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ORDER SCHAUDER BASES IN BANACH LATTICES
ANNA GUMENCHUK, OLENA KARLOVA, AND MIKHAIL POPOV
Abstract. We introduce and study the notion of an order Schauder
basis of a vector lattice E by replacing the norm convergence in the
definition of a Schauder basis with the order convergence in E. By
a bibasis of a Banach lattice E we mean a sequence which is both a
Schauder basis and an order Schauder basis of E. We find necessary
and sufficient conditions for a system to be a bibasis, and extend some
known theorems on Schauder bases to the setting of bibases. Then we
show that the Haar system is a bibasis of Lp with 1 < p < ∞ and
an order Schauder basis of L∞ . One of the results asserts that L1 is
not lattice embedded with σ-order continuous inverse into a σ-order
continuous Banach lattice with a bibasis. However, we do not know if
L1 admits an order Schauder basis.
1. Introduction
We follow the standard terminology and notation on Banach spaces from
the Lindenstrauss-Tzafriri books [7], [8] of the Albiac-Kalton book [3], and
on vector lattices from the Aliprantis-Burkinshaw book [5]. All vector spaces
are considered over the reals only.
By Lp we mean Lp [0, 1] with the Lebesgue measure µ on the Lebesgue
σ-algebra Σ on [0, 1]. The disjoint union A = B ⊔ C for A, B, C ∈ Σ means
that A = B ∪ C and B ∩ C = ∅.
Let E be a vector lattice, (xα ) be a net in E. The notation xα ↓ 0 is
used to mean that the net (xα ) is decreasing (in the non-strict sense) and
inf α xα = 0. A net (xα ) in E order converges to an element x ∈ E (notation
o
xα −→ x) if there exists a net (uα ) with the same indices such that uα ↓ 0 in
E and |xα − x| ≤ uα for all α. One can easily show that an order convergent
sequence must be order bounded. The same is true for general nets having
well ordered index sets. However, similar assertion for general nets does not
hold. On the other hand, it is clear that in most interesting cases we can
restrict ourselves to order bounded nets only, which is necessary to prove
some fundamental results on the order convergence. The order convergence
possesses many usual properties, like limit uniqueness, limit of the sum
equals the sum of limits, taking off the scalar multiple, passing to a limit in
inequalities and others, see [4, p. 322], [2].
2010 Mathematics Subject Classification. Primary 46A35; secondary 46B15; 46A40;
46B42.
Key words and phrases. Vector lattice; Banach lattice; order convergence; basis.
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ANNA GUMENCHUK, OLENA KARLOVA, AND MIKHAIL POPOV
It is also interesting to note that the order convergence of sequences need
not have a topological origin. For example, the linear space L0 of equivalence classes of measurable functions x : [0, 1] → R with the natural order
x ≤ y whenever x(t) ≤ y(t) for almost all t ∈ [0, 1] is a Dedekind complete
vector lattice, however the order convergence in L0 , which is equivalent to
the convergence a.e. on [0, 1], coincides with the convergence in no Hausdorff topology on L0 . Indeed, let (xn ) be a sequence in L0 converging in
measure to zero which is divergent at each point. Then (xn ) has the following additional property: every subsequence (yn ) of (xn ) contains a further
subsequence (zn ) which converges a.e. to zero. If there were a topology on
L0 generating the convergence a.e. then the latter property of (xn ) would
imply its convergence a.e. to zero, a contradiction. So, the order convergence in a vector lattice may lead to new unexpected results in the theory
of bases.
Definition 1.1. Let F be a linear subspace of a vector lattice E. A sequence
(xn ) in F is called an order Schauder basis of F if for every x ∈ F there
exists a unique sequence of scalars (an ) such that
n
X
o
ak xk −→ x.
k=1
A sequence (xn ) in a vector lattice E is called an order Schauder basic
sequence if it is an order Schauder basis of the sequential order closure of
the linear span of (xn ).
For example, the unit vectors (en ) form an order Schauder basis in the
Banach lattices c0 and ℓp with 1 ≤ p ≤ ∞ (the same system in both c0 and
ℓ∞ lattices).
We remark that the notion of an order basis studied in [9] is completely
different.
To distinguish the norm and the sequential order closure of the linear
span of a sequence (xn ) in a Banach lattice E, we will denote them by [xn ]
and [xn ]O respectively.
Definition 1.2. An order Schauder basic sequence (xn ) in a Banach lattice
E is called a strong order Schauder basic sequence if [xn ] = [xn ]O .
Recall that a Banach lattice E is said to be order continuous (σ-order
continuous) if for any net (sequence) (xα ) in E the condition xα ↓ 0 implies
that kxα k → 0. Observe that, in an order continuous (σ-order continuous)
Banach lattice the order convergence of a net (sequence) implies its norm
convergence.
We do not know the answer to the following problem.
Problem 1.3. Let (xn ) be an order Schauder basis of an order continuous
Banach lattice E. Is then (xn ) a Schauder basis of E? What about E = Lp
with 1 ≤ p < ∞?
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ORDER SCHAUDER BASES IN BANACH LATTICES
3
Actually, Problem 1.3 asks of whether an order Schauder basis (xn ) of an
order continuous Banach
lattice E possesses the uniqueness of an expansion
P
of an element x = ∞
a
n=1 n xn in the sense of the norm convergence of E.
Acknowledgements. The authors are very grateful to the referee for huge
work during the reviewing process, whose patience together with high qualification made the paper as it is.
2. Bibases
Our results below concern order Schauder bases which are also Schauder
bases.
Definition 2.1. An order Schauder basis (strong order Schauder basic sequence) (xn ) in a Banach lattice E is called a bibasis (bibasic sequence) if it
is a Schauder basis (basic sequence).
The next proposition easily follows from the definitions and the well
known properties of the order convergence.
Proposition 2.2. Let E be a σ-order continuous Banach lattice. Then
(1) [xn ]O ⊆ [xn ] holds for any sequence (xn ) in E;
(2) every order Schauder basis of E is a strong order Schauder basic
sequence;
(3) if (xn ) is a bibasic
P∞sequence in E then for any sequence of scalars
(an ) the series
n=1 an xn order converges to some x ∈ E if and
only if it norm converges to x.
Theorem 2.3. Let (xn ) be a bibasic sequence in a σ-order continuous Banach lattice E. Then there is a constant M < ∞ such that for each m ∈ N
and each collection of scalars (ak )m
1 one has
i
m
m X
_
X
≤
M
a
x
a
x
k k k k .
(2.1)
i=1 k=1
k=1
Proof. Denote by K the Schauder basis constant of (xn ). Observe that for
any fixed m ∈ N and scalars (ak )m
1 one has
(2.2)
m
i
i
m X
m X
X
_
X
≤
mK
a
x
a
x
≤
a
x
k k .
k k
k k i=1 k=1
i=1 k=1
k=1
Now suppose to the contrary, (2.1) is not satisfied for any number M .
Using (2.2) and the negation of (2.1) we are going to construct a block basis
(un )∞
n=1
mn
X
un =
ak xk
k=mn−1 +1
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ANNA GUMENCHUK, OLENA KARLOVA, AND MIKHAIL POPOV
of (xn ) such that 0 = m0 < m1 < . . . , kun k ≤ 2−n and for every n ∈ N
(2.3)
m
_n
i
X
i=mn−1 +1 k=mn−1 +1
For n = 1 we choose any u1 =
n−1
mX
ak xk .
ak xk > n + k=1
m
P1
ak xk with ku1 k = 2−1 and
k=1
m1 X
i
_
ak xk > 1.
i=1 k=1
Now suppose that u1 , . . . , un−1 are constructed with the desired properties.
n
so that
We choose a number mn > mn−1 and scalars (bk )m
1
mn
X
b
x
k k =
(2.4)
1
2n (K + 1)
k=1
and
n−1
n i
mX
m
mn−1 K + 1
_ X
ak xk + K n
bk x k > n + .
2 (K + 1)
(2.5)
i=1 k=1
k=1
On the other hand,
n−1 i
n i
_
m_
m
X
_ X
bk x k bk x k = i=1
i=1 k=1
n−1 i
m_
X
≤
i=1
k=1
bk x k + mn−1
i
X
i=mn−1 +1 k=1
X
≤ mn−1 K bk x k + k=1
k=1
m
_n
m
_n
i=mn−1 +1
m
_n
i
X
bk x k i=mn−1 +1 k=1
bk x k n−1
mX
bk x k + i
X
k=mn−1 +1
k=1
bk x k .
Then using the equality (x+ y)∨ (x+ z) = x+ (y ∨ z) [5, p. 3], we continue
our previous estimation
n−1
n−1
mX
mX
bk x k +
bk x k + = mn−1 K k=1
k=1
mn−1
X
≤ (mn−1 K + 1)
bk x k + k=1
mn−1 K + 1
≤K n
+
2 (K + 1)
m
_n
m
_n
i=mn−1 +1 k=mn−1 +1
m
_n
i
X
i=mn−1 +1 k=mn−1 +1
i
X
i
X
i=mn−1 +1 k=mn−1 +1
bk xk .
bk xk bk xk 1
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ORDER SCHAUDER BASES IN BANACH LATTICES
Thus, we have
m
_n
(2.6) 5
proved that
i
n i
m
X
mn−1 K + 1
_ X
.
bk x k − K n
bk xk ≥ 2 (K + 1)
i=mn−1 +1 k=mn−1 +1
i=1 k=1
Combining (2.6) with (2.5), we obtain
m
_n
i
X
i=mn−1 +1 k=mn−1 +1
n−1
mX
ak xk .
bk xk > n + k=1
It is left to estimate using (2.4)
mn
mn
mn
n−1
mX
X
X
X
1
bk x k ≤ n .
bk x k + bk xk ≤ (1 + K)
bk x k ≤ 2
k=mn−1 +1
k=1
k=1
k=1
Then we set ak = bk for mn−1 + 1 ≤ k ≤ mn , and our construction is
completed.
P∞
P∞
Now set u =
n=1 un =
k=1 ak xk . Since (xn ) is an order Schauder
basis, by Proposition 2.2, the series order converges as well. In particular,
the sequence of partial sums is order bounded and thus there is a constant
C < ∞ such that
i
n X
_
ak xk ≤ C.
(2.7)
sup
n
i=1
k=1
On the other hand, taking into account (2.7)
≤
=
≤
m
_n
i
X
i=mn−1 +1 k=mn−1 +1
m
_n
i=mn−1 +1
m
_n
m
_n
k=1
mn−1
i
X
X
ak x k −
ak x k i=mn−1 +1 k=1
i
n−1
mX
X
ak xk ak xk + k=1
k=1
i
n−1
X
mX
ak xk + ak xk i=mn−1 +1 k=1
m
_n
ak xk = k=1
i
n−1
n−1
X
mX
mX
ak xk + ak x k ≤ C + ak xk ,
i=mn−1 +1 k=1
what contradicts (2.3).
k=1
k=1
Definition 2.4. Let (xn ) be a bibasic sequence in a σ-order continuous
Banach lattice E. The least constant M < ∞ such that (2.1) is fulfilled will
be called the bibasis constant of (xn ).
Remark 2.5. Obviously, the Schauder basis constant K and the bibasis
constant M of a bibasic sequence satisfy K ≤ M , and for some obvious
cases, these constants are equal. However, as we will see below, the Haar
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ANNA GUMENCHUK, OLENA KARLOVA, AND MIKHAIL POPOV
system, the Schauder basis constant of which equals one, is a bibasis of
Lp for p > 1 with the bibasis constant tending to infinity as p → 1 (see
Proposition 4.7).
According to Lindenstrauss-Tzafriri [8, p. 30], a Banach lattice E has the
+
Fatou property if any
W sequence xn ↑ in E with supn kxn k < ∞ has the exact
upper bound x = n xn ∈ E with kxk = limn kxn k. In another terminology
W
(see e.g. Abramovich-Aliprantis [1, p. 65]), the condition x = n xn ∈ E
moves to the assumptions of the σ-Fatou property, making the definition less
restrictive, and the class of Banach lattices possessing this property wider.
For our purpose the definition of Lindenstrauss-Tzafriri works, so we use this
definition. For example, the classical Banach lattices Lp (µ) with 1 ≤ p ≤ ∞
have the Fatou property.
Theorem 2.6. Let E be a σ-order continuous Banach lattice with the Fatou
property. A sequence (xn ) of nonzero elements is a bibasic sequence if and
only if (2.1) is satisfied.
Proof. In view of Theorem 2.3 we need to prove the sufficiency of (2.1) only.
So, assume (2.1) is fulfilled. Then for each i < m and each collection of
scalars (ak )m
1 one has
m
i
X
X
ak xk .
ak x k ≤ M k=1
k=1
By the well known Schauder basis criterion [7, p. 2], (xn ) is a Schauder
basic sequence with a basis constant ≤ M . Denote by (x∗n ) the biorthogonal
functionals to (xn ) and consider any x ∈ [xn ]. We are going to prove that
n
P
o
x∗k (x)xk −→ x. For each n < m we set
k=1
zn,m =
i
m X
_
x∗k (x)xk − x
i=n k=1
and observe that zn,m ↑ with respect to m. Moreover, using (2.1) we obtain
i
i
m X
m X
_
_
∗
∗
xk (x)xk |x| + xk (x)xk = |x| +
kzn,m k ≤ i=n
i=n k=1
k=1
m
_
≤ kxk + ≤ kxk +
i
X
x∗k (x)xk i=1 k=1
m
X
M
x∗k (x)xk k=1
≤ (1 + M 2 )kxk.
By the Fatou property of E, for each n ∈ N there exists
i
∞ X
∞
_
_
x∗k (x)xk − x ∈ E,
zn,m =
yn =
m=n+1
i=n k=1
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ORDER SCHAUDER BASES IN BANACH LATTICES
7
and moreover,
kyn k = lim kzn,m k.
(2.8)
m→∞
P
n ∗
Obviously, xk (x)xk − x ≤ yn ↓ . It remains to prove that inf yn = 0.
n
k=1
Let z ≤ yn for each n ∈ N. For z + = 0 ∨ z we have that 0 ≤ z + ≤ yn .
Our goal is to prove that z ≤ 0, i.e. that z + = 0. Consider the following
elements for each n < m < ℓ
ℓ
m X
_
x∗k (x)xk .
un,m,ℓ =
i=n k=i+1
Using (2.1) we obtain
ℓ
i
m X
_
X
x∗k (x)xk −
x∗k (x)xk kun,m,ℓ k = i=n k=n
ℓ
m X
_
≤
i=n
k=n
k=n
x∗k (x)xk i
X
+
x∗k (x)xk k=n
i
m X
ℓ
_
X
∗
∗
xk (x)xk xk (x)xk +
= i=n k=n
k=n
ℓ
m
ℓ
X
X
X
≤
x∗k (x)xk + M x∗k (x)xk ≤ (1 + M 2 )
x∗k (x)xk .
k=n
k=n
k=n
Since zn,m = limℓ→∞ un,m,ℓ , for any n < m one has that
∞
X
2 kzn,m k = lim kun,m,ℓ k ≤ (1 + M )
x∗k (x)xk .
ℓ→∞
k=n
Thus, by (2.8), we obtain
∞
X
kz + k ≤ kyn k ≤ (1 + M 2 )
x∗k (x)xk .
k=n
By the arbitrariness of n ∈ N, this yields that z + = 0. In particular,
it follows that [xn ] ⊆ [xn ]O . By (1) of Proposition 2.2, (xn ) is a bibasic
sequence.
3. Stability of bibases
Proposition 3.1. Let (xn ) be a bibasic sequence in a σ-order continuous
Banach lattice E. Then every block basis
un =
mn
X
k=mn−1 +1
ak xk , 0 = m0 < m1 < . . . , un 6= 0
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ANNA GUMENCHUK, OLENA KARLOVA, AND MIKHAIL POPOV
is a bibasic sequence, the bibasis constant of which does not exceed that of
(xn ). In particular, every subsequence of (xn ) is a bibasic sequence.
Proof. By (1) of Proposition 2.2, [un ]O ⊆ [un ]. And by (2) of Proposition 2.2,
(xn ) and hence, (un ) is a Schauder basic sequence. Let x ∈ [un ]. Then there
are constants (bn ) such that
(3.1)
x=
∞
X
bn un =
∞
X
mn
X
a k bn x k .
n=1 k=mn−1 +1
n=1
By proving that
s
X
(3.2)
o
bn un −→ x,
n=1
we show simultaneously the converse inclusion [un ] ⊆ [un ]O and that (un )
is an order Schauder basic sequence, that is, (un ) is a bibasic sequence.
P
o
To prove (3.2), choose scalars (ck ) so that sj=1 cj xj −→ x. Hence, x =
P∞
j=1 cj xj in norm. Since (xn ) is a basic sequence, from the last expansion
of x and (3.1) we obtain that ck = ak bn for all pairs (k, n) ∈ N2 such that
mn−1 + 1 ≤ k ≤ mn . Hence,
s
X
n=1
bn un =
s
X
mn
X
ak bn xk =
n=1 k=mn−1 +1
s
X
mn
X
ck xk =
n=1 k=mn−1 +1
ms
X
o
ck xk −→ x.
k=1
So, (3.2) is proved. It remains to estimate its bibasis constant. For any
scalars (bn )s1 let
s
X
bn un =
n=1
s
X
mn
X
ak bn xk =
n=1 k=mn−1 +1
ms
X
ck xk
k=1
where (ck ) are suitable constants. Then
i
s X
i
s X
_
_
bn u n = i=1 n=1
mn
X
i=1 n=1 k=mn−1 +1
ak bn xk j
ms X
s
X
_
b
u
≤
M
c
x
≤
n n .
k k j=1 k=1
n=1
The following theorem extends the well known Krein-Milman-Rutman
theorem on stability of Schauder basic sequences to the setting of bibases.
Theorem 3.2. Let (xn ) be a normalized bibasic sequence
a σ-order conPin
∞
tinuous Banach lattice E, (yn ) a sequence in E with
n=1 kxn − yn k <
(2K)−1 where K is the Schauder basis constant of (xn ). Then (yn ) is a
bibasic sequence. If, moreover, (xn ) is a bibasis of E then so is (yn ).
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Proof. Set F = [xn ] and G = [yn ]. By [7, Proposition 1.a.9], (yn ) is a
Schauder basis of G. We show that it is also an order Schauder basis of G.
This, in particular, imply
P∞ that G ⊆ [yn ]O , and hence, G = [yn ]O . Indeed,
let y ∈ G, say, y =
n=1 an yn where the series is norm convergent. By
[7, PropositionP1.a.9], the sequences (xn ) and (yn ) are equivalent. Hence,
the series x = ∞
n=1 an xn norm converges. Since (xn ) is a bibasic sequence,
by (3) of Proposition 2.2, the latter series
Pnis orderconvergent to x as well.
Choose a sequence (un ) in E with x − k=1 ak xk ≤ un ↓ 0. Then
∞
n
X
X
ak yk ak y k = y −
k=1
k=n+1
∞
∞
X
X
ak (xk − yk )
≤
ak xk + k=n+1
k=n+1
∞
n
X
X
|ak ||xk − yk |
ak x k +
≤ x −
k=1
≤ un + 2Kkxk
k=n+1
∞
X
|xk − yk |
k=n+1
P
for every n ∈ N. Now observe that vn = ∞
vn ↓
k=n+1 |x k − yk | ↓ 0, because
P
Pn
and kvn k = k>n kxk −yk k → 0 as n → ∞. Thus, y − k=1 ak yk ≤ wn ↓ 0
where wn = un + 2Kkxkvn .
4. The Haar system in Lp with 1 ≤ p ≤ ∞
Now we show that the Haar system is an order Schauder basis of Lp
with 1 < p ≤ ∞, and is not a bibasis of L1 . To this concern, it is worth
mentioning that in Lp for 1 ≤ p < ∞ the order convergence is stronger than
the norm convergence, however, in L∞ the order convergence is weaker than
the norm convergence. More precisely, the following lemma characterizes
the order convergence in these Banach lattices (for a proof, which is an easy
technical exercise, see [4, Lemma 8.17]).
Lemma 4.1. A sequence (xn ) in Lp with 1 ≤ p ≤ ∞ order converges to an
element x ∈ Lp if and only if xn → x a.e. on [0, 1] and it is order bounded
in Lp .
Let (xn ) be a minimal system in a Banach space X, that is, xm ∈
/ [xn ]n6=m
for all m ∈ N. By [12, Theorem 6.1], there are functionals fn ∈ X ∗ , n ∈ N,
such that (xn , fn ) is a biorthogonal system, that is, fn (xm ) = δn,m . If,
moreover, (xn ) is complete in X, that is, [xn ] = X, then obviously, the
functionals fP
n are uniquely defined. Anyway, for every x ∈ [xn ] the terms
of the series ∞
n=1 fn (x) xn , which is called the Fourier series of an element
x ∈ [xn ] with respect to (xn ), do not depend on the choice of functionals fn .
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ANNA GUMENCHUK, OLENA KARLOVA, AND MIKHAIL POPOV
P
The partial sums of the Fourier series are defined as Sn (x) = nk=1 fk (x) xk .
Obviously, if (xn ) is a Schauder basic sequence then the Fourier series of any
x ∈ [xn ] converges to x in norm.
As consequences of Lemma 4.1, we obtain the following characterizations
of order bases in Lp .
Corollary 4.2. Let 1 ≤ p < ∞. A complete minimal system (xn ) in Lp is
a bibasis of Lp if and only if the partial sums Sn (x) of the Fourier series of
any x ∈ Lp with respect to (xn ) have the following properties
(i) Sn (x) tends to x a.e.
on [0, 1];
(ii) the sequence Sn (x) is order bounded in Lp .
Proof. Let a complete minimal system (xn ) be a bibasis of Lp . Given any x ∈
P
P
o
Lp , we find coefficients (an ) so that nk=1 ak xk −→ x. Then ∞
n=1 an xn = x
Pn
o
in Lp , hence Sn (x) = k=1 ak xk −→ x. Then use Lemma 4.1.
Let a complete minimal system (xn ) be such that the Fourier series of
any x ∈ Lp with respect to (xn ) satisfy (i) and (ii). Given any x ∈ Lp , by
o
Lemma 4.1, Sn (x) −→ x. Since the order convergence of a sequence in Lp
implies its norm convergence, every x ∈ Lp has an expansion with respect to
(xn ) converging to x in both norm and order senses. So, if we suppose that
(xn ) is not a bibasis then some x ∈ Lp has two distinct expansions converging
to x in norm (we use here again that the order convergence yields the norm
convergence), that contradicts the minimality of (xn ).
Before the next corollary, observe that if a system of elements (xn ) in L∞
is a bibasis of Lp for some p ∈ [1, +∞) then the partial sums Sn (x) of the
Fourier series of any x ∈ L∞ with respect to (xn ) are well defined. Indeed,
being a Schauder basis (xn ) is a complete minimal system in Lp , and the
partial sums Sn (x) of the Fourier series of any x ∈ L∞ ⊆ Lp with respect to
(xn ) are well defined.
Corollary 4.3. Let a sequence (xn ) in L∞ have the following properties:
(1) it is a bibasis of Lp for some p ∈ [1, +∞);
∞
(2) the sequence Sn (x) n=1 of partial sums of the Fourier series of any
x ∈ L∞ with respect to (xn ) is order bounded in L∞ .
Then (xn ) is an order Schauder basis in L∞ .
o
Proof. Fix any x ∈ L∞ . By Corollary 4.2, Sn (x) −→ x in Lp . By Lemma 4.1
P
o
o
and (2), Sn (x) −→ x in L∞ . Assume that nk=1 ak xk −→ x in L∞ , where
Pn
o
(ak ) are some scalars. Hence, k=1 ak xk −→ x in Lp by Lemma 4.1. Being
a Schauder basis, (xn ) is a complete minimal system in Lp . By Corollary 4.2,
P
n
k=1 ak xk = Sn (x) for all n.
n
∞ 2
A sequence (Gn,k )n=0,k=1
of measurable subsets Gn,k ⊆ [0, 1] is called a
tree of sets if Gn,k = Gn+1,2k−1 ⊔ Gn+1,2k and µ(Gn,k ) = 2−n for n = 0, 1, . . .
and k = 1, . . . , 2n . The corresponding system of functions (g i )∞
i=1 defined
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ORDER SCHAUDER BASES IN BANACH LATTICES
11
by g1 = 1G0,1 and g 2n +k = 1Gn+1,2k−1 − 1Gn+1,2k for n = 0, 1, . . . and k =
1, . . . , 2n is called a Haar type system. The Haar system (hn ) on [0, 1] is the
k
Haar type system with respect to the dyadic intervals In,k = k−1
2n , 2n . We
n
will also consider an L1 -normalized Haar type system g2n +k = 2 g 2n +k , and
the L1 -normalized Haar system h2n +k = 2n h2n +k .
For every x ∈ L1 , by Sn (x) we denote the n-th partial sum of the Fourier
series for x with respect to the Haar system:
n Z
X
Sn (x) =
x hk dµ hk .
k=1
[0,1]
Since (hn ) is a Schauder basis of Lp for any p ∈ [1, +∞), one has that
kx − Sn (x)kp → 0.
Theorem 4.4. The Haar system is a bibasis of Lp with 1 < p < ∞, and an
order Schauder basis of L∞ .
Theorem 4.4 is a consequence of corollaries 4.2 and 4.3, and item (2) of
the following result from [6, p. 83]).
Theorem 4.5. The Fourier series of any function x ∈ L1 converges to x
a.e. on [0, 1]. Moreover, for the function
S ∗ (x)(t) = sup |Sn (x)(t)|, t ∈ [0, 1]
n
one has the following estimates
(1) µ t ∈ [0, 1] : S ∗ (x)(t) > a ≤ Ca kxk1 ;
(2) kxkp ≤ kS ∗ (x)kp ≤ Cp kxkp , x ∈ Lp , 1 < p ≤ ∞,
where C and Cp are some constants.
Theorem 4.6. A Haar type system fails to be a bibasic sequence in L1 .
Proof. Let (gn ) be the Haar type system with respect to a tree of sets
2n
(Gn,k )∞
n=0,k=1 . We show that (gn ) fails (2.1), and this will do by Theorem 2.6. Observe that for each n ∈ N the function zn = 2n 1Gn,1 has the
following expansion with respect to (gn )
zn = g1 + g2 + g3 + g5 + . . . + g2n−1 +1 .
Thus, we have that kzn k = 1 for each n ∈ N and, on the other hand,
|g1 | ∨ |g1 + g2 | ∨ |g1 + g2 + g3 | ∨ . . . ∨ |g1 + g2 + . . . + g2n−1 +1 | = 1[0,1] ∨ z1 ∨ z2 ∨ . . . ∨ zn n
= 1G1,2 + 2 · 1G2,2 + 4 · 1G3,2 + . . . + 2n−1 · 1Gn,2 + 2n · 1Gn,1 = + 1,
2
so (2.1) is not satisfied for each M .
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ANNA GUMENCHUK, OLENA KARLOVA, AND MIKHAIL POPOV
Using arguments of the above proof, one can obtain an estimate for the
bibasis constant Mp of the Haar system (hn ) in Lp for 1 < p < ∞. Indeed,
for the functions zn we obtain kzn kpp = 2n(p−1) and
p
|h1 | ∨ |h1 + h2 | ∨ |h1 + h2 + h3 | ∨ . . . ∨ |h1 + h2 + . . . + h2n−1 +1 | p
p
= 1I1,2 + 2 · 1I2,2 + 4 · 1I3,2 + . . . + 2n−1 · 1In,2 + 2n · 1In,1 p
2p
22p
2(n−1)p
2np
2n(p−1)
1
−1
+ 2 + 3 + ... +
+ n =
+ 2n(p−1) .
2
2
2
2n
2
2p − 2
Dividing the above expression by kzn kpp , we obtain the following estimate
for Mpp for each n ∈ N
!
1 − 2−n(p−1)
2n(p−1) − 1
n(p−1)
p
−n(p−1)
+2
.
=1+
Mp ≥ 2
p
2 −2
2p − 2
=
Thus, we obtain the following statement.
Proposition 4.7. The bibasis constant Mp of the Haar system in Lp , 1 <
p < ∞ satisfies
1
.
Mpp ≥ 1 + p
2 −2
5. L1 is not embedded into a Banach lattice with a bibasis
A linear bounded operator T : X → Y between Banach spaces X and
Y is called an isomorphic embedding, or an into isomorphism, if there is
δ > 0 such that kT xk ≥ δkxk for each x ∈ X. If, moreover, X and Y are
Banach lattices and T is also a lattice homomorphism then T is called a
lattice embedding [5, p. 215]. In this case the image T (X) is a norm closed
vector sublattice of Y . If such an embedding exists, we say that a Banach
lattice X is lattice embedded in a Banach lattice Y .
The well known Pelczy´
nski theorem [10] asserts that L1 is not isomorphically embedded into a Banach space with an unconditional basis. A similar
result is true for bibases.
Theorem 5.1. The Banach lattice L1 is not lattice embedded by means of
a lattice embedding with σ-order continuous inverse map, into a σ-order
continuous Banach lattice with a bibasis.
To prove the theorem, it is convenient to use strictly rich subspaces. A
subspace X of Lp , 1 ≤ p < ∞, is called strictly rich if for every A ∈ Σ there
exists a decomposition A = B ⊔C with µ(B) = µ(C) such that 1B −1C ∈ X.
For example, every subspace X ⊆ Lp of finite codimension in Lp is strictly
rich [11, Theorem 2.15 (a)]. It is interesting to remark that there is a strictly
rich subspace of Lp isometrically isomorphic to Lp [11, Corollary 4.16 (ii)].
Proof of Theorem 5.1. Assume on the contrary that there are a σ-order continuous Banach lattice E with a bibasis (xn ), and a lattice isomorphism
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T from L1 into E with σ-order continuous inverse map T −1 : F → L1
where F = T (L1 ). Let (Pn ) be the basis projections on E, and K the
Schauder basis constant of (xn ). Our goal is to construct an L1 -normalized
Haar type system (gn ), and a normalized
block basis (un ) of (xn ) such that
P∞
for yn = kT gn k−1 T gn one has
ku
− yn k < (2K)−1 . By Proposin
n=1
tion 3.1, (un ) is a bibasic sequence. By the well known and easy fact [7,
p. 6], the Schauder basis constant of (un ) does not exceed K. So, by Theorem 3.2, (yn ), and hence, (T gn ) is a bibasic sequence. By the σ-order
continuity of T −1 , this yields that (gn ) is a bibasic sequence, which contradicts Theorem
P 4.6 (indeed, for each x ∈ [gn ] there exist scalars (an ) such
that T x = ∞
order converges, and hence, by the
n=1 an T gn , where the series P
−1
σ-order continuity of T , the series x = ∞
n=1 an gn order converges in L1 ).
We use the following notation: given x ∈ E \ {0}, we denote x = kxk−1 x.
Set g1 = 1[0,1] and choose n1 ∈ N so that for u1 = Pn1 T g1 one has
ku1 − T g1 k < 2−2 K −1 (this is possible because Pn T g1 → T g1 as n → ∞).
Since the subspace X1 = T −1 ([xk ]k>n1 ) of L1 has finite codimension, it is
strictly rich. Let G0,1 = [0, 1] = G1,1 ⊔ G1,2 be a decomposition such that
µ(G1,i ) = 1/2 for i = 1, 2 and g2 = 1G1,1 − 1G1,2 ∈ X1 . Then T g2 ∈ [xk ]k>n1
and hence Pn1 T g2 = 0.
Then choose n2 > n1 so that for u2 = Pn2 T g2 = (Pn2 − Pn1 )T g2 one has
ku2 − T g2 k < 2−3 K −1 . Analogously, the subspace X2 = T −1 ([xk ]k>n2 ) of L1
is rich. Let G1,1 = G2,1 ⊔ G2,2 be a decomposition such that µ(G2,i ) = 1/4
for i = 1, 2 and g3 = 1G2,1 − 1G2,2 ∈ X2 , hence, Pn2 T g3 = 0.
Now choose n3 > n2 so that for u3 = Pn3 T g3 = (Pn3 − Pn2 )T g3 one has
ku3 − T g3 k < 2−4 K −1 . The subspace X3 = T −1 ([xk ]k>n3 ) of L1 is rich. Let
G1,2 = G2,3 ⊔ G2,4 be a decomposition such that µ(G2,i ) = 1/4 for i = 3, 4
and g4 = 1G2,3 − 1G2,4 ∈ X3 , hence, Pn3 T g4 = 0.
Continuing the procedure in the obvious manner, we complete the construction.
By Theorem 4.6, the Haar system is not an order Schauder basis of L1 .
It is well known that in several senses the Haar system has the best basic
properties. So, we conjecture that L1 does not admit an order Schauder
basis.
Problem 5.2. Does there exist an order Schauder basis of L1 ?
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Chernivtsi Medical College, str. Heroiv Majdanu 60, Chernivtsi, 58000
(Ukraine)
E-mail address: anna [email protected]
Department of Mathematics and Informatics, Chernivtsi National University, str. Kotsyubyns’koho 2, Chernivtsi, 58012 (Ukraine)
E-mail address: [email protected]
Department of Mathematics and Informatics, Chernivtsi National University, str. Kotsyubyns’koho 2, Chernivtsi, 58012 (Ukraine)
E-mail address: [email protected]