Piecewise Linear Vector Optimization Problems on Locally Convex Hausdorff Topological Vector Spaces

TL;DR

This paper studies piecewise linear vector optimization problems in locally convex Hausdorff topological vector spaces, showing their efficient solutions form unions of finitely many semi-closed generalized polyhedral convex sets, extending previous results from normed spaces.

Contribution

It extends the characterization of solution sets of piecewise linear vector optimization problems to locally convex Hausdorff topological vector spaces, generalizing prior normed space results.

Findings

Efficient solution sets are unions of finitely many semi-closed generalized polyhedral convex sets.

Convex problems have solution sets that are unions of finitely many generalized polyhedral convex sets.

Solution sets are connected by line segments in convex cases.

Abstract

Piecewise linear vector optimization problems in a locally convex Hausdorff topological vector spaces setting are considered in this paper. The efficient solution set of these problems are shown to be the unions of finitely many semi-closed generalized polyhedral convex sets. If, in addition, the problem is convex, then the efficient solution set and the weakly efficient solution set are the unions of finitely many generalized polyhedral convex sets and they are connected by line segments. Our results develop the preceding ones of Zheng and Yang [Sci. China Ser. A. 51, 1243--1256 (2008)], and Yang and Yen [J. Optim. Theory Appl. 147, 113--124 (2010)], which were established in a normed spaces setting.

Equations

$${\rm(VP)}\qquad\quad{\rm Min}_{K}\big{\{}f(x)\mid x\in D\big{\}}.$$

$$(1-\lambda )f(x_1)+\lambda f(x_2)-f((1-\lambda )x_1+\lambda x_2)\in K$$

$$D=\big{\{}x\in X\mid x\in L,\ \langle x^{*}_{i},x\rangle\leq\alpha_{i},\ i=1,\dots,p\big{\}}.$$

$$D=\big{\{}x\in X\mid Ax=z,\ \langle x^{*}_{i},x\rangle\leq\alpha_{i},\ i=1,\dots,p\big{\}}.$$

$$\displaystyle D\!=\!\Bigg{\{}\sum\limits_{i=1}^{k}\lambda_{i}u_{i}+\sum\limits_{j=1}^{\ell}\mu_{j}v_{j}\mid$$

$$\displaystyle\sum\limits_{i=1}^{k}\lambda_{i}=1,\,\mu_{j}\geq 0,\,\forall j=1,\dots,\ell\Bigg{\}}+X_{0}.$$

$$D\!=\!\big{\{}x\in L\mid\langle x^{*}_{i},x\rangle\leq\alpha_{i},\,i=1,\dots,p;\,\langle x^{*}_{i},x\rangle<\alpha_{i},\,i=p+1,\dots,q\big{\}}.$$

$$K=\big{\{}y\in Y\mid\langle y^{*}_{j},y\rangle\leq 0,\ j=1,\dots,q\big{\}},$$

$$\ell(K)=Y_{0}:=\big{\{}y\in Y\mid\langle y^{*}_{j},y\rangle=0,\ j=1,\dots,q\big{\}}$$

$$K=Y_0+K_1.$$

$${\rm int}\,K=\big{\{}y\in Y\mid\langle y^{*}_{j},y\rangle<0,\ j=1,\dots,q\big{\}}.$$

$$K\setminus{\ell(K)}=\Big{\{}y\in K\mid\text{there exists $j\in\{1,\dots,q\}$ such that }\langle y^{*}_{j},y\rangle<0\Big{\}}.$$

$$\mathrm{epi}f=\{(x,y)\in X\times Y\mid y\in f(x)+K\}.$$

$$M_k=T_k(D\cap P_k)+b_k.$$

$$\displaystyle D\cap P_{k}\!=\!\Big{\{}\sum\limits_{i=1}^{r_{k}}\lambda_{i}u_{k,i}+\sum\limits_{j=1}^{s_{k}}\mu_{j}v_{k,j}\mid$$

$$\displaystyle\,\sum\limits_{i=1}^{r_{k}}\lambda_{i}=1,\,\mu_{j}\geq 0,\,j=1,\dots,s_{k}\Big{\}}\!+\!X_{0,k}.$$

$$\displaystyle M^{\prime}_{k}\!=\!\Big{\{}\sum\limits_{i=1}^{r_{k}}\lambda_{i}(Tu_{k,i}+b_{k})+\sum\limits_{j=1}^{s_{k}}\mu_{j}Tv_{k,j}\mid$$

$$\displaystyle\sum\limits_{i=1}^{r_{k}}\lambda_{i}=1,\,\mu_{j}\geq 0,\,j=1,\dots,s_{k}\Big{\}}.$$

$$M_k+K=M_k^{\prime }+T_k(X_{0,k})+Y_0+K_1.$$

$$(\mathrm{VLP}){\mathrm{Min}}_K\left\{Gy\mid y\in M+K\right\}$$

$$E(M|K)=M\cap E(M+K|K),E^w(M|K)=M\cap E^w(M+K|K).$$

$$\mathrm{Sol}(\mathrm{VP})\cap P_k$$

$$\displaystyle=\big{\{}u\in D\cap P_{k}\mid f(u)\in E(M+K|K)\big{\}}$$

$$\displaystyle=\bigcup\limits\limits_{j=1}^{d}\big{\{}u\in D\cap P_{k}\mid f(u)\in Q_{j}\big{\}}$$

$$\displaystyle=\bigcup\limits\limits_{j=1}^{d}\big{\{}u\in D\cap P_{k}\mid T_{k}u+b_{k}\in Q_{j}\big{\}}$$

$$\displaystyle=\bigcup\limits\limits_{j=1}^{d}\Big{(}D\cap P_{k}\cap T_{k}^{-1}(-b_{k}+Q_{j})\Big{)}.$$

$${\rm Sol(VP)}=\bigcup\limits_{k=1}^{m}\bigcup\limits\limits_{j=1}^{d}\Big{(}D\cap P_{k}\cap T_{k}^{-1}(-b_{k}+Q_{j})\Big{)}.$$

$$[y_i,y_{i+1}]\subset E(M+K|K)(i=1,\dots ,r-1).$$

$$(1-\lambda )f(u_i)+\lambda f(u_{i+1})=f((1-\lambda )u_i+\lambda u_{i+1})+w.$$

$$y_{\lambda }=$$

$$=$$

$$y_{\lambda }$$

$$=[f(u_{\lambda })+w]+(1-\lambda )w_i+\lambda w_{i+1}$$

$$=f(\bar{x})+\bar{w}+(w+(1-\lambda )w_i+\lambda w_{i+1}).$$

$$⟨x_i^{\ast },x⟩=\underset{-1}{\overset{1}{\int }}{t}^{i}x(t)dt(i=1,2),$$

Full text

1 Introduction

The intersection of a finite number of closed half-spaces of a finite-dimensional Euclidean space is called a polyhedral convex set (a convex polyhedron in brief). Following Bonnans and Shapiro (Bonnans_Shapiro_2000, , Definition 2.195), we call a subset of a locally convex Hausdorff topological vector space (lcHtvs) a generalized polyhedral convex set (or a generalized convex polyhedron) if it is the intersection of finitely many closed half-spaces and a closed affine subspace of that topological vector space. If the affine subspace can be chosen as the whole space, the generalized polyhedral convex set (gpcs) is said to be a polyhedral convex set (or a convex polyhedron).

From now on, if not otherwise stated, $X$ and $Y$ are locally convex Hausdorff topological vector spaces. Similarly as in ZhengYang_2008 , we say that a mapping $f:X\to Y$ is a piecewise linear function (or a piecewise affine function) if there exist polyhedral convex sets $P_{1},\dots,P_{m}$ in $X$, continuous linear mappings $T_{1},\dots,T_{m}$ from $X$ to $Y$, and vectors $b_{1},\dots,b_{m}$ in $Y$ such that $X=\bigcup\limits_{k=1}^{m}P_{k}$ and $f(x)=T_{k}x+b_{k}$ for all $x\in P_{k}$, $k=1,\dots,m$.

We call a subset $K\subset Y$ a cone if $tK\subset K$ for all $t>0$.

Given a piecewise linear function $f:X\rightarrow Y$, a generalized polyhedral convex set $D\subset X$, and a polyhedral convex cone $K\subset Y$ with $K\neq Y$, we consider the piecewise linear vector optimization problem

[TABLE]

In the terminology of (Giannessi_1984, , p. 341), one says that $f$ is a $K$-function111Some authors use the term “$K$-convex function” instead of $K$-function. on $D$ if

[TABLE]

for any $x_{1},x_{2}$ in $D$ and $\lambda\in[0,1]$. It is clear that if $f$ is linear, then it is a $K$-function on $D$. In the case where $Y=\mathbb{R}$ and $K=\mathbb{R}_{+}$, $f$ is a $K$-function on $D$ if and only if $f$ is convex on $D$. When $f$ is a $K$-function on $D$, we say that (VP) is a convex problem.

A point $y$ from a subset $Q\subset Y$ is called an efficient point (resp., a weakly efficient point) of $Q$ if there is no $y^{\prime}\in Q$ such that $y-y^{\prime}\in K\setminus{\ell(K)}$ (resp., $y-y^{\prime}\in{\rm int}\,K$), where $\ell(K):=K\cap(-K)$. The efficient point set and the weakly efficient point set of $Q$ are denoted respectively by $E(Q|K)$ and $E^{w}(Q|K)$. Clearly, in the case where $K$ is a pointed cone, i.e., $\ell(K)=\left\{0\right\}$, a point $y\in Q$ belongs to $E(Q|K)$ if and only if $y-y^{\prime}\notin K\setminus\{0\}$ for every $y^{\prime}\in Q$. Since ${\rm int}\,K\subset K\setminus{\ell(K)}$ by (Luan_2016, , Proposition 2.2), one gets $E(Q|K)\subset E^{w}(Q|K)$.

A vector $u\in D$ is called an efficient solution (resp., a weakly efficient solution) of (VP) if $f(u)\in E(f(D)|K)$ (resp., $f(u)\in E^{w}(f(D)|K)$). The efficient solution set and the weakly efficient solution set of (VP) are denoted respectively by ${\rm Sol(VP)}$ and ${\rm Sol}^{w}{\rm(VP)}$. Since $E(Q|K)\subset E^{w}(Q|K)$, one has ${\rm Sol(VP)}\subset{\rm Sol}^{w}{\rm(VP)}$.

The study of the structures and characteristic properties of these solution sets is useful in the design of efficient algorithms for solving (VP).

Zheng and Yang ZhengYang_2008 have proved that if $X,\,Y$ are normed spaces and $D$ is a polyhedral convex set, then ${\rm Sol}^{w}{\rm(VP)}$ is the union of finitely many polyhedral convex sets. If $f$ is a $K$-function, then ${\rm Sol}^{w}{\rm(VP)}$ is connected by line segments.

In order to describe the structure of ${\rm Sol(VP)}$ and obtain sufficient conditions for its connectedness, Yang and Yen Yang_Yen_2010 have applied the image space approach Giannessi_1984 ; Giannessi_1987 to optimization problems and variational systems and proposed the notion of semi-closed polyhedral convex set. On account of (Yang_Yen_2010, , Theorem 2.1), if $X,\,Y$ are normed spaces, $Y$ is of finite dimension, $K\subset Y$ is a pointed cone, and $D\subset X$ is a polyhedral convex set, then ${\rm Sol(VP)}$ is the union of finitely many semi-closed polyhedra. In this setting, if $f$ is a $K$-function on $D$, then ${\rm Sol(VP)}$ is the union of finitely many polyhedra and it is connected by line segments; see (Yang_Yen_2010, , Theorem 2.2). Observe that the main tool for proving the latter results is the representation formula for convex polyhedra in $\mathbb{R}^{n}$ via a finite number of points and a finite number of directions (see, e.g., (Klee_1959, , Theorem 2.12) and (Rockafellar_1970, , Theorem 19.1)). This celebrated result is attributed (Rockafellar_1970, , p. 427) primarily to Minkowski Minkowski_1910 and Weyl Weyl_1935 ; Weyl_1953 .

Theorem 2.3 of Yang_Yen_2010 is an infinite-dimensional version of the classic Arrow, Barankin and Blackwell theorem (the ABB theorem; see ABB_1953 ; Luc_1989 ; Luc_2016 ), which says that the efficient solution set of a finite-dimensional linear vector optimization problem is the union of finitely many polyhedral convex sets and it is connected by line segments.

In Fang_Meng_Yang_2012 , Fang, Meng, and Yang studied multiobjective optimization problems with either continuous or discontinuous piecewise linear objective functions and polyhedral convex constraint sets. They obtained an algebraic representation of a semi-closed polyhedron and apply it to show that the image of a semi-closed polyhedron under a continuous linear function is always a semi-closed polyhedron. They proposed an algorithm for finding the Pareto point set of a continuous piecewise linear bi-criteria program and generalized it to the discontinuous case. The authors applied that algorithm to solve discontinuous bi-criteria portfolio selection problems with an $\ell_{\infty}$ risk measure and transaction costs. Some examples with the historical data of the Hong Kong Stock Exchange are discussed. Other results in this direction were given in Fang_Huang_Yang_2012 and Fang_Meng_Yang_2015 . Later, Zheng and Ng Zheng_Ng_2014 have studied the metric subregularity of piecewise polyhedral multifunctions and applied this property to piecewise linear multiobjective optimization.

Very recently, in a locally convex Hausdorff topological vector spaces setting, by the use of a representation formula for generalized polyhedral convex sets, Luan and Yen Luan_Yen_2015 have obtained solution existence theorems for generalized linear programming problems, a scalarization formula for the weakly efficient solution set of a generalized linear vector optimization problem, and proved that the latter is the union of finitely many generalized polyhedral convex sets. In Luan_2016 , where the relative interior of the dual cone of a polyhedral convex cone is described, a similar result is given for the corresponding efficient solution set. Moreover, it is shown that both efficient solution set and weakly efficient solution set of a generalized linear vector optimization problem are connected by line segments. Thus, the ABB theorem in linear vector optimization has been extended to the lcHtvs setting.

Various generalized polyhedral convex constructions in locally convex Hausdorff topological vector spaces can be found in the new paper of Luan, Yao, and Yen Luan_Yao_Yen_2016 .

It is well known that any infinite-dimensional normed space equipped with the weak topology is not metrizable, but it is a locally convex Hausdorff topological vector space. Similarly, the dual space of any infinite-dimensional normed space equipped with the weak∗-topology is not metrizable, but it is a locally convex Hausdorff topological vector space. Actually, the just mentioned two models provide us with the most typical examples of locally convex Hausdorff topological vector space, whose topologies cannot be given by norms.

The aim of the present paper is to revisit the results of Zheng and Yang ZhengYang_2008 , Yang and Yen Yang_Yen_2010 in a broader setting. Namely, instead of normed spaces, we will consider locally convex Hausdorff topological vector spaces. Apart from using some ideas and schemes of Yang_Yen_2010 , our investigation is based on the results of Luan_2016 ; Luan_Yao_Yen_2016 ; Luan_Yen_2015 .

The paper organization is as follows. The next section gives some preliminaries. In Section 3, we prove that if the problem (VP) is convex, then the solution sets are the unions of finitely many generalized polyhedral convex sets and they are connected by line segments. In Section 4, without imposing the convexity assumption on (VP), we show that ${\rm Sol}{\rm(VP)}$ is the union of finitely many semi-closed generalized polyhedral convex sets, while ${\rm Sol}^{w}{\rm(VP)}$ is the union of finitely many generalized polyhedral convex sets. Two illustrative examples are given in Sections 3 and 4.

2 Preliminaries

We now recall some concepts and results on generalized polyhedral convex sets and polyhedral convex cones. Let $X$ be a locally convex Hausdorff topological vector space. Denote by $X^{*}$ the dual space of $X$ and by $\langle x^{*},x\rangle$ the value of $x^{*}\in X^{*}$ at $x\in X$.

Definition 1

(See (Bonnans_Shapiro_2000, , p. 133)) A subset $D\subset X$ is said to be a generalized polyhedral convex set, or a generalized convex polyhedron, if there exist $x^{*}_{i}\in X^{*}$, $\alpha_{i}\in\mathbb{R}$, $i=~{}1,2,\dots,p$, and a closed affine subspace $L\subset X$, such that

[TABLE]

If $D$ can be represented in the form (2.1) with $L=X$, then we say that it is a polyhedral convex set (pcs), or a convex polyhedron.

If $L$ is a closed affine subspace of $X$ then, by (Bonnans_Shapiro_2000, , Remark 2.196), there exists a continuous surjective linear mapping $A$ from $X$ to a lcHtvs $Z$ and a vector $z\in Z$ such that L=\big{\{}x\in X\mid Ax=z\big{\}}. So, one can rewrite (2.1) as follows

[TABLE]

Next lemmas, which were obtained in Luan_2016 ; Luan_Yao_Yen_2016 ; Luan_Yen_2015 , will be useful for our subsequent investigations.

Lemma 1

(See (Luan_Yen_2015, , Theorem 2.7) and (Luan_Yao_Yen_2016, , Lemma 2.12))* Suppose that $D$ is a nonempty subset of $X$. The set $D$ is generalized polyhedral convex if and only if there exist $u_{1},\dots,u_{k}\in X$, $v_{1},\dots,v_{\ell}\in X$, and a closed linear subspace $X_{0}\subset X$ such that*

[TABLE]

In particular, $D$ is polyhedral convex if and only if $D$ admits a representation of the form (2.3), where $X_{0}$ is a closed linear subspace of finite codimension.

Lemma 2

(See (Luan_Yao_Yen_2016, , Propositions 2.10–2.11))* Suppose that $D_{1},D_{2}$ are generalized polyhedral convex sets of $X$. If ${\rm aff}D_{1}$ is finite-dimensional or $D_{1}$ is polyhedral convex, then $D_{1}+D_{2}$ is a generalized polyhedral convex set.*

Lemma 3

(See (Luan_Yao_Yen_2016, , Lemma 2.13))* If $X_{1}$ and $X_{2}$ are linear subspaces of $X$ with $X_{1}$ being closed and finite-codimensional, then $X_{1}+X_{2}$ is closed and finite-codimensional.*

Lemma 4

(See (Luan_Yao_Yen_2016, , Corollary 2.16))* If a convex subset $D\subset X$ is the union of a finite number of generalized polyhedral convex sets (resp., of polyhedral convex sets) in $X$, then $D$ is generalized polyhedral convex (resp., polyhedral convex).*

Lemma 5

(See (Luan_Yao_Yen_2016, , Proposition 2.7))* Let $T:X\rightarrow Y$ be a continuous linear mapping between locally convex Hausdorff topological vector spaces. If $Q\subset Y$ is a nonempty generalized polyhedral convex set, then $T^{-1}(Q)$ is a generalized polyhedral convex set.*

Lemma 6

(See (Luan_Yen_2015, , Lemma 2.5))* If $W$ and $Z$ are Hausdorff finite-dimensional topological vector spaces of dimension $n$ and if $g:W\rightarrow Z$ is a linear bijective mapping, then $g$ is a homeomorphism.*

According to Yang and Yen Yang_Yen_2010 , a subset of a normed space is called a semi-closed polyhedron if it is the intersection of a finite family of (closed or open) half-spaces. The following definition appears naturally in that spirit.

Definition 2

A subset $D\subset X$ is said to be a semi-closed generalized polyhedral convex set, or a semi-closed generalized convex polyhedron, if there exist $x^{*}_{i}\in~{}X^{*}$, $\alpha_{i}\in\mathbb{R}$, $i=~{}1,2,\dots,q$, with a positive integer $p\leq q$, and a closed affine subspace $L\subset X$, such that

[TABLE]

If $D$ can be represented in the form (2.4) with $L=X$, then we say that it is a semi-closed polyhedral convex set, or a semi-closed convex polyhedron.

Remark 1

From Definitions 1 and 2 it follows that every polyhedral convex set is a semi-closed polyhedral convex set. It is not difficult to show that, if $D_{1}$ and $D_{2}$ are semi-closed polyhedral convex sets, then $D_{1}\setminus D_{2}$ is the union of finitely many semi-closed polyhedral convex sets.

Suppose that $Y$ is a locally convex Hausdorff topological vector space and $K\subset Y$ is a polyhedral convex cone defined by

[TABLE]

where $y^{*}_{j}\in Y^{*}\setminus\{0\}$ for $j=1,\dots,q$. Note that

[TABLE]

is a closed linear subspace of $Y$ and the codimension of $Y_{0}$, denoted by ${\rm codim}Y_{0}$, is finite. Moreover, there exists a finite-dimensional linear subspace $Y_{1}$ of $Y$, such that $Y=Y_{0}+Y_{1}$ and $Y_{0}\cap Y_{1}=\{0\}$. According to (Rudin_1991, , Theorem 1.21(b)), $Y_{1}$ is closed. Clearly, K_{1}:=\big{\{}y\in Y_{1}\mid\langle y^{*}_{j},y\rangle\leq 0,\;j=1,\dots,q\big{\}} is a pointed polyhedral convex cone in $Y_{1}$ and

[TABLE]

The assertions of two following lemmas have been proved in Luan_2016 .

Lemma 7

(See (Luan_2016, , Lemma 2.1))* It holds that $K\setminus\ell(K)=Y_{0}+K_{1}\setminus\{0\}$.*

Lemma 8

(See (Luan_2016, , Proposition 2.2))* Let $K\subset Y$ be a polyhedral convex cone of the form (2.5). The following are valid:*

•

(a)* The interior of $K$ has the representation*

[TABLE]

•

(b)* The set $K\setminus{\ell(K)}$ is a convex cone and*

[TABLE]

The invariance of $K\setminus{\ell(K)}$ and ${\rm int}\,K$ w.r.t. a translation by a vector from $K$ is described by the forthcoming lemma, which can be proved similarly as (Luc_1989, , Proposition 4.3, p. 19) (see also (ChenHuangYang_2005, , Lemma 1.2 (iii))).

Lemma 9

We have

•

(a)* $K\setminus{\ell(K)}+K\subset K\setminus{\ell(K)}$;*

•

(b)* ${\rm int}\,K+K\subset{\rm int}\,K$.*

3 Structure of the Solution Sets in the Convex Case

One says that a subset $A\subset X$ is connected by line segments if for any points $u,u^{\prime}$ in $A$, there are some points $u_{1},\dots,u_{r}$ in $A$ with $u_{1}=u$ and $u_{r}=u^{\prime}$ such that $[u_{i},u_{i+1}]\subset A$ for $i=1,2,\dots,r-1$.

The next result is an extension of (Yang_Yen_2010, , Theorem 2.2) and (ZhengYang_2008, , Theorems 3.2 and 3.3) to a locally convex Hausdorff topological vector spaces setting.

Theorem 3.1

If $f$ is a $K$-function on $D$, the efficient solution set and the weakly efficient solution set of (VP) are the unions of finitely many generalized polyhedral convex sets and they are connected by line segments.

The proof of this result is based on the next lemma.

Lemma 10

(See (Luan_2016, , Theorems 3.2–3.4) and (Luan_Yen_2015, , Theorem 4.5))* If the function $f$ is linear, then ${\rm Sol(VP)}$ (resp., ${\rm Sol}^{w}{\rm(VP)}$) is the union of finitely many generalized polyhedral convex sets and it is connected by line segments.*

Proof of Theorem 3.1  Without loss of generality, we may assume that $D$ is nonempty.

Claim 1. The sum $f(D)+K$ is a polyhedral convex set.

Following (Luc_1989, , p. 18), we define the epigraph of $f$ by the formula

[TABLE]

Since $f$ is a $K$-function, ${\rm epi}\,f$ is convex by (Luc_1989, , Proposition 6.2, p. 29). As $f(D)+K$ is the projection of ${\rm epi}\,f\cap(D\times Y)$ on $Y$, it is convex.

For $k=1,\dots,m$, set $M_{k}:=f(D\cap P_{k})$ and note that

[TABLE]

We will show that $M_{k}+K$, $k=1,\dots,m$, are pcs. Clearly, $D\cap P_{k}$ is a gpcs in $X$. If $D\cap P_{k}$ is empty, then $M_{k}=\emptyset$; hence $M_{k}+K=\emptyset$ is a special polyhedral convex set. In the case where $D\cap P_{k}$ is nonempty, by the representation for gpcs (see Lemma 1) one can find $u_{k,1},\dots,u_{k,r_{k}}$, $v_{k,1},\dots,v_{k,s_{k}}$ in $X$ and a closed linear subspace $X_{0,k}\subset X$ such that

[TABLE]

It is not difficult to show that $M_{k}=M^{\prime}_{k}+T_{k}(X_{0,k})$, where

[TABLE]

Combining this with the equality (2.6), one has

[TABLE]

Since $T_{k}(X_{0,k})$ is a linear subspace of $Y$ and $Y_{0}$ is a closed finite-codimensional linear subspace of $Y$, Lemma 3 shows that $Y_{0}+T_{k}(X_{0,k})$ is a closed subspace of $Y$ and {\rm codim}\big{(}Y_{0}+T_{k}(X_{0,k})\big{)} is finite. Hence, by Lemma 1, M^{\prime}_{k}+\big{(}Y_{0}+T_{k}(X_{0,k})\big{)} is a pcs in $Y$. As $K_{1}$ is a gpcs in $Y$, Lemma 2 yields $K_{1}+M^{\prime}_{k}+Y_{0}+T_{k}(X_{0,k})$ is a polyhedral convex set. So, $M_{k}+K$ is a pcs in $Y$.

In accordance with Lemma 4, since the convex set $f(D)+K$ is the union of polyhedral convex sets $M_{1}+K,\dots,M_{m}+K$, we may conclude that $f(D)+K$ is a polyhedral convex set.

Claim 2. The sets $E(M+K|K)$ and $E^{w}(M+K|K)$, where $M:=f(D)$, are the unions of finitely many generalized polyhedral convex sets and they are connected by line segments.

It suffices to apply Lemma 10 to the problem

[TABLE]

with $G(y):=y$ for all $y\in Y$.

Claim 3. A vector $u\in D$ belongs to Sol(VP) (resp., belongs to ${\rm Sol}^{w}{\rm(VP)}$) if and only if $f(u)\in E(M+K|K)$ (resp., $f(u)\in E^{w}(M+K|K)$).

Arguing similarly as in the proof of (Luc_Ratiu_2014, , Proposition 7.10), we can show that

[TABLE]

For any $u\in D$, one has $f(u)\in M$. By definition, $u$ belongs to ${\rm Sol(VP)}$ (resp., to ${\rm Sol}^{w}{\rm(VP)}$) if and only if $f(u)\in E(M|K)$ (resp., $f(u)\in E^{w}(M|K)$). Hence, the assertions follow from (3.1).

Claim 4. The efficient solution set and the weakly efficient solution set of (VP) are the unions of finitely many generalized polyhedral convex sets.

According to Claim 2, one can find generalized polyhedral convex sets $Q_{1},\dots,Q_{d}$ in $Y$ such that $E(M+K|K)=\bigcup\limits\limits_{j=1}^{d}Q_{j}$. For each $k\in\{1,\dots,m\}$, by using Claim 3, one has

[TABLE]

Since {\rm Sol(VP)}=\bigcup\limits_{k=1}^{m}\big{(}{\rm Sol(VP)}\cap P_{k}\big{)}, we obtain

[TABLE]

For any $k\in\{1,\dots,m\}$ and $j\in\{1,\dots,d\}$, the set $-b_{k}+Q_{j}$ is generalized polyhedral convex by Lemma 2. Hence, since $T_{k}$ is a continuous linear mapping, by Lemma 5 we can assert that $T_{k}^{-1}(-b_{k}+Q_{j})$ is a gpcs. This implies that $D\cap P_{k}\cap T_{k}^{-1}(-b_{k}+Q_{j})$ is a gpcs. Then, formula (3.2) justifies the fact that ${\rm Sol(VP)}$ is the union of finitely many generalized polyhedral convex sets. The assertion concerning ${\rm Sol}^{w}{\rm(VP)}$ can be proved similarly.

Claim 5. The efficient solution set and the weakly efficient solution set of (VP) are connected by line segments.

By analogy, it suffices to prove the assertion about the efficient solution set. Take any $u,u^{\prime}$ in ${\rm Sol(VP)}$. By Claim 3, $f(u)$ and $f(u^{\prime})$ are contained in $E(M+K|K)$. Since $E(M+K|K)$ is connected by line segments (see Claim 2), there exist $y_{1},\dots,y_{r}$ in $Y$ such that $y_{1}=f(u)$, $y_{r}=f(u^{\prime})$ and

[TABLE]

Let $u_{i}\in D$ and $w_{i}\in K$ be such that $y_{i}=f(u_{i})+w_{i}$ for $i=2,\dots,r-1$. Setting $u_{1}=u$, $w_{1}=0$, $u_{r}=u^{\prime}$, and $w_{r}=0$, we have $y_{i}=f(u_{i})+w_{i}$ for $i=1,\dots,r$. Let $i\in\{1,\dots,r-1\}$ be chosen arbitrarily. To show that $[u_{i},u_{i+1}]\subset{\rm Sol(VP)}$, we suppose the contrary: There is a $\lambda\in[0,1]$ such that the vector $u_{\lambda}:=(1-\lambda)u_{i}+\lambda u_{i+1}$ does not belong to ${\rm Sol(VP)}$. Then one can find $\bar{x}\in D$ and $\bar{w}\in K\setminus{\ell(K)}$ satisfying $f(u_{\lambda})=f(\bar{x})+\bar{w}$. As $f$ is a $K$-function on $D$, there is $w\in K$ with

[TABLE]

Set $y_{\lambda}:=(1-\lambda)y_{i}+\lambda y_{i+1}$. On one hand, the inclusion $[y_{i},y_{i+1}]\subset E(M+K|K)$ gives $y_{\lambda}\in E(M+K|K)$. On the other hand,

[TABLE]

Combining (3.3), (3.4) with the equality $f(u_{\lambda})=f(\bar{x})+\bar{w}$, we obtain

[TABLE]

By the convexity of the cone $K$, one has $w+(1-\lambda)w_{i}+\lambda w_{i+1}\in K$. Therefore, as $\bar{w}\in K\setminus{\ell(K)}$, Lemma 9 shows that $\bar{w}+(w+(1-\lambda)w_{i}+\lambda w_{i+1})\in K\setminus{\ell(K)}$. It follows that $y_{\lambda}-f(\bar{x})\in K\setminus{\ell(K)}$. Hence, $y_{\lambda}\notin E(M+K|K)$, a contradiction. We have proved that the line segments $[u_{i},u_{i+1}],i=1,\dots,r-1,$ which connect $u$ with $u^{\prime}$, lie wholly in Sol(VP).

The proof is complete.$\hfill\Box$

Remark 2

In the proof of Theorem 3.1, we have used some ideas of Yang and Yen Yang_Yen_2010 . For the case where $X,Y$ are normed spaces, $Y$ is finite-dimensional, $K\subset Y$ is a pointed cone, and $D$ is a polyhedral convex set, the assertions of Theorem 3.1 about ${\rm Sol(VP)}$ recover (Yang_Yen_2010, , Theorem 2.2). For the case where $X,Y$ are normed spaces and $D$ is a pcs, the assertions of Theorem 3.1 about ${\rm Sol}^{w}{\rm(VP)}$ have been established by Zheng and Yang (ZhengYang_2008, , Theorems 3.2 and 3.3). For the special case where $f$ is linear, Theorem 3.1 expresses several recent results in (Luan_2016, , Theorems 3.2–3.4) and (Luan_Yen_2015, , Theorem 4.5).

The following example is designed as an illustration for Theorem 3.1.

Example 1

Let $X=C[-1,1]$ be the Banach space of continuous real-valued functions defined on $[-1,1]$ with the norm given by $||x||=\max\limits_{t\in[-1,1]}|x(t)|.$ The Riesz representation theorem (see, e.g., (Kolmogorov_Fomin_1975, , Theorem 6, p. 374) and (Luenberger_1969, , Theorem 1, p. 113)) asserts that the dual space of $X$ is $X^{*}=NBV[-1,1]$, the normalized space of functions of bounded variation on $[-1,1]$, i.e., functions $y:[-1,1]\rightarrow\mathbb{R}$ of bounded variation, $y(-1)=0$, and $y(\cdot)$ is continuous from the right at every point of $(-1,1)$. We define two elements $x^{*}_{1},\,x^{*}_{2}\in X^{*}$ by setting

[TABLE]

where the integrals are Riemannian. They equal respectively to the Riemann-Stieltjes integrals (see (Kolmogorov_Fomin_1975, , p. 367)) $\int\limits_{-1}^{1}x(t)dy_{i}(t)$, $i=1,2$, which are given by the $C^{1}$-smooth functions $y_{i}(t)=\int\limits_{-1}^{t}\tau^{i}d\tau$ for $i=1,2$. Now, we consider $X$ with the weak topology. Then $X$ is a locally convex Hausdorff topological vector space whose topology is much weaker than the norm topology. It is clear that $X_{0}:={\rm ker}\,x^{*}_{1}\cap{\rm ker}\,x^{*}_{2}$ is a closed linear subspace of $X$. Let

[TABLE]

and

[TABLE]

We have $e_{1},e_{2}\in X$, $\langle x^{*}_{1},e_{1}\rangle=\langle x^{*}_{2},e_{2}\rangle=1,$ and $\langle x^{*}_{1},e_{2}\rangle=\langle x^{*}_{2},e_{1}\rangle=0$. For any $x\in X$, put $t_{i}=\langle x^{*}_{i},x\rangle$ for $i=1,2$, and observe that the vector $x_{0}:=x-t_{1}e_{1}-t_{2}e_{2}$ belongs to $X_{0}$. Conversely, if $x=x_{0}+t_{1}e_{1}+t_{2}e_{2}$, with $x_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$, then

[TABLE]

Therefore, for any $x\in X$, there exists a unique element $(x_{0},t_{1},t_{2})\in X_{0}\times\mathbb{R}\times\mathbb{R}$ such that $x=x_{0}+t_{1}e_{1}+t_{2}e_{2}$.

Let $Y=X$, $y^{*}_{i}=x^{*}_{i}$ for $i=1,2,$ and $K:=\left\{y\in Y\mid\langle y^{*}_{i},y\rangle\leq 0,\,i=1,2\right\}.$ Note that $K$ is a polyhedral convex cone. Clearly,

[TABLE]

By Lemma 8, ${\rm int}\,K=\left\{y\in Y\mid\langle y^{*}_{i},y\rangle<0,\,i=1,2\right\}$, and

[TABLE]

An easy computation shows that

[TABLE]

and

[TABLE]

Given any $e_{0}\in X_{0}$ and put $L=\left\{x\in X\mid x(t)=e_{0}(t),\,t\in[-1,0]\right\}$. Clearly, $L$ is a closed affine subspace of $X$. Therefore, the set

[TABLE]

is generalized polyhedral convex. Observe that $e_{0}+e_{2}\in D$ and $D$ is not a polyhedral convex set in $X$.

Let $X_{2}$ be the linear subspace of $X$ generated by $e_{2}$. It is clear that ${\rm dim}\,X_{2}=1$, $X={\rm ker}\,x^{*}_{2}+X_{2}$, and ${\rm ker}\,x^{*}_{2}\cap X_{2}=\{0\}$. Let

[TABLE]

be the canonical projection from $X$ on the quotient space $X/{\rm ker}\,x^{*}_{2}$. According to (Rudin_1991, , Theorem 1.41(a)), the linear mapping $\pi$ is continuous. Since the operator

[TABLE]

is a linear bijective mapping, $\Phi$ is a homeomorphism by Lemma 6. So,

[TABLE]

is a linear continuous mapping. Define the map $\varphi:X_{2}\rightarrow Y$ by $\varphi(te_{2})=te_{1}$ for all $t\in\mathbb{R}$. In accordance with Theorem 3.4 of (Schaefer_1971, , p. 22), since ${\rm dim}\,X_{2}=1$, the linear mapping $\varphi$ is continuous. Set $T=\varphi\circ\Phi\circ\pi$, and observe that $T:X\rightarrow X$ is a linear continuous mapping. It is easy to check that if $x=x_{0}+t_{1}e_{1}+t_{2}e_{2}$ with $x_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$, then $Tx=t_{2}e_{1}$. Moreover, ${\rm ker}\,T={\rm ker}\,x^{*}_{2}$.

Obviously, the space $X$ is the union of the pcs $P_{1}:=\{x\in X\mid\langle x^{*}_{2},x\rangle\geq 0\}$ and $P_{2}:=\{x\in X\mid\langle x^{*}_{2},x\rangle\leq 0\}.$ We see at once that if $x=x_{0}+t_{1}e_{1}+t_{2}e_{2}$, where $x_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$, then $x$ belongs to $P_{1}$ (resp., belongs to $P_{2}$) if and only if $t_{2}\geq 0$ (resp., $t_{2}\leq 0$). Let $f:X\rightarrow Y$ be given by

[TABLE]

Since $x-Tx=X+Tx=x$ for any $x\in P_{1}\cap P_{2}={\rm ker}\,T$, the values of $f$ is well defined. Moreover, $f$ is a piecewise linear vector-valued function.

Claim 1. The mapping $f$ is a $K$-function on $X$.

Indeed, take any vectors $x_{1},\,x_{2}\in X$, a number $\lambda\in[0,1]$, and set

[TABLE]

If there exists an index $i\in\{1,2\}$ such that $x_{1},x_{2}\in P_{i}$, then $y=0$ because $f$ is linear on $P_{i}$; so $y\in K$. If $x_{1}\in P_{1}$ and $x_{2}\in P_{2}$, then $x_{i}=x_{i,0}+t_{i,1}e_{1}+t_{i,2}e_{2}$, $i=1,2$, with $x_{i,0}\in X_{0}$, $t_{i,1}\in\mathbb{R}$, $i=1,2$, and $t_{1,2}\geq 0\geq t_{2,2}$. Therefore, $Tx_{i}=t_{i,2}e_{1}$ for $i=1,2$. In the case where $(1-\lambda)x_{1}+\lambda x_{2}\in P_{1}$, we obtain

[TABLE]

Since $2\lambda t_{2,2}\leq 0$, one has $y\in K$ by (3.5). If $(1-\lambda)x_{1}+\lambda x_{2}\in P_{2}$, then

[TABLE]

From (3.5), since $-2(1-\lambda)t_{1,2}\leq 0$, one gets $y\in K$. With $x_{1}\in P_{2}$ and $x_{2}\in P_{1}$, a similar conclusion is obtained. It follows that $f$ is a $K$-function on $X$.

Claim 2. It holds that

[TABLE]

First, take any $u\in{\rm Sol(VP)}$. Let $u_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$ be such that $u=u_{0}+t_{1}e_{1}+t_{2}e_{2}$. Since $u\in D$, $t_{1}\leq 0$ and $t_{2}\leq 1$. Moreover, $u_{0}(t)=e_{0}(t)$ for all $t\in[-1,0]$; so, $u_{0}\in D$. If $t_{2}<0$, then $u\in P_{2}$. Therefore,

[TABLE]

Since $f(u)-f(u_{0})=(t_{1}+t_{2})e_{1}+t_{2}e_{2}$ with $t_{1}+t_{2}<0$ and $t_{2}<0$, by (3.7), $f(u)-f(u_{0})\in K\setminus\ell(K)$. The fact contradicts the assumption $u\in{\rm Sol(VP)}$. This clearly forces $0\leq t_{2}\leq 1$; hence $u\in P_{1}$. If $t_{1}<0$, then we choose $x=u_{0}+t_{2}e_{2}$. Since $x\in D\cap P_{1}$,

[TABLE]

Since $t_{1}<0$, by (3.7), $f(u)-f(x)\in K\setminus\ell(K)$. This inclusion contradicts the assumption $u\in{\rm Sol(VP)}$. We thus get $t_{1}=0$ and $0\leq t_{2}\leq 1$. Consequently, $u\in S$, where $S$ is the set on the right-hand side of (3.8). We have proved that ${\rm Sol(VP)}\subset S$. To obtain the opposite inclusion, take any $u\in S$. Let us show that $f(u)-f(x)\notin K\setminus\ell(K)$ for all $x\in D$. Suppose that $u=u_{0}+t_{1}e_{1}+t_{2}e_{2}$ with $u_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$. Of course, $t_{1}=0$ and $0\leq t_{2}\leq 1$. Since $u\in P_{1}$,

[TABLE]

For any $x\in D$, there exist a vector $x_{0}\in X_{0}$, numbers $\tau_{1}\leq 0$ and $\tau_{2}\leq 1$ satisfying $x=x_{0}+\tau_{1}e_{1}+\tau_{2}e_{2}$.

If $\tau_{2}<0$, then $x\in P_{2}$; so $f(x)=x+Tx=x_{0}+(\tau_{1}+\tau_{2})e_{1}+\tau_{2}e_{2}$. According to (3.7), since $f(u)-f(x)=(u_{0}-x_{0})+(-t_{2}-\tau_{1}-\tau_{2})e_{1}+(t_{2}-\tau_{2})e_{2}$ with $t_{2}-\tau_{2}>0$, we can assert that $f(u)-f(x)\notin K\setminus\ell(K)$.

If $\tau_{2}\geq 0$ then $x\in P_{1}$. Since $f(x)=x-Tx=x_{0}+(\tau_{1}-\tau_{2})e_{1}+\tau_{2}e_{2}$,

[TABLE]

As $(-t_{2}-\tau_{1}+\tau_{2})+(t_{2}-\tau_{2})=-\tau_{1}\geq 0,$ one gets $f(u)-f(x)\notin K\setminus\ell(K)$ by (3.7). From what that has already been said, we obtain $u\in{\rm Sol}{\rm(VP)}$. We have proved that ${\rm Sol(VP)}=S$.

Claim 3. It holds that

[TABLE]

First, take any $u\in{\rm Sol}^{w}{\rm(VP)}$. Suppose that $u=u_{0}+t_{1}e_{1}+t_{2}e_{2}$ with $u_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$. Since $u\in D$, $t_{1}\leq 0$ and $t_{2}\leq 1$. Moreover, $u_{0}(t)=e_{0}(t)$ for all $t\in[-1,0]$. Therefore, $u_{0}\in D$. If $t_{2}<0$ then $u\in P_{2}$; so

[TABLE]

Since $f(u)-f(u_{0})=(t_{1}+t_{2})e_{1}+t_{2}e_{2}$ with $t_{1}+t_{2}<0$ and $t_{2}<0$, by (3.6), $f(u)-f(u_{0})\in{\rm int}\,K$. This contradicts the assumption $u\in{\rm Sol}^{w}{\rm(VP)}$. We thus get $0\leq t_{2}\leq 1$. Therefore, $u\in P_{1}$ and

[TABLE]

If $t_{1}<0$ and $t_{2}<1$, one can find a positive number $\varepsilon$ such that $t_{1}+\varepsilon<0$ and $t_{2}+\varepsilon<1$. Take $x=u_{0}+(t_{2}+\varepsilon)e_{2}$, and observe that $x\in D\cap P_{1}$. Since $f(x)=x-Tx=u_{0}-(t_{2}+\varepsilon)e_{1}+(t_{2}+\varepsilon)e_{2}$,

[TABLE]

As $t_{1}+\varepsilon<0$ and $-\varepsilon<0$, formula (3.6) shows that $f(u)-f(x)\in{\rm int}\,K$, which is impossible because $u\in{\rm Sol}^{w}{\rm(VP)}$. We thus get $t_{1}=0$ or $t_{2}=1$. Consequently, $u$ belongs to $S^{w}$, where $S^{w}$ is the set on the right-hand side of (3.9). We have proved that ${\rm Sol}^{w}{\rm(VP)}\subset S^{w}$. To obtain the opposite inclusion, take any $u\in S^{w}$. Let $u_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$ be such that $u=u_{0}+t_{1}e_{1}+t_{2}e_{2}$.

Case 1: $t_{1}=0$ and $0\leq t_{2}\leq 1$. Clearly, $\langle x^{*}_{1},u\rangle=0$ and $0\leq\langle x^{*}_{2},u\rangle\leq 1$. According to (3.8), $u\in{\rm Sol}{\rm(VP)}$; hence $u\in{\rm Sol}^{w}{\rm(VP)}$.

Case 2: $t_{1}<0$ and $t_{2}=1$. Since $u\in P_{1}$, one has

[TABLE]

For any $x\in D$, there are $x_{0}\in X_{0}$ and $\tau_{1},\tau_{2}\in\mathbb{R}$ satisfying $x=x_{0}+\tau_{1}e_{1}+\tau_{2}e_{2}$. Clearly, $\tau_{1}\leq 0$ and $\tau_{2}\leq 1$. If $\tau_{2}<0$, then $x\in P_{2}$ and

[TABLE]

According to (3.6), since $f(u)-f(x)=(u_{0}-x_{0})+(t_{1}-1-\tau_{1}-\tau_{2})e_{1}+(1-\tau_{2})e_{2}$ with $1-\tau_{2}>0$, $f(u)-f(x)\notin{\rm int}\,K$. If $0\leq\tau_{2}\leq 1$, then $x\in P_{1}$ and

[TABLE]

Since $f(u)-f(x)=(u_{0}-x_{0})+(t_{1}-1-\tau_{1}+\tau_{2})e_{1}+(1-\tau_{2})e_{2}$ with $1-\tau_{2}\geq 0$, formula (3.6) gives $f(u)-f(x)\notin{\rm int}\,K$. It follows that $f(u)-f(x)\notin{\rm int}\,K$ for all $x\in D$. Hence, $u\in{\rm Sol}^{w}{\rm(VP)}$. We have proved that ${\rm Sol}^{w}{\rm(VP)}=S^{w}$.

Observe that ${\rm Sol}^{w}{\rm(VP)}\neq{\rm Sol}{\rm(VP)}$. Indeed, let $e_{3}$ in $X$ be given by

[TABLE]

Since $e_{0}+e_{3}\in L$, $\langle x^{*}_{1},e_{0}+e_{3}\rangle=-\frac{5}{4}$, and $\langle x^{*}_{2},e_{0}+e_{3}\rangle=1$, we can see that $e_{0}+e_{3}$ belongs to ${\rm Sol}^{w}{\rm(VP)}$ but $e_{0}+e_{3}\notin{\rm Sol}{\rm(VP)}$.

4 Structure of the Solution Sets in the Nonconvex Case

In Theorem 3.1, the assumption $f$ is a $K$-function on $D$ cannot be dropped (see (Yang_Yen_2010, , Example 2.1) for an example about the efficient solution set, (ZhengYang_2008, , p. 1252) for an example about the weakly efficient solution set). For the case where $X,\,Y$ are normed spaces, $Y$ is of finite dimension, $K\subset Y$ is a pointed cone, and $D\subset X$ is a polyhedral convex set, the efficient solution set of (VP) is shown to be the union of finitely many semi-closed polyhedral convex sets (see (Yang_Yen_2010, , Theorem 2.1)). This result can be extended to the lcHtvs setting we are studying as follows.

Theorem 4.1

The efficient solution set of (VP) is the union of finitely many semi-closed generalized polyhedral convex sets.

Proof   We may assume, without loss of generality, that $D$ is a nonempty set.

Claim 1. There exists a closed linear subspace $X_{1}$ of $X$ such that $D$ is a polyhedral convex set in $X_{1}$.

Suppose that $D$ is given by (2.2). Set $X_{0}:={\rm ker}\,A$ and observe that $X_{0}$ is a closed linear subspace of $X$.

If $z=0$, then $D=\{x\in X_{0}\mid\langle x_{i}^{*},x\rangle\leq\alpha_{i},\,i=1,\dots,p\}$. For $i=1,\dots,p$, let $x^{*}_{i,0}$ be the restriction to $X_{0}$ of the functional $x^{*}_{i}$. Clearly, $x^{*}_{i,0}\in X^{*}_{0}$ for $i=1,\dots,p$. Since

[TABLE]

we can assert that $D$ is a polyhedral convex set in $X_{0}$.

If $z\neq 0$, then fix any vector $\bar{x}\in D$. Clearly, $\bar{x}$ does not belong to $X_{0}$. Set $X_{1}:=X_{0}+\{t\bar{x}\mid t\in\mathbb{R}\}.$ Since $\{t\bar{x}\mid t\in\mathbb{R}\}$ is a linear subspace of one-dimension, the linear subspace $X_{1}$ is closed by (Rudin_1991, , Theorem 1.42). According to (Rudin_1991, , Theorem 3.5), there exists $x_{1,0}^{*}\in X_{1}^{*}$ such that $\langle x_{1,0}^{*},\bar{x}\rangle=1$ and $\langle x_{1,0}^{*},x\rangle=0$ for all $x\in X_{0}$. It is not difficult to show that $X_{0}={\rm ker}\,x_{1,0}^{*}$. For $i=1,\dots,p$, let $x^{*}_{1,i}$ be the restriction to $X_{1}$ of the functional $x^{*}_{i}$. Of course, $x^{*}_{1,i}\in X^{*}_{1}$ for $i=1,\dots,p$. Set

[TABLE]

and observe that $D^{\prime}$ is a polyhedral convex set in $X_{1}$. Let us show that $D=D^{\prime}$. To obtain the inclusion $D\subset D^{\prime}$, take any $x\in D$. As $x_{0}:=x-\bar{x}$ belongs to $X_{0}$, one has $x\in X_{1}$. Therefore,

[TABLE]

For each $i=1,\dots,p$, since $\langle x^{*}_{1,i},x\rangle=\langle x^{*}_{i},x\rangle$, one gets $\langle x_{1,i}^{*},x\rangle\leq\alpha_{i}$. It follows that $x\in D^{\prime}$. We have proved that $D\subset D^{\prime}$. To obtain the opposite inclusion, take any $x\in D^{\prime}$. Let $x_{0}\in X_{0}$ and $t\in\mathbb{R}$ be such that $x=x_{0}+t\bar{x}$. Then

[TABLE]

Since $x\in D^{\prime}$, $\langle x_{1,0}^{*},x\rangle=1$; so $t=1$. Therefore,

[TABLE]

For each $i=1,\dots,p$, the inequality $\langle x_{1,i}^{*},x\rangle\leq\alpha_{i}$ implies $\langle x^{*}_{i},x\rangle\leq\alpha_{i}$. The inclusion $D^{\prime}\subset D$ has been proved. Thus $D=D^{\prime}$; hence $D$ is a polyhedral convex set in $X_{1}$.

By Claim 1, we may assume that $D$ is a pcs in $X_{1}$, where $X_{1}$ is a closed linear subspace of $X$. For each $k=1,\dots,m,$ set $P_{1,k}:=P_{k}\cap X_{1}$ and observe that $P_{1,k}$ is a pcs in $X_{1}$. Of course, $X_{1}=\bigcup\limits_{k=1}^{m}P_{1,k}$. Let

[TABLE]

be the canonical projection from $Y$ on the quotient space $Y/Y_{0}$. By (Rudin_1991, , Theorem 1.41(a)), $\pi_{1}$ is a linear continuous mapping. Since the operator

[TABLE]

is a linear bijective mapping, $\Phi_{1}$ is a homeomorphism by Lemma 6. So, the operator $\pi:=\Phi_{1}\circ\pi_{1}:Y\rightarrow Y_{1}$ is linear and continuous.

Claim 2. For any $y\in Y$, we have $\pi(y)\in K_{1}\setminus\{0\}$ if and only if $y\in K\setminus\ell(K)$.

Indeed, suppose that $y\in K\setminus\ell(K)$. By Lemma 7, one can find $y_{0}\in Y_{0}$ and $y_{1}\in K_{1}\setminus\{0\}$ such that $y=y_{0}+y_{1}$. Then $\pi(y)=y_{1}\in K_{1}\setminus\{0\}$. Now, suppose that $\pi(y)\in K_{1}\setminus\{0\}$, i.e., there exists $y_{1}\in K_{1}\setminus\{0\}$ satisfying $\pi(y)=y_{1}$. This implies that $y-y_{1}\in Y_{0}$; hence $y\in y_{1}+Y_{0}\subset K_{1}\setminus\{0\}+Y_{0}$. In accordance with Lemma 7, $y\in K\setminus\ell(K)$.

Let $f_{1}$ be the restriction to $X_{1}$ of the mapping $\pi\circ f$. Clearly, $f_{1}$ is a piecewise linear vector-valued function from $X_{1}$ to $Y_{1}$. Let us consider a piecewise linear vector optimization problem

[TABLE]

Claim 3. It holds that ${\rm Sol(VP)}={\rm Sol(VP_{1})}$.

First, to show that ${\rm Sol(VP)}\subset{\rm Sol(VP_{1})}$, we suppose the contrary: There exists $u\in{\rm Sol(VP)}$ not belonging to ${\rm Sol(VP_{1})}$. Then one can find $x\in D$ such that $f_{1}(u)-f_{1}(x)\in K_{1}\setminus\{0\}.$ Since $\pi(f(u)-f(x))\in K_{1}\setminus\{0\}$, by Claim 2, one has $f(u)-f(x)\in K\setminus\ell(K)$. This contradicts the assumption $u\in{\rm Sol(VP)}$.

Now, to obtain the inclusion ${\rm Sol(VP_{1})}\subset{\rm Sol(VP)}$, take any $u\notin{\rm Sol(VP)}$. Then there exists $x\in D$ such that $f(u)-f(x)\in K\setminus\ell(K)$. Combining this with Claim 2, one gets $\pi(f(u)-f(x))\in K_{1}\setminus\{0\}$, i.e., $f_{1}(u)-f_{1}(x)\in K_{1}\setminus\{0\}.$ Therefore, $u\notin{\rm Sol(VP_{1})}$. We have thus proved that ${\rm Sol(VP)}={\rm Sol(VP_{1})}$.

Since $Y_{1}$ is finite-dimensional, $K_{1}$ is a pointed cone, $D$ is a polyhedral convex set in $X_{1}$, arguing similarly as in the proof of (Yang_Yen_2010, , Theorem 2.1), we can assert that ${\rm Sol(VP_{1})}$ is the union of finitely many semi-closed polyhedral convex sets in $X_{1}$. As ${\rm Sol(VP)}={\rm Sol(VP_{1})}$ by Claim 3, the assertion of the theorem has been proved. $\hfill\Box$

The next result is a generalization of (ZhengYang_2008, , Theorem 3.1).

Theorem 4.2

If ${\rm int}\,K$ is nonempty, then the weakly efficient solution set of (VP) is the union of finitely many generalized polyhedral convex sets.

Proof   We may assume, without loss of generality, that $D$ is nonempty. Set

[TABLE]

Invoking Lemma 8, we have ${\rm int}\,H_{j}=\{y\in Y\mid\langle y^{*}_{j},y\rangle<0\}$ for $j=1,\dots,q,$ and ${\rm int}\,K=\bigcap\limits_{j=1}^{q}{\rm int}\,H_{j}$. By Claim 1 in the proof of Theorem 3.1, one sees $M_{k}+K$ is a polyhedral convex set in $Y$, where $M_{k}:=f(D\cap P_{k})$ for all $k=1,\dots,m.$ Then, one can find $y^{*}_{k,1},\dots,y^{*}_{k,\ell_{k}}$ in $Y^{*}$, $\beta_{k,1},\dots,\beta_{k,\ell_{k}}$ in $\mathbb{R}$ such that

[TABLE]

with $H_{k,i}:=\{y\in Y\mid\langle y^{*}_{k,i},y\rangle\leq\beta_{k,i}\}.$ Put $Q=f(D)+K$ and observe that $Q=\bigcup\limits_{k=1}^{m}(M_{k}+K)$. Since $E^{w}(Q|K)=Q\setminus(Q+{\rm int}\,K),$ one has

[TABLE]

For any $k\in\{1,\dots,m\}$ and $i\in\{1,\dots,\ell_{k}\}$,

[TABLE]

Observe that, for any $k,\,k_{1}\in\{1,\dots,m\}$, $i\in\{1,\dots,\ell_{k}\}$, and $j\in\{1,\dots,q\}$, the set (M_{k_{1}}+K)\setminus\big{(}H_{k,i}+{\rm int}\,H_{j}\big{)} is polyhedral convex.

Indeed, let us show that there exist $y^{*}_{k,i,j}\in Y^{*}$ and $\beta_{k,i,j}\in\mathbb{R}$ such that

[TABLE]

First, consider the case where $y^{*}_{k,i}=0$. If $\beta_{k,i}<0$, then $H_{k,i}$ is empty. So, one can choose $y^{*}_{k,i,j}=0$ and $\beta_{k,i,j}=\beta_{k,i}$ because $H_{k,i}+{\rm int}\,H_{j}=\emptyset$. If $\beta_{k,i}\geq 0$, then $H_{k,i}=Y$. Since $H_{k,i}+{\rm int}\,H_{j}=Y$, (4.3) is fulfilled with $y^{*}_{k,i,j}=0$ and $\beta_{k,i,j}=\beta_{k,i}.$ Now, we consider the case where $y^{*}_{k,i}\neq 0$. One can find $\bar{y}_{k,i},w_{k,i}$ in $Y$ satisfying $\langle y^{*}_{k,i},\bar{y}_{k,i}\rangle=\beta_{k,i}$ and $\langle y^{*}_{k,i},w_{k,i}\rangle=1$. It is not difficult to show that

[TABLE]

Since $y^{*}_{j}\neq 0$, there exists $w_{j}\in Y$ such that $\langle y^{*}_{j},w_{j}\rangle=1$. We see at once that $H_{j}=\{t_{j}w_{j}\mid t_{j}\leq 0\}+{\rm ker}\,y^{*}_{j}$ and ${\rm int}\,H_{j}=\{t_{j}w_{j}\mid t_{j}<0\}+{\rm ker}\,y^{*}_{j}$. It follows that

[TABLE]

If ${\rm ker}\,y^{*}_{k,i}\neq{\rm ker}\,y^{*}_{j}$, then ${\rm ker}\,y^{*}_{k,i}+{\rm ker}\,y^{*}_{j}=Y$ by ${\rm codim}({\rm ker}\,y^{*}_{k,i})=1.$ Therefore, (4.4) shows that $H_{k,i}+{\rm int}\,H_{j}=Y$. So, one can choose $y^{*}_{k,i,j}=0$ and $\beta_{k,i,j}=1$.

If ${\rm ker}\,y^{*}_{k,i}={\rm ker}\,y^{*}_{j}$, then we take $\lambda_{k,i,j}=\langle y^{*}_{j},w_{k,i}\rangle$. For every $y\in Y$, put $t_{k,i}=\langle y^{*}_{k,i},y\rangle$. Clearly, the vector $y_{0}:=y-t_{k,i}w_{k,i}$ belongs to ${\rm ker}\,y^{*}_{k,i}$. Therefore,

[TABLE]

We thus get $y^{*}_{j}=\lambda_{k,i,j}y^{*}_{k,i,j}$. Since $y^{*}_{j}\neq 0$, one has $\lambda_{k,i,j}\neq 0$. If $\lambda_{k,i,j}>0$, then ${\rm int}\,H_{j}=\{y\in Y\mid\langle y^{*}_{k,i},y\rangle<0\}$. So,

[TABLE]

Of course, the formula (4.3) is fulfilled with $y^{*}_{k,i,j}=y^{*}_{k,i}$ and $\beta_{k,i,j}=\beta_{k,i}.$ If $\lambda_{k,i,j}<0$, then ${\rm int}\,H_{j}=\{y\in Y\mid\langle y^{*}_{k,i},y\rangle>0\}$. Therefore, $H_{k,i}+{\rm int}\,H_{j}=Y$. Hence, one can choose $y^{*}_{k,i,j}=0$ and $\beta_{k,i,j}=1$.

From (4.3) we see that

[TABLE]

Therefore, (M_{k_{1}}+K)\setminus\big{(}H_{k,i}+{\rm int}\,H_{j}\big{)} is a polyhedral convex set in $Y$.

From (4.2) it follows that, for all $k\in\{1,\dots,m\}$ and $i\in\{1,\dots,\ell_{k}\}$, Q\setminus\big{(}H_{k,i}+{\rm int}\,K\big{)} is the union of finitely many gpcs. Therefore, by (4.1), $E^{w}(Q|K)$ is the union of finitely many gpcs. Hence, using the same argument for getting Claim 4 in the proof of Theorem 3.1, we can prove that ${\rm Sol}^{w}{\rm(VP)}$ is the union of finitely many gpcs. $\hfill\Box$

Remark 3

For the case where $X,\,Y$ are normed spaces and $D$ is a polyhedral convex set, the result in Theorem 4.2 is due to Zheng and Yang (ZhengYang_2008, , Theorem 3.1).

Let us consider an illustrative example for Theorems 4.1 and 4.2.

Example 2

Keeping the notations of Example 1, we redefine the piecewise linear function $f$ by

[TABLE]

Claim 1. It holds that

[TABLE]

First, to show that ${\rm Sol(VP)}\subset S$, where $S$ is the set on the right-hand side of (4.5), take any $u\in{\rm Sol(VP)}$. Then one can find a vector $u_{0}\in X_{0}$ and numbers $t_{1},t_{2}$ satisfying $u=u_{0}+t_{1}e_{1}+t_{2}e_{2}$. Since $u\in D$, we have $t_{1}\leq 0$ and $t_{2}\leq 1$. Moreover, $u_{0}(t)=e_{0}(t)$ for all $t\in[-1,0]$; so, $u_{0}\in D$. The condition $u\in{\rm Sol(VP)}$ yields $f(u)-f(x)\notin K\setminus\ell(K)$ for every $x\in D$.

If $t_{2}\geq 0$, then $u\in P_{1}$; so $f(u)=u+Tu=u_{0}+(t_{1}+t_{2})e_{1}+t_{2}e_{2}$. Observe that $x:=u_{0}+e_{2}$ belongs to $D\cap P_{1}$. It is clear that $f(x)=x+Tx=u_{0}+e_{1}+e_{2}$; hence

[TABLE]

Since $t_{1}+t_{2}-1\leq 0$ and $t_{2}-1\leq 0$, (3.5) shows that $f(u)-f(x)\in K$. Combining this with the inclusion $f(u)-f(x)\notin K\setminus\ell(K)$, one gets $f(u)-f(x)\in\ell(K)$. This implies that $t_{1}+t_{2}-1=0$ and $t_{2}-1=0$, i.e., $t_{1}=0$ and $t_{2}=1$. Therefore, $u\in S$.

If $t_{2}<0$, then $u\in P_{2}$ and $f(u)=u-Tu=u_{0}+(t_{1}-t_{2})e_{1}+t_{2}e_{2}$. If we take $x=u_{0}+e_{2}$, then $x\in D\cap P_{1}$. As $f(x)=x+Tx=u_{0}+e_{1}+e_{2}$, one has

[TABLE]

Since $f(u)-f(x)\notin K\setminus\ell(K)$, (3.5) shows that $t_{1}-t_{2}-1>0$, by $t_{2}-1<0$. Therefore, $t_{2}<-1$ as $t_{1}\leq 0$. If $t_{1}<0$, then one can find a positive number $\varepsilon$ such that $t_{1}+\varepsilon<0$ and $t_{2}+\varepsilon<-1$. Set $x=u_{0}+(t_{2}+\varepsilon)e_{2}$, and observe that $x\in D\cap P_{2}$. Since $f(x)=x-Tx=u_{0}-(t_{2}+\varepsilon)e_{1}+(t_{2}+\varepsilon)e_{2}$,

[TABLE]

As $t_{1}+\varepsilon<0$ and $-\varepsilon<0$, (3.7) yields $f(u)-f(x)\in K\setminus\ell(K)$. This contradicts the assumption $u\in{\rm Sol}{\rm(VP)}$. We thus get $t_{1}=0$. Consequently, $u\in S$.

We have proved that ${\rm Sol(VP)}\subset S$. To obtain the opposite inclusion, take any $u\in S$. Let $u_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$ be such that $u=u_{0}+t_{1}e_{1}+t_{2}e_{2}$. Of course, $t_{1}=0$. Given any $x\in D$, one can find a vector $x_{0}\in X_{0}$, numbers $\tau_{1}\leq 0$ and $\tau_{2}\leq 1$ such that $x=x_{0}+\tau_{1}e_{1}+\tau_{2}e_{2}$.

If $t_{2}=1$, then $u\in P_{1}$ and $f(u)=u+Tu=u_{0}+e_{1}+e_{2}$. If $0\leq\tau_{2}\leq 1$, then $x\in P_{1}$ and $f(x)=x_{0}+(\tau_{1}+\tau_{2})e_{1}+\tau_{2}e_{2}$. Since

[TABLE]

with $1-\tau_{1}-\tau_{2}\geq 0$ and $1-\tau_{2}\geq 0$, (3.7) shows that $f(u)-f(x)\notin K\setminus\ell(K)$. If $\tau_{2}<0$, then $x\in P_{2}$; so $f(x)=x_{0}+(\tau_{1}-\tau_{2})e_{1}+\tau_{2}e_{2}$. According to (3.7), since $f(u)-f(x)=(u_{0}-x_{0})+(1-\tau_{1}+\tau_{2})e_{1}+(1-\tau_{2})e_{2}$ with

[TABLE]

one gets $f(u)-f(x)\notin K\setminus\ell(K)$. It follows that $f(u)-f(x)\notin K\setminus\ell(K)$ for all $x\in D$; hence, $u\in{\rm Sol}{\rm(VP)}$.

If $t_{2}<-1$, then $u\in P_{2}$. Therefore, $f(u)=u-Tu=u_{0}-t_{2}e_{1}+t_{2}e_{2}$. If $0\leq\tau_{2}\leq 1$, then $x\in P_{1}$ and $f(x)=x_{0}+(\tau_{1}+\tau_{2})e_{1}+\tau_{2}e_{2}$. As

[TABLE]

with $-t_{2}-\tau_{1}-\tau_{2}>1-\tau_{1}-\tau_{2}\geq 0$, one has $f(u)-f(x)\notin K\setminus\ell(K)$ by (3.7). If $\tau_{2}<0$, then $x\in P_{2}$ and $f(x)=x_{0}+(\tau_{1}-\tau_{2})e_{1}+\tau_{2}e_{2}$. Observe that

[TABLE]

with $(-t_{2}-\tau_{1}+\tau_{2})+(t_{2}-\tau_{2})=-\tau_{1}\geq 0$. So, $f(u)-f(x)\notin K\setminus\ell(K)$ by (3.7). Therefore, $f(u)-f(x)\notin K\setminus\ell(K)$ for all $x\in D$. Hence, $u\in{\rm Sol}{\rm(VP)}$. We have proved that ${\rm Sol}{\rm(VP)}=S$.

Observe that ${\rm Sol}{\rm(VP)}$ is the union of two semi-closed generalized polyhedral convex sets. Furthermore, ${\rm Sol}{\rm(VP)}$ is disconnected and non-closed.

Claim 2. It holds that

[TABLE]

First, to clear that ${\rm Sol}^{w}{\rm(VP)}\subset S^{w}$, where $S^{w}$ is the set on the right-hand side of (4.6), take any $u\in{\rm Sol}^{w}{\rm(VP)}$. Suppose that $u=u_{0}+t_{1}e_{1}+t_{2}e_{2}$ with $u_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$. Since $u\in D$, $t_{1}\leq 0$ and $t_{2}\leq 1$. Observe that $u_{0}(t)=e_{0}(t)$ for all $t\in[-1,0]$; so, $u_{0}\in D$. The inclusion $u\in{\rm Sol}^{w}{\rm(VP)}$ implies that $f(u)-f(x)\notin{\rm int}\,K$ for all $x\in D$.

If $t_{2}\geq 0$, then $u\in P_{1}$ and $f(u)=u+Tu=u_{0}+(t_{1}+t_{2})e_{1}+t_{2}e_{2}$. Since the vector $x:=u_{0}+e_{2}$ belongs to $D\cap P_{1}$, $f(x)=x+Tx=u_{0}+e_{1}+e_{2}$. Therefore,

[TABLE]

If $t_{2}<1$, then $t_{1}+t_{2}-1<0$ and $t_{2}-1<0$. So, $f(u)-f(x)\in{\rm int}\,K$ by (3.6). This contradicts the assumption $u\in{\rm Sol}^{w}{\rm(VP)}$. We thus get $t_{2}=1$. Consequently, $u\in S^{w}$.

If $t_{2}<0$, then $u\in P_{2}$ and $f(u)=u-Tu=u_{0}+(t_{1}-t_{2})e_{1}+t_{2}e_{2}$. Clearly, the vector $x:=u_{0}+e_{2}$ belongs to $D\cap P_{1}$. Since $f(x)=x+Tx=u_{0}+e_{1}+e_{2}$,

[TABLE]

If $t_{1}-t_{2}-1<0$, then $f(u)-f(x)\in{\rm int}\,K$ by (3.6). This contradicts the assumption $u\in{\rm Sol}^{w}{\rm(VP)}$. It follows that $t_{1}-t_{2}-1\geq 0$. Hence, $t_{2}\leq-1$ as $t_{1}\leq 0$. Therefore, $u\in S^{w}$.

We have proved that ${\rm Sol}^{w}{\rm(VP)}\subset S^{w}$. To obtain the opposite inclusion, take any $u\in S^{w}$. Let $u_{0}\in X_{0}$ and $t_{1},t_{2}\in\mathbb{R}$ be such that $u=u_{0}+t_{1}e_{1}+t_{2}e_{2}$. Given any $x\in D$, one can find a vector $x_{0}\in X_{0}$, numbers $\tau_{1}\leq 0$ and $\tau_{2}\leq 1$ such that $x=x_{0}+\tau_{1}e_{1}+\tau_{2}e_{2}$.

If $t_{2}=1$ and $t_{1}\leq 0$, then $f(u)=u+Tu=u_{0}+(t_{1}+1)e_{1}+e_{2}$ by $u\in P_{1}$. If $0\leq\tau_{2}\leq 1$, then $x\in P_{1}$ and $f(x)=x_{0}+(\tau_{1}+\tau_{2})e_{1}+\tau_{2}e_{2}$. Since

[TABLE]

with $1-\tau_{2}\geq 0$, one gets $f(u)-f(x)\notin{\rm int}\,K$ by (3.6). If $\tau_{2}<0$, then $x\in P_{2}$ and $f(x)=x_{0}+(\tau_{1}-\tau_{2})e_{1}+\tau_{2}e_{2}$. In accordance with (3.6), since

[TABLE]

with $1-\tau_{2}\geq 0$, one gets $f(u)-f(x)\notin{\rm int}\,K$. It follows that $f(u)-f(x)\notin{\rm int}\,K$ for all $x\in D$. Hence, $u\in{\rm Sol}^{w}{\rm(VP)}$.

If $t_{2}\leq-1$ and $t_{1}=0$, then $f(u)=u-Tu=u_{0}-t_{2}e_{1}+t_{2}e_{2}$ by $u\in P_{2}$. If $0\leq\tau_{2}\leq 1$, then $x\in P_{1}$ and $f(x)=x_{0}+(\tau_{1}+\tau_{2})e_{1}+\tau_{2}e_{2}$. Therefore,

[TABLE]

By (3.6), since $-t_{2}-\tau_{1}-\tau_{2}\geq 1-\tau_{1}-\tau_{2}\geq 0$, one has $f(u)-f(x)\notin{\rm int}\,K$. If $\tau_{2}<0$, then $x\in P_{2}$ and $f(x)=x_{0}+(\tau_{1}-\tau_{2})e_{1}+\tau_{2}e_{2}$. As

[TABLE]

with $(-t_{2}-\tau_{1}+\tau_{2})+(t_{2}-\tau_{2})=-\tau_{1}\geq 0$, by (3.6), $f(u)-f(x)\notin{\rm int}\,K$. It follows that $f(u)-f(x)\notin{\rm int}\,K$ for all $x\in D$. Hence, $u\in{\rm Sol}^{w}{\rm(VP)}$. We have proved that ${\rm Sol}^{w}{\rm(VP)}=S^{w}$.

Clearly, ${\rm Sol}^{w}{\rm(VP)}$ is disconnected and it is the union of two generalized polyhedral convex sets.

Acknowledgements.

This research was supported by the Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant number 101.01-2014.37. The author would like to thank Professor Nguyen Dong Yen for his guidance and the anonymous referees for valuable suggestions.