目标优化作为最优化理论和应用的重要分支, 其理论研究涉及到凸分析、非光滑分析、非线性分析等学科, 同时, 它在国民生活中的众多领域均有着重要应用.我们知道, 多目标优化问题的有效解在非紧条件下往往不存在, 在实际应用数值方法求解优化问题时, 所得到的解常常是近似解, 而且近似解在很弱的情况下都可能存在.故研究多目标优化问题的近似解更加符合实际应用和理论研究的需要.近些年, 众多学者对此类问题开展了深入的研究, 获得了丰富的成果^[1-10].

本文中我们考虑如下带一般约束的多目标优化问题:

$\mathbf{ (VP)}\quad \left\{ \begin{array}{cl}{\min} & f(x)=(f_1 (x), f_2 (x), \cdots, f_k (x))^T, \\ \textrm{s.t.}&h(x)=(h_1 (x), h_2 (x), \cdots, h_m (x))^T\le 0, \\ & g(x)=(g_1 (x), g_2 (x), \cdots, g_l (x))^T=0, \\ & x\in X, \end{array} \right.$

其中$X\subset R^n$是非空开集, $f_i, h_j, g_t :R^n\to R(i=1, \cdots, k, j=1, \cdots, m, t=1, \cdots, l)$是$X$上的局部Lipschitz连续函数.问题(VP)的可行集记为$S=\left\{x\in X:h(x)\le 0, g(x)=0\right\}$, $I(x)=\left\{j\in \{1, \cdots, m\}:h_j (x)=0\right\}.$

设$R^n$是$n$维欧氏空间, 记$R_+^n =\left\{x\in R^n:x_i \ge 0, i=1, \cdots, n\right\}$, $\textrm{int} R_+^n =\{x\in R^n: x_i >0, i=1, \cdots, n\}$.定义$R^n$中的序关系:

$ x<y\Leftrightarrow y-x\in \textrm{int}R_+^n, \qquad x\le y\Leftrightarrow y-x\in R_+^n. $

设$\varepsilon=(\varepsilon _1, \varepsilon _2, \cdots, \varepsilon_k)^T\in R_+^k $, 以下是$\varepsilon$ -拟弱有效解的概念.

定义1.1 ^[4] 称$\overline x \in S $为问题(VP)的$\varepsilon$ -拟弱有效解, 如果

$ f(S)-f(\overline x )+\varepsilon \left\| {x-\overline x }\right\|\;\bigcap\;(-\textrm{int}R_+^k )=\emptyset. $

若$\varepsilon =0, $则以上定义退化为问题(VP)的弱有效解.

定义1.2 ^[5] $f(x):R^n\to R$在点$\overline x$处是Lipschitz的, 则$f(x)$在点$\overline x$处沿方向$d\in R^n$的Clarke方向导数、Clarke广义方向导数、Clarke广义次梯度分别定义为如下形式:

$\begin{aligned} f'\left(\overline x, d\right) & =\lim\limits_{t\downarrow 0}\dfrac{f(\overline x +td)-f(\overline x)}{t}, \\ f^\circ(\overline x, d) & =\lim\limits_{t\downarrow 0}\sup\limits_{y\to x} \dfrac{f(y+td)-f(y)}{t}\\ & =\lim\limits_{\delta\downarrow 0}\sup\left\{\left.\dfrac{f(y+td)-f(y)}{t}\right| 0<t<\delta, \left\|y-x\right\|<\delta\right\}, \\ \partial f(\overline x ) & =\left\{\xi\in R^n\left| f^\circ (\overline x, d)\ge \langle \xi, d\rangle, \forall d\in R^n\right.\right\}. \end{aligned}$

定义1.3 ^[6] 称$f(x)$在点$x$处是正则的, 如果满足:对任意$d\in R^n$, 有$f'(x, d)$存在, 并且$f'(x, d)=f^\circ (x, d)$.

定义1.4 ^[7] 对任意给定的$\varepsilon \ge 0$, 称函数$\varphi\to R$在集合$X$上是$\varepsilon$ -凸函数, 如果对所有的$x, y\in K$和$\lambda \in (0, 1)$, 有

$\varphi(\lambda x+(1-\lambda)y)\le \lambda\varphi (x)+(1-\lambda )\varphi (y)+\varepsilon \lambda (1-\lambda )\left\| x-y \right\|.$

定义1.5 ^[8] 称函数$f:X\to R$在$x\in K$处是$\varepsilon$ -半凸的, 对$\emptyset \ne K\subset X, \varepsilon>0$, 如果下列条件成立:

ⅰ) $f$在点$x$的球形邻域内是Lipschitz的;

ⅱ)对任意$d\in X$, 有${f}'(x, d)$存在, 且满足$0\le f^\circ (x, d)-{f}'(x, d)\le \sqrt \varepsilon \left\| d \right\|$;

ⅲ)由$x+d\in K$和$f'(x, d)+\sqrt \varepsilon \left\| d \right\|\ge0$可推出$f(x+d)+\sqrt \varepsilon \left\| d \right\|\ge f(x)$成立.

引理1.1 ^[4] 如果$\overline x \in S$是问题(VP)的一个$\varepsilon $ -拟弱有效解, 则$\overline x$是如下向量优化问题的弱有效解.

$\mathbf{(VP1)}\qquad \mathop {\min }\limits_{x\in S} f(x)+\varepsilon \left\| {x-\overline x }\right\|.$

引理1.2 ^[9] 设$X$是Asplund空间, $f_i :X\to \overline R, i=1, 2$是下半连续函数且其中一个函数在$\overline x$处是局部Lipschitz的, 则有如下关系成立:

$\partial (f_1 +f_2 )(\overline x )\subset \partial f_1 (\overline x )+\partial f_2 (\overline x ).$

受文献[4]等工作的启发, 我们在相关文献考虑MP问题的基础上, 增加了等式约束条件, 即本文考虑的VP问题, 并将文献[4]中定理3.1、定理3.2中的凸性假设改为半凸性假设, 得到VP问题$\varepsilon$-拟弱有效解的相应最优性条件(见本文定理2.1和定理2.2).

类似的, 我们定义了VP问题的拉格朗日函数及其$\varepsilon$ -拟弱鞍点, 得到VP问题$\varepsilon$ -拟弱鞍点的相应定理(见本文定理3.1和定理3.2).接着, 我们定义了VP问题的对偶问题VD问题, 得到VP问题的弱对偶和强对偶定理(见本文定理3.3和定理3.4).

定理2.1 设存在$\overline x \in S$, $\lambda =(\lambda _1, \cdots, \lambda _k )^T\in R_+^k \backslash \left\{ 0\right\}$, 使得

ⅰ) $\sum\limits_{i=1}^k {\lambda _i f_i } $在$\overline x $处是$(\lambda^T\varepsilon )^2$ -半凸；

ⅱ) $\sum\limits_{i=1}^k {\lambda _i f_i ' (\overline x , x-\overline x)} +\lambda ^T\varepsilon \left\| {x-\overline x } \right\|\ge 0, \quad \forall x\in S, $

那么$\overline x $是问题(VP)的$\varepsilon$ -拟弱有效解.

证 由条件ⅱ)和ⅰ), 可得

$\sum\limits_{i=1}^k {\lambda _i f_i (x)} +\lambda ^T\varepsilon \left\| {x-\overline x } \right\|\ge \sum\limits_{i=1}^k {\lambda _i f_i (\overline x )}, \quad \forall x\in S.$

(2.1)

假设$\overline x $不是问题(VP)的$\varepsilon$ -拟弱有效解.那么存在$\hat{x}\in S$使得

$f(\hat{x})-f(\overline x )+\varepsilon \left\| {\hat{x}-\overline x } \right\|\in -\textrm{int}R_+^k.$

因此, 由$\lambda \in R_+^k \backslash \left\{ 0 \right\}$, 可得

$ \lambda ^T f(\hat{x})+\lambda ^T \varepsilon \left\| {\hat{x}-\overline x } \right\|\in \lambda ^T f(\overline x ). $

这与(2.1) 式矛盾.因此, $\overline x $是问题(VP)的$\varepsilon$ -拟弱有效解.

定理2.2 假设$f_i, h_j, v_t(i=1, \cdots, k; j=1, \cdots, m; t=1, \cdots, l)$在$\overline x \in S$处是正则的, 并且满足

$\sum\limits_{i=1}^k{\overline \lambda _i f_i +} \sum\limits_{j=1}^m {\overline u _j h_j +}\sum\limits_{t=1}^l {\overline v _t g_t } $

在$\overline x $处是$(\lambda^T\varepsilon )^2$ -半凸的, 则$\overline x$是问题(VP)的$\varepsilon$ -拟弱有效解当且仅当存在$\lambda \in R_+^k \backslash \left\{ 0 \right\}$, $\mu \in R_+^m$, $v\in R_+^l$, 使得

ⅰ) $0\in \sum\limits_{i=1}^k {\lambda _i \partial f_i (\overline x )+} \sum\limits_{j=1}^m {\mu _j \partial h_j (\overline x )+} \sum\limits_{t=1}^l {v_t \partial g_t (\overline x )+} \lambda ^T\varepsilon B^\ast$;

ⅱ) $\sum\limits_{j=1}^m {\mu _j h_j (\overline x )} =0$;

ⅲ) $\sum\limits_{i=1}^k {\lambda _i f_i +} \sum\limits_{j=1}^m {\mu _j h_j+} \sum\limits_{t=1}^l {v_t g_t } $在$\overline x $处是$\left(\lambda^T\varepsilon \right)^2$ -半凸,

这里$B^\ast $是$R^n$中的单位球.

证 $(\Longrightarrow)$由引理1.1可知, $\overline x$是问题(VP1) 的弱有效解.利用Fritz-John型必要性条件可知, 存在$\lambda\in R_+^k $和$u\in R_+^m$, $v\in R_+^l $使得

$0\in \sum\limits_{i=1}^k {\lambda _i \partial (f_i (\overline x )} +\varepsilon _i \left\| {x-\overline x } \right\|)+\sum\limits_{j=1}^m {u_j \partial } h_j (\overline x )+\sum\limits_{t=1}^l {v_t } \partial g_t (\overline x ), \\ \quad\; \sum\limits_{j=1}^m {u_j } h_j (\overline x )=0.$

由引理1.2有

$\sum\limits_{i=1}^k {\lambda _i \partial (f_i (\overline x )} + \varepsilon_i \left\| {x-\overline x } \right\|) \subseteq \sum\limits_{i=1}^k {\lambda_i } \partial f_i (\overline x )+ \sum\limits_{i=1}^k {\lambda _i } \partial\varepsilon _i \left\| {x-\overline x } \right\|. $

故

$0\in \sum\limits_{i=1}^k {\lambda _i \partial f_i (\overline x )} +\sum\limits_{j=1}^m {u_j \partial h_j (\overline x )}+ \sum\limits_{t=1}^l{v_t } \partial g_t (\overline x )+ \lambda ^T\varepsilon B^\ast .$

$( \Longleftarrow )$由ⅰ)知, 存在

$x_i^\ast \in \partial f_i (\overline x ), \; y_j^\ast \in \partial h_j(\overline x ), \; z_t^\ast \in \partial g_t (\overline x ), \; b^\ast \in B^\ast, $

其中$i=1, \cdots, k; j=1, \cdots, m; t=1, \cdots, l$, 使得

$ 0=\sum\limits_{i=1}^k {\lambda _i x_i^\ast +} \sum\limits_{j=1}^m {\mu _j y_j^\ast +} \sum\limits_{t=1}^l {v_t z_t^\ast +} \lambda ^T\varepsilon b^\ast. $

由Clarke方向导数的性质和定理2.2的假设, 可知

$\begin{aligned} \left(\sum\limits_{i=1}^k{\lambda _i f_i }+ \sum\limits_{j=1}^m {\mu _j h_j +} \sum\limits_{t=1}^l {v_t g_t }\right)'\big(\overline x, d\big)&=\sum\limits_{i=1}^k{\lambda_i f'_i(\overline x, d)}+ \sum\limits_{j=1}^m{\mu _j h'_j(\overline x, d)} + \sum\limits_{t=1}^l{v_t g'_t (\overline x, d)}\\&=\sum\limits_{i=1}^k{\lambda _i f^\circ_i(\overline x, d)}+ \sum\limits_{j=1}^m{\mu _j h^\circ_j(\overline x, d)} + \sum\limits_{t=1}^l{v_t g^\circ_t (\overline x, d)} \\&\ge\sum\limits_{i=1}^k {\lambda _i } \langle x_i^\ast, d\rangle +\sum\limits_{j=1}^m {\mu _j \langle y_j^\ast, d\rangle} +\sum\limits_{t=1}^l {v_t \langle z_t^\ast, d \rangle } \\&=\left\langle \sum\limits_{i=1}^k {\lambda _i x_i^\ast }+ \sum\limits_{j=1}^m {\mu _j y_j^\ast } + \sum\limits_{t=1}^l {v_t z_t^\ast }, d\right\rangle \\&=\langle -\lambda ^T\varepsilon b^\ast, d\rangle =-\lambda^T\varepsilon \langle b^\ast, d\rangle \ge -\lambda ^T\varepsilon \left\| d\right\|. \end{aligned}$

即成立

$\left(\sum\limits_{i=1}^k {\lambda _i f_i +} \sum\limits_{j=1}^m {\mu _j h_j+} \sum\limits_{t=1}^l {v_t g_t } \right)'\big(\overline x, d\big)+ \lambda^T\varepsilon \left\| d \right\|\ge 0 .$

由$\sum\limits_{i=1}^k{\overline \lambda _i f_i +} \sum\limits_{j=1}^m {\overline u _j h_j +}\sum\limits_{t=1}^l {\overline v _t g_t } $在$\overline x $处是$(\lambda^T\varepsilon )^2$ -半凸的, 取$d=x-\overline x , x\in X$, 可得

$\quad\sum\limits_{i=1}^k {\lambda _i f_i (x)+} \sum\limits_{j=1}^m \mu _j h_j(x)+\sum\limits_{t=1}^l {v_t g_t (x)} +\lambda ^T\varepsilon \left\|{x-\overline x } \right\|\\ \ge \sum\limits_{i=1}^k {\lambda _i f_i (\overline x )+} \sum\limits_{j=1}^m {\mu _j h_j (\overline x )+} \sum\limits_{t=1}^l {v_t g_t (\overline x )}.$

由ⅱ)及$\sum\limits_{t=1}^l {v_t g_t (\overline x)} =0$, 得

$ \sum\limits_{i=1}^k {\lambda _i f_i (x)+} \sum\limits_{j=1}^m {\mu _j h_j (x)+} \sum\limits_{t=1}^l {v_t g_t (x)} +\lambda ^T\varepsilon \left\| {x-\overline x } \right\|\ge \sum\limits_{i=1}^k {\lambda _i f_i (\overline x )} . $

又由于

$\sum\limits_{j=1}^m {\mu _j h_j (x)\le 0, } \quad \sum\limits_{t=1}^l {v_tg_t (x)} =0, \forall x\in S, $

从而

$ \sum\limits_{i=1}^k {\lambda _i f_i (x)} +\lambda ^T\varepsilon \left\| {x-\overline x } \right\|\ge \sum\limits_{i=1}^k {\lambda _i f_i (\overline x ), \quad \forall x\in S}, $

即$\overline x $是问题(VP)的$\varepsilon$ -拟弱有效解.

定义3.1 设$e=(1, \cdots, 1)^T\in R^k$.对任意的$x\in X$和$u\in R_+^m$, $v\in R_+^l$, 定义拉格朗日函数形式如下

$ L(x, u, v)=f(x)+[u^Th(x)+v^Tg(x)]e. $

定义3.2 称$(\overline x, \overline u, \overline v )\in X\times R_+^m\times R_+^l $是拉格朗日函数的$\varepsilon $ -拟弱鞍点, 如果

$\begin{aligned}&L(\overline x, u, v)-\varepsilon \left\| {u-\overline u }\right\|- \varepsilon \left\| {v-\overline v } \right\| \le L(\overline x, \overline u, \overline v ), && \forall u\in R_+^m, v\in R_+^l, \\&L(x, \overline u, \overline v )+\varepsilon \left\| {x-\overline x }\right\|- L(\overline x, \overline u, \overline v ) \notin -\textrm{int}R_+^k, && \forall x\in X. \end{aligned}$

注 若$\varepsilon =0$, $\varepsilon$ -拟弱鞍点退化到文[10]中研究的广义拉格朗日鞍点.

定理3.1 设$(\overline x, \overline u, \overline v )\in X\times R_+^m\times R_+^l $是$\varepsilon$ -拟弱鞍点, $\overline x $是问题

$\mathop {\max}\limits_{x\in X} \overline u ^T h(x), \quad \mathop {\max }\limits_{x\in X}\overline v ^T g(x)$

的最优解, 则$\overline x $是问题(VP)的$\varepsilon$ -拟弱有效解.

证 由$\varepsilon$ -拟弱鞍点的定义可得, 对任意$ x\in X$, 有

$\bigg(f(x)-f(\overline x )+\varepsilon \left\| {x-\overline x }\right\| +\overline u ^T\big(h(x)-h(\overline x )\big)e +\overline v^T\big(g(x)-g(\overline x )\big)e\bigg) \bigcap \bigg(-\mathrm{int}R_+^k\bigg )=\emptyset.$

再由$\overline x $是问题$\mathop {\max }\limits_{x\in X} \overline u^Th(x)$, $\mathop {\max }\limits_{x\in X} \overline v ^Tg(x)$的最优解可知, 对任意$ x\in X$, 有

$\;\;\overline u ^T(h(x)-h(\overline x ))e\le 0, \\ \;\; \overline v ^T(g(x)-g(\overline x ))e\le 0, \\ 0\le f(x)-f(\overline x )+\varepsilon \left\| {x-\overline x } \right\|+\overline u ^T(h(x)-h(\overline x ))e+\overline v ^T(g(x)-g(\overline x ))e\\ \;\; \le f(x)-f(\overline x )+\varepsilon \left\| {x-\overline x } \right\|,$

故

$\big(f(x)-f(\overline x )+\varepsilon \left\| {x-\overline x } \right\|\big)\bigcap \big(-\textrm{int}R_+^k \big)=\emptyset, $

因此结论成立.

定理3.2 设$\overline x $是问题(VP)的$\varepsilon$ -拟弱有效解.若定理2.2中的假设条件成立, 则存在$\overline u \in R_+^m, \overline v \in R_+^l $使得$(\overline x, \overline u, \overline v )$是$\varepsilon$ -拟弱鞍点.

证 由定理2.2, 存在$\overline \lambda \in R_+^k $和$\overline u \in R_+^m$和$\overline v \in R_+^l $, 使得

$0\in \sum\limits_{i=1}^k {\overline \lambda _i \partial f_i (\overline x )} +\sum\limits_{j=1}^m {\overline u _j \partial } h_j (\overline x )+\sum\limits_{t=1}^l {\overline v _t } \partial g_t (\overline x )+\varepsilon ^T\overline \lambda B^\ast,$

(3.1)

$\sum\limits_{j=1}^m {\overline u _j } h_j (\overline x )=0.$

(3.2)

易证

$L(\overline x, u, v)-\varepsilon \left\| {u-\overline u } \right\|-\varepsilon \left\| {v-\overline v } \right\|\le L(\overline x, \overline u, \overline v ), \quad\forall u\in R_+^m, v\in R_+^l.$

下面证明

$L(x, \overline u, \overline v )+\varepsilon \left\| {x-\overline x } \right\|-L(\overline x, \overline u, \overline v )\notin -\textrm{int}R_+^k, \quad\forall x\in X.$

假设不成立, 则存在$x\in X$使得

$f(\overline x )+\overline u ^Th(\overline x )e+\overline v ^Tg(\overline x )e>f(x)+\overline u ^T h(x)e+\overline v ^Tg(x)e+\varepsilon \left\| {x-\overline x } \right\|, $

即

$ f(x)-f(\overline x )+\overline u ^T(h(x)-h(\overline x ))e+\overline v ^T(g(x)-g(\overline x ))e+\varepsilon \left\| {x-\overline x } \right\|<0. $

结合$\sum\limits_{i=1}^k {\overline {\lambda _i } } =1$有

$\sum\limits_{i=1}^k {\overline {\lambda _i } } (f_i (x)-f_i (\overline x ))+\sum\limits_{j=1}^m {\overline {u_j } ^T(h_j (x)-h_j (\overline x ))} +\sum\limits_{t=1}^l {\overline {v_t } ^T(g_t (x)-g_t (\overline x ))} +\varepsilon ^T\overline \lambda \left\| {x-\overline x } \right\|<0.$

(3.3)

再由定理2.2中的假设条件: $f_i, h_j, v_t .(i=1, \ldots , k;j=1, \ldots, m;t=1, \ldots, l)$在$\overline x \in S$处是正则的, 和

$\sum\limits_{i=1}^k {\overline \lambda _i f_i +} \sum\limits_{j=1}^m {\overline u _j h_j +} \sum\limits_{t=1}^l {\overline v _t g_t }$

在$\overline x $处是$(\lambda^T\varepsilon )^2$ -半凸的, 类似定理2.2的证明可得到

$\quad \sum\limits_{i=1}^k {\overline \lambda _i f_i (x)+} \sum\limits_{j=1}^m {\overline u _j h_j(x)} + \sum\limits_{t=1}^l {\overline v _t g_t (x)} +\overline \lambda ^T\varepsilon \left\| {x-\overline x } \right\|\\ \ge \sum\limits_{i=1}^k {\overline \lambda _i f_i (\overline x )+} \sum\limits_{j=1}^m {\overline \mu _j h_j (\overline x )+} \sum\limits_{t=1}^l {\overline v _t g_t (\overline x )}, \quad \forall x\in X.$

这与式(3.3) 矛盾, 因此结论成立.

下面我们考虑(VP)的$\varepsilon$拉格朗日对偶问题.

对$u\in R^m, v\in R^l$考虑如下问题

$\mathbf{ (VP)}_{(u, v)}\quad \left\{ \begin{array}{cl} {\min} & L(x, u, v), \\ \textrm{s.t.}&x\in S. \end{array} \right.$

设$W(u, v)$为问题(VP)$_{(u, v)}$的所有$\varepsilon$-拟弱有效解的集合.并设$\Omega _w (u, v)=\{L(x, u, v):x\in W(u, v)\}$.下面考虑如下形式的对偶问题

$\mathbf{ (VD)}\quad \left\{ \begin{array}{cl} {\max}&\Omega _w (u, v), \\ \textrm{s.t.}&u\in R_+^m, v\in R_+^l. \end{array} \right.$

定理3.3 (弱对偶定理) 设$x$是问题(VP)的任意可行点,

$\overline y\in \bigcup\limits_{u\in R_+^m, v\in R_+^l } {\Omega _w (u, v)}, $

即存在$\overline u \in R_+^m$, $\overline v \in R_+^l $和$\overline x \in W(\overline u, \overline v )$, 使得$\overline y =L(\overline x, \overline u, \overline v )$, 则

$(f(x)+\varepsilon \left\| {x-\overline x } \right\|-\overline y )\;\bigcap\; (-\textrm{int}R_+^k )=\emptyset.$

证 由$\overline x \in W(\overline u, \overline v)$可知, 不存在$\hat{x}\in S$使得

$L(\hat{x}, \overline u, \overline v )+\varepsilon \left\| {\hat{x}-\overline x } \right\|<L(\overline x, \overline u , \overline v )=\overline y .$

由于$\overline u \in R_+^m$, $\overline v \in R_+^l$, $h(\hat{x})\le 0$, $g(\hat{x})=0$, 故不存在$\hat{x}\in S$使得

$f(\hat{x})<L(\hat{x}, \overline u, \overline v) =f(\widehat{x})+\left[\overline u ^T h(\widehat{x})+\overline v^T g(\hat{x})\right]e, \\ f(\hat{x})+\varepsilon \left\| {\hat{x}-\overline x }\right\| <L(\hat{x}, \overline u, \overline v )+\varepsilon \left\| {\hat{x}-\overline x } \right\|<\overline y .$

故$f(\hat{x})+\varepsilon \left\| {\hat{x}-\overline x } \right\|<\overline y$, 因此结论成立.

定理3.4 (强对偶定理) 设$\overline x $是问题(VP)的$\varepsilon$ -拟弱有效解.定理2.2中的条件成立.则存在$\overline u \in R_+^m$, $\overline v\in R_+^l $, 使得$(\overline x, \overline u, \overline v)$是问题(VD)的$\varepsilon$-拟弱有效解.

证 由定理3.2可知, 存在$\overline u \in R_+^m, \overline v \in R_+^l$使得$(\overline x, \overline u, \overline v )$是$L(x, u, v)$的$\varepsilon$ -拟弱鞍点, 且$\sum\limits_{j=1}^m {\overline u _j } h_j (\overline x )=0.$

下面证明$\overline x $是问题(VP)$_{(\overline u, \overline v )}$的$\varepsilon$ -拟弱有效解.若不然, 存在$\hat{x}\in S$使得

$L(\hat{x}, \overline u, \overline v )+\varepsilon \left\| {\hat{x}-\overline x } \right\|<L(\overline x, \overline u, \overline v).$

这与$\varepsilon$ -拟弱鞍点的定义矛盾.

下面用反证法证明结论.假设$(\overline x, \overline u, \overline v)$不是问题(VD)的$\varepsilon$ -拟弱有效解, 则存在$\overline u \in R_+^m$, $\overline v \in R_+^l $, $\hat{x}\in W(\hat{u}, \hat{v})$和$\hat{y}\in \Omega _w(\hat{u}, \hat{v})$使得

$L(\overline x, \overline u, \overline v )+\varepsilon \left\| {(\overline u, \overline v, \overline x )-(\hat{u}, \hat{v}, \hat{x})} \right\|<L(\hat{x}, \hat{u}, \hat{v})=\hat{y}.$

因此$L(\overline x , \overline u, \overline v )+\varepsilon \left\| {\overline x -\hat{x}} \right\|<L(\hat{x}, \hat{u}, \hat{v})=\hat{y}.$由$\sum\limits_{j=1}^m {\overline u _j } h_j (\overline x )=0$, $\sum\limits_{t=1}^l {\overline v _t } g_t (\overline x )=0$，可得$f(\overline x )+\varepsilon \left\| {\overline x -\hat{x}} \right\|<\hat{y}.$这与定理3.3矛盾.