SOOCHOW JOURNAL OF MATHEMATICS
Volume 29, No. 3, pp. 313-326, July 2003
VECTOR VALUED PARANORMED BOUNDED AND NULL
SEQUENCE SPACES ASSOCIATED WITH MULTIPLIER
SEQUENCES
BY
BINOD CHANDRA TRIPATHY AND MAUSUMI SEN
Abstract. In this article we introduce the multiplier vector valued sequence spaces
∞ {Ek , Λ, p} and c0 {Ek , Λ, p}, where Λ = (γk ) is an associated multiplier sequence
of non-zero complex numbers and the terms of the sequence are chosen from the
seminormed spaces Ek , k∈N . This generalizes the scalar sequence spaces ∞ {p}
and c0 {p}. We study some properties of these spaces like solidity, completeness and
prove some inclusion results. We characterize the multiplier problem and obtain
their duals.
1. Introduction
Let w denote the set of all sequences of complex terms. Then w is a linear
space under co-ordinatewise addition and scalar multiplication. Any subspace of
w is called as a sequence space. For example ∞ the set of all bounded complex
sequences is a sequence space.
The notion of paranormed sequence space was introduced by Nakano [10]
and Simons [15]. It was further investigated from sequence space point of view
and linked with summability theory by Maddox [8, 9], Lascarides [6], Nanda [11],
Ratha [13], Rath and Tripathy [12] and many others.
The studies on vector valued sequence spaces is done by Gupta [3], Ratha
and Srivastava [14], Das and Choudhury [1], Leonard [7], Tripathy and Sarma
[16] and many others.
Received May 31, 2002; revised November 7, 2002.
AMS Subject Classification. 40A05, 46A45.
Key words. paranormed sequence spaces, solid spaces, multiplier sequence.
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BINOD CHANDRA TRIPATHY AND MAUSUMI SEN
The scope for the studies on sequence spaces was extended by using the
notion of associated multiplier sequences. Goes et al. [2] defined the differentiated
sequence space dE and integrated sequence space
E, using the multiplier sequences
(k−1 )
E for a given sequence space
and (k) respectively. Kamthan [4] used
the multiplier sequence (k!). In the present article we shall consider a general
multiplier sequence Λ = (γk ) of non-zero scalars.
Let Λ = (γk ) be a sequence of non-zero scalars. Then for E a sequence space,
the multiplier sequence space E(Λ), associated with the multiplier sequence Λ is
defined as
E(Λ) = {(xk ) ∈ w : (γk xk ) ∈ E}.
Example 1. Let γk = k−1 for all k ∈ N . Then
c0 (k−1 ) = {(xk ) ∈ w : k−1 xk → 0, as k → ∞}.
From the above example it is clear that c0 (k−1 ) contains some unbounded
sequences too. Further it accelerates the convergence of the sequences in c0 . It
also covers a larger class of sequences for the study. Some times the associated
multiplier sequence delays the rate of convergence of a sequence.
During a chemical reaction, a catalyst is used to accelerate the process of
reaction. For example AIBN is used as a catalyst for polumarization.
2. Definitions and Preliminaries
Let X be a linear space and g : X → R is such that
(i) g(x) ≥ 0.
(ii) x = θ implies g(x) = 0.
(iii) g(x + y) ≤ g(x) + g(y).
(iv) g(−x) = g(x), for all x ∈ X.
(v) g(λn xn − λx) → 0, as n → ∞, for scalars λn and λ and xn , x ∈ X, whenever
λn → λ and xn → x, as n → ∞.
Then g is said to be a paranorm on X and (X, g) is called as a paranormed
space.
VECTOR VALUED PARANORMED BOUNDED AND NULL SEQUENCE SPACES
315
Example 2. Let p = (pk ) be such that p ∈ ∞ and M = max(1, sup pk ).
Then
∞ (p) = {(xk ) ∈ w : sup |xk |pk < ∞}
k
pk
is a paranormed space, paranormed by g(xk ) = supk |xk | M .
A vector valued sequence space E is called solid (or normal) if αx = (αk xk ) ∈
E, whenever x = (xk ) ∈ E and α = (αk ) is a sequence of scalars such that
|αk | ≤ 1 for all k ∈ N . A sequence space E is said to be monotone if E contains
the canonical preimages of all its subspaces (one may refer to Kamthan and Gupta
[5], p.48).
Throughout the article Ek will denote a seminormed sequence space, seminormed by fk for all k ∈ N , defined over C, the field of complex numbers.
Throughout p = (pk ) is a sequence of strictly positive numbers and tk =
all k ∈ N .
1
pk ,
for
We define the following vector valued multiplier sequence spaces:
∞ (Ek , Λ, p) = {(xk ) : xk ∈ Ek for all k ∈ N and sup(fk (γk xk ))pk < ∞},
k
c0 (Ek , Λ, p) = {(xk ) : xk ∈ Ek for all k ∈ N and (fk (γk xk ))pk → 0, as k → ∞},
∞ {Ek , γ, p} = {(xk ) : xk ∈ Ek for all k ∈ N and there exists r > 0
such that sup(fk (rγk xk ))pk tk < ∞},
k
c0 {Ek , Λ, p} = {(xk ) ∈ Ek for all k ∈ N and there exists r > 0
such that (fk (rγk xk ))pk tk → 0, as k → ∞}.
Two sequence spaces E and F are said to be equivalent if there exists a
sequence u = (uk ) of strictly positive numbers such that the mapping
u : E → F defined by y = ux = (uk xk ) ∈ F,
whenever (xk ) ∈ E,
is one-to-one correspondence between E and F . It is denoted by E ∼
= F (u) or
∼ F (see for instance Nakano [10]).
simply E =
It is remarked by Lascarides [6] (Remark 3) that “If E is a sequence space
paranormed (or normed) by g and E ∼
= F (u), then F is a sequence space paranormed (or normed) by gu defined by gu (y) = g(u−1 y), y ∈ F ”.
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BINOD CHANDRA TRIPATHY AND MAUSUMI SEN
Further it is noted by Lascarides [6] that “if (pk ) ∈ ∞ , then c0 (p) ∼
= c0 {p}(u),
tk
∼
(as well as ∞ (p) = ∞ {p}(u)), where u = (pk )”.
For E, F two sequence spaces we define M (F, E) as follows:
M (F, E) = {(γk ) : (γk xk ) ∈ E for all (xk ) ∈ F },
where Λ = (γk ) is a multiplier sequence.
Lemma 1. A sequence space E is solid implies E is monotone.
Lemma 2. (Lascarides [6], Proposition 1) Let h = inf pk and H = sup pk .
Then the following conditions are equivalent:
(i) H < ∞ and h > 0.
(ii) c0 (p) = c0 or ∞ (p) = ∞ .
(iii) ∞ {p} = ∞ (p).
(iv) c0 {p} = c0 (p).
(v) {p} = (p).
3. Main Results
In this section we prove the results involving ∞ {Ek , Λ, p} and c0 {Ek , Λ, p}.
First we prove the following properties of the spaces ∞ {Ek , Λ, p} and
c0 {Ek , Λ, p}.
Property 1. c0 {Ek , Λ, p} is a linear space for any sequence p = (pk ).
Proof. Let x ∈ c0 {Ek , Λ, p}.
Then there exists r > 0 such that
(fk (rγk xk ))pk tk → 0, as k → ∞. Let λ ∈ C. Without loss of generality we can
take λ = 0. Let ρ = r(|λ|)−1 > 0, then we have
(fk (γk (λxk )ρ))pk tk = (fk (rγk xk ))pk tk → 0,
as
k → ∞.
Therefore λx ∈ c0 {Ek , Λ, p}, for all λ ∈ C and for all x ∈ c0 {Ek , Λ, p}.
Next we suppose that x, y ∈ c0 {Ek , Λ, p}. Then there exist r1 , r2 > 0 such
that
(fk (γk xk r1 ))pk tk → 0,
as
k→∞
VECTOR VALUED PARANORMED BOUNDED AND NULL SEQUENCE SPACES
317
and
(fk (γk yk r2 ))pk tk → 0,
as
k → ∞.
Thus given ε > 0, there exists k1 , k2 > 0 such that
(fk (γk xk r1 ))pk tk < pk ,
for all k ≥ k1
(1)
(fk (γk yk r2 ))pk tk < pk ,
for all k ≥ k2 .
(2)
and
Let r = r1 r2 (r1 + r2 )−1 and k0 = max(k1 , k2 ).
Then we have for all k ≥ k0 ,
[fk (γk (xk +yk )r)]pk ≤ [fk (γk xk r1 )r2 (r1 +r2 )−1 +fk (γk yk r2 )r1 (r1 +r2 )−1 ]pk < pk .
Hence x + y ∈ c0 {Ek , Λ, p}.
Thus c0 {Ek , Λ, p} is a linear space.
In view of the techniques applied in proving Property 1, the proof of the
following is a routine work.
Property 2. ∞ {Ek , Λ, p} is linear for any sequence p = (pk ).
Theorem 1. If p ∈ ∞ and each Ek is complete, then c0 {Ek , Λ, p} is a
complete paranormed space, paranormed by
pk
k
M
g(x) = sup(fk (γk xk p−t
k )) ,
k
where M = max(1, H), H = supk pk .
Proof. Clearly, for any x ∈ c0 {Ek , Λ, p}
g(x) ≥ 0,
g(θ) = 0
and g(−x) = g(x).
Now, let x, y ∈ c0 {Ek , Λ, p}. Then clearly g(x + y) ≤ g(x) + g(y).
Now we check the continuity of scalar multiplication. It can be shown by
standard techniques that α → 0, x → θ imply αx → θ and α fixed, x → θ imply
αx → θ.
Next we show that α → 0 and x fixed imply αx → θ.
pk
k
M
We have g(αx) = supk (fk (γk (αxk )p−t
k )) .
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BINOD CHANDRA TRIPATHY AND MAUSUMI SEN
Now, (xk ) ∈ c0 {Ek , Λ, p}. Then there exists r > 0 such that
min(1, r H )(fk (γk xk ))pk tk ≤ (fk (γk xk r))pk tk → 0
k → ∞.
as
(since min(1, r H ) ≤ r pk for all k ∈ N )
pk
k
M → 0
as
⇒ (fk (γk xk p−t
k ))
k → ∞.
⇒ Given > 0, there exists k0 ∈ N such that
pk
k
M < ,
(fk (γk xk p−t
k ))
for all k > k0 .
Since α → 0, without loss of generality let |α| < 1, then we have
pk
k
M < ,
(fk (γk (αxk )p−t
k ))
for all k > k0 .
Again x ∈ c0 {Ek , Λ, p} implies there exists L < ∞ such that
pk
k
M ≤ L,
(fk (γk xk p−t
k ))
pk
For k = 1, 2, 3, . . . , k0 , let |α| M < δ(=
L ),
pk
k
M < ,
(fk γk ((αxk )p−t
k ))
for k = 1, 2, 3, . . . , k0 .
then we have
for all k = 1, 2, 3, . . . , k0 .
Taking α small enough, we have
pk
k
M < ,
(fk (γk (αxk )p−t
k ))
for all k ∈ N.
So, g(αx) → 0 as α → 0.
Hence g is a paranorm on c0 {Ek , Λ, p}.
We now show that c0 {Ek , Λ, p} is complete under the paranorm g. For this,
let (x(n) ) be a Cauchy sequence in c0 {Ek , Λ, p}. Then given ε > 0, there exists
n0 ∈ N such that
g(x(n) − x(m )) < ,
for all n, m ≥ n0
M
⇒ fk (γk xk (n) pk −tk − γk xk (m) pk −tk ) < pk < ,
for all n, m ≥ n0
and for all
k ∈ N.
(3)
We write zk (n) = γk xk n) pk −tk , for all k ∈ N .
Then (zk (n) )n∈N is a Cauchy sequence in Ek for each k ∈ N . Since Ek ’s are
(n)
complete there exists zk ∈ Ek such that zk → zk as n → ∞, for all k ∈ N .
VECTOR VALUED PARANORMED BOUNDED AND NULL SEQUENCE SPACES
319
Since Ek ’s are linear, we can express zk as zk = γk xk pk −tk where xk ∈ Ek . Let
x = (xk ).
Taking m → ∞ in (3), we get
pk
(n)
k
M < ,
sup(fk (γk (xk − xk )p−t
k ))
k
⇒ g(x(n) − x) < ,
⇒x
(n)
→ x,
as
for all n ≥ n0 .
for all n ≥ n0 .
n → ∞.
(4)
Also (4) implies that x(n) − x ∈ c0 {Ek , Λ, p}. Since c0 {Ek , Λ, p} is linear, we
have x = (x(n) − x) + x(n) ∈ c0 {Ek , Λ, p}.
Hence c0 {Ek , Λ, p} is complete under the paranorm g.
Corollary 1. If 0 < inf pk ≤ sup pk < ∞, then ∞ {Ek , Λ, p} is paranormed
by g.
Proof. In view of Theorem 1, we need only to check that α → 0 and x fixed
imply αx → 0.
We have, (g(αx))M ≤ supk |α|pk (g(x))M .
Since α → 0, without loss of generality we can take |α| < 1. Now if inf pk = 0,
then the right side of the above inequality will be independent of α.
Hence the result.
Using Corollary 1 and following Theorem 1 one will get the following result.
Theorem 2. If 0 < inf pk ≤ sup pk < ∞ and each Ek is complete, then
∞ {Ek , Λ, p} is a complete paranormed space paranormed by g.
Proposition 3. The space c0 {Ek , Λ, p} and ∞ {Ek , Λ, p} are normal.
Proof. Let x be an element of either of the spaces and |αk | ≤ 1, for k =
1, 2, 3, . . ..
Since |αk |pk ≤ max(1, |αk |H ) ≤ 1, for all k ∈ N , so
(fk (γk (αk xk )r))pk tk ≤ (fk (γk xk r))pk tk .
Thus x ∈ X and |αk | ≤ 1 for all k ∈ N imply αx ∈ X where X is either
c0 {Ek , Λ, p} or ∞ {Ek , Λ, p}.
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BINOD CHANDRA TRIPATHY AND MAUSUMI SEN
Hence the result.
The next result follows immediately from Lemma 1 and Proposition 3.
Proposition 4. The spaces c0 {Ek , Λ, p} and ∞ {Ek , Λ, p} are monotone.
Theorem 5. Let (pk ) be a given sequence of strictly positive real numbers. Then (γk ) ∈ (E, E) if and only if ((γk )pk ) ∈ ∞ , where E = c0 {Ek , p} or
∞ {Ek , p} or c0 (Ek , p) or ∞ (Ek , p).
Proof. The sufficiency for all the cases is obvious.
/ ∞ .
For the necessity, for the case E = ∞ {Ek , p}, suppose that ((γk )pk ) ∈
Then there exists a subsequence ((γki )pki ) of ((γk )pk ) such that
(γki )pki → ∞,
as
i → ∞.
(5)
Then
[(fki (rγki xki ))pki tki ] → ∞,
as
i → ∞.
/ (∞ {Ek , p}, ∞ {Ek , p}).
Hence (γk ) ∈
pk
Thus ((γk ) ) ∈ ∞ is necessary for (γk ) ∈ (∞ {Ek , p}, ∞ {Ek , p}).
Next for E = c0 {Ek , p}, for the necessity, suppose that (5) holds. Then one
can find a sequence (xk ) ∈ c0 {Ek , p} such that
|(γki )pki | ≥ (fki (rxki ))pki tki
for all i ∈ N.
Then [(fki (rγki xki ))pki tki ] ≥ 1 for all i ∈ N.
/ (c0 {Ek , p}, c0 {Ek , p}). As such we arrive at a contradiction.
Thus (γk ) ∈
Hence ((γk )pk ) ∈ ∞ is necessary for (γk ) ∈ (c0 {Ek , p}, c0 {Ek , p}).
The other cases can be proved similarly.
The following result follows from Lemma 2 and Theorem 5.
Corollary 2. M (E, E) = ∞ , for E = c0 {Ek , p} or ∞ {Ek , p} or c0 (Ek , p)
or ∞ (Ek , p) if and only if h > 0 and H < ∞.
From Lemma 2 and Corollary 2, the following result follows.
Corollary 3. Let h = inf pk and H = sup pk . Then the following are
equivalent:
VECTOR VALUED PARANORMED BOUNDED AND NULL SEQUENCE SPACES
321
(i) H < ∞ and h > 0.
(ii) ∞ {Ek , Λ, p} = ∞ (Ek , Λ, p).
(iii) c0 {Ek , Λ, p} = c0 (Ek , Λ, p).
Theorem 6. If p, q ∈ ∞ . Then ∞ {Ek , Λ, p} ⊆ ∞ {Ek , Λ, q} if and only if
q
− pk
lim inf qk (M pk )
>0
k
k
(6)
for every integer M > 1.
Proof. Sufficiency. Let x ∈ ∞ {Ek , Λ, p}. Then there exists r > 0 such
that supk fk (γk xk r))pk tk < ∞. So there exists M = M (r) > 1 such that
fk (γk xk r))pk ≤ M pk for all k ∈ N.
(7)
Also (6) implies there exists α > 0 such that
q
− pk
qk (M pk )
>α
k
(8)
for all sufficiently large k.
Now,
qk
(fk (γk xk r))qk qk −1 ≤ (M pk ) pk qk−1
< α−1 ,
[by (7)]
for sufficiently large k [by (8)].
So, (xk ) ∈ ∞ {Ek , Λ, q}.
Hence ∞ {Ek , Λ, p} ⊆ ∞ {Ek , Λ, q}.
Necessity. Suppose ∞ {Ek , Λ, p} ⊆ ∞ {Ek , Λ, q} but there exists M0 > 1
such that
q
− pk
lim inf qk (M0 pk )
k
k
= 0.
Then there exists k1 < k2 < k3 < · · · such that
qk
i
−p
qki (M0 pki )
ki
< i−1 , i = 1, 2, 3, . . . .
Let H = supk qk < ∞. Define the sequence x = (xk ) as follows:
xk =

1
 (M0 pk ) pk

|γk |
θk ,
Ik ,
k = ki ,
otherwise,
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BINOD CHANDRA TRIPATHY AND MAUSUMI SEN
where θk is the zero element and Ik the identity of Ek for all k ∈ N.
Then (fk (γk xk 1))pk p−1
k = M0 , for every k = ki , i = 1, 2, 3, . . . .
Now (xk ) ∈ ∞ {Ek , Λ, p}, but
qk
(fk (γk xk r))qk qk−1 = (M0 pk ) pk r pk qk −1 ,
for every k = ki , i = 1, 2, 3, . . .
and for every r > 0,
> i min(1, r H ).
/ ∞ {Ek , Λ, q}. Hence we arrive at a contradiction.
Therefore (xk ) ∈
Thus (6) holds.
Theorem 7. Let p, q ∈ ∞ . Then c0 {Ek , Λ, p} ⊆ c0 {Ek , Λ, q} if and only if
qk
lim lim sup qk−1 (M −1 pk ) pk = 0.
M
k
(9)
qk
Proof. Let I(M ) = lim supk qk−1 (M −1 pk ) pk , for all M > 1 and I(M, k) =
qk
qk−1 (M −1 pk ) pk .
By (9) given > 0, there exists M0 > 1 such that
I(M ) < , for all M > M0 .
(10)
Let x ∈ c0 {Ek , Λ, p} and M ∗ be fixed with M ∗ > M0 . Then there exists
r > 0 and k0 ∈ N such that
∗ −1
, for all k > k0 .
(fk (γk xk r))pk p−1
k <M
qk
Now, (fk (γk xk r))qk qk−1 = qk−1 (M ∗ −1 pk ) pk < I(M ∗ ) < , for all k > k0 . [by (10)]
So, x ∈ c0 {Ek , Λ, p}.
Hence c0 {Ek , Λ, p} ⊆ c0 {Ek , Λ, q}.
Conversely suppose that c0 {Ek , Λ, p} ⊆ c0 {Ek , Λ, q} but (9) does not hold.
Then we have two cases:
(i) I(M ) = ∞ for every integer M > 1.
(ii) There exists M0 > 1 such that I(M ) < ∞, for all M > M0 and limM I(M ) >
0.
VECTOR VALUED PARANORMED BOUNDED AND NULL SEQUENCE SPACES
323
Case (i) Consider a strictly increasing sequence (ki ) of positive integers such
that
I(i + 1, ki ) > i, i = 1, 2, 3, . . . .
Define (xk ) as follows:

 (i+1)−tk pk tk I ,
k
|γk |
xk =
θ ,
k = ki ,
otherwise.
k
−1
So, (fk (γk xk 1))pk p−1
k = (i + 1) , for all k = ki , i = 1, 2, 3, . . . .
Hence (xk ) ∈ c0 {Ek , Λ, p}.
qk
q
− pk
Also, (fk (γk xk r))qk qk−1 = r qk qk −1 pk pk (i + 1)
k
≥ I(i + 1, ki ) min(1, r H ) , for k = ki and for all r > 0
(H = sup qk < ∞)
> i min(1, r
H
k
).
/ c0 {Ek , Λ, q}. Hence we arrive at a contradiction.
Therefore (xk ) ∈
Case (ii) Suppose limM >M0 I(M ) = 2a > 0. Therefore there exists a strictly
increasing sequence (ki ) of positive integers such that
I(M0 + i − 1, ki ) > a, i = 1, 2, 3, . . . .
We define a sequence x = (xk ) as follows:

t
 (M0 +i−1)−tk pkk
xk =
θ ,
k
|γk |
Ik ,
k = ki ,
otherwise.
−1
Then (fk (γk xk 1))pk p−1
k = (M0 + i − 1) , for all k = ki , i = 1, 2, 3, . . . .
So, (xk ) ∈ c0 {Ek , Λ, p}.
Also for k = ki , i = 1, 2, 3, . . ., we have
(fk (γk xk r))qk qk−1
qk
=
p
r qk qk−1 pk k
q
− pk
(M0 + i − 1)
k
≥ I(M0 + i − 1, ki ) min(1, r H ), for all r > 0
> a min(1, r H ).
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BINOD CHANDRA TRIPATHY AND MAUSUMI SEN
Thus (xk ) ∈
/ c0 {Ek , Λ, q}. Once again we arrive at a contradiction.
Therefore limM I(M ) = 0.
This completes the proof of the theorem.
Remark. Taking Ek ’s as normed linear spaces, normed by .
Ek , one will
get the analogue of the above results for ∞ {Ek , Λ, p} and c0 {Ek , Λ, p}. These
spaces are paranormed by
pk
k M
f (x) = sup{
γk xk Ek p−t
k } .
k
4. Duals of the Above Sequence Spaces
By 1 we denote the class of absolutely summable sequence spaces of complex
terms. The α-dual i.e. the Köthe-Toeplitz dual of a sequence space E of complex
terms is defined as
E α = {(yk ) ∈ w : (xk yk ) ∈ 1 for all (xk ) ∈ E}.
For any normed space E, the set of all continuous linear functions on E is
called its continuous dual, denoted by E ∗ . Clearly E ∗ is a linear space.
Throughout this section we take Ek ’s to be normed linear spaces, normed by
.
Ek for all k ∈ N . Then the Köthe-Toeplitz dual of Z(Ek ) is defined as
[Z(Ek )]α = {(yk ) : yk ∈ Ek∗ for all k ∈ N and (
xk Ek yk Ek∗ ) ∈ 1 }.
The following well-known result will be used for establishing the result of
this section.
Lemma 3. Let pk > 0 for all k ∈ N . Then
(i) [c0 (p)]α = M0 (p) = ∪J∈N {(ak ) :
∞
|ak |J −tk < ∞}.
k=1
∞
(ii) [∞ (p)]α = M∞ (p) = ∩J∈N {(ak ) :
|ak |J tk < ∞}.
k=1
Theorem 8. Let pk > 0 for all k ∈ N . Then
α
(i) [c0 (Ek , Λ, p)] = ∪J∈N {(ak ) :
∞
γk −1 ak Ek∗ J −tk < ∞}.
k=1
VECTOR VALUED PARANORMED BOUNDED AND NULL SEQUENCE SPACES
∞
(ii) [∞ (Ek , Λ, p)]α = ∩J∈N {(ak ) :
325
γk −1 ak Ek ∗ J tk < ∞}.
k=1
(iii) [c0 {Ek , Λ, p}]α = ∪J∈N {(ak ) :
∞
r −1 γk −1 ptkk ak Ek ∗ J −tk < ∞, for r > 0}.
k=1
∞
(iv) [∞ {Ek , Λ, p}]α = ∩J∈N {(ak ) :
t
∗ J k < ∞, for r > 0}.
r −1 γk −1 pk tk ak EK
k=1
Proof. The proof follows from Lemma 3 and the following expressions.
∞
xk Ek ak Ek ∗ =
k=1
=
∞
γk xk Ek γk −1 ak Ek ∗
k=1
∞
rγk xk tk tk Ek r −1 γk −1 pk tk ak Ek ∗ .
k=1
Acknowledgment
The authors thank the referees for their constructive comments on the first
draft of the article.
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Mathematical Sciences Division, Institute of Advanced Study in Science and Technology, Khanapara, Guwahati - 781 022, ASSAM, INDIA.
E-mail: [email protected]
E-mail: tripathybc@rediffmail.com