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A UNIFIED APPROACH TO ZERO DUALITY GAP FOR CONVEX OPTIMIZATION PROBLEMS. Dong Thap University Journal of Science, Vol 11, No 5, 2022, 09 18 9 A UNIFIED APPROACH TO ZERO DUALITY GAP FOR CONVEX OPTIMIZATION PROBLEMS Dang Hai Long and Tran Hong Mo Faculty of Education and Ba.
Dong Thap University Journal of Science, Vol 11, No 5, 2022, 09-18 A UNIFIED APPROACH TO ZERO DUALITY GAP FOR CONVEX OPTIMIZATION PROBLEMS Dang Hai Long and Tran Hong Mo* Faculty of Education and Basic Sciences, Tien Giang University * Corresponding author: tranhongmo@tgu.edu.vn Article history Received: 13/5/2021; Received in revised form: 26/7/2021; Accepted: 08/9/2021 Abstract In this paper we establish necessary and sufficient condition for zero duality gap of the optimization problem involving the general perturbation mapping via characteringsetunder the convex setting An application to the class of composite optimization problems will also be given to show that our general results can be applied to various classes of optimization problems Keywords: Characterizing set, composite optimization problem, perturbation function, zero duality gap * , * : tranhongmo@tgu.edu.vn ịch sử b 13/5/2021 bá 26/7/2021 08/9/2021 óm tắt liên quan khóa: DOI: https://doi.org/10.52714/dthu.11.5.2022.975 Cite: Dang Hai Long and Tran Hong Mo (2022) A unified approach to zero duality gap for convex optimization problems Dong Thap University Journal of Science, 11(5), 9-18 Natural Sciences issue Introduction It is well known that duality theory plays an important role in optimization For a primal problem, there are different ways to define its dual problems (Feizollahi et al., 2017, Huang and Yang, 2003, Li, 1995, Yang and Huang, 2001) The zero duality gap is known as the state in which the optimal values of the primal problem and that of its dual problem are equal Many attempts have been made to study the zero duality gap for various classes of optimization problems in recent decades (Feizollahi et al., 2017, Huang and Yang, 2003, Jeyakumar and Li, 2009a, Jeyakumar and Li, 2009b, Jeyakumar and Wolkowicz, 1990, Li, 1995, Huang and Yang, 2003, Yang and Huang, 2001, Li, 1999, Long and Zeng, 2020, Rubinov et al., 2002) In this paper, we establish characterizations of zero duality gap property for the general optimization problem which can then be applied to many different specific classes optimization problems We are concerned with the so-called perturbation function : X Y {} and the optimization problem (P) The paper is organized as follows: In Section we recall some notation and introduce some preliminary results which will be used in the sequel Characterizing set and Lagrange dual problems of the problem (P) are introduced in Section with related basic properties Section is devoted to establish the main results of this paper, that is, the characterization of zero duality gap for the problem (P) under the convex setting As an illustrative example, in Section 5, we show how to apply generalized results to the classes of composite optimization problems Preliminaries Throughout the paper, we consider X and Y the locally convex Hausdorff topological vector spaces with topological dual spaces X and Y , respectively Y is a non-empty convex cones in Y while Y aims the set of positive functionals on Y with respect to Y , i.e., Y := { y* Y * : y* , k for all k Y } inf ( x,0Y ), xX Where X , Y are locally convex Hausdorff topological vector spaces, Y is non-empty convex cone in Y We assume in this paper that dom (.,0Y ) , or in other words, the problem (P) is feasible, meaning that (P) < It is worth commenting that many classes of optimization problems can be written in the form of (P) (see Boţ, 2010) So, investigating the problem (P) gives us a unified approach to all optimization problems In this paper, we study characterizations of the zero duality gap property for the problem (P) via its characterizing set which is inspired by the concept of characterizing set introduced by Dinh et al (2020) for the vector optimization with geometric and cone constrains It is worth observing that the characterizing set is rather simpler than those sets in the form of epigraph of conjugate mapping Therefore, the conditions imposed on the characterizing set will be easier to handle than the ones related to the epigraph of conjugate mapping proposed recently to examine 10 the zero duality gap property (see, e.g., Jeyakumar and Li, 2009a) f : X : {, ) Domain, Let epigraph, and hypograph of f are defined by, respectively, dom f := {x X : f ( x) }, epi f := {( x, ) X : f ( x) }, hyp f := {( x, ) X : f ( x) } f is said to be proper if f ( x) for all x X and dom f We say that f is convex if the following condition holds for all x1 , x2 X and (0,1) f ( x1 (1 ) x2 ) f ( x1 ) (1 ) f ( x2 ) It is easy to see that f is convex if and only if epi f is a convex subset of X The conjugate function of f is defined as f : X such that f ( x ) = sup[ x , x f ( x)] xX Y , We consider in Y the partial order induced by Y , defined as y1 Y y2 if and only if y2 y1 Y Dong Thap University Journal of Science, Vol 11, No 5, 2022, 09-18 ( y, r ) C x X : ( y, r ) epi ( x,.) We also enlarge Y by attaching a greatest element Y and a smallest element Y , which not x X : r ( x, y) belong to Y , and define Y := Y {Y , Y } Let H : X Y We say that H is a Y -convex mapping if, for all x1 , x2 X and (0,1), H ( x1 (1 ) x2 ) Y H ( x1 ) (1 ) H ( x2 ) We H define the domain of as dom H := {x X : H ( x) Y } and say that H is proper if Y H ( X ) and dom H When H is a proper mapping, the image and the graph of H are defined by, respectively, im H := {H ( x) : x dom H }, gr H := {( x, H ( x)) : x dom H } xX 3.1 Characterizing set (P), we (3.1) xX Proposition 3.1 Under the current assumption dom (.,0Y ) , one has (0Y , r ) C for some r In particular, C Proof As dom (.,0Y ) , there exists ( x ,0Y ) xX such that Take r := ( x ,0Y ) , one has (0Y , r ) epi ( x ,.) C, and we are done Proof Let us denote := {r : (0Y , r ) C} We will prove that (P) = inf (P) = inf ( x,0Y ) x dom H else epi ( x,.) Y The next proposition gives a presentation of the value of the problem (P) via its characterizing set C Firstly, recall that Characterizing set and Lagrange dual problems C := So, if is a convex function then epi is a convex subset of X Y which yields that C = Y epi is convex, as well (P) = inf {r : (0Y , r ) C} the meantime, for y Y , we convention that Corresponding to the problem consider the characterizing set ( y, r ) Y epi Proposition 3.3 It holds We say that g : Y is a Y -nondecreasing function if g ( y1 ) g ( y2 ) whenever y1 Y y2 In y , H ( x) if ( y H )( x) = x X : ( x, y, r ) The convexity of C is shown in the following proposition Proposition 3.2 If is convex then C is a convex subset of Y Proof We begin by proving that C is image of the set epi by the conical projection Y : X Y Y , Y ( x, y, r ) = ( y, r ) for all ( x, y, r ) X Y Indeed, for all ( y, r ) Y , (3.2) Take arbitrarily r Then, there exists a net (ri )iI such that (0Y , ri )iI C and ri r For each i I , as (0Y , ri ) C, there is xi X such that ( xi ,0Y ) ri By (3.2), ( xi ,0Y ) (P) , and hence, (P) ri for all i I Letting ri r , we get (P) r Take > (P) It follows from (3.2) that there is x X satisfying > ( x ,0Y ) := r Note that (0Y , r ) epi ( x ,.) C , which leads to r {r : (0Y , r ) C} Briefly, we have just shown that, for all > (P) , there exists r such that that > r So, (P) = inf and we are done 3.2 Lagrange dual problems The Lagrange dual problem and the loose Lagrange dual problem of (P) are defined as follows, respectively, (D) (D ) sup inf [ ( x, y ) y , y ], yY ( x , y )X Y sup inf [ ( x, y ) y , y ] yY* ( x , y )X Y 11 Natural Sciences issue ri ( xi , yi ) It is worth noting that y * in the dual problem (D) can be considered as the Lagrange multiplier while the one in (D ) also can be understood as a positive Lagrange multiplier Next, taking arbitrarily y Y , one has D := y Combining (3.4) and (3.5) gives D ri y , yi Proof The first inequality follows immediately from the property of supremum For the second inequality, taking arbitrarily x X , we will prove that y for all i I Proceeding to the limit, we obtain D r (recall that ( yi , ri ) (0Y , r ) ) So, y (D) = sup Dy r yY (3.3) Step Taking such that (D) , we will show that On the contrary, suppose that Then, it follows from this that Indeed, for all y Y , one has y inf { ( x, y ) y , y} ( x , y )X Y ( x ,0Y ) y ,0Y = ( x ,0Y ) (0Y , ) C As is a convex function, the set C is (D) = sup Dy ( x ,0Y ) convex (see Proposition 3.2), and hence, C is convex as well So, according to the separation theorem (see Rudin, 1991, Theorem 3.4), there are y Y , and such that Hence, yY We have just shown that (3.3) holds for any x X This leads to the fact that (D) inf ( x ,0Y ) = (P) x X The last one comes from the fact that (P) is feasible, and the proof is complete Theorem 3.1 Assume that is convex Then, one has Moreover, if (D) < < y , y r, ( y, r ) C then We next prove that > Fix x dom (.,0Y ) (it is possible as dom(.,0Y ) ) Then we have ( x ,0Y ) Set r = max{ , ( x ,0Y )} Then, one has r ( x ,0Y ) , hence, (0Y , r ) epi ( x ,.) C which, together with (3.6), yields < r , or of r ) we obtain > Consequently, it follows from this and (3.6) that (D) = min{r : (0Y , r ) C} < < y , y r, ( y, r ) C, Proof Denote := {r : (0Y , r ) C} It follows from Proposition 3.1 that Let us divide the proof into three steps (3.6) (r ) > Combining this equivalently, inequality with the fact that r (by the definition (D) = inf{r : (0Y , r ) C} Step Take arbitrarily r We claim that (D) r As r , one has (0Y , r ) C , and hence, there exists a net (( yi , ri ))iI C such that ( yi , ri ) (0Y , r ) For each i I , as ( yi , ri ) C , there is xi X such that ( yi , ri ) epi ( xi ,.), or equivalently, 12 (3.5) ( xi , yi ) y , yi (D ) (D) (P) < D := inf { ( x, y ) y , y} ( x , y )X Y Proposition 3.4 (Weak duality) (D) ( x ,0Y ) (3.4) where y := y and := (3.7) It is clear that for any ( x, y) dom , one has ( y, ( x, y)) epi ( x,.) C , and hence, (3.7) entails < < y , y ( x, y) Thus, Dong Thap University Journal of Science, Vol 11, No 5, 2022, 09-18 such that ( xˆ, k ) Mˆ for all k Y , which yields ( xˆ, k ) Mˆ for all > Hence, for any ( x , y )X Y This implies that < sup inf { ( x, y) y , y} = (D) ( x , y )X Y yY which contradicts the assumption (D) So, as desired Step Conclusion We have just shown that: (i) (D) r , r (Step 1) (ii) Take > (D) Then, there exists such that > (D) (recall that (D) < , see Proposition 3.4) According to Step 2, one has Briefly, for all > (D), there is such that > We (D) = inf thus get from (i) and (ii) that We now assume further that (D) Then, it is obvious that (D) (D) Replacing by (D) in Step 2, we get (D) This, together with (i), yields that (D) = Theorem 3.2 Assume that is convex and the following condition holds (C0 ) ( xˆ,.) is bounded from above on Y for some xˆ X if (D ) , then (D ) = min{r : (0Y , r ) C} Proof Let us set or equivalently, y , k > Mˆ , > Letting , one gets y , k 0, which implies y Y It is obvious that ( y, ( x, y)) epi ( x,.) C for all ( x, y) dom So, it follows from (3.8) that < < y , y ( x, y) for any ( x, y ) dom , and hence, 0, This contradicts our assumption (D ) Consequently, we arrive at Then, (D ) = inf{r : (0Y , r ) C} Moreover, > , ( k , Mˆ ) C, and then, (3.8) leads to We are in the position to establish the main results of this paper, that is characterizing zero duality gap for general vector optimization problem (P) in convex setting We assume throughout this section that is a convex function Definition 4.1 We say that the problem (P) has zero duality gap if (P) = (D) and that (P) has zero loose duality gap if (P) = (D ) (3.8) 13 Natural Sciences issue According to Proposition 3.4, one has (D ) (D) (P) So, if (P) = (D ) then (P) = (D), or in the other words, if (P) has zero loose duality gap then it has zero duality gap For this purpose, we take (0Y , r ) C ({0Y } ) We now show that (0Y , r ) C ({0Y } ) Indeed, as (D) = inf{r : (0Y , r ) C} It is easy to see that C ({0Y } ) C ({0Y } ) = C ({0Y } ), (4.1) where the last equality follows from the fact that is a closed subset of Y Let us 0Y introduce the qualifying condition: (CQ) C ({0Y } ) = C ({0Y } ), (4.2) which also means that the converse inclusion of (4.1) holds It is observing that the condition (CQ) is a general type of the one introduced recently by Khanh et al (2019) when they studied zero duality gap for linear programming problems Theorem 4.1 (Characterization of zero duality gap) The following statements are equivalent to each other: (i) (CQ) holds (ii) (P) has zero duality gap Proof (i) (ii) Let :Y be the conical projection from Y to (i.e., ( y, r ) = r for all ( y, r ) Y ) According to Proposition 3.3 and Theorem 3.1, one has (P) = inf {r : (0Y , r ) C} = inf C {0Y } The last n inequality (4.4) implies that rn > inf xX ( x,0Y ) for For each n any n , we set rn := r , which leads to the existence of xn X such that rn > ( xn ,0Y ) for any n Hence, (0Y , rn ) epi ( xn ,.) C, giving rise to (0Y , rn ) C (0Y ) This, together with the fact that (0Y , rn ) (0Y , r ), yields (0Y , r ) C ({0Y } ), which completes the proof Example 4.1 Let X be a non-empty convex cone in X We consider the equality constrained linear programming problem of the form: inf , x (EP) s.t Ax b x X Let us introduce the perturbation mapping {} such that C ({0 } ) Y So, if (CQ) holds then (P) = (D) , which is nothing else but (ii ) (ii) (i) Assume that (ii ) holds, i.e., (P) = (D) The proof is completed by showing that (i ) holds According to (4.1), it is sufficient to show that (4.3) , x if Ax y b and x X else ( x, y ) Then, (EP) can be rewritten as inf ( x,0Y ) in the xX form of (P) The characterizing set C now reduces to the set M b Ax, , x r : x X , r while the dual problem (D) becomes (ED) sup y* , b s.t A# y* X * y* Y * 14 (4.4) xX : X Y (D) = inf{r : (0Y , r ) C} C ({0Y } ) C ({0Y } ) r (P) = inf ( x,0Y ) where X * , b Y , and A being a continuous linear function from X to Y and = inf (see Theorem 3.1), one has r (D) Consequently, by assumption that (ii ) holds, we obtain: Dong Thap University Journal of Science, Vol 11, No 5, 2022, 09-18 In this case, the condition (CQ) collapses to M ({0Y } ) M ({0Y } ) According to Theorem 4.1, one has inf (EP) sup(ED) if and Theorem 4.2 (Characterization of zero loose duality gap) Assume that the condition (C0 ) in Theorem 3.2 is fulfilled Then, the following statements are equivalent to each other: (i) (CQ) holds (ii) (P) has zero loose duality gap Proof Similar to the proof of Theorem 4.1, using Theorem 3.2 instead of Theorem 3.1 (CQR) consider the new cone qualifying We say that C is closed regarding the set 0Y if (CQR) holds It is worth observing that if (CQR) holds, then (CQR) does, too immediate Corollary 4.1 Assume that (CQR) holds Then, it holds: (i) (P) has zero duality gap (ii) If (C0 ) in Theorem 3.2 holds then (P) has zero loose duality gap Proof As (CQR) holds, one has C ({0Y } ) = C ({0Y } ) = C ({0Y } ), In the rest of this section, we will establish various characterizations of zero duality gap for the problem (CP) due to different choices of the perturbation function introduced in Section 5.1 The first way of transforming Consider Y = Z , Y = Z , and 1 : X Z defined by 1 ( x, z) = f ( x) g ( H ( x) z) Application: Zero duality gap for composite optimization problems In this last section, we apply the general results established in the previous sections to derive zero duality gap for the composite optimization problem We are concerned with the composite optimization problems, of the form (Boţ, 2010, Boţ et al., 2005, Dinh and Mo, 2012) (CP) inf [ f ( x) ( g H )( x)], (5.1) It is easy to see that dom 1 (.,0Z ) = dom( f g H ) = dom f H 1 (dom g ), and hence, by above assumption, dom1 (.,0Z ) It is worth noting that when taking = 1 , the problem (P) collapses to the problem (CP) In this case, characterizations of zero duality gap for the problem (P) are also the ones for the problem (CP) The next lemma gives us specific forms of the characterization set C and dual problems (D) and (D ) in this setting Lemma 5.1 With Y = Z , Y = Z , and = 1 given by (5.1), the set C , the problems (D) and (D ) become, respectively, which means that (CQ) holds The conclusion now follows from Theorems 4.1 and 4.2 xX and H : X Z are proper mappings such that dom f H 1 (dom g ) and we adopt the convention g (Z ) = C ({0Y } ) = C ({0Y } ) The next corollary is an consequence of the above theorems Z, f : X , g : Z , in only if M ({0Y } ) M ({0Y } ) We now condition where X , Z are locally convex Hausdorff topological vector spaces, Z is non-empty convex C1 := im( H , f ) hyp( g ), (5.2) and (CD1 ) sup { g ( z ) inf [ f ( x) z , H ( x) ]}, x X z dom g (CD ) sup z dom g Z { g ( z ) inf [ f ( x) z , H ( x) ]}, x X where im( H , f ) ( H ( x), f ( x)) : x dom H dom f Proof See Appendix A We now establish the first characterization of zero duality gap for the problem (CP) and the one of zero loose duality gap for the problem (CP) 15 Natural Sciences issue Corollary 5.1 (Characterization of zero duality gap 1) Assume that f is convex, that g is convex and Y -nondecreasing, and that H is a Y -convex mapping Then, the following statements are equivalent: (i) C1 ({0Z } ) = C1 ({0Z } ), (ii) (CP) = (CD1 ) Proof The convexity of 1 implies directly from the above assumption Then, the conclusion follows from Theorem 4.1 and Lemma 5.1 Corollary 5.2 (Characterization of zero loose duality gap 1) Assume that the assumption of Corollary 5.1 holds Assume further that the following condition holds xˆ X : g ( H ( xˆ ) Z ) is bounded (C1 ) from above Then, the following statements are equivalent: (CD2 ) sup { f (x )g (z )(z H )(x)} x dom f z dom g Y Proof See Appendix B By combining Lemma 5.2 to Theorem 4.1 and to Theorem 4.2, respectively, we get directly the consequences as follows: Corollary 5.3 (Characterization of zero duality gap 2) Assume all the assumption of Corollary 5.1 Then, the following statements are equivalent: (i) C2 ({0 X } {0Z }) = C2 ({0 X } {0Z }) , (ii) (CP) = (CD2 ) Corollary 5.4 (Characterization of zero loose duality gap 2) Assume all the assumption of Corollary 5.1 Assume further that the condition (C1 ) in Corollary 5.2 holds Then, the following statements are equivalent: (i) C1 ({0Z } ) = C1 ({0Z } ) , (i) C2 ({0 X } {0Z }) = C2 ({0 X } {0Z }) , (ii) (CP) = (CD1 ) (ii) (CP) = (CD2 ) Proof It follows from Theorem 4.2 and Lemma 5.1 5.2 The second way of transforming We now take Y = X Z , Y = {0 X } Z , and the perturbation 2 : X X Z defined by References Anderson, E.J (1983) A review of duality theory for linear programming over topological vector spaces J Math Anal Appl., 97(2), 380-392 (5.3) Boţ, R.I (2010) Conjugate Duality in Convex Optimization Berlin: Springer It is easy to check that dom2 (.,0 X ,0Z ) It is worth observing that in this case, taking = 2 , the problem (P) collapses to the problem (CP) Boţ, R.I., Hodrea, I B., and Wanka, G (2005) Composed convex programming: duality and Farkas-type results Proceeding of the International Conference In Memoriam Gyula Farkas, 23-26 The formulas of characterization set C and dual problems (D) and (D ) in this the case are given by the following lemma Feizollahi, M.J., Ahmed, S., and Sun, A (2017) Exact augmented Lagrangian duality for mixed integer linear programming Math Program.,161, 365-387 2 ( x, x, z) = f ( x x) g ( H ( x) z) Lemma 5.2 With Y = X Z , Y = {0 X } Z and = 2 given by (5.3), the set C becomes C2 := gr(0Z , f ) gr(H ,0) {0 X } hyp( g ), (5.4) while the problems (D) and (D ) become, respectively, (CD ) sup { f (x)g (z )(z H )(x)}, (x , z )dom f dom g 16 Huang, X.X and Yang, X.Q (2003) A unified augmented Lagrangian approach to duality and exact penalization Math Oper Res., 3, 533-552 Huang, X.X and Yang, X.Q (2005) Further study on augmented Lagrangian duality theory.J Global Optim., 31, 193-210 Jeyakumar, V and Li, G.Y (2009a) Stable zero duality gaps in convex programming: Dong Thap University Journal of Science, Vol 11, No 5, 2022, 09-18 Complete dual characterizations with application to semidefinite programs, J Math Anal Appl., 360, 156-167 Appendix Proof of Lemma 5.1 Jeyakumar, V and Li, G.Y (2009b) New dual constraint qualifications characterizing zero duality gaps of convex programs and semidefinite programs Nonlinear Anal., 71, 2239-2249 (i ) Prove that C = C1 Take ( z, r ) C Then, there exists x X such that ( z, r ) epi 1 ( x,.), which means r 1 ( x, z) = f ( x) g ( H ( x) z), or equivalently, So, f ( x) r g ( H ( x) z) ( H ( x) z, f ( x) r ) hyp( g ), and hence, Jeyakumar, V and Wolkowicz, H (1990) Zero duality gaps in infinite-dimensional programming J Optim Theory Appl., 67(1), 87-108 ( z, r ) = ( H ( x), f ( x)) ( H ( x) z, f ( x) r ) ( H ( x), f ( x)) hyp( g ) Li, D (1995) Zero duality gap for a class of nonconvex optimization problems J Optim Theory Appl., 85(2), 309-324 Li, D (1999) Zero duality gap in integer programming: P-norm surrogate constraint method Oper Res Lett., 25(2), 89-96 Long, F and Zeng, B (2020) The zero duality gap property for an optimal control problem governed by a multivalued hemivariational inequality Applied Mathematics & Optimization, 10.1007/s00245-020-09721-z Nguyen Dinh, Dang Hai Long, Tran Hong Mo, and Yao, J.-C (2020) Approximate Farkas lemmas for vector systems with applications to convex vector optimization problems J Nonlinear Convex Anal., 21(5), 1225-1246 Nguyen Dinh and Tran Hong Mo (2012) Qualification conditions and Farkas-type results for systems involving composite functions.Vietnam J Math., 40(4), 407-437 Pham Duy Khanh, Tran Hong Mo, and Tran Thi Tu Trinh (2019) Necessary and sufficient conditions for qualitative properties of infinite dimensional linear programming problems Numer Func Anal Opt., 40, 924-943 Moreover, the inequality also leads to r f ( x ) g ( H ( x) z ) x dom f dom H = dom( H , f ) So, one gets ( z, r ) im( H , f ) hyp( g ) Then, there are x dom( H , f ) = dom f dom H and (u, ) hyp( g ) such that ( z, r ) = ( H ( x), f ( x)) (u, ) , which means z = H ( x) u and r = f ( x) Rudin, W (1991) Functional Analysis (2nd Edition) New York: McGraw-Hill Yang, X.Q and Huang, X.X (2001) A nonlinear Lagrangian approach to constrained optimization problems SIAM J Optim., 11(4), 1119-1144 (5.5) As (u, ) hyp( g ) , one has g (u) , or equivalently, g (u) , and hence, by (5.5), r f ( x) g (u) = f ( x) g ( H ( x) z) = 1 ( x, z) This yields ( z, r ) epi 1 ( x,.) C1 and we are done (ii ) Prove that sup(D) = sup(CD1 ) By the definition of the Lagrange dual problem (D) (see Subsection 3.2), one has sup(D) = sup zZ D z where D := z inf [1 ( x, z ) z , z] ( x , z )X Z (recall that, at this time, Y = Z and = 1 ) For each z Z , according to (5.1), we have D = z Rubinov, A.M., Huang, X.X., and Yang, X.Q (2002) The zero duality gap property and lower semicontinuity of the perturbation function Math Oper Res., 27(4), 775-791 ( z, r ) C1 Take = inf [ f ( x) g ( H ( x) z ) z , z ] ( x , z ) X Z inf [ f ( x) g (u ) z , H ( x) u ] ( x ,u ) X Z = sup[ z , u g(u)] inf [ f (x) z ,H (x) ] x X uZ = g ( z ) inf [ f ( x) z , H ( x) ] x X So, we get 17 Natural Sciences issue f ( x x) z g ( H ( x) z) sup(D) = sup { g ( z )inf [ f ( x) z , H ( x) ]} zZ xX = sup { g (z )inf [ f (x) z ,H (x) ]} z dom g x X = sup(CD1 ) where the third equality follows from the fact that g (u ) = whenever u dom g (iii) Similar arguments apply to the problem (D ) to obtain sup(D ) = sup(CD ), and the proof is complete Proof of Lemma 5.2 Prove that C = C Take ( x, z, r ) C Then, there is x X such that ( x, x, z, r ) epi 2 ( x,.,.) , i.e., r 2 ( x, x, z) = f ( x x) g ( H ( x) z) (5.6) On the other hand, we can rewrite ( x, z, r ) as ( x, z, r ) = ( x x,0Z , f ( x x)) ( x, H ( x),0) (0 X , H ( x) z, f ( x x) r ) (5.7) It follows from (5.6) that x x dom f = dom(0Z , f ), x dom H = dom(H ,0), and 18 This, together with (5.6), yields ( x, z, r ) gr(0Z , f ) gr(H ,0) {0 X } hyp( g ) Conversely, take ( x, z, r ) C Then, there are u dom(0Z , f ) = dom f , v dom(H ,0) = dom H , and (w, ) hyp( g ) such that ( x, z, r ) = (u,0Z , f (u)) (v, H (v),0) (0 X , w, ), and hence x = u v, z = H (v) w and r = f (u) (5.8) As (w, ) hyp( g ), we have g (u) Combining this with (5.8) g (u) , we get r f (u ) g ( w) = f (v x) g ( H (v) z ) = 2 (u, x, z ) Consequently, ( x, z, r ) epi 2 (u,.) C2 and we are done The proof of equalities sup(D) = sup(CD2 ) and sup(D) = sup(CD2 ) is similar as in that of Lemma 5.1 ... are equal Many attempts have been made to study the zero duality gap for various classes of optimization problems in recent decades (Feizollahi et al., 2017, Huang and Yang, 2003, Jeyakumar and... duality gap for linear programming problems Theorem 4.1 (Characterization of zero duality gap) The following statements are equivalent to each other: (i) (CQ) holds (ii) (P) has zero duality gap Proof... 16 Huang, X.X and Yang, X.Q (2003) A unified augmented Lagrangian approach to duality and exact penalization Math Oper Res., 3, 533-552 Huang, X.X and Yang, X.Q (2005) Further study on augmented