PROCEEDINGS OF THE AMERICAN MATHEMATICALSOCIETY Volume 116, Number 2, October 1992
ON ARCHIMEDEAN ORDERED VECTOR SPACES AND A CHARACTERIZATION OF SIMPLICES
GERHARD GIERZ AND BORIS SHEKHTMAN
(Communicated by William J. Davis)
Abstract. We show that a convex subset K of a linear space is a simplex if and only if it is line compact and every nonempty intersection of two translates of K is a homothet of K . This answers a problem posed by Rosenthal. The proof uses a reformulation of this problem in terms of Archimedean ordered spaces
Introduction Let K be a convex subset of a linear space E. If K x {1} is the base for a lattice cone in X x 5R, then K is called a simplex (see [7] for the definitions). A remarkable result of Kendall [4] shows that K is a simplex if and only if K is line-compact1 and the nonempty intersection of two homothets2 of K is a homothet of K. Moreover, every simplex has the property that it is line compact and every nonempty intersection of two translates of ATis a homothet of K. It is an open problem whether this last condition fully characterizes simplices, but a result of Rosenthal [7] shows that this is the case at least for a- convex subsets of topological vector spaces. Using Archimedean ordered spaces and their known relation to simplices, Rosenthal [7] reformulated this open problem as follows. Let E be an ordered vector space. Given a positive function p: E —»SR, we say that E has a /¿-lattice structure if p(u\) — p(u2) implies that u\ V u2 exists in E. Note that in an ordered vector space the existence of —u\ V H - u2 implies that u\ A u2 exists and is equal to -(-u\ V -u2). Hence in a p-lattice the infimum U\ A u2 exists whenever p(u\) = p(u2). Recall that a positive functional p is called strictly positive if u > 0 and p(u) —0 implies that u = 0. Rosenthal has shown that an affirmative answer to the following question leads to a positive solution of the open problem concerning simplices as stated above. Received by the editors December 4, 1990 and, in revised form, February 21, 1991. 1991 Mathematics Subject Classification. Primary 46A55, 46A40. Key words and phrases. Simplices, Archimedean ordered spaces, sublattices of C(K). 1 K is line-compact if the intersection of every line with K is compact. 2A homothet of K is a set of the form a + r • K , where r > 0 is a positive constant.
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Problem. Let E be an Archimedean ordered space. If E has a //-lattice struc- ture for a strictly positive linear functional p on E, is it necessarily a vector lattice? In this note we give an affirmative answer to this problem (Theorem 3.5). For definitions and results concerning vector lattices and Banach lattices, see [5, 3].
2. /Z-SUBLATTICES OF C(K) In the following, let E C C([0, 1]) be a subspace such that (i) I, x £ E and (ii) there is a positive measure p on [0, 1] such that 0, 1 G supp(/í) and such that for all f, g £ E with p(f) = p(g) we have sup(/, g) £ E. We would like to show that E contains all piecewise linear functions. We need some notation. For a point a £ [0,1], define maps Xa, pa £ C([0, 1]) by ( a -x if x < a, Xa{x) = { 0 else " and f x - a if x > a, *(jC)sS\q else." Since Xa(x) - pa(x) = a-x = a-l-x,vie have Xa £ E if and only if pa £ E. Also, since Xa(x) = sup{(a • 1 -x), 0), condition (ii) implies that there is a number a0 with 0 < ao < 1 such that X^ , pao £ e . 2.1. Proposition. Let E ç C([0, 1]) be a subspace such that (i) 1, x£E; (ii) there is a positive measure p on [0,1] such that 0, 1 G supp(/¿) and such that E is a p-lattice. Then E contains all piecewise linear functions. Proof. Every piecewise linear function is a linear combination of functions of the forms Xb and pb, 0 < b < 1 . Hence, we have to show that Xb, pb g E for all b with 0 < b < 1. Let a0 be a number with 0 < a0 < 1 such that X^ , pao £ E, and let 0 < b < 1 be given. We shall assume that a0 < b; the case where b < ao is treated similarly. Consider the function / given by f(x) — x - b and let r = J f(x) dp. Then, since p is positive and 0 G supp(/¿), it follows that ¡Xao(x)dp > 0, and we can find a number s such that
■j Xao(x)dp.
From our assumptions on E it follows that
gl = SUp(/, 5 • Xao) - S • Xa<¡£ E . We will consider two cases: (i) -b/ao < s . In this case, pb = g\ £ E. (ii) 5 < -b/ao . In this case, g\ = r{ • Xa¡ + pb where
/1 n b + sao ^ r\ — -( 1 + s) and a\ = —-< üq .
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Let R= pbdp, S = Xaodp. Since pb < gi, we have / g\dp> R. Again, we can find a number S\ so that si / K dp= g\dp,
and it follows that s\ > R/S. It follows that ao • R/S < s\ • ao = Si • Xao(0). Define g2 = SUp(g[, Si -Xao)-Si -Xao. Then g2 £ E, and either g2 = pb £ E (in the case where si • ao > r\ • a\) or g2 = r2- Xa2+ pb where a2 r„ ■an , and therefore gn+\ = pb £ E . Note that this procedure has to terminate after finitely many steps since rn- o-n < f„-\ • an-\ - (R/S) • ao and sn > R/S. D In the following result, we generalize the domain of the functions slightly. Before we do this, let us make some remarks concerning positive measures. Let K and K' be compact Hausdorff spaces, and let
Then v' = TUv), where Tl is the adjoint to the operator Tf. C(K') ^ C(K), f^f°4>- The supports of v and v' are related by the equation SUpp(i/) = 0(SUpp(l/)) . 2.2. Proposition. Let E c C([a, b]) be a subspace such that (i) 1,xg£; (ii) there is a positive measure p on [0,1] such that a, b £ supp(p) and such that E is a p-lattice. Then E contains all piecewise linear functions. Proof. Define a map
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p = T^(v), where v is a positive measure on [a,b]. Furthermore, since supp(p) — 0(supp(/^)), we have 0, 1 G supp(/z) if and only if a, b £ supp(i/). Hence Proposition 2.2 follows from 2.1. D 2.3. Theorem. Let K be a compact Hausdorff space, let E ç C(K) be a linear subspace containing a strictly positive function e, and let p be a strictly positive measure on K such that E is a p-lattice. Then E is a sublattice of C(K). Proof. First, consider the map Te-r. C(K) ^ C(K), f^e-'-f. This map is linear, bijective, and preserves the lattice structure. Hence, we may replace E by Te-,(E), e by l,and p by e-p in the statement of the theorem. It follows that we may assume w.l.o.g. that 1 G E. Now let 0 G E be arbitrary. We have to show that \ \= |7^(z')| —7^(1/1) £ E. D 2.4. Proposition. Let E ç C(K) be a subspace such that I £ E. Assume that E separates the points of K. Further, let License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ON ARCHIMEDEAN ORDERED VECTOR SPACES 373 to show that 4>' is strictly positive, we have to show that supp(/z) = K. Assume not, and let Xo G K \ supp(A"). We will construct a positive element fo £ E such that fo(xo) > 0 and fo(y) = 0 for all y £ supp(p). This will lead to the contradiction 0 < >{f0)= cp'(f) = ¡f0dp = /suppMf0dp = 0. As a first step, we show that for every x G K there is an open neighborhood U of x and a positive element f £ E such that 0 jé f and f(y) = 0 for all y £ U. In order to construct /, pick any nonconstant function g £ E and consider the element g - (g)• 1)+ = (g - 4>(g)• 1) V 0 belongs to E and is 0 on a neighborhood of U. Moreover, since g is not constant, g - cf>(g)-1^0. Since g - (g)• 1 is not negative, hence there is at least one point xx for which f(x\) = (g- 3. ARCHIMEDEAN ORDERED VECTOR SPACES In this section, we will show that every Archimedean ordered vector space that is also a /¿-lattice for a certain strictly positive functional p is actually a vector lattice. License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 374 GERHARD GIERZ AND BORIS SHEKHTMAN We will start our discussion with order unit spaces. This special case can be reduced to 2.5. The arguments are to a large extend standard in the theory of compact convex sets and boundary integrals, see also [1]. If (E, e) is an Archimedean ordered vector space with an order unit e, let S(E) denote the state space of E, that is S = {4>:E —>SR|0 is positive, and (p(e) — 1}. When equipped with the weak-*-topology, S is a compact convex set. The following result is a consequence of Herve's Theorem (see also Proposition 1.4.1 in [1]); the proof is a variation of the proof of Theorem II. 1.9 in [1]. 3.1. Theorem. Let (E, e) be an Archimedean ordered vector space with order unit e, and let p g dS(E) be an extreme point of S(E). If a, b £ E and aM b exists, then p(a V b) = max{p(a), p(b)}. Moreover, if License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ON ARCHIMEDEAN ORDERED VECTOR SPACES 375 If p(x) < 0, then p(-x) > 0, hence -x £ C - C by the previous argument, and thus x £ C - C, since C - C is a linear subspace. D We are now able to show 3.5. Theorem. Let E be an Archimedean ordered vector space, and assume that the positive cone of E is generating. If there exists a strictly positive functional p on E such that E is a p-lattice, then E is a vector lattice. Proof. For each positive u £ E let Eu = U«>o{x £ E: nu < x < nu} be the order ideal generated by u . Then (Eu, u) is an Archimedean ordered vector space with order unit u. Moreover, the restriction of p to Eu defines a strictly positive functional on Eu , and Eu is a //-lattice with respect to this functional. Hence each Eu is a vector lattice by 3.3. Since E = \JU>0Eu , it can be expected that E is also a vector lattice; we just have to show that the supremum of the element x £ E with 0 does not depend on the order ideal Eu in which it is computed. Let us denote the supremum of x and 0 in Eu by xu . Let u and v be two positive elements such that x G Eu n Ev . If u < v , then Eu ç Ev , and hence it follows that xv < xu since xu g Ev is an upper bound of x and 0. Because Eu is an order ideal, it follows that xv £ Eu, and since xu is the least upper bound of x and 0 in Eu, we conclude that xu < xv , i.e., xu = xv . If u and v are arbitrary positive elements such that x £ Euf) Ev , then also x G Eu+V , and we have just argued that xu = xu+v = xv . Therefore the supremum of x and 0 exists in E, and F is a vector lattice. D In connection with Rosenthal's result in [7] we now obtain 3.6. Corollary. Let K be a line-compact convex set such that the nonempty intersection of two translates of K is a homothet of K. Then K is a simplex. References 1. E. M. Alfsen, Compact convex sets and boundary integrals, Springer-Verlag, New York, Heidelberg, and Berlin, 1971. 2. H. H. Schaefer, Topological vector spaces, Springer-Verlag, New York, Heidelberg, and Berlin, 1971. 3. _, Banach lattices and positive operator, Springer-Verlag, New York, Heidelberg, and Berlin, 1974. 4. D. G. Kendall, Simplexes and vector lattices, 3. London Math. Soc. (2) 37 (1962), 365-371. 5. W. A. J. Luxemburg and A. C. Zaanen, Riesz spaces. I, North-Holland, Amsterdam and Berlin, 1971. 6. H. Rosenthal, O -convexity, Functional Analysis Proceedings, University of Texas, Lecture Notes in Math., vol. 1332, Springer-Verlag, Berlin, Heidelberg, and New York, 1988, pp. 156-174. 7. _, On the Choquet representation theorem, Functional Analysis Proceedings, University of Texas, Lecture Notes in Math., vol. 1332, Springer-Verlag, Berlin, Heidelberg, and New York, 1988, pp. 1-32. 8. K. Yoshida, Functional analysis, Springer-Verlag, New York, Heidelberg, and Berlin, 1965. Department of Mathematics, University of California, Riverside, California 92521 E-mail address : [email protected] Department of Mathematics, University of South Florida, Tampa, Flordia 33620 E-mail address : [email protected] License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use