Unbounded Norm Convergence in Banach Lattices
aa r X i v : . [ m a t h . F A ] M a y UNBOUNDED NORM CONVERGENCEIN BANACH LATTICES
Y. DENG, M. O’BRIEN, AND V.G. TROITSKY
Abstract.
A net ( x α ) in a vector lattice X is unbounded orderconvergent to x ∈ X if | x α − x | ∧ u converges to 0 in order forall u ∈ X + . This convergence has been investigated and appliedin several recent papers by Gao et al. It may be viewed as ageneralization of almost everywhere convergence to general vectorlattices. In this paper, we study a variation of this convergence forBanach lattices. A net ( x α ) in a Banach lattice X is unboundednorm convergent to x if (cid:13)(cid:13) | x α − x | ∧ u (cid:13)(cid:13) → u ∈ X + . Weshow that this convergence may be viewed as a generalization ofconvergence in measure. We also investigate its relationship withother convergences. Introduction
We begin by recalling a few definitions. A net ( x α ) α ∈ A in a vectorlattice X is said to be order convergent to x ∈ X if there is a net( z β ) β ∈ B in X such that z β ↓ β ∈ B , there exists α ∈ A such that | x α − x | ≤ z β whenever α ≥ α . For short, we will denotethis convergence by x α o −→ x and write that x α is o-convergent to x . Anet ( x α ) α ∈ A in a vector lattice X is unbounded order convergent to x ∈ X if | x α − x | ∧ u o −→ u ∈ X + . We denote this convergenceby x α uo −→ x and write that x α uo-converges to x . We refer the readerto [GTX] for a detailed exposition on uo-convergence and further ref-erences. In particular, for sequences in L p ( µ ) spaces, uo-convergenceagrees with almost everywhere (a.e.) convergence. Furthermore, if X can be represented as an ideal (or, more generally, a regular sublattice)in L ( µ ), then the uo-convergence of sequences in X agrees with thea.e. convergence in L ( µ ). Thus, uo-convergence may be viewed as ageneralization of a.e. convergence to general vector lattices. Date : May 12, 2016.2010
Mathematics Subject Classification.
Primary: 46B42. Secondary: 46A40.
Key words and phrases.
Banach lattice, un-convergence, uo-convergence, orderconvergence, AL-representation.The third author was supported by an NSERC grant.
Throughout this paper, X will stand for a Banach lattice. For anet ( x α ) in X we write x α → x if ( x α ) converges to x in norm, i.e., k x α − x k →
0. A net ( x α ) in X is unbounded norm convergent or un-convergent to x if | x α − x | ∧ u → u ∈ X + ; we then write x α un −→ x . Un-convergence was introduced in [Tro04] as a tool to studymeasures of non-compactness. In this paper, we study properties of un-convergence and its relationship to other convergences. In particular,we show that un-convergence may be viewed as a generalization ofconvergence in measure to general Banach lattices.2. Basic properties of un-convergence
Unless stated otherwise, we will assume that X is a Banach lat-tice and all nets and vectors lie in X . We routinely use the followinginequality: ( x + y ) ∧ u ≤ x ∧ u + y ∧ u for all x, y, u ∈ X + . Lemma 2.1. (i) x α un −→ x iff ( x α − x ) un −→ ; (ii) If x α un −→ x , then y β un −→ x for any subnet ( y β ) of ( x α ) . (iii) Suppose x α un −→ x and y α un −→ y . Then ax α + by α un −→ ax + by forany a, b ∈ R . (iv) If x α un −→ x and x α un −→ y , then x = y . (v) If x α un −→ x , then | x α | un −→ | x | .Proof. (i), (ii), and (iii) are straightforward. To prove (iv), observethat | x − y | ≤ | x − x α | + | y − x α | for every α . Put u = | x − y | ; it followsthat | x − y | = | x − y | ∧ u ≤ | x − x α | ∧ u + | y − x α | ∧ u → . Finally, (v) follows from (cid:12)(cid:12) | x α | − | x | (cid:12)(cid:12) ≤ | x α − x | . (cid:3) Remark 2.2.
In particular, x α un −→ x iff | x α − x | un −→
0. This often allowsone to reduce general un-convergence to un-convergence of positive netsto zero.
Example 2.3.
It was observed in Examples 21 and 22 in [Tro04] thaton c un-convergence agrees with coordinate-wise convergence and on C (Ω) un-convergence agrees with uniform convergence on compacta.The proofs of the following two facts are straightforward. Proposition 2.4. If x α → then x α un −→ . For order bounded nets,un-convergence and norm convergence agree. This justifies the name unbounded norm convergence.
Proposition 2.5.
In an order continuous Banach lattice, uo-convergenceimplies un-convergence.
NBOUNDED NORM CONVERGENCE 3
Example 2.6.
Let ( e n ) be the standard unit sequence ( e n ) in ℓ ∞ . Itfollows immediately from Proposition 2.4 that it is not un-null. Thisserves as a counterexample to several “natural” statements. First,while [GTX, Corollary 3.6] asserts that every disjoint sequence is uo-null (see also [Gao14, Lemma 1.1]), this example shows that a disjointsequence need not be un-null. Second, since ( e n ) is uo-null, this ex-ample shows that the order continuity assumption in Proposition 2.5cannot be dropped.Third, it was observed in [GTX, Theorem 3.2] that for a net ( x α ) ina regular sublattice F of a vector lattice E , x α uo −→ F iff x α uo −→ E . This fails for un-convergence: indeed, ( e n ) is un-null as a sequencein c , but not in ℓ ∞ . However, it is easy to see that if x α un −→ X then x α un −→ X . Example 2.7.
The next example shows that un-convergence in a sub-lattice does not imply un-convergence in the entire space even whenthe sublattice is a lattice copy of ℓ . Let X = ℓ ⊕ ∞ ℓ ∞ ; let ( f n ) be thestandard unit basis of ℓ and ( g n ) the standard unit sequence in ℓ ∞ .Put x n = f n ⊕ g n . Let Y be the closed span of ( x n ) in X . Since ( x n ) isa disjoint sequence in X , Y is exactly the closed sublattice generatedby ( x n ). Observe that (cid:13)(cid:13)(cid:13) n X k =1 α k x k (cid:13)(cid:13)(cid:13) = (cid:13)(cid:13)(cid:13) n X k =1 α k f k (cid:13)(cid:13)(cid:13) ∨ (cid:13)(cid:13)(cid:13) n X k =1 α k g k (cid:13)(cid:13)(cid:13) = (cid:16) n X k =1 | α k | (cid:17) ∨ (cid:16) n _ k =1 | α k | (cid:17) = n X k =1 | α k | . for any n and any scalars α , . . . , α n . It follows that the basic sequence( x n ) in X is 1-equivalent to ( f n ) in ℓ and, therefore, Y is an isometriclattice copy of ℓ in X . It is easy to see that f n un −→ ℓ ; hence x n un −→ Y .However, we claim that x n un −→ X . Indeed, let u = 0 ⊕ = W ∞ k =1 g k . Then x n ∧ u = g n for every n , hence ( x n ∧ u ) does not convergeto zero in X .The following three results are similar to Lemmas 3.6 and 3.7, andProposition 3.9 of [GX14]; we replace uo-convergence with un-convergence,and we do not require the space be order continuous in this case. Theproofs are similar. Lemma 2.8. If x α un −→ x then | x α | ∧ | x | → | x | and k x k ≤ lim inf α k x α k . Recall that a subset A of X is almost order bounded if for every ε > u ∈ X + such that A ⊆ [ − u, u ] + εB X . Equivalently, (cid:13)(cid:13)(cid:0) | x | − u ) + (cid:13)(cid:13) < ε for all x ∈ A . Y. DENG, M. O’BRIEN, AND V.G. TROITSKY
Lemma 2.9. If x α un −→ x and ( x α ) is almost order bounded then x α → x . Proposition 2.10. If ( x α ) is relatively weakly compact and x α un −→ x then ( x α ) converges to x in | σ | ( X, X ∗ ) .Proof. Without loss of generality, x = 0. Let f ∈ X ∗ + . Fix ε >
0. By[AB06, Theorem 4.37], there exists u ∈ X + such that f (cid:16)(cid:0) | x α | − u (cid:1) + (cid:17) <ε for every α . It follows from | x α | ∧ u → f (cid:0) | x α | ∧ u (cid:1) → , sothat f (cid:0) | x α | (cid:1) = f (cid:0) | x α | ∧ u (cid:1) + f (cid:16)(cid:0) | x α | − u (cid:1) + (cid:17) < ε for all sufficiently large α . It follows that f (cid:0) | x α | (cid:1) → . (cid:3) If ( x α ) is a net in a vector lattice with a weak unit e then x α uo −→ | x α | ∧ e o −→
0; see, e.g., [GTX, Lemma 3.5]. Analogously, the next resultlimits the task of checking un-convergence to a single quasi-interiorpoint, if one exists; cf [Tro04, Lemma 24].
Lemma 2.11.
Let X be a Banach lattice with a quasi-interior point e .Then x α un −→ iff | x α | ∧ e → .Proof. The forward implication is immediate. For the reverse implica-tion, let u ∈ X + be arbitrary and fix ε >
0. Note that | x α |∧ u ≤ | x α |∧ ( u − u ∧ me )+ | x α |∧ ( u ∧ me ) ≤ ( u − u ∧ me )+ m (cid:0) | x α |∧ e (cid:1) and, therefore, (cid:13)(cid:13) | x α | ∧ u (cid:13)(cid:13) ≤ k u − u ∧ me k + m (cid:13)(cid:13) | x α | ∧ e (cid:13)(cid:13) for all α and all m ∈ N . Since e is quasi-interior, we can find m suchthat k u − u ∧ me k < ε . Furthermore, it follows from | x α | ∧ e → α such that (cid:13)(cid:13) | x α | ∧ e (cid:13)(cid:13) < εm whenever α ≥ α . It followsthat (cid:13)(cid:13) | x α | ∧ u (cid:13)(cid:13) < ε + m εm = 2 ε. Therefore, | x α | ∧ u → (cid:3) Corollary 2.12.
Let X be an order continuous Banach lattice with aweak unit e . Then x α un −→ iff | x α | ∧ e → .Proof. If X is order continuous, then e is a weak unit iff e is a quasi-interior point. (cid:3) Recall that norm convergence is sequential in nature. In particular,given a net ( x α ) in a normed space, if x α → x then there exists an in-creasing sequence of indices ( α n ) such that x α n → x . This often allowsone to reduce nets to sequences when dealing with norm convergence.In view of Lemma 2.11, we can do the same with the un-convergence NBOUNDED NORM CONVERGENCE 5 as long as the space has a quasi-interior point (in particular, when X is separable): Corollary 2.13.
Suppose that X has a quasi-interior point and x α un −→ for some net ( x α ) in X . Then there exists an increasing sequence ofindices ( α n ) such that x α n un −→ . Question 2.14.
Does Corollary 2.13 remain valid without a quasi-interior point?We will show in Corollary 3.5 that the answer is affirmative for ordercontinuous spaces. 3.
Disjoint subsequences
The following lemma is standard; we provide the proof for the con-venience of the reader.
Lemma 3.1.
Let | x | = u + v for some vector x and some positivevectors u and v in a vector lattice. Then there exist y and z such that x = y + z , | y | = u , and | z | = v .Proof. Applying the Riesz Decomposition Property [AB06, Theorem 1.20]to the equality x + + x − = u + v , we find four positive vectors vectors a , b , c , and d such that u = a + b , v = c + d , x + = a + c , and x − = b + d .Put y = a − b and z = c − d . Then y + z = x + − x − = x . It fol-lows from 0 ≤ a ≤ x + and 0 ≤ b ≤ x − that a ⊥ b and, therefore, | y | = | a − b | = a + b = u . Similarly, c ⊥ d , and, therefore, | z | = v . (cid:3) Theorem 3.2.
Let ( x α ) be a net in X such that x α un −→ . Then thereexists an increasing sequence of indices ( α k ) and a disjoint sequence ( d k ) such that x α k − d k → .Proof. Assume first that x α ≥ α . Pick any α . Supposethat α , . . . , α k − have been constructed. Note that x α ∧ x α i → i = 1 , . . . , k −
1. Choose α k > α k − so that (cid:13)(cid:13) x α k ∧ x α i (cid:13)(cid:13) ≤ k + i forevery i = 1 , . . . , k −
1. This produces an increasing sequence of indices( α k ) such that k z ik k ≤ k + i where z ik = x α i ∧ x α k , 1 ≤ i < k .For every k , put v k = P k − i =1 z ik + P ∞ j = k +1 z kj . Clearly, v k is definedand k v k k < k . Put d k = ( x α k − v k ) + . It is easy to see that 0 ≤ x α k − d k ≤ v k , so that k x α k − d k k → k → ∞ . It is left to showthat the sequence ( d k ) is disjoint. Let k < m . Then d k = ( x α k − v k ) + ≤ ( x α k − z km ) + = x α k − x α k ∧ x α m , and d m = ( x α m − v m ) + ≤ ( x α m − z km ) + = x α m − x α k ∧ x α m . It follows that d k ⊥ d m . Y. DENG, M. O’BRIEN, AND V.G. TROITSKY
For the general case, we first apply the first part of the proof to thenet (cid:0) | x α | (cid:1) and produce an increasing sequence of indices ( α k ) and twopositive sequences ( w k ) and ( h k ) such that | x α k | = w k + h k , ( w k ) isdisjoint, and h k →
0. By Lemma 3.1, we can find sequences ( d k ) and( g k ) in X with | d k | = w k , | g k | = h k and x α k = d k + g k . It follows that( d k ) is a disjoint sequence and g k →
0. Thus, x α k − d k → (cid:3) Remark 3.3.
Theorem 3.2 is a variant of the Kadeˇc-Pe lczy´nski di-chotomy theorem; cf [LT79, p.38]. Theorem 3.2 clearly implies [GTX,Lemma 6.7]; unlike in [GTX, Lemma 6.7], we do not require the spaceto be order continuous or the net be norm bounded. Also, we startwith a net instead of a sequence.Recall the following standard fact; see, e.g., Exercise 13 in [AA02,p. 25].
Proposition 3.4.
Every norm convergent sequence in a Banach latticehas a subsequence which converges in order to the same limit.
Corollary 3.5.
Let ( x α ) be a net in an order continuous Banach lattice X such that x α un −→ . Then there exists an increasing sequence ofindices ( α k ) a such that x α k uo −→ and x α k un −→ .Proof. Let ( α k ) and ( d k ) be as in Theorem 3.2. Since ( d k ) is disjoint, wehave d k uo −→ d k un −→
0. It now follows from x α k − d k → x α k − d k un −→ x α k un −→
0. Furthermore, since x α k − d k →
0, passing to a further subsequence, we may assume that x α k − d k o −→ x α k − d k uo −→
0. This yields x α k uo −→ (cid:3) Note that Corollary 3.5 provides a partial answer to 2.14.4.
Uo-convergent subsequences and convergence inmeasure
We now have an analogue of Proposition 3.4 for un- and uo-convergences.
Proposition 4.1. If x n un −→ then there is a subsequence ( x n k ) of ( x n ) such that x n k uo −→ .Proof. Define e := P ∞ n =1 | x n | n k x n k . Let B e be the band generated by e in X . It follows from x n un −→ | x n | ∧ e → X and, therefore, in B e . There exists a subsequence ( x n k ) of ( x n ) such that | x n k | ∧ e o −→ B e . Since e is a weak unit in B e , we have x n k uo −→ B e . Finally, since B e is an ideal in X , it follows from [GTX, Corollary 3.8] that x n k uo −→ X . (cid:3) NBOUNDED NORM CONVERGENCE 7
It was observed in [Tro04, Example 23] that for sequences in L p ( µ ),where µ is a finite measure, un-convergence agrees with convergence inmeasure. We now provide an alternative proof of this fact. Corollary 4.2. ( [Tro04] ) Let ( f n ) be a sequence in L p ( µ ) where ≤ p < ∞ and µ is a finite measure. Then f n un −→ iff f n µ −→ .Proof. Without loss of generality, f n ≥ n . Suppose f n µ −→
0. Itis easy to see that f n ∧ → L p ( µ ). It follows fromLemma 2.11 that f n un −→ f n un −→
0. Then every subsequence ( f n k )is still un-null and, therefore, by Proposition 4.1, has a further subse-quence f n ki such that f n ki uo −→ f n ki a . e . −−→
0. This yields f n µ −→ (cid:3) Remark 4.3.
In the last step of the preceding proof, we used the factthat given a sequence of measurable functions over a measure spacewith a finite measure, the sequence converges in measure iff every sub-sequence has a further subsequence which converges a.e. (to the samelimit); see, e.g., Exercise 22 in [Roy88, p. 96]. Note that Proposition 4.1may be viewed as an extension of one of the directions of this equiva-lence to general Banach lattices. The next result shows that for ordercontinuous Banach lattices the other direction extends as well.
Theorem 4.4.
A sequence in an order continuous Banach lattice X is un-null iff every subsequence has a further subsequence which uo-converges to zero.Proof. The forward implication is Proposition 4.1. To show the con-verse, assume that x n un −→
0. Then there exist δ > u ∈ X + , and asubsequence ( x n k ) such that (cid:13)(cid:13) | x n k | ∧ u (cid:13)(cid:13) > δ for all k . By assumption,there is a subsequence ( x n ki ) of ( x n k ) such that x n ki uo −→
0, and, there-fore, x n ki un −→ | x n ki | ∧ u →
0, which isa contradiction. (cid:3)
Remark 4.5.
Again, Example 2.6 shows that the order continuityassumption cannot be removed.Suppose that X is an order continuous Banach lattice with a weakunit e . It is known that X can be represented as an order and normdense ideal in L ( µ ) for some finite measure µ . That is, there is avector lattice isomorphism T : X → L ( µ ) such that Range T is anorder and norm dense ideal of L ( µ ). Note that T need not be a normisomorphism, though T may be chosen to be continuous and T e = . Y. DENG, M. O’BRIEN, AND V.G. TROITSKY
Moreover, Range T contains L ∞ ( µ ) as a norm and order dense ideal.It is common to identify X with Range T and just view X as an idealof L ( µ ); we also identify e with . We call such an inclusion of X intoan L ( µ ) space an AL-representation of X . We refer the readerto [LT79, Theorem 1.b.14] or [GTX, Section 4] for more details onAL-representations.It was observed in [GTX, Remark 4.6] that for a sequence ( x n ) in X , x n uo −→ X iff x n a . e . −−→ L ( µ ). We prove an analogous resultfor un-convergence. Theorem 4.6.
Let X be an order continuous Banach lattice with aweak unit; let L ( µ ) be an AL-representation for X . For a sequence ( x n ) in X , we have x n un −→ in X iff x n µ −→ in L ( µ ) .Proof. By Theorem 4.4, x n un −→ X iff for every subsequence ( x n k )there is a further subsequence ( x n ki ) such that x n ki uo −→
0. The latter isequivalent to x n ki a . e . −−→
0. Now apply Remark 4.3. (cid:3) When do un- and uo-convergences agree?
Our next goal is to prove that uo- and un-convergences for sequencesagree iff X is order continuous and atomic. Recall that a non-zeroelement a ∈ X + is an atom iff the ideal I a consists only of the scalarmultiples of a . In this case, I a is a projection band. Let P a be thecorresponding band projection. We say that X is atomic or discrete if it equals the band generated by all the atoms in it. Suppose that X is atomic, and fix a maximal disjoint collection A of atoms in X .For every x ∈ X + , we have x = W a ∈ A x a a , where x a a = P a x . Onecan also write this as x = P a ∈ A x a a , where the sum is understoodas the order limit (or the supremum) of sums over finite subsets of A . This sum may be viewed as a coordinate expansion of x over A .Furthermore, if we are also given a positive vector y = P a ∈ A y a a , then x ≤ y iff x a ≤ y a for every a ∈ A . Suppose now that, in addition, X is order continuous. Then it can be shown that only countablymany of coefficients x a are non-zero. Enumerating them, we get x = P ∞ i =1 x a i a i , where the series converges in order and, therefore, in norm.For details, see [AB03], Exercise 7 in [Sch74, p. 143], and the proof ofProposition 1.a.9 in [LT79]. We will need two standard lemmas. Lemma 5.1.
Suppose that X is atomic and order continuous, and ( x n ) is an order bounded sequence in X . If x n → then x n o −→ .Proof. Without loss of generality, x n ≥ n . Let u ∈ X + suchthat x n ≤ u for every n . There is a sequence of distinct atoms ( a i ) in A NBOUNDED NORM CONVERGENCE 9 such that u = P ∞ i =1 u i a i for some coefficients ( u i ); the series convergesin norm and in order. Given n ∈ N , it follows from 0 ≤ x n ≤ u that wecan write x n = P ∞ i =1 x ni a i for some sequence of coefficients ( x ni ). Notethat 0 ≤ x ni a i ≤ x n for every n and i ; it follows that lim n x ni = 0 forevery i ; that is, the sequence ( x n ) converges to zero “coordinate-wise”.For each k ∈ N , define v k = k X i =1 ( k ∧ u i ) a i + ∞ X i = k +1 u i a i . It is easy to see that v k ↓
0. On the other hand, since x n ≤ u and ( x n )converges to zero coordinate-wise, for every k we can find n k such that x n ≤ v k whenever n ≥ n k . It follows that x n o −→ (cid:3) Lemma 5.2. If µ is a finite non-atomic measure then there exists asequence ( f n ) in L ∞ ( µ ) which converges to zero in measure but not a.e..Proof. In the special case when µ is the Lebesgue measure on the unitinterval, we take ( f n ) to be the “typewriter” sequence f n = χ [ n − k k , n − k +12 k ] where k ≥ k ≤ n < k +1 . In the general case, we produce a similar sequence using Exercise 2 in[Hal70, p. 174]. (cid:3)
Theorem 5.3.
The following are equivalent: (i) x n uo −→ ⇐⇒ x n un −→ for every sequence ( x n ) in X ; (ii) X is order continuous and atomic.Proof. (ii) ⇒ (i) Suppose X is order continuous and atomic. The im-plication x n uo −→ ⇒ x n un −→ ⇒ (ii) Let ( x n ) be a disjoint order bounded sequence in X . Then x n uo −→ x n un −→
0. Sincethe sequence is order bounded, this yields x n →
0. Hence, X is ordercontinuous. It follows that every closed ideal in X is a projection band.It remains to show that X is atomic. Suppose not; then the band X generated by all the atoms in X is a proper subset of X . Let X be thecomplementary band. Fix a non-zero w ∈ X +2 , and let Y = B w , theband generated by w in X . Clearly, Y ⊆ X , Y is order continuous, w is a weak unit in Y , and Y has no atoms. We can find an AL-representation for Y such that L ∞ ( µ ) ⊆ Y ⊆ L ( µ ). Since Y has noatoms, it is easy to see that µ is a non-atomic measure. By Lemma 5.2, there is a sequence ( x n ) in L ∞ ( µ ) such that x n µ −→ x n a . e . −−→
0. It follows that x n un −→ x n uo −→ Y and, therefore,in X ; a contradiction. (cid:3) Un-convergence and weak convergence
In this section, we consider the relationship between un- and weakconvergences. For a mononote net, weak convergence implies norm con-vergence, and, therefore, un-convergence; however, weak convergencedoes not imply un-convergence in general. For example, the followingfact was observed in [CW98, Theorem 2.2]:
Lemma 6.1. ( [CW98] ) If X is non-atomic and order continuous, and x ∈ X + , then there exists a sequence ( x n ) such that x n w −→ yet | x n | = x for all n . Clearly, ( x n ) is not un-null. Proposition 6.2.
The following are equivalent: (i) x n w −→ implies x n un −→ for every sequence ( x n ) in X ; (ii) X is order continuous and atomic.Proof. (i) ⇒ (ii) The proof is similar to that of Theorem 5.3. Let ( x n ) bea disjoint order bounded sequence in X . Then x n w −→
0. By assumption, x n un −→
0. Since ( x n ) is order bounded, this yields x n →
0. Therefore, X is order continuous.Now suppose that X is not atomic. Then X = X ⊕ X where X isthe band generated by the atoms, and X is the complementary band.Since X is not atomic, we have X = { } . Now apply Lemma 6.1 to X .(ii) ⇒ (i) Lemma 6.14 in [GTX] asserts that if X is atomic and x n w −→ x n uo −→
0. If, in addition, X is order continuous, this yields x n un −→ (cid:3) In the previous result, the conditions for a weakly-null sequence tobe un-null are quite strong. On the other hand, in an arbitrary Banachlattice, this fact is true for monotone nets. If we remove the constraintthat X is atomic, then we obtain the following result. Recall that for anorder bounded positive net ( x α ) in an order continuous Banach lattice, x α w −→ x α → Proposition 6.3.
Let ( x α ) be a positive net in an order continuousBanach lattice X . If x α w −→ then x α un −→ . NBOUNDED NORM CONVERGENCE 11
Proof.
For every u ∈ X + we have 0 ≤ x α ∧ u ≤ x α for every α . Thisyields x α ∧ u w −→
0. Since this net is positive and order bounded, x α ∧ u →
0. Hence, x α un −→ (cid:3) We have now seen several conditions on X that yield x n w −→ x n un −→
0. For the converse, we have the following.
Theorem 6.4. If X ∗ is order continuous then x α un −→ implies x α w −→ for every norm bounded net ( x α ) in X + .Proof. Suppose that X ∗ is order continuous and ( x α ) is a norm boundednet in X + with x α un −→
0. Without loss of generality, k x α k ≤ α . Since X ∗ is order continuous, it follows from [AB06, Theorem 4.19]that for every ε > f ∈ X ∗ + there exists u ∈ X + such that f (cid:0) | x | −| x | ∧ u (cid:1) < ε whenever k x k ≤
1. In particular, f (cid:0) x α − x α ∧ u (cid:1) < ε for every α . It follows from x α ∧ u → f ( x α ) < ε for all sufficientlylarge α . Hence, f ( x α ) → x α w −→ (cid:3) We do not know whether the converse is true. If X ∗ is not order con-tinuous then ℓ is lattice embeddable in X ; see, e.g., [AB06, Theorem4.69]. Let ( e n ) be the standard basis of ℓ viewed as a sequence in X .It is easy to see that e n w −→ ℓ and, therefore, in X . It is also easyto see that e n un −→ ℓ . However, this does not imply that e n un −→ X ; see Example 2.7.7. Un-convergence is topological
It is well known that a.e. convergence is not topological; see, e.g.,[Ord66]. That is, this convergence is not given by a topology. It fol-lows that uo-convergence need not be topological in general. We showthat un-convergence is topological. Moreover, we explicitly define theneighborhoods of this topology.Given an ε > u ∈ X + , we put V u,ε = (cid:8) x ∈ X : (cid:13)(cid:13) | x | ∧ u (cid:13)(cid:13) < ε (cid:9) . Let N be the collection of all the sets of this form. We claim that N is a base of neighborhoods of zero for some Hausdorff linear topology.Once we establish this, we will be able to define arbitrary neighbor-hoods as follows: a subset U of X is a neighbourhood of y if y + V ⊆ U for some V ∈ N . It follows immediately from the definition of un-convergence that x α un −→ N contains a tail of thisnet, hence the un-convergence is exactly the convergence given by thistopology. We have to verify that N is indeed a base of neighborhoods of zero;cf. [KN76, Theorem 5.1] or [Run05, Theorem 3.1.10].First, every set in N trivially contains zero.Second, we need to show that the intersection of any two sets in N contains another set in N . Take V u ,ε and V u ,ε in N . Put ε = ε ∧ ε and u = u ∨ u . We claim that V u,ε ⊆ V u ,ε ∩ V u ,ε . Indeed, take any x ∈ V u,ε . Then (cid:13)(cid:13) | x | ∧ u (cid:13)(cid:13) < ε . It follows from | x | ∧ u ≤ | x | ∧ u that (cid:13)(cid:13) | x | ∧ u (cid:13)(cid:13) ≤ (cid:13)(cid:13) | x | ∧ u (cid:13)(cid:13) < ε ≤ ε , so that x ∈ V u ,ε . Similarly, x ∈ V u ,ε .It is easy to see that V u,ε + V u,ε ⊆ V u, ε . This immediately impliesthat for every U in N there exists V ∈ N such that V + V ⊆ U . It isalso easy to see that for every U ∈ N and every scalar λ with | λ | ≤ λU ⊆ U .Next, we need to show that for every U ∈ N and every y ∈ U ,there exists V ∈ N such that y + V ⊆ U . Let y ∈ V u,ε for some ε > u ∈ X + . We need to find δ > v ∈ X + such that y + V v,δ ⊆ V u,ε . Put v := u . It follows from y ∈ V u,ε that (cid:13)(cid:13) | y | ∧ u (cid:13)(cid:13) < ε ; take δ := ε − (cid:13)(cid:13) | y | ∧ u (cid:13)(cid:13) . We claim that y + V v,δ ⊆ V u,ε . Let x ∈ V v,δ ; it suffices show that y + x ∈ V u,ε . Indeed, | y + x | ∧ u ≤ | y | ∧ u + | x | ∧ u, so that (cid:13)(cid:13) | y + x | ∧ u (cid:13)(cid:13) ≤ (cid:13)(cid:13) | y | ∧ u (cid:13)(cid:13) + (cid:13)(cid:13) | x | ∧ u (cid:13)(cid:13) < (cid:13)(cid:13) | y | ∧ u (cid:13)(cid:13) + δ = ε. Finally, in order to show that the topology is Hausdorff, we needto verify that T N = { } . Indeed, suppose that 0 = x ∈ V u,ε forall non-zero u ∈ X + and ε >
0. In particular, x ∈ V | x | ,ε , so that k x k = (cid:13)(cid:13) | x | ∧ | x | (cid:13)(cid:13) < ε for every ε >
0; a contradiction.Note that we could also conclude that the topology is linear andHausdorff from Lemma 2.1.
Acknowledgement.
We would like to thank Niushan Gao for valu-able discussions.
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School of Mathematics, Southwest Jiaotong University, Chengdu,Sichuan, 610000, China.
E-mail address : Department of Mathematical and Statistical Sciences, Universityof Alberta, Edmonton, AB, T6G 2G1, Canada.
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