Open Access

Size of Convergence Domains for Generalized Hausdorff Prime Matrices

Journal of Inequalities and Applications20112011:131240

Received: 8 December 2010

Accepted: 2 March 2011

Published: 14 March 2011


We show that there exit E-J generalized Hausdorff matrices and unbounded sequences such that each matrix has convergence domain .

1. Introduction

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, .

The E-J generalized Hausdorff matrices under consideration were defined independently by Endl ([1, 2]) and Jakimovski [3]. Each matrix is a lower triangular matrix with nonzero entries

where is real number, is a real or complex sequence and is forward difference operator defined by . We will consider here only nonnegative . For , one obtains an ordinary Hausdorff matrix.

From [1] or [3] a E-J generalized Hausdorff matrix (for ) is regular if and only if there exists a function with such that

in which case is called the moment generating function, or mass function, for and is called moment sequence.

For ordinary Hausdorff summability [4], the necessary and sufficient conditions, for regularity are that function , , , and (1.2) is satisfied with .

As noted in [5], the set of all multiplicative Hausdorff matrices forms a commutative Banach algebra that is also an integral domain, making it possible to define the concepts of unit, prime, divisibility, associate, multiple, and factor. Hille and Tamarkin ([6, 7]), using some techniques from [8], showed that every Hausdorff matrix with moment function

is prime. In 1967, Rhoades [9] 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 [10] constructed a regular matrix with convergence domain . It has been shown by Parameswaran [11] that if is any unbounded sequence such that is bounded, divergent, and Borel summable, then no Hausdorff matrix exists with .

The main result of this paper is to show that there exist E-J generalized Hausdorff matrices whose moment sequences are

and unbounded sequences such that each matrix has convergent domain .

Define the sequences by

where it is understood that if is positive integer, then for .

If is the moment sequence defined by , , then it is clear that . Hence, it will be sufficient to prove the theorem by using , in (1.4). To have the convenience of regularity, we will use the sequence

since the constant does not affect the size of the convergence domain of .

2. Auxiliary Results

In order to prove the main theorem of this paper, we will need the following results.

Lemma 2.1.

Let , , , , . Then, formally, for any ,


Lemma 2.1 appears as formula 12 on page 138 of [12].

Lemma 2.2.

For integers , as in (1.5),


Using Lemma 2.1,

Lemma 2.3.



can be written as
so that, for , . From Lemma 2.1 and (3.11),

3. Main Result

Theorem 3.1.

If for fixed and the matrix is defined by (1.4) and a sequence by (1.5), then .


We will first show that .

We can write the matrix , where the diagonal entries of are
For each and ,
Define , where
From Lemma 2.1,

This argument is valid provided is not a positive integer. If is a positive integer, then for .

Then, for , and for , from Lemma 2.1, we get

Since is regular, . Thus, .

To prove the converse, we will use Zeller's technique to construct a regular matrix with and then show that .

Set and define a sequence inductively by selecting to be smallest integer such that . (Such a construction is clearly possible, since is not bounded.) Let , . Define a matrix by
Now, define the matrix as follows:
If for any integer , then there exists an integer such that . For this , define

Set otherwise. From [10], 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.

If is real and not a positive integer, the E-J generalized Hausdorff matrix generated by (1.6) has a unique two sided inverse with generating sequence
For ,

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.

If for any integer , and denotes the integer such that , then from the definition of ,
If for , then
For ,
For ,
For for any , if we now let denote the integer such that , then for
For ,
For ,
For ,
The quantity in brackets is equal to , giving
For ,
and finally,
By using (3.13)–(3.17),
By using Lemma 2.2, and noting that
Note that
For for any , the integer such that , and using (3.18)–(3.24), we have
Writing and using Lemma 2.2, the quantity in brackets, which we call , takes the form
The sum

Clearly, has null columns. It remains to show that has finite norm.

For all integers, , is positive and . From (3.25),
Since , then, , and the above sum is bounded by . From (3.29),
From choice of , . Again, using the fact that , we have

Since there are only a finite number of rows of with , has finite norm and is regular.

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.

For ,

Since and . By induction it is showed that , where is a function of .

For , for any integer , , and the integer such that ,

Proof of Case  III.

If is complex, then none of the vanish, and we may use the matrix of Case I. It will be sufficient to show that has finite norm. From (3.25),
Again, . It can be shown that . Since

the first two and last terms of (3.40), are clearly bounded in .

For , using (3.16),
where . for all sufficiently large. From Lemma 2.1, we can write

and the sum is uniformly bounded in , since is bounded away from zero.

If , for any , then from (3.29),
Terms , and 6 of (3.44) are clearly bounded in . Recalling that , the first summation may be written in the form

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.

Authors’ Affiliations

Department of Mathematics, Mimar Sinan Fine Arts University
Department of Mathematics, Istanbul Commerce University
Department of Mathematics, Indiana University


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© T. Selmanogullari et al. 2011

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