calculations on some sequence spaces - European Mathematical ...

IJMMS 2004:31, 1653–1670
PII. S0161171204209401
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CALCULATIONS ON SOME SEQUENCE SPACES
BRUNO DE MALAFOSSE
Received 16 September 2002
We deal with space of sequences generalizing the well-known spaces w p ∞(λ), c ∞ (λ,µ), replacing
the operators C(λ) and ∆(µ) by their transposes. We get generalizations of results
concerning the strong matrix domain of an infinite matrix A.
2000 Mathematics Subject Classification: 46A45, 40C05.
1. Notations and preliminary results. For a given infinite matrix A = (a nm ) n,m≥1 ,
the operators A n are defined, for any integer n ≥ 1, by
A n (X) =
∞∑
a nm x m , (1.1)
m=1
where X = (x n ) n≥1 , the series intervening in the second member being convergent. So
we are led to the study of the infinite linear system
A n (X) = b n , n= 1,2,..., (1.2)
where B = (b n ) n≥1 is a one-column matrix and X the unknown, see [1, 2, 3, 4, 5, 6, 7,
8, 10]. Equation (1.2) can be written in the form AX = B, where AX = (A n (X)) n≥1 .In
this paper, we will also consider A an operator from a sequence space into another
sequence space.
A Banach space E of complex sequences with the norm ‖‖ E is a BK space if each
projection P n X = x n is continuous for all X ∈ E. A BK space E is said to have AK, (see
[12, 13]), if B = ∑ ∞
m=1 b m e m , for every B = (b n ) n≥1 ∈ E, (withe n = (0,...,1,...),1being
in the nth position), that is,
∥
∞∑
m=N+1
b m e m
∥ ∥∥∥∥E
→ 0 (n →∞). (1.3)
We will write s for the set of all complex sequences, l ∞ , c, c 0 for the sets of bounded,
convergent, and null sequences, respectively. We will denote by cs and l 1 the sets of
convergent and absolutely convergent series, respectively.
In all that follows we will use the set
U +∗ = {( u n
)n≥1 ∈ s | u n > 0 ∀n } . (1.4)

1654 BRUNO DE MALAFOSSE
From Wilansky’s notations [15], we define for any sequence
α = ( α n
)n≥1 ∈ U +∗ , (1.5)
and for any set of sequences E, the set
( 1
α
) −1 { ∣ ( }
(xn ) ∣∣∣
∗E =
n≥1 ∈ s xn
∈ E . (1.6)
α n
)n
We will write α∗E instead of (1/α) −1 ∗E forshort.Soweget
⎧
s ⎪⎨
α ◦ if E = c 0 ,
α∗E = s α (c) if E = c,
⎪⎩
s α if E = l ∞ .
(1.7)
We have for instance
α∗c 0 = s ◦ α = {( x n
)n≥1 ∈ s | x n = o ( α n
)
n →∞
}
. (1.8)
Each of the spaces α∗E, where E ∈{c 0 ,c,l ∞ }, is a BK space normed by
( ∣ ∣ ) ∣xn ‖X‖ sα = sup , (1.9)
n≥1 α n
and s ◦ α has AK.
Now let α = (α n ) n≥1 and β = (β n ) n≥1 ∈ U +∗ . S α,β is the set of infinite matrices
A = (a nm ) n,m≥1 such that
(
anm α m
)
m≥1 ∈ l1 ∀n ≥ 1,
S α,β is a Banach space with the norm
∞∑ (∣ ∣ ) ( )
∣anm∣αm = O βn
m=1
(n →∞). (1.10)
⎛
‖A‖ Sα,β = sup⎝
n≥1
⎞
∞∑
∣
∣ anm α m ⎠ . (1.11)
β n
m=1
Let E and F be any subsets of s.WhenA maps E into F,wewillwriteA ∈ (E,F),see[11].
So for every X ∈ E, AX ∈ F, (AX ∈ F will mean that for each n ≥ 1 the series defined
by y n = ∑ ∞
m=1 a nm x m is convergent and (y n ) n≥1 ∈ F). It has been proved in [9] that
A ∈ (s α ,s β ) if and only if A ∈ S α,β . So we can write that (s α ,s β ) = S α,β .
When s α = s β , we obtain the unital Banach algebra S α,β = S α ,(see[1, 2, 3, 5, 6, 10])
normed by ‖A‖ Sα =‖A‖ Sα,α .
We also have A ∈ (s α ,s α ) if and only if A ∈ S α .If‖I −A‖ Sα < 1, we will say that A ∈ Γ α .
Since the set S α is a unital algebra, we have the useful result that if A ∈ Γ α , A is bijective
from s α into itself.

CALCULATIONS ON SOME SEQUENCE SPACES 1655
If α = (r n ) n≥1 , Γ α , S α , s α , sα ◦ ,ands(c) α are replaced by Γ r , S r , s r , sr ◦ , and s(c) r , respectively,
(see [1, 2, 3, 5, 6, 10]). When r = 1, we obtain s 1 = l ∞ , s1 ◦ = c 0, and s (c)
1 = c, and
putting e = (1,1,...), we have S 1 = S e . It is well known, see [11], that
(
s1 ,s 1
)
=
(
c0 ,s 1
)
=
(
c,s1
)
= S1 . (1.12)
For any subset E of s, weput
AE ={Y ∈ s |∃X ∈ E, Y = AX}. (1.13)
If F is a subset of s, wewilldenote
F(A)= F A ={X ∈ s | Y = AX ∈ F}. (1.14)
We can see that F(A)= A −1 F.
2. Some properties of the operators ∆ + and Σ + . Here we will deal with the operators
represented by C + (λ) and ∆ + (λ).
Let
U = {( u n
)n≥1 ∈ s | u n ≠ 0 ∀n } . (2.1)
We define C(λ) = (c nm ) n,m≥1 ,forλ = (λ n ) n≥1 ∈ U, by
⎧
⎪⎨
1
if m ≤ n,
c nm = λ n
⎪⎩
0 otherwise.
(2.2)
So, we put C + (λ) = C(λ) t . It can be proved that the matrix ∆(λ) = (c ′ nm) n,m≥1 with
⎧
λ ⎪⎨ n if m = n,
c nm ′ = −λ n−1 if m = n−1, n≥ 2,
⎪⎩
0 otherwise,
(2.3)
is the inverse of C(λ),see[12, 14]. Similarly, we put ∆ + (λ) = ∆(λ) t .Ifλ = e, weget
the well-known operator of first difference represented by ∆(e) = ∆ and it is usually
written as Σ = C(e). Note that ∆ = Σ −1 and Σ belong to any given space S R with R>1.
Writing D λ = (λ n δ nm ) n,m≥1 , (where δ nm = 0forn ≠ m and δ nn = 1 otherwise), we have
∆ + (λ) = D λ ∆ + .Soforanygivenα ∈ U +∗ , we see that if (α n−1 /α n )|λ n /λ n−1 |=O(1),
then ∆ + (λ) ∈ (s (α/|λ|) ,s α ).SinceKer∆ + (λ) ≠ 0, we are lead to define the set
sα
∗ (
∆ + (λ) ) (
= s α ∆ + (λ) ) ⋂
s(α/|λ|) = { X = ( )
x n n≥1 ∈ s (α/|λ|) | ∆ + }
(λ)X ∈ s α . (2.4)

1656 BRUNO DE MALAFOSSE
It can easily be seen that
s ∗ (α/|λ|)(
∆ + (e) ) = s ∗ (α/|λ|)(
∆
+ ) = s ∗ α
(
∆ + (λ) ) . (2.5)
2.1. Properties of the sequence C(α)α. We will use the following sets:
⎧
⎛
⎨ ∣ ∣∣∣
Ĉ 1 =
⎩ α ∈ U +∗ 1
n∑
⎝
α n
k=1
⎧
⎛
⎨ ∣ ∣∣∣
Ĉ =
⎩ α ∈ U +∗ 1
n∑
⎝
α n
⎧
⎨
Ĉ 1 + = ⎩ α ∈ U ⋂ +∗ cs
∣
k=1
1
α n
⎛
⎝
α k
⎞
⎠ = O(1) (n →∞)
⎫
⎬
⎭ ,
α k
⎞
⎠ ∈ c
⎫
⎬
⎭ ,
∞∑
k=n
{ ∣ ( ) }
∣∣∣
Γ = α ∈ U +∗ αn−1
lim < 1 ,
n→∞ α n
{ ∣ ( ) }
∣∣∣
Γ + = α ∈ U +∗ αn+1
lim < 1 .
n→∞ α n
α k
⎞
⎠ = O(1) (n →∞)
⎫
⎬
⎭ ,
(2.6)
Note that α ∈ Γ + if and only if 1/α ∈ Γ . We will see in Proposition 2.1 that if α ∈ Ĉ 1 , α
tends to infinity. On the other hand, we see that ∆ ∈ Γ α implies α ∈ Γ and α ∈ Γ if and
only if there is an integer q ≥ 1 such that
( )
αn−1
γ q (α) = sup < 1. (2.7)
n≥q+1 α n
We obtain the following results in which we put [C(α)α] n = ( ∑ n
k=1 α k)/α n .
Proposition 2.1. Let α ∈ U +∗ .Then
(i) α n−1 /α n → 0 if and only if [C(α)α] n → 1,
(ii) (a) α ∈ Ĉ implies that (α n−1 /α n ) n≥1 ∈ c,
(b) [C(α)α] n → l implies that α n−1 /α n → 1−1/l,
(iii) if α ∈ Ĉ 1 , there are K>0 and γ>1 such that
α n ≥ Kγ n ∀n, (2.8)
(iv) the condition α ∈ Γ implies that α ∈ Ĉ 1 and there exists a real b>0 such that
[
C(α)α
]n ≤ 1
1−χ +bχn for n ≥ q +1, χ= γ q (α) ∈ ]0,1[, (2.9)
(v) the condition α ∈ Γ + implies that α ∈ Ĉ+ 1 .

CALCULATIONS ON SOME SEQUENCE SPACES 1657
Proof.
Assume that α n−1 /α n → 0. Then there is an integer N such that
n ≥ N +1 ⇒ α n−1
α n
≤ 1 2 . (2.10)
So there exists a real K>0 such that α n ≥ K2 n for all n and
Then
⎛
n−1
1 ∑
⎝
α n
k=1
α k
=
α ( )
k
··· αn−1 1 n−k
≤ for N ≤ k ≤ n−1. (2.11)
α n α k+1 α n 2
⎞ ⎛
N−1
α k
⎠
1 ∑
= ⎝
α n
k=1
⎞
n−1
∑
α k
⎠ +
k=N
⎛
α k
≤ 1
N−1
∑
⎝
α n K2 n
k=1
⎞
n−1
∑ ( )
α k
⎠ 1 n−k
+ , (2.12)
2
k=N
and since ∑ n−1
k=N(1/2) n−k = 1−(1/2) n−N → 1, (n →∞), wededucethat
⎛ ⎞
n−1
1 ∑
⎝ α k
⎠ = O(1) (2.13)
α n
k=1
and ([C(α)α] n ) ∈ l ∞ . Using the identity
[
C(α)α
]n = α 1 +···+α n−1
α n−1
α n−1
α n
we get [C(α)α] n → 1. This proves the necessity.
Conversely, if [C(α)α] n → 1, then
[ ]
α n−1 C(α)α
n
= [ ]
−1
α n C(α)α n−1
+1 = [ C(α)α ] n−1
(
αn−1
)
+1, (2.14)
α n
→ 0. (2.15)
(ii) is a direct consequence of the identity (2.14).
(iii) We put Σ n = ∑ n
k=1 α k .ThenforarealM>1,
[
C(α)α
]n = Σ n
Σ n −Σ n−1
≤ M ∀n. (2.16)
So Σ n ≥ (M/(M −1))Σ n−1 and Σ n ≥ ( M/(M−1) ) n−1
α1 ∀n. Therefore, from
( )
α 1 M n−1
≤ [ C(α)α ] n
α n M −1
= Σ n
≤ M, (2.17)
α n
we conclude that α n ≥ Kγ n for all n, withK = (M −1)α 1 /M 2 and γ = M/(M−1)>1.

1658 BRUNO DE MALAFOSSE
(iv) If α ∈ Γ , there is an integer q ≥ 1 for which
So there is a real M ′ > 0 for which
k ≥ q +1 implies α k−1
α k
≤ χ

CALCULATIONS ON SOME SEQUENCE SPACES 1659
X = (x n,0 ) n≥1 in s α . Then there is a unique scalar x 1 such that
n−1
∑
x n,0 = x 1 − α k . (2.25)
k=1
So
∣ ⎛
⎞
∣ xni , 0 n i −1
=
1 ∑
⎝
α x1 − α k
⎠
ni ∣ α →∞ as i →∞, (2.26)
ni ∣
k=1
and X ∉ s α , which is contradictory.
We conclude that each of the properties e ∈ Ker∆ + ⋂ s α and e ∉ Ker∆ + ⋂ s α is impossible
and ∆ + is not bijective from s α into itself. This proves the lemma.
Lemma 2.6. For every X ∈ c 0 , Σ + (∆ + X) = X and for every X ∈ cs, ∆ + (Σ + X) = X.
Proof.
It can easily be seen that
[
Σ
+ ( ∆ + X )] n = ∞ ∑
(
xm −x m+1
)
= xn ∀X ∈ c 0 ,
m=n
[
∆
+ ( Σ + X )] ∞ n = ∑
∞∑
x m − x m = x n
m=n m=n+1
∀X ∈ cs.
(2.27)
We can assert the following result, in which we put α + = (α n+1 ) n≥1 and s ◦∗
α (∆ + ) =
s ◦ α(∆ + ) ⋂ s ◦ α. Note that from (2.5) we have
s ∗ α
(
∆ + (e) ) = s ∗ α
(
∆
+ ) = s α
(
∆
+ ) ⋂
sα . (2.28)
Theorem 2.7. (i) (a) s α (∆) = s α if and only if α ∈ Ĉ 1 ,
(b) sα(∆) ◦ = sα ◦ if and only if α ∈ Ĉ 1 ,
(c) s α
(c) (∆) = s α (c) if and only if α ∈ Ĉ.
(ii) (a) α ∈ Ĉ 1 if and only if s α +(∆ + ) = s α and ∆ + is surjective from s α into s α +,
(b) α ∈ Ĉ+ 1 if and only if s∗ α (∆ + ) = s α and ∆ + is bijective from s α into s α ,
(c) α ∈ Ĉ+ 1 implies that s◦∗ α (∆ + ) = sα ◦ and ∆ + is bijective from sα ◦ into sα.
◦
(iii) α ∈ Ĉ+ 1 if and only if s α(Σ + ) = s α and s α (Σ + ) = s α implies sα ◦ (Σ+ ) = sα ◦ .
Proof. (i) has been proved in [8].
(ii)(a) Sufficiency. If ∆ + is surjective from s α into s α +, then for every B ∈ s α +
solutions of ∆ + X = B in s α are given by
the
n∑
x n+1 = x 1 − b k n = 1,2,..., (2.29)
k=1

1660 BRUNO DE MALAFOSSE
where x 1 is arbitrary. If we take B = α + ,wegetx n = x 1 − ∑ n
k=2 α k.So
⎛
x n
= x 1
− 1 ⎝
α n α n α n
n∑
k=2
α k
⎞
⎠ = O(1). (2.30)
Taking x 1 =−α 1 , we conclude that ( ∑ n−1
k=1 α k )/α n = O(1) and α ∈ Ĉ 1 .
Conversely, assume that α ∈ Ĉ 1 . From the inequality
α n−1
α n
≤ 1 α n
⎛
⎝
n∑
k=1
α k
⎞
⎠ = O(1), (2.31)
we deduce that α n−1 /α n = O(1) and ∆ + ∈ (s α ,s α +). ThenforanygivenB ∈ s α +,the
solutions of the equation ∆ + X = B are given by x 1 =−u and
n−1
∑
−x n = u+ b k for n ≥ 2, (2.32)
where u is an arbitrary scalar. So there exists a real K>0 such that
∣ ∣ xn ∣
∑ n−1 u+
=
α n
k=1 b k
α n
∣
k=1
≤ |u|+K(∑ n
k=2 α )
k
= O(1) (2.33)
α n
and X ∈ s α . We conclude that ∆ + is surjective from s α into s α +.
(ii)(b) Necessity. Assume that α ∈ Ĉ+ 1 .Then∆+ ∈ (s α ,s α ),since
α n+1
α n
≤ 1 α n
⎛
⎝
∞∑
k=n
α k
⎞
⎠ = O(1) (n →∞). (2.34)
Further, from s α ⊂ cs, we deduce, using Lemma 2.4, that for any given B ∈ s α ,
∆ +( Σ + B ) = B. (2.35)
On the other hand, Σ + B = ( ∑ ∞
k=n b k ) n≥1 ∈ s α ,sinceα ∈ Ĉ+ 1 .So∆+ is surjective from s α
into s α . Finally, ∆ + is injective because the equation
∆ + X = O (2.36)
admits the unique solution X = O in s α ,since
Ker∆ + = { ue t | u ∈ C } (2.37)
and e t ∉ s α .

CALCULATIONS ON SOME SEQUENCE SPACES 1661
Sufficiency. For every B ∈ s α , the equation ∆ + X = B admits a unique solution in
s α .ThenfromLemma 2.5, α ∈ cs and since s α ⊂ cs, we deduce from Lemma 2.6 that
X = Σ + B ∈ s α is the unique solution of ∆ + X = B. TakingB = α, wegetΣ + α ∈ s α ,thatis,
α ∈ Ĉ+ 1 .
(ii)(c) If α ∈ Ĉ+ 1 , ∆+ is bijective from sα ◦ into itself. Indeed, we have D 1/α ∆ + D α ∈ (c 0 ,c 0 )
from (2.34) andLemma 2.4. Furthermore, since α ∈ Ĉ+ 1 we have s◦ α ⊂ cs and for every
B ∈ sα,
◦
∆ +( Σ + B ) = B. (2.38)
From Lemma 2.4, we have Σ + ∈ (sα,s ◦ α), ◦ so the equation ∆ + X = B admits the solution
X 0 = Σ + B in sα ◦ and we have proved that ∆ + is surjective from sα ◦ into itself. Finally,
α ∈ Ĉ+ 1 implies that et ∉ sα,soKer∆ ⋂ ◦ + sα ◦ ={0} and we conclude that ∆ + is bijective
from sα ◦ into itself.
(iii) comes from (ii), since α ∈ Ĉ+ 1 if and only if ∆+ is bijective from s α into itself and
Σ +( ∆ + X ) = ∆ +( Σ + X ) = X ∀X ∈ s α . (2.39)
As a direct consequence of Theorem 2.7 we obtain the following results.
Corollary 2.8.
Let R be any real > 0. Then
R>1 ⇐⇒ s R (∆) = s R ⇐⇒ s ◦ R (∆) = s◦ R ⇐⇒ s R(
∆
+ ) = s R . (2.40)
Proof. From (i) and (ii) in Theorem 2.7, we see that it is enough to prove that α =
(R n ) n≥1 ∈ Ĉ 1 if and only if R>1. We have (R n ) n≥1 ∈ Ĉ 1 if and only if R ≠ 1 and
R −n ⎛
⎝
n∑
k=1
⎞
R k ⎠
1 =
1−R R−n+1 −
This means that R>1 and the corollary is proved.
R = O(1)
1−R
as n →∞. (2.41)
Using the notation α − = (1,α 1 ,α 2 ,...,α n−1 ,...) we get the next result.
Corollary 2.9. Let α ∈ U +∗ and µ ∈ U. Then
(i) α/|µ|∈Ĉ 1 if and only if
s α
(
∆ + (µ) ) = s (α/|µ|) −, (2.42)
(ii) α/|µ|∈Ĉ+ 1
if and only if sα
∗ (
∆ + (µ) ) = s (α/|µ|) . (2.43)
Proof.
First we have
s α
(
∆ + (µ) ) = s (α/|µ|)
(
∆
+ ) . (2.44)

1662 BRUNO DE MALAFOSSE
Indeed,
X ∈ s α
(
∆ + (µ) ) ⇐⇒ D µ ∆ + X ∈ s α ⇐⇒ ∆ + X ∈ s (α/|µ|) ⇐⇒ X ∈ s (α/|µ|)
(
∆
+ ) . (2.45)
Now, if α/|µ| ∈ Ĉ 1 , from (i) in Theorem 2.7, we have s (α/|µ|) (∆ + ) = s (α/|µ|) − and
s α (∆ + (µ)) = s (α/|µ|) −. Conversely, assume s α (∆ + (µ)) = s (α/|µ|) −. Reasoning as above,
we get s (α/|µ|) (∆ + ) = s (α/|µ|) −, and using (i) in Theorem 2.7 we conclude that α/|µ|∈Ĉ 1
and (i) holds.
(ii) α/|µ|∈Ĉ+ 1 implies that ∆+ is bijective from s (α/|µ|) into itself. Thus
s ∗ α
(
∆ + (µ) ) = s ∗ (α/|µ|)(
∆
+ ) = s (α/|µ|) . (2.46)
This proves the necessity. Conversely, assume that sα ∗ (∆ + (µ))=s (α/|µ|) .Thens ∗ (α/|µ|) (∆+ )
= s (α/|µ|) and from Theorem 2.7(ii)(b), α/|µ|∈Ĉ+ 1 and (ii) holds.
2.3. Spaces w p α(λ) and w +p
α (λ) for given real p>0. Here we will define sets generalizing
the well-known sets
w p ∞(λ) = { X ∈ s | C(λ) ( |X| p) ∈ l ∞
}
,
w p 0 (λ) = { X ∈ s | C(λ) ( |X| p) ∈ c 0
}
,
(2.47)
see [9, 12, 13, 14, 15]. It is proved that each of the sets w p 0 = wp 0 ((n) n) and w∞ p =
w∞((n) p n ) is a p-normed FK space for 0

CALCULATIONS ON SOME SEQUENCE SPACES 1663
Theorem 2.10. (i) (a) The condition α ∈ Ĉ+ 1
is equivalent to
w +p
α (λ) = s (αλ) 1/p . (2.51)
(b) If α ∈ Ĉ+ 1 , then
w ◦p
α (λ) = s ◦ (αλ) 1/p . (2.52)
(ii) (a) The condition αλ ∈ Ĉ 1 is equivalent to
w p α(λ) = s (αλ) 1/p . (2.53)
(b) If αλ ∈ Ĉ 1 , then
w ◦+p
α (λ) = s ◦ (αλ) 1/p . (2.54)
Proof.
Assume that α ∈ Ĉ+ 1 .SinceC+ (λ) = Σ + D 1/λ , we have
w +p
α (λ) = { X | ( Σ + D 1/λ
)(
|X|
p ) ∈ s α
}
=
{
X | D1/λ
(
|X|
p ) ∈ s α
(
Σ
+ )} , (2.55)
and since α ∈ Ĉ+ 1
implies s α(Σ + ) = s α , we conclude that
w +p
α (λ) = { X ||X| p ∈ D λ s α = s αλ
}
= s(αλ) 1/p . (2.56)
Conversely, we have (αλ) 1/p ∈ s (αλ) 1/p = w α
+p
(λ). So
⎛
C + (λ) [ (αλ) 1/p] ∞∑
p
= ⎝
k=n
α k λ k
λ k
⎞
⎠
n≥1
∈ s α , (2.57)
that is, α ∈ Ĉ+ 1 and we have proved (i). We obtain (i)(b) by reasoning as above.
(ii) Assume that αλ ∈ Ĉ 1 .Then
w p α(λ) = { X ||X| p ∈ ∆(λ)s α
}
. (2.58)

1664 BRUNO DE MALAFOSSE
Since ∆(λ) = ∆D λ ,weget∆(λ)s α = ∆s αλ . Now, from αλ ∈ Ĉ 1 we deduce that ∆ is
bijective from s αλ into itself and w p α(λ) = s (αλ) 1/p . Conversely, assume that w p α(λ) =
s (αλ) 1/p .Then(αλ) 1/p ∈ s (αλ) 1/p implies that
C(λ)(αλ) ∈ s α , (2.59)
and since D 1/α C(λ)(αλ) ∈ s 1 = l ∞ , we conclude that C(αλ)(αλ) ∈ l ∞ . The proof of
(ii)(b) follows the same lines as in the proof of the necessity in (ii) replacing s αλ by s ◦ αλ .
3. New sets of sequences of the form [A 1 ,A 2 ]. In this section, we will deal with the
sets
[
A1 (λ),A 2 (µ) ] = { X ∈ s | A 1 (λ) (∣ ∣ A2 (µ)X ∣ ) }
∈ s α , (3.1)
where A 1 and A 2 are of the form C(ξ), C + (ξ), ∆(ξ), or∆ + (ξ) and we give necessary
conditions to get [A 1 (λ),A 2 (µ)] in the form s γ .
Let λ and µ ∈ U +∗ . For simplification, we will write throughout this section
[ ] [
A1 ,A 2 = A1 (λ),A 2 (µ) ] = { X ∈ s | A 1 (λ) (∣ ∣ A2 (µ)X ∣ ) }
∈ s α (3.2)
for any matrices
A 1 (λ) ∈ { ∆(λ),∆ + (λ),C(λ),C + (λ) } ,
A 2 (µ) ∈ { ∆(µ),∆ + (µ),C(µ),C + (µ) } .
(3.3)
So we have for instance
[C,∆] = { X ∈ s | C(λ) (∣ ∣) } (
∣ ∆(µ)X ∈ sα = wα (λ) ) ∆(µ) ,.... (3.4)
In all that follows, the conditions ξ ∈ Γ ,or1/η ∈ Γ for any given sequences ξ and η can
be replaced by the conditions ξ ∈ Ĉ 1 and η ∈ Ĉ+ 1 .
3.1. Spaces [C,C], [C,∆], [∆,C],and[∆,∆]. For the convenience of the reader we
will write the following identities, where A 1 (λ) and A 2 (µ) are lower triangles and we
will use the convention µ 0 = 0:
⎧
⎛
⎛ ⎞
⎞
⎫
⎨ ∣ ∣∣∣
[C,C] =
⎩ X ∈ s 1
n∑
⎝
1
m∑
⎬
⎝
λ x k
⎠
⎠
n ∣ µ = αn O(1)
m=1 m ∣
⎭ ,
k=1
⎧
⎛
⎞
⎫
⎨ ∣ ∣∣∣
[C,∆] =
⎩ X ∈ s 1
n∑
∣ ⎬
⎝ ∣ µk x k −µ k−1 x k−1 ⎠ = αn O(1)
λ n
⎭ ,
k=1
⎧
∣ ⎛ ⎞
⎛ ⎞
⎫
⎨ ∣ ∣∣∣∣∣ ∣∣∣
n−1
[∆,C]=
⎩ X ∈ s 1 ∑
−λ n−1
⎝ x i
⎠
µ n−1 ∣ +λ 1
n∑
⎬
⎝
n
x i
⎠
∣ µ n ∣ = α nO(1)
⎭ ,
k=1
[∆,∆] = { X ∈ s |−λ n−1
∣ ∣µn−1
x n−1 −µ n−2 x n−2
∣ ∣+λn
∣ ∣µn x n −µ n−1 x n−1
∣ ∣ = αn O(1) } .
(3.5)
k=1

CALCULATIONS ON SOME SEQUENCE SPACES 1665
Note that for α = e and λ = µ, [C,∆] is the well-known set of sequences that are strongly
bounded, denoted by c ∞ (λ), see[9, 12, 13, 14, 15]. We get the following result.
Theorem 3.1.
(i) If αλ and αλµ ∈ Γ , then
[C,C] = s (αλµ) , (3.6)
(ii) if αλ ∈ Γ , then
[C,∆] = s (α(λ/µ)) , (3.7)
(iii) if α and αµ/λ ∈ Γ , then
[∆,C]= s (α(µ/λ)) , (3.8)
(iv) if α and α/λ ∈ Γ , then
[∆,∆] = s (α(µ/λ)) . (3.9)
Proof.
We have for any given X
C(λ) (∣ ∣ C(µ)X
∣ ∣
)
∈ sα (3.10)
if and only if C(µ)X ∈ s α
(
C(λ)
)
= s(αλ) ,sinceαλ ∈ Γ .Soweget
X ∈ ∆(µ)s αλ (3.11)
and the condition αλµ ∈ Γ implies ∆(µ)s αλ = s (αλµ) , which permits us to conclude (i).
(ii) Now, for any given X, the condition C(λ)(|∆(µ)X|) ∈ s α is equivalent to
∣
∣ ∆(µ)X ∈ ∆(λ)sα = ∆s αλ = s αλ , (3.12)
since αλ ∈ Γ . Thus
X ∈ C(µ)s αλ = D 1/µ Σs αλ = s (α(λ/µ)) . (3.13)
(iii) Similarly, ∆(λ)(|C(µ)X|) ∈ s α if and only if
∣ ( )
∣ C(µ)X ∈ sα ∆(λ) = C(λ)sα = D 1/λ Σs α = s (α/λ) , (3.14)
since α ∈ Γ .So
X ∈ ∆(µ)s (α/λ) = ∆s (αµ/λ) . (3.15)
We conclude since αµ/λ ∈ Γ implies that ∆s (αµ/λ) = s (αµ/λ) . (iv) Here,
if α ∈ Γ . Thus we have
∆(λ) (∣ ∣ ∆(µ)X
∣ ∣
)
∈ sα if and only if ∆(µ)X ∈ C(λ)s α = s (α/λ) , (3.16)
since α/λ ∈ Γ . So (iv) holds.
X ∈ C(µ)s (α/λ) = s (α/λµ) (3.17)

1666 BRUNO DE MALAFOSSE
Remark 3.2.
If we define
[
A1 ,A 2
]
0 = { X ∈ s | A 1 (λ) (∣ ∣ A2 (µ)X ∣ ∣ ) ∈ s ◦ α}
, (3.18)
we get the same results as in Theorem 3.1, replacing in each case (i), (ii), (iii), and (iv) s ξ
by s ◦ ξ .
3.2. Sets [∆,∆ + ], [∆,C + ], [C,∆ + ], [∆ + ∆], [∆ + ,C], [∆ + ∆ + ], [C + ,C], [C + ,∆], [C + ,∆ + ],
and [C + ,C + ]. We get immediately from the definitions of the operators ∆(ξ), ∆ + (η),
C(ξ), and C + (η), the following:
[
∆,∆
+ ] = { ∣ ∣ ∣ ∣
X | λ ∣µn n x n −µ n+1 x ∣−λn−1∣µn−1 n+1 x n−1 −µ n x n = αn O(1) } ,
⎧
∣ ∣ ∣ ⎫
[
∆,C
+ ] ⎨
=
⎩ X | λ ∞∑ ∣∣∣∣∣ ∣∣∣∣∣ x i
∑
∞ ∣∣∣∣∣
x
⎬
i
n
−λ n−1 = α n O(1)
∣ µ
i=n i µ
i=n−1 i
⎭ ,
⎧ ⎛
⎞
⎫
[
C,∆
+ ] ⎨ ∣ ∣∣∣
=
⎩ X 1
n∑
∣ ⎬
⎝ ∣ µk x k −µ k+1 x k+1 ⎠ = αn O(1)
λ n
⎭ ,
k=1
[
∆ + ,∆ ] = { ∣ ∣ ∣ ∣
X | λ ∣µn n x n −µ n−1 x ∣−λn+1∣µn+1 n−1 x n+1 −µ n x n = αn O(1) } ,
⎧ ∣ ∣ ∣ ∣ ⎫
[
∆ + ,C ] ⎨ ∣ ∣∣∣∣∣ ∣∣∣
=
⎩ X λ n n
∑ ∣∣∣∣∣ ∣∣∣∣∣
x i − λ n+1
∑ ∣∣∣∣∣ ⎬
n+1
x i = α n O(1)
µ n µ
i=1
n+1
⎭ ,
i=1
[
∆ + ,∆ +] = { ∣ ∣ ∣ ∣
X | λ ∣µn n x n −µ n+1 x ∣−λn+1∣µn+1 n+1 x n+1 −µ n+2 x n+2 = αn O(1) } ,
⎧ ⎛ ∣ ∣⎞
⎫
[
C + ,C ] ⎨ ∣ ∣∣∣
∞
=
⎩ X ∑ ∣∣∣∣∣
⎝ 1 1 ∑
k ∣∣∣∣∣ ⎬
x i
⎠ = αn O(1)
λ
k=n k µ k
⎭ ,
i=1
⎧
⎫
[
C + ,∆ ] ⎨ ∣ ∣∣∣
∞
=
⎩ X ∑ ( )
1 ∣ ∣ ⎬
∣µk x k −µ k−1 x k−1 = α n O(1)
λ
k=n k
⎭ ,
⎧
⎫
[
C + ,∆ +] ⎨ ∣ ∣∣∣
∞
=
⎩ X ∑ ( )
1 ∣ ∣ ⎬
∣µk x k −µ k+1 x k+1 = α n O(1)
λ
k=n k
⎭ ,
⎧ ⎛ ∣ ∣⎞
⎫
[
C + ,C +] ⎨ ∣ ∣∣∣
∞
=
⎩ X ∑ ∣∣∣∣∣
⎝ 1 ∑
∞ ∣∣∣∣∣
x
⎬
i ⎠ = αn O(1)
λ
k=n k µ
i=k i
⎭ . (3.19)
We can assert the following result, in which we do the convention α n = 1forn ≤ 0.
Theorem 3.3.
(i) Assume that α ∈ Γ .Then
[
∆,∆
+ ] = s (α/λµ) − if α λµ ∈ Γ ,
[
∆,C
+ ] = s (α(µ/λ)) if λ (3.20)
α ∈ Γ .
(ii) The conditions αλ ∈ Γ and αλ/µ ∈ Γ together imply
[
C,∆
+ ] = s (α(λ/µ)) −. (3.21)

CALCULATIONS ON SOME SEQUENCE SPACES 1667
(iii) The condition α/λ ∈ Γ implies
[
∆ + ,∆ ] = s (αn−1 /µ nλ n−1 ) n
= s (1/µ(α/λ) − ). (3.22)
(iv) If α/λ and µ(α/λ) − = (µ n (α n−1 /λ n−1 )) n ∈ Γ , then
[
∆ + ,C ] = s µ(α/λ) −. (3.23)
(v) If α/λ and 1/µ(α/λ) − = (α n−1 /µ n λ n−1 ) n ∈ Γ , then
[
∆ + ,∆ +] = s ((α/λ) − /µ) − = s (α n−2 /λ n−2 µ n−1 ) n
. (3.24)
(vi) If 1/α and αλµ ∈ Γ , then
[
C + ,C ] = s (αλµ) . (3.25)
(vii) If 1/α and αλ ∈ Γ , then
[
C + ,∆ ] = s (α(λ/µ)) . (3.26)
(viii) If 1/α and α(λ/µ) ∈ Γ , then
[
C + ,∆ +] = s (α(λ/µ)) −. (3.27)
(ix) If 1/α and 1/αλ ∈ Γ , then
[
C + ,C +] = s (αλµ) . (3.28)
Proof.
(i) First, for any given X, the condition ∆(λ)(|∆ + (µ)X|) ∈ s α is equivalent to
∣ ∆ + (µ)X ∣ ( ) ∈ sα ∆(λ) = s(α/λ) , (3.29)
since α ∈ Γ .SoX ∈ s αλ (∆ + (µ)) and applying Corollary 2.9, we conclude the first part
of the proof of (i).
We have ∆(λ)(|C + (µ)X|) ∈ s α if and only if
∣ C + (µ)X ∣ ∈ C(λ)sα = D 1/λ Σs α . (3.30)
Since α ∈ Γ , we have Σs α = s α and D 1/λ Σs α = s (α/λ) . Then, for α/λ ∈ Γ + , X ∈ [∆,C + ] if
and only if
X ∈ w +1
(α/λ) (µ) = s (α(µ/λ)). (3.31)
(ii) For any given X, C(λ)(|∆ + (µ)X|) ∈ s α is equivalent to
and since αλ ∈ Γ we have w 1 α(λ) = s αλ .So
if αλ/µ ∈ Γ . Then (ii) is proved.
∆ + (µ)X ∈ w 1 α(λ), (3.32)
X ∈ s αλ
(
∆ + (µ) ) = s (α(λ/µ)) − (3.33)

1668 BRUNO DE MALAFOSSE
(iii) Here, ∆ + (λ)(|∆(µ)X|) ∈ s α if and only if
∣ (
∣ ∆(µ)X ∈ sα ∆ + (λ) ) = s (α/λ) −, (3.34)
since α/λ ∈ Γ . Thus
X ∈ C(µ)s (α/λ) − = D 1/µ Σs (α/λ) − = s (αn−1 /λ n−1 µ n) (3.35)
if (α/λ) − ∈ Γ , that is, α/λ ∈ Γ .
(iv) If α/λ ∈ Γ ,weget
∆ + (λ) (∣ ∣ C(µ)X
∣ ∣
)
∈ sα ⇐⇒ ∣ ∣ C(µ)X
∣ ∣ ∈ sα
(
∆ + (λ) )
= s (α/λ) − ⇐⇒ X ∈ ∆(µ)s (α/λ) −.
(3.36)
Since µ(α/λ) − ∈ Γ , we conclude that [∆ + ,C]= s (µ(α/λ) − ).
(v) One has
[
∆ + ,∆ +] = { X | ∆ + (
(µ)X ∈ s α ∆ + (λ) )} , (3.37)
and since α/λ ∈ Γ ,weget
We deduce that if α/λ ∈ Γ ,
s α
(
∆ + (λ) ) = s (α/λ) −. (3.38)
[
∆ + ,∆ +] (
= s (α/λ) − ∆ + (µ) ) . (3.39)
Then, from Corollary 2.9, ifα/λ ∈ Γ and (α/λ) − /µ = (α n−1 /λ n−1 µ n ) n ∈ Γ ,
(vi) We have
s (α/λ) −(
∆ + (µ) ) = s ((α/λ) − /µ) − = s (α n−2 /λ n−2 µ n−1 ) n
. (3.40)
C + (λ) (∣ ∣ C(µ)X
∣ ∣
)
∈ sα ⇐⇒ C(µ)X ∈ w +1
α (λ), (3.41)
and since α ∈ Γ + , we have w α
+1 (λ) = s αλ .Thenforαλµ ∈ Γ , X ∈ [C + ,C] if and only if
(vii) The condition C + (λ)(|∆(µ)X|) ∈ s α is equivalent to
and since α ∈ Γ + , we have w α
+1 (λ) = s αλ . Thus
since αλ ∈ Γ . So (vii) holds.
X ∈ ∆(µ)s αλ = s (αλµ) . (3.42)
∆(µ)X ∈ w +1
α (λ), (3.43)
X ∈ s αλ
(
∆(µ)
)
= D1/µ Σs αλ = s (α(λ/µ)) , (3.44)

CALCULATIONS ON SOME SEQUENCE SPACES 1669
(viii) First, we have
[
C + ,∆ +] = { X | ∆ + (µ)X ∈ w +1
α (λ) } , (3.45)
and the condition α ∈ Γ + implies that w α
+1 (λ) = s αλ . Thus
[
C + ,∆ +] = { X | ∆ + } (
(µ)X ∈ s αλ = sαλ ∆ + (µ) ) , (3.46)
and we conclude since
s αλ
(
∆ + (µ) ) = s (αλ/µ) −
for αλ
µ ∈ Γ . (3.47)
(ix) If α ∈ Γ + ,
[
C + ,C +] = { X | C + (µ)X ∈ w α +1 }
(λ) = s αλ = w
+1
αλ
(µ). (3.48)
We conclude that w +1
αλ (µ) = s (αλµ), sinceαλ ∈ Γ + .
Remark 3.4. Note that in Theorem 3.3, we have [A 1 ,A 2 ] = s α (A 1 A 2 ) = (s α (A 1 )) A2
for A 1 ∈{∆(λ),∆ + (λ),C(λ),C + (λ)} and A 2 ∈{∆(µ),∆ + (µ),C(µ),C + (µ)}. For instance,
we have
⎧
⎫
⎨ ∣ ( ∣∣∣
[∆,C]=
⎩ X λn
− λ ) n−1
n−1 ∑
x i + λ ⎬
n
αµ
x n = α n O(1) for
µ n µ n−1 µ
i=1 n
⎭ λ ∈ Γ . (3.49)
Similarly, under the corresponding conditions given in Theorems 3.1 and 3.3, weget
[∆,∆] = { ( )
X |−λ n−1 µ n−2 x n−2 +µ n−1 λn +λ n−1 xn−1 −λ n µ n x n = α n O(1) } ,
⎧
⎫
[
∆,C
+ ] ⎨ ∣ ∣∣∣
=
⎩ X λ n
x n + ( )
∞∑ x
⎬
m
λ n −λ n−1 = α n O(1)
µ n µ
m=n−1 m
⎭ ,
[
∆,∆
+ ] = { ( )
X |−λ n−1 µ n−1 x n−1 +µ n λn +λ n−1 xn −λ n µ n+1 x n+1 = α n O(1) } ,
(3.50)
[
∆ + ,∆ ] = { X |−λ n µ n−1 x n−1 + ( λ n +λ n+1
)
µn x n −λ n+1 µ n+1 x n+1 = α n O(1) } .
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Bruno de Malafosse: Laboratoire de Mathématiques Appliquées du Havre (LMAH), Université
du Havre, Institut Universitaire de Technologie du Havre, 76610 Le Havre, France
E-mail address: bdemalaf@wanadoo.fr