Vector optimization problems with weakened convex and weakened affine constraints in linear topological spaces

Renying Zeng

doi:10.1515/math-2024-0073

Article Open Access

Vector optimization problems with weakened convex and weakened affine constraints in linear topological spaces

Renying Zeng

Published/Copyright: October 17, 2024

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$Open Mathematics$

From the journal Open Mathematics Volume 22 Issue 1

Abstract

In this article, we work on vector optimization problems in linear topological spaces. Our vector optimization problems have weakened convex inequality constraints and weakened affine equality constraints. Our inequalities are given by partial orders that are induced by pointed convex cones. We prove a Farkas–Minkowski-type theorem of alternative and obtain some optimality conditions through the discussions of vector saddle points and scalar saddle points.

Keywords: topological vector spaces; optimization problem; theorem of alternative; vector saddle point; scalar saddle point

MSC 2010: 90C26; 90C30; 90C29; 90C48

1 Introduction and preliminary

There were many discussions about constraint optimization. For optimization problems that have equality constraints, inequality constraints, and/or abstract constraints, authors usually required that the equality constraints are affine-linear, e.g., see [1], locally Lipschitz [2,3,4], lower semi-continuous [5], or continuous differentiable [6,7]. In this article, we assume that our inequality constraints are generalized convex functions, and our equality constraints are generalized affine functions. Our inequalities are given by partial orders that are induced by pointed convex cones. We obtain a Fakas alternative theorem and some vector saddle point theorems as well as some scalar saddle point theorems.

Let Y be a linear topological space. A subset Y ₊ of Y is a cone if λ y ∈ Y + for ∀ y ∈ Y + and ∀ λ ≥ 0 . We denote by 0 Y the zero element in the topological vector space Y and simply by 0 if there is no confusion. A convex cone is one for which λ 1 y 1 + λ 2 y 2 ∈ Y + for ∀ y 1 , y 2 ∈ Y + and ∀ λ 1 , λ 2 ≥ 0 . A cone Y ₊ of Y is a pointed one if Y + ∩ ( − Y + ) = { 0 } .

A functional on the vector space Y is a real-valued continuous linear function on Y. The set Y ^* of all functionals on Y is called the dual space of Y. The dual cone Y + ⁎ of Y ⁎ is defined as follows:

Y + ⁎ = { ξ ∈ Y ⁎ : ξ ( y ) ≥ 0 , ∀ y ∈ Y + } .

Let Y be a linear topological space with pointed convex cone Y ₊. We denote the partial order induced by Y ₊ as follows:

y 1 ≻ y 2 iff y 1 − y 2 ∈ Y + , or , y 1 ≺ y 2 iff y 1 − y 2 ∈ − Y + , y 1 ≻ ≻ y 2 iff y 1 − y 2 ∈ int Y + , or , y 1 ≺ ≺ y 2 iff y 1 − y 2 ∈ − int Y + ,

where int Y ₊ denotes the topological interior of the set Y ₊.

A function f : X → Y is said to be a linear function on D if

f ( α x 1 + β x 2 ) = α f ( x 1 ) + β f ( x 2 )

for ∀ x 1 , x 2 ∈ D ⊆ X , ∀ α , β ∈ R (the set of all real numbers; f: X → Y is said to be an affine function on D if

f ( α x 1 + ( 1 − α ) x 2 ) = α f ( x 1 ) + ( 1 − α ) f ( x 2 )

for ∀ x 1 , x 2 ∈ D ⊆ X , ∀ α ∈ R ; and f is said to be Y ₊-convex on D if

α f ( x 1 ) + ( 1 − α ) f ( x 2 ) ≺ f ( α x 1 + ( 1 − α ) x 2 )

for ∀ x 1 , x 2 ∈ D ⊆ X , ∀ α ∈ ( 0 , 1 ) , where the inequality is induced by the pointed convex cone Y + of Y.

It is known that f is a linear function if and only if it is in the form of f ( x ) = a x ; f is an affine function if and only if it is in the form of f ( x ) = a x + b , where a and b are scalars.

We define the following concept of affine cone.

Definition 1.1. If there exists a pointed convex cone Y + such that Y + = Y + ∪ ( − Y + ) , then Y + is said to be an affine cone.

By use of the definition of affine cones, we introduce the definition of generalized affine function as follows.

Definition 1.2. Let M be an affine cone of Y. A function f : X → Y is said to be generalized M-affine on D ⊆ X , ∀ x 1 , x 2 ∈ D , ∀ α ∈ R , ∃ x 3 ∈ D , ∃ v ∈ M , ∃ t ≠ 0 such that

v + α f ( x 1 ) + ( 1 − α ) f ( x 2 ) = t f ( x 3 ) ,

where t ≠ 0 is a scalar.

Example 1.1 shows that the generalized affineness does not implies affineness.

Example 1.1. Given the function f ( x ) = x 2 + 1 , x ∈ R (the set of all real numbers), and take M = R .

A function is an affine function if and only if it is in the form of f ( x ) = a x + b . Therefore, f ( x ) = x 2 + 1 , x ∈ R is not an affine function.

But f is a generalized M-affine function. ∀ v ∈ M , ∀ x 1 , x 2 ∈ R , ∀ α ∈ R , taking t = 1 if v + α f ( x 1 ) + ( 1 − α ) f ( x 2 ) ≥ 0 , t = − 1 if v + α f ( x 1 ) + ( 1 − α ) f ( x 2 ) < 0 , then

v + α f ( x 1 ) + ( 1 − α ) f ( x 2 ) = t f ( x 3 ) ,

where x 3 = ∣ v + α f ( x 1 ) + ( 1 − α ) f ( x 2 ) ∣ 1 / 2 .

The following example shows that there does exist non generalized affine functions, and so the definition of generalized affineness is not trivial.

Example 1.2. Given the function f ( x , y , z ) = ( x 2 , y 2 , z 2 ) , x , y , z ∈ R , and assume that M = { ( x , y , 0 ) : x , y ∈ R } .

Take α = − 4 , ( x 1 , y 1 , z 1 ) = ( 0 , 0 , 1 ) , ( x 2 , y 2 , z 2 ) = ( 1 , 1 , 0 ) , then

α f ( x 1 , y 1 , z 1 ) + ( 1 − α ) f ( x 2 , y 2 , z 2 ) = ( 5 , 5 , − 4 ) .

Therefore, ∀ v = ( x , y , 0 ) ∈ M , one has

v + α f ( x 1 , y 1 , z 1 ) + ( 1 − α ) f ( x 2 , y 2 , z 2 ) = ( x + 5 , y + 5 , − 4 ) .

Due to the fact that f ( x 3 , y 3 , z 3 ) = ( x 3 2 , y 3 2 , z 3 2 ) ≻ 0 , ∀ x 3 , y 3 , z 3 ∈ R and x + 5 > 0 , y + 5 > 0 , one obtains

v + α f ( x 1 , y 1 , z 1 ) + ( 1 − α ) f ( x 2 , y 2 , z 2 ) = ( x + 5 , y + 5 , − 4 ) ≠ t f ( x 3 , y 3 , z 3 ) , ∀ t ≠ 0 .

Therefore, f ( x , y , z ) = ( x 2 , y 2 , z 2 ) , x , y , z ∈ R is not a generalized M-affine function.

2 Alternative theorems

Definition 2.1. A function f : X → Y is said to be generalized Y ₊ -convex on D if ∀ u ∈ int Y + , ∀ x 1 , x 2 ∈ D , ∀ α ∈ [ 0 , 1 ] , ∃ x 3 ∈ D , ∃ τ > 0 such that

u + α f ( x 1 ) + ( 1 − α ) f ( x 2 ) ≺ τ f ( x 3 ) ,

where τ > 0 is a scalar.

Theorem 2.1

(Alternative theorem) Suppose that X, Y, Z, and W are real topological linear spaces, D ⊆ X . Y + , Z + are pointed convex cones of Y and Z, respectively, and M is an affine cone of W. Assume that functions f : D → Y , g : D → Z , h : D → W satisfy that

f and g are generalized convex functions on D, i.e., ∀ u 1 ∈ int Y + , ∀ u 2 ∈ int Z + , ∀ α ∈ ( 0 , 1 ) , ∀ x 1 , x 2 ∈ D , ∃ τ 1 , τ 2 > 0 , ∃ x ′ , x ″ ∈ D such that
u 1 + α f ( x 1 ) + ( 1 − α ) f ( x 2 ) ≺ τ 1 f ( x ′ ) , u 2 + α g ( x 1 ) + ( 1 − α ) g ( x 2 ) ≺ τ 2 g ( x ″ ) ;
1. h is a generalized affine function on D, i.e., ∀ α ∈ ( − ∞ , + ∞ ) , ∀ x 1 , x 2 ∈ D , ∃ t ≠ 0 , ∃ x ' ' ' ∈ D , ∃ v ∈ M such that
v + α h ( x 1 ) + ( 1 − α ) h ( x 2 ) = t h ( x ' ' ' ) ;
int h ( D ) ≠ ∅ ;

and let the items (i) and (ii) be defined as follows:

(i) ∃ x ∈ D , s . t . , f ( x ) ≺ ≺ 0 , g ( x ) ≺ 0 , h ( x ) = 0 ;

(ii) ∃ ( ξ , η , ζ ) ∈ ( Y + ⁎ × Z + ⁎ × W ⁎ ) \ { ( 0 Y , 0 Z , 0 W ) } such that

ξ ( f ( x ) ) + η ( g ( x ) ) + ς ( h ( x ) ) ≥ 0 , ∀ x ∈ D .

If (i) has no solution, then (ii) has solutions.

Moreover, if (ii) has a solution ( ξ , η , ς ) with ξ ≠ 0 Y ⁎ , then (i) has no solutions.

Proof. ∀ z 1 , z 2 ∈ ⋃ t ≠ 0 t h ( D ) + M , ∀ α ∈ R , ∃ x 1 , x 2 ∈ D , ∃ b 1 , b 2 ∈ M , ∃ t 1 , t 2 ≠ 0 such that

α y 1 + ( 1 − α ) y 2 = α t 1 h ( x 1 ) + ( 1 − α ) t 2 h ( x 2 ) + α b 1 + ( 1 − α ) b 2 = ( α t 1 + ( 1 − α ) t 2 ) [ α t 1 α t 1 + ( 1 − α ) t 2 h ( x 1 ) + ( 1 − α ) t 2 α t 1 + ( 1 − α ) t 2 h ( x 2 ) ] + α b 1 + ( 1 − α ) b 2 .

By the assumption (a), ∃ x 3 ∈ D , ∃ t 3 > 0 , ∃ v ∈ M , such that

α t 1 α t 1 + ( 1 − α ) t 2 h ( x 1 ) + ( 1 − α ) t 2 α t 1 + ( 1 − α ) t 2 h ( x 2 ) = t 3 h ( x 3 ) − v .

Therefore,

α z 1 + ( 1 − α ) z 2 = α t 1 h ( x 1 ) + ( 1 − α ) t 2 h ( x 2 ) + α b 1 + ( 1 − α ) b 2 = ( α t 1 + ( 1 − α ) t 2 ) α t 1 α t 1 + ( 1 − α ) t 2 h ( x 1 ) + ( 1 − α ) t 2 α t 1 + ( 1 − α ) t 2 h ( x 2 ) + α b 1 + ( 1 − α ) b 2 = ( α t 1 + ( 1 − α ) t 2 ) t 3 h ( x 3 ) + α b 1 + ( 1 − α ) b 2 − ( α t 1 + ( 1 − α ) t 2 ) v ⊆ ∪ t ≠ 0 t h ( D ) + M .

So, ⋃ t ≠ 0 t h ( D ) + M is a convex set.

Similarly, ∪ t > 0 ( t f ( D ) + int Y + and ∪ t > 0 t g ( D ) + Z + are convex as well. Therefore, the set B = ( ∪ t > 0 ( t f ( D ) + int Y + ) × ( ∪ t > 0 ( t g ( D ) + Z + ) × ( ∪ t ≠ 0 t h ( D ) + M ) is convex.

From the assumption (c), int B ≠ ∅ . We also have ( 0 Y , 0 Z , 0 W ) ∉ B since (i) has no solution. Therefore, according to the separation theorem of convex sets of topological linear space, there exists a nonzero vector ( ξ , η , ς ) ∈ Y ⁎ × Z ⁎ × W ⁎ such that

ξ ( τ 1 f ( x ) + y 0 ) + η ( τ 2 g ( x ) + z 0 ) + ς ( t h ( x ) + w 0 ) ≥ 0 ,

for all ∀ x ∈ D , ∀ y 0 ∈ int Y + , ∀ z 0 ∈ Z + , ∀ w 0 ∈ M , ∀ τ 1 , τ 2 > 0 , ∀ t ≠ 0 ,

Since int Y ₊ and int Z ₊ are convex cones, one obtains

ξ ( τ 1 f ( x ) + λ 1 y 0 ) + η ( τ 2 g ( x ) + λ 2 z 0 ) + ς ( t h ( x ) + k w 0 ) ≥ 0 ,

∀ x ∈ D , ∀ y 0 ∈ int Y + , ∀ z 0 ∈ int Z , + ∀ w 0 ∈ M , ∀ τ i > 0 , ∀ λ i > 0 , ( i = 1 , 2 ) , ∀ t ≠ 0 , ∀ k ∈ R .

Let τ i → 0 ( i = 1 , 2 ) , λ 2 → 0 , t → 0 , k → 0 , one has

ξ ( y 0 ) ≥ 0 , ∀ y 0 ∈ int Y + .

Therefore, ξ ( y ) ≥ 0 , ∀ y ∈ Y + . Hence, ξ ∈ Y + ⁎ .

Similarly, let τ i → 0 ( i = 1 , 2 ) , λ 1 → 0 , t → 0 , k → 0 , we obtain η ∈ Z + ⁎ . Thus,

( ξ , η , ς ) ∈ Y + ⁎ × Z + ⁎ × W ⁎ .

Letting λ i → 0 ( i = 1 , 2 ) , k → 0 , one has

ξ ( f ( x ) ) + η ( g ( x ) ) + ς ( h ( x ) ) ≥ 0 , x ∈ D .

This means that (ii) has solutions.

On the other hand, suppose that (ii) has a solution ( ξ , η , ς ) with ξ ≠ 0 Y + ⁎ , i.e.,

ξ ( f ( x ) ) + η ( g ( x ) ) + ς ( h ( x ) ) ≥ 0 , x ∈ D .

We are going to prove that (i) has no solution.

Otherwise, if (i) has a solution x ˜ ∈ D , then f ( x ˜ ) ≺ ≺ 0 , g ( x ˜ ) ≤ 0 , h ( x ˜ ) = 0 .

Therefore, we would have

ξ ( f ( x ˜ ) ) + η ( g ( x ˜ ) ) + ς ( h ( x ˜ ) ) < 0 ,

which is a contradiction. The proof is completed.

For a finite dimensional space, we may assume that its pointed convex cone is composed by all elements whose components are nonnegative real numbers. Then one has the following Theorem 2.2.

Theorem 2.2

Let Y, Z, and W are finite dimensional. Assume that the functions f : X → Y , g : X → Z , g : X → W satisfy the conditions in Theorem 2.1, and (i) and (ii) denote the systems

(i) ∃ x ∈ D , s . t . , f ( x ) ≺ ≺ 0 , g ( x ) ≺ 0 , h ( x ) = 0 ;

(ii) ∃ ( ξ , η , ζ ) ∈ ( Y + ⁎ × Z + ⁎ × W ⁎ ) \ { ( 0 Y , 0 Z , 0 W ) } , such that

〈 f ( x ) , ξ 〉 + 〈 g ( x ) , η 〉 + 〈 h ( x ) , ς 〉 ≥ 0 , ∀ x ∈ D .

If (i) has no solution, then (ii) has solutions. Moreover, if (ii) has a solution ( ξ , η , ς ) with ξ ≠ 0 Y ⁎ , then (i) has no solutions.

3 Vector saddle point theorems

Consider the following vector optimization problem:

( V P ) Y + − min f ( x ) , g ( x ) ≺ 0 , h ( x ) = 0 , x ∈ D ,

where f : X → Y , g : X → Z , h : X → W , Y ₊ is a pointed convex cones in Y, and D is a nonempty subset of X.

From now on, we assume that f, g, and h satisfy the conditions in Theorem 2.1.

Let F be the feasible set of (VP), i.e.,

F ≔ { x ∈ D : g ( x ) ≺ 0 , h ( x ) = 0 } .

Definition 3.1. A point x ¯ ∈ F is said to be a weakly efficient solution of (VP) if there exists no x ∈ D satisfying f ( x ¯ ) ≻ ≻ f ( x ) .

Let

P min [ A , Y + ] = { y ∈ A : ( y − A ) ∩ int Y + = ∅ } ,

P max [ A , Y + ] = { y ∈ A : ( A − y ) ∩ int Y + = ∅ } .

In the sequel, B ( W , Y ) denotes the set of all continuous linear mappings T from W to Y; B + ( Z , Y ) denotes the set of all nonnegative and continuous linear mappings S from Z to Y, where nonnegative mapping S means that S ( z ) ∈ Y + , ∀ z ∈ Z .

Definition 3.2. A triple ( x ¯ , S ¯ , T ¯ ) ∈ X × B + ( Z , Y ) × B ( W , Y ) is said to be a vector saddle point of L ( x ¯ , S ¯ , T ¯ ) if

L ( x ¯ , S ¯ , T ¯ ) ∈ P min [ L ( X , S ¯ , T ¯ ) , Y + ] ∩ P max [ L ( x ¯ , B + ( Z , Y ) , B ( W , Y ) ) , Y + ] .

where

L ( x ¯ , S ¯ , T ¯ ) = f ( x ¯ ) + S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ) ,

and

P max [ L ( x ¯ , B + ( Z , Y ) , B ( W , Y ) ) , Y + ] = { μ : μ = P max [ L ( x ¯ , S , T ) , Y + ] , ( S , T ) ∈ B + ( Z , Y ) × B ( W , Y ) } .

Theorem 3.1

A triple ( x ¯ , S ¯ , T ¯ ) ∈ X × B + ( Z , Y ) × B ( W , Y ) is a vector saddle point of L ( x ¯ , S ¯ , T ¯ ) if and only if

(i) f ( x ¯ ) ∈ P min [ L ( X , S ¯ , T ¯ ) , Y + ] ,

(ii) g ( x ¯ ) ≺ 0 , h ( x ¯ ) = 0 ,

where L ( X , S ¯ , T ¯ ) = { σ : σ = L ( x ¯ , S ¯ , T ¯ ) , x ¯ ∈ X } .

Proof. The necessity.

Assume that ( x ¯ , S ¯ , T ¯ ) ∈ X × B + ( Z , Y ) × B ( W , Y ) is a vector saddle point of L ( x ¯ , S ¯ , T ¯ ) , i.e.,

L ( x ¯ , S ¯ , T ¯ ) ∈ P min [ L ( X , S ¯ , T ¯ ) , Y + ] ∩ P max [ L ( x ¯ , B + ( Z , Y ) × B ( W , Y ) ) , Y + ] .

So,

{ S ( g ( x ¯ ) ) + T ( h ( x ¯ ) ) − [ S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ) ] } ∩ int Y + = ∅ ∀ ( S , T ) ∈ B + ( Z , Y ) × B ( W , Y ) ,

and ∃ x ∈ X such that

{ f ( x ¯ ) + S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ) − [ f ( x ) + S ¯ ( g ( x ) ) + T ¯ ( h ( x ) ) ] } ∩ int Y + = ∅ .

Let T = T ¯ , then

S ( g ( x ¯ ) ) − S ¯ ( g ( x ¯ ) ) ∉ int Y + , ∀ S ∈ B + ( Z , Y ) .

Let S = S ¯ , then

T ( h ( x ¯ ) ) − T ¯ ( h ( x ¯ ) ) ∉ int Y + , ∀ T ∈ B ( W , Y ) .

Aim to show that g ( x ¯ ) ≺ 0 .

Otherwise, we would have − g ( x ¯ ) ≠ 0 .

By the separate theorem ∃ η ∈ Z ⁎ \ { 0 } ,

η ( t z ) > η ( − g ( x ¯ ) ) , ∀ z ∈ Z + , ∀ t > 0 .

i.e.,

η ( z ) > 1 t η ( − g ( x ¯ ) ) , ∀ z ∈ Z + , ∀ t > 0 .

Let t → ∞ and we obtain η ( z ) ≥ 0 , ∀ z ∈ Z + . This means that η ∈ Z + ⁎ \ { 0 } . Therefore, η ( g ( x ¯ ) ) > 0 . Given z ˜ ∈ int Z + and let

S ( z ) = η ( z ) η ( ( g ( x ¯ ) ) z ˜ + S ¯ ( z ) .

Then S ¯ ∈ B + ( Z , Y ) and

S ( g ( x ¯ ) ) − S ¯ ( g ( x ¯ ) ) = z ˜ ∈ int Y + .

This implies that − z ¯ ∈ Z + , i.e.,

g ( x ¯ ) ≺ 0 .

Write h ( x ¯ ) = w ¯ . By the separation theorem, ∃ ς ∈ W ⁎ such that

ς ( w ) < ς ( w ¯ ) , ∀ w ∈ W + .

If h ( x ¯ ) = w ¯ ≠ 0 , then ς ( w ¯ ) ≠ 0 since 0 ∈ W + . Taking y 0 ∈ int Y + and define T 0 ∈ B + ( W , Y ) by

T 0 ( w ) = ς ( w ) ς ( w ¯ ) y 0 + T ¯ ( w ) .

Then

T 0 ( w ¯ ) − T ¯ ( w ¯ ) = y 0 ∈ int Y + .

This is contradicting to T ( h ( x ¯ ) ) − T ¯ ( h ( x ¯ ) ) ∉ int Y + , ∀ T ∈ B ( W , Y ) . Therefore,

w ¯ = h ( x ¯ ) = 0 .

The sufficiency.

Suppose that the conditions (i) and (ii) are satisfied.

First, g ( x ¯ ) ≺ 0 and h ( x ¯ ) = 0 imply that

S ( g ( x ¯ ) ) ≺ 0 , T ( h ( x ¯ ) ) = 0 , ∀ ( S , T ) ∈ B + ( Z , Y ) × B ( W , Y ) .

Condition (i) states that

T ( h ( x ¯ ) ) − T ¯ ( h ( x ¯ ) ) ∉ int Y + , ∀ T ∈ B ( W , Y ) .

So, Y + + int Y + ⊆ Y + and S ( h ( x ¯ ) ) ⊆ Y + together imply that

{ f ( x ¯ ) + S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ) − [ f ( X ) + S ¯ ( g ( X ) ) + T ¯ ( h ( X ) ) ] } ∩ int Y + = ∅ .

Hence,

f ( x ¯ ) + S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ) ∈ P min [ L ( X , S ¯ , T ¯ ) , Y + ] .

On the other hand, from S ( z ¯ ) ∩ int Y + = ∅ and int Y + + Y + ⊆ int Y + , we conclude that

{ ⋃ ( S , T ) ∈ B + ( Z , Y ) × B ( W , Y ) [ f ( x ¯ ) + S ( g ( x ¯ ) ) + T ( h ( x ¯ ) ) ] − [ y ¯ + S ¯ ( z ¯ ) + T ¯ ( w ¯ ) ] } ∩ int Y + = ∅ .

Hence,

f ( x ¯ ) + S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ) ∈ P max [ L ( x ¯ , B + ( Z , Y ) , B ( W , Y ) ) , Y + ] .

Consequently,

L ( x ¯ , S ¯ , T ¯ ) ∩ P min [ L ( X , S ¯ , T ¯ ) , Y + ] ∩ P max [ L ( x ¯ , B + ( Z , Y ) , B ( W , Y ) ) , Y + ] ≠ ∅ .

Therefore, ( x ¯ , S ¯ , T ¯ ) ∈ X × B + ( Z , Y ) × B ( W , Y ) is a vector saddle point of L ( x ¯ , S ¯ , T ¯ ) .

Theorem 3.2

If ( x ¯ , S ¯ , T ¯ ) ∈ X × B + ( Z , Y ) × B ( W , Y ) is a vector saddle point of L ( x ¯ , S ¯ , T ¯ ) , and if S ¯ ( g ( x ¯ ) ) = 0 , then x ¯ is a weak efficient solution of (VP).

Proof. Assume that ( x ¯ , S ¯ , T ¯ ) ∈ D × B + ( Z , Y ) × B ( W , Y ) is a vector saddle point of L ( x ¯ , S ¯ , T ¯ ) . From Theorem 3.1, we have

S ( g ( x ¯ ) ) ≺ 0 , h ( x ¯ ) = 0 .

So x ¯ ∈ D (the feasible solution of (VP)). And

f ( x ¯ ) ∈ P min [ L ( X , S ¯ , T ¯ ) , Y + ] ,

i.e.,

( f ( x ¯ ) − [ f ( X ) + S ¯ ( g ( X ) ) + T ¯ ( h ( X ) ] ) ∩ int Y + = ∅ .

Thus,

( f ( x ¯ ) − [ f ( D ) + S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ] ) ∩ int Y + = ∅ .

Therefore,

( f ( x ¯ ) − f ( D ) ) ∩ int Y + = ∅ ,

which means x ¯ is a weakly efficient solution of (VP).

Theorem 3.3

If x ¯ ∈ D is a weakly efficient solution of (VP), then ∃ ( S ¯ , T ¯ ) ∈ B + ( Z , Y ) × B ( W , Y ) such that ( x ¯ , S ¯ , T ¯ ) ∈ X × B + ( Z , Y ) × B ( W , Y ) is a vector saddle point of L ( x ¯ , S ¯ , T ¯ ) .

Proof. Assume that x ¯ ∈ D is a weakly efficient solution of (VP), then there is no x ∈ X such that

f ( x ) − f ( x ¯ ) ≺ ≺ 0 , g ( x ) ≺ 0 , h ( x ) = 0 .

By Theorem 2.1, ∃ ( ξ , η , ς ) ∈ Y + ⁎ × Z + ⁎ × W ⁎ \ { 0 } such that

ξ ( f ( x ) − f ( x ¯ ) ) + η ( g ( x ) ) + ς ( h ( x ) ) ≥ 0 , ∀ x ∈ D .

Take x = x ¯ from the aforementioned equation, and we obtain η ( g ( x ¯ ) ) ≥ 0 . But x ¯ ∈ D and η ∈ Z + ⁎ imply that η ( g ( x ¯ ) ) ≤ 0 . Therefore,

η ( g ( x ¯ ) ) = 0 .

And so,

ξ ( f ( x ) − f ( x ¯ ) ) ≥ 0 , ∀ x ∈ D .

Here, ξ ≠ 0 . So we may take y 0 ∈ int Y + such that ξ ( y 0 ) = 1 . Define the operator S : Z → Y and T : W → Y by

S ( z ) = η ( z ) y 0 , T ( w ) = ς ( w ) y 0 .

It is easy to see that

S ∈ B + ( Z , Y ) , S ( Z + ) = η ( Z + ) y 0 ⊆ Y + , T ∈ B ( W , Y ) .

Hence,

S ( g ( x ¯ ) ) = η ( g ( x ¯ ) ) y 0 ∈ 0 ⋅ Y + = 0 .

Therefore,

f ( x ¯ ) = f ( x ¯ ) + S ( g ( x ¯ ) ) + T ( h ( x ¯ ) ) .

Then

ξ [ f ( x ) + S ( g ( x ) ) + T ( h ( x ) ) ] = ξ ( f ( x ) ) + η ( ( g ( x ) ) ξ ( y 0 ) + ς ( h ( x ) ) ξ ( y 0 ) = ξ ( f ( x ) ) + η ( g ( x ) ) + ς ( h ( x ) ) ≥ ξ ( f ( x ¯ ) ) , ∀ x ∈ D ,

i.e.,

ξ [ f ( x ) − f ( x ¯ ) ) + S ( g ( x ) ) + T ( h ( x ) ) ] ≥ 0 , ∀ x ∈ D .

Taking F ( x ) = f ( x ) + S ( g ( x ) ) + T ( h ( x ) ) , G ( x ) = 0 , and H ( x ) = 0 , and applying Theorem 2.1 to the functions F ( x ) − y ¯ , G ( x ) , H ( x ) , then one has

( f ( x ¯ ) − [ f ( D ) + S ( g ( D ) ) + T ( h ( D ) ) ] ∩ int Y + = ∅ .

Hence,

f ( x ¯ ) = f ( x ¯ ) + S ( g ( x ¯ ) ) + T ( h ( x ¯ ) ) ∈ P min [ L ( X , S ¯ , T ¯ ) , Y + ] .

Similarly, together η ( g ( x ¯ ) ) = 0 and h ( x ¯ ) = 0 deduce that

f ( x ¯ ) = f ( x ¯ ) + S ( g ( x ¯ ) ) + T ( h ( x ¯ ) ) ∈ P max [ L ( x ¯ , B + ( Z , Y ) , B ( W , Y ) ) , Y + ] .

Therefore, ( x ¯ , S ¯ , T ¯ ) ∈ X × B + ( Z , Y ) × B ( W , Y ) is a vector saddle point of L ( x ¯ , S ¯ , T ¯ ) .

4 Scalar saddle point theorems

Definition 4.1. Given ξ ¯ ∈ Y + ⁎ \ { 0 } . The real-valued Lagrangian function of (VP) l ξ ¯ ( x , η , ς ) : X × Z + ⁎ × W ⁎ → R is defined by

l ξ ¯ ( x , η , ς ) = ξ ¯ ( f ( x ) ) + η ( g ( x ) ) + ς ( h ( x ) ) .

Definition 4.2. Given ξ ¯ ∈ Y + ⁎ \ { 0 } . A triple ( x ¯ , η ¯ , ς ¯ ) is said to be a scalar saddle point of the Lagrangian function l ξ ¯ ( x , η , ς ) , if

l ξ ¯ ( x ¯ , η , ς ) ≤ l ξ ¯ ( x ¯ , η ¯ , ς ¯ ) ≤ l ξ ¯ ( x , η ¯ , ς ¯ ) , ∀ x ∈ D , ∀ ( η , ς ) ∈ Z + ⁎ × W ⁎ .

Theorem 4.1

If x ¯ ∈ D is a weakly efficient solution of (VP), then ∃ ( ξ ¯ , η ¯ , ς ¯ ) ∈ ( Y + ⁎ \ { 0 } ) × Z + ⁎ × W ⁎ such that ( x ¯ , η ¯ , ς ¯ ) is a scalar saddle point of the Lagrangian function l ξ ¯ ( x , η , ς ) .

Proof. Suppose that x ¯ ∈ D is a weakly efficient solution of (VP). Similar to the proof of Theorem 3.3, ∃ ( S ¯ , T ¯ ) ∈ B + ( Z , Y ) × B ( W , Y ) such that

S ¯ ( g ( x ¯ ) ) = 0 ,

and noting that g ( x ¯ ) ≺ 0 , h ( x ¯ ) = 0 one has

ξ ¯ ( f ( x ¯ ) ) + S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ) ) ≤ ξ ¯ ( f ( x ¯ ) ) ≤ ξ ¯ [ f ( x ) ) + S ¯ ( g ( x ) ) + T ¯ ( h ( x ) ) ] = ξ ¯ ( f ( x ) ) + ξ ¯ ∘ S ¯ ( g ( x ) ) + ξ ¯ ∘ T ¯ ( h ( x ) ) , ∀ x ∈ D .

Take η ¯ = ξ ¯ ∘ S ¯ , ς ¯ = ξ ¯ ∘ T ¯ , then ( η ¯ , ς ¯ ) ∈ Z + ⁎ × W ⁎ . And so

l ξ ¯ ( x ¯ , η ¯ , ς ¯ ) ≤ ξ ¯ ( f ( x ¯ ) ) + η ¯ ( g ( x ¯ ) ) + ς ¯ ( h ( x ¯ ) ) = ξ ¯ [ f ( x ¯ ) + S ¯ ( g ( x ¯ ) ) + T ¯ ( h ( x ¯ ) ] ≤ ξ ¯ [ f ( x ) + S ¯ ( g ( x ) ) + T ¯ ( h ( x ) ] = ξ ¯ ( f ( x ) ) + η ¯ ( g ( x ) ) + ς ¯ ( h ( x ) ) = l ξ ¯ ( x , η ¯ , ς ¯ ) , ∀ x ∈ D .

And, similar to the proof of Theorem 3.3

η ¯ ( g ( x ¯ ) ) = 0 .

Hence,

l ξ ¯ ( x ¯ , η , ς ) = ξ ¯ ( f ( x ¯ ) ) + η ( g ( x ¯ ) ) + ς ( h ( x ¯ ) ) ≤ ξ ¯ ( f ( x ¯ ) ) + η ¯ ( g ( x ¯ ) ) + ς ¯ ( h ( x ¯ ) ) = l ξ ¯ ( x ¯ , , η ¯ , ς ¯ ) , ∀ ( η , ς ) ∈ Y + ⁎ × W ⁎ .

Therefore, ( x ¯ , η ¯ , ς ¯ ) is a scalar saddle point of the Lagrangian function l ξ ¯ ( x , η , ς ) .

Theorem 4.2

Let x ¯ ∈ D . If ∃ ( ξ ¯ , η ¯ , ς ¯ ) ∈ ( Y + ⁎ \ { 0 } ) × Z + ⁎ × W ⁎ , such that ( x ¯ , η ¯ , ς ¯ ) is a scalar saddle point of the Lagrangian function l ξ ¯ ( x , η , ς ) , then x ¯ ∈ D is a weakly efficient solution of (VP) and η ¯ ( g ( x ¯ ) ) = 0 .

Proof. Suppose ∃ ( ξ ¯ , η ¯ , ς ¯ ) ∈ ( Y + ⁎ \ { 0 } ) × Z + ⁎ × W ⁎ such that ( x ¯ , η ¯ , ς ¯ ) is a scalar saddle point of the Lagrangian function l ξ ¯ ( x , η , ς ) , i.e.,

l ξ ¯ ( x ¯ , η , ς ) ≤ l ξ ¯ ( x ¯ , η ¯ , ς ¯ ) ≤ l ξ ¯ ( x , η ¯ , ς ¯ ) , ∀ ( η , ς ) ∈ Z + ⁎ × W ⁎ , ∀ x ∈ D .

That is to say

ξ ¯ ( f ( x ¯ ) ) + η ( g ( x ¯ ) ) + ς ( h ( x ¯ ) ) ) ≤ ξ ¯ ( f ( x ¯ ) ) + η ¯ ( g ( x ¯ ) ) + ς ¯ ( h ( x ¯ ) ) ) , ∀ x ∈ D , ∀ ( η , ς ) ∈ Z + ⁎ × W ⁎ .

Then

η ( g ( x ¯ ) ) + ς ( h ( x ¯ ) ) ) ≤ η ¯ ( g ( x ¯ ) ) + ς ¯ ( h ( x ¯ ) ) , ∀ ( η , ς ) ∈ Z + ⁎ × W ⁎ .

Take η = η ¯ , or ς = ς ¯ we have

η ( g ( x ¯ ) ) ≤ η ¯ ( g ( x ¯ ) ) , ∀ η ∈ Z + ⁎ , ς ( h ( x ¯ ) ) ) ≤ ς ¯ ( h ( x ¯ ) ) , ∀ ς ∈ W ⁎ .

Therefore, taking η = 0 , we obtain η ¯ ( g ( x ¯ ) ) ≥ 0 , but taking η = 2 η ¯ , we obtain η ¯ ( g ( x ¯ ) ) ≤ 0 . Hence,

η ¯ ( g ( x ¯ ) ) = 0 .

Since g ( x ) ≺ 0 and ς ¯ ( h ( x ) ) = 0 , ∀ x ∈ D , one has

ξ ¯ ( f ( x ¯ ) ) ≤ ξ ¯ ( f ( x ) ) , ∀ x ∈ D .

Therefore, x ¯ is a weakly efficient solution of (VP).

5 Conclusion

A function f : X → Y is said to be linear on D if ∀ x 1 , x 2 ∈ D ⊆ X , ∀ α , β ∈ ( − ∞ , + ∞ ) , there holds

f ( α x 1 + β x 2 ) = α f ( x 1 ) + β f ( x 2 ) ;

f is said to be affine on D if ∀ x 1 , x 2 ∈ D ⊆ X , ∀ α ∈ ( − ∞ , + ∞ ) , there holds

f ( α x 1 + ( 1 − α ) x 2 ) = α f ( x 1 ) + ( 1 − α ) f ( x 2 ) ;

and f is said to be Y ₊-convex on D if ∀ x 1 , x 2 ∈ D ⊆ X , ∀ α ∈ ( 0 , 1 ) , there holds

α f ( x 1 ) + ( 1 − α ) f ( x 2 ) ≺ f ( α x 1 + ( 1 − α ) x 2 ) ,

where the inequality is induced by the pointed convex cone Y + of Y.

We assume that our inequality constraints g : X → Z of the optimization problem is generalized Y ₊ -convex, i.e., ∀ u ∈ int Y + , ∀ x 1 , x 2 ∈ D , ∀ α ∈ [ 0 , 1 ] , ∃ x 3 ∈ D , ∃ τ > 0 such that

u + α f ( x 1 ) + ( 1 − α ) f ( x 2 ) ≺ τ f ( x 3 ) ,

where τ > 0 is a scalar.

In this article, we also assume that the equality constraint h: X → W of the optimization problem is a generalized affine function, i.e., ∀ x 1 , x 2 ∈ D , ∀ α ∈ ( − ∞ , + ∞ ) , ∃ x 3 ∈ D , ∃ v ∈ M , ∃ t ≠ 0 such that

v + α h ( x 1 ) + ( 1 − α ) h ( x 2 ) = t h ( x 3 ) ,

where M is a affine cone of W.

We remark that the similar assumptions before this author for the equality constraint is as follows, e.g., see [8], ∀ x 1 , x 2 ∈ D , ∀ α ∈ [ 0 , 1 ] , ∃ x 3 ∈ D , ∃ t > 0 such that

α h ( x 1 ) + ( 1 − α ) h ( x 2 ) = t h ( x 3 ) .

For the equality constraints, we note that the condition is reduced from “ ∃ t > 0 ” to “ ∃ t ≠ 0 .”

Our theorems of alternative are generalizations or modifications of the theorems of alternative in [8,9,10,11,12,13,14,15], and our saddle point theorems are generalizations or modifications of the saddle point theorems in [8,16,17,18,19].

This article uses “affine cone” to define the generalized affineness for vector-valued functions, and a discussion for set-valued functions can be found in [20]. We may also use some other “auxiliary sets” to replace the affine cone to define generalized affine functions and properties, and applications would be similar, e.g., see [21,22].

We note that [23] proposed definitions of α-affine-connected and weakly α-affine-connected family of functions, which were about relationships between functions, while our definition of generalized affineness is a property of an individual function.

Funding information: The author receives no external funding.
Author contributions: The author confirms the sole responsibility for the conception of the study and presented results and manuscript preparation.
Conflict of interest: The author states no conflicts of interest.

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Received: 2023-09-05

Revised: 2024-09-11

Accepted: 2024-09-12

Published Online: 2024-10-17

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/math-2024-0073

Keywords for this article

topological vector spaces; optimization problem; theorem of alternative; vector saddle point; scalar saddle point

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