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Владикавказский математический журнал 2009, Том 11, выпуск 3, С. 44-50
UDC 517.983.27:517.972.8
BOOLEAN MODELS AND SIMULTANEOUS INEQUALITIES
To Yuri G. Reshetnyak on His 80th Birthday
S. S. Kutateladze
Boolean valued analysis is applied to deriving operator versions of the classical Farkas Lemma in the
theory of simultaneous linear inequalities.
Mathematics Subject Classification (2000): 03H10, 47B60, 90C48.
Key words: Farkas lemma, theorem of the alternative, interval equations.
1. Agenda
The Farkas Lemma, also known as the Farkas-Minkowski Lemma, plays a key role in linear programming and the relevant areas of optimization [1]. The aim of this article is to demonstrate how Boolean valued analysis [2] may be applied to simultaneous linear inequalities with operators. This particular theme is another illustration of the deep and powerful technique of «stratified validity» which is characteristic of Boolean valued analysis. We only outline the main scheme of proof and announce the main results. The complete exposition will appear elsewhere.
2. Environment
Assume that X is a real vector space, Y is a Kantorovich space also known as a complete vector lattice or a complete Riesz space. Let B := B(Y) be the base of Y, i. e., the complete Boolean algebras of positive projections in Y; and let m(Y) be the universal completion of Y. Denote by L(X,Y) the space of linear operators from X to Y. In case X is furnished with some Y-seminorm on X, by L(m)(X, Y) we mean the space of dominated operators from X to Y. As usual, {T < y} := {x G X | Tx < y} and ker(T) = T-1(0) for T : X ^ Y.
3. Inequalities: explicit dominance
The following are well known:
(1): (3X) XA = B ^ ker(A) С ker(B);
(2):1 If W is ordered by W+ and A(X) - W+ = W+ - A(X) = W, then
(3X ^ 0) XA = B ^ {A < 0} С {B < 0}.
© 2009 Kutateladze S. S.
1The Kantorovich Theorem [3].
4. Farkas: explicit dominance
Theorem 1. Assume that A\,... ,AN and B belong to L(m)(X, Y). The following are equivalent:
(1) Given b G B, the operator inequality bBx ^ 0 is a consequence of the simultaneous linear operator inequalities bA\X ^ 0,..., bAnx ^ 0, i. e.,
{bB < 0} D {bAi < 0} n ■ ■ ■ n [bAN < 0}.
(2) There are positive orthomorphisms ai,..., aN G Orth(m(Y)) such that
N
B = ^ ak Ak; k=i
i. e., B lies in the operator convex conic hull of Ai,..., An.
5. Farkas: hidden dominance
Lemma 1. Let X be a vector space over some subfield R of the reals R. Assume that f and g are R-linear functionals on X; in symbols, f,g G X # := L(X, R).
For the inclusion {g ^ 0} D {f ^ 0} to hold it is necessary and sufficient that there be a G R+ satisfying g = af. < Sufficiency is obvious.
Necessity; The case of f = 0 is trivial. If f = 0 then there is some x G X such that f (x) G R and f (x) > 0. Denote the image f (X) of X under f by R0. Put h := g o f-i, i .e. h G R# is the only solution for h o f = g. By hypothesis, h is a positive R-linear functional on Ro. By the Bigard Theorem [3, p. 108] h can be extended to a positive homomorphism h : R ^ R, since Ro — R+ = R+ — Ro = R. Each positive automorphism of R is multiplication by a positive real. As the sought a we may take h(1). >
6. Reals: explicit dominance
Lemma 2. Let X be an R-seminormed vector space over some subfield R of R that fi,..., fN and g are bounded R-linear functionals on X; in symbols, fi,... X* := L(m) (X, R). For the inclusion
N
{g < 0}D f|{fk < 0}
k=i
to hold it is necessary and sufficient that there be ai,... ,aN G R+ satisfying
N
g = ^2 ak fk. k=i
7. Origins
Cohen's final solution of the problem of the cardinality of the continuum within ZFC gave rise to the Boolean valued models Scott forecasted in 1969 [4]:
We must ask whether there is any interest in these nonstandard models aside from the independence 'proof; that is, do they have any mathematical interest? The answer must be yes, but we cannot yet give a really good argument.
. Assume
,fN,g G
Takeuti coined the term «Boolean valued analysis» for applications of the models to analysis [5].
8. Boolean valued universe
Let B be a complete Boolean algebra. Given an ordinal a, put
Va(B) := {x | (3£ € a) x : dom(x) ^ B & dom(x) C Vjf^ . The Boolean valued universe V(B) is
V(B) := y pKB), agOn
with On the class of all ordinals. The truth value € B is assigned to each formula ^ of ZFC relativized to V(B).
9. Descending and ascending
Given a formula of ZFC, and y, a member of VB; put Av : = A^, y) : = {x | ^(x, y)}. The descent of a class A^ is
A^I := {t | t € V(B) & [<p(t,
=
If t € A^I, then it is said that t satisfies y) inside V(B). The descent xI of x € V(B) is defined as
xI := {t | t € V(B) & [t € x] = l} ,
i. e. xI = A^exI. The class xI is a set. If x is a nonempty set inside V(B) then
(3z € xI)[(3t € x) p(t)] = Mz)]. The ascent functor acts in the opposite direction.
10. The reals within
There is an object R inside V(B) modeling R, i. e.,
[R is the reals ] = 1. Let RI be the descent of the carrier |R| of the algebraic system
R := (|R|, +, •, 0,1, <)
inside V(B). Implement the descent of the structures on |R| to RI as follows:
x + y = z ^ [x + y = z] = 1; xy = z ^ [xy = z] = 1;
x ^ y ^ [x ^ y ] = 1; Ax = y ^ [AAx = y] = 1 (x,y,z € RI, A € R).
Gordon Theorem. R j with the descended structures is a universally complete vector lattice with base B(R j) isomorphic to B.
< Proof of Theorem 1.
(2)- (1): If B = £N= i ak Ak for some positive ai,..., an in Orth(m(Y )) while bAk x ^ 0 for b G B and x G X, then
N N
bBx = b ^ ak Akx = ^ ak bAkx ^ 0
k=i k=i
since orthomorphisms commute and projections are orthomorphisms of m(Y).
(1)^ (2): Consider the separated Boolean valued universe V(B) over the base B of Y. By the Gordon Theorem the ascent Yj of Y is R, the reals inside V(B).
Using the canonical embedding, we see that XA is an R-seminormed vector space over the standard name RA of the reals R. Moreover, RA is a subfield and sublattice of R = Yj inside V(B).
Put fk := Ak j for all k := 1,... ,N and g := Bj. Clearly, all fi,..., fN ,g belong to (XA)* inside VB.
Define the finite sequence
f : {1,...,N}A ^ (XA)* as the ascent of (fi,..., fN). In other words, the truth values are as follows:
[fkA (xA) = Akx] = 1, [g(xA) = Bx] = 1
for all x G X and k := 1 , . . . , N. Put
b := [Aix ^ 0A] A ■ ■ ■ A [Anx ^ 0A].
Then bAkx ^ 0 for all k := 1,... ,N and bBx ^ 0 by (1). Therefore,
[Aix ^ 0A] A ■ ■ ■ A [Anx ^ 0A] ^ [Bx ^ 0A].
In other words,
[(V k := 1A,..., NA)fk(xA) < 0A] = / [fkA (xA) < 0A] < [g(xA) < 0A].
k:=i,...,N
Using Lemma 2 inside V(B) and appealing to the maximum principle of Boolean valued analysis, we infer that there is a finite sequence a : {1A,..., NA} ^ R+ inside V(B) satisfying
N
A
(Vx G XA)g(x) = £ a(k)fk(x) k=1A
=.
Put ak := a(kA) G R+ j for k := 1,...,N. Multiplication by an element in Rj is an orthomorphism of m(Y). Moreover,
N
B = ^2 ak Ak,
k=i
which completes the proof. >
11. Counterexample: no dominance
Lemma 1, describing the consequences of a single inequality, does not restrict the class of functionals under consideration. The analogous version of the Farkas Lemma simply fails for two simultaneous inequalities in general. Indeed, the inclusion {/ = 0} C {g ^ 0} equivalent to the inclusion {/ = 0} C {g = 0} does not imply that / and g are proportional in the case of an arbitrary subfield of R. It suffices to look at R over the rationals Q, take some discontinuous Q-linear functional on Q and the identity automorphism of Q. This gives grounds for the next result.
12. Reconstruction: no dominance
Theorem 2. Take A and B in L(X, Y). The following are equivalent:
(1) (3a € Orth(m(Y))) B = aA;
(2) There is a projection k € B such that
{k6B < 0} D {K6A < 0}; {-KbB < 0} D {-KbA ^ 0}
for all b € B.
< Boolean valued analysis reduces the claim to the scalar case. Applying Lemma 1 twice and writing down the truth values, complete the proof. >
13. Interval operators
Let X be a vector lattice. An interval operator T1 from X to Y is an order interval [T, T] in L(r)(X, Y), with T < T.
The interval equation B1 = XA1 has a weak interval solution [6] provided that (3X)(3A € A1 )(3B € B1) B = XA.
Given an interval operator T1 and x € X, put
Pti (x) = Tx+ - Tx-.
Call T1 is adapted in case T — T is the sum of finitely many disjoint addends, and put ~ (x) := —x for all x € X.
14. Interval equations
Theorem 3. Let X be a vector lattice, and let Y be a Kantorovich space. Assume that A{,..., AN are adapted interval operators and B1 is an arbitrary interval operator in the space of order bounded operators L(r)(X, Y). The following are equivalent:
(1) The interval equation
N
B1 = £ ak Ak fc=i
has a weak interval solution ai,..., aN € Orth(Y)+.
(2) For all b € B we have
{bB ^ 0} D {bA^ < 0} n • • • n {bAN < 0},
where A^ := Pai o ~ for k := 1,..., N and B := Pbi .
15. Inhomogeneous inequalities
Theorem 4. Let X be a Y-seminormed space, with Y a Kantorovich space. Assume given some dominated operators A\,..., An, B G L(m (X, Y) and elements u\,..., un, v G Y. The following are equivalent:
(1) For all b G B the operator inhomogeneous inequality bBx ^ bv is a consequence of the consistent simultaneous inhomogeneous operator inequalities bA\x ^ bui,... ,JjAnx ^ buN, i. e.,
{bB ^ bv} D {bAi ^ bui} n ■ ■ ■ n {bAN ^ buN}.
(2) There are positive orthomorphisms ai,..., aN G Orth(m(Y)) satisfying
NN
B = ^ ak Ak; v >^2 ak uk. k=1 k=1
16. Inhomogeneous matrix inequalities
In applications we encounter inhomogeneous matrix inequalities over various finite-dimensional spaces [7].
Theorem 5. Let X be a real Y-seminormed space, with Y a Kantorovich space. Assume that A G L(m) (X,Ys), A G L(m) (X,Ys), u G Yf and v G Ym, where s and t are some naturals.
The following are equivalent:
(1) For all b G B the inhomogeneous operator inequality bBx ^ bv is a consequemce of the consistent inhomogeneous inequality bAx ^ bu, i. e., {bB ^ bv} D {bA ^ bu}.
(2) There is some s x t matrix with entries positive orthomorphisms of m(Y) such that B = XA and Xu ^ v for the corresponding linear operator X G L+(Ys ,Yt).
17. Complex scalars
Theorem 6. Let X be a complex Y-seminormed space, with Y a Kantorovich space. Assume given ui,... ,un ,v G Y and dominated operators Ai,..., An ,B G L(m)(X,Yc) from X into the complexification YC := Y ® iY of Y. The following are equivalent:
(1) For all b G B and x G X the inhomogeneous inequality b\Bx\ ^ bv is a consequence of the consistent simultaneous inhomogeneous inequalities b\Aix\ ^ bui,..., J\Anx\ ^ Jjun, i. e.,
{b\B\ < bv} D {b\A\i < bui} n ■ ■ ■ n {J\A\n < Jun}.
(2) There are complex orthomorphisms ci,..., cn G Orth(m(Y)c) satisfying
N N
B = ^2 CkAk; v >^2 \Ck\uk. k=i k=i
18. Theorem of the alternative
Theorem 7. Let X be a Y-seminormed real vector space, with Y a Kantorovich space. Assume that A1,...,An and B belong to L(m)(X, Y).
Then one and only one of the following holds:
(1) There are x € X and b, b' € B such that b' ^ b and
b'Bx > 0, bAix < 0,..., bANx < 0.
(2) There are a1,..., aN € Orth(m(Y))+ such that B = J2N=1 akAk.
19. All is number
The above results, although curious to some extent, are nothing more than simple illustrations of the powerful technique of model theory shedding new light at the Pythagorean Thesis. The theory of the reals enriches mathematics, demonstrating the liberating role of logic.
References
1. Floudas C. A., Pardalos P. M. (eds.) Encyclopedia of Optimization.—Berlin etc.: Springer, 2009.— 4626 p.
2. Kusraev A. G., Kutateladze S. S. Introduction to Boolean Valued Analysis.—Moscow: Nauka, 2005.— 526 p.
3. Kusraev A. G., Kutateladze S. S. Subdifferential Calculus: Theory and Applications.—Moscow: Nauka, 2007.—560 p.
4. Scott D. Boolean Models and Nonstandard Analysis // In: Luxemburg W. (ed.) Applications of Model Theory to Algebra, Analysis, and Probability.— N. Y.: Holt, Rinehart, and Winston, 1969.—P. 87-92.
5. Takeuti G. Two Applications of Logic to Mathematics.—Tokyo; Princeton: Iwanami and Princeton Univ. Press, 1978.—137 p.
6. Fiedler M. (eds.) Linear Optimization Problems with Inexact Data.—N. Y.: Springer, 2006.—214 p.
7. Mangasarian O. L. Set containment characterization // J. Glob. 0ptim.—2002.—Vol. 24, № 4.—P. 473480.
Received August 3, 2009.
Kutateladze Semen Samsonovich Sobolev Institute of Mathematics, senior staff scientist 4 Koptyug Avenue, Novosibirsk, 630090, Russia E-mail: [email protected]
