dm  ()=sin 2 d,forevery1<p<1andofweaktype(1,1).Thesamebound- edness propertieshold for the maximaloperator of the trun ated operators

ULTRASPHERICAL POLYNOMIALS

DARIUSZBURACZEWSKI,TERESAMARTINEZ,JOS



EL.TORREA,ANDROMANURBAN

Abstra t. We dene and investigate the Riesz transform asso iated to the dif-

ferential operator L



f() = f 00

() 2 otf 0

()+ 2

f(). We prove that it

an bedened as a prin ipal value, and that it isbounded on L p

([0;℄;dm

 ()),

dm



()=sin 2

d,forevery1<p<1andofweaktype(1,1).Thesamebound-

edness propertieshold for the maximaloperator of the trun ated operators. The

speedof onvergen eofthetrun atedoperatorsismeasuredintermsofthebound-

ednessinL p

(dm



),1<p<1andweaktype(1;1)oftheos illationand-variation

asso iated to them. Also, a multiplier theoremis proved toget the boundedness

ofthe onjugate fun tion studiedbyMu kenhoupt and Steinfor 1 <p <1asa

orollaryoftheresultsfortheRiesztransform. Moreover,wenda onditiononthe

weightvwhi hisne essaryand suÆ ientfortheexisten eofaweightusu hthat

theRiesztransformisboundedfromL p

(vdm

 )intoL

p

(udm

 ):

1. Introdu tion

We onsider the ultraspheri al polynomials of type , with any  > 0, P



n (x) de-

nedasthe oeÆ ientsintheexpansionofthegenerating fun tion(1 2x!+! 2

)



=

P

1

n=0

! n

P



n

(x)(see[17℄forfurtherdetails). ItisknownthatthesetfP



n

( os):n2Ng

isorthogonaland omplete inL 2

[0;℄ withrespe tto themeasure dm



():=sin 2

d:

Morepre isely,wehave

(1.1) Z



0 P



n

( os)P



m

( os)dm

 ()=Æ

n;m 2

1 2

 () 2

(n+2)

(n+)n!

=Æ

n;m

=

n :

Itisalsoknownthat thefun tionsP



n

( os)areeigenfun tionsoftheoperatorL



(1.2) L



f()= f 00

() 2 otf 0

()+ 2

f();

withL

 P



n

( os)=(n+) 2

P



n ( os).

In this paper we develop a general method for studying the properties of ertain

operators asso iated to the dierential operator L



, that appear similarly to the way

that lassi aloperatorsdowhen onsidering theLapla ianoperator 2



in[0;2℄. Given

anoperatorTweprodu eapartitionofT intoits\lo al"andits\global"partsa ording

totworegions: the\lo alregion",whereroughlyspeakingwe anusethattheLebesgue

measure d and dm



() are equivalent, and its omplementary, that we all \global

region"(see(3.2)and(3.3)). Thismethodwaswidelyused by manyauthors invarious

ontexts (see e.g. [4, 5, 9, 11, 14℄). The pro edure for the lo al partof the operators

2000Mathemati sSubje tClassi ation. 42C05,42C15.

Keywordsandphrases. Ultraspheri alpolynomials,Poissonintegral,Riesztransform,maximalop-

erator,weightedinequalities.

Theauthorswere partiallysupportedbyRTN Harmoni Analysisand Related Problems ontra t

HPRN-CT-2001-00273-HARP. Therst and fourth authors weresupported inpart byKBN grant1

P03A01826. These ondandthirdauthorswerepartiallysupportedbyBFMgrant2002-04013-C02-02.

onsistsin omparingthekerneloftheoperatorwiththekernelofthe lassi alanalogous

operatorinthetorus. Weshowthatthedieren ebetweenthemisboundedbya\good"

kernel,i.e.,akernelwhi hdenesaboundedoperatoronL p

(d)foreverypintherange

1 p  1. By a\heritage"theorem (Theorem 3.8), we provethat the lo al part of

the lassi al operator in the torus inherits the boundedness in L p

with respe t to the

Lebesguemeasure. Next, foroperators withkernelssupported in thelo al part(\lo al

operators"),weprove(seeLemma 3.9) that theboundedness withrespe tto Lebesgue

measureisequivalenttotheboundednesswithrespe ttodm



(). Thereforewe on lude

that thelo alpartofT is bounded in L p

(dm



):Intheglobalpartwejust usethat the

kerneloftheoperatorT isboundedbyapositiveandni ekernel.

InthispaperweapplytheabovemethodinordertostudytheRiesztransform(The-

orem2.14),themaximaloperator asso iatedtoit(Theorem 7.1)anditsos illationand

variation(Theorem 8.2). We alsoprovesomemultiplier theorem in order to showthat

theresults in [10℄ anbe obtainedfrom ours,see se tion 6. Wewould liketo thankS.

Medaforsomehelpful suggestions on erningthistheorem.

Forevery\ni e"fun tionf dened on[0;℄weasso iateitsultraspheri alexpansion

(1.3) f() 1

X

n=0 a

n P



n

( os); where a

n

=

n Z



0 f()P



n

( os)dm

 ():

Inthepaper[10℄ Mu kenhouptandStein studied ultraspheri alexpansionsfromahar-

moni analysis point of view. Forany fun tion asin (1.3)they dene the \harmoni "

extensionoff as

(1.4) f(r;)=

1

X

n=0 a

n r

n

P



n ( os)

andthe\ onjugateharmoni "seriesas

(1.5)

e

f(r;)= 1

X

n=1 2

n+2

a

n r

n

sinP

+1

n 1 ( os):

Then,bothfun tions satisfyCau hy{Riemannequations:



r

((rsin) 2

e

f) = r

2 1

(sin) 2



 f;





((rsin) 2

e

f) = r 2+1

(sin) 2



r f:

Oneofthemainresultsin[10℄isthatiff 2L p

(dm



)for1<p<1,thenthereexists

theboundaryfun tion

~

f()of e

f(r;)su hthat

~

f()=lim

r!1 e

f(r;) in L p

(dm

 ) norm

andk

~

fk

L p

(dm)

A

p kfk

L p

(dm) .

~

f()is alledthe onjugatefun tion off. Con erning

the asep=1,theyshowthatthe\ onjugateharmoni "fun tionf()!

~

f(r;)satises

an appropriate substitute relation (see Corollary 2, page 43, in [10℄). As far as we

understand, this result doesnot lead in an obvious wayto the weak type (1;1) of the

onjugatefun tion

~

f().

The onne tion between the results in [10℄ and the ones ontained in this paper is

as follows. In se tion 6 we prove a multiplier theorem that allows us to obtain the

boundedness properties in L p

forp in the range1<p<1, of the onjugatefun tion

~

f()as a orollaryofourresultsfortheRiesz transformdened spe trallyin (1.7),and

a posteriori as a prin ipal value operator. Although we obtain indeed the weak type

f()!

~

f()ofMu khenhouptand Stein,whi hisdened onlyforf 2L p

,satises also

aweaktype(1;1) inequality.

Itiseasyto he kthatL



isformallyself-adjointonthespa eL 2

(dm



). Ouroperator

L



fa torizesasL



f() =( 







 +

2

)f(),see se tion 2,where 





=



+2 ot is

formallyadjointto



;i.e.,h



 f;gi

L 2

(m

 )

= hf;

 gi

L 2

(m

 )

. Forafun tionf asin(1.3)

wedeneitsPoissonintegralasso iatedtoouroperatorL



asfollows

(1.6) Pf(e

t

;)=e t

p

L

f()= 1

X

n=0 a

n e

t(n+)

P



n ( os):

Wepro eednowtodeneoneofthemainobje tswearegoingtoinvestigateinthispaper,

theRiesz transform. Let us allpolynomialfun tion to anynitelinear ombination of

ultraspheri al polynomials,that is, any f of the form f = P

N

n=0 a

n P



n

. We dene the

Riesz transformof any polynomialfun tion f byspe tralte hniques,asit issuggested

in[15℄,asfollows:

R



f() = 

 (L

 )

1=2

f()=

 N

X

n=0 a

n

n+ P



n ( os)

!

(1.7)

= 2

N

X

n=1 a

n

n+ sinP

+1

n 1 ( os):

Theaboveequationfollowsfromtheformula



t P



n

(t)=2P

+1

n 1

(t), see[17℄.

It turns out that if for a given fun tion f and its Poisson integral Pf, we dene

analogouslyto(1.5)the onjugatePoissonintegral

R



f(r;)= 2

N

X

n=1 a

n

n+ r

n+

sinP

+1

n 1 ( os):

ThefollowingCau hy{Riemannequationsaresatised



 Pf(e

t

;)= 

t R

 f(e

t

;);



t Pf(e

t

;)+ 2

Z

1

t

Pf(e s

;)ds=



 R

 f(e

t

;):

(1.8)

Ourrstmain result,Theorem2.13,saysthatinfa t ourRiesztransformonpolyno-

mialfun tions isgivenbyaprin ipalvalue,i.e.,foranyf apolynomialfun tion,

R



f()=p:v:

Z



0 R



(;)f()dm



()=lim

"!0 Z

j j>"

R



(;)f()dm

 ()a.e:

Also, we provethat theRiesz transform R



is bounded on L p

(dm



) and of weak type

(1,1)(seeTheorem2.14). Asa orollaryweobtainthatR

 f(e

t

;)tendstoR

 f in L

p

normfor1<p<1,ast goesto 0.

These ondobje tweareinterestedinisamaximalfun tion

R





f():=sup

">0 





 Z

j j>"

R



(;)f()dm

 ()







 :

Theorem7.1saysthatR





isboundedonL p

(dm



)for1<p<1;andofweaktype(1,1).

Asa onsequen eofthis result,wegetthattheRiesztransformisaprin ipalvaluealso

for fun tions in L 1

(dm



) (Theorem 7.2). Given this, our next aim is measuring the

speed of onvergen e, something whi h has lassi ally been done by means of ertain

use the so alled os illation operator and -variation operator, and obtain that they

bothare bounded in L p

(dm



) for1<p<1;and that they satisfyaweak type(1;1)

inequalitywithrespe ttodm



(seese tion8forthepre isedenitionoftheoperatorsand

statementoftheresults). Asanintermediatestepintheproof ofthisresultsweobtain

the boundedness in L p

(d) for 1<p <1 and theweak type(1;1) for the os illation

and -variationasso iated to the lassi al onjugate fun tion in the torus. This result,

that webelieveofintrinsi interest, isobtainedby omparisonwith thesameoperators

fortheHilberttransformintheline.

Thelastpartofthispaperisdevotedtothefollowingproblem. Givenaxedoperator

T (inour asewe onsidertheRiesztransformR



)boundedinL p

(dm



)forsome1<p<

1;ndne essaryandsuÆ ient onditionsforaweightv in ordertohavetheexisten e

of anontrivialweight usu h that T be omes bounded from L p

(vdm



) into L p

(udm

 ):

Inthe aseof theRiesz transform onsidered in thispaperweprove(seeTheorem 9.3)

thatthe onditionisthat

Z



0 v(x)

1

p 1

dm



(x)<1:

Toobtainthisresult,we onsiderave tor-valuedextensionofR



totheoperatortaking

valuesintheBana hspa el p

. Thenweapplytheanalogousve tor-valuedversionofthe

te hniqueofsplittingtheoperatorintoitslo alandglobalparts,togettheboundedness

inL p

` p

oftheextension(seese tion3forthedenitions).

Guide to the paper. Thestru tureof thepaperisasfollows. Inse tion2westate

pre iselythenotationandallthene essarydenitions inorder tointrodu etheobje ts

ofourinterest. Inparti ular,theRiesztransformR



isdened. Inse tion3weintrodu e

a\generalma hinery"inordertohandlethe\lo al"and\global"partsoftheoperators.

Inse tion4westudythebehavioroftheRieszkernela ordingtothispartition,andin

se tion5weshowthattheRiesztransformR



isgivenasaprin ipalvalueforpolynomial

fun tions anditsboundednessproperties. Inse tion6weproveamultipliertheorem.

Se tion7isdevotedtostudythemaximalfun tionR





f()andprovethatthemaximal

operator f 7! R





f is bounded on every L p

(dm



), for 1 < p < 1 and of weak type

(1;1). Thenextse tion,se tion8dealswiththestudyoftheos illationand-variation

operators. Finally,inse tion9we onsiderweightedinequalitiesandanswerthequestion

mentionedabove(Theorem9.3).

2. The Riesz transform

2.1. Poissonkernel. Wearegoingtodes ribeheresomefundamentalpropertiesofour

PoissonkernelP. Byusing(1.4)and(1.6),observethat

(2.1) Pf(e

t

;)=e t

f(e t

;):

Theresults in [10℄ forf(r;) leadeasily to provethat thePoisson integralPf is given

byakernel,whi h anbe omputedexpli itly. Namely,forany2[0;℄wehave

(2.2) Pf(r;)=r

 Z



0

P(r;;)f()dm

 ()

wherer=e t

and

(2.3) P(r;;)=



 (1 r

2

) Z



0

sin 2 1

t

2

+1 dt:

Anothereasy onsequen eof (2.1)andTheorem2,se tion6,p.31in[10℄isthefollowing

theorem.

Theorem2.4. [Theorem 2,[10℄℄ Letf 2L p

(dm



);1p1. Then

(a) kPf(r;
)k

L p

(dm

 )

r



kfk

L p

(dm

 )

;

(b) kPf(r;
) fk

L p

(dm

 )

!0asr!1;

( ) lim

r!1

Pf(r;)=f() almosteverywhere,

(d) ksup

r<1

jPf(r;
)jk

L p

(dm

 )

A

p kfk

L p

(dm

 )

:

2.2. The Riesz Transform. Inthisse tion we omputethekernelof theRiesztrans-

form. Byusing(1.1),oneshowsthat

(2.5) kR

 fk

L 2

(dm)

Ckfk

L 2

(dm) :

Withthesameproof,wegettheboundednessoftheoperatorforanyf 2L 2

(dm



)andso

thedenition(1.7)makessenseforeveryf 2L 2

(dm



). Observethatasimilarargument

givesthattheoperator(L

 )

1=2

isalsowelldenedinL 2

(dm



)withtheparallelformula

to theoneabove. Inthefollowinglemmaweshowthat (L

 )

1=2

is denedby akernel

almosteverywhere,andthattheRiesztransformhasanasso iatedkernelinthesenseof

Calderon-Zygmund. Thislaststatementmeans,forT alinearoperator,thatthereexists

afun tionk(;)su hthatforeveryf 2L 2

([0;℄;dm



)andoutsidethesupportoff,

Tf()= R

k(;)f()dm

 ().

Lemma2.6. Given f 2L 1

(dm



), for almost every2[0;℄,we havethat

(2.7) (L

 )

1

2

f()= Z



0 W



(;)f()dm



(); where W



(;)= Z

1

0 r

 1

P(r;;)dr:

Given f 2L 1

(dm



)and outsidethe supportof f,we havethat

(2.8) R

 f()=

Z



0 R



(;)f()dm



(); where R



(;)= Z

1

0 r

 1

P



(r;;)dr;

whereP



standsfor the partialderivative ofP with respe tto.

Proof. Toprovetherstformula,weusetheintegralformula

s a

= 1

(a) Z

1

0 e

ts

t a

dt

t

todenethenegativesquarerootofL

 :

L 1=2



f() = 1

(1) Z

1

0 e

t p

L

f()t dt

t

= Z

1

0 e

t

f(e t

;)dt

= Z

1

0 Z



0 r

 1

P(r;;)f()dm

 ()dr;

(2.9)

after applying the hange of variables r=e t

. We gethere(2.7) ifwe an hange the

order of integration in the formerintegral. Before we pro eed further let us introdu e

somenotation:

a = os os+sinsin ost= os( ) sinsin(1 ost);

(2.10)

D

r

= 1 2ra+r 2

:

Observe that a  os( ), and that for negative a, D

r

 1, therefore we always

have that D

r

 1 os( ) 2

 0. In parti ular, this implies that r

 1

P(r;;)

 1 2 +1

observethattheexpressionf(r;)= R



0

P(r;;)f()dm



() iswelldenedforalmost

every2[0;℄sin ebytheresultsin[10℄, namelyLemmas5.2 and5.3,pg. 28,wehave

thatjf(r;)j(jfj)



(), where

f



()= sup

h6=0;0+h

R

+h



jf()jdm

 ()

R

+h

 dm

 ()

andthisfun tionisboundedin L p

(dm



)foreveryp2(1;1℄andofweaktype(1;1). In

parti ular, for f 2L 1

(dm



), it is nite foralmost every . Thus, we ansee that the

integrandin(2.9)belongstoL 1

(dm



dr)forf 2L 1

([0;℄;dm

 ).

On e we hange the order of integration in (2.9), to get the expression of R

 (;)

we just have to be able to put the derivative 



inside the integral. We prove it by

showingthatr

 1

P



(r;;)f()2L 1

(dm



dr). Observethatj



aj2andthat then

jr

 1

P



(r;;)j anbeboundedabovebya onstantforoutsidethesupportoff,what

givestheresult. 

Remark2.11. One analsousetheCau hy{Riemannequations (1.8)toprovetheabove

lemma. Namely, inviewofequations

 Pf(e

t

;)= 

t R

 f(e

t

;)andR



f(0;)=0,

weobtain

R

 f(e

t

;)= Z

1

t



 P

t

f()dt= Z

1

t Z



0 e

t

P

 (e

t

;;)f()ddt:

Further,if does notbelongtothesupportoff,we anmake r=e t

andthenargueas

beforetojustifymaking r going to1toobtain (2.8).

Remark2.12. Observethatforanyboundedfun tionf,f



()kfk

L

1 isawelldened

fun tion,nitefor every2[0;℄. Thus,byrepeatingthe proofof (2.7), wehavethatit

holdsfor every 2[0;℄.

Next,letus stateherethemain results on erningtheRiesz transform,namelythat

itisaprin ipalvalue(Theorem2.13)anditsboundedness propertiesinL p

, 1<p<1,

andin weakL 1

(Theorem2.14). Wewillprovethemin se tion5.

Theorem 2.13. The Riesz transform R



is given as a prin ipal value on polynomial

fun tions, i.e.,for any polynomial fun tion f andfor almostevery 2[0;℄

R



f()=lim

"!0 Z

j j>"

R



(;)f()dm

 ():

Theorem 2.14. The Riesztransform R



isbounded on L p

(dm



), for 1<p<1 and

fromL 1

(dm



)intoL 1;1

(dm

 ).

Remark2.15. Givenp2(1;1)andf 2L p

(dm



)su hthatithasanasso iatedFourier

seriesexpansionf  P

a

n P



n

( os), then

R

 f()

X

2

a

n

n+ sinP

+1

n 1 ( os):

Indeed, onsider thelinear operators fromL p

(dm



) intoC given by

U

n

f =

2a

n

n+

;

V

n

f = ksinP

+1

n 1

( os)k 2

L 2

(dm) Z



0 R



f()sinP

+1

n 1

( os)dm

 ():

Let us note that U

n

is trivially bounded by (1.3) and Theorem 2.14, and that V

n is

boundedalsobyTheorem2.14 andthefa tthat sinP

+1

( os)arein L p

0

(dm



). Now,

observe that fsinP

+1

n 1

( os)g forman orthogonal systemin L 2

(dm



), and thatthere-

fore,for every polynomialfun tion f,U

n f =V

n

f. Sin epolynomial fun tionsaredense

inL p

(dm



)for everyp2(1;1), ourthesis holds.

3. General ma hinery

Westartthis se tionwithanappropriateforourpurposes overinglemma. First,we

onstru tafamilyofballsontheinterval[0;



2

℄. LetusxaverysmallÆ>0throughout

thepaper,anddeneasequen eof pointsandballsasfollows



0

=



2

; 

i+1

=



i

1+Æ

; B

i

=



: 1

1+Æ

<



i



<1+Æ



;

Letus also onsider ballsB(

i

;Æ 0

)=f: (1+Æ 0

) 1

<

i

= <1+Æ 0

g. One an easily

extend(bysymmetrywith respe tto



2

)theabovefamilyto aset of balls overingthe

interval(0;),thatwillbedenotedin thesamewayfB

i g

1

i=0 .

Lemma3.1. Forthe olle tion of ballsfB

i g

1

i=0

denedabove wehave

(1) The olle tion fB

i g

1

i=0

of losedballs overs(0;).

(2) ThereexistsÆ

0

su hthatthe ballsB(

i

;Æ

0

)arepairwisedisjoint.

(3) Foreveryn1,if1+Æ

0

=(1+Æ) n

;thenthe olle tion fB(

i

;Æ

0

)g hasbounded

overlap.

(4) For every n1, if 1+Æ

0

=(1+Æ) n

; then there exist a onstant C depending

onlyon Æ,n,su hthatfor every measurable setEB(

i

;Æ

0

), wehavethat

1

C

 2

i

jEjm



(E)C

2

i jEj:

Proof. By the symmetry of the onstru tion it is enough to prove this properties for

theballs with enter in [0;=2℄: Inorder to prove(2) itis enoughto takeÆ

0

satisfying

(1+Æ

0 )

2

<1+Æ. Infa t,wehave

(1+Æ

0 )

i+1

<

(1+Æ)

i+1

1+Æ

0

=



i

1+Æ

0 :

Weleavethedetails oftheremainingproofto thereader. 

ConsideralinearoperatorT mappingthespa eofpolynomialfun tionsintothespa e

ofmeasurablefun tions on[0;℄,satisfyingthefollowingassumptions:

(a) T either extendsto abounded operator on L q

(d) for some1< q <1; or of

weaktype(1,1)withrespe ttod,

(b) thereexistsameasurablefun tionK,denedinthe omplementofthediagonal

in[0;℄[0;℄;su hthatforeveryfun tionf andalloutsidethesupportoff,

Tf()= Z



0

K(;)f()d;

( ) forall(;);6=,KveriesjK(;)jCj j 1

.

Let us denote by N

t

the region onstru ted in the following way: onsider the part of

theregionM

t

=f(;): 1

1+t

<





<1+tgunderthediagonalofthesquare[0;℄

2

with

negativeslope,anddeneN

t

tobethisregiontogetherwithitsre e tedwithrespe tto

thepoint(=2;=2). ItisnotdiÆ ultto seethat theregionsN

t1 andN

t2 , witht

1

=Æ

and1+t =(1+Æ) 3

havetheshapeshowninthenextpi ture.

6

-







2



2





=



(1+Æ ) 3

=



1+Æ

j

s o

6

=(1+Æ)

=(1+Æ) 3

Observethatwegetaregion ontainedinM

t

andsymmetri withrespe ttothepoint

(=2;=2). Thisregionwilldenethe\lo al"and \global"parts oftheoperators.

GivenanoperatorT satisfying(b),wedeneitsglobalandlo alpartsby

T

glob f()=

Z



0

K(;)(1 1

N

t

2

(;))f()d;

(3.2)

T

lo

f()=Tf() T

glob f():

(3.3)

Wewill need thefollowingtwolemmas, whose proofsare areformulationofthe proofs

ofLemmas3.2and3.3in [5℄usingonlypropertiesofthe overingfrom Lemma3.1.

Lemma 3.4. Let d be any positive measure in [0;℄, ff

j

gbe asequen e of fun tions

and dene f = P

j 1

B

j f

j

; where fB

j

: j 2 Ng is the olle tion of balls in Lemma 3.1.

Then,for all >0;

(3.5) f:jf()j>g X

j

f2B

j :jf

j

()j>=2g;

andfor any 1q<1

(3.6) kfk

L q

(d)

2



X

j Z

B

j jf

j ()j

q

d()



1=q

;

Proof. Observe that any point  2 [0;℄ belongs to at most two balls of the overing

fB

j

g. Then, the levelset f : jf()j > gis a subset of S

j f 2 B

j : jf

j

()j > =2g,

whi hgives(3.5). Toprove(3.6),just observethat

Z



0 jf()j

q

d() Z



0

(2max

j jf

j ()j1

B

j ())

q

d()2 q

Z



0 X

j jf

j ()j

q

1

B

j

()d():



Lemma3.7. LetT bealinearoperator mappingpolynomialfun tionsintothe spa eof

measurable fun tionson [0;℄ verifying (a). Given the overing fB

j g

1

, wedene the

operator

T 1

f()= X

j 1

Bj ()jT(1

B 0

j

f)()j; where B 0

j

=B(

j

;Æ 0

); 1+Æ 0

=(1+Æ) 2

:

Then, T 1

isalso bounded from L 1

(d) intoL 1;1

(d) or fromL q

(d) into L q

(d),1<

q<1; asthe asemightbe.

Proof. Toprovethat T 1

isofweaktype(1;1)withrespe tto difT is,observethat

jf2B

j : jT(1

B 0

j

f)()j>=2gj C

 Z

B

j

jf()jd:

We nish if we apply Lemma 3.4 and the bounded overlapproperty from Lemma 3.1.

Theproofofstrongtype(q;q)isanalogous. 

Thefollowingtheoremisoneofthemaintoolsintheproofsoftheresultsinthepaper.

Theorem3.8. Under the assumptions (a), (b) and ( )made on T above, the operator

T

lo

inherits fromT either L q

(d)-boundednessortheweaktype(1,1)withrespe ttothe

Lebesguemeasure, asthe asemight be.

Proof. Assumethat isin theballB

j

anddeneB 0

j

asinLemma 3.7. Then

T

lo

f()=Tf() T

glob f()

=T(f1

B 0

j

)()+T(f(1 1

B 0

j ))()

Z



0 (1 1

N

t

2

(;))K(;)f()d

=T(f1

B 0

j )()+

Z



0 (1

N

t

2

(;) 1

B 0

j

())K(;)f()d:

Bymultiplyingby1

Bj

andsummingoverj,wegetthatforany2[0;℄

jT

lo

f()jjT 1

(f)()j+T 2

(f)();

where T 1

is asin Lemma 3.7, and by that lemma itis bounded in L q

orof weak type

(1;1)as orresponds. Therefore,weonlyneedtoprovethatT 2

isbounded, where

T 2

(f)()= Z



0 X

j 1

Bj ()j1

Nt

2

(;) 1

B 0

j

()jjK(;)jjf()jd:

Thisoperatoris givenbythekernel

H(;)= X

j 1

Bj ()j1

Nt

2

(;) 1

B 0

j

()jjK(;)j;

whi hgivesaboundedoperator.Wewillstudythiskernelinea hofthesquares[0;=2℄

2

,

[0;=2℄[=2;℄,[=2;℄[0;=2℄and[=2;℄

2

. Intherstandfourthones,H(;)is

supportedin N

t2 nN

t1

. Indeed,letusassumerst(;)2[0;=2℄

2

. If(;)62N

t2 then

1

(1+Æ) 3

>



 or





>(1+Æ) 3

:

Hen e, if  2 B

j

, then  annot belong to B 0

j

and then j1

Nt

2

(;) 1

B 0

j

()j = 0. If

(;)2N

t1

and2B

j then

1

<



<1+Æ and

1

2

<



j

<(1+Æ) 2

;

whi himpliesthat2B 0

j andj1

N

t

2

(;) 1

Æ 0

B

j

()j=0. For(;)2[=2;℄

2

, weuse

that ea hterm 1

Bj ()j1

Nt

2

(;) 1

B 0

j

()j is symmetri with respe t to (=2;=2) and

wegetinthisregionthesameresult.

In theregion where (;) 2[0;=2℄[=2;℄, one an see bysimilar argumentsas

beforethat thesupport ofH(;)is ontainedin N

t1

. Sin ein that regionthereexists

a onstant " su h that j j >", there jK(;)j C

"

and by thenite overlapping

propertyfromLemma3.1,H(;)C1

N

t

1

(;). Bysymmetry,wegetthesamebound

intheregion[=2;℄[0;=2℄.

Sin ethepartoftheoperatorintheregions[0;=2℄[=2;℄and[=2;℄[0;=2℄is

triviallybounded inL p

(d) foreveryp(1p1),to provethat T 2

alsosatisesthis

property,itisenoughtoprovethat

sup

2[0;=2℄

Z

=2

0 1

Nt

2 nNt

1 (;)

j j

d<1; sup

2[0;=2℄

Z

=2

0 1

Nt

2 nNt

1 (;)

j j

d<1:

Letusprovetherstinequality,sin ethese ondonefollowsinanidenti alway. Tothis

end,write:

sup

 Z



0 1

Nt

2 nNt

1 (;)

j j

dsup

 Z



1+t

1



1+t

2 1

 

d+sup

 Z

(1+t2)

(1+t1) 1

  d

=log 1

1

1+t

1

1 1

1+t2 +log

1+t

2

1+t

1

<1:



Similarargumentsas theones shownintheaboveproofleadtothefollowinglemma.

Lemma 3.9. Assume that an operator S dened on polynomial fun tionssatises (b),

with a kernel supported in N

t2

. Then, strong type (p,p) for Lebesgue measure and m



measureareequivalent. Thesame holdsfor weak type(p,p), 1p1.

Proof. One an see that jSf()j P

j 1

B

j jS(f1

~

B

j

)()j, where B

j

are asin Lemma 3.1

and

~

B

j

=B(

j

;

~

Æ),for1+

~

Æ=(1+Æ) 4

,just pro eeding asin theproofofTheorem3.8.

Then, by using Lemma 3.4, and parts (3) and (4) from Lemma 3.1, one aneasily get

theresult. 

In se tions 7 and 9 wewill have to onsider linear operators taking valuesin some

Bana h spa es. We denote by L p

B

() = L p

B

(;[0;℄), p < 1, the Bo hner-Lebesgue

spa e onsistingofallB-valued(strongly)measurablefun tions f denedon[0;℄su h

that kfk p

L p

B ()

= R



0 kf()k

p

B

d() < 1. For p =1 wewrite L 1

B

for the spa e of all

f su h thatkfk

L 1

B

=supesskf()k

B

<1. Similarly,thespa e L p;1

B

()=weak-L p

B ()

is formed by all B-valued fun tions f su h that kfk

L p;1

B ()

= sup

t>0

t(f 2 [0;℄ :

kf()k

B

>tg) 1=p

<1.

We note here that the \heritage" Theorem 3.8 remains valid in the ve tor valued

ase with the same proof. More pre isely, given B

1

; B

2

Bana h spa es, let T be a

linearoperator dened in thespa e ofpolynomialfun tions with B

1

-valued oeÆ ients

andtakingvaluesinthespa eofB

2

-valuedandstronglymeasurablefun tionson[0;℄;

satisfying onditions(a'),(b')and( ')whi hare,infa t,(a),(b)and( )withappropriate

(a') T extends to a bounded operator either from L q

B

1

(d) into L q

B

2

(d) for some

1<q<1; orfromL 1

B

1

(d)into L 1;1

B2 (d),

(b') there exists a L(B

1

;B

2

)-valued measurable fun tion K, dened in the omple-

mentof thediagonalin [0;℄[0;℄; su hthat foreveryfun tion f 2L 1

B1 and

outsidethesupportoff,

Tf()= Z



0

K(;)f()d:

( ') thefun tionK satiseskK(;)kCj j 1

. forall6=.

In this\ve torvaluedlanguage" we an obtainLemmas 3.4 and 3.7, Theorem 3.8 and

Lemma 3.9in asimilar way (just hangingabsolutevaluesbynorms). Forfurther use,

letusstatetheve torvaluedversionofTheorem3.8asfollows.

Theorem3.10. If T isan operator satisfying(a'), (b')and( '). Dene T

lo andT

glob

as in 3.3 and 3.2. Then T

lo

inherits from T either the L q

-boundedness or the weak

type (1,1) as the ase might be. Besides, the orresponding boundedness holds for both

Lebesgueandm



measures.

4. Estimates onthe Riesz kernel

4.1. The lo al part. In this se tion we study the behaviorof the kernel of theRiesz

transform,R



(;)relatedto thekernelofthe lassi al onjugatefun tionin thetorus,

C(;)= 1

2

1

tan(( )=2)

,when onsidered onthelo alpart. Wearegoingto provethat

thedieren e

jR



(;)(sin) 2

C

0

C(;)j;

where C

0

is appropriately hosenreal number, anbeestimated in thelo al region by

some\ni e"fun tionM(;)su hthatthelo aloperator R



0

M(;)f()disbounded

oneveryL p

(d); for1p1.

We will usethe notationfor a and D

r

introdu ed in (2.10), as well asthe following

ones

b = 



a= sin os+ ossin ost= sin( ) ossin(1 ost);

r

= 1 2r os( )+r 2

;



r

= rsinsin:

Also,weshallwriteD,
, insteadofD

1 ,

1 ,

1

. Asin[16℄wearegoingto makeuse

ofthefollowingsimpleidentity, whi hisvalidforeveryreal> 1

(4.1)

Z

1

0 r



(1 r 2

)

D +2

r

dr=

1

(+1)D +1

:

BytheargumentsintheproofofLemma2.6,(4.1)and(2.3),wemaysimplifythekernel

oftheRiesztransformasfollows

(4.2) R



(;)= 2

 Z



0 bsin

2 1

t

D

+1 dt:

Dene

(4.3) N(;)=1+log

+



sinsin

1 os( )



:

Lemma4.4. There exist onstants C

0

andC>0su hthat in the lo al region we have

the following estimate

jR



(;)(sin) 2

C

0

C(;)jCM(;);

whereM(;)=N(;)(sin) 1

.

Proof. Observe that R



( ; ) = R



(;), C( ; ) = C(;) and

M( ; )=M(;). Therefore,ifwegetthe omparisonintheregion(;)2[0;=

2℄[0;℄wegetthesameinequalityintheregion[=2;℄[0;℄. Thus,wewillrestri t

ourselvestoprovetheinequalityfor(;)2[0;=2℄[0;℄. Inthelo alregionforthis

rangeof (;), weget thefollowingestimates, that will beused throughout theproof:

thereexistsa onstantC su h that

1

C

sinsinCsin; jsin( )jCsin:

Theproofoftheseinequalitiesistrivial,althoughtheargumentdierswhen2[0;=2℄

thanwhen2[=2;℄. Wewillalsohandleintegralsoftheform

I 



= Z 

2

0 t



(
+

t 2

2 )

 dt:

Byusingthe hangeofvariablest=
1

2

 1

2

u,andtheformerinequalities,forany>0,

if=2+1and =+1,theaboveintegral anbeestimatedasfollows

I 2+1

+1

C(sinsin) (+1)

N(;):

Firstwegetrid ofapartofthekernel(4.2),byobservingthat







 (sin)

2

2

 Z



=2 b(sint)

2 1

D

+1 dt









C(sin) 2+1

Z



=2 (sint)

2 1



+1

t 2+2

dt C

sin

Next,letuspro eedwiththe omparisonof thekernelsin severalsteps.

Step1. Dene

R 1



(;)= 2

 Z

=2

0 bt

2 1

D

+1 dt:

Thenusingthat jsin 2 1

t t 2 1

j=O(t 2+1

),weobtain

jR



(;) R 1



(;)j(sin) 2

C(sin) 2+1

I 2+1

+1

CM(;):

Step2. Dene

R 2



(;)= 2

 Z

=2

0 bt

2 1

D

+1 dt;

where

D=2(1 os( )+sinsin t

2

2 ):

ThenbytheTaylorexpansion,fort2[0;℄.





 1

D

+1 1

D

+1 



C

sinsint 4

D

+2 :

Thus,wehave

jR 1



(;) R 2



(;)j(sin) 2

 (sin) 2+3

I 2+3

+2

CM(;):

Step3. Dene

R 3



(;)= 2

 Z

=2

0

sin( )t 2 1

+1

dt:

Then

jR 2



(;) R 3



(;)j(sin) 2

C(sin) 2+1

I 2+1

+1

CM(;):

Step4. Compute:

R 3



(;)(sin) 2

= (sin) 2

2

 Z



2

0

sin( )t 2 1

D

+1

dt

= 2



sin( )(sin) 2

1







C Z

1



2

 1

2

1

2 u

2 1

(1+u 2

)

+1 du



= C

0



sin

sin





C(;)+I;

where

jIjC

sin( )

Z

1



2

 1

2

1

2 1

u 3

duCjsin( )j

1

 C

sin

;

andnallynoti e











sin

sin





C(;) C(;) 







 C

sin :



Lemma4.5. The lo al operator denedfor all f 2L 1

(d) and2[0;℄as

Mf()= Z



0

M(;)f()d;

whereM(;)isdenedinLemma4.4isboundedon L p

(d) for 1p1.

Proof. Toprovethelemmaitisenoughto he kthefollowing onditions,for =(1+Æ) 3

sup

2[0;℄

Z





M(;)d<1; sup

2[0;℄

Z





M(;)d<1:

Bythe hangeofvariablesx=,wehavethat

sup

2[0;℄

Z





N(;)

sin

dC



1+ sup

2[0;℄

Z





log

+



( ) 2

d





C+C Z

1

log

+ x

(1 x) 2

dx

x

<1:



4.2. The global part. Using the symmetry of the Riesz kernel R



(;) = R

 (

; ), wemay restri t ourselvesto onsider the global partof theRiesz transform

onlyfor2[0;=2℄. Observe(seethepi tureinpage8)thatfor2[0;=2℄,we angive

abound aboveforR

;glob

withthree integralsasfollows,where =(1+Æ) 3

:

jR

;glob

f()j  Z



0 jR



(;)j(1 1

Nt

2

(;))jf()jdm

 () (4.6)

= Z

f> g +

Z

f



>> g +

Z

[=2;℄

=I+II+III:

For the last integral, observe that there exists a " > 0 su h that in the interse tion

[0;=2℄[=2;℄ withtheglobalregion, j j>". Therefore, jR



(;)jC

"

as an

beeasilyseenfrom(4.2),andweget

III C

"

Z

[=2;℄

jf()jdm



()C

Æ kfk

L 1

(dm

 )

;

thatis, thispartisboundedin allL p

(dm



),1p1. Letusnowhandletheintegral

I.If> and,2[0;=2℄,thensin( )sinandjbjsin( ). Wewilluseas

wellthatforandinthisregion,D=1 a=1 os( )C( ) 2

Csin( ) 2

.

Hen e,inthispartweobtainthat

jR



(;)j

Csin( )

(sin ( )) 2+2

= C

(sin) 2+1

:

Wewillstatetheboundednessofthisoperator inthefollowinglemma.

Lemma4.7. The operator

M 1

f()= Z



0 M

1

(;)f()dm



(); whereM 1

(;)= 1

(sin) 2+1

1

f> g\[0;=2℄

2(;

)

for >1,is of weak type(1,1) and strong type(p;p) for 1<p<1 with respe t tothe

measuredm

 .

Proof. Firstweprovetheweaktype. Sin etheoperatorislinear,we anrestri tourselves

to provethe weak type (1;1) for positive fun tions with kfk

L 1

(dm

 )

=1. In this ase,

jM 1

f()j 1

(sin) 2+1

. Takeany2. Then,

m



(f: M 1

f()>g)m



([0;℄) C

 :

For>2,observethat 1

(sin) 2+1

>ifandonly if<ar sin (1=) 1=(2+1)

,sin ein

thatregionsinhasawelldenedin reasinginverse. Also,inthisregion(=2 "),

thereexistsC su hthat0<C< os. Therefore,

m



(f: M 1

f()>g) = m



(f: <ar sin (1=) 1=(2+1)

g)

 1

C Z

ar sin (1=) 1=(2+1)

0

os(sin) 2

d= 1

C

Nowwehavetoprovethat M 1

is bounded onL 1

(dm



),andthentheresultwillfollow

fromMar inkiewi zinterpolationtheorem. Thisboundednessholds,sin e

sup

0=2 Z



2

0

(sin) 2

(sin) 2+1

1

f> g

(;)d<1:



To giveabound for II, observethat in theregion where > and , 2[0;=2℄,

wehavethat bCsin( )andalsojsin( )jsin. Sin ehereitalsoholdsthat

DCsin( ) 2

weobtain

jR



(;)j C

(sin) 2+1

:

Lemma4.8. The operator

M 2

f()= Z



0 M

2

(;)f()dm



(); whereM 2

(;)= 1

(sin) 2+1

1

f> g\[0;=2℄

2(;

)

for >1 isof weak type (1;1) and strong type (p;p) for 1<p<1 with respe t tothe

measuredm

 .

Proof. Toproveweaktype(1,1)weargueasin thepreviouslemma,sin ein thisregion

sinCsin. These ondpartholds,be auseM 2

istheadjointofM 1

inL 2

(dm

 ). 

5. Proofs of Theorems2.13 and2.14

5.1. Proof of Theorem 2.13. Letf beapolynomialfun tion. Then by thespe tral

denition ofR



(1.7)andRemark 2.12,we anwrite

R



f()=(

 (L

 )

1=2

)f()=(

 [(L

 )

1=2

f℄)()=

 Z



0 W



(;)f()d

for every  2 [0;℄, sin e the integral exists for every  2 [0;℄ (by Lemma 2.6 and

Remark2.12).

Next step is adding and subtra ting the fun tion W(;) = 1



logjsin(( )=2)j,

thekernelof theoperator(

2

 )

1

2

in thetorus. LetC(;)bethekernelofthe lassi al

onjugate fun tion in thetorus. Sin e W(;) is anintegrable fun tion in , wehave

that R



0

W(;)f()dm



() iswelldened forevery, andwe anwrite

R



f() = 

 Z



0 (W



(;)sin 2

 C

0

W(;))f()d

(5.1)

+C

0



 Z



0

W(;)f()d:

These ond termisaprin ipalvalue,sin eit isthe onjugatefun tion in thetorus. To

handle the rst term, observethat for 6=, 

 W



(;)= R



(;) (sin e for  6= ,

jr

 1

P



(r;;)jCr



=(1 os( ) 2

)

+1

andthis fun tion is dr-integrable). Thus,

foralmostevery2[0;℄,wehavethat





(W(;)sin 2

 C

0

W(;))=R



(;)sin 2

 C

0 C(;):

Now,byusingLemma4.4and theargumentsofsubse tion4.2, wehavethat

jR



(;)sin 2

 C

0 C(;)j

 M(;)1

Nt

2

(;)+M 1

(;)(sin) 2

+M 2

(;)(sin) 2

+C1

N

t

2 (;);

wherethe onstantC omes fromtheboundedness forIII in(4.6)andtheboundedness

ofC(;)intheglobalregion. BytheproofsofLemmas4.5,4.7 and4.8, theright-hand

fun tionisd-integrable. Thus,byLebesgue'sdominated onvergen etheorem,therst

term in (5.1) is also the limitof the trun ated integrals, and thereforewe getthat for

almostevery2[0;℄,

R

 f()=

= lim

"!0 Z

j j>"

(R



(;)sin 2

 C

0

C(;))f()d+C

0 lim

"!0 Z

j j>"

C(;)f()d

= lim

"!0 Z

j j>"

R



(;)f()dm

 ():

5.2. ProofofTheorem2.14. Letuspointoutthatitisenoughtoprovethebounded-

nessinL p

orinweakL 1

foradensesub lassoffun tionsinthosespa es. Inthisse tion,

we onsider polynomialfun tionsf 2L p

forany1p<1,whi hformadensesubset

of L p

and for whi h Theorem 2.13holds. Next,wedene theglobal and lo al parts of

theRiesztransformR



a ordingto(3.2)and(3.3). Theboundednessoftheglobalpart

wasobtainedin subse tion4.2.

Considerthelo alpartR

;lo

. Wewrite

(5.2) R

;lo

=(R

;lo C

0 C

lo )+C

0 C

lo

;

where C denotesthe lassi al onjugatefun tion in thetorus. Letusobservethat ifan

operator T is given by a prin ipal value, then its lo al part is also a prin ipal value.

Thus, by Lemmas 4.4 and 4.5it followsthat R

;lo C

0 C

lo

is bounded on L p

(d) for

1p1. Letusre allthatby lassi alresultsandTheorem3.8,C

lo

isalsoboundedin

L p

(d),1<p<1andofweaktype(1;1). Thus,R

;lo

isboundedinL p

(d)1<p<1

andof weaktype(1;1),and byLemma 3.9it followsthatthesamestatementholds for

L p

(dm



())for1<p<1andfortheweaktype(1,1)withrespe ttodm

 .

6. Comparison of the results forMu kenhoupt-Stein's onjugate

fun tion andthe Riesz transform

Theaimof thisse tionis givinganargumenttoderivetheboundednessin L p

(dm

 )

for p 2 (1;1) obtained by Mu kenhoupt and Stein in [10℄ for the onjugate fun tion

dened there,asa orollaryof theboundednessobtainedin Theorem2.14fortheRiesz

transform in the same range of p's. First step is observing that Mu kenhoupt-Stein's

onjugate fun tion is dened, for a polynomial fun tion f() = P

N

n=0 a

n P



n

( os), as

~

f() = lim

r!1

~

f(r;). By using (1.5) and the expression for the Riesz transform of f

(1.7), we get that

~

f()=R

 (T

m

f)(), where T

m

isthe multiplier operator dened for

polynomialfun tions as

T

m f()=

N

X

n=0 a

n m

n P



n

( os); m

0

=0; m

n

=

n+

n+2

for n1:

Therefore,ifweobtainthatthemultiplierT

m

isboundedinL p

(dm



)foreveryp2(1;1),

wewouldgettriviallytheboundednessofthe onjugatefun tioninL p

(sin epolynomial

fun tionsaredensein allL p

(dm



)). Thismultiplierfallsintothes opeofthemultiplier

theorem proved by Mu kenhoupt and Stein in [10℄, Theorem 10 in page 71. But this

theoremgivestheboundednessofT

m

onlyinarestri tedrangeofp's,soweneedanother

multipliertheorem. Wewillrestri tourselvestostatethemultipliertheoremjustforT

m ,

sin eitisnotourinterestat thispointtoproveageneraltheorem.

Lemma6.1. The multiplieroperator T

m

isboundedinL p

(dm



)for every p2(1;1).

Proof. Dene,forp2(1;1)thespa e

X

p

=(I 

0 )L

p

(dm

 );

where 

0

is the ontinuousproje tion(orthogonal in L 2

(dm



)) onto the subspa egen-

erated by P



0

( os), whi h is a onstant (see[17℄). X

p

is a losed subspa e of L p

, the

subspa eoffun tions ofnullmean,andin parti ularaBana hspa eitself, wherepoly-

nomial fun tions of the form f() = P

K

k =1 a

k P



k

( os) are dense. Now, observe that

foranypolynomialfun tionf = P

K

k =0 a

k P



k

( os),T

m

f()=T

m (f a

0 P



0

( os)),and

thereforeitisenoughtogettheboundednessofT forthepolynomialfun tionsin X .

Let us onsider rst the ase p  2. To this end, let us observe that by spe tral

te hniques,wehavethat

T

m f()=

N

X

n=1 a

n



1

1+



n+



P



n

( os)=

1

1+L 1=2



f()=F(L 1=2



)f();

whereF(z)= 1

1+z

. Now,observethatthespe trumofL 1=2



inX

p

forp>2is

(L 1=2

 )=



1

n+

: n1



$



z: jzj<

1



;

and that the fun tion F is holomorphi and admits a power series expansion F(z) =

P

1

k =0 a

k z

k

intheright-handset.Byusingwellknownresultsonoperatortheory,seefor

exampleTheoremVII.3.10in[3℄,weget

(6.2) T

m

=F(L 1=2

 )=

1

X

k =0 a

k (L

1=2

 )

k

:

Next step is giving a good bound for the norm of L 1=2



in X

p

. This is given by the

followinglemma

Lemma6.3. Foreveryp,2p<1,thereexists"

p

2(0;1)su hthatforeveryf 2X

p ,

kL 1=2

 fk

L p

(dm)

 C

+"

p kfk

L p

(dm) :

Letus postponethe proofof thislemma, in order to larifythereading ofthe proof

ofLemma6.1. By(6.2),wehavethat foranyf 2X

p

,2p<1,

kT

m fk

L p

(dm

 )

 1

X

k =0 ja

k jk(L

1=2

 )

k

fk

L p

(dm

 )

 1

X

k =0 ja

k j



C

+"

p



k

kfk

L p

(dm)

Ckfk

L p

(dm)

;

sin etheseries onvergesabsolutelyforjzj<1=.

To prove the boundedness of T

m in X

p

for p 2 (1;2), we use the standard duality

argument. 

Proofof Lemma 6.3. First,observethat foranyf 2X

2

polynomialfun tion,

ke t

p

L



fk 2

L 2

(dm)

=

K

X

k =1 a

k e

t(k +)

P



k ( os)

2

L 2

(dm)

 e

2t(+1) K

X

k =1 ja

k j

2

=e

2t(+1)

kfk 2

L 2

(dm

 )

:

Next, for any q > 2 and any fun tion in X

q

, by (a) in Theorem 2.4, we have that

ke t

p

L

fk

L q

(dm)

e t

kfk

L q

(dm)

. Takeanypsu hthat2<p<q,1=p="

p

=2+(1

"

p

)=q,thenbytheinterpolationtheorem,wehavethat

e t

p

L

Xp!L p

(dm)

C

e t

p

L

"p

X2!L 2

(dm)

e t

p

L

(1 "p)

Xq!L q

(dm)

Ce

t(+"p)

:

Finally,wegetthat

kL 1=2

 k

X

p

!L p

(dm

 )

 Z

1

0

e t

p

L



X

p

!L p

(dm

 )

dtC Z

1

0 e

t(+"

p )

dt= C

+"

p :



7. The maximal operator R



 .

Denethefollowingmaximaloperators

R





f() = sup

"

jR

;"

f()j; where R

;"

f()= Z

j j>"

R



(;)f()d;

C



f() = sup

"

jC

"

f()j; where C

"

f()= Z

j j>"

C(;)f()d:

Themainresultof thisse tionistheboundednessof therstone. Theboundedness of

these ondoneisawellknownfa t,see[18℄.

Theorem7.1. R





is boundedonL p

(dm



)for 1<p<1, andof weak type (1,1) with

respe ttodm

 .

Proof. TherststepissplittingR





asfollows,

R





f()=sup

"

jR

;"

f()jsup

"

Z

j j>"

jR

;lo

(;)(sin) 2

C

0 C

lo

(;)jjf()jd

+C

0 sup

"





 Z

j j>"

C

lo

(;)f()d





 +sup

"

Z

j j>"

jR

;glob

(;)jjf()jdm

 ():

The rst term is bounded aboveby the operator Mf(), as follows from Lemma 4.4.

Then,Lemma4.5givestheboundednessofthisterm. Thethirdfa tor anbeestimated

by the same arguments as in the proof of Theorem 2.14, see subse tion 4.2, namely,

gettinga parallel inequality to (4.6). Then, Lemmas 4.7 and 4.8 givethe boundedness

ofthis term. Thelo al partpartofthe lassi almaximaloperatorC



lo

isbounded asa

onsequen eofTheorem 3.10andtheboundedness ofC

lo

. 

Asa orollaryoftheformerresult,by lassi alargumentswe anextendthedenition

oftheRiesztransformasaprin ipalvaluetoallfun tionsinL 1

(dm

 ).

Theorem7.2. Forany fun tion f 2L 1

(dm



)andalmost every 2[0;℄

R



f()=lim

"!0 Z

j j>"

R



(;)f()dm

 ():

8. Os illations

Bytheresultsinthepreviousse tions,weknowthatourRiesztransformisthelimitof

thetrun ated operatorsR

;"

almosteverywhere. Ingeneral,for everyfamily T =fT

"

g

su h that there exists Tf = lim

"!0 T

"

f, it is lassi al to measure the speed of this

onvergen ebymeansofexpressionsinvolvingdieren esofthetypejT

"

f T

"

0fj. Some

oftheoperatorsusedforthispurposearetheso alled os illationoperator,denedas

O(T)f(x)=



1

X

sup

t

i+1

"

i+1

<"

i

<t

i jT

"i+1

f(x) T

"i f(x)j

2



1=2

;

foraxedsequen eft

i

gde reasingtozero. Forin therange2<<1;wehavethe

so alled -variationoperator,

V



(T)f(x)=sup

f"ig



1

X

i=1 jT

"i+1

f(x) T

"i f(x)j





1=

;

where the supis takenoverall sequen es f"

i

gde reasing to zero. Classi alresults are

that when these operators are dened for the family of the trun ated operators of the

Hilbert transformor for thefamily of the Poisson integrals in R, they are bounded in

L p

(R ), 1<p<1andofweaktype(1;1)withrespe ttotheLebesguemeasure(see[2℄

and[7℄andthereferen estherein). Ouraim isprovingsimilarresultsfortheos illation

and -variation ofthefamily ofthetrun ated operatorsof theRiesz transform. Letus

all R



= fR

;"

g and C = fC

"

g to the families of the trun ated operators asso iated

to the Riesz transform and the lassi al onjugate fun tion, respe tively. First of all,

letus pointoutthat with theve torvaluedte hniques shown in [7℄, one anget easily

the boundedness of the os illationand -variationasso iated to the lassi al onjugate

fun tioninthetorusfromtheknownresultsfortheHilberttransformintheline. Letus

denote by H=fH

t

gthefamilyof thetrun atedoperators fortheHilbert transformin

R, thatis,

H

t f(x)=

Z

jx yj>t f(y)

x y dy:

Before pro eeding further, let us observe that for any family T = fT

"

g, O(T)f(x) 

O 0

(T)f(x)foralmosteveryx,where

O 0

(T)f(x)=



1

X

i=1 sup

t

i+1

<st

i jT

ti+1

f(x) T

s f(x)j

2



1=2

:

Thus,itisenoughtoprovethedesiredboundednesspropertiesforthisoperatorinstead

of O. Denote by B = ` 2

L 1

(0;1)

the spa e of sequen es u = fu

i

g su h that for every

i, u

i

= u

i

(s)is afun tion in L 1

(0;1), with kuk

B

=kfku

i k

L 1gk

`

2. Then, if we put

U(T)f(x)=f(T

ti+1

f(x) T

s f(x))1

(t

i+1

;t

i

℄

(s)g,wehavethat

O 0

(T)f(x)=kf(T

ti+1

f(x) T

s f(x))1

(t

i+1

;t

i

℄ (s)gk

B

=kU(T)fk

B :

Also, let us all  =



" = f"

i g : "

i

# 0

the set of sequen es de reasing to zero,

and B



= B

`

();

with  in the range 2 <  < 1; the spa e of bounded fun tions

v =v(")dened on ,whose valuesaresequen es in `



, v(")=fv

i

(")g, and su hthat

kvk

B

= sup

"2

kfv

i (")gk

`



. Then, if we put V(T)f(x) =fT

"i+1

f(x) T

"i

f(x)g, we

havethat

V



(T)f(x)=kfT

"i+1

f(x) T

"i f(x)gk

B

=kV(T)fk

B

:

Inthe asethat ea hT

t

hasan asso iatedkernelK

t

, thenU(T)andV(T)havealsoan

asso iatedkernel:

U(T)f(x) = f(T

ti+1

f(x) T

s f(x))1

(t

i+1

;t

i

℄ (s)g

= Z



(K

t

i+1

(x;y) K

s (x;y))1

(ti+1;ti℄

(s)

f(y)d(y);

and

V(T)f(x) = Z



K

"i+1

(x;y) K

"i (x;y)

f(y)d(y):

With all these preliminaries, it is lear that the boundedness of the os illation and -

variationoperators asso iatedto ertain family, namely, C, is equivalentto the bound-

ednessofU(C)andV(C). We anprovethisboundednessby omparisonwithU(H ) and

V(H ).

Proposition 8.1. Let us onsider f : [0;℄ ! R and extend it as 0 to the whole R.

Then, thereexistsa positive onstant C su hthat for every 2[0;℄ andp2[1;1℄ we

havethat

kU(C)f() 1

 1

[0;℄

()U(H )f()k

B

Ckfk

L p

(d)

;

kV(C)f() 1

 1

[0;℄

()V(H )f()k

B

Ckfk

L p

(d) :

Proof. WewillprovetheresultonlyforU,sin etheproofforV is ompletelyanalogous.

Inthis ase,wehavethat

U(C)f() 1

 1

[0;℄

()U(H )f()

=



(R

ti+1

f() R

s f())1

(t

i+1

;t

i

℄ (s)

1





(H

ti+1

f() H

s f())1

(t

i+1

;t

i

℄ (s)

= Z



0 1

2



1

tan(( )=2)

1

( )=2



1

fti+1<j jsg

(j j)1

(ti+1;ti℄

(s)



f()d:

Noti ethat



1

tan(( )=2)

1

( )=2



1

fti+1<j jsg

(j j)1

(ti+1;ti℄

(s)



B

= 







1

tan(( )=2)

1

( )=2 







;

thereforebyMinkowski'sinequality,wehave

kU(C)f() 1

 1

[0;℄

()U(H )f()k

B

C Z



0 







1

tan(( )=2)

1

( )=2 







jf()jd:

And now, for ; 2[0;℄,  2 [ ;℄ but both fun tions are odd. Therefore, we

onlyhaveto give abound for j1=tan(x=2) 1=(x=2)j forx 2[0;℄, orequivalently, for

j1=tanx 1=xj forx 2[0;=2℄. Inthis range, learlyj1=tanx 1=xjC, and we get

theresult. 

Thefollowingtheoremstatesthemainresultsofthisse tion.

Theorem8.2. Forevery p2(1;1)thereexistsa onstantC su hthat

kO(R

 )fk

L p

(dm

 )

Ckfk

L p

(dm

 )

; kV

 (R

 )fk

L p

(dm

 )

Ckfk

L p

(dm

 )

andalso

m



f: O(R



)f()>g

 C

 kfk

L 1

(dm

 )

m

 f: V

 (R



)f()>g

 C

 kfk

L 1

(dm

 )

Proof. Theproofofthistheorem anbedevelopedbyusingthesamete hniquesapplied

in theproofsofTheorems2.13and2.14,i.e.,splittingtheoperatorsintotheirlo al and

globalparts. Tothisend,deneforafamilyT =fT

t

gasabove,a tingonfun tionsover

[0;℄, itslo alandglobalpartsasT

lo

=fT

t;lo

gandT

glob

=fT

t;glob g.

Lemma8.3. Given afamily of operatorsT =fT

t

gon fun tionsdenedon [0;℄ su h

thatea hoperator T

t

isgiven by akernelK

t

,thenwe have

U(T)

lo

=U(T

lo

); U(T)

glob

=U(T

glob

) V(T)

lo

=V(T

lo

); V(T)

glob

=V(T

glob ):

Proof. Letusseetheprooffortheglobalpart,sin eforthelo alpartitfollowsfromits

denitionandthelinearityin T oftheoperators U andV:

U(T)

lo

=U(T) U(T)

glob

=U(T) U(T

glob

)=U(T T

glob

)=U(T

lo ):

Bydenition oftheglobalpartofanoperator,wehave

U(T)

glob

f() = Z



0



(L

ti+1

(;) L

s (;))1

(t

i+1

;t

i

℄ (s)

(1 1

Nt

2

(;))f()d;

= f(T

t

i+1

;glob

f() T

s;glob f())1

(ti+1;ti℄

(s)g=U(T

glob )f():

Asimilarproofholds forV. 

Now,wewillprovethatthelo alpartofU(R



)andV(R



)dierfromthelo alpartof

U(C)andV(C),respe tively,inanoperatorwhi hisboundedinL p

(dm



)for1<p<1

andofweaktype(1;1).

Lemma 8.4. There exists apositive onstant C su hthat for every  2[0;℄ we have

that

kU(R

 )

lo

f() C

0 U(C)

lo f()k

B

C Z



0

M(;)1

Nt

2

(;)jf()jd;

kV(R

 )

lo

f() C

0 V(C)

lo f()k

B

C Z



0

M(;)1

Nt

2

(;)jf()jd

whereC

0

andM(;)areasin Lemma 4.4.

Proof. Sin e foreverytrun ated operatorweare awayfrom thediagonal, observethat

we anwrite

U(R

 )

lo

f() C

0 U(C)

lo

f()=U(R

;lo

)f() C

0 U(R

lo )f()

=



(R

;ti+1;lo

f() R

;s;lo f())1

(ti+1;ti℄

(s)

C

0



(C

t

i+1

;lo

f() C

s;lo f())1

(ti+1;ti℄

(s)

= Z



0



[R



(;)(sin) 2

C

0 C(;)℄

1

fs<j jt

i+1 g

(j j)1

(t

i+1

;t

i

℄ (s)

1

Nt

2

(;)f()d:

Sin e



[R



(;)(sin) 2

C

0

C(;)℄1

fs<j jt

i+1 g

(j j)1

(t

i+1

;t

i

℄ (s)

B

=jR



(;)(sin) 2

C

0

C(;)j;

byMinkowski'sinequalityand Lemma4.4,wehavethat

kU(R

 )

lo

f() C

0 U(C)

lo f()k

B

 Z



0

M(;)1

N

t

2

(;)jf()jd:

ByProposition8.1,weknowtheboundednessofU(C)andV(C)inL p

(d)for1<p<

1 and weak type(1;1) with respe t to d. Now, one ansee that U(C)and V(C) are

ve torvaluedoperatorsasdenedinse tion3. Thus,bytheve torvaluedversionofthe

heritagetheorem,Theorem3.10,thelo alpartofbothoftheminheritstheboundedness

properties mentionedabovewithrespe tto d. ByLemmas 8.4 and4.5, this givesthe

boundedness ofthe lo al part ofU(R



) and V(R



)with respe tto d. Hen e,by the

ve tor valued version of Lemma 3.9, we obtain the boundedness of the lo al part of

U(R



)and V(R



)alsofor dm



. Inorder to proveTheorem 8.2, itis onlyleft toprove

thattheglobalpartsarealsoboundedinL p

(dm



). This anbeobtainedbyusingsimilar

argumentsasinLemma8.4tohandletheve torvaluedkernels,andthenpro eedingasin

theproofofTheorem2.14fortheglobalpart, on retelyobtainingtheparallelinequality

to4.6. 

9. Weightedinequalities

Let us all weight to anystri tly positive measurable fun tion v. In this se tion we

will answerthe following question: nd ane essaryand suÆ ient onditionona given

weightv, fortheexisten eofaweightusu hthat theRiesztransformisboundedfrom

L p

(vdm



) intoL p

(udm



)for1<p<1. Wewill useawell knownte hniquebasedin

ideasbyRubiodeFran ia. Inparti ular,weshallneedthefollowingtheoremduetothis

author(see[12℄).

Theorem 9.1. Let (X;) be a measure spa e, B a Bana h spa e and T a sublinear

operator fromB intoL s

(X)su hthatthe following inequalityissatisedforsomes<p

andeverysequen eff

j gB



X

j jTf

j j

p



1=p

L s

(X)

C

p;s



X

j kf

j k

p

B



1=p

;

where C

p;s

isa onstant depending on pand s. Then there exists apositive fun tion u

su hthat u 1

2L s

p s

(X)and



Z

X

jTf()j p

u()d()



1=p

kfk

B :

ConsidertheextensionoftheRiesztransformR



totheoperatordenedforintegrable

fun tions f = ff

j g

j

taking values in ` p

, given by R



f = fR

 f

j g

j

. For this extended

operator,wehavethefollowinglemma.

Lemma9.2. Letpbein therange1<p<1;then

R

 : L

1

l p(m

 )!L

1;1

l p

(m

 )

Proof. Itisknownthatthe` p

-valuedextensionofthe lassi al onjugatefun tion inthe

torus,Cf =fCf

j g

j

;maps L 1

l

p(d)intoL 1;1

l p

(d) (see[1℄). Bytheve tor-valuedversion

oftheheritagetheorem,Theorem3.10,thelo alpartofthisextensionofCisalsoofweak

type (1;1). Therefore by using identity (5.2), Lemma 4.4 and Lemma 4.5 weget that

R

;lo

maps L 1

l

p(d)into L 1;1

l p

(d): Wehaveusedherethat theoperatorin Lemma 4.5

isgivenbyapositivekernelandthereforeit anbeextended to ` p

-valuedfun tions for

1p1. Anstraightforward` p

-valuedextensionofLemma3.9givestheboundedness

ofR

;lo

fromL 1

l p(m



)intoL 1;1

l p

(m

 ):

Asfortheglobalpartweobservethatsin eitismajorizedbyapositiveoperator(see