MINIMIZERS OF GERSTEWITZ FUNCTIONALS
by
PETRA WEIDNER111HAWK Hochschule für angewandte Wissenschaft und Kunst Hildesheim/Holzminden/ Göttingen University of Applied Sciences and Arts, Faculty of Natural Sciences and Technology,
D-37085 Göttingen, Germany, [email protected].
Research Report
April 25, 2017
**Abstract:
Scalarization in vector optimization is essentially based on the minimization of Gerstewitz functionals. In this paper, the minimizer sets of Gerstewitz functionals are investigated. Conditions are given under which such a set is nonempty and compact. Interdependencies between solutions of problems with different parameters or with different feasible point sets are shown. Consequences for the parameter control in scalarization methods are derived. It is pointed out that the minimization of Gerstewitz functionals is equivalent to an optimization problem which generalizes the scalarization by Pascoletti and Serafini.
**
**Keywords:
nonlinear programming; vector optimization; scalarization; optimization in general spaces
**
Mathematics Subject Classification (2010):
90C30, 90C48, 90C29
1. Introduction
Gerstewitz functionals were introduced by Gerstewitz (later Gerth, now Tammer) in the context of vector optimization [1]. They have been investigated in [2] and [3], later followed by
[4], [5] and [6]. Necessary and sufficient conditions for its basic properties under more general assumptions are given in [7] and [8]. In [8], it is shown that Gerstewitz functionals can represent orders, preference relations and other binary relations and thus act as a general tool for scalarization. This tool is used in multicriteria optimization, decision theory, mathematical finance, production theory and operator theory. In many cases, it is not obvious or not known that the function which is minimized is a Gerstewitz functional. A complete characterization of solutions in vector optimization by minimizers of Gerstewitz functionals has been given in [9] and [10].
The original version of the results presented in this paper was developed by the author in [3]. It has been extended using the properties of Gerstewitz functionals proved in [7] and [8].
In Section 2, we will give the basic definitions and some preliminaries.
Section 3 contains the formulation of the problem and connects it with a generalization of an optimization problem introduced by Pascoletti and Serafini [11].
Moreover, relationships to problems with an altered feasible point set are proved.
In Section 4, the existence of optimal solutions and properties of the solution set are studied.
Section 5 deals with the question for which varying parameters the optimization problems have the same solution set. Finally, further interdependencies between solution sets for different parameters are shown in Section 6.
2. Preliminaries
Throughout this paper, Y is assumed to be a real topological vector space.
From now on, R and N will denote the set of real numbers and of nonnegative integers, respectively.
We define
N> as the set of positive integers,
R+:={x∈R∣x≥0}, R>:={x∈R∣x>0},
R+n:={(x1,…,xn)T∈Rn∣xi≥0∀i∈{1,…,n}} for each n∈N>. R:=R∪{−∞,+∞} denotes the extended real-valued set.
A set C in Y is said to be a cone iff λc∈C\mboxforallλ∈R+,c∈C. The cone C is called nontrivial iff C=∅, C={0} and C=Y hold. It is said to be pointed iff C∩(−C)={0}.
Let A be a subset of Y.
0+A:={u∈Y∣A+R+u⊆A} denotes the recession cone of A.
coreA stands for the algebraic interior of A.
clA, intA, bdA and convA denote the closure, the interior, the boundary and the convex hull, respectively, of A.
Given some set B⊆R, d∈Y, and D⊆Y, we write Bd:={b⋅d∣b∈B} and BD:={b⋅d∣b∈B,d∈D}.
Consider a functional φ:Y→R. We use its effective domain
domφ:={y∈Y∣φ(y)∈R∪{−∞}}.
For some binary relation R given on R, the sets levφ,R(t):={y∈Y∣φ(y)Rt} are defined for t∈R. In this way, the sublevel sets of φ are given as levφ,≤(t).
φ is said to be finite-valued on A iff it attains only real values on A. It is called finite-valued iff it is finite-valued on Y. φ is said to be proper iff domφ=∅ and φ is finite-valued on domφ. According to the rules of convex analysis, inf∅=+∞.
From now on, we suppose in this section
(H1A,k): A is a proper closed subset of Y and k∈−0+A∖{0}.
Definition 1**.**
The function φA,k:Y→R defined by
[TABLE]
is called a Gerstewitz functional.
We will use the following statements from [7] and [8].
Lemma 1**.**
φA,k* is lower semicontinuous on domφA,k,*
[TABLE]
Moreover,
φA,k(y)=−∞⟺y+Rk⊆A.
φA,k* is finite-valued on domφA,k∖A.*
φA,k* is proper if and only if
A does not contain lines parallel to k, i.e.,*
[TABLE]
If k∈−core0+A, then φA,k is finite-valued.
If k∈(−0+A)∩0+A, then φA,k does not attain any real value.
If k∈(−0+A)∖0+A and A is convex, then φA,k is proper.
Lemma 2**.**
We have
[TABLE]
for each y∈Y with φA,k(y)∈R.
Proof.
φA,k(y)∈R implies
φA,k(y)=min{t∈R∣φA,k(y)≤t}=min{t∈R∣y∈A+tk} by (2.1). Thus, we get the first equation. The second one results from (2.2).
∎
[7, Theorem 3.1] contains the following results.
Lemma 3**.**
Assume A−R>k⊆intA.
Then
[TABLE]
Moreover,
φA,k(y)=−∞⟺y+Rk⊆intA.
bdA+Rk* is the subset of Y on which φA,k is finite-valued, and*
domφA,k∖intA⊆bdA+R+k.
The following conditions are equivalent:
[TABLE]
φA,k* is finite-valued if and only if
Y=bdA+Rk.*
We will use two other lemmata from [8].
Lemma 4**.**
Consider some arbitrary λ∈R>.
Then (H1A,λk) holds, domφA,λk=domφA,k and
φA,λk(y)=λ1φA,k(y) for all y∈Y.
Lemma 5**.**
Consider some arbitrary c∈R.
Then (H1A+ck,k) holds, domφA+ck,k=domφA,k and
φA+ck,k(y)=φA,k(y)−c for all y∈Y.
In [9] and [10], Gerstewitz functionals are used for the scalarization of vector optimization problems. There, a vector optimization problem is given by a function f:S→Y defined on a nonempty set S and by a set D⊂Y which defines the solution concept. D is called the domination set of the problem. A solution of the vector optimization problem is each s∈S for which f(s) is an efficient element of F:=f(S) with respect to D.
An element y0∈F is called an efficient element of F
w.r.t. D iff
[TABLE]
We denote the set of efficient elements of F with respect to D by Eff(F,D).
In our paper, we need the following statement [10, Lemma 11].
Lemma 6**.**
Assume F,D⊆Y and F⊆F~⊆F+(D∪{0}).
Then Eff(F~,D)⊆Eff(F,D).
If, additionally, D∩(−D)⊆{0} and D+D⊆D, then Eff(F~,D)=Eff(F,D).
Furthermore, the following lemmata will be used in the proofs of the next sections.
We get from [12, Theorem 2.43]:
Lemma 7**.**
A real-valued lower semicontinuous function on a nonempty compact subset of a topological space attains a minimum value, and the nonempty set of minimizers is compact.
This implies the next statement by the use of the topology which is induced on a subset of the space.
Lemma 8**.**
Assume that the following conditions hold:
φ:Y→R* is finite-valued on M⊆domφ and lower semicontinuous,*
M* is a nonempty compact set.*
Then the set of minimizers of φ on M is nonempty and compact.
The next lemma is due to [3].
Lemma 9**.**
Assume that D⊆Y is a cone with intD=∅ and d∈intD. Then
[TABLE]
Proof.
V:=intD−d is a neighbourhood of [math]. ⇒tV=t(intD−d)⊆intD−td∀t∈R> since D is a cone. ⇒R>V⊆intD−R>d. ⇒Y=R+V⊆intD−R>d since d∈intD.
∎
We will also need the following lemmata, which were proved in [13].
Lemma 10**.**
Assume that M and D are closed convex sets in Rℓ. If there exists some a∈Rℓ such that M∩(a−D) is not bounded, then 0+(M∩(b−D))={0} for each
b∈Rℓ with M∩(b−D)=∅.
Lemma 11**.**
Assume that D⊆Rℓ is a closed convex set with 0∈D and that M⊂Rℓ is a set for which there exist a polyhedral cone C⊂Rℓ and some u∈Rℓ such that −D∖{0}⊂intC and (M−u)∩intC=∅.
Then each set M∩(b−D) with b∈Rℓ is bounded.
3. Problem formulation and related problems
From now on, we assume
[TABLE]
unless not stated otherwise.
In this paper, we study the optimization problem
[TABLE]
Since φa−H,k has been defined as an extended real-valued functional
and F is not necessarily contained in the effective domain of φa−H,k, the feasible range F∩domφa−H,k of the optimization problem does not always coincide with F.
Theorem 1**.**
The optimization problem (3.1)
is equivalent to the optimization problem
[TABLE]
i.e., both problems have the same feasible point set F∩domφa−H,k, each problem has an optimal solution if and only if the other one has an optimal solution, and the optimal solutions y as well as the optimal value of both problems coincide.
The optimization problem (3.1) has an optimal solution if and only if the optimization problem
[TABLE]
has an optimal solution. In this case, the optimal solutions y as well as the optimal value of both problems coincide.
If (PF,a,H,k) has a feasible solution (y0,t0)∈F×R, then the set of minimizers y of (PF,a,H,k) and its optimal value coincide with those of the problem
[TABLE]
Proof.
By Lemma 2, φa−H,k(y)∈R implies φa−H,k(y)=min{t∈R∣y∈a−H+tk}=min{t∈R∣y∈a−bdH+tk}. Thus, we get (a) and (b).
Assume now that (PF,a,H,k) has a feasible solution (y0,t0)∈F×R.
Since k∈0+H, a−H+tk⊆a−H+t0k∀t≤t0.
Hence, (PF,a,H,k) has the same optimal solutions and the same optimal value as the problem (3.2).
∎
Geometrical interpretation of (PF,a,H,k):
Stick the set −H to the point a and shift the set a−H along the line {a+tk∣t∈R} until the smallest
t is reached for which the intersection of the set with F is not empty,
i.e., for which F∩(a−H+tk)=∅.
Then t is the optimal value of (PF,a,H,k), and the set of points in which F and a−H+tk intersect is the set of optimal solutions y of (PF,a,H,k).
We illustrate this by an example.
Example 1**.**
Choose Y=R2, F={(y1,y2)T∈R2∣0≤y2≤y1≤1},
H=R+2, a=(−1,0)T, k=(1,1)T.
(PF,a,H,k) attains an optimal value topt. y=(0,0)T is the unique optimal solution of
(PF,a,H,k).
In applications, we will mainly work with (PF,a,H,k), where the equivalence to (PF,a,bdH,k) can be used for restricting the attention in certain steps to the boundary of H. The equivalence to problem (3.1) and problem (3.2) is useful for proving properties of (PF,a,H,k). Scalarization results for solution sets of vector optimization problems have been formulated in [9] and [10] using problem (3.1).
From now on, let MF,a,H,k denote the set of optimal solutions y of problem (PF,a,H,k). As we have just shown, this set coincides with y∈Fargminφa−H,k(y) under assumption (H1–OPF,a,H,k).
Sometimes, F+H may be closed or convex though F does not have this property.
In these cases, the following proposition can be useful. Here, we also consider the efficient elements of the minimizer set. As pointed out in [9] and [10], only these minimizers are efficient elements of a vector optimization problem with respect to the same domination set.
Proposition 1**.**
Assume, additionally, 0∈H and H+H⊆H.
We get:
(PF,a,H,k)* has a solution if and only if (PF+H,a,H,k) has a solution. In this case, both optimal values coincide. Moreover,*
[TABLE]
If F+H is closed, then
[TABLE]
If, moreover, H∩(−H)={0} holds, then
[TABLE]
Proof.
First, assume that φa−H,k attains on F the minimal value t. Then φa−H,k attains the function value t also on F+H.
Supposition: ∃tˉ<t,y∈F,h∈H:φa−H,k(y+h)=tˉ.
⇒y+h∈a−H+tˉk because of (2.1). ⇒y∈a−h−H+tˉk⊆a−H+tˉk.
Then (2.1) implies φa−H,k(y)≤tˉ, a contradiction to the definition
of t.
Hence, t is also the minimal value of φa−H,k on F+H.
This implies the first inclusion.
Assume now that t is the minimum of φa−H,k on F+H.
Then φa−H,k does not attain any smaller value than t on F.
Take any y∈F and h∈H with φa−H,k(y+h)=t.
⇒y+h∈a−H+tk.
⇒y∈a−H+tk.
⇒φa−H,k(y)=t.
y is a minimal solution of φa−H,k on F. This yields the second inclusion.
For MF,a,H,k=∅, the statement is obvious.
Assume now MF,a,H,k=∅ and
t to be the minimum of φa−H,k on F.
⇒MF,a,H,k=F∩(a−H+tk).
Consider an arbitrary yˉ∈clMF,a,H,k.
clMF,a,H,k⊆clF⊆cl(F+H)=F+H.
⇒∃y∈F,h∈H:yˉ=y+h.
clMF,a,H,k⊆a−H+tk.⇒y+h∈a−H+tk.
⇒y∈a−h−H+tk⊆a−H+tk.⇒y∈MF,a,H,k.
Thus, yˉ∈MF,a,H,k+H.
Hence, clMF,a,H,k⊆MF,a,H,k+H, and Lemma 6 yields the assertion.
∎
4. Existence of optimal solutions and properties of the solution set
Proposition 2**.**
If F is closed, then MF,a,H,k is closed.
If F and H are convex, then MF,a,H,k as well as the feasible range of (PF,a,H,k)
are convex.
Proof.
If MF,a,H,k=∅, then the statement of the proposition is fulfilled.
Otherwise, MF,a,H,k=F∩(a−H+t0k), where t0 denotes the optimal value of (PF,a,H,k). This yields the assertion.
∎
Let us now investigate the existence of optimal solutions for the considered optimization problems and the compactness of the solution set.
Immediately from (PF,a,H,k), we get necessary conditions for the existence of optimal solutions.
Proposition 3**.**
If MF,a,H,k=∅, the following conditions are fulfilled:
[TABLE]
Condition (4.1) expresses the existence of a feasible solution and is equivalent to F∩domφa−H,k=∅. It holds, if F is nonempty and φa−H,k is finite-valued on F.
Condition (4.2) is fulfilled if and only if φa−H,k is bounded below on F. In this case, φa−H,k is finite-valued on F∩domφa−H,k.
Lemma 12**.**
The following conditions are equivalent to each other:
φa−H,k* is finite-valued on F∩domφa−H,k.*
φa−H,k* is finite-valued on F∩(a−H+tk) for some t∈R.*
Proof.
Obviously, (a) implies (b). Assume now that there exists some y∈F∩domφa−H,k with φa−H,k(y)=−∞.
Then y+Rk⊆a−H by Lemma 1. ⇒∀t∈R:y−tk∈a−H, hence, y∈a−H+tk. Then (b) is not fulfilled.
∎
We will now study sufficient conditions for the existence of optimal solutions, based on the following theorem.
Theorem 2**.**
Assume
[TABLE]
If (4.2) holds, then MF,a,H,k is nonempty and compact.
Assumption (4.2) holds if and only if
φa−H,k is finite-valued onF∩domφa−H,k.
Proof.
(4.2) implies by (2.1) that φa−H,k is finite-valued on F∩domφa−H,k.
Assume now that φa−H,k is finite-valued on F∩domφa−H,k.
There exists some t1∈R such that B:=F∩(a−H+t1k) is nonempty and compact. By Theorem 1, MF,a,H,k=MB,a,H,k. Lemma 8 yields that MB,a,H,k is nonempty and compact. This proves (a) and, because of (2.1), also (b).
∎
Lemma 7 implies by Lemma 1:
Proposition 4**.**
MF,a,H,k* is nonempty and compact if
F is a nonempty compact set and φa−H,k is finite-valued.*
We get especially by Lemma 1(d):
Corollary 1**.**
Assume that F is a nonempty compact set, and k∈core0+H.
Then MF,a,H,k is nonempty and compact.
Let us now investigate the set of minimizers for the case that φa−H,k is not necessarily finite-valued.
Proposition 5**.**
*Assume that F is a compact set and that the problem (PF,a,H,k) has a feasible solution.
Then MF,a,H,k is nonempty and compact under each of the following conditions:*
F∩(a−H)=∅,
F∩(a−intH)=∅* and H+R>k⊂intH,*
H* does not contain a line in direction k,*
H* is convex and k∈−0+H.*
Proof.
(PF,a,H,k) has a feasible solution. ⇒∃t0∈R:B:=F∩(a−H+t0k)=∅. By Theorem 1, MF,a,H,k=MB,a,H,k.
B is compact since F is compact and H is closed.
φa−H,k is finite-valued on B
in the cases (a) and (c) by Lemma 1(b) and (c), respectively, in case (b) by Lemma 3(b), in case (d) by Lemma 1(f).
Lemma 8 yields the assertion.
∎
For Y=Rℓ, the existence of optimal solutions of (PF,a,H,k) can be guaranteed without the assumption that F is compact.
Proposition 6**.**
Assume that the following conditions are fulfilled:
a∈Rℓ, F⊆Rℓ is a nonempty closed set, and there exists some u∈Rℓ with F⊆u+R+ℓ.
H* is a proper closed subset of Rℓ with R+ℓ⊆0+H and k∈core0+H that, for each j∈{1,…,ℓ}, does not contain any line in the direction ej.*
Then MF,a,H,k is nonempty and compact.
Proof.
By Lemma 1(d), φa−H,k is finite-valued. Hence,
(PF,a,H,k) has a feasible solution. ⇒∃t0∈R:B:=F∩(a−H+t0k)=∅.
For each j∈{1,…,ℓ}, H does not contain any line in the direction ej. Thus, ∀j∈{1,…,ℓ}∃sj∈R:
[TABLE]
zj:=uj+sj∀j∈{1,…,ℓ}. Suppose that there exists some y∈B and some i∈{1,…,ℓ} with yi>zi.
y∈F⊆u+R+ℓ implies u+siei∈y−R+ℓ⊆(a−H+t0k)−R+ℓ⊆a−H+t0k, a contradiction to (4.3).
Thus, B⊆z−R+ℓ. Hence, B is compact.
By Corollary 1, MB,a,H,k is nonempty and compact.
Theorem 1 yields the assertion.
∎
Note that R+ℓ⊆0+H and k∈intR+ℓ imply k∈core0+H.
The assumption in Proposition 6 that refers to lines in directions ej is not superfluous, even for convex sets H.
Example 2**.**
Y=R2, F={(y1,y2)T∈R2∣y1>0,y2=y11},
a=(0,0)T, k=(1,1)T
and H={(y1,y2)T∈R2∣y1≥0} fulfill all assumptions of Proposition 6 but the condition that H does not contain any line in direction (0,1)T.
F∩(a−H+tk)=∅∀t≤0 and
F∩(a−H+tk)=∅∀t>0.
Thus, (PF,a,H,k) does not have any optimal solution.
Furthermore, the statement of Proposition 6 is not true any more without the assumption that F is bounded below.
Example 3**.**
Consider Y=R2, F={(y1,y2)T∈R2∣y2=0},
H={(y1,y2)T∈R2∣y1>0,y2≥y11},
k=(1,1)T.
For each a∈R2, (PF,a,H,k) does not have any optimal solution.
Proposition 6 implies a special statement for convex cones H, which results from the following lemma.
Lemma 13**.**
Assume that H⊂Rℓ is a nontrivial closed pointed convex cone with R+ℓ⊆H.
Then, for each j∈{1,…,ℓ}, H does not contain any line in the direction ej.
Proof.
Consider an arbitrary j∈{1,…,ℓ}. ej∈H by R+ℓ⊆H, and ej∈H∖(−H) since H is pointed.
By Lemma 1(f), φ−H,ej is proper. Thus, by Lemma 1(c), H does not contain any line in direction ej.
∎
Thus, we get by Proposition 6:
Corollary 2**.**
Assume that the following conditions are fulfilled:
a∈Rℓ, F⊆Rℓ is a nonempty closed set, and there exists some u∈Rℓ with F⊆u+R+ℓ.
H⊆Rℓ* is a nontrivial closed pointed convex cone with R+ℓ⊆H and k∈intH.*
Then MF,a,H,k is nonempty and compact.
Let us now consider the optimization problems on Y=Rℓ for the case that k is not necessarily an element of core0+H.
Proposition 7**.**
Assume that the following conditions hold:
a∈Rℓ, F⊆Rℓ is closed, and there exists some u∈Rℓ with F⊆u+R+ℓ.
H* is a proper closed subset of Rℓ with R+ℓ⊆0+H and k∈0+H∖{0} that, for each j∈{1,…,ℓ}, does not contain any line in the direction ej.*
The problem (PF,a,H,k) has a feasible solution.
One of the following assumptions is fulfilled:
F∩(a−H)=∅,
F∩(a−intH)=∅* and H+R>k⊂intH,*
H* does not contain a line in direction k,*
H* is convex and k∈−0+H.*
Then MF,a,H,k is nonempty and compact.
Proof.
(PF,a,H,k) has a feasible solution. ⇒∃t0∈R:B:=F∩(a−H+t0k)=∅.
By the same arguments as in the proof of Proposition 6, B is compact.
By Proposition 5, MB,a,H,k is nonempty and compact.
Theorem 1 yields the assertion.
∎
For convex sets H, Proposition 7 implies a statement which can be proved by means of the following lemma.
Lemma 14**.**
Assume that H⊂Rℓ is a proper closed convex subset of Rℓ with H+(R+ℓ∖{0})⊆intH, H+R>k⊆intH and k∈−0+H.
Then, for each j∈{1,…,ℓ}, H does not contain any line in direction ej.
Proof.
Supposition: ∃i∈{1,…,ℓ}:H contains some line in direction ei.
⇒∃u∈Rℓ:{u+sei∣s∈R}⊆H.
⇒{u+sei∣s∈R}={u+(s−1)ei∣s∈R}+ei⊆H+(R+ℓ∖{0})⊆intH.
By Lemma 1(f), φ−H,k is proper. Hence, Lemma 3(c) implies intH=bdH+R>k.
u∈intH⇒∃h∈bdH,t0>0:h=u−t0k.
h+ei∈intH since H+(R+ℓ∖{0})⊆intH.
⇒∃t1>0:h+ei−t1k∈intH.
t2:=t0+t1t0∈(0,1).s0:=−1−t2t2.
⇒t2+(1−t2)s0=0,t2(t0+t1)=t0.
⇒h=u−t0k=u+(t2+(1−t2)s0)ei−t2(t0+t1)k=t2(u+ei−(t0+t1)k)+(1−t2)(u+s0ei)=t2(h+ei−t1k)+(1−t2)(u+s0ei)∈intH since intH
is convex.
This contradicts h∈bdH.
∎
Thus, we get by Proposition 7:
Corollary 3**.**
Assume that the following conditions hold:
a∈Rℓ, F⊆Rℓ is closed and there exists some u∈Rℓ with F⊆u+R+ℓ.
H* is a proper closed convex subset of Rℓ with H+(R+ℓ∖{0})⊆intH, H+R>k⊆intH and k∈−0+H.*
The problem (PF,a,H,k) has a feasible solution.
Then MF,a,H,k is nonempty and compact.
Without the assumption F⊆u+R+ℓ in Corollary 3, the existence of optimal solutions for (PF,a,H,k) can depend on the choice of a,
even if H is a closed convex cone.
Example 4**.**
Assume Y=R3 and that H is the convex cone generated by the vectors (2,0,−1)T, (0,2,−1)T
and (−1,0,2)T. Choose k=(1,1,1)T, a=(0,0,0)T and b=(1,−1,−21)T.
The set F:={(y1,y2,y3)T∈R3∣2y1+y2+2y3=0,y1>0,y2≤−y11} is contained in the same hyperplane as the points
(0,0,0)T,(2,0,−2)T und (0,−2,1)T.
It is closed and convex.
Since (a−H)∩F=∅ and (a−intH+tk)∩F=∅∀t>0, (PF,a,H,k) does not have any optimal solution.
But there exist optimal solutions of the problem (PF,b,H,k). These are
the elements of the set
(b−bdH)∩F=(b−H)∩F={(y1,y2,y3)T∈R3∣y1=1,y2=−2λ−1,y3=λ−21,λ∈R+}.
Proposition 7 and Lemma 13 imply for convex cones H:
Corollary 4**.**
Assume that the following conditions hold:
a∈Rℓ, F⊆Rℓ is closed, and there exists some u∈Rℓ with F⊆u+R+ℓ.
H⊂Rℓ* is a nontrivial closed convex pointed cone with R+ℓ⊆H and k∈H∖(−H).*
The problem (PF,a,H,k) has a feasible solution.
Then MF,a,H,k is nonempty and compact.
Without the assumption F⊆u+R+ℓ in Corollary 4, the existence of optimal solutions for (PF,a,H,k) can depend on the choice of a, even if H is the nonnegative orthant.
Example 5**.**
Consider Y=R3, H=R+3, k=(1,1,1)T, a=(0,0,0)T, b=(1,−1,−1)T.
F:={(y1,y2,y3)T∈R3∣y1+y3=0,y2≤−y11,y1>0} is a closed convex set, which is contained in the hyperplane {(y1,y2,y3)T∈R3∣y1+y3=0}.
(a−H)∩F=∅, but (a−intH+tk)∩F=∅∀t>0.
Thus, (PF,a,H,k) does not have any optimal solution.
But (b−bdH)∩F=(b−H)∩F={(y1,y2,y3)T∈R3∣y1=1,y2≤−1,y3=−1}
is the set of minimizers y of (PF,b,H,k).
We now turn to statements in Y=Rℓ without the assumption that F is bounded below.
Proposition 8**.**
Assume that the following conditions are fulfilled:
a∈Rℓ, F⊆Rℓ is a nonempty closed set.
H* is a proper closed subset of Rℓ with k∈core0+H.*
There exist a polyhedral cone C⊂Rℓ, some z∈clconvH and u∈Rℓ such that
[TABLE]
Then MF,a,H,k is nonempty and compact.
Proof.
By Lemma 1(d), φa−H,k is finite-valued. Hence,
(PF,a,H,k) has a feasible solution. ⇒∃t0∈R:B:=F∩(a−H+t0k)=∅.
B⊆F∩(a+t0k−z+(z−clconvH)) is bounded by Lemma 11, hence compact.
By Corollary 1, MB,a,H,k is nonempty and compact.
Theorem 1 yields the assertion.
∎
The assumptions of Proposition 8 do not imply that F is bounded below or that H is a cone .
Example 6**.**
The sets H:={(y1,y2)T∈R2∣y1≥−1,y2≥−1,(y1+1)(y2+1)≥1} and F:={(y1,y2)T∈R2∣y1≥0,y2≥−2y1} fulfill
the conditions in Proposition 8 with C:={(y1,y2)T∈R2∣y1+y2≤0} and u=z=(0,0)T.
For the case that k is not necessarily an element of core0+H, we can prove the following statement.
Proposition 9**.**
Assume that the following conditions are satisfied:
a∈Rℓ, F⊆Rℓ is a nonempty closed set.
H* is a proper closed subset of Rℓ with k∈0+H∖{0}.*
There exist a polyhedral cone C⊂Rℓ, some z∈clconvH and u∈Rℓ such that
[TABLE]
The problem (PF,a,H,k) has a feasible solution.
One of the following assumptions is fulfilled:
F∩(a−H)=∅,
F∩(a−intH)=∅* and H+R>k⊂intH,*
H* does not contain a line in direction k.*
Then MF,a,H,k is nonempty and compact.
Proof.
(PF,a,H,k) has a feasible solution. ⇒∃t0∈R:B:=F∩(a−H+t0k)=∅.
B⊆F∩(a+t0k−z+(z−clconvH)) is bounded by Lemma 11, hence compact.
By Proposition 5, MB,a,H,k is nonempty and compact.
Theorem 1 yields the assertion.
∎
Apply now the idea of the previous proposition to convex sets H.
Proposition 10**.**
Assume that the following conditions are fulfilled:
a∈Rℓ, F⊆Rℓ is a nonempty closed set.
H* is a proper closed convex subset of Rℓ with k∈0+H∖(−0+H).*
There exist a polyhedral cone C⊂Rℓ, some h∈H and u∈Rℓ such that
[TABLE]
The problem (PF,a,H,k) has a feasible solution.
Then MF,a,H,k is nonempty and compact.
Proof.
(PF,a,H,k) has a feasible solution. Then, there exists some t0∈R with B:=F∩(a−H+t0k)=∅.
B=F∩(a−h+t0k+(h−H)) is bounded by Lemma 11, hence compact.
By Proposition 5, MB,a,H,k is nonempty and compact.
Theorem 1 yields the assertion.
∎
Condition (d) in Proposition 10 is fulfilled under the assumptions (a)-(c) if k∈core0+H.
5. Parameter control
Problems (PF,a,H,k) with varying parameters a and k are used in vector optimization procedures.
Proposition 11**.**
*Assume that H is a proper closed convex subset of Y with 0+H={0} and that F is a nonempty subset of Y for which
F+0+H is convex.
Then
{(a,k)∈Y×(0+H∖{0})∣(PF,a,H,k)\mboxhasafeasiblesolution}
is a convex set.*
Proof.
Assume (a1,k1),(a2,k2)∈Y×(0+H∖{0}) such that
(PF,a1,H,k1) and (PF,a2,H,k2) have feasible solutions.
⇒∃y1,y2∈F,t1,t2∈R:y1∈a1−H+t1k1,y2∈a2−H+t2k2.
Take any λ∈(0,1).
λy1+(1−λ)y2∈λa1+(1−λ)a2−H+λt1k1+(1−λ)t2k2
since H is convex.
t:=max(t1,t2).
⇒h:=λ(t−t1)k1+(1−λ)(t−t2)k2∈0+H
since 0+H is a convex cone.
λt1k1+(1−λ)t2k2=t(λk1+(1−λ)k2)−h implies
λy1+(1−λ)y2∈λa1+(1−λ)a2−H+t(λk1+(1−λ)k2).
Since F+0+H is convex, there exist y3∈F and
h1∈0+H such that
λy1+(1−λ)y2=y3+h1.
Hence, y3 is a feasible solution of (PF,λa1+(1−λ)a2,H,λk1+(1−λ)k2).
∎
Note that F+0+H is convex if F is convex.
Lemma 1(e) and Lemma 4 imply restrictions to the set of parameters k which have to be considered.
Proposition 12**.**
Assume (H1–OPF,a,H,k).
If k∈−0+H, then (PF,a,H,k) does not have an optimal solution.
For each λ∈R>, the problem (PF,a,H,λk) has the same feasible vectors y and the same optimal solutions y as (PF,a,H,k).
The proposition underlines that replacing k by another vector in the same direction does not alter the optimal solutions. Hence, it is sufficient to consider only one vector k per direction, e.g., to restrict k to unit vectors if Y is a normed space. In Y=Rℓ, the range for k can also be restricted by Proposition 15(d).
We now investigate whether the set of parameters a can be restricted.
We get from Lemma 5:
Proposition 13**.**
Assume (H1–OPF,a,H,k).
Then MF,a,H,k=MF,a+ck,H,k for each c∈R.
This implies:
Proposition 14**.**
If Λ⊂Y with Y={Λ+βk∣β∈R}, then
[TABLE]
Λ can be chosen as the complementary space of the linear subspace
{βk∣β∈R}. Thus, in Y=Rℓ, the dimension of the parameter set for a can be reduced by one.
Proposition 15**.**
Assume F⊆Rℓ and H is a proper closed subset of Rℓ with 0+H={0}.
For each k∈0+H∖{0} and j∈{1,…,ℓ} with
kj=0, we have
[TABLE]
If k∈0+H∖{0} and k∈intR+ℓ∪(−intR+ℓ), we have
[TABLE]
If k∈0+H∖{0} with i=1∑ℓki=0, then
[TABLE]
For a∈Rℓ being fixed, we get
[TABLE]
Proof.
Apply first Proposition 13.
For c=−kjaj, we get (a+ck)j=0.
Choose n∈{1,…,ℓ} with
knan=i∈{1,…,ℓ}maxkiai,
m∈{1,…,ℓ} withkmam=i∈{1,…,ℓ}minkiai.
If k∈intR+ℓ, we get for c=−knan
a+ck∈−R+ℓ and for c=−kmam
a+ck∈R+ℓ.
If k∈−intR+ℓ, the two vectors have the opposite sign.
For c=−i=1∑ℓkii=1∑ℓai,
i=1∑ℓ(a+ck)i=0 holds.
MF,a,H,k=MF,a,H,λk∀λ∈R> implies with λ:=i=1∑ℓki1
the assertion by Proposition 12.
∎
6. Solution sets for varying parameters
The solution of a problem (PF,a,H,k) can deliver some information about solution sets for problems with altered parameters a and k.
Lemma 15**.**
Assume (H1–OPF,a,H,k) and k∈int0+H. Then:
Each function φb−H,k0 with b∈Y and k0∈int0+H is finite-valued.
If φa−H,k is bounded below on F, then φb−H,k0 is bounded below on F for all b∈Y and k0∈int0+H.
Proof.
results from Lemma 1(d).
φa−H,k is bounded below on F.
⇒∃s∈R∀y∈F:φa−H,k(y)≥s.
Then (2.3) implies (a−intH+sk)∩F=∅.
By Lemma 9, Y=α∈R⋃(int0+H+αk0)=α∈R⋃(a+sk−int0+H−αk0).
⇒∃t∈R:b∈a+sk−int0+H−tk0.
⇒b+tk0∈a+sk−int0+H.
⇒b−intH+tk0⊆a+sk−(intH+int0+H)⊆a+sk−intH.
⇒(b−intH+tk0)∩F=∅. (2.3) implies
φb−H,k0≥t∀y∈F,
i.e., φb−H,k0 is bounded below on F.
∎
In Lemma 15, it is essential that k and k0 belong to the interior of 0+H.
Example 7**.**
H={(y1,y2)T∈R2∣y2≥y12}* fulfills the assumptions of Lemma 15 without the condition that the interior of 0+H is nonempty.
k=(0,1)T∈0+H.
Consider F={(y1,y2)T∈R2∣y1≥0,y2=−y12}∪{(−1,0)T}.
Then F∩−intH=∅, hence,
φ−H,k(y)≥0∀y∈F,
i.e., φ−H,k is bounded below on F.
Let t be an arbitrary negative number and b=(1,0)T.
y1:=43−2t>0, y2:=−y12. Then y∈F.
yˉ:=y−b−tk=(−2t−41,−4t2−4t−169)T∈−intH since yˉ2<−yˉ12.
Hence, y∈b−intH+tk and φb−H,k(y)<t.
Since t can be arbitrarily small, φb−H,k is not bounded below on F.*
Lemma 15 implies by Theorem 1:
Proposition 16**.**
Assume (H1–OPF,a,H,k) and k∈int0+H. Then:
F* is the feasible range of each problem (PF,b,H,k0) with b∈Y, k0∈int0+H.*
If the objective function of (PF,a,H,k) is not bounded below on F, then none of the problems (PF,b,H,k0) with b∈Y and k0∈int0+H has an optimal solution.
Proposition 17**.**
Assume F⊆Rℓ is a nonempty closed convex set, a∈Rℓ, H is a proper closed convex subset of Rℓ, and k∈0+H∖(−0+H).
If MF,a,H,k=∅ and F∩(a−H+Rk)=∅ or if MF,a,H,k is not bounded, then, for each b∈Rℓ and k0∈0+H∖(−0+H), MF,b,H,k0=∅ or
0+MF,b,H,k0={0}.
If MF,a,H,k is nonempty and bounded, then, for each b∈Rℓ and k0∈0+H∖(−0+H) with F∩(b−H+Rk0)=∅, MF,b,H,k0 is nonempty and compact.
Proof.
By Lemma 1(f), each function φb−H,k0 with b∈Y and k0∈0+H∖(−0+H) is proper.
Consider first the case that MF,a,H,k=∅ and that
there exists some t∈R with F∩(a+tk−H)=∅.
F∩(a+tk−H) cannot be compact, since otherwise φa−H,k would attain a minimum on F∩(a+tk−H) und thus on F∩domφa−H,k. Since F∩(a+tk−H) is closed, it is not bounded.
Consider now the case that MF,a,H,k is not bounded.
Let t be the minimum of φa−H,k on F.
⇒F∩(a+tk−H) is not bounded.
For b∈Rℓ, k0∈0+H∖(−0+H), if the minimum tˉ of φb−H,k0 on F exists, then
MF,b,H,k0=F∩(b+tˉk0−H), and 0+MF,b,H,k0={0} by Lemma 10.
follows from (a).
∎
Note that the condition F∩(a−H+Rk)=∅ is equivalent to the existence of feasible solutions of problem (PF,a,H,k).
Proposition 18**.**
Assume F⊆Rℓ is a nonempty closed convex set, a∈Rℓ, H is a proper closed convex subset of Rℓ
and k∈int0+H.
If MF,a,H,k=∅ or if MF,a,H,k is not bounded, then, for each b∈Rℓ and k0∈0+H∖(−0+H), MF,b,H,k0=∅ or
0+MF,b,H,k0={0}.
If MF,a,H,k is nonempty and bounded, then, for each b∈Rℓ and k0∈int0+H, MF,b,H,k0 is nonempty and compact.
Proof.
By Lemma 1(d), each function φb−H,k0 with b∈Y and k0∈int0+H is finite-valued. Apply Proposition 17.
∎
If (PF,a,H,k) in Proposition 18(b) has a unique minimal solution, then this is not necessarily the case for the problems (PF,b,H,k0) considered.
Example 8**.**
Choose F=H=R+2, a=(0,0)T, b=(1,0)T,
k=(1,1)T.
Then F∩(a−H)=(0,0)T and
F∩(b−H)=F∩(b−bdH)={(y1,y2)T∈R2∣0≤y1≤1,y2=0}.
Thus, φa−H,k attains its minimum on F only at (0,0)T, whereas
each element of F∩(b−H) is a minimizer of φb−H,k on F.
Further sensitivity results for the studied problems (PF,a,H,k) in Y=Rℓ were proved by Pascoletti and Serafini [11] for polyhedral cones H, later by Helbig for closed convex cones H [14] as well as for closed convex sets [15] and by Eichfelder [16] for closed convex pointed cones H.
The way in which the solution set of problems (PF,a,H,k) depends on alternations in a and k had also been studied for convex cones H in more general spaces Y by Sterna-Karwat in [17] and [18].
In [16], further references related to sensitivity are given.