On the sets of sequences 19
2.3. Properties of some new sets of sequences.
In this subsection we shall characterize the sets E 1 µ, E 1
+
µ for E ∈
s
α
, s
◦
α
, s
• α
, and the sets w
p α
λ, w
+ p
α
λ and w
◦
+ p
α
λ. In order to state some new results we need the following lemmas. First recall the
well known result. L
EMMA
1. A ∈ c ,
c if and only if
A ∈ S
1
, lim
n
a
nm
= for each m ≥ 1.
The next result has been shown in [11]. L
EMMA
2. If 1
+
is bijective from s
α
into itself, then α ∈ cs. We also need to state the following elementary result.
L
EMMA
3. We have 6
+
1
+
X =
X ∀
X ∈ c and
1
+
6
+
X =
X ∀
X ∈ cs. Put now
w
+ p
α
λ =
X ∈ s C
+
λ |X |
p
∈ s
α
, w
◦
+ p
α
λ =
n X ∈ s C
+
λ |X |
p
∈ s
◦
α
o ,
see [11]. Letting β
−
= β
n−1 n≥1
, with β =
1, for any β = β
n n≥1
∈ U
+∗
, we can state the following results.
T
HEOREM
2. Let α ∈ U
+∗
, λ, µ ∈ U and p 0. We successively have i a s
α
1 µ = s
α |µ|
if and only if α ∈ c C
1
; b s
◦
α
1 µ = s
◦ α
|µ|
if and only if α ∈ c C
1
; c s
• α
1 µ = s
•
α |µ|
if and only if α ∈ b C.
ii a s
α
1
+
µ =
s
α |µ|
−
if and only if α
|µ| ∈ c
C
1
; b
α
n−1
α
n
µ
n
µ
n−1
= o 1 implies s
◦
α
1
+
µ =
s
◦ α
|µ| −
; c
α |µ|
∈ c C
+ 1
if and only if s
∗ α
1
+
µ =
s
α |µ|
; d
α |µ|
∈ c C
+ 1
if and only if s
◦
∗ α
1
+
µ =
s
◦ α
|µ|
.
20 B. de Malafosse
iii a s
α
6
+
= s
α
if and only if α ∈ c C
+ 1
and s
◦
α
6
+
= s
◦
α
if and only if α ∈ c C
+ 1
. b α ∈ c
C
+ 1
if and only if w
+ p
α
λ = s
α|λ| 1
p
, c if α ∈ c
C
+ 1
, then w
◦
+ p
α
λ = s
◦ α|λ|
1 p
, d α |λ| ∈ c
C
1
if and only if w
p α
λ = s
α|λ| 1
p
. e If α |λ| ∈ c
C
1
, then w
◦
p α
λ = s
◦ α|λ|
1 p
. Proof. Assertion i has been proved in [10]. Throughout the proof of part ii we shall
put β =
α |µ|
. Assertion ii a. First we have s
α
1
+
µ =
s
β
1
+
. Indeed, X ∈ s
α
1
+
µ ⇔
D
µ
1
+
X ∈ s
α
⇔ 1
+
X ∈ s
β
⇔ X ∈ s
β
1
+
. To get a, it is enough to show that β ∈ c
C
1
if and only if s
β
1
+
= s
β
−
. We assume that β ∈ c
C
1
. From the inequality β
n−1
β
n
≤ 1
β
n n
X
k=1
β
k
= O 1 ,
we deduce that β
n−1
β
n
= O 1 and 1
+
∈ s
β
−
, s
β
. Then for any given B ∈ s
β
the solutions of the equation 1
+
X = B are given by x
1
= − u and
3 −
x
n
= u +
n−1
X
k=1
b
k
, for n ≥ 2, where u is an arbitrary scalar. So there exists a real K 0, such that
|x
n
| β
n−1
= u +
n−1
P
k=1
b
k
β
n−1
≤ |u| + K
n−1
P
k=1
β
k
β
n−1
= O 1 ,
since iii in Proposition 1 implies
|u| β
n−1
= O 1. So X ∈ s
α
and we conclude that 1
+
is surjective from s
β
−
into s
β
. Then β =
α |µ|
∈ c C
1
implies s
α
1
+
µ =
s
α |µ|
−
. Conversely, assume that s
α
1
+
µ =
s
α |µ|
−
. If we take B = β, we get x
n
= x
1
−
n−1
P
k=1
β
k
, where x
1
is an arbitrary scalar and x
n
β
n−1
= x
1
β
n−1
− 1
β
n−1 n−1
X
k=1
β
k
= O 1 .
On the sets of sequences 21
Putting x
1
= 0, we conclude that β ∈ c
C
1
. ii b First we have 1
+
∈ s
◦
β
−
, s
◦
β
because β
n−1
β
n
= O 1. Let us show that
1
+
is surjective from s
◦
β
−
into s
◦
β
. For this, let B = b
n n≥1
∈ s
◦
β
. The solutions X = x
n n≥1
of the equation 1
+
X = B are given by 3. We have x
n
β
n−1
= o 1 −
n−1
P
k=1
b
k
β
n−1
, because from Proposition 2.1, the condition
β
n−1
β
n
= o 1 implies β ∈ c
C
1
and β → ∞
. Since B ∈ s
◦
β
there is a sequence ν = ν
n n≥1
∈ c
, such that b
n
= β
n
ν
n
. Then we have for a real M 0
n−1
P
k=1
b
k
β
n−1
≤ 1
β
n−1 n−1
X
k=1
β
k
ν
k
for all n ≥ 2.
It remains to show that
1 β
n−1
n−1
P
k=1
β
k
ν
k
= o 1. For this consider any given ε 0. Since
β → ∞ there is an integer N such that
S
n
= 1
β
n−1 N
X
k=1
β
k
ν
k
≤ ε
2 for n N , and
sup
k≥N +1
|ν
k
| ≤ ε
2 sup
n≥2
[C β β]
n−1
. Writing R
n
= 1
β
n−1 n−1
P
k=N +1
β
k
ν
k
for n N + 2, we deduce that
R
n
≤ sup
N +1≤k≤n−1
|ν
k
| [C β β]
n−1
≤ ε
2 .
Finally, we obtain |x
n
| β
n−1
= 1
β
n−1 N
X
k=1
β
k
ν
k
+ 1
β
n−1
n−1
X
k=N +1
β
k
ν
k
≤ S
n
+ R
n
≤ ε for
n ≥ N, and X ∈ s
◦
β−
. So we have proved ii b.
22 B. de Malafosse
Assertion ii c. Necessity. Assume that β =
α |µ|
∈ c C
+ 1
. Since we have
s
∗ α
1
+
µ =
s
∗ β
1
+
= s
β
, it is enough to show that 1
+
is bijective from s
β
to s
β
. We can write that 1
+
∈ s
β
, s
β
, since 4
β
n+1
β
n
≤ 1
β
n ∞
X
k=n
β
k
= O 1 n → ∞ .
Further, from s
β
⊂ cs, we deduce using Lemma 3 that for any given B ∈ s
β
, 1
+
6
+
B =
B. On the other hand 6
+
B =
∞
P
k=n
b
k n≥1
∈ s
β
, since β ∈ c C
+ 1
. So 1
+
is surjective from s
β
into s
β
. Finally, 1
+
is injective because the equation 1
+
X = O admits the unique solution X = O in s
β
, since K er 1
+
= ue
t
u ∈ C and e
t
∈ s
β
. Sufficiency. For every B ∈ s
β
the equation 1
+
X = B admits a unique solution in s
β
. Then from Lemma 2, β ∈ cs and since s
β
⊂ cs we deduce from Lemma 3 that
X = 6
+
B ∈ s
β
is the unique solution of 1
+
X = B. Taking B = β, we get 6
+
β ∈ s
α
that is β ∈ c C
+ 1
. As above to prove ii d it is enough to verify that β =
α |µ|
∈ c C
+ 1
if and only if s
◦
∗ β
1
+
= s
β
. If β ∈ c C
+ 1
, 1
+
is bijective from s
◦
β
into itself. Indeed, we have D
1 β
1
+
D
β
∈ c ,
c from 4 and Lemma 1. Furthermore, since β ∈ c
C
+ 1
we have s
◦
β
⊂ cs and for every B ∈ s
◦
β
, 1
+
6
+
B =
B. From Lemma 1, we have 6
+
∈ s
◦
β
, s
◦
β
, so the equation 1
+
X = B admits in s
◦
β
the solution X = 6
+
B and we have proved that 1
+
is surjective from s
◦
β
into itself. Finally, β ∈ c C
+ 1
implies that e
t
∈ s
◦
β
, so K er 1
+
T s
◦
β
= { 0} and we conclude that 1
+
is bijective from s
◦
β
into itself. iii a comes from ii, since α ∈ c
C
+ 1
if and only if 1
+
is bijective from s
α
into itself and is also bijective from s
◦
α
into itself, and 6
+
1
+
X = 1
+
6
+
X =
X for all X ∈ s
α
. b Assume that α ∈ c
C
+ 1
. Since C
+
λ = 6
+
D
1 λ
, we have w
+ p
α
λ = n
X ∈ s 6
+
D
1 λ
|X |
p
∈ s
α
o =
n X D
1 λ
|X |
p
∈ s
α
6
+
o ;
and since α ∈ c C
+ 1
implies s
α
6
+
= s
α
, we conclude that w
+ p
α
λ = X ∈ s |X |
p
∈ D
λ
s
α
= s
α|λ|
= s
α|λ| 1
p
.
On the sets of sequences 23
Conversely, we have α |λ|
1 p
∈ s
α|λ| 1
p
= w
+ p
α
λ. So C
+
λ h
α |λ|
1 p
i
p
=
∞
X
k=n
α
k
|λ
k
| |λ
k
|
n≥1
∈ s
α
, i.e. α ∈ c
C
+ 1
and we have proved i. We get iii c reasoning as above. iii d has been proved in [4]. iii e Assume that α |λ| ∈ c
C
1
. Then w
◦
p α
λ = n
X ∈ s |X |
p
∈ 1 λ s
◦
α
o .
Since 1 λ = 1D
λ
, we get 1 λ s
◦
α
= 1 s
◦
α|λ|
. Now, from i b we deduce that α |λ| ∈ c
C
1
implies that 1 is bijective from s
◦
α|λ|
into itself and w
α
λ = s
◦
α|λ|
1 p
. We get e reasoning as above.
As a direct consequence of Theorem 2 we obtain the following results given in [11]. C
OROLLARY
1. Let r 0 be any real. We get r 1 ⇔ s
r
1 = s
r
⇔ s
◦
r
1 = s
◦
r
⇔ s
r
1
+
= s
r
. We deduce from the previous section the following.
3. Sets of sequences that are strongly α−bounded and α−convergent to zero with index p and generalizations.