---

Calder´on couples of re-arrangement invariant spaces

arXiv:math/9209217v1 [math.FA] 25 Sep 1992Calder´on couples of re-arrangement invariant spacesN.J. Kalton1Department of MathematicsUniversity of Missouri-ColumbiaColumbia, Missouri 65211.Abstract.

We examine conditions under which a pair of re-arrangement invariant functionspaces on [0, 1] or [0, ∞) form a Calder´on couple. A very general criterion is developed todetermine whether such a pair is a Calder´on couple, with numerous applications.

We give,for example, a complete classification of those spaces X which form a Calder´on couple withL∞. We specialize our results to Orlicz spaces and are able to give necessary and sufficientconditions on an Orlicz function F so that the pair (LF , L∞) forms a Calder´on pair.1.

Introduction.Suppose (X, Y ) is a compatible pair of Banach spaces (see [4] or [5]). We denote byK(t, f) = K(t, f; X, Y ) the Peetre K-functional on X + Y i.e.K(t, f) = inf{∥x∥X + t∥y∥Y : x + y = f}.Then (X, Y ) is called a Calder´on couple (or a Calder´on-Mityagin couple) if whenever f, gsatisfyK(t, f) ≤K(t, g)for all t then there is a bounded operator T : X +Y →X +Y such that ∥T∥X, ∥T∥Y < ∞and Tg = f. We will say that (X, Y ) is a uniform Calder´on couple (with constant C)if we can further insist that max(∥T∥X, ∥T∥Y ) ≤C.

Calder´on couples are particularlyimportant in interpolation theory because it is possible to give a complete description ofall interpolation spaces for such a couple. Indeed, for such a couple, it is easy to showthat a space Z is an interpolation space if and only if it is K-monotone, i.e.

if f ∈Z andg ∈X + Y with K(t, g) ≤K(t, f) then g ∈Z. It follows from the K-divisibility theoremof Brudnyi and Krugljak [7] that if Z is a normed K-monotone space then allows ∥f∥Z onZ is equivalent to a norm ∥K(t, f)∥Φ where Φ is an appropriate lattice norm on functionson (0, ∞).

Thus, for Calder´on couples, one has a complete description of all interpolation1Research supported by NSF-grants DMS-8901636 and DMS-9201357; the author alsoacknowledges support from US-Israel BSF-grant 87-002441

spaces. We remark also at this point that there are apparently no known examples ofCalder´on couples which are not uniform.There has been a considerable amount of subsequent effort devoted to classifyingCalder´on couples of rearrangement-invariant spaces on [0, 1] or [0, ∞).

It is a classicalresult of Calder´on and Mityagin ([9],[32]) that the pair (L1, L∞) is a uniform Calder´oncouple with constant 1. It is now known that any pair (Lp, Lq) is a Calder´on couple (andindeed weighted versions of these theorems are valid); we refer the reader to Lorentz andShimogaki [27], Sparr [36], Arazy and Cwikel [1], Sedaev and Semenov [35] and Cwikel[13], [15].Subsequent work has shown that under certain hypotheses pairs of Lorentzspaces or Marcinkiewicz spaces are Calder´on couples; see Cwikel [14], Merucci [30], [31]and Cwikel-Nilsson [16], [17].

For further positive results on Calder´on couples see [18] (forweighted Banach lattices), and [21] and [38] (for Hardy spaces).On the negative side, Ovchinnikov [34] showed that on [0, ∞) the pair (L1 + L∞, L1 ∩L∞) is not a Calder´on couple; indeed Maligranda and Ovchinnikov show that if p ̸= 2 thenLp ∩Lq and Lp + Lq ( 1p + 1q = 1) are interpolation spaces not obtainable by the K-method[29].The general problem we consider in this article is that of providing necessary andsufficient conditions on a pair (X, Y ) of r.i. spaces (always assumed to have the so-calledFatou property) on either [0, 1] or [0, ∞) so that (X, Y ) is a Calder´on couple. Althoughwe cannot provide a complete answer to this problem, we can resolve it in certain casesand this enables us to settle some open questions in the area (see, e.g.

Maligranda [28],Problems 1-3 or Brudnyi-Krugljak [8] p.685 [g], [i]).For example, we give a completeclassification of all r.i. spaces X so that (X, L∞) is a Calder´on couple and hence giveexamples of r.i. spaces (even Orlicz spaces) X so that (X, L∞) is not a Calder´on couple.Our methods give fairly precise information in the problem of classifying pairs of Orliczspaces which form Calder´on couples. It should also be mentioned that our results applyequally to symmetric sequence spaces.We now describe our results in more detail.

Let X be an r.i. space on [0, 1] or [0, ∞)or a symmetric sequence space. Let en = χ[2n,2n+1) for n ∈J where J = Z−= −N orJ = Z or J = N ∪{0}.

We associate to X a K¨othe sequence space EX on J by defining∥ξ∥EX = ∥Xn∈Jξ(n)en∥X.We then say that X is stretchable if the sequence space EX has the right-shift property(RSP) i.e. there is a constant C so that if (xn, yn)Nn=1 is any pair of finite normalizedsequences in EX so that supp x1 < supp y1 < supp x2 < · · · < supp yn then for any2

α1, . .

. , αN we have∥NXn=1αnyn∥EX ≤C∥NXn=1αnxn∥EX.Thus EX has (RSP) if the right-shift operator is uniformly bounded on the closed linearspan of every block basic sequence with respect to the canonical basis.

We similarly saythat X is compressible if EX has the corresponding left-shift property (LSP). Finally we saythat X is elastic if it is both stretchable and compressible.

It is easy to see that Lp−spacesand more generally Lorentz spaces with finite Boyd indices are elastic because EX in thiscase is a weighted ℓp−space (in fact this property characterizes Lorentz spaces when theBoyd indices are finite). On the other hand it is not difficult to give examples of r.i. spaceswhich are neither compressible nor stretchable.

Curiously, however, we have no exampleof a space which is either stretchable or compressible and not elastic.The significance of these ideas is illustrated by Theorem 5.4. The pair (X, L∞) is aCalder´on couple if and only if X is stretchable.

Dually if we assume that X has nontrivialconcavity then (X, L1) is a Calder´on couple if and only if X is compressible (Theorem5.5). More generally if (X, Y ) is any pair of r.i. spaces such that either the Boyd indicessatisfy pY > qX or there exists p so that X is p-concave and Y is p-convex and hasnontrivial concavity then (X, Y ) is a Calder´on couple if and only if X is stretchable andY is compressible.In Section 6 we study these concepts for Orlicz spaces.

We show that for an Orliczspace to be compressible it is necessary and sufficent that it is stretchable; thus we needonly consider elastic Orlicz spaces. We show for example that LF [0, 1] (where F satisfiesthe ∆2−condition) is elastic if and only if there is a constant C and a bounded monotoneincreasing function w(t) so that for any 0 < x ≤1 and any 1 ≤s ≤t < ∞we haveF(tx)F(t) ≤C F(sx)F(s) + w(t) −w(s).This condition implies that the Boyd indices (or Orlicz-Matuszewska indices) pF and qFof LF coincide.

In fact it implies the stronger condition that F must be equivalent to afunction which is regularly varying in the sense of Karamata (see [6]). We give examples toshow that F can be regularly varying with LF inelastic and that LF can be elastic withoutcoinciding with a Lorentz space (cf.

[26], [33]).Brudnyi (cf. [8]) has conjectured that if a pair of (distinct) Orlicz spaces (LF [0, 1],LG[0, 1]) is a Calder´on couple then pF = qF and pG = qG.

We show by example that thisis false. However we also show that either pF = pG and qF = qG or both LF and LG areelastic and hence pF = qF and pG = qG.Let us now introduce some notation and conventions.

Let Ωbe a Polish space and letµ be a σ−finite Borel measure on Ω. Let L0(µ) denote the space of all real-valued Borel3

functions on Ω(where functions differing on a set of measure zero are identified), equippedwith the topology of convergence in µ−measure on sets of finite measure. By a K¨othefunction space on Ωwe shall mean a Banach space X which is a subspace of L0 containingthe characteristic function χB whenever µ(B) < ∞and such that the norm ∥∥X satisfiesthe conditions:(a) ∥f∥X ≤∥g∥X whenever |f| ≤|g| a.e.

(b) BX = {f : ∥f∥X ≤1} is closed in L0.Condition (b) is usually called the Fatou property; note here that we include the Fatouproperty in our definition and so it is an implicit assumption throughout the paper. It issometimes convenient to extend the definition of ∥f∥X by setting ∥f∥X = ∞if f /∈X.

Wewill also write PBf = fB = fχB when B is a Borel subset of Ω. We let supp f = {ω :f(ω) ̸= 0}.If X is a K¨othe function space then we say that X is p-convex (1 ≤p ≤∞) if thereis a constant M so that for any f1, .

. .fn ∈X we have∥(nXk=1|fk|p)1/p∥X ≤M(nXk=1∥fk∥pX)1/pand p-concave if there exists M so that(nXk=1∥fk∥pX)1/p ≤M∥(nXk=1|fk|p)1/p∥X.Similarly we say that X has an upper p-estimate if there is a constant M so that if f1, .

. ., fnare disjoint in X then∥nXk=1fk∥X ≤M(nXk=1∥fk∥pX)1/pand X has a lower p-estimate if there exists M so that if f1, .

. ., fn are disjoint then(nXk=1∥fk∥pX)1/p ≤M∥nXk=1fk∥X.See Lindenstrauss-Tzafriri [25] for a fuller discussion.We will sometimes use ⟨f, g⟩forRΩfg dµ.

With this notion of pairing we will also useX∗for the K¨othe dual of X (which will coincide with the full dual if X is separable).If (X, Y ) are two K¨othe function spaces on (Ω, µ) then the pair (X, Y ) is necessarilyGagliardo complete (cf. [4]).

We denote by A(X, Y ) the space of admissible operators i.e.operators T : X+Y →X+Y such that ∥T∥X = sup{∥Tf∥X : ∥f∥X ≤1} < ∞and ∥T∥Y =sup{∥Tf∥Y : ∥f∥Y ≤1} < ∞. We norm A(X, Y ) by ∥T∥(X,Y ) = max(∥T∥X, ∥T∥Y ).4

In the special case when Ω= J is a subset of Z and µ is counting measure we writeω(J) = L0(µ) and a K¨othe function space X is called a K¨othe sequence space modelled onJ. An operator T on X is then called a matrix if it takes the formTx(n) =Xk∈Jankx(k)for some (ank)n,k∈J.

We remark here that the assumption that T is a matrix forces theexistence of an adjoint operator T ∗: X∗→X∗even in the nonseparable situation whenX∗is not the full dual of X.If Ω= [0, 1] or [0, ∞) (with µ Lebesgue measure) or if Ω= N (with µ countingmeasure) then for f ∈L0(µ) we define the decreasing rearrangement f ∗of f byf ∗(t) =supB:µ(B)≤tinfs∈B |f(s)|,for 0 < t < ∞. We say that X is a rearrangement-invariant space (or a symmetric sequencespace if Ω= N) if ∥f∥X = ∥f ∗∥X for all f ∈L0.

If we definef ∗∗(t) = 1tZ t0f ∗(s)dsthen it is well-known that if f, g ∈L0 with f ∗∗≤g∗∗then ∥f∥X ≤∥g∥X.If X is an r.i. space on [0, 1] or [0, ∞) then the dilation operators Da on X are thendefined by Daf(t) = f(t/a) (where we regard f as vanishing outside [0, 1] in the formercase). We can then define the Boyd indices pX and qX of X bypX = lima→∞log alog ∥Da∥XqX = lima→0log alog ∥Da∥X.In the case when X is a symmetric sequence space we define pX and qX in the same waybut we define Da by the nonlinear formulaDaf(n) = f ∗(n/a)where f ∗is well-defined on [0, ∞).Finally let us mention two special classes of r.i. spaces.

If 1 ≤p < ∞we will say thatan r.i. space X on Ω= [0, 1] or [0, ∞) is a Lorentz space of order p if there is a positivemonotone increasing weight function w : Ω→(0, ∞) such that supt,2t∈Ωw(2t)/w(t) < ∞and ∥f∥X is equivalent to the quasinorm∥f∥w,p = (ZΩf ∗(t)pw(t)p dtt )1/p.5

We can then write X = Lw,p. If we take w(t) = t1/q we obtain the standard Lorentz spacesL(q, p).

It is easy to compute that the Lorentz space X = Lw,p has Boyd indices pX, qXwhere1pX= lima→∞supt,at∈Ωlog w(at) −log w(t)log a1qX= lima→∞inft,at∈Ωlog w(at) −log w(t)log a.If we impose the additional restriction that qX < ∞then it can easily be seen that wemay suppose that w satisfies inft,2t∈Ωw(2t)/w(t) > 1.We will also be interested in Orlicz function spaces and sequence spaces. By an Orliczfunction we shall mean a continuous strictly increasing convex function F : [0, ∞) →[0, ∞)such that F(0) = 0.

F is said to satisfy the ∆2−condition if there is a constant ∆suchthat F(2x) ≤∆F(x) for all x ≥0.The Orlicz function space LF (Ω, µ) is defined by∥f∥LF = inf{α > 0 :ZΩF(α−1f(t))dt ≤1}so that LF = {f : ∥f∥LF < ∞}.In this case the Boyd indices pF = pLF and qF = qLF are closely related to the Orlicz-Matuszewska indices of F (see Lindenstrauss-Tzafriri [25] p.139).More precisely letα∞(F) (resp. α0(F)) be the supremum of all p so that for some C we have F(st) ≤CspF(t)for all 0 ≤s ≤1 and all t ≥1 (resp.

t ≤1). Similarly let β∞(F) (resp.

β0(F)) be theinfimum of all q so that for some C we have spF(t) ≤CF(st) for all 0 ≤s ≤1 and all t ≥1(resp. t ≤1).

Then if Ω= [0, 1] we have pF = α∞(F) and qF = β∞(F). If Ω= [0, ∞) thenpF = min(α∞(F), α0(F)) and qF = max(β∞(F), β0(F)).

If we assume the ∆2−condition(and we always will) then qF < ∞.2. The shift properties.Let J be one of the three sets Z, Z+ = {n ∈Z : n ≥0} or Z−= Z \ Z+.

Let ω(J)denote the space of all sequences modelled on J. If x = {x(k)}k∈J is a sequence (modelledon J) we write supp x = {k : x(k) ̸= 0}.

If A, B are subsets of J we write A < B if a < bfor every a ∈A, b ∈B. If I is any interval of Z and (xn, yn)n∈I is a pair of sequences inω(J) we say (xn, yn) is interlaced if each xn, yn has finite support and supp xn < suppyn (n ∈I) and supp yn < supp xn+1 whenever n, n + 1 ∈I.6

Let E be a K¨othe sequence space modelled on J. We will say that E has the right-shiftproperty (RSP) if there is a constant C such that whenever (xn, yn)n∈I is an interlacedpair with ∥yn∥E ≤∥xn∥E = 1 (n ∈I) then for every finitely nonzero sequence of scalars(αn)n∈I we have∥Xn∈Iαnyn∥E ≤C∥Xn∈Iαnxn∥E.Conversely we will say that E has the left-shift property (LSP) if there is a constant C′so that for every interlaced pair (xn, yn)n∈I with ∥xn∥E ≤∥yn∥E = 1, and every finitelynonzero (αn)n∈I we have∥Xn∈Iαnxn∥E ≤C′∥Xn∈Iαnyn∥E.Proposition 2.1.

E has (LSP) if and only if E∗has (RSP).Proof: We will only prove one direction. Let us assume E∗has (RSP) with constantC.

Let (xn, yn)n∈I, be an interlaced pair with ∥yn∥E ≤∥xn∥E = 1. We may assume eachxn, yn is positive (i.e.

xn(k), yn(k) ≥0 for every k). Suppose (αn)n∈I is a finitely nonzerosequence of nonnegative reals.

Let f = P αnyn. Then there exists positive g ∈E∗withsupp g ⊂supp f and so that ⟨f, g⟩= ∥f∥E while ∥g∥E∗= 1.

We can writeg =Xβnvnwhere each vn is positive, ∥vn∥E∗= 1 and supp vn ⊂supp yn.Next pick positive un with supp un ⊂supp xn, ⟨xn, un⟩= 1 and ∥un∥E∗= 1. Weconclude from the fact that E∗has (RSP) that∥Xn∈Iβnun∥E∗≤C.Thus∥Xn∈Iαnyn∥E =Xn∈Iαnβn⟨yn, vn⟩≤Xαnβn≤⟨(Xαnxn), (Xβnun)⟩≤C∥Xαnxn∥E.Thus the proposition is proved.7

Proposition 2.2. Suppose E is a K¨othe sequence space modelled on Z. Define E+ =E(Z+) and E−= E(Z−).

Then E has (RSP) (resp. (LSP)) if and only both E+ and E−have (RSP) (resp.

(LSP)).Proof: One direction is obvious. For the other, suppose both E+ and E−have (RSP)with constant C, say.

Suppose (xn, yn)n∈I is an interlaced pair of sequences with ∥yn∥E ≤∥xn∥E = 1 and that (αn)n∈I is finitely nonzero. Then there exists m ∈I so that supp(xn + yn) ⊂Z−for n < m and supp (xn + yn) ⊂Z+ for n > m. Now∥Xn∈Iαnyn∥E ≤|αm| + ∥Xnmαnyn∥E≤(2C + 1)∥Xn∈Iαnxn∥E.Thus E has (RSP) with constant at most 2C + 1.To simplify our discussion we introduce the idea of an order-reversal.

Let E = E(J)be a K¨othe sequence space.We let ˜J = {−(n + 1) : n ∈J} and if x ∈ω(J) we set˜x(n) = x(−(n + 1)) for n ∈J. Let ˜E(˜J) be defined by ∥x∥˜E = ∥˜x∥E; then ˜E is the order-reversal of E. Clearly (LSP) (resp.

(RSP)) for E is equivalent to (RSP) (resp. (LSP)) for˜E.Next observe that if (wn)n∈J satisfy wn > 0 for all n then the weighted sequence spaceE(w) = {x : xw ∈E} normed by ∥x∥E(w) = ∥xw∥E satisfies (LSP) (resp.

(RSP)) if andonly if E satisfies (LSP) (resp. (RSP)).Proposition 2.3.

Let E = E(J) be a symmetric sequence space. Suppose E has either(LSP) or (RSP).

Then E = ℓp(J) for some 1 ≤p ≤∞.Proof: For convenience of notation we consider only the case J = Z+ and (LSP) andleave the reader to make the minor adjustments necessary for the other cases. Let (un)n∈Nbe any normalized positive block basic sequence in E(J).Select an ∈supp un.

Then(u2n, ea2n+1)n∈N is an interlaced pair. Thus∥Xn∈Nαnen∥E ≤C∥Xn∈Nαnu2n∥E.Similarly (ea2n−1, u2n)n∈N is an interlaced pair and so∥Xn∈Nαnu2n∥E ≤C∥Xn∈Nαnen∥E.Thus (u2n) is C2−equivalent to (en) and similarly so is (u2n−1)n∈N.

It then follows thatthe basis (or basic sequence) (en) is perfectly homogeneous and by a theorem of Zippin [39](see Lindenstrauss-Tzafriri [24]) this implies that it is equivalent to the ℓp-basis for somep or the c0-basis; in the latter case we deduce that E = ℓ∞(J). The result then follows.8

Proposition 2.4. Let E = E(Z+) be a K¨othe sequence space with (LSP) or (RSP).

IfE contains a symmetric basic sequence then there exists 1 ≤p ≤∞and an increasingsequence (ak)k≥0 with a0 = 0 so that E = ℓp(E[ak, ak+1)). In particular, when E isseparable, we have p < ∞and any symmetric basic sequence in E is equivalent to thecanonical ℓp−basis.Remark: Of course there is a similar result if J = Z−.

However in the two-ended settingJ = Z we recall that E has (RSP) (resp. (LSP)) if and only if both E(Z+) and E(Z−)have (RSP) (resp.

(LSP)). In particular, ℓp(Z−) ⊕ℓr(Z+) has (LSP) and (RSP) even ifr ̸= p.Proof: We can suppose that (un) is a normalized symmetric block basic sequence.

Byan interlacing argument as in Proposition 2.3 it will follow that a subsequence (eak) ofthe unit vectors is symmetric. Since the restriction of E to this subsequence has (LSP)or (RSP) it follows that it is equivalent to the ℓp-basis for some 1 ≤p < ∞or to thec0−basis by Proposition 2.3.

For convenience we suppose the former case and fix p. LetIk = Ik = [ak, ak+1). Then, for suitable C, by an interlacing argument any normalizedsequence (vk) supported on I2k is C−equivalent to the ℓp−basis; similarly any normalizedsequence supported on I2k+1 is C−equivalent to the ℓp−basis and the first part of theresult follows.

For the last part, if E is separable then obviously p < ∞and a simpleblocking argument gives the result.Remark: It is possible that E contains no symmetric basic sequence. Indeed, Tsirelsonspace T [37] and its convexifications provide examples of such spaces with (RSP) and(LSP) (see [10] and [12]).

It is not difficult to use Krivine’s theorem [22] to show that ifE = E(Z+) has (LSP) (or (RSP)) then there is a subsequence (ean) of the unit vectorbasis so that for some C, p we have for all k and every k vectors x1, x2, . .

., xk with suppx1 < supp x2 < . .

. < supp xk and supp (x1 + · · · + xk) ⊂{an}n≥1 thenC−1(kXn=1∥xn∥pE)1/p ≤∥kXn=1xn∥E ≤C(∞Xn=1∥xn∥pE)1/pwith appropriate modifications when p = ∞.

Thus any space E having either (LSP) or(RSP) and no symmetric basic sequence has a “Tsirelson-like” subspace.Problem: Does there exist a K¨othe sequence space with (LSP) and not (RSP)?Let us remark that this is probably non-trivial. Indeed the corresponding question forsimple shifts has been considered [3] and Bellenot has only recently given an example [2].Lemma 2.5.

Let E be a K¨othe sequence space on J with (RSP); then there is a constantC so that whenever (xn, yn)n∈I is an interlaced pair of sequences with ∥yn∥E ≤∥xn∥E = 19

and (x∗n)n∈I is a sequence in E∗with supp x∗n ⊂supp xn and x∗n(xn) = ∥xn∥E∗= 1 thenthe operator T defined by Tx = Pn∈I⟨x, x∗n⟩yn is bounded on E with ∥T∥E ≤C.Proof: For any x ∈E with finite support, Tx has finite support and we can defineg∗n ∈E∗and a finitely nonzero sequence (αn)n∈I so that ∥g∗n∥= 1(n ∈I), supp g∗n ⊂suppyn, ∥P αng∗n∥E∗= 1 and⟨Tx,Xn∈Iαng∗n⟩= ∥Tx∥E.Thus∥Tx∥E =Xn∈Iαnx∗n(x)= ⟨x,Xn∈Iαnx∗n⟩≤∥x∥E∥Xn∈Iαnx∗n∥E∗≤C∥x∥E∥Xn∈Iαng∗n∥E∗≤C∥x∥Ewhere C is the (LSP) constant of E∗(which actually is the (RSP) constant of E byProposition 2.1 and its proof). The result follows.Lemma 2.6.

Under the hypotheses of Lemma 2.5, there is a constant C1 so that (Jn)n∈Iis a sequence of intervals in J with Jn < Jn+1 whenever n, n + 1 ∈I, (xn)n∈I, (yn)n∈I aretwo normalized sequences with supp xn, supp yn ⊂Jn and (x∗n) is any sequence with suppx∗n ⊂supp xn and x∗n(xn) = 1 = ∥x∗n∥E∗then the operatorTx =Xn∈I⟨x, x∗n⟩yn+1(where yn+1 = 0 if n + 1 /∈I) is bounded on E with ∥T∥E ≤C1.Proof: The sequence pairs (x2n, y2n+1)2n,2n+1∈I and (x2n−1, y2n)2n−1,2n∈I are interlacedand the lemma follows from 2.5 with C1 = 2C by simply adding.Remark: If E is separable and has both (LSP) and (RSP) then Lemma 2.6 quickly showsthat every normalized block basic sequence in E spans a complemented subspace; thisproperty is, of course, enjoyed by Tsirelson space [12] (see also Casazza-Lin [11] for anearlier similar example). If this property holds for a symmetric sequence space then it isisomorphic to ℓp for some 1 ≤p < ∞(see Lindenstrauss-Tzafriri [23]).10

3. The shift properties for pairs.We next consider a pair of K¨othe sequences spaces (E, F) modelled on J.

We will saythat (E, F) has the (RSP) if there is a constant C so that whenever {xn, yn}n∈I is aninterlaced pair with ∥yn∥E ≤∥xn∥E = 1 and xn, yn ≥0 then there is a positive matrix Twith ∥T∥(E,F ) ≤C and Txn = yn. We will say that the pair (E, F) has (LSP) if ( ˜F, ˜E)has (RSP).

If (E, F) has both (LSP) and (RSP) then we say that it has the shift property(SP).We first note that if (E, F) has (RSP) then E has (RSP). Conversely it follows fromLemma 2.5 that if E has (RSP) then (E, E) has (RSP).In this section, we show that, under certain hypotheses, one can deduce (RSP) forthe couple (E, F) from the property (RSP) for E alone.

We will need some definitions.We define the shift operators τn for n ∈Z on ω(J) by τn(x)(k) = x(k −n), where weinterpret x(j) = 0 when j /∈J. We define κ+(E) = limn→∞∥τn∥1/nE(which can be ∞inthe case when τ1 is unbounded on E) and κ−(E) = limn→∞∥τ−n∥1/nE .

We will also letρ(n) = ρ(n; E, F) = ∥en∥E/∥en∥F . We will say that (E, F) is exponentially separated ifthere exists β > 0 and C0 so that if m, m + n ∈J then ρ(m + n) ≥C−10 2nβρ(m).Lemma 3.1.

If κ−(E)κ+(F) < 1 then (E, F) is exponentially separated.Proof: Here we have ρ(m + n)/ρ(m) ≥(∥τ−n∥E∥τn∥F )−1. The hypothesis then impliesthat for some β > 0 we have ∥τ−n∥E∥τn∥F ≤C2−nβ for some C > 0.

The result thenfollows.Lemma 3.2. Suppose (E, F) is exponentially separated.

Then there is a constant C1 sothat if supp x ⊂[a, b] thenC−11 ρ(a)∥x∥F ≤∥x∥E ≤C1ρ(b)∥x∥F.Proof: Suppose x = Pbk=a x(k)ek. Then∥x∥E ≤X|x(k)|∥ek∥E≤X|x(k)|ρ(k)∥ek∥F≤C0Xρ(b)2−β(b−k)|x(k)|∥ek∥F≤C0ρ(b)(∞Xk=02−βk) maxj|x(j)|∥ej∥F≤Cρ(b)∥x∥Ffor a suitable constant C. The other inequality is similar.11

Lemma 3.3. Let E, F be a pair of K¨othe sequence spaces satisfying κ−(E)κ+(F) < 1.Suppose E has (RSP).

Then (E, F) has the (RSP). Similarly if F has (LSP) then (E, F)has (LSP).Proof: We first note that it is only necessary to prove the first statement since ( ˜F, ˜E)will satisfy the same hypothesis κ−( ˜F)κ+( ˜E) < 1 and ˜F will have (RSP) if F has (LSP).We refer back to Lemma 2.5; it is clear that there exists C0 so that if {xn, yn}n∈I isa positive interlaced pair with ∥yn∥E ≤∥xn∥E = 1 then if we pick x∗n ≥0 with supp x∗n ⊂supp xn and ⟨xn, x∗n⟩= ∥x∗n∥E∗= 1 for n ∈I then ∥T∥E ≤C0 whereTx =Xn∈A⟨x, x∗n⟩yn.Obviously T is a positive matrix.

We now compute ∥T∥F . Suppose k ∈supp yn wheren ∈I.

Then, since supp x∗n ⊂(−∞, k) and yn(k)ek ≤yn,|Tx(k)| = |x∗n(x)|yn(k) ≤∥x(−∞,k)∥E∥ek∥−1E .Now∥x(−∞,k)∥E ≤Xj 0. Hence:∥Tx∥F ≤∞Xj=1Mδj∥x∥Fso that ∥T∥F ≤C1 for some constant C1 depending only on E, F.Although Lemma 3.3 is enough for most of our purposes, there are some possiblemodifications.

First we give a simple argument in the case F = ℓ∞, which will be usefullater.12

Lemma 3.4. Suppose E is a K¨othe sequence space with (RSP) and that F = ℓ∞(J).Suppose (E, F) is exponentially separated.

Then (E, F) has (RSP).Proof: We may assume that for some C0, β > 0 we have ∥em∥E ≤C02−βn∥em+n∥E. Inthis case we proceed as in Lemma 3.3 but note that∥Tx∥F = supk∈J|Tx(k)|.If k ∈supp yn,|Tx(k)| = |⟨x, x∗n⟩|yn(k)≤∥x(−∞,k)∥E∥ek∥−1E≤Xj

Suppose (E, F) is exponentially separated, E has (RSP) and that either (a)there exists 1 ≤p ≤∞so that E has a lower p-estimate and F has an upper p-estimateor (b) E is r-concave for some r < ∞and there exists 1 < q < ∞so that E has an upperq-estimate and F has a lower q-estimate. Then (E, F) has (RSP).Proof: For (a) we note that the case p = ∞is essentially covered in Lemma 3.4.

Supposep < ∞. By Lemma 3.2 there is a constant C1 so that if supp x < supp y then ∥x∥E∥y∥F ≤C1∥x∥F ∥y∥E.

There is also a constant C2 so that if u1, . .

., un are disjoint vectors in E orF,(nXj=1∥uj∥pE)1/p ≤C2∥nXj=1uj∥E∥nXj=1uj∥F ≤C2(nXj=1∥uj∥pF )1/p.We suppose (xn, yn)n∈I is a positive interlaced pair with ∥yn∥E ≤∥xn∥E = 1. Define T as13

in Lemma 3.3, and set Jk = supp x∗k. Now if x ∈F,∥Tx∥F = ∥Xk∈Ax∗k(x)yk+1∥F≤C2(Xk∈A∥xJk∥pE∥yk+1∥pF )1/p≤C2C1(Xk∈A∥xJk∥pF )1/p≤C22C1∥x∥Fand (a) follows.We turn to the proof of (b).

Let 1 < p < ∞be conjugate to q. Then ˜E∗has (RSP) byProposition 2.1.

Also ( ˜E∗, ˜F ∗) is exponentially separated, and ˜E∗has a lower p-estimatewhile ˜F ∗has an upper p-estimate; thus by (a) the couple ( ˜E∗, ˜F ∗) has (RSP). It followsthat (F ∗, E∗) has (LSP).

We further can assume, by renorming, that E has an upper p-estimate with constant 1 and an r-concavity constant 1 (apply Lindenstrauss-Tzafriri [25]p.88 Lemma 1.f.11 to E∗.) Let C1 be the associated (LSP) constant for this couple.

Wefirst prove a claim:Claim. There exist constants C2 and δ < 1 depending only on (E, F) with the followingproperty.

Suppose {xn, yn}n∈I is a positive interlaced pair of sequences with ∥xn∥E =∥yn∥E = 1. Then there is a subset D of J, and a positive matrix operator S with ∥S∥(E,F ) ≤C2 so that Sxn = PDyn and ∥yn −PDyn∥E ≤δ, whenever n ∈I.Choose x∗k ≥0 with supp x∗k ⊂supp xk and ∥x∗k∥E = ∥xk∥E = ⟨xk, x∗k⟩= 1.

Similarlychoose y∗k ≥0 with supp y∗k ⊂supp yk and ∥y∗k∥E = ∥yk∥E = ⟨yk, y∗k⟩. We begin by usingthe (LSP) property of (F ∗, E∗) to produce a positive matrix V on E with ∥V ∥(E,F ) ≤C1and V ∗y∗k = x∗k whenever k ∈I.

Thus ⟨V xk, y∗k⟩= 1.Fix τ > 0 small enough so that 12τ −1pCp1τ p = γ > 0. Let Dk be the set of j ∈supp ykso that 2V xk(j) ≥yk(j).

Let D = ∪k∈IDk. Clearly there is a positive matrix S with∥S∥(E,F ) ≤2C1 = C2 and Sxk = PDyk.

Now observe that ⟨V xk −PDV xk, y∗k⟩≤12 so that⟨τPDV xk + yk −PDyk, y∗k⟩≥1 + 12τ −∥PDyk∥E.Thus1 + 12τ −∥PDyk∥E ≤(Cp1τ p + 1)1/p ≤1 + 1pCp1τ p.Upon reorganization this yields:∥PDyk∥E ≥12τ −1pCp1τ p = γ.14

This in turn implies∥yk −PDyk∥E ≤(1 −γr)1/r = δ < 1.This establishes the claim.To complete the proof from the claim is quite easy by an inductive argument. We mayclearly construct a disjoint sequence of subsets (Dn)n≥1 of J and a sequence of positivematrix operators (Sn)n≥1 with ∥Sn∥(E,F ) ≤2C1δn−1 and so that Snxk = PDnyk and∥yk −Pnj=1 PDjyk∥E ≤δn.

The operator T = P∞n=1 Sn is a positive matrix and Txk = yk;further ∥T∥(E,F ) ≤2C1(1 −δ)−1.Proposition 3.6. Suppose (E, F) is a pair of K¨othe sequence spaces.

Suppose either :(a) (E, F) is exponentially separated, F is r-concave for some r < ∞, and there exists1 < p < ∞so that E has a lower p-estimate and F has an upper p-estimate.or(b) κ−(E)κ+(F) < 1.Then (E, F) has (SP) if and only if E has (RSP) and F has (LSP).Proof: (a) We use Lemma 3.5 to show that (E, F) has (RSP) and ( ˜F, ˜E) has (RSP) andthe result follows. (b) This is immediate from Lemma 3.3.4.

Calder´on couples of sequence spaces.We now turn to calculating the K-functional for an exponentially separated pair.Lemma 4.1. Suppose (E, F) is exponentially separated.

Then there is a constant C2 sothat if ρ(a) ≤t ≤ρ(a + 1)K(t, x) ≤∥x(−∞,a]∥E + t∥x(a,∞)∥F ≤C2K(t, x).In particular,∥x(−∞,a]∥E + ρ(a)∥x[a,∞)∥F ≤C2K(ρ(a), x).Similarly if t ≤ρ(a) for all a (in the case J = Z+) thent∥x∥F ≤C2K(t, x)while if t ≥ρ(a) for all a (when J = Z−) then∥x∥E ≤C2K(t, x).Proof: If supp x ⊂(−∞, a] then it follows from Lemma 3.2 that C1K(ρ(a), x) ≥∥x∥E.Similarly if supp x ⊂[a, ∞) then C1K(ρ(a), x) ≥ρ(a)∥x∥F. Combining these statementsgives the results.15

Theorem 4.2. Suppose (E, F) is exponentially separated and forms a Calder´on pair.Then E satisfies (RSP) and F satisfies (LSP).Proof: First we remark that it suffices to prove the result for E. Once this is establishedwe can apply an order-reversal argument to get the result for F. Indeed ( ˜F, ˜E) is alsoexponentially separated and a Calder´on pair; thus ˜F has (RSP) and F has (LSP).We will suppose that ρ(m) ≤C02−nβρ(m + n) for m, m + n ∈J and that C1 and C2are the constants given in Lemmas 3.2 and 4.1.We now introduce a notion which helps in the argument.

An admissible pair is a pair(x, I) where I is a finite interval in J and x is a positive vector with supp x ⊂I, max(supp x) < max I, and ∥x∥E = 1. An admissible family is a finite collection F = (xk, Ik)nk=1of admissible pairs so that (Ik) are pairwise disjoint.

We define supp F = ∪Ik. If F is anadmissible family then we define Γ(F) to be the least constant M so that if (yk)nk=1 satisfy∥yk∥E ≤1, supp yk ⊂Ik and supp xk < supp yk, then there exists T ∈A(E, F) with∥T∥(E,F ) ≤M and Txk = yk for 1 ≤k ≤n.

Notice that since max(supp xk) < max Ikthere is “room” for some yk satisfying our hypotheses. It is not difficult to show thatsuch a Γ(F) is well-defined since we can restrict the problem for each such family to afinite-dimensional space.We next make the remark that if T is such an optimal choice of operator then T can bereplaced without altering its norm by Pnk=1 PIkTPIk.

Thus it can be assumed that Tx = 0for any x whose support is disjoint from ∪Ik. Now suppose F and G are two admissiblefamilies with disjoint supports so that their union F ∪G is also admissible.

Then usingthe above remark it is clear that we can simply add optimal operators to obtain that(1)Γ(F ∪G) ≤Γ(F) + Γ(G).Next suppose F is a single admissible pair (x, I). Suppose y is supported on I andsatisfies ∥y∥E ≤1, and supp x < supp y.

Then we can choose x∗∈E∗with ∥x∗∥E∗= 1supp x∗⊂supp x and ⟨x, x∗⟩= 1. Consider the operator S defined by Sξ = ⟨ξ, x∗⟩y.

Ofcourse ∥S∥E ≤1. Now suppose the maximum of supp x is a. Then∥Sξ∥F ≤∥y∥F∥ξ(−∞,a]∥E≤C21∥ξ∥Fwhere C1 is the constant of Lemma 3.2.

Hence Γ(F) ≤C21. It then follows by the additionprinciple (1) that if |F| = n then Γ(F) ≤nC21.Now we seek to prove that Γ(F) is bounded over all admissible families.Let ussuppose on the contrary that it is not.

We then can construct inductively a sequence of16

admissible families (Fn) for n ∈N and an increasing sequence of integers (mn) so thatsupp Fn ⊂[−mn, mn] and Γ(Fn+1) ≥n(n + 1) + C21(2mn + 2n + 1).Now refine Fn by deleting all pairs (x, I) so that I intersects [−mn −n, mn + n].This removes at most 2mn + 1 pairs and creates a new admissible family F′n so thatΓ(F′n) ≥n(n + 1). The families F′n are now disjoint.

If we write the members of F′n inincreasing order of support as (xk, Ik)Nk=1 then we can define Fn,r for 0 ≤r ≤n to be thefamily of all (xk, Ik) where k ≡r mod (n+1)). At least one of Fn,r satisfies Γ(Fn,r) ≥n by(1).

Call this family Gn. We note that if (x, I) and (y, J) are two consecutive members of Gnthen I + n < J (since n nontrivial intervals in Fn lie between I and J.) Furthermore thereis a gap of at least n between any interval represented in Gn and any interval representedin Gk for some k < n.Finally let us consider the union of all Gn for n ≥1.

This may be written as a sequenceof admissible pairs (xk, Ik)k∈A where A is one of the sets Z, Z−, Z+ and Ik < Ik+1 for allk, k+1 ∈A. Let us write Ik = [ak, bk].

Then bk < ak+1 whenever k, k+1 ∈A. Furthermorethe gaps between the intervals tend to infinity as |k| →∞.

Precisely, if σk = (ak+1 −bk)then lim|k|→∞σk = ∞. Now let dk =max (supp xk) so that ak ≤dk < bk.

Let Jk = (dk, bk]for k ∈A.We now claim:Claim. There exists a finite subset A0 of A and a constant M so that if A1 = A \ A0,and (yk)k∈A1 is any sequence satisfying ∥yk∥E = 1 and supp yk ⊂Jk, then there existsT ∈A(E, F) with ∥T∥(E,F ) ≤M and ∥Txk −yk∥E ≤12.Let us first assume the claim is established and show how the proof is completed.Under these hypotheses we consider the space Y = ℓ∞(E(Jk))k∈A1 and the map S :A(E, F) →Y defined by S(T) = (PJkTkxk)k∈A1.

Clearly ∥S∥≤1 and it follows fromthe claim that if y = (yk)k∈A1 ∈Y there exists T ∈A(E, F) with ∥T∥(E,F ) ≤M∥y∥and ∥S(T) −y∥≤12∥y∥. By a well-known argument from the Open Mapping Theoremthis is enough to show that S is onto and indeed if ∥y∥≤1 then there exists T with∥T∥(E,F ) ≤2M and S(T) = y.Now suppose Gn = {(xk, Ik)}k∈Bn where Bn ⊂A1.

Then if (yk)k∈Bn satisfy ∥yk∥E ≤1, and supp xk < supp yk ⊂Ik it follows that there is an operator T ∈A(E, F) with∥T∥(E,F ) ≤2M and PJkTxk = yk. If we set T0 = Pk∈Bn PJkTPIk then ∥T0∥(E,F ) ≤2Mand T0xk = yk.

Thus Γ(Gn) ≤2M. Now since A0 is finite we conclude that Γ(Gn) ≤2M forall but finitely many n. This contradicts the original construction of Gn.

The contradictionshows that there is a constant M0 so that Γ(F) ≤M0 for all admissible families F. Inparticular if we have a finite set of finitely supported vectors x1, x2, . .

., xn, y1, . .

., yn sothat supp x1 < supp y1 < . .

. < supp xn < supp yn and ∥xk∥E = 1 for all k and ∥yk∥E ≤117

then there is an operator T : E →E with ∥T∥E ≤2M0 and Txk = yk. Hence for anyα1, .

. .

, αn we would have∥nXk=1αkyk∥E ≤2M0∥nXk=1αkxk∥and this means that E has (RSP).Thus it only remains to prove the claim. We start by defining a sequence (λk)k∈Asuch that λk+1 −λk = 12βσk.

We next make some initial observations. Let us suppose thatsupp uk ⊂Ik for k ∈A and ∥uk∥E = 1 for all k. We claim that there exists a constant C3independent of the choice of (uk) so that if k ∈A then(2)∥Xj≤k2λjuj∥E ≤C32λkand(3)∥Xj≥k2λjuj∥F ≤C32λk∥uk∥F .In fact (2) follows easily from the fact that if j ≤k thenλj = λk −12βk−1Xi=jσi ≤λk −12(k −j)β.For (3) we note that if j ≥k,∥uj∥F ≤C1ρ(aj)−1≤C1C02−β(aj−bk)ρ(bk)−1≤C21C02−β(aj−bk)∥uk∥F≤C21C02−2(λj−λk)∥uk∥Fso that2λj∥uj∥F ≤C21C02−12 β(j−k)2λk∥uk∥Ffrom which (3) will follow.In particular let us define z = Pk∈A 2λkxk.

The above calculations show that z ∈E+F. Since (E, F) is a Calder´on couple there is a constant M0 = M0(z) so that if u ∈E+Fand K(t, u) ≤K(t, z) for all t then there exists T ∈A(E, F) with ∥T∥(E,F ) ≤M0 andTz = u.18

Now suppose (yk)k∈A is any sequence with ∥yk∥E = 1 and supp yk ⊂Jk. We setv = P 2λkyk ∈E + F. We turn to comparing K(t, v) with K(t, z).

Let us note first thatfor every k ∈A we have∥yk∥F ≤C1ρ(dk)−1 ≤C21∥xk∥F .If t satisfies t ≤ρ(ak) for all k then we must have A = Z+ and we make the estimateK(t, v) ≤t∥v∥F ≤C3t2λ0∥y0∥F ≤C3C21t2λ0∥x0∥Fso that by Lemma 3.3,K(t, v) ≤C3C2C21K(t, z).Similarly if t ≥ρ(bk) for all k ∈A then we can have A = Z−and we make a similarestimateK(t, v) ≤∥v∥E ≤C32λ−1 ≤C3∥z[a−1,b−1]∥E ≤C2C3K(t, z).In the other cases we first consider the case when ρ(n) ≤t ≤ρ(n + 1) for some n inan interval [ak, dk). Then K(t, xk) ≥tC−11 ρ(dk)−1 by Lemma 3.2.HenceK(t, v[ak,∞)) ≤C3t∥v[ak,ak+1]∥F= C3t2λk∥yk∥F≤C3C1t2λkρ(dk)−1≤C3C212λkK(t, xk)≤C3C21K(t, z).If k is the initial element of A we are done.

Otherwise:K(t, v(−∞,ak)) ≤C32λk−1≤C3∥z(−∞,ak)∥E≤C3K(t, z).Combining in this case we have K(t, v) ≤CK(t, z) for some constant C depending onlyon C1, C2 and C3.For the final case, we can suppose there exists n not in any interval [ak, dk) and suchthat ρ(n) ≤t ≤ρ(n + 1); it may also be assumed that there exists k ∈A with k + 1 ∈Aand dk ≤n ≤ak+1. Then by Lemma 3.3,∥z(−∞,n]∥E + t∥z(n,∞)∥F ≤C2K(t, z).19

Now∥v(−∞,bk]∥E ≤C32λk ≤C3∥z(−∞,n]∥E.Also∥v(bk,∞)∥F ≤C32λk+1∥yk+1∥F ≤C21C3∥z(n,∞)∥F .Thus combining all the cases there exists C4 independent of (yk) so that K(t, v) ≤C4K(t, z). Hence there is an operator T ∈A(E, F) with Tz = v and ∥T∥(E,F ) ≤C4M0.Now for fixed k ∈A assume first that k is not the initial element of A. Then∥(z(−∞,ak)∥E ≤C32λk−1 ≤C32−12 βσk−12λk.Thus we have that∥T(z(−∞,ak))∥E ≤C4C3M02−12 βσk−12λk.If k is not the final element,∥z(bk,∞)∥F ≤C32λk+1∥yk+1∥F≤C1C32λk+1ρ(ak+1)−1≤C0C1C32λk+1−βσkρ(bk)−1.Thus if f = T(z(bk,∞))(dk,bk] then∥f∥E ≤C1ρ(bk)∥f∥F≤C1C4M0ρ(bk)∥z(bk,∞)∥F≤C0C21C3C4M02−12 βσk2λk.It follows that if k is not an initial or final element of A,∥yk −PJkTxk∥E ≤C5M02−12 βτkwhere C5 is a constant depending only on E and F and τk = min(σk−1, σk).

Now if we setS = Pk∈A PJkTPIk then ∥S∥(E,F ) ≤C4M0. Further if we let A1 be the set of k ∈A sothat k is not an initial or final element and C5M 20 2−12 βτk < 1/2 then A0 = A \ A1 is finiteand ∥Sxk −yk∥< 1/2 for k ∈A1.

Thus the claim is established and the proof is complete.Lemma 4.3. Suppose (E, F) satisfies (RSP).

Then there is a constant C so that if 0 ≤x, y ∈E + F and ∥y(−∞,a]∥E ≤∥x(−∞,a]∥E for all a ∈J then there exists a positiveT ∈A(E, F) with ∥T∥(E,F ) ≤C and Tx = y.Proof: By applying the argument of Lemma 2.6 we deduce from (RSP) the existence ofa constant C0 so that if A is an interval in Z, (Jk)k∈A is a collection of finite intervals in J20

with Jk < Jk+1 whenever k, k + 1 ∈A and (xk)k∈A, (yk)k∈A are positive and satisfy suppxk, supp yk ⊂Jk and ∥yk+1∥E ≤∥xk∥E for k, k + 1 ∈A then there is a positive matrixoperator T with ∥T∥(E,F ) ≤C0 and Txk = yk+1 whenever k, k + 1 ∈A.Let us prove the lemma when x, y have disjoint supports. We first define a functionσ : Z →Z ∪{±∞} by setting σ(k) = −∞if k < J, σ(k) = ∞if k > J and otherwise σ(k)is the greatest j ∈Z ∪{−∞} so that ∥x(−∞,k]∥E ≥4j.Let I0 = {k ∈J : σ(k) > σ(k −1)}.

We then let I be the subset of I0 of all k so thatfor every n ∈I0 with n < k then ∥x(−∞,n]∥E ≤12∥x(−∞,k]∥E.We can now index I as (an)n∈A where A is an interval in Z which can be assumed tohave 0 as its initial element if I is bounded below.We now define B to be Z−when infk∈Z σ(k) > −∞and to be empty otherwise.We will only need to introduce B in the case when lima→−∞∥x(−∞,a]∥E > 0. If B isnonempty then I is bounded below and there exists a greatest λ so that ∥x(−∞,k]∥E ≥4λfor every k (in this case J cannot be bounded below).

We must have ∥x(−∞,k]∥E < 4λ+1whenever k < a0. It follows that we may pick a−1 so that ∥x(a−1,a0]∥E ≥4σ(a0)−1 and theninductively a−n so that ∥x(a−n,a−(n−1)]∥E ≥4λ−1.In this way we define (an)n∈A∪B.

We now let xn = x(an−1,an] and yn = y(an−1an]if n ∈A ∪B is not the initial element of A ∪B; if n = 0 is the initial element we letx0 = x(−∞,a0] and y0 = y(−∞,a0]. If n is the final element of A ∪B we set yn+1 = y(an,∞].We may now verify that Pn∈A∪B xn ≤x.

We also claim that Pn∈A∪B yn+1 = y. IfA ∪B = Z this is clear. If A ∪B = (−∞, n] for some n it follows from our definition ofyn+1.

If A ∪B is bounded below (by 0) then B is empty and hence σ(a0 −1) = −∞. Thusy0 = 0 and we obtain our claim easily.We first prove that if n, n + 1 ∈A then σ(an+1 −1) ≤σ(an) + 1.

If not there exists afirst k1 so that σ(k1) ≥σ(an) + 1 and a first k2 so that σ(k2) ≥σ(an) + 2 and an < k1 ≤k2 ≤an+1 −1. Then k1, k2 are in I0 \ I.

Thus ∥x(−∞,an]∥E > 12∥x(−∞,k1]∥E. The equalityk1 = k2 would entail ∥x(−∞,k1]∥E ≥4σ(an)+2 and thus ∥x(−∞,an]∥E > 4σ(an)+1 whichcontradicts the definition of σ(an).

Thus k1 < k2 and we conclude also that ∥x(−∞,k1]∥E >12∥x(−∞,k2]∥E so that ∥x(−∞,an]∥E > 14∥x(−∞,k2]∥E which implies the absurd conclusionσ(k2) ≤σ(an) + 1. Thus, as claimed, σ(an+1 −1) ≤σ(an) + 1.The same argument shows that if A is bounded above then if k > an we must haveσ(k) ≤σ(an) + 1.Now if n, n + 1 ∈A we can argue that since x, y have disjoint supports that yn+1 issupported on (an, an+1) and thus ∥yn+1∥E ≤∥x(−∞,an+1−1]∥E ≤4σ(an)+2.

Similarly, letn be the last element of A. Then for all k > an ∥y(−∞,k]∥E ≤∥x(−∞,k]∥E ≤4σ(k)+1 ≤4σ(an)+2.

Thus ∥yn+1∥E ≤4σ(an)+2.21

On the other hand, if n is not the initial element of A,∥xn∥E ≥∥x(−∞,an]∥E −∥x(−∞,an−1]∥E ≥124σ(an).If n = 0 is the initial element, we either have, if B = ∅, x0 = x(−∞,a0] so that ∥x0∥≥4σ(a0)or if B ̸= ∅then ∥x1∥≥4σ(a0)−1. In all such cases, if n ∈A we have ∥yn+1∥E ≤43∥xn∥E.Next suppose n, n + 1 ∈B.

Then ∥yn+1∥E ≤∥x(−∞,an+1)∥E ≤4λ+1 while ∥xn∥≥4λ−1. Thus ∥yn+1∥E ≤42∥xn∥E.Finally, consider the case n = −1 ∈B and n + 1 = 0 ∈A.

Then since a0 is in thesupport of x we have ∥yn+1∥E ≤∥x(−∞,a0−1)∥E < 4λ+1. However ∥xn∥E ≥4λ−1 so that∥yn+1∥E ≤42∥xn∥E.Combining all cases, we conclude that there is a positive operator T with ∥T∥(E,F ) ≤43C0 so that Txn = yn+1.

Now it is clear that Pn∈A∪B xn ≤x while Pn∈A∪B yn+1 = y.Thus if S = Pn∈A∪B Psupp yn+1TPsupp xn then ∥S∥(E,F ) ≤64C0 and Sx = y. Thus thelemma is established in the case when x and y have disjoint supports.For the general case we let I = {n ∈J : yn > 2xn}.

Let J = J \ I. Then set u = xJand v = yI.

For any k ∈J we have ∥xI∩(−∞,k]∥E ≤12∥y(−∞,k]∥E ≤12∥x(−∞,k]∥E. Thus∥u(−∞,k]∥E ≥12∥x(−∞,k]∥E.

Hence there is a positive operator S in A(E, F) with Su = vand ∥S∥(E,F ) ≤128C0. On the other hand yJ ≤2x and so there is a multiplication operatorV ∈(E, F) with ∥V ∥(E,F ) ≤2 and V x = yJ.

Finally the operator T = SPJ +V establishesthe lemma.Lemma 4.4. Suppose (E, F) is exponentially separated and satisfies (SP).

Then thereexists a constant C so that if 0 ≤x, y ∈E + F and K(t, y) ≤K(t, x) for all t ≥0 thenthere exists a positive matrix T ∈A(E, F) with ∥T∥(E,F ) ≤C and Tx = y.Proof: It follows from Lemma 4.1 that there is a constant C2 so that for all a ∈J wehave, whenever K(t, y) ≤K(t, x) for all t ≥0,max(∥y(−∞,a]∥E, ρ(a)∥y[a,∞)∥F ) ≤2C2 max(∥x(−∞,a]∥E, ρ(a)∥x(a,∞)∥F ).Thus for every a either(4)∥y−∞,a]∥E ≤2C2∥x(−∞,a]∥Eor(5)∥y[a,∞)∥F ≤2C2∥x[a,∞)∥F .Let J1 be the set of a so that (4) holds and let J2 = J \ J1. Since (E, F) has (RSP) wecan apply Lemma 4.3 to deduce the existence of a positive matrix T1 with ∥T1∥(E,F ) ≤C3where C3 depends only (E, F) and T1x = yJ1.

Similarly since (E, F) has (LSP) we canfind a positive matrix T2 with ∥T2∥(E,F ) ≤C3 so that T2x = yJ2. Then (T1 + T2)(x) = y.22

Theorem 4.5. Let (E, F) be a pair of K¨othe sequence spaces.

Suppose either:(a) κ−(E)κ+(F) < 1.or(b) (E, F) is exponentially separated, F is r-concave for some r < ∞and there exists pwith 1 ≤p < ∞so that E has a lower p-estimate and F has an upper p-estimate.Then (E, F) is a (uniform) Calder´on couple if and only if E has (RSP) and F has (LSP).Proof: This is an immediate deduction from Proposition 3.6, Theorem 4.2 and Lemma4.4.Remark: Note that in fact Lemma 4.4 implies that under these circumstances if K(t, y) ≤K(t, x) for all t and x, y ≥0 then there is a positive operator T with ∥T∥(E,F ) < ∞andTx = y.The following theorem is similar to results of Cwikel and Nilsson [18].Theorem 4.6. Let E, F be symmetric sequence spaces on Z+ and suppose (E, F(w)) isa Calder´on pair for a weight sequence w = (wn).

Then either F(w) = F (i.e. 0 < inf wn ≤sup wn < ∞) or E = ℓp, F = ℓq for some 1 ≤p, q ≤∞.Proof: If (wn) is unbounded we can pass to a subsequence satisfying wnk > 2wnk−1.Then the pair (E, F(wnk)) is a Calder´on pair and we can apply Theorem 4.2 to get that Fhas (RSP) and E has (LSP).

An application of Proposition 2.3 gives the result. If (w−1n )is unbounded we can argue similarly.5.

Calder´on couples of r.i. spaces.Let Ωdenote one of the sets [0, ∞), [0, 1] and N. Let J be the set Z, Z−, or Z+respectively. If X is an r.i. space on Ω(or a symmetric sequence space if Ω= N) we willassociate to X a K¨othe sequence space EX on J.

To do this let en, n ∈J be defined byen = χ[2n,2(n+1)). We then define for x ∈ω(J),∥x∥EX = ∥Xk∈Jx(k)ek∥X.

(Here we use ek with a dual meaning as both the canonical basis element of ω(J) and as anelement of X(Ω).) We observe that EX regarded as a subspace of X is 1-complemented bythe natural averaging operator.

Notice also that EX∗= E∗X(2n) is a weighted version ofE∗X. We also note that on EX we can compute ∥τn∥EX ≤∥D2n∥X where Ds is the naturaldilation operator.

Furthermore it is easy to see that for f ∈X we have D2nf ∗≤τn+1Pf ∗where P is the natural averaging projection of X onto EX; thus ∥D2n∥X ≤∥τn+1∥EX.Thus κ+(EX) = 21/pX and κ−(EX) = 2−1/qX where pX and qX are the Boyd indices ofX.23

We now show how to build examples of r.i. spaces from sequence spaces. To keepthe notation straight we prove our results for the case of function spaces Ω= [0, 1] orΩ= [0, ∞).

However simple modifications give the analogous results for sequence spaces.Proposition 5.1. Let E be a K¨othe sequence space on J.

Then:(1) If κ+(E) < 2 there is an r.i.space X = X(Ω) so that ∥f∥X is equivalent to∥Pn∈J f ∗(2n)en∥E. (2) If κ−(E) < 1 ≤κ+(E) < 2, and X is an r.i. space so that ∥f∥X is equivalent to∥Pn∈J f ∗(2n)en∥E then EX = E (up to equivalence of norm).Proof: (1) We define X to be the set of measurable functions on Ωsuch that∥f∥X = ∥Xn∈Jf ∗(2n)en∥E < ∞.We show that the functional ∥f∥X is equivalent to a norm by computing ∥f ∗∗∥X wheref ∗∗(t) = 1tR t0 f ∗(s)ds.

Thenf ∗∗(2n) ≤2−n Xk

space. (2) Let ∥∥X denote the quasinorm induced by E. We remark that it follows from(1) that there exists a constant C2 so that for f ∈X we have ∥f ∗∗∥X ≤C2∥f∥X.

Now,considering the EX-quasinorm induced on ω(J) it is clear that if x is a nonincreasingsequence then ∥x∥EX = ∥x∥E. In general we note that if f ∈EX then for some C3 =Pj≥0 ∥τ−j∥E since κ−(E) < 1,∥maxj≥0 |τ−jf|∥E ≤C3∥f∥Eso that∥f∥EX ≤C1∥f∥E.For the converse direction we observe that if f ∈X it is trivial that ∥D2jf∥X ≤∥τj∥E∥f∥X.24

Then∥f∥E ≤∥maxj≥0 |τ−jf|∥E= ∥maxj≥0 |τ−jf|∥X≤∥Xj≥0D2−j|f|∥X≤Xj≥0∥D2−jf ∗∗∥X≤C3∥f ∗∗∥X≤C2C3∥f∥X.Thus EX is (up to equivalence of norm) identical with E.Remark: It follows from the above Proposition that there is a natural one-one corre-spondence between r.i. spaces X with Boyd indices satisfying 1 < pX ≤qX < ∞andsequence spaces E on J with κ−(E) < 1 ≤κ+(E) < 2 determined by E = EX. Under thiscorrespondence if 1 < p < ∞an r.i. space X with qX < ∞is a Lorentz space (of order p)if and only if EX is a weighted ℓp−space.

For if∥f∥X = (Z ∞0(f ∗(t)w(t))p dtt )1/pwhere w is an increasing function satisfying 1 < inf w(2t)/w(t) ≤sup w(2t)/w(t) < ∞then the above Proposition shows that EX = ℓp(wn), where wn = w(2n). Conversely ifEX is an ℓp−space then EX = ℓp(wn) where the assumption that qX < ∞enables us toassume inf wn+1/wn > 1.

If we define w(t) = wn whenever 2n−1 < t ≤2n then it is easyto see that X is a Lorentz space.We now prove the elementary:Proposition 5.2. Let (X, Y ) be a pair of r.i. spaces on Ω.

Then (X, Y ) is a Calder´oncouple if and only if (EX, EY ) is a Calder´on couple.Proof: By using the averaging projection it is clear that if (X, Y ) is a Calder´on couplethen so is (EX, EY ). In fact it is trivial to see that for f ∈EX +EY we have K(t, f; X, Y ) =K(t, f; EX, EY ).

Thus if K(t, g; EX, EY ) ≤K(t, f; EX, EY ) for all t there exists T ∈A(X, Y ) so that Tf = g. If P is the averaging projection then PT ∈A(EX, EY ) andPTf = g.Conversely suppose (EX, EY ) is a Calder´on couple.Suppose f, g ∈X + Y andK(t, g; X, Y ) ≤K(t, f; X, Y ) for all t ≥0. We then observe that if G = Pn∈J g∗(2n)enand F = Pn∈J f ∗(2n)en then g∗≤G ≤D2g∗and f ∗≤F ≤D2f ∗.

andK(t, G; X, Y ) ≤K(t, D2g∗; X, Y ) ≤2K(t, f; X, Y ) ≤2K(t, F; X, Y ).25

Since F, G are in EX + EY we can deduce the existence of T ∈A(EX, EY ) with TF = G.Now since F ≤D2f ∗and g∗≤G it is clear that there exists S ∈A(X, Y ) with Sf = g.Remarks: It now follows that every pair of Lorentz spaces whose Boyd indices are finiteis a Calder´on couple, since every pair of weighted ℓp−spaces is a Calder´on couple (cf. [36],[13]); this result is due to Cwikel [14] and Merucci [30] for certain special cases.We introduce the following definitions.

We say X is stretchable if EX has (RSP) andwe say that X is compressible if EX has (LSP). If X both stretchable and compressible, wesay that X is elastic.

It is immediate from Proposition 2.1 that X is stretchable if and onlyif X∗is compressible and vice versa; thus elasticity is a self-dual property. We remark thatwe have no example of a stretchable (or compressible) space which is not already elastic.In fact we shall see that for Orlicz spaces these concepts do indeed coincide.Theorem 5.3.

Let (X, Y ) be a pair of r.i. spaces on Ωwhose Boyd indices satisfy pY > qX.Then (X, Y ) is a Calder´on couple if and only if X is stretchable and Y is compressible.Proof: As κ−(EX) = 2−1/qX and κ+(EY ) = 21/pY we have κ−(EX)κ+(EY ) < 1 and sothe theorem is immediate from Theorems 4.5 and 5.2.If one space is L∞we can do rather better.Theorem 5.4. Let X be an r.i. space on Ω= [0, 1] or Ω= [0, ∞).

Then (X, L∞) is a(uniform) Calder´on couple if and only if X is stretchable. Similarly if X is a symmetricsequence space then (ℓ∞, X) is a (uniform) Calder´on couple if and only if X is stretchable.Before proving Theorem 5.4 we state a result which has a very similar proof.

Weremark that Theorem 5.5 only improves on Theorem 5.3 under the assumption that pY =p = qX since the case pY < qX is already covered.Theorem 5.5. Suppose (X, Y ) is a couple of r.i. spaces on Ωso that for some 1 ≤p < ∞X is p-concave and Y is p-convex and suppose also that Y is r-concave for some r < ∞.Then (X, Y ) is a (uniform) Calder´on couple if and only if X is stretchable and Y iscompressible.Proofs of Theorems 5.4 and 5.5: Theorem 5.4 corresponds to the case p = ∞, andY = L∞.

We can and do assume that the p-convexity constant of Y and the p-concavityconstant of X are both equal to one. Under this hypothesis it is easy to see that, whenp < ∞2−k/p∥ek∥X is increasing and 2−k/p∥ek∥Y is decreasing.

Thus for p ≤∞, ρ(k) =∥ek∥X/∥ek∥Y is an increasing function and ρ(k + 1) ≤2ρ(k) whenever k, k + 1 ∈J. Thenfor k ∈J we let Ik = {n ∈J : 2k < ρ(n) ≤2k+1}.Before continuing let us make remark which we use several times in the proof.

As-suming p < ∞suppose f, g are two finitely supported functions in EX which satisfy26

∥f∥p = ∥g∥p andZ t0(f ∗(s))pds ≤Z t0(g∗(s))pdsfor every t ≥0. Then we have the inequalities ∥f∥Y ≤∥g∥Y and ∥f∥X ≤∥g∥X.

In fact itfollows from a well-known lemma of Hardy, Littlewood and Polya, [19], [25], p.124, that|f ∗|p is in the convex hull of the set of all rearrangements of |g∗|p; this can be proved bypartitioning the supports of f ∗, g∗into finitely many sets of equal measure. The assertionis then a direct consequence of the definitions of p-convexity and p-concavity.We make some initial remarks which will be needed in both directions of the proof.Each set Ik is an interval (possibly infinite) or is empty.

The set of k so that Ik is nonemptyis an interval A. Let E(Ik) be the linear span of (en : n ∈Ik) when k ∈A.

We state thefollowing Lemma.Lemma 5.6. If f, g ∈E(Ik) then, under the hypotheses of Theorem 5.5,∥f∥X∥g∥p ≤2∥f∥p∥g∥X∥f∥Y ∥g∥p ≤2∥f∥p∥g∥Xwhere ∥∥p denotes the usual Lp-norm, so that ∥P αkek∥p = (P 2k|αk|p)1/p.Under the hypotheses of Theorem 5.4, we have∥f∥X∥g∥∞≤4∥f∥∞∥g∥X.Proof: In fact suppose f, g ∈[en : a ≤n ≤b] where a, b ∈Ik, and that neither is zero.We may observe that for all t ≥0 we have2−aZ t0(e∗a(s))p)ds ≥∥f∥−ppZ t0(f ∗(s))pds ≥2−bZ t0(e∗b(s))pdswith similar inequalities for g. It thus follows from the remarks above that2−a/p∥ea∥Y ≥∥f∥−1p ∥f∥Y ≥2−b/p∥eb∥Y .Similarly2−a/p∥ea∥X ≤∥f∥−1p ∥f∥X ≤2−b/p∥eb∥X.There are similar inequalities for g. Since 2k < ρ(a) ≤ρ(b) ≤2k+12−b/p∥eb∥X ≤2k+1−b/p∥eb∥Y ≤2k+1−a/p∥ea∥Y ≤2.2−a/p∥ea∥X.27

Combining these we see that∥f∥−1p ∥f∥X ≤2∥g∥−1p ∥g∥Xand∥f∥−1p ∥f∥Y ≤2∥g∥−1p ∥g∥Ywhence the claimed inequalities follow. For the last part, we observe that∥f∥∞∥ea∥X ≤∥f∥X ≤∥f∥∞∥ea + · · · + eb∥X ≤2∥f∥∞∥eb∥Xand proceed similarly.We draw immediately the conclusion that if A is finite (so that ρ is bounded) thenboth X and Y coincide with Lp(µ) and there is nothing to prove.

In other cases at most oneIk is infinite. We write Ik = [ak, bk] if Ik is finite and Ik = [ak, ∞) or Ik = (−∞, bk] if Ik isinfinite.

Let A0 be the set of k so that k−1 and k+1 ∈A. We define a set J by taking onepoint dk from each Ik for k ∈A.

We introduce the sequence spaces FX and FY modelledon A by setting ∥x∥FX = ∥Pk∈A 2−dk/px(k)edk∥X and ∥x∥FY = ∥Pk∈A 2−dk/px(k)edk∥Y .In the case p = ∞we define ∥x∥FX = ∥Pk∈A x(k)edk∥X.Lemma 5.7. Under the hypotheses of Theorem 5.5, suppose EY (J) has (LSP).

Then thereis a constant C0 so that if f ∈EY then ∥f∥Y is C0−equivalent to ∥(∥fIk∥p)∥FY .Proof: It suffices to prove such an equivalence if f ∈EY satisfies fIk = 0 for k /∈A0,since there are most two values of k /∈A0 and Lemma 5.6 shows that the Y -normon each such E(Ik) is equivalent to the Lp−norm.Next observe that for such f ifg = Pk∈A0 2−dk+1/p∥fIk∥pedk+1 then for all t ≥0,R t0(g∗(s))pds ≤R t0 (f ∗(s))pds. Thuswe have immediately by the p-convexity and rearrangement-invariance of Y, ∥g∥Y≤∥f∥Y .

Similarly if h = Pk∈A0 2−dk−1/p∥fIk∥pedk−1 then ∥h∥Y ≥∥f∥Y . Next let ˜f =Pk∈A0 2−dk/p∥fIk∥pedk.

We complete the proof by showing that for some C, ∥h∥Y ≤C∥˜f∥Y and ∥˜f∥Y ≤C∥g∥Y . Once this is done it will be clear that ∥f∥Y is actually equiv-alent to ∥˜f∥Y as claimed.The proofs of these statements are essentially the same, so we concentrate on the first.Note that2−dk−1/p∥edk−1∥Y ≤2−(k−1)−dk−1/p∥edk−1∥X≤2−(k−1)−dk/p∥edk∥X≤21−dk/p∥edk∥Yand so if C is the (LSP) constant of EY (J) we have ∥h∥Y ≤2C∥˜f∥Y .

Similarly ∥˜f∥Y ≤2C∥g∥Y .In a very similar way, exploiting the p-concavity of X one has,28

Lemma 5.8. Suppose EX(J) has (RSP).

Then there is a constant (which we also nameC0) so that if f ∈EX then ∥f∥X is C0−equivalent to ∥(∥fIk∥p)∥FX.Sketch: First consider the case of Theorem 5.5. We assume f ∈EX is finitely supported.Proceed as in Lemma 5.7, defining g, h, ˜f as before.

In this case we have that ∥g∥X ≥∥f∥X ≥∥h∥X. The remainder of the argument mirrors that of Lemma 5.7.Let us also sketch the argument when p = ∞(i.e.

for Theorem 5.4). Analogouslyto Lemma 5.7 we note that ∥g∥X ≥∥f∥X ≥∥h∥X where g = Pk∈A0 ∥fIk∥∞edk+1 andh = Pk∈A0 ∥fIk∥∞edk.

The remainder of the argument is the same.Now let us turn to the proofs of Theorems 5.4 and 5.5. Suppose first that the couple(X, Y ) is a Calder´on couple.

Then the couple (EX(J), EY (J)) must also be a Calder´oncouple since there is a common averaging projection from (X, Y ) onto (EX, EY ). Now it isclear that (EX(J), EY (J)) is exponentially separated (when J is indexed as a sequence).We can thus apply Theorem 4.2 to obtain that EY (J) has (LSP) and EX(J) has (LSP).We conclude this direction of the proof by showing that if EY (J) (and hence FY ) has(LSP) then EY has (LSP) and so Y is compressible.

A very similar argument shows thatX is stretchable.To prove this we suppose that {fj, gj}j∈B is an interlaced pair of positive sequencesin EY with ∥fj∥Y ≤∥gj∥Y = 1. For given j let l(j) be the largest k so that PIkfj ̸= 0.

(Note here if such a largest k does not exist then j is the maximal element of B and gj = 0;hence this case can be ignored.) We then split fj = f ′j + f ′′j where f ′′j = PIl(j)fj.

Similarlywe let gj = g′j + g′′j where g′j = PIl(j)gj. Let B0 = {j : ∥f ′j∥Y ≥1/2}, and let B1 = B \ B0.For j ∈B0 we set vj = (∥PIkf ′j∥p)k∈A ∈FY ; for j ∈B1 we set vj = (∥PIkf ′′j ∥p)k∈A.For all j ∈B we set w′j = (∥PIkg′j∥p) and w′′j = (∥PIkg′′j ∥p).Let (αj)j∈B be positive and finitely nonzero.

First observe that j in B0 we must havesupp vj < supp (w′j + w′′j ). Further ∥vj∥FY ≥(2C0)−1 while ∥wj + w′′j ∥FY ≤C0.

Thus,since FY has (LSP) applying Lemma 2.6 we get the existence of a constant C1 dependingonly on (E, F) so that∥Xj∈B0αj(w′j + w′′j )∥FY ≤C1∥Xj∈B0αjvj∥FY .Notice also that (w′j + w′′j )j∈B0 have disjoint supports so that we can conclude that∥Xj∈B0αjgj∥Y ≤C0∥Xj∈B0αj(w′j + w′′j )∥FY .Similarly∥Xj∈B0αjvj∥FY ≤C0∥Xj∈B0αjfj∥Y .29

Combining we have∥Xj∈B0αjgj∥Y ≤C20C1∥Xj∈B0αjfj∥Y .We now obtain a similar estimate on B1. In fact, if we set B2 = {j ∈B1 : w′′j ̸= 0}then we can argue as above to show that∥Xj∈B2αjw′′j ∥FY ≤C1∥Xj∈B2αjvj∥FYand hence obtain an estimate∥Xj∈B1αjg′′j ∥Y ≤C20C1∥Xj∈B1αjfj∥Y .Finally we observe that for j ∈B1, ∥Pl(j)gj∥p ≤4∥Pl(j)fj∥p by Lemma 5.6.

Thus forany k∥PIkXj∈B1αjg′j∥p = (Xl(j)=k|αj|p∥PIkg′j∥pp)1/p≤4(Xl(j)=k|αj|p∥PIkf ′′j ∥pp)1/p= 4∥PIkXj∈B1αjf ′′j ∥p.Thus∥Xj∈B1αjg′j∥Y ≤4C20∥Xj∈B1αjf ′′j ∥Y .Combining these estimates gives that∥Xj∈Bαjgj∥Y ≤C∥Xj∈Bαjfj∥Yfor a suitable constant C. This completes the proof that Y is compressible and, as explainedabove a similar argument shows that X is stretchable.We now consider the other direction in Theorems 5.4 and 5.5.We suppose X isstretchable and Y is compressible. It follows that EX has (RSP) and EY has (LSP) andwe can apply both Lemmas 5.7 and 5.8.

We can immediately deduce:Lemma 5.9. There exists C so that if 0 ≤f, g ∈EX + EY and ∥fIk∥p ≥∥gIk∥p for allk ∈A then there exists 0 ≤T ∈A(EX, EY ) with ∥T∥(EX,EY ) ≤C and Tf = g.Now suppose f, g ≥0 in EX + EY and that K(t, g) ≤K(t, f) for all t ≥0.

We definef ′ = Pk∈A 2−dk/p∥fIk∥pedk and g′ = Pk∈A 2−dk/p∥gIk∥pedk. Then Lemma 5.9 yields30

the conclusion that K(t, g′) ≤CK(t, g) ≤K(t, f) ≤C2K(t, f ′). Now (EX(J), EY (J)) isexponentially separated.Now for Theorem 5.5 we quote Theorem 4.5 to give that (EX(J), EY (J)) is a Calder´oncouple and hence there exists S ∈A(EX(J), EY (J)) with ∥S∥(EX(J),EY (J)) ≤C2, whereC2 depends only on (E, F), and Sf ′ = g′.

It follows easily from Lemma 5.9 that (EX, EY )and hence (X, Y ) is a uniform Calder´on couple.In the case of Theorem 5.4 we note that it suffices to consider the case when f and gare decreasing functions; then f ′ and g′ are also decreasing. Then K(t, g′) ≤C2K(t, f ′)for all t implies that∥g′χ[0,t]∥X ≤C2∥f ′χ[0,t]∥X.We further note that (EX(J), ℓ∞(J)) has (RSP) by Lemma 3.4 and then apply Lemma4.3 to obtain a positive S ∈A(EX(J), ℓ∞(J)) with ∥S∥EX(J),ℓ∞(J) ≤C2 and Sf ′ = g′.This leads to the desired conclusion.Corollary 5.10.

Let X be an r.i. space on [0, 1] or [0, ∞). Suppose X is r-concave forsome r < ∞.

In order that both (L1, X) and (L∞, X) be Calder´on couples it is necessaryand sufficient that X be elastic.Examples: We begin with the obvious remark that the spaces Lp for 1 ≤p ≤∞areelastic and so our results include the classical results cited in the introduction. On thespace [0, ∞) one can basically separate behavior at ∞from behavior at 0 so that spacesof the form Lp + Lq and Lp ∩Lq are also elastic.

Note however that we cannot applyTheorems 5.3 or 5.5 unless we have appropriate assumptions on either the Boyd indicesor convexity/concavity assumptions; thus pairs of of such spaces are not always Calder´oncouples.Let us now specialize to [0, 1]. In certain special cases we can easily see that an r.i.space is elastic.

For example, suppose X is the Lorentz space on [0, 1], for which qX < ∞.Then it is immediately clear that X is elastic since EX is a weighted ℓp−space. Rathermore obscure elastic spaces can be built using a weighted Tsirelson space for EX.On the other hand, it is possible to give easy examples where EX fails (RSP) or (LSP).Indeed if one takes any symmetric sequence space E on J which is not an ℓp-space andconsiders E(wn) where 1 < w < 2 then there is an r.i. space X for which EX = E(wn).By Proposition 2.3 EX fails (RSP) and (LSP).

In this case we note that since κ+(EX) = wand κ−(EX) = w−1, we have pX = qX = (log2 w)−1. If say E = ℓF (Z−) for some Orliczfunction F satisfying the ∆2−condition then X is an “Orlicz-Lorentz space” given by∥f∥X ∼Z 10F(f ∗(t)t−1/p)dtt31

where p = pX = qX. Note that for such a space the pair (L∞, X) fails to be a Calder´oncouple.

This answers a well-known question (cf. [8],[28]).In the next section we will investigate Orlicz spaces in more detail.

We will also giveexamples of Orlicz spaces LF for which (L∞, LF) is not a Calder´on couple.We will conclude this section by considering a situation suggested by the example ofOvchinnikov [34] (cf. [29]).Theorem 5.11.

Suppose 1 < p < ∞and that X is an r.i. space on [0, ∞) whose Boydindices satisfy either qX < p or p < pX ≤qX < ∞. Then (X ∩Lp, X + Lp) is a Calder´oncouple if and only if X is a Lorentz space of order p.Proof: If X is a Lorentz space of order p, then both X +Lp and X ∩Lp are also Lorentzspaces of order p, and so form a Calder´on couple.

Conversely suppose (X ∩Lp, X + Lp)is a Calder´on couple; then so is (EX∩Lp, EX+Lp). Let us consider the case qX < p; theother case is similar.

Then E0 = EX∩Lp = EX(Z−) ⊕ELp(Z+) and E1 = EX+Lp =ELp(Z−) ⊕EX(Z+). Note that for all n we have ∥en∥X∩Lp ≥∥en∥X+Lp; further if werearrange the sequence (en)n∈Z so that ∥en∥X∩Lp/∥en∥X+Lp increases, it is not difficult tosee that (E0, E1) is exponentially separated.

Thus E0 has have (RSP) and E1 has (LSP)for this ordering. It also follows easily form our assumptions on the Boyd indices that thereexists k so that the gap in the new ordering for E0 between two consecutive elements of Z+is at most k. Indeed the ratio ∥en∥X∩Lp/∥en∥X+Lp behaves like 2−n/p∥en∥X for n < 0 andlike 2n/p∥en∥−1X for n ≥0 and we have an estimate for k > 0, C−12k/r ≤∥en+k∥X/∥en∥X ≤C2k for suitable C and r with qX < r < p. Thus E0 must be a weighted ℓp-space by theargument of Proposition 2.3.

It follows that EX(Z−) is a weighted ℓp−space. SimilarlyEX(Z+) is a weighted ℓp−space and so X is a Lorentz space of order p.6.

Orlicz spaces.Let F be an Orlicz function, i.e. a strictly increasing convex function F : [0, ∞) →[0, ∞) satisfying F(0) = 0.

We will also assume that F satisfies the ∆2−condition withconstant ∆i.e.F(2x) ≤∆F(x) for every x > 0. We will use the notation Ft(x) =F(tx)/F(x).We recall first that F is said to be regularly varying at ∞(resp.

at 0), in the senseof Karamata, if the limit limt→∞Ft(x) (resp. limt→0 Ft(x)) exists for all x (in fact, itsuffices that the limit exists when x ≤1.) In this case there exists p, 1 ≤p < ∞so thatlimt→∞Ft(x) = xp (resp.

limt→0 Ft(x) = xp); F is then said to be regularly varying withorder p. See [6] for details.32

Lemma 6.1. The following conditions are equivalent:(1) F is equivalent to an Orlicz function G which is regularly varying with order p at ∞(resp.

0). (2) There exists a constant C so that if x0 ≤1 there exists 0 < t0 < ∞so that if t ≥t0(resp.

t ≤t0) and x0 ≤x ≤1,C−1xp ≤Ft(x) ≤Cxp. (3) There exists a constant C so that if x ≤1 lim supt→∞Ft(x) ≤C lim inft→∞Ft(x).(resp.

lim supt→0 Ft(x) ≤C lim inft→0 Ft(x). )Proof: The implication (1) ⇒(3) is immediate and (3) ⇒(2) is a simple compactnessargument.

We indicate the details of (2) ⇒(1). Let f(x) = log F(ex) for x ∈R.

Thefunction f(x) −x is then increasing. Then it is easy to translate (2) as:(2)′ : there exists c so that if y0 ≥0 there exists x0 so that if 0 ≤y ≤y0, then|f(x) −f(x −y) −py| ≤c,whenever x ≥x0.Now we can pick a function u = u(x) for x ∈R so that u(x) = 0 for x ≤0, u isdifferentiable, increasing, u′(x) ≤1, limx→∞u(x) = ∞, and |f(x) −f(x −y) −py| ≤c for0 ≤y ≤u(x).

Now define g(x) = f(x) if u(x) = 0 andg(x) = 1uZ xx−u(f(s) + p(x −s))dsif u > 0. It is easy to show that f −g is bounded.

Further if u > 0,g′(x) = −u′u (g(x) −f(x −u) −pu) + 1u(f(x) −f(x −u) −pu) + p.Sinceu′(x)u(x) ≤1uit is easy to see that lim g′(x) = p and so if G0(x) = exp g(log x) then G0 is regularlyvarying and equivalent to F.It remains to construct a convex G with the same properties. First note that sincef(x) −x is increasing we have if u > 0,g(x) −f(x) ≤1uZ xx−u(p −1)(x −s)ds ≤p −12u.33

Henceg′(x) ≥p + 1 −u′u(f(x) −f(x −u) −pu) −p −12u′≥p −(p −1)(1 −u′) −(p −1)u′≥1.It now follows that G0(x)/x is increasing.The proof is completed by setting G(x) =R x0 G0(x)/xdx and it is then easy to verify that G has the desired properties.If F is an Orlicz function, 0 < x ≤1, and C > 1 we can define Ψ∞p (x, C) (resp.Ψ0p(x, C)) to be the supremum (possibly ∞) of all N so that there exist a1 < a2 < · · · < aNwith ak/ak−1 ≥2 for k ≤N −1 and a1 ≥1 (resp. aN ≤1) so that for all k eitherFak(x) ≥Cxp or xp ≥CFak(x).

It is easy to show thatProposition 6.2. F is equivalent to a regularly varying function of order p at ∞(resp.

at0) if and only if for some C and all 0 < x ≤1 we have Ψ∞p (x, C) < ∞(resp. Ψ0p(x, C) < ∞).We omit the proof which is immediate.However we can now state the result ofMontgomery-Smith [33] which characterizes Orlicz spaces which are Lorentz spaces (seeLorentz [26]).Theorem 6.3.

In order that LF [0, 1] coincides with a Lorentz space of order p it isnecessary and sufficient that there exist C0, C1 and r > 0 so that for every x with 0 < x ≤1we have Ψp(x, C0) ≤C1x−r.This is a somewhat disguised restatement of Montgomery-Smith’s result. However wewill not pause in our exposition to derive this result as a proof is implicit in our approachto elastic Orlicz spaces.

Further, we state the result in order to motivate the followingdefinition.For C > 1 and 0 < x ≤1 let us define Φ∞+ (x, C) (resp. Φ0+(x, C) to be the supremum ofall n so that there exist a1 < b1 ≤a2 < b2 · · · ≤an < bn with a1 ≥1 (resp.

bn ≤1) so thatFbk(x) ≥CFak(x), for 1 ≤k ≤n. For C > 1 and 0 < x ≤1 let us define Φ∞−(x, C) (resp.Φ0−(x, C)) to be the supremum of all n so that there exist a1 < b1 ≤a2 < b2 · · · ≤an < bnwith a1 ≥1 (resp.

bn ≤1) so that Fak(x) ≥CFbk(x), for 1 ≤k ≤n. We say that F iselastic at ∞(resp.

at 0) if there exist C0, C1 > 1 and r > 0 so that for 0 < x ≤1 we haveΦ∞+ (x, C0) + Φ∞−(x, C0) ≤C1x−r (resp. Φ0+(x, C0) + Φ0−(x, C0) ≤C1x−r).

From now on,we will consider only the case at ∞although similar results can always be proved at 0.Lemma 6.3. F is elastic at ∞if and only if there exist constants C0, C1 > 1 and r > 0 sothat if 0 < x ≤1, Φ∞+ (x, C0) ≤C1x−r (resp.

Φ∞−(x, C0) ≤C1x−r).Proof: Assume Φ∞+ (x, C0) ≤C1x−r. Suppose 1 ≤a1 < b1 ≤· · · ≤an < bn withFak(x) ≥eC0Fbk(x) for 1 ≤k ≤n.

Consider an interval [bk, ak+1] where 1 ≤k ≤n −1.34

Let ν = νk be the integer part of (log C0)−1(log Fak+1(x) −log Fbk(x)). Then we can findbk = c0 < c1 < · · · < cν ≤ak+1 so that log Fck(x) −log Fck−1(x) = log C0.

It follows thatn−1Xk=1νk ≤Φ+(x, C0)and hence thatn−1Xk=1(log Fak+1(x) −log Fbk(x)) ≤(log C0)(Φ+(x, C0) + n −1)and thuslog Fbn(x) −log Fa1(x) ≤(log C0)(C1x−r −1) −n.Nowlog Fbn(x) −log Fa1(x) ≥log Fbn(x) ≥−C2| log x| −C3for suitable C2, C3 by the ∆2 condition. Hencen ≤(log C0)(C1x−r −1) + C2| log x| + C3and soΦ∞−(x, eC0) ≤C4x−2rfor a suitable C4.

The other case is similar.Proposition 6.4. The following conditions on F are equivalent:(1) F is elastic at ∞.

(2) There exist constants C0, C1 > 1 so that if 1 ≤a1 < b1 ≤· · · ≤an < bn and 0 ≤x ≤1then:nXk=1(Fbk(x) −C0Fak(x)) ≤C1. (3) There exist constants C0, C1 > 1 so that if 1 ≤a1 < b1 ≤· · · ≤an < bn and 0 ≤x ≤1then:nXk=1(Fak(x) −C0Fbk(x)) ≤C1.

(4) There exists a bounded monotone increasing function w : [1, ∞) →R and a constantC0 so that if 1 ≤s ≤t and 0 ≤x ≤1 thenFt(x) ≤CFs(x) + w(t) −w(s).35

(5) There exists a bounded monotone increasing function w : [1, ∞) →R and a constantC0 so that if 1 ≤s ≤t and 0 ≤x ≤1 thenFs(x) ≤CFt(x) + w(t) −w(s).Proof: (1) ⇒(2). We assume that for suitable constants C2, C3 > 1 and r > 0, we haveΦ∞+ (x, C2) ≤C3x−r.We will assume that C2 > ∆from which it follows easily Fb(x)/Fa(x) ≥C2 implies thatb > 2a.

First suppose m is an integer with m > r. We will estimate Φ∞+ (x, Cm2 ). Suppose1 ≤a1 < b1 ≤· · · ≤an < bn and Fbk(x) > Cm2 Fak(x).

Let s be the smallest integer greaterthan | log2 x| + 1. Then as ≥x−1 and aks > x−1b(k−1)s for 2 ≤k ≤[n/s].

Let ξ = x1/m.Now for each 1 ≤k ≤[n/s] there exists σk with 0 ≤σk ≤m −1 so that Fξσk bsk(ξ) ≥C2Fξσk ask(ξ) and the intervals [ξσkask, ξσkbsk] are disjoint in [1, ∞). Hence we have anestimate thatΦ∞+ (ξ, C2) ≥[n/s]and this means that[n/s] ≤C3ξ−r.Thusn ≤C3(s + 1)x−r/m ≤C4 + C5| log x|x−r/mfor suitable constants C4, C5.

This leads to an estimateΦ∞+ (x, Cm2 ) ≤C6x−αwhere 0 < α < 1.Now suppose C0 = ∆Cm2 . Suppose 1 ≤a1 < b1 ≤· · · ≤an < bn and that 0 ≤xk ≤1for 1 ≤k ≤n.

For j ∈N let Ij be the set of k such that 2−j < xk ≤2.2−j. ThenXk∈Ij(Fbk(xk) −C0Fak(xk)) ≤Xk∈Ij(∆Fbk(2−j) −C0Fak(2−j))≤∆Φ∞+ (2−j, Cm2 ) maxkFbk(2−j)≤C6∆2−(1−α)j.ThusnXk=1(Fbk(x) −C0Fak(x)) ≤C6∆∞Xj=12−(1−α)j.36

This establishes (2). (2) ⇒(4).

We define w(t) for 1 ≤t < ∞by setting w(t) to be the supremum ofPnk=1(Fbk(xk) −C0Fak(xk)) over all n and all 1 ≤a1 < b1 ≤· · · ≤an < bn ≤t and all0 ≤xk ≤1 for 1 ≤k ≤n. Clearly w(t) is increasing and bounded above by C1.

Condition(4) is immediate from the definition. (4) ⇒(1).

Suppose 0 < x ≤1 and that 1 ≤a1 < b1 ≤· · · ≤an < bn, are such thatFbk(x) > 2C0Fak(x). Then we haveC0nXk=1Fbk(x) ≤nXk=1(w(bk) −w(ak)) ≤C1where C1 = limx→∞w(x) −w(1).

Now Ft(x) ≥C2xr for all t, for a suitable C2, by the∆2−condition. ThusΦ∞+ (x, 2C0) ≤C1(C0C2)−1x−r.The implication now follows from Lemma 6.3.The remaining implications are similar.Lemma 6.5.

If F is elastic at ∞then F is equivalent to an Orlicz function which isregularly varying at ∞.Proof: It follows immediately from (4) above thatlim supt→∞Ft(x) ≤C0 lim inft→∞Ft(x)for 0 < x ≤1. Apply Lemma 6.1.We now come to our main theorem on elastic Orlicz functions.Theorem 6.6.

Let F be an Orlicz function satisfying the ∆2−condition. Then the fol-lowing are equivalent:(1) F is elastic at ∞.

(2) LF [0, 1] is stretchable. (3) LF [0, 1] is compressible.

(4) LF [0, 1] is elastic.Proof: We will only show (1) ⇒(2) and (2) ⇒(1). The other implications will then beclear.

We will write E = EF for EX where X = LF [0, 1]. Then EF is the modular sequencespace of Z−defined by ∥x∥EF = 1 if and only if Pn∈Z−F(x(n))2n = 1.

Let us define λnfor n ∈Z−by F(λn) = 2−n. Then (λn)n∈Z−is strictly decreasing and λn−1 ≤2λn forn < 0.37

(1) ⇒(2). We must show that E has (RSP).

To show this it suffices to show theexistence of a constant C so that if a1 < b1 < c1 ≤a2 < b2 < c2 ≤· · · ≤an < bn < cn ≤0and supp xk ⊂[ak, bk), supp yk ⊂[bk, ck) and ∥yk∥E ≤∥xk∥E = 1 then∥nXk=1αkyk∥E ≤C∥nXk=1αkxk∥E.To do this let us suppose n, ak, bk, ck, xk as fixed and let Γ be the least constant C forwhich this inequality holds. We show a uniform bound on Γ.

We can suppose the existenceof constants C0, C1 and an increasing function w : [1, ∞) →R with limx→∞w(x) =w(1) + C1 so that if 1 ≤s ≤tFs(x) ≤C0Ft(x) + w(t) −w(s)for 0 ≤x ≤1.Let us define x′k = P|xk(j)|≥12 λbk xk(j)ej and also y′k = P|yk(j)|≥12 λck yk(j)ej.ThenX2jF(2|xk(j) −x′k(j)|) ≤2−bk Xj

Then ∥x′′k∥E, ∥y′′k∥E ≤2.Now for any α1, . .

., αn such that ∥Pnk=1 αkxk∥E = 1 we set z = Pnk=1 αky′k andv = Pnk=1 αkx′k. We also let u = Pnk=1 αkλbkebk.

Then for fixed k,Xj∈[bk,ck)2jF(|αky′k(j)|) ≤X2jF(|αky′′k(j)|)≤C02bkF(|αk|λbk)Xy′′k (j)̸=02jF(y′′k(j))+Xy′′k (j)̸=02jF(y′′k(j))(w(λbk) −w(y′′k(j)))≤C0∆2bkF(|αk|λbk) + ∆(w(λbk) −w(λck)).On summing, we getXj2jF(|z(j)|) ≤C0∆Xj2jF(|u(j)|) + ∆C1.Now in the other direction, for fixed k,2bkF(|αk|λbk) ≤C0F(|αk|x′′k(j))F(x′′k(j))−1 + w(λak) −w(λbk)38

whenever x′′k(j) ̸= 0.Thus2bkF(|αk|λbk)(Xj2jF(x′′k(j))) ≤C0Xj2jF(|αkx′′k(j)|)++ (Xj2jF(x′′k(j)))(w(λak −w(λbk)).Now we observe that 1/2 ≤∥x′k∥E ≤1 so that 1/2 ≤∥x′′k∥E ≤2. Hence 1/2 ≤Pj 2jF(x′′k(j)) ≤∆.

Thus we have:2bkF(|αk|λbk) ≤2C0∆Xj2jF(|αkx′k(j)|) + 2∆(w(λak) −w(λbk)).Summing as beforeXj2jF(|u(j)|) ≤2C0∆Xj2jF(|v(j)|) + 2C1∆.We thus have an estimateXj2jF(|z(j)|) ≤C2Xj2jF(|v(j)|) + C3for constants C2, C3 depending only on F. This in turn implies an estimate ∥z∥E ≤C4∥v∥Efor some constant C4.Now we conclude by noting that:∥nXk=1αkyk∥E ≤∥z∥E + ∥kXj=1αk(yk −y′k)∥E≤C4∥v∥E + Γ maxj∥yj −y′j∥E∥nXk=1αkxk∥E≤(C4 + Γ2 )Thus Γ ≤C4 + Γ/2 and so Γ ≤2C4 and E has (RSP). (2) ⇒(1).

Suppose E has (RSP). This implies that for some C0, if a1 < b1

We also note from the ∆2-condition that we can supposeC1Ft(x) ≥xr for some C1 and r and all t > 0, 0 ≤x ≤1,For any constant C > 2C0∆3 and 0 < x ≤1 suppose now that 1 ≤c1 < d1 ≤c2 < · · · < dn−1 ≤cn < dn and Fck(x) ≥CFdk(x). Then we must have dk > 23ck.

Now39

choose bn−k+1 ∈Z−to be the largest integer so that λbn−k+1 > ck and let an−k+1 be thesmallest integer so that λan−k+1 < dk. It is clear that a1 < b1 < a2 · · · < an < bn.

Furtherλbn−k+1 ≤2ck and λan−k+1 ≥dk/2. It follows that for every k with 1 ≤k ≤n we have2bkF(λbkx) ≥C∆−22akF(λakx).Now suppose n > C1x−r.

Then we can select a subset J of {1, 2, . .

., n} so that1/2 ≤Pk∈J 2akF(λakx) ≤1. Then we can conclude thatC∆−22≤Xk∈J2bkF(λbkx) ≤C0.Since C > 2C0∆2 we reach a contradiction and conclude that n ≤C1x−r.

ThusΦ∞−(x, C) ≤C1x−rand F is elastic by Lemma 6.3.Of course there are corresponding results for sequence spaces and Orlicz spaces on[0, ∞). We will omit the proofs.Theorem 6.7.

Suppose F is an Orlicz function satisfying the ∆2−condition. Then:(1) In order that ℓF be elastic (resp.

compressible, resp. stretchable) it is necessary andsufficient that F be elastic at 0.

(2) In order that LF [0, ∞) be elastic (resp. compressible, resp.

stretchable) it is necessaryand sufficient that F be elastic at both 0 and ∞.Remark: It is perhaps worth pointing out at this point that the theorem of Montgomery-Smith (Theorem 6.3) cited above can be proved in much the same manner as Theorem6.6; the problem in this case is to show that E is a weighted ℓp−space. In fact our proofof Theorem 6.6 is derived from the arguments used by Montgomery-Smith [33].Returning to the case of [0, 1] we note the following simple deduction.Proposition 6.8.

If the Orlicz space LF [0, 1] is elastic then its Boyd indices pF = pLFand qF = qLF coincide.Proof: In fact we can suppose F is regularly varying by Lemma 6.5 and so the conclusionis immediate.Remark: The analogous result holds for sequence spaces, but not for LF [0, ∞) where onemust consider behavior at both 0 and ∞. Thus Lp ∩Lq is elastic for any p, q.

Let us alsomention at this point that Proposition 6.8 allows us very easily to give examples of Orlicz40

function spaces LF [0, 1] so that (L∞, LF) is not a Calder´on couple by simply ensuring thatpF ̸= qF .Examples: We now give two examples to separate the concepts implicit in our discussionabove. We first construct a regularly varying Orlicz function which is not elastic.

To dothis first suppose (ξn) is a positive sequence, bounded by one and tending monotonicallyto zero. We define φ(x) = 2 if x ≤1 and then φ(x) = 2 + (−1)nξn if 2n−1 < x ≤2n.Define f(x) =R x0 φ(t)dt and F(x) = exp(f(log x)).

Then F(x)/x is increasing and henceF1(x) =R x0 F(t)/t dt is an Orlicz function equivalent to F. Further F and F1 are regularlyvarying of order 2.It remains to show that F1 or equivalently F is not elastic at ∞.Suppose C > 1 and that 0 < x ≤1. If 2n−1 > log x−1 then| log(F(e2nx)F(e2n) ) −log F(e2n+1x)F(e2n+1) | ≥(ξn + ξn+1) log x−1.If we assume that ξn goes to zero slowly enough, say ξn ∼(log log n)−1 this will exceedlog C, O(exp(x−r)) times for some r > 0 and so F1 cannot be elastic.Our second construction is of an elastic Orlicz space which is not a Lorentz space.

Itis of course clear that conversely that every Lorentz space is elastic. We note first that ifF(x) = exp(f(log x)) where f is convex then F is elastic at ∞, by applying Proposition 6.4(4) (equally the same conclusion holds when f is concave).

We thus consider a functionφ(t) = 2 + ψ(t) where ψ(t) is bounded by one and decreases monotonically to 0. Letf(x) =R t0 φ(t)dt as above.

As usual it may be necessary to convexify F by constructingF1; however this is equivalent to F. Now we show that for LF [0, 1] to be a Lorentz space itis necessary that ψ tends to zero at a certain rate. In fact if Ψ∞2 (x, C0) ≤C1x−r it followsthat ψ(2C1x−r+1) < log C0/ log x−1 and hence that ψ(u) = O((log log u)−1).

Thus if wechoose ψ converging to zero slowly enough then LF [0, 1] is an elastic non-Lorentz space.We now turn to the general problem of determining when a pair of Orlicz spacesLF [0, 1] and LG[0, 1] forms a Calder´on couple. Of course if the Boyd indices satisfy qF < pGthis can only happen if both F and G are elastic at ∞in which case pF = qF and pG = qG.Brudnyi [8] has conjectured that if LF and LG are distinct then if (LF , LG) forms aCalder´on couple then we must have pF = qF and pG = qG.

The next theorem shows thatthat if either pF ̸= qF or pG ̸= qG then F and G must in some sense be similar functions.However following the theorem we will give a counterexample to Brudnyi’s conjecture.Theorem 6.9. Suppose F and G are Orlicz functions satisfying the ∆2−condition andsuch that (LF [0, 1], LG[0, 1]) forms a Calder´on pair.

Then either F and G are both elasticor pF = pG and qF = qG.Proof: Let us assume that qF > qG. The other case is similar.

It will be convenient to41

pick q0, q1 so that qG < q0 < qF < q1 and to suppose (by passing to equivalent functions)that F(x)/xq1 and G(x)/xq0 are decreasing.Let F be the closure of the set of functions {Ft : t ≥1} in C[0, 1].This set isrelatively compact. For each M > 1 let FM be the closure of the set of functions {Ft : t ≥1, F(t)/G(t) ≥M} and let F∞= ∩MFM.

Similarly if a < 1 we let Fa be the closure ofthe set of functions {Ft : t ≥1, F(t)/G(t) ≤a} and set F0 = ∩aFa.Now suppose t ≥1 and At is a measurable subset of [0, 1] such that µ(At) = F(t)−1.Then ∥χAt∥LF = t−1 while ∥χAt∥LG = s−1 where G(s) = F(t). If F(t) > G(t) we con-clude that s > t and further from the ∆2−condition for G we have that ∥χAt∥LG ≥φ(F(t)/G(t))∥χAt∥LF where φ is a function satisfying limu→∞φ(u) = ∞.Suppose F∞is nonempty and H1, H2 ∈F∞.

Then we can find a sequence (tn)n≥1such that t1 ≥2, tn > 2tn−1 and F(tn)/G(tn) →∞,∥χAtn ∥LG∥χAtn ∥LF≥2∥χAtn−1∥LG∥χAtn−1 ∥LFfor n ≥1 and such that Ft2n →H1 while and Ft2n−1 →H2. Since µAtn ≤2−n we cansuppose these sets are disjoint.

If we restrict to the sub−σ−algebra A of the Borel setsgenerated by (Atn) then (LF (A), LG(A)) forms a Calder´on couple. Regarded as a couple ofsequence spaces it is exponentially separated and hence the Orlicz modular space ℓFtn has(LSP) by Theorem 4.2.

By passing to a subsequence of the unit vectors it follows that boththe Orlicz sequence spaces ℓH1 and ℓH2 have (LSP) and further that the space obtainedby interlacing their bases has (LSP). Hence from Proposition 2.4 H1(x) and H2(x) areboth equivalent to some (common) xp0.

We thus conclude that there exists p0 so that anyH ∈F∞is equivalent to xp0.By similar reasoning, if F0 is nonempty there exists p1 so that every H ∈F0 isequivalent to xp1.Now suppose q0 < r1 < r2 < qF . We pick m an integer large enough so that (m−2)r1+2q1 < mr2.

Then for any ξ < 1 the function F0(x) = max{xr2t−r2F(t) : ξmx ≤t ≤x}is equivalent to F and therefore F0(x)/xr2 cannot be decreasing everywhere. Thus forany x0 there exists x ≥x0 such that for some δ > 0 we have F0(u)/ur2 < F0(x)/xr2 ifx−δ < u < x.

It follows that F0(x) = F(x) and hence F(x) ≥xr2t−r2F(t) if ξmx ≤t ≤x.Next define F1(y) = max{yr1t−r1F(t) : ξy ≤t ≤y}. Notice that F1(y) ≤ξq1−r1F(ξy).We will argue that F1(x)/xr1 cannot be decreasing on (ξm−1x, ξx).

If it is then F1(ξx) ≤42

F1(ξm−1x) and henceF(x) ≤ξ−q1F(ξx)≤ξ−q1F1(ξx)≤ξ−(m−2)r1−q1F1(ξm−1x)≤ξ−(m−2)r1−2q1F(ξmx)and hence (m −2)r1 + 2q1 > mr2 contrary to assumption.We now argue as above and conclude similarly that there exists u with ξm−1x ≤u ≤ξxsuch that F(u) ≥ur1t−r1F(t) for ξu ≤t ≤u.Now notice thatF(u)/G(u) ≤ur2−q0xq0−r2F(x)/G(x)and soF(u)/G(u) ≤ξr2−q0F(x)/G(x).It follows that given any ξ < 1 any x0 we can pick x ≥x0 so that F(x) ≥xr1t−r1F(t)for ξx ≤t ≤x and either F(x)/G(x) ≥ξ−(r2−q0)/2 or F(x)/G(x) ≤ξ(q0−r2)/2. Thuswe can find a sequence tn →∞such that Ftn(x) ≥xr1 for n−1 ≤x ≤1 and eitherF(tn)/G(tn) →∞or F(tn)/G(tn) →0.Consider the former case.

Then there exists H ∈F∞with H(x) ≥xr1. Hence p0 ≥r1.In the latter case p1 ≥r1.

Since r1 < qF is arbitrary we conclude that either p0 = qF orp1 = qF .Consider the case p0 = qF ; in particular F(t)/G(t) is unbounded for t ≥1. Wewill argue that F(t)/G(t) tends to infinity.

For any ξ < 1 consider the function h(x) =min{F(t)/G(t) : ξx ≤t ≤x}. If h does not converge to ∞then given any M and x0 thereexists x > x0 and δ > 0 so that h(x) = M and h(x) < h(u) for x −δ < u < x and thisimplies that F(t)/G(t) ≤F(x)/G(x) for ξx ≤t ≤x.

Thus we can construct tn →∞sothat F(tn)/G(tn) →∞and Ftn(x) ≤Gtn(x) ≤xq0 for n−1 ≤x ≤1. Thus F∞contains afunction H with H(x) ≤xq0.

This contradicts the fact that q0 < pF . Thus F(t)/G(t) →∞and it follows easily that since F∞= F that pF = qF .

We can invoke Theorem 5.3 toobtain that both F and G must be elastic.The case p1 = qF is similar. In this case G(t)/F(t) is unbounded and we use the sameargument as above to show that G(t)/F(t) →∞.We omit the case pF < pG; the reasoning is much the same.Example: It remains to construct an example of a Calder´on couple (LF [0, 1], LG[0, 1])with F and G non-equivalent and pF = pG < qF = qG.

Such an example is a counterex-ample to the previously mentioned conjecture of Brudnyi [8]. Our construction dependson the following lemma:43

Lemma 6.10. Let (Y0, Y1) be a Calder´on couple and let X be a Banach space.

Then thepair (X ⊕Y0, X ⊕Y1) also forms a Calder´on couple.Proof: We suppose the direct sums are ℓ1−sums. Suppose (x0, y0), (x1, y1) ∈X ⊕(Y0 +Y1) satisfiesK(t, (x0, y0), X ⊕Y0, X ⊕Y1) ≤K(t, (x1, y1), X ⊕Y0, X ⊕Y1).Then we observe that∥x0∥X ≤∥x1∥X + K(1, y1, Y0, Y1).Thus there is an operator S : X ⊕(Y0 + Y1) →X with ∥S0∥≤1 and S(x1, y1) = x0.

Onthe other hand:K(t, y0, Y0, Y1) ≤min(1, t)∥x1∥+ K(t, y1, Y0, Y1)Now (Y0, Y1) is Gagliardo complete ([13], Lemma 3) so by K-divisibility ([4],[7], [15]) wecan write y0 = u + v whereK(t, u, Y0, Y1) ≤γ min(1, t)∥x1∥andK(t, v, Y0, Y1) ≤γK(t, y, Y0, Y1)and γ is an absolute constant. The former inequality implies that max(∥u∥Y0, ∥u∥Y1) ≤γ∥x1∥X and hence that there exists S1 : X →Y0 ∩Y1 with ∥S1∥≤γ and S1x1 = u. Thelatter inequality yiels the existence of S2 : Y0 + Y1 →Y0 + Y1 with S2 ∈A(Y0, Y1) andS2y1 = v. Let S(x, y) = (S0x, S1x + S2y).

Then S is bounded on each X ⊕Yi and maps(x1, y1) to (x0, y0).We now construct the example. We suppose q > p > 1; we set r = 12(p+q), α = p−1and β = q −r.

We next define a1 = 1 and then inductively (bn)n≥1, (cn)n≥1, (dn)n≥1 and(an)n≥2 by letting bn = 2nan, cn = 4bn, dn = cn + 2n and an+1 = 4dn.We then can construct an unbounded nonnegative Lipschitz function φ : R →Rso that supp φ ⊂∪n≥1[an, bn] and |φ′(x)| ≤αx−1 a.e. (or equivalently |φ(x) −φ(y)| ≤| log x−log y| for x, y ≥1.) We then also define a nonnegative Lipschitz function ψ : R →Rwith supp ψ ⊂∪n≥1[cn, dn] by defining ψ(x) = β(x −cn) for cn ≤x ≤cn + n andψ(x) = β(dn −x) for cn + n ≤x ≤dn.

Finally we put F(x) = xr exp(ψ(log x) andG(x) = xr exp(ψ(log x) + φ(log x)).Now observe that F and G both satisfy the ∆2-condition and both F(x)/x and G(x)/xare increasing functions so that F and G are equivalent to convex Orlicz functions. Weprefer to work directly with F and G. We consider the pair (EF , EG).

For n < 0 let λn44

be the unique solution of F(λn) = 2−n and let νn be the unique solution of G(νn) = 2−n.We split Z−into two disjoint sets J0, J1 by setting J0 = {n : log λn ∈∪k[ck/2, 2dk]} andJ1 = Z−\ J0.We claim that on ω(J0) the norms ∥∥LF and ∥∥LG are equivalent. In fact sinceF ≤G we need only bound Pn∈J0 2nG(ξn) subject to Pn∈J0 2nF(ξn) = 1.

To do thisobserve that if n ∈J0 and 0 ≤ξn ≤λn then F(ξn) = G(ξn) unless log ξn < log λn/2. ThusXn∈J0G(ξn) ≤1 +Xn∈J02nG(√λn)≤1 +Xn∈J0λ−1/2nand this establishes the required estimate since λn increases geometrically.On ω(J1) we claim both ∥∥F and ∥∥G are equivalent to weighted ℓr−norms andhence form a Calder´on couple by the result of Sparr [36].

Let us do this for the caseof ∥∥G which we claim is equivalent to (Pn∈J1 |ξn/νn|r)1/r. It suffices to (a) boundPn∈J1 2nG(ξn) subject to Pn∈J1 ξrnν−rn= 1 and (b) conversely bound Pn∈J1 ξrnν−rnsub-ject to Pn∈J1 2nG(ξn) = 1.For (a) note that if 0 ≤ξn ≤νn then| log G(ξn) −log G(νn) −r log( ξnνn)| ≤α log(log νnlog ξn)as long as log ξn > log νn/2.

HenceG(ξn) ≤2α2−nν−rn ξrn + G(√νn)for n ∈J1. ThusXn∈J12nG(ξn) ≤2α +Xn∈J12nG(√νn) ≤2α +Xn∈J1ν−1/2nand this gives the required estimate.

(b) is similar. The argument that ∥∥F is equivalentto (Pn∈J1 |ξn|rλ−rn )1/r is slightly simpler and we omit it.This completes the construction of the example.

It is clear from Lemma 6.10 that(EF , EG) and hence (LF [0, 1], LG[0, 1]) is a Calder´on couple with pF = pG = p and qF =qG = q but that F and G are non-equivalent.We remark in closing that it is possible to find Orlicz function spaces LF [0, 1] so thatif (LF , LG) forms a Calder´on pair then LF = LG. (We assume the ∆2−condition for bothF and G.) We sketch the details.

The argument of Theorem 6.9 can be used to establishthat if F and G are not equivalent at ∞then there exists p with 1 ≤p < ∞so that xp is45

equivalent, for 0 ≤x ≤1, to a function of the form lim Ftn(x) where tn →∞. Now thereare many examples of functions F which fail this property; for example one can take theminimal Orlicz function:F(x) = x2 exp(α∞Xn=0(1 −cos(2π(log t)/2n)).See [20].Acknowledgements.

The author wishes to thank Michael Cwikel for introducing himto the problems studied in this paper and for many stimulating discussions and helpfulcomments. He would also like to thank E. Pustylnik for some valuable comments on anearlier draft of the paper.References.1.

J. Arazy and M. Cwikel, A new characterization of the interpolation spaces betwenLp and Lq, Math. Scand.

55 (1984) 253-270.2. S.F.

Bellenot, The Banach spaces of Maurey and Rosenthal and totally incomparablebases, J. Functional Analysis 95 (1991) 96-105.3. S.F.

Bellenot, R. Haydon and E. Odell, Quasi-reflexive and tree spaces constructed inthe spirit of R.C. James, Contemporary Math.

85 (1987) 19-43.4. C. Bennett and R. Sharpley, Interpolation of Operators, Academic Press, New York1988.5.

J. Bergh and J. L¨ofstr¨om, Interpolation spaces. An Introduction, Springer, Berlin1976.6.

N.H. Bingham, C.M. Goldie and J.L.

Teugels, Regular Variation, Cambridge Univer-sity Press, Cambridge 1987.7. Y. Brudnyi and N. Krugljak, Real interpolation functors, Soviet Math.

Doklady 23(1981) 5-8.8. Y. Brudnyi and N. Krugljak, Interpolation functors and interpolation spaces NorthHolland 1991.9.

A.P. Calder´on, Spaces between L1 and L∞and the theorems of Marcinkiewicz, StudiaMath.

26 (1966) 273-299.10. P.G.

Casazza, W.B. Johnson and L. Tzafriri, On Tsirelson’s space, Israel J.

Math. 47(1984) 81-98.11.

P.G. Casazza and B.L.

Lin, On symmetric basic sequences in Lorentz sequence spaces,II, Israel J. Math.

17 (1974) 191-218.12. P.G.

Casazza and T.J. Shura, Tsirelson’s space, Springer Lecture Notes 1363, Berlin198946

13. M. Cwikel, Monotonicity properties of interpolation spaces, Ark.

Mat.14 (1976)213-236.14. M. Cwikel, Monotonicity properties of interpolation spaces II, Ark.

Mat. 19 (1981)123-136.15.

M. Cwikel, K-divisibility of the K-functional and Calder´on couples, Ark. Mat.

22(1984) 39-62.16. M. Cwikel and P. Nilsson, On Calder´on-Mityagin couples of Banach lattices, Proc.Conf.Constructive Theory of Functions, Varna, 1984, Bulgarian Acad.Sciences1984, 232-236.17.

M. Cwikel and P. Nilsson, Interpolation of Marcinkiewicz spaces, Math. Scand.

56(1985) 29-42.18. M. Cwikel and P. Nilsson, Interpolation of weighted Banach lattices, Memoirs Amer.Math.

Soc. to appear.19.

G.H. Hardy, J.E.

Littlewood and G. Polya, Inequalities, Cambridge University Press,1934.20. F.L.

Hernandez and B. Rodriguez-Salinas, On ℓp-complemented copies in Orlicz spacesII, Israel J. Math.

68 (1989) 27-55.21. P.W.

Jones, On interpolation between H1 and H∞, 143-151 in Interpolation spacesand allied topics in analysis, Proceedings, Lund Conference 1983, M. Cwikel and J.Peetre, editors, Springer Lecture Notes 1070, 1984.22. J.L.

Krivine, Sous espaces de dimension finite des espaces de Banach reticul´es, Ann.Math. 104 (1976) 1-29.23.

J. Lindenstrauss and L. Tzafriri, On the complemented subspaces problem, Israel J.Math. 9 (1971) 263-269.24.

J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces I Springer, Berlin 1977.25. J. Lindenstrauss and L. Tzafriri, Classical Banach Spaces II Springer, Berlin 1979.26.

G.G. Lorentz, Relations between function spaces, Proc.

Amer. Math.

Soc. 12 (1961)127-132.27.

G.G. Lorentz and T. Shimogaki, Interpolation theorems for the pairs of spaces (L1, Lp)and (Lp, L∞), Trans.

Amer. Math.

Soc. 159 (1971) 207-222.28.

L. Maligranda, On Orlicz results in interpolation theory, Proc. Orlicz Memorial Con-ference, Univ.

of Mississippi, 1990.29. L. Maligranda and V.I.

Ovchinnikov, On interpolation between L1 +L∞and L1 ∩L∞,J. Functional Analysis 107 (1992) 342-351.30.

C. Merucci, Interpolation r´eelle avec fonction param`etre: r´eit´eration et applicationsaux espaces Λp(φ), (0 < p ≤+∞), C.R. Acad.

Sci. (Paris) I, 295 (1982) 427-430.47

31. C. Merucci, Applications of interpolation with a function parameter to Lorentz, Sob-olev and Besov spaces, 183-201 in Interpolation spaces and allied topics in analysis,Proceedings, Lund Conference 1983, M. Cwikel and J. Peetre, editors, Springer LectureNotes 1070, 1984.32.

B.S. Mityagin, An interpolation theorem for modular spaces, Mat.

Sbornik 66 (1965)473-482.33. S.J.

Montgomery-Smith, Comparison of Orlicz-Lorentz spaces, Studia Math. to ap-pear.34.

V.I. Ovchinnikov, On the estimates of interpolation orbits, Mat.Sb.115 (1981)642-652 (= Math.

USSR Sbornik 43 (1982) 573-583.)35. A.A. Sedaev and E.M. Semenov, On the possibility of describing interpolation spacesin terms of Peetre’s K-method, Optimizaciya 4 (1971) 98-114.36.

G. Sparr, Interpolation of weighted Lp−spaces, Studia Math. 62 (1978) 229-271.37.

B.S. Tsirelson, Not every Banach space contains an embedding of ℓp or c0, FunctionalAnal.

Appl. 8 (1974) 138-141.38.

Q. Xu, Notes on interpolation of Hardy spaces, preprint, 1991.39. M. Zippin, On perfectly homogeneous bases in Banach spaces, Israel J.

Math. 4 (1966)265-272.48

---

출처: arXiv:9209.217 • 원문 보기