A note on “Higher-order optimality conditions in set-valued optimization using Studniarski derivatives and applications to duality” [Positivity. 18, 449–473(2014)]

Abstract

The note points out that the sufficiency of proposition 2.1 in Anh (Positivity 18:449–473, 2014) is erroneous and we provide an example to illustrate it. Also the proof of proposition 2.2 in Anh (Positivity 18:449–473, 2014) is incorrect and we give a new proof.

1 Introduction

The concept of generalized convexity plays an important role in operations research and applied mathematics. Yang et al. [1, 2] introduced the concepts of generalized cone subconvexlike set-valued map and nearly cone-subconvexlike set-valued map. Sach [3] introduced a new convexity notion for set-valued maps, called ic-cone-convexlikeness. Xu and Song [4] obtained the following results: (a) when the ordering cone has nonempty interior, ic-cone-convexness is equivalent to near cone-subconvexlikeness; (b) when the ordering cone has empty interior, ic-cone-convexness implies near cone-subconvexlikeness, a counter example is given to show that the converse implication is not true.

Anh [5] gave an equivalent characterization of generalized cone subconvexlikeness in Proposition 2.1 and applied it to obtain Proposition 2.2, under the assumption of generalized cone subconvexlikeness, some higher-order optimality conditions were established.

In this paper, we will point out that the sufficiency of [5, Proposition 2.1] is invalid and the proof of [5, Proposition 2.2] is incorrect.

2 Preliminaries

Throughout the paper, suppose X, Y are two normed spaces, $0_{X}$ and $0_{Y}$ denote the original points of X and Y, respectively. $C\subset Y$ is a convex cone with nonempty interior $\mathrm{int}C$ such that $0_{Y}\in C$.

Definition 2.1

(See [1, 5]) Suppose S is a nonempty set in X and $F:S\rightarrow 2^{Y}$ is a set-valued map. F is said to be generalized $C-$subconvexlike on S if $\exists v\in \mathrm{int}C$, $\forall x_{1}, x_{2}\in S$, $\forall \lambda \in (0,1)$, $\forall \epsilon >0$, $\exists x_{3}\in S$ and $\exists r>0$ such that

$$\begin{aligned} \epsilon v+\lambda F(x_{1})+(1-\lambda )F(x_{2})\subseteq rF(x_{3})+C. \end{aligned}$$

3 Main result

By virtue of generalized $C-$subconvexlike of set-valued maps, Anh obtained an equivalent characterization for generalized $C-$subconvexlikeness, as is shown in the following proposition.

Proposition 3.1

(See [5, Proposition 2.1]) The map $F:S\rightarrow 2^Y$ is generalized $C-$subconvexlike on S if and only if $\mathrm{cone}_{+}(F(S))+\mathrm{int}C$ is convex, where $\mathrm{cone_+}(F(S)):=\{ry:r>0, y\in F(S)\}$.

Remark 3.1

The sufficiency of [5, Proposition 2.1] is incorrect, to illustrate the point, we need the following Lemma.

Lemma 3.1

The following statements are equivalent for the set-valued map F:

• (i) $\forall \hat{u}\in \mathrm{int}C$, $\forall x_{1},x_{2}\in S$, $\forall \alpha \in [0,1]$, $\exists x_{3}\in S$ and $\exists \rho >0$, such that

  $$\begin{aligned} \hat{u}+\alpha F(x_{1})+(1-\alpha )F(x_{2})\subseteq \rho F(x_{3})+C; \end{aligned}$$

• (ii) F is generalized $C-$subconvexlike on S;

• (iii) $\forall x_{1},x_{2}\in S$, $\forall \alpha \in [0,1]$, $\exists u=u(x_{1},x_{2},\alpha )\in \mathrm{int}C$, and $\forall \varepsilon >0$, $\exists x_{3}=x_{3}(u,\varepsilon )\in S$ and $\exists \rho =\rho (u,\varepsilon )>0$ such that

  $$\begin{aligned} \varepsilon u+\alpha F(x_{1})+(1-\alpha )F(x_{2})\subseteq \rho F(x_{3})+C. \end{aligned}$$

Proof

By Theorem 2.1 of [1], (i) implies (ii) and (ii) implies (iii).

In what follows, we show that (iii) implies (i). Let

$$\begin{aligned} \hat{u}\in \mathrm{int}C, x_{1},x_{2}\in S, \alpha \in [0,1]. \end{aligned}$$

Then, from (iii), $\exists u=u(x_{1},x_{2},\alpha )\in \mathrm{int}C$, and $\forall \varepsilon >0$, $\exists \bar{x}_{3}=\bar{x}_{3}(u,\varepsilon )\in S$ and $\exists \bar{\rho }=\bar{\rho }(u,\varepsilon )>0$ such that

$$\begin{aligned} \varepsilon u+\alpha F(x_{1})+(1-\alpha )F(x_{2})\subseteq \bar{\rho } F(\bar{x}_{3})+C. \end{aligned}$$

Since $\hat{u}\in \mathrm{int}C$, one can find $\varepsilon _{0}>0$ and $u_{0}\in \mathrm{int}C$ such that

$$\begin{aligned} \hat{u}-\varepsilon _{0}u=u_{0}. \end{aligned}$$

From (iii), $\exists x_{3}=x_{3}(u,\varepsilon _{0})\in S$ and $\rho =\rho (u,\varepsilon _{0})>0$ such that

$$\begin{aligned} \varepsilon _{0}u+\alpha F(x_{1})+(1-\alpha )F(x_{2})\subseteq \rho F(x_{3})+C. \end{aligned}$$

Hence

$$\begin{aligned} \hat{u}+\alpha F(x_{1})+(1-\alpha )F(x_{2})= & {} \left[ \varepsilon _{0}u+\alpha F(x_{1})+(1-\alpha )F(x_{2})\right] +u_{0}\\\subseteq & {} \rho F(x_{3})+C+u_{0}\\\subseteq & {} \rho F(x_{3})+\mathrm{int}C\\\subseteq & {} \rho F(x_{3})+C. \end{aligned}$$

Thus we complete the proof. $\square$

Example 3.1

Let us set

$$\begin{aligned}&S\!=\!\left\{ (x_{1},x_{2})\in R^{2}:x_{1}+x_{2}=1\right\} , \quad C=R^{2}_{+}\!=\!\left\{ (x_{1},x_{2})\in R^{2}:x_{1}\!\ge \!0,x_{2}\ge 0\right\} ,\\&F(x_{1},x_{2})=\left\{ (x_{1},x_{2}),(1/2,1/2)\right\} ,\forall (x_{1},x_{2})\in S. \end{aligned}$$

A direct calculation gives

$$\begin{aligned} \mathrm{cone}_{+}(F(S))+\mathrm{int}C=\left\{ (y_{1},y_{2})\in R^{2}:y_{1}+y_{2}>0\right\} . \end{aligned}$$

Then $\mathrm{cone}_{+}(F(S))+\mathrm{int}C$ is convex.

However, F is not generalized $C-$subconvexlike on S, as is illustrated in the following.

Let $z_{1}=(-1,2), z_{2}=(2,-1), \lambda _{0}=1/2,\hat{u}=(1/8,1/8)$. Then

$$\begin{aligned} \hat{u}+\lambda _{0}F(z_{1})+(1-\lambda _{0})F(z_{2})= \{(-1/8,11/8),(11/8,-1/8),(5/8,5/8)\}. \end{aligned}$$

In what follows, we show that

$$\begin{aligned} \hat{u}+\lambda _{0}F(z_{1})+(1-\lambda _{0})F(z_{2})\not \subseteq \rho F(z)+C, \forall z\in S,\forall \rho >0. \end{aligned}$$

In fact, $\forall z=(x_{1},x_{2})\in \left\{ (x_{1},x_{2}):x_{1}+x_{2}=1\right\}$,$\forall \rho >0$,

• (1) If $x_{1}\ge 0$, $x_{2}<0$, then $(-1/8,11/8)\notin \rho F(z)+C$;

• (2) If $x_{1}\ge 0$, $x_{2}\ge 0$, then $(-1/8,11/8)\notin \rho F(z)+C$;

• (3) If $x_{1}<0$, $x_{2}>0$, then $(11/8,-1/8)\notin \rho F(z)+C$.

From above discussions, we deduce that $\exists \hat{u}\in \mathrm{int}C$, $\exists z_{1}, z_{2}\in S$, $\exists \lambda _{0}=1/2$ such that $\forall z\in S,\forall \rho >0$,

$$\begin{aligned} \hat{u}+\lambda _{0}F(z_{1})+(1-\lambda _{0})F(z_{2})\not \subseteq \rho F(z)+C. \end{aligned}$$

From Lemma 3.1, it follows that F is not generalized $C-$subconvexlike on S.

Remark 3.2

Since the sufficient condition of Proposition 2.1 in Ref. [5]. was applied to the proof of Proposition 2.2 in Ref. [5]., the proof is erroneous. However, Proposition 2.2 in Ref. [5]. is true. In the following, we give the proposition and new proof.

Proposition 3.2

(See [5, Proposition 2.2]) Suppose that the map $F:S\rightarrow 2^Y$ is generalized $C-$subconvexlike on S. Then F is also generalized $K-$subconvexlike on S, where K is a convex cone satisfying $C\subseteq K$.

Proof

Since F is generalized $C-$subconvexlike on S, there exists $v\in \mathrm{int}C$, for any $x_{1}, x_{2}\in S$, $\forall \lambda \in (0,1)$, $\forall \epsilon >0$, $\exists x_{3}\in S$ and $\exists r>0$ such that

$$\begin{aligned} \epsilon v+\lambda F(x_{1})+(1-\lambda )F(x_{2})\subseteq rF(x_{3})+C. \end{aligned}$$

From $C\subseteq K$, it follows that

$$\begin{aligned} \epsilon v+\lambda F(x_{1})+(1-\lambda )F(x_{2})\subseteq rF(x_{3})+C\subseteq rF(x_{3})+K. \end{aligned}$$

Then F is generalized $K-$subconvexlike on S. $\square$