Positivity (2018) 22:837–843 https://doi.org/10.1007/s11117-017-0548-z
Positivity
Unbounded order continuous operators on Riesz spaces Akbar Bahramnezhad1 · Kazem Haghnejad Azar1
Received: 31 October 2017 / Accepted: 2 December 2017 / Published online: 7 December 2017 © Springer International Publishing AG, part of Springer Nature 2017
Abstract In this paper, using the concept of unbounded order convergence in Riesz spaces, we define new classes of operators, named unbounded order continuous (uocontinuous, for short) and boundedly unbounded order continuous operators. We give some conditions under which uo-continuity will be equivalent to order continuity of some operators on Riesz spaces. We show that the collection of all uo-continuous linear functionals on a Riesz space E is a band of E ∼ . Keywords Riesz space · Order convergence · Unbounded order convergence · Unbounded order continuous operator Mathematics Subject Classification 46B42 · 47B60
1 Introduction The notion of unbounded order convergence (uo-convergence, in brief) was initially introduced in [9] and “uo-convergence” is proposed firstly in [3]. Many papers are devoted to this concept (see [4–8]). Unbounded convergence net in term of weak convergence, uaw-convergence, was introduced and considered in [10]. Let E be a Riesz space. For a net (xα )α∈A in E, if there is a net (yβ )β∈B , possibly over a different index set, with yβ ↓ 0 and for every β ∈ B there exists α0 ∈ A such that |xα − x| ≤ yβ for all α ≥ α0 , we say that (xα )α∈A converges to x in order, denoted
B
Akbar Bahramnezhad
[email protected] Kazem Haghnejad Azar
[email protected]
1
Department of Mathematics, University of Mohaghegh Ardabili, Ardabil, Iran
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A. Bahramnezhad, K. H. Azar o
by xα − → x. A net (xα ) in a Riesz space E is unbounded order convergent to x ∈ E if uo o |xα − x| ∧ u − → 0 for all u ∈ E + . We denote this convergence by xα −→ x and write that xα uo-convergent to x. If there exists a net (yα )α∈A (with the same index set) in o → x. a Riesz space E such that yα ↓ 0 and |xα − x| ≤ yα for each α ∈ A, then xα − Conversely, if E is a Dedekind complete Riesz space and (xα )α∈A is order bounded, o then xα − → x in E implies that there exists a net (yα )α∈A (with the same index set) such that yα ↓ 0 and |xα − x| ≤ yα for each α ∈ A. For sequences in a Riesz space E, o → x if and only if there exists a sequence (yn ) such that yn ↓ 0 and |xn − x| ≤ yn xn − for each n ∈ N (see [1, Page 17–18]). For unexplained terminology and facts on Riesz spaces and positive operators, we refer the reader to the excellent book of [2]. We start by identifying unbounded order continuous operators. Definition 1 An operator T : E → F between two Riesz spaces is said to be: uo
1. Unbounded order continuous (or, uo-continuous for short), if xα −→ 0 in E implies uo T xα −→ 0 in F. uo 2. Unbounded σ -order continuous (or, uσ o-continuous for short), if xn −→ 0 in E uo implies T xn −→ 0 in F. Definition 2 Let E be a normed Riesz space and F be a Riesz space. An operator T : E → F is said to be uo
1. Boundedly unbounded order continuous (or, buo-continuous for short) if T xα −→ 0 in F for any norm bounded uo-null net (xα ) in E. uo 2. Boundedly unbounded σ -continuous (or, buσ o-continuous for short) if T xn −→ 0 in F for any norm bounded uo-null sequence (xn ) in E. The collection of all uo-continuous operators of L b (E, F) will be denoted by L uo (E, F), that is L uo (E, F) := {T ∈ L b (E, F) : T is uo-continuous}. Similarly, L uσ o (E, F) will denote the collection of all order bounded operators from E to F that are uσ o-continuous. That is, L uσ o (E, F) := {T ∈ L b (E, F) : T is uσ o-continuous}. Clearly, we have L uo (E, F) ⊂ L uσ o (E, F) and L uo (E, F), L uσ o (E, F) are both vector subspaces of L b (E, F). The collection of all boundedly uo-continuous (resp. uσ o-continuous) operators from a normed Riesz space E into a Riesz space F will be denoted by L buo (E, F) (resp. L buσ o (E, F)). Example 1 Let B be a projection band and P the corresponding band projection. Then P is uo-continuous (see [6, Lemma 3.3]).
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Example 2 Let (, , μ) be a measure space. We write L 0 (μ) for the Riesz space of real-valued measurable functions on modulo almost everywhere (a.e.) equality equipped with the a.e. order: f ≥ g if and only if f (t) ≥ g(t) for a.e. t ∈ . By [7, Proposition 3.1], if T : L 0 (μ) → L 0 (μ) is an operator, then T is uσ o-continuous if and only if T is σ -order continuous. Example 3 Consider the Lozanovsky-like example T : C[0, 1] → c0 in [4], given by
1
T( f ) = 0
1
f (x) sinx d x,
f (x) sin2x d x, . . .
0
for each f ∈ C[0, 1]. Note that T is not order bounded and hence not order continuous. uo Let ( f n ) be a norm bounded sequence in C[0, 1] for which f n −→ 0 holds. Since uaw C[0, 1] is an AM-space, we have f n −−→ 0. It follows from [10, Theorem 7] that w w → 0 and that T ( f n ) − → 0. Since T ( f n ) is a norm bounded sequence in c0 , fn − uo T ( f n ) −→ 0. Hence, T is buσ o-continuous.
2 Main results We first remark useful facts about uo-convergence. Lemma 1 ([6, Lemma 3.1]) In a Riesz space we have the following: uo
uo
1. If xα −→ x and xα −→ y, then x = y. In other hands, unbounded order limits are uniquely determined. uo uo uo 2. If xα −→ x, yα −→ y and k, r are real numbers, then kxα + r yα −→ kx + r y. uo uo uo Furthermore xα ∨ yα −→ x ∨ y and xα ∧ yα −→ x ∧ y. In particular xα+ −→ x + , uo uo xα− −→ x − and |xα | −→ |x|. uo 3. If xα −→ x and xα ≥ y holds for all α, then x ≥ y. uo 4. If xα ↑ or xα ↓ and xα −→ x , then xα ↑ x or xα ↓ x, respectively. Proposition 1 ([2, Exercise 11, P.57]) Let E be a Dedekind complete Riesz space o and (xα ) an order bounded net in E. Then xα − → x if and only if x = limsupα xα = liminf α xα . Proof Assume first that x = limsupα xα = liminf α xα . Note that limsupα xα = inf α (supβ≥α xβ ) = ∧α (∨β≥α xβ ) and liminf α xα = supα (inf β≥α xβ ) = ∨α (∧β≥α xβ ). Clearly, ∨β≥α xβ ↓ x and ∧β≥α xβ ↑ x and xα − x ≤ ∨β≥α xβ − ∧β≥α xβ and x − xα ≤ ∨β≥α xβ − ∧β≥α xβ . o Thus, |xα − x| ≤ (∨β≥α xβ − ∧β≥α xβ ) ↓ 0. Therefore, xα − → x. Conversely, assume o that xα − → x. Then there exists a net (yα ) such that |xα − x| ≤ yα ↓ 0 for each
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α. For each β ≥ α we have, −yα ≤ −yβ ≤ xβ − x ≤ yβ ≤ yα . Therefore, −yα ≤ ∧β≥α xβ − x ≤ ∨β≥α xβ − x ≤ yα holds for all α. So, x = supα (x − yα ) ≤ ∨α (∧β≥α xβ ) ≤ ∧α (∨β≥α xβ ) ≤ inf α (x + yα ) = x.
(1) (2) (3)
Hence, x = limsupα xα = liminf α xα . This completes the proof. Corollary 1 Let E be a Dedekind complete Riesz space and (xα ) a net in E. Then uo xα −→ x for some x ∈ E if and only if inf α supβ≥α |xβ − x| ∧ u = supα inf β≥α |xβ − x| ∧ u = 0 for all u ∈ E + . It should be mentioned that the uo-convergence in a Riesz space E is not topological and does not necessarily correspond to a topology on E in the sense that order and topological nets or sequences are the same. Example 4 Let E = c0 , the Riesz space of all real valued null sequences. Then uoconvergence in E is equivalent to pointwise convergence which is not topological. Example 5 Let E = c, the Banach lattice of real valued convergent sequences. Put n e , where (e ) is the standard basis. Then (x ) is uo-convergent to x = xn = k=1 k n n (1, 1, 1, . . .), but it is not norm convergent. Example 6 Assume that (, , μ) is a measure space and let E = L p (μ) for some 1 ≤ p < ∞. Then uo-convergence of sequences in L p (μ) is the same as almost everywhere convergence. Note that order convergence implies uo-convergence but the converse is not true in general. Example 7 Let E = c0 , the Riesz space of all real valued null sequences and (en ) be the standard basis of E. Then (en ) uo-converges to 0 but it is not order convergent to zero, since E has order continuous norm. The most explicit property of the class of uo-continuous operators is that is satisfies the domination property, that is, if an uo-continuous operator T dominates S, then S is also uo-continuous. But, unfortunately we don’t know the full analogue of Ogasawara’s result in [2, Theorem 1.57], for this class of operators is false or true. So, in this paper the following problem remains as open. Problem 1 Let E, F be Riesz spaces. 1. Assume that a uo-continuous operator T : E → F has a modulus |T |. Is |T | uo-continuous?
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2. Are L uo (E, F) and L uσ o (E, F) both bands of L b (E, F)? In this part, we show that the collection of all order bounded uo-continuous linear functionals on a Riesz space E is a Band of E ∼ . Let T : E → F be a positive operator between Riesz spaces. We say that an operator S : E → F is dominated by T (or that T dominates S ) whenever |Sx| ≤ T |x| holds for each x ∈ E. Proposition 2 If a positive uo-continuous operator T : E → F dominates S, then S is uo-continuous. Proof Follows immediately from part (2) of Lemma 1 and |Sx| ∧ u ≤ (T |x|) ∧ u for all u ∈ F + . Theorem 1 For an order bounded linear functional f on a Riesz space E the following statements are equivalent. 1. f is uo-continuous. 2. f + and f − are both uo-continuous. 3. | f | is uo-continuous. uo
Proof (1) ⇒ (2) By Lemma 1, we may assume that (xα ) ⊂ E + . Let xα −→ 0 and let (rα ) be a net in R such that rα ↓ 0. In view of f + x = sup{ f y : 0 ≤ y ≤ x}, for each α there exists a net (yα ) in E with 0 ≤ yα ≤ xα such that f + xα − rα ≤ f yα . uo uo So, f + xα ≤ f yα + rα . Since xα −→ 0, we have yα −→ 0. Thus, by assumption, uo uo uo f yα −→ 0. It follows from f + xα ≤ ( f yα + rα ) −→ 0 that f + xα −→ 0. Hence, f + is uo-continuous. Now, as f − = (− f )+ , we conclude that f − is also uo-continuous. (2) ⇒ (3) Follows from the identity | f | = f + + f − . (3) ⇒ (1) Follows immediately from Proposition 2 by observing that | f | dominates f. Theorem 2 If E is a Riesz space, then L uo (E, R) and L uσ o (E, R) are both bands of E ∼. Proof We only show that L uo (E, R) is a band of E ∼ . Note first that by Theorem 1, L uo (E, R) is an ideal of E ∼ . To see that the ideal L uo (E, R) is a band, let 0 ≤ f λ ↑ f uo in E ∼ with ( f λ ) ⊂ L uo (E, R), and let 0 ≤ xα −→ 0 in E. Then for each fixed λ and each w ∈ R+ we have o
0 ≤ f (xα ) ∧ w ≤ (( f − f λ )(xα ) ∧ w + f λ (xα ) ∧ w) − → 0. o
uo
So, f (xα ) ∧ w − → 0 and hence, f (xα ) −→ 0. Thus, f ∈ L uo (E, R), and the proof is finished. Now, we give some conditions under which uo-continuity (resp. uσ o-continuity) will be equivalent to order continuity (resp. σ -order continuity) of some operators on Riesz spaces. Recall that a Riesz space is said to be σ -laterally complete if every disjoint sequence has a supremum. For a set A, R A is an example of σ -laterally complete Riesz space.
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Proposition 3 Let E, F be Riesz spaces such that E is σ -Dedekind complete and σ -laterally complete and let T : E → F be an order bounded operator. Then T is uσ o-continuous if and only if T is σ -order continuous. Proof Suppose T is uσ o-continuous. Since order convergent sequences in a Riesz space are automatically order bounded and for order bounded nets, uo-convergence is equivalent to o-convergence, T is σ -order continuous. For the converse assume that T is σ -order continuous. Since E is a σ -Dedekind complete and σ -laterally complete Riesz space, using from Theorem 3.9 of [7], yields the desired result. Recall from [2, P. 46] that a positive operator T : E → F between two Riesz spaces is order continuous if and only if xα ↓ 0 in E implies T xα ↓ 0 in F. Proposition 4 Let E, F be Riesz spaces. Then we have the following assertions: 1. If 0 ≤ T ∈ L uo (E, F), then xα ↓ 0 in E implies T xα ↓ 0 in F. 2. If T : E → F is an onto lattice homomorphism and xα ↓ 0 in E implies T xα ↓ 0 in F, then T ∈ L uo (E, F). uo
Proof 1. Let 0 ≤ T ∈ L uo (E, F) and xα ↓ 0. Clearly, xα −→ 0 and by assumption, uo T xα −→ 0. It follows from T xα ↓ and part (4) of Lemma 1 that T xα ↓ 0. uo 2. Let xα −→ 0. Then for each u ∈ E + there exists a net (yβ ) in E satisfying: (1) yβ ↓ 0; (2) for any β there exists α0 such that |xα | ∧ u ≤ yβ for all α ≥ α0 . Thus by assumption, T (|xα | ∧ u) = |T xα | ∧ T u ≤ T yβ ↓ 0. o
So, for each u ∈ E + , |T xα | ∧ T u − → 0. Since T is onto, for each z ∈ F + there exists a vector u ∈ E + such that T u = z. Thus by above argument, for each uo o z ∈ F + , |T xα | ∧ z − → 0. Therefore T xα −→ 0. Hence, T is uo-continuous. Remark 1 One can easily formulate by himself the analogue of Proposition 4 for uσ o-continuous operators. Corollary 2 Let T : E → F be an onto lattice homomorphism between Riesz spaces. Then we have the following: 1. T is uσ o-continuous if and only if T is σ -order continuous. 2. T is uo-continuous if and only if T is order continuous. As a consequence of [2, Theorem 2.40] and preceding Corollary we obtain the following result. Corollary 3 Let E, F be Archimedean Riesz spaces. For an order continuous disjoint preserving operator T from E onto F, the modulus |T | is uo-continuous. Recall that a functional on a Riesz space is order continuous if it maps order null nets to order null nets. The collection of all order continuous linear functionals on a Riesz space E will be denoted by E n∼ .
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Remark 2 Let E and F be normed Riesz spaces. Clearly, we have L uo (E, F) ⊂ L buo (E, F). Since each order convergent net has a tail which is order bounded, and therefore, norm bounded, it is easy to see that L uo (E, R) ⊂ L buo (E, R) ⊂ E n∼ . Proposition 5 Let E, F be Banach lattices with order continuous norms and F be atomic. If T : E → F is a continuous operator, then T : F → E is buo-continuous. Proof Let (xα ) be a norm bounded uo-null net in F . Since F has order continuous w∗
w∗
norm, by [5, Theorem 2.1], xα −→ 0 in F . As T is w ∗ -to-w ∗ continuous, T xα −→ 0. uo Now, by [5, Theorem 3.4], we have T xα −→ 0. Hence, T is buo-continuous. For the details of next example see [2, Page 289, Exercise 10]. Example 8 The operator T : L 1 [0, 1] → c0 defined by T( f ) = 0
1
1
f (x) sinx d x,
f (x) sin2x d x, . . . ,
0
is a continuous operator which is not order continuous. Since the norms of Banach lattices L 1 [0, 1] and c0 are order continuous and c0 is atomic, by preceding Theorem, T is buo-continuous. Acknowledgements We would like to thank the referee for her/his valuable suggestions.
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