aa r X i v : . [ m a t h . L O ] J a n On Farkas’ Lemma and Related Propositionsin BISH
Josef Berger ∗ and Gregor Svindland † January 12, 2021
Abstract
In this paper we analyse in the framework of constructive mathe-matics (BISH) the validity of Farkas’ lemma and related propositions,namely the Fredholm alternative for solvability of systems of linearequations, optimality criteria in linear programming, Stiemke’s lemmaand the Superhedging Duality from mathematical finance, and vonNeumann’s minimax theorem with application to constructive gametheory.
Keywords:
Farkas’ lemma, constructive mathematics, Fredholm alternative, Stiemke’slemma, Superhedging Duality, von Neumann minimax theorem, constructive game theory.
MSC2010 classification:
In this paper we analyse in the framework of constructive mathematics thevalidity of Farkas’ lemma and related propositions, namely • the Fredholm alternative for solvability of systems of linear equations, • optimality criteria in linear programming, • Stiemke’s lemma and the Superhedging Duality, ∗ Mathematisches Institut, Ludwig-Maximilians-Universit¨at M¨unchen, email:[email protected] † Institut f¨ur Mathematische Stochastik & House of Insurance, Leibniz Universit¨at Han-nover, email: [email protected] von Neumann’s minimax theorem and existence of solutions to two-person zero-sum games.The latter two lines of results are fundamental in mathematical finance andeconomics. Constructive mathematics refers to mathematics in the traditionof Errett Bishop [4, 6], also denoted (BISH). Farkas’ lemma [7] in a formula-tion as two conflicting alternatives is the following proposition: For any real m × n -matrix A and b ∈ R m we haveFAR( A, b ) Exactly one of the following statements is true.i) ∃ ξ ∈ R m ( ξ · A ≥ ∧ ξ · b < ∃ q = ( q , . . . q n ) ∈ R n ( q i ≥ i = 1 , . . . , n ) ∧ A · q = b )Obviously, i ) and ii ) cannot hold simultaneously. It is clear that Farkas’lemma cannot be proved in (BISH), and in fact we show that it is equivalentto the limited principle of omniscience (LPO) which is a strong instance ofthe law of excluded middle (LEM). LPO may be stated as ∀ x ∈ R ( x > ∨ x ≤ . However, our main focus lies on deriving useful constructively valid versionsof Farkas’ lemma. The first type of such results replace the alternatives inFAR(
A, b ) by equivalences and are useful in applications such as solvabilitycriteria for systems of linear equations, see Propositions 4, 8 and Corollary 2.The second type of constructively valid versions of Farkas’ lemma concludesFAR(
A, b ) in the original formulation as alternatives from the detachabilityof a suited set from { , . . . , k } for some k ∈ N , see Proposition 5. We thensay that FAR( A, b ) is conditionally constructive . The rule of intuitionisticpropositional logic (( ϕ ∨ ¬ ϕ ) ⇒ ¬ ψ ) ⇒ ¬ ψ implies that conditionally constructive formulas ν such as FAR( A, b ) may beused to prove negated statements:( ν ⇒ ¬ ψ ) ⇒ ¬ ψ, see Proposition 6. This observation is very useful because Farkas’ lemmaoften comes into play when we wish to derive falsum. Indeed, based onthe fact that FAR( A, b ) is conditionally constructive we provide short proofsof constructive versions of classically well-known results such as optimalitycriteria in linear programming, see Section 5.2, Stiemke’s lemma and theSuperhedging Duality from mathematical finance, see Section 5.3, and von2eumann’s minimax theorem with application to constructive game theory,see Section 5.4. The constructive von Neumann minimax theorem was al-ready proved differently in [5]. In Section 5.4 we also combine our resultswith some recent findings in [3] to verify a conjecture stated in [5] as regardsthe existence of solutions to two-person zero-sum games.
Let k, n ∈ N . We set I k := { , . . . , k } and for x, y ∈ R n x ≤ y : ⇔ ∀ i ∈ I n ( x i ≤ y i ) , y ≥ x : ⇔ x ≤ y. Also we will need the following sets: X n := { p ∈ R n | ≤ p } and S n := ( p ∈ X n | X i ∈ I n p i = 1 ) . For any vector x ∈ R n we write x i for its i th component, that is x =( x , . . . , x n ). Given x, y ∈ R n , z ∈ R m and A = ( a ij ) i ∈ I m ,j ∈ I n ∈ R m × n wewrite x · y := X i ∈ I n x i y i for the Euclidean scalar product, A · x for the element of R m with i th com-ponent ( A · x ) i = X j ∈ I n a ij x j , i = 1 , . . . , m, and z · A for the element of R n with j th component( z · A ) j = ( X i ∈ I m a ij z i ) , j = 1 , . . . , n. A subset K ⊆ R n is a cone if it is inhabited, that is ∃ x ∈ R n ( x ∈ K ), and if ∀ x ∈ K ∀ t ≥ tx ∈ K ) . A subset C ⊆ R n is convex if it is inhabited and if ∀ x, y ∈ C ∀ λ ∈ [0 ,
1] ( λx + (1 − λ ) y ∈ C ) . Let C ⊆ R n and let f : C → R . f is convex if C is convex and if ∀ x, y ∈ C ∀ λ ∈ [0 ,
1] ( f ( λx + (1 − λ ) y ) ≤ λf ( x ) + (1 − λ ) f ( y )) . y , . . . , y k ∈ R n . We denote the span, convex hull, and convex conegenerated by y , . . . , y k byspan( y , . . . , y k ) = span(( y i ) i ∈ I k ) = ( k X i =1 λ i · y i | λ ∈ R k ) , hull( y , . . . , y k ) = hull(( y i ) i ∈ I k ) = ( k X i =1 λ i · y i | λ ∈ S k ) , cone( y , . . . , y k ) = cone(( y i ) i ∈ I k ) = ( k X i =1 λ i · y i | λ ∈ X k ) , respectively.A set U ⊆ R n is located if it is inhabited and if for all x ∈ R n the distance d ( x, U ) = inf {k x − y k | y ∈ U } exists, where throughout this paper k · k denotes the Euclidean norm on R n .There are a number of sufficient conditions ensuring locatedness such as thefollowing variation of [6, Lemma 5.2.3] which we believe has not been statedin the literature yet as it follows from a quite recent result in [2] on infimaof positive convex functions: Lemma 1.
Fix vectors y , . . . , y k ∈ R n such that each element of the convexhull has positive norm, that is ∀ x ∈ hull( y , . . . , y k ) ( k x k > . Then the convex cone cone( y , . . . , y k ) is located.Proof. By [2, Corollary 1] the value µ := inf (cid:8) k x k | x ∈ hull( y , . . . , y k ) (cid:9) is defined and positive. Hence, the assertion follows from [6, Lemma 5.2.3]. Corollary 1.
Suppose that the vectors y , . . . , y k ∈ R n are linearly indepen-dent, that is ∀ λ ∈ R k ( k λ k > ⇒ k P i ∈ I k λ i y i k > . Then cone( y , . . . , y k ) is closed and located. roof. Note that k λ k > λ ∈ S k . Hence, by linear independence k P i ∈ I k λ i y i k >
0. Apply Lemma 1 to conclude locatedness of cone( y , . . . , y k ).As for closedness, note that linear independence implies that the mapping R k ∋ ( λ , . . . , λ k ) X i ∈ I k λ i y i is a bounded linear injection and the same is true for its inverse, see [6,Corollary 4.1.5]. Proposition 1.
Let K ⊆ R m be a located convex cone and fix b ∈ R m . Thefollowing statements are equivalent.i) ∃ ξ ∈ R m ∀ x ∈ K ( ξ · x ≥ ∧ ξ · b < ii) d ( b, K ) > .Proof. i ) ⇒ ii ): As R m ∋ x ξ · x is continuous and ξ · b <
0, there exists δ > ∀ x ∈ R m ( k b − x k < δ ⇒ ξ · x < . Fix x ∈ K . If k b − x k < δ , we can conclude that ξ · x <
0, a contradiction.Thus, ∀ x ∈ K ( k b − x k ≥ δ ) . This implies ii ). ii ) ⇒ i ): Set d := d ( b, K ). By [1, Lemma 6], there exists ξ ∈ R n such that ∀ x ∈ K (cid:0) ξ · ( x − b ) ≥ d (cid:1) . Thus, ∀ x ∈ K (cid:0) ξ · x ≥ d + ξ · b (cid:1) . Since 0 ∈ K , we conclude that ξ · b <
0. Finally, K being a cone implies ∀ x ∈ K ( ξ · x ≥ . Farkas’ Lemma
Proposition 2.
Equivalent are:i)
FAR : ∀ A ∈ R m × n ∀ b ∈ R m FAR(
A, b ) ii) LPO
Proof.
For the moment we only prove that Farkas’ lemma implies LPO, theconverse implication is shown in Lemma 7 below. Consider x ∈ R and let A = ( x ) and b = 1. By FAR( A, b ) either there is ξ ∈ R such that ξ < ξx ≥ x ≤ q ≥ xq = 1 which implies x > A ∈ R m × n we henceforth denote by a , a , . . . , a n ∈ R m the columns of A , and we write cone( A ) := cone(( a i ) i ∈ I n ),and similarly for the span and convex hull. Proposition 3.
Fix a matrix A ∈ R m × n and b ∈ R m . If cone( A ) is located,the following are equivalent:i) ∃ ξ ∈ R m ( ξ · A ≥ ∧ ξ · b < ii) d ( b, cone( A )) > Proof.
Apply Proposition 1.Note that locatedness of cone( A ) cannot be dropped from Proposition 3. Infact, an inspection of the proof of Proposition 1 shows that i) always implies ii ) ′ : ∃ δ > ∀ x ∈ cone( A ) ( k b − x k ≥ δ )which is equivalent to ii) in case cone( A ) is located. However, without re-quiring locatedness of cone( A ) ii ) ′ ⇒ i ) would imply the (constructively notvalid) lesser limited principle of omniscience (LLPO): ∀ x ∈ R ( x ≥ ∨ x ≤ . Indeed, for x ∈ R let A = (cid:18) | x | x (cid:19) and set b = (cid:18) (cid:19) . Then ii ) ′ is satisfied, and i ) would provide a vector ξ = ( ξ , ξ ) such that ξ < ξ | x | + ξ x ≥
0. Either ξ < ξ > ξ . In the first case weobtain x ≤
0, and in the second it follows that x ≥ roposition 4. Fix A ∈ R m × n and b ∈ R m . If cone( A ) is located and closed,then the following are equivalent:i) ∀ ξ ∈ R m ( ξ · A ≥ ⇒ ξ · b ≥ ii) ∃ q ∈ X n ( A · q = b ) Proof.
Since cone( A ) is located and closed, the statement ∃ q ∈ X n ( A · q = b )is equivalent to d ( b, cone( A )) = 0, that is ¬ ( d ( b, cone( A )) > A ) in Proposition 4 is not possible sincethat would imply LPO: For x ∈ R let A = (cid:18) | x | | x | (cid:19) and b = (cid:18) (cid:19) . Then i) of Proposition 4 holds. Indeed, suppose that ξ · A ≥ ξ <
0. Then ξ | x | ≥ − ξ > | x | >
0. As also ξ | x | ≥ , we conclude that ξ ≥ ξ ≥ ∃ q ∈ X ( A · q = b ) . By q | x | + q = 1 we must have that either q | x | > q >
0. In the firstcase | x | >
0. If q >
0, then | x | q = 0 implies | x | = 0. Hence, we have shownthat either x = 0 or | x | > M of a set N is said tobe detachable from N if ∀ x ∈ N ( x ∈ M ∨ x M ) . Definition 1.
A formula ϕ is conditionally constructive if there exists a k ∈ N and a subset M of I k such that the detachability of M from I k implies ϕ . One verifies that conditionally constructive formulas are closed under con-junction and implication:
Lemma 2.
Let the formulas ϕ and ψ be conditionally constructive. Theni) if ϕ ⇒ ν , then ν is conditionally constructive, i) ϕ ∧ ψ is conditionally constructive.Proof. i) is obvious. As for ii), let k, k ′ ∈ N and M ⊆ I k and M ′ ⊆ I k ′ suchthat the detachability of M from I k implies ϕ and the detachability of M ′ from I k ′ implies ψ . Set M ′′ := { l + k | l ∈ M ′ } . Then the detachability of M ′′ from { k + 1 , . . . , k + k ′ } implies ψ . Hence, thedetachability of M ∪ M ′′ from I k + k ′ implies ϕ ∧ ψ . Proposition 5.
Fix A ∈ R m × n and b ∈ R m . Then the formula FAR(A , b) is conditionally constructive. Proposition 5 will be proved throughout the following auxiliary results and isthen a direct consequence of Lemma 6. To this end, fix a matrix A ∈ R m × n .Consider the formulaIND( A ) Exactly one of the following statements is true:i) a , . . . , a n are linearly independent, that is ∀ λ ∈ R n ( k λ k > ⇒ k X i ∈ I n λ i a i k > , ii) a , . . . , a n are linearly dependent, ∃ λ ∈ R n ( k λ k > ∧ X i ∈ I n λ i a i = 0) . Let IND : ∀ A ∈ R m × n IND( A ) . IND is equivalent LPO. Indeed, let x ∈ R and A = ( x ). On the one hand, x is linearly independent if and only if | x | >
0, so either x > x <
0. Onthe other hand x is linearly dependent if and only if x = 0. That is we haveLPO. The fact that LPO implies IND follows from Lemma 7 below.For each inhabited subset J of I n set A J = ( a i ) i ∈ J , i.e. the matrix consisting of columns a i , i ∈ J . We will write cone( A J )for cone(( a i ) i ∈ J ), and similarly for the span and convex hull. Moreover,we say that A J is linearly independent if the vectors a j , j ∈ J are linearlyindependent, and A J is linearly dependent if the vectors a j , j ∈ J are linearly8ependent. We call A linearly independent if and only if A I n is and similarlyfor the linear dependent case. Set L := { J ∈ P ( I n ) | J is inhabited and A J is linearly independent } where P ( I n ) denotes the power set of I n . Lemma 3.
Fix A ∈ R m × n . Suppose that L is detachable from P ( I n ) , then IND( A ) . Hence, IND( A ) is conditionally constructive.Proof. If L = ∅ , then in particular { a i } is not linear independent for all i ∈ I n which implies ¬ ( k a i k > k a i k = 0 for all i ∈ I n . In that case A isthe zero matrix which is linearly dependent. Suppose now that L is inhabitedand pick J ∈ L with a maximal cardinality. If J = I n , then A is linearlyindependent. Otherwise, if J ( I n , note that span( A J ) is located and closedby [6, Lemma 4.1.2, Proposition 4.1.6]. Let j ∈ I n \ J . If d ( a j , span( A J )) > J ∪ { j } ∈ L ; see [6, Lemma 4.1.10], which contradicts maximality of J .Hence, d ( a j , span( A J )) = 0 which implies that A is linearly dependent. Lemma 4.
Fix a subset J of I n and suppose that | J | ≥ . Moreover, supposethat A J is linearly dependent. Let x ∈ cone( A J ) and ε > . Then there exist j ∈ J and y ∈ cone( A J \{ j } ) such that k x − y k < ε. Proof. As x ∈ cone( A J ) there is q ∈ R J with coordinates q i ≥ i ∈ J , suchthat x = A J · q . Fix M > max {k a i k | i ∈ J } . Let J ′ ⊆ J such that i ∈ J ′ implies q i > i J ′ implies q i < ε/M . If J \ J ′ is inhabited, pick j ∈ J \ J ′ and set y = X i ∈ J \{ j } q i a i ∈ cone( A J \{ j } ) . Then k x − y k = q j k a j k < ε . Therefore in the following we may assume that J ′ = J .Since A J is linearly dependent there is λ ∈ R J with k λ k > A J λ = 0. Switching to − λ if necessary, we may assume that λ i > i ∈ J . Set β := max (cid:26) λ i q i | i ∈ J (cid:27) . Then β > j ∈ J such that λ j > (cid:12)(cid:12)(cid:12)(cid:12) q j λ j − β (cid:12)(cid:12)(cid:12)(cid:12) < ε ˜ M M >
M > max {k P i ∈ J \{ k } λ i a i k | k ∈ J } . Set y = X i ∈ J \{ j } ( q i − β λ i ) a i . As q i − β λ i ≥ i ∈ J \ { j } we have that y ∈ cone( A J \{ j } ). Note that x = X i ∈ J \{ j } ( q i − q j λ j λ i ) a i . Hence, k x − y k = (cid:12)(cid:12)(cid:12)(cid:12) q j λ j − β (cid:12)(cid:12)(cid:12)(cid:12) k X i ∈ J \{ j } λ i a i k < ε. Lemma 5.
Suppose that L is detachable from P ( I n ) and inhabited. Choosearbitrary x ∈ cone( A ) and ε > . Then there exists J ∈ L such that d ( x, cone( A J )) < ε . In particular cone( A ) is located and for all z ∈ R m we have d ( z, cone( A )) = min J ∈L d ( z, cone( A J )) .Proof. Note that cone( A ˜ J ) is located and closed for any ˜ J ∈ L by Corollary 1.Let q ∈ X n such that x = A · q . Since { i } ∈ L ⇔ k a i k > L is detachable from P ( I n ), the set J := { i ∈ I n | k a i k > } is detachable from I n , and i J implies k a i k = 0, that is a i = 0. Hence, wehave x = X i ∈ J q i a i ∈ cone( A J ) . If J ∈ L , then set J = J . Otherwise, A J is linearly dependent by Lemma 3,so we may apply Lemma 4 to find j ∈ J and y ∈ cone( A J \{ j } ) such that k x − y k < εn . If J := J \ { j } ∈ L , set J = J , and note that d ( x, cone( A J )) ≤ k x − y k < ε. Otherwise A J is linearly dependent by Lemma 3, so we may apply Lemma 4to find j ∈ J and y ∈ cone( A J \{ j } ) such that k y − y k < εn . If J := J \ { j } ∈ L , set J = J and note that d ( x, cone( A J )) ≤ k x − y k < k x − y k + k y − y k < ε { i } ∈ L for all i ∈ J , after at most | J | − J ⊆ I n and y | J |−| J | ∈ cone( A J ), where y := x , such that J ∈ L and d ( x, cone( A J )) ≤ k x − y | J |−| J | k≤ k x − y k + . . . + k y | J |−| J |− − y | J |−| J | k < ( n − | J | ) εn < ε. Finally, we prove that cone( A ) is located. Let z ∈ R m be arbitrary and set d := min { d ( z, A J ) | J ∈ L} . We prove that for all y ∈ cone( A ) we have k z − y k ≥ d which implies thatinf {k z − y k | y ∈ cone( A ) } exists and equals d . To this end, let y ∈ cone( A )and suppose that k z − y k < d . Then, according to what we have shownabove, there exists J ∈ L such that d ( y, cone( A J )) < d − k z − y k . Thisimplies that d ( z, cone( A J )) ≤ k z − y k + d ( y, cone( A J )) < d which is absurd. Lemma 6.
Fix A ∈ R m × n and b ∈ R m . Define subsets Ω , Ω , Ω , and Ω of P ( I n ) × I by ( J, ∈ Ω ⇔ J ∈ L , ( J, ∈ Ω ⇔ J ∈ L ∧ d ( b, cone( A J )) > , ( J, ∈ Ω ⇔ J ∈ L ∧ d ( b, cone( A J )) = 0 , ( I n , ∈ Ω ⇔ k b k > . Assume that the set Ω ∪ Ω ∪ Ω ∪ Ω is detachable from P ( I n ) × I , then FAR(
A, b ) .Proof. The assumption in particular implies that L is detachable from P ( I n ).If Ω = ∅ and Ω is inhabited, alternative i) of FAR( A, b ) holds. Indeed, sincein that case ( { i } , Ω for each i ∈ I n , it follows that k a i k = 0 for all i ∈ I n .Hence, A is the matrix in which all entries are 0. Therefore i) of FAR( A, b )is satisfied by ξ = − b . If Ω = Ω = ∅ , then b = 0 and alternative ii) ofFAR( A, b ) holds with q = 0. Therefore, from now on we may assume that L is inhabited. We show that ∀ J ∈ L ( d ( b, cone( A J )) > ∨ d ( b, cone( A J )) = 0) . (1)Fix J ∈ L . Consider the following cases:11 ( J, ∈ Ω and ( J, ∈ Ω • ( J, ∈ Ω and ( J, / ∈ Ω • ( J, / ∈ Ω and ( J, ∈ Ω • ( J, / ∈ Ω and ( J, / ∈ Ω The first and the last case are absurd. The remaining cases both imply (1).Recall that cone( A J ) is closed for all J ∈ L according to Corollary 1. Hence,if ( J, ∈ Ω for some J ∈ P ( I n ), then there is q ∈ X n with q i = 0 for all i ∈ I n \ J such that Aq = b . That is alternative ii) of FAR( A, b ) holds. Itremains to consider the case ∀ J ∈ L ( d ( b, cone( A J )) > . In view of Lemma 5 we can conclude that cone( A ) is located and that d ( b, cone( A )) >
0. Thus Proposition 3 implies that alternative i) of FAR(
A, b )holds.
Lemma 7.
Assume
LPO . Then
IND and
FAR .Proof.
Let A ∈ R m × n , b ∈ R m , and let J ∈ P ( I n ) be inhabited. The unitball S = { λ ∈ R J | k λ k = 1 } is compact and thus α := inf {k A J λ k | λ ∈ S } exists, see [6, Corollary 2.2.7]. LPO implies that either α > α = 0. If α >
0, then A J is linearly independent. If α = 0, then, as LPO implies theminimum principle (see [10]), there exists λ ∈ S such that A J λ = 0 whichimplies that A J is linearly dependent. In particular, letting J = I n , we haveshown IND( A ). Also, as J ∈ P ( I n ) was arbitrary, we have that L is detach-able from P ( I n ). Moreover, cone( A J ) is located for any J ∈ L by Corollary 1and LPO implies that either d ( b, cone( A J )) > d ( b, cone( A J )) = 0. Againby LPO we have either k b k > k b k = 0. Thus the setΩ ∪ Ω ∪ Ω ∪ Ω from Lemma 6 is detachable from P ( I n ) × I . This implies FAR( A, b ) ac-cording to Lemma 6. 12
Conditionally Constructive Formulas andProofs of Negated Statements
Consider the following rule of intuitionistic propositional logic:(( ϕ ∨ ¬ ϕ ) ⇒ ¬ ψ ) ⇒ ¬ ψ, (2)see also [11]. (2) allows to prove a negated statement ¬ ψ by assuming a finitenumber of case distinctions ϕ ∨ ¬ ϕ and proving ¬ ψ in each resulting case:([( ϕ ∨ ¬ ϕ ) ∧ . . . ∧ ( ϕ k ∨ ¬ ϕ k )] ⇒ ¬ ψ ) ⇒ ¬ ψ, or equivalently, if we prove ν ∧ . . . ∧ ν k ⇒ ¬ ψ (3)for all 2 k possible combinations ν i ∈ { ϕ i , ¬ ϕ i } , i ∈ I k , then ¬ ψ . As a resultwe obtain the following proposition: Proposition 6.
Suppose that the formula ϕ is conditionally constructive.Then ( ϕ ⇒ ¬ ψ ) ⇒ ¬ ψ. Proof.
Since ϕ is conditionally constructive, there is k ∈ N and a subset M of I k such that M being detachable from I k implies ϕ . For each i ∈ I k considerthe cases i ∈ M or i M . This gives 2 k instances of type ν ∧ . . . ∧ ν k where ν i ∈ { i ∈ M, i M } , i ∈ I k , as in (3). In each such instance M is detachablefrom I k and thus we obtain ϕ . Hence if ϕ ⇒ ¬ ψ , then we may conclude ¬ ψ .Clearly, for any formula ϕ the formula ϕ ∨ ¬ ϕ is conditionally constructive,simply by choosing k = 1 and M ⊆ I given by 1 ∈ M if and only if ϕ .Hence, in that case Proposition 6 is nothing but (2). A basic solvability theorem from Linear Algebra is the so-called Fredholmalternative theorem (FRED): For all A ∈ R m × n and b ∈ R m FRED(
A, b ) Exactly one of the following statements is true:13) ∃ ξ ∈ R m ( ξ · A = 0 ∧ | ξ · b | > ∃ x ∈ R n ( A · x = b )In fact, like FAR, also FRED is equivalent to LPO: Let a ∈ R and set A = ( a )and b = 1. Then FRED( A, b ) yields either a = 0 or ax = 1 for some x ∈ R .The latter implies | a | >
0, so either a < a >
0. Hence, we have LPO.Conversely, as LPO implies FAR (Proposition 2), the following propositionalso implies that LPO ⇒ FRED.
Proposition 7.
Fix A ∈ R m × n and b ∈ R m . Let B = ( A − A ) ∈ R m × n .Then FAR(
B, b ) ⇒ FRED(
A, b ) . Hence, FRED(
A, b ) is conditionally con-structive.Proof. By FAR(
B, b ) there is either ξ ∈ R m such that ξ · B ≥ ξ · b < q ∈ X n such that B · q = b . In the latter case, letting x be givenby x i = q i − q n + i , i ∈ I n , yields x ∈ R n with A · x = b . In the first case ξ · A ≥ − ξ · A ≥ ξ · A = 0.Proposition 5 and Lemma 2 imply that FRED( A, b ) is conditionally con-structive.We now prove a constructive version of FRED.
Proposition 8.
Let A ∈ R m × n and b ∈ R m . Suppose that span( A ) is locatedand closed. Equivalent are:i) ∀ ξ ∈ R m ( ξ · A = 0 ⇒ ξ · b = 0) ,ii) ∃ x ∈ R n ( A · x = b ) .Proof. Again consider the matrix B := ( A − A ), then cone( B ) = span( A ) isclosed and located. Hence, by Proposition 4 the following are equivalent1) ∀ ξ ∈ R m ( ξ · B ≥ ⇒ ξ · b ≥ ∃ q ∈ X n ( B · q ) = b .Now i) is equivalent to 1) and ii) is equivalent to 2).As a consequence we obtain the following constructive version of the Fred-holm alternative for solvability of systems of linear equations. Corollary 2.
Let A ∈ R m × n and b ∈ R m . Suppose span( A ) is located andclosed. If the homogeneous equation ξ · A = 0 admits a unique solution, thenthere exists a solution to the system of linear equations A · x = b .Proof. The unique solution to ξ · A = 0 is of course ξ = 0, so i) of Proposition 8is satisfied which implies ii). 14 .2 Optimality Criteria of Linear Programming Consider the following linear optimisation problems: Let A ∈ R m × n , b ∈ R m ,and c ∈ R n . The primal problem is( P ) minimise c · x subject to x ∈ P := { y ∈ X n | A · y = b } , whereas the dual problem is( D ) maximise b · u subject to u ∈ D := { v ∈ R m | v · A ≤ c } . Before we state constructive versions of optimality criteria in linear program-ming in Propositions 9 and 10, we briefly recall the following well-knownresult.
Lemma 8.
Fix x ∈ P and u ∈ D such that c · x = b · u . Then x solves ( P ) and u solves ( D ) .Proof. This follows immediately once we observe that for all y ∈ P and all v ∈ D we have b · v = v · A · y ≤ c · y. Proposition 9.
Suppose that there exists a solution u to ( D ) . The followingstatement is conditionally constructive:there exists a solution x to (P) and c · x = b · u . For the proof we need the following auxiliary lemma:
Lemma 9.
Let u be a solution to ( D ) . Define J ⊆ I n by ∀ i ∈ I n ( i ∈ J ⇔ ( u · A ) i < c i ) . Consider ϕ : ( k b k = 0) ∨ ( k b k > , J is detachable from I n , | J | < n , FAR( A I n \ J , b )) .ϕ is conditionally constructive.Proof. By Lemma 2 and Proposition 5 ψ : k b k = 0 ∨ k b k > ,ψ : ^ i ∈ I n (( u · A ) i < c i ∨ ( u · A ) i = c i ) , ψ : ^ J ′ ∈P ( I n ) , J ′ inhabited FAR( A J ′ , b )are conditionally constructive, and thus also ψ ∧ ψ ∧ ψ . ψ ∧ ψ ∧ ψ implies that either k b k = 0 or k b k > J is detachable from I n . Incase k b k > u solves ( D ), we have that | J | < n , because otherwise u + tb ∈ D for small t >
0, and b · ( u + tb ) = b · u + t k b k > b · u which isabsurd. Now ψ implies FAR( A I n \ J , b ). Hence, we have ψ ∧ ψ ∧ ψ ⇒ ϕ ,so ϕ is conditionally constructive by Lemma 2. Proof of Proposition 9.
Recall ϕ from lemma 9. We show that ϕ implies thatthere exists a solution x to (P) and c · x = b · u . To this end, consider A ′ = (cid:18) Ac (cid:19) ∈ R ( m +1) × n and b ′ := (cid:18) bb · u (cid:19) ∈ R m +1 . We show that b ′ ∈ cone( A ′ ), because in that case there is x ∈ X n such that A · x = b and c · x = b · u , so x solves (P) according to Lemma 8.If k b k = 0, then b ′ = 0 ∈ cone( A ′ ).If k b k >
0, then FAR( A I n \ J , b ), with J as in Lemma 9, yields the followingcases:Case 1: There is ξ ∈ R m such that ξ · A I n \ J ≥ ξ · b <
0. Then thereis t > u − tξ ) · A ≤ c and ( u − tξ ) · b > u · b which contradictsoptimality of u .Case 2: There is x ∈ R I n \ J with x ≥ A I n \ J · x = b . In that case x ′ ∈ R n given by x ′ i = x i , i ∈ I n \ J , and x ′ i = 0 otherwise satisfies x ′ ∈ X n , A · x ′ = A I n \ J · x = b , and c · x ′ = c I n \ J · x = ( u · A I n \ J ) · x = u · b, so A ′ · x ′ = b ′ . Here we used that ( u · A ) i = c i for all i ∈ I n \ J .Now one readily finds the following version of the optimality criteria in lin-ear programming, replacing the requirement ‘conditionally constructive’ inProposition 9 by a sufficiently strong condition on the input A, b, c such thatproving x to be a solution to (P) boils down to proving a negated statement: Proposition 10.
Consider the ( m + 1) × n -matrix A ′ = (cid:18) Ac (cid:19) and suppose that cone( A ′ ) is closed and located. If there is a solution u to ( D ) , then there exists a solution x to (P) and c · x = b · u . roof. Again set b ′ := (cid:18) bb · u (cid:19) ∈ R m +1 . As in the proof of Proposition 9 we need to show that b ′ ∈ cone( A ′ ). Notethat cone( A ′ ) being closed and located implies that b ′ ∈ cone( A ′ ) is equivalentto d (cone( A ′ ) , b ′ ) = 0, that is ¬ ( d (cone( A ′ ) , b ′ ) > ¬ ( d (cone( A ′ ) , b ′ ) >
0) under the assumption thatthere exists a solution x to (P) and c · x = b · u. But the latter obviously implies that b ′ ∈ cone( A ′ ). In the following for x, y ∈ R k we write x < y : ⇔ ∀ i ∈ I k ( x i < y i ) , y > x : ⇔ x < y and x (cid:12) y : ⇔ x ≤ y ∧ ∃ i ∈ I k ( x i < y i ) , x (cid:13) y : ⇔ y (cid:12) x. Let P n = { q ∈ S n | q > } . Stiemke’s lemma (STI) states that for all A ∈ R m × n we haveSTI(A) Exactly one of the following alternatives is true:i) ∃ ξ ∈ R m ( ξ · A (cid:13) ∃ p ∈ P n ( A · p = 0)Like FAR and FRED also STI is equivalent to LPO. Indeed, for x ∈ R let A = ( | x | ). Then STI( A ) implies that either there exists ξ ∈ R such that ξ | x | >
0, that is | x | >
0, or | x | = 0. Hence, we have LPO. The implicationLPO ⇒ STI follows from LPO ⇒ FAR (Proposition 2) and the proof of thefollowing proposition.
Proposition 11.
Fix A ∈ R m × n . Then STI(A) is conditionally constructive. roof. First, assume that n = 1. k a k > ∨ k a k = 0 is conditionallyconstructive. If k a k >
0, alternative i ) of STI(A) holds. If k a k = 0,alternative ii ) of STI(A) holds.Now assume that n ≥
2. For each i ∈ I n , let A i be the matrix which resultsfrom removing the column a i from A . By Proposition 5 and Lemma 2 ϕ : FAR( A , − a ) ∧ FAR( A , − a ) ∧ . . . ∧ FAR( A n , − a n )is conditionally constructive. We prove that ϕ ⇒ STI(A). Note that ϕ implies that the sets N = (cid:8) i ∈ I n | alternative i ) of FAR( A i , − a i ) holds (cid:9) and N = (cid:8) i ∈ I n | alternative ii ) of FAR( A i , − a i ) holds (cid:9) . are detachable from I n and that I n = N ∪ N . If N is inhabited, there exist i ∈ I n and ξ ∈ R m such that ξ · A i ≥ ξ · ( − a i ) < . This implies that ξ · A (cid:13)
0. Thus alternative i ) of STI(A) holds.Now assume that N = ∅ and therefore N = I n . For each i ∈ I n there exists q i ∈ X n − such that A i · q i = − a i , which yields the existence of p i ∈ X n with( p i ) i = 1 and A · p i = 0. Then˜ p i := 1 P j ∈ I n p ij p i ∈ S n , i ∈ I n , and p := 1 n X i ∈ I n ˜ p i ∈ P n satisfies A · p = 0. Thus, alternative ii ) of STI(A) holds.Let us now briefly consider a simple stochastic one-period financial marketmodel. For further details and explanations we refer to [2, 8]. The matrix A ∈ R m × n represents the discounted price changes between time 0 (today)18nd time t = 1 (tomorrow). More precisely, we assume that the marketconsists of m financial assets and that there are n possible states of theworld tomorrow. Thus a ij is the discounted price change between times 0and 1 of asset j ∈ I m in state i ∈ I n . A so-called equivalent martingalemeasure for the market is a p ∈ P n such that A · p = 0. We denote the set ofequivalent martingale measures by P . If P is inhabited, the market model iscalled arbitragefree. A contingent claim is a financial contract which pays acertain amount c i ≥ i ∈ I n at time 1. We assume that c i is alreadydiscounted, that is c = ( c , . . . , c n ) ∈ X n is the discounted payoff profile ofthe claim c . For any p ∈ P the price c · p is a fair (arbitragefree) price of theclaim c . Trading strategies are given by vectors ξ ∈ R m , where ξ i representsthe amount of shares of asset i which are bought. Shortselling, that is ξ i < ξ are thus given by ξ · A . Assuming we have availablecapital x ∈ R at time 0, a superhedge of the claim c given the capital x isa trading strategy ξ such that x + ξ · A ≥ c . Here := (1 , , . . . , ∈ R n represents the bank account in which the investor keeps her capital. Indeed,assuming that the investor buys ξ at time 0, she has x minus the price of ξ leftin the bank account. At time 1 the discounted value of the investment is x minus the price of ξ at time 0 plus the price of ξ at time 1 which correspondsto x + ξ · A . Thus a superhedge of c given the capital x is an investmentwhich outperforms c in any possible future state of the world. The so-calledSuperhedging Duality in classical financial mathematics states thatsup p ∈P c · p = min { x ∈ R | ∃ ξ ∈ R m ( x + ξ · A ≥ c ) } . We now prove a constructive version of this Superhedging Duality:
Proposition 12.
Suppose that P is inhabited and that sup p ∈P c · p and inf { x ∈ R | ∃ ξ ∈ R m ( x + ξ · A ≥ c ) } exist. Then sup p ∈P c · p = inf { x ∈ R | ∃ ξ ∈ R m ( x + ξ · A ≥ c ) } . Proof.
Consider x ∈ R such that there exists ξ ∈ R m with x + ξ · A ≥ c. For any p ∈ P we obtain x = ( x + ξ · A ) · p ≥ c · p . Hence, we have thatsup p ∈P c · p ≤ inf { x ∈ R | ∃ ξ ∈ R m ( x + ξ · A ≥ c ) } .
19t remains to prove that ¬ (sup p ∈P c · p < inf { x ∈ R | ∃ ξ ∈ R m ( x + ξ · A ≥ c ) } ) . To this end, assume that there is y ∈ R such thatsup p ∈P c · p < y < inf { x ∈ R | ∃ ξ ∈ R m ( x + ξ · A ≥ c ) } , and consider the extended market B = (cid:18) Ac − y (cid:19) ∈ R ( m +1) × n . Since we are proving a negated statement, according to Proposition 6, itsuffices to prove this under the assumption of STI( B ). Note that ii) inSTI( B ) is absurd because for any p ∈ P n with B · p = 0 we have p ∈ P and c · p − y = 0 which contradicts the assumption c · p < y . Hence, we mayassume i) in STI( B ), that is there exists ξ ∈ R m and η ∈ R such that ξ · A + η ( c − y ) (cid:13) . (4)Pick any p ∈ P . Then η ( c · p − y ) = ( ξ · A + η ( c − y )) · p > η < . Thus deviding both sides in (4) by | η | and rearrangingwe obtain y + 1 | η | ξ · A ≥ c which contradicts y < inf { x ∈ R | ∃ ξ ∈ R m ( x + ξ · A ≥ c ) } . The discussion in this section is based on the lemma on alternatives (ALT):For all A ∈ R m × n we haveALT( A ) Exactly one of the following statements is true:20) ∃ p ∈ S m ( p · A ≥ ∃ q ∈ S n ( A · q < x ∈ R and A = ( x ) by ALT( A )we either have x ≥ x < ⇒ ALT.
Proposition 13.
Let A ∈ R m × n . Define B = ( A E m ) ∈ R m × ( n + m ) , where E m ∈ R m × m denotes the identity matrix, i.e. the matrix with diagonal entriesall equal to and all other entries equal to . Set b = ( − , . . . , − ∈ R m .Then FAR(
B, b ) ⇒ ALT( A ) . Hence, ALT( A ) is conditionally constructive.Proof. By FAR(
B, b ) either there is ξ ∈ R m such that ξ · B ≥ ξ · b < q ∈ X n + m such that B · q = b . In the first case we must have ξ ≥
0, since 0 ≤ ξ · E m = ξ , and P i ∈ I m ξ i = − ξ · b >
0. Hence, p := 1 P i ∈ I m ξ i ξ ∈ S m satisfies p · A ≥
0. In the second case ˆ q := ( q , . . . , q n ) ∈ X n satisfies A · ˆ q ≤ B · q = b < . In particular, min { a i | i ∈ I n } X i ∈ I n ˆ q i < , which implies | P i ∈ I n ˆ q i | > P i ∈ I n ˆ q i > q ∈ X n . Hence,˜ q := 1 P i ∈ I n ˆ q i ˆ q ∈ S n satisfies A · ˜ q < A ∈ R m × n max p ∈ S m min q ∈ S n p · A · q = min q ∈ S n max p ∈ S m p · A · q. A thorough discussion of this result in (BISH) is given in [5]. In that articlealso the following constructive version of von Neumann’s minimax theoremwas introduced, see [5, Theorem 2.3]. Here we provide a short proof of thisresult based on Propositions 6 and 13.
Proposition 14.
Let A ∈ R m × n . Then sup p ∈ S m inf q ∈ S n p · A · q = inf q ∈ S n sup p ∈ S m p · A · q. roof. Note that inf q ∈ S n p · A · q = min i ∈ I n ( p · A ) i and sup p ∈ S m p · A · q =max j ∈ I m ( A · q ) j , and the functions S m ∋ p min i ∈ I n ( p · A ) i and S n ∋ q max j ∈ I m ( A · q ) j are uniformly continuous, whencesup p ∈ S m inf q ∈ S n p · A · q and inf q ∈ S n sup p ∈ S m p · A · q exist, see [6, Corollary 2.2.7]. Clearly,sup p ∈ S m inf q ∈ S n p · A · q ≤ inf q ∈ S n sup p ∈ S m p · A · q, so it remains to show that ¬ ( sup p ∈ S m inf q ∈ S n p · A · q < inf q ∈ S n sup p ∈ S m p · A · q ) . Suppose sup p ∈ S m inf q ∈ S n p · A · q < inf q ∈ S n sup p ∈ S m p · A · q. Without loss of generality, by suitable translation, we may assume that thereexists ι > p ∈ S m inf q ∈ S n p · A · q ≤ − ι and ι ≤ inf q ∈ S n sup p ∈ S m p · A · q. (5)As we aim at proving falsum, by Propositions 6 and 13 it suffices to considerthe casesi) ∃ p ∈ S m ( p · A ≥ ∃ q ∈ S n ( A · q < p ∈ S m inf q ∈ S n p · A · q ≥ > − ι, a contradiction, and in the second caseinf q ∈ S n sup p ∈ S m p · A · q ≤ < ι, also a contradiction. 22s a consequence of a recent result on the minimum principle for convexfunctions, see [3, Theorem 1], we obtain the following existence result forsolutions to two-person zero-sum games; see for instance [9] for a classicaldiscussion of such games. To this end, note that a function f : C → R ,where C ⊆ R k , such that α := inf x ∈ C f ( x ) exists is said to admit at mostone minimum, if ∀ x, y ∈ C ( k x − y k > ⇒ ( f ( x ) > α ∨ f ( y ) > α )) . Proposition 15.
Let A ∈ R m × n . Suppose that f A : S n ∋ q sup p ∈ S m p · A · q admits at most one minimum, and that g A : S m ∋ p inf q ∈ S n p · A · q admits at most one maximum, that is − g A admits at most one minimum.Then there exists (ˆ p, ˆ q ) ∈ S m × S n such that ˆ p · A · ˆ q = sup p ∈ S m inf q ∈ S n p · A · q = inf q ∈ S n sup p ∈ S m p · A · q. Proof.
Note that S n and S m are compact and that f A is convex whereas g A is concave, that is − g A is convex. Hence, according to [3, Theorem 1] thereexists a minimiser ˆ q ∈ S n of f A and a minimiser ˆ p ∈ S m of − g A , i.e. ˆ p is amaximiser of g A . We havesup p ∈ S m inf q ∈ S n p · A · q = inf q ∈ S n ˆ p · A · q ≤ ˆ p · A · ˆ q ≤ sup p ∈ S m p · A · ˆ q = inf q ∈ S n sup p ∈ S m p · A · q. Now apply Proposition 14.Saddle points (ˆ p, ˆ q ) as in Proposition 15 are called solutions to the two-personzero-sum game given by A . The following Corollary 3 generalises [5, Theorem3.2] and verifies the conjecture as regards existence of solutions to two-personzero-sum games made at the end of [5]. Corollary 3.
Let A ∈ R m × n , and suppose that the associated two-personzero-sum game has at most one solution in the sense of [5], that is, denoting α := sup p ∈ S m inf q ∈ S n p · A · q = inf q ∈ S n sup p ∈ S m p · A · q, we have for any pairs ( p, q ) , ( p ′ , q ′ ) ∈ S m × S n with k p − p ′ k + k q − q ′ k > that either | p · A · q − α | > or | p ′ · A · q ′ − α | > . Then the game has aunique solution, that is there exists a unique (ˆ p, ˆ q ) ∈ S m × S n such that ˆ p · A · ˆ q = α. roof. For uniqueness, assume that ( p, q ) , ( p ′ , q ′ ) ∈ S m × S n are two solutionsto the game. Then, as the game has at most one solution, k p − p ′ k + k q − q ′ k > p, q ) = ( p ′ , q ′ ).As regards existence of solutions, we show that the function f A defined inProposition 15 admits at most one minimum. Note that inf q ∈ S n f A ( q ) = α and ∀ δ > ∀ q ∈ S n ∃ p ∈ S m ( | p · A · q − f A ( q ) | < δ ) . (6)Fix q, q ′ ∈ S n and suppose that k q − q ′ k >
0. The function h : S m × S m → R ( p, p ′ )
7→ | p · A · q − α | + | p ′ · A · q ′ − α | is uniformly continuous, convex, and positive-valued. The latter follows fromthe assumption that the game has at most one solution. Thus, according to[2, Proposition 1] there exists ε > ( p,p ′ ) ∈ S m × S m h ( p, p ′ ) > ε. (7)We have that either f A ( q ) < α + ε/ f A ( q ) > α and either f A ( q ′ ) < α + ε/ f A ( q ′ ) > α . Assume that f A ( q ) < α + ε f A ( q ′ ) < α + ε . Then there are p, p ′ ∈ S m such that | p · A · q − α | < ε | p ′ · A · q ′ − α | < ε . This is a contradiction to (7). Thus, either f A ( q ) > α or f A ( q ′ ) > α. Similarly, one verifies that g A defined in Proposition 15 admits at most onemaximum. Hence, the assertion follows from Proposition 15.24 eferences [1] Josef Berger and Gregor Svindland. Convexity and constructive infima. Arch. Math. Logic , 55, 2016.[2] Josef Berger and Gregor Svindland. A separating hyperplane theorem,the fundamental theorem of asset pricing, and Markov’s principle.
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