On second-order duality for a class of nondifferentiable minimax fractional programming problem with...

$: On second-order duality for a class of nondifferentiable minimax fractional programming problem with (C,α,ρ,d)-convexity$
JAMCJ Appl Math ComputDOI 10.1007/s12190-013-0649-1

O R I G I NA L R E S E A R C H

On second-order duality for a class of nondifferentiableminimax fractional programming problemwith (C,α,ρ,d)-convexity

D. Dangar · S.K. Gupta

Received: 26 November 2012© Korean Society for Computational and Applied Mathematics 2013

Abstract In this paper, we derive appropriate duality theorems for three second-order dual models of a nondifferentiable minimax fractional programming prob-lem under second-order (C,α,ρ, d)-convexity assumptions. A nontrivial examplehas also been exemplified to show the existence of second-order (C,α,ρ, d)-convexfunctions. Several known results including many recent works are obtained as specialcases.

Keywords Minimax fractional programming · Nondifferentiable programming ·Second-order duality · (C,α,ρ, d)-convexity

Mathematics Subject Classification 49J35 · 90C32 · 49N15

1 Introduction

Schmitendorf [14] first developed necessary and sufficient conditions for a general-ized minimax programming problem. Tanimoto [16] defined a dual problem to theminimax problem discussed in [14] and proved usual duality results under convexityassumptions. After that, several researchers have worked to develop optimality con-ditions and duality relations for differentiable/nondifferentiable minimax fractionalprogramming problems.

Chandra and Kumar [3] modified two dual problems discussed in [17] and studiedduality theorems under convexity assumptions. Motivated by Chandra and Kumar

D. DangarDepartment of Mathematics, Indian Institute of Technology, Patna 800 013, Indiae-mail: [email protected]

S.K. Gupta (�)Department of Mathematics, Indian Institute of Technology, Roorkee 247 667, Indiae-mail: [email protected]

mailto:[email protected]

mailto:[email protected]

D. Dangar, S.K. Gupta

[3], Liu and Wu [11, 12] established sufficient optimality conditions for the mini-max fractional programming problem and derived usual duality results for three dualmodels under η-pseudo-invexity/quasi-invexity/(F,ρ)-convexity assumptions. Lateron, Yang and Hou [18] extended the work in [3, 11, 12] in the framework of gen-eralized convexity. In the spirit of [9–12], Yuan et al. [19] introduced the concept of(C,α,ρ, d)-convexity and obtained necessary and sufficient optimality conditions fora nondifferentiable minimax fractional programming problem. Long [13] has provedsufficient optimality conditions and usual duality relations for a nondifferentiablemulti-objective fractional programming problem under (C,α,ρ, d)-convexity.

This paper deals with the following nondifferentiable minimax fractional program-ming problem:

(P) Minimize ξ(x) = supy∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2

subject to g(x) ≤ 0,

where Y is a compact subset of Rl , f : Rn × Rl → R, h : Rn × Rl → R aretwice continuously differentiable on Rn × Rl and g : Rn → Rm is twice con-tinuously differentiable on Rn, B and D are n × n positive semidefinite matrix,f (x, y) + (xT Bx)1/2 ≥ 0 and h(x, y) − (xT Dx)1/2 > 0 for each (x, y) ∈ J × Y ,where J = {x ∈ Rn : g(x) ≤ 0}.

Ahmad and Husain [1] derived sufficient optimality conditions for the problem(P) using (F,α,ρ, d)-convexity and discussed appropriate duality results for the twodual models of (P). Introducing the concept of α-univexity, Jayswal [6] proved suf-ficient optimality conditions for (P) and obtained duality relations for the three dualproblems of (P).

Second-order duality for a minimax fractional programming problem was firstintroduced by Husain et al. [5]. This work was later extended in Hu et al. [4] by in-cluding an additional vector r to the dual models considered in [5] and establishedduality relations involving η-bonvex functions. Recently, Sharma and Gulati [15]have studied second-order duality for a minimax fractional programming problemunder second order generalized α-type I univexity. In this paper, we have introducedthe concept of second-order (C,α,ρ, d)-convexity and discussed weak, strong, strictconverse duality theorems connecting (P) and its three second-order dual models.An example has been illustrated to show the existence of second-order (C,α,ρ, d)-convex functions. Special cases have also been discussed to show that the resultsobtained in this paper generalizes some existing known works in the literature.

2 Notations and preliminaries

For each (x, y) ∈ J × Y , we define

J (x) = {j ∈ M = {1,2, . . . ,m} : gj (x) = 0

},

Y (x) ={y ∈ Y : f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2= sup

z∈Y

f (x, z) + (xT Bx)1/2

h(x, z) − (xT Dx)1/2

},

On second-order duality

K(x) ={(s, t, y) ∈ N × Rs+ × Rls : 1 ≤ s ≤ n + 1, t = (t1, t2, . . . , ts) ∈ Rs+,

s∑

i=1

ti = 1, y = (y1, y2, . . . , ys), yi ∈ Y(x), i = 1,2, . . . , s

}.

Now, we need the following definitions in sequel:Let X ⊂ Rn and α : X × X → R+\{0}, ρ ∈ R and let d : X × X → R+ be a

function with the property that d(x, x0) = 0 ⇔ x = x0. Let C : X × X × Rn → R bea function such that Cx,x0(0) = 0 for any (x, x0) ∈ X × X.

Definition 2.1 ([19]) A function C : X ×X ×Rn → R is said to be convex on Rn ifffor any fixed (x, x0) ∈ X × X and for any y1, y2 ∈ Rn,

Cx,x0

(λy1 + (1 − λ)y2

) ≤ λCx,x0(y1) + (1 − λ)Cx,x0(y2), ∀λ ∈ (0,1).

The following definition requires the condition of convexity of C:

Definition 2.2 A twice differentiable function h : X → R is said to be second-order(C,α,ρ, d)-convex at x0 ∈ X iff for any x ∈ X, there exists vector p ∈ Rn such that

1

α(x, x0)

[h(x) − h(x0) + 1

2pT ∇xxh(x0)p

]

≥ Cx,x0

(∇xh(x0) + ∇xxh(x0)p) + ρd(x, x0)

α(x, x0).

Definition 2.3 A twice differentiable function h : X → R is said to be strictlysecond-order (C,α,ρ, d)-convex at x0 ∈ X iff for any x ∈ X, x = x0, there existsvector p ∈ Rn such that

1

α(x, x0)

[h(x) − h(x0) + 1

2pT ∇xxh(x0)p

]

> Cx,x0

(∇xh(x0) + ∇xxh(x0)p) + ρd(x, x0)

α(x, x0).

The function h is said to be second-order (strictly second-order) (C,α,ρ, d)-convexover X iff it is second-order (strictly second-order) (C,α,ρ, d)-convex at every pointin X. For p to be zero vector Definition 2.2 reduces to the definition considered in[13, 19].

Example 2.1 Let X = [0,∞) ⊂ R. Let h : X → R be defined as h(x) = x2 −2 sin2 x.Suppose α : X × X → R+\{0} and C : X × X × Rn → R are given by α(x, x0) =0.50, Cx,x0(a) = 3ax and d : X × X → R+ be defined by d(x, x0) = (x − x0)

2. Forp = −2, ρ = −3 and x0 = 0.5π , we have

A = 1

α(x, x0)

[h(x) − h(x0) + 1

2pT ∇xxh(x0)p

]


Fig. 1 Graph of A versus x

− Cx,x0

(∇xh(x0) + ∇xxh(x0)p) − ρd(x, x0)

α(x, x0)

= 2x2 + 4 cos2 x − 0.5π2 − 3πx + 36x + 6(x − 0.5π)2 + 24

= 4 cos2 x + 36x + 8(x − 9π/16)2 + 24 − 392π2/256

≥ 0 ∀x ∈ [0,∞).

This can also be seen from Fig. 1. Hence A ≥ 0. Therefore, the function h is second-order (C,α,ρ, d)-convex but h is not (C,α,ρ, d)-convex since for x = 0.5π , wehave

1

α(x, x0)

[h(x) − h(x0)

] − Cx,x0

(∇xh(x0)) − ρd(x, x0)

α(x, x0)

= 2x2 + 4 cos2 x − 0.5π2 − 3πx + 6(x − 0.5π)2

= −1.5π2 < 0.

Lemma 2.1 (Generalized Schwarz inequality) Let B be a positive semidefinite ma-trix of order n. Then, for all x,w ∈ Rn,

xT Bw ≤ (xT Bx

)1/2(wT Bw

)1/2.

The equality holds if Bx = λBw for some λ ≥ 0.

The following Theorem 2.1 ([8], Theorem 3.1) will be required to prove strongduality theorems:


Theorem 2.1 (Necessary condition) If x∗ is an optimal solution of problem (P) satis-fying x∗T Bx∗ > 0, x∗T Dx∗ > 0, and ∇gj (x

∗), j ∈ J (x∗) are linearly independent,then there exist (s, t∗, y) ∈ K(x∗), λ∗ ∈ R+, w,v ∈ Rn and μ∗ ∈ Rm+ such that

s∑

i=1

t∗i{∇f

(x∗, yi

) + Bw − λ∗(∇h(x∗, yi

) − Dv)} +

m∑

j=1

μ∗j∇gj

(x∗) = 0, (1)

f(x∗, yi

) + (x∗T Bx∗)1/2 − λ∗(h

(x∗, yi

) − (x∗T Dx∗)1/2) = 0,

i = 1,2, . . . , s, (2)

m∑

j=1

μ∗j gj

(x∗) = 0, (3)

t∗i ≥ 0 (i = 1,2, . . . , s),

s∑

i=1

t∗i = 1, (4)

wT Bw ≤ 1, vT Dv ≤ 1,(5)(

x∗T Bx∗)1/2 = x∗T Bw,(x∗T Dx∗)1/2 = x∗T Dv.

In the above theorem, both matrices B and D are positive semidefinite at the so-lution x∗. If either x∗T Bx∗ or x∗T Dx∗ is zero, then the functions involved in theobjective function of problem (P) are not differentiable. To derive necessary condi-tions under this situation, for (s, t∗, y) ∈ K(x∗), define

Zy

(x∗) = {

z ∈ Rn : zT ∇gj

(x∗) ≤ 0, j ∈ J

(x∗),

and any one of the next conditions (i)–(iii) holds}.

(i) x∗T Bx∗ > 0, x∗T Dx∗ = 0

⇒ zT

(s∑

i=1

t∗i{∇f

(x∗, yi

) + Bx∗

(x∗T Bx∗)1/2− λ∗∇h

(x∗, yi

)})

+ (zT

(λ∗2D

)z)1/2

< 0,

(ii) x∗T Bx∗ = 0, x∗T Dx∗ > 0

⇒ zT

(s∑

i=1

t∗i{∇f

(x∗, yi

) − λ∗(

∇h(x∗, yi

) − Dx∗

(x∗T Dx∗)1/2

)})

+ (zT Bz

)1/2< 0,


(iii) x∗T Bx∗ = 0, x∗T Dx∗ = 0

⇒ zT

(s∑

i=1

t∗i{∇f

(x∗, yi

) − λ∗∇h(x∗, yi

)})

+ (zT

(λ∗2D

)z)1/2

+ (zT Bz

)1/2< 0.

If in addition, Zy(x∗) = φ, then the result of Theorem 2.1 still holds.

Throughout this paper, we use

ψ1(.) =s∑

i=1

ti[(

h(z, yi ) − zT Dv)(

f (., yi ) + (.)T Bw)

− (f (z, yi ) + zT Bw

)(h(., yi ) − (.)T Dv

)],

ψ2(.) =[

s∑

i=1

ti(h(z, yi ) − zT Dv

)][

s∑

i=1

ti(f (., yi ) + (.)T Bw

) +m∑

j=1

μjgj (.)

]

−[

s∑

i=1

ti(f (z, yi) + zT Bw

) +m∑

j=1

μjgj (z)

][s∑

i=1

ti(h(., yi ) − (.)T Dv

)]

and

ψ3(.) =[

s∑

i=1

ti(f (., yi) + (.)T Bw − λ

(h(., yi ) − (.)T Dv

))]

.

3 Duality Model I

In this section, we consider the following dual problem to (P):

(DM1) max(s,t,y)∈K(z)

sup(z,μ,w,v,p,r)∈H1(s,t,y)

F (z),

where F(z) = supy∈Y

f (z,y)+(zT Bz)1/2

h(z,y)−(zT Dz)1/2 and H1(s, t, y) denotes the set of all (z,μ,

w,v,p, r) ∈ Rn × Rm+ × Rn × Rn × Rn × Rn satisfying

∇ψ1(z) + ∇2ψ1(z)p +m∑

j=1

μj∇gj (z) + ∇2m∑

j=1

μjgj (z)p = 0, (6)

m∑

j=1

μjgj (z) +(

m∑

j=1

μj∇gj (z)

)T

r − 1

2pT ∇2

m∑

j=1

μjgj (z)p ≥ 0, (7)


(∇ψ1(z))T

r − 1

2pT ∇2ψ1(z)p ≥ 0, (8)

(∇ψ1(z))T

r +(

m∑

j=1

μj∇gj (z)

)T

r ≤ 0, (9)

wT Bw ≤ 1, vT Dv ≤ 1,

(10)(zT Bz

)1/2 = zT Bw,(zT Dz

)1/2 = zT Dv.

If the set H1(s, t, y) = φ, we define the supremum of F(z) over H1(s, t, y) equal to−∞.

Theorem 3.1 (Weak duality) Let x and (z,μ,w,v, s, t, y,p, r) be feasible solutionsof (P) and (DM1), respectively. Assume that

(i) ψ1(.) and gj (.) are second-order (C,α,ρ, d) and (C,β, γj , cj )-convex, j =1,2, . . . ,m, respectively at z with α = β and

(ii) ρd(x, z) + ∑mj=1 μjγj cj (x, z) ≥ 0.

Then

supy∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2≥ F(z).

Proof Contrary to the result suppose

supy∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2< F(z). (11)

For any yi ∈ Y(z), i = 1,2, . . . , s, we have

F(z) = f (z, yi) + (zT Bz)1/2

h(z, yi ) − (zT Dz)1/2. (12)

From (11) and (12), we get

f (x, yi) + (xT Bx)1/2

h(x, yi ) − (xT Dx)1/2≤ sup

y∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2

<f (z, yi) + (zT Bz)1/2

h(z, yi ) − (zT Dz)1/2, i = 1,2, . . . , s.

As h(., yi ) − ((.)T D(.))1/2 > 0, therefore, we have

[(h(z, yi ) − (

zT Dz)1/2)(

f (x, yi) + (xT Bx

)1/2)

− (f (z, yi ) + (

zT Bz)1/2)(

h(x, yi) − (xT Dx

)1/2)]< 0, i = 1,2, . . . , s


which from ti ≥ 0, i = 1,2, . . . , s and t = 0 yields

s∑

i=1

ti[(

h(z, yi ) − (zT Dz

)1/2)(f (x, yi) + (

xT Bx)1/2)

− (f (z, yi) + (

zT Bz)1/2)(

h(x, yi) − (xT Dx

)1/2)]< 0. (13)

Now,

ψ1(x) =s∑

i=1

ti[(

h(z, yi ) − zT Dv)(

f (x, yi) + xT Bw)

− (f (z, yi) + zT Bw

)(h(x, yi) − xT Dv

)].

This further using Lemma 2.1 and (10) yields

ψ1(x) ≤s∑

i=1

ti[(


)1/2)(f (x, yi) + (

xT Bx)1/2)

− (f (z, yi ) + (

zT Bz)1/2)(

h(x, yi) − (xT Dx

)1/2)].

In view of (13), the above inequality gives

ψ1(x) < 0 = ψ1(z). (14)

Now, by the hypothesis (i), we have

1

α(x, z)

[ψ1(x) − ψ1(z) + 1

2pT ∇2ψ1(z)p

]

≥ Cx,z

(∇ψ1(z) + ∇2ψ1(z)p) + ρd(x, z)

α(x, z)(15)

and

1

β(x, z)

[gj (x) − gj (z) + 1

2pT ∇2gj (z)p

]

≥ Cx,z

(∇gj (z) + ∇2gj (z)p) + γj cj (x, z)

β(x, z), j = 1,2, . . . ,m. (16)

Denote τ = 1 +∑mj=1 μj . It is easy to see that τ > 0. Multiplying (15) by 1

τand (16)

byμj

τ, j = 1,2, . . . ,m and using feasibility of x for (P) in (16), we have

1

τα(x, z)

[ψ1(x) − ψ1(z) + 1

2pT ∇2ψ1(z)p

]

≥ 1

τCx,z

(∇ψ1(z) + ∇2ψ1(z)p) + ρd(x, z)

τα(x, z)(17)


and

μj

τβ(x, z)

[−gj (z) + 1

2pT ∇2gj (z)p

]

≥ μj

τCx,z

(∇gj (z) + ∇2gj (z)p) + μjγj cj (x, z)

τβ(x, z), j = 1,2, . . . ,m. (18)

Summing (18) for all j = 1,2, . . . ,m and then adding with (17), using α = β and theconvexity of C, we obtain

1

τα(x, z)

[

ψ1(x) − ψ1(z) + 1

2pT ∇2

(

ψ1(z) +m∑

j=1

μjgj (z)

)

p −m∑

j=1

μjgj (z)

]

≥ Cx,z

(1

τ

[

∇ψ1(z) + ∇2ψ1(z)p +m∑

j=1

μj∇gj (z) + ∇2m∑

j=1

μjgj (z)p

])

+ 1

τα(x, z)

{

ρd(x, z) +m∑

j=1

μjγj cj (x, z)

}

. (19)

Now, from dual constraints (7)–(9), we obtain

−m∑

j=1

μjgj (z) + 1

2pT ∇2

(

ψ1(z) +m∑

j=1

μjgj (z)

)

p ≤ 0.

This together with (6), (19), hypothesis (ii) and Cx,z(0) = 0 yields

ψ1(x) ≥ ψ1(z) (20)

which contradicts (14). Hence proved. �


(i) f (., yi )+ (.)T Bw, −h(., yi )+ (.)T Dv and gj (.) are second-order (C,α,ρi, di),(C,α′, ρ′

i , d′i ) and (C,β, γj , cj )-convex, i = 1,2, . . . , s, j = 1,2, . . . ,m, respec-

tively at z with α = α′ = β and

(ii)s∑

i=1

ti{(


)1/2)ρidi(x, z) + (

f (z, yi) + (zT Bz

)1/2)ρ′

id′i (x, z)

}

+m∑

j=1

μjγj cj (x, z) ≥ 0

.

Then

supy∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2≥ F(z).


Proof Following the lines of proof of Theorem 3.1, we have

ψ1(x) < ψ1(z). (21)

From hypotheses (i), we get

1

α(x, z)

[f (x, yi) + xT Bw − (

f (z, yi ) + zT Bw) + 1

2pT ∇2f (z, yi )p

]

≥ Cx,z

(∇f (z, yi) + Bw + ∇2f (z, yi )p) + ρidi(x, z)

α(x, z), (22)

1

α′(x, z)

[−h(x, yi) + xT Dv + h(z, yi ) − zT Dv − 1

2pT ∇2h(z, yi )p

]

≥ Cx,z

(−∇h(z, yi ) + Dv − ∇2h(z, yi )p) + ρ′

id′i (x, z)

α′(x, z), (23)

1

β(x, z)

[gj (x) − gj (z) + 1

2pT ∇2gj (z)p

]

≥ Cx,z

(∇gj (z) + ∇2gj (z)p) + γj cj (x, z)

β(x, z), j = 1,2, . . . ,m. (24)

Denote Γ = ∑si=1 ti (h(z, yi ) − (zT Dz)1/2 + f (z, yi ) + (zT Bz)1/2) + ∑m

j=1 μj .

Therefore, Γ > 0 as ti ≥ 0, t = 0, μj ≥ 0, h(z, yi ) − (zT Dz)1/2 > 0 and f (z, yi) +(zT Bz)1/2 ≥ 0 for i = 1,2, . . . , s, j = 1,2, . . . ,m. Now, from the inequality (24),μj

Γ≥ 0, j = 1,2, . . . ,m and feasibility of x for (P) it follows that

μj

Γβ(x, z)

[−gj (z) + 1

2pT ∇2gj (z)p

]

≥ μj

ΓCx,z

(∇gj (z) + ∇2gj (z)p) + μjγj cj (x, z)

Γβ(x, z), j = 1,2, . . . ,m. (25)

Applying α = α′ = β and convexity of C and using (10) in the addition of the ex-pression obtained by multiplying and summing (22) by ti

Γ[h(z, yi )− (zT Dz)1/2] ≥ 0

and (23) by tiΓ

[f (z, yi ) + (zT Bz)1/2] ≥ 0 for i = 1,2, . . . , s and summing (25) forj = 1,2, . . . ,m, we get

1

Γ α(x, z)

[

ψ1(x) − ψ1(z) + 1

2pT ∇2

(

ψ1(z) +m∑

j=1

μjgj (z)

)

p −m∑

j=1

μjgj (z)

]

≥ Cx,z

(1

Γ

{

∇ψ1(z) + ∇2ψ1(z)p +m∑

j=1

μj∇gj (z) + ∇2m∑

j=1

μjgj (z)p

})


+ 1

α(x, z)

[s∑

i=1

ti{(

h(z, yi) − (zT Dz

)1/2)ρidi(x, z)

+ (f (z, yi ) + (

zT Bz)1/2)

ρ′id

′i (x, z)

} +m∑

j=1

μjγj cj (x, z)

]

which using (6) and hypothesis (ii) yields

ψ1(x) − ψ1(z) + 1

2pT ∇2

(

ψ1(z) +m∑

j=1

μjgj (z)

)

p −m∑

j=1

μjgj (z) ≥ 0.

This further from (7)–(9) implies

ψ1(x) ≥ ψ1(z)

which contradicts (21). This completes the proof. �

Theorem 3.3 (Strong duality) Let x∗ be an optimal solution for (P) and let ∇gj (x∗),

j ∈ J (x∗) be linearly independent. Then, there exist (s∗, t∗, y∗) ∈ K(x∗) and(x∗,μ∗,w∗, v∗,p∗ = 0, r∗ = 0) ∈ H1(s

∗, t∗, y∗) such that (x∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗ = 0, r∗ = 0) is feasible solution of (DM1) and the two objectives have same val-ues. If, in addition, the assumption of weak duality hold for all feasible solutions(x,μ,w,v, s, t, y,p, r) of (DM1), then (x∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗ = 0, r∗ = 0)

is an optimal solution of (DM1).

Proof Since x∗ is an optimal solution of (P) and ∇gj (x∗), j ∈ J (x∗) are lin-

early independent, then by Theorem 2.1, there exist (s∗, t∗, y∗) ∈ K(x∗) and(x∗,μ∗,w∗, v∗,p∗ = 0, r∗ = 0) ∈ H1(s

∗, t∗, y∗) such that (x∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗ = 0, r∗ = 0) is feasible solution of (DM1) and the two objectives have same val-ues. Optimality of (x∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗ = 0, r∗ = 0) for (DM1), thus fol-lows from Theorem 3.1 or 3.2. �

Theorem 3.4 (Strict converse duality) Let x∗ be an optimal solution to (P) and(z∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗, r∗) be an optimal solution to (DM1). Assume that

(i) ψ1(.) and gj (.) are strictly second-order (C,α,ρ, d) and second-order (C,β,

γj , cj )-convex, j = 1,2, . . . ,m, respectively at z∗ with α = β ,(ii) {∇gj (x

∗), j ∈ J (x∗)} are linearly independent and(iii) ρd(x∗, z∗) + ∑m

j=1 μ∗j γj cj (x

∗, z∗) ≥ 0.

Then z∗ = x∗.

Proof Suppose z∗ = x∗. Since x∗ and (z∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗, r∗) are optimalsolutions to (P) and (DM1), respectively and {∇gj (x

∗), j ∈ J (x∗)} are linearly inde-pendent, by Theorem 3.3, we have

supy∗∈Y

f (x∗, y∗) + (x∗T Bx∗)1/2

h(x∗, y∗) − (x∗T Dx∗)1/2= F

(z∗).


Then, for y∗i ∈ Y(z∗), we get

f (x∗, y∗i ) + (x∗T Bx∗)1/2

h(x∗, y∗i ) − (x∗T Dx∗)1/2

≤ supy∗∈Y

f (x∗, y∗) + (x∗T Bx∗)1/2

h(x∗, y∗) − (x∗T Dx∗)1/2

= f (z∗, y∗i ) + (z∗T Bz∗)1/2

h(z∗, y∗i ) − (z∗T Dz∗)1/2

.

This together with t∗i ≥ 0, i = 1,2, . . . , s∗ and t∗ = 0 implies

s∗∑

i=1

t∗i[(

h(z∗, y∗

i

) − (z∗T Dz∗)1/2)(

f(x∗, y∗

i

) + (x∗T Bx∗)1/2)

− (f

(z∗, y∗

i

) + (z∗T Bz∗)1/2)(

h(x∗, y∗

i

) − (x∗T Dx∗)1/2)] ≤ 0. (26)

Now,

ψ1(x∗) =

s∗∑

i=1

t∗i[(

h(z∗, y∗

i

) − z∗T Dv∗)(f(x∗, y∗

i

) + x∗T Bw∗)

− (f

(z∗, y∗

i

) + z∗T Bw∗)(h(x∗, y∗

i

) − x∗T Dv∗)]

≤s∗∑

i=1

t∗i[(

h(z∗, y∗

i

) − (z∗T Dz∗)1/2)(

f(x∗, y∗

i

) + (x∗T Bx∗)1/2)

− (f

(z∗, y∗

i

) + (z∗T Bz∗)1/2)(

h(x∗, y∗

i

) − (x∗T Dx∗)1/2)]

(using Lemma 2.1 and (10)

)

≤ 0(from (26)

).

Hence,

ψ1(x∗) ≤ 0 = ψ1

(z∗). (27)

By hypothesis (i), we have

1

α(x∗, z∗)

[ψ1

(x∗) − ψ1

(z∗) + 1

2p∗T ∇2ψ1

(z∗)p∗

]

> Cx∗,z∗(∇ψ1

(z∗) + ∇2ψ1

(z∗)p∗) + ρd(x∗, z∗)

α(x∗, z∗)(28)

and

1

β(x∗, z∗)

[gj

(x∗) − gj

(z∗) + 1

2p∗T ∇2gj

(z∗)p∗

]

≥ Cx∗,z∗(∇gj

(z∗) + ∇2gj

(z∗)p∗) + γj cj (x

∗, z∗)β(x∗, z∗)

, j = 1,2, . . . ,m. (29)


Denote Γ ∗ = 1 +∑mj=1 μ∗

j > 0. Therefore, employingμ∗

j

Γ ∗ in (29) and then summingover j = 1,2, . . . ,m with the feasibility of x for (P), we have

1

Γ ∗β(x∗, z∗)

[

−m∑

j=1

μ∗j gj

(z∗) + 1

2p∗T ∇2

m∑

j=1

μ∗j gj

(z∗)p∗

]

≥m∑

j=1

μ∗j

Γ ∗ Cx∗,z∗(∇gj

(z∗) + ∇2gj

(z∗)p∗) +

m∑

j=1

μ∗j γj

cj (x∗, z∗)

β(x∗, z∗). (30)

Multiplying (28) by 1Γ ∗ and then adding with (30), using the convexity of C, (7)–(9)

and hypothesis (iii), we obtain

ψ1(x∗) − ψ1(z

∗)Γ ∗α(x∗, z∗)

> Cx∗,z∗

(1

Γ ∗

[

∇ψ1(z∗) + ∇2ψ1

(z∗)p∗

+m∑

j=1

μ∗j∇gj

(z∗) + ∇2

m∑

j=1

μ∗j gj

(z∗)p∗

])

which using (6) and Cx∗,z∗(0) = 0 gives

ψ1(x∗) > ψ1

(z∗).

This contradicts (27). Hence the result. �

Theorem 3.5 (Strict converse duality) Let x∗ be an optimal solution to (P) and(z∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗, r∗) be an optimal solution to (DM1). Assume that

(i) f (., yi ) + (.)T Bw∗ and −h(., yi ) + (.)T Dv∗ are strictly second-order (C,α,

ρi, di) and (C,α′, ρ′i , d

′i )-convex, i = 1,2, . . . , s∗ at z∗, respectively and gj (.)

be second-order (C,β, γj , cj )-convex at z∗, j = 1,2, . . . ,m, with α = α′ = β ,(ii) {∇gj (x

∗), j ∈ J (x∗)} are linearly independent and

(iii)s∗∑

i=1

t∗i{(

h(z∗, yi

) − (z∗T Dz∗)1/2)

ρidi

(x∗, z∗)

+ (f

(z∗, yi

) + (z∗T Bz∗)1/2)

ρ′id

′i

(x∗, z∗)} +

m∑

j=1

μ∗j γj cj

(x∗, z∗) ≥ 0.

Then z∗ = x∗.

Proof Follows on the lines of Theorems 3.2 and 3.4. �


4 Duality Model II

In this section, we consider the following dual problem to (P):


sup(z,μ,w,v,p,r)∈H2(s,t,y)

∑si=1 ti (f (z, yi ) + (zT Bz)1/2) + ∑m

j=1 μj gj (z)∑s

i=1 ti (h(z, yi ) − (zT Dz)1/2),

where H2(s, t, y) denotes the set of all (z,μ,w,v,p, r) ∈ Rn × Rm+ × R+ × Rn ×Rn × Rn × Rn satisfying

∇ψ2(z) + ∇2ψ2(z)p = 0, (31)

(∇ψ2(z))T

r + 1

2pT ∇2ψ2(z)p ≤ 0, (32)

(∇ψ2(z))T

r ≥ 0, (33)

wT Bw ≤ 1, vT Dv ≤ 1,

(zT Bz

)1/2 = zT Bw,(zT Dz

)1/2 = zT Dv.

(34)

If the set H2(s, t, y) is empty, we define the supremum in (DM2) over H2(s, t, y)

equal to −∞.


(i) ψ2(.) is second-order (C,α,ρ, d)-convex at z and(ii) ρ1 ≥ 0.

Then

supy∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2≥

∑si=1 ti (f (z, yi ) + (zT Bz)1/2) + ∑m

j=1 μjgj (z)∑s

i=1 ti (h(z, yi ) − (zT Dz)1/2).

Proof By the hypothesis (i), we get

1

α(x, z)

[ψ2(x) − ψ2(z) + 1

2pT ∇2ψ2(z)p

]

≥ Cx,z

(∇ψ2(z) + ∇2ψ2(z)p) + ρ1d(x, z)

α(x, z).

This further from hypothesis (ii), (31)–(33) and Cx,z(0) = 0 implies

ψ2(x) − ψ2(z)

α(x, z)≥ 0


i.e.

ψ2(x) ≥ ψ2(z). (35)

Now, contrary to the result suppose

supy∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2<

∑si=1 ti (f (z, yi ) + (zT Bz)1/2) + ∑m

j=1 μjgj (z)∑s

i=1 ti (h(z, yi ) − (zT Dz)1/2)

or

(f (x, yi) + (

xT Bx)1/2)

[s∑

i=1

ti(h(z, yi) − (

zT Dz)1/2)

]

<(h(x, yi) − (

xT Dx)1/2)

[s∑

i=1

ti(f (z, yi ) + (

zT Bz)1/2) +

m∑

j=1

μjgj (z)

]

,

∀yi ∈ Y(z), i = 1,2, . . . , s

which from ti ≥ 0, i = 1,2, . . . , s, t = 0 and (34) yields

s∑

i=1

ti(f (x, yi) + (

xT Bx)1/2)

[s∑

i=1

ti(h(z, yi) − zT Dv

)]

<

s∑

i=1

ti(h(x, yi) − (

xT Dx)1/2)

[s∑

i=1

ti(f (z, yi ) + zT Bw

) +m∑

j=1

μjgj (z)

]

.

From Lemma 2.1 and (34) it follows that

ψ2(x) ≤[

s∑

i=1

ti(f (x, yi) + (

xT Bx)1/2) +

m∑

j=1

μjgj (x)

]s∑

i=1


)

−s∑

i=1

ti(h(x, yi) − (

xT Dx)1/2)

[s∑

i=1

ti(f (z, yi ) + zT Bw

) +m∑

j=1

μjgj (z)

]

<

s∑

i=1


) m∑

j=1

μjgj (x)

≤ 0

(

sinces∑

i=1


)> 0 and

m∑

j=1

μjgj (x) ≤ 0

)

Therefore, we have

ψ2(x) < 0 = ψ2(z)

which contradicts (35). This proves the theorem. �

In a similar way, we can prove the following theorems between (P) and (DM2):


Theorem 4.2 (Strong duality) Let x∗ be an optimal solution for (P) and let∇gj (x

∗), j ∈ J (x∗) be linearly independent. Then, there exist (s∗, t∗, y∗) ∈ K(x∗)and (x∗,μ∗,w∗, v∗,p∗ = 0, r∗ = 0) ∈ H2(s

∗, t∗, y∗) such that (x∗,μ∗,w∗, v∗, s∗,t∗, y∗,p∗ = 0, r∗ = 0) is feasible solution of (DM2) and the two objectives havesame values. If, in addition, the assumption of weak duality hold for all feasible so-lutions (x,μ,w,v, s, t, y,p, r) of (DM2), then (x∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗ = 0,

r∗ = 0) is an optimal solution of (DM2).

Theorem 4.3 (Strict converse duality) Let x∗ be an optimal solution to (P) and(z∗,μ∗,w∗, v∗, s∗, t∗, y∗,p∗, r∗) be an optimal solution to (DM2). Let

(i) ψ2(.) be strictly second-order (C,α,ρ, d)-convex at z∗,(ii) {∇gj (x

∗), j ∈ J (x∗)} be linearly independent and(iii) ρ2 ≥ 0.

Then z∗ = x∗, that is z∗ is an optimal solution to (P).

5 Duality Model III

In this section, we consider another dual problem to (P) which is as follows:


sup(z,μ,λ,w,v,p)∈H2(s,t,y)

λ,

where H2(s, t, y) denotes the set of all (z,μ,λ,w,v,p) ∈ Rn × Rm+ × R+ × Rn ×Rn × Rn satisfying

∇ψ3(z) + ∇2ψ3(z)p +m∑

j=1

μj∇gj (z) + ∇2m∑

j=1

μjgj (z)p = 0, (36)

ψ3(z) + 1

2pT ∇2ψ3(z)p +

∑

j∈J0

μjgj (z) − 1

2pT ∇2

∑

j∈J0

μjgj (z)p ≥ 0, (37)

∑

j∈Jα

μjgj (z) − 1

2pT ∇2

∑

j∈Jα

μjgj (z)p ≥ 0, α = 1,2, . . . , k, (38)

wT Bw ≤ 1, vT Dv ≤ 1, (39)

where Jα ⊆ M , α = 0,1,2, . . . , k with⋃k

α=0 Jα = M and Jα ∩ Jβ = φ, if α = β . If,for a triplet (s, t, y) ∈ K(z), the set H2(s, t, y) = φ, then we define the supremumover it to be −∞.

Theorem 5.1 (Weak duality) Let x and (z,μ,λ,w,v, s, t, y,p) be feasible solutionsto (P) and (DM3), respectively. Assume that


(i) (ψ3(.) + ∑j∈J0

μjgj (.)) and gj (.), j ∈ Jα , α = 1,2, . . . , k are second-order

(C,α1, ρ1, d1)-convex and (C,α2, ρ2, d2)-convex at z, respectively and

(ii)ρ1d1(x, z)

α1(x, z)+

∑kα=1

∑j∈Jα

μjρ2αd2(x, z)

α2(x, z)≥ 0.

Then

supy∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2≥ λ.

Proof Suppose

supy∈Y

f (x, y) + (xT Bx)1/2

h(x, y) − (xT Dx)1/2< λ.

Therefore, we have

f (x, yi) + (xT Bx

)1/2 − λ(h(x, yi) − (

xT Dx)1/2)

< 0 ∀yi ∈ Y(x).

From ti ≥ 0, i = 1,2, . . . , s, t = 0, Lemma 2.1 and dual constraint (39) it follows that

s∑

i=1

ti[f (x, yi) + xT Bw − λ

(h(x, yi) − xT Dv

)]< 0

i.e.

ψ3(x) < 0

which by (37), feasibility at x of (P) and μj ≥ 0 yields

ψ3(x) − ψ3(z) +∑

j∈J0

μjgj (x) −∑

j∈J0

μjgj (z)

− 1

2pT ∇2

(ψ3(z) +

∑

j∈J0

μjgj (z)

)p < 0.

Then using second-order (C,α1, ρ1, d1)-convexity of (ψ3(.) + ∑j∈J0

μjgj (.)) at z,we have

Cx,z

(∇ψ3(z) + ∇2ψ3(z)p +

∑

j∈J0

μj∇gj (z) + ∇2∑

j∈J0

μjgj (z)p

)< −ρ1d1(x, z)

α1(x, z).

(40)Now, from the second-order (C,α2, ρ2, d2)-convexity of gj (.), j ∈ Jα , α = 1,2,

. . . , k at z, we get

gj (x) − gj (z) + 1

2pT ∇2gj (z)p ≥ Cx,z

(∇gj (z) + ∇2gj (z)p) + ρ2

αd2(x, z)

α2(x, z).


This implies forμj

τ≥ 0, j ∈ Jα (α = 1,2, . . . , k), where τ = 1 + ∑k

α=1∑

j∈Jαμj ,

with feasibility of x for (P) and the inequality (38) that

k∑

α=1

∑

j∈Jα

μj

τCx,z

(∇gj (z) + ∇2gj (z)p)< −

∑kα=1

∑j∈Jα

μjρ2αd2(x, z)

τα2(x, z). (41)

Multiplying (40) by τ > 0, we have

1

τCx,z

(∇ψ3(z)+∇2ψ3(z)p+

∑

j∈J0

μj∇gj (z)+∇2∑

j∈J0

μjgj (z)p

)< −ρ1d1(x, z)

τα1(x, z).

(42)Hence, using the convexity of C and hypothesis (ii) in the addition of (41) and (42),we obtain

Cx,z

(

∇ψ3(z) + ∇2ψ3(z)p +m∑

j=1

μj∇gj (z) + ∇2m∑

j=1

μjgj (z)p

)

< − 1

τ

(ρ1d1(x, z)

α1(x, z)+

∑kα=1

∑j∈Jα

μjρ2αd2(x, z)

α2(x, z)

)≤ 0.

This contradicts (36) as Cx,z(0) = 0. �

The proof of the following theorems can be obtained in a similar way:

Theorem 5.2 (Strong duality) Let x∗ be an optimal solution for (P) and let ∇gj (x∗),

j ∈ J (x∗) be linearly independent. Then, there exist (s∗, t∗, y∗) ∈ K(x∗) and(x∗,μ∗, λ∗,w∗, v∗,p∗ = 0) ∈ H2(s

∗, t∗, y∗) such that (x∗,μ∗, λ∗,w∗, v∗, s∗, t∗, y∗,p∗ = 0) is feasible solution of (DM3) and the two objectives have same val-ues. If, in addition, the assumption of weak duality hold for all feasible solutions(x,μ,λ,w,v, s, t, y,p) of (DM3), then (x∗,μ∗, λ∗,w∗, v∗, s∗, t∗, y∗,p∗ = 0) is anoptimal solution of (DM3).

Theorem 5.3 (Strict converse duality) Let x∗ and (z∗,μ∗, λ∗,w∗, v∗, s∗, t∗, y∗,p∗)be feasible solutions to (P) and (DM3) respectively. Assume that

(i) (ψ3(.) + ∑j∈J0

μ∗j gj (.)) and gj (.) are strictly second-order (C,α1, ρ1, d1)-

convex and second-order (C,α2, ρ2, d2)-convex at z∗, respectively,(ii) {∇gj (x

∗), j ∈ J (x∗)} are linearly independent and

(iii)ρ1d1(x∗, z∗)α1(x∗, z∗)

+∑k

α=1∑

j∈Jαμ∗

j ρ2αd2(x∗, z∗)

α2(x∗, z∗)≥ 0.

Then z∗ = x∗.

6 Special cases

(i) If p = 0 and r = 0, then the dual model (DM1) reduce to the problems studiedin [1, 6, 7].


(iii) If p = 0 and r = 0, then the dual model (DM2) becomes the dual model consid-ered in [2, 7].

(iv) If p = 0, r = 0 and B and D are zero matrices of order n, then (DM1) and(DM2) reduce to the problem studied in [11, 12].

(v) If B and D are zero matrices of order n, then (DM3) reduce to the problem(GMD) studied in [5].

7 Conclusion

In this paper, we have introduced the concept of second-order (C,α,ρ, d)-convexfunctions and formulated three types of second order dual models for a nondifferen-tiable minimax fractional programming problem. Further, we have established usualduality relations under this assumption. A nontrivial example has also been given toshow the existence second-order (C,α,ρ, d)-convex functions. It will be interestingto see whether or not the work in the present paper can be further extended to non-differentiable minimax fractional programming problem with the square root term inboth numerator and denominator of the objective function replaced by support func-tion of a compact convex set.

Acknowledgements The authors wish to thank anonymous reviewers for their constructive and valu-able suggestions which have considerably improved the presentation of the paper. The first author is alsothankful to MHRD, Government of India, New Delhi (India) for financial support.

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