Seria I: PRACE MATEMATYCZNE XLV (1) (2005), 23-32
Mohamed Morsli, Fatiha Boulahia
Uniformly non-l
1nBesicovitch-Orlicz space of almost periodic functions
Abstract. We consider uniformly non-l1nproperty of the Besicovitch-Orlicz space of almost periodic functions. It is shown that this property is equivalent to the reflexiv- ity of this space.
2000 Mathematics Subject Classification: 46B20.
Key words and phrases: Uniformly non-ln1, B-convexity, reflexivity, Besicovitch- Orlicz, almost periodic functions.
1. Introduction. Recall at the beginning that a Banach space (X, k.k) is said to be uniformly non-l1n, ( n ∈ N , n ≥ 2) or ((n, δ)−convex) if there is a δ ∈ (0, 1) such that for each choice of x1, x2, ...xnfrom the unit ball of X we have :
kx1± x2± ... ± xnk ≤ n(1 − δ)
for some choice of the signs ±. The definition remains true if we replace kxik ≤ 1 by kxik = 1 for i = 1, 2, ..., n (see [5]).
This geometric property has been introduced by R. C. James in [6]. Banach spaces that are uniformly non - l21 are called uniformly non-square (see [6]). Any uniformly non -square Banach space is reflexive (see [6]). Banach spaces that are uniformly non-l1nfor some integer n ≥ 2 are called B−convex. This notion is useful in the probability theory (see [1], [2], [4]. B-convexity is a topological notion i.e., it is invariant under equivalent renorming (see [4]). Any B-convex Banach space which has an unconditional Schauder bases or which is a Banach lattice is reflexive.
In the class of Orlicz spaces, B-convexity coincides with reflexivity and also with super reflexivity (see [3]).
The purpose of this paper is to study these questions in the Besicovitch-Orlicz space of almost periodic functions.
Now, we shall give some notations and definitions.
2. Preliminaries.
2.1. Orlicz functions. A function φ : R → R+is said to be an Orlicz function if it is even, convex, φ (0) = 0, φ (u) > 0 iff u 6= 0 and lim
u→0 φ(u)
u = 0, lim
u→∞
φ(u) u = +∞.
An Orlicz function admits a derivative φ0 except perhaps on a countable set of points. It satisfies φ0(0) = 0, φ0(|u|) > 0 whenever u > 0 and lim
|u|→∞
φ0(|u|) = +∞, so that is increasing to infinity (see [3], [11])
The function ψ (y) = sup {x |y| − φ (x) , x ≥ 0} is called conjugate to φ. It is an Orlicz function when φ is. The pair (φ, ψ) satisfies the Young inequality
xy ≤ φ (x) + ψ (y) , x ∈ R, y ∈ R.
Let us note that equality holds in the Young’s inequality iff x = ψ0(y) or y = φ0(x) .
We say that an Orlicz function satisfies ∆2-condition if there exist K > 2 and u0≥ 0 for which
φ (2u) ≤ Kφ (u) , ∀u ≥ u0.
In this case we write φ ∈ 42. If φ ∈ 42and ψ ∈ 42we write φ ∈ 42∩ ∇2. When φ /∈ 42 then there exists a sequence (an)n≥1 of positive reals numbers increasing to infinity for which (see [3])
φ
1 + 1 n
an
≥ 2nφ (an) , ∀ n ≥ 1.
2.2. The Besicovitch-Orlicz space of almost periodic functions. Let M (R, E) be the set of all real Lebesgue measurable functions with value in a Banach space (E, k.k).
The functional,
ρBφ: M (R, E) → [0, +∞] , ρBφ(f ) = lim
T →+∞
1 2T
+T
Z
−T
φ (kf (t)k) dt
is a pseudomodular (see [9], [10]).
The associated modular space,
Bφ(R, E) = n
f ∈ M (R, E) , lim
α→0ρBφ(αf ) = 0o
= {f ∈ M (R, E) , ρBφ(λf ) < +∞, for some λ > 0}
is called the Besicovitch-Orlicz space.
This space is endowed with the Luxemberg pseudonorm kf kBφ= inf
k > 0, ρBφ f k
≤ 1
, f ∈ Bφ(R, E) called the Luxemburg norm (see [9], [10]).
Let now A be the linear set of generalized trigonometric polynomials, i.e
A =
P (t) =
n
X
j=1
αjexp (iλjt) , λj∈ R , αj∈ E, n ∈ N
.
The Besicovitch-Orlicz space of almost periodic functions denoted by Bφa.p. (R, E) ( resp. eBφa.p. (R, E) ) is the closure of A in Bφ(R, E) with respect to the pseudonorm k.kBφ ( resp. to the modular convergence ), more exactly :
Bφa.p. (R, E) =
f ∈ Bφ, ∃ (Pn)n≥1⊂ A, s.t. lim
n→+∞kf − PnkBφ= 0 ff
=
f ∈ Bφ, ∃ (Pn)n≥1⊂ A, s.t. ∀k > 0, lim
n→+∞ρBφ(k (f − Pn)) = 0 ff
Beφa.p. (R, E) =
f ∈ Bφ, ∃ (Pn)n≥1⊂ A, s.t. ∃k > 0, lim
n→+∞ρBφ(k (f − Pn)) = 0 ff
. Clearly, Bφa.p. (R, E) ⊂ eBφa.p. (R, E) and equality holds when φ ∈ ∆2.
Some structural and topological properties of these spaces are considered in [7],[9]
and [10].
From [9], we know that when f ∈ Bφa.p. (R, E) and φ ∈ ∆2the limits exists in the expression of ρBφ, i.e :
ρBφ(f ) = lim
T →+∞
1 2T
+T
Z
−T
φ (kf (t)k) dt, f ∈ Bφa.p. (R, E) .
This fact is very useful in our computations.
We now state some fundamental results that will be used below.
From now on, we always denote by φ, ψ a pair of Orlicz functions complementary to each other.
Lemma 2.1 (see [9]). Let f ∈ Bφa.p. (R, E), and suppose the Orlicz function φ satisfies the ∆2-condition. Then :
1. kf kBφ≤ 1 if and only if ρBφ(f ) ≤ 1.
2. kf kBφ= 1 if and only if ρBφ(f ) = 1.
3. ∀ε > 0, ∃η > 0 such that kf kBφ > ε ⇒ ρBφ(f ) > η and the converse is also true.
4. ∀ε > 0, ∃δ (ε) > 0 such that ρBφ(f ) ≤ 1 − ε ⇒ kf kBφ ≤ 1 − δ (ε) .
Lemma 2.2 (see [8]). Let E be a uniformly non-l1nnormed linear space, if ψ satisfies the 42−condition, there exists 0 < r < 1 such that for all x1, x2, ..., xn in E
X
±
φ
x1± x2± ... ± xn
n
≤ 2n−1r n
n
X
i=1
φ (kxik) ,
where the first sum is taken over the 2n−1 choices of signs ±.
Lemma 2.3 (see [9]). Let {an}n≥1, an > 0, be a sequence of real numbers. With every n ≥ 1, we associate a measurable set An⊂ [0, 1] such that :
a) Ai∩ Aj= ∅ if i 6= j and S
n≥1
An⊂ [0, α] , where 0 < α < 1.
b) P
n≥1
φ (an) µ (An) < +∞.
Consider the function f = P
n≥1
anχAn on [0, 1] and let ef be the periodic extension of f to the whole R, with period τ = 1. Then ef ∈ eBφa.p. (R, R).
3. Auxiliary results.
Lemma 3.1 Let φ satisfies the ∆2-condition, then the mapping I : (Lφ([0, 1]) , k.kφ) → Bφa.p. (R, R)
f → fe
(where ef is the periodic extension of f ) is an isometry for the respective modulars and also an isometry for the respective norms. Lφ([0, 1]) is the classical Orlicz space on [0, 1] and k.kφdenotes it’s usual Luxemburg norm.
The results of this lemma remains true if we take an interval [a, b] , a, b ∈ R.
Proof We have only to prove that ef ∈ Bφa.p. (R, R) . Indeed, if f ∈ Lφ([0, 1]) , then for every ε > 0, there exists a step function fε=
n
P
i=1
aiχAi d´efined on [0, 1] such that kf − fεkφ≤ ε4 (see [3], [11]). Since f is absolutely continuous (cf.[3]), we may choose δ > 0 such that µ(A) ≤ δ ⇒ kf χAk ≤ε4. We take α > 0 with 1 − α ≤ δ and put Aαi = Ai∩ [0, α], i = 1, 2, · · · , n. Let fεα=
n
P
i=1
aiχAα
i, then fεα∈ Lφ([0, 1]).
If ef and efεα are the respective periodic extensions of f and fεα to the whole R (with the same period τ = 1), we will have
f − ee fεα
Bφ
≤ kf − fεαkφ
≤
(f − fεα)χ[0,α]
φ+
(f − fεα)χ[α,1]
φ
≤ kf − fεkφ+ f χ[α,1]
φ
≤ ε
4+ε 4= ε
2
From Lemma 2.3 since φ ∈ ∆2we know that efεα∈ Bφa.p. (R, R) , and then there exists a trigonometric polynomial Pεsuch that
feεα− Pε Bφ
≤ ε2, hence,
f − Pe ε
Bφ
≤ f − ee fεα
Bφ
+
feεα− Pε Bφ
≤ ε 2+ε
2 = ε which means that ef ∈ Bφa.p. (R, R) .
The mapping is clearly a modular isometry and also an isometry for the norms. The following Lemma is a direct consequence of Theorem 1.13 in [3]:
Lemma 3.2 Suppose that there exist two constants η, ξ ∈ (0, 1) and δ ≥ 0 such that φ (ηu) ≤ ηξφ (u) ∀u ≥ δ.
Then ψ satisfies the 42−-condition.
Proposition 3.3 We suppose that φ doesn’t satisfies the ∆2-condition. Then Beφa.p. (R, E) contains an isometric copy of C0.
(C0is the classical Banach space of sequences converging to 0).
Proof Let Sn, n ≥ 1, be a family of disjoints subsets of [0, 1[ , such that µ (Sn) = 21n
and S
n≥1
Sn⊂ [0, 1[ .
Let an,1, n ≥ 1, be a sequence of real numbers with an,1> 2n. Since φ does’nt satisfies the 42−condition, for every n ≥ 1, we may find a sequence (an,k)k≥1 increasing to infinity and such that
φ
1 +1 k
an,k
> 2n+kφ (an,k) , k ≥ 1.
Since the measure µ is nonatomic, by the Laponov’s theorem {µ (A) , A is Lebesgue measurable} = [0. + ∞[ .
Consequently we can find disjoint sets {Sn} of [0, 1] with µ (Sn) = 21n and for each n a sequence (Sn,k)k≥1a disjoint subsets of Snwith µ (Sn,k) = 2n+kφ(a1 n,k).
We define fn= P
k≥1
an,kχSn,k, n = 1, 2, ... on [0, 1[ and fn= efn, n = 1, 2, ... with e ∈ E, kek = 1.
Let ffn be the periodic extension of fn to the whole R, with period τ = 1. From Lemma 2.3 we know that ffn∈ eBφa.p. (R, E) . Moreover,
ρBφ
ffn
=
1
Z
0
φ (|fn(t)|) dt
= X
k≥1
φ (an,k) µ (Sn,k) =X
k≥1
1
2n+k < 1, ∀n = 1, 2, ... . Now if λ > 1 there exists k0∈ N such that 1 +1k < λ, ∀k ≥ k0and thus
ρBφ
λffn
=
1
Z
0
φ (λ |fn(t)|) dt =X
k≥1
φ (λan,k) µ (Sn,k)
≥ X
k≥k0
φ
1 +1 k
an,k
µ (Sn,k)
≥ X
k≥k0
2n+kφ (an,k) µ (Sn,k) ≥ X
k≥k0
1 = +∞, ∀n ≥ 1.
So that ffn
Bφ
= 1, ∀n ≥ 1.
Consider now the mapping,
P : C0 −→ X
c = (cn) 7−→ fec= P
n≥1
cnffn,
where X = (
P
n≥1
cnffn, (cn)n≥1∈ C0 )
.
First, we prove that efc∈ eBφa.p. (R, E) . Indeed, if efN =
N
P
n=1
cnffnwe have clearly feN ∈ eBφa.p. (R, E) and then it is sufficient to show that the sequencen
feNo
N ≥1
is norm convergent to efc or that for each λ > 0,
lim
N →+∞ρBφ
λ
fec− efN
= 0.
We have
ρBφ
λ
fec− efN
= ρBφ
X
n≥N
λcnffn
=
1
Z
0
φ
X
n≥N
λcn|fn(t)|
dt = X
n≥N 1
Z
0
φ (λcn|fn(t)|) dt
Taking N such that λcn≤ 1; ∀ n ≥ N, we will have,
ρBφ
λ
fec− efN
= X
n≥N
X
k≥1
φ (λcnan,k) µ (Sn,k)
≤ X
n≥N
X
k≥1
φ (an,k) µ (Sn,k)
≤ X
n≥N
X
k≥1
1 2n+k
≤ 1
2N −1
which tends to zero when N → +∞. Hence the space being complete (see [10]), we get efc∈ eBφa.p. (R, E) .
Now, we show that P is an isometry.
Let λ < kck∞, then there exists n0such that cn0λ = α0> 1.
Consequently :
ρBφ
fec λ
!
=
1
Z
0
φ
X
n≥1
cn λfn(t)
dt ≥
1
Z
0
φ (α0|fn0(t)|) dt = +∞
If λ ≥ kck∞, we will have,
ρBφ
fec λ
!
=
1
Z
0
φ
X
n≥1
cn λfn(t)
dt
≤ X
n≥1 1
Z
0
φ (|fn(t)|) dt ≤X
n≥1
X
k≥1
1 2n+k
≤ 1.
Finally, fec
Bφ= kck∞ and P is an isometry.
4. Uniform non-l1n of eBφa.p. (R, E).
Theorem 4.1 The space eBφa.p. (R, E) is uniformly non-l1nif and only if E is uni- formly non-l1nand both φ and ψ satisfy the 42−condition.
Proof Sufficiency. Suppose E is uniformly non-l1n and both φ and ψ satisfy the 42−condition. Let n be the natural number from the definition of uniformly non-ln1 of E and let f1, f2, ..., fnbe norm one elements in eBφa.p. (R, E) . Since φ ∈ 42we have also ρBφ(fi) = 1 for all i = 1, 2, · · · , n. By Lemma 2.2, there exists 0 < r < 1 such that
(1) X
±
φ
f1(t) ± f2(t) ± ... ± fn(t) n
≤2n−1r n
n
X
i=1
φ (kfi(t)k) . Integrating inequality (1) over [−T, T ] yields
X
± T
Z
−T
φ
f1(t) ± f2(t) ± ... ± fn(t) n
dt ≤ 2n−1r n
n
X
i=1 T
Z
−T
φ (kfi(t)k) dt
and since the limits exist, letting T → +∞ we get:
X
±
lim
T →+∞
1 2T
T
Z
−T
φ
f1(t) ± f2(t) ± ... ± fn(t) n
dt
≤ 2n−1r n
n
X
i=1
lim
T →+∞
1 2T
T
Z
−T
φ (kfi(t)k) dt
≤ 2n−1r n
n
X
i=1
ρBφ(fi) = 2n−1r Consequently,
ρBφ
f1± f2± ... ± fn
n
= lim
T →+∞
1 2T
T
Z
−T
φ
f1(t) ± f2(t) ± ... ± fn(t) n
dt
≤ r < 1
for some choice of signs ±. Hence there exists δ ∈ (0, 1) (δ depends on r) such that
f1± f2± ... ± fn n
Bφ
≤ 1 − δ for some choice of signs ±.
Therefore Bφa.p. (R, E) is uniformly non-ln1.
Necessity. Suppose φ does not satisfy the 42−condition, then by Proposition 3.3, eBφa.p. (R, E) will contain an isometric copy of C0and so eBφa.p. (R, E) can’t be uniformly non- ln1 (it is well known (cf.[4]) that a B-convex Banach space contains no subspace isomorphic to C0).
Now, assume that φ satisfies the 42-condition and ψ does not satisfy the 42- condition, then by Lemma 3.2 we have :
∀η, ξ ∈ (0, 1) and δ ≥ 0, there exists u0≥ δ such that φ (ηu0) > ηξφ (u0) Putting η = 1n. we get
φ 1 nu0
> 1 nξφ (u0) In particular we can write :
∀ξ ∈ (0, 1) , ∃uξ such that φ (uξ) > n and φuξ n
> ξ nφ (uξ) .
By Lapunov’s theorem, there are pairwise disjoint sets A1, A2, ..., An such that µ (Ai) = φ(u1
ξ) i = 1, 2, ..., n;
n
S
i=1
Ai⊂ [0, 1[ , this implies from φ (uξ) > n, µ (Ai) <
1 n.
Define fi = uξ χAi on [0, 1[ , i = 1, 2, ..., n. Let efi be the periodic extension of fi to the whole R, with period τ = 1. From Lemma 2.3, we know that efi ∈ Beφa.p. (R, R) , ∀i = 1, 2, ..., n.
Now let fi = e efiwhere e ∈ E with kek = 1 for i = 1, 2, ..., n. We have : ρBφ
f1± f2± ... ± fn
nξ
≥ 1
ξρBφ
f1± f2± ... ± fn
n
≥ 1
ξ lim
T →+∞
1 2T
T
Z
−T
φ
f1(t) ± f2(t) ± ... ± fn(t) n
! dt
≥ 1
ξ
1
Z
0
φ |f1(t) ± f2(t) ± ... ± fn(t)|
n
dt
≥ 1
ξ
n
X
i=1
φuξ n
µ (Ai)
≥ n
ξφuξ n
1
φ (uξ) ≥ n ξ ξ n = 1.
Thus,
f1± f2± ... ± fn
Bφ ≥ nξ, for any choice of signs ±. ξ ∈ (0, 1) being arbitrary, this means that eBφa.p. (R, E) is not uniformly non-l1n.
The necessity of the uniform non-ln1of E is a direct consequence of the following injective isometry
i : E −→ Beφa.p. (R, E)
x 7−→ xf
where f ∈ eBφa.p. (R, E) satisfies kf kBφ = 1 i.e. E is embedded isometrically into
Beφa.p. (R, E) .
Remark 4.2 The necessity of the 42−condition on ψ may be proved as follows:
Since φ satisfies the 42−condition, the mapping I : (Lφ([0, 1]) , k.kφ) → Bφa.p. (R, R) is a modular isometry for the respective norms (see Lemma 3.1). Then, from the uniformly non-l1nof Lφ([0, 1]), it follows that ψ must be of 42−type (see [3]).
Remark 4.3 If E is a Hilbert space the space eBφa.p. (R, E) is reflexive iff φ ∈
∆2∩ ∇2.
We have only to show the necessity.
Suppose eBφa.p. (R, E) is reflexive, then φ ∈ ∆2since in the contrary case eBφa.p. (R, E) will contain an isometric copy of C0.
To show that we have also ψ ∈ ∆2, remark that since considering the canonical isometry of Lemma 3.1, we deduce that Lφ[0, 1] is also reflexive and so that ψ ∈ ∆2.
Remark 4.4 When E = R, uniform non-l1nproperties are all equivalent in eBφa.p.(R).
In fact, they are equivalent to the reflexivity of the space.
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Mohamed Morsli University of Tizi-Ouzou
Department of Mathematics. Faculty of Sciences.University of Tizi-Ouzou Oued-Aissi, Algeria E-mail: morsli@ifrance.com
Fatiha Boulahia University of Tizi-Ouzou
Department of Mathematics. Faculty of Sciences.University of Tizi-Ouzou Oued-Aissi, Algeria E-mail: Boulahia@yahoo.fr
(Received: 15.06.2003; revised: 22.03.2004)
