2003
DOI: 10.2298/fil0317059j
|View full text |Cite
|
Sign up to set email alerts
|

Ordinary, absolute and strong summability and matrix transformations

Abstract: Abstract. Many important sequence spaces arise in a natural way from various concepts of summability, namely ordinary, absolute and strong summability. In the first two cases they may be considered as the domains of the matrices that define the respective methods of summability; the situation, however, is different and more complicated in the case of strong summability.Given sequence spaces X and Y , we find necessary and sufficient conditions for the entries of a matrix to map X into Y , and characterize the … Show more

Help me understand this report

Search citation statements

Order By: Relevance

Paper Sections

Select...
1
1
1
1

Citation Types

0
38
0

Year Published

2004
2004
2022
2022

Publication Types

Select...
7
3

Relationship

0
10

Authors

Journals

citations
Cited by 101 publications
(42 citation statements)
references
References 8 publications
0
38
0
Order By: Relevance
“…A Banach space X ⊂ s is a BK space if the projection P n : x → x n from X into C is continuous for all n. A BK space X ⊃ ϕ is said to have AK if for every x ∈ X, x = lim p→∞ p k=1 x k e k , where e k = (0,...,1,...), 1 being in the kth position. It is well known that if X has AK, then Ꮾ(X) = (X,X), see [9,13,19].…”
Section: Recall Of Some Results In Summabilitymentioning
confidence: 99%
“…A Banach space X ⊂ s is a BK space if the projection P n : x → x n from X into C is continuous for all n. A BK space X ⊃ ϕ is said to have AK if for every x ∈ X, x = lim p→∞ p k=1 x k e k , where e k = (0,...,1,...), 1 being in the kth position. It is well known that if X has AK, then Ꮾ(X) = (X,X), see [9,13,19].…”
Section: Recall Of Some Results In Summabilitymentioning
confidence: 99%
“…A BK space E is said to have AK if every sequence X = (x n ) ∞ n=1 ∈ E has a unique representation X = ∞ n=1 x n e n , where e n is the sequence with 1 in the nth position and 0 otherwise. It is well known [9] that if E has AK, then Ꮾ(E) = (E, E).…”
Section: Preliminary Results a = (A Nm ) Nm≥1mentioning
confidence: 99%
“…Since A(r, s, t) is a triangle, it has a unique inverse and the inverse is also a triangle [7]. Take…”
Section: Preliminariesmentioning
confidence: 99%