数学物理学报, 2024, 44(4): 1052-1065

基于非 Lipschitz 步长策略的临近分裂可行问题的强收敛性研究

马小军1, 陈富1, 贾芝福,2,*

1山西大同大学 山西大同 037009

2宿迁学院 江苏宿迁 223800

Research on a Strong Convergence Theorem for Proximal Split Feasibility Problems with Non-Lipschitz Stepsizes

Ma Xiaojun1, Chen Fu1, Jia Zhifu,2,*

1School of Mathematics and Statistics, Shanxi Datong University, Shanxi Datong 037009

2Suqian College, Jiangsu Suqian 223800

通讯作者: *贾芝福, E-mail: fzhuangmaxj@163.com

收稿日期: 2023-07-20   修回日期: 2024-02-25  

基金资助: 山西大同大学人才引进科研启动(2023-B-06)
山西大同大学人才引进科研启动(202303021222208)
宿迁市科技计划项目(K202332)
国家自然科学基金(12172266)
国家自然科学基金(61803241)

Received: 2023-07-20   Revised: 2024-02-25  

Fund supported: Startup Foundation for Newly Recruited Employee(2023-B-06)
Startup Foundation for Newly Recruited Employee(202303021222208)
Suqian Sci Tech Program(K202332)
National Natural Science Foundation of China(12172266)
National Natural Science Foundation of China(61803241)

摘要

针对 Hilbert 空间中的临近分裂可行问题, 该文提出了一种惯性粘滞类算法. 其中主要引入了一种非 Lipschitz 步长策略, 其克服了原步长远离零的缺点. 另外, 通过弱化临近映射的完全非扩张性, 证明了修正后算法的强收敛性. 进一步, 将所得的结论应用于分裂均衡问题. 最后, 列举实例充分说明了修正后算法的有效性.

关键词: 临近分裂可行问题; 分裂均衡问题; 非 Lipschitz 连续映射; 粘滞类算法; 强收敛性

Abstract

In this paper, An inertial viscosity-type algorithm is proposed to solve proximal split feasibility problems in Hilbert spaces. In this algorithm, a non-Lipschitz stepsize rule is given, which overcomes the drawback that the stepsize tends to zero. Further, a strong convergence theorem for our proposed algorithm is established without Lipschitz continuity of the gradient operators. As theoretical applications, the split equilibrium problem is investigated. Finally, numerical experiments are provided for demonstration and comparison.

Keywords: Proximal split feasibility problem; Split equilibrium problem; Non-Lipschitz continuity; Viscosity-type algorithm; Strong convergence

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本文引用格式

马小军, 陈富, 贾芝福. 基于非 Lipschitz 步长策略的临近分裂可行问题的强收敛性研究[J]. 数学物理学报, 2024, 44(4): 1052-1065

Ma Xiaojun, Chen Fu, Jia Zhifu. Research on a Strong Convergence Theorem for Proximal Split Feasibility Problems with Non-Lipschitz Stepsizes[J]. Acta Mathematica Scientia, 2024, 44(4): 1052-1065

1 引言

分裂可行问题 (Split Feasibility Problem, 记作 SFP) 是最优化领域的一个研究热点, 并在 LASSO 问题、信号处理、图像去噪问题等领域有着广泛的应用[2,9,10,16,19,27,32,37]. 因此, 该问题的理论算法及应用得到了广泛的关注和研究[3,4,6,11,14,17,20,25,27,28,31]. 1994 年, Censor 和 Elfving[5] 首次介绍了分裂可行问题模型, 其用于模拟反问题. 此后, 文献 [3,4] 建立了求解该模型的经典算法称之为 Byrne CQ 算法, 并将其推广到求解分裂可行问题更广义的临近分裂可行问题[22] (Proximal Split Feasibility Problem, 记作 PSFP). 针对 PSFP, 文献 [1,22,29-31,35] 给出了相关的算法及其收敛性的推证. 即展开来讲, Moudafi 和 Thakur[22]构造了无需 Armijo 搜索[12,13,18,24,28,34] (该搜索常常计算额外的临近算子) 的自适应步长策略, 并讨论了分裂临近算法的弱收敛性. 基于惯性思想 (其用于加速算法的收敛性[2,16,20,27,28,32]), Shehu 等[31]修正了 Moudafi 等人的算法, 并证明了其弱收敛性. 为了研究其强收敛性, 众多学者给出了经典的强收敛算法, 例如, Abbas 等[1]设计了两种不同的一步迭代算法;随后, Shehu 等[29,30]结合 Mann-type, 加速混合粘滞类算法[21], 并提出了最速下降法; Wang 和 Xu[35]提出了分裂临近梯度法; Ma 和 Liu[20]构造了一种惯性分裂临近算法. 以上算法存在一个共同点: 在单调算子的 Lipschitz 连续的条件下, 步长趋近于零, 导致算法的效率较差. 此外, 算法的强收敛性证明需要较强的条件如临近算子的完全非扩张性. 针对这些因素, 本文提出了如下问题.

如何在算法[31]的基础上, 提出一个有界远离零的步长策略, 并通过弱化临近算子的完全非扩张性和单调算子的 Lipschitz 连续性来获得算法的强收敛性?

受文献 [20,31] 的启发, 本文给出了一种非 Lipschitz 步长策略, 并结合惯性技巧、粘滞类算法及分裂临近算法, 提出了一种惯性粘滞类算法, 在无需单调算子的 Lipschitz 连续性和临近映射的完全非扩张性的条件下, 证明所修正算法的强收敛性. 在算法的应用方面, 主要考虑了分裂均衡问题. 最后数值结果验证了算法的有效性.

文章的结构如下. 第 2 节给出了一些基本概念和结论; 第 3 节给出了本文的主要结论; 第 4 节考虑了算法的应用; 第 5 节提供了数值实验.

2 预备知识

符号“$\rightharpoonup$”代表弱收敛性, “$\to$”代表强收敛性. 令 $H_{1}$$H_{2}$ 是实 Hilbert 空间. 对于真、弱下半连续函数 $F:H_{1}\to \left] {-\infty, \infty} \right]$, 其有效域记作 $\rm{dom}F$, 即, ${\rm{dom}F}:=\{x\in H_{1}: F(x)<\infty\}$. 临近算子 ${\rm{prox}}_{\tau G}: H_{2} \to H_{2}$定义为

${\rm{pro}}{{\rm{x}}_{\tau G}}\left( x \right)\mathop { := \arg \min }\limits_{y \in H_{2}} \left\{ {G\left( y \right) + \frac{1}{{2\tau }}{{\left\| {x -y } \right\|}^2}} \right\}, $

其中 $\tau>0$, $G:H_{2}\to \mathbb{R}\cup \left\{ { + \infty } \right\}$是真凸下半连续函数 (lsc). 文献 [9] 证明了临近映射 $\rm{prox}_{\tau G}$ 是完全非扩张性. 令 $C\subset H_{1}$ 为非空闭凸集, 则 $x$$C$ 上的正交投影定义为

$P_{C}x\mathop { := \arg \min }\left\{\left\| {x -y } \right\| | y \in C \right\}, \forall x\in H_{1}.$

另外, 示性函数定义为

$\begin{equation*}\ \delta_{C}(x)= \begin{cases} 0, & \text{若}\ x\in C, \\ \infty, & \text{否则}. \\ \end{cases} \end{equation*}$

在其作用下, 临近算子 ${\rm{prox}}_{\tau}$ 等价于正交投影算子 $P_{C}$, 详细内容参见文献 [7,33].

定义 2.1$h:H_{1}\to H_{1}$ 为一映射, 若

(i) $h$ 称之为非扩张性, 则即有

$\|hx-hy\|\leq\|x-y\|, \forall x, y \in H_{1}.$

(ii) $h$ 称之为完全非扩张性, 则即有

$\langle hx-hy, x-y\rangle\geq\|hx-hy\|^2, \forall x, y \in H_{1}.$

(iii) 令 $D\subset H_{1}$ 为一集合, $h:D\to \mathbb{R}\cup \left\{ { + \infty }\right\}$ 称之为弱下半连续函数, 如果$x_{n}\rightharpoonup x, $ 蕴含如下关系:

$\mathop {\liminf}\limits_{n \to \infty }h(x_{n})\geq h(x). $

引理 2.1$\nu \in \left[ {0, 1} \right]$, 对所有的 $x, y, z\in H_{1}$, 则

(i) $\|\nu x+(1-\nu)y\|^2=\nu\|x\|^2+(1-\nu)\|y\|^2-\nu(1-\nu)\|x-y\|^2;$

(ii) $\|x+y\|^2\leq \|x\|^2+2\langle y, x+y\rangle;$

(iii) $\langle x-y, x-z\rangle=\frac{1}{2}\|x-y\|^2+\frac{1}{2}\|x-z\|^2-\frac{1}{2}\|y-z\|^2. $

引理 2.2[15]$C \subset H_{1}$ 为非空闭凸集, 则 $H_{1}$$C$上的投影 $P_{C}$性质概述如下

(i) $\langle x-P_{C}x, y-P_{C}x\rangle\leq 0 \forall x \in H_{1}, y \in C;$

(ii) $\| P_{C}x-P_{C}y\|\leq \|x-y\| \forall x, y \in H_{1};$

引理 2.3[26]$\{s_{n} \}_{n=1}^{\infty}$ 是一个非负实数序列, 使得

$\begin{array}{*{20}{l}} \begin{array}{l} \quad \;s_{n+1}\leq (1-\alpha_{n})s_{n}+\alpha_{n}\delta_{n}, \forall n\geq 1, \end{array} \end{array}$

其中

(i) $\{ \alpha_{n} \}_{n=1}^{\infty}\subset\left( {0, 1} \right)$, 并满足 $ \sum\limits_{n = 1}^\infty { \alpha_{n}=\infty }$,

(ii) 若对于序列 $\{n\}$ 的子列 $\{n_{k}\}$, 都有 $\mathop {\limsup}\limits_{k\rightarrow\infty}\delta_{n_{k}}\leq0$. 另外, 序列 $\{s_{n}\}$ 的子列 $\{s_{n_{k}}\}$ 满足

$\mathop {\liminf}\limits_{k\to\infty}\|s_{n_{k}+1}-s_{n_{k}}\|\geq0$.

$ \mathop {\lim}\limits_{n\to\infty} s_{n}=0. $

引理 2.4[23]$\{ \varphi_{n}\}$ 是一个非负实数序列, 并满足

$ \varphi_{n+1}\leq \phi_{n}\varphi_{n}+\psi_{n}, \forall n \in \mathbb{N},$

其中 $\{\phi_{n}\}$$\{\psi_{n}\}$ 是非负实数列, 并满足$\{\phi_{n}\} \subset [1, +\infty), $$\sum\limits_{n = 1}^\infty ({\phi_{n}-1)<\infty }$$\sum\limits_{n = 1}^\infty {\psi_{n}<\infty }$.$ \underset{n\rightarrow \infty}\lim \varphi_{n}$存在.

3 惯性粘滞类算法及其强收敛性

在本节, 首先, 令 $H_{1}$$H_{2}$ 为实 Hilbert 空间, $A:H_{1}\rightarrow H_{2}$ 是有界线性算子及其共轭算子为 $A^*, $$F:H_{1}\to \mathbb{R}\cup \left\{ { + \infty } \right\}$$G:H_{2}\to \mathbb{R}\cup \left\{ { + \infty } \right\}$ 是真凸下半连续函数.

现在, 本节引入如下 PSFP[22]

$ \begin{array}{l} \mbox{寻找一点} z^* \in H_{1} \mbox{使得} \mathop {\min }\left\{F(x)+G_{\tau}(Ax):x\in H_{1}\right\}, \end{array} $

其中 $\tau>0$$G_{\tau}(x):=\mathop {\min }\limits_{y \in H_{2}}\left\{ {G\left( y\right) + \frac{1}{{2\tau }}{{\left\| {y -x } \right\|}^2}} \right\}$ 看作参数 $\tau$ 和函数 $G$ 的 Moreau-Yosida 近似函数. 若点$z^{*}$存在, PSFP 的解集记作 $\Gamma$. 接下来, 文献 [1,22,31] 给出了本节需要的定义.

任给 $\tau>0$$x\in H_{1}$, 定义

$\begin{align*} &E(x)=\frac{1}{2}\|(I-\mbox{prox}_{\tau G})Ax\|^2;\\ &L(x)=\frac{1}{2}\|(I-\mbox{prox}_{\tau F})x\|^2;\\ &\theta(x)=\sqrt{\|\nabla E(x)+\nabla L(x)\|^2}. \end{align*}$

则上述函数的 Lipschitz 梯度分别为

$\nabla E(x)=A^*(I-{\rm{prox}}_{\tau G})Ax {\mbox{和}} \nabla L(x)=(I-{\rm{prox}}_{\tau F})x,$

其 Lipschitz 常数分别为 $\|A\|^2$$1$, $I$ 是恒等映射. 在阐述算法之前, 本节首先给出了收敛性分析需要的条件.

(A1) 问题 PSFP 的解集是非空的, 即 $\Gamma\neq \emptyset. $

(A2) 令$ \{\tilde{\tau}_{n}\} \subset \big[{0, \tilde{\theta}}\big)$$\tilde{\theta}>0$ 是正的序列并满足 $ \tilde{\tau}_{n}=o(\gamma_{n})$, 即, $\mathop {\lim }\limits_{n\to\infty} \frac{\tilde{\tau}_{n}}{\gamma_{n}}=0$, 其中序列 $\{\gamma_{n}\}\subset\left({0, 1}\right)$ 满足$\sum_{n=1}^{\infty} \gamma_{n}=\infty, \lim _{n \rightarrow \infty} \gamma_{n}=0$

(A3) 映射 $f:H_{1}\to H_{1}$$\tilde{\rho}$-压缩的及其常数为 $\tilde{\rho}\in \left[{0, 1}\right). $

本节的算法框架如下:

算法 1

步骤 0 固定 $x_{0}, x_{1}\in H_{1}$. 选择序列 $\{\sigma_{n}\}\subset\left[{0, \sigma}\right)\subset\left[{0, 1}\right)$$0<\gamma_{n}<1$.

步骤 1 给定 $x_{n-1}$, $x_{n} (n\geq 1)$ 并计算

$\begin{equation} \begin{array}{l} w_{n}=x_{n}+\sigma_{n}\left(x_{n}-x_{n-1}\right), \\ y_{n}=w_{n}-\lambda_{n}(\nabla E(w_{n})+\nabla L(w_{n})), \\ x_{n+1}=\gamma_{n}f(x_{n})+(1-\gamma_{n})y_{n}, \end{array} \end{equation}$

其中步长 $\lambda_{n}$ 更新如下

$\begin{equation} \ \lambda_{n+1}= \begin{cases} \mbox{min}\left\{{\min\left\{\frac{\delta L(w_{n})}{ \|\nabla L(w_{n})\|^2}, \frac{\delta E(w_{n})}{ \|\nabla E(w_{n})\|^2}\right\}, \phi_{n}\lambda_{n}+\psi_{n} }\right\}, \\[1mm] \hskip 4cm \mbox{若}\ \nabla E(w_{n})\neq0 \mbox{和} \nabla L(w_{n})\neq0, \\ \phi_{n}\lambda_{n}+\psi_{n},\hskip 2cm \ \, \mbox{否则}, \end{cases} \end{equation}$

其中 $\theta(w_{n})=\sqrt{\|\nabla E(w_{n})+\nabla L(w_{n})\|^2}$.

步骤 2$\nabla E(w_{n})=\nabla L(w_{n})=0$$w_{n}=x_{n}$, 则停止, $x_{n}$ 是 PSFP 的解. 否则, 设 $n: = n + 1$并返回步骤 1.

注 3.1$\Gamma\neq \emptyset$, $\nabla E(w_{n})=\nabla L(w_{n})=0$$w_{n}=x_{n}$, 则 $x_{n}\in \Gamma$. 这是由于 $A^*(I-\mbox{prox}_{\tau G})Aw_{n}=(I-\mbox{prox}_{\tau F})w_{n}=0$, 进一步推证 $w_{n}\in \Gamma$. 此外, $A^*(I-\mbox{prox}_{\tau G})Aw_{n}=(I-\mbox{prox}_{\tau F})w_{n}=0$ 结合算法 1 的 (3.1) 式可推出 $y_{n}=w_{n}. $ 再由 $w_{n}=x_{n} $ 可得 $y_{n}=x_{n} \in \Gamma$. 因此, $x_{n}\in \Gamma. $

注 3.2 在算法 1 中, 惯性参数 $\sigma_{n}$ 的选取方式如下

$\begin{equation} \sigma_{n}= \begin{cases} \min\left\{{\frac{\tilde{\tau}_{n}}{\|x_{n}-x_{n-1}\|}, \sigma }\right\}, & \mbox{若}\ x_{n}\neq x_{n-1}, \\[1mm] \sigma, &{\mbox{否则}. } \end{cases} \end{equation}$

接下来, 本节的引理和定理的证明无需映射 $I-\mbox{prox}_{\tau}$ 的完全非扩张性.

引理 3.1 令步长序列 $\{\lambda_{n}\}$ 是由算法 1 生成的. 则有

$\underset{n\rightarrow\infty}\lim \lambda_{n}=\lambda \mbox{和} \lambda \in \left[\min\left\{\min\left\{\frac{\delta}{2\|A^*A\|}, \frac{\delta}{2}\right\}, \lambda_{1}\right\}, \infty\right).$

假设情形 $\nabla E(w_{n})\neq0$$\nabla L(w_{n})\neq0$, 则有

$\begin{align*} \|\nabla E(w_{n})\|^2&=\left\langle\nabla E(w_{n}), \nabla E(w_{n})\right\rangle\\ &= \left\langle A^*\left(I-\mbox{prox}_{\tau G}\right)Aw_{n}, A^*(I-\mbox{prox}_{\tau G})Aw_{n}\right\rangle\\ &= \langle (I-\mbox{prox}_{\tau G})Aw_{n}, AA^*(I-\mbox{prox}_{\tau G})Aw_{n}\rangle\\ & \leq \|A^*A\|\langle (I-\mbox{prox}_{\tau G})Aw_{n}, (I-\mbox{prox}_{\tau G})Aw_{n}\rangle\\ &\leq \|A^*A\|\left\|\left(I-\mbox{prox}_{\tau G}\right)Aw_{n}\right\|^2\\ &= 2\|A^*A\|E(w_{n}), \end{align*}$

进一步推出

$\frac{\delta E(w_{n})}{\|\nabla E(w_{n})\|^2} \geq\frac{\delta}{2\|A^*A\|}.$

同理, 可得

$ \|\nabla L(w_{n})\|^2=2L(w_{n}) \mbox{和} \frac{\delta L(w_{n})}{\|\nabla L(w_{n})\|^2}=\frac{\delta}{2}.$

上式整理后, 即得

$\min\left\{\frac{\delta E(w_{n})}{\|\nabla E(w_{n})\|^2}, \frac{\delta L(w_{n})}{\|\nabla L(w_{n})\|^2}\right\}\geq \min\left\{\frac{\delta}{2\|A^*A\|}, \frac{\delta}{2}\right\}.$

由数学归纳法知, 序列 $\{\lambda_{n}\}$ 有下界 $\mbox{min}\left\{\min\left\{\frac{\delta}{2\|A^*A\|}, \frac{\delta}{2}\right\}, \lambda_{1}\right\}$. 根据 $\lambda_{n+1}$ 的定义, 则有

$\begin{equation} \nonumber \lambda_{n+1}\leq \phi_{n}\lambda_{n}+\psi_{n}. \end{equation}$

由引理 2.4 知, $\underset{n\rightarrow\infty}\lim \lambda_{n}$ 存在, 并记作

$\underset{n\rightarrow\infty}\lim \lambda_{n}=\lambda.$

由于序列 $\{\lambda_{n}\}$ 有下界 $\mbox{min}\left\{\min\left\{\frac{\delta}{2\|A^*A\|}, \frac{\delta}{2}\right\}, \lambda_{1}\right\}$, 则 $\lambda>0$.

现在, 在无需梯度算子 $\nabla E$, $\nabla L$ 的 Lipschitz 连续性和映射 $I-\mbox{prox}_{\tau}$ 完全非扩张性的条件下, 证明本节的主要定理.

定理 3.1 假设条件 (A1)-(A3) 成立. 则由算法 {1} 产生的序列 $\{x_{n}\}$ 强收敛于一点 $z \in \Gamma $, 其中 $z=P_{\Gamma}o f(z)$.

$z\in \Gamma$. 由于 $\mbox{prox}_{\tau}$ 是非扩张映射, 元素 $z$ 是 PSFP 的解, 并且, 任意函数的极小值点是临近映射的不动点, 结合引理 2.1(iii), 可得

$\begin{align*} {\langle w_{n}-z, -\nabla E(w_{n})\rangle} &= \langle w_{n}-z, A^*(\mbox{prox}_{\tau G}-I)Aw_{n}\rangle\\ & = \langle Aw_{n}-Az, (\mbox{prox}_{\tau G}-I)Aw_{n}\rangle\\ &= \langle Aw_{n}-\mbox{prox}_{\tau G}Aw_{n}+\mbox{prox}_{\tau G}Aw_{n}-Az, (\mbox{prox}_{\tau G}-I)Aw_{n}\rangle\\ &= \langle \mbox{prox}_{\tau G}Aw_{n}-Az, \mbox{prox}_{\tau G}Aw_{n}-Aw_{n}\rangle -\|\mbox{prox}_{\tau G}Aw_{n}-Aw_{n}\|^2\\ & = \frac{1}{2}\left(\|\mbox{prox}_{\tau G}Aw_{n}-Az\|^2+\|\mbox{prox}_{\tau G}Aw_{n}-Aw_{n}\|^2 -\|Aw_{n}-Az\|^2 \right)\\ &\quad\ - \|\mbox{prox}_{\tau G}Aw_{n}-Aw_{n}\|^2\\ &\le -\frac{1}{2}\|\mbox{prox}_{\tau G}Aw_{n}-Aw_{n}\|^2 = -E(w_{n}), \end{align*}$

同理, 可得${\langle w_{n}-z, -\nabla L(w_{n})\rangle}\leq -L(w_{n}),$ 结合 (3.1}) 和 (3.2) 式, 可证

$\begin{aligned} \left\|y_{n}-z\right\|^{2}= & \left\|w_{n}-\lambda_{n}\left(\nabla E\left(w_{n}\right)+\nabla L\left(w_{n}\right)\right)-z\right\|^{2} \\ = & \left\|w_{n}-z\right\|^{2}+\lambda_{n}^{2}\left\|\nabla E\left(w_{n}\right)+\nabla L\left(w_{n}\right)\right\|^{2}+2 \lambda_{n}\left\langle w_{n}-z,-\left(\nabla E\left(w_{n}\right)+\nabla L\left(w_{n}\right)\right)\right\rangle \\ = & \left\|w_{n}-z\right\|^{2}+\lambda_{n}^{2}\left\|\nabla E\left(w_{n}\right)+\nabla L\left(w_{n}\right)\right\|^{2} \\ & +2 \lambda_{n}\left\langle w_{n}-z,-\nabla E\left(w_{n}\right)\right\rangle+2 \lambda_{n}\left\langle w_{n}-z,-\nabla L\left(w_{n}\right)\right\rangle \\ \leq & \left\|w_{n}-z\right\|^{2}+2 \lambda_{n}^{2}\left\|\nabla E\left(w_{n}\right)\right\|^{2}-2 \lambda_{n} E\left(w_{n}\right)+2 \lambda_{n}^{2}\left\|\nabla L\left(w_{n}\right)\right\|^{2}-2 \lambda_{n} L\left(w_{n}\right) \\ = & \left\|w_{n}-z\right\|^{2}+2 \lambda_{n}\left(\frac{\lambda_{n}}{\lambda_{n+1}} \lambda_{n+1}\left\|\nabla E\left(w_{n}\right)\right\|^{2}-E\left(w_{n}\right)\right) \\ & +2 \lambda_{n}\left(\frac{\lambda_{n}}{\lambda_{n+1}} \lambda_{n+1}\left\|\nabla L\left(w_{n}\right)\right\|^{2}-L\left(w_{n}\right)\right) \\ \leq & \left\|w_{n}-z\right\|^{2}+2 \lambda_{n}\left(\frac{\delta \lambda_{n}}{\lambda_{n+1}}-1\right)\left(E\left(w_{n}\right)+L\left(w_{n}\right)\right), \end{aligned}$

其中 $\|\nabla E(w_{n})+\nabla L(w_{n})\|^2\leq 2\|\nabla E(w_{n})\|^2+2\|\nabla L(w_{n})\|^2$. 由于

$\underset{n\rightarrow\infty}\lim\left(1-\frac{\delta\lambda_{n}}{\lambda_{n+1}}\right)=1-\delta>1-\mu,$

其中 $\delta \in (0, \mu)\subset(0, 1)$.$ \exists N\geq 0$ 使得 $ \forall n\geq N$, $1-\frac{\delta\lambda_{n}}{\lambda_{n+1}}> 1-\mu$. 因此, $ \forall n\geq N$,

$\| y_{n}-z\|^2\leq \|w_{n}-z\|^2+2\lambda_{n}(\mu-1)(E(w_{n})+L(w_{n})).$

由上式可知, $\forall n\geq N$,

$\begin{equation} \|y_{n}-z\|\leq \|w_{n}-z\|. \end{equation}$

根据 $w_{n}$ 的表达式, 可知

$\begin{align*} {\|w_{n}-z\|} &= \| x_{n}+\sigma_{n}(x_{n}-x_{n-1})-z\|\\ &\leq \|x_{n}-z\|+\sigma_{n}\| x_{n}-x_{n-1}\| \\ & = \|x_{n}-z\|+\gamma_{n} \cdot\frac{\sigma_{n}}{\gamma_{n}}\| x_{n}-x_{n-1}\|. \end{align*}$

依据 (3.3) 式, 则有$\sigma_{n}\|x_{n}-x_{n-1}\|\leq\tilde{\tau}_{n}\forall n\geq 1$, 进一步结合 $ \mathop {\lim }\limits_{n\to\infty}\frac{\tilde{\tau}_{n}}{\gamma_{n}}=0$, 即证

$ \mathop {\lim }\limits_{n\rightarrow\infty}\frac{\sigma_{n}}{\gamma_{n}}\| x_{n}-x_{n-1}\|\leq \mathop {\lim }\limits_{n\rightarrow\infty}\frac{\tilde{\tau}_{n}}{\gamma_{n}}=0.$

那么存在一个常数 $M_{1}>0$ 使得

$ \frac{\sigma_{n}}{\gamma_{n}}\| x_{n}-x_{n-1}\|\leq M_{1}, \forall n\geq 1, $

上式结合 (3.5) 和 (3.6) 式, 可知

$\begin{equation} \|y_{n}-z\|\leq \|w_{n}-z\|\leq \|x_{n}-z\|+\gamma_{n}M_{1}. \end{equation}$

上式结合 (3.1) 式, 可知 $ \forall n\geq N$,

$\begin{align*} {\|x_{n+1}-z\|} &= \|\gamma_{n}f(x_{n})+(1-\gamma_{n})y_{n}-z\|\\ &= \|\gamma_{n}(f(x_{n})-z)+(1-\gamma_{n})(y_{n}-z)\|\\ &\leq \gamma_{n}\|f(x_{n})-z\|+(1-\gamma_{n})\|y_{n}-z\|\\ &= \gamma_{n}\|f(x_{n})-f(z)+f(z)-z\|+(1-\gamma_{n})\|y_{n}-z\|\\ & \leq \gamma_{n}\tilde{\rho}\|x_{n}-z\|+\gamma_{n}\|f(z)-z\|+(1-\gamma_{n})\|y_{n}-z\|\\ &\leq \gamma_{n}\tilde{\rho}\|x_{n}-z\|+(1-\gamma_{n})(\|x_{n}-z\|+\gamma_{n}M_{1}\|)+\gamma_{n}\|f(z)-z\|\\ &\leq (1-\gamma_{n}(1-\tilde{\rho}))\|x_{n}-z\|+\gamma_{n}(1-\tilde{\rho})\frac{\|f(z)-z\|+M_{1}}{1-\tilde{\rho}}\\ &\leq \max \left\{\|x_{n}-z\|, \frac{ M_{1}+ \|f(z)-z\| }{1-\tilde{\rho}}\right\}\\ &\leq\cdots\leq \max \left\{\|x_{N}-z\|, \frac{ M_{1}+ \|f(z)-z\| }{1-\tilde{\rho}}\right\}. \end{align*}$

上式表明序列 $\{x_{n}\}$ 是有界的. 因此, 序列 $\{y_{n}\}$, $\{ f(x_{n})\}$$\{w_{n}\}$ 也是有界的. 现在, 结合 $\|\cdot\|^2$ 的凸性和 (3.1) 式, 可知存在 $M_{2}>0$,

$\begin{align*} \begin{array}{l} \quad\;{\|x_{n+1}-z\|^2} = \| \gamma_{n}f(x_{n})+(1-\gamma_{n})y_{n}-z\|^2\\ \quad\quad \quad \quad \;\; = \| \gamma_{n}(f(x_{n})-z)+(1-\gamma_{n})(y_{n}-z)\|^2\\ \quad\quad \quad \quad \;\; \leq \gamma_{n}\|(f(x_{n})-z)\|^2+(1-\gamma_{n})\|y_{n}-z\|^2\\ \quad\quad \quad \quad \;\; \leq \gamma_{n}(\|f(x_{n})-f(z)\|+\|f(z)-z\|)^2+(1-\gamma_{n})\|y_{n}-z\|^2\\ \quad\quad \quad \quad \;\; \leq \gamma_{n}(\tilde{\rho}\|x_{n}-z\|+\|f(z)-z\|)^2+(1-\gamma_{n})\|y_{n}-z\|^2\\ \quad\quad \quad \quad \;\; \leq \gamma_{n}(\|x_{n}-z\|+\|f(z)-z\|)^2+(1-\gamma_{n})\|y_{n}-z\|^2\\ \quad\quad \quad \quad \;\; \leq \gamma_{n}\|x_{n}-z\|^2+\gamma_{n}(\|f(z)-z\|^2+2\|x_{n}-z\|\|f(z)-z\|) +(1-\gamma_{n})\|y_{n}-z\|^2\\ \quad\quad \quad \quad \;\; \leq \gamma_{n}\|x_{n}-z\|^2+(1-\gamma_{n})\|y_{n}-z\|^2 +\gamma_{n}M_{2}. \end{array} \end{align*}$

再结合 (3.5) 式上面的式子, 可得 $\forall n\geq N$,

$\begin{aligned} \left\|x_{n+1}-z\right\|^{2} \leq & \gamma_{n}\left\|x_{n}-z\right\|^{2}+\left(1-\gamma_{n}\right)\left\|w_{n}-z\right\|^{2} \\ & -\left(1-\gamma_{n}\right) 2 \lambda_{n}(1-\mu)\left(E\left(w_{n}\right)+L\left(w_{n}\right)\right)+\gamma_{n} M_{2}. \end{aligned}$

把 (3.7) 式代入上式, 则存在 $ M_{3}>0$ 使得 $ \forall n\geq N$,

$\begin{align*} {\|x_{n+1}-z\|^2}& \leq \gamma_{n}\|x_{n}-z\|^2+(1-\gamma_{n})(\|x_{n}-z\|+\gamma_{n}M_{1})^2\\ &\ - (1-\gamma_{n})2\lambda_{n}(1-\mu)(E(w_{n})+L(w_{n})) +\gamma_{n}M_{2}\\ &= \gamma_{n}\|x_{n}-z\|^2+(1-\gamma_{n})\|x_{n}-z\|^2+(1-\gamma_{n})(\gamma_{n}M_{1})^2+ 2(1-\gamma_{n})\gamma_{n}M_{1}\|x_{n}-z\|\\ &\ - (1-\gamma_{n})2\lambda_{n}(1-\mu)(E(w_{n})+L(w_{n}))+\gamma_{n}M_{2}\\ & \leq \|x_{n}-z\|^2-(1-\gamma_{n})2\lambda_{n}(1-\mu)(E(w_{n})+L(w_{n}))+\gamma_{n}M_{3}. \end{align*}$

即, $\forall n\geq N$,

$\begin{equation} \nonumber (1-\gamma_{n})2\lambda_{n}(1-\mu)(E(w_{n})+L(w_{n})) \leq \|x_{n}-z\|^2-\|x_{n+1}-z\|^2+\gamma_{n}M_{3}. \end{equation}$

运用引理 2.1(i)-(ii) 和 (3.7) 式, 即证 $ \forall n\geq N$,

$\begin{align*} \|x_{n+1}-z\|^2&= \|\gamma_{n}f(x_{n})+(1-\gamma_{n})y_{n}-z\|^2\\ & = \| \gamma_{n}(f(x_{n})-f(z))+(1-\gamma_{n})(y_{n}-z)+\gamma_{n}(f(z)-z)\|^2\\ &\leq \| \gamma_{n}(f(x_{n})-f(z))+(1-\gamma_{n})(y_{n}-z)\|^2+ 2 \gamma_{n}\langle f(z)-z, x_{n+1}-z\rangle\\ & \leq \gamma_{n}\|f(x_{n})-f(z)\|^2+(1-\gamma_{n})\|y_{n}-z\|^2 +2\gamma_{n}\langle f(z)-z, x_{n+1}-z\rangle\\ & \leq \gamma_{n}\tilde{\rho}^2\|x_{n}-z\|^2+(1-\gamma_{n})\|y_{n}-z\|^2 +2\gamma_{n}\langle f(z)-z, x_{n+1}-z\rangle\\ & \leq \gamma_{n}\tilde{\rho}\|x_{n}-z\|^2+(1-\gamma_{n})\|w_{n}-z\|^2 +2\gamma_{n}\langle f(z)-z, x_{n+1}-z\rangle. \end{align*}$

$w_{n}$ 的定义, 可知

$\begin{aligned} \left\|w_{n}-z\right\|^{2} & =\left\|x_{n}+\sigma_{n}\left(x_{n}-x_{n-1}\right)-z\right\|^{2} \\ & =\left\|x_{n}-z\right\|^{2}+\sigma_{n}^{2}\left\|x_{n}-x_{n-1}\right\|^{2}+2 \sigma_{n}\left\langle x_{n}-z, x_{n}-x_{n-1}\right\rangle \\ & \leq\left\|x_{n}-z\right\|^{2}+\sigma_{n}^{2}\left\|x_{n}-x_{n-1}\right\|^{2}+2 \sigma_{n}\left\|x_{n}-z\right\|\left\|x_{n}-x_{n-1}\right\|. \end{aligned}$

$M=\underset{n\geq N}\sup \{ \sigma\|x_{n}-x_{n-1}\|, 2\|x_{n}-z\| \}$. 结合 (3.9) 式和 (3.10) 式, 可得 $ \forall n\geq N$,

$\begin{align*} \|x_{n+1}-z\|^2 \leq\ & (1-\gamma_{n}(1-\tilde{\rho}))\|x_{n}-z\|^2+\sigma_{n}^{2}\| x_{n}-x_{n-1}\|^2+2\sigma_{n}\|x_{n}-z\| \| x_{n}-x_{n-1}\|\\ &+ 2\gamma_{n}\langle f(z)-z, x_{n+1}-z\rangle\\ \leq\ & (1- \gamma_{n}(1-\tilde{\rho}))\|x_{n}-z\|^2+(1-\tilde{\rho})\frac{2}{1-\tilde{\rho}}\gamma_{n}\langle f(z)-z, x_{n+1}-z\rangle\\ &+ \sigma_{n} \| x_{n}-x_{n-1}\|( \sigma_{n} \| x_{n}-x_{n-1}\|+2\|x_{n}-z\|)\\ \leq\ & (1- \gamma_{n}(1-\tilde{\rho}))\|x_{n}-z\|^2+(1-\tilde{\rho})\frac{2}{1-\tilde{\rho}}\gamma_{n}\langle f(z)-z, x_{n+1}-z\rangle\\ & + \sigma_{n} \| x_{n}-x_{n-1}\|( \sigma \| x_{n}-x_{n-1}\|+2\|x_{n}-z\|)\\ \leq\ & (1- \gamma_{n}(1-\tilde{\rho}))\|x_{n}-z\|^2+(1-\tilde{\rho})\frac{2}{1-\tilde{\rho}}\gamma_{n}\langle f(z)-z, x_{n+1}-z\rangle\\ & + 2M\sigma_{n} \| x_{n}-x_{n-1}\|\\ \leq\ & (1- \gamma_{n}(1-\tilde{\rho}))\|x_{n}-z\|^2+(1-\tilde{\rho})\gamma_{n}(\frac{2}{1-\tilde{\rho}}\langle f(z)-z, x_{n+1}-z\rangle \\ &+ \frac{2M\sigma_{n}}{(1-\tilde{\rho})\gamma_{n}}\| x_{n}-x_{n-1}\|). \end{align*}$

接下来, 需证 $ \{ \|x_{n}-z\|^2 \}$ 趋近于零.

假设 $ \{ \|x_{n}-z\|\}$ 的子列为 $ \{ \|x_{n_{k}}-z\|\}$, 并满足$\underset{k\rightarrow \infty}\liminf (\|x_{n_{k}+1}-z\|-\|x_{n_{k}}-z\|)\geq0. $ 则有

$\begin{aligned} & \liminf _{k \rightarrow \infty}\left(\left\|x_{n_{k}+1}-z\right\|^{2}-\left\|x_{n_{k}}-z\right\|^{2}\right) \\ = & \liminf _{k \rightarrow \infty}\left(\left(\left\|x_{n_{k}+1}-z\right\|-\left\|x_{n_{k}}-z\right\|\right)\left(\left\|x_{n_{k}+1}-z\right\|+\left\|x_{n_{k}}-z\right\|\right)\right) \geq 0. \end{aligned}$

上式结合 $\underset{k\rightarrow \infty}\lim \gamma_{n_{k}}=0$$\underset{k\rightarrow \infty}\lim \lambda_{n_{k}}=\lambda>0$, 有

$\begin{align*} &\ \underset{k\rightarrow \infty}\limsup 2(1-\mu)(1-\gamma_{n_{k}})\lambda_{n_{k}}(E(w_{n_{k}})+L(w_{n_{k}}))\\ &\leq \underset{k\rightarrow \infty}\limsup (\|x_{n_{k}+1}-z\|^2-\|x_{n_{k}}-z\|^2+ \gamma_{n_{k}}M_{3} )\\ & \leq \underset{k\rightarrow \infty}\limsup (\|x_{n_{k}+1}-z\|^2-\|x_{n_{k}}-z\|^2)+\underset{k\rightarrow \infty}\limsup \gamma_{n_{k}}M_{3}\\ & = -\underset{k\rightarrow \infty}\liminf (\|x_{n_{k}+1}-z\|^2-\|x_{n_{k}}-z\|^2)\leq0. \end{align*}$

$\begin{equation} \underset{k\rightarrow \infty}\lim(E(w_{n_{k}})+L(w_{n_{k}}))=0 \Leftrightarrow \underset{k\rightarrow \infty}\lim E(w_{n_{k}})=0 \mbox{和} \underset{k\rightarrow \infty}\lim L(w_{n_{k}})=0. \end{equation}$

由迭代公式 (3.1) 和 (3.2) 可知

$\begin{align*} \|y_{n_{k}}-w_{n_{k}}\|^2&=\|\lambda_{n_{k}}(\nabla E(w_{n_{k}})+\nabla L(w_{n_{k}}))\|^2 \\ &\leq 2\lambda_{n_{k}}^{2}\left(\|\nabla E(w_{n_{k}}\|^2+\|\nabla L(w_{n_{k}}\|^2\right)\\ &\leq 2\lambda_{n_{k}}^{2}\left(\frac{1}{\lambda_{n_{k}+1}} \lambda_{n_{k}+1}\|\nabla E(w_{n_{k}})\|^2+ \frac{1}{\lambda_{n_{k}+1}} \lambda_{n_{k}+1}\|\nabla L(w_{n_{k}})\|^2\right)\\ &\leq 2\lambda_{n_{k}}^{2}\left(\frac{\delta E(w_{n_{k}})}{\lambda_{n_{k}+1}}+\frac{\delta L(w_{n_{k}})}{\lambda_{n_{k}+1}}\right) = \frac{2\delta\lambda_{n_{k}}^{2}}{\lambda_{n_{k}+1}}(E(w_{n_{k}})+L(w_{n_{k}})), \end{align*}$

其中 $\|\nabla E(w_{n_{k}})+\nabla L(w_{n_{k}})\|^2\leq 2\|\nabla E(w_{n_{k}})\|^2+2\|\nabla L(w_{n_{k}})\|^2. $ 上式结合$\underset{k\rightarrow \infty}\lim \lambda_{n_{k}}=\lambda>0$, 有

$\begin{equation} \underset{k\rightarrow \infty}\lim\|y_{n_{k}}-w_{n_{k}}\|=0. \end{equation}$

依据 $\underset{k\rightarrow\infty}\lim \gamma_{n_{k}}=0$ 和 (3.1) 式, 可得

$\begin{equation} \|x_{n_{k}}-w_{n_{k}}\|= \sigma_{n_{k}}\|x_{n_{k}}-x_{n_{k}-1}\|=\gamma_{n_{k}} \frac{ \sigma_{n_{k}}}{\gamma_{n_{k}}}\|x_{n_{k}}-x_{n_{k}-1}\|\rightarrow 0, \mbox{当} k\rightarrow \infty. \end{equation}$

利用 (3.13) 和 (3.14) 式, 可知

$\begin{equation} \underset{k\rightarrow \infty}\lim\|y_{n_{k}}-x_{n_{k}}\|=0. \end{equation}$

由 (3.1) 式和 $\underset{k\rightarrow\infty}\lim \gamma_{n_{k}}=0$, 可推出

$\begin{equation} \|x_{n_{k}+1}-y_{n_{k}}\|=\gamma_{n_{k}}\| f(x_{n_{k}})-y_{n_{k}}\|\rightarrow 0, \mbox{当} k\rightarrow \infty. \end{equation}$

上式可推出

$\begin{equation} \|x_{n_{k}+1}-x_{n_{k}}\| \leq \|x_{n_{k}+1}-y_{n_{k}}\|+\|y_{n_{k}}-x_{n_{k}}\|\rightarrow 0, \mbox{当} k\rightarrow \infty. \end{equation}$

由于序列 $\{ x_{n_{k}}\}$ 是有界的, 则存在 $\{ x_{n_{k}}\}$ 的子列 $\{ x_{n_{k_{i}}}\}$ 弱收敛于一点 $z_{0} \in H$ 使得

$\begin{equation} \nonumber \underset{k\rightarrow \infty}\limsup \langle f(z)-z, x_{n_{k}}-z\rangle=\underset{i\rightarrow \infty}\lim \langle f(z)-z, x_{n_{k_{i}}}-z\rangle = \langle f(z)-z, z_{0}-z\rangle. \end{equation}$

此外, 由 (3.14) 式, 可得

$\begin{equation} \nonumber w_{n_{k_{i}}}\rightharpoonup z_{0}, \mbox{当} i\rightarrow \infty. \end{equation}$

利用函数 $E$ 的弱下半连续性, 所得的结论是

$\begin{equation} \nonumber 0\leq E(z_{0})\leq \underset{i\rightarrow\infty}\liminf E(w_{n_{k_{i}}})=\underset{i\rightarrow\infty}\lim E(w_{n_{k_{i}}})=0. \end{equation}$

上式意味着 $E(z_{0})=\frac{1}{2}\|(I-\mbox{prox}_{\tau G})Az_{0}\|^2=0, $ 即, $Az_{0}$ 是函数 $G$ 的临近映射的不动点, 或者等价于 $0\in \partial G(Az_{0}). $ 换句话说, $Az_{0}$$G$ 的极小点. 同理, 根据 $L$ 的弱下半连续性, 可得

$\begin{equation} \nonumber 0\leq L(z_{0})\leq \underset{i\rightarrow\infty}\liminf L(w_{n_{k_{i}}})=\underset{i\rightarrow\infty}\lim L(w_{n_{k_{i}}})=0. \end{equation}$

进一步可知, $L(z_{0})=\frac{1}{2}\left\|(I-\mbox{prox}_{\tau F})z_{0}\right\|^2=0, $ 即, $z_{0}$ 是函数 $F$ 的临近映射的不动点, 或者等价于 $0 \in \partial F(z_{0}). $ 也就是说, $z_{0}$$F$ 的极小点. 因此, $z_{0}\in \Gamma. $ 那么由投影性质对 $z=P_{\Gamma}of(z)$ 展开得

$\underset{k\rightarrow \infty}\limsup \langle f(z)-z, x_{n_{k}}-z\rangle=\underset{i\rightarrow \infty}\lim \langle f(z)-z, x_{n_{k_{i}}}-z\rangle=\langle f(z)-z, z_{0}-z\rangle\leq 0,$

结合 (3.16) 式, 即证

$\begin{align*} \underset{k\rightarrow \infty}\limsup\langle f(z)-z, x_{n_{k}+1}-z\rangle &\leq \underset{k\rightarrow \infty}\limsup \langle f(z)-z, x_{n_{k+1}}-x_{n_{k}}\rangle+\underset{k\rightarrow \infty}\limsup \langle f(z)-z, x_{n_{k}}-z\rangle\\ & \leq 0. \end{align*}$

因此

$\underset{k\rightarrow \infty}\limsup \tilde{\delta}_{n_{k}}=\underset{k\rightarrow \infty}\limsup \left[\frac{2}{1-\tilde{\rho}}\langle f(z)-z, x_{n_{k}+1}-z\rangle +\frac{2M\sigma_{n_{k}}}{(1-\tilde{\rho})\gamma_{n_{k}}}\| x_{n_{k}}-x_{n_{k}-1}\|\right]\leq 0.$

运用引理 2.3, 可证 $ \underset{n\rightarrow \infty}\lim\|x_{n}-z\|=0$.

$F\equiv\delta_{C}$ (若 $x \in C$, 则 $\delta_{C}(x)=0$, 否则, $+\infty$), $G\equiv\delta_{Q}$, 分别为非空闭凸集 $C\subset H_{1}$$Q\subset H_{2}$ 上的示性函数, 则 PSFP退化为如下的 SFP.

$ \begin{array}{l} \mbox{寻找一点} z^* \in C \mbox{使得} Az^* \in Q, \end{array} $

进一步, 由定理 3.1 获得如下强收敛推论.

推论 3.1$\Gamma\neq\emptyset$, $\lambda_{1}>0$, $\delta \in (0, \mu)\subset(0, 1)$, $\{\sigma_{n}\}\subset[0, \sigma)\subset[0, 1)$, $\{\phi_{n}\}$$\{\psi_{n}\}$ 是满足引理 2.4 的两个序列, 条件 (A2) 和 (A3) 成立. 令 $x_{0}, x_{1}\in H_{1}$, 则序列 $\{x_{n}\}$ 生成的迭代格式如下

$\begin{equation}\nonumber\ \begin{cases} w_{n}=x_{n}+\sigma_{n}(x_{n}-x_{n-1}), \\ y_{n}=w_{n}-\lambda_{n}(\nabla E(w_{n})+\nabla L(w_{n})), \\ x_{n+1}=\gamma_{n}f(x_{n})+(1-\gamma_{n})y_{n}. \end{cases} \end{equation}$

其中 $\sigma_{n}$ 定义在 (3.3) 式, $\nabla E(w_{n})=A^*(I-P_{Q})Aw_{n}$, 步长 $\lambda_{n}$ 的更新方式如下

$\lambda_{n+1}= \begin{cases} \mbox{min}\bigg\{{\min\bigg\{\frac{\delta L(w_{n})}{ \|\nabla L(w_{n})\|^2}, \frac{\delta E(w_{n})}{ \|\nabla E(w_{n})\|^2}\bigg\}, \phi_{n}\lambda_{n}+\psi_{n} }\bigg\}, \\[1mm] \hskip 4cm \mbox{若}\ \nabla E(w_{n})\neq0 \mbox{和} \nabla L(w_{n})\neq0, \\ \phi_{n}\lambda_{n}+\psi_{n}, \hskip 2cm\ \, \mbox{否则}, \end{cases}$

则上式生成的迭代序列 $\{x_{n}\}$ 强收敛于一点 $z \in \Gamma$, 其中 $z=P_{\Gamma}of(z)$.

4 应用于分裂均衡问题

文献 [31] 介绍了分裂均衡问题属于变分包含问题, 并且, 其是临近分裂可行问题的特例. 接下来, 本节给出的分裂均衡问题 (Split Equilibrium Problem, 记作 SEP) 是指

$\begin{align*} \begin{array}{l} \mbox{寻找一点} z^* \in H_{1} \mbox{使得} g_{1}(z^*, x) \geq 0, \forall x \in H_{1}, \\ \mbox{寻找一点} y^*=Az^* \in H_{2} \mbox{使得} g_{2}(y^*, z) \geq 0, \forall z \in H_{2}, \\ \end{array} \end{align*}$

其中 $H_{1}$$H_{2}$ 是实 Hilbert 空间, $g_{1}: H_{1}\times H_{1}\rightarrow \mathbb{R}$$g_{2}: H_{2}\times H_{2} \rightarrow \mathbb{R}$ 是双边函数. 该问题的解集记作 $\Gamma$$\Gamma\neq\emptyset. $ 在给出算法之前, 现给出如下假设

对于双边函数 $g_{i}: H_{i}\times H_{i}\rightarrow \mathbb{R}, i=1, 2, $ 并给出相关条件

(i) $g_{i}(x, x)=0, \forall x\in H_{i};$

(ii) $g_{i}$ 是单调的, 即, $g_{i}(x, y)+g_{i}(y, x)\leq 0,$ for all $x, y \in H_{i};$

(iii) 对每一个 $ x, y \in H_{i}$, $\underset{\alpha\rightarrow 0}\lim g_{i}(\alpha z+(1-\alpha)x, y)\leq g_{i}(x, y)$;

(iv) 对每一个 $ x \in H_{i}, $$y\mapsto g_{i}(x, y)$ 是凸的且为下半连续函数. 对每一个满足上述条件的双边函数, 其临近算子 $J_{i}^{\tau}: H_{i}\rightarrow H_{i}$ 定义为

$J_{i}^{\tau}(x):=\{z\in H_{i}: g_{i}(z, y)+\frac{1}{\tau}\langle y-z, z-x \rangle, \forall y \in H_{i} \mbox{和} \tau>0\}.$

另外, 文献 [8,36] 证明了 $J_{i}^{\tau}$是单值的且为完全非扩张性映射. 即可获得如下推论.

推论 4.1$\Gamma\neq\emptyset$, $\lambda_{1}>0$, $\delta \in (0, \mu)\subset(0, 1)$, $\{\sigma_{n}\}\subset[0, \sigma)\subset[0, 1)$, $\{\phi_{n}\}$$\{\psi_{n}\}$ 是满足引理 2.4 的序列, 条件 (A2) 和 (A3) 成立. 令$x_{0}, x_{1}\in H_{1}$, 序列 $\{x_{n}\}$ 生成的迭代格式如下

$\begin{equation}\nonumber\ \begin{cases} w_{n}=x_{n}+\sigma_{n}(x_{n}-x_{n-1}), \\ y_{n}=w_{n}-\lambda_{n}(\nabla E(w_{n})+\nabla L(w_{n})), \\ x_{n+1}=\gamma_{n}f(x_{n})+(1-\gamma_{n})y_{n}. \end{cases} \end{equation}$

其中 $\sigma_{n}$ 的定义见 (3.3) 式, $\nabla E(w_{n})=A^*(I-J_{2}^{\tau})Aw_{n}$, $\nabla L(w_{n})=(I-J_{1}^{\tau})w_{n}$, 步长 $\lambda_{n}$ 更新如下

$\lambda_{n+1}= \begin{cases} \mbox{min}\left\{{\min\left\{\frac{\delta L(w_{n})}{ \|\nabla L(w_{n})\|^2}, \frac{\delta E(w_{n})}{ \|\nabla E(w_{n})\|^2}\right\}, \phi_{n}\lambda_{n}+\psi_{n} }\right\}, \\[1mm] \hskip 4cm \mbox{若}\ \nabla E(w_{n})\neq0 \mbox{和} \nabla L(w_{n})\neq0, \\ \phi_{n}\lambda_{n}+\psi_{n}, \hskip 2cm\ \, \mbox{否则}, \end{cases}$

则如上方案生成的迭代序列 $\{x_{n}\}$ 强收敛于一点 $z \in \Gamma$, 其中 $z=P_{\Gamma}of(z)$.

5 数值实验

本节提供与 PSFP 相关的数值算例. 在第一个算例中, 算法 1 与算法 3.1-3.2[1], 算法 3.1[29]及算法[30]进行比较. 所有的实验执行于 MATLAB R2017a, 其安装在运行内存为 RAM 8.00 GB, 型号为 PC Desktop Intel(R) Core(TM) i7-6700 CPU @3.40 GHZ 的计算机上.

在下面算例中, 本节主要研究 PSFP 的情形为 $\mbox{arg}\min F\cap A^{-1}(\mbox{arg}\min G) \neq\emptyset$, 换句话说, 寻找函数 $F$ 的一个极小点 $z^*$ 使得 $Az^*$ 是函数 $G$ 的极小点, 即

$\begin{align*} \mbox{寻找一点} z^*\in H_{1} \mbox{使得} z^*\in\mbox{arg}\underset{x \in H_{1}}\min F(x) \mbox{和} Az^*\in\mbox{arg}\underset{y \in H_{2}}\min G(y), \end{align*}$

其中 $F:H_{1}\rightarrow \mathbb{R}$$G:H_{2}\rightarrow \mathbb{R}$ 是真凸下半连续函数, $\mbox{arg}\min F=\{ z^*\in H_{1} : F(z^*)\leq F(x), \forall x\in H_{1}\}$$\mbox{arg}\min G=\{ y^*\in H_{2} : G(y^*)\leq G(x), \forall x\in H_{2}\}$, 该问题的解集记为 $\Gamma$.

例 5.1[9]$H_{1}=H_{2}=\mathbb{R}^{N}$$F(x)=\frac{1}{2}d_{C}^{2}(x)$,

其中 $C\subset \mathbb{R}^{N}$ 是一个单位球, $ d_{C}(x)=\inf_{z\in C}\|x-z\|. $

函数 $G(x)=\frac{1}{2}\|x\|^2$.$Ax=x$, $x\in \mathbb{R}^{N}$. 观察到 $0\in \Gamma$, 则 $\Gamma\neq\emptyset$.

对于算法 3.1-3.2[1]和算法 3.1[29], 选取 $\kappa_{n}=1. 9$, $\alpha_{n}=\frac{1}{n+1}$$\alpha_{n}=\frac{1}{10^4(n+1)}$.

对于算法[30], 设参数 $\tau_{n}=10^{-4}$, $\alpha_{n}=\frac{1. 99}{n+1}$, $\mu=1$, $\tilde{F}=I$ (在文献 [30] 中指出, $\tilde{F}$ 是压缩映射, $I$$H_{1}$ 上的恒等映射)和 $\beta_{n}=\frac{0. 001}{(n+1)^2}$.

对于算法 1, 固定 $\kappa_{n}=1. 9$, $\alpha_{n}=\frac{1}{n^2}$, $\sigma=0. 3$$\tilde{\tau}_{n}=\frac{1}{n^3}$.

对于所有算法, 选择初始点 $x_{0}=[\cdots,0]$$x_{1}=[\cdots,1] \in \mathbb{R}^{N}$$\tau=5$. 所有算法的准止条件是 $ \|x_{n}-\mbox{prox}_{\tau, F}(x_{n})\|+\|Ax_{n}- \mbox{prox}_{\tau, G}(Ax_{n})\|<\epsilon. $ 为了保证所对比算法有相同的收敛点, 则设 $f(x)=0$.

数值结果参见表1表2.

表1   例 5.1 的数值结果

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表2   例 5.1 的数值结果

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注 5.1表1-2可知, 算法 1 数值效果较好, 主要体现在两个方面: 一是迭代次数较少, 二是 CPU 时间较短.

对于问题 (SFP), 我们列举了如下无限维空间的数值算例, 其用于算法 1 与算法 3.1[32]、算法 3.1[14] 进行对比.

例 5.2[9]$H_{1}=H_{2}=L^2([0,1])$, 则范数$\|x\|_{L^2}=\big( \int_{0}^{1} x(t)^2{\rm d}t\big)^\frac{1}{2}$ 和内积 $\langle x, y\rangle=\int_{0}^{1} x(t)y(t){\rm d}t, x, y \in L^2([0,1])$.$C=\{ x\in L^2([0,1]): \|x\|_{L^2}\leq 1 \}$$Q=\{ x\in L^2([0,1]):\langle x, t\rangle=0\}$, $Ax(t)=\frac{x(t)}{2}$. 观察到 $0 \in \Gamma$. 因此, $\Gamma\neq\emptyset$.

对于算法 3.1[32] 和算法 3.1[14], 固定$\sigma=0. 3$, $\tilde{\tau}_{n}=\frac{1}{n^2}$, $\alpha_{n}=\frac{1}{10^4n}$, $\kappa_{n}=1. 6$, $f(x)=0. 01x$, $\beta_{n}=0. 7$$\chi_{n}=\max(0, \chi_{n}-0. 1)$.

对于算法 1, 选取 $\kappa_{n}=1. 6$, $\sigma=0. 3$, $f(x)=0. 01x$, $\alpha_{n}=\frac{1}{10^4n}$, $\tilde{\tau}_{n}=\frac{1}{n^2}$.

对于 LópezAlg. 5.1, 采取$\kappa_{n}=9\times10^{-5}$$\alpha_{n}=\frac{10^{-5}}{n}. $

所有算法的准止条件为 $ \|x_{n+1}-x_{n}\|_{L^2}<\epsilon$. 初始点的选取有以下两种情形

情形 1 $x_{0}=t^4, x_{1}=t+1$;

情形 2 $x_{0}={\rm e}^t, x_{1}=3{\rm e}^t$.

接下来, 集合 $C$$Q$ 上的投影公式为

$\begin{equation}\ P_{C}(x)= \begin{cases} \frac{x}{\|x\|_{L^2}}, & \mbox{如果}\ \|x\|_{L^2}>1, \nonumber \\ x, &\mbox{如果}\ \|x\|_{L^2}\leq1. \nonumber \end{cases} \mbox{和} \ P_{Q}(x)= x-\frac{\langle t, x\rangle}{\|t\|_{L^2}}t. \end{equation}$

实验结果详见表3表4.

表3   例 5.2 的实验结果

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表4   例 5.2 的实验结果

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注 5.2表3-4的实验结果可知, 算法 1 的收敛速度较快于其它算法, 主要体现在算法 1 的迭代次数较少及执行时间较短.

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