SHARPESTIMATESINHARMONICANALYSIS
By
GuillermoRey
ADISSERTATION
Submittedto
MichiganStateUniversity
inpartialentoftherequirements
forthedegreeof
Mathematics{DoctorofPhilosophy
2015
ABSTRACT
SHARPESTIMATESINHARMONICANALYSIS
By
GuillermoRey
Weinvestigatecertainsharpestimatesrelatedtosingularintegrals.Inparticularwegive
sharplevelsetestimatesforsparseoperators,weshowhowtoreducetheproblemof
estimatingon-Zygmundoperatorsbysparseoperators,andwestudysomeweighted
inequalitiesfortheseoperators.
ACKNOWLEDGEMENTS
Todoesto,nosoloestatesis,habrsidoimposibledenoserpormifamilia.Gracias,
a,porense~narmeaserfeliz,yporaguantarmetodosestosa~nos;todavmeacuerdo
cuandoentedlasaticasporprimeravezcuandoabamosenElkridge.Gracias,
a,portodasesasdiscusionessobreciencia,informatica,ydetodoengeneral;
portenersiempeunarespuestaamispreguntas,ypordejarmejugarcontodoslos
aparatosqueencontraba.GraciasalosdosporestarsiempreahYgraciasaMarta,por
ayudarmeentodaslasgrandesdecisionesquehetomado;contigoheaprendidomuch
yesoqueestosdostellevanmasdeveintea~nosdeventaja.Gracias,Jaime,porhacerme
reirtantveces:asesimposiblenotenerbuenasmemorias.Y,Pablo,graciaspor
querermetanto;eresdelasprimeraspersonasconlasquequieroestarcadavezquehago
unavisita.Yporsupuesto,graciasalrestodemifamilia,queesdemasiadograndecomo
paraenumeraraquQueaisqueospongocomoejemplodelasuertequetengo.
AMichayAntonioporesosinterrailes,yaLuis

Angel.Porserdelospocosbuenos
recuerdosquetengodemionantesdeiralauniversidad.AJorgeporquecontigo
siempresepuedehablardeloqueasnosgusta.ARita,porserunaamigatanchachiy
porense~narmetantaspalabrasguays.ACarmenportenerconversacionesinterminables
conmigosobreomoarreglarelmundo.AMarporqueyonoibaalacafetercontigo
porlacalidaddeloscafes,sinoporqueaspodestarmastiempocontigo;porqueeras,y
siguessiendo,absolutamenteimprescindible.AAnonporquedespuesdetantotiempoyo
creoqueyapensamosaticamente.Porense~narmequelosproblemassinonno
existen,yporquecontigopuedoestarhablandohastaelamanecer.YaMiranda,porque
aunmeacuerdodecuandoteconocydelasagujetasquetuvealdsiguientedetanto
reirme.
AJose,porquetegustanlasaticasylacomidatantocomoamPorvisitarme,
viajarconmigo,yayudarmeenelproyectodelqueestoymasorgulloso.ABeatrizporque
siempretepreocupaspormyporqueabrazasfenomenal.Aanporense~narmetantas
iii
cosas,yportodaslasqueaunnomehasense~nado.YaMarporquepuedoconartelo
todoysiempreasahcuandotenecesito.Porquehablarcontigomealegraeldaunque
estemuylejos.
AlosprofesoresdelaUAMquemeayudarontantodesdeelprincipio.GraciasaAntonio
ordoba,JoseGarca,yFernandoSoria.Sobretodo,graciasaAnaVargaspor
enmyense~narmetanto.
TothegraduatestudentsatMSUwhichmademystayheremuchmoretolerable;thank
youAlexforteachingmeBellmanfunctions,thankyouTylerandBenforallthoseBeggars
nights,andthankyouCharlotteandSamiforallthelaughsinyournew
Andlast,butnotleast.Thankyoutomytwowonderfuladvisors:IgnacioUriarte-Tuero
andAlexanderVolberg.Thankyouforyourconstantsupportandforteachingmesomuch
aboutmathematicsandlife.This,obviously,wouldnothavehappenedwithoutyourhelp.
GuillermoRey
iv
TABLEOFCONTENTS
LISTOFFIGURES
vi
1Introduction
1
2Sharpweak-typeboundsforpositivedyadicshifts
5
2.1Introduction
5
2.2TheBellmanfunctiontechnique
8
2.3FindingtheBellmanfunctioncandidate
12
2.4Optimality
16
3Dyadicmodelsforsingularintegrals
27
3.1Introduction
27
3.2Pointwisedomination
33
3.3Applications
48
3.3.1Multilinear
A
2
theorem
48
3.3.2SharpapertureweightedLittlewood-Paleytheorem
49
3.4Theweak-typeestimateformultilinear
m
-shifts
51
4Ontheembeddingof
A
1
into
A
1
57
4.1Introduction
57
4.1.1Organization
62
4.2TheBellmanfunctionapproach
63
4.3FindingtheBellmanfunction
69
4.3.1Explicitextremizers
71
4.4VerifyingtheMainInequality
76
5Borderlineweak-typeboundsforsingularintegrals
82
5.1Introduction
82
5.2Proof
83
BIBLIOGRAPHY
89
v
LISTOFFIGURES
Figure4.1
Plotsof
f
and
e
f






































62
Figure4.2
Domains
k









































79
vi
1
Introduction
Inanalysis,oneoftenneedstocommutealimitwithanoperator.Afamousexampleof
suchsituationconcernstheFouriertransform:
b
f
(
˘
):=
Z
1

e

2
ˇix˘
f
(
x
)
dx:
Wewouldliketorecoverthefunction
f
fromitsFouriertransform
b
f
,andthisiscertainly
possibleinsomecases,butoneshouldbecarefulwithwhatexactlydowemeanby
\recover".Inparticular,eventhoughtheFouriertransformmaybewellthe
inverseFouriertransformmaynot.
Atypicalwaytoresolvethisissueistonotinvertthewholefunction
b
f
,butatruncationof
it:
f
R
(
x
):=
Z
R

R
e
2
ˇix˘
b
f
(
˘
)
d˘:
Nowthequestionis:Does
f
R
tendto
f
as
R
!1
?Inwhatsense?
Thisisaveryoldquestionandtherearemanywaystoanswerit.Perhaps,thebestknown
answeristhefollowing:if
f
2
L
2
(
R
)then
f
R
!
f
in
L
2
(
R
),thatis
lim
R
!1
k
f
R

f
k
L
2
(
R
)
=0
:
(1.1)
1
ThemaintoolusedtoprovethisresultisPlancherel'stheorem:
k
b
f
k
L
2
(
R
)
=
k
f
k
L
2
(
R
)
:
Ifwetakethistheoremforgranted,thenwecangiveashortproofof(
1.1
):
lim
R
!1
k
f
R

f
k
2
L
2
=lim
R
!1
k
c
f
R

b
f
k
2
L
2
=lim
R
!1
Z
R
j
f
(
x
)
j
2
(1

1
[

R;R
]
(
x
))
dx;
=0
:
wherewehaveusedtheDominatedConvergenceTheoreminthesecondtolastline.
Onecouldaskwhatissospecialabout
L
2
,apartfromPlancherel'stheorem.Isthistruein,
say,
L
3
?
Thisisindeedtruefor
f
2
L
p
(
R
)forall1
<p<
1
,buttheproofisconsiderablymore
involved.OnewaytoproveitistostudytheFouriermultiplieroperator
d
Hf
(
˘
)=sign(
˘
)
b
f
(
˘
)
:
Ifonecouldshowthatthisoperatorisboundedin
L
p
,thenonewouldimmediatelyget
k
f
R

f
k
L
p
!
0
sincewecanreconstructthemultiplier
1
[

R;R
]
bytranslated,modulated,anddilated
versionsofthemultipliersign.
So,theproblemisreducedtostudyingtheboundednessof
H
,whichisalsoknownasthe
Hilberttransform(uptoamultiplicativeconstant),andwhichcanalsobewrittenas
Hf
(
x
)=
c
p.v.
Z
R
f
(
y
)
x

y
dy;
2
forsomeconstant
c
6
=0
:
Hence,wehavereducedtheproblemofstudyingtheconvergencepropertiesoftheFourier
transformtostudyingacertain
singularintegral
operator.Andmoreover,quantitative
boundsforthesingularintegralyieldquantitativeinformationabouttheoriginalproblem.
Singularintegralscanbefurthergeneralizedinmanytdirections,andhave
numerousapplications.Thereisaparticularlycommonclass,calledon-Zygmund
operators,whichwillbethemainthemeinthiswork.Theirrigorouswillbe
postponeduntilChapter
3
,butessentiallytheyareoperatorsoftheform
Tf
(
x
)=p.v.
Z
R
d
K
(
x;y
)
f
(
y
)
dy;
wherethekernel
K
certaingrowthconditionslike
j
K
(
x;y
)
j
.
1
j
x

y
j
d
;
andcertainregularityconditionslike
jr
K
(
x;y
)
j
.
1
j
x

y
j
d
+1
:
TheseoperatorsgeneralizetheHilberttransformtohigherdimensions,wheremore
applicationsappear.Forexample,three-dimensionalsingularintegralsareofspecial
interestindynamics.
InChapter
2
westudytheweak-typeboundednessofacertaindyadicmodelof
on-Zygmundoperatorscalleddyadicshifts.InthisChapterweobtainthebest
possibleupper-levelsetestimatesforsuchoperatorsusingtheBellmanfunctiontechnique.
InChapter
3
weshowhowtoreducetheproblemofobtainingestimatesfor
on-Zygmundoperators(aswellasothersimilaroperators)tostudyingthesimple
dyadicshiftsintroducedinthepreviousChapter.InChapter
4
westudytheembeddingof
A
1
weightsinto
A
1
;theseareclassesofabsolutelycontinuousmeasureswhichplayakey
3
roleinthestudyofweightedinequalitiesforsingularintegrals.InChapter
5
wegiveshort
proofsoftwoweak-typebounds(forsingularintegralsandforsquarefunctions)usingthe
machineryintroducedinthepreviouschapters.
4
2
Sharpweak-typeboundsforpositive
dyadicshifts
GuillermoReyandAlexanderReznikov
AdvancesinMathematics,Vol254,March2014.
2.1Introduction
Thepurposeofthisarticleistostudytheweak-type(1
;
1)boundednessoftheoperator
A
f
=
X
Q
2D
(
I
)

Q
h
f
i
Q
1
Q
:
Here
I
denotesanyintervalin
R
,
h
f
i
I
=
1
j
I
j
R
I
f
,
D
(
I
)denotesthedyadicgrid
consistingofdyadicsubintervalsof
I
and
f

J
g
J
2D
(
I
)
isaCarlesonsequenceadaptedto
I
,
i.e.:

J

0forall
J
2D
(
I
)and
sup
J
2D
(
I
)
1
j
J
j
X
K
2D
(
J
)

K
j
K
j
=
C<
1
:
TheseoperatorshaverecentlyappearedintheworksofA.K.Lerner[
20
]and[
21
],where
5

K
wasabinarysequence,althoughtheideasgobackto[
9
].Hence,wewillcallthem
Lerneroperators
inthesequel.HerewetheexactBellmanfunctiondescribingthe
localboundednessof
A
from
L
1
to
L
1
;
1
.
Itiseasytoseethattheoperator
A
isboundedin
L
2
.This,togetherwithadecomposition
ofon-Zygmundtype,canbeusedtoproveanestimateoftheform



n
x
2
I
:



X
J
2D
(
I
)

J
h
f
i
J
1
J
(
x
)



>
o




C

Z
I
j
f
j
:
However,herewepreciselydescribehowthebestconstantintheaboveinequalitychanges
withrespecttotheparametersoftheproblem.
Themainresultofthearticleisthefollowingtheorem:
Theorem2.1.
Let
A
,

and
t
bepositivenumbersand
I
anintervalin
R
,then
sup
1
j
I
j



n
x
2
I
:
X
J
2D
(
I
)

J
h
f
i
J
1
J
(
x
)
>
o



=
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
2
At

+
t
if
0

t




q
At

if
0

A

min

t

;

t

;
1
otherwise.
Wherethesupremumistakenoverallnonnegativefunctions
f
with
h
f
i
I
=
t
andallnon-
negativesequences
f

J
g
J
2D
(
I
)
withCarlesonconstantatmost
1
whichsatisfy
1
j
I
j
X
J
2D
(
I
)

J
j
J
j
=
A:
Wealsoprovideasequenceofexampleswhich,inthelimit,attainthesupremumofthe
previousresult.Seethelastsectionfordetailsonthestructureofsuchexamples.
Asanimmediatecorollarywehavethefollowinglocalweak-type(1,1)estimate:
Corollary2.2.
Foranynonnegative
f
2
L
1
([0
;
1))
andforanyCarlesonsequence
f

J
g
J
2D
([0
;
1))
6
withconstantatmost
1
wehavethesharpbound



x
2
[0
;
1):
A
f
(
x
)
>




8
>
>
<
>
>
:
2
k
f
k
L
1

+
k
f
k
L
1
if
k
f
k
L
1


1
if
k
f
k
L
1


whichinparticularimpliesthat
kA
f
k
L
1
;
1
([0
;
1))

2
k
f
k
L
1
([0
;
1))
;
andthattheconstant
2
issharp.
Operatorssimilartothesewererecentlystudiedin[
29
],[
33
],[
31
]and[
32
],howevertheir
resultsareslightlytfromours.Theyconsiderthesupremumtakenoverall
functions
f
satisfying
Z
I
f
=
s
and
Z
I
G
(
f
)=
t;
where
G
isastrictlyconvexfunctionsatisfying
G
(
x
)
=x
!1
as
x
!1
.Thisdoesnot
includethequestionofboundednessfrom
L
1
to
L
1
;
1
.Ourmethodofproofistthan
theoneusedinthearticlescitedabove,wheretheyusethedeepcombinatorialproperties
oftheseoperators.Seealsothemonograph[
40
]byA.Os
`
ekowskiforrelatedresults.We
insteadfollowtheideasin[
45
]and[
46
]tosolvetheBellmanPDEandproveitssharpness.
ThisproblemisalsocloselyrelatedtostudyingHaarshifts,themainbeingthat
Haarshiftsarenotpositiveoperators.Ithasbeenshownhowever,see[
5
],thatLerner-type
operatorscanbeusedtoboundHaarshifts.Thereadercanresultssimilartooursin
[
44
],[
34
]and[
38
].
Thearticleisorganizedasfollows.InSection2weexplainhowtheBellmanfunction
techniqueisusedtocomputethesupremuminTheorem
2.1
.InSection3wegivea
supersolutiontotheBellmanvariationalproblemwhichservesasanupperboundforthe
7
exactBellmanfunction.Finally,inSection4weshowthatthefunctionwefoundinthe
previoussectionistheexactBellmanfunction,wealsogiveasequenceofexampleswhich,
inthelimit,extremizetheinequalityofTheorem
2.1
.
2.2TheBellmanfunctiontechnique
Considerthefunctionin=
f
(
t;A;
):0

t;
0

A

1
;
2
R
g
B
(
t;A;
)=sup
n
1
j
I
j




x
2
I
:
X
J
2D
(
I
)

J
h
f
i
J
1
J
>




o
;
wherethesupremumistakenoverallallnonnegativefunctions
f
on
I
with
h
f
i
I
=
t
and
allCarlesonsequences
f

J
g
j
2D
(
I
)
withconstantatmost1and
A
=
1
j
I
j
X
J

I

J
j
J
j
:
Notethat
I
is
not
aparameterin
B
,thisisbecausethesupremumisinvariantunder
dilationsandtranslationsin
I
,andhenceindependentof
I
.
TheBellmanfunctiontechnique,whichappearedinthe1995preprintversionof[
36
],
isbasedonshowingthat
B
solvesacertainminimizationproblem.Oneshowsthat
B
akindofconcavitypropertyandexplicitlycomputes
B
inasubdomainnaturalto
theproblem(thisisusuallyeasy).Thenoneshowsthatanycontinuouspositivefunction
satisfyingtheseconditionsmajorizes
B
,whichreducestheproblemtoingthesmallest
functionwhichtheseproperties.Finallyonehastoactuallysuchafunction,
thisisusuallythehardestpart.Thereadercaninsightfulintroductionsin[
37
]and
[
39
],seealso[
36
],[
45
],and[
46
]formoreexamplesofthistechnique.
Letusbeginbydescribingmorepreciselytheconcavitypropertywhich
B

8
Lemma2.3
(Maininequality)
.
B
(
t;A;
)

1
2

B
(
t
1
;A
1
;
0
)+
B
(
t
2
;A
2
;
0
)

(2.1)
whenever
t
=
t
1
+
t
2
2
;A
=
A
1
+
A
2
2
+

and

=

0
+
t
and


0
.
Proof.
Consideranydyadicinterval
I
,anyfunction
f

0satisfying
h
f
i
I

=
t
1
and
h
f
i
I
+
=
t
2
andanyCarlesonsequence
f

J
g
J
2D
(
I
)
withconstantatmost1on
I
satisfying
1
j
I

j
X
J
2D
(
I

)

J
j
J
j
=
A
1
;
1
j
I

j
X
J
2D
(
I
+
)

J
j
J
j
=
A
2
and

I
=
:
Supposealsothat

=

0
+
t
.
Since
h
f
i
I
=
t
thenwemusthave
B
(
t;A;
)

1
j
I
j



n
x
2
I
:
X
J
2D
(
I
)

J
h
f
i
J
1
J
(
x
)
>
o



sincethesupremum
B
istakenoveralargerspace.
9
Observenowthat
1
j
I
j



n
x
2
I
:
X
J
2D
(
I
)

J
h
f
i
J
1
J
(
x
)
>
o



=
1
2
j
I

j



n
x
2
I

:
X
J
2D
(
I
)

J
h
f
i
J
1
J
(
x
)
>
o



+
1
2
j
I
+
j



n
x
2
I
+
:
X
J
2D
(
I
)

J
h
f
i
J
1
J
(
x
)
>
o



=
1
2
j
I

j



n
x
2
I

:
X
J
2D
(
I

)

J
h
f
i
J
1
J
(
x
)
>


I
t
o



+
1
2
j
I
+
j



n
x
2
I
+
:
X
J
2D
(
I
+
)

J
h
f
i
J
1
J
(
x
)
>


I
t
o



=
1
2
j
I

j



n
x
2
I

:
X
J
2D
(
I

)

J
h
f
i
J
1
J
(
x
)
>
0
o



+
1
2
j
I
+
j



n
x
2
I
+
:
X
J
2D
(
I
+
)

J
h
f
i
J
1
J
(
x
)
>
0
o



andthustheclaimfollows.
Also,wetriviallyseethat
B
mustsatisfythefollowing\obstacle"condition:
B
(
t;A;
)=1whenever
<
0
:
(2.2)
Aswedescribedinthebeginningofthesection,thefunction
B
isaminimizerinthespace
ofpositivefunctionswhichsatisfytheseproperties.Thefollowingpropositionmakesthis
precise:
Proposition2.4.
Supposeacontinuousfunction
F
inequality
(
2.1
)
togetherwith
theobstaclecondition
(
2.2
)
,thenwemusthave
B
(
t;A;
)

F
(
t;A;
)
:
Proof.
Let
f

0beanintegrablefunctiononaninterval
I
andlet
f

J
g
J
2D
(
I
)
beaCarleson
10
sequencewithconstantatmost1,thenforall

wehave(by(
2.1
))
F
(
h
f
i
I
;A;
)=
F

h
f
i
I

+
h
f
i
I
+
2
;
A

+
A
+
2
+

I
;


1
2

F
(
h
f
i
I

;A

;


I
h
f
i
I
)+
F
(
h
f
i
I
+
;A
+
;


I
h
f
i
I
)

;
where
A
=
1
j
I
j
P
J

I

J
j
J
j
and
A

isanalogouslyfor
I

and
I
+
.
Ifweiteratethisinequalityweobtain
F
(
h
f
i
I
;A;
)

1
2
N
X
J
ˆ
I;
j
J
j
=2

N
j
I
j
F
(
h
f
i
J
;A
J
;

N
X
k
=1

J
(
k
)
h
f
i
J
(
k
)
1
J
(
k
)
(
c
J
))
;
where
A
J
=
1
j
J
j
P
P

J

P
j
P
j
.
IfweassumeapriorithattheCarlesonsequence

isthenwecanlet
N
!1
and
obtain
F
(
h
f
i
I
;A;
)

1
j
I
j
Z
I
F
(
f
(
x
)
;A
(
x
)
;
A
f
(
x
))
dx

1
j
I
j
Z
f
x
2
I
:


f
(
x
)
<
0
g
1
dx
by(
2.2
)
=
1
j
I
j
jf
x
2
I
:
A
f
(
x
)
>
gj
:
Here
A
(
x
)isalmostevasthelimitof
A
(
J
)as
J
!
x
,thisiseasilyseento
existalmosteverywherebytheLebesguedtiationtheorem.
Lettingthenumberofnon-zeroelementsof
f

J
g
J
2D
(
I
)
tendtoyandthentakingthe
supremumintheof
B
weobtain
F
(
h
f
i
I
;A;
)

B
(
h
f
i
I
;A;
)
:
Remark
2.5
.
Notethatwedon'tknowyetifthefunction
B
iscontinuous,thusa
11
minimizerinthespaceofcontinuousfunctionsmightnotgiveusthetrueBellmanfunction.
Itturnsout,however,thatassumingcontinuity(actually
C
1
smoothness)weareableto
apositivefunctionsatisfying(
2.1
)and(
2.2
)whichmoreoverisbestpossiblewithoutthea
prioriassumptionofsmoothness.Weshowthisinthelastsection.
Wehavethereforeseenthatanypositivecontinuousfunction
F
satisfying(
2.1
)and
(
2.2
)willgiveusanupperboundfor
B
.Inthenextsectionwesuchafunction.
2.3FindingtheBellmanfunctioncandidate
Ourgoalnowistothesmallestcontinuousfunction
F
satisfying(
2.1
)and(
2.2
).As
weremarkedafterProposition
2.4
,wewillassumeapriorithat
F
is
C
1
.Moreover,wewill
restricttheminimizationspaceevenmorebyrequiring
F
tohavethesamekindof
homogeneitythatthetrue
B
musthave,i.e.:
B
(
t;A;
)=
B
(
t;A;
)
8
>
0
;>
0
:
ThisinprinciplemightmakeourcandidateforBellmanfunctionlargerthantheonewe
couldwithoutrequiringsuchhomogeneity.However,theoptimalBellmanfunction
thisidentity,sorequiring
F
toalsosatisfyitwillnotpreventusfromingit.
AssumingsmoothnesswecanwritetheMainInequality(
2.1
)asaconcavitycondition,
togetherwithamonotonicitypropertyalongcertaincharacteristics.Moreprecisely,if
F
is
asmoothpositivefunction,then(
2.1
)togetherwith(
2.2
)andtheabovehomogeneityis
equivalenttothefollowingconditions:
1.
F
isnonnegative,andconcaveinthetwovariables.
2.
F
(
t;A;
)isincreasinginthedirection(0
;
1
;t
).
3.
F
(
st;A;
)=
F
(
t;A;
)forall
s>
0.
4.
F
(
t;A;
)=1whenever
<
0
12
Indeed,ifwelet

=0in(
2.1
)weseethat
B
isconcaveinthevariables(
t;A
).Ifweset
A
1
=
A
2
=
A
and
t
1
=
t
2
=
t
thenwesee,byvarying

,that
B
(
t;A;
)isincreasinginthe
direction(0
;
1
;t
).Thisshowsthatanysmooth
F
satisfying(
2.1
)and(
2.2
),andwhichis
alsohomogeneousintheabovesense,mustalsosatisfyproperties(1)through(4).
Moreover,if
F
isanysmoothfunctionsatisfyingproperties(1)through(4),thenitalso
mustsatisfythemaininequality(
2.1
)andtheobstaclecondition(
2.2
).Toseethisobserve
thatusingproperty(1)weobtain(
2.1
)butwith

=0,nowproperty(2)allowsusto
insertan

asinthehypothesesforthemaininequalitysinceitdescribesthepathalong
which
F
isincreasing.Thehomogeneityandobstacleconditionsareexactly(3)and(4)
respectively,sothisprovestheequivalence.
Usingthehomogeneityproperty,wecanreduceto
M
:(0
;
1
)

[0
;
1]
!
[0
;
1
)such
thatif
F
(
x;y;z
)=
8
>
>
<
>
>
:
M
(
x=z;y
)if
z>
0
1if
z

0
;
then
F
(1)through(4).Theseproperties,whentranslatedtothefunction
M
,
become:
1.
M
isconcave.
2.
M
y

x
2
M
x

0.
3.
M
(
x;y
)
!
1when
x
!1
.
Thesecondofthesepropertiestellsusthat
M
isincreasingalongthecharacteristics
8
>
>
<
>
>
:
_
x
(
t
)=

x
2
_
y
(
t
)=1
:
Observethatthesecharacteristicsfoliate[0
;
1
)

[0
;
1].Also,ifwemovebackwardsintime
13
alongacharacteristicwhichstartsat(
x
0
;
1)with
x
0

1,thenthischaracteristicisabove
thecurve
y
=
1
x
andfurthermorethecharacteristictendsto(
1
;y
f
)forsome0
<y
f
<
1.
Usingthefactthat
M
(
x;y
)
!
1as
x
!1
andthatweshoulddecreaseifwemove
backwardsalongthesecharacteristics,wemusthave
M
(
x;y
)

1whenever
y

1
x
:
However,wemayassume(ifourgoalistothetrueBellmanfunction)that
M

1since
thetrueBellmanfunction
B
obviouslycannotbelargerthan1,sowewillactuallyimpose
M
(
x;y
)=1whenever
y

1
x
:
Observethat
B
(0
;
0
;
1)is0andconsiderthestraightlinejoiningthepoint(0
;
0)with
(
x
1
;y
1
),where
x
1
y
1
=1.Observealsothatthepointwiseminimumofanytwopositive
continuousfunctionssatisfying(
2.1
)and(
2.2
)willgiveusasmallerfunctionwhichalso
theseproperties.
Weknowthatthefunction
M
shouldbe1at(
x
1
;y
1
)andthat,alongthisline,
M
should
beconcave.Thesmallestconcavecurvejoiningthesetwopointsisobviouslyastraight
line,soif
M
inthiswayproducesasmoothconcavefunctionsatisfyingthe
monotonicityproperty(2)thentheoptimal
M
shouldbesuchafunction.Joiningthe
point(0
;
0)withthepoints(
x
1
;y
1
)
2
[0
;
1
)

[0
;
1]satisfying
x
1
y
1
=1coverseverythingin
thesubdomain0

y

min(
x;x

1
),soletus
M
hereby
M
(
x;y
)=
p
xy:
Thisfunctionislinearalongstraightlinesjoining(0
;
0)withtheboundarycurve
xy
=1
andis1atthisboundary.Itisfurthermoreconcaveandthemonotonicityproperty
(2),soifweknewthat
B
iscontinuousthen
B
(
x;y;
1)mustbeasaboveinthis
subdomain.
14
Wearethereforeleftwith
M
intheuppertriangle
T
=
f
0

x

y

1
g
.
Inspiredbythelinearbehaviorof
M
inthedomain,wemaketheansatzthat
M
is
actually1-homogeneousinthewholedomain.
Let
f
(
x
)=
M
(
x;
1)for0

x

1,thenif
M
is1-homogeneousweshouldhave
M
(
x;y
)=
yf
(
x=y
)
:
Ifwewantcondition(2)toholdthenweshouldhave
f
(
x=y
)

(
x=y
)
f
0
(
x=y
)

x
2
f
0
(
x=y
)

0
:
Weexpectthistobeanequalityontheboundary,whichiswhen
y
=1,sowewillassume
that
f
(
x
)

f
0
(
x
)(
x
+
x
2
)=0
:
Thisordinaryerentialequationhasthesolutions
f
(
x
)=
C
x
1+
x
;
andweshouldfurthermorehave
f
(1)=
M
(1
;
1)=1.So
C
=2andtherefore
f
(
x
)=
2
x
x
+1
=
)
M
(
x;y
)=
2
xy
x
+
y
whenever1

y

x

0.Oneeasilyvesthat
M
alltherequirementsinthis
subdomain,sowejusthavetoshowthatthewholefunction
M
isconcave,butthis
immediatelyfollowsfromthefactthat
M
isconcaveineachsubdomainandthat
M
is
C
1
(ascanbeeasilyseen).
15
Thisgivesusthat
M
(
x;y
)=
8
>
>
<
>
>
:
2
xy
x
+
y
if0

x

y

1
p
xy
if0

y

min(
x;x

1
)
;
whichusingthehomogeneitygivesusthefullfunctionofTheorem
2.1
.
2.4Optimality
Inthissectionweshowthatthefunctionfoundintheprevioussectionisactuallytheexact
Bellmanfunction.Weweneedasimpletechnicallemmawhichwillallowustodeduce
that
B
(

;

;
1)mustbesuperlinearalonglinesjoining(0
;
0
;
1)to(
x;
1
;
1).
Lemma2.6.
Let
f
:[0
;
1]
!
[0
;
1
)
beafunctionwhich
f

x
+
y
2


1
2
f
(
x
)+
1
2
f
(
y
)(2.3)
forall
0

x

y

1
.Thenwemusthave
f
(
x
)

f
(1)
x
forall
x
2
[0
;
1]
:
Proof.
Wecanassumewithoutlossofgeneralitythat
x
2
(0
;
1)andthat
f
(1)=1.Using
(
2.3
)wehave
f
(
x
0
+

(1

x
0
))


(2.4)
foralldyadicrationals

2
[0
;
1],i.e.:numbersoftheform

=
k
2

N
for0

k

2
N
.
Forevery
N
2
N
let
k
N
betheuniqueintegerin0

k

2
N
x
which



x

k
2
N



<
1
2
N
16
(thisexistsbecausethesequence
k
7!
k
2

N
isanarithmeticsequenceofstep2

N
).
Observethatthen,ifwe
x
N
:=
2
N
x

k
N
2
N

k
N
=
x

k
N
2
N
1

k
N
2
N
;
wemusthave0

x
N

1
2
N
(1

x
)
,soinparticular
x
N
!
0as
N
!1
.
Butthen

:=
k
N
2
N
=
x

x
N
1

x
N
isadyadicrationalandpluggingitinto(
2.4
),with
x
N
playingtheroleof
x
0
,yields
f
(
x
)

x

x
N
1

x
N
;
soletting
N
!1
completestheproof.
Usingthislemma,togetherwiththeMainInequality(
2.1
)weimmediatelyhavethe
followingcorollary:
Corollary2.7.
Wehavethefollowingidentity:
B
(
x;y;
1)=
M
(
x;y
)
forall
x;y
inthesubdomain
0

y

min(
x;x

1
)
.
Proof.
Weshowedintheprevioussectionthat
B
(
x;y;
1)

M
(
x;y
)forall(
x;y
)
2

0
.To
showthereverseinequalitynoticethattheMainInequality(
2.1
)togetherwithLemma
2.6
imply
B
(
x;y;
1)


B

x

;
y

;
1

:
(2.5)
Wewouldbedoneifwecanshowthat
B
(
x;y;
1)=1whenever
xy
=1.Indeed,thenwecan
justuseequation(
2.5
)with

=
p
xy
.
17
Fix(
x;y
)
2

0
with
xy
=1andconsiderthefunction
f
n
=
2
n
x
2
n

1
1
[0
;
1

2

n
)
:
If
I
istheinterval[0
;
1)thenobviously
h
f
n
i
I
=
x
.ConsideralsotheCarlesonsequence
f

J
g
J
2D
(
I
)
by

J
=
8
>
>
<
>
>
:
y
1

2

n
if
J
=[2

n
(
k

1)
;
2

n
k
)and
k
2f
1
;:::;
2
n

1
g
0otherwise.
Thenwehave
1
j
I
j
X
J
2D
(
I
)

J
j
J
j
=
y
2
n

1
X
k
=1
2

n
1

2

n
=
y:
Also,
A
f
n
(
t
)=
8
>
>
<
>
>
:

2
n
2
n

1

2
if0

t<
1

2

n
0otherwise,
hence
B
(
x;y;
1)

1

2

n
forall
n

1.Letting
n
!1
yieldstheclaim.
Remark
2.8
.
Observethatusingtheconstantfunction
f
(
t
)=
x
1
I
(
t
)andtheone-term
Carlesonsequencewhichis
y
on
I
and0everywhereelse,oneobtainsthat
A
f
=
xy
1
I
,
hence
B
(
x;y;
1)=1forall
xy>
1.
UsingLemma
2.6
inthesameway,wejusthavetoshowthat
B
(
x;
1
;
1)=
2
x
x
+1
toprovethat
B
(
x;y;
1)=
M
(
x;y
)intherestofthedomain,howeverthisturnsouttobeharder.
Theorem2.9.
Fix
x
2
(0
;
1)
andlet
>
0
.Foranyinterval
I
thereexistsanonnegative
function
f
on
I
with
h
f
i
I
=
x
andaCarlesonsequence
f

J
g
J
2D
(
I
)
withCarlesonconstant
18
atmostoneandverifying
1
j
I
j
X
J
2D
(
I
)

J
j
J
j
=1
suchthat
1
j
I
j



I
\
n
X
J
2
D
(
I
)

J
h
f
i
J
1
J
>
1
o



=
2
x
x
+1
+
O
(

)
:
ToprovethiswewillusetheMainInequality(
2.1
)iterativelytogiveadecompositionof
f
consistingofconstantfunctionsoncertaindyadicintervals,thisalsogivesusthe
constructionofthesequence
f

J
g
J
2D
(
I
)
.Thebasicideaisto,startingwithapoint(
x;
1)in

0
,use(
2.1
)tosplitthispointintoanotherpoint(
x
+
;
1)ontheboundaryandsomepoint
(
x

;A

).Thepoint(
x

;A

)isthenabsorbedbackintotheinitialpointandweapplythe
sameproceduretothepoint(
x
+
;
1)untilwegettoapointpasttheobstacle
xy

1(where
extremizersconsistofconstantfunctionstogetherwithone-termCarlesonsequencesasin
theRemarkafterCorollary
2.7
).
Inordertoillustratetheideawewillprovethelowerboundfor
B
withoutexplicitly
constructingtheexample.Thewayinwhichweprovethelowerboundwillmakethe
constructionmoreintuitive.
Theorem2.10.
TheBellmanfunction
B

B
(
x;
1
;
1)=
2
x
x
+1
forall
x
2
[0
;
1]
.
Proof.
Let
E
(
x;y
)=
B
(
x;y;
1),thenusingtheMainInequality(
2.1
)weseethatwehave
thefollowingbehavior:
E
(
t;
1)

1
2
E

t
1
1

t
;A
1

+
1
2
E

t
2
1

t
;A
2

19
whenever
t
=
t
1
+
t
2
2
and1=
A
1
+
A
2
2
+

.Letting
>
0,
x
=
t
and
A
2
=1weget
E
(
x;
1)

1
2

E

x

2

1

2

x

+
E

x
+
;
1

;
where
x
+
=
x
1+

1


+2

Since
B
issuperlinearinthetwovariablesand
B
(0
;
0
;
1)=0,wemusthave
E

x

2

1

2

x



1

2

x

E
(
x;
1)
soputtingeverythingtogetherweobtain
E
(
x;
1)

x
x
+2

E
(
x
+
;
1)
:
(2.6)
Ifwedeinductively
x
n
+1
=
x
n
1+

1


+2

and
x
0
=
x
,thenweeasilyseethat
x
n
=

n

1
1


+
x


1
1


;
where

=
1+

1


.
Wewanttostoptheiterationonce
x
n

1,andthishappenswhen

n

2


1+
x
(1


)
;
let
N
=
N
(
x
)bethesmallestintegerforwhichtheaboveinequalitydoesnothold.Then
iterating(
2.6
)
N
timesweget(since
E
(1
;
1)=1)
E
(
x;
1)

N
Y
j
=0
x
j
x
j
+2

;
itjusttogivealowerboundfortherighthandside.
20
Tothisendobservethat
N
Y
j
=0
x
j
x
j
+2


exp


N
X
j
=0
log

1+
2

x
j


exp


N
X
j
=0
2

x
j

=exp


2

N
X
j
=0
1
x
j

:
Letusestimate

2

P
N
j
=0
1
x
j
.Usingtheexplicitformulafor
x
n
wehave

2

N
X
j
=0
1
x
j
=

2

N
X
j
=0
1

j

1
1


+
x


1
1


=

2

N
X
j
=0

1

j

1
1


+
x


1
1



1

j

1+
x


1

+
N
X
j
=0
2

1


j
(1+
x
)
:
Thesttermtendsto0as

!
0andthesecondisaRiemannsum,indeed(recallingthe
of
N
=
N
(
x;
):
N
X
j
=0
2

1


j
(1+
x
)
=(1


)
N
X
j
=0

j
2

1



j
(1


j
(1+
x
))
=(1


)
N
X
j
=0
f
(

j
)(

j
+1


j
)
=
Z
2
1+
x
1
f
(
y
)
dy
+
O
(

)
;
as

!
0andwhere
f
(
y
)=
1
y
(1

y
(
x
+1))
:
Itiseasytoseethat
Z
2
1+
x
1
1
y
(1

y
(
x
+1))
dy
=log

2
x
x
+1

;
whichcompletestheproofofthelowerbound.
21
Letusnowusetheseideastoconstructtheexample.Therearetwobasicstepsinthe
iteration:wesplitthepoint(
x;
1)into(
x

;A

)and(
x
+
;
1),thenweabsorb(
x

;A

)
into(
x;
1)andobtainalowerboundfor
E
(
x;
1)intermsof
E
(
x
+
;
1),wetheniteratethis
until
x
+
>
1,wherewestopbecauseweknowthat
E
(
x
+
;
1)mustbe1there.Thesetwo
stepsareimposingacertainself-similarityon
f
andtheCarlesonsequence

intermsof
(
f
+
;
+
).ThefollowingLemma,whichisbasedontheideasfrom[
46
],makesthisprecise.
Lemma2.11.
Fixaninterval
I
andlet
g
+
beanonnegativefunctionon
I
+
.Supposealso
that

+
isaCarlesonsequenceadaptedto
I
+
withconstantatmost
1
andsuchthat
1
j
I
+
j
X
J
2D
(
I
+
)

+
J
j
J
j
=1
:
If
h
g
+
i
I
+
=
x
1+

1


+2

forsome
x
2
(0
;
1)
andasmall
>
0
,thenwecanconstruct
afunction
f
on
I
andaCarlesonsequence

adaptedto
I
withconstantatmost
1
suchthat
h
f
i
I
=
x
,
1
j
I
j
X
J
2D
(
I
)

J
j
J
j
=1(2.7)
and
1
j
I
j



I
\
n
X
J
2D
(
I
)

J
h
f
i
J
1
J
>
1
o





x
x
+2


1
j
I
+
j



I
+
\
n
X
J
2D
(
I
+
)

+
J
h
g
+
i
J
1
J
>
1
o



:
(2.8)
Proof.
Wewillassumewithoutlossofgeneralitythat
I
=[0
;
1),alsodenote

=

x
.

J
tobe

if
J
=
I
and

+
J
for
J
2D
(
I
+
).
22

f
tobe(1


)
g
+
on
I
+
anddenote
g

=(1


)

1
f
1
I

,then
1
j
I
j



y
2
I
:
X
J
2D
(
I
)

J
h
f
i
J
1
J
(
y
)
>
1



=
1
2
j
I

j



y
2
I

:
X
J
2D
(
I

)

J
h
f
i
J
1
J
(
y
)
>
1





+
1
2
j
I
+
j



y
2
I
+
:
X
J
2D
(
I
+
)

J
h
f
i
J
1
J
(
y
)
>
1





=
1
2
j
I

j



y
2
I

:
X
J
2D
(
I

)

J
h
g

i
J
1
J
(
y
)
>
1



+
1
2
j
I
+
j



y
2
I
+
:
X
J
2D
(
I
+
)

J
h
g
+
i
J
1
J
(
y
)
>
1



:
Let
I
j
=[
e
j
;e
j
+1
),where
e
j
=
1
2

2

j
,andsupposethat

b
I
j
=0for
j

1and

I

=0,then
1
2
j
I

j



y
2
I

:
X
J
2D
(
I

)

J
h
g

i
J
1
J
(
y
)
>
1



=
1
2
1
X
j
=1
2

j
1
j
I
j
j



y
2
I
j
:
A
(
g

1
j
)(
y
)
>
1



:
(2.9)
Let

=1

2

andwrite

=
1
X
j
=1
2

j
b
j
forsomebinarysequence
f
b
j
g
j
2
N
(i.e.:write

inbinary).
Foragiveninterval
J
let
S
J
f
bethescaledversionof
f
adaptedto
J
,i.e.:if
J
=[
a;b
)then
S
J
f
(
x
)=
f

x

a
b

a

:
Abusingnotation,letusalsodenoteby
S
J

thescaledversionoftheCarlesonsequence

tothedyadicsubinterval
J
of
I
,thenwehave
1
j
J
j



n
y
2
J
:
X
K
2D
(
J
)
(
S
J

)
K
h
S
J
f
i
K
1
K
(
y
)
>
1
o



=
1
j
I
j



n
y
2
I
:
X
K
2D
(
I
)

K
h
f
i
K
1
K
(
y
)
>
1
o



:
Supposethat(1


)
f
,whenrestrictedto
I
j
,agreeswith
S
I
j
f
forall
j

1suchthat
b
j
=1
23
andis0otherwise.SupposefurthermorethattheCarlesonsequence

alsothesame
similarity,i.e.:ifwescaleto
I
therestrictionof

to
I
j
weobtain

again.Ifwedenoteby
theleft-handsidein(
2.8
)thenwecoulduse(
2.9
)toobtain
=
1
2
1
X
j
=1
2

j
b
j
+
1
2
j
I
+
j



y
2
I
+
:
X
J
2D
(
I
+
)

J
h
g
+
i
J
1
J
(
y
)
>
1



;
hence
=

1
1+2


1
j
I
+
j



y
2
I
+
:
X
J
2D
(
I
+
)

J
h
g
+
i
J
1
J
(
y
)
>
1



=

x
x
+2


1
j
I
+
j



y
2
I
+
:
X
J
2D
(
I
+
)

J
h
g
+
i
J
1
J
(
y
)
>
1



;
whichiswhatwewanted.Notealsothatwecouldusethesamemethodtocomputethe
averageof
f
andityieldspreciselytherightamount:
x
.
Thereforewejusthavetoshowthatwecanafunction
f
andaCarlesonsequence

satisfyingtheseself-similarityconditions.Letusstartwith
f
:theoperator
T
by
Tf
=(1


)
1
X
j
=1
b
j
1
I
j
S
I
j
f
+(1


)
1
I
+
g
+
:
Weneedtoshowthat
T
hasapointin
L
1
(
I
);wewilldothisfollowingthestepsofthe
proofoftheBanachpointtheorem.Let
f
0
=(1


)
g
+
1
I
+
andinductively
f
n
+1
=
Tf
n
:
24
Weshouldshowthat
f
n
isaCauchysequencein
L
1
(
I
),butobservethat
k
f
n
+1

f
n
k
L
1
(
I
)
=(1


)
Z
I




1
X
j
=1
b
j
1
I
j
S
I
j
(
f
n
)

1
X
j
=1
b
j
1
I
j
S
I
j
(
f
n

1
)



=(1


)
1
X
j
=1
b
j
Z
I
j
j
S
I
j
(
f
n
)

S
I
j
(
f
n

1
)
j
=(1


)
1
X
j
=1
b
j
j
I
j
j
Z
I
j
f
n

f
n

1
j
=(1


)
Z
I
j
f
n

f
n

1
j
1
X
j
=1
b
j
2

j

1
=
(1


)(1

2

)
2
Z
I
j
f
n

f
n

1
j
:
Theconstant
˘
:=
(1


)(1

2

)
2
isstrictlylessthan1andbyinductionwehave
k
f
n
+1

f
n
k
L
1
(
I
)
.
˘
n
;
hencethesequenceisCauchy.Thistheproofofexistencefor
f
sincewecanjust

f
tobethelimitin
L
1
ofthesequence
f
n
above.
ToshowtheexistenceoftheCarlesonsequencewecanfollowthesamestepsasabove,but
nowwedon'thavetodealwithconvergenceissues.Indeed,startwithasequenceasinthe
beginningoftheproofandinductivelythe(
n
+1)-thsequence

n
+1
byinsertingthe
entiredyadictreeof

n
ateach
I
j
.Ateachstepweareonlychangingthevalueofthe
sequenceatdeeperanddeeperlevels,sowecanjust

K
asthethevalueof

n
K
,where
n
istheintegeratwhichthesequence

n
K
stabilizes.
WearenowreadytoproveTheorem
2.9
,wewillusethesameideasandnotationasinthe
proofofTheorem
2.10
.Given

,
I
and
x
2
(0
;
1)let
N
bethesmallestintegersuchthat

n

2


1+
x
(1


)
:
Let
f
1
betheconstantfunction
x
on
I
+
andlet

1
betheone-termCarlesonsequence
25
whichis1at
I
+
.Nowthefunction
f
n
+1
andCarlesonsequence

n
+1
inductivelyby
applyingLemma
2.11
tothefunction
g
+
:=
S
I
+
(
f
n
)andtheCarlesonsequence
S
I
+
(

n
);let
f
=
f
N
and

=

N
.Then,asintheproofofTheorem
2.10
,wehave
1
j
I
j



n
y
2
I
:
X
J
2D
(
I
)

J
h
f
i
J
1
J
>
1
o




exp

N
X
j
=0
2

1


j
(1+
x
)

;
whichweshowedtobe
2
x
x
+1
+
O
(

)
;
andthisiswhatwewantedtoprove.
26
3
Dyadicmodelsforsingularintegrals
JoseM.Conde-AlonsoandGuillermoRey
MathematischeAnnalen,October2015.
3.1Introduction
OneparticularlyusefulwaytostudymanyimportantoperatorsinHarmonicAnalysisis
thatofdecomposingthemintosumsofsimplerdyadicoperators.Anexampleofarecent
strikingresultusingthisstrategyistheproofofthesharpweightedestimateforthe
HilberttransformbyS.Petermichl[
42
].Thiswasakeysteptowardsthefull
A
2
theorem
forgeneralCalon-Zygmundoperators,provenbyT.onenin[
14
].Ofcourse
therearemanyinstancesofthisusefultechnique,butwewillnottrytogiveathorough
historicaloverviewhere.
Theproofin[
14
]wasa
tourdeforce
whichwastheculminationofmanypreviouspartial
byothers,see[
14
]andthereferencestherein.onendidnotonlyprovethe
A
2
theorem,buthealsoshowedthatgeneralon-Zygmundoperatorscouldbe
representedasaveragesofcertainsimpler\Haarshifts"inthespiritof[
42
].Thesharp
weightedboundthenfollowedfromthecorrespondingoneforthesesimpleroperators.
Later,A.Lernergaveaofthe
A
2
theoremin[
22
]whichavoidedtheuseof
27
mostofthecomplicatedmachineryin[
14
];itmainlyreliedonageneralpointwiseestimate
forfunctionsintermsofpositivedyadicoperatorswhichhadalreadybeenprovenin[
20
].
Theweightedresultforthepositivedyadicshiftsthatthiscontributionreducedthe
problemtohadalreadybeenshownbeforein[
18
],seealso[
4
]and[
5
].Moreprecisely,the
proofofLerner(essentially)gavethefollowingpointwiseestimateforgeneral
on-Zygmundoperators
T
:foreverydyadiccube
Q
j
Tf
(
x
)
j
.
1
X
m
=0
2

m
A
m
S
j
f
j
(
x
)fora.e.
x
2
Q;
(3.1)
where
>
0dependsontheoperator
T
,
S
arecollectionsofdyadiccubes(belongingto
samedyadicgridforeach
S
)whichdependon
f
,
T
and
m
,and
A
m
S
arepositive
dyadicoperatorsby
A
m
S
f
(
x
)=
X
Q
2S
h
f
i
Q
(
m
)
1
Q
(
x
)
;
where
Q
(
m
)
denotesthe
m
-thdyadicparentof
Q
.Moreover,thecollections
S
in(
3.1
)are
sparse
intheusualsense:given0
<<
1,wesaythatacollectionofcubes
S
belongingto
thesamedyadicgridis

-sparseifforallcubes
Q
2S
thereexistmeasurablesubsets
E
(
Q
)
ˆ
Q
with
j
E
(
Q
)
j

j
Q
j
and
E
(
Q
)
\
E
(
Q
0
)=
;
unless
Q
=
Q
0
.Acollectioniscalled
simplysparseifitis
1
2
-sparse.
FromthispointwiseestimateLernercontinuestheproofbyshowingthatboundingthe
operatornormofeach
A
m
S
canbereducedtojustestimatingtheoperatornormof
A
0
S
0
in
thesamespaceforallpossiblesparsecollections
S
0
.Moreprecisely,heshowsthat
kA
m
S
f
k
X
.
(
m
+1)sup
D
;
S
0
kA
0
S
0
f
k
X
;
(3.2)
wherethesupremumistakenoveralldyadicgrids
D
andallsparsecollection
S
0
ˆ
D
,and
where
X
isanyBanachfunctionspace,inthesenseof[
1
],Chapter1.
28
Itisatthispointwherethedualityof
X
isneededintheargument;theoperators
A
m
S
do
notlendthemselvestoLerner'spointwiseformula,whiletheiradjointsdo.Consequently,
thequestionofwhattodowhennodualityispresentwasleftopen.Ourmainresult
answersthisquestionbyprovingastronger(thoughlocalized)statement:theoperators
A
m
S
areactuallypointwiseboundedbypositivedyadic0-shifts:
Theorem3.1.
Let
P
beacubeand
S
asparsecollectionofdyadicsubcubes
Q
suchthat
Q
(
m
)

P
,thenforallnonnegativeintegrablefunctions
f
on
P
thereexistsanothersparse
collection
S
0
ofdyadicsubcubesof
P
suchthat
A
m
S
f
(
x
)
.
(
m
+1)
A
0
S
0
f
(
x
)
8
x
2
P
(3.3)
Infact,weproveTheorem
3.1
inaslightlymoregeneralsetting:thestatementis
provenforacertainnaturalmultilineargeneralizationoftheoperators
A
m
S
.Second,the
sparsecollection
S
isreplacedbyamoregeneralCarlesonsequence.Therelevantdetails
aregiveninthenextsection.
Thenoveltyinourapproachistwo-fold:wedirectlyattackthepointwiseestimateforthe
operators
A
m
,insteadofboundingtheirnorminvariousspaces.Also,inprovingthe
pointwiseboundwedevelopanalgorithmthatconstructivelyselectsthosecubeswhichwill
formthefamily
S
0
.Thisalgorithmhas\memory"inacertainsense:eachiterationtakes
intoaccounttheprevioussteps,afeaturewhichiscrucialinourmethodtoensurethat
S
is
sparse.
AsacorollaryofTheorem
3.1
,weananalogueof(
3.1
)forCon-Zygmund
operatorswithmoregeneralmoduliofcontinuity(seethenextsectionfortheprecise
Inparticular,weobtainthefollowingpointwiseestimatefor
on-Zygmundoperators:
Corollary3.2.
If
P
isadyadiccube,
f
isanintegrablefunctionsupportedon
P
and
T
is
29
aon-Zygmundoperatorwhosekernelhasmodulusofcontinuity
!
,then
j
Tf
(
x
)
j
.
1
X
m
=0
!
(2

m
)(
m
+1)
A
0
S
m
j
f
j
(
x
)
fora.e.
x
2
P;
(3.4)
where
S
m
aresparsecollectionsbelongingtoatmost
3
d
entdyadicgrids.
Moreover,ifweknowthat
!
thelogarithmicDinicondition:
Z
1
0
!
(
t
)

1+log

1
t

dt
t
<
1
;
(3.5)
thenwecansparsecollections
fS
0
1
;:::;
S
0
3
d
g
,belongingtopossiblyentdyadicgrids,
suchthat
j
Tf
(
x
)
j
.
3
d
X
i
=1
A
0
S
0
i
j
f
j
(
x
)
fora.e.
x
2
P:
(3.6)
Thefactor
m
in(
3.2
)precludedanaiveadaptationoftheproofin[
23
]toan
A
2
theorem
withtheusualDinicondition:
Z
1
0
!
(
t
)
dt
t
<
1
;
(3.7)
sincethesum
1
X
m
=0
!
(2

m
)(
m
+1)
'
Z
1
0
!
(
t
)

1+log
1
t

dt
t
(3.8)
coulddivergeforsomemoduli
!
satisfyingonly(
3.7
).Moreover,itwasshownin[
12
]that
theweak-type(1
;
1)normoftheadjointsoftheoperators
A
m
S
wasatleastlinearin
m
,even
intheunweightedcase,sousingdualitypreventedanextensionofthistype.However,
althoughourargumentdoesnotquitegivean
A
2
theoremforon-Zygmundoperators
satisfyingtheDinicondition(westillneed(
3.8
)tobeourproofavoidstheuseof
dualityandthestudyoftheadjointoperators(
A
m
S
)

.Itthusremovesatleastoneofthe
obstructionstopossibleproofsofthe
A
2
theoremwiththeDiniconditionwhichfollowthis
30
strategy.Hence,removingthelinearfactorof
m
inTheorem
3.1
remainsasanopen
problem.
Apartfrombeinginterestinginitsownright,aboundforon-Zygmundoperatorsby
thesesumsofpositive0-shiftsincaseswherethereisnodualityhasinterestingapplications,
someofwhichwedescribelater.Before,letusstateasecondcorollarytoTheorem
3.2
:
Corollary3.3.
Let
kk
X
beafunctionquasi-norm(seesection
3.2
)and
T
aon-
ZygmundoperatorsatisfyingthelogarithmicDinicondition,then
k
Tf
k
X
.
sup
D
;
S
kA
0
S
j
f
jk
X
;
(3.9)
wherethesupremumistakenoveralldyadicgrids
D
andallsparsecollections
Sˆ
D
.
Wenowdescribetwoimmediateapplicationsofourresult.Firstwecancontinuethe
program,initiatedin[
6
]andextendedin[
28
],whichaimstoextendthesharpweighted
estimatesforCaldeon-Zygmundoperatorstotheirmultilinearanalogues(asin[
10
]).In
particularweobtain
Theorem3.4.
Let
T
beamultilinearon-Zygmundoperator.Suppose
1
<p
1
;:::;p
k
<
1
,
1
p
=
1
p
1
+

+
1
p
k
and
~w
2
A
~
P
.Then
k
T
~
f
k
L
p
(
v
~w
)
.
[
~w
]
max(1
;
p
0
1
p
;:::;
p
0
k
p
)
A
~
P
k
Y
i
=1
k
f
i
k
L
p
(
w
i
)
:
(3.10)
Thesametheoremwasprovenin[
28
]butwiththeadditionalhypothesisthat
p
hadtobe
atleast1.Theproofofthistheoremisanapplicationoftheresultin[
28
]whichprovedthe
sameestimate(withoutthecondition
p

1)butforamultilinearanalogueoftheoperators
A
m
S
,togetherwithTheorem
3.1
.Infact,wewillneedamultilinearversionofTheorem
3.1
whichwestateandproveinthenextsection.
Oursecondapplicationisasharpapertureweightedestimateforsquarefunctionswhich
extendsaresultin[
24
].Inparticular:
31
Theorem3.5.
Let
>
0
,thenthesquarefunction
S

fortheconein
R
d
+1
+
ofapperture

andthestandardkernel
 

k
S

f
k
L
p;
1
(
R
d
;w
)
.

d
[
w
]
1
=p
A
p
k
f
k
L
p
(
R
d
;w
)
for
1
<p<
2
and
k
S

f
k
L
2
;
1
(
R
d
;w
)
.

d
[
w
]
1
=
2
A
2
(1+log[
w
]
A
2
)
k
f
k
L
2
(
R
d
;w
)
:
(3.11)
Ananalogousresultwasshownin[
24
]for2
<p<
3:
k
S

f
k
L
p;
1
(
R
d
;w
)
.

d
[
w
]
1
=
2
A
p
(1+log[
w
]
A
p
)
k
f
k
L
p
(
R
d
;w
)
:
TheproofreliesontheuseofLerner'spointwiseformulaandpreviousresultsbyLaceyand
Scurry[
19
].However,in[
24
]therequirementof
p>
2wasnecessaryforthesamereason
whytheproofofthemultilinearweightedestimatesrequired
p

1(acertainspacehadno
satisfactorydualityproperties).Theorem
3.1
canbeusedinalmostthesamewayaswith
theweightedmultilinearestimatestoproveTheorem
3.5
.Indeed,theproofsin[
19
]and[
24
]
reducetheproblemtoestimatingcertaindiscretepositiveoperatorswhichcanbeseento
beparticularinstancesofthepositivemultilinear
m
-shiftsusedintheproofofTheorem
3.4
.
Aswasnotedin[
19
],estimate(
3.11
)canbeseenasananalogueoftheresultin[
26
]
stablishingtheendpointweightedweak-typeestimateforon-Zygmundoperators
k
Tf
k
L
1
;
1
(
w
)
.
[
w
]
A
1
(1+log[
w
]
A
1
)
k
f
k
L
1
(
w
)
:
Seealso[
34
]forasimilarestimatefrombelowandmoreinformationonthesharpnessof
thisestimate,knownastheweak
A
1
conjecture.Inthisdirection,itseemsreasonablethat
LaceyandScurry'sproofin[
19
]couldbeadaptedtothemultilinearsetting,howeverwe
willnotpursuethisproblemhere.
Finally,asathirdapplicationofourresults,itispossibletogiveamoredirectproofofthe
32
resultin[
15
]forthe
q
-variationofon-Zygmundoperatorssatisfyingthelogarithmic
Diniconditionbyusingthepointwiseestimateanalogousto(
3.1
)in[
15
]andthenapplying
Theorem
3.1
.However,wewillnotpursuethisargumentationeither.
Shortlybeforeuploadingthispreprint,AndreiLernerkindlycommunicatedtotheauthors
thathe,jointlywithFedorNazarov,hadindependentlyprovenatheoremverysimilarto
Corollary
3.2
[
25
].Thoughthehypothesisarethesame,theirresultfromtheonein
thisnoteinthatwegivealocalizedpointwiseestimatewhiletheirpointwiseestimateis
validforallof
R
d
.However,ourresultseemstobeaspowerfulintheapplications.
3.2Pointwisedomination
ThegoalofthissectionistheproofofTheoremAanditsconsequencesasstatedinthe
introduction.Wewillprovetheresultinthelevelofgeneralityofmultilinearoperators.
Givenacube
P
0
on
R
d
,wewilldenoteby
D
(
P
0
)thedyadiclatticeobtainedbysuccessive
dyadicsubdivisionsof
P
0
.Byadyadicgridwewilldenoteanydyadiclatticecomposedof
cubeswithsidesparalleltotheaxis.A
k
-linearpositivedyadicshiftofcomplexity
m
isan
operatoroftheform
A
m
P
0

~
f
(
x
)=
A
m
P
0

(
f
1
;f
2
;

;f
k
)(
x
):=
X
Q
2
D
(
P
0
)
Q
(
m
)

P
0

Q

k
Y
i
=1
h
f
i
i
Q
(
m
)

1
Q
(
x
)
:
AsasteptowardstheproofofTheoremA,itisconvenienttoseparatethescalesof(or
Sinceweuploadedthisdocumentto
arXiv
,twootherarticleshaveappeared:[
17
]and[
11
],inwhich
similarestimatesareobtained.
33
slice
)
A
m
P
0

asfollows:
A
m
P
0

~
f
(
x
)=
m

1
X
n
=0
1
X
j
=1
X
Q
2
D
jm
+
n
(
P
0
)

Q

k
Y
i
=1
h
f
i
i
Q
(
m
)

1
Q
(
x
)
=:
m

1
X
n
=0
A
m;n
P
0

~
f
(
x
)
:
Notethat
D
k
(
P
0
)denotesthe
k
-thgenerationofthelattice
D
(
P
0
).Nowwerewrite
A
m
;
n
P
0

asasumofdisjointlysupportedoperatorsoftheform
A
m
;0
P;
.Indeed,
A
m
;
n
P
0

~
f
(
x
)=
1
X
j
=1
X
Q
2
D
jm
+
n
(
P
0
)

Q

k
Y
i
=1
h
f
i
i
Q
(
m
)

1
Q
(
x
)
=
X
P
2
D
n
(
P
0
)
1
X
j
=1
X
Q
2
D
jm
(
P
)

Q

k
Y
i
=1
h
f
i
i
Q
(
m
)

1
Q
(
x
)
=
X
P
2
D
n
(
P
0
)
A
m
;0
P
~
f
(
x
)
;
whichleadstotheexpression
A
m

0
~
f
(
x
)=
m

1
X
n
=0
X
P
2
D
n
(
P
0
)
A
m
;0
P
~
f
(
x
)
:
Wesaythatasequence
f

Q
g
Q
2
D
(
P
0
)
isCarlesonifitsCarlesonconstant
k

k
Car(
P
0
)
<
1
,
where
k

k
Car(
P
0
)
=sup
P
2
D
(
P
0
)
1
j
P
j
X
Q
2
D
(
P
)

Q
j
Q
j
:
Thefollowingintermediatestepisthekeytoourapproach:
Proposition3.6.
Let
m

1
and

beaCarlesonsequence.Forintegrablefunctions
34
f
1
;:::;f
k

0
on
P
0
thereexistsasparsecollection
S
ofcubesin
D
(
P
0
)
suchthat
A
m
;0
P
0

~
f
(
x
)

C
1
k

k
Car(
P
0
)
X
Q
2S

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
;
where
C
1
onlydependson
k
and
d
,andinparticularisindependentof
m
.
ToproveProposition
3.6
wewillproceedinthreesteps:wewillconstructthecollection
S
,thenshowthatwehavetherequiredpointwisebound,andthat
S
issparse.By
homogeneity,wewillassumethat
k

k
Car(
P
0
)
=1.Also,wewillassumethatthesequence

isbutourconstantswillbeindependentofthenumberofelementsinthesequence.
Let
P
0
=0and,foreach
Q
2
D
mj
(
P
0
)with
j

0,thesequence
f

Q
g
Q
by

Q
=max
R
2
D
m
(
Q
)

R
:
Foreach
Q
2
D
mj
(
P
0
)with
j

0,wewillinductivelythequantities
Q
and

Q
as
follows:

Q
=
8
>
>
<
>
>
:
0if
Q


Q
k
i
=1
h
f
i
i
Q


Q

0
2
2(
k
+1)
C
W
otherwise
;
where
C
W
istheboundednessconstantoftheunweightedendpointweak-typeofthe
operators
A
m
provedinTheorem
3.16
inthelastsection.Also,forevery
R
2
D
m
(
Q
)we


R
=
Q
+(

Q


R
)

k
Y
i
=1
h
f
i
i
Q

:
Notethattheonlyappliestocubesin
D
mj
(
P
0
)forsome
j
.Forallothercubesin
D
P
0
,weset

Q
=
Q
=0.Thecollection
S
consistsofthosecubes
Q
2
D
(
P
0
)forwhich

Q
6
=0.Notethat,since2
2(
k
+1)
C
W
>
1=
k

k
Car(
P
0
)


R
forall
R
andbytheof

Q
,wemusthave
Q

0forall
Q
.Thiscanbeeasilyseenbyinduction.
35
Remark
3.7
.
Wearetryingtoconstructasparseoperatorofcomplexity0whichdominates
A
m
;0
P
0

.Onewaytoachievethisistolet
S
bethecollectionofalldyadicsubcubesof
P
0
,but
ofcoursethisdoesnotyieldasparsecollection.Abetterwaywouldbetolet
S
consistofall
dyadiccubesin
P
0
forwhichatleastoneofits
m
-thgenerationchildren
R


R
>
0;
unfortunatelythisyieldsacollection
S
whichisnotsparse,andinfactitcanbeseenthat
theCarlesonsequence

associatedwiththiscollectioncanhaveaCarlesonnorm
k

k
Car(
P
0
)
whichgrowsexponentiallyin
m
.
Themainproblemwiththisapproachisthat,whenthetimecomestodecidewhethera
cubeshouldbein
S
ornot,wedonottakeintoaccountwhichcubeshavebeenselectedin
theprevioussteps.Notethatwheneverweaddacube
Q
to
S
wearenotonly\helping"
todominatetheportionof
A
m
;0
P
0

comingfrom
Q
,butalsowhatmaycomefromanyofits
descendants.
Onecanaccountforthisbyhavingthealgorithmuseasortof\memory"to,essentially,
keeptrackofhowmanycubesin
S
(appropriatelyweightedwiththeaveragesof
~
f
)lieabove
anyparticularcube.Thisisthepurposeof
Q
.Thiscanalsobeseenasthestoppingtime
algorithmwhichselectsacubewheneverthepreviouslyselectedcubesdonotprovideenough
heighttodominatetheoperatoruntilthatpoint.
Lemma3.8.
Wehavethepointwisebound
A
m
;0
P
0

~
f
(
x
)

X
Q
2
D
(
P
0
)

Q

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
:
(3.12)
Proof.
Wewillprovebyinductionthefollowingclaim:if
P
2
D
jm
(
P
0
)forsome
j

0,then
A
m
;0
P;
~
f
(
x
)


P
+
X
Q
2
D
(
P
)

Q

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
:
(3.13)
Notethat,when
P
=
P
0
,thisisexactly(
3.12
).Since

isthereisasmallest
j
0
2
N
suchthat

Q
=0forallcubes
Q
2
D

j
0
m
(
P
0
)
.Let
Q
beanycubein
D
j
0
m
(
P
0
),weobviously
Weuse
D

k
(
P
)todenotethosecubes
Q
in
D
(
P
)ofgenerationatleast
k
,so
j
Q
j
2

dk
j
P
j
.
36
have
A
m
;0

~
f

0in
Q:
Since
Q

0,theclaim(
3.13
)istrivialfor
P
2
D
j
0
m
(
P
0
).Now,assumebyinductionthat
wehaveproved(
3.13
)forallcubes
P
2
D
jm
(
P
0
)with1

j
1

j
andlet
P
beanycubein
D
(
j
1

1)
m
(
P
0
).By
A
m
;0
P
~
f
(
x
)=
X
Q
2
D
m
(
P
)


Q

k
Y
i
=1
h
f
i
i
P

1
Q
(
x
)+
A
m
;0

~
f
(
x
)

:
Let
x
2
Q
2
D
m
(
P
),thenbytheinductionhypothesisandtheof
Q
:
A
m
;0
P
~
f
(
x
)


Q

k
Y
i
=1
h
f
i
i
P

+
Q
+
X
R
2
D
(
Q
)

R

k
Y
i
=1
h
f
i
i
R

1
R
(
x
)
=

Q

k
Y
i
=1
h
f
i
i
P

+
P
+(

P


Q
)

k
Y
i
=1
h
f
i
i
P

+
X
R
2
D
(
Q
)

R

k
Y
i
=1
h
f
i
i
R

1
R
(
x
)
=
P
+

P

k
Y
i
=1
h
f
i
i
P

+
X
R
2
D
(
Q
)

R

k
Y
i
=1
h
f
i
i
R

1
R
(
x
)
=
P
+
X
R
2
D
(
P
)

R

k
Y
i
=1
h
f
i
i
R

1
R
(
x
)
;
whichiswhatwewantedtoshow.
Lemma3.9.
Thecollection
S
issparse.
Proof.
Let
P
2S
,wehavetoshowthattheset
F
:=
[
Q
(
P;Q
2S
Q

j
F
j
1
2
j
P
j
.Tothisend,let
R
bethecollectionofmaximal(strict)subcubesof
P
whicharein
S
,Notethatforall
R
2R
wehave
R
2
D
N
R
m
(
P
)forsome
N
R

1.Wethus
37
have
F
=
G
R
2R
R:
Bymaximality,forall
R
2R
anddyadiccubes
Q
with
R
(
Q
(
P
wehave

Q
=0.Forall
R
2R
and1

j

N
R
wenowclaimthat

R
((
N
R

j
)
m
)


P

k
Y
i
=1
h
f
i
i
P


j
X

=1

R
((
N
R


)
m
)

k
Y
i
=1
h
f
i
i
R
((
N
R


+1)
m
)

:
(3.14)
Indeed,onecanprovethisbyinductionon
j
.If
j
=1thenbydwehave

R
((
N
R

1)
m
)
=
P
+(

P


R
((
N
R

1)
m
)
)

k
Y
i
=1
h
f
i
i
P



P

k
Y
i
=1
h
f
i
i
P



R
((
N
R

1)
m
)

k
Y
i
=1
h
f
i
i
P

;
since
P

0.
Toprovetheinductionstep,observethat(bytheinductionhypothesis)for
j>
1

R
((
N
R

j
)
m
)
=
R
((
N
R

j
+1)
m
)
+(

R
((
N
R

j
+1)
m
)


R
((
N
R

j
)
m
)
)

k
Y
i
=1
h
f
i
i
R
((
N
R

j
+1)
m
)

=
R
((
N
R

j
+1)
m
)


R
((
N
R

j
)
m
)

k
Y
i
=1
h
f
i
i
R
((
N
R

j
+1)
m
)



P

k
Y
i
=1
h
f
i
i
P


j
X

=1

R
((
N
R


)
m
)

k
Y
i
=1
h
f
i
i
R
((
N
R


+1)
m
)

:
From(
3.14
)with
j
=
N
R
,wehave(sincethetermsarenonnegative)

R


P

k
Y
i
=1
h
f
i
i
P

A
m
;0
P
~
f
(
x
)
forall
x
2
R
.Since

R
6
=0,wemusthave

k
Y
i
=1
h
f
i
i
R


R


R
>
0
;
38
i.e.:

k
Y
i
=1
h
f
i
i
R


R
+
A
m
;0
P
~
f
(
x
)
>
2
2(
k
+1)
C
W

k
Y
i
=1
h
f
i
i
P

forall
x
2
R
.Let
G
P
~
f
=
P
R
2R

R

Q
k
i
=1
h
f
i
i
R

1
R
,thenforall
x
2
R
wehave
G
P
f
(
x
)+
A
m
;0
P
~
f
(
x
)
>
2
2(
k
+1)
C
W

k
Y
i
=1
h
f
i
i
P

;
hence
j
F
j





(
x
2
P
:
G
P
~
f
(
x
)+
A
m
;0
P
~
f
(
x
)
>
2
2(
k
+1)
C
W

k
Y
i
=1
h
f
i
i
P

)






kG
P
+
A
m
;0
P;
k
1
=k
L
1
(
P
)

L
1
(
P
)
!
L
1
=k;
1
(
P
)

2
2(
k
+1)
C
W

Q
k
i
=1
h
f
i
i
P

1
=k

k
Y
i
=1
k
f
i
k
L
1
(
P
)

1
=k
=
kG
P
+
A
m
;0
P
k
1
=k
L
1
(
P
)

L
1
(
P
)
!
L
1
=k;
1
(
P
)
(2
2(
k
+1)
C
W
)
1
=k
j
P
j
Letuscomputetheoperatornorm
kG
P
k
L
1
(
P
)

L
1
(
P
)
!
L
1
=k;
1
(
P
)
.Observethat,since

Q

1
forall
Q
,theoperator
G
ispointwiseboundedbythemulti-linearprojection
P
P
~
f
(
x
)=
X
R
2R

k
Y
i
=1
h
f
i
i
R

1
R
(
x
)=
k
Y
i
=1

X
R
2R
h
f
i
i
R
1
R
(
x
)

:
Foreach1

i

k
,wehave
k
P
R
2R
h
f
i
i
R
1
R
k
L
1
(
P
)
k
f
i
k
L
1
(
P
)
.Therefore,byolder's
inequalityweget
kP
P
~
f
k
L
1
=k;
1
(
P
)

k
Y
i
=1



X
R
2R
h
f
i
i
R
1
R



L
1
(
P
)

k
Y
i
=1
k
f
i
k
L
1
(
P
)
:
Ontheotherhandwehave
kA
m
;0
P
~
f
k
L
1
=k;
1
(
P
)

C
W
k
Y
i
=1
k
f
i
k
L
1
(
P
)
39
byTheoremW.1.Combiningtheseestimatesweget
kG
P
+
A
m
;0
P
k
L
1
(
P
)

L
1
(
P
)
!
L
1
=k;
1
(
P
)

2
k
+1
(1+
C
W
)

2
k
+2
C
W
andtheresultfollows.
Fromlemmas
3.8
and
3.9
Proposition
3.6
followsatonce.Theproofshowsthatonecan
actuallytake
C
1
=2
2+
k
(7+
d
(2
k

1))
.WearenowreadytotheproofofTheoremA,
whichwestatehereinfullgenerality:
Theorem3.10.
Let

beaCarlesonsequenceandlet
P
0
beadyadiccube.Forevery
k
-tuple
ofnonnegativeintegrablefunctions
f
1
;:::;f
k
on
P
thereexistsasparsecollection
S
ofcubes
in
D
(
P
)
suchthat
A
m
P
~
f
(
x
)

C
2
X
Q
2S

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
:
Proof.
If
m
=0wecanjustapplyProposition
3.6
afternotingthat
A
0
P
0

canbewrittenas
A
1;0
P
0

,where

Q
=

Q
(1)
:
Oneeasilyseesthat
k

k
Car(
P
0
)
=
k

k
Car(
P
0
)
.Hence,wemayassumethat
m

1.Recallthe
expression
A
m
P
0

~
f
(
x
)=
m

1
X
n
=0
X
P
2
D
n
(
P
0
)
A
m
;0
P
~
f
(
x
)
:
fromthebeginningofthesection.ByProposition
3.6
,foreach0

n

m

1andeach
P
2
D
n
(
P
0
)wecanasparsecollectionofcubes
S
n
P
ˆ
D
(
P
)suchthat
A
m
;0
P;
~
f
(
x
)

C
1
k

k
Car(
P
0
)
X
Q
2S
n
P

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
:
40
Observethatthecollection
S
n
=
[
P
2
D
n
(
P
0
)
S
n
P
isalsosparse,so
A
m
P
0

~
f
(
x
)

C
1
k

k
Car(
P
0
)
m

1
X
n
=0
X
Q
2S
n

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
:
(3.15)
For0

n

m

1

n
Q
=
8
>
>
<
>
>
:
1if
Q
2S
n
0otherwise.
Sincethecollections
S
n
aresparse,thesequences

n
areCarlesonsequenceswith
k

n
k
Car(
P
0
)

2,thereforethesequence

Q
:=
m

1
X
n
=0

n
Q
isalsoCarlesonwith
k

k
Car(
P
0
)

2
m
.
Withthiswecancontinuetheargumentusingestimate(
3.15
)andthecase
m
=0:
A
m
P
0

~
f
(
x
)

C
1
k

k
Car(
P
0
)
m

1
X
n
=0
X
Q
2S
n

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
=
C
1
k

k
Car(
P
0
)
m

1
X
n
=0
X
Q
2
D
(
P
0
)

n
Q

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
=
C
1
k

k
Car(
P
0
)
X
Q
2
D
(
P
0
)

Q

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
=
C
1
k

k
Car(
P
0
)
A
0
P
0

~
f
(
x
)

C
1
k

k
Car(
P
0
)
C
1
2
m
X
Q
2S

k
Y
i
=1
h
f
i
i
Q

1
Q
(
x
)
;
whichyieldstheresultwith
C
2
=2
C
2
1
.
Remark
3.11
.
TheaboveproceduredoesnotrelyonanysppropertyoftheLebesgue
measure.Infact,Theorem
3.1
alsoholdswhenwereplaceallaverages{bothincomplexity
0andcomplexity
m
operators{byaverageswithrespecttoanyotherlocallyBorel
measure,becausetheproofis
41
WenowdetailhowtouseTheorem
3.1
toderivethemultilinearversionofcorollaries
3.2
and
3.3
.Forus,amultilinearon-Zygmundoperatorwillbeanoperator
T
satisfying
T
(
f
1
;:::;f
k
)=
Z
R
dk
K
(
x;y
1
;:::;y
k
)
f
1
(
y
1
)

f
k
(
y
k
)
dy
1
:::dy
k
forall
x=
2\
k
i
=1
supp
f
i
forappropriate
f
i
.Alsowewillrequirethat
T
extendstoabounded
operatorfrom
L
q
1

:::L
q
k
to
L
q
where
1
q
=
1
q
1
+

+
1
q
k
;
andthatitthesizeestimate
j
K
(
y
0
;:::;y
k
)
j
A

P
k
i;j
=0
j
y
i

y
j
j

kd
:
!
willbethemodulusofcontinuityofthekerneloftheoperatori.e.apositive
nondecreasingcontinuousanddoublingfunctionthat
j
K
(
y
0
;:::;y
j
;:::;y
k
)

K
(
y
0
;:::;y
0
j
;:::;y
k
)
j
C!
 
j
y
j

y
0
j
j
P
k
i;j
=0
j
y
i

y
j
j
!
1

P
k
i;j
=0
j
y
i

y
j
j

kd
forall0

j

k
,whenever
j
y
j

y
0
j
j
1
2
max
0

i

k
j
y
j

y
i
j
.WecannowproveCorollary
3.2
:
ProofofCorollary
3.2
.
Fixameasurable
f
,andacube
Q
0
ˆ
R
d
.Ourstartingpointisthe
formula
j
T
~
f
(
x
)

m
T
~
f
(
Q
0
)
j
.
X
Q
2S
1
X
m
=0
!
(2

m
)
m
Y
i
=1
hj
f
i
ji
2
m
Q
1
Q
(
x
)
;
whichholdsforasparsesubcollection
Sˆ
D
(
Q
0
)(see[
6
]and[
15
],weareimplicitlyusing
aslightimprovementofLerner'sformulawhichcanbefoundin[
12
],Theorem2.3).Here
m
f
(
Q
)denotesthemedianofameasurablefunction
f
overacube
Q
(see[
23
]fortheprecise
42
which
j
m
f
(
Q
)
j
.
k
f
k
L
1
;
1
(
Q
)
j
Q
j
:
Hencewecanjustwrite
j
T
~
f
(
x
)
j
.
1
X
m
=0
!
(2

m
)
X
Q
2S
m
Y
i
=1
hj
f
i
ji
2
m
Q
1
Q
(
x
)
;
(3.16)
Byanelaborationofthewell-knownone-thirdtrick,itwasprovenin[
15
]thatthereexist
dyadicsystems
f
D
ˆ
g
ˆ
2f
0
;
1
=
3
;
2
=
3
g
d
suchthatforeverycube
Q
in
R
d
andevery
m

1,there
exists
ˆ
2f
0
;
1
=
3
;
2
=
3
g
d
and
R
Q;m
2
D
ˆ
suchthat
Q
ˆ
R
Q;m
;
2
m
Q
ˆ
Q
(
m
)
;
3
`
(
Q
)
<`
(
R
Q;m
)

6
`
(
Q
)
:
Also,wemayassumethatforeach
ˆ
2f
0
;
1
=
3
;
2
=
3
g
d
thereexistsacube
P
(
ˆ
)suchthat
Q
0
ˆ
P
(
ˆ
)
ˆ
c
d
P
(
ˆ
)forsomedimensionalconstant
c
d
.Usingthis,wecanfurtherwrite
(
3.16
)as
j
T
~
f
(
x
)
j
.
X
ˆ
2f
0
;
1
3
;
2
3
g
d
1
X
m
=0
!
(2

m
)
X
Q
2S
R
Q;m
2
D
ˆ

k
Y
i
=1
hj
f
i
ji
R
(
m
)
Q;m

1
R
Q
:
Let
F
ˆ
m
=
f
R
Q;m
:
R
Q
2
D
ˆ
gˆ
D
(
P
(
ˆ
)).Then,wecanestimate
j
T
~
f
(
x
)
j
.
6
d
X
ˆ
2f
0
;
1
3
;
2
3
g
d
1
X
m
=0
!
(2

m
)
X
R
2F
ˆ
m

k
Y
i
=1
hj
f
i
ji
R
(
m
)

1
R
;
sinceatmost6
d
cubes
Q
in
D
aremappedtothesamecube
R
Q;m
.thesequence

ˆ
Q
=
8
>
>
<
>
>
:
1if
Q
2F
ˆ
m
0otherwise
:
43
Thecollections
F
ˆ
m
are2

1

6

d
-sparse,andhenceCarlesonwithconstant2

6
d
.Inorderto
applyTheorem
3.1
,foreach
ˆ
2f
0
;
1
3
;
2
3
g
d
,
m

0,wenowsplitthesumasfollows:
X
Q
2
D
ˆ

ˆ
Q

k
Y
i
=1
hj
f
i
ji
Q
(
m
)

1
Q
(
x
)=
X
Q
2
D

m
(
P
(
ˆ
))

ˆ
Q

k
Y
i
=1
hj
f
i
ji
Q
(
m
)

1
Q
(
x
)
+
1
X
`
=1
X
Q
2
D
m

`
(
P
(
ˆ
))

ˆ
Q

k
Y
i
=1
hj
f
i
ji
Q
(
m
)

1
Q
(
x
)
=I+II
:
Now,since
f
i
issupportedon
Q
0
ˆ
P
(
ˆ
)for1

i

k
andall
ˆ
2f
0
;
1
3
;
2
3
g
d
,weclaimthat
II

I.Indeed,compute
1
X
`
=1
X
Q
2
D
m

`
(
P
(
ˆ
))

ˆ
Q

k
Y
i
=1
hj
f
i
ji
Q
(
m
)

1
Q
(
x
)

1
X
`
=1
X
Q
2
D
m

`
(
P
(
ˆ
))

k
Y
i
=1
hj
f
i
ji
Q
(
m
)

1
Q
(
x
)
=
1
X
`
=1

k
Y
i
=1
hj
f
i
ji
P
(
ˆ
)
(
`
)

:
Nowobservethat,bythesupportconditiononthetuple
~
f
,
k
Y
i
=1
hj
f
i
ji
P
(
ˆ
)
(
`
)
=2

dk`
k
Y
i
=1
hj
f
i
ji
P
(
ˆ
)
;
whichisenoughtoprovetheclaim.Therefore,weonlyneedtoworkinthelocalizedcubes
P
(
ˆ
),
ˆ
2f
0
;
1
3
;
2
3
g
d
.Therefore,wecanobtaintheassertionofCorollary
3.2
applying
Theorem
3.1
:
44
j
T
~
f
(
x
)
j
.
X
ˆ
2f
0
;
1
3
;
2
3
g
d
1
X
m
=0
!
(2

m
)
X
Q
2
D
ˆ
;Q
ˆ
P
(
ˆ
)
(
m
)

ˆ
Q

k
Y
i
=1
hj
f
i
ji
Q
(
m
)

1
Q
(
x
)
.
X
ˆ
2f
0
;
1
3
;
2
3
g
d
1
X
m
=0
!
(2

m
)(
m
+1)
X
Q
2S
m;
~
f

k
Y
i
=1
hj
f
i
ji
Q

1
Q
=
X
ˆ
2f
0
;
1
3
;
2
3
g
d
1
X
m
=0
!
(2

m
)(
m
+1)
A
S
m;
~
f
~
f
(
x
)
;
forsparsecollections
S
m;
~
f
thatmaydependbothon
m
and
~
f
(andwhicharesubfamiliesof
D
(
P
(
ˆ
))foreachvalueof
ˆ
).Now,reorganizingthesumaboveweobtain
j
T
~
f
(
x
)
j
.
X
ˆ
2f
0
;
1
3
;
2
3
g
d
X
S
m;
~
f
ˆ
D
ˆ
!
(2

m
)(
m
+1)
A
S
m;
~
f
~
f
(
x
)
=:
X
ˆ
2f
0
;
1
3
;
2
3
g
d
A
ˆ
~
f
(
x
)
:
Now,bythelogarithmicDinicondition,eachoftheoperators
A
ˆ
isboundedabovebysome
absoluteconstanttimesa0-shiftwhoseassociatedsequenceis1-Carleson(andlocalizedin
P
(
ˆ
))towhichwecanapplyagainTheorem
3.1
.Therefore,weobtain
j
T
~
f
(
x
)
j
.
X
ˆ
2f
0
;
1
3
;
2
3
g
d
A
S
ˆ
~
f
(
x
)
;
forsomesparsefamilies
S
ˆ
ˆ
D
ˆ
whichdependon
~
f
.
Wenowintroducethenotionoffunctionquasi-norm.Wesaythat
kk
X
,ontheset
ofmeasurablefunctions,isafunctionquasi-normif:
(P1)
Thereexistsaconstant
C>
0suchthat
k
f
+
g
k
X

C

k
f
k
X
+
k
g
k
X

;
45
(P2)
k

k
X
=
j

jk
f
k
X
forall

2
C
.
(P3)
If
j
f
(
x
)
jj
g
(
x
)
j
almost-everywherethen
k
f
k
X
k
g
k
X
.
(P4)
k
liminf
n
!1
f
n
k
X

liminf
n
!1
k
f
n
k
X
Fixsomedyadicsystem
D
suchthatthereexistsanincreasingsequenceofdyadiccubes
f
P
`
g
`
ˆ
D
whoseunionisthewholespace
R
d
,anddenote
1
P
`
~
f
=(
1
P
`
f
1
;:::;
1
P
`
f
k
).Now,
takingintoaccountproperties
(P1)
and
(P3)
,ifwetakequasi-normsinthesecond
assertionofCorollary
3.2
,wehave
k
1
P
`
T
(
1
P
`
~
f
)
k
X
.
sup
D
;
S
kA
S
(
1
P
`
~
f
)
k
X
8
`:
Ontheonehand,since
~
f
isintegrable,
T
(
1
P
`
~
f
)convergespointwiseto
T
(
~
f
).Therefore,we
have
1
P
`
T
(
1
P
`
~
f
)
!
T
(
~
f
)
pointwise.Finally,weapplyproperty
(P4)
andweget
k
T
~
f
k
X
=



liminf
`
1
P
`
T
(
1
P
`
~
f
)



X

liminf
`



1
P
`
T
(
1
P
`
~
f
)



X
.
sup
D
;
S



A
S
~
f



X
:
ThisisexactlyCorollary
3.3
.
Remark
3.12
.
Wenotethatthedependenceon
m
inthepointwiseestimateofshiftsof
complexity
m
mustbeatleastlinearin
m
.Toseethis,letusworkindimensiononeand
alargeinteger
m
.Foranyinterval
I
=[
a;b
)let
I
j
bethe
j
-thintervalof
D
m
(
I
):
I
j
=
a
+
j
I
j
[
j
2

m
;
(
j
+1)2

m
)
:
46
atoweroveraninterval
I
tobethecollectionofintervals
T
I
=
f
[
a;a
+2

k
j
I
j
):
k
2
N
g
:
Thecollectionofintervals
S
=
S
J
2
D
m
(
I
)
T
J
isasparsecollection.Nowconsiderafunction
f
on
I
whichisby
f
(
x
)=
8
>
>
<
>
>
:
0if
x
2
I
j
with
j
even
;
2otherwise
:
Denotegen(
J
)=log
2
(
`
(
I
)
`
(
J
)

1
)forcubes
J
2
D
(
I
).Observethatforanydyadicinterval
J

I
withgen(
J
)

m

1wehave
h
f
i
J
=1
:
Considernowtheactionof
A
m
S
on
f
.If
x
2
(
I
j
)
0
with
j
eventhen
A
m
S
f
(
x
)=
m:
Inordertoconstructacollection
S
0
ofintervalsin
I
forwhichwehave
A
m
S
f
(
x
)

C
A
0
S
0
f
(
x
)
;
wewouldneedtoselecteveryinterval
J
ˆ
I
withgen(
J
)

m

1.Indeed,let
I
k
(
x
)bethe
intervalin
D
k
(
I
)whichcontains
x
andlet

J
be1if
J
2S
0
and0otherwise.Then
C
A
0
S
0
f
(
x
)=
C
m

1
X
k
=0

I
k
(
x
)

m
forall
x
2
(
I
j
)
0
with
j
even.Thisimpliesthatatleast
m=C
oftheseintervalsmustbein
47
S
0
.Butthisimpliesthattheheight
X
J
2S
0

J
1
J
(
x
)

m=C
onhalfoftheinterval
I
,whichcontradictsthehypothesisof
S
0
beingsparseif
m
islarge
enough.
3.3Applications
Wearenowreadytofullystateandprovetheapplicationsofthepointwiseboundas
statedintheintroduction.Webeginwiththemultilinearsharpweightedestimates:
3.3.1Multilinear
A
2
theorem
WeneedsomemoresThesewereintroducedin[
27
].
3.13
(
A
~
P
weights)
.
Let
~
P
=(
p
1
;:::;p
k
)with1

p
1
;:::;p
k
<
1
and
1
p
=
1
p
1
+

+
1
p
k
.Given
~w
=(
w
1
;:::;w
k
),set
v
~w
=
k
Y
i
=1
w
p=p
i
i
:
Wesaythat
~w
the
k
-linear
A
~
P
conditionif
[
~w
]
A
~
P
:=sup
Q

1
j
Q
j
Z
Q
v
~w

k
Y
i
=1

1
j
Q
j
Z
Q
w
1

p
0
i
i

p=p
i
:
Wecall[
~w
]
A
~
P
the
A
~
P
constantof
~w
.Asusual,if
p
i
=1thenweinterpret
1
j
Q
j
R
Q
w
1

p
0
i
i
tobe
(essinf
Q
w
i
)

1
.
Thefollowingtheoremwasprovedin[
28
]:
48
Theorem3.14.
Suppose
1
<p
1
;:::;p
k
<
1
,
1
p
=
1
p
1
+

+
1
p
k
and
~w
2
A
~
P
.Then
kA
S
~
f
k
L
p
(
v
~w
)
.
[
w
]
max(1
;
p
0
1
p
;:::;
p
0
k
p
)
A
~
P
k
Y
i
=1
k
f
i
k
L
p
(
w
i
)
;
whenever
S
issparse.
WecannowuseCorollary
3.3
toextendtheaboveresulttogeneral
k
-linear
on-Zygmundoperators:
Theorem3.15.
UndertheconditionsofTheorem
3.14
,forany
k
-linearon-Zygmund
operator
T
,wehave
k
T
~
f
k
L
p
(
v
~w
)
.
[
~w
]
max(1
;
p
0
1
p
;:::;
p
0
k
p
)
A
~
P
k
Y
i
=1
k
f
i
k
L
p
(
w
i
)
:
Proof.
WejustneedtoapplyCorollary
3.3
with
kk
X
:=
kk
L
p
(
v
~w
)
,whichclearlyisafunction
quasi-norm.Theassumptionof
~
f
beingintegrableisaqualitativeoneandcanbetrivially
removedbytheusualdensityarguments.
3.3.2SharpapertureweightedLittlewood-Paleytheorem
HerewefollowLerner[
24
],thereadercananiceintroductionandsomereferences
there.Webeginwithsome
Let
 
2
L
1
(
R
d
)with
R
R
d
 
(
x
)
dx
=0satisfy
j
 
(
x
)
j
.
1
(1+
j
x
j
)
d
+

(3.17)
Z
R
d
j
 
(
x
+
h
)

 
(
x
)
j
dx
.
j
h
j

:
(3.18)
Wewilldenotetheupperhalf-space
R
d

R
by
R
d
+1
+
andthe

-coneat
x
by


(
x
)=
f
(
y;t
)
2
R
d
+1
+
:
j
y

x
j
t
g
:
49
Let
 
t
bethedilationof
 
whichpreservesthe
L
1
norm,i.e.:
 
t
(
x
)=
t

d
 
(
x=t
),thenwe
canthesquarefunction
S

f
by
S

f
(
x
)=

Z


(
x
)
j
(
f

 
t
)(
y
)
j
2
dydt
t
d
+1

1
=
2
:
Wewillalsoneedaregularizedversion.LetbeaSchwartzfunctionsuchthat
1
B
(0
;
1)
(
x
)


x
)

1
B
(0
;
2)
(
x
)
:
Wetheregularizedsquarefunction
e
S

by
e
S

f
(
x
)=

Z
R
d
+1
+


x

y


j
(
f

 
t
)(
y
)
j
2
dydt
t
d
+1

1
=
2
:
Theregularizedversioncanbeusedinsteadof
S

inmostcasessincewehave
S

f
(
x
)

e
S

f
(
x
)

S
2

f
(
x
)
:
Itwasprovedin[
24
]that
j
(
e
S

f
(
x
))
2

(
m
Q
0
(
e
S

f
)
2
)
j
.

2
d
1
X
m
=0
2

m
X
Q
2S
hj
f
ji
2
2
m
Q
1
Q
(
x
)
BythesameTheorem
3.1
initsbilinearformulation(with
f
1
=
f
2
=
f
),thelastexpression
canbebounded,uptoaconstant,byanexpressionoftheform

2
d
X
ˆ
2f
0
;
1
3
;
2
3
g
d
1
X
m
=0
2

m
(
m
+1)
X
Q
2S
ˆ;m
hj
f
ji
2
Q
1
Q
(
x
)
:
Asin[
24
],weknow(apriori)that
m
Q
0
(
e
S

f
)
!
0as
j
Q
j!1
sobythetriangle
inequalityandFatou'slemmawecanignorethatterm(orbyarguingaswedidinthe
50
previoussection).Finally,arguingasintheproofofcorollaries
3.2
and
3.3
,wearriveat
k
e
S

f
k
L
p;
1
(
w
)
.

d
sup
D
;
S
kA
0
S
(
f;f
)
1
=
2
k
L
p;
1
(
w
)
;
wherethesupremumistakenoveralldyadicgrids
D
andallsparsecollections
Sˆ
D
.To
theargumentwerecallthefollowingresult,whichwasshownin[
19
]:
kA
0
S
(
f;f
)
1
=
2
k
L
p;
1
(
w
)
.
[
w
]
max(
1
2
;
1
p
)
A
p

p
([
w
]
A
p
)
k
f
k
L
p
(
w
)
(3.19)
for1
<p<
3,where

p
(
t
)=
8
>
>
<
>
>
:
1if1
<p<
2
1+log
t
if2

p<
3
:
WearethusabletoextendLerner'sestimateto1
<p

2,obtaining
k
S

f
k
L
p;
1
(
w
)
.

d
[
w
]
1
=p
A
p
k
f
k
L
p
(
w
)
for1
<p<
2
and
k
S

f
k
L
2
;
1
(
w
)
.

d
[
w
]
1
=
2
A
2
(1+log[
w
]
A
2
)
k
f
k
L
2
(
w
)
:
3.4Theweak-typeestimateformultilinear
m
-shifts
Hereweprovetheweak-typeestimatefor
k
-linear
m
-shiftsneededinsection
3.2
.Notice
thattheonlyimportantpointoftheestimatesbelowistheindependenceoftheconstants
fromtheparameter
m
.Theproofcouldbemoreorlessstandardbynow,buttheauthors
havenotbeenabletoitelsewhere.Thereforeweincludeitforcompleteness.
51
Theorem3.16.
sup

0

jf
x
2
P
0
:
A
m
P
0

~
f
(
x
)
>
gj
k

C
W
k

k
Car(
P
0
)
k
Y
i
=1
k
f
i
k
L
1
(
P
0
)
;
(3.20)
where
C
W
>
0
onlydependson
k
and
d
,andinparticularisindependentof
m
.
WewillessentiallyfollowGrafakos-Torres[
10
]and[
14
].Weprovean
L
2
boundand
thenapplyaon-Zygmunddecomposition.Forthe
L
2
boundwewillusea
multilinearCarlesonembeddingtheorembyW.ChenandW.an[
2
],fromwhichwe
onlyneedtheunweightedresult:

X
Q
2
D
(
P
0
)

Q

k
Y
i
=1
h
f
i
i
Q

p

1
p
k

k
Car(
P
0
)
k
Y
i
=1
p
0
i
k
f
i
k
L
p
i
(
P
0
)
(3.21)
whenever
1
p
=
1
p
1
+

+
1
p
k
:
Nowwecanprove
Proposition3.17.
kA
m
P
0

~
f
k
L
2
(
P
0
)

4
k

k
Car(
P
0
)
k
Y
i
=1
k
f
i
k
L
2
k
(
P
0
)
Proof.
Webeginbyusingdualityandhomogeneitytoreducetoshowing
Z
P
0
g
(
x
)
A
m
P
0

~
f
(
x
)
dx

4
assumingthat
k
f
i
k
L
2
k
(
P
0
)
=
k
g
k
L
2
(
P
0
)
=
k

k
Car(
P
0
)
=1and
g

0.Byand
Cauchy-Schwarz,thisisequivalentto

X
Q
2
D

m
(
P
0
)

Q

k
Y
i
=1
h
f
i
i
Q
(
m
)

2
j
Q
j

1
=
2

X
Q
2
D

m
(
P
0
)

Q
h
g
i
2
Q
j
Q
j

1
=
2
:
52
Thesecondtermcanbeestimated,using(
3.21
)inthelinearcase,by

X
Q
2
D

m
(
P
0
)

Q
h
g
i
2
Q
j
Q
j

1
=
2

2
:
Forthetermobservethatthesequence

Q
by

Q
=
1
2
dm
X
R
2
D
m
(
Q
)

R
isaCarlesonsequenceadaptedto
P
0
ofthesameconstant.Indeed:
1
j
Q
j
X
R
2
D
(
Q
)

R
j
R
j
=
1
j
Q
j
X
R
2
D
(
Q
)
j
R
j
1
2
dm
X
T
2
D
m
(
R
)

T
=
1
j
Q
j
X
R
2
D
(
Q
)
X
T
2
D
m
(
R
)

T
j
T
j
=
1
j
Q
j
X
R
2
D

m
(
Q
)

R
j
R
j
k

k
Car(
I
)
=1
:
Therefore,wecanwritethetermas

X
Q
2
D
(
P
0
)

Q

k
X
i
=1
h
f
i
i
Q

2
j
Q
j

1
=
2
;
whichcanalsobeestimatedby(
3.21
)asfollows:

X
Q
2
D
(
P
0
)

Q

k
X
i
=1
h
f
i
i
Q

2
j
Q
j

1
=
2


2
k
2
k

1

k

2
:
Combiningbothtermswearriveat
Z
P
0
g
(
x
)
A
m
P
0

~
f
(
x
)
dx

4
53
whichiswhatwewanted.
NowwecanproveTheorem
3.16
.
Proof.
Byhomogeneitywecanassume
k

k
Car(
P
0
)
=
k
f
i
k
L
1
(
P
0
)
=1.Wenowfollowthe
classicalschemewhichusesthe
L
2
boundandastandardon-Zygmunddecomposition,
seeforexampleGrafakos-Torres[
10
].However,weneedtobecarefulwiththedependence
on
m
,sowewilladapttheproofin[
14
]toouroperators.
Assumewithoutlossofgeneralitythat
f
i

0.

i
=
f
x
2
P
0
:
M
d
f
i
(
x
)
>
1
=k
g
:
If
h
f
i
i
P
0
>
1
=k
thenbythehomogeneityassumption
j
P
0
j
<

1
=k
andtheestimatefollows.Therefore,wecanassume
h
f
i
i
P
0


1
=k
forall1

i

k
andhence
wecanwrite
i
asaunionthecubesinacollection
R
i
consistingofpairwisedisjointdyadic
(strict)subcubesof
P
0
withtheproperty
h
f
i
i
R
>
1
=k
and
h
f
i
i
R
(1)


1
=k
:
Foreach1

i

k
let
b
i
=
P
R
2R
i
b
R
i
,where
b
R
i
(
x
):=

f
i
(
x
)
h
f
i
i
R

1
R
(
x
)
:
Wenowlet
g
i
=
f
i

b
i
.
Observethatwehave
j
g
i
(
x
)
j
2
d

1
=k
;
54
aswellas
j

i
j
=
X
R
2R
i
j
R
j


1
=k
:
=
[
k
i
=1

i
,thenwehave
jf
x
2
P
0
:
A
m
P
0

~
f
(
x
)
>
gjj

j
+
jf
x
2
P
0
n
:
A
m
P
0

~
f
(
x
)
>
gj

k

1
=k
+
jf
x
2
P
0
n
:
A
m
P
0

~
f
(
x
)
>
gj
:
(3.22)
Toestimatethesecondtermobservethat
A
m
P
0

~
f
(
x
)=
A
m
P
0

(
~g
+
~
b
)(
x
)
=
A
m
P
0

~g
(
x
)+
2
k

1
X
j
=1
A
m
P
0

(
h
j
1
;:::;h
j
k
)(
x
)
;
wherethefunctions
h
j
i
areeither
g
i
or
b
i
and,furthermore,foreach1

j

2
k

1thereis
atleastone1

i

k
suchthat
h
j
i
=
b
i
.Fix
j
andlet
i
j
besuchthat
h
j
i
j
=
b
i
j
,then
A
m
P
0
(
h
j
1
;h
j
2
;:::;h
j
i
j
;:::;h
j
k
)(
x
)=
X
Q
2
D

m
(
P
0
)

Q

k
Y
i
=1
h
h
j
i
i
Q
(
m
)

1
Q
(
x
)
=
X
Q
2
D

m
(
P
0
)

Q
h
b
i
j
i
Q
(
m
)

Y
1

i

k;i
6
=
i
j
h
h
j
i
i
Q
(
m
)

1
Q
(
x
)
=
X
R
2R
i
j
X
Q
2
D

m
(
P
0
)

Q
h
b
R
i
j
i
Q
(
m
)

Y
1

i

k;i
6
=
i
j
h
h
j
i
i
Q
(
m
)

1
Q
(
x
)
=
X
R
2R
i
j
X
Q
2
D
>m
(
R
)

Q
h
b
R
i
j
i
Q
(
m
)

Y
1

i

k;i
6
=
i
j
h
h
j
i
i
Q
(
m
)

1
Q
(
x
)
:
Sowededucethat
A
m
P
0

(
h
j
1
;:::;h
j
k
)(
x
)=0forall
x=
2

i
j
.Withthisfactwecanseethat
thesecondtermin(
3.22
)isactuallyidenticalto
jf
x
2
P
0
n
:
A
m
P
0

~g
(
x
)
>
gj
:
55
Nowwecanusethe
L
2
boundasfollows:
jf
x
2
P
0
n
:
A
m
P
0

~g
(
x
)
>
gj
1

2
kA
m
P
0

~g
k
2
L
2
(
P
0
)

16

2
k
Y
i
=1
k
g
i
k
2
L
2
k
(
P
0
)

16

2
k
Y
i
=1

2
d

1
=k

2
k

1
k
k
g
i
k
1
=k
L
1
(
P
0
)
=
16

2
2
d
(2
k

1)

2

1
=k
=2
4+
d
(2
k

1)


1
=k
:
Puttingbothestimatestogetherwearriveat
jf
x
2
P
0
:
A
m
P
0

~
f
(
x
)
>
gj
2
5+
d
(2
k

1)


1
=k
whichyieldstheresultwith
C
W
=2
k
(5+
d
(2
k

1))
.
56
4
Ontheembeddingof
A
1
into
A
1
GuillermoRey
Submitted.
4.1Introduction
Thepurposeofthisarticleistogiveaquantitativeversionoftheclassicalembedding
betweenMuckenhouptclasses
A
1
,
!
A
1
:
(4.1)
Theclass
A
1
istobeallweights
w

0forwhich
Mw

Cw
forsome
C
,where
Mf
(
x
)=sup
P
3
x
1
j
P
j
Z
P
j
f
(
y
)
j
dy
istheuncenteredHardy-Littlewoodmaximaloperator(herethesupremumistakenover
cubeswithsidesparalleltothecoordinateaxes).
Theclass
A
1
istobeallweights
w

0forwhichthereexistsaconstant
C
andan
57
exponent
>
0suchthat
w
(
E
)
j
P
j

C

j
E
j
j
P
j


forallcubes
P
andallsubsets
E

P
.See[
8
]formoreequivalent
Itisawell-knownfactthateveryweightin
A
1
isalsoin
A
1
;herewegiveaquantitative
versionofthisembedding.
Wewillactuallyworkwithawiderclassofweights,thedyadic
A
p
weights.Tostatethe
result,letusawaytoquantifyexactlyhowaweightliesindyadic
A
1
.Let
P
beacube
in
R
d
,wethe
A
d
1
(
P
)characteristicofaweight
w

0tobe
[
w
]
A
d
1
(
P
)
:=esssup
x
2
P
M
dyadic
P
w
(
x
)
w
(
x
)
;
where
M
dyadic
P
isthedyadicmaximaloperatorlocalizedto
P
:
M
d
P
f
(
x
)=sup
R
2D
(
P
)
hj
f
ji
R
1
R
(
x
)
:
Herewearedenotingby
D
(
P
)thecollectionofalldyadicsubcubesof
P
,andtheaverage
ofafunction
f
overaset
E
by
h
f
i
E
:=
1
j
E
j
Z
E
f
(
x
)
dx:
Also,wedenotethecharacteristicfunctionofaset
E
by
1
E
.
Wethe(non-dyadic)
A
1
characteristicsimilarly:
[
w
]
A
1
(
P
)
:=esssup
x
2
P
M
P
w
(
x
)
w
(
x
)
;
where
M
P
istheuncenteredHardy-Littlewoodmaximaloperatorwherethecubesare
constrainedtolieinside
P
.
58
Theclassicalwaytoprove(
4.1
)proceedsbyusingthe
reverseolderinequality
of
Coifman-F[
3
](see[
13
]forarecentsharpreverseolderinequalityvalidinavery
generalcontext):foranyweight
w
2
A
p
wehave
h
w
q
i
P

C
h
w
i
q
P
;
forsomeexponent
q>
1dependingon
w
.Indeed,let
C
RH
bethebestconstantinthe
aboveinequality(whichwilldependon
q
andonhow
w
liesin
A
p
),then:
w
(
E
)=
Z
P
w
1
E


Z
P
w
q

1
=q
j
E
j
1
=q
0

C
1
=q
RH
w
(
P
)

j
E
j
j
P
j

1
=q
0
:
For(non-dyadic)
A
1
weightsthemostquantitativeversionofthereverseolderinequality
wasgivenby[
47
]indimensionone.Usingtheresultsof[
47
]oneobtains
w
(
E
)
w
(
P
)

a
a

1

j
E
j
j
P
j

1
a
[
w
]
A
1
(
P
)
forall
a>
1,soonecangetarbitrarilyclosetotheexponent
1
[
w
]
A
1
atthecostofa
multiplicativeconstant.Theresultsin[
47
]are,however,validonlyfornon-dyadic
A
p
weights,whichbehavemuchbetterintermsofsharpconstants;also[
47
]isvalidonlyin
dimension1.
In[
30
]A.Melasshowedthat,fordyadic
A
1
weights,onehas
D
(
M
dyadic
w
)
p
E
P

C
(
p;
[
w
]
A
d
1
)
h
w
i
p
P
;
59
forall
p
suchthat
1

p<
log(2
d
)
log

2
d

2
d

1
[
w
]
A
d
1

;
andwhere
C
(
p;
[
w
]
A
d
1
)isaconstantwhichblows-upas
p
tendstotheendpointabove.
Followingthesamestepsasbefore,thisimpliesaninequalityoftheform
w
(
E
)
w
(
P
)

C


j
E
j
j
P
j


forall

suchthat
0

<

log

1

2
d

1
2
d
[
w
]
A
d
1

d
log2
:=

([
w
]
A
d
1
;d
)
;
andwhere
C

isaconstantwhichblows-upas

tendstotheendpoint

([
w
]
A
d
1
;d
).
Itwasofinterestwhetheronecouldachieveanestimatewiththeendpoint

([
w
]
A
d
1
;d
),and
thiswasansweredpositivelybyA.Os˘ekowskiin[
41
],whereheprovedthefollowing
weak-typeestimate:
1
j
P
j



n
x
2
P
:
M
dyadic
w
(
x
)
>
1
o



h
w
i
p
P
(4.2)
forall
p
suchthat
1

p

log(2
d
)
log

2
d

2
d

1
[
w
]
A
d
1

:
Thisestimate,coupledwitholder'sinequalityforLorentzspacesyields
w
(
E
)
w
(
P
)

C

(
Q;d
)

j
E
j
j
P
j


(
Q;d
)
forallweights
w
with[
w
]
A
d
1

Q
,thussettlingtheendpointquestionofwhetheradecay
60
rateof(
j
E
j
=
j
P
j
)

(
Q;d
)
couldbeachieved.However,notethatolder'sinequalityforLorentz
spaces(whenusedinthisway)hasaconstantwhichexplodeswhen
p
!
1whichinthis
caseimpliesthattheconstant
C

(
Q;d
)
willblow-upas
Q
!1
.
Inthisarticleweimprovethisconclusionbydirectlycomputingthefunction
B
(
x;y;m
)=sup
w
(
E
)
j
P
j
;
wherethesupremumistakenoverallsets
E

P
with
j
E
j
=
j
P
j
=
x
,andalldyadic
A
1
weights
w
with[
w
]
A
d
1
(
P
)

Q
,
h
w
i
P
=
y
andessinf
z
2
P
w
(
z
)=
m
.
Theexpressionfor
B
isalittleinvolvedandwereferthereadertosection
4.3
foritsfull
form,butwecanalreadygiveanupperboundfor
B
(

;Q;
1):
B
(
x;Q;
1)

e
f
(
x
):=
Qx

(
Q;d
)
:
(4.3)
ThisshowsthatthedecayratededucedfromOs˘ekowski'sestimatecanbeachievedwitha
uniformconstantas
Q
!1
(notethattheconstant
Q
cancelswhenestimating
w
(
E
)
w
(
P
)
).
ObservealsothatthisrecoverstheresultofOs˘ekowskiwhenonetakes
w
insteadofits
maximalfunctionin(
4.2
),whichcanbeinterpretedasaweak-typereverseolder
inequality.Indeed,assumewithoutlossofgeneralitythat
j
P
j
=essinf
w
=1andlet
E

=
f
x
2
P
:
w
(
x
)
>
g
,thenourestimatewillshow(see(
4.12
))that
w
(
E

)

Q

w
(
P
)

1
Q

1

j
E

j
Q

1
w
(
P
)

1


(
Q;d
)
:
Sointegrating
w
overthissetyields

j
E

j
1


(
Q;d
)

(
h
w
i
P

1)
1


(
Q;d
)

Q
(
Q

1)

(
Q;d
)

h
w
i
P
:
61
Or,inotherwords,
k
w
k
L
p;
1

Z
P
w
(
x
)
dx
forthesame
p
'sasin(
4.2
).
However,thefunction
B
(

;Q;
1)is,surprisingly,slightlybetter.Indeedifwe
f
(
x
)=
B
(
x;Q;
1),thenourmainresultshowsthat
f
isthepiecewise-linearinterpolationof
thefunction
e
f
evaluatedatthepoints2

dk
for
k
2
N
.
Figure4.1Plotsof
f
and
e
f
In
Figure4.1
weshowanormalizedsectionoftheplot(thevaluesaredividedby
Q
)ofthe
functions
f
and
e
f
with
Q
=10andindimensiontwo.
4.1.1Organization
Thearticleisorganizedasfollows:insection
4.2
wecasttheproblemasoneofa
certainBellmanfunction,theninsection
4.3
wegivealowerboundfortheBellman
function;wealsodescribethestructureofthemaximizers.Insection
4.4
weshowthatthe
lowerboundfoundintheprevioussectionisalsoanupperbound,henceshowingthatthe
functionfoundistheactualBellmanfunction.
62
4.2TheBellmanfunctionapproach
asintheintroduction,thefunction
B
(
x;y;m
)=sup
n
w
(
E
)
j
P
j
:
E

P;
[
w
]
A
d
1
(
P
)

Q
suchthat
j
E
j
=
x
j
P
j
;
h
w
i
P
=
y;m
=essinf
w
o
:
Bytranslationanddilationinvariance,thefunction
B
isindependentof
P
.
Thedomain,whichwillbedenotedby
B
is:
0

x

1
0
<m

y

Qm:
Inthissectionwecastnding
B
asaminimizationproblem.WewillfollowtheBellman
functionmethod,seeforexample[
36
],[
47
]or[
45
],and[
41
]foranapproachclosertoours.
Thefunction
B
thefollowing
MainInequality
B
(
x;y;m
)

D
B
(
x
i
;y
i
;m
i
)
E
;
(4.4)
where
h
x
i
i
=
x
,
h
y
i
i
=
y
,min
m
i
=
m
,(
x
i
;y
i
;m
i
)
2
and(
x;y;m
)
2
Ininequality
(
4.4
),andfortherestofthearticle,weusethenotation
h
˘
i
i
:=
1
n
n
X
i
=1
˘
i
;
whenever
f
˘
i
g
isadiscretesequenceof
n
numbers;usually
n
willbeobviousfromthe
contextsowewillomititsdependence.
Wecansee(
4.4
)bycombiningalmost-extremizersonthedyadic
subcubesof
P
intooneonthewholecube
P
.
63
Wealsohavethe
obstaclecondition
B
(1
;y;y
)=
y;
whichisjusttheobservationthatif
E
=
P
almosteverywhere,then
h
1
E
w
i
P
=
h
w
i
P
.
Fromtheof
B
wehavethehomogeneityproperty
B
(
x;;
)=

B
(
x;y;m
)
:
(4.5)
Ifwendanonnegativefunction
B
in
B
andwhichthemaininequality
andtheobstacleconditionabove,then
B

B
.Thisisatypicalfactwhoseproofweomit,
butthereadercanconsult[
41
]foraproofinasimilarcase.
Thehomogeneityconditionwillletusassumethat
m
=1in(
4.4
):
Proposition4.1.
Ifafunction
B
don

B
themaininequality
(
4.4
)
with
m
=1
andthehomogeneityproperty
(
4.5
)
,thenitmustalsosatisfythemaininequalityfor
all
m>
0
.
Proof.
Thisisjusttheobservationthatthedomainof
B
isinvariantundersimultaneous
dilationsofthevariables
y
and
m
.
Wewanttoasetofnecessaryandtconditionsfor
B
tosatisfythemain
inequality,butwhicharesimplertoverify.Tothisend,letusprove
necessary
conditionsthatanysuch
B
mustsatisfy.
ThefollowingLemmaissimplebutimportantinwhatfollows.Ittellsusthat,inorderto
exploit(
4.4
),weshouldstrivetominimizethevariables
m
i
asmuchaspossible.Wewilllet
N
:=2
d
fortherestofthearticle.
Lemma4.2.
Anyfunction
B
satisfying
(
4.4
)
isdecreasingin
m
.Moreprecisely:assume
(
x;y;m
1
)
and
(
x;y;m
2
)
aretwopointsin

B
with
m
1

m
2
,then
B
(
x;y;m
1
)

B
(
x;y;m
2
)
:
(4.6)
64
Proof.
Let
x
i
=
x
and
y
i
=
y
forall1

i

2
d
:=
N
.Also,let
e
m
i
=
8
>
>
<
>
>
:
m
1
if
i
=1
m
2
if
i>
1
:
Thenthepoints(
x
i
;y
i
;
e
m
i
)areallinAlso,
h
x
i
i
=
x
and
h
y
i
i
=
y
.Since
m
1

m
2
wealso
havethatmin(
e
m
i
)=
m
1
,sousing(
4.4
):
B
(
x;y;m
1
)

1
N
B
(
x;y;m
1
)+
N

1
N
B
(
x;y;m
2
)
;
whichafterrearrangingyields(
4.6
).
ThefollowingLemmafollowsdirectlyfromthemaininequality(
4.4
).
Lemma4.3.
Foranyd
m>
0
,thefunction
(
x;y
)
7!
B
(
x;y;m
)
isconcave.
Proof.
Thisisjusttheobservationthatthedomainisconvex,togetherwith(
4.4
)with
m
i
=
m
.
Nowweareabletomakethereductionin(
4.4
)(afterthetrivialoneofsetting
m
=1):
Proposition4.4.
Suppose
B
isanonnegativefunctiondin

B
andwhich
theobstaclecondition,
(
4.5
)
,and
(
4.6
)
.If
B

B
(
x;y;
1)

D
B

x
i
;y
i
;
max

1
;
y
i
Q

(4.7)
forall
N
-tuplesofpoints
(
x
i
;y
i
)
satisfying
0

x
i

1
;
and
h
x
i
i
=
x;
(4.8)
1

y
i
;
min(
y
i
)

Q;
and
h
y
i
i
=
y;
(4.9)
thenwemusthavethat
B
=
B
.
65
Proof.
Theaboveconditionsmake(
4.7
)certainlynecessary.Toseethatitist,take
any
N
-tuple(
x
i
;y
i
;m
i
)ofpointsin
B
satisfying
h
x
i
i
=
x;
h
y
i
i
=
y
andmin(
m
i
)=1
:
Considernowthealternative
N
-tupleformedby(
x
i
;y
i
;
e
m
i
),where
e
m
i
=
8
>
>
<
>
>
:
y
i
Q
if
y
i

Q
1otherwise
:
=max

1
;
y
i
Q

:
Thesepointsallliein
B
andmoreovertheystillsatisfythecondition
min(
e
m
i
)=1
:
However,byinequality(
4.6
)wehave
B

x
i
;y
i
;
max

1
;
y
i
Q


B
(
x
i
;y
i
;m
i
)
:
Thispropositionisusefulbecauseitallowsusto\almost"eliminatethethirdvariablefrom
ouranalysis.Thereasonthatweusedtheword\almost"comesfromthefactthatwestill
havetheextraneousconditionthatmin(
y
i
)

Q
,whichisanofhavingmin(
m
i
)=1.
Wenowproceedtoeliminatethisconditiontoo.
Supposethatofthe
N
points(
x
i
;y
i
),thereareexactly
N

k
ofthemforwhich
y
i

Q
.
Then,afterpossiblyreorderingtheinequality(whichwecandowithoutlossofgenerality),
66
therighthandsideof(
4.7
)becomes
1
N

k
X
i
=1
B
(
x
i
;y
i
;
1)+
N
X
i
=
k
+1
B

x
i
;y
i
;
max

y
i
Q

whichcanbewritten,afterapplyingthehomogeneityproperty(
4.5
),as
1
N

k
X
i
=1
B
(
x
i
;y
i
;
1)+
N
X
i
=
k
+1
y
i
Q
B
(
x
i
;Q;
1)

:
So,verifying(
4.7
)reducestojustshowingthat
B
isconcavein(
x;y
),decreasingin
m
,and
thatforeach1

k

N

1
B
(
x;y;
1)

1
N

k
X
i
=1
B
(
x
i
;y
i
;
1)+
N
X
i
=
k
+1
y
i
Q
B
(
x
i
;Q;
1)

(4.10)
forall(
x;y
)andall(
x
i
;y
i
)asinProposition(
4.4
),withtheadditionalassumptionthat
y
i

Q
for
k

k
+1.
Thenextpropositionallowsustojustconsiderthecasewhere
k
=
N

1intheabove
inequality.
Proposition4.5.
Let
M
beanonnegativefunctiondon

andwhichthat
1.
M
isconcave.
2.
Thefunction
t
7!
tM
(
x;y=t
)
isdecreasing.
3.
Forall
(
x;y
)
andall
(
e
x;
e
y
)
in

wehave
M
(
x;y
)

N

1
N
M
(
e
x;
e
y
)+
Ny

(
N

1)
e
y
QN
M
(
Nx

(
N

1)
e
x;Q
)(4.11)
whenever
Nx

(
N

1)
e
x

0
and
Ny

(
N

1)
e
y

Q
.
Then,
B
byhomogeneityasin
(
4.5
)
:
B
(
x;y;m
)=
mM
(
x;y=m
)
;
67
yieldsafunctionwhichtheconditionsofProposition
4.4
Proof.
Firstofallnotethat,bytheabovediscussion,wejustneedto
M
satisfyingthe
conditions(1),(2)and
M
(
x;y
)

1
N

k
X
i
=1
M
(
x
i
;y
i
;
1)+
N
X
i
=
k
+1
y
i
Q
M
(
x
i
;Q;
1)

;
wheretheaverageof
x
i
is
x
,theaverageof
y
i
is
y
andall
y
i

Q
for
i

k
+1.
Also,notethat(
4.11
)isjustthecaseof(
4.7
)with
k
=
N

1.So,inwhatfollowsweassume
k<N

1.
Fixallpoints(
x
i
;y
i
)for
i

k
andconsiderthecollection
V
ofallvectors
~y
=(
y
k
+1
;:::;y
N
)
with
y
i

Q
for
k

k
+1andsatisfying.
1
N
N
X
i
=
k
+1
y
i
+
1
N
k
X
i
=1
y
i
=
y:
Wecanwritethisconditionas
b
y
:=
1
N

K
N
X
i
=
k
+1
y
i
=
Ny

P
k
i
=1
y
i
N

k
=
Ny

k
e
y
N

k
;
wherewehave
e
y
=
1
k
P
k
i
=1
y
i
.
Itisaneasyexercisetoverifythat
1
N
N
X
i
=
k
+1
y
i
Q
M
(
x
i
;Q
)

1
QN
N
X
i
=
k
+1
b
i
M
(
x
i
;Q
)
;
where
b
i
areby
b
i
=
8
>
>
<
>
>
:
Q
if
i
6
=
i
max
(
N

k
)
b
y

Q
(
N

k

1)if
i
=
i
max
;
andwhere
i
max
istobetheindexwhichmaximizes
M
(
x
i
;Q
)for
i

k
+1.
68
Observethatthevector(
b
k
+1
;:::;b
N
)
2V
,sowecanassumethat
y
i
=
b
i
for
i

k
+1.But
then,wecanreorganizetheinequalitytoputallofthetermsexceptone(theonewith
i
max
)
onthesummation.Writingitthiswaymakesitevidentthatitreallywasaparticular
exampleoftheinequalitywith
k
=
N

1.
4.3FindingtheBellmanfunction
Inthissectionwegivealowerbound
M
for
M
,andinthenextsectionwewillshowthat
thislowerboundisalsoanupperboundandhencethat
M
=
M
.
Firstrecallthat
t
7!
M
(
x;y=t
)
isnon-increasingandthereforethat
M
(1
;y
)

y
(hereweareusingtheobstacle
M
(1
;
1)=1.Since
M
(0
;
1)

0,wenowcanextendthisboundtothesubdomain
y

1+(
Q

1)
x
toget:
M
(
x;y
)

x
+
y

1
8
(
x;y
)
2
:
y

1+(
Q

1)
x:
Wewillnowgivealowerboundfor
M
intherestofthedomain.Theideaistouse
inequality(
4.11
)settingthenumber
Nx

(
N

1)
e
x
tobeaslargeaspossible,withinthe
domainthatweknow,andtheniterate.
Morepreciselylet
x
0
=1,observethatif
Nx

(
N

1)
e
x
=
x
0
,then
e
x
=
Nx

x
0
N

1
:
Clearlyweneed
x

1
=N
for
e
x
tobeinthedomain,soweset
x
=
1
N
.Wewillalsomake
e
y
assmallaspossible,whichmeans
e
y
=1.
69
Puttingitalltogetherweobtain,using(
4.11
)with
x
=
1
N
and
y
=
Q
:
M

1
N
;Q


NQ

(
N

1)
NQ
M
(
x
0
;Q
)=
Q

1

N

1
NQ

:
Nowweiteratethisprocedure.Set
x
=
x
k
+1
=
x
k
N
,
y
=
Q
,
e
y
=1and
e
x
=0,then(
4.11
)
gives
M
(
x
k
+1
;Q
)

(1

N

1
NQ

M
(
x
k
;Q
)
;
so
M
(
N

k
;Q
)

Q

1

N

1
NQ

k
:
Between
x
k
+1
and
x
k
weknowthat
M
(

;Q
)isconcave,so
M
mustcertainlybeatleast
linearintheseintervals.Now,since
M
(0
;
1)

0,wecanalsoextendthisboundby
homogeneityandgettheupperbound
M
(
x;y
)

y

1
Q

1
M

x
Q

1
y

1
;Q


y

1
Q

1
f

x
Q

1
y

1

for
y

1

x
.Here,
f
isthepiecewiselinearfunctionon[0
;
1]bylinearly
interpolatingthepoints
f
(
x
k
)=
Q

1

N

1
NQ

k
between
x
k
+1
and
x
k
,
Figure4.1
showswhat
f
typicallylookslike.
Puttingitalltogether,weget
M
(
x;y
)

8
>
>
<
>
>
:
x
+
y

1if
y

1+(
Q

1)
x
y

1
Q

1
f

x
Q

1
y

1

if
y

1+(
Q

1)
x:
)
=:
M
(
x;y
)
:
(4.12)
70
Thewayweprovedtheseboundsalsoshowshowonewouldconstructpairsofweights
w
andsets
E
showingthat
M
isatleastthepromisedlowerbound.Wenowgiveadetailed
descriptionoftheseexamples.
4.3.1Explicitextremizers
Let'sstartwithexamplescorrespondingtotheline(1
;y
)with
y
2
[1
;Q
].Togetthebound
M
(1
;y
)

y
weusedthemaininequalitykeepingalltheparametersexceptoneofthe
m
i
's.Soletusrepeattheproof,butnowwithactualweights.Fixacube
P
andlet
P
1
;:::P
N
beitsdyadicchildren.
w
i
(
x
)=1forall
i
andall
x
2
P
i
exceptfor
i
=
N
,
forwhichwene
w
i
(
x
)=1+
N
(
y

1)forall
x
2
P
N
.Now
w
(
x
)=
w
i
(
x
)forall
x
2
P
i
;clearlyessinf
x
2
P
w
(
x
)=1and
h
w
i
P
=
y
.Now,since
x
=1,weshouldset
E
=
P
.
Thepair(
w;E
)isclearlycontainedinthesupremumintheof
B
(1
;y;
1)=
M
(1
;y
)andso
M
(1
;y
)

w
(
P
)
j
P
j
=
y
(4.13)
forthisparticularchoiceof
w
.Ofcourse,anyweightwith
h
w
i
P
=
y
wouldalsohavebeen
tsince
x
=1.
Examplesforweightsandsetscorrespondingtopoints(
x;y
)ontherestofthedomainare
morecomplicated.Wewillstartbyconstructingexamplesalongtheline
y
=
Q
.
Thewayweprovedthat
M
(
1
N
;Q
)

Q
(1

N

1
NQ
)wasbyusing(
4.11
)with
e
x
=0,
e
y
=1,
x
=
1
N
and
y
=
Q
.Similarly,wegotthebound
M
(
x
k
+1
;Q
)

(1

N

1
NQ
)
M
(
x
k
;Q
)byusing
(
4.11
)with
e
x
=0,
e
y
=1,
x
=
1
N
k
+1
and
y
=
Q
.Lookingbackathowwegot(
4.11
),wesee
thatwecombined
N

1trivialweight-setpairs(thepairs(
w

1
;E
=
;
))withanexample
comingfrom
B

1
N
k
;N
(
Q

1)+1
;N

N

1
Q

:
71
Wethenusedhomogeneitytotranslatethistoanexamplewhichwouldextremize
M

1
N
k
;Q

;
buthavinglostafactorslightlylargerthanone.
Wecantracebackthesestepswiththefollowinglemma:
Lemma4.6.
Let
P
beacubein
R
d
.Givenapair
(
w;E
)
where
w
isadyadic
A
1
weightwith
[
w
]
A
1

Q
andwith
h
w
i
P
=
Q
,
essinf
z
2
P
w
(
z
)=1
,and
h
1
E
i
P
=
x
,thereexistsapair
(
e
w;
e
E
)
where
e
w
isanotherdyadic
A
1
weightwith
[
w
]
A
d
1

Q
andwith
h
e
w
i
P
,
essinf
z
2
P
e
w
(
z
)=1
,
and
h
1
E
i
P
=
x=N
forwhich
e
w
(
e
E
)
j
P
j


1

N

1
NQ

w
(
E
)
j
P
j
:
Moreover,theset
e
E
isentirelycontainedinoneofthedyadicsubcubesof
P
and
e
w
is
identically
1
onthecomplementof
e
E
.
Proof.
Asbefore,enumeratethechildrenof
P
by
P
1
;:::;P
N
.Westartbytranslatingand
dilating(
w;E
)tothesubcube
P
1
,wedothiswiththeobviouslinearchangeofvariables.
Wethenmultiplytheweightwejustconstructedbytheconstant
NQ

(
N

1)
Q
.Letuscallthis
newweight
w
1
.Clearlyessinf
z
2
P
1
w
1
(
z
)=
NQ

(
N

1)
Q

1and
h
w
1
i
P
1
=
NQ

(
N

1).Now

w
i
(
z
)=1forall
z
2
P
i
andeach
i

2andcombinealloftheseweightsintoone:
e
w
(
z
)=
w
i
(
z
),forall
z
2
P
i
.Thisnewweightisadyadic
A
1
weightwith[
e
w
]
A
d
1

Q
.
With
E
wedothesame:wetranslateanddilateitto
P
1
;letuscallthisnewset
E
1
.This
newsethas
h
1
E
1
i
=
x
.
e
E
tobejust
E
1
(so
1
e
E
(
z
)=
1
E
1
(
z
)).
Weassertthatthisnewpair(
e
w;
e
E
)thepromisedestimate.Indeed(assumingwith-
72
outlossofgeneralitythat
j
P
j
=1):
e
w
(
e
E
)=
1
N

(
N

1)
w
2
(
e
E
)+
w
1
(
e
E
)

=
1
N
w
1
(
e
E
)
=

1

N

1
NQ

w
(
E
)
;
whichiswhatwewanted.
Givenacube
P
andapair(
w;E
)asinLemma
4.6
,we
T
(
w
)=
e
w;
where
e
w
istheweightconstructedintheproofofLemma
4.6
.Similarly,we
S
(
E
)=
e
E
.
Withthislemmaathandwecannowdescribethestructureoftheexampleswhichshow
that
M
(
N

k
;Q
)

Q
(1

N

1
NQ
)
k
.
Lemma4.7.
Let
P
beanycubeandlet
w
0
betheweightconstructedwhenproving
(
4.13
)
(butanyweightwith
h
w
0
i
P
=
Q
,
essinf
z
2
P
w
0
(
z
)=1
,andwith
[
w
]
A
d
1
=
Q
willworkaswell).
theweights
w
k
andthesets
E
k
inductivelyby
w
k
+1
=
Tw
k
and
E
k
+1
=
SE
k
;
where
E
0
=
P
.
Then
w
k
isan
A
d
1
weightwith
[
w
]
A
d
1
=
Q
,
h
w
k
i
P
=
Q
,
essinf
z
2
P
w
k
(
z
)=1
,
h
1
E
k
i
P
=
N

k
and
w
k
(
E
k
)
j
P
j
=
Q

1

N

1
NQ

k
:
Proof.
TheproofisjusttoiterativelyapplyLemma
4.6
.
Itremainstoextendtheexamplestotherestofthedomain.Butrecallthattheboundwe
73
gavefor
M
ontherestofthedomainwasobtainedbylinearinterpolation,sowejustneed
tocombineexamplesthathavealreadybeenconstructed.
Thefollowinglemmashowshowtocombinetwopairs(
w
0
;E
0
)and(
w
1
;E
1
)intoone:
Lemma4.8.
Let
P
beacubeandlet
(
w
0
;E
0
)
and
(
w
1
;E
1
)
betwopairs.Assume
w
0
and
w
1
arebothdyadic
A
1
weightswith
[
w
i
]
A
d
1

Q
,andalso:
h
1
E
i
i
P
=
x
i
;
h
w
i
i
P
=
y
i
;
essinf
z
2
P
w
i
(
z
)=1
:
Then,forany

2
[0
;
1]
wecanconstructapair
C

((
w
0
;E
1
)
;
(
w
1
;E
1
))=(
w;E
)
,where
w
is
adyadic
A
1
weightwith
[
w
]
A
d
1

Q
,
h
1
E
i
P
=
x;
h
w
i
P
=
y;
essinf
z
2
P
w
(
z
)=1
;
and
w
(
E
)
j
P
j
=(1


)
w
0
(
E
0
)
j
P
j
+

w
1
(
E
1
)
j
P
j
;
where
x
=(1


)
x
0
+

1
and
y
=(1


)
y
0
+

1
:
Proof.
Notethat,atleastwhen

isadyadicrational,repeatedapplicationsoftheMain
Inequalitygiveexactlythesedynamics.SoweshouldfollowtheproofoftheMainInequality,
whosemeaningistoshowwhathappenswhenonecombinespairs(
w
i
;E
i
)onthe
dyadicchildrenofacubeintoonepair(
w;E
)onthewholecube.
Thereisaslighttechnicality:ifoneappliesthiscombinationprocedureanumberof
times,onecanonlyprovethislemmainthecasewhere

isadyadicrational,butwecan
stillprovethislemmawithalimitingargument.
Let
b
i
bethedigitsof

whenwritteninbinary:

=
1
X
i
=1
b
i
2

i
74
(itdoesnotmatterwhichofthepossiblebinaryrepresentationsoneuses).
Fixthecube
P
andlet
R
beanyofitsdyadicsubcubes.
S
P
!
R
tobethelinearchange
ofvariableswhichmaps
P
to
R
.
Givenacube
P
let
P
1
;:::P
N
beaenumerationofitschildren,this
orderingwillbethroughouttheproof(inthesensethatwewillusethesameordering
oneveryothercube,whichweobtainbytranslatinganddilatingtheoriginalordering).
Theideaistosplitthesubcubesof
P
andonhalfofthemputatranslatedanddilatedcopy
ofeither(
w
0
;E
0
)or(
w
1
;E
1
),dependingonthebinarydigitofthecurrentstep.Weapply
thesameprocedureoneachoftheremainingcubes(butnowwiththenextdigit).
Moreprecisely,letch(
P
)bethedyadicsubcubesof
P
and
H
1

(
P
)to
bethesubsetofch(
P
)consistingoftheorsecondhalfthedyadicchildren,i.e.:
H
1

(
P
)=
f
P
1
;:::;P
2
d

1
g
and
H
1
+
(
P
)=
f
P
2
d

1
+1
;:::;P
2
d
g
:
Weinductively
H
j
+1

(
P
)asfollows:
H
j
+1

(
P
)=
[
R
2H
j
+
(
P
)
H

(
R
)
:
Wetheweight
w
by
w
(
x
)=
1
X
j
=1
X
R
2H
j

(
P
)

(1

b
j
)
S
P
!
R
w
0
(
x
)+
b
j
S
P
!
R
w
1
(
x
)

:
Similarly,wetheset
E
by
1
E
(
x
)=
1
X
j
=1
X
R
2H
j

(
P
)

(1

b
j
)
S
P
!
R
1
E
0
(
x
)+
b
j
S
P
!
R
1
E
1
(
x
)

:
Onecannowcheckthatthispairtherequiredproperties;see[
43
]foraverysimilar
construction.
75
WiththisLemma,wecannowexpressthestructureoftheexamplesontheline
y
=
Q
of
whichliebetweenthepointswithcoordinates
x
=
N

k
.Indeed,let(
w
k
;E
k
)bethe
weight-setpairconstructedbyLemma
4.7
.Thenforany
x
2
(
N

k

1
;N

k
)wehave
(
w
x
;E
x
):=
C

((
w
k
+1
;E
k
+1
)
;
(
w
k
;E
k
))
;
where
x
=(1


)
N

k

1
+


k
:
Toextendtotherestoflet(
x;y
)
2
with
y<Q
.Firstassumethat
y

1+(
Q

1)
x
;
thenweshouldusethepreviousLemmawithboundaryon
x
=1.Indeedlet
(
w;E
)=
C

((
1
;
;
)
;
(
w
y
;P
))
;
where

=1+
y

1
x
andwhere
w
y
isanydyadic
A
1
weightwith[
w
]
A
d
1

Q
,
h
w
y
i
P
=
y
and
essinf
z
2
P
w
(
z
)=1.Thispairclearlysatisesalltherequiredestimates.
Nowsupposethat
y

1+(
Q

1)
x
andlet(
w

;E

)bethepairwejustconstructedonthe
line
y
=
Q
with
x
-coordinate
x
Q

1
y

1
.Then
(
w;E
)=
C


(
1
;
;
)
;
(
w

;E

)

;
with

=
x
Q

1
y

1
alsoalltherequiredestimates.
4.4VerifyingtheMainInequality
Wenowhavetoshowthatthefunction
M
thatwefoundintheprevioussection
alltherequiredconditionswhich,werecall,are:
1.
M
isconcave.
76
2.
Thefunction
t
7!
tM
(
x;y=t
)isnonincreasing.
3.
Forall(
x;y
)
2
andall(
e
x;
e
y
)inwith
e
x

x
and
Ny

(
N

1)
e
y

Q
,wehave
M
(
x;y
)

N

1
N
M
(
e
x;
e
y
)+
Ny

(
N

1)
e
y
NQ
M
(
Nx

(
N

1)
e
x;Q
)
:
(4.14)
Itwillbeconvenienttoexaminethefunction
f
,inparticularobservethat
f
0
(
x
)=(
N
)
k
;
where

=1

N

1
NQ
,whenever
x
2
(
N

k

1
;N

k
).
Theratio
N>
1whenever
Q>
(
N

1)
=N
,whichisalwaysthecasesince
Q

1,hence
f
isconcave.Since
f
isconcave,itfollowsthat
M
mustalsobeconcave,since
M
isjustthe
extensionof
f
byhomogeneitytothesubdomainofwhichliesabovethediagonal
y
=1+(
Q

1)
x
,andbelowthislinethefunctionisjusttheplane
z
=
x
+
y

1.Abrief
checknowshowsthat
M
isindeedconcaveinThisproves(1).
Nowwewillshowthatthefunction
t
7!
tM
(
x;y=t
)
isdecreasing,thusproving(2).
Toshowthis,notethatwejustneedtoprove
yM
y

M
wherever
M
istiable.This
obviouslyholdsfor
y<
1+(
Q

1)
x
,soittoassume
y>
1+(
Q

1)
x
.By
homogeneity,wecantranslatethisconditiontoonefor
f
:
y
Q

1
f

x
Q

1
y

1


xy
y

1
f
0

x
Q

1
y

1


y

1
Q

1
f

x
Q

1
y

1

:
Let
u
=
x
Q

1
y

1
,thenthisinequalitybecomes
1
u
f
(
u
)

yf
0
(
u
)

0
77
forall
u
2
[0
;
1]andall
y
2
[1
;Q
].Since
f
isincreasing,thisinequalityisstrongestwhen
y
=
Q
,soittoshow
f
(
u
)

Quf
0
(
u
)
:
Recallthat
f
ispiecewiselinear,solet
u
0
=
N

k

1
and
u
1
=
N

k
andassume
u
2
(
u
0
;u
1
).
Theaboveinequalitynowbecomes
f
(
u
0
)+(
u

u
0
)
f
0
(
u
0
+)

Quf
0
(
u
0
+)
:
Thus,wecanreducetoshowing
f
(
u
0
)
f
0
(
u
0
+)

u
0
+(
Q

1)
u
1
:
Butaneasycomputation,usingthevalueof
f
0
computedbefore,yieldsthatthisinequality
isequivalentto


1

N

1
NQ
;
whichispreciselythevalueof

sowearedone.Thisshows(2).
Finally,weareleftwithverifying(3).Todothiswewillconstructasequenceoffunctions
M
k
onallofwhichsatisfy(3)onaspsubsetof

k
=
f
(
x;y
)
2
:
y

1+(
Q

1)
N
k
x
g
:
Figure4.2
representsthethreeofthesedomains(again,thediagramisnottoscale).
Forexample
2
isthesubdomainofwhichliestotherightofthelinejoining
O
and
C
.
78
Figure4.2Domains
k
We
M
k
tobethewedgeformedbythe
k
-thplaneof
M
on
n

k

1
andthe
(
k

1)-thplaneof
M
on
k

1
,thatis:
M
k
(
x;y
)=
8
>
>
<
>
>
:
a
k
x
+
b
k
(
y

1)if(
x;y
)
2

n

k

1
a
k

1
x
+
b
k

1
(
y

1)if(
x;y
)
2

k

1
:
where
M
(
x;y
)=
a
k
x
+
b
k
(
y

1)on
k
n

k

1
.Onecangivetheexplicitformulasfor
a
k
and
b
k
:
a
k
=(
N
)
k
;b
k
=

k
:
Obviously
M
0
(3).
Fixany(
x;y
)
2
wecanassumewithoutlossofgeneralitythat(
x;y
)
2

k
forsome
k
.
Introducethenotation
x
=
N

1
N
e
x
+
1
N
b
x
and
y
=
N

1
N
e
y
+
1
N
b
y:
79
Since
M
isconcave,wehavethat
M
k

M
on(
M
k
isa\supportingwedge"ofthegraph
of
M
).Insteadof(3)wewillprove(underthesamehypotheses)
M
k
(
x;y
)

N

1
N
M
k
(
e
x;
e
y
)+
1
N
b
y
Q
M
k
(
b
x;Q
)
;
(4.15)
which,bytheaboveremark,isastrongerstatement.
Wewillshowthatwecanassumethepoint(
b
x;Q
)tobein
k
.Indeed,supposethat
e
x
issosmallthat(
b
x;Q
)
=
2

k
,then
@
@
e
x

Righthandsideof(
4.15
)

=
N

1
N
a
k

N

1
N
b
y
Q
a
k

1
=

N

1
N

a
k

b
y
Q
a
k

1



N

1
N

a
k

Ny

(
N

1)
Q
a
k

1



N

1
N

a
k

NQ

(
N

1)
Q
a
k

1

:
Nowrecallthat
a
k
=(
N
)
k
,sothepartialderivativeoftherighthandsideofequation
(
4.15
)isatleast
N

1
N
(
N
)
k

1

N

NQ

(
N

1)
Q

=0
;
sotherighthandsideisincreasing,atleastaslongas(
b
x;Q
)
2

k

1
.
Thisallowsustoassumethat
e
x
islargeenoughtomake(
b
x;Q
)
2

k
(bycontinuity).
Underthisassumptiontheinequalitybecomesmucheasiersince
M
k
isnowbeingevaluated
alwayson
k
,andhencewecanassumethat
M
k
itselfisaplane.Nowitiseasytocheck
thattheinequalityisindeedtrueundertheseconditions.
Toseethis,observethatinequality(
4.15
)canbewrittenas:
ax
+
b
(
y

1)

N

1
N

a
e
x
+
b
(
e
y

1)

+
1
N
b
y
Q

a
b
x
+
b
(
Q

1)

:
80
Wecanreorganizethisas:
a

x

N

1
N
e
x

1
N
b
y
Q
b
x

+
b

y

1

N

1
N
e
y
+
N

1
N

1
N
b
y
Q
(
Q

1)


0
:
Thistoshowing
a

b
x
N

b
x
N
b
y
Q

+
b

b
y
NQ

1
N


0
;
whichisequivalentto

b
y
Q

1


b

a
b
x


0
:
Sincetheassumptionsforce
b
y
tobeatleast
Q
,wejustneedtocheckthat
b
x

b
a
.Butthis
isexactlytheboundthatisguaranteedfromtheconsiderationsabovesince
b
a
=
N

k
.
81
5
Borderlineweak-typeboundsfor
singularintegrals
CarlosDomingo-Salazar,MichaelLacey,andGuillermoRey
BulletinoftheLondonMathematicalSociety,October2015.
5.1Introduction
Thepurposeofthischapteristoshowsomeapplicaitonsofthetechniquesdevelopedso
far.Theresultsinthischapterusethepointwisedominationofsingularintegralsfrom
Chapter
3
,aswellasseveralfactsaboutsparsefamiliesandMuckenhouptweights.
Thetheoremisinthecontextoflinearsingularintegraloperators:
Theorem5.1.
Let
T
beaon-Zygmundoperatoron
R
d
and
w
an
A
1
weight,then
k
Tf
k
L
1
;
1
(
w
)
.
T;d
[
w
]
A
1

1+log[
w
]
A
1

k
f
k
L
1
(
w
)
:
Itisunknownwhetherthelogarithmictermissharp,butapowerisnecessary,see[
35
].
Wecanalsostateaverysimilartheoremforsquarefunctions:
82
Theorem5.2.
Let
G
beasquarefunctionasin
3.5
,then
k
Gf
k
L
2
;
1
(
w
)
.
G;d
p
[
w
]
A
2
(1+log[
w
]
A
1
)
k
f
k
L
2
(
w
)
:
Theorem
5.1
wasalreadyknown,seeforexample[
26
],butherewegiveanelementary
proofwhichusesthemachinerydevelopedinthepreviouschapters.Theorem
5.2
was
obtainedin[
7
].Sincetheproofsareverysimilar,herewejustproveTheorem
5.1
.The
prooffollowsthestepsin[
19
],asdevelopedin[
7
].
5.2Proof
Bythepointwisedominationofon-ZygmundoperatorsprovedinChapter
3
it
toprove
kA
S
f
k
L
1
;
1
(
w
)
.
d
[
w
]
A
1

1+log[
w
]
A
1

k
f
k
L
1
(
w
)
;
(5.1)
where
S
isasparsefamilyofcubes,and
f
isnonnegative.SeeChapter
4
forprecise
of
A
1
and
A
1
.
Afterpossiblysplittingthefamilyintoseveralsubfamilies,wecanassumethat
S
is
1
4
-sparse,thatis:


[
R
(
Q;R
2S
R



1
4
j
Q
j8
Q
2S
:
Now,byhomogeneity,ittoshow
w

x
:
A
S
f
(
x
)
>
3

.
d
[
w
]
A
1

1+log[
w
]
A
1

;
forallnonnegativefunctions
f
with
k
f
k
L
1
(
w
)
=1,andforallweights
w
.
83
Itwillbeconvenienttosplitthefamily
S
intobetter-behavedsubfamilies:
S
m
=
f
Q
2S
:2

m

1
<
h
f
i
Q

2

m
g
and
S

=
f
Q
2S
:
h
f
i
Q
>
1
g
:
Wehave
w

x
:
A
S
f
(
x
)
>
3


w

x
:
A
S

f
(
x
)
>
1

+
w

x
:
A
S
+
f
(
x
)
>
2

;
wherewehave
S
+
tobetheunionofallthefamilies
S
m
for
m

0.
Webeginestimatingtheterm.Notethat
A
S

f
issupportedon
f
x
:
M
d
f
(
x
)
>
1
g
,
where
M
d
isthedyadicmaximalfunction,so
w

x
:
A
S

f
(
x
)
>
1

k
M
d
k
L
1
(
w
)
!
L
1
;
1
(
w
)

[
w
]
A
1
;
sothisdealswiththesummand.
Thesecondsummandwillbesplitintotwo:
w

x
:
A
S
+
f
(
x
)
>
2


w

x
:
m
0

1
X
m
=0
A
S
m
f
(
x
)
>
1

+
w

x
:
1
X
m
=
m
0
A
S
m
f
(
x
)
>
1

=
I
+
II:
Thewaywethesubfamilies
S
m
givesusverygoodcontroloftheaveragesof
f
.
Indeed:
84
Lemma5.3.
Foreach
m
2
N

E
m
(
Q
)=
Q
n
[
R
(
Q;R
2S
m
R;
then
h
f
1
E
m
(
Q
)
i
Q
˘h
f
i
Q
(5.2)
forall
Q
2S
m
.
Proof.
Indeed:ifwelet
R
1
;R
2
;:::
bethemaximalsubcubesof
Q
en
S
m
then
1
j
Q
j
Z
Q
f
1
E
m
(
Q
)
=
h
f
i
Q

X
i
1
j
Q
j
Z
R
i
f
>
2

m

1

X
i
j
R
i
j
j
Q
j
1
j
R
i
j
Z
R
i
f

2

m

1

2

m
X
i
j
R
i
j
j
Q
j

2

m

1

2

m
1
4
&
h
f
i
Q
:
Thereasonthisisusefulisbecausethesets
f
E
m
(
Q
)
g
arepairwisedisjointwhen
Q
runs
85
over
S
m
.Wecanusethistodealwith
I
:
w

x
:
m
0

1
X
m
=0
A
S
m
f
(
x
)
>
1


Z
m
0

1
X
m
=0
A
S
m
f
(
x
)
w
(
x
)
dx
=
m
0

1
X
m
=0
Z
X
Q
2S
m
h
f
i
Q
1
Q
w
(
x
)
dx
.
m
0

1
X
m
=0
X
Q
2S
m
h
f
1
E
m
(
Q
)
i
Q
1
Q
(
x
)
w
(
x
)
dx
=
m
0

1
X
m
=0
X
Q
2S
m
Z
f
1
E
m
(
Q
)
w
(
Q
)
j
Q
j

[
w
]
A
1
m
0

1
X
m
=0
Z
fw
=
m
0
[
w
]
A
1
:
Finally,toestimate
II
,let
f
a
m
g
1
m
=
m
0
beasequenceofnonnegativenumberssuchthat
1
X
m
=
m
0
a
m
=1
:
Then
II
=
w

x
:
1
X
m
=
m
0
A
S
m
f
(
x
)
>
1
X
m
=
m
0
a
m


1
X
m
=
m
0
w

x
:
A
S
m
f
(
x
)
>a
m


1
X
m
=
m
0
w

x
:
X
Q
2S
m
h
f
i
Q
1
Q
(
x
)
>a
m


1
X
m
=
m
0
w

x
:
X
Q
2S
m
1
Q
(
x
)
>
2
m
a
m

:
Call
b
m
(
x
):=
X
Q
2S
m
1
Q
(
x
)
:
86
Since
S
m
issparse,thisfunctionisalmost,butnotquite,uniformlybounded;itisactually
inBMO.Infact,foreach
m
thereexistacollectionofmaximaldyadiccubes
Q
m
1
;Q
m
2
;
2S
m
suchthat
b
m
issupportedintheunionofthesecubesand
1
j
Q
m
i
j



n
x
2
Q
m
i
:
b
m
(
x
)
>
o




e

C
forall


1andall
i

1.
Nowwecanusethatevery
A
1
weightsisalsoin
A
1
(seealsoChapter
4
)toobtain
w

x
2
Q
m
i
:
b
m
(
x
)
>

w
(
Q
m
i
)

exp


c
[
w
]
A
1


:
Aftersummingin
i
,thisyields
w

x
:
b
m
(
x
)
>


exp


c
[
w
]
A
1


X
i
w
(
Q
m
i
)
:
Sinceallofthecubes
Q
m
i
arecontainedintheset
f
x
:
M
d
f
(
x
)
>
2

m
g
;
wecanusethe(weighted)weak-typeboundednessofthemaximalfunctiontogivethe
estimate
w

x
:
b
m
(
x
)
>


[
w
]
A
1
2
m
exp


c
[
w
]
A
1


:
Now,pluggingthisestimateback,wehave
II

[
w
]
A
1
1
X
m
=
m
0
2
m
exp


c
[
w
]
A
1
2
m
a
m

:
Weshouldchoose
a
m
sothatit\looses"against2
m
(sincewewantexponentialgrowth),
87
whilestillsummingto1.Apossiblechoiceis
a
m
=
˘

m
(1

˘

1
)
˘
m
0
forsome1
<˘<
2,likeforexample3
=
2.
Pluggingthisinthepreviousinequality,andestimatingthesumbyanintegralwe
obtain
II
.
[
w
]
A
1
Z
1
m
0
2
x
exp


c
[
w
]
A
1
2
x
˘

x
(1

˘

1
)
˘
m
0

dx:
Calling

=2
=˘
weseethat1
<<
2and:
II
.
[
w
]
A
1
Z
1
0
2
x
exp


c
0
[
w
]
A
1

x
˘
m
0

dx
.
[
w
]
A
1
Z
1
0
y

e

y
dy
y
.
[
w
]
A
1
providedwechoose
m
0
˘
log[
w
]
A
1
,andwhere
>
0.
Withthischoiceof
m
0
weobtain
w

A
S
f>
3

.
[
w
]
A
1
(1+log[
w
]
A
1
)
;
whichiswhatweneededtoprovethetheorem.
88
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