数学物理学报, 2026, 46(1): 94-107

研究论文

Hilbert 空间中关于分层变分不等式问题与不动点问题的投影算法

刘佳宇,, 高兴慧,*, 淡鹭涵

延安大学数学与计算机科学学院 陕西延安 716000

Projection Algorithm for Hierarchical Variational Inequality Problem and Fixed Point Problem in Hilbert Space

Liu Jiayu,, Gao Xinghui,*, Dan Luhan

College of Mathematics and Computer Science, Yan'an University, Shaanxi Yan'an 716000

通讯作者: *高兴慧, Email: yadxgaoxinghui@163.com

收稿日期: 2025-03-10   修回日期: 2025-06-29  

基金资助: 国家自然科学基金(61866038)
陕西省特支计划人才项目(2021年)

Received: 2025-03-10   Revised: 2025-06-29  

Fund supported: NSFC(61866038)
Shaanxi Special Support Plan for High-level Talent Program (2021)

作者简介 About authors

刘佳宇,Email:3438956880@qq.com

摘要

该文在 Hilbert 空间中为了求解分层变分不等式问题与拟非扩张映射不动点问题的公共元给出了一种新的多步惯性正则化算法, 一定条件下, 得出了该算法生成的迭代序列的强收敛定理. 最后, 给出数值例子说明该算法的有效性.

关键词: 分层变分不等式; 多步惯性正则化算法; 强收敛定理

Abstract

This paper introduces a novel multi-step inertial regularization algorithm in Hilbert spaces, designed to solve the common solution of hierarchical variational inequality problems and quasi-non expansive mapping fixed-point problems. Under certain conditions, the strong convergence theorem for this iterative sequence generated by the algorithm is established. Finally, a numerical example is provided to demonstrate that this algorithm is more efficient than the existing ones.

Keywords: hierarchical variational inequalities; multi-step inertial regularization algorithm; strong convergence theorem

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本文引用格式

刘佳宇, 高兴慧, 淡鹭涵. Hilbert 空间中关于分层变分不等式问题与不动点问题的投影算法[J]. 数学物理学报, 2026, 46(1): 94-107

Liu Jiayu, Gao Xinghui, Dan Luhan. Projection Algorithm for Hierarchical Variational Inequality Problem and Fixed Point Problem in Hilbert Space[J]. Acta Mathematica Scientia, 2026, 46(1): 94-107

1 引言

$C$ 是实 Hilbert 空间 $H$ 的一个非空闭凸子集, $\langle\cdot, \cdot\rangle$$\|\cdot\|$ 表示 $H$ 中的内积与范数. $A: H \rightarrow H$ 为一映射, 经典的变分不等式 (VIP) 为寻找一点 $\hat{x} \in C$ 使得

$\langle A \hat{x}, y-\hat{x}\rangle \geq 0, \forall y \in C.$

记 (1.1) 式的解集为 $\operatorname{VI}(C, A)$. 此外, 不动点问题为寻找一点 $p \in H$, 使得 $T(p)=p$, 即若 $\operatorname{Fix}(T)=\{p \in H \mid T(p)=p\}$, 则 ${\rm Fix}(T)$ 称为 $T$ 的不动点集.

该问题已经被广泛应用于控制论、信号处理、经济学、最优控制理论等领域[1-3], 变分不等式和不动点问题受到学者们的关注和研究[4-7]. 为了求解变分不等式问题, 1976 年, Korpelevich[8]提出双投影外梯度算法, 该算法需计算两次到可行集 $C$ 上的投影, 当 $C$ 的结构复杂时, 数值实验中投影算子的计算会更加复杂. 因此, Censor[9]等提出次梯度外梯度算法, 将第二次在 $C$ 上的投影替换为半空间上的投影, 形式如下

$\left\{\begin{array}{l}y_{n}=P_{C}\left(x_{n}-\lambda A x_{n}\right), \\x_{n+1}=P_{T_{n}}\left(x_{n}-\lambda A y_{n}\right).\end{array}\right.$

近年来, 很多学者对分层变分不等式 (HVIP) 问题产生了很大的兴趣. 设 $F: H \rightarrow H$ 是一个强单调且 $L$-Lipschitz 映射, HVIP 问题为寻找一点 $x^{*} \in \operatorname{VI}(C, A)$ 使得

$\left\langle F x^{*}, x-x^{*}\right\rangle \geq 0, \forall x \in \mathrm{VI}(C, A).$

为了求解分层变分不等式问题, 2021 年, Jiang 等[10]结合次梯度外梯度外梯度法和多步惯性提出了一种新的投影算法, 形式如下

$\left\{\begin{array}{l}w_{n}=x_{n}+\alpha_{i, n} \sum_{i=1}^{N}\left(x_{n-i+1}-x_{n-i}\right), \\y_{n}=P_{C}\left[w_{n}-\lambda_{n}\left(A w_{n}+\beta_{n} F w_{n}\right)\right], \\x_{n+1}=P_{T_{n}}\left[w_{n}-\lambda_{n}\left(A y_{n}+\beta_{n} F w_{n}\right)\right],\end{array}\right.$

其中 $A: H \rightarrow H$ 为单调且 $L$-Lipschitz 映射, $F: H \rightarrow H$ 为强单调, 半连续且 $K$-广义 Lipschitz 映射.

2022 年, Song[11]在 Jiang 等[10]的基础上增加一个逆强单调映射, 提出一种新的分层变分不等式问题, 该问题为寻找一点 $x^{*} \in \mathrm{VI}(C, A)$ 使得

$\left\langle F x^{*}, x-x^{*}\right\rangle \geq 0, \forall x \in \mathrm{VI}(C, A) \cap \operatorname{Zer}(B). $

其中 ${\rm Zer}(B)=\{x \in H: B x=0\}$, 他们提出的算法具体形式如下

$\left\{\begin{array}{l}\omega_{n}=x_{n}+\alpha_{n}\left(x_{n}-x_{n-1}\right), \\y_{n}=P_{C}\left[\omega_{n}-\lambda_{n}\left(A \omega_{n}+\beta_{n}^{w} B \omega_{n}+\beta_{n} F \omega_{n}\right)\right], \\\lambda_{n+1}=\left\{\begin{array}{l}\min \left\{\lambda_{n}+\gamma_{n}, \frac{\mu\left\|\omega_{n}-y_{n}\right\|}{\left\|A \omega_{n}-A y_{n}\right\|}\right\}, \text { 若 } A \omega_{n} \neq A y_{n},\\\lambda_{n}+\gamma_{n},\text { 其他. }\end{array}\right. \\x_{n+1}=P_{T_{n}}\left[\omega_{n}-\lambda_{n}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)\right],\end{array}\right.$

其中 $B: H \rightarrow H$$v$-逆强单调映射, $F: H \rightarrow H$ 为强单调, 半连续且 $K$-广义 Lipschitz 映射. 2025 年姜炳南等[12]为了加快算法的收敛速度, 提出新的正则化次梯度外梯度法求解上述分层变分不等式问题, 该算法与文献[11] 的算法相比对于映射 $A, B$ 只需要计算 $A y_{n}$$B y_{n}$, 从而减少了计算时间, 具体形式如下

$\left\{\begin{array}{l}\omega_{n}=x_{n}+\alpha_{i, n} \sum_{i=1}^{N}\left(x_{n-i+1}-x_{n-i}\right), \\y_{n}=P_{C}\left[\omega_{n}-\lambda_{n}\left(A y_{n-1}+\beta_{n}^{w} B y_{n-1}+\beta_{n} F \omega_{n}\right)\right], \\\lambda_{n+1}=\left\{\begin{array}{l}\min \left\{\lambda_{n}+\gamma_{n}, \frac{\mu\left\|y_{n-1}-y_{n}\right\|}{\left\|A y_{n-1}-A y_{n}\right\|}\right\},\text { 若 } A y_{n-1} \neq A y_{n}, \\\lambda_{n}+\gamma_{n},\text { 其他. }\end{array}\right. \\\quad x_{n+1}=P_{T_{n}\left[\omega_{n}-\lambda_{n}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)\right].}\end{array}\right.$

在上述文献的基础上, 我们考虑将文献 [12] 的分层变分不等式问题与不动点问题结合, 并试图加快文献 [12] 中算法的收敛速度. 为此, 本文提出了一种带最大值自适应步长的多步惯性正则化次梯度外梯度算法, 在一定条件下证明了该算法生成的迭代序列强收敛于分层变分不等式解集与拟非扩张映射不动点集的公共元. 并给出数值例子说明该算法的收敛速度更快. 本文的结果改进和推广了文献 [10-12] 的相关结果.

2 预备知识

在本文中我们设 $H$ 是一个实 Hilbert 空间, $C$$H$ 上的非空闭凸子集.

定义 2.1[13,14]$T: H \rightarrow H$ 是一个映射.

(i) $T$ 被称为 $L$-Lipschitz 映射 $(L>0)$, 若

$\|T x-T y\| \leq L\|x-y\|, \forall x, y \in H;$

(ii) $T$ 被称为 $K$ 广义-Lipschitz 映射 $(K>0)$, 若

$\|T x-T y\| \leq K(\|x-y\|+1), \forall x, y \in H;$

(iii) $T$ 被称为单调的, 若

$\langle T x-T y, x-y\rangle \geq 0, \forall x, y \in H;$

(iv) $T$ 被称为 $\eta-$ 强单调映射 $\eta>0$, 若

$\langle T x-T y, x-y\rangle \geq \eta\|x-y\|^{2}, \forall x, y \in H;$

(v) $T$ 被称为 $v$-逆强单调映射 $v>0$, 若

$\langle T x-T y, x-y\rangle \geq v\|T x-T y\|^{2}, \forall x, y \in H;$

(vi) $T$ 被称为拟非扩张映射 $(\operatorname{Fix}(T) \neq \varnothing)$, 若

$\|T x-p\| \leq\|x-p\|, \forall x \in H, p \in \operatorname{Fix}(T).$

引理 2.1[15] 对任意 $x, y \in H, \alpha \in R$, 有

(i) $\|x+y\|^{2}=\|x\|^{2}+\|y\|^{2}+2\langle x, y\rangle$.

(ii) $ \|\alpha x+(1-\alpha) y\|^{2}=\alpha\|x\|^{2}+(1-\alpha)\|y\|^{2}-\alpha(1-\alpha)\|x-y\|^{2}.$

引理 2.2[16]$C$ 是 Hilbert 空间上的非空闭凸子集, 称 $T: C \rightarrow H$ 为拟非扩张映射, 则 Fix$(T)$ 是 C 中的非空闭凸子集.

引理 2.3[17]$C$ 是 Hilbert 空间上的非空闭凸子集, 对于 $x \in H, z \in C$

(i) $z=P_{C} x \Leftrightarrow\langle x-z, y-z\rangle \leq 0, \forall y \in C$;

(ii) $\left\|P_{C} x-P_{C} y\right\| \leq\left\langle P_{C} x-P_{C} y, x-y\right\rangle, \forall y \in C.$

引理 2.4[18]$A: H \rightarrow H$ 是一个强单调且半连续映射, 则 $\mathrm{VI}(C, A)$ 是单点集.

引理 2.5[19]$\left\{a_{n}\right\}$ 是非负的实数序列, $\left\{\alpha_{n}\right\},\left\{b_{n}\right\}$ 为实数列. 且满足

$a_{n+1} \leq\left(1-\alpha_{n}\right) a_{n}+\alpha_{n} b_{n}.$

其中 $\left\{\alpha_{n}\right\} \subset(0,1), \sum_{n=1}^{\infty} \alpha_{n}=\infty, \lim _{n \rightarrow \infty} \sup b_{n} \leq 0$, 那么 $\lim _{n \rightarrow \infty} a_{n}=0$.

3 主要结果

${\bf(C1)} A: H \rightarrow H$ 是单调且 $L$-Lipschitz 映射;

${\bf(C2)} T: H \rightarrow H$ 是闭的拟非扩张映射, ${\rm Fix}(T) \cap \mathrm{VI}\left(C, A+\beta_{n}^{w} B+\beta_{n} F\right) \cap {\rm Zer}(B) \neq \varnothing$;

${\bf(C3)} B: H \rightarrow H$$v$-逆强单调映射, ${\rm Fix}(T) \cap \mathrm{VI}(C, A) \cap {\rm Zer}(B) \neq \varnothing$;

${\bf(C4)} F: H \rightarrow H$$\eta$-强单调, 半连续且 $K$-广义 Lipschitz 映射.

${\bf 算法 1}$

选取 $x_{0}, x_{1} \in H,k\in(0,\frac{1}{2}), \theta_{n} \in[0,1]$ 且满足 $ \sum_{n=1}^{\infty} \theta_{n}=\infty$, $\lim _{n \rightarrow \infty} \theta_{n}=\theta \in\left(0, \frac{1}{3}\right)$, $ \mu \in\left(0, \frac{3-\sqrt{2}}{5}\right)$. 实数列 $\left\{\beta_{n}\right\} \subset(0, \infty)$ 满足 $\lim _{n \rightarrow \infty}\beta_{n}=0$, $\lim _{n \rightarrow \infty}\frac{\beta_{n+1}-\beta_n}{\beta_n^2\beta_{n+1}}=0$. 对每个 $i=1,2, \cdots, N$, 选取 $\alpha_{i} \geq 0$, 实数列 $\left\{\sigma_{i, n}\right\} \subset(0, \infty)$ 始终满足 $\lim _{n \rightarrow 0} \frac{\sigma_{i, n}}{\beta_{n}}=0$.$n=1$.

${\bf 第一步}$ 计算

$\omega_{n}=x_{n}+\alpha_{i, n} \sum_{i=1}^{N}\left(x_{n-i+1}-x_{n-i}\right),$

其中

$\alpha_{i, n}= \begin{cases}\min \left\{\alpha_{i}, \frac{\sigma_{i, n}}{\left\|x_{n-i+1}-x_{n-i}\right\|}\right\}, & \text { 若 } x_{n-i+1} \neq x_{n-i}, \\ \alpha_{i}, & \text { 其他. }\end{cases}$

${\bf 第二步}$ 计算

$y_{n}=P_{C}\left[\omega_{n}-\frac{k}{\lambda_{n}}\left(A y_{n-1}+\beta_{n}^{w} B y_{n-1}+\beta_{n} F \omega_{n}\right)\right].$

${\bf 第三步}$ 计算

$z_{n}=P_{T_{n}}\left[\omega_{n}-\frac{k}{\lambda_{n}}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)\right]$

$\lambda_{n+1}=\left\{\begin{array}{ll}\max \left\{\lambda_{n}, \frac{\left\|A y_{n-1}-A y_{n}\right\|}{\mu\left\|y_{n-1}-y_{n}\right\|}\right\},& \text { 若 } y_{n-1} \neq y_{n}, \\\lambda_{n}, & \text { 其他, }\end{array}\right.$

其中 $T_{n}=\left\{z \in H:\left\langle\omega_{n}-\frac{k}{\lambda_{n}}\left(A y_{n-1}+\beta_{n}^{w} B y_{n-1}+\beta_{n} F \omega_{n}\right)-y_{n}, z-y_{n}\right\rangle \leq 0\right\}$.

${\bf 第四步}$ 计算

$x_{n+1}=\left(1-\theta_{n}\right) \omega_{n}+\theta_{n} T z_{n}.$

$n=n+1$ 并返回第一步.

首先, 我们在条件 (C1)-(C4) 成立时, 给出以下变分不等式问题, 该问题为寻找一点 $\bar{x} \in C$, 满足

$\left\langle\left(A+\beta^{w} B+\beta F\right) \bar{x}, x-\bar{x}\right\rangle \geq 0, \forall x \in C,$

其中 $w \in(0,1), \beta>0$. 显然 $A+\beta^{w} B+\beta F$ 为强单调映射, 由引理 2.4 可得该问题存在唯一解 $x_{\beta}$.

引理 3.1 假设条件 (C1)-(C4) 成立, 由算法 1 生成的数列 $\left\{\lambda_{n}\right\}$ 满足

$0 \leq \lim _{n \rightarrow \infty} \lambda_{n}=\lambda \leq \max \left\{\lambda_{1}, \frac{L}{\mu}\right\}.$

这里的 $\lambda$ 为常数.

证明 由 $A$$H$ 上的 $L$-Lipschitz 映射, 我们有

$\left\|A y_{n-1}-A y_{n}\right\| \leq L\left\|y_{n-1}-y_{n}\right\|.$

$y_{n-1} \neq y_{n}$, 可以得到

$\frac{\left\|A y_{n-1}-A y_{n}\right\|}{\mu\left\|y_{n-1}-y_{n}\right\|} \leq \frac{L}{\mu}.$

$\lambda_{n+1}$ 的构造可知 $\left\{\lambda_{n}\right\}$ 非递减序列, 结合数学归纳法可得

$\lambda_{n} \leq \max \left\{\lambda_{n-1}, \frac{L}{\mu}\right\} \leq \cdots \leq \max \left\{\lambda_{1}, \frac{L}{\mu}\right\}.$

所以 $\left\{\lambda_{n}\right\}$ 有上界, 故存在 $\lambda>0$, 有 $\lim _{n \rightarrow \infty} \lambda_{n}=\lambda \leq \max \left\{\lambda_{1}, \frac{L}{\mu}\right\}$.

引理 3.2[12] 假设条件 (C1)-(C4) 成立, 任意 $\omega, \alpha, \beta \in(0,1)$, 都有 $\left\|x_{\alpha}-x_{\beta}\right\| \leq \frac{|\alpha-\beta|}{\alpha \beta} M$, 其中 $M=\frac{1}{\eta}\left[\left(1+\frac{K}{\eta}+\frac{1}{v \eta}\right)\left\|F x^{*}\right\|+\left\|B x^{*}\right\|+\left(2 K+\frac{2}{v}\right)\left\|x^{*}\right\|+K\right]$.

引理 3.3[12] 假设条件 (C1)-(C4) 成立, 任意 $\omega \in(0,1)$, 都有 $\lim _{\beta \rightarrow 0} x_{\beta}=x^{*}$.

定理 3.1 假设 $k \leq 1$, 条件 (C1)-(C4) 成立, 由算法 1 生成的序列 $\left\{x_{n}\right\}$ 强收敛到一点 $x^{*} \in \operatorname{VI}(C, A) \cap \operatorname{Fix}(T) \cap \operatorname{Zer}(B)$, 这里的 $x^{*}$ 是 (1.3) 式的解.

根据引理 2.4 可知对于任意 $n \in \mathbb{N}$, 存在唯一解 $x_{\beta_{n}} \in C$, 满足

$\left\langle\left(A+\beta_{n}^{w} B+\beta_{n} F\right) x_{\beta_{n}}, x-x_{\beta_{n}}\right) \geq 0, \forall x \in C.$

$x_{\beta_{n}} \in C \subseteq T_{n}$, 根据引理 2.3 可得

$\begin{aligned}& \left\|z_{n}-x_{\beta_{n}}\right\|^{2} \\= & \left\|P_{T_{n}}\left[\omega_{n}-\frac{k}{\lambda_{n}}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)\right]-P_{T_{n}} x_{\beta_{n}}\right\|^{2} \\\leq & \left\langle z_{n}-x_{\beta_{n}}, \omega_{n}-\frac{k}{\lambda_{n}}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)-x_{\beta_{n}}\right\rangle \\= & \frac{1}{2}\left\|z_{n}-x_{\beta_{n}}\right\|^{2}+\frac{1}{2}\left\|\omega_{n}-\frac{k}{\lambda_{n}}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)-x_{\beta_n}\right\|^{2}\\-&\frac{1}{2}\left\|z_{n}-\omega_{n}+\frac{k}{\lambda_{n}}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)\right\|^{2} \\= & \frac{1}{2}\left\|z_{n}\!-\!x_{\beta_{n}}\right\|^{2}+\frac{1}{2}\left\|\omega_{n}\!-\!x_{\beta_{n}}\right\|^{2}-\frac{1}{2}\left\|z_{n}\!-\!\omega_{n}\right\|^{2}\!-\!\left\langle z_{n}-x_{\beta_{n}}, \frac{k}{\lambda_{n}}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)\right\rangle.\end{aligned}$

于是有

$\begin{aligned} \left\|z_n-x_{\beta_n}\right\|^2 \leq & \left\|\omega_n-x_{\beta_n}\right\|^2-\left\|z_n-\omega_n\right\|^2-2\left\langle z_n-x_{\beta_n}, \frac{k}{\lambda_n}\left(A y_n+\beta_n^w B y_n+\beta_n F \omega_n\right)\right\rangle \\ = & \left\|\omega_n-x_{\beta_n}\right\|^2-\left\|z_n-\omega_n\right\|^2-2\left\langle z_n-y_n, \frac{k}{\lambda_n}\left(A y_n+\beta_n^w B y_n+\beta_n F \omega_n\right)\right\rangle \\ & -2\left\langle y_n-x_{\beta_n}, \frac{k}{\lambda_n}\left(A y_n+\beta_n^w B y_n+\beta_n F \omega_n\right)\right\rangle. \end{aligned}$

接下来我们考虑

$\begin{aligned}& 2\left\langle y_{n}-z_{n}, \frac{k}{\lambda_{n}}\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)\right\rangle \\= & 2 \frac{k}{\lambda_{n}}\left\langle y_{n}-z_{n},\left(A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right)-\left(A y_{n-1}+\beta_{n}^{w} B y_{n-1}+\beta_{n} F \omega_{n}\right)\right\rangle \\& +2 \frac{k}{\lambda_{n}}\left\langle y_{n}-z_{n}, A y_{n-1}+\beta_{n}^{w} B y_{n-1}+\beta_{n} F \omega_{n}\right\rangle \\= & 2 \frac{k}{\lambda_{n}}\left\langle y_{n}-z_{n}, A y_{n}-A y_{n-1}\right\rangle+2 \frac{k \beta_{n}^{w}}{\lambda_{n}}\left\langle y_{n}-z_{n}, B y_{n}-B y_{n-1}\right\rangle\\&+2 \frac{k}{\lambda_{n}}\left\langle y_{n}-z_{n}, A y_{n-1}+\beta_{n}^{w} B y_{n-1}+ \beta_{n} F \omega_{n}\right\rangle.\end{aligned}$

由均值不等式和定义 2.1 可得

$\begin{aligned} & 2 \frac{k}{\lambda_{n}}\left\langle y_{n}-z_{n}, A y_{n}-A y_{n-1}\right\rangle \\ \leq & \frac{2 \mu k \lambda_{n+1}}{\lambda_{n}}\left\|y_{n}-y_{n-1}\right\|\left\|\left\|y_{n}-z_{n}\right\|\right. \\ \leq & \frac{\mu k \lambda_{n+1}}{\lambda_{n}}\left(\frac{1}{\sqrt{2}}\left\|y_{n}-y_{n-1}\right\|^{2}+\sqrt{2}\left\|y_{n}-z_{n}\right\|^{2}\right) \\ \leq & \frac{\mu k \lambda_{n+1}}{\lambda_{n}}\left((1+\sqrt{2})\left\|y_{n}-\omega_{n}\right\|^{2}+\left\|\omega_{n}-y_{n-1}\right\|^{2}+\sqrt{2}\left\|y_{n}-z_{n}\right\|^{2}\right). \end{aligned}$

类似地, 因为 $B$$v$-逆强单调映射, 所以是 $\frac{1}{v}$-Lipschitz 映射

$\begin{aligned}& \quad \frac{2 k \beta_{n}^{w}}{\lambda_{n}}\left\langle y_{n}-z_{n}, B y_{n}-B y_{n-1}\right\rangle \\& \leq \frac{2 k \beta_{n}^{w}}{v \lambda_{n}}\left\|y_{n}-y_{n-1}\left|\left\|\mid y_{n}-z_{n}\right\|\right.\right. \\& \leq \frac{k \beta_{n}^{w}}{v \lambda_{n}}\left(\frac{1}{\sqrt{2}}\left\|y_{n}-y_{n-1}\right\|^{2}+\sqrt{2}\left\|y_{n}-z_{n}\right\|^{2}\right) \\& \leq \frac{k \beta_{n}^{w}}{v \lambda_{n}}\left((1+\sqrt{2})\left\|y_{n}-\omega_{n}\right\|^{2}+\left\|\omega_{n}-y_{n-1}\right\|^{2}+\sqrt{2}\left\|y_{n}-z_{n}\right\|^{2}\right).\end{aligned}$

$\psi_{n}\!=\!\frac{\mu k \lambda_{n+1}}{\lambda_{n}}+\frac{k \beta_{n}^{w}}{v \lambda_{n}}$, 根据引理 3.1 得 $\lim _{n \rightarrow 0} \psi_{n}\!=\!\mu k$.$\varepsilon \!\in\!\left(0, \min \left\{\frac{2 \eta \theta_{n}}{K \theta_{n}+2}, 1\!-\!(3\!+\!\sqrt{2}) \mu k\right\}\right)$, 则有

$\begin{align*} &\lim _{n \rightarrow \infty}\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}\right)=1-(1+\sqrt{2}) \mu k>\varepsilon,\\ \quad &\lim _{n \rightarrow \infty} 1-\sqrt{2} \psi_{n}=1-\sqrt{2} \mu k>0, \\ & \lim _{n \rightarrow \infty} \frac{\left(1-\theta_{n}\right)}{\theta_{n}}\left(1-\frac{k \varepsilon \beta_{n}}{\lambda_{n}}\right)-\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\\ &=\frac{(1-\theta)}{\theta}-1+(1+\sqrt{2}) \mu k+\varepsilon>0. \end{align*}$

根据数列极限的保号性, 存在正整数 $n_{0} \geq N$, 当 $n \geq n_{0}$ 时有

$\begin{aligned}& 1-\sqrt{2} \psi_{n}>0, \quad 1-(1+\sqrt{2}) \psi_{n}-\frac{ k\beta_{n} K}{\varepsilon\lambda_{n}}-\varepsilon>0, \quad \sum_{i=1}^{N} \sigma_{i, n} \leq \frac{\varepsilon k \beta_{n}}{\lambda_{n}}, \\& \frac{\left(1-\theta_{n}\right)}{\theta_{n}}\left(1-\frac{k \varepsilon \beta_{n}}{\lambda_{n}}\right)-\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)>0.\end{aligned}$

在后面证明中均假设 $n \geq n_{0}$.

利用引理 2.3, 由 $z_{n} \in T_{n}$$T_{n}$ 的定义有

$\left\langle z_{n}-y_{n}, \omega_{n}-\frac{k}{\lambda_{n}}\left(A y_{n-1}+\beta_{n}^{w} B y_{n-1}+\beta_{n} F \omega_{n}\right)-y_{n}\right\rangle \leq 0,$

即有

$\begin{aligned}& \frac{2 k}{\lambda_{n}}\left\langle y_{n}-z_{n},\left(A y_{n-1}+\beta_{n}^{w} B y_{n-1}+\beta_{n} F \omega_{n}\right)\right\rangle \\\leq & 2\left\langle z_{n}-y_{n}, y_{n}-\omega_{n}\right\rangle=\left\|z_{n}-\omega_{n}\right\|^{2}-\left\|y_{n}-\omega_{n}\right\|^{2}-\left\|z_{n}-y_{n}\right\|^{2}.\end{aligned}$

由均值不等式和映射 $A, B, F$ 的单调性, 有

$\begin{aligned}& \frac{2 k}{\lambda_{n}}\left\langle x_{\beta_{n}}-y_{n}, A y_{n}+\beta_{n}^{w} B y_{n}+\beta_{n} F \omega_{n}\right\rangle \\\leq & \frac{2 k \beta_{n}}{\lambda_{n}}\left\langle x_{\beta_{n}}-y_{n}, F \omega_{n}-F x_{\beta_{n}}\right\rangle \\\leq & -\frac{2 k \beta_{n} \eta}{\lambda_{n}}\left\|\omega_{n}-x_{\beta_{n}}\right\|^{2}+\frac{2 k \beta_{n} K}{\lambda_{n}}\left\|\omega_{n}-x_{\beta_{n}}\right\|\left\|\omega_{n}-y_{n}\right\|+\frac{2 k \beta_{n} K}{\lambda_{n}}\left\|\omega_{n}-y_{n}\right\| \\\leq & -\frac{2 k \beta_{n} \eta}{\lambda_{n}}\left\|\omega_{n}-x_{\beta_{n}}\right\|^{2}+\frac{k \beta_{n} K \varepsilon}{\lambda_{n}}\left\|\omega_{n}-x_{\beta_{n}}\right\|^{2}+\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}\left\|\omega_{n}-y_{n}\right\|^{2}+\varepsilon\left\|\omega_{n}-y_{n}\right\|\\&+\frac{k^{2} \beta_{n}^{2} K^{2}}{\varepsilon \lambda_{n}^{2}} \\= & -(2 \eta-K \varepsilon) \frac{k \beta_{n}}{\lambda_{n}}\left\|\omega_{n}-x_{\beta_{n}}\right\|^{2}+\left(\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}+\varepsilon\right)\left\|\omega_{n}-y_{n}\right\|^{2}+\frac{k^{2} \beta_{n}^{2} K^{2}}{\varepsilon \lambda_{n}^{2}}.\end{aligned}$

将 (3.3)-(3.6) 式代入 (3.2) 式可得

$\begin{aligned} &\left\|z_{n}-x_{\beta_{n}}\right\|^{2}\\ \leq&\left(1-(2 \eta-K \varepsilon) \frac{k \beta_{n}}{\lambda_{n}}\right)\left\|\omega_{n}-x_{\beta_{n}}\right\|^{2}-\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left\|y_{n}-\omega_{n}\right\|^{2} \\ &+\psi_{n}\left\|\omega_{n}-y_{n-1}\right\|^{2}-\left(1-\sqrt{2} \psi_{n}\right)\left\|z_{n}-y_{n}\right\|^{2}+\frac{k^{2} \beta_{n}^{2} K^{2}}{\varepsilon \lambda_{n}^{2}} \\ \leq&\left(1-(2 \eta-K \varepsilon) \frac{k \beta_{n}}{\lambda_{n}}\right)\left\|\omega_{n}-x_{\beta_{n}}\right\|^{2}-\frac{1}{2}\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left\|y_{n}-\omega_{n+1}\right\|^{2} \\ &+\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left\|\omega_{n}-\omega_{n+1}\right\|^{2}+\psi_{n}\left\|\omega_{n}-y_{n-1}\right\|^{2}+\frac{k^{2} \beta_{n}^{2} K^{2}}{\varepsilon \lambda_{n}^{2}}. \end{aligned}$

$\omega_{n}$ 的构造和均值不等式可得

$\begin{aligned}& \left\|\omega_{n+1}-\omega_{n}\right\|^{2} \\= & \left\|x_{n+1}-x_{n}+\alpha_{i, n+1} \sum_{i=1}^{N}\left(x_{n-i+2}-x_{n-i+1}\right)\right\|^{2} \\\leq & \left\|x_{n+1}-x_{n}\right\|^{2}+\sum_{i=1}^{N} \sigma_{i, n+1}^{2}+\left(\left\|x_{n+1}-x_{n}\right\|^{2}+1\right) \sum_{i=1}^{N} \sigma_{i, n+1}+2 \sum_{1 \leq i<j} \sigma_{i, n+1} \sigma_{j, n+1} \\\leq & \left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n+1}-x_{n}\right\|^{2}+\tau_{n+1},\end{aligned}$

其中 $\tau_{n+1}=\sum_{i=1}^{N} \sigma_{i, n+1}^{2}+\sum_{i=1}^{N} \sigma_{i, n+1}+2 \sum_{1 \leq i<j} \sigma_{i, n+1} \sigma_{j, n+1}$, 将 (3.8) 式代入 (3.7) 式, 有

$\begin{aligned} & \left\|z_{n}-x_{\beta_{n}}\right\|^{2} \\ \leq& \left(1-(2 \eta-K \varepsilon) \frac{k \beta_{n}}{\lambda_{n}}\right)\left\|\omega_{n}-x_{\beta_{n}}\right\|^{2}+\psi_{n}\left\|\omega_{n}-y_{n-1}\right\|^{2}+\frac{k^{2} \beta_{n}^{2} K^{2}}{\varepsilon \lambda_{n}^{2}} \\ &-\frac{1}{2}\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left\|y_{n}-\omega_{n+1}\right\|^{2}+\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}- \varepsilon\right) \tau_{n+1} \\ &+ \left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n+1}-x_{n}\right\|^{2}. \end{aligned}$

由引理 2.4 可知对于任意 $n \in \mathbb{N}$, $x_{\beta_{n}}$为单点集, 设 $x_{\beta_{n}} \in \mathrm{VI}\left(C, A+\beta_{n}^{w} B+\beta_{n} F\right) \cap {\rm F i x}(T)$, 由 $x_{n+1}$ 的定义有

$\begin{aligned}\left\|x_{n+1}-x_{\beta_{n}}\right\|^{2} & =\left\|\left(1-\theta_{n}\right) \omega_{n}+\theta_{n} T z_{n}-x_{\beta_{n}}\right\|^{2} \\& \leq\left(1-\theta_{n}\right)\left\|\omega_{n}-x_{\beta_{n}}\right\|^{2}+\theta_{n}\left\|z_{n}-x_{\beta_{n}}\right\|^{2}-\left(1-\theta_{n}\right) \theta_{n}\left\|T z_{n}-\omega_{n}\right\|^{2}.\end{aligned}$

另外, 由 $x_{n+1}$ 的构造可知

$T z_{n}-\omega_{n}=\frac{1}{\theta_{n}}\left(x_{n+1}-\omega_{n}\right).$

于是有

$\begin{aligned}\left\|x_{n+1}-\omega_{n}\right\|^{2}& =\left\|x_{n+1}-x_{n}-\alpha_{i, n} \sum_{i=1}^{N}\left(x_{n-i+1}-x_{n-i}\right)\right\|^{2} \\& =\left\|x_{n+1}-x_{n}\right\|^{2}+\left\|\alpha_{i, n} \sum_{i=1}^{N}\left(x_{n-i+1}-x_{n-i}\right)\right\|^{2}\\& -2 \alpha_{i, n} \sum_{i=1}^{N}\left\langle x_{n-i+1}-x_{n-i}, x_{n+1}-x_{n}\right\rangle \\& \geq\left\|x_{n+1}-x_{n}\right\|^{2}+\left\|\alpha_{i, n} \sum_{i=1}^{N}\left(x_{n-i+1}-x_{n-i}\right)\right\|^{2}\\& -2 \alpha_{i, n} \sum_{i=1}^{N}\left\|x_{n-i+1}-x_{n-i}\right\|\left\|\mid x_{n+1}-x_{n}\right\| \\& \geq\left\|x_{n+1}-x_{n}\right\|^{2}+\left\|\alpha_{i, n} \sum_{i=1}^{N}\left(x_{n-i+1}-x_{n-i}\right)\right\|^{2}\\& -2 \sum_{i=1}^{N} \sigma_{i, n}\left\|x_{n+1}-x_{n}\right\|.\end{aligned}$

类似地, 由 $\omega_{n}$ 的定义有

$\begin{aligned}&\left\|y_{n}-\omega_{n+1}\right\|^{2} \\=&\left\|y_{n}-x_{n+1}-\alpha_{i, n+1} \sum_{i=1}^{N}\left(x_{n-i+2}-x_{n-i+1}\right)\right\|^{2} \\\geq&\left\|y_{n}-x_{n+1}\right\|^{2}+\left\|\alpha_{i, n+1} \sum_{i=1}^{N}\left(x_{n-i+2}-x_{n-i+1}\right)\right\|^{2}-2 \sum_{i=1}^{N} \sigma_{i, n+1}\left\|y_{n}-x_{n+1}\right\|.\end{aligned}$

将 (3.11) 式与 (3.12) 式代入 (3.10) 式, 有

$\begin{aligned}&\left\|x_{n+1}-x_{\beta_n}\right\|^2 \\\leq&\left(1-\theta_n\right)\left\|\omega_n-x_{\beta_n}\right\|^2+\theta_n\left\|z_n-x_{\beta_n}\right\|^2-\frac{\left(1-\theta_n\right)}{\theta_n}\left\|x_{n+1}-\omega_n\right\|^2 \\\leq& \left(1-\theta_n\right)\left\|\omega_n-x_{\beta_n}\right\|^2+\theta_n\left\|z_n-x_{\beta_n}\right\|^2-\frac{\left(1-\theta_n\right)}{\theta_n}\left(1-\frac{\varepsilon k \beta_n}{\lambda_n}\right)\left\|x_{n+1}-x_n\right\|^2\\&+\frac{\lambda_n\left(1-\theta_n\right)}{\theta_n} \frac{\sum_{i=1}^N \sigma_{i, n}^2}{\varepsilon k \beta_n}.\end{aligned}$

$\omega_{n}$ 的定义与均值不等式可得

$\begin{aligned}& \left\|\omega_{n}-x_{\beta_{n}}\right\|^{2} \\\leq & \left(\left\|x_{n}-x_{\beta_{n}}\right\|+\alpha_{i, n}\left\|\sum_{i=1}^{N}\left(x_{n-i+1}-x_{n-i}\right)\right\|\right)^{2} \\\leq & \left\|x_{n}-x_{\beta_{n}}\right\|^{2}+\sum_{i=1}^{n} \sigma_{i, n}^{2}+\left(\left\|x_{n}-x_{\beta_{n}}\right\|^{2}+1\right) \sum_{i=1}^{n} \sigma_{i, n}+2 \sum_{1 \leq i<j} \sigma_{i, n} \sigma_{j, n} \\\leq & \left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n}-x_{\beta_{n}}\right\|^{2}+\tau_{n}.\end{aligned}$

类似地, 由均值不等式可得

$\left\|\omega_{n}-y_{n-1}\right\|^{2} \leq\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n}-y_{n-1}\right\|^{2}+\tau_{n}.$

将 (3.9) 式, (3.13) 式, (3.15) 式与 (3.16) 式代入 (3.14) 式, 可得

$\begin{aligned} & \left\|x_{n+1}-x_{\beta_{n}}\right\|^{2} \\ \leq&\left(1-\theta_{n}\right)\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n}-x_{\beta_{n}}\right\|^{2}+\left(1-(2 \eta-K \varepsilon) \frac{k \beta_{n}}{\lambda_{n}}\right) \theta_{n}\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n}-x_{\beta_{n}}\right\|^{2} \\ &+\psi_{n} \theta_{n}\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n}-y_{n-1}\right\|^{2} \\ &-\frac{1}{2}\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left(1-\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \theta_{n}\left\|y_{n}-x_{n+1}\right\|^{2} \\ &-\left(\frac{\left(1-\theta_{n}\right)}{\theta_{n}}\left(1\!-\!\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\!-\!\left(1\!-\!(1+\sqrt{2}) \psi_{n}\!-\!\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}\!-\!\varepsilon\right)\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\right)\left\|x_{n+1}\!-\!x_{n}\right\|^{2}+\phi_{n} \\ \leq&\left(1-\left((2 \eta-K \varepsilon) \theta_{n}-\varepsilon\right) \frac{k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n}-x_{\beta_{n}}\right\|^{2}+\psi_{n} \theta_{n}\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n}-y_{n-1}\right\|^{2} \\ &-\frac{1}{2}\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left(1-\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \theta_{n}\left\|y_{n}-x_{n+1}\right\|^{2}+\phi_{n}, \end{aligned}$

其中 $\phi_{n}=\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right) \theta_{n} \tau_{n+1}+\left(1-(2 \eta-K \varepsilon) \frac{k \beta_{n}}{\lambda_{n}}\right) \theta_{n} \tau_{n}+\left(1-\theta_{n}\right) \tau_{n}+\psi_{n} \theta_{n} \tau_{n}+\frac{k^{2} \beta_{n}^{2} K^{2} \theta_{n}}{\varepsilon \lambda_{n}^{2}}+\frac{\lambda_{n}\left(1-\theta_{n}\right)}{\theta_{n}} \frac{\sum_{i=1}^{N} \sigma_{i, n}^{2}}{\varepsilon k \beta_{n}}+\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right) \theta_{n} \lambda_{n} \sum_{i=1}^{N} \sigma_{i, n}^{2}$, 由 $\left\{\beta_{n}\right\}$ 的条件可得 $\lim _{n \rightarrow \infty} \frac{\phi_{n}}{\beta_{n}}=0, \sum_{n=1}^{\infty} \phi_{n}=\infty$. 由引理 3.2 和均值不等式可得

$\begin{aligned}& \left\|x_{n+1}-x_{\beta_{n+1}}\right\|^{2} \\\leq & \left\|x_{n+1}-x_{\beta_{n}}\right\|^{2}+\left\|x_{\beta_{n+1}}-x_{\beta_{n}}\right\|^{2}+2\left\|x_{n+1}-x_{\beta_{n}}\right\|\left\|x_{\beta_{n+1}}-x_{\beta_{n}}\right\| \\\leq & \left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n+1}-x_{\beta_{n}}\right\|^{2}+\frac{\lambda_{n}+\varepsilon k \beta_{n}}{\varepsilon k \beta_{n}}\left\|x_{\beta_{n+1}}-x_{\beta_{n}}\right\|^{2} \\= & \left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n+1}-x_{\beta_{n}}\right\|^{2}+\frac{M^{2}\left(\lambda_{n}+\varepsilon k \beta_{n}\right)\left(\beta_{n+1}-\beta_{n}\right)^{2}}{\varepsilon k \beta_{n}^{3} \beta_{n+1}^{2}},\end{aligned}$

其中 $M=\frac{1}{\eta}\left[\left(1+\frac{K}{\eta}+\frac{1}{v \eta}\right)\left\|F x^{*}\right\|+\left\|B x^{*}\right\|+\left(2 K+\frac{2}{v}\right)\left\|x^{*}\right\|+K\right]$. 将 (3.17) 式代入 (3.18) 式有

$\begin{aligned} & \left\|x_{n+1}-x_{\beta_{n+1}}\right\|^{2} \\ \leq&\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left(1-\left((2 \eta-K \varepsilon) \theta_{n}-\varepsilon\right) \frac{k}{\lambda_{n}} \beta_{n}\right)\left\|x_{n}-x_{\beta_{n}}\right\|^{2} \\ &+\psi_{n} \theta_{n}\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)^{2}\left\|x_{n}-y_{n-1}\right\|^{2} \\ &-\frac{1}{2}\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left(1-\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \theta_{n}\left\|y_{n}-x_{n+1}\right\|^{2} \\ &+\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \phi_{n}+\frac{M^{2}\left(\lambda_{n}+\varepsilon k \beta_{n}\right)\left(\beta_{n+1}-\beta_{n}\right)^{2}}{\varepsilon k \beta_{n}^{3} \beta_{n+1}^{2}} \\ \leq&\left(1-\left((2 \eta-K \varepsilon) \theta_{n}-2 \varepsilon\right) \frac{k \beta_{n}}{\lambda_{n}}\right)\left\|x_{n}-x_{\beta_{n}}\right\|^{2}+\psi_{n} \theta_{n}\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)^{2}\left\|x_{n}-y_{n-1}\right\|^{2} \\ &-\frac{1}{2}\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left(1-\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \theta_{n}\left\|y_{n}-x_{n+1}\right\|^{2} \\ &+\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \phi_{n}+\frac{M^{2}\left(\lambda_{n}+\varepsilon k \beta_{n}\right)\left(\beta_{n+1}-\beta_{n}\right)^{2}}{\varepsilon k \beta_{n}^{3} \beta_{n+1}^{2}}. \end{aligned}$

$\varsigma_n=\left((2 \eta-K \varepsilon) \theta_n-2 \varepsilon\right) \frac{k \beta_n}{\lambda_n}$, 由引理 3.1 和 $\left\{\beta_n\right\}$ 的定义可得 $\lim _{n \rightarrow \infty} \varsigma_n=0, \sum_{n=1}^{\infty} \varsigma_n=\infty$.

$\Phi_{n}=\left\|x_{n}-x_{\beta_{n}}\right\|^{2}+\frac{\psi_{n} \theta_{n}}{1-\varsigma_{n}}\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)^{2}\left\|x_{n}-y_{n-1}\right\|^{2}.$

由 (3.19) 式, 有

$\begin{aligned} \Phi_{n+1} \leq&\left(1-\varsigma_n\right) \Phi_n+\left(1+\frac{\varepsilon k \beta_n}{\lambda_n}\right) \phi_n+\frac{M^2\left(\lambda_n+\varepsilon k \beta_n\right)\left(\beta_{n+1}-\beta_n\right)^2}{\varepsilon k \beta_n^3 \beta_{n+1}^2} \\ & -\frac{1}{2}\left(1-(1+\sqrt{2}) \psi_n-\frac{k \beta_n K}{\varepsilon \lambda_n}-\varepsilon\right)\left(1-\frac{\varepsilon k \beta_n}{\lambda_n}\right)\left(1+\frac{\varepsilon k \beta_n}{\lambda_n}\right) \theta_n\left\|x_{n+1}-y_n\right\|^2 \\ & +\frac{\psi_{n+1} \theta_{n+1}}{1-\varsigma_{n+1}}\left(1+\frac{\varepsilon k \beta_{n+1}}{\lambda_{n+1}}\right)^2\left\|x_{n+1}-y_n\right\|^2. \end{aligned}$

根据引理 3.1 和 $\left\{\beta_{n}\right\}$ 的定义可得

$\begin{align*} &\lim _{n \rightarrow \infty} \frac{1}{2}\left(1-(1+\sqrt{2}) \psi_{n}-\frac{k \beta_{n} K}{\varepsilon \lambda_{n}}-\varepsilon\right)\left(1-\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right)\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \theta_{n}\\ &-\frac{\psi_{n+1} \theta_{n+1}}{1-\varsigma_{n+1}}\left(1+\frac{\varepsilon k \beta_{n+1}}{\lambda_{n+1}}\right)^{2} \\& = \frac{1}{2}(1-(3+\sqrt{2}) \mu k-\varepsilon) \theta_{n}>0. \end{align*}$

于是依据极限的保号性当 $n \geq n_{0}$ 时有

$\begin{aligned} &\frac{1}{2}\left(1-(1+\sqrt{2}) \psi_n-\frac{k \beta_n K}{\varepsilon \lambda_n}-\varepsilon\right)\left(1-\frac{\varepsilon k \beta_n}{\lambda_n}\right)\left(1+\frac{\varepsilon k \beta_n}{\lambda_n}\right) \theta_n \\ &-\frac{\psi_{n+1} \theta_{n+1}}{1-\varsigma_{n+1}}\left(1+\frac{\varepsilon k \beta_{n+1}}{\lambda_{n+1}}\right)^2>0. \end{aligned}$

将 (3.21) 式代入 (3.20) 式有

$\begin{aligned}\Phi_{n+1}& \leq\left(1-\varsigma_{n}\right) \Phi_{n}+\varsigma_{n}\left(\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \frac{\phi_{n}}{\varsigma_{n}}+\frac{M^{2}\left(\lambda_{n}+\varepsilon k \beta_{n}\right)\left(\beta_{n+1}-\beta_{n}\right)^{2}}{\left((2 \eta-K \varepsilon) \theta_{n}-2 \varepsilon\right) \varepsilon k \lambda_{n} \beta_{n}^{4} \beta_{n+1}^{2}}\right) \\& \leq\left(1-\varsigma_{n}\right) \Phi_{n}+\varsigma_{n} \chi_{n},\end{aligned}$

其中 $\chi_{n}=\left(1+\frac{\varepsilon k \beta_{n}}{\lambda_{n}}\right) \frac{\phi_{n}}{\varsigma_{n}}+\frac{M^{2}\left(\lambda_{n}+\varepsilon k \beta_{n}\right)\left(\beta_{n+1}-\beta_{n}\right)^{2}}{\left((2 \eta-K \varepsilon) \theta_{n}-2 \varepsilon\right) \varepsilon k \lambda_{n} \beta_{n}^{4} \beta_{n+1}^{2}}$. 由引理 3.1 和 $\left\{\beta_{n}\right\}$ 的定义得 $\lim _{n \rightarrow \infty} \chi_{n}$$=0$, 由引理 2.5 可知 $\lim _{n \rightarrow \infty} \Phi_{n}=0$. 即有 $\lim _{n \rightarrow \infty}\left\|x_{n}-x_{\beta_{n}}\right\|=0 ; \lim _{n \rightarrow \infty}\left\|x_{n}-y_{n-1}\right\|=0$. 由引理 3.3 有 $\lim _{\beta \rightarrow 0} x_{\beta}=x^{*}$, 所以 $x_{n} \rightarrow x^{*}$, 由 (3.15) 式可得 $\lim _{n \rightarrow \infty}\left\|\omega_{n}-x_{\beta_{n}}\right\|=0, \omega_{n} \rightarrow x^{*}$, 由 (3.9) 式可得 $\lim _{n \rightarrow \infty}\left\|z_{n}-x_{\beta_{n}}\right\|=0, z_{n} \rightarrow x^{*}$, 由 $x_{n+1}$ 的构造可知

$\left(1-\theta_{n}\right)\left\|x_{n+1}-\omega_{n}\right\|=\theta_{n}\left\|T z_{n}-x_{n+1}\right\|.$

所以 $\lim _{n \rightarrow \infty}\left\|T z_{n}-x_{n+1}\right\|=0$. 由此可得

$\lim _{n \rightarrow \infty}\left\|T z_{n}-z_{n}\right\| \leq \lim _{n \rightarrow \infty}\left(\left\|T z_{n}-x_{n+1}\right\|+\left\|x_{n+1}-x_{\beta_{n}}\right\|+\left\|x_{\beta_{n}}-z_{n}\right\|\right)=0.$

故由 $T$ 是闭的拟非扩张映射与引理 2.2 可得 $x^{*} \in \operatorname{Fix}(T)$, 由 (1.3) 式有 $x^{*} \in \operatorname{VI}(C, A) \cap \operatorname{Zer}(B)$, 即 $x^{*} \in \operatorname{VI}(C, A) \cap F(T) \cap \operatorname{Zer}(B)$.

注 3.1 本文所得结果从下列几个方面对文献 [12] 进行了改进

1) 本文在文献 [12] 的基础上与拟非扩张映射不动点问题结合;

2) 将本文算法 1 与文献 [12] 的算法相比, 将最小值自适应步长改为最大值自适应步长并加入变量 $k$, 改进了固定的自适应步长并使迭代序列收敛速度加快;

3) 本文在与拟非扩张映射不动点结合的证明过程中去掉了文献 [7] 中映像满足次闭原理的条件.

4 数值实验

本节将通过两个例子说明算法 1 的有效性. 所有数值实验均在 MATLAB-R2022a 和 Windows11 中运行, 用 $\left\|x_{n}-x^{*}\right\|$ 测量第 $n$ 步迭代的误差, 当 $\left\|x_{n}-x^{*}\right\| \leq $ eps 时迭代停止.

${\bf 例 4.1}$$C=[-2,5], H=\mathbb{R}, A: \mathbb{R} \rightarrow \mathbb{R}, B: \mathbb{R} \rightarrow \mathbb{R}$, 满足

$\begin{gathered}Ax=x+\sin x, Bx=x-\sin x, \quad F=I, T x=x / 99\end{gathered}$

其中 $I$ 为恒等映射, 易知算子 $A$ 为单调 $L$-Lipschitz 映射, 算子 $F$ 为强单调映射, 算子 $B$ 为逆强单调映射, 算子 $T$ 为拟非扩张算子. 可以得出 $\mathrm{VI}(C, A) \cap \operatorname{Zer}(B) \cap F(T)=\{0\}$, 即 $x^{*}=0$.

我们比较本文的算法 1, 文献 [算法 4], 文献 [算法 4]以及文献 [算法 2], 选取共同的参数 $x_{0}=x_{1}=1 \quad \lambda_{1}=0.3, \sigma_{n}=n^{-2}, \beta_{n}=n^{-1 / 3}, \mu=0.3, \alpha_{i}=0.08$. 本文的算法 1 中选取 $\omega=0.005, k=0.2, \theta_{n}=0.33$, 文献 [算法 4] 选取 $\omega=0.005$, 文献 [算法 2] 选取 $\gamma_{n}=1 / n, \omega=0.005$. 在该条件下, 当 $N=1$ 时, 取 eps $=10^{-8}$. 实验结果如表1图1 所示. 由表1图1 可以看出本文的算法迭代步数更少, 速度更快.

表1   例 4.1 不同 eps 值下五种算法的比较

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图1

图1   例 4.1 中五种算法的比较


${\bf 例 4.2}$$H=L^{2}[0,1]$, 分别定义内积和范数为

$\begin{align*} \langle x, y\rangle=\int_{0}^{1} x(t) y(t) {\rm d} t, \forall x, y \in L^{2}[0,1]; \quad \|x\|=\sqrt{\int_{0}^{1}(x(t))^{2} {\rm d} t}, \forall x \in L^{2}[0,1]. \end{align*}$

$\begin{align*} &C=\left\{x \in L^{2}[0,1]:\|x\| \leq 1\right\}. \\[3mm] &A: L^{2}[0,1] \rightarrow L^{2}[0,1], \\[3mm] &F: L^{2}[0,1] \rightarrow L^{2}[0,1], \end{align*}$

满足

$\begin{align*} &(A x)(t)=\max \{x(t), 0\}, B=I / 3, \\[3mm] &(F x)(t)=x(t)+\max \{x(t), 0\}, T x(t)=x(t) / 99. \end{align*}$

这里 $A$ 为单调且 Lipschitz 映射, $F$ 为强单调映射, 算子 $B$ 为逆强单调映射, 算子 $T$ 为拟非扩张算子. 容易得出 $\mathrm{VI}(C, A) \cap \operatorname{Zer}(B) \cap F(T)=\{0\}$, 即 $x^{*}=0$.

我们比较本文的算法 1 与文献 [算法 2], 选取共同的参数 $x_{0}=x_{1}=1$, $\lambda_{1}=0.2, \sigma_{n}=n^{-2}, \beta_{n}=n^{-1 / 3}, \mu=0.1, \omega=0.005, \alpha_{i}=0.08$. 本文的算法 1 中选取 $k=0.2, \theta_{n}=0.33$. 分别选取 $x_{1}(t)=t^{3}, {\rm e}^{t}$ 并当 eps $=10^{-4}$ 时, 实验结果见图2图3. 由图2图3 可以看出本文的算法更优.

图2

图2   例 4.2 $x_{1}(t)=t^{3}$ 时两算法的对比图


图3

图3   例 4.2 $x_{1}(t)={\rm e}^{t}$ 时两算法的对比图


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