# On the Geometry of Müntz Spaces

Authors: Sergey V. Ludkovsky; Wolfgang Lusky
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

LetΛ
=
{
λ
k
}
k
=
1
∞ satisfy 0
<
λ
1
<
λ
2
<
⋯, ∑
k
=
1
∞
‍
1
/
λ
k
<
∞ and i
n
f
k
(
λ
k
+
1
-
λ
k
)
>
0. We investigate the Müntz spaces M
p
Λ
=
s
p
a
n
¯
{
t
λ
k
:
k
=
1,2
,
…
}
⊂
L
p
(
0,1
) for 1
≤
p
≤
∞. We show that, for each p, there is a Müntz space F
p which contains isomorphic copies of all Müntz spaces as complemented subspaces. F
p is uniquely determined up to isomorphisms by this maximality property. We discuss explicit descriptions of F
p. In particular F
p is isomorphic to a Müntz space M
p
(
Λ
^
) where Λ
^ consists of positive integers. Finally we show that the Banach spaces (
∑
n
‍
⊕
F
n
)
p for 1
≤
p
<
∞ and (
∑
n
‍
⊕
F
n
)
0 for p
=
∞ are always isomorphic to suitable Müntz spaces M
p
(
Λ
) if the F
n are the spans of arbitrary finitely many monomials over [
0,1
].


---

## Body

## 1. Introduction

Our paper is concerned with the Banach space geometry of Müntz spaces which is largely unexplored.LetΛ
=
{
λ
k
:
k
=
1,2
,
…
} be a sequence of real numbers with 0
<
λ
1
<
λ
2
<
⋯ and(1)
∑
k
=
1
∞
1
λ
k
<
∞
,
inf
k
⁡
λ
k
+
1
-
λ
k
>
0
.For 1
≤
p
<
∞ and f
:
[
0,1
]
→
R, put(2)
f
p
=
∫
0
1
f
t
p
d
t
1
/
p
,
f
∞
=
sup
t
∈
0,1
⁡
f
t
.For λ
>
0, let t
λ be the function f
:
[
0,1
]
→
R with f
(
t
)
=
t
λ. In this paper, we consider the Müntz spaces(3)
M
p
Λ
=
span
¯
⁡
t
λ
k
:
k
=
1,2
,
…
⊂
L
p
0,1
for
1
≤
p
≤
∞
.By the Müntz theorem [1, 2], it is clear that M
p
(
Λ
)
≠
L
p
(
0,1
) if 1
≤
p
<
∞ and M
∞
(
Λ
)
≠
C
0
(
0,1
)
:
=
{
f
:
[
0,1
]
→
R
:
f
c
o
n
t
i
n
u
o
u
s
,
f
(
0
)
=
0
}. It is known that M
p
(
Λ
) is isomorphic to l
p for 1
≤
p
<
∞ and M
∞
(
Λ
) is isomorphic to c
0 provided that Λ satisfies in addition the lacunary condition i
n
f
k
(
λ
k
+
1
/
λ
k
)
>
1 [3]. On the other hand, there are examples of Müntz spaces M
∞
(
Λ
) which are not isomorphic to c
0 ([4] or [5, Corollary 10.2
.
4.]).We want to show that, for fixedp, all Müntz spaces are isomorphic to complemented subspaces of one special function space which itself is a Müntz space M
p
(
Λ
^
) where Λ
^ consists of positive integers.For two Banach spacesX and Y, we write X
~
Y if X is isomorphic to Y.For a sequence of Banach spacesE
n, we put(4)
∑
n
=
1
∞
⊕
E
n
p
=
e
n
:
e
n
∈
E
n
,
n
=
1,2
,
…
,
∑
n
=
1
∞
e
n
p
1
/
p
<
∞
,if 1
≤
p
<
∞, and(5)
∑
n
=
1
∞
⊕
E
n
0
=
e
n
:
e
n
∈
E
n
,
n
=
1,2
,
…
,
lim
n
→
∞
⁡
e
n
=
0with the norm (
e
n
)
=
s
u
p
n
e
n.Theorem 1.
Let(
Ω
k
) be a sequence of finite subsets of N containing each finite subset of N infinitely many times. Moreover let F
p
(
Ω
k
) be the linear s
p
a
n of the functions t
λ, λ
∈
Ω
k, in L
p
(
0,1
). Put(6)
F
p
=
∑
k
∈
N
⊕
F
p
Ω
k
p
i
f
1
≤
p
<
∞
,
F
∞
=
∑
k
∈
N
⊕
F
∞
Ω
k
0
.
(a)
For eachp
∈
[
1
,
∞
], there is a set Λ
^
⊂
N satisfying (1) such that F
p is isomorphic to M
p
(
Λ
^
).
(b)
IfΛ is any set of positive real numbers satisfying (1), then M
p
(
Λ
) is isomorphic to a complemented subspace of F
p.On the other hand, we do not know if any complemented subspace ofF
p is isomorphic to a Müntz space.We prove Theorem1 in Section 4. It turns out that the “maximality” condition of Theorem 1(b) implies uniqueness up to isomorphism. Indeed we have the following.Theorem 2.
Let1
≤
p
≤
∞ and assume that Λ is a set of real positive numbers satisfying (1). If M
p
(
Λ
) contains a complemented isomorphic copy of any Müntz space M
p
(
Λ
~
) then M
p
(
Λ
) is isomorphic to the space F
p of Theorem 1.Proof.
Consider the spacesF
p
(
Ω
k
) of Theorem 1. By the choice of the Ω
k and the definition of F
p, we have F
p
~
F
p
⊕
F
p
⊕
⋯.
Now letΛ satisfy the assumptions of Theorem 2. Since F
p is isomorphic to a Müntz space, by Theorem 1 there is a complemented subspace A
⊂
M
p
(
Λ
) with M
p
(
Λ
)
~
A
⊕
F
p. Moreover, by Theorem 1(b) there is a complemented subspace B
⊂
F
p with F
p
~
B
⊕
M
p
(
Λ
). Using Pelczynski’s decomposition method [6, Theorem 2.2.3.], we obtain F
p
~
M
p
(
Λ
).Corollary 3.
F
∞ is not isomorphic to c
0.Proof.
There is a Müntz spaceM
∞
(
Λ
) which is not isomorphic to c
0 ([4] or [5, Corollary 10.2.4.]). But, according to Theorem 1, M
∞
(
Λ
) is isomorphic to a complemented subspace of F
∞. Since all infinite dimensional complemented subspaces of c
0 are isomorphic to c
0 [7, Theorem I.2.a.3.], the space F
∞ cannot be isomorphic to c
0.In connection with the proof of Theorem1, we also get the following.Theorem 4.
LetΩ
n, n
=
1,2
,
…, be arbitrary finite subsets of positive real numbers and put F
p
(
Ω
n
)
=
s
p
a
n
{
t
λ
:
λ
∈
Ω
n
}
⊂
L
p
(
0,1
). Then, for each p
∈
[
1
,
∞
], there is a subset Λ
⊂
N satisfying (1) such that (
∑
n
⊕
F
p
(
Ω
n
)
)
p
~
M
p
(
Λ
), if 1
≤
p
<
∞, and (
∑
n
⊕
F
∞
(
Ω
n
)
)
0
~
M
∞
(
Λ
), if p
=
∞.We do not know if, conversely, any Müntz spaceM
p
(
Λ
) is isomorphic to the direct sum of the spaces F
p
(
Ω
n
) for suitable Ω
n.


## 2. Finite Dimensional Subspaces ofL
p
(
0,1
)

PutC
0,0
(
0,1
)
=
{
f
∈
C
0
(
0,1
)
:
f
(
1
)
=
0
}. For a finite dimensional subspace E
⊂
L
p
(
0,1
), 1
≤
p
<
∞, or E
⊂
C
0
(
0,1
), p
=
∞, and ϵ
>
0, put(7)
α
E
,
p
,
ϵ
=
sup
⁡
α
>
0
:
e
·
1
0
,
α
p
≤
ϵ
e
p
∀
e
∈
E
.If E
⊂
C
0
(
0,1
), p
=
∞, also put(8)
γ
E
,
ϵ
=
inf
⁡
γ
<
1
:
e
1
-
e
·
1
γ
,
1
∞
≤
ϵ
e
∞
∀
e
∈
E
.Similarly, for E
⊂
L
p
(
0,1
), 1
≤
p
<
∞, or E
⊂
C
0,0
(
0,1
), p
=
∞, define(9)
β
E
,
p
,
ϵ
=
inf
⁡
β
>
0
:
e
·
1
β
,
1
p
≤
ϵ
e
p
∀
e
∈
E
.We collect a few properties of these parameters in the following.Lemma 5.
One has(a)
α
(
E
,
p
,
ϵ
)
≤
β
(
E
,
p
,
ϵ
) for all p
<
∞ if ϵ
<
1
/
2,
(b)
α
(
E
,
∞
,
ϵ
)
≤
β
(
E
,
∞
,
ϵ
) if ϵ
<
1,
(c)
α
(
E
,
∞
,
ϵ
)
≤
γ
(
E
,
ϵ
) if ϵ
<
1
/
3.Proof.
We prove (c). The proofs of (a) and (b) are similar. Lete
∈
E and let γ
<
1 be such that e
1
-
e
·
1
γ
,
1
∞
≤
ϵ
e
∞. If t
∈
[
γ
,
1
], we obtain e
t
-
e
γ
≤
2
ϵ
e
∞. This yields(10)
e
∞
=
max
⁡
e
·
1
0
,
γ
∞
,
e
·
1
γ
,
1
∞
≤
max
⁡
e
·
1
0
,
γ
∞
,
e
γ
+
2
ϵ
e
∞
.Hence, 1
-
2
ϵ
e
∞
≤
e
·
1
0
,
γ
∞. For ϵ
<
1
/
3, we obtain (
1
-
2
ϵ
)
>
ϵ and therefore ϵ
e
∞
<
e
·
1
0
,
γ
∞. This implies α
(
E
,
∞
,
ϵ
)
≤
γ
(
E
,
ϵ
).Lemma 6.
ConsiderE
⊂
L
p
(
0,1
), 1
≤
p
<
∞, or E
⊂
C
0,0
(
0,1
), p
=
∞. Fix 0
<
ϵ
<
1
/
2 and let α
=
α
(
E
,
p
,
ϵ
) and β
=
β
(
E
,
p
,
ϵ
). Then one has(11)
1
-
2
ϵ
e
p
≤
e
·
1
α
,
β
p
≤
e
p
∀
e
∈
E
.For p
=
∞, one even has e
·
1
α
,
β
∞
=
e
∞.Proof.
We obtain(12)
e
p
≤
e
·
1
α
,
β
p
+
e
·
1
0
,
α
p
+
e
·
1
β
,
1
p
≤
e
·
1
α
,
β
p
+
2
ϵ
e
p
.which yields the first inequality. The second inequality is trivial. The last assertion of Lemma 6 follows directly from the definitions of α and β.Proposition 7.
Fixϵ
n
>
0 such that ∑
n
=
1
∞
ϵ
n
<
1
/
8. Let E
n
⊂
L
p
(
0,1
), 1
≤
p
<
∞, or E
n
⊂
C
0,0
(
0,1
), p
=
∞, be finite dimensional subspaces with(13)
β
n
≔
β
E
n
,
p
,
ϵ
n
<
α
E
n
+
1
,
p
,
ϵ
n
+
1
≕
α
n
+
1
.Then, for any e
n
∈
E
n and N
∈
N, one has (14)
1
2
∑
n
=
1
N
e
n
p
p
1
/
p
≤
∑
n
=
1
N
e
n
p
≤
4
3
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
≤
4
3
∑
n
=
1
N
e
n
p
p
1
/
p
,if 1
≤
p
<
∞, and(15)
7
8
sup
n
≤
N
⁡
e
n
∞
≤
∑
n
=
1
N
e
n
∞
≤
9
8
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
∞
≤
9
8
sup
n
≤
N
⁡
e
n
∞
,if p
=
∞.Proof.
Let1
≤
p
<
∞. Put e
=
∑
n
=
1
N
e
n. We obtain with Lemma 6 the following: (16)
e
-
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
=
∑
n
=
1
N
e
n
-
e
n
·
1
α
n
,
β
n
p
≤
∑
n
=
1
N
2
ϵ
n
e
n
p
≤
∑
n
=
1
N
2
ϵ
n
1
-
2
ϵ
n
e
n
·
1
α
n
,
β
n
p
≤
1
3
sup
n
≤
N
⁡
e
n
·
1
α
n
,
β
n
p
≤
1
3
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
p
1
/
p
.Here we used ∑
n
=
1
∞
ϵ
n
<
1
/
8. Hence, 1
-
2
ϵ
n
≥
3
/
4 for all n and ∑
n
=
1
∞
2
ϵ
n
/
(
1
-
2
ϵ
n
)
<
1
/
3.
This yields, in view of (13), (17)
e
p
≤
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
+
1
3
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
p
1
/
p
=
4
3
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
p
1
/
p
≤
4
3
∑
n
=
1
N
e
n
p
p
1
/
p
,
e
p
≥
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
-
1
3
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
p
1
/
p
=
2
3
∑
n
=
1
N
e
n
·
1
α
n
,
β
n
p
p
1
/
p
≥
1
2
∑
n
=
1
N
e
n
p
p
1
/
p
.Here we used Lemma 6 and 1
-
2
ϵ
n
≥
3
/
4. The proof for p
=
∞ is similar.Proposition 8.
Fixϵ
n
>
0 such that ∑
n
=
1
∞
ϵ
n
<
1
/
8. Let E
n
⊂
C
0
(
0,1
) be finite dimensional subspaces with(18)
γ
n
≔
γ
E
n
,
ϵ
n
<
α
E
n
+
1
,
∞
,
ϵ
n
+
1
≕
α
n
+
1
.Then, for any e
k
∈
E
k and N
∈
N, one has(19)
1
2
sup
n
≤
N
⁡
∑
k
=
1
n
e
k
∞
≤
∑
k
=
1
N
e
k
∞
≤
s
u
p
n
≤
N
∑
k
=
1
n
e
k
∞
.Proof.
By (18) we have, for any n
≤
N, (20)
∑
k
=
1
N
e
k
·
1
0
,
γ
n
∞
-
∑
k
=
1
n
e
k
·
1
0
,
γ
n
∞
≤
∑
k
=
n
+
1
N
e
k
·
1
0
,
α
k
∞
≤
∑
k
=
n
+
1
N
ϵ
k
e
k
∞
.Equation (18) and Lemma 5 also imply that γ
n are monotonically increasing. Hence, if t
∈
[
γ
n
,
1
], we obtain (21)
∑
k
=
1
n
e
k
t
-
∑
k
=
1
n
e
k
γ
n
≤
∑
k
=
1
n
e
k
t
-
e
k
1
+
e
k
1
-
e
k
γ
n
≤
∑
k
=
1
n
2
e
k
1
-
e
k
·
1
γ
n
,
1
∞
≤
2
∑
k
=
1
n
ϵ
k
e
k
∞
.Here we used that γ
k
≤
γ
n and hence [
γ
n
,
1
]
⊂
[
γ
k
,
1
] for all k
=
1
,
…
,
n. The preceding estimate implies (22)
∑
k
=
1
n
e
k
∞
≤
max
⁡
∑
k
=
1
n
e
k
·
1
0
,
γ
n
∞
,
∑
k
=
1
n
e
k
γ
n
+
2
∑
k
=
1
n
ϵ
k
e
k
∞
≤
∑
k
=
1
n
e
k
·
1
0
,
γ
n
∞
+
2
∑
k
=
1
n
ϵ
k
e
k
∞
.We conclude(23)
0
≤
∑
k
=
1
n
e
k
∞
-
∑
k
=
1
n
e
k
·
1
0
,
γ
n
∞
≤
2
∑
k
=
1
n
ϵ
k
e
k
∞
,
(24)
∑
k
=
1
N
e
k
∞
≥
∑
k
=
1
N
e
k
·
1
0
,
γ
n
∞
≥
∑
k
=
1
n
e
k
·
1
0
,
γ
n
∞
-
∑
k
=
n
+
1
N
ϵ
k
e
k
∞
≥
∑
k
=
1
n
e
k
∞
-
∑
k
=
1
N
2
ϵ
k
e
k
∞
≥
∑
k
=
1
n
e
k
∞
-
1
4
sup
k
≤
N
⁡
e
k
∞
.Applying this estimate for n and n
-
1 yields (25)
2
∑
k
=
1
N
e
k
∞
≥
∑
k
=
1
n
e
k
∞
+
∑
k
=
1
n
-
1
e
k
∞
-
1
2
sup
k
≤
N
⁡
e
k
∞
≥
e
n
∞
-
1
2
sup
k
≤
N
⁡
e
k
∞
,from which we obtain s
u
p
k
e
k
∞
≤
4
∑
k
=
1
N
e
k
∞. Moreover, using (24) we see that s
u
p
n
∑
k
=
1
n
e
k
∞
≤
2
∑
k
=
1
N
e
k
∞ which proves the first inequality of the proposition. The second inequality is trivial.Now consider, for fixedp and N
>
0, the maps(26)
T
N
f
t
=
N
1
/
p
t
N
-
1
/
p
f
t
N
,
S
N
f
t
=
N
1
/
p
t
N
-
1
/
p
f
t
N
,
t
∈
[
0,1
]. (Here ⌈
r
⌉ is the smallest integer ≥
r.) We include the case p
=
∞ by putting 1
/
∞
=
0. We clearly have |
T
N
f
|
≥
|
S
N
f
| and we obtain T
N
f
p
=
f
p for f
∈
L
p
(
0,1
). If p
=
∞, then S
N
=
T
N.Lemma 9.
One has
(a)(27)
α
T
N
E
,
p
,
ϵ
=
α
1
/
N
E
,
p
,
ϵ
,
β
T
N
E
,
p
,
ϵ
=
β
1
/
N
E
,
p
,
ϵ
,
γ
T
N
E
,
ϵ
=
γ
1
/
N
E
,
ϵ
,
(b)(28)
α
S
N
E
,
p
,
ϵ
1
-
2
ϵ
≥
α
1
/
N
E
,
p
,
ϵ
,
β
S
N
E
,
p
,
ϵ
1
-
2
ϵ
α
1
/
N
E
,
p
,
ϵ
≤
β
1
/
N
E
,
p
,
ϵ
,
(c)(29)
α
1
/
N
E
,
p
,
ϵ
1
-
2
ϵ
e
p
≤
S
N
e
p
≤
e
p
,
e
∈
E
,for 0
<
ϵ
<
1
/
2.Proof.
(a) follows from simple substitution. (Recall thatγ is only defined for p
=
∞.)
(b): Letα
0
=
α
1
/
N
(
E
,
p
,
ϵ
)
=
α
(
T
N
E
,
p
,
ϵ
) and β
0
=
β
1
/
N
(
E
,
p
,
ϵ
)
=
β
(
T
N
E
,
p
,
ϵ
). At first, let p
<
∞. Consider e
∈
E. Put θ
=
⌈
(
N
-
1
)
/
p
⌉
-
(
N
-
1
)
/
p. Since 0
≤
θ
≤
1 and (
S
N
e
)
(
t
)
=
t
θ
(
T
N
e
)
(
t
) for t
∈
[
0,1
], we obtain, with Lemma 6, (30)
S
N
e
·
1
0
,
α
0
p
≤
α
0
θ
T
N
e
·
1
0
,
α
0
p
≤
α
0
θ
ϵ
T
N
e
p
≤
α
0
θ
ϵ
1
-
2
ϵ
T
N
e
·
1
α
0
,
β
0
p
≤
ϵ
1
-
2
ϵ
S
N
e
p
.This yields the first inequality if p
<
∞.
Similarly, we obtain(31)
S
N
e
·
1
β
0
,
1
p
≤
T
N
e
·
1
β
0
,
1
p
≤
ϵ
T
N
e
p
≤
ϵ
1
-
2
ϵ
T
N
e
·
1
α
0
,
β
0
p
≤
α
0
-
θ
ϵ
1
-
2
ϵ
S
N
e
·
1
α
0
,
β
0
p
≤
ϵ
1
-
2
ϵ
α
0
S
N
e
p
.This yields the second inequality of (c) if p
<
∞.
Ifp
=
∞, then the two inequalities follow from (a) and the fact that α is increasing and β is decreasing with respect to ϵ. (Recall that here is S
N
=
T
N).
(c): Ifp
=
∞, then (c) is trivial. Now let p
<
∞. By definition, S
N
e
p
≤
T
N
e
p
=
e
p. On the other hand, with α
0, β
0 of (b), and Lemma 6, (32)
S
N
e
p
≥
S
N
e
·
1
α
0
,
β
0
p
≥
α
0
N
-
1
/
p
-
N
-
1
/
p
T
N
e
·
1
α
0
,
β
0
p
≥
α
0
1
-
2
ϵ
T
N
e
p
=
α
0
1
-
2
ϵ
e
psince 0
≤
⌈
(
N
-
1
)
/
p
⌉
-
(
N
-
1
)
/
p
≤
1.


## 3. Some Facts about Müntz Spaces

IfΛ satisfies (1), then any f
∈
M
p
(
Λ
) can be represented as f
(
t
)
=
∑
n
=
1
∞
α
k
t
λ
k where the series converges pointwise on [
0,1
[. There are universal constants c
k
>
0 with |
α
k
|
≤
c
k
f
p for all k [8]. Moreover, it is easily seen that for any λ
>
0 we have(33)
lim
μ
→
λ
⁡
t
λ
-
t
μ
∞
=
lim
μ
→
λ
⁡
λ
μ
λ
/
μ
-
λ
1
-
λ
μ
=
0
.Lemma 10.
Letϵ
k
>
0 such that ∑
k
=
1
∞
ϵ
k
c
k
<
1
/
2. Moreover, let μ
k satisfy t
λ
k
-
t
μ
k
∞
≤
ϵ
k for all k. Then, T with T
(
∑
n
=
1
∞
α
k
t
λ
k
)
=
∑
n
=
1
∞
α
k
t
μ
k is an isomorphism between M
p
(
Λ
) and M
p
(
{
μ
k
:
k
=
1,2
,
…
}
).Proof.
Forf
(
t
)
=
∑
k
=
1
∞
α
k
t
λ
k we obtain(34)
T
f
-
f
p
≤
∑
k
=
1
∞
c
k
ϵ
k
f
p
≤
1
2
f
p
.This proves 2
-
1
f
p
≤
T
f
p
≤
3
·
2
-
1
f
p.Using Lemma10, we can always assume in the following that Λ
⊂
Q.Put(35)
M
∞
0
Λ
=
f
∈
M
∞
Λ
:
f
1
=
0
=
span
¯
⁡
t
λ
k
-
t
λ
k
+
1
:
k
=
1,2
,
…
.Since M
∞
0
(
Λ
) is 1-codimensional in M
∞
(
Λ
) and M
∞
(
Λ
) contains a complemented subspace isomorphic to c
0, we obtain M
∞
0
(
Λ
)
~
M
∞
(
Λ
).Proposition 11.
FixX
=
M
p
(
Λ
), 1
≤
p
<
∞, or X
=
M
∞
0
(
Λ
), p
=
∞. Then, there are linear bounded finite rank operators R
n
:
X
→
X with l
i
m
n
→
∞
R
n
f
=
f, f
∈
X, R
n
R
m
=
R
m
i
n
(
n
,
m
) if n
≠
m and(36)
c
1
∑
k
=
1
∞
R
k
-
R
k
-
1
f
p
p
1
/
p
≤
f
p
≤
c
2
∑
k
=
1
∞
R
k
-
R
k
-
1
f
p
p
1
/
p
,if 1
≤
p
<
∞, or(37)
c
1
sup
k
⁡
R
k
-
R
k
-
1
f
∞
≤
f
∞
≤
c
2
sup
k
⁡
R
k
-
R
k
-
1
f
∞
,if p
=
∞. Moreover, there are indices m
k
<
n
k, k
=
1,2
,
…, with(38)
R
k
-
R
k
-
1
X
=
span
⁡
t
λ
j
-
t
λ
j
+
1
:
m
k
≤
j
≤
n
k
,if p
=
∞, or(39)
R
k
-
R
k
-
1
X
=
span
⁡
t
λ
j
:
m
k
≤
j
≤
n
k
,if 1
≤
p
<
∞.For a proof see [5, Proposition 9.1
.
4].The conditionl
i
m
n
→
∞
R
n
f
=
f implies the following.Corollary 12.
For the operatorsR
k of Proposition 11 one has f
=
∑
k
=
1
∞
(
R
k
-
R
k
-
1
)
f (with R
0
=
0), f
∈
X, where the series converges in norm.However,R
k
-
R
k
-
1 are no projections and the “summands overlap”; that is, (
R
k
-
R
k
-
1
)
2
≠
R
k
-
R
k
-
1 and (
R
k
-
R
k
-
1
)
(
R
k
+
1
-
R
k
)
≠
0 in general.


## 4. Final Proofs

LetΛ satisfy (1). Fix 1
≤
p
≤
∞. Take R
k, m
k, and n
k as in Proposition 11 and put E
k
=
(
R
k
-
R
k
-
1
)
M
p
(
Λ
), if 1
≤
p
<
∞, or E
k
=
(
R
k
-
R
k
-
1
)
M
∞
0
(
Λ
), if p
=
∞. Fix 0
<
ϵ
k
<
1
/
4 and let 0
<
N
1
<
N
2
<
⋯ be such that(40)
α
1
/
N
k
E
k
,
p
,
ϵ
k
≥
1
2
∀
k
.Let F
j be finite dimensional Banach spaces and put X
=
(
∑
j
=
1
∞
⊕
F
j
)
p, if 1
≤
p
<
∞, or X
=
(
∑
j
=
1
∞
⊕
F
j
)
0, if p
=
∞.Lemma 13.
Assume thatF
j
k
=
S
N
k
E
k for some j
k, k
=
1,2
,
…. Then, M
p
(
Λ
) is isomorphic to a complemented subspace of X.Proof.
Fore
∈
M
p
(
Λ
), 1
≤
p
<
∞, or e
∈
M
∞
0
(
Λ
), p
=
∞, put U
e
=
(
f
j
) where f
j
k
=
S
N
k
(
R
k
-
R
k
-
1
)
e and f
j
=
0 if j
≠
j
k, k
=
1,2
,
…. Using Proposition 11, Corollary 12, and Lemma 9(c), we see that U
e is a well-defined element in X and U is a bounded linear operator M
p
(
Λ
)
→
X if p
<
∞ or M
∞
0
(
Λ
)
→
X. By Lemma 9(c) and condition (40), U is even an into-isomorphism. Hence, there are universal constants c
1
>
0 and c
2
>
0 such that c
1
e
p
≤
U
e
p
≤
c
2
e
p for all e.
It remains to prove thatU
M
p
(
Λ
) and U
M
∞
0
(
Λ
), respectively, are complemented in X. To this end, let f
∈
X. At first assume that, for some N
∈
N, f
=
(
f
1
,
…
,
f
N
,
0,0
,
…
) with f
j
∈
F
j and f
j
k
=
S
N
k
e
k for some e
k
∈
E
k if j
k
≤
N. Put V
f
=
∑
j
k
≤
N
e
k. The fact that R
m
R
n
=
R
m
i
n
(
m
,
n
) if m
≠
n implies (
R
n
-
R
n
-
1
)
(
R
k
-
R
k
-
1
)
=
0 if n
-
k
≥
2. Hence,(41)
R
n
-
R
n
-
1
V
f
=
R
n
-
R
n
-
1
e
n
-
1
+
e
n
+
e
n
+
1
.We obtain a universal constant c
3
>
0 with (42)
R
n
-
R
n
-
1
V
f
p
≤
c
3
e
n
-
1
p
p
+
e
n
p
p
+
e
n
+
1
p
p
1
/
p
≤
4
c
3
S
N
n
-
1
e
n
-
1
p
p
+
S
N
n
e
n
p
p
+
S
N
n
+
1
e
n
+
1
p
p
1
/
pif 1
≤
p
<
∞. Here again, (40) and Lemma 9(c) were used. In view of Proposition 11, this implies that V is a bounded linear operator on the dense subspace of X consisting of finite sequences. Hence, V can be uniquely extended to a linear bounded operator V
:
X
→
M
p
(
Λ
), called V again. The same argument shows that V is also bounded if p
=
∞.
We obtain by definitionV
U
e
=
e if e
∈
M
p
(
Λ
), 1
≤
p
<
∞, or e
∈
M
∞
0
(
Λ
), p
=
∞. This follows from Corollary 12. Indeed, we have U
e
=
(
f
j
) with f
j
k
=
S
N
k
e
k and f
l
=
0 for l
∉
{
j
k
:
k
=
1,2
,
…
}. Hence,(43)
V
U
e
=
V
f
j
=
∑
k
=
1
∞
R
k
-
R
k
-
1
e
=
e
.(Here we used R
0
=
0.) This shows that U
V is a bounded projection from X onto U
M
p
(
Λ
), if 1
≤
p
<
∞, or onto U
M
∞
0
(
Λ
), if p
=
∞. Since M
∞
0
(
Λ
)
~
M
∞
(
Λ
), this completes the proof.Proposition 14.
Let1
≤
p
<
∞. Assume that Ω
k, k
=
1,2
,
…, are finite subsets of positive numbers and put(44)
F
k
=
span
⁡
t
λ
:
λ
∈
Ω
k
⊂
L
p
0,1
.Then, there is a Müntz space M
p
(
Λ
) which is isomorphic to ∑
k
=
1
∞
⊕
F
k
p.
HereΛ
=
∪
k
=
1
∞
⌈
N
k
-
1
/
p
⌉
+
N
k
Ω
k for suitable integers N
k
>
0. If all Ω
k consist of rational numbers, then N
k can be taken such that in addition Λ
⊂
N.Proof.
Fixϵ
k
>
0 such that(45)
∑
k
=
1
∞
ϵ
k
1
-
2
ϵ
k
<
1
16
.Find integers 1
<
N
1
<
N
2
<
⋯ such that m
a
x
N
k
Ω
k
<
m
i
n
N
k
+
1
Ω
k
+
1 for all k and(46)
1
2
<
α
1
/
N
k
F
k
,
p
,
ϵ
k
≤
β
1
/
N
k
F
k
,
p
,
ϵ
k
<
α
1
/
N
k
+
1
F
k
+
1
,
p
,
ϵ
k
+
1for all k. Put α
k
=
α
1
/
N
k
(
F
k
,
p
,
ϵ
k
). Then, we obtain(47)
∑
k
=
1
∞
ϵ
k
1
-
2
ϵ
k
α
k
<
1
8
.Define Λ
=
∪
k
=
1
∞
⌈
N
k
-
1
/
p
⌉
+
N
k
Ω
k. (Of course we can take N
k so large that Λ satisfies (1) and consists of integers if all Ω
k consist of rational numbers.)
We obtain, by Lemma5, Lemma 9 ((a) and (b)), and (46),(48)
α
S
N
k
+
1
F
k
+
1
,
p
,
ϵ
k
+
1
1
-
2
ϵ
k
+
1
α
k
+
1
≥
α
S
N
k
+
1
F
k
+
1
,
p
,
ϵ
k
+
1
1
-
2
ϵ
k
+
1
≥
α
1
/
N
k
+
1
F
k
+
1
,
p
,
ϵ
k
+
1
≥
β
1
/
N
k
F
k
,
p
,
ϵ
k
≥
β
S
N
k
F
k
,
p
,
ϵ
k
1
-
2
ϵ
k
α
k
.Here we used that α
(
·
,
·
,
ϵ
) is increasing with respect to ϵ. Now Proposition 7 yields(49)
M
p
Λ
=
span
¯
⁡
⋃
k
=
1
∞
S
N
k
F
k
~
∑
k
=
1
∞
⊕
S
N
k
F
k
p
.Finally, with Lemma 9(c) we obtain(50)
∑
k
=
1
∞
⊕
S
N
k
F
k
p
~
∑
k
=
1
∞
⊕
F
k
p
.Now we turn to the casep
=
∞.Proposition 15.
LetΩ
k, k
=
1,2
,
…, be finite subsets of positive numbers and put (51)
F
k
=
span
⁡
t
λ
:
λ
∈
Ω
k
⊂
C
0
0,1
,
G
k
=
span
⁡
t
λ
-
t
μ
:
λ
,
μ
∈
Ω
k
⊂
C
0,0
0,1
.Then, there is a Müntz space M
∞
(
Λ
), where Λ
=
∪
k
=
1
∞
N
k
Ω
k for suitable integers N
k, such that(52)
∑
k
=
1
∞
⊕
F
k
0
~
M
∞
Λ
~
∑
k
=
1
∞
⊕
G
k
0
.If all Ω
k consist of rational numbers, then N
k can be taken such that in addition Λ
⊂
N.Proof.
Recall that forp
=
∞ we have S
N
=
T
N and(53)
S
N
f
∞
=
T
N
f
∞
=
f
∞
.Fix λ
¯
k
∈
Ω
k and put H
k
=
s
p
a
n
{
t
λ
¯
k
}. For any f
∈
F
k, we have f
=
(
f
-
f
(
1
)
t
λ
¯
k
)
+
f
(
1
)
t
λ
¯
k which implies(54)
F
k
=
G
k
⊕
H
k
.
Fixϵ
k
>
0 with ∑
k
=
1
∞
ϵ
k
<
1
/
8. Then, we find integers 1
<
N
1
<
N
2
<
⋯ with N
k
λ
¯
k
<
N
k
+
1
λ
¯
k
+
1, (55)
1
2
<
β
1
/
N
k
G
k
,
∞
,
ϵ
k
<
α
1
/
N
k
+
1
G
k
+
1
,
∞
,
ϵ
k
+
1
,
1
2
<
γ
1
/
N
k
F
k
,
ϵ
k
<
α
1
/
N
k
+
1
F
k
+
1
,
∞
,
ϵ
k
+
1
.Put(56)
Y
=
span
¯
⁡
⋃
k
=
1
∞
T
N
k
G
k
,
Z
=
span
¯
⁡
⋃
k
=
1
∞
T
N
k
H
k
,
Λ
=
⋃
k
=
1
∞
N
k
Ω
k
.(The integers N
k should be taken so large that Λ satisfies (1) and that Λ
⊂
N if all Ω
k consist of rational numbers.) Hence Y
⊂
M
∞
0
(
Λ
) and Z
⊂
M
∞
(
Λ
).
Letg
k
∈
G
k and fix M
∈
N. Lemma 9(a) and Proposition 7 imply(57)
7
8
sup
k
≤
M
⁡
T
N
k
g
k
∞
≤
∑
k
=
1
M
T
N
k
g
k
∞
≤
9
8
sup
k
≤
M
⁡
T
N
k
g
k
∞
.Moreover, let f
k
∈
F
k. Then, Lemma 9(a) and Proposition 8 show(58)
1
2
sup
n
≤
M
⁡
∑
k
=
1
n
T
N
k
f
k
∞
≤
∑
k
=
1
M
T
N
k
f
k
∞
≤
sup
n
≤
M
⁡
∑
k
=
1
n
T
N
k
f
k
∞
.Let(59)
P
∑
k
=
1
M
T
N
k
f
k
=
∑
j
=
1
M
f
j
1
t
N
j
λ
¯
j
.Then, P is a projection. (Recall that f
j
(
1
)
=
T
N
j
f
j
(
1
)).
We haveP
≤
2. Indeed, we obtain, in view of (58), (60)
P
∑
k
=
1
M
T
N
k
f
k
∞
=
∑
j
=
1
M
f
j
1
t
N
j
λ
¯
j
∞
=
∑
j
=
1
M
-
1
∑
k
=
1
j
f
k
1
t
N
j
λ
¯
j
-
t
N
j
+
1
λ
¯
j
+
1
+
∑
k
=
1
M
f
k
1
t
N
M
λ
¯
M
∞
≤
sup
j
≤
M
⁡
∑
k
=
1
j
f
k
1
=
sup
j
≤
M
⁡
∑
k
=
1
j
T
N
k
f
k
1
≤
2
∑
k
=
1
M
T
N
k
f
k
∞
.Here we used that t
N
j
λ
¯
j
-
t
N
j
+
1
λ
¯
j
+
1
≥
0 and(61)
∑
j
=
1
M
-
1
t
N
j
λ
¯
j
-
t
N
j
+
1
λ
¯
j
+
1
+
t
N
M
λ
¯
M
≤
1
.Hence, P can be extended to a bounded projection from M
∞
(
Λ
)
=
s
p
a
n
¯
∪
k
=
1
∞
T
N
k
F
k onto Z whose kernel is Y. Equation (58) implies that H
~
c
~
c
0 since dim T
N
k
H
k
=
1 for all k. Thus, we have M
∞
(
Λ
)
~
Y
⊕
c
0. Since Y contains a complemented isomorphic copy of c
0, we obtain M
∞
(
Λ
)
~
Y and (53) and (57) show(62)
Y
~
∑
k
=
1
∞
⊕
T
N
k
G
k
0
~
∑
k
=
1
∞
⊕
G
k
0
.Finally, put X
=
∑
k
=
1
∞
⊕
F
k
0. As a consequence of (54), we have X
~
∑
k
=
1
∞
⊕
G
k
0
⊕
c
0 which implies by the preceding M
∞
(
Λ
)
~
X.Propositions14 and 15 prove Theorem 4. Indeed, in view of (33) and the fact that ·
p
≤
·
∞ for all p, we may assume that all Ω
k consist of rational numbers.We conclude with the following.Proof of Theorem1.
LetΩ
k
⊂
N, k
=
1,2
,
…, be a family of finite subsets such that each finite subset of N occurs infinitely many times among the Ω
k. Put F
k
=
s
p
a
n
{
t
λ
:
λ
∈
Ω
k
}, F
p
=
(
∑
k
=
1
∞
⊕
F
k
)
p, if 1
≤
p
<
∞, and F
∞
=
(
∑
k
=
1
∞
⊕
F
k
)
0. According to Propositions 14 and 15, these spaces are isomorphic to Müntz spaces M
p
(
Λ
^
) with Λ
^
⊂
N. This proves Theorem 1(a).
To prove (b), letΛ be a given set of positive numbers satisfying (1). According to Lemma 10, we may assume that Λ consists of rational numbers. As in the beginning of this section, consider E
k
=
s
p
a
n
{
t
λ
j
:
m
k
≤
j
≤
n
k
}, if 1
≤
p
<
∞, and E
k
=
s
p
a
n
{
t
λ
j
-
t
λ
j
+
1
:
m
k
≤
j
≤
n
k
}, if p
=
∞. Moreover, let N
k be integers fulfilling inequality (40). Put Λ
k
=
{
⌈
N
k
-
1
/
p
⌉
+
N
k
λ
j
:
m
k
≤
j
≤
n
k
} if 1
≤
p
<
∞ and Λ
k
=
{
N
k
λ
j
:
m
k
≤
j
≤
n
k
+
1
} if p
=
∞. At the same time we can take N
k so large that all Λ
k consist of positive integers. Then, Lemma 13 shows that M
p
(
Λ
) is isomorphic to a complemented subspace of (
∑
k
=
1
∞
⊕
S
N
k
E
k
)
p if 1
≤
p
<
∞ which itself is complemented in F
p since the sets Λ
k appear among the sets Ω
n introduced in the proof of part (a). For p
=
∞, Lemma 13 implies that M
∞
(
Λ
) is isomorphic to a complemented subspace of (
∑
k
=
1
∞
⊕
S
N
k
E
k
)
0 which, in view of Proposition 15, is isomorphic to (
∑
k
=
1
∞
⊕
B
k
)
0 where B
k
=
s
p
a
n
{
t
N
k
λ
j
:
m
k
≤
j
≤
n
k
}. (
∑
k
=
1
∞
⊕
B
k
)
0, of course, is complemented in F
∞.



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*Source: 787291-2015-06-16.xml*