On optimality conditions and duality theorems for robust semi-infinite multiobjective optimization problems

Abstract

In this paper, we consider a semi-infinite multiobjective optimization problem with more than two differentiable objective functions and uncertain constraint functions, which is called a robust semi-infinite multiobjective optimization problem and give its robust counterpart ${\mathrm{(RSIMP)}}$ of the problem, which is regarded as the worst case of the uncertain semi-infinite multiobjective optimization problem. We prove a necessary optimality theorem for a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$, and then give a sufficient optimality theorem for a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$. We formulate a Wolfe type dual problem of ${\mathrm{(RSIMP)}}$ and give duality results which hold between ${\mathrm{(RSIMP)}}$ and its dual problem.

1 Introduction

Mathematical programming problems in the face of data uncertainty have been treated by the worst case approach or the stochastic approach. The worst case approach for optimization problems, which has emerged as a powerful deterministic approach for studying optimization problems with data uncertainty, associates an uncertain optimization problem with its robust counterpart. Recently, the study of convex programs that are affected by data uncertainty is becoming increasingly important in optimization (Beck and Ben-Tal 2009; Ben-Tal et al. 2009; Ben-Tal and Nemirovski 1999, 2002, 2008; Jeyakumar et al. 2012, 2015; Jeyakumar and Li 2010a, b; Kuroiwa and Lee 2012, 2014; Lee and Jiao 2016; Lee and Lee 2012, 2015b). Many researchers (Beck and Ben-Tal 2009; Jeyakumar and Li 2010b, 2014; Li et al. 2011; Suzuki et al. 2013) have investigated optimality and duality theories for linear or convex programming problems under uncertainty with the worst-case approach(the robust approach). It was shown that the value of the robust counterpart of primal problem is equal to the value of the optimistic counterpart of the dual primal (“primal worst equals dual best”) (Beck and Ben-Tal 2009; Jeyakumar and Li 2010b; Jeyakumar et al. 2012). Recently, many researchers (Chuong 2016; Chuong and Kim 2014; Kuroiwa and Lee 2012, 2014; Lee and Lee 2015a, b) have studied optimality and duality theories for robust multiobjective optimization programming problems under different suitable constrained qualifications. Especially, Goberna et al. (2013a, b, 2014), have established optimality conditions and duality theorems for robust multiobjective linear semi-infinite programming problems under uncertainty. In this paper, we investigate optimality conditions and duality theorems for robust multiobjective nonlinear semi-infinite programming problems under uncertainty.

Consider the following semi-infinite multiobjective optimization problem in the absence of data uncertainty

where $f_i$, $i=1,\ldots ,l$, and $g_t: \mathbb {R}^n \rightarrow \mathbb {R}$, $t\in T$, are continuously differentiable and T is an arbitrary (possible infinite) index set.

The semi-infinite multiobjective optimization problem (SIMP) in the face of data uncertainty in the constraints can be captured by the problem

where $f_i:\mathbb {R}^n \rightarrow \mathbb {R}$, $i=1,\ldots ,l$ and $g_t : \mathbb {R}^n \times \mathbb {R}^q \rightarrow \mathbb {R}$ are continuously differentiable and $v_t \in \mathbb {R}^q$ is an uncertain parameter which belongs to the convex compact set $\mathcal {V}_t\subset \mathbb {R}^q$, $t\in T$. The uncertainty set-valued mapping $\mathcal {V}:T\rightrightarrows \mathbb {R}^q$ is defined as $\mathcal {V}(t):=\mathcal {V}_t$ for all $t\in T$. So, ${\mathrm{gph}}\mathcal {V}=\{(t,v_t) \ | \ v_t\in \mathcal {V}_t, \ t\in T\}$ and $v\in \mathcal {V}$ means that v is a selection of $\mathcal {V}$, i.e., $v:T\rightarrow \mathbb {R}^q$ and $v_t\in \mathcal {V}_t$ for all $t\in T$.

The robust counterpart of (USIMP) is

The robust feasible set F of ${\mathrm{(RSIMP)}}$ is defined by

$$\begin{aligned} F:=\left\{ x\in \mathbb {R}^n \ | \ g_t(x,v_t)\le 0, \ \forall {t}\in T, \ \forall {v_t}\in \mathcal {V}_t\right\} . \end{aligned}$$

Then $\bar{x} \in F$ is called a weakly robust efficient solution of (RSIMP) if there does not exist a robust feasible solution x of (RSIMP) such that

$$\begin{aligned} f_i (x) < f_i (\bar{x}), \ i = 1, \ldots , l. \end{aligned}$$

In this paper, we prove a necessary optimality theorem for a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$, and then give a sufficient optimality theorem for a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$. We formulate a Wolfe type dual problem of ${\mathrm{(RSIMP)}}$ and give duality results which holds between ${\mathrm{(RSIMP)}}$ and its dual problem.

2 Preliminaries

Let us first recall some notation and preliminary results which will be used throughout this paper. We denote by $\mathbb {R}^n$ the Euclidean space of dimension n. The nonnegative orthant of $\mathbb {R}^n$ is denoted by $\mathbb {R}_{+}^n$. We say that the set A is convex whenever $\mu a_1 + (1-\mu )a_2 \in A$ for all $\mu \in [0, 1]$, $a_1, a_2\in A$. A function $f : \mathbb {R}^n \rightarrow \mathbb {R}$ is said to be convex if for all $\mu \in [0, 1]$,

$$\begin{aligned} f((1-\mu )x + \mu y)\le (1-\mu )f(x) + \mu f(y) \end{aligned}$$

for all $x, y \in \mathbb {R}^n$. The function f is said to be concave whenever $-f$ is convex. For a set $E\subset \mathbb {R}^n$, $\mathrm{co}E$ and $\mathrm{cone}E$ denote the convex hull of E and the convex conical hull of E, respectively. For a convex cone $K\subset \mathbb {R}^n$, its positive polar cone is defined as $K^+:=\{y\in \mathbb {R}^n \ | \ x^Ty\ge 0 \ \forall x\in K\}$. Let $\mathbb {R}^{(T)}$ be the linear space of mappings $\mu \in \mathbb {R}^T$ such that $\{t\in T \ | \ \mu _t\ne 0\}$ is finite and let $\mathbb {R}_+^{(T)}$ be the positive cone in $\mathbb {R}^{(T)}$. Let X, Y be topological spaces and let $S : X \rightrightarrows Y$ be a multifunction. Then the multifunction S is said to be upper semi-continuous function at $x_0\in X$ if for any open neighborhood $N(S(x_0))$ of $S(x_0)$, there exists an open neighborhood $N(x_0)$ of $x_0$ such that for all $x\in N(x_0)$, $S(x)\subset N(S(x_0))$. We say that S is upper semi-continuous on X if it is upper semi-continuous at any point of X.

Proposition 1

(Aubin 1979) Suppose that a multifunction S is upper semi-continuous with non-empty compact images. Then S is closed.

Proposition 2

(Aubin 1979) Suppose that a multifunction S is upper semi-continuous with non-empty compact images. If X is compact, then $S(X)=\bigcup _{x\in X}S(x)$ is also compact.

Throughout this paper we assume that the following assumptions hold:

• (A1) T is a compact metric space.

• (A2) $\mathcal {V}$ is compact-valued and upper semi-continuous on T.

• (A3) $g_{t_n}(x_n,v_{t_n})\rightarrow g_{t}(x,v_{t})$, whenever $t_n\in T\rightarrow t\in T$, $v_{t_n}\in \mathcal {V}_{t_n}\rightarrow v_{t}\in \mathcal {V}_t$ and $x_n\in \mathbb {R}^n\rightarrow x\in \mathbb {R}^n$ as $n\rightarrow \infty$.

• (A4) $\nabla g_{t_n}(x_n,v_{t_n})\rightarrow \nabla g_{t}(x,v_{t})$, whenever $t_n\in T\rightarrow t\in T$, $v_{t_n}\in \mathcal {V}_{t_n}\rightarrow v_{t}\in \mathcal {V}_t$ and $x_n\in \mathbb {R}^n\rightarrow x\in \mathbb {R}^n$ as $n\rightarrow \infty$.

Remark 1

Under the conditions (A1) and (A2), by Proposition 1 and 2, $\mathcal {V}(T)$ is compact and ${\mathrm{gph}}\mathcal {V}$ is closed in $T\times \mathbb {R}^q$.

3 Robust necessary and sufficient optimality conditions

Let $x\in F$. Let us decompose T into two index sets $T=T_1(x)\cup T_2(x)$, where $T_1(x)=\{t\in T \ | \ \exists v_t\in \mathcal {V}_t \ {\mathrm{s.t.}} \ g_t(x,v_t)=0\}$ and $T_2(x)=T\backslash T_1(x)$. Let $\mathcal {V}_t(x)=\{v_t\in \mathcal {V}_t \ | \ g_t(x, v_t)=0\}$.

Definition 1

(Bonnans and Shapiro 2000; Jeyakumar et al. 2015) We say that the extended Mangasarian–Fromovitz constraint qualification (EMFCQ) is satisfied at $x \in F$ iff

$$\begin{aligned} \exists d \in \mathbb {R}^n {\text { s.t. }} \nabla _x g_t( x , v_t)^Td < 0 \ \forall t \in T_1 ( x), \ \forall v_t \in \mathcal {V}_t( x). \end{aligned}$$

Remark 2

(i) Let $x\in F$ and let $T:=\{t_1,\ldots , t_p\}$. Since $\mathrm{co}\{\nabla _xg_{t_i}(x,v_{t_i}) \ | \ v_{t_i}\in \mathcal {V}_{t_i}(x), \ i=1,\ldots ,p\}$ is convex and compact, ${\mathrm{(EMFCQ)}}$ is equivalent to that $0\notin \mathrm{co}\{\nabla _xg_{t_i}(x,v_{t_i}) \ | \ v_{t_i}\in \mathcal {V}_{t_i}(x), \ i=1,\ldots ,p\}$. The latter condition is the special case of Definition 3.2 in Chuong (2016).

(ii) If ${\mathrm{(EMFCQ)}}$ holds at $x\in F$, then

$$\begin{aligned} \mathbf{N}_F(x)\subset \bigcup _{\lambda \in A(x)}\left\{ \sum _{t\in T}\lambda _t\nabla _x g_t(x,v_t)\right\} , \end{aligned}$$

where $\mathbf{N}_F(x)$ is the normal cone to F at $x\in F$ and $A(x):=\{\lambda \in \mathbb {R}_+^{(T)} \ | \ \lambda _tg_t(x,v_t)=0 \ \forall t\in T, \ \forall v_t\in \mathcal {V}_t(x)\}$. The latter condition is a kind of the robust versions of Definition 3.2 in Chuong and Kim (2014).

Now we prove a robust necessary optimality theorem for a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$.

Theorem 1

(Robust Necessary Optimality Condition) Let $\bar{x}$ be a weakly robust efficient solution of (RSIMP). Suppose that $g_t(x,\cdot )$ is concave on $\mathcal {V}_t$, for each $x\in \mathbb {R}^n$ and for each $t\in T$. Then there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, $\left( \bar{\lambda }_t\right) _{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\bar{\mu }_i+\sum _{t\in T}\bar{\lambda }_t=1$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)=0 \ \ {\text {and}} \ \ \bar{\lambda }_tg_t(\bar{x},\bar{v}_t)=0, \ t\in T. \end{aligned}$$

(1)

Moreover, if we further assume that the (EMFCQ) holds at $\bar{x}$, then there exist $\hat{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero, $(\hat{\lambda }_t)_{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\hat{\mu }_i=1$,

$$\begin{aligned} \sum _{i=1}^l\hat{\mu }_i\nabla f_i(\bar{x})+\sum _{t\in T}\hat{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)=0 \ \ {\text {and}} \ \ \hat{\lambda }_tg_t(\bar{x},\bar{v}_t)=0, \ t\in T. \end{aligned}$$

(2)

Proof

Let $\bar{x}$ be a weakly efficient solution of (RSIMP). Then there does not exist $x\in F$ such that

$$\begin{aligned} f_i(x)<f_i(\bar{x}), \ i=1,\ldots ,l. \end{aligned}$$

(3)

[Step 1] Since $\mathcal {V}_t(\bar{x})$ is a closed subset of the compact set $\mathcal {V}_t$, $\mathcal {V}_t(\bar{x})$ is compact. Now, we may assume, without loss of generality, that $T_1(\bar{x})\ne \emptyset$. We first show that the following system:

$$\begin{aligned} \nabla f_i(\bar{x})^Td<0, \ i=1,\ldots ,l, \ {\mathrm{and }} \ \nabla _x g_t(\bar{x}, v_t)^Td<0 \ \forall t\in T_1(\bar{x}), \ \forall v_t\in \mathcal {V}_t(\bar{x}) \end{aligned}$$

(4)

has no solution. Otherwise, there exists $d\in \mathbb {R}^n$ such that (4) holds. Then we will show that there exists $\bar{\tau }>0$ such that

$$\begin{aligned} g_t(\bar{x} + \tau d, v_t)<0 \ \forall \tau \in (0,\bar{\tau }], \ \forall t\in T, \ \forall v_t\in \mathcal {V}_t. \end{aligned}$$

(5)

Now, we suppose that (5) does not hold. Then, for each natural number m, there exists $\tau _m\in (0, \textstyle {\frac{1}{m}})$, $t_m\in T$ and $v_{t_m}\in \mathcal {V}_{t_m}$ such that

$$\begin{aligned} g_{t_m}(\bar{x} +\tau _nd,v_{t_m})\ge 0. \end{aligned}$$

(6)

Since T is compact, there exist some subsequences $\{\tau _{m_k}\}$ and $\{t_{m_k}\}$ such that $\tau _{m_k}\rightarrow 0$ and $t_{m_k}\rightarrow t^*\in T$. Since, by Remark 1, $\mathcal {V}(T)$ is compact, we may assume that $v_{t_{m_k}}\rightarrow v_{t^*}$. Moreover, since, by Remark 1, ${\mathrm{gph\mathcal {V}}}$ is closed and $(t_{m_k}, v_{t_{m_k}})\rightarrow (t^*,v_{t^*})$, $(t^*,v_{t^*})\in {\mathrm{gph}}\mathcal {V}$, that is, $v_{t^*}\in \mathcal {V}_{t^*}$. From (6) and (A3), we have $g_{t_{m_k}}(\bar{x}+\tau _{m_k}d, v_{t_{m_k}})\rightarrow g_{t^*}(\bar{x},v_{t^*})\ge 0$. Moreover, since $\bar{x}\in F$, we have $g_{t^*}(\bar{x},v_{t^*})=0$, and so, $t^*\in T_1(\bar{x})$ and $v_{t^*}\in \mathcal {V}_{t^*}(\bar{x})$. On the other hand, by the Mean-Value Theorem, there exists $\alpha _{m_k}\in (0, \tau _{m_k})$ such that

$$\begin{aligned} 0 \le g_{t_{m_k}}\left( \bar{x}+\tau _{m_k}d, v_{t_{m_k}}\right) -g_{t_{m_k}}(\bar{x}, v_{t_{m_k}})=\tau _{m_k}\nabla _x g_{t_{m_k}}\left( \bar{x}+\alpha _{m_k}d, v_{t_{m_k}}\right) ^Td, \end{aligned}$$

and hence, $\nabla _x g_{t_{m_k}}\left( \bar{x}+\alpha _{m_k}d, v_{t_{m_k}}\right) d\ge 0$. So, by (A4), we have

$$\begin{aligned} \nabla _x g_{t^*}(\bar{x}, v_{t^*})^Td=\lim _{k\rightarrow \infty }\nabla _x g_{t_{m_k}}\left( \bar{x}+\alpha _{m_k}d, v_{t_{m_k}}\right) ^Td\ge 0, \end{aligned}$$

which contradicts the choice of d. So, (5) holds. Moreover, since $\nabla f_i(\bar{x})^Td<0$, $i=1,\ldots ,l$, there exists $\tilde{\tau }>0$ such that

$$\begin{aligned} f_i(\bar{x}+\tau d)<f_i(\bar{x}), \ i=1,\ldots ,l, \quad \forall \tau \in (0,\tilde{\tau }]. \end{aligned}$$

So, whenever $\tau \in (0, \min \{\bar{\tau },\tilde{\tau }\}]$, we have

$$\begin{aligned} g_t(\bar{x}+\tau d,v_t)\le 0, \ \forall t\in T, \ \forall v_t\in \mathcal {V}_t \ {\mathrm{and }} \ f_i(\bar{x}+\tau d)<f_i(\bar{x}), \quad i=1,\ldots ,l. \end{aligned}$$

This contradicts (3).

[Step 2] Now we consider the set $S=\{\nabla f_i(\bar{x}) \ | \ i=1,\ldots ,l\}\cup \{\nabla _x g_t(\bar{x},v_t) \ | \ v_t\in \mathcal {V}_t(\bar{x}), \ t\in T_1(\bar{x})\}$. Let $\tilde{S}=\{\nabla _x g_t(\bar{x},v_t) \ | \ v_t\in \mathcal {V}_t(\bar{x}), \ t\in T_1(\bar{x})\}$. Then we will show that $\tilde{S}$ is compact. First, we assume that $\tilde{S}$ is not bounded. Then there exist $t_m\in T_1(\bar{x})$ and $v_{t_m}\in \mathcal {V}_t(\bar{x})$ such that

$$\begin{aligned} \Vert \nabla _xg_{t_m}(\bar{x}, v_{t_m})\Vert \rightarrow \infty . \end{aligned}$$

(7)

Since $t_m\in T_1(\bar{x})\subset T$, there exists $\tilde{v}_{t_m}\in \mathcal {V}_{t_m}$ such that $g_{t_m}(\bar{x},\tilde{v}_{t_m})=0$. Since T is compact, we may assume that $t_m\rightarrow \tilde{t}$. Moreover, since $\mathcal {V}_{t_m}\subset \mathcal {V}(T)$ and $\mathcal {V}(T)$ is compact, we may assume that $\tilde{v}_{t_m}\rightarrow \tilde{v}$. Since ${\mathrm{gph}}\mathcal {V}$ is closed and $(t_m,\tilde{v}_{t_m})\in {\mathrm{gph}}\mathcal {V}\rightarrow (\tilde{t}, \tilde{v})$, $(\tilde{t}, \tilde{v})\in {\mathrm{gph}}\mathcal {V}$, that is, $\tilde{v}\in \mathcal {V}_{\tilde{t}}$. So, by (A4),

$$\begin{aligned} \lim _{n\rightarrow \infty }\nabla _x g_{t_m}(\bar{x}, v_{t_m})=\nabla _x g_{\tilde{t}}(\bar{x}, \tilde{v}). \end{aligned}$$

This contradicts (7). Hence, $\tilde{S}$ is bounded. Now we will show that $\tilde{S}$ is closed. Let $t_m\in T_1(\bar{x})$, $v_{t_m}\in \mathcal {V}_t(\bar{x})$ and $\nabla _xg_{t_m}(\bar{x},v_{t_m})\rightarrow a$. Since T is compact and $\mathcal {V}(T)$ is compact, we may assume that $t_m\rightarrow t^*$ and $v_{t_m}\rightarrow v^*$. Moreover, since $\mathrm{gph}\mathcal {V}$ is closed, $v^*\in \mathcal {V}_{t^*}$. So, by (A4),

$$\begin{aligned} \lim _{n\rightarrow \infty }\nabla _x g_{t_m}(\bar{x}, v_{t_m})=\nabla _x g_{t^*}(\bar{x}, v^*). \end{aligned}$$

Hence, $\nabla _xg_{t^*}(\bar{x},v^*)=a$. Thus, $\tilde{S}$ is closed. Hence $\tilde{S}$ is compact and consequently, S is compact.

[Step 3] Now we will prove that $0\in \mathrm{co}S$. Otherwise, suppose that $0\notin \mathrm{co}S$. Then, by the separation theorem, there exists $d^*\in \mathbb {R}^n$ such that for any $v\in \mathrm{co}S$, $v^Td^*<0$. Thus $\nabla f_i(\bar{x})^Td^*<0$, $i=1,\ldots ,l$, and $\nabla _x g_t(\bar{x},v_t)^Td^*<0$ for all $t\in T_1(\bar{x})$ and all $v_t\in \mathcal {V}_t(\bar{x})$. This contradicts the result obtained in [Step 1].

[Step 4] Since $0\in \mathrm{co}S$, there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, $\lambda _{t_j}\ge 0$, $j=1,\ldots ,p$, not all zero, $v_{t_j}\in \mathcal {V}_{t_j}(\bar{x})$ and $t_j\in T_1(\bar{x})$, $j=1,\ldots ,p$, such that $\sum _{i=1}^l\bar{\mu }_i+\sum _{j=1}^p\lambda _{t_j}=1$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum _{j=1}^p\lambda _{t_j}\nabla _xg_{t_j}(\bar{x},v_{t_j})=0. \end{aligned}$$

We should notice that many $v_{t_j}$’s may in the same $\mathcal {V}_{t_j}$. But (1) implies that only one $v_t$ is in $\mathcal {V}_t$. If $\sum _{j=1}^p\lambda _{t_j}=0$, then the result of this Theorem trivially holds. We can assume without loss of generality $\lambda _{t_j}>0$, $j=1,\ldots ,p$. Now we consider that there exist $\mu _i\ge 0$, $i=1,\ldots ,l$, $\lambda _{t_j}^k\ge 0$, $j=1,\ldots ,q$, $k=1,\ldots ,s_j$, not all zero, $v_{t_j}^k\in \mathcal {V}_{t_j}(\bar{x})$ and $t_j\in T_1(\bar{x})$, $j=1,\ldots ,p$, such that $\sum _{j=1}^qs_j=p$, $\sum _{i=1}^l\bar{\mu }_i+\sum _{j=1}^q\sum _{k=1}^{s_j}\lambda _{t_j}^k=1$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum _{j=1}^q\sum _{k=1}^{s_j}\lambda _{t_j}^k\nabla _xg_{t_j}\left( \bar{x},v_{t_j}^k\right) =0. \end{aligned}$$

So, we have, for any $d\in \mathbb {R}^n$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})^Td+\sum _{j=1}^q\sum _{k=1}^{s_j}\lambda _{t_j}^k\nabla _xg_{t_j}\left( \bar{x},v_{t_j}^k\right) ^Td=0. \end{aligned}$$

(8)

Moreover, for each $j=1,\ldots ,q$,

$$\begin{aligned}&\sum _{k=1}^{s_j}\lambda _{t_j}^k\nabla _xg_{t_j}\left( \bar{x},v_{t_j}^k\right) ^Td =\sum _{k=1}^{s_j}\lambda _{t_j}^k\lim \limits _{\alpha \rightarrow 0} \frac{g_{t_j}\left( \bar{x}+\alpha d,v_{t_j}^k\right) -g_{t_j}\left( \bar{x},v_{t_j}^k\right) }{\alpha }\\&\quad =\sum \limits _{k=1}^{s_j}\lambda _{t_j}^k\lim \limits _{\alpha \rightarrow 0}\frac{\sum \nolimits _{k=1}^{s_j}\lambda _{t_j}^k\left( g_{t_j}(\bar{x} +\alpha d,v_{t_j}^k)-g_{t_j}(\bar{x},v_{t_j}^k)\right) }{\sum \nolimits _{k=1}^{s_j} \lambda _{t_j}^k\alpha }\\&\quad \quad \le \sum _{k=1}^{s_j}\lambda _{t_j}^k\lim _{\alpha \rightarrow 0} \frac{g_{t_j}\left( \bar{x}+\alpha d,{\sum \nolimits _{k=1}^{s_j}\lambda _{t_j}^kv_{t_j}^k}\big /{\sum \nolimits _{k=1}^{s_j} \lambda _{t_j}^k}\right) -g_{t_j}\left( \bar{x},{\sum \nolimits _{k=1}^{s_j}\lambda _{t_j}^kv_{t_j}^k}\big / {\sum \nolimits _{k=1}^{s_j}\lambda _{t_j}^k}\right) }{\alpha }\\&\quad =\sum _{k=1}^{s_j}\lambda _{t_j}^k\nabla _xg_{t_j} \left( \bar{x},{\sum _{k=1}^{s_j}\lambda _{t_j}^kv_{t_j}^k}\bigg / {\sum _{k=1}^{s_j}\lambda _{t_j}^k}\right) ^Td, \end{aligned}$$

where the last inequality follows from the fact that the concavity of $g_{t_j}(x,\cdot )$, $v_{t_j}^k\in \mathcal {V}_{t_j}(\bar{x})$, $t_j\in T_1(\bar{x})$ and the convexity of $\mathcal {V}_{t_j}(\bar{x})$, $j=1,\ldots ,p$ (i.e, ${\sum _{k=1}^{s_j}\lambda _{t_j}^kv_{t_j}^k}/{\sum _{k=1}^{s_j}\lambda _{t_j}^k}\in \mathcal {V}_{t_j}(\bar{x})$, and so, $g_{t_j}(\bar{x},{\sum _{k=1}^{s_j}\lambda _{t_j}^kv_{t_j}^k}/{\sum _{k=1}^{s_j}\lambda _{t_j}^k})=0$). So, from (8), for any $d\in \mathbb {R}^n$,

$$\begin{aligned} 0= & {} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})^Td +\sum \limits _{j=1}^q\sum \limits _{k=1}^{s_j}\lambda _{t_j}^k \nabla _xg_{t_j}\left( \bar{x},v_{t_j}^k\right) ^Td\\&\quad \le \left( \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x}) +\sum \limits _{j=1}^q\sum _{k=1}^{s_j}\lambda _{t_j}^k\nabla _xg_{t_j} \left( \bar{x},\frac{\sum _{k=1}^{s_j}\lambda _{t_j}^kv_{t_j}^k}{\sum _{k=1}^{s_j}\lambda _{t_j}^k}\right) \right) ^Td. \end{aligned}$$

Hence, we have

$$\begin{aligned} 0=\sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum _{j=1}^p\sum _{k=1}^{s_j} \lambda _{t_j}^k\nabla _xg_{t_j}\left( \bar{x},\frac{\sum _{k=1}^{s_j} \lambda _{t_j}^kv_{t_j}^k}{\sum _{k=1}^{s_j}\lambda _{t_j}^k}\right) . \end{aligned}$$

Since $\mathcal {V}_{t_j}(\bar{x})$ is convex, ${\sum _{k=1}^{s_j}\lambda _{t_j}^kv_{t_j}^k}/{\sum _{k=1}^{s_j}\lambda _{t_j}^k}\in \mathcal {V}_{t_j}(\bar{x})$. Thus, there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, $\bar{\lambda }_{t_j}\ge 0$, $j=1,\ldots ,q$, not all zero, $\bar{v}_{t_j}\in \mathcal {V}_{t_j}(\bar{x})$ and $t_j\in T_1(\bar{x})$, $j=1,\ldots ,p$, such that $\sum _{i=1}^l\bar{\mu }_i+\sum _{j=1}^q\bar{\lambda }_{t_j}=1$,

$$\begin{aligned} 0=\sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum _{j=1}^q\bar{\lambda }_{t_j}\nabla _xg_{t_j}(\bar{x},\bar{v}_{t_j}) \ \ {\mathrm{and }} \ \ \bar{\lambda }_{t_j}g_{t_j}(\bar{x},\bar{v}_{t_j})=0, \quad j=1,\ldots ,p. \end{aligned}$$

Let $\lambda _t=0$, for all $t\in T_2(\bar{x})$. Then there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, $\left( \bar{\lambda }_t\right) _{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\bar{\mu }_i+\sum _{t\in T}\bar{\lambda }_t=1$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)=0 \ \ {\mathrm{and }} \ \ \bar{\lambda }_tg_t(\bar{x},\bar{v}_t)=0, \ t\in T. \end{aligned}$$

Now we assume that the (EMFCQ) holds at $\bar{x}$, that is, there exists $d \in \mathbb {R}^n$ such that $\nabla _x g_t(\bar{x}, \bar{v}_t)^Td<0$ for all $t\in T_1 (\bar{x})$ and all $\bar{v}_t\in \mathcal {V}_t(\bar{x})$. Then $\sum _{i=1}^l\bar{\mu }_i>0$. Otherwise, $\bar{\mu }_i=0$, $i=1,\ldots ,l$, then $\sum _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)=0$, which is a contradiction. Thus we may assume that $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero. Let $\hat{\mu }_i=\bar{\mu }_i/\sum _{i=1}^l\bar{\mu }_i$, $i=1,\ldots ,l$ and let $\hat{\lambda }_t=\bar{\lambda }_t/\sum _{i=1}^l\bar{\mu }_i$, $t\in T$. Then, we have $\sum _{i=1}^l\hat{\mu }_i=1$. Moreover, we see that

$$\begin{aligned} 0= & {} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum \limits _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)\\= & {} \sum _{i=1}^l\bar{\mu }_i\left( \sum _{i=1}^l\hat{\mu }_i\nabla f_i(\bar{x})+\sum \limits _{t\in T}\hat{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)\right) \\ \end{aligned}$$

and $0=\bar{\lambda }_{t}g_t(\bar{x},\bar{v}_{t})=\sum _{i=1}^l\bar{\mu }_i\hat{\lambda }_{t}g_{t}(\bar{x},\bar{v}_{t}), \ t\in T$. Since $\sum _{i=1}^l\bar{\mu }_i>0$, we have $\sum _{i=1}^l\hat{\mu }_i\nabla f_i(\bar{x})+\sum _{t\in T}\hat{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)=0$ and $\hat{\lambda }_{t}g_{t}(\bar{x},\bar{v}_{t})=0$, $t\in T$. So, (2) holds. $\square$

Now we present an example which illustrates Theorem 3.1.

Example 1

Consider the following uncertain semi-infinite multiobjective problem, with $n=1$ and $l=2$, (RSIMP):

where the data $v_t$ is uncertain, $v_t\in \mathcal {V}_t=[-t+2,t+2]$ and $T=[0,1]$. The robust counterpart is

Let $f_1(x)=x$, $f_2(x)=x^2-5x$ and $g_t(x,v_t)=tx^2-2v_tx$. The multifunction $\mathcal {V}$, defined by for any $t\in T$, $\mathcal {V}(t)=\{v_t \ | \ v_t\in \mathcal {V}_t\}$, is upper semi-continuous and the conditions (A3) and (A4) hold for $g_t(x,v_t)$. Moreover, we can easily see that $\mathrm{gph}\mathcal {V}=\mathrm{co}\{(0,2),(1,1),(1,3)\}$. So, it can be verified that the feasible set is [0, 2] and the set of all weakly robust efficient solutions is [0, 2]. Let $\bar{x}=2$, $\bar{\mu }_1=0$ and $\bar{\mu }_2=\frac{2}{3}$. Let $\bar{\lambda }\in \mathbb {R}_+^{(T)}$ be such that $\bar{\lambda }_t=0$ for all $t\in [0,1)$ and $\bar{\lambda }_1=\frac{1}{3}$. Let $\bar{v}_t\in [-t+2,t+2]$ for all $t\in [0,1)$ and $\bar{v}_1=1$. Then, we see that

$$\begin{aligned} \bar{\mu }_1\nabla f_1(\bar{x})+\bar{\mu }_2\nabla f_2(\bar{x})+\sum _{t\in T}\bar{\lambda }_t\nabla _x g_t(\bar{x},\bar{v}_t)=0 \ \ {\mathrm{and }} \ \ \bar{\lambda }_tg_t(\bar{x},\bar{v}_t)=0, \ t\in T. \end{aligned}$$

Moreover, $T_1(\bar{x})=\{1\}$ and $\mathcal {V}_1(\bar{x})=\{1\}$. For $d=-1$, $\nabla _xg_1(\bar{x},1)^Td=-2$. So, the (EMFCQ) holds at $\bar{x}$. Let $\hat{\mu }_i=\frac{\bar{\mu }_i}{\bar{\mu }_1+\bar{\mu }_2}$, $i=1,2$, and $\hat{\lambda }_t=\frac{\bar{\lambda }_t}{\bar{\mu }_1+\bar{\mu }_2}$, $t\in [0,1]$. Then $\hat{\mu }_1=0$, $\hat{\mu }_2=1$, $\hat{\lambda }_1=\frac{1}{2}$ and $\hat{\lambda }_t=0$ for all $t\in [0,1)$. So, we have

$$\begin{aligned} \hat{\mu }_1\nabla f_1(\bar{x})+\hat{\mu }_2\nabla f_2(\bar{x})+\sum _{t\in T}\hat{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)=0 \ \ {\mathrm{and }} \ \ \hat{\lambda }_tg_t(\bar{x},\bar{v}_t)=0, \ t\in T. \end{aligned}$$

So, Theorem 1 holds.

From Theorem 1, we get the following corollary which will be used to establish the dual problem to ${\mathrm{(RSIMP)}}$.

Corollary 1

(Robust Necessary Optimality Condition) Let $\bar{x}$ be a weakly robust efficient solution of (RSIMP). Suppose that $g_t(x,\cdot )$ is concave on $\mathcal {V}_t$, for each $x\in \mathbb {R}^n$ and for each $t\in T$. Assume that the (EMFCQ) holds at $\bar{x}$. Then there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero, $\left( \bar{\lambda }_t^i\right) _{t\in T}\in \mathbb {R}_+^{(T)}$, $i=1,\ldots ,l$, and $\bar{v}_t\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\bar{\mu }_i=1$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\left( \nabla f_i(\bar{x})+\sum _{t\in T}\bar{\lambda }_t^i\nabla _xg_t(\bar{x},\bar{v}_t)\right) =0 \ \ {\mathrm{and }} \ \ \bar{\lambda }_t^ig_t(\bar{x},\bar{v}_t)=0, \ i=1,\ldots ,l, \ t\in T. \end{aligned}$$

(9)

Proof

Let $\bar{x}$ be a weakly robust efficient solution of (RSIMP). Assume that the (EMFCQ) holds at $\bar{x}$. Then, by Theorem 1, there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero, $(\bar{\lambda }_t)_{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\bar{\mu }_i=1$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)=0 \ \ {\mathrm{and }} \ \ \bar{\lambda }_tg_t(\bar{x},\bar{v}_t)=0, \ t\in T. \end{aligned}$$

(10)

Let $I_1=\{i\in I \ | \ \mu _i>0\}$ and $I_2=I\backslash I_1$, where $I=\{1,\ldots ,l\}$. Define

$$\begin{aligned} \bar{\lambda }_t^i=\left\{ \begin{array}{ll} \bar{\lambda }_t\bar{\mu }_i, &{} \quad i\in I_1, \\ 0, &{} \quad i\in I_2. \end{array} \right. \end{aligned}$$

Then we see that $\bar{\lambda }^i=(\bar{\lambda }_t^i)_{t\in T}\in \mathbb {R}_+^{(T)}$, $i=1,\ldots ,l$, and $\sum _{i=1}^l\bar{\lambda }_t^i=\bar{\lambda }_t$. Moreover, from (10), we see that

$$\begin{aligned} 0=\sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum \limits _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)= & {} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum \limits _{i=1}^l\bar{\mu }_i\sum \limits _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)\\= & {} \sum _{i=1}^l\bar{\mu }_i\left( \nabla f_i(\bar{x})+\sum \limits _{t\in T}\bar{\lambda }_t^i\nabla _xg_t(\bar{x},\bar{v}_t)\right) . \end{aligned}$$

So, (9) holds. Also, for $i\in I$,

$$\begin{aligned} \bar{\lambda }_t^ig_t(\bar{x},\bar{v}_t)=\bar{\mu }_i\bar{\lambda }_tg_t(\bar{x},\bar{v}_t) \end{aligned}$$

So, $\bar{\lambda }_t^ig_t(\bar{x},\bar{v}_t)=0$, $i=1,\ldots ,l$, $t\in T$. Thus, (13) holds. $\square$

Now we give a robust sufficient optimality theorem for a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$.

Theorem 2

(Robust Sufficient Optimality Condition) Let $\bar{x}\in F$. Suppose that $f_i : \mathbb {R}^n\rightarrow \mathbb {R}$, $i=1,\ldots ,l$, are convex functions. Let $g_t(\cdot ,v_t)$ be convex on $\mathbb {R}^n$, for each $v_t\in \mathcal {V}_t$ and for each $t\in T$. Suppose that there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero, $\left( \bar{\lambda }_t\right) _{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t\in \mathcal {V}_t$, $t\in T$, such that

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})+\sum _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)=0 \ \ {\text {and}} \ \ \bar{\lambda }_tg_t(\bar{x},\bar{v}_t)=0, \quad t\in T. \end{aligned}$$

Then $\bar{x}$ is a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$.

Proof

Suppose that $\bar{x}$ is not a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$. Then there exists $x^*\in F$ such that $f_i(x^*)<f_i(\bar{x})$, $i=1,\ldots ,l$. By the convexity assumption of $f_i$, $i=1,\ldots ,l$, and $g_t(\cdot ,v_t)$, $t\in T$, for any $x\in \mathbb {R}^n$,

$$\begin{aligned}&f_i (x)-f_i (\bar{x})\ge \nabla f_i(\bar{x})^T(x-\bar{x}), \quad i = 1, \ldots , l,\\&g_t(x, \bar{v}_t)-g_t (\bar{x},\bar{v}_t)\ge \nabla _x g_t (\bar{x},\bar{v}_t)^T(x-\bar{x}), \quad t\in T. \end{aligned}$$

So, by assumption, for any $x\in F$,

$$\begin{aligned}&\bar{\mu }_if_i (x)-\bar{\mu }_if_i (\bar{x})\ge \bar{\mu }_i\nabla f_i(\bar{x})^T(x-\bar{x}), \quad i = 1, \ldots , l,\\&0\ge \bar{\lambda }_tg_t(x, \bar{v}_t)=\bar{\lambda }_tg_t(x, \bar{v}_t)-\bar{\lambda }_tg_t (\bar{x},\bar{v}_t)\ge \bar{\lambda }_t\nabla _x g_t (\bar{x},\bar{v}_t)^T(x-\bar{x}), \ t\in T. \end{aligned}$$

Since $\sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})=-\sum _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_if_i (x)-\sum _{i=1}^l\bar{\mu }_if_i (\bar{x})\ge \sum _{i=1}^l\bar{\mu }_i\nabla f_i(\bar{x})^T(x-\bar{x})=-\sum _{t\in T}\bar{\lambda }_t\nabla _xg_t(\bar{x},\bar{v}_t)^T(x-\bar{x})\ge 0. \end{aligned}$$

So, for any $x\in F$, $\sum _{i=1}^l\bar{\mu }_if_i (x)\ge \sum _{i=1}^l\bar{\mu }_if_i (\bar{x})$, which contradicts the fact $f_i(x^*)<f_i(\bar{x})$, $i=1,\ldots ,l$. Thus, $\bar{x}$ is a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$. $\square$

Remark 3

Theorem 2 can be extended when the following generalized convexity conditions for $f_i(\cdot )$ and $g_t(\cdot ,\bar{v}_t)$ holds at $\bar{x}$ and $\bar{v}_t$; there exists a function $\eta :\mathbb {R}^n\times \mathbb {R}^n\rightarrow \mathbb {R}^n$ such that for any $x\in \mathbb {R}^n$,

$$\begin{aligned}&f_i(x)-f_i(\bar{x}) \ge \nabla f_i(\bar{x})^T\eta (x,\bar{x}), \ i=1,\ldots , l \\&{\text {and }} g_t(x,v_t)-g_t(\bar{x},\bar{v}_t) \ge \nabla _x g_t(\bar{x},\bar{v}_t)^T\eta (x,\bar{x}), \ t\in T. \end{aligned}$$

3.1 Applications to linear semi-infinite optimization problem

Now we apply Theorem 3.1 to robust multiobjective linear semi-infinite programming problems under uncertainty, which was studied by Goberna et al. (2013a, b, 2014).

Consider the following linear semi-infinite multiobjective optimization problem in the absence of data uncertainty:

where $c_i$, $i=1,\ldots ,l$, $a_t\in \mathbb {R}^n$ and $b_t\in \mathbb {R}$, $t\in T$. The semi-infinite optimization problem in the face of data uncertainty in the linear constraints can be captured by the problem

where $a_t$ and $b_t$ are uncertain parameters, and $(a_t,b_t)$ belongs to the set $\mathcal {V}_t\subset \mathbb {R}^{n+1}$ for all $t\in T$. The set-valued mapping $\mathcal {V}:T\rightrightarrows \mathbb {R}^{n+1}$ is defined as $\mathcal {V}(t):=\mathcal {V}_t$ for all $t\in T$.

The robust counterpart of (ULSIMP) is

Clearly, $F^L:=\{x\in \mathbb {R}^n \ | \ a_t^Tx\ge b_t,\ \forall (a_t,b_t)\in \mathcal {V}_t, \ \forall t\in T\}$ is the feasible set of (RLSIMP).

The specification of (EMFCQ) in Definition 1 to $F^L$ is as follows:

Definition 2

We say that the extended linear Mangasarian–Fromovitz constraint qualification (ELMFCQ) is satisfied at $x \in F^L$ iff

$$\begin{aligned} \exists d \in \mathbb {R}^n {\text { s.t. }} a_t^Td > 0\; \;\forall t \in T_1 (x), \quad \forall (a_t,b_t) \in \mathcal {V}_t(x). \end{aligned}$$

Remark 4

Goberna et al. (2014) have established characterizations of robust solutions under the locally Farkas–Minkowski constraint qualification ${\mathrm{(LFMCQ)}}$ at $\bar{x}\in F^L$, that is, $D(F^L;\bar{x})^+=A(\bar{x})$, where

$$\begin{aligned}&A(\bar{x}):=\mathrm{cone}\left\{ a \ | \ (a,b)\in \bigcup _{t\in T}\mathcal {V}_t \ {\mathrm{and }} \ a^T\bar{x}=b\right\} \subset \mathbb {R}^n\\&{\text {and }} D(F^L;\bar{x}):=\left\{ d\in \mathbb {R}^n \ | \ \exists \eta >0 \ {\mathrm{s.t.}} \ \bar{x}+\eta d\in F^L\right\} . \end{aligned}$$

In linear programming with finite uncertain linear constraints, generally, even if the extended $\mathrm{(ELMFCQ)}$ does not hold, $\mathrm{(LFMCQ)}$ may hold as $\mathrm{(LFMCQ)}$ is weaker than $\mathrm{(ELMFCQ)}$.

The following example shows that the (ELMFCQ) may not hold for a linear programming problem with finite uncertain linear constraints.

Example 2

Consider the following uncertain linear multiobjective optimization problem, with $n=2$ and $l=2$, (ULMP):

where $(a_1,b_1)\in \mathcal {V}_1:=[1,2]\times [1,2]\times \{-1\}$, $(a_2,b_2)\in \mathcal {V}_2:=[-3,-2]\times [-3,-2]\times \{1\}$. The robust counterpart of (ULMP) is

Let $h_i(x):=\inf _{(a_i,b_i)\in \mathcal {V}_i}\left( a_i^Tx-b_i\right)$, $i=1,2$. Then we can easily see that

$$\begin{aligned} h_1(x_1,x_2)= & {} \left\{ \begin{array}{ll} x_1+2x_2+1 &{} (x_1>0),\\ 2x_1+2x_2+1 &{} \left( -\frac{1}{2}\le x_1\le 0\right) , \\ 2x_1+x_2+1 &{} \left( x_1<-\frac{1}{2}\right) \end{array} \right. \\ {\text { and }}\;\; h_2(x_1,x_2)= & {} \left\{ \begin{array}{ll} -3x_1-2x_2-1 &{} (x_1>0),\\ -2x_1-2x_2-1 &{} \left( -\frac{1}{2}\le x_1\le 0\right) , \\ -2x_1-3x_2-1 &{} \left( x_1<-\frac{1}{2}\right) . \end{array} \right. \end{aligned}$$

So, the feasible set of (RLMP) is as follows:

$$\begin{aligned}&\left\{ x\in \mathbb {R}^2 \ | \ h_1(x)\ge 0, \ h_2(x)\ge 0\right\} =\left\{ (x_1,x_2)\in \mathbb {R}^2 \ | \ 2x_1+2x_2+1=0, \ -\textstyle {\frac{1}{2}}\le x_1\le 0 \right\} \\&\quad =\mathrm{co}\left\{ \left( \textstyle -\frac{1}{2},0\right) ,\left( 0,\textstyle -\frac{1}{2}\right) \right\} . \end{aligned}$$

Also, it can be verified that the set of all weakly robust efficient solutions of (RLMP) is $\mathrm{co}\{(\textstyle -\frac{1}{2},0),(0,\textstyle -\frac{1}{2})\}$. Let $(\bar{x}_1,\bar{x}_2)=(\textstyle -\frac{1}{4},\textstyle -\frac{1}{4})$. Then $T_1(\bar{x}_1,\bar{x}_2)=\{1,2\}$, $\mathcal {V}_1(\bar{x}_1,\bar{x}_2)=\{2\}\times \{2\}\times \{-1\}$ and $\mathcal {V}_2(\bar{x}_1,\bar{x}_2)=\{-2\}\times \{-2\}\times \{1\}$. Let $d=(d_1,d_2)\in \mathbb {R}^2$ be any given vector and let $(a_1,b_1)\in \mathcal {V}_1(\bar{x}_1,\bar{x}_2)$ and $(a_2,b_2)\in \mathcal {V}_2(\bar{x}_1,\bar{x}_2)$. Then the following inequalities do not hold simultaneously: $a_1^Td=2d_1+2d_2>0$ and $a_2^Td=-2d_1-2d_2>0$. So, (ELMFCQ) does not hold at $\bar{x}$. Otherwise, (RLMP) satisfies (LFMCQ) because its characteristic cone is finitely generated (by the extreme points of $\mathcal {V}_1$ and $\mathcal {V}_2$), so that the Farkas–Minkowski property holds (For details see Goberna and Lopez 1998).

Theorem 3

(Robust Necessary Optimality Condition) Assume that the conditions (A1) and (A2) hold. Let $\bar{x}$ be a weakly robust efficient solution of (ULSIMP). Then there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, $\left( \bar{\lambda }_t\right) _{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t=(\bar{a}_t,\bar{b}_t)\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\bar{\mu }_i+\sum _{t\in T}\bar{\lambda }_t=1$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_ic_i -\sum _{t\in T}\bar{\lambda }_t\bar{a}_t=0 \ \ {\mathrm{and }} \ \ \bar{\lambda }_t(\bar{a}_t^T\bar{x}-\bar{b}_t)=0, \ t\in T. \end{aligned}$$

(11)

Moreover, if we further assume that the (ELMFCQ) holds at $\bar{x}$, then there exist $\hat{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero, $(\hat{\lambda }_t)_{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t=(\bar{a}_t,\bar{b}_t)\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\hat{\mu }_i=1$,

$$\begin{aligned} \sum _{i=1}^l\hat{\mu }_ic_i -\sum _{t\in T}\hat{\lambda }_t\bar{a}_t=0 \ \ {\mathrm{and }} \ \ \hat{\lambda }_t(\bar{a}_t^T\bar{x}-\bar{b}_t)=0, \ t\in T. \end{aligned}$$

(12)

Proof

Let $f_i(x)=c_i^Tx$ and $g_t(x,v_t)=b_t-a_t^Tx$, where $v_t=(a_t,b_t)\in \mathcal {V}_t\subset \mathbb {R}^{n+1}$, $t\in T$. Clearly, $g_t(x, \cdot )$ is affine. So, we only need to show that (A3) and (A4) hold. Let $x_{n_k}\rightarrow \bar{x}\in \mathbb {R}^n$. Since T is compact, there exists some subsequence $\{t_{n_k}\}$ such that $t_{n_k}\rightarrow t^*\in T$. Since, by Remark 1, $\mathcal {V}(T)$ is compact, we may assume that $v_{t_{n_k}}=(a_{t_{n_k}},b_{t_{n_k}})\rightarrow v_{t^*}=(\bar{a}_{t^*},\bar{b}_{t^*})\in \mathcal {V}_{t^*}$. Moreover, since, by Remark 1, ${\mathrm{gph\mathcal {V}}}$ is closed and $(t_{n_k}, v_{t_{n_k}})\rightarrow (t^*,v_{t^*})$, $(t^*,v_{t^*})\in {\mathrm{gph}}\mathcal {V}$, that is, $v_{t^*}\in \mathcal {V}_{t^*}$. So, we have

$$\begin{aligned} \lim _{k\rightarrow \infty }g_{t_{n_k}}\left( x_{n_k},v_{t_{n_k}}\right) =\lim _{k\rightarrow \infty }\left( b_{t_{n_k}}-a_{t_{n_k}}^Tx_{n_k}\right) =\bar{b}_{t^*}-{\bar{a}_{t^*}}^T\bar{x}. \end{aligned}$$

Hence, (A3) holds. Moreover, since

$$\begin{aligned} \lim _{k\rightarrow \infty }\nabla _x g_{t_{n_k}}\left( x_{n_k},v_{t_{n_k}}\right) =\lim _{k\rightarrow \infty }\left( -a_{t_{n_k}}\right) =-{\bar{a}_{t^*}}, \end{aligned}$$

(A4) holds. Thus, by Theorem 1, (11) and (12) hold. $\square$

Remark 5

We note that the ${\mathrm{(ELMFCQ)}}$ could be replace by ${\mathrm{(LFMCQ)}}$ in Theorem 3. In fact, Goberna et al. (2014) have established a characterization for robust weakly efficient solutions for ${\mathrm{(ULSIMP)}}$ under the ${\mathrm{(LFMCQ)}}$ [For detail, see Theorem 3.3 in Goberna et al. (2014)], and our Theorem 3 could be obtained by applying their result under the ${\mathrm{(LFMCQ)}}$.

From Corollary 1, we get the following corollary which will be used to establish the dual problem to (RLSIMP).

Corollary 2

Assume that the conditions (A1) and (A2) hold. Let $\bar{x}$ be a weakly robust efficient solution of (ULSIMP). Assume that the (ELMFCQ) holds at $\bar{x}$. Then there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero, $\left( \bar{\lambda }_t^i\right) _{t\in T}\in \mathbb {R}_+^{(T)}$, $i=1,\ldots ,l$, and $(\bar{a}_t,\bar{b}_t)\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\bar{\mu }_i=1$,

$$\begin{aligned}&\sum _{i=1}^l\bar{\mu }_i\left( c_i -\sum \limits _{t\in T} \bar{\lambda }_t^i\bar{a}_t\right) =0\\&{\mathrm{and }}\; \bar{\lambda }_t^i\left( \bar{a}_t^T\bar{x}-\bar{b}_t\right) =0, \ i=1,\ldots ,l, \ t\in T. \end{aligned}$$

The following example illustrates how Theorem 3 applies.

Example 3

Consider the following uncertain linear semi-infinite problem, with $n=2$ and $l=2$, (ULSIMP):

where the data $a_t$, $b_t$ are uncertain, $a_t\in [-1-2t,-1+2t]\times [-1-2t,-1+2t]$, $b_t\equiv -1$ and $T=[0,1]$. The robust counterpart is

As the boxes $[-1-2t,-1 + 2t]^2$ expand with $t\in [0, 1]$, their union is $[-3, 1]^2$ and the feasible set of ${\mathrm{(RLSIMP)}}$ is as follows:

$$\begin{aligned} F^L=\{(x_1,x_2) \ | \ -3x_1-3x_2\ge -1, \ -3x_1+x_2\ge -1, \ x_1-3x_2\ge -1, \ x_1+x_2\ge -1\} \end{aligned}$$

and the set of all weakly robust efficient solution is $\mathrm{co}\{(-1,0),(0,-1)\}$. Let $(\bar{x}_1,\bar{x}_2)=(-1,0)$, $\bar{\mu }_1=\textstyle \frac{1}{3}$ and $\bar{\mu }_2=\textstyle \frac{1}{3}$. Let $(\bar{\lambda }_t)_{t\in T}\in \mathbb {R}_+^{(T)}$ be such that $\bar{\lambda }_t=0$ for all $t\in [0,1)$ and $\bar{\lambda }_1=\textstyle \frac{1}{3}$. Let $(\bar{a}_t,\bar{b}_t)\in [-1-2t,-1+2t]\times [-1-2t,-1+2t]\times \{-1\}$ for all $t\in [0,1)$ and $(\bar{a}_1,\bar{b}_1)=((1,1),-1)$. Then we see that

$$\begin{aligned} \bar{\mu }_1(1,0)+\bar{\mu }_2(0,1)-\sum _{t\in T}\bar{\lambda }_t\bar{a}_t=0 \ \ {\mathrm{and }} \ \ \bar{\lambda }_t(\bar{a}_t\bar{x}-\bar{b}_t)=0, \ t\in T. \end{aligned}$$

Moreover, $T_1(\bar{x}_1,\bar{x}_2)=\{1\}$ and $\mathcal {V}_1(\bar{x})=\{1\}\times [-3,1]\times \{-1\}$. Let $(d_1,d_2)=(1,0)$ and $(\bar{a}_t,\bar{b}_t)=((1,0),-1)$. Then $\bar{a}_t^Td > 0$ $\forall (\bar{a}_t,\bar{b}_t)\in \mathcal {V}_t(\bar{x}_1,\bar{x}_2)$, $\forall t \in T_1 (\bar{x}_1,\bar{x}_2)$. So, the (ELMFCQ) holds at $(\bar{x}_1,\bar{x}_2)$. Let $\hat{\mu }_i=\frac{\bar{\mu }_i}{\bar{\mu }_1+\bar{\mu }_2}$, $i=1,2$ and $\hat{\lambda }_t=\frac{\bar{\lambda }_t}{\bar{\mu }_1+\bar{\mu }_2}$, $t\in [0,1]$. Then $\hat{\mu }_1=\frac{1}{2}$, $\hat{\mu }_2=\frac{1}{2}$, $\hat{\lambda }_1=\frac{1}{2}$ and $\hat{\lambda }_t=0$ for all $t\in [0,1)$. So, we have

$$\begin{aligned} \hat{\mu }_1(1,0)+\hat{\mu }_2(0,1)-\sum _{t\in T}\hat{\lambda }_t\bar{a}_t=0 \ \ {\mathrm{and }} \ \ \hat{\lambda }_t(\bar{a}_t\bar{x}-\bar{b}_t)=0, \ t\in T. \end{aligned}$$

Thus, Theorem 3 is satisfied.

Theorem 4

(Robust Sufficient Optimality Condition) Let $\bar{x}\in F$. Suppose that there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero, $\left( \bar{\lambda }_t\right) _{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t=(\bar{a}_t,\bar{b}_t)\in \mathcal {V}_t$, $t\in T$, such that

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_ic_i -\sum _{t\in T}\bar{\lambda }_t\bar{a}_t=0 \ \ {\mathrm{and }} \ \ \bar{\lambda }_t(\bar{a}_t^T\bar{x}-\bar{b}_t)=0, \ t\in T. \end{aligned}$$

Then $\bar{x}$ is a weakly robust efficient solution of ${\mathrm{(RLSIMP)}}$.

Proof

Suppose that $\bar{x}$ is not a weakly robust efficient solution of ${\mathrm{(RLSIMP)}}$. Then there exists $x^*\in F$ such that $c_i^Tx^*< c_i^T\bar{x}$, $i=1,\ldots ,l$. So, by assumption, there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, $\left( \bar{\lambda }_t\right) _{t\in T}\in \mathbb {R}_+^{(T)}$ and $\bar{v}_t=(\bar{a}_t,\bar{b}_t)\in \mathcal {V}_t$, $t\in T$, such that

$$\begin{aligned}&\sum _{i=1}^l\bar{\mu }_ic_i^Tx^*-\sum _{i=1}^l\bar{\mu }_ic_i^T\bar{x}=\sum _{i=1}^l\bar{\mu }_ic_i^T(x^*-\bar{x})\\&\quad =\sum \limits _{t\in T}\bar{\lambda }_t\bar{a}_t^T(x^*-\bar{x})\\&\quad =\sum \limits _{t\in T}\bar{\lambda }_t\bar{a}_t^Tx^*-\sum \limits _{t\in T}\bar{\lambda }_t\bar{a}_t^T\bar{x}\\&\quad =\sum \limits _{t\in T}\bar{\lambda }_t\bar{a}_t^Tx^*-\sum \limits _{t\in T}\bar{\lambda }_t\bar{b}_t\\&\quad \ge \sum \limits _{t\in T}\bar{\lambda }_t\bar{b}_t-\sum \limits _{t\in T}\bar{\lambda }_t\bar{b}_t\\&\quad =0 \end{aligned}$$

This contradicts the fact that $c_i^Tx^*< c_i^T\bar{x}$, $i=1,\ldots ,l$. Thus, $\bar{x}$ is a weakly robust efficient solution of ${\mathrm{(RLSIMP)}}$. $\square$

4 Duality theorems

Now, we formulate a Wolfe type dual problem ${\mathrm{(RSIMD)}}$ for ${\mathrm{(RSIMP)}}$ as follows:

Denote that $F_D$ is the robust feasible set of ${\mathrm{(RSIMD)}}$. Now we define weakly robust efficient solutions of ${\mathrm{(RSIMD)}}$ in ways similar to the case of ${\mathrm{(RSIMP)}}$. $(\bar{y},\bar{v},\bar{\mu },\bar{\lambda }) \in F_D$ is called a weakly robust efficient solution of ${\mathrm{(RSIMD)}}$ if there does not exist a robust feasible solution $(y,v,\mu ,\lambda )$ of ${\mathrm{(RSIMD)}}$ such that

$$\begin{aligned} f_i (y)+\sum _{t\in T}\lambda _t^ig_t(y,v_t) > f_i (\bar{y})+\sum _{t\in T}\bar{\lambda }_t^ig_t(\bar{y},\bar{v}_t), \ i = 1, \ldots , l. \end{aligned}$$

Now we establish a weak duality result which holds between ${\mathrm{(RSIMP)}}$ and ${\mathrm{(RSIMD)}}$.

Theorem 5

(Weak Duality) Let $f_i : \mathbb {R}^n\rightarrow \mathbb {R}$, $i=1,\ldots ,l$, be convex functions and let $g_t(\cdot ,v_t)$ be convex on $\mathbb {R}^n$, for each $v_t\in \mathcal {V}_t$ and for each $t\in T$. Let x and $(y,v,\mu ,\lambda )$ be feasible solutions of ${\mathrm{(RSIMP)}}$ and ${\mathrm{(RSIMD)}}$, respectively. Then the following inequalities cannot hold simultaneously:

$$\begin{aligned} f_i(x)< f_i(y)+\sum _{t\in T}\lambda _t^ig_t(x,v_t), \ i=1,\ldots ,l. \end{aligned}$$

Proof

Suppose to the contrary that there exist a feasible x of ${\mathrm{(RSIMP)}}$ and a feasible $(y,v,\mu ,\lambda )$ of ${\mathrm{(RSIMD)}}$ such that

$$\begin{aligned} f_i(x)<f_i(y)+\sum _{t\in T}\lambda _t^ig_t(y,v_t), \end{aligned}$$

for all $i=1,\ldots ,l$. Let $\mu _i\ge 0$, $i=1,\ldots ,l$. Then, by multiplying both sides of above inequalities by $\mu _i$, we have

$$\begin{aligned} \sum _{i=1}^l\mu _if_i(x)<\sum _{i=1}^l\mu _if_i(y)+\sum _{i=1}^l\mu _i\sum _{t\in T}\lambda _t^ig_t(y,v_t). \end{aligned}$$

(13)

By the convexity assumption on $f_i$, $i=1,\ldots ,l$, and $g_t(\cdot ,v_t)$, $t\in T$,

$$\begin{aligned}&\sum _{i=1}^l\mu _if_i (x)-\sum _{i=1}^l\mu _if_i (y)\ge \sum _{i=1}^l\mu _i\nabla f_i(y)^T(x-y)\\&\quad {\mathrm{and }} \ \ \quad \sum _{t\in T}\lambda _t^ig_t(x, v_t)-\sum _{t\in T}\lambda _t^ig_t (y,v_t)\ge \sum _{t\in T}\lambda _t^i\nabla _x g_t (y,v_t)^T(x-y). \end{aligned}$$

Since $\sum _{i=1}^l \mu _i\nabla f_i(y)=-\sum _{i=1}^l \mu _i\sum _{t\in T}\lambda _t^i\nabla _xg_t(y, v_t)$, we see that

$$\begin{aligned}&\sum _{i=1}^l\mu _if_i (x)-\sum _{i=1}^l\mu _i\left( f_i (y)+\sum _{t\in T}\lambda _t^ig_t(y,v_t)\right) \\&\quad \ge \sum _{i=1}^l\mu _i\nabla f_i(y)^T(x-y)-\sum _{i=1}^l\mu _i\sum _{t\in T}\lambda _t^ig_t(y,v_t)\\&\quad =-\sum _{i=1}^l\mu _i\sum _{t\in T}\lambda _t^i\nabla _x g_t (y,v_t)^T(x-y)-\sum _{i=1}^l\mu _i\sum _{t\in T}\lambda _t^ig_t(y,v_t)\\&\quad \ge -\sum _{i=1}^l\mu _i\left( \sum _{t\in T}\lambda _t^ig_t (x,v_t)-\sum _{t\in T}\lambda _t^ig_t(y, v_t)\right) -\sum _{i=1}^l\mu _i\sum _{t\in T}\lambda _t^ig_t(y,v_t)\\&\quad =-\sum _{i=1}^l\mu _i\sum _{t\in T}\lambda _t^ig_t (x,v_t)\\&\quad \ge 0. \end{aligned}$$

So, $\sum _{i=1}^l\mu _if_i (x)\ge \sum _{i=1}^l\mu _i(f_i (y)+\sum _{t\in T}\lambda _t^ig_t(y,v_t))$, which contradicts (13). $\square$

Now we establish a strong duality result which holds between ${\mathrm{(RSIMP)}}$ and ${\mathrm{(RSIMD)}}$.

Theorem 6

(Strong Duality) Let $f_i : \mathbb {R}^n\rightarrow \mathbb {R}$, $i=1,\ldots ,l$, be convex functions and let for each $v_t\in \mathcal {V}_t$ and for each $t\in T$, $g_t(\cdot ,v_t)$ be convex on $\mathbb {R}^n$ and for each $x\in \mathbb {R}^n$ and each $t\in T$, $g_t(x,\cdot )$ be concave on $\mathcal {V}_t$. Assume that the (EMFCQ) holds at $\bar{x}$. Let $\bar{x}$ be a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$. Then there exists $(\bar{\mu },\bar{\lambda },\bar{v})\in \mathbb {R}_+^l\times \mathbb {R}_+^{(T)}\times \mathcal {V}$ such that $(\bar{x},\bar{v},\bar{\mu },\bar{\lambda })$ is a weakly robust efficient solution of ${\mathrm{(RSIMD)}}$.

Proof

Let $\bar{x}$ be a weakly robust efficient solution of ${\mathrm{(RSIMP)}}$. Then, by Corollary 1, there exist $\bar{\mu }_i\ge 0$, $i=1,\ldots ,l$, not all zero, $\left( \bar{\lambda }_t^i\right) _{t\in T}\in \mathbb {R}_+^{(T)}$, $i=1,\ldots ,l$, and $\bar{v}_t\in \mathcal {V}_t$, $t\in T$, such that $\sum _{i=1}^l\bar{\mu }_i=1$,

$$\begin{aligned} \sum _{i=1}^l\bar{\mu }_i\left( \nabla f_i(\bar{x})+\sum _{t\in T}\bar{\lambda }_t^i\nabla _xg_t(\bar{x},\bar{v}_t)\right) =0 \ \ {\mathrm{and }} \ \ \bar{\lambda }_t^ig_t(\bar{x},\bar{v}_t)=0, \quad i=1,\ldots ,l, \ t\in T. \end{aligned}$$

So, $(\bar{x},\bar{v},\bar{\mu },\bar{\lambda })$ is feasible for ${\mathrm{(RSIMD)}}$. Now we suppose that $(\bar{x},\bar{v},\bar{\mu },\bar{\lambda })$ is not a weakly robust efficient solution of ${\mathrm{(RSIMD)}}$. Then there exists a robust feasible solution $(\tilde{y},\tilde{v},\tilde{\mu },\tilde{\lambda })$ of ${\mathrm{(RSIMD)}}$ such that

$$\begin{aligned} f_i (\tilde{y})+\sum _{t\in T}\tilde{\lambda }_t^ig_t(\tilde{y},\tilde{v}_t) > f_i (\bar{x})+\sum _{t\in T}\bar{\lambda }_t^ig_t(\bar{x},\bar{v}_t), \quad i = 1, \ldots , l. \end{aligned}$$

Since $\bar{\lambda }_t^ig_t(\bar{x},\bar{v}_t)=0$,   $i=1,\ldots ,l$, $t\in T$, we have

$$\begin{aligned} f_i (\tilde{y})+\sum _{t\in T}\tilde{\lambda }_t^ig_t(\tilde{y},\tilde{v}_t) > f_i (\bar{x}) , \ i = 1, \ldots , l, \end{aligned}$$

which contradicts weak duality. Thus, $(\bar{x},\bar{v},\bar{\mu },\bar{\lambda })$ is a weakly robust efficient solution of ${\mathrm{(RSIMD)}}$. $\square$

We now revisit Example 1 to illustrate Theorems 5 and 6.

Example 4

Consider the following uncertain semi-infinite multiobjective problem, with $n=1$ and $l=2$, (USIMP):

where the data $v_t$ is uncertain, $v_t\in \mathcal {V}_t=[-t+2,t+2]$ and $t=[0,1]$. The robust counterpart is

Let $f_1$, $f_2$ and $g_t$ be as in Example 1. Recall that the feasible set and the set of weakly robust efficient solutions are $F=[0,2]$.

Now we consider a Wolfe type dual problem ${\mathrm{(RSIMD)}}$ for ${\mathrm{(RSIMP)}}$ as follows:

where $v_t\in \mathcal {V}_t=[-t+2,t+2]$ and $t=[0,1]$. Let x and $(y,v,\mu ,\lambda )$ be feasible solutions of ${\mathrm{(RSIMP)}}$ and ${\mathrm{(RSIMD)}}$, respectively. Then, we see that

$$\begin{aligned} x-y= & {} f_1 (x)-f_1 (y)\ge f_1'(y)(x-y)=x-y,\\ x^2-5x-(y^2-5y)= & {} f_2 (x)-f_2 (y)\ge f_2'(y)(x-y)=(2y-5)(x-y) \end{aligned}$$

and for all $t\in T$,

$$\begin{aligned} tx^2-2v_tx-\left( ty^2-2v_ty\right) =g_t(x, v_t)-g_t (y,v_t)\ge g_t'(y,v_t)(x-y)=(2ty-2v_t)(x-y) \end{aligned}$$

Let $\mu _i\ge 0$, $i=1,2$ with $\mu _1+\mu _2=1$ and $(\lambda _t^i)_{t\in T}\in \mathbb {R}_+^{(T)}$, $i=1,2$. Then we have

$$\begin{aligned}&\mu _1(x-y)\ge \mu _1(x-y), \ \mu _2\left( x^2-5x-(y^2-5y)\right) \ge \mu _2(2y-5)(x-y) \\&\quad {\mathrm{and }} \ \ \quad \mu _i\sum _{t\in T}\lambda _t^i\left( tx^2-2v_tx-(ty^2-2v_ty)\right) \ge \mu _i\sum _{t\in T}\lambda _t^i(2ty-2v_t)(x-y), \ i=1,2. \end{aligned}$$

Since x and $(y,v,\mu ,\lambda )$ are feasible solutions of ${\mathrm{(RSIMP)}}$ and ${\mathrm{(RSIMD)}}$, respectively, $\sum _{t\in T}(tx^2-2v_tx)\le 0$ and $\mu _1+\mu _2(2y-5)+\sum _{i=1}^2\mu _i\sum _{t\in T} \lambda _t^i(2ty-2v_t)=0$. So, we have

$$\begin{aligned}&\mu _1f_1 (x)+\mu _2f_2 (x)-\left( \mu _1f_1 (y)+\mu _2f_2(y)+\sum _{i=1}^2\mu _i\sum _{t\in T}\lambda _t^ig_t(y,v_t)\right) \\&\quad =\mu _1(x-y)+\mu _2\left( x^2-5x-(y^2-5y)\right) -\sum _{i=1}^2\mu _i\sum _{t\in T}\lambda _t^i\left( ty^2-2v_ty\right) \\&\quad \ge \mu _1(x-y)+\mu _2\left( x^2-5x-(y^2-5y)\right) +\sum _{i=1}^2\mu _i\sum _{t\in T}\lambda _t^i\left( tx^2-2v_tx\right) \\&\quad -\sum _{i=1}^2\mu _i\sum _{t\in T}\lambda _t^i\left( ty^2-2v_ty\right) \\&\quad \ge \mu _1(x-y)+\mu _2(2y-5)(x-y)+\sum _{i=1}^2\mu _i\sum _{t\in T}\lambda _t^i(2ty-2v_t)(x-y)\\&\quad =\left( \mu _1+\mu _2(2y-5)+\sum _{i=1}^2\mu _i\sum _{t\in T}\lambda _t^i(2ty-2v_t)\right) (x-y)\\&\quad =0. \end{aligned}$$

Hence, $\mu _1f_1 (x)+\mu _2f_2 (x)\ge \mu _1f_1 (y)+\mu _2f_2(y)+\sum _{i=1}^2\mu _i\sum _{t\in T}\lambda _t^ig_t(y,v_t)$. Thus, Theorem 5 holds.

Let $\bar{x}=2$. Then $\bar{x}$ is a weakly efficient solution of ${\mathrm{(RSIMP)}}$. Moreover, $T_1(\bar{x})=\{1\}$ and $\mathcal {V}_1(\bar{x})=\{1\}$. We already checked that the (EMFCQ) holds at $\bar{x}$ in Example 1. Let $\bar{v}_1=1$, $\bar{\mu }_1=0$, $\bar{\mu }_2=1$, $\bar{\lambda }_1^i=\textstyle \frac{1}{2}$ and $\bar{\lambda }_t^i=0$, $t\in [0,1)$, $i=1,2$. Since $\bar{\mu }_1\nabla f_1(\bar{x})+\bar{\mu }_2\nabla f_2(\bar{x})+\sum _{i=1}^2\bar{\mu }_i\sum _{t\in T}\bar{\lambda }_t^i\nabla _xg_t(\bar{x},\bar{v}_t)=0$, $\mu _1+\mu _2=1$, $(\bar{\lambda }_t^i)_{t\in T}\in \mathbb {R}_+^{(T)}$, $i=1,2$, $\bar{v}_t\in \mathcal {V}_t$, $t\in T$, $(\bar{y},\bar{v},\bar{\mu },\bar{\lambda })$ is feasible for ${\mathrm{(RSIMD)}}$. Since $\sum _{t\in T}\bar{\lambda }_t^ig_t(\bar{x},\bar{v}_t)=0$, $i=1,2$, by Theorem 5, the following inequalities cannot hold simultaneously:

$$\begin{aligned} f_i(y)+\sum _{t\in T}\lambda _tg_t(x,v_t)>f_i(\bar{x})=f_i(\bar{x})+\sum _{t\in T}\bar{\lambda }_t^ig_t(\bar{x},\bar{v}_t), \quad i=1,2. \end{aligned}$$

So, $(\bar{y},\bar{v},\bar{\mu },\bar{\lambda })$ is a weakly robust efficient solution of ${\mathrm{(RSIMD)}}$. Thus Theorem 6 holds.

4.1 Duality theorems for linear semi-infinite optimization problem

Now we apply Theorems 5 and 6 to robust multiobjective linear semi-infinite programming problems under uncertainty, which was studied by Goberna et al. (2013a, b, 2014). We formulate a dual problem ${\mathrm{(RLSIMD)}}$ for ${\mathrm{(RLSIMP)}}$ as follows:

Denote that $F_D^L$ is the robust feasible set of ${\mathrm{(RLSIMD)}}$. Now we define weakly robust efficient solutions of ${\mathrm{(RLSIMD)}}$. $(\bar{y}, (\bar{a},\bar{b}),\bar{\mu },\bar{\lambda }) \in F_D^L$ is called a weakly robust efficient solution of ${\mathrm{(RLSIMD)}}$ if there does not exist a robust feasible solution $(y,(a,b),\mu ,\lambda )$ of ${\mathrm{(RLSIMD)}}$ such that

$$\begin{aligned} c_i^Ty+\sum _{t\in T}\lambda _t^i\left( b_t-a_t^Ty\right) > c_i^T\bar{y}+\sum _{t\in T}\bar{\lambda }_t^i\left( \bar{b}_t-\bar{a}_t^T\bar{y}\right) , \ i = 1, \ldots , l. \end{aligned}$$

Now we give a weak duality theorem for ${\mathrm{(RLSIMP)}}$ and ${\mathrm{(RLSIMD)}}$.

Theorem 7

(Weak Duality) Let x and $(y,(a,b),\mu ,\lambda )$ be feasible solutions of ${\mathrm{(RLSIMP)}}$ and ${\mathrm{(RLSIMD)}}$, respectively. Then the following cannot hold:

$$\begin{aligned} {c_i}^Tx<c_i^Ty+\sum _{t\in T}\lambda _t^i\left( b_t-a_t^Ty\right) , \ i=1,\ldots ,l. \end{aligned}$$

Proof

Define functions $f_i : \mathbb {R}^n\rightarrow \mathbb {R}$, $i=1,\ldots ,l$, and $g_t:\mathbb {R}^n\times \mathbb {R}^{n+1}\rightarrow \mathbb {R}$, $t\in T$ by for any $x\in \mathbb {R}^n$ and for all $(a_t,b_t)\in \mathcal {V}_t$, $t\in T$, $f_i(x)={c_i}^Tx$, $i=1,\ldots ,l$ and $g_t(x,a_t,b_t)=b_t-a_t^Tx$, $t\in T$. Then $f_i$, $i=1,\ldots ,l$, is convex, for each $(a_t,b_t)\in \mathcal {V}_t$ and for each $t\in T$, $g_t(\cdot ,(a_t,b_t))$ is convex on $\mathbb {R}^n$ and for each $x\in \mathbb {R}^n$ and each $t\in T$, $g_t(x,\cdot )$ is concave on $\mathcal {V}_t$. So, by Theorem 5, we obtain the desired result. $\square$

Now we give a strong duality theorem for ${\mathrm{(RLSIMP)}}$ and ${\mathrm{(RLSIMD)}}$.

Theorem 8

(Strong Duality) Let $\bar{x}\in F^L$. Assume that the (ELMFCQ) holds at $\bar{x}$. Let $\bar{x}$ be a weakly robust efficient solution of ${\mathrm{(RLSIMP)}}$. Then there exists $(\bar{y},(\bar{a},\bar{b}),\bar{\mu },\bar{\lambda })\in \mathbb {R}^n\times \mathcal {V}\times \mathbb {R}_+^l\times \mathbb {R}_+^{(T)}$ such that $(\bar{y},(\bar{a},\bar{b}), \bar{\mu },\bar{\lambda })$ is a weakly robust efficient solution of ${\mathrm{(RLSIMD)}}$.

Proof

Define functions $f_i : \mathbb {R}^n\rightarrow \mathbb {R}$, $i=1,\ldots ,l$, and $g_t:\mathbb {R}^n\times \mathbb {R}^{n+1}\rightarrow \mathbb {R}$, $t\in T$ by for any $x\in \mathbb {R}^n$ and for all $(a_t,b_t)\in \mathcal {V}_t$, $t\in T$, $f_i(x)={c_i}^Tx$, $i=1,\ldots ,l$ and $g_t(x,a_t,b_t)=b_t-a_t^Tx$, $t\in T$. Then $f_i$, $i=1,\ldots ,l$, is convex, for each $(a_t,b_t)\in \mathcal {V}_t$ and for each $t\in T$, $g_t(\cdot ,(a_t,b_t))$ is convex on $\mathbb {R}^n$ and for each $x\in \mathbb {R}^n$ and each $t\in T$, $g_t(x,\cdot )$ is concave on $\mathcal {V}_t$. So, by Theorem 6, we obtain the desired result. $\square$

We now revisit Example 3 to illustrate Theorems 7 and 8.

Example 5

Consider the following uncertain linear semi-infinite problem, with $n=2$ and $l=2$, (ULSIMP):

where the data $a_t$, $b_t$ are uncertain, $a_t\in [-1-2t,-1+2t]\times [-1-2t,-1+2t]$, $b_t\equiv -1$ and $T=[0,1]$.The robust counterpart is

Recall that the feasible set and the set of weakly robust efficient solutions are

$$\begin{aligned} F^L=\{(x_1,x_2) \ | \ -3x_1-3x_2\ge -1, \ -3x_1+x_2\ge -1, \ x_1-3x_2\ge -1, \ x_1+x_2\ge -1\} \end{aligned}$$

and $\mathrm{co}\{(-1,0),(0,-1)\}$, respectively.

Now we consider a dual problem ${\mathrm{(RLSIMD)}}$ for ${\mathrm{(RLSIMP)}}$ as follows:

where $\mathcal {V}_t=[-1-2t,-1+2t]\times [-1-2t,-1+2t]\times \{-1\}$ and $T=[0,1]$. Let x and $(y,(a,b),\mu ,\lambda )$ be feasible solutions of ${\mathrm{(RLSIMP)}}$ and ${\mathrm{(RLSIMD)}}$, respectively. Since $\sum _{t\in T}a_t^Tx+1\ge 0$ and $(\mu _1,\mu _2)=\mu _1\sum _{t\in T}\lambda _t^1a_t+\mu _2\sum _{t\in T}\lambda _t^2a_t$,

$$\begin{aligned}&\mu _1x_1+\mu _2x_2-\mu _1y_1-\mu _2y_2-\mu _1\sum _{t\in T}\lambda _t^1\left( b_t-a_t^Ty\right) -\mu _2\sum _{t\in T}\lambda _t^2\left( b_t-a_t^Ty\right) \\&\quad =(\mu _1,\mu _2)^T(x-y)-\mu _1\sum _{t\in T}\lambda _t^1\left( b_t-a_t^Ty\right) -\mu _2\sum _{t\in T}\lambda _t^2\left( b_t-a_t^Ty\right) \\&\quad =\left( \mu _1\sum _{t\in T}\lambda _t^1a_t+\mu _2\sum _{t\in T}\lambda _t^2a_t\right) ^T(x-y)-\mu _1\sum _{t\in T}\lambda _t^1\left( b_t-a_t^Ty\right) \\&\quad \quad -\mu _2\sum _{t\in T}\lambda _t^2\left( b_t-a_t^Ty\right) \\&\quad =\left( \mu _1\sum _{t\in T}\lambda _t^1a_t+\mu _2\sum _{t\in T}\lambda _t^2a_t\right) ^Tx-\mu _1\sum _{t\in T}\lambda _t^1b_t-\mu _2\sum _{t\in T}\lambda _t^2b_t \\&\quad \ge \mu _1\sum _{t\in T}\lambda _t^1b_t+\mu _2\sum _{t\in T}\lambda _t^2b_t-\mu _1\sum _{t\in T}\lambda _t^1b_t-\mu _2\sum _{t\in T}\lambda _t^2b_t \\&\quad =0. \end{aligned}$$

Hence $\mu _1x_1+\mu _2x_2\ge \mu _1y_1+\mu _2y_2+\mu _1\sum _{t\in T}\lambda _t^1(b_t-a_t^Ty)+\mu _2\sum _{t\in T}\lambda _t^2(b_t-a_t^Ty)$. Thus, Theorem 7 holds.

Let $(\bar{x}_1,\bar{x}_2)=(-1,0)$. Then $(\bar{x}_1,\bar{x}_2)$ is a weakly efficient solution of ${\mathrm{(RLSIMP)}}$. Moreover, $T_1(\bar{x}_1,\bar{x}_2)=\{1\}$ and $\mathcal {V}_1(\bar{x})=\{1\}\times [-3,1]\times \{-1\}$. We already checked that the $\mathrm{(ELMFCQ)}$ holds at $(\bar{x}_1,\bar{x}_2)$ in Example 3. Let $\bar{a}_1=(1,1)$, $\bar{\mu }_1=1$, $\bar{\mu }_2=0$, $\bar{\lambda }_1^1=1$, $\bar{\lambda }_1^2=0$ and $\bar{\lambda }_t^1=\bar{\lambda }_t^2=0$ for all $t\in [0,1)$. Then $\bar{\mu }_1(1,0)+\bar{\mu }_2(0,1)-\bar{\mu }_1\sum _{t\in T}\bar{\lambda }_t^1a_t-\bar{\mu }_2\sum _{t\in T}\bar{\lambda }_t^2a_t=0$, $\bar{\mu }_1+\bar{\mu }_2=1$, $(\bar{\lambda }_t^i)_{t\in T}\in \mathbb {R}_+^{(T)}$, $i=1,2$ and $(\bar{a}_t,\bar{b}_t)\in \mathcal {V}_t$, $t\in T$. So, $(\bar{x},(\bar{a},\bar{b}), \bar{\mu },\bar{\lambda })$ is feasible for ${\mathrm{(RLSIMD)}}$. Since $\sum _{t\in T}\bar{\lambda }_t^i(\bar{b}_t-{\bar{a}_t}^T\bar{x})=0$, $i=1,2$, by Theorem 7, the following inequality cannot hold:

$$\begin{aligned} c_i^Ty+\sum _{t\in T}\lambda _t^i\left( b_t-a_t^Ty\right) > c_i^T\bar{x}=c_i^T\bar{x}+\sum _{t\in T}\bar{\lambda }_t^i\left( \bar{b}_t-\bar{a}_t^T\bar{y}\right) , \ i=1,2. \end{aligned}$$

So, $(\bar{\mu },\bar{\lambda },(\bar{a},\bar{b}))$ is a weakly robust efficient solution of ${\mathrm{(RLSIMD)}}$. Thus, Theorem 8 holds.

Remark 6

Putting $c_i^Ty+\sum _{t\in T}\lambda _t^i(b_t-a_t^Ty)=r_i$, $i=1,\ldots ,l$ in ${\mathrm{(RLSIMD)}}$, ${\mathrm{(RLSIMD)}}$ becomes the following problem (Goberna et al. 2014):

Goberna et al. (2014) have established weak duality and strong duality theorems for ${\mathrm{(RLSIMP)}}$ and ${\mathrm{(RLSIMD)}}_{1}$.