THE ABEL-TYPE TRANSFORMATIONS INTO

Let t be a sequence in (0,1) that converges to 1, and define the Abel-type matrix Aα,t by ank = ( k+α k ) tk+1 n (1−tn)α+1 for α>−1. The matrix Aα,t determines a sequenceto-sequence variant of the Abel-type power series method of summability introduced by Borwein in [1]. The purpose of this paper is to study these matrices as mappings into . Necessary and sufficient conditions for Aα,t to be , G, and Gw are established. Also, the strength of Aα,t in the setting is investigated.

1. Introduction and background.The Abel-type power series method [1], denoted by A α ,α > −1, is the following sequence-to-function transformation: if then we say that u is A α -summable to L. In order to study this summability method as a mapping into , we must modify it into a sequence to sequence transformation.This is achieved by replacing the continuous parameter x with a sequence t such that 0 < t n < 1 for all n and lim t n = 1.Thus, the sequence u is transformed into the sequence A α,t u whose nth term is given by This transformation is determined by the matrix A α,t whose nkth entry is given by (1.4) The matrix A α,t is called the Abel-type matrix.The case α = 0 is the Abel matrix introduced by Fridy in [5].It is easy to see that the A α,t matrix is regular and, indeed, totally regular.

Basic notations.
Let A = (a nk ) be an infinite matrix defining a sequence-tosequence summability transformation given by where (Ax) n denotes the nth term of the image sequence Ax.The sequence Ax is called the A-transform of the sequence x.If X and Z are sets of complex number sequence, then the matrix A is called an X-Z matrix if the image Au of u under the transformation A is in Z whenever u is in X.
Let y be a complex number sequence.Throughout this paper, we use the following basic notations: (2.2)

The main results.
Our first result gives a necessary and sufficient condition for A α,t to be -.[6] guarantees that A α,t is anmatrix.Also, if A α,t is an -matrix, then by Knopp-Lorentz theorem, we have and this yields (1 − t) α+1 ∈ .
Proof.The corollary follows easily by Theorem 1.
Before considering our next theorem, we recall the following result which follows as a consequence of the familiar Hölder's inequality for summation.The result states that if x and y are real number sequences such that x ∈ p ,y ∈ q , p > 1, and If A α,t is an -matrix, then, by Theorem 1, we have ∈ , then we have (1 − t) ∈ and, consequently, the assertion follows.If α > 0, then the theorem follows using the preceding result by letting Proof.If A α,t is an -matrix, then, by Theorem 1, it follows that (1 − t) α+1 ∈ and this yields α > 0. Conversely, suppose that α > 0. Then we have for some M > 0. This is possible as both the summation and the integral are finite since α > 0. Now, we let and we compute g(k) using integration by parts repeatedly.We have where and (3.9) It follows that where Continuing this process, we get It is easy to see that g(k) can be written using summation notation as Consequently, we get Thus by the Knopp-Lorentz theorem [6], A α,t is an -matrix.α+1 ∈ , then we prove that A α,t is not -by showing that the condition of the Knopp-Lorentz theorem [6] fails to hold.For convenience, we let q = 1/p and 2 1/p = R, where p > 1.Then we have for some M > 0. This is possible as both the summation and integral are finite since (1 − t) α+1 ∈ .Now, let us define Using integration by parts repeatedly, we can easily deduce that (3.17) This implies that and hence A α,t is not -.
In case t n = 1 − (n + 2) −q , it is natural to ask whether A α,t is an -matrix.For −1 < α ≤ 0, it is easy to see that A α,t is -if and only if α > (1 − q)/q, by Theorem 1.
For α > 0, the answer to this question is given by the next theorem, which gives a necessary and sufficient condition for the matrix to be -.
Proof.Suppose that q ≥ 1.Let q = 1/p, 2 1/p = R and (R−1)/R = S, where 0 < p ≤ 1.Then we have for some M > 0. This is possible as both the summation and the integral are finite since (1 − t) α+1 ∈ for α > 0. Now, let us define Using integration by parts repeatedly, we can easily deduce that . (3.23) Now, from the hypotheses that q ≥ 1 and α > 0, it follows that By writing the right-hand side of the preceding inequality using the summation notation, we obtain Consequently, we have Thus, by Knopp-Lorentz theorem [6], A α,t is an -matrix .Conversely, if A α,t is an -matrix, then it follows, by Theorems 3 and 4, that q ≥ 1.
Proof.The result follows immediately from Theorems 1 and 5.
Then both A α,t and A α,w are -matrices.
Corollary 10.Suppose that t n = 1 − (n + 2) −q .Then A α,t is a G-matrix if and only if α > (1 − q)/q.For q = 1, A α,t is a G-matrix if and only if it is an -matrix.
Proof.The proof follows using Theorems 3 and 6.
Our next few results suggest that the Abel-type matrix A α,t is -stronger than the identity matrix (see [7,Def. 3]).The results indicate how large the sizes of (A α,t ) and d(A α,t ) are.Theorem 8. Suppose that −1 < α ≤ 0, A α,t is an -matrix, and the series ∞ k=0 x k has bounded partial sums.Then it follows that x ∈ (A α,t ).
Proof.Since, for −1 < α ≤ 0, k+α k is decreasing, the theorem is proved by following the same steps used in the proof of [7,Thm. 4].
Remark 2. Although the preceding theorem is stated for −1 < α ≤ 0, the conclusion is also true for α > 0 for some sequences.This is demonstrated as follows: let x be the bounded sequence given by (3.28) Let Y be the A α,t -transform of the sequence x.Then it follows that the sequence Y is given by Hence, if A α,t is an -matrix, then by Theorem 1, (1 − t) α+1 ∈ , and so x ∈ (A α,t ).
Corollary 12. Suppose that −1 < α ≤ 0, A α,t is an -matrix.Then (A α,t ) contains the class of all sequences x such that ∞ k=0 x k is conditionally convergent.Remark 3. In fact, we can give a further indication of the size of (A α,t ) by showing that if A α,t is an -matrix, then it also contains an unbounded sequence.To verify this, consider the sequence x given by

.31)
Let Y be the A -transform of the sequence x.Then we have and, consequently, Hence, if A α,t is an -matrix, then by Theorem 1, (1 − t) α+1 ∈ , and so x ∈ (A α,t ).This example clearly indicates that A α,t is a rather strong method in the -setting for any α > −1.
The -strength of the A α,t matrices can also be demonstrated by comparing them with the familiar Norland matrices (N p ) [3].By using the same techniques used in the proof of [3,Thm. 8], we can show that the class of the A α,t matrix summability methods is -stronger than the class of N p matrix summability methods for some p.
When discussing the -strength of A α,t , or the size of (A α,t ), it is very important that we also determine the domain of A α,t .The following proposition, which can be easily proved, gives a characterization of the domain of A α,t .Example 2. The A α,t matrix is not -stronger than the familiar Euler-Knopp matrix E r for r ∈ (0, 1).Also, E r is not -stronger than A α,t .To demonstrate consider the sequence x defined by x k = (−q) k and r = 1 q , (3.37)

Proposition 1. The complex number sequence x is in the domain of the matrix A α,t if and only if
where q > 1.Let Y be the E r -transform of the sequence x .Then it is easy to see that the sequence Y is defined by (3.38) Since q > 1, we have Y ∈ and hence x ∈ (E r ), but x ∉ (A α,t ) by Proposition 1.Hence, A α,t is not -stronger than E r .To show that E r is not -stronger than A α,t , we let −1 < α ≤ 0 and consider the sequence x that was constructed by Fridy in his example of [5, p. 424] .Here, we have x ∉ (E r ), but x ∈ (A α,t ) by Theorem 8. Thus, E r is not -stronger than A α,t .

.34) Remark 4 .
Proposition 1 can be used as a powerful tool in making a comparison between the -strength of the A α,t matrices and some other matrices as shown by the following examples.Example 1.The A α,t matrix is not -stronger than the Borel matrix B[8, p. 53].To demonstrate this, consider the sequence x given byx k = (−3) k .k = e −4n .(3.36) Thus, we have Bx ∈ and hence x ∈ (B), but by Proposition 1, x ∉ (A α,t ) .Hence, A α,t is not -stronger than B.