When are the nonstandard hulls of normed lattices discrete or continuous?

In Functional Analysis, the ultrapower and the nonstandard analysis approaches are equivalent: results obtained by one of these two methods can usually be translated into the other. In this short note, we present nonstandard analysis versions of the main results of [5], where they were originally presented in the ultrapower language. We believe that in this new form the ideas of the proofs are more transparent.

Suppose that E is a Archimedean vector lattice. Recall that an element 0 < e G E is said to be discrete if 0 6 x 6 e implies that x is a scalar multiple of e or, equivalently, the interval [0, e] doesn’t contain two non-zero disjoint vectors (see [3, Theorem 26.4]). We say that E is continuous if it contains no discrete elements and discrete if every non-zero positive vector dominates a discrete element or, equivalently, E has a complete disjoint system consisting of discrete elements (see [1, p. 40]).

If E is a normed space. We will write ^∗ E for the nonstandard extension of E and E for the nonstandard hull of E . We refer the reader to [2, 6] for terminology and details on nonstandard hulls of normed spaces and normed lattices. We will use the following standard fact (see, e. g., [4, Remark 4]).

Lemma 1. Suppose that E is a normed lattice and a, x, b ∈ ^∗ E such that a 6 b and a 6 x 6 b . Then there exists y G * E such that y ~ x and a 6 y 6 b .

The following is a variant of Theorem 2.2 of [5]:

Theorem 2. Let E be a normed lattice. Then the following are equivalent.

• (i)    E is continuous;

• (ii)    3 e > 0 V x G E + 3 a, b G [0, x] a ± b and ||a k Л ||b || >  e||x||.

C (i) ⇒ (ii) Suppose that E fails (ii). Let ε be a positive infinitesimal. Then there exists a vector x G * E + such that for all a,b G * [0,x] with a ± b we have | a | Л | b | < e k x | . Without loss of generality, | x | = 1 . Let a,b G [0,x] and a ± b . By Lemma 1, we may assume that

(° 2009 Troitsky V. G.

• a, b G * [0, x] . Furthermore, a ± b implies that a Л b to 0 . Let u = a — a Л b and v = b — a Л b , then u, v G * [0, x] and u ± v , so that ||u k Л ||v k < e . It follows that either ||u k or || v k is infinitesimal. Say, u to 0 . Then a = u + a Л b is infinitesimal as well, so that a = 0 . Thus, x is discrete in E .

• (ii)    ^ (i) Suppose that (ii) holds for some (standard) e > 0 . Let x G E + , show that x is not discrete. Without loss of generality, x G * E + and ||x || = 1 . By (ii), we can find a, b G * [0, x] such that a ⊥ b and k a k ∧ k b k >  ε . It follows that neither a nor b is infinitesimal, so that a, b are two non-zero disjoint elements of [0, x] . Hence, x is not discrete. B

Recall that a normed lattice satisfies the Fatou property if 0 6 x _a f x implies | x _a | ^ | x | , and the ст -Fatou property if 0 6 x _n f x implies | x n k ^ | x | , see, e. g., [1]. We will use the following simple lemma.

Lemma 3. Suppose that E is a normed lattice with the Fatou property and S ⊆ E + such that x = sup S exists. Then for every e >  0 there is a finite subset y of S such that || sup y k >  (1 — e) | x | . The same is true for countable families if E satisfies the ст -Fatou property.

C Let Л be the collection of all finite subsets of S , ordered by inclusion. Clearly, sup _{a G} л sup a = x . Let x _a = sup a , then (x _a ) _{a G} л is an increasing net and 0 6 x _a f x . It follows from the Fatou property that | x _a | ^ k x k , so that there exists 7 G Л with ||x ₇ k >  (1 — e) | x | .

Now suppose that E satisfies ст -Fatou property and x = V £ i x i . Let Z k = V i =i x i , then x k 6 Z k 6 x , so that x = Vfc =i Z k . Now ст -Fatou property guarantees that k z k k ^ k x k , so that (1 — e) k x k 6 ||z _m k = k x i V • • • V x _m | for some m . B

The following is a variant of Theorem 3.1 of [5].

Theorem 4. Let E be a discrete normed lattice, and D the set of all discrete elements of norm one in E . If E satisfies the Fatou property (or the ст -Fatou property if D is countable) then the discrete elements of E are exactly the positive scalar multiples of the elements of { e | e G * D } .

C It suffices to show that given x G * E with k x k = 1 , then x is discrete in E if and only if x = e for some e G * D . Suppose that x = e for some e G * D . Take any a G * E such that 0 6 a 6 x . By Lemma 1, we may assume that 0 6 a 6 e . It follows that a is a scalar multiple of e , hence a is a scalar multiple of x .

Conversely, suppose that x is discrete in E . Note that the set D is a complete disjoint system in E . By [1, Theorem 1.75], we have x = sup { P _e x | e G * D } . For every e G * D , the vector P e x is a scalar multiple of e , and 0 6 P e x 6 x , hence 0 6 P e x 6 x . Therefore, if P e x is not infinitesimal for some e G * D then x is a scalar multiple of P e x , hence of e .

Suppose now that Pex is infinitesimal for every e G *D. It follows from x = sup{Pex | e ∈ ∗D} and Lemma 3 that there exist n ∈ ∗N and e1, . . . , en ∈ ∗D such that kzk > 34 , where z = ||Pe1 x V • • • V Penxk- Choose k 6 n in *N so that °Pe1 x V • • • V Pek-1 x° < 4, while kPeix V • • • V Pei,xk > 1. Put u = Peix V • • • V Pe,x = Peix + • • • + Pe.x. Then e1 ek 4 e1 ek e1 ek

4 ^{6 k} u ^{k 6} ° P e i ^{x V} • •• ^V P e k - 1 ^x ° ^{+ k} P e k ^{x |} . 4

hence | u | ~ 4 . Put v = z — u , then u ± v , 0 6 u, v 6 z , and | u | , | v | >  4 . Therefore, u and v are non-zero and disjoint elements of [0, x] ; a contradiction. B

Corollary 5. Suppose that E is an AM-space with a strong unit, and H is a discrete regular sublattice of E . Then H is discrete.

C Let D be a complete disjoint system of discrete elements of norm one in H . Suppose ---- that x G H+. We will show that x majorizes a discrete vector. Without loss of generality,

When are hulls discrete or continuous?

x G *H+ with kxk = 1. Then x = sup{Pex | e G *D} by [1, Theorem 1.75]. Since E is an AM-space, we can apply Lemma 3 with e to 0 and find n G *N and ei,..., en G *D such that kPe1 x V • • • V Penx| > (1 — e)kxk ~ 1. Again, since E is an AM-space, we have kPe1 x V • •• V Penxk = kPe1 xk V ••• V ||Penxk, so that ||Pekxk to 1 for some k 6 n. Then [ is non-zero. It is discrete by Theorem 4 because Pek x is a multiple of ek . Finally, notice that Pekx 6 x. B