Strongly Order Continuous Operators on Riesz Spaces

Home , Band (order theory)

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uo 1. Strongly order continuous (or, so-continuous for short), if xα −→ 0 in E implies o T xα −→ 0 in F .

uo 2. Strongly σ-order continuous (or, sσo-continuous for short), if xn −→ 0 in E implies o T xn −→ 0 in F .

The collection of all so-continuous operators of Lb(E, F ) will be denoted by Lso(E, F ), that is Lso(E, F ) := {T ∈ Lb(E, F ) : T is so-continuous}.

Similarly, Lsσo(E, F ) will denote the collection of all order bounded operators from E to F that are sσo-continuous. That is,

Lsσo(E, F ) := {T ∈ Lb(E, F ) : T is sσo-continuous}.

Clearly, we have Lso(E, F ) ⊂ Lsσo(E, F ) and Lso(E, F ) and Lsσo(E, F ) are both vector subspaces of Lb(E, F ). Recall that an operator T : E → F between two Riesz spaces is said to be order o o o continuous (resp. σ-order continuous) if xα −→ 0 (resp. xn −→ 0) in E implies T xα −→ 0 o (resp. T xn −→ 0) in F . The collection of all order continuous operators of Lb(E, F ) will be denoted by Ln(E, F ), that is

Ln(E, F ) := {T ∈ Lb(E, F ) : T is order continuous}.

Similarly, Lc(E, F ) will denote the collection of all order bounded operators from E to F that are σ-order continuous. That is,

Lc(E, F ) := {T ∈ Lb(E, F ) : T is σ-order continuous}.

Note that every so-continuous (resp. sσo-continuous ) operator is order (resp. σ-order) continuous. The converse is not true in general. For example the identity operator I : c0 → c0 is order continuous but is not so-continuous. Indeed, the standard basis sequence of c0 is uo-converges to 0 but is not order convergent.

2 Main Results

Lemma 1. ([5, 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 + ryα −→ kx + ry. 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. A. Bahramnezhad and K. Haghnejad Azar 3

Note that the uo-convergence in a Riesz space E does not necessarily correspond to a topology on E. For example, let E = c, the Banach lattice of real valued convergent n sequences. Put xn =Σk=1ek, where (en) is the standard basis. Then (xn) is uo-convergent to x = (1, 1, 1, ...), but it is not norm convergent.

Proposition 1. Let E, F be Riesz spaces such that E is ﬁnite-dimensional. Then Lso(E, F )= Ln(E, F ) and Lsσo(E, F )= Lc(E, F ).

Proof. Follows immediately if we observe that in a ﬁnite-dimensional Riesz space order convergence is equivalent to uo-convergence.

Recall that a Riesz space is said to be σ-laterally complete if every disjoint sequence has a supremum. For a set A, RA is an example of σ-laterally complete Riesz space.

Proposition 2. Let E, F , G be Riesz spaces. Then we have the following:

1. If T ∈ Lso(E, F ) and S ∈ Ln(F, G), then ST ∈ Lso(E, G). As a consequence, Lso(E) is a left ideal for Ln(E). Similarly, Lsσo(E) is a left ideal for Lc(E).

2. If E is σ-Dedekind complete and σ-laterally complete and S ∈ Lc(E, F ) and T ∈ Lsσo(F, G), then TS ∈ Lsσo(E, G). In this case, Lsσo(E, F )= Lc(E, F ).

uo o Proof. 1. Let (xα) be a net in E such that xα −→ 0. By assumption, T xα −→ 0. So, o ST xα −→ 0. Hence, ST ∈ Lso(E, G).

2. Let E be a σ-Dedekind complete and σ-laterally complete Riesz space. By Theorem 3.9 of [6], we see that a sequence (xn) in E is uo-null if and only if it is order null. uo o o So, if (xn) be a sequence in E such that xn −→ 0, then xn −→ 0. Thus, Sxn −→ 0 and o then T Sxn −→ 0. Hence, TS ∈ Lsσo(E, G). Clearly, we have Lsσo(E, F )= Lc(E, F ). This ends the proof.

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 3. If a positive so-continuous operator T : E → F dominates S, then S is so-continuous.

Proof. Let T : E → F be a positive so-continuous operator between Riesz spaces such that uo uo T dominates S : E → F and let xα −→ 0 in E. By part (2) of Lemma 1, |xα| −→ 0. So, o by assumption, T |xα| −→ 0 and from the inequality |Sx| ≤ T |x| and part (2) of Lemma 1 o again, we have Sxα −→ 0. Hence, S is so-continuous. 4 Strongly order continuous operators

For an operator T : E → F between two Riesz spaces we shall say that its modulus |T | exists ( or that T possesses a modulus) whenever |T | := T ∨ (−T ) exists in the sense that |T | is the supremum of the set {−T,T } in L(E, F ). If E and F are Riesz spaces with F Dedekind complete, then every order bounded operator T : E → F possesses a modulus [2, Theorem 1.18]. From this discussion it follows that when E and F are Riesz spaces with F Dedekind complete, then each order bounded operator T : E → F satisﬁes T +(x) = sup{Ty :0 ≤ y ≤ x}, and T −(x) = sup{−Ty :0 ≤ y ≤ x} for each x ∈ E+. Theorem 1. For an order bounded linear functional f on a Riesz space E the following statements are equivalent. 1. f is so-continuous. 2. f + and f − are both so-continuous. 3. |f| is so-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{fy : 0 ≤ y ≤ x}, for + each α there exists a net (yα) in E with 0 ≤ yα ≤ xα such that f xα − rα ≤ fyα. So, + uo uo o f xα ≤ fyα + rα. Since xα −→ 0, we have yα −→ 0. Thus, by assumption, fyα −→ 0. It + o + o + follows from f xα ≤ (fyα + rα) −→ 0 that f xα −→ 0. Hence, f is so-continuous. Now, as f − = (−f)+, we conclude that f − is also so-continuous. (2) ⇒ (3) Follows from the identity |f| = f + + f −. (3) ⇒ (1) Follows immediately from Proposition 3 by observing that |f| dominates f.

Remark 1. One can easily formulate by himself the analogue of Theorem 1 for sσo- continuous operators.

Recall that a subset A of a Riesz space is said to be order closed whenever (xα) ⊂ A o and xα −→ x imply x ∈ A. An order closed ideal is referred to as a band. Thus, an ideal A is a band if and only if (xα) ⊂ A and 0 ≤ xα ↑ x imply x ∈ A. In the next theorem we ∼ show that Lso(E, R) and Lsσo(E, R) are both bands of E . The details follow. ∼ Theorem 2. If E is a Riesz space, then Lso(E, R) and Lsσo(E, R) are both bands of E . ∼ Proof. We only show that Lso(E, R) is a band of E . That Lsσo(E, R) is a band can be ∼ proven in a similar manner. Note first that if |g| ≤ |f| holds in E with f ∈ Lso(E, R), ∼ then from Theorem 1 it follows that g ∈ Lso(E, R). That is Lso(E, R) is an ideal of E . ∼ To see that the ideal Lso(E, R) is a band, let 0 ≤ fλ ↑ f in E with (fλ) ⊂ Lso(E, R), and uo let 0 ≤ xα −→ 0 in E. Then for each fixed λ we have o 0 ≤ f(xα) = ((f − fλ)(xα)+ fλ(xα)) −→ 0. o So, f(xα) −→ 0. Thus, f ∈ Lso(E, R), and the proof is finished. A. Bahramnezhad and K. Haghnejad Azar 5

Problem 1. Can we ﬁnd a so-continuous operator T : E → F between Riesz spaces whose modulus is not so-continuous?

References

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