# Multipliers on Generalized Mixed Norm Sequence Spaces

Authors: Oscar Blasco; Carme Zaragoza-Berzosa
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Given1
≤
p
,
q
≤
∞ and sequences of integers (
n
k
)
k and (
n
k
′
)
k such that n
k
≤
n
k
′
≤
n
k
+
1, the generalized mixed norm space ℓ
ℐ
(
p
,
q
) is defined as those sequences (
a
j
)
j such that (
(
∑
j
∈
I
k
‍
|
a
j
|
p
)
1
/
p
)
k
∈
ℓ
q where I
k
=
{
j
∈
ℕ
0
s
.
t
.
n
k
≤
j
<
n
k
′
}
,
k
∈
ℕ
0. The necessary and sufficient conditions for a sequence λ
=
(
λ
j
)
j to belong to the space of multipliers (
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
), for different sequences ℐ and 𝒥 of intervals in ℕ
0, are determined.


---

## Body

## 1. Introduction

Let𝒮 be the space of complex valued sequences with the locally convex vector topology given by means of the seminorms p
j
(
λ
)
=
|
λ
j
| where λ
=
(
λ
j
)
j
∈
ℕ
0. Given two Banach spaces A
,
B continuously contained in 𝒮, we write (
A
,
B
) for the space of multipliers from A into B. More precisely,
(1)
(
A
,
B
)
=
{
λ
=
(
λ
j
)
j
:
(
λ
j
a
j
)
j
∈
B
;
∀
(
a
j
)
j
∈
A
}
.
We will use the notation supp
⁡
(
a
)
=
{
n
∈
ℕ
0
:
a
n
≠
0
} and λ
*
a for the sequence (
λ
j
a
j
)
j
∈
ℕ
0 where λ
=
(
λ
j
)
j
∈
ℕ
0 and a
=
(
a
j
)
j
∈
ℕ
0.Of course for the classicalℓ
p spaces, one easily sees that (
ℓ
p
1
,
ℓ
p
2
)
=
ℓ
p where 1
/
p
=
(
1
/
p
2
-
1
/
p
1
)
+. We use the notation p
2
⊖
p
1
=
p to mean 1
/
p
2
⊖
p
1
=
1
/
p
2
-
1
/
p
1 whenever p
1
>
p
2 and p
=
∞ whenever p
1
≤
p
2.The above result can be extended (see [1]) to the class of mixed norm sequence spaces, denoted ℓ
(
p
,
q
), which are defined by the condition
(2)
(
∑
n
=
0
∞
(
∑
2
n
-
1
≤
k
<
2
n
+
1
-
1
|
a
k
|
p
)
q
/
p
)
1
/
q
<
∞
.Theorem 1.
Let1
≤
r
,
s
,
u
,
v
≤
∞. Then
(3)
(
ℓ
(
r
,
s
)
,
ℓ
(
u
,
v
)
)
=
ℓ
(
u
⊖
r
,
v
⊖
s
)
.In particular, the Köthe dual ofℓ
(
p
,
q
), defined by (
ℓ
(
p
,
q
)
,
ℓ
1
), becomes ℓ
(
p
′
,
q
′
) for 1
≤
p
,
q
<
∞ and 1
/
p
+
1
/
p
′
=
1
/
q
+
1
/
q
′
=
1.Also multipliers between sequence spaces given by Taylor coefficients of holomorphic functions in the disk have been deeply studied in the literature. Since the time of Hardy and Littlewood, mixed norm and related spaces have been used to study function spaces on the unit disk and later to study multipliers between such spaces. Special emphasis has been put on the case where the spacesA and B correspond to the sequence space of Taylor coefficient of analytic functions such as Hardy spaces, Bergman spaces, mixed norm spaces of analytic functions, and so forth. The theory of Hardy spaces and mixed norm spaces of analytic functions was originated in the work of Hardy and Littlewood (see [2, 3]) who implicitly considered the spaces H
(
p
,
q
,
α
) of functions f
∈
ℋ
(
𝔻
) such that
(4)
(
∫
0
1
‍
(
1
-
r
)
q
α
-
1
M
p
q
(
f
,
r
)
d
r
)
1
/
q
<
∞
.
Their work on these spaces was continued by Flett and Sledd (see [4–8]) and later on by Pavlović (see [9, 10]). Multipliers on Hardy spaces were in fashion for a long time and much work was done on them and related spaces. However nowadays complete descriptions of multipliers between Hardy spaces (
H
p
,
H
q
) for certain values of p and q remain still open. The reader is referred to the surveys (see [11, 12]) for lots of results and references. Also many results on multipliers between mixed norm spaces of analytic functions have been established in the last decades (see [13–15] and references thereby). For such a purpose, the use of solid spaces (sequence spaces whose norm depends only on the size of the coefficients), and in particular ℓ
(
p
,
q
) spaces, is a rather important tool. It is worth mentioning that the smallest solid space contained or which contains one of classical Hardy, Bergman, and H
(
p
,
q
,
α
) is actually H
(
2
,
q
,
α
) for some values p,  q, and α (see [14, 15]) and this last space can be identified with certain weighted ℓ
(
2
,
q
), due to Plancherel’s theorem.Another appearance of mixed norm spaces comes with the use of lacunary sequences, that is,a
=
(
a
n
), such that supp
⁡
(
a
)
⊂
{
n
k
:
k
∈
ℕ
0
} for a sequence of integers satisfying inf
⁡
n
k
+
1
/
n
k
=
λ
>
1. Recently (see [16]) the description of the Taylor coefficient of analytic functions F
(
z
)
=
∑
k
=
0
∞
‍
b
k
z
n
k, where n
k is a lacunary sequence, belonging to the weighted Bergman-Besov space B
1
(
ρ
) has been achieved under certain conditions on the weight. It corresponds again with certain weighted ℓ
(
2,1
).In this paper, we consider families of intervalsℐ
=
{
I
k
:
k
∈
ℕ
0
} where I
k
=
{
j
∈
ℕ
0
s
.
t
.
n
k
≤
j
<
n
k
′
} for some increasing sequences (
n
k
)
k and (
n
k
′
) such that n
k
<
n
k
′
≤
n
k
+
1 and we use the notation Λ
ℐ
=
∪
I
k. We will introduce the spaces ℓ
ℐ
(
p
,
q
) given by sequences a
=
(
a
j
)
j
∈
Λ
ℐ verifying
(5)
(
(
∑
j
∈
I
k
‍
|
a
j
|
p
)
1
/
p
)
k
∈
ℓ
q
and the obvious modifications for p
=
∞ or q
=
∞.In particular,ℓ
(
p
,
q
)
=
ℓ
ℐ
(
p
,
q
) for I
k
=
[
2
k
-
1
,
2
k
+
1
-
1
)
∩
ℕ
0. Also a lacunary sequence a
=
(
a
n
)
n corresponds to supp
⁡
(
a
)
⊆
Λ
ℐ where ℐ
=
{
I
k
:
k
∈
ℕ
0
} with I
k
=
{
n
k
} (that is, n
k
′
=
n
k
+
1) for some inf
⁡
k
n
k
+
1
/
n
k
=
λ
>
1.We will give the necessary and sufficient conditions for a sequenceλ
=
(
λ
j
)
j to belong to the multiplier space (
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) whenever Λ
ℐ
=
Λ
𝒥. We also get some applications to multipliers between certain weighted mixed norm spaces of analytic functions. The paper is organized as follows. Section 2 contains the definitions and first properties of the spaces ℓ
ℐ
(
p
,
q
), studying inclusions between them and conditions for coincidence results ℓ
ℐ
(
p
,
q
)
=
ℓ
𝒥
(
p
,
q
). Section 3 contains the main result, which is split into three subsections: the case where intervals in 𝒥 are union of intervals in ℐ, to be denoted ℐ
≤
𝒥, the case where for each I
∈
ℐ there exists J
∈
𝒥 such that either I
⊆
J or J
⊆
I, and finally the case where there exists (
I
,
J
)
∈
ℐ
×
𝒥 such that I
∩
J
≠
∅ and I
∩
J
∉
ℐ
∪
𝒥. In Section 4, we include some application to multipliers on spaces of analytic functions and extend some recent result on weighted Bergman-Besov classes.From now on, we will writeA
≈
B whenever there exists C
>
0 such that C
-
1
A
≤
B
≤
C
A and, as usual, #
I stands for the cardinal of I, 1
/
p
+
1
/
p
′
=
1 for 1
≤
p
≤
∞ and also C denotes a constant that may vary from line to line.


## 2. Generalized Mixed Norm Spaces

Definition 2.
Let1
≤
p
,
q
≤
∞ and let ℐ be a collection of disjoint intervals in ℕ
0, say I
k
=
ℕ
0
∩
[
n
k
,
n
k
′
), where n
k
<
n
k
′
≤
n
k
+
1. One sets Λ
ℐ
=
∪
k
∈
ℕ
0
I
k. One writes ℓ
ℐ
(
p
,
q
) for the space of sequences (
a
n
)
n
∈
Λ
ℐ verifying
(6)
(
(
∑
j
∈
I
k
‍
|
a
j
|
p
)
1
/
p
)
k
∈
ℓ
q
.
This space becomes a Banach space under the norm
(7)
∥
a
∥
p
,
q
ℐ
=
(
∑
k
=
0
∞
‍
(
∑
j
∈
I
k
‍
|
a
j
|
p
)
q
/
p
)
1
/
q
with the obvious modifications for p
=
∞ or q
=
∞.Remark 3.
Of courseℓ
ℐ
(
p
,
p
)
=
{
(
a
n
)
n
∈
Λ
ℐ
:
(
∑
n
‍
|
a
n
|
p
)
1
/
p
<
∞
}. In particular, ℓ
ℐ
(
p
,
p
)
=
ℓ
p whenever Λ
ℐ
=
ℕ
0.
An elementary approach, using Hölder’s inequality, leads to the duality(8)
ℓ
ℐ
(
p
,
q
)
*
=
ℓ
ℐ
(
p
′
,
q
′
)
for 1
≤
p
,
q
<
∞ and 1
/
p
+
1
/
p
′
=
1
/
q
+
1
/
q
′
=
1.Remark 4.
It is clear that(
a
j
)
j
∈
ℓ
ℐ
(
p
,
q
)
⇔
(
a
j
p
)
j
∈
ℓ
ℐ
(
1
,
q
/
p
) in the case p
<
q and also (
a
j
)
j
∈
ℓ
ℐ
(
p
,
q
)
⇔
(
a
j
q
)
j
∈
ℓ
ℐ
(
(
p
/
q
)
,
1
) in the case p
>
q.
Moreover, fora
p
=
(
a
j
p
)
j,
(9)
∥
a
∥
p
,
q
ℐ
=
(
∥
a
p
∥
1
,
q
/
p
ℐ
)
1
/
p
=
(
∥
a
q
∥
p
/
q
,
1
ℐ
)
1
/
q
.Remark 5.
Leta
∈
ℓ
ℐ
(
p
,
q
).(i)
Ifℐ
′ is a subcollection of intervals in ℐ, then ∥
a
∥
p
,
q
ℐ
′
≤
∥
a
∥
p
,
q
ℐ.
(ii)
Ifℐ
=
ℐ
′
∪
ℐ
′′ for two disjoint collections ℐ
′ and ℐ
′′, then ∥
a
∥
p
,
q
ℐ
=
(
(
∥
a
∥
p
,
q
ℐ
′
)
q
+
(
∥
a
∥
p
,
q
ℐ
′′
)
q
)
1
/
q.We would like to analyze the embedding betweenℓ
ℐ
(
p
1
,
q
1
) and ℓ
ℐ
(
p
2
,
q
2
).Proposition 6.
Letℐ be a collection of disjoint intervals in ℕ
0 and let 1
≤
p
1
,
p
2
,
q
≤
∞ with p
1
≠
p
2. Then ℓ
ℐ
(
p
1
,
q
)
=
ℓ
ℐ
(
p
2
,
q
) (with equivalent norms) if and only if
(10)
sup
⁡
k
∈
ℕ
0
#
I
k
<
∞
.
In particular, ifsup
⁡
k
∈
ℕ
0
#
I
k
<
∞, then
(11)
ℓ
ℐ
(
p
,
q
)
=
{
(
a
n
)
n
∈
Λ
ℐ
:
(
∑
n
‍
|
a
n
|
q
)
1
/
q
<
∞
}
.Proof.
⇒) Assume, for instance, p
1
<
p
2 and that ∥
a
∥
p
1
,
q
ℐ
≈
∥
a
∥
p
2
,
q
ℐ for all a supported in Λ
ℐ. Hence, taking a
=
χ
I
k, one concludes that (
n
k
′
-
n
k
)
1
/
p
1
-
1
/
p
2
≤
C for any k. Hence sup
⁡
k
#
I
k
<
∞.
⇐) Note that #
I
k
=
(
n
k
′
-
n
k
) and assume M
=
sup
⁡
k
(
n
k
-
n
k
′
). Then
(12)
∥
a
∥
p
1
,
q
ℐ
=
(
∑
k
=
0
∞
‍
(
∑
j
∈
I
k
‍
|
a
j
|
p
1
)
q
/
p
1
)
1
/
q
≈
(
∑
k
=
0
∞
(
∑
j
∈
I
k
|
a
j
|
p
2
)
q
/
p
2
)
1
/
q
=
∥
a
∥
p
2
,
q
ℐ
since ∥
·
∥
p
1
≈
∥
·
∥
p
2 in ℂ
M.Proposition 7.
Let1
≤
p
1
,
q
1
,
p
2
,
q
2
≤
∞ and let ℐ be a collection of disjoint intervals in ℕ
0 with sup
⁡
k
#
I
k
=
∞.
Thenℓ
ℐ
(
p
1
,
q
1
)
⊆
ℓ
ℐ
(
p
2
,
q
2
) if and only if p
1
≤
p
2 and q
1
≤
q
2.Proof.
⇒) Assume that there exists C
>
0 such that ∥
a
∥
p
2
,
q
2
ℐ
≤
C
∥
a
∥
p
1
,
q
1
ℐ for all a supported in Λ
ℐ. Hence, taking k
∈
ℕ
0 and a
=
χ
I
k, one concludes that (
#
I
k
)
1
/
p
2
-
1
/
p
1
≤
C. Hence p
1
≤
p
2. Let N
∈
ℕ
0 and consider a
=
∑
k
=
1
N
‍
χ
n
k. Applying the above inequality, we obtain N
1
/
q
2
-
1
/
q
1
≤
C. Therefore, q
1
≤
q
2.
⇐) Let us denote
(13)
ℓ
q
(
ℓ
p
)
=
{
(
x
k
)
k
∈
ℕ
0
:
x
k
∈
ℓ
p
,
(
∑
k
=
0
∞
‍
∥
x
k
∥
ℓ
p
q
)
1
/
q
<
∞
}
.
Hence the mapping
(14)
(
a
n
)
n
∈
ℕ
0
⟶
(
(
a
j
)
j
∈
I
k
)
k
∈
ℕ
0
is an isometric embedding from ℓ
ℐ
(
p
,
q
) into ℓ
q
(
ℓ
p
). Taking into account that ℓ
r
1
(
E
)
⊆
ℓ
r
2
(
E
) for any Banach space E and r
1
≤
r
2, we conclude that
(15)
ℓ
ℐ
(
p
,
q
1
)
⊆
ℓ
ℐ
(
p
,
q
2
)
,
ℓ
ℐ
(
p
1
,
q
)
⊆
ℓ
ℐ
(
p
2
,
q
)
.
Therefore
(16)
ℓ
ℐ
(
p
1
,
q
1
)
⊆
ℓ
ℐ
(
p
2
,
q
1
)
⊆
ℓ
ℐ
(
p
2
,
q
2
)
.We would like to analyze the embedding betweenℓ
ℐ
(
p
,
q
) and ℓ
𝒥
(
p
,
q
) for ℐ
≠
𝒥 whenever Λ
ℐ
=
Λ
𝒥.Proposition 8.
Letℐ
=
{
I
l
:
l
∈
ℕ
0
} and 𝒥
=
{
J
k
:
k
∈
ℕ
0
}. If Λ
ℐ
=
Λ
𝒥, p
≤
q (respect. q
≤
p), and sup
⁡
k
#
J
k
<
∞ (respect. sup
⁡
l
#
I
l
<
∞), then
(17)
ℓ
ℐ
(
p
,
q
)
⊆
ℓ
𝒥
(
p
,
q
)
(
r
e
s
p
e
c
t
.
ℓ
𝒥
(
p
,
q
)
⊆
ℓ
ℐ
(
p
,
q
)
)
.Proof.
Proposition6 gives ℓ
𝒥
(
p
,
q
)
=
ℓ
𝒥
(
q
,
q
), and clearly ℓ
𝒥
(
q
,
q
)
=
ℓ
ℐ
(
q
,
q
). Then the result follows using ℓ
ℐ
(
p
,
q
)
⊆
ℓ
ℐ
(
q
,
q
) whenever p
≤
q.Let us mention another particular case where they coincide.Proposition 9.
Letℐ be such that I
k
=
[
n
k
,
n
k
′
)
∩
ℕ
0 with n
2
k
′
=
n
2
k
+
1 and define
(18)
𝒥
=
{
J
k
=
I
2
k
∪
I
2
k
+
1
:
k
∈
ℕ
0
}
.
Then ℓ
ℐ
(
p
,
q
)
=
ℓ
𝒥
(
p
,
q
).Proof.
Note thatJ
k
=
I
2
k
∪
I
2
k
+
1 is again an interval in ℕ
0. Using that (
a
+
b
)
α
≤
C
α
(
a
α
+
b
α
) for a
,
b
,
α
>
0, then
(19)
∥
a
∥
p
,
q
𝒥
=
(
∑
k
=
0
∞
‍
(
∑
j
∈
J
k
‍
|
a
j
|
p
)
q
/
p
)
1
/
q
=
(
∑
k
=
0
∞
‍
(
∑
j
∈
I
2
k
‍
|
a
j
|
p
+
∑
j
∈
I
2
k
+
1
‍
|
a
j
|
p
)
q
/
p
)
1
/
q
≤
C
(
∑
k
=
0
∞
‍
(
∑
j
∈
I
2
k
‍
|
a
j
|
p
)
q
/
p
+
∑
k
=
0
∞
‍
(
∑
j
∈
I
2
k
+
1
‍
|
a
j
|
p
)
q
/
p
)
1
/
q
≤
C
∥
a
∥
p
,
q
ℐ
.
On the other hand, using now(
a
β
+
b
β
)
≤
C
β
(
a
+
b
)
β for a
,
b
,
β
>
0,
(20)
∥
a
∥
p
,
q
ℐ
=
(
∑
k
=
0
∞
‍
(
∑
j
∈
I
2
k
‍
|
a
j
|
p
)
q
/
p
+
(
∑
j
∈
I
2
k
+
1
‍
|
a
j
|
p
)
q
/
p
)
1
/
q
≤
C
′
(
∑
k
=
0
∞
‍
(
∑
j
∈
I
2
k
∪
I
2
k
+
1
‍
|
a
j
|
p
)
q
/
p
)
1
/
q
≤
C
′
∥
a
∥
p
,
q
𝒥
.The previous idea is easily generalized using the following definition.Definition 10.
Letℐ
:
=
{
I
l
:
l
∈
ℕ
0
} and 𝒥
:
=
{
J
k
:
k
∈
ℕ
0
}. One says that ℐ
≤
𝒥 if the following conditions hold:(i)
Λ
ℐ
=
Λ
𝒥;
(ii)
F
k
=
F
k
(
ℐ
,
𝒥
)
:
=
{
l
∈
ℕ
0
:
I
l
⊆
J
k
}
≠
∅ for all k
∈
ℕ
0;
(iii)
J
k
=
∪
l
∈
F
k
I
l for all k
∈
ℕ
0.Proposition 11.
Let1
≤
p
,
q
≤
∞ and ℐ
≤
𝒥. Then(i)
ℓ
𝒥
(
p
,
q
)
⊆
ℓ
ℐ
(
p
,
q
)
for
p
≤
q;
(ii)
ℓ
ℐ
(
p
,
q
)
⊆
ℓ
𝒥
(
p
,
q
)
for
q
≤
p.
Moreover, the embeddings above are of norm1.Proof.
(i) Caseq
=
∞: let a
∈
ℓ
𝒥
(
p
,
∞
) and l
∈
ℕ
0. We know that there is k such that I
l
⊆
J
k. Hence
(21)
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
≤
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
≤
∥
a
∥
p
,
∞
𝒥
.
This gives ∥
a
∥
p
,
∞
ℐ
≤
∥
a
∥
p
,
∞
𝒥.
The casep
=
1: let a
∈
ℓ
𝒥
(
1
,
q
) and q
≥
1. Therefore
(22)
(
∥
a
∥
1
,
q
𝒥
)
q
=
∑
k
‍
(
∑
l
∈
F
k
‍
∑
n
∈
I
l
‍
|
a
n
|
)
q
≥
∑
k
∑
l
∈
F
k
(
∑
n
∈
I
l
|
a
n
|
)
q
=
(
∥
a
∥
1
,
q
ℐ
)
q
.
The case 1
<
p
≤
q
<
∞ follows using (9) and the previous one.
(ii) The casep
=
∞: let a
∈
ℓ
ℐ
(
∞
,
q
). Then
(23)
∥
a
∥
∞
,
q
𝒥
=
(
∑
k
‍
sup
⁡
l
∈
F
k
(
sup
⁡
n
∈
I
l
|
a
n
|
)
q
)
1
/
q
≤
(
∑
k
‍
∑
l
∈
F
k
‍
(
sup
⁡
n
∈
I
l
|
a
n
|
)
q
)
1
/
q
=
∥
a
∥
∞
,
q
ℐ
.
To cover the remaining cases, from (9), we simply need to show that ℓ
ℐ
(
p
,
1
)
⊆
ℓ
𝒥
(
p
,
1
) for p
≥
1. Now observe that
(24)
∥
a
∥
p
,
1
𝒥
=
∑
k
‍
(
∑
l
∈
F
k
‍
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
=
∑
k
‍
(
∑
l
∈
F
k
‍
∥
a
χ
I
l
∥
p
p
)
1
/
p
≤
∑
k
‍
∑
l
∈
F
k
‍
∥
a
χ
I
l
∥
p
=
∑
l
‍
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
=
∥
a
∥
p
,
1
ℐ
.Theorem 12.
Letℐ
≤
𝒥 and 1
≤
p
,
q
≤
∞ with p
≠
q.
ℓ
ℐ
(
p
,
q
)
=
ℓ
𝒥
(
p
,
q
) (with equivalent norms) if and only if sup
⁡
k
#
F
k
<
∞.Proof.
⇒) Assume that ∥
a
∥
p
,
q
𝒥
≈
∥
a
∥
p
,
q
ℐ for all a finitely supported. Let k
∈
ℕ
0 and define
(25)
a
(
k
)
=
∑
l
∈
F
k
‍
(
#
I
l
)
-
1
/
p
χ
I
l
.
Then ∥
a
∥
p
,
q
𝒥
=
(
#
F
k
)
1
/
p and ∥
a
∥
p
,
q
ℐ
=
(
#
F
k
)
1
/
q.
One concludes thatC
2
≤
(
#
F
k
)
1
/
p
-
1
/
q
≤
C
1 which implies, in the case p
≠
q, sup
⁡
k
∈
ℕ
0
(
#
F
k
)
<
∞.
⇐)Case
p
<
q. From Proposition 11, we only need to show ℓ
ℐ
(
p
,
q
)
⊆
ℓ
𝒥
(
p
,
q
). Using now Hölder’s inequality for q
/
p
>
1,
(26)
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
≤
(
∑
l
∈
F
k
‍
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
≤
(
∑
l
∈
F
k
‍
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
q
/
p
)
1
/
q
(
#
F
k
)
1
/
p
⊖
q
.
Therefore, if M
=
sup
⁡
k
F
k, we have
(27)
∥
a
∥
p
,
q
𝒥
=
(
∑
k
=
0
∞
‍
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
q
/
p
)
1
/
q
≤
M
1
/
p
⊖
q
(
∑
k
∈
ℕ
0
‍
∑
l
∈
F
k
‍
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
q
/
p
)
1
/
q
=
M
1
/
p
⊖
q
(
∑
l
∈
ℕ
0
‍
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
q
/
p
)
1
/
q
=
M
1
/
p
⊖
q
∥
a
∥
p
,
q
ℐ
.
Case
p
>
q. Using again Proposition 11, we will show ℓ
𝒥
(
p
,
q
)
⊆
ℓ
ℐ
(
p
,
q
). Using 1
/
q
=
1
/
q
⊖
p
+
1
/
p,
(28)
∥
a
∥
p
,
q
ℐ
=
(
∑
l
‍
∥
a
χ
I
l
∥
p
q
)
1
/
q
=
(
∑
k
‍
∑
l
∈
F
k
‍
∥
a
χ
I
l
∥
p
q
)
1
/
q
≤
(
∑
k
‍
(
∑
l
∈
F
k
‍
∥
a
χ
I
l
∥
p
p
)
q
/
p
(
#
F
k
)
q
/
q
⊖
p
)
1
/
q
≤
M
1
/
q
⊖
p
(
∑
k
‍
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
q
/
p
)
1
/
q
≤
M
1
/
q
⊖
p
∥
a
∥
p
,
q
𝒥
.Let us now exhibit an example where neitherℓ
(
p
,
q
)
ℐ
⊆
ℓ
𝒥
(
p
,
q
) nor ℓ
𝒥
(
p
,
q
)
⊆
ℓ
ℐ
(
p
,
q
).Example 13.
Let1
≤
p
<
q
<
∞ and take ℐ
,
𝒥 as shown in Figure 1 with(29)
card
(
I
0
)
=
n
1
card
(
I
1
)
=
⋯
=
card
(
I
n
1
)
=
1
card
(
I
n
1
+
1
)
=
n
2
card
(
I
n
1
+
2
)
=
⋯
=
card
(
I
n
1
+
n
2
+
1
)
=
1
card
(
I
n
1
+
n
2
+
2
)
=
n
3
,
…
card
(
J
0
)
=
⋯
=
card
(
J
n
1
)
=
1
card
(
J
n
1
)
=
n
1
card
(
J
n
1
+
1
)
=
⋯
=
card
(
J
n
1
+
n
2
)
=
1
card
(
J
n
1
+
n
2
+
1
)
=
n
2
,
…
.Figure 1Let us see that neitherℓ
𝒥
(
p
,
q
)
⊂
ℓ
ℐ
(
p
,
q
) nor ℓ
ℐ
(
p
,
q
)
⊂
ℓ
𝒥
(
p
,
q
).Taking(30)
a
=
(
β
1
,
…
,
β
1
︷
n
1
,
0
,
…
,
0
︷
n
1
,
β
2
,
…
,
β
2
︷
n
2
,
0
,
…
)
,
b
=
(
0
,
…
,
0
︸
n
1
,
β
1
,
…
,
β
1
︸
n
1
,
0
,
…
,
0
︸
n
2
,
β
2
,
…
)
we have
(31)
∥
a
∥
p
,
q
ℐ
=
∥
b
∥
p
,
q
𝒥
=
(
∑
j
‍
β
j
q
n
j
q
/
p
)
1
/
q
,
∥
a
∥
p
,
q
𝒥
=
∥
b
∥
p
,
q
ℐ
=
(
∑
j
‍
β
j
q
n
j
)
1
/
q
.Then it is enough to considerq
>
p and β
j
=
n
j
-
1
/
p
j
-
1
/
q. Now
(32)
(
∑
j
‍
β
j
q
n
j
q
/
p
)
1
/
q
=
(
∑
j
‍
j
-
1
)
1
/
q
=
∞
and, since n
j
≥
j,
(33)
(
∑
j
‍
β
j
q
n
j
)
1
/
q
=
(
∑
j
‍
j
-
1
n
j
1
-
q
/
p
)
1
/
q
≤
(
∑
j
j
-
q
/
p
)
1
/
q
<
∞
.
Hence we have a
∈
ℓ
𝒥
(
p
,
q
)
∖
ℓ
ℐ
(
p
,
q
) and b
∈
ℓ
ℐ
(
p
,
q
)
∖
ℓ
𝒥
(
p
,
q
).We would like to explain a procedure to analyze the general caseΛ
ℐ
=
Λ
𝒥.Definition 14.
Letℐ and 𝒥 be families of disjoint intervals in ℕ
0 with Λ
ℐ
=
Λ
𝒥. For each k
∈
ℕ
0, one uses the notation, as above, F
k
=
{
l
∈
ℕ
0
:
I
l
⊆
J
k
} which now might be empty. One also defines
(34)
F
~
k
=
{
l
∈
ℕ
0
:
J
k
∩
I
l
≠
∅
}
.
We write ϕ and Φ for the mappings given by
(35)
ϕ
(
k
)
=
min
⁡
F
~
k
,
Φ
(
k
)
=
max
⁡
F
~
k
.Similarly, interchangingℐ and 𝒥, we define G
l, G
~
l, ψ
(
l
), and ψ
(
l
).Definition 15.
One defines the “left” and “right” part of the intervalJ
k by
(36)
J
ˇ
k
=
J
k
∩
I
ϕ
(
k
)
,
J
^
k
=
J
k
∩
I
Φ
(
k
)
and, denoting J
k
′
=
∪
l
∈
F
k
I
l and J
~
k
=
∪
l
∈
F
~
k
I
l, one has
(37)
J
k
′
⊆
J
k
⊆
J
~
k
,
(38)
J
k
=
J
k
′
∪
J
^
k
∪
J
ˇ
k
,
where J
k
′
=
∅ whenever F
k
=
∅.Similarly, interchangingℐ and 𝒥, we consider I
ˇ
l,  I
^
l,  I
l
′, and I
~
l.With this notation out of the way we can classify intervals in𝒥 into four different types (according to ℐ). Note that for each interval J
∈
𝒥 there are four possibilities: J coincides with I for some I
∈
ℐ, J can be written as a union of at least two intervals in ℐ, J is strictly contained into some interval I
∈
ℐ, or there exists I
∈
ℐ which overlaps with J and its complement J
c.Therefore we decomposeℕ
0 into four disjoint sets defined as follows.Definition 16.
Letℐ and 𝒥 be families of disjoint intervals in ℕ
0 with Λ
ℐ
=
Λ
𝒥. One introduces
(39)
N
equal
𝒥
=
{
k
∈
ℕ
0
:
#
(
F
~
k
∖
F
k
)
=
0
,
#
F
~
k
=
1
}
,
hi
N
big
𝒥
=
{
k
∈
ℕ
0
:
#
(
F
~
k
∖
F
k
)
=
0
,
#
F
~
k
≥
2
}
,
N
small
𝒥
=
{
k
∈
ℕ
0
:
#
(
F
~
k
∖
F
k
)
>
0
,
#
F
~
k
=
1
}
,
i
N
inter
𝒥
=
{
k
∈
ℕ
0
:
#
(
F
~
k
∖
F
k
)
>
0
,
#
F
~
k
≥
2
}
.
We define the setsN
equal
ℐ,  N
big
ℐ,  N
small
ℐ, and N
inter
ℐ similarly.Remark 17.
Using (38), we can also give a description of the sets above in terms of ϕ and Φ:
(40)
N
equal
𝒥
=
{
k
:
ϕ
(
k
)
=
Φ
(
k
)
,
J
k
=
I
ϕ
(
k
)
}
.
hi
N
big
𝒥
=
{
k
:
ϕ
(
k
)
<
Φ
(
k
)
,
J
k
=
J
~
k
}
.
N
small
𝒥
=
{
k
:
ϕ
(
k
)
=
Φ
(
k
)
,
J
k
⊊
I
ϕ
(
k
)
}
.
i
N
inter
𝒥
=
{
k
:
ϕ
(
k
)
<
Φ
(
k
)
,
J
k
⊊
J
~
k
}
.Using the above decomposition, we can generalize Propositions8 and 11 and Theorem 12. Note that sup
⁡
k
#
J
k
<
∞ implies sup
⁡
k
#
F
~
k
<
∞ and also that ℐ
≤
𝒥 corresponds to the case where N
inter
𝒥
∪
N
small
𝒥
=
∅ or equivalently #
G
~
l
=
1 for any l
∈
ℕ
0.Theorem 18.
Let1
≤
p
<
q
≤
∞ and ℐ
,
𝒥 be collections of intervals such that
(41)
ℓ
ℐ
(
p
,
q
)
⊆
ℓ
𝒥
(
p
,
q
)
⟺
sup
⁡
{
#
F
~
k
;
k
∈
ℕ
0
}
<
∞
.Proof.
⇒) Arguing as in Theorem 12, for k
∈
ℕ
0 we consider
(42)
a
(
k
)
=
∑
l
∈
F
~
k
‍
(
#
(
I
l
∩
J
k
)
)
-
1
/
p
χ
I
l
∩
J
k
.
Hence
(43)
∥
a
(
k
)
∥
p
,
q
𝒥
=
(
∑
n
∈
J
k
|
a
n
|
p
)
1
/
p
=
(
∑
l
∈
F
~
k
∑
n
∈
I
l
∩
J
k
|
a
n
|
p
)
1
/
p
=
(
#
F
~
k
)
1
/
p
,
∥
a
(
k
)
∥
p
,
q
ℐ
=
(
∑
l
∈
F
~
k
‍
(
∑
n
∈
I
l
∩
J
k
‍
|
a
n
|
p
)
q
/
p
)
1
/
q
=
(
#
F
~
k
)
1
/
q
.
Therefore, using that ∥
a
(
k
)
∥
p
,
q
𝒥
≤
C
∥
a
(
k
)
∥
p
,
q
ℐ and p
<
q, we conclude that sup
⁡
{
#
F
~
k
;
k
∈
ℕ
0
}
<
∞.
⇐) Denote sup
⁡
k
(
#
F
k
)
=
M
≥
0 and let k
∈
ℕ
0.
Case
q
=
∞
. If k
∈
N
small
𝒥
∪
N
equal
𝒥, then
(44)
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
≤
(
∑
n
∈
I
ϕ
(
k
)
‍
|
a
n
|
p
)
1
/
p
≤
∥
a
∥
p
,
∞
ℐ
.
If k
∈
N
big
𝒥
∪
N
inter
𝒥, we have
(45)
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
=
(
∑
l
∈
F
k
‍
∑
n
∈
I
l
‍
|
a
n
|
p
+
∑
n
∈
J
˘
k
∪
J
^
k
‍
|
a
n
|
p
)
1
/
p
≤
(
∑
l
∈
F
k
‍
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
+
(
∑
n
∈
I
ϕ
(
k
)
‍
|
a
n
|
p
)
1
/
p
+
(
∑
n
∈
I
Φ
(
k
)
‍
|
a
n
|
p
)
1
/
p
≤
C
(
sup
⁡
l
∈
F
k
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
×
(
#
F
k
)
1
/
p
+
2
∥
a
∥
p
,
∞
ℐ
sup
⁡
l
∈
F
k
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
)
.
This showsℓ
ℐ
(
p
,
∞
)
⊆
ℓ
𝒥
(
p
,
∞
).
Caseq
<
∞: arguing as in Proposition 11, we simply show that ℓ
ℐ
(
1
,
q
)
⊆
ℓ
𝒥
(
1
,
q
) for q
>
1.
Observe that(46)
∑
k
∈
N
small
𝒥
(
∑
n
∈
J
k
‍
|
a
n
|
)
q
‍
≤
∑
l
∈
N
big
ℐ
∪
N
inter
ℐ
‍
∑
ϕ
(
k
)
=
l
‍
(
∑
n
∈
J
k
‍
|
a
n
|
)
q
≤
∑
l
∈
N
big
ℐ
∪
N
inter
ℐ
‍
(
∑
ϕ
(
k
)
=
l
‍
∑
n
∈
J
k
‍
|
a
n
|
)
q
=
∑
l
∈
N
big
ℐ
∪
N
inter
ℐ
‍
(
∑
n
∈
I
l
‍
|
a
n
|
)
q
≤
(
∥
a
∥
1
,
q
ℐ
)
q
.
Also we have(47)
∑
k
∈
N
equal
𝒥
∪
N
big
𝒥
‍
(
∑
n
∈
J
k
‍
|
a
n
|
)
q
≤
∑
k
∈
N
equal
𝒥
∪
N
big
𝒥
‍
(
∑
l
∈
F
k
‍
∑
n
∈
I
l
‍
|
a
n
|
)
q
≤
∑
k
∈
N
equal
𝒥
∪
N
big
𝒥
‍
(
#
F
k
)
q
-
1
∑
l
∈
F
k
‍
(
∑
n
∈
I
l
‍
|
a
n
|
)
q
≤
M
q
-
1
∑
k
∈
N
equal
𝒥
∪
N
big
𝒥
‍
∑
l
∈
F
k
‍
(
∑
n
∈
I
l
‍
|
a
n
|
)
q
≤
M
q
-
1
(
∥
a
∥
1
,
q
ℐ
)
q
.
Finally
(48)
∑
k
∈
N
inter
𝒥
‍
(
∑
n
∈
J
k
‍
|
a
n
|
)
q
≤
∑
k
∈
N
inter
𝒥
‍
(
∑
l
∈
F
k
‍
∑
n
∈
I
l
‍
|
a
n
|
+
∑
n
∈
J
ˇ
k
‍
|
a
n
|
+
∑
n
∈
J
^
k
‍
|
a
n
|
)
q
≤
C
∑
k
∈
N
inter
𝒥
‍
(
#
F
k
)
q
-
1
∑
l
∈
F
k
‍
(
∑
n
∈
I
l
‍
|
a
n
|
)
q
+
C
∑
k
∈
N
inter
𝒥
‍
(
∑
n
∈
J
ˇ
k
‍
|
a
n
|
)
q
+
C
∑
k
∈
N
inter
𝒥
‍
(
∑
n
∈
J
^
k
‍
|
a
n
|
)
q
≤
C
M
q
-
1
∑
l
∈
N
inter
ℐ
∪
N
small
ℐ
‍
(
∑
n
∈
I
l
‍
|
a
n
|
)
q
+
C
∑
k
∈
N
inter
𝒥
‍
(
∑
n
∈
I
ϕ
(
k
)
‍
|
a
n
|
)
q
+
∑
k
∈
N
inter
𝒥
‍
(
∑
n
∈
I
Φ
(
k
)
‍
|
a
n
|
)
q
≤
C
(
∥
a
∥
1
,
q
ℐ
)
q
.
Combining the above estimates, we conclude this implication.Corollary 19.
Let1
≤
p
<
q
≤
∞ and ℐ
,
𝒥 be such that
(49)
ℓ
𝒥
(
p
,
q
)
⊆
ℓ
ℐ
(
p
,
q
)
⟺
sup
⁡
{
#
G
~
l
;
l
∈
ℕ
0
}
<
∞
.The next result can be achieved using duality, but we include a direct proof.Theorem 20.
Let1
≤
q
<
p
≤
∞ and ℐ
,
𝒥 be such that
(50)
ℓ
𝒥
(
p
,
q
)
⊆
ℓ
ℐ
(
p
,
q
)
⟺
sup
⁡
{
#
F
~
k
;
k
∈
ℕ
0
}
<
∞
.Proof.
⇒) Repeat the argument presented in the direct implication of Theorem 18.
⇐) Denote again sup
⁡
k
(
#
F
k
)
=
M.
Case
p
=
∞. Observe first that if l
∈
N
big
ℐ
∪
N
equal
ℐ, we have
(51)
(
sup
⁡
n
∈
I
l
|
a
n
|
)
q
=
|
a
n
(
l
)
|
q
≤
(
sup
⁡
n
∈
J
k
|
a
n
|
)
q
for some k
=
k
(
l
)
∈
N
small
𝒥
∪
N
equal
𝒥. Since k
(
l
)
≠
k
(
l
′
) for l
≠
l
′
∈
N
big
ℐ
∪
N
equal
ℐ, we obtain
(52)
∑
l
∈
N
big
ℐ
∪
N
equal
ℐ
‍
(
sup
⁡
n
∈
I
l
|
a
n
|
)
q
≤
∑
k
∈
N
small
𝒥
∪
N
equal
𝒥
‍
(
sup
⁡
n
∈
J
k
|
a
n
|
)
q
.
Also if l
∈
N
inter
ℐ, then (
sup
⁡
n
∈
I
l
|
a
n
|
)
q
=
|
a
n
(
l
)
|
q where n
(
l
)
∈
I
l
′
∪
I
^
l
∪
I
ˇ
l. Note that n
(
l
)
∈
J
k for some k
∈
N
small
𝒥
∪
N
inter
𝒥 and
(53)
1
≤
#
(
{
l
∈
N
inter
ℐ
:
n
(
l
)
∈
J
k
}
)
≤
2
.
Hence
(54)
∑
l
∈
N
inter
ℐ
‍
(
sup
⁡
n
∈
I
l
|
a
n
|
)
q
≤
2
∑
k
∈
N
small
𝒥
∪
N
inter
𝒥
‍
(
sup
⁡
n
∈
J
k
|
a
n
|
)
q
.
On the other hand,
(55)
∑
l
∈
N
small
ℐ
‍
(
sup
⁡
n
∈
I
l
|
a
n
|
)
q
≤
∑
k
∈
N
big
𝒥
∪
N
inter
𝒥
‍
∑
ψ
(
l
)
=
k
‍
(
sup
⁡
n
∈
I
l
|
a
n
|
)
q
≤
∑
k
∈
N
big
𝒥
∪
N
inter
𝒥
‍
(
sup
⁡
n
∈
J
k
|
a
n
|
)
q
(
#
F
k
)
q
≤
M
q
(
∥
a
∥
p
,
∞
𝒥
)
q
.
Combining the previous cases, we getℓ
𝒥
(
∞
,
q
)
⊆
ℓ
ℐ
(
∞
,
q
).
Case
p
<
∞. Arguing as in Proposition 11, we simply show that ℓ
𝒥
(
p
,
1
)
⊆
ℓ
ℐ
(
p
,
1
) for p
>
1. Consider
(56)
∥
a
∥
p
,
1
ℐ
=
∑
l
‍
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
≤
∑
l
∈
N
small
ℐ
‍
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
+
∑
l
∈
N
equal
ℐ
∪
N
big
ℐ
‍
(
∑
k
∈
G
l
‍
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
+
∑
l
∈
N
inter
ℐ
‍
(
+
∑
n
∈
I
˘
l
‍
|
a
n
|
p
+
∑
n
∈
I
^
l
‍
|
a
n
|
p
∑
k
∈
G
l
‍
∑
n
∈
J
k
‍
|
a
n
|
p
+
∑
n
∈
I
˘
l
‍
|
a
n
|
p
+
∑
n
∈
I
^
l
‍
|
a
n
|
p
)
1
/
p
=
I
1
+
I
2
+
I
3
.
Now observe that
(57)
I
1
≤
∑
k
∈
N
big
𝒥
∪
N
inter
𝒥
‍
∑
l
∈
F
k
‍
(
∑
n
∈
I
l
‍
|
a
n
|
p
)
1
/
p
≤
∑
k
∈
N
big
𝒥
∪
N
inter
𝒥
‍
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
#
(
F
k
)
≤
M
∥
a
∥
p
,
1
𝒥
.
Also note, since p
>
1,
(58)
I
2
≤
∑
l
∈
N
equal
ℐ
∪
N
big
ℐ
‍
∑
k
∈
G
l
‍
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
≤
∥
a
∥
p
,
1
𝒥
.
Finally
(59)
I
3
≤
∑
l
∈
N
inter
ℐ
‍
(
∑
k
∈
G
l
‍
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
+
(
∑
n
∈
I
ˇ
l
‍
|
a
n
|
p
)
1
/
p
+
(
∑
n
∈
I
^
l
‍
|
a
n
|
p
)
1
/
p
≤
∑
k
∈
N
inter
𝒥
∪
N
small
𝒥
‍
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
+
∑
l
∈
N
inter
ℐ
‍
(
∑
n
∈
J
ψ
(
l
)
‍
|
a
n
|
p
)
1
/
p
+
∑
l
∈
N
inter
ℐ
‍
(
∑
n
∈
J
Ψ
(
l
)
‍
|
a
n
|
p
)
1
/
p
≤
C
∑
k
‍
(
∑
n
∈
J
k
‍
|
a
n
|
p
)
1
/
p
=
C
∥
a
∥
p
,
1
𝒥
.
The converse implication is now complete.Corollary 21.
Let1
≤
q
<
p
≤
∞ and ℐ
,
𝒥 be collections of intervals such that
(60)
ℓ
ℐ
(
p
,
q
)
⊆
ℓ
𝒥
(
p
,
q
)
⟺
sup
⁡
{
#
G
~
l
;
l
∈
ℕ
0
}
<
∞
.Corollary 22.
Let1
≤
p
,
q
≤
∞ with p
≠
q and ℐ
,
𝒥 be collections of intervals such that
(61)
ℓ
𝒥
(
p
,
q
)
=
ℓ
ℐ
(
p
,
q
)
⟺
sup
⁡
{
(
#
F
~
k
)
(
#
G
~
l
)
;
k
,
l
∈
ℕ
0
}
<
∞
.Proof.
It suffices to show the casep
<
q. Note that ℓ
ℐ
(
p
,
q
)
⊆
ℓ
𝒥
(
p
,
q
) and ℓ
𝒥
(
p
,
q
)
⊆
ℓ
ℐ
(
p
,
q
) are equivalent, due to Theorem 18 and Corollary 19, to the facts su
p
k
(
#
F
~
k
)
<
∞ and su
p
l
(
#
G
~
l
)
<
∞, or equivalently
(62)
sup
⁡
{
(
#
F
~
k
)
(
#
G
~
l
)
;
k
,
l
∈
ℕ
0
}
=
sup
⁡
k
(
#
F
~
k
)
sup
⁡
l
(
#
G
~
l
)
<
∞
.


## 3. Multipliers on Generalized Mixed Norm Spaces

In this section, we consider1
≤
r
,
s
,
u
,
v
≤
∞ and ℐ
,
𝒥 such that Λ
ℐ
=
Λ
𝒥. We define
(63)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
{
λ
=
(
λ
n
)
n
∈
Λ
ℐ
∩
Λ
𝒥
:
∥
(
λ
n
a
n
)
n
∈
Λ
𝒥
∥
u
,
v
𝒥
≤
C
∥
(
a
n
)
n
∈
Λ
ℐ
∥
r
,
s
ℐ
λ
=
(
λ
n
)
n
∈
Λ
ℐ
∩
Λ
𝒥
:
∥
(
λ
n
a
n
)
n
∈
Λ
𝒥
∥
u
,
v
𝒥
}
.The caseℐ
=
𝒥 can be shown repeating the proof for ℐ
=
{
I
k
:
k
∈
ℕ
0
} where I
k
=
[
2
k
-
1
,
2
k
+
1
-
1
)
∩
ℕ
0 (see [1, Theorem 1]).Theorem 23.
Consider1
≤
r
,
s
,
u
,
v
≤
∞. Then
(64)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
ℐ
(
u
,
v
)
)
=
ℓ
ℐ
(
u
⊖
r
,
v
⊖
s
)
.We define the Köthe dualℓ
ℐ
(
p
,
q
)
K
=
(
ℓ
ℐ
(
p
,
q
)
,
ℓ
ℐ
(
1,1
)
).Corollary 24.
Consider1
≤
r
,
s
,
u
,
v
≤
∞. Then
(65)
ℓ
ℐ
(
r
,
s
)
K
=
ℓ
ℐ
(
r
′
,
s
′
)
.
There are some other cases where the set of multipliers can be easily determined. Using Proposition 6 and Corollary 22, one easily obtains the following results.Proposition 25.
(i) Ifsup
⁡
k
∈
ℕ
0
#
J
k
<
∞, then (
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
ℓ
ℐ
(
v
⊖
r
,
v
⊖
s
).
(ii) Ifsup
⁡
l
∈
ℕ
0
#
I
l
<
∞, then (
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
ℓ
𝒥
(
u
⊖
s
,
v
⊖
s
).
(iii) Ifsup
⁡
{
(
#
F
~
k
)
(
#
G
~
l
)
;
k
,
l
∈
ℕ
0
}
<
∞, then
(66)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
ℓ
𝒥
(
u
⊖
r
,
v
⊖
s
)
=
ℓ
ℐ
(
u
⊖
r
,
v
⊖
s
)
.Also as a direct consequence of Theorem18 we obtain the following.Proposition 26.
Ifr
≤
u, s
≤
v, and u
<
v and sup
⁡
{
#
F
~
k
;
k
∈
ℕ
0
}
<
∞, then
(67)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
{
(
λ
n
)
n
∈
Λ
ℐ
:
sup
⁡
n
|
λ
n
|
<
∞
}
.Proof.
Ifλ
=
(
λ
n
)
n is a multiplier then the sequence (
λ
n
) is necessarily bounded. 							(68)
{
(
λ
n
)
n
∈
Λ
ℐ
:
sup
⁡
n
|
λ
n
|
<
∞
}
⊆
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
is equivalent to ℓ
ℐ
(
r
,
s
)
⊆
ℓ
𝒥
(
u
,
v
). Now use the embedding ℓ
ℐ
(
r
,
s
)
⊆
ℓ
ℐ
(
u
,
v
) and Theorem 18 to conclude the result.Definition 27.
Given two families of intervals,ℐ
,
𝒥, such that Λ
ℐ
=
Λ
𝒥, one defines the collection of pairwise disjoint intervals in ℕ
0 as
(69)
ℐ
∩
𝒥
~
=
{
I
l
∩
J
k
:
k
∈
ℕ
0
,
l
∈
F
~
k
}
.
It coincides with {
I
l
∩
J
k
:
l
∈
ℕ
0
,
k
∈
G
~
l
}.Proposition 28.
Let1
≤
r
,
s
,
u
,
v
≤
∞.(i)
If
r
≤
s
,
v
≤
u,
then
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
⊆
ℓ
ℐ
∩
𝒥
~
(
u
⊖
r
,
v
⊖
s
).
In particular, ifsup
⁡
k
#
F
~
k
<
∞, then
(70)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
⊆
ℓ
𝒥
(
u
⊖
r
,
v
⊖
s
)
.
(ii)
If
s
≤
r
,
u
≤
v,
then
ℓ
ℐ
∩
𝒥
~
(
u
⊖
r
,
v
⊖
s
)
⊆
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
).
In particular, ifsup
⁡
l
#
G
~
l
<
∞, then
(71)
ℓ
ℐ
(
u
⊖
r
,
v
⊖
s
)
⊆
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
.Proof.
(i) Note thatℐ
∩
𝒥
~
≤
ℐ and ℐ
∩
𝒥
~
≤
𝒥. Hence, from Proposition 11,
(72)
ℓ
ℐ
∩
𝒥
~
(
p
,
q
)
⊆
ℓ
ℐ
(
p
,
q
)
,
p
≥
q
,
hi
ℓ
𝒥
(
p
,
q
)
⊆
ℓ
ℐ
∩
𝒥
~
(
p
,
q
)
,
p
≤
q
.
Now using (72) and Theorem 23, we obtain
(73)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
⊆
(
ℓ
ℐ
∩
𝒥
~
(
r
,
s
)
,
ℓ
ℐ
∩
𝒥
~
(
u
,
v
)
)
=
ℓ
ℐ
∩
𝒥
~
(
u
⊖
r
,
v
⊖
s
)
.
Also we have(74)
F
k
(
ℐ
∩
𝒥
~
,
𝒥
)
=
{
l
∈
ℕ
0
:
I
l
∩
J
k
⊆
J
k
}
=
F
~
k
,
G
l
(
ℐ
∩
𝒥
~
,
ℐ
)
=
{
k
∈
ℕ
0
:
I
l
∩
J
k
⊆
I
l
}
=
G
~
l
.
Using now Theorem12,
(75)
ℓ
ℐ
∩
𝒥
~
(
p
,
q
)
=
ℓ
𝒥
(
p
,
q
)
⟺
sup
⁡
k
#
F
~
k
<
∞
,
(76)
ℓ
ℐ
∩
𝒥
~
(
p
,
q
)
=
ℓ
ℐ
(
p
,
q
)
⟺
sup
⁡
l
#
G
~
l
<
∞
.
The particular case follows now applying (75).
(ii) is similar to the previous one and left to the reader.Our purpose is to get a final description of multipliers(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
). We will deal first with the case ℐ
≤
𝒥 and get a reduction to this situation in the remaining cases.


### 3.1. The Caseℐ
≤
𝒥

In this section, we considerℐ and 𝒥 such that ℕ
0
=
N
big
𝒥
∪
N
equal
𝒥. This means that F
~
k
=
F
k
≠
∅ and J
k
=
∪
l
∈
F
k
I
l for all k. Notice that l
∈
F
k means I
l
⊆
J
k and we have
(77)
F
k
=
{
l
∈
ℕ
0
:
ϕ
(
k
)
≤
l
≤
Φ
(
k
)
}
.We use the notation𝒥
/
ℐ
=
{
F
k
:
k
∈
ℕ
0
}.We will need the following well-known fact.Lemma 29.
Let0
<
u
,
r
≤
∞, A
⊆
ℕ
0, and (
λ
i
)
i
∈
A. There exists (
a
i
)
i
∈
A such that
(78)
hhh
(
∑
i
∈
A
‍
|
a
i
|
r
)
1
/
r
=
1
,
(
∑
i
∈
A
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
=
(
∑
i
∈
A
‍
|
a
i
λ
i
|
u
)
1
/
u
(with the obvious modifications whenever u
,
r or u
⊖
r equals ∞).Proof.
Forr
=
∞ (then u
⊖
r
=
u), it suffices to take a
i
=
1
,
i
∈
A.
Ifr
<
∞ and u
≥
r (hence u
⊖
r
=
∞), it suffices to take
(79)
a
i
=
{
1
i
=
i
(
A
)
0
otherwise
for i
(
A
) such that sup
⁡
i
∈
A
|
λ
i
|
=
|
λ
i
(
A
)
|.
Ifu
<
r
<
∞, take
(80)
a
i
=
(
∑
i
∈
A
‍
|
λ
i
|
u
⊖
r
)
-
1
/
r
λ
i
u
⊖
r
/
r
,
i
∈
A
.
Using1
+
u
⊖
r
/
r
=
u
⊖
r
/
u, one shows the result.Theorem 30.
Ifℐ
≤
𝒥, then
(81)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
{
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
u
⊖
r
)
l
(
λ
n
)
n
:
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
l
∈
ℓ
𝒥
/
ℐ
(
u
⊖
s
,
v
⊖
s
)
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
u
⊖
r
)
l
}
.Proof.
⊆) Assume that (
λ
n
)
n
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
).
We use Lemma29 with A
=
I
l to select for each l
∈
ℕ
0 a sequence (
a
i
(
l
)
)
i
∈
I
l such that (
∑
i
∈
I
l
‍
|
a
i
(
l
)
|
r
)
1
/
r
=
1 and β
l
=
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
=
(
∑
i
∈
I
l
‍
|
λ
i
a
i
(
l
)
|
u
)
1
/
u.
Now, again use Lemma29 with A
=
F
k for each k
∈
ℕ
0 to choose (
α
l
)
l
∈
F
k verifying (
∑
l
∈
F
k
‍
|
α
l
|
s
)
1
/
s
=
1 and (
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
1
/
u
⊖
s
=
(
∑
l
∈
F
k
‍
|
β
l
α
l
|
u
)
1
/
u.
Finally, using Lemma29 for A
=
ℕ
0, one more time, take γ
=
(
γ
k
)
k verifying (
∑
k
‍
|
γ
k
|
s
)
1
/
s
=
1 and
(
∑
k
‍
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
⊖
s
/
u
⊖
s
)
1
/
v
⊖
s
=
(
∑
k
‍
γ
k
v
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
/
u
⊖
s
)
1
/
v.
This procedure allows us to obtain the sequencea
=
(
a
i
)
i, a
i
=
γ
k
α
l
a
i
(
l
) where i
∈
I
l, l
∈
F
k, and k
∈
ℕ
0. With this choice, we get that ∥
a
∥
r
,
s
ℐ
=
1 and
(82)
∥
β
∥
u
⊖
s
,
v
⊖
s
𝒥
/
ℐ
=
∥
λ
*
a
∥
u
,
v
𝒥
≤
∥
λ
∥
.
⊇) Let a
=
(
a
i
)
i
∈
ℓ
ℐ
(
r
,
s
) and λ
=
(
λ
i
)
i such that (
β
l
)
l
∈
ℓ
𝒥
/
ℐ
(
u
⊖
s
,
v
⊖
s
) where
(83)
β
l
=
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
.
Fix k
∈
ℕ
0:
(84)
(
∑
i
∈
J
k
‍
|
λ
i
a
i
|
u
)
1
/
u
=
(
∑
l
∈
F
k
‍
∑
i
∈
I
l
‍
|
λ
i
a
i
|
u
)
1
/
u
≤
(
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
u
/
u
⊖
r
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
u
/
r
)
1
/
u
≤
(
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
u
⊖
s
/
u
⊖
r
)
1
/
u
⊖
s
×
(
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
s
/
r
)
1
/
s
.
Taking the v-norm, we get
(85)
(
∑
k
‍
(
∑
i
∈
J
k
‍
|
λ
i
a
i
|
u
)
v
/
u
)
1
/
v
≤
(
∑
k
‍
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
/
u
⊖
s
(
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
s
/
r
)
v
/
s
)
1
/
v
≤
(
∑
k
‍
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
⊖
s
/
u
⊖
s
)
1
/
v
⊖
s
×
(
∑
k
‍
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
s
/
r
)
1
/
s
=
(
∑
k
‍
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
⊖
s
/
u
⊖
s
)
1
/
v
⊖
s
×
(
∑
l
‍
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
s
/
r
)
1
/
s
.
Hence (
λ
n
)
n
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) and ∥
λ
∥
≤
∥
β
∥
u
⊖
s
,
v
⊖
s
𝒥
/
ℐ.Corollary 31.
Let𝒥
≤
ℐ and 1
≤
r
,
s
,
u
,
v
≤
∞. Then
(86)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
{
(
λ
n
)
n
:
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
ℓ
ℐ
/
𝒥
(
v
⊖
r
,
v
⊖
s
)
(
λ
n
)
n
:
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
}
.Proof.
Recall thatG
~
l
=
G
l
=
{
k
∈
ℕ
0
:
J
k
⊆
I
l
} and I
l
=
∪
k
∈
G
l
J
k. We now denote ℐ
/
𝒥
=
{
G
l
:
l
∈
ℕ
0
}. Using Köthe duals. we actually have
(87)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
(
ℓ
𝒥
(
u
′
,
v
′
)
,
ℓ
ℐ
(
r
′
,
s
′
)
)
.
Taking into account that p
′
⊖
q
′
=
q
⊖
p for all p
,
q, the result follows from Theorem 30.


### 3.2.  The Caseℐ
∩
𝒥
~
⊆
ℐ
∪
𝒥

Letℐ
=
{
I
l
:
l
∈
ℕ
0
} and 𝒥
=
{
J
k
:
k
∈
ℕ
0
} be such that Λ
ℐ
=
Λ
𝒥. We assume in this section that N
inter
ℐ
=
∅ and N
inter
𝒥
=
∅, that is to say, for a given l
∈
ℕ
0 either there exists k such that I
l
⊆
J
k or there exists k
′ such that J
k
′
⊆
I
l. In other words, each interval in ℐ
∩
𝒥
~ belongs either to ℐ or to 𝒥.To extend the result on multipliers to this setting we will use the following lemma whose easy proof is left to the reader.Lemma 32.
Letℐ
=
{
I
l
:
l
∈
ℕ
0
} and 𝒥
=
{
J
k
:
k
∈
ℕ
0
} be such that Λ
ℐ
=
Λ
𝒥 and let ℐ
i (respect. 𝒥
i) be subcollections of ℐ (respect. 𝒥) for i
=
1
,
…
,
m where ℐ
=
∪
i
=
1
m
ℐ
i (respect. 𝒥
=
∪
i
=
1
m
𝒥
i) satisfying Λ
ℐ
i
=
Λ
𝒥
i for i
=
1
,
…
,
m. Then
(88)
λ
=
(
λ
n
)
n
∈
Λ
ℐ
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
if and only if
(89)
λ
(
i
)
=
(
λ
n
)
n
∈
Λ
ℐ
i
∈
(
ℓ
ℐ
i
(
r
,
s
)
,
ℓ
𝒥
i
(
u
,
v
)
)
,
i
=
1
,
…
,
m
.
Moreover∥
λ
∥
≈
∑
i
=
1
m
‍
∥
λ
(
i
)
∥.Theorem 33.
Letℐ
∩
𝒥
~
⊆
ℐ
∪
𝒥. Then (
λ
n
)
n
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) if and only if it satisfies the conditions
(90)
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
N
e
q
u
a
l
𝒥
∈
ℓ
v
⊖
s
,
(91)
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
l
∈
N
s
m
a
l
l
ℐ
∈
ℓ
ℱ
(
u
⊖
s
,
v
⊖
s
)
,
(92)
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
N
s
m
a
l
l
𝒥
∈
ℓ
𝒢
(
v
⊖
r
,
v
⊖
s
)
,
where ℱ
=
{
F
k
:
k
∈
N
big
𝒥
} and 𝒢
=
{
G
l
:
l
∈
N
big
ℐ
}.Proof.
Let us consider the following collection of intervals:(93)
𝒥
b
=
{
J
k
:
k
∈
N
big
𝒥
}
,
𝒥
e
=
{
J
k
:
k
∈
N
equal
𝒥
}
,
𝒥
s
=
{
J
k
:
k
∈
N
small
𝒥
}
and similarly for ℐ.
IfJ
k
∈
𝒥
b (respect.  I
l
∈
ℐ
b), we have F
k
=
{
l
∈
ℕ
0
:
I
l
⊊
J
k
}
≠
∅ (respect.  G
l
=
{
k
∈
ℕ
0
:
J
k
⊊
I
l
}
≠
∅) and
(94)
J
k
=
∪
l
∈
F
k
I
l
,
I
l
∈
ℐ
s
(
respect
.
I
l
=
∪
l
∈
G
l
J
k
,
J
k
∈
𝒥
s
)
.
Hence𝒥
=
𝒥
e
∪
𝒥
b
∪
𝒥
s, ℐ
=
ℐ
e
∪
ℐ
b
∪
ℐ
s, and
(95)
𝒥
e
=
{
J
k
:
k
∈
N
equal
𝒥
}
=
{
I
l
:
l
∈
N
equal
ℐ
}
=
ℐ
e
.
Observe that ℐ
s
≤
𝒥
b and 𝒥
s
≤
ℐ
b and, in particular, 𝒢
=
ℐ
b
/
𝒥
s
and
ℱ
=
𝒥
b
/
ℐ
s.
We use Lemma32 and observe that, denoting Λ
0
=
Λ
𝒥
e, Λ
1
=
Λ
𝒥
b
=
Λ
ℐ
s, and Λ
2
=
Λ
𝒥
s
=
Λ
ℐ
b,
(96)
(
λ
n
)
n
∈
Λ
0
∈
(
ℓ
𝒥
e
(
r
,
s
)
,
ℓ
ℐ
e
(
u
,
v
)
)
corresponds to (90) invoking Theorem 23, also that
(97)
(
λ
n
)
n
∈
Λ
1
∈
(
ℓ
ℐ
s
(
r
,
s
)
,
ℓ
𝒥
b
(
u
,
v
)
)
corresponds to (91) invoking Theorem 30, and, finally, that
(98)
(
λ
n
)
n
∈
Λ
2
∈
(
ℓ
ℐ
b
(
r
,
s
)
,
ℓ
𝒥
s
(
u
,
v
)
)
corresponds to (92) invoking Corollary 31.


### 3.3.  The General Case

In this section, we assume that there existk
∈
ℕ
0 and l
∈
F
~
k such that I
l
∩
J
k
∈
ℐ
∩
𝒥
~ and I
l
∩
J
k
∉
ℐ
∪
𝒥.Since the notation may be a bit confusing, we will illustrate the idea. Letℐ
,
𝒥 be different partitions of ℕ
0 as shown in Figure 2.Figure 2The situation we are handling now corresponds toN
inter
𝒥
≠
∅ (and hence N
inter
ℐ
≠
∅).Definition 34.
Consider(99)
𝒥
′
=
{
J
k
′
=
∪
l
∈
F
k
I
l
:
k
∈
ℕ
0
,
#
F
k
>
0
}
,
ℋ
=
ℐ
∩
𝒥
~
∖
(
ℐ
∪
𝒥
)
,
𝒥
s
=
{
J
k
:
k
∈
N
small
𝒥
}
.
Denote𝒥
′′
=
𝒥
′
∪
𝒥
s and 𝒥
new
=
𝒥
′′
∪
ℋ.
We use similar notations forℐ.Recalling thatϕ
(
k
)
=
min
⁡
F
~
k and Φ
(
k
)
=
max
⁡
F
~
k for k
∈
ℕ
0, we easily observe that ϕ
(
N
equal
𝒥
)
⊆
N
equal
ℐ, ϕ
(
N
big
𝒥
)
⊆
N
small
ℐ, ϕ
(
N
small
𝒥
)
⊆
N
big
ℐ
∪
N
inter
ℐ, and ϕ
(
N
inter
𝒥
)
⊆
N
small
ℐ
∪
N
inter
ℐ. Same results hold for Φ.Lemma 35.
Consider the families defined in the previous definition. Then:(100)
ℋ
=
{
J
^
k
:
k
∈
N
i
n
t
e
r
𝒥
,
ϕ
(
k
)
∈
N
i
n
t
e
r
ℐ
}
∪
{
J
ˇ
k
:
k
∈
N
i
n
t
e
r
𝒥
,
Φ
(
k
)
∈
N
i
n
t
e
r
ℐ
}
.Proof.
⊆) Let I
∈
ℋ. Since I
∈
ℐ
∩
𝒥
~, then there exist k
∈
ℕ
0 and l
∈
F
~
k such that I
=
I
l
∩
J
k. On the other hand, since I
∉
ℐ
∪
𝒥, we have that I
⊊
I
l and I
⊊
J
k. Hence, either ϕ
(
k
)
=
l and ψ
(
l
)
=
k or Φ
(
k
)
=
l and ψ
(
l
)
=
k. This gives either k
∈
N
inter
𝒥 and ϕ
(
k
)
∈
N
inter
ℐ (and hence I
=
J
^
k) or k
∈
N
inter
𝒥 and Φ
(
k
)
∈
N
inter
ℐ (and hence I
=
J
ˇ
k).
⊇) Let k
∈
N
inter
𝒥 with ϕ
(
k
)
∈
N
inter
ℐ and consider J
^
k
=
J
k
∩
I
ϕ
(
k
)
∈
ℐ
∩
𝒥
~. Then J
^
k
⊊
J
k (hence J
^
k
∉
𝒥) and J
^
k
⊊
I
ϕ
(
k
) (hence J
^
k
∉
ℐ). One may proceed similarly for J
˘
k in the case k
∈
N
inter
𝒥 with Φ
(
k
)
∈
N
inter
ℐ.Remark 36.
Note thatJ
^
k
=
J
k
∩
I
l if and only if I
ˇ
l
=
I
l
∩
J
k. Therefore,
(101)
ℋ
=
{
I
^
l
:
l
∈
N
inter
ℐ
,
ψ
(
l
)
∈
N
inter
𝒥
}
∪
{
I
ˇ
l
:
k
∈
N
inter
ℐ
,
Ψ
(
l
)
∈
N
inter
𝒥
}
.Lemma 37.
Consider the different families defined in Definition34. Then:
(102)
ℐ
′′
∩
𝒥
′′
~
⊆
ℐ
s
∪
𝒥
s
∪
ℐ
e
⊆
ℐ
′′
∪
𝒥
′′
.Proof.
LetI
∈
ℐ
′
∪
ℐ
s and J
∈
𝒥
′
∪
𝒥
s with I
∩
J
≠
∅. The case I
∈
ℐ
s and J
∈
𝒥
s can not hold. If I
∈
ℐ
s and J
∈
𝒥
′, then I
∩
J
=
I
∈
ℐ
s. Similarly if I
∈
ℐ
′ and J
∈
𝒥
s, then I
∩
J
=
J
∈
𝒥
s. Finally if I
∈
ℐ
′ and J
∈
𝒥
′, then I
=
J
∈
ℐ
e
=
𝒥
e.Theorem 38.
λ
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) if and only if (
λ
n
)
n satisfies(103)
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
N
e
q
u
a
l
𝒥
∈
ℓ
v
⊖
s
(104)
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
l
∈
N
s
m
a
l
l
ℐ
∈
ℓ
ℱ
(
u
⊖
s
,
v
⊖
s
)
(105)
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
N
s
m
a
l
l
𝒥
∈
ℓ
𝒢
(
v
⊖
r
,
v
⊖
s
)
(106)
(
(
∑
i
∈
J
ˇ
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
Λ
r
+
(
(
∑
i
∈
J
^
k
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
Λ
l
∈
ℓ
v
⊖
s
,
where
(107)
Λ
r
=
{
k
∈
N
i
n
t
e
r
𝒥
,
Φ
(
k
)
∈
N
i
n
t
e
r
ℐ
}
,
Λ
l
=
{
k
∈
N
i
n
t
e
r
𝒥
,
ϕ
(
k
)
∈
N
i
n
t
e
r
ℐ
}
,
i
𝒢
=
{
G
l
:
l
∈
N
b
i
g
ℐ
∪
N
i
n
t
e
r
ℐ
,
#
G
l
>
0
}
,
ℱ
=
{
F
k
:
k
∈
N
b
i
g
𝒥
∪
N
i
n
t
e
r
𝒥
,
#
F
k
>
0
}
.Proof.
UsingJ
k
=
J
k
′
∪
J
^
k
∪
J
ˇ
k and Lemma 35, one obtains 𝒥
new
≤
𝒥 and ℐ
new
≤
ℐ. Clearly, #
F
l
(
ℐ
new
,
ℐ
)
≤
3 and #
F
k
(
𝒥
new
,
𝒥
)
≤
3 for all k. Therefore, using Theorem 12, we have ℓ
𝒥
new
(
p
,
q
)
=
ℓ
𝒥
(
p
,
q
) and ℓ
ℐ
new
(
p
,
q
)
=
ℓ
ℐ
(
p
,
q
), which gives
(108)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
(
ℓ
ℐ
new
(
r
,
s
)
,
ℓ
𝒥
new
(
u
,
v
)
)
.
Taking into account Lemma35 and Remark 36, we observe that Λ
ℋ
=
Λ
r
∪
Λ
l and Λ
ℐ
′′
=
Λ
𝒥
′′.
Since𝒥
new
=
𝒥
′′
∪
ℋ and ℐ
new
=
ℐ
′′
∪
ℋ, we can apply Lemma 32 to conclude that λ
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) if and only if (
λ
n
)
n
∈
Λ
ℋ
∈
(
ℓ
ℋ
(
r
,
s
)
,
ℓ
ℋ
(
u
,
v
)
) and (
λ
n
)
n
∉
Λ
ℋ
∈
(
ℓ
ℐ
′′
(
r
,
s
)
,
ℓ
𝒥
′′
(
u
,
v
)
).
Now apply Theorem23 to obtain (
λ
n
)
n
∈
Λ
ℋ
∈
ℓ
ℋ
(
u
⊖
r
,
v
⊖
s
) which corresponds to (106).
On the other hand, comparingℐ
′′ and 𝒥
′′, we notice that I
∈
ℐ
big
′′ corresponds to I
=
I
l
′ for some l
∈
N
big
ℐ
∪
N
inter
ℐ and #
G
l
≥
1. Hence we obtain that 𝒢
=
{
G
I
:
I
∈
ℐ
big
′′
} and similarly ℱ
=
{
F
J
:
J
∈
𝒥
big
′′
}.
We now use Lemma37 together with Theorem 33 to obtain the equivalence with (103), (104), and (105) and (
λ
n
)
n
∉
Λ
ℋ
∈
(
ℓ
ℐ
′′
(
r
,
s
)
,
ℓ
𝒥
′′
(
u
,
v
)
).


## 3.1. The Caseℐ
≤
𝒥

In this section, we considerℐ and 𝒥 such that ℕ
0
=
N
big
𝒥
∪
N
equal
𝒥. This means that F
~
k
=
F
k
≠
∅ and J
k
=
∪
l
∈
F
k
I
l for all k. Notice that l
∈
F
k means I
l
⊆
J
k and we have
(77)
F
k
=
{
l
∈
ℕ
0
:
ϕ
(
k
)
≤
l
≤
Φ
(
k
)
}
.We use the notation𝒥
/
ℐ
=
{
F
k
:
k
∈
ℕ
0
}.We will need the following well-known fact.Lemma 29.
Let0
<
u
,
r
≤
∞, A
⊆
ℕ
0, and (
λ
i
)
i
∈
A. There exists (
a
i
)
i
∈
A such that
(78)
hhh
(
∑
i
∈
A
‍
|
a
i
|
r
)
1
/
r
=
1
,
(
∑
i
∈
A
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
=
(
∑
i
∈
A
‍
|
a
i
λ
i
|
u
)
1
/
u
(with the obvious modifications whenever u
,
r or u
⊖
r equals ∞).Proof.
Forr
=
∞ (then u
⊖
r
=
u), it suffices to take a
i
=
1
,
i
∈
A.
Ifr
<
∞ and u
≥
r (hence u
⊖
r
=
∞), it suffices to take
(79)
a
i
=
{
1
i
=
i
(
A
)
0
otherwise
for i
(
A
) such that sup
⁡
i
∈
A
|
λ
i
|
=
|
λ
i
(
A
)
|.
Ifu
<
r
<
∞, take
(80)
a
i
=
(
∑
i
∈
A
‍
|
λ
i
|
u
⊖
r
)
-
1
/
r
λ
i
u
⊖
r
/
r
,
i
∈
A
.
Using1
+
u
⊖
r
/
r
=
u
⊖
r
/
u, one shows the result.Theorem 30.
Ifℐ
≤
𝒥, then
(81)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
{
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
u
⊖
r
)
l
(
λ
n
)
n
:
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
l
∈
ℓ
𝒥
/
ℐ
(
u
⊖
s
,
v
⊖
s
)
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
u
⊖
r
)
l
}
.Proof.
⊆) Assume that (
λ
n
)
n
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
).
We use Lemma29 with A
=
I
l to select for each l
∈
ℕ
0 a sequence (
a
i
(
l
)
)
i
∈
I
l such that (
∑
i
∈
I
l
‍
|
a
i
(
l
)
|
r
)
1
/
r
=
1 and β
l
=
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
=
(
∑
i
∈
I
l
‍
|
λ
i
a
i
(
l
)
|
u
)
1
/
u.
Now, again use Lemma29 with A
=
F
k for each k
∈
ℕ
0 to choose (
α
l
)
l
∈
F
k verifying (
∑
l
∈
F
k
‍
|
α
l
|
s
)
1
/
s
=
1 and (
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
1
/
u
⊖
s
=
(
∑
l
∈
F
k
‍
|
β
l
α
l
|
u
)
1
/
u.
Finally, using Lemma29 for A
=
ℕ
0, one more time, take γ
=
(
γ
k
)
k verifying (
∑
k
‍
|
γ
k
|
s
)
1
/
s
=
1 and
(
∑
k
‍
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
⊖
s
/
u
⊖
s
)
1
/
v
⊖
s
=
(
∑
k
‍
γ
k
v
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
/
u
⊖
s
)
1
/
v.
This procedure allows us to obtain the sequencea
=
(
a
i
)
i, a
i
=
γ
k
α
l
a
i
(
l
) where i
∈
I
l, l
∈
F
k, and k
∈
ℕ
0. With this choice, we get that ∥
a
∥
r
,
s
ℐ
=
1 and
(82)
∥
β
∥
u
⊖
s
,
v
⊖
s
𝒥
/
ℐ
=
∥
λ
*
a
∥
u
,
v
𝒥
≤
∥
λ
∥
.
⊇) Let a
=
(
a
i
)
i
∈
ℓ
ℐ
(
r
,
s
) and λ
=
(
λ
i
)
i such that (
β
l
)
l
∈
ℓ
𝒥
/
ℐ
(
u
⊖
s
,
v
⊖
s
) where
(83)
β
l
=
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
.
Fix k
∈
ℕ
0:
(84)
(
∑
i
∈
J
k
‍
|
λ
i
a
i
|
u
)
1
/
u
=
(
∑
l
∈
F
k
‍
∑
i
∈
I
l
‍
|
λ
i
a
i
|
u
)
1
/
u
≤
(
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
u
/
u
⊖
r
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
u
/
r
)
1
/
u
≤
(
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
u
⊖
s
/
u
⊖
r
)
1
/
u
⊖
s
×
(
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
s
/
r
)
1
/
s
.
Taking the v-norm, we get
(85)
(
∑
k
‍
(
∑
i
∈
J
k
‍
|
λ
i
a
i
|
u
)
v
/
u
)
1
/
v
≤
(
∑
k
‍
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
/
u
⊖
s
(
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
s
/
r
)
v
/
s
)
1
/
v
≤
(
∑
k
‍
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
⊖
s
/
u
⊖
s
)
1
/
v
⊖
s
×
(
∑
k
‍
∑
l
∈
F
k
‍
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
s
/
r
)
1
/
s
=
(
∑
k
‍
(
∑
l
∈
F
k
‍
β
l
u
⊖
s
)
v
⊖
s
/
u
⊖
s
)
1
/
v
⊖
s
×
(
∑
l
‍
(
∑
i
∈
I
l
‍
|
a
i
|
r
)
s
/
r
)
1
/
s
.
Hence (
λ
n
)
n
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) and ∥
λ
∥
≤
∥
β
∥
u
⊖
s
,
v
⊖
s
𝒥
/
ℐ.Corollary 31.
Let𝒥
≤
ℐ and 1
≤
r
,
s
,
u
,
v
≤
∞. Then
(86)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
{
(
λ
n
)
n
:
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
ℓ
ℐ
/
𝒥
(
v
⊖
r
,
v
⊖
s
)
(
λ
n
)
n
:
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
}
.Proof.
Recall thatG
~
l
=
G
l
=
{
k
∈
ℕ
0
:
J
k
⊆
I
l
} and I
l
=
∪
k
∈
G
l
J
k. We now denote ℐ
/
𝒥
=
{
G
l
:
l
∈
ℕ
0
}. Using Köthe duals. we actually have
(87)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
(
ℓ
𝒥
(
u
′
,
v
′
)
,
ℓ
ℐ
(
r
′
,
s
′
)
)
.
Taking into account that p
′
⊖
q
′
=
q
⊖
p for all p
,
q, the result follows from Theorem 30.


## 3.2.  The Caseℐ
∩
𝒥
~
⊆
ℐ
∪
𝒥

Letℐ
=
{
I
l
:
l
∈
ℕ
0
} and 𝒥
=
{
J
k
:
k
∈
ℕ
0
} be such that Λ
ℐ
=
Λ
𝒥. We assume in this section that N
inter
ℐ
=
∅ and N
inter
𝒥
=
∅, that is to say, for a given l
∈
ℕ
0 either there exists k such that I
l
⊆
J
k or there exists k
′ such that J
k
′
⊆
I
l. In other words, each interval in ℐ
∩
𝒥
~ belongs either to ℐ or to 𝒥.To extend the result on multipliers to this setting we will use the following lemma whose easy proof is left to the reader.Lemma 32.
Letℐ
=
{
I
l
:
l
∈
ℕ
0
} and 𝒥
=
{
J
k
:
k
∈
ℕ
0
} be such that Λ
ℐ
=
Λ
𝒥 and let ℐ
i (respect. 𝒥
i) be subcollections of ℐ (respect. 𝒥) for i
=
1
,
…
,
m where ℐ
=
∪
i
=
1
m
ℐ
i (respect. 𝒥
=
∪
i
=
1
m
𝒥
i) satisfying Λ
ℐ
i
=
Λ
𝒥
i for i
=
1
,
…
,
m. Then
(88)
λ
=
(
λ
n
)
n
∈
Λ
ℐ
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
if and only if
(89)
λ
(
i
)
=
(
λ
n
)
n
∈
Λ
ℐ
i
∈
(
ℓ
ℐ
i
(
r
,
s
)
,
ℓ
𝒥
i
(
u
,
v
)
)
,
i
=
1
,
…
,
m
.
Moreover∥
λ
∥
≈
∑
i
=
1
m
‍
∥
λ
(
i
)
∥.Theorem 33.
Letℐ
∩
𝒥
~
⊆
ℐ
∪
𝒥. Then (
λ
n
)
n
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) if and only if it satisfies the conditions
(90)
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
N
e
q
u
a
l
𝒥
∈
ℓ
v
⊖
s
,
(91)
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
l
∈
N
s
m
a
l
l
ℐ
∈
ℓ
ℱ
(
u
⊖
s
,
v
⊖
s
)
,
(92)
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
N
s
m
a
l
l
𝒥
∈
ℓ
𝒢
(
v
⊖
r
,
v
⊖
s
)
,
where ℱ
=
{
F
k
:
k
∈
N
big
𝒥
} and 𝒢
=
{
G
l
:
l
∈
N
big
ℐ
}.Proof.
Let us consider the following collection of intervals:(93)
𝒥
b
=
{
J
k
:
k
∈
N
big
𝒥
}
,
𝒥
e
=
{
J
k
:
k
∈
N
equal
𝒥
}
,
𝒥
s
=
{
J
k
:
k
∈
N
small
𝒥
}
and similarly for ℐ.
IfJ
k
∈
𝒥
b (respect.  I
l
∈
ℐ
b), we have F
k
=
{
l
∈
ℕ
0
:
I
l
⊊
J
k
}
≠
∅ (respect.  G
l
=
{
k
∈
ℕ
0
:
J
k
⊊
I
l
}
≠
∅) and
(94)
J
k
=
∪
l
∈
F
k
I
l
,
I
l
∈
ℐ
s
(
respect
.
I
l
=
∪
l
∈
G
l
J
k
,
J
k
∈
𝒥
s
)
.
Hence𝒥
=
𝒥
e
∪
𝒥
b
∪
𝒥
s, ℐ
=
ℐ
e
∪
ℐ
b
∪
ℐ
s, and
(95)
𝒥
e
=
{
J
k
:
k
∈
N
equal
𝒥
}
=
{
I
l
:
l
∈
N
equal
ℐ
}
=
ℐ
e
.
Observe that ℐ
s
≤
𝒥
b and 𝒥
s
≤
ℐ
b and, in particular, 𝒢
=
ℐ
b
/
𝒥
s
and
ℱ
=
𝒥
b
/
ℐ
s.
We use Lemma32 and observe that, denoting Λ
0
=
Λ
𝒥
e, Λ
1
=
Λ
𝒥
b
=
Λ
ℐ
s, and Λ
2
=
Λ
𝒥
s
=
Λ
ℐ
b,
(96)
(
λ
n
)
n
∈
Λ
0
∈
(
ℓ
𝒥
e
(
r
,
s
)
,
ℓ
ℐ
e
(
u
,
v
)
)
corresponds to (90) invoking Theorem 23, also that
(97)
(
λ
n
)
n
∈
Λ
1
∈
(
ℓ
ℐ
s
(
r
,
s
)
,
ℓ
𝒥
b
(
u
,
v
)
)
corresponds to (91) invoking Theorem 30, and, finally, that
(98)
(
λ
n
)
n
∈
Λ
2
∈
(
ℓ
ℐ
b
(
r
,
s
)
,
ℓ
𝒥
s
(
u
,
v
)
)
corresponds to (92) invoking Corollary 31.


## 3.3.  The General Case

In this section, we assume that there existk
∈
ℕ
0 and l
∈
F
~
k such that I
l
∩
J
k
∈
ℐ
∩
𝒥
~ and I
l
∩
J
k
∉
ℐ
∪
𝒥.Since the notation may be a bit confusing, we will illustrate the idea. Letℐ
,
𝒥 be different partitions of ℕ
0 as shown in Figure 2.Figure 2The situation we are handling now corresponds toN
inter
𝒥
≠
∅ (and hence N
inter
ℐ
≠
∅).Definition 34.
Consider(99)
𝒥
′
=
{
J
k
′
=
∪
l
∈
F
k
I
l
:
k
∈
ℕ
0
,
#
F
k
>
0
}
,
ℋ
=
ℐ
∩
𝒥
~
∖
(
ℐ
∪
𝒥
)
,
𝒥
s
=
{
J
k
:
k
∈
N
small
𝒥
}
.
Denote𝒥
′′
=
𝒥
′
∪
𝒥
s and 𝒥
new
=
𝒥
′′
∪
ℋ.
We use similar notations forℐ.Recalling thatϕ
(
k
)
=
min
⁡
F
~
k and Φ
(
k
)
=
max
⁡
F
~
k for k
∈
ℕ
0, we easily observe that ϕ
(
N
equal
𝒥
)
⊆
N
equal
ℐ, ϕ
(
N
big
𝒥
)
⊆
N
small
ℐ, ϕ
(
N
small
𝒥
)
⊆
N
big
ℐ
∪
N
inter
ℐ, and ϕ
(
N
inter
𝒥
)
⊆
N
small
ℐ
∪
N
inter
ℐ. Same results hold for Φ.Lemma 35.
Consider the families defined in the previous definition. Then:(100)
ℋ
=
{
J
^
k
:
k
∈
N
i
n
t
e
r
𝒥
,
ϕ
(
k
)
∈
N
i
n
t
e
r
ℐ
}
∪
{
J
ˇ
k
:
k
∈
N
i
n
t
e
r
𝒥
,
Φ
(
k
)
∈
N
i
n
t
e
r
ℐ
}
.Proof.
⊆) Let I
∈
ℋ. Since I
∈
ℐ
∩
𝒥
~, then there exist k
∈
ℕ
0 and l
∈
F
~
k such that I
=
I
l
∩
J
k. On the other hand, since I
∉
ℐ
∪
𝒥, we have that I
⊊
I
l and I
⊊
J
k. Hence, either ϕ
(
k
)
=
l and ψ
(
l
)
=
k or Φ
(
k
)
=
l and ψ
(
l
)
=
k. This gives either k
∈
N
inter
𝒥 and ϕ
(
k
)
∈
N
inter
ℐ (and hence I
=
J
^
k) or k
∈
N
inter
𝒥 and Φ
(
k
)
∈
N
inter
ℐ (and hence I
=
J
ˇ
k).
⊇) Let k
∈
N
inter
𝒥 with ϕ
(
k
)
∈
N
inter
ℐ and consider J
^
k
=
J
k
∩
I
ϕ
(
k
)
∈
ℐ
∩
𝒥
~. Then J
^
k
⊊
J
k (hence J
^
k
∉
𝒥) and J
^
k
⊊
I
ϕ
(
k
) (hence J
^
k
∉
ℐ). One may proceed similarly for J
˘
k in the case k
∈
N
inter
𝒥 with Φ
(
k
)
∈
N
inter
ℐ.Remark 36.
Note thatJ
^
k
=
J
k
∩
I
l if and only if I
ˇ
l
=
I
l
∩
J
k. Therefore,
(101)
ℋ
=
{
I
^
l
:
l
∈
N
inter
ℐ
,
ψ
(
l
)
∈
N
inter
𝒥
}
∪
{
I
ˇ
l
:
k
∈
N
inter
ℐ
,
Ψ
(
l
)
∈
N
inter
𝒥
}
.Lemma 37.
Consider the different families defined in Definition34. Then:
(102)
ℐ
′′
∩
𝒥
′′
~
⊆
ℐ
s
∪
𝒥
s
∪
ℐ
e
⊆
ℐ
′′
∪
𝒥
′′
.Proof.
LetI
∈
ℐ
′
∪
ℐ
s and J
∈
𝒥
′
∪
𝒥
s with I
∩
J
≠
∅. The case I
∈
ℐ
s and J
∈
𝒥
s can not hold. If I
∈
ℐ
s and J
∈
𝒥
′, then I
∩
J
=
I
∈
ℐ
s. Similarly if I
∈
ℐ
′ and J
∈
𝒥
s, then I
∩
J
=
J
∈
𝒥
s. Finally if I
∈
ℐ
′ and J
∈
𝒥
′, then I
=
J
∈
ℐ
e
=
𝒥
e.Theorem 38.
λ
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) if and only if (
λ
n
)
n satisfies(103)
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
N
e
q
u
a
l
𝒥
∈
ℓ
v
⊖
s
(104)
(
(
∑
i
∈
I
l
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
l
∈
N
s
m
a
l
l
ℐ
∈
ℓ
ℱ
(
u
⊖
s
,
v
⊖
s
)
(105)
(
(
∑
i
∈
J
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
N
s
m
a
l
l
𝒥
∈
ℓ
𝒢
(
v
⊖
r
,
v
⊖
s
)
(106)
(
(
∑
i
∈
J
ˇ
k
‍
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
Λ
r
+
(
(
∑
i
∈
J
^
k
|
λ
i
|
u
⊖
r
)
1
/
u
⊖
r
)
k
∈
Λ
l
∈
ℓ
v
⊖
s
,
where
(107)
Λ
r
=
{
k
∈
N
i
n
t
e
r
𝒥
,
Φ
(
k
)
∈
N
i
n
t
e
r
ℐ
}
,
Λ
l
=
{
k
∈
N
i
n
t
e
r
𝒥
,
ϕ
(
k
)
∈
N
i
n
t
e
r
ℐ
}
,
i
𝒢
=
{
G
l
:
l
∈
N
b
i
g
ℐ
∪
N
i
n
t
e
r
ℐ
,
#
G
l
>
0
}
,
ℱ
=
{
F
k
:
k
∈
N
b
i
g
𝒥
∪
N
i
n
t
e
r
𝒥
,
#
F
k
>
0
}
.Proof.
UsingJ
k
=
J
k
′
∪
J
^
k
∪
J
ˇ
k and Lemma 35, one obtains 𝒥
new
≤
𝒥 and ℐ
new
≤
ℐ. Clearly, #
F
l
(
ℐ
new
,
ℐ
)
≤
3 and #
F
k
(
𝒥
new
,
𝒥
)
≤
3 for all k. Therefore, using Theorem 12, we have ℓ
𝒥
new
(
p
,
q
)
=
ℓ
𝒥
(
p
,
q
) and ℓ
ℐ
new
(
p
,
q
)
=
ℓ
ℐ
(
p
,
q
), which gives
(108)
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
)
=
(
ℓ
ℐ
new
(
r
,
s
)
,
ℓ
𝒥
new
(
u
,
v
)
)
.
Taking into account Lemma35 and Remark 36, we observe that Λ
ℋ
=
Λ
r
∪
Λ
l and Λ
ℐ
′′
=
Λ
𝒥
′′.
Since𝒥
new
=
𝒥
′′
∪
ℋ and ℐ
new
=
ℐ
′′
∪
ℋ, we can apply Lemma 32 to conclude that λ
∈
(
ℓ
ℐ
(
r
,
s
)
,
ℓ
𝒥
(
u
,
v
)
) if and only if (
λ
n
)
n
∈
Λ
ℋ
∈
(
ℓ
ℋ
(
r
,
s
)
,
ℓ
ℋ
(
u
,
v
)
) and (
λ
n
)
n
∉
Λ
ℋ
∈
(
ℓ
ℐ
′′
(
r
,
s
)
,
ℓ
𝒥
′′
(
u
,
v
)
).
Now apply Theorem23 to obtain (
λ
n
)
n
∈
Λ
ℋ
∈
ℓ
ℋ
(
u
⊖
r
,
v
⊖
s
) which corresponds to (106).
On the other hand, comparingℐ
′′ and 𝒥
′′, we notice that I
∈
ℐ
big
′′ corresponds to I
=
I
l
′ for some l
∈
N
big
ℐ
∪
N
inter
ℐ and #
G
l
≥
1. Hence we obtain that 𝒢
=
{
G
I
:
I
∈
ℐ
big
′′
} and similarly ℱ
=
{
F
J
:
J
∈
𝒥
big
′′
}.
We now use Lemma37 together with Theorem 33 to obtain the equivalence with (103), (104), and (105) and (
λ
n
)
n
∉
Λ
ℋ
∈
(
ℓ
ℐ
′′
(
r
,
s
)
,
ℓ
𝒥
′′
(
u
,
v
)
).


## 4. An Application

Letρ
:
[
0,1
)
→
[
0
,
∞
) be a nondecreasing function such that ρ
(
0
)
=
0 and ρ
(
t
)
/
t
∈
L
1
(
[
0,1
)
); we define the weighted Bergman-Besov space B
1
(
ρ
) as those analytic functions F in the unit disk such that
(109)
∫
𝔻
‍
|
F
′
(
z
)
|
ρ
(
1
-
|
z
|
)
1
-
|
z
|
d
A
(
z
)
<
∞
.An analytic functionF is called lacunary if F
(
z
)
=
∑
n
∈
Λ
ℒ
‍
a
n
z
n where ℒ
=
{
{
n
k
}
:
k
∈
ℕ
0
} for some (
n
k
) such that inf
⁡
k
n
k
+
1
/
n
k
>
1.Recently, weights with the following condition had been considered in [16]: there exist C
1
,
C
2
>
0 and K
(
n
,
ρ
) such that
(110)
C
1
∫
0
1
‍
r
2
n
-
1
ρ
(
1
-
r
)
1
-
r
d
r
≤
K
(
n
,
ρ
)
≤
C
2
∫
1
-
2
-
n
1
-
2
-
(
n
+
1
)
‍
r
2
n
+
1
-
1
ρ
(
1
-
r
)
1
-
r
d
r
,
and the following result has been shown.Theorem 39 (see [16]).
LetF
(
z
)
=
∑
n
∈
Λ
ℒ
‍
a
n
z
n be a lacunary function and let ρ be a weight satisfying (110). Then F belongs to B
1
(
ρ
) if and only if
(111)
∑
k
=
0
∞
‍
(
∑
n
∈
J
k
‍
|
a
n
|
2
)
1
/
2
2
k
K
(
k
,
ρ
)
<
∞
,
where J
k
=
{
n
:
2
k
-
1
≤
n
<
2
k
+
1
-
1
}.We will extend the previous result for more general classes of weight functions and families of intervals𝒥.Definition 40.
Let0
<
q
<
∞; let 𝒥 be a collection of disjoint intervals in ℕ
0, say J
k
=
ℕ
0
∩
[
m
k
,
m
k
+
1
), where m
0
=
0 and (
m
k
) is some increasing sequence in ℕ
0. And let ρ
:
[
0,1
)
→
[
0
,
∞
) be a measurable function such that ρ
(
t
)
/
t
∈
L
1
(
[
0,1
)
).We say thatρ is q-adapted to 𝒥 whenever there exists C
>
0 depending on m
n
,
q and ρ such that
(112)
∫
0
1
‍
r
q
m
n
ρ
(
1
-
r
)
1
-
r
d
r
≤
C
∫
A
n
‍
r
q
m
n
+
1
ρ
(
1
-
r
)
1
-
r
d
r
for all n
≥
0 where A
0
=
[
0,1
-
1
/
m
1
)  and  A
n
=
[
1
-
1
/
m
n
,
1
-
1
/
m
n
+
1
) for n
≥
1.We denote(113)
μ
ρ
(
s
)
=
∫
0
1
‍
r
s
ρ
(
1
-
r
)
1
-
r
d
r
,
s
≥
0
.
In particular, from condition (112) if ρ is q-adapted to 𝒥, we get that
(114)
μ
ρ
(
q
m
n
)
≈
μ
ρ
(
q
m
n
+
1
)
.
Note also that condition (110) means that ρ is 1
/
2-adapted for 𝒥 where m
n
=
2
n
-
1.Proposition 41.
Letρ
α
(
t
)
=
t
α with α
>
0 and 𝒥
=
{
[
m
n
,
m
n
+
1
)
∩
ℕ
0
:
n
∈
ℕ
0
}. The following statements are equivalent:(i)
ρ
α
is
q-adapted to
𝒥
for all
q
>
0,
(ii)
ρ
α
is
q-adapted to
𝒥
for some
q
>
0,
(iii)
sup
⁡
n
m
n
+
1
/
m
n
<
∞.Proof.
(
i
)
⇒
(
ii
) It is trivial.
(
ii
)
⇒
(
iii
) It is well known that B
(
n
+
1
,
α
)
=
∫
0
1
‍
r
n
(
1
-
r
)
α
-
1
d
r
≈
n
-
α and therefore μ
ρ
α
(
q
m
n
)
≈
m
n
-
α.
Hence it follows from (114) that m
n
+
1
≈
m
n. Therefore sup
⁡
m
n
+
1
/
m
n
<
∞.
(
iii
)
⇒
(
i
) Let sup
⁡
m
n
+
1
/
m
n
=
δ and take q
>
0. Now observe that
(115)
∫
1
-
1
/
m
n
1
-
1
/
m
n
+
1
‍
r
q
m
n
+
1
(
1
-
r
)
α
-
1
d
r
≥
(
1
-
1
m
n
)
q
m
n
+
1
∫
1
/
m
n
+
1
1
/
m
n
‍
s
α
-
1
d
s
≥
1
α
(
1
-
1
m
n
)
q
m
n
+
1
m
n
-
α
(
1
-
(
m
n
m
n
+
1
)
α
)
≥
1
α
(
(
1
-
1
m
n
)
m
n
)
δ
q
m
n
-
α
(
1
-
1
δ
α
)
≥
C
μ
ρ
α
(
q
m
n
)
.We now modify the proof of Lemma 3 in [14] to obtain the following result.Lemma 42.
Let0
<
q
≤
1, let 𝒥 be a collection of disjoint intervals in ℕ
0, and assume ρ is a weight q-adapted to 𝒥. If (
α
n
)
≥
0, then
(116)
∫
0
1
‍
(
∑
n
=
0
∞
‍
α
n
r
n
)
q
ρ
(
1
-
r
)
1
-
r
d
r
≈
∑
n
=
0
∞
‍
(
∑
k
∈
J
n
‍
α
k
)
q
μ
ρ
(
q
m
n
)
,
where J
n
=
{
k
:
m
n
≤
k
<
m
n
+
1
}.Proof.
As above,A
0
=
[
0,1
-
1
/
m
1
)
and
A
n
=
[
1
-
1
/
m
n
,
1
-
1
/
m
n
+
1
) for n
≥
1. Then
(117)
∫
0
1
‍
(
∑
n
=
0
∞
‍
α
n
r
n
)
q
ρ
(
1
-
r
)
1
-
r
d
r
=
∑
n
=
0
∞
‍
∫
A
n
‍
(
∑
n
=
0
∞
‍
α
n
r
n
)
q
ρ
(
1
-
r
)
1
-
r
d
r
≥
∑
n
=
0
∞
‍
∫
A
n
‍
(
∑
k
∈
J
n
‍
α
k
r
k
)
q
ρ
(
1
-
r
)
1
-
r
d
r
≥
∑
n
=
0
∞
‍
∫
A
n
‍
(
∑
k
∈
J
n
‍
α
k
)
q
r
q
m
n
+
1
ρ
(
1
-
r
)
1
-
r
d
r
≥
C
-
1
∑
n
=
0
∞
‍
(
∑
k
∈
J
n
‍
α
k
)
q
μ
ρ
(
q
m
n
)
.
Conversely, since q
≤
1,
(118)
∫
0
1
‍
(
∑
n
=
0
∞
‍
α
n
r
n
)
q
ρ
(
1
-
r
)
1
-
r
d
r
≤
∫
0
1
‍
∑
n
=
0
∞
‍
(
∑
k
∈
J
n
‍
α
k
r
k
)
q
ρ
(
1
-
r
)
1
-
r
d
r
≤
∑
n
=
0
∞
‍
(
∑
k
∈
J
n
‍
α
k
)
q
(
∫
0
1
‍
r
q
m
n
ρ
(
1
-
r
)
1
-
r
d
r
)
≤
∑
n
=
0
∞
‍
(
∑
k
∈
J
n
‍
α
k
)
q
μ
ρ
(
q
m
n
)
.We first note that for lacunary functionsF and 0
<
p
<
∞, we have (see [17])
(119)
M
p
(
F
,
r
)
=
(
∫
0
2
π
‍
|
F
(
r
e
i
θ
)
|
p
d
θ
2
π
)
1
/
p
≈
M
2
(
F
,
r
)
=
(
∫
0
2
π
‍
|
F
(
r
e
i
θ
)
|
2
d
θ
2
π
)
1
/
2
.
Therefore, for lacunary functions F, one has that F
∈
B
1
(
ρ
) if and only if
(120)
∫
0
1
‍
M
2
(
F
′
,
r
)
ρ
(
1
-
r
)
1
-
r
d
r
<
∞
.
Therefore, invoking Plancherel’s theorem and Lemma 42, we recover Theorem 39.Recall that an analytic functionF
:
𝔻
→
ℂ with F
(
z
)
=
∑
n
=
0
∞
‍
a
n
z
n is said to belong to H
(
p
,
q
,
ρ
) (see [14, Definition 2]) whenever
(121)
∥
F
∥
H
(
p
,
q
,
ρ
)
=
(
∫
0
1
‍
M
p
q
(
F
,
r
)
ρ
(
1
-
r
2
)
1
-
r
2
r
d
r
)
1
/
q
<
∞
.
We use the notation H
(
p
,
q
,
α
) if ρ
(
t
)
=
t
α.A consequence of Lemma42 is the following result.Corollary 43.
Let0
<
q
≤
2, let 𝒥 be a collection of disjoint intervals in ℕ
0, and let ρ be a weight q
/
2-adapted to 𝒥. Then
(122)
∥
F
∥
H
(
2
,
q
,
ρ
)
≈
(
∑
n
=
0
∞
‍
(
∑
k
∈
J
n
‍
|
a
k
|
2
)
q
/
2
μ
ρ
(
q
m
n
2
)
)
1
/
q
.
Moreover, if F is lacunary and 0
<
p
<
∞, then
(123)
∥
F
∥
H
(
p
,
q
,
ρ
)
≈
(
∑
n
=
0
∞
‍
(
∑
k
∈
J
n
∩
Λ
ℒ
‍
|
a
k
|
2
)
q
/
2
μ
ρ
(
q
m
n
2
)
)
1
/
q
.Theorem 44.
Let0
<
q
<
∞, let 𝒥 be a collection of disjoint intervals in ℕ
0, and assume ρ is a weight q-adapted to 𝒥. Define λ
=
(
λ
k
)
k such that
(124)
λ
k
=
(
∫
0
1
‍
r
q
m
n
ρ
(
1
-
r
)
1
-
r
)
1
/
q
,
k
∈
J
n
and λ
k
=
0 otherwise. Then (
λ
k
)
k
∈
(
H
(
1
,
q
,
ρ
)
,
ℓ
𝒥
(
∞
,
q
)
).Proof.
We will show that(125)
(
∑
n
=
0
∞
‍
(
sup
⁡
k
∈
J
n
|
a
k
|
)
q
μ
ρ
(
q
m
n
)
)
1
/
q
≤
C
∥
F
∥
H
(
1
,
q
,
ρ
)
.
Recall that
(126)
sup
⁡
k
∈
J
n
-
1
|
a
k
|
r
k
≤
M
1
(
F
,
r
)
and therefore, if A
0
=
[
0,1
-
1
/
m
1
)  and  A
n
=
[
1
-
1
/
m
n
,
1
-
1
/
m
n
+
1
) for n
≥
1, then
(127)
∑
n
=
0
∞
‍
(
sup
⁡
k
∈
J
n
|
a
k
|
)
q
μ
ρ
(
q
m
n
)
≤
C
∑
n
=
0
∞
‍
(
sup
⁡
k
∈
J
n
|
a
k
|
)
q
∫
A
n
‍
r
q
m
n
+
1
ρ
(
1
-
r
)
1
-
r
d
r
≤
C
∑
n
=
0
∞
‍
∫
A
n
‍
(
sup
⁡
k
∈
J
n
|
a
k
|
r
k
)
q
ρ
(
1
-
r
)
1
-
r
d
r
≤
C
∑
n
=
0
∞
‍
∫
A
n
‍
M
1
q
(
F
,
r
)
ρ
(
1
-
r
)
1
-
r
d
r
=
C
∥
F
∥
H
(
1
,
q
,
ρ
)
q
.Theorem 45.
Let1
≤
q
2
<
q
1
≤
2 and let 𝒥 and ℐ be collections of disjoint intervals in ℕ
0, generated by sequences m
k and n
k, respectively, such that ℐ
≤
𝒥. Assume that ρ
1 is a weight q
1
/
2-adapted to ℐ and ρ
2 is a weight q
2
/
2-adapted to 𝒥. Denote
(128)
μ
ρ
1
,
ρ
2
(
k
)
=
(
(
μ
ρ
2
(
q
2
m
k
2
)
)
1
/
q
2
(
μ
ρ
1
(
q
1
n
k
2
)
)
-
1
/
q
1
)
1
/
q
2
⊖
q
1
;
then
(129)
(
H
(
2
,
q
1
,
ρ
1
)
,
H
(
2
,
q
2
,
ρ
2
)
)
=
{
(
λ
n
)
n
;
(
sup
⁡
k
∈
I
n
μ
ρ
1
,
ρ
2
(
k
)
|
λ
k
|
)
∈
ℓ
𝒥
/
ℐ
(
∞
,
q
2
⊖
q
1
)
}
.Proof.
Let(130)
F
ℐ
(
z
)
=
∑
k
=
0
∞
‍
(
μ
ρ
1
(
q
1
n
k
2
)
)
1
/
q
1
(
∑
j
∈
I
k
‍
z
j
)
,
F
~
ℐ
(
z
)
=
∑
k
=
0
∞
‍
(
μ
ρ
1
(
q
1
n
k
2
)
)
-
1
/
q
1
(
∑
j
∈
I
k
‍
z
j
)
,
G
𝒥
(
z
)
=
∑
k
=
0
∞
‍
(
μ
ρ
2
(
q
2
m
k
2
)
)
1
/
q
2
(
∑
j
∈
J
k
‍
z
j
)
.
Using Corollary 43, one has that f
∈
H
(
2
,
q
1
,
ρ
1
) if and only if f
*
F
ℐ
∈
ℓ
ℐ
(
2
,
q
1
) and g
∈
H
(
2
,
q
2
,
ρ
2
) if and only if g
*
G
𝒥
∈
ℓ
𝒥
(
2
,
q
2
).
We use thatλ
∈
(
H
(
2
,
q
1
,
ρ
1
)
,
H
(
2
,
q
2
,
ρ
2
)
) is equivalent to λ
*
G
𝒥
∈
(
H
(
2
,
q
1
,
ρ
1
)
,
ℓ
𝒥
(
2
,
q
2
)
) and also equivalent to λ
*
G
𝒥
*
F
~
ℐ
∈
(
ℓ
ℐ
(
2
,
q
1
)
,
ℓ
𝒥
(
2
,
q
2
)
).
Making use of Theorem30, we have
(131)
(
ℓ
ℐ
(
2
,
q
1
)
,
ℓ
𝒥
(
2
,
q
2
)
)
=
{
(
γ
n
)
n
;
(
sup
⁡
k
∈
I
n
|
γ
k
|
)
n
∈
ℓ
𝒥
/
ℐ
(
∞
,
q
2
⊖
q
1
)
}
.
This concludes the result.Let us finish by observing some examples to apply the above results.Example 46.
Letλ
>
1 and denote m
0
(
λ
)
=
0 and m
k
(
λ
)
=
[
λ
k
] for k
∈
ℕ
0 and 𝒥
(
λ
) the partition of intervals J
k
(
λ
)
=
[
m
k
(
λ
)
,
m
k
+
1
(
λ
)
)
∩
ℕ
0. In this case, μ
ρ
α
(
q
m
n
)
≈
λ
-
α
n, and then, from Proposition 41, ρ
α is q-adapted to 𝒥
(
λ
) for any value of q
>
0.
Letλ
>
γ
>
1 with λ
=
γ
N with N
∈
ℕ
0. Then 𝒥
(
γ
)
≤
𝒥
(
λ
) because
(132)
m
k
(
λ
)
=
[
λ
k
]
=
[
γ
N
k
]
=
m
N
k
(
γ
)
and therefore
(133)
J
k
(
λ
)
=
∪
l
∈
F
k
J
l
(
γ
)
,
where F
k
=
{
l
:
N
k
≤
l
<
N
k
+
N
}. Hence 𝒥
(
λ
)
/
𝒥
(
γ
)
=
ℐ where I
k
=
[
N
k
,
N
(
k
+
1
)
)
∩
ℕ
0; that is, m
k
(
ℐ
)
=
N
k.



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*Source: 983273-2014-04-03.xml*