设$E$和$E^*$分别为实Banach空间和它的对偶空间.广义对偶映射$J_q:E\rightarrow 2^{E^*}$定义为

$ {{J}_{q}}(x)=\left\{ {{x}^{*}}\in {{E}^{*}}:\ \left\langle x, {{x}^{*}} \right\rangle ={{\left\| x \right\|}^{q}}, \ \left\| {{x}^{*}} \right\|={{\left\| x \right\|}^{q-1}} \right\}, \forall \ x\in E. $

易知${{J}_{q}}={{\left\| x \right\|}^{q-2}}{{J}_{2}}(x), \forall \ x\ne 0$, 其中$J_2=J$称为正规对偶映射.众所周知, 若$E$光滑, 则$J_q$是单值的, 记为$j_q$.现在回忆文献[1-2]中一些基本概念.

Banach空间$E$称为严格凸的, 若对任意$x, y\in E$, $x\neq y$且$\left\| x \right\|=\left\| y \right\|=1$, 有$\left\| {\frac{{x + y}}{2}{\rm{ }}} \right\| < 1$. $E$中的凸模$\delta_E:[0, 2]\rightarrow[0, 1]$定义为

$ {\delta _E}() = \inf \left\{ {1-\frac{{\left\| {x + y} \right\|}}{2}:\;\left\| x \right\| = 1, \left\| y \right\| = 1且\left\| {x-y} \right\| \ge } \right\}, \;0 \le 2. $

Banach空间$E$称为一致凸的, 若对任意$0 < \epsilon\leq2$, 有$\delta_E(\epsilon)>0$. Banach空间$E$称为$p$ -一致凸的, 若存在常数$c>0$使得$\delta_E(\epsilon)\geq c\epsilon^p$.众所周知, Hilbert空间是$2$ -一致凸的, $L^p$空间是$\max \left\{ p, 2\right\}$-一致凸的, 其中$p>1$.若$1 < p < 2$, 则任何Banach空间都不是$p$ -一致凸的.

Banach空间$E$中的范数称为Gâteaux可微的, 若对$E$中单位球面$S(E)$上的任意点$x, y$, 极限$\mathop {\lim }\limits_{t \to 0} \frac{{\left\| {x + ty} \right\|-\left\| x \right\|}}{t}$存在.此时, $E$也称为光滑的. $E$中范数称为一致Gâteaux可微的, 若任取$S(E)$中元素$y$, 上述极限能够对$S(E)$中$x$一致的取到. $E$中范数称为Fréchet可微的, 若任取$S(E)$中元素$x$, 上述极限能够对$S(E)$中$y$一致的取到. $E$中范数称为一致Fréchet可微的, 若上述极限能够对$S(E)$中$x, y$能一致的取到.

$E$的光滑模$\rho_E:[0, \infty)\rightarrow [0, \infty)$定义为

$ {\rho _E}(t) = \sup \left\{ {\frac{1}{2}(\left\| {x + y} \right\| + \left\| {x-y} \right\|)-1:{\mkern 1mu} x \in S(E), {\mkern 1mu} \left\| y \right\| \le t} \right\}. $

Banach空间$E$称为一致光滑的, 若$\frac{\rho_E(t)}{t}\rightarrow 0$, 其中$t\rightarrow 0$. $E$称为$q$ -一致光滑的, 若存在常数$c>0$使得$\rho_E(t)\leq ct^q$.众所周知, $E$是一致光滑的当且仅当$E$中的范数是一致Fréchet可微的.若$E$是$q$ -一致光滑的, 则$q\leq 2$且$E$是一致光滑的. $L^p$是$\min \left\{ p,2 \right\}$ -一致光滑的, 其中$p>1$. Hilbert空间是2 -一致光滑的.

设$C$和$D$为Banach空间$E$的两个非空子集使得$C$是非空闭凸的且$D\subset C$, 映射$P:C\rightarrow D$称为向阳的^[1], 若当$x+t(x-P(x))\in C$时, 有

$ P(x+t(x-P(x)))=P(x), \quad \forall \, x\in C, \, t\geq0. $

映射$P:C\rightarrow D$称为拉回, 若$Px=x, \ \forall\ x\in D$. $P$称为从$C$到$D$上的向阳非扩张拉回, 若$P$是$C$到$D$上的拉回且是非扩张的. $C$中子集$D$称为$C$的向阳非扩张拉回, 若存在一个从$C$到$D$上的向阳非扩张拉回映射.

命题 1.1^[3] 设$C$为Banach空间$E$的闭凸子集, $D$为$C$的子集.设$P:C\rightarrow D$是拉回映射且$J$为$E$中正规对偶映射.则下面命题等价:

(a) $P$是向阳非扩张的;

(b) ${{\left\| Px-Py \right\|}^{2}}\le \left\langle x-y,J(Px-Py) \right\rangle ,\forall x,y\in C$;

(c) $\left\langle x-Px,J(y-Px) \right\rangle \le 0,\forall \in C,y\in D$.

命题 1.2^[4] 若$E$是严格凸的且是一致光滑的, $T:C\rightarrow C$为非扩张映射且其不动点集为$F(T)$, 则$F(T)$是$C$中向阳非扩张拉回集.

映射$T:C\rightarrow C$称为李普希兹的, 若存在常数$L>0$使得

$ \left\| {Tx-Ty} \right\| \le L\left\| {x-y} \right\|, \forall {\mkern 1mu} {\mkern 1mu} x, y \in C. $

(1.1)

映射$T:C\rightarrow C$称为非扩张的, 若

$ \left\| {Tx - Ty} \right\| \le \left\| {x - y} \right\|,\forall x,y \in C $

(1.2)

映射$A:C\rightarrow E$称为增生的, 若存在$j_q(x-y)\in J_q(x-y)$满足

$ \left\langle {Ax-Ay, {j_q}(x-y)} \right\rangle \ge 0, \forall {\mkern 1mu} {\mkern 1mu} x, y \in C. $

(1.3)

映射$A:C\rightarrow E$称为$\alpha$ -逆强增生的, 若存在$j_q(x-y)\in J_q(x-y)$与$\alpha>0$满足

$ \left\langle {Ax-Ay, {j_q}(x-y)} \right\rangle \ge \alpha {\left\| {Ax-Ay} \right\|^q}, \forall {\mkern 1mu} {\mkern 1mu} x, y \in C. $

(1.4)

映射$T:C\rightarrow C$称为$\lambda$ -严格伪压缩的[5], 若对于任意$x, y\in C$和某个$j_q(x-y)\in J_q(x-y)$, 存在常数$\lambda>0$满足

$ \left\| {Tx-Ty} \right\|, {j_q}(x-y) \le {\left\| {x-y} \right\|^q} - \lambda {\left\| {(I - T)x - (I - T)y} \right\|^q}. $

(1.5)

注 1.1 根据$(1.5)$式, 可以证明, 若$T$是$\lambda$ -严格伪压缩的, 则$T$是李普希兹的, 其中李普希兹常数$L=\frac{1+\lambda^{\frac{1}{q-1}}}{\lambda^{\frac{1}{q-1}}}$.事实上, 由(1.5)式得

$ \lambda {\left\| {(I-T)x-(I-T)y} \right\|^q} \le \left\langle {x - y - (Tx - Ty), {j_q}(x - y)} \right\rangle, $

于是

$ \lambda {\left\| {(I-T)x-(I-T)y} \right\|^q} \le \left\| {x - y - (Tx - Ty)} \right\|{\left\| {x - y} \right\|^{q - 1}}. $

则

$ {\lambda ^{\frac{1}{{q-1}}}}\left\| {x-y-(Tx - Ty)} \right\| \le \left\| {x - y} \right\|. $

故有

$ {\lambda ^{\frac{1}{{q-1}}}}(\left\| {Tx-Ty} \right\|-\left\| {x - y} \right\|) \le {\lambda ^{\frac{1}{{q - 1}}}}\left\| {x - y - (Tx - Ty)} \right\| \le \left\| {x - y} \right\|. $

于是

$ \left\| {Tx-Ty} \right\| \le \frac{{1 + {\lambda ^{\frac{1}{{q-1}}}}}}{{{\lambda ^{\frac{1}{{q-1}}}}}}\left\| {x - y} \right\|. $

设${\Bbb N}$和${\Bbb R}^+$分别为正整数集与正实数集.映射$\varphi:{\Bbb R}^+\rightarrow {\Bbb R}^+$称为$L$ -函数, 若$\varphi(0)=0, $ $\varphi(t)>0, \forall\ t>0$且对于任意$s>0$存在$u>s$满足$\varphi(t)\leq s, \forall\ t\in[s, u]$.

设$(E, d)$为距离空间.映射$f:E\rightarrow E$称为$(\varphi, L)$ -压缩映射, 若$\varphi:{\Bbb R}^+\rightarrow {\Bbb R}^+$为$L$ -函数且有$d(f(x), f(y)) < \varphi(d(x, y)), \forall\ x, y\in E, x\neq y$.映射$f:E\rightarrow E$称为Meir-Keeler型映射, 若对于任意$\epsilon>0$, 存在$\delta=\delta(\epsilon)>0$使得对于任意$x, y\in E$, $\epsilon\leq d(d, y) < \epsilon+\delta$, 有$d(f(x), f(y)) < \epsilon$.

命题 1.3^[6] 设$(E, d)$为距离空间, $f:E\rightarrow E$为映射.则下述论断等价:

(ⅰ) $f$为Meir-Keeler型映射;

(ⅱ) 存在$L$ -函数$\varphi:{\Bbb R}^+\rightarrow {\Bbb R}^+$, 使得$f$为$(\varphi, L)$ -压缩映射.

命题1.4^[7] 设$C$为Banach空间$E$中闭凸子集, $f:C\rightarrow C$为Meir-Keeler型映射.则对任意$\epsilon>0$, 存在$r\in(0, 1)$使得

$ 若\left\| x-y \right\|\ge \epsilon, 则有\text{ }\left\|\text{ }f(x)-f(y) \right\|\le r\left\| x-y \right\|. $

本文中, Meir-Keeler型映射或$(\varphi, L)$ -压缩映射都称为广义压缩映射.假定$L$ -函数是连续的, 严格增的且满足$\lim\limits_{t\rightarrow\infty}\eta(t)=\infty$, 其中$\eta(t)=t-\varphi(t), \forall\ t\in{\Bbb R}^+$.

变分不等式理论成为研究纯理论和应用数学中许多问题的一个重要工具, 已经发展了一些解决变分不等式问题和不动点问题的迭代算法, 见文献[8-15, 27-32].

设$C$为实Hilbert空间$H$中的非空闭凸子集, $A:C\rightarrow H$为非线性映射.经典的变分不等式就是找$x^*$满足

$ \left\langle {A{x^*}, x-{x^*}} \right\rangle \ge 0, \;\forall \;x \in C. $

(1.6)

设$A, B:C\rightarrow H$为两个非线性映射. Ceng^[8]研究如下问题:找$(x^*, y^*)\in C\times C$满足

$ \left\{ {\begin{array}{*{20}{l}} {{\rm{ }}\left\langle {\lambda A{y^*} + {x^*}-{y^*}, x-{x^*}} \right\rangle \ge 0, {\mkern 1mu} {\mkern 1mu} \forall {\mkern 1mu} x \in C, }\\ {{\rm{ }}\left\langle {\mu B{x^*} + {y^*}-{x^*}, x - {y^*}} \right\rangle \ge 0, {\mkern 1mu} {\mkern 1mu} \forall {\mkern 1mu} x \in C, } \end{array}} \right. $

(1.7)

此问题称为一般变分不等式, 其中$\lambda>0$, $\mu>0$为常数.特别地, 若$A=B, \ x^*=y^*$, 则问题(1.7)变为问题(1.6).他们也研究了下面算法

$ \left\{ \begin{array}{l} x_1=u\in C, \\ y_n=P_C(x_n-\mu Bx_n), \\ x_{n+1}=\alpha_nu+\beta_nx_n+\gamma_nTP_C(y_n-\lambda Ay_n), \\ \end{array} \right. $

(1.8)

并证明了由算法(1.8)生成的序列$\left\{ {{x_n}} \right\}$强收敛到问题(1.7)的解集与一个非扩张映射的不动点集的公共元.

设$C$为实Banach空间$E$中非空闭凸子集, $A, B:C\rightarrow E$为非线性映射. Yao^[9]考虑了下面问题:找$(x^*, y^*)\in C\times C$满足

$ \left\{ {\begin{array}{*{20}{l}} {\left\langle {A{y^*} + {x^*}-{y^*}, j(x-{x^*})} \right\rangle \ge 0, {\mkern 1mu} {\mkern 1mu} \forall {\mkern 1mu} x \in C, }\\ {\left\langle {B{x^*} + {y^*}-{x^*}, j(x - {y^*})} \right\rangle \ge 0, {\mkern 1mu} {\mkern 1mu} \forall {\mkern 1mu} x \in C, } \end{array}} \right. $

(1.9)

此问题称为Banach空间中的一般变分不等式.他们也引入下面算法

$ \left\{ \begin{array}{l} x_0, u\in C, \\ y_n=Q_C(x_n- Bx_n), \\ x_{n+1}=\alpha_nu+\beta_nx_n+\gamma_nQ_C(y_n- Ay_n), \end{array} \right. $

(1.10)

并得到了强收敛定理.

本文在$q$-一致Banach空间$E$里考虑下面问题:找$(x^*, y^*)\in C\times C$满足

$ \left\{ {\begin{array}{*{20}{l}} {\;\left\langle {\lambda A{y^*} + {x^*}-{y^*}, {j_q}(x-{x^*})} \right\rangle \ge 0, {\mkern 1mu} {\mkern 1mu} \forall {\mkern 1mu} x \in C, }\\ {\;\left\langle {\mu B{x^*} + {y^*}-{x^*}, {j_q}(x - {y^*})} \right\rangle \ge 0, {\mkern 1mu} {\mkern 1mu} \forall {\mkern 1mu} x \in C, } \end{array}} \right. $

(1.11)

这里$\lambda>0, \ \mu>0$为两个固定常数, $A, B:C\rightarrow E$为两个非线性映射.特别地, 若$\lambda=\mu=1, \ q=2$, 则问题(1.11)成为问题(1.9).若$E$是Hilbert空间, 则问题(1.11)变为问题(1.7).因此问题(1.11)包括(1.7)与(1.9)作为特殊情况.进一步, 在$q$ -一致光滑, 一致凸Banach空间里, 引入关于广义压缩映射的粘性迭代算法, 找到了关于两个逆强增生算子的变分不等式问题解集与无限个严格伪压缩映射的公共不动点集的公共元.

为了证明主要结果, 需要下面引理.

引理 2.1^[16] 设$\left\{ {{a_n}} \right\}$为非负实数列满足$a_{n+1}\leq(1-\alpha_n)a_n+\delta_n, \, \, n\geq0$, 其中$\left\{ {{\alpha _n}} \right\}$为(0, 1)中序列且$\left\{ {{\delta _n}} \right\}$是${\Bbb R}$中序列满足

(ⅰ) $\sum\limits_{n=0}^\infty \alpha_n=\infty$;

(ⅱ) $\limsup\limits_{n\rightarrow\infty}\frac{\delta_n}{\alpha_n}\leq0$或$\sum\limits_{n=0}^{\infty }{\left| {{\delta }_{n}} \right|}\text{ }<\infty$.

则$\lim\limits_{n\rightarrow\infty}a_n=0$.

引理 2.2^[17] 设$\left\{ {{x_n}} \right\}$和$\left\{ {{z_n}} \right\}$为Banach空间$E$中有界序列, $\left\{ {{\beta _n}} \right\}$为[0, 1]中序列满足: $0 < \liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n < 1$.假设$x_{n+1}=\beta_nx_n+(1-\beta_n)z_n$, $n\geq0$且$\underset{n\to \infty }{\mathop{\lim \sup }}\,(\left\| {{z}_{n+1}}-{{z}_{n}} \right\|-\left\| {{x}_{n+1}}-{{x}_{n}} \right\|)\le 0$.则$\mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{z_n} - {x_n}} \right\| = 0$.

引理 2.3^[18] 设$E$为实$q$ -一致光滑Banach空间, 则存在常数$C_q>0$满足

$ {\left\| {x + y} \right\|^q} \le {\left\| x \right\|^q} + q\left\langle {y, {j_q}x} \right\rangle + {C_q}{\left\| y \right\|^q}, \forall \;x, y \in E. $

特别地, 若$E$是2-一致光滑Banach空间, 则存在最佳光滑常数$K>0$满足

$ {\left\| {x + y} \right\|^2} \le {\left\| x \right\|^2} + 2\left\langle {y, jx} \right\rangle + 2{\left\| {Ky} \right\|^2}, \forall \;x, y \in E. $

引理 2.4^{[19, p63]} 设$q>1$.则对于任意正数$a, b$, 下面不等式成立

$ ab\leq\frac{1}{q}a^q+\frac{q-1}{q}b^{\frac{q}{q-1}}. $

引理 2.5^[20] 设$C$为严格凸Banach空间$E$中闭凸子集. $T_1$和$T_2$为$C$中非扩张自映射满足$F(T_1)\cap F(T_2)\neq\emptyset$.定义映射$S$为

$ Sx=\lambda T_1x+(1-\lambda)T_2x, \ \forall\ x\in C, $

其中$\lambda$为$(0, 1)$中常数.则$S$是非扩张的且$F(S)=F(T_1)\cap F(T_2)$.

引理 2.6^[21] 设$E$为实光滑、一致凸Banach空间, $r>0$.则存在严格增、连续的凸函数$g:[0, 2r]\rightarrow{\Bbb R}$使得$g(0)=0$且$g(\left\| x-y \right\|)\le {{\left\| x \right\|}^{2}}-2\left\langle x,jy \right\rangle +{{\left\| y \right\|}^{2}},\forall \ x,y\in {{B}_{r}}=\left\{ z\in E:\left\| x \right\|\le r \right\}$.

引理 2.7^[22] 设$C$为实$q$-一致光滑Banach空间$E$中非空闭凸子集, $T:C\rightarrow C$为$\lambda$ -严格伪压缩映射.给定$\alpha\in (0, 1)$, 定义$T_\alpha x=(1-\alpha)x+\alpha Tx$.则当$\alpha\in (0, \mu]$, $\mu =\min \left\{ 1,{{\left\{ \frac{q\lambda }{{{C}_{q}}} \right\}}^{\frac{1}{q-1}}} \right\}$, 有$T_\alpha:C\rightarrow C$是非扩张映射且$F(T_\alpha)=F(T)$.

引理 2.8^[23] 在Banach空间$E$中, 下面不等式成立

$ {\left\| {x + y} \right\|^2} \le {\left\| x \right\|^2} + 2\left\langle {y, j(x + y)} \right\rangle, \forall {\mkern 1mu} x, y \in E, $

其中$j(x+y)\in J(x+y)$.

引理 2.9^{[24, 引理3.1]} 设$C$为Banach空间$E$中非空子集, $\left\{ {{T_n}} \right\}$为一族从$C$到$E$的映射.假设对于$C$中任意有界集$D$, 存在从${\Bbb R}^{+}$到${\Bbb R}^{+}$上的连续的单调增函数$h_D$使得$h_D(0)=0$且$\lim\limits_{k, l\rightarrow\infty}\rho_l^k=0$, 其中$\rho _{l}^{k}:=\sup \left\{ {{h}_{D}}(\left\| {{T}_{k}}z-{{T}_{l}}z) \right\|:z\in D \right\} < \infty, \forall \ k, l\in \mathbb{N}$.则$\underset{n\to \infty }{\mathop{\lim \sup }}\, \left\{ {{h}_{D}}\left\| (Tz-{{T}_{n}}z) \right\|:z\in D \right\}=0$.

注 2.1^{[24, 注3.2]} 假设$\sum\limits_{n=1}^{\infty }{\sup }\left\{ \left\| {{T}_{n+1}}z-{{T}_{n}}z \right\|:z\in D \right\}<\infty$且$h_D:{\Bbb R}^{+}\rightarrow {\Bbb R}^{+}$是连续的, 增函数满足$h_D(0)=0$, 则$\mathop {\lim \sup }\limits_{k, l \to \infty } {\rm{ }}\left\{ {{h_D}(\left\| {{T_k}z-{T_l}z} \right\|)} \right\}:z \in D=0$.

引理 2.10^[25] 设$C$为一致光滑Banach空间$E$中非空闭凸子集. $T:C\rightarrow C$为非扩张映射满足$F(T)\neq\emptyset$且$f:C\rightarrow C$为广义压缩映射.则由$x_t=tf(x_t)+(1-t)Tx_t, t\in (0, 1)$定义的序列$\left\{ {{x}_{t}} \right\}$强收敛到$\hat{x}\in F(T)$, 且解决了下面变分不等式

$ \left\langle {f(\hat x)-\hat x, j(z-\hat x)} \right\rangle \le 0, \forall \;z \in F(T). $

引理 2.11^[25] 设$C$为一致光滑Banach空间$E$中非空闭凸子集. $T:C\rightarrow C$为非扩张映射满足$F(T)\neq\emptyset$, $f:C\rightarrow C$为广义压缩映射.假设当$t\rightarrow0$时, 由$x_t=tf(x_t)+(1-t)Tx_t, t\in (0, 1)$定义的序列$\left\{ {{x}_{t}} \right\}$强收敛到$\hat{x}\in F(T)$且$\left\{ {{x}_{n}} \right\}$为有界序列满足当$n\rightarrow\infty$时, 有$x_n-Tx_n\rightarrow0$.则

$ \mathop {\lim \sup }\limits_{n \to \infty } \left\langle {f(\hat x)-\hat x, j({x_n}-\hat x)} \right\rangle \le 0. $

引理 2.12^[26] 设$C$为实$q$-一致光滑Banach空间$E$中非空闭凸子集. $A:C\rightarrow E$为$\alpha$-逆强增生算子.则下面不等式成立

$ {\left\| {(I-\lambda A)x-(I-\lambda A)y} \right\|^q} \le {\left\| {x - y} \right\|^q} - \lambda (q\alpha - {C_q}{\lambda ^{q - 1}}){\left\| {Ax - Ay} \right\|^q}. $

特别地, 若$0 < \lambda\leq(\frac{q\alpha}{C_q})^{\frac{1}{q-1}}$, 则$I-\lambda A$是非扩张的.

引理 2.13^[26] 设$C$为实$q$-一致光滑Banach空间$E$中非空闭凸子集. $P_C$为$E$到$C$上向阳非扩张拉回. $A:C\rightarrow E$为$\alpha$ -逆强增生的且$B:C\rightarrow E$为$\beta$ -逆强增生的.映射$G:C\rightarrow C$定义为

$ G(x) = {P_C}\left[{{P_C}(x-\mu Bx)-\lambda A{P_C}(x-\mu Bx)} \right], {\mkern 1mu} {\mkern 1mu} \forall {\mkern 1mu} x \in C. $

若$0 < \lambda\leq(\frac{q\alpha}{C_q})^{\frac{1}{q-1}}$且$0 < \mu\leq(\frac{q\beta}{C_q})^{\frac{1}{q-1}}$, 则$G:C\rightarrow C$是非扩张的.

引理 2.14^[26] 设$C$为实$q$-一致光滑Banach空间$E$中非空闭凸子集. $P_C$为$E$到$C$上向阳非扩张拉回. $A, B:C\rightarrow E$为非线性算子.给定$x^*, y^*\in C$, 则$(x^*, y^*)$为问题(1.11)的解当且仅当$x^*=P_C(y^*-\lambda Ay^*)$, 其中$y^*=P_C(x^*-\mu Bx^*)$.

引理 2.15 设$C$为实$q$-一致Banach空间$E$中非空闭凸子集. $S:C\rightarrow C$为非扩张映射且$T:C\rightarrow C$为$\delta$ -严格伪压缩映射满足$F(S)\cap F(T)\neq\emptyset$.定义映射$Wx=[(1-\alpha)I+\alpha T]Sx, \forall\ x\in C$, 其中$\alpha \in (0,\mu ),\mu =\min {{\left\{ 1,\left\{ \frac{q\delta }{{{C}_{q}}} \right\} \right\}}^{\frac{1}{q-1}}}$.则$F(W)=F(S)\cap F(T)$.

证 首先证明$F(S)\cap F(T)\subseteq F(W)$.事实上, 任取$x\in F(S)\cap F(T)$, 有

$ [(1-\alpha)I+\alpha T]Sx=[(1-\alpha)I+\alpha T]x=(1-\alpha)x+\alpha x=x, $

故$x\in F(W)$.因此$F(S)\cap F(T)\subseteq F(W)$.下面证明$F(W)\subseteq F(S)\cap F(T)$.任给$x\in F(W)$且$y\in F(S)\cap F(T)$, 由引理2.3得

$ {{\left\| x-y \right\|}^{q}}\\ ={{\left\| [(1-\alpha )I+\alpha T]Sx-y \right\|}^{q}}\\ ={{\left\| Sx-y+\alpha (TSx-Sx) \right\|}^{q}}\\ \le {{\left\| Sx-y \right\|}^{q}}+q\alpha \left\langle TSx-Sx,{{j}_{q}}(Sx-y) \right\rangle +{{C}_{q}}{{\alpha }^{q}}{{\left\| TSx-Sx \right\|}^{q}}\\ ={{\left\| Sx-y \right\|}^{q}}+q\alpha \left\langle TSx-y,{{j}_{q}}(Sx-y) \right\rangle +q\alpha \left\langle y-Sx,{{j}_{q}}(Sx-y) \right\rangle +{{C}_{q}}{{\alpha }^{q}}{{\left\| TSx-Sx \right\|}^{q}}\\ \le {{\left\| x-y \right\|}^{q}}+q\alpha ({{\left\| Sx-y \right\|}^{q}}-\delta {{\left\| TSx-Sx \right\|}^{q}})-q\alpha {{\left\| Sx-y \right\|}^{q}}+{{C}_{q}}{{\alpha }^{q}}{{\left\| TSx-Sx \right\|}^{q}}\\ ={{\left\| x-y \right\|}^{q}}-(q\alpha \delta -{{C}_{q}}{{\alpha }^{q}}){{\left\| TSx-Sx \right\|}^{q}}, $

从而

$ (q\alpha \delta-{C_q}{\alpha ^q}){\left\| {TSx-Sx} \right\|^q} \le 0. $

因此

$ TSx = Sx. $

(2.1)

于是

$ x=[(1-\alpha)I+\alpha T]Sx=(1-\alpha)Sx+\alpha Sx=Sx. $

(2.2)

故$x\in F(S)$.根据$(2.1)$和$(2.2)$式, 有$x=Sx=TSx=Tx$, 因此$x\in F(T)$.故$x\in F(S)\cap F(T)$.于是$F(W)\subseteq F(S)\cap F(T)$.证毕.

定理 3.1 设$C$为实$q$-一致光滑Banach空间$E$中非空闭凸子集. $Q_C$为$E$到$C$上向阳非扩张拉回.映射$A, B:C\rightarrow E$分别为$\alpha$-逆强增生的和$\beta$-逆强增生的. $f:C\rightarrow C$为广义压缩映射. $W:C\rightarrow C$为非扩张映射且$\left\{ {{S}_{i}}:C\to C \right\}_{i=0}^{\infty }$为一族$\lambda_i$ -严格伪压缩映射满足$F:=\bigcap\limits_{i=0}^{\infty} F(S_i)\cap F(W)\cap F(G)\neq\emptyset$, 其中$G$的定义见引理2.13.假设$\lambda=\inf \left\{ {{\lambda }_{i}}:i=0, 1, 2, \cdots \right\}>0$.给定$x_0\in C$, 序列$\left\{ {{x}_{n}} \right\}$定义为

$ \begin{equation} \left\{ \begin{array}{l} y_n=Q_C(x_n-\mu Bx_n), \\ z_n=Q_C(y_n-\lambda Ay_n), \\ x_{n+1}=\alpha_nf(x_n)+\beta_nx_n+\gamma_nT_nWz_n, \end{array} \right. \end{equation} $

(3.1)

这里$0 < \lambda < (\frac{q\alpha}{C_q})^{\frac{1}{q-1}}$, $0 < \mu < (\frac{q\beta}{C_q})^{\frac{1}{q-1}}$, $T_n=(1-\delta_n)I+\delta_n S_n$.假设$\left\{ {{\alpha }_{n}} \right\},\left\{ {{\beta }_{n}} \right\}$和$\left\{ {{\gamma }_{n}} \right\}$为[0, 1]中序列满足

(ⅰ) $\alpha_n+\beta_n+\gamma_n=1$;

(ⅱ) $\lim\limits_{n\rightarrow\infty}\alpha_n=0, \sum\limits_{n=1}^\infty\alpha_n=\infty$;

(ⅲ) $0 < \liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n < 1$;

(ⅳ) $\delta_n\in(0, \rho), \delta_n\rightarrow\delta\in(0, \rho)$, 其中$\rho =\min {{\left\{ 1,\left\{ \frac{q\lambda }{{{C}_{q}}} \right\} \right\}}^{\frac{1}{q-1}}}$.

假设对于$C$中任意有界集$D$, 存在单调增、连续的凸函数$h_D:{\Bbb R}^+\rightarrow {\Bbb R}^+$满足$h_D(0)$且$\underset{k,l\to \infty }{\mathop{\lim \sup }}\,\left\{ {{h}_{D}}(\left\| {{S}_{k}}z-{{S}_{l}}z \right\|) \right\}:z\in D=0$.定义$Sx=\lim\limits_{n\rightarrow\infty}S_nx, \forall\ x\in C$且假设$F(S)=\bigcap\limits_{i=0}^\infty F(S_i)$.则$\left\{ {{x}_{n}} \right\}$强收敛到$z\in F$且解决下面变分不等式

$ \left\langle {z-f(z), {j_q}(z-p)} \right\rangle \le 0, \forall {\mkern 1mu} p \in F. $

证 首先证明$\left\{ {{x}_{n}} \right\}$有界.事实上, 任取$x^*\in F$和$y^*=Q_C(x^*-\mu Bx^*)$, 由引理2.13得

$ \left\| {{z_n}-{x^*}} \right\| = \left\| {G{x_n}-G{x^*}} \right\| \le \left\| {{x_n}-p} \right\|. $

(3.2)

根据引理2.7, 有$T_n$是非扩张的且$T_nx^*=x^*$.由(3.1)式得

$ \left\|{x_{n+1}-x^*}\right\|=\left\|{\alpha_n(f(x_n)-x^*)+\beta_n(x_n-x^*)+\gamma_n(T_nWz_n-x^*)}\right\|\\ \leq\alpha_n\left\|{f(x_n)-x^*}\right\|+\beta_n\left\|{x_n-x^*}\right\|+\gamma_n\left\|{T_nWz_n-x^*}\right\|\\ \leq\alpha_n\left\|{f(x_n)-f(x^*)}\right\|+\alpha_n\left\|{f(x^*)-x^*}\right\|+\beta_n\left\|{x_n-x^*}\right\|+\gamma_n\left\|{x_n-x^*}\right\|\\ \leq\alpha_n\varphi(\left\|{x_n-x^*}\right\|)+(1-\alpha_n)\left\|{x_n-x^*}\right\|+\alpha_n\left\|{f(x^*)-x^*}\right\|\\ =(1-\alpha_n\eta)\left\|{x_n-x^*}\right\|+\alpha_n\left\|{f(x^*)-x^*}\right\|\\ \leq\max\left\{ {\left\| {{x_n} - {x^*}} \right\|, {\eta ^{ - 1}}(\left\| {f({x^*}) - {x^*}} \right\|)} \right\}. $

由归纳法得

$ \left\| {{x_n}-{x^*}} \right\| \le \max \left\{ {\left\| {{x_0}-{x^*}} \right\|, {\eta ^{-1}}(\left\| {f({x^*}) - {x^*}} \right\|)} \right\}, \forall \;n \ge 1, $

从而$\left\{ {{x}_{n}} \right\}$有界.同样地, $\left\{ {{z}_{n}} \right\}$和$\left\{ W{{z}_{n}} \right\}$也有界.不失一般性, 可以假设存在$C$中的有界集$B'$包含序列$\left\{ {{x}_{n}} \right\},\left\{ {{z}_{n}} \right\}$和$\left\{ W{{z}_{n}} \right\}$.

下面证明$\mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{x_{n + 1}} - {x_n}} \right\| = 0$.

因为

$ \left\| {{z_{n + 1}}-{z_n}} \right\| = \left\| {G{x_{n + 1}}-G{x_n}} \right\| \le \left\| {{x_{n + 1}}-{x_n}} \right\|. $

(3.3)

故

$ \left\|{T_{n+1}Wz_{n+1}-T_nWz_n}\right\|\\ \leq\left\|{T_{n+1}Wz_{n+1}-T_{n+1}Wz_{n}}\right\|+\left\|{T_{n+1}Wz_{n}-T_nWz_n}\right\|\\ \leq\left\|{z_{n+1}-z_n}\right\|+\left\|{(1-\delta_{n+1})Wz_n+\delta_{n+1}S_{n+1}Wz_n-(1-\delta_n)Wz_n-\delta_n S_nWz_n}\right\|\\ \leq\left\|{x_{n+1}-x_n}\right\|+\delta_n\left\|{S_{n+1}Wz_n-S_nWz_n}\right\|+{\rm{|}}{\delta _{n + 1}} - {\delta _n}|\left\|{S_{n+1}Wz_n-Wz_n}\right\|. $

(3.4)

定义$x_{n+1}=\beta_nx_n+(1-\beta_n)l_n, \forall\ n\geq0$.

于是

$ l_{n+1}-l_n=\frac{x_{n+2}-\beta_{n+1}x_{n+1}}{1-\beta_{n+1}}-\frac{x_{n+1}-\beta_nx_n}{1-\beta_n} \\ =\frac{\alpha_{n+1}f(x_{n+1})+\gamma_{n+1}T_{n+1}Wz_{n+1}}{1-\beta_{n+1}}-\frac{\alpha_nf(x_n)+\gamma_nT_nWz_n}{1-\beta_n} \\ =\frac{\alpha_{n+1}(f(x_{n+1})-T_{n+1}Wz_{n+1})}{1-\beta_{n+1}} -\frac{\alpha_n(f(x_n)-T_nWz_n)}{1-\beta_n}\\ +T_{n+1}Wz_{n+1}-T_nWz_n. $

(3.5)

结合(3.4)与(3.5)式, 有

$ \left\| {{l}_{n+1}}-{{l}_{n}} \right\|\le \frac{{{\alpha }_{n+1}}}{1-{{\beta }_{n+1}}}\left\| f({{x}_{n+1}})-{{T}_{n+1}}W{{z}_{n+1}} \right\|+\frac{{{\alpha }_{n}}}{1-{{\beta }_{n}}}\left\| f({{x}_{n}})-{{T}_{n}}W{{z}_{n}} \right\|\\ +\left\| {{T}_{n+1}}W{{z}_{n+1}}-{{T}_{n}}W{{z}_{n}} \right\|\\ \le \frac{{{\alpha }_{n+1}}}{1-{{\beta }_{n+1}}}\left\| f({{x}_{n+1}})-{{T}_{n+1}}W{{z}_{n+1}} \right\|+\frac{{{\alpha }_{n}}}{1-{{\beta }_{n}}}\left\| f({{x}_{n}})-{{T}_{n}}W{{z}_{n}} \right\|\\ +\left\| {{x}_{n+1}}-{{x}_{n}} \right\|+{{\delta }_{n}}\left\| {{S}_{n+1}}W{{z}_{n}}-{{S}_{n}}W{{z}_{n}} \right\|\\ +\left| {{\delta }_{n+1}}-{{\delta }_{n}} \right|\left\| {{S}_{n+1}}W{{z}_{n}}-W{{z}_{n}} \right\|, $

从而

$ \left\| {{l}_{n+1}}-{{l}_{n}} \right\|-\left\| {{x}_{n+1}}-{{x}_{n}} \right\|\\ \le \frac{{{\alpha }_{n+1}}}{1-{{\beta }_{n+1}}}\left\| f({{x}_{n+1}})-{{T}_{n+1}}W{{z}_{n+1}} \right\|+\frac{{{\alpha }_{n}}}{1-{{\beta }_{n}}}\left\| f({{x}_{n}})-{{T}_{n}}W{{z}_{n}} \right\|\\ +{{\delta }_{n}}\left\| {{S}_{n+1}}W{{z}_{n}}-{{S}_{n}}W{{z}_{n}} \right\|+\left| {{\delta }_{n+1}}-{{\delta }_{n}} \right|\left\| {{S}_{n+1}}W{{z}_{n}}-W{{z}_{n}} \right\|. $

(3.6)

注意假设$\mathop {\lim }\limits_{k,l \to \infty } {\mkern 1mu} \sup \left\{ {{h_{B'}}(\left\| {{S_k}z - {S_l}z} \right\|)} \right\}:z \in B' = 0$.故当$n\rightarrow\infty$时

$ {{h}_{{{B}'}}}(\left\| {{S}_{n+1}}W{{z}_{n}}-{{S}_{n}}W{{z}_{n}} \right\|)\le \sup \left\{ {{h}_{{{B}'}}}(\left\| {{S}_{n+1}}z-{{S}_{n}}z \right\|):z\in {B}' \right\}\\\to 0. $

根据$h_{B'}$的性质得

$ \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{S_{n + 1}}W{z_n} - {S_n}W{z_n}} \right\| = 0. $

(3.7)

注意条件(ⅱ), (ⅳ)与(3.7)式, 根据(3.6)式得

$ \underset{n\to \infty }{\mathop{\lim \sup }}\, (\left\| {{l}_{n+1}}-{{l}_{n}} \right\|-\left\| {{x}_{n+1}}-{{x}_{n}} \right\|)\le 0. $

由引理2.2得$\mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{l_n} - {x_n}} \right\| = 0$.因此

$ \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{x_{n + 1}} - {x_n}} \right\| = \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} (1 - {\beta _n})\left\| {{l_n} - {x_n}} \right\| = 0. $

(3.8)

下面证明$\mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{x_n} - {z_n}} \right\| = 0$与$\underset{n\to \infty }{\mathop{\lim }}\,\left\| {{x}_{n}}-{{T}_{n}}W{{z}_{n}} \right\|=0$.

由引理2.12得

$ {{\left\| {{y}_{n}}-{{y}^{*}} \right\|}^{q}}={{\left\| {{Q}_{C}}({{x}_{n}}-\mu B{{x}_{n}})-{{Q}_{C}}({{x}^{*}}-\mu B{{x}^{*}}) \right\|}^{q}}\\ \le {{\left\| (I-\mu B){{x}_{n}}-(I-\mu B){{x}^{*}} \right\|}^{q}}\\ \le {{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}-\mu (q\beta-{{C}_{q}}{{\mu }^{q-1}}){{\left\| B{{x}_{n}}-B{{x}^{*}} \right\|}^{q}} $

(3.9)

与

$ {{\left\| {{z}_{n}}-{{x}^{*}} \right\|}^{q}}={{\left\| {{Q}_{C}}({{y}_{n}}-\lambda A{{y}_{n}})-{{Q}_{C}}({{y}^{*}}-\lambda A{{y}^{*}}) \right\|}^{q}}\\ \le {{\left\| (I-\lambda A){{y}_{n}}-(I-\lambda A){{y}^{*}} \right\|}^{q}}\\ \le {{\left\| {{y}_{n}}-{{y}^{*}} \right\|}^{q}}-\lambda (q\alpha-{{C}_{q}}{{\lambda }^{q-1}}){{\left\| A{{y}_{n}}-A{{y}^{*}} \right\|}^{q}}. $

(3.10)

将(3.9)式代入(3.10)式, 有

$ {{\left\| {{z}_{n}}-{{x}^{*}} \right\|}^{q}}\le \text{ }{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}-\mu (q\beta-{{C}_{q}}{{\mu }^{q-1}}){{\left\| B{{x}_{n}}-B{{x}^{*}} \right\|}^{q}}\\ -\lambda (q\alpha-{{C}_{q}}{{\lambda }^{q-1}}){{\left\| A{{y}_{n}}-A{{y}^{*}} \right\|}^{q}}. $

(3.11)

根据${\left\| \cdot \right\|^q}$的凸性得

$ {{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}={{\left\| {{\alpha }_{n}}(f({{x}_{n}})-{{x}^{*}})+{{\beta }_{n}}({{x}_{n}}-{{x}^{*}})+{{\gamma }_{n}}({{T}_{n}}W{{z}_{n}}-{{x}^{*}}) \right\|}^{q}}\\ \le {{\alpha }_{n}}{{\left\| f({{x}_{n}})-{{x}^{*}} \right\|}^{q}}+{{\beta }_{n}}{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}+{{\gamma }_{n}}{{\left\| {{T}_{n}}W{{z}_{n}}-{{x}^{*}} \right\|}^{q}}\\ \le {{\alpha }_{n}}{{\left\| f({{x}_{n}})-{{x}^{*}} \right\|}^{q}}+{{\beta }_{n}}{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}+{{\gamma }_{n}}{{\left\| {{z}_{n}}-{{x}^{*}} \right\|}^{q}}. $

(3.12)

结合(3.11)与(3.12)式, 有

$ {{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}\\ \le {{\alpha }_{n}}{{\left\| f({{x}_{n}})-{{x}^{*}} \right\|}^{q}}+{{\beta }_{n}}{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}\\ +{{\gamma }_{n}}[{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}-\mu (q\beta-{{C}_{q}}{{\mu }^{q-1}}){{\left\| B{{x}_{n}}-B{{x}^{*}} \right\|}^{q}}-\lambda (q\alpha-{{C}_{q}}{{\lambda }^{q-1}}){{\left\| A{{y}_{n}}-A{{y}^{*}} \right\|}^{q}}]\\ \le {{\alpha }_{n}}{{\left\| f({{x}_{n}})-{{x}^{*}} \right\|}^{q}}+{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}-{{\gamma }_{n}}\mu (q\beta -{{C}_{q}}{{\mu }^{q-1}}){{\left\| B{{x}_{n}}-B{{x}^{*}} \right\|}^{q}}\\ -{{\gamma }_{n}}\lambda (q\alpha -{{C}_{q}}{{\lambda }^{q-1}}){{\left\| A{{y}_{n}}-A{{y}^{*}} \right\|}^{q}}. $

(3.13)

注意到$x^r-y^r\leq rx^{r-1}(x-y), \ \forall\ r\geq 1$.于是根据(3.13)式得

$ {{\gamma }_{n}}\mu (q\beta-{{C}_{q}}{{\mu }^{q-1}}){{\left\| B{{x}_{n}}-B{{x}^{*}} \right\|}^{q}}+{{\gamma }_{n}}\lambda (q\alpha-{{C}_{q}}{{\lambda }^{q-1}}){{\left\| A{{y}_{n}}-A{{y}^{*}} \right\|}^{q}}\\ \le {{\alpha }_{n}}{{\left\| f({{x}_{n}})-{{x}^{*}} \right\|}^{q}}+{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q}}-{{\left\| {{x}_{n+1}}-{{x}^{*}} \right\|}^{q}}\\ \le {{\alpha }_{n}}{{\left\| f({{x}_{n}})-{{x}^{*}} \right\|}^{q}}+q{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q-1}}(\left\| {{x}_{n}}-{{x}^{*}} \right\|-\left\| {{x}_{n+1}}-{{x}^{*}} \right\|)\\ \le {{\alpha }_{n}}{{\left\| f({{x}_{n}})-{{x}^{*}} \right\|}^{q}}+q{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{q-1}}\left\| {{x}_{n}}-{{x}_{n+1}} \right\|. $

因为$0 < \lambda < (\frac{q\alpha}{C_q})^{\frac{1}{q-1}}$, $0 < \mu < (\frac{q\beta}{C_q})^{\frac{1}{q-1}}$, $\lim\limits_{n\rightarrow\infty}\alpha_n=0$, $\limsup\limits_{n\rightarrow\infty}\beta_n < 1$和(3.8)式, 有

$ \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {A{y_n} - A{y^*}} \right\| = \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {B{x_n} - B{x^*}} \right\| = 0. $

(3.14)

令${r_1}=\mathop {\sup }\limits_{n \ge 0} {\mkern 1mu} \left\{ {\left\| {{y_n} -{y^*}} \right\|, \left\| {{z_n} -{x^*}} \right\|} \right\}$.根据命题1.1与引理2.6得

$ {{\left\| {{z}_{n}}-{{x}^{*}} \right\|}^{2}}={{\left\| {{P}_{C}}({{y}_{n}}-\lambda A{{y}_{n}})-{{P}_{C}}({{y}^{*}}-\lambda A{{y}^{*}}) \right\|}^{2}}\\ \le \left\langle {{y}_{n}}-\lambda A{{y}_{n}}-({{y}^{*}}-\lambda A{{y}^{*}}), j({{z}_{n}}-{{x}^{*}}) \right\rangle\\ =\left\langle {{y}_{n}}-{{y}^{*}}, j({{z}_{n}}-{{x}^{*}}) \right\rangle-\lambda \left\langle A{{y}_{n}}-A{{y}^{*}}, j({{z}_{n}}-{{x}^{*}}) \right\rangle\\ \le \frac{1}{2}[{{\left\| {{y}_{n}}-{{y}^{*}} \right\|}^{2}}+{{\left\| {{z}_{n}}-{{x}^{*}} \right\|}^{2}}-{{g}_{1}}(\left\| {{y}_{n}}-{{z}_{n}}+{{x}^{*}}-{{y}^{*}} \right\|)]\\ +\lambda \left\langle A{{y}^{*}}-A{{y}_{n}}, j({{z}_{n}}-{{x}^{*}}) \right\rangle, $

从而

$ {{\left\| {{z}_{n}}-{{x}^{*}} \right\|}^{2}}\le {{\left\| {{y}_{n}}-{{y}^{*}} \right\|}^{2}}-{{g}_{1}}(\left\| {{y}_{n}}-{{z}_{n}}+{{x}^{*}}-{{y}^{*}} \right\|)+2\lambda \left\langle A{{y}^{*}}-A{{y}_{n}}, j({{z}_{n}}-{{x}^{*}}) \right\rangle\\ \le {{\left\| {{y}_{n}}-{{y}^{*}} \right\|}^{2}}-{{g}_{1}}(\left\| {{y}_{n}}-{{z}_{n}}+{{x}^{*}}-{{y}^{*}} \right\|)+2\lambda \left\| A{{y}_{n}}-A{{y}^{*}} \right\|\left\| {{z}_{n}}-{{x}^{*}} \right\|. $

(3.15)

令${r_2} = \mathop {\sup }\limits_{n \ge 0} {\mkern 1mu} \left\{ {\left\| {{x_n} - {x^*}} \right\|,\left\| {{y_n} - {y^*}} \right\|} \right\}$.再次根据命题1.1与引理2.6得

$ {{\left\| {{y}_{n}}-{{y}^{*}} \right\|}^{2}}={{\left\| {{P}_{C}}({{x}_{n}}-\mu B{{x}_{n}})-{{P}_{C}}({{x}^{*}}-\mu B{{x}^{*}}) \right\|}^{2}}\\ \le \left\langle {{x}_{n}}-\mu B{{x}_{n}}-({{x}^{*}}-\mu B{{x}^{*}}), j({{y}_{n}}-{{y}^{*}}) \right\rangle\\ =\left\langle {{x}_{n}}-{{x}^{*}}, j({{y}_{n}}-{{y}^{*}}) \right\rangle-\mu \left\langle B{{x}_{n}}-B{{x}^{*}}, j({{y}_{n}}-{{y}^{*}}) \right\rangle\\ \le \frac{1}{2}[{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}+{{\left\| {{y}_{n}}-{{y}^{*}} \right\|}^{2}}-{{g}_{2}}(\left\| {{x}_{n}}-{{y}_{n}}-({{x}^{*}}-{{y}^{*}}) \right\|)]\\ +\mu \left\langle B{{x}^{*}}-B{{x}_{n}}, j({{y}_{n}}-{{y}^{*}}) \right\rangle, $

从而

$ {{\left\| {{y}_{n}}-{{y}^{*}} \right\|}^{2}}\le {{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}-{{g}_{2}}(\left\| {{x}_{n}}-{{y}_{n}}-({{x}^{*}}-{{y}^{*}}) \right\|)+2\mu \left\langle B{{x}^{*}}-B{{x}_{n}}, j({{y}_{n}}-{{y}^{*}}) \right\rangle\\ \le {{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}-{{g}_{2}}(\left\| {{x}_{n}}-{{y}_{n}}-({{x}^{*}}-{{y}^{*}}) \right\|)+2\mu \left\| B{{x}_{n}}-B{{x}^{*}} \right\|\left\| {{y}_{n}}-{{y}^{*}} \right\|. $

(3.16)

将(3.16)代入(3.15)式得

$ {{\left\| {{z}_{n}}-{{x}^{*}} \right\|}^{2}}\le {{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}-{{g}_{2}}(\left\| {{x}_{n}}-{{y}_{n}}-({{x}^{*}}-{{y}^{*}}) \right\|)-{{g}_{1}}(\left\| {{y}_{n}}-{{z}_{n}}+{{x}^{*}}-{{y}^{*}} \right\|)\\ +2\mu \left\| B{{x}_{n}}-B{{x}^{*}} \right\|\left\| {{y}_{n}}-{{y}^{*}} \right\|+2\lambda {{\left\| A{{y}_{n}}-Ay \right\|}^{*}}\left\| {{z}_{n}}-{{x}^{*}} \right\|. $

(3.17)

根据${\left\| \cdot \right\|^2}$的凸性和引理2.8得

$ {{\left\| {{x}_{n+1}}-{{x}^{*}} \right\|}^{2}}\\ ={{\left\| {{\beta }_{n}}({{x}_{n}}-{{x}^{*}})+(1-{{\beta }_{n}})({{T}_{n}}W{{z}_{n}}-{{x}^{*}})+{{\alpha }_{n}}(f({{x}_{n}})-{{T}_{n}}W{{z}_{n}}) \right\|}^{2}}\\ \le {{\left\| {{\beta }_{n}}({{x}_{n}}-{{x}^{*}})+(1-{{\beta }_{n}})({{T}_{n}}W{{z}_{n}}-{{x}^{*}}) \right\|}^{2}}+2{{\alpha }_{n}}\left\langle f({{x}_{n}})-{{T}_{n}}W{{z}_{n}}, j({{x}_{n+1}}-{{x}^{*}}) \right\rangle \\ \le {{\beta }_{n}}{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}+(1-{{\beta }_{n}}){{\left\| {{T}_{n}}W{{z}_{n}}-{{x}^{*}} \right\|}^{2}}+2{{\alpha }_{n}}\left\| f({{x}_{n}})-{{T}_{n}}W{{z}_{n}} \right\|\left\| {{x}_{n+1}}-{{x}^{*}} \right\|\\ \le {{\beta }_{n}}{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}+(1-{{\beta }_{n}}){{\left\| {{z}_{n}}-{{x}^{*}} \right\|}^{2}}+2{{\alpha }_{n}}{{M}_{1}}, $

(3.18)

其中${M_1} = \mathop {\sup }\limits_{n \ge 0} {\mkern 1mu} \left\| {f({x_n}) - {T_n}W{z_n}} \right\|\left\| {{x_{n + 1}} - {x^*}} \right\|$.

由(3.17)和(3.18)式得

$ {{\left\| {{x}_{n+1}}-{{x}^{*}} \right\|}^{2}}\\ \le {{\beta }_{n}}{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}+(1-{{\beta }_{n}})[{{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}-{{g}_{2}}(\left\| {{x}_{n}}-{{y}_{n}}-({{x}^{*}}-{{y}^{*}}) \right\|)\\ \quad-{{g}_{1}}(\left\| {{y}_{n}}-{{z}_{n}}+{{x}^{*}}-{{y}^{*}} \right\|)+2\mu \left\| B{{x}_{n}}-B{{x}^{*}} \right\|\left\| {{y}_{n}}-{{y}^{*}} \right\|\\ \quad +2\lambda \left\| A{{y}_{n}}-A{{y}^{*}} \right\|\left\| {{z}_{n}}-{{x}^{*}} \right\|]+2{{\alpha }_{n}}{{M}_{1}}\\ \le {{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}-(1-{{\beta }_{n}}){{g}_{2}}(\left\| {{x}_{n}}-{{y}_{n}}-({{x}^{*}}-{{y}^{*}}) \right\|)-(1-{{\beta }_{n}}){{g}_{1}}(\left\| {{y}_{n}}-{{z}_{n}}+{{x}^{*}}-{{y}^{*}} \right\|)\\ +2\mu \left\| B{{x}_{n}}-B{{x}^{*}} \right\|\left\| {{y}_{n}}-{{y}^{*}} \right\|+2\lambda \left\| A{{y}_{n}}-A{{y}^{*}} \right\|\left\| {{z}_{n}}-{{x}^{*}} \right\|+2{{\alpha }_{n}}{{M}_{1}}, $

从而

$ (1-{{\beta }_{n}}){{g}_{2}}(\left\| {{x}_{n}}-{{y}_{n}}-({{x}^{*}}-{{y}^{*}}) \right\|)+(1-{{\beta }_{n}}){{g}_{1}}(\left\| {{y}_{n}}-{{z}_{n}}+{{x}^{*}}-{{y}^{*}} \right\|)\\ \le {{\left\| {{x}_{n}}-{{x}^{*}} \right\|}^{2}}-{{\left\| {{x}_{n+1}}-{{x}^{*}} \right\|}^{2}}+2\mu \left\| B{{x}_{n}}-B{{x}^{*}} \right\|\left\| {{y}_{n}}-{{y}^{*}} \right\|\\ \quad +2\lambda \left\| A{{y}_{n}}-A{{y}^{*}} \right\|\left\| {{z}_{n}}-{{x}^{*}} \right\|+2{{\alpha }_{n}}{{M}_{1}}\\ \le (\left\| {{x}_{n}}-{{x}_{n+1}} \right\|)(\left\| {{x}_{n}}-{{x}^{*}} \right\|+\left\| {{x}_{n+1}}-{{x}^{*}} \right\|)+2\mu \left\| B{{x}_{n}}-B{{x}^{*}} \right\|\left\| {{y}_{n}}-{{y}^{*}} \right\|\\ \quad +2\lambda {{\left\| A{{y}_{n}}-Ay \right\|}^{*}}\left\| {{z}_{n}}-{{x}^{*}} \right\|+2{{\alpha }_{n}}{{M}_{1}}. $

因为$\lim\limits_{n\rightarrow\infty}\alpha_n=0$, $\limsup\limits_{n\rightarrow\infty}\beta_n < 1$, (3.8)和(3.14)式, 有

$ {g_2}(\left\| {{x_n} - {y_n} - ({x^*} - {y^*})} \right\|) = 0,\mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} {g_1}(\left\| {{y_n} - {z_n} + {x^*} - {y^*}} \right\|) = 0. $

根据$g_1$和$g_2$的性质, 有

$ \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} {\left\| {{x_n} - {y_n} - ({x^*} - y} \right\|^*}) = 0,\mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{y_n} - {z_n} + {x^*} - {y^*}} \right\| = 0. $

于是, 当$n\rightarrow\infty$时

$ \left\| {{x}_{n}}-{{z}_{n}} \right\|\le \text{ }\left\| {{x}_{n}}-{{y}_{n}}-({{x}^{*}}-{{y}^{*}}) \right\|+\left\| {{y}_{n}}-{{z}_{n}}+{{x}^{*}}-{{y}^{*}} \right\|\to 0. $

(3.19)

由(3.1)式得

$ \left\| {{x}_{n+1}}-{{x}_{n}} \right\|=\left\| {{\alpha }_{n}}(f({{x}_{n}})-{{x}_{n}})+{{\gamma }_{n}}({{T}_{n}}W{{z}_{n}}-{{x}_{n}}) \right\|\\ \ge {{\gamma }_{n}}\left\| {{T}_{n}}W{{z}_{n}}-{{x}_{n}} \right\|-{{\alpha }_{n}}\left\| f({{x}_{n}})-{{x}_{n}} \right\|, $

从而

$ \left\| {{T}_{n}}W{{z}_{n}}-{{x}_{n}} \right\|\le \frac{1}{{{\gamma }_{n}}}[{{\alpha }_{n}}\left\| f({{x}_{n}})-{{x}_{n}} \right\|+\left\| {{x}_{n+1}}-{{x}_{n}} \right\|]. $

根据$\lim\limits_{n\rightarrow\infty}\alpha_n=0$和(3.8)式得

$ \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{T_n}W{z_n} - {x_n}} \right\| = 0. $

(3.20)

由(3.19)与(3.20)式得:当$n\rightarrow\infty$时

$ \left\| {{x}_{n}}-{{T}_{n}}W{{x}_{n}} \right\|\le \left\| {{x}_{n}}-{{T}_{n}}W{{z}_{n}} \right\|+\left\| {{T}_{n}}W{{z}_{n}}-{{T}_{n}}W{{x}_{n}} \right\|\\ \le \left\| {{x}_{n}}-{{T}_{n}}W{{z}_{n}} \right\|+\left\| {{z}_{n}}-{{x}_{n}} \right\|\\ \to 0. $

(3.21)

定义映射$T_\delta:C\rightarrow C$为$T_\delta x=(1-\delta)x+\delta Sx$, 观察

$ \left\| {{x}_{n}}-{{T}_{\delta }}W{{x}_{n}} \right\|\le \left\| {{x}_{n}}-{{T}_{n}}W{{x}_{n}} \right\|+\left\| {{T}_{n}}W{{x}_{n}}-{{T}_{\delta }}W{{x}_{n}} \right\|\\ =\left\| {{x}_{n}}-{{T}_{n}}W{{x}_{n}} \right\|+\left\| (1-{{\delta }_{n}})W{{x}_{n}}+{{\delta }_{n}}{{S}_{n}}W{{x}_{n}}-(1-\delta )W{{x}_{n}}-\delta SW{{x}_{n}} \right\|\\ =\left\| {{x}_{n}}-{{T}_{n}}W{{x}_{n}} \right\|+\left| {{\delta }_{n}}-\delta \right|\left\| {{S}_{n}}W{{x}_{n}}-W{{x}_{n}} \right\|+\delta \left\| {{S}_{n}}W{{x}_{n}}-SW{{x}_{n}} \right\|. $

(3.22)

注意到

$ {{h}_{{{B}'}}}(\left\| {{S}_{n}}W{{x}_{n}}-SW{{x}_{n}} \right\|)\le \sup \left\{ {{h}_{{{B}'}}}(\left\| {{S}_{n}}x-Sx \right\|:x\in {B}') \right\}. $

由引理2.9和$h_{B'}$的连续性, 知$\mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} {h_{B'}}(\left\| {{S_n}W{x_n} - SW{x_n}} \right\|) = 0$.根据$h_{B'}$的性质得

$ \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{S_n}W{x_n} - SW{x_n}} \right\| = 0. $

(3.23)

因此由(3.22)式得

$ \mathop {\lim }\limits_{n \to \infty } {\mkern 1mu} \left\| {{x_n} - {T_\delta }W{x_n}} \right\| = 0. $

(3.24)

根据引理2.15得$F(S)\cap F(W)=F(T_\delta W)$.定义映射$U:C\rightarrow C$为$Ux=(1-\theta)T_\delta Wx+\theta Gx$, 其中$G$的定义见引理2.13, $\theta\in(0, 1)$为固定常数.由引理2.5得

$ F(G)=F(S)\cap F(W)\cap F(G)=F. $

根据(3.19)和(3.24)式得, 当$n\rightarrow\infty$时

$ \left\| {{x}_{n}}-U{{x}_{n}} \right\|\le (1-\theta )\left\| {{x}_{n}}-{{T}_{\delta }}W{{x}_{n}} \right\|+\theta \left\| {{x}_{n}}-G{{x}_{n}} \right\|\\ =(1-\theta )\left\| {{x}_{n}}-{{T}_{\delta }}W{{x}_{n}} \right\|+\theta \left\| {{x}_{n}}-{{z}_{n}} \right\|\\ \to 0. $

(3.25)

定义$z=\lim\limits_{t\rightarrow0}x_t$, 其中$x_t$定义为$x_t=tf(x_t)+(1-t)Ux_t$.由引理2.11得$z\in F(U)=F$且

$ \left\langle (I-f)z, j(z-p) \right\rangle \le 0, \forall \ p\in F. $

(3.26)

因为${j_q}(x) = {\left\| x \right\|^{q - 1}}j(x),x \ne 0$, 则

$ \left\langle (I-f)z, {{j}_{q}}(z-p) \right\rangle \le 0, \forall \ p\in F. $

(3.27)

根据(3.25)式和引理2.11得

$ \left\langle f(z)-z, j({{x}_{n}}-z) \right\rangle \le 0. $

(3.28)

再次根据${j_q}(x) = {\left\| x \right\|^{q - 1}}j(x),x \ne 0$得

$ \left\langle f(z)-z, {{j}_{q}}({{x}_{n}}-z) \right\rangle \le 0. $

(3.29)

最后我们证明当$n\rightarrow\infty$时, $x_n\rightarrow z$.假设$\left\{ {{x_n}} \right\}$不强收敛到$z$.则存在$\epsilon>0$和$\left\{ {{x_n}} \right\}$中子列$\left\{ {{x_{{n_j}}}} \right\}$满足$\left\| {{x_{{n_j}}} - z} \right\| \ge ,\forall \;j \ge 0$.根据命题1.4, 对上述$\epsilon$, 存在$\alpha'\in(0, 1)$满足$\left\| {f({x_{{n_j}}}) - f(z)} \right\| \le \alpha '\left\| {{x_{{n_j}}} - z} \right\|$.由(3.1)式和引理2.4得

$ {{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q}}=\left\langle {{\alpha }_{{{n}_{j}}}}(f({{x}_{{{n}_{j}}}})-z)+{{\beta }_{{{n}_{j}}}}({{x}_{{{n}_{j}}}}-z)+{{\gamma }_{{{n}_{j}}}}({{T}_{{{n}_{j}}}}{{z}_{{{n}_{j}}}}-z), {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle \\ \le {{\alpha }_{{{n}_{j}}}}\left\langle f({{x}_{{{n}_{j}}}})-f(z), {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle +{{\beta }_{{{n}_{j}}}}\left\langle {{x}_{{{n}_{j}}}}-z, {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle \\ \quad +{{\gamma }_{{{n}_{j}}}}\left\langle {{T}_{{{n}_{j}}}}{{z}_{{{n}_{j}}}}-z, {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle +{{\alpha }_{{{n}_{j}}}}\left\langle f(z)-z, {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle \\ \le {{\alpha }_{{{n}_{j}}}}{\alpha }'\left\| {{x}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}+{{\beta }_{{{n}_{j}}}}\left\| {{x}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}\\ \quad +{{\gamma }_{{{n}_{j}}}}\left\| {{T}_{{{n}_{j}}}}{{z}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}+{{\alpha }_{{{n}_{j}}}}\left\langle f(z)-z, {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle \\ \le {{\alpha }_{{{n}_{j}}}}{\alpha }'\left\| {{x}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}+{{\beta }_{{{n}_{j}}}}\left\| {{x}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}\\ \quad +{{\gamma }_{{{n}_{j}}}}\left\| {{z}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}+{{\alpha }_{{{n}_{j}}}}\left\langle f(z)-z, {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle \\ \le {{\alpha }_{{{n}_{j}}}}{\alpha }'\left\| {{x}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}+{{\beta }_{{{n}_{j}}}}\left\| {{x}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}\\ \quad +{{\gamma }_{{{n}_{j}}}}\left\| {{x}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}+{{\alpha }_{{{n}_{j}}}}\left\langle f(z)-z, {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle \\ =(1-{{\alpha }_{{{n}_{j}}}}(1-{\alpha }'))\left\| {{x}_{{{n}_{j}}}}-z \right\|{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q-1}}\\ +{{\alpha }_{{{n}_{j}}}}\left\langle f(z)-z, {{j}_{q}}({{x}_{{{n}_{j}}+1}}-z) \right\rangle \\ \le [1-{{\alpha }_{{{n}_{j}}}}(1-{\alpha }')](\frac{1}{q}{{\left\| {{x}_{{{n}_{j}}}}-z \right\|}^{q}}+\frac{q-1}{q}{{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q}})\\ +{{\alpha }_{{{n}_{j}}}}\left\langle f(z)-z, j({{x}_{{{n}_{j}}+1}}-z) \right\rangle, $

从而

$ {{\left\| {{x}_{{{n}_{j}}+1}}-z \right\|}^{q}}\le [1-{{\alpha }_{{{n}_{j}}}}(1-{\alpha }')]{{\left\| {{x}_{{{n}_{j}}}}-z \right\|}^{q}}+q{{\alpha }_{{{n}_{j}}}}(1-{\alpha }')\frac{\left\langle f(z)-z, j({{x}_{{{n}_{j}}+1}}-z) \right\rangle }{1-{\alpha }'}. $

(3.30)

注意到条件(ⅱ)与(3.29)式.应用引理2.1到(3.30)式, 于是当$j\rightarrow\infty$时, 有$x_{n_j}\rightarrow z$.矛盾.因此当$n\rightarrow\infty$时, $x_n\rightarrow z$.证毕.

现在给一个关于参数$\alpha_n$, $\beta_n$和$\gamma_n$的例子.

例 3.1 令$\alpha_n=\frac{1}{4n}, \ \beta_n=\frac{1}{3}+\frac{1}{4n}$且$\gamma_n=\frac{2}{3}-\frac{1}{2n}$, 则易知它们满足定理3.1中的条件(ⅰ)和(ⅱ).

由定理3.1, 易得下面结果, 故略去证明.

定理 3.2 设$C$为实$q$-一致光滑Banach空间$E$中非空闭凸子集. $Q_C$为$E$到$C$上向阳非扩张拉回.映射$A, B:C\rightarrow E$分别为$\alpha$ -逆强增生的和$\beta$ -逆强增生的. $f:C\rightarrow C$为广义压缩映射. $W:C\rightarrow C$为非扩张映射且$S:C\rightarrow C$为$\lambda$ -严格伪压缩映射满足$F:=F(S)\cap F(W)\cap F(G)\neq\emptyset$, 其中$G$的定义见引理2.13.给定$x_0\in C$, 序列$\left\{ {{x_n}} \right\}$定义为

$ \left\{ \begin{array}{l} y_n=Q_C(x_n-\mu Bx_n), \\ z_n=Q_C(y_n-\lambda Ay_n), \\ x_{n+1}=\alpha_nf(x_n)+\beta_nx_n+\gamma_nTWz_n, \end{array} \right. $

(3.31)

这里$0 < \lambda < (\frac{q\alpha}{C_q})^{\frac{1}{q-1}}$, $0 < \mu < (\frac{q\beta}{C_q})^{\frac{1}{q-1}}$, $T=(1-\delta)I+\delta S$, $\delta\in(0, \rho)$, 其中$\rho = \min \left\{ {1,{{\left\{ {\frac{{q\lambda }}{{{C_q}}}} \right\}}^{\frac{1}{{q - 1}}}}} \right\}$.假设$\left\{ {{\alpha _n}} \right\},\left\{ {{\beta _n}} \right\}$和$\left\{ {{\gamma _n}} \right\}$为[0, 1]中序列满足

(ⅰ) $\alpha_n+\beta_n+\gamma_n=1$;

(ⅱ) $\lim\limits_{n\rightarrow\infty}\alpha_n=0, \sum\limits_{n=1}^\infty\alpha_n=\infty$;

(ⅲ) $0 < \liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n < 1$.

则$\left\{ {{x_n}} \right\}$强收敛到$z\in F$且解决下面变分不等式

$ \left\langle {z-f(z), {j_q}(z-p)} \right\rangle \le 0, \forall {\mkern 1mu} p \in F. $

定理 3.3 设$C$为实$2$-一致光滑Banach空间$E$中非空闭凸子集. $Q_C$为$E$到$C$上向阳非扩张拉回.映射$A, B:C\rightarrow E$分别为$\alpha$ -逆强增生的和$\beta$ -逆强增生的. $f:C\rightarrow C$为广义压缩映射. $W:C\rightarrow C$为非扩张映射且$\left\{ {{S_i}:C \to C} \right\}_{i = 0}^\infty $为一族$\lambda_i$ -严格伪压缩映射满足$F:=\bigcap\limits_{i=0}^{\infty} F(S_i)\cap F(W)\cap F(G)\neq\emptyset$, 其中$G$的定义见引理2.13.假设$\lambda = \inf \left\{ {{\lambda _i}:i = 0,1,2, \cdots } \right\} > 0$.给定$x_0\in C$, 序列$\left\{ {{x_n}} \right\}$定义为

$ \left\{ \begin{array}{l} y_n=Q_C(x_n-\mu Bx_n), \\ z_n=Q_C(y_n-\lambda Ay_n), \\ x_{n+1}=\alpha_nf(x_n)+\beta_nx_n+\gamma_nT_nWz_n, \end{array} \right. $

(3.32)

这里$0 < \lambda < \frac{\alpha}{K^2}$, $0 < \mu < \frac{\beta}{K^2}$, $T_n=(1-\delta_n)I+\delta_n S_n$.假设$\left\{ {{\alpha _n}} \right\},\left\{ {{\beta _n}} \right\}$和$\left\{ {{\gamma _n}} \right\}$为[0, 1]中序列满足

(ⅰ) $\alpha_n+\beta_n+\gamma_n=1$;

(ⅱ) $\lim\limits_{n\rightarrow\infty}\alpha_n=0, \sum\limits_{n=1}^\infty\alpha_n=\infty$;

(ⅲ) $0 < \liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n < 1$;

(ⅳ) $\delta_n\in(0, \frac{\lambda}{K^2}), \delta_n\rightarrow\delta\in(0, \frac{\lambda}{K^2})$.

假设对于$C$中任意有界集$D$, 存在单调增、连续的凸函数$h_D:{\Bbb R}^+\rightarrow {\Bbb R}^+$满足$h_D(0)$且$\mathop {\lim \sup }\limits_{k,l \to \infty } \left\{ {{h_D}(\left\| {{S_k}z - {S_l}z} \right\|):z \in D} \right\} = 0$.定义$Sx=\lim\limits_{n\rightarrow\infty}S_nx, \forall\ x\in C$且假设$F(S)=\bigcap\limits_{i=0}^\infty F(S_i)$.则$\left\{ {{x_n}} \right\}$强收敛到$z\in F$且解决下面变分不等式

$ \left\langle {z-f(z), {j_q}(z-p)} \right\rangle \le 0, \forall {\mkern 1mu} p \in F. $

定理 3.4 设$C$为实$2$-一致光滑Banach空间$E$中非空闭凸子集. $Q_C$为$E$到$C$上向阳非扩张拉回.映射$A, B:C\rightarrow E$分别为$\alpha$ -逆强增生的和$\beta$ -逆强增生的. $f:C\rightarrow C$为广义压缩映射. $W:C\rightarrow C$为非扩张映射且$S:C\rightarrow C$为$\lambda$ -严格伪压缩映射满足$F:=F(S)\cap F(W)\cap F(G)\neq\emptyset$, 其中$G$的定义见引理2.13.给定$x_0\in C$, 序列$\left\{ {{x_n}} \right\}$定义为

$ \left\{ \begin{array}{l} y_n=Q_C(x_n-\mu Bx_n), \\ z_n=Q_C(y_n-\lambda Ay_n), \\ x_{n+1}=\alpha_nf(x_n)+\beta_nx_n+\gamma_nTWz_n, \end{array} \right. $

(3.33)

这里$0 < \lambda < \frac{\alpha}{K^2}$, $0 < \mu < \frac{\beta}{K^2}$, $T=(1-\delta_n)I+\delta_n S$, $\delta\in(0, \frac{\lambda}{K^2})$.假设$\left\{ {{\alpha _n}} \right\},\left\{ {{\beta _n}} \right\}$和$\left\{ {{\gamma _n}} \right\}$为[0, 1]中序列满足

(ⅰ) $\alpha_n+\beta_n+\gamma_n=1$;

(ⅱ) $\lim\limits_{n\rightarrow\infty}\alpha_n=0, \sum\limits_{n=1}^\infty\alpha_n=\infty$;

(ⅲ) $0 < \liminf\limits_{n\rightarrow\infty}\beta_n\leq\limsup\limits_{n\rightarrow\infty}\beta_n < 1$.

则$\left\{ {{x_n}} \right\}$强收敛到$z\in F$且解决下面变分不等式

$ \left\langle {z-f(z), {j_q}(z-p)} \right\rangle \le 0, \forall {\mkern 1mu} p \in F. $