< ∞ in Musielak-Orlicz spaces

Criteria for k

< ∞ in Musielak-Orlicz spaces

Lianying Cao, Tingfu Wang

Abstract. In this paper, some necessary and sufficient conditions for sup{kx :kxk⁰ = 1}<∞in Musielak-Orlicz function spaces as well as in Musielak-Orlicz sequence spaces are given.

Keywords: Musielak-Orlicz space, Orlicz norm Classification: 46B20, 46B30

In Orlicz spaces endowed with the Orlicz norm, denote k_M = sup

kxk⁰_M=1

nk >0 :kxk⁰= 1

k(1 +ρ_M(kx))o .

Since the study of many geometric properties in Orlicz spaces is related to whether k_M < ∞ is true, the criterion for k_M < ∞ has been discussed extensively. In 1986 S. Chen obtained a concise result in classical Orlicz spaces:

k_M <∞ ⇐⇒M ∈ ∇₂ (i.e. N ∈∆₂).

But because of the fact that the Musielak-Orlicz functions and condition ∆ are more complicated, the corresponding problem in Musielak-Orlicz spaces has not been solved. And it has become to be an obstacle for the study of many geometric properties in these spaces. In this paper, we shall generalize the result of S. Chen to Musielak-Orlicz function and sequence spaces.

The triple (T,Σ, µ) stands for a finite nonatomic measurable space. A mapping M :T×[0,∞)→[0,∞] is said to be Musielak-Orlicz function if it satisfies:

(*) for eachu∈[0,∞),M(t, u) is aµ-measurable function oft onT; (**) fort∈T (a.e.), M(t, u) is convex and left-continuous with respect tou;

(***) fort∈T (a.e.),M(t,0) = 0, limu→∞M(t, u) =∞ andM(t, u^′)<∞for someu^′ >0.

The subject supported by NSFC (19871020).

We denote byN(t, v) the complementary function ofM(t, u), where N(t, v) = sup

u≥0

{uv−M(t, u)} (t∈T, v≥0).

It is easy to see thatN is also a Musielak-Orlicz function.

Letx(t) :T →(−∞,∞) be aµ-measurable function. The linear set {x(t);∃λ >0 such that ρ_M(λx) =

Z

T

M(t, λx(t))dµ <∞}

equipped with Orlicz norm kxk⁰= sup

ρX(y)≤1

Z

T

x(t)y(t)dµ= inf

k>0

1

k(1 +ρM(kx))

forms a Banach space denoted byL⁰_M. It is called the Musielak-Orlicz function space. For 06=x∈L⁰_M,kxk⁰=_k¹(1 +ρ_M(kx)) iffk∈k(x) = [k^∗_x, k^∗∗_x ], where

k^∗_x= infn

k >0 :ρ_N(ρ(k|x|)) = Z

T

N(t, p(t, k|x(t)|))dµ≥1o , k_x^∗∗= supn

k >0 :ρ_N(ρ(k|x|)) = Z

T

N(t, p(t, k|x(t)|))dµ≤1o .

We say thatM(t, u) satisfies condition ∆ (M ∈∆ for short) if there existλ >1 and a measurable nonnegative functionδdefined on T withR

T δ(t)dµ <∞such that

M(t,2u)≤λM(t, u) +δ(t) (t∈T a.e., − ∞< u <+∞).

The right derivative of M(t, u) (N(t, v)) at u(v) is denoted by p(t, u) (q(t, v), respectively).

We start with the following lemmas.

Lemma 1. The following statements are equivalent:

(1) N ∈∆, i.e. there existλ >1and 0≤δ(t)∈L¹ such that N(t,2v)≤λN(t, v) +δ(t) (t∈T a.e., v∈R);

(2) for anyε >0 there existλ >1and0≤δ(t)∈L¹such that N(t,v

ε)≤λN(t, v) +δ(t) (t∈T a.e., v∈R);

(3) for anyε∈(0,1) there existθ∈(0,1)and0≤δ(t)∈L¹ such that M(t, εu)≤θεM(t, u) +δ(t) (t∈T a.e., v∈R);

(4) there existε, θ∈(0,1)and0≤δ(t)∈L¹ such that M(t, εu)≤θεM(t, u) +δ(t) (t∈T a.e., v∈R).

Proof: See Theorem 1.13 in [2].

Lemma 2. For any06=x∈L⁰_M, if R

{t∈T:x(t)6=0}N(t, B(t))dµ >1thenK(x)6=

∅, where B(t) = sup{v≥0 :N(t, v)<∞}.

Proof: See Theorem 1.35 in [2].

In the following, we always denote k_M = sup

kxk⁰=1

nk >0 :kxk⁰ = 1

k(1 +ρ_M(kx))o .

Theorem 1. The necessary and sufficient condition forkM <∞isN ∈∆.

Proof: Necessity. Suppose thatN /∈∆. For anyε >0, define δ(t) = supn

u≥0 :M(t, εu)> ε

1 +εM(t, u)o . Then

Z

T

M(t, δ(t))dµ=∞.

(Otherwise

M(t, εu)≤ ε

1 +εM(t, u) +M(t, δ(t)) (t∈T a.e., u∈R).

This shows thatN ∈∆.)

Thus we can takeu(t)≥0 satisfying M(t, εu(t))> ε

1 +εM(t, u(t)) and

Z

T

M(t, u(t))dµ >1 +ε ε . Then

Z

T

M(t, εu(t))dµ >1.

Sokεuk⁰>1. And recallingM(t, εu(t))<∞(t∈T a.e.), it is easy to check that there exists Ω⊂T such thatkεu|_Ωk⁰= 1. Takek∈K(εu|_Ω), i.e.

1 =kεu|_Ωk⁰= 1

k(1 +ρ_M(kεu|_Ω)).

Then 1

k +ρ_M(εu|_Ω)≤ 1

k(1 +ρ_M(kεu|_Ω)) =kεu|_Ωk⁰

≤ε(1 +ρ_M(1

ε·εu|_Ω)) =ε(1 +ρ_M(u|_Ω))

≤ε(1 +1 +ε ε

Z

Ω

M(t, εu(t))dµ)

=ε+ (1 +ε)ρM(εu|_Ω).

So _k¹ ≤ ε+ερ_M(εu|_Ω) ≤ 2ε. By the arbitrariness of ε > 0, we obtain the contradictionk_M =∞.

Sufficiency. Since N ∈ ∆, according to Lemma 1, there exist η > 0 and 0 ≤ δ(t)∈L¹ such that

M(t,2u)≥2(1 + 2η)M(t, u)−δ(t) (t∈T a.e., v∈R).

So forusatisfyingM(t, u)> ^δ(t)_2η , we have

(1) M(t,2u)≥2(1 +η)M(t, u).

TakeD >0 such thatD−1−_2η¹ R

Tδ(t)dµ≥1. For anyx∈L⁰_M withkxk⁰= 1, denote

Hx=n

t∈T :M(t, D|x(t)|)> δ(t) 2η

o. Since 1 =kxk⁰≤ _D¹(1 +ρM(Dx)), we get

ρ_M(Dx)≥D−1.

It follows that

(2)

Z

Hx

M(t, Dx(t))dµ=ρM(Dx)− Z

T\Hx

M(t, Dx(t))dµ

≥D−1− 1 2η

Z

T\Hx

δ(t)dµ

≥D−1− 1 2η

Z

T

δ(t)dµ≥1.

ByN ∈∆, we getB(t) =∞(a.e.). Consequently, by virtue of Lemma 2, for any x∈L⁰_M with kxk⁰ = 1,K(x)6=∅. Ifk ∈K(x), then k≤D or k > D. If k > Dthere existsj≥1 such that

2^j−1D < k≤2^jD.

From (1) and (2), we have

2^jD≥k= 1 +ρ_M(kx)>

Z

Hx

M(t, kx(t))dµ

≥ Z

Hx

M(t,2^j−1D|x(t)|)dµ

≥2^j−1(1 +η)^j−1 Z

Hx

M(t, D|x(t)|)dµ

≥(1 +η)^j−12^j−1.

It implies thatj−1 ≤log_1+η2D. Thus k ≤D·2^log1+η2D+1. This shows that kM <∞.

Next we present the criterion fork_M <∞in the Musielak-Orlicz sequence spa- cel_M⁰ . LetM ={Mi}^∞_i=1 be a sequence of functions. For eachi,Mi(u) is convex and left continuous with respect tou, and satisfiesMi(0) = 0, limu→∞Mi(u) =∞ andMi(u^′)<∞for someu^′>0. We denote byNi(v) the complementary function ofMi(u), whereNi(v) = Sup_u≥0{uv−M_i(u)}. pi(u) andqi(v) denote their right derivatives, respectively. p⁻_i (u) denotes the left derivative ofMi(u). We say that M satisfies conditionδif there existλ >1,i0, ci≥0 (i > i0) withP

i>i0ci<∞, anda >0 such that

Mi(2u)≤λMi(u) +ci (i≥i₀, Mi(u)≤a).

In what follows, denote byx= (x(i))^∞_i=1 a real sequence, and define a modular ofxwith respect to M by

ρM(x) =

∞

X

i=1

Mi(|x(i)|).

The linear set

nx:∃c >0, ρ_Mx c

<∞o equipped with the Orlicz norm

kxk⁰= inf

k>0

1

k(1 +ρ_M(kx)) = sup

ρ_N(y)≤1

∞

X

i=1

x(i)y(i)

forms a Banach space denoted by l_M⁰ and called the Musielak-Orlicz sequence space. For any 06=x∈l_M⁰ we define K(x),k_x^∗, k_x^∗∗ andk_M in the same way as in the function case. For eachi, denote

b_i= sup{v≥0 :N_i(v)<∞}.

We obtain the following theorem.

Theorem 2. k_M <∞if and only if (1) N ∈δ;

(2) for anya >0 there existsλ >1such that if Ni(pi(u))≥athen N_i(p_i(λu))>1 (i= 1,2, . . .).

Proof: Necessity. The necessity of (1) can be verified analogously to Theorem 1.

If (2) does not hold, there exista >0,un>0 andin such that Nin(pin(un))≥a, Nin(pin(nun))≤1, (n= 1,2, . . .).

Letxn:xn(in) =un,xn(i) = 0 (i6=in) (n= 1,2, . . .). Thenxn∈l⁰_M and kxk⁰ ≥unpin(un)≥Nin(pin(un))≥a.

Since

ρ_N(p(nxn)) =Nin(pin(nun))≤1, by the definition ofk^∗∗_xn we havek_x^∗∗_n ≥n. Consequently

k^∗∗xn

kxnk0 =kxnk⁰k_x^∗∗_n≥na.

This shows thatkM =∞.

Sufficiency. Since N ∈ δ, there exist λ^′ > 1, i^′₀, c_i ≥ 0 (i > i^′₀) with P

i>i^′₀ci<∞, anda^′ >0 such that

Ni(2v)≤λ^′Ni(v) +ci (i > i^′₀, Ni(v)≤a^′).

Takei₀> i^′₀ satisfyingP

i>i0ci <1 anda < a^′ satisfying M_i(q_i(N_i⁻¹(a)))≤ 1

i₀ (i= 1,2, . . . i₀).

Next takeλ > λ^′ such that

(3) Ni(pi(u))≥a=⇒Ni(pi(λu))>1 (i−1,2, . . .).

It is easy to prove that

(4) Ni(2v)≤λNi(v) +ci (i > i₀, Ni(v)≤a).

Notice that _λ¹N_i(2v) and ¹_λM_i(^λ₂u) are complementary to each other. So for i > i₀,N_i(p_i(u))≤a, we have

Mi(u) +Ni(p(u)) =upi(u)≤ 1 λMi

λu 2

+1

λNi(2pi(u))

≤ 1 λMi

λu 2

+Ni(pi(u)) +ci

λ

≤ 1

2λMi(λu) +Ni(pi(u)) + ci

λ. Then

(5) Mi(λu)≥2λMi(u)−2ci (i > i₀, Ni(pi(u))≤a).

Condition (2) impliesNi(bi)>1 (i= 1,2, . . .). Applying Theorem 1 in [3], for any x∈l_M⁰ withkxk⁰ = 1, we get K(x)6=∅. Then for any givenk ∈K(x), we havek≤5λork >5λ. Ifk >5λ, then

N_i(p_i(5λ|x(i)|))≤ρ_N(p(5λ|x|))≤ρ_N(p⁻(kx))≤1 (i= 1,2, . . .).

Applying (3) we have

Ni(pi(5|x(i)|))< a (i= 1,2, . . .), i.e.

5|x(i)| ≤q_i(N_i⁻¹(a)) (i= 1,2, . . .).

Consequently

i0

X

i=1

M_i(5|x(i)|)≤

i0

X

i=1

M_i(q_i(N_i⁻¹(a)))≤1.

From 1 =kxk⁰≤ ¹₅(1 +ρM(5x)), we deduce thatρM(5x)≥4. So

(6) X

i>i0

M_i(5|x(i)|)≥3.

Takej≥1 such that 5λ^j< k <5λ^j+1. Combining (3) with N_i(p_i(λ5λ^j−1|x(i)|))≤ρ_N(p(5λ^j|x|))≤1, we obtain

Ni(pi(5λ^j−1|x(i)|))< a (i= 1,2, . . .).

From (5) and (6), we conclude that 5λ^j+1> k= 1 +ρM(kx)> X

i>i0

Mi(k|x(i)|)

≥ X

i>i0

M_i(5λ^j|x(i)|)

≥ X

i>i0

n(2λ)^jM_i(5|x(i)|)−(2λ)^j−1(2c_i)−(2λ)^j−2(2c_i)− · · · −2c_io

= X

i>i0

(2λ)^jn

M_i(5|x(i)|)− 1 2λ+ 1

(2λ)² +· · ·+ 1 (2λ)^j

·2c_io

≥(2λ)^jX

i>i0

nM_i(5|x(i)|)−2c_io

≥(2λ)^j.

Thusj≤log₂5λ, i.e.k≤5λ^log25λ+1. Hencek_M <∞.

References

[1] Chen S.,Some rotundities of Orlicz space with Orlicz norm, Bull. Polish Acad. Sci. Math.

34(1986), 585–596.

[2] Chen S.,Geometry of Orlicz space, Dissertationes Math.356(1996), 1-204.

[3] Wu C., Sun H.,Norm calculation and complex convexity of the Musielak-Orlicz sequence space, Chinese Ann. Math.12A(1991), suppl., 98–102.

Harbin University of Science and Technology, P.O. Box 123, 150080 Harbin, China E-mail: [email protected]

(Received July 3, 2000,revised January 2, 2001)