© Hindawi Publishing Corp.
OPTIMIZATION PROBLEMS FOR SET FUNCTIONS
SLAWOMIR DOROSIEWICZ
(Received 16 June 2000)
Abstract.This paper gives the formal definition of a class of optimization problems, that is, problems of finding conditional extrema of given set-measurable functions. It also formulates the generalization of Lyapunov convexity theorem which is used in the proof of first-order optimality conditions for the mentioned class of optimization problems.
2000 Mathematics Subject Classification. 90C48.
1. Introduction. This paper concentrates on the analysis of certain class of opti-mization problems, namely the problems of finding conditional extrema of set func-tions. This type of problems has been shown to arise in interval and pointwise estima-tion of probability distribuestima-tion, testing hypothesis (in particular during the construc-tion of the strongest tests)—proof of Neyman-Pearson lemma (cf. [3, 4, 10]). These problems also appear in analysis of some transportation problems, especially com-putations of trafficassignment in transportation networks including finding optimal systems of regular lines (see [7]). In future, one can expect the widening possibilities of applications of this class of optimization problems.
The considerations are divided into four parts. The first part consists of general re-marks on set functions and optimization problems for these functions. Sections3and
4have an auxiliary character. They include some special properties of bounded mea-sures (the generalization of Lapunov convexity theorem (Lemma 3.1andTheorem 3.2) and the definition of differentiability of set functions). The most important results are presented inSection 5. There are theorems and some corollaries which formulate the properties of solutions of optimization problems for set functions. These necessary conditions for optimality seem to be the generalization of the results formulated in [5,6,9,11,12]. The mentioned theorems are similar to the well-known Kuhner-Tucker first-order optimality conditions. The main difficulty to obtain these results is caused by the structure of the union of feasible solutions—this family does not usually have the useful structures: compactness, convexity, the structure of a Banach space.
2. Optimization problems for set functions. Let(X,M,µ)be a measurable space with bounded measureµ:M→R.
Definition2.1. Every map (for a givenk∈NMk denotes the cartesian product
M×···×M
nfactors
Example2.2. We consider theµ-integrable functionsfi:X→R(i=1,...,k). An
important family consists of set functions
F:Mk →R, F(S)=k i=1
Si
fidµ. (2.1)
The obvious generalization is the following:
F:Mk →R, F(S)=u
S1f1dµ,...,
Sk
fkdµ
, (2.2)
whereu:Rk→Ris given.
We formulate the general form of optimization problem for a set function. This problem relies on finding extrema (without loss of generality—minima) of a given set functionF0:Mk→R:
F0(S) →min, (2.3)
under some constraints (conditions). They will have two main forms:
• constraints defined by other set functionsFi:Mk→R, wherei=1,...,s
Fi(S)≤0 (i=1,...,s); (2.4)
• constraints imposed directly on the sets, or equivalently on characteristic func-tions of these sets
χS(x)∈v(x) forµ-almost all (a.a.)x∈X, (2.5)
wherev:X→2{0,1}k
is a given function taking values in the family of all subsets of{0,1}k,χ
Sdenotes the vector of characteristic functions of the elementS∈
Mk, ifS=(S1,...,S
k), thenχS=(χS1,...,χSk)and
χSj(x)=
1 if0 ifxx∈∈SXj−, Sj. (2.6)
The notions: feasible solution and optimal one can be defined in a standard way. The feasible solution is an elementS∈Mksatisfying conditions (2.4) and (2.5); an optimal
solution is feasible which minimizes the value of (2.3) among other feasible solutions. To investigate the properties of optimal solutions of (2.3), (2.4), and (2.5) we need some special properties of measures and set functions.
3. Some properties of bounded measure. This part provides the proof of some properties of bounded measures. It follows that a wide class of optimization prob-lems with set functions is equivalent to the probprob-lems of convex mathematical pro-gramming in Euclidean space and can be examined using appropriate methods of convex analysis.
Let(X,M,µ)be a space with bounded measure. ForS=(S1,...,Sk)∈MkandU=
Lemma3.1. Suppose that
(1) µis bounded and nonatomic;
(2) Cis a (totally) unimodular matrix;
(3) for allx∈Xthe system of equations
Cy=χU(x) (3.1)
is consistent. The set
W=µS1,...,µSk
:S∈Mk, Cχ
S=χU (3.2)
is compact and convex. Ifµis bounded (but not necessarily nonatomic), then the setW
is compact (but it may not be connected).
Proof. Suppose first thatµis nonatomic. This part of the proof is similar to the proof of Lapunov convexity theorem and uses the ideas of Lindenstrauss (see [8]).
Without loss of generality, we assume thatµis scalar and nonnegative. Let
ᏸ= k
i=1
L∞(X,M,µ),
K=g=g1,...,gk∈ᏸ: 0≤gi≤1, Cg=χU,
J:ᏸ →Rk, J(g)=
Xg1dµ,...,
Xgkdµ
.
(3.3)
The set ᏸ with the standard norm g =ki=1gi∞, where · ∞ is the norm in L∞(X,M,µ)such thath∞=infA:µ(A)=0sup
x∈X−A|h(x)|is a Banach space.
The setKis an intersection of cartesian product ofksets{h∈L∞(X,M,µ): 0≤h≤ 1}and the set of functions satisfying the equationCy=χU. From Alaoglu theorem
(cf. [1,2]) and Tichonov theorem, we obtain that
k
i=1
g∈L∞(X,M,µ): 0≤g≤1 (3.4)
is compact in∗-weak topology ofᏸ. Because the set of solutions of (3.1) is closed, the setKis compact.
LetJi fori=1,...,kdenotes theith components of the mapJ. The functionsJi
are linear and continuous in the topology ofL∞(X,M,µ)defined by the norm · ∞. Dunford-Schwartz theorem implies thatJiare continuous in the∗-weak topology of
L∞(X,M,µ). Therefore the mapJis continuous in the weak∗-topology ofᏸ. The set J(K)⊂Rkis the image of a compact and convex set in the linear map, therefore it is
also compact and convex.
We show thatKis nonempty. We denote byx, for any givenx∈X, the set of
non-negative solutions of the system (3.1). Under the assumptions, forµ-a.a.x∈Xwe have x≠∅. Because{(x,z):x∈X, z∈x} ∈M⊗β(Rk), then there exists a measurable
selectorv0of the family{x:x∈X}(i.e., the mapv0:X→Rk, such that forµ-a.a.
x∈X,v0(x)∈xholds). Definition ofKimplies thatv0∈K, thusKis nonempty. Of
andW⊂J(K), then it suffices to show that either the setJ−1({a})∩Kis empty or it includes the mapχS for anyS∈Mk.
Fixa∈J(K). The setJ−1({a})∩Kis nonempty, compact, and convex. Krein-Millman theorem follows that this set has at least one extreme point. Letf=(f1,...,fm)be
such a point. We will show that each component offtakes only values 0,1 (µ-a.e. onX). LetXC denote the set{y:Cy=0}. If dimXC=0, then for everyx∈X,x
con-tains one element. This unique element is obviously the extreme point ofx. As the
matrixC is unimodular we have that every component ofr (x)∈x is either 0 or
1. The selector v0 of the family {x :x ∈X} (the mapv0(x)=r (x) for x∈ X)
can be written as v0 = χS for any S ∈ Mk. This means that Lemma 3.1 holds in
this case.
Suppose that for anyi∈ {1,...,k}, fi is not a characteristic function of any
mea-surable set. In this case, dimXC>0 and there exist* >0 and the setE0∈Msuch that
µ(E0) >0 and* < fi(x) <1−*for everyx∈E0. The unimodularity ofCimplies that
every component of the extreme point ofxis equal to 0 or 1. Therefore, for every
x∈E0, the vectorf (x)is not an extreme point ofx. Hence there exist a setE0⊂E0, a vectorc∈XC,c≠0, and a numberδ >0, such that
µE 0
>0, ∀x∈E 0,
λ∈[0,δ]⇒fi(x)+λc∈ +x−x,ex, (3.5)
wherex,exdenotes the set of extreme points ofx.
Becauseµis nonatomic, there exists a measurable setE1⊂E
0such that 0< µ(E1) < µ(E0). Choose the measurable sets G1⊂E1andG2⊂E
0−E1with positive measures and the numberss1,s2∈R(s1+s2>0), such that
maxs1,s2< δ;
s1µE1−2µG1+s2µE
0−E1−2µG2=0.
(3.6)
Consider the function
h=s1χE1−2χG1
+s2χE0−E1−2χG2
. (3.7)
It is easy to check thatXhdµ=0 and for sufficiently smalls1,s2functionsf+=f+hc,
f−=f−hcbelong toJ−1({a})∩K. Becausef=1/2(f++f−), thenf is not an ex-treme point ofJ−1({a})∩K. This contradicts the definition off. Fori=1,...,kand µ-a.a.x∈X, we havefi(x)∈ {0,1}, which implies the existence of an elementS∈Mk
such thatχS=f. This finishes the proof of the first part ofLemma 3.1.
Suppose that the measureµis only bounded (not necessarily nonatomic). It is easy to see that the following decomposition holds:
µ=µna+µa, (3.8)
where the measuresµna,µaare singular and the first of them is nonatomic. Of course
atoms andaµ the sum of sets of this family. For S =(S1,...,Sk)∈Mk, we denote
S−aµ=(S1−aµ,...,Sk−aµ)andS∩aµ=(S1∩aµ,...,Sk∩aµ). For everyA∈Mwe
have an obvious decompositionA=(A−aµ)∪(A∩aµ), soχU=χU−aµ+χU∩aµ and
χS=χS−aµ+χS∩aµ. This means that
W=WnaWa, (3.9)
where
Wna=µnaS1,...,µnaSk:S∈Mk, CχS=χU−aµ
,
Wa=µaS1,...,µaSk:S∈Mk, CχS=χU∩aµ
. (3.10)
From the first part ofLemma 3.1, it follows thatWnais compact. It remains to prove
thatWaalso has this property. Note thatWacan be rewritten as follows:
Wa=
b∈Aµ
µS1∩b,...,µSk∩b:S∈Mk, CχS=χU∩aµ
=
b∈Aµ
µ(b)x1b,...,µ(b)xbk:xb1,...,xbk∈ {0,1}k∩b, b∈Aµ ,
(3.11)
whereb, forb∈Aµ, satisfies the condition
µx∈b:x=b=µ(b). (3.12)
The setWa is equalφ(U), where (denotes the cartesian product of set)
U=
b∈Aµ
{0,1}k∩ b,
φ:[0,1]k|Aµ| →Rk, φxbi
b∈Aµ,i=1,...,k
=
b∈Aµ
µ(b)xb1,...,µ(b)xbk.
(3.13)
Continuity ofφand the compactness ofUimply thatWais compact. This completes
the proof.
Note that inLemma 3.1, the systemCχS=χUwas rewritten as the conditionχS(x)∈
(x)forµ-a.a.x∈X. Unimodularity of matrixCand other assumptions guarantee that the sets(x)were nonempty subsets of{0,1}k. Taking the measurable map
v:X→2{0,1}k
− {∅}(the symbol 2{0,1}k
denotes the family of all subsets of{0,1}k),
measurability ofvmeans thatv−1(B)∈M for everyB⊂ {0,1}k, and replacing (3.1)
by (the mapχS satisfying such condition will be called the selection ofv)
χS(x)∈v(x) forµ-a.a.x∈X, (3.14)
Theorem3.2. Letµbe the bounded, nonatomic measure and letv:X→2{0,1}k be the measurable map, such that forµ-a.a.x∈Xv(x)≠∅. Then the set
W=µS1,...,µSk:S=S1,...,Sk∈Mk, χS is selection ofv (3.15)
is compact and convex.
Proof. The collection{v−1({B}):B⊂ {0,1}k}is the partition ofX. It follows that
W=B⊂{0,1}kWB, where
W=µS1,...,µSk:S=S1,...,Sk∈Mk,
Sj⊂v−1(B), χS|v−1(B)is a selection ofv|v−1(B)
, (3.16)
seeLemma 3.1. The setWis nonempty and compact since anyWBhas these properties
(seeLemma 3.1). This completes the proof.
4. Differentiability of set functions. The definition presented below seems to be a generalization (on the case of bounded measure with any family of atoms) of the notion of differentiability introduced in [5,9].
Before defining a derivative of set functions, we note a few facts. Note that any set function can be equivalently defined on the family of characteristic functions of measurable subsets of X. We denote ˆM = {χS :S ∈M}. Any element S ∈Mk
cor-responds tokcharacteristic functionsχS∈L∞(X,M,µ,Rk)(orχS∈L∞(X,M,µ,Rk),
whereL∞(X,M,µ,Rk)denotes the space of essentially bounded functions), this
im-plies equivalence between set functions and maps defined on ˆMk, set functionF
cor-responds to ˆF: ˆMk→R, ˆF(χ
A)=F(A). Additionally, if we can identify the sets whose
symmetricdifference has measure zero, thenMkcan be viewed as a metric space—the
distance can be defined as follows: ifA=(A1,...,Ak)∈Mk,B=(B1,...,Bk)∈Mk, then
ρ:Mk×Mk →R, ρ(A,B)=k i=1
µAiBi, (4.1)
wheredenotes the symmetricdifference of sets.
Of course, the derivative of the function can be easily defined, if its domain has the structure of a linear and metric space. Unfortunately, Mk (or equivalently ˆMk)
has no “natural” linear structure. Additionally ˆMkis not a convex nor a closed
sub-set of the space of integrable or essentially bounded functions. These facts do not make impossible defining differentiability, but they require making the appropriate modifications.
Definition4.1. We say that a set functionF is differentiable at S0∈Mk if for
any S ∈Mk there exist an integrable functionf
S0,S∩aµ :X−aµ→Rk and the map φS0:Mk→R, such that the following decomposition holds:
F(S)−FS0=
XfS0,S∩aµ
χS−χS0dµ+φS0S∩aµ+RS0(S),
RS0(S)=oρS−aµ,S0−aµ.
(4.2)
Example4.2. The set functions (2.1) are differentiable. Moreover,
fS0,S∩aµ=f|X−aµ, φS0(S)=
Xf
χS∩aµ−χS0∩aµdµ. (4.3)
The notion of differentiability has some “good” properties. For example, the unique-ness property: any set functionF:Mk→Rhas, at givenS0∈Mk, no more than one
derivative. The proof can be found in [7].
In the special case whenµis nonatomic, we haveaµ= ∅and for everyS∈MkS∩aµ = ∅def=(∅,...,∅)∈Mkwhich means thatf
S0,S∩aµ=fS0,∅andφS0≡0 forS∈Mk.
5. Properties of solutions of optimization problems withset functions. In this section, we formulate the necessary conditions for optimality in the problems (2.3), (2.4), and (2.5) with differentiable set functionsFi (i=0,1,...,s). The derivatives of
these functions are denoted by(fS,i,0), respectively, the components offS,iare
de-noted byfS,ij, wherej=1,...,k.
We begin by considering the case with nonatomic measure.
Theorem5.1. LetS∗∈Mkbe an optimal solution of (2.3), (2.4), and (2.5).
(1)Suppose that
(a) µis bounded and nonatomic;
(b) Fi(i=0,1,...,s)are differentiable atS∗in the sense ofDefinition 4.1;
(c) the mapv:X→2{0,1}k
−{∅}is measurable; then there exist the nonnegative numbersλ∗
0,λ∗1,...,λ∗s, not simultaneously equal to zero, such that for every
S∈Mksatisfyingχ
S(x)∈v(x)forx∈X, the following inequality holds: s
i=0 λ∗
i
XfS∗,i
χS−χS∗dµ≥0. (5.1)
Moreover, fori=1,...,s,
λ∗
iFiS∗=0. (5.2)
(2)If there existsSˆ∈Mk, such thatχˆ
S is measurable selection ofv satisfying, for
i=1,...,s, the inequality
FiSˆ+
XfS∗,i
χˆS−χS∗dµ <0, (5.3)
then in (5.1) we may putλ∗ 0=1.
Proof. The presented conditions are similar to the Lusternik theorem describing necessary conditions for conditional extrema of functionals in Banach spaces. The proof is divided into a sequence of steps.
Step1. We show that
V=x0,...,vs∈Rs+1:∃S∈Mk, χS selection ofv,
∀i=1,...,s, xi≥
XfS∗,i
χS−χS∗dµ+1−δi0FiS∗,
(5.4)
whereδ denotes Kronecker delta, is convex. Fix p1,p2∈V, a∈[0,1]. There exist S1,S2∈Mksuch that, form=1,2,i=0,1,...,s,
pm i ≥
XfS∗,i
andχSm(x)∈v(x)forµ-a.a.x∈X.Lemma 3.1applied to the measure
MA→ν(A)=
XfS∗,iχAdµ
i=0,1,...,s
∈Rs+1 (5.6)
shows that the set
XfS∗,0χSdµ,...,
XfS∗,sχSdµ
∈Rs+1:S∈Mk, χ
S(x)∈v(x)forx∈X
(5.7)
is convex. For anya∈[0,1]there exists an elementSa∈Mksuch that
XfSa,0χSadµ,...,
XfSa,sχSadµ
=a
XfS1,0χS1dµ,...,
XfS1,sχS1dµ
+(1−a)
XfS2,0χS2dµ,...,
XfS2,sχS2dµ (5.8)
and forµ-a.a.x∈X,χSa(x)∈v(x).
This implies that
ap1+(1−a)p2≥
XfSa,0
χSa−χS∗dµ,...,
XfSa,s
χSa−χS∗dµ
+1−δi0FiS∗,
(5.9) which means thatap1+(1−a)p2∈V. This finishes the proof of convexity.
Step2. We show thatVis disjoint with]−∞,0[s+1. If forµ-a.a.x∈Xthe setv(x) has only one element, then forS∈Mksatisfying the conditionχ
S(x)∈v(x)we have
S=S∗. The setVis disjoint with]−∞,0[s+1and the inequality (5.1) obviously holds. Assume that there existsS0≠S∗such thatχ
S0(x)∈v(x)forµ-a.a.x∈X. Suppose thatV∩]−∞,0[s+1≠∅. Hence there existsS0∈Mk, such that for alli=0,1,...,sthe
following inequality holds:
XfS∗,i
χS0−χS∗dµ+1−δi0FiS∗≤0. (5.10)
Lapunov convexity theorem applied to the measure
MA→
A
χS0−χS∗dµ,
AfS∗,0
χS0−χS∗dµ,...,
AfS∗,s
χS0−χS∗dµ
∈Rs+2,
(5.11) where |χS0−χS∗| =kj=1|χS0
j −χS∗j|, implies that for any α∈[0,1]there exists a
µ-measurable set
Ωα⊂x∈X:χ
S∗(x)≠χS0(x) (5.12)
such that
α
X
χS0−χS∗dµ=
ΩαχS0−χS∗dµ, (5.13)
and for everyi=0,...,s,
α
XfS∗,i
χS0−χS∗dµ=
ΩαfS∗,i
Consider for fixedα∈[0,1], the map
uα:X →Rk, uα(x)=χ
S∗(x)+χS0(x)−χS∗(x)χΩα(x), (5.15)
and we denote bySα the elementMkwhich corresponds touα, that is,χ
Sα=uα. It
is easy to check thatχSα(x)∈v(x)forµ-a.a.x∈X. For sufficiently smallα >0 the
elementSαis closed toS∗with respect to the metricρ(see (4.1)):
ρSα,S∗=
X
χSα−χS∗dµ=
ΩαχSα−χS∗dµ=α
X
χS0−χS∗dµ, (5.16)
thereforeρ(Sα,S∗)→0 ifα→0+. Differentiability ofFiimplies that
FiSα−FiS∗=
XfS∗,i
χSα−χS∗dµ+Ri,S∗Sα, (5.17)
and hence
FiSα−FiS∗=α
XfS∗,i
χS0−χS∗dµ+Ri,S∗Sα, (5.18)
whereRi,S∗(Sα)=o(ρ(Sα,S∗))=o(α)if α→0. Equation (5.10) gives that the first component in (5.18) is negative, therefore for sufficiently smallα >0 its absolute value is greater than the second one. This implies thatF0(Sα) < F0(S∗)andF
i(Sα) <
Fi(S∗)≤0 fori=1,...,s. ElementSαis the feasible solution of (2.3), (2.4), and (2.5);
the value of the objective function forSα is less than forS∗. This contradicts the definition ofS∗.
Step 3. This step finishes the proof of inequality (5.1), the separation theorem implies that there exist nonnegative numbersλ∗
0,...,λ∗s not all zero such that, for all
(t0,t1,...,ts)∈V,
s
i=0 λ∗
iti≥0. (5.19)
ForS∈Mksatisfyingχ
S(x)∈v(x)forµ-a.a.x∈X, we have
s
i=0 λ∗
i
XfS∗,i
χS−χS∗dµ+1−δ0,iFiS∗
≥0. (5.20)
LettingS =S∗ we obtain s
i=1λ∗iFi(S∗)≥0. Becauseλ∗i ≥0 andFi(S∗)≤0 fori=
1,...,s, thenλ∗
iFi(S∗)=0. This finishes the first part of the proof.
Now suppose that there exists an element ˆS∈Mk, which satisfies the inequalities
(5.3) withλ∗
0=0. The inequality (5.20) gives
s
i=1 λ∗
i
XfS∗,i
χˆS−χS∗dµ+FiS∗
≥0. (5.21)
Each component of this sum is nonpositive, thenλ∗
i =0 fori=1,...,s. We obtain the
contradiction with the fact thatλ∗
0,λ∗1,...,λ∗s are not all zero. Henceλ∗0>0. Dividing both sides of (5.1) byλ∗
Corollary5.2. Suppose thatS∗=(S∗
1,...,Sk∗)∈Mkis an optimal solution of (2.3), (2.4), and (2.5) and this problem does not have the conditions (2.5). If the assumptions ofTheorem 5.1are satisfied, then there exist nonnegative numbersλ∗
0,λ∗1,...,λ∗s, not simultaneously equal to zero, such that forj=1,...,k,
s
i=0 λ∗
ifS∗,i(x)
≤0 ifx∈S
∗
j, ≥0 ifx∈X−S∗
j.
(5.22)
Proof. According toTheorem 5.1, for everyS∈Mkwe have s
i=0 λ∗
i
XfS∗,i
χS−χS∗dµ≥0. (5.23)
Fixj ∈ {1,...,k}. Inequality (5.23) must hold for every measurableu=(u1,...,uk)
such thatX→ {0,1}k, whereu
i=χS∗i fori≠j. This gives (the symbolfS∗,ij denotes thejth component offS∗,i)
s
i=0 λ∗
i
XfS∗,ij
uj−χSj∗
dµ≥0, (5.24)
which is equivalent to (5.22). This completes the proof.
Corollary5.3. LetS∗be the optimal solution of (2.3), (2.4), and (2.5) in which the
condition (2.5) can be written asχS(x)∈Vforµ-a.a.x∈X, whereV is a given subset of{0,1}k.Theorem 5.1shows that there exist nonnegative numbersλ∗
i (i=0,1,...,s), which do not vanish simultaneously, such that, forv¯∈Vandx∈χ−1
S∗(v)¯ ,
∀v∈V s
i=0 λ∗
ifS∗,i(v−v)¯ ≥0. (5.25)
Example5.4. Let(X,M,µ)be a probabilistic space with nonatomic, bounded mea-sureµ,Fi(i=0,1,...,s), the differentiable set functions onM3. Consider the problem
F0S1,S2,S3→min, (5.26)
subject to
S1,S2,S3∈M3, F
iS1,S2,S3≤0 (i=1,...,s), S1∪S2⊂S3. (5.27)
Note that the last constraint can be written, using the characteristic functions of S1,S2,S3, in the following form:
χS1∪S2=max
χS1,χS2
≤χS3. (5.28)
This inequality has five solutions:
v1=(0,0,0), v2=(0,0,1), v3=(1,0,1), v4=(0,1,1), v5=(1,1,1), (5.29)
thus we have
χS1(x),χS2(x),χS3(x)
∈V forx∈X, (5.30)
Suppose thatFiare differentiable. We denote the derivative ofFiatS=(S1,S2,S3)
byfS,i, and its components byfS,i,1,fS,i,2,fS,i,3, respectively. Assume that the
prob-lem (5.26) and (5.27) has the optimal solutionS∗=(S∗
1,S2∗,S3∗).Corollary 5.3implies that there exist nonnegative numbersλ∗
i (i=0,...,s)(which do not vanish
simulta-neously), such that for anyv∈V,
XᏲS∗
v−χS∗dµ≥0, (5.31)
whereᏲS∗=si=0λ∗ifS∗,i. The functionᏲS∗, likefS∗,i, has three components which correspond toS1,S2,S3. We denote them byᏲS1∗,ᏲS2∗,ᏲS3∗, respectively.
Consider the inequality (5.31). Putting the functions
v(k)(x)= χS
∗(x) forx∈S1∗∪S2∗∪S3∗, vk forx∈S1∗∪S2∗∪S3∗,
(5.32)
wherek=1,...,5, we obtain the system of inequalities which holdsµfor a.a.X−S∗ 1∪ S∗
2∪S3∗(i.e., forxsuch thatχS∗(x)=v1), ᏲS∗
1(x)+ᏲS3∗(x)≥0, ᏲS2∗(x)+ᏲS∗3(x)≥0, ᏲS∗1(x)+ᏲS2∗(x)+ᏲS3∗(x)≥0.
(5.33)
Forx∈S∗ 3−
S∗ 1∪S2∗
(ifχS∗(x)=v2),
ᏲS3∗(x)≤0, ᏲS1∗(x)≥0, ᏲS2∗(x)≥0. (5.34)
Ifx∈S∗
1−S2∗(χS∗(x)=v3), then
ᏲS1∗(x)≤0, ᏲS2∗(x)≥0, ᏲS3∗(x)≤0. (5.35)
Ifx∈S∗
2−S1∗(χS∗(x)=v4), then
ᏲS1∗(x)≥0, ᏲS2∗(x)≤0, ᏲS3∗(x)≤0. (5.36)
In the case whenx∈S∗
1∩S2∗∩S3∗(χS∗(x)=v5), then
ᏲS1∗(x)≤0, ᏲS2∗(x)≤0, ᏲS3∗(x)≤0. (5.37)
Inequalities (5.33), (5.34), (5.35), (5.36), and (5.37) allow us to determine the optimal solution(−s)S∗=(S∗
1,S2∗,S3∗)(if exist) of the problem (5.26) and (5.27).
The considerations presented so far concerned the special case when the measure µ was nonatomic. It is easy to prove some generalizations ofTheorem 5.1. We have the following corollary.
Corollary5.5. LetS∗be an optimal solution of (2.3), (2.4), and (2.5), with
differ-entiable set functionsFi (i=0,...,s). There exist the nonnegative numbers λ∗i (i=
0,...,s), which do not vanish simultaneously, such that any feasible solutionSsatisfies the following inequality:
s
i=0 λ∗
i
X−aµfS
∗,S∗∩aµ,iχS−χS∗dµ≥0, (5.38)
Proof. The elementS∗is the optimal solution of (2.3), (2.4), and (2.5) if and only ifS∗−a
µ=(S1∗−aµ,...,Sk∗−aµ)is the optimal solution of the problem (with decision
variablesA=(A1,...,Ak))
F0A∪S∗∩a
µ →min, (5.39)
subject to
Aj∈M, Aj⊂X−aµ forj=1,...,k;
FiA∪S∗∩aµ≤0 fori=1,...,s;
χA(x)∈v(x) forµ-a.a.x∈X−aµ.
(5.40)
Measure µ restricted to the family of measurable subsets of X−aµ is obviously
nonatomic. ApplyingTheorem 5.1to the problem (5.39) and (5.40) finishes the proof.
Corollary 5.7concentrates on ordinary (unconditional) extrema of set function. This kind of extreme can be viewed as the optimal solutions of (2.3), (2.4), and (2.5), in which the constraints (2.4) do not appear explicitly.
Definition5.6. We say that the set functionF:Mk→Rhas inS∗∈Mkminimum
(local minimum), ifS∗is the optimal solution of the problem
F(S) →min, (5.41)
subject to
S∈Mkthere exists* >0, such thatρS∗,S< *. (5.42)
The necessary condition for optimality gives the following corollary.
Corollary5.7. IfFhas minimum (respectively, local minimum) inS∗=(S∗ 1,...,Sk∗) and((fS∗,S∩aµ)S∈Mk,φS∗)is the derivative ofFinS∗, then for anyS∈Mk(respectively,
there exists* >0, such thatρ(S,S∗) < *) andj=1,...,k,
x∈S∗
j −aµ ⇒fS∗,S∗∩aµ,j(x)≤0;
x∈S∗
j −aµ ⇒fS∗,S∗∩aµ,j(x)≥0, (5.43)
φS∗S∩aµ≥0. (5.44)
Proof. Without loss of generality, it is sufficient to consider the case of local min-imum. Minimum ofF corresponds to the situation with*≥µ(X). Formulas (5.43) follows directly fromTheorem 5.1. To finish the proof, we should show the inequality (5.44). Differentiability ofF and optimality ofS∗imply that ifρ(S∗,S) < *, then
0≤F(S)−F(S∗)=
X−aµfS
∗,S∗∩aµχS−χS∗dµ
+φS∗S∩aµ+oρS−aµ,S∗−aµ.
(5.45)
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