More Examples of Hereditarily ℓp Banach Spaces
Extending our previous result, we construct a class of hereditarily ℓp for 1 ≤ p < ∞ (c₀) Banach spaces, investigate their properties, and show that the classical Pitt theorem on compactness of operators from ℓs to ℓp for 1 ≤ p < s < ∞ is false in the general setting of hereditarily ℓs and...
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nasplib_isofts_kiev_ua-123456789-1245852025-02-23T18:59:34Z More Examples of Hereditarily ℓp Banach Spaces Popov, M.M. Extending our previous result, we construct a class of hereditarily ℓp for 1 ≤ p < ∞ (c₀) Banach spaces, investigate their properties, and show that the classical Pitt theorem on compactness of operators from ℓs to ℓp for 1 ≤ p < s < ∞ is false in the general setting of hereditarily ℓs and ℓp spaces. 2005 Article More Examples of Hereditarily ℓp Banach Spaces / M.M. Popov // Український математичний вісник. — 2005. — Т. 2, № 1. — С. 92-108. — Бібліогр.: 10 назв. — англ. 1810-3200 2000 MSC. 46B20, 46E30. https://nasplib.isofts.kiev.ua/handle/123456789/124585 en Український математичний вісник application/pdf Інститут прикладної математики і механіки НАН України |
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Extending our previous result, we construct a class of hereditarily ℓp for 1 ≤ p < ∞ (c₀) Banach spaces, investigate their properties, and show that the classical Pitt theorem on compactness of operators from ℓs to ℓp for 1 ≤ p < s < ∞ is false in the general setting of hereditarily ℓs and ℓp spaces. |
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Popov, M.M. More Examples of Hereditarily ℓp Banach Spaces Український математичний вісник |
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Popov, M.M. |
| author_sort |
Popov, M.M. |
| title |
More Examples of Hereditarily ℓp Banach Spaces |
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More Examples of Hereditarily ℓp Banach Spaces |
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More Examples of Hereditarily ℓp Banach Spaces |
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More Examples of Hereditarily ℓp Banach Spaces |
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More Examples of Hereditarily ℓp Banach Spaces |
| title_sort |
more examples of hereditarily ℓp banach spaces |
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Інститут прикладної математики і механіки НАН України |
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2005 |
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More Examples of Hereditarily ℓp Banach Spaces / M.M. Popov // Український математичний вісник. — 2005. — Т. 2, № 1. — С. 92-108. — Бібліогр.: 10 назв. — англ. |
| series |
Український математичний вісник |
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AT popovmm moreexamplesofhereditarilylpbanachspaces |
| first_indexed |
2025-11-24T12:27:03Z |
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1849674667336400896 |
| fulltext |
Український математичний вiсник
Том 2 (2005), № 1, 92 – 108
More Examples of Hereditarily ℓp
Banach Spaces
Mikhail M. Popov
(Presented by O. I. Stepanets)
Abstract. Extending our previous result, we construct a class of
hereditarily ℓp for 1 ≤ p < ∞ (c0) Banach spaces, investigate their
properties, and show that the classical Pitt theorem on compactness of
operators from ℓs to ℓp for 1 ≤ p < s < ∞ is false in the general setting
of hereditarily ℓs and ℓp spaces.
2000 MSC. 46B20, 46E30.
Key words and phrases. Hereditarily, the spaces ℓp and c0.
1. Preliminaries and Introduction
We use the standard terminology and usual notations as in [5-7]. By
[xi]
∞
i=1 we denote the closed linear span of a sequence {xi}∞i=1 in a Banach
space X. S(X) stands for the unit sphere of a Banach space X. By a
“subspace” of a Banach space we mean a closed linear subspace.
Recall that an infinite dimensional Banach space X is said to be
hereditarily Y (Y is a Banach space), if each infinite dimensional sub-
space X0 of X contains a further subspace Y0 ⊆ X0 which is isomorphic
to Y . Thus, if X is hereditarily Y then we naturally expect to have the
interior properties of X to be close to that of Y . Any exception may be
of interest. So, it is well known that ℓ1 possesses the Schur property (a
Banach space X is said to have the Schur property provided weak con-
vergence of sequences in X implies their norm convergence), while there
are hereditarily ℓ1 Banach spaces without the Schur property [3], [2], [9].
A hereditarily ℓ2 Banach space need not be reflexive, a counterexample
is the James tree JT [4]. See also a recent paper of P. Azimi [1], where
the author makes an attempt to generalize the idea of [2] for constructing
new hereditarily ℓp Banach spaces.
Received 15.03.2004
ISSN 1810 – 3200. c© Iнститут прикладної математики i механiки НАН України
M. M. Popov 93
Using the main idea of [9], we construct classes of hereditarily ℓp, 1 ≤
p < ∞ (and respectively, c0) Banach sequence spaces Zp. Section 3 is
devoted to a proof that ℓp (resp., c0) is isomorphic to a complemented
subspace of Zp, which is used below. Section 4 is devoted to a study of
the question whether the classical Pitt theorem (that if 1 ≤ p < s < ∞
then every continuous linear operator from ℓs to ℓp is compact) remains
true for the setting of hereditarily ℓs and ℓp spaces instead of the ℓs
and ℓp themselves. We show that in general, the answer is negative,
but nevertheless not everything is clear in this emphasis. We state some
open questions in the last section and do some historical comments on
them. We note also that for some values of parameters our spaces Zp are
isometrically embedded in the classical function spaces Lp = Lp[0, 1].
The author would like to thank A. M. Plichko and the participants
of V. K. Maslyuchenko’s seminar (Chernivtsi) for valuable remarks.
We recall that for an arbitrary sequence of Banach spaces {Xn}∞n=1
and any number p ∈ [1,∞) the direct sum of these spaces in the sense of
ℓp is defined as the linear space
X =
( ∞∑
n=1
⊕Xn
)
p
of all sequences x = (x1, x2, · · · ), xn ∈ Xn, n = 1, 2, · · · for which
‖x‖ =
( ∞∑
n=1
‖xn‖p
) 1
p
<∞,
where the norm ‖xn‖ is considered in the corresponding space Xn. Anal-
ogously, the direct sum of the spaces {Xn}∞n=1 in the sense of c0 is defined
as the linear space
X =
( ∞∑
n=1
⊕Xn
)
0
of all sequences x = (x1, x2, · · · ), xn ∈ Xn, n = 1, 2, · · · for which
limn ‖xn‖ = 0 with the norm
‖x‖ = max
n
‖xn‖.
Fix any decreasing sequence P of reals p1 > p2 > · · · > 1 (note that
we do not care if pn tends to 1 or not). Consider any fixed value of p
from the set p ∈ {0} ∪ [1,∞) and the following corresponding sequence
space
XP
p =
( ∞∑
n=1
⊕ℓpn
)
p
,
94 More Examples of Hereditarily ℓp Banach Spaces
where the direct sum is considered in the sense of ℓp, p ≥ 1 or c0 if p = 0.
For each n ≥ 1, denote by {ei,n}∞i=1 the unit vector basis of ℓpn and
by {ei,n}∞i=1 its natural copy in XP
p :
ei,n =
(
0, · · · , 0︸ ︷︷ ︸
n−1
, ei,n, 0, · · ·
)
∈ XP
p .
Let δn > 0 be such reals that for ∆ = (δ1, δ2, · · · ) we have ‖∆‖p = 1
(i.e.
∞∑
n=1
δp
n = 1 if p ≥ 1, and lim
n
δn = 0 and max
n
δn = 1 if p = 0.
For i ≥ 1 put zi =
∞∑
n=1
δnei,n. Evidently, ‖zi‖ = 1 for each i.
Denote by Zp = Zp(P) the closed linear span of {zi}∞i=1 (formally, Zp
depends also on ∆, but actually nothing would change if we replace one
value of ∆ by another and hence we fix ∆ from now on). We show that
Zp is hereditarily ℓp if p ≥ 1 and c0 if p = 0. Note that this construction
is a generalized version of [9] and that this fact is actually proved for
p = 1 in [9].
There is an essential difference between the cases p = 0, 1 and 1 <
p < ∞. For X = ℓ1 or X = c0, every Banach space isomorphic to X
for arbitrary ε > 0 contains a subspace which is (1 + ε)-isomorphic to
X [6,p.97], while this is false for X = ℓp when 1 < p < ∞ [8,p.1348]
(recall that Banach spaces X and Y are said to be λ-isomorphic provided
there exists an isomorphism T : X → Y with ‖T‖ · ‖T−1‖ ≤ λ; evidently,
λ ≥ 1 in this case). Thus, when speaking of hereditarily ℓ1 or c0 spaces,
it is enough to say “subspace isomorphic to X” and by “X is hereditarily
ℓp” we mean the strongest (1 + ε)-isomorphic version, i.e. each infinite
dimensional subspace X0 of X for every ε > 0 contains a further subspace
Y0 ⊆ X0 which is (1 + ε)-isomorphic to ℓp.
Now we recall some notions on bases in Banach spaces. A sequence
{xn}∞n=1 in a Banach spaceX is called a basis forX if for each x ∈ X there
is a unique sequence of scalars {an}∞n=1 such that x =
∞∑
n=1
anxn in the
sense of the norm convergence in X. By a theorem of S. Banach [6,p.1],
the following so-called projections associated with the basis {xn}∞n=1
Pn
( ∞∑
k=1
akxk
)
=
n∑
k=1
akxk,
are uniformly bounded, and the number K = supn ‖Pn‖ is called the basis
constant of the basis {xn}∞n=1. A sequence which is a basis for its closed
linear span is called a basic sequence. A block basis of a basic sequence
M. M. Popov 95
{xn}∞n=1 is any sequence of non-zero elements of the form
uk =
nk+1∑
j=nk+1
ajxj , k = 1, 2, · · · ,
where 0 = n1 < n2 < · · · — some increasing sequence of integers. Evi-
dently, a block basis is also a basic sequence whose basis constant is less
or equal to that of the basic sequence. Two basic sequences {xn}∞n=1 in
X and {yn}∞n=1 in Y are said to be λ-equivalent if there exists an isomor-
phism T : [xi]
∞
i=1 → [yi]
∞
i=1 with ‖T‖ · ‖T−1‖ ≤ λ. Basic sequences are
called equivalent if they are λ-equivalent for some λ ≥ 1. A basis {xn}∞n=1
in a Banach space X is said to be symmetric if for any permutation π
of integers the sequence {xπ(n)}∞n=1 is equivalent to {xn}∞n=1. If they are
1-equivalent for any permutation π then the basis is called 1-symmetric.
2. The Proof that Zp is Hereditarily ℓp
For each I ⊆ N by PI we denote the natural projection of XP
p onto
[ei,n : i ∈ N, n ∈ I] (i.e. with the kernel [ei,n : i ∈ N, n /∈ I]). Of
course, ‖PI‖ = ‖Id − PI‖ = 1. Given an infinite dimensional subspace
E0 of Zp, we find a sequence {xs}∞s=1 in E0 and a block basic subsequence
{us}∞s=1 of {zi}∞i=1 having “almost disjoint supports” and which is close
enough to {xs}∞s=1. (Here by “almost disjoint supports” we mean that for
each ε > 0 there are disjoint subsets Is of N with ‖PIsus‖ ≥ (1− ε)‖us‖).
Hence {xs}∞s=1 contains a subsequence equivalent to the unit vector basis
of ℓp.
Lemma 2.1. For all scalars {ai}m
i=1 and each permutation of integers
τ : N → N one has
∥∥∥∥
m∑
i=1
aizτ(i)
∥∥∥∥
p
=
∞∑
n=1
δp
n
( m∑
i=1
|ai|pn
) p
pn
, if 1 ≤ p <∞
and ∥∥∥∥
m∑
i=1
aizτ(i)
∥∥∥∥ = sup
n∈N
δn
( m∑
i=1
|ai|pn
) 1
pn
, if p = 0.
Hence, {zi}∞i=1 is a 1-symmetric basic sequence.
Proof. The proof is straightforward:
∥∥∥∥
m∑
i=1
aizτ(i)
∥∥∥∥
p
=
∞∑
n=1
δp
n
∥∥∥∥
m∑
i=1
aieτ(i),n
∥∥∥∥
p
=
∞∑
n=1
δp
n
( m∑
i=1
|ai|pn
) p
pn
96 More Examples of Hereditarily ℓp Banach Spaces
for 1 ≤ p <∞ and
∥∥∥∥
m∑
i=1
aizτ(i)
∥∥∥∥ = sup
n∈N
δn
∥∥∥∥
m∑
i=1
aieτ(i),n
∥∥∥∥ = sup
n∈N
δn
( m∑
i=1
|ai|pn
) 1
pn
for p = 0.
Thus, if a series
∞∑
i=1
aizi converges then
∞∑
i=1
|ai|pn <∞ for each n and
lim
n
δn
(
m∑
i=1
|ai|pn
) 1
pn
= 0.
The following lemma as well as its proof exactly coincides with the
corresponding lemma from [9]. To make our note self-contained, we pro-
vide it with a complete proof.
Lemma 2.2. Let E0 be an infinite dimensional subspace of Zp, n,m, j ∈
N (n > 1) and ε > 0. Then there are {xi}m
i=1 ⊂ E0 and {ui}m
i=1 ⊂ Zp of
the form
ui =
ji+1∑
s=ji+1
ai,szs where j = j1 < j2 < ... < jm+1
such that
ji+1∑
s=ji+1
|ai,s|pn−1 = 1 and ‖ui − xi‖ <
ε
m
‖ui‖
for each i = 1, ...,m.
Proof. Put E1 = E0
⋂
[zi]
∞
i=j+1. Since E0 is infinite dimensional and
[zi]
∞
i=j+1 has finite codimension in Zp, E1 is infinite dimensional as well.
Put j1 = j and choose any
x1 =
∞∑
s=j1+1
a1,szs ∈ E1 \ {0}.
Without lost of generality we may assume that
∞∑
s=j1+1
|a1,s|pn−1 = 1
(otherwise we multiply x1 by a suitable number). Then choose j2 > j1
so that for
u1 =
j2∑
s=j1+1
a1,szs
M. M. Popov 97
we have
‖u1 − x1‖ <
ε‖x1‖
4m
, λ1 =
( j2∑
s=j1+1
|a1,s|pn−1
) 1
pn−1 ≥ 1
2
and
‖u1‖ ≥ ‖x1‖
2
.
Hence,
‖u1 − x1‖ <
ε‖u1‖
2m
Now put a1,s = λ−1
1 a1,s, x1 = λ−1
1 x1 and u1 = λ−1
1 u1 . Then
j2∑
s=j1+1
|a1,s|pn−1 =
1
λ
pn−1
1
j2∑
s=j1+1
|a1,s|pn−1 = 1
and
‖u1 − x1‖ =
1
λ1
‖u1 − x1‖ <
ε ‖u1‖
2λ1m
≤ ε ‖u1‖
m
≤ ε ‖u1‖
m
.
Continuing the procedure in the obvious manner, we construct the
desired sequences.
For n ∈ N denote Qn = P{n, n+1, ··· }.
Lemma 2.3. Let E0 be an infinite dimensional subspace of Zp, j, n ∈ N
and ε > 0. There exist an x ∈ E0, x 6= 0 and a u ∈ Zp of the form
u =
l∑
i=j+1
aizi, where l > j
such that
(i) ‖Qnu‖ ≥ (1 − ε) ‖u‖;
(ii) ‖x− u‖ < ε ‖u‖.
Proof. Choose m ∈ N so that
1
δp
n
m
1
pn−1
− 1
pn < ε or
1
δn
m
1
pn−1
− 1
pn < ε if p = 0.
Using Lemma 2.2, choose {xi}m
i=1 ⊂ E0 and {ui}m
i=1 ⊂ Zp to satisfy
the claims of the lemma and put
x =
m∑
i=1
xi and u =
m∑
i=1
ui.
98 More Examples of Hereditarily ℓp Banach Spaces
First, we prove (ii). Since {zs}∞s=1 is 1-symmetric then ‖ui‖ ≤ ‖u‖
for i = 1, ...,m and
‖x− u‖ ≤
m∑
i=1
‖xi − ui‖ <
m∑
i=1
ε ‖ui‖
m
≤
m∑
i=1
ε ‖u‖
m
= ε ‖u‖.
To prove (i), we first show that
‖u‖ − ‖Qnu‖ < m
1
pn−1 .
Anyway, ‖u‖ − ‖Qnu‖ ≤ ‖P{1,··· ,n−1}u‖. Hence, for p ≥ 1 one has
(
‖u‖ − ‖Qnu‖
)p
≤
n−1∑
k=1
δp
k
∥∥∥∥
m∑
i=1
ji+1∑
s=ji+1
ai,ses,k
∥∥∥∥
p
=
n−1∑
k=1
δp
k
( m∑
i=1
ji+1∑
s=ji+1
|ai,s|pk
) p
pk ≤
n−1∑
k=1
δp
k
( m∑
i=1
ji+1∑
s=ji+1
|ai,s|pn−1
) p
pn−1
=
n−1∑
k=1
δp
k
( m∑
i=1
1
) p
pn−1
= m
p
pn−1
n−1∑
k=1
δp
k < m
p
pn−1
and for p = 0
‖u‖ − ‖Qnu‖ ≤ max
1≤k<n
δk
∥∥∥∥
m∑
i=1
ji+1∑
s=ji+1
ai,ses,k
∥∥∥∥
= max
1≤k<n
δk
( m∑
i=1
ji+1∑
s=ji+1
|ai,s|pk
) 1
pk ≤ max
1≤k<n
δk
( m∑
i=1
ji+1∑
s=ji+1
|ai,s|pn−1
) 1
pn−1
= max
1≤k<n
δk
( m∑
i=1
1
) 1
pn−1
= m
1
pn−1 max
1≤k<n
δk ≤ m
1
pn−1 .
On the other hand, for p ≥ 1
‖u‖p =
∞∑
k=1
δp
k
∥∥∥∥
m∑
i=1
ji+1∑
s=ji+1
ai,ses,k
∥∥∥∥
p
≥ δp
n
∥∥∥∥
m∑
i=1
ji+1∑
s=ji+1
ai,ses,n
∥∥∥∥
p
= δp
n
( m∑
i=1
ji+1∑
s=ji+1
|ai,s|pn
) p
pn
≥ δp
n
( m∑
i=1
( ji+1∑
s=ji+1
|ai,s|pn−1
) pn
pn−1
) p
pn
= δp
n
( m∑
i=1
1
) p
pn
= δp
nm
p
pn
M. M. Popov 99
and for p = 0
‖u‖ = max
k∈N
δk
∥∥∥∥
m∑
i=1
ji+1∑
s=ji+1
ai,ses,k
∥∥∥∥ ≥ δn
∥∥∥∥
m∑
i=1
ji+1∑
s=ji+1
ai,ses,n
∥∥∥∥
= δn
( m∑
i=1
ji+1∑
s=ji+1
|ai,s|pn
) 1
pn
≥ δn
( m∑
i=1
( ji+1∑
s=ji+1
|ai,s|pn−1
) pn
pn−1
) 1
pn
= δn
( m∑
i=1
1
) 1
pn
= δnm
1
pn .
Thus, anyway ‖u‖ ≥ δnm
1
pn and hence
1 − ‖Qnu‖
‖u‖ ≤ 1
δn
m
1
pn−1
− 1
pn < ε
and ‖Qnu‖ ≥ (1 − ε) ‖u‖.
Lemma 2.4. Suppose ε > 0 and εs for s ∈ N are such that:
2εs ≤ ε if p = 1;
∞∑
s=1
(2εs)
q ≤ εq if 1 < p <∞ where 1
p + 1
q = 1;
∞∑
s=1
2εs ≤ ε if p = 0.
If for given vectors {us}∞s=1 ⊂ S(Zp) where Zp = Zp(P), there is
a sequence of integers 1 ≤ n1 < n2 < · · · such that the following two
conditions hold
(i) ‖us −Qnsus‖ ≤ εs,
(ii) ‖Qns+1us‖ ≤ εs
for each s ∈ N then {us}∞s=1 is (1 + ε)(1 − 3ε)−1-equivalent to the unit
vector basis of ℓp (respectively, c0).
Proof. Put vs = Qnsus − Qns+1us for s ∈ N. Since vs = us − (us −
Qnsus +Qns+1us), then ‖vs‖ ≥ 1 − 2εs > 1 − 2ε. On the other hand, by
definitions of Qi and the norm on Zp one has ‖vs‖ ≤ ‖us‖ = 1. Thus,
1− 2ε < ‖vs‖ ≤ 1 for each s ∈ N. Then for each scalars {as}m
s=1 one has
(1 − 2ε)p
m∑
s=1
|as|p ≤
m∑
s=1
|as|p‖vs‖p =
∥∥∥∥
m∑
s=1
asvs
∥∥∥∥
p
≤
m∑
s=1
|as|p (1)
100 More Examples of Hereditarily ℓp Banach Spaces
for 1 ≤ p <∞ and
(1 − 2ε) max
1≤s≤m
|as| ≤ max
1≤s≤m
|as|‖vs‖ =
∥∥∥∥
m∑
s=1
asvs
∥∥∥∥ ≤ max
1≤s≤m
|as| (2)
for p = 0. By the lemma conditions
∥∥∥∥
m∑
s=1
as(us − vs)
∥∥∥∥ ≤
∥∥∥∥
m∑
s=1
as(us −Qnsus)
∥∥∥∥+
∥∥∥∥
m∑
s=1
asQns+1us
∥∥∥∥
≤
m∑
s=1
|as|‖us −Qnsus‖ +
m∑
s=1
|as|‖Qns+1us‖ ≤
m∑
s=1
|as| 2εs
then depending on p:
≤
( m∑
s=1
|as|p
) 1
p
·
( m∑
s=1
(2εs)
q
) 1
q
< ε
( m∑
s=1
|as|p
) 1
p
(3)
if 1 < p <∞,
≤
m∑
s=1
|as|p · max
1≤s≤m
2εs ≤ ε
m∑
s=1
|as| (4)
if p = 1 and
≤ max
1≤s≤m
|as| ·
m∑
s=1
2εs < ε max
1≤s≤m
|as| (5)
if p = 0. Using (1) − (5) we obtain
∥∥∥∥
m∑
s=1
asus
∥∥∥∥ ≥
∥∥∥∥
m∑
s=1
asvs
∥∥∥∥−
∥∥∥∥
m∑
s=1
as(us − vs)
∥∥∥∥
depending on p:
≥ (1 − 2ε)
( m∑
s=1
|as|p
) 1
p
− ε
( m∑
s=1
|as|p
) 1
p
= (1 − 3ε)
( m∑
s=1
|as|p
) 1
p
(6)
if 1 ≤ p <∞ and
≥ (1 − 2ε) max
1≤s≤m
|as| − ε max
1≤s≤m
|as| = (1 − 3ε) max
1≤s≤m
|as| (7)
if p = 0. On the other hand,
∥∥∥∥
m∑
s=1
asus
∥∥∥∥ ≤
∥∥∥∥
m∑
s=1
asvs
∥∥∥∥+
∥∥∥∥
m∑
s=1
as(us − vs)
∥∥∥∥
M. M. Popov 101
depending on p:
≤
( m∑
s=1
|as|p
) 1
p
+ ε
( m∑
s=1
|as|p
) 1
p
= (1 + ε)
( m∑
s=1
|as|p
) 1
p
(8)
if 1 ≤ p <∞ and
≤ max
1≤s≤m
|as| + ε max
1≤s≤m
|as| = (1 + ε) max
1≤s≤m
|as| (9)
if p = 0. Combining (6)–(9) we obtain the desired inequalities
(1 − 3ε)
( m∑
s=1
|as|p
) 1
p
≤
∥∥∥∥
m∑
s=1
asus
∥∥∥∥ ≤ (1 + ε)
( m∑
s=1
|as|p
) 1
p
for 1 ≤ p <∞ and
(1 − 3ε) max
1≤s≤m
|as| ≤
∥∥∥∥
m∑
s=1
asus
∥∥∥∥ ≤ (1 + ε) max
1≤s≤m
|as|
for p = 0.
Theorem 2.1. The Banach space Zp = Zp(P) is hereditarily ℓp if 1≤
p <∞ and is hereditarily c0 if p = 0.
Proof. Let E0 be an infinite dimensional subspace of Zp and fix an ε > 0,
quite enough small to satisfy (1+ε)(1−3ε)−1 ≤ 2. Choose any sequence
of positive numbers εs to satisfy the conditions of Lemma 2.4. Then
choose by the Krein-Milman-Rutman stability of basic sequences theorem
[6,p.5] numbers ηs > 0, s ∈ N such that if {xn} is a basic sequence in
a Banach space X with the basis constant ≤ K and ys are vectors in
X with ‖xs − ys‖ < (2K)−1ηs then {ys} is also a basic sequence which
is (1 + ε)-equivalent to {xs}. Using Lemma 2.3, construct inductively
sequences {xs}∞s=1 ⊂ E0, {us}∞s=1 ⊂ Zp of the form
us =
js+1∑
i=js+1
aizi,
where j1 < j2 < ... and ‖us‖ = 1 and a sequence 1 ≤ n1 < n2 < · · · so
that
(i) ‖Qnsus‖ ≥ 1 − εs ,
(ii) ‖us − xs‖ ≤ ηs
4
,
102 More Examples of Hereditarily ℓp Banach Spaces
(iii) ‖Qns+1us‖ < 1 − εs .
To see that this can be done, let j1 = n1 = 1. Choose by Lemma 2.3
an x1 ∈ Zp \ {0} and
u1 =
j2∑
i=j1+1
aizi
such that ‖u1‖ = 1, ‖Qn1u1‖ ≥ 1 − ε1 and ‖x1 − u1‖ < 4−1δ1. Then
choose n2 > n1 so that ‖Qn2u1‖ < ε1. Continuing the procedure in the
obvious manner, we construct the desired sequences.
Evidently, (i) yields
(i′) ‖us −Qnsus‖ ≤ εs.
Conditions (i′) and (iii) imply that {us}∞s=1 is (1 + ε)(1 − 3ε)−1-
equivalent to the unit vector basis of ℓp (respectively, c0), by Lemma
2.4. Then by the choice of {ηs}∞s=1, {xs}∞s=1 is a basic sequence (1 + ε)-
equivalent to {us}∞s=1. Thus, {xs}∞s=1 is (1 + ε)2(1 − 3ε)−1-equivalent to
the unit vector basis of ℓp (respectively, c0).
3. Zp(P) Contains a Complemented Copy of ℓp
Recall that a subspace X of a Banach space Z is called complemented
if there exists a subspace Y of Z such that Z can be decomposed into a
direct sum Z = X ⊕ Y . Of course, for each subspace X of Z there are a
lot of linear subspaces Y ⊆ Z such that Z = X ⊕ Y , but it may happen
that all of them are not closed. In other words, a subspace X of Z is
complemented if it is the range of some linear bounded projection of Z
onto X.
Theorem 3.1. 1. The space Zp = Zp(P) contains a complemented sub-
space isomorphic to ℓp (resp., c0) for each p and P.
2. The space Zp(P)⊕ℓp is isomorphic to Zp (respectively, Z0(P)⊕c0).
Proof. For j,m ∈ N we set ũj,m = zj+1 + · · · + zj+m and uj,m =
‖ũj,m‖−1ũj,m.
We prove the following statement (A): for each n ∈ N and each ε > 0
there is an m0 such that for every j ∈ N and every m ≥ m0 we have
‖uj,m −Qnuj,m‖ < ε. Indeed, for 1 ≤ p <∞
‖uj,m −Qnuj,m‖p =
‖ũj,m −Qnũj,m‖p
‖ũ‖p
=
n−1∑
s=1
δp
s m
p
ps
∞∑
s=1
δp
s m
p
ps
≤
n−1∑
s=1
δp
s m
p
pn−1
δp
n m
p
pn
M. M. Popov 103
<
m
p
pn−1
δp
n m
p
pn
= δ−p
n m
p
pn−1
−
p
pn → 0 as m→ ∞
and for p = 0
‖uj,m −Qnuj,m‖ =
max
1≤s<n
δs m
1
ps
sup
1≤s<∞
δs m
1
ps
≤ m
1
pn−1
δn m
1
pn
= δ−1
n m
1
pn−1
− 1
pn → 0
as m→ ∞ and (A) is proved.
Then, using an inductive procedure, prove the following fact (B):
given a sequence of positive numbers {εs}∞s=1, there exist sequences of
integers 1 = j1 < j2 < · · · and 1 = n1 < n2 < · · · such that for
ũs = ũjs, js+1−js = zjs+1 + · · · + zjs+1 and us =
ũj
‖ũj‖
we have
(i) ‖us −Qnsus‖ ≤ εs,
(ii) ‖Qns+1us‖ ≤ εs
for each s ∈ N.
Indeed, put j1 = n1 = 1 and j2 = 2. Then we have u1 = z2 and
Qn1u1 = u1 and hence (i) is trivially satisfied for s = 1. Then choose
n2 > n1 to satisfy (ii) for s = 1, i.e. so that ‖Qn2u1‖ < ε1. Then
using (A), choose j2 > j1 to satisfy (i). Continuing the procedure in the
obvious manner, we construct the desired sequences.
Now applying to (B) Lemma 2.4, we obtain the following statement
(C): for each ε > 0 there exists a sequence {σj}∞j=1 of disjoint nonempty
finite subsets of N with maxσj < minσj+1 such that the corresponding
block basis with constant coefficients of the basis {zi}∞i=1
us =
∑
n∈σs
zn
spans a subspace E, (1 + ε)-isomorphic to ℓp (resp., c0). By [6,p.116],
E is complemented and the claim 1 of the theorem is proved. Claim 2
follows from [6,p.117].
104 More Examples of Hereditarily ℓp Banach Spaces
4. Operators between Zp1(P1) and Zp2(P2)
Definition 4.1. Let X and Y be any of the spaces ℓp(1 ≤ p < ∞), c0,
Zp (1 ≤ p < ∞, p = 0) with their natural bases {xn}∞n=1 and {yn}∞n=1
respectively. The formal (maybe, unbounded) operator T : X → Y which
extends by linearity and continuity the equality Txn = yn we shall call
the natural operator from X to Y .
Proposition 4.1. Let p ∈ {0} ∪ [1,+∞), P be arbitrary, as above.
(i) If inf
n
pn < p then the natural operator from ℓp to Zp is unbounded.
(ii) If inf
n
pn ≥ p then the natural operator from Zp to ℓp is unbounded.
Proof. For constant scalars a1 = a2 = · · · = am = 1 we have by
Lemma 2.1 ∥∥∥
m∑
i=1
zi
∥∥∥
p
=
∞∑
n=1
δp
n m
p
pn , if 1 ≤ p <∞
and ∥∥∥
m∑
i=1
zi
∥∥∥ = sup
n∈N
δn m
1
pn , if p = 0.
On the other hand,
∥∥∥∥
m∑
i=1
e
(p)
i
∥∥∥∥
p
= m, if 1 ≤ p <∞ and
∥∥∥∥
m∑
i=1
e
(p)
i
∥∥∥∥ = 1, if p = 0.
Consider the case 1 ≤ p <∞ and put
λ(p)
m =
∥∥∥
m∑
i=1
zi
∥∥∥
p
∥∥∥
m∑
i=1
e
(p)
i
∥∥∥
p
=
∞∑
n=1
δp
n m
p
pn
− 1
.
If inf
n
pn < p then there exists an n0 such that pn0 < p and hence
λ(p)
m ≥ δp
n0
m
p
pn0
− 1
→ ∞ as m→ ∞.
Suppose now that inf
n
pn ≥ p. In this case p
pn
− 1 < 0 for each n. Given
ε > 0, choose n0 so that
∞∑
n=n0
δp
n <
ε
2 . Then choose m0 so that
(
max
1≤i≤n0
δi
)p
m
p
pn0
− 1
<
ε
2n0
M. M. Popov 105
for m ≥ m0. Then for such m we have
λ(p)
m =
n0∑
n=1
δp
n m
p
pn
− 1
+
∞∑
n=n0+1
δp
n m
p
pn
− 1
≤
n0∑
n=1
(
max
1≤i≤n0
δi
)p
m
p
pn0
− 1
+
∞∑
n=n0
δp
n <
ε
2
+
ε
2
= ε.
The case p = 0 is quite trivial: λ
(p)
m → ∞ as m→ ∞ anyway.
Thus, we have shown that the basis {zi}∞i=1 of Zp which is normalized
and symmetric (by Lemma 2.1) is not equivalent to the unit vector basis
of ℓp (resp., c0) (which is also normalized and symmetric), for any value
of p. Note that the spaces ℓp, 1 ≤ p <∞ and c0 have, up to equivalence,
a unique symmetric basis [6, p.129]. Therefore we obtain
Corollary 4.1. Let p ∈ {0} ∪ [1,+∞), P be arbitrary. Then the spaces
ℓp and Zp are not isomorphic.
Of course, for distinct indices p 6= s the spaces ℓs and Zp cannot be
isomorphic (see Proposition 4.2 below).
Recall that a linear bounded operator T : X → Y between Banach
spaces (denoted as T ∈ L(X,Y ) is called compact if TB(X) is a relatively
compact set in Y , and is called strictly singular provided the restriction
T |X0 of T to any infinite dimensional subspace X0 ⊆ X is not an isomor-
phic embedding. Of course, each compact operator is strictly singular,
but the converse does not hold, for example for the embedding operators
Ip,s : ℓp → ℓs when 1 ≤ p < s <∞.
Two infinite dimensional Banach spaces are said to be totally incom-
parable if they do not contain isomorphic infinite dimensional subspaces.
For example, each two spaces from the class {c0, ℓp : 1 ≤ p < ∞} are
totally incomparable [6, p.54]. Evidently, if X and Y are totally incom-
parable and X1, Y1 are hereditarily X and respectively, Y then X1 and
Y1 are totally incomparable too. On the other hand, evidently if X and
Y are totally incomparable then each operator T ∈ L(X,Y ) is strictly
singular. Thus we have the following
Proposition 4.2. Let s, p ∈ {0} ∪ [1,+∞), X ∈ {ℓs, Zs} and Y ∈
{ℓp, Zp} (if s = 0 or p = 0 then we mean c0 instead of ℓs or ℓp respec-
tively). If s 6= p then every operator T ∈ L(X,Y ) is strictly singular.
Now we prove that the Pitt theorem does not hold in general for
hereditarily ℓp spaces.
106 More Examples of Hereditarily ℓp Banach Spaces
Example 4.1. Suppose that inf
n
pn ≥ s where P2 = {p1, p2, · · · }. Then
for any p and P1 there exist non-compact operators
T ∈ L(ℓs, Zp(P2)) and T1 ∈ L(Zs(P1), Zp(P2)).
Certainly, the example may be of interest if s > p.
Proof. By Theorem 3.1, it is enough to construct a noncompact operator
T ∈ L(ℓs, Zp) where Zp = Zp(P2). We show that the natural operator
from ℓs to Zp which cannot be compact is bounded. Indeed, let x =
m∑
i=1
aie
(s)
i ∈ ℓs. Since inf
n
pn ≥ s, then
( m∑
i=1
|ai|
pn
) 1
pn
≤ ‖x‖
for each n ∈ N and hence by Lemma 2.1
‖Tx‖ =
( ∞∑
n=1
δp
n
( m∑
i=1
|ai|
pn
) p
pn ) 1
p
≤ ‖x‖
and T can be extended to the whole space ℓs.
5. Remarks and Open Problems
From [7, p.212] we easily deduce
Remark 5.1. If for the set P we have 1 ≤ p ≤ inf
n
pn < p1 ≤ 2 then the
space XP
p and hence its subspace Zp is isometric to a subspace of Ls for
any s ∈ [2, p].
We do not know whether the condition inf
n
pn ≥ s is essential in
Example 4.1. In a view of Proposition 4.1 (i), it looks very likely. More-
over, note that from a result of H. P. Rosenthal (Theorem A2) [20] and
Remark 5.1 we obtain
Corollary 5.1. (1) Let 1 ≤ p < · · · < p2 < p1 ≤ 2 < s < ∞. Then
every operator T ∈ L(ℓs, Zp) is compact.
(2) Let 1 ≤ p < s < · · · < p2 < p1 ≤ 2. Then every operator
T ∈ L(Zs, ℓp) is compact.
Thus, we have
M. M. Popov 107
Problem 1. Suppose that p < s, inf
n
pn < s but the condition in Corol-
lary 5.1 (i) is not fulfilled. Does there exist a non-compact operator
T ∈ L(ℓs, Zp)?
We do not know whether we can replace the range space Zp by ℓp in
Example 4.1. More exactly
Problem 2. Suppose that p < s but the condition in Corollary 5.1 (ii)
is not fulfilled. Does there exist a non-compact operator T ∈ L(Zs, ℓp)?
Or, more general
Problem 3. Let 1 ≤ p < s < ∞ and let X be a hereditarily ℓs Banach
space. Does there exist a non-compact operator T ∈ L(X, ℓp)?
We are not interested in the case when the domain space is c0 because
Remark 4 of [20] yields
Remark 5.2. If a Banach space Y contains no subspace isomorphic to
c0 then every operator T ∈ L(c0, Y ) is compact.
We would like to ask in general:
Problem 4. What properties of the spaces c0 and ℓp for 1 ≤ p < ∞
remain true for hereditarily c0 and respectively ℓp spaces and what are
not true (otherwise trivial cases)?
Some more questions concern the geometric structure of the spaces
Zp. Recall that a Banach space X is said to be primary if for every
decompositions of X onto complemented subspaces X = Y ⊕Z either Y
or Z is isomorphic to X.
Problem 5. Is Zp primary?
Problem 6. How many (finite, countable or uncountable) non-equivalent
normalized symmetric bases does Zp have?
References
[1] P. Azimi, A new class of Banach sequence spaces // Bull. Iran. Math. Soc. 28
(2)(2002), 57–68.
[2] P. Azimi and J. N. Hagler, Examples of hereditarily ℓ1 Banach spaces failing the
Schur property // Pacif. J. Math. 122 (2)(1986), 287–297.
[3] J. Bourgain, ℓ1-subspaces of Banach spaces. Lecture notes. Free University of
Brussels.
[4] R. C. James, A separable somewhat reflexive Banach space with non-separable
dual // Bull. Amer. Math. Soc. 80 (1974), 738–743.
108 More Examples of Hereditarily ℓp Banach Spaces
[5] W. B. Johnson and J. Lindenstrauss, Basic concepts in the geometry of Banach
spaces. Handbook of the geometry of Banach spaces. Vol.I. (W. B. Johnson and
J. Lindenstrauss eds.), Elsevier, Amsterdam. 2001, 1–84.
[6] J. Lindenstrauss and L. Tzafriri, Classical Banach spaces. I. Springer-Verlag,
Berlin-Heidelberg-New York, 1977.
[7] J. Lindenstrauss and L. Tzafriri, Classical Banach spaces. II. Springer-Verlag,
Berlin-Heidelberg-New York, 1979.
[8] E. Odell and Th. Shlumprecht, Distortion and asymptotic structure, Handbook
of the geometry of Banach spaces. Vol.II. (W. B. Johnson and J. Lindenstrauss
eds.) Elsevier, Amsterdam. 2003, 1333–1360.
[9] M. M. Popov, A hereditarily ℓ1 subspace of L1 without the Schur property //
Proc. Amer. Math. Soc. (2005), (to appear).
[10] H. P. Rosenthal, On quasi-complemented subspaces of Banach spaces, with an
appendix on compactness of operators from Lp(µ) to Lr(ν) // J. Funct. Anal. 4
(2)(1969), 176–214.
Contact information
Mikhail M. Popov Department of Mathematics
Chernivtsi National University
str. Kotsjubyn’skoho 2,
58012 Chernivtsi,
Ukraine
E-Mail: popov@chv.ukrpack.net
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