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数学物理学报, 2018, 38(6): 1144-1152 doi:

论文

自反巴拿赫空间中方向扰动的广义混合变分不等式的可解性

罗雪萍,, 崔梦天

Solvability of Directional Perturbed Generalized Mixed Variational Inequalities in Reflexive Banach Spaces

Luo Xueping,, Cui Mengtian

通讯作者: 罗雪萍, E-mail: xuepingluo2014@outlook.com

收稿日期: 2017-10-20  

基金资助: 国家自然科学基金.  11701480
西南民族大学中央高校基本科研业务费基金.  2018HQZZ23
四川省教育厅重点项目.  18ZA0511
西南民族大学创新团队基金.  14CXTD03
四川省教育厅创新团队项目.  15TD0050
四川省青年科技创新研究团队项目.  2017TD0028

Received: 2017-10-20  

Fund supported: the NSFC.  11701480
the Fundamental Research Funds for the Central Universities of Southwest University for Nationalities.  2018HQZZ23
the Key Projects of the Education Department of Sichuan Province.  18ZA0511
the Innovation Team Funds of Southwest University for Nationalities.  14CXTD03
the Innovative Research Team of the Education Department of Sichuan Province.  15TD0050
the Sichuan Youth Science and Technology Innovation Research Team.  2017TD0028

摘要

该文给出了在自反巴拿赫空间中,一个强制条件下,方向扰动的广义混合变分不等式的可解性.其中,关于集合受方向扰动的研究结果是全新的.该文改进与推广了一些已有的结果(数学物理学报,2016,36A(3):473-480).

关键词: 广义混合变分不等式 ; 可解性 ; 方向扰动 ; 强制条件

Abstract

Solvability of directional perturbed generalized mixed variational inequalities is discussed in reflexive Banach spaces, under a coercivity condition. In particular, the result we present that the set is directional perturbed is new. Our results generalize and extend some known results in this area (Acta Math Sci, 2016, 36A(3):473-480).

Keywords: Generalized mixed variational inequality ; Existence and boundedness ; Directional perturbation ; Coercivity condition

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本文引用格式

罗雪萍, 崔梦天. 自反巴拿赫空间中方向扰动的广义混合变分不等式的可解性. 数学物理学报[J], 2018, 38(6): 1144-1152 doi:

Luo Xueping, Cui Mengtian. Solvability of Directional Perturbed Generalized Mixed Variational Inequalities in Reflexive Banach Spaces. Acta Mathematica Scientia[J], 2018, 38(6): 1144-1152 doi:

1 引言

X是自反巴拿赫空间, X是其对偶空间, x\in X的范数, \left\langle {\phi, x} \right\rangle\phi\in X^*x\in X的对偶积.设K\subset X是非空闭凸子集, F:K\to 2^{X^*}是具有非空值的集值映射和f:K\to {\Bbb R}\cup\{+\infty\}是函数.本文将讨论以下广义混合变分不等式:求x\in Kx^*\in F(x)使得

\begin{equation}\label{eq:mvi}\left\langle {x^*}{y-x} \right\rangle +f(y)-f(x)\ge 0, \quad \forall y\in K.\end{equation}
(1.1)

上述问题引起了众多学者的广泛关注[1-6].下面的讨论中, GMVIP(F, f, K)和S(F, f, K)分别表示问题(1.1)及其解集.

GMVIP(F, f, K)包括大量其他问题作为其特例:

(ⅰ)如果f\equiv0,则(1.1)式退化为以下广义变分不等式GVIP(F, K)[7-14]:求x\in Kx^*\in F(x)使得

\left\langle {x^*}{y-x} \right\rangle\ge 0, \quad \forall y\in K.

(ⅱ)如果F是单值且f\equiv0,则(1.1)式退化为经典的变分不等式VIP(F, K)[15-17]:求x\in K使得

\label{ineq2} \left\langle {F(x)}{y-x} \right\rangle\ge 0, \quad \forall y\in K.

数学规划中,研究带扰动项的变分不等式的一些性质是重要课题之一[18-24].特别地,文献[15,推论5.5.12]利用拓扑度为理论工具研究了X={\Bbb R}^nF是单值时,强制条件蕴含了扰动变分不等式VIP(F+q, K)有解.

数学规划中,研究带扰动项的变分不等式的一些性质是重要课题之一[18-24].特别地,文献[15,推论5.5.12]利用拓扑度为理论工具研究了X={\Bbb R}^nF是单值时,强制条件蕴含了扰动变分不等式VIP(F+q, K)有解.

目前,扰动变分不等式的稳定性分析被推广到了空间是无限维, F是集值的情形. He[10]在自反巴拿赫空间中,研究了当F是伪单调时,映射或约束集扰动的情形下, GVIP(F, K)的稳定性.在文献[9]中, Fan和Zhong推广了文献[10]的结果到映射与约束集同时被扰动的情形. Zhong和Huang[25]进一步将文献[9-10]的结果推广到含f -伪单调映射的混合变分不等式.最近, Li和He[12]推广了文献[15]的结果,在一个较弱的强制条件下,讨论了F是集值时,扰动广义变分不等式的可解性.随后, Li和Sun[26]推广了文献[12]的主要结果到无限维空间,还给出了比文献[12,定理1.1]更强的结论.那么,研究广义混合变分不等式的可解性结果是一个自然的问题.

受到文献[12, 26]中研究工作的启发,本文讨论了在自反巴拿赫空间中,映射和集合同时被方向扰动的情形下,广义混合变分不等式的可解性结果.本文改进与推广了文献[12, 26]的结果.值得一提的是集合受方向扰动的情形是首次被讨论,是一个全新的结果.

2 预备知识

X, X^*K与第一部分的假设相同. "\rightarrow"和"\rightharpoonup"分别表示强收敛和弱收敛.定义

barr(K):=\{x^*\in X^*:\mathop {\sup }\limits_{x \in K}\left\langle {x^*}{x} \right\rangle<\infty\}

K的匣锥. K的回收锥是闭凸锥,被定义为

K_\infty:=\{d\in X:\exists t_n\downarrow 0, \exists x_n\in K, t_nx_n\rightharpoonup d\}.

众所周知,给定x_0\in K,则

K_\infty=\{d\inX:x_0+\lambda d\in K \mbox{ for all }\lambda>0\}.

对于X中的非空子集D, D^-:=\{x^*\in X^*:\left\langle {x^*}{x} \right\rangle\le0, \forall x\in D\}int(D)表示D的内部.由文献[27,命题3.10],可知

\begin{eqnarray}\label{eq:barr}barr(K)^-=K_\infty.\end{eqnarray}
(2.1)

对于正整数m\in\mathbb N, K_m:=\{x\in K:\left\| x \right\|\leq m\}, {\Bbb B}(0, m):=\{x\in X:\left\| x \right\| < m\}\bar{{\Bbb B}}(0, m):=\{x\in X:\left\| x \right\|\leq m\}.

f:K\to {\Bbb R}\cup\{+\infty\}是真凸下半连续函数.定义f的上境图为

epi f:=\{(x, \lambda)\in K\times {\Bbb R}: f(x)\leq \lambda\}.

epi f的回收锥是回收函数f_{\infty}的上境图,被定义为

f_{\infty}(d):=\mathop {\lim }\limits_{t \to + \infty } \frac{f(x_0+t d)-f(x_0)}{t},

其中,任意的x_0\in dom f.可知,回收函数f_{\infty}是真凸下半连续的,即是弱下半连续的.

定义2.1 设F:K\to 2^{ X^{*}}是具有非空值的集值映射和f:K\to {\Bbb R}\cup\{+\infty\}是真凸下半连续函数.称F

(i)在K上单调,如果对于任意的x, y\in K及任意的x^{*}\in F(x)y^{*}\in F(y),

\left\langle {y^*-x^*}{y-x} \right\rangle\ge 0;

(ⅱ)在K上拟单调,如果对于任意的x, y\in K及任意的x^{*}\in F(x)y^{*}\in F(y),

\left\langle {x^{*}, y-x} \right\rangle>0\Rightarrow\langle y^{*}, y-x\rangle\geq0;

(ⅲ)在Kf -拟单调,如果对于任意的x, y\in K及任意的x^{*}\in F(x)y^{*}\in F(y),

\langle x^{*}, y-x\rangle+f(y)-f(x)>0\Rightarrow\langle y^{*}, y-x\rangle+f(y)-f(x)\geq0;

(ⅳ)在Kf -伪单调,如果对于任意的x, y\in K及任意的x^{*}\in F(x)y^{*}\in F(y),

\langle x^{*}, y-x\rangle+f(y)-f(x) \geq0\Rightarrow\langle y^{*}, y-x\rangle+f(y)-f(x)\geq0;

(ⅴ)在K上关于U\subset X^*稳定f -拟单调,如果FF(\cdot)-u对于所有的u\in U是拟单调的;

(ⅵ)在x_{0}\in K处上半连续,如果对于F(x_{0})的任意邻域 {\cal N}(F(x_{0})),存在x_{0}的邻域{\cal N}(x_{0}), 使得

F(x)\subset{\cal N}(F(x_{0})), \quad \forall\ x\in{\cal N}(x_{0})\cap K;

(ⅶ)在K上沿线节上半连续,如果FX^*中关于弱拓扑在K中的任意线段上都是上半连续的.

注2.1 (ⅰ)显然地,单调映射必定是拟单调和f -拟单调的;

(ⅱ)如果f\equiv0, f -拟单调退化为拟单调的;

(ⅲ) f -伪单调映射是f -拟单调的,然而f -拟单调映射不一定是f -伪单调的.

以下例子说明了如果f\neq0, f -拟单调映射不一定是拟单调.

例2.1 设X={\Bbb R}=(-\infty, +\infty)K=[1, 2].设

f(x)=x^2, \ \forall x\in K \quad \mbox{和} \quadF(x)\equiv[-2, 2], \ \forall x\in K.

易知F(\cdot)不是拟单调的.然而, F(\cdot)K上是f -拟单调.事实上,设x^*\in F(x)y^*\in F(y),

\left\langle {x^*}{y-x} \right\rangle+f(y)-f(x)=(x^*+y+x)(y-x)>0,

x^*\in[-2, 2]和x, y\in[1, 2]蕴含x^*+y+x>0.所以y-x>0.这样,有

\left\langle {y^*}{y-x} \right\rangle+f(y)-f(x)=(y^*+y+x)(y-x)\geq0,

因为y^*+y+x\in[0, 6].因此, F(\cdot)f -拟单调的.

以下例子说明了如果f\neq0, 拟单调映射不一定是f -拟单调的.

例2.2 设X={\Bbb R}=(-\infty, +\infty)K=[-2, 2].

f(x)=x^2, \ \forall x\in K \quad \mbox{和}\quadF(x)\equiv[0, 3], \ \forall x\in K.

易知F(\cdot)K上是拟单调的.取(x, x^*)=(-2, 3)(y, y^*)=(-\frac{1}{2}, 2).

\left\langle {x^*}{y-x} \right\rangle+f(y)-f(x)=(3-\frac{1}{2}-2)(-\frac{1}{2}+2)>0,

然而

\left\langle {y^*}{y-x} \right\rangle+f(y)-f(x)=(2-\frac{1}{2}-2)(-\frac{1}{2}+2)<0.

因此, F(\cdot)不是f -拟单调的.

以下例子说明了如果f\neq0, f -拟单调映射不一定是f -伪单调的.

例2.3 设X={\Bbb R}=(-\infty, +\infty)K=[1, 2].设f:K\rightarrow{\Bbb R}是可微凸函数,使得对于所有x\in K, |f'(x)|\leq r, 其中r>0是常数.设

F(x)\equiv[r, r+e^x], \ \forall x\in K.

接下来证明F(\cdot)K上是f -拟单调.设x^*\in F(x)y^*\in F(y)

\left\langle {x^*}{y-x} \right\rangle+f(y)-f(x)>0,

由中值定理,存在\xi\in(1, 2)使得f(y)-f(x)=f'(\xi)(y-x), 所以

(x^*+f'(\xi))(y-x)>0.

这样, x^*\geq r蕴含x^*+f'(\xi)>0, 所以y-x>0.因为{y^*} \in \left[ {r, r + {e^x}} \right],

\left\langle {y^*}{y-x} \right\rangle+f(y)-f(x)=(y^*+f'(\xi))(y-x)\geq0.

因此, F(\cdot)f -拟单调.另一方面,取(x, x^*)=(2, r)(y, y^*)=(1, r+e),

\left\langle {x^*}{y-x} \right\rangle+f(y)-f(x)=(r+f'(\xi))(1-2)=0,

易知r+f'(\xi)=0.然而

\left\langle {y^*}{y-x} \right\rangle+f(y)-f(x)=(r+e+f'(\xi))(1-2)=-e<0.

因此, F(\cdot)不是f -伪单调的.

引理2.1[10] 设KX中的非空闭凸集且int(barr K)\ne\emptyset, 则不存在\{x_n\}\subset K\left\| {{x_n}} \right\| \to \infty使得\frac{{{x_n}}}{{\left\| {{x_n}} \right\|}} ⇀ 0.如果K是锥,则不存在\{d_n\}\subset K\left\| {{d_n}} \right\|=1使得d_n\rightharpoonup 0.

引理2.2[9] 设KX中的非空闭凸集且int(barr K)\ne\emptyset, 则不存在\{d_n\}\subset K_{\infty}\left\| {{d_n}} \right\| = 1使得d_n\rightharpoonup0.

命题2.1  (ⅰ)如果DK中是非空有界闭凸集, F: K\rightarrow 2^{X^*}是具有弱^*紧凸值的沿线节上半连续和拟单调映射,则GVIP(F, D)有解.

(ⅱ)如果DK中是非空有界闭凸集, f:K\to {\Bbb R}\cup\{+\infty\}是真凸下半连续函数, F: K\rightarrow 2^{X^*}是具有弱^*紧凸值的沿线节上半连续和稳定f -拟单调映射,则对于任意的\zeta\in X^*, GMVIP ({F-\zeta}, {f}, {D})有解.

  (ⅰ)的证明参见文献[13]. (ⅱ)利用与(ⅰ)中类似的方法即可得证.

3 可解性

本部分为本文的主要结果,给出了扰动混合变分不等式在扰动项为映射F和集合K时的可解性结果. I表示集合K-tK_\infty\setminus\{0\}, 对于任意的t>0.

引理3.1 如果KX中的非空闭凸集,则K_\infty=I_{\infty}barr(K)=barr(I).

 设d\in I_{\infty}.给定x_0\in K, 因为K:=I+tK_\infty\setminus\{0\}, 存在y_0\in I\bar{d}\in K_\infty\setminus\{0\}使得x_0=y_0+t\bar{d}.对于任意的\lambda>0,

x_0+\lambda d=(y_0+t\bar{d})+\lambda d=(y_0+\lambda d)+t\bar{d}\in I+tK_\infty\setminus\{0\}=K,

因此, d\in K_\infty.这样, I_{\infty}\subset K_\infty得证.

另一方面,设d_1\in K_{\infty}.给定x_1\in I, 因为I:=K-tK_\infty\setminus\{0\}, 存在y_1\in K\bar{d_1}\in K_\infty\setminus\{0\}使得x_1=y_1-t\bar{d_1}.对于任意的\lambda_1>0,

x_1+\lambda_1 d_1=(y_1-t\bar{d_1})+\lambda_1 d_1=(y_1+\lambda d_1)-t\bar{d_1}\in K-tK_\infty\setminus\{0\}=I,

因此, d_1\in I_\infty.这样, K_\infty\subset I_{\infty}.因此, K_\infty=I_{\infty}.

再由(2.1)式,有barr(K)=barr(I).

定理3.1 设KX中的非空闭凸集且int(barr K)\ne\emptyset.f:I\to {\Bbb R}\cup\{+\infty\}是真凸下半连续函数和F:I\to 2^{X^*}是具有非空弱^*-紧凸值的沿线节上半连续和稳定f -拟单调映射.如果以下强制条件成立: \exists r>0, \forall x\in K\setminus K_r, \exists y\in K_r使得

\begin{eqnarray}\label{eq:2220}&&\nonumber\exists y^*\inF(y)\cap\{y^*:\left\langle {y^*}{d} \right\rangle+f_{\infty}(d)<0, \forall d\inK_\infty\setminus\{0\}\}, \\&&\left\langle { y^*, y-x} \right\rangle+f(y)-f(x)<0, \end{eqnarray}
(3.1)

则对于任意的q\in int(barr K), 存在k>r, 使得

\emptyset\neq{\rm S}(F-\varepsilon q, f, K-tK_\infty\setminus\{0\})\subset\overline{\mathbb{B}}(0, \frac{1}{\varepsilon}), \quad\forall\varepsilon\in(0, \frac{1}{k}), \forall t>0.

  (1)首先,需要证明对于任意的q\in int(barr K), 存在m>r使得对所有的\varepsilon\in(0, \frac{1}{m}), S(F-\varepsilon q, f, K-tK_{\infty}\setminus\{0\})\neq\emptyset.假设存在q\in int(barr K), 对所有的m>r, 存在\varepsilon_m\in(0, \frac{1}{m})使得S(F-\varepsilon_m q, f, K-tK_{\infty}\setminus\{0\})=\emptyset.

E_m:=\{x\in I:\left\| x \right\|\leq \frac{1}{\varepsilon_m}\}.E_m是有界闭凸的,利用命题2.1(ⅱ),可知GMVIP ({F-\varepsilon_{m}q}, {f}, {E_m})有解.对于每一m>r, 存在x_m\in E_m使得

\begin{eqnarray}\label{eq:11x}\sup\limits_{\xi\in F(x_m)}\left\langle {\xi-\varepsilon_{m}q}{y-x_m} \right\rangle+f(y)-f(x_m)\geq0, \quad \forall y\in E_m.\end{eqnarray}
(3.2)

(ⅰ)对某一m>r, \left\| {{x}_{m}} \right\| < \frac{1}{\varepsilon_m}, 则对于任意的y\in K-tK_\infty\setminus\{0\}, 存在\lambda\in(0, 1)使得z_{\lambda}:=x_m+\lambda(y-x_m)\in E_m, 因为KK_{\infty}的凸性蕴含K-tK_{\infty}\setminus\{0\}是凸的.所以,由(3.2)式和f是凸函数,有

\begin{eqnarray*}0&\leq&\sup\limits_{\xi\in F(x_m)}\left\langle {\xi-\varepsilon_{m}q}{z_{\lambda}-x_m} \right\rangle+f(z_{\lambda})-f(x_m)\\&\leq&\lambda[\sup\limits_{\xi\inF(x_m)}\left\langle {\xi-\varepsilon_{m}q} {y-x_m} \right\rangle+f(y)-f(x_m)].\end{eqnarray*}

y\in K-tK_\infty\setminus\{0\}的任意性,可知x_m是GMVIP({F-\varepsilon_{m}q}, {f}, {K-tK_{\infty}\setminus\{0\}})的解.这样, x_m\in {\rm S}(F-\varepsilon_{m}q, f, K-tK_\infty\setminus\{0\})\neq\emptyset.

(ⅱ)对每一m>r, \left\| {{x}_{m}} \right\|=\frac{1}{\varepsilon_m}>m.因为x_m\in K-tK_\infty\setminus\{0\}, 存在\widetilde{x_m}\in Kp_m\in K_\infty\setminus\{0\}使得x_m=\widetilde{x_m}-tp_m.下面需要证明\widetilde{x_m}\not\in K_r.事实上,若不然,则\left\| {\widetilde{x_m}} \right\|\leq r.这样

t||p_m||\leq\left\| {{x}_{m}} \right\|+||\widetilde{x_m}||\leq\frac{1}{\varepsilon_m}+r<\frac{2}{\varepsilon_m}.

所以,对于所有的m, t||p_m|| < +\infty.然而,由p_m\in K_\infty\setminus\{0\}, 可得||p_m||>0.t\to+\infty时, t||p_m||没有上界,则产生矛盾.

由强制条件(3.1)可知存在y_m\in K_r使得存在y^*_m\in F(y_m)\cap\{y^*:\left\langle {y^*}, {d} \right\rangle+f_{\infty}(d) < 0, \forall d\in K_\infty\}, \left\langle {y^*_m}{y_m-\widetilde{x_m}} \right\rangle+f(y_m)-f(\widetilde{x_m}) < 0.因此

\begin{eqnarray*}\label{eq:xx}&&\left\langle {y^*_m}{y_m-x_m} \right\rangle+f(y_m)-f(x_m)\\&=&\left\langle {y^*_m}{y_m-\widetilde{x_m}} \right\rangle+f(y_m)-f(\widetilde{x_m})+\left\langle {y^*_m}{\widetilde{x_m}-x_m} \right\rangle+f(\widetilde{x_m})-f(x_m)\\&<&\left\langle {y^*_m}, {tp_m} \right\rangle+f(x_m+tp_m)-f(x_m), \quad \forall t>0.\end{eqnarray*}

进而

\begin{eqnarray*}\label{eq:YY}\frac{\left\langle {y^*_m}{y_m-x_m} \right\rangle+f(y_m)-f(x_m)}{t}&<&\left\langle {y^*_m}, {p_m} \right\rangle+\frac{f(x_m+tp_m)-f(x_m)}{t}\\&\rightarrow&\left\langle {y^*_m}, {p_m} \right\rangle+f_{\infty}(p_m), \quad \mbox{ as } t\rightarrow\infty.\end{eqnarray*}

所以

\begin{eqnarray*}\label{eq:YY1}\lim\limits_{t\rightarrow\infty}\frac{\left\langle {y^*_m}{y_m-x_m} \right\rangle+f(y_m)-f(x_m)}{t}\leq\left\langle {y^*_m}, {p_m} \right\rangle+f_{\infty}(p_m)<0.\end{eqnarray*}

t>0,

\mathop {\inf }\limits_{y_m^* \in F({y_m})} \left\langle {y^*_m}{y_m-x_m} \right\rangle+f(y_m)-f(x_m)<0.

Ff -拟单调性,可得

\begin{eqnarray}\label{eq:x}\mathop {\sup }\limits_{\xi \in F({x_m})} \left\langle {\xi}{y_m-x_m} \right\rangle+f(y_m)-f(x_m)\leq0.\end{eqnarray}
(3.3)

不失一般性,假设\frac{x_m}{\left\| {{x}_{m}} \right\|}\rightharpoonup d\in I_{\infty}.由引理3.1,有d\in K_{\infty}.因为int(barr I)=int(barr K)\ne\emptyset, 引理2.2蕴含d\neq0.q\in int(barr K)和(2.1)式,可知\left\langle {q}, {d} \right\rangle\leq0.接下来证明\left\langle {q}, {d} \right\rangle < 0.事实上,若不然,则\left\langle {q}, {d} \right\rangle=0.因为q\in int(barr K), 所以对于所有的x^*\in X^*, 存在\delta\in(0, 1)使得\delta q+(1-\delta)x^*\in barr K.所以, \left\langle {\delta q+(1-\delta)x^*}, {d} \right\rangle\leq0,进而, \left\langle {x^*}, {d} \right\rangle\leq0.类似地,可证得-\left\langle {x^*}, {d} \right\rangle\leq0.这样, \left\langle {x^*}, {d} \right\rangle=0, d\neq0产生矛盾.因此, \left\langle {q}, {d} \right\rangle < 0.

因为\left\| {{y_m}} \right\| < r < \frac{1}{\varepsilon_m}, 则对任意的y\in K-tK_\infty\setminus\{0\}, 存在\theta\in(0, 1)使得z_{\theta}:=y_m+\theta(y-y_m)\left\| {{z_\theta }} \right\|\leq \frac{1}{\varepsilon_m}.现在,将证明z_{\theta}\in I.事实上,由I:=K-tK_\infty\setminus\{0\}y_m\in K, 存在\widetilde{y_m}\in I\widetilde{p_m}\in K_{\infty}, y_m=\widetilde{y_m}+t\widetilde{p_m}, 因此, z_{\theta}=\widetilde{y_m}+\theta(y-\widetilde{y_m})+(1-\theta)t\widetilde{p_m}.因为I是凸的, \widetilde{y_m}+\theta(y-\widetilde{y_m})\in I.由引理3.1,可知\widetilde{p_m}\in K_{\infty}=I_{\infty}.由回收锥的定义和(1-\theta)t>0, z_{\theta}\in I.因此, z_{\theta}\in E_m.

由(3.2)和(3.3)式,有

\begin{eqnarray*}0&\leq&\sup\limits_{\xi\in F(x_m)}\left\langle {\xi-\varepsilon_{m}q}, {z_{\theta}-x_m} \right\rangle+f(z_{\theta})-f(x_m)\\&=&\theta[\sup\limits_{\xi\inF(x_m)}\left\langle {\xi-\varepsilon_{m}q}, {y-x_m} \right\rangle+f(y)-f(x_m)]\\&&+(1-\theta)[\sup\limits_{\xi\in F(x_m)}\left\langle {\xi}, {y_m-x_m} \right\rangle+f(y_m)-f(x_m)]+\varepsilon_{m}(1-\theta)(\left\langle {q}, {x_m} \right\rangle-\left\langle {q}, {y_m} \right\rangle)\\&\leq&\theta[\sup\limits_{\xi\in F(x_m)}\left\langle {\xi-\varepsilon_{m}q}, {y-x_m} \right\rangle+f(y)-f(x_m)]+\varepsilon_{m}(1-\theta)(\left\langle {q}, {x_m} \right\rangle-\left\langle {q}, {y_m} \right\rangle).\end{eqnarray*}

因为\varepsilon_{m}=\frac{1}{\left\| {{x}_{m}} \right\|}, 所以{\varepsilon _m}\left\langle {q, {x_m}} \right\rangle = \left\langle {q, \frac{{{x_m}}}{{\left\| {{x_m}} \right\|}}} \right\rangle \to \left\langle {q, d} \right\rangle < 0.由极限的保号性定理,存在M>0, m>M, \varepsilon_{m} q. {x_m} < \frac{1}{2} q. {d} < 0.

又因为\{y_m\}\subset K_r, \{y_m\}是有界的,有\varepsilon_{m}\left\langle {q}, {y_m} \right\rangle\rightarrow0.因此,对于足够大的m,

\mathop {\sup }\limits_{\xi \in F({x_m})} \left\langle {\xi-\varepsilon_{m}q}{y-x_m} \right\rangle+f(y)-f(x_m)\geq0.

y\in K-tK_\infty\setminus\{0\}的任意性, x_m是GMVIP{F-\varepsilon_{m}q}, {f}, {K-tK_\infty\setminus\{0\}}的解.这样, x_m\in {\rm S}(F-\varepsilon_{m}q, f, K-tK_\infty\setminus\{0\})\neq\emptyset.

上述两种情况均与{\rm S}(F-\varepsilon_{m}q, f, K-tK_\infty\setminus\{0\})=\emptyset产生矛盾.

(2)下面需要证明对于任意的q\in int(barr K), 存在k\geq m>r使得对每一\varepsilon\in(0, \frac{1}{k}), {\rm S}(F-\varepsilon q, f, K-tK_{\infty}\setminus\{0\}) \subset\bar{{\Bbb B}}(0, \frac{1}{\varepsilon}).鉴于第一部分的证明,存在m>r, 对任意的k\geq m\varepsilon\in(0, \frac{1}{k}), {\rm S}(F-\varepsilon q, f, K-tK_{\infty}\setminus\{0\})\neq\emptyset.假设存在q\in int(barr K), 对任意的k\geq m, 存在\varepsilon_k\in(0, \frac{1}{k})使得x_k\in {\rm S}(F-\varepsilon_k q, f, K-tK_{\infty}\setminus\{0\}), 然而, x_k\not\in\bar{{\Bbb B}}(0, \frac{1}{\varepsilon_k}).

因为x_k\in I||x_k||>\frac{1}{\varepsilon_k}>k\geq m>r, 运用第一部分(1)的证明方法,总能找到\widetilde{x_k}\in K||\widetilde{x_k}||>r.由强制条件(3.1),存在y_k\in K_r使得存在y_k^*\in F(y_k)\cap\{y^*:\left\langle { y^*}, {d} \right\rangle+f_{\infty}(d) < 0, \forall d\in K_\infty\}, \left\langle {y_k^*, {y_k} - \widetilde {{x_k}}} \right\rangle + f({y_k}) - f(\widetilde {{x_k}}) < 0.类似地,可得

\mathop {\sup }\limits_{{\xi _k} \in F({x_k})} \left\langle {\xi_k}, {y_k-x_k} \right\rangle+f(y_k)-f(x_k)\leq0.

同时, y_k\in K蕴含y_k\in I, 因为K=I+tK_\infty=I+tI_\infty\subset I.x_k\in{\rm S}(F-\varepsilon_k q, f, K-tK_{\infty}\setminus\{0\}),

\begin{eqnarray}\label{eq:j1}0&\nonumber\leq&\sup\limits_{\xi^*_k\in F(x_k)}\left\langle {\xi^*_k-\varepsilon_k q}, {y_k-x_k} \right\rangle+f(y_k)-f(x_k)\\&\nonumber=&\sup\limits_{\xi^*_k\inF(x_k)}\left\langle {\xi^*_k}, {y_k-x_k} \right\rangle+f(y_k)-f(x_k)+\varepsilon_{k}(\left\langle {q}, {x_k} \right\rangle-\left\langle {q}, {y_k} \right\rangle)\\&\leq&\varepsilon_{k}(\left\langle {q}, {x_k} \right\rangle-\left\langle {q}, {y_k} \right\rangle).\end{eqnarray}
(3.4)

类似地,可以假设\frac{x_k}{||x_k||}⇀ e.这样, \left\langle {q}, {\frac{x_k}{||x_k||}} \right\rangle\rightarrow\left\langle {q}, {e} \right\rangle < 0.由极限的保号性定理,存在K>0, k>K时, \left\langle {q}, {\frac{x_k}{||x_k||}} \right\rangle < 0.因为||x_k||>\frac{1}{\varepsilon_k}, ||x_k||\varepsilon_k>1.k>K时,有

\varepsilon_k\left\langle {q}, {x_k} \right\rangle=||x_k||\varepsilon_k\left\langle {q}{\frac{x_k}{||x_k||}} \right\rangle<0.

类似地, \{y_k\}\subset K_r蕴含\{y_k\}是有界的, \varepsilon_{k}\left\langle {q}, {y_k} \right\rangle\rightarrow0.因此,对足够大的k, \varepsilon_{k}(\left\langle {q}, {x_k} \right\rangle-\left\langle {q}, {y_k} \right\rangle) < 0.这样,与(3.4)式产生矛盾.证毕.

注3.1 Li和He[12]{\Bbb R}^n中给出了仅仅映射被扰动的GVIP({F}{K})结果(参见文献[12,定理3.2]).因此,定理3.1是文献[12]中定理3.2的推广.定理3.1也改进与推广了文献[26]中的定理3.1到广义混合变分不等式.

如果扰动项仅仅是映射,则有如下推论3.1.

推论3.1 设KX中的非空闭凸集且int(barr K)\ne\emptyset.f:K\to {\Bbb R}\cup\{+\infty\}是真凸下半连续函数和F:K\to 2^{X^*}是具有非空弱^*-紧凸值的沿线节上半连续和稳定f -拟单调映射.如果以下强制条件成立: \exists r>0, \forall x\in K\setminus K_r, \exists y\in K_r使得

\begin{eqnarray*}\inf\limits_{y^*\in F(y)}\left\langle {y^*, y-x} \right\rangle+f(y)-f(x)<0, \end{eqnarray*}

则对任意的q\in int(barr K), 存在k>r, 使得

\emptyset\neq{\rm S}(F-\varepsilon q, f, K)\subset\bar{{\Bbb B}}(0, \frac{1}{\varepsilon}), \quad\forall\varepsilon\in(0, \frac{1}{k}).

如果扰动项仅仅是集合,则有如下推论3.2.

推论3.2 设KX中的非空闭凸集且int(barr K)\ne\emptyset.f:I\to {\Bbb R}\cup\{+\infty\}是真凸下半连续函数和F:I\to 2^{X^*}是具有非空弱^* -紧凸值的沿线节上半连续和稳定f -拟单调映射.如果以下强制条件成立: \exists r>0, \forall x\in K\setminus K_r, \exists y\in K_r使得

\begin{eqnarray*}&&\nonumber\exists y^*\inF(y)\cap\{y^*:\left\langle {y^*}, {d} \right\rangle+f_{\infty}(d)<0, \forall d\inK_{\infty}\setminus\{0\}\}, \\&&\left\langle {y^*, y-x} \right\rangle+f(y)-f(x)<0, \end{eqnarray*}

则存在k>r, 使得

\emptyset\neq{\rm S}(F, f, K-tK_\infty\setminus\{0\})\subset\bar{{\Bbb B}}(0, \frac{1}{\varepsilon}), \quad\forall\varepsilon\in(0, \frac{1}{k}), \forall t>0.

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