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On a Theorem of Von NeumannAuthor(s): Lloyd L. DinesSource: Proceedings of the National Academy of Sciences of the United States of America,Vol. 33, No. 11 (Nov. 15, 1947), pp. 329-331Published by: National Academy of SciencesStable URL: http://www.jstor.org/stable/87659 .

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VOL.33, 1947 MA THEMATICS: L. L. DINES 329ON A THEOREM OF VON NEUMANN

BY LLOYD L. DINESDEPARTMENT OF MATHEMATICS, SMITH COLLEGE

CommunicatedAugust 13, 1947In these PROCEEDINGS,ugust 15, 1946, Loomis' presented a simpleproof of a theorem dud to J. von Neumann2' 3 and formulated by Loomisas follows:Let aij and bj be two rectangular matrices (i = 1, ..., n), (j = 1, ..., m).such that aij > 0 for all i, j. Then there exists a unique X and vectorsx =

(X,..., xm), y = (y1, ..., Yn)subject to Xj_ O,0 O,m, xj = 1 and Zlyir = 1such that m mX E aijxj E bijxj, i = 1, ..., n, (1)j=l j=ln nX E aijy < bijyz,j = 1, .. w . (2)i=l i=1

The present note gives an alternative simple proof, the essence of whichis a reduction of the question to consideration of a single system of linearequations. This reduction is accomplished in two steps: 1st, replacementof (1) and (2) by an equivalent pair of adjoint linear systemsm

ajx >O , i= 1, ...,n + m, (1')j=1n+m

E Coyi = 0, j =1,...,m; (2')i-1and 2nd, utilization of a known relationship4 between adjoint systems ofform (1'), (2'). This relationship is simply that the system of inequalities(1') admits a solution x which does not annul all the left members f, and onlyif, thesystem of equations (2') admits no positive solution y (everyyi positive).The system (1') consists of n + m inequalities, the first n of which arethose in (1) after obvious transpositions, and the last m are the conditionsxj _ 0, j = 1, ..., m, which occur as collateral conditions in the statementof the theorem.The system (2') consists of m equations, which are equivalent to the minequalities of (2) by virtue of the introduction of m new and essentiallynon-negative variables yn+l, ..., yn+m.The explicit definitions of the coefficients aio are

aij = Xaij - bi, i = 1, ...;, n; 1, ..., m,ai = a,j i = n + 1, ..., n + m; j 1, ...,m,where bij = 1 if i = n + j; otherwise it is zero.

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330 MA THEMA TICS: L. L. DINES PROC.N. A. S.

In view of the relationships which have been indicated above, thevon Neumann theorem will be justified if we show the existence of a uniquevalue of the parameter X for which the system of equations (2') admitsno positive solution y = (yl, ..., y,+m), but does admit a non-negativesolution with at least one yk positive.To this end, we consider the explicit form of the system (2'), whichmay be writtenn

(Xa, - bij)yi + yn+j = 0, j = ..., m. (2")i=1Since aij > 0 for every i, j, it is obvious that for X sufficiently small the

system will admit positive solutions y, while for X sufficiently large it willadmit no such solution. Furthermore if it admits a positive solution forany particular value of X, it will admit such a solution for any smallervalue. We denote by Xo he least upper bound of the values of X for which(2") admits a positive solution y. Since to every positive solution y therecorresponds by simple normalization a solution for which Zln+myk = 1,it follows from the compactness of the closed and bounded sub-spacedefined by yk > 0, Zln+myk = 1 that to the value X =Xo there correspondsa solution of (2") in this subspace; call it y? = (y, , . , y+m).At least one of the co6rdinates yk? must be zero. For if they were allpositive, a sufficiently small e > 0 would make correspondto the parametervalue Xo + e a positive solution y' = (yl', .. ., yn+m') with Yi' = yi?, (i = 1,

.., n), and y,,+.' = yn+j? - e Zlnaijyi0, (i = 1, ..., m), thus contradictingthe definition of Xo. In Xowe have therefore one value of the parameter Xwhich affords an admissible solution.No second value of X does. For if Xi > Xodid'afford an admissiblesolution y', then 2"l(Xila, - boj)yi' _ O, (j = 1, . .., m), and consequently,for any e > 0

nZ [(x - e)ai - bi]yi' < 0 j = 1, ...,m,i=1

and from the continuity of the linear functions these inequalities remain'strictly valid if each yi' be replaced by yi" = yi' + qi with 71i> 0 andsufficiently small. Defining y,+ j", (j = 1, ...m) by the equations

E [(Xi - e)ai - bi]yi" + y,+j" = 0 j = 1,. .., m,i=1we would then have a positive solution corresponding to the parametervalue X1- e > X0, hus contradicting the definition of X0.This completes the proof of the theorem. The method of proof definitelysuggests the possibility of generalization, since the vital relationshipbetween the adjoint systems (1') and (2') persists when the finite matrix


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VOL.33, 1947 MATHEMATICS: L. GARDING 331

aij is replaced by a function a(p, q) and the summations are replaced bymore general linear operators.

1Loomis, L. H., "On a Theorem of von Neumann," Proc. Nat. Acad. Sci., 32, 213-215(1946).2von Neumann, J., "Uber ein okonomisches Gleichungssystem, etc.," Ergebnisseeines Mathematischen Kolloquiums, 8, 73-83 (1937).3 von Neumann, J., and Morgenstern, O., "Theory of Games and Economic Behavior,"Princeton University Press (1944).4Dines, L. L., "Convex Extension and Linear Inequalities,"Bull. Amer. Math. Soc.,42, 353-365 (1936).

NOTE ON CONTINUOUS REPRESENTATIONS OF LIE GROUPSBY LARS GARDING

INSTITUTE OF MATHEMATICS, LUND, SWEDEN, AND PRINCETON UNIVERSITYCommunicated August 28, 1947

Let G be an analytic Lie group with unit element e and let a -* T(a),(a e G), be a representation of G as bounded operators T(a) on a Banachspace B such that for every x e B, T(e)x = x and T(a)x -> T(b)x whena -> b. Let A = (a(s)), (s real), be a one-parameter subgroup in G anddefine

TAX = lim,_es-l((T(a(s)) - )xwhenever the limit exists in which case x is said to be in the domain ofTA. I. Gelfand proved that every such domain is dense in B.1 By a slightextension of his method we will prove that they have a dense intersectionwhen A varies, thus answering a question left open by W. Wigner2 and V.Bargmann.3Let r be a not-negative integer or + co, let Cr be the class of real func-tions defined on G with continuous derivatives of order < r, (< r if r = oo),and let Cr?be the subset of Cr whose elements vanish outside a compactset and let Br be the set of elements of B of the form

x(h) = cGh(b)T(b)xdb (h e Cr?, x e B),where db is a left invariant volume element on G.

THEOREM. Every Br+1 is dense in B, it is in the domain of any TA andTAB,+l is contained in Br for any A.Proof: One has with s = 0

s-'(T(a(s)) - l)x(h) = s-l fG h(b)(T(a(s)b) - T(b))xdb =JG s-l(h(a-(s)b) - h(b))T(b)xdb.

VOL.33, 1947 MATHEMATICS: L. GARDING 331

aij is replaced by a function a(p, q) and the summations are replaced bymore general linear operators.

1Loomis, L. H., "On a Theorem of von Neumann," Proc. Nat. Acad. Sci., 32, 213-215(1946).2von Neumann, J., "Uber ein okonomisches Gleichungssystem, etc.," Ergebnisseeines Mathematischen Kolloquiums, 8, 73-83 (1937).3 von Neumann, J., and Morgenstern, O., "Theory of Games and Economic Behavior,"Princeton University Press (1944).4Dines, L. L., "Convex Extension and Linear Inequalities,"Bull. Amer. Math. Soc.,42, 353-365 (1936).

NOTE ON CONTINUOUS REPRESENTATIONS OF LIE GROUPSBY LARS GARDING

INSTITUTE OF MATHEMATICS, LUND, SWEDEN, AND PRINCETON UNIVERSITYCommunicated August 28, 1947

Let G be an analytic Lie group with unit element e and let a -* T(a),(a e G), be a representation of G as bounded operators T(a) on a Banachspace B such that for every x e B, T(e)x = x and T(a)x -> T(b)x whena -> b. Let A = (a(s)), (s real), be a one-parameter subgroup in G anddefine

TAX = lim,_es-l((T(a(s)) - )xwhenever the limit exists in which case x is said to be in the domain ofTA. I. Gelfand proved that every such domain is dense in B.1 By a slightextension of his method we will prove that they have a dense intersectionwhen A varies, thus answering a question left open by W. Wigner2 and V.Bargmann.3Let r be a not-negative integer or + co, let Cr be the class of real func-tions defined on G with continuous derivatives of order < r, (< r if r = oo),and let Cr?be the subset of Cr whose elements vanish outside a compactset and let Br be the set of elements of B of the form

x(h) = cGh(b)T(b)xdb (h e Cr?, x e B),where db is a left invariant volume element on G.

THEOREM. Every Br+1 is dense in B, it is in the domain of any TA andTAB,+l is contained in Br for any A.Proof: One has with s = 0

s-'(T(a(s)) - l)x(h) = s-l fG h(b)(T(a(s)b) - T(b))xdb =JG s-l(h(a-(s)b) - h(b))T(b)xdb.


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