2019
DOI: 10.1016/j.jfa.2018.09.002
|View full text |Cite
|
Sign up to set email alerts
|

ξ-completely continuous operators and ξ-Schur Banach spaces

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
2
1
1
1

Citation Types

0
18
0

Year Published

2019
2019
2023
2023

Publication Types

Select...
7
1

Relationship

2
6

Authors

Journals

citations
Cited by 12 publications
(18 citation statements)
references
References 29 publications
0
18
0
Order By: Relevance
“…. < r s such that ∪ s i=1 supp(P M,ri ) is an initial segment of N , then P N,i = P M,ri for all 1 i s. This was proved in [10].…”
Section: Proofmentioning
confidence: 75%
See 1 more Smart Citation
“…. < r s such that ∪ s i=1 supp(P M,ri ) is an initial segment of N , then P N,i = P M,ri for all 1 i s. This was proved in [10].…”
Section: Proofmentioning
confidence: 75%
“…The concepts behind these classes are that weakly null sequences are mapped by the operator to sequences which are "very" weakly null (completely continuous operators send weakly null sequences to 0-weakly null sequences, and weak Banach-Saks operators send weakly null sequences to 1-weakly null sequences). In [10], the notions of ξ-completely continuous operators and ξ-Schur Banach spaces were introduced. These notions are weakenings of the notions of completely continuous operators and Schur Banach spaces, respectively.…”
Section: Introductionmentioning
confidence: 99%
“…For 0 < ξ < ω 1 , fix a sequence ξ n ↑ ω ξ and a sequence ϑ It was shown in [6] that Z ξ is ζ -Schur for each ζ < ω ξ , but Z ξ is not ω ξ -Schur. Thus for every ξ < ω 1 , we have found a Banach space which is ζ -Schur for every ζ < ω ξ and which fails to be ω ξ -Schur.…”
Section: Examples Of ξ -Schur Banach Spacesmentioning
confidence: 99%
“…In the terminology of [11], the ξ th level of the repeated averages hierarchy considered as a family of probability measures on S ξ are a ξ-sufficient, and therefore ξ-regulatory probability block (see [11,Corollary 4.8]). For 0 < ε 1 < ε, there exists N ∈ [N] such that for each F ∈ S ξ , there exists E) ε 1 (see the definition of ξ-full and the definitions of the sets E(f, P, P, ε) and G(f, P, M, ε) preceding [11,Lemma 4.4]). Since f (N(i), E) ε 1 > 0, then N(i) ∈ E.…”
Section: Properties Of Interestmentioning
confidence: 99%