众所周知,非光滑约束向量优化问题在经济理论研究,工程设计,管理科学等众多领域中发挥着关键的作用,见文献[1-10].当所讨论的空间是无穷维时,约束向量优化问题的精确解并不一定存在.因此,在无穷维空间中,考虑优化问题的某种意义下的近似解是有必要和有价值的.事实上,以各种方式定义的约束向量优化问题的近似解,以及相关这些解的必要条件已经出现在各种文献中,参见文献[11-21].

因为在有限维空间中,拉格朗日乘数法是求解标量优化问题最优解的重要方法.这也正是许多学者研究约束向量优化问题的拉格朗日乘数法的动机和原因所在,见文献[19-21].在文献[20]中,根据Clarke法锥以及目标多值映射和约束多值映射的coderivatives,并利用fuzzy分离性定理及Ekeland变分原则,郑教授和他的合作者证明了巴拿赫空间中关于弱近似帕雷托解的多目标约束向量优化问题的类拉格朗日乘数法.

受文献[20]中的结果所启发,在文献[21]中,我们考虑实希尔伯特空间中含有一个约束条件的向量优化问题,利用proximal法锥及目标多值映射和约束多值映射的有关proximal法锥的coderivatives,我们给出了该类优化问题的弱近似帕雷托解存在的拉格朗日乘数法,见文献[21].然而,通过进一步地研究,我们发现文献[21]中的结果并不能自然地延拓到含有两个及两个以上约束条件的优化问题上.其中的原因是当$F$被定义为有限个多值映射$F_{i}\ (i=1, 2, \cdots, m)$ ($m>1$)的乘积时, $F$相关于proximate法锥的coderivative不一定包含于每个$F_{i}$关于proximate法锥的coderivative的和中.下面我们结合文献[21]中的结论对以上所述作详细地说明.

设$X$, $Y_{0}$和$Y$是实希尔伯特空间, $F_{0}: X \rightrightarrows Y_{0}$和$F: X \rightrightarrows Y$是闭多值映射, $A$是$X$中的闭子集, $K_{0}$和$K$分别是空间$Y_{0}$和$Y$中的闭凸锥.文献[21]中考虑如下向量优化问题

(1.1) $K_{0}-\mbox{min}F_{0}(x)\ \ \ \mbox{subject to}\ \ \ F(x)\cap-K\neq\emptyset\ \ \mbox{and}\ \ x\in A, $

并建立了问题(1.1)的弱近似帕雷托解存在的必要条件-拉格朗日乘数法^[21].

定理1.1 设$\varepsilon>0$和$(\bar{x}, \bar{y}_{0})$是$(1.1)$的弱$\varepsilon$ -帕雷托解.假设$\bar{y}\in F(\bar{x})\cap(-K)$.则任意$\lambda>0$,存在

$x_{0}, \ x\in B(\bar{x}, \lambda), \ \ y_{0}\in F_{0}(x_{0})\cap B(\bar{y}_{0}, \lambda), \ \ y\in F(x)\cap B(\bar{y}, \lambda), $

$a\in A\cap B(\bar{x}, \lambda), \ \ y_{0}^{*}\in K_{0}^{+}, \ \ y^{*}\in K^{+}, \ \ \gamma\in N_{P}(A, a)+\frac{\varepsilon}{\lambda}B_{X}, $

$\zeta_{0}\in D_{P}^{*}F_{0}(x_{0}, y_{0})\big(y_{0}^{*}+\frac{\varepsilon}{\lambda}B_{Y_{0}}\big)+\frac{\varepsilon}{\lambda}B_{X} $

和

$\zeta\in D_{P}^{*}F(x, y)\big(y^{*}+\frac{\varepsilon}{\lambda}B_{Y}\big)+\frac{\varepsilon}{\lambda}B_{X}, $

使得

$\zeta_{0}+\zeta+\gamma=0, $

并且

$\frac{1}{48}-\frac{\varepsilon^{2}}{4\lambda^{2}}-\frac{\varepsilon}{2\lambda} \leq\| \zeta_{0}\| ^{2}+\| \zeta\| ^{2}+\| y_{0}^{*}\| ^{2}+\| y^{*}\| ^{2} \leq\frac{(1+\frac{\varepsilon}{\lambda})^{2}}{4}, $

其中$N_{P}(A, a)$和$D_{P}^{*}F_{0}(x_{0}, y_{0})$分别表示proximal法锥和$F_{0}$在点$(x_{0}, y_{0})$处的proximal法锥意义下的coderivative (具体定义见预备知识部分).

设$X$, $Y_{0}$, $Y_{1}, \cdots, Y_{m}$是实希尔伯特空间, $\Phi_{i} : X \rightrightarrows Y_{i}\ (i=0, 1, \cdots, m)$是闭多值映射, $A$是$X$的闭子集,并且$K_{i}$是$Y_{i}\ (i=0, 1, \cdots, m)$中的闭凸锥.考虑如下多目标向量优化问题

(1.2) $\begin{array}{l} K_{0}-\min\Phi_{0}(x), \\ \Phi_{i}(x)\bigcap-K_{i}\neq\emptyset, i=1, \cdots, m, \\ x\in A. \end{array}$

当$Y_{0}=\cdots=Y_{m}=\mathbb{R} $, $K_{0}=\cdots=K_{n}=\mathbb{R} _{+}$, $K_{n+1}=\cdots=K_{m}=\{0\}$且$\Phi_{i}$都是单值映射时,问题(1.2)将变为实数域上的标量优化问题.本文中用$Z$表示问题(1.2)的可行集,即

$Z:= A\cap\bigg(\bigcap\limits_{i=1}^{m}\Phi_{i}^{-1}(-K_{i})\bigg).$

事实上,问题(1.1)和(1.2)是等价的.只需要令$Y=Y_{1}\times\cdots\times Y_{m}$, $F_{0}(x)=\Phi_{0}(x)$, $F(x)=\Phi_{1}(x)\times\cdots\times \Phi_{m}(x)$和$K=K_{1}\times\cdots\times K_{m}$,则问题(1.2)就变成了问题(1.1).由此,自然地想到这样一个问题:直接利用定理1.1,我们是否能得到问题(1.2)的相应拉格朗日乘数法.也就是,下面的定理1.2是否能通过把问题(1.2)变换成问题(1.1)后,直接应用定理1.1得到.

定理1.2 设$\varepsilon>0$和$(\bar{x}, \bar{y}_{0})$是$(1.2)$的弱$\varepsilon$ -帕雷托解.假设$\bar{y}_{i}\in\Phi_{i}(\bar{x})\cap(-K_{i})\ (i=1, \cdots, m)$.则对任意的$\lambda>0$,存在$\xi>0$, $\eta>0$,

$x_{i}\in B(\bar{x}, \lambda), \ \ \ y_{i}\in\Phi_{i}(x_{i})\cap B(\bar{y}_{i}, \lambda), \ \ \ x_{m+1}\in A\cap B(\bar{x}, \lambda), $

$c_{i}\in K_{i}^{+}, \ \ \ \zeta_{m+1}\in N_{P}(A, x_{m+1})+\frac{\varepsilon}{\lambda}B_{X}, $

及

$\zeta_{i}\in D_{P}^{*}\Phi_{i}(x_{i}, y_{i})(c_{i}+\frac{\varepsilon}{\lambda}B_{Y_{i}})+ \frac{\varepsilon}{\lambda}B_{X}\ \ (0\leq i\leq m), $

使得

$\sum\limits_{i=0}^{m+1}\zeta_{i}=0, $

且

$\xi\leq \sum\limits_{i=0}^{m}(\| \zeta_{i}\| ^{2}+\| c_{i}\| ^{2}) \leq\eta.$

接下来,我们回答以上疑问.首先把问题(1.2)转换为问题(1.1),设$Y=Y_{1}\times\cdots\times Y_{m}$, $F_{0}(x)=\Phi_{0}(x)$, $F(x)=\Phi_{1}(x)\times\cdots\times \Phi_{m}(x)$, $K=K_{1}\times\cdots\times K_{m}$.根据定理1.1,对任意的$\lambda>0$,存在$y_{0}^{*}\in K_{0}^{+}$, $y^{*}=(c_{1}\times\cdots\times c_{m})\in K^{+}$, $\gamma\in N_{P}(A, a)+\frac{\varepsilon}{\lambda}B_{X}$, $\zeta_{0}\in D_{P}^{*}\Phi_{0}(x_{0}, y_{0})\big(y_{0}^{*}+\frac{\varepsilon}{\lambda}B_{Y_{0}}\big)+\frac{\varepsilon}{\lambda}B_{X}$和$\zeta\in D_{P}^{*}(\Phi_{1}\times\cdots\times\Phi_{m})\big(x, (y_{1}, \cdots, y_{m})\big)\big((c_{1}, \cdots, c_{m})+\frac{\varepsilon}{\lambda}B_{Y_{1}\times\cdots\times Y_{m}}\big)+\frac{\varepsilon}{\lambda}B_{X}, $使得$ \zeta_{0}+\zeta+\gamma=0. $显然,关键在于

$\begin{eqnarray*} &&D_{P}^{*}(\Phi_{1}\times\cdots\times\Phi_{m})\big(x, (y_{1}, \cdots, y_{m})\big)\big((c_{1}, \cdots, c_{m}) +\frac{\varepsilon}{\lambda}B_{Y_{1}\times\cdots\times Y_{m}}\big)+\frac{\varepsilon}{\lambda}B_{X}\\ &\subset&\sum\limits_{i=1}^{m}\big(D_{p}^{*}\Phi_{i}(x, y_{i})(c_{i}+\frac{\varepsilon}{\lambda}B_{Y_{i}}) +\frac{\varepsilon}{\lambda}B_{X}\big) \end{eqnarray*}$

是否成立.本文预备知识最后部分的例子告诉我们上述包含关系并不一定成立.

受此启发,本文对上述问题进行了研究,并在文中第三部分给出了定理1.2的证明.此外,我们将由定义看到, proximal法锥体现二阶变分性质并且有明显的几何意义,而Clarke法锥和Fréchet法锥只体现一阶变分性质,并且在非光滑的情况下, Clarke法锥难以计算,见文献[19-20, 22-24]及预备知识.因此,在处理实际问题时,我们的结论可能比郑教授们的成果^[20]更加有用.

设$X$是实希尔伯特空间,以$B_{X}$和$B(x, r)$记$X$中的闭单位球和半径为$r$的开球.给定$X$中的闭集$S$和点$s\in S$,我们用$N_{P}(S, s)$表示在点$s$处集合$S$的proximal法锥,即$v\in N_{P}(S, s)$当且仅当存在$\sigma$, $\delta\in (0, +\infty)$,使得

$\langle v, x-s\rangle\leq\sigma\| x-s\| ^{2}, \ \ \forall \ x\in S\cap B(s, \delta).$

下面关于proximal法锥的引理在我们后面的分析中将多次用到(参见文献[22,命题1.3,命题1.5]).

命题2.1 设$S$是实希尔伯特空间$X$中的闭集, $s\in S$, $v\in X$.则下述结论等价

(ⅰ) $v\in N_{P}(S, s)$;

(ⅱ)存在$\sigma\in (0, +\infty)$,使得

$\langle v, x-s\rangle\leq\sigma\| x-s\| ^{2}, \ \ \ \forall \ x\in S;$

(ⅲ)存在$t\in (0, +\infty)$,使得$d(s+tv, S)=t\| v\| $,其中$d(s+tv, S)$表示点$s+tv$到$S$的距离.

设$\Phi$是实希尔伯特空间$X$到$Y$中的多值映射,用$\mbox{gph}(\Phi)$记它的图,即

$\mbox{gph}(\Phi):=\left\{(x, y)\in X\times Y: y\in \Phi(x)\right\}.$

称$\Phi$是闭的,当且仅当$\mbox{gph}(\Phi)$是$X\times Y$中的闭子集.给定$x\in X$和$y\in\Phi(x)$,本文用$D_{p}\Phi(x, y): Y\rightrightarrows X$表示映射$\Phi$在$(x, y)$处关于proximal法锥的coderivative

$D_{p}^{*}\Phi(x, y)(\eta):=\left\{\xi\in X: (\xi, -\eta)\in N_{P}\big(\mbox{gph}(\Phi), (x, y)\big)\right\}, \ \ \ \forall\ \eta\in Y.$

假设$\phi:X\rightarrow\mathbb{R} $是一个真下半连续函数, $x\in\mbox{dom}(\phi)$.我们用$\partial_{P}\phi(x)$表示$\phi$在点$x$处的proximal次微分,即如果$\xi\in\partial_{P}\phi(x)$,当且仅当存在$\sigma, \delta\in (0, +\infty)$,满足

$\langle\xi, x'-x\rangle\leq\phi\big(x'\big)-\phi(x)+\sigma\| x'-x\| ^{2}, \ \ \ \forall \ x'\in B(x, \delta).$

特别地,对任意$x\in X$,当$\Phi(x)=[\phi(x), +\infty)$时,容易推出下式成立.

$D_{p}^{*}\Phi\big(x, \phi(x)\big)(1)=\partial_{P}\phi(x), \ \ \ \forall \ x\in\mbox{dom}(\phi).$

称$\Phi$在$(\bar{x}, \bar{y})\in \mbox{gph}(\Phi)$处是pseudo-Lipschitz的(参见文献[21, 23-24]),当且仅当存在$L, r\in (0, +\infty)$,使得

(2.1) $\Phi(x_{1})\cap B(\bar{y}, r)\subset\Phi(x_{2})+\| x_{1}-x_{2}\| LB_{Y}, \ \ \ \forall\ \ x_{1}, x_{2}\in B(\bar{x}, r).$

以下结论见于文献[21,推论2.1].

命题2.2 设$\Phi: X\rightrightarrows Y$是一个闭多值映射,并假设存在$L, r\in (0, +\infty)$满足$(2.1)$式.则对任意的$x\in B(\bar{x}, r)$, $y\in\Phi(x)\cap B(\bar{y}, r)$及$\eta\in Y$,有

$\sup\{\| \xi\| : \xi\in D_{P}^{*}\Phi(\bar{x}, \bar{y})(\eta)\}\leq L\| \eta\| .$

设$K$是实希尔伯特空间$X$中的闭凸锥.可定义$X$上的偏序关系$\leq_{K}$为^[13]:任意$x_{1}, x_{2} \in X$,

$x_{1}\leq_{K} x_{2}\ \ \ \mbox{当且仅当}\ \ \ x_{2}-x_{1}\in K.$

本文用$K^{+}$记$K$的对偶锥,即

$K^{+}:=\{v\in X:0\leq \langle v, x\rangle, \ \ \forall \ x\in K\}.$

设$\bar{x}\in Z$, $\bar{y}\in\Phi_{0}(\bar{x})$,称$(\bar{x}, \bar{y})$是优化问题(1.2)的一个帕雷托有效解,当且仅当

$\Phi_{0}(Z)\cap(\bar{y}-K_{0})=\{\bar{y}\}.$

当$\bar{x}\in Z$, $\bar{y}\in\Phi_{0}(\bar{x})$及$\mbox{int}(K_{0})\neq\emptyset$的情况下,称点$(\bar{x}, \bar{y})$是问题(1.2)的一个弱帕雷托有效解,当且仅当

$\Phi_{0}(Z)\cap\big(\bar{y}-\mbox{int}(K_{0})\big)=\emptyset.$

当我们假设$\mbox{int}(K_{0})\neq0$时, $(\bar{x}, \bar{y}_{0})$即为问题(1.2)的一个弱帕雷托有效解.以上定义见文献[13-14, 16-17].郑教授及其合作者在文献[20]中定义了问题(1.2)的一种更弱的有效解,他们称之为弱$\varepsilon$ -帕雷托解:如果$\bar{x}\in Z$, $\bar{y}_{0}\in\Phi_{0}(\bar{x})$,并且存在$e\in Y_{0}$满足$\| e\| <\varepsilon$,使得

$\Phi_{0}(Z)\cap(\bar{y}_{0}+e-K_{0})=\emptyset.$

则称$(\bar{x}, \bar{y}_{0})$是优化问题(1.2)的一个弱$\varepsilon$ -帕雷托解.以下结论给出了优化问题(1.2)存在弱$\varepsilon$ -帕雷托解的必要条件.

命题2.3 设$\Phi_{0}$在可行集$Z$上关于$K_{0}$是下有界的,即存在$b\in Y_{0}$,使得

$b\leq_{k_{0}}y, \ \ \ \ \forall \ y \in \Phi_{0}(Z).$

则任取$\varepsilon>0$,问题$(1.2)$存在弱$\varepsilon$ -帕雷托解.

为证明我们的主要结果,我们需要下面的一些引理.

引理2.1 设$X$是实希尔伯特空间, $A\subset X$, $B=\{(x, \cdots, x)\in X^{n}: x\in A\}$, $s=(\bar{s}, \cdots, \bar{s})\in X^{n}, $$ \bar{s}\in A$.则

(2.2) $N_{P}(B, s)=\{(\eta_{1}, \cdots, \eta_{n})\in X^{n}: \eta_{1}+\cdots+\eta_{n}\in N_{P}(A, \bar{s})\}.$

证 由命题2.1 (ⅱ)知,对于$\eta=(\eta_{1}, \cdots, \eta_{n})\in N_{p}(B, s)$,存在$\sigma>0$使得

(2.3) $\begin{array}[b]{rl}&\langle\eta, s'-s\rangle \leq\sigma\| s'-s\| ^{2}\\ \Leftrightarrow& \langle(\eta_{1}, \cdots, \eta_{n}), (x-\bar{s}\cdots, x-\bar{s})\rangle \leq\sigma\| (x-\bar{s}, \cdots, x-\bar{s})\| ^{2}\\ \Leftrightarrow&\langle \eta_{1}+\cdots+\eta_{n}, x-\bar{s}\rangle\leq \sigma\sqrt{n} \| x-\bar{s}\| ^{2}, \ \ \ \forall \ s'\in B\ {\rm且}\ x\in A. \end{array}$

故由命题2.1 (ⅱ),有

$\eta_{1}+\cdots+\eta_{n}\in N_{P}(A, \bar{s}).$

(2.2)式的反包含关系容易由(2.3)式得到.证毕.

引理2.2 设$(E, d)$是距离空间, $A$是$E$中的一个完备子集.设$\varphi: E\rightarrow\mathbb{R} \cup\{+\infty\}$是非负下半连续函数, $s_{0}\in A\cap{\rm{dom}}(\varphi)$, $\alpha_{i}\in(0, +\infty)\ (i=0, 1, \cdots)$满足$\sum\limits_{i=0}^{\infty}\alpha_{i}<+\infty$.则对于任意的$\sigma>0$和$\varepsilon'>0$,如下结论成立

(ⅰ)存在$s_{i}\in A(i=0, 1, \cdots)$和$\hat{s}\in A$,使得

$\hat{s}\in B_{E}(s_{0}, \sqrt{\frac{1}{\sigma}\alpha_{0}\varphi(s_{0})}), $

$d(\hat{s}, s_{k+1})\leq\frac{\varepsilon'}{\sqrt{\sigma}}\alpha_{k}, \ \ \ \forall \ k=0, 1, \cdots.$

(ⅱ) $\tilde{\varphi}(\cdot):=\varphi(\cdot)+\sigma\sum\limits_{i=0}^{\infty}\alpha_{i}d^{2}(\cdot, s_{i})$是$E$上的下半连续函数,并且

$\tilde{\varphi}(x)\geq\tilde{\varphi}(\hat{s}), \ \ \ \ \forall \ x\in A.$

证 设

$W(s_{0}):=\{x\in A: \varphi(x)+\sigma\alpha_{0}d^{2}(x, s_{0})\leq\varphi(s_{0})\}, $

显然$W(s_{0})$是非空闭集.任取$x\in W(s_{0})$,有

$\sigma\alpha_{0}d^{2}(x, s_{0})\leq \varphi(s_{0}), $

因此

(2.4) $W(s_{0})\subset B_{E}\bigg(s_{0}, \sqrt{\frac{1}{\sigma}\alpha_{0}\varphi(s_{0})}\; \bigg) .$

对于$k=0, 1, \cdots$,我们按如下方法归纳地构造点列$s_{k}\in A$和集合$W(s_{k})$:当给定$s_{k}$和$W(s_{k})$后,取$s_{k+1}\in W(S_{k})$使得

$ \begin{eqnarray*} \varphi(s_{k+1})+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(s_{k+1}, s_{i})\leq\inf\limits_{x\in W(s_{k})}\bigg\{\varphi(x)+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(x, s_{i})\bigg\} +(\varepsilon')^{2}\alpha_{k+1}(\alpha_{k})^{2}, \end{eqnarray*} $

然后定义

$W(s_{k+1}):=\bigg\{x\in A: \varphi(x)+\sigma\sum\limits_{i=0}^{k+1}\alpha_{i}d^{2}(x, s_{i}) \leq\varphi(s_{k+1})+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(s_{k+1}, s_{i})\bigg\}.$

由此易知$k$, $s_{k}\in W(s_{k})$,并且$W(s_{k})$是闭集,此外$W(s_{k+1})\subset W(s_{k})$.下面我们说明当$k\rightarrow \infty$时, $\mbox{diam}W(s_{k})\rightarrow0$.设$x\in W(s_{k+1})$,因为

$\begin{eqnarray*} \sigma\alpha_{k+1}d^{2}(x, s_{k+1})&\leq&\varphi(s_{k+1})+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(s_{k+1}, s_{i})-\bigg(\varphi(x)+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(x, s_{i})\bigg)\\ &\leq&\varphi(s_{k+1})+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(s_{k+1}, s_{i})-\inf\limits_{x\in W(s_{k})}\bigg\{\varphi(x)+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(x, s_{i})\bigg\} \\&<&(\varepsilon')^{2}\alpha_{k+1}(\alpha_{k})^{2}. \end{eqnarray*}$

所以

(2.5) $d(x, s_{k+1})\leq \frac{\varepsilon'}{\sqrt{\sigma}}\alpha_{k}, \ \ \ \forall \ k=0, 1, \cdots.$

故当$k\rightarrow\infty$时, $\mbox{diam}\big(W(s_{k+1})\big)\leq 2\frac{\varepsilon'}{\sqrt{\sigma}}\alpha_{k}\rightarrow0$.此时由康托定理,我们知道存在$\bigcap\limits_{k=0}^{\infty}W(s_{k})$使得$\bigcap\limits_{k=0}^{\infty}W(s_{k})=\{\hat{s}\}$.又注意到$W(s_{k+1})\subset W(s_{k})$,因此我们得到

$\varphi(\hat{s})+\sigma\sum\limits_{i=0}^{\infty}\alpha_{i}d^{2}(\hat{s}, s_{i})\leq\varphi(s_{0})<+\infty.$

任取$x\in E$,由三角不等式,可得

$d(x, s_{i})\leq d(x, \hat{s})+d(\hat{s}, s_{i})\leqd(x, \hat{s})+\mbox{diam}(W(s_{i})).$

由此推出

(2.6) $\sum\limits_{i=0}^{\infty}\alpha_{i}d(x, s_{i})<+\infty, \ \ \ \forall \ x\in E.$

事实上$\hat{s}$是下半连续函数

$\tilde{\varphi}(x):=\varphi(x)+\sigma\sum\limits_{i=0}^{\infty}\alpha_{i}d^{2}(x, s_{i})$

在$A$上的最小值点.这是因为$\bigcap\limits_{k=0}^{\infty}W(s_{k})=\{\hat{s}\}$,因此任取$x\in A$, $x\neq\hat{s}$,存在$k_{0}\in {\Bbb N}$满足$x\not\in W(s_{k_{0}})$,故由$W(s_{k})$的构造,易知当$k\geq k_{0}$时,有

$\begin{eqnarray*} \varphi(x)+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(x, s_{i}) >\varphi(s_{k})+\sigma\sum\limits_{i=0}^{k-1}\alpha_{i}d^{2}(s_{k}, s_{i})\geq \varphi(\hat{s})+\sigma\sum\limits_{i=0}^{k}\alpha_{i}d^{2}(s_{k}, s_{i}). \end{eqnarray*}$

令$k\rightarrow\infty$,则

(2.7) $\tilde{\varphi}(x) \geq \tilde{\varphi}(\hat{s}).$

结合(2.4)-(2.7)式结论得证.证毕.

进一步,考虑以下两个二元函数

$f(x_{1}, x_{2}):=\| \alpha x_{1}-\beta x_{2}-\gamma e\|, \ \ \ \forall \ (x_{1}, x_{2})\in X\times X, $

$g(x_{1}, x_{2}):=\sum\limits_{i=0}^{\infty}\alpha_{i}\| (x_{1}, x_{2})-(s_{1i}, s_{2i})\| ^{2}, \ \ \ \forall \ (x_{1}, x_{2})\in X\times X, $

其中$\alpha, \beta, \gamma\in \mathbb{R} $, $e\in X$, $(s_{1i}, s_{2i})\in X\times X (i=0, 1, \cdots)$且$\alpha_{i}\in (0, +\infty)(i=0, 1, \cdots)$满足$\sum\limits_{i=0}^{\infty}\alpha_{i}<\infty$.下面的两个引理给出了这两个函数在某些点处的Fréchet导数.

引理2.3 设$X$是实希尔伯特空间, $(\bar{x}_{1}, \bar{x}_{2})\in X\times X$满足$f(\bar{x}_{1}, \bar{x}_{2})\neq0$.则$f$在点$(\bar{x}_{1}, \bar{x}_{2})$的某领域内二阶连续可微,并且

$f'(\bar{x}_{1}, \bar{x}_{2})=\bigg(\frac{\alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e} {\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| }, -\frac{\alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e} {\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| }\bigg).$

证 因为$\| \cdot\| $在希尔伯特空间中的任意非零点处是无穷次可微,所以存在$(\bar{x}_{1}, \bar{x}_{2})$的某个领域$U$,使得$f\in C^{2}(U)$.任取$(u_{1}, u_{2})\in X\times X$,由定义知

$ \begin{eqnarray*} f((\bar{x}_{1}, \bar{x}_{2})+(u_{1}, u_{2}))-f(\bar{x}_{1}, \bar{x}_{2})&=&\| \alpha \bar{x}_{1}+u_{1}-\beta \bar{x}_{2}-u_{2}-\gamma e\| - \| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| \\&=&\frac{\| \alpha \bar{x}_{1}+u_{1}-\beta \bar{x}_{2}-u_{2}-\gamma e\| ^{2}-\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| ^{2}} {\| \alpha \bar{x}_{1}+u_{1}-\beta \bar{x}_{2}-u_{2}-\gamma e\| +\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| }\\&=&\frac{2\langle\alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e, u_{1}-u_{2}\rangle+\| u_{1}-u_{2}\| ^{2}} {\| \alpha \bar{x}_{1}+u_{1}-\beta \bar{x}_{2}-u_{2}-\gamma e\| +\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| }. \end{eqnarray*}$

令$\xi=\frac{\alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e} {\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| }$,于是

$ \begin{eqnarray*} && \mid f((\bar{x}_{1}, \bar{x}_{2})+(u_{1}, u_{2}))-f(\bar{x}_{1}, \bar{x}_{2})-\langle(\xi, -\xi), (u_{1}, u_{2})\rangle\mid\\& \leq& \bigg|\bigg \langle \xi, \bigg(\frac{2\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| } {\| \alpha \bar{x}_{1}+u_{1}-\beta \bar{x}_{2}-u_{2}-\gamma e\| +\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| } -1\bigg)(u_{1}, u_{2})\bigg\rangle\bigg|\\&&{}+\frac{\| u_{1}-u_{2}\| ^{2}} {\| \alpha \bar{x}_{1}+u_{1}-\beta \bar{x}_{2}-u_{2}-\gamma e\| +\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| }\\& \leq &\bigg| \frac{2\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| } {\| \alpha \bar{x}_{1}+u_{1}-\beta \bar{x}_{2}-u_{2}-\gamma e\| +\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| }-1\bigg| (\sqrt{2}\| (u_{1}, u_{2})\| )\\&&{}+\frac{2\| (u_{1}, u_{2})\| ^{2}} {\| \alpha \bar{x}_{1}+u_{1}-\beta \bar{x}_{2}-u_{2}-\gamma e\| +\| \alpha \bar{x}_{1}-\beta \bar{x}_{2}-\gamma e\| }. \end{eqnarray*}$

由此

$\begin{eqnarray*} \lim\limits_{(u_{1}, u_{2})\rightarrow(0, 0)} \frac{f((\bar{x}_{1}, \bar{x}_{2})+(u_{1}, u_{2}))-f(\bar{x}_{1}, \bar{x}_{2})-\langle(\xi, -\xi), (u_{1}, u_{2})\rangle} {\| (u_{1}, u_{2})\| }=0. \end{eqnarray*}$

证毕.

引理2.4 设$X$是实希尔伯特空间.假设存在$\eta>0$和$(s_{1}, s_{2})\in X\times X$满足

$g(x_{1}, x_{2})=\sum\limits_{i=0}^{\infty}\alpha_{i}\| (x_{1}, x_{2})-(s_{1i}, s_{2i})\| ^{2}<+\infty, \ \ \ \forall \ (x_{1}, x_{2})\in B((s_{1}, s_{2}), \eta)$

及

(2.8) $\sum\limits_{i=0}^{\infty}\alpha_{i}\| (x_{1}, x_{2})-(s_{1i}, s_{2i})\| <+\infty.$

则对任意的$(x_{1}, x_{2})\in B((s_{1}, s_{2}), \eta)$,有

(2.9) $g'(x_{1}, x_{2})=\bigg(\sum\limits_{i=0}^{\infty}2\alpha_{i}(x_{1}-s_{1i}), \sum\limits_{i=0}^{\infty}2\alpha_{i}(x_{2}-s_{2i})\bigg), $

并且

(2.10) $g''(x_{1}, x_{2})=\sum\limits_{i=0}^{\infty}2\alpha_{i}(I, I),$

其中$I$是$X$上的恒等映射.

证 设$(x_{1}, x_{2})\in B((s_{1}, s_{2}), \eta)$和$\delta:=\eta-\| (x_{1}, x_{2})-(s_{1}, s_{2})\| $.则对任意的$(u_{1}, u_{2})\in B((0, 0), \delta)$,有

$ \begin{eqnarray*} &&g((x_{1}, x_{2})+(u_{1}, u_{2}))-g(x_{1}, x_{2})\\ &=&\sum\limits_{i=0}^{\infty}\alpha_{i}(\| x_{1}+u_{1}-s_{1i}\| ^{2}+\| x_{2}+u_{2}-s_{2i}\| ^{2})-\sum\limits_{i=0}^{\infty}\alpha_{i}(\| x_{1}-s_{1i}\| ^{2}+\| x_{2}-s_{2i}\| ^{2})\\ &=&\sum\limits_{i=0}^{\infty}\alpha_{i}[(\| x_{1}+u_{1}-s_{1i}\| ^{2}-\| x_{1}-s_{1i}\| ^{2})\ +(\| x_{2}+u_{2}-s_{2i}\| ^{2}-\| x_{2}-s_{2i}\| ^{2})]\\ &=&\sum\limits_{i=0}^{\infty}\alpha_{i}(2\langle x_{1}-s_{1i}, u_{1}\rangle+\| u_{1}\| ^{2} +2\langle x_{2}-s_{2i}, u_{2}\rangle+\| u_{2}\| ^{2}). \end{eqnarray*} $

利用(2.8)式及三角不等式性,可得

$\sum\limits_{i=0}^{\infty}\alpha_{i}\| (x_{1}, x_{2})+(s_{1i}, s_{2i})\| <+\infty. $

再结合$X\times X$的完备性,推出

$\bigg(\sum\limits_{i=0}^{\infty}\alpha_{i}(x_{1}-s_{1i}), \sum\limits_{i=0}^{\infty}\alpha_{i}(x_{2}-s_{2i})\bigg) \in X\times X.$

我们令$(\xi_{1}, \xi_{2}):= (\sum\limits_{i=0}^{\infty}2\alpha_{i}(x_{1}-s_{1i}), \sum\limits_{i=0}^{\infty}2\alpha_{i}(x_{2}-s_{2i}))$,故

$\begin{eqnarray*} &&\lim\limits_{(u_{1}, u_{2})\rightarrow(0, 0)}\frac{g((x_{1}, x_{2})+(u_{1}, u_{2}))-g(x_{1}, x_{2})- \langle(\xi_{1}, \xi_{2}), (u_{1}, u_{2})\rangle} {\| (u_{1}, u_{2})\| }\\&=&\lim\limits_{(u_{1}, u_{2})\rightarrow(0, 0)}\frac {\sum\limits_{i=0}^{\infty}\alpha_{i}(\| u_{1}\| ^{2}+\| u_{1}\| ^{2})} {\| (u_{1}, u_{2})\| }=0. \end{eqnarray*}$

这说明(2.9)式成立. (2.10)式容易由(2.9)式简单推得.证毕.

在这部分的最后,我们给出一个例子对第一部分最后的疑问加以说明.

例2.1 设$F_{1}, F_{2}: \mathbb{R} \rightrightarrows\mathbb{R} $定义如下

$ \begin{eqnarray*} F_{1}(x)= \left\{\begin{array}{lll} \{y:y=4x\},&x>0, \\ \{y:y\geq6x\},&x\leq0, \end{array}\right. \ \ \ \ \ F_{2}(x)= \left\{\begin{array}{lll} \{0\},&x>0, \\ \{y:y\leq2x\},&x\leq0. \end{array}\right. \end{eqnarray*}$

由命题2.1 (ⅲ)易知

$\begin{eqnarray*} D_{p}^{*}F_{1}(0, 0)(v)= \left\{\begin{array}{lll} \{0\},&v=0, \\ \emptyset,&v\neq0, \end{array}\right. \end{eqnarray*}$

并且对任意的$v\in\mathbb{R} $, $D_{p}^{*}F_{2}(0, 0)(v)$满足如下性质

(ⅰ)若$v>0$,则$D_{p}^{*}F_{2}(0, 0)(v)=\emptyset$.

(ⅱ)若$v\leq0$且$u\in D_{p}^{*}F_{2}(0, 0)(v)$,则$u\leq0$.

下面说明$1\in D_{p}^{*}(F_{1}\times F_{2})\big(0, (0, 0)\big)\big(\frac{1}{4}, -\frac{1}{4}\big)$.事实上,对任意的$x\in\mathbb{R} $和$(y_{1}, y_{2})\in F_{1}(x)\times F_{2}(x)$,容易得到

(ⅲ)如果$x>0$,则

$\langle(1, (-\frac{1}{4}, \frac{1}{4})), (x, (y_{1}, y_{2}))\rangle=x-\frac{1}{4}y_{1}+\frac{1}{4}y_{2}=0.$

(ⅳ)如果$x\leq0$,则

$\begin{eqnarray*}\langle(1, (-\frac{1}{4}, \frac{1}{4})), (x, (y_{1}, y_{2}))\rangle&=&(x-\frac{1}{6}y_{1})-\frac{1}{12}y_{1}+\frac{1}{4}y_{2}\\&\leq&-\frac{1}{12}y_{1}+\frac{1}{4}y_{2}\\&\leq&-\frac{1}{2}x+\frac{1}{4}y_{2}\leq0.\end{eqnarray*}$

由(ⅲ), (ⅳ)和命题2.1 (ⅱ),我们知道

$1\in D_{p}^{*}(F_{1}\times F_{2})\big(0, (0, 0)\big)\big(\frac{1}{4}, -\frac{1}{4}\big).$

因此, $1\not\in D_{p}^{*}F_{1}(0, 0)(\frac{1}{4})+D_{p}^{*}F_{2}(0, 0)(-\frac{1}{4})=\emptyset$.而

$1\in D_{p}^{*}(F_{1}\times F_{2})\big(0, (0, 0)\big)\big(\frac{\sqrt{2}}{4}B_{\mathbb{R} ^2}\big)+\frac{\sqrt{2}}{4}B_{\mathbb{R} }, $

并且

$1\not\in\bigg\{D_{p}^{*}F_{1}(0, 0)\big(\frac{\sqrt{2}}{4}B_{\mathbb{R} }\big)+\frac{\sqrt{2}}{4}B_{\mathbb{R} }\bigg\}+\bigg\{D_{p}^{*}F_{2}(0, 0)\big(\frac{\sqrt{2}}{4}B_{\mathbb{R} }\big)+\frac{\sqrt{2}}{4}B_{\mathbb{R} }\bigg\}.$

此例说明定理1.1不能自然地延拓到有多个约束的优化问题中.

这一节的主要目标是给出定理1.2的证明.证明的思想借鉴了文献[24-25]中Borwein和Presis非光滑变分原则及文献[21]中的定理1.1的证明思想.

在整个这部分中,我们假设$X$, $Y$, $Y_{0}$, $Y_{1}, \cdots, Y_{m}$是实希尔伯特空间, $A$是$X$中的闭子集, $K_{i}$是$Y_{i}\ (i=0, 1, \cdots, m)$中的闭凸锥,以$\langle\cdot, \cdot\rangle$表示内积,并设$\Phi_{i} : X\rightrightarrows Y_{i}\ (i=0, 1, \cdots, m)$是闭多值映射.

定义空间$X\times\prod\limits_{i=0}^{m}Y_{i}$的内积和范数为：任取$(x_{1}, y_{0}^{(1)}, \cdots, y_{m}^{(1)})$, $(x_{2}, y_{0}^{(2)}, \cdots, y_{m}^{(2)})\in X\times\prod\limits_{i=0}^{m}Y_{i}$,

$ \langle (x_{1}, y_{0}^{(1)}, \cdots, y_{m}^{(1)}), (x_{2}, y_{0}^{(2)}, \cdots, y_{m}^{(2)})\rangle= \langle x_{1}, x_{2}\rangle+\sum\limits_{i=0}^{m}\langle y_{i}^{(1)}, y_{i}^{(2)}\rangle, $

$\| (x_{1}, y_{0}^{(1)}, \cdots, y_{m}^{(1)})\| =\bigg(\| x_{1}\| ^{2}+ \sum\limits_{i=0}^{m}\| y_{i}^{(1)}\| ^{2}\bigg)^{\frac12}.$

定理3.1 设$\varepsilon>0$, $(\bar{x}, \bar{y}_{0})$是多目标优化问题$(1.2)$的一个弱$\varepsilon$-帕雷托解.假设$\bar{y}_{i}\in\Phi_{i}(\bar{x})\cap(-K_{i})\ (i=1, \cdots, m)$.则对任意的$\lambda>0$,存在

$x_{i}\in B(\bar{x}, \lambda), \ \ \ y_{i}\in\Phi_{i}(x_{i})\cap B(\bar{y}_{i}, \lambda), \ \ \ x_{m+1}\in A\cap B(\bar{x}, \lambda), $

$c_{i}\in K_{i}^{+}, \ \ \ \zeta_{i}\in D_{P}^{*}\Phi_{i}(x_{i}, y_{i})(c_{i}+\frac{\varepsilon}{\lambda}B_{Y_{i}})+ \frac{\varepsilon}{\lambda}B_{X}\ \ (0\leq i\leq m)$

及$\zeta_{m+1}\in N_{P}(A, x_{m+1})+\frac{\varepsilon}{\lambda}B_{X}$,使得$\sum\limits_{i=0}^{m+1}\zeta_{i}=0$且

(3.1) $\begin{equation}\frac{1}{64(m+3)(m+1)^{2}}-(\frac{\varepsilon}{\lambda})^{2} \leq \sum\limits_{i=0}^{m}(\| \zeta_{i}\| ^{2}+\| c_{i}\| ^{2}) \leq\frac{1}{32(m+1)}+\frac{1}{2}(\frac{\varepsilon}{\lambda})^{2}.\end{equation}$

证 因为$(\bar{x}, \bar{y}_{0})$是优化问题(1.2)的弱$\varepsilon$ -帕雷托解,所以存在$e\in Y_{0}$满足$\| e\| <\varepsilon$,使得

(3.2) $\begin{equation}\Phi_{0}(Z)\cap(\bar{y}_{0}+e-K_{0})=\emptyset, \end{equation}$

其中$Z:=A\cap(\bigcap\limits_{i=1}^{m}\Phi_{i}^{-1}(-K_{i}))$.考虑积空间$X\times \prod\limits_{i=0}^{m}Y_{i}$,并设

$A_{i}:=\bigg\{(x, y_{0}, \cdots, y_{m})\in X\times\prod\limits_{i=0}^{m}Y_{i}:(x, y_{i})\in\mbox{gph}(\Phi_{i})\bigg\}\ \ (i=0, \cdots, m), $

$A_{m+1}:=A\times(y_{0}-K_{0})\times\prod\limits_{i=1}^{m}(y_{i}-K_{i}), \ \ \bar{e}:=(0, e, 0, \cdots, 0)\in X\times \prod\limits_{i=0}^{m}Y_{i}.$

记

$E:=\bigg(X\times \prod\limits_{i=0}^{m}Y_{i}\bigg)^{m+1}, \ \ \ \tilde{e}:=(\bar{e}, \cdots, \bar{e})\in E, $

$B_{1}:=A_{0}\times A_{1}\times \cdots \times A_{m}\ \ \mbox{及}\ \ B_{2}:=\{(x, \cdots, x)\in E:x\in A_{m+1}\}.$

定义$\varphi: E\times E\rightarrow\mathbb{R} $为

$\varphi(\textbf{x})=\frac{1}{8(m+1)^{\frac{3}{2}}}\| x_{1}-x_{2}-\tilde{e}\|, \ \ \ \ \mbox{其中}\ \ \textbf{x}:=(x_{1}, x_{2})\in E\times E.$

令$a:=(\bar{x}, \bar{y}_{0}, \cdots, \bar{y}_{m})$, $s:=(a, \cdots, a)\in E$及$\textbf{s}_{0}:=(s, s)$.显然有$\textbf{s}_{0}\in B_{1}\times B_{2}$,并且$\varphi(\textbf{s}_{0})=\frac{1}{8(m+1)^{\frac{3}{2}}}\| \tilde{e}\| =\frac{1}{8(m+1)}\| e\| $.

设$\sigma:=\frac{\varepsilon}{2\lambda^{2}(m+1)}$, $\alpha_{i}:=\frac{1}{2^{i+2}} (i=0, 1, \cdots)$和$0<\varepsilon'<\sqrt{\frac{\varepsilon}{2(m+1)}}$.则由引理2.2,存在$\hat{\textbf{s}}=(\hat{s}_{1}, \hat{s}_{2}) \in B_{1}\times B_{2}$和$\textbf{s}_{i}=(s_{1i}, s_{2i})\in B_{1}\times B_{2}\ (i=1, \cdots, )$,使得

(3.3) $\begin{equation} \hat{\textbf{s}}\in B_{E\times E}\bigg(\textbf{s}_{0}, \lambda\sqrt{\frac{\| e\| }{\varepsilon}}\bigg) \subset B_{E\times E}(\textbf{s}_{0}, \lambda), \end{equation}$

(3.4) $\begin{equation}\| \hat{\textbf{s}}-\textbf{s}_{k+1}\| \leq\frac{\varepsilon'}{\sqrt{\sigma}}\cdot\frac{1}{2^{k+2}} \leq \lambda, \ \ \ \forall \ k=0, 1, \cdots, \end{equation}$

且有

(3.5) $ \begin{equation}\tilde{\varphi}(\textbf{x}):=\varphi(\textbf{x})+\frac{\varepsilon}{2\lambda^{2}(m+1)} \sum\limits_{i=0}^{\infty}\frac{\| \textbf{x}-\textbf{s}_{i}\| ^{2}}{2^{i+2}} \geq \tilde{\varphi}(\hat{\textbf{s}}), \ \ \ \forall \ \textbf{x}\in B_{1}\times B_{2}.\end{equation}$

下面我们用反证法证明$\varphi(\hat{\textbf{s}})\neq 0$.为此,假设$\hat{s}_{1}=\hat{s}_{2}+\tilde{e}$.由$B_{1}$和$B_{2}$的定义知,存在$x\in A_{m+1}$,使得

$\hat{s}_{2}=(x+\bar{e}, \cdots, x+\bar{e})\ \ \mbox{且}\ \ \ x+\bar{e}\in \bigcap\limits_{i=0}^{m}A_{i}.$

于是由$A_{m+1}$的构造知,存在$\tilde{x}\in A$和$k_{i}\in K_{i}$$ (i=0, \cdots, m)$,满足

$(\tilde{x}, \bar{y}_{0}+e-k_{0}, \bar{y}_{1}-k_{1}, \cdots, \bar{y}_{m}-k_{m})\in \bigcap\limits_{i=0}^{m}A_{i}.$

由此说明$\bar{y}_{0}+e-k_{0}\in \Phi_{0}(\tilde{x})\cap(\bar{y}_{0}+e-k_{0})$和$\bar{y}_{i}-k_{i}\in \Phi_{i}(\tilde{x})\cap(\bar{y}_{i}-k_{i})\subset \Phi_{i}(\tilde{x})\cap(-k_{i})$$ (i=1, \cdots, m)$成立.故

$\tilde{x}\in A\cap\bigg(\bigcap\limits_{i=0}^{m}\Phi_{i}^{-1}(-K_{i})\bigg)=Z.$

所以

$\bar{y}_{0}+e-k_{0}\in \Phi_{0}(Z)\cap(\bar{y}_{0}+e-k_{0}), $

和(3.2)式矛盾.因此$\varphi(\hat{\textbf{s}})\neq 0$.

容易由(3.4)式得到$\sum\limits_{i=0}^{\infty}\frac{\| \hat{\textbf{s}}-\textbf{s}_{i}\| }{2^{i+2}}<+\infty$.故由引理2.3和引理2.4可知$\tilde{\varphi}$在点$\hat{\textbf{s}}$的某领域内二阶连续可微,即存在领域$U:=B_{E\times E}(\hat{\textbf{s}}, r)$和常数$\rho>0$使得

$\tilde{\varphi}\in C^{2}(U) \ \ \ \mbox{和}\ \ \ \| \tilde{\varphi}''(\textbf{x})\| \leq2\rho, \ \ \ \forall \ \textbf{x}\in U.$

于是,在点$\textbf{z}\in (B_{1}\times B_{2})\cap U$处$\tilde{\varphi}$能二阶泰勒展开,并根据(3.5)式可知

$\begin{eqnarray*} 0\leq\tilde{\varphi}(\textbf{z})-\tilde{\varphi}(\hat{\textbf{s}}) &=&\langle\tilde{\varphi}'(\hat{\textbf{s}}), \textbf{z}-\hat{\textbf{s}}\rangle +\frac{1}{2}\langle\tilde{\varphi}''(\theta), (\textbf{z}-\hat{\textbf{s}})^{2}\rangle\\&\leq& \langle\tilde{\varphi}'(\hat{\textbf{s}}), \textbf{z}- \hat{\textbf{s}}\rangle + \rho\| \textbf{z}-\hat{\textbf{s}}\| ^{2}, \end{eqnarray*}$

其中$\theta$是$\textbf{z}$和$\hat{\textbf{s}}$连线上的某个点.这表明

(3.6) $\begin{equation} -\tilde{\varphi}'(\hat{\textbf{s}})\in N_{p}(B_{1}\times B_{2}, \hat{\textbf{s}}) =N_{P}(B_{1}, \hat{s}_{1})\times N_{P}(B_{2}, \hat{s}_{2}).\end{equation}$

又根据引理2.3和引理2.4,我们知道

(3.7) $ \begin{eqnarray}\tilde{\varphi}'(\hat{\textbf{s}})&=&\bigg(\frac{1}{8(m+1)^{\frac{3}{2}}}\cdot \frac{\hat{s}_{1}-\hat{s}_{2}-\tilde{e}}{\| \hat{s}_{1}-\hat{s}_{2}-\tilde{e}\| } +\frac{\varepsilon}{2\lambda^{2}(m+1)}\bigg(\frac{\hat{s}_{1}-s}{2}+\sum\limits_{i=1}^{\infty}\frac{\hat{s}_{1}-s_{1i}}{2^{i+1}}\bigg), \\ &&-\frac{1}{8(m+1)^{\frac{3}{2}}}\cdot \frac{\hat{s}_{1}-\hat{s}_{2}-\tilde{e}}{\| \hat{s}_{1}-\hat{s}_{2}-\tilde{e}\| } +\frac{\varepsilon}{2\lambda^{2}(m+1)} \bigg(\frac{\hat{s}_{2}-s}{2}+\sum\limits_{i=1}^{\infty}\frac{\hat{s}_{2}-s_{2i}}{2^{i+1}}\bigg)\bigg). \end{eqnarray}$

而(3.3)和(3.4)式表明

$\bigg\|\frac{\hat{s}_{1}-s}{2}+\sum\limits_{i=1}^{\infty}\frac{\hat{s}_{1}-s_{1i}}{2^{i+1}}\bigg\|\leq\lambda \ \ \ \mbox{和}\ \ \ \bigg\| \frac{\hat{s}_{2}-s}{2}+\sum\limits_{i=1}^{\infty}\frac{\hat{s}_{2}-s_{2i}}{2^{i+1}}\bigg\| \leq\lambda.$

于是,由(3.6)和(3.7)式得

$-\frac{1}{8(m+1)^{\frac{3}{2}}}\cdot \frac{\hat{s}_{1}-\hat{s}_{2}-\tilde{e}}{\| \hat{s}_{1}-\hat{s}_{2}-\tilde{e}\| } \in N_{P}(B_{1}, \hat{s}_{1})+\frac{\varepsilon}{2\lambda(m+1)}B_{E} $

且

(3.8) $\begin{equation} \frac{1}{8(m+1)^{\frac{3}{2}}}\cdot \frac{\hat{s}_{1}-\hat{s}_{2}-\tilde{e}}{\| \hat{s}_{1}-\hat{s}_{2}-\tilde{e}\| } \in N_{P}(B_{2}, \hat{s}_{2})+\frac{\varepsilon}{2\lambda(m+1)}B_{E}.\end{equation}$

注意到$\hat{s}_{1}\in B_{1}$, $\hat{s}_{2}\in B_{2}$,则存在$\bar{s}_{i}=(x_{i}, y_{i, 0}, \cdots, y_{i, m})\in A_{i}\ (i=0, \cdots, m)$和$\bar{s}=(x_{m+1}, y_{m+1, 0}, \cdots, y_{m+1, m})\in A_{m+1}$,使得

$(\bar{s}_{0}, \cdots, \bar{s}_{m})=\hat{s}_{1}\ \ \ \mbox{and}\ \ \ (\bar{s}, \cdots, \bar{s})=\hat{s}_{2}.$

于是结合(3.3)式,可得

(3.9) $\begin{equation}\sum\limits_{i=0}^{m}\| x_{i}-\bar{x}\| ^{2} +\sum\limits_{i=0}^{m}\sum\limits_{j=0}^{m}\| y_{i, j}-\bar{y}_{j}\| ^{2} +(m+1)(\| x_{m+1}-\bar{x}\| ^{2} +\sum\limits_{j=0}^{m}\| y_{m+1, j}-\bar{y}_{j}\| ^{2}) <\lambda^{2}. \end{equation}$

接下来,令$t:=\| \hat{s}_{1}-\hat{s}_{2}-\tilde{e}\| $,则(3.8)式表明

(3.10) $ \begin{eqnarray}&&-\frac{1}{8(m+1)^{\frac{3}{2}}t}(\bar{s}_{0}-\bar{s}-\bar{e}, \cdots, \bar{s}_{m}-\bar{s}-\bar{e})\\&=&-\frac{1}{8(m+1)^{\frac{3}{2}}t}\big((x_{0}-x_{m+1}, y_{0, 0}-y_{m+1, 0}-e, y_{0, 1}-y_{m+1, 1}, \cdots, y_{0, m}-y_{m+1, m}), \\&& \cdots, (x_{m}-x_{m+1}, y_{m, 0}-y_{m+1, 0}-e, y_{m, 1}-y_{m+1, 1}, \cdots, y_{m, m}-y_{m+1, m})\big)\\&& \in N_{P}(A_{0}, \bar{s}_{0})\times \cdots \times N_{P}(A_{m}, \bar{s}_{m})+\frac{\varepsilon}{2\lambda(m+1)}B_{E}. \end{eqnarray}$

再由$A_{i}$的定义,我们又能得到当$0\leq i\leq m$时,有

$ \begin{eqnarray*} N_{P}(A_{i}, \bar{s}_{i})=\{(\xi, \eta_{0}, \cdots, \eta_{m}): (\xi, \eta_{i})\inN_{P}\big(\mbox{gph}(\Phi_{i}), (x_{i}, y_{i, i})\big), \eta_{k}=0\ \forall \ k\neq i\}, \end{eqnarray*}$

这和(3.10)式推出存在

$(\xi_{i}, \eta_{i})\in N_{P}\big(\mbox{gph}(\Phi_{i}), (x_{i}, y_{i, i})\big)\ (0\leq i\leq m), $

满足

(3.11) $\begin{eqnarray}&&\sum\limits_{i=0}^{m}\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}(x_{i}-x_{m+1})+\xi_{i}\bigg\| ^{2} +\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}(y_{0, 0}-y_{m+1, 0}-e)+\eta_{0}\bigg\| ^{2}\\&&{}+\sum\limits_{i=1}^{m}\bigg(\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}(y_{i, i}-y_{m+1, i})+\eta_{i}\bigg\| ^{2} +\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}(y_{i, 0}-y_{m+1, 0}-e)\bigg\| ^{2}\bigg)\\&&{}+\sum\limits_{i, k=0, i\neq k}^{m}\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}(y_{i, k}-y_{m+1, k})\bigg\| ^{2}< \bigg(\frac{\varepsilon}{2\lambda(m+1)}\bigg)^{2}. \end{eqnarray} $

应用引理2.1得

$N_{P}(B_{2}, \hat{s}_{2})=\{(\eta_{1}, \cdots, \eta_{m+1})\in E: \eta_{1}+\cdots+\eta_{m+1}\in N_{P}(A_{m+1}, \bar{s})\}.$

这和(3.8)式表明存在

$(\omega_{0}, \cdots, \omega_{m})\in \frac{\varepsilon}{2\lambda(m+1)}B_{E}, $

使得

(3.12) $\begin{equation}\frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(\bar{s}_{i}-\bar{s}-\bar{e})-(\omega_{0}+\cdots+\omega_{m}) \in N_{P}(A_{m+1}, \bar{s}).\end{equation}$

因为

$ \begin{eqnarray*} \| \omega_{0}+\cdots+\omega_{m}\| ^{2} &\leq&(\| \omega_{0}\| +\cdots+\| \omega_{m}\| )^{2}\\&\leq&(m+1)\sum\limits_{i=0}^{m}\| \omega_{i}\| ^{2}=(m+1)\| (\omega_{0}, \cdots, \omega_{m})\| ^{2}\\&\leq& \frac{1}{4(m+1)}\cdot(\frac{\varepsilon}{\lambda})^{2}, \end{eqnarray*}$

所以由(3.12)式可得

(3.13) $\begin{equation}\frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(\bar{s}_{i}-\bar{s}-\bar{e}) \in N_{P}(A_{m+1}, \bar{s})+\frac{\varepsilon}{2\lambda(m+1)^\frac12}B_{E}.\end{equation}$

通过$A_{m+1}$的定义,容易证明如下结论

$N_{P}(A_{m+1}, \bar{s})\subset N_{P}(A, x_{m+1})\times \prod\limits_{i=0}^{m}K_{i}^{+}.$

这又和(3.13)式表明存在

$\zeta \in N_{P}(A, x_{m+1})\ \ \ \mbox{且}\ \ \ (c_{0}', \cdots, c_{m}')\in \prod\limits_{i=0}^{m}K_{i}^{+}, $

使得

(3.14) $\begin{eqnarray}&&\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(x_{i}-x_{m+1})-\zeta\bigg\| ^{2}+\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(y_{i, 0}-y_{m+1, 0}-e)-c_{0}'\bigg\| ^{2}\\&&+\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(y_{i, 1}-y_{m+1, 1})-c_{1}'\bigg\| ^{2}+\cdots+\bigg\| \frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(y_{i, m}-y_{m+1, m})-c_{m}'\bigg\| ^{2}\\& <&\frac{1}{4(m+1)}\cdot\bigg(\frac{\varepsilon}{\lambda}\bigg)^{2}. \end{eqnarray}$

又注意到

$ \begin{eqnarray*} -\eta_{0}&=&c_{0}'-\bigg(\frac{1}{8(m+1)^{\frac{3}{2}}t}(y_{0, 0}-y_{m+1, 0}-e)+\eta_{0}\bigg)\\ &&+\bigg(\frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(y_{i, 0}-y_{m+1, 0}-e)-c_{0}'\bigg)\\ &&-\sum\limits_{i=1}^{m}\frac{1}{8(m+1)^{\frac{3}{2}}t}(y_{i, 0}-y_{m+1, 0}-e) \end{eqnarray*}$

及

$ \begin{eqnarray*} -\eta_{k}&=&c_{k}'-\bigg(\frac{1}{8(m+1)^{\frac{3}{2}}t}(y_{k, k}-y_{m+1, k})+\eta_{k}\bigg)\\ &&+\bigg(\frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(y_{i, k}-y_{m+1, k}-e)-c_{k}'\bigg)\\ &&-\sum\limits_{i=0, i\neq k}^{m}\frac{1}{8(m+1)^{\frac{3}{2}}t}(y_{i, k}-y_{m+1, k})\ (k=1, \cdots, m), \end{eqnarray*}$

再结合(3.11)和(3.14)式,可知

$-\eta_{0}\in c_{0}'+\frac{\varepsilon}{\lambda\sqrt{m+1}}B_{Y_{0}}$

和

$-\eta_{k}\in c_{k}'+\frac{\varepsilon}{\lambda\sqrt{m+1}}B_{Y_{k}}\ (k=1, \cdots, m)$

成立.因此

$\xi_{i}\in D_{P}^{*}\Phi_{i}(x_{i}, y_{i, i})(-\eta_{i})\subset D_{P}^{*}\Phi_{i}(x_{i}, y_{i, i})(c_{i}'+\frac{\varepsilon}{\lambda\sqrt{m+1}}B_{Y_{i}}) \ (i=0, \cdots, m).$

令$\xi_{i}':=-\frac{1}{8(m+1)^{\frac{3}{2}}t}(x_{i}-x_{m+1})\ (0\leq i\leq m)$及$\xi_{m+1}':=\frac{1}{8(m+1)^{\frac{3}{2}}t}\sum\limits_{i=0}^{m}(x_{i}-x_{m+1})$,则由(3.11)式和(3.14)式得

$\xi_{i}'\in D_{P}^{*}\Phi_{i}(x_{i}, y_{i, i})(c_{i}'+\frac{\varepsilon}{\lambda\sqrt{m+1}}B_{Y_{i}})+\frac{\varepsilon}{2\lambda(m+1)}B_{X} \ (i=0, \cdots, m), $

并且

$\xi_{m+1}'\in N_{P}(A, x_{m+1})+\frac{\varepsilon}{2\lambda\sqrt{m+1}}B_{X}.$

又记$\zeta_{i}:=\sqrt{m+1}\xi_{i}', \ c_{i}:=\sqrt{m+1}c_{i}' \ \ (i=0, \cdots, m)$和$\zeta_{m+1}:=\sqrt{m+1}\xi_{m+1}'$,则

$\zeta_{i}\in D_{P}^{*}\Phi_{i}(x_{i}, y_{i, i})(c_{i}+\frac{\varepsilon}{\lambda}B_{Y_{i}})+\frac{\varepsilon}{\lambda}B_{X} \ (i=0, \cdots, m), $

$\zeta_{m+1}\in N_{P}(A, x_{m+1})+\frac{\varepsilon}{\lambda}B_{X}\ \ \ \mbox{且}\ \ \ \sum\limits_{i=0}^{m+1}\zeta_{i}=0.$

至此,由(3.9)式可知我们还需要说明(3.1)式成立即可.下面我们证明(3.1)式.首先注意到$\frac{\| \hat{s}_{1}-\hat{s}_{2}-\tilde{e}\| }{t}=1$,即有

(3.15) $ \begin{eqnarray}&&\frac{1}{t^{2}}\bigg(\sum\limits_{i=0}^{m}\| x_{i}-x_{m+1}\| ^{2} +\sum\limits_{i=0}^{m}\| y_{i, 0}-y_{m+1, 0}-e\| ^{2} +\sum\limits_{i=1}^{m}\| y_{i, i}-y_{m+1, i}\| ^{2}\\&& +\sum\limits_{i=0, k=1, i\neq k}^{m}\| y_{i, k}-y_{m+1, k}\| ^{2}\bigg) =1. \end{eqnarray}$

类似地,因为

$y_{0, 0}-y_{m+1, 0}-e=\sum\limits_{i=0}^{m}(y_{i, 0}-y_{m+1, 0}-e)-\sum\limits_{i=1}^{m}(y_{i, 0}-y_{m+1, 0}-e), $

并且

$y_{k, k}-y_{m+1, k}=\sum\limits_{i=0}^{m}(y_{i, k}-y_{m+1, k})-\sum\limits_{i=0, i\neq k}^{m}(y_{i, k}-y_{m+1, k})\ (1\leq k\leq m), $

所以对于$(1\leq k\leq m)$时,有

$\begin{eqnarray*} \| y_{0, 0}-y_{m+1, 0}-e\| ^{2} &\leq& (m+2)\bigg(\bigg\| \sum\limits_{i=0}^{m}(y_{i, 0}-y_{m+1, 0}-e)-8(m+1)^{\frac{3}{2}}t c_{0}'\bigg\| ^{2}\\ &&+64(m+1)^{3}\cdot t^{2}\cdot \| c_{0}'\| ^{2} +\sum\limits_{i=1}^{m}\| y_{i, 0}-y_{m+1, 0}-e\| ^{2}\bigg) \end{eqnarray*}$

和

$\begin{eqnarray*} \| y_{k, k}-y_{m+1, k}-e\| ^{2} &\leq& (m+2)\bigg(\bigg\| \sum\limits_{i=0}^{m}(y_{i, k}-y_{m+1, k})-8(m+1)^{\frac{3}{2}}t c_{k}'\bigg\| ^{2}\\ &&+64(m+1)^{3}\cdot t^{2}\cdot \| c_{k}'\| ^{2} +\sum\limits_{i=0, i\neq k}^{m}\| y_{i, k}-y_{m+1, k}\| ^{2}\bigg) \end{eqnarray*}$

成立.这说明

$\begin{eqnarray*}& &\sum\limits_{i=0}^{m}\| y_{i, 0}-y_{m+1, 0}-e\| ^{2} +\sum\limits_{i=1}^{m}\| y_{i, i}-y_{m+1, i}\| ^{2}+\sum\limits_{i=0, k=1, i\neq k}^{m}\| y_{i, k}-y_{m+1, k}\| ^{2}\\ &{}\leq &(m+3)\bigg(\sum\limits_{i=1}^{m}\| y_{i, 0}-y_{m+1, 0}-e\| ^{2}+\sum\limits_{i=0, k=1, i\neq k}^{m}\| y_{i, k}-y_{m+1, k}\| ^{2}\bigg)\\ &&{}+(m+2)\bigg(\bigg\| \sum\limits_{i=0}^{m}(y_{i, 0}-y_{m+1, 0}-e)-8(m+1)^{\frac{3}{2}}t c_{0}'\bigg\| ^{2}\\ &&{}+\sum\limits_{k=1}^{m}\bigg\| \sum\limits_{i=0}^{m}(y_{i, k}-y_{m+1, k})-8(m+1)^{\frac{3}{2}}t c_{k}'\bigg\| ^{2}\bigg)+64(m+2)(m+1)^{3}\cdot t^{2}\sum\limits_{i=0}^{m}\| c_{i}'\| ^{2}. \end{eqnarray*} $

此时,应用(3.11), (3.14)和(3.15)式可得

$\begin{eqnarray*} \frac{1}{64(m+1)^{2}} &\leq& \sum\limits_{i=0}^{m}\bigg\| -\frac{1}{8(m+1)t}(x_{i}-x_{m+1})\bigg\| ^{2} +(m+3)\cdot(\frac{\varepsilon}{\lambda})^{2}\cdot \frac{1}{4(m+1)} \\&&+\frac{(m+2)}{4}\cdot(\frac{\varepsilon}{\lambda})^{2}+(m+2)\sum\limits_{i=0}^{m}\| \sqrt{m+1}c_{i}'\| ^{2}, \end{eqnarray*}$

故

$\frac{1}{64(m+3)(m+1)^{2}}-(\frac{\varepsilon}{\lambda})^{2} \leq \sum\limits_{i=0}^{m}(\| \zeta_{i}\| ^{2}+\| c_{i}\| ^{2}).$

进一步,因为

$\begin{eqnarray*} \sum\limits_{i=0}^{m}\| c_{i}\| ^{2} &\leq&\bigg(\bigg\| \frac{1}{8(m+1)t}\sum\limits_{i=0}^{m}(y_{i, 0}-y_{m+1, 0}-e)-\sqrt{m+1}c_{0}'\bigg\| \\ &&{}+\bigg\| \frac{1}{8(m+1)t}\sum\limits_{i=0}^{m}(y_{i, 0}-y_{m+1, 0}-e)\bigg\| \bigg)^{2}\\&&+\bigg(\bigg\| \frac{1}{8(m+1)t}\sum\limits_{i=0}^{m}(y_{i, 1}-y_{m+1, 1})-\sqrt{m+1}c_{1}'\bigg\| \\&&{}+\bigg\| \frac{1}{8(m+1)t}\sum\limits_{i=0}^{m}(y_{i, 1}-y_{m+1, 1})\bigg\| \bigg)^{2}\\&&+\cdots+\bigg(\bigg\| \frac{1}{8(m+1)t}\sum\limits_{i=0}^{m}(y_{i, m}-y_{m+1, m})-\sqrt{m+1}c_{m}'\bigg\| \\ &&{}+\bigg\| \frac{1}{8(m+1)t}\sum\limits_{i=0}^{m}(y_{i, m}-y_{m+1, m})\bigg\| \bigg)^{2}\\&\leq& 2(m+1)\bigg(\sum\limits_{i=0}^{m}\bigg\| \frac{1}{8(m+1)t}(y_{i, 0}-y_{m+1, 0}-e)\bigg\| ^{2}\\&&{}+\sum\limits_{k=1}^{m}\sum\limits_{i=0}^{m}\bigg\| \frac{1}{8(m+1)t}(y_{i, k}-y_{m+1, k})\bigg\| ^{2}\bigg)\\&&+2\bigg(\bigg\| \frac{1}{8(m+1)t}\sum\limits_{i=0}^{m}(y_{i, 0}-y_{m+1, 0}-e)-\sqrt{m+1}c_{0}'\bigg\| ^{2}\\&&{}+\sum\limits_{k=1}^{m}\bigg\| \frac{1}{8(m+1)t}\sum\limits_{i=0}^{m}(y_{i, k}-y_{m+1, k})-\sqrt{m+1}c_{k}'\bigg\| ^{2}\bigg), \end{eqnarray*}$

由此, (3.14)和(3.15)式表明(3.1)式的第二个不等式成立.证毕.

我们知道拉格朗日乘数法是求解条件极值问题的有效方法.当$X=Y_{0}=\cdots=Y_{m}=\mathbb{R} $, $K_{0}=K_{1}=\cdots=K_{m}={\Bbb R_{+}}$和$\Phi_{i}\ (i=0, \cdots, m)$是pseudo-Lipschitz映射时,拉格朗日乘数法能表示为:如果$\bar{x}$是优化问题(1.2)的解,则存在$\lambda_{i}\in[0, 1]\ (i=0, \cdots, m)$,使得

$\sum\limits_{i=0}^{m}\lambda_{i}=1\ \ \ \mbox{且}\ \ \ 0\in\sum\limits_{i=0}^{m}\lambda_{i}\partial\Phi_{i}(\bar{x})+N(A, \bar{x}), $

其中$\partial\Phi_{i}(\bar{x})$和$N(A, \bar{x})$分别表示$\Phi_{i}$在点$\bar{x}$处的Clarke次微分和Clarke法锥.如果$\Phi_{i}\ (i=0, \cdots, m)$是光滑的,则Clarke次微分能用导数代替.因为proximal次微分体现二阶变分性质,所以就算是光滑标量优化问题也不能得到其关于proximal次微分的拉格朗日乘数法^[21].为此,文献[21]给出了问题(1.1)的$\varepsilon$-proximal拉格朗日点和fuzzy proximal拉格朗日点的定义.自然地,这两个定义能推广到多目标优化问题(1.2).

定义3.1 设$\varepsilon>0$, $\bar{x}$是问题$(1.2)$的一个可行点,并且$\bar{y}_{0}\in \Phi_{0}(\bar{x})$.称$(\bar{x}, \bar{y}_{0})$是

(ⅰ)问题$(1.2)$的$\varepsilon$-proximal拉格朗日点,当且仅当$\bar{y}_{i}\in\Phi_{i}(\bar{x})\cap(-K_{i})\ (i=1, \cdots, m)$,且存在$x_{0}\in B(\bar{x}, \varepsilon)$, $y_{0}\in\Phi_{0}(x_{0})\cap B(\bar{y}_{0}, \varepsilon)$, $x_{i}\in B(\bar{x}, \varepsilon)$, $y_{i}\in\Phi_{i}(x_{i})\cap B(\bar{y}_{i}, \varepsilon)\ (i=1, \cdots, m)$, $a\in A\cap B(\bar{x}, \varepsilon))$和$c_{i}\in K_{i}^{+}\ (i=1, \cdots, m)$满足以下性质

$\sum\limits_{i=0}^{m}\| c_{i}\| =1\ \ \ \mbox{和} \ \ \ 0\in\sum\limits_{i=0}^{m}D_{P}^{*}\Phi_{i}(x_{i}, y_{i})(c_{i}+\varepsilon B_{Y_{i}}) +N_{P}(A, a)+\varepsilon B_{X}.$

(ⅱ)问题$(1.2)$的fuzzy proximal拉格朗日点,当且仅当对任意的$\varepsilon>0$, $(\bar{x}, \bar{y}_{0})$都是问题$(1.2)$的$\varepsilon$-proximal拉格朗日点.

定理3.2 设$\Phi_{0}$在可行集$Z$上关于序锥$K_{0}$下有界.则下述结论有一个成立.

(ⅰ)对任意的$\varepsilon>0$,存在$\bar{x}\in Z$和$\bar{y}_{0}\in\Phi_{0}(\bar{x})$,使得$(\bar{x}, \bar{y}_{0})$是问题$(1.2)$的一个$\varepsilon$-帕雷托解,并且$(\bar{x}, \bar{y}_{0})$是问题$(1.2)$的一个$\varepsilon$-proximal拉格朗日点.

(ⅱ)对任意的$\varepsilon>0$,存在$\bar{x}\in Z$和$\bar{y}_{0}\in \Phi_{0}(\bar{x})$,使得$(\bar{x}, \bar{y}_{0})$是问题$(1.2)$的一个$\varepsilon$ -帕雷托解,并且存在$x_{0}\in B(\bar{x}, \varepsilon)$, $y_{0}\in\Phi_{0}(x_{0})\cap B(\bar{y}_{0}, \varepsilon)$, $x_{i}\in B(\bar{x}, \varepsilon)$, $y_{i}\in \Phi_{i}(x_{i})\cap (-K_{i}+\varepsilon B_{Y_{i}})\ (1\leq i\leq m)$, $a\in A\cap B(\bar{x}, \varepsilon))$,

$\zeta_{i}\in D_{P}^{*}\Phi_{i}(x_{i}, y_{i})(\varepsilon B_{Y_{i}})\ (0\leq i\leq m)\ \ \ \mbox{和}\ \ \ \xi\in N_{P}(A, a)+\varepsilon B_{X}$

满足

$\sum\limits_{i=0}^{m}\zeta_{i}+\xi=0\ \ \ \mbox{和}\ \ \ \sum\limits_{i=0}^{m}\| \zeta_{i}\| +\| \xi\| =1.$

证 由命题2.3得,任意$n\in {\Bbb N}$,存在$\bar{x}_{n}\in Z$和$\bar{y}_{n}\in\Phi_{0}(\bar{x}_{n})$,使得$(\bar{x}_{n}, \bar{y}_{n})$是问题(1.2)的$\frac{1}{n^{2}}$ -帕雷托解.应用定理3.1 (取$\varepsilon=\frac{1}{n^{2}}$, $\lambda=\frac{1}{n}$),存在$x_{0}(n)\in B(\bar{x}_{n}, \frac{1}{n})$, $y_{0}(n)\in \Phi_{0}(x_{0}(n))\cap B(\bar{y}_{n}, \frac{1}{n})$, $x_{i}(n)\in B(\bar{x}_{n}, \frac{1}{n})$$y_{i}(n)\in \Phi_{i}(x_{i}(n))\cap (-K_{i}+\frac{1}{n}B_{Y_{i}})\ (1\leq i\leq m)$, $x_{m+1}(n)\in A\cap B(\bar{x}_{n}, \frac{1}{n})$, $c_{i}(n)\in K_{i}^{+}$,

(3.16) $ \begin{equation}\zeta_{i}(n)\in D_{P}^{*}\Phi_{i}(x_{i}(n), y_{i}(n))(c_{i}(n)+\frac{1}{n}B_{Y_{i}})+\frac{1}{n}B_{X}\ \ \ 0\leq i\leq m\end{equation} $

和

(3.17) $\begin{equation} \zeta_{m+1}(n)\in N_{P}(A, x_{m+1}(n))+\frac{1}{n}B_{X}, \end{equation}$

使得$\sum\limits_{i=0}^{m+1}\zeta_{i}(n)=0$和

(3.18) $\begin{equation}\frac{1}{64(m+3)(m+1)^{2}}-\frac{1}{n^{2}} \leq \sum\limits_{i=0}^{m}(\| \zeta_{i}(n)\| ^{2}+\| c_{i}(n)\| ^{2}) \leq\frac{1}{32(m+1)}+\frac{1}{2n^{2}}\end{equation}$

成立.对任意的$n\in {\Bbb N}$,设$r_{n}:=\sum\limits_{i=0}^{m}\| c_{i}(n)\| $.我们首先考虑$\{r_{n}\}$不收敛到$0$的情况.不失一般性,假设存在$r$,使得$r_{n}>r, \forall \ n\in{\Bbb N}$ (如果必要可取子列).设$\tilde{c}_{i}(n):=\frac{c_{i}(n)}{r_{n}}$.则$\tilde{c}_{i}(n)\in K_{i}^{+}$, $\sum\limits_{i=0}^{m}\| \tilde{c}_{i}(n)\| =1\ (n\in {\Bbb N})$,结合(3.16)-(3.18)式,知

$0\in\sum\limits_{i=0}^{m}D_{P}^{*}\Phi_{i}(x_{i}(n), y_{i}(n))(\tilde{c}_{i}(n)+\frac{1}{nr}B_{Y_{i}}) +N_{P}(A, x_{m+1}(n))+\frac{m+2}{nr}B_{X}\ \ \forall \ n\in{\Bbb N}.$

由此可得(ⅰ)成立.下面假设$r_{n}\rightarrow0$.则由(3.18)式,对充分大的$n$,有$l_{n}:=\sum\limits_{i=0}^{m}\| \zeta_{i}(n)\| \geq \frac{1}{16(m+1)(m+3)}$.应用(3.16)-(3.18)式,得

$\frac{\zeta_{i}(n)}{l_{n}}D_{P}^{*}\Phi_{i}(x_{i}(n), y_{i}(n))\bigg(\frac{r_{n}}{l_{n}}+\frac{1}{nl_{n}}B_{Y_{i}}\bigg)+\frac{1}{nl_{n}}B_{X}\ \ \ 0\leq i\leq m, $

$\frac{\zeta_{m+1}(n)}{l_{n}}\in N_{P}(A, x_{m+1}(n))+\frac{1}{nl_{n}}B_{X}, $$ $$ \sum\limits_{i=0}^{m+1}\frac{\zeta_{i}(n)}{l_{n}}=0\ \ \ \mbox{和}\ \ \ \sum\limits_{i=0}^{m+1}\bigg\| \frac{\zeta_{i}(n)}{l_{n}}\bigg\| =1.$

于是(ⅱ)成立.证毕.

若$\Phi_{i}\ (0\leq i\leq m)$是pseudo-Lipschitz映射,由命题2.2和定理3.3容易推出以下结论.

推论3.1 设$\Phi_{0}$在可行集$Z$上关于序锥$K_{0}$下有界.假设$\Phi_{i}$都是pseudo-Lipschitz映射.则定理$3.3$中的(i)成立.即对任意的$\varepsilon>0$,存在$\bar{x}\in Z$和$\bar{y}_{0}\in\Phi_{0}(\bar{x})$,使得$(\bar{x}, \bar{y}_{0})$是问题$(1.2)$的一个$\varepsilon$-proximal拉格朗日点.

注3.1 如果$(\bar{x}, \bar{y}_{0})$是问题$(1.2)$的一个帕雷托有效解,则$\Phi_{0}(Z)\cap(\bar{y}_{0}-K_{0})=\{\bar{y}_{0}\}$,故

${\rm diam}\big(\Phi_{0}(Z)\cap(\bar{y}_{0}-K_{0})\big)=0.$

这说明对于任意的$\varepsilon'>0$, $(\bar{x}, \bar{y}_{0})$都是问题$(1.2)$的$\varepsilon'$ -帕雷托解.因此,依据定理$3.3 $和推论$3.4$的证明思想,容易推出问题$(1.2)$的每一个帕雷托解都是该问题的fuzzy proximal拉格朗日点.

注3.2 当int$(K_{0})\neq0$,通常会考虑问题$(1.2)$的弱帕雷托解.假设$(\bar{x}, \bar{y}_{0})$是问题$(1.2)$的弱帕雷托解,则

$\Phi_{0}(Z)\cap\big(\bar{y}_{0}-{\rm int}(K_{0})\big)=\emptyset.$

因为$K_{0}$是凸锥, int$(K_{0})+K_{0}\subset {\rm int}(K_{0})$,所以任取$e\in-{\rm int}(K_{0})$,有$\Phi_{0}(Z)\cap(\bar{y}_{0}+e-K_{0})=\emptyset$.说明,问题$(1.2)$的任意弱帕雷托解都是它的弱$\varepsilon'$ -帕雷托解(任意$\varepsilon'>0$).因此,问题$(1.2)$的每一个帕雷托解也都是它的fuzzy proximal拉格朗日点.