Rhoades3 1 Department of Mathematics, Mimar Sinan Fine Arts University, Besiktas, 34349 Istanbul, Turkey 2 Department of Mathematics, Istanbul Commerce University, Uskudar, 34672 Istanbu
Trang 1Volume 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 Savas¸,2 and B E Rhoades3
1 Department of Mathematics, Mimar Sinan Fine Arts University, Besiktas, 34349 Istanbul, Turkey
2 Department of Mathematics, Istanbul Commerce University, Uskudar, 34672 Istanbul, Turkey
3 Department of Mathematics, Indiana University, Bloomington, IN 47405-7106, USA
Correspondence should be addressed to T Selmanogullari,tugcenmat@yahoo.com
Received 8 December 2010; Accepted 2 March 2011
Academic Editor: Q Lan
Copyrightq 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 a nk n, k 0, 1, will be denoted by
A and is defined by A : {x {x n } | A n x ∈ c}, where c denotes the space of convergence sequences, A n x : k a nk x k The necessary and sufficient conditions of Silverman and Toeplitz for a matrix to be conservative are limn a nk a k exists for each k, lim n∞
k0 a nk t
exists, and||A|| : sup n∞
k0 |a nk | < ∞ A conservative matrix A is called multiplicative if each a k 0 and regular if, in addition, t 1.
The E-J generalized Hausdorff matrices under consideration were defined indepen-dently by Endl1,2 and Jakimovski 3 Each matrix H μ αis a lower triangular matrix with nonzero entries
h α nk
⎛
⎝n α
n − k
⎞
where α is real number, {μ n} is a real or complex sequence and Δ is forward difference operator defined by Δμ k μ k − μ k 1 , Δ n 1 μ k ΔΔn μ k We will consider here only
nonnegative α For α 0, one obtains an ordinary Hausdorff matrix.
Trang 2From1 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
μ α n
1
0
in which case χ is called the moment generating function, or mass function, for H μ α and μ α n
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 in5, 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 from8, showed that every Hausdorff matrix with moment function
μ z z − a
z b , R a > 0, R b > 0 1.3
is prime In 1967, Rhoades9 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 domainA c ⊕ x It has been shown by Parameswaran 11 that if x is any
unbounded sequence such that{x n − x n−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
matrices H μ αwhose moment sequences are
μ α n n − a
n b α , R a > 0, R b > 0, 1.4
and unbounded sequences x α such that each matrix has convergent domain c ⊕ x α
Define the sequences x αby
x n α Γn α 1
Γn − a 1 for Ra > 0, 1.5
where it is understood that if a is positive integer, then x α n 0 for n 0, 1, , a − 1.
If λ α n is the moment sequence defined byn − a/n 1 α, Ra > 0, then it is clear
thatH μ α H λ α 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
μ α n n − a
−a αn 1 α , 1.6
since the constant−1/a α does not affect the size of the convergence domain of H μ α
Trang 32 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
kd
ΓA k
ΓB k
1
A − B 1
ΓA n 1
ΓB n −ΓB − 1 d ΓA d . 2.1
Proof. Lemma 2.1appears as formula 12 on page 138 of12
Lemma 2.2 For m, n integers n > m 1 > a, x α n as in1.5,
n
km 1
1
x α k k − a
1
a α
1
x α m
− 1
x α n
Proof UsingLemma 2.1,
n
km 1
1
x α k k − a
n
km 1
Γk − a
Γk α 1
−a − α 1 1
Γn 1 − a
Γn 1 α−Γm 1 − a Γm 1 α
1
a α
1
x α m
− 1
x α n
.
2.3
Lemma 2.3 For 0 ≤ r ≤ a
−n−1
ja 1
h α nj−1
h α jr x α n
Γa α 1 −
a α 1
n − a . 2.4
Proof μ α n can be written as
μ α n −1
a α a αn α 1 a α 1 , 2.5
Trang 4so that, for 0≤ k < n, h α nk a α 1/a αn 1 α FromLemma 2.1and3.11,
−n−1
ja 1
h α nj −1
h α jr −n−1
ja 1
−x α n a αa α 12
x α j
j − a
a αj α 1
a α 12x α n
n−1
ja 1
Γj − a
Γj 2 α
a α 12x n α
a α 1
1
Γa α 2−Γn 1 α Γn − a
x α n
Γa α 1−
a α 1
n − a .
2.6
3 Main Result
Theorem 3.1 If for fixed a and b the matrix H μ α is defined by1.4 and a sequence x α by1.5,
then H μ α c ⊕ x α
Proof We will first show that c ⊕ x α ⊆ H μ α
We can write the matrix H μ α −1/a αI − H λ α, where the diagonal entries of
H λ αare
λ α n a α 1
n α 1 . 3.1
For each n and k,
Δn−k λ α k a α 1
1
0
t k α 1 − t n−k
dt
a α 1Γk α 1Γn − k 1
Γn α 2 .
3.2
Therefore,
H λ α
n,k
n α
n − k
Δn−k λ α k
n α
n − k
a α 1Γk α 1Γn − k 1
Γn α 2
a α 1
n α 1 .
3.3
Trang 5Define y n −u n /a α, where
u n x α n −n
k0
h α n,k x α k 3.4 FromLemma 2.1,
u n Γn α 1
Γn − a 1 −
a α 1
n α 1α a 1
Γn α 2
Γn − a 1 −
Γα 1
Γ−a
Γn α 1
Γn − a 11 − 1 Γ−an α 1 Γα 1
Γα 1
Γ−an α 1 −→ 0 as n −→ ∞.
3.5
This argument is valid provided a is not a positive integer If a is a positive integer, then
x α k 0 for 0 ≤ k ≤ a − 1.
Then, u n 0 for 0 ≤ n ≤ a − 1, and for n ≥ a, fromLemma 2.1, we get
u n x α n −a α 1 n α 1n
ka
Γk α 1
Γk − a 1
Γn α 1
Γn − a 1 −
a α 1
n α 1
Γa α 1
Γ1 −
1
n α 1
Γn α 2
Γn − a 1 −Γa α 2Γ1
Γn α 1
Γn − a 1 −
Γa α 2
n 1 α −
Γn α 2
n 1 αΓn − a 1
Γa α 2
n 1 α
Γn α 1
Γn − a 1
1−n 1 α n 1 α
0 −→ 0 as n −→ ∞.
3.6
Since H μ α is regular, c ⊆ H μ α Thus, c ⊕ x α ⊆ H μ α
To prove the converse, we will use Zeller’s technique to construct a regular matrix A
withA c ⊕ x αand then show thatH μ α ⊆ A.
Set P0 0 and define a sequence {P n } inductively by selecting P n 1 to be smallest
integer P > P nsuch that|x P α | ≥ 2 |x α P n | Such a construction is clearly possible, since x αis not bounded. Let qα n 1 − x α P n−1 /x α P n , n 1, 2, Define a matrix B by
b00 1,
b n,n−1 1
q α n
, n ≥ 1,
b n,n − x
α
P n−1
q α n x P α n , n ≥ 1,
b n,k 0 otherwise.
3.7
Trang 6Now, define the matrix A as follows:
a P n ,P k b nk ,
a P n ,k 0, k / P i for any integer i,
a nn 1, n / P i for any integer i.
3.8
If n / P i for any integer i, then there exists an integer r such that P r < n < P r 1 For this r,
define
a n,P r−1 x α n
x α P r − x α P
r−1
,
a n,P r −x α n
x α P r − x α P
r−1
.
3.9
Set a nk 0 otherwise From 10, A is regular and A c ⊕ x α 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
H μ αgenerated by1.6 has a unique two sided inverse H μ α−1 h α nk−1 with generating sequence
1
μ α n
−a αn 1 α
n − a −a α − a αa α 1
n − a . 3.10
For k < n,
h α nk−1
n α
n − k
Δn−k 1
μ α k
−a αa α 1Γn α 1Γk − a
Γk α 1Γn − a 1
−x n α a αa α 1
x α k k − a ,
3.11
h α nn
−1
−a αn 1 α
To show thatH μ α ⊆ A, it will be sufficent to show that D AH μ α−1is a regular matrix Each column of H μ α−1 is essentially a scalar multiple of1.5, so it is obvious that each
Trang 7column ofH μ α−1belongs 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 / P i for any integer i, and r denotes the integer such that P r−1 < k < P r, then from
the definition of A,
d P n ,k n
jr
a P n ,P j h α P j ,k−1
b n,n−1 h α P n−1 ,k−1
b nn h α P n ,k−1
0.
3.13
If k P r for r < n − 1, then
d P n ,P r n
jr
a P n ,P j h α P j ,P r−1
For k P n−1,
d P n ,P n−1 a P n ,P n−1 h α P n−1 ,P n−1−1
a P n ,P n h α P n ,P n−1−1
1
q α n
−a αP n−1 α 1
P n−1 − a
⎛
⎝ −x P α n−1
q α n x α P n
⎞
⎠
⎛
⎝−a αa α 1x α P n
x P α n−1 P n−1 − a
⎞
⎠
−a α
q α n
.
3.15
For P n−1 < k < P n,
d P n ,k a P n ,P n h α P n ,k−1
a α1 a αx
α
P n−1
k − aq α n x α k , 3.16
d P n ,P n a P n ,P n h α P n ,P n−1
a α1 α P n x
α
P n−1
P n − aq n α x α P n . 3.17
For n / P i for any i, if we now let r denote the integer such that P r < n < P r 1, then for
0 < k < P r−1,
d nkn
jk
a n,j h α j,k−1
a n,P r−1 h α P ,k−1
a n,P r h α P ,k−1
a n,n h α n,k−1
0.
3.18
Trang 8For k P r−1,
d n,P r−1 a n,P r−1 h α P r−1 ,P r−1−1
a n,P r h α P r ,P r−1−1
a n,n h α n,P r−1−1
−a αx α n
P r−1 − a
⎛
⎝P r−1 α 1
x P α r − x α P r−1 −
a α 1
x P α r − x P α r−1
x α P r
x α P r−1 a α 1
x α P r−1
⎞
⎠
−a αx α n
P r−1 − a
⎛
⎝P r−1 α 1
x P α r − x α P r−1 −
a α 1
x P α r − x P α r−1
⎞
⎠
−a αx α n
x α P r − x α P r−1 .
3.19
For P r−1 < k < P r,
d nk a n,P r h α P r ,k−1
a n,n h α n,k−1
a αa α 1x
α
n x P α r−1
x α k x α P r − x P α r−1k − a .
3.20
For k P r,
d n,P r a n,P r h α P r ,P r−1
a n,n h α n,P r−1
−a αx α n
P r − a
⎡
⎣−P r α 1
x α P r − x α P r−1
a α 1
x α P r
⎤
⎦
−a αx α n
P r − ax α P r x P α r − x P α
r−1
−P r α 1x α P r x α P r − x α P r−1a α 1.
3.21
The quantity in brackets is equal to−P r − ax α P r − x α P r−1 a α 1, giving
d n,P r a αx α n
x P α
r−x α P
r−1
x n α
x P α
r
a αa α 1x α P r−1
P r − a x α P
r − x α P
r−1
For P r < k < n,
d n,k a n,n h α n,k−1
−x α n
x α k
a αa α 1
Trang 9and finally,
d n,n −a αn 1 α
By using3.13–3.17,
P n
k0
d P n ,k d P n ,P n−1 Pn−1
kP n−1 1
d P n ,k d P n ,P n
a α
q α n
⎡
⎣−1 x α
Pn−1 a α 1 Pn−1
kP n−1 1
1
x α k k − a
P n α 1x α P n−1
P n − ax P α n
⎤
⎦.
3.25
By usingLemma 2.2, and noting that
P n α 1
P n − a 1
1 a α
P n − a ,
P n
k0
d P n ,k a α
q n α
⎡
⎣−1 x α
P n−1
a α 1
a α
⎛
⎝ 1
x α P
n−1
− 1
x α P
n−1
⎞
⎠ 1 a α
P n − a
x P α n−1
x α P
n
x
α
P n−1
x α P
n
⎤
⎦
a α
q n α
⎡
⎣−1 a α 1
a α −a α 1
a α
x α P n−1
x α P n−1 1 a α
P n − a
x P α n−1
x α P n x
α
P n−1
x α P n
⎤
⎦.
3.26 Note that
−a α 1
a α
x α P n−1
x α P n−1 1 a α
P n − a
x α P n−1
x α P n a α 1
⎡
⎣− x
α
P n−1
a αx α P n−1
x α P n−1
P n − ax α P n
⎤
⎦
a α 1
⎡
⎣− x α P n−1 ΓP n − a
a αΓP n α
x α P n−1 ΓP n − a 1
P n − aΓP n α 1
⎤
⎦
a α 1 ΓP n − a
ΓP n α
⎡
⎣−x P α n−1
a α
x α P n−1
P n α
⎤
⎦
−a α 1ΓP n − a 1 x α P n−1
ΓP n α 1a α .
3.27
Trang 10P n
k0
d P n ,k a α
q α n
⎡
⎣−1 a α 1
a α x
α
P n−1
x P α n − a α 1ΓP n − a 1x α P n−1
a αΓP n α 1
⎤
⎦
a α
q α n
⎡
⎣−1 a α 1
a α
⎛
⎝1 −x P α n−1
x α P n
⎞
⎠ x α P n−1
x α P n
⎤
⎦
a α
q α n
⎡
⎣−1 1 a α
q α n
a α
x
α
P n−1
x P α n
⎤
⎦
−a α
q α n
⎛
⎝1 −x α P n−1
x α P n
⎞
⎠ 1 a α 1.
3.28
For n / P i for any i, r the integer such that P r < n < P r 1, and using3.18–3.24, we have
n
k0
d nk a αx α n
x α P r − x α P r−1
⎡
⎣−1 x α
P r−1 a α 1 Pr−1
kP r−1 1
1
x k α k − a 1
x α P r−1 a α 1
x α P r P r − a
⎤
⎦
n−1
kP r 1
−x α n a αa α 1
x α k k − a −
a αn 1 α
n − a .
3.29
Writingn 1 α/n−a 1 x n α 1 a α/x α n n−a and usingLemma 2.2, the quantity
in brackets, which we call I1, takes the form
I1 x α P r−1 a α 1 a α
⎛
⎝ 1
x α P
r−1
− 1
x α P
r−1
⎞
⎠ x α P r−1 a α 1
x α P
r P r − a
x α P
r−1 a α 1
⎛
a αx α P r−1 −
1
a αx P α r−1
1
x α P r P r − a
⎞
⎠.
3.30
The sum
a αx P α r−1
1
x α P r P r − a −
ΓP r − a
a αΓP r α ΓP
r − a 1
P r − aΓP r α 1
− 1
a αx α P .
3.31
Trang 11a αx α n
x α P r − x α P
r−1
I1 a αx α n
x α P r − x α P
r−1
x α P r−1 a α 1
⎛
a αx α P
r−1
a αx P α
r
⎞
⎠
x n α a α 1
x P α r .
3.32
Finally,
n
k0
d nk x α n a α 1
x α P r − a α1 a αx α n
⎡
⎣ n−1
kP r 1
1
x α k k − a
1
x α n n − a
⎤
⎦ − a α
a α 1x α n
x α P r − a α1 a αx α n
⎡
⎣ 1
a α
⎛
⎝ 1
x α P r − 1
x n α
⎞
⎠
⎤
⎦ − a α
a α 1
x α P
r
x α n − x α n − x α P r − a α
1.
3.33
Clearly, D has null columns It remains to show that D has finite norm.
For all integers, n ≥ a 1, x α n is positive and1/2 ≤ q α n ≤ 1 From 3.25,
P n
k0
d P
n, k d P n, P n−1 Pn−1
kP n−1 1
d P
n, k d P n, P n
a α
q α n
⎡
⎣1 x α P n−1
x α P n
⎤
⎦ 1 a α.
3.34
Since|x α Pn | ≥ 2|x α Pn−1 |, then, x α P n−1 /x α P n ≤ 1/2, and the above sum is bounded by 4α 4a 1.
From3.29,
n
k0
|d nk| 2a αxα n
x α P r − x α P r−1 a α 2a α 1xn α
x α P r − a α 1. 3.35
From choice of n, |x α n | < 2|x P α r | Again, using the fact that |x α P r | ≥ 2|x α P r−1|, we have
n
k0
|d nk | < 2a α 2
x α P r
x α P −x α
P /2 4a α 1 2a α 1 14a α 5. 3.36
...Proof of Case I If a is real and not a positive integer, the E-J generalized Hausdorff matrix
H μ αgenerated by1.6 has a unique two sided inverse... otherwise From 10, A is regular and A c ⊕ x α There are three cases to
consider, based on whether a is real number and not a positive integer, a is positive... data-page="7">
column of< i>H μ α−1belongs to the convergence domain of A However, it will be necessary to calculate the terms of D explicitly, since