# The Distance toL∞ in the Grand Orlicz Spaces

Authors: Fernando Farroni; Raffaella Giova
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

We establish a formula for the distance toL∞ from the grand Orlicz space LΦ)Ω  introduced in Capone et al. (2008). A new formula for the distance to L∞ from the grand Lebesgue space Ln)Ω introduced in Iwaniec and Sbordone (1992) is also provided.


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## Body

## 1. Introduction

LetΩ be a bounded open subset of ℝn and let 1<p<∞. The grandLp-space, denoted by Lp)=Lp)(Ω), consists of functions h∈⋂0<ε≤p-1‍Lp-ε(Ω) such that
(1)∥h∥p)=sup0<ε<p-1(12ε⨍Ω|h(x)|p-εdx)1/(p-ε)<∞,where⨍Ω=(1/|Ω|)∫Ω‍ denotes the average over Ω. Note that ∥·∥p) is a norm and Lp)(Ω) is a Banach space. This space was introduced by  Iwaniec and Sbordone in connection with the integrability of the Jacobian [1], and it comes into play in a various number of problems (see, e.g., [2–15]).It is worth pointing out thatL∞(Ω) is not a dense subspace of Lp)(Ω) (see [9]); it is proved in [16] that the distance to L∞ in Lp) is given by
(2)distLp)(h,L∞)=lim  supε→0+(12ε⨍Ω|h(x)|p-εdx)1/(p-ε).A generalization of the grand Lebesgue space is thegrand Orlicz spaceLΦ)(Ω), introduced by Capone et al.  in [17]. Let us recall that Φ:[0,∞)→[0,∞) is called an Orlicz function if it is continuous, strictly increasing, and satisfies Φ(0)=0 and limt→∞Φ(t)=∞. The Orlicz spaceLΦ(Ω) associated with Φ consists of all measurable functions u:Ω→ℝ for which there exists λ>0 such that
(3)∫Ω‍Φ(|u(x)|λ)dx<∞.Let us introduce the Luxemburg functional defined as(4)∥u∥LΦ(Ω)=inf{λ>0:⨍ΩΦ(|u(x)|λ)dx≤1}.Because of the monotonicity ofΦ we have
(5)Φ((1-α)s+αt)≤[Φ(s)+Φ(t)]foreverys,t∈(0,∞),α∈(0,1),and among Orlicz functions we will consider the ones satisfying the following condition:(6)Φ(αt)≤C(α)Φ(t)foreveryt∈(0,∞),α∈(0,1),
for some constant C(α) such that C(α)→0 as α→0. This will be done in order to ensure that the functional in (4) is a quasinorm. In what follows, we will lose no generality in assuming that
(7)Φ(1)=1.Suppose that(8)∫1∞‍Φ(t)tn+1dt=∞,
and let NΦ:[0,1]→[0,∞) be the increasing weight defined as
(9)NΦ(σ)=1∫1∞‍([Φ(t)]1/(1+σ)/tn+1)dt.Following a definition given in [18], we suppose that NΦ is tempered; that is,
(10)c1NΦ(σ)≤NΦ(2σ)≤c2NΦ(σ)foreveryσ∈(0,σ0),σ0∈(0,12)
for some c1,c2>0.An example of functionΦ satisfying (6)–(10) is Φ(t)=tn(1+log(1+t))-α for 0≤α≤1, and in this case NΦ(σ)≈σ1-α as σ→0+ when 0≤α<1 and NΦ(σ)≈|logσ|-1 as σ→0+ when  α=1 (see Section 5 for details).Thegrand Orlicz spaceLΦ)(Ω) consists of all measurable functions u:Ω→ℝ for which there exists λ>0 such that
(11)sup0<σ<σ0NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt<∞,whereu* is the decreasing rearrangement of u(12)u*(t)=inf{h≥0:μu(h)≤t}foreveryt∈[0,|Ω|]
and μu is the distribution function of u(13)μu(h)=|{x∈Ω:|u(x)|>h}|foreveryh≥0.The quasinorm denoted by∥·∥LΦ)(Ω) is defined as follows:
(14)∥u∥LΦ)(Ω)=inf{12(u*(t)λ)(u*(t)λ)λ>0:sup0<σ<σ0NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt≤1}.We address that if we takeΦ(t)=tn also the grand Orlicz space LΦ)(Ω) reduces to the grand Lebesgue space Ln)(Ω) (see [17, Proposition 2.6], [6]).Our main result provides a formula for the distance of a functionu∈LΦ)(Ω) to L∞(Ω), defined by
(15)distLΦ)(Ω)(u,L∞(Ω))=infφ∈L∞(Ω)∥u-φ∥LΦ)(Ω).Theorem 1.
LetΩ be a bounded open set of ℝn. Assume that Φ:[0,∞)→[0,∞) is an Orlicz function verifying (6)–(10). For every function u∈LΦ)(Ω), one has
(16)distLΦ)(Ω)(u,L∞(Ω))=inf{limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt≤1λ>0:limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt≤1}.Our theorem is in the framework of the results of paper [19], which cannot be directly applied to our context, without a preliminary check that the grand Orlicz spaces LΦ) can be characterized as interpolation or extrapolation spaces. We also refer to [5, 16, 20–24] for the problem of finding formulae for the distance to a subspace in a given function space.Theorem1 gives, as byproduct, a characterization of the closure of L∞(Ω) in LΦ)(Ω) with respect to the norm∥·∥LΦ)(Ω), which will be denoted by LbΦ)(Ω).Theorem 2.
A functionu belongs to LbΦ)(Ω) if and only if
(17)limσ→0+NΦ(σ)⨍0|Ω|tσΦ(βu*(t))dt=0∀β>0.For the special choiceΦ(t)=tn, Theorem 1 also provides new formula for the distance to L∞ in Ln) (see Theorem 5).


## 2. The Main Result

We start this section recalling few basic properties of the decreasing rearrangementu* of a measurable function u:Ω→ℝ defined in a bounded open set Ω of ℝn. We refer the reader to [25, Propositions 1.7and 1.8] for details.Lemma 3.
Letu,v:Ω→ℝ be measurable functions defined in a bounded open set Ω of ℝn:
(18)(u+v)*(t1+t2)≤u*(t1)+v*(t2)foreveryt1,t2≥0witht1+t2≤|Ω|,(19)(cv)*=cv*foreveryc>0,(20)v*(0)=∥v∥L∞(Ω)ifv∈L∞(Ω).We need a technical result providing a useful property of the quantity(21)(u)LΦ)(Ω)=inf{lim  supσ→0+NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt≤1λ>0:lim  supσ→0+NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt≤1}.We recall that the goal of Theorem1 consists in proving that (u)LΦ)(Ω) is equal to distLΦ)(Ω)(u,L∞(Ω)). We notice that if φ∈L∞(Ω), then (φ)LΦ)(Ω)=0, because from (8) and (10) it is NΦ(σ)→0 as σ→0+ and the average remains bounded for every σ>0.Lemma 4.
LetΦ:[0,∞)→[0,∞) be an Orlicz function satisfying the assumptions of Theorem 1. Assume that u∈LΦ)(Ω) and φ∈L∞(Ω). Then
(22)(u)LΦ)(Ω)=(u-φ)LΦ)(Ω).Proof.
Letλ>(u)LΦ)(Ω) and let ε∈(0,λ) so that
(23)λ>λ-ε>(u)LΦ)(Ω).
We use (18) with t1=t∈[0,|Ω|] and t2=0 and (20), and we get
(24)(u-φ)*(t)≤u*(t)+∥φ∥L∞(Ω).
We use (5) to get
(25)Φ((u-φ)*(t)λ)≤Φ(u*(t)λ+∥φ∥L∞(Ω)λ)=Φ(λ-ελu*(t)λ-ε+ελ∥φ∥L∞(Ω)ε)≤Φ(u*(t)λ-ε)+Φ(∥φ∥L∞(Ω)ε).
We multiply bytσ and we integrate over [0,|Ω|] to get
(26)⨍0|Ω|tσΦ((u-φ)*(t)λ)dt≤⨍0|Ω|tσΦ(u*(t)λ-ε)dt+Φ(∥φ∥L∞(Ω)ε)⨍0|Ω|tσdt=⨍0|Ω|tσΦ(u*(t)λ-ε)dt+Φ(∥φ∥L∞(Ω)ε)|Ω|σ1+σ.
We multiply byNΦ(σ), and since NΦ(σ)→0 as σ→0+, we have
(27)limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ((u-φ)*(t)λ)dt≤limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ-ε)dt.
From (23) we get
(28)limsupσ→0+NΦ(σ)Φ⨍0|Ω|tσ((u-φ)*(t)λ)dt≤1.
We apply the definition of(u-φ)LΩ)(Φ),  and we have
(29)(u-φ)LΦ)(Ω)≤λ,
and then, passing to the limit as λ→(u)LΦ)(Ω), we have
(30)(u-φ)LΦ)(Ω)≤(u)LΦ)(Ω).
By replacingu with u-φ and φ with -φ in (30), we obtain the converse inequality
(31)(u)LΦ)(Ω)≤(u-φ)LΦ)(Ω).
Equality (22) is finally proved.Now, we are in a position to prove Theorem1.Proof of Theorem1.
From Lemma4 we know that
(32)(u)LΦ)(Ω)=(u-φ)LΦ)(Ω),
for every φ∈L∞(Ω). This clearly proves that
(33)distLΦ)(Ω)(u,L∞(Ω))≥(u)LΦ)(Ω)
since (u)LΦ)(Ω)=(u-φ)LΦ)(Ω)≤∥u-φ∥LΦ)(Ω) for every φ∈L∞(Ω).
Now, we want to show that(34)distLΦ)(Ω)(u,L∞(Ω))≤(u)LΦ)(Ω).
In order to achieve the claimed inequality, we prove that if(35)λ0>(u)LΦ)(Ω),
then
(36)distLΦ)(Ω)(u,L∞(Ω))<λ0.
Without loss of generality we may assume that u∉L∞(Ω). From (35) we find λ∈(0,λ0) such that
(37)limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt≤1.
For each ε>0 there exists σε∈(0,σ0) such that
(38)NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt≤1+εforeveryσ∈(0,σε].
Let h*>0 be such that
(39)μu(h*)<1,
and let σ∈(σε,σ0). From (38), we find some constant hλ (depending on λ), with hλ>h*, such that
(40)NΦ(1)|Ω|∫0μu(hλ)‍tσεΦ(u*(t)λ)dt≤1.
Using the monotonicity of weight NΦ, the fact that hλ>h*, and (39), we deduce from (40) that
(41)NΦ(σ)|Ω|∫0μu(hλ)‍tσΦ(u*(t)λ)dt≤1foreveryσ∈(σε,σ0).
We setuλ(x)=u(x) if |u(x)|≤hλ and,  uλ(x)=0 if |u(x)|>hλ, and we show that
(42)(u-uλ)*(t)=u*(t)fort∈[0,μ(hλ)),(u-uλ)*(t)=0fort∈[μ(hλ),|Ω|].
Let us observe that(43)μu-uλ(h)=μu(hλ)if0<h≤hλ,
while
(44)μu-uλ(h)=μu(h)ifh≥hλ.
Using the fact that the distribution function is decreasing, we easily see that(45)μu-uλ(h)≤μu(hλ)foreveryh≥0.
Therefore, if we lett∈[μ(hλ),|Ω|], we see that condition
(46)μu-uλ(h)≤t
is verified for all h≥0. Thus (u-uλ)*(t)=0 for t∈[μ(hλ),|Ω|]. On the other hand, if we let t∈[0,μ(hλ)), we see that condition (46) is the same as requiring
(47)μu(h)≤t.
Thus(u-uλ)*(t)=u*(t) holds if t∈[0,μ(hλ)), and (42) is proved.
It follows directly from (42) that
(48)NΦ(σ)⨍0|Ω|tσΦ((u-uλ)*(t)λ)dt=NΦ(σ)|Ω|∫0μu(hλ)‍tσΦ(u*(t)λ)dt,
for every σ∈(0,σ0). Hence, we make use of (38) if σ∈(0,σε] and of (41) if σ∈(σε,σ0)  to conclude that
(49)NΦ(σ)⨍0|Ω|tσΦ((u-uλ)*(t)λ)dt≤1+εforeveryσ∈(0,σ0).
In particular,(50)sup0<σ<σ0NΦ(σ)⨍0|Ω|tσΦ((u-uλ)*(t)λ)dt≤1+ε.
Since (50) holds for every ε>0, we obtain that its left-hand side is smaller than 1, and therefore ∥u-uλ∥LΦ)(Ω)≤λ. We get
(51)distLΦ)(Ω)(u,L∞(Ω))≤∥u-uλ∥LΦ)(Ω)≤λ<λ0.
Hence (36) is established. Since λ0 is any arbitrary number for which (35) holds, we may pass to the limit as λ0 approaches (u)LΦ)(Ω) in (36) to get
(52)distLΦ)(Ω)(u,L∞(Ω))≤(u)LΦ)(Ω).
Combining (52) with (33) we obtain (16) as desired.Proof of Theorem2.
As a consequence of Theorem1, it is clear that u∈LbΦ)(Ω) if and only if
(53)limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt≤1∀λ>0.
We fix an arbitraryα∈(0,1) and we set λ=α/β. Using (6) we have
(54)NΦ(σ)⨍0|Ω|tσΦ(βu*(t))dt=NΦ(σ)⨍0|Ω|tσΦ(αu*(t)λ)dt≤C(α)NΦ(σ)⨍0|Ω|tσΦ(u*(t)λ)dt.
Hence, using (53) we have
(55)lim supσ→0+NΦ(σ)⨍0|Ω|tσΦ(βu*(t))dt≤C(α),
and (17) follows since C(α)→0 as α→0.


## 3. The Case of the Grand Lebesgue SpaceLn)

We denote by(u)Ln)(Ω) the functional (u)LΦ)(Ω) as in (21) when Φ(t)=tn. In this case, (u)Ln)(Ω) takes the form
(56)(u)Ln)(Ω)=limsupσ→0+(nσ1+σ⨍0|Ω|tσ|u*(t)|ndt)1/n.Our next result proves that the distance given by formula (2) reduces to (u)Ln)(Ω).Theorem 5.
LetΩ be a bounded open set of ℝn. For every function u∈Ln)(Ω), one has
(57)distLΦ)(Ω)(u,L∞(Ω))=lim supσ→0+(nσ1+σ⨍0|Ω|tσ|u*(t)|ndt)1/n.Proof.
First we prove that(58)lim supε→0+(ε⨍Ω|u(x)|n-εdx)1/(n-ε)≤lim supσ→0+(nσ1+σ⨍0|Ω|tσ|u*(t)|ndt)1/n.
To this aim, we considerε,σ>0 and k>1 such that
(59)n-ε=n1+kσ.
Using Hölder's inequality we have(60)ε⨍Ω|u(x)|n-εdx=nkσ1+kσ⨍0|Ω|tσ/(1+kσ)|u*(t)|n/(1+kσ)t-σ/(1+kσ)dt≤nkσ1+kσ(⨍0|Ω|tσ|u*(t)|ndt)1/(1+kσ)(⨍0|Ω|t-1/kdt)kσ/(1+kσ)
which in turn implies that
(61)(ε⨍Ω|u(x)|n-εdx)1/(n-ε)≤(k(k-1)|Ω|1/k)kσ/n(nkσ1+kσ)(1+kσ)/n×(⨍0|Ω|tσ|u*(t)|ndt)1/n=(k(k-1)|Ω|1/k)kσ/n×(nkσ1+kσ)(1+kσ)/n(1+σnσ)1/n×(nσ1+σ⨍0|Ω|tσ|u*(t)|ndt)1/n=(k(k-1)|Ω|1/k)kσ/n×(k1+kσ)(1+kσ)/n(1+σ)1/n(nσ)kσ/n×(nσ1+σ⨍0|Ω|tσ|u*(t)|ndt)1/n.
Since (nσ)kσ/n→1 as σ→0+, we deduce from (61) that
(62)lim  supε→0+(ε⨍Ω|u(x)|n-εdx)1/(n-ε)≤k1/nlim  supσ→0+(nσ1+σ⨍0|Ω|tσ|u*(t)|ndt)1/n.
Sincek is any number strictly greater than 1, (62) immediately implies (58).
We wish to prove the converse inequality(63)lim  supε→0+(ε⨍Ω|u(x)|n-εdx)1/(n-ε)≥lim  supσ→0+(nσ1+σ⨍0|Ω|tσ|u*(t)|ndt)1/n.
For eacht∈[0,|Ω|], we have
(64)⨍Ω|u(x)|n/(1+σ)dx=⨍0|Ω||u*(s)|n/(1+σ)ds≥1|Ω|∫0t‍|u*(s)|n/(1+σ)ds≥1|Ω|t|u*(t)|n/(1+σ).
Thus(65)⨍0|Ω|tσ|u*(t)|ndt=⨍0|Ω|tσ|u*(t)|nσ/(1+σ)|u*(t)|n/(1+σ)dt≤|Ω|σ(⨍0|Ω||u*(t)|n/(1+σ)dt)1+σ.
We considerε,σ>0 such that
(66)n-ε=n1+σ.
Then(67)(nσ1+σ⨍0|Ω|tσ|u*(t)|ndt)1/n≤[(nσ1+σ)-σ|Ω|σ(nσ1+σ⨍0|Ω||u*(t)|n/(1+σ)dt)1+σ]1/n=(nσ1+σ)-σ/n|Ω|σ/n(nσ1+σ⨍0|Ω||u*(t)|n/(1+σ)dt)(1+σ)/n=(nσ1+σ)-σ/n|Ω|σ/n(ε⨍Ω|u(x)|n-εdx)1/(n-ε),
which proves (63).


## 4. Few Properties of the Distance

In this concluding section we provide certain properties of the functional(·)LΦ)(Ω).Lemma 6.
LetΦ:[0,∞)→[0,∞) be an Orlicz function satisfying the assumptions of Theorem 1 and let v∈LΦ)(Ω). Assume that
(68)limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ(v*(t)λ)dt≤L,for some constants positive λ and L. Then, there exists a positive constant C0 depending only on L such that
(69)(v)LΦ)(Ω)≤C0λ.Proof.
LetC(α) be the constant appearing in (6). We may take α0∈(0,1) such that
(70)C(α0)L≤1
since C(α)→0 as α→0. We use (6) to get
(71)NΦ(σ)⨍0|Ω|tσΦ(α0v*(t)λ)dt≤C(α0)NΦ(σ)⨍0|Ω|tσΦ(v*(t)λ)dt.
We take thelimsup as σ→0+ and use (68) to get
(72)limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ(α0v*(t)λ)dt≤C(α0)L.
Therefore, from (72) and (70) we have
(73)limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ(α0v*(t)λ)dt≤1.
The desired constantC0 is obtained by setting C0=1/α0. We address that C0 is independent of v, and thus the proof is completed.Remark 7.
It is clear from the definition of(u)LΦ)(Ω) that we can pick C0=1 if L=1.Our next lemma provides a sort of triangle inequality involving the functional(·)LΦ)(Ω).Lemma 8.
LetΦ:[0,∞)→[0,∞) be an Orlicz function satisfying the assumptions of Theorem 1 and let u,v∈LΦ)(Ω). Then, there exists a constant C1 depending only on Φ such that
(74)(u+v)LΦ)(Ω)≤C1[(u)LΦ)(Ω)+(v)LΦ)(Ω)].Proof.
Take(75)λ1>(u)LΦ)(Ω),λ2>(v)LΦ)(Ω).
Lett∈[0,|Ω|]. We use (18) with t1=t2=t/2, the monotonicity of Φ, to obtain
(76)Φ((u+v)*(t)λ1+λ2)≤Φ(λ1λ1+λ2u*(t/2)λ1+λ2λ1+λ2v*(t/2)λ2)≤[Φ(u*(t/2)λ1)+Φ(v*(t/2)λ2)].
Fixσ∈(0,1). We multiply by tσ and we integrate over [0,|Ω|] to get
(77)∫0|Ω|‍tσΦ((u+v)*(t)λ1+λ2)dt≤[∫0|Ω|‍tσΦ(u*(t/2)λ1)dt+∫0|Ω|‍tσΦ(v*(t/2)λ2)dt].
With the aid of two changes of variables in the integrals appearing at the right-hand side of (77) we have
(78)∫0|Ω|‍tσΦ((u+v)*(t)λ1+λ2)dt≤2σ+1[∫0|Ω|/2‍sσΦ(u*(s)λ1)ds+∫0|Ω|/2‍rσΦ(v*(r)λ2)dr],
which in turn implies(79)⨍0|Ω|tσΦ((u+v)*(t)λ1+λ2)dt≤4[⨍0|Ω|sσΦ(u*(s)λ1)ds+⨍0|Ω|rσΦ(v*(r)λ2)dr].
We multiply both sides of (79) by NΦ(σ), and we take the limsup as σ→0+ and use (75) to get
(80)limsupσ→0+NΦ(σ)⨍0|Ω|tσΦ((u+v)*(t)λ1+λ2)dt≤8.
We appeal to Lemma6 to conclude that there exists a constant C1 such that
(81)(u+v)LΦ)(Ω)≤C1(λ1+λ2).
Finally, (74) follows letting λ1→(u)LΦ)(Ω) and λ2→(v)LΦ)(Ω), respectively.


## 5. An Example

In this section we study the behaviour of weightNΦ(σ) as σ→0+ when Φ(t)=tn(1+log(1+t))-α with 0≤α≤1. We follow closely the lines of Example 3.6  in [17].Example 9.
LetΦ(t)=tn(1+log(1+t))-α and let  0≤α≤1. We start by proving that
(82)NΦ(σ)≈σ1-αasσ→0+when0≤α<1.
To see this, letθ>0. Then
(83)∫1∞‍Φ(t)1-θtn+1dt≈∫1∞‍t-nθ-1(1+logt)-α(1-θ)dt≈∫1∞‍e-nθuu-α(1-θ)du≈∫θ∞‍e-nv(vθ)-α(1-θ)dvθ≈θα-1∫θ∞‍e-nvv-α(1-θ)dv≈θα-1(1+∫θ1‍v-α(1-θ)dv).
We pickθ=σ/(1+σ) in such a way that
(84)NΦ(σ)=1∫1∞‍([Φ(t)]1/(1+σ)/tn+1)dt≈σ1-α.
A similar argument leads to(85)NΦ(σ)≈|logσ|-1asσ→0+whenα=1.



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