Optimality and mixed duality in multiobjective E-convex programming

Abstract

In this paper, we consider a class of multiobjective E-convex programming problems with inequality constraints, where the objective and constraint functions are E-convex functions which were firstly introduced by Youness (J. Optim. Theory Appl. 102:439-450, 1999). Fritz-John and Kuhn-Tucker necessary and sufficient optimality theorems for the multiobjective E-convex programming are established under the weakened assumption of the theorems in Megahed et al. (J. Inequal. Appl. 2013:246, 2013) and Youness (Chaos Solitons Fractals 12:1737-1745, 2001). A mixed duality for the primal problem is formulated and weak and strong duality theorems between primal and dual problems are explored. Illustrative examples are given to explain the obtained results.

1 Introduction

The concepts of an E-convex set and an E-convex function were introduced first by Youness [1]. Subsequently, necessary and sufficient optimality criteria for a class of E-convex programming problems were discussed by Youness [2], and E-Fritz-John and E-Kuhn-Tucker problems, which modified the Fritz-John and Kuhn-Tucker problems, were also presented. In Megahed et al. [3], the concept of an E-differentiable convex function which transforms a non-differentiable convex function to a differentiable function under an operator E: $\Bbb{R}^{n} \rightarrow\Bbb{R}^{n}$ was presented, then a solution of mathematical programming with a non-differentiable function could be found by applying the Fritz-John and Kuhn-Tucker conditions due to Mangasarian [4].

However, on the other hand, the results on E-convex programming in Youness [1] were not correct, and some counterexamples were given by Yang [5]. The results concerning the characterization of an E-convex function f in terms of its E-epigraph in Youness [1] were also not correct, and some characterizations of E-convex functions using a different notion of epigraph were given by Duca et al. [6].

Based on the correct results in Youness [1], a class of semi-E-convex functions was introduced by Chen [7], the concepts of E-quasiconvex functions and strictly E-quasiconvex functions were introduced by Syau and Stanley Lee [8], respectively.

In fact, after defining the E-convex function in 1999, Youness [1] pointed out that the E-convex function that he defined had more generalized results than a convex function. He dealt mainly with some properties of an E-convex set and an E-convex function, a programming problem without E in both objective functions or constrained functions, and the relation between solutions of objective and constrained functions with and without E. He then drew the conclusion that the E-convex set and E-convex function were more generalized than the convex set and function proposed by Hanson [9], Hanson and Mond [10], and Kaul and Kaur [11].

This paper also addresses a counterexample of Theorem 4.1 in Youness [1]. Characterization of efficient solutions based on the modification of Theorem 4.2 in Youness [1] is presented. A sufficient optimality theorem is given by using this characterization and E-convexity conditions. We obtain the scalarization method due to Chankong and Haimes [12] for multiobjective E-convex programming. By employing this scalarization method, Fritz-John and Kuhn-Tucker necessary theorems for the multiobjective case are established under the weakened assumption of the theorems in Megahed et al. [3] and Youness [2]. Moreover, a mixed type dual for the primal problem is given. Under the assumption of the E-convex conditions, weak and strong duality theorems between the primal and dual problems are established, and we also propose some examples to illustrate our results.

2 Preliminaries

Let $\Bbb{R}^{n}$ denote the n-dimensional Euclidean space. The following conventions for a vector in $\Bbb{R}^{n}$ will be used in this paper:

$$\begin{aligned} &x< y\quad \mbox{if and only if}\quad x_{i}< y_{i} \quad \mbox{for all } i=1,2, \ldots,p, \\ &x\leqq y \quad\mbox{if and only if}\quad x_{i} \leqq y_{i} \quad\mbox{for all } i=1,2, \ldots, p, \\ &x\leq y \quad\mbox{if and only if}\quad x_{i} \leqq y_{i} \quad\mbox{for all } i=1,2, \ldots,p \mbox{ but } x\neq y. \end{aligned}$$

We present some concepts of E-convex set and E-convex function; for convenience, we recall the definition of E-convex set first.

Definition 2.1

[1]

A set $M\subset{\Bbb{R}^{n}}$ is said to be E-convex iff there is a map $E:{\Bbb{R}^{n}} \to{\Bbb{R}^{n}}$ such that $(1-\lambda)E(x)+ \lambda E(y) \in M$, for each $x,y \in M$, and $\lambda\in[0, 1]$.

It is clear that if $M\subset\Bbb{R}^{n}$ is convex, then M is E-convex by taking a map $E:{\Bbb{R}^{n}} \to{\Bbb{R}^{n}}$ as the identity map, but the converse may not be true; see the following example.

Example 2.1

Consider the set $S_{1}= \{(x,y)\in\Bbb{R}^{2}\mid y\leqq x, 0\leqq x\leqq1 \}$. Let $E(x,y)=(\sqrt{x},y)$, it is clear that $S_{1}$ is E-convex (since $S_{1}$ is convex). It is easy to check that $E(S_{1})$ is E-convex by taking the map $E(x,y)=(\sqrt{x},y)$, while $E(S_{1})$ is not convex, where $E(S_{1})= \{(x,y)\in\Bbb{R}^{2}\mid y\leqq x^{2}, 0\leqq x\leqq1 \}$.

However, if $E: \Bbb{R}^{n} \to\Bbb{R}^{n}$ is a surjective map, it is easy to check that the converse also holds. Note that E is said to be surjective if there exists $x\in M$ such that $E(x)=y$, $\forall y \in E(M)$.

Definition 2.2

[1]

A function $f: {\Bbb{R}^{n}}\to{\Bbb{R}}$ is said to be E-convex on $M \in{\Bbb{R}^{n}}$ iff there is a map $E: {\Bbb{R}^{n}} \to{\Bbb{R}^{n}}$ such that M is an E-convex set and

$$f\bigl(\lambda E(x) + (1-\lambda)E(y)\bigr)\leqq\lambda f\bigl(E(x)\bigr) + (1- \lambda) f\bigl(E(y)\bigr) $$

for each $x,y \in M$ and $0 \leqq\lambda\leqq1$. Moreover, if

$$f\bigl(\lambda E(x) + (1-\lambda)E(y)\bigr)\geqq\lambda f\bigl(E(x)\bigr) + (1- \lambda) f\bigl(E(y)\bigr) $$

then f is called E-concave on M. If the inequality signs in the above two inequalities are strict, then f is called strictly E-convex and strictly E-concave, respectively.

Remark 2.1

Let f, g be E-convex on M. Then $f + g$, αf ($\alpha\geqq 0$) are E-convex on the set M.

It is easy to check that every convex function f on a convex set M is an E-convex function, where E is the identity map. But the converse may not hold, we recall the example from [1].

Example 2.2

Define the function $f: \Bbb{R}\to\Bbb{R}$ as

$$f(x)= \left \{ \textstyle\begin{array}{@{}l@{\quad}l} 1, & \mbox{if } x> 0, \\ -x, & \mbox{if } x\leqq0, \end{array}\displaystyle \right . $$

and let $E:{\Bbb{R}} \to{\Bbb{R}}$ be defined as $E(x)=-x^{2}$. Then $\Bbb{R}$ is an E-convex set and f is E-convex but not convex.

Obviously, if f is a real-valued differentiable function on an E-convex set $M \subset{\Bbb{R}^{n}}$, we can define a differentiable E-convex function in the following.

Definition 2.3

f is E-convex on M if and only if for each $x,y \in M$

$$f\bigl(E(x)\bigr)-f\bigl(E(y)\bigr) \geqq\nabla f\bigl(E(y)\bigr) \bigl(E(x)-E(y)\bigr). $$

3 Optimality criteria

In this section, we suppose that $E:M\to M$ ($M\subset{\Bbb{R}^{n}}$) is a surjective map. In addition, as we know if a set $M\subset\Bbb{R}^{n}$ is E-convex with respect to a mapping $E: \Bbb{R}^{n} \to\Bbb{R}^{n}$, then $E(M)\subset M$ (see [1], Proposition 2.2). For an E-convex function f, we say that the function $(f\circ E): M\to\Bbb{R}$ defined by $(f\circ E)(x)=f(E(x))$ for all $x\in M$ is well defined (see [8]).

Consider the following multiobjective nonlinear program:

$$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l@{}} (\mbox{MP}) &\mbox{Maximize}& f(x)=\bigl(f_{1}(x),f_{2}(x), \ldots, f_{p}(x)\bigr),\\ &\mbox{subject to}& x\in M= \bigl\{ x \in\Bbb{R}^{n}\mid g_{j}(x) \leqq0 , j=1,2, \ldots, m \bigr\} , \end{array}\displaystyle \end{aligned}$$

where $f_{i}:{\Bbb{R}^{n}}\to{\Bbb{R}}$ , $i\in P=\{1,2,\ldots,p\}$ and $g_{j}:{\Bbb{R}^{n}}\to{\Bbb{R}}$ , $j\in Q=\{1,2,\ldots,m\}$ are E-convex functions.

Then we give the following E-convex program related to (MP):

$$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l} (\mathbf{MP}_{\mathbf{E}}) &\mbox{Maximize} &(f\circ E) (x)= \bigl((f_{1}\circ E) (x),f_{2}\circ E\bigr) (x), \ldots, f_{p}\circ E) (x)),\\ &\mbox{subject to}& x\in E(M)= \bigl\{ x\in\Bbb{R}^{n}\mid (g_{j} \circ E) (x) \leqq0, j=1,2, \ldots,m \bigr\} , \end{array}\displaystyle \end{aligned}$$

where $f_{i}\circ E$ , $i\in P$ and $g_{j}\circ E$ , $j\in Q$ are differentiable on M.

It states that, for a surjective map E, if f is E-convex, then $f\circ E$ is obviously convex.

Definition 3.1

A point $\bar{x} \in E(M)$ is said to be an efficient solution of ${(\mathrm{MP}_{\mathrm{E}})}$ if and only if there is no other $x\in E(M)$ such that

$$(f_{i_{0}}\circ E) (x) < (f_{i_{0}}\circ E) (\bar{x}) \quad \mbox{for some } {i_{0}}\in P $$

and

$$(f_{i}\circ E) (x) \leqq(f_{i}\circ E) (\bar{x}) \quad \mbox{for all } i\in P, $$

where $P= \{1,2,\ldots,p \}$, that is

$$(f\circ E) (x) \leq(f\circ E) (\bar{x}). $$

Now we give a counterexample which is easier to understand than the one in [5], to show that Theorem 4.1 (In (MP), the set M is an E-convex set.) in Youness [1] is incorrect.

Example 3.1

In (MP), $g_{j}$, $j\in Q$ are E-convex, but M does not always need to be E-convex set.

Let $g(x)=x\in\Bbb{R}$ and define the map E as $E(x)=|x|$. Then $g(x)$ is E-convex. Take $x=-1$, $y=-1/2$. Then $g(-1)=-1$, $g(-1/2)=-1/2$.

So, $-1,-1/2 \in M= \{x\in\Bbb{R}\mid g(x) \leqq0 \}$. But, for all $\lambda\in[0,1]$,

$$g\bigl(\lambda E(x)+(1-\lambda)E(y)\bigr)=g\bigl(\lambda|x|+(1-\lambda)|y| \bigr)= {\frac{1+\lambda}{2}} >0. $$

Hence, M is not E-convex set.

Also, Theorem 4.2 in Youness [1] is incorrect. The counterexample was given by Yang [5].

Now we would like to present the characterization of efficient solutions modifying Theorem 4.2 in Youness [1] by using only surjective assumption of the mapping E as follows.

Theorem 3.1

Let $E: M\to M$ be a surjective map. Then x̄ is an efficient solution of (MP_E) if and only if $E(\bar{x})$ is an efficient solution of (MP).

Proof

Suppose that $E(\bar{x})$ is not an efficient solution of (MP). Then there exists $\bar{z}\in M$ such that $f(\bar{z}) \leq f(E(\bar{x}))$. Since E is surjective, we have $E(M)=M$, then there exists $\bar{y} \in M$ such that $\bar{z}=E(\bar{y})$, that is, $(f\circ E)(\bar{y}) \leq(f\circ E)(\bar{x})$, which contradicts that x̄ is an efficient solution of ($\mathrm{MP}_{\mathrm{E}}$).

Conversely, suppose that x̄ is not an efficient solution of ${(\mathrm{MP}_{\mathrm{E}})}$, then there exists $y^{*}\in E(M)$ such that $(f\circ E)(y^{*}) \leq(f\circ E)(\bar{x})$. Since E is surjective, there exists $z^{*}\in M$ such that $E(y^{*})=z^{*}$. Hence $f(z^{*}) \leq f(E(\bar{x}))$, which contradicts that $E(\bar{x})$ is an efficient solution of (MP). □

With the help of Theorem 3.1 and the E-convexity assumption, we now give the sufficient optimality condition.

Theorem 3.2

(Sufficient optimality condition)

Assume that $(\bar{x},\bar{\lambda},\bar{\mu})$ satisfies the following conditions:

$$\begin{aligned} &\bar{\lambda}\nabla(f\circ E) (\bar{x})+\bar{\mu}\nabla(g\circ E) (\bar{x})=0, \\ &\bar{\mu}(g\circ E) (\bar{x})=0, \\ &(g\circ E) (\bar{x}) \leqq0, \\ &\bar{\lambda}> 0,\qquad\bar{\mu}\geqq0, \end{aligned}$$

where $\bar{\lambda}\in\Bbb{R}^{p}$, $\bar{\mu}\in\Bbb{R}^{m}$.

Then x̄ is an efficient solution of ($\mathrm{MP}_{\mathrm{E}}$).

Proof

Suppose that x̄ is not an efficient solution of ($\mathrm{MP}_{\mathrm{E}}$), then there exists $x^{*}\in E(M)$ such that

$$\begin{aligned} (f\circ E) \bigl(x^{*}\bigr) \leq(f\circ E) (\bar{x}). \end{aligned}$$

(3.1)

Since $f_{i}$ and $g_{j}$ are E-convex and $f_{i}\circ E$ and $g_{j}\circ E$ are differentiable on M, for any $x\in E(M)$, we have

$$\begin{aligned} &(f_{i}\circ E) (x)-(f_{i}\circ E) (\bar{x}) \geqq(x-\bar{x})\nabla(f_{i}\circ E) (\bar{x}), \end{aligned}$$

(3.2)

$$\begin{aligned} &(g_{j}\circ E) (x)-(g_{j}\circ E) (\bar{x}) \geqq(x-\bar{x})\nabla(g_{j}\circ E) (\bar{x}). \end{aligned}$$

(3.3)

Since $\bar{\lambda}>0$, $\bar{\mu}\geqq0$, from (3.2) and (3.3), for each $i\in P$ and $j\in Q$, we have

$$\begin{aligned} &\bar{\lambda}(f\circ E) (x)-\bar{\lambda}(f\circ E) (\bar{x})+\bar{\mu}(g\circ E) (x)-\bar{\mu}(g\circ E) (\bar{x}) \\ &\quad\geqq(x-\bar{x})\bigl[\bar{\lambda}\nabla(f\circ E) (\bar{x})+\bar{\mu}\nabla (g \circ E) (\bar{x})\bigr]. \end{aligned}$$

Since $\bar{\lambda}\nabla(f\circ E)(\bar{x})+\bar{\mu}\nabla(g\circ E)(\bar{x})=0$, $\bar{\mu}(g\circ E)(\bar{x})=0$ and $(g\circ E)(\bar{x}) \leqq0$, we get

$$(f\circ E) (x) \geqq(f\circ E) (\bar{x}), $$

which contradicts (3.1). □

Remark 3.1

If we replace the E-convexity of $f_{i}$ and $\bar{\lambda}> 0$ by the strictly E-convexity of $f_{i}$ and $\bar{\lambda}\geq0$, respectively, then Theorem 3.2 also holds.

Now we present the following lemma due to Chankong and Haimes [12] to deal with the relationship between the scalar and multiobjective programming problems.

Lemma 3.1

x̄ is an efficient solution for ${(\mathrm{MP}_{\mathrm{E}})}$ if and only if x̄ solves

$$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l@{}} (\mathbf{MP}_{\mathbf{E}})_{\boldsymbol {k}} &\textit{Minimize} & (f_{k}\circ E) (x)\\ &\textit{subject to} &(f_{i}\circ E) (x) \leqq(f_{i} \circ E) (\bar{x}),\quad i\in P^{k}:=P\setminus\{k\},\\ &&(g\circ E) (x) \leqq0, \end{array}\displaystyle \end{aligned}$$

for each $k=1,2,\ldots,p$.

Proof

Suppose that x̄ is not a solution of $(\mathrm{MP}_{\mathrm{E}})_{k}$. Then there exists $x\in E(M)$ such that

$$\begin{aligned} &(f_{k}\circ E) (x)< (f_{k}\circ E) (\bar{x}), \quad k\in P, \end{aligned}$$

(3.4)

$$\begin{aligned} &(f_{i}\circ E) (x)\leqq(f_{i}\circ E) (\bar{x}) , \quad i\neq k. \end{aligned}$$

(3.5)

From (3.4) and (3.5), we conclude that x̄ is not efficient for ${(\mathrm{MP}_{\mathrm{E}})}$.

Conversely, assume that x̄ is a solution of $(\mathrm{MP}_{\mathrm{E}})_{k}$ for every $k \in P$, then for all $x\in E(M)$ with $(f_{i}\circ E)(x)\leqq (f_{i}\circ E)(\bar{x})$, $i\neq k$, we have $(f_{k}\circ E)(\bar{x}) \leqq(f_{k}\circ E)(x)$. Then there exists no other $x\in E(M)$ such that $(f_{i}\circ E)(x) \leqq(f_{i}\circ E )(\bar{x})$, $i\in P$, with strict inequality holding for at least one i. This implies that x̄ is efficient for ($\mathrm{MP}_{\mathrm{E}}$). □

Remark 3.2

Without loss of generality, we assume that $P\cap Q= \emptyset$. Set

$$(G_{t}\circ E) (x)=\left \{ \textstyle\begin{array}{@{}l@{\quad}l} (f_{t}\circ E)(x)-(f_{t}\circ E)({\bar{x}}),& t \in P^{k},\\ (g_{t}\circ E)(x),& t\in Q, \end{array}\displaystyle \right . $$

and $T=P^{k} \cup Q$. Then $(\mathrm{MP}_{\mathrm{E}})_{k}$ is equivalent to the following problem:

$$\min (f_{k}\circ E) (x) \quad\mbox{subject to}\quad (G_{t} \circ E) (x) \leqq0 ,\quad t\in T , \mbox{for each } k\in P. $$

In order to obtain the necessary optimality condition, we employ the following generalized linearization lemma due to Mangasarian [4].

Lemma 3.2

Let x̄ be a local solution of $(\mathrm{MP}_{\mathrm{E}})_{k}$, let $f_{k}\circ E$, for each $k\in P$ and $G_{t}\circ E$, $t\in T$ be differentiable at x̄. Then the system

$$\begin{aligned}& \nabla(f_{k}\circ E) (\bar{x})z< 0, \\& \nabla(G_{W}\circ E) (\bar{x})z< 0, \\& \nabla(G_{V}\circ E) (\bar{x})z\leqq0, \end{aligned}$$

has no solution $z\in\Bbb{R}^{n}$, for each $k\in P$, where we denote

$$\begin{aligned} &I=\bigl\{ t \mid (G_{t}\circ E) (\bar{x})=0\bigr\} ,\qquad J=\bigl\{ t \mid (G_{t}\circ E) (\bar{x})< 0\bigr\} , \qquad I\cup J=T, \\ &V=\bigl\{ t \mid (G_{t}\circ E) (\bar{x})=0 \textit{ and } (G_{t}\circ E) \textit{ is } E\textit{-concave at } \bar{x}\bigr\} , \\ &W=\bigl\{ t\mid (G_{t}\circ E) (\bar{x})=0 \textit{ and } (G_{t}\circ E) \textit{ is not } E\textit{-concave at } \bar{x}\bigr\} , \\ &I=V\cup W. \end{aligned}$$

We establish the following Fritz-John necessary optimality criteria by using Lemma 3.2.

Theorem 3.3

(Fritz-John necessary condition)

Assume that $\bar{x}\in E(M)$ is an efficient solution of ($\mathrm{MP}_{\mathrm{E}}$), then there exist $\bar{\lambda}\in\Bbb{R}^{p}$, $\bar{\mu}\in \Bbb{R}^{m}$ such that

$$\begin{aligned}& \bar{\lambda}\nabla(f\circ E) (\bar{x})+ \bar{\mu}\nabla(g\circ E) (\bar{x})=0, \\& \bar{\mu}(g\circ E) (\bar{x})=0, \\& (g\circ E) (\bar{x}) \leqq0, \\& (\bar{\lambda},\bar{\mu}) \geq0. \end{aligned}$$

Proof

Since x̄ is an efficient solution of ($\mathrm{MP}_{\mathrm{E}}$), then from Lemma 3.1, x̄ solves $(\mathrm{MP}_{\mathrm{E}})_{k}$ for each $k\in P$. By Lemma 3.2 and Remark 3.2, we see that the system

$$\begin{aligned}& \nabla(f_{k}\circ E) (\bar{x})z< 0, \\& \nabla(G_{W}\circ E) (\bar{x})z< 0, \\& \nabla(G_{V}\circ E) (\bar{x})z\leqq0, \end{aligned}$$

has no solution $z\in\Bbb{R}^{n}$. Hence by Motzin’s theorem [4], there exist $\bar{\lambda}_{k}$, $\bar{\mu}_{W}$, $\bar{\mu}_{V}$ such that

$$\begin{aligned}& \bar{\lambda}_{k}\nabla(f_{k}\circ E) (\bar{x})+\bar{\mu}_{W}\nabla(G_{W}\circ E) (\bar{x})+\bar{\mu}_{V} \nabla(G_{V}\circ E) (\bar{x})=0, \\& (\bar{\lambda}_{k}, \bar{\mu}_{W}) \geq0, \\& \bar{\mu}_{V} \geqq0. \end{aligned}$$

Since $(G_{W}\circ E)(\bar{x})=0$ and $(G_{V}\circ E)(\bar{x})=0$, it follows that if we define $\bar{\mu}_{J}=0$ and $\bar{\mu}=(\bar{\mu}_{W}, \bar{\mu}_{V}, \bar{\mu}_{J})$, then

$$\bar{\mu}(G\circ E) (\bar{x})=\bar{\mu}_{W}(G_{W}\circ E) (\bar{x})+\bar{\mu}_{V}(G_{V}\circ E) (\bar{x})+\bar{\mu}_{J}(G_{J}\circ E) (\bar{x})=0, $$

here, we can reduce $\bar{\mu}(g\circ E)(\bar{x})=0$. Thus $\bar{\lambda}_{k}\nabla(f_{k}\circ E)(\bar{x})+\bar{\mu}(g\circ E)(\bar{x})=0$ and $(\bar{\lambda}_{k}, \bar{\mu}) \geq0$. Then, for each $k\in P$, we have

$$\begin{aligned}& \bar{\lambda}\nabla(f\circ E) (\bar{x})+ \bar{\mu}\nabla(g\circ E) (\bar{x})=0, \\& (\bar{\lambda},\bar{\mu}) \geq0. \end{aligned}$$

Since $x^{*}\in E(M)$, $(g\circ E)(x^{*}) \leqq0$.

The proof is complete. □

Theorem 3.4

(Kuhn-Tucker necessary condition)

If $\bar{x}\in E(M)$ is an efficient solution of ($\mathrm{MP}_{\mathrm{E}}$) and $G_{t}\circ E$, $t\in T$ satisfies a constraint qualification [4] for $(\mathrm{MP}_{\mathrm{E}})_{k}$ for at least one $k\in P$. Then there exist $\bar{\lambda}\in\Bbb{R}^{p}$ and $\bar{\mu}\in\Bbb{R}^{m}$ such that

$$\begin{aligned}& \bar{\lambda}\nabla(f\circ E) (\bar{x})+\bar{\mu}\nabla(g\circ E) (\bar{x})=0, \\& \bar{\mu}(g\circ E) (\bar{x})=0, \\& (g\circ E) (\bar{x}) \leqq0, \\& \bar{\lambda}\geq0, \qquad\bar{\mu}\geqq0. \end{aligned}$$

Proof

Since x̄ is an efficient solution of ($\mathrm{MP}_{\mathrm{E}}$), then by Theorem 3.3 there exist $\bar{\lambda}\in\Bbb{R}^{p}$, $\bar{\mu}\in\Bbb{R}^{m}$ such that $(\bar{x}, \bar{\lambda}, \bar{\mu})$ satisfies

$$\begin{aligned}& \bar{\lambda}\nabla(f\circ E) (\bar{x})+\bar{\mu}\nabla(g\circ E) (\bar{x})=0, \\& \bar{\mu}(g\circ E) (\bar{x})=0, \\& (g\circ E) (\bar{x}) \leqq0, \\& (\bar{\lambda},\bar{\mu})\geq0. \end{aligned}$$

We only have to show that $\bar{\lambda}\geq0$, that is, $\bar{\lambda}_{k} >0$ for at least one $k\in P$.

Since $(\bar{\lambda},\bar{\mu})\geq0$, $(\bar{\lambda}, \bar{\mu}_{W}) \geq0$, we have $\bar{\lambda}_{k} > 0$ for at least one $k\in P$ if W is empty. Now, we show that $\bar{\lambda}_{k} > 0$ for at least one $k\in P$ if W is nonempty by contradiction.

Suppose that $\bar{\lambda}_{k} = 0$ for all $k\in P$. Since $\bar{\mu}_{J} =0$ as we define in the proof of Theorem 3.3, we have $\bar{\mu}_{W}\nabla(G_{W}\circ E)(\bar{x}) + \bar{\mu}_{V}\nabla (G_{V}\circ E)(\bar{x})=0$, $\bar{\mu}_{W} \geq0$, $\bar{\mu}_{V} \geqq0$. Since $G_{t}\circ E$ satisfies the Arrow-Hurwicz-Uzawa constraint qualification [4] at x̄ for ${(\mathrm{MP}_{\mathrm{E}})_{k}}$ for at least one $k\in P$, there exists $\bar{z}\in\Bbb{R}^{n}$ such that

$$\begin{aligned} \nabla(G_{W}\circ E) (\bar{x})\bar{z}>0, \end{aligned}$$

(3.6)

$$\begin{aligned} \nabla(G_{V}\circ E) (\bar{x})\bar{z}\geqq0. \end{aligned}$$

(3.7)

Multiplying (3.6) and (3.7) by $\bar{\mu}_{W}$ and $\bar{\mu}_{V}$, respectively, then yields

$$\bar{\mu}_{W} \nabla(G_{W}\circ E) (\bar{x})\bar{z}+\bar{\mu}_{V} \nabla (G_{W}\circ E) (\bar{x})\bar{z}>0, $$

which contradicts the fact that

$$\bar{\mu}_{W} \nabla(G_{W}\circ E) (\bar{x})\bar{z}+\bar{\mu}_{V} \nabla (G_{W}\circ E) (\bar{x})\bar{z}=0. $$

Hence $\bar{\lambda}_{k} > 0$ for at least one $k\in P$. Then we obtain $\bar{\lambda}\geq0$. □

Remark 3.3

If we replace our surjective assumption of E by bijection (or linearity) of E, then our Fritz-John and Kuhn-Tucker necessary optimality results reduce to the ones in Megahed et al. [3] (or Youness [2]).

Example 3.2

Consider the following problem:

$$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l@{}} \widehat{(\mathbf{MP})} &\mbox{Minimize}& \bigl(f_{1}(x),f_{2}(x) \bigr),\\ &\mbox{subject to}& x\in M= \bigl\{ x \in\Bbb{R} \mid g_{1}(x) \leqq0, g_{2}(x)\leqq0 \bigr\} , \end{array}\displaystyle \end{aligned}$$

where $f_{1}(x)=x$, $f_{2}(x)=x^{2}$, $g_{1}(x)=x-1$, and $g_{2}(x)=-x$.

Let $E:M\to E(M)$ defined by $E(x)=x+1$ be the surjective map, then we get the following E-convex programming problem related to $\widehat{(\mathrm{MP})}$:

$$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l@{}} \widehat{(\mathbf{MP}_{\mathbf{E}})} &\mbox{Minimize}& \bigl((f_{1}\circ E) (x), (f_{2}\circ E) (x)\bigr),\\ &\mbox{subject to}& x\in E(M)= \bigl\{ x \in\Bbb{R} \mid (g_{1}\circ E) (x) \leqq0, (g_{2}\circ E) (x)\leqq0 \bigr\} , \end{array}\displaystyle \end{aligned}$$

where $(f_{1}\circ E)(x)=x-1$, $(f_{2}\circ E)(x)=x^{2}-2x+1$, $(g_{1}\circ E)(x)=x-2$, and $(g_{2}\circ E)(x)=-x+1$.

• (a) It is easy to check that the feasible sets of ${\widehat {(\mathrm{MP})}}$ and ${\widehat{(\mathrm{MP}_{\mathrm{E}})}}$ are $M=[0, 1]$ and $E(M)=[1,2]$, respectively.

• (b) By the definition of an efficient solution, we see that $x^{*}=0 \in M$ is the efficient solution of ${\widehat{(\mathrm{MP})}}$ and $\bar{x}=E(x^{*})=1\in E(M)$ is the efficient solution of ${\widehat {(\mathrm{MP}_{\mathrm{E}})}}$, hence Theorem 3.1 holds.

• (c) We can easily check that $(\bar{x}, (\bar{\lambda}_{1}, \bar{\lambda}_{2}), (\bar{\mu}_{1}, \bar{\mu}_{2}))=(1, ({1\over 2}, 1), (0, {1\over 2}))$ satisfy the conditions in Theorem 3.2, and $\bar{x}=1$ is the efficient solution of ${\widehat{(\mathrm{MP}_{\mathrm{E}})}}$, hence Theorem 3.2 holds.

• (d) Since the efficient solution $\bar{x}=1$ for ${\widehat {(\mathrm{MP}_{\mathrm{E}})}}$, also solves both ${\widehat{(\mathrm{MP}_{\mathrm{E}})}_{1}}$ and ${\widehat{(\mathrm{MP}_{\mathrm{E}})}_{2}}$, Lemma 3.1 holds, where

  $$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l@{}} \widehat{(\mathbf{MP}_{\mathbf{E}})}_{\mathbf{1}} &\mbox{Mimimize}& (f_{1}\circ E) (x),\\ &\mbox{subject to}& (f_{2}\circ E) (x)\leqq(f_{2}\circ E) ( \bar{x}),\\ &&x\in E(M), \end{array}\displaystyle \end{aligned}$$

  and

  $$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l@{}} \widehat{(\mathbf{MP}_{\mathbf{E}})}_{\mathbf{2}} &\mbox{Mimimize}& (f_{2}\circ E) (x),\\ &\mbox{subject to}& (f_{1}\circ E) (x)\leqq(f_{1}\circ E) (\bar{x}),\\ &&x\in E(M). \end{array}\displaystyle \end{aligned}$$

• (e) As $\bar{x}=1$ is the efficient solution of $\widehat {(\mathrm{MP}_{\mathrm{E}})}$, then there exist $\bar{\lambda}=(\frac{1}{2}, 1)$ and $\bar{\mu}= (0, \frac{1}{2})$ satisfy the conditions in Theorem 3.3, hence Theorem 3.3 holds.

• (f) $\bar{x}= 1$ is the efficient solution of $\widehat{(\mathrm{MP}_{\mathrm{E}})}$ and it is easy to check the problem $\widehat{(\mathrm{MP}_{\mathrm{E}})}_{1}$ satisfies the Kuhn-Tucker constraint qualification [4], and there exist $\bar{\lambda}=(\frac{1}{2}, 1)$ and $\bar{\mu}= (0, \frac{1}{2})$ satisfying the conditions in Theorem 3.4, hence Theorem 3.4 holds.

4 Duality

Recently, several researchers found some results on mixed dual model under some generalized convexity; see [13–15], for example. In this section, first we establish the following mixed dual problem (MD) to (MP):

$$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l@{}} (\mathbf{MD}) &\mbox{Maximize} &\displaystyle\biggl(f_{1}(u)+\sum _{j\in J_{0}}\mu_{j}^{T} g_{j}(u), \ldots ,f_{p}(u)+\sum_{j\in J_{0}} \mu_{j}^{T} g_{j}(u)\biggr)\\ &\mbox{subject to} &\displaystyle \sum_{i=1}^{p} \lambda_{i}^{T} \nabla f_{i}^{T}(u)+\sum _{j=1}^{q}\mu_{j}^{T} \nabla g_{j}(u)=0,\\ &&\displaystyle\sum_{j\in J_{\alpha}}\mu_{j}^{T} g_{j}(u)\geqq0 , \quad\alpha=1,2,\ldots ,r,\\ &&\displaystyle \lambda=(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{p})\in\Lambda ^{+},\\ &&\displaystyle \mu_{j} \geqq0,\quad j\in Q=\{1,2,\ldots, q\}, \end{array}\displaystyle \end{aligned}$$

where $J_{\alpha}\subset Q=\{1,2,\ldots,q\}$, $\alpha=0,1,\ldots,r$ with $\bigcup_{\alpha=0}^{r} J_{\alpha}=Q$ and $J_{\alpha}\cap J_{\beta}=\emptyset$ if $\alpha\neq\beta$. $\Lambda^{+}=\{\lambda\in\Bbb{R}^{p} \mid\lambda\geqq0, \lambda^{T}e=1, e=(1,\ldots, 1)^{T}\in\Bbb{R}^{p}\}$.

Then we formulate the following mixed dual problem ($\mathrm{MD}_{\mathrm{E}}$) to ($\mathrm{MP}_{\mathrm{E}}$):

$$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l@{}} (\mathbf{MD}_{\mathbf{E}}) &\mbox{Maximize} & \displaystyle\biggl((f_{1} \circ E) (u)+\sum_{j\in J_{0}} \mu_{j}^{T}(g_{j} \circ E) (u),\ldots,\\ &&\displaystyle(f_{p}\circ E) (u)+\sum _{j\in J_{0}} \mu_{j}^{T}(g_{j}\circ E) (u)\biggr)\\ &\mbox{subject to} &\displaystyle \sum_{i=1}^{p} \lambda_{i}^{T} \nabla(f_{i}\circ E) (u)+\sum _{j=1}^{q}\mu_{j}^{T} \nabla(g_{j}\circ E) (u)=0 ,\\ &&\displaystyle\sum_{j\in J_{\alpha}}\mu_{j}^{T} (g_{j}\circ E) (u)\geqq0 ,\quad \alpha =1,2,\ldots,r,\\ &&\displaystyle \lambda=(\lambda_{1}, \lambda_{2}, \ldots, \lambda_{p})\in\Lambda ^{+},\\ &&\displaystyle \mu_{j} \geqq0,\quad j\in Q=\{1,2,\ldots, q\}, \end{array}\displaystyle \end{aligned}$$

where $J_{\alpha}\subset Q=\{1,2,\ldots,q\}$, $\alpha=0,1,\ldots,r$ with $\bigcup_{\alpha=0}^{r} J_{\alpha}=Q$ and $J_{\alpha}\cap J_{\beta}=\emptyset$ if $\alpha\neq\beta$; $\Lambda^{+}=\{\lambda\in\Bbb{R}^{p} \mid\lambda\geqq0, \lambda^{T}e=1, e=(1,\ldots, 1)^{T}\in\Bbb{R}^{p}\}$.

• (1) If $J_{0}=Q$, then our mixed dual type ($\mathrm{MD}_{\mathrm{E}}$) (or (MD)) reduces to the Wolfe dual type.

• (2) If $J_{0}=\emptyset$, then our mixed dual type ($\mathrm{MD}_{\mathrm{E}}$) (or (MD)) reduces to the Mond-Weir dual type.

Theorem 4.1

Let $E:M\to M$ be a surjective map. Then ū is an efficient solution of ($\mathrm{MD}_{\mathrm{E}}$) if and only if $E(\bar{u})$ is an efficient solution of (MD).

Proof

By Lemma 3.1, we can obtain this theorem. □

Assume that f is an E-convex function and $E:M\to M$ ($M\subset\Bbb{R}^{n}$) is a surjective map, by Lemma 3.1, we can study dual problem between (MP) and (MD). Here, we would like to study the dual problem between ${(\mathrm{MP}_{\mathrm{E}})}$ and ${(\mathrm{MD}_{\mathrm{E}})}$.

Theorem 4.2

(Weak duality)

Assume that for all feasible x of ${(\mathrm{MP}_{\mathrm{E}})}$ and all feasible $(u,\lambda,\mu)$ of ${(\mathrm{MD}_{\mathrm{E}})}$, $f_{i}$, $g_{j}$ are E-convex functions. If also either

• (a) $\lambda_{i}>0$ for all $i=1,2,\ldots,p$, or

• (b) $\sum_{i=1}^{p}\lambda_{i}f_{i}(\cdot)+\sum_{j=1}^{q}\mu_{j}g_{j}(\cdot)$ is strictly E-convex at u,

then the following cannot hold:

$$\begin{aligned} &(f_{i_{0}}\circ E) (x) \leqq(f_{i_{0}}\circ E) (u)+ \sum_{j\in J_{0}}\mu _{j}^{T}(g_{j} \circ E) (u) \quad\textit{for all } i_{0}\in P, \end{aligned}$$

(4.1)

$$\begin{aligned} &(f_{i}\circ E) (x) < (f_{i}\circ E) (u)+ \sum_{j\in J_{0}}\mu_{j}^{T}(g_{j} \circ E) (u) \quad\textit{for some } i\in P . \end{aligned}$$

(4.2)

Proof

Suppose to the contrary that (4.1) and (4.2) hold. Since x is feasible for ${(\mathrm{MP}_{\mathrm{E}})}$ and $\mu\geqq0$, from (4.1) and (4.2), we imply

$$\begin{aligned} &(f_{i_{0}}\circ E) (x)+\sum_{j\in J_{0}} \mu_{j}^{T}(g_{j}\circ E) (x)\leqq (f_{i_{0}}\circ E) (u)+\sum_{j\in J_{0}} \mu_{j}^{T}(g_{j}\circ E) (u) \quad\mbox{for all } i_{0}\in P, \end{aligned}$$

(4.3)

$$\begin{aligned} &(f_{i}\circ E) (x)+\sum_{j\in J_{0}} \mu_{j}^{T}(g_{j}\circ E) (x)< (f_{i} \circ E) (u)+\sum_{j\in J_{0}}\mu_{j}^{T}(g_{j} \circ E) (u) \quad \mbox{for some } i\in P. \end{aligned}$$

(4.4)

If hypothesis (a) holds, then with $\sum_{i=1}^{p}\lambda_{i}=1$, one has

$$\begin{aligned} \sum_{i=1}^{p}\lambda_{i}(f_{i} \circ E) (x)+\sum_{j\in J_{0}}\mu _{j}^{T}(g_{j} \circ E) (x)< \sum_{i=1}^{p} \lambda_{i}(f_{i}\circ E) (u)+\sum _{j\in J_{0}}\mu_{j}^{T}(g_{j}\circ E) (u) \end{aligned}$$

(4.5)

and since $f_{i}$, $g_{j}$ are E-convex and $\lambda_{i}>0$, $i=1,2,\ldots ,p$, $\mu\geqq0$, it now follows from (4.5) that

$$\bigl(E(x)-E(u)\bigr)^{T}\Biggl(\sum_{i=1}^{p} \lambda_{i}\nabla(f_{i}\circ E) (u)+\sum _{j\in J_{0}}\mu_{j}^{T}(g_{j}\circ E) (u)\Biggr)< 0, $$

which contradicts the fact that

$$\sum_{i=1}^{p} \lambda_{i} \nabla(f_{i}\circ E) (u)+\sum_{j=1}^{q} \mu _{j}^{T}\nabla(g_{j}\circ E) (u)=0. $$

On the other hand, since $\lambda_{i}\geqq0$, $i=1,2,\ldots,p$ and $\sum_{i=1}^{p}\lambda_{i}=1$, (4.3) and (4.4) imply

$$\begin{aligned} \sum_{i=1}^{p}\lambda_{i}(f_{i} \circ E) (x)+\sum_{j\in J_{0}}\mu _{j}^{T}(g_{j} \circ E) (x) \leqq\sum_{i=1}^{p} \lambda_{i}(f_{i}\circ E) (u)+\sum _{j\in J_{0}}\mu_{j}^{T}(g_{j}\circ E) (u) . \end{aligned}$$

(4.6)

Now (4.6) and hypothesis (b) imply (4.5), which also contradicts the fact that

$$\sum_{i=1}^{p} \lambda_{i} \nabla(f_{i}\circ E) (u)+\sum_{j=1}^{q} \mu _{j}^{T}\nabla(g_{j}\circ E) (u)=0. $$

 □

Corollary 4.1

Assume that weak duality (Theorem  4.2) holds between ($\mathrm{MP}_{\mathrm{E}}$) and ($\mathrm{MD}_{\mathrm{E}}$). If $(\bar{u},\bar{\lambda},\bar{\mu})$ is feasible for ($\mathrm{MD}_{\mathrm{E}}$) with $\bar{\mu}^{T}(g\circ E)(\bar{u})=0$ and if ū is feasible for ($\mathrm{MP}_{\mathrm{E}}$), then ū is efficient for ($\mathrm{MP}_{\mathrm{E}}$) and $(\bar{u},\bar{\lambda},\bar{\mu})$ is efficient for ($\mathrm{MD}_{\mathrm{E}}$).

Proof

Suppose that ū is not efficient for ($\mathrm{MP}_{\mathrm{E}}$). Then there exists a feasible x for ($\mathrm{MP}_{\mathrm{E}}$) such that

$$\begin{aligned} &(f_{i_{0}}\circ E) (x)\leqq(f_{i_{0}}\circ E) (\bar{u}) \quad\mbox{for all } i_{0}\in P, \end{aligned}$$

(4.7)

$$\begin{aligned} &(f_{i}\circ E) (x)< (f_{i}\circ E) (\bar{u}) \quad\mbox{for some } i\in P. \end{aligned}$$

(4.8)

By hypothesis $\bar{\mu}^{T}(g\circ E)(\bar{u})=0$, so (4.7) and (4.8) can be written as

$$\begin{aligned} &(f_{i_{0}}\circ E) (x)\leqq(f_{i_{0}}\circ E) (\bar{u})+\bar{\mu}^{T}(g\circ E) (\bar{u}) \quad\mbox{for all } i_{0}\in P, \\ &(f_{i}\circ E) (x)< (f_{i}\circ E) (\bar{u})+\bar{\mu}^{T}(g\circ E) (\bar{u}) \quad\mbox{for some } i \in P. \end{aligned}$$

Since $(\bar{u},\bar{\lambda}, \bar{\mu})$ is feasible for ($\mathrm{MD}_{\mathrm{E}}$) and x is feasible for ($\mathrm{MP}_{\mathrm{E}}$), these inequalities contradict weak duality (Theorem 4.2).

Also, suppose that $(\bar{u},\bar{\lambda},\bar{\mu})$ is not efficient for ($\mathrm{MD}_{\mathrm{E}}$), then there exists a feasible solution $(u,\lambda ,\mu)$ for ($\mathrm{MD}_{\mathrm{E}}$) such that

$$\begin{aligned} &(f_{j}\circ E) (u)+\mu^{T}(g\circ E) (u) \geqq(f_{j}\circ E) (\bar{u})+\bar{\mu}^{T}(g\circ E) (\bar{u}) \quad\mbox{for all } j\in P, \end{aligned}$$

(4.9)

$$\begin{aligned} &(f_{i}\circ E) (u)+\mu^{T}(g\circ E) (u)>(f_{i}\circ E) (\bar{u})+\bar{\mu}^{T}(g\circ E) (\bar{u}) \quad\mbox{for some } i\in P. \end{aligned}$$

(4.10)

Since $\bar{\mu}^{T}(g\circ E)(\bar{u})=0$, (4.9) and (4.10) reduce to

$$\begin{aligned} &(f_{j}\circ E) (u)+\mu^{T}(g\circ E) (u) \geq(f_{j}\circ E) (\bar{u}) \quad\mbox{for all } j\in P, \\ &(f_{i}\circ E) (u)+\mu^{T}(g\circ E) (u)>(f_{i} \circ E) (\bar{u}) \quad\mbox{for some } i\in P. \end{aligned}$$

Since ū is feasible for ($\mathrm{MP}_{\mathrm{E}}$), these inequalities contradict weak duality (Theorem 4.2). Therefore ū and $(\bar{u},\bar{\lambda},\bar{\mu})$ are efficient for their respective problems. □

Theorem 4.3

(Strong duality)

Let x̄ be an efficient solution for ${(\mathrm{MP}_{\mathrm{E}})}$ and assume that x̄ satisfies a constraint qualification [4] for ${(\mathrm{MP}_{\mathrm{E}})_{k}}$ for at least one $k=1,2,\ldots,p$. Then there exist $\bar{\lambda}\in\Bbb{R}^{p}$ and $\bar{\mu}\in\Bbb{R}^{q}$ such that $(\bar{x},\bar{\lambda},\bar{\mu})$ is feasible for ${(\mathrm{MD}_{\mathrm{E}})}$. Moreover, if weak duality (Theorem  4.2) holds between ${(\mathrm{MP}_{\mathrm{E}})}$ and ${(\mathrm{MD}_{\mathrm{E}})}$, then $(\bar{x},\bar{\lambda},\bar{\mu})$ is efficient for ${(\mathrm{MD}_{\mathrm{E}})}$.

Proof

Since x̄ is efficient for ($\mathrm{MP}_{\mathrm{E}}$), by Lemma 3.1, x̄ solves $(\mathrm{MP}_{\mathrm{E}})_{k}$ for all $k=1,2,\ldots,p$. By hypothesis, there exists a $k\in P= \{ 1,2,\ldots,p \}$ for which x̄ satisfies a constraint qualification of $(\mathrm{MP}_{\mathrm{E}})_{k}$.

From the Kuhn-Tucker necessary conditions [4], there exist $\lambda_{i}\geqq0$ such that, for all $i\neq k$ and $\mu\geqq0$, $\mu\in\Bbb{R}^{m}$,

$$\begin{aligned} &(f_{k}\circ E) (\bar{x})+\sum _{i\neq k}\lambda_{i}\nabla(f_{i}\circ E) ( \bar{x})+\nabla\mu^{T}(g\circ E) (\bar{x})=0, \end{aligned}$$

(4.11)

$$\begin{aligned} &\mu^{T}(g\circ E) (\bar{x})=0. \end{aligned}$$

(4.12)

Now we divide all terms in (4.11) and (4.12) by $1+\sum_{i\neq k} \lambda_{i}$ and set $\bar{\lambda}_{k}={\frac{1}{1+\sum_{i\neq k} \lambda_{i}}} > 0$, $\bar{\lambda}_{j}={\frac{\lambda_{i}}{1+\sum_{i\neq k} \lambda_{i}}} \geqq0$, $\bar{\mu}={\frac{\mu}{1+\sum_{i\neq k} \lambda_{i}}} \geqq0$. Since weak duality (Theorem 4.2) holds, from Corollary 4.1, we conclude that $(\bar{x},\bar{\lambda},\bar{\mu})$ is feasible as well as efficient for ($\mathrm{MD}_{\mathrm{E}}$). □

Example 4.1

Recall the problem in Example 3.2, and we now give the mixed dual problem to $\widehat{(\mathrm{MP}_{\mathrm{E}})}$.

$$\begin{aligned} \textstyle\begin{array}{@{}l@{\quad}l@{\quad}l} \widehat{(\mathbf{MD}_{\mathbf{E}})} &\mbox{Maximize}& \bigl((f_{1}\circ E) (u)+\mu _{1}(g_{1}\circ E) (u), (f_{2}\circ E) (u)+\mu_{1}(g_{1}\circ E) (u) \bigr)\\ &\mbox{subject to}& \lambda^{T}\nabla(f\circ E) (u)+\mu^{T} \nabla (g\circ E) (u)=0,\\ && \mu_{2}(g_{2}\circ E) (u)\geqq0,\\ && \lambda_{1}+\lambda_{2}=1,\qquad \lambda\geqq0,\qquad \mu \geqq0, \end{array}\displaystyle \end{aligned}$$

where $\lambda=(\lambda_{1}, \lambda_{2})$ and $\mu=(\mu_{1}, \mu_{2})$.

As we know the feasible set of $\widehat{(\mathrm{MP}_{\mathrm{E}})}$ is $E(M)=[1, 2]$ and it is easy to check that the feasible set of $\widehat {(\mathrm{MD}_{\mathrm{E}})}$ denoted by G is $G=\{ (u, \lambda, \mu)\in\Bbb{R}\times\Bbb{R}^{2}\times\Bbb{R}^{2}\mid \lambda_{2}(2u-3)+1+\mu_{1}-\mu_{2}=0, \mu_{2}(-u+1)\geqq0, 0\leqq \lambda_{2} \leqq1, \mu_{1}\geqq0, \mu_{2}\geqq0\}$.

Now we check the validity of weak duality, say Theorem 4.2, that is, for any feasible point $x\in E(M)$ and $(u, \lambda, \mu)\in G$ with positive $\lambda_{1}$ and $\lambda_{2}$,

$$ \begin{pmatrix} x-1\\ x^{2}-2x+1 \end{pmatrix} \leq \begin{pmatrix} u-1+\mu_{1}(u-2)\\ u^{2}-2u+1+\mu_{1}(u-2) \end{pmatrix} $$

(4.13)

cannot hold. In fact, by the positivity of $\lambda_{2}$, we have $G=\{ (u, \lambda, \mu)\in\Bbb{R}\times\Bbb{R}^{2}\times\Bbb{R}^{2}\vert\ 1\leqq u\leqq{3\over 2}-{{1+\mu_{1}}\over {2\lambda_{2}}}, 0< \lambda _{2} < 1, \mu_{1}\geqq0\}$, and

$$\min(x-1) =0 > \max\bigl(u-1+\mu_{1}(u-2)\bigr)={ {{1}\over {2}}-{{1}\over {2 \lambda_{2}}}}, $$

which implies (4.13) cannot hold, and we conclude that weak duality (Theorem 4.2) holds.

Finally we turn to strong duality (Theorem 4.3), as we know $\bar{x}=1$ is an efficient solution of $\widehat{(\mathrm{MP}_{\mathrm{E}})}$, and with the satisfy of Kuhn-Tucker constraint qualification [4], it is easy to check that there exist $\bar{\lambda}=(1, 0)$ and $\bar{\mu}=(0, 1)$ such that $(\bar{x}, \bar{\lambda}, \bar{\mu})=(1, (1,0), (0, 1))$ is a feasible solution of $\widehat{(\mathrm{MD}_{\mathrm{E}})}$. Moreover, if weak duality (Theorem 4.2) holds, $(\bar{x}, \bar{\lambda}, \bar{\mu})=(1, (1,0), (0, 1))$ is efficient for $\widehat{(\mathrm{MD}_{\mathrm{E}})}$, hence strong duality (Theorem 4.3) holds.