Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2011, Article ID 131240, 14 pages
doi:10.1155/2011/131240
Research Article
Size of Convergence Domains for Generalized
Hausdorff Prime Matrices
T. Selmanogullari,1 E. Savaş,2 and B. E. Rhoades3
1
Department of Mathematics, Mimar Sinan Fine Arts University, Besiktas, 34349 Istanbul, Turkey
Department of Mathematics, Istanbul Commerce University, Uskudar, 34672 Istanbul, Turkey
3
Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
2
Correspondence should be addressed to T. Selmanogullari, tugcenmat@yahoo.com
Received 8 December 2010; Accepted 2 March 2011
Academic Editor: Q. Lan
Copyright q 2011 T. Selmanogullari et al. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
We show that there exit E-J generalized Hausdorff matrices and unbounded sequences x such that
each matrix has convergence domain c ⊕ x.
1. Introduction
The convergence domain of an infinite matrix A ank n, k 0, 1, . . . will be denoted by
A and is defined by A : {x {xn } | An x ∈ c}, where c denotes the space of convergence
and
sequences, An x :
k ank xk . The necessary and sufficient conditions of Silverman
a
Toeplitz for a matrix to be conservative are limn ank ak exists for each k, limn ∞
k0 nk t
∞
exists, and ||A|| : supn k0 |ank | < ∞. A conservative matrix A is called multiplicative if
each ak 0 and regular if, in addition, t 1.
The E-J generalized Hausdorff matrices under consideration were defined indepenα
dently by Endl 1, 2 and Jakimovski 3. Each matrix Hμ is a lower triangular matrix with
nonzero entries
⎛
⎞
n
α
α
⎠Δn−k μk ,
1.1
hnk ⎝
n−k
where α is real number, {μn } is a real or complex sequence and Δ is forward difference
operator defined by Δμk μk − μk1 , Δn1 μk ΔΔn μk . We will consider here only
nonnegative α. For α 0, one obtains an ordinary Hausdorff matrix.
2
Journal of Inequalities and Applications
From 1 or 3 a E-J generalized Hausdorff matrix for α > 0 is regular if and only if
there exists a function χ ∈ BV 0, 1 with χ1 − χ0 1 such that
α
μn
1
1.2
tnα dχt,
0
α
α
in which case χ is called the moment generating function, or mass function, for Hμ and μn
is called moment sequence.
For ordinary Hausdorff summability 4, the necessary and sufficient conditions, for
regularity are that function χ ∈ BV 0, 1, χ1 − χ0 1, χ0 χ0, and 1.2 is satisfied
with α 0.
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
μz z−a
,
zb
Ra > 0,
1.3
Rb > 0
is prime. In 1967, Rhoades 9 showed that the convergence domain of every known prime
Hausdorff matrix is of the form c ⊕ x for a particular unbounded sequence x.
Given any unbounded sequence x, Zeller 10 constructed a regular matrix A with
convergence domain A c ⊕ x. It has been shown by Parameswaran 11 that if x is any
unbounded sequence such that {xn − xn−1 } is bounded, divergent, and Borel summable, then
no Hausdorff matrix H exists with H c ⊕ x.
The main result of this paper is to show that there exist E-J generalized Hausdorff
α
matrices Hμ whose moment sequences are
α
μn n−a
,
nbα
Ra > 0,
1.4
Rb > 0,
and unbounded sequences xα such that each matrix has convergent domain c ⊕ xα .
Define the sequences xα by
α
xn Γn α 1
Γn − a 1
1.5
for Ra > 0,
α
where it is understood that if a is positive integer, then xn 0 for n 0, 1, . . . , a − 1.
α
If λn is the moment sequence defined by n − a/n 1 α, Ra > 0, then it is clear
α
α
that Hμ Hλ . Hence, it will be sufficient to prove the theorem by using b 1, in 1.4.
To have the convenience of regularity, we will use the sequence
α
μn n−a
,
−a αn 1 α
1.6
α
since the constant −1/a α does not affect the size of the convergence domain of Hμ .
Journal of Inequalities and Applications
3
2. Auxiliary Results
In order to prove the main theorem of this paper, we will need the following results.
Lemma 2.1. Let A, B ∈ C, d, n ∈ N ∪ {0}, d ≤ n. Then, formally, for any n,
n
ΓA k
kd
ΓB k
1
ΓA n 1
ΓA d
−
.
A−B1
ΓB n
ΓB − 1 d
2.1
Proof. Lemma 2.1 appears as formula 12 on page 138 of 12.
α
Lemma 2.2. For m, n integers n > m 1 > a, xn as in 1.5,
n
1
1
α
a
α
km1 xk k − a
1
α
xm
−
1
α
xn
.
2.2
Proof. Using Lemma 2.1,
n
1
α
km1 xk k
− a
n
Γk − a
Γk α 1
km1
1
Γn 1 − a Γm 1 − a
−
−a − α 1 1 Γn 1 α Γm 1 α
1
aα
1
α
xm
−
1
α
xn
2.3
.
Lemma 2.3. For 0 ≤ r ≤ a
−
n−1 ja1
α −1 α
hjr
hnj
α
xn
a α 1
−
.
Γa α 1
n−a
2.4
α
Proof. μn can be written as
α
μn aα1
−1
,
a α a αn α 1
2.5
4
Journal of Inequalities and Applications
α
so that, for 0 ≤ k < n, hnk a α 1/a αn 1 α. From Lemma 2.1 and 3.11,
−
n−1 ja1
α −1 α
hjr
hnj
−
α
n−1
−xn a αa α 12
α j − a a α j α 1
ja1 xj
a α α
12 xn
Γ j −a
ja1 Γ j 2 α
n−1
α a α 12 xn
a α 1
xn
a α 1
−
.
Γa α 1
n−a
Γn − a
1
−
Γa α 2 Γn 1 α
2.6
α
3. Main Result
α
Theorem 3.1. If for fixed a and b the matrix Hμ is defined by 1.4 and a sequence xα by 1.5,
α
then Hμ c ⊕ xα .
α
Proof. We will first show that c ⊕ xα ⊆ Hμ .
α
α
We can write the matrix Hμ
α
−1/a αI − Hλ , where the diagonal entries of
Hλ are
α
λn aα1
.
nα1
3.1
For each n and k,
α
Δn−k λk a α 1
1
tkα 1 − tn−k dt
0
a α 1Γk α 1Γn − k 1
.
Γn α 2
3.2
Therefore,
α
Hλ
n,k
nα
n−k
α
Δn−k λk
n α a α 1Γk α 1Γn − k 1
n−k
aα1
.
nα1
Γn α 2
3.3
Journal of Inequalities and Applications
5
Define yn −un /a α, where
α
un xn −
n
k0
α α
hn,k xk .
3.4
From Lemma 2.1,
un aα1
Γn α 2 Γα 1
Γn α 1
−
−
Γn − a 1 n α 1α a 1 Γn − a 1
Γ−a
Γn α 1
Γα 1
1 − 1 Γn − a 1
Γ−an α 1
Γα 1
−→ 0 as n −→ ∞.
Γ−an α 1
3.5
This argument is valid provided a is not a positive integer. If a is a positive integer, then
α
xk 0 for 0 ≤ k ≤ a − 1.
Then, un 0 for 0 ≤ n ≤ a − 1, and for n ≥ a, from Lemma 2.1, we get
α
un xn −
n
Γk α 1
a α 1 n α 1 ka Γk − a 1
1
Γn α 2 Γa α 2
Γn α 1 a α 1 Γa α 1
−
−
−
Γn − a 1 n α 1
Γ1
n α 1 Γn − a 1
Γ1
Γn α 1 Γa α 2
Γn α 2
Γa α 2
−
−
Γn − a 1
n 1 α
n 1 αΓn − a 1
n 1 α
Γn α 1
n 1 α
1−
Γn − a 1
n 1 α
3.6
0 −→ 0 as n −→ ∞.
α
α
α
Since Hμ is regular, c ⊆ Hμ . Thus, c ⊕ xα ⊆ Hμ .
To prove the converse, we will use Zeller’s technique to construct a regular matrix A
α
with A c ⊕ xα and then show that Hμ ⊆ A.
Set P0 0 and define a sequence {Pn } inductively by selecting Pn1 to be smallest
α
α
integer P > Pn such that |xP | ≥ 2 |xPn |. Such a construction is clearly possible, since xα is
α
α
α
not bounded. Let qn 1 − xPn−1 /xPn , n 1, 2, . . .. Define a matrix B by
b00 1,
bn,n−1 1
α
qn
,
n ≥ 1,
3.7
α
bn,n −
bn,k 0
xPn−1
α α
qn xPn
,
n ≥ 1,
otherwise.
6
Journal of Inequalities and Applications
Now, define the matrix A as follows:
aPn ,Pk bnk ,
aPn ,k 0,
ann 1,
k/
Pi for any integer i,
3.8
n/
Pi for any integer i.
If n /
Pi for any integer i, then there exists an integer r such that Pr < n < Pr1 . For this r,
define
α
an,Pr−1 an,Pr xn
α
α
xPr − xPr−1
α
−xn
α
α
xPr − xPr−1
,
3.9
.
Set ank 0 otherwise. From 10, A is regular and A c ⊕ xα . There are three cases to
consider, based on whether a is real number and not a positive integer, a is positive integer,
or a is complex.
Proof of Case I. If a is real and not a positive integer, the E-J generalized Hausdorff matrix
α
α
α
Hμ generated by 1.6 has a unique two sided inverse Hμ −1 hnk −1 with generating
sequence
1
α
μn
−a αn 1 α
a αa α 1
.
−a α −
n−a
n − a
3.10
For k < n,
α −1
hnk
nα
1
Δn−k α
n−k
μk
−a αa α 1Γn α 1Γk − a
Γk α 1Γn − a 1
3.11
α
α −1
hnn
α
−xn a αa α 1
α
xk k − a
,
−a αn 1 α
.
n−a
3.12
α −1
To show that Hμ ⊆ A, it will be sufficent to show that D AHμ Each column of
α −1
Hμ is a regular matrix.
is essentially a scalar multiple of 1.5, so it is obvious that each
Journal of Inequalities and Applications
7
α −1
column of Hμ belongs to the convergence domain of A. However, it will be necessary to
calculate the terms of D explicitly, since we must show that t 1 and that D has finite norm.
If k /
Pi for any integer i, and r denotes the integer such that Pr−1 < k < Pr , then from
the definition of A,
dPn ,k n
jr
α −1
aPn ,Pj hPj ,k
−1
α
α −1
bn,n−1 hPn−1 ,k
bnn hPn ,k
3.13
0.
If k Pr for r < n − 1, then
dPn ,Pr n
jr
−1
α
aPn ,Pj hPj ,Pr
0.
3.14
For k Pn−1 ,
−1
−1
α
α
dPn ,Pn−1 aPn ,Pn−1 hPn−1 ,Pn−1
aPn ,Pn hPn ,Pn−1
1
α
qn
−a αPn−1 α 1
Pn−1 − a
−a α
α
qn
⎛
⎝
α
−xPn−1
α α
qn xPn
⎞⎛
⎠⎝
α
−a αa α 1xPn
α
xPn−1 Pn−1
− a
⎞
⎠
3.15
.
For Pn−1 < k < Pn ,
dPn ,k aPn ,Pn
α −1
hPn ,k
α
a α1 a αxPn−1
α α
k − aqn xk
3.16
,
−1 a α1 α Pn xPα
α
n−1
dPn ,Pn aPn ,Pn hPn ,Pn
.
α α
Pn − aqn xPn
3.17
For n /
Pi for any i, if we now let r denote the integer such that Pr < n < Pr1 , then for
0 < k < Pr−1,
dnk n
α −1
an,j hj,k
jk
α
an,Pr−1 hPr−1 ,k
−1
α
an,Pr hPr ,k
−1
α
an,n hn,k
−1
3.18
0.
8
Journal of Inequalities and Applications
For k Pr−1 ,
−1
−1
−1
α
α
α
an,Pr hPr ,Pr−1
an,n hn,Pr−1
dn,Pr−1 an,Pr−1 hPr−1 ,Pr−1
⎛
α
−a αxn ⎝ Pr−1
α
Pr−1 − a
xPr
α
−a αxn ⎝ Pr−1
α
Pr−1 − a
xPr
⎛
α1
−
α
xPr−1
α1
α
− xPr−1
⎞
α
−
a α 1 xPr
α
xPr
−
α
xPr−1
α
xPr−1
a α 1⎠
α
xPr−1
⎞
−
3.19
a α 1 ⎠
α
α
xPr − xPr−1
α
−a αxn
α
α
xPr − xPr−1
.
For Pr−1 < k < Pr ,
α −1
α −1
dnk an,Pr hPr ,k
an,n hn,k
3.20
α α
a αa α 1xn xPr−1
.
α
α
α
xk xPr − xPr−1 k − a
For k Pr ,
−1
α
α −1
dn,Pr an,Pr hPr ,Pr
an,n hn,Pr
⎤
⎡
α
−a αxn ⎣ −Pr α 1 a α 1 ⎦
α
α
α
Pr − a
xPr − xPr−1
xPr
3.21
−a αxn
α
α
α
−Pr α 1xPr xPr − xPr−1 a α 1 .
α
α
α
Pr − axPr xPr − xPr−1
α
α
α
The quantity in brackets is equal to −Pr − axPr − xPr−1 a α 1, giving
α
dn,Pr a αxn
α
α
xPr − xPr−1
α
α
x a αa α 1xPr−1
nα
.
xPr Pr − a xPα − xPα
r
r−1
3.22
For Pr < k < n,
α
−xn a αa α 1
α −1
dn,k an,n hn,k
,
α
k − a
xk
3.23
Journal of Inequalities and Applications
9
and finally,
dn,n −a αn 1 α
.
n−a
3.24
By using 3.13–3.17,
Pn
dPn ,k dPn ,Pn−1 k0
P
n −1
dPn ,k dPn ,Pn
kPn−1 1
⎡
P
n −1
1
a α ⎣
α
−1
x
α
1
a
P
α
α
n−1
qn
kPn−1 1 xk k − a
α
Pn α 1xPn−1
α
Pn − axPn
⎤
3.25
⎦.
By using Lemma 2.2, and noting that
Pn α 1
1aα
1
,
Pn − a
Pn − a
⎞
⎛
⎡
⎤
α
α
Pn
x
x
1
a
α
1
α
1
1
a
a α ⎣
Pn−1
Pn−1
α
⎝
⎦
−1 xPn−1
dPn ,k − α ⎠ α
α
α
α
a
α
P
−
a
n
qn
xPn−1 xPn −1
xPn
xPn
k0
⎡
⎤
α
α
α
a α 1 a α 1 xPn−1
1 a α xPn−1 xPn−1 ⎦
a α ⎣
−1 −
α .
α
aα
a α xα
Pn − a xα
qn
xPn
Pn −1
Pn
3.26
Note that
⎡
⎤
α
α
α
α
xPn−1
xPn−1
a α 1 xPn−1
1 a α xPn−1
⎦
−
a α 1⎣−
α
α
a α xα
Pn − a xα
−
ax
αx
a
P
n
Pn −1
Pn
Pn −1
Pn
⎡
a α 1⎣−
α
xPn−1 ΓPn − a
a αΓPn α
α
xPn−1 ΓPn − a 1
Pn − aΓPn α 1
⎤
⎦
⎡
⎤
α
α
xPn−1
ΓPn − a ⎣ −xPn−1
⎦
a α 1
ΓPn α a α Pn α
α
−
a α 1ΓPn − a 1 xPn−1
ΓPn α 1a α
.
3.27
10
Journal of Inequalities and Applications
Finally,
⎡
⎤
α
α
Pn
x
−
a
1x
α
1ΓP
a
n
a α 1
a α ⎣
Pn−1
Pn−1
⎦
−1 α
dPn ,k −
α
a
α
α
1
αΓP
a
n
qn
xPn
k0
⎞
⎛
⎡
⎤
α
α
xPn−1
xPn−1
a α 1 ⎝
a α ⎣
−1 1 − α ⎠ α ⎦
α
aα
qn
xPn
xPn
⎡
a α ⎣
−1 1 a α
α
qn
−
⎛
α
a
⎝
α
qn
α
1−
xPn−1
α
xPn
α
qn
aα
α
xPn−1
α
xPn
3.28
⎤
⎦
⎞
⎠ 1 a α 1.
For n /
Pi for any i, r the integer such that Pr < n < Pr1 , and using 3.18–3.24, we have
⎤
⎡
α
α
P
n
r −1
xPr−1 a α 1
1
a αxn ⎣
α
⎦
dnk α
1
−1 xPr−1 a α 1
α
α
α
xPr − xPr−1
xPr Pr − a
k0
kPr−1 1 xk k − a
α
−xn a αa α
α
xk k − a
kPr 1
n−1
1
−
3.29
a αn 1 α
.
n − a
α
α
Writing n 1 α/n − a 1 xn 1 a α/xn n − a and using Lemma 2.2, the quantity
in brackets, which we call I1 , takes the form
I1 α
xPr−1
⎞
⎛
α
1 ⎠ xPr−1 a α 1
a α 1 ⎝ 1
−
α
α
α
a α
xPr−1 xPr −1
xPr Pr − a
⎞
⎛
xPr−1 a α 1⎝
1
α
a α
αxPr−1
−
1
a α
αxPr −1
1
α
xPr Pr
− a
3.30
⎠.
The sum
−
1
a α
αxPr −1
1
α
xPr Pr
− a
−
ΓPr − a 1
ΓPr − a
a αΓPr α Pr − aΓPr α 1
3.31
−
1
α
a αxPr
.
Journal of Inequalities and Applications
11
Thus,
α
a αxn
α
α
xPr − xPr−1
α
I1 a αxn
α
α
xPr − xPr−1
α
xn a
⎛
xPr−1 a α 1⎝
⎞
1
α
α 1
α
xPr
α
a αxPr−1
−
1
α
a αxPr
⎠
3.32
.
Finally,
n
dnk α
xPr
k0
α
a α 1xn
α
xPr
⎤
⎡
α
xn a α 1
− a α1 a n−1
1
α
αxn ⎣
α
kPr 1 xk k
− a
1
α
xn n − a
⎦ − a α
⎡
⎞⎤
⎛
1
1
1
α
⎠⎦ − a α
⎝
− a α1 a αxn ⎣
−
a α xα xnα
Pr
a α 1 α α
α
xn − xn − xPr
− a α
α
xPr
1.
3.33
Clearly, D has null columns. It remains to show that D has finite norm.
α
α
For all integers, n ≥ a 1, xn is positive and 1/2 ≤ qn ≤ 1. From 3.25,
Pn dP k dP P n,
n, n−1
k0
P
n −1
kPn−1 1
⎡
α
α
α
dP k dP P n,
n, n
α
xPn−1
3.34
⎤
a α ⎣
1 α ⎦ 1 a α.
α
qn
xPn
α
Since |x Pn | ≥ 2|x Pn−1 |, then, xPn−1 /xPn ≤ 1/2, and the above sum is bounded by 4α 4a 1.
From 3.29,
n
α
|dnk | k0
α
2a αxn
α
α
xPr − xPr−1
α
α
a α 2a α 1xn
α
xPr
α
− a α 1.
3.35
α
From choice of n, |xn | < 2|xPr |. Again, using the fact that |xPr | ≥ 2|xPr−1 |, we have
α 2a α 2xPr 4a α 1 2a α 1 14a α 5.
|dnk | < α α xPr − xPr /2
k0
n
3.36
12
Journal of Inequalities and Applications
Since there are only a finite number of rows of D with n < a 1, D has finite norm and is
regular.
α
α
Proof of Case II. If a is a positive integer, μa 0, and Hμ fails to have a two-sided inverse.
α
However, if we define a new matrix F fnk with fa,a 1 and which agrees with Hμ
α
elsewhere, then F does possess a unique two-sided inverse. Morever, F Hμ and, for
α −1
α
−1
hnk , where the hnk −1 are computed using 3.11 and 3.12.
k > a, fnk
α
xn
From 1.5,
0 for 0 ≤ n < a. Consequently, P0 0, P1 a and P2 a 1. Now, let
E : AF −1 enk . To prove that E is regular, we are concerned with the behavior of the enk
α
−1
for all n sufficiently large. We will restrict our attention to n > a 1. Since fnk
hnk −1 for
all k > a, it is clear that enk dnk for k > a. If we can show that enk 0 for all 0 ≤ k ≤ a and
n > a 1, then it will follow that E is regular, since D is.
For n > a 1,
n
−1
fnj
fja 0,
ja
−1
fna
faa
n
−1
−
fnj
fja
ja1
−
n−1 ja1
α −1 α
hnj
hja
−
α −1 α
α −1 α
hnn
hna .
3.37
α
−1
Since fa,a 1 and hnn hna −aα1/n−a, fna
xn /Γaα1. By induction
α
α
α
−1
it is showed that fn,a−r kr axn , where kr a is a function of a.
For n > a 1, Pn−1 > Pa ≥ P1 a ≥ r
ePn ,r Pn
aPn ,j fjr−1 bn,n−1 fP−1n−1 ,r bn,n fP−1n ,r
jr
⎞
⎛
α
x
1 ⎝ α
Pn−1
α
α α
xPn ka−r a ⎠ 0.
α ka−r axPn−1 − α
qn
xPn
3.38
For n > a 1, n /
Pi for any integer i, 0 ≤ r ≤ a, and s the integer such that Ps < n < Ps1 ,
en,r n
jr
−1
an,j fjr−1 an,Ps−1 fP−1s−1 ,r an,Ps fP−1s ,r an,n fn,r
α
xn
α
α
xPs − xPs−1
α
α
α
α
ka−r axPs−1 − ka−r axPs
3.39
α
α
ka−r axn 0.
α
Proof of Case III. If a is complex, then none of the μn vanish, and we may use the matrix D
of Case I. It will be sufficient to show that D has finite norm. From 3.25,
Pn
P
−a α a αPn α 1xα n −1
Pn−1 |dPn ,k | |dPn ,k | |dPn ,Pn−1 1 |.
α
α
qα kP 2
q P − ax
k0
n
n
n
Pn
n−1
3.40
Journal of Inequalities and Applications
α
13
α
α
Again, |xPn | ≥ 2|xPn−1 |. It can be shown that 1/2 ≤ |qn | ≤ 3/2. Since
a αa α 1 |dPn ,Pn−1 1 | α
,
q P 1 α 3.41
n−1
n
the first two and last terms of 3.40, are clearly bounded in n.
For Pn−1 1 < k < Pn , using 3.16,
a αa α 1ΓP 1 αΓk − a n−1
|dPn ,k | α
ΓP 1 − aΓk 1 αq
n−1
n
a αa α 1k − a − 1 · · · P − a 1 n−1
ΓPn−1 1 α
α
Γk 1 αq
n
a αa α 1ΓP 1 α n−1
≤
α
q
3.42
n
|Pn−1 1 − a||Pn−1 1 − a| 1 · · · |Pn−1 1 − a| k − Pn−1 − 2 ·
Γk 1 α
a αa α 1ΓP 1 α Γk w
n−1
,
α
Γk 1 α
qn Γ|Pn−1 1 − a|
where w |Pn−1 1 − a| − Pn−1 − 1. w < 0 for all n sufficiently large. From Lemma 2.1, we can
write
Pn
a αa α 1ΓP 1 α Γx w
n−1
|dPn ,k | ≤ α
w − αΓx α Pn−1 2
q Γ|P 1 − a|
2
P
n −1
kPn−1
n
n−1
n
n−1
a αa α 1ΓP 1 α ΓPn−1 2 w
n−1
,
<
α
α − wΓPn−1 2 α
q Γ|P 1 − a|
3.43
and the sum is uniformly bounded in n, since −w is bounded away from zero.
If n /
Pi , for any i, then from 3.29,
α α α P
−1
n
r
αa
α
1x
x
a
−a αxn Pr−1 n |dnk | α
xP − xPα kPr−1 1 xα − xα xα k − a k0
r
Pr
r−1
Pr−1
k
α α α a αxn a αa α 1xPr−1 xn α
xP − xPα xα − xα xα Pr − a Pr
Pr−1
Pr
r
r−1
n−1 a αa α 1xnα −a αn 1 α .
α
n−a
x k − a
kP 1
r
k
3.44
14
Journal of Inequalities and Applications
α
Terms 1, 3, 4, and 6 of 3.44 are clearly bounded in n. Recalling that qr
first summation may be written in the form
r −1 xα
a αa α 1xnα P
P
r−1
.
α
α α
kPr−1 1 xk k − a qr xP
α
α
1 − xPr−1 /xPr , the
3.45
r
The summation is identical with the one in 3.40, and the above expression is uniformly
α
α
bounded, since |xn | < 2|xPr |. 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.
Acknowledgment
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.
References
1 K. Endl, “Abstracts of short communications and scientific program,” International Congress of
Mathematicians, vol. 73, p. 46, 1958.
2 K. Endl, “Untersuchungen über Momentenprobleme bei Verfahren vom Hausdorffschen Typus,”
Mathematische Annalen, vol. 139, pp. 403–432, 1960.
3 A. Jakimovski, “The product of summability methods; new classes of transformations and their
properties,” Tech. Note, Contract no. AF61 052-187, 1959.
4 G. H. Hardy, Divergent Series, Clarendon Press, Oxford, UK, 1949.
5 E. Hille and R. S. Phillips, Functional Analysis and Semi-Groups, vol. 31 of American Mathematical Society
Colloquium Publications, American Mathematical Society, Providence, RI, USA, 1957.
6 E. Hille and J. D. Tamarkin, “On moment functions,” Proceedings of the National Academy of Sciences of
the United States of America, vol. 19, pp. 902–908, 1933.
7 E. Hille and J. D. Tamarkin, “Questions of relative inclusion in the domain of Hausdorff means,”
Proceedings of the National Academy of Sciences of the United States of America, vol. 19, pp. 573–577, 1933.
8 L. L. Silvermann and J. D. Tamarkin, “On the generalization of Abel’s theorem for certain definitions
of summability,” Mathematische Zeitschrift, vol. 29, pp. 161–170, 1928.
9 B. E. Rhoades, “Size of convergence domains for known Hausdorff prime matrices,” Journal of
Mathematical Analysis and Applications, vol. 19, pp. 457–468, 1967.
10 K. Zeller, “Merkwürdigkeiten bei Matrixverfahren; Einfolgenverfahren,” Archiv für Mathematische,
vol. 4, pp. 1–5, 1953.
11 M. R. Parameswaran, “Remark on the structure of the summability field of a Hausdorff matrix,”
Proceedings of the National Institute of Sciences of India. Part A, vol. 27, pp. 175–177, 1961.
12 H. T. Davis, The Summation of Series, The Principia Press of Trinity University, San Antonio, Texas,
USA, 1962.