Matrix Mappings on the Domains of Invertible Matrices

Abstract

We focus on sequence spaces which are matrix domains of Banach sequence spaces. We show that the characterization of a random matrix operator 𝑇=(𝑡_𝑛𝑘)∈(𝐸_𝐴,𝐹_𝐵), where 𝐸_𝐴 and 𝐹_𝐵 are matrix domains with invertible matrices 𝐴 and 𝐵, can be reduced to the characterization of the operator 𝑆=𝐵∘𝑇∘𝐴⁻¹∈(𝐸,𝐹). As an application, the necessary and sufficient conditions for the matrix operators between invertible matrix domains of the classical sequence spaces and norms of these operators are given.

1. Introduction and Preliminaries

Let 𝜔 denote the set of all complex sequences. Any subspace of 𝜔 is called as a sequence space. We will write ℓ_∞, 𝑐, and 𝑐₀ for the spaces of all bounded, convergent, and null sequences, respectively. By ℓ_𝑝, we denote the space of all 𝑝 absolutely summable sequences, where 1≤𝑝<∞.

Let 𝑋 and 𝑌 be Banach spaces. Then, (𝑋,𝑌) is the set of all operators 𝐿∶𝑋→𝑌 and ℬ(𝑋,𝑌) is the set of all continuous linear operators 𝐿∶𝑋→𝑌. ℬ(𝑋,𝑌) is a Banach space with the operator norm defined by ‖𝐿‖=sup{‖𝐿(𝑥)‖/‖𝑥‖∶0≠𝑥∈𝑋}(𝐿∈ℬ(𝑋,𝑌)).

Let 𝐸 and 𝐹 be two sequence spaces and 𝐴=(𝑎_𝑛𝑘) an infinite matrix of real or complex numbers 𝑎_𝑛𝑘, where 𝑛, 𝑘∈ℕ={1,2,3,…}. Then, we say that 𝐴 defines a matrix mapping from 𝐸 into 𝐹, and we denote it by writing 𝐴∶𝐸→𝐹, if for every sequence 𝑥=(𝑥_𝑘)∈𝐸 the sequence 𝐴𝑥={(𝐴𝑥)_𝑛}, the 𝐴-transform of 𝑥, is in 𝐹, where (𝐴𝑥)_𝑛=_𝑘𝑎_𝑛𝑘𝑥_𝑘(𝑛∈ℕ). (1) Thus, 𝐴∈(𝐸,𝐹) if and only if the series on the right side of (1) converges for each 𝑛∈ℕ and every 𝑥∈𝐸, and we have 𝐴𝑥={(𝐴𝑥)_𝑛}_𝑛∈ℕ∈𝐹 for all 𝑥∈𝐸.

Let 𝐴=(𝑎_𝑛𝑘) and 𝐵=(𝑏_𝑛𝑘). Suppose the sums 𝑐_𝑛𝑘=_𝑗𝑎_𝑛𝑗𝑏_𝑗𝑘 (2) exist for all 𝑛, 𝑘∈ℕ. Then, the product of 𝐴 and 𝐵 is defined by 𝐴𝐵=(𝑐_𝑛𝑘).

If 𝐸 is a subset of 𝜔, then 𝐸_𝐴={𝑥∈𝜔∶𝐴𝑥∈𝐸} is the matrix domain of 𝐴 in 𝐸. We will say that a matrix 𝐴 is invertible over a sequence space  𝐸 if the operator 𝐴∶𝐸_𝐴→𝐸 is bijective; that is, there exists an operator 𝐴⁻¹ such that 𝐴⁻¹(𝐴𝑥)=𝑥 for all 𝑥∈𝐸_𝐴 and 𝐴(𝐴⁻¹𝑦)=𝑦 for all 𝑦∈𝐸. We will say 𝐴 is invertible if 𝐴 is invertible over 𝜔.

A BK space is a Banach sequence space with continuous coordinates. The sequence spaces 𝑐₀, 𝑐, ℓ_𝑝, and ℓ_∞ are the well-known examples of BK spaces.

Several authors studied matrix mappings on sequence spaces that are matrix domains of the difference operator or of the matrices of some classical methods of summability in spaces such as ℓ_𝑝, 𝑐₀, 𝑐, or ℓ_∞. For instance, some matrix domains of the difference operator were studied in [1, 2], of the Cesàro matrices in [3, 4], of the Euler matrices in [5–7], and of the Nörlund matrices in [8]. A general approach was done in [9] reducing the characterizations of the classes (𝑋_𝑇,𝑌) for arbitrary FK spaces 𝑋 with AK and 𝑌⊂𝜔 to those of the classes (𝑋,𝑌) and (𝑋,𝑐), where 𝑇 is a triangle. Compact operators on matrix domains of triangles were examined in [10]. The gliding hump properties of matrix domains were examined in [11].

In this work, our aim is to give some general results for matrix mappings between sequence spaces, which are matrix domains of invertible matrices of sequence spaces. Also, we give some applications of the results.

Theorem 1 (see [12, Theorem 4.2.8, page 57]). Matrix operators between BK spaces are continuous.

An infinite matrix 𝑇=(𝑡_𝑛𝑘) is said to be a triangle if 𝑡_𝑛𝑘=0 for 𝑛<𝑘 and 𝑡_𝑛𝑛≠0 for 𝑛∈ℕ. The following is a well-known result about triangles.

Theorem 2 (see [12, 1.4.8, page 9]). Every triangle 𝑇 has an inverse 𝑇⁻¹ which also is a triangle, and 𝑥=𝑇(𝑇⁻¹𝑥)=𝑇⁻¹(𝑇𝑥) for all 𝑥∈𝜔.

Let 𝐴₁,𝐴₂ be two matrices and 𝜇 a sequence space. If 𝐴₁(𝐴₂𝑥)=(𝐴₁𝐴₂)𝑥 holds for all 𝑥∈𝜇, then we will say 𝐴₁ and 𝐴₂ are associative over the space  𝜇. If 𝐴₁ and 𝐴₂ are associative over the space 𝜔, we will shortly say 𝐴₁ and 𝐴₂ are associative. For row finite matrices, we do not generally have an inverse. But we have associativity; that is, for any two row finite matrices 𝑅₁ and 𝑅₂ we have 𝑅₁𝑅₂𝑥=𝑅₁𝑅₂𝑥∀𝑥∈𝜔. (3) Moreover, we have the following result.

Theorem 3 (see [12, Theorem 1.4.4, page 8]). Let 𝐴₁,𝐴₂ be two matrices.(i)If rows of 𝐴₁ are in ℓ₁ and 𝐴₂∈(ℓ_∞,ℓ_∞), then 𝐴₁ and 𝐴₂ are associative over ℓ_∞.(ii)If 𝐴₁ is row finite, then 𝐴₁ and 𝐴₂ are associative over 𝜔_𝐴2.

Corollary 4. The set of matrices (ℓ_∞,ℓ_∞) are associative over ℓ_∞.

Remark 5. The associativity property does not hold in general (see, e.g., [12, Example 1.4.6]).

Theorem 6. Let 𝐴 be an invertible matrix over a normed sequence space 𝐸 with norm ‖⋅‖_𝐸. Then, 𝐸_𝐴 is a normed sequence space with norm ‖𝑧‖_𝐸𝐴=‖𝐴𝑧‖_𝐸.

Proof. Let 𝛼∈ℂ and 𝑧∈𝐸_𝐴. Then, ‖𝛼𝑧‖_𝐸𝐴=‖𝐴(𝛼𝑧)‖_𝐸=‖𝛼𝐴𝑧‖_𝐸=|𝛼|‖𝐴𝑧‖_𝐸=|𝛼|‖𝑧‖_𝐸𝐴. (4) Secondly, for 𝑦,𝑧∈𝐸_𝐴 we have 𝑦+𝑧_𝐸𝐴=𝐴𝑦+𝑧_𝐸=𝐴𝑦+𝐴𝑧_𝐸≤𝐴𝑦_𝐸+‖𝐴𝑧‖_𝐸=𝑦_𝐸𝐴+‖𝑧‖_𝐸𝐴, (5) so the triangle inequality holds.
Now, suppose that ‖𝑧‖_𝐸𝐴=0. Then, ‖𝐴𝑧‖_𝐸=0 and since ‖⋅‖_𝐸 is a norm we have 𝐴𝑧=𝜃. Since 𝐴 is invertible, we have 𝑧=𝜃.

Theorem 7. Let 𝐸 be a Banach sequence space and 𝐴 a matrix. Then, the operator 𝐴∶𝐸_𝐴→𝐸 is linear and continuous.

Proof. Let 𝐴_𝑛 be the operator that corresponds to the 𝑛th row of the matrix operator 𝐴; that is, 𝐴_𝑛𝑥=∑^∞_𝑘=1𝑎_𝑛𝑘𝑥_𝑘 for all 𝑥=(𝑥_𝑘)∈𝐸_𝐴. Let 𝑥=(𝑥_𝑘)∈𝐸_𝐴 and 𝑦=(𝑦_𝑘)∈𝐸_𝐴. Clearly, we have 𝐴_𝑛𝑐𝑥=𝑐𝐴_𝑛𝑥 for any 𝑐∈ℂ. Let 𝑥^′=𝐴_𝑛𝑥=∑^∞_𝑘=1𝑎_𝑛𝑘𝑥_𝑘 and 𝑦^′=𝐴_𝑛𝑦=∑^∞_𝑘=1𝑎_𝑛𝑘𝑦_𝑘. We have 𝐴_𝑛(𝑥+𝑦)=∑^∞_𝑘=1𝑎_𝑛𝑘(𝑥_𝑘+𝑦_𝑘)=lim_𝑚∑^𝑚_𝑘=1𝑎_𝑛𝑘(𝑥_𝑘+𝑦_𝑘)=lim_𝑚∑^𝑚_𝑘=1𝑎_𝑛𝑘𝑥_𝑘+lim_𝑚∑^𝑚_𝑘=1𝑎_𝑛𝑘𝑦_𝑘=𝑥^′+𝑦^′=𝐴_𝑛𝑥+𝐴_𝑛𝑦.  That means the operator 𝐴_𝑛 is linear for arbitrary 𝑛, which implies the linearity of 𝐴.
𝐴 is continuous since ‖𝐴‖‖‖=sup𝐴𝑥_𝐸‖𝑥‖_𝐸𝐴∶0≠𝑥∈𝐸_𝐴‖‖=sup𝐴𝑥_𝐸‖𝐴𝑥‖_𝐸∶0≠𝑥∈𝐸_𝐴=1. (6)

2. Main Results

Theorem 8. If (𝐸,‖⋅‖) is a Banach sequence space and 𝐴 is an invertible matrix over 𝐸, then (𝐸_𝐴,‖⋅‖_𝐴) is a Banach space.

Proof. Let (𝐸,‖⋅‖) be a Banach sequence space. Then, (𝐸_𝐴,‖⋅‖_𝐴) is a normed space by the previous theorem. Now, let (𝑥_𝑛) be a Cauchy sequence in 𝐸_𝐴. Let 𝑦_𝑛=𝐴𝑥_𝑛. Then (𝑦_𝑛) is a sequence in 𝐸 and is Cauchy in 𝐸 since 𝑥_𝑛−𝑥_𝑚𝐸𝐴=𝐴𝑥_𝑛−𝑥_𝑚𝐸=𝐴𝑥_𝑛−𝐴𝑥_𝑚𝐸=𝑦_𝑛−𝑦_𝑚𝐸. (7) Let 𝑦∈𝐸 such that lim𝑦_𝑛=𝑦 in 𝐸. Let 𝑥=𝐴⁻¹𝑦. Then, lim_𝑛→∞𝑥_𝑛−𝑥_𝐸𝐴=lim_𝑛→∞𝐴𝑥_𝑛−𝑥_𝐸=lim_𝑛→∞𝑦_𝑛−𝑦_𝐸=0. (8)

Theorem 9. Let 𝐴 and 𝐵 be invertible matrices over the BK spaces 𝐸 and 𝐹, respectively. Suppose that 𝑇∈(𝐸_𝐴,𝐹_𝐵). If 𝐵∘𝑇∘𝐴⁻¹=𝐵𝑇𝐴⁻¹ over 𝐸, then the operator 𝑇∈(𝐸_𝐴,𝐹_𝐵) is continuous.

Proof. Let 𝑇∈(𝐸_𝐴,𝐹_𝐵) and let ‖⋅‖ be the norm of 𝐸. Since 𝐴 is invertible over 𝐸, 𝐴∶𝐸_𝐴→𝐸 is bijective. Also, 𝐴 is continuous by Theorem 7.
By the bounded inverse theorem 𝐴⁻¹ is continuous. Similarly the inverse 𝐵⁻¹ of 𝐵∶𝐹_𝐵→𝐹 is continuous.
Now, we have that the matrix operator 𝑅=𝐵∘𝑇∘𝐴⁻¹∈(𝐸,𝐹) and is continuous by Theorem 1. Since 𝐴 and 𝐵⁻¹ are continuous, the operator 𝑇=𝐵⁻¹∘𝑅∘𝐴 is continuous.

Theorem 10. Let 𝐴 and 𝐵 be invertible matrices over the sequence spaces 𝐸 and 𝐹, respectively. Then, for an operator 𝑇, 𝑇∈(𝐸_𝐴,𝐹_𝐵) if and only if 𝐵∘𝑇∘𝐴⁻¹∈(𝐸,𝐹).

Proof. Suppose that 𝑇∈(𝐸_𝐴,𝐹_𝐵), and let 𝑧=(𝑧_𝑘)∈𝐸. Then, clearly 𝐴⁻¹𝑧∈𝐸_𝐴. Hence, we have (𝑇∘𝐴⁻¹)𝑧∈𝐹_𝐵 and so (𝐵∘𝑇∘𝐴⁻¹)𝑧∈𝐹.
For the inverse implication, suppose that 𝐵∘𝑇∘𝐴⁻¹∈(𝐸,𝐹), and let 𝑧∈𝐸_𝐴. Then, 𝐴𝑧∈𝐸 and so (𝐵∘𝑇∘𝐴⁻¹∘𝐴)𝑧=(𝐵∘𝑇)𝑧∈𝐹. Since 𝐵 is invertible, we have 𝑇𝑧∈𝐹_𝐵.

Corollary 11. Let 𝐸 and 𝐹 be sequence spaces and 𝐴 and 𝐵 triangles. Then, for an operator 𝑇, 𝑇∈(𝐸_𝐴,𝐹_𝐵) if and only if 𝐵∘𝑇∘𝐴⁻¹∈(𝐸,𝐹).

Theorem 12. Let 𝐸 and 𝐹 be two normed sequence spaces and 𝐴 and 𝐵 invertible matrices over 𝐸 and 𝐹, respectively. Then, for an operator 𝑇, 𝑇∈ℬ(𝐸_𝐴,𝐹_𝐵) if and only if 𝐵∘𝑇∘𝐴⁻¹∈ℬ(𝐸,𝐹). In this case, one has ‖𝑇‖_{(𝐸𝐴,𝐹𝐵)}=𝐵∘𝑇∘𝐴⁻¹_(𝐸,𝐹). (9)

Proof. It is enough to show the following equality: ‖𝑇‖_{(𝐸𝐴,𝐹𝐵)}=sup_{𝜃≠𝑧∈𝐸𝐴}‖𝑇𝑧‖_𝐹𝐵‖𝑧‖_𝐸𝐴=sup_{𝜃≠𝐴𝑧∈𝐸}‖(𝐵∘𝑇)𝑧‖_𝐹‖𝐴𝑧‖_𝐸=sup_{𝜃≠𝐴𝑧∈𝐸}𝐵∘𝑇∘𝐴⁻¹∘𝐴𝑧_𝐹‖𝐴𝑧‖_𝐸=𝐵∘𝑇∘𝐴⁻¹_𝐸,𝐹. (10)

3. Examples and Applications

Theorem 10, Corollary 11, and Theorem 12 have many applications, especially in the subject of characterization of matrices which act as operators between certain sequence spaces. We just give a taste by the following examples and theorems.

Example 13. Let 𝜆=(𝜆_𝑘) be a sequence of nonzero scalars in ℂ. For any 𝑧=(𝑧_𝑘)∈𝜔, let 𝜆𝑧=(𝜆_𝑘𝑧_𝑘). Then, for a sequence space 𝐸, the multiplier sequence space 𝐸(𝜆), associated with the multiplier sequence 𝜆, is defined as 𝐸(𝜆)={𝑧∈𝜔∶𝜆𝑧∈𝐸}. (11) Let 𝜆Λ=¹000⋯0𝜆₂00⋯00𝜆₃0⋯000𝜆₄⋯⋮⋮⋮⋮⋱. (12) Then, the matrix Λ is invertible with Λ⁻¹=𝜆₁⁻¹000⋯0𝜆₂⁻¹00⋯00𝜆₃⁻¹0⋯000𝜆₄⁻¹⋯⋮⋮⋮⋮⋱, (13) and 𝐸(𝜆)=𝐸_Λ. So, all the Theorems in Section 2 are applicable for multiplier sequence spaces (see [13] for detailed applications and examples).

Example 14. Let 𝑇=(𝑡_𝑛𝑘) be a row-finite matrix and 𝐵 the operator with matrix representation 𝐵=1000⋯1100⋯1110⋯1111⋯⋮⋮⋮⋮⋱. (14)𝐵 is invertible and the inverse operator has the matrix representation 𝐵⁻¹=1000⋯−1100⋯0−110⋯00−11⋯⋮⋮⋮⋮⋱. (15)𝐵 and 𝐵⁻¹ are triangles and so they are one to one. 𝐵⁻¹ is the operator 𝐵(𝑟,𝑠) of [14] with 𝑟=1 and 𝑠=−1. Since both 𝑇 and 𝐵⁻¹ are row finite, we have 𝑇∘𝐵⁻¹=𝑇𝐵⁻¹. Now, using Corollary 11 we have that 𝑇 is in (𝑐_𝐵,𝑐) if and only if 𝑇𝐵⁻¹ is in (𝑐,𝑐):𝑇𝐵⁻¹_𝑛𝑘=𝑡_𝑛𝑘−𝑡_𝑛(𝑘+1), (16) so by the Kojima-Schur Theorem (see, e.g., Theorem 2.7 of [15]) we have that 𝑇∈(𝑐_𝐵,𝑐) if and only if the following three conditions hold:(i)lim_𝑛𝑡_𝑛𝑘−𝑡_𝑛(𝑘+1) exists for each 𝑘;(ii)lim_𝑛∑_𝑘𝑡_𝑛𝑘−𝑡_𝑛(𝑘+1) exists;(iii)sup_𝑛∑_𝑘|𝑡_𝑛𝑘−𝑡_𝑛(𝑘+1)|<∞.
Now, let us examine (𝑐,𝑐_𝐵). 𝑇 is in (𝑐,𝑐_𝐵) if and only if 𝐵𝑇 is in (𝑐,𝑐). We have [𝐵𝑇]_𝑛𝑘=𝑠_𝑛𝑘, (17) where 𝑠_𝑛𝑘 is defined as the sum of the first 𝑛 terms of the 𝑘th column of the matrix 𝑇; that is, 𝑠_𝑛𝑘=𝑡_1𝑘+𝑡_2𝑘+⋯+𝑡_𝑛𝑘. (18) Now, by using the Kojima-Schur Theorem, we have, 𝑇∈(𝑐,𝑐_𝐵) if and only if the following three conditions hold:(i)lim_𝑛𝑠_𝑛𝑘 exists for each 𝑘;(ii)lim_𝑛∑_𝑘𝑠_𝑛𝑘 exists;(iii)sup_𝑛∑_𝑘|𝑠_𝑛𝑘|<∞.Now, let us examine (𝑐_𝐵,𝑐_𝐵). 𝑇 is in (𝑐_𝐵,𝑐_𝐵) if and only if 𝐵𝑇𝐵⁻¹ is in (𝑐,𝑐). We have 𝐵𝑇𝐵⁻¹_𝑛𝑘=𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1). (19) Now, by using the Kojima-Schur Theorem we have that 𝑇∈(𝑐_𝐵,𝑐_𝐵) if and only if the following three conditions hold:(i)lim_𝑛𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1) exists for each 𝑘;(ii)lim_𝑛∑_𝑘𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1) exists;(iii)sup_𝑛∑_𝑘|𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1)|<∞.

Theorem 15. An infinite matrix 𝐴=(𝑎_𝑛𝑘)∈(ℓ_∞,𝑏𝑣) if and only if sup_{(ℕ)𝑁∈𝑓𝑘𝑛∈𝑁}𝑎_𝑛𝑘−𝑎_{(𝑛−1)𝑘}<∞, (20) where 𝑓(ℕ) denotes the collection of all finite subsets of ℕ.

Proof. First, we observe that 𝑏𝑣=ℓ_1Δ, where Δ is the difference operator 𝐵⁻¹ of Example 14. The inverse of Δ is 𝐵. By Corollary 11, 𝐴∈(ℓ_∞,ℓ_1Δ) if and only if Δ∘𝐴∈(ℓ_∞,ℓ₁). Since 𝑏𝑣⊂𝑐, we have 𝐴∈(ℓ_∞,𝑐) and so by the Schur Theorem (see, e.g., Theorem 2.11 of [15]) rows of 𝐴 are in ℓ₁. By Theorem 3 part (ii), Δ and 𝐴 are associative over ℓ_∞. So, Δ∘𝐴=Δ𝐴 over ℓ_∞. We have [Δ𝐴]_𝑛𝑘=𝑎_𝑛𝑘−𝑎_{(𝑛−1)𝑘}, (21) where 𝑎_0𝑘=0 for all 𝑘∈ℕ. Now, using Theorem 2.14 of [15], we have 𝐴∈(ℓ_∞,𝑏𝑣) if and only if sup_{𝑁∈𝑓(ℕ)𝑘𝑛∈𝑁}𝑎_𝑛𝑘−𝑎_{(𝑛−1)𝑘}<∞. (22)

Using Theorem 12, we have the following.

Corollary 16. If 𝐴=(𝑎_𝑛𝑘)∈(ℓ_∞,𝑏𝑣), then 𝐴=(𝑎_𝑛𝑘)∈ℬ(ℓ_∞,𝑏𝑣) and ‖𝐴‖_{(ℓ∞,𝑏𝑣)}=‖Δ𝐴‖_{(ℓ∞,ℓ1)}. (23)

Example 17. Let 𝑇=(𝑡_𝑛𝑘) be a row finite matrix and 𝐶 the Cesàro operator which has a matrix representation 𝐶=1000⋯1/21/200⋯1/31/31/30⋯1/41/41/41/4⋯⋮⋮⋮⋮⋱. (24) Then, 𝑐_𝐶 is the well-known space of Cesàro summable sequences. This is a sequence space which includes the space of convergent sequences 𝑐. 𝐶 is invertible and the inverse operator has the matrix representation 𝐶⁻¹=1000⋯−1200⋯0−230⋯00−34⋯⋮⋮⋮⋮⋱. (25) Since 𝑇,𝐶, and 𝐶⁻¹ are all row finite, the operator 𝐶∘𝑇∘𝐶⁻¹ is represented by the matrix 𝐶𝑇𝐶⁻¹. Now, using Corollary 11, we have that 𝑇 is in ((ℓ_∞)_𝐶,(ℓ_∞)_𝐶) if and only if 𝐶𝑇𝐶⁻¹ is in (ℓ_∞,ℓ_∞), where (ℓ_∞)_𝐶 denotes the space, introduced by Ng and Lee [3], of all sequences whose 𝐶-transforms are in the space ℓ_∞.
Let 𝑠_𝑛𝑘 be defined as in the previous example. Then, after some calculations, we have 𝐶𝑇𝐶⁻¹_𝑛𝑘=𝑘𝑛𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1). (26) Then, 𝐶𝑇𝐶⁻¹ is in (ℓ_∞,ℓ_∞) if and only if sup_𝑛𝑘𝑘𝑛𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1)<∞ (27) by Theorem 2.6 of [15]. So, we have 𝑇∈((ℓ_∞)_𝐶,(ℓ_∞)_𝐶) if and only if (27) holds. Similarly, we have 𝑇∈(𝑐_𝐶,(ℓ_∞)_𝐶) or 𝑇∈((𝑐₀)_𝐶,(ℓ_∞)_𝐶) if and only if (27) holds, where (𝑐₀)_𝐶 and 𝑐_𝐶 denote the spaces 𝑐₀ and 𝑐̃, studied by Şengönül and Başar [4], of all sequences whose 𝐶-transforms are in the spaces 𝑐₀ and 𝑐, respectively.

Theorem 18. Let C be the Cesàro matrix. Then, 𝐴=(𝑎_𝑛𝑘)∈(𝑐,𝑐_𝐶) if and only if sup_𝑛1𝑛_𝑘𝑠_𝑛𝑘<∞,lim_𝑛𝑠_𝑛𝑘𝑛𝑒𝑥𝑖𝑠𝑡𝑠𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑘,lim_𝑛1𝑛_𝑘𝑠_𝑛𝑘𝑒𝑥𝑖𝑠𝑡𝑠, (28) where 𝑠_𝑛𝑘=𝑎_1𝑘+𝑎_2𝑘+⋯+𝑎_𝑛𝑘.

Proof. By Corollary 11, 𝐴∈(𝑐,𝑐_𝐶) if and only if 𝐶∘𝐴∈(𝑐,𝑐). Since 𝐴∈(𝑐,𝑐_𝐶)⊂(𝑐,𝜔), we have that rows of 𝐴 are in ℓ₁. By Theorem 3 part (ii), 𝐶 and 𝐴 are associative over 𝑐. So, 𝐶∘𝐴=𝐶𝐴 over 𝑐. We have [𝐶𝐴]_𝑛𝑘=𝑠_𝑛𝑘𝑛. (29) Now, using Kojima-Schur Theorem, we have 𝐴∈(𝑐,𝑐_𝐶) if and only if conditions (28) hold.

Lemma 19. If ∑^∞_𝑘=1𝑎_𝑘𝑐_𝑘 is convergent for all Cesàro summable sequences (𝑐_𝑘), then lim𝑘𝑎_𝑘=0.

Proof. Suppose that lim𝑘𝑎_𝑘≠0. Then, there exist 𝜖>0 and a subsequence (𝑎_𝑘𝑛) of (𝑎_𝑘) such that 𝑘_𝑛|𝑎_𝑘𝑛|>𝜖. Without loss of generality, we can choose this subsequence (𝑎_𝑘𝑛) such that 𝑘_𝑛≥𝑛²𝑘_𝑛−1. Let 𝑐_𝑘=𝑘_𝑛𝑎_𝑘𝑛𝑛𝑎_𝑘𝑛if𝑘=𝑘_𝑛0otherwise. (30) Then, the sequence (𝑐_𝑘) is Cesàro summable because for 𝑘_𝑛≤𝑚<𝑘_𝑛+1 we have 𝑐₁+𝑐₂+⋯+𝑐_𝑚𝑚≤𝑐₁+𝑐₂+⋯+𝑐_𝑚𝑚=𝑐_𝑘1+𝑐_𝑘2+⋯+𝑐_𝑘𝑛𝑚=𝑘₁/1+𝑘₁/2+⋯+𝑘_𝑛/𝑛𝑚≤𝑘₁/1+𝑘₁/2+⋯+𝑘_𝑛−1/(𝑛−1)+𝑘_𝑛/𝑛𝑘_𝑛=𝑘₁𝑘_𝑛+𝑘₂2𝑘_𝑛𝑘+⋯+^𝑛−1(𝑛−1)𝑘_𝑛+1𝑛≤1𝑛²+12𝑛²1+⋯+(𝑛−1)𝑛²+1𝑛≤()1𝑛−1𝑛²+1𝑛≤2𝑛. (31) On the other hand, ^∞_𝑘=1𝑎_𝑘𝑐_𝑘=^∞_𝑛=1𝑎_𝑘𝑛𝑐_𝑘𝑛=^∞_𝑛=1𝑘_𝑛𝑎_𝑘𝑛𝑛>𝜖^∞_𝑛=11𝑛, (32) and so we get to the contradiction ∑^∞_𝑘=1𝑎_𝑘𝑐_𝑘=∞ for the Cesàro summable sequence (𝑐_𝑘). So, our assumption lim𝑘𝑎_𝑘≠0 is not true.

Theorem 20. Let C be the Cesàro matrix. Then, 𝐴=(𝑎_𝑛𝑘)∈(𝑐_𝐶,𝑐_𝐶) if and only if lim_𝑘𝑘𝑎_𝑛𝑘=0𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑛,sup_𝑛𝑘𝑘𝑛𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1)<∞,lim_𝑛𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1)𝑛𝑒𝑥𝑖𝑠𝑡𝑠𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑘,lim_𝑛1𝑛_𝑘𝑘𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1)𝑒𝑥𝑖𝑠𝑡𝑠, (33)(34)(35)(36) where 𝑠_𝑛𝑘=𝑎_1𝑘+𝑎_2𝑘+⋯+𝑎_𝑛𝑘.

Proof. Suppose that 𝐴∈(𝑐_𝐶,𝑐_𝐶). Then, by Lemma 19, we have (33). By Corollary 11, 𝐴∈(𝑐_𝐶,𝑐_𝐶) if and only if 𝐶∘𝐴∘𝐶⁻¹∈(𝑐,𝑐). First let us show that 𝐴 and 𝐶⁻¹ are associative over 𝑐. Let 𝑥=(𝑥_𝑘)∈𝑐. We have that 𝐴𝐶⁻¹𝑥_𝑛=^∞_𝑘=1𝑎_𝑛𝑘𝑘𝑥_𝑘−(𝑘−1)𝑥_𝑘−1 (37) is convergent in ℂ. Let 𝑆_𝑚 be the 𝑚th partial sum of this series. Then, 𝑆_𝑚=^𝑚−1_𝑘=1𝑘𝑎_𝑛𝑘−𝑎_𝑛(𝑘+1)𝑥_𝑘+𝑚𝑎_𝑛𝑚𝑥_𝑚; (38)lim_𝑚𝑚𝑎_𝑛𝑚=0, so 𝐴𝐶⁻¹𝑥_𝑛=lim_𝑚𝑆_𝑚=^∞_𝑘=1𝑘𝑎_𝑛𝑘−𝑎_𝑛(𝑘+1)𝑥_𝑘=𝐴𝐶⁻¹𝑥_𝑛; (39)𝐴𝐶⁻¹∈(𝑐,𝜔) and 𝐶 is row finite. Then, by Theorem 3 part (ii), 𝐶 and 𝐴𝐶⁻¹ are associative over 𝑐. So, 𝐶∘𝐴∘𝐶⁻¹=𝐶𝐴𝐶⁻¹ over 𝑐. We have 𝐶𝐴𝐶⁻¹_𝑛𝑘=𝑘𝑛𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1). (40) Now, using Kojima-Schur Theorem, 𝐴∈(𝑐_𝐶,𝑐_𝐶) implies conditions (34)–(36).
For the reverse implication, suppose that conditions (33)–(36) hold. Then, by the Kojima-Schur Theorem, conditions (34)–(36) imply 𝐶𝐴𝐶⁻¹∈(𝑐,𝑐). By (33), 𝐴∘𝐶⁻¹=𝐴𝐶⁻¹ over 𝑐. By Part (ii) of Theorem 3, 𝐶 and 𝐴𝐶⁻¹ are associative over 𝑐, and so we have 𝐶∘𝐴∘𝐶⁻¹=𝐶𝐴𝐶⁻¹ over 𝑐. Hence, 𝐶∘𝐴∘𝐶⁻¹∈(𝑐,𝑐), and now by Corollary 11, 𝐴∈(𝑐_𝐶,𝑐_𝐶).

Corollary 21. If 𝐴=(𝑎_𝑛𝑘)∈(𝑐_𝐶,𝑐_𝐶), then 𝐴=(𝑎_𝑛𝑘)∈ℬ(𝑐_𝐶,𝑐_𝐶) and ‖𝐴‖_{(𝑐𝐶,𝑐𝐶)}=𝐶𝐴𝐶⁻¹_(𝑐,𝑐)=sup_𝑛𝑘𝑘𝑛𝑠_𝑛𝑘−𝑠_𝑛(𝑘+1). (41) Finally, let one give an application on a matrix operator which is not a triangle.

Theorem 22. Let 𝛾Γ=¹𝛾₂𝛾₃𝛾₄⋯0𝛾₂𝛾₃𝛾₄⋯00𝛾₃𝛾₄⋯000𝛾₄⋯⋮⋮⋮⋮⋱, (42) where 𝛾_𝑘≠0 for all 𝑘∈ℕ. Then, a matrix 𝐴∈(𝑐_0Γ,𝑐₀) if and only if the following three conditions hold: 𝑎_𝑛𝑘𝛾_𝑘^∞_𝑘=1𝑖𝑠𝑏𝑜𝑢𝑛𝑑𝑒𝑑𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑛,sup_𝑛^∞_𝑘=1𝑎_𝑛𝑘𝛾_𝑘−𝑎_{𝑛(𝑘−1)}𝛾_𝑘−1<∞,lim_𝑛𝑎_𝑛𝑘𝛾_𝑘−𝑎_{𝑛(𝑘−1)}𝛾_𝑘−1=0𝑓𝑜𝑟𝑒𝑎𝑐ℎ𝑘, (43) where 𝑎_𝑛0/𝛾₀=0.

Proof. Suppose that 𝐴∈(𝑐_0Γ,𝑐₀). The operator Γ∶𝜔_Γ→𝜔 is one to one, because for Γ𝑥=0 with 𝑥=(𝑥_𝑘)∈𝜔 we have the system of equations 0=^∞_𝑗=𝑘𝛾_𝑗𝑥_𝑗for𝑘=1,2,3,…,0=𝛾_𝑘𝑥_𝑘+^∞_𝑗=𝑘+1𝛾_𝑗𝑥_𝑗=𝛾_𝑘𝑥_𝑘+0=𝛾_𝑘𝑥_𝑘for𝑘=1,2,3,…. (44)
Now, let us show that Γ∶𝑐_0Γ→𝑐₀ is onto. Let 𝑦=(𝑦_𝑘)∈𝑐₀. Then Γ𝑥=𝑦 for 𝑥=(𝑥_𝑘) with 𝑥_𝑘=𝛾_𝑘⁻¹(𝑦_𝑘−𝑦_𝑘+1) since (Γ𝑥)_𝑛=𝛾_𝑛𝑥_𝑛+𝛾_𝑛+1𝑥_𝑛+1=𝑦+⋯_𝑛−𝑦_𝑛+1+𝑦_𝑛+1−𝑦_𝑛+2+⋯=𝑦_𝑛. (45) Hence, Γ∶𝑐_0Γ→𝑐₀ is onto. So, Γ∶𝑐_0Γ→𝑐₀ is bijective. The inverse operator is then Γ⁻¹=𝛾₁⁻¹−𝛾₁⁻¹00⋯0𝛾₂⁻¹−𝛾₂⁻¹0⋯00𝛾₃⁻¹−𝛾₃⁻¹⋯000𝛾₄⁻¹⋯⋮⋮⋮⋮⋱. (46) Now, by Theorem 10, 𝐴∈(𝑐_0Γ,𝑐₀) if and only if 𝐴∘Γ⁻¹∈(𝑐₀,𝑐₀).
For 𝑥=(𝑥_𝑘)∈𝑐₀ we have Γ⁻¹𝑥=(𝛾₁⁻¹[𝑥₁−𝑥₂],𝛾₂⁻¹[𝑥₂−𝑥₃],…), and then for 𝐴=(𝑎_𝑛𝑘) we have (𝐴(Γ⁻¹𝑥))_𝑛=∑_𝑘𝑎_𝑛𝑘𝛾_𝑘⁻¹(𝑥_𝑘−𝑥_𝑘+1). Let 𝑆_𝑚 be the 𝑚th partial sum of this series. Then, 𝑆_𝑚=^𝑚_𝑘=1𝑎_𝑛𝑘𝛾_𝑘⁻¹−𝑎_{𝑛(𝑘−1)}𝛾⁻¹_𝑘−1𝑥_𝑘−𝑎_𝑛𝑚𝛾_𝑚⁻¹𝑥_𝑚+1, (47) where 𝑎_𝑛0=0.
By a similar proof to the proof of Lemma 19, we can see that the sequence (𝑎_𝑛𝑚𝛾_𝑚⁻¹)^∞_𝑚=1 is bounded for each 𝑛, when 𝐴∈(𝑐_0Γ,𝜔). Then, 𝐴Γ⁻¹𝑥_𝑛=lim_𝑚𝑆_𝑚=^∞_𝑘=1𝑎_𝑛𝑘𝛾_𝑘⁻¹−𝑎_{𝑛(𝑘−1)}𝛾⁻¹_𝑘−1𝑥_𝑘=𝐴Γ⁻¹𝑥_𝑛. (48) The conditions for 𝐴Γ⁻¹ to be in (𝑐₀,𝑐₀), by Example 5A page 129 of [12], are sup_𝑛^∞_𝑘=1𝑎_𝑛𝑘𝛾_𝑘⁻¹−𝑎_{𝑛(𝑘−1)}𝛾⁻¹_𝑘−1<∞,lim_𝑛𝑎_𝑛𝑘𝛾_𝑘⁻¹−𝑎_{𝑛(𝑘−1)}𝛾⁻¹_𝑘−1=0foreach𝑘. (49) Hence, we have all conditions (43). We leave the reverse implication part to the reader.

Acknowledgment

The author thanks the referees for their valuable comments and suggestions.