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On Archimedean Ordered Vector Spaces and a Characterization of Simplices 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. © 1992 American Mathematical Society 0002-9939/92 $1.00+ $.25 per page 369 License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 370 GERHARD GIERZ AND BORIS SHEKHTMAN 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 . License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use ON ARCHIMEDEAN ORDERED VECTOR SPACES 371 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<a\< a0 and r2 • a2 = rx • a\ - s\ • ao < r\ • a\ - (R/S) • ao. We continue in this way with r2 and a2 in the place of r¡ and a2 until we finally find an index n such that s„ • ao > 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 <f>: K -> K' be a continuous map. If v is a measure on K, then define a measure u' on K' by /ddv'= f(fo<f>)dv. 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 <P-[0> 1] -►[a, b], x^a + (b-a)-x. Then the map 7VC([a,ô])-C([0,l]), f^focp is a (norm-preserving) linear bijection that respects the lattice structure of C([a, b]) and that also sends the space of all piecewise linear functions on C([a, b]) onto the space of all piecewise linear functions on C([0, 1]). More- over, p is a positive measure on C([0, 1]) if and only if it is of the form License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 372 GERHARD GIERZ AND BORIS SHEKHTMAN 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 \<j)\£ E. Let a = min <b(x), b = max 6(x). 0<max<l 0<x<l Consider the linear operator Tf.C([a,b])^C(K) f^focj). Again, this operator is linear and a lattice homomorphism (in the sense that T*(\f\) = IW)I for all continuous / G C([a, b])). Let F = T~X(E) and let p' — T^(p). Then F is a linear subspace of C([a, b]) with 1 G F (since 7^(1) = 1 o (p = 1), and we have x £ F ; indeed, x £ F means that i £ F, where i(x) = x. But i £ F is equivalent to T$(i) = i o0 — cf>g E, and the last statement is true by our assumptions.
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