# Admissible Differential Bases in Banach Spaces

Authors: Evgenii I. Berezhnoj
Journal: International Journal of Mathematics and Mathematical Sciences
 (2013)

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## Abstract

In the present work, a new method for constructing differential bases is presented. The bases constructed by this method allow us to distinguish symmetric spaces with different behaviour of fundamental functions at zero, as in the case of Hayes and Stokolos, and even allow us to distinguish Lorentz and Marzinkiewicz spaces or Lebesgue and Marzinkiewicz spaces whose fundamental functions are the same.


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## Body

## 1. Introduction

We recall certain definitions connected with differential bases [1]. By a differential basisB(t)at a pointt∈ℝn is meant a family of bounded measurable sets with positive measure which contain t and are such that there is at least one sequence {Bk∈B(t)} satisfying the condition diamBk→0. A union of such families B={⋃‍B(t):t∈ℝn} is called a differential basis in Rn. We note that the bulk of the problems and assertions in the theory of differentiation of integrals in ℝn consists of testing the validity almost everywhere of some fact. Therefore, it is natural to consider differential bases which are defined, not for all points in ℝn, but only almost everywhere. In what follows, we will make use of these observations.The classical examples of differential bases are the bases inℝn, usually denoted by Bs(ℝn)  (s=1,2,…,n) and consisting of all rectangular parallelepipeds B of the form
(1)B={(t1,…,tn)∈ℝn:αi<ti<αi+γi,i=1,…,n},
which satisfy the condition γj=γ1 for j=1,…,s and αi∈ℝ, γi∈ℝ+.A basisB made up of open sets is called a Busemann-Feller basis (BF-basis) if it follows from the conditions B∈B and t∈B such that B∈B(t). The significance of the introduction of BF-bases lies in the fact that questions arising in the theory of differentiation with respect to bases can be easily resolved for BF bases. We now define the upper and lower derivatives of the integral of a locally integrable function f at a point t∈ℝn with respect to a basis B by means of the identities
(2)DB(f,t)¯=sup{lim¯k→∞1μ(Bk)∫Bk‍fds:Bk∈B(t),diamBk→0},DB(f,t)_=inf{lim_k→∞1μ(Bk)∫Bk‍fds:Bk∈B(t),diamBk→0}.We note that the upper and lower derivatives are variants of the functionalsμ* and μ* which are the subject of investigation in several sections of the chapter 1 in [2].Following [1], we say that a basis B  differentiates the integral off if the identities
(3)DB(f,t)¯=DB(f,t)_=f(t)
hold almost everywhere. If B differentiates the integral of any function in the space X, then we say that the basis B  differentiates the spaceX. If the basis B differentiates L∞, then it is said to be a density basis [1].One of the fundamental problems of the theory of differentiation of integrals has the following form: given two function spacesX, Y which are different in some sense, is it possible to distinguish these two spaces with the help of differential bases? In other words, does there exist a differential basis, which differentiates all integrals obtained from functions in X, but a function can be found in Y whose integral is not differentiated by the given basis? This problem is still far from its final solution. The first result related to this problem is due to Hayes [1, page 155], who showed that it is possible to distinguish the spaces Lp and Lq for p≠q by constructing an appropriate differential basis. Then, Stokolos [3, 4] showed that it is possible to distinguish Orlicz spaces by constructing an appropriate differential basis. Apparently, everything that has been achieved on the problem of distinguishing spaces with the help of differential bases is exhausted by these two results.The following classical results are fundamental for the theory of differentiation of integrals inℝn: Bs differentiates L(Log+L)n-s (the case s=n is due to Lebesgue [5], the case s=1 to Jessen et al. [6], and the case of arbitrary s to Zygmund [7]). In 1935, Saks [8] showed that B1 does not differentiate o(L(Log+L)n-1). It was proved later in [9] that Bs does not differentiate the space o(L(Log+L)n-s).LetQ0 be the unit cube in ℝn with the usual Lebesgue measure. For f:Q0→ℝ, we will denote by λ(f,γ) the distribution function of f:
(4)λ(f,γ)=μ({t:|f(t)|>γ});
its rearrangement in nonincreasing order will be denoted by f*:
(5)f*(t)=inf{s>0:λ(f,s)<t}.
A Banach space X consisting of measurable functions is said to be ideal [10] if, given y∈X, measurability of x and |x(t)|≤|y(t)| almost everywhere, it follows that x∈X and ||x∣X||≤||y∣X||. An ideal space is said to be symmetric [10] if for x,y∈X, given that the inequality λ(x,γ)≤λ(y,γ) holds for all γ∈ℝ+, it follows that ||x∣X||≤||y∣X||. As usual we denote by X′ the space of integral functionals on X with norm
(6)∥g∣X′∥=sup{∫Q0‍g(t)x(t)dt:∥x∣X∥≤1}.For each symmetric spaceX, its fundamental function is defined by means of the identity
(7)ψ(X,t)={∥χ(D)∣X∥:μ(D)=t},
where χ(D) denotes the characteristic function of a set D. This function is nondecreasing and t-1ψ(X,t) does not increase. The spaces Lp of Lebesgue, Lh of Orlicz, M(ψ) of Marzinkiewicz, and Λ(ψ) of Lorentz are the classical examples of symmetric spaces. The theory of symmetric spaces of functions is described in detail in [10] and that of symmetric spaces of sequences in [11]. Since Lorentz and Marzinkiewicz spaces will be used below, we will describe M(ψ) and Λ(ψ) in greater detail. Let us denote by W the set of concave functions φ:[0,1]→ℝ+, each of which is continuous with φ(0)=0 and limt→0t-1φ(t)=∞. Then, the Lorentz space Λ(φ) (resp., the Marzinkiewicz space M(φ)) consists of all those f for which the following expression
(8)∥f∣Λ(φ)∥=∫01‍f*(s)dφ(s)=∫0∞‍φ(λ(f,s))ds.
Respectively,
(9)∥f∣M(φ)∥=sup0<t<11φ(t)∫0t‍f*(s)ds=supD⊂Q01φ(μ(D))∫D‍|f(s)|ds,
is a finite norm. It is clear that for the fundamental functions of the Lebesgue, Orlicz, Lorentz, and Marzinkiewicz spaces we have the identities
(10)ψ(Lp,t)=t1/p,ψ(Lh,t)=1h-1(t-1),ψ(Λ(φ),t)=φ(t),ψ(M(φ),t)=tφ(t).The duality between Lorentz and Marzinkiewicz spaces,(11)Λ(φ)′=M(φ),M(φ)′=Λ(φ),
is well known, as is the embedding theorem of Krejn et al. [10]: let X be a symmetric space and ψ(X,t)=φ(t) its fundamental function; then we have the continuous embeddings
(12)Λ(φ)⊆X⊆M(tφ(t)).Ifφ(t)=tθ with 0<θ<1, then in this case the Lorentz (Marzinkiewicz) space will be denoted by Λθ(Mθ).


## 2. The Construction of Differential Bases

Now we pass to the construction of differential bases. For simplicity and clarity, we will demonstrate the construction forn=2. We begin the construction of the corresponding bases from the construction of a certain collection U of sets in the square Q0=[0,1]×[0,1]; we will refer to the construction process as the basic construction.Suppose that we are given a sequence of triples of natural numbers(mi,pi,li), for which mi+1>mi, pi+1>pi, and li+1>li. First we divide the square Q0 into m12 equal squares Q(1,i)(i=1,2,…,m12). These we will call squares of the first level. Then again we divide each of the squares Q(1,i) into m22 equal squares and obtain the squares Q(2,i)(i=1,2,…,m12m22). These we will call squares of the second level, and so on. Each of the squares of the kth level Q(k,i) is divided in turn into pklk equal rectangles Π(k,i,j)(j=1,2,…,pklk), which we will take as open. For the sake of definiteness, we will agree that the vertical side of the square is divided into pk equal parts, while the horizontal side of the square is divided into lk equal parts. Now, we will construct a collection of sets U(k) relating to the kth level.We take the square of thekth level with the first index. The sets of the collection U(k) which are contained in the square Q(k,1) are formed in the following manner. By construction the square is split up into lk vertical strips and pk horizontal strips which form pklk equal rectangles. We take the lower left rectangle Π(k,1,1), which lies at the intersection of the first horizontal and the first vertical. All the other rectangles from the square Q(k,1) lying on the first horizontal do not take part in the continuation of the construction and can be discarded. The set U(k,1,1) consists of the rectangle Π(k,1,1) to which have been added the pk-1 rectangles from the first vertical. The set U(k,1,2) consists of the rectangle Π(k,1,1) to which have been added the pk-1 rectangles from the second vertical. Then, we again take the rectangle Π(k,1,1), adding to it the pk-1 rectangles from the third vertical, and we will form the set U(k,1,3). We continue this process and thus obtain the collection of sets
(13){U(k,1,1),…,U(k,1,lk)}=U(k,1).The sets of the collectionU(k,1) have three important properties:
(14)U(k,1,j)∩U(k,1,l)=Π(k,1,1)(j≠l),μ(U(k,1,j))=pkμ(Π(k,1,1))=μ(Q(k,1))lk,diam(U(k,1,j+1))≥diam(U(k,1,j))(j=1,…,lk-1).Now, we repeat this construction for all the remaining squaresQ(k,i) of the kth level (i=1,2,…,m12m22⋯mk2). We will obtain the collections of sets U(k,i). The sets of the collections U(k,i) have the following important properties:
(15)U(k,i1,j)∩U(k,i2,l)=∅(i1≠i2),U(k,i,j)∩U(k,i,l)=Π(k,i,1)(j≠l),μ(U(k,i,j))=pkμ(Π(k,1,1))=pkμ(Π(k,i,1))=μ(Q(k,1))lk,limk→∞diam(U(k,i,j))=0.We now define the collectionU(k) as the union over i of all the collections U(k,i). Now on combining the collections of sets U(1),U(2),…,U(i),… constructed for each level, we obtain the desired collection U. We note that the measure of those points of Q0 which are not included in the sets of the collection U(k) does not exceed pk-1. Therefore, from the definition and (15), we may take U as a differential BF basis.Corresponding to the collectionU(k), we construct the maximal operator
(16)Mkx(t)=sup{1μ(U(k,i,j))×∫U(k,i,j)‍|x(t)|dt:t∈U(k,i,j)∈U(k)},
and for the collection U and δ>0, we construct its maximal operator
(17)Mδx(t)=sup{1μ(B)∫B‍|x(t)|dt:t∈B∈U,diamB<δ}.Forδ=diamQ0, we will write simply M in place of Mδ. We denote by Γ(U(k)) the union of the boundaries of all the sets U(k,i,j) appearing in the collection U(k). It is easy to see that the identity μ(Γ(U(k)))=0 holds.


## 3. An Effective Bound for Maximal Operator

Our immediate goal is to obtain an effective bound for the maximal operator. The following lemma can be obtained easily from well-known facts of the metric theory of functions and the identityμ(Γ(U(k)))=0.Lemma 1.
Suppose that the collections of setsU(1),…,U(n) have been constructed relative to the triples of natural numbers (mi,pi,li)(i=1,…,n). Then, for any ϵ>0, there exists a number N=N(ϵ) such that for all m>N(ϵ) the following condition is satisfied: if the square Q0 is divided into m2 equal nonintersecting squares {Qi}, then one has the inequality
(18)∑1m2‍{μ(Qi):Qi∩(⋃1nΓ(U(k)))≠∅}<ϵ.In other words, the area of those squares which intersect the boundary of at least one set of the collectionsU(1),…,U(n) will be less than ϵ.Now from all the differential bases obtained by means of the basic construction we extract a certain subclass for which it is possible to verify its differential properties.Definition 2.
We will say that a sequence of triples of natural numbers(mk,pk,lk) is admissible if the conditions
(19)mk+1>mk,pk+1>pk,lk+1>lk,mk+1>N(μ(Π(k,1,1))),
are satisfied, where the number N(·) is taken from Lemma 1 for ϵ=μ(Π(k,1,1)).Definition 3.
We will say that a differential basisU={⋃‍U(k)} is admissible if the families {U(k)} have been constructed by means of the basic construction applied with an admissible sequence of triples.The following lemma is a key tool for the characterization of differential properties of admissible bases.Lemma 4.
Letx∈L1 with x(t)>0 almost everywhere, and let U be an admissible differential basis. Suppose that α>0 and δ>0 are given and choose k(δ) such that
(20)diam(Π(k(δ),1,1))≤δ,butdiam(Π(k(δ)-1,1,1))>δ.
Then, one has the following relation:(21){t:Mδx(t)>α}⊂E0∪E1∪E2,
where
(22)μ(E0)<2α-1∫E0‍x(t)dt,μ(E1)<∑k≥k(δ)‍μ(Π(k,1,1)),
and the set E2 consists of the union of nonintersecting sets Vk with k≥k(δ), each of which can be represented in the form
(23)Vk={⋃Q(k,i):i∈J(k)}
for some J(k), and moreover for each square Q(k,i)∈Vk the condition
(24)μ(Q(k,i))<2α-1lk∫Π(k,i,1)‍x(t)dt
is satisfied.Proof.
We will establish the lemma forδ=diamQ0. We will indicate all the changes required for other δ at the end. Let
(25)D1=D(x,α)={t:Mx(t)>α}={⋃U(k,i,j):1μ(U(k,i,j))∫U(k,i,j)‍x(t)dt>α},I1={(k,i,j):U(k,i,j)⊂D1},k(1)=min{k:(k,i,j)∈I1}.
We introduce the set(26)W1={⋃U(k(1),i,j):(k(1),i,j)∈I1}.
Then, as it is not difficult to verify, we have the identity(27)W1={t:Mk(1)x(t)>α}.
Let us put (recall that (15) holds)
(28)J(k(1))={αμ(Q(k(1),i))2lk(1)i:∫Π(k(1),i,1)‍x(t)dt>α2μ(U(k(1),i,1))=αμ(Q(k(1),i))2lk(1)},W1+={⋃Q(k(1),i):i∈J(k(1))},W1-=W1∖W1+.
IfU(k(1),i,j)⊂W1-, then from the inequality
(29)∫U(k(1),i,j)∖Π(k(1),i,1)‍x(t)dt=∫U(k(1),i,j)‍x(t)dt-∫Π(k(1),i,1)‍x(t)dt>αμ(U(k(1),i,j))-α2μ(U(k(1),i,j))=α2μ(U(k(1),i,j)),
we obtain
(30)μ(U(k(1),i,j))<2α∫U(k(1),i,j)∖Π(k(1),i,1)‍x(t)dt.
Now the setsU(k(1),i,j1)∖Π(k(1),i,1) and U(k(1),i,j2)∖Π(k(1),i,1) are disjoint for j1≠j2 according to (15). Therefore, we have the relation
(31)μ(W1-)≤∑U(k(1),I,j)⊂W1-‍μ(U(k(1),i,j))≤∑U(k(1),i,j)⊂W1-‍2α∫U(k(1),i,j)∖Π(k(1),i,1)‍x(t)dt≤2α∫W1-‍x(t)dt.
Then, ifQ(k(1),i)⊂W1+, according to (15) and the construction, we will obtain
(32)μ(Q(k(1),i))=lk(1)μ(U(k(1),i,1))<2lk(1)α∫Π(k(1),i,1)‍x(t)dt,
and for the set W1+ we will have according to the construction
(33)μ(W1+)=∑i∈J(k(1))‍μ(Q(k(1),i)).
Next, among the sets{U(k,i,j):U(k,i,j)⊂D1,k>k(1)} we reject those which lie entirely in W1-∪W1+ and put
(34)D2={⋃U(k,i,j):U(k,i,j)⊂D1,U(k,i,j)∩(W1-∪W1+)≠U(k,i,j)⋃‍},I2={(k,i,j):U(k,i,j)⊂D2},k(2)=min{k:(k,i,j)∈I2}.
We divide the setD2 into two parts:
(35)D2γ={⋃U(k(2),i,j)⊂D2:U(k(2),i,j)∩(W1-∪W1+)≠∅⋃U(k(2),i,j)},W2={U(k(2),i,j)⊂D2:U(k(2),i,j)∩(W1-∪W1+)=∅}.
Let us denote byMk(2)¯ the operator, analogous to Mk(2) from (16), but with the upper bound taken only over those sets U(k(2),i,j) which are disjoint from W1-∪W1+. Then, as is not difficult to verify, we have the identity
(36)W2={⋃U(k(2),i,j):(k(2),i,j)∈I2}={t:Mk(2)¯x(t)>α}.
Therefore, we can put(37)J(k(2))={αμ(Q(k(2),i))2lk(2)i:∫Π(k(2),i,1)‍x(t)dt>α2μ(U(k(2),i,1))=αμ(Q(k(2),i))2lk(2)},W2+={⋃Q(k(2),i):i∈J(k(2))},W2-=W2∖W2+.
IfU(k(2),i,j)⊂W2-, then analogous to (30) we obtain
(38)μ(U(k(2),i,j))<2α∫U(k(2),i,j)∖Π(k(2),i,1)‍x(t)dt.
Because the setsU(k(2),i,j1)∖Π(k(2),i,1) and U(k(2),i,j2)∖Π(k(2),i,1) are disjoint for j1≠j2, we have the relations
(39)μ(W2-)≤∑U(k(2),i,j)⊂W2-‍μ(U(k(2),i,j))≤∑U(k(2),i,j)⊂W2-‍2α∫U(k(2),i,j)∖Π(k(2),i,1)‍x(t)dt≤2α∫W2-‍x(t)dt.
And for the setW2+, we have in analogy to W1+ the following relations: if Q(k(2),i)⊂W2+, then
(40)μ(Q(k(2),i))=lk(2)μ(U(k(2),i,1))<2lk(2)α∫Π(k(2),i,1)‍x(t)dt,μ(W2+)=∑i∈J(k(2))‍μ(Q(k(2),i)).
The setD2γ is made up of those U(k(2),i,j) which intersect the boundaries of the sets U(k(1),i,j) contained in W1-∪W1+. Since the basis is admissible, we can write the bound
(41)μ(D2γ)<μ(Π(k(1),1,1)).
Now we reject those among the sets{U(k,i,j):U(k,i,j)⊂D2,k>k(2)}, which lie entirely in (W1-∪W1+)∪(W2-∪W2+), and put
(42)D3={⋃U(k,i,j):U(k,i,j)⊂D1,U(k,i,j)∩((W1-∪W1+)∪(W2-∪W2+))≠U(k,i,j)⋃‍},I3={(k,i,j):U(k,i,j)⊂D3},k(3)=min{k:(k,i,j)∈I3}.
We divide the setD3 into two parts:
(43)D3γ={U(k(3),i,j)⊂D3:U(k(3),i,j)∩((W1-∪W1+)∪(W2-∪W2+))≠∅},W3={U(k(3),i,j)⊂D3:U(k(3),i,j)∩((W1-∪W1+)∪(W2-∪W2+))=∅},
construct the sets W3-, W3+ and the operator Mk(3)¯, and write analogues of (39), (40), (41), and so on. We continue this process to infinity and obtain as a result the collections of sets
(44)W1-,W2-,…,Wk-,…W1+,W2+,…,Wk+,…D2γ,D3γ,…,Dkγ,….
Moreover,(45)Wi+∩Wj+=∅,Wi-∩Wj-=∅(i≠j),Wi-∩Wj+=∅,
and the following relations hold:
(46)μ(Wk-)≤2α-1∫Wk-‍x(t)dt,
and if Q(k(i),j)⊂Wi+1, then
(47)μ(Q(k(i),i))=lk(i)μ(U(k(i),j,1))<2lk(i)α∫Π(k(i),j,1)‍x(t)dt,μ(Wi+)=∑j∈J(k(i))‍μ(Q(k(i),j)),μ(Diγ)<μ(Π(k(i-1),1,1)).
To complete the proof, it is enough to put(48)E0={⋃Wi-:i=1,2,…},E1={⋃Diγ:i=2,3,…},E2={⋃Wi+:i=1,2,…},Vk(i)={⋃Q(k(i),j):j∈J(k(i))}.
Then by (47), all the relations (22) and (24) will be satisfied for δ=diamQ0 and k(δ)=1. But if δ<diamQ0, then all the arguments remain in force, only k(1)≥k(δ), and therefore (22) and (24) are always true. The lemma is proved.It is immediately useful to note that for admissible differential basesμ(E1)→0 as δ→0.Lemma4 has a series of important corollaries.Corollary 5.
Each admissible differential basis is a density basis; that is, it differentiatesL∞.Proof.
Letx(t)=χ(A) and fix α>0. We first choose k0 such that the inequality 0.5α>pk-1 holds for k≥k0. Then by (15), we have for such k the inequality
(49)1μ(U(k,i,j))∫Π(k,i,1)‍χ(A)dt≤μ(Π(k,i,1))μ(U(k,i,j))=pk-1<0.5α.
Therefore, if we putδ=0.5diamΠ(k0,1,1), then it will follow from (49) the definition of δ and the definition of the sets Vk that E2=∅, and, consequently,
(50)μ({t:Mδχ(A)>α})≤∑i>k0‍μ(Π(i,1,1))+2α-1μ(A).
It follows from (50) that μ({t:Mδχ(A)>α})→0 as δ→0 and μ(A)→0. According to the criterion given in Theorem 1.1 of the book [1], the last observation implies that U is a density basis.Corollary 6.
LetX=Lp(1<p<∞). If the relation
(51)supiliq-1pi=C<∞
holds, where 1/p+1/q=1, then U differentiates Lp.Proof.
Letx∈Lp with x≥0 almost everywhere. We will bound the measure of the set {t:Mδx(t)>α} in accordance with Lemma 4. We obtain from duality and (22) for any symmetric space X the following:
(52)μ(E0)≤2α∥x∣X∥min{ψ(X′,μ(E0)),ψ(X′,μ(suppx))}.
In the caseX=Lp, we have
(53)μ(E0)≤2α∥x∣Lp∥μ(E0)1/q.
To obtain the bound forE2, we can put
(54)g(x,δ,t)=∑k≥k(δ)‍∑i∈J(k)‍lkχ(Π(k,i,1))
and by (24), we can write the relation,
(55)μ(E2)≤∑k≥k(δ)‍∑i∈J(k)‍μ(Q(k,i))≤2α-1∑k≥k(δ)‍∑i∈J(k)‍lk∫Π(k,i,1)‍x(s)χ(Π(k,i,1))ds≤2α-1∥x∣X∥∥g(x,δ,·)∣X′∥.
But in our case(56)∥g(x,δ,·)∣Lq∥=(∑k≥k(δ)‍∑i∈J(k)‍lkqμ(Π(k,i,1)))1/q=(∑k≥k(δ)‍∑i∈J(k)‍lkqμ(Q(k,i))lkpk)1/q.
Taking into account (15), (51), and (55) along with the fact that E2 consists of nonintersecting sets Vk, and consequently of squares Q(k,i), we obtain
(57)μ(E2)≤2Cα-1μ(E2)1/q∥x∣Lq∥.
If we put(58)xk(t)={x(t)forx(t)≥k,0forx(t)<k,
and apply the bounds (53) and (57) to xk(t), then we will obtain μ({t:Mδxk(t)>α})→0 as δ→0 and k→∞. For density bases, the last observation is equivalent by the criterion given in Theorem 2.2 of the book [1] to the differentiability of the integral ∫‍x(t)dt. Since this result is true for any x∈Lp, x≥0 almost everywhere, it follows that U differentiates Lp. The corollary is proved.Corollary 7.
Letφ∈W and X=M(φ). If the relations
(59)∑i≥1‍liφ(1pili)<∞,∑i=k∞‍1pili≤Cpklk,
are satisfied and C does not depend on k∈ℕ, then U differentiates M(φ).Proof.
The proof of this corollary follows the scheme of the proof of Corollary6. Let x∈M(φ) with x(t)≥0 almost everywhere. We find a bound for the measure of the set {t:Mδx(t)>α} by using Lemma 4 and (52) and (55). From (52) and duality, we obtain
(60)μ(E0)≤2α-1∥x∣M(φ)∥ψ(Λ(φ),μ(suppx))=2α-1∥x∣M(φ)∥φ(μ(suppx)),
and we will determine directly the norm of g(x,δ,t) in Λ(φ). According to (59) for γ>ln≥lk(δ), we have the inequality
(61)μ({t:g(x,δ,t)>γ})≤∑k>n‍∑i∈J(k)‍μ(Π(k,i,1))≤∑k>n‍1pili≤Cpn+1ln+1,
and for γ≤lk(δ), there is the bound
(62)μ({t:g(x,δ,t)>γ})≤∑k≥k(δ)‍∑i∈J(k)‍μ(Π(k,i,1))≤∑k≥k(δ)‍1pili≤Cpk(δ)lk(δ).
Therefore,(63)∥g(x,δ,t)∣Λ(φ)∥=∫0∞‍φλ(g(x,δ,·),γ)dγ={∫0lk(δ)‍+∑j=k(δ)∞‍∫ljlj+1‍}φλ(g(x,δ,·),γ)dγ≤lk(δ)φ(Cpk(δ)lk(δ))+∑j≥k(δ)‍(lj+1-lj)φ(Cpj+1lj+1)≤C∑j≥k(δ)‍ljφ(1pjlj).
From this, we obtain(64)μ(E2)≤2Cα∥x∣M(φ)∥∑j≥k(δ)‍ljφ(1pjlj).
The final part of the proof repeats word for word the proof of the corresponding part in Corollary6.Corollary 8.
Letφ∈W and X=Λ(φ). If the conditions
(65)supn1pnφ(1/pn+1ln+1)<∞,∑i=n∞‍pi-1<Cpn-1,
are satisfied and C does not depend on n∈ℕ, then U differentiates Λ(φ).Proof.
Letx∈Λ(φ) with x(t)≥0 almost everywhere. Using Lemma 4 and (52), and (55) we find a bound for the measure of the set {t:Mδx(t)>α}. From duality and (52), we obtain
(66)μ(E0)≤2α-1∥x∣Λ(φ)∥ψ(M(φ),μ(E0))=2α-1∥x∣Λ(φ)∥μ(E0)φ(μ(E0)),
and we will arrive at a bound for the norm of the function g(x,δ,t) in M(φ) by use of (61) and (62). If C(pn+1ln+1)-1≤t<C(pnln)-1, then
(67)∫0t‍g*(x,δ,s)ds≤∑j=n∞‍∫C(pj+1lj+1)-1C(pjlj)-1‍g*(x,δ,s)ds≤∑j=n∞‍C(1pjlj-1pj+1lj+1)g*(x,δ,Cpj+1lj+1).
According to the definition,(68)g*(x,δ,Cpj+1lj+1)=inf{s>0:λ(g(x,δ,·),s)<Cpj+1lj+1}.
Therefore, from (61), (62), and (65), we obtain
(69)μ({s>0:g(x,δ,s)>lj})≤∑i>j‍1pi+1li+1≤1lj+1∑i>j‍1pi<Cpj+1lj+1,
and, consequently,
(70)g*(x,δ,Cpj+1lj+1)≤lj.
Therefore, forC(pn+1ln+1)-1≤t<C(pnln)-1, we will have
(71)∫0t‍g*(x,δ,s)ds≤∑j=n∞‍C(1pjlj-1pj+1lj+1)lj≤∑j=n∞‍C(1pj-ljpj+1lj+1)≤C∑j=n∞‍1pj≤C2pn.
Therefore,(72)∥g(x,δ,t)∣M(φ)∥≤C1supn{1pnφ(1/pn+1ln+1)}.
The final part of the proof is similar to that in the proof of the preceding corollary.


## 4. The Admissible Differential Bases

We now construct for each admissible differential basisU a certain function which will play one of the key roles in the resolution of the problems formulated at the beginning of this section. Suppose that we are given a differential basis U which has been constructed relative to an admissible sequence of triples (mi,pi,li). Suppose that we are also given a sequence of numbers {di}. We define the function f0(t) by means of the identity
(73)f0(t)=∑k=1∞‍dkχ(⋃j≥1‍Π(k,j,1)).It turns out that with a suitable choice of numbers{di} the basis U does not differentiate ∫‍f0.Lemma 9.
LetU be a differential basis constructed relative to (mi,pi,li). Suppose that the following relation holds for the sequence {di}:
(74)lim¯i→∞dipi-1=C>0.
Then, the differential basisU does not differentiate ∫f0.Proof.
Choose a positive integerk and put Ak=⋃j≥1‍Π(k,j,1). Then, in accordance with the construction and condition (15), we will have for any set U(k,i,j)(75)1μ(U(k,i,j))∫U(k,i,j)‍f0(t)χ(Ak)dt≥1μ(U(k,i,j))∫Π(k,i,1)‍dkχ(Π(k,i,1))dt≥dkμ(Π(k,i,1))μ(U(k,i,j))=dkpk-1.
Therefore, from (15) and (74), we have the identity
(76)lim¯k→∞μ({t:Mkf0(t)χ(Ak)>0.5C})≥limk→∞μ(⋃i,j≥1‍U(k,i,j))=1.
If we note thatμ(Ak)=(pklk)-1→0 as k→∞, we can deduce from Theorem 3.2.2 in [1] that the basis U does not differentiate ∫‍f0. The lemma is proved.Now, we will produce certain conditions on the function (73) which should guarantee its membership of certain spaces.Lemma 10.
Let(mi,pi,li) be an admissible collection of triples and put
(77)γj=∑i=1j‍di,βj=∑i=j∞‍(pili)-1,Aj=∑i=j∞‍(pili)-1γi.
Then, the following inequalities hold:(78)∥f0∣Λ(φ)∥≤C+∑i=1∞‍γiφ(βi),(79)∥f0∣M(φ)∥≤CsupjAjϕ(βj+1).Proof.
We have from the definition the inequality(80)λ(f0,γj)≤∑k=j+1∞‍μ(⋃i≥1‍Π(k,i,1))≤∑k=j+1∞‍1pklk=βj+1.
Therefore,(81)∥f0∣Λ(φ)∥=∫0∞‍φ(λ(f0,γ))dγ≤{∫0γ1‍+∑i=1∞‍∫γiγi+1‍}φ(λ(f0,γ))dγ≤γ1φ(λ(f0,0))+∑i=1∞‍(γi+1-γi)φ(λ(f0,γi))≤γ1φ(λ(f0,0))+∑i=1∞‍(γi+1-γi)φ(βi+1)≤γ1φ(λ(f0,0))+∑i=1∞‍γi+1φ(βi+1),
which also establishes (78). Let us verify (79). If βj+1≤t<βj, then
(82)∫0t‍f0*(t)dt≤∑i=j∞‍∫βi+1βi‍f0*(t)dt≤∑i=j∞‍(βi-βi+1)f0*(βi+1).
According to the definition off* and (80)
(83)f*(βi+1)=inf{s>0:λ(f0,s)<βi+1}≤γi.
Therefore, from the monotonicity of the functionφ(t), we obtain
(84)∥f0∣M(φ)∥=supt1φ(t)∫0t‍f0*(s)ds≤Csupj1φ(βj+1)∑i=j∞‍(βi-βi+1)γi=CsupjAjφ(βj+1).
The lemma is proved.Now, we have everything ready for the proof of the following theorem. We note that the fundamental functions of the spaces involved in Theorem11 are the same.Theorem 11.
Let1<p<∞. Then, there exists an admissible differential basis U which differentiates Lp but not M1-θ(p=1/θ, 1/p+1/q=1).Proof.
We will construct an admissible sequence of triples(mi,pi,li) in the following way. First, we put pi=2i and choose a monotone increasing sequence of natural numbers {li} such that for sufficiently large i we have the inequality
(85)1≤liq-1pi<2.
This is possible because of the inequality1<p<∞. Then, we choose a sequence {mi} such that the sequence of triples (mi,pi,li) is admissible. The possibility of such a choice of {mi} is also obvious. Now, we construct the differential basis U corresponding to (mi,pi,li). It follows from (85) and Corollary 6 that U differentiates Lp. Let us show that U does not differentiate the Marzinkiewicz space M1-θ. With the help of formula (73), we construct the function f0(t), putting di=pi. It then follows from Lemma 9 that the basis U does not differentiate ∫f0. We show that f0∈M1-θ. For this, we make use of Lemma 10. From the definition of pi and (85), we obtain
(86)γi=pj+1-2≃pj,Aj≃1lj,βj≃1pjlj.
Therefore,(87)∥f0∣M(φ)∥≤CsupjAjβj+11-θ≤C1supj(pjlj)1-θlj≤C2supj(pjlj1-q)1-θ<∞.
Thus, by Lemma10, f0∈M1-θ but U does not differentiate ∫‍f0, and, consequently, U does not differentiate the Marzinkiewicz space M1-θ. The theorem is proved.Now, we show that with the help of the bases which have been set up it is possible to distinguish Lorentz and Marzinkiewicz spaces with exactly the same quasiexponential fundamental functions.Theorem 12.
Letφ∈W and suppose that the following conditions are satisfied:
(88)supi≥1φ(2-(i+1))φ(2-i)=q1<1,supi≥1φ(2-i)φ(2-(i+1))=q2<2.
Then, there exists a differential basis which differentiatesΛ(φ) but does not differentiate M(t/φ(t)).Proof.
We define the sequences{ai}1∞, {bi}1∞ by means of the equations ai=1/φ(2-i), bi=2iφ(2-i) and put pi=[ai], li=[bi], di=ai ([a] denotes the integer part of a). Now, we choose a sequence {mi} such that the sequence of triples (mi,pi,li) is admissible and construct the corresponding basis U. Then, in accordance with (88), we have the following relations:
(89)∑i=n∞‍pi-1≃∑i=n∞‍ai-1=∑i=n∞‍φ(2-i)≃φ(2-n)1-q1≃pn-11-q1.
Sincepnln≃anbn≃2n, then
(90)supn1pnφ(1/pnln)≃supn1anφ(2-n)≤C.
We therefore obtain from this inequality and (89) by Corollary 8 that U differentiates Λ(φ). Let us verify that U does not differentiate M(t/φ(t)). For this, we construct the function f0 according to (73). Since di=ai and pi=[ai], then by Lemma 9  U does not differentiate ∫‍f0. We show that f0∈M(t/φ(t)), for which we make use of Lemma 10. Since pnln≃2n, we have by (88) the following relations:
(91)γj=∑i=1j‍di=∑i=1j‍ai=∑i=1j‍1φ(2-i)≤1φ(2-j)(1-q1),βj=∑i=j∞‍1pili≃∑i=j∞‍2-i≤C1pjlj,Aj=∑i=j∞‍γipili≤11-q1∑i=j∞‍aipili≤C21-q1∑i=j∞‍1lj≤C31-q1∑i=j∞‍1bi=C41-q1∑i=j∞‍2-iφ(2-i)≤C5(1-q1)(1-q2)lj.
We therefore obtain from Lemma10 and the fact that φ(t)/t does not increase that
(92)∥f0∣M(tφ(t))∥≤C6supjAjφ(βj+1)βj+1≤C7supj1ljφ(1pj+1lj+1)pj+1lj+1≤C8supjφ(1pjlj)pj≤C9supjφ(2-j)pj≤C10.
Thus,f0∈M(t/φ(t)), and it follows from Lemma 9 that U does not differentiate ∫‍f0. The theorem is proved.Now, we show that if there is a pair of symmetric spaces with nonequivalent fundamental functions, then there exists a differential basis which differentiates one but not the other. For this, we consider first the case of Lorentz and Marzinkiewicz spaces.Theorem 13.
Letϕ,ψ∈W, let Λ(ψ) be the Lorentz space and let M(t/φ(t)) be the Marzinkiewicz space. Suppose that
(93)limt→0ψ(t)φ(t)=0.
Then, there exists a differential basisU which differentiates M(t/φ(t)) but not Λ(ψ).Proof.
It follows from (93) that ψ(t)≤a(t)φ(t), and it can be assumed that a(t)↓ (as t→0+), a(0)=0 and a(t) is continuous. We choose a sequence {δi} such that the conditions of the following two inequalities are satisfied:
(94)ψ(δi+1)<0.25ψ(δi),a(δi)≤10-i,
and put
(95)ai=12iψ(δi),bi=2iψ(δi)δi,pi=[ai]+1,li=[bi]+1,di=ai.
Then from (94) and the concavity of ψ follow the inequalities
(96)ai+1=12i+1ψ(δi+1)≥212iψ(δi)=2ai,bi+1=2i+1ψ(δi+1)δi+1≥2×2iψ(δi)δi=2bi.
Let us construct the functionf0 according to (73). We will verify that f0∈Λ(ψ). For this, we make use of Lemma 10. According to (96), we will have
(97)γj=∑i=1j‍di=∑i=1j‍ai≃aj,βj=∑i=j∞‍1pili≃1pjlj≃1ajbj≃δj.
Therefore,(98)∥f0∣Λ(ψ)∥≤C+∑j=1∞‍γjψ(βj)≤C+∑j=1∞‍γjψ(δj)≤C+∑j=1∞‍ψ(δj)2jψ(δj)<∞.
Thus,f0∈Λ(ψ). Now, we choose a sequence {mi} in such a way that the sequence of triples (mi,pi,li) is admissible. It follows from the definition of pi and di in (95) that U does not differentiate ∫‍f0. Therefore, according to (98), U does not differentiate the space Λ(ψ).
It remains to be shown thatU differentiates the Marzinkiewicz space M(t/φ(t)). According to the definitions (95) and (96),
(99)∑i=k∞‍1pili≃1pklk≃δk.
On the other hand we obtain from the inequalityψ(t)≤a(t)φ(t), (94), and (95), that
(100)∑i=1∞‍li1pili1φ(1/pili)≤C∑i=1∞‍1pia(1/pili)ψ(1/pili)≤C1∑i=1∞‍1pia(δi)ψ(δi)≤C2∑i=1∞‍2iψ(δi)a(δi)ψ(δi)≤C3∑i=1∞‍2ia(δi)<∞.
Thus, condition (59) for Corollary 7 is satisfied, and so U differentiates M(t/φ(t)). The theorem is proved.This theorem has an important corollary.Corollary 14.
LetX, Y be a pair of symmetric spaces and φ(t), ψ(t) their fundamental functions. If the condition (93) is satisfied, then there exists a differential basis U which differentiates the space X but not the space Y.Proof.
We apply the embedding theorem (12) for the spaces X and Y. Then,
(101)Λ(φ)⊆X⊆M(tφ(t)),Λ(ψ)⊆Y⊆M(tψ(t)).
Now, we apply Theorem13 to the spaces Λ(ψ) and M(t/φ(t)). As a result of this, we find a basis U which differentiates M(t/φ(t)), and consequently X, but does not differentiate Λ(ψ), and consequently U does not differentiate Y.



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*Source: 748645-2013-08-05.xml*