show that
\(\bar{x}\in \operatorname{VI}(C,A)\). Let
T be a maximal monotone mapping defined by
$$Tx= \begin{cases} Ax+N_{C}x, & x\in C, \\ \emptyset, & x\notin C. \end{cases} $$
For any given
\((x,y)\in \operatorname{Graph}(T)\), we have
\(y-Ax\in N_{C}x\). Since
\(y_{n}\in C\), we have
\(\langle x-y_{n}, y-Ax\rangle\geq0\). Since
\(y_{n}=P_{C}(z_{n}-s_{n}Az_{n}+e_{n})\), we see that
\(\langle x-y_{n}, y_{n}-(I-s_{n}A)z_{n}-e_{n}\rangle\geq0 \) and hence
$$\biggl\langle x-y_{n},\frac{y_{n}-z_{n}-e_{n}}{s_{n}}+Az_{n}\biggr\rangle \geq 0. $$
It follows that
$$\begin{aligned} \langle x-y_{n_{i}},y\rangle \geq&\langle x-y_{n_{i}},Ax\rangle \\ \geq&\langle x-y_{n_{i}},Ax\rangle-\biggl\langle x-y_{n_{i}}, \frac {y_{n_{i}}-z_{n_{i}}-e_{n_{i}}}{s_{n_{i}}}+Az_{n_{i}}\biggr\rangle \\ =&\langle x-y_{n_{i}},Ax-Ay_{n_{i}}\rangle+\langle x-y_{n_{i}},Ay_{n_{i}}-Az_{n_{i}}\rangle \\ &{} -\biggl\langle x-y_{n_{i}},\frac{y_{n_{i}}-z_{n_{i}}-e_{n_{i}}}{s_{n_{i}}}\biggr\rangle \\ \geq&\langle x-y_{n_{i}},Ay_{n_{i}}-Az_{n_{i}}\rangle- \biggl\langle x-y_{n_{i}},\frac {y_{n_{i}}-z_{n_{i}}-e_{n_{i}}}{s_{n_{i}}}\biggr\rangle . \end{aligned}$$
Since
A is Lipschitz continuous, we see that
\(\langle x-\bar{x},y\rangle\geq0\). Notice that
T is maximal monotone and hence
\(0\in T\bar{x}\). This shows that
\(\bar{x}\in \operatorname{VI}(C,A)\). By using Lemma
1.2, we find that
\(\bar {x}\in F(S)\). It follows that
\(\limsup_{n\rightarrow\infty}\langle f(q)-q,x_{n}-q\rangle\leq0\),
$$\begin{aligned}& \| x_{n+1}-q\|^{2} \\& \quad \leq\alpha_{n} \bigl\langle f(x_{n})-q,x_{n+1}-q \bigr\rangle +\beta_{n}\|x_{n}-q\|\| x_{n+1}-q\|+ \gamma_{n}\| S_{n}y_{n}-q\|\|x_{n+1}-q\| \\& \quad \leq\alpha_{n} \bigl\langle f(x_{n})-f(q),x_{n+1}-q \bigr\rangle +\alpha_{n} \bigl\langle f(q)-q,x_{n+1}-q\bigr\rangle +\beta_{n}\|x_{n}-\bar{x}\|\|x_{n+1}-q\| \\& \qquad {} +\gamma_{n}\bigl(\Vert x_{n}-q\Vert +e_{n}\bigr)\|x_{n+1}-q\| \\& \quad \leq\frac{\alpha_{n}\beta+\beta_{n}+\gamma_{n}}{2}\bigl(\|x_{n}-q\|^{2}+ \|x_{n+1}-q\| ^{2}\bigr)+\alpha_{n} \bigl\langle f(q)-q,x_{n+1}-q\bigr\rangle +e_{n}\|x_{n+1}-q\|. \end{aligned}$$
It follows that
$$\| x_{n+1}-q\|^{2} \leq \bigl(1-\alpha_{n}(1-\beta) \bigr)\|x_{n}-q\|^{2}+2\alpha_{n} \bigl\langle f(q)-q,x_{n+1}-q\bigr\rangle +2K_{n}e_{n}, $$
where
\(K=\sup_{n\geq1}\{\|x_{n}-q\|\}\). By using Lemma
1.6, we find that
\(\lim_{n\rightarrow\infty}\|x_{n}-q\| =0\). This completes the proof. □