Lagrange multipliers theorem and saddle point optimality criteria in mathematical programming

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1 J. Math. Anal. Appl. 323 (2006) Lagrange multipliers theorem and saddle point optimality criteria in mathematical programming Duong Minh Duc, Nguyen Dinh Hoang, Lam Hoang Nguyen National University of Hochiminh City, Viet Nam Received 16 July 2005 Available online 28 November 2005 Submitted by B.S. Mordukhovich Abstract We prove a version of Lagrange multipliers theorem for nonsmooth functionals defined on normed spaces. Applying these results, we extend some results about saddle point optimality criteria in mathematical programming Elsevier Inc. All rights reserved. Keywords: Lagrange multipliers theorem; Saddle point; Mathematical programming 1. Introduction Let U be an open neighborhood of a vector x in a normed space E,g 1,...,g N and f be real functions on U. We consider the following optimization problem with inequality constraints: (PI) min f(y) s.t. g i (y) 0 i = 1,...,N. If E is a Banach space, x is a solution of (PI), g 1,...,g N and f are Fréchet differentiable at x and D((g 1,...,g N ))(x)(e) = R N, Ioffe and Tihomirov in [7] proved that there are a real number a 0 and nonpositive real numbers a 1,...,a N such that (a 0,...,a N ) 0 R N+1 and a 0 Df (x) = a 1 Dg 1 (x) + +a N Dg N (x). * Corresponding author. address: dmduc@hcmc.netnam.vn (D.M. Duc) X/$ see front matter 2005 Elsevier Inc. All rights reserved. doi: /j.jmaa

2 442 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) In [4], Halkin extended this result to the following optimization problem: (PIE) min f(y) s.t. g i (y) 0 i = 1,...,n, g n+j (y) = 0 j = 1,...,N n. Halkin also proved that a 1,...,a n are nonpositive. Furthermore, if g 1,...,g N are not only Fréchet differentiable but also C 1 at x, that is the Fréchet derivatives of g 1,...,g N exist on a neighborhood of x and are continuous at x, Ioffe and Tihomirov in [7, p. 73] proved that a 0 in their cited result is not equal to 0. Using the notation of generalized gradients, Clarke, Ioffe, Michel and Penot, Mordukhovich, Rockafellar and Treiman have extended the above results for Lipschitz constraint functions in [3,5,6,10,11,17,19,20]. In [21,22], Ye considered the problem (PIE) with mixed assumptions of Gâteaux, Fréchet differentiability and Lipschitz continuity of constraint functions. In [8,9,12 16], using the extremal principle, Kruger, Mordukhovich and Wang consider the problem (PIE) for locally Lipschitzian constraint functions on subsets in Asplund spaces. If N = 1, we reduced the Lagrange multipliers rule to a two-dimensional problem in [1] and obtained a version of Lagrange multipliers theorem, in which we only required the smoothness of the restrictions of f and g on F U, where F is any two-dimensional vector subspace containing x of E. This smoothness is very weak and may not imply the continuity of the functions. In the next section of the present paper, we prove a discrete implicit mapping theorem (see Lemma 2.1) and apply it to extend the results in [1] to the case N>1forf and g, whose restrictions on F U are smooth for any (N + 1)-dimensional vector subspace F containing x of E. We note that our results can be applied to functions which are not C 1 -Fréchet differentiable neither Lipschitz continuous, even they are not continuous at x (see Remark 2.3). Applying these results, we extend some results of Bector et al. [2] in the last section. 2. Lagrange multipliers rule Let U be a nonempty open subset of a normed space (E, E ), X be a linear subspace of E, Z be a finite-dimensional linear subspace of X and J be a mapping from U into a normed space Y. We consider Z as a normed subspace of X.Letv be a vector in X and x be in U. Denote by Z(v) and Z(x,v) the vector subspaces of E generated by Z {v} and Z {x,v}, respectively. We put U x,z ={y Z: x + y U}, J x,z (y) = J(x+ y) y U x,z. Then U x,z is an open subset of Z. Wesay (i) J is (X, Z)-continuous at x on U if and only if for every v in X, there is a positive real number η v such that J u,z is continuous at 0 for any u in Z(x,v) B E (x, η v ); (ii) J is X-differentiable at x if and only if there exists a linear mapping DJ(x) from X into Y such that J(x+ th) J(x) = DJ (x)(h) h X; t 0 t

3 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) (iii) J is (X, Z)-differentiable at x if and only if J is X-differentiable at x and for any v in X,if the sequence {(h m,t m )} m N Z(v) R converges to (h, 0) in Z(v) R then J(x+ t m h m ) J(x) = DJ (x)(h). m t m Remark 2.1. Let J be a linear mapping from E into R n, X be a linear subspace of E and Z be a finite-dimensional vector subspace of X. It is clear that J is (X, Z)-continuous at any x in E and (X, Z)-differentiable at any x in E although it may not be continuous on E. Remark 2.2. If J is (X, F )-differentiable at x then there are a positive real number ɛ v and a mapping φ v from B E (0,ɛ v ) Z(v) into Y such that B E (x, ɛ v ) U, z 0 φ v (z) = 0 and J(x+ z) = J(x)+ DJ (x)(z) + z E φ v (z) z B E (0,ɛ v ) Z(v). Indeed, it is sufficient to prove that J(x+ z) J(x) DJ (x)(z) = 0. z Z(v),z 0 z E We assume by contradiction that there exist a sequence {z m } m N in Z(v) and a positive real number ɛ such that 0 < z m E <m 1 and J(x+ z m ) J(x) DJ (x)(z m ) z m >ɛ n N. ( ) E Y We put s m = z m 1 E z m Z(v). Because Z(v) is a finite-dimensional subspace of X, there exists a subsequence {s mk } k N of {s m } m N such that k s mk = s in Z(v). Because J is (X, F )-differentiable at x then J(x+ z mk ) J(x) J(x+ z mk E s mk ) J(x) = = DJ (x)(s). k z mk E k z mk E Since Z(v) is finite-dimensional, Df (x) is continuous on Z(v) and ( ) DJ(x) zmk = k z mk DJ (x)(s m k ) = DJ (x)(s). E k Then we have J(x+ z mk ) J(x) DJ (x)(z mk ) = 0, k z mk E we get the contradiction with ( ) and we get the result. We have the following result: Lemma 2.1 (Discrete Implicit Mapping Theorem). Let U be an open neighborhood of a vector x in a normed linear space E, X be a vector subspace of E, F be a n-dimensional vector subspace of X and g be a mapping from U into R n with g = (g 1,...,g n ). Assume that g is (X, F )-continuous and (X, F )-differentiable at x. Put M ={y U: g(y) = g(x)} and e i = (δ 1 i,...,δn i ) for any i in {1,...,n}, where δj i is the Kronecker number. Let v be in X and h 1,...,h n be n vectors in F such that Dg(x)(v) = 0 and Dg(x)(h i ) = e i for any i in {1,...,n}.

4 444 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) Then there exist a sequence {s m = (s 1 m,...,sn m )} m N converging to 0 in R n and a sequence of positive real numbers {α m } m N converging to 0 in R such that u m α m ( s 1 m h 1 + +s n m h n) + αm v + x M m N. Proof. We can assume without loss of generality that x = 0 and g(x) = 0. Fix a vector v in X such that Dg(0)(v) = 0. By Remark 2.1 about the (X, F )-continuity and (X, F )-differentiability of g at 0, there is a positive real number ɛ v and a mapping φ v from B F(v) (0,ɛ v ) F(v) B E (0,ɛ v ) to R n such that: B E (x, ɛ v ) U, g y,f is continuous at 0 for every y B F(v) (0,ɛ v ), y 0 φ v (y) = 0 and g(z) = Dg(0)(z) + z E φ v (z) z B F(v) (0,ɛ v ). There is a real positive number r such that α(s 1 h 1 + +s n h n + v) belongs to B F(v) (0,ɛ v ) for any (α, s 1,...,s n ) (0,r) B n (0,r), where B k (0,p) ={t = (t 1,...,t k ) R k : t R k (t 1 ) 2 + +(t k ) 2 <p}. We put η(s) = s 1 h 1 + +s n h n s = ( s 1,...,s n) B n (0,r), G α (s) = α 1 g ( α ( η(s) + v )) (α, s) (0,r) B n (0,r). Because g y,f is continuous at 0 for every y in B F(v) (0,ɛ v ), we see that G α is continuous on B n (0,r)for any fixed α in (0,r)and G α (s) = α 1 g ( α ( s 1 h 1 + +s n h n + v )) = α 1 g ( α ( η(s) + v )) = α 1 Dg(0) ( α ( η(s) + v )) + α 1 α ( η(s) + v ) ( ( )) E φ v α η(s) + v = Dg(0) ( η(s) ) + ( ( )) η(s) + v E φ v α η(s) + v = ( s 1,...,s n) + η(s) + v ( ( )) E φ v α η(s) + v s = ( s 1,...,s n) B n (0,r). Note that there is a positive real number M such that η(s) + v E M for any s = (s 1,...,s n ) in B n (0,r). Thus [ { ( ( )) sup φv α η(s) + v E : ( s 1,...,s n) B n (0,r) }] = 0 and α 0 Gα (s), s = s,s + η(s) + v ( ( )) φv E α η(s) + v,s m 2{ 1 mm ( ( )) } ( φ v α η(s) + v E s B n 0,m 1 ), where m is an integer greater than r 1. Thus there is a real number α m in (0,m 1 ) such that Gαm (s), s > 0 s B n ( 0,m 1 ). By Lemma 4.1 in [18, p. 14], we have a solution s m in B n (0,m 1 ) to the equation G αm (s) = 0 for any integer m greater than r 1, which yields the lemma. Remark 2.3. If E is a Banach space, U is an open subset of E, g is Fréchet differentiable on U and Dg is continuous at x, then by the Ljusternik theorem (see [7, p. 41]), M is a manifold and the tangent space TM x = Dg(x) 1 ({0}). Moreover, M is a C 1 -manifold if g is C 1 -Fréchet

5 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) differentiable on U (see [23, Theorem 43.C]). The smoothness of g in Lemma 2.1 is very weak so that M may not be a smooth manifold and we cannot define the tangent space of M at x. Note that the sequence {αm 1(u m x)} m N converges to v in E. Therefore, we can admit v as a generalized tangent vector of M at x for any v in Dg(x) 1 ({0}), if we consider {u m } m N as a discrete curve passing through x. This idea is illustrated by the following example. Let E = X = U = R 3 and Z = R {0} {0}. We define P 1 = { (s, t) R 2 : (α, β) Q Q, αs+ βt = 0 }, P 2 = { (s, t) R 2 : s 2 + t 2 Q }, P =[P 1 P 2 ] [( R 2 ) ( \ P 1 R 2 )] \ P 2, A = { (r,s,t) R 3 : (s, t) P }, B = { (r,s,t) R 3 : r = 0, s>0, t= s 2}, g(r,s,t) = r + ( s 2 + t 2) χ A (r,s,t)+ χ B (r,s,t) (r,s,t) R 3, where χ C is the characteristic function of the set C. Let v = (0,s,t)in R 3 such that st 0. Put δ v = 1 2 s 2 t 2 + s 4 t 4,wehaveB E ((0, 0, 0), δ v ) Z(v) B =. Therefore, for any v in {0} R 2, there is a positive real number δ v such that B E ((0, 0, 0), δ v ) Z(v) B =. Thus we see that g is (X, Z)-continuous at (0, 0, 0) and (X, Z)-differentiable at (0, 0, 0). Butg is not Fréchet differentiable at (0, 0, 0) because g is not continuous at (0, 0, 0). Put M ={(r,s,t) R 3 : g(r,s,t) = g(0, 0, 0)},wehave Dg ( (0, 0, 0) ) (Z) = R, M = { (0,s,t): (s, t) / (P B) } { (r,s,t): (s, t) (P \ B), r = s 2 t 2}. Note that (0,s,t)is a generalized tangent vector of M at (0, 0, 0) for any (s, t) in R 2. The results in [3 7,17,19 22] cannot be applied to this case. It is easy to derive this example to the case of vector functions. The idea of generalized tangent vectors is essential to get the following generalized Lagrange multipliers theorem. Theorem 2.1. Let U be an open subset of normed vector space E, X be a vector subspace of E, F be a n-dimensional vector subspace of X, u be in U, r be in R n, f be a mapping from U into R, g = (g 1,...,g n ) be a mapping from U into R n, M ={x U: g(x) = r} and u M. Assume that (i) f(u)is the minimum (or maximum) of f(m), (ii) f is (X, F )-differentiable at u, (iii) g is (X, F )-continuous at u and (X, F )-differentiable at u, (iv) Dg(u)(F ) = R n. Then there exists a unique mapping Λ L(R n, R) such that Df (u)(k) = Λ ( Dg(u)(k) ) k X.

6 446 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) Proof. Assume f(u) is the minimum of f(m). Choose n vectors h 1,...,h n in F such that Dg(u)(h i ) = e i (δi 1,...,δn i ) for any i in {1,...,n}. We define a real linear mapping Λ on Rn as follows: Λ(e i ) = Df (u)(h i ) i = 1,...,n. Now fix a vector k in X. Put n v = k Dg i (u)(k)h i X. Then Dg(u)(v) = 0. By Lemma 2.1, there are a sequence {α m } m N of positive real numbers and a sequence {s m = (s 1 m,...,sn m )} m N in R n such that {α m } m N and {s m } m N converge to 0 in R and R n respectively, u m U, and g(u m ) = g(u), where u m = u + α m ( s 1 m h 1 + +s n m h n) + αm v m N, or u m M for every m N. Since f(u)is the minimum of f(m)and f is (X, F )-differentiable at u,wehave Df (u)(k) Λ ( Dg(u)(k) ) ( n = Df (u)(k) Λ = Df (u)(k) Dg i (u)(k)e i ) n D i (u)(k)λ(e i ) = Df (u)(k) n Dg i (u)(k)df (u)(h i ) ( n ) ( = Df (u)(k) Df (u) Dg i (u)(k)h i = Df (u) k ) n Dg i (u)(k)h i [f(u+ α m (sm 1 = Df (u)(v) = h 1 + +sm n h n + v)) f(u)] 0. m α m Therefore, Df (u)(k) Λ(Dg(u)(k)) for any k in X. Replacing k in the above inequality by k, we get the theorem. Since Dg(u)(F ) = R n, we can get the uniqueness of Λ. The proof for the case f(u)= max f(m)is similar and omitted. Remark 2.4. If E is a Banach space, g is Fréchet differentiable on U and Dg is continuous at u, then Theorem 2.1 has been proved in [7, p. 73]. Here we only need the differentiability of f and g at u. Ifn = 1, Theorem 2.1 has been proved in [1]. Theorem 2.2. Let U be an open subset of normed vector space E, X be a vector subspace of E, F be a (n + m)-dimensional vector subspace of X, u be in U, f be a mapping from U into R, g = (g 1,...,g n+m ) be a mapping from U into R n+m and M = { x U: g i (x) 0,g n+j (x) = 0 i {1,...,n}, j {1,...,m} }. Assume that (i) u M and f(u)is the minimum of f(m),

7 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) (ii) f is (X, F )-differentiable at u, (iii) g is (X, F )-continuous at u and (X, F )-differentiable at u, (iv) Dg(u)(F ) = R n+m. Then g(u) = 0 and there exists a unique (a 1,...,a n+m ) in R n+m such that a 1,...,a n are negative and n+m Df (u)(k) = a i Dg i (u)(k) k X. Proof. We prove the theorem by the following steps. Step 1. Put α = (α 1,...,α n+m ) = g(u) and S ={x U: g(x) = α} we see that S M. Since f(u)= min f(s), by Theorem 2.1, there exists a unique (a 1,...,a n+m ) in R n+m such that n+m Df (u)(k) = a i Dg i (u)(k) k X. Step 2. We prove that a i < 0 for every i {1,...,n}. First, we prove that a 1 < 0. Since Dg(u)(F ) = R n+m, there is k in F such that Dg 1 (u)(k) = 1 and Dg 2 (u)(k) = = Dg n (u)(k) = ε<0 and Dg n+j (u)(k) = 0 for any j in {1,...,m}. Let {h i },...,n+m be in F such that Dg(u)(h i ) = (δi 1,...,δn+m i ) for any i {1,...,n+ m}, we have D(g n+1,...,g n+m )(u)(h i ) = ( δi n+1,...,δi n+m ) i {n + 1,...,n+ m}. By Lemma 2.1, there are a sequence {s l = (sl n+1,...,sl n+m )} l N converging to 0 in R m and a sequence of positive real numbers {α l } l N converging to 0 in R such that ( u l α l s n+1 l h n+1 + +sl n+m ) h n+m + αl k + u, (g n+1,...,g n+m )(u l ) = (g n+1,...,g n+m )(u) = 0 l N. Note that g i (u l ) g i (u) = l α l l i=2 g i (α l (s n+1 l h n+1 + +sl n+m h n+m ) + α l k + u) g i (u) α l = Dg i (u)(k) < 0 i {1,...,n}. Thus there exists l 0 N such that g i (u l )<g i (u) i {1,...,n}, l l 0, which implies that u l is in M for any l l 0 and n f(u l ) f(u) a 1 ε a i = Df (u)(k) = 0 ε>0. l α l Let ε tend to 0, we have a 1 0ora 1 0. Since Dg(u)(F ) = R n+m,wehavea 1 < 0. Similarly, we have a i < 0 i = 1,...,n.

8 448 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) Step 3. We shall prove g(u) = 0. Put I = { i {1,...,n}: g i (u) < 0 } and J = { j {1,...,n}: g j (u) = 0 }. We assume by contradiction that I is not empty. In this case, we see that a i < 0. i I On the other hand, there is k in F such that (1) Dg i (u)(k) = 1 i I, Dg j (u)(k) = ε j J, Dg n+l (u)(k) = 0 l = 1,...,m. By Lemma 2.1, there are a sequence {s l = (sl n+1,...,sl n+m )} l N converging to 0 in R m and a sequence of positive real numbers {α l } l N converging to 0 in R such that ( u l α l s n+1 l h n+1 + +sl n+m ) h n+m + αl k + u and 0 = (g n+1,...,g n+m )(u l ) = (g n+1,...,g n+m )(u) l N. We have g i (u l ) g i (u) = l α l l g i (α l (s n+1 l = Dg i (u)(k) = 1 i I. h n+1 + +sl n+m h n+m ) + α l k + u) g i (u) α l Thus l g i (u l ) = g i (u) < 0. Then there exists an integer l 1 such that We have g i (u l )<0 i I, l l 1. g j (u l ) g j (u) = l α l l Thus there exists l 2 N such that g j (α l (s n+1 l = Dg j (u)(k) = ε<0 j J. g j (u l )<g j (u) = 0 j J, l l 2. h n+1 + +sl n+m h n+m ) + α l k + u) g j (u) α l Therefore u l is in M for any l max{l 1,l 2 } and a i ε f(u l ) f(u) a j = Df (u)(k) = 0 ε>0. l α i I j J l It implies that i I a i 0, which contradicts to (1) and I should be empty and we get the result.

9 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) Applications in programming problems Let X R n denote the n-dimensional Euclidean space and let R n + be its nonnegative orthant. Let f,h 1,...,h m be real functions on an open subset S of R n. We consider the following nonlinear programming problem min f(x) (P ) subject to h j (x) 0 j {1, 2,...,m}. Put J ={1,...,r} and K ={r + 1,...,m}, h J (x) = (h 1 (x),..., h r (x)) and h K (x) = (h r+1 (x),...,h m (x)) for any x in S. Leth(x) denote the column vector (h 1 (x), h 2 (x),..., h m (x)) T and be partitioned as h(x) = (h J (x), h K (x)) T. We denote S h = { x S: h j (x) 0, j {1,...,m} }, S K = { x S: h k (x) 0, k K }. Let a = (a 1,...,a N ) R N with N N. Wesaya 0(or 0) if and only if a i 0(or 0) for every i {1,...,N}. Definition 3.1. Let S be an open subset of R n, u be in S, g be a real function on S and Gâteaux differentiable at u, and η is a function from S S to R n.wesay (i) the function g is said to be invex at u with respect to the function η, if for every x S, we have g(x) g(u) η(x,u) T g(u); (ii) the function g is said to be pseudo-invex at u with respect to the function η, if for every x S,wehave η(x,u) T g(u) 0 g(x) g(u); (iii) the function g is said to be quasi-invex at u with respect to the function η, if for every x S, we have g(x) g(u) η(x,u) T g(u) 0. Applying results in the first section, we study the following programming problems. Problem 1: Nonlinear programming problem Let f, g, h 1,...,h m be real functions on an open subset S of R n.letj, K, h J and h K be as in the beginning of this section. Put L(x, λ J ) = f(x)+ (λ J ) T h J (x) for any (x, λ J ) in S K R J +.ThemapL is called the incomplete Lagrange function of the problem (P ). Definition 3.2. A point ( x, λ J ) S K R J + is called a saddle point of the incomplete Lagrange function L if L( x,λ J ) L( x, λ J ) L(x, λ J ) (x, λ J ) S K R J +.

10 450 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) Theorem 3.1. Let S be an open subset of X R n, f and h 1,...,h m be real functions on S and Gâteaux differentiable at x with x S. F be a m-dimensional vector subspace of X and x be optimal for (P ). Assume that (i) L(,λ J ) is pseudo invex and (λ K ) T h K ( ) is quasi invex at x with respect to a function η for any (λ J,λ K ) in R m + ; (ii) f is (X, F )- differentiable at x; (iii) h is (X, F )-differentiable at x and (X, F )-continuous at x; (iv) Dh( x)(f ) = R m. Then there exists λ J R J + such that ( x, λ J ) is a saddle point of the incomplete Lagrange function L. Proof. Applying Theorem 2.2 in case E = X = R n, U = S for f and {h j } j=1,m, there exists λ = ( λ J, λ K ) R m, λ J R J +, λ K R K +, such that [ f( x) + (λ J ) T h J ( x) + (λ K ) T h K ( x) ] = 0, (λ J ) T h J ( x) = 0, (λ K ) T h K ( x) = 0, λ = ( λ J, λ K ) 0. Now arguing as in the proof of Theorem 2.1 in [2], we get the theorem. We consider the following propositions. (A1) x be optimal for (P ). (A2) There are positive real numbers λ 1,...,λ m such that h i ( x) = 0 for any i in {1,...,m} and Df ( x) + λ 1 Dh 1 ( x) + +λ m Dh m ( x) = 0. We have the following result. Theorem 3.2. Let S be an open subset of X R n, f and h 1,...,h m be real functions on S. Let x be in S. Then (i) if f, h 1,...,h m satisfy the conditions (ii), (iii) and (iv) of Theorem 3.1, then (A1) implies (A2); (ii) if f( ) + m λ i h i ( ) is Gâteaux differentiable at x and invex with respect to the function η at x then (A2) implies (A1). Proof. Applying Theorem 2.2 in case E = X = R n, U = S for f and {h j } j=1,m, we get (i). Consider (ii). By the invexity property of f( ) + m λ i h i ( ) with respect to the function η at x,wehave [ ] m m f(x)+ λ i h i (x) f( x) + λ i h i ( x) [ m η(x, x) T f + λ i h i ]( x) = 0 x S h.

11 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) It follows that m m f(x) f(x)+ λ i h i (x) f( x) + λ i h i ( x) = f( x) x S h, and we get (ii). Problem 2: Fractional programming problem Let f, g,h 1,...,h m be real functions on an open subset S of R n.letj, K, h J and h K be as in the beginning of this section. Put S h = { x S: h j (x) 0, j {1,...,m} }, S K = { x S: h k (x) 0, k K }. Assume that g(x) 0 for any x in S and g(x) > 0 for any x in S K. We now consider the fractional programming problem min f(x) (FP) g(x) subject to h j (x) 0 j {1, 2,...,m}. We note that if x is optimal for the problem (FP) then x is also optimal for the following problem R J + (FP1) min f(x) g(x) subject to h j (x) g(x) 0 j J and h k(x) 0 k K. This form of (FP1) suggests the choice of the incomplete Lagrange function L F : S K R as follows: L F (x, λ J ) = f(x)+ (λ J ) T h J (x). g(x) Definition 3.3. A point ( x, λ J ) S K R J + is called a saddle point of the incomplete Lagrange function L F if L F ( x,λ J ) L F ( x, λ J ) L F (x, λ J ) (x, λ J ) S K R J +. Theorem 3.3. Let S be an open subset of X R n, f and h 1,...,h m be real functions on S and Gâteaux differentiable at x with x S. F be a m-dimensional vector subspace of X and x be optimal for (FP). Assume that (i) L F (,λ J ) is pseudo invex and (λ K ) T h K ( ) is quasi invex at x with respect to a function η for any (λ J,λ K ) in R m + ; (ii) f g is (X, F )-differentiable at x; (iii) h 1 g,..., h r g,h r+1,...,h m are (X, F )-differentiable at x and (X, F )-continuous at x; (iv) D(( h 1 g,..., h r g,h r+1,...,h m ))( x)(f ) = R m.

12 452 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) Then there exists λ J R J + such that ( x, λ J ) is a saddle point of the incomplete Lagrange function L F. Proof. Since x is optimal for (FP) (and hence for (FP1)), applying Theorem 2.2 in case E = X = R n, U = S for f g and h 1 g,..., h r g,h r+1,...,h m, we can find λ = ( λ J, λ K ) in R J + R K +, such that [ f( x) g( x) + ( λ J ) T h ] J ( x) g( x) + ( λ K ) T h K ( x) = 0, (λ J ) T h J ( x) = 0 and (λ K ) T h K ( x) = 0. Fix x in S K,wehave( λ K ) T h K (x) 0 = ( λ K ) T h K ( x). By the quasi-invexity of ( λ K ) T h K ( ) with respect to the function η at x, we obtain η(x, x) T [ ( λ K ) T h K ( x) ] 0. Thus [ f( x) η(x, x) T g( x) + ( λ J ) T h ] J ( x) 0. g( x) By the pseudo-invexity of f( )+( λ J ) T h J ( ) g( ) with respect to the function η at x, we see that f(x) g(x) + ( λ J ) T h J (x) f( x) g(x) g( x) + ( λ J ) T h J ( x) g( x). And we get L F (x, λ J ) L F ( x, λ J ). If λ J is in R J +,wehave L F ( x, λ J ) = f( x) g( x) + ( λ J ) T h J ( x) g( x) = f( x) f( x) g( x) g( x) + (λ J ) T h J ( x) g( x) = L F ( x,λ J ). Thus we get the theorem. Problem 3: Generalized fractional programming problem Let f 1,...,f p,g 1,...,g p,h 1,...,h m be real functions on an open subset S of R n. Put S h = { x S: h j (x) 0, j {1,...,m} }, S K = { x S: h k (x) 0, k K }. Assume that g 1 (x),...,g p (x) 0 for any x in S and g 1 (x),..., g p (x) are positive for every x in S h. We consider the generalized fractional programming problem min (GFP) max f i (x) x S 1 i p g i (x) subject to h j (x) 0 j {1, 2,...,m}. Put Y ={y R p +, p y i = 1}. The incomplete Lagrange function L G : S K Y R J + R for the problem (GFP) can be chosen as L G (x,y,λ J ) = yt f(x)+ (λ J ) T h J (x) y T. g(x)

13 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) Definition 3.4. A point ( x,ȳ, λ J ) S K Y R J + is called a saddle point of the incomplete Lagrange function L G if L G ( x,y,λ J ) L G ( x,ȳ, λ J ) L G (x, ȳ, λ J ) (x,y,λ J ) S K Y R J +. We consider the problem (EGFP) v min q subject to f i (x) vg i (x) q and h j (x) 0 (i, j) {1,...,p} {1,...,m}. We have the following lemma (see [2, Lemma 2.3, p. 9]). Lemma 3.1. The point x is (GFP) optimal with corresponding optimal value of the (GFP) objective equal to v if and only if (x,q ) is (EGFP) v optimal with corresponding optimal value of the (EGFP) v objective equal to 0, i.e., q = 0. Now we define H v (x, q) = q (x, q) S R, G v 1i (x, q) = f i(x) v g i (x) q (x,q,i) S R {1,...,p}, G v 2j (x, q) = h j (x) (x,q,j) S R {1,...,m}. Theorem 3.4. Let S be an open subset of X R n, f 1,...,f p,g 1,...,g p,h 1,...,h m be real functions on S and Gâteaux differentiable at x with x S. F be a (p + m)-dimensional vector subspace of X R R n R and x be optimal for (GFP) with corresponding optimal value of the (GFP) objective equal to v. Assume that (i) L G (,y,λ J ) is pseudo invex and (λ K ) T h K ( ) is quasi invex at x with respect to a function η for any (y, λ J,λ K ) in Y R m + ; (ii) G v 11,...,Gv 1p, Gv 21,...,Gv 2m are (Rn R,F)-differentiable at ( x,0) and (R n R,F)- continuous at ( x,0); (iii) D((G v 11,...,Gv 1p, Gv 21,...,Gv 2m ))( x,0)(f ) = Rp+m. Then there exists (ȳ, λ J ) Y R J + such that ( x,ȳ, λ J ) is a saddle point of the incomplete Lagrange function L G. Proof. Because x is an optimal value of (GFP) with corresponding optimal value of the (GFP) objective equal to v, by Lemma 3.1, (x,q ) is (EGFP) v optimal with corresponding optimal value of the (EGFP) v objective equal to 0. Then the problem (L) min H v (x, q) subject to G v 1i (x, q) 0 and Gv 2j (x, q) 0 has an optimal solution (x,q ) with q = 0. (i, j) {1,...,p} {1,...,m}

14 454 D.M. Duc et al. / J. Math. Anal. Appl. 323 (2006) Applying Theorem 2.2 in case E = X = R n R, U = S R for H v and G v 11,...,Gv 1p,Gv 21,...,G v 2m, there exists (ȳ, λ J, λ K ) R p + R J + R K + such that DH v ( x,0)(k, l) + p m ȳ i DG v 1i ( x,0)(k, l) + for every (k, l) R n R and j=1 λ j DG v 2j ( x,0)(k, l) = 0, ( ) [ ȳ i fi ( x) v g i ( x) ] = 0 i {1,...,p}, (λ J ) T h J ( x) = 0, (λ K ) T h K ( x) = 0, λ = ( λ J, λ K ) 0, ȳ 0. If l in ( ) is equal to 0, we have [ ȳ T f( x) v ȳ T g( x) + ( λ J ) T h J ( x) + ( λ K ) T h K ( x) ] = 0. If k in ( ) is equal to 0, we have p ȳi = 1orȳ Y. Now arguing as in the proof of Theorem 2.5 in [2], we get the theorem. We consider the following propositions: (A3) x be optimal for (GFP) with corresponding optimal value of the (GFP) objective equal to v. (A4) x has the following properties: [ ȳ T f( x) v ȳ T g( x) + ( λ J ) T h J ( x) + ( λ K ) T h K ( x) ] = 0, [ ȳ i fi ( x) v g i ( x) ] = 0 i {1,...,p}, (λ J ) T h J ( x) = 0, (λ K ) T h K ( x) = 0, λ = ( λ J, λ K ) 0, ȳ 0, p ȳ i = 1. Theorem 3.5. Let S be an open subset of X R n, f 1,...,f p,g 1,...,g p,h 1,...,h m be real functions on S. Let x be in S. (i) If the conditions (ii), (iii) of Theorem 3.4 are fulfilled then (A3) implies (A4). (ii) If ȳ T f( ) v ȳ T g( ) + ( λ J ) T h J ( ) + ( λ K ) T h K ( ) is Gâteaux differentiable at x and invex with respect to the function η at x then (A4) implies (A3). Proof. Applying Theorem 2.2 in case E = X = R n R, U = S R for H v and G v 11,...,Gv G v 21,...,Gv 2m (we define in p. 453) and with the same proof of Theorem 3.4, we get (i). Consider (ii). Since the map ȳ T f( ) v ȳ T g( ) + ( λ J ) T h J ( ) + ( λ K ) T h K ( ) is invex with respect to the function η at x, for every x S h,wehave ȳ T f(x) v ȳ T g(x) + ( λ J ) T h J (x) + ( λ K ) T h K (x) [ ȳ T f( x) v ȳ T g( x) + ( λ J ) T h J ( x) + ( λ K ) T h K ( x) ] η(x, x) T [ ȳ T f( x) v ȳ T g( x) + ( λ J ) T h J ( x) + ( λ K ) T h K ( x) ] = 0. 1p,

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