## The Finite Termination of Feasible Solution Sequence for Optimization and Variational Inequality

Wang Ruyu, Zhao Wenling,*, Song Daojin

School of Mathematics and Statistics, Shandong University of Technology, Shandong Zibo 255000

 基金资助: 山东省自然科学基金(ZR2021MA066)

 Fund supported: Shandong Provincial Natural Science Foundation(ZR2021MA066)

Abstract

To provide a finite characterization of feasible solution sequences for optimization problems (OP) and variational inequality problems (VIP), an augmented set value map is introduced for the solution sets of these problems. Additionally, the concepts of augmented weak sharpness with respect to feasible solution sequences are established. These novel notions extend the traditional concepts of weak sharpness and strongly non-degeneracy relative to feasible solution sequences, addressing the limitation that solution sets often lack weak sharpness or strongly non-degeneracy in many cases. When the feasible solution sets of optimization problems and variational inequality problems exhibit augmented weak sharpness, the necessary and sufficient conditions for the finite termination of feasible solution sequences are provided for each problem. These conditions extend the corresponding results found in existing literature, where solution sets are weak sharp or strongly non-degenerate. Furthermore, sufficient conditions with fewer restrictions are provided for the finite termination of various optimization algorithms.

Keywords： Optimization problem; Variational inequality problem; Feasible solution sequence; Augmented weak sharpness; Finite termination

Wang Ruyu, Zhao Wenling, Song Daojin. The Finite Termination of Feasible Solution Sequence for Optimization and Variational Inequality[J]. Acta Mathematica Scientia, 2024, 44(4): 1037-1051

## 1 引言

$\begin{equation*} \min_{x\in S} f(x), \end{equation*}$

$\begin{equation*} \hat{S}=\{x\in S\mid\exists u\in\partial f(x),\quad \text{使得}\quad \langle u, y-x\rangle\geq0,\quad \forall y\in S\}, \end{equation*}$

$$$\bar{S}\subseteq \hat{S}$$$

$N\subseteq \{1, 2, \cdots\}$ 是一个无限子序列, $C^{k}\subset \mathbb{R}^{n}(k=1,2,\cdots)$. 定义

$\begin{equation*} \limsup_{k\rightarrow \infty }C^{k} =\{x \in \mathbb{R}^{n}\mid\ \text{存在}\ N\ \text{与}\ x^{k}\in C^k,\quad\text{使得}\ \lim_{k\in N,k\rightarrow \infty }x^{k} = x\}, \end{equation*}$
$\begin{equation*} \liminf_{k\rightarrow \infty }C^{k} =\{x \in \mathbb{R}^{n}\mid\ \text{存在}\ x^{k}\in C^{k},\quad\text{使得}\ \lim_{k\rightarrow \infty} x_{k}=x\}. \end{equation*}$

$\begin{equation*} \liminf_{k\rightarrow \infty }C^{k} \subseteq\limsup_{k\rightarrow \infty }C^{k}. \end{equation*}$

$C\subset \mathbb{R}^{n}$, $\bar{x}\in C$, $C$ 在点 $\bar{x}$ 的切锥定义为

$\begin{equation*} T_{C}(\bar{x})=\{d\in \mathbb{R}^{n}\mid\ \text{存在}\ x^{k}\in C,x^{k}\rightarrow \bar{x}, \tau^{k}\downarrow 0, (k\rightarrow \infty), \text{使得}\ \lim_{k\rightarrow \infty}(x^{k}-\bar{x})/\tau^{k}=d\}. \end{equation*}$

$\begin{equation*} \hat{N}_{C}(\bar{x})=\{d\in \mathbb{R}^{n} \mid \langle d,x-\bar{x}\rangle\leq o(\|x-\bar{x}\|),x\in C\}. \end{equation*}$

$\begin{equation*} N_{C}(\bar{x})=\limsup_{x^k\in C,x^k\rightarrow \bar{x}}\hat{N}_C(x^k). \end{equation*}$

$\begin{equation*} C^{\circ}=\{y\in \mathbb{R}^{n}\mid \langle y,x\rangle\leq 0,\forall x\in C\}. \end{equation*}$

$\begin{equation*} T_{C}(x)^{\circ}=\hat{N}_C(x). \end{equation*}$

$C$ 是凸集时, 据文献 [1,定理 6.9] 有

$\begin{equation*} N_{C}(\bar{x})=\hat{N}_C(\bar{x})=\{d\mid \langle d,x-\bar{x}\rangle\leq 0,\forall x\in C\}. \end{equation*}$

$x\in \mathbb{R}^{n}$, $x$ 在闭集 $C$ 上的投影定义为

$\begin{equation*} P_C(x)=\arg\min_{y\in C}\|y-x\|. \end{equation*}$

$x$ 到集合 $C$ 的距离定义为

$\text{ dist}(x,C)=\inf_{y\in C}\|y-x\|.$

$\text{ dist}(x,C)=\|P_{C}(x)-x\|.$

$\begin{equation*} P_{T_{C}(x)}(-\partial\psi(x))=\left\{P_{T_{C}(x)}(-u)\mid u\in \partial\psi(x)\right\}. \end{equation*}$

$$$\alpha B\subset \partial f(x)+[T_{S}(x)\cap N_{\bar{S}}(x)]^{\circ},\quad\forall x\in \bar{S},$$$

$$$-\nabla f (\bar{x})\in \text{ int} \bigcap_{x\in\bar{S}}[T_{S}(x)\cap N_{\bar{S}}(x)]^{\circ},\quad\forall \bar{x}\in \bar{S}.$$$

$\begin{equation*} \min_{x\in S}f(x)=\frac{1}{2}x^2, \end{equation*}$

$\begin{equation*} G:=\bigcap_{x\in \bar{S}}[T_S(x)\cap\hat{N}_{\bar{S}}(x)]^{\circ}. \end{equation*}$

$$$-\partial f(x)\cap(-N_S(x))\subset {\rm int} G, \forall x\in \bar{S},$$$

$\begin{equation*} \alpha B-\partial f(x)\cap(-N_S(x))\subset[T_S(x)\cap N_{\bar{S}}(x)]^{\circ}. \end{equation*}$

$\begin{eqnarray*} \ \alpha B&\subset&\partial f(x)\cap(-N_S(x))+[T_S(x)\cap N_{\bar{S}}(x)]^{\circ}\\ &\subset&\partial f(x)+[T_S(x)\cap N_{\bar{S}}(x)]^{\circ}, \forall x\in \bar{S}, \end{eqnarray*}$

$\bar{S}$ 是一个弱尖锐集, 从而由命题 2.1 即得证明.

### 2.2 (OP) 是光滑的情况

$$$-\nabla f(x)\in {\rm int} G, \forall x\in \bar{S},$$$

$\bar{S}$ 是弱尖锐集, 则对任意 $\{x^k\}\subset S, \bar{S}$ 是广义弱尖锐性的.

$$$\lim_{k\rightarrow\infty}\|\nabla f(x^k)-\nabla f(z^k)\|=0.$$$

$K=\{k\mid x^k\notin \bar{S}\}$ 是一无穷序列. 令 $H(z)=\nabla f(z)$, $\forall z\in \bar{S}$. 由假设 $\bar{S}$ 是弱尖锐的, 因此, 由定义 2.2 知定义 2.5 中 (a) 成立. 再据 (2.8) 式即得

$\begin{equation*} \limsup_{k\in K, k \rightarrow\infty}\psi_{k}=\limsup_{k\in K, k \rightarrow\infty} \frac{1}{\|x^{k}-z^{k}\|}\langle\nabla f(x^{k})-\nabla f(z^{k}), x^{k}-z^{k}\rangle=0, \end{equation*}$

$K=\{k\mid x^k\notin \bar{S}\}$ 是一无穷序列. 由假设 $\{\nabla f(x^k)\}$ 有界, 因此 $\{\nabla f(x^k)\}_{k\in K}$ 必有一聚点 $\bar{p}$, 满足 $-\bar{p}\in \text{ int} G$, 即存在常数 $\alpha>0$, 使得

$$$\alpha B\subset\bar{p}+[T_{S}(z)\cap\hat{N}_{\bar{S}}(z)]^{\circ}, \forall z\in\bar{S}.$$$

$H(z)=\bar{p}$, $\forall z\in\bar{S}$, 由 (2.9) 式知, 定义 2.5 中 (a) 成立. 设 $K_{0} \subset K$ 使得

$$$\lim_{k\in K_{0}, k\rightarrow\infty}\nabla f(x^{k})=\bar{p}.$$$

$\begin{eqnarray*} \limsup_{k\in K, k \rightarrow\infty}\psi_{k}\geq\limsup_{k\in K_{0}, k \rightarrow\infty}\psi_{k} =\lim_{k\in K_{0}, k\rightarrow\infty}\frac{1}{\| x^{k}-P_{\bar{s}}(x^{k})\|}\langle\nabla f(x^{k}) -\bar{p}, x^{k}-P_{\bar{S}}(x^{k})\rangle=0, \end{eqnarray*}$

$K=\{k\mid x^k\notin \bar{S}\}$ 是一无穷序列. 则由假设 $\{x^k\}_{k\in K}$ 必有一聚点 $\bar{x}$.$K_0\subset K$ 使得

$$$\lim_{k\in K_0, k\rightarrow\infty}x^k=\bar{x}.$$$

$\begin{eqnarray*} \limsup_{k\in K,k \rightarrow\infty}\psi_{k}&\geq&\limsup_{k\in K_{0},k \rightarrow\infty}\psi_{k} =\lim_{k\in K_{0},k\rightarrow\infty}\frac{1}{\| x^{k}-\bar{x}\|}\langle\nabla f(x^{k}) -\nabla f(\bar{x}),x^{k}-\bar{x}\rangle=0, \end{eqnarray*}$

### 2.3 变分不等式的情况

$$$\lim_{k\in K_{0},k\rightarrow\infty}x^{k}=(\frac{\pi}{2},0).$$$

$\begin{eqnarray*} \ \psi_{k}&=&\frac{1}{\|x^{k}-\bar{x}\|}[(\cos x_{1}^{k}+\lambda)(x_{1}^{k}-\frac{\pi}{2})+({\rm e}^{x_{2}^{k}}-\lambda)x_{2}^{k}]\\ &\geq&\frac{1}{\|x^{k}-\bar{x}\|}[\cos x_{1}^{k}(x_{1}^{k}-\frac{\pi}{2})+({\rm e}^{x_{2}^{k}}-2\lambda)x_{2}^{k}] { (\text{由} (i))}\\ &\geq&\frac{1}{\|x^{k}-\bar{x}\|}(x_{1}^{k}-\frac{\pi}{2})\cos x_{1}^{k} {(\text{由} \lambda\in(0,\frac{1}{2}))}. \end{eqnarray*}$

$\begin{equation*} \limsup_{k\in K,k\rightarrow\infty}\psi_{k}\geq\lim_{k\in K_{0},k\rightarrow\infty}\frac{1}{\|x^{k}-\bar{x}\|}(x_{1}^{k}-\frac{\pi}{2})\cos x_{1}^{k}=0, \end{equation*}$

$\begin{equation*} \min_{x\in S}f(x)=\sin x_1 \cos x_2, \end{equation*}$

$\begin{equation*} S=\{x\in \mathbb{R}^{2}\mid 0\leq x_1\leq \frac{\pi}{2},\ 0\leq x_2\leq \frac{\pi}{2}\}, \end{equation*}$
$\begin{equation*} \bar{S}=\{x\in \mathbb{R}^{2}\mid x_1=0, 0\leq x_2\leq \frac{\pi}{2}\}\cup\{x\in \mathbb{R}^{2}\mid 0\leq x_1\leq \frac{\pi}{2}, x_2=\frac{\pi}{2}\}, \end{equation*}$
$\begin{equation*} \nabla f(x)=(\cos x_1\cos x_2,-\sin x_1\sin x_2), \end{equation*}$
$$$\nabla f(x)=\left\{\begin{array}{ll} (0, 0),& x=(0,\frac{\pi}{2}),\\[3mm] (\cos x_2, 0),& x_1=0,\ 0\leq x_2< \frac{\pi}{2},\\[3mm] (0,-\sin x_1),& 0< x_1\leq \frac{\pi}{2},\ x_2=\frac{\pi}{2}, \end{array} \right.$$$

$$$[T_S(x)\cap \hat{N}_{\bar{S}}(x)]^\circ =\left\{\begin{array}{ll} \mathbb{R}^2,& x=(0,\frac{\pi}{2}),\\[3mm] \{x\in \mathbb{R}^2\mid x_1\leq0,-\infty<x_2<+\infty\},& x_1=0,\ 0\leq x_2< \frac{\pi}{2},\\[3mm] \{x\in \mathbb{R}^2\mid -\infty<x_1<+\infty,x_2\geq0\},& 0< x_1\leq \frac{\pi}{2},\ x_2=\frac{\pi}{2}. \end{array} \right.$$$

$\begin{equation*} \nabla f(x)\!+\![T_S(x)\cap \hat{N}_{\bar{S}}(x)]^\circ =\left\{\begin{array}{ll} \mathbb{R}^2,& x=(0,\frac{\pi}{2}),\\[3mm] \{x\in \mathbb{R}^2\mid x_1\leq \cos x_2,-\infty\!<\!x_2\!<\!+\infty\},& x_1=0,\ 0\leq x_2< \frac{\pi}{2},\\[3mm] \{x\in \mathbb{R}^2\mid -\infty\!<\!x_1\!<\!+\infty,x_2\geq -\sin x_1\},& 0<x_1\leq \frac{\pi}{2},\ x_2\!=\!\frac{\pi}{2}. \end{array}\right. \end{equation*}$

(1) $P_{\bar{S}}(x^k)=(0,x^k_2) $$0\leq x_2^k<\frac{\pi}{2} ; (2) P_{\bar{S}}(x^k)=(x^k_1,\frac{\pi}{2})$$ 0< x_1^k\leq\frac{\pi}{2}$.

$\begin{eqnarray*} \ \Psi_k&=&\frac{1}{\|x^k-P_{\bar{S}}(x^k)\|}\langle\nabla f(x^k)-H(P_{\bar{S}}(x^k)),x^k-P_{\bar{S}}(x^k)\rangle\\ &=&\cos x_1^k\cos x_2^k-\lambda \geq \cos ^2(\frac{\pi}{2}-\varepsilon)-\lambda\geq0. \end{eqnarray*}$

$\begin{eqnarray*} \Psi_k&=&\frac{x_2^k-\frac{\pi}{2}}{|x_2^k-\frac{\pi}{2}|}(-\sin x_1^k\sin x_2^k+\lambda) =\sin x_1^k\sin x_2^k-\lambda \geq \sin ^2\varepsilon-\lambda\geq0. \end{eqnarray*}$

$\begin{equation*} \min_{x\in S}f(x)=x_1\log (1+x_2)+\frac{1}{2}(x_1-1)^2(x_2-1)^2, \end{equation*}$

$\begin{equation*} S=\{x\in \mathbb{R}^2\mid x_1\geq0,x_2\geq0,x_1^2+x_2^2\leq1\}, \bar{S}=\{(1,0),(0,1)\}. \end{equation*}$

$\begin{equation*} F(x)=(-x_1{\rm e}^{1-x_2^2},x_2(x_2^2-1)), \end{equation*}$

$\begin{equation*} S=\{x\in \mathbb{R}^2\mid -1\leq x_1\leq 1,-1\leq x_2\leq1\}, \end{equation*}$
$\begin{equation*} \bar{S}=\{((-1)^i,(-1)^j)\mid i=1,2;\ j=1,2\}. \end{equation*}$

### 3.1 最优化问题情况

(1) 设 (1.1) 式成立. 如果 $\{x^k\}$ 有限终止于 $\bar{S}$, 则有

$$$0\in\liminf_{k\rightarrow\infty}P_{T_{S}(x^{k})}(-\partial f(x^{k}));$$$

(2) 如果 (3.1) 式成立, 则 $\{x^k\}$ 有限终止于 $\bar{S}$.

(1) 如果 $x^k\in \bar{S}$, 则根据假设 (1.1), 可知 $\exists u^k\in\partial f(x^k)$, 使得 $-u^k\in N_S(x^k)$. 因此据 $S$ 的凸性与投影分解, 即知 $P_{T_S(x^k)}(-u^k)=0$, 即 (3.1) 式成立.

(2) 设 (3.1) 式成立.下面证明 $\{x^{k}\}$ 有限终止于 $\bar{S}$, 否则 $K=\{k\mid x^{k}\notin\bar{S}\}$ 将是一个无穷序列. 这样据 $\bar{S}$ 关于 $\{x^{k}\}\subset S$ 的广义弱尖锐性, 存在映射 $H:\bar{S}\rightarrow 2^{R^{n}}$与常数 $\alpha>0$ 使得

$$$\alpha B\subset H(z)+[T_{S}(z)\cap\hat{N}_{\bar{S}}(z)]^{\circ}, \forall z\in\bar{S},$$$

$\begin{equation*} \langle x^{k}-z^{k}, z-z^{k}\rangle\leq \frac{1}{2}\|z-z^{k}\|^{2}= \circ(\|z-z^{k}\|). \end{equation*}$

$$$x^{k}-z^{k}\in\hat{N}_{\bar{S}}(z^{k}).$$$

$$$x^{k}-z^{k}\in T_{S}(z^{k}), z^{k}-x^{k}\in T_{S}(x^{k}),$$$

$$$x^{k}-z^{k}\in T_{S}(z^{k})\cap\hat{N}_{\bar{S}}(z^{k}).$$$

$g_{k}=\frac{x^{k}-z^{k}}{\|x^{k}-z^{k}\|}(k\in K)$. 据 (3.2) 式可知存在 $\bar{v}^{k}\in H(z^{k})$, $\bar{\xi}^k\in[T_{S}(z^{k})\cap\hat{N}_{\bar{S}}(z^{k})]^{\circ}$, 使得

$$$\alpha g_{k}=\bar{v}^{k}+\bar{\xi}^{k}.$$$

$$$\alpha=\langle\bar{v}^{k},g_{k}\rangle+\langle\bar{\xi}^{k},g_{k}\rangle \leq \langle\bar{v}^{k},g_{k}\rangle.$$$

$\begin{equation*} 0\in\liminf_{k\in K,k\rightarrow\infty}P_{T_{S}(x^{k})}(-\partial f(x^{k}). \end{equation*}$

$$$\lim_{k\in K, k\rightarrow\infty}P_{T_{S}(x^{k})}(-\bar{u}^{k})=0.$$$

$\begin{eqnarray*} \alpha&\leq &\langle\bar{v}^{k},g_{k}\rangle\\ &=&\langle-\bar{u}^{k},-g_{k}\rangle-\langle\bar{u}^{k}-\bar{v}^{k},g_{k}\rangle\\ &\leq &\max\{\langle-\bar{u}^{k},d\rangle\mid d\in T_{S}(x^{k}),\|d\|\leq 1\}-\langle\bar{u}^{k}-\bar{v}^{k},g_{k}\rangle\\ &=&\|P_{T_{S}(x^{k})}(-\bar{u}^{k})\|-\langle\bar{u}^{k}-\bar{v}^{k},g_{k}\rangle. \end{eqnarray*}$

$\begin{eqnarray*} \alpha&\leq &\liminf_{k\in K,k\rightarrow\infty}\{\|P_{T_{S}(x^{k})}(-\bar{u}^{k})\|-\langle\bar{u}^{k}-\bar{v}^{k},g_{k}\rangle\}\\ &=&-\limsup_{k\in K,k\rightarrow\infty}\langle\bar{u}^{k}-\bar{v}^{k},g_{k}\rangle\leq 0, \end{eqnarray*}$

$f(\cdot) $$\mathbb{R}^n 上是局部 Lipschitz 的, 则 \partial f(x)\neq \emptyset 且 (1.1) 式成立. 因此由定理 3.1 即得下面推论. 推论 3.1 设在规划 (OP) 中, f(\cdot)$$ \mathbb{R}^n$ 上是局部 Lipschitz 的, $\bar{S}$ 关于 $\{x^k\}\subset S$ 是广义弱尖锐性的. 则 $\{x^k\}$ 有限终止于 $\bar{S}$, 当且仅当 (3.1) 式成立.

### 3.3 实例

$\begin{equation*} \min_{x\in S} f(x)=\log (1+x_1)+\frac{1}{2}(x_2)^2, \end{equation*}$

$\begin{equation*} S=\{x\in \mathbb{R}^2\mid 0\leq x_1\leq1, -1\leq x_2\leq1\}, \bar{S}=\{(0,0)\}. \end{equation*}$

$$$[T_S(0,0)\cap {\hat{N}}_{\bar{S}}(0,0)]^\circ=N_{S}(0,0)=\{\xi\in \mathbb{R}^2\mid\xi_1\leq0,\xi_2=0\},$$$

$$$\alpha B\subset H(z)+[T_S(z)\cap {\hat{N}}_{\bar{S}}(z)]^\circ, \forall z\in \bar{S}.$$$

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