Tv’eST‘TGTOT‘TE NORMS AND SEMTNORMS ON AN ARCHTMEDEAN VECTOR LATTICE Thesis for the Degree of Ph. D. MICHIGAN STATE UNTVERSTTY ROSAUND REICHARD 1970 LIBRARY " ‘lVlitillizgetzl Eitzatza ' University This is to certify that the thesis entitled MONOTONE NORMS AND SEMINORMS ON AN ARCHIMEDEAN VECTOR LATTICE presented by Rosalind Reichard has been accepted towards fulfillment of the requirements for thD-_. degree in_Ma_themaJ;ics ivgtx~k7bée 74% flajor professor ( /" (7' vi) (I , -, .\ Date ,7 {727/2 “67/ [17) /; 75 0-169 Y BIN‘DING BY nuns & snus' enmT/2 ) + +-TTE:\§(n)|. We shall show that (E, s, Hon) is not norm de:;: in (E, s, n-HM) but is norm dense in (E, s, H.H1) where Haul = n§1(\§c(n)|/2“) + L(\$E\) and L is any Banach limit. This not only indicates that u-HM and H-Hl are not equivalent, but also answers a question of Nishiura [15, Problem 1]. In order to prove that (E, s, H-H) is norm dense in (E, s, H-Hl) we note that for any i E E = La and any positive number s > 0, there exists a positive integer N such that ; 2M/2n < e where M is a bound for the sequence x. New n=N+l let x (x(l),x(2),...) E c be the element such that x(n) = x(n) for n = 1,2,...,N and x(n) = L(\§‘) for n >>N. Then ” n “x - RH]. S 2‘, ZM/Z + 0 < e. n=N+l However, we can show that (E, s, H-H) is not norm dense in (E, s, “'“M) by choosing R = (1,0,1.O.---) 6 La and noting that L(§) = 1/2. If x is any element in c, it is easy to see that “x - xHM 2 1/2. As Example 0.2 illustrates, the normiextensions of a normed vector lattice need not be equivalent. However, the example given is not a Banach lattice. Therefore, we ask whether all extensions of a complete, monotone norm on a vector lattice E must be equivalent on E. Chapter One of this paper is involved with an attempt at answering this question. Nishiura [15, Theorem A] has shown that if (E, s, H-H) is a Banach lattice then “on" is complete on E. Since all monotone, complete norms on a vector lattice are equivalent [14, Theorem 30.28] the above question asks whether every extension of a complete norm on E must be complete on E. In Section one we consider conditions under which norm extensions of a normed vector lattice are unique and apply these results to the question of equivalence of norm extensions for Banach lattices. The boundedness property and some of its con- sequences on normed vector lattices is the subject of Section two since this property plays an important role in the equivalence of norm extensions. In Section three we discuss the Egoroff property on normed vector lattices and we obtain a Riesz-Fischer Theorem as well as some results on norm extensions which aid us in answering the above question. We shall need the following definitions in our discussion: Definitions 0.3. A subvector lattice M C E of a vector lattice E is a linear subspace of E such that sup(x,y) is in M whenever x,y 6 M. A linear subspace M of E is an ideal in E if y is an element of M whenever \y‘ s x and x E M. An ideal M in E is a band in E whenever sup(B) = x0 in E and B C M implies that xo 6 M. A subvector lattice M of E is said to be order dense in E whenever for all 0 s x E E there exists a directed system 0 s x t (given a any a1, 02’ there exists an 03 such that x 2 x and 0’3 011 x 2 x ) of elements of M such that sup(x : a 6 a0 = x. 03 02 0 Definitions 0.4. A vector lattice E is universally complete if it is Dedekind complete and sup(xa: a E 60 = x0 exists in E whenever [x : a E 6% c E and inf(x ,x ) = 0 for all a1 f a2. The vector lattice E# is said to be a universal completion of E whenever E# is universally complete and E can be isomorphically embedded as an order dense subvector lattice of E#. A vector lattice E has a universal completion if and only if E is Archimedean and any two universal completions are isomorphic [13, Theorem 34.4]. In Theorem 39.1 of Modern Spectral Theory[l3], H. Nakano characterizes a normed vector lattice (E, s, H-H) having the property that whenever {x0} is a set of elements in E which is bounded above in E then sup(Hxa“: a E 60 = inf(HyH: y is an upper bound for {xa}). Such a normed vector lattice must be a norm dense subvector lattice of a space C(K) of continuous, bounded functions, vanishing at infinity, and defined on a locally compact, Hausdorff Space K. The first part of the proof of this theorem can be used to prove the following very useful result. Theorem 0.5. Let (E, s, H-H) be a normed vector lattice such that whenever {xa} c E+ (E+ = the positive elements of E) and sup(xa: a E 60 = x0 in E then sup(“xaH: a 6 a0 = HXOH. If {xa} C E+’ and Hxau s M for all a, then sup(xa: a E 60 = x exists in E#. O The proof of this theorem involves decomposing each x a into the supremum of a disjoint collection of elements in E# and in this way forcing an element of E# to be above each of the x . We do not reproduce the proof here because it would a involve introducing several terms that we shall not need else- where in this paper. However, for the convenience of the reader, the proof of Theorem 0.8, below, can be found in the Appendix. The importance of Theorems 0.5 and 0.8 will become evident in Sections one and three when we apply these results to the problem of norm extensions. Before stating Theorem 0.8, we shall need the following definitions. Definitions 0.6. A net [xaz a 6 a9 in a vector lattice E is said to decrease (increase) £2. x0 in E (notation: i = ° : = : xa 1 x0, xa 1 x0) f xO inf(xa a 6 a0 (xo 8up(xa a E 60) and x z x whenever 2 a x s x whenever 2 a . A as B (as a) net {x : a E GQ is said to order converge to xO in E a __ (notation: x 3 x0) if [x } is an order bounded subset of a a E and there exists a net (ya: a 6 a0 in E that decreases to o and lxa - xo| 3 ya for all a e 4. Definitions 0.7. Let (E, S, H°H) be a normed vector lattice. no“ is semi-continuous (gfsemi-continuous) whenever O S xa 1 x0 in E (0 s xn 1 x0 in E) implies that sup(“xau: a’E a0 = “x0“ (sup(“xnnz n E N) = “x0H). A vector lattice E is said to be order separable whenever given any subset A C E such that sup(A) = x0 exists in E, then there exists a countable subset A' c A such that sup(A') = x0 in E. Remark. Whenever (E, S, No“) is c-semi-continuous and E is order separable it follows that “-H is semi-con- tinuous. Examples of semi-continuous norms are frequently found; e.g., let E = c and “x“ = sup(x(n): n E N). The norm prOperty stated in Theorem 0.5 is used very strongly at one point of the proof. However, we can obtain the following theorem using a slight variation. Theorem 0.8. Let (E, s, H'H) be a normed vector lattice such that “o“ is semi-continuous (a-semi-continuous). If 0 s x“: (o s xn 1) in E and uxau s M for all a e a (Hxnu S M for all n E N) then sup(xa: a E G? = x0 exists 7 in E# (sup(xn: n E N) = x0 exists in E1). As noted above, applications of Theorems 0.5 and 0.8 can be found in Sections one and three. However, one immediate reSult is the following theorem which was proved by W.A.J. Luxemburg [7, Theorem 61.7] in a different manner. Definition 0.9. Let (E, S, H'H) be a normed vector lattice. “-u is said to be locally norm complete whenever [xn} C E, ‘xn| S x0 6 E for n = 1,2,..., and Hxn - xm“ a O as n,m a¢n implies there exists an element x in E such that “xn - xula 0 as n ~»m. Theorem 0.10. Let (E, S, “'H) be a normed vector lattice such that E is Dedekind o-complete and .H is a-semi-continuous. Then (E, S, H-H) must be locally norm complete. Proof. Suppose {xn} C E is Hon-Cauchy and \xn‘ S xo E E for all n E N. We can assume without loss of n generality that uxn - xn+1H S 1/2 for n 8 1,2,... . Con- sider the sequence 8n = ‘XZ - xl‘ +"x3 - xz‘ +-o-+"xn+1 ~ xn] * and note that by Theorem 0.8 sn t x in E# since “Sun . “‘32 ‘ xl‘ +"'+"Xn+1 - xn‘u S 1 for all n and H.” is a-semi-continuous. But then we must have # x +(x2 - x +---+-(xn - X“) 3 x E E . (Absolute convergence 1 l) of a series implies convergence of the series in a Dedekind o-complete vector lattice [23, p. 101]). o - # . Therefore, we have xn « x in E . However, Since A ‘xn‘ S xo E E C E and E is an ideal in E# we must have 0 _p [1” Q E E and xn i in [23, p. 64]. Now let y = inf (sup (x )) which exists in E since E is Dedekind n kzn k o-complete and {xk} is bounded above in E. We note that by definition of y we must have xn 3 y in E [16, p. 44]. o - . a - a x in E we most also have x = inf (sup (Xk)) However, since xn n kzn where the suprema and infima are taken in E. But suprema and infima are preserved under the embedding E w E. Hence, we to most have i = y and xn E in E, Q E E. To complete the proof of the theorem*we need only show that xnia Q in norm. Q = x1 + (x2 - x1)‘+--~ implies that ‘x - xn‘ = ‘(xl +(x2 - x1) +----+-(xn - xn_1) +'--) - a) - (x1 +-(x2 - x1) +~--+-(xn - xn_1))‘ S k2n|xk+1 - xk‘. It now - m “ n-l follows that “x - xnu S H 2 ‘xk+1 - xk]u S. 2 “x1“,1 - ka S 1/2 , k=n k=n . with the second inequality holding because “-H is a-semi- continuous. Therefore, “x - an a 0 as n a m. The purpose of presenting the above theorem is to demon- strate some of the techniques used in proving theorems in the theory of vector lattices, and also to indicate the strength of Nakano's result. As one reads Chapter One, he or she should keep in mind the main problem (i.e., when are norm extensions of a Banach lattice complete) as it is the motivation for the material pre- sented here. Chapter Two discusses a problem of a different nature and it is presented in this paper because it is an interesting result. In this last chapter we prove that every universally complete vector lattice with a countable collection of continuous, ‘monotone seminorms defining a locally convex, Hausdorff topology must be the space of all real sequences or a finite dimensional space. CHAPTER ONE ARE THE NORM EXTENSIONS OF A BANACH LATTICE EQUIVALENT? Section One. Uniqueness of Norm Extensions Given a Banach lattice (E, s, u-H), we know that the maximal extension is always complete [15, Theorem A]. There- fore, whenever norm extensions are unique, the question of the completeness of all norm extensions can be easily answered. For this reason, it becomes important to determine conditions under which “.HM is the only extension of a monotone norm H'H defined on E. Definition 1.1. A norm H-H defined on a vector lattice E is said to be continuous (o-continuous) whenever 0 S xa 1 0 (0 S xn l 0) in E implies that inf(“xau: a 6 a0 = O (inf(“an: n E N) = 0). Remark. A o-continuous norm is continuous whenever E is order separable. Also, a continuous norm is always semi- continuous. However, if E = La and “x“ = sup(\x(n)\: n E N) then “on is semi-continuous but not continuous. We shall prove that whenever (E, S, H-H) is a normed vector lattice and “c“ is continuous, “'HM is the only extension of “-H to E. This result was also proved by V.A. Solovev 10 ll [19, Theorem.4], but he obtained the result as a corollary to a combination of several theorems. We shall give a short direct proof. Once we have the fact that continuous norms extend uniquely, it becomes a natural question to ask.whether semi-continuous, a-semi-continuous, or o-continuous norms extend uniquely. The answer to all three questions is negative and several examples will be given to illustrate these points. At all times, we shall apply our results to the question concerning the equivalence of norm extensions. Theorem 1.2. If (E, S, MD is a normed vector lattice and H“ is a continuous norm, then “H extends uniquely to E. 3529;. Let (p6!) = sup(uyn: o S y S m, y e E) for all E E E. If u-Hl is any extension of u-H to E we must have on?) S “1“:“1 S HxHM for all 5‘: E E. Hence, to show that “-H extends uniquely to E we need only prove that m(x) = uxuM for all s e E. Choose SEEE and let A={w-y:w2 m, OSysm, w,y e E}. A is a directed set in E so we can let A = {ua: o e d} where a is a directed set. Then 0 s ua i 0 since sup(y: 0 S y S hi], y E E) = inf(w: w 2 fit], w E E) = ‘5“. Using the hypothesis that “on is continuous we must have inf(“uaH: a 6 a0 = 0. Hence, given any e > 0 there exists an element w 2 \fi| and an element 0 S y S \fi‘ with w,y E E such that “w - y“ < e. However, “w“ - “Y“ S “w - y“. Therefore, uwu - Hy” < a and it follows that (not) = HEM- This proves that u.“ extends uniquely to E. 12 Corollary 1.3. Let E be a vector lattice having any continuous, monotone, complete norm “'“c defined on it. If (E, S, “-H) is a Banach lattice, every extension of H‘H to E must be complete. m3. Since “'Hc and W“ are both complete they must be equivalent [14, Theorem 30.28]. Therefore, there exist real numbers a,b > 0 such that fix“ s auxuc S bnxu for all x in E. It easily follows that “-H is continuous because “one is continuous. By the theorem, “'H extends uniquely to E and since H-“M is complete, every extension of “-u to E is comp 1e te . This corollary is also true if we assume that E is any Archimedean vector lattice such that E has a continuous, complete norm defined on it. For example, if E is any order dense subvector lattice of co (the real sequences which converge to zero) and E = c0 then if H.“ is any complete norm on E every extension of H-H to E = C0 must be complete. This follows because “xuc = sup(‘x(n)‘: n E N) is a continuous, complete norm on co. Then since H-HM is a complete norm on co, H'HM must be equivalent to H-Hc. This forces “OHM to be continuous on i = co. Then Ho“ must be continuous on E and the result follows. Although we shall see that, in general, a semi-continuous, monotone norm on a vector lattice E need not extend uniquely to E, for a very large class of semi-continuous norms we do get unique extensions. 13 + Definition 1.4. If E is a vector lattice, e E E is said to be a strong unit for E if for every x 6 E there exists a real number k > 0 such that ‘xl S xe. As in [23] we shall call a vector lattice with a strong unit a Space of bounded elements. If E is a space of bounded elements with strong unit e we can define a monotone seminorm on E by: ”X“ = inf(x: Xe 2 ‘xl). If E is Archimedean this seminorm is a norm and H.” is semi- continuous. To see this latter fact we choose a net {xat a G aq C E such that 0 S xa 1 X0 in E. By definition of “x0“ we have that xa S uxaH-e for all a 6 an This implies that sup(uxau: a E GO-e = sup(“xaH-e: a E 60 2 sup(xa: a 6 a0 = x0. Then “sup(“xan: a E gooe“ = sup(“xaH: a E 00 2 “x0“. Since we always have that sup(“xau: a 6 a5 S “x0“ it follows that H°H is semi-continuous. . If E is an Archimedean space of bounded elements with strong unit e then E is a space of bounded elements with strong unit e and qus = inf(x: Xe 2 ‘§\) is a semi-con- tinuous norm on E. Using a characterization of E as a Space of continuous, bounded functions defined on a compact set, V.A. Solovev [19, Theorem 5] proved that H-N extends uniquely to E. We shall give a direct proof of the fact that semi-continuous norms of this type extend uniquely. However, first we need the following lemma and remark. 14 Lemma 1.5. Let (E, S, H-H) be a normed vector lattice. If H-H1 and H-H2 are extensions of H°H to E and both H-H1 and H-HZ are semi-continuous on E then HitH1 = Hi‘tH2 for all i e E. Proof. Choose any § 6 E and let {xaz a E G? be a net in E such that 0Sx01|§|in E. Since HH1 and H-H2 (are semi-continuous we must have .sup(HxaH1: a 6 a0 = “in, and sup<\\x,n2= new =nszu2. nut M1 and M, agree on E. Therefore, HSEH2 a sup(HxaHZ: a 6 a0 = sup(HxaHl: a Q d) = HEHI 339355. It is known ([20, Lemma 1] and [6, Theorem 1.51) that if (E, S, H‘H) is a normed vector lattice and H-H is semi-continuOus then HfiH* = sup(HyH: 0 S y S ‘x‘) is a semi- continuous norm extension of H-H to E. As the example E = c and HxH I sup(‘x(n)‘: n E N) + lim |x(n)‘ demonstrates, this n—m is not necessarily true when H-H is not semi-continuous. In this case H§H* = sup(‘i(n)‘) + 122 |x(n)\, which is not a norm nan because it is not Sub-additive. Also whenever H-H is semi- . 6+ continuous and we choose any x E E and any two nets {xa: a 6 fig, {yB: B 618} in E such that O S Xa 1 fl and O S yB 1 E then H§H* = sup(HxaH: a E d0 = sup(HyBH: B 6‘8). We can prove this latter statement by fixing some so and noting that xa A y.BO 1 x A yB a (The notation xa A y8 = y . 0 80 means inf(xa , yB o)). The semi-continuity of H-H implies that sup(Hxa A y a 06 a0 = HyB oH. Then xa 2 xa A yBo for all H' so a implies that sup(Hxa H: a 06 a0 2 sup(Hxa A yBo H: a E a0 = = HyB H for all 30. This shows that sup(Han a 6 d0: 0 15 2 sup(HyBH: B 616). A similar argument proves the other inequality. Theorem 1.6. Let E be an Archimedean vector lattice with strong unit e. Define HxH = inf(x: xe e ‘XW). Then H-H extends uniquely to E. “Egggft We note that H’H is semi-continuous on E so A that H-H* is a semi-continuous norm on E. Hence, using Lemma 1.5 we need only show that H-HM is semi-continuous on E. We shall prove that HxHM = inf(x: Xe 2 \fi‘) for all a in E because the right hand side is a semi-continuous norm on E. Choose A > 0 such that Xe 2 \x‘. Then by letting w = le 6 E and noting that erH = xHeH = x we have that {l: he 2 \a|} : {HwH: w 2 \a‘, w 6 E}. Furthermore, if we choose w e E such that w 2 \i‘ it follows that HwH-e 2 ‘w| = w 2 ‘fi‘. And then {HwH: w 2 \sl, w E E} c c {1: is 2 ‘2‘}. Therefore, HfiHM = inf(x: is r \fi‘) and this completes the proof of the theorem. Using Theorem 1.6 we know that whenever GE, S, H-H) is a Banach lattice of bounded elements every extension of H°H to E must be complete (because H-HM is the only extension). However, it is also true if E is a vector lattice of bounded elements and H-H is any monotone norm which is complete on E, then every extension of H-H to E must be complete. This result will follow from Corollary 1.12. Having the results that a continuous norm on E must extend uniquely to E and a special class of semi-continuous 16 norms extend uniquely to E we ask.whether all semi-continuous norms extend uniquely to E. We shall give an example of a normed vector lattice (E, S, H-H) such that H-H satisfies the pro- perty that if {xa: a E aq is any subset of E+ (not necessarily directed upward) and sup(xa: a’E a0 ==xo then sup(HxaH; o 6 a9 = = onH. But H-H does not extend uniquely to E. This answers our question negatively since the stated property is stronger than semi-continuity. This example also shows that Theorem 3 in [20] is incorrect. Further, we shall prove that with an A added condition on E, H-H does extend uniquely to E. Example 1.7. Let E = c and e = (1,2,1,2,...) e E = L00. 3 is a strong unit for Lm. Therefore, for any fi 6 La HfiHS ‘ inf(x: Ag 2 ‘9‘) defines a semi-continuous norm on La. It follows that H'H = H-HSWE is a semi-continuous norm on E. We note that H-H also has the property that if {x0} C E+ and sup(xa: a E d0 = x0 then sup(HxaH: a E d§ = HXOH- EV=swmm=osyssLyen>=mh since both H-H* and WHS are semi-continuous norms on E which agree on E. (See Lemma 1.5). We shall show that HitHM ' inf(HwH: w 2 ‘fi‘, w E E) # H9H* for some E E E. Con- sider E = (1,2,l,2,...). HéH* = 1. However, given any w E E = c such that w 2 E we must have lim W(n) 2 2. It follows that HwH 2 2 since HwH = inf(x: :ng w). Therefore, HéHM ' 2 f 1 = HéH*. This proves that H-H* and H-HM are A two distinct extensions of H-H to E. 17 Our previous example is one in which there exists a non semi-continuous extension, namely H-HM, to E of a semi-con- tinuous norm H-H on E. However, H-H’c is an extension of A * H'H to E. Of course, it is not always true that H-H is an extension of H-H to E. As noted in a previous remark, if E = c and HxH = sup(‘x(n)‘: n E N) + lim ‘x(n)‘ then 7': W HfiH = sup(\x(n)‘: n E N) + lim ‘x(n)\ is not a norm on nam E = La. But in this example H-H is not o-semi-continuous. Therefore, we ask whether there exists a normed vector lattice * (E, S, H-H) such that H'H is c-semi-continuous but H-H is not an extension of H-H to E. Such an example would have to be non order separable. The following example answers the above-mentioned question as well as shows that o-semi-continuous norms need not extend uniquely. Example 1.8. Let X be any uncountable set of points and let E {f: f is a real-valued, bounded function on X and f is constant except possibly on a finite set of points in X}. Then E = {f: f is a real-valued, bounded function on X}. For every f E E let c(f) = that real number such that {x: f(x) = c(f)} is uncountable. Also let Hle = Sup(‘f(x)‘: x e X) and define “f“ = ”le + 2‘c(f)|. (i) (E, S, H-H) is a normed vector lattice and H'H is o-semi-continuous on E. It is easy to see that H-H is a monotone norm on E. To show the o-semi-continuity of H-H we choose any sequence {fn} in E such that 0 S fn 1 f in E. If c(f) = 0 then c(fn) = 0 for n = 1,2,... . And 18 sup(anH: n E N) = sup(anHI: n E N) = Hle = HfH since H-H1 is o-semi-continuous. If c(f) # O and e > 0 is any real number, then there exists a positive integer N such that c(f) - c(fN) S g. (It is crucial here that {fn} is a sequence and not an arbitrary net.) Hence, 2c(fn) 2 2c(f) - 23 for all n n N. Now, sup(anH: n e N) = sup(sup(fn(x)) + 2c(fn)) 2 - sup(sup(fn(x)) + 2c(f) - 2s) = Hle : Zeif) - 23 = HfH - 26. 1t follZws that sup(anH: n E N) 2 HfH. Since the other in- equality always holds we must have Sup(anH: n e N) = HfH and H-H is o-semi-continuous. (ii) H§H* = sup(HyH: 0 S y S |m\) does not define a norm on E. Let x = x1 U x2 where x1 n x2 = ¢ and card(X1) = card(X2) > R0. Let xx, 1x1, Ixz denote the char- acteristic functions of X, X1, and X2, reapectively. Then IX, IX , IX are elements of E and HIXH* = 3. However, 1 2 HIX1H* = HIX2H* = 1 because both 1x1 and 1X2 are zero on a tabl s t as HI + \\* 3 > 2 u n" + n uncoun e e . nce, = = * Xl Ix2 IX1 + HIX H which shows that H-H* is not a norm. 2 (iii) HxHM = inf(HwH: w 2 |x\, w e E) is not o-semi- continuous. Let {xnz n E N} be any countable set of points in X and define f(x) = {:1 if x = xn for n = 1,2,... 0 otherwise. Then f E E. For n = 1,2,..., define fn(x) = {I if x = Xk’ k = 1,2,...,n 0 otherwise. Then 0 S fn 1 f in E and sup(anHM: n e N) = sup(anH: n e N) = 1. However, HfHM = 3. Therefore, HfHM > sup(anHM: n E N). This l9 proves that “on“ is not o-semi-continuous. (iv) We now exhibit an extension of “-u distinct grom H°HM. Define HxHML = inf(:::.HanM: 0 s xn 1 ‘x‘ in E). Given any norm “on, H-HL is always a seminorm. (See Section three for a thorough discussion of H°HL.) For our particular example we shall show that H-HML agrees with H-H on E. This fact forces H HML to be a norm extension of H'“ to E. Obviously, H HML cannot be equal to “.“M on S because H°HM is not o-semi-continuous on E. Hence, H-HM and H-HML [11) are two distinct extensions of u-u to Choose any f in E+. If c(f) 0 and 0 s f; 1 f in S then we must have f; E E for n 1,2,... . And we 5 know that sup(HfSflM: n E‘N) = sup(“fguz e N) = “f“ since “on is o-semi-continuous on E. Suppose c(f) # 0. Then if e > O is any real number there exists a positive integer N such that {x: ffi(x) 2 f(x) - s} is an uncountable subset of x. Hence, f§(x) 2 c(f) - e for uncountably many x in x. Then, HffiHM 2 Sup(f§(x): x e X) + 2c(f) - ze. We now have sup(“fRHM: n e N) 2 S:p(s:p(f;(x)) + 2c(f) - 2.) = Hrul + + 2c(f) - 26 = Hf“ - 26. It follows that for any sequence {£3} in E with o S i; i f we must have sup(HERHM: n e N) = = Hf“. Hence, uanL = Hf“ for all f in E. We note that the norm defined in Example 1.8 is o-continuous as well as o-semi-continuous and “.“M is not a-continuous. 20 The following example is one in which H'H is o-semi- * continuous, “-H is not a norm, but “'HM is o-semi-continuous. We shall then indicate an interesting generalization of Examples 1.8 and 1.9. Example 1.9. Let X be an uncountable set of points and E = {f: f is a real-valued, bounded function on X and f is constant except possibly on a countable set of points in X}. Define H-H as in Example 1.8 and note that E is also as in Example 1.8. We merely remark that H'H is a-semi-con- tinuous and H'H* is not a norm on E as the proofs are exactly like those in the previous example. However, “- is o-semi- HM continuous. To show that u-HM is o-Semi-continuous we first note that E is Dedekind o-complete. Then we show that whenever (E, s, “'H) is a Dedekind o-complete normed vector lattice and H-u is a-semi-continuous, “OHM must be o-semi-continuous. The proof is due to V.A. Solovev [20, Theorem 1]. Suppose 0 5 En 1 f in E and choose any real number s > 0. By defini- tion of H°HM, there exists an element xn E E such that Xu 2 f“ and “xnu S Hf;“M + e for n = 1,2,... . We also choose an element x in E such that x 2 f and we can assume without loss of generality that xn s x for all n 6 N. Define the element yn = inf(xk: k 2 n) in E, n = 1,2,... . Since yn s x for all n there exists an element y in E such that 0 s yn t y. By the o-semi-continuity of H'H we know that sup(“ynfl: n 6 N) I “y“. Also, y 2 f and yn s xn for n = 1,2,... 21 implies that “E“M s Hy” = sup(HynH: n E N) s sup(anHM: n E N) +.€. Since a > 0 was arbitrary we have “E“M s SUP(Hf;HM: n 6 N). It follows that “.HM is a-semi-continuous. The previous two examples can be generalized in the following manner. Example 1.10. Let N1 be any cardinal number such that 81 > RC. Choose a cardinal number 32 and a cardinal number 33 such that R3 > R2 > RI. Let X be a set Such that card(X) = R3 and let E = {f: f is a real—valued, bounded function on X and f is constant except possibly on a subset of X having cardinality less than or equal to 81}. Then E = {f: f is real-valued and bounded on X}. For each f in E, let c(f) = that real number such that {x: f(x) # c(f)} has cardinality at most R1. For f in E let Hle = = sup(\f(x)\; x e x) and define Hf“ =HfH1 + 2\c(£)|. Again, it is easy to see that (E, s, H°H) is a normed vector lattice. (i) H -“ is Rz-semi-continuous on E. That is, if {a: o E 6% is any directed set such that its cardinality is less than or equal to R and O 3 Ed 1 f in E then 2 sup(Hf “: o 6 a0 = “f“. Suppose O s fo 1 f in E as just ' o described. If c(f) = 0 then c(fa) B 0 for all a e d’ and sup(“faH: a E d0 = “f“. If c(f) # 0 and e > O is any real number, there exists an index do such that c(f) - c(fo ) S e 0 (because card({a: a E a?) 5 R2 < 83). Now, as in part (i) of Example 1.8, it follows that sup(“fau: a E 00 = SUP($UP(f (X)) + 26(f )) 2 SUP(SUP(f (X)) +‘26(f) - 26) = a x a a a x a 22 = Hle + 2c(f) - 26 = “f“ - 26. Hence, sup(ufyu; o c G) = Hf“. (ii) H-H* is not an extension of n-H to E. Choose X1, X2 subsets of X such that X = X1 U X2, X1 0 X2 = ¢ and card(Xl) = card(XZ) = N3. The fact that u.u* is not a norm follows just as in Example 1.8. (iii) “.“M is not RZ-semi-continuous on E. Let {x8: B 616} be a set of points in X having cardinality equal to R2 and define f(x) = .’1 if x = x6 for any B 615 \0 otherwise. Then f E E and we shall define a net {fx: A E A} in E such that card({x: k E A}) = 32 and 0 s fx 1 f in E. For each 8 E 8 define fB(X) = fl if x = x8 to otherwise. Then sup(fB(x): B 618} = f(x) for every x E X. Consider the collection {f8: B 618} of elements in E and define {fx: l E A} = {collection of all finite supreme of the elements of {fB}}. Then {fx: A E A} is a directed net in E and 0 S f) 1 f in E. Also, card({x: x e A}) = N2. It now follows that H-HM is not Rz-semi-continuous because Sup(nfoM: l E A) = = sup(ufiu: i e A) = 1 but “f“M = 3. If we let E = {f: f is a real-valued, bounded function on X and f is constant except possibly on a subset of X having cardinality at most 82} we would have an example of a normed vector lattice (E, s, “-H) such that “-H is Rz-semi- continuous H-“* is not a norm and “’H is R -semi-continuous ’ 1 | ’ M ' 2 (noting that E is Dedekind Rz-complete). 23 Although Examples 1.8 and 1.9 are special cases of Example 1.10 we present them all here so that we can see how the generaliza- tion in Example 1.10 can be motivated. As we have seen by the previous examples, semi-continuity or a-semi-continuity does not insure uniqueness of norm extensions. As Example 1.7 indicates, even the stronger condition that {xa} is any collection of elements in E+' (not necessarily directed upward) and sup(xa: a 6 a0 = x0 implies 3“P(“xa“: a 6 d0 = on“ does not force norm extensions to be unique. However, when we add the condition that e A,w E E whenever w 6 E and e = sup(x E E+l “x“* s 1) (which exists in E# by applying Theorem 0.5) we shall have that “-u extends uniquely to E. We note that this is precisely what fails to hold in Example 1.7. However, this condition is not a heavy restriction as several examples, including any space of bounded elements, will clearly indicate. After proving the following theorem, we shall apply it to the question of completeness of norm extensions. Theorem 1.11. Let (E, S, H-H) be a normed vector lattice such that {xa} C E+I and sup(xa: a 6 d0 = x0 in E implies that sup(uxa“: a 6 a0 3 “x0“. In addition, letting 5 = sup(& 6 13+: “5211* S 1) (which exists in E#), we shall assume that e A,w 6 E whenever w E E. Then H-H extends uniquely to E. 12322;. Since “-H is semi-continuous we can extend “'u '30 E by MW = BUNNY“: 0 s y s 1&1, y e E). Also, 24 it is easy to see that for every {§a: a E a0 C E+ such that sup(fia: a E a0 - EC in E we must have sup(“fiau*; o e a9 = = u§o“*' Hence, applying Theorem 0.5 we know that g = sup(& 6 E+} “§“* 5 1) exists in E#. We extend “-H* to E# (allowing u-H* to be infinite # on some elements) as follows: for i 6 E , define qu* = * - a * = sup(HyH : 0 S y s htLy E E). Since H°H is semi-continuous on E it follows that if {y;: a E afl is any net in E and 0 s y; 1 ‘5‘ in E# then “xu* = Sup(“y;u*: a E d0. - * A .+ * (1) “en 3 1. For, let A = {x 6 E : Hi“ 5 1}. A is a directed subset of E because given any s,y E A we nust have us y y\\* = “in" v \\y\\* s 1. Then 5‘: v y e A and E V y 2‘fi, R v y 2 9. Let A = {eaz a E 60 G E where 47 is a directed set. Then 0 5 fig 1 a in E# and since “-H* is * - * semi-continuous, sup(ne;u : o 6 a0 = “a“ . Now since “can* 3 1 for all a and some elements of A must have H-H* - * * equal to one we have that Me“ = sup(“éau : a 6 a0 = l. . - # - - - * (ii) For any y E E , \yl s e if and only if Hy“ s 1. By the monotonicity of H-H* on E# we have that ‘y‘ s 5 - * implies that Hy“ s 1. On the other hand, suppose Hyu* S l and let 0 s Ea 1 \yl, {x;} C E. Then “x;“* s 1 for all a so that id 6 A for all a. Therefore, 5 2 x; and we have that |§1 = sup(xé: a E as S 5. Now let E'= {§ E E#: “xH* < +-m}. Then (E, s, H-H*) * is a normed vector lattice with “on semi-continuous, E is _ _, _. * an ideal in E# and e 6 E. Also, we can show that (E, s, H-H ) is a space of bounded elements with strong unit 5 and 25 u;u* - inf(x: )5 2 1&1). To prove this last statement we choose E E E' and let a - “;“*. Then “Ila ‘§\H* = l and (ii) implies that 1/a ‘5‘ S E. Therefore, \i\ S aé. This shows that E is a strong unit for El Also, u§u* = inf(x: )5 2 ‘§\) since \i1 S u£u*.é, which follows easily from (ii). In order to show that “0“ extends uniquely to E we need only show that for every n e E uiu* = inf(x: )é 2 \i\) 2 2 inf(HwH: w 2 |fi‘, w E E) = ufiuM. Choose any x > 0 such that is 2 |§|. By hypothesis, w A E E E whenever w e E. Hence, if w 2 ‘%| and w E E then w' = w A )5 E E and w' 2 \fi‘. Since uw'“ - uw'u* S H15H* = x it follows that for all i such that x8 2 121 there exists an element w' 2 1§| with w' E B such that “w'“ S x. Hence, inf(Hw'“: w' 2 ‘fi‘, w' E E) S S inf(x: A; 2 ‘§‘). This completes the proof of the theorem. Corollary 1.12. Let E be a vector lattice having any monotone, complete norm H-H defined on it such that (E, 2, Ho”) satisfies the hypotheses of the theorem. Then if H‘on‘ is any monotone, complete norm on E every extension of H|~H\ to E must be complete. ‘ggggg. Let (E, S, H'H) be any Banach lattice satisfying the hypotheses of the theorem and Hl'U1 any complete, monotone norm on E. Then H'“ and HI'Ul are equivalent. Hence, therc exist constants a,b > 0 such that “x“ S a“\xu\ S beH for all x in E. Let “i'U‘l be any extension of u|oh| to E. Then usu* S aH‘fiu‘l s bHEHM for all E e E. But uiu* = uiuM so that “1'“‘1 is equivalent to u-HM. This implies that “1°“11 is complete on E. 26 Example 1.13. Let E = c and “x“ = sup(‘x(n)‘: n E N). Then (E, S, H'“) satisfies the hypotheses of Theorem 1.11 and H'H is complete on E. Now defining u|xn\ = sup(|x(n)\: n e N) + + lim |x(n)‘ we know that (E, S, HM”) is a Banach lattice angd:everal distinct extensions of “‘-“| to E exist. (See Example 0.1.) However, according to Corollary 1.12, every extension of H‘ou‘ to E must be complete. Remark. Corollary 1.12 can be applied to any Archimedean vector lattice with strong unit e and any complete, monotone norm “"H' defined on E. Let “x“ = inf(x: he 2 ‘x|). Then H‘-“‘ and H.“ are equivalent on E. To prove this latter fact we note that H°H-convergence always coincides with (r)- convergence (xn 5 x if there exists a z in E and a sequence of real numbers X“ i 0 such that \x - xn‘ S xnz for In = 1,2,...) [23, Theorem‘VII.4.l]. And xn a x in norm H|-H\ and “‘-H| complete implies every subsequence of {xn} has a subsequence which (r)-converges to x [22, Theorem‘VII.2.1]. Therefore, (E, S, H'H) satisfies the hypotheses of Theorem 1.11. This forces every extension of H‘°“| to E to be complete. As illustrated in the foregoing paragraphs, if (E, S H-H) ..‘. ’ H1) force “-H to extend uniquely to E. As indicated, these pro- is a normed vector lattice various properties on (E, S, \ perties have applications to the problem of when every extension of a complete, monotone norm on E must be complete on E. Therefore, we consider relationships among the following properties 27 in order to clarify the preceding results: (Pl). Every set of elements {xa} : E+ which is bounded above in E has an upper bound X0 in E Such that sup satisfies (23), then (E, S, H'H ) also satisfies (P3). This fact was actually used in the beginning of the proof of Theorem 1.11. We shall now prove it for the convenience of the reader. Let (Ea: a E afl be a collection 28 of elements in E+ such that sup(fia: a E d0 = E0 in E. Since E is a Dedekind completion of E, for each a 5.7 there exists a net x : E A c E such that x 1 E . { KEG x } X’Q’ )i 0’ We know that “-H* is semi-continuous on E. Hence, sup(“x H: l G A) = “5? H* for all a E 4. Also, )ua 0’ su an x )) = E in E. Now let : 66 = O11> ( xv ( l.a 0 {Y8 B l 8 {the collection of all finite supreme of the elements x :aEdlEA-Then 1?! inEad { lea ’ i} ya 0 “ sup(“yeu: B 6.6) = sup(sup ”XX a“) since (E, S, u-H) satisfies 3 a (P3). Also, “-H* semi-continuous on E implies that n * .. it when: sea =ux.n. mm, hon -sup<\\y,u= as em = s sup(sup “x H) = sup(ux H*: o 6 a0. It follows that A a l x,a a a (E, S, “-H*) satisfies (P3). Since (P3), (P2) and (P1) are all equivalent on a Dedekind complete vector lattice it follows that whenever (E, S, H'H> satisfies (p3), (E, S, u-n*) satisfies (P1), (P2) and (P3). (v). If (E, s, “-u) satisfies (P2) then “-H extends uniquely to E. However, as the example in (iii) indicates, (E, s, “'“) such that “'H extends uniquely to E does not necessarily imply (P2). Most of the properties we have discussed up to this point have been restrictions on the norm rather than on the vector lattice. We shall now see that whenever E is "close to" E (in a sense that we shall define) and (E, S, “'H) is a normed vector lattice then “-H extends uniquely to E. 29 Definitions 1.14. A band B in E is a projection pgpg.whenever given any x E E+y sup(y: y 6 B and y S X) exists in E. sup(y: y E B and y S x) is an element of B and is called the projection p§_ x gpgg B. A vector lattice E is said to have the projection property whenever every band in E is a projection band. Whenever a vector lattice E has the projection pro- perty, E is "close to" E in the sense that many properties which hold on E also hold on E. (See [22] for a discussion of this.) We note that c does not have the projection pro- perty. The band B generated by the set [xn = = (1,0,1,0,...,1,0,0,0,...): n 6 N} and the element x = (l,l,1,...) in c demonstrate this fact since the pro- jection of x onto B does not exist in c. (sup(y E B: y S x) = = (l,0,l,0,1,0,...) 4 c.) Theorem 1.15. If (E, S, H-H) is a normed vector lattice and E has the projection property, H-H extends uniquely to E- Proof. A.I. Veksler has shown [22, Lemma 3] that when- «4- ever E has the projection property and E E B there exists a monotone sequence {xn} C E+ such that 0 S xn t E in E n . c (g) a . . and xn (r)-converges to x in E. (xn x if and only 1f .+ there exists an element E E E and a sequence of positive real numbers An 1 0 such that ‘xn - E\ S Kn? for n = 1,2,... Let “-“1 be any extension of H°H to E and choose A x 6 E. Using Veksler's result we know there exists a monotone + (r) n h d a . sequence [xn} : E suc that Xu 1 |fi‘ an xn ‘x‘ 30 .. ~+ Hence, there exists an element a E E and a sequence kn l 0 such that ‘ ‘E‘ - x ‘ S 1 E for n = 1,2,... . Then n n H ‘fil ' xnul " O as n ” °° Hence, “EH1 - “Kn“ 4 0 as n a m. But then sup(uan: n E N) = “i“l where H-Hl is any extension of “-H to E. This implies that “fiul = “EHM for all E E E A and “.HM is the unique extension of “o“ to E. In [21, Theorem 3] Veksler proved that if E is an Archimedean vector lattice of bounded elements not having the projection preperty there exists a positive, order bounded (order bounded = bounded on order intervals) linear functional m on E such that m has at least two distinct, positive, order bounded extensions m1 and m2 to E. Then “x“ = = inf(x: xe 2 ‘x‘) + ¢(‘x‘) is a monotone norm on E and “9H1 - inf(x: )(e 2 |E\) + (“dip and “EH2 - inf(x: )(e 2 M) + + m2(‘§‘) are two distinct extensions of “'H to E. However, as noted in the remark following Example 1.13, if H.” is complete on E, both “.“1 and “.H2 are complete on E. All of the previous reSults have aided us in determining when every norm.extension of a Banach lattice is complete. In concluding this section we shall prove that under a specific restriction on E, if (E, S, “-H) is a Banach lattice, every extension of “-H to E must be locally norm complete. Definitions 1.16. The principgl ideal in E generated by an element x in E is the ideal [y E E: ‘y| S a ‘x‘ for some real 0 2 0}. The ppincipal band in E generated by an element x in E is the smallest band in E containing x. 31 In [25, Theorem 6.3], A.C. Zaanen considers an Archimedean vector lattice E such that every principal band in E has a strong unit. If we restrict E to satisfy this condition we obtain the following theorem. Theorem 1.17. Let (E, S, H-n) be a Banach lattice and suppose that every principal band in E has a strong unit. Then every extension of “-H to E must be locally norm complete. 2:22;. Suppose H-Hl is an extension of “on to E 3 {En} : E such that HE“ - Emul a O as n,m » m and 33' m x “o E E for all n E N. Choose an element x in E such that x 2 20. Let IX be the ideal in E generated by x; 1.8., Ix = [u E E: \u‘ S ax for some real number q 2 0]. Let Bx be the order closure of IX in E. Then Bx is the band in E generated by x. Since Bx is order closed in E, BX is a norm closed subset of E [9, Theorem 35.5], and, therefore, (Bx, S, H'“) is a Banach lattice. Also, since BX has a strong unit, every extension of H‘H to B; must be complete. (See the remark following Example 1.13.) Bx an ideal in E implies that B; C E and B is ’00. an ideal in E. Hence, E e B“ and 9 e B“ for n = 1,2 0 X n X A Also, “-“1‘8; is an extension of “-H from Bx to Bx. Therefore, (3;, S, “-Hl) must be a Banach lattice. This implies that there exists an element i E B; such that Hi - finul « 0 as n «to. It follows that (E, S, n-Hl) is a locally norm complete vector lattice. 32 Remark. If H°H is semi-continuous, (E, g, H-H*) is a Dedekind complete normed vector lattice with H-H* semi-con- tinuous. This implies that “-u* is locally norm complete (Theorem 0.10). However, we do not necessarily have that every extension of “-H to E is locally norm complete. That is, qul S “KHZ for all x and “-HI locally norm complete does not necessarily imply that H'Hz is locally norm complete. As a counterexample to the latter statement let E = L”, Hxnl I §l(‘x(n)‘/zn) and “XHZ = §1(\x(n)\/2n) +-TT; [x(n)]. Il-m 33 Section Two. The Boundedness PrOperty. In any normed Space the norm boundedness of a Set B can be characterized by the property that for any sequence [xn} C B and any sequence of positive real numbers kn l O we must have {xnxn] converging to zero in norm as n a an The question of whether an analogous property in an Archimedean vector lattice E can characterize the order boundedness of a subset B was first considered by L.V. Kantovovich [5]; i.e. given any Archimedean vector lattice E can we say that a Sub- set B in E is order bounded if and only if given any sequence [xn} : B and any sequence of positive real numbers kn i 0 then {xnxn} order converges to zero as n 4 m. Of course, if B is order bounded, {xn} c B and En L 0 then {xnxn} order converges to zero because E is Archimedean. However, the subset B = [xn = (l,l,1,...,1,0,...): n E N} is not order bounded in co (the real sequences converging to zero) and given any sequence of positive real numbers kn l O, {Knxn} = = [(xn,xn,...,xn,0,0,...)} does order converge to zero as n a m. Therefore, in general, we cannot characterize the order boundedness of a set in this manner. As in [16, p. 51] we shall say that E has the EEEEQEE' pggg grapgrty if a subset B of E is order bounded whenever xnxn 3 0 as n dim for every sequence {xn} c B and every sequence {Kn} of positive real numbers such that kn i 0. The importance of the boundedness property in the present paper centers around one of the main results of this section: If (E, S, H-H) is a Banach lattice and E has the boundedness 34 property, then all extensions of “o“ to E must be complete. Definitions 2.1. A normed vector lattice (E, S, H'H) is said to be monotone complete (monotone bounded) whenever {xa: a E a? in E+ and 0 S x 1 in E with Hxan S M for a all 0 implies sup(xa: a E a0 = x0 exists in E ([xa} is a bounded set in E). Replacing arbitrary nets by sequences gives us the definitions for monotone o-complete and monotone g-bounded. Remark. A normed vector lattice (E, S, H-H) is monotone bounded (monotone c-bounded) if and only if 0 S Xa ? and {x : a E a] an unbounded subset of E (0 S xn t and a {xnz n E N} an unbounded subset of E) implies that “x0“ t +-m (Hxnu 1 +-m). Also, if H'H is monotone complete (monotone o-complete), E must be Dedekind complete (Dedekind o-complete). I. Amemiya [1] has proved that whenever (E, S, H’H) is a normed vector lattice and H’H is monotone c-complete then “-H is norm complete. S. Yamamuro [24, Theorem 1.1] introduced a property similar to the boundedness property (i.e., ro rt ‘5: if 0 S xn 1 and [xn} is unbounded in E, then there exists a sequence of positive real numbers in 1 0 Such that {xfixn} is unbounded in E) and proved that whenever (E, S, “°“) is a Banach lattice and E has property K then “'H must be monotone o-bounded. It is not difficult to show [19, Lemma 5] that if (E, S, “-H) is monotone o-bounded and 35 E is order separable then every extension of “-H to E must be monotone a-complete. Combining these three results we know that whenever (E, S, H-“) is a Banach lattice and E is order separable and has property K then every extension of “-H to E must be complete. We have indicated the importance of the boundedness property and property K to the question of equivalence of norm extensions of a Banach lattice. Before obtaining the results mentioned above we shall discuss the relationships among property K, the boundedness property and the following two properties on an Archimedean vector lattice E: (A): Given any unbounded set B C E B must have a countable unbounded Subset. (B): If [An] is a countable collection of unbounded subsets in E, there exist finite subsets A; c An n = 1,2,..., such that {sup(Aé): n = 1,2,...} is an unbounded set in E. If E satisfies this property we say that E is finitely unbounded [17]. Remark. The following are some well-known results about the above-mentioned properties: (1). Any Archimedean vector lattice E with a countable subset [hn: n E N} C E+' such that for any x in E there exists a positive integer N(x) and a real number x(x) > 0 such that ‘x| S th must have the boundedness property [17, Pr0pos it ion 1 .8] . (ii). Any perfect sequence Space must have the bounded- ness property [17, Proposition 1.7]. 36 (iii). Any Archimedean vector lattice which is finitely unbounded has the boundedness property [12, Proposition 2.5]. (iv). Any Archimedean vector lattice with the boundedness prOperty satisfies property (A) [23, Theorem VI.6.3]. We shall now prove that the boundedness prOperty implies property K. However, property K does not imply the bounded- ness prOperty. Theorem 2.2. If E has the boundedness prOperty then E must have property K. EEEQE, Let [xn] C E+ such that xn 1 +-m (i.e. xn 1 and [xn} is unbounded in E). We shall Show that there exists a sequence of positive real numbers an i 0 Such that [anxn} is unbounded in E. E having the boundedness property implies that there exists a subsequence {xnk} C [xn] and a sequence of positive numbers kk l 0 such that xkxn -;4+0. k i let “k = (xk)2. Then ak l O and [akxn } is unbounded in k E because if akxnk S 2 G E for all k E N then xkxnk S akz which implies that xkxn 2 0 as k.» m. k Now for n = 1,2,... define an = ak if nk_1 < n S nk (n = 1). Then a l 0 and [a x } is unbounded in E. O n nn Corollary 2.3. If (E, S, H-“) is a Banach lattice and E has the boundedness property then “’H must be monotone o-bounded. 12322;. This corollary follows directly from the theorem and Yamamuro's result [24, Theorem 1.1]. 37 Example 2.4. Let E = {f: f is a real-valued, bounded function on [0,1], f(0) = 0, and f(O) # f(x) for at most countably many x in [0,1]}. We shall show that (i) E does not satisfy the boundedness prOperty, but (ii) E does have property K. (i). Let B=[f EE:f(t)=x if t=x and x x fx(t) = 0 if t # x: x E [0,1]}. B is an unbounded subset of E with no countable unbounded subset; i.e., E does not satisfy property (A). But the boundedness prOperty implies property (A). (See Remark (iv)). Therefore, E does not have the boundedness property. (ii). In order to Show that E has property K we note that “f“ = sup(‘f(t)\: t E [0,1]) is a monotone a-complete, monotone norm on E and we shall show that any normed vector lattice (E, S, H'H) which is monotone a-bounded must have property K: Let 0 S xn 1 +~m in E. Since ”-H is monotone o- bounded we must have Hxn“ 1 +-m. Let in = l/(Han)% which is a sequence of positive real numbers decreasing to 0. Then “xnan = (“xn“)% 1 +-m. Hence, {xnxn} is unbounded in E. This proves that E has property K. We also note that E is Dedekind complete and order separable, but (E, S, H°H) is not monotone bounded. Since Example 2.4 does not satisfy preperty (A) we ask whether property K and property (A) together imply the bounded- ness property. 38 Theorem 2.5. An Archimedean vector lattice E has the boundedness property if and only if E has property K and property (A). Pr_oo£. We already know that prOperty K and property (A) are implied by the boundedness property. Therefore, we need only prove that property K and property (A) together imply the boundedness property. Choose a set B C E such that B is unbounded in E. E having property (A) implies there exists a countable un- + bounded subset {Kn} C B . Let {yk} = [x1 V x v---v xk} 2 for k = 1,2,... . Then 0 S yk 1 +'m in E. Therefore, applying property K, there exists a sequence of positive real numbers xk 1 0 such that {xkyk} is unbounded in E. Then {Xh*k} is unbounded in E because if xfixk S 2 E E for k = 1,2,... then xkyk = ikxl v ikxz v'--v ikxk S S Alxl V 12x2 V°°~V Akxk S 2 for k = 1,2,... . Now [xkxk] unbounded in E implies that xkxk-3490 in E. Therefore, E has the boundedness property. Since we are interested in extending norms from E to E we shall ask questions concerning the extension of the above- mentioned properties from E to E. In the following we shall prove that E has the boundedness property if and only if E has the boundedness property. Lemma 2.6. If E is any Archimedean vector lattice having the boundedness prOperty, then E must have property K. 39 choose a net {xa’nz a E aq in B such that 0 5 x0,“ ‘0 Xn' Let {ya: B 616} = {collection of all finite suprema of the elements a 6 an n e N}} C E. Then 0 S y8 1 + m in E. {xa,n: E having the boundedness prOperty and {y8: B 613} being un- bounded in E implies there exists a sequence {yB } C {ye} 'k which is unbounded in E. Then, there exists a sequence of positive numbers xk l 0 such that {xkyB } is unbounded in E. R For each k, choose xnk 2 yak. Then {xkxnk} is un- bounded in E. If we let Kn = xk for nk_1 < n S nk (nO = 1) we shall have N. i o and {inxg} is unbounded in E. This completes the proof of the lemma. Remark. If E has property K it easily follows that E has property K. Lemma 2.7. An Archimedean vector lattice E has pro- perty (A) if and only if E has property (A). 2:22;, Suppose B C E and B is unbounded in E. let A'= {i 6 E: \s‘ S 9 for some 9 6 B}. A. is unbounded in E. Let A = A]? E. It then follows that A is unbounded in E because every n 6 A’ is such that a = Sup(xa: Xa S x, xq E A) Assuming that E has property (A), there exists a sequence {xn} C A such that {xn} is unbounded in E. For n = 1,2,..., choose y; E B such that y; 2 X“. Then {y;} is a countable subset of B which is unbounded in E. It follows that E has property (A) whenever E does. "Q a he Unbc 40 The fact that E has property (A) implies that E has property (A) is a direct result of the definition. Theorem 2.8. An Archimedean vector lattice E has the boundedness property if and only if E has the boundedness property. ‘Egggg. Recall that if E has the boundedness property then E has preperty K (Lemma 2.6). Also, property K and property (A) together imply the boundedness property (Theorem 2.5). Assuming that E has the boundedness property, it must have property (A) (Remark (iv)) and by the above lemma E has property (A). Therefore, E has property K and property (A) so it must have the boundedness property. On the other hand, assume E has the boundedness pro- perty and let B be an unbounded subset of E. Let B'= {x E E: \fi‘ S ‘y‘ for some y E B} C E. Then B is un- bounded in E. Therefore, there exists a sequence {fin} C B and a sequence of positive numbers kn i 0 such that lnxn_;450. For n = 1,2,..., choose yn E B such that \yn‘ 2 \fin‘. Then xnyn.;z;o because if kn‘yn‘ S 2n 6 E and zn i 0 then in|§n\ S xn‘yn‘ S zn which would imply that infin g 0. Therefore, we have proved that E has the bounded- ness property. Given an Archimedean vector lattice E which is finitely unbounded, it is easy to see that it also has property (A). How- ever, 8 (the Space of all real sequences) is not finitely un- bounded [17, p. 324] but 8 does satisfy property (A). In Al fact, 8 has the boundedness property since 8 is a regular, extended space. (See [22, p. l65-172]). Hence, although every Archimedean vector lattice which is finitely unbounded must have the boundedness property (Remark (iii)), the implication does not reverse. We now recall the example of a normed vector lattice (E, S, H°H) such that H.” was monotone o-bounded but not monotone-bounded (Example 2.4). A natural question to ask at this point is which property (or properties) on E will force every monotone o-bounded, monotone norm to be monotone bounded. We note that order separability of E will not do it since Example 2.4 is order separable. Theorem 2.9. It E has property (A) and (E, S, H-H) is a monotone o-bounded normed vector lattice, then ”on is monotone bounded. .1. Proof. Suppose {xa} is a net in E and O S Xa 1 +'m in E. E having property (A) implies there exists a sequence {x : n E N} C {x : a G a0 such that {x } is unbounded in an a an E. If the set {xa } is directed upward we must have “x H t +.@ a n because “‘H is monotone g-bounded and it would then follow that “x01“ 1 +m. If {xa : n 6 N] is not directed upward we define n {yn: n 6 N} as follows: Y1 = x y2 = x v x 42 Then 0 g y 1 +oo in E which implies that Hymn t + as. n For n = 1,2,..., choose zan E {xa} such that 2an 2 yn. za exists because {xa} is directed upward. Now n {zan} : {x0} and sup(“zanH: n E N) = +'m. It follows that “x0“ 1 +-m and (E, S, “°“) is monotone bounded. V.A. Solovev proved [19, Theorem 5] that if (E, S, H-H) is a normed vector lattice and "-H is monotone bounded then every extension of H°H to E must be monotone complete. Combining this result with the other theorems quoted or proved in the preceding paragraphs we can state the main conclusions of this section. Theorem 2.10. If E has the boundedness prOperty and (E, S, “-H) is a Banach lattice then every extension of H.“ to E must be complete. Corollary 2.11. If E is finitely unbounded and (E: 59 “'u) is a Banach lattice then every extension of H.“ to E must be complete. Theorem 2.12. If E is order separable and has pro- perty K and (E, S, H-H) is a Banach lattice, then every extension of “'H to E must be complete. Corollary 2.13. If E is order separable, E has pro- perty K and (E, S, No“) is a Banach lattice then every extension of “-H to E must be complete. 43 Example 2.14. Let E = c and “x“ = snp(\x(n)\) + + lim ‘x(n)‘. E has the boundedness property because it has a 2::ong unit. (See Remark (i)). Then since H.“ is complete on c, by Theorem 2.10 we have that every extension of H-W to E must be complete. (We proved the same result in Example 1.13, but in a different manner.) Remark. Let E be any Archimedean vector lattice with a strong unit e. E having the boundedness property implies that any complete, monotone norm on E has equivalent norm extensions. (Again, this was noted in the remark following Example 1.13;) Also, “X” = inf(x: Xe 2 \x‘) is a monotone bounded norm on E. To show this we choose any net {xa} C E such that 0 S xa 1 and Hxa“ S M for all a 6 an Then since ‘xa‘ S “xau°e for all a we have that ‘xa‘ S M-e which proves that H-H is monotone bounded. The fact that H'H is monotone bounded shows that whenever E is Dedekind o-complete, H°H is complete. One example of an Archimedean vector lattice not having property K (or the boundedness property) is co. As noted in Section one, however, if E is any Archimedean vector lattice such that E = C0 then every extension of a complete norm on E must be a complete norm on E. Other examples of vector lattices not having the bounded- ness property are c (9 co, L” @ co or x 6-) c0 where x is any sequence Space. 44 Section Three. The Egoroff PrOperty In [9], W.A.J. luxemburg and A.C. Zaanen discuss some results concerning Banach function Spaces. Given a set X, a o-complete Boolean algebra ‘8 of subsets of X which is a sub— algebra of the power set of X, and a countably additive, non- negative measure p, on 6, let M(X,B,p.) denote the space of all extended, real-valued functions on X, measurable with reSpect to [B with identification of almost everywhere equal functions. Let “-H be an extended, monotone seminorm defined on M and let E = {x 6 M: “x” < +-m}. If M is complete with reSpect to “on, then E is said to be a Banach function Space. Luxemburg and Zaanen show that whenever n is g-finitE, “qu a inf {It}: Hxnu: 0 s xn 1 ‘X‘} is a a-semi-continuous, extended, monotone seminorm on M, “XML s “x“ for all x E M, and “.HL is the maximal seminorm on M having these preperties. The condition on M that appears to be essential to their proofs is the Egoroff property. Hence, we shall show that if (E, s, H-H) is any normed vector lattice and E has the Egoroff property then H-HL as defined above satisfies the same properties and H-HL is a norm. ‘We shall study “.HL in detail because the functional mCx) - HxH - HxHL will aid us in determining when all extensions of “-H to E are equivalent. Combining Theorem 0.8 in the introduction with some results we have about “ HL’ we shall prove that completeness of (E, S, “'H) is equivalent to a Riesz-Fischer property. Although this Riesz-Fischer Theorem does not bear directly on 45 the problem of equivalence of norm extensions of a Banach lattice, we present it here as an interesting application of Theorem.0.8. The main results of this section are: (1). If (E, s, H-H) is a Banach lattice, E is order separable and has the Egoroff property and m(&) = “QHM - “xHML is a con- A tinuous functional at x = O with reSpect to on E, H ° “ML then every extension of H-H to E must be complete. (2). If (E, S, H-H) is a Banach lattice which has the Egoroff property and ¢(x) = HxH - “xHL is a continuous functional on E at x 8 0 with respect to H'HL, then every extension of “~H to E is complete, provided that either “.HL is con- tinuous on E or (E, s, H-HL) satisfies the hypotheses of Theorem 1.11. Definition 3.1. A vector lattice E has the Egoroff property if for each x G E and for any double sequence {xn k: n,k.E N} of elements of E+ such that for n = 1,2,..., , 0 s xn,k 11“x| in E, there exists a sequence {xm} : E such that 0 s xm t ‘x‘ and for every m,n E N there exists a positive integer k(n,m) such that xm s xn,k(n’m). For a discussion of the Egoroff property and its rela- tionship with the classical Egoroff theorem, we refer the reader to [4]. Theorem 3.;. If (E, s, H-H) is a normed vector lattice and E has the Egoroff property, then HXHL = inf{lim “xnuz ““03 0 s xn 1 'x‘}is a o-semi-continuous norm on E, HXHL s “x“ 46 for all x E E and h'HL is the largest such norm on E. 2592;, One can easily show that “-HL is a seminorm as the triangle inequality can be proved by the usual method. (1) We shall Show that “.“L is a norm on E. Choose x E E and for n = 1,2,..., by definition of HxHL we can find a sequence {vn’kz k E N} such that 0 s vn’k 1k ‘x‘ and uvmku s ML + l/n for all k e n. E having the Egoroff property implies there exists a sequence {xm} : E+ such that xm 1 ‘x‘ and for every m,n 6 N there exists an integer k(n,m) such that xm < Hence, ’ vn,k(n,m)' meu s “vn,k“ s “XML + l/n for all n,m. Therefore, uxmu S “XHL for all m and it follows that lim “xmn S HXHL. man We can now prove that “.HL is a norm on E because if “x“L = 0 then there exists a sequence 0 s xm 1 \x‘ with lim “x H = 0. But since H-u is a norm we must have x = 0 m—Iaa m m for all m E N and then x = 0. (ii) “.HL is a-semiecontinuous. Let 0 s xn 1 x in E. We shall Show that ”XML 5 ii: HanL and this will imply that “qu '- lim “anL' As in (i), for n = 1,2,..., choose a sequence {vnT;: k E N} such that O s vn,k 1k Xn and “v“,kn S “xnnL + 1/n for all k E N. Let u = n,k vn,k + (X - Xn) for all k,n 6 N. Then 0 s u x for all n. E having n,k 1k + the Egoroff prOperty implies there exists a sequence {2m} ; E such that 0 s 2m 1 x and for all n,m 6 N there exists a positive integer k(n,m) such that zm s Let un,k' + ym (zm +'xm - x) . Then 0 S ym 1 x and for each n,m E N there exists a k(n,m) such that ym = (2m +~xm - x)+ s 47 S un,k +-xm - x = vn,k. Now, Hymn S “v“,k“ S “XnHL + l/n for all n,m E N. Therefore, ii: Hym“ S ii: “xnuL' It follows that “xHL S :32.uanL which proves that “'HL is a-semi- continuous. The fact that “XML S “x“ for all x E E and H°HL is the maximal norm on E satisfying the above properties follows easily, completing the proof of the theorem. J.A.R. Holbrook has shown that if H'U is a monotone seminorm on an Archimedean vector lattice E and u-HK is the maximal a-semi-continuous seminorm below H-u, then “-HL - u-HK on E if and only if E has the Egoroff property [4, Theorem 5.2]. The above theorem shows that “.HL is a norm if and only if u-H is a norm. We shall now apply Theorem 0.8. Theoremi3.3. Let (E, S, H-u) be a normed vector lattice such that E has the Egoroff property. If 0 S xn 1 in E and Hxnu S M < +'m for all n E N, then sup(xn: n E N) = = x exists in E#. 0 Proof. Using Theorem 3.2 we know there exists a o-semi- continuous norm H°HL on E such that “x“L S “x“ for all x E E. Therefore, 0 S “anL S M < +-« for all n 6 N. Apply- ing Theorem 0.8 to (E, S, “-HL) it follows that sup(xn: n E N) = # I=x inE. 0 We are now motivated to make the following definition. 48 Definition 3.4. Let (E, S, H-H) be a normed vector lattice such that E has the Egoroff property. (E, S, H-H) is said to have the Riesz-Fischer property whenever G on “xnu < +-¢. and {xn} C E+ implies that 2 xn is an n31 , n=l element in E. (Note that 2 xn exists in E# whenever a n-l Z Hxnn < +-o by applying Theorem 3.3.) n: In [9, Theorem 4.8] Luxemburg and Zaanen have shown that a normed function space E c: M(X,B,u.) is complete if and only if 0 Sxn 6 E for n = 1,2,..., and ugluxnn < +'m Q implies that H 2 an < +-m. Of course, since E is a normed n= m function space, 2 xn always exists in M. This is not n-l necessarily true for a general normed vector lattice (E, S, H-H). + However, whenever E has the Egoroff property and {xn} : E a a with z “xnn < +-a we know that 2 xn exists in E#. (Apply n-l n=l Theorem 3.3.) The proof of the following theorem is completely analogous to that of Luxemburg and Zaanen. Theorem 3.5. Let (E, S, “°“) be a normed vector lattice such that E has the Egoroff property. (E, S, H-H) is complete if and only if (E, S, “-N) has the Riesz-Fischer property. Proof. (1) If (E, s, \HD is complete and {xn} c E+ a such that galuxn“ < +-m, let Sn = x1 +-x2 +...+.xn, n a 1,2,... Then “Sn - Smu‘“ 0 as n,m‘~ a. Hence, there exists an element So E E such that “Sn - So“ a 0 as n q.¢, But 0 S Sn 1 and “Sn - so“ a 0 implies that So ' sup(Sn: n E N); i.e., O 80" anéE. n-l 49 (ii). We now assume that (E, S, H‘H) has the Riesz- Fischer property. We shall first show that whenever Z Hx H < +'a» then H g x H S E Hx H. Assume that for some n=1 n n=1 n n=1 mu m positive number s > 0 we have H 2 an > e +- 2 Han. For n— n=l k = 1,2,..., obtain a sequence {bn k: n E N} in E such that ’ Hzn bn kH > k +'anbn kH. Then for every positive integer p , ’ 2 H 2 bn,k“ > k +- 2 an,k” and for some p = pk 2 an,k“ < 1/k . n2p n2P n2p Then by reindexing we obtain for every k = 1,2,..., 3 sequence . 2 {2“,k. n E N} in E such that ZnHZn,kH < l/k and Hznzn’kH > k. Let {un: n E N} be the double sequence {zn,k} . 2 arranged in single order. Then anunH < 2k 1/k < +-m and HznunH > “Enzn,ku > k for all k e N. But ZnHunH < +-m and HznunH = +-m implies that Enun is not in E which contradicts the Riesz-Fischer property. Using the above result we can prove that (E, S, H-H) is complete. Let {xn} be H-H-Cauchy in E and assume without n loss of generality that Hxn+1 - an < 1/2 for n = 1,2,... . 0 Then x = 2 ‘xn+1 - xn‘ is an element of E. Let yn = xn - x1 + n-1 n-1 +- g ‘xk*1 - xk| for n = 1,2,..., and note that y1 = O and 0 S yn 1. Since yn+1 - yn S Z‘xn+1 - xn‘ we have an E Hyn+1 - ynH < +-m. Then sup(yn: n E N) = y is an element of E. Let Q = y - x + x1. Then i - x = y - x +-x - Xn = n l n-l ““1 glxkn'xk‘ Wi glx...-xkn= a CO - on “E (yk+1 ' yk) ":3 ‘xk-l-l " “k“ Hence’ ‘x " Xn‘ SE (VH1 ' yk) ‘ m m - - E ‘xk*1 - ka S.3 E ‘xk+1 - xk‘. Now, Hx - an S 50 m S 3H2 ‘xk+1 - xk‘H S 3 E ka+1 - ka the last inequality n n holding because of the result proved in the preceding paragraph. Hence, HE - an‘» 0 as n ~»m which proves that (E, S, H.H) is complete in norm. As noted in Section two, the monotone completeness of a normed vector lattice is a relatively strong property. For this reason, we ask whether a normed vector lattice (B, S, H’H) can be isomorphically and isometrically embedded in a monotone complete normed vector lattice. Although this question does not directly involve the problem of equivalence of norm extensions, the proof of the theoreulwhich answers this question is a nice application of some of the results of this section as well as Theorem 0.8. Theorem 3.6. Let (E, S, HoH) be a normed vector lattice such that E has the Egoroff property and E is order separable. If 0 S xa 1 in E and HxaH S M for all a 6(7, then sup(xa: o E 60 = x0 exists in E#. Furthermore, if H-H is a—semi-continuous (E, S, H-H) can be isomorphically and isometrically embedded in a monotone complete normed vector lattice. Igzggg. As in Theorem 3.2, H-HL is a g-semi-continuous norm on E such that HxHL S HxH for all x in E. However, since E is order separable it follows that H-HL is semi- continuous. We can now apply Theorem 0.8 to H-HL and we have that whenever 0 S xa t and HxaH S M for all a E a' then sup(xa: a 6 a0 = x0 in E#. 51 If H‘H is o-semi-continuous, H-H = H'HL and we can 1 extend H-H to EH (allowing H-H to be infinite) by defining HxH = sup(HyH: O S y S §,y E E). Let E'= {x E E#: HxH < + m}. One can easily show that (E, S, H-H) is a monotone complete normed vector lattice and (E, S, H-H) is isomorphically and isometrically embedded in (E, S, H°H). A If (E, S, H-H) is a normed vector lattice and E has the Egoroff property, H-HML can be defined on E where A H-HM is the maximal extension of H-H to E and H-HML is A m) the maximal o-semi-continuous norm below HHM on natural question to ask is whether is equal to H-HL n-nML when HoHML is restricted to E. If E is order separable then E has the Egoroff property if and only if E has the Egoroff property [18, Chapter 1, Theorem 1.3]. Therefore, we consider the above-mentioned question for (E, S, H-H) where E is order separable and has the Egoroff property. The impact of the following two theorems on the prdblem of equivalence of norm extensions of a Banach lattice will become evident in the corollary following them. Theorem 3.7. Let (E, S, H-H) be a normed vector lattice such that E is order separable and has the Egoroff property. Then HxHL a inf(iim Han: O S xn 1 x, {xn} C E) is equal to ~aao HxHKL = inf(lim HfinHK: O S in 1 x, {2n} C E) for all x in ndm . E+ where H-HK is any extension of H-H to E. .££22£° By the definition of “'HL and H-HKL we can see that HxHKL S HxHL for all x in E. To Show that 52 HxHKL 2 HxHL we prove that given any sequence {2“} c E such A that 0 S 2n 1 x in E+ there exists a sequence {xn} C E such A that x S 2 for n = 1,2,..., and 0 S x 1 x. n n n Let An = {y E E: 0 S y S 2“}, n = 1,2,..., and let G A = U Ah. Then A is a directed set in E and sup(A) = x. n-l Since E is order separable, there exists a sequence {yk} C A such that 1 x. ({yk} can be chosen directed upward beacuse yk A is a directed set.) By definition of A there exists a sub- sequence {fin } C {2“} such that z; 2 yk for k = 1,2,... . k k yk for n such that nk S n < nk+1, For 1 S nr< n let x = x where x E E and O S x S 2 . l n o o o 1 Then for n = 1,2,... we must have 0 S xn S En. And Since Choose xn ' k = 1,2,... . yk 1 x it follows that xn 1 x. This completes the proof of the theorem. Theorem 3.8. Let (E, S, H-H) be order separable and n * A A e have the Egoroff property. Then HxHL = HxHKL for all x in E where H-HK is any extension of H-H to E. Proof. Using the fact that E is order separable it follows that H-HL is semi-continuous on E and then H.H: is a semi-continuous extension of H-HL to E. Also, since E is order separable whenever E is [19, Lemma 3] H-HKL is A * also a semi-continuous norm on E. Now, H-HL and are H'HKL semi-continuous norms on E and by Theorem 3.7 H-H: and H.HKL agree on E. By Lemma 1.5, we have that H.H: and H-HKL agree on E. 53 Corollary 3.9. Let CE, S, H-H) be a Banach lattice such that E is order separable and has the Egoroff property. If H'HK is equivalent to n-HKL on i}. where \\-\\K is any extension of H-H to E, every extension of H-H to E must be complete. Before examining conditions under which H-HK is equi- valent to H-HKL we present the following examples. The first example uses corollary 3.9 and the others illustrate what happens when E does not have the Egoroff property. Example 3.10. Let E = c and HxH = sup(‘x(n)‘: n E N) + +~lim ‘x(n)‘. Then if H-HK is any extension of H-H to n—m A E -.L” ‘we must have HxHKL = sup(|x(n)‘: n E N) since * H§HL ' 3UP(‘*(D)‘= n E N). Furthermore, since H- is Hm. equivalent to H-HM (HxHM = sup(\x(n)‘: n E N) +'liml\x(n)|) n—m on E we must have that every extension of H-H to E is complete. Note that we have now proved this fact in three different ways. (Compare with Examples 1.13 and 2.14.) Example 3.11. Theorems 3.7 and 3.8 were proved for E order separable and having the Egoroff prOperty. We shall now give an example of a normed vector lattice (E, S, HoH) such that E has the Egoroff property but E does not. Let X = the set of all sequences x (X(1).X(2),---) with positive integer coordinates and let E (f: f is a real- valued, bounded function on X such that f is constant except possibly on a finite number of points in X}. J.J. Masterson 54 and G. Crofts have shown [12, Example 1] that E is diagonaliz- able for any uncountable set X. (A vector lattice E is diagonalizable if whenever {xn k: n,k E N} is any double + o o . sequence in E , x a x for each n and x a x in E, n,k k n n then there exists a strictly increasing sequence (kn: n = 1,2,...} of positive integers such that 3,x as n a m.) It is xn,k n easy to see that whenever E is diagonalizable it must have the Egoroff property [10, p. 185]. Therefore, the vector lattice E which we have defined above has the Egoroff property. We shall now prove that E = {f: f is a real-valued, bounded function on X} does not have the Egoroff prOperty. The proof is due to J.A.R. Holbrook [4, Example 4.2]. Define subsets An,k of X by letting An,k = {x E X: x(n) S k} and let fn,k = IA , for n,k = 1,2,... . Then 0 S fn,k 1k f A n,k in E for n = 1,2,..., where f = Ix. However, for no choice of k(n) do we have sup(f ° n E N) = f. To prove this n,k(n)' latter statement we note that if we choose any sequence {k(n): n E N} of positive integers and let i = (x(l),x(2),...) be the element in x such that £(n) - k(n) + 1 for n = 1,2,... then i does not belong to any of the sets An,k(n)° It now follows that E does not have the Egoroff property for if we assume there exists a sequence {um} C E such that for all n,m E N there exists a positive integer k(n,m) with um S fn k then there exists a k(m) for each m such that u S f . m m,k(m) m,k(m): m E N) i f. The element f = xx not having the Egoroff property implies Hence, sup(um: m E N) S sup(f there exists a monotone seminorm H-H on E such that 55 HfHL * HfHLL [4, Theorem 5.2]; i.e. H’H is not o-semi-con- tinuous on E. Then ng' = sup(‘g(t)\: t 6 X) + Hg“ is a monotone norm on E and HfH£ = “Ix“i = 1 + HEHL # 1 + “EHLL = aniL; i.e., “-H' is not o-semi-continuous on E. However, H-H"E = H-Hl is a monotone norm on E such that H H1L is o-semi-continuous on E because E has the Egoroff property. Although H.HL need not be a-semi-continuous whenever E does not have the Egoroff property, we can still consider H.HL and H HML and ask some questions about these seminorms. Example 3.12. We shall give an example of a normed vector lattice (E, S, u-H) where is not equal to H'HML M... on ’é. Let X be an uncountable set of points and let E = {f: f is a real-valued, bounded function on X and f is constant except possibly on a finite set of points in X}. Then E = {f: f is real-valued and bounded on X}. Assuming the continuum hypothesis, W.A.J. Luxemburg and A.C. Zaanen have shown that E does not have the Egoroff prOperty [10, Theorem 43.3]. Let g E E be the element g = 1x1 + 21X Where 2 X = X1 U x2, x1.” X2 = ¢ and card(Xl) = card(XZ) > 80. Then g is a strong unit for E. Hence, we can introduce a semi- continuous norm on E by defining Hf” = inf(x: kg 2 \f‘) for every f in E. (i). H°H* is a semi-continuous extension of Ho] to A E and “hn* = inf(x: xg 2 ‘h‘) for all h in E. S6 (11). H°H* # u-HM on i because n§n* = 1, but if w is any element of E such that w 2 g then W(x) 2 2 except possibly for a finite number of points in X. This A implies that Hg = inf(HwH: w 2 g, w E E) = 2. HM (iii). H HML # H'“* on E. We know that ng* = 1. However, we shall Show that ugHML = 2. Let {f;} C E be a sequence of elements such that O S En t g. Then for some positive integer N, ffi must be greater than one on an un- countable set in X (i.e., £8 4 E). Also, given any 6 > 0 there exists a positive integer K > 0 such that {x: ‘ffi(x) - §(x)‘ < e} is an uncountable set in X. This implies that “fifiM 7 2 - e. Hence, s:p(uf;uM) = 2 and we have that “EHML = 2. A (iv). H-HML # H-HLM on E. Since H'H = H.HL on E we must show that “OHML # H-HM on E; i.e., H-HM is not o-semi-continuous on E. Define h E E by choosing {xk: k E N} a countable subset of X and letting 2 h(x) I {1 for x =xk, k = 1,2,... 0 otherwise. Then HhHM = HEHIMZ= 1. However, taking fn(x) = ['1 for x = xk, k = 1,2,...,n I\O otherwise, we have 0 S fn 1 h in E, {fn} C E and anHM Hence, “BHML = inf(ii: Hug“: 0 S u; 1 h in E) S 1/2. This =1fnn = 1/2- proves that “EHML # HhHLM. Note that the previous example is similar to Example 1.7 in Section one in the sense that it is an example of a normed 57 vector lattice with a semi-continuous norm which does not extend uniquely to E. Although our present example is net order separable, Example 1.7 provides us with a normed vector lattice which is order separable and has the Egoroff property but “HIM f H'HML on E. In Example 1.7, “HML = “H: on E. However, in our present example “'HML f H-H: on E. In Example 3.12 if we had chosen E = {f: f is real- valued and bounded on X and f is constant except possibly on a countable set of points in X} with the norm defined in the same way, (E, S, H'H) would be a normed vector lattice with H°H semi-continuous, “-H* a semi-continuous norm extension, “-H* # H'HM’ but “'HM would be a-semi-continuous on E. (See Example 1.9.) Hence, we would have “xHML = = 1152“,. = Wm on E. For the case when E is order separable and has the Egoroff property we know that H-HML = “'H: on E. However, as the previous example illustrates, this is not necessarily true in other cases. We now ask whether H'HML is an extension of “.HL (i.e., does H HML = H-HL on E) even though H HML i may not be equal to H-H: on E. * Example 3.12 (continued). We recall that H°HML # H°HL on E. However, we shall show that H-HML = H-HI on E. By definition of H-HL and H-HML we always have HEHML S “fuL for every f E E. Let {f;: n E N} be a sequence in E such that o s f; 1 f in E. For n = 1,2,..., let m An = {y E E: 0 S y S fn} and let A = lJ An' Then A is a n=1 directed set in E and sup(A) = f. Since A is directed we 58 can let A = {yi: i e A} c E where 0 s yx 1 f. Since H'H is semi-continuous on E we have sup(“yiu: x E A) = Hf“. But, by definition of A, sup(nyiu: i e A) s sup(HfQHM: n c N). Hence, “f“ s sup(“ngM: n e N) which implies that HfH = = HfHL = ufum. Obviously, if E = {f: f is real-valued and bounded on X and f is constant except possibly on a countable set of points in X} then H HML n-HML = MM and M, = M. is an extension of “.HL since Example 3.13. We now consider Examples 1.8 and 1.9 in Section one. In both of those examples we proved that * * c * “.HL a n.“ was not a norm on E. Hence, “.HL # “ HML on E. However, we also showed in both cases that H-HML agreed with “-H = H-HL on E. As noted in Corollary 3.9, whenever (E, S, H-H) is order separable and has the Egoroff property and H-H is complete, if we can show that is equivalent to “.H H-HML then every extension of H'“ to E must be complete. We note that H'H is not always equivalent to H-HL. (Let Q _ n ‘-r- E = La and let “x“ - E (}x(n)}/2 ) + 11m }x(n)}. Then a: n=1 n—m “xHL * n§1(}x(n)|/2n) and “.HL is locally norm complete while “-H is not. Therefore, H°HL cannot be equivalent to “0“.) However, in all of the examples that we presently have at hand, whenever H-H is complete, “.“L is equivalent to H'H. Proving that this is true in general depends on prepcrties of the functional m(x) = “x“ - “XHIK W.A.J. Luxemburg and 59 A.C. Zaanen considered this functional (defined on normed function Spaces) in [9, Note IV, p. 262]. They noted that in all of their examples m is a monotone seminorm. However, they could not prove that m is even monotone, nor did they have a counterexample. We can use Example 1.7 from Section one and Theorem 3.8 of the present section to answer Luxemburg and Zaanen's question. Example 3.14. Let E = La and for each x in E let “x“ = inf(uwuo: w2 M, w e c) where “""Ho = inf(x: i’e M) and a = (1,2,1,2,...) E Loo. Note that (Loo, S, H”) is a Banach function space. Recalling the form in which we stated Example 1.7 and applying Theorem 3.8 to this example it follows that uqu - inf(x: it 2 |x|). We can now easily show that @CX) = HxH - HxHL is not monotone on E. Let y = (2,2,...) 6 c S gm. Then y 2 § = (1,2,1,2,...) and ¢(y) = Hy“ - HyHL = 0 Since H'H is equal to H'HL on c. However, m(e) = He“ — HEHL = 2-1 = 1. (Recall that “e“ = 2 and “eHL = He“* = l as noted in Example 1.7.) Therefore, ¢(e) > ¢(y). Also, m is not sub- additive on E. Let x = (1,1,1,...) and y = (O,1,0,1,...). Then x +ty = e and m(x +-y) = 1. But mCx) = 0 while aw=uuM-hW=1-e=e. We are motivated by Example 3.14 to prove the following theorem. 60 Theorem 3.15. Let (E, S, H'“) be a normed vector lattice such that E has the Egoroff property. Then if gp(x) - “x“ - “qu is monotone on E, D = {x e E: “x” = MM is a solid subset of E. (D is said to be a £21£§.SUb3et of E whenever 0 S y S |x} and x E D implies that y E D). 2122;. Suppose D is not solid in E and choose x in D and y e E such that 0 s y s |x| but y 4 D. Then met) a tp(‘x}) = 0 but q)(y) = Hy” - HyHL > 0 which shows that m is not monotone. In [4, Section 6, Comment b], J.A.R. Holbrook asks if E having the almost Egoroff property (almost Egoroff = Egoroff whenever E is Archimedean) implies that “-H is equal to H-HL on some super order dense ideal in E. (A subset A C E is said to be super order dense in_ §_ whenever given any element x in E there exists a sequence {xn} : A such that sup(xn: n E A) = x). Example 3.14 gives us an Egoroff vector lattice E and a monotone norm H-H such that D {x: “x“ = “x“L} is not a super order dense ideal in E because D is not solid in E. (An ideal is always a solid subset.) However, co is a super order dense ideal in E and co C D. Holbrook notes [4, p. 77] that if “-H = H-HL on a - _ . . . A super order dense ideal in E then “.“L — H HLL’ i.e., H AL must be o-semi-continuous. This, of course, implies that E has the Egoroff prOperty (provided that E is Archimedean) [4, Theorem 5.2]. Using Example 3.14 as motivation we can prove the following generalization of this latter statement. 61 Theorem 3.16. Let E be an order separable, Archimedean vector lattice. If H-HL = “-H on E where “-H is a monotone seminorm on is, then n.nLL=\\.uL on E. tea HfiHLL = mug: nannp o s ant m) \an - inns: “an“: o a an t m). Let {in} be any sequence in E such that o s an t \&|. As in the proof of Theorem 3.7 we can find a sequence {2“} c E such that 0 S zn S fin for n = 1,2,..., and zn 1 |fi|. Now, lim “2““ = :im “anL s -—ioo new . S 1im “fin“L. Therefore, “E“LL 2 “EHL ,which completes the proof n a” of the theorem. Corollary 3.17. Let E be an order separable, Archimedean vector lattice and E its Dedekind completion. Then if for every monotone seminorm “-H on E, either H-H = H.HL on a super order dense ideal in E or H'“ = H-HL on E, then E must have the Egoroff preperty. Although m(x) = “-H - “.“L need not be monotone or subadditive, if we place a different kind of restriction on m we can prove that H.“ is equivalent to “.HL Theorem 3.18. Let (E, S, H'H) be a normed vector lat- tice having the Egoroff property. Then H'“ is equivalent to “.“L if and only if m(x) = “x“ - HxHL is continuous at x = 0 with respect to H-HL. Proof. Suppose first that “-H is equivalent to H-HL and let {xn} be a sequence in E such that “XnHL « O as n a m. Then “an a O as n film so that 62 Win) " “an ' anHL -* 0 as n -' °°- 0n the other hand, assuming that m is continuous at x = 0 with respect to “'HL and choosing a sequence {xn} in E such that “anL a 0 as n a m, we must have m(xn) « 0 as n a m. Hence, “an a O as n a m. Since ”X“ 2 “X“L for all x 6 E we have that “-H and H'HL are equivalent. Example 3.19. Let E = L and “x“ = E (}x(n)\/2n) + m n=1 +‘TEE |x(n)}. m(x) = lim }x(n)} is not continuous at x = 0 new “Sm with reapect to H-HL. To show this let xk = (0,0,...,O,l,1,...). R Then uxkuL -+ o as k —. a but cp(xk) = 1 for all k e N. However, H'“ is not a complete norm. In all of the examples which we presently have at hand, whenever H-“ is complete m is continuous at x = O with respect to H-HL. Note that even in Example 3.14, although m is not monotone or subadditive m is continuous with reSpect to “.HL at x = 0. In this example “.HL -is a complete norm on L”. (See the remark following Example 2.14 where we show that if E is Dedekind o-complete and “X“L = inf(x: is 2 \x}) and e is a strong unit for E then “'HL is complete.) Also, HXHL S “x” for all x in E forces “-H to be complete [14, Theorem 30.28] as well. Hence, by our previous theorem m is continuous at x = 0 with respect to H°HL‘ We can now state the main results of this section. Theorem 3.20. Let (E, S, H'H) be a Banach lattice such that E is order separable and has the Egoroff property. 63 If q)(§) = “EHM - HEHML is a continuous functional at x = O with respect to H-HML on E then every extension of H-H to E must be complete. Theorem 3.21. Let (E, S, H-H) be a Banach lattice such that E has the Egoroff property. Also, suppose m(x) - HxH - HxHL is a continuous functional at x = 0 with respect to “-HL. Then if either HOHL is continuous or (E, s, n-HL) satisfies the hypotheses of Theorem 1.11, every extension of H-H to E must be complete. In order to apply the results of this section to Specific examples it would be nice to have several Archimedean vector lattices having the Egoroff property. We note that if A is any Archimedean vector lattice which is also a sequence Space (i.e., A c S and A 2 m where S is the space of all real sequences and m is the Space of all eventually zero sequences) then A must have the Egoroff property. In order to see this we first note that l# = S since S is universally complete and A is order dense in S. (The fact that A is order dense in S follows because A 2 m.) Obviously, S has the Egoroff property because it is the Space of all measurable functions on the countable discrete Space 80. Also, whenever E# has the Egoroff property and is order separable it follows that E must be order separable and have the Egoroff property. To prove this latter statement recall that whenever E has the Egoroff property and is order separable E also has these 64 two properties [18, Chapter 1, Theorem 1.3]. Therefore, we need only show that E has the Egoroff property and is order separable. Obviously, if E# is order separable then E must be order separable. Choose a double sequence .+ . {xn,k° n,k E N} in E such that O S xn,k 1k x in E for n = 1,2,... . Since E is an ideal in E# we have # O S x 1 x in E . Hence, there exists a sequence n,k k - — # [th m E N} C E# such that O S xm 1 x in E and for each n,m E N there exists a positive integer k(n,m) such that xSx . m n,k Using a construction similar to the one used in the proof of Theorem.3.7 we can find a sequence {ym} C E+ such that ym 1 x in E and ym S Qm for all m E N. Therefore, given any n,m E N there exists a k(n,m) such that ym a xn,k' It now follows that A has the Egoroff property. In concluding this section we shall prove the following theorem which relates some of the properties of Section two with the present section. Theorem 3.22. Let (E, S, H-H) be any Dedekind o- complete Banach lattice which has the Egoroff property. If E also has property K, qu) = HxH - “xHL is continuous at x = 0 with reapect to “-HL. EEQQE. By Yamamuro's result [24, Theorem 1.1], (E, S, “-H) must be monotone o-complete. Amemiya [1, Theorem 1] has shown that whenever H°H is monotone o-complete, there exists a real number k > 0 such that for every x in E and for every sequence 0 Sxn 1 |x| we must have 65 “X“ S k sup(“xnfl: n E N). Therefore, it follows that “x” S knan for all x in E. Hence, “-H and H-HL are equivalent. The conclusion follows by applying Theorem 3.18. Corollary 3.23. Let (E, S, H-H) be any order separable Banach lattice which has the Egoroff property. If E also has property K, m(x) = “X“ - “XHL is continuous at x = O with reSpect to n-HL. Proof. Note that (E, S, H-HM) satisfies the hypotheses of the theorem. Therefore, H-HM is equivalent to H-“ML. Also, since H HML is an extension of H-HL (i.e., H‘HML = H-H:) it follows that u-HL is equivalent to H'H on E. CHAPTER TWO A CHARACTERIZATION OF THE SPACE OF ALL REAL SEDUENCES While the subject of this chapter is not directly related to the problems discussed in the preceeding chapter, it is interesting in its own right and does involve the pro- perties of monotone seminorms on an Archimedean vector lattice. We shall characterize a universally complete vector lattice E having a countable collection of continuous, monotone semi- norms defining a locally convex, Hausdorff topology as either a finite dimensional Space or the space S of all real sequences. Every vector lattice E always has a collection {xaz 0 6(7} of mutually disjoint (i.e., x0!1 A xa,2 = O for o1 # a2) elements in E such that xa A y = 0 for all a C¢7 implies that y = 0. ({xa} is said to be complete system in E.) [13, Theorem 4.6]. If E is also universally complete and {Xe} has at least countably many elements then E cannot have a monotone norm defined on it. For, if H'H is a norm on E and {xa } is a countable subset of {xa} n then Hxann > O for all n E N. Letting yn = (n/HxanH)-xo[n we must have sup(yn: n E N) = y in E because E is universally complete. But “ynH = n for n = 1,2,..., implies that fly“ = +~w, which is impossible if y E E. Hence, no such monotone norm exists. Of course, if E is any Archimedean 66 67 vector lattice such that any system of mutually disjoint non- zero elements is finite, then E is of finite dimension [25, Theorem 4.3]. Hence, a universally complete vector lattice which is not a finite dimensional Space cannot have a monotone norm defined on it. Also, many universally complete spaces do not even have monotone seminorms defined on them. (All of our norms and semi- norms are assumed to be real-valued, i.e., they do not take on +-m at any points.) For example, the space of all equivalence classes of Lebesgue measurable functions defined on [0,1] has no non-trivial monotone seminorm defined on it [2, Corollary 1, p. 119]. However, pn(x) = x(n) for n = 1,2,..., and x = (x(l),x(2),...) defines a continuous, monotone semi- norm on S and {pn: n E N} also defines a locally convex, Hausdorff topology on S. Therefore, we ask whether S and the finite dimensional Spaces are the only universally complete vector lattices having such a topology. We shall need the following preliminary results. Lemma 1.1. Let E be an Archimedean vector lattice with a countable collection of monotone seminorms [H-Hn: n E N} defining a locally convex, Hausdorff topology on E. If [fn: n E N} c E, fn 1 and there exists an element g in E such that “fn - gnk a o (k = 1,2,...) as n a a, then sup(fn: n E N) = g. .EEEEER For n > m we have ‘fm - inf(g,fm)\ = a ‘inf(fn,fm) - inf(g,fm)| s ‘fn - g}. Therefore, 68 Hfm - inf(g,fm)“k = O for k = 1,2,... . But the collection {H “k: k E N} defining a Hausdorff topology on E implies that fm = inf(g,fm) S g for all m E N. Hence, g is an upper bound for [fn}. Let h be another upper bound for {fn}. Then \in - inf(g,h)} = ‘inf(fn,h) - inf(g,h)‘ s \fn — g} and we must have urn - inf(g,h)uka o (k = 1,2,...) as n a m. But since we also know that Hf“ - ng a O (k = 1,2,...) as n a m, it follows that g = inf(g,h) S h. Hence, g = sup(fn: n E N). Theorem 1.2. Let E be an Archimedean vector lattice with [H-Hk: k E N} a countable collection of monotone, con- tinuous seminorms defining a locally convex, Hausdorff t0pology on E. Then if {x } is any net in E such that O S x 1 x, o o at there exists a sequence {xa } C [xa} such that O n x 1 x n n in E (i.e., E is order separable). Proof. If {x } is any net in E such that O S x 1 _____. o a in E, then “X - onk 1 O for k = 1,2,... . For n = 1,2,... choose x E [x } Such that 0 S x 1 and fix - x a a ' n on n onHk for k = 1,2,...,n. Then for any k E N and any 6 > 0 we < 1/n can choose N > 0 Such that 1/N < e and N 2 R. Then “X - Xanuk < l/N < e for all n 2 N. Hence, “x - xanHk for k = 1,2,... . Then, by the above lemma, 0 S xa 1 x. n We shall need the following definition concerning linear functionals on an Archimedean vector lattice E. 69 Definition 1.3. A linear functional f defined on E is said to be order continuous if inf(f(xa): o E a) = 0 . 0 whenever {x0} is a net in E such that xa a 0. Letting 5(E) = the set of all order continuous linear functionals on E, it is known that 5(E) is a Dedekind complete vector lattice [23, p. 247]. We shall denote the topological dual Space of a locally convex Space (E, S, {H-Ha: a E<7}) by E! Theorem 1.4. Let E be an Archimedean vector lattice with a collection {H.Ha: 0 E17} of monotone seminorms defining a locally convex, Hausdorff topology on E. Then “-Ha is continuous for all a 6¢7 if and only if E' C:5(E). 2522;. Assume every H-Hq is continuous and let f E E', f 2 0. Suppose O S x 2 0. Then “XBHG E 0 for each 8 a e 4. Therefore f(xB) —. o and we have f e 5(a). On the other hand, if 5(E):D E' and we choose any net {x6: 8 E B} in E such that O S x6 9, O and any f E E'1C15(E) we must have f(xB) a 0. Applying proposition 3.4 in [16, p. 91] we have that “xfiua E O for all a 6(7. Therefore, “-Ha is continuous for each a €<7. John J. Masterson [11, Corollary 2] has proved that an Archimedean vector lattice E is isomorphic (algebraically and lattice) to a Space M(X,B,u) of equivalence classes of u-measurable, almost everywhere finite-valued functions on the completely additive, a-finite measure space (X,B,u) if 70 and only if E is order separable and universally complete and F(E) is separating on E, where F(E) is an extended dual Space for E. For our purposes, all we need know about F(E) is that 5(E) separating on E implies that F(E) is separating on E. (For a discussion of F(E), see [8].) Therefore, if E is a universally complete vector lattice with a countable collection of continuous, monotone seminorms defining a locally convex, Hausdorff topology on E we know that E : M(X,/3,p.) where M(X,B,u) is described as above. (This result follows from Theorems 1.2 and 1.4 and Masterson's result.) In order to show that under our assumptions on E, M(X,B,p.) is either a finite-dimensional Space or the space of all real sequences we need some results of C. Goffman [2]. We shall refer the reader to Goffman's paper for the definitions of carriers of a lattice and the space of Caratheodory functions generated by a lattice. We note that if (X,B,u) is any localizable measure Space (i.e., the lattice of equivalence classes of u-measurable subsets is complete) having the finite subset property (i.e., any set of u-positive measure has a subset of u-finite measure) and ‘Bfi is the Boolean algebra of equivalence classes of measurable subsets of X then Er iS the lattice of carriers of M(X,B,p.) and M(X,/3,u) is the * Space of Caratheodory functions generated by .6 . x * Goffman's Criterion 2. Suppose x 616 may be split as follows: '1: *U h -k -0* x x1 x2, w ere x1 n x2 - ~k * h * _ 0* 1 x1 ' x11 U xiZ’ w ere x11 n x12 _ (i — ’2) x* =x* U x* 1112°°°1n-1 1112...1n_11 1112...1n_12 * * * where X n X ' = 0 (i :i >°~°:i,- = 1,2) 1112...in-11 1112...1 _12 1 2 a 1 and so that for every subsequence i1,i2,...,in,... = 1,2 * * * we have that xi n xi 1 n...n x 1 1 2 i112 * i n... =0 . Then * 0.. n x is said to satisfy criterion 2. ’ ~k * Goffman's Theorem. If x E B is the carrier of * x E M(X,B,p.) and x satisfies criterion 2, then there exists no monotone seminorm p on M(X,/3,u) Such that p(x) > O. * We shall Show that if x does not contain an atom * C (an atom e in a Boolean algebra 1s a non-zero element such * * it 9.- * * 'c that whenever e 2y 20 then c =y or y =0?) and ~k (X,B,p.) is a-finite then we can Split x as in criterion 2. Lemma 1.5. If (X,B,p.) is a a-finite measure Space, * 9: * . x E B such that x contains no atom, and O s 01 is any 1: extended real number with O s 01 s “(x ) then there exists * * ~k * *— an element x 6 B such that 0 c. x s: x and ”(X1) = (y. l 1 Proof. See [3], p. 174, exercise 2. Theorem 1.6. If (X,B,u) is a o-finite measure space * 'k * 7% :‘: and x is an element of 8 Such that x 3‘ 0 then x 9: does not contain an atom if and only if x satisfies criterion 2. 72 2322;. The fact that x* satisfying criterion 2 implies that x* does not contain an atom follows directly from some of Goffman's results. (See criterion 1 and some of the following discussion.) Therefore, we need only show that if x* does not con- tain an atom then x* satisfies criterion 2. We consider two cases. * * * * Case 1: “(x )‘< +~m. Choose x ,x2 €18 such that 1 * * * * * * * * * x = x1 U 32, x1 0 x2 = O and “(x1) = u(x2) = 1/2p(x ). (We can do this using the previous lemma.) Continue splitting * * x so that for n = 1,2,..., we choose x. . . 1, 111.2...ln-1 * . h * * U * X. o ... 2 W1C X. . =X. , 1 X. o 2, 1112 in-l 1112...1n_1 11...1 _1 1 .1n_1 * fl * 0* d ( * ) x. . x = an x . = 1 - 2 P - 1 1112 1mi 11 1n-1 11 1n-i * . —u(x 2) . 1/2 ”(x. ) (1 9000,]. - = 1,2). 11'..1n-1 11...in_1 1 n 1 Now for every sequence i1,i2,...,in,... (in = 1 or 2, * * n = 1,2,...) we get a chain x. . 2 x . 2°°°2 x. . . 2' m * * n * Therefore, u( n x ) = lim u(x. . ) = lim 1/2 u(x ) = O. - i 0.01 1 0.01 n-l l n nan l n new Hence, for any sequence 11,...,in,... (in = 1,2, n = 1,2,...) m * * * we get that 0 xi 1 i = 0 and this proves that x n=1 1 2 n satisfies criterion 2. 'k * * Case 2: u(x ) = +-m. Let X - C38 * * * X where X. n X, = 0 n 1 J * n 1 whenever j i i and suppose u(Xn) < +-m for n = 1,2,... . Let {nkz k = 1,2,...} be the subsequence of positive integers * * * * * * such that x n Xn # 0 for k = 1,2,..., and x n Xm = 0 k * if m * nk for each k. We split x as follows: Let 73 * 7 * * 'k * x* = x d x* where x* = x? n U X , x = x n X . Then * * * x: 0 x2 = 0 and p(xZ) < +-m. ‘We continue by Splitting x2 d o * o * * * as in case 1 an we split x1 into x1 x11 U x12 where * * * * 'k = x n U Xn and x12 = x1 0 X . AS m = 1,2,..., we ki1,2 k n2 * * have at the mth step: x111 1 = x11 1 U X11 ..12 where * x11 (m-l) times m times * * U X* d * 7': x = x 0 an X = x 11...1 k#1,2,...’m nk 11.0.12 nm * . and x . is split as in case 1 for any i ,1 i i ...1 1 2 m-l which are not all ones. Now we consider any sequence i1,iz,...,i ,... n . * * (1 = 1 or 2 n = 1,2,...). We note that x, . x 33 n ’ i i 11 1 2 * 7% i i i .3-- 3 xi i in iD--- forms a strictly decreasing l 2 311 2 * chain. If for any n we have that p(xi i i n) < + m then 1 2' * x as in case 1 fl xi i i = 0 . Hence, we need only consider n=111—[-112... ._)X. 7': ‘k * * the case i1,12,ooo,in,ooo : 1. X :)X 3x D"°:_)x ,)o¢o 1 11 111* * 1': 'k 1 11 11...1 m=l k>m nk ll = x* n i. q ( u x: )1 = 0*. m= 1 k>m “R We are ready to prove the following theorem. Theorem 1.7. If (Xn6,n) is a o—finite meaSure Space and x E M(Xy6,p) then there exists a monotone seminorm p on M(Xy8,u) with p(x) > 0 if and only if x* contains an atom. Egggf. Again, the proof that x* contains an atom implies that there exists a monotone seminorm p on M(X,5§u) such that p(x) > 0 follows directly from criterion 1 in Goffman's paper. 74 * To prove the other direction we note that if x does * not contain an atom then x must satisfy criterion 2 by the previous theorem. But then Goffman's Theorem shows that there exists no monotone seminorm p on M such that p(x) > 0. Before proving the characterization theorem we give the following definitions. Definitions 1.8. Let E be an Archimedean vector lattice. An element x in E is said to be discrete whenever there do not exist mutually disjoint elements y > O and z > 0 such that y,z s ‘x‘. Let x be any element in E+ and Bx the band in E generated by x. Then if y is any element of E+ the projection of, y. gg£g_ Bx is defined to be sup(y A (nx): n E N) whenever this supremum exists. (Notation: [x]y = sup(y A (nx): n E N).) A discrete element x always satisfies the property that whenever O < z < \xl then 2 = Xx for some real number x > 0 [23, p. 73]. Also, note that whenever E is Dedekind o-complete all projections onto principal bands must exist. Theorem 1.8. E is a universally complete vector lattice having a countable collection {u-un: n E N} of monotone, continuous seminorms defining a locally convex, Hausdorff topology on E if and only if E is a finite dimensional space or E is the Space of all real sequences. Egggf, We have already shown that E must be order separable and E : M(X,B,u) where (X,B,p.) is o-finite. 75 Since {H°Hn: n E N} defines a Hausdorff topology on E, given any 0 # x E E there exists a positive integer n(x) such that HxHn > 0. Hence, by the previous theorem, every * 0 # x E E is Such that x (the carrier of x) contains an * atom. (X,B,p.) o-finite implies that B can have at most * countably many atoms. Let {en} be the collection (possibly * finite) of distinct, and, therefore, disjoint atoms in [B . For each n let en = I * (the characteristic function of e * en) and note that en E M2Xn6,u) = E. Then {en} forms a complete system of pairwise disjoint, discrete elements in E. * * * Obviously, e A e = 0 for n ¥ m because e n e = 0 . n m n m To show that en is discrete for each n, choose any elements x,y E E such that O s x,y s en and show that either one of * x and y must be zero or x A y is not zero. Since en * * * * . * * is an atom and 0 c x , y c en either one of x and y * * must be zero or both x and y are equal to e * . If one of n * * * * * x or y is zero we are done, so we assume x = y = e But then x A y cannot be zero. Also, the collection {en} is complete in E because if f A en = O for all n where * * it f E E = M(X,B,p.) then f n e = 0 for all n. But this n is possible only if f* = 0* because if f* i 0* then f* must contain an atom; i.e., one of the e:. Hence, f = 0. Now, E is a universally complete vector lattice and {en} is a complete, countable (possibly finite) system of pairwise disjoint, discrete elements. We define a mapping R F: E -0 R , a 1-1, onto, lattice and algebraic isomorphism 76 where R = R0 or R = No (some finite integer) as follows: For every x in E note that [en]x = anen for each n Since en is a discrete element which forces B8 to be a one dimensional universally complete vector lattiZe [23, Theorem IV.12.1]. Define F(x) = fx € RR where fx(n) = an. One can easily see that this is a lattice and algebraic isomorphism. Also, the mapping is onto since E is universally complete. Hence, we have proved that every universally complete vector lattice satisfying the hypotheses of the theorem must be S or a finite dimensional space. The other direction of the theorem follows easily because pn(x) = x(n) for x =(x(1),x(2),...) in S (or x =(x(1),x(2),...,x(N)) for the case of N-dimensional Space) defines a continuous, monotone seminorm for n = 1,2,... (n = 1,2,...,N) and {pn} defines a locally convex, Hausdorff topology on E. BIBLIOGRAPHY 10. ll. BIBLIOGRAPHY Amemiya, I., "A generalization of Riesz-Fischer's theorem,” J, Math. Soc. of Japan 5 (1953), 353-354. Goffman, CaSper, "Remarks on lattice ordered groups and vector lattices. I. Caratheodory functions," Trans. Amer. Math. Soc. 88, No. 1 (1958), 107-120. Halmos, Paul, Measure Theory, Van Nostrand, New York, 1950. Holbrook, John A.R., "Seminorms and the Egoroff property in Riesz Spaces," Trans. Amer. Math. Soc. 132 (1968), 67 -77 o Kantorovitch, L., "Lineare halbgeordnete Rafime,” Mat. Sbornik 2 (1937), 121-168. Kawaii, I., "Locally convex lattices," J, Math. Soc. Japan 9 (1957), 281-314. Luxemburg, W.A.J., "Notes on Banach function Spaces,” Proc. Acad. Sci. Amsterdam: Note XIV, A 68 (1965), 229-248; Note XV, A 68, 415-446; Note XVI, A 68, 646-667. Luxemburg, W.A.J. and Masterson, John J., "An extension of the concept of the order dual of a Riesz space," Can. J. Math. 19 (1967), 488-498. Luxemburg, W.A.J. and Zaanen, A.C., "Notes on Banach function Spaces," Proc. Acad. Sci. Amsterdam: Note I, A 66 (1963), 135-147; Note II, A 66, 148-153; Note III, A 66, 239-250; Note IV, A 66, 251-263; Note V, A 66, 496-504; Note VI, A 66, 655-668; Note VII, A 66, 669-681; Note VIII, A 67 (1964), 104-119; Note IX, A 67, 360-376; Note X, A 67, 493-506; Note XI, A 67, 507-518; Note XII, A 67, 519-529; Note XIII, A 67, 530-543. Inxemburg,'W.A.J. and Zaanen, A.C., Riesz Spaces, Part I (Preprint of Book). Masterson, John J., "A characterization of the Riesz space of measurable functions," Trans. Amer. Math. Soc. 135 (1969), 193-197. 77 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 78 Masterson, John J. and Crofts, G., "Vector lattices and sequence spaces," Illinois J, of Math. (to be published). Nakano, H., Modern Spectral Theory, Maruzen Co., Tokyo, 1950. Nakano, H., Modulared Semi-Ordered Linear Spaces, Maruzen Co., Tokyo, 1950. Nishiura, Togo, "Completions of normed linear lattices," Colloq. Math. 19 (1968), 271-275. Peressini, A.L., Ordered Topological Vector Spaces, Harper and Row, New York, 1967. Peressini, A.L. and Sherbert, D.R., "Order properties of linear mappings on sequence Spaces," Math. Ann. 165 (1966), 318-332. Quinn, Joseph, Measure Riesz Spaces, Michigan State University Ph.D. dissertation, August, 1970. Solovev, V.A., "On the extension of a monotone norm from a normed vector lattice to its Dedekind completion," Sibirsk. Mat. g. 7 (1966), 1360-1369. Solovev, V.A., "The extension of a semi-continuous norm and a monotonically complete norm from a KN-lineal to its K-completion," Vestnik Leningrad. Univ. 23 (1968), 52-570 Veksler, A.I., "On the homomorphism between the class of regular operators on a K-lineal and its completion," Izv. Vyss. Ucebn. Zaved. Matematika No. 1 (1960), 48-570 Veksler, A.I., "Concerning a new construction of the Dedekind completion of some lattices and quotient Spaces of L-groups," Sibirsk. Mat. a. 10 (1969), 1206-1213. Vulich, B.Z., Introduction to the Theory of Partially Ordered Spaces, Wolters-Noordhoff, Groningen, Netherlands, 1967. Yamamuro, S., ”Monotone completeness of normed semi-ordered linear Spaces," Pacific ;. ngath. 7 (1957), 1715-1725. Zaanen, A.C., "Stability of order convergence and regularity in Riesz Spaces," Studia Math. 31 (1968), 159-172. APPENDIX APPENDIX Before proving Theorem 0.8 of the introduction we shall introduce some terms and state some results. The following discussion comes directly from [13, Sections 7 and 8]. Let E be a Dedekind o-complete vector lattice. For any two elements a and p in E+ define [213 = sup(a A (np): n E N). For arbitrary elements a and p in E define [p]a = [‘p\]a+ - [‘p\]a-. [p] is an Operator on E and is called the projector of p. Following are some results con- cerning the projector operator. Any of the proofs can be found in [13]. (1)- [131(ch + By) = aEPlx + BIPJy for all x,y.p in E and 0,8 real numbers. (2). \[p]x\ g ‘x\ for all x and p in E. (3). [p]x = 0 if and only if p A x = 0. (4). [p] = 0 if and only if p = 0. Definition. An element e # 0 of a vector lattice E is said to be a complete element in E if for any x in E e A x = 0 implies x = 0. A universally complete vector lattice always has a complete element [13, Theorem 32.4]. '79 80 1 if and only if p is a complete element (5). [p] of E. (6). [p] [up] for all real a i 0. (7). Given any a in E, [a+]a = a+, <8). tPJEq] = [[qu1 = [\p\ A \q\]. For any two projectors [p] and [q] we shall write + [p] 2 [q] if and only if [p]x 2 [q]x for every x in E . (9). [p] 2 [q] is equivalent to each of IPJIq] = [Q] and [plq = q- (10). ‘p‘ 2 \q‘ implies [P] 2 [q]. Since [p] 2 [q] as defined above defines an order re- lation on the set of projectors in a vector lattice we can de- fine Such terms as upper bound and least Upper bound of a set of projectors just as in a vector lattice. (11). If lp\ = SUP(\px\: x E A) then [P18 = sup([px]: 1 e A)a. Theorem 0.8. Let (E, s, H'H) be a normed vector lattice such that u-H is semi-continuous. If {xa: a E a? + is any'increasing net in E such that “Kg“ 5 M for all a E 63 then # sup x = X exists in E . a 0 Proof. Let e be a complete element in E#. For n = O,l,2,..., define S;'= sup([(xa - n5)+]e: a 6 GO in E#. Then 0 s'E; 1. And by (10) it follows that 0 s [3;] 1. We shall Show that o s [3;] 1 o. 81 Suppose there exists an element a in E such that [p]e 2 a 2 0 where [p] = inf([pg]: n E N). We shall show that a = 0. Since [(xa - nE)+](xa - he) 2 o for all a in <7 and all n (see (7)) we must have [(Xo - né)+]xa 2 2 [(xa - n5)+](ne). But, using (2), we then have xa 2 2 [(xa - n5)+](ne). Since (2) also implies that e 2 [p]e we also have ne 2 n[p]e 2 na. Combining these last two statements we obtain xa 2 [(xo - n5)+](na) and it now follows that sup(xa A (na): a Q G? 2 sup([(xa - ne)+](na): o 6 GB for n = O,l,2,... . On the other hand, using (11), (8), and (5), we have sup([(xa - n5)+](na): a G a0 = [3;](na). Also, [3;](na) 2 2 [p](na). Applying (9) to the inequality [p] 2 [na] which holds because [p] 2 [[535] 2 [a] we obtain [p](na) = na. Combining these inequalities we have sup([(xa - ne)+](na): a 6 a0 2 [p](na) = (na) 2 sup(xa A (na): a 6 d9. It now follows that (na) = sup(xa A (na): 0 E d9 for n = O,l,2,... . Applying the semi-continuity of H'H to the net xa A (na) To (na) we have “nau = sup(Hxa A (na)“: a 6 d9 S M. But HnaH S M for all n implies a = 0. Hence, [5]; = 0 because 5 = sup(ax: A 6 A) for some set {ax} C E. We now apply (4), (3) and the fact that e is a complete element to obtain [p] = 0. . #. Since E is universally complete the supremum of any ; collection of mutually disjoint elements in E1 must exist in E#. Hence. 30 = n§1n([p“'1] - [3;3)é - sup(n<[6n_1] - [En1>é> exists in E#. Also, since [3;] l 0 and [50] 2 [xa][5] = [X0] 82 for all or e a, we must have xa = n:1[E1-:] - Grand. Note that ([pn_11 - [5n]>(x, - né> = = [En_1]<1 - [in] + [En][*j - [(xa - né>+j> = = [5“,11<1 - [5n]><1 - [(xa - n5>+j)