On linear systems containing strict inequalities

Transcription

1 On linear systems containing strict inequalities M.A Goberna, V. Jornet and M.M.L. Rodriguez Dep. of stistics and Oper. Res., University of Alicante Abstract This paper deals with systems of an arbitrary (possibly infinite) number of both weak and strict linear inequalities. We analize the existence of solutions for such kind of systems and show th the large class of convex sets which admit this kind of linear representions (i.e., the so-called evenly convex sets) enjoys most of the well-known properties of the subclass of the closed convex sets. We also show th it is possible to obtain geometrical informion on these sets from a given linear represention. Finally, we discuss the theory and methods for those linear optimizion problems which contain strict inequalities as constraints. 1. INTRODUCTION This paper deals with linear inequality systems in R n containing an arbitrary number of either weak or strict inequalities. Such kind of systems can be written as σ = {a 0 t x,t W ; a 0 t x>,t S}, (1.1) where W S 6=, W S =, a t R n and R for all t T := W S. We shall denote by F the solution set of σ. There exists an extensive literure on ordinary linear inequality systems (S =, W < ) as far as they are closely reled to Linear Programming theory and methods. Concerning linear semi-infinite systems (S =, W 6= arbitrary), whose analysis provides the theoretical foundions for Linear Semi-Infinite Programming (LSIP), different conditions for F 6= (existence theorems) and many results characterizing the geometrical properties of F in terms of the coefficients of σ are well-known (see Part II of 7 and references therein). Linear systems containing strict inequalities (S 6= ) nurally arise in separion problems, optimizion, stability analysis and other fields. In fact, a family of m 2 non-empty sets in R n, A 1,...,A m is said to be strictly separable 1 if there exist m closed halfspaces Σ 1,...,Σ m such th A j int Σ j, j =1,...,m,and m j=1 int Σ j =, i.e., if, for each j = 1,...,m,

2 µ c j there exists a solution of σ j = {a 0 x x n+1 > 0, a A j }, R n+1, with c j 6=0 n, such th the system σ 0 = (c j ) 0 x>d j,j=1,...,m ª is inconsistent. Moreover, if m A k 6=, µ P j =1,...,m, then the inconsistency of σ 0 can be replaced by m c j =0 n+1 (by Theorem 2 in 1). In particular, the search for a hyperplane separing strictly a pair of disjoint sets in R n, Y and Z, can be formuled as the system of strict inequalities d j j=1 d j k=1 k6=j where the unknown hyperplane. {y 0 x>x n+1,y Y; z 0 x> x n+1,z Z}, µ x R n+1 determines the vector of coefficients of the separing x n+1 On the other hand, if A 0,A 1,...,A n are given square symmetric mrices and the model building of a certain optimizion problem requires th a linear combinion of them, say A (x) =A 0 + P x j A j, must be positive definite, then this constraint can be formuled as {s 0 A (x) s>0, s S n },wheres n stands for the unit sphere in R n. Finally, a continuous linear semi-infinite system {a 0 tx,t W} is stable (in the different senses specified in Theorem 6.9 of 7) if and only if there exists a solution of the corresponding system of strict inequalities {a 0 tx>,t W}. Despite the many potential applicions of the linear systems containing strict inequalities, only existence theorems for particular cases have been given up to now ( T < in 2, 4 and 12 and homogeneous systems in 7). All these results are subsumed by the existence theorems in Section 2, where the numerical compution of a solution is also discussed. Section 3 shows th the solution setsoflinearinequality systemssuch as σ in (1.1) enjoy nice geometrical properties. Indeed, this family of convex sets (called evenly convex in 6) captures the most outstanding properties of a subclass, the closed convex sets, which plays a crucial role in optimizion theory and practice. Evenly convex sets were introduced by Fenchel in 1952 to extend the polarity theory. Given an evenly convex set C containing 0 n (0 n represents the zero vector in R n ), its modified (negive) polar was defined by Fenchel as C o = {y R n x 0 y<1, x C}, proving th C oo = C. On the other hand, given a function f : R n R {± }, f is said to be evenly quasiconvex if {x R n f (x) α} is evenly convex for all α R. New characterizions 2

3 of these functions have been given recently in 3. This class of functions has applicions in quasiconvex programming (duality and conjugacy, 9, 10, 13 and 14) and mhemical economy (consumer theory, 11). Section 4 analyzes the geometrical properties of F intermsofthecoefficientsofitsgiven represention σ. Finally, in Section 5 we show th it is possible to handle linear optimizion problems involving strict constraints in an effective way, extending to this class of problems LSIP results on optimality, strong uniqueness and boundedness of the optimal set. Now let us introduce the necessary notion. Given a non-empty set X R p, p N, we denote by conv X, cone X, aff X and span X the convex hull of X, the convex cone genered by X, the affine hull of X and the linear subspace of R n spanned by X, respectively. Moreover, we define cone =span = {0 p }. Some of the above sets can be described by means of the space of generalized finite sequences, R (T ), whose elements are the functions λ : T R such th λ t 6=0only on a finite subset of T. The convex cone, in R (T ), of the nonnegive finite sequences is R (T ) +.IfX 6= is convex, we denote by O + X the recession cone of X, bydim X the dimension of X and by D (X; x) the cone of feasible directions x X. IfX 6= is a convex cone, X o denotes the positive polar cone of X. Moreover, from the topological side, we denote by cl X, int X, rbd X and rint X the closure, the interior, the relive boundary and the relive interior of X, respectively. We shall exploit throughout the paper the existing relionship between σ, in(1.1),and its relaxed system σ = {a 0 tx,t T } (obtained by replacing a 0 tx> with a 0 tx for all t S). Obviously, the consistency of σ does not entail the consistency of σ (consider, e.g., the system σ = {0 <x<0} in R). PROPOSITION 1.1. Let σ be the relaxed system of σ and let F be the solution set of σ. Then the following conditions hold: (i) If F 6=, then F =clf. (ii) If F = and σ does not contain trivial inequalities (i.e., then either F = or dim F<n. µ 6= 0 n+1 for all t T ) Proof. (i) Let x 1 F.Ifx F,then (1 λ) x + λx 1 F, 0 <λ<1, (1.2) 3

4 so th x = lim (1 λ) x + λx 1 cl F. λ 0 Hence F cl F. The reverse inclusion is trivial. µ (ii) Assume F =, 6= 0 n+1 for all t T and F 6=. Since σ is consistent and {a 0 tx>,t T } has no solution (i.e., there is no Sler point for σ), there exists a t T such th a 0 tx = for all x F (Corollary µ in 7), with a t 6=0 n (otherwise, taking an arbitrary x F,weget =0 0 nx =0, so th =0 n+1 ). Hence a 0 tx = defines a hyperplane containing F. Observe th, for σ = {0 0 nx>0}, F = and F = R n. Thus, stement (ii) in Proposition 1.1 could fail for systems containing trivial inequalities. 2. EXISTENCE OF SOLUTIONS PROPOSITION 2.1. Let σ be the system in (1.1). (i) If σ is consistent, then µ 0n / cl cone 1 ½µ Moreover, if S 6=, then the following stement also holds: ½µ ¾ ½µ 0 n+1 / conv,t S +cone ¾,t T. (2.1),t W (ii) Each of the following conditions guarantees the consistency of σ: (ii.a) S = and (2.1) holds. (ii.b) S 6=, (2.1) and (2.2) hold and the set in (2.2) is closed. ¾. (2.2) Proof. (i) Assume th Then there exists a sequence µ 0n cl cone 1 ½µ ¾ d r δ r µ d r = X λ r t t T δ r ½µ ¾,t T. µ d R n+1 r such th lim r µ δ r =,λ r R (T ) +,r=1, 2,... µ 0n and 1 4

5 Since σ is consistent, we can take x 0 F. Then, we have µ d r 0 µ x 0 = X λ r t a 0 1 t x 0 0, r=1, 2,..., (2.3) t T δ r and, taking limits in (2.3), we get the contradiction µ 0 µ 0n x Hence, (2.1) holds. Now assume th S 6=. If 0 n+1 conv ½µ ¾,t S then there exists λ R (T ) + such th P t Sλ t =1and 0 n+1 = X t S +cone ½µ µ λ t + X µ λ t. b t W t ¾,t W, Taking an arbitrary solution of σ, x 0,wehave µ x 0=0 0 0 n+1 = X X λ t a 0 1 t x 0 + λ t a 0 t x 0 > 0. t S t W Hence, (2.2) holds. (ii.a) Let us suppose µ th S = and (2.1) holds. µ By Corollary in 15, µ there exists a x c hyperplane in R n+1, c 0 = γ, with c = R n R, such th c 0 0n = γ and 1 all v cone x n+1 ½µ c 0 v>γfor all v cl cone ½µ,t W,t W c n+1 ¾. The last condition implies th c 0 v 0 >γfor ¾,inparticular,a 0 tc + c n+1 0 for all t W. Since c n+1 = γ<0, definingx 1 = c n+1 1 c,wehavea 0 tx 1 for all t W,sothσ is consistent. (ii.b) Now, let us assume th S 6=, (2.1) and (2.2) hold and the set ½µ ¾ ½µ ¾ A := conv,t S +cone,t W is closed. Since 0 n+1 / A, by Corollary in 15, there exists a vector µ c c n+1 R n+1 such 5

6 µ c th µ + µ c n+1 µ as 0 µ a>0for all a A. Since A for all t S, a 0 b tc + c n+1 > 0. Since t A for all t S, s W and µ>0, wehave b s Hence, a 0 sc + b s c n+1 0 for all s W. a 0 s c + b sc n+1 > a0 tc + c n+1. µ Let x be a solution of σ (system sisfying (ii.a) and, so, consistent) and consider the following point of R n : bx := c c n+1, if c n+1 < 0 c + x, if c n+1 =0 2x + c, if c n+1 > 0 c n+1 Simple algebraic calculions show th bx F,sothσ is consistent. REMARKS 2.1. Conditions (2.1) and (2.2) can be interpreted in terms of consequence relions, defined by Kuhn 8 as those linear inequalities which would hold true for all solutions of the given system. To do this, we shall consider again part (i) in Proposition 2.1. Assume th σ is consistent. Obviously, every non-negive linear combinion of the weak inequalities a 0 tx, t T, is consequence of σ. Moreover, given a sequence of consequence relions of σ such th the n +1sequences of coefficients are convergent, the limit inequality µ is also a consequence of σ. Identifying each inequality a 0 x b with the vector of coefficients a R n+1, (2.1) means th 0 0 b nx 1 (which cannot be consequence of σ) cannot be obtained from σ through non-negive linear combinions followed by limits. Since the aggregion of 0 0 nx 1 to σ does ½µ not change itsµ solution ¾ set, F, we can replace the right hand side 0n cone in (2.1) by cl cone,t T;, which actually represents all the consequence 1 relions of σ by µ the non-homogeneous Farkas Lemma (see, for example, Corollary in 7). a Similarly, if belongs to the right hand side set in (2.2), then a 0 x>bis non-negive b linear combinion of the inequalities of σ, with least a positive multiplier for a certain strict inequality a 0 tx, t S, and so it is a consequence of σ. Thus (2.2) means th 0 0 nx>0 cannot be obtained from σ in this way. Observe µ again th, ½µ since the aggregion ¾ ½µ of 0 0 nx> 1to ¾ 0n σ does not modify F, we can aggrege to either,t S or,t W 1 in (2.2). 6.

7 2.2. If S =, Proposition 2.1 coincides with the existence theorem of Fan 5 or, equivalently (taking into account Remark 2.1), the existence theorem of Zhu 16 for linear semi-infinite systems If W =, (2.2) reads 0 n+1 / conv ½µ ¾,t T, and this set is closed if T is compact and a t and are continuous functions (in particular if T < ). This version provides an easy proof of Carver s Theorem 2: {A 0 x<b} (A (m n), b R m ) is consistent if and only if the unique solution of {A 0 y =0 n,b 0 y 0, y 0 m } is 0 m If σ is homogeneous (i.e., =0for all t T )ands 6=, then Proposition 2.1 becomes the extended Motzkin s Alternive Theorem (see Theorem 3.5 in 7) If W = and σ is homogeneous, then Proposition 2.1 coincides with the extended Gordan s Theorem (Theorem 3.2 in 7) If T <, the closedness condition in (ii.b) holds. From here it is easy to prove Motzkin s Transposition Theorem 12: the system {Ax < b; Cx d} (A (m n), C (p n), b R m and d R p ) is consistent if and only if (i) if y 0 A + z 0 C =0 0 n, y 0 m and z 0 p, then y 0 b + z 0 d 0; and (ii) if y 0 A + z 0 C =0 0 n, y 0 m, y 6= 0 m and z 0 p, then y 0 b + z 0 d>0. (if (2.1) fails, then (i) fails; if (2.2) fails, then (ii) fails) The finite version of Proposition 2.1 is an implicit consequence of Theorems I-III in 8. The following example shows th the closedness assumption in condition (ii.b) of Proposition 2.1 is not superfluous. EXAMPLE 2.1. Consider σ = {tx 1 + x 2 > t 2,t R\{0} ; x 2 > 0(t =0)}. (2.1) holds because 0 2 is solution of σ. If (2.2) fails, then we can write 0 0 = λ X λ t t 1, (2.4) 0 0 t6=0 t 2 where λ t 0 for all t R, {t R λ t 6=0} < and P t Rλ t =1. Comparing the third components in both sides of (2.4), we get λ t =0for all t 6= 0, so th λ 0 =0and P t Rλ t =0 (contradiction). Now, assume th x R n sisfies tx 1 + x 2 > t 2 for all t 6= 0. Taking limits as t 0 we get x 2 0,sothσ is inconsistent. 7

8 The following result provides the nural way to decide whether σ is consistent or not, and to compute a solution of σ in the first case. To do this we associe with σ the LSIP problem (P σ ) Inf x n+1 s.t. a 0 tx + x n+1, t S a 0 t x, t W. PROPOSITION 2.2. (i) If v (P σ ) < 0, then σ is consistent. (ii) If v (P σ ) > 0, then σ is inconsistent. (iii) If v (P σ )=0and (P σ ) is not solvable, then σ is inconsistent. µ bx Proof. (i) Let bx n+1 R n+1 such th a 0 tbx + bx n+1 for all t S, a 0 tbx for all t W and bx n+1 < 0. Then bx is solution of σ. µ x (ii) If x is a solution of σ, then is a feasible solution of (P σ ), so th v (P σ ) 0. 0 Hence, v (P σ ) > 0 entails the inconsistency of σ (and σ). µ bx (iii) If v (P σ )=0and bx is a solution of σ, then is an optimal solution of (P σ ). 0 If v (Pµ σ )=0and (P σ ) is solvable, there exists an optimal solution of (P σ ) which can be x written as. Then x is solution of σ. Nevertheless, σ is not necessarily consistent. 0 The system in Example 2.1 illustres the dubious case: v (P σ )=0with (P σ ) solvable. In fact, taking limits as t 0 in tx 1 + x 2 + x 3 t 2, t 6= 0,givesx 2 + x 3 0. The remaining constraint is x 2 + x 3 0, so th x 3 0 for all feasible solution of (P σ ).Since0 3 is feasible solution, v (P σ )=0and 0 3 is an optimal solution of (P σ ). In this case σ is inconsistent. µ x Observe th, given ε>0, if is a solution of x n+1 σ ε := {a 0 tx x n+1 ε, t S; x n+1 ε; a 0 tx x n+1 0, t W }, µ x then (x n+1 ) 1 is a feasible solution of (P σ ), so th (as observed in 4) the consistency ε of σ ε entails the consistency of σ, according to Proposition µ 2.2. The converse stement x holds if S < (since, given x F, then εδ 1 is solution of σ ε for δ := 1 min {1; a 0 tx,t S}), but it can fail for infinite systems. In fact, for the system in R σ = {tx > t 2,t6= 0}, F = {0} whereas σ ε is inconsistent for all ε>0. 8

9 3. EVENLY CONVEX SETS REVISITED AsetC R n is said to be evenly convex (in 6) if it is the intersection of a family of open halfspaces. Since this family can be empty, R n and are evenly convex sets. On the other hand, since any closed halfspace is the intersection of infinitely many open halfspaces, C is evenly convex if and only if C is the solution set of a certain linear inequality system such as (1.1). In particular, any closed convex set is evenly convex. According to (1.2), if C is an evenly convex set, x 1 C and x 2 cl C, then ]x 1,x 2 [ C (compare with the proof of stement 3.5 in 6). The next result provides two characterizions of evenly convexity. PROPOSITION 3.1. Given C R n such th 6= C 6= R n, the following conditions are equivalent to each other: (i) C is evenly convex; (ii) C is a convex set and for each x R n \C there exists a hyperplane H such th x H and H C = ; and (iii) C is the result of elimining from a closed convex set the union of a certain family of its exposed faces. Proof. (i) (ii) Let C = t T {x R n a 0 tx> },witha t 6=0 n for all t T.Ifx/ C there exists a t T such th a 0 tx. Then H := {x R n a 0 t (x x) =0} sisfies the desired conditions. (ii) (i) Given t T := R n \C, there exists a vector a t 6=0 n such th a 0 t (x t) > 0 for all x C. Defining = a 0 tt, wehavec {x R n a 0 tx>,t T }. On the other hand, if x/ C, takingt := x, wehavea 0 tx =,sothx/ {x R n a 0 tx>,t T }. Hence,C is the solution set of {a 0 tx>,t T }. (i) (iii) Let C be the solution set of σ = {a 0 tx>,t T } and let σ be the relaxed system of σ. The solution set of σ is F =clc (by Proposition 1.1). Given t T, the set X t := ª x F a 0 tx = is an exposed face of F (maybe empty). Moreover, C = F \ X t. t T (iii) (i) Let X be a closed convex set and let {X t,t S} be a family of exposed faces 9

10 of X such th C = X\ X t. t S Since X t 6= X for all t S (otherwise C = ), t S X t rbd X and we get rint X = X\ (rbd X) C X. (3.1) Taking closures in (3.1) we conclude th X =clc, soth C =(clc) \ X t. (3.2) t S Since cl C is a closed convex set, it is the solution set of a certain linear semi-infinite system {a 0 tx,t U} and, so, cl C is evenly convex. If X t = for all t S, then, from (3.2), C =clc is evenly convex. So, we can assume without loss of generality X t 6= for all t S 6= (since we can elimine in (3.2) those X t, t S, which are empty). Given t S, there exist a t 6=0 n and R such th a 0 tx for all x cl C and X t = {x cl C a 0 tx = }. Defining T := U S, it is clear th cl C = {x R n a 0 tx,t T }. Let W = T \S. We shall prove th σ, as in (1.1), is a linear represention of C. If x C, sincex/ (cl C) \C = t S X t, according to (3.2), we must have x/ X t for all t S. Hence a 0 tx for all t T (since x cl C), with a 0 tx> for all t S, so th x is solution of σ. Conversely, if x is solution of σ, then x cl C (since a 0 tx for all t T )andx/ t S X t (since a 0 tx> for all t S). Then x C, again by (3.2). We conclude th C is the solution set of σ. REMARK 3.1. The following characterizions of evenly convex sets have been proposed by Dr. Martínez-Legaz in a prive communicion: (i) C is the intersection of a non empty collection of non empty open convex sets; (ii) C is a convex set and is the intersection of a collection of complements of hyperplanes; (iii) C is a convex set and for any convex set K contained in (cl C) \C, there exists a hyperplane containing K and not intersecting C; (iv) C is a convex set and for any convex set K (cl C) \C, the minimal exposed face (in cl C) containing K does not intersect C; (v) C is a convex set and for any x (cl C) \C, the minimal exposed face (in cl C) 10

11 containing x does not intersect C; and (vi) C is a convex set and for any x (cl C) \C, there exists a supporting hyperplane of cl C x not intersecting C. As a consequence of Proposition 3.1 and Theorem 11.2 in 15, any relively open convex set is evenly convex. Analogously, any strictly convex set C (i.e., a convex set C such th bd cl C does not contain segments) is evenly convex since the exposed faces of cl C are the singleton sets determined by its boundary points. Observe th any convex set X 6= can be fitted from inside by rint X and from outside by clx, rint X and cl X being evenly convex sets. COROLLARY 3.1. If C is an evenly convex non-closed set, then (cl C) \C cannot be the union of a family of non-exposed faces of cl C. Proof. According to Proposition 3.1, it will be sufficient to prove th two disjoint families of faces have different unions. So, in particular, no non-empty union of non-exposed faces of a closed convex set is equal to a union of exposed faces. Let X beaclosedconvexsetandlet{x u,u U} and {X v,v V } be two disjoint families of non-empty faces of X. Assume th X u = X v. u U v V Given u 1 U, thereexistsx rint X u1 X v,soththereexistsv 1 V such th v V x X v1 and (rint X u1 ) X v1 6=. Then X u1 X v1 (by Theorem 18.1 in 15), the inclusion being strict since {X u,u U} {X v,v V} =. Hence dim X u1 < dim X v1. Similarly, given v 1 V,thereexistsu 2 U such th dim X v1 < dim X u2. By induction, there exists sequences {u k } U and {v k } V such th dim X uk < dim X vk < dim X uk+1,k=1, 2,... Hence, lim k dim X uk =+, contradicting dim X u n for all u U. EXAMPLE 3.1. Consider the closed convex set X = x R 2 tx 1 +(1 t)x 2 t t 2,t [0, 1] ª (3.3) represented in Figure 3.1. The non-trivial faces of X are X u = {(u, 1+u 2 u)}, 0 u 1, X 2 = {0} [1, + [ and X 3 =[1, + [ {0}. All these faces are exposed, except X 0 and X 1. 11

12 X Figure 3.1 C = X\ X u is evenly convex if U ]0, 1[ {2, 3} (by Proposition 3.1 and Corollary u U 3.1), and it is not evenly convex if 6= U {0, 1} (by Corollary 3.1). The next two results compare different elements of C and cl C. PROPOSITION 3.2. If C 6= is evenly convex, then (i) D (C, x) =D (cl C, x) for all x C. (ii) The extreme points of C are those extreme points of cl C belonging to C. Proof. (i) We shall prove the non-trivial inclusion D (cl C, x) D (C, x) for all x C. We assume the contrary. Let x C and u D (cl C, x) \D (C, x). Letε>0 such th x + εu cl C. Sinceu/ D (C, x), x + ε u (cl C) \C and there exists an exposed face of 2 cl C,sayX, such th X C = and x + ε u X (by Proposition 3.1). 2 Let a 6= 0 n and b R such th a 0 x b for all x cl C and X = {x cl C a 0 x = b}. ³ Since a 0 x + ε u = b and a 0 x b, wehavea 0 u 0. We shall obtain a contradiction in 2 the two possible cases. If a 0 u<0, then a 0 (x + εu) <a 0 ³ x + ε 2 u = b, sothx + εu / cl C (contradiction). If a 0 u =0, then a 0 x = a 0 ³ x + ε 2 u = b, sothx X. Hencex/ C andthisisagaina contradiction. (ii) Let x C. If x is not an extreme point of cl C, then there exists u 6= 0 n such th ±u D (cl C, x) =D (C, x), according to part (i), so th x cannot be an extreme point of C. The converse stement is trivial. 12

13 PROPOSITION 3.3. If C 6= is evenly convex, then O + C = O + (cl C). Consequently, C is bounded if and only if O + C = {0 n }. Proof. Let C = {x R n a 0 tx>,t T }. Then O + C = {y R n a 0 ty 0, t T } = O + (cl C), (3.4) since cl C = {x R n a 0 tx,t T}(Proposition 1.1). Finally, C is bounded if and only if cl C is bounded if and only if O + (cl C) = {0 n } (Theorem 8.4 in 15). The last two results could fail or not for general convex sets. EXAMPLE 3.1 (revisited). Neither C 1 := X\ (1, 0) 0ª nor C 2 = X\ (]2, + [ {0}) is evenly convex. C 1 sisfies both Proposition 3.2 and 3.3 whereas none of them is sisfied by C 2 (consider the point (2, 0) 0 in Figure 3.2). C 2 Figure 3.2 PROPOSITION 3.4. Let C 6= be an evenly convex set and let y 6= 0 n. If there exists x C such th {x + λy λ 0} C, then y O + C. Proof. Let C = {x R n a 0 tx>,t T} and assume the existence of t T such th a 0 ty<0. Then a 0 t (x + λy) < for λ sufficiently large, so th x + λy / C. Therefore a 0 ty 0 for all t T, and this entails y O + C according to (3.4). Proposition 3.4 is a direct extension of Theorem 8.3 in 15 to evenly convex sets, and 13

14 implies the same consequence: such y belongs to O + (rint C). Similarly, the next result extends Corollary in 15. PROPOSITION 3.5. Let 6= C R n an evenly convex set and let A : R m R n be a linear transformion such th A 1 C 6=. Then A 1 C is evenly convex and O + (A 1 C)= A 1 (O + C). Proof. Let C = {x R n a 0 tx>,t T }. Then it can be realized th A 1 C = z R m (A 0 a t ) 0 z>,t T ª, so th A 1 C is evenly convex. A 1 (O + C) O + (A 1 C). Infact,ify A 1 (O + C), taking an arbitrary z A 1 C,we have for each λ 0, sinceay O + C, A (z + λy) =Az + λay C, so th z + λy A 1 Ċ. Hence y O + (A 1 C). Conversely, assume th y O + (A 1 C). Let z A 1 C and λ 0 arbitrarily chosen. Since z + λy A 1 C,wehaveAz + λay C. Applying Proposition 3.4, we conclude th Ay O + C,sothy A 1 (O + C). PROPOSITION 3.6. The cartesian product of two evenly convex sets is also evenly convex. Proof. If C 1 = {x R n a 0 ux>b u,u U} and C 2 = {x R m c 0 vx>d v,v V }, then C 1 C 2 = ½ x R n+m µ au 0 m Hence C 1 C 2 is evenly convex. 0 x>b u,u U; µ 0n c v 0 x>d v,v V ¾. Concerning the sum of closed convex sets, we know th it is not necessarily closed unless a certain recession condition holds which guarantees th the recession cone of the sum is the sum of the corresponding recession cones. Next we show th the second stement remains true for evenly convex sets, but their sum is not necessarily evenly convex (even though one of the two sets is bounded). PROPOSITION 3.7. Let C 1 and C 2 be non-empty evenly convex sets in R n such th 14

15 (O + C 1 ) ( O + C 2 )={0 n }. Then, O + (C 1 + C 2 )=O + C 1 + O + C 2. (3.5) Proof. According to Proposition 3.3, [O + (cl C 1 )] [ O + (cl C 2 )] = {0 n },sothwecan apply Corollary in 15 to conclude th O + [cl (C 1 + C 2 )] = O + (cl C 1 )+O + (cl C 2 )=O + C 1 + O + C 2. Then, we have O + C 1 + O + C 2 O + (C 1 + C 2 ) O + [cl (C 1 + C 2 )] = O + C 1 + O + C 2, so th (3.5) holds. EXAMPLE 3.1 (revisited). Consider the evenly convex set X in (3.3). The compact convex set C 1 := {x X x 1 + x 2 1} (see Figure 3.3) and the set C 2 = {x R 2 x 1 0; x 2 0; x 1 + x 2 > 0} (see Figure 3.4) are obviously evenly convex and sisfy (3.5). Nevertheless, C 1 + C 2 is not evenly convex (see Figure 3.5). C 1 Figure

16 C 2 Figure 3.4 C 1 + C 2 Figure 3.5 Observe also th C 1 + C 2 = A (C 1 C 2 ) if we define A : R 2n R n as A (x, z) =x + z. This shows th the image of an evenly convex set through a linear transformion may fail to be evenly convex (as it happens with the closed convex sets). In contrast, the linear transformion of a relively open convex set is another relively open convex set (Theorem 6.6 in 15). The next two results are the extensions to evenly convex sets of two well-known properties of the closed convex sets (Corollary and Corollary in 15, respectively). PROPOSITION 3.8. If {C i i I} is a family of evenly convex sets such th C i 6=, i I then µ O + C i = O + C i. i I i I 16

17 Proof. Obviously the set C = i I C i is evenly convex. Take x C arbitrarily. If y O + C i for all i I, then we have {x + λy λ 0} C i for all i I, so th {x + λy λ 0} C. Hence y O + C and O + C i i I trivial. O + C. The reverse inclusion is COROLLARY 3.2 Let C be an evenly convex set and let M be an affine manifold such th C M is a non-empty bounded set. Then M 0 C is bounded for each affine manifold M 0 which is parallel to M. Proof. Obviously, since any affine manifold is a closed convex set, M 0 is evenly convex. Moreover, since M 0 is assumed to be parallel to M, O + M 0 = O + M (a linear subspace). If M 0 C 6=, from Propositions 3.8 and 3.3, O + (M 0 C) =O + M 0 O + C = O + M O + C = O + (M C) ={0 n }, so th M 0 C is bounded. 4. GEOMETRY Along this section we show th it is possible to obtain geometrical informion about the solution set F of a consistent system σ = {a 0 tx,t W; a 0 tx>,t S}. To do this we appeal to well-known relionships between the corresponding relaxed system σ = {a 0 tx,t T := W S} and its solution set F. Recall th σ is said to be locally Farkas-Minkowski (LFM) if each linear consequence of σ defining a supporting halfspace to F is also the consequence of a finite subsystem of σ. PROPOSITION 4.1. Let F 6= be the solution set of σ and let T c be its set of carrier indices (i.e., T c = {t T a 0 tx = for all x F } W ). Then, rint F {x R n a 0 t x =,t T c ; a 0 t x>,t T \T c }. (4.1) Moreover, if σ is LFM, then both members of (4.1) are equal, aff F = {x R n a 0 t x =,t T c } 17

18 and dim F = n dim span {a t,t T c }. Proof. Observe th the carrier indices of σ are those of σ since a hyperplane contains a convex set if and only if contains its closure and F =clf (Proposition 1.1). On the other hand, rint F =rintf, aff F =aff F and dim F =dimf, so th it is sufficient to prove all the stements above for σ and F instead of σ and F. Hence, the conclusion follows from Theorem 5.1and5.9in7. PROPOSITION 4.2. equivalent to each other: (i) Fµ is bounded; ½µ 0n (ii) int cone,t T; 1 (iii) cone {a t,t T} = R n ; and If the solution set of σ is F 6=, then the following stements are µ ¾ 0n ; 1 (iv) there exists a finite subsystem of σ whose solution set is bounded. Proof. Since the solution set of an arbitrary consistent system is bounded if and only if the solution set of its corresponding relaxed system is bounded (by Proposition 1.1), each of the stements from (i) to (iv) holds if and only if it holds for σ. The conclusion is a straightforward consequence of Theorem 9.3 in 7. The active cone x F (with respect to σ)is A (x) =cone{a t a 0 tx =,t W }. Analogously, the active cone x F (with respect to σ)is A (x) =cone{a t a 0 tx =,t T }. σ is said to be locally polyhedral (LOP) if D F,x = A (x) o for all x F. PROPOSITION 4.3. Let x F.Ifdim A (x) =n, then x is an extreme point of F.The converse stement holds if σ is LOP. Proof. It can be easily seen th A (x) =A (x) for all x F. Moreover, x is an extreme point of F if and only if x is an extreme point of F (by Proposition 3.2). The conclusion follows from Theorem 9.1 in 7. 18

19 The following example shows th the additional conditions for the converse stements in Propositions 4.1 and 4.3 are not superfluous. EXAMPLE 3.1. (revisited) Let σ = {tx 1 +(1 t)x 2 >t t 2,t ]0, 1[} whose solution set, F, is represented in Figure 4.1. F Figure 4.1 Here T c =, so th (4.1) becomes rint F F. Nevertheless, rint F 6= F since σ is not LFM (observe th x 1 0 and x 2 0 are not the consequence of finite subsystems of σ). Despite the failure of the LFM property, aff F = R 2 and dim F =2, as prescribed by Proposition 4.1. On the other hand, cone {a t,t T } =cone ½µ ¾ t,t ]0, 1[ 6= R 2, 1 t so th F is unbounded. Finally, observe th A (x) ={0 2 } for all x F, even the extreme points of F, (1, 0) 0 and (0, 1) 0 (in fact, any LOP system is LFM). 5. LINEAR OPTIMIZATION We associe with the linear optimizion problem with strict inequalities (P ) Inf c 0 x s.t. a 0 tx, t W a 0 tx>, t S 6=, 19

20 where c 6= 0 n,thelsip P Inf c 0 x s.t. a 0 tx, t T = W S. Obviously, the values of the above problems are reled by v P v (P ) (the inequality can be strict: e.g., for (P )Minx s.t. 0 < x < 0, v (P ) = + and v P = 0), with v (P ) = v P if (P ) is consistent (by Proposition 1.1). Hence, any outer approximion method for P (as grid and cutting plane discretizion methods) is an outer approximion method for (P ). On the other hand, any feasible directions method for P (as the simplexlike methods) provides a sequence of feasible solutions for (P ) approaching v (P ). Infact,if {x r } F sisfies lim c 0 x r = v P,takingbx F and a sequence {λ r } [0, 1] such th r lim λ r =0,wehave{(1 λ r ) x r + λ r bx} F and lim c 0 [(1 λ r ) x r + λ r bx] =v (P ). r r Although (P ) will be usually unsolvable (even though F is bounded), we can ste a KKT condition which provides an exact stopping rule for any LSIP method adapted to (P ). PROPOSITION 5.1. Let bx F.Ifc A (bx) (c int A (bx)), then bx is an optimal solution of (P ) (a strongly unique optimal solution of (P ), respectively). The converse stements are true if σ is LFM. Proof. If c A (bx), with bx F,thenwehavebx F and c A (bx), sothbx is an optimal solution of P (Theorem 7.1 in 7). Similarly, if c int A (bx), then c int A (bx) and bx is a strongly unique optimal solution of P (Theorem 10.6 in 7). Conversely, if bx is an optimal solution of (P ), then it is also an optimal solution of P (since v P = v (P )). Then Theorem 7.1 of 7 applies again to conclude th c A (bx) =A (bx) under the LFM assumption. The argument is similar for the other converse stement, taking into account Theorem 10.6 in 7. The last result deals with the boundedness of the optimal set of (P ), th we denote by F (the boundedness of F can be seen as a well-posedness condition for (P )). PROPOSITION 5.2 If (P ) is solvable, the following conditions are equivalent to each other: (i) F is a bounded set; 20

21 (ii) all the non-empty sublevel sets of (P ) (either {x F c 0 x<α} or {x F c 0 x α}, with α R) are bounded; (iii) there exists a finite subproblem of (P ) whose non-empty sublevel sets are bounded; and (iv) c int cone {a t,t T}. Proof. First, observe th the sublevel sets of (P ) (in particular, F ) are evenly convex. (i) (ii) Let α R such th {x F c 0 x α} 6= (the argument applies for strict sublevel sets). According to Proposition 3.8, we have O + F = O + (F {x R n c 0 x v (P )}) =O + F {y R n c 0 y 0} = = O + (F {x R n c 0 x α}) =O + ({x F c 0 x α}). Hence, since O + F = {0 n } if and only if O + ({x F c 0 x α}) ={0 n }, (by Proposition 3.3) F is bounded if and only if {x F c 0 x α} is bounded. Now let us show th all the non-empty sublevel sets of (P ) are bounded if and only if all the non-empty sublevel sets of P are bounded. This is the consequence of the double inclusion {x F c 0 x α} x F c 0 x α ª cl {x F c 0 x<α+ ε} for all α R and for all ε>0 (if x = lim r x r,with{x r } F and c 0 x α, then c 0 x r <α+ ε for r sufficiently large, so th x cl {x F c 0 x<α+ ε}). The same argument applies for the non-empty sublevel sets of the subproblems obtained replacing W and S by the finite sets W 0 W and S 0 S in (P ) and P. Hence we conclude th (ii) (iii) (iv) by straightforward applicion of Corollary in 7. ACKNOWLEDGMENTS The authors are indebted to Dr. suggestions. J.-E. Martínez-Legaz for his valuable comments and [1] J. Bair, Strict separion of several convex sets, Bull. Soc. Mh. Bel., Ser. B 32 (1980) [2] W. B. Carver, Systems of linear inequalities, Annals of Mhemics 23 ( ) [3] A. Daniilidis and J.-E. Martínez-Legaz, Characterizions of evenly convex sets and evenly quasiconvex functions, submitted to Journal of Mhemical Analysis and Applicions. 21

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