# On Second-Order Duality for Minimax Fractional Programming Problems with Generalized Convexity

Authors: Izhar Ahmad
Journal: Abstract and Applied Analysis
 (2011)

---

## Abstract

We focus our study on a discussion of duality relationships of a minimax fractional programming problem with its two types of second-order dual models under the second-order generalized convexity type assumptions. Results obtained in this paper naturally unify and extend some previously known results on minimax fractional programming in the literature.


---

## Body

## 1. Introduction

Fractional programming is an interesting subject applicable to many types of optimization problems such as portfolio selection, production, and information theory and numerous decision making problems in management science. More specifically, it can be used in engineering and economics to minimize a ratio of physical or economical functions, or both, such as cost/time, cost/volume, and cost/benefit, in order to measure the efficiency or productivity of the system (see Stancu-Minasian [1]).Minimax type functions arise in the design of electronic circuits; however, minimax fractional problems appear in the formulation of discrete and continuous rational approximation problems with respect to the Chebyshev norm [2], continuous rational games [3], multiobjective programming [4, 5], and engineering design as well as some portfolio selection problems discussed by Bajona-Xandri and Martinez-Legaz [6].In this paper, we consider the minimax fractional programming problem(1.1)minimizeϕ(x)=sup⁡y∈Yf(x,y)h(x,y),subjecttog(x)≤0,x∈Rn,
where Y is a compact subset of Rl and f(·,·):Rn×Rl→R,  h(·,·):Rn×Rl→R, and g(·):Rn→Rm are twice continuously differentiable functions on Rn×Rl, Rn×Rl, and Rn, respectively. It is assumed that, for each (x,y) in Rn×Rl, f(x,y)≥0 and h(x,y)>0.For the case of convex differentiable minimax fractional programming, Yadav and Mukherjee [7] formulated two dual models for (1.1) and derived duality theorems. Chandra and Kumar [8] pointed out certain omissions and inconsistencies in the dual formulation of Yadav and Mukherjee [7]; they constructed two modified dual problems for (1.1) and proved appropriate duality results. Liu and Wu [9, 10] and Ahmad [11] obtained sufficient optimality conditions and duality theorems for (1.1) assuming the functions involved to be generalized convex.Second-order duality provides tighter bounds for the value of the objective function when approximations are used. For more details, one can consult ([12, page 93]). One more advantage of second-order duality, when applicable, is that, if a feasible point in the primal is given and first-order duality does not apply, then we can use second order duality to provide a lower bound of the value of the primal (see [13]).Mangasarian [14] first formulated the second-order dual for a nonlinear programming problem and established second-order duality results under certain inequalities. Mond [12] reproved second-order duality results assuming rather simple inequalities. Subsequently, Bector and Chandra [15] formulated a second-order dual for a fractional programming problem and obtained usual duality results under the assumptions [14] by naming these as convex/concave functions.Based upon the ideas of Bector et al. [16] and Rueda et al. [17], Yang and Hou [18] proposed a new concept of generalized convexity and discussed sufficient optimality conditions for (1.1) and duality results for its corresponding dual. Recently, Husain et al. [19] formulated two types of second-order dual models to (1.1) and discussed appropriate duality results involving η-convexity/generalized η-convexity assumptions.In this paper, we are inspired by Chandra and Kumar [8], Bector et al. [16], Liu [20], and Husain et al. [19] to discuss weak, strong, and strict converse duality theorems connecting (1.1) with its two types of second-order duals by using second-order generalized convexity type assumptions [21].


## 2. Notations and Preliminaries

LetS={x∈Rn:g(x)≤0} denote the set of all feasible solutions of (1.1). For each (x,y)∈Rn×Rl, we define
(2.1)J(x)={j∈M:gj(x)=0},
where M={1,2,…,m},(2.2)Y(x)={y∈Y:f(x,y)=sup⁡z∈Yf(x,z)},K(x)={∑i=1sti=1,(s,t,ỹ)∈N×R+s×Rls:1≤s≤n+1,t=(t1,t2,…,ts)∈R+swith∑i=1sti=1,ỹ=(y̅1,y̅2,….y̅s)withy̅i∈Y(x),i=1,2,…,s}.Definition 2.1.
A functionalℱ:X×X×Rn→R, where X⊆Rn is said to be sublinear in its third argument, if ∀x,x¯∈X,(i)
ℱ(x,x¯;a1+a2)≤ℱ(x,x¯;a1)+ℱ(x,x¯;a2)∀a1,a2∈Rn,(ii)
ℱ(x,x¯;αa)=αℱ(x,x¯;a)∀α∈R+,a∈Rn. By (ii), it is clear that ℱ(x,x¯;0a)=0.Definition 2.2.
A pointx¯∈S is said to optimal solution of (1.1) if ϕ(x)≥ϕ(x¯) for each x∈S.The following theorem [8] will be needed in the subsequent analysis.Theorem 2.3 (necessary conditions).
Letx* be a solution (local or global) of (1.1), and let  ∇gj(x*),j∈J(x*) be linearly independent. Then there exist  (s*,t*,y¯*)∈K(x*),λ*∈R+, and  μ*∈R+m such that
(2.3)∇∑i=1s*ti*(f(x*,y¯i*)-λ*h(x*,y¯i*))+∇∑j=1mμj*gj(x*)=0,f(x*,y¯i*)-λ*h(x*,y¯i*)=0,i=1,2,…,s*,∑j=1mμj*gj(x*)=0,ti*≥0,∑i=1s*ti*=1,y¯i*∈Y(x*),i=1,2,…,s*.
Throughout the paper, we assume that ℱ is a sublinear functional. For β=1,2,…,r let b,b0,bβ:X×X→R+,ϕ,ϕ0,ϕβ:R→R,ρ,ρ0,ρβ be real numbers, and let θ:Rn×Rn→R.


## 3. First Duality Model

In this section, we discuss usual duality results for the following dual [19]:(3.1)max⁡(s,t,y¯)∈K(z)sup⁡(z,μ,λ,p)∈H1(s,t,y¯)λ,
where H1(s,t,y¯) denotes the set of all (z,μ,λ,p)∈Rn×R+m×R+×Rn satisfying(3.2)∇∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+∇∑j=1mμjgj(z)+∇2∑j=1mμjgj(z)p=0,(3.3)∑i=1sti(f(z,y¯i)-λh(z,y¯i))-12pT∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p≥0,(3.4)∑j=1mμjgj(z)-12pT∇2∑j=1mμjgj(z)p≥0.
If, for a triplet (s,t,y¯)∈K(z), the set H1(s,t,y¯)=∅, then we define the supremum over it to be -∞.Remark 3.1.
IfP=0, then (3.1) becomes the dual considered in [9].Theorem 3.2 (weak duality).
Letx and  (z,μ,λ,s,t,y¯,p) be the feasible solutions of (1.1) and (3.1), respectively. Suppose that there exist  ℱ,θ,ϕ,b and  ρ such that
(3.5)b(x,z)ϕ[∑i=1sti(f(x,y¯i)-λh(x,y¯i))-∑i=1sti(f(z,y¯i)-λh(z,y¯i))-∑j=1mμjgj(z)+12pT∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+12pT∑j=1mμjgj(z)p]<0⟹F(x,z;∇∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∇∑j=1mμjgj(z)+∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+∇2∑j=1mμjgj(z)p)<-ρ‖θ(x,z)‖2.
Further assume that
(3.6)a<0⟹ϕ(a)<0,(3.7)b(x,z)>0,(3.8)ρ≥0.
Then
(3.9)sup⁡y∈Yf(x,y)h(x,y)≥λ.Proof.
Suppose contrary to the result that(3.10)sup⁡y∈Yf(x,y)h(x,y)<λ.
Thus, we have
(3.11)f(x,y¯i)-λh(x,y¯i)<0,∀y¯i∈Y(x),i=1,2,…,s.
It follows from ti≥0,i=1,2,…,s, that
(3.12)ti[f(x,y¯i)-λh(x,y¯i)]≤0,
with at least one strict inequality since t=(t1,t2,…,ts)≠0. Taking summation over i, we have
(3.13)∑i=1sti(f(x,y¯i)-λh(x,y¯i))<0,
which together with (3.3) gives
(3.14)∑i=1sti(f(x,y¯i)-λh(x,y¯i))<0≤∑i=1sti(f(z,y¯i)-λh(z,y¯i))-12pT∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p.
The above inequality along with (3.4) implies
(3.15)∑i=1sti(f(x,y¯i)-λh(x,y¯i))-∑i=1sti(f(z,y¯i)-λh(z,y¯i))-∑j=1mμjgj(z)+12pT∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+12pT∇2∑j=1mμjgj(z)p<0.
Using (3.6) and (3.7), it follows from (3.15) that
(3.16)b(x,z)ϕ[∑i=1sti(f(x,y¯i)-λh(x,y¯i))-∑i=1sti(f(z,y¯i)-λh(z,y¯i))-∑j=1mμjgj(z)+12pT∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+12pT∇2∑j=1mμjgj(z)p]<0,
which along with (3.5) and (3.8) yields
(3.17)F(x,z;∇∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∇∑j=1mμjgj(z)+∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+∇2∑j=1mμjgj(z)p)<0,
which contradicts (3.2) since ℱ(x,z;0)=0.Theorem 3.3 (strong duality).
Assume thatx* is an optimal solution of (1.1) and  ∇gj(x*),j∈J(x*) are linearly independent. Then there exist  (s*,t*,y¯*)∈K(x*) and  (x*,μ*,λ*,p*=0)∈H1(s*,t*,y¯*) such that  (x*,μ*,λ*,s*,t*,y¯*,p*=0) is a feasible solution of (3.1) and the two objectives have the same values. Further, if the assumptions of weak duality (Theorem 3.2) hold for all feasible solutions  (z,μ,λ,s,t,y¯,p) of (3.1), then  (x*,μ*,λ*,s*,t*,y¯*,p*=0) is an optimal solution of (3.1).Proof.
Sincex* is an optimal solution of (1.1) and ∇gj(x*),j∈J(x*) are linearly independent, then, by Theorem 2.3, there exist (s*,t*,y¯*)∈K(x*) and (x*,μ*,λ*,p*=0)∈H1(s*,t*,y¯*) such that (x*,μ*,λ*,s*,t*,y¯*,p*=0) is a feasible solution of (3.1) and the two objectives have the same values. Optimality of (x*,μ*,λ*,s*,t*,y¯*,p*=0) for (3.1) thus follows from weak duality (Theorem 3.2).Theorem 3.4 (Strict converse duality).
Letx* and  (z*,μ*,λ*,s*,t*,y¯*,p*) be the optimal solutions of (1.1) and (3.1), respectively. Suppose that  ∇gj(x*),j∈J(x*) are linearly independent and there exist  ℱ,θ,ϕ,b and  ρ such that
(3.18)b(x*,z*)ϕ[∑i=1s*ti*(f(x*,y¯i*)-λ*h(x*,y¯i*))-∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))-∑j=1mμj*gj(z*)+12p*T∇2∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))p*+12p*T∇2∑j=1mμj*gj(z*)p*]≤0⟹F(x*,z*;∇∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∇∑j=1mμj*gj(z*)+∇2∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))p*+∇2∑j=1mμj*gj(z*)p*)<-ρ‖θ(x*,z*)‖2.
Further Assume
(3.19)a<0⟹ϕ(a)≤0,(3.20)b(x*,z*)>0,(3.21)ρ≥0.
Then z*=x*, that is, z* is an optimal solution of (1.1).Proof.
Suppose contrary to the result thatz*≠x*. Since x* and (z*,μ*,λ*,s*,t*,y¯*,p*) are optimal solutions of (1.1) and (3.1), respectively, and ∇gj(x*),j∈J(x*) are linearly independent, therefore, from strong duality (Theorem 3.3), we reach
(3.22)sup⁡y*∈Yf(x*,y*)h(x*,y*)=λ*.
Thus, we have
(3.23)f(x*,y¯i*)-λ*h(x*,y¯i*)≤0,∀y¯i*∈Y(x*),i=1,2,…,s*.
Now, proceeding as in Theorem 3.2, we get
(3.24)∑i=1s*ti*(f(x*,y¯i*)-λ*h(x*,y¯i*))-∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))-∑j=1mμj*gj(z*)+12p*T∇2∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))p*+12p*T∇2∑j=1mμj*gj(z*)p*<0.
Using (3.19) and (3.20), it follows from (3.24) that
(3.25)b(x*,z*)ϕ[∑i=1s*ti*(f(x*,y¯i*)-λ*h(x*,y¯i*))-∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))-∑j=1mμj*gj(z*)+12p*T∇2∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))p*+12p*T∇2∑j=1mμj*gj(z*)p*]≤0,
which along with (3.18) and (3.21) implies
(3.26)F(x*,z*;∇∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∇∑j=1mμj*gj(z*)+∇2∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))p*+∇2∑j=1mμj*gj(z*)p*)<0,
which contradicts (3.2) since ℱ(x*,z*;0)=0.


## 4. Second Duality Model

This section deals with duality theorems for the following second-order dual to (1.1):(4.1)max⁡(s,t,y¯)∈K(z)sup⁡(z,μ,λ,p)∈H2(s,t,y¯)λ,
where H2(s,t,y¯) denotes the set of all (z,μ,λ,p)∈Rn×R+m×R+×Rn satisfying(4.2)∇∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+∇∑j=1mμjgj(z)+∇2∑j=1mμjgj(z)p=0,(4.3)∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∑j∈J∘μjgj(z)-12pT∇2[∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∑j∈J∘μjgj(z)]p≥0,(4.4)∑j∈Jαμjgj(z)-12pT∇2∑j∈Jαμjgj(z)p≥0,α=1,2,…,r,
where Jα⊆M,α=0,1,2,…,r, with ⋃α=0rJα=M and Jα∩Jβ=∅, if α≠β.If, for a triplet(s,t,y¯)∈K(z), the set H2(s,t,y¯)=∅, then we define the supremum over it to be -∞.Theorem 4.1 (weak duality).
Letx and  (z,μ,λ,s,t,y¯,p) be the feasible solutions of (1.1) and (4.1), respectively. Suppose that there exist  ℱ,θ,ϕ0,b0,ρ0 and  ϕβ,bβ,ρβ,β=1,2,…,r such that
(4.5)b0(x,z)ϕ0[∑i=1sti(f(x,y¯i)-λh(x,y¯i))-∑i=1sti(f(z,y¯i)-λh(z,y¯i))-∑j∈J0μjgj(z)+12pT∇2(∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∑j∈J0μjgj(z))p]<0⟹F(x,z;∇∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+∇∑j∈J0μjgj(z)+∇2∑j∈J0μjgj(z)p)<-ρ0‖θ(x,z)‖2,(4.6)-bα(x,z)ϕα[∑j∈Jαμjgj(z)-12pT∇2∑j∈Jαμjgj(z)p]≤0⟹F(x,z;∇∑j∈Jαμjgj(z)+∇2∑j∈Jαμjgj(z)p)≤-ρα‖θ(x,z)‖2,α=1,2,…,r.
Further assume that
(4.7)a≥0⟹ϕα(a)≥0,α=1,2,…,r,(4.8)a<0⟹ϕ0(a)<0,(4.9)b0(x,z)>0,bα(x,z)≥0,α=1,2,…,r,(4.10)ρ0+∑α=1rρα≥0.s
Then
(4.11)sup⁡y∈Yf(x,y)h(x,y)≥λ.Proof.
Suppose contrary to the result that(4.12)sup⁡y∈Yf(x,y)h(x,y)<λ.
Thus, we have
(4.13)f(x,y¯i)-λh(x,y¯i)<0,∀y¯i∈Y(x),i=1,2,…,s.
It follows from ti≥0,i=1,2,…,s, that
(4.14)ti[f(x,y¯i)-λh(x,y¯i)]≤0,
with at least one strict inequality since t=(t1,t2,…,ts)≠0. Taking summation over i, we have
(4.15)∑i=1sti(f(x,y¯i)-λh(x,y¯i))<0,
which together with (4.3) implies
(4.16)∑i=1sti(f(x,y¯i)-λh(x,y¯i))<0≤∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∑j∈J0μjgj(z)-12pT∇2[∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∑j∈J0μjgj(z)]p.
Using (4.8) and (4.9), it follows from (4.16) that
(4.17)b0(x,z)ϕ0[∑i=1sti(f(x,y¯i)-λh(x,y¯i))-∑i=1sti(f(z,y¯i)-λh(z,y¯i))-∑j∈J0μjgj(z)+12pT∇2[∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∑j∈J0μjgj(z)]p]<0,
which by (4.5) implies
(4.18)F(x,z;∇∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+∇∑j∈J0μjgj(z)+∇2∑j∈J0μjgj(z)p)<-ρ0‖θ(x,z)‖2.
Also, inequality (4.4) along with (4.7) and (4.9) yields
(4.19)-bα(x,z)ϕα[∑j∈Jαμjgj(z)-12pT∇2∑j∈Jαμjgj(z)p]≤0,α=1,2,…,r.
From (4.6) and the above inequality, we have
(4.20)F(x,z;∇∑j∈Jαμjgj(z)+∇2∑j∈Jαμjgj(z)p)≤-ρα‖θ(x,z)‖2,α=1,2,…,r.
On adding (4.18) and (4.20) and making use of the sublinearity of ℱ with (4.10), we obtain
(4.21)F(x,z;∇∑i=1sti(f(z,y¯i)-λh(z,y¯i))+∇2∑i=1sti(f(z,y¯i)-λh(z,y¯i))p+∇∑j=1mμjgj(z)+∇2∑j=1mμjgj(z)p)<0,
which contradicts (4.2) since ℱ(x,z;0)=0.The proof of the following theorem is similar to that of Theorem3.3 and, hence, is omitted.Theorem 4.2 (strong duality).
Assume thatx* is an optimal solution of (1.1) and  ∇gj(x*),j∈J(x*), are linearly independent. Then there exist  (s*,t*,y¯*)∈K(x*) and  (x*,μ*,λ*,p*=0)∈H2(s*,t*,y¯*) such that  (x*,μ*,λ*,s*,t*,y¯*,p*=0) is a feasible solution of (4.1) and the two objectives have the same values. Further, if the assumptions of weak duality (Theorem 4.1) hold for all feasible solutions  (z,μ,λ,s,t,y¯,p) of (4.1), then  (x*,μ*,λ*,s*,t*,y¯*,p*=0) is an optimal solution of (4.1).Theorem 4.3 (strict converse duality).
Letx* and  (z*,μ*,λ*,s*,t*,y¯*,p*) be the optimal solutions of (1.1) and (4.1), respectively. Suppose that  ∇gj(x*),j∈J(x*) are linearly independent and there exist  ℱ,θ,ϕ0,b0,ρ0 and  ϕβ,bβ,ρβ,β=1,2,…,r such that
(4.22)b0(x*,z*)ϕ0[(∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∑j∈J0μj*gj(z*))∑i=1s*ti*(f(x*,y¯i*)-λ*h(x*,y¯i*))-∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))-∑j∈J0μj*gj(z*)+12p*T∇2(∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∑j∈J0μj*gj(z*))p*]≤0⟹F(x*,z*;∇∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∇2∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))p*+∇∑j∈J0μj*gj(z*)+∇2∑j∈J0μj*gj(z*)p*)<-ρ0‖θ(x*,z*)‖2(4.23)-bα(x*,z*)ϕα[∑j∈Jαμj*gj(z*)-12p*T∇2∑j∈Jαμj*gj(z*)p*]≤0⟹F(x*,z*;∇∑j∈Jαμj*gj(z*)+∇2∑j∈Jαμj*gj(z*)p*)≤-ρα‖θ(x*,z*)‖2,α=1,2,…,r.
Further assume that
(4.24)a≥0⟹ϕα(a)≥0,α=1,2,…,r,(4.25)a<0⟹ϕ0(a)≤0,(4.26)b0(x*,z*)>0,bα(x*,z*)≥0,α=1,2,…,r,(4.27)ρ0+∑α=1rρα≥0.
Then z*=x*, that is, z* is an optimal solution of (1.1).Proof.
Suppose contrary to the result thatz*≠x*. Since x* and (z*,μ*,λ*,s*,t*,y¯*,p*) are optimal solutions of (1.1) and (4.1), respectively, and ∇gj(x*),j∈J(x*) are linearly independent, therefore, from strong duality (Theorem 4.2), we reach
(4.28)sup⁡y*∈Yf(x*,y*)h(x*,y*)=λ*.
Thus, we have
(4.29)f(x*,y¯i*)-λ*h(x*,y¯i*)≤0,∀y¯i*∈Y(x*),i=1,2,…,s*.
Now, proceeding as in Theorem 4.1, we get
(4.30)∑i=1s*ti*(f(x*,y¯i*)-λ*h(x*,y¯i*))-∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))-∑j∈J0μj*gj(z*)+12p*T∇2(∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∑j∈J0μj*gj(z*))p*<0.
Using (4.25) and (4.26), it follows from (4.30) that
(4.31)b0(x*,z*)ϕ0[∑i=1s*ti*(f(x*,y¯i*)-λ*h(x*,y¯i*))-∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))-∑j∈J0μj*gj(z*)+12p*T∇2(∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∑j∈J0μj*gj(z*))p*]≤0,
which by (4.22) implies
(4.32)F(x*,z*;∇∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∇2∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))p*+∇∑j∈J0μj*gj(z*)+∇2∑j∈J0μj*gj(z*)p*)<-ρ0‖θ(x*,z*)‖2.
Also, inequality (4.4) along with (4.24) and (4.26) yields
(4.33)-bα(x*,z*)ϕα[∑j∈Jαμj*gj(z*)-12p*T∇2∑j∈Jαμj*gj(z*)p*]≤0,α=1,2,…,r.
From (4.23) and the above inequality, we have
(4.34)F(x*,z*;∇∑j∈Jαμj*gj(z*)+∇2∑j∈Jαμj*gj(z*)p*)≤-ρα‖θ(x*,z*)‖2,α=1,2,…,r.
On adding (4.32) and (4.34) and making use of the sublinearity of ℱ with (4.27), we obtain
(4.35)F(x*,z*;∇∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))+∇2∑i=1s*ti*(f(z*,y¯i*)-λ*h(z*,y¯i*))p*+∇∑j=1mμj*gj(z*)+∇2∑j=1mμj*gj(z*)p*)<0,
which contradicts (4.2) since ℱ(x*,z*;0)=0.


## 5. Conclusion and Further Developments

In this paper, we have established weak, strong, and strict converse duality theorems for a class of minimax fractional programming problems in the frame work of second-order generalized convexity. The second-order duality results developed in this paper can be further extended for the following nondifferentiable minimax fractional programming problem [22, 23]:(5.1)minimizeψ(x)=sup⁡y∈Yf(x,y)+(xTBx)1/2h(x,y)-(xTDx)1/2,subjecttog(x)≤0,x∈Rn,
where Y is a compact subset of Rl, B and D are n×n positive semidefinite symmetric matrices, and f(·,·):Rn×Rl→R,   h(·,·):Rn×Rl→R, and g(·):Rn→Rm are twice continuously differentiable functions on Rn×Rl, Rn×Rl, and Rn, respectively.The question arises as to whether the second-order fractional duality results developed in this paper hold for the following complex nondifferentiable minimax fractional problem:(5.2)minimizeΨ(ξ)=sup⁡ν∈WRe[f(ξ,ν)+(zTBz)1/2]Re[h(ξ,ν)-(zTDz)1/2],subjectto-g(ξ)∈S,ξ∈C2n,
where ξ=(z,z¯),ν=(ω,ω) for z∈𝒞n,ω∈𝒞l,f(·,·):𝒞2n×𝒞2l→𝒞 and h(·,·):𝒞2n×𝒞2l→𝒞 are analytic with respect to ω, W ia a specified compact subset in 𝒞2l,S is a polyhedral cone in 𝒞m, and g:𝒞2n→𝒞m is analytic. Also B,D∈𝒞n×n are positive semidefinite Hermitian matrices.



---

*Source: 563924-2011-10-18.xml*