Sergey Astashkin and Konstantin Lykov- Extrapolation description of rearrangement invariant spaces...

8/3/2019 Sergey Astashkin and Konstantin Lykov- Extrapolation description of rearrangement invariant spaces and related pr

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Extrapolation description

of rearrangement invariant spacesand related problems

Sergey Astashkin and Konstantin Lykov

Abstract. Problems of the extrapolation theory of operators

in the scale of Lpspaces are considered. Mainly we focus on the

problem of an extrapolation description of rearrangement invariant

(r.i.) spaces on [0, 1]. Generalizing JawerthMilmans approach we

introduce and study a new class SE of extrapolation r.i. spaceswith respect to the scale of Lpspaces. In spite of the fact that the

class SE contains almost all known by now extrapolation spacesclose to L it has a rather simple description. We completely

characterize Marcinkiewicz and Lorentz spaces belonging to this

class and present some examples of not extrapolation r.i. spaces

X such that X Lp for every p < . Under mild assumptions itis proved that Peetres Kfunctional of a couple (X, L) has anextrapolation description if and only if X SE. Moreover, it isinvestigated the problem when an Orlicz space is representable as

an intersection of Lpspaces with scalar weights. As an applica-

tion, a sharp Yano type theorem for operators bounded in Lp isestablished.

1 Introduction and preliminaries

The starting point of the extrapolation theory of operators is the fol-lowing remarkable theorem proved by Japanese mathematician S.Yanoin 1951.

2000 subject classification: Primary 46M35, 46E30

Key words and phrases: Extrapolation spaces, rearrangement invariant spaces,

extrapolation of operators, Yanos theorem, Orlicz spaces, Peetres K-functional

The first author was partially supported by the RFBR grant 09-01-08236

1


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2 S. Astashkin and K.Lykov

Theorem 1.1 ([45]). (a) Suppose that a linear operator T is bounded

in Lp for every p (1, p0), p0 > 1, and

(1) TLpLp C(p 1), p (1, p0),

for some > 0, with a constant C > 0 independent of p. Then,

T : L(log L) L1,

where the Zygmund space L(log L)

consists of all measurable on [0,1]functions x(t) such that

xL(logL) =1

0

log(e/t)x(t) dt < .

Here and next, x(t) stands for the non-increasing rearrangement of

|x(t)|.(b) If T is a linear operator bounded in Lp for all p > p0 and

(2) TLpLp Cp1/, p (p0, ),

for some > 0, with a constant C > 0 independent of p, then

T : L Exp L.

The Zygmund space Exp L consists of all measurable functions x(t)

such thatxExpL = sup

0


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Extrapolation description of rearrangement invariant spaces 3

In fact, the paper [45] contains only the part (a) of Theorem 1.1

but in the classical Zygmunds book Trigonometric series both partswere published (see [47, Theorem 12.4.41]). Yanos theorem at once hasfound valuable applications to studying classical operators of analysissuch that the maximal Hardy-Littlewood operator and the operator oftrigonometric conjugate function [45, 47]. Their norms in Lpspacessatisfy estimates of the form (1) or (2) and therefore, by Theorem 1.1, weimmediately obtain the appropriate estimates in Zygmund spaces. ThusYanos theorem can be treated as a converse statement with respect tointerpolation theorems: estimates of operator norms in Lp-spaces implythese in appropriate limiting spaces of this scale. So, Yanos result hasbeen called an extrapolation theorem.

Soon after that some more papers devoted to extrapolation problemswere appeared (see, for instance, [46, 43, 19]). In particular, in [46] and[43] Yanos result was really rediscovered.

Later, in nineties, B. Jawerth and M. Milman laid down the founda-tions of the general extrapolation theory, which studies natural limitingspaces associated with various interpolation scales and provides esti-mates for norms of appropriate operators [16, 17]. Mainly, they considerscales generated by the real method of interpolation.

Recall that for any Banach couple (A0, A1) (i.e., A0 and A1 are Ba-nach spaces linearly and continuously embedded in a common Hausdorfftopological vector space) and t > 0, the Peetre

Kfunctional is defined

as follows:

K(t, a; A0, A1) = inf{a0A0 + ta1A1 : a = a0 + a1, ai Ai, i = 0, 1}.Then, for every 0 < < 1 and 1 q , the spaceA,q consists of alla A0 + A1 such that

a A,q :=

0

tK(t, a; A0, A1)

q dtt

1/q

< if q <

and

a

A,

:= sup

0


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As extrapolation methods (functors) B.Jawerth and M.Milman in-

troduce the sum and the intersection of families of Banach spaces. Let{A}0


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where

A,q

:= ((1

)q)1/q A,q . In particular, for the Banach couple

A = (L1, L) of the spaces on the segment [0, 1] we have

A,1 Lp = A,p A, for = 1 1p

uniformly with respect to 1 p . Therefore, assuming that (p) (2p) as p we obtain

M() :=

1

1

= (p) (2p) = M

1 +

2

as 1,

whence

1p


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with the more general space Lr (1 r

) which allows to obtain

new extrapolation spaces. In [6] and [9], we focused on the following ina sense opposite problem: which rearrangement invariant spaces suchthat X Lp for every p < have an extrapolation description withrespect to the scale of Lp-spaces or, in other words, when the norm ofan r.i. space X can be expressed by Lpnorms?

First of all, recall basic definitions of the theory of r.i. spaces on [0, 1](its detailed exposition can be found in [21, 22]). A Banach functionspace X is called a Banach lattice if the conditions y = y(t) X and|x(t)| |y(t)| imply that x = x(t) X and x y. A Banachlattice X is said to be an rearrangement invariant (r.i.) space if fromy = y(t)

X and x(t) y(t) it follows that x = x(t)

X and

xX yX .The simplest and most important example of an r.i. space is given

by Lpspace (1 p ), with the usual norm

xp = 10

|x(t)|p dt1/p

(1 p < ) and x = ess sup0t1

|x(t)|.

Note that L is the smallest space among all r.i. spaces on [0, 1] andL1 is the largest one [21, Theorem 2.4.1]. Let 1 < p < , 1 q .Then, the space Lp,q is defined as the set of all functions for which the

following quasi-norm is finite:

xp,q :=q

p

10

t1/px(t)

q dtt

1/q

(1 q < )

and

xp, = sup0


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on [0,

) such that M(0) = 0. The Orlicz space LM consists of all

measurable functions x(t) on [0, 1] such that the function M |x(t)|

u

L1for some u > 0, with the norm

xLM = infu > 0 :

10

M

|x(t)|u

dt 1

.

In the case of the power function M(u) = up (1 p < ) we obtainLpspace; if M(u) is equivalent to the function e

u ( > 0) for all ularge enough, LM coincides with the Zygmund space Exp L

appearedin Yanos theorem. Moreover, the exponential Orlicz space Exp LN is

generated by the function eN(t) 1 (N(t) is an Orlicz function).Other examples of r.i. spaces are Lorentz and Marcinkiewicz spaces.

Let (t) be an increasing concave function on [0, 1], with (0) = 0. TheMarcinkiewicz space M() consists of all measurable functions x(t) suchthat

xM() = sup0 0, (t) increases, and (t)/t


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decreases). Every such function is equivalent to its least concave ma-

jorant (t); more exactly, 1/2(t) (t) (t) (0 t 1) [21,Theorem 4.1.1]. An r.i. space X is said to have the Fatou property if theconditions 0 xn x a.e. on [0, 1] and supnN xn < imply x Xand xn x.

An Orlicz space LM close to L coincides with the Marcinkiewiczspace with the same fundamental function . More precisely [23, 42],LM = M() if and only if 1/ LM. In view of the expression forthe fundamental function of LM, the last condition is equivalent to thefollowing: there exists a > 0 such that

(7)

1

0

M1

M11

t

dt < .

In particular, a Zygmund space Exp L and an exponential Orlicz spaceExp LN coincide with the Marcinkiewicz space M() generated by thefunction (t) log1/(e/t) and (t) = 1/N1(log(e/t)), respectively.

Let us recall the definition from [6], where instead of weighted Lr-parameters arbitrary Banach lattices are considered. Let F be a Banachlattice of measurable functions on [1, ) (it will be called an extrapola-tion parameter). The set LF consists of all functions x(t) for which thefunction fx(p) := xp F, with the norm

xLF := fxF .It is not hard to check that LF is an r.i. space such that LF Lp forevery p < .

Definition 1.2 ([6]). We will say that an r.i. space X on [0, 1] is an

extrapolation space with respect to the scale ofLpspaces (X E) if thereis a Banach lattice F on [1, ) such that X = LF (with equivalence ofnorms).

The simplest example of such a space is the space L. Since

fL = limp+ fp,


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then L

=

LL[1,

). Other examples are Zygmund spaces Exp L

( >

0) (see (6)) and exponential Orlicz spaces Exp LN since, by [4],

(8) xExpLN suppp0

xpN1(p)

.

Note that in both cases we can take for a parameter F an appropriateweighted Lspace.

Some basic properties of Banach spaces LF were studied in [27].Moreover, in [6] necessary and sufficient conditions under which impor-tant in applications Marcinkiewicz and Lorentz spaces belong to theclass Ehave been found. Next, in the paper [9] the following interestingsubclass of strong or nice extrapolation spaces was singled out. If X isan r.i. space on [0, 1], then by X we denote the Banach lattice of allmeasurable on [1, ) functions f such that

fX := f(

log2(2/t))

X< .

Definition 1.3 ([9]). An r.i. space X is called a strong extrapolation

space (X SE) if X = LX (with equivalence of norms).In particular, relations (6) and (8) demonstrate that all Zygmund

spaces Exp L ( > 0) and exponential Orlicz spaces Exp LN are strong

extrapolation spaces.Thus, for a strong extrapolation space X the corresponding extrap-

olation parameter X is completely and explicitly determined by X. Weobserve the following surprised phenomenon: the norm of any functionin every strong extrapolation r.i. space X is equivalent to the norm inX of its Lpnorm. More precisely, xX

xlog2(2/t) X .In [9], a rather simple characterization of strong extrapolation spaces

has been found. Namely, an r.i. space X SE if and only if theoperator Sx(t) = x(t2) is bounded in X. Moreover, in a partial caseincluding Marcinkiewicz and Lorentz spaces a simpler criterion holds. InSection 2, we review these results. Moreover, we introduce the concept of

a tempered extrapolation parameter generalizing the Jawerth-Milmansidea of a tempered weight. Then, we prove that an r.i. space X SE if and only if X = LF with a tempered extrapolation parameter


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F (see Theorem 2.2). This implies, in particular, that every limiting

space of the Lpscale near to L whose norm can be calculated byJawerth-Milmans approach is a strong extrapolation space. At the sametime, in this section some examples of extrapolation but not strongextrapolation Marcinkiewicz spaces are presented. By Theorem 2.2,every such space is generated by a not tempered parameter and thereforecannot be obtained by Jawerth-Milmans approach. Section 3 containsa review of results from [9] about an extrapolation description of the

K-functional K(t, x; X, L) provided that X is a strong extrapolationspace (Theorem 3.1). In Section 4, it is investigated the problem whenan Orlicz space LM is representable in the form:

(9) LM = 1p


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(X0, X1), a Banach space X is an interpolation space with respect to

(X0, X1) if X0

X1 X X0 + X1 and, for every linear operator Tbounded in X0 and in X1, T is bounded in X.

2 Strong extrapolation spaces

The next definition is a generalization of the Jawerth-Milman idea of atempered weight.

Definition 2.1. An extrapolation parameter F will be called tempered

if the operator Df(p) := f(2p) is bounded in F.

Recall that the concepts of extrapolation and strong extrapolationr.i. spaces (with respect to the scale of Lpspaces) were been intro-duced in [6] and [9], respectively (see Definitions 1.2 and 1.3 in theIntroduction). One of the main results of this article is the followingcharacterization of strong extrapolation spaces.

Theorem 2.2. For any r.i. space X the following conditions are equiv-

alent:

(a) X SE;(b) the operator Sf(t) = f(t2) is bounded in X.

(c) X = LF with a tempered extrapolation parameter F;To prove Theorem 2.2 and some other statements we need the fol-

lowing lemma.

Lemma 2.3. The operator Sx(t) = x(t2) is bounded in an r.i. space X

if and only ifSxX CxX for some C > 0 and all x X.

Proof. Since necessity is obvious, we should prove only sufficiency. Let

x E and let > 0. IfA = {t [0, 1] : | x(t) |> } and B = {t [0, 1] : | x(t2) |> },


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then B = (A) with (u) =

u. Hence (see, for instance, [36, Lem-

ma 9.5.1]),

{t : |Sx(t)| > } = (B) =A

du

2

u

{t: x(t)>}0

du

2

u= {t : Sx(t) > } .

Thereby (Sx)(t) Sx(t) and, by assumption, we obtain

Sx X and SxX SxX CxX .

Proof of Theorem 2.2. Firstly, show that the conditions (a) and (b) are

equivalent.

Let x = x(t) be a measurable function on [0, 1]. Consider the func-

tion x(t) := xlog2 2/t (0 < t 1). Since

xp x[0,2p+1]p 12 x(2p+1) (p 1),

then xX 12xX . Therefore, by the definition of a strong extrapola-tion space, condition (a) of the theorem is equivalent to the inequality

(10) xX CxX ,

which holds for all x X and some C > 0.First, we assume S to be bounded in X. Since SnxX SnxX


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(n = 1, 2, . . . ), we infer

x(2p+1) = xp 2x [0,2p)p 2

n=0

x [22n+1p,22np)p

2

n=0

22n

x

22n+1p

,

for all p 1, or equivalently

x(2t) 2

n=0

22n

Sn+1x(t) (0 < t 1/2).

Hence,

x(2t)X 2

n=0

22nSn+1xX C1xX .

Since the dilation operator 2x(t) = x(t/2) is bounded in every r.i. space

and 2XX 2 [21, Theorem 2.4.4], the last inequality implies (10)with C = 2C1.

Let now conversely X = LX , i.e., x X whenever x X andinequality (10) holds. If t = 2p+1 (p 1), then

x(t2) x(22p) 4x [0,22p]p 4xp = 4x(t).

Therefore, SxX 4xX 4CxX . Applying Lemma 2.3, we con-clude that S is bounded in X.

Next, prove that conditions (a) and (b) imply (c). In fact, with (a)

at hand we may take F = X. Then, by condition (b) and again by

the boundedness of the dilation operator 2x(t) = x(t/2) in X, for an

arbitrary f = f(p) F we havef(2p)F = f(2 log2(2/t))X = f(log2(4/t2))X


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C

f(log2(4/t))

X 2C

f(log2(2/t))

X = 2C

f

F.

At last, we check that (c) implies (b). Suppose that X = LF, whereF is a tempered extrapolation parameter. Applying the HolderRogers

inequality, we obtain

10

|Sx(t)|p dt =1

0

x(t2)p dt =1

0

|x(s)|p ds2

s

10

|x(s)|4p ds

14 1

0

1(2

s)

43

ds

34

,

whence Sxp C1x4p. Since F is a tempered parameter, then

SxX C2Sxp

F C1C2

x4pF C3

xpF C4xX ,

which implies that the operator S is bounded in X.

Corollary 2.4. An r.i. space X SE

if and only if there exists a

constant C > 0 such that

n=1

x[22n,22n+1)n[2n,2n+1)

X CxX

for allx X.

Proof. First of all, using the boundedness of the dilation operator 2 in

X [21, Theorem 2.4.4] once more, we obtain

1

2

k=1

x(2k)[2k,2k+1)

X xX 2

k=1

x(2k+1)[2k,2k+1)

X


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This together with the following elementary inequality

x(2k+1) 2x[2k,2k+1)

k x(2k) (k = 1, 2, . . . )

imply that k=1

x[2k,2k+1)k

[2k,2k+1)

X xX

4

k=1

x[2k,2k+1)k

[2k,2k+1)

X

.

Applying Theorem 2.2, we complete the proof.

Theorem 2.2 implies that every r.i. space being an interpolationspace with respect to a couple of strong extrapolation spaces is a strongextrapolation space as well. In particular, we obtain

Corollary 2.5. Let N1 and N2 be two Orlicz functions and X be an

interpolation r.i. space with respect to the couple (Exp LN1, Exp LN2).

Then X is a strong extrapolation space.

Remark 2.6. As it is easily seen, instead of the norm in X we can use

in Theorem 2.2 the seminorm

fX,p0 = f [p0,)X ,

with some fixed p0 > 1.

Remark 2.7. Recall that using Jawerth-Milmans theory [18] we may

identify an extrapolation space (with respect to the Lp-scale) of the form

(r)((p)Lp) with the norm

x(r)

p0

(p)xp

rdp

1

r

ifr <


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and

x() suppp0

(p)xp

whenever

(11) (p) (2p) as p .

Without loss of generality, we may assume that (p/2) C(p) if p

2p0. Thus,

p0

((p)f(2p))r dp

1/r =

2p0

((p/2)f(p))rdp

2

1/r

C

p0

((p)f(p))r dp

1/r

,

which implies that the extrapolation parameter F = Lr() is tempered.

Therefore, all r.i. spaces of the form (r)((p)Lp), with (p) satisfying

condition (11), possess property (c) from Theorem 2.2 and, according to

this theorem, are strong extrapolation spaces.

Remark 2.8. The construction resulting in strong extrapolation spaces

as well as the (r)-functor with a tempered weight is in a sense sta-

ble. Consider instead of the scale of Lpspaces the scale of Lp,q-spaces

(1 < p < ), where a q [1, ] is fixed, and a given Banach latticeF on [1,

) define the space

LF,q consisting of all functions x = x(t)

measurable on [0, 1] such that fx,q(p) := xp,q F. Arguing in thesame way as in the proof of Theorem 2.2, it can be showed that either


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of its conditions (a), (b), and (c) is equivalent to either of the following

ones:

(d) for every q [1, ] X = LX,q , i.e., xX xlog2(2/t),qX

(x X);(e) there is a q [1, ] such that X = LX,q .In some important cases condition (b) under which an r.i. space

belongs to the class SEis equivalent to a much simpler condition [6].

Definition 2.9. We say that a nonnegative function (t) on [0, 1] sat-

isfies the 2condition ( 2) whenever there exists a C > 0 suchthat

(12) (t) C(t2), 0 t 1.

Proposition 2.10. Given an r.i. space X on [0, 1], consider the follow-

ing conditions:

(i) the operator Sx(t) = x(t2) is bounded in X;

(ii) the fundamental function X(t) of X satisfies the 2condition.

Then (i) implies (ii). Conversely, if X is an interpolation space with

respect to a couple ((), M()) of Lorentz and Marcinkiewicz spaces

with the same fundamental function (t) (in particular, if X coincides

with () or with M()), then (ii) implies (i).

Proof. Since the implication (i) (ii) is obvious it suffices to prove theopposite one.

Let X = (). It is well known that the boundedness in a Lorentz

space of a linear operator that is continuous in the space of all finite


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a.e. measurable functions with respect to convergence in measure is

equivalent to its boundedness on the set of characteristic functions [21,

Lemma 2.5.2]. Since for the operator S the last is equivalent to (12),

then in this case we done.

Examine now the case of a Marcinkiewicz space. Firstly, if 2,then for every x M()

(t)Sx(t) C(t2)x(t2) C sup0


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Corollary 2.12. Suppose

2 and 1 r n.

Clearly, tn 0. We put

x(t) =

n=1

1

(t2n1) (t2n,t2n1](t)

and

X0 := {y M() : y(t) Bx(t) for some B, > 0and all t (0, 1]} .

It is easy to see that X0 is a linear subset of the Marcinkiewicz space

M(). Define X as the closure of X0 in M(). Then X is an r.i.

space endowed with the norm of M(). In particular, the fundamental

function of this space X(t) coincides with (t) and therefore satisfies

the 2condition.


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Show now that the function x1(t) := x(t2)

X. It suffices to check

that

(16) inf yX0

x1 yM() > 0.

Let y X0 be arbitrary. Applying the well-known inequality [21, The-orem 2.3.1]

a bM() a bM(),

we obtain

(17) x1 yM() x1 yM() (x1 y)(t2n,tn]M().

Choose a sufficiently large n N so that

1)(t2n1)

(tn)> 2B and

1 +

1

t2n tn,

where B > 0 and (0, 1) are such that y(t) Bx(t). In this case

(x1 y)(t2n,tn]M() (x1(tn) Bx(t))(t2n/,tn]M() =

=(

x1(tn) Bx(t2n1))

(tn t2n/) (

x(t2n) Bx(t2n1))

(t2n) =

=

1

(t2n) B

(t2n1)

(t2n)

1

(tn) B

(t2n1)

(t2n) =

=1

(tn)

1 B(tn)

(t2n1)

(t2n) >

(t2n)

(tn) 1

2

1

2C.

Hence, (17) implies that for every y X0

x1 yM() 1

2C,

and thus (16) is proved. Finally, Theorem 2.2 guarantees that the r.i.

space X is not a strong extrapolation space.


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By [6, Theorem 1], a Marcinkiewicz space M() is an extrapolation

space if and only if

(18) (t) C supp1

t1/p

1/p , 0 < t 1.

Moreover, the last condition implies that

(19) xM() supp1

xp1/p .

Now, assume that 2. Then, by [6, Lemma 1],

1/p C

(2p) (p

1),

and therefore we have (18). Moreover, it is obvious that in this case (19)implies (13). In this connection, a natural question arises: Is there an ex-trapolation Marcinkiewicz space M() generated by a function 2? In other words, is there an extrapolation space which cannot be ob-tained by Jawerth-Milmans theory ? We give an example showing thatthe answer is affirmative (other examples of this sort will be presentedin Section 4).

Example 2.14. Consider an increasing concave function (t) such that

(t) (t) = e1ln e/t for some (0, 1). Such a function exists since(t) > 0, (t) > 0, (t) < 0 for t (0, t0) (where t0 is sufficientlysmall) and limt0+ (t) = 0.

Show that M() E. It suffices to prove (18) with instead of .Firstly,

1/pp =1

0

ep ln(e/t)pdt = ep+1

+1

epss ds.

Let us represent the integrand in the form: epss = epss e(1)s,

where (0, 1) is a parameter, and consider the function f(s) =


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ps

s (s 1). It is not hard to check that maxs1 f(s) = (1

)

p 11 (/) 1 . Consequently,

1/pp ep+1 exp

(1 )p 11

1

0

e(1)sds

= ep+1 exp

(1 )p 11

1

1

1 ,

whence

1/p

exp

(1 ) p

1

1

1 1/p

.

Put = p/(1 +p), where = /(1 ). Then

(1 )1/p = (1 +p)1/p 2p/p < C

for some C > 0 and all p 1. Moreover, since (1 + x) 1 + Cx for

all x [0, 1], we have

(1 ) p

1

= p 1 +1

p

(1 ) 1 (1 ) 1 (p+ C),

and thus

(20) 1/p C1 exp((1 )

1 p 1 ).

To prove (18), verify that the inequality

(21) (t) 1/p C2t1/p

holds for p = 1 ln1 e/t with the constant C2 = C1 e. In fact, on the

one hand, by (20),

(t) 1/p C1 exp((1 ln e/t)+(1) ln e/t) = C2 exp( ln e/t).


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On the other hand,

t1/p = exp((ln t)/p) = exp((ln1 e/t)(1 ln e/t)) exp( ln e/t).

Thus, (21) is proved and M() E. At the same time, it is not hardto check that 2.

3 Extrapolation description of Peetres Kfunc-tional

The extrapolation theory developed in [16, 17, 18, 34] allows us to reversein a sense the interpolation process. In particular, in the case of scalesgenerated by the real method of interpolation it is possible to recover theKfunctional corresponding to the initial Banach couple. More precisely,for any Banach coupleA = (A0, A1) we have

a0 0;


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(ii) If there is a Banach lattice F on [1,

), F

L

, such that

(23) K(t, x; X, L) K(t, xp; F, L[1, )),

with constants independent of x X and t > 0, then the fundamentalfunction of X satisfies the 2condition.

To prove this theorem, we need several auxiliary assertions. The firstof them is proved in [28, Example 3]. Since the proof of the second oneis similar, it is omitted as well.

Let X be an r.i. space on [0, 1], X = L. Then its fundamentalfunction (t) satisfies the condition limt

0+ (t) = 0. Without loss of

generality, we may assume that (t) is strictly increasing on [0, 1] and(1) = 1.

Lemma 3.2. For every x X and 0 < t 1 we have

x[0,1(t)]X K(t, x; X, L) 2 x[0,1(t)]X .Let now X E, i.e. X = LF, with a Banach lattice F on [1, ). If

X = L, then there exists x X such that limp+ xp = . Hence,the function

(24) (u) := [u,)F (u 1)satisfies the condition limu+ (u) = 0. Without loss of generality, itmay be assumed that is strictly decreasing on [1, ) and (1) = 1.

Lemma 3.3. Leta F be an increasing nonnegative function on [1, )and 0 < t 1. Then

a[1(t),)F K(t, a; F, L[1, )) 2 a[1(t),)F.In what follows, a connection between the fundamental function

of an extrapolation space X =L

F and the function defined by (24)plays an important role. In particular, if X SE, then

(u) = [u,)X = [u,)(log2(2/s))X = (21u) (u 1),


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whence

(25) (t) = (log2(2/t)) (0 < t 1).

In the general case, we have the following: for every h (0, 1) there isA = A(h) > 0 such that

(26)

h log2

2

t

A(t) if 0 < t 211/h.

In fact, ifp h log2(2/t), then we have t1/p t

1h log2 2/t 21/h, and the

definitions of and imply that (t) C121/h(

h log22t

), where C

is the constant of equivalence of norms in X and LF. Thereby we obtain(26), with A = 21/hC.

In the proof of the second part of Theorem 3.1 we will use the nextstatement.

Lemma 3.4. Let X = LF, where F is a Banach lattice on [1, ), andlet be the fundamental function of X. Then the following conditions

are equivalent:

(a) there exists C1 > 0 such that

(27) (t) C1 t1/p[1((t)),)F

for all0 < t 1;

(b) there exist C2 > 0 and 0 < h < 1 such that

(28) (t) C2

h log2

2

t

for all0 < t 211/h.

Proof. First, we assume that (b) holds. In the case when 0 < t 211/h

the definition of yields

(t) C2[h log2 2/t,)F 21/hC2 t 1h log2 2/t [h log2 2/t,)F

21/hC2

t1/p[h log2 2/t,)

F.


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Moreover, since the function 1 decreases, then (26) implies that

1(A(t)) h log2(2/t) if 0 < t 211/h.

Combining this with the previous inequality, we infer

(t) 21/hC2

t1/p[1(A(t)),)F

.

At the same time, Lemma 3.3 and the concavity of the Kfunctionalyield

t1/p[1(A(t)),)F K(A(t), t1/p; F, L[1, ))

AK((t), t1/p; F, L[1, )) 2At1/p[1((t)),)

F.

Thus, (27) is proved for 0 < t 211/h and hence for all 0 < t 1.

Prove now the opposite implication (a) (b). Let h be for nowan arbitrary number from the interval (0, 1/2). If 0 < t 211/h then

t < 1/2. Thereby log2 t1log2 t 12 , which implies t

h1 log12 2/t 21/(2h).

Therefore, assuming that h log2(2/t) > 1((t)), we obtain thatt1/p[1((t)),h log2(2/t))

F th

1 log12 2/t[1((t)),)F 21/(2h)(t).

Thus, choosing h (0, 1/2) so that 2C121/(2h) < 1, by (27), we have

(t) C1

t1/p[1((t)),)

F

C1t1/p[1((t)),h log2(2/t))F + C1 t1/p[h log2(2/t),)F

C121/(2h)(t) + C1 (h log2(2/t)) (1/2)(t) + C1 (h log2(2/t)) ,


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whence

(t) 2C1 (h log2(2/t)) (0 < t 211/h).

Since in the case when h log2 2/t 1((t)) the last inequality is a

straightforward consequence of (27), the proof is completed.

Proof of Theorem 3.1. Since for X = L both statements of the the-

orem are obvious, we confine exposition to the case when X = L.

Therefore, we may assume that the fundamental function of X sat-

isfies the conditions mentioned before Lemma 3.2. Since (1) = 1, we

have that xX x (x L), whence

(29) K(t, x; X, L) = xX (t 1).

Moreover, aX aL[1,) (a L[1, )) and, therefore, if a Xthen

K(t, a; X, L

[1,

)) =

a

X (t 1).

The last equality and (29) imply that under the condition X = LX wehave

K(t, x; X, L) K(t, xp; X, L[1, )) (t 1).

Thus, taking into account Lemmas 3.2 and 3.3 and equality (25), we see

that (i) will be proved as soon as we will show that

(30)

1(t)

0 x(s)p

ds1/p

X xp[V(t),)X (0 < t

1),

where V(t) := log2(

2/1(t))

.


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First of all, in the case when p V(t)

1(t)0

x(s)p ds1/p

1(t)1/p xp 1(t)1/V(t) xp.

Since log2 z1log2 z 1 (0 < z 1), the definition of V(t) and the previousinequality yield

1(t)0

x(s)p ds1/p

1

2xp (p V(t)),

whence

(31)1(t)

0x(s)p ds

1/pX

1

2xp[V(t),)X (0 < t 1).

To prove the opposite inequality, we may assume that

(32) {s [0, 1] : x(s) > 0} [0, 1(t)].

If 21p 1(t) (which is equivalent to the inequality p V(t)), then

x(21p) 2

21p

0

x(s)p ds1/p 2

x

p[V(t),

)(p).

From this and (32) it follows

xX = x(21p)X 2 xp[V(t),)X .

On the other hand, since X is a strong extrapolation space, then

1(t)0

x(s)p ds1/p

X CxX .

The last two relations yield

1(t)0

x(s)p ds1/p

X 2Cxp[V(t),)X ,


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which combined with (31) implies (30). Thus, the proof of (i) is com-

pleted.

Proceed with the proof of (ii). Firstly, since F L[1, ), we have

K(t, a; F, L[1, )) aF (t 1)

for all a F. Combining this with (29) and (23) we obtain that X = LF.Hence, applying Lemmas 3.2 and 3.3 (we may assume that the function

defined by (24) satisfies the conditions stated before the last of them)

and hypothesis (23), we infer1(t)0

x(s)p ds1/p

F xp[1(t),)F (0 < t 1).

Choosing x = [0,u) and t = (u), where 0 < u 1, we conclude that

(u) u1/pF Cu1/p[1((u)),)

F.

Thus, Lemma 3.4 yields (28). This fact and (26) imply that

(u) C2 (h log2(2/u)) = C2 ((h/2)log2(2/u)2) C2

((h/2)log2(2/u

2)) C2A(h/2)(u

2)

for some h (0, 1) and all 0 < u 21/21/h, i.e., 2. The proof iscompleted.

As it was showed there are Marcinkiewicz spaces M() E suchthat 2 (see Example 2.14). At the same time, from Theorems 2.11and 3.1 it follows the following result.

Corollary 3.5. If X is an interpolation space with respect to a Banach

couple ((), M()) (in particular, if X = () or X = M()), then

the following conditions are equivalent:


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(i) there exists a Banach lattice F on [1,

) such that F

L

and

equivalence (23) is valid with constants independent ofx X and t > 0;(ii) X is a strong extrapolation space.

In both cases relation (23) is fulfilled for F = X.

Example 3.6. Consider an exponential Orlicz space ExpL generated

by the function e(u) 1, where is an increasimg convex functionon the half-line (0, ) and (0) = 0. As it is well known [23, 42],we have ExpL = M() (with equivalence of norms), where (u) =

1/1 (log2(2/u)) . Since 1 is a concave function, then

1(

log2(2/u2)) 1 (2log2(2/u)) 2

1 (log2(2/u)) .

Therefore, (u) 2(u2), which implies that 2. Thus, ExpL SEand

K(t, f;ExpL, L) fp[1(t),)

F(0 < t 1).

In this case, 1(t) = (1/t) (0 < t 1) and, by (13), F = L((2p)).

Thereby

K(t, f;ExpL, L) supp(1/t)

fp1(p)

(0 < t 1),

whence

K(t, f; L, ExpL) tK(1/t,f;ExpL, L) t supp(t)

fp1(p)

(t 1).

The last relation coincides with the statement of Proposition 1 in [4]

(see also Theorem 2 in [1]) to within change of variables.


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In the power case, i.e., (t) = t, we arrive at the classical Zygmund

space ExpL and the equivalence

K(t, f; L, ExpL) t suppt

fpp1/

= t suppt

fpp

is valid for every > 0 [2, Theorem 2].

4 Extrapolation description of Orlicz spaces.

In this section we look for conditions under which an Orlicz space isa limiting space of the Lpscale or, in other words, when it is possi-ble to describe the norm of an Orlicz space by Lpnorms. Analogousproblems and various applications to differential equations, imbeddingtheorems, probability theory, orthogonal series have been considered in[1, 2, 4, 6, 12-14, 26, 29-33, 37-40]. In particular, Mamontov [29] [33]has developed the techniques of special integral transformations andused them to prove some imbedding theorems for Orlicz spaces and in-tersections of Lpspaces.

Note that our exposition is close to that by Mamontov and Ostrovsky[38] [40]. The main idea is to look at Lpspaces as Orlicz spacesand a key observation is the following: an intersection of Orlicz spacesgenerated by functions M ( I) often coincides with the Orlicz space,which is generated by the envelope of the family {M}I, i.e., by thefunction M(u) = sup M(u).

Lemma 4.1. Let M(u) ( I) be arbitrary Orlicz functions and letL := ILM , where LM are Orlicz spaces generated by the functions

M. Then

(33) LM L and xL xLM,

where M(u) := sup M(u) (u > 0).

Moreover, LM(t) = L(t) (0 t 1).


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Proof. Since (33) is obvious, we should prove only the equality of funda-

mental functions. By the definition of the space L, we have LM (t)

L(t). Therefore,1

0

M

[0,t](s)

L(t)

ds 1

for all t (0, 1] and I. On the other hand, for any > 0 andt (0, 1] there is = (, t) such that

M1

L

(t) (1 + )M1

L

(t) .Thus,

10

M

[0,t](s)

L(t)

ds (1 + )

10

M

[0,t](s)

L(t)

ds 1 + ,

which implies LM(t) L(t). The opposite inequality is a consequence

of (33).

Let (Lp) := 1p 1).

Setting M(u) = Mp(u) = (p)pup, we see that the space (Lp)

coincides with the space L from Lemma 4.1 with equality of norms. Wewill examine under which conditions L = LM, where M is the envelopeof the family {Mp}1p 1, or, equivalently, (Lp) = L. We begin withthe following result.

Theorem 4.2. Let Mp(u) = (p)pup and let

M(u) := supp1

Mp(u) 1

Mp(u)R(p) dp, for u > u0,


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where a nonnegative function R(p) satisfies the condition

(34)

1

R(p)p dp < ,

with some > 0. Then (Lp) = LM.

Proof. Suppose that x (Lp) and x(Lp) < 1. By the definitionof the functions Mp, we have

1

0

Mp|x(t)|

dt

1

p

,

with from hypothesis (34). Therefore,

10

M

|x(t)|

dt M(u0) +

10

1

Mp

|x(t)|

R(p) dpdt =

= M(u0)+

1

1

0

Mp

|x(t)|

dt

R(p) dp M(u0)+

1

R(p)

pdp < ,

which implies that x

LM. The opposite imbedding follows from

Lemma 4.1.

Next, we want to clarify when an Orlicz function may be representedin the form

(35) M(u) = supp1

(p)up,

with some function (p).

Lemma 4.3. A function M(u) may be represented in the form (35) if

and only if

(36) supp1

infv>0

M(v)

vp

up = M(u).


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Proof. Firstly, if equality (36) holds then, setting (p) := infv>0M(v)

vp ,

we obtain (35).

Conversely, assume that (35) is fulfilled with some (p). Then

(p) M(u)up , for all u > 0 and p 1. Therefore, (p) infv>0M(v)

vp

and, finally,

M(u) = supp1

(p)up supp1

infv>0

M(v)

vp

up sup

p1

M(u)

up

up = M(u).

Suppose that a function M(u) may be represented in the form (35).Putting (p) := supqp (q), we see that

M(u) = supp1

(p)up

for all u 1. Since Orlicz spaces generated by equivalent functions haveequivalent norms we may assume that (p) decreases and (1) = 1. Inthis case M(u) = u if 0 u 1.

After change of variables u = et, v = es and taking the logarithm ofboth sides of equality (36) we obtain that

log M(et) = supp1

{pt sup

sR{sp log M(es)}

}, t R.

Since log M(et) = t if t < 0, we may put in the right hand side ofthe last equality the supremum over all p R. By the well-knownFenchelMoreau theorem [44, Remark 1.63, p. 26], the last means thatthe function N(t) := log M(et) is convex. Recall that the Legendreconjugate function N for a function N is defined by

N(p) = supt

R

{pt N(t)}.

In our case N(p) increases for p > 1, N(1) = 0, and N(p) = + ifp < 1. Thus, Lemma 4.3 implies the following statement.


36/60


Theorem 4.4. An Orlicz functionM(u) may be represented in the form

(35) if and only if

(37) M(u) = eN(log u),

with some convex function N such that N(t) = t if t < 0. Moreover,

(p) = eN(p).

Remark 4.5. In the case when N(t) = qt + C (t > 0, q 1) we have

M(u) = C1uq (u > 1) and (p) = 0 for p > q .

The following results follow from Theorems 4.2, 4.4 and Lemma 4.1.

Theorem 4.6. Let a function M be of the form (37) with some convex

function N(t) such that limt+ N(t)/t = . If

(38) eN(t)

1

eptN(p)+Cp dp, for t t0,

with some C > 0, then LM = (Lp), where

(39) (p) = eN(p)

p .

Proof. According to Theorem 4.4

M(u) = supp1

Mp(u), where Mp(u) = (p)pup = ep log uN

(p).

After change of variables t = log u in (38) we see that conditions of

Theorem 4.2 hold for R(p) = eCp , = eC+1, and u0 = et0.

Theorem 4.7. Let a function M be of the form (37) with some convex

functionN(t) and let the Orlicz space LM coincide with a Marcinkiewicz

space. Then LM = (Lp), where is defined by (39).


37/60


Proof. By Theorem 4.4 and Lemma 4.1, LM

(Lp) and the fun-

damental functions of these spaces are equal. Since LM is a Marcin-

kiewicz space being the largest space among all r.i. spaces with the

same fundamental function [21, Theorem 2.5.7], we deduce the opposite

imbedding.

Remark 4.8. In [38], a somewhat similar result is proved. Moreover,

in the same article its applications can be found as well.

Remark 4.9. Note that the function N(p) for p large enough is defined

by values of N(p) for large p. Therefore, by the standard extending

argument, it is not hard to check that Theorems 4.6 and 4.7 will be still

valid if equality (37) is assumed to be only for u u0 (u0 > 0), where

N(t) is a function convex for t t0.

Remark 4.10. Recall that an Orlicz space LM coincides with a Mar-

cinkiewicz space if and only if the function M satisfies condition (7).

Suppose M can be represented in the form (37). Then after logarithmic

change of variables we see that (7) is equivalent to the following:

(40)

s0

eN(N1(s)C)s ds < ,

with some constant C > 0. Moreover, the change N1(s) = x allows to

rewrite condition (40) as follows

(41)

x0

eN(xC)N(x)N(x) dx < .


38/60


Next, using the convexity of the function N together with (40) and (41),

we obtain the following conditions guaranteeing the coincidence of the

corresponding Orlicz and Marcinkiewicz spaces:

(42)

s0

exp

Cs

N1(s)

ds <

and

(43)

x0

exp

CN(x)

x

N(x) dx < .

Analogously, by the inequality

CN(x) < N(x C) N(x),

we conclude that the Orlicz and Marcinkiewicz spaces do not coincide

whenever for arbitrary C > 0 we have

(44)

x0

eCN(x)N(x) dx = .

Let us state a result dual with respect to Theorem 4.7.

Theorem 4.11. Let be an increasing concave function on [0, 1] such

that(t) teL(log(1/t)), with a concave positive functionL(u) on[0, +)satisfying the condition limu+ L(u)/u = 0. If the Lorentz space ()

coincides with an Orlicz space, then

() = 1


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Proof. Firstly, the Marcinkiewicz space

M() , where (t) = t/(t),

coincides with the Orlicz space LM, M(u) = eN(log u). According to

Theorem 4.7

(45) yM() supp0


40/60


then

(t) = (0,t) inf1


41/60


= 1/q, we see that M()

M(2) as

0. Since

A,1 Lp = A,p A,, for = 1 1p

=1

p,

with uniformly bounded constants, then the previous relations implythat

1


42/60


Example 4.12. Let > 0 and M(u) = eu

if u u0, where u0 > 0

is large enough. This function is representable in the form (37), with

N(t) = et. It is not hard to verify that this function satisfies condition

(40) and hence the corresponding Orlicz space Exp L coincides with the

Marcinkiewicz space M(), where (t) = log1/(e/t). Therefore, byTheorem 4.7, the corresponding Orlicz space Exp L is an extrapolation

one. Moreover, elementary calculations show that

N(p) =

p

logp

p

,

and we infer

xExpL supp1

xpp1/.

Example 4.13. Let > 1 and > 0. Consider the Orlicz space LM,

generated by the function M,(u) = elog(u) if u u,, where u, >

0 is large enough. This function is of the form (37), with N,(t) = t.

Moreover, since condition (42) holds, LM, is a Marcinkiewicz space.

Taking into account that

N,(p) =p

()1if p p0, where

1

+

1

= 1,

and using Theorem 4.7, we infer

LM, =

exp

p

1

()1

Lp

,

which means that

xLM, supp1 xp exp

p1

()1

.


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In the case when = 1/ these expressions are simplified and for the

function M(u) = e1 log(u) (u u) we have the following extrapola-

tion formula

xLM supp1 xp exp

p1

.

Example 4.14. Let 1. Consider the Orlicz space generated by the

function of the form (37), with N(t) = N(t) = t logt if t t, where

t is large enough. Since N(t) grows at infinity essentially more slowly

than the function N,(t) from the previous example, then the Orlicz

space LM , where M(u) = eN(log u), is wider than the Orlicz space

LM, from Example 4.13. However, LM is an extrapolation space as

well. In fact, since

x0

e2log x logx dx < if 1,

then (43) holds. Therefore, LM coincides with the corresponding Mar-cinkiewicz space.

It is not hard to find the Legendre conjugate function to an Orlicz

function which is equivalent to N (see, for example, [20]). But it is

insufficient because the corresponding Orlicz spaces may be different.

At the same time, to find the Legendre conjugate function N in the

general case is not an easy thing. However, if = 1 and = 2 elementary

calculations show that

N1 (p) = ep1 and N2 (p) = pe

p+11/

p + 1,


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respectively. Therefore, by Theorem 4.7, we have

xLM1 supp1 xp exp

ep1

p

and

xLM2 supp1 xp exp

e

p+11

p + 1

,

where

M1(u) = exp (log u log log u) and M2(u) = exp(

log u log2 log u)

,

respectively.Note that in the case = 1 we may make use of Theorem 4.6 as

well, without exploiting the fact of the coincidence of the Orlicz and

Marcinkiewicz spaces. We must check only condition (38), which is a

consequence of the following inequality

et log t

0

eptep1+p dp if t t0.

In fact, the integrand attains its maximum at the point p0 = log(t+1)+1.

Therefore,0

eptep1+p dp

p0+1t

p0

eptep1+p dp

1

texp

p0 +

1

t

t ep0+ 1t1 +p0 + 1

t

=

= exp

(t + 1) log(t + 1) + t + 2 +

1

t (t + 1)e 1t log t

et log t

if t is large enough, because of

limt+

(t + 1) log(t + 1) + t + 2 +

1

t (t + 1)e1t log t t log t

= 1.


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The last example shows that there are Orlicz spaces generated by

functions of the form (37) that do not satisfy the condition of Theorem4.7.

Example 4.15. Let N be of the same form as in the previous example

but now (0, 1). Since for arbitrary C > 0

x0

eClog x logx dx = ,

then (44) holds, and the corresponding Orlicz space doesnt coincide

with any Marcinkiewicz space. Analogously, Theorem 4.7 cannot be

used for the function N(t) = t log log t if t t0.

5 Concluding remarks and some applications.

5.1 Discrete extrapolation relations.

Here we confine to consideration of Lpspaces with integer p. Firstly,

show that in contrast to the case of the continuous scale any extrapola-tion functor of the type always gives on the scale {Lk}k=1 an Orliczspace. In other words,

xd(Lk) := supkN

(k)xk

is an Orlicz norm, for an arbitrary nonnegative sequence {(k)}k=1.Taking into account Lemma 4.1, it is sufficient to check that

L := k=1LMk LM,

where

Mk(u) := (k)kuk (k N), M(u) = sup

k=1,2,...(k)kuk.


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Consider the function G(u) :=k=1 Mk(u). If x L and xL < 1,then 10 Mk(|x(t)|) dt 1 for all k N. Therefore,1

0

G

|x(t)|2

dt =

k=1

10

Mk

|x(t)|2

dt

k=1

1

2k= 1,

whence xLM xLG 2, i.e., x LM.

Remark 5.1. Analogous arguments show that x = supk>2

(k)xlog kis an Orlicz norm, for an arbitrary nonnegative sequence {(k)}k=1.

Now, compare a discrete extrapolation method with its correspond-ing continuous version. The following example shows that in the casewhen (p) tends to zero (as p ) not too fast these methods areequivalent, i.e., give the same extrapolation spaces.

Example 5.2. Let be an arbitrary quasi-concave function on [0, 1].

Then, it follows easily that

supkN

(2k)xk supp1

(2p)xp.

In particular, this implies that for the Orlicz spaces LM from Example

4.13 if 2 we obtain

LM = d(exp(k1/)Lk), where 1

+

1

= 1,

i.e.,

xLM supkN

xkek1/.

Remark 5.3. It is not hard to prove that the analogous result holds

for every extrapolation method generated by a tempered Banach lat-

tice F (see Section 2). However, Example 5.2 shows that discrete and

continuous methods may coincide by essentially weaker assumptions.


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In general, a discrete extrapolation method and its corresponding

continuous version give, of course, different spaces.

Example 5.4. Let x(t) = elog3/4 1/t (0 < t 1). Using asymptotic

Laplaces method, it can be proved that

(47) xp ep3

, where =27

256.

In fact, after change of variables log 1/t = s, we have

x

pp =

1

eps3/4s ds.

The integrand increases on the interval [1, sp) and decreases on (sp, ),where sp := (3p/4)

4. Therefore,

xpp sp

sp1eps

3/4s ds exp

p

(3p/4)4 13/4 (3p/4)4 + 1 ,

whence xp cep3

. On the other hand, if p is large enough, then

xpp ep+p4 +

ep

e s2 ds 2ep+p4.

Combining this with the previous inequality, we obtain (47). Thus,

(48) limp

xp+1/2xp

= .

Setting

(p) :=

k=1

1

xk[k,k+1)(p)

and taking into account (48), we see that

(p) 0 as p , x d(Lk), but x (Lp).


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5.2 A sharp extrapolation Yano type theorem.

As usual, let x(t) = 1tt0

x(s) ds. Given > 0 and an r.i. space

X define the space X(log) as follows: it consists of all functionsx(t) such that x(t)log2 (2/t) X and it is endowed with the normxX(log) = x(t)log2 (2/t)X . Such spaces arise in the theory ofinterpolation of weak type operators [21, 2.6] as well as in studyinggeometry of r.i. spaces (for instance, see [5, Theorem 2.17]).

First, we show that the replacement of two stars with one star inthe norm ofX(log) leads to an equivalent functional provided that Xis a strong extrapolation space.

Proposition 5.5. LetX SE. ThenX(log) consists of all functionsx(t) such that x(t)log2 (2/t) X and

(49) xX(log) x(t)log2 (2/t)X .

Proof. Since x(t) decreases, we obtain the inequality

x(t)log2 (2/t) [1/2,1]X 2x(t) [1/2,3/4]X 2

3

x(t)log

2

(2/t)

[1/4,1/2]X

.

Hence,

xX(log) x(t)log2 (2/t) [0,1/2]X+ x(t)log2 (2/t) [1/2,1]X (1 + 2 3)x(t)log2 (2/t) [0,1/2]X .

Further, assume that t [0, 1/2]. Then,

x(t) =1

t

n=0

t2n

t2n+1

x(s) ds

n=0

x

t2n+1

t2n1.


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Therefore,

x(t)log2 (2/t)

n=0

x

t2n+1

log2

2/t2n+1

log2

2/t2

n+1

log2 (2/t)t2n1

n=0

Sn+1(

x(t)log2 (2/t)) 2(n+1) 1

2

2n1,

where, as above, Sx(t) = x(t2). By Theorem 2.2, the operator S is

bounded in X and hence

xX(log) = x(t)log2 (2/t)X

(1 + 2 3)

n=0

Sn+1 2(n+1)+12nx(t)log2 (2/t)X

= Cx(t)log2 (2/t)X .

Since x(t) x(t), the opposite inequality is obvious.

Corollary 5.6. Under the conditions of the previous proposition, X(log)is a strong extrapolation space. Moreover, we have that X(log) =

LX(p), wherefX(p) := f(p) pX .

Proof. Formula (49) indicates that the boundedness of S in X implies

its boundedness in X(log). Therefore, by Theorem 2.2, X(log) is a

strong extrapolation space as well, i.e., X(log) = L X(log). The lastassertion follows now from the isometric equality L X(log) = LX(p),which can be checked directly.


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Theorem 5.7. Let T be a bounded linear operator in Lp for all p

p0 1 and for some > 0 and C > 0

(50) TLpLp Cp (p p0).

Then, for every strong extrapolation r.i. space X the operator T is

bounded from X in X(log).

Moreover, there exists a linear operator T0 satisfying condition (50)

with the following property: if T0 is bounded from X SE into an r.i.space Y, then

(51) X(log) Y.

Proof. Estimate (50), Corollary 5.6, and Remark 2.6 yield

T xX(log) C1 T xp p X,p0 C2 xp X,p0 C3xX

and the first assertion of the theorem is proved.

Given > 0 we consider the operator

T0x(t) =

1t

log1(s/t)s

x(s) ds.

It is not hard to check that the associated operator to T0 is defined as

follows:

T0x(t) =1

t

t0

log1(t/s)x(s) ds.

Let us estimate

T0xq 10

log1(1/s)x(st)Lq(t)ds 10

s1/q log1(1/s) dsxq.


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Therefore, by the equality

10

s1/q log1(1/s) ds =0

et(1/q1)t1 dt

=

q

q 1

0evv1 dv =

q

q 1

(),

where (x) is the Euler gamma-function, we obtain that

T0LpLp = T0LpLp () p, where

1

p+

1

p= 1,

for every 1 < p < . Thereby T0 satisfies condition (50). Suppose thatX SEand that T0 is bounded from X into an r.i. space Y.

Let x X(log). Then, for some y X we have

x(t) y(t)log2 (2/t) (0 < t 1)

and, by [21, p. 93], we obtain that

(52) x(t) (y(t)log2 (2/t)) y(t/2)log2 (4/t) z

(t)log2 (2/t),

with some z X. Put z1(t) = z(t2). By Theorem 2.2 and by theassumption X SE, we have that z1 X and

T0z1(t)

tt

log1(s/t)s

z1(s) ds z1(

t)

tt

log1(s/t)s

ds

= 1z(t)log(1/

t) = 12z(t)log(1/t)

122 log 2 z(t)log2 (2/t) [0,1/2](t).

If T0 is bounded from X into Y then T0z1 Y and from the last in-equality and (52) it follows x Y, i.e., we have imbedding (51).


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Remark 5.8. If X = L

then the space X(log) coincides with the

Zygmund space Exp L1/. In this case, from Theorem 5.7 it follows the

assertion of the second part of classical Yanos theorem, i.e., if a linear

operator T satisfies (50) then T is bounded from L into Exp L1/.

Using Theorem 5.7, we obtain the following sufficient conditions forthe convergence of orthogonal series in r.i. spaces close to L.

Corollary 5.9. Let X be a strong extrapolation space and let {n}n=1be a complete in X orthonormal system of functions such that for every

x X and for all positive integers N we have N

n=1

cn(x)n

p Cpxp if p p0,

where cn(x) are Fourier coefficients of x with respect to {n}n=1. Then, for every x X the series

n=1

cn(x)n converges to x in the space

X(log).

In particular, by [47, Theorems 4.3.16 and 7.6.4] and [15, Theo-rem 5.3.2], using the interpolation Marcinkiewicz theorem in the sameway as in [26], we deduce that Lpnorms of trigonometric Fourier andWalshFourier partial sums satisfy inequality (50) if = 1. Therefore,we obtain

Corollary 5.10. Suppose X is a separable strong extrapolation r.i.

space. Then the Fourier trigonometric and the FourierWalsh series

of a function x X converge to x in the space X(log1).

In the case of Lorentz spaces the last corollary was proved by Lukom-skii [24, 25]. Moreover, he showed that this result is sharp. For Marcin-kiewicz spaces the analogous result was proved in [26].


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5.3 Rearrangement invariance of Rademacher multipli-

cator spaces.

The Rademacher functions are rn(t) := sign sin(2nt), t [0, 1], n N.

Let R denote the set of all functions of the form n=1 anrn, where theseries converges a.e. For an r.i. space X on [0,1], let R(X) be the closedlinear subspace of X given by R X. The Rademacher multiplicatorspace ofX is the space (R, X) of all measurable functions x : [0, 1] Rsuch that x n=1 anrn X, for every n=1 anrn R(X). It is aBanach function lattice on [0,1] when endowed with the norm

x(R,X) = supx n=1 anrn

X :

n=1 anrn X,

n=1 anrn

X

1

.

The space (R, X) can be viewed as the space of operators from R(X)into the whole space X given by multiplication by a measurable function.

The Rademacher multiplicator space (R, X) was firstly consideredin [11] where it was shown that for a broad class of classical r.i. spaces Xthe space (R, X) is not r.i. This result was extended in [5] to includeall r.i. spaces such that the lower dilation index X of their fundamentalfunction X satisfies X > 0. This result motivated the study of thesymmetric kernel Sym (R, X) of the space (R, X), that is, the largestr.i. space embedded into (

R, X). The space Sym (

R, X) was studied

in [5], where it was shown that, if X is an r.i. space satisfying the Fatouproperty and X ExpL2, then Sym(R, X) is the r.i. space with thenorm x := x(t)log1/2(2/t)X . It was also shown that any spaceX that has the Fatou property and that is an interpolation space forthe couple (Llog1/2L, L) can be realized as the symmetric kernel ofa certain r.i. space. The opposite situation is when the Rademachermultiplicator space (R, X) is r.i. The simplest case of this situationis when (R, X) = L. In [3], it was shown that (R, X) = Lholds for all r.i. spaces X that are interpolation spaces for the couple(L, ExpL2). More exactly, in [7], it was proved that (R, X) = Lif and only if the function log1/2(2/t) does not belong to the closure

of L in X, which will be denoted by X0.. In the paper [8], the casewhen (R, X) is an r.i. space different from L was investigated. Inparticular, the following sufficient condition for (R, X) to be r.i. has


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been obtained [8, Theorem 3.4].

Theorem 5.11. If X is an r.i. space on [0, 1] such that the operator

Sx(s) = x(s2) is bounded in X, then (R, X) is an r.i. space.The proof in [8] is based on some distribution function estimates. An

extrapolation proof presented here is essentially shorter showing therebyusefulness of the concept of a strong extrapolation space.

Proof. Note that if log1/2(2/t) X0 then, as it was proved in [7],(R, X) = Sym (R, X) = L. Thus, we may assume that log1/2(2/t) X0. Let kn := [ k12n , k2n ), where n N and k = 1, , 2n. In view of[8, Proposition 3.3], it is sufficient to prove the following: there exists

A > 0 such that for all n N and every c1 c2 c2n 0 we have(53) 2

nk=1

ckkn log1/22

t

X A 2

nk=1

ckkn log1/2 2

2nt + 1 k

X.

Firstly, let x be a measurable function such that x Lp for all p < .For every 0 < t 1, setting p := log(2/t), we have

xp = xlog(2/t) t

0(x(s))log(2/t) ds

1/ log(2/t) x(t)t1/ log(2/t)

1

ex(t),

since t1/ log(2/t) 1/e for 0 < t 1. Therefore,

(54)x(t) log1/22

t

X e xlog(2/t) log1/22t

X

.

If 1 p < , then, by [11, Lemma 1], there exists a Rademacher

sum hp =

mk=n+1 bkrk such that (bk)l2 = 1 and(55) xp 3p1/2x hpp.


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By [41] and the relation

xLN sup0


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Sergey Astashkin

Department of Mathematics and Mechanics, Samara State University,Akad.Pavlova 1, 443011 Samara, RussiaE-mail: [email protected]

Konstantin LykovDepartment of Mathematics and Mechanics, Samara State University,Akad.Pavlova 1, 443011 Samara, RussiaE-mail: [email protected]

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