A note on infinite systems of linear inequalities in Rn

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1 Carnegie Mellon University Research CMU Department of Mathematical Sciences Mellon College of Science 1973 A note on infinite systems of linear inequalities in Rn Charles E. Blair Carnegie Mellon University Follow this and additional works at: This Technical Report is brought to you for free and open access by the Mellon College of Science at Research CMU. It has been accepted for inclusion in Department of Mathematical Sciences by an authorized administrator of Research CMU. For more information, please contact

2 A Note on Infinite Systems of Linear Inequalities in by R n C. E. Blair Research Report March, This work was done while the author was supported by NSF Grant GJ-28457X1. NUT UUABY tftmemmeum UNIVERSITY

3 ABSTRACT A Note on Infinite Systems of Linear Inequalities in R n by C. E. Blair A necessary and sufficient condition for infinite systems of linear inequalities to have solutions is given, using the Kuhn-Fourier Theorem, which deals with finite systems.

4 A Note on Infinite Systems of Linear Inequalities in R by C. E. Blair Consider a system of inequalities A given by: a^x ^ b i5 iel where a., xer n and b^r, and I is arbitrary. We shall say that an inequality c*x ^ d can be linearly deduced (abbreviated: deduced) from A iff there is a finite subset J = { j.,..., j, } of I and non-negative reals m.,...,iru such that (i) m.a. = c and i J x x (ii) L m i b i ^ cl. iej (Compare with the notion of legal linear combination in [2], p. 8.) Clearly, if x satisfies the system A, x must satisfy any inequality deduced from A. Notice that if B is a set of inequalities, each of which can be deduced from A, then any inequality deducible from B is also deducible from A. (See [2], Chapter 1.) For I finite the Kuhn-Fourier Theorem [2, p. 10] gives a useful necessary and sufficient condition for the system A to have a solution: A has no solution iff the inequality 0*x ^ d can be deduced from A for some positive d. In this note, we prove the following result for infinite systems.

5 Theorem; Suppose that every finite subsystem of A has a solution. Then A has no solution iff there is a ver n such that vx ^ N can be deduced from A for every N. Proof; The "if" part is immediate: If there were an x satisfying A, then x would satisfy vx ^ N for every N, which is impossible. To prove the "only if" part we first show: Lemma: If A has no solution but every finite subsystem of A has a solution, then there is a sequence of w. R n with w- = 1 such that, for all i, w.^x ^> i can be deduced from A. Proof of Lemma; Let e. be the vector in R whose ith coordinate is 1 with all other coordinates 0. If, for some N, every finite subsystem of A had a solution x such that e.*x N (i = l,...,n) then A itself would have a solution. (For jel, let C. be the set of x whose coordinates are between -N and N and a.^x ^> b.. The C. are compact and, by assumption, every finite set of C. has nonempty intersection, so PI C. T* 0-) jel D Therefore we know that, for every N, there is a finite subsystem F c A, F = (a^-x ^> b^jiej}, j finite, such that the system has no solution. F U {e^x ;> -N i = 1,...,n} U {-e^x ;> -N i = 1,...,n} By the Kuhn-Fourier Theorem, there are nonnegative m.,...,m v, r.,...,r, s.,...,s such that i K. I n i n (iii) m i a i + 2 r.e. + s.(-e.) = 0 and iej

6 (iv) 2 m.b. + r i (-N) + 2 s^-n) > 0. iej L Let u = 2m.a. = 2 (s. - r.)e.. We can use the m. to deduce from F (hence, from A) u N «x =( 2 m i a i ) -x 2 2 m i b i > N(2(r i + s^ ) 2 N I U NI U N We now obtain the sequence needed by setting w = -i =V. Q.E.D. W U N' Next we obtain a v such that v x ^ N can be deduced, for all N, from the inequalities {w.»x^ i i = 1,2,. } given by the Lemma. Since each of the w-inequalities can be deduced from A, this will complete the proof. Since the w. R n, {w. i = 1,2,3,...) contains a finite linearly independent set {x.,...,x } such that every w. can be written as a linear combination of the x.. It is a trivial exercise in linear algebra to show that there exists a bound R > 0, such that for every vector w = t 2 a.x. i=l x x in the span of the (x i,...,x t ) with w = 1, we have a. < R for i = 1,..., t. t t Now let v = 2 x.. For any w, we have w = 2 a.x., i=l x N N i=l L x where a i < R, (with the a^ depending on N). But can be deduced from A. v x = w x + ( S ( l - a./ R) x.)- x^^ R JN i=]_ 1 1 R m mm mmt-mm

7 The Kuhn-Fourier Theorem is valid in any ordered field, but our theorem is not (recall our use of compactness arguments in the preceding proof). The following infinite system of linear inequalities in Q (= the rationals) provides a counterexample to our theorem for the rational field: (v) x^q, for each q < -x / q, for each q < - In fact, if our theorem held for Q, the vector veq would exist with the properties stated, and (v) would not be solvable in R, by the argument for the trivial direction of our theorem. However, x = /2 solves (v) in R. Acknowledgement: This problem was suggested to me by Robert Jeroslow, who also provided encouragement and advice. This work was done while being supported by NSF Grant GJ-28457X1. References [1] R. J. Duffin, "Infinite Programs", in Linear Inequalities and Related Systems, Kuhn and Tucker, eds., Princeton University Press, Princeton, New Jersey, [2] J. Stoer and C. Witzgall, Convexity and Optimization in Finite Dimensions 1^, Springer-Verlag, New York, /nlc 3/14/7 3

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