# Sufficient Conditions and Duality  Theorems for Nondifferentiable Minimax Fractional Programming

Authors: Shun-Chin Ho
Journal: ISRN Mathematical Analysis
 (2011)

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## Abstract

We consider nondifferentiable minimax fractional programming problems
involvingB-(p, r)-invex functions with respect to η and b. Sufficient optimality conditions
and duality results for a class of nondifferentiable minimax fractional programming problems are obtained undr B-(p, r)-invexity assumption on objective and constraint functions.
Parametric duality, Mond-Weir duality, and Wolfe duality problems may be formulated,
and duality results are derived under B-(p, r)-invex functions.


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## Body

## 1. Introduction

Convexity plays an important role in many aspects of mathematical programming including sufficient optimality conditions and duality theorems. In general, we use the invex function to replace convexity on sufficient optimality conditions and duality theorems (see, e.g., [1–6]).Many authors investigated the optimality conditions and duality theorems for minimax (fractional) programming problems. For details, one can consult [1–14]. In particular, Lai et al. [10] have established the theorems of necessary and sufficient optimality conditions for nondifferentiable minimax fractional problem under the conditions of convexity. In [11], Lai and Lee employed the optimality conditions to construct two parameter-free dual models of nondifferentiable minimax fractional programming problem which involve pseudoconvex and quasiconvex functions, and derived weak and strong duality theorems. In the formulation of the dual models in [11] optimality conditions given in [10] are used. Mishra et al. [4] derived a Kuhn-Tucker-type sufficient optimality condition for an optimal solution to the nondifferentiable minimax fractional programming problem and established weak, strong, and converse duality theorems for the problem and its three different forms of dual problems under generalized univexity. Mishra et al. [5, 13] considered the nondifferentiable minimax fractional programming problem and obtain optimality and duality results under generalized α-invexity [5] and generalized α-unvexity [13]. Recently, Antczak [15] defined a new class of functions, named B-(p,r)-invex, which is an extension of invex function. In [1], parametric and nonparametric sufficient optimality conditions and several parametric and parameter-free duality models for the generalized fractional minimax programs are obtained under B-(p,r)-invexity assumption on objective and constraint functions.In this paper, we are inspired to extend the result of Lai et al. [10] to B-(p,r)-invexity and organize this paper as follows. In Section 2 we introduce some basic results. We establish sufficient optimality conditions for nondifferentiable minimax fractional programming problem under B-(p,r)-invex with respect to the same function η and with respect to, not necessarily, the same function b in Section 3. Employing these results, we construct three dual problems in Sections 4–6. Here we investigate weak, strong, and strict converse duality theorems under the framework of B-(p,r)-invex with respect to the same function η and with respect to, not necessarily, the same function b.


## 2. Some Notations and Preliminary Results

Letℝn be the n-dimensional Euclidean space and ℝ+n its nonnegative orthant. Throughout the paper, let X be a nonempty open set of ℝ𝕟.The following definition can be found in [15].Definition 2.1 (see [15]).
Letp and r be any real numbers. The differentiable function f:X→ℝ is said to be (strictly) B-(p,r)-invex with respect to η and b at u∈X on a nonempty set X⊂ℝn if, there exist a function η:X×X→ℝn and a function b:X×X→ℝ+∖{0} such that, for all x∈X, the inequalities
(2.1)1rb(x,u)(er(f(x)-f(u))-1)≥{1p∇f(u)(epη(x,u)-1)(>ifx≠u)forp≠0,r≠0,∇f(u)η(x,u)(>ifx≠u)forp=0,r≠0,b(x,u)(f(x)-f(u))≥{1p∇f(u)(epη(x,u)-1)(>ifx≠u)forp≠0,r=0,∇f(u)η(x,u)(>ifx≠u)forp=0,r=0
hold.f is said to be B-(p,r)-invex (strictly B-(p,r)-invex) with respect to η and b on X if it is B-(p,r)-invex with respect to the same η and b at each u∈X.It should be pointed out that exponentials appearing on the right-hand sides of the above inequalities are understood to be taken componentwise and1=(1,1,…,1)∈ℝn.We consider the following nondifferentiable minimax fractional programming problem:(P)minimizeF(x)=sup⁡⁡y∈Yf(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2subjecttog(x)≤0,
where Y is a compact subset of ℝm,   f(·,·):ℝn×ℝm↦ℝ, h(·,·):ℝn×ℝm↦ℝ, and g(·):ℝn↦ℝp are C1-functions, B and D are n×n positive semidefinite matrices, f(x,y)+(x⊤Bx)1/2≥0, and h(x,y)-(x⊤Dx)1/2>0 for each (x,y) in X∘×Y, where X∘ is the set of feasible solutions of problem (P); that is, X∘={x∈X:g(x)≤0}. This is a nondifferentiable programming problem if either B or D is nonzero. If B and D are null matrices, then problem (P) is a minimax fractional programming problem.For each(x,y)∈ℝn×ℝm define (2.2)ϕ(x,y)=f(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2.
We let(2.3)J={1,2,…,p},J(x)={j∈J∣gj(x)=0},Y(x)={y∈Y∣ϕ(x,y)=sup⁡z∈Y⁡ϕ(x,z)},K(x)={(s,t,y)∈N×R+s×Rms∣1≤s≤n+1,t=(t1,t2,…,ts)∈R+swith∑i=1sti=1,y=(y1,y2,…,ys)withyi∈Y(x),i=1,2,…,s∑i=1s}.Becausef and h are contionuous differentiable and Y is compact subset of ℝm, we see that for each x0∈X∘, Y(x0)≠∅, and for any y¯i∈Y(x0), we have a postive constant (2.4)k0=ϕ(x0,y¯i)=f(x0,y¯i)+(x0⊤Bx0)1/2h(x0,y¯i)-(x0⊤Dx0)1/2.We will use the generalized Schwarz inequality(2.5)x⊤Bv≤(x⊤Bx)1/2(v⊤Bv)1/2,forx,v∈Rn;
the equality holds when Bx=λBv, for some λ≥0.Hence ifv⊤Bv≤1, we have (2.6)x⊤Bv≤(x⊤Bx)1/2.In [10] Lai et al. derived the following necessary conditions for optimality (P).Theorem 2.2 ((necessary conditions) see, [10]).
Letx0 be a (P)-optimal solution and satisfying x0⊤Bx0>0, x0⊤Dx0>0, and ∇gj(x0),j∈J(x0) is linearly independent. Then there exist (s,t*,y¯)∈K(x0), k0∈ℝ+, w,v∈ℝn, and μ*∈ℝ+p such that
(2.7)∑i=1sti*{∇f(x0,y¯i)+Bw-k0(∇h(x0,y¯i)-Dv)}+∇∑j=1pμj*gj(x0)=0,(2.8)f(x0,y¯i)+(x0⊤Bx0)1/2-k0(h(x0,y¯i)-(x0⊤Dx0)1/2)=0,i=1,2,…,s,(2.9)∑j=1pμj*gj(x0)=0,(2.10)ti*≥0,∑i=1sti*=1,(2.11)w⊤Bw≤1,v⊤Dv≤1,x0⊤Bw=(x0⊤Bx0)1/2,x0⊤Dv=(x0⊤Bx0)1/2.It should be noted that both the matricesB and D are positive definite at the solution x0 in the above theorem. If one of x0⊤Bx0 and x0⊤Dx0 is zero, or both B and D are singular at x0, then, for (s,t*,y¯)∈K(x0), we define a set Zy¯(x0) by (2.12)Zy¯(x0)={z∈Rn∣z⊤∇gj(x0)≤0,j∈J(x0),withanyoneofthenextconditions(i)–(iii)holds}.Here conditions (i)–(iii) are given as follows:(i)
ifx0⊤Bx0>0 and x0⊤Dx0=0, then
(2.13)z⊤(∑i=1sti*(∇f(x0,y¯i)+Bx0(x0⊤Bx0)1/2-k0∇h(x0,y¯i)))+(z⊤(k02D)z)1/2<0,(ii)
ifx0⊤Bx0=0 and x0⊤Dx0>0, then
(2.14)z⊤(∑i=1sti*(∇f(x0,y¯i)-k0(∇h(x0,y¯i)-Dx0(x0⊤Dx0)1/2)))+(z⊤Bz)1/2<0,(iii)
ifx0⊤Bx0=0 and x0⊤Dx0=0, then
(2.15)z⊤(∑i=1sti*(∇f(x0,y¯i)-k0∇h(x0,y¯i)))+(z⊤(k0D)z)1/2+(z⊤Bz)1/2<0.If we take conditionZy¯(x0)=∅ in Theorem 2.2, then the result of Theorem 2.2 still holds.


## 3. Optimality Conditions

In this section we derive sufficient conditions for optimality of (P) under the assumpition of a particular form of generalized B-(p,r)-invexity. All theorems in this work will be proved only in the case when p≠0, r≠0 (other cases can be dealt with by similarity since the only difference is arised from the form of the inequality defining the class of the B-(p,r)-invex functions with respect to η and b for given p and r). The proofs of the other cases are easier than this one.We would establish the sufficient conditions under theB-(p,r)-invex function.Theorem 3.1 (sufficient optimality conditions).
Letx0∈X∘ be a feasible solution of (P). There exist a positive interger s, 1≤s≤n+1, t*∈ℝs, y¯i∈Y(x0)(i=1,2,…,s), k0∈ℝ+, (w,v)∈ℝn×ℝn, and μ*∈ℝ+p to satisfy relations (2.7)~(2.11). Furthermore suppose that any one of conditions (a) and (b) holds:  (a)
Υ(·)=∑i=1sti*((f(·,y¯i)+(·)⊤Bw)-k0(h(·,y¯i)-(·)⊤Dv)) is B-(p,r)-invex with respect to η and b at x0, and ∑j=1pμj*gj(·) is B-(p,r)-invex with respect to the same function η and another function b1 at x0 on X∘, not necessarily, equal to b,(b)
Υ1(·)=∑i=1sti*((f(·,y¯i)+(·)⊤Bw)-k0(h(·,y¯i)-(·)⊤Dv))+∑j=1pμj*gj(·) is B-(p,r)-invex with respect to η and b at x0 on X∘,then x0 is an optimal solution of (P).Proof.
Suppose thatx0 is not an optimal solution of (P). Then there exists a (P)-feasible solution x1 such that
(3.1)sup⁡⁡y∈Yf(x1,y)+(x1⊤Bx1)1/2h(x1,y)-(x1⊤Dx1)1/2<sup⁡⁡y∈Yf(x0,y)+(x0⊤Bx0)1/2h(x0,y)-(x0⊤Dx0)1/2.
We note that
(3.2)sup⁡⁡y∈Yf(x0,y)+(x0⊤Bx0)1/2h(x0,y)-(x0⊤Dx0)1/2=f(x0,y¯i)+(x0⊤Bx0)1/2h(x0,y¯i)-(x0⊤Dx0)1/2=k0,
for y¯i∈Y(x0), i=1,2,…,s, and
(3.3)f(x1,y¯i)+(x1⊤Bx1)1/2h(x1,y¯i)-(x1⊤Dx1)1/2≤sup⁡⁡y∈Yf(x1,y)+(x1⊤Bx1)1/2h(x1,y)-(x1⊤Dx1)1/2.
Then, we obtain
(3.4)f(x1,y¯i)+(x1⊤Bx1)1/2h(x1,y¯i)-(x1⊤Dx1)1/2<k0,fori=1,2,…,s.
It follows that
(3.5)f(x1,y¯i)+(x1⊤Bx1)1/2-k0(h(x1,y¯i)-(x1⊤Dx1)1/2)<0,fori=1,2,…,s.
From relations (2.6), (2.11), (2.8), (2.10), and (3.5), we have
(3.6)Υ(x1)=∑i=1sti*(f(x1,y¯i)+(x1⊤Bw)-k0(h(x1,y¯i)-(x1⊤Dv)))≤∑i=1sti*(f(x1,y¯i)+(x1⊤Bx1)1/2-k0(h(x1,y¯i)-(x1⊤Dx1)1/2))<0=∑i=1sti*(f(x0,y¯i)+(x0⊤Bx0)1/2-k0(h(x0,y¯i)-(x0⊤Dx0)1/2))=∑i=1sti*(f(x0,y¯i)+(x0⊤Bw)-k0(h(x0,y¯i)-(x0⊤Dv)))=Υ(x0).
That is,
(3.7)Υ(x1)<Υ(x0).
From relations (P) and (2.9), we obtain
(3.8)∑j=1pμj*gj(x1)≤0=∑j=1pμj*gj(x0).
If hypothesis (a) holds, from the B-(p,r)-invexity with respect to η and b1 at x0 of ∑j=1pμj*gj(·), we have
(3.9)1rb1(x1,x0)(er(∑j=1pμj*gj(x1)-∑j=1pμj*gj(x0))-1)≥1p∇∑j=1pμj*gj(x0)(epη(x1,x0)-1).
From the inequalities (3.8) and (3.9), we get
(3.10)1p∇(∑j=1pμj*gj(x0))(epη(x1,x0)-1)≤0.
Now, multiplying equality (2.7) by (1/p)(epη(x1,x0)-1), we know
(3.11)1p(epη(x1,x0)-1)(∑i=1sti*{∇f(x0,y¯i)+Bw-k0(∇h(x0,y¯i)-Dv)}+∇∑j=1pμj*gj(x0))=0.
From relations (3.10) and (3.11), we have
(3.12)1p(epη(x1,x0)-1)(∑i=1sti*{∇f(x0,y¯i)+Bw-k0(∇h(x0,y¯i)-Dv)})≥0.
From the B-(p,r)-invexity with respect to the same function η and the function b at x0 of Υ(·),
(3.13)1rb(x,u)(er(Υ(x1)-Υ(x0))-1)≥1p∇Υ(x0)(epη(x1,x0)-1).
From inequality (3.12) and the above inequality, we obtain
(3.14)Υ(x1)≥Υ(x0),
which contradicts (3.7), and proves that x0 is an optimal solution to (P).
If hypothesis (b) holds, from theB-(p,r)-invexity with respect to η and b at x0 of Υ1(·), then
(3.15)1rb(x,u)(er(Υ1(x1)-Υ1(x0))-1)≥1p∇Υ1(x0)(epη(x1,x0)-1).
The above inequality along with (2.7) yields
(3.16)Υ1(x1)≥Υ1(x0),
which contradicts (3.7). Hence, the proof is completed.


## 4. Parametric Dual-Type Model

We use the optimality conditions of the preceding section and show that the following formation is a dual (D) to the minimax problem (P):(D)max⁡(s,t,y)∈K(z)⁡sup⁡(z,μ,k,w,v)∈H1(s,t,y)⁡k,
where H1(s,t,y) denotes the set of (z,μ,k,w,v)∈ℝn×ℝ+p×ℝ+×ℝn×ℝn satisfying (4.1)∑i=1sti{∇f(z,yi)+Bw-k(∇h(z,yi)-Dv)}+∇∑j=1pμjgj(z)=0,(4.2)∑i=1sti{f(z,yi)+z⊤Bw-k(h(z,yi)-z⊤Dv)}≥0,(4.3)∑j=1pμjgj(z)≥0,(4.4)w⊤Bw≤1,v⊤Dv≤1,(s,t,y)∈K(z).
If for a triplet (s,t,y)∈K(z) the set H1(s,t,y) is empty, then we define the supremum over it to be -∞.LetΓ denote the set of all feasible points of (D). Moreover, we denote prXΓ={z∈X∣(z,μ,s,t,y,w,v,k)∈Γ}.We can derive the following weak duality theorem between (P) and (D).Theorem 4.1 (weak duality).
Letx and (z,μ,s,t,y,w,v,k) be (P)-feasible and (D)-feasible, respectively. Suppose that any one of the following conditions (a) and (b) holds:  (a)
Υ2(·)=∑i=1sti((f(·,yi)+(·)⊤Bw)-k(h(·,yi)-(·)⊤Dv)) is B-(p,r)-invex with respect to η and b at z and ∑j=1pμjgj(·) is B-(p,r)-invex with respect to the same function η and another function b2 at z on X∘∪prXΓ, not necessarily, equal to b,(b)
Υ3(·)=∑i=1sti((f(·,yi)+(·)⊤Bw)-k(h(·,yi)-(·)⊤Dv))+∑j=1pμjgj(·) is (p,r)-invex with respect to η and b at z on X∘∪prXΓ.Then
(4.5)sup⁡⁡y∈Yf(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2≥k.Proof.
Suppose on the contrary that(4.6)sup⁡⁡y∈Yf(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2<k.
Then, we have an inequality
(4.7)f(x,y)+(x⊤Bx)1/2-k(h(x,y)-(x⊤Dx)1/2)<0∀y∈Y.
It follows that for ti≥0, i=1,2,…,s with ∑i=1sti=1, we have
(4.8)ti[f(x,yi)+(x⊤Bx)1/2-k(h(x,yi)-(x⊤Dx)1/2)]≤0,
with at least one strict inequality because t=(t1,t2,…,ts)≠0. From relations (2.6), (4.4), (4.8), and (4.2), we obtain
(4.9)Υ2(x)=∑i=1sti(f(x,yi)+x⊤Bw-k(h(x,yi)-x⊤Dv))≤∑i=1sti[f(x,yi)+(x⊤Bx)1/2-k(h(x,yi)-(x⊤Dx)1/2)]<0≤∑i=1sti(f(z,yi)+z⊤Bw-k(h(z,yi)-z⊤Dv))=Υ2(z).
That is,
(4.10)Υ2(x)<Υ2(z).
From relations (P) and (4.3), we have
(4.11)∑j=1pμjgj(x)≤0≤∑j=1pμjgj(z),
If hypothesis (a) holds, from B-(p,r)-invexity with respect to η and b at z of Υ2(·), we get
(4.12)1rb(x,z)(er(Υ2(x)-Υ2(z))-1)≥1p∇Υ2(z)(epη(x,z)-1).
From the above inequality together with relation (4.10), we have
(4.13)1p∇Υ2(z)(epη(x,z)-1)<0.
Multiplying (4.1) by (1/p)(epη(x,z)-1), we obtain
(4.14)1p(epη(x,z)-1)(∑i=1sti{∇f(z,yi)+Bw-k(∇h(z,yi)-Dv)}+∇∑j=1pμjgj(z))=0.
From the above equality and inequality (4.13), we get
(4.15)1p(epη(x,z)-1)∑j=1pμj∇gj(z)>0.
Using the B-(p,r)-invexity of ∑j=1pμjgj(·) with respect to the same function η and the function b2 at z and inequality (4.15), we get
(4.16)∑j=1pμjgj(x)>∑j=1pμjgj(z).
which contradicts (4.11) and proves that supy∈Y((f(x,y)+(x⊤Bx)1/2)/(h(x,y)-(x⊤Dx)1/2))≥k.
If hypothesis (b) holds, from theB-(p,r)-invexity with respect to η and b at z of Υ3(·), then
(4.17)1rb(x,z)(er(Υ3(x)-Υ3(z))-1)≥1p∇Υ3(z)(epη(x,z)-1).
By the above inequality and equality (4.1), we have
(4.18)Υ3(x)≥Υ3(z).
From relations (4.10) and (4.11), we obtain
(4.19)Υ3(x)<Υ3(z),
which contradicts inequality (4.18). Thus, the proof is complete.Theorem 4.2 (strong duality).
Letx* be an optimal solution of (P), and let x* satisfy a constraint qualification for (P). Then there exist (s*,t*,y*)∈K(x*) and (x*,μ*,k*,w*,v*)∈H1(s*,t*,y*) such that (x*,μ*,s*,t*,y*,k*,w*,v*) is a feasible solution of (D). If in addition the hypothesis of Theorem 4.1 holds, then (x*,μ*,s*,t*,y*,k*,w*,v*) is an optimal solution of (D) and the two problems (P) and (D) have the same optimal value.Proof.
By Theorem2.2, there exist (s*,t*,y*)∈K(x*) and (x*,μ*,k*,w*,v*)∈H1(s*,t*,y*) such that (x*,μ*,s*,t*,y*,k*,w*,v*) is feasible for (D), and
(4.20)k*=f(x*,yi*)+(x*⊤Bx*)1/2h(x*,yi*)-(x*⊤Dx*)1/2.
The optimality of this feasible solution for (D) follows from Theorem 4.1.Theorem 4.3 (strict converse duality).
Letx¯ and (z*,μ*,s*,t*,y*,k*,w*v*) be optimal solutions of (P) and (D), respectively, and assume that the hypothesis of Theorem 4.2 is fulfilled. Suppose that any one of the following conditions (a) and (b) holds:  (a)
Υ4(·)=∑i=1s*ti*{f(·,yi*)+(·)⊤Bw*-k*(h(·,yi*)-(·)⊤Dv*)} is strictly B-(p,r)-invex with respect to η and b at z* and ∑j=1pμj*gj(·) is (p,r)-invex with respect to the same function η and another function b4 at z* on X∘∪prXΓ, not necessarily, equal to b,(b)
Υ5(·)=∑i=1s*ti*{f(·,yi*)+(·)⊤Bw*-k*(h(·,yi*)-(·)⊤Dv*)}+∑j=1pμj*gj(·) is strictly (p,r)-invex with respect to η and b at z* on X∘∪prXΓ.Then x¯=z*, that is, z* solves (P) and supy∈Y((f(z*,y)+((z*)⊤Bz*)1/2)/(h(z*,y)-((z*)⊤Dz*)1/2))=k*.Proof.
We shall assume thatx¯≠z* and reach a contradiction. From Theorem 4.2, we know that there exist (s¯,t¯,y¯)∈K(x¯) and (x¯,μ¯,k¯,w¯,v¯)∈H1(s¯,t¯,y¯) such that (x¯,μ¯,s¯,t¯,y¯,k¯,w¯,v¯) is an optimal solution for (D) with the optimal value
(4.21)sup⁡⁡y∈Yf(x¯,y)+(x¯⊤Bx¯)1/2h(x¯,y)-(x¯⊤Dx¯)1/2=k¯.
Now like the proof of Theorem 4.1 by x replacing by x¯ and (z,μ,s,t,y,w,v,k) by (z*,μ*,s*,t*,y*,k*,w*,v*), we obtain
(4.22)sup⁡⁡y∈Yf(x¯,y)+(x¯⊤Bx¯)1/2h(x¯,y)-(x¯⊤Dx¯)1/2>k*.
The above inequality contradicts
(4.23)sup⁡⁡y∈Yf(x¯,y)+(x¯⊤Bx¯)1/2h(x¯,y)-(x¯⊤Dx¯)1/2=k*=k¯.
Therefore, we conclude that x¯=z*. Here, the proof of the theorem is complete.Remark 4.4.
In Theorem4.3, if ∑j=1pμj*gj(·) is a strictly B-(p,r)-invex function with respect to η and b4 and Υ4(·) is a B-(p,r)-invex function with respect to the same function η and the function b, not necessarily, equal to b4, then Theorem 4.3 also holds.


## 5. Mond-Weir Dual-Type Model

In this section, we formulate the Mond-Weir-type dual model to the problem(P) as follows:(MWD)max⁡(s,t,y)∈K(z)⁡sup⁡(z,μ,w,v)∈H2(s,t,y)⁡F(z),
where H2(s,t,y) denotes the set of (z,μ,w,v)∈ℝn×ℝ+p×ℝn×ℝn satisfying(5.1)∑i=1sti{(h(z,yi)-(z⊤Dz)1/2)(∇f(z,yi)+Bw)-(f(z,yi)+(z⊤Bz)1/2)(∇h(z,yi)-Dv)}+∑j=1pμj∇gj(z)=0,(5.2)∑j=1pμjgj(z)≥0,(5.3)w⊤Bw≤1,v⊤Dv≤1,zBw=(z⊤Bz)1/2zDv=(z⊤Bz)1/2,
where (5.4)F(z)=sup⁡⁡y∈Yf(z,y)+(z⊤Bz)1/2h(z,y)-(z⊤Dz)1/2=sup⁡⁡y∈Yϕ(z,y).If for a triplet(s,t,y)∈K(z) the set H2(s,t,y) is empty, then we define the supremum over it to be -∞.LetΓ̃ denote the set of all feasible points of (MWD). Moreover, we denote prXΓ̃={z∈X∣(z,μ,s,t,y,w,v)∈Γ̃}.We establish the weak, strong, and strict converse duality theorems for (MWD) with respect to the primal problem (P).Theorem 5.1 (weak duality).
Letx and (z,μ,s,t,y,w,v) be (P)-feasible and (MWD)-feasible, respectively. Suppose that any one of the following conditions (a) and (b) holds:  (a)
Υ6(·)=∑i=1sti{(h(z,yi)-z⊤Dv)(f(·,yi)+(·)⊤Bw)-(f(z,yi)+z⊤Bw)(h(·,yi)-(·)⊤Dv)} is B-(p,r)-invex, respect to η and b at z and ∑j=1pμjgj(·) is B-(p,r)-invex with respect to the same function η and another function b6 at z on X∘∪Γ̃, not necessarily, equal to b,(b)
Υ7=∑i=1sti{(h(z,yi)-z⊤Dv)(f(·,yi)+(·)⊤Bw)-(f(z,yi)+z⊤Bw)(h(·,yi)-(·)⊤Dv)}+∑j=1pμjgj(·) is B-(p,r)-invex with respect to η and b at z on X∘∪Γ̃.Then
(5.5)sup⁡⁡y∈Yf(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2≥F(z).Proof.
On the contrary, if possible, suppose that for eachx∈X∘,
(5.6)sup⁡⁡y∈Yf(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2<F(z).
From the above inequality and yi∈Y(z), i=1,2,…,s, we obtain
(5.7)f(x,yi)+(x⊤Bx)1/2h(x,yi)-(x⊤Dx)1/2<F(z)=f(z,yi)+(z⊤Bz)1/2h(z,yi)-(z⊤Dz)1/2,i=1,2,…,s.
By the above inequality, we know that
(5.8)(f(x,yi)+(x⊤Bx)1/2)(h(z,yi)-(z⊤Dz)1/2)-(h(x,yi)-(x⊤Dx)1/2)(f(z,yi)+(z⊤Bz)1/2)<0,
for all i=1,2,…,s and yi∈Y(z).
Multiplying the above inequality byti∈ℝ+s with ∑i=1sti=1, we have
(5.9)∑i=1sti{(f(x,yi)+(x⊤Bx)1/2)(h(z,yi)-(z⊤Dz)1/2)-(h(x,yi)-(x⊤Dx)1/2)(f(z,yi)+(z⊤Bz)1/2)}<0.
From relations (2.6), (5.3), and (5.9), we get
(5.10)Υ6(x)=∑i=1sti{(f(x,yi)+x⊤Bw)(h(z,yi)-z⊤Dv)-(h(x,yi)-x⊤Dv)(f(z,yi)+z⊤Bw)}≤∑i=1sti{(f(x,yi)+(x⊤Bx)1/2)(h(z,yi)-(z⊤Dz)1/2)-(h(x,yi)-(x⊤Dx)1/2)(f(z,yi)+(z⊤Bz)1/2)}<0=Υ6(z).
By relations (P) and (5.2), we have
(5.11)∑j=1pμjgj(x)≤0≤∑j=1pμjgj(z).
Now, if condition (a) holds, from B-(p,r)-invexity with respect to η and b6 at z of ∑j=1pμjgj(·), we get
(5.12)1rb1(x1,x0)(er(∑j=1pμj*gj(x)-∑j=1pμj*gj(z))-1)≥1p∇∑j=1pμj*gj(z)(epη(x,z)-1).
From the above inequality and inequality (5.11), we obtain
(5.13)1p∇∑j=1p(μjgj(z))(epη(x,z)-1)≤0.
Multiplying (5.1) by (1/p)(epη(x,z)-1), we have
(5.14)1p(epη(x,z)-1){∑i=1sti{(h(z,yi)-(z⊤Dz)1/2)(∇f(z,yi)+Bw)-(f(z,yi)+(z⊤Bz)1/2)(∇h(z,yi)-Dv)}+∑j=1pμj∇gj(z)}=0.
By the above equality and inequality (5.13), we obtain
(5.15)1p(epη(x,z)-1){∑i=1sti{(h(z,yi)-(z⊤Dz)1/2)(∇f(z,yi)+Bw)-(f(z,yi)+(z⊤Bz)1/2)(∇h(z,yi)-Dv)}∑i=1s}≥0.
Using the B-(p,r)-invexity with respect to the same function η and the function b at z of Υ6(·) and the above inequality, we have
(5.16)Υ6(x)≥Υ6(z),
which contradicts (5.10) and proves that supy∈Y((f(x,y)+(x⊤Bx)1/2)/(h(x,y)-(x⊤Dx)1/2))≥ℱ(z).
If hypothesis (b) holds, from theB-(p,r)-invexity with respect to η and b at z of Υ7(·) and the equality (5.1), then
(5.17)Υ7(x)≥Υ7(z).
From relations (5.10) and (5.11), we have
(5.18)∑i=1sti{(h(z,yi)-z⊤Dv)(f(x,yi)+x⊤Bw)-(f(z,yi)+z⊤Bw)(h(x,yi)-x⊤Dv)}+∑j=1pμjgj(x)<∑i=1sti{(h(z,yi)-z⊤Dv)(f(z,yi)+z⊤Bw)-(f(z,yi)+z⊤Bw)(h(z,yi)-z⊤Dv)}+∑j=1pμjgj(z),
which contradicts inequality (5.17). Hence, the proof is complete.Similar to the proof of Theorem4.2, we can establish Theorem 5.2.Theorem 5.2 (strong duality).
Letx* be an optimal solution of (P) and let x* satisfy a constraint qualification for (P). Then there exist (s*,t*,y*)∈K(x*) and (x*,μ*,w*,v*)∈H1(s*,t*,y*) such that (x*,μ*,s*,t*,y*,w*,v*) is a feasible solution of (MWD). If in addition the hypothesis of Theorem 5.1 holds, then (x*,μ*,s*,t*,y*,w*,v*) is an optimal solution of (MWD) and the two problems (P) and (MWD) have the same optimal value.Theorem 5.3 (strict converse duality).
Letx¯ and (z*,μ*,s*,t*,y*,w*,v*) be optimal solutions of (P) and (MWD), respectively, and assume that the hypothesis of Theorem 5.2 is fulfilled. Suppose that any one of the following conditions (a) and (b) holds:  (a)
Υ8(·)=∑i=1s*ti*{(h(z*,yi*)-(z*)⊤Dv)(f(·,yi*)+(·)⊤Bw)-(f(z*,yi*)+(z*)⊤Bw)(h(·,yi*)-(·)⊤Dv)} is strictly B-(p,r)-invex with respect to η and b at z* and ∑j=1pμj*gj(·) is B-(p,r)-invex with respect to the same function η and another function b8 at z* on X∘∪Γ̃, not necessarily, equal to b,(b)
Υ9(·)=∑i=1s*ti*{(h(z*,yi*)-(z*)⊤Dv)(f(·,yi*)+(·)⊤Bw)-(f(z*,yi*)+(z*)⊤Bw)(h(·,yi*)-(·)⊤Dv)}+∑j=1pμj*gj(·) is strictly B-(p,r)-invex with respect to η and b at z* on X∘∪Γ̃.Then x¯=z*, that is, z* solves (P) and supy∈Y((f(z*,y)+((z*)⊤Bz*)1/2)/(h(z*,y)-((z*)⊤Dz*)1/2))=ℱ(z*).Proof.
We suppose on the contrary that ifx¯≠z*, then
(5.19)sup⁡⁡y∈Yf(x¯,y)+(x¯⊤Bx¯)1/2h(x¯,y)-(x¯⊤Dx¯)1/2≤F(z*).
Since yi*∈Y(z*), for i=1,2,…,s*,
(5.20)f(z*,yi*)+((z*)⊤Bz*)1/2h(z*,yi*)-((z*)⊤Dz*)1/2=F(z*),fori=1,2,…,s*.
It follows that
(5.21)(f(x¯,y)+(x¯⊤Bx¯)1/2)(h(z*,yi*)-((z*)⊤Dz*)1/2)-(h(x¯,y)-(x¯⊤Dx¯)1/2)(f(z*,yi*)+(z*)⊤Bz*)1/2≤0,fori=1,2,…,s*,y∈Y.
From yi*∈Y(z*)⊂Y and t∈ℝ+s* with ∑i=1s*ti*=1, we have
(5.22)∑i=1s*ti*{(h(z*,yi*)-((z*)⊤Dz*)1/2)(f(x¯,yi*)+(x¯⊤Bx¯)1/2)-(h(x¯,yi*)-(x¯⊤Dx¯)1/2)(f(z*,yi*)+((z*)⊤Bz*)1/2)}≤0.
From relations (2.6), (5.3), and inequality (5.22), we obtain that
(5.23)Υ8(x¯)=∑i=1s*ti*{(h(z*,yi*)-(z*)⊤Dv)(f(x¯,yi*)+x¯⊤Bw)-(f(z*,yi*)+(z*)⊤Bw)(h(x¯,yi*)-x¯⊤Dv)}≤∑i=1s*ti*{(h(z*,yi*)-(x¯⊤Dx¯)1/2)(f(z*,yi*)+((z*)⊤Bz*)1/2)-(h(x¯,yi*)-(x¯⊤Dx¯)1/2)(f(z*,yi*)+((z*)⊤Bz*)1/2)}≤0=Υ8(z*).
By relations (P) and (5.2), we get
(5.24)∑j=1pμj*gj(x¯)≤0≤∑j=1pμj*gj(z*).
Now, if condition (a) holds, from B-(p,r)-invexity with respect to η and b8 at z of ∑j=1pμj*gj(·), then
(5.25)1rb1(x¯,z*)(er(∑j=1pμj*gj(x¯)-∑j=1pμj*gj(z*))-1)≥1p∇∑j=1pμj*gj(z*)(epη(x¯,z*)-1).
From the above inequality and inequality (5.24), we know that
(5.26)1p∇∑j=1pμj*gj(z*)(epη(x¯,z*)-1)≤0.
Multiplying (MWD) by (1/p)(epη(x¯,z*)-1), we have
(5.27)1p(epη(x¯,z*)-1){∑j=1p∑i=1s*ti*{(h(z*,yi*)-((z*)⊤Dz*)1/2)(∇f(z*,yi*)+Bw*)-(f(z*,yi*)+((z*)⊤Bz*)1/2)(∇h(z*,yi*)-Dv)}+∑j=1pμj*∇gj(z*)}=0.
Basing on the above inequality and (5.26), we get the inequality
(5.28)1p(epη(x¯,z*)-1){∑i=1s*ti*{(h(z*,yi*)-((z*)⊤Dz*)1/2)(∇f(z*,yi*)+Bw*)-(f(z*,yi*)+((z*)⊤Bz*)1/2)(∇h(z*,yi*)-Dv)}∑i=1s*}≥0.
Since Υ8(·) is strictly B-(p,r)-invex with respect to the same function η and the function b at z* and the above inequality, we obtain
(5.29)Υ8(z*)<Υ8(x¯),
which contradicts relations (5.23). Hence x¯=z* and z* is an optimal solution of (P).
If hypothesis (b) holds, from the strictB-(p,r)-invexity with respect to η and b at z* of Υ9(·) and equality (5.1), then
(5.30)Υ9(x¯)>Υ9(z*).
Relations (5.23) together with (5.24) yield
(5.31)∑i=1s*ti*{(h(z*,yi*)-(z*)⊤Dv)(f(x¯,yi*)+x¯⊤Bw)-(f(z*,yi*)+z⊤Bw)(h(x¯,yi*)-x¯⊤Dv)}+∑j=1pμj*gj(x¯)≤∑i=1s*ti*{(h(z*,yi*)-(z*)⊤Dv)(f(z*,yi*)+(z*)⊤Bw)-(f(z*,yi*)+(z*)⊤Bw)(h(z*,yi*)-(z*)⊤Dv)}+∑j=1pμj*gj(z*),
which contradicts (5.30). Hence z*=x¯ is also an optimal solution of (P). The proof is complete.Remark 5.4.
In Theorem5.3, if ∑j=1pμj*gj(·) is a strictly B-(p,r)-invex function with respect to η and b8 and Υ8(·) is a B-(p,r)-invex function with respect to the same function η and the function b, not necessarily, equal to b8, then Theorem 5.3 also holds.


## 6. Wolfe Dual-Type Model

Based on the result of Theorem2.2, we can rewrite Theorem 2.2 as follows.Theorem 6.1 ((necessary conditions) Lai and Lee [11, Theorem 4]).
Letx0 be a (P)-optimal solution, and let ∇gj(x0), j∈J(x0) be linearly independent. Then there exist (s*,t*,y*)∈K(x0),(w,v)∈ℝn×ℝn, and μ*∈ℝ+p such that
(6.1)∇(∑i=1s*ti*f(x0,yi*)+x0⊤Bw+∑j=1pμj*gj(x0)∑i=1s*ti*(h(x0,yi*)-x0⊤Dv))=0,∑j=1pμj*gj(x0)=0,ti*≥0,∑i=1s*ti*=1,w⊤Bw≤1,v⊤Dv≤1,x0⊤Bw=(x0⊤Bx0)1/2,x0⊤Dv=(x0⊤Bx0)1/2.In this section, we present the Wolfe dual (WD) to the minimax program (P):
(WD)max⁡(s,t,y)∈K(z)⁡sup⁡(z,μ,w,v)∈H3(s,t,y)⁡A(z),
where H3(s,t,y) denotes the set of (z,μ,w,v)∈ℝn×ℝ+p×ℝn×ℝn satisfying
(6.2)∇(∑i=1sti(f(z,yi)+z⊤Bw)+∑j=1pμjgj(z)∑i=1sti(h(z,yi)-z⊤Dv))=0,(s,t,y)∈K,(6.3)w⊤Bw≤1,v⊤Dv≤1,z⊤Bw=(z⊤Bz)1/2,z⊤Dv=(z⊤Bz)1/2,(6.4)A(z)=∑i=1sti(f(z,yi)+z⊤Bw)+∑j=1pμjgj(z)∑i=1sti(h(z,yi)-z⊤Dv).
If for a triplet (s,t,y)∈K(z) the set H3(s,t,y) is empty, then we define the supremum over it to be -∞.LetΓ¯ denote the set of all feasible points of (WD). Moreover, we denote prXΓ¯={z∈X∣(z,μ,s,t,y,w,v)∈Γ¯}.We assume throughout this section that∑i=1sti(f(z,yi)+z⊤Bw)+∑j=1pμjgj(z)≥0 and ∑i=1sti(h(z,yi)-z⊤Dv)>0.We establish the weak, strong, and strict converse duality theorems for(WD) with respect to the primal problem (P).Theorem 6.2 (weak duality).
Letx and (z,μ,s,t,y,w,v) be (P)-feasible and (WD)-feasible, respectively, and assume that
(6.5)Υ10(⋅)=[∑i=1sti(h(z,yi)-z⊤Dv)][∑i=1sti(f(⋅,yi)+(⋅)⊤Bw)+∑j=1pμjgj(⋅)]-[∑i=1sti(f(z,yi)+z⊤Bw)+∑j=1pμjgj(z)][∑i=1sti(h(⋅,yi)-(⋅)⊤Dv)]
is B-(p,r)-invex with respect to η and b at z on X∘⋃prXΓ¯; then
(6.6)sup⁡⁡y∈Yf(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2≥A(z).Proof.
Suppose on the contrary that(6.7)sup⁡⁡y∈Yf(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2<A(z).
Hence, we have an inequality
(6.8)f(x,y)+(x⊤Bx)1/2h(x,y)-(x⊤Dx)1/2<A(z)∀y∈Y.
Then, we get
(6.9)(f(x,y)+(x⊤Bx)1/2){∑i=1sti(h(z,yi)-z⊤Dv)}-(h(x,y)-(x⊤Dx)1/2){∑i=1sti(f(z,yi)+z⊤Bw)+∑i=1pμjgj(z)}<0,
for all y∈Y.
Ify is replaced by yi in the above inequality and is multiplied by ti, then summing up, we get
(6.10)∑i=1sti(f(x,yi)+(x⊤Bx)1/2){∑i=1sti(h(z,yi)-z⊤Dv)}-{∑i=1sti(h(x,yi)-(x⊤Dx)1/2)}{∑i=1sti(f(z,yi)+z⊤Bw)+∑i=1pμjgj(z)}<0.
By (2.6) and (6.3), we have
(6.11)Υ10(x)≤{∑i=1sti(f(x,yi)+(x⊤Bx)1/2)+∑i=1pμjgj(x)}{∑i=1sti(h(z,yi)-z⊤Dv)}-{∑i=1sti(h(x,yi)-(x⊤Dx)1/2)}{∑i=1sti(f(z,yi)+z⊤Bw)+∑i=1pμjgj(z)}<∑i=1pμjgj(x){∑i=1sti(h(z,yi)-z⊤Dv)}.
Since ∑i=1sti(h(z,yi)-z⊤Dv)>0 and ∑i=1pμjgj(x)≤0, it follows that
(6.12)Υ10(x)<0=Υ10(z).
Using the B-(p,r)-invexity with respect to η and b at z of Υ10(·), we have the inequality
(6.13)1p∇Υ10(z)(epη(x,z)-1)<0.
This contradicts the equality of (6.2) ×(1/p)(epη(x,z)-1). Here the proof is complete.As a consequence of Theorems6.1 and 6.2, we obtain Theorem 6.3. By a similar way, we can prove the strong duality and strictly converse duality theorems with respect to (P) and (WD) which we state as follows.Theorem 6.3 (strong duality).
Letx* be an optimal solution of (P) satisfying the hypothesis of Theorem 6.2. Then there exist (s*,t*,y*)∈K(x*) and (x*,μ*,w*,v*)∈H3(s*,t*,y*) such that (x*,μ*,s*,t*,y*,w*,v*) is feasible for (WD). If any of the conditions of Theorem 6.2 hold, then (x*,μ*,s*,t*,y*,w*,v*) is an optimal solution of (WD) and the two problems (P) and (WD) have the same extremal values.Theorem 6.4 (strict converse duality).
Letx¯ and (z*,μ*,s*,t*,y*,w*,v*) be optimal solutions of (P) and (WD), respectively. Suppose that the assumptions of Theorem 6.3 are fulfilled and
(6.14)Υ11(⋅)=[∑i=1s*ti*(h(z*,yi*)-(z*)⊤Dv*)][∑i=1s*ti*(f(⋅,yi*)+(⋅)⊤Bw*)+∑j=1pμj*gj(⋅)]-[∑i=1s*ti*(f(z*,yi*)+(z*)⊤Bw*)+∑j=1pμj*gj(z*)][∑i=1sti*(h(⋅,yi*)-(⋅)⊤Dv*)]
is strictly B-(p,r)-invex with respect to η and b at z* on X∘∪prXΓ¯.Thenx¯=z*, that is, z* is an optimal solution of (P) and supy∈Y((f(z*,y)+((z*)⊤Bz*)1/2)/(h(z*,y)-((z*)⊤Dz*)1/2))=𝒜(z*).



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