space of almost periodic functions

Characterization of the strict convexity of the Besicovitch-Musielak-Orlicz

space of almost periodic functions

Mohamed Morsli, Mannal Smaali

Abstract. We introduce the new class of Besicovitch-Musielak-Orlicz almost periodic functions and consider its strict convexity with respect to the Luxemburg norm.

Keywords: Besicovitch-Orlicz space, almost periodic functions, strict convexity Classification: 46B20, 42A75

1. Introduction

We denote byC⁰a.p.the linear set of all continuous almost periodic functions (u.a.p.). Let A be the subspace of C⁰a.p. whose elements are the generalized trigonometric polynomials i.e.,

A=



Pn(t) = Xn j=1

aje^iλjt, aj ∈C, λj ∈R, n∈N



.

The classC⁰a.p.is in fact the closure ofAin the uniform norm ofCb(R) (the space of continuous and bounded functions onR).

This topological characterization is used to define widest classes of almost periodic functions as the closure of the linear setAwith respect to some specific norms.

The first extension was obtained by A.S. Besicovitch (cf. [2]) in the context of L^p spaces. Namely he defined the Sa.p.^q ,Wa.p.^q and Ba.p.^q spaces (resp. Stepanoff, Weyl and Besicovitch spaces of almost periodic functions). Later on, T.R. Hill- mann (cf. [5]) used a similar approach to obtain an extension in the context of Orlicz spaces.

Most of the Hillmann’s work concerns topological and structural properties of the new spaces.

In [9], [10], [11], there are considered the fundamental geometric properties of the Besicovitch-Orlicz spaces of almost periodic functions.

In this paper, we consider the natural extension of almost periodicity to the context of Besicovitch-Musielak-Orlicz spaces, in particular the case when the functionϕgenerating the space depends on a parameter.

The theory of spaces of generalized almost periodic functions was since its be- ginning a subject of great interest. This was essentially motivated by the devel- opment of the theory of differential and partial differential equations with almost periodic terms (cf. [1], [8], [13]).

Actually this interest is still in growth and is enlarged to cover new domains of applications.

2. Preliminaries

In the sequel ϕ : R×[0,+∞[ → [0,+∞[ will be a continuous function on R×[0,+∞[ satisfying:

(i) For everyt∈R, ϕ(t,0) = 0.

(ii) For eacht∈R, ϕ(t, u) is convex with respect tou∈[0,+∞[.

(iii) For everyu∈[0,+∞[,ϕ(t, u) is periodic with respect tot∈R, the period τ being fixed and independent ofu∈[0,+∞[. Without loss of generality we may suppose thatτ= 1.

(iv) For eachα >0, we have inf_t∈R ϕ(t, α) =φ(α)>0.

We denote byM(R) the space of all real valued Lebesgue measurable functions.

The functional

ρϕ:M(R) → [0,+∞]

f 7→ ρϕ(f) = lim

T→+∞

1 2T

Z +T

−T

ϕ(t,|f(t)|)dt is a convex pseudomodular (cf. [10], [12]).

We define the Besicovitch-Musielak-Orlicz space associated to this pseudomod- ular by

B^ϕ(R) =

f ∈M(R) : lim

α→0ρϕ(αf) = 0

=

f ∈M(R) :ρϕ(αf)<+∞, for some α >0 . The spaceB^ϕ(R) is naturally endowed with the pseudonorm

kfkϕ= inf

k >0 :ρϕ

f k

≤1

, f ∈B^ϕ(R). LetAbe the set of all generalized trigonometric polynomials, i.e.,

A=



Pn(t) = Xn j=1

a_je^iλjt, a_j ∈C, λ_j ∈R, n∈N



.

We denote by ˜Ba.p.^ϕ (R) (resp.B^ϕa.p.(R)) the closure of A with respect to the pseudomodularρϕ (resp. with respect to the pseudonormk.kϕ), more precisely:

B˜_a.p.^ϕ (R) =

f ∈B^ϕ(R) :∃fn∈A,∃k0>0, lim

n→+∞ρϕ(k0(fn−f)) = 0

, B_a.p.^ϕ (R) =

f ∈B^ϕ(R) :∃fn∈A,∀k >0, lim

n→+∞ρϕ(k(fn−f)) = 0

=

f ∈B^ϕ(R) :∃fn∈A, lim

n→+∞kfn−fk_ϕ= 0

.

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

It is clear that

B_a.p.^ϕ (R) ⊆B˜^ϕ_a.p.(R)⊆B^ϕ(R).

Whenϕ(t,|x|) =|x|, we denote byB¹(R) andB¹a.p.(R) the respective spaces.

The notationρ₁ is used for the associated pseudomodular.

Recall that the functionϕis said to be strictly convex ifϕ(t, λu+ (1−λ)v)<

λϕ(t, u) + (1−λ)ϕ(t, v) for almost all t ∈ R and for every 0≤ u < v < +∞, 0< λ <1.

A normed linear space (X,k.k) is strictly convex if^x+y₂

<1 wheneverkxk= kyk= 1 andkx−yk>0.

We say thatϕ satisfies the ∆₂-condition (ϕ∈∆₂) if there existk >1 and a measurable nonnegative functionhsuch thatρϕ(h)<+∞andϕ(t,2u)≤kϕ(t, u) for almost allt∈Rand allu≥h(t).

3. Auxiliary results

The spaceBa.p.^ϕ (R) can be regarded as a subspace of measurable functions on Rwith respect to Lebesgue measure. However, the theory of B^ϕa.p.(R) spaces is different from that ofL^ϕ(R) spaces: the usual convergence results of the Lebesgue measure theory are not valid in theB^ϕa.p.(R) spaces (see [11]).

To handleBa.p.^ϕ (R) spaces asL^ϕ(R) ones, we introduce the set function ¯µ.

Let Σ = Σ(R) be the σ-algebra of all Lebesgue measurable subsets ofR. We denote by ¯µthe set function defined on Σ by

¯

µ(A) = lim

T→+∞

1 2T

Z +T

−T

χ_A(t)dt= lim

T→+∞

1

2Tµ(A∩[−T,+T]), whereµdenotes the Lebesgue measure onR.

It is easily seen that the set function ¯µis notσ-additive.

A sequence{fn} ⊂B^ϕ(R) is said to be ¯µ-convergent to somef ∈B^ϕ(R) (in symbolfn µ¯

−−→f) when, for everyα >0, we have

n→+∞lim µ¯{x∈R:|fn(x)−f(x)|> α}= 0.

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

Lemma 1. Letν(A) = limT→+∞ 1 2T

R+T

−T ϕ(t, χA(t))dt. Then the set function

¯

µis absolutely continuous with respect toν, i.e., for everyε >0there existsδ >0 such that

(3.1) (A∈Σ, ν(A)< δ)⇒(¯µ(A)< ε).

Proof: Suppose that (3.1) is false. Then for some ε0 > 0 we will have the following:

for eachn∈N, there existsEn∈Σ s.t.ν(En)< ₂¹n and ¯µ(En)> ε0. Thus ν(En) = lim

T→+∞

1 2T

Z +T

−T

ϕ(t, χEn(t))dt

= lim

T→+∞

1 2T

Z _+T

−T

ϕ(t,1)χ_En(t)dt

≥φ(1)¯µ(En)≥φ(1)ε₀,

a contradiction.

Lemma 2. Let{fn}_n≥1 ⊂ B_a.p.^ϕ (R) be a sequence modular convergent to f ∈ Ba.p.^ϕ (R), i.e.,limn→+∞ρϕ(fn−f) = 0. Thenfn

¯

−−→µ f.

Proof: Notice first that we have also lim_n→+∞ρ_φ(fn−f) = 0. Then from a similar result for functions without parameter (cf. [10]) it follows thatfn µ¯

−−→f. Lemma 3. Leth∈B^ϕ(R)be such thatρϕ(h) =a >0. Then for everyθ¯∈(0,1) there exist constantsβ >0,T₀>0 and a setG¯={t∈R,|h(t)| ≤β}such that (3.2) µG¯∩[−T,+T] ≥θ2T,¯ for T ≥T0.

Proof: It is clear thath∈B^φ(R). Then ifρφ(h)>0 the conclusion follows from a similar result for the functionφwithout parameter (cf. [10]). The conclusion is

immediate ifρφ(h) = 0.

Lemma 4. Letg ∈B^ϕa.p.(R). Then for all ε >0 there exist δ > 0 andT₀ >0 such thatρϕ(gχ_Q)≤ε, for allQ∈Σsatisfyingµ{Q∩[−T,+T]} ≤2δT,T ≥T₀. Proof: We may suppose ρϕ(g)>0.

Letε >0 andPε∈Abe such thatρϕ(2(g−Pε))<₂^ε. Using the properties ofϕ we haveϕ(t,2|Pε(t)|)∈C⁰a.p.(cf. [4]). We then putMε= sup_t∈Rϕ(t,2|Pε(t)|).

We choose ¯θ∈(0,1) satisfyingMε(1−θ)¯ < ^ε₂. Then by Lemma 3 there exist β > 0 and a set ¯G = {t ∈ R,|g(t)| ≤ β} for which µ{G¯∩[−T,+T]} ≥ 2¯θT,

∀T ≥T₀, for someT₀ >0. Hence, denoting by ¯G^′ the complement of ¯G, we will have for allT ≥T₀,

(3.3) 1 2T

Z

G¯^′∩[−T,+T]

ϕ(t,|g(t)|)dt

≤ 1 2

1 2T

Z

G¯^′∩[−T,+T]

[ϕ(t,2|g(t)−Pε(t)|) +ϕ(t,2|Pε(t)|)]dt

!

≤ ε 4+ 1

4TMε 1−θ¯ 2T ≤ ε

2. We putδ=_{2 sup} ^ε

t∈Rϕ(t,β) and let Q⊂Rbe such thatµ{Q∩[−T,+T]} ≤2δT forT ≥T0.

Then ifQ₁ =Q∩G¯ and Q₂ =Q∩G¯^′, we will have 1

2T Z

Q1∩[−T,T]ϕ(t,|g(t)|)dt≤ 1 2T

Z

Q1∩[−T,T]ϕ(t, β)dt

≤ 1

2Tµ(Q₁) sup

t∈R

ϕ(t, β)

≤δsup

t∈R

ϕ(t, β)≤ ε 2. Similarly using (3.3) we get

1 2T

Z

Q2

ϕ(t,|g(t)|)dt≤ 1 2T

Z

G¯^′∩[−T,+T]ϕ(t,|g(t)|)dt≤ ε 2. Finally for allT ≥T₀, we have

1 2T

Z

Q∩[−T,+T]

ϕ(t,|g(t)|)dt≤ε,

which means thatρϕ(gχ_Q)≤ε.

Proposition 1. Letf ∈Ba.p.^ϕ (R). Thenϕ(t,|f(t)|)∈B¹_a.p.(R)and consequently the limitlim_{T→+∞ 2T}¹ R+T

−T ϕ(t,|f(t)|)dtexists and is finite.

Proof: Let {fn} be a sequence of trigonometric polynomials such that kfn− fkϕ →0. Then using Lemma 2 we have alsofn µ¯

−−→f.

Let ¯θ ∈ (0,1). In view of Lemma 3, there existβ > 0 and T₀ > 0 for which

¯

µ( ¯G)≥θ¯with ¯G={t∈R:|f(t)| ≤β}.

Let α > 0 and A^α_n = {t ∈ R : |fn(t)−f(t)| > α}. It is easily seen that

|fn(t)| ≤β+α,∀t∈G¯∩(A^α_n)^′.

Now, the functionϕbeing continuous onR×[0,+∞[, is also uniformly contin- uous on [0,1]×[0, α+β]. Moreover, using the periodicity ofϕ(t, u) with respect tot∈R, it follows thatϕis uniformly continuous onR×[0, α+β].

Then for everyη >0 there existsαη>0 such that

∀t∈G¯∩(A^α_n)^′ :|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)| ≥η=⇒ |fn(t)−f(t)|> αη. Hence, sincefn µ¯

−−→f we get also

n→+∞lim µ¯n

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

= 0.

Consequently,

¯

µ{t∈R:|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)| ≥η}

≤µ¯n

t∈G¯∩(A^α_n)^′ :|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)| ≥ηo + ¯µn

t∈ G¯′

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

≤µ¯n

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

G¯′

+ ¯µ(A^α_n)

≤µ¯n

t∈G¯∩(A^α_n)^′ :|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)| ≥ηo + 1−θ¯

+ ¯µ(A^α_n). Lettingntend to infinity, we will have

n→+∞lim µ¯{t∈R:|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)| ≥η} ≤ 1−θ¯ .

Finally, since ¯θ∈(0,1) is arbitrary, we deduce that for allη >0

(3.4) lim

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

On the other hand, using Lemma 4, it is easy to see that given ε > 0 there existδ >0 andn₀∈Nsuch that for alln≥n₀ the following implication holds

(Q∈Σ,µ(Q)¯ ≤δ) =⇒max ρϕ f χ_Q

, ρϕ fnχ_Q

≤ε.

Let E_n^ε = {t ∈ R : |ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)| ≥ ε}. Then since by (3.3),

¯

µ(E^ε_n)≤δforn≥n0, we get

T→+∞lim 1 2T

Z _+T

−T

|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)|dt

≤ lim

T→+∞

1 2T

Z

E^ε_n∩[−T,T]|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)|dt + lim

T→+∞

1 2T

Z

(E_n^ε)^′∩[−T,T]

|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)|dt

≤2ε+ε= 3ε.

Finally byε >0 being arbitrary we deduce that

n→+∞lim lim

T→+∞

1 2T

Z _+T

−T

|ϕ(t,|fn(t)|)−ϕ(t,|f(t)|)|dt= 0.

It remains to see thatϕ(t,|fn(t)|)∈C⁰a.p. This follows from the properties of the functionϕand the fact thatfn∈A(see [4]).

Lemma 5. Let{fn}n ⊂B_a.p.¹ (R) be such thatfn µ¯

−−→f ∈ B_a.p.¹ (R). Suppose there exists g ∈ B_a.p.¹ (R) for which max(|fn(t)|,|f(t)|) ≤ g(t), t ∈ R. Then ρ₁(fn)→ρ₁(f).

Proof: Take ε > 0 and let δ >0 be associated to g as in Lemma 4. We put A^ε_n={t∈R:|fn(t)−f(t)| ≥ ^ε₂}. Then sincefn ¯µ

−−→f it follows that ¯µ(A^ε_n)≤δ for alln≥n0 and then by Lemma 4

ρ₁ |fn−f|χ_A^ε_n

≤ρ₁ 2gχ_A^ε_n

≤ε 2. Consequently, for alln≥n₀ we have

ρ₁(|fn−f|)≤ρ₁

|fn−f|χ_Aε

n

+ρ₁

|fn−f|χ

(^Aεn)^′

≤ ε 2 +ε

2 =ε

i.e., limn→+∞ρ1(fn) =ρ1(f).

Lemma 6. Letf ∈B_a.p.^ϕ (R). Then the functionalλ7→ρϕ

_f

λ

is continuous on ]0,+∞[.

Proof: First, notice that since f ∈ Ba.p.^ϕ (R) we have ρϕ(αf) < +∞ for each α >0. Indeed,f being in Ba.p.^ϕ (R) there exists a sequence{fn}n⊂A such that limn→∞kf−fnkϕ= 0 or equivalently limn→∞ρϕ(α(f−fn)) = 0 for everyα >0.

Letα >0 andn₀∈Nsuch thatρϕ(2α(f−fn0))≤1. Then ρϕ(αf)≤1

2ρϕ(2α(f−fn0)) +1

2ρϕ(2αfn0),

consequently, using the fact that the trigonometric polynomialfn0 is uniformly bounded, it follows thatρϕ(αf)<+∞.

Let nowλ₀∈]0,+∞[ and{λn}be a sequence of real numbers which converges toλ₀. We have

ρϕ

f λn − f

λ₀

≤

1 λn − 1

λ₀

ρϕ(f) for every n≥n₀. Then lim_n→+∞ρϕ

_f

λn −_λ^f₀

= 0.

Now, using Lemma 2 we get _λ^f

n

¯

−−→µ _λ^f₀ and thenϕ t,^|f(t)|_λ

n

_µ¯

−−→ϕ

t,^|f(t)|_λ0 (see the proof of Proposition 1). Furthermore

max

ϕ

t,|f(t)|

λn

, ϕ

t,|f(t)|

λ₀

≤ϕ

t, 2 λ₀|f(t)|

and by Proposition 1 we have ϕ

t,_λ²₀|f(t)|

∈ B_a.p.¹ (R). Consequently, using Lemma 5 we deduce

ρϕ

f λn

→ρϕ

f λ₀

. This means thatλ7→ρϕ

_f

λ

is continuous on ]0,+∞[.

Corollary 1. Letf ∈B^ϕa.p.(R). Then (1) kfkϕ≤1if and only if ρϕ(f)≤1;

(2) kfkϕ= 1if and only if ρϕ(f) = 1.

Proof: We prove briefly (2), the assertion (1) follows then easily.

Letf ∈B^ϕa.p.(R) with kfkϕ = 1. Then for ε >0 we will haveρϕ

_f

1+ε

≤1 and using Lemma 6 it follows thatρϕ(f)≤1.

We have alsoρϕ

_f

1−ε

≥1 and again by Lemma 6 we getρϕ(f)≥1. Finally, ρϕ(f) = 1.

The converse implication is known for a general modular space.

Remark 1. We recall that a similar result holds in classical Musielak-Orlicz spaces under the additional ∆₂-condition. This condition is not necessary in our case since Lemma 6 holds with the restrictionf ∈Ba.p.^ϕ (R).

Lemma 7. Let f ∈ Ba.p.^ϕ (R) with kfkϕ = 1. Then there exist real numbers 0 < α < β and θ ∈(0,1) such that if G₁ ={t ∈R : α≤ |f(t)| ≤ β} we have

¯

µ(G1)≥θ.

Proof: Let ¯θ ∈(0,1). Then from Lemma 3 there existβ >0 andT0 >0 such thatµ{G¯∩[−T,+T]} ≥θ2T,¯ ∀T ≥T0, where ¯G={t∈R:|f(t)| ≤β}.

We claim that the following is also true:

• for each δ∈ (0,1) there exist ˜θ ∈ (0,1), T₀ >0 and a set ˜G ={t ∈ R, ϕ(t,|f(t)|)≤1−δ}such that forT ≥T₀

(3.5) µn

G˜∩[−T,+T]o

<θ2T.˜

For, letδ∈(0,1) andPn be a sequence of trigonometric polynomials approxi- matingf, i.e.,kf−Pnkϕ→0. We takeP_δ such thatρϕ(2|f−P_δ|)<₄^δ and put M = sup_t∈Rϕ(t,2Pδ(t)).

Letε >0 be such that_δ

4+M ε

< δand suppose that (3.5) is not satisfied.

Then taking ˜θ= 1−ε, there will exists a sequence{Tn}increasing to infinity for whichµ{G˜∩[−Tn,+Tn]} ≥θ2T˜ n. We then get

1 2Tn

Z _+Tn

−Tn

ϕ(t,|f(t)|)dt= 1 2Tn

Z

G∩[−T˜ n,+Tn]

ϕ(t,|f(t)|)dt + 1

2Tn

Z

(^G˜)^{′∩[−Tn,+Tn]}ϕ(t,|f(t)|)dt

≤(1−δ) + 1 2Tn

Z

(^G˜)^′∩[−Tn,+Tn]

ϕ(t,|f(t)|)dt.

Moreover, we have 1 2Tn

Z

(^G˜)^′∩[−Tn,+Tn]

ϕ(t,|f(t)|)dt

≤ 1 2

"

1 2Tn

Z

(^G˜)^{′∩[−Tn,+Tn]}ϕ(t,2|f(t)−P_δ(t)|)dt

+ 1

2Tn

Z

(^G˜)^{′∩[−Tn,+Tn]}ϕ(t,2|P_δ(t)|)dt

#

≤ 1 2

δ 4 +M ε

≤δ 2.

Then

1 2Tn

Z +Tn

−Tn

ϕ(t,|f(t)|)dt≤1−δ+δ

2 ≤1−δ 2.

Hence, letting n tend to infinity we will have ρϕ(f) ≤ 1− ₂^δ. Finally, using Corollary 1 it followskfkϕ<1. This contradicts the fact thatkfkϕ= 1.

We now show the statement of the lemma. Letδ∈(0,1) andα >0 be such that sup_t∈Rϕ(t, α)≤1−δ. We choose ˜θas in (3.5) and then take ¯θ >θ˜as in Lemma 3.

Ifβ > αis a fixed number we define the setG₁ ={t∈R:α≤ |f(t)| ≤β}. Then since

(G₁)^′∩[−T, T] ={t∈[−T, T] :|f(t)| ≤α} ∪ {t∈[−T, T] :f(t)≥β} ⊂G˜∪ G¯′

,

it follows that forT ≥T₀ we have µ (G₁)^′∩[−T, T]

≤µ

G˜∩[−T, T] +µ

G¯′

∩[−T, T]

≤θ2T˜ + 1−θ¯ 2T =

1−

θ¯−θ˜ 2T, or equivalently

µ(G₁∩[−T, T])≥ θ¯−θ˜

2T, for T ≥T₀.

Lemma 8. Let{an}n, an>0 be a sequence of real numbers andα∈(0,1). To eachn we associate a measurable setAn such that

(i) Ai∩Aj =φ, fori6=j andS

n≥1An⊂[0, α[,α <1;

(ii) P

n≥0

R₁

0 ϕ(t, anχ_An(t))dt <+∞.

Consider the functionf =P

n≥1anχ_An on[0,1]and letf˜be the periodic exten- sion off to the wholeR(with periodτ= 1). Thenf˜∈B˜a.p.^ϕ .

Proof: Let us first remark that sinceR1

0 ϕ(t, an)dt <+∞, forn≥1 there exists a set An ⊂ [0, α[ for which R1

0 ϕ(t, anχ_An(t))dt < _n¹2. It is also clear that we may choose theAn’s so that the conditions of the lemma are satisfied. Now, for an arbitraryε >0 we fix n₀ such thatP

n≥n⁰

R₁

0 ϕ(t, anχ_An(t))dt≤ ^ε₃ and put f₁=Pn0

i=1aiχ_Aion [0,1[. Let thenM = max_i≤n0sup_t∈[0,1]ϕ(t,2ai) andδ≤_3M^ε (remark that we may suppose 1−α > δ).

Letf₁^rdenote the restriction off₁ to [0,1−δ]. Then by Luzin’s theorem there exists a continuous functiong_ε^ron [0,1−δ] such that

µ{t∈[0,1−δ] :ϕ(t,|f₁^r(t)−g_ε^r(t)|)>0} ≤ ε 3M .

Moreover sincef₁ is bounded so isg_ε^r(with the same bound).

Let nowgεbe a linear extension ofg^r_ε to [0,1], more preciselygε is such that gε=g^r_ε on [0,1−δ],gε is linear between 1−δand 1 and satisfiesgε(1) =g^r_ε(0).

We then get Z 1

0

ϕ

t,|f(t)−gε(t)|

2

dt

≤ Z ₁

0

ϕ

t,|f(t)−f₁(t)|+|f₁(t)−gε(t)|

2

dt

≤1 2

Z ₁

0

ϕ(t,|f(t)−f₁(t)|)dt+1 2

Z ₁

0

ϕ(t,|f₁(t)−gε(t)|)dt

≤1 2

Z 1 0

ϕ t, X

n≥n0

anχ_An(t)

! dt

+1 2

Z 1−δ 0

ϕ(t,|f₁^r(t)−g^r_ε(t)|)dt+1 2

Z 1 1−δ

ϕ(t,|f1(t)−gε(t)|)dt

≤1 2

X

n≥n⁰

Z ₁

0

ϕ(t, anχ_An(t))dt+1 2M ε

3M +1 2M ε

3M

≤ ε 2.

Finally, the continuous functiongε: [0,1]→Rsatisfies gε(0) =gε(1) and

Z ₁

0

ϕ

t,|f(t)−gε(t)|

2

dt≤ ε

2.

Let now ˜f and ˜gεbe the respective periodic extensions off andgεto the whole R(with the periodτ= 1). Clearly ˜gε isu.a.p.and then it is also inBa.p.^ϕ (R).

Consequently, there existsPε∈A for whichρϕ

˜gε−Pε

2

≤ ^ε₂.

On the other hand ˜f and ˜g being periodic with periodτ = 1, using the peri- odicity ofϕ(withτ= 1), we get

ρϕ

f˜−g˜ε

2

!

= lim

T→+∞

1 2T

Z +T

−T ϕ



t,

f˜(t)−˜gε(t) 2



dt

= Z ₁

0

ϕ

t,|f(t)−gε(t)|

2

dt≤ ε

2. Finally,

ρϕ

f˜−Pε

4

!

≤ 1 2

"

ρϕ

f˜−g˜ε

2

! +ρϕ

˜gε−Pε

2

#

≤ε,

i.e., ˜f ∈B˜^ϕa.p..

4. Results

Lemma 9. Let ϕ(t, u) be strictly convex with respect to u ≥ 0 and fn, gn ∈ Ba.p.^ϕ (R)be sequences such that, for somer >0, we have

ρϕ(fn)≤r, ρϕ(gn)≤r and lim

n→∞ρϕ

fn+gn

2

=r.

Then(fn−gn)−−→^µ¯ 0.

Proof: Suppose that lim_n→∞(fn−gn)6= 0 in the ¯µ-convergence sense. Then there existε >0,σ >0 andn_kր ∞such that ifE_k={t∈R:|fn_k(t)−gn_k(t)| ≥ σ}we have ¯µ(E_k)> ε.

Take a numberkε>1 such that (see Lemma 1) there holds

¯

µ(E)≥ ε

4 ⇒ρϕ(χ_E)> r kε, wherer >0 is the constant from the lemma.

Then putting

A_k={t∈R:|fnk(t)|> kε}, B_k={t∈R:|gnk(t)|> kε} we obtain

r≥ρϕ(fn_k)

= lim

T→+∞

1 2T

Z _+T

−T

ϕ(t,|fnk(t)|)dt

≥ lim

T→+∞

1 2T

Z

Ak∩[−T,T]ϕ(t, kε)dt

≥kε lim

T→+∞

1 2T

Z

Ak∩[−T,T]

ϕ(t,1)dt=kερϕ χ_Ak . It follows thatρϕ(χA_k)≤ _k^r_ε and then ¯µ(Ak)≤ ^ε₄.

In the same way we show that ¯µ(B_k)≤ ^ε₄. Now, define the set

Q={(u, v)∈R²/|u| ≤kε,|v| ≤kε,|u−v| ≥σ}, and consider the function

F(t, u, v) = 2ϕ t,^u+v₂ ϕ(t, u) +ϕ(t, v).

Sinceϕ is strictly convex we haveF(t, u, v)<1, for all (t, u, v)∈R×Q. Then using the continuity of ϕ on R×Q (where Q is a compact set of R²) and its periodicity with respect tot, it follows that

sup

R×QF(t, u, v) = 1−δ for some δ >0.

More precisely, for (t, u, v)∈R×Qwe have ϕ

t,u+v

2

≤(1−δ)ϕ(t, u) +ϕ(t, v)

2 .

Let nowt∈E_k\(A_k∪B_k). Thenfnk(t), gnk(t)∈Qand consequently ϕ

t,|fn_k(t) +gn_k(t)|

2

≤(1−δ)ϕ(t,|fn_k(t)|) +ϕ(t,|gn_k(t)|)

2 .

Hence r−ρϕ

fn_k+gn_k

2

≥ ρϕ(fn_k) +ρϕ(gn_k)

2 −ρϕ

fn_k+gn_k

2

≥ lim

T→+∞

1 2T

Z

[Ek\(Ak∪Bk)]∩[−T,+T]

ϕ(t,|fnk(t)|) +ϕ(t,|gnk(t)|)

2 −ϕ

t,|fnk(t) +gnk(t)|

2

dt

≥ δ 2 lim

T→+∞

1 2T

Z

[Ek\(Ak∪Bk)]∩[−T,+T]

[ϕ(t,|fn_k(t)|) +ϕ(t,|gn_k(t)|)]dt

≥δ lim

T→+∞

1 2T

Z

[Ek\(Ak∪Bk)]∩[−T,+T]

ϕ

t,|fn_k(t)−gn_k(t)|

2

dt

≥δϕσ

2 ε−ε 4 −ε

4

=δε 2ϕσ

2

.

Finally,

r−ρϕ

fn+gn

2

≥δε 2φσ

2 >0,

a contradiction with the hypothesisρϕ

_f

n+gn

2

→r.

Theorem 1. B˜a.p.^ϕ (R)is strictly convex if and only if ϕis strictly convex andϕ satisfies the∆₂-condition.

Proof: Sufficiency. Suppose that ϕ is strictly convex and satisfies the ∆₂- condition but ˜B^ϕ_a.p.(R) is not strictly convex. Then for somef and g∈B˜^ϕ_a.p.(R) we will havekfkϕ =kgkϕ= 1 and kf−gkϕ>0 but^f+g2

_ϕ= 1. From Corol- lary 1 we will have alsoρϕ(f) =ρϕ(g) =ρϕ

_f+g

2

= 1. Then from Lemma 9 it follows that for eachα >0, ¯µ{t∈R:|f −g|> α}= 0. Finally, using Lemma 7 we getρϕ(f −g) = 0. Contradiction.

Necessity. Let L^ϕ = L^ϕ([0,1]) = {f ∈ M(R) : R₁

0 ϕ(t, λ|f(t)|)dt < +∞

for some λ > 0} be the usual Musielak-Orlicz space and k.k_L^ϕ its associated Luxemburg norm.

We consider the injection map

i:L^ϕ ֒→B˜^ϕ_a.p.(R), i(f) =f ,e

where feis the periodic extension (with period τ = 1) off toR. We show first thati(L^ϕ)⊂B˜^ϕa.p.(R).

Letf ∈L^ϕ([0,1]). Then there existsλ >0 such thatϕ(t, λ|f(t)|)∈L¹([0,1]).

From usual arguments of Lebesgue theory we have lim_N→+∞µ(V_N) = 0, where V_N ={t∈[0,1] :ϕ(t, λ|f(t)|)≥N}.

LetEN ={t∈[0,1] :|f(t)| ≥N}. Then fort∈EN we have ϕ(t, λ|f(t)|)≥ϕ(t, λN)≥λN ϕ(t,1)≥λN φ(1),

where φ(1) = inf_t∈[0,1]ϕ(t,1), φ(1) >0 (we may suppose φ(1) = 1). It follows thatE_N ⊂V_λN and then we get lim_N→+∞µ(E_N) = 0.

Consider the following functions forN ∈N, f_N(t) =

f(t) if f(t)≤N N if f(t)≥N.

It is clear that the sequence {f_N} is increasing andf_N ≤f. Moreover, since lim_N→+∞µ(E_N) = 0 we have lim_N→+∞R

ENϕ(t, λ|f(t)|)dt= 0.

Then for a givenε >0 there is anNε∈Nsuch that Z ₁

0

ϕ(t, λ|f(t)−f_Nε(t)|)dt≤ Z

E_{N ε}

ϕ(t, λ|f(t)|)dt≤ε.

Now forf_Nε being bounded there exists a sequence of simple functions (S_Nε)n

uniformly convergent to f_Nε. In particular, there exists a simple function S_Nε such that sup_t∈[0,1]|λ(f_Nε(t)−S_Nε(t))| ≤εand then

Z ₁

0

ϕ

t,λ

2|f(t)−S_Nε(t)|

dt

≤1 2

Z 1 0

ϕ(t, λ|f(t)−f_Nε(t)|)dt+1 2

Z 1 0

ϕ(t, λ|f_Nε(t)−S_Nε(t)|)dt≤ε.

We denote byfe,fe_Nε and Se_Nε the respective periodic extensions (with period τ= 1) of the functions f, f_Nε andS_Nε. We have from the periodicity properties ofϕ,fe,fe_Nε andSe_Nε:

ρϕ

λ 2

fe−Se_Nε

= lim

T→+∞

1 2T

Z _+T

−T

ϕ

t,λ 2

ef(t)−Se_Nε(t)

dt

= Z ₁

0

ϕ

t,λ

2 |f(t)−S_Nε(t)|

dt≤ε.

Moreover, from Lemma 8 we haveSe_Nε ∈Be^ϕa.p.(R). Then there exists Pε∈A for whichρϕ

1

4(SeNε−Pε)

≤ε (see the proof of Lemma 8).

Finally, puttingα= min λ,¹₄ we get ρϕ

α 2

fe−Pε

≤1 2

ρϕ

λ 2

fe−Se_Nε +ρϕ

1 4

Se_Nε−Pε

≤ε.

This means thatfe∈B˜^ϕ_a.p.(R).

Now, since i : L^ϕ([0,1]) ֒→ B˜a.p.^ϕ (R) is an isometry, the strict convexity of B˜a.p.^ϕ (R) implies the strict convexity ofL^ϕ([0,1]).

Consequentlyϕ(t, u), t ∈[0,1],u≥0 is strictly convex and satisfies the ∆₂- condition for Musielak-Orlicz spaces (see [6], [7]) i.e., there existk≥1 andh≥0 withR₁

0 h(t)dt <∞such thatϕ(t,2u)≤kϕ(t, u)+h(t) for allu≥0 and almost all t∈[0,1]. The periodically (withτ = 1) extended functionsϕ(t, u),t ∈R,u≥0 andeh(t), t ∈R satisfy the conditions eh∈B¹(R) andϕ(t,2u)≤kϕ(t, u) +eh(t) foru≥0 and almost allt∈R.

Now, puttingf(t) = sup{u≥0 :ϕ(t, u)≤eh(t)}it follows thatf is measurable andϕ(t, f(t)) =eh(t) fort∈R. Finally, we get

ϕ(t,2u)≤kϕ(t, u) +eh(t)≤(k+ 1)ϕ(t, u)

foru≥f(t) and almost allt∈R, i.e.,ϕsatisfies the ∆₂-condition for Besicovitch- Musielak-Orlicz spaces.

References

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D´epartement de Math´ematiques, Faculte des Sciences, Universit´e de Tizi Ouzou, Algeria

E-mail: [email protected]

D´epartement de Math´ematiques, Faculte des Sciences, Universit´e de Tizi Ouzou, Algeria

E-mail: Smaali [email protected]

(Received October 4, 2005,revised April 11, 2007)