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Lecture Notes in Economics and Mathematical Systems
Managing Editors: M. Beckmann and W. Krelle
319
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H. Albach M. Beckmann (Managing Editor) P. Dhrymes G.Fandel G. Feichtinger J. Green W. Hi!denbrand W. Krclle (Managing Editor) H.P. KOnzi K. Ritter R. Sata U. 8chittko P. Schonfeld R. Selten
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Author
Dr. Dinh The Luc JnstHute of Mathematjcs RBox 631 Boho 10000 Hanoi, Vietnam
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© Springor.Ver:ag Berlin Heidc!berg 1989 Printed in Germany
C...·n+inn ":Int4 hil"'ni ....n· n ... ,,...lth~ll~ RAI·,. H"'m~Mch/Rpr~crtr
To My Mother and
Preface
These notes grc,v out of a series of lectures given by the author at the Univer­ sity of Budapest during 1985-1986. Additional results have been included which \vere obtained \V'hile the author "\vas at the University of Erlangen-)Jiirnberg under a grant of the Alexander .von !Iumboldt Foundation.
Vector optimization has two main sources coming from economic equilibrium and ,velfare theories of EGge,vorth (1881) a.ud Pareto (1906) and from mathemat­ ical backgrounds of ordered spaces of Cantor (1897) and Iiausdorff (1906). Later, game theory of Borel (1921) and von Neumann (1926) and production theory of I(oopmans (1951) have also contributed to this area. Ho\vever, only in the fifties, after the publication of I(uhu·-Tucker's paper (1951) on the necessary and sufficient .conditions for efficiency, and of Deubreu's paper (1954) on valuation equilibrium and Pareto optimum, has vector optimization been recognized as a mathematical discipline. The stretching development of this field began later in the seventies and eighties. Today there are a number of books on vector optimization. Most of them arc concerned with the rnethodology and the applications. Fev"," of them offer a systematic study of the theoretical aspects.The aim of these notes is to pro­ vide a tmified background of vector optinuzation,"\vith the emphasis on nonCOllvex problems in infinite dimensional spaces ordered by convex cones.
The notes are arranged into six chapters. The first chapter presents prelim­ inary material. It cOlltains a study of nonconvex analysis \vith respect to convex cones. In chapter 2, we introduce main concepts of vector optimization such as preference order, efficiencies. vector optimality etc. and the existence of efficient points. Chapter 3 deals "\vith vector optimization problems ,vith set.· ·valued objec­ tives and constraints. Necessary and sufficient conditions are established in terms of generalized derivatives. In chapter 4, \Ve present a scalarizatiolllTIcthod to con­ vert a vector problem to a scalar problem. Stability properties of solution sets-of vector problems arc addressed. Chapter 5 is devot.ed to duality. Together ,vith the classical approaches to dualit.~· sneh a~ Lagrangcan and conjugate duality we also provide axiomatic duality and au approach via thcoreUls of the alternative. These approaches arc especially appropriate for llonconvex problc1l1s. In the last chap­ tel", the structure of efficient point. sets of linear, convex and quasiconyex problems is investigated. In the references \ve include only the papers \\·hich are directly
VI
involved with the topics of our consideration and those of recent publication& the subject. .
A cknov.dcdgernen ts. The author ,vould like to express his deep thanks to Professor A.1:lrekopa \vno was his PhD supervisor at the Hungarian Academy of Sciences and \vho suggested giving a course of lectures on multiobjective optimiza­ tion at the University of Budapcst~ ITurther thanks go to Professor J~Jahn of the University of Erlangen-Xiirnberg for reading the manuscript and for useful sug­ gestions. turthermore, the author is grateful to the Computer and Automation Institute of the I-Iungarian AcadenlY of Sciences and the Institute for Applied Math ematics of the University of Erlangen-Nurnberg for the hospitality and the \:vork­ ing facilities he received during his stay there. (These notes would never have been completed ,vithout a grant of the Alexander von Jlumboldt Foundation to \vhom the author is especially indebtcd~The last) but not least thanks are addressed to the Institute of Mathernatics, Hanoi, for the permission and support to carry out the research abroad..
Erlangen, West Germany lvIay 1988
Contents
Chapter 1: Analysis over Cones 1
1~ Convex cones ~. 6 • ~ • ~ •• ~ ~ •• ~ •• ~ • ~ ~ •• ~ • ~ ~ 6 • ~ •• ~ • 6 •••••••••••••••••••••••• 1 2.Recession cones 6 ••••••••••••• I ••• I •• I • ~ I ~ I ••• ~ I •••• 8 3.Cone closed sets .. 6 ••••••••••••••• ~ •••••• ~ • ~ ••••••••••••••••••• 6 • • • • • • 13 '1. Cone monotonic fUIlctions ..... I •• I •••• I •••• I • I •••••••••••••••••••••••• 18 5.Cone continuous functions I •••••••• 6 •••••••••••• 6 • 6 •••••• I •••••••••••• 22 6.Cone conve..x functions 6 •••• ~ •• ~ • 6 ••• ~ • 6 ••••• 6 ••••••• ~ ~ ••••• 29 7~Sct-valued maps ~ ~ .. ~ .. ~ ~ ~ "' I •••••••••••• I •••••••••••••• 33
Chapter 2: Efficient Points and Vector Optimization Problems 37
. I.Binary relations and p"artial o~dcrs 6 •• I ••••••••••••• 37 2.Efficient points ~ .. ~ ~ ~ . ~ ~ I •••••••••••••••••••••••••••••• ~ ~ 39 3~Existcncc of efficient points ~ ~ . ~ .. ~ ~ . ~ .. ~ ~ ~ .. ~ ~ I I I I I • 46 4.Domination property . ~ ~ "' ~ . ~ .. ~ . ~ ~ . ~ ~ . 53 51Vector optimization problems I ••••••••••••••' •••••••••••••••••• ~ •• 57
Chapter 3: Nonsmooth Vector Optimization Problems 62
1.Contingent cierivatives ~ ... I • I I • I ••••••••••••••••••••••••••• 6 •••••••••• 63 2.Unconstrained problems ~. ~ ~ .. ~ "' . "' .. I •••••••••• I. "' ••••••••••• 67 3.Constrained problems 6 ••• ~ ••••••••• ~ •••• ~ ~ ~ • ~ ~ ~ •• ~ ~ ••• I •• I •• 6 .70 4.Differentiable case ~ ~ ~ ~ . ~ . ". 74 5.Convex case ~ I' •• I I I I •••••• 6 •••••••••••••••••••••••••••••••••••••••••• 77
Chapter 4: Scalarization and Stability so
I.Separation by monotonic functions ~ ~ .. ~ I •• I' I •••• I' •••• ~ ~ •• 81 2.Scalar representations . ~ .. ~ ~ . I I • 6 ••••••••• I ••••••••••••••••••••••• I •••• 86
VI1l
Chapter 5: Duality 109
1.Lagrangcan duality ~ 110 2~Conjtlgate duality ~ . ~ ~ ~ 117 3.Axiomatic duality " ~ ~ ~ 120 4.Dllality and alternative ~ ~ 129
Chapter 6: Structure of Optimal Solution Sets 135
I.General ca..'iC •••••• ~ ~ ..•••••••. ~ ~ .•..•••••••••.•••••••••••••••••.•• ~ •. 135 2.Lineal· case " 137 3.Convcx case 139 4.Quasiconvex case 148
Comments
References
Index
155
151
171
Analysis over Cones
This chapter is of preliminary character. It contains a study of sets and functions with respect to cones in infinite dimensional spaces. First, we give defi­ nitions concerning convex cones and properties of cones with special structure such as correct cones and cones with conve.x bounded base. In Section 2, we introduce the concept of recession cones of nonconvex sets which plays an important role in nonconvex optimization. The next four sections deal with sets and functions in the spaces with the presence of convex cones. 1"\he last section provides some defi.i:l..itions and results about set-valued maps which will be needed in the study of optimization problems \vith set-valued data.
I .. CONVEX CONES
Let E be a topological vector space over the field of real numbers. For a set C ~ E, the following notations '\vill be used throughout: elC, intC, riC, cc , conv(C) stay for the closure, interior, relative interior, complement and convex hull of C in E, respectivelY6 Besides, I(C) denotes the set C n -C~
Definition 1.1 A subset C of E is said to be a cone if tc E C for every c E C and every nonnegative number t. It is said to be a convex set if for any c, dEC,
. the line segment [c,d] = {tc+ (1 - t)d: 0 :5 t 5 I} belongs to C. Further, suppose
2
that C is a convex cone in E, then we say that it is 1) pointed if I(C) = {O}, 2) acute if its closure is pointed, 3) strictly supported if C \ l(C) is contained in an open homogeneous half
space, 4) correct if (cIC) + C \ I(C) ~ c.
Example 1.2 Below we give some examples to· clarify the definition.
1. Let Rn be the n-dimensional Euclidean space, then the nonnegative orthant R+. consisting of all vectors of Rn with nonnegative coordinates, is convex closed acute strictly Stlpprorted and correct as welL
The set {O} is also such a cone, but it is a trivial cone.
1"he set composed of zero and of the vectors with the first coordinates being positive, is a pointed strictly supported correct cone, but it is not acute.
Any closed homogeneous half space is a correct strictly supported cone, but it is not painted.
2. Let
c = {(x, y, z) E R3 : x > 0, Y > 0, Z > O} U {(x, y) z) : x ;::: y ;:: 0, z = O}.
Then C is convex, acute but not correct~
3. Let n be the vector space of all sequences x = {xn } of real numbers. Let
c = {x = {xn } En: X n ?: 0 for all n}.
Then C is a convex pointed cone~ We cannot say whether it is correct or acute because no topology has been given in the space.
4. Ubiquitous cone. Let 0 1 be the subspace of n consisting of the sequences x = {x n } such that X n = 0 for all but a finite number of choises for n~ It is a normed space if we provide it with the norm
I]xll = max{lxnl : n = 1, 2, ~ .. }~
Let C be the cone composed of zero and of sequences whose last nonzero term is positive~ Then C is pointed. It is called a ubiquitous cone (Holmes-1975) because the linear space spanned by C is the whole 01 . This cone is neither strictly supported, nor correct.
5.. Lcxicographic cone. Let
3
and let C be the set composed of zero and of seqnences ,vhose first nonzero term is positive~ This is a convex cone, called lexicographic~ It is pointed, but neither correct nor strictly supported.
6. IJet Lp[O, 1], 0 < P < 1, be the space of functions x(~) on [0,1] ,vhich are
integrable with respect to Lebesgue measure IJ. and I0 1 ]xl PdJl < oc ~ A mctrizable
topology of this space is determined by the basis of neighborhoods of zero:
{x E Lp[O, 1] : (Id IxI PdJ.L)l/p < lin}, n = 1,2, ....
Let C be the set of functions which are nonnegative almost everywhere. l"'hen the cone C is convex closed, hence correct (Proposition 1.4). Later on we shall see that this cone is not strictly supported.
Correct cones will play an important role in the next chapter, therefore \ve provide here some criteria for a cone to be correct. In the sequel C is presumed to be convex.
Proposition 1.3 C is correct if and only if (elC) + C \ l(C) ~ C \ I(C).
Proof. Since C\Z(C) ~ C, if the relation stated in the proposition holds, then the cone is obviously correct~ No\v, suppose that C is a convex correct cone. Observe that since 1(C) is a linear subspace and C is convex, for each a, bEe the relation a + b E I(C) implies that a, b E l(C). By this, the following relations hold:
C \ l(C) + C \ l(C) = C \ l(C);
G + C \ I(C) ~ C \ l(C). With these 've can rewrite the inclusion in the defmition of correct cones as
(ciC) + C \ I(C) = (cIC) + C \ l(C) +C \ Z(C) ~ C+C\l(C) ~ C\l(C),
completing the proof.•
Proposition 1.4 C is correct if one of the following conditions holds: i) C is closed;
ii) C \ I(C) is nonempty open; iii) C is composed of zero and the intersection of certain open and closed
homogeneous half spaces in E.
Proof. The first case is obvious. Now, if C \ I(C) is nonempty open, then the interior intO of C is nonempty and intC = C \ I(C)~ Hence,
(ciC) + C \ l(G) = (ciC) + (intC) ~ c.
4
Finally, assume that
C = {O} u (n{H>.: A E A}), \\There H).. is either a closed or open half space of E~ If all of HJ+.. arc closed, then this is equivalent to the first case. Therefore we may assume that at least one of the half spaces is open. In that case, I(C) = {OJ and a vector bEE belongs to C \ I(C) if and only if it belongs to every l!).., A E A. Further, it is clear that a E clC if and only if a E clH).. for all A E A. Now, since
(clH>J + H).. ~ H).., whatever H).. be open or closed, we conclude that a + bEe \vhenever a E clC, bEe \ l(C), completing the proof.-
Definition 1.5 Given a cone C in the space E. We say that a set B ~ E generates the cone C and write C ::::: cone(B) if
C = {tb : ~ E B, t ~ OJ. If in addition B does not contain zeTO and JOT each c E C, c f= 0, there are unique b E B, t > 0 such that c = tb, then we say that B is a base of C. Whenever B is a finite set, cone(conv (B)) is called a polyhedral cone.
In the literature (Pcressini-1967) sometimes a base is required to be convex closed. According to our definition, every nontrivial cone has a base. Later on \ve shall impose other requirements on the base if needed.
Remark 1.6 It is clear that in finite dimensional spaces a cone has a closed convex bounded base if and only if it is pointed closed. This fact, however is not true in infinite dimensions as it will be demonstrated by the example of Remark 4.6.
Proposition 1.7 (Jamcson-1970) If the space E is Hausdorff, then a cone with a closed convex bounded base is closed pointed, hence correc.t.
Proof. We sho\v first that C is closed. :Ei'or, let {c~} be a net from C converging to c. Since B is a base, there exist a net {ba} from B and a net {tal of positive numbers such that Co = to:ba: 6 vVe claim that {to:} is bounded. In fact, if this is not the case, t4at is, we suppose that lim to: = 00. Then, since the space is Hausdorff, the net {ba = cOo/ta} converges to O. Moreover, since B is closed, we arrive at the contradiction: 0 = limbo: E B. In this way, we may assume that {ta:} converges to some to 2:: o. If to = 0, then by the boundeness of B ~ lim lobo; = O. Hence· c = 0 and of course c E C. If to > 0, we may assume that tQ > c for
5
all a and some positive c. No\v, bo: = cex./ta converges to clio and again by the closedness of B, the vector clto E B. Hence c E C and C is closed.
1'he pointedness of C is obvious.•
Below \ve furnish t\VO other characterizations of cones ,vith closed convex bounded base. frhe first one describes a local property; for t~~o vectors x and y, the intersection (x + C) n (y - C) must be small enough if they are sufficiently closed to each other.The second one describes a global property: if two vectors of the cone are far from the origin, then so is their sum. The space E is presumed to be separated. I
Proposition 1.8 Assume that C has a closed convex bounded base. Then in any neighborhood W of the origin in E, there exists another neighborhood, say V , such thai
(V + C) n (V - C) ~ w. (1.1)
Proof. Let B be a base meeting the requirements of the proposition. First ,\ve prove that there is a balanced absorbing neighborhood U of zero in E such that
B n (U - C) = 0. (1.2)
In fact, since B is closed and does not contain zero, there is a neighborhood U of zero which may be assumed to be balanced absorbing symmetric such that
B n U = 0. (1.3)
We show that this neighborhood U will yield the relation (1~2). Suppose to the contrary that that relation does not hold, i.e. there are some b E B, u E U and c E C with b=u - c. Since B is a base, one can find a nonnegative number t and a vector b' E B such that c = tb' . We have
u = b+ c = b+ tb' .
Consider the vector u/(l + t). On one hand it belongs to U since the latter set is balanced. On the other hand it belongs to B because it is a convex combination of the vectors band b' of the convex set B. In other words,
u/{l + t) E Un B,
contradicting the relation (1.3)~ Thus, the relation (1.2) holds. Now, let W be an arhitrary neighborhood of zero in E. We construct a set V with the property (1.1). Let to be a positive number, which is smaller than 1 and such that the follo\ving relation holds,
toB S; W/2. (1.4)
Such a number exists because B is bounded. Further, we may assume that
U ~ W/2~ (1.5)
v = (t o /2)U, (1~6)
and verify that this neighborhood yields the relation (1.1). Indeed,let a = v + c, \vhcre v E V,c E C. Supposing that a rf. W we show that .
a¢V-C,
and by this the lemma will be proven. In virtue of the relations (1.5) and (1~6):
c ~ W/2. (1.7)
Since the set B is a base of C, hence so is toB. Moreover, it follows from (1.4) and (1.7) that there is a positive number t1 > 1 \vith c = t1tob, for some b E B. This together with (1~2) implies the fact that
c ~ toU - C~
In particular,
The proof is complete.•
Proposition 1.9 Suppose that C has a closed convex bounde.d base. Then for every bounded neighborhood V of zero, there is another one, say U, such that
x, y E en UC implies x + y E V e • (1~8)
ProoE Let B be a closed convex bounded base or-C and suppose to the contrary that (1~8) does not hold, i.e. there exists a bounded neighborhood ~ of zero such that for every neighborhood U of zero, one can find x, y E en UC with x +y E 1;;: Since B is bounded, we fix a neighborhood U which contains B and consider the family of neighborhoods {nU : n = 1, 2, .~.}. For every n > 0, there axe some :tn, Yn E C n (nU)C \vith X n + Yn E Vo~ As B is a baf-ie of C, there are some an, bn E B and positive numbers tn, Sn such that X n = tnan , Yn = snbn . It can be assumed that tn and Sn are greater than n. Indeed, since
xn/n E en uc,
50 that if xn/n = ta for some a E B, t ~ 0, then t must be greater than 1 because B ~ u. Hence ~n = nta where the quantity nt is greater that n. Further, the points (tna n + snbn)/(tn + sn) belong to V:/(tn + sn), n = 1,2, .1' • On the other hand, they belong to B due to the convexity of B I It is clear that
lim(tnan + snbn)/(tn + 511.) = O~
This and the closedness of B imply 0 E B, a contradiction~ •
7
Xext, let B* and E' denote the algebraic and topological dual spaces of E~
The algebraic and topological polar cones C"« and C' of C arc:
C* = {~ E E»- : ~(x) C 0, for all x E C},
C' = {~ E E' = {(x) ~ 0 , for all x E e}.
Denote also
c*+ = {~ E E* : ~(x) > 0, for all x E C \ l(C)},
G1+ = {~ E E' : ~(x) > 0, for all x E C \ l(C)}.
It should be noted that the first two cones are noncmpty convex, for instance they contain zero, but the last t,vo cones are not necessarily noncmpty.
In Example 1.2 (1), (R:t.)* = (R+)' = R+" In Example 1.2 (3), C" = en (21.
In Example 1.2 (4)(6), c*- = {O}, which shows that the cone is not strictly supported.
Belov" we provide a condition under ,vhich polar cones are nontriviaL
Proposition 1.10 (Pcrcssini-1967, Borwein-1980) In a vector space E, a convex set B is a base of C if and only if there exists a vector €E C*+ such that
B={CEC:~(c)=l}.
F'lLTthermore, if E is locally convex .separate.d and C has a convex weakly compact base, then C'+ is nonempty, and if in addii1.on E' is metrizable or barreled, then intC' is also nonempiy.
ProoE It is clear that the set B = {c E C : ~(c) = I} with ~ E C-+ is a convex base of the cone C. Conversely, for a given convex base B, consider the family of a1l1inear manifolds in E containing B but not zero. By Zorn's lernma there exists a maximal one which is a hyperplane, i.e. it is the set c;-l(l) for some ~ E EIt<. Since B ~ €-1(1), it follows that ~ E C*+ and indeed B = {c E C : ~(c) = 1}.
Further, if E is locally convex separated and B is a convex weakly compact base of C, then in view of a sepaxation theorem there is a vector €o E E' such that
~o(b) > 0, for all b E B. (1.9)
The relation above implies that ~o E G1+. If in addition, E' is metrizable or barreled, by Proposition 36.3 of Treves (1967), the topology of E' is the same as the Mackey's topology T(EJ
, E) and consequently (1.9) gives the relation: ~(b) > 0, for all b E B and for all ~ belonging to some neighborhood of (0 in E'. In other words, ~o E intC'.-
8
"'vVe recall that ~ E C' is an extreme vector if there are no two linearly inde­ pendent vectors ~1, ~2 E C' so that ~ = ~1 + ~2·
Proposition 1 .. 11 Assume that E is a separated locally convex space, C is a closed convex cone with C' having weakly compact convex base. Then x ¢ C if and only if there is an extreme vector ~ of C' such that ~(x) < o.
Proof. If ~(x) < 0, some ~ E C', then obviously x cannot belong to C. Conversely, if x ~ C, then one can separate{x} and C by a nonzero vector ~ E C' ,i.e. ~(x) < O. Let B' be a weakly compact convex base of C' . The inequality obtained above shows that
inf{~(x) : ~ E B' } < O.
Since B' is weakly compact convex, the function f(f,) = ~(x) attains its infimum at an extreme point of B' , which is also an extreme vector of C' .•
2.RECESSION CONES
Recession cones, or sometimes called asymptotic cones, were first introduced for convex sets (Steinitz-1913, 1914, 1916, Fenchel-1951, Dieudonne-1966 and Rockafcllar-1970), and then they were extended for arbitrary sets in infinite di­ mensional spaces (Dedicu-1978) .. Here we give a new definition of recession cones and provide several properties of these.
Let us add the point 00 to the space E which is, as in the previous section, a separated topological vector space ~ver rcals.
Definition 2.1 A nonempty set·V in E is said to be an open (resp., closed) neighborhood of 00 if it is open (resp.,closed) and its complement VC in E is a bounded set.
. In the sequel let B denote the filter of neighborhoods of 00.
Definition 2.2 11he recession cone of a.nonempty set X ~ E is the cone
Rec(X) = n{clcone(X nV) : VEE}.
In this definition we set
9
clcone(X n V) = {OJ
if the intersection X n V is empty~ If the space is normable, then there exists a bounded balanced absorbing neighborhood W of zero and in this case direct verification shows that
Rec(X) = n{clcone(X n (nW)C) : n =.1, 2, .~.}.
Relative to the set X, we also define two other cones:
X CXJ = n{cl(O, o:]X : Q > O}, where (0, l1']X = {tX ~ 0 < t ~ a}; ·
As(X) = {a E E : there axe a net {xoe } from X and a net {tal of positive numbers converging to 0 such that a = lim taxa}.
Remark 2.3 It is not diflictilt to see that in normablc spaces the cones X oo )
As(X) and Rec(X) coincide. In other cases, we have the relation:
X oo = As(X) ~ Rec(X).
In Dedieu (1978), the cone X~ is called the asymptotic cone of J\.
Lemma 2.4 A vector a E E belongs to Rec(X) if and only if for each V E 13 and each neighborhood U of zero in B,
cone(a + U) n X n V i- 0.
Proof. By definition, a E Rec(X) if and only if a E clcone(XnV) for each V E 13, which is equivalent to the relation:
(a + U) n cone(X nV) -# 0 or cquiva1ently~
cone(a + U) n (X nV) =F 0, for each neighborhood U of zero and each V E B.•
Proposition 295 The following properties are true:
1) Rec(X) = {OJ if X is bounded; 2) Rec(X) ~ Rec(Y) if X ~ y ~ E; 3) Rec(tX) = sign(t)Rec(X), each scalar t, where sign(t) is 1,0 and-1 if
t is positive, zero and negative, respectively; 4) Rec(X) = Rec(clX); 5) Rec(X) = clX if X is a cone; 6) Rec(X U Y) =Rec(X) U Rec(Y) for each X, Y ~ E; 7) Rec(X nY) ~ Rec(X) n Rec(Y) for each X, Y ~ E}~
8) conv(Rec(X)) ~ Rec(conv(X)).
Proof. Invoke to Definition 2.2 and Lemma 2.4.•
Remark 2.6 In the case ,vhere X and Yare convex closed with X nY being nonempty, the inclusion in 7) of Proposition 2.3 becomes the equality:
Rec(X n Y) = Rec(X) n Rec(Y).
However, this is not the case when the sets are arbitrary. The opposite inclusion of 8), in general, docs not hold even in finite dimensional spaces. Furthermore, it is not difficult to prove that a subset of a finite dimensional space is unbounded if and only if its recession cone is nontrivial. In infinite dimensions, the recession cone of an unbounded set is not necessarily nontrivial unless the set happens to have special structure at the infinity or to be convex.
Definition 2.7 We say that X satisfies the condition (CB) if there exists a neighborhood v;, of 00 such that the cone clcone(X n Vo ) has a compact base; and it satisfies the condition (CD) if for each a E R~c(X), there is a bounded set A ~ E S'l.Lch that (ta + A) n X is nonempty for all t ~ o.
Remark 2.8 Direct verification shows that the condition (OB) holds for every set in finite dimensional spaces. In infinite dimensional spaces if a set is convex, locally compact, then both of the conditions (CB) and (CD) are satisfied.
Lemma 2.9 Assume that X satisfies the condition (CB). Then X is bounded if and only if Rec(X) consists of the zero vector alone.
Proof. The "only if' part is the property 1) of J:>roposition 2.5. As for the converse assertion suppose that X is unbounded. Then for each neighborhood V E B, there is a point Xv E X n V. Without loss of generality we may assume that V ~ Vo , where Va is as in Definition 2.7. Let B be a compact base of the cone clcone(X n Vo)~ Then there exists a positive number t and a point bv E B such t~at Xv = tbv • Since B is compact, the set {bv : V E B} h&.~ at least one point of accumulation, say a E B. 11he vector a is nonzero since B does not contain zero and using Lemma 2.4 we can verify that a E Rec(X), completing the proof.-
Lemma 2.10 Assume that X is unbounded and it satisfies the condition (CB). Then for every filter U on X which is finer than the filter generated by the basis
{XnV:VEB,V~Vo},
where Va is defined by the condition (CB), there exists a nonzero vector v E Rec(X) such that v E c[cone(U), for every U E U.
11
Proof. Let B be a compact base of the cone clcone(X n Yo). Since for each clement U of the :filter U,
clcone(U) ~ clcone(X n Vo ),
B u = B n clcone(U)
is a compact base of clcone(U). Indeed, it is compact as the intersection of a compact set and a closed set; further, it does not contain zero since neither docs B. Moreover, for each x,
x E clcone(U) ~ clcone(X n Vo )
there are some b E B and a nonnegative number t so that x = tb. The vector b obviously belongs to B u , Le~ cone(Bu ) = clcone(U). The family {B-u : U E U} forms a basis of filter on B. Since B is compact, this family has at least onc accumulation point, say v. It is clear that v E B and v # o. lvlorcover,
v E clcone(U) for each U E U,
completing the proof.•
In vie\v of Remark 2.8, a useful case of Lemma 2.10 occurs v;hcn the set is closed convex, locally compact. The rcsult of this case was established by Dieudonnc (1966).
Lemma 2.11 If a set A ~ E is bounded, then
Rec(X + A) = Rec(X).
Proof. Invoke to Defmition 2.2 and Lemma 2.4.-
In the remainder of this section for the sake of simplicity we assume that E is a normable space.
Theorem 2 ..12 FOT every nonempty subsets X and Y of the space E, we have the following re lations:
1) Rec(X),Rec(Y) ~ Rec(X + Y); 2) Rec(X) + Rec(Y) ~ Rec(X +- Y) if the condition (C1) holds for at least
one of the two sets; 3) Rec(X +Y) ~ Rec(X) +Rec(Y} if the condition (CB) holds for at least
one of the two sets and if Rec(X) n -Rec(Y) = {O}.
Proof. For the first assertion, let x be a fixed point of X. Then
x+Y~X+Y.
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The inclusion:
is established by the same way~
For the second assertion~let a E Rec(X), b E Rec(Y). We have to prove the relation
a + b E Rec(X + Y)~ (2.1)
Suppose that X satisfies the condition (CD). By Remark 2~3, there are some Yec E Y, i Q > 0 ,vith lim ta = 0 and lim toyo = b. By the condition (CD), there exists a bounded net {ao } such that
X Q = (l/io.)a + aa EX.
I t is clear that
a + b = limta(xa + Yet),
and (2~1) is proven.
For the last assertion, suppose that X satisfies the condition (CB) and let a E Rec(X + Y) ~ i.e. a = lim to:(xo: + Ycx), for some Xo: E X, Ya. E Y and to > 0, limtQ = o. If the net {xa } is bounded, then
a = limto.Y", E Rec(X) + Rec(Y),
completing the proof. If not, since the filter generated by that net is finer tha:L the filter generated by the basis
{X n V : V E B, V ~ Vol, in virtue of Lemma 2.9, there exists a nonzero vector z E Rec(X) such that
z = lim A/3Xo:", \vhere A{3 > O,limA,B = 0 and {xO"IJ} is a subnet of {XCi}. Denoting
Pap = to.~ /)../3 ,
we may assume by taking a subnet if necessary, that limpCtp = Po , where Po may be infinite or a nonnegative number. If Po is finite, we have that
a = PaZ + limPap )..[jYo.~ ,
o= lima/pOop = limA.e(xQI3 + YQfJ)·
13
:
B = {(O~ 22n+1) E R2 : n = 0, 1~ IU}'
It is clear that the vectors (1,0) -and (0,1) belong to Rec(A) and Rec{B), respectively, nevertheless their sum does not belong to Rec(A +B). l"'hese sets do not satisfy the condition (CD).
Further, let x(n) be a sequence· in 0 1 (Example 1.2{4)) \vhose terrllS arc all zero except for the first and the n-th ones being n, while y(n) is a sequence whose terms are all zero except for the first one being n and the n-th one being -n. Set
A = {x(n) : n = 2,3, .. ~},
B = {y(n) : n = 2,3, ~ ..}.
Then Rec(A) = Rec(B) = {O} and meanwhile Rec(A + B) -I {O} , for instance it contains the sequence with the unique nonzero term being the first one. The two sets above do not satisfy the condition (CB).
Corollary 2.13 Assume that X and Yare nonempty with
Rec(X) n -Rec(Y) = {OJ, and one of them is convex, locally compact. Then
Rec(X + Y) = Rec(X) + Rec(Y).
Proof. If one of the two sets is convex~ locally compact, then it yields the condi­ tions (CB) and (CD). The corollary is then deduced from Theorem 2.12.•
3loCONE CLOSED SETS
As in the previous section, let C be a nonempty convex cone in a separated topological space E. We shall examine the sets which are closed or compact not in the usual sense, but with respect to the cone C.
Definition 3.1 Let X be a subset of E. We say that it is 1) C·-bounded if for each neighborhood U of zero in E, there is some positive
number t such that X ~ tU + C, 2) C -closed if X + clC is closed,
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3) C-compact if any cover of X of the f01m
{Ua +C : Q E I, Uo arc open} admits a finite subcover,
4) C-semicompact if any cover of X of the form
{(xo: - cIC)C : 0: E I, Xo. E X} admits a finite subcover.
The last definition was first given by Corley (1980)Jt is clear that whenever C = {O}, every C-notion becomes the corresponding ordinary one.
Lemma 3.2 If X ~ A +C for some bounded set A ~ E, then X is C-bounded. Conversely, if the space is normable and X is C-bounded, then there exists a bounded set A such that
X ~ A+C.
Proof. The proof is straighfo!'\vard. We omit it.•
Proposition 3.3 Every C-compact set is C-semicompact, C-closed and G'­ bounded.
Proof. Let X ~ E be a C-compact set~ That it is C-scmicompact follows from the fact that (x - clC)C is an open set and it is the same as (x - clC)C + C. We prove now that X is C-closed. For, let x be a clustcr point of the set X +clC.. We have to show that x E X +cIC. Suppose to the contrary that x does not belong to that set. Let U be the filter of neighborhoods of zero in E. Consider the family
G = {(cl(x - clC + U))C + C : U E U}.
It forms an open cover of X. Indeed, let y E X. Then y ¢ x - clC. lIenee, thcrc is a neighborhood V of zero such that
Y E (cl(x - ciG + V))C ~ (cl(x - clC + V))C +c. Since y is arbitrary, G is actually an open cover of JY. We show now that this cover has no finite subcovers. In fact, if that is not the case, Lc. thcre are some U1 , ... , Un from U such that the family
{(cl(x - cIG + Ui))C +C : i = 1, ... , n}
covers X, then taking
U = n{Ui : i = 1, ... , n} \ve have the incIusion
X ~ (x - etC + U)C + C
and arrive at the contradiction: x cannot be a cluster point of X + etC.
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For the C-boundedness) let U E U. Then the family {x + U + C : x E X} forms an open cover of X. There are a finite number of points, say Xi, ••• , X n of X such that the family {Xi + U + C : i = 1, ... , n} still covers X. Since the set {Xl' .~., x n } is bounded, there is a positive t such that Xi E tU, for i = 1, ... ,11.. By this \ve get the relation
X ~ U{Xi + U + C : i = 1, .. ~, n} ~ (t + l)U + C,
completing the proof.•
We must confess here that any compact set is C-compact \vhatever the convex cone C be, ho\vcver, a closed set is not necessarily C-closed unless C happens to be {OJ. ::vI:oreover, in finite dimensional spaces, not every C-closed, C'-bounded set is c- compact~ The unboundedness of C destroys the nice property of the usual compactncss~
Proposition 3.4 Let L be a linear map from E into another separated topological vector space. Then we have
1) L(X) is L(C)-convex if X is C-convex (i.e. X +C is convex)J 2) L(X) is L(C)-bounded if X is C-bounded.
Proof. This is immediate from the definition.•
Proposition 3 .. 5 Assume that !( is a convex cone contained in C. 'l'hen X is C-bounded if it is !(-·bounded. The conclusion remains valid if instead of bounded we write compact or semicompact.
Proof. If X is K-bounded, then for any neighborhood U of zero in E, there is a positive number t such that
X ~ tU + !(.
and by this, X is C-bounded.
1'ow,assume that X is K-compact, and let {Uce + C = a: E I} be a cover of X as in Definition 3.1, then {Ua + C + K : 0:' E I} is also a cover, here the sets Uo. +C are open. By the !(-compactness of X, there is a finite set [0 from I such that {Ua +C +!( : 0' E [o} still covers X. Observing that !( ~ C we see that the latter family is the same as {Ua +C ; Q' E [o} and X is C-·compact.
ltbr the C-semicompactness the proof is similar.-
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Proposition 3.6 If the space is normable and X is C -bounded, then
Rec(X) ~ eiC..
Proof. By Lemma 3.2, there exists a bounded set A ~ E such that ...Y £; A + C. Applying Proposition 2.5 we get the relation
Rec(X) ~ Rec(A + C) = Rec(C) = elG,
completing the proof.-
Theorem 3.7 Assume that X and Yare two nonempty C-closed sets in a normable space E and the following conditions hold,
i) X yields the condition (CD) and any bounded subset of X is relatively compact,
ii) Rec(X) n -Rec(Y +cIC) = {O}. Then X + Y is C -closed.
Proof. I.lct p be a cluster point of X + Y + cIG. 'fhe aim is to establish the relation
P EX +Y+clC.
For this purpose, let U be a neighborhood of zero in E. Let us consider the set
X V = X n (p - y - U - cIC). (3.1)
It is obvious that the family {Xu: U E U}, where U is the filter of neighborhoods of zero in E, forms a basis of filter on X. If one of Xu is bounded, then by condition i), that filter has an accumulation point, say X o• Since X is C:.... closed, X o E X + cIC. This and (3.1) sho\v that
X o E P - Y - 2U - ciC.
Since Y is C-closed and U is arbitrary, X o must be in p - Y - cleo Consequently, p belongs to X + Y + ciC. Further, if none of Xu is bounded, then in virtue of Lemma 2~lO there exists a nonzero vector v E Rec(X), such that
v E clcone(Xu nV), (3.2)
for every U E U and V E B, where B is the filter of neighborhoods of 00. By (3.1), relation (3.2) gives
v E clcone((p - Y - U - cIG) n V), for each V E B.
This and Proposition 2.5 show that
v E Rec(-Y - U - ciC), for every U.
In particular, when U is bounded, we get the relation
v E Rec(-Y - cIC),
contradicting condition ii) of the theorem. The proof is complete.a
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Remark 3.8 It is clear that the theorem above is not true if condition ii) docs not hold~ We furnish a simple example to show that the result of the theorem may fail if condition i) is violated. We are in a Hilbert space and let
X = {ei : i = 1,2, ...}
be an orthonormal set in the space. Further, let C = {O} and let
Y = {(-l + 1/2i)ei: i = 1,2, ...}.
The sets .)( and Y arc closed with
Rec(X) n -Rec(Y) = '{o}.
Theorem 3.9 Assume that
Rec(X) n -elC = {O} and either X or C satisfies condition i) of Theorem 3.7. Then X is C-closed if and only if X + ciA is closed for some subset A of C which contains the origin of the space.
Proof. Suppose that X is C~-closed. Take A to be C to get the closcdness of the set X +A.
Assume now that X + A is closed for some A ~ C with 0 E A and p is a cluster point of X + cleo The aim is to establish that
p EX +cIC.
Let W be a bounded neighborhood of zero in E. Supposing that X satisfies con­ dition i) of the preceding theorem, we consider the set
Xw = {x EX: (p+ W) n (x +cIC) =#= 0}~
It is obvious that the family {Xw : W E U} forms a basis of filter on X. If one of the sets of the basis is bounded, then that filter has an accumulation point,say x. Since X + A is closed and X ~ X + A, 'vc have that x E X + A~ This also shows that p - x is a cluster point of C~ hence it belongs to cIC. In this way p E X + ciC. Further, if none of the sets of the basis is bounded. Applying the technique developed in the proof of the previous theorem one can find a nonzero vector v fromRec(X)n-Rec(cIC)~ Hence, Rec(X)n-clC is nODzcro,contradicting the assumption. For the case where C yields condition i), instead of X \ve have to consider the set
Cw = {c E C : (p + W) n (X +c) f:. 0} and repeat the procedure described above for this set to ensure· the relation-
pEX+C.
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Remark 3.10 We give here t\VO e"xaroplcs to show that the assumptions of the above theorem cannot be v/eakened~ Let the space be as in Remark 3.8. Consider the following sets and cone:
X = {(l - 1/2n )el - nen : n = 1,2, UI},
A = {OJ, and C = cone(conv{en : n = 2,3, ~.~}).
The cone Rec(X) = {O} , the set X + A is closed, although the set X + C is not closed. Ko\v, if
x = {en: n = 1,2, UI},
A = cone(conv(X)),
C = cone(conv{(l/n)el - en = n = 1, 2, .. ~}),
then X + A is closed while X + C is not it.
4.CONE MONOTONIC FUNCTIONS
Let E 1 and E2 be two real topological vector spaces and let K and C be t\VO
convex cones in E 1 and E 2 , respectively. Let further f be a function from X s: E 1
to E2 • Denote the epigraph of f by epi f,i.e.
epi f = {(x~ Y) E E 1 X E2 : Y E f(x) + C, x EX},
and the level set of f at a point y E E2 by lev(y),Le.
lev(y) = {x E E1 : f(x) E y - C,x EX}.
Besides, we shall use also two other notations:
levo(Y) = {x E E1 : f(x) E y - c \ I(C)}, \vhcre l(C) = C n -~C, and
levl{Y) = {x EEl": f(x) E y - intC}
\vhen intC is noncmpty.
Definition 4.1 FOT a given junction !, we say that 1) it is nondecreasing (or monotonic) at X o E LY with respect to (1<, C) if
x E X n (xo - !() implies f(x) E f(x o ) - C;
2) it is increasing at x E X with respect to (K, C) if it is nondecreasing at that point and
19
x E X n (xo - I( \ l(I()) implies J(x) E f(x o ) - C \ l(C); Whenever intI( and intC are nonempty, we say that f is strictly increasing
at x 0 E X with respect to (!(, C) if it is nondecreasing with respeet to (!(, C) and increasing with respect to ({O} U intI(, {O} U intC).
Further, if f is nondecreasing (resp., increasing~ .. ) at every point of X with respect to (]{~ C), we say that it is nondecreasing (resp., increasing ~ ~.) on X with respect to (1(, C) or even say that it is nondecreasing if it is clear where and which cones it is with respect to.
In a special case where the spaces coincide with the field of real numbers, R) and the cones are the set of nonnegativcs numbers, R+, 've have everything in the usual sense, for instance, I is nondecreasing if f(x) 2: fey) for every x, y with x ~ y and f is increasing if f(x) > f(y) for every x > y. In this case, incrcasingness and strict increasingness are the same.
Proposition 4.2 We have the following 1) f is nondecreasing at x E X if and only if
X n (x - I() ~ lev(f(x)); (4~1)
2) f is increasing at x E X if and only if in addition to (4.1),
X n (x - !( \ I(I()) ~ levo(f(x)); (4~2)
3) f is strictly increasing at x E X if and only if in addition to (4.1)
X n (x - intI<) ~ levl (f(x))~ (4.3)
Proof. This is immediate from the definition.•
~ow,lct E3 be another real topological vector space and a convex cone D be given in E3 •
Proposition 4.3 Suppose that f and 9 are functions from X to E2 and h is a function from f(X) to E3 • Then
1) if is nondecreasing (resp~,increasingor strie:tly increasing) for each t > 0 if so is I;
2) f + 9 is nondecreasing if so are J and g; 3) f + 9 is increasing (resp.,strictly increasing) if they are nondecreasing
and at least one of them is increasing (resp.,strictly increasing); 4) h 0 f is nondecreasing (resp~,increasing or strictly increasing) if so are f
and h.
Proof. Assertions 1), 2) and 4) arc immediate from the defintion. For 3) it suffices to observe that by the convexity of C,
20
C + intC ~ intC~.
Proposition 4.4 Let T be a nonempty set and g(x, t) is a function from X x T to R, and let the cone C be R+~ Assume that the following conditions hold:
i) g(~, t) is nondecreasing (resp~,increasing or strictly increasing) on X for every fixed t E T;
ii) f(x) = max{g(x, t) : t E T} exists for every fixed x E- X. Then I(x) is a nondecreasing (resp.,increasing Qr strictly increasing) function on X"~
Proof. Let first x, y E X ,vith y E x - I(~ By condition ii), there are some t3;' t y E T so that
f(x) = g(x, t x ) and
f(y) = g(y, t y ).
In view of condition i) for the fixed t y ,
g(x, tv) c g(y, ty).
Consequently,
Below are some examples of monotonic functions
1. Positive linear operators:
Let L(E1 ) E2) be the space of linear operators from E 1 to E2 . An operator A E L(E1 , E 2 ) is said to be positive if A(I<) ~ C. It is clear that A is nondecreasing if and only if it is positive. If in addition
A(K \ 1(1<)) ~ C \ I(C) or
A(intK) ~ intC,
then it is increasing or strictly increasing and vice versa.
2 .. Positive linear functionals:
Assume that in the previous example E2 = R, C = R+ and we write E instead of E 1 • Then L(E, R) is the space of linear functionals on E called the algebraic dual of E which we have denoted in Section 1 by E>Ic . We recall that K* is the algebraic polar cone of K.
21
Proposition 4.5 FOT every junctional ~ E E* , we have that 1) ~ is nondecreasing if and only if ~ E J(*; 2) ~ is increasing if and only if ~ E [(*+; 3) ~ is strictly increasing if and only if ~ E K* \ {O} ~
Proof. This is immediate from the definition. •
3. The smallest strictly monotonic funcqons:
Assume that intK is noncmpty. 1."'he spaces and cones are as in the previous cxamplc~ Let e E intI( be a fixed vector and a E E, define a function he,a on E as follu\vs;
he,a(x) = min{t : x E a + te - !(, t E R}.
It is obvious that this function is strictly increasing on E. It is the smallest in the sense that if f is a strictly increasing function at a, then the level set of f at f(a) must contain that of he,a at O.
4. Cherbysbev norm:
Let the spaces and cones be as before.Assume further that 1< has a convex bounded base and intI( is nonempty. Let e E intI( be fixed. The ::vIinkowski functional corresponding to e is defined by
f(x) = inf{t : t > 0 and (l/t)x E (e - !() n (-e + !()}, for every x E E. One can verify direct that J gives a norm on E :
llxll = f(x) which is called a generalized Cherbyshev norm~ The word "generalized" falls down when E = Rn , !( = R+ and e is the vector with all components equal to 1. In this case
IIxU = max{lxil : i = 1, ... , n} where Xi are the ith components of x ..
Proposition 4.6 For every e E intI(, the Minkowski functional corresponding to e is strictly increasing on K and -1(.
Proof. Direct verification completes the proof.•
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5.CONE CONTINUOUS FUNCTIONS
In this section we give a definition of cone continuity of vector valued functions and using the concept of equiscmicontinuity of scalar valued functions we establish SOlue criteria for a function to be conc-~·continuous.Let E 1 and E2 be real normable spaces and a convex cone C be given in E2• l,ct f be a function from a nonempty set. X ~ E1 to E2.
Definition 5.1 For a given junction f from X to E 2 , we say that 1) J is C··continuous at Xo E X if for any neighborhood V of f(x o ) in E2
there is a ne ighborhood U of x 0 in }J1 such that
f(x) E V + C, for all x E un X, (561) and f is C -continuous on X if it is C -continuous at any point of X.
Further, assuming that C is closed we say that 2) f is epi-closed if epij is a closed set in the product space E 1 x E2; 3) f is level-closed if the level set of f at any point of E 2 is closed~
In the Iitcrature sometimes epi-closed functions are called closed and 1evcl­ closed ones are called C-·semicontinuous (Corlcy-1980). 'vVe recall that a scalar valued function h from X into R is lower semicontinuous at x E X if for each positive e , there is a neighborhood U of X o in E 1 such that
h(x) ~ h(xo ) - c, for all x E Un x. (5~2)
Whenever E2 = R and C = R+ (the cone of nonnegative numbercs), C -continuity is the same as lovler scmicontinuity. In this case three concepts: C-continuity, cpi­ closedness and level-closedncss coincide. In other cases they are different from each other as this will be seen later.
Definition 5.2 Let {h(x, t) : t E T} be a family of scalar-valued functions on X, where T is a nonempty parameter set. We say that this family is lower equisemicontinuous at X o E X if for every c > 0, there is a neighborhood U of Xo
in E 1 such that
h(x,"t) ~ h(xo , t) - c, for all x E U n X and t E T. (5.3)
Theorem 5.3 Assume that C has a closed convex bounded base. Then in order that f be continuous it is necessary and sufficient that it be C - and (- C)­ cantinuous simultaneously.
23
ProoE It is obvious that if f is continuous,Lc {O}-continuous, then it is D~­
continuous for any cone D in E2 • Suppose now that f is C- and (-C)-continuous at a point X o E X and let W be a neighborhood of f(x o ) in £2- \"Ve have to show that there is a neighborhood U of $0 in E 1 such that
f(x) E W, for all x E X n U. (5.4)
For the neighborhood W, due to Proposition 1.7, one can find a neighborhood V of zero in E2 such that (1.1) holds. By the assumption of the theorem, for V, there arc two neighborhoods U1 and U2 of X o i'Q E1 such that
f(x) E V + C, for x E U1 n X and
f(x) E V - C, for x E U2 n X.
This and (1.1) imply (5.4) for U = U1 n U2 .•
Remark 5.4 If the cone C is merely convex closed pointed, then the result above is not always true. To see this, let us consider the following sets and functions: the space is as in Remark 3.8,
C = cone(conv{ei, bi : i = 1,2, ...}),
where bi = (1/2 i - 1)el - ei, X = [0,1]
and the function f is given by the rule:
f(O) = 0,
f(t) = 2(1 - 2i )bi+1 + (2 i+1t - l)bi ,
for t, 1/2i+1 ~ t ~ 1/2i , i = 0,1, ... with bo = O.
It is clear that f is C- and (-C)-continuous at 0, but not continuous thcrc~ The cone C also serVes an example clarifying Remark 1.5.
For the sake of simplicity we assume that it is given a norm 11.11 in E2 and the norm in the topological dual space is denoted by the same~
Theorem 5.5 f is C-continuoU8 at a point X o E X if and only if the family G = {~o f : ~ E C', Ilell = I} is lower equisemicontinuous at that point.
Proof. Assume first that f is C-continuous at X o E X~ Then for every c > 0) there is a neighborhood U of :to in E 1 so that
f(x) E f(x o ) + B(O, e) + C, for every x E Un X,
where B(O,e) is the ball of center 0 and radius c in E2 ~ Let ~ be a unit normed vector from C'. We have that
~ 0 f(x) ~ ~ 0 f(x o ) +inf{~(y) : y E B(O, c:)} == ~ 0 f(x o ) - £~
24
This relation sho'vs the lo\ver equisernicontinuity of the family G at X O • As for the converse assertion of the theorem, suppose that f is not C··continuous at X o E X, i.e. there exist () > 0 and a net {xo; = Q' E I} from X ,vith limxo := X o such that
f(x a ) ¢ f(x o ) +B(O, 28) +C, for all G E [.
Since the set cl(f(xo)+B(O~8)+C) is convex closed, applying a separation theorem, one can find some ~Q from the topological dual of E1 with unit norm such that
~a(f(x~) ::; ~a(Y), (5.5) for all y E f(x o ) + B(O, 8) + C. It follo"\vs from (5.5) that ~a E C' and
~Q(f(xQ) ~ ~Ct(f(xo))+inf{~(y) : y E B(O, 8)} = ~~(f(xo)) - o. In this "\\o"ay G docs not yield (5.3) for e = 8/2, completing the proof. -
Corollary 5.6 S'llppose that C' is a polyhedral cone. Then f is C··continuous at X o E X if and only if every function of G is lower semicontinuous at that point.
Proof. Assume that
0' = cone(conv{~i: i = 1, ... ,n}).
It can easily be proven that G is lower cquisemicontinuous if and only if the family {~i 0 f : i = 1, ... , n} is it. But the latter family is finite and it is lower equisemicontinuous if and only if every element of it is lower semicontinuous. II
Remark 5.7 The fact that the result of the corollary above may fail when C is not polyhedral is shown by the following example in R3 . Let
ai = (1,1 - 1/22i , 1/2 i ) E R3) i = 0,1) ... ,
C = cone(conv({ai : i = 0,1, ...} U (1,0,0) U (1,1,0»)). Denote
Pi =(1, -1/(1 + 1/22i+1), -3/(2i+1 + 1/2i )).
Then it can be verified that the cone C consists of the vectors a E 1~3 \vhich solve the following system:
(a~Pi) ;::: 0, i = 0,1, u.
(a, (0,0, 1)) ~ 0,
{a, (0, 1,0)) 2: 0,
\vhere (.) is the inner prodliet. Further,denote bi = 2iai -.. Pi. We construct a function f from [0,1] into R3
as follows:
1(0) = 0,
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The cone C' is obviously generated by the set
G = {Pi: i = 0, 1, ..~} U (0,0,1) U (0,1,0).
It is clear that f is not C-continuous at O. Nevertheless we shall prove that any composition ~ 0 f, € E C' , is lower semicontinuous at o. Indeed, it suffices to verify that fact for the vectors from G. Direct computation shows that
{bi, (0,0, 1)) = 1+ 1/(2i+1 + 1/2i );
(bi, (0, 1,0)) =2i (1 - 1/22i ) + 1/(1 + 1/22i+1
),
(bi,pj) = 2i (ailPj) - (Pi,Pj) ~ 2i /(1 + 22i+1 ) - 14.
lIenee, when j is fixed and i is large enough, we have
(bi,pj) ~ O. (5.8) Cornbining (5~6), (5.7) with (5.8) and taking (5.5) into account we obtain that for each ~ E G, if t is sufficiently small, then (,(f(t)) 2: ~(f(O)), i.e. ~ 0 f is lo,vcr scmicontinuous at O.
Theorem 5.8 Every epi-closed function is level-closed. Conversely, if intC is nonemptYJ then every level-closed function is epi~~closed.
Proof. Suppose that epif is closed and let x be a cluster point of L(y), for some y E E. We have to show that x E L(Y)6 If that is not the c&Clc,Lc. f(x) <t y - C, then by the closedness of the epigraph, there is a neighborhood (U, V) of (x, y) in E 1 x E2 such that
(U, V) n epif = 0. In particular,
(U, y) n epif = 0, for every x E E.
In other ,vords,
Un L(y) = 0, that is, x cannot be a cluster point of L(y).
Assuming that iniC is nonempty, we no\v demonstrate the converse assertion. For, suppose that L(y) is closed for each y E E2 • vVe have to prove the closedness of epif. Let (x,y) E E 1 X E2 and (x,y) ¢ epif. The latter relation means that y rt f (x) +C. Since C is closed, there is a neighborhood of zero , say W, such that
(y + W) n (f(x) + C) = 0. Taking a vector e E W n intC, we get
y t/. f(x) - e + C,
which means that x tt L(y + e). The set L(y + e) is closed, hence there is a neighborhood U of x in E 1 such that
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y ¢ f(U) - e + C. (5.9)
J:urthcr, as the vector e belongs to intC) there exists a neighborhood V of zero in E2 such that
e- V ~ C. (5.10)
\Ve are going to establish the relation
(U, y + V) n epif = 0, and by this the theorem will be proven. If that relation is not true,Le. for some x' E U and v E V, (x', y + v) E epif, then
f(x') E y + v-C.
In virtue of (5.10),
contradicting (5~g).1"he proof is complete.•
The following simple example shows the need for the condition iniC t= 0 in Theorem 5.8 .. Let f be a function from R to R2 defined by the relations:
f(x) = (1,0), for x $ 0, f(x) = (0,1), for x > 0;
C = {(t, 0) E R2 : t ~ OJ.
Then f is level-closed, although epif is not closed.
Theorem 5.9 Assume that ~ 0 f is lower semicontinuous for each ~ E C' . 7'hen f is epi-closed~
Proof. We suppose to the contrary that f is not epi-closcd, i.e. there is a cluster point (xo,Yo) of epif for which Yo rt f(xo) + C. Let t be a positive number with the property:
(Yo -~ B(O, t)) n (f(xo) + C) = 0~
Separating these convex sets by a unit normed vector ~ E E 1 , we obtain
(c;, f(xo ) + c) ~ (~, V), for all c E C and y E Yo + B(G, t)~
It is obvious that ~ E C and
(~, f(x o )) ~ ({, y) + (~, V'), for all y E Yo +B(O, t/2) and Y' E B(O, t/2). Consequently,
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(f" f(x o )) ~ (~, y) + sup{ {c;, y'} : y' E B(O, l/2)}
~ (~, y) +t/2, for all y E Yo +B(O, t/2). Remembering that (xo' Yo) is a cluster point of epij \vhich means that there are points x as closed to X o as we want so that f(x) E yo+B(o, t/2), we conclude that the function ~ 0 f cannot be lower scrnicontinuous at X O ' •
Corollary 5.10 Every C-continuOU5 fun~tion is epi-closed, hence level-closed.
Proof. Invoke the corollary to 'I'heorems 5.8,5.9 and Theorem 5.12 below.•
We now study the compositions of C-continuous functions. Let 9 be a function from E2 to a normed space E3 and let D be a convex cone in E3 I
Theorem 5.11 Assume that X is a subset of E 1 with at least one accumulation point, say XO ' The composition 9 0 f is D-continuous at Xo for eve1~ function f from X to E 2 , being C-continuous at Xo if and only if g is D-contin1l0Us and nondecreasing on E 2 •
Proof. Suppose that 9 is D-continuous and nondecreasing on E 2 and f is C­ continuous at XOI Let W be a neighborhood of g<f(xo ) in Ea. By the D-continuity of g, one can find a neighborhood V of f(x o ) in Ez such that
g(y) E W +D, for all y E V. (5.11) By the Ch·continuity of !, there is a neighborhood U of X o in E 1 such that
j(x) E V + C, for all x E U nx~ (5.12)
Since g is nondecrcasing, (5.11) and (5.12) imply the relation
g(f(x)) E g(V + C) ~ W + D, for all x E Un x. Thus, go f is D-continuous at xo~
Conversely, suppose first that 9 is not nondecreasing on E2 , i. eO. there are a point Yo E E 2 and a nonzero vector c E C such that
g(y + c) 95 g(y) + D.
Since the set g(y) +D is closed, there exists a neighborhood ltV of zero in Ea such that
(g(y + c) +W) n (g(y) +D) = 0. (5.13) We construct a fvnction f from X into E2 as follows:
/(xo ) = y,
j(x) = y +c, for all x E X, x # X o '
It is clear that f is C-continuous at X o• However, 9 0 f is not D-continuous at that point as (5.13) sho\vs.
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Further, suppose that 9 is not D· 'continuous, say at Yo E E, , Le. there are a neighborhood W of g(yo) in E3 and a sequence {Yn} from E2 ,vith limYn = Yo such that
g(Yn) fJ. W +D, for all n = 1,2, ... . (5.14)
Since X o is an accumulation point of X, there is a scquence {xn }, X n i: xo , from X \vhich converges to X o • The aim at the moment is to construct a continuous function f from X to E2 for which go f is not D-continuous at X o • Without loss of generality, we may assume that {llYn - yoll} is decreasing ,vith IIYl - Yo!1 = 1. First we construct a function 11 from {xn } to the interval [0,1] by the formula:
fl(xn) = nYn - Yolr·
This function is continuous on the closed subset of the space E1 • Apply Tietze extension theorem (Kurato,vski-1972) to get a continuous function /2 from E1 to [0, I). Now we construct a continuous function is from [0, 1J to E2 as follows. First we note that for every t from that interval, there exists exactly one n such that
r1Yn-l - Yo II > t 2:: llYn - yoll· In other words,
t = SUYn-l - YoH + (1 - s)IIYn - YoU, for some $,0 < S ::; 1. We set
f3( t) = SYn-l + (1 - S)Yn. It is clear that fa is continuous on [0) 1]. Hence, the composition f = fa 0 12 is continuous on E1 • Moreover,
f(xn) = Yn, n = 1,2, ....
The composition 9 0 f is obviously not D-continuous at Xo••
Theorem 5.12 Assume that the cone D does not coinside with the whole space. The.n 9 0 f is D-continuous at a point X o E X for every function 9 being D­ continuous at f (x0) if and only if f is cantinuous at x0.
Proof. Supposc that f is continuous at xo~ Then it is {O}-continuous at that point. Since any function from E2 . to E3 is nonincreasing with respect to the cones {O} ~ E2 and D, applying Theorem 5.11 ,ve get the D-continuity of 9 0 f for each 9 which is D-continuous at f(x o). Suppose now that f is not continuous at the point X O • Then there are a neighborhood V of f(x o) and a sequence {xn }
from X with limxn = X o such that f(xn ) E V, for each n. Define a function 9 from E2 to E3 as follows~ Let v be a nonzero vector of E3 which docs not belong to D, and let t be a positive number with the property:
B(O, t) + f(x o ) C V.
Set
g(y) = (Jly - f(xo)lI/t)v, for every y E E2 6
It is obvious that 9 is continuous, hence D-continuous. Despite of this, go! is not D-continuous at X O ' The theorem is proven.•
6.CONE CONVEX FUNCTIONS
In this section E 1 and E2 are real topological vector spaces, X is a nonempty convex set in E 1 and in E2 it is given a convex cone C.
Definition 6.1 Let f be a function from X to E2 • We say that 1) f is C-·convex on X if for Xl,X2 EX, t E [0,1],
f(tXl + (1 - t)X2) E tf(Xl) + (1 - t)f(X2J - C; (6.1) 2) f is strictly C-·convex on X, when intC is nonempty, if
fOT Xl, X2 E X, Xl i=- X2, t E (0,1), i.e. 0 < t < 1,
f(txl + (1 - t)X2) E tf(Xl) + (1 - t)f(X2) - intC; 3) f is C-·quasiconvex on X if for y E E 2 ? Xl, X2 E J"l() t E [0, 1],
f(Xl),!(X2) E y - C implies f(tXl + (1 - t)X2) E Y - C~
4) f is strictly C-guasiconvex, when intC is nonempty, if for y E E2 , Xl, X2 E E3 , Xl ~ X2, t E (0,1),
f(Xl), f(X2) E y - C implies !(tXt + (1- t)X2) E Y - intC.
In a particular case \vhere E2 = R, C = R+, we get the definition of convex and quasiconvex functions in the usual sense.
Here are some simple properties of C-conve.."{ and C-quasiconvex functions.
Proposition 6.2 f is q-convex if and only if epif is a convea; set. Moreover, if E 2 is separated and C is closed, then f is C-convex if and only if ~ 0 f is convex for every €E C/.
PrOOL The first part of the proposition is immediate from the definition.
For the second part, suppose that f is C-convcx. Then the relation 1) of Definition 6.1 holds. We already know that any functional ~ E C' is nondccrcasing and linear, therefore applying ~ to the relation of 1), \ve obtain
~f(txl + (1 - t)X2) $ t(f(Xl) + (1 - t)~f(X2)'
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\vhich sho\vs that ~ 0 f is convex as a scalar valued function.
Conversely, if relation (6.1) docs not hold for some Xl, X2 E X, t E [0,1]. By a separation theorem, there is a functional ~ E E' separating the point f(tXl + (1 - t)X2) and the convex closed set in the right hand side of (6.1). It is clear that {E C' and
€f(txl + (1 - t)X2) > t~f(Xl) + (1 - t)~f(X2),
which sho'vs that ~ 0 f is not convex, completing the proof. •
Proposition 6.3 We have the following: 1) f is C-quasiconvex if and only if lev(y) is convex for each y E E2 ;
2) f is C -quasiconvex if and only if he,a () f is quasiconvex for every a. E E2 and a fixed e E intC, where he,a is the smallest strictly monotonic function at the point a (Sec.4), whene'uer iniC is nonempty.
Proof. The first assertion is obvious. For the second OIle, suppose that f is not C-quan."iiconvcx,Le. the relation in 3) of Definition 6.1 docs not hold. Take a = y and consider the function hc,y. By the definition of this function, we have
he,y(f(Xl)) ~ 0 and
which shows that hc~y 0 f is not quasiconvex.
1'hc converse assertion follows from Proposition 6.8 below.•
Proposition 6~4 Assume that E 2 is locally convex separated and C t has a weakly compact convex base. If ~ 0 f is quasiconvex"!oT every extreme vector ~ of C' , then f is C-quasicon·vex.
Proof. Suppose ~o the contrary that f is not C-quasiconvex which means that the relation in 3) of Definition 6.1 docs not hold. In virtue of Proposition 1.10, there is an extreme vector ~ E C' such that
~(f(txl + (1 - t)X2) - y) > o. This and the fact
c;(f(x2) - y) ~ 0, i = 1,2
sho\v that ~ 0 f is not quasiconvex, completing the proof. II
Proposition 6.5 Assume that E2 = RR and C is a polyhedral cone generated by n linearly independent 'Vectors. Then f is C-quasiconvex if and only if ~ 0 f is quasiconvex for every extreme vector ~ of ct.
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Proof. By Proposition 6.4, it suffices to prove that the C-quasiconvexity of f implies the quasiconvexity of ~ 0 f for every extreme vector ~ of C'. First 1,ve note that if aI, ... , an generate C, then C' is generated by the nonzero vectors b1 , ••• , bn
\vhich are the only extreme vectors of C' and defined by (bi,aj) = 0, i =f:. j, (6.2)
(bi, ai) = 1.
Suppose that ~ 0 f is not quasiconvcx for say ~ = b1 • Then there exist some Xl,
X2 E X, t E (0,1) such that ~f(tXl + (1 - t)X2) > max{~J(xl); ~f(X2)},
\vIDch means that ~ strictly separates f(tXl +(1- t)X2) and f(Xl) U f(X2). Assume that
~f(Xl) ~ ~f(X2).
Consider the hyperplane H generated by ~ and passing through f(Xl)' By (6.2), \ve have that
H = leXt) + lin{a2' ... , an}, where lin denotes the linear subspace stretched on a2, ... , an' Consequently)
(/(x!) + C) nH = {!(X2) +Zin{al}} nH
+cone(conv{a2' .~., an})~ (6.3) Let c be the point which yields the relation
(J(Xl) + C) n (f(X2) + C) = c + C (such a point exists because C is generated by n linearly independent vectors, see the Choquet-Kendall rfheorem in Pcrcssini-1967). We prove that c E H. Indeed, consider the (n - I)-space H - !(Xl) and the cone
Co = cone(conv{a2, ... ,an})
in it. One can easily verify that there is a unique point Co such that
Co n {{f(X2) + lin(al)} n H + Co} ~ Co + Co.
It follows from (6.3) and from the definition of Co that
c = Co + f(xl)~
j(Xl), /(:£2) E C- C, meanwhile
f(tXl + (1 - t)X2) (j. H - C.
In this 1,vay, f is not C-quasiconvex.•
Corollary 6.6 Under the assumptions of the previous proposition, if in addition C == R+ ' then f is C···quasiconvex if and only if every component junction of f is quasiconvex. .
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Proof. This follows from Proposition 6.5 and the fact that the polar cone of C is itself...
Proposition 6.7 Let f and 9 be two functions from X to E 2• 11hen 1) if is C -~convex (resp.,strictly C··convex,~~.) for each t > 0 if so is ji 2) f +9 is C-convex (resp.,C-quasiconvex) if 80 are f and g; 3) f +9 is strictly C -convex (resp., strictly C -~quasiconvex) if they are G­
convex and at leas t one of them is strictly C -con1)ex (resp., strictly C­ quasiconvex).
Proof. This is immediate from the definition.•
Proposition 6.8 Let f be a junction from X to E 2 and 9 be a, junction from Ez to another space E:~ in which a convex cone D is given. Then
1) go f is D-convex if f is C-convex, g is D-convex and nondecreasing; 2) 9 0 f is strictly D-convcx if f is strictly C-convex, 9 is D-~convex and
strictly increasing j
3) g 0 f is D-quasiconvex if f is C-convex and if g is D-~quasiconvex and nondecreasing,.
4) go f is strictly D-~quasiconvex if f is strictly C-convex) and if g is D­ quasiconvex and strictly increasing.
Proof. We prove l)w Let Xl, X2 E X, and 0 < t < 1. Since f is C·-convex~
f(tXl + (1 - t)X2) E tf(Xl) + (1 :- t)J(X2) - c. rrhis combines with the nondecreasingncss of 9 to yield the relation:
go J(tXl + (1 - t)X2) E g(tf(Xl) + (1 - t)f(X2)) - Dw (6.4) By the D-convexity of 9 we have that
g(tf(Xl) + (1 - t)f(X2)) E tg 0 J(Xl) +(1 - t)g 0 !(X2) - D.
The latter relation and (6.4) sho\v that 9 0 f is D-convex.
Other parts of the proposition are proven similarly.•
Remark 6.9 In the proposition above, in every case f must be C-convex or strictly C-convex. If it is merely C·-quasiconvex or strictly C-quasiconvcx, then the last two assertions may fail. Below we give an example where" f is C­ quasiconvex, 9 is D-convcx and increasing, but go f is not D-convex.
Let X = [-1, 1], E2 = R2, C = R~ ~ Es = R, D = Rt; Let f be defined by
f(x) = (-x,O), if x E 10, 1J f(x) = (O,x), if X E [-1,0],
33
and let 9 be defined by g((x, y)) = x + y, for (x, y) E R2~
It is easy to verify that f is C-quasiconvcx, 9 is linear increasing, although the composition go f(x) = -Ix] is not quasiconvex.
7.SET-VALUED MAPS
Suppose that E 1 and E2 are t\VO real topological vector spaces and it is given a convex cone C in E2 • Let }' be a set-valued map from E1 to E2 v{hich means that }?(x) is a set in E2 for each x EEl. The following notations will be used for set~· valued maps:
domF = {x E E1 : F(x) f 0} grafF = {(x,y) EEl X E 2 : y E F(x),x E domF}
epiF = {(x, y) E El X E2 : y E F(x) + C, ~ E domF}
Definition 7.1 Let X be a subset of domF. We say that 1) F is upper C-continuous at X o E X if for each neighborhood V of F(xo )
in E2 ) there is a neighborhood U of X o in E 1 such that
F(x) ~ V + C, for all x E Un dam}';
2) F is lower C-continuous at X o E X if for any y E F(xo ), any neighbor· hood V of y in E z, there is a neighborhood U of X o in E 1 such that
F(x) n (V + C) -:f 0, for each x E Un domF;
3) F is C-continuous at X o if it is upper and lower C-continuous at that . point; and F is upper (resp.,lower, ...) C-continuOU8 on X if it is upper
(resp.,lower, ...) C-continuous at every point of X; 4) F is C-closed if epiF is closed; 5) whenever "N" denotes some property of sets in E 2 , we say that F is
'tN'~ ··-valued on X if F(x) has the property "N", for every x E x.
In the above definition, setting C = {O} we get the definitions in the usual sense which we meet in the literature with adding "semi" to "continuous'~. Some­ times we say simply upper continuous instead of upper {OJ-continuous. There are a lot of books dealing with set-valued maps ( see for instance Aubin and Ekeland-1984; Berge-1962). vVe develop here only what we need in the chapters to come.
34
Theorem 7.2 Assume that X is a compact set in E 1 and F is an L'lVn-valued, upper C-coniinuous map from E: to E 2 with X ~ domF, where "lV)' may be C­ closed, G-bounded, C-compact or C-semicompact. Then F(X) has the property "lV" in E2 .
f)roof. First let ,roN" be C-closcd and let {aa = (} E I} be a net from F(X) + etC 1,vith limaQ = Q. 'ATe have to prove that there is some x E X such that
a E F(x) + clG.
Let Xo: E X, Yet E F(xa ) and Co: E clC be such that ao: = YQ +co. We may assume that
limxa = x EX.
~br any neighborhood V of }'(x) in E2 , there is some f3 E I such that
F(xaJ S;; V + cIC, for all a ~ {3.
In particular,
aa E V + clC, for all Q ~ (3.
Since V is arbitrary and F' is C-closcd-valued, we conclude that a E F(x) + cleo
Now,lct "lV" be C~·boundcd and let V be an arbitrary neighborhood of zero in E2. "'rVe have to show that there is some i > 0 such that
F(X) ~ tV +C.
'1.'0 this end, for every x E X, consider the set
U(x) = {y EX: F(y) ~ F(x) + V + C} which is open in X due to the upper C-continuity of F. By the compactness of X, there are a finite number of points from X, say Xl, ... , X n , such that {U(Xi) : i = 1, ... ,n} coversX. Thus,
F(X) ~ U{F(Xi) :i= 1, ...~n}+nV:tC. Remember that F is C-bounded-valued, which means that there are some ii ~ a so that
F(Xi) ~ tiV + C.
Take t = n + t 1 + -.. + "tn to get the inclusion F(X) ~ tV +G.
furthcr,lct "N': be C-compact and suppose that {Va + C : a E I} where Vo; are open is a cover of F(X). We have to draw a finite subcover from that cover. For x E X, denote by I(x) a fInite index set from I which exists by the C···compactness of F(x) such that {Va + C : a E I(x)} covers F(x). Again, the set
U(x) = {y E X = F(x) ~ U{Vc:e : a E I(x)} + C} is open and we can obtain a finite cover of X, say {U(Xj) : i = 1, ... , n}. Then the family {Va + C : a E I(Xl) U ~ .. U I(xn)} forms a finite subcover of F(X).
35
{(a Q - clC)C : a E I, a~ E F(X)}
be a cover of F(X). For each x E X, since F(x) is C-scmicompact, it can be seen that there is a finite index set I(x) ~ I such that {(aQ - cIC)C : 0' E I(x)} covers F(x), where a el may lie outside of F(x). Further, it is obvious that
(a - cIC)C + C = (a - clC)C, for each a E E2 ,
therefore the set
U(x) = {y EX: F(y}S; U{(aQ - cIC)C : a E lex)}} is open in X. No,v the argument of the previous part can be applied without any change.•
Definition 7.3 (Penot-1984) F is said to be compact at x E damEl if any net {(xQ' Va)} from grafF possesses a convergent subnet with the limit belonging to grafF as soon as {xo:} converges to x.
Whenever this is true for each x E X s;; damF., we say that F is compact on the set X.
Definition 7.4 Let now E1 and E 2 be metric spaces and let F be a compact­ valued on E 1 • We say that F is Lipschitz at x E domF if there is a neighborhood U of x in E 1 and a positive number k, called a Lipschitz constant, such that
h(F(x), F(y)) ~ kd(x, y), for each y E U,
where d(.,.) is the metric in E 1 , and h(.,.) is the Haus40rff distance between two compact sets in E2 •
Proposition 7.5 If F is Lipschitz at x E damP, then it is compact and contin­ uous at that point
Proof. The continuity of the map at x is obvious. We prove that the map is compact .. For, let {(xo:' Yo.)} be a net from grafF \vith {xa:} converging to x E domP. Consider the net d(yo.' F(x)) of real numbers. Since F is Lipschitz at the point x,
limd(Ya, F(x)) = O.
By the compactness of F(x), there is a net {za:} from F(x) such that
d(yo., F(x») = d(yco zoJ and this net may be assumed to converge to some z E F(x). VVc have then,
d(Ya., z) ~ d(Ya' zce) + d(zce, z) and {Yell converges to z, completing the proof.•
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No,,, suppose that E a is a real topological vector space, G is a set-valued map from E2 to E3 . The composition G 0 F is the map from E 1 to E2 defined as follows:
(G 0 F) (x) = U{G(y) : y E F(x)}. Further, for t,vo set-valued maps F1 , F2 from E 1 to E2, their sum and the multiplication with a scalar t are defined by the rules:
(FI + F2 )(x) = F1(x) + F2 (x); (iF} )(x) = tF1(x), for each x E E 1 •
Proposition 7~6 Suppose that the maps F, F I and F2 are compact at x E E 1 and G is compact on F(x). Then the map tF, PI + F2 and Go F are compact at the point x.
Proof. This is immediate from the definitions ~ •
Chapter 2
Efficient Points and Vector Optimization Problems
rrhis chapter is devoted to the basic concepts of efficiency in topological vec­ tor spaces. We deal with partial orders generated by convex cones over all, after having introduced them in Section 1. l'he next three sections contain the def­ initions, properties and· existence conditions of efficient points. In Section 5 \ve define vector optimization problems, their solution concepts and investigate the existence of optimal solutions.
1.BINARY RELATIONS AND PARTIAL ORDERS
Given an arbitrary set E, a binary relation on E is, by definition, a subset B of the product set E X E. This means that an element x E E is in relation ,vith Y E E, if (x,.y) E B.
Definition 1.1 Let B be a binary relation on E. We say that it is 1) reflexive if (x, x) E B for every x E E ; otherwise it is irreftexive 2) symmetric if (x, y) E B implies (y, x) E B for each x, y E E, otherwise it
is asymmetric; 3) transitive if (x, y) E B, (y, z) E B imply (x, z) E B for every x, y, z E B,
oiherwise it is nontransitive;
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4) complete or connected if (x, y) E B or (y, x) E B for each x, y E E,
x:f:. Yi 5) linear in the case where E is a real vector space, if (x, y) E B implies
that (tx +z, ty +z) E B for every x, y, z E E, t > 0, 6) closed in the case where E is a topological vector space, if it is clo.ged as
a subset of the product space E x E.
To clarify this definition, let us consider the following classical example: let E be a community of inhabitants of a city and we define binary relations as follo\vs.The inhabitants are named by x, y, z, ....
1) (x, y) E B 1 if x is older or as aged as y. 2) (x, y) E B2 if x and yare of different sex. 3) (x,y) E B 3 if x and yare relatives (they come from one family tree). 4) (x, y) E B 4 jf x and y are relatives of somebody 5) (x, Y) E Bs if x and y axe more weighted than any cocitizcns.
It can be seen that B 1 is re.flexive, transitive, asymmetric, complete; B 2 is ir­ re.flexive~ symmetric, nontransitivc, noncomplete ; B3 is reflexive,symmetric, non­ transitive, noncomplcte; B4 is refLexive,symmetric,transitivc)complctc, while Bs is irreflexive and noncomplete.Thc t\VO last relations are extreme cases:B4 = E X E and Bs = 0.
Definition 1.2 A binary relation is said to be a partial order if it is reflexive, transitive.
It is known that if B is a partial order which is linear in a vector space, then the set
c = {x E E : (x) 0) E B}
is a convex cone. If in addition B is asymmetric, then C is pointed. Conversely, every convex cone C in E gives a binary relation
Be = {(x, y) E E x E = x - y E C} which is reflexive, transitive and linear. If in addition C is pointed, then Be is asymmetric.
From now on we shall consider only orders generated by convex cones. We wri te sometimes .
x?:.cY instead of x - y E C,
or simply x ~ y if it is clear that the binary relation is defined by C;
x >c Y if x~cy·and not Y?cx, in other words, x E y + C \ I(C);
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When intC is nonempty,
x>Cy means that x >I( Y ,vith ]( = {O} U intC.
Here are some examples:
1.Let us be in Rn and let C = Ri.o Then Be is reflexive transitive linear closed asymmetric but not complete. For x = (Xl' ~.~, Xn), Y = (Yl' ... , y,.J E Rn :
x?acY if and only if Xi.~ Yi for i = 1, ... , n;
x >e Y if and only if Xi ~ Yi, for i = 1, ... , n and at least «;>uc of the inequalities is strict;
x >c Y if and only if Xi > Yi for all i = 1, ... , n.
2. In R2 , if C = (Rl, 0) , then Be is reflexive transitive linear closed and
syrnmetric. In this case, x~cY if and only if the second components of these vectors coincide. The order is not complete.
3. The ubiquitous cone (Example 1.2(4))Chaptcr 1) gives a reflexive transitive linear, but not complete relation in 0 1,
4. The lexicographic cone (Example 1.2(5),Chapter 1) provides a reflexive transi tive linear complete relation in l'P.
2.EFFICIEKT POINTS
Let E be a real topological vector space with partial order (~) generated by a convex cone C.
Definition 2.1 Let A be a nonempty subset of E. We say that 1) x E A is an ideal efficient (or ideal minimal) point of A with respect to
C if Y ~ x for every yEA; The set of ideal minimal points of A is denoted by IMin(AfC);
2) x E A is an efficient ( OT Pareto-minimal, or nondominated) point of A with respect to C if x ~ y, for some yEA, then y ~ x; The set of efficient points of A is denoted by lVlin(AIC);
3) x E A is a (global) properly efficient point of A with respect to C if there exists a convex cone ]( which is not the whole space and contains C \ I(C) in its interior so that x'E Min(AII{); The set of proper efficient points of A is denoted by Pr.lvlin(AIC);
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4) supposing that intC is nonempty) x E A is a weakly efficient point of A with respect to C if x E l\1in(A]{O} U inter); The set of weakly efficient points of A is denoted by Wlv[in(A~C).
In the literature some authors exclude indifferent points from the set of effi­ cient points (two points x, YEA, x ;f. Y arc indifferent \vith respect to C if they satisfy simultaneously the relations: x ~c Y and y ~c X4 In other "\vords, a point x E A is said to be efficient if there is no yEA, x =f:. y such that x ~C Y~ 0 bviously~
this definition coincides with ours only in the case where C is a pointed cone (sec Proposition 2.3 below). In the sequel, sometimes, if no confusion occurs, we omit "with respect to C" and ,t Ie" in the definition above. The notions
[Max, lVfax, PrMax, l-VJvlax
are defined dually. When \ve restrict ourselves in a neighborhood of x in E; we get the local ideal efficient, local efficient etc. points notions and denote them by the same with the lower index "1": [Mini, MinI etc.. When speaking of weakly efficient points we always nlean that C is assumed to have nonempty interior.
IIere are some simple examples;­
1. "Vo are in the 2-space R2 ~ Let
A = {(x, Y) E R2 : x2 + y2 :5 1, Y ::; O} U {(x, y) : x ~ 0, 0 ~ Y 2 -l};
B = AU {(-2, -2)}. For C = R~, ,vc have,
IMin(B) = PrlVlin(B) = Min(B) = WMin(B) = {(-2, -2)}; Il\1Iin(A) = 0, PrMin(A) = {(x,y) E R2 : x 2 +y2 = 1,0 > x,D> y},
Min(A) = PrMin(A) U{(a, -l)} U {(-1, O)}, W Min(A) = i\1"in(A) U {(x, y) : y = -1, x ~ O}~
Now, for C = (Rl,O) S; R2 , we have,
IMin(B) = 0, PrMin(B) = Min(D) = WMin(B) = B,
IMin(A) = .0, Prl\1Iin(A) = Min(A) = Wl11in(A) = A.
2. Let.us be in 0 1 and denote
B(O, 1) = {x E rl1 : Hxll ::; 1}. It is easy to see that for C being as in Example 1.2(3) of Chapter 1,
Min(BIC) = {x E ~1 : IIxH = l)x ~ O}. For C being the ubiquitous cone,
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PrMin(A) ~ Min(A) ~ WMin{A).
I1Vlin(A) = Min(A)
Proof. We prove first the inclusion
PrMin(A) ~ 1\!Iin(A).
Let x E PrMin(A). Posit to the contrary that it is not an efficient point of A, Le. there is some yEA such that
x E y+G\l(C). Hence x E y + intK, where K is the cone in the definition of the proper efficiency. Since !( is not the whole space, intK belongs to K \ l(I(). Consequcntly, x >K y~
contradicting the fact that x is efficient with respect to [(.
Further, to prove the inclusion
Min(A) ~ WMin(A),
let x E Min(A) and let [( be the cone composed of zero and of intC. Suppose that yEA and x 2:K y. We have to show that y ~K x which implies that
x E WMin(AIC).
Indeed, if x = y, nothing to prove. If x =F y, X ~K Y means that
x - y E intC. . (2.1)
Since x E Min(A) and [( ~ C, x ~K y implies that y ~c x. In othcr words, y - x E C. This and (2.1) show that 0 E intC, Le. C = E and hence y ~J( x as well.
Finally, it is clear that
IMin(A) ~ Min(A).
If IMin(A) is nonempty, say x is one of its elements, then for each Y E Min(A), y ~ x implies x 2:: y. The transitivity of the order gives us the relation: z 2: Y for every z E A. This means that y E IMin(A) and hence IMin(A) and i\lIin(A) coincide. Whenever C is pointed, x ~ y and y ~ x axe possible only in the case x = y. Thus,IMin(A) is a point.•
Proposition 2.3 An equivalent definition of ejjiciency.~
1) x E IMin(A) if and only if x E A and A ~ x + c;
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2) x E j\1in(A) if and only if An (x - C) S; x +I(C), or equiv~lently, there is no yEA such that x > y. In particular, when G is pointed, x E i\1in(A) if and only if A n (x - C) = {x};
3) when C is not the whole space, x E Wlvlin(A) if and only if A n (x -- intC') = 0, or equivalently, there. is no yEA such that x > y.
Proof. This is immediate from the definition.•
Proposition 2.4 Suppose that there exists a convex pointed cone ]( containing C. Then
1) IJ1fin(AII() = I1W"in(AIC) in case I1Vin(AIC) exists,
2) PrMin(ArK) ~ PrMin(ArC), 3) 1\din(AII() ~ 1\din(AJC),
1) Wl¥in(AII() ~ Wl\lin(AIC).
Proof. To prove the first assertion we observe that C is pointed~ Hence by Propo­ sition 2~2, IlvIin(AiC) is a point,say x E A, if it exists~ By Proposition 2.3,
A ~ x+C ~ x+I(. Consequently, x E Ii\lIin(ArK) and actually we have the equality since 1< is pointed~
The second assertion is triviaL .
For the third onc,let x E Min(AIK), by Proposition 2.3,
A n (x - K) = {x}. Since C ~ !<,
A n (x - C) ~ A n (x - J<). Conscquently,
A n (x - C) = {x} and
x E Min(AIC). To prove the last assertion it suffices to note that intI( is noncmpty whenever
intC is noncmpty , and in this case intC k intK.•
A counterexample for Proposition 2.4, in the case where the pointedness of !< is violated, is obtained \vhen [( is the whole space. In that case every point of a set, in particular, the points which are not efficient with respect to C, is efficient \vith respect to ](. Howevcr,the following proposition provides a useful exception.
Proposition 2.5 Assume that there is a closed homogeneous half space H which contains C \ l(C) in its interior. Then
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1) IMin(ArH) ~ IMin(AIC) in case the right hand side set is nonempty,
2) Min(AJH) ~ Min(AJC), 3) W ..i\1in(AIH) ~ W.l\IIin(AIC).
Proof. For the first assertion, supposing x E IMin(AIII), ,\ve prove that
x E IMin(ArC). By Proposition 2.3, it suffices to show th~t