Characterization of the uniform convexity of the Besicovitch-Musielak-Orlicz spaces of almost

Seria I: PRACE MATEMATYCZNE XLVI (2) (2006), 215-231

Mohamed Morsli, Mannal Smaali

Characterization of the uniform convexity of the Besicovitch-Musielak-Orlicz spaces of almost

periodic functions.

Abstract. We introduce the new class of Besicovitch-Musielak-Orlicz spaces of al- most periodic functions Ba.p.^ϕ .The uniform convexity of this space is characterized in terms of it’s generating functional ϕ.

2000 Mathematics Subject Classification: 46B20, 42A75.

Key words and phrases: Besicovitch-Orlicz space, Musielak-Orlicz space, almost pe- riodic functions, uniform convexity .

1. Preliminaries. In the sequel the notation ϕ stands for a function ϕ : R× [0, +∞[ → [0, +∞[ satisfying the following conditions:

• ϕ (t, u) is convex on [0, +∞[ with respect to u.

• ϕ (t, u) is periodic with respect to t ∈ R.

• ϕ (t, u) is continuous on R× [0, +∞[ .

• lim

u→+∞ϕ (t, u) = +∞; lim

u→+∞

ϕ(t,u)

u = +∞ and lim

u→0 ϕ(t,u)

u = 0.

Moreover it is supposed that inf {ϕ (t, α) , t ∈ R} = C (α) > 0, ∀α > 0 and ϕ (t, 0) = 0, ∀t ∈ R.

We denote by M (R) the space of all Lebesgue measurable functions on R and by L^ϕ_loc (R) it’s subspace of ϕ−locally integrable functions, i.e. the subspace of functions f ∈ M (R) such that for each compact K ⊂ R there exists a λK > 0 for

whichR

K

ϕ (t, λK|f (t)|) dt < +∞.

The functional

ρ_ϕ: L^ϕ_loc(R) → [0, +∞]

f 7→ ρ_ϕ(f ) = lim_{T →+∞2T}¹

+T

R

−T

φ (t, |f (t)|) dt

is a convex pseudomodular on L^ϕ_loc(R) .

The linear space associated to this pseudomodular,

B^ϕ(R) = n

f ∈ L^ϕ_loc(R) : lim_α→0ρϕ(αf ) = 0o

= {f ∈ L^ϕ_loc(R) : ρϕ(αf ) < +∞, for some α > 0} ,

is called the Besicovitch-Musielak-Orlicz space and is endowed with the usual Luxemburg’s pseudonorm

kf k_ϕ= inf



k > 0/ρϕ

 f k



≤ 1



, f ∈ B^ϕ(R) .

Let A be the linear set of all generalized trigonometric polynomials, i.e.:

A =







Pn(t) =

n

X

j=1

aje^iλjt, aj∈ C, λj ∈ R, n ∈ N





 .

We denote by ˜B_a.p.^ϕ (R) (resp. B^ϕa.p.(R)) the closure of A with respect to the pseudomodular ρϕ (resp. with respect to the pseudonorm k.k_ϕ), more precisely:

B˜_a.p.^ϕ (R) =



f ∈ B^ϕ(R) : ∃fn∈ A, ∃k0> 0 s.t. lim

n→+∞ρ_ϕ(k₀(f_n− f )) = 0



B_a.p.^ϕ (R) =



f ∈ B^ϕ(R) : ∃fn∈ A, s.t.∀k > 0 lim

n→+∞ρ_ϕ(k (f_n− f )) = 0



=



f ∈ B^ϕ(R) : ∃fn∈ A s.t. lim

n→+∞kf_n− f k_ϕ= 0

 .

B˜_a.p.^ϕ (R) and Ba.p.^ϕ (R) will be called Besicovitch-Musielak-Orlicz spaces of almost periodic functions.

As usual {u.a.p.} denotes the algebra of Bohr’s almost periodic functions i.e., the closure of the set A in the uniform norm of Cb(R) (the space of continuous and bounded functions on R).

The following inclusions hold :

B_a.p.^ϕ (R) ⊆ ˜B_a.p.^ϕ (R) ⊆ B^ϕ(R) .

The function ϕ (t, u) is called uniformly convex if : ∀ε ∈ ]0, 1[ , ∃f_ε∈ M (R) with ρ_ϕ(f_ε) = ε and ∃p (ε) ∈ ]0, 1[ s.t. ∀x, y ∈ [0, +∞[ : if max (x, y) ≥ |f_ε(t)| and

|x − y| ≥ ε max (x, y) then

ϕ

 t,x + y

2



≤ 1 − p (ε)

2 (ϕ (t, x) + ϕ (t, y)) , for almost all t ∈ R.

We say that ϕ (t, u) satisfies the condition ∆2(ϕ ∈ ∆2) if there exist k > 1 and a measurable nonnegative function h with ρϕ(h) < +∞ such that ϕ (t, 2u) ≤ kϕ (t, u) for almost all t ∈ R and all u ≥ h (t) .

Note that this definition of condition ∆₂is similar to that for classical Musielak- Orlicz spaces, the condition of integrability of h being replaced by ρ_ϕ(h) < +∞ (see [2]) .

2. Auxiliary results. The structure of the class B_a.p.^ϕ (R) is not adapted to the methods of Lebesgue measure theory. In fact, the usual convergence results are not valid in the B^ϕ_a.p.(R) spaces (see [9]).

To handle B_a.p.^ϕ (R) spaces as L^ϕ(R) ones, we introduce the set function µ.

Let Σ (R) denotes the σ−algebra of Lebesgue measurable subsets of R. We denote by µ the set function defined on Σ (R) by:

¯

µ (A) = limT →+∞ 1 2T

+T

R

−T

χA(t) dt = limT →+∞ 1

2Tµ (A ∩ [−T, +T ]) where µ is the Lebesgue’s measure on R.

As usual a sequence {fn} from B^ϕ(R) is said to be ¯µ−convergent to some f ∈ B^ϕ(R) (in symbol fⁿ−→ f ) when, ∀α > 0^µ

n→+∞lim µ {x ∈ R, |f¯ ⁿ(x) − f (x)| > α} = 0.

We give here some technical results that are the key arguments in the proof of the main theorem.

Lemma 2.1 Let {fn}_n≥1 ⊂ B_a.p.^ϕ (R) . If lim

n→+∞ρ_ϕ(f_n− f ) = 0 for some f ∈ B^ϕ_a.p.(R) , then fⁿ−→ f.^µ

Proof Put A^αn = {t ∈ R, |fn(t) − f (t)| ≥ α} , then

ρϕ(fn− f ) = limT →+∞

1 2T

+T

Z

−T

ϕ (t, |fn(t) − f (t)|) dt

≥ limT →+∞

1 2T

+T

Z

−T

ϕ t, αχA^α_n(t)
dt

≥ C (α) ¯µ (A^α_n) . It follows directly that lim

n→+∞µ (A¯ ^α_n) = 0, i.e. f_n−→ f.^µ _

Remark 2.2 The condition inf {ϕ (t, α) , t ∈ R} = C (α) > 0, ∀α > 0 is necessary in Lemma 2.1, as it is shown by the following example:

We define ϕ : R× [0, +∞[ → [0, +∞[ by ϕ (t, u) = f (t).u² where f : R → [0, +∞[

is a continuous and periodic function (with period T = 1) defined by:

f (t) =











0 if t ∈0,³₈ 8t − 3 if t ∈₃

8,¹₂

−8t + 5 if t ∈₁

2,⁵₈

0 if t ∈₅

8, 1

Consider now the continuous and periodic function (with period T = 1) defined as follows

h(t) =















1 if t ∈0,¹₄

−8t + 3 if t ∈1 4,³₈

0 if t ∈₃

8,⁵₈ 8t − 5 if t ∈₅

8,³₄

1 if t ∈₃

4, 1

It is easily seen that ϕ(t, |h(t)|) = f (t) h²(t) = 0 ∀t ∈ R and then ρϕ(h) = 0.

But clearly ¯µt ∈ R, |h (t)| ≥ ¹2 ≥ ¹₂.

Lemma 2.3 Let h ∈ B^ϕ(R) with ρ^ϕ(h) = a > 0. Then

(1) ∀θ ∈ (0, 1) , ∃β > 0 and G = {t ∈ R, |h (t)| ≤ β} such that ¯µ (G) ≥ θ.

Proof For θ ∈ (0, 1) take β > 0 such that (1 − θ) C (β) > 2ρϕ(h) = 2a and suppose that (1) is false. Then, for some sequence {Tn} increasing to infinity we will have

µ{G∩[−T_n,+T_n]}

2Tn < θ. Hence denoting by G⁰ the complementary of G in R we get

1 2Tn

+T_n

Z

−Tn

ϕ (t, |h (t)|) dt ≥ 1 2Tn

Z

G⁰∩[−Tn,+Tn]

ϕ (t, |h (t)|) dt

≥ C (β) 1

2T_nµ {G⁰∩ [−T_n, +T_n]}

≥ C (β) (1 − θ) > 2a,

and letting n tends to infinity, it follows ρϕ(h) > 2a, a contradiction. _

Lemma 2.4 Let g ∈ Ba.p^ϕ (R) , then

∀ε > 0 ∃δ > 0 s.t.∀Q ∈X

(R) , ¯µ (Q) ≤ δ ⇒ ρϕ(gχQ) ≤ ε.

Proof We may suppose ρ^ϕ(g) > 0.

For ε > 0 take Pε∈ A such that ρϕ(2 (g − Pε)) < ^ε₂and put Mε= sup

t∈R

ϕ (t, 2 |Pε(t)|).

Let θ ∈ (0, 1) satisfying Mε(1 − θ) < ^ε₂. If β and G are as in Lemma 2.3 we will have

lim_{T →+∞} 1 2T

Z

G⁰∩[−T,+T ]

ϕ (t, |g (t)|) dt (2)

≤ 1

2limT →+∞

1 2T

Z

G⁰∩[−T,+T ]

[ϕ (t, 2 |g (t) − Pε(t)|) + ϕ (t, 2 |Pε(t)|)] dt

≤ ε

4 +1

2Mε(1 − θ) ≤ ε 2 Let δ = _{2 sup}^ε

t∈R

ϕ(t,β) and Q ∈ P (R) with ¯µ (Q) ≤ δ. Putting Q1 = Q ∩ G and Q2= Q ∩ G⁰ it follows

limT →+∞

1 2T

Z

Q₁∩[−T,T ]

ϕ (t, |g (t)|) dt ≤ limT →+∞

1 2T

Z

Q₁∩[−T,T ]

ϕ (t, β) dt

≤ sup

t∈R

ϕ (t, β) ¯µ (Q₁)

≤ δ sup

t∈R

ϕ (t, β) ≤ ε 2. Similarly, in view of (2) we have

limT →+∞

1 2T

Z

Q₂∩[−T,T ]

ϕ (t, |g (t)|) dt ≤ limT →+∞

1 2T

Z

G⁰∩[−T,+T ]

ϕ (t, |g (t)|) dt ≤ ε 2. Finally,

limT →+∞

1 2T

Z

Q∩[−T ,+T ]

ϕ (t, |g (t)|) dt ≤ ε,

which means that ρ_ϕ(gχ_Q) ≤ ε. _

Proposition 2.5 Let f ∈ Ba.p^ϕ (R) . Then ϕ (t, |f (t)|) ∈ Ba.p¹ (R) and consequently lim

T →+∞

1 2T

+T

R

−T

ϕ (t, |f (t)|) dt exists (and is finite).

Proof Let {fn} be a sequence of trigonometric polynomials such that kfn− f k_ϕ→ 0, then using Lemma 2.1 we have also fn

−→ f.µ

Let θ ∈ (0, 1), in view of Lemma 2.3 there exist β > 0 and a set G = {t ∈ R, |f (t)| ≤ β}

for witch ¯µ (G) ≥ θ.

Take α > 0 and A^α_n = {t ∈ R, |fn(t) − f (t)| > α}, then it is easily seen that

|f_n(t)| ≤ β + α, ∀t ∈ G ∩ (A^α_n)⁰ and since ϕ is continuous on R× [0, +∞[, periodic

with respect to t ∈ R, using the fact that |fⁿ(t)| , |f (t)| ∈ [0, β + α] for t ∈ G∩(A^α_n)⁰, we claim that

∀η > 0 ∃α_η> 0 ∀t ∈ G∩(A^α_n)⁰, |ϕ (t, |f_n(t)|) − ϕ (t, |f (t)|)| ≥ η =⇒ |f_n(t) − f (t)| > α_η then, since fn

−→ f we get alsoµ n→+∞lim µ¯n

t ∈ G ∩ (A^α_n)⁰, |ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| ≥ ηo

= 0.

Consequently,

µ {t ∈ R, |ϕ (t, |f¯ ⁿ(t)|) − ϕ (t, |f (t)|)| ≥ η}

≤ µ¯n

t ∈ G ∩ (A^α_n)⁰, |ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| ≥ ηo +¯µ {t ∈ G⁰, |ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| ≥ η}

+¯µ {t ∈ A^α_n, |ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| ≥ η}

≤ µ¯n

t ∈ G ∩ (A^α_n)⁰, |ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| ≥ ηo +¯µ (G⁰) + ¯µ (A^α_n)

≤ µ¯n

t ∈ G ∩ (A^α_n)

0

, |ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| ≥ ηo + (1 − θ) + ¯µ (A^α_n) ,

hence, letting n tends to infinity we will have

limn→+∞µ {t ∈ R, |ϕ (t, |f¯ ⁿ(t)|) − ϕ (t, |f (t)|)| ≥ η} ≤ (1 − θ) . Finally, since θ ∈ (0, 1) is arbitrary, we deduce the following

(3) ∀η > 0, lim

n→+∞µ {t ∈ R, |ϕ (t, |f¯ n(t)|) − ϕ (t, |f (t)|)| ≥ η} = 0.

Let now ε > 0, then there exists δ = δ (f, ε) > 0 such that for all Q ∈P (R) with

¯

µ (Q) ≤ δ we have ρϕ(fnχQ) ≤ ε for sufficiently large n. Indeed, since ρϕ(fnχQ) ≤

1

2[ρϕ(2 (f − fn) χQ+ ρϕ(f χQ))] , using lemma 2.4 the result is immediate.

On the other hand from (3) there is n0∈ N such that for n ≥ n0

µ {t ∈ R, |ϕ (t, |f¯ n(t)|) − ϕ (t, |f (t)|)| ≥ ε} ≤ δ.

Let A^ε_n= {t ∈ R, |ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| ≥ ε} . Then

limT →+∞

1 2T

+T

Z

−T

|ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| dt

≤ limT →+∞

1 2T

Z

Aⁿ_ε∩[−T,T ]

|ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| dt

+limT →+∞

1 2T

Z

(Aⁿ_ε)⁰∩[−T,T ]

|ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| dt,

and since ¯µ (Aⁿ_ε) ≤ δ for some δ > 0 and n ≥ n0 we get

limT →+∞

1 2T

Z

Aⁿ_ε∩[−T,T ]

|ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| dt

≤ limT →+∞

1 2T

Z

Aⁿ_ε∩[−T,T ]

ϕ (t, |fn(t)|) dt + limT →+∞

1 2T

Z

Aⁿ_ε∩[−T,T ]

ϕ (t, |f (t)|) dt

≤ 2ε.

On the other hand we have also limT →+∞

1 2T

Z

(Aⁿ_ε)⁰∩[−T,T ]

|ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| dt ≤ ε

then, since ε > 0 is arbitrary, we get

(4) lim

n→+∞lim_{T →+∞} 1 2T

+T

Z

−T

|ϕ (t, |fn(t)|) − ϕ (t, |f (t)|)| dt = 0.

Now, the continuous function ϕ : R× [0, +∞[ → [0, +∞[ being periodic with re- spect to t ∈ R and since for each n ∈ N, fⁿ∈ {u.a.p} , we deduce that ϕ (t, |fn(t)|) ∈ {u.a.p} (see[3], p.61).

Finally from (4) it follows that ϕ (t, |f (t)|) ∈ B_a.p¹ (R) and by a classical result (see[1]) the limit lim

T →+∞

1 2T

+T

R

−T

ϕ (t, |f (t)|) dt exists (and is finite). _

Lemma 2.6 Let {fn} ⊂ B¹_a.p(R) be such that fⁿ −→ f ∈ B^{µ 1}_a.p(R) . Suppose there exists g ∈ B_a.p¹ (R) for which max (|fⁿ(t)| , |f (t)|) ≤ g (t) , ∀t ∈ R, then ρ¹(fn) → ρ1(f ) .

Remark 2.7 The notation Ba.p¹ (R) is used for the space corresponding to the func- tion ϕ (t, x) = |x| and ρ₁ is the associated modular.

Proof Take ε > 0 and put En^ε =t ∈ R/ |fn(t) − f (t)| ≥ ^ε₂ . Let δ > 0 be asso- ciated to the function 2g and ^ε₂ as in Lemma 2.4. Since f_n −→ f it follows that^µ

¯

µ (E_n^ε) ≤ δ for n ≥ n₀ and then by Lemma 2.4 ρ₁ (f_n(t) − f (t)) χ_Eε

n
≤ ρ1 2gχ_Eε n
≤ ε

2.

Finally we get for n ≥ n0

|ρ1(fn(t)) − ρ1(f (t))| ≤ ρ1((fn(t) − f (t)))

≤ ρ1



(fn(t) − f (t)) χ_Eε

n

 + ρ1



(fn(t) − f (t)) χ

(Eεn)⁰



≤ ε

2 +ε 2 = ε i.e. lim

n→+∞ρ₁(f_n) = ρ₁(f ) _

Lemma 2.8 Let f ∈ Ba.p^ϕ (R), then the functional λ 7→ ρ^ϕ_f

λ



is continuous on ]0, +∞[ .

Proof Let λ⁰ ∈ ]0, +∞[ and{λn} be a sequence of real numbers which converges to λ0.We have

ρ_ϕ f λn

− f λ0



≤    

1 λn

− 1 λ0

ρ_ϕ(f ) , ∀n ≥ n₀ and then lim

n→+∞ρϕ

_f

λ_n −_λ^f

0



= 0.

From Lemma 2.1, it follows_λ^f

n

−→µ _λ^f

0 and then ϕ t,^{|f (t)|}_λ

n

 _µ

−→ ϕ t,^{|f (t)|}_λ

0

 (see (3)) . Furthermore, max

ϕ t,^{|f (t)|}_λ

n

, ϕ t,^{|f (t)|}_λ

0

≤ ϕ t,_λ²

0|f (t)|

∈ B¹_a.p(R)(see prop. 2.5), consequently, using Lemma 2.6 we deduce

ρ_ϕ f λn



→ ρϕ

 f λ0



this means that λ 7→ ρϕ

_f

λ



is continuous at λ0. _

Corollary 2.9 Let f ∈ B^ϕa.p(R) , then 1. kf k_ϕ≤ 1 if and only if ρϕ(f ) ≤ 1.

2. kf k_ϕ= 1 if and only if ρϕ(f ) = 1.

3. ∀ε ∈ ]0, 1[ , ∃δ ∈ ]0, 1[ such that ρϕ(f ) ≤ δ implies kf k_ϕ≤ ε.

Proof This follows from Lemma 2.8 and usual arguments of Orlicz spaces theory.

Note that a similar result holds in classical Musielak-Orlicz spaces with the ad- ditional condition ∆₂ on the function ϕ. In fact, this condition is necessary for the continuity of the function λ 7→ ρϕ

_f

λ

 .

Here the continuity holds without condition ∆₂, but with the restriction f ∈

B^ϕ_a.p(R) . 

Lemma 2.10 Let ϕ (t, u) ∈ ∆² then

∀θ ∈ (0, 1) ∀ε > 0∃hε∈ B^ϕ(R) and k⁰> 1 such that:

(5) ϕ (t, 2u) ≤ k⁰ϕ (t, u) ∀u ≥ hε(t) , ∀t ∈ G with ¯µ (G⁰) < θ and ρϕ(hε) < ε.

Proof Since ϕ (t, u) ∈ ∆2, by definition there exist k > 1 and a nonnegative function h ∈ B^ϕ(R) such that ϕ (t, 2u) ≤ kϕ (t, u) for almost all t ∈ R and all u ≥ h(t).

Let us remark that the later holds if we replace h (t) by the function h1(t) =

 δ if 0 ≤ h(t) ≤ δ h(t) if h(t) ≥ δ .

Then we may assume that h(t) ≥ δ, ∀t ∈ R for some δ > 0. Moreover for ε > 0 there exists η0> 0 such that ρϕ

h η0



< ε.

We take θ ∈ (0, 1) and β > 0 as in Lemma 2.3 then ¯µ (G⁰) ≤ θ where G = {t ∈ R, |h (t)| ≤ β} .

Now if we put hε= _η^h

0 and k⁰= max (k, k1) where k1= max ϕ (t, 2u)

ϕ (t, u) , u ∈ δ η₀, β

 , t ∈ R

 ,

we obtain (5) . Note that k1< +∞ due to the periodicity of ϕ (., u) . _

Lemma 2.11 Let f ∈ ˜B^ϕ_a.p(R) and ϕ ∈ ∆2, then

∀ε ∈ ]0, 1[ ∃δ (ε) ∈ ]0, 1[ s.t. ρϕ(f ) ≤ 1 − ε ⇒ kf k_ϕ≤ 1 − δ (ε) Proof In view of Lemma 2.10 we have

∆⁰₂: ∀θ ∈ (0, 1) ∀ε > 0, ∃hε∈ B^ϕ(R) and k⁰> 1 such that ϕ (t, 2u) ≤ k⁰ϕ (t, u) ∀u ≥ hε(t) , ∀t ∈ G with ¯µ (G⁰) < θ and ρ_ϕ(h_ε) < ε.

On the other hand, if p (t, u) is the right derivative of ϕ (t, u) with respect to u, we have (see [2], [10])

up (t, u) ≤ ϕ (t, 2u) ≤ 2up (t, 2u) ∀u ∈ R+, ∀t ∈ R.

Let λ ∈ 0,¹₂
, then

ϕ t,_1−λ^u  ϕ (t, u) = 1 +

u 1−λ

Z

u

p (t, s) ϕ (t, u)ds,

hence, since p (t, .) is non-decreasing we obtain 

ϕ t,_1−λ^u  ϕ (t, u) − 1

≤

   

u 1 − λ − u

/ϕ (t, u)

 .p

 t, u

1 − λ



≤    

1 1 − λ − 1

u. p (t, 2u) ϕ (t, u)



≤ 1 2

λ

1 − λ.ϕ (t, 4u) ϕ (t, u)

≤ λϕ (t, 4u) ϕ (t, u) . (6)

Consider the function

fλ(t, u) = ϕ

t,_1−λ^u  ϕ (t, u) . From (6), we have

f_λ(t, u) ≤ 1 + λϕ (t, 4u)

ϕ (t, u) , ∀u ∈ R+, ∀t ∈ R and using the condition ∆⁰₂, we get

fλ(t, u) ≤ 1 + (k⁰)²λ, ∀u ≥ hε(t) , ∀t ∈ G with ¯µ (G⁰) < θ and ρϕ(hε) < ε.

Moreover it is easily seen that given ε > 0 there exists δ1(ε) > 0 s.t.

|λ| ≤ δ1(ε) ⇒ fλ(t, u) ≤ 1 + (k⁰)²λ ≤ 1 1 −₂^ε (take for example δ1(ε) = ^ε

(2k⁰²)(^1−ε₂)), so we have f_δ1(ε)(t, u) ≤ ₁₋¹ε 2

, i.e.

 1 −ε

2

 ϕ



t, u

1 − δ1(ε)



≤ ϕ (t, u) , ∀u ≥ hε(t) , ∀t ∈ G

with ¯µ (G⁰) < θ and ρ_ϕ(h_ε) < ε. This inequality remains valid for each δ ≤ δ₁(ε) (we may assume δ₁(ε) ≤ ¹₂).

Suppose now that ρ_ϕ(f ) ≤ 1 − ε and put E = {t ∈ R, f (t) ≥ hε(t)} , then 1

2T

+T

Z

−T

 1 − ε

2

 ϕ



t, |f (t)|

1 − δ1(ε)

 dt

= 1

2T Z

G∩[−T ,+T ]

1 −ε 2

ϕ



t, |f (t)|

1 − δ1(ε)

 dt

+ 1 2T

Z

G⁰∩[−T,+T ]

1 − ε 2

ϕ



t, |f (t)|

1 − δ1(ε)

 dt.

In view of Lemma 2.4, we choose θ such that ¯µ (G⁰) ≤ θ implies ρϕ(2f χG⁰) ≤ ^ε₄, then

1 2T

Z

G⁰∩[−T,+T ]

 1 −ε

2

 ϕ



t, |f (t)|

1 − δ1(ε)

 dt

≤ 1

2T Z

G⁰∩[−T,+T ]

 1 −ε

2



ϕ (t, 2 |f (t)|) dt

≤ ε

4.

On the other hand

1 2T

Z

G∩[−T ,+T ]

 1 − ε

2

 ϕ



t, |f (t)|

1 − δ1(ε)

 dt

= 1

2T Z

G∩E∩[−T ,+T ]

 1 − ε

2

 ϕ



t, |f (t)|

1 − δ₁(ε)

 dt

+ 1 2T

Z

G∩E⁰∩[−T,+T ]

 1 − ε

2

 ϕ



t, |f (t)|

1 − δ₁(ε)

 dt

≤ 1

2T Z

G∩E∩[−T ,+T ]

ϕ (t, |f (t)|) dt

+ 1 2T

 1 − ε

2

 Z

G∩E⁰∩[−T,+T ]

ϕ



t, |f (t)|

1 − δ1(ε)

 dt

≤ 1

2T Z

G∩[−T ,+T ]

ϕ (t, |f (t)|) dt − 1 2T

Z

G∩E⁰∩[−T,+T ]

ϕ (t, |f (t)|) dt

+ 1 2T

1 − ε 2

 Z

G∩E⁰∩[−T,+T ]

ϕ



t, |f (t)|

1 − δ1(ε)

 dt

≤ 1

2T Z

G∩[−T ,+T ]

ϕ (t, |f (t)|) dt

+ 1 2T

Z

G∩E⁰∩[−T,+T ]

 ϕ



t, |f (t)|

1 − δ₁(ε)



− ϕ (t, |f (t)|)

 dt

−ε 2

1 2T

Z

G∩E⁰∩[−T,+T ]

ϕ



t, |f (t)|

1 − δ₁(ε)

 dt

≤ 1 2T

Z

G∩[−T ,+T ]

ϕ (t, |f (t)|) dt

+ 1 2T

Z

G∩E⁰∩[−T,+T ]

 ϕ



t, |f (t)|

1 − δ1(ε)



− ϕ (t, |f (t)|)

 dt.

Let now t ∈ E⁰∩ G, then |f (t)| ≤ hε(t) ≤ β and _1−δ^{|f (t)|}

1(ε) ≤ 2β. Using the uniform continuity of ϕ (while taking δ₁(ε) small enough) and it’s periodicity with respect to t ∈ R, we obtain

    ϕ



t, |f (t)|

1 − δ₁(ε)



− ϕ (t, |f (t)|)    

≤ε 4, it follows

1 2T

+T

Z

−T

 1 − ε

2

 ϕ



t, |f (t)|

1 − δ₁(ε)

 dt ≤ 1

2T

+T

Z

−T

ϕ (t, |f (t)|) dt +ε

2, ∀T ≥ T0

then letting T tends to infinity, we obtain

 1 − ε

2

 ρϕ

 f

1 − δ1(ε)



≤ ρϕ(f ) + ε

2 ≤ 1 − ε + ε

2 = 1 − ε 2

and finally kf k_ϕ≤ 1 − δ1(ε) . _

Lemma 2.12 Let (aⁿ)_n≥1 be a sequence of positive real numbers. For each n, we associate a measurable set An such that

i. Ai∩ Aj= φ, for i 6= j and S

n≥1

An ⊂ [0, α[ , α < 1.

ii. P

n≥0 1

R

0

ϕ (t, anχA_n) dt < +∞.

Consider the function f = P

n≥1

anχA_n on [0, 1] and let ˜f be the periodic exten- sion of f to the whole R (with period τ = 1). Then ˜f ∈ ˜B_a.p.^ϕ .

Proof Let us first remark that in view of Lemma 2.4, for each n ≥ 1 there exists a set An⊂ [0, α[ for which

1

R

0

ϕ (t, anχA_n) dt < _n¹2. It is also clear that we may choose the A_n’s so that the conditions of the Lemma are satisfied. Now, for an arbitrary ε > 0, fix n0 such that P

n≥n₀ 1

R

0

ϕ (t, anχA_n) dt ≤ ^ε₃ and put f1=

n0

P

i=1

aiχA_i on [0, 1[ . Let then M = max

i≤n₀sup

t

ϕ (t, 2a_i) and δ ≤ _3M^ε . We can suppose 1 − α > δ.

If f₁^r is the restriction of f1 to [0, 1 − δ], by Luzin’s theorem there exists a con- tinuous function g^r_ε on [0, 1 − δ] such that

µ {t ∈ [0, 1 − δ] /ϕ (t, |f₁^r(t) − g_ε^r(t)|) > 0} ≤ ε 3M. Moreover since f₁ is bounded so is g_ε^r(with the same bound).

Let gεbe a linear extension of g_ε^rto [0, 1] , more precisely gεis such that gε= g_ε^r on [0, 1 − δ], gεis linear between 1 − δ and 1 and satisfies gε(1) = g_ε^r(0) .

We then get

1

Z

0

ϕ



t,|f (t) − gε(t)|

2

 dt

≤

1

Z

0

ϕ



t,|f (t) − f1(t)| + |f1(t) − gε(t)|

2

 dt

≤ 1

2

1

Z

0

ϕ (t, |f (t) − f1(t)|) dt +1 2

1

Z

0

ϕ (t, |f1(t) − gε(t)|) dt

≤ 1

2

1

Z

0

ϕ



t, X

n≥n0

anχAn



dt +1 2

1−δ

Z

0

ϕ (t, |f₁^r(t) − g_ε^r(t)|) dt

+ 1

2

1

Z

1−δ

ϕ (t, |f1(t) − gε(t)|) dt

≤ 1

2 X

n≥n0

1

Z

0

ϕ (t, anχA_n) dt + 1 2M ε

3M +1 2M ε

3M

≤ ε

2.

Finally the continuous function gε: [0, 1] → R satisfies

gε(0) = gε(1) and

1

Z

0

ϕ



t,|f (t) − gε(t)|

2

 dt ≤ ε

2.

Let now ˜f be the periodic extension of f to the whole R and ˜gε be the periodic extension of gε. Clearly ˜gε is u.a.p. and then it is also in B_a.p.^ϕ (R) .

Consequently, there exists Pε∈ A for which ρϕ

_g˜

ε−Pε

2

≤^ε₂.

On the other hand ˜f and ˜g being periodic with period T = 1, we have

ρ_ϕ ˜f − ˜g_ε 2

!

= lim_{T →+∞} 1 2T

+T

Z

−T

ϕ



t,   

f (t) − ˜˜ gε(t)   2



dt

=

1

Z

0

ϕ



t,|f (t) − g_ε(t)|

2

 dt ≤ ε

2. Finally,

ρϕ

˜f − Pε

4

!

≤1 2

"

ρϕ

˜f − ˜gε

2

! + ρϕ

 ˜gε− Pε

2

#

≤ ε.

i.e. ˜f ∈ ˜B_a.p.^ϕ . _

3. Result.

Theorem 3.1 ˜B_a.p.^ϕ is uniformly convex if and only if ϕ is uniformly convex and satisfies the condition ∆2.

Proof sufficiency:

Let ε ∈ ]0, 1[ and f ,g in ˜B_a.p.^ϕ be such that kf k_ϕ= kgk_ϕ= 1 and

f −g 2

_ϕ≥ ε.

From corollary 2.9 we have also ρϕ(f ) = ρϕ(g) = 1 and ρϕ

_{f −g}

2

≥ δ for some δ = δ (ε) ∈ ]0, 1[ .

Let h_δ be a measurable function such that ρ_ϕ(h_δ) = ^δ₄, then from the uniform convexity of ϕ there exists p (δ) ∈ ]0, 1[ for which the following implication holds



|hδ(t)| ≤ max (|f (t)| , |g (t)|) ≤ 4

δ|f (t) − g (t)|



=⇒

 ϕ



t,|f (t) + g (t)|

2



≤1 − p (δ)

2 [ϕ (t, |f (t)|) + ϕ (t, |g (t)|)]

 . Put

B =



t ∈ R, |h^δ(t)| ≤ max (|f (t)| , |g (t)|) ≤ 4

δ|f (t) − g (t)|

 , then

1 2T

+T

Z

−T

ϕ



t,|f (t) + g (t)|

2

 χ_Bdt

≤ 1 − p (δ) 2



 1 2T

+T

Z

−T

ϕ (t, |f (t)|) χ_Bdt + 1 2T

+T

Z

−T

ϕ (t, |g (t)|) χ_Bdt



.

It follows

ρ_ϕ f + g 2 χ_B



≤ 1 − p (δ)

2 [ρ_ϕ(f χ_B) + ρ_ϕ(gχ_B)]

and then,

1 − ρ_ϕ f + g 2



= 1

2(ρ_ϕ(f ) + ρ_ϕ(g)) − ρ_ϕ f + g 2



≥ 1

2(ρ_ϕ(f χ_B) + ρ_ϕ(gχ_B)) − ρ_ϕ f + g 2 χ_B



≥ p (δ)

2 (ρϕ(f χB) + ρϕ(gχB)) . (7)

We define the sets

C = {t ∈ B⁰ s.t. max (|f (t)| , |g (t)|) < |h_δ(t)|}

D =



t ∈ B⁰ s.t. |f (t) − g (t)| < δ

4max (|f (t)| , |g (t)|)

 . Then B⁰= CU D and

ρϕ

 f − g 2 χC



≤ 1

2[ρϕ(f χC) + ρϕ(gχC)] ≤δ 4

ρϕ

 f − g 2 χD



≤ δ

8[ρϕ(f χD) + ρϕ(gχD)] ≤ δ 4 hence ρϕ

_{f −g}

2 χB⁰

≤₂^δ and consequently ρϕ

_{f −g}

2 χB

≥ ^δ₂. Now, from the inequality

δ

2 ≤ ρ_ϕ f − g 2 χ_B



≤1

2(ρ_ϕ(f χ_B) + ρ_ϕ(gχ_B)) it follows

(ρ_ϕ(f χ_B) + ρ_ϕ(gχ_B)) ≥ δ and using (7) we deduce

ρϕ

 f + g 2



≤ 1 −δp (δ) 2 .

Finally, since ϕ ∈ ∆2 and recalling the fact that δ = δ (ε) depends on ε, by lemma 2.11 there exists q (ε) ∈ ]0, 1[ such that

f +g 2

≤ 1 − q (ε) .