S.V. Astashkin, L. Maligranda and E.M. Semenov- Multiplicator space and complemented subspaces of rearrangement invariant space

8/3/2019 S.V. Astashkin, L. Maligranda and E.M. Semenov- Multiplicator space and complemented subspaces of rearrangeme

1/30

Journal of Functional Analysis 202 (2003) 247276

Multiplicator space and complemented subspacesof rearrangement invariant space$

S.V. Astashkin,a L. Maligranda,b,* and E.M. Semenovc

a

Department of Mathematics, Samara State University, Akad. Pavlova 1, 443011 Samara, RussiabDepartment of Mathematics, Lulea University of Technology, 971 87 Lulea, SwedencDepartment of Mathematics, Voronezh State University, Universitetskaya pl.1, 394693 Voronezh, Russia

Received 20 May 2002; revised 23 August 2002; accepted 7 October 2002

Communicated by G. Pisier

Abstract

We show that the multiplicator space MX of an rearrangement invariant (r.i.) space X on0; 1 and the nice part N0X ofX; that is, the set of all aAX for which the subspaces generatedby sequences of dilations and translations of a are uniformly complemented, coincide when the

space X is separable. In the general case, the nice part is larger than the multiplicator space.

Several examples of descriptions ofMX and N0X for concrete X are presented.r 2002 Elsevier Inc. All rights reserved.

MSC: 46E30; 46B42; 46B70

Keywords: Rearrangement invariant spaces; Multiplicator spaces; Lorentz spaces; Orlicz spaces;

Marcinkiewicz spaces; Subspaces; Complemented subspaces; Projections

0. Introduction

For rearrangement invariant (r.i.) function space X on I 0; 1; we will considerthe multiplicator space MX and the nice part N0X of the space X: The space

ARTICLE IN PRESS

$Research supported by a grant from the Royal Swedish Academy of Sciences for cooperation between

Sweden and the former Soviet Union (Project 35147). The second author was also supported in part by the

Swedish Natural Science Research Council (NFR)-Grant M5105-20005228/2000 and the third author by

the Russian Fund of Fundamental Research (RFFI)-Grant 02-01-00146 and the Universities of Russia

Fund (UR)-Grant 04.01.051.

*Corresponding author.E-mail addresses: [email protected] (S.V. Astashkin), [email protected] (L. Maligranda),

[email protected] (E.M. Semenov).

URL: http://www.sm.luth.se/~lech/.

0022-1236/03/$ - see front matterr 2002 Elsevier Inc. All rights reserved.

doi:10.1016/S0022-1236(02)00094-0


2/30

MX is connected with the tensor product of two functions xsyt; s; tA0; 1; andN0X is the space given by uniformly bounded sequence in X of projections intoQa;n generated by the dilations and translations of the non-zero, decreasing function

aAX on dyadic intervals k12n ; k2n in I; k 1; 2;y; 2n; n 0; 1; 2;y . Thesefunctions are given by

an;kt a2nt k 1 if tAk1

2n; k

2n;

0 elsewhere:

(

The spaces MX and N0X coincide when X is a separable space but in the non-separable case the nice part can be larger than the multiplicator space. Such a

description is helpful in the proofs of properties of N0

X

and it motivates us to

investigate more the multiplicator spaceMX: We will describe MX for concreter.i. spaces X as Lorentz, Orlicz and Marcinkiewicz spaces. Suitable results on N0X;especially when X is a Marcinkiewicz space Mj; are given.

The paper is organized as follows. In Section 1 we collect some necessary

definitions and notations.

Section 2 contains results on the multiplicator space MX of a r.i. space X on0; 1: At first we collect its properties. After that the multiplicator space MX isdescribed for concrete spaces like Lorentz Lp;j spaces, Orlicz LF spaces and

Marcinkiewicz Mj spaces. The main result here is Theorem 1 which gives necessary

and sufficient condition for the tensor product operator to be bounded between

Marcinkiewicz spaces Mj:In Section 3, we consider a subspace N0X of X generated by dilations and

translations in r.i. space on 0; 1 of a decreasing function from X: The main result ofthe paper is Theorem 2 showing that the multiplicator space MX is a subset of thenice part N0X of X and that they are equal when a space X is separable. In thegeneral case, the nice part is larger than the multiplicator space (cf. Example 2). Here

we apply results on multiplicators from Section 2 to the description of N0X:Special attention is taken about N0X when X is a Marcinkiewicz space Mj (seeCorollary 5 and Theorem 3). Stability properties of the class N0 with respect to the

complex and real interpolation methods are presented. There is also given, inTheorem 7, a characterization of Lp-spaces among the r.i. spaces on 0; 1; which issaying that r.i. space X on 0; 1 coincides with Lp0; 1 for some 1pppN if and onlyif X and its associated space X0 belong to the class N0:

Finally, in Section 4 we show that, in general, you cannot compare the results on

the interval 0; 1 with the results on 0;N and vice versa.

1. Definitions and notations

We first recall some basic definitions. A Banach function space X on I 0; 1 issaid to be a rearrangement invariant (r.i.) space provided xntpynt for everytA0; 1 and yAX imply xAX and jjxjjXpjjyjjX; where xn denotes the decreasing

ARTICLE IN PRESS

S.V. Astashkin et al. / Journal of Functional Analysis 202 (2003) 247276248


3/30

rearrangement of jxj: Always we have imbeddings LN0; 1CXCL10; 1: By X0 wewill denote the closure of LN0; 1 in X:

An r.i. space X with a norm

jj jjX has the Fatou property if for any increasing

positive sequence xn in X with supn jjxnjjXoN we have that supn xnAX andjjsupn xnjjX supn jjxnjjX:

We will assume that the r.i. space X is either separable or it has the Fatou

property. Then, as follows from the Caldero nMityagin theorem [BS,KPS],

the space X is an interpolation space with respect to L1 and LN; i.e., if a linearoperator T is bounded in L1 and LN; then T is bounded in X andjjTjjX-XpCmaxjjTjjL1-L1 ; jjTjjLN-LN for some CX1:

If wA denotes the characteristic function of a measurable set A in I; then clearlyjjwAjjX depends only on mA: The function jXt jjwAjjX; where mA t; tAI; iscalled the fundamental function of X:For s40; the dilation operator ss given by ssxt xt=swIt=s; tAI is welldefined in every r.i. space X and jjssjjX-Xpmax1; s: The classical Boyd indices ofX are defined by (cf. [BS,KPS,LT])

aX lims-0

lnjjssjjX-Xln s

; bX lims-N

lnjjssjjX-Xln s

:

In general, 0paXpbXp1: It is easy to see that %jXtpjjstjjX-X for any t40; where%jXt sup0oso1;0osto1 jXstjX

s

:

The associated space X0 to X is the space of all (classes of) measurable functionsxt such that R1

0jxtytjdtoN for every yAX endowed with the norm

jjxjjX0 supZ1

0

jxtytj dt : jjyjjXp1& '

:

For every r.i. space X the embedding XCX00 is isometric. If an r.i. space X isseparable, then X0 Xn:

Let us recall some classical examples of r.i. spaces. Denote by C the set of

increasing concave functions jt on 0; 1 with j0 j0 0: Each functionjAC generates the Lorentz space Lj endowed with the norm

jjxjjLj Z1

0

xnt djt

and the Marcinkiewicz space Mj endowed with the norm

jjxjjMj sup0otp1

1

j

t

Zt

0

xns ds:

If F is a positive convex function on 0;N with F0 0; then the Orlicz spaceLF LF0; 1 (cf. [KR,M89]) consists of all measurable functions xt on 0; 1 for

ARTICLE IN PRESS

S.V. Astashkin et al. / Journal of Functional Analysis 202 (2003) 247276 249
http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


4/30

which the functional jjxjjLF is finite, where

jjxjjLF inf l40 : IFx

l p1

n owith IFx : Z

1

0 Fjxtj dt:An Orlicz space LF is separable if and only if the function F satisfies the D2-condition

(i.e. F2upCFu for every uXu0 and some constants u040 and C40).The Lorentz space Lp;q; 1opoN; 1pqpN; is the space generated by the

functionals (quasi-norms)

jjxjjp;q Z1

0

t1=pxntq dtt

1=qif qoN

and

jjxjjp;N sup0oto1

t1=pxnt:

For 1ppoN and jAC the Lorentz space Lp;j is the space generated by the norm

jjxjjp;j Z1

0

xntp djt 1=p

:

We will use the CalderonLozanovski construction (see [C,M89]). Let X0; X1 be apair of r.i. spaces on 0; 1 and rAU (rAU means that rs; t srt=s for s40 withan increasing, concave function r on 0;N such that r0 0 and r0; t 0: ByrX0; X1 we mean the space of all measurable functions xt on 0; 1 for which

jxtjplrjx0tj; jx1tj a:e: on 0; 1

for some xiAXi with jjxijjXip1; i 0; 1; and with the infimum of these l as the normjjxjjr: In the case of the power function rys; t s1yty with 0pyp1; ryX0; X1is the Calderon construction X1y0 X

y1 (see [C,LT,M89]). The particular case

X1=pLN11=p Xp for 1opoN is known as the p-convexification of X definedas Xp fx is measurable on I : jxjpAXg with the norm jjxjjXp jjjxjpjj1=pX (see[LT,M89]).

For other general properties of lattices of measurable functions and r.i. spaces we

refer to books [BS,KPS,LT].

2. Multiplicator space of an r.i. space

Let X XI be an r.i. space on I 0; 1: Then the corresponding r.i. spaceXI I on I I is the space of measurable functions xs; t on I I such thatx,tAXI with the norm jjxjjXII jjx,jjXI; where x, denotes the decreasing

ARTICLE IN PRESS

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


5/30

rearrangement of jxj with respect to the Lebesgue measure m2 on I I: For twomeasurable functions x xs;y yt on I 0; 1 we define the bilinear operatorof the tensor product # by

x#ys; t xsyt; s; tAI:

Definition 1. The multiplicator space MX of an r.i. space X on I 0; 1 is the setof all measurable functions x xs such that x#yAXI I for arbitrary yAXwith the norm

jjxjjMX supfjjx#yjjXII : jjyjjXp1g: 1

The multiplicator space MX is an r.i. space on 0; 1 because for the productmeasure we have

m2fs; tAI I : jxsytj4lg Z1

0

mfsAI : jxsytj4lg dt:

Let us collect some properties ofMX: First note that for any measurable setA in I the functions wA#x and smAx are equimeasurable, i.e., their distributions areequal

dwA#xl m2fs; tAI I : wAsjxtj4lg

mftAI : jsmAxtj4lg dsmAxl

for all l40: In particular, jjxjjMXXjjx#w0;1=jX1jjX jjxjjX=jX1 gives the

imbedding

MXCX and jjxjjXpjX1jjxjjMX for xAMX: 2

Moreover,M

X

X if and only if the operator# : X

X-X

I

I

is bounded.

In particular, MLp;q Lp;q for 1opoN and 1pqpp since from the ONeiltheorem (see [O, Theorem 7.4]) the tensor product # : Lp;q Lp;q-Lp;qI I isbounded.

From the equality w0;u#w0;v,t w0;uvt we obtain that if XCMX; thenfundamental function jX is submultiplicative on 0; 1; i.e., there exists a constantc40 such that jXstpcjXsjXt for all s; tA0; 1:

Some other properties of the multiplicator space MX (cf. [A97] for theproofs):

(a) jMX

t jj

stjjX-X

;jj

stjjMX-MX jj

stjjX-X

for 0otp1 and

jjs1=tjj1X-XpjjstjjMX-MXpjjstjjX-X for t41: In particular, aMX aXand bMXpbX:

ARTICLE IN PRESS

http://-/?-http://-/?-http://-/?-http://-/?-


6/30

(b) We have imbeddings LcCMXCLp; where ct jjstjjX-X; 0otp1;p 1=aX and the constants of imbeddings are independent of X: In

particular,

b1 MX LN if and only if aX 0:b2 If the operator # : X X-XI I is bounded, then XCL1=aX:

(c) If X is an interpolation space between Lp and Lp;N for some 1opoN; then

MX Lp: In particular, MLp;q Lp for 1opoN and ppqpN:

Note that the operationMX is not monotone, i.e., ifX; Y are r.i. spaces on 0; 1such that XCY then, in general, it is not true that MXCMY: Namely, considerthe r.i. space X on 0; 1 constructed by Shimogaki [S]. This space has Boyd lowerindex aX

0 with jX

t

t1=2 and L2CX: On the other hand, M

L2

L2 but

MX LN by b1:Proposition 1. We have MMX MX with equal norms.

Proof. It is enough to show the imbedding MXCMMX: Let xAMX withthe norm jjxjj

MXpC: Then

jjx#ujjXIIpCjjujjXI 8uAX:

In particular, for u y#z,

with fixed yAMX and any zAX with jjzjjXIp1 wehave

jjx#y#z,jjXIIpCjjy#z,jjXI Cjjy#zjjXII:

Since

m2fs; tAI I : jxsy#z,tj4lg

Z10

mftAI : jxsy#z,tj4lg ds

Z1

0

m2ft; aAI I : jxsytzaj4lg ds

m3fs; t; aAI I I : jxsytzaj4lg

Z1

0

m2fs; tAI I : jxsytzaj4lg da

Z1

0 mftAI : jx#y,

tzaj4lg da m2ft; aAI I : x#y,tzaj4lg

ARTICLE IN PRESS

http://-/?-http://-/?-http://-/?-


7/30

for any l40 it follows that

jjx#

y#z

,

jjX

I

I jj

x#y

,#z

jjX

I

I:

Taking the supremum over all zAX with jjzjjXIp1 we obtain

jjx#y,jjMX supfjjx#y,#zjjXII : jjzjjXIp1g

supfjjx#y#z,jjXII : jjzjjXIp1g

pCsupfjjy#zjjXII : jjzjjXIp1g CjjyjjMX:

This means that xAMMX and its norm is pC: &

Note that if X MY for some r.i. space Y; then X MX: Indeed, MX MMY MY X:

For concrete r.i. spaces, like Lorentz, Orlicz and Marcinkiewicz, we have the

following results about multiplicator space. From the above discussion we have that

if 1opoN and 1pqpN; then

MLp;q Lp;minp;q: 3

Proposition 2 (cf. Astashkin [A97] for p 1). Let jAC and 1ppoN: Then(i) Lp; %jCMLp;jCLp;j:

(ii) MLp;j Lp;j if and only if j is submultiplicative on 0; 1:(iii) If %jt lims-0 jstjs ; then MLp;j Lp; %j:

The proof follows from [A97] (cf. also [Mi76,Mi78]), property (b) and the fact that

MXp MXp; where Xp is the p-convexification of X .

Proposition 3. For the Orlicz space LF LF0; 1 we have the following:(i) If FeD2; then MLF LN:

(ii) If FAD2; then L %FCMLFCLF; where %Fu supvX1 Fuv

Fv ; uX1:

(iii) If FAD2; then MLF LF if and only if F is a submultiplicative function forlarge u; i.e., FuvpCFuFv for some positive C; u0 and all u; vXu0:

Proof. (i) It follows from property b1 and the fact that Boyd index aLF 0:(ii) The imbedding L %FCMLF follows from Ando theorem [A, Theorem 6] on

boundedness of tensor product between Orlicz spaces. In fact, if xsAL %F and

ARTICLE IN PRESS

http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-http://-/?-


8/30

ytALF; then I%Fx=l IFy=loN for some lX1; and so

F

l2

jx

s

jjy

t

jp

1

%F

jx

s

j=l

F

1

F

jy

t

j=l

;

from which

IFl2x#yp1 I%Fx=lF1 IFy=loN;

that is, x#yALFI I: Therefore, L %FCMLF: The second imbedding followsfrom (2).

(iii) It follows directly from (ii) and it was also proved in [A,A82,Mi81,O]. &

The situation is different in the case of Marcinkiewicz spaces.

Theorem 1. Let jAC: The following statements are equivalent:

(i) MMj Mj:(ii) The tensor product # : Mj Mj-MjI I is bounded.

(iii) j0#j0AMj:(iv) There exists a constant C40 such that the inequality

Xn

i1j

ui

j

i

n j

i 1n

pCj 1nX

n

i1ui ! 4

is valid for any uiA0; 1; i 1; 2;y; n and every nAN:

Proof. The equivalence i3ii is true for any r.i. space, in particular also for theMarcinkiewicz space Mj:

Implication ii ) iii follows from the fact that j0AMj:iii ) iv: Given an integer n and a sequence u1; u2;y; unA0; 1; consider the

set

A [ni1

i 1n

; in

0; uiC0; 1 0; 1:

Then

ZA

j0#j0 dm2pCjm2A;

where m2 is the Lebesque measure on 0; 1 0; 1: SinceZ

A

j0tj0s dt ds Xni1

jui j in

j i 1

n

ARTICLE IN PRESS

http://-/?-http://-/?-


9/30

and

m2

A

Xn

i1

1

n

ui

it follows that estimate (4) holds.

iv ) ii: Assume that (4) is valid. It is sufficient to prove that the inequality

jjx#yjjMjpC

holds for x;yAMj; jjxjjMj jjyjjMj 1 and x xn; y yn: Given x xnAMj withjjxjjMj 1 and e40 there exists a strictly decreasing function z znAMj such that

jjz

jjMjp1

e and zXx: Therefore, we can assume in addition that x and y are

strictly decreasing and continuous on 0; 1: We must prove the inequalityZAt

xtys dt dspCm2At

for any set

At ft; sA0; 1 0; 1 : xtysXtg; t40:

Given t40; there exists a continuous decreasing function g

t

gt

t

such that

At ft; s : gsXtg:Put

Pn [ni1

0; gi

n

i 1

n;

i

n

and

Qn [n

i1

0; gi 1

n i 1

n;

i

n :Then

PnCAtCQn:

The continuity of the function g implies that

limn-N

mQn\Pn limn-N

Xni1

1

ng

i 1n

g i

n

p limn-N

max1pipn

gi 1

n

g i

n

0:

ARTICLE IN PRESS



10/30

The function xtys belongs to L1m2: Hence

limn-N

ZQn\Pn

x

t

y

s

dt ds

0

and ZAt

xtys dt ds limn-N

ZPn

xtys dt ds

limn-N

Xni1

Zg in

0

xt dtZi

n

i1n

ys ds

p limn-NXni1

j g

i

n Zin

0ys ds Z

i1n

0ys ds !

limn-N

Xni1

j gi

n

j g i 1

n

Zin

0

ys ds:

Since

j gi

n

j g i 1

n

40

and

Zin

0

ys dspj in

it follows thatZAt

xtys dt dsp limn-N

Xnk1

j gi

n

j g i 1

n

j

i

n

limn-N

Xni1

j g in

j i

n

j i 1n

:

Denoting gi

n

ui and applying (4) we get

ZAt

xtys dt dsp limn-N

Xni1

jui j in

j i 1

n

pC limn-N

j1

nXni1

ui

! C limn-N jm2Pn Cm2At;and the proof is complete. &

ARTICLE IN PRESS



11/30

Observe that we have even proved that the tensor multiplicator norm in the space

Mj is equal

sup0ouip1;i1;2;y;n;nAN

Pni1 juij in ji1n

j1n

Pni1 ui

:

The concavity of j implies that the supremum is attained on the set of decreasing

sequences 1Xu1Xu2X?Xun40:

Remark 1. Theorem 1 can be formulated in a more general form. Let j1; j2; j3AC:Then the tensor product# : Mj1 Mj2-Mj3I I is bounded if and only if thereexists a constant C40 such that the inequality

Xni1

j1ui j2i

n

j2

i 1n

pCj3

1

n

Xni1

ui

!5

is true for every integer n and every uiA0; 1; i 1; 2;y; n:Condition (5) can be also written in the integral formZ1

0

j1utj02t dtpCj3Z1

0

ut dt

for all functions u

t

on0; 1

such that 0pu

tp1: The last integral condition is

satisfied when for example Z10

j1s

t

j02t dtpCj3s

for all s in 0; 1: A similar assumption appeared in papers [Mi76,Mi78].

We will find a condition on jAC under which estimate (4) will be true.

Lemma 1. Let jAC and j

t

pKj

t2

for some positive number K and for every

tA0; 1: ThenXni1

jui j in

j i 1

n

pK 1j1j 1

n

Xni1

ui

!

for every integer n and every uiA0; 1; i 1; 2;y; n:

Proof. The concavity of j implies that we can suppose the monotonicity

1Xu1Xu2X?XunX0: Denote

s 1n

Xni1

ui:

ARTICLE IN PRESS

http://-/?-http://-/?-


12/30

There exists a natural number m; 1pmpn; such that umpffiffi

sp

and ui4ffiffi

sp

for ipm:Since

ns X

n

i1uiX

Xm

i1uiXm

ffiffisp

it yields that mpnffiffi

sp

and

Xmi1

jui j in

j i 1

n

pj1

Xmi1

ji

n

j i 1

n

j1j m

n

pj

1

j

n ffiffisp

n j1j ffiffis

ppKj

1

j

s

:

If i4m; then juipjs and soXn

im1jui j i

n

j i 1

n

pjs

Xnim1

ji

n

j i 1

n

pj1js:

Hence

Xni1

j

ui

ji

n j

i

1

n pK 1j1js K 1j1j

1

nXni1

ui

!: &Immediately from Theorem 1 and Lemma 1 we have the following assertion.

Corollary 1. Let jAC and jtpKjt2 for some positive number K and for everytA0; 1: Then

MMj Mj:

Let us note that the power function j

t

ta with 0oao1 does not satisfy

inequality (4) but there are functions jAC which satisfy the estimate jtpKjt2for some positive number K and for every tA0; 1: This estimate gives, of course, thesupermultiplicativity of j on 0; 1:

Example 1. For each l40 there exists a alA0; 1 such that the function

jlt 0 if t 0;lnl 1

tif 0otpal;

linear if tA

a

l

; 1

;

8>:

belongs to C: Clearly, jltp2ljlt2 for every tA0; al: Consequently, jlsatisfies the conditions of Lemma 1.

ARTICLE IN PRESS



13/30

Remark 2. There exists a function jAC such that the tensor product acts from

Mj Mj into Mj and j does not satisfy the condition jtpKjt2 for somepositive number K and for every tA

0; 1

: It is enough to take j

t

ta lnb at

for

0oao1; b41 and a4e2b=1a:

We finish this part with the imbeddings of Caldero nLozanovski construction on

multiplicator spaces.

Proposition 4. Let X0; X1 be r.i. spaces on 0; 1: Then(i) MX01yMX1yCMX1y0 Xy1 :

(ii) If rAU is a supermultiplicative function on 0;N; i.e., there exists a constantc40 such that rstXcrsrt for all s; tA0;N; then

rMX0;MX1CMrX0; X1:

Proof. (i) Observe first that YCMX if and only if the operator # : Y X-XI I is bounded.

Since# :MXi Xi-XiI I; i 0; 1; is bounded with the normp1 and theCaldero n construction is an interpolation method for positive bilinear operators

(cf. [C]) it follows that

# :M

X0

1yM

X1

y

X1y0 X

y1-X0

I

I

1y

X1

I

I

y

X1y0 X

y1

I

I

is bounded with the norm p1: Therefore, MX01yMX1yCMX1y0 Xy1 .

(ii) When r is a supermultiplicative function the Caldero nLozanovski construc-

tion is an interpolation method for positive bilinear operators (see [As97,M]

Theorem 2]) and the proof of the imbedding is similar as in (i). &

Note that the inclusions in Proposition 4 can be strict. For the spaces X0 Lp;q;X1 Lp;N with 1pqopoN we have

M

X0

1yM

X1

y

M

Lp;q

1yM

Lp;N

y

L1yp;q L

yp

Lp;r;

where 1=r 1 y=q y=p andMX1y0 Xy1 ML1yp;q Lyp;N MLp;s Lp;minp;s;

where 1=s 1 y=q: The strict imbedding Lp;rD! Lp;minp;s gives then thecorresponding example.

3. Subspaces generated by dilations and translations in r.i. spaces

Given an r.i. space X on I 0; 1 let us denote byV0X faAX : aa0; a ang:

ARTICLE IN PRESS

http://-/?-http://-/?-http://-/?-http://-/?-


14/30

For a fixed function aAV0X and dyadic intervals Dn;k k12n ; k2n; k 1; 2;y; 2n;n 0; 1; 2;y; let us consider the dilations and translations of a function a

an;kt a2n

t k 1 if tADn;k;0 elsewhere:

(Then supp an;kCDn;k and

mftADn;k : jan;ktj4lg 2nmftAI : jatj4lg for all l40:

For aAV0X and n 0; 1; 2;y we denote by Qa;n the linear span fan;kg2n

k1generated by functions an;k in X:

Definition 2. For an r.i. function space X on

0; 1

the nice part N0

X

ofX is defined

by

N0X aAV0X : there exists a sequence of projections fPngNn0 on X such that

ImPn Qa;n and supn0;1;y jjPnjjX-XoN:

We say that X is a nice space (or shortly XAN0) if an belongs to N0X for every a

from X:

We are using here similar notions as in the paper [HS99]. They were consid-

ering r.i. space X X0;N on 0;N; the corresponding set VX faAX : aa0; supp aC0; 1; a ang and the set NX of all aAVX such that Qais a complemented subspace of X X0;N; where Qa is the linear closed spangenerated by the sequence akNk1 with

akt at k 1 for tAk 1; k and akt 0 elsewhere:

If NX X; then they write that XAN (or say that X is a nice space).We are putting sub-zero notions, that is, V0X and N0X; so that we have

difference between of our case of r.i. spaces on 0; 1 and their case 0;N:

Theorem 2. Let X be an r.i. space on 0; 1 and let X0 denote the closure of LN0; 1 inX: Then we have embeddings

(i) MXCN0X;(ii) N0X0CMX:

Before the proof of this theorem we will need some auxiliary results.

Let aAV0

X

and fAV0

X0

be such thatZ1

0

ftat dt 1: 6

ARTICLE IN PRESS

http://-/?-http://-/?-http://-/?-


15/30

Define the sequence of natural projections (averaging operators)

Pnxt Pn;a;fxt X2n

k1 2n Z

Dn;kfn;ksxsdsan;kt; n 0; 1; 2;y : 7

Lemma 2. The sequence of norms fjjPn;a;fjjX-XgNn0 is a non-decreasing sequence.

Proof. For x xt with supp xCDn;k we define

Rn;kxt x 2t k 12n

; Sn;kxt x 2t k

2n

:

Then

supp Rn;kxCDn1;2k1; supp Sn;kxCDn1;2k

and

mftADn1;2k1 : jRn;kxtj4lg mftADn1;2k : jSn;kxtj4lg

12

mftAI : jxtj4lg

for all l40: Therefore,ZDn1;2k1

Rn;kxt dt Z

Dn1;2kSn;kxt dt 12

ZDn;k

xt dt:

Moreover,

Rn;kfn;kxwDn;kt fn1;2k1Rn;kxwDn;kt;

Sn;kfn;kxwDn;kt fn1;2kSn;kxwDn;kt

and

mftADn1;j : an1;jt4lg 12 mftADn;i : an;i4lg

for all l40 and any i 1; 2;y; 2n; j 1; 2;y; 2n1:Denote Pn Pn;a;f: The last equality and the equality of integrals give that the

function Pnxt is equimeasurable with the function

Pn1yt X

2n

k12n1

ZDn1;2k1

fn1;2k1sRn;kxwDn;ks ds !

an1;2k1t

X2nk1

2n1Z

Dn1;2kfn1;2ksSn;kxwDn;ks ds

!an1;2kt;

ARTICLE IN PRESS



16/30

where

yt X2n

k1 Rn;kxwDn;kt Sn;kxwDn;kt:

From the above we can see that y is equimeasurable with x and so

jjPnxjjX jjPn1yjjXpjjPn1jj jjyjjX jjPn1jj jjxjjX;

that is, jjPnjjpjjPn1jj: &

Lemma 3. Let X be a separable r.i. space. If aAN0X; then there exists a functionfAN0

X0

such that (6) is fulfilled and for the sequence of projections

fPn;a;f

gdefined

by (7) we have

supn0;1;2;y

jjPn;a;fjjX-XoN:

Proof. Since X is a separable space and aAN0X it follows that there are functionsgn;kAX

0k 1; 2;y; 2n; n 0; 1; 2;y such that

Z1

0

gn;k

s

an;k

s

ds

1 and Z

1

0

gn;k

s

an;j

s

ds

0; jak;

and for the projections

Tnxt X2nk1

Z10

gn;ksxs ds

an;kt

we have supn0;1;y jjTnjjX-XoN: Let frigni1 be the first n Rademacher functions onthe segment 0; 1: Since X is an r.i. space it follows that for every uA0; 1 the normsof the operators

Tn;uxt X2nk1

rkuX2ni1

riuZ

Dn;i

gn;ksxs ds !

an;kt

are the same as the norms of Tn: Let us consider the operators

Snxt Z1

0

Tn;uxt du X2nk1

ZDn;k

gn;ksxs ds !

an;kt:

Then

jjSnjjp supuA0;1

jjTn;ujj jjTnjjpC:

ARTICLE IN PRESS



17/30

Therefore, we can assume that

supp gn;kCDn;k and gn;k is decreasing on Dn;k:

Moreover, we shift supports of the functions gn;kk 1; 2;y; 2n to the segment0; 2n and consider the averages

Gnt 2nX2nk1

tk12n gn;kt;

where tsgt gt s:Then the shifts hn;kt 2ntk12n Gnt generate operators

Unxt X2nk1

2nZ

Dn;k

hn;ksxs ds !

an;kt

and we can show that jjUnjjX-XpC:Since hn;kt Fnn;kt; where Fnt 2nGn2nt for 0ptp1; it follows that

Z10

Fntat dt Z2n

0

Gntan;1t dt

2nX2nk1

Z2n

0

tk12n gn;ktan;1t dt

2nX2nk1

ZDn;k

gn;ktan;kt dt 1:

Let us show that there exists a subsequence fFnktg of Fnt which converges atevery tA0; 1:

Lemma 2 gives that the norm of the one-dimensional operator

Lnxt Z1

0

Fnsxs ds

at

does not exceed jjUnjjX-X; and consequently also not C: Therefore,

jjFnjjX0pC

jjajjX: 8

By the definition of Fn we have Fn

n t Fnt and

1 Z1

0

Fnsas dsXFntZt

0

as ds

ARTICLE IN PRESS



18/30

or

Fntp Zt

0 as ds 1

for all tA0; 1:Applying Helly selection theorem (see [N]) we can choose subsequences

fFng*fFn;1g*fFn;2g*y*fFn;mg*y

that converge on the intervals

1

2; 1

;

1

3; 1

;y;

1

m

1; 1

;y;

respectively. Then the diagonal sequence fnt Fn;nt converges at every tA0; 1 toa function ft fnt: Using estimate (8) we obtainZs

0

fntat dtpjjfnjjX0 jjaw0;sjjXpC

jjajjXjjaw0;sjjX:

Since X is a separable r.i. space it follows that jjaw0;sjjX-0 as s-0: Therefore, theequalities fnn fn and an a imply that ffnag is an equi-integrable sequence offunctions on

0; 1

: Hence (see [E, Theorem 1.21], or [HM, Theorem 6, Chapter V])

Z10

ftat dt limn-N

Z10

fntat dt 1:

Let mAN be fixed. By the estimate jjUnjjX-XpC; the definition offn; and Lemma 2we have

X2mk1

2mZ

Dm;k

fnm;ktxt dt !

am;k

XpCjjxjjX 9

for all nXm and all xAX:Suppose that xt is a non-negative and non-increasing function on every interval

Dm;k; k 1; 2;y; 2m: As above, from (8) it follows that ffnm;kxwDm;kgNm1 is an equi-integrable sequence on Dm;k: Hence

limn-N

ZDm;k

fnm;ktxt dt Z

Dm;k

fm;ktxt dt

and for all such functions xt estimate (9) impliesX2mk1

2mZ

Dm;k

fm;ktxt dt !

am;k

X

pCjjxjjX:

ARTICLE IN PRESS



19/30

Since X is an r.i. space we can prove that the above estimate holds for all xAX: Theproof is complete. &

Proof of Theorem 2. (i) At first by the result in [A97, Theorem 1.14], we have thataAMX if and only if there exists a constant C40 such that

X2nk1

cn;kan;k

X

pCX2nk1

cn;kwDn;k

X

10

for all cn;kAR and k 1; 2;y; 2n; n 0; 1; 2;y .Suppose that aAV0X-MX; that is, estimate (10) holds. If et 1; then the

operators

Pn;ext X2nk1

2nZ

Dn;k

xs ds !

wDn;kt n 1; 2;y 11

are bounded projections in every r.i. space X and jjPn;ejjX-Xp1 (see [KPS, Theorem4.3]).

Define operators Rn;a : Qe;n Im Pn;e-Qa;n as follows:

Rn;aX2nk1

cn;kwDn;k ! X2n

k1cn;kan;k:

By the assumption aAMX or equivalently by estimate (10) we havejjRn;ajjQe;n-XpC: Therefore, for the operators

Pn;a 1jjajjL1Rn;aPn;e:

We have

jjPn;ajjX-XpCjjajj1L1 ; n 1; 2;y :

It is easy to check that Pn;a are projections and Im Pn;a Qa;n: Therefore, aAN0X:(ii) If X LN; then MX N0X0 LN:If XaLN; then X

0 is a separable r.i. space. In this case, by Lemma 3, for any

aAN0X0 there exists a function fAV0X00 such that (6) is fulfilled and for theprojections Pn;a;f defined as in (7) we have

C supn0;1;2;y

jjPn;a;fjjX0-X0oN: &

ARTICLE IN PRESS



20/30

If x is a function of the form xs P2nk1 cn;kwDn;ks; thenPn;a;fx jjfjjL1 X

2n

k1 cn;kan;k:

Therefore,

X2nk1

cn;kan;k

X0

pCjjfjj1L1X2nk1

cn;kwDn;k

X0

;

and we obtain (10) for X0:

IfX0 X; that is, X is a separable r.i. space, then the theorem is proved. If X is anon-separable r.i. space, then X0 is an isometric subspace of X

X00: By using the

Fatou property, we can extend the above inequality to the whole space X and obtain(10), which gives that aAMX: The proof of Theorem 2 is complete. &

Immediately from Theorem 2 and the properties of the multiplicator space we

obtain the following corollaries.

Corollary 2. If X is a separable r.i. space, then MX N0X:

Corollary 3. If 1opoN; 1pqpN; then N0Lp;q Lp;q for 1pqp p andN0Lp;q N0L

0p;N Lp for poqoN:

Corollary 4. Let X 0 and X 1 be separable r.i. spaces. If X0; X1AN0; then

X1y0 Xy1AN0:

Corollaries 3 and 4 show that the class of nice spaces N0 is stable with respect to

the complex method of interpolation but it is not stable with respect to the real

interpolation method.

Corollary 5. If jAC and j

tpKj

t2

for some positive number K and for every

tA0; 1; then N0Mj Mj:

By jAC0 we mean jAC such that limt-0t

jt 0:

Theorem 3. Let jAC0:

(i) If

lim supt-0

j2tj

t

2 12

then

LNCN0MjCLN,Mj\M0j: 13

ARTICLE IN PRESS



21/30

(ii) If

lim

t-

0

j2tjt

2

then

N0Mj LN,Mj\M0j:

Proof. (i) By Theorem 2 the left part of (13) is valid for any r.i. space. Assumption

(12) implies

lim supt-0

jtj

ts

1

s; 0oso1

and so

jjssjjMj-MjXs lim supt-0

jtjst 1

for every 0osp1: This means that aMj 0:Let xAN0Mj-M0j: By using Corollary 2 and property b1 we get

xAN0M0j MM0j LN:

This proves the right part of (13).

(ii) We must only prove the inclusion

Mj\M0jCN0Mj:

Let aAMj\M0j; jjajjMj 1 and ct

Rt

0ans ds: It is well known that

dista; M0j lim supt-0

1jt

Zt

0

ans ds:

Therefore,

g lim supt-0

ctjt40

and there exists a sequence ftmg tending to 0 such that

limm-N

ctmjtm g:

ARTICLE IN PRESS



22/30

Since

limm-

N

jtm2

n

jtm2n

1

for every natural n; it follows that

jjan;kjjMjX lim supm-N

ctm2njtm

2n lim supm-N

ctmjtm limm-N

jtm2njtm

2n g

for every 1pkp2n; n 1; 2;y .Consider the subspaces Hn;k=closed span fan;igiak: These subspaces are closed

subspaces of Mj and an;keHn;k: Thus, by the HahnBanach theorem, there are

bn;kAMjn such that bn;kjHn;k 0; bn;kan;k 1 and jjbn;kjj 1jjan;kjjp1g: Then theoperators

Pnx X2nk1

bn;kxan;k

are projections from Mj onto Qa;n: Moreover, Pn are uniformly bounded since

jjPnxjjMjp1

gX2nk1 jjxjjMj an;k

Mj

1

g jjxjjMj :

Therefore, aAN0Mj: The proof is complete. &

Example 2. There exists a non-separable r.i. space X such that MXaN0X:

Take X Mj with jt t ln et on 0; 1: Since aMj 0 it follows that MMj LN: The function at ln et for tA0; 1 as unbounded is not in MX but it is inMj\M

0j and by Theorem 3(ii) it shows that aAN0

X

: Therefore, N0

X

aM

X

:

Corollary 6. If jAC0 and lim supt-0

j2tjt 2; then j0AN0Mj and consequently

N0Mj is neither a linear space nor a lattice.

Problem 1. For 1opoN describe N0Lp;N:

Note that MLp;N Lp and N0L0p;N ML0p;N Lp:

Theorem 4. Let X be an r.i. space X on0; 1

: The following conditions are equivalent:

(i) # : X X-XI I is bounded.(ii) MX X:

ARTICLE IN PRESS



23/30

(iii) XAN0; i.e., N0X X:(iv) There exists a constant C40 such that

X2

n

k1cn;kan;k

X

pCjjajjXX

2

n

k1cn;kwDn;k

X

14

for all aAX and all cn;kAR; k 1; 2;y; 2n; n 0; 1; 2;y .

Proof. Implication i ) ii follows by definition ii ) iii from Theorem 2 andiv ) i by the result in [A97, Theorem 1.14]. Therefore, it only remains to provethat (iii) implies (iv).

First, assume additionally that X is separable. If XAN0; then, similarly as in theproof of Theorem 2, for any aAX there exist C

140 and fAV

0X

0such thatR1

0ftat dt 1 and

X2nk1

cn;kan;k

X

pC1jjfjj1L1X2nk1

cn;kwDn;k

X

:

Let us introduce a new norm on X defined by

jjajj1 supP2n

k1 cn;kan;k

XP2nk1 cn;kwDn;k

X

: cn;kAR; k 1; 2;y; 2n; n 0; 1; 2;y

8>:

9>=>;:

Then jjajjXpjjajj1 and jjajj1oN for all aAX: By the closed graph theorem we obtainthat jjajj1pCjjajjX and (14) is proved.

Now, let X be a non-separable r.i. space. In the case X LN both conditions (iii)

and (iv) are fulfilled. Therefore, consider the case Xa

LN: Then X0

is a separable r.i.space. The canonical isometric imbedding X0CX X00 gives that X0AN0: LetaAV0X: The separability of X0 implies

X2nk1

cn;kamn;k

X0

pCjjamjjX0X2nk1

cn;kwDn;k

X0

CjjamjjX0 X2n

k

1

cn;kwDn;k

X

;

where amt minat; m; m 1; 2;y .

ARTICLE IN PRESS

http://-/?-http://-/?-


24/30

Since X X00 has the Fatou property and amn;k an;km it follows that

X2nk1

cn;kan;k

X

limm-N

X2nk1

cn;kan;k" #m

X

pCjjajjXX2nk1

cn;kwDn;k

X

and Theorem 4 is proved. &

Theorem 5. Let X be an r.i. space X on 0; 1: Then XAN0 if and only if X00AN0:

Proof. The proof is similar to that of Theorem 4. The essential part is the proof of

the estimate (14). We leave the details to the reader.

Theorem 6. Let X be a separable r.i. space X on 0; 1: Then the following conditionsare equivalent:

(i) aAN0X:(ii) The operators Rn;a : Qe;n-Qa;n given by

Rn;aX2nk1

cn;kwDn;k ! X2

n

k1cn;kan;k

are uniformly bounded.

(iii) The operators Rn;a and their inverses are uniformly bounded.

Proof. i ) iii: Let aAN0X: Then jjRn;ajjpC for all n 0; 1; 2;y; by Theorem2. Next, since aa0 there exists n0AN and e en040 such that atXut ew0;2n0 t: Therefore, for all cn;kAR;

X2nk1

cn;kan;k

X

X

X2nk1

cn;kun;k

X

eX2nk1

cn;kwk12n;k12n0 2n

X

e s2n0X2nk1

cn;kwDn;k

!

X

Xejjs2n0 jj1X-XX2nk1

cn;kwDn;k

X

;

which shows that the inverses Rn;a1 are uniformly bounded.ii ) i: If the operators Rn;a are uniformly bounded, then we have estimate (14)

or equivalently aAM

X

and Theorem 2(i) gives that aAN0

X

: &

Now, we present a characterization of Lp spaces among all r.i. spaces on

0; 1:

ARTICLE IN PRESS



25/30

Theorem 7. Let X be an r.i. space X on 0; 1: The following conditions are equivalent:

(i) XAN0 and X0AN0:

(ii) There exists a constant C40 such that

C1X2nk1

cn;kwDn;k

X

p

X2nk1

cn;kan;k

X

pCX2nk1

cn;kwDn;k

X

15

for all aAV0X with jjajjX 1 and all cn;kAR with k 1; 2;y; 2n; n 0; 1; 2;y .

(iii) For any pair of functions a;f such that aAV0X; fAV0X0 satisfying (6) theoperators Pn;a;f defined in (7) are uniformly bounded in X:

(iv) The operator of the tensor product# is bounded from X X into XI I and from X0 X0 into X0I I:(v) There exists a pA1;N such that X Lp:

Proof. i ) iv: This follows from Theorem 4.iv ) ii: Let aAV0X; jjajjX 1: Assumption (iv) implies, by Theorem 4, that

X2n

k1cn;kan;k

XpC1

X2n

k1cn;kwDn;k

X

for some constant C140:Therefore it only remains to prove left estimate in (15). For arbitrary bAV0X0

such that

Z10

atbt dt 1 andX2nk1

dn;kbn;k

X0p1dn;kAR

we obtainR

Dn;k an;ktbn;kt dt 2n and

X2nk1

cn;kan;k

X

X

Z10

X2nk1

cn;kan;kt ! X2n

k1dn;kbn;kt

!dt 2n

X2nk1

cn;kdn;k:

Since# is bounded from X0 X0 into X0I I it follows, again by Theorem 4 usedto X0; that

X2nk1

dn;kbn;k

X0pC2

X2nk1

dn;kwDn;k

X0

ARTICLE IN PRESS



26/30

for some constant C240; from which we conclude that

X2nk1 c

n;kan;k

XXC

1

2 supdZ1

0

X2nk1 c

n;kan;kt ! X2n

k1 dn;kbn;kt ! dt

2nC12 supd

X2nk1

cn;kdn;k;

where the supremum is taken over all d dn;k2n

k1 such thatP2n

k1 dn;kwDn;k

X0p1:

The operator Pn;e defined as in (11) by

Pn;ex

t

X2n

k12n Z

Dn;k

x

s

ds !wDn;kt n 1; 2;y

satisfies jjPn;ejjX0-X0p1: Therefore,X2nk1

cn;kwDn;k

X

supjjyjjX0p1

Z10

X2nk1

cn;kwDn;kt !

yt dt

supjjyjjX0p1

Z10

X2nk1

cn;kwDn;kt !

Pn;eyt dt

p supjjPn;eyjjX0p1

Z10

X2n

k1c

n;kw

Dn;kt !Pn;eyt dt

p 2n supd

X2nk1

cn;kdn;k:

Hence X2nk1

cn;kwDn;k

X

pC2X2nk1

cn;kan;k

X

:

ii )

v

: By Krivines theorem [K,LT, p. 141] for every r.i. space X there exists

pA1;N with the following property:for any n 0; 1; 2;y and e40 there exist disjoint and equimeasurable functions

vkAX; k 1; 2;y; 2n; such that

1 ejjcjjppX2nk1

cn;kvk

X

p1 ejjcjjp 16

for any c cn;k2n

k1; where jjcjjp P2n

k1 jcn;kjp 1=p

: Hence, in particular, it follows

(with the notion nN

0

that

1 e2n=ppX2nk1

vk

X

p1 e2n=p:

ARTICLE IN PRESS

http://-/?-http://-/?-


27/30

Let

a

t

r1 X

2n

k1vk !

n

t

; where r

X2n

k1vk

X

:

Then jjajjX 1 and an;k are equimeasurable functions with r1vk for everyk 1; 2;y; 2n: Therefore,

1 e1 e 2

n=pjjcjjppX2nk1

cn;kan;k

X

p1 e1 e 2

n=pjjcjjp;

that is,

1 e1 e X

2n

k1cn;kwD

n;k

p

p X2n

k1cn;kan;k

X

p1 e1 e X

2n

k1cn;kwD

n;k

p

:

Hence, by assumption (15), we have

C1eX2nk1

cn;kwDn;k

p

p

X2nk1

cn;kwDn;k

X

pCeX2nk1

cn;kwDn;k

p

; 17

where Ce C1 e=1 e:Let 1ppoN: If X is a separable r.i. space, then (17) implies that X Lp: In the

case when X

X00 it is sufficient to consider r.i. spaces XaLN: Then X0 satisfies

15 and so X0 Lp: Hence, X0 X00 Lp0 and X X00 Lp00 Lp:Let p N: Suppose that there is a function xAX\LN: Then from (17) we obtain

jjxjjXXX2nk1

xnk2nwDn;k

X

XC1eX2nk1

xnk2nwDn;k

N

C1e xn2n:

Since xeLN it follows that limn-N xn2n N: This contradiction shows that

XCLN: The reverse imbedding is always true.v ) iii: This follows from the estimate of the norm of natural projections in Lp

space

jjPn;a;fjjLp-Lppjjajjpjjfjjp0 :

iii ) i: By definition of the operators Pn;a;f we have that XAN0: We want toshow that also X0AN0: For all xAX and yAX0

Z10

Pn;a;fxtyt dt X2nk1

2nZ

Dn;k

fn;ksxs dsZ

Dn;k

an;ktyt dt Z1

0

Pn;f;aysxs ds:

Therefore, the conjugate operator Pn;a;fn

to Pn;a;f is Pn;f;a on the space X0: Since X0is isometrically imbedded in Xn the last equality implies that the operators Pn;f;a are

uniformly bounded, and so X0AN0: The proof is complete. &

ARTICLE IN PRESS



28/30

4. Additional remarks and results

First we describe the difference between the cases on

0; 1

and

0;N

: Let X

0;N

denote an r.i. space on 0;N and X fxAX0;N : xt 0 for t41g thecorresponding r.i. space on 0; 1: We use here also the notion X0;NAN from thepaper [HS99, p. 56] (cf. also our explanation after Definition 2). Let us present

examples showing that no one of the following statements:

(i) X0;NAN;(ii) XAN0

implies the other one, in general.

Example 3. The Orlicz space LFp 0;N; where Fpu eup

1; 1opoN; belongsto the classN: On the other hand, the lower Boyd index aLFp ofLFp on 0; 1 equals 0and so MLFp LN: Therefore, by Theorem 4, LFpeN0 .

Example 4. Consider the function

jt ta if 0ptp1;

ta lnbt e 1 if 1ptoN;

(

where 0obpao1: Then j is a quasi-concave function on0;N

; i.e., j

t

is

increasing on 0;N and jt=t is decreasing on 0;N: Let *j be the smallestconcave majorant of j: Then

sup0otp1;nAN

*jnt*jn *jt N:

In fact, for every n 1; 2;y; we can choose tA0; 1 such that nto1: Then*jnt

*j

n

*j

t

X

1

4

jntj

n

j

t

lnbn e 1-N as n-N:

This implies that the Lorentz space L *j0;NeN (see [HS99, Theorem 4.1]). At thesame time for L *j Lta Lp;1 with p 1=a on 0; 1 we have, by (3), that ML *j MLp;1 Lp;1 L *j; and, by Theorem 4, L *jAN0:

The reason of non-equivalences (i) and (ii) is coming from the fact that the dilation

operator st in r.i. spaces on 0;N does not satisfy an equation of the formjjstxjjX0;N ftjjxjjX0;N; for xAX0;N and for all t40:

If this equation is satisfied, then the function ft is a power function ft ta forsome aA0; 1 and then the above statements (i) and (ii) are equivalent. Thisobservation allows us to improve, for example, Theorem 4.2 from [HS99]: if

ARTICLE IN PRESS



29/30

1opoN and 1pqpN; then

NLp;q0;N Lp;q31pqpp:

We can characterize NLp;q0;N for 1oppqoN:

Theorem 8. If 1oppqoN; then NLp;q0;N Lp:

Proof. Let aANLp;q0;N: The spaces Lp;q0;N are separable for qoN:Therefore, similarly as in the proof of Theorem 2, we can show that

Xn

k

1

ckak

Lp;q0;N

pC Xn

k

1

ck wk1;k

Lp;q0;N

18

for all ckAR; k 1; 2;y; n; n 1; 2;y . Since

jjstxjjLp;q0;N t1=pjjxjjLp;q0;N for xALp;q0;N and all t40; 19

it follows that

Xnk1

ckank

Lp;qpC

Xnk1

ckw k1n

;kn

Lp;q

;

and, by Theorem 1.14 in [A97] together with property (c), we obtain aALp:Conversely, if aALp then, by using property (c), Theorem 1.14 in [A97] and

equality (19), we get (18) for all n of the form 2m; m 1; 2;y . The space Lp;q has theFatou property, thus passing to the limit, we obtain

XNk1

ckak

Lp;q0;Np

XNk1

ckwk1;k

Lp;q0;N:

Next, arguing as in the proof of Theorem 2 (see also [HS99, Theorem 2.3]) we obtain

aANLp;q0;N:

References

[A] T. Ando, On products of Orlicz spaces, Math. Ann. 140 (1960) 174186.

[A82] S.V. Astashkin, On a bilinear multiplicative operator, Issled. Teor. Funkts. Mnogikh Veshchestv.

Perem. Yaroslavl (1982) 315 (in Russian).

[A97] S.V. Astashkin, Tensor product in symmetric function spaces, Collect. Math. 48 (1997) 375391.

[As97] S.V. Astashkin, Interpolation of positive polylinear operators in Caldero nLozanovski

spaces,

Sibirsk. Mat. Zh. 38 (1997) 12111218 (English translation in Siberian Math. J. 38 (1997)

10471053).

[BS] C. Bennett, R. Sharpley, Interpolation of Operators, Academic Press, Boston, 1988.

ARTICLE IN PRESS



30/30

[C] A.P. Caldero n, Intermediate spaces and interpolation, Studia Math. 24 (1964) 113190.

[E] R.J. Elliott, Stochastic Calculus and Applications, Springer, New York, 1982.

[HM] S. Hartman, J. Mikusin ski, The Theory of Lebesgue Measure and Integration, Pergamon Press,

New York, Oxford, London, Paris, 1961.[HS99] F.L. Hernandez, E.M. Semenov, Subspaces generated by translations in rearrangement invariant

spaces, J. Funct. Anal. 169 (1999) 5280.

[KR] M.A. Krasnoselskii, Ya.B. Rutickii, Convex Functions and Orlicz Spaces, Noordhoff Ltd.,

Groningen, 1961.

[KPS] S.G. Krein, Yu.I. Petunin, E.M. Semenov, Interpolation of Linear Operators, Nauka, Moscow,

1978 (in Russian) (English translation in Amer. Math. Soc., Providence, 1982).

[K] J.L. Krivine, Sous-espaces de dimension finie des espaces de Banach re ticule s, Ann. of Math. 104

(1976) 129.

[LT] J. Lindenstrauss, L. Tzafriri, Classical Banach Spaces, II. Function Spaces, Springer, Berlin, New

York, 1979.

[M89] L. Maligranda, Orlicz Spaces and Interpolation, Seminars in Mathematics, Vol. 5, University of

Campinas, Campinas SP, Brazil, 1989.[M] L. Maligranda, Positive bilinear operators in Caldero nLozanovski spaces, Lulea University of

Technology, Department of Mathematics, Research Report, No. 11, 2001, pp. 117.

[Mi76] M. Milman, Tensor products of function spaces, Bull. Amer. Math. Soc. 82 (1976) 626628.

[Mi78] M. Milman, Embeddings of LorentzMarcinkiewicz spaces with mixed norms, Anal. Math. 4

(1978) 215223.

[Mi81] M. Milman, A note on Lp; q spaces and Orlicz spaces with mixed norms, Proc. Amer. Math.Soc. 83 (1981) 743746.

[N] I.P. Natanson, Theory of Functions of a Real Variable, Vol. II, Frederick Ungar Publishing Co.,

New York, 1961.

[O] R. ONeil, Integral transforms and tensor products on Orlicz spaces and Lp; q spaces,J. Analyse Math. 21 (1968) 1276.

[S] T. Shimogaki, A note on norms of compression operators on function spaces, Proc. Japan Acad.46 (1970) 239242.

ARTICLE IN PRESS


Documents

S.V. Astashkin, L. Maligranda and E.M. Semenov- Multiplicator space and complemented subspaces of rearrangement invariant space