Size of Convergence Domains for Generalized Hausdorff Prime Matrices
© T. Selmanogullari et al. 2011
Received: 8 December 2010
Accepted: 2 March 2011
Published: 14 March 2011
The convergence domain of an infinite matrix will be denoted by and is defined by , where denotes the space of convergence sequences, . The necessary and sufficient conditions of Silverman and Toeplitz for a matrix to be conservative are exists for each , exists, and . A conservative matrix is called multiplicative if each and regular if, in addition, .
For ordinary Hausdorff summability , the necessary and sufficient conditions, for regularity are that function , , , and (1.2) is satisfied with .
is prime. In 1967, Rhoades  showed that the convergence domain of every known prime Hausdorff matrix is of the form for a particular unbounded sequence .
Given any unbounded sequence , Zeller  constructed a regular matrix with convergence domain . It has been shown by Parameswaran  that if is any unbounded sequence such that is bounded, divergent, and Borel summable, then no Hausdorff matrix exists with .
2. Auxiliary Results
In order to prove the main theorem of this paper, we will need the following results.
Lemma 2.1 appears as formula 12 on page 138 of .
3. Main Result
Set otherwise. From , is regular and . There are three cases to consider, based on whether is real number and not a positive integer, is positive integer, or is complex.
Proof of Case I.
To show that , it will be sufficent to show that is a regular matrix. Each column of is essentially a scalar multiple of (1.5), so it is obvious that each column of belongs to the convergence domain of . However, it will be necessary to calculate the terms of explicitly, since we must show that and that has finite norm.
Proof of Case II.
If is a positive integer, , and fails to have a two-sided inverse. However, if we define a new matrix with and which agrees with elsewhere, then does possess a unique two-sided inverse. Morever, and, for , , where the are computed using (3.11) and (3.12).
From (1.5), for . Consequently, and . Now, let . To prove that is regular, we are concerned with the behavior of the for all sufficiently large. We will restrict our attention to . Since for all , it is clear that for . If we can show that for all and , then it will follow that is regular, since is.
Proof of Case III.
The summation is identical with the one in (3.40), and the above expression is uniformly bounded, since . Using an argument similar to the one used in establishing (3.40), the second summation of (3.44) can be shown to be uniformly bounded.
The first author acknowledges support from the Scientific and Technical Research Council of Turkey in the preparation of this paper, the authors wish to thank the referee for his careful reading of the manuscript and for his helpful suggestions.
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