# Some Reiteration Theorems forR, L, RR, RL, LR, and LL Limiting Interpolation Spaces

Authors: Leo R. Ya. Doktorski
Journal: Journal of Function Spaces
 (2021)

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## Abstract

We consider theK-interpolation methods involving slowly varying functions. We establish some reiteration formulae including so-called L or R limiting interpolation spaces as well as the RR, RL, LR, and LL extremal interpolation spaces. These spaces arise in the limiting situations. The proofs of most reiteration formulae are based on Holmstedt-type formulae. Applications to grand and small Lorentz spaces in critical cases are given.


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## Body

## 1. Introduction

LetA¯≔A0,A1 be a compatible couple of Banach or quasi-Banach spaces such that A0⋂A1≠0. For t>0, Peetre’s K-functional is given by(1)Kt,f;A¯≡Kt,f≔inff0A0+tf1A1:f=f0+f1,fi∈Ai.We refer to, e.g., [1] for basic concepts on interpolation theory and properties of the K-functional. In recent years, the scale of interpolation spaces A¯θ,q,a has been intensively studied, which is defined via the (quasi-)norms(2)fθ,q,a≔u−θ−1/qauKu,fq,0,∞,where 0≤θ≤1, 0<q≤∞, a is a slowly varying function (see Definition 5 below), and ∗q,0,∞ is the usual (quasi-)norm in the Lebesgue space Lq on the interval 0,∞.An important property of this scale is that the reiteration spacesA¯θ0,∗,A¯θ1,∗θ,r,a with 0≤θ0<θ1≤1 and 0<θ<1 belong to the same scale. (Here and below, “∗” stands for further parameters.) The consideration of the limiting cases θ∈0,1 leads normally to some new limiting interpolation spaces A¯θ,∗L and A¯θ,∗R. Following [2], we call them R and L spaces (see Definition 7 below). These limiting interpolation spaces occur naturally in many fields of analysis. For example, it is shown that the so-called grand and small Lorentz spaces can be described in terms of the R and L spaces [3, 4].Naturally, the question arises about the description of the reiteration spaces∗,∗θ,r,a for couples where one or both of the operands are R or L spaces. Such reiteration formulae under condition 0<θ<1 are established in the previous paper [5]. Some limiting cases θ∈0,1, but not all, are also considered in [5]. For example, if 0<θ0<θ1<1, it has been shown in [5] that A¯θ0,∗,A¯θ1,∗Lθ,∗=A¯η,∗ if 0<θ<1 and A¯θ0,∗,A¯θ1,∗L0,∗=A¯θ0,∗L. However, the formula for A¯θ0,∗,A¯θ1,∗L1,∗ has not been derived. The principal aim of this paper is to establish reiteration formulae for combinations of parameters, which are not included in [5].The motivation for this work was the articles [6–8], where it has been shown that for some limiting combinations of parameters, new interpolation spaces are required. Following [6, 7], we call them LL, LR, RL, and RR extremal interpolation spaces (see Definition 8 below). For example, A¯θ0,∗,A¯θ1,∗L1,∗=A¯θ1,∗L,R (see Theorem 19 below).Our paper [5] and the articles [6–8] appeared practically at the same time. The results of the papers [6–8] demonstrate—mathematically speaking—a large intersection and a symmetric difference with the ones in [5] and in this paper. Fernández-Martínez and Signes work with the scale A0,A1θ,E,b of interpolation spaces based on the family of rearrangement invariant Banach function spaces E. They have introduced and studied this rich scale in [2, 9]. If E=Lq and 1≤q≤∞, the space A0,A1θ,E,b coincides with the space A0,A1θ,q;b. However, the spaces A0,A1θ,q;b with 0<q<1 do not belong to the scale A0,A1θ,E,b. The same remark holds about R and L limiting interpolation spaces, too.Note that in [5] and in this paper, the main parameters θ0 and θ1 for considered R und L spaces lie between 0 and 1. Limiting cases θ0,θ1∈0,1 will be the topic of a future paper.The paper is organized as follows. Sections2 and 3 contain necessary notations, definitions, and technical results. In Sections 4–6, we establish reiteration formulae for the couples, where one of the operands is either L or R space. Reiteration formulae for the couples, where both operands are the L or R space, are established in Section 7. In Section 8, we consider couples where one of the operands is LL, LR, RL, or RR space. Note that latter combinations are not studied in [6–8]. Finally, in Section 9, we present interpolation results for the grand and small Lorentz spaces in critical cases as applications of our general reiteration theorems.


## 2. Preliminaries

Throughout the paper, we writeX⊂Y for two (quasi-)normed spaces X and Y to indicate that X is continuously embedded in Y. We write X=Y if X⊂Y and Y⊂X. For f and g being positive functions, we write f≺g if f≤Cg, where the constant C is independent of all significant quantities. Two functions f and g are considered equivalent (f≈g) if f≺g and g≺f. We adopt the conventions 1/∞=0 and 1/0=∞. The abbreviation LHS(∗) (RHS(∗)) will be used for the left- (right-) hand side of the relation (∗). By χa,b, we denote the characteristic function on an interval a,b.


### 2.1. Slowly Varying Functions

In this section, we summarize some of the properties of slowly varying functions, which will be required later. For more details, we refer to, e.g., [2, 9–11].Definition 1.
We say that a positive Lebesgue measurable functionb is slowly varying on 0,∞, notation b∈SV, if, for each ε>0, the function tεbt is equivalent to an increasing function while the function t−εbt is equivalent to a decreasing function.Lemma 2.
Letb,b1,b2∈SV, α>0, 0<q≤∞, λ∈−∞,∞, and t∈0,∞.(i)
Then,bλ∈SV, b1/t∈SV, btαb1t∈SV, and b1b2∈SV(ii)
Iff≈g, then b∘f≈b∘g(iii)
uα−1/qbuq,0,t≈tαbt and u−α−1/qbuq,t,∞≈t−αbt(iv)
The functionsu−1/qbuq,0,t and u−1/qbuq,t,∞ (if exist) belong to SV and bt≺u−1/qbuq,0,t and bt≺u−1/qbuq,t,∞(v)
uλ−1/qbuq,t,2t≈tλbt
We will often use these properties without referencing explicitly every time.


### 2.2. Hardy-Type Inequalities

The following Hardy-type inequalities will be applied later.Lemma 3 (see [10], Lemma 2.7).
Let1≤P≤∞ and b∈SV.(i)
The inequality(3)tν−1/Pbt∫0tguduP,0,∞≺tν+1−1/PbtgtP,0,∞,holds for all (Lebesgue) measurable nonnegative functions g on 0,∞ if and only if ν<0.(ii)
The inequality(4)tν−1/Pbt∫t∞guduP,0,∞≺tν+1−1/PbtgtP,0,∞,holds for all (Lebesgue) measurable nonnegative functions g on 0,∞ if and only if ν>0.These inequalities will be used in the form of the following corollary.Corollary 4.
Let0<q≤r≤∞, b∈SV, and μ>0. Then, for all (Lebesgue) measurable nonnegative functions h on 0,∞,(5)t−μ−1/rbtu−1/qhuq,0,tr,0,∞≺t−μ−1/rbthtr,0,∞,tμ−1/rbtu−1/qhuq,t,∞r,0,∞≺tμ−1/rbthtr,0,∞.Proof.
We begin with the first estimate. If0<q<r<∞, using Lemma 3 with gu=u−1/qhuq, ν=−qμ, and P=r/q, we get(6)t−μ−1/rbtu−1/qhuq,0,tr,0,∞=∫0∞t−qμ−1/Pbtq∫0tguduPdt1/Pq≺∫0∞t−qμ+1−1/PbtqgtPdt1/Pq=t−μ−1/rbthtr,0,∞.
The case0<q<r=∞ can be considered similarly with P=∞. The case 0<q=r≤∞ can be proved by Fubini’s theorem or by the exchange of the essential suprema (if p=∞).
The second estimate follows from the first one if we takeu=1/v and t=1/s.☐


## 2.1. Slowly Varying Functions

In this section, we summarize some of the properties of slowly varying functions, which will be required later. For more details, we refer to, e.g., [2, 9–11].Definition 1.
We say that a positive Lebesgue measurable functionb is slowly varying on 0,∞, notation b∈SV, if, for each ε>0, the function tεbt is equivalent to an increasing function while the function t−εbt is equivalent to a decreasing function.Lemma 2.
Letb,b1,b2∈SV, α>0, 0<q≤∞, λ∈−∞,∞, and t∈0,∞.(i)
Then,bλ∈SV, b1/t∈SV, btαb1t∈SV, and b1b2∈SV(ii)
Iff≈g, then b∘f≈b∘g(iii)
uα−1/qbuq,0,t≈tαbt and u−α−1/qbuq,t,∞≈t−αbt(iv)
The functionsu−1/qbuq,0,t and u−1/qbuq,t,∞ (if exist) belong to SV and bt≺u−1/qbuq,0,t and bt≺u−1/qbuq,t,∞(v)
uλ−1/qbuq,t,2t≈tλbt
We will often use these properties without referencing explicitly every time.


## 2.2. Hardy-Type Inequalities

The following Hardy-type inequalities will be applied later.Lemma 3 (see [10], Lemma 2.7).
Let1≤P≤∞ and b∈SV.(i)
The inequality(3)tν−1/Pbt∫0tguduP,0,∞≺tν+1−1/PbtgtP,0,∞,holds for all (Lebesgue) measurable nonnegative functions g on 0,∞ if and only if ν<0.(ii)
The inequality(4)tν−1/Pbt∫t∞guduP,0,∞≺tν+1−1/PbtgtP,0,∞,holds for all (Lebesgue) measurable nonnegative functions g on 0,∞ if and only if ν>0.These inequalities will be used in the form of the following corollary.Corollary 4.
Let0<q≤r≤∞, b∈SV, and μ>0. Then, for all (Lebesgue) measurable nonnegative functions h on 0,∞,(5)t−μ−1/rbtu−1/qhuq,0,tr,0,∞≺t−μ−1/rbthtr,0,∞,tμ−1/rbtu−1/qhuq,t,∞r,0,∞≺tμ−1/rbthtr,0,∞.Proof.
We begin with the first estimate. If0<q<r<∞, using Lemma 3 with gu=u−1/qhuq, ν=−qμ, and P=r/q, we get(6)t−μ−1/rbtu−1/qhuq,0,tr,0,∞=∫0∞t−qμ−1/Pbtq∫0tguduPdt1/Pq≺∫0∞t−qμ+1−1/PbtqgtPdt1/Pq=t−μ−1/rbthtr,0,∞.
The case0<q<r=∞ can be considered similarly with P=∞. The case 0<q=r≤∞ can be proved by Fubini’s theorem or by the exchange of the essential suprema (if p=∞).
The second estimate follows from the first one if we takeu=1/v and t=1/s.☐


## 3. Interpolation Methods

Below is a collection of crucial definitions and statements building a base of the real interpolation methods involving slowly varying functions. In the following, letA¯≡A0,A1 be a compatible couple of (quasi-)Banach spaces such that A0⋂A1≠0. Peetre’s K-functional on A0+A1 is given by (1).


### 3.1. Standard Interpolation Spaces

Definition 5 (see [10]).
Let0≤θ≤1, 0<q≤∞, and b∈SV. We put(7)A¯θ,q;b≡A0,A1θ,q;b≔f∈A0+A1:fθ,q;b=u−θ−1/qbuKu,fq,0,∞<∞.Lemma 6 (see [10], Proposition 2.5).
A0⋂A1⊂A¯θ,q;b⊂A0+A1 if and only if one of the following conditions is satisfied:(i)
0<θ<1(ii)
θ=0 and u−1/qbuq,1,∞<∞(iii)
θ=1 and u−1/qbuq,0,1<∞
Moreover, if none of these conditions holds, thenA¯θ,q;b=0.
SpacesA¯θ,q;b are referred to as standard interpolation spaces.


### 3.2.L and R Spaces

Definition 7 (see [10]).
Let0<r,q≤∞, 0≤θ≤1, and a,b∈SV. We put(8)A¯θ,r,b,q,aL≡A0,A1θ,r,b,q,aL≔f∈A0+A1:fL;θ,r,b,q,a≔t−1/rbtu−θ−1/qauKu,fq,0,tr,0,∞<∞.
Similarly,(9)A¯θ,r,b,q,aR≡A0,A1θ,r,b,q,aR≔f∈A0+A1:fR;θ,r,b,q,a≔t−1/rbtu−θ−1/qauKu,fq,t,∞r,0,∞<∞.
In the literature (e.g., [2, 5, 9, 12–14]), similar definitions are given alongside with properties of these spaces. As proposed in [2], we refer to the spaces A¯∗L and A¯∗R as L and R limiting interpolation spaces.


### 3.3.LL, LR, RL, and RR Spaces

We introduce here four interpolation spaces. We follow [6–8] where it has been shown that they appear in relation to the extreme reiteration results.Definition 8.
Let0<p,q,r≤∞, 0≤θ≤1, and a,b,c∈SV. The space A¯θ,p,c,r,b,q,aL,L≡A0,A1θ,p,c,r,b,q,aL,L is the set of all f∈A0+A1 for which(10)fL,L;θ,p,c,r,b,q,a≔t−1/pcts−1/rbsu−θ−1/qauKu,fq,0,sr,0,tp,0,∞<∞.
The spaceA¯θ,p,c,r,b,q,aL,R≡A0,A1θ,p,c,r,b,q,aL,R is the set of all f∈A0+A1 for which(11)fL,R;θ,p,c,r,b,q,a≔t−1/pcts−1/rbsu−θ−1/qauKu,fq,t,sr,t,∞p,0,∞<∞.
The spaceA¯θ,p,c,r,b,q,aR,L≡A0,A1θ,p,c,r,b,q,aR,L is the set of all f∈A0+A1 for which(12)fR,L;θ,p,c,r,b,q,a≔t−1/pcts−1/rbsu−θ−1/qauKu,fq,s,tr,0,tp,0,∞<∞.
The spaceA¯θ,p,c,r,b,q,aR,R≡A0,A1θ,p,c,r,b,q,aR,R is the set of all f∈A0+A1 for which(13)fR,R;θ,p,c,r,b,q,a≔t−1/pcts−1/rbsu−θ−1/qauKu,fq,s,∞r,t,∞p,0,∞<∞.
We refer to these spaces asLL, LR, RL, and RR extremal interpolation spaces.


### 3.4. Some Known Formulae, Auxiliary Lemmas, and Reiteration Theorems

LetC≥1, t>0, and b~t≔bt−1. Under suitable conditions, the following formulae hold ([1], Chap. 5, Proposition 1.2) [6–8, 10]:(14)Kt,f≤KCt,f≤CKt,f,Kt,f;A0,A1=tKt−1,f;A1,A0,A0,A1θ,q,b=A1,A01−θ,q,b~,(15)A0,A1σ,r,b,q,aL=A1,A01−σ,r,b~,q,a~R,(16)A0,A1θ,p,c,r,b,q,aL,L=A1,A01−θ,p,c~,r,b~,q,a~R,R,A0,A1θ,p,c,r,b,q,aL,R=A1,A01−θ,p,c~,r,b~,q,a~R,L.Lemma 9 (cf. [5]; [6], Remark 2.6).
Leta∈SV and 0<r<q≤∞. Then, for all f∈A0+A1 and t>0,(17)u−θ−1/qauKu,fq,0,t≺u−θ−1/rauKu,fr,0,t,θ<1,(18)u−θ−1/qauKu,fq,t,∞≺u−θ−1/rauKu,fr,t,∞,0<θ.
Moreover,(19)t−1u−1/raur,0,tKt,f≤u−1−1/rauKu,fr,0,t,(20)u−1/raur,t,∞Kt,f≤u−1/rauKu,fr,t,∞.Proof.
The estimates (19) and (20) follow immediately from the monotonicity of the K-functional. We prove the estimate (18). The estimate (17) can be proved similarly. For u>t, because θ>0 and Ku,f is nondecreasing, we have(21)u−θauKu,f≈Ku,fs−θ−1/rasr,u,∞≤s−θ−1/rasKs,fr,u,∞≤s−θ−1/rasKs,fr,t,∞.
Thus, (18) is proved for q=∞:(22)supt<u<∞u−θauKu,f≺u−θ−1/rauKu,fr,t,∞.
Ifr<q<∞, using the last estimate, we get(23)u−θ−1/qauKu,fq,t,∞≤supt<u<∞u−θauKu,fq−r/q∫t∞u−θauKu,frduu1/q≺u−θ−1/rauKu,fr,t,∞.
This completes the proof.☐The next lemma can be proved by repeating the proofs of [6] (Lemma 2.12) and [7] (Lemma 2.10).Lemma 10.
Let0≤θ≤1, 0<q,r≤∞, and a,b∈SV. Then, for all f∈A0+A1 and t>0,(24)t−θats−1/rbsr,0,tKt,f≺s−1/rbsu−θ−1/qauKu,fq,s,tr,0,t,(25)t−θats−1/rbsr,t,∞Kt,f≺s−1/rbsu−θ−1/qauKu,fq,t,sr,t,∞,(26)t−θatbtKt,f≺s−1/rbsu−θ−1/qauKu,fq,s,∞r,t,∞,(27)t−θatbtKt,f≺s−1/rbsu−θ−1/qauKu,fq,0,sr,0,t.Lemma 11.
Let0<θ<1, a,b∈SV, 0<p≤∞, and 0<r<q≤∞. Then, for all f∈A0+A1 and t>0,(28)s−1/qbsu−θ−1/pauKu,fp,0,sq,0,t≺s−1/rbsu−θ−1/pauKu,fp,0,sr,0,t,s−1/qbsu−θ−1/pauKu,fp,s,∞q,t,∞≺s−1/rbsu−θ−1/pauKu,fp,s,∞r,t∞.Proof.
We prove the first estimate. The second one can be proved analogously. Because0<θ<1, the function s1−θas is equivalent to an increasing function. It is known ([10], Theorem 3.1) that for all f∈A0+A1 and t>0,(29)Kt1−θat,f;A¯θ,p,a,A1≈u−θ−1/pauKu,fp,0,t.
Therefore, the function1/t1−θatu−θ−1/pauKu,fp,0,t is equivalent to a decreasing function. Hence, for v<t, we get(30)bvu−θ−1/pauKu,fp,0,v≈s1−θ−1/rasbsr,0,v1v1−θavu−θ−1/pauKu,fp,0,v≺s−1/rbsu−θ−1/pauKu,fp,0,sr,0,v≤s−1/rbsu−θ−1/pauKu,fp,0,sr,0,t.
Thus, (28) is proved for q=∞:(31)sup0<s<tbsu−θ−1/pauKu,fp,0,s≺s−1/rbsu−θ−1/pauKu,fp,0,sr,0,t.
Denoteds=bsu−θ−1/pauKu,fp,0,s. If r<q<∞, using the last estimate, we get(32)s−1/qbsu−θ−1/pauKu,fp,0,sq,0,t=∫0tdsqduu1/q≤sup0<u<tdsq−r/q∫0tdsrduu1/q≺s−1/rbsu−θ−1/pauKu,fp,0,sr,0,t.
This completes the proof.☐It is worth remarking that the estimates from Lemmas9–11 can be used to establish different embedding theorems.Lemma 12 (cf. [8], Lemma 4.2).
Let0<q,r≤∞, a,b∈SV, and β>0. Then, for all f∈A0+A1,(33)t−β−1/rbtuα−1/qauKu,fq,0,tr,0,∞≈tα−β−1/ratbtKt,fr,0,∞,provided that α>−1, and(34)tβ−1/rbtuα−1/qauKu,fq,t,∞r,0,∞≈tα+β−1/ratbtKt,fr,0,∞,provided that α<0.Proof.
We prove the first equivalence. The second one can be proved analogously. We have(35)uα−1/qauKu,fq,0,t≥uα−1/qauKu,fq,t/2,t≥uα−1/qauq,t/2,tKt2,f≈tαatKt,f.
Hence,(36)t−β−1/rbtuα−1/qauKu,fq,0,tr,0,∞≻tα−β−1/ratbtKt,fr,0,∞.
Let us prove the converse inequality.
Caseq≤r. Because β>0, by Corollary 4 (taking hu=uαauKu,f), we get(37)t−β−1/rbtuα−1/qauKu,fq,0,tr,0,∞≺tα−β−1/ratbtKt,fr,0,∞.
In particular,(38)t−β−1/rbtuα−1/rauKu,fr,0,tr,0,∞≺tα−β−1/ratbtKt,fr,0,∞.
Caseq>r. Because α>−1, using Lemma 9 and the last estimate, we get(39)t−β−1/rbtuα−1/qauKu,fq,0,tr,0,∞≺t−β−1/rbtuα−1/rauKu,fr,0,tr,0,∞≺tα−β−1/ratbtKt,fr,0,∞.
This completes the proof.☐Lemma 13.
Let0≤θ0<θ1≤1, 0<p,q,r≤∞, and a,b,c∈SV. Then, for all f∈A0+A1,(40)tθ0−θ1−1/pcts−1/rbsu−θ0−1/qauKu,fq,0,sr,0,tp,0,∞≈t−θ1−1/patbtctKt,fp,0,∞,tθ1−θ0−1/pcts−1/rbsu−θ1−1/qauKu,fq,s,∞r,t,∞p,0,∞≈t−θ0−1/patbtctKu,fp,0,∞.Proof.
We prove the first equivalence. The second one can be proved analogously or using the first equivalence and symmetry argument. DenoteIr≔LHS40.
Caser≤p. Because θ0−θ1<0 and θ0<1, by Corollary 4 (taking hs=bsu−θ0−1/qauKu,fq,0,s) and by Lemma 12, we get(41)Ir≺tθ0−θ1−1/rcthtp,0,∞=tθ0−θ1−1/pctbtu−θ0−1/qauKu,fq,0,tp,0,∞≈t−θ1−1/patctbtKt,fp,0,∞.
In particular, it was shown that(42)Ip≺t−θ1−1/patctbtKt,fp,0,∞.
Caser>p. By Lemma 11 and (42), we get(43)Ir≺tθ0−θ1−1/pcts−1/pbsu−θ0−1/qauKu,fq,0,sp,0,tp,0,∞=Ip≺t−θ1−1/patctbtKt,fp,0,∞.
Let us prove the inverse estimate. By (27), we have(44)t−θ0atbtKt,f≺s−1/rbsu−θ0−1/qauKu,fq,0,sr,0,t.
Hence,(45)RHS40≺tθ0−θ1−1/pcts−1/rbsu−θ0−1/qauKu,fq,0,sr,0,tp,0,∞.
This completes the proof.☐Lemma 14.
Let0≤θ0<θ1≤1, 0<p,q,r≤∞, and a,b∈SV.(i)
Ifs−1/rbsr,0,1<∞, then for all f∈A0+A1,(46)tθ0−θ1−1/pcts−1/rbsu−θ0−1/qauKu,fq,s,tr,0,tp,0,∞≈t−θ1−1/pats−1/rbsr,0,tctKt,fp,0,∞.(ii)
Ifs−1/rbsr,1,∞<∞, then for all f∈A0+A1,(47)tθ1−θ0−1/pcts−1/rbsu−θ1−1/qauKu,fq,t,sr,t,∞p,0,∞≈t−θ0−1/pats−1/rbsr,t,∞ctKt,fp,0,∞.Proof.
We prove the first equivalence. The second one can be proved analogously or using the symmetry argument. For0≤θ0≤1, by (24), we have(48)t−θ0ats−1/rbsr,0,tKt,f≺s−1/rbsu−θ0−1/qauKu,fq,s,tr,0,t.
Hence, for eachθ1,(49)RHS46≺tθ0−θ1−1/pcss−1/rbsu−θ0−1/qauKu,fq,s,tr,0,tp,0,∞=LHS46.
Let us prove the inverse estimate. Becauseθ0−θ1<0 and θ0<1 by Lemma 12, we get(50)LHS46≤tθ0−θ1−1/pcts−1/rbsr,0,tu−θ0−1/qauKu,fq,0,tp,0,∞≈t−θ1−1/pcts−1/rbsr,0,tatKt,fp,0,∞=RHS46.
This completes the proof.☐The following lemma deals with the quasi-norms on the spacesA¯0,∗;∗ and A¯1,∗;∗.Lemma 15.
Let0<r,p,q≤∞ and a,b,c∈SV.(i)
Ifs−1/rcsr,1,∞<∞ and s−1/pbsp,1,∞<∞, then for all f∈A¯0,r;c,(51)f0,r;c≈tθ−1/rctats−1/pbsp,t,∞s−1/pbsu−θ−1/qauKu,fq,t,sp,t,∞r,0,∞,provided that 0<θ≤1. Furthermore,(52)f0,r;c≺t−1/rctats−1/pbsp,t,∞s−1/pbsu−1/qauKu,fq,t,sp,t,∞r,0,∞.(ii)
Ifs−1/rcsr,0,1<∞ and s−1/pbsp,0,1<∞, then for all f∈A¯1,r;c,(53)f1,r;c≈tθ−1−1/rctats−1/pbsp,0,ts−1/pbsu−θ−1/qauKu,fq,s,tp,0,tr,0,∞,provided that 0≤θ<1. Furthermore,(54)f1,r;c≺t−1/rctats−1/r0bsr0,0,ts−1/pbsu−1−1/qauKu,fq,s,tp,0,tr,0,∞.(iii)
Ifs−1/rcsr,1,∞<∞ and 0<θ≤1, then for all f∈A¯0,r;c,(55)f0,r,c≈tθ−1/rctatu−θ−1/qauKu,fq,t,∞r,0,∞.(iv)
Ifs−1/rcsr,0,1<∞ and 0≤θ<1, then for all f∈A¯1,r;c,(56)f1,r,c≈tθ−1−1/rctatu−θ−1/qauKu,fq,0,tr,0,∞.Proof.
We prove the first assertion. The second one can be proved analogously or using the symmetry argument. By Lemma10, for 0≤θ≤1, we have(57)f0,r;c=tθ−1/rctats−1/pbsp,t,∞t−θats−1/pbsp,t,∞Kt,fr,0,∞≺tθ−1/rctats−1/pbsp,t,∞u−1/pbus−θ−1/qasKs,fq,t,up,t,∞r,0,∞.
Thus,LHS51≺RHS51 and the estimate (52) is proved. Let 0<θ≤1. Note that(58)s−1/r1bsu−θ−1/qauKu,fq,t,sr1,t,∞≤s−1/r1bsr1,t,∞u−θ−1/qauKu,fq,t,∞.
Hence, by Lemma12, we get(59)RHS51≤tθ−1/rctatu−θ−1/qauKu,fq,t,∞r,0,∞≈t−1/rctKt,fr,0,∞=LHS51.
The equivalences (iii) and (iv) follow immediately from Lemma12.☐In Section8, we will use the following reiteration theorems. See [9, 14].Theorem 16.
Let0<r,r0≤∞, a,b0∈SV, and s−1/r0b0sr0,1,∞<∞. Put ρt=ts−1/r0b0sr0,t,∞.(i)
If0<θ<1, then(60)A¯0,r0,b0,A1θ,r,a=A¯θ,r,a#,where a#t=s−1/r0b0sr0,t,∞1−θaρt.(ii)
Ift−1/ratr,1,∞<∞, then(61)A¯0,r0,b0,A10,r,a=A¯0,r,a#∩A¯0,r,a∘ρ,r0,b0L,where a#t=s−1/r0b0sr0,t,∞aρt.(iii)
Ift−1/ratr,0,1<∞, then(62)A¯0,r0,b0,A11,r,a=A¯1,r,a∘ρ.Theorem 17.
Let0<r,r1≤∞, a,b1∈SV, and s−1/r1b1sr1,0,1<∞. Put ρt=t/t−1/r1b1tr1,0,t.(i)
If0<θ<1, then(63)A0,A¯1,r1,b1θ,r,a=A¯θ,r,a#,where a#t=s−1/r1b1sr1,0,tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(64)A0,A¯1,r1,b10,r,a=A¯0,r,a∘ρ.(iii)
Ift−1/ratr,0,1<∞, then(65)A0,A¯1,r1,b11,r,a=A¯1,r,a#∩A¯1,r,a∘ρ,r1,b1R,where a#t=s−1/r1b1sr1,0,taρt.


## 3.1. Standard Interpolation Spaces

Definition 5 (see [10]).
Let0≤θ≤1, 0<q≤∞, and b∈SV. We put(7)A¯θ,q;b≡A0,A1θ,q;b≔f∈A0+A1:fθ,q;b=u−θ−1/qbuKu,fq,0,∞<∞.Lemma 6 (see [10], Proposition 2.5).
A0⋂A1⊂A¯θ,q;b⊂A0+A1 if and only if one of the following conditions is satisfied:(i)
0<θ<1(ii)
θ=0 and u−1/qbuq,1,∞<∞(iii)
θ=1 and u−1/qbuq,0,1<∞
Moreover, if none of these conditions holds, thenA¯θ,q;b=0.
SpacesA¯θ,q;b are referred to as standard interpolation spaces.


## 3.2.L and R Spaces

Definition 7 (see [10]).
Let0<r,q≤∞, 0≤θ≤1, and a,b∈SV. We put(8)A¯θ,r,b,q,aL≡A0,A1θ,r,b,q,aL≔f∈A0+A1:fL;θ,r,b,q,a≔t−1/rbtu−θ−1/qauKu,fq,0,tr,0,∞<∞.
Similarly,(9)A¯θ,r,b,q,aR≡A0,A1θ,r,b,q,aR≔f∈A0+A1:fR;θ,r,b,q,a≔t−1/rbtu−θ−1/qauKu,fq,t,∞r,0,∞<∞.
In the literature (e.g., [2, 5, 9, 12–14]), similar definitions are given alongside with properties of these spaces. As proposed in [2], we refer to the spaces A¯∗L and A¯∗R as L and R limiting interpolation spaces.


## 3.3.LL, LR, RL, and RR Spaces

We introduce here four interpolation spaces. We follow [6–8] where it has been shown that they appear in relation to the extreme reiteration results.Definition 8.
Let0<p,q,r≤∞, 0≤θ≤1, and a,b,c∈SV. The space A¯θ,p,c,r,b,q,aL,L≡A0,A1θ,p,c,r,b,q,aL,L is the set of all f∈A0+A1 for which(10)fL,L;θ,p,c,r,b,q,a≔t−1/pcts−1/rbsu−θ−1/qauKu,fq,0,sr,0,tp,0,∞<∞.
The spaceA¯θ,p,c,r,b,q,aL,R≡A0,A1θ,p,c,r,b,q,aL,R is the set of all f∈A0+A1 for which(11)fL,R;θ,p,c,r,b,q,a≔t−1/pcts−1/rbsu−θ−1/qauKu,fq,t,sr,t,∞p,0,∞<∞.
The spaceA¯θ,p,c,r,b,q,aR,L≡A0,A1θ,p,c,r,b,q,aR,L is the set of all f∈A0+A1 for which(12)fR,L;θ,p,c,r,b,q,a≔t−1/pcts−1/rbsu−θ−1/qauKu,fq,s,tr,0,tp,0,∞<∞.
The spaceA¯θ,p,c,r,b,q,aR,R≡A0,A1θ,p,c,r,b,q,aR,R is the set of all f∈A0+A1 for which(13)fR,R;θ,p,c,r,b,q,a≔t−1/pcts−1/rbsu−θ−1/qauKu,fq,s,∞r,t,∞p,0,∞<∞.
We refer to these spaces asLL, LR, RL, and RR extremal interpolation spaces.


## 3.4. Some Known Formulae, Auxiliary Lemmas, and Reiteration Theorems

LetC≥1, t>0, and b~t≔bt−1. Under suitable conditions, the following formulae hold ([1], Chap. 5, Proposition 1.2) [6–8, 10]:(14)Kt,f≤KCt,f≤CKt,f,Kt,f;A0,A1=tKt−1,f;A1,A0,A0,A1θ,q,b=A1,A01−θ,q,b~,(15)A0,A1σ,r,b,q,aL=A1,A01−σ,r,b~,q,a~R,(16)A0,A1θ,p,c,r,b,q,aL,L=A1,A01−θ,p,c~,r,b~,q,a~R,R,A0,A1θ,p,c,r,b,q,aL,R=A1,A01−θ,p,c~,r,b~,q,a~R,L.Lemma 9 (cf. [5]; [6], Remark 2.6).
Leta∈SV and 0<r<q≤∞. Then, for all f∈A0+A1 and t>0,(17)u−θ−1/qauKu,fq,0,t≺u−θ−1/rauKu,fr,0,t,θ<1,(18)u−θ−1/qauKu,fq,t,∞≺u−θ−1/rauKu,fr,t,∞,0<θ.
Moreover,(19)t−1u−1/raur,0,tKt,f≤u−1−1/rauKu,fr,0,t,(20)u−1/raur,t,∞Kt,f≤u−1/rauKu,fr,t,∞.Proof.
The estimates (19) and (20) follow immediately from the monotonicity of the K-functional. We prove the estimate (18). The estimate (17) can be proved similarly. For u>t, because θ>0 and Ku,f is nondecreasing, we have(21)u−θauKu,f≈Ku,fs−θ−1/rasr,u,∞≤s−θ−1/rasKs,fr,u,∞≤s−θ−1/rasKs,fr,t,∞.
Thus, (18) is proved for q=∞:(22)supt<u<∞u−θauKu,f≺u−θ−1/rauKu,fr,t,∞.
Ifr<q<∞, using the last estimate, we get(23)u−θ−1/qauKu,fq,t,∞≤supt<u<∞u−θauKu,fq−r/q∫t∞u−θauKu,frduu1/q≺u−θ−1/rauKu,fr,t,∞.
This completes the proof.☐The next lemma can be proved by repeating the proofs of [6] (Lemma 2.12) and [7] (Lemma 2.10).Lemma 10.
Let0≤θ≤1, 0<q,r≤∞, and a,b∈SV. Then, for all f∈A0+A1 and t>0,(24)t−θats−1/rbsr,0,tKt,f≺s−1/rbsu−θ−1/qauKu,fq,s,tr,0,t,(25)t−θats−1/rbsr,t,∞Kt,f≺s−1/rbsu−θ−1/qauKu,fq,t,sr,t,∞,(26)t−θatbtKt,f≺s−1/rbsu−θ−1/qauKu,fq,s,∞r,t,∞,(27)t−θatbtKt,f≺s−1/rbsu−θ−1/qauKu,fq,0,sr,0,t.Lemma 11.
Let0<θ<1, a,b∈SV, 0<p≤∞, and 0<r<q≤∞. Then, for all f∈A0+A1 and t>0,(28)s−1/qbsu−θ−1/pauKu,fp,0,sq,0,t≺s−1/rbsu−θ−1/pauKu,fp,0,sr,0,t,s−1/qbsu−θ−1/pauKu,fp,s,∞q,t,∞≺s−1/rbsu−θ−1/pauKu,fp,s,∞r,t∞.Proof.
We prove the first estimate. The second one can be proved analogously. Because0<θ<1, the function s1−θas is equivalent to an increasing function. It is known ([10], Theorem 3.1) that for all f∈A0+A1 and t>0,(29)Kt1−θat,f;A¯θ,p,a,A1≈u−θ−1/pauKu,fp,0,t.
Therefore, the function1/t1−θatu−θ−1/pauKu,fp,0,t is equivalent to a decreasing function. Hence, for v<t, we get(30)bvu−θ−1/pauKu,fp,0,v≈s1−θ−1/rasbsr,0,v1v1−θavu−θ−1/pauKu,fp,0,v≺s−1/rbsu−θ−1/pauKu,fp,0,sr,0,v≤s−1/rbsu−θ−1/pauKu,fp,0,sr,0,t.
Thus, (28) is proved for q=∞:(31)sup0<s<tbsu−θ−1/pauKu,fp,0,s≺s−1/rbsu−θ−1/pauKu,fp,0,sr,0,t.
Denoteds=bsu−θ−1/pauKu,fp,0,s. If r<q<∞, using the last estimate, we get(32)s−1/qbsu−θ−1/pauKu,fp,0,sq,0,t=∫0tdsqduu1/q≤sup0<u<tdsq−r/q∫0tdsrduu1/q≺s−1/rbsu−θ−1/pauKu,fp,0,sr,0,t.
This completes the proof.☐It is worth remarking that the estimates from Lemmas9–11 can be used to establish different embedding theorems.Lemma 12 (cf. [8], Lemma 4.2).
Let0<q,r≤∞, a,b∈SV, and β>0. Then, for all f∈A0+A1,(33)t−β−1/rbtuα−1/qauKu,fq,0,tr,0,∞≈tα−β−1/ratbtKt,fr,0,∞,provided that α>−1, and(34)tβ−1/rbtuα−1/qauKu,fq,t,∞r,0,∞≈tα+β−1/ratbtKt,fr,0,∞,provided that α<0.Proof.
We prove the first equivalence. The second one can be proved analogously. We have(35)uα−1/qauKu,fq,0,t≥uα−1/qauKu,fq,t/2,t≥uα−1/qauq,t/2,tKt2,f≈tαatKt,f.
Hence,(36)t−β−1/rbtuα−1/qauKu,fq,0,tr,0,∞≻tα−β−1/ratbtKt,fr,0,∞.
Let us prove the converse inequality.
Caseq≤r. Because β>0, by Corollary 4 (taking hu=uαauKu,f), we get(37)t−β−1/rbtuα−1/qauKu,fq,0,tr,0,∞≺tα−β−1/ratbtKt,fr,0,∞.
In particular,(38)t−β−1/rbtuα−1/rauKu,fr,0,tr,0,∞≺tα−β−1/ratbtKt,fr,0,∞.
Caseq>r. Because α>−1, using Lemma 9 and the last estimate, we get(39)t−β−1/rbtuα−1/qauKu,fq,0,tr,0,∞≺t−β−1/rbtuα−1/rauKu,fr,0,tr,0,∞≺tα−β−1/ratbtKt,fr,0,∞.
This completes the proof.☐Lemma 13.
Let0≤θ0<θ1≤1, 0<p,q,r≤∞, and a,b,c∈SV. Then, for all f∈A0+A1,(40)tθ0−θ1−1/pcts−1/rbsu−θ0−1/qauKu,fq,0,sr,0,tp,0,∞≈t−θ1−1/patbtctKt,fp,0,∞,tθ1−θ0−1/pcts−1/rbsu−θ1−1/qauKu,fq,s,∞r,t,∞p,0,∞≈t−θ0−1/patbtctKu,fp,0,∞.Proof.
We prove the first equivalence. The second one can be proved analogously or using the first equivalence and symmetry argument. DenoteIr≔LHS40.
Caser≤p. Because θ0−θ1<0 and θ0<1, by Corollary 4 (taking hs=bsu−θ0−1/qauKu,fq,0,s) and by Lemma 12, we get(41)Ir≺tθ0−θ1−1/rcthtp,0,∞=tθ0−θ1−1/pctbtu−θ0−1/qauKu,fq,0,tp,0,∞≈t−θ1−1/patctbtKt,fp,0,∞.
In particular, it was shown that(42)Ip≺t−θ1−1/patctbtKt,fp,0,∞.
Caser>p. By Lemma 11 and (42), we get(43)Ir≺tθ0−θ1−1/pcts−1/pbsu−θ0−1/qauKu,fq,0,sp,0,tp,0,∞=Ip≺t−θ1−1/patctbtKt,fp,0,∞.
Let us prove the inverse estimate. By (27), we have(44)t−θ0atbtKt,f≺s−1/rbsu−θ0−1/qauKu,fq,0,sr,0,t.
Hence,(45)RHS40≺tθ0−θ1−1/pcts−1/rbsu−θ0−1/qauKu,fq,0,sr,0,tp,0,∞.
This completes the proof.☐Lemma 14.
Let0≤θ0<θ1≤1, 0<p,q,r≤∞, and a,b∈SV.(i)
Ifs−1/rbsr,0,1<∞, then for all f∈A0+A1,(46)tθ0−θ1−1/pcts−1/rbsu−θ0−1/qauKu,fq,s,tr,0,tp,0,∞≈t−θ1−1/pats−1/rbsr,0,tctKt,fp,0,∞.(ii)
Ifs−1/rbsr,1,∞<∞, then for all f∈A0+A1,(47)tθ1−θ0−1/pcts−1/rbsu−θ1−1/qauKu,fq,t,sr,t,∞p,0,∞≈t−θ0−1/pats−1/rbsr,t,∞ctKt,fp,0,∞.Proof.
We prove the first equivalence. The second one can be proved analogously or using the symmetry argument. For0≤θ0≤1, by (24), we have(48)t−θ0ats−1/rbsr,0,tKt,f≺s−1/rbsu−θ0−1/qauKu,fq,s,tr,0,t.
Hence, for eachθ1,(49)RHS46≺tθ0−θ1−1/pcss−1/rbsu−θ0−1/qauKu,fq,s,tr,0,tp,0,∞=LHS46.
Let us prove the inverse estimate. Becauseθ0−θ1<0 and θ0<1 by Lemma 12, we get(50)LHS46≤tθ0−θ1−1/pcts−1/rbsr,0,tu−θ0−1/qauKu,fq,0,tp,0,∞≈t−θ1−1/pcts−1/rbsr,0,tatKt,fp,0,∞=RHS46.
This completes the proof.☐The following lemma deals with the quasi-norms on the spacesA¯0,∗;∗ and A¯1,∗;∗.Lemma 15.
Let0<r,p,q≤∞ and a,b,c∈SV.(i)
Ifs−1/rcsr,1,∞<∞ and s−1/pbsp,1,∞<∞, then for all f∈A¯0,r;c,(51)f0,r;c≈tθ−1/rctats−1/pbsp,t,∞s−1/pbsu−θ−1/qauKu,fq,t,sp,t,∞r,0,∞,provided that 0<θ≤1. Furthermore,(52)f0,r;c≺t−1/rctats−1/pbsp,t,∞s−1/pbsu−1/qauKu,fq,t,sp,t,∞r,0,∞.(ii)
Ifs−1/rcsr,0,1<∞ and s−1/pbsp,0,1<∞, then for all f∈A¯1,r;c,(53)f1,r;c≈tθ−1−1/rctats−1/pbsp,0,ts−1/pbsu−θ−1/qauKu,fq,s,tp,0,tr,0,∞,provided that 0≤θ<1. Furthermore,(54)f1,r;c≺t−1/rctats−1/r0bsr0,0,ts−1/pbsu−1−1/qauKu,fq,s,tp,0,tr,0,∞.(iii)
Ifs−1/rcsr,1,∞<∞ and 0<θ≤1, then for all f∈A¯0,r;c,(55)f0,r,c≈tθ−1/rctatu−θ−1/qauKu,fq,t,∞r,0,∞.(iv)
Ifs−1/rcsr,0,1<∞ and 0≤θ<1, then for all f∈A¯1,r;c,(56)f1,r,c≈tθ−1−1/rctatu−θ−1/qauKu,fq,0,tr,0,∞.Proof.
We prove the first assertion. The second one can be proved analogously or using the symmetry argument. By Lemma10, for 0≤θ≤1, we have(57)f0,r;c=tθ−1/rctats−1/pbsp,t,∞t−θats−1/pbsp,t,∞Kt,fr,0,∞≺tθ−1/rctats−1/pbsp,t,∞u−1/pbus−θ−1/qasKs,fq,t,up,t,∞r,0,∞.
Thus,LHS51≺RHS51 and the estimate (52) is proved. Let 0<θ≤1. Note that(58)s−1/r1bsu−θ−1/qauKu,fq,t,sr1,t,∞≤s−1/r1bsr1,t,∞u−θ−1/qauKu,fq,t,∞.
Hence, by Lemma12, we get(59)RHS51≤tθ−1/rctatu−θ−1/qauKu,fq,t,∞r,0,∞≈t−1/rctKt,fr,0,∞=LHS51.
The equivalences (iii) and (iv) follow immediately from Lemma12.☐In Section8, we will use the following reiteration theorems. See [9, 14].Theorem 16.
Let0<r,r0≤∞, a,b0∈SV, and s−1/r0b0sr0,1,∞<∞. Put ρt=ts−1/r0b0sr0,t,∞.(i)
If0<θ<1, then(60)A¯0,r0,b0,A1θ,r,a=A¯θ,r,a#,where a#t=s−1/r0b0sr0,t,∞1−θaρt.(ii)
Ift−1/ratr,1,∞<∞, then(61)A¯0,r0,b0,A10,r,a=A¯0,r,a#∩A¯0,r,a∘ρ,r0,b0L,where a#t=s−1/r0b0sr0,t,∞aρt.(iii)
Ift−1/ratr,0,1<∞, then(62)A¯0,r0,b0,A11,r,a=A¯1,r,a∘ρ.Theorem 17.
Let0<r,r1≤∞, a,b1∈SV, and s−1/r1b1sr1,0,1<∞. Put ρt=t/t−1/r1b1tr1,0,t.(i)
If0<θ<1, then(63)A0,A¯1,r1,b1θ,r,a=A¯θ,r,a#,where a#t=s−1/r1b1sr1,0,tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(64)A0,A¯1,r1,b10,r,a=A¯0,r,a∘ρ.(iii)
Ift−1/ratr,0,1<∞, then(65)A0,A¯1,r1,b11,r,a=A¯1,r,a#∩A¯1,r,a∘ρ,r1,b1R,where a#t=s−1/r1b1sr1,0,taρt.


## 4. Limiting Interpolation between theL Spaces on the Right and the Standard Interpolation Spaces

In this section, we establish some limiting reiteration formulae for the couples of the formA¯θ0,∗,A¯θ1,∗L (0≤θ0<θ1<1). They amend Theorems 16, 17, and 19 from [5]. Here, we need the relevant Holmstedt-type formulae. The proofs of all Holmstedt-type formulae in this paper are based on the paper [15]. Therefore, we adopt the notation from it.Theorem 18 (cf. [6], Theorem 3.4).
Let0<θ0<θ1<1, 0<r0,q1,r1≤∞, b0,a1,b1∈SV, and s−1/r1b1sr1,1,∞<∞. Put χ1t=a1ts−1/r1b1sr1,t,∞.(i)
Letρt=tθ1−θ0b0t/χ1t. Then, for all f∈A¯θ0,r0,b0+A¯θ1,r1,b1,q1,a1L and t>0,(66)Kρt,f;A¯θ0,r0,b0,A¯θ1,r1,b1,q1,a1L≈u−θ0−1/r0b0uKu,fr0,0,t+ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞.(ii)
Lets−1/r0b0sr0,1,∞<∞ and ρt=tθ1s−1/r0b0sr0,t,∞/χ1t. Then, for all f∈A¯0,r0,b0+A¯θ1,r1,b1,q1,a1L and t>0,(67)Kρt,f;A¯0,r0,b0,A¯θ1,r1,b1,q1,a1L≈u−1/r0b0uKu,fr0,0,t+ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞.(iii)
Letρt=tθ11/χ1t. Then, for all f∈A0+A¯θ1,r1,b1,q1,a1L and t>0,(68)Kρt,f;A0,A¯θ1,r1,b1,q1,a1L≈ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞.Proof.
The formulae (i) and (iii) have been proved in [5]. Here, we prove the formula (ii). Let Φ0 be the function space corresponding to A¯0,r0,b0:(69)F∗Φ0=u−1/r0b0uFur0,0,∞.
Consider the functionsIt,f, g0t, and h0t.(70)It,f≔χ0,t∗K∗,fΦ0=u−1/r0b0uKu,fr0,0,t,g0t≔tχt,∞∗Φ0=tu−1/r0b0ur0,t,∞,h0t≔∗χ0,t∗Φ0=u1−1/r0b0ur0,0,t≈tb0t.
Thus,(71)h0t≺g0t≈g0t+h0t≈tu−1/r0b0ur0,t,∞.
LetΦ1 be the function space corresponding to A¯θ1,r1,b1,q1,a1L:(72)F∗Φ1=s−1/r1b1su−θ1−1/q1a1uFuq1,0,sr1,0,∞.
Consider the functionsJt,f, g1t, and h1t.(73)Jt,f≔χt,∞∗K∗,fΦ1=s−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞,g1t≔tχt,∞∗Φ1=ts−1/r1b1su−θ1−1/q1a1uq1,t,sr1,t,∞≺t1−θ1s−1/r1b1sr1,t,∞a1t=t1−θ1χ1t,h1t≔∗χ0,t∗Φ1=s−1/r1b1su1−θ1−1/q1a1uq1,0,mint,sr1,0,∞≈s−1/r1b1smint,s1−θ1a1mint,sr1,0,∞≈s1−θ1−1/r1a1sb1sr1,0,t+t1−θ1a1ts−1/r1b1sr1,t,∞≈t1−θ1a1tb1t+t1−θ1χ1t.
Becausea1tb1t≺a1ts−1/r1b1sr1,t,∞=χ1t, we arrive at(74)g1t≺h1t≈g1t+h1t≈t1−θ1χ1t.
Now, we estimate the functionsgt and ht.(75)1gt≔χt,∞uug0u+h0uΦ1≈s−1/r1b1su−θ1−1/q1a1uv−1/r0b0vr0,u,∞q1,t,sr1,t,∞≤u−θ1−1/q1a1uv−1/r0b0vr0,u,∞q1,t,∞s−1/r1b1sr1,t,∞≈t−θ1a1tv−1/r0b0vr0,t,∞s−1/r1b1sr1,t,∞=1ρt,ht≔χ0,tuug1u+h1uΦ0≈uθ1−1/r0b0uχ1ur0,0,t≈tθ1b0tχ1t≺tθ1s−1/r0b0sr0,t,∞χ1t=ρt.
We see thatht≺ρt≺gtand g0t/g1t+h1t≈g0t+h0t/h1t≈ρt. Note that(76)g0t+ρth1t≈ts−1/r0b0sr0,t,∞.
Hence, by [15] (Theorem 4, Case 1), we arrive at(77)Kρt,f;A¯0,r0,b0,A¯θ1,r1,b1,q1,a1L≈u−1/r0b0uKu,fr0,0,t+ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞+s−1/r0b0sr0,t,∞Kt,f.
Using Lemma10, we get(78)s−1/r0b0sr0,t,∞Kt,f≺ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞.
Therefore,(79)Kρt,f;A¯0,r0,b0,A¯θ1,r1,b1,q1,a1L≈u−1/r0b0uKu,fr0,0,t+ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞.
This completes the proof.☐Theorem 19 (cf. [6], Theorem 4.7 (c)).
Let0<θ0<θ1<1, 0<r,r0,q1,r1≤∞, a,b0,a1,b1∈SV, s−1/r1b1sr1,1,∞<∞, and s−1/rasr,0,1<∞. Put χ1t=a1ts−1/r1b1sr1,t,∞ and ρt=tθ1−θ0b0t/χ1t. Then,(80)A¯θ0,r0,b0,A¯θ1,r1,b1,q1,a1L1,r,a=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A¯θ0,r0,b0, X1=A¯θ1,r1,b1,q1,a1L, K¯t,f=Kt,f;X¯, and Z=X¯1,r,a. Using the change of variables u=ρt (see, e.g., [5], Remark 3) and Theorem 18 (i), we can write(81)fZ=u−1−1/rauK¯u,fr,0,∞≈ρt−1t−1/raρtK¯ρt,fr,0,∞≈I1+I2,where(82)I1≔tθ0−θ1−1/raρtχ1tb0tu−θ0−1/r0b0uKu,fr0,0,tr,0,∞,I2≔t−1/raρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞=fL,R;θ1,r,a∘ρ,r1,b1,q1,a1.
Lemmas12 and 10 imply that(83)I1≈t−θ1−1/raρta1ts−1/r1b1sr1,t,∞Kt,fr,0,∞≺I2.
This completes the proof.☐Theorem 20 (cf. [6], Theorem 4.8 (b, c)).
Let0<θ1<1, 0<r,r0,q1,r1≤∞, a,b0,a1,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,1,∞<∞. Put χ1t=a1ts−1/r1b1sr1,t,∞ and ρt=tθ1s−1/r0b0sr0,t,∞/χ1t.(i)
Ifs−1/rasr0,1,∞<∞, then(84)A¯0,r0,b0,A¯θ1,r1,b1,q1,a1L0,r,a=A¯0,r,a#⋂A¯0,r,a∘ρ,r0,b0L,where a#t=s−1/r0b0sr0,t,∞aρt.(ii)
Ifs−1/rasr0,0,1<∞, then(85)A¯0,r0,b0,A¯θ1,r1,b1,q1,a1L1,r,a=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A¯0,r0,b0, X1=A¯θ1,r1,b1,q1,a1L, and K¯t,f=Kt,f;X¯. First, we prove (i). Let Z=X¯0,r,a. Using the change of variables u=ρt and Theorem 18 (ii), we can write(86)fZ=u−1/rauK¯u,fr,0,∞≈t−1/raρtK¯ρt,fr,0,∞≈I1+I2,where(87)I1≔t−1/raρtu−1/r0b0uKu,fr0,0,tr,0,∞=fL;0,r,a∘ρ,r0,b0,I2≔t−1/raρtρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞=tθ1−1/ra#tχ1ts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞.
By Lemma15 (i), we get I2≈f0,r;a#. The formula (i) is proved.
Now, we prove (ii). LetZ=X¯1,r,a. Using the change of variables and Theorem 18 (ii), we can write(88)fZ=u−1−1/rauK¯u,fr,0,∞≈ρt−1t−1/raρtK¯ρt,fr,0,∞≈I1+I2,where(89)I1≔t−θ1−1/rχ1ts−1/r0b0sr0,t,∞aρtu−1/r0b0uKu,fr0,0,tr,0,∞,I2≔t−1/raρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞=fL,R;θ1,r,a∘ρ,r1,b1,q1,a1.
By Lemmas12 and 10, we get(90)I1≈t−θ1−1/rχ1tb0ts−1/r0b0sr0,t,∞aρtKt,fr,0,∞≺t−θ1−1/ra1ts−1/r1b1sr1,t,∞aρtKt,fr,0,∞≺t−1/raρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞=I2.
This completes the proof.☐Theorem 21 (cf. [6], Theorem 4.9 (b)).
Let0<θ1<1, 0<r,q1,r1≤∞, a,a1,b1∈SV, s−1/r1b1sr1,1,∞<∞, and s−1/rasr,0,1<∞. Put χ1t=a1ts−1/r1b1sr1,t,∞ and ρt=tθ11/χ1t. Then,(91)A0,A¯θ1,r1,b1,q1,a1L1,r,a=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A0, X1=A¯θ1,r1,b1,q1,a1L, K¯t,f=Kt,f;X¯, and Z=X¯1,r,a. Using the change of variables and Theorem 18 (iii), we can write(92)fZ=u−1−1/rauK¯u,fr,0,∞≈ρt−1t−1/raρtK¯ρt,fr,0,∞≈t−1/raρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞=fL,R;θ1,r,a∘ρ,r1,b1,q1,a1.
This completes the proof.☐


## 5. Limiting Interpolation between theL Spaces on the Left and the Standard Interpolation Spaces

In this section, we establish some limiting reiteration formulae for the couples of the formA¯θ0,∗L,A¯θ1,∗0<θ0<θ1≤1. They amend Theorems 11–13 from [5]. Here, we also need the relevant Holmstedt-type formulae.Theorem 22 (cf. [7], Theorem 3.2).
Let0<θ0<θ1<1, 0<q0,r0,r1≤∞, a0,b0,b1∈SV, and s−1/r0b0sr0,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞.(i)
Letρt=tθ1−θ0χ0t/b1t. Then, for all f∈A¯θ0,r0,b0,q0,a0L+A¯θ1,r1,b1 and t>0,(93)Kρt,f;A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t+ρtu−θ1−1/r1b1uKu,fr1,t,∞.(ii)
Lets−1/r1b1sr1,0,1<∞ and ρt=t1−θ0χ0t/s−1/r1b1sr1,0,t. Then, for all f∈A¯θ0,r0,b0,q0,a0L+A¯1,r1,b1 and t>0,(94)Kρt,f;A¯θ0,r0,b0,q0,a0L,A¯1,r1,b1≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t+ρtu−1−1/r1b1uKu,fr1,t,∞.(iii)
Letρt=t1−θ0χ0t. Then, for all f∈A¯θ0,r0,b0,q0,a0L+A¯1 and t>0,(95)Kρt,f;A¯θ0,r0,b0,q0,a0L,A1≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t.Proof.
Recall that we adopt the notation from the paper [15]. Let Φ0 be the function space corresponding to A¯θ0,r0,b0,q0,a0L:(96)F∗Φ0=s−1/r0b0su−θ0−1/q0a0uFuq0,0,sr0,0,∞.
Consider the functionsIt,f, g0t, and h0t.(97)It,f≔χ0,t∗K∗,fΦ0=s−1/r0b0su−θ0−1/q0a0uχ0,tuKu,fq0,0,sr0,0,∞≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t,g0t≔tχt,∞∗Φ0=ts−1/r0b0su−θ0−1/q0a0uq0,t,sr0,t,∞≤tu−θ0−1/q0a0uq0,t,∞s−1/r0b0sr0,t,∞≈t1−θ0a0ts−1/r0b0sr0,t,∞=t1−θ0χ0t,h0t≔∗χ0,t∗Φ0=s−1/r0b0su1−θ0−1/q0a0uq0,0,mint,sr0,0,∞≈s−1/r0b0smint,s1−θ0a0mint,sr0,0,∞≈s1−θ0−1/r0b0sa0sr0,0,t+t1−θ0a0ts−1/r0b0sr0,t,∞≈t1−θ0b0ta0t+t1−θ0a0ts−1/r0b0sr0,t,∞≈t1−θ0a0ts−1/r0b0sr0,t,∞=t1−θ0χ0t.
Thus,(98)g0t≺h0t≈g0t+h0t≈t1−θ0χ0t.
First, we prove (i). LetΦ1 be the function space corresponding to A¯θ1,r1,b1:(99)F∗Φ1=u−θ1−1/r1b1uFur1,0,∞.
Consider the functionsJt,f, g1t, and h1t.(100)Jt,f≔χt,∞∗K∗,fΦ1=u−θ1−1/r1b1uKu,fr1,t,∞,g1t≔tχt,∞∗Φ1=tu−θ1−1/r1b1ur1,t,∞≈t1−θ1b1t,h1t≔∗χ0,t∗Φ1=u1−θ1−1/r1b1ur1,0,t≈t1−θ1b1t.
Thus,(101)g1t≈h1t≈g1t+h1t≈t1−θ1b1t.
Now, we estimate the functionsgt and ht. Because θ0−θ1<0, we get(102)1gt≔χt,∞uug0u+h0uΦ1≈uθ0−θ1−1/r1b1uχ0ur1,t,∞≈tθ0−θ1b1tχ0t=1ρt,ht≔χ0,tuug1u+h1uΦ0≈s−1/r0b0suθ1−θ0−1/q0a0ub1uq0,0,mint,sr0,0,∞≈sθ1−θ0−1/r0b0sa0sb1sr0,0,t+tθ1−θ0a0tb1ts−1/r0b0sr0,t,∞≈tθ1−θ0a0tb0tb1t+tθ1−θ0a0tb1ts−1/r0b0sr0,t,∞≈tθ1−θ0a0tb1ts−1/r0b0sr0,t,∞=ρt.
Thus,(103)ht≈ρt≈gt.
Using [15] (Theorem 4, Case 4), we arrive at (93).
Proof of statement (ii). Put χ1t=s−1/r1b1sr1,0,t. Let Φ1 be the function space corresponding to A¯1,r1,b1:(104)F∗Φ1=u−1−1/r1b1uFur1,0,∞.
Consider the functionsJt,f, g1t, and h1t.(105)Jt,f≔χt,∞∗K∗,fΦ1=u−1−1/r1b1uKu,fr1,t,∞,g1t≔tχt,∞∗Φ1=tu−1−1/r1b1ur1,t,∞≈b1t,h1t≔∗χ0,t∗Φ1=u−1/r1b1ur1,0,t=χ1t.
Thus,(106)g1t≺h1t≈g1t+h1t≈χ1t.
Now, we estimate the functionsgt and ht. Because θ0<1, we get(107)1gt≔χt,∞uug0u+h0uΦ1≈uθ0−1−1/r1b1uχ0ur1,t,∞≈tθ0−1b1tχ0t≺tθ0−1χ1tχ0t=1ρt,ht≔χ0,tuug1u+h1uΦ0≈s−1/r0b0su1−θ0−1/q0a0uχ1tq0,0,mint,sr0,0,∞≈s1−θ0−1/r0b0sa0sχ1sr0,0,t+t1−θ0a0tχ1ts−1/r0b0sr0,t,∞≈t1−θ0b0ta0tχ1t+t1−θ0a0tχ1ts−1/r0b0sr0,t,∞≈t1−θ0a0tχ1ts−1/r0b0sr0,t,∞=ρt.
Thus,(108)ht≈ρt≈gt.
Using [15] (Theorem 4, Case 4), we arrive at (94).
Proof of assertion (iii). Here, we use [15] (Theorem 1). Because g0t+h0t≈ρt, we get(109)Kρt,f;A¯θ0,r0,b0,q0,a0L,A1≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t+t−θ0χ0tKt,f.
By Lemma9, we know that(110)t−θ0χ0tKt,f=s−1/r0b0sr0,t,∞t−θ0a0tKt,f≺s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t.
So, we arrive at (95).☐Theorem 23 (cf. [7], Theorem 4.8).
Let0<θ0<θ1<1, 0<r,q0,r0,r1≤∞, a,a0,b0,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/rasr,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞ and ρt=tθ1−θ0χ0t/b1t. Then,(111)A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b10,r,a=A¯θ0,r,a#,q0,a0L⋂A¯θ0,r,a∘ρ,r0,b0,q0,a0L,L,where a#t=s−1/r0b0sr0,t,∞aρt.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0L, X1=A¯θ1,r1,b1, K¯t,f=Kt,f;X¯, and Z=X¯0,r,a. Using the change of variables and Theorem 22 (i), we can write(112)fZ=u−1/rauK¯u,fr,0,∞≈t−1/raρtK¯ρt,fr,0,∞≈I1+I2+I3,where(113)I1≔t−1/raρts−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,tr,0,∞=fL,L;θ0,r,a∘ρ,r0,b0,q0,a0,I2≔t−1/raρts−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,tr,0,∞=fL;θ0,r,a#,q0,a0,I3≔tθ1−θ0−1/raρtχ0tb1tu−θ1−1/r1b1uKu,fr1,t,∞r,0,∞.
It is enough to show thatI3≺I2. By Lemmas 12 and 9, we get(114)I3≈t−θ0−1/raρta0ts−1/r0b0sr0,t,∞Kt,fr,0,∞≺t−1/raρts−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,tr,0,∞=I2.
This completes the proof.☐Theorem 24 (cf. [7], Theorem 4.9).
Let0<θ0<1, 0<r,q0,r0,r1≤∞, a,a0,b0,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞ and ρt=t1−θ0χ0t/s−1/r1b1sr1,0,t.(i)
Ifs−1/rasr,1,∞<∞, then(115)A¯θ0,r0,b0,q0,a0L,A¯1,r1,b10,r,a=A¯θ0,r,a#,q0,a0L⋂A¯θ0,r,a∘ρ,r0,b0,q0,a0L,L,where a#t=s−1/r0b0sr0,t,∞aρt.(ii)
Ifs−1/rasr,0,1<∞, then(116)A¯θ0,r0,b0,q0,a0L,A¯1,r1,b11,r,a=A¯1,r,a#⋂A¯1,r,a∘ρ,r1,b1R,where a#t=s−1/r1b1sr1,0,taρt.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0L, X1=A¯1,r1,b1, and K¯t,f=Kt,f;X¯. First, we prove (i). Let Z=X¯0,r,a. Using the change of variables and Theorem 22 (ii), we can write(117)fZ=u−1/rauK¯u,fr,0,∞≈t−1/raρtK¯ρt,fr,0,∞≈I1+I2+I3,where(118)I1≔t−1/raρts−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,tr,0,∞=fL,L;θ0,r,a∘ρ,r0,b0,q0,a0,I2≔t−1/raρts−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,tr,0,∞=fL;θ0,r,a#,q0,a0,I3≔t1−θ0−1/raρtχ0ts−1/r1b1sr1,0,tu−1−1/r1b1uKu,fr1,t,∞r,0,∞.
The estimateI3≺I2 can be proved by repeating the corresponding part of the proof of Theorem 23.
Now, we prove (ii). LetZ=X¯1,r,a. Using the change of variables and Theorem 22 (ii), we can write(119)fZ=u−1−1/rauK¯u,fr,0,∞≈ρt−1t−1/raρtK¯ρt,fr,0,∞≈I1+I2+I3,where(120)I1≔tθ0−1−1/ra#tχ0ts−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,tr,0,∞,I2≔tθ0−1−1/ra#ta0tu−θ0−1/q0a0uKu,fq0,0,tr,0,∞,I3≔t−1/raρtu−1−1/r1b1uKu,fr1,t,∞r,0,∞=fR;1,r,a∘ρ,r1,b1.
Using Lemma15 (iv), we get I2≈f1,r,a#. Hence, it is enough to show that I1≺f1,r,a#. By Lemma 13, we get(121)I1≈t−1−1/ra#ts−1/r0b0sr0,t,∞b0sKt,fr,0,∞≺t−1−1/ra#tKt,fr,0,∞=f1,r,a#.
This completes the proof.☐Theorem 25 (cf. [7], Theorem 4.10).
Let0<θ0<1, 0<r,q0,r0≤∞, a,a0,b0∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/rasr,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞ and ρt=t1−θ0χ0t. Then,(122)A¯θ0,r0,b0,q0,a0L,A10,r,a=A¯θ0,r,a#,q0,a0L⋂A¯θ0,r,a∘ρ,r0,b0,q0,a0L,L,where a#t=s−1/r0b0sr0,t,∞aρt.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0L, X1=A1, K¯t,f=Kt,f;X¯, and Z=X¯0,r,a. Using the change of variables and Theorem 22 (iii), we can write(123)fZ=u−1/rauK¯u,fr,0,∞≈t−1/raρtK¯ρt,fr,0,∞≈I1+I2,where(124)I1≔t−1/raρts−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,tr,0,∞=fL,L;θ0,r,a∘ρ,r0,b0,q0,a0,I2≔t−1/raρts−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,tr,0,∞=fL;θ0,r,a#,q0,a0.
This completes the proof.☐


## 6. Limiting Interpolation between theR Spaces and the Standard Interpolation Spaces

The theorems in this section amend theorems from [5] (Section 5). The proofs are based on the corresponding reiteration theorems from the previous sections and on the formulae (14)–(16). They are left to the reader. See, as an example, the proof of Theorem 34.Theorem 26 (cf. [6], Theorem 4.4).
Let0<θ0<θ1<1, 0<r,q0,r0,r1≤∞, a,b0,a1,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/rasr,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,0,t and ρt=tθ1−θ0χ0t/b1t. Then,(125)A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b10,r,a=A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.Theorem 27 (cf. [6], Theorem 4.5).
Let0<θ0<1, 0<r,q0,r0,r1≤∞, a,a0,b0,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,0,t and ρt=t1−θ0χ0t/s−1/r1b1sr1,0,t.(i)
Ifs−1/rasr0,1,∞<∞, then(126)A¯θ0,r0,b0,q0,a0R,A¯1,r1,b10,r,a=A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(ii)
Ifs−1/rasr0,0,1<∞, then(127)A¯θ0,r0,b0,q0,a0R,A¯1,r1,b11,r,a=A¯1,r,a#⋂A¯1,r,a∘ρ,r1,b1R,where a#t=s−1/r1b1sr1,0,taρt.Theorem 28 (cf. [6], Theorem 4.6).
Let0<θ0<1, 0<r,q0,r0≤∞, a,a0,b0∈SV, s−1/r0b0sr0,0,1<∞, and s−1/rasr,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,0,t and ρt=t1−θ0χ0t. Then,(128)A¯θ0,r0,b0,q0,a0R,A10,r,a=A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.Theorem 29 (cf. [7], Theorem 4.5).
Let0<θ0<θ1<1, 0<r,r0,q1,r1≤∞, a,b0,a1,b1∈SV, s−1/r1b1sr1,0,1<∞, and s−1/rasr,0,1<∞. Put χ1t=a1ts−1/r1b1sr1,0,t and ρt=tθ1−θ0b0t/χ1t. Then,(129)A¯θ0,r0,b0,A¯θ1,r1,b1,q1,a1R1,r,a=A¯θ1,r,a#,q1,a1R⋂A¯θ1,r,a∘ρ,r1,b1,q1,a1R,R,where a#t=s−1/r1b1sr1,0,taρt.Theorem 30 (cf. [7], Theorem 4.6).
Let0<θ1<1, 0<r,r0,q1,r1≤∞, a,b0,a1,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,0,1<∞. Put χ1t=a1ts−1/r1b1sr1,0,t and ρt=tθ1s−1/r0b0sr0,t,∞/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(130)A¯0,r0,b0,A¯θ1,r1,b1,q1,a1R0,r,a=A¯0,r,a#⋂A¯0,r,a∘ρ,r1,b1L,where a#t=s−1/r0b0sr0,t,∞aρt.(ii)
Ifs−1/rasr,0,1<∞, then(131)A¯0,r0,b0,A¯θ1,r1,b1,q1,a1R1,r,a=A¯θ1,r,a#,q1,a1R⋂A¯θ1,r,a∘ρ,r1,b1,q1,a1R,R,where a#t=s−1/r1b1sr1,0,taρt.Theorem 31 (cf. [7], Theorem 4.7).
Let0<θ1<1, 0<r,q1,r1≤∞, a,a1,b1∈SV, s−1/r1b1sr1,0,1<∞, and s−1/rasr,0,1<∞. Put χ1t=a1ts−1/r1b1sr1,0,t and ρt=tθ11/χ1t. Then,(132)A0,A¯θ1,r1,b1,q1,a1R1,r,a=A¯θ1,r,a#,q1,a1R⋂A¯θ1,r,a∘ρ,r1,b1,q1,a1R,R,where a#t=s−1/r1b1sr1,0,taρt.


## 7. Limiting Reiteration Formulae for Couples Formed Only byL and R Spaces

The reiteration theorems in this section amend theorems from [5] (Section 6). Here, we also need the relevant Holmstedt-type formulae.Theorem 32 (cf. [8], Theorem 3.2).
Let0<θ0<θ1<1, 0<q0,r0,q1,r1≤∞, a0,b0,a1,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞, χ1t=a1ts−1/r1b1sr1,t,∞, and ρt=tθ1−θ0χ0t/χ1t. Then, for all f∈A¯θ0,r0,b0,q0,a0L+A¯θ1,r1,b1,q1,a1L and t>0,(133)Kρt,f;A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1L≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t+ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞.Proof.
Recall that we adopt the notation from the paper [15]. Let Φ0 be the function space corresponding to A¯θ0,r0,b0,q0,a0L:(134)F∗Φ0=s−1/r0b0su−θ0−1/q0a0uFuq0,0,sr0,0,∞.
From the proof of Theorem22, we know that(135)It,f≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t,g0t≺h0t≈g0t+h0t≈t1−θ0χ0t.
LetΦ1 be the function space corresponding to A¯θ1,r1,b1,q1,a1L:(136)F∗Φ1=s−1/r1b1su−θ1−1/q1a1uFuq1,0,sr1,0,∞.
From the proof of Theorem18, we know that(137)Jt,f=s−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞,g1t≺h1t≈g1t+h1t≈t1−θ1χ1t.
Now, we estimate the functionsgt and ht. Because θ0−θ1<0, we get(138)1gt≔χt,∞uug0u+h0uΦ1≈s−1/r1b1suθ0−θ1−1/q1a1uχ0uq1,t,sr1,t,∞≤s−1/r1b1sr1,t,∞uθ0−θ1−1/q1a1uχ0uq1,t,∞≈s−1/r1b1sr1,t,∞tθ0−θ1a1tχ0t=1ρt,ht≔χ0,tuug1u+h1uΦ0≈s−1/r0b0suθ1−θ0−1/q0a0uχ1uq0,0,mint,sr0,0,∞≈sθ1−θ0−1/r0b0sa0sχ1sr0,0,t+s−1/r0b0sr0,t,∞uθ1−θ0−1/q0a0uχ1uq0,0,t≈tθ1−θ0b0ta0tχ1t+s−1/r0b0sr0,t,∞tθ1−θ0a0tχ1t≈s−1/r0b0sr0,t,∞tθ1−θ0a0tχ1t=ρt.
Thus,(139)ht≈ρt≺gt.
Therefore, using [15] (Theorem 4, Case 4), we arrive at (133).☐Theorem 33 (cf. [8], Theorem 4.4).
Let0<θ0<θ1<1, 0<r,q0,r0,q1,r1≤∞, a,a0,b0,a1,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞, χ1t=a1ts−1/r1b1sr1,t,∞, and ρt=tθ1−θ0χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(140)A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1L0,r,a=A¯θ0,r,a#,q0,a0L⋂A¯θ0,r,a∘ρ,r0,b0,q0,a0L,L,where a#t=s−1/r0b0sr0,t,∞aρt.(ii)
Ifs−1/rasr,0,1<∞, then(141)A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1L1,r,a=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0L, X1=A¯θ1,r1,b1,q1,a1L, and K¯t,f=Kt,f;X¯. First, we prove (i). Let Z=X¯0,r,a. Using the change of variables and Theorem 32, we can write(142)fZ=u−1/rauK¯u,fr,0,∞≈t−1/raρtK¯ρt,fr,0,∞≈I1+I2+I3,where(143)I1≔t−1/raρts−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,tr,0,∞=fL,L;θ0,r,a∘ρ,r0,b0,q0,a0,I2≔t−1/raρts−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,tr,0,∞=fL;θ0,r,a#,q0,a0,I3≔tθ1−θ0−1/raρtχ0tχ1ts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞.
It is enough to show thatI3≺I2. Using Lemmas 12 and 9, we get(144)I3≤tθ1−θ0−1/raρtχ0tχ1ts−1/r1b1sr1,t,∞u−θ1−1/q1a1uKu,fq1,t,∞r,0,∞≈t−θ0−1/raρta0ts−1/r0b0sr0,t,∞Kt,fr,0,∞≺t−1/raρts−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,tr,0,∞=I2.
Statement (i) is proved.
Now, we prove (ii). LetZ=X¯1,r,a. Using the change of variables and Theorem 32, we can write(145)fZ=u−1−1/rauK¯u,fr,0,∞≈ρt−1t−1/raρtK¯ρt,fr,0,∞≈I1+I2+I3,where(146)I1≔tθ0−θ1−1/raρtχ1tχ0ts−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,tr,0,∞,I2≔tθ0−θ1−1/raρtχ1ta0tu−θ0−1/q0a0uKu,fq0,0,tr,0,∞,I3≔t−1/raρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞=fL,R;θ1,r,a∘ρ,r1,b1,q1,a1.
Note that due to Lemma10, we have(147)I3≻t−θ1−1/raρta1ts−1/r1b1sr1,t,∞Kt,fr,0,∞≔I4.
Hence, it is enough to show thatI1≺I4 and I2≺I4. Consider I1. By Lemma 13, we have(148)I1≈t−θ1−1/raρtχ1tχ0tb0ta0tKt,fr,0,∞=t−θ1−1/raρta1ts−1/r1b1sr1,t,∞s−1/r0b0sr0,t,∞b0tKt,fr,0,∞≺t−θ1−1/raρta1ts−1/r1b1sr1,t,∞Kt,fr,0,∞=I4.
ConsiderI2. By Lemma 12, we get(149)I2≈t−θ1−1/raρtχ1tKt,fr,0,∞=I4.
This completes the proof.☐The next assertion is a symmetric counterpart of Theorem33.Theorem 34 (cf. [8], Theorem 4.3).
Let0<θ0<θ1<1, 0<r,q0,r0,q1,r1≤∞, a,a0,b0,a1,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,0,t, χ1t=a1ts−1/r1b1sr1,0,t, and ρt=tθ1−θ0χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(150)A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1R0,r,a=A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(ii)
Ifs−1/rasr,0,1<∞, then(151)A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1R1,r,a=A¯θ1,r,a#,q1,a1R⋂A¯θ1,r,a∘ρ,r1,b1,q1,a1R,R,where a#t=s−1/r1b1sr1,0,taρt.Proof.
We prove the second formula. The first one can be proved similarly. Recall thatA¯=A0,A1, and we use the denotation b~t=bt−1. Note that(152)s−1/rib~isri,1,∞=s−1/ribisri,0,1<∞i=0,1,s−1/ra~sr,1,∞=s−1/rasr,0,1<∞.
Using Theorem33 (i) and formulae (14) and (15), we get(153)A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1R1,r,a=A1,A01−θ0,r0,b~0,q0,a~0L,A1,A01−θ1,r1,b~1,q1,a~1L1,r,a=A1,A01−θ1,r1,b~1,q1,a~1L,A1,A01−θ1,r0,b~0,q0,a~0L0,r,a~=A1,A01−θ1,r,c,q1,a~1L⋂A1,A01−θ1,r,a~∘ρ1,r0,b~0,q0,a~0L,L,where(154)ρ1t=tθ1−θ0a~1ts−1/r1b~1sr1,t,∞a~0ts−1/r0b~0sr0,t,∞=tθ1−θ0a1t−1s−1/r1b1sr1,0,t−1a0t−1s−1/r0b0sr0,0,t−1=1ρt−1=1ρ~t,ct=s−1/r0b~1sr0,t,∞a~ρ1t=s−1/r0b1sr0,0,t−1a~1ρ~t=s−1/r0b1sr0,0,t−1aρ~t=s−1/r0b1sr0,0,t−1aρt−1=a#t−1=a#~t.
Additionally, we have(155)a~∘ρ1t=a1ρ1t=aρ~t=a∘ρt−1.
Hence,a~∘ρ1~=a∘ρ. Therefore,(156)A1,A01−θ1,r,c,q1,a~1L=A¯θ1,r,a#,q1,a1R,
and by (16),(157)A1,A01−θ1,r,a~∘ρ1,r0,b~0,q0,a~0L,L=A¯θ1,r,a~∘ρ1~,r0,b0,q0,a0R,R=A¯θ1,r,a∘ρ,r0,b0,q0,a0R,R.
This completes the proof.☐Theorem 35 (cf. [8], Theorem 3.4).
Let0<θ0<θ1<1, 0<q0,r0,q1,r1≤∞, a0,b0,a1,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞, χ1t=a1ts−1/r1b1sr1,0,t, and ρt=tθ1−θ0χ0t/χ1t. Then, for all f∈A¯θ0,r0,b0,q0,a0L+A¯θ1,r1,b1,q1,a1R and t>0,(158)Kρt,f;A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1R≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t+ρts−1/r1b1sr1,0,tu−θ1−1/q1a1uKu,fq1,t,∞+ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,s,∞r1,t,∞.Proof.
As usual, we use the notation from the paper [15]. Let Φ0 be the function space corresponding to A¯θ0,r0,b0,q0,a0L:(159)F∗Φ0=s−1/r0b0su−θ0−1/q0a0uFuq0,0,sr0,0,∞.
From the proof of Theorem22, we know that(160)It,f≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,t+s−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,t,g0t≺h0t≈g0t+h0t≈t1−θ0χ0t.
LetΦ1 be the function space corresponding to A¯θ1,r1,b1,q1,a1R:(161)F∗Φ1=s−1/r1b1su−θ1−1/q1a1uFuq1,s,∞r1,0,∞.
Consider the functionsJt,f, g1t, and h1t.(162)Jt,f≔χt,∞∗K∗,fΦ1=s−1/r1b1su−θ1−1/q1a1uKu,fq1,maxs,t,∞r1,0,∞≈s−1/r1b1sr1,0,tu−θ1−1/q1a1uKu,fq1,t,∞+s−1/r1b1su−θ1−1/q1a1uKu,fq1,s,∞r1,t,∞,g1t≔tχt,∞∗Φ1=ts−1/r1b1su−θ1−1/q1a1uq1,maxs,t,∞r1,0,∞≈ts−1/r1b1sr1,0,tu−θ1−1/q1a1uq1,t,∞+s−1/r1b1su−θ1−1/q1a1uq1,s,∞r1,t,∞≈ts−1/r1b1sr1,0,tt−θ1a1t+s−θ1−1/r1b1sa1sr1,t,∞≈t1−θ1a1ts−1/r1b1sr1,0,t+b1t≈t1−θ1a1ts−1/r1b1sr1,0,t=t1−θ1χ1t,h1t≔∗χ0,t∗Φ1=s−1/r1b1su1−θ1−1/q1a1uq1,s,tr1,0,t≤s−1/r1b1sr1,0,tu1−θ1−1/q1a1uq1,0,t≈s−1/r1b1sr1,0,tt1−θ1a1t=t1−θ1χ1t.
Thus,(163)h1t≺g1t≈g1t+h1t≈t1−θ1χ1t.
Now, we estimate the functionsgt and ht. Because θ0−θ1<0, we get(164)1gt≔χt,∞uug0u+h0uΦ1≈s−1/r1b1suθ0−θ1−1/q1a1uχ0uq1,maxs,t,∞r1,0,∞≈s−1/r1b1sr1,0,tuθ0−θ1−1/q1a1uχ0uq1,t,∞+s−1/r1b1suθ0−θ1−1/q1a1uχ0uq1,s,∞r1,t,∞≈s−1/r1b1sr1,0,ttθ0−θ1a1tχ0t+sθ0−θ1−1/r1b1sa1sχ0sr1,t,∞≈tθ0−θ1a1tχ0ts−1/r1b1sr1,0,t+b1t≈tθ0−θ1a1tχ0ts−1/r1b1sr1,0,t=1ρt,ht≔χ0,tuug1u+h1uΦ0≈s−1/r0b0suθ1−θ0−1/q0a0uχ1uq0,0,mins,tr0,0,∞≈s−1/r0b0suθ1−θ0−1/q0a0uχ1uq0,0,sr0,0,t+s−1/r0b0sr0,t,∞uθ1−θ0−1/q0a0uχ1uq0,0,t≈sθ1−θ0−1/r0b0sa0sχ1sr0,0,t+s−1/r0b0sr0,t,∞tθ1−θ0a0tχ1t≈tθ1−θ0a0tχ1tb0t+s−1/r0b0sr0,t,∞≈tθ1−θ0a0tχ1ts−1/r0b0sr0,t,∞=ρt.
Thus,ht≈ρt≈gt. Therefore, using [15] (Theorem 4, Case 4), we arrive at (158).☐Theorem 36 (cf. [8], Theorem 4.6).
Let0<θ0<θ1<1, 0<r,q0,r0,q1,r1≤∞, a,a0,b0,a1,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞, χ1t=a1ts−1/r1b1sr1,0,t, and ρt=tθ1−θ0χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(165)A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1R0,r,a=A¯θ0,r,a#,q0,a0L⋂A¯θ0,r,a∘ρ,r0,b0,q0,a0L,L,where a#t=s−1/r0b0sr0,t,∞aρt.(ii)
Ifs−1/rasr,0,1<∞, then(166)A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1R1,r,a=A¯θ1,r,a#,q1,a1R⋂A¯θ1,r,a∘ρ,r1,b1,q1,a1R,R,where a#t=s−1/r1b1sr1,0,taρt.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0L, X1=A¯θ1,r1,b1,q1,a1R, and K¯t,f=Kt,f;X¯. We prove (i). Statement (ii) can be proved analogously. Let Z=X¯0,r,a. Using the change of variables and Theorem 35, we can write(167)fZ=u−1/rauK¯u,fr,0,∞≈t−1/raρtK¯ρt,fr,0,∞≈I1+I2+I3+I4,where(168)I1≔t−1/raρts−1/r0b0su−θ0−1/q0a0uKu,fq0,0,sr0,0,tr,0,∞=fL,L;θ0,r,a∘ρ,r0,b0,q0,a0,I2≔t−1/raρts−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,tr,0,∞=fL;θ0,r,a#,q0,a0,I3≔tθ1−θ0−1/raρtχ0ta1tu−θ1−1/q1a1uKu,fq1,t,∞r,0,∞,I4≔tθ1−θ0−1/raρtχ0tχ1ts−1/r1b1su−θ1−1/q1a1uKu,fq1,s,∞r1,t,∞r,0,∞.
It is enough to show thatI3≺I2 and I4≺I1. By Lemmas 12 and 9, we get(169)I3≈t−θ0−1/raρta0ts−1/r0b0sr0,t,∞Kt,fr,0,∞≺t−1/raρts−1/r0b0sr0,t,∞u−θ0−1/q0a0uKu,fq0,0,tr,0,∞=I2.
Using Lemmas13 and 10, we arrive at(170)I4≈t−θ0−1/raρtχ0tχ1tb1ta1tKt,fr,0,∞=t−θ0−1/raρtχ0tb1ts−1/r1b1sr1,0,tKt,fr,0,∞≺t−θ0−1/raρta0ts−1/r0b0sr0,t,∞Ku,fr,0,∞≺t−1/raρts−1/r0b0su−θ0−1/qa0uKu,fq,t,sr0,t,∞r,0,∞≤t−1/raρts−1/r0b0su−θ0−1/qa0uKu,fq,0,sr0,t,∞r,0,∞=I1.
This completes the proof.☐Theorem 37 (cf. [8], Theorem 3.3).
Let0<θ0<θ1<1, 0<q0,r0,q1,r1≤∞, a0,b0,a1,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,0,t, χ1t=a1ts−1/r1b1sr1,t,∞, and ρt=tθ1−θ0χ0t/χ1t. Then, for all f∈A¯θ0,r0,b0,q0,a0R+A¯θ1,r1,b1,q1,a1L and t>0,(171)Kρt,f;A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1L≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,s,tr0,0,t+ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞.Proof.
LetΦ0 be the function space corresponding to A¯θ0,r0,b0,q0,a0R:(172)F∗Φ0=s−1/r0b0su−θ0−1/q0a0uFuq0,s,∞r0,0,∞.
Consider the functionsIt,f, g0t, and h0t.(173)It,f≔χ0,t∗K∗,fΦ0=s−1/r0b0su−θ0−1/q0a0uKu,fq0,s,tr0,0,t,g0t≔tχt,∞∗Φ0=ts−1/r0b0su−θ0−1/q0a0uq0,maxt,s,∞r0,0,∞≈ts−1/r0b0sr0,0,tu−θ0−1/q0a0uq0,t,∞+s−1/r0b0su−θ0−1/q0a0uq0,s,∞r0,t,∞≈ts−1/r0b0sr0,0,tt1−θ0a0t+s−θ0−1/r0a0sb0sr0,t,∞≈t1−θ0a0ts−1/r0b0sr0,0,t+b0t≈t1−θ0a0ts−1/r0b0sr0,0,t=t1−θ0χ0t,(174)h0t≔∗χ0,t∗Φ0=s−1/r0b0su1−θ0−1/q0a0uq0,s,tr0,0,∞≤u1−θ0−1/q0a0uq0,0,ts−1/r0b0sr0,0,∞≈t1−θ0a0ts−1/r0b0sr0,0,∞=t1−θ0χ0t.
Thus,(175)h0t≺g0t≈g0t+h0t≈t1−θ0χ0t.
LetΦ1 be the function space corresponding to A¯θ1,r1,b1,q1,a1L:(176)F∗Φ1=s−1/r1b1su−θ1−1/q1a1uFuq1,0,sr1,0,∞.
From the proof of Theorem18, we know that(177)Jt,f=s−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞,g1t≺h1t≈g1t+h1t≈t1−θ1χ1t.
Now, we estimate the functionsgt and ht.(178)1gt≔χt,∞uug0u+h0uΦ1≈s−1/r1b1suθ0−θ1−1/q1a1uχ0uq1,t,sr1,t,∞≤uθ0−θ1−1/q1a1uχ0uq1,t,∞s−1/r1b1sr1,t,∞≈tθ0−θ1a1tχ0ts−1/r1b1sr1,t,∞=1ρt,ht≔χ0,tuug1u+h1uΦ0≈s−1/r0b0suθ1−θ0−1/q0a0uχ1uq0,s,tr0,0,∞≤uθ1−θ0−1/q0a0uχ1uq0,0,ts−1/r0b0sr0,0,∞≈tθ1−θ0a0tχ1ts−1/r0b0sr0,0,∞=ρt.
Becauseht≺ρt≺gt, using [15] (Theorem 4, Case 2), we arrive at(179)Kρt,f;A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1L≈s−1/r0b0su−θ0−1/q0a0uKu,fq0,s,tr0,0,t+ρts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞+t−θ0χ0tKt,f.
By Lemma10, we get(180)t−θ0χ0tKt,f≺s−1/r0b0su−θ0−1/q0a0uKu,fq0,s,tr0,0,t.
Thus, (171) is proved.☐Theorem 38 (cf. [8], Theorem 4.5).
Let0<θ0<θ1<1, 0<r,q0,r0,q1,r1≤∞, a,a0,b0,a1,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,0,t, χ1t=a1ts−1/r1b1sr1,t,∞, and ρt=tθ1−θ0χ0t/χ1t.(i)
If0<θ<1, then(181)A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1Lθ,r,a=A¯η,r,a#,where η=1−θθ0+θθ1 and a#t=χ0t1−θχ1tθaρt.(ii)
Ifs−1/rasr,1,∞<∞, then(182)A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1L0,r,a=A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(iii)
Ifs−1/rasr,0,1<∞, then(183)A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1L1,r,a=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0R, X1=A¯θ1,r1,b1,q1,a1L, K¯t,f=Kt,f;X¯, and Z=X¯θ,r,a (0≤θ≤1). Using the change of variables and Theorem 37, we can write(184)fZ=u−θ−1/rauK¯u,fr,0,∞≈t−1/rρt−θaρtK¯ρt,fr,0,∞≈I1+I2,where(185)I1≔t−θθ1−θ0−1/rχ1tχ0tθaρts−1/r0b0su−θ0−1/q0a0uKu,fq0,s,tr0,0,tr,0,∞,I2≔t1−θθ1−θ0−1/rχ1tχ0tθ−1aρt∗s−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞.
First, let0<θ<1. In this case, −θθ1−θ0=θ0−η and 1−θθ1−θ0=θ1−η. By Lemma 14 (i), we get(186)I1≈t−η−1/rχ1tχ0tθaρta0ts−1/r0b0sr0,0,tKt,fr,0,∞=t−η−1/ra#tKt,fr,0,∞=fη,r,a#.
So, it is enough to show thatI2≺fη,r,a#. Note that(187)s−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞≤s−1/r1b1sr1,t,∞u−θ1−1/q1a1uKu,fq1,t,∞.
Thus, by Lemma12, we get(188)I2≤tθ1−η−1/rχ1tχ0tθ−1aρts−1/r1b1sr1,t,∞u−θ1−1/q1a1uKu,fq1,t,∞r,0,∞≈t−η−1/rχ1tχ0tθ−1aρta1ts−1/r1b1sr1,t,∞Kt,fr,0,∞=t−η−1/ra#tKt,fr,0,∞=fη,r,a#.
Now, we prove (ii). In this case,(189)I1=t−1/raρts−1/r0b0su−θ0−1/q0a0uKu,fq0,s,tr0,0,tr,0,∞=fR,L;θ0,r,a∘ρ,r0,b0,q0,a0,I2=tθ1−θ0−1/raρtχ0tχ1ts−1/r1b1su−θ1−1/q1a1uKu,fq1,t,sr1,t,∞r,0,∞.
It is enough to show thatI2≺I1. Using Lemmas 12 and 10, we have(190)I2≤tθ1−θ0−1/raρtχ0tχ1ts−1/r1b1sr1,t,∞u−θ1−1/q1a1uKu,fq1,t,∞r,0,∞≈t−θ0−1/raρta0ts−1/r0b0sr0,0,tKt,fr,0,∞≺t−1/raρts−1/r0b0su−θ0−1/q0a0uKu,fq0,s,tr0,0,tr,0,∞=I1.
The proof of (iii) follows similar steps.☐Remark 39.
Assertion (i) of Theorem38 coincides with [5] (Theorem 29). But there it has been wrongly stated that it can be proved based on symmetry arguments. The correct proof is presented above.


## 8. Reiteration Formulae for Couples Where One of the Operands IsLL, LR, RL, or RR Space

In this section, we establish reiteration formulae for the couples in which one of the operands isLL, LR, RL, or RR space. Here, we have chosen an indirect method of the proofs. Compared to the direct method, i.e., calculation of the corresponding Holmstedt-type formulae, we subsequently apply known formulae but unfortunately often arrive at weaker results (embeddings instead of isomorphisms).Theorem 40.
Let0<θ0<θ1<1, 0<r,r0,p1,q1,r1≤∞, a,b0,a1,b1,c1∈SV, s−1/r1b1sr1,1,∞<∞, and u−1/p1c1up1,0,1<∞. Put χ1t=a1ts−1/r1b1sr1,t,∞u−1/p1c1up1,0,t and ρt=tθ1−θ0b0t/χ1t.(i)
If0<θ<1, then(191)A¯θ0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a=A¯η,r,a#,where η=1−θθ0+θθ1 and a#t=b0t1−θχ1tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(192)A¯θ0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1L,R0,r,a=A¯θ0,r,a∘ρ,r0,b0L.(iii)
Ift−1/ratr,0,1<∞, then(193)A¯θ0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1L,R1,r,a⊂A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A¯θ0,r0,b0, X1=A¯θ1,r1,b1,q1,a1L, and ψt=a1ts−1/r1b1sr1,t,∞. Let σ be a strongly increasing, differentiable function such that σt≈tθ1−θ0b0t/ψt and σ1=1. Additionally, denote x=c1∘σ−1, Z=A¯θ0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a, and yt=s−1/p1xsp1,0,tθat/s−1/p1xsp1,0,t (0≤θ≤1). By the change of variables u=σ−1s, we get s−1/p1xsp1,0,σt≈u−1/p1c1up1,0,t. Hence, s−1/p1xsp1,0,1<∞ and(194)yσt≈u−1/p1c1up1,0,tθaρt.
By Theorem19, we obtain(195)A¯θ1,p1,c1,r1,b1,q1,a1L,R=A¯θ0,r0,b0,A¯θ1,r1,b1,q1,a1L1,p1,x.
So,Z=X0,X¯1,p1,xθ,r,a. Using Theorem 17, we conclude that for 0≤θ<1, it holds(196)Z=X¯θ,r,y=A¯θ0,r0,b0,A¯θ1,r1,b1,q1,a1Lθ,r,y.
If0<θ<1, by [5] (Theorem 19) (i) and (194), we get Z=A¯η,r,a#, where(197)a#t=b0t1−θa1ts−1/r1b1sr1,t,∞θyσt≈b0t1−θa1ts−1/r1b1sr1,t,∞u−1/p1c1up1,0,tθaρt=b0t1−θχ1tθaρt.
Ifθ=0, by [5] (Theorem 19) (ii), we get(198)Z=A¯θ0,r0,b0,A¯θ1,r1,b1,q1,a1L0,r,y=A¯θ0,r,a∘ρ,r0,b0L,while due to (194), y∘σ≈a∘ρ. Note that t−1/rytr,1,∞<∞.
Ifθ=1, using Theorems 17 and 19, we conclude that(199)Z=X0,X¯1,p1,x1,r,a⊂X¯1,r,y=A¯θ0,r0,b0,A¯θ1,r1,b1,q1,a1L1,r,y=A¯θ1,r,y∘σ,r1,b1,q1,a1L,R=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.
This completes the proof.☐Theorem 41.
Let0<θ1<1, 0<r,r0,p1,q1,r1≤∞, a,b0,a1,b1,c1∈SV, s−1/r0b0sr0,1,∞<∞, s−1/r1b1sr1,1,∞<∞, and u−1/p1c1up1,0,1<∞. Put χ0t=s−1/r0b0sr0,t,∞, χ1t=a1ts−1/r1b1sr1,t,∞u−1/p1c1up1,0,t, and ρt=tθ1χ0t/χ1t.(i)
If0<θ<1, then(200)A¯0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a=A¯η,r,a#,where η=θθ1 and a#t=χ0t1−θχ1tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(201)A¯0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1L,R0,r,a=A¯0,r,a#⋂A¯0,r,a∘ρ,r0,b0L,where a#t=χ0taρt.(iii)
Ift−1/ratr,0,1<∞, then(202)A¯0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1L,R1,r,a⊂A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A¯0,r0,b0, X1=A¯θ1,r1,b1,q1,a1L, and ψt=a1ts−1/r1b1sr1,t,∞. Let σ be a strongly increasing, differentiable function such that σt≈tθ1χ0t/ψt and σ1=1. Additionally, denote x=c1∘σ−1, Z=A¯0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a, and yt=s−1/p1xsp1,0,tθat/s−1/p1xsp1,0,t (0≤θ≤1). By the change of variables u=σ−1s, we get s−1/p1xsp1,0,σt≈u−1/p1c1up1,0,t and(203)yσt≈u−1/p1c1up1,0,tθaρt.
By Theorem20 (ii), we obtain(204)A¯θ1,p1,c1,r1,b1,q1,a1L,R≅A¯0,r0,b0,A¯θ1,r1,b1,q1,a1L1,p1,x.
Hence,Z=X0,X¯1,p1,xθ,r,a. Using Theorem 17, we conclude that for 0≤θ<1, it holds(205)Z=X¯θ,r,y=A¯0,r0,b0,A¯θ1,r1,b1,q1,a1Lθ,r,y.
If0<θ<1, by [5] (Theorem 17) and (203), we get(206)Z=A¯0,r0,b0,A¯θ1,r1,b1,q1,a1Lθ,r,y≅A¯η,r,a#,where η=θθ1 and a#t=χ0t1−θψtθyσt≈χ0t1−θχ1tθaρt.
Ifθ=0, using Theorem 20 (i), we get(207)Z=A¯0,r0,b0,A¯θ1,r1,b1,q1,a1L0,r,y=A¯0,r,a#⋂A¯0,r,a∘ρ,r0,b0L,where a#t=χ0tyσt≈χ0taρt.
Ifθ=1, using Theorems 17 and 20, we conclude that(208)Z=X0,X¯1,p1,x1,r,a⊂X¯1,r,y=A¯0,r0,b0,A¯θ1,r1,b1,q1,a1L1,r,y=A¯θ1,r,y∘σ,r1,b1,q1,a1L,R=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.
This completes the proof.☐Theorem 42.
Let0<θ1<1, 0<r,p1,q1,r1≤∞, a,a1,b1,c1∈SV, s−1/r1b1sr1,1,∞<∞, and u−1/p1c1up1,0,1<∞. Put χ1t=a1ts−1/r1b1sr1,t,∞u−1/p1c1up1,0,t and ρt=tθ11/χ1t.(i)
If0<θ<1, then(209)A0,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a=A¯η,r,a#,where η=θθ1 and a#t=χ1tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(210)A0,A¯θ1,p1,c1,r1,b1,q1,a1L,R0,r,a=A¯0,r,a∘ρ.(iii)
Ift−1/ratr,0,1<∞, then(211)A0,A¯θ1,p1,c1,r1,b1,q1,a1L,R1,r,a⊂A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A0, X1=A¯θ1,r1,b1,q1,a1L, and ψt=a1ts−1/r1b1sr1,t,∞. Let σ be a strongly increasing, differentiable function such that σt≈tθ1/ψt andσ1=1. Additionally, denote x=c1∘σ−1, Z=A0,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a, and yt=s−1/p1xsp1,0,tθat/s−1/p1xsp1,0,t (0≤θ<1). By the change of variables u=σ−1s, we get s−1/p1xsp1,0,σt≈u−1/p1c1up1,0,t and(212)yσt≈u−1/p1c1up1,0,tθaρt.
By Theorem21, we obtain(213)A¯θ1,p1,c1,r1,b1,q1,a1L,R=A0,A¯θ1,r1,b1,q1,a1L1,p1,x.
Hence,Z=X0,X¯1,p1,xθ,r,a.
If0≤θ<1, by Theorem 17, [5] (Theorem 16), and (212), we get(214)Z=X¯θ,r,y=A0,A¯θ1,r1,b1,q1,a1Lθ,r,y=A¯η,r,a#,where η=θθ1 and a#t=ψtθyσt≈χ1tθaρt.
Ifθ=1, using Theorems 17 and 21, we conclude that(215)Z=X0,X¯1,p1,x1,r,a⊂X¯1,r,y=A0,A¯θ1,r1,b1,q1,a1L1,r,y=A¯θ1,r,y∘σ,r1,b1,q1,a1L,R=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.
This completes the proof.☐Theorem 43.
Let0<θ0<θ1<1, 0<r,q0,r0,p1,q1,r1≤∞, a,a0,b0,a1,b1,c1∈SV, s−1/r0b0sr0,1,∞<∞, s−1/r1b1sr1,1,∞<∞, and u−1/p1c1up1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞, χ1t=a1ts−1/r1b1sr1,t,∞u−1/p1c1up1,0,t, and ρt=tθ1−θ0χ0t/χ1t.(i)
If0<θ<1, then(216)A¯θ0,r0,b0,q0,a0L,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a=A¯η,r,a#,where η=1−θθ0+θθ1 and a#t=χ0t1−θχ1tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(217)A¯θ0,r0,b0,q0,a0L,A¯θ1,p1,c1,r1,b1,q1,a1L,R0,r,a=A¯θ0,r,a#,q0,a0L⋂A¯θ0,r,a∘ρ,r0,b0,q0,a0L,L,where a#t=s−1/r0b0sr0,t,∞aρt.(iii)
Ift−1/ratr,0,1<∞, then(218)A¯θ0,r0,b0,q0,a0L,A¯θ1,p1,c1,r1,b1,q1,a1L,R1,r,a⊂A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0L, X1=A¯θ1,r1,b1,q1,a1L, and ψt=a1ts−1/r1b1sr1,t,∞. Let σ be a strongly increasing, differentiable function such that σt≈tθ1−θ0χ0t/ψt and σ1=1. Additionally, denote x=c1∘σ−1, Z=A¯θ0,r0,b0,q0,a0L,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a, and yt=s−1/p1xsp1,0,tθat/s−1/p1xsp1,0,t. By the change of variables u=σ−1s, we get s−1/p1xsp1,0,σt≈u−1/p1c1up1,0,t and(219)yσt≈u−1/p1c1up1,0,tθaρt.
By Theorem33 (ii), we obtain(220)A¯θ1,p1,c1,r1,b1,q1,a1L,R=A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1L1,p1,x.
Hence,Z=X0,X¯1,p1,xθ,r,a. Using Theorem 17, we conclude that for 0≤θ<1, it holds(221)Z=X¯θ,r,y=A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1Lθ,r,y.
If0<θ<1, by [5] (Theorem 26) (i) and (219), we get Z=A¯η,r,a#, where a#t=χ0t1−θψtθyσt≈χ0t1−θχ1tθaρt.
Ifθ=0, by Theorem 33 (i), we get(222)Z=A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1L0,r,y=A¯θ0,r,a#,q0,a0L⋂A¯θ0,r,a∘ρ,r0,b0,q0,a0L,L,where a#t=s−1/r0b0sr0,t,∞yσt≈s−1/r0b0sr0,t,∞aρt.
Ifθ=1, using Theorems 17 and 33 (ii), we conclude that(223)Z=X0,X¯1,p1,x1,r,a⊂X¯1,r,y=A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1,q1,a1L1,r,y=A¯θ1,r,y∘σ,r1,b1,q1,a1L,R=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.
This completes the proof.☐Theorem 44.
Let0<θ0<θ1<1, 0<r,r0,q0,p1,q1,r1≤∞, a,a0,a1,b0,b1∈SV, s−1/r0b0sr0,0,1<∞, s−1/r1b1sr1,1,∞<∞, and u−1/p1c1up1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,0,t, χ1t=a1ts−1/r1b1sr1,t,∞u−1/p1c1up1,0,t, and ρt=tθ1−θ0χ0t/χ1t.(i)
If0<θ<1, then(224)A¯θ0,r0,b0,q0,a0R,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a=A¯η,r,a#,where η=1−θθ0+θθ1 and a#t=χ0t1−θχ1tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(225)A¯θ0,r0,b0,q0,a0R,A¯θ1,p1,c1,r1,b1,q1,a1L,R0,r,a=A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(iii)
Ift−1/ratr,0,1<∞, then(226)A¯θ0,r0,b0,q0,a0R,A¯θ1,p1,c1,r1,b1,q1,a1L,R1,r,a⊂A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0R, X1=A¯θ1,r1,b1,q1,a1L, and ψt=a1ts−1/r1b1sr1,t,∞. Let σ be a strongly increasing, differentiable function such that σt≈tθ1−θ0χ0t/ψt and σ1=1. Additionally, denote x=c1∘σ−1, Z=A¯θ0,r0,b0,q0,a0R,A¯θ1,p1,c1,r1,b1,q1,a1L,Rθ,r,a, and yt=s−1/p1xsp1,0,tθat/s−1/p1xsp1,0,t. By the change of variables u=σ−1s, we get s−1/p1xsp1,0,σt≈u−1/p1c1up1,0,t and(227)yσt≈u−1/p1c1up1,0,tθaρt.
By Theorem38 (iii), we obtain(228)A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1L1,p1,x=A¯θ1,p1,c1,r1,b1,q1,a1L,R.
Hence,Z=X0,X¯1,p1,xθ,r,a. Using Theorem 17, we conclude that for 0≤θ<1, it holds(229)Z=X¯θ,r,y=A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1Lθ,r,y.
If0<θ<1, by Theorem 38 (i) and (227), we get Z=A¯η,r,a#, where a#t=χ0t1−θψtθyσt≈χ0t1−θχ1tθaρt.
Ifθ=0, by Theorem 38 (ii), we get(230)Z=A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1L0,r,y=A¯θ0,r,y∘σ,r0,b0,q0,a0R,L=A¯θ0,r,a#,r0,b0,q0,a0R,L,where a#t=yσt≈aρt.
Ifθ=1, using Theorems 17 and 38 (iii), we conclude that(231)Z=X0,X¯1,p1,x1,r,a⊂X¯1,r,y=A¯θ0,r0,b0,q0,a0R,A¯θ1,r1,b1,q1,a1L1,r,y=A¯θ1,r,y∘σ,r1,b1,q1,a1L,R=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.
This completes the proof.☐From the previous theorems in this section, using the symmetry arguments given by the formulae (14)–(16), the following five theorems can be proved.Theorem 45.
Let0<θ0<θ1<1, 0<r,p0,q0,r0,r1≤∞, a,b0,a0,c0,b1∈SV, s−1/r0b0sr0,0,1<∞, and u−1/p0c0up0,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,0,tu−1/p0c0up0,t,∞ and ρt=tθ1−θ0χ0t/b1t.(i)
If0<θ<1, then(232)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b1θ,r,a=A¯η,r,a#,where η=1−θθ0+θθ1 and a#t=χ0t1−θb1tθaρt.(ii)
Ift−1/ratr,0,1<∞, then(233)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b10,r,a⊂A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(iii)
Ift−1/ratr,0,1<∞, then(234)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b11,r,a=A¯θ1,r,a∘ρ,r1,b1R.Theorem 46.
Let0<θ0<1, 0<r,p0,q0,r0,r1≤∞, a,a0,b0,b1,c0∈SV, s−1/r0b0sr0,0,1<∞, u−1/p0c0up0,1,∞<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,0,tu−1/p0c0up0,t,∞, χ1t=s−1/r1b1sr1,0,t, and ρt=t1−θ0χ0t/χ1t.(i)
If0<θ<1, then(235)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯1,r1,b1θ,r,a=A¯η,r,a#,where η=1−θθ0+θ and a#t=χ0t1−θχ1tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(236)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯1,r1,b10,r,a⊂A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(iii)
Ift−1/ratr,0,1<∞, then(237)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯1,r1,b11,r,a=A¯1,r,a#⋂A¯1,r,a∘ρ,r1,b1R,where a#t=χ1taρt.Theorem 47.
Let0<θ0<1, 0<r,p0,q0,r0≤∞, a,a0,b0,c0∈SV, s−1/r0b0sr0,0,1<∞, and u−1/p0c0up0,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,0,tu−1/p0c0up0,t,∞ and ρt=t1−θ0χ0t.(i)
If0<θ<1, then(238)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A1θ,r,a=A¯η,r,a#,where η=1−θθ0+θ and a#t=χ0t1−θaρt.(ii)
Ift−1/ratr,1,∞<∞, then(239)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A10,r,a⊂A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(iii)
Ift−1/ratr,0,1<∞, then(240)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A11,r,a=A¯1,r,a∘ρ.Theorem 48.
Let0<θ0<θ1<1, 0<r,p0,q0,r0,r1,q1≤∞, a,a0,b0,a1,b1,c0∈SV, s−1/r0b0sr0,0,1<∞, u−1/p0c0up0,1,∞<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=a0ts−1/r0b0sr0,0,tu−1/p0c0up0,t,∞, χ1t=a1ts−1/r1b1sr1,0,t, and ρt=tθ1−θ0χ0t/χ1t.(i)
If0<θ<1, then(241)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b1,q1,a1Rθ,r,a=A¯η,r,a#,where η=1−θθ0+θθ1 and a#t=χ0t1−θχ1tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(242)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b1,q1,a1R0,r,a⊂A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(iii)
Ift−1/ratr,0,1<∞, then(243)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b1,q1,a1R1,r,a=A¯θ1,r,a#,q1,a1R⋂A¯θ1,r,a∘ρ,r1,b1,q1,a1R,R,where a#t=s−1/r1b1sr1,0,taρt.Theorem 49.
Let0<θ0<θ1<1, 0<r,p0,q0,r0,r1,q1≤∞, a0,a1,b0,b1∈SV, s−1/r0b0sr0,0,1<∞, u−1/p0c0up0,1,∞<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,0,tu−1/p0c0up0,t,∞, χ1t=a1ts−1/r1b1sr1,t,∞, and ρt=tθ1−θ0χ0t/χ1t.(i)
If0<θ<1, then(244)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b1,q1,a1Lθ,r,a=A¯η,r,a#,where η=1−θθ0+θθ1 and a#t=χ0t1−θχ1tθaρt.(ii)
Ift−1/ratr,1,∞<∞, then(245)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b1,q1,a1L0,r,a⊂A¯θ0,r,a∘ρ,r0,b0,q0,a0R,L.(iii)
Ift−1/ratr,0,1<∞, then(246)A¯θ0,p0,c0,r0,b0,q0,a0R,L,A¯θ1,r1,b1,q1,a1L1,r,a=A¯θ1,r,a∘ρ,r1,b1,q1,a1L,R.The technique used in the section allows considering also some couples, where one of the operands isLL or RR space. In these cases, though, only “reiteration embedding” can be proved.Theorem 50.
Let0<θ0<θ1<1, 0<r,p0,q0,r0,r1≤∞, a,a0,b0,c0,b1∈SV, s−1/r0b0sr0,1,∞<∞, u−1/p0c0up0,1,∞<∞, and s−1/rasr,1,∞<∞. Put χ0t=a0ts−1/r0b0sr0,t,∞u−1/p0c0up0,t,∞ and ρt=tθ1−θ0χ0t/b1t.(i)
If0<θ<1, then(247)A¯θ0,p0,c0,r0,b0,q0,a0L,L,A¯θ1,r1,b1θ,r,a⊃A¯η,r,a#,where η=1−θθ0+θθ1 and a#t=χ0t1−θb1tθaρt.(ii)
Ift−1/ratr,0,1<∞, then(248)A¯θ0,p0,c0,r0,b0,q0,a0L,L,A¯θ1,r1,b11,r,a⊃A¯θ1,r,a∘ρ,r1,b1R.Proof.
DenoteX0=A¯θ0,r0,b0,q0,a0L, X1=A¯θ1,r1,b1, and ψt=a0ts−1/r0b0sr0,t,∞. Let σ be a strongly increasing, differentiable function such that σt≈tθ1−θ0ψt/b1t and σ1=1. Additionally, denote x=c0∘σ−1, Z=A¯θ0,p0,c0,r0,b0,q0,a0L,L,A¯θ1,r1,b1θ,r,a, and yt=s−1/p0xsp0,t,∞1−θats−1/p0xsp0,t,∞. By the change of variables u=σ−1s, we get s−1/p0xsp0,σt,∞≈u−1/p0c0up0,t,∞ and(249)yσt≈u−1/p0c0up0,t,∞1−θaρt.
By Theorem23, we obtain(250)A¯θ0,p0,c0,r0,b0,q0,a0L,L⊃A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b10,p0,x.
Thus,Z⊃X¯0,p0,x,X1θ,r,a. Using Theorem 16, we conclude that for 0<θ≤1, it holds(251)X¯0,p0,x,X1θ,r,a=X¯θ,r,y=A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b1θ,r,y.
If0<θ<1, by [5] (Theorem 19) (i) and (249), we get Z⊃A¯η,r,a#, where a#t=ψt1−θb1tθyσt≈χ0t1−θb1tθaρt.
Letθ=1. Using [5] (Theorem 11) (ii), we arrive at(252)Z⊃A¯θ0,r0,b0,q0,a0L,A¯θ1,r1,b11,r,y=A¯θ1,r,y∘σ,r1,b1R=A¯θ1,r,a∘ρ,r1,b1R.☐Similar theorems can be formulated and proved also for the couplesA¯θ0,p0,c0,r0,b0,q0,a0L,L,A¯1,r1,b1, A¯θ0,p0,c0,r0,b0,q0,a0L,L,A1, A¯θ0,p0,c0,r0,b0,q0,a0L,L,A¯θ1,r1,b1,q1,a1L, A¯θ0,p0,c0,r0,b0,q0,a0L,L,A¯θ1,r1,b1,q1,a1R, A¯θ0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1R,Rθ,r,a, A¯0,r0,b0,A¯θ1,p1,c1,r1,b1,q1,a1R,R, A0,A¯θ1,p1,c1,r1,b1,q1,a1R,R, A¯θ0,r0,b0,q0,a0R,A¯θ1,p1,c1,r1,b1,q1,a1R,R, and A¯θ0,r0,b0,q0,a0L,A¯θ1,p1,c1,r1,b1,q1,a1R,R. We leave this to the reader.


## 9. Applications

Here, we demonstrate how our general reiteration theorems can be used to establish limiting interpolation results for the grand and small Lorentz spaces. For the sake of shortness, we present only some possible results.LetΩ,μ denote a σ-finite measure space with a nonatomic measure μ. We consider functions f from the set MΩ,μ of all μ-measurable functions on Ω. As conventional (see, e.g., [1]), f∗t (t>0) denotes the nonincreasing rearrangement of f and the maximal function of f∗ is defined by(253)f∗∗t≔1t∫0tf∗udu.For simplicity, we consider below only interpolation spaces betweenL1 and L∞. For functions from these spaces, Peetre has shown that ([1], Theorem V.1.6)(254)Kt,f;L1,L∞=∫0tf∗sds=tf∗∗t.The following assertion is a modification of [10] (Lemma 5.2) and can be proved similarly.Lemma 51.
Let0<θ≤1, 0<q≤∞, b∈SV, and p=1/1−θ. If θ=1, we additionally suppose that u−1/qbuq,0,1<∞. Then, for all f∈L1+L∞ and t>0,(255)u−θ−1/qbuKu,f;L1,L∞q,0,t=u1/p−1/qbuf∗∗uq,0,t≈u1/p−1/qbuf∗uq,0,t.


### 9.1. Function Spaces

First, we define function spaces under consideration.Definition 52.
Let0<p,q≤∞ and b∈SV. The Lorentz–Karamata space Lp,q;b is the set of all f∈MΩ,μ such that(256)fp,q;α≔t1/p−1/qbtf∗tq,0,∞<∞.
The Lorentz–Karamata spaces form an important scale of spaces. It contains, e.g., the Lebesgue spaceLp, Lorentz space Lp,q, Lorentz–Zygmund, and the generalized Lorentz–Zygmund spaces. For further information about Lorentz–Karamata spaces, we refer to, e.g., [10, 11, 16]. They have found many different important applications in analysis (see, e.g., [1, 4, 10–12, 14, 16–19] and the references therein).Lemma51 implies the following interpolation result.Lemma 53 (cf. [10], Corollary 5.3; [9]).
Let0<θ≤1, 0<q≤∞, b∈SV, and p=1/1−θ. Then, L1,L∞θ,q,b=Lp,q,b.Definition 54 (cf. [10], (5.21), (5.33)).
Let1<p<∞, 0<q,r≤∞, and a,b∈SV. The spaces Lp,r,b,q,aL and Lp,r,b,q,aR are the sets of all f∈MΩ,μ such that(257)fLp,r,b,q,aL≔t−1/rbtu1/p−1/qauf∗uq,0,tr,0,∞<∞,or(258)fLp,r,b,q,aR≔t−1/rbtu1/p−1/qauf∗∗uq,t,∞r,0,∞<∞,correspondingly.
We will require thatt−1/rbtr,1,∞<∞ for Lp,r,b,q,aL and t−1/rbtr,0,1<∞ for Lp,r,b,q,aR. Otherwise, the corresponding space consists only of the null element. Similar definitions can be found in [8, 9, 17]. We refer to the spaces L∗L and L∗R as LL and LR spaces, respectively. Note that the spaces L∗L are a special case of the generalized gamma space with double weights [3].In order to be able to compare our results with the results from [3, 4, 7–9], we introduce the generalized grand and small Lorentz spaces. Note that the interpolation results in [3, 4] do not contain limiting cases θ∈0,1.Definition 55.
Let1<p<∞, 0<q,r≤∞, and b∈SV. We define the small Lorentz space Lbp,q,r as the set of all f∈MΩ,μ such that(259)fLbp,q,r≔t−1/rbtu1/p−1/qf∗uq,0,tr,0,∞<∞.
We define the grand Lorentz spaceLbp,q,r as the set of all f∈MΩ,μ such that(260)fLbp,q,r≔t−1/rbtu1/p−1/qf∗∗uq,t,∞r,0,∞<∞.
It is clear thatLbp,q,r=Lp,r,b,q,1L and Lbp,q,r=Lp,r,b,q,1R. Grand and small Lebesgue and Lorentz spaces find many different important applications and have been intensively studied by many authors (see [3, 4, 6–9] and the references therein). Normally, they are defined on a bounded domain Ω in Rn with measure 1. In [4], it is required that the functions are real-valued. We do not require that μΩ=1 and neither that the functions are real-valued.The following lemma characterizes theLL and LR spaces as appropriate L and R limiting interpolation spaces. It follows from the corresponding definitions, formula (254), and Lemma 51.Lemma 56 (see [10], Lemmas 5.4 and 5.9).
Let1<p<∞, 0<q,r≤∞, a,b∈SV, and θ=1−1/p. Then, Lp,r,b,q,aL=L1,L∞θ,r,b,q,aL and Lp,r,b,q,aR=L1,L∞θ,r,b,q,aR.
In particular,Lbp,q,r=L1,L∞θ,r,b,q,1L and Lbp,q,r=L1,L∞θ,r,b,q,1R.Definition 57.
Let1<p<∞, 0<q,r,s≤∞, and a,b,c∈SV. The spaces Lp,s,c,r,b,q,aL,L, Lp,s,c,r,b,q,aL,R, Lp,s,c,r,b,q,aR,L, and Lp,s,c,r,b,q,aR,R are the sets of all f∈MΩ,μ such that(261)fLp,s,c,r,b,q,aL,L≔u−1/scut−1/rbtv1/p−1/qavf∗vq,0,tr,0,us,0,∞<∞,(262)fLp,s,c,r,b,q,aL,R≔u−1/scut−1/rbtv1/p−1/qavf∗∗vq,u,tr,u,∞s,0,∞<∞,(263)fLp,s,c,r,b,q,aR,L≔u−1/scut−1/rbtv1/p−1/qavf∗∗vq,t,ur,0,us,0,∞<∞,or(264)fLp,s,c,r,b,q,aR,R≔u−1/scut−1/rbtv1/p−1/qavf∗∗vq,t,∞r,u,∞s,0,∞<∞,correspondingly.Remark 58.
Because of Lemma51, if in the formulae (257), (259), and (261) f∗ will be replaced with f∗∗, one gets equivalent (quasi-)norms for corresponding spaces.In view of Lemma51, Remark 58, and (254), the following analogue of Lemma 56 holds.Lemma 59.
Let1<p<∞, 0<q,r,s≤∞, a,b∈SV, and θ=1−1/p. Then,(265)Lp,s,c,r,b,q,aL,L=L1,L∞θ,s,c,r,b,q,aL,L,Lp,s,c,r,b,q,aL,R=L1,L∞θ,s,c,r,b,q,aL,R,Lp,s,c,r,b,q,aR,L=L1,L∞θ,s,c,r,b,q,aR,L,Lp,s,c,r,b,q,aR,R=L1,L∞θ,s,c,r,b,q,aR,R.


### 9.2. Limiting Interpolation between the Small or Great Lorentz Spaces and the Lorentz–Karamata Spaces

Using results from Sections4–6 and Lemmas 56 and 59, we are able to characterize some limiting interpolation spaces lying between LL or LR spaces and Lorentz–Karamata spaces. In the corollaries below, we restrict ourselves to the grand and small Lorentz spaces.Corollary 60 (cf. [9], Corollary 7.8; [7], Corollary 5.12).
Let1<p0<∞, 0<r,r0,q0≤∞, a,b0∈SV, and s−1/r0b0sr0,1,∞<∞. Put χ0t=s−1/r0b0sr0,t,∞ and ρt=t1/p0χ0t.(i)
Ifs−1/rasr,1,∞<∞, then(266)Lb0p0,q0,r0,L∞0,r,a=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L,where a#t=χ0taρt.(ii)
Ifs−1/rasr,0,1<∞, then(267)Lb0p0,q0,r0,L∞1,r,a≅L∞,r,a∘ρ.Proof.
Letθ0=1−1/p0. By Lemmas 56 and 59 and Theorem 25, we have(268)Lb0p0,q0,r0,L∞0,r,a=L1,L∞θ0,r0,b0,q0,1L,L∞0,r,a=L1,L∞θ0,r,a#,q0,1L⋂L1,L∞θ0,r,a∘ρ,r0,b0,q0,1L,L=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L.
Similarly, by [5] (Theorem 13) and Lemma 53, we get(269)Lb0p0,q0,r0,L∞1,r,a=L1,L∞θ0,r0,b0,q0,1L,L∞1,r,a=L1,L∞∞,r,a∘ρ=L∞,r,a∘ρ.
This completes the proof.☐Corollary 61 (cf. [7], Corollary 5.10).
Let1<p0<p1<∞, 0<r,r0,r1,q0≤∞, a,b0,b1∈SV, and s−1/r0b0sr0,1,∞<∞. Put χ0t=s−1/r0b0sr0,t,∞ and ρt=t1/p0−1/p1χ0t/b1t. Then,(i)
Ifs−1/rasr,1,∞<∞, then(270)Lb0p0,q0,r0,Lp1,r1,b10,r,a=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L,where a#t=χ0taρt.(ii)
Ifs−1/rasr,0,1<∞, then(271)Lb0p0,q0,r0,Lp1,r1,b11,r,a≅La∘ρp1,r1,r.Proof.
Letθ0=1−1/p0 and θ1=1−1/p1. By Lemmas 56, 53, and 59 and Theorem 23, we have(272)Lb0p0,q0,r0,Lp1,r1,b10,r,a=L1,L∞θ0,r0,b0,q0,1L,L1,L∞θ1,r1,b10,r,a=L1,L∞θ0,r,a#,q0,1L⋂L1,L∞θ0,r,a∘ρ,r0,b0,q0,1L,L=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L.
Similarly, by [5] (Theorem 11), we get(273)Lb0p0,q0,r0,Lp1,r1,b11,r,a=L1,L∞θ0,r0,b0,q0,1L,L1,L∞θ1,r1,b11,r,a=L1,L∞θ1,r,a∘ρ,r1,1R=La∘ρp1,r1,r.
This completes the proof.☐Analogously, by using Theorems28 and 26 and [5] (Theorem 23), the next two corollaries can be proved.Corollary 62. (cf. [6], Corollary 5.9).
Let1<p0<∞, 0<r,r0,q0≤∞, a,b0∈SV, and s−1/r0b0sr0,0,1<∞. Put ρt=t1/p0s−1/r0b0sr0,0,t. Then,(i)
Ifs−1/rasr,1,∞<∞, then(274)Lb0p0,q0,r0,L∞0,r,a=Lp0,r,a∘ρ,r0,b0,q0,1R,L.(ii)
Ifs−1/rasr,0,1<∞, then(275)Lb0p0,q0,r0,L∞1,r,a≅L∞,r,a∘ρ.Corollary 63 (cf. [6], Corollary 5.5).
Let1<p0<p1<∞, 0<r,r0,r1,q0≤∞, a,b0,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/rasr,1,∞<∞. Put ρt=t1/p0−1/p1s−1/r0b0sr0,0,t/b1t. Then,(276)Lb0p0,q0,r0,Lp1,r1,b10,r,a=Lp0,r,a∘ρ,r0,b0,q0,1R,L.Four corollaries above amend Corollaries 38–40 and 47–49 from [5]. Other corollaries from Sections 8.2 and 8.3 of [5] can be similarly amended.


### 9.3. Limiting Interpolation Formulae for Couples Where Both Operands Are Small or Grand Lorentz Spaces

Four corollaries below amend corollaries from Section 8.4 of [5]. We prove only the first one. Three further corollaries can be proved analogously using theorems from Section 7.Corollary 64 (cf. [6], Corollary 5.12; [8], Corollary 5.7).
Let1<p0<p1<∞, 0<r,r0,r1,q0,q1≤∞, b0,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=s−1/r0b0sr0,t,∞, χ1t=s−1/r1b1sr1,t,∞, and ρt=t1/p0−1/p1χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(277)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L,where a#t=χ0taρt.(ii)
Ifs−1/rasr,0,1<∞, then(278)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=Lp1,r,a∘ρ,r1,b1,q1,1L,R.Proof.
Letθ0=1−1/p0 and θ1=1−1/p1. Using Lemmas 56 and 59 and Theorem 33, we get(279)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=L1,L∞θ0,r0,b0,q0,1L,L1,L∞θ1,r1,b1,q1,1L0,r,a=L1,L∞θ0,r,a#,q0,1L⋂L1,L∞θ0,r,a∘ρ,r0,b0,q0,1L,L=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L.
Similarly,(280)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=L1,L∞θ1,r,a∘ρ,r1,b1,q1,1L,R=Lp1,r,a∘ρ,r1,b1,q1,1L,R.
This completes the proof.☐Corollary 65 (cf. [6], Corollary 5.7; [8], Corollary 5.5).
Let1<p0<p1<∞, 0<r,r0,r1,q0,q1≤∞, b0,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=s−1/r0b0sr0,0,t, χ1t=s−1/r1b1sr1,0,t, and ρt=t1/p0−1/p1χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(281)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=Lp0,r,a∘ρ,r0,b0,q0,1R,L.(ii)
Ifs−1/rasr,0,1<∞, then(282)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=La#p1,q1,r1⋂Lp1,r,a∘ρ,r1,b1,q1,1R,R,where a#t=χ1taρt.Corollary 66 (cf. [7], Theorem 5.7; [8], Corollary 5.10).
Let1<p0<p1<∞, 0<r,r0,r1,q0,q1≤∞, b0,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=s−1/r0b0sr0,t,∞, χ1t=s−1/r1b1sr1,0,t, and ρt=t1/p0−1/p1χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(283)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L,where a#t=χ0taρt.(ii)
Ifs−1/rasr,0,1<∞, then(284)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=La#p1,q1,r⋂Lp1,r,a∘ρ,r1,b1,q1,1R,R,where a#t=χ1taρt.Corollary 67 (cf. [8], Corollary 5.9).
Let1<p0<p1<∞, 0<r,r0,r1,q0,q1≤∞, b0,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=s−1/r0b0sr0,0,t, χ1t=s−1/r1b1sr1,t,∞, and ρt=t1/p0−1/p1χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(285)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=Lp0,r,a∘ρ,r0,b0,q0,1R,L.(ii)
Ifs−1/rasr,0,1<∞, then(286)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=Lp1,r,a∘ρ,r1,b1,q1,1L,R.


## 9.1. Function Spaces

First, we define function spaces under consideration.Definition 52.
Let0<p,q≤∞ and b∈SV. The Lorentz–Karamata space Lp,q;b is the set of all f∈MΩ,μ such that(256)fp,q;α≔t1/p−1/qbtf∗tq,0,∞<∞.
The Lorentz–Karamata spaces form an important scale of spaces. It contains, e.g., the Lebesgue spaceLp, Lorentz space Lp,q, Lorentz–Zygmund, and the generalized Lorentz–Zygmund spaces. For further information about Lorentz–Karamata spaces, we refer to, e.g., [10, 11, 16]. They have found many different important applications in analysis (see, e.g., [1, 4, 10–12, 14, 16–19] and the references therein).Lemma51 implies the following interpolation result.Lemma 53 (cf. [10], Corollary 5.3; [9]).
Let0<θ≤1, 0<q≤∞, b∈SV, and p=1/1−θ. Then, L1,L∞θ,q,b=Lp,q,b.Definition 54 (cf. [10], (5.21), (5.33)).
Let1<p<∞, 0<q,r≤∞, and a,b∈SV. The spaces Lp,r,b,q,aL and Lp,r,b,q,aR are the sets of all f∈MΩ,μ such that(257)fLp,r,b,q,aL≔t−1/rbtu1/p−1/qauf∗uq,0,tr,0,∞<∞,or(258)fLp,r,b,q,aR≔t−1/rbtu1/p−1/qauf∗∗uq,t,∞r,0,∞<∞,correspondingly.
We will require thatt−1/rbtr,1,∞<∞ for Lp,r,b,q,aL and t−1/rbtr,0,1<∞ for Lp,r,b,q,aR. Otherwise, the corresponding space consists only of the null element. Similar definitions can be found in [8, 9, 17]. We refer to the spaces L∗L and L∗R as LL and LR spaces, respectively. Note that the spaces L∗L are a special case of the generalized gamma space with double weights [3].In order to be able to compare our results with the results from [3, 4, 7–9], we introduce the generalized grand and small Lorentz spaces. Note that the interpolation results in [3, 4] do not contain limiting cases θ∈0,1.Definition 55.
Let1<p<∞, 0<q,r≤∞, and b∈SV. We define the small Lorentz space Lbp,q,r as the set of all f∈MΩ,μ such that(259)fLbp,q,r≔t−1/rbtu1/p−1/qf∗uq,0,tr,0,∞<∞.
We define the grand Lorentz spaceLbp,q,r as the set of all f∈MΩ,μ such that(260)fLbp,q,r≔t−1/rbtu1/p−1/qf∗∗uq,t,∞r,0,∞<∞.
It is clear thatLbp,q,r=Lp,r,b,q,1L and Lbp,q,r=Lp,r,b,q,1R. Grand and small Lebesgue and Lorentz spaces find many different important applications and have been intensively studied by many authors (see [3, 4, 6–9] and the references therein). Normally, they are defined on a bounded domain Ω in Rn with measure 1. In [4], it is required that the functions are real-valued. We do not require that μΩ=1 and neither that the functions are real-valued.The following lemma characterizes theLL and LR spaces as appropriate L and R limiting interpolation spaces. It follows from the corresponding definitions, formula (254), and Lemma 51.Lemma 56 (see [10], Lemmas 5.4 and 5.9).
Let1<p<∞, 0<q,r≤∞, a,b∈SV, and θ=1−1/p. Then, Lp,r,b,q,aL=L1,L∞θ,r,b,q,aL and Lp,r,b,q,aR=L1,L∞θ,r,b,q,aR.
In particular,Lbp,q,r=L1,L∞θ,r,b,q,1L and Lbp,q,r=L1,L∞θ,r,b,q,1R.Definition 57.
Let1<p<∞, 0<q,r,s≤∞, and a,b,c∈SV. The spaces Lp,s,c,r,b,q,aL,L, Lp,s,c,r,b,q,aL,R, Lp,s,c,r,b,q,aR,L, and Lp,s,c,r,b,q,aR,R are the sets of all f∈MΩ,μ such that(261)fLp,s,c,r,b,q,aL,L≔u−1/scut−1/rbtv1/p−1/qavf∗vq,0,tr,0,us,0,∞<∞,(262)fLp,s,c,r,b,q,aL,R≔u−1/scut−1/rbtv1/p−1/qavf∗∗vq,u,tr,u,∞s,0,∞<∞,(263)fLp,s,c,r,b,q,aR,L≔u−1/scut−1/rbtv1/p−1/qavf∗∗vq,t,ur,0,us,0,∞<∞,or(264)fLp,s,c,r,b,q,aR,R≔u−1/scut−1/rbtv1/p−1/qavf∗∗vq,t,∞r,u,∞s,0,∞<∞,correspondingly.Remark 58.
Because of Lemma51, if in the formulae (257), (259), and (261) f∗ will be replaced with f∗∗, one gets equivalent (quasi-)norms for corresponding spaces.In view of Lemma51, Remark 58, and (254), the following analogue of Lemma 56 holds.Lemma 59.
Let1<p<∞, 0<q,r,s≤∞, a,b∈SV, and θ=1−1/p. Then,(265)Lp,s,c,r,b,q,aL,L=L1,L∞θ,s,c,r,b,q,aL,L,Lp,s,c,r,b,q,aL,R=L1,L∞θ,s,c,r,b,q,aL,R,Lp,s,c,r,b,q,aR,L=L1,L∞θ,s,c,r,b,q,aR,L,Lp,s,c,r,b,q,aR,R=L1,L∞θ,s,c,r,b,q,aR,R.


## 9.2. Limiting Interpolation between the Small or Great Lorentz Spaces and the Lorentz–Karamata Spaces

Using results from Sections4–6 and Lemmas 56 and 59, we are able to characterize some limiting interpolation spaces lying between LL or LR spaces and Lorentz–Karamata spaces. In the corollaries below, we restrict ourselves to the grand and small Lorentz spaces.Corollary 60 (cf. [9], Corollary 7.8; [7], Corollary 5.12).
Let1<p0<∞, 0<r,r0,q0≤∞, a,b0∈SV, and s−1/r0b0sr0,1,∞<∞. Put χ0t=s−1/r0b0sr0,t,∞ and ρt=t1/p0χ0t.(i)
Ifs−1/rasr,1,∞<∞, then(266)Lb0p0,q0,r0,L∞0,r,a=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L,where a#t=χ0taρt.(ii)
Ifs−1/rasr,0,1<∞, then(267)Lb0p0,q0,r0,L∞1,r,a≅L∞,r,a∘ρ.Proof.
Letθ0=1−1/p0. By Lemmas 56 and 59 and Theorem 25, we have(268)Lb0p0,q0,r0,L∞0,r,a=L1,L∞θ0,r0,b0,q0,1L,L∞0,r,a=L1,L∞θ0,r,a#,q0,1L⋂L1,L∞θ0,r,a∘ρ,r0,b0,q0,1L,L=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L.
Similarly, by [5] (Theorem 13) and Lemma 53, we get(269)Lb0p0,q0,r0,L∞1,r,a=L1,L∞θ0,r0,b0,q0,1L,L∞1,r,a=L1,L∞∞,r,a∘ρ=L∞,r,a∘ρ.
This completes the proof.☐Corollary 61 (cf. [7], Corollary 5.10).
Let1<p0<p1<∞, 0<r,r0,r1,q0≤∞, a,b0,b1∈SV, and s−1/r0b0sr0,1,∞<∞. Put χ0t=s−1/r0b0sr0,t,∞ and ρt=t1/p0−1/p1χ0t/b1t. Then,(i)
Ifs−1/rasr,1,∞<∞, then(270)Lb0p0,q0,r0,Lp1,r1,b10,r,a=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L,where a#t=χ0taρt.(ii)
Ifs−1/rasr,0,1<∞, then(271)Lb0p0,q0,r0,Lp1,r1,b11,r,a≅La∘ρp1,r1,r.Proof.
Letθ0=1−1/p0 and θ1=1−1/p1. By Lemmas 56, 53, and 59 and Theorem 23, we have(272)Lb0p0,q0,r0,Lp1,r1,b10,r,a=L1,L∞θ0,r0,b0,q0,1L,L1,L∞θ1,r1,b10,r,a=L1,L∞θ0,r,a#,q0,1L⋂L1,L∞θ0,r,a∘ρ,r0,b0,q0,1L,L=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L.
Similarly, by [5] (Theorem 11), we get(273)Lb0p0,q0,r0,Lp1,r1,b11,r,a=L1,L∞θ0,r0,b0,q0,1L,L1,L∞θ1,r1,b11,r,a=L1,L∞θ1,r,a∘ρ,r1,1R=La∘ρp1,r1,r.
This completes the proof.☐Analogously, by using Theorems28 and 26 and [5] (Theorem 23), the next two corollaries can be proved.Corollary 62. (cf. [6], Corollary 5.9).
Let1<p0<∞, 0<r,r0,q0≤∞, a,b0∈SV, and s−1/r0b0sr0,0,1<∞. Put ρt=t1/p0s−1/r0b0sr0,0,t. Then,(i)
Ifs−1/rasr,1,∞<∞, then(274)Lb0p0,q0,r0,L∞0,r,a=Lp0,r,a∘ρ,r0,b0,q0,1R,L.(ii)
Ifs−1/rasr,0,1<∞, then(275)Lb0p0,q0,r0,L∞1,r,a≅L∞,r,a∘ρ.Corollary 63 (cf. [6], Corollary 5.5).
Let1<p0<p1<∞, 0<r,r0,r1,q0≤∞, a,b0,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/rasr,1,∞<∞. Put ρt=t1/p0−1/p1s−1/r0b0sr0,0,t/b1t. Then,(276)Lb0p0,q0,r0,Lp1,r1,b10,r,a=Lp0,r,a∘ρ,r0,b0,q0,1R,L.Four corollaries above amend Corollaries 38–40 and 47–49 from [5]. Other corollaries from Sections 8.2 and 8.3 of [5] can be similarly amended.


## 9.3. Limiting Interpolation Formulae for Couples Where Both Operands Are Small or Grand Lorentz Spaces

Four corollaries below amend corollaries from Section 8.4 of [5]. We prove only the first one. Three further corollaries can be proved analogously using theorems from Section 7.Corollary 64 (cf. [6], Corollary 5.12; [8], Corollary 5.7).
Let1<p0<p1<∞, 0<r,r0,r1,q0,q1≤∞, b0,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=s−1/r0b0sr0,t,∞, χ1t=s−1/r1b1sr1,t,∞, and ρt=t1/p0−1/p1χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(277)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L,where a#t=χ0taρt.(ii)
Ifs−1/rasr,0,1<∞, then(278)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=Lp1,r,a∘ρ,r1,b1,q1,1L,R.Proof.
Letθ0=1−1/p0 and θ1=1−1/p1. Using Lemmas 56 and 59 and Theorem 33, we get(279)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=L1,L∞θ0,r0,b0,q0,1L,L1,L∞θ1,r1,b1,q1,1L0,r,a=L1,L∞θ0,r,a#,q0,1L⋂L1,L∞θ0,r,a∘ρ,r0,b0,q0,1L,L=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L.
Similarly,(280)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=L1,L∞θ1,r,a∘ρ,r1,b1,q1,1L,R=Lp1,r,a∘ρ,r1,b1,q1,1L,R.
This completes the proof.☐Corollary 65 (cf. [6], Corollary 5.7; [8], Corollary 5.5).
Let1<p0<p1<∞, 0<r,r0,r1,q0,q1≤∞, b0,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=s−1/r0b0sr0,0,t, χ1t=s−1/r1b1sr1,0,t, and ρt=t1/p0−1/p1χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(281)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=Lp0,r,a∘ρ,r0,b0,q0,1R,L.(ii)
Ifs−1/rasr,0,1<∞, then(282)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=La#p1,q1,r1⋂Lp1,r,a∘ρ,r1,b1,q1,1R,R,where a#t=χ1taρt.Corollary 66 (cf. [7], Theorem 5.7; [8], Corollary 5.10).
Let1<p0<p1<∞, 0<r,r0,r1,q0,q1≤∞, b0,b1∈SV, s−1/r0b0sr0,1,∞<∞, and s−1/r1b1sr1,0,1<∞. Put χ0t=s−1/r0b0sr0,t,∞, χ1t=s−1/r1b1sr1,0,t, and ρt=t1/p0−1/p1χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(283)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=La#p0,q0,r⋂Lp0,r,a∘ρ,r0,b0,q0,1L,L,where a#t=χ0taρt.(ii)
Ifs−1/rasr,0,1<∞, then(284)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=La#p1,q1,r⋂Lp1,r,a∘ρ,r1,b1,q1,1R,R,where a#t=χ1taρt.Corollary 67 (cf. [8], Corollary 5.9).
Let1<p0<p1<∞, 0<r,r0,r1,q0,q1≤∞, b0,b1∈SV, s−1/r0b0sr0,0,1<∞, and s−1/r1b1sr1,1,∞<∞. Put χ0t=s−1/r0b0sr0,0,t, χ1t=s−1/r1b1sr1,t,∞, and ρt=t1/p0−1/p1χ0t/χ1t.(i)
Ifs−1/rasr,1,∞<∞, then(285)Lb0p0,q0,r0,Lb1p1,q1,r10,r,a=Lp0,r,a∘ρ,r0,b0,q0,1R,L.(ii)
Ifs−1/rasr,0,1<∞, then(286)Lb0p0,q0,r0,Lb1p1,q1,r11,r,a=Lp1,r,a∘ρ,r1,b1,q1,1L,R.



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*Source: 8513304-2021-08-02.xml*