全局优化问题的最优性条件一直是学者们研究的热点^{[6-7, 10]}.凸函数差的全局最小值问题作为全局优化问题的典型问题,其研究也备受关注.一般情况下,可用抽象凸函数替换凸函数,则凸函数差的全局最小值问题转变为抽象凸函数差的全局最小值问题.目前关于抽象凸函数差问题的研究,主要集中在ICAR函数差^[8-9]、ICR函数差^[3-4]以及Topical函数差^[5]等几方面.

ICR函数在数理经济学领域有着广泛的应用^{[1-2, 11]}.此前关于ICR函数的性质定理的研究日益完善^[4],关于DICR函数性质定理的研究也逐渐展开^[12].本论文将继续在拓扑向量空间中研究DICR函数,引入该函数关于支撑集及次微分的概念,研究该函数支撑集与次微分之间的关系,结合该函数的抽象凸性及相关性质定理,得到严格DICR函数差的全局最小值的充要条件.

本文结构如下:第一章说明了DICR函数的研究背景.第二章回顾了DICR函数的定义及部分性质定理.第三章给出了DICR函数关于支撑集及次微分的概念,并研究该函数支撑集与次微分之间的关系.第四章研究了严格DICR函数与最大元的关系,给出了严格DICR函数差的全局最小值的充要条件.

设$X$为拓扑向量空间, $S$为$X$中的闭凸点锥,且$S\bigcap(-S)=\{0\}$.假设int$S \neq \emptyset$,在$X$上定义序关系$S:x\geq y\Leftrightarrow x-y\in S( x, y\in X )$.定义$R_{+}=[0 , +\infty)$和$R_{++}=(0 , +\infty)$.我们先回顾一些相关的定义.

定义2.1  设$g:X\rightarrow[0 , +\infty]$,

(ⅰ)称$g$为$X$上的co-radiant函数,如果对于任意的$x\in X$, $\lambda_{1} \in(0 , 1]$,有$ g(\lambda_{1}x)\geq \lambda_{1} g(x)$(易证该定义也等价于对于任意的$x\in X$, $ ~\lambda_{2}\in[1, +\infty)$,有$g(\lambda_{2}x)\leq \lambda_{2}g(x)$);

(ⅱ)称$g$为$X$上的inverse co-radiant函数,如果对于任意的$x \in X$, $\lambda_{1} \in(0, 1]$,有$g(\lambda_{1}x)\leq\frac{1}{\lambda_{1}}g(x)$ (易证该定义也等价于对于任意的$x\in X$, $\lambda_{2}\in[1, +\infty)$,有$g(\lambda_{2}x)\geq\frac{1}{\lambda_{2}}g(x)$);

(ⅲ)称$g$为$X$上递增函数,如果对于任意的$x, y\in X$,当$x\geq y$时,有$g(x)\geq g(y)$;

(ⅳ)称$g$为$X$上递减函数,如果对于任意的$x, y\in X$,当$x\geq y$时,有$g(x)\leq g(y)$;

(ⅴ)称$g$为$X$上严格递增函数,如果对于任意的$x, y\in X$,当$x>y$时,有$g(x)>g(y)$;

(ⅵ)称$g$为$X$上严格递减函数,如果对于任意的$x, y\in X$,当$x<y$时,有$g(x)<g(y)$.

定义2.2  设$g:X\rightarrow[0, +\infty]$,

(ⅰ)称$g$为ICR函数,如果$g$是递增的co-radiant函数;

(ⅱ)称$g$为DICR函数,如果$g$是递减的inverse co-radiant函数.

定义2.3  设$K\subseteq X$, $K\neq\emptyset$, $P=\{p:K\rightarrow[0, +\infty], p $为函数$\}$,如果存在集合$P_{0}\subseteq P$,对任意的$k\in K$,有

$ g(k)=\sup\{p(k):p\in P_{0}\}, $

称函数$g:K\rightarrow[0, +\infty]$是抽象凸的(或$P$-凸的).

定义2.4  设$P=\{p:K\rightarrow[0, +\infty], p$为函数$\}$为定义在集合$K$上的函数全体,在$P$上定义一般的序关系.任取$p, p_{0}\in P$,对任意的$k\in K$,有

$p_{0}(k)\geq p(k) \Rightarrow p_{0}=p, $

称函数$p$称为集合$P$上的最大元.

考虑函数$\varphi:X\times X\times R_{++}\rightarrow[0, +\infty]$,对任意的$x, y\in X, \alpha\in R_{++}$,定义

$\varphi(x, y, \alpha)=\sup\{\lambda:0\leq\lambda\leq \alpha, \lambda x \leq y\}.$

约定sup$\emptyset=0$.显然,对任意的$x, y\in X, \alpha\in R_{++}$,有$0\leq\varphi(x, y, \alpha)\leq\alpha$.

函数$\varphi(x, y, \alpha)$有以下性质^[4]

(2.1) $\varphi(\mu x, y, \alpha)=\frac{1}{\mu}\varphi(x, y, \mu \alpha);\\\varphi(x, \mu y, \alpha)=\mu\varphi(x, y, \frac{\alpha}{\mu});$

(2.2) $x\leq x'\Rightarrow\varphi(x, y, \alpha)\geq\varphi(x', y, \alpha);\\y\leq y'\Rightarrow\varphi(x, y, \alpha)\leq\varphi(x, y', \alpha);\\\alpha\leq\alpha'\Rightarrow\varphi(x, y, \alpha)\leq\varphi(x, y, \alpha');$

(2.3) $\varphi(\gamma x, y, \alpha)\leq\frac{1}{\gamma}\varphi(x, y, \alpha);\\\varphi(x, \gamma y, \alpha)\geq\gamma\varphi(x, y, \alpha);\\\varphi(x, y, \alpha)=\alpha\Longleftrightarrow\alpha x \leq y.$

任意$x, y\in X, \alpha\in R_{++}$,定义函数$\varphi_{(x, \alpha)}:X\rightarrow[0, +\infty]$表达式为$\varphi_{(x, \alpha)}(y)=\varphi(x, y, \alpha)$;函数$\varphi_{(y, \alpha)}:X\rightarrow[0, +\infty]$表达式为$\varphi_{(y, \alpha)}(x)=\varphi(x, y, \alpha)$.

所有$\varphi_{(x, \alpha)}$组成的集合记为$Y_{\varphi}=\{\varphi_{(x, \alpha)}:x\in X, \alpha\in R_{++}\}$.类似的, $X_{\varphi}=\{\varphi_{(y, \alpha)}:y\in X, \alpha\in R_{++}\}$.

以下关于DICR函数的性质定理可参阅文献[12].

命题2.1  函数$\varphi(x, y, \alpha)$关于第一变元为DICR函数,关于第二变元为ICR函数.

定理2.1  设$g:X\rightarrow[0, +\infty]$,则以下条件等价:

(ⅰ) $g$是DICR函数;

(ⅱ)对任意的$x, y\in X, \lambda \in(0, 1]$,当$\lambda x\leq y$时, $g(x)\geq \lambda g(y)$;

(ⅲ)对任意的$x, y\in X, \alpha \in R_{++}$, $\alpha g(x)\geq\varphi(x, y, \alpha)g(\frac{y}{\alpha})$,约定$0\times(+\infty)=0$.

定理2.2  设$L:X\rightarrow[0, +\infty]$,则$L$为DICR函数的充要条件是存在集合$X_{\varphi_{0}}\subseteq X_{\varphi}$,对任意的$y\in X, \alpha \in R_{++}$,都有

$ L(x)=\sup\limits_{\varphi_{(y, \alpha)}\in X_{\varphi_{0}}}\varphi_{(y, \alpha)}(x), $

其中

$X_{\varphi_{0}}=\Big\{\varphi_{(y, \alpha)}\in X_{\varphi}:L(\frac{y}{\alpha})\geq\alpha\Big\}. $

因此, $L$为DICR函数当且仅当$L$为$X_{\varphi}$-凸函数.

本章主要在拓扑向量空间中研究DICR函数的支撑集与次微分及相关结果.

定义3.1  设$L:X\rightarrow [0, +\infty]$为DICR函数,则函数$L$的$X_{\varphi}$-支撑集为

${\rm supp}_{\varphi}(L, X_{\varphi})=\{\varphi_{(y, \alpha)}\in X_{\varphi}:\varphi_{(y, \alpha)}(x)\leq L(x), \forall x \in X\}. $

根据定理2.2,对每一个DICR函数$L:X\rightarrow [0, +\infty]$,其$X_{\varphi}$-支撑集supp$_{\varphi}(L, X_{\varphi})\neq\emptyset$.

命题3.1  设$L:X\rightarrow [0, +\infty]$为DICR函数,则可得

${\rm supp}_{\varphi}(L, X_{\varphi})=\{\varphi_{(y, \alpha)}\in X_{\varphi}:L(\frac{y}{\alpha})\geq\alpha\}. $

证  任取$\varphi_{(y, \alpha)}\in {\rm supp}_{\varphi}(L, X_{\varphi})$,由定义3.1有$\varphi_{(y, \alpha)}(x)\leq L(x)$对任意的$x\in X$都成立.取$x=\frac{y}{\alpha}$,由(2.3)式可得$\alpha=\varphi_{(y, \alpha)}(\frac{y}{\alpha})\leq L(\frac{y}{\alpha})$.

反之,假设$\varphi_{(y, \alpha)}\in X_{\varphi}$且满足条件$L(\frac{y}{\alpha})\geq\alpha$,由定理2.1(ⅲ)有$\varphi_{(y, \alpha)}(x)\leq L(x)$.结合定义3.1即得$\varphi_{(y, \alpha)}\in {\rm supp}_{\varphi}(L, X_{\varphi})$.

综上所述,即得证.

定义3.2  设$L:X\rightarrow [0, +\infty]$为DICR函数,则函数$L$在点$x_{0}\in X$($L(x_{0})\in R_{++}$)处的$X_{\varphi}$-次微分为

$ \partial _{X_{\varphi}}L(x_{0})=\{\varphi_{(y, \alpha)}\in X_{\varphi}:L(x)-L(x_{0})\geq\varphi_{(y, \alpha)}(x)-\varphi_{(y, \alpha)}(x_{0}), \forall x\in X\}. $

命题3.2  设$L:X\rightarrow [0, +\infty]$是DICR函数,任取$x_{0}\in X$($L(x_{0})\in R_{++}$),则

$\{\varphi_{(y, \alpha)}\in X_{\varphi}:L(\frac{y}{\alpha})\geq\alpha, \varphi_{(y, \alpha)}(x_{0})=L(x_{0})\}\subseteq\partial _{X_{\varphi}}L(x_{0}).$

此外, $\partial _{X_{\varphi}}L(x_{0})\neq\emptyset$.

证  任取$\varphi_{(y, \alpha)}\in \{\varphi_{(y, \alpha)}\in X_{\varphi}:L(\frac{y}{\alpha})\geq\alpha\}$.由命题3.1有$\varphi_{(y, \alpha)}(x)\leq L(x)$对任意的$x\in X$都成立.结合已知条件$\varphi_{(y, \alpha)}(x_{0})=L(x_{0})$由定义3.2可得$\varphi_{(y, \alpha)}\in\partial _{X_{\varphi}}L(x_{0})$.

定理3.1  设$L:X\rightarrow [0, +\infty]$是DICR函数,任取$x_{0}\in X$($L(x_{0})\in R_{+}$),则

$\{\varphi_{(y, \alpha)}\in X_{\varphi}:L(x_{0})\leq\varphi_{(y, \alpha)}(x_{0}), \alpha-\varphi_{(y, \alpha)}(x_{0})\leq L(\frac{y}{\alpha})-L(x_{0})\}\subseteq\partial _{X_{\varphi}}L(x_{0}).$

特别地,当$\inf \limits_{x\in X}L(x)=0$时上式取等.

证  令${\bf D}=\{\varphi_{(y, \alpha)}\in X_{\varphi}:L(x_{0})\leq\varphi_{(y, \alpha)}(x_{0}), \alpha-\varphi_{(y, \alpha)}(x_{0})\leq L(\frac{y}{\alpha})-L(x_{0})\}$.根据$\varphi$的定义有$0\leq\frac{\varphi_{(y, \alpha)}(x)}{\alpha}\leq 1$,又因为$\varphi_{(y, \alpha)}(x_{0})-L(x_{0})\geq0$与$\alpha-L(\frac{y}{\alpha})\leq \varphi_{(y, \alpha)}(x_{0})-L(x_{0})$对任意的$\varphi_{(y, \alpha)}\in {\bf D}$都成立,即得

(3.1) $ \begin{equation}\label{equ:4} \frac{\varphi_{(y, \alpha)}(x)}{\alpha}\cdot(\alpha-L(\frac{y}{\alpha}))\leq\frac{\varphi_{(y, \alpha)}(x)}{\alpha}\cdot(\varphi_{(y, \alpha)}(x_{0})-L(x_{0}))\leq\varphi_{(y, \alpha)}(x_{0})-L(x_{0}). \end{equation} $

再根据定理2.1(ⅲ)有$\frac{\varphi_{(y, \alpha)}(x)}{\alpha}\cdot L(\frac{y}{\alpha})\leq L(x)$对任意$x\in X$都成立,结合(3.1)式可得

$\varphi_{(y, \alpha)}(x)-L(x)=\frac{\varphi_{(y, \alpha)}(x)}{\alpha}\cdot\alpha-L(x)\leq\frac{\varphi_{(y, \alpha)}(x)}{\alpha}\cdot(\alpha-L(\frac{y}{\alpha}))\leq\varphi_{(y, \alpha)}(x_{0})-L(x_{0}).$

由定义3.2可知$\varphi_{(y, \alpha)}\in\partial _{X_{\varphi}}L(x_{0})$,即证${\bf D}\subseteq\partial _{X_{\varphi}}L(x_{0})$.

当$\inf \limits_{x\in X}L(x)=0$时,任取$\varphi_{(y, \alpha)}\in\partial _{X_{\varphi}}L(x_{0})$.对任意的$x\in X$有

(3.2) $\begin{equation}\label{equ:5} \varphi_{(y, \alpha)}(x)-\varphi_{(y, \alpha)}(x_{0})\leq L(x)-L(x_{0}), \end{equation} $

结合$\varphi$的定义有

(3.3) $\begin{equation}\label{equ:15} -\varphi_{(y, \alpha)}(x_{0})\leq\varphi_{(y, \alpha)}(x)-\varphi_{(y, \alpha)}(x_{0})\leq L(x)-L(x_{0}). \end{equation} $

由(3.3)式化简可得$L(x_{0})-\varphi_{(y, \alpha)}(x_{0})\leq L(x)\leq\inf \limits_{x\in X}L(x)=0$,即$L(x_{0})\leq\varphi_{(y, \alpha)}(x_{0})$成立.再取(3.2)式中$x=\frac{y}{\alpha}$,由$\varphi$的定义可得

$\alpha-\varphi_{(y, \alpha)}(x_{0})\leq L(\frac{y}{\alpha})-L(x_{0}). $

因此, $\varphi_{(y, \alpha)}\in {\bf D}$,即证$\partial _{X_{\varphi}}L(x_{0})\subseteq{\bf D}$.

综上所述,当$\inf \limits_{x\in X}L(x)=0$时,有$\partial _{X_{\varphi}}L(x_{0})={\bf D}$.

本章主要研究严格DICR函数与最大元之间的关系,给出严格DICR函数差的全局最小值的一个充要条件.

假设$p, q:X\rightarrow [0, +\infty]$为恰当的DICR函数,且$D(q)\subseteq D(p)$.假设$(+\infty)-(+\infty)=(+\infty)$.令$f=q-p$,即

$ f(x)= \left\{ \begin{array}{ll} q(x)-p(x), &x\in D(q), \\ +\infty,&x\notin D(q). \end{array} \right.$

考虑以下优化问题

(4.1) $\begin{equation}\label{ques:9}\mbox{min}~~f(x), ~{\rm s.t.}~~x\in X, \end{equation}$

其中$\inf \limits_{x\in X}f(x)>-\infty$.

假设$\eta=\inf \limits_{x\in X}f(x)>-\infty$且满足$\gamma\leq\eta$,即有

$q(x)\geq p(x)+\gamma, \quad\forall x\in X .$

取$\widetilde{p}(x)= p(x)+\gamma$.对任意的$x\in X$,有$q(x)\geq p(x)+\gamma$当且仅当supp$_{\varphi}(\widetilde{p}, X_{\varphi})\subseteq {\rm supp}_{\varphi}(q, X_{\varphi})$.反之,假设supp$_{\varphi}(\widetilde{p}, X_{\varphi})= {\rm supp}_{\varphi}(p, X_{\varphi})+\gamma$,有

$\mbox{supp}_{\varphi}(p, X_{\varphi})+\gamma\subseteq \mbox{supp}_{\varphi}(q, X_{\varphi})$

当且仅当有$q(x)\geq p(x)+\gamma$对任意的$x\in X$都成立.

对任意的$y\in X, \alpha\in R_{++}$,定义$\varphi_{p}=\{\varphi_{(y, \alpha)}:\varphi_{(y, \alpha)}\in {\rm supp}_{\varphi}(p, X_{\varphi})\}$, $\varphi_{q}=\{\varphi_{(y, \alpha)}:\varphi_{(y, \alpha)}\in {\rm supp}_{\varphi}(q, X_{\varphi})\}$.

命题4.1  设$p:X\rightarrow [0, +\infty]$为DICR函数且$\varphi_{(y, \alpha)}\in\varphi_{p}$.若$\varphi_{(y, \alpha)}$为$\varphi_{p}$的最大元,则$p(\frac{y}{\alpha})=\alpha$.

证  任取$\varphi_{(y, \alpha)}\in\varphi_{p}$,由命题3.1可得$p(\frac{y}{\alpha})\geq\alpha$,即$\frac{p(\frac{y}{\alpha})}{\alpha}\geq1$成立.又因为$p(\frac{y}{\alpha})\geq\alpha$,由命题3.1有$\varphi_{((\frac{y}{\alpha})\cdot p(\frac{y}{\alpha}), p(\frac{y}{\alpha}))}\in\varphi_{p}$.已知$\frac{p(\frac{y}{\alpha})}{\alpha}\geq1$,根据(2.1)式,函数$\varphi_{((\frac{y}{\alpha})\cdot p(\frac{y}{\alpha}), p(\frac{y}{\alpha}))}$对任意的$x\in X$都有

(4.2) $\begin{equation}\label{equ:10} \varphi_{(y, \alpha)}(x)\leq\frac{p(\frac{y}{\alpha})}{\alpha}\cdot\varphi_{(y, \alpha)}(x)=\varphi_{((\frac{y}{\alpha})\cdot p(\frac{y}{\alpha}), p(\frac{y}{\alpha}))}(x). \end{equation} $

若$\varphi_{(y, \alpha)}$为$\varphi_{p}$的最大元,结合(4.2)式知对任意$x\in X$,有

(4.3) $\begin{equation}\label{equ:11} \varphi_{(y, \alpha)}(x)=\varphi_{((\frac{y}{\alpha})\cdot p(\frac{y}{\alpha}), p(\frac{y}{\alpha}))}(x). \end{equation} $

取(4.3)式中$x=\frac{y}{\alpha}$,根据$\varphi$的定义可得

$p(\frac{y}{\alpha})=\varphi_{((\frac{y}{\alpha})\cdot p(\frac{y}{\alpha}), p(\frac{y}{\alpha}))}(\frac{y}{\alpha})=\varphi_{(y, \alpha)}(\frac{y}{\alpha})=\alpha.$

即证$p(\frac{y}{\alpha})=\alpha$.

命题4.2  设$p:X\rightarrow[0, +\infty]$为严格DICR函数,对任意的$y\in X$,取$\varepsilon=$ max$\{\alpha:p(\frac{y}{\alpha})\geq\alpha\}<+\infty$.则$\varphi_{(y, \alpha)}$为$\varphi_{p}$的最大元的充分必要条件是

$p(\frac{y}{\varepsilon})=\varepsilon.$

证  由命题4.1证明过程,以下仅证当$p(\frac{y}{\varepsilon})=\varepsilon$时, $\varphi_{(y, \varepsilon)}$为$\varphi_{p}$的最大元.当$p(\frac{y}{\varepsilon})=\varepsilon$时,由命题3.1可得$\varphi_{(y, \varepsilon)}\in\varphi_{p}$.假设存在$\varphi_{p}$的最大元为$\varphi_{(y', \varepsilon')}$,则$\varphi_{(y, \varepsilon)}(x)\leq\varphi_{(y', \varepsilon')}(x)$对任意的$x\in X$成立,以下证明$\varphi_{(y, \varepsilon)}=\varphi_{(y', \varepsilon')}$.于是,由(2.3)式以及$\varphi$的定义知

(4.4) $\begin{equation}\label{equ:12} \varepsilon=\varphi_{(y, \varepsilon)}(\frac{y}{\varepsilon})\leq\varphi_{(y', \varepsilon')}(\frac{y}{\varepsilon})\leq\varepsilon'. \end{equation} $

结合(2.3)式与定义3.1可得$\varepsilon=p(\frac{y}{\varepsilon})=\varphi_{(y, \varepsilon)}(\frac{y}{\varepsilon})\leq\varphi_{(y', \varepsilon')}(\frac{y}{\varepsilon})\leq p(\frac{y}{\varepsilon})=\varepsilon$,即得$\varphi_{(y', \varepsilon')}(\frac{y}{\varepsilon})=\varepsilon$.再根据(2.3)和(4.4)式,有$\frac{y'}{\varepsilon}\geq\frac{y'}{\varepsilon'}\geq\frac{y}{\varepsilon}$成立.已知$p$为递减的,有$p(\frac{y'}{\varepsilon})\leq p(\frac{y}{\varepsilon})=\varepsilon$.由(2.2)与(4.4)式可得$\varphi_{(y', \varepsilon)}\leq\varphi_{(y', \varepsilon')}$,由定义3.1即得$\varphi_{(y', \varepsilon)}\in\varphi_{p}$,结合命题3.1有$p(\frac{y'}{\varepsilon})\geq\varepsilon$成立.因此$p(\frac{y'}{\varepsilon})= p(\frac{y}{\varepsilon})=\varepsilon$.又已知$p$为严格的DICR函数,故$y=y'$.

既然$\varepsilon=\max\{\alpha:p(\frac{y}{\alpha})\geq\alpha\}$且$p(\frac{y}{\varepsilon'})\geq\varepsilon'$,那么$\varepsilon\geq\varepsilon'$成立.结合(4.4)式,有$\varepsilon=\varepsilon'$.

综上所述, $\varphi_{(y, \varepsilon)}=\varphi_{(y', \varepsilon')}$.

命题4.3  设$p:X\rightarrow[0, +\infty]$为严格DICR函数.对任意的$y\in X$,取$\varepsilon_{y}=\max\{\alpha:p(\frac{y}{\alpha})\geq\alpha\}<+\infty$.则对任意的$\varphi_{(y, \alpha)}\in \varphi_{p}$,存在最大元$\varphi_{(\widetilde{y}, \widetilde{\alpha})}\in\varphi_{p}$,使得$\varphi_{(y, \alpha)}\leq\varphi_{(\widetilde{y}, \widetilde{\alpha})}$成立,其中取$\widetilde{\alpha}=p(\frac{y}{\varepsilon_{y}})$, $\widetilde{y}=(\frac{y}{\varepsilon_{y}}\cdot p(\frac{y}{\varepsilon_{y}}))$.

证  因为$p(\frac{\widetilde{y}}{\widetilde{\alpha}})=\widetilde{\alpha}$,由命题3.1可得$\varphi_{(\widetilde{y}, \widetilde{\alpha})}\in \varphi_{p}$.任取$\varphi_{(y, \alpha)}\in \varphi_{p}$,结合(2.1)与(2.2)式可得

$\varphi_{(\widetilde{y}, \widetilde{\alpha})}=\varphi_{((\frac{y}{\varepsilon_{y}})\cdotp p(\frac{y}{\varepsilon_{y}}), p(\frac{y}{\varepsilon_{y}}))}=\frac{p(\frac{y}{\varepsilon_{y}})}{\varepsilon_{y}}\cdot\varphi_{(y, \varepsilon_{y})}\geq\varphi_{(y, \varepsilon_{y})}\geq\varphi_{(y, \alpha)}, $

即存在$\varphi_{(\widetilde{y}, \widetilde{\alpha})}\in \varphi_{p}$,使得$\varphi_{(y, \alpha)}\leq\varphi_{(\widetilde{y}, \widetilde{\alpha})}$成立.接下来证明$\varphi_{(\widetilde{y}, \widetilde{\alpha})}$为$\varphi_{p}$的最大元.

对任意的$\delta>0$,当$\widetilde{\alpha}\geq p(\frac{\widetilde{y}}{\delta})\geq\delta$时,有$\widetilde{\alpha}=\max \{\delta:p(\frac{\widetilde{y}}{\delta})\geq\delta\}<+\infty$.当$p(\frac{\widetilde{y}}{\delta})\geq\delta\geq\widetilde{\alpha}$时,已知$\widetilde{\alpha}=p(\frac{y}{\varepsilon_{y}})$,由(2.2)式有

$\varphi_{((\frac{y}{\varepsilon_{y}})\cdot p(\frac{y}{\varepsilon_{y}}), \delta)}\geq\varphi_{((\frac{y}{\varepsilon_{y}})\cdot p(\frac{y}{\varepsilon_{y}}), \widetilde{\alpha})}\geq\varphi_{((\frac{y}{\varepsilon_{y}})\cdot p(\frac{y}{\varepsilon_{y}}), p(\frac{y}{\varepsilon_{y}}))}, $

再结合(2.1)式可得

$\varphi_{(y, \frac{\varepsilon_{y}\cdot\delta}{p(\frac{y}{\varepsilon_{y}})})}=\varphi_{((\frac{y}{\varepsilon_{y}})\cdot p(\frac{y}{\varepsilon_{y}}), \delta)}\geq\varphi_{((\frac{y}{\varepsilon_{y}})\cdot p(\frac{y}{\varepsilon_{y}}), p(\frac{y}{\varepsilon_{y}}))}=\varphi_{(y, \varepsilon_{y})}. $

已知$p(\frac{\widetilde{y}}{\delta})\geq\delta$, $p(\frac{y}{\varepsilon_{y}})\geq\varepsilon_{y}$及$\widetilde{y}=(\frac{y}{\varepsilon_{y}}\cdot p(\frac{y}{\varepsilon_{y}}))$,可得

$p(\frac{1}{\delta}\cdot\frac{y}{\varepsilon_{y}}\cdot p(\frac{y}{\varepsilon_{y}}))=p(\frac{\widetilde{y}}{\delta})\geq\delta\geq\delta \cdot \frac{\varepsilon_{y}}{p(\frac{y}{\varepsilon_{y}})}, $

即有$\frac{\varepsilon_{y}\cdot\delta}{p(\frac{y}{\varepsilon_{y}})}\in\{\alpha:p(\frac{y}{\alpha})\geq\alpha\}$,因此$\frac{\varepsilon_{y}\cdot\delta}{p(\frac{y}{\varepsilon_{y}})}\leq\varepsilon_{y}$.既然$\delta\leq p(\frac{y}{\varepsilon_{y}})=\widetilde{\alpha}$,则$\widetilde{\alpha}=$max$\{\delta:p(\frac{\widetilde{y}}{\delta})\geq\delta\}$.

综上所述,对任意的$\widetilde{y}\in X$,有$\widetilde{\alpha}=$max$\{\delta:p(\frac{\widetilde{y}}{\delta})\geq\delta\}$成立,且$p(\frac{\widetilde{y}}{\widetilde{\alpha}})=\widetilde{\alpha}$.由命题证明过程可知$\varphi_{(\widetilde{y}, \widetilde{\alpha})}$为$\varphi_{p}$的最大元.

命题4.4  设$p, q:X\rightarrow[0, +\infty]$为严格DICR函数,对任意的$y\in X$,取$\varepsilon_{y}=\max\{\alpha:p(\frac{y}{\alpha})\geq\alpha\}<+\infty$,对任意的$z\in X$,取$\eta_{z}=\max\{\alpha:p(\frac{z}{\alpha})\geq\alpha\}<+\infty$.有以下条件等价:

(ⅰ) $\varphi_{p}\subseteq\varphi_{q}$;

(ⅱ)关于$\varphi_{p}$中的最大元$\varphi_{(y, \varepsilon_{y})}$,总存在$\varphi_{q}$中的最大元$\varphi_{(z, \eta_{z})}$,使得$\varphi_{(y, \varepsilon_{y})}(x)\leq\varphi_{(z, \eta_{z})}(x)$对任意的$x\in X$都成立.

证  (ⅰ)$\Rightarrow$(ⅱ)假设$\varphi_{(y, \varepsilon_{y})}$为$\varphi_{p}$的最大元,因为$\varphi_{p}\subseteq\varphi_{q}$,可得$\varphi_{(y, \varepsilon_{y})}\in\varphi_{q}$.又因为$\varphi_{(y, \varepsilon_{y})}\in\varphi_{q}$,由命题4.3知总存在$\varphi_{q}$的最大元$\varphi_{(z, \eta_{z})}$,使得$\varphi_{(y, \varepsilon_{y})}\leq\varphi_{(z, \eta_{z})}$成立.

(ⅱ)$\Rightarrow$(ⅰ)任取$\varphi_{(y^{*}, \alpha)}\in\varphi_{p}$,由命题4.3知存在$\varphi_{p}$的最大元为$\varphi_{(y, \varepsilon_{y})}$,使得$\varphi_{(y^{*}, \alpha)}\leq\varphi_{(y, \varepsilon_{y})}$成立.假设$\varphi_{(z, \eta_{z})}$为$\varphi_{q}$最大元,结合已知条件$\varphi_{(y, \varepsilon_{y})}\leq\varphi_{(z, \eta_{z})}$,有$\varphi_{(y^{*}, \alpha)}\leq\varphi_{(z, \eta_{z})}$,可得$\varphi_{(y^{*}, \alpha)}\in\varphi_{q}$.即证$\varphi_{p}\subseteq\varphi_{q}$.

定理4.1  设$p, q:X\rightarrow[0, +\infty]$为严格DICR函数.对任意的$x\in X$,有$p(x)\leq q(x)$且$\widetilde{p}(x)=p(x)+f(x_{0})$.对任意的$y, z\in X$,取$\varepsilon_{y}=\max\{\alpha:\widetilde{p}(\frac{y}{\alpha}) \geq\alpha\}<+\infty$, $\eta_{z}=\max \{\alpha:q(\frac{z}{\alpha})\geq\alpha\}<+\infty$.则$x_{0}$为问题(4.1)所定义函数的全局最小值当且仅当对任意满足条件$\widetilde{p}(\frac{y}{\varepsilon_{y}})= \varepsilon_{y}\neq0$的$y\in X$,存在满足条件$q(\frac{z}{\eta_{z}})=\eta_{z}$的$z\in X$,使得

$\varphi_{(y, \varepsilon_{y})}\leq\varphi_{(z, \eta_{z})}.$

证  $x_{0}$为$f(x)$的全局最小值当且仅当$\varphi_{\widetilde{p}}\subseteq\varphi_{q}$,可直接由命题4.4及命题4.3证明.