WEIGHTEDNORMINEQUALITIESFORCALDER

ON{ZYGMUND
OPERATORS
By
AleksandrB.Reznikov
ADISSERTATION
Submittedto
MichiganStateUniversity
inpartialentoftherequirements
forthedegreeof
Mathematics|DoctorofPhilosophy
2014
ABSTRACT
WEIGHTEDNORMINEQUALITIESFORCALDER

ON{ZYGMUND
OPERATORS
By
AleksandrB.Reznikov
Givenaon-Zygmundoperator
T
andtwoweights
u
and
v
,westudyt
conditionsforthisoperatortobeboundedfromthespace
L
p
(
u
)to
L
p
(
v
).Wealsostudy
sharpboundsforthecorrespondingnorm.Further,westudyaquestionaboutconditions
forboundednessofallon-Zygmundoperatorsfrom
L
p
(
u
)to
L
p
(
v
).Wedoitinthe
Euclidiansettingandinmetricspaces.
Finally,westudythelimitingcase
p
=1andthecase
u
=
v
,whentheoperatorhasa
chansetobeweaklybounded,i.e.boundedfromthespace
L
1
(
u
)tothespace
L
1
;
1
(
u
).
Inparticular,wedisprovethe\
A
1
conjecture",provethe\
A
2
conjecture"inthemetric
spacesetting,provethe\bumpconjecture"for
p
=2;moreover,westatethe\separated
bumpconjecture"andproveitinseveralparticularcases.
Tomymom,wifeanddaughter|threemostwonderfulwomen.
iii
ACKNOWLEDGMENTS
Thissectionisthemostpleasanttowrite.DuringmygraduatestudyatMSUIgot,besides
someniceresults,alotofgreatfriends,colleguesand,mostimportant,gotmarried.Though
someproblemsremainunsolvedandsomethingsremainundone,Icouldnotwishbettere
yearsformyself.
Firstofall,IwouldliketoexpressmyutmostgratitudetoAlexanderVolberg,who,in
theseeyears,becameformemorethanjustanacademicfather.Hishelpful,respectful
andfriendlyattitudetowardsgraduatestudentsareabsolutelyunique,andIneverfeltsorry
formychoiceoftheuniversityandoftheadvisor.Theproblemsthathegaveme,andthe
conferencesandworkshopsthatheencouragedmetoattend,wereextremelyusefulformy
mathematicalcareer.
IwouldalsoliketomentionotherMSUfacultyandthatmentoredandhelped
meduringtheseeyears.OursecretariesBarbaraMillerandLeslieAitcheson,graduate
directorsZhengfangZhouandCasimAbbas,professorsAaronLevin,SheldonNewhouse,
VladimirPeller,MichaelShapiro,IgnacioUriarte-Tuero,KirillVanninskii,JonWolfson,
DapengZhanandJaneZimmermanmademyacademiclifealotmorepleasant,challenging
andinteresing.
MyresearchwasalsogreatlybycolleguesoutsideMSU.Itwasagreatpleasure
tointeractwithDavidCruz-Uribe,PaulHagelstein,ViktorHavin,SteveAlex
Iosevich,CarlosKenig,MichaelLacey,FedorNazarov,CarlosPerez,StefaniePetermichl,
MariaReguera,SergeiTreil,CristophThiele,VasiliiVasyunin,IgorVerbitskii.
ItwouldbeapitynottomentionmycolleguesandfriendsAlyaBeznosova,MattBond,
NickBoros,LipingChen,AsyaDiop,MishaDubashinskii,PaataIvanishvili,MishaIvanov,
iv
BoryaKrylov,SeryozhaMatveenko,SeryozhaNikitin,NickPattakos,GuillermoRey,Asya
Slutskaya,SashaVallander,LiangminZhou.
IwouldliketothankVladimirShlapentokhfororganizingalotofinterestingseminars
andconversations.Theystimulatedmealot,andIdothinkthattheywereveryimportant
formydevelopment.Specialthankstoparticipantsofthesemeetings:LyubaandVladimir
Shlapentokhs,ViktoriyaandMarkDykmans,ZhannaandYuriyGandelsmans,Irinaand
MichaelShapiro,OlgaandAlexanderVolbergs.Thelevelofthesemeetingsisextremely
high,andtheywereinspiringbothfor\academic"and\real"life.
Finally,itismyutmostpleasuretothankforallsupportmyfamily:mygrandfather
GrishaMints,mygrandmotherInnaMints,mymotherAnnaReznikova,myparentsinlaw
TatianaandViktorShushkovs,andmywifeNataliaShushkova.
v
TABLEOFCONTENTS
Chapter1Preliminarynotation,andtheorems
..........
1
1.1Preliminarynotationandnitions......................1
1.2Someknowntheorems..............................7
Chapter2Twoweightestimates
..........................
9
2.1Mainresults....................................9
2.2Thestoppingtimeapproachtotheseparatedbumpconjecture........11
2.2.1ProofofTheorem2.2.4..........................23
2.2.2Prooffortheloglog-bumps.......................24
2.3TheBellmanfunctionapproachtothebumpconjecture............27
2.3.1Mainresult................................27
2.3.2Orlicznormsanddistributionfunctions.................27
2.3.2.1AlowerboundfortheOrlicznorm..............27
2.3.2.2Examples............................32
2.3.3Mainresultinnewlanguage.......................33
2.3.4Generalsetup...............................33
2.3.4.1Haarshifts...........................34
2.3.4.2Paraproducts..........................36
2.3.5Reductiontothemartingalecase.....................37
2.3.6Proofof(thetialEmbedding)Theorem2.3.6:Bellmanfunc-
tionandmaintialinequality...................41
2.3.7Maininequalityintheform..............45
2.3.7.1Dyadiccase...........................45
2.3.7.2Generalcase..........................46
2.3.8Proofof(theEmbedding)Theorem2.3.7................47
2.3.8.1Anauxiliaryfunction......................47
2.3.8.2Bellmanfunctionandthemaindtialinequality....48
2.3.9Finiteformofthemaininequality..............51
2.3.9.1Frommaininequality(2.48)toTheorem2.3.7........52
2.3.10Frommaininequality(2.43)toTheorem2.3.6.............56
2.4TheBellmanfunctionapproachtotheseparatedbumpconjecture......57
2.4.1Aquickreminder.............................57
2.4.2Aconstructionoffunction.......................58
2.4.3Themainresults.Boundednessandweakboundedness.........59
2.4.4Examplesofsatisfyingtherestrictionsofthemainresults:thecases
from[CURV]...............................61
2.4.5SelfimprovementsofOrlicznorms....................62
2.4.6Examples.................................64
2.4.6.1Example1:log-bumps.....................64
vi
2.4.6.2Example2:loglog-bumps...................64
2.4.7Proofofthemainresult:notationandthereduction.......65
2.4.8Bellmanproofof(2.55):introducingthe\maininequality"......66
2.4.9Fourthstep:buildingthefunction
B
2
..................74
2.4.10Fifthstep:buildingthefunction
B
1
...................78
Chapter3Oneweightestimateforthelimitingcase:the
A
1
conjecture
.
79
3.1Themainresult..................................79
3.2TheBellmanapproach..............................80
3.2.1Homogeneity...............................80
3.2.2Themaininequality...........................81
3.3Theunweightedestimate:theexactBellmanfunction.............85
3.3.1
B
>
B
...................................87
3.3.2
B
(
g;f;F
)
6
B
(
g;f;F
)..........................89
3.3.2.1Changeofvariables.......................92
3.3.2.2Theproofof
B
>
B
......................92
3.3.2.3Thecase
F
>
2.........................94
3.3.2.4Thecase
F
6
2.........................94
3.3.3Buildingtheextremalsequenseforpoints(1
;
1
;F
)...........95
3.3.3.1Thecase
F
>
2.........................101
3.3.3.2Thecase
F<
2.........................102
3.3.4HowtotheBellmanfunction
B
...................103
3.3.4.1Theboundary
F
=
y
1

y
2
..................104
3.3.4.2Thedomain.........................106
3.4Theweightedestimate..............................110
3.4.1tialpropertiesof
B
translatedtorentialpropertiesof
B
.110
3.4.2Obstacleconditionsfor
B
.........................125
Chapter4Oneweightestimatesinmetricspaces
................
126
4.1Mainresults....................................126
4.2Themainconstruction..............................127
4.2.1FirstStep.................................127
4.2.2Secondstep:technicallemmata.....................131
4.2.3Mainandtheorem.......................135
4.2.4Probabilitytobe\good"isthesameforeverycube..........138
4.3TheHaarshiftdecomposition..........................139
4.3.1Paraproducts...............................144
4.3.2Estimatesof
˙
1
..............................150
4.3.3Estimateof
in
..............................151
4.3.4Estimatesfor
out
............................153
4.4Therestoftheproof...............................156
4.4.1ParaproductsandBellmanfunction...................156
4.4.2WeightedestimatesfordyadicshiftsviaBellmanfunction.......158
vii
BIBLIOGRAPHY
..............................
170
viii
Chapter1
Preliminarynotation,and
theorems
1.1Preliminarynotationand
Webeginwithsomenotationandthatareneededthroughoutthisthesis.
1
(Weight)
.
Byaweightin
R
n
wecallafuntion
w
,whichispositivealmost
everywhereandlocallyintegrablewithrespecttotheLebesguemeasure
dx
.
2
on-Zygmundkernel)
.
Afunction
K
:
R
n

R
n
!
R
iscalledaon-
Zygmundkernelifthereexistpositivenumbers
C
and
"
suchthatthefollowingconditions
are
j
K
(
x;y
)
j
6
C
j
x

y
j
n
(1.1)
j
K
(
x;y
)

K
(
x
0
;y
)
j
6
C
j
x

x
0
j
"
j
x

y
j
n
+
"
if
j
x

x
0
j
<
1
2
j
x

y
j
(1.2)
j
K
(
x;y
)

K
(
x;y
0
)
j
6
C
j
y

y
0
j
"
j
x

y
j
n
+
"
if
j
y

y
0
j
<
1
2
j
x

y
j
:
(1.3)
3
on-ZygmundOperator)
.
Anoperator
T
iscalledaon-Zygmund
1
operator,ifthereexistsaon-ZygmundKernel
K
,suchthat
Forany
f
2
C
1
0
;
andany
x
62
supp
f
:
Tf
(
x
)=
Z
K
(
x;y
)
f
(
y
)
dy
(1.4)
T
isaboundedoperatorfrom
L
2
(
dx
)to
L
2
(
dx
)
:
(1.5)
Anexampleofaon-ZygmundoperatorindimensiononeisHilberttransform
H
withkernel
K
(
x;y
)=
1
x

y
.Anexampleofaon-Zygmundoperatorindimension
n
is
Rietztransformwithvector-valuedkernel
K
(
x;y
)=
x

y
j
x

y
j
n
+1
.
Notation1.
Foraset
Q
ˆ
R
n
andafunction
'
wedenote
h
'
i
Q
=
1
j
Q
j
Z
Q
'
(
x
)
dx;
where
j
Q
j
istheLebesguemeasureoftheset
Q
.
4
(The
A
p
class)
.
Let1
<p<
1
.Aweight
w
belongstoaclass
A
p
onacube
I
(where
I
isallowedtobeequalto
R
n
)ifthefollowingholds:
Foreverycube
J
ˆ
I
:
h
w
i
J
h
w

1
p

1
i
p

1
J
6
Q:
Thebestconstant
Q
iscalledthe
A
p
characteristic
of
w
andisdenotedby[
w
]
p
.
5
(The
RH
p
class)
.
Let1
<p<
1
.Aweight
w
belongstoaclass
RH
p
ona
cube
I
(where
I
isallowedtobeequalto
R
n
)ifthefollowingholds:
Foreverycube
J
ˆ
I
:
h
w
p
i
1
p
J
6
R
h
w
i
J
:
2
Thebestconstant
R
iscalledthe
RH
p
characteristic
of
w
andisdenotedby[
w
]
RH
p
.
Asonecansee,theseionsdonotworkfor
p
=1and
p
=
1
.However,ifwe
carefullyconsiderthelimitoftheleft-handside,wegetthefollowing.
6
(Thelimitingcases)
.
1.Aweight
w
belongstoaclass
A
1
onacube
I
(where
I
isallowedtobeequalto
R
n
)ifthefollowingholds:
Foreverycube
J
ˆ
I
:
h
w
i
J
6
Q
inf
J
w:
Thebestconstant
Q
iscalledthe
A
1
characteristic
of
w
andisdenotedby[
w
]
1
.
2.Aweight
w
belongstoaclass
A
1
onacube
I
(where
I
isallowedtobeequalto
R
n
)
ifthefollowingholds:
Foreverycube
J
ˆ
I
:
h
w
i
J
6
Qe
h
log(
w
)
i
J
:
Thebestconstant
Q
iscalledthe
A
1
characteristic
of
w
andisdenotedby[
w
]
1
.
3.Aweight
w
belongstoaclass
RH
1
onacube
I
(where
I
isallowedtobeequalto
R
n
)
ifthefollowingholds:
Foreverycube
J
ˆ
I
:
h
w
h
w
i
J
log
w
h
w
i
J
i
J
6
R:
Thebestconstant
R
iscalledthe
RH
1
characteristic
of
w
andisdenotedby[
w
]
RH
1
.
Nextwillconcernso-called\dyadicmodels"foron-Zygmundoperators.
Wemakeuseofthesemodelsforweightedestimatesoftheoperators.
3
Notation2.
If
I
isacubein
R
n
thenby
`
(
I
)wedenotethesidelengthof
I
.
7
(Dyadicgrid)
.
Adyadicgrid
D
isaunionofcollectionsofcubes
D
k
,such
that:
1.Fora
k
cubesin
D
k
donotintersect,andfor
I
2D
k
:
`
(
I
)=2

k
.
2.Foranycube
I
2D
k
+1
thereexistsauniquecube
J
2D
k
,suchthat
I
ˆ
J
.Inthis
situation
I
iscalledadyadicchildof
J
;
J
iscalledthedyadicfatherof
I
.
Elementsof
D
arecalleddyadiccubes.
8
(Carlesonsequense)
.
Fixadyadicgrin
D
.Asequense
f
a
I
g
I
2D
iscalled
C
-Carleson,ifforanyinterval
J
2D
thefollowinginequalityholds:
X
I
ˆ
J
a
I
j
I
j
6
C
j
J
j
:
9
(GeneralizedHaarfunction)
.
Givenadyadiccube
J
,
h
J
isa(generalized)
Haarfunctionassociatedtoacube
J
if
h
J
(
x
)=
X
I
2
ch
(
J
)
c
I
˜
J
(
x
)
;
where
ch
(
J
)isthesetofdyadicchildrenof
J
and
j
c
I
j
1.
10
(GeneralizedHaarshift)
.
Wesaythatanoperator
S
hasaHaarshiftkernel
ofcomplexity(
m;n
)if
Sf
(
x
)=
X
J
S
J
(
f
)
;
4
where
S
J
(
f
)=
1
j
J
j
X
I;I
0
ˆ
J
`
(
I
)=2

n
`
(
J
)
`
(
I
0
)=2

m
`
(
J
)
(
f;h
I
)
h
I
0
and
h
I
and
h
I
0
aregeneralizedHaarfunctionsassociatedtothecubes
I
and
I
0
respectively.
Wesaythat
S
isaHaarshiftofcomplexity(
m;n
)ifithasaHaarshiftkernelofcomplexity
(
m;n
),anditisboundedon
L
2
(
dx
).
11
(Positiveshift)
.
Wesaythatanoperator
S
isapositiveshift,ifthereexists
a2-Carlesonsequense
f
a
I
g
I
2D
,suchthat
Sf
(
x
)=
X
J
2D
a
J
h
f
i
J
˜
J
:
12
(Orlitznorm)
.
GivenaYoungfunctionwethecorrespondingOrlitz
normonacube
J
by
k
f
k
J;

=
k
f
k
L

(
J
)
=
k
f
k
L

J
=inf
ˆ
>
0:
1
j
J
j
Z
J


j
f
(
x
)
j


dx

1
˙
:
Finally,wetranslateseveralnotionstothemetricspacesetting.
13
(Doublingmetricspace)
.
Aspace
X
withametric
ˆ
iscalleddoublingifthere
existsameasure

on
X
andaconstant
C
,suchthatforanyball
B
(
x;r
)=
f
y
:
ˆ
(
x;y
)
<r
g
thefollowingholds:

(
B
(
x;
2
r
))
6
C
(
B
(
x;r
))
:
Inwhatfollowsallmetricspacesareassumedtobedoubling.
14
(Aon-Zygmundkernel)
.
Let

(
x;r
)beapositivefunction,increas-
5
inganddoublingin
r
,i.e.

(
x;
2
r
)
6
C
(
x;r
),where
C
doesnotdependon
x
and
r
.
K
(
x;y
):
X

X
!
R
isaon-Zygmundkernel,associatedtoafunction

,ifthere
existpositivenumbers
C;"
,suchthat
j
K
(
x;y
)
j
6
C
min

1

(
x;ˆ
(
x;y
))
;
1

(
y;ˆ
(
x;y
))

;
(1.6)
j
K
(
x;y
)

K
(
x
0
;y
)
j
6
C
ˆ
(
x;x
0
)
j
"
ˆ
(
x;y
)
"

(
x;ˆ
(
x;y
))
;ˆ
(
x;y
)
>
Cˆ
(
x;x
0
)
;
(1.7)
j
K
(
x;y
)

K
(
x;y
0
)
j
6
C
ˆ
(
y;y
0
)
"
ˆ
(
x;y
)
"

(
y;ˆ
(
x;y
))
;ˆ
(
x;y
)
>
Cˆ
(
y;y
0
)
:
(1.8)
15
on-Zygmundoperator)
.
Let

and
K
beasintheprevious
Let

beameasureon
X
,suchthat

(
B
(
x;r
))
6
C
(
x;r
),where
C
doesnotdependon
x
and
r
.Wesaythat
T
isaon-Zygmundoperatorwithkernel
K
if
T
isbounded
L
2
(

)
!
L
2
(

)
;
(1.9)
Tf
(
x
)=
Z
K
(
x;y
)
f
(
y
)

(
y
)
;
8
x
62
supp
f
(1.10)
16
(
A
2
weightsinmetricspaces)
.
Let

beadoublingmeasure.Wesaythata
weight
w
belongsto
A
2

if
[
w
]
2

=sup
x;r
1

(
B
(
x;r
))
Z
B
(
x;r
)
w

1

(
B
(
x;r
))
Z
B
(
x;r
)
w

1
<
1
:
Themeasure

willalwaysbeandwewillsupressthesubindex

.
6
1.2Someknowntheorems
Inthissectionwecollectthetheoremsthatareknownandthatwewillusewithoutproofs.
Theorem1.2.1
onen'sdecomposition)
.
Inaspace
R
d
,set
D
0
=
f
2

k
([0
;
1)
d
+
m
):
k
2
Z
;m
2
Z
d
g
.Forabinaryfamily
!
=(
!
j
)
j
2
Z
,
!
j
2f
0
;
1
g
,set
I
+
!
=
I
+
P
j
:2

j
<`
(
I
)
2

j
!
j
.
Denote
D
!
=
f
I
+
!
:
I
2D
0
g
.Forthecanonicalprobabilityonsetofbinarysequences,and
foraon-Zygmundoperator
T
itistruethat
(
Tf;g
)=
c
(
T
)

E
!
X
i
>
0
;j
>
0
˝
ij
(
S
ij
!
f;g
)
;f;g
2
C
1
0
;
where
˝
ij
6
c
2

i
+
j
,and
S
ij
!
isageneralizeddyadicshiftofcomplexity
(
i;j
)
.
IntheSection4.3weproveaversionofthistheoremformetricspaces.Thenexttheorem
isanotherwaytoestimateaon-Zygmundoperator.
Theorem1.2.2
(Lerner'sdecomposition)
.
Foraon-Zygmundoperator
T
andafunc-
tionBanachspace
X
itistruethat
k
T
k
X
6
C
(
T;X
)

sup
k
S
k
X
;
where
S
isapositivedyadicshiftwitha
2
-Carlesonsequense
f
a
I
g
.Thesupremumistaken
overalldyadicgridson
R
d
andall
2
-Carlesonsequences.
Thenexttheoremshowshowtoestimatesuchoperators.
7
Theorem1.2.3.
Let
S
beapositiveHaarshiftofcomplexity
(
m;n
)
.Then
k
S
(

˙
)
k
L
p
(
˙
)
!
L
p
(
u
)
6
˝
k
M
(

˙
)
k
L
p
(
˙
)
!
L
p
(
u
)
+sup
Q
k
˜
Q
S
(
˜
Q
˙
)
k
L
p
(
u
)
˙
(
Q
)
1
p
+sup
Q
k
˜
Q
S

(
˜
Q
u
)
k
L
p
0
(
˙
)
u
(
Q
)
1
p
0
;
(1.11)
where
M
istheHardy-Littlewoodmaximalfunction.
8
Chapter2
Twoweightestimates
2.1Mainresults
Suppose
T
isaCaon-Zygmundoperatoron
R
,and
u;v
areweights.By
T
u
wedenote
theoperatorthatactsas
T
u
(
f
)=
T
(
fu
).Thequestiononeasksitthefollowing:
Whatconditionsshouldweights
u
and
v
havetoassurethattheoperator
T
u
isbounded
from
L
2
(
u
)to
L
2
(
˙
)?
ThusquestionsgotsomeattentionrecentlyintheworksofF.Nazarov,S.Treil,A.
Volberg,A.Lerner,M.Lacey,E.Sawyer,C.Y.Shen,I.Uriarte-Tuero,D.Cruz-Uribe,C.
Perez,J.M.Martell.
Thisquestionwasconsideredforindividualoperators:Haarshift,see[NTV2,NTV3]
andHilbertTransform,see[NTV4,LSUT,LSSUT,Lac1,Lac2].Theconditionsinthese
questionswereformulatedintermsoftheseindividualoperators.
Inourpapers[CURV,NRTV1,NRTV2,NRV1]weconsiderthe\bump"approachtothis
problem.Wegiveaconditiononweights
u;˙
whichassuresthatfor
any
on-Zygmund
OperatorT,
T
u
isboundedfrom
L
2
(
u
)to
L
2
(
˙
).In[CURV]wealsohavesome
L
p
results.
The\bump"conditionappearsfromthefamousSarasonconjecture.
Conjecture.
Ifthereexistsanumber
R
,suchthatforanyinterval
I
wehave
P
u
(
z
)

P
v
(
z
)
6
9
R
,thentheHilbertTransform
H
u
isboundedfrom
L
2
(
u
)to
L
2
(
˙
).Here
P
u
(
z
)=
Z
R
Im
z
(Re
z

t
)
2
+(Im
z
)
2
u
(
t
)
dt:
Thisconjectureisknowntobe
false
,see[NV]or[LSUT]foracounterexample.
ThebumpapproachappearedinworksofC.F[F],Chang-Wilson-Wolf,[CWW].
Formorehistorywereferthereadertothebook[CUMP1].Infact,the
A
2
conditionfor
aweight
w
readsas
h
w
i
I
h
w

1
i
I
6
Q
.Inoursettingwehavetwoweights
u;v
,andthe
condition
h
u
i
I
h
v
i
I
6
Q
isevenweakerthantheoneintheSarasonconjecture(justtake
z
=
c
I
+
j
I
j
i
,where
c
I
isthecenteroftheinterval
I
).Wenoticethat
h
u
i
I
isthesquared
L
2
(
dx
j
I
j
)normofthefunction
u
1
2
ontheinterval
I
.Wetrytoconsiderastrongernorm.
Precisely,let
A
:[0
;
1
)
!
[0
;
1
)beaYoungfunction.In[CURV]weprovethefollowing
theorems.
Theorem2.1.1
(Separatedbumpconjecture)
.
Suppose
A
(
t
)=
t
2
log
1+
"
(
t
)
.Theniffor
anyinterval
I
k
u
1
2
k
I;A
h
˙
i
1
2
I
+
k
˙
1
2
k
I;A
h
u
i
1
2
I
6
Q;
thentheoperator
T
u
isboundedfrom
L
2
(
u
)
to
L
2
(
˙
)
.
Thesameistruefor
A
(
t
)=
t
2
log(
t
)(loglog(
t
))
1+
"
forbig
"
.
Theorem2.1.2
(Weakseparatedbumpconjecture)
.
Suppose
A
(
t
)=
t
2
log
1+
"
(
t
)
.Thenif
foranyinterval
I
k
v
1
2
k
I;A
h
u
i
1
2
I
6
Q;
thentheoperator
T
u
isboundedfrom
L
2
(
u
)
to
L
2
;
1
(
v
)
.
Thesameistruefor
A
(
t
)=
t
2
log(
t
)(loglog(
t
))
1+
"
forbig
"
.
10
Theorem2.1.3
(Bumpconjecture)
.
Suppose

t
)
isayoungfunction,suchthat
1

t
)
is
integrableat
1
.Thenifforanyinterval
I
k
u
k
I;

k
v
k
I;

6
Q;
thentheoperator
T
w
isboundedfrom
L
2
(
u
)
to
L
2
(
v
)
.
Noticethatwhilethebumpconjectureisproveninfull,theseparatedbumpconjecture
isprovenonlyforsomefunctions
A
.Inwhatfollowswegivethepreciseconditionson
A
for
whichwecanprovetheresult.
2.2Thestoppingtimeapproachtotheseparatedbumpconjecture
Westartwiththefollowing
17.
WewillsaythataYoungfunction

A
the
B
p
0
condition,1
<p<
1
,
ifforsome
c>
0,
Z
1
c

A
(
t
)
t
p
0
dt
t
<
1
:
If
A
and

A
aredoubling(i.e.,if
A
(2
t
)

CA
(
t
),andsimilarlyfor

A
),then

A
2
B
p
ifand
onlyif
Z
1
c

t
p
A
(
t
)

p
0

1
dt
t
<
1
:
Examplesofsuchbumpsare
A
(
t
)=
t
p
log(
e
+
t
)
p

1+

;

A
(
t
)
ˇ
t
p
0
log(
e
+
t
)
1+

0
;
(2.1)
11
B
(
t
)=
t
p
0
log(
e
+
t
)
p
0

1+

;

B
(
t
)
ˇ
t
p
log(
e
+
t
)
1+

00
;
(2.2)
where
>
0,

0
=
=
(
p

1),

00
=
=
(
p
0

1)Given
p
,1
<p<
1
,let
A
and
B
beYoung
functionssuchthat

A
2
B
p
0
and

B
2
B
p
.Ourmainconditiononweights
u
and
˙
willbe
sup
Q
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q
<
1
;
(2.3)
sup
Q
k
u
1
=p
k
A;Q
h
˙
i
1
=p
0
Q
<
1
:
(2.4)
Remark1.
BythepropertiesoftheLuxemburgnormwehavethateitherconditionimplies
thetwo-weight
A
p
condition:
sup
Q
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
<
1
:
(2.5)
RecallthattheHardy-Littlewoodmaximaloperatoristobe
Mf
(
x
)=sup
Q
3
x
hj
f
ji
Q
=sup
Q
3
x
k
f
k
1
;Q
:
GivenaYoungfunction
A
,wetheOrliczmaximaloperator
M
A
by
M
A
f
(
x
):=sup
Q
3
x
k
f
k
A;Q
:
ThefollowingresultisduetoPerez[Pe](seealso[1]).
Theorem2.2.1.
Fix
p
,
1
<p<
1
,andlet
A
beaYoungfunctionsuchthat
A
2
B
p
.Then
M
A
:
L
p
!
L
p
.
The
B
p
conditionisalsotforatwo-weightnorminequalityfortheHardy-
Littlewoodmaximaloperator.ThisresultisalsoduetoPerez[Pe,1].
12
Theorem2.2.2.
Fix
p
,
1
<p<
1
,andlet
B
beaYoungfunctionsuchthat

B
2
B
p
.If
thepairofweights
(
u;˙
)

sup
Q
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q
<
1
;
(2.6)
then
k
M
(
f˙
)
k
L
p
(
u
)

C
k
f
k
L
p
(
˙
)
:
(2.7)
BythedecompositiontheoremofLerner,toprovetheseparatedbumpconjectureitwill
toprovethattheyholdforapositivedyadicshift.Morepreciselywewillprovethe
following.
Theorem2.2.3.
Given
p
,
1
<p<
1
,suppose
A
and
B
arelog-bumpsoftheform
(2.1)
,
(2.2)
,andthepairofweights
(
u;˙
)

(2.3)
and
(2.4)
.Givenanypositivedyadicshift
S
itholdsthat
k
S
(
f˙
)
k
L
p
(
u
)

C
k
f
k
L
p
(
˙
)
:
Theorem2.2.4.
Given
p
,
1
<p<
1
,suppose
A
isalog-bumpoftheform
(2.1)
,andthe
pairofweights
(
u;˙
)

(2.4)
.Givenanypositivedyadicshift
S
itholdsthat
k
S
(
f˙
)
k
L
p;
1
(
u
)

C
k
f
k
L
p
(
˙
)
:
BytheTheorem1.2.3istoestimate
k
˜
I
S
(
˜
I
˙
)
k
L
p
(
u
)
.Thedualestimatewillbe
thesame.
13
Fixacube
Q
0
;usingthenotationfromtheofaHaarshift,wehavethat
˜
Q
0
S
(
˜
Q
0
˙
)=
X
R
ˆ
Q
0
S
R
(
˙
)+
˜
Q
0
X
R;Q
0
(
R
S
R
(
˜
Q
0
˙
)
6
X
R
ˆ
Q
0
S
R
(
˙
)+
˜
Q
0
h
˙
i
Q
0
:
(2.8)
Thesecondinequalityisstraightforward:see,forinstance,[H,HyLa,HLM+,HPTV].As
wenotedabove,thepair(
u;˙
)thetwo-weight
A
p
condition(2.5).Therefore,the
L
p
(
u
)normofthesecondtermisboundedby
k
˜
Q
0
k
L
p
(
u
)
h
˙
i
Q
0
=
h
u
i
1
=p
Q
0
h
˙
i
1
=p
0
Q
0
˙
(
Q
0
)
1
=p

C˙
(
Q
0
)
1
=p
:
Toestimatethe
L
p
(
u
)normoftheterm,weformthefollowingdecomposition
(see[HyLa]):
K
=
K
i
=
f
Q
ˆ
Q
0
:
`
(
Q
)=2
i
+
˝n
g
;n
2
Z
;
K
a
=
f
Q
2K
:2
a
6
h
u
i
1
p
Q
h
˙
i
1
p
0
Q
<
2
a
+1
g
;
P
a
0
=allmaximalcubesin
K
a
;
P
a
n
=
n
maximalcubes
P
0
ˆ
P
2P
a
n

1
,suchthat
P
0
2K
a
;
h
˙
i
P
0
>
2
h
˙
i
P
o
;
P
a
=
[
n
>
0
P
a
n
:
Followingtheterminologyfrom[LPR],wecallmembersof
P
a
principalcubes
.
Hereafterwesuppresstheindex
i
;thiswillgiveusasumwith
˝
+1terms.Given
Q
2K
a
,
let
Q
)denotetheminimalprincipalcubethatcontainsit,and
K
a
(
P
)=
f
Q
2K
a
:
Q
)=
P
g
:
14
Wewillestimatethe
L
p
(
u
)normofthesumontheright-handsideof(2.8)using
theexponentialdecaydistributionalinequalityoriginatedin[LPR].Below,
S
isanypositive
generalizedHaarshiftthatisboundedonunweighted
L
2
.Inparticular,wewilltake
S
to
beoneofthepositiveHaarshifts
S
L
fromabove.
Weneedthefollowingnotation.Forafamily
S
wedenote
S
Q
=
X
Q
2Q
S
Q
:
Thefollowingdistributionalinequalityholds.
Theorem2.2.5.
Thereexistsaconstant
c
,dependingonlyonthedimensionandtheun-
weighted
L
2
normoftheshift,suchthatforany
P
2P
a
,
u

x
2
P
:
j
S
K
a
(
P
)
(
˙
)
j
>t
˙
(
P
)
j
P
j

.
e

ct
u
(
P
)
:
ItfollowsfromTheorem2.2.5thatforsomepositiveconstant
c
,
k
X
R
(
Q
S
R
(
˙
)
k
L
p
(
u
)
6
C˝
X
a
0
@
X
P
2P
a
u
(
P
)

˙
(
P
)
j
P
j

p
1
A
1
p
:
(2.9)
Wesketchtheproofof(2.9)followingthebeautifulcalculationsin[HyLa]:
X
R
(
Q
S
R
(
˙
)=
˝
X
i
=0
X
a
X
P
2P
a
S
K
a
(
P
)
(
˙
)
;
15
andso
k
X
R
(
Q
S
R
(
˙
)
k
L
p
(
u
)
6
(
˝
+1)
X
a
k
X
P
2P
a
S
K
a
(
P
)
(
˙
)
k
L
p
(
u
)
:
Fix
a
.UsingFubini'stheoremwewrite
k
X
P
2P
a
S
K
a
(
P
)
(
˙
)
k
L
p
(
u
)
=

Z

X
j
X
P
2P
a
˜
f
S
K
a
(
P
)
(
˙
)
2
(
j;j
+1]
˙
(
P
)
j
P
j
g
S
K
a
(
P
)
(
˙
)(
x
)

p
u
(
x
)
dx

1
=p

X
j
(
j
+1)

Z

X
P
2P
a
˜
f
S
K
a
(
P
)
(
˙
)
2
(
j;j
+1]
˙
(
P
)
j
P
j
g
˙
(
P
)
j
P
j

p
u
(
x
)
dx

1
=p
:
Bythechoiceofthestoppingcubes
P
2P
a
wehavethat

X
P
2P
a
˜
f
S
K
a
(
P
)
(
˙
)
2
(
j;j
+1]
˙
(
P
)
j
P
j
g
˙
(
P
)
j
P
j

p
.
X
P
2P
a
˜
f
S
K
a
(
P
)
(
˙
)
2
(
j;j
+1]
˙
(
P
)
j
P
j
g

˙
(
P
)
j
P
j

p
:
Letusexplainit.Takeapoint
x
andnestedcubes
P
0
˙
P
1
˙
:::
˙
P
N
,
P
k
2P
a
n
k
,
x
2
P
k
,
and
x
2f
S
K
a
(
P
)
(
˙
)
2
(
j;j
+1]
˙
(
P
)
j
P
j
g
(i.e.,forwhichthetermsinsumsabovearenon-zero).
Bynitionof
P
a
n
k
wehave
h
˙
i
P
k
>
2
h
˙
i
P
k

1
.Theinequalitymaybemuchbetterifwe
skipseveralgenerations,but2intheright-handsideisguaranteed.Wehaveasequense
y
k
=
˙
(
P
k
)
j
P
k
j
.Theinequalitysaysthat
 
X
k
y
k
!
p
6
C
X
k
y
p
k
:
Noticethat
y
k
>
2
y
k

1
.Thus,
X
k
y
k
6
N
X
k
=0
1
2
N

k
y
N
.
y
N
:
16
Therefore,
 
X
k
y
k
!
p
.
y
p
N
6
X
k
y
p
k
:
Thisbeautifulobservationfrom[HyLa]letsuswrite
k
X
P
2P
a
S
K
a
(
P
)
(
˙
)
k
L
p
(
u
)
.
X
j
(
j
+1)

X
P
2P
a

˙
(
P
)
j
P
j

p
u
(
S
K
a
(
P
)
(
˙
)
2
(
j;j
+1]
˙
(
P
)
j
P
j
)

1
=p
:
ThenbythedistributionalinequalityfromTheorem2.2.5:
k
X
P
2P
a
S
K
a
(
P
)
(
˙
)
k
L
p
(
u
)
.
X
j
(
j
+1)
e

cj=p

X
P
2P
a

˙
(
P
)
j
P
j

p
u
(
P
)

1
=p
:
Thisgivesus(2.9).
Itisatthispointintheproofthatwecannolongerassumethatourpairofweights
(
u;˙
)thegeneral
A
p
bumpconditionandwemustinsteadmakethemorerestrictive
assumptionthatwehavelogbumps.Beforedoingso,however,wewanttoshowhowtheproof
goesandwheretheproblemarisesforgeneralbumps.Wewillthengivethemo
necessarytomakethisargumentworkforlogbumps.
thesequence

Q
=
8
>
>
>
<
>
>
>
:
j
P
j
;Q
=
P;
forsomecube
P
2P
a
0
;
otherwise;
17
thentheinnersumin(2.9)becomes
X
Q
ˆ
Q
0
u
(
Q
)
j
Q
j

˙
(
Q
)
j
Q
j

p

Q
:
Butbyolder'sinequalityinthescaleofOrliczspaces,
˙
(
Q
)
j
Q
j
=
h
˙
1
p
˙
1
p
0
i
Q
6
C
k
˙
1
p
0
k
Q;B
k
˙
1
p
k
Q;

B
6
k
˙
1
p
0
k
Q;B
inf
x
2
Q
M

B
(
˙
1
p
˜
Q
)
:
(2.10)
Therefore,by(2.3),
X
Q
ˆ
Q
0
u
(
Q
)
j
Q
j

˙
(
Q
)
j
Q
j

p

Q
6
K
p
X
Q
ˆ
Q
0

Q
inf
x
2
Q
M

B
(
˙
1
p
˜
Q
)
p
:
(2.11)
Tocompletetheproofweneedtwolemmas.Thecanbefoundin[LPR],formula
(3
:
3).
Lemma2.2.6.
f

Q
g
isaCarlesonsequence.
Thesecondisafolktheorem;aproofcanbefoundin[MP].
Lemma2.2.7.
If
f

Q
g
isaCarlesonsequence,then
X
Q
ˆ
Q
0

Q
inf
Q
˜
Q
0
F
(
x
)
.
Z
Q
0
F
(
x
)
dx:
18
CombiningthesetwolemmaswithTheorem2.2.1(since

B
2
B
p
)weseethat
X
Q
u
(
Q
)
j
Q
j

˙
(
Q
)
j
Q
j

p

Q
6
K
p
X
Q;Q
ˆ
Q
0

Q
inf
x
2
Q
M

B
(
˙
1
p
˜
Q
0
)
p
.
K
p
k
M

B
(
˙
1
p
˜
Q
0
)
k
p
L
p
(
dx
)
.
K
p
k
˙
1
p
˜
Q
0
k
p
L
p
(
dx
)
=
K
p
˙
(
Q
0
)
:
Thiswouldcompletetheproofexceptthatwemustnowsumover
a
,andin(2.9)thissum
goesfrom

tothelogarithmofthetwo-weight
A
p
constantofthepair(
u;˙
).Wecannot
evaluatethissumunlesswecanmodifytheaboveargumenttoyieldadecayconstantin
a
.
Intheone-weightargumentin[HyLa]theauthorscouldusethefactthattheparameter
a
runfrom0tothelogarithmof
A
p
constant:thisfollowssincebyolder'sinequalitythe
A
p
constantofanyweightisatleast1.Inthetwo-weightcasethe
A
p
constantcanbearbitrarily
small,andthereforewemustsumovermanyvaluesof
a
.Weareabletogetthe
desireddecayconstantonlybyassumingthatweareworkingwithlogbumps.
Wemodifytheaboveargumentasfollows.Essentially,wewillusethepropertiesoflog
bumpstoreplace

B
withaslightlylargerYoungfunction.
B
0
(
t
)=
t
p
0
log(
e
+
t
)
p
0

1+

2
;
thenweagainhavethat

B
0
2
B
p
.Insteadof(2.11)wewillprovethatthereexists

,
0
<<
1,suchthat
X
Q
ˆ
Q
0
u
(
Q
)
j
Q
j

˙
(
Q
)
j
Q
j

p

Q
6
C

K
(1


)
p
2
p
X
Q
ˆ
Q
0

Q
inf
x
2
Q
M

B
0
(
˙
1
p
˜
Q
0
)
p
:
(2.12)
Giveninequality(2.12),wecanrepeattheargumentabove,butwenowhavethedecayterm
2
p
whichallowsustosumin
a
andgetthedesiredestimate.
19
Toprove(2.12)supposeforthemomentthatthereexists

suchthat
k
˙
1
p
0
k
Q;B
0

C
1
k
˙
1
p
0
k
1


Q;B
k
˙
1
p
0
k

L
p
0
(
Q;dx=
j
Q
j
)
:
(2.13)
Giventhis,acube
Q
2P
a
|wecandothissinceotherwise

Q
=0.Then
u
(
Q
)
j
Q
j

˙
(
Q
)
j
Q
j

p
.
h
u
i
Q
k
˙
1
=p
0
k
p
B
0
;Q
k
˙
1
=p
k
p

B
0
;Q
.
h
u
i
Q
k
˙
1
=p
0
k
(1


)
p
B;Q
k
˙
1
=p
0
k
p
L
p
0
(
Q;dx=
j
Q
j
)
k
˙
1
=p
k
p

B
0
;Q
=(
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q
)
(1


)
p

(
h
u
i
1
=p
Q
k
˙
1
=p
0
k
L
p
0
(
Q;dx=
j
Q
j
)
)
p
k
˙
1
=p
k
p

B
0
;Q
.
K
(1


)
p

(
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
)
p
k
˙
1
=p
k
p

B
0
;Q
.
K
(1


)
p

2
p
k
˙
1
=p
k
p

B
0
;Q
.
K
(1


)
p

2
p

inf
x
2
Q
M

B
0
(
˙
1
p
˜
Q
0
)
p
:
Inequality(2.12)nowfollowsimmediately.
Therefore,tocompletetheproofwemustestablish(2.13).Bytherescalingpropertiesof
theLuxemburgnorm[1,Section5.1],theright-handsideofthisinequalityisequalto
k
˙
1


p
0
k
C;Q
k
˙

p
0
k
p
0
;Q
;
where
C
(
t
)=
B
(
t
1
1


).Therefore,bythegeneralizedolder'sinequalityinOrliczspaces
([1,Lemma5.2]),inequality(2.13)holdsifforall
t>
1,
C

1
(
t
)
t

p
0
.
B

1
0
(
t
)
:
(2.14)
20
Astraightforwardcalculation(see[1,Section5.4])showsthat
C

1
(
t
)=
B

1
(
t
)
1


ˇ
t
1


p
0
log(
e
+
t
)
1


p
+

(1


)
p
0
;B

1
0
(
t
)
ˇ
t
1
p
0
log(
e
+
t
)
1
p
+

2
p
0
:
Byequatingtheexponentsonthelogarithmterms,weseethat(2.14)holdsifwetake

=

2(
p
0

1+

)
:
Therefore,withthisvalueof

inequality(2.13)holds,andthiscompletesourproof.
Fortheconvenienceofthereaderwegiveadirectproofof(2.13);thiscomputationwill
alsobeusedbelowinSection2.2.2.Thedesireinequalityobviouslyfollowsfromthefollowing
lemma.
Lemma2.2.8.
Givenaprobabilitymeasure

,let
f
beanon-negativemeasurablefunction.
Let
B;B
0
belogarithmicbumpsasin
(2.2)
with

=
˝
and

=
˝
2
respectively.Thenthere
existsanabsoluteconstant
C
and

=

(
p
0
;˝
)
>
0
suchthat
k
f
k
B
0


C
k
f
k
1


B
k
f
k

L
p
0
(

)
:
(2.15)
Proof.
Wewillactuallyshowthat

=
1
2+(
p
0

1)
2
˝
.:=
R
j
f
j
p
0

.Sinceinequal-
ity(2.15)ishomogeneous,wemayassumewithoutlossofgeneralitythat
k
f
k
B
=1
:
(2.16)
Moreover,wemayassumethat

1:otherwise(2.15)canbeachievedbychoosing
C
21
tlylarge.Let
<
1and
K
beconstants;wewilldeterminetheirprecisevalue(in
thisorder)below.Thenwehavethat
Z
f
p
0

p
0
log

e
+
f


p
0

1+
˝
2


Z
f
f

K
g

+
Z
f
f

K
g
:::



p
0
[log(
e
+
K
)]
p
0

1+
˝
2
+
Z
f
f

K
g
f
p
0

p
0
log(
e
+
f

)
p
0

1+
˝
[log(
e
+
K
)]
˝
2




p
0
[log(
e
+
K
)]
p
0

1+
˝
2
+
Z
f
p
0

p
0
log(
e

+
f

)
p
0

1+
˝
[log(
e
+
K
)]
˝
2




p
0
[log(
e
+
K
)]
p
0

1+
˝
2
+
1

p
0
[log(
e
+
K
)]
˝=
2

Z
f
p
0
log
p
0

1+
˝
(
e
+
f
)

+
Z
f
p
0
log(


1
)
p
0

1+
˝





p
0
[log(
e
+
K
)]
p
0

1+
˝
2
+
1

p
0
[log(
e
+
K
)]
˝=
2

1+log(


1
)
p
0

1+
˝

:
Inthelastlineweused(2.16).Fix

sothat
=(

p
0
)
1+
c
;
where
c
=1+(
p
0

1)
2
˝
.Inotherwords,

=
1
=p
0
)

=
k
f
k

L
p
0
(

)
;
=
1
1+
c
:
Nowchoose
K
sothat
[log(
e
+
K
)]
˝=
2
ˇ


p
0
;
22
then
[log(
e
+
K
)]
p
0

1+
˝=
2
ˇ
(


p
0
)
1+(
p
0

1)
2
˝
=:(


p
0
)
c
:
Ifwesubstitutethesevaluesintotheabovecalculation,weseethattherighthandsideis
dominatedbyaconstant.Hence,bytheoftheLuxemburgnorm,
k
f
k
B
0


C
=
C
k
f
k

L
p
0
(

)
:
Thiscompletestheproof.
Remark2.
Theconjugatetestingconditioncanbevsimilarly.Theadjoint
S

isalso
aHaarshift,andsowecanapplythedistributioninequalityfromTheorem2.9toit.Also,
thesecondsumin(2.8)willhavethesamepointwiseestimate(exchanging
˙
and
v
)ifwe
replace
S
with
S

.
Remark3.
Intheproofofthetestingconditionweonlyusedthebumpcondition(2.3);
toprovethesecondtestingconditionweusethesecondbumpcondition(2.4).
2.2.1ProofofTheorem2.2.4
Theproofoftheweak-typeinequalityusesessentiallythesameargumentasabove;herewe
sketchthechangesrequired.Werepeattheargument,replacingthe
L
p
(
u
)normwiththe
L
p;
1
(
u
)norm.Sincethepair(
u;˙
)thetwo-weight
A
p
conditionwehavethewell
knowninequalitythat
k
M
(
f˙
)
k
L
p;
1
(
u
)

C
k
f
k
L
p
(
˙
)
;
wheretheconstant
C
dependsonlyonthe
A
p
constantandthedimension.Thereforeit
remainstoestimatethe
L
p;
1
(
u
)normof
S
L
(
j
f
j
˙
).However,fromonen,
etal.
[HLM+,
23
Theorem4.3]wehavethefollowinganalogofTheorem1.2.3.
Theorem2.2.9.
Let
S
beapositiveHaarshiftofcomplexity
(
m;n
)
.Then
k
S
(

˙
)
k
L
p
(
˙
)
!
L
p;
1
(
u
)
6
˝
k
M
(

˙
)
k
L
p
(
˙
)
!
L
p;
1
(
u
)
+sup
Q
k
˜
Q
S

(
˜
Q
u
)
k
L
p
0
(
˙
)
u
(
Q
)
1
p
0
:
GivenTheorem2.2.9theargumentnowproceedsexactlyasbefore,usingthebump
condition(2.4)toboundthetestingcondition.Thiscompletestheproof.
2.2.2Prooffortheloglog-bumps
Inthissectionweconsiderbumpsofthefollowingform.
A
(
t
)=
t
p
log(
e
+
t
)
p

1
loglog(
e
e
+
t
)
p

1+


A
(
t
)
ˇ
t
p
0
log(
e
+
t
)loglog(
e
e
+
t
)
1+

0
;
(2.17)
B
(
t
)=
t
p
0
log(
e
+
t
)
p
0

1
loglog(
e
e
+
t
)
p
0

1+

;

B
(
t
)
ˇ
t
p
log(
e
+
t
)loglog(
e
e
+
t
)
1+

00
;
(2.18)
where
>
0.Ourproofsareverysimilartotheproofsgiveninprevioussections,sowe
willdescribetheprinciplechanges.Asbefore,weneedtoprovethefollowingtheoremsfor
positivedyadicshifts.
Theorem2.2.10.
Given
p
,
1
<p<
1
,suppose
A
and
B
areloglog-bumpsoftheform
(2.17)
,
(2.18)
with
>
0
tlylarge
,andthepairofweights
(
u;˙
)

(2.3)
and
(2.4)
.Givenanypositivedyadicshift
S
,
k
S
(
f˙
)
k
L
p
(
u
)

C
j
f
k
L
p
(
˙
)
.
Theorem2.2.11.
Given
p
,
1
<p<
1
,suppose
A
isaloglog-bumpoftheform
(2.17)
with
>
0
tlylarge
,andthepairofweights
(
u;˙
)

(2.4)
.Givenanypositive
dyadicshift
S
,
k
S
(
f˙
)
k
L
p;
1
(
u
)

C
j
f
k
L
p
(
˙
)
.
24
WewillproveTheorem2.2.10bymodifyingtheproofofTheorem2.2.3above;The-
orem2.2.11isprovedsimilarly.ThemainstepistoadaptLemma2.2.8toworkwith
loglog-bumps.Let
B
beasin(2.18),and
B
0
similarlybutwith

replacedby
=
2.
ThenarguingalmostexactlyaswedidintheproofofLemma2.2.8,wehavethat
k
f
k
B
0


C
k
f
k
B
"

k
f
k
L
p
0
(

)
k
f
k
B

;
(2.19)
where
"
(
t
)=(log
C
t
)


,
C
=
C
(
p;
),and

=

(
p;
)with
>
1if

islargeenough.For
thedetailedproofforwiderrangeofbumpssee[NRV],Section5.2.2.
Given(2.19)wehavethat
u
(
Q
)
j
Q
j

˙
(
Q
)
j
Q
j

p

C
h
u
i
Q
k
˙
1
=p
0
k
p
B
0
;Q
k
˙
1
=p
k
p

B
0
;Q

C
h
u
i
Q
k
˙
1
=p
0
k
p
B;Q
"

h
˙
i
1
=p
0
Q
k
˙
1
=p
0
k
B;Q

p
k
˙
1
=p
k
p

B
0
;Q

C
0
@
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
1
A
p
"

h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q

p
(
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
)
p
k
˙
1
=p
k
p

B
0
;Q
:
Tocompletetheproof,weneedagoodboundin
a
fortheproductofthreeterms.
Moreover,itisenoughtogetagoodboundfornegativeandverybiginabsolutevalue
a
.
Thus,wecanthinkthat
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
isverysmall.
Considerafunction
'
(
t
)=
t"
(
1
t
)
:
25
Then
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
"

h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q

=
'
0
@
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
1
A
:
Set
t
0
=
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
.Then
c
6
t
0
6
C
1
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
.Theright-handsideofthelast
inequalityisverybig.Moreover,thefunction
'
(
t
)=
t
(log(
Ct
))

{
isincreasingnear
1
.
Therefore,
'
(
t
0
)
6
C'
0
@
1
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
1
A
6
C
1
'
(2

a
)
:
Therefore,sinceonallcubes
P
2P
a
wehave
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
˘
2
a
,weget
C
0
@
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
1
A
p
"

h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
h
u
i
1
=p
Q
k
˙
1
=p
0
k
B;Q

p
(
h
u
i
1
=p
Q
h
˙
i
1
=p
0
Q
)
p
k
˙
1
=p
k
p

B
0
;Q
6
C
2
"
(2
a
)
p
k
˙
1
=p
k
p

B
0
;Q
:
UsingtheCarlesonpropertyofthesequence

Q
,weget
X
Q
ˆ
Q
0

Q
u
(
Q
)
j
Q
j

˙
(
Q
)
j
Q
j

p
6
C
2
"
(2
a
)
p
Z
M

B
0
(
˙
1
=p
˜
Q
0
)
p
dx
6
C
3
"
(2
a
)
p
˙
(
Q
0
)
:
Thus,returningtotheformula(2.9),weget
X
a
0
@
X
P
2P
a
u
(
P
)

˙
(
P
)
j
P
j

p
1
A
1
p
6
˙
1
p
(
Q
0
)
X
a<
0
"
(2
a
)
:
Bytheformulator
"
,theseriesconvergeif
{
>
1.
26
2.3TheBellmanfunctionapproachtothebumpconjecture
2.3.1Mainresult
InthissectionwewillworkwithOrlitzfunctionssuchthat
1

isintegrableneary.
Weaimtoprovethefollowingtheorem.
Theorem2.3.1.
Letfunctions

1
and

2
beasabove.Lettheweights
v
,
w
satisfy
sup
I
k
v
k

1
;Q
k
w
k

2
;Q
<
1
;(2.20)
herethesupremumistakenoverallcubes
I
.
Thenforanyboundedon-Zygmundoperator
T
theoperator
T
v
isboundedfrom
L
2
(
v
)
to
L
2
(
w
)
.
2.3.2Orlicznormsanddistributionfunctions
Orlicznormisnotveryconvenienttoworkwith,sowewouldliketoreplaceitbysomething
moretractable.
2.3.2.1AlowerboundfortheOrlicznorm
Westartwiththeremarkthatnotation
R
0
f
(
t
)
dt<
1
meansthatthefunctionisintegrable
nearzero.Similarly,
R
1
f
(
t
)
dt<
1
meansthatthefunctionisintegrableneary.
Letbeacontinuousnon-negativeincreasingconvexfunctionsuchthat=0and
R
+
1
dt

t
)
<
+
1
.
s
)parametricallyby
s
)=
0
(
t
)when
s
=
1

t

0
(
t
)
(
t>
0).
Then
s
)ispositiveanddecreasingfor
s>
0and
s

s
)isincreasing.Moreover
R
0
ds
s

s
)
<
27
+
1
.Indeed,usingourparametrizationwecanrewritethelastintegralas
Z
+
1

1

t
)
+

00
(
t
)

0
(
t
)
2

dt:
Theintegralconvergesbyourassumptionandthesecondintegrandhasaboundednear
+
1
antiderivative

1

0
(
t
)
.
Let
w

0on
I
ˆ
R
n
.thenormalizeddistributionfunction
N
of
w
by
N
(
t
)=
N
w
I
(
t
)=
1
j
I
j
jf
x
2
I
:
w
(
x
)
>t
gj
(2.21)
Lemma2.3.2.
Let
:(0
;
1]
!
R
+
beadecreasingfunctionsuchthatthefunction
s
7!
s

s
)
isincreasing.Let

beaYoungfunctionandlet

s
)

C

0
(
t
)
where
s
=
1

t

0
(
t
)
foralllarge
t
.Thenfor
N
=
N
w
I
n

(
N
):=
Z
1
0
N
(
t

N
(
t
))
dt

C
k
w
k
L

(
I
)
:
(2.22)
Proof.
Thelefthandsidescaleslikeanormundermultiplicationbyconstants,soitisenough
toshowthatif
k
w
k
L

(
I
)

1,i.e.,
1
j
I
j
Z
I

w
)=
Z
1
0
N
(
t

0
(
t
)
dt

1
then
n

(
N
)isboundedbyaconstant.Since
s

s
)increases,wemayhavetroubleonlyat
28
+
1
Itisclearthatittoestimatetheintegraloverthesetwhere
N
(
t
))
>

0
(
t
)but
sinceisdecreasingthismeansthat
N
(
t
)

C=

t

0
(
t
)),sowegetatmost
R
+
1

t
)

1
dt
andwearedone.
Remark4.
Infact,fortlyregulartheconverseinequality
k
w
k
L

(
I
)

C
Z
1
0
N
(
t

N
(
t
))
dt
holdsforanypositivedecreasingintegrable
N
.Toseethis,letusconsiderthefamilyof
suchthat
t
)=
tˆ
(
t
)and
ˆ
ismonotonicallyincreasingand\logarithmicallyconcave"in
thesensethat
tˆ
0
(
t
)
ˆ
(
t
)
decreasesmonotonicallywhen
t
!1
.Wealsoassumeofcoursethat
lim
t
!1
ˆ
(
t
)=
1
andthat
ˆ
(
t
)

1.Let
G
(
t
):=
N
(
t

N
(
t
)).Whentgoestoy,
N
ismonotonicallydecreasingtozero,andhence
G
isalsomonotonicallydecreasing(as
s

s
)
increasesnearzero).
Put
s

1
=
t

0
(
t
)

tˆ
2
(
t
)(justbecause
0
(
t
)

ˆ
(
t
)byour\logarithmicconcavity"
of
ˆ
assumption).Hence
s

c
1
(
tˆ
2
(
t
))

1
.Nowisdecreasingbyandthis
implies

c
1
(
tˆ
2
(
t
))

1
)


0
(
t
)

ˆ
(
t
)

c
2
ˆ
(
t
)
:
(2.23)
Wenowaskanadditionto\logarithmicconcavity",namely:
tˆ
0
(
t
)
ˆ
(
t
)
log
ˆ
(
t
)
!
0as
t
!1
:
(2.24)
Denote
r
(
x
)=log(
ˆ
(
e
x
)).Werequiredatthebeginningthatlim
x
!1
r
(
x
)=
1
.The
lastinequalitysaysinparticularthat
r
0
(
x
)=
o
(1)
r
(
x
)

1
,andtherefore,
r
0
tendstozeroat
29
y.Thus
r
(
x
)

x
3
foralllarge
x
.Keepingthisinmindwecontinue.
Set
u
=
tˆ
2
(
t
)
c
1
:
Then
t
=
c
1
u
ˆ
2
(
t
)
:
Thus,since
ˆ
isanincreasingtoyfunctionandweassumethat
t
iscientlybig,we
get
ˆ
2
(
t
)

c
1
.Therefore,
t

u;
and,thus,
t
=
c
1
u
ˆ
2
(
t
)

c
1
u
ˆ
2
(
u
)
:
Hence,using(2.23),weget

u

1
)

c
2
ˆ
(
t
)

c
2
ˆ
(
c
1
u
ˆ
2
(
u
)
)
:
(2.25)
Next,wewillprovethefollowinginequality.Recallthat
r
(
x
)=log(
ˆ
(
e
x
)).Weclaim
that

x
)=
r
(
x
)

r
(
x

2
r
(
x
)

c
0
)

C:
Infact,bythemeanvaluetheoremwehaveforcertain
˘
2
(
x

2
r
(
x
)

c
0
;x
)

x
)=(2
r
(
x
)+
c
0
)
r
0
(
˘
)=(2
r
(
x
)+
c
0
)
ˆ
0
(
e
˘
)
ˆ
(
e
˘
)
e
˘
:
Sinceweassumedthat
t
ˆ
0
(
t
)
ˆ
(
t
)
ismonotonicallydecreasing,weget(nowusing(2.24)inthe
30
secondcomparisonbelow):

x
)

(2
r
(
x
)+
c
0
)
ˆ
0
(
e
x

2
r
(
x
)

c
0
)
ˆ
(
e
x

2
r
(
x
)

c
0
)
e
ˆ
(
e
x

2
r
(
x
)

c
0
)
=
=(2
r
(
x
)+
c
0
)
o
(1)
log
ˆ
(
e
x

2
r
(
x
)

c
0
)
=(2
r
(
x
)+
c
0
)
o
(1)
r
(
x

2
r
(
x
)

c
0
)
=
=
o
(1)

2
r
(
x
)+
c
0
r
(
x

2
r
(
x
)

c
0
)

2+2

=
o
(1)

x
)+
c
0
r
(
x

2
r
(
x
)

c
0
)
+
o
(1)
:
Finally,weusethat
r
(
x

2
r
(
x
)

c
0
)isseparatedfromzerowhen
x
isbig.Thus

x
)


x
)
o
(1)+
o
(1)
:
Thisimmediatelyimplies
x
)=
o
(1)when
x
!1
,andthus

x
)

C:
(2.26)
Letusnowwritewhatdoesitmean.Infact,bytheof
r
andby(2.26),wecan
concludethat
C

r
(
x
)

r
(
x

2
r
(
x
)

c
0
)=log
ˆ
(
e
x
)
ˆ
(
e
x

2log
ˆ
(
e
x
)

c
0
)
=log
ˆ
(
e
x
)
ˆ
(
e
x
c
3
ˆ
2
(
e
x
)
)
:
Thus,wegetforalllarge
u
:
ˆ
(
u
)

c
4
ˆ
(
u
c
3
ˆ
2
(
u
)
)
:
Wechose
c
3
=
c

1
1
andplugtheaboveinequalityinto(2.25).Thenwellyget

u

1
)

c
5
ˆ
(
u
)
31
If
N

N
)=
G
then
c
6
G

Nˆ
(
1
N
)bythepreviousinequality.Therefore,
N

c
6
G
(we
assumedthat
ˆ

1),and
N

c
6
G
ˆ
(
1
N
)

c
6
G
ˆ
(
1
c
6
G
)
.Andwecancontinuethepreviousestimate:
N
(
t
)

c
6
G
(
t
)
ˆ
(
1
c
6
G
(
t
)
)

c
6
G
(
t
)
ˆ
(
t
)
.Weusedherethefactthattheintegrabilityandmonotonicityof
G
impliesthat
G
(
t
)=
o
(
1
t
),inparticular,
G
(
t
)
<
1
c
6
t
forlarge
t
.Butwealreadymentioned
that
0
(
t
)

c
7
ˆ
(
t
).Combiningthelasttwoinequalities,weget
N
(
t

0
(
t
)

c
6
c
7
G
(
t
),and
wejustobtainedthat
R
1
0
N
(
t

0
(
t
)
dt

c
4
c
5
.
2.3.2.2Examples
Intheabovesectiononlythebehaviorofat+
1
(equivalently,thebehaviorofnear0)
wasimportant,sowewillconcentrateourattentionthere.
Let
t
)=
t
(ln
t
)

,
>
1near
1
.Then

0
(
t
)
˘
(ln
t
)

;

t

0
(
t
)
˘
t
(ln
t
)
2

;
so
s
):=(ln(1
=s
))

theassumptionsofLemma2.3.2:toseethatwenotice

t

0
(
t
))
˘
ln
t:
If
t
)=
t
ln
t
(lnln
t
)

,
>
1,then

0
(
t
)
˘
ln
t
(lnln
t
)

;

t

0
(
t
)
˘
t
(ln
t
)
2
(lnln
t
)
2

and
s
)=ln(1
=s
)(lnln(1
=s
))

works.becauseagain
t

0
(
t
))
˘
ln
t
.
Notethatinbothexamples
R
0
(
s

s
))

1
ds<
1
.
32
TheexamplesofYoungfunctionswithhigherorderlogarithmsaretreatedsimilarly.
2.3.3Mainresultinnewlanguage
Werestateourmainresultusingthenewlybuiltfunctions
1
;
2
.Let
1
;

2
:(0
;
1]
!
R
+
beasabove,i.e.for
i
=1
;
2,
i
isdecreasing,
s
7!
s

i
(
s
)isincreasingand
Z
1
0
ds
s

i
(
s
)
<
1
:
Recallthatforaweight
w
thenormalizeddistributionfunction
N
w
I
isby(2.21)
Theorem2.3.3.
Lettheweights
v
,
w
satisfy
sup
I
n

1
(
N
v
I
)
n

2
(
N
w
I
)
<
1
;(2.27)
herethesupremumistakenoverallcubes
I
,and
n

isdby
(2.22)
.
Thenforanyon-Zygmundoperator
T
theoperator
T
v
isboundedfrom
L
2
(
v
)
to
L
2
(
w
)
.
2.3.4Generalsetup
Considerameasurespace
X
with
˙
measure

let
L
k
=
f
I
k
j
g
j
,
k
2
Z
(or
k
2
Z
+
)be
partitionsof
X
intodisjointsets
I
k
j
,0
<
(
I
k
j
)
<
1
.
Weassumethatthepartition
L
k
+1
isatof
L
k
.
Let
A
bethe
˙
-algebrageneratedbyallthepartitions
L
k
.Inwhatfollowsallfunctions
on
X
weconsiderwillbeassumedtobe
A
-measurable.
Withrespecttothis
˙
-algebraswecanmartingaleaveragingoperators
E
k
,and
33
martingaleoperators
n
k
:=

E
k
+
E
k
+
n
.
Weadaptthefollowingnotation.
ch
I
Thecollectionofchildrenof
I
2L
,i.e.if
I
2L
n
thench
I
=
f
J
2L
n
+1
:
J
ˆ
I
g
.
ch
k
I
Thecollectionofchildrenoftheorder
k
of
I
2L
;ch
0
(
I
)=
f
I
g
,ch
k
+1
(
I
)=
f
ch(
J
):
J
2
ch
k
(
I
)
g
.
h
f
i
I
Theaverageof
f
over
I
,
h
f
i
I
=

(
I
)

1
R
I
f
(
x
)

(
x
);
E
I
Theaveragingoperator,
E
I
f
:=
h
f
i
I
1
I
;notethat
E
k
=
P
I
2L
k
E
I
.

I
Martingaleoperator,
I
:=

E
I
+
P
J
2
ch(
I
)
E
J
;notethat
k
=
P
I
2L
k

I
.

n
I
Martingaleoperatoroforder
n
,

n
I
:=

E
I
+
X
J
2
ch
n
(
I
)
E
J
:
Sincethemeasure

isassumedtobewesometimeswillbeusing
j
E
j
for

(
E
)and
dx
for

(
x
).Wealsowillbeusing
L
2
for
L
2
(

)
Theprototypicalexampleis
X
=
R
or
R
d
with
L
beingadyadiclattice
D
.
2.3.4.1Haarshifts
WewilluseaslightlymoregeneralofaHaarshift.
34
18.
AHaarshift
S
ofcomplexity
n
isgivenby
S
f
=
X
I
2D
S
I

n
I
f;
wheretheoperators
S
I
acton
n
I
L
2
andcanberepresentedasintegraloperatorswith
kernels
a
I
,
k
a
I
k
1
j
I
j

1
.Thelattermeansthatforall
f;g
2

n
I
L
2
h
S
I
f;g
i
=
Z
I
Z
I
a
I
(
x;y
)
f
(
y
)
g
(
x
)
dxdy:
Thisisaslightlymoregeneralnitionthantheonein[HPTV],butonlytheestimate
k
a
I
k
1
j
I
j

1
isessentialforourconstruction.Notealsothataccordingtothe
in[HPTV]thecomplexityofthecorrespondingshiftis
n

1,not
n
,whichreallydoesnot
matter;wejustourofcomplexityabitmoreconvenient.
Theestimate
k
a
I
k
1
j
I
j

1
meansthattheoperators
S
I
are\
L
1

L
1
normalized",
meaningthat
jh
S
I
f;g
ijj
I
j
k
f
k
1
j
I
j
k
g
k
1
j
I
j
8
f;g
2

n
I
L
2
(2.28)
Haarshiftsofcomplexity1aresimply\
L
1

L
1
normalized"martingaletransforms;mar-
tingaletransformheremeansinparticularthatthesubspaces
I
areorthogonal,and
S
can
berepresentedasanorthogonalsumoftheoperators
S
I
.
AHaarshiftofcomplexity
n

2isnotgenerallyamartingaletransform,meaningthat
thesubspaces
n
I
generallyintersect,so
S
doesnotsplitintodirectsumof
S
I
.
However,ifonegoeswithstep
n
,thenthecorrespondingoperatorisamartingaletrans-
form,soaHaarshiftofcomplexity
n
canberepresentedasasumof
n
Haarshiftsof
35
complexity1.Namely,for
k
=1
;
2
;:::;n

1
L
k
=
f
I
:
I
2L
k
+
nj
;j
2
Z
g
;
andlet
S
k
=
X
I
2L
k
S
I
:
Then
S
=
P
n

1
k
=0
S
k
andeach
S
k
isaHaarshiftofcomplexity1withrespecttothelattice
L
k
.
Remark5.
Therefore,uniformestimatefortheHaarshiftsofcomplexity1(i.e.forthe
\
L
1

L
1
normalized"martingaletransforms)givesthelinearincomplexityestimateforthe
generalHaarshifts.Noticethattheestimatedoesnotdependonthenumberofchildren.
2.3.4.2Paraproducts
Giventhelattice
L
andalocallyintegrablefunction
b
,theparaproduct=
b
=
b
(
L
)is
as

f
:=
X
I
2L
(
E
I
f

I
b
)
:
Thenecessaryandtconditionfortheparaproducttobeboundedisthat
sup
J
2L
j
J
j

1
X
I
2L
:
I
ˆ
J
k

I
b
k
2
2
<
1
:
Inthecaseofdyadiclatticein
R
d
or,moregenerallyinthehomogeneoussituation,when
inf
J
2L
inf
I
2
ch(
J
)
j
I
j
j
J
j
>
0
36
thisconditionisequivalentto
b
belongingtothecorrespondingmartingaleBMOspace
BMO
L
2.3.5Reductiontothemartingalecase.
Toreducetheproblemtothemartingalecaseweusethefollowingresultthatcanbefound
in[H]and[HPTV]:
Theorem2.3.4.
Let
T
beaon{Zygmundoperatorin
R
d
.Thereexistsaprobability
space

;
P
)
ofdyadiclattices
D
!
,suchthat
T
=
C
 
Z

1
X
n
=1
2

"n
S
n
(
!
)
d
P
(
!
)+
Z


1
(
!
)+
2
(
!
))

)
d
P
(
!
)
!
;
where
S
n
(
!
)
areHaarshiftsofcomplexity
n
withrespecttothelattice
D
!
,

1
;
2
(
!
)
arethe
paraproductswithrespecttothelattice
D
!
,
k

1
;
2
(
!
)
k
1
.
Theconstants
C
and
"
dependon
d
,
k
T
k
andon{Zygmundparametersofthekernel
of
T
.
Theorem2.3.4impliesimmediatelythatthemaintheorem(Theorem2.3.3)followsfrom
thetheorembelow.
Theorem2.3.5.
Lettheweights
v
,
w
satisfytheassumptionsofTheorem2.3.3.Then
1.ForallHaarshifts
S
oforder
1
theoperators
S
(

v
)
areuniformlyboundedfrom
L
2
(
v
)
to
L
2
(
w
)
,
k
S
(

v
)
k
L
2
(
v
)
!
L
2
(
w
)

C
,where
C
dependson

2
,

2
,thesupremumin
(2.27)
,butnotonthelattice
L
.
2.ForallHaarshifts
S
n
theoperators
S
n
(

v
)
areuniformlyboundedfrom
L
2
(
v
)
to
L
2
(
w
)
by
Cn
,where
C
.
37
3.Let
=
b
beaparaproductsuchthat
j
J
j

1
X
I
2L
:
I
ˆ
J
k

I
b
k
2
1
j
I
j
1
8
J
2L
:
(2.29)
Thentheoperator


v
)
isboundedfrom
L
2
(
v
)
to
L
2
(
w
)
by
C
,whereagain
C
depends
on

1
,

2
,thesupremumin
(2.27)
,butnotonthelattice
L
.
Remark6.
Forthehomogeneouslattices,i.e.forlatticessatisfying
inf
J
2L
inf
I
2
ch(
J
)
j
I
j
j
J
j
=:
>
0
allthenormalized
L
p
norms
j
I
j

1
=p
k

I
g
k
p
,
p
2
[1
;
1
]areequivalentinthesenseoftwo
sidedestimates.Soforsuchlatticescondition(2.29)meansthat
k

k
C
(

).SoTheorem
2.3.5givestheestimatesthatbeingfedtoTheorem2.3.4implyTheorem2.3.3.
Theorem2.3.6.
Let

beasabove.Thenforanyweight
w
on
X
suchthat
n

(
N
w
I
)
<
1
forall
I
2L
X
I
2L
n

(
N
w
I
)

1

j
I
j

1
Z
X
j

I
(
fw
1
=
2
)
j
dx

2
j
I
j
C
k
f
k
2
L
2
(
dx
)
(2.30)
forall
f
2
L
2
(
dx
)
;here
C
=
C

andinthesummationweskip
I
onwhich
w

0
.
Letusseethatthistheoremimpliesthecondition1ofTheorem2.3.5.Assume,multi-
plyingtheweightsbyappropriateconstantsthattheinequality
n

1
(
N
w
I
)
n

2
(
N
v
I
)

1(2.31)
38
holdsforall
I
2L
.Then
jh
S
(
fw
1
=
2
)
;gv
1
=
2
ij
X
I
2L
jh
S
I

I
(
fw
1
=
2
)
;

I
(
gv
1
=
2
)
ij

X
I
2L
j
I
j

1
k

I
(
fw
1
=
2
)
k
1
k

I
(
gv
1
=
2
)
k
1

X
I
2L
j
I
j

1
k

I
(
fw
1
=
2
)
k
1
k

I
(
gv
1
=
2
)
k
1

n

1
(
N
w
I
)
n

2
(
N
v
I
)

1
=
2

1
2
X
I
2L
j
I
j

1
k

I
(
fw
1
=
2
)
k
2
1
n

1
(
N
w
I
)
+
1
2
X
I
2L
j
I
j

1
k

I
(
gv
1
=
2
)
k
2
1
n

2
(
N
v
I
)
:
Thesecondinequalityherefollowsfrom\
L
1

L
1
normalization"condition(2.28),thesecond
onefrom(2.31)andthelastoneisjustthetrivialinequality2
xy

x
2
+
y
2
.
ApplyingTheorem2.3.6toeachsumwegetthat
jh
S
(
fw
1
=
2
)
;gv
1
=
2
ij
1
2

C

1
)
k
f
k
2
2
+
C

2
)
k
g
k
2
2

:
Replacing
f
7!
tf
,
g
7!
t

1
g
,
t>
0weget
jh
S
(
fw
1
=
2
)
;gv
1
=
2
ij
1
2

t
2
C

1
)
k
f
k
2
2
+
t

2
C

2
)
k
g
k
2
2

:
Takingumoverall
t>
0andrecallingthat2
ab
=inf
t>
0
(
t
2
a
+
t

2
b
)for
a;b

0we
obtain
jh
S
(
fw
1
=
2
)
;gv
1
=
2
ij
(
C

1
)
C

2
))
1
=
2
k
f
k
2
k
g
k
2
;
whichisexactlystatement1ofTheorem2.3.5.
Forthestatement3ofTheorem2.3.5
wealsoneedanotherembeddingtheorem.
39
Theorem2.3.7.
Let

beasabove.ThenforanynormalizedCarlesonsequence
f
a
I
g
I
2D
(
a
I

0
),i.e.foranysequencesatisfying
sup
I
2D
j
I
j

1
X
I
0
2D
:
I
0
ˆ
I
a
I
0
j
I
0
j
1
weget
X
I
2D
h
fw
1
=
2
i
2
I
n

(
N
w
I
)
a
I
j
I
j
C
k
f
k
2
L
2
(
dx
)
;
whereagain
C
=
C

.
LetusshowthatthistheoremtogetherwithTheorem2.3.6impliesstatement3ofThe-
orem2.3.5.Let
a
I
=
k

I
b
k
2
1
.
Again,multiplyingifnecessarytheweights
v
and
w
byappropriateconstantswecan
assume(2.31).Thenwecanwrite
jh

b
(
fw
1
=
2
)
;gv
1
=
2
ij
X
I
2D
jh
fw
1
=
2
i
I
jjh

I
b;

I
(
gv
1
=
2
)
ij

X
I
2D
jh
fw
1
=
2
i
I
j
(
a
I
)
1
=
2
j
I
j
1
=
2

n

1
(
N
w
I
)

1
=
2

k

I
(
gv
1
=
2
)
k
1

n

2
(
N
v
I
)

1
=
2
j
I
j
1
=
2

0
@
X
I
2D
jh
fw
1
=
2
i
I
j
2
a
I
n

1
(
N
w
I
)
j
I
j
1
A
1
=
2
0
@
X
I
2D
k

I
(
gv
1
=
2
)
k
2
1
n

2
(
N
v
I
)
j
I
j
;
1
A
1
=
2
thesecondinequalityholdsbecauseof(2.31),andthelastoneisjusttheCauchy{Schwarz
inequality.
EstimatingthesumsinparenthesesbyTheorem2.3.7and2.3.6respectivelyweget
statement3ofTheorem2.3.5.
40
2.3.6Proofof(thentialEmbedding)Theorem2.3.6:Bellmanfunc-
tionandmaintialinequality
Let
'
(
s
):=
s

s
).Multiplyingbyanappropriateconstantwecanassumewithoutloss
ofgeneralitythat
Z
1
0
1
'
(
s
)
ds
=1
:
(2.32)

m
(
s
)on[0
;
1]by
m
(0)=
m
0
(0)=0,
m
00
(
s
)=1
='
(
s
).Identity(2.32)impliesthat
m
iswand0

m
0
(
s
)

1,0

m
(
s
)

s
.Foradistributionfunction
N
=
N
w
I

u
(
N
)=
Z
1
0
(2
N
(
t
)

m
(
N
(
t
)))
dt
=2
h
w
i
I

Z
1
0
m
(
N
(
t
))
dt
;(2.33)
Notethattheinequality
m
(
s
)

s
impliesthat
u
(
N
w
I
)
h
w
i
I
.
Thefunctional
u
isontheconvexsetofdistributionfunctions,i.e.onthesetof
decreasingfunctions
N
:[0
;
1
)
!
[0
;
1]suchthat
R
1
0
N
(
t
)
dt<
1
.
Inwhatfollowswecanconsideronlysupportedfunctions
N
,andthenuse
standardapproximationreasoning.Considertwodistributionfunctions
N
and
N
1
andlet

N
=
N
1

N
.Denotealso
w
:=
Z
1
0
N
(
t
)
dt;
w
1
:=
Z
1
0
N
1
(
t
)
dt;
andlet

w
:=
w
1

w
=
Z
1
0

N
(
t
)
dt
;
themotivationforthisnotationisthatif
N
and
N
1
arethedistributionfunctionsofthe
41
weights
w
and
w
1
,thentheintegralsaretheaveragesonthecorrespondingweights.Denote
also
w

:=
Z
1
0
j

N
(
t
)
j
dt
;(2.34)
clearly
j

w
j
w

.
Letuscomputederivativesof
u
inthedirectionof

N
.Therstderivativeisgivenby
u
0

N
(
N
)=
d
d˝
u
(
N
+
˝

N
)



˝
=0
=
Z
1
0

2

m
0
(
N
(
t
))


N
(
t
)
dt;
so,inparticular
j
u
0

N
j
2
w

:
Thereforewecanwrite
u
0

N
=

w

;
=

(

N
)
;
j

j
2
:
(2.35)
Thesecondderivativeinthedirection

N
=
N
1

N
isgivenby

u
00

N
(
N
)=

d
2
d˝
2
u
(
N
+
˝

N
)



˝
=0
=
Z
1
0
'
(
N
(
t
))

1
(

N
(
t
))
2
dt
42
ByCauchy-Schwarz,theintegralintherightsideisatleast
h
Z
1
0
N
(
t

N
(
t
))
dt
i

1
h
Z
1
0
j

N
(
t
)
j
dt
i
2
=
n
(
N
)

1
h
Z
1
0
j

N
(
t
)
j
dt
i
2
=
n
(
N
)

1
(
w

)
2
;
so

u
00

N
(
N
)

(
w

)
2
n
(
N
)
(2.36)
Forthescalarvariable
f
2
R
andthedistributionfunction
N
theBellmanfunction
e
B
(
f;N
)=
B
(
f
;
u
(
N
))where
B
(
f
;
u
)=
f
2
u
:
Computingsecondderivativeof
e
B
inthedirection

=(

f
;

N
)weget
e
B
00

=
0
B
@

f
u
0

N
1
C
A
T
0
B
@
B

B
fu
B
fu
B
uu
1
C
A
0
B
@

f
u
0

N
1
C
A
+
B
u
u
00

N
Inthelastformulathederivativeof
e
B
isevaluatedatthepoint(
f;N
),andderivatives
of
B
areevaluatedat(
f
;
u
(
N
)).
TheHessianiseasytocompute
0
B
@
B

B
fu
B
fu
B
uu
1
C
A
=
0
B
@
2
u

2
f
u
2

2
f
u
2
2
f
2
u
3
1
C
A
;(2.37)
notethatthismatrixispositive
43
Since
B
u
=

f
2
=
u
2
,wegetusing(2.36)
B
u
u
00

N

f
2
u
2
n
(
w

)
2
:
Thus,gatheringeverythingandusing(2.35)weget
e
B
00


0
B
@

f

w

1
C
A
T
0
B
@
2
u

2
f
u
2

2
f
u
2
2
f
2
u
3
(1+
u
2

2
n
)
1
C
A
0
B
@

f

w

1
C
A
(2.38)
ThematrixhereisobtainedfromtheHessianin(2.37)bymultiplyingthelowerrightentry
by1+
u
2

2
n

1,soithasmorepositivitythantheHessian.Inparticular,ifwedividethe
upperleftentryofthematrixin(2.38)bythesamequantity1+
u
2

2
n
,thematrixstillbe
positivee.Butourmatrixin(2.38)hassomethingbiggerintheupper-leftcorner!
Therefore,since
1


1+
u
2

2
n


1
=
u
2

2
n
+
u
wegetthat
e
B
00


2(

f
)
2
2

2
n
+
u

2(

f
)
2
2

2
2
n
+
u

c
(

f
)
2
n
;(2.39)
thelastinequalityholdsforsome
c>
0because
u

2
w

C
n
.
Letusexplainit.Infact,wewant
Z
N
I
(
t
)
dt
=
h
w
i
I

C
Z
N
I
(
t

N
I
(
t
))
dt:
Clearly,itisenoughtoconsidertheset
B
=
f
t
:
N
I
(
t
))

1
g
.Sinceisdecreasing,for
44
t
2
B
wegetthat
N
I
(
t
)



1
(1).Since
s
7!
s

s
)isincreasing,weget
N
I
(
t

N
I
(
t
))



1
(1)



1
(1)
N
I
(
t
)(thelastisbecause
N
I
isnormalized).Wearedone.
Inequality(2.39)isexactlywhatwewillusetoobtaintheMaininequalityin
forminthenextsection.
2.3.7Maininequalityintheform
2.3.7.1Dyadiccase
Lemma2.3.8.
Let
f
=
f
1
+
f
2
2
;N
(
t
)=
N
1
(
t
)+
N
2
(
t
)
2
:
Then
1
2

B
(
f
1
;
u
(
N
1
))+
B
(
f
2
;
u
(
N
2
))

B
(
f
;
u
(
N
))

c
4

(
f
1

f
)
2
n
(
N
)
:
(2.40)
where
c
istheconstantfrom
(2.39)
.(Notethat
f
1

f
=
f

f
2
,sowecanreplace
(
f
1

f
)
2
intherightsideby
(
f
2

f
)
2
)
Proof.
Noticethat
s
1
+
s
2
2


s
1
+
s
2
2


s
1
+
s
2
2
(
s
1
+
s
2
)

1
2
s
1

s
1
);(2.41)
heretheinequalityholdsbecauseisdecreasingandthesecondonebecause
s

s
)is
increasing.Ofcourse,wecaninterchange
s
1
and
s
2
intheaboveinequality.
45
Let

f
:=
f
1

f
,

N
:=
N
1

N
.
F
(
˝
)=
B
(
f
+
˝

f
;
u
(
N
+
˝

N
))+
B
(
f

˝

f
;
u
(
N

˝

N
))
Taylor'sformulatogetherwiththeestimate(2.39)implythat
F
(1)

F
(0)

c
2
(

f
)
2

1
n
(
N
+
˝

N
)
+
1
n
(
N

˝

N
)

(2.42)
forsome
˝
2
(0
;
1).
Estimate(2.41)impliesthat
n
(
N
)

1
2
n
(
N

˝

N
)
;
so

1
n
(
N
+
˝

N
)
+
1
n
(
N

˝

N
)


1
n
(
N
)
:
Thenitfollowsfrom(2.42)that
F
(1)

F
(0)

c
2

(

f
)
2
n
(
N
)
:
Recallingthenitionof
F
anddividingthisinequalityby2weget(2.40).
2.3.7.2Generalcase
Let
'
and
e
B
beasabove.
Lemma2.3.9.
Let
f
;
f
k
2
R
,

k
2
R
+
andthedistributionfunctions
N
,
N
k
,
k
=1
;
2
;:::;n
46
satisfy
f
=
n
X
k
=1

k
f
k
;N
=
n
X
k
=1

k
N
k
;
n
X
k
=1

k
=1
:
Then

e
B
(
f
;N
)+
n
X
k
=1

k
e
B
(
f
k
;N
k
)

c
16

1
n
(
N
)
 
n
X
k
=1

k
j
f
k

f
j
!
2
(2.43)
2.3.8Proofof(theEmbedding)Theorem2.3.7.
2.3.8.1Anauxiliaryfunction
LetbethefunctionfromTheorem2.3.7.
'
(
s
):=
s

s
).
Forthenumbers
A
2
[1
;
2],
N
2
R
+

T
(
A;N
):=
N
Z
N=A
0
1
'
(
s
)
ds
Lemma2.3.10.
Thefunction
T
isconvexandtheentialinequality

@T
@A

1
4

N
2
'
(
N
)
:
Proof.
tiatingtheintegralweget

@T
@A
=
N
2
A
2
'
(
N=A
)

1
4

N
2
'
(
N
)
:
(2.44)
since
'
isincreasingand1

A

2.
Toprovetheconvexitynoticethat
T
islinearonthelines
N
=
kA
,sotheHessian
d
2
T
47
degenerates.
tiating(2.44)weget
@
2
T
@A
2
=
N
2
2
A'
(
N=A
)

N'
0
(
N=A
)
(
A
2
'
(
N=A
))
2
Notethattherightsideispositiveif
s'
0
(
s
)
<
2
'
(
s
)(because
'
(
s
)
>
0).
Butforourfunctionevenastrongerinequality
s'
0
(
s
)

'
(
s
)isIndeed,since
'
(
s
)=
s

s
)isincreasingandisdecreasing,then
0

(
s

s
))
0
=
s
)+
s

0
(
s
)


s
)
(thesecondinequalityholdsbecauseisdecreasing).Multiplyingthisinequalityby
s
we
get
s'
0
(
s
)

'
(
s
).
Therefore,since
'
(
s
)
>
0,weconcludethat
@
2
T
@A
2
>
0.
ButtheHessian
d
2
T
issingular,anditisaneasyexerciseinlinearalgebratoshowthat
asingularHermitian2

2matrixwithapositiveentryonthemaindiagonalispositive

2.3.8.2Bellmanfunctionandthemaintialinequality.
Letnow
N
beadistributionfunction,andlet
T
(
A;N
)=
Z
1
0
T
(
A;N
(
t
))
dt:
48
AsinSection2.3.6assume,multiplyingbyanappropriateconstant,that
Z
1
0
1
'
(
s
)
ds
=1
:
Then
T
(
A;N
(
t
))

N
(
t
),so
T
(
A;N
)

Z
1
0
N
(
t
)
dt
=:
w
=
w
(
N
)
:
For
f
2
R
,
M
2
[0
;
1]andforadistributionfunction
N
thefunction
e
B
(
f
;N;M
):=
B
(
f
;
u
(
M;N
)),where
B
(
f
;
u
)=
f
2
u
and
u
=
u
(
M;N
)=2
Z
1
0
N
(
t
)
dt

T
(
M
+1
;N
)
=:2
w
(
N
)

T
(
M
+1
;N
)
:
Notethat2
w
(
N
)

u
(
M;N
)

w
(
N
).
Weclaimthatthefunction
e
B
isconvex.Indeed,adirection

:=(

f
;

N;

M
)
T
and
computethesecondderivative
e
B
00

inthisdirection
e
B
00

=
d
2
d˝
2
e
B
(
f
+
˝

f
;N
+
˝

N;M
+
˝

M
)



˝
=0
:
49
Weget
e
B
00

=
0
B
@

f
u
0

1
C
A
T
0
B
@
B

B
fu
B
fu
B
uu
1
C
A
0
B
@

f
u
0

1
C
A
+
B
u
u
00

:
TheHessian
0
B
@
B

B
fu
B
fu
B
uu
1
C
A
=
0
B
@
2
u

2
f
u
2

2
f
u
2
2
f
2
u
3
1
C
A
isclearlypositivesothetermisnonnegative.Forthesecondtermnotice
that
B
u
=

f
2
u
2
;
u
00

=

T
00


0(2.45)
because
T
,andtherefore
T
isconvex.Thus
e
B
00


0,so
e
B
isconvex.Letuscomputethe
partialderivative

@
e
B
@M
=

u
@
u
@M
=
f
2
u
2



@
T
@M

(2.46)
ByLemma2.3.10

@
T
@M

1
4

Z
1
0
N
(
t
)
2
'
(
N
(
t
))
dt

1
4

Z
1
0
N
(
t
)
dt

2

Z
1
0
'
(
N
(
t
))
dt


1
=
1
4

w
(
N
)
2
n
(
N
)
;
thesecondinequalityhereisjusttheCauchy{Schwarzinequality.Combiningwith(2.46)
50
andrecallingthat
u

2
w
weget

@
e
B
@M

1
16

f
2
n
(2.47)
Thisinequality(togetherwiththeconvexityof
e
B
)isthemaintialinequalityforour
function.
2.3.9Finiteformofthemaininequality
Let
X
=(
f
;N;M
),
X
k
=(
f
k
;N
k
;M
k
),(
f
;
f
k
2
R
,
M;M
k
2
[0
;
1],
N
,
N
k
arethedistribution
functions)satisfy
f
=
n
X
k
=1

k
f
k
;N
=
n
X
k
=1

k
N
k
;M
=
a
+
n
X
k
=1

k
M
k
;a

0
;
where
n
X
k
=1

k
=1
;
k

0
:
Then

e
B
(
X
)+
n
X
k
=1

k
e
B
(
X
k
)

1
16

a
f
2
n
(2.48)
where
n
=
n
(
N
).
Indeed,for
M
0
:=
P
n
k
=1

k
M
k
themaininequality(2.47)implies
e
B
(
f
;N;M
0
)

e
B
(
f
;N;M
)

1
16

a
f
2
n
:
51
Theconvexityof
e
B
impliesthat
e
B
(
f
;N;M
0
)

n
X
k
=1

k
e
B
(
X
k
)
whichtogetherwiththepreviousinequalitygivesus(2.48).
2.3.9.1Frommaininequality(2.48)toTheorem2.3.7.
ThereasoninghereisalmostverbatimthesameasinSection2.3.10.
Foracube
I
2L
letusdenote
f
I
=
h
fw
1
=
2
i
I
,
N
I
=
N
w
I
,
M
I
=
j
I
j

1
P
I
0
ˆ
I
a
I
0
,
w
I
=
h
w
i
I
,
u
I
=
u
(
M
I
;N
I
).
Fix
I
0
2L
,andlet
I
k
beitschildren.Applyingtheinequality(2.48)with

k
=
j
I
k
j
=
j
I
0
j
,
f
k
=
f
I
k
,
N
k
=
N
w
I
k
,
M
k
=
M
I
k
wegetthat
1
16

a
I
0
f
2
I
0
n
(
N
w
I
0
)
j
I
0
j
I
0
j
e
B
(
X
I
0
)+
X
I
2
ch(
I
0
)
j
I
j
e
B
(
X
I
)
Writingthecorrespondingestimatesforthechildrenof
I
0
,thenfortheirchildren,weget
aftergoing
n
generationsdownandusingthetelescopingsumintherightside
1
16
X
I
2
ch
k
(
I
0
)
0

k<n
a
I
f
2
I
n
(
N
w
I
)
j
I
jj
I
0
j
e
B
(
X
I
0
)+
X
I
2
ch
n
(
I
0
)
j
I
j
e
B
(
X
I
)

X
I
2
ch
n
(
I
0
)
j
I
j
e
B
(
X
I
);
thelastinequalityholdsbecause
e
B
0.
52
Since
e
B
(
X
I
)

f
2
I
=
u
I

f
2
I
=
w
I
(thelastinequalityholdsbecause
u

w
)andbyCauchy{Schwarz
jh
fw
1
=
2
i
I
j
2
hj
f
j
2
i
I
h
w
i
I
;
weconclude,exactlyasinSection2.3.10that
j
I
j
e
B
(
X
I
)
j
I
jhj
f
j
2
i
I
=
Z
I
j
f
j
2

so
1
16
X
I
2
ch
k
(
I
0
)
0

k<n
a
I
f
2
I
n
(
N
w
I
)
j
I
j
Z
I
0
j
f
j
2

ConclusionoftheproofisexactlyasinSection2.3.10:welet
n
!1
,andthentaking
thesumover
I
0
2L

m
andletting
m
!1
getthedesiredestimate.
Proof.
Thereasoningbelowisa\babyversion"ofthereasoningusedtoprovethemain
estimate(Lemma6.1)in[T].
Foraweight

=
f

k
g
n
k
=1
,

k

0,let
`
p
(

)betheweighted
`
p
spaces,
k
x
k
p
`
p
(

)
=
P
n
k
=1

k
j
x
k
j
p
(
`
1
(

)isjusttheusuall
`
1
).
Let
h
;
i

bethestandardduality
h
x;y
i

=
P
n
k
=1

k
x
k
y
k
.

e
2
`
p
(

),
e
=(1
;
1
;:::;
1).
53
Considerthequotientspace
X
=
`
1
(

)
=
span
f
e
g
.For
x
2
`
1
(

)let
x
0
:=
x
k
e
k

1
`
1
(

)
h
x;
e
i

e
;
so
P
n
k
=1

k
x
0
k
=0.Then
k
x
k
X
k
x
0
k
`
1
(

)

2
k
x
k
X
:
(2.49)
Indeed,theinequalityistrivial(followsfromtheofthenorminthequotient
space).Asforthesecondone,
jh
x;
e
i

jk
x
k
`
1
(

)
,soitfollowsfromthetriangleinequality
that
k
x
0
k
`
1
(

)
k
x
k
`
1
(

)
+
k
e
k

1
`
1
(

)
jh
x;
e
ijk
e
k
`
1
(

)

2
k
x
k
`
1
(

)
:
Thisinequalityremainstrueifonereplaces
x
by
x


e
,

2
R
,sothesecondinequalityin
(2.49)isproved.
Thedualspace
X

canbeidenwithssubspaceof
`
1
=
`
1
(

)consistingof
x

2
`
1
(

)suchthat
h
e
;x

i

=0(withtheusual
`
1
-norm).
So,forthevector
x
=(
x
1
;x
2
;:::x
n
),
x
k
=
f
k

f
(noticethat
h
x;
e
i

=0thereis

=
f

k
g
n
k
=1
,
j

k
j
1suchthat
P
n
k
=1

k

k
=0and
n
X
k
=1

k

k
(
f
k

f
)=
k
x
k
X

1
2
k
x
k
`
1
(

)
=
1
2
n
X
k
=1

k
j
f
k

f
j
:

f
+
,
f

,
N
+
,
N

by
f

=
n
X
k
=1

k
(1


k
)
f
k
;N

:=
n
X
k
=1

k
(1


k
)
N
k
:
54
ByLemma2.3.8
1
2

e
B
(
f
+
;N
+
))+
e
B
(
f

;N

))


e
B
(
f
;N
)

c
4

(
f
+

f
)
2
n
(
N
)
(2.50)
Weknowthat
f
+

f
=
n
X
k
=1

k

k
f
k
=
n
X
k
=1

k

k
(
f
k

f
)

1
2
n
X
k
=1

k
j
f
k

f
j
(thesecondequalityholdsbecause
P
n
k
=1

k

k
=0),sotherightsideof(2.50)isestimated
belowby
c
16

1
n
(
N
)
 
n
X
k
=1

k
j
f
k

f
j
!
2
Sincethefunction
e
B
isconvex
e
B
(
f
+
;N
+
)

n
X
k
=1

k
(1+

k
)
e
B
(
f
k
;N
k
)
;
e
B
(
f

;N

)

n
X
k
=1

k
(1


k
)
e
B
(
f
k
;N
k
)
andaddingtheseinequalitieswecanestimateabovetheleftsideof(2.50)by

e
B
(
f
;N
)+
n
X
k
=1

k
e
B
(
f
k
;N
k
)
:
55
2.3.10Frommaininequality(2.43)toTheorem2.3.6.
Fixaninterval
I
0
andlet
I
k
beitschildren.ApplyingLemma2.3.9with
f
k
=
h
fw
1
=
2
i
I
k
,
N
k
=
N
w
I
k
and

k
=
j
I
k
j
=
j
I
0
j
wegetdenoting
~
f
:=
fw
1
=
2
c
16

k

I
0
~
f
k
2
1
n
(
N
w
I
0
)
j
I
0
j

I
0
j
e
B
(
h
~
f
i
I
0
;N
w
I
0
)+
X
I
2
ch(
I
0
)
j
I
j
e
B
(
h
~
f
i
I
;N
w
I
)
Applyingthisformulatoallchildrenof
I
0
,thentotheirchildrenandaddinguptheinequal-
itieswegetaftergoing
n
generationsdownthat
c
16
X
I
2
ch
k
(
I
0
)
0

k<n
k

I
~
f
k
2
1
n
(
N
w
I
)
j
I
j

I
0
j
e
B
(
h
~
f
i
I
0
;N
w
I
0
)+
X
I
2
ch
n
(
I
0
)
j
I
j
e
B
(
h
~
f
i
I
;N
w
I
)

X
I
2
ch
n
(
I
0
)
j
I
j
e
B
(
h
~
f
i
I
;N
w
I
)
:
Weknowthat
e
B
(
f
;N
)

C
f
2
u
(
N
)
,andsince(see(2.33))
u
(
N
w
I
)
h
w
i
I
weconcludeusing
theCauchy{Schwarzestimate
jh
fw
1
=
2
i
I
j
2
hj
f
j
2
i
I
h
w
i
I
that
j
I
j
e
B
(
h
~
f
i
I
;N
w
I
)

C
j
I
j
h
fw
1
=
2
i
2
I
h
w
i
I
=
C
j
I
jhj
f
j
2
i
I
=
C
Z
I
j
f
j
2

Therefore,estimatingtherightsideweget
c
16
X
I
2
ch
k
(
I
0
)
0

k<n
k

I
~
f
k
2
1
n
(
N
w
I
)
j
I
j

C
Z
I
0
j
f
j
2

56
Sincetherightsidedoesnotdependon
n
wecanmake
n
!1
,andhavethesuminthe
leftsideoverall
I
2L
,
I
ˆ
I
0
.
Takingthesumoverall
I
0
2L

m
andletting
m
!1
wegetconclusionofthetheorem.
2.4TheBellmanfunctionapproachtotheseparatedbumpconjec-
ture
2.4.1Aquickreminder
WeagainworkwithYoungfunctionsthatsatisfy
isconvexincreasingfunctionsuchthat
Z
1
1
dt

t
)
<
1
:
(2.51)
Aswehaveseen,intheseparatedbumpconjectureweworkwiththefollowingquitenatural
one-sidedbumpassumption:
Thereexistsaconstant
C
,suchthatforanyinterval
I
thefollowingholds:
k
u
k
L

I
k
v
k
L
1
(
I;
dx
j
I
j
)
6
C;
and
k
u
k
L
1
(
I;
dx
j
I
j
)
k
v
k
L

I
6
C:
(2.52)
Andnowone-sidedbumpconjectureisthefollowingstatement:suppose(2.52)holds
forallintervals(cubes),andsupposeintegrabilitycondition(2.51),thenany
57
on{Zygmundoperatorisboundedfrom
L
2
(
u
)into
L
2
(
v
)inthesense
8
f
2
C
1
0
k
T
(
fu
)
k
L
2
(
v
)
6
C
k
f
k
L
2
(
u
)
:
(2.53)
2.4.2Aconstructionoffunction
Toformulatethemainresultweuseacertainlanguage.
Forthatweneedthefollowingconstruction.afunctioninthefollowingpara-
metricway:
8
>
>
>
<
>
>
>
:
s
=
1

t

0
(
t
)

s
):=
0
(
t
)
:
Ofcourse,weinthiswaynear
s
=0.
Wegivethefollowing
19.
Afunctioniscalled
regularbump
,ifforanyfunction
u
thereholds
k
u
k
L

I
>
C
Z
N
I
(
t

N
I
(
t
))
dt:
Remark7.
Anexampleofregularbumpisthefollowing:
t
)=
tˆ
(
t
),and
t
ˆ
0
(
t
)
ˆ
(
t
)
log
ˆ
(
t
)
!
0
;
as
t
!1
:
Theimportantresultisthefollowing.
Lemma2.4.1.
Thefunction
s
7!

s
)
isdecreasing;thefunction
s
7!
s

s
)
isincreasing;
thefunction
1
s

s
)
isintegrablenear
0
.Moreover,thefollowinginequalityistruewitha
58
uniformconstant
C
(whichmaydependonlyon

):
C
k
u
k
L

I
>
Z
N
I
(
t

N
I
(
t
))
dt;
where
N
I
(
t
)=
1
j
I
j
jf
x
2
I
:
u
(
x
)
>
t
gj
:
Further,for\regular"functions

wehavethat
k
u
k
L

I
˘
Z
N
I
(
t

N
I
(
t
))
dt:
2.4.3Themainresults.Boundednessandweakboundedness.
Givenafunctionsatisfying(2.51),buildthecorrespondingfunctionasinSection2.4.2.
Weprovethefollowingtheorems.Regularityconditionsarenotveryimportant,butthelast
conditioninthestatementofthetheoremisactuallyanimportantrestriction.Thisisthe
restrictiononewouldwishtogetridof.Ortoprovethatitisactuallyneeded.Latelywe
believethatonecannotgetridofit.Wegiveanon-standardn.
20.
Afunction
f
is\weaklyconcave"onitsdomain,ifforanynumbers
x
1
;:::;x
n
and

1
;:::;
n
,suchthat0
6

j
6
1,and
P

j
=1,thefollowinginequalityholds:
f
(
X

j
x
j
)
>
C
X

j
f
(
x
j
)
;
wheretheconstant
C
doesnotdependon
n
.
Theorem2.4.2.
Supposethereexistsafunction

0
withcorresponding

0
,suchthat:
59


0

(2.51)
;


and

0
areregularbumps;

Thereisafunction
"
,suchthat

0
(
s
)
6
C

s
)
"

s
))
;

Thefunction
t
7!
t"
(
t
)
isweaklyconcave,inthesenseofthe20;

Thefunction
t
7!
t"
(
t
)
isstrictlyincreasingnear
1
;

Thefunction
t
7!
t"
(
t
)
isconcavenear
1
;

Thefunction
t
7!
"
(
t
)
t
isintegrableat
1
.
Supposethatthereexistsaconstant
C
,suchthataone-sidedbumpcondition
(2.52)
holds.
Thenanyon{Zygmundoperatorisboundedfrom
L
2
(
u
)
into
L
2
(
v
)
inthesenseof
(2.53)
.
Theorem2.4.3.
Supposethefunction

allconditionsfromthetheoremabove.
Supposethatthereexistsaconstant
C
,suchthat
k
u
k
L
1
(
I;
dx
j
I
j
)
k
v
k
L

I
6
C:
ThenanyCalderon-Zygmundoperatorisweaklyboundedfrom
L
2
(
u
)
into
L
2
;
1
(
v
)
,i.e.there
existsaconstant
C
,suchthatforanyfunction
f
2
C
1
0
thereholds
k
T
(
fu
)
k
L
2
;
1
(
v
)
6
C
k
f
k
L
2
(
u
)
:
(2.54)
60
2.4.4Examplesofsatisfyingtherestrictionsofthemainresults:thecases
from[CURV]
Thebiggestoftheaboveresultswiththoseof[CURV]isthatherewegavethe
integralconditiononthecorrespondingbumpfunctionTocomparewith[CURV]we
noticethatin[CURV]theoremsabovewereprovedintwocases:
1.
t
)=
t
log
1+
˙
(
t
);
2.
t
)=
t
log(
t
)loglog
1+
˙
(
t
),fortlybig
˙
.
Weshowthattheseresultsarecoveredbyourtheorems.
First,suppose
t
)=
t
log
1+
˙
(
t
).Then
s
)

log
1+
˙
(
1
s
).Weput
0
(
s
)=
t
log
1+
˙
2
(
t
),
andthen
"
(
t
)=
t

˙
2(1+
˙
)
.Then,clearly,allpropertiesof
"
fromourtheoremare
Next,suppose
t
)=
t
log(
t
)loglog
1+
˙
(
t
).Then
s
)

log(
1
s
)loglog
1+
˙
(
1
s
).Weput

0
(
t
)=
t
log(
t
)loglog
1+
˙
(
t
),
<
1whichgives
"
(
t
)=log

(1


)
˙
(
t
).Then,theintegral
R
1
"
(
t
)
t
dt
convergesif
˙>
1,andwechoose

tobeverysmall.
Moreover,examiningtheproofofTheorem5.1from[CURV],wegettheresultfromour
paperbutwithacondition
Thefunction
t
7!
p
"
(
t
)
t
isintegrableat
1
.
Wenoticethatforregularfunctionswehave
"
(
t
)
!
0when
t
!1
,andso
"
(
t
)
<
p
"
(
t
).
Thus,ourresultsworkformorefunction
"
and,thus,bumps
61
2.4.5SelfimprovementsofOrlicznorms.
Inthissectionweproveatechnicalresult,whichhasthefollowing\hand-waving"explana-
tion:supposewetakeafunctionandasmallerfunction
0
.Weexplainhowsmallcanbe
thequotient
k
u
k
L

0
I
k
u
k
L

I
intermsofsmallnessof

0

.Inwhatfollowsweconsideronly\regular
bumps"functionsinthesenseofthe19.
Supposewehavetwofunctionsand
0
,andwehavebuiltfunctionsand
0
.We
supposethat

0
(
s
)
6
C

s
)
"

s
))
:
Thefollowingtheoremholds.
Theorem2.4.4.
Let
I
beanarbitraryinterval(cube).Ifafunction
t
7!
t"
(
t
)
isweakly
concave,then
k
u
k
L

0
I
6
C
k
u
k
L

I
"

k
u
k
L

I
h
u
i
I

:
Todothatweneedthefollowingeasylemma:
Lemma2.4.5.
ForweaklyconcavefunctionstheJenseninequalityholdswithaconstant:
Z
f
(
g
(
t
))

(
t
)
6
Cf
(
Z
g
(
t
)

(
t
))
:
Proof.
Thisistruesinceif
g
isastepfunction,thenthisisjustaThenwepass
tothelimit.Hereweessentiallyusedthatwecantakeaconvexcombinationof
n
points,
andtheconstantintheabovedoesnotdependon
n
.
ProofoftheTheorem.
Intheproofweomittheindex
I
.Sinceforregularbumpsweknow
62
that
k
u
k
L

˘
Z

N
(
t
))
N
(
t
)
dt;
wesimplyneedtoprovethat
Z

0
(
N
(
t
))
N
(
t
)
dt
6
C
Z

N
(
t
))
N
(
t
)
dt"

R

N
(
t
))
N
(
t
)
dt
R
N
(
t
)
dt

Ourstepistheobviousestimateoftheleft-handside:
Z

0
(
N
(
t
))
N
(
t
)
dt
6
C
Z

N
(
t
))
"

N
(
t
))
N
(
t
)
dt:
Denote
a
(
t
)=
t"
(
t
).Thenweneedtoprovethat
Z
a

N
(
t
))
N
(
t
)
dt
6
C
Z
N
(
t
)
dta

R

N
(
t
))
N
(
t
)
dt
R
N
(
t
)
dt

:
Wedenote

=
N
(
t
)
R
N
(
t
)
dt
dt;
itisaprobabilitymeasure.Moreover,byassumption,
t
7!
a
(
t
)isconcave.Therefore,by
Jensen'sinequality(fromtheLemma),
Z
a
(
f
(
t
))

(
t
)
6
Ca

Z
f
(
t
)

(
t
)

:
Take
f
(
t
)=(
N
(
t
)),andtheresultfollows.
63
2.4.6Examples
2.4.6.1Example1:log-bumps
First,if
t
)=
t
log
1+
˙
(
t
),then
s
)=log
1+
˙
(1
=s
),and

0
(
s
)

s
)
=log

˙
2
(1
=s
)=

˙
2(1+
˙
)
:
Thus,
"
(
t
)=
t

˙
2(1+
˙
)
,andeverythingis
2.4.6.2Example2:loglog-bumps
Nextexampleiswithdoublelogs.Infact,when

s
)=
log
(1
=s
)(loglog(1
=s
))
1+
˙
;

0
(
s
)=
log
(1
=s
)(loglog(1
=s
))
1+
˙=
2
then

0
(
s
)

s
)
=loglog

˙=
2
(1
=s
)
˘
(log
s
)))

˙=
2
:
Thus,
"
(
t
)=(log
t
)

˙
2
.Everythingwouldbealsoexceptforonelittlething:the
function
t
7!
t"
(
t
)isconcaveony,butnotnear1.However,
t
7!
t"
(
t
)isweakly
concaveon[2
;
1
),andthisisenoughforourgoalsaswithoutlossofgenerality,
s
)
>
2.
Soletusprovethat
a
(
t
)=
t"
(
t
)isweaklyconcaveon[2
;
1
).
Let
{
:=
˙
2
.Thefunction
a
hasalocalminimumat
e
{
anditsconcavitychangesat
e
{
+1
.Wenowtake
x
j
,

j
and
x
=
P

j
x
j
.Wenoticethatif
x>e
{
+1
,theweare
done,becausethen(
x;
P

j
a
(
x
j
))liesunderthegraphof
a
.
64
If2

x<e
{
+1
,then
a
(
x
)
>
min
[2
;e
{
+1
]
a
=
c
(
{
).Moreover,if
`
isalinetan-
genttographof
a
,startingat(2
;a
(2)),and
`
\kisses"thegraphatapoint(
r;a
(
r
)),then
P

j
a
(
x
j
)
6
a
(
r
)=
c
1
(
{
).Thisfollowsfromthepicture:aconvexcombinationof
a
(
x
j
)
cannotbehigherthanthisline.
Therefore,
a
(
X

j
x
j
)
>
c
(
{
)
>
Cc
1
(
{
)
>
X

j
a
(
x
j
)
:
Thisourproof.
2.4.7Proofofthemainresult:notationandthereduction.
Weadyadicgrid
D
.Toproveourmainresultsitisenoughtoshowthatthefollowing
implicationholds:
ifforall
I
k
u
k
L

I
k
v
k
L
1
(
I;
dx
j
I
j
)
6
B
u;v
then
k
˜
J
T
D
;
f
a
I
g
(
u˜
J
)
k
2
L
2
(
v
)
6
Cu
(
J
)
;
where
C
doesnotdependneitheronthegrid,noronthesequense
f
a
I
g
.Itcan,ofcourse,
dependon
B
u;v
.Thiswillprovetheweakbound
T
:
L
2
(
v
)
!
L
2
;
1
(
u
).Forsimplicity,we
denote
T
a
=
T
D
;
f
a
I
g
.Itisaneasycalculationthat,underthejoint
A
2
condition(whichis
underthebumpcondition),itisenoughtogetanestimateofthefollowing
form:
1
j
J
j
X
J
ˆ
I
a
I
h
u
i
I

1
j
I
j
X
K
ˆ
I
a
K
h
u
i
K
h
v
i
K
j
K
jj
I
j
6
Cu
(
J
)
:
(2.55)
Remark8.
Bytherescalingargumentitisclearthatwecanassume
B
u;v
assmallaswe
need(where\smallness",ofcourse,dependsonlyonthefunctionWeneedthisremark,
sinceallbehaviorsofourfunction
"
arestudiednear0.
65
Remark9.
Everythingisreducedto(2.55).Weconcentrateonproving(2.55).Clearly,by
scaleinvariance,itlooksverytemptingtomake(2.55)aBellmanfunctionstatement.This
willbeexactlyourplanfromnowon.
2.4.8Bellmanproofof(2.55):introducingthe\maininequality"
WestartthisSectionwiththefollowingnotation.Wetwoweights
u
and
v
,andaCarleson
sequense
f
a
I
g
.Wedenote
u
I
=
h
u
i
I
;v
I
=
h
v
i
I
;
N
I
(
t
)=
1
j
I
j
jf
x
:
u
(
x
)
>
t
gj
;
A
I
=
1
j
I
j
X
J
ˆ
I
a
J
j
J
j
;
L
I
=
1
j
I
j
X
J
ˆ
I
a
J
h
u
i
J
h
v
i
J
j
J
j
:
Weproceedwithtwotheoremsthatproveourmainresult.Everywhereinthefuturewe
usethat
h
u
i
I
h
v
i
I
=
u
I
v
I
6
<
1forany
I
.Wecandoitduetosimplerescaling.
Theorem2.4.6.
Supposethat

0

6
"

;
where
"
propertiesofTheorem2.9,fromwhichthemainoneis
Z
1
"
(
t
)
t
dt<
1
:
(2.56)
Let

besmallenough,and

1
=
f
(
N;A
):0
6
N
6
1;0
6
A
6
1
g
66
andforsomeconstant
P

2
=
f
(
u;v;L;A
):0
6
A
6
1;
u;v;L
>
0;
uv
6

;
L
6
P

p
uv
g
:
Supposewehavefoundafunction
B
1
,don

1
,andafunction
B
2
,don

2
,
suchthat:
0
6
B
1
6
N
;(2.57)
(
B
1
)
0
A
>
10
N

0
(
N
)
;(2.58)

d
2
B
1
>
0;(2.59)
0
6
B
2
6
u
;(2.60)
(
B
2
)
0
A
>
0(2.61)
(
B
2
)
0
A
>
c

u

L;
when
P

p
uv

L
>
uv
"
(
1
uv
)
;(2.62)
uv
(
B
2
)
0
L
>


1
uL;
forsmall

1
inthewholeof

2
;(2.63)

d
2
B
2
>
0
:
(2.64)
Thenforthefunctionofaninterval
B
(
I
):=
B
2
(
u
I
;v
I
;L
I
;A
I
)+
1
R
0
B
1
(
N
I
(
t
)
;A
I
)
dt
the
followingholds:
0
6
B
(
I
)
6
2
u
I
(2.65)
B
(
I
)

B
(
I
+
)+
B
(
I

)
2
>
Ca
I

u
I

L
I
:
(2.66)
Next,westate
67
Theorem2.4.7.
Ifsuchtwofunctions
B
1
and
B
2
exist,then
(2.55)
holds,namely
1
j
I
j
X
J
ˆ
I
a
I
h
u
i
J

1
j
J
j
X
K
ˆ
J
a
K
h
u
i
K
h
v
i
K
j
K
jj
I
j
6
R
2
Z
I
u:
ProofoftheTheorem2.4.7.
ThisisastandardGreen'sformulaappliedtofunction
B
(
I
)on
thetreeofdyadicintervals.Letusexplainthedetails.
Sincethefunction
B
isnon-negative,wehavethat
2
j
I
j
u
I
>
j
I
jB
(
I
)
>
j
I
jB
(
I
)

2
n
X
k
=1
j
I
n;k
jB
(
I
n;k
)
:
Here
n
ised,and
I
n;k
are
n

th
generationdescendantsof
I
.Clearly,all
j
I
n;k
j
areequal
to2

n
.
Letusdenote
J
)=
j
J
jB
(
J
)
j
J
+
jB
(
J
+
)
j
J

jB
(
J

),where
J

arechildrenof
J
.By
theproperty(2.66)weknowthat
J
)
>
C
j
J
j
a
J
u
J
L
J
.Bythetelescopiccancellation,we
getthat
j
I
jB
(
I
)

2
n
X
k
=1
j
I
n;k
jB
(
I
n;k
)=
n

1
X
m
=0
2
m
X
k
=1

I
m;k
)
:
Combiningourestimates,weget
2
j
I
j
u
I
>
C
n

1
X
m
=0
2
m
X
k
=1
j
I
m;k
j
a
I
m;k
u
I
m;k
L
I
m;k
=
C
X
J
ˆ
I;
j
J
j
>
2

n
j
I
j
j
J
j
a
J
u
J
L
J
:
Thisistrueforevery
n
,withtheconstant
C
independentof
n
.Thus,
u
I
>
C
1
j
I
j
X
J
ˆ
I
a
J
u
J
L
J
j
J
j
:
68
Theresultfollowsfromtheof
L
J
.
InthefutureweusethefollowingvariantofSylvestercriterionofpositivityofmatrix.
Lemma2.4.8.
Let
M
=(
m
ij
)
3
i;j
=1
bea
3

3
realsymmetricmatrixsuchthat
m
11
<
0
,
m
11
m
22

m
12
m
21
>
0
,and
det
M
=0
.Then
M
isnonpositive
Proof.
Let
E
beamatrixwithallentriesbeing0exceptfor
e
33
=1.Consider
t>
0
and
A
:=
A
(
t
):=
M
+
tE
.Itiseasytoseethat
a
11
<
0,
a
11
a
22

a
12
a
21
>
0,and
det
A
=
t

(
m
11
m
22

m
12
m
21
)
>
0when
t>
0.BySylvestercriterion,matrices
A
(
t
),
t>
0,areallnegativelyTherefore,tending
t
to0+,weobtain,that
M
isnonpositive

Weneedthefollowinglemma,whichisinspiritof[VaVo].
Lemma2.4.9.
Let
L
I
begivenby
L
I
=
1
j
I
j
X
J
ˆ
I
a
J
h
u
i
J
h
v
i
J
j
J
j
:
Let
A
I
givenby
A
I
=
1
j
I
j
P
J
ˆ
I
a
J
j
J
j
.Supposethatitisboundedby
1
foranydyadic
I
(Carlesoncondition).Ifforanydyadicinterval
I
wehavethat
h
u
i
I
h
v
i
I
6
1
,thenitholds
thatforanydyadicinterval
I
wehave
L
I
6
P
q
h
u
i
I
h
v
i
I
.
Proof.
Itistruesincethefunction
T
(
u;v;A
)=100
p
uv

uv
A
+1
isconcaveenoughinthe
domain
G
:=
f
0
6
A
6
1
;uv<
1
;u;v
>
0
g
.Onecanadapttheprooffrom[VaVo].
First,weneedtocheckthatthefunction
T
(
x;y;A
)isconcavein
G
.Clearly,
T
00
A;A
<
0.
Next,
det
0
B
@
T
00
A;A
T
00
A;v
T
00
A;v
T
00
v;v
1
C
A
=
x
y
(
A
+1)
4

(50(
A
+1)
p
xy

xy
)
>
0
:
(2.67)
69
Thisexpressionisnon-negative,because
A
+1
>
1,and
p
uv
6
1.Finally,
det
0
B
B
B
B
B
@
T
00
A;A
T
00
A;v
T
00
A;u
T
00
A;v
T
00
v;v
T
00
v;u
T
00
A;u
T
00
v;u
T
00
u;u
1
C
C
C
C
C
A
=0
:
Therefore,byLemma2.4.8weconcludethat
T
(
u;v;A
)isaconcavefunction.
Next,
T
0
A
=
uv
(
A
+1)
2
>
1
4
uv:
Thus,ifwethreepoints(
u;v;A
),(
u

;v

;A

),suchthat
u
=
u
+
+
u

2
,
v
=
v
+
+
v

2
,
and
A
=
A
+
+
A

2
+
a
,wegetbytheTaylorformula:
T
(
u;v;A
)

T
(
u
+
;v
+
;A
+
)+
T
(
u

;v

;A

)
2
>
aT
0
A
(
u;v;A
)
>
Ca

uv:
Thisrequirestheexplanation.TheTaylorformulaweusedhasaremainderwiththesec-
ondderivativeattheintermediatepoint
P

onsegments
S
+
:=[(
u;v;
A

+
A
+
2
)
;
(
u;v;A
+
)]
;
S

:=[(
u;v;
A

+
A
+
2
)
;
(
u;v;A

)].Oneofthissegmentsliesinsidedomain
G
,
where
T
isconcave,andthisremainderwillhavetherightsign.Howeverthesecondseg-
mentcaneasilystickoutofdomain
G
,because
G
itselfisnotconvex.Butnoticethatif,for
example,
S
+
isnotinside
G
,still(
x;y;B
)
2
S
+
impliesthatoneofthecoordinates,say
x
,
mustbesmallerthan
u
.Then
y
canbebiggerthan
v
,butnotmuch.Infact,
v
+

v
=
v

v

)
v
+

2
v

v


2
v:
Therefore,
y

v
+

2
v
.Thenwehavethat
xy

2
uv

2.Letusconsider
e
G
:=
f
(
x;y;A
):
70
0

A

1
;x;y

0
;
0

xy

2
g
.Nowcomebacktotheproofthat
T
isconcavein
G
.
In(2.67)weusedthatif(
x;y;A
)
2
G
,then
xy

1andthecorrespondingdeterminantis
non-negative.Butthesamenon-negativityin(2.67)holdsunderslightlyrelaxedassumption
(
x;y;A
)
2
e
G
.
Wenoticethatour
u
I
=
h
u
i
I
,
v
I
=
h
v
i
I
,and
A
I
=
1
j
I
j
P
J
ˆ
I
a
I
j
I
j
havethedynamics
above.TherestoftheproofreadsexactlyastheproofoftheTheorem2.4.7.
ProofoftheTheorem2.4.6.
WestartwiththefollowingcorollaryfromtheTaylorexpansion.
Supposewehavethreetuples(
N;A
),(
N

;A

),suchthat:
N
=
N
+
+
N

2
;
A
=
A
+
+
A

2
+
m:
Moreover,supposethereare(
u;v;L
),(
u

;v

;L

),suchthat
u
=
u
+
+
u

2
;
v
=
v
+
+
v

2
;
L
=
L
+
+
L

2
+
m

uv:
Then,since
d
2
B
1
6
0,wewrite
B
1
(
N
+
;A
+
)
6
B
1
(
N;A
)+(
B
1
)
0
N
(
N;A
)(
N
+

N
)+(
B
1
)
0
A
(
N;A
)(
A
+

A
)
:
Thus,
B
1
(
N;A
)

B
1
(
N
+
;A
+
)+
B
1
(
N

;A

)
2
>
(
B
1
)
0
A
(
N;A
)

(
A

A
+
+
A

2
)=
m

(
B
1
)
0
A
(
N;A
)
>
m
N

0
(
N
)
:
71
Similarly,
B
2
(
u;v;L;A
)

B
2
(
u
+
;v
+
;L
+
;A
+
)+
B
2
(
u

;v

;L

;A

)
2
>
m

((
B
2
)
0
A
(
u;v;L;A
)
+
uv
(
B
2
)
0
L
)
First,supposethat
L
I
6
u
I
v
I
"
(
1
u
I
v
I
)
.Then,using
m
=
a
I
weget
B
(
I
)

B
(
I
+
)+
B
(
I

)
2
>
>
Z
 
B
1
(
N
I
(
t
)
;A
I
)

B
1
(
N
I
+
(
t
)
;A
I
+
)+
B
1
(
N
I

(
t
)
;A
I

)
2
!
dt
+
+
 
B
2
(
u
I
;v
I
;L
I
;A
I
)

B
2
(
u
I
+
;v
I
+
;L
I
+
;A
I
+
)+
B
2
(
u
I

;v
I

;L
I

;A
I

)
2
!
>
a
I

(
B
2
)
0
A
(
u
I
;v
I
;L
I
;A
I
)+
u
I
v
I
(
B
2
)
0
L
(
u
I
;v
I
;L
I
;A
I
)

+
a
I

Z
(
B
1
)
0
A
(
N
I
(
t
)
;A
I
)
dt

>
a
I

Z
N
I
(
t
)

0
(
N
I
(
t
))
dt


1
u
I
L
I

:
(2.68)
Thelastinequalityistrue,since(
B
2
)
0
A
>
0and
uv
(
B
2
)
0
L
>


1
uL
onthedomainof
B
2
.
Weuseolder'sinequality(andthat
R
N
I
(
t
)
dt
=
u
I
)toget:
Z
N
I
(
t
)

0
(
N
I
(
t
))
dt
>
u
2
I
R
N
I
(
t

0
(
N
I
(
t
))
dt
>
C
u
2
I
R
N
I
(
t

N
I
(
t
))
dt"

R
N
I
(
t

N
I
(
t
))
dt
u
I

:
(2.69)
72
LastinequalityisTheorem2.4.4.Therefore,wegetthat
Z
N
I
(
t
)

0
(
N
I
(
t
))
dt
>
u
I

u
I
k
u
k
L

I

1
"
0
@
k
u
k
L

I
u
I
1
A
=
u
I
u
I
v
I
k
u
k
L

I
v
I

1
"
0
@
k
u
k
L

I
v
I
u
I
v
I
1
A
:
(2.70)
Wearegoingtousetheone-sidedbumpcondition
k
u
k
L

I
v
I
6
B
u;v
6
1.Thus,
u
I
v
I
6
u
I
v
I
k
u
k
L

I
v
I
:
Sincethefunction
x
7!
x
"
(
1
x
)
isincreasingnear0(on[0
;c
"
])andboundedfrombelowbetween
c
"
and1,weget
u
I
v
I
v
I
k
u
k
L

I

1
"
0
@
v
I
k
u
k
L

I
u
I
v
I
1
A
>
C

u
I
v
I
1
"
(
1
u
I
v
I
)
:
Therefore,
Z
N
I
(
t
)

0
(
N
I
(
t
))
dt
>
Cu
I
u
I
v
I
"
(
1
u
I
v
I
)
>
Cu
I
L
I
:
Thelastinequalityfollowsfromourassumptionthat
L
I
6
u
I
v
I
"
(
1
u
I
v
I
)
.Puttingeverything
together,weget
B
(
I
)

B
(
I
+
)+
B
(
I

)
2
>
a
I
u
I
L
I
(
C


1
)
>
C
1

a
I
u
I
L
I
:
73
Weproceedtothecase
L
I
>
u
I
v
I
"
(
1
u
I
v
I
)
.Thenwewrite
B
(
I
)

B
(
I
+
)+
B
(
I

)
2
>
B
2
(
u
I
;v
I
;L
I
;A
I
)

B
2
(
u
I
+
;v
I
+
;L
I
+
;A
I
+
)+
B
2
(
u
I

;v
I

;L
I

;A
I

)
2
:
Thisisobviouslytrue,since(
B
1
)
0
A
>
0everywhereand
B
1
isaconcavefunction.Next,we
use
B
2
(
u
I
;v
I
;L
I
;A
I
)

B
2
(
u
I
+
;v
I
+
;L
I
+
;A
I
+
)+
B
2
(
u
I

;v
I

;L
I

;A
I

)
2
>
a
I

(
B
2
)
0
A
+
uv
(
B
2
)
0
L

>
ca
I

u
I
L
I
;
(2.71)
bythepropertyof
B
2
.Therefore,wearedone.
2.4.9Fourthstep:buildingthefunction
B
2
Inordertotheproof,weneedtobuildfunctions
B
1
and
B
2
.Inthissectionwewill
presentthefunction
B
2
.Denote
'
(
x
)=
x
"
(
1
x
)
:
Thisfunctionisincreasing(byregularityassumptionson
"
inTheorem2.9),therefore,there
exists
'

1
.Weintroduce
B
2
(
u;v;L;A
)=
Cu

L
2
v
1
Z
A
+1
L
'

1

1
x

dx:
74
Letusexplainwhytheintegralisconvergent.Infact,usingchangeofvariables,weget
1
Z
1
'

1

1
x

dx
=
'

1
(1)
Z
0
"
(
1
t
)

t
d
dt
(
"
(
1
t
))
t
dt;
whichconvergesat0byassumption(2.56).
Therefore,since
L
I
6
C
p
u
I
v
I
,weget
0
6
B
(
u
I
;v
I
;L
I
;A
I
)
6
Cu
I
:
Next,
(
B
2
)
0
A
+
uv
(
B
2
)
0
L
=
L
v
'

1

L
A
+1


u
(
A
+1)
'

1

L
A
+1


2
uL
1
Z
A
+1
L
'

1

1
x

dx
=
uL

0
B
B
B
@
1
uv
'

1

L
A
+1


A
+1
L
'

1

L
A
+1


2
1
Z
A
+1
L
'

1

1
x

dx
1
C
C
C
A
(2.72)
Weusethat
L
>
uv
"

1
uv

=
'
(
uv
).Then
'

1
(
L
)
>
uv
,and,since
A
+1
˘
1,weget
1
uv
'

1

L
A
+1

>
C
1
:
Moreover,since
uv
6

isasmallnumber,wegetthat
L
issmallenoughfortheintegral
1
R
A
+1
L
'

1
(
1
x
)
dx
tobelessthanasmallnumber
c
2
.Finally,letuscompare
A
+1
L
'

1
(
L
A
+1
)
75
withasmallnumber
c
3
.Since
L
issmall,wecanwrite
"

1
c
3
L

6
c
3
:
Wedoit,since
c
3
isfromthebeginning(say,
c
3
=
1
10
).Thus,
L
6
'
(
c
3
L
)
:
Thisimplies
'

1
(
L
)
6
c
3
L;
thus
1
L
'

1
(
L
)
6
c
3
:
Since
A
+1
˘
1,wegetthedesired.Therefore,if
L
>
uv
"
(
1
uv
)
=
'
(
uv
)then(
B
2
)
0
A
+
uv
(
B
2
)
0
L
>
cuL
.
Moreover,inthewholedomainof
B
2
weget,since(
B
2
)
0
A
>
0,
(
B
2
)
0
A
+
uv
(
B
2
)
0
L
>
uv
(
B
2
)
0
L
>

(
c
2
+
c
3
)
uL
withsmall
c
2
+
c
3
.ThisisapenultimateinequalityinthestatementofTheorem2.4.6.
Nowweshallprovetheconcavityof
B
2
.Forthisitisenoughtoprovetheconcavityof
thefunctionofthreevariables:
B
(
v;L;A
):=
B
2
(
u;v;L;A
)

Cu
.Clearly,(
B
)
00
vv
<
0,which
76
isobvious.Also,itisacalculationthat
det
0
B
B
B
B
B
@
(
B
)
00
vv
(
B
)
00
vA
(
B
)
00
vL
(
B
)
00
vA
(
B
)
00
AA
(
B
)
00
AL
(
B
)
00
vL
(
B
)
00
AL
(
B
)
00
LL
1
C
C
C
C
C
A
=0
:
Thus,weneedtoconsiderthematrix
0
B
@
(
B
)
00
vv
(
B
)
00
vA
(
B
)
00
vA
(
B
)
00
AA
1
C
A
andtoprovethatitsdeterminantispositive.Wedenote
f
(
t
)=
'

1
(
t
),tosimplifythenext
formula.Thecalculationshowsthatthedeterminantaboveisequalto
g

L
A
+1

:=

f

L
A
+1

2
+2

L
A
+1

2

f
0

L
A
+1

1
Z
A
+1
L
f

1
x

dx:
Weneedtoprovethat
g
ispositivenear0.First,
g
(0)=0.Next,
g
0
(
s
)=

2
f
(
s
)
f
0
(
s
)+4
sf
0
(
s
)
1
Z
1
s
f

1
x

dx
+2
s
2
f
00
(
s
)
1
Z
1
s
f

1
x

dx
+2
f
0
(
s
)
f
(
s
)=
4
sf
0
(
s
)
1
Z
1
s
f

1
x

dx
+2
s
2
f
00
(
s
)
1
Z
1
s
f

1
x

dx:
(2.73)
Wenoticethat
f
0
ispositive,since
'

1
isincreasingnear0.Moreover,bythefactthat
'
isstrictlymonotonous,andbyconcavityof
t"
(
t
)(seeTheorem2.4.2),wegetthat
'
is
strictly
convex,hence
'

1
isstrictlyconvexnear0aswell.Thatis,
f
00
isalsopositive.
77
Therefore,
g
0
(
s
)
>
0,andso
g
(
s
)
>g
(0)=0.TheapplicationofLemma2.4.8
theproofofconcavityof
B
(andthereforeoftheconcavityof
B
2
).Wearedone.
Remark10.
Wecanalwaysthinkthatthebumpconstant
B
u;v
6
C
"
,where
C
"
issuch
that
L
I
6
c
"
.Thenwecanusethemonotonicityandconcavityofthefunction
'
near0.
2.4.10Fifthstep:buildingthefunction
B
1
Wepresentthefunctionfrom2.3.8.1.
B
1
(
N;A
)=
CN

N
N
A
Z
0
ds
s

0
(
s
)
78
Chapter3
Oneweightestimateforthelimiting
case:the
A
1
conjecture
3.1Themainresult
Weareon
I
0
:=[0
;
1].Asalways
D
denotethedyadiclattice.Inthischapterweusethe
usualHaarsystem
f
h
I
g
:
h
I
(
x
):=
8
>
>
>
<
>
>
>
:
1
p
j
I
j
;x
2
I
+

1
p
j
I
j
;x
2
I

Theweightedweaknormofanoperator
T
isby
k
T
k
L
1
;
1
(
w
)
=sup
t>
0
;
k
f
k
L
1
(
w
)
=1
t

w
f
x
:
j
Tf
(
x
)
j
>
t
g
Weconsidertheoperator
T
"
:
'
!
X
I

I
0
;I
2D
"
I
(
';h
I
)
h
I
;
where
"
I
=

1.Noticethatthesumdoesnotcontaintheconstantterm.
Ourmaintheoremisthefollowing.
Theorem3.1.1.
Forany
p<
1
5
andforanylarge
Q
thereexistsaweight
w
,suchthat
79
[
w
]
1
=
Q
,and
sup
"
=
f
"
I
g
k
T
"
k
L
1
;
1
(
w
)
>
Q
log
p
Q:
3.2TheBellmanapproach
Put
F
=
hj
f
j
w
i
I
;f
=
h
f;
i
I
;
=
w
=
h
w
i
I
;m
=inf
I
w:
Weareinthedomain
:=
f
(
F;w;m;f;
):
F
j
f
j
m;m

w

Qm
g
:
(3.1)
Introduce
B
(
F;w;m;f;
):=sup
1
j
I
j
w
f
x
2
I
:
X
J

I;J
2
D
"
J
(
';h
J
)
h
J
(
x
)
>
g
;
(3.2)
wherethe
sup
istakenoverall
"
J
;
j
"
J
j
1
;J
2
D;J

I
,andoverall
f
2
L
1
(
I;wdx
)
suchthat
F
:=
hj
f
j
w
i
I
;f
:=
h
f
i
I
,
w
=
h
w
i
I
;m

inf
I
w
,and
w
aredyadic
A
1
weights,
suchthat
8
I
2
D
h
w
i
I

Q
inf
I
w
,and
Q
beingthebestsuchconstant.Inotherwords
Q
:=[
w
]
dyadic
A
1
.
3.2.1Homogeneity
Byitisclearthat
s
B
(
F=s;w=s;m=s;f;
)=
B
(
F;w;m;f;
)
;
80
B
(
tF;w;m;tf;
)=
B
(
F;w;m;f;
)
:
Choosing
s
=
m
and
t
=


1
,wecanseethat
B
(
F;w;m;f;
)=
mB
(
F

;
w
m
;
f

)(3.3)
foracertainfunction
B
.Introducingnewvariables

=
F

;
=
w
m
;
=
f

wewritethat
B
isin
G
:=
f
(
;;
):
j

j
;
1



Q
g
:
(3.4)
3.2.2Themaininequality
Theorem3.2.1.
Let
P;P
+
;P

2

;P
=(
F;w;
min(
m
+
;m

)
;f;
)
,
P
+
=(
F
+
;w
+
;m
+
;f
+
;
+

)
,
P

=(
F

;w

;m

;f

;


)
.Then
B
(
P
)

1
2
(
B
(
P
+
)+
B
(
P

))

0
:
(3.5)
Atthesametime,if
P;P
+
;P

2

;P
=(
F;w;
min(
m
+
;m

)
;f;
)
,
P
+
=(
F
+
;w
+
;m
+
;f
+
;


)
,
P

=(
F

;w
+
;m
+
;f

;
+

)
.Then
B
(
P
)

1
2
(
B
(
P
+
)+
B
(
P

))

0
:
(3.6)
Inparticular,withd
m
,andwithallpointsbeinginside

weget
B
(
F;w;m;f;
)

1
4
(
B
(
F

dF;w

dw;m;f




)+
B
(
F

dF;w

dw;m;f
+



)+
B
(
F
+
dF;w
+
dw;m;f


+

)+
B
(
F
+
dF;w
+
dw;m;f
+

+

))

0
:
(3.7)
81
Remark.
1)tialnotations
dF;dw;
justmeansmallnumbers.2)In(3.7)weloose
abitofinformation(incomparisonto(3.5),(3.6)),butthisisexactly(3.7)thatwearegoing
touseinthefuture.
Proof.
Fix
P;P
+
;P

2
Let
'
+
;'

,
w
+
;w

befunctionsandweightsgivingthesupre-
mumin
B
(
P
+
)
;B
(
P

)respectivelyuptoasmallnumber
>
0.Usingthefactthat
B
does
notdependon
I
,wethinkthat
'
+
;w
+
ison
I
+
and
'

;w

ison
I

.Consider
'
(
x
):=
8
>
>
>
<
>
>
>
:
'
+
(
x
)
;x
2
I
+
'

(
x
)
;x
2
I

!
(
x
):=
8
>
>
>
<
>
>
>
:
w
+
(
x
)
;x
2
I
+
w

(
x
)
;x
2
I

Noticethatthen
(
';h
I
)

1
p
j
I
j
=
:
(3.8)
Thenitiseasytoseethat
hj
'
j
!
i
I
=
F
=
P
1
;
h
'
i
I
=
f
=
P
4
:
(3.9)
82
Noticethatfor
x
2
I
+
using(3.8),wegetif
"
I
=

1
1
j
I
j
w
+
f
x
2
I
+
:
X
J

I
+
;J
2
D
"
J
(
';h
J
)
h
J
(
x
)
>
g
=
1
j
I
j
w
+
f
x
2
I
+
:
X
J

I
+
;J
2
D
"
J
(
';h
J
)
h
J
(
x
)
>
+

g
=
1
2
j
I
+
j
w
+
f
x
2
I
+
:
X
J

I
+
;J
2
D
"
J
(
'
+
;h
J
)
h
J
(
x
)
>P
+
;
3
g
1
2
B
(
P
+
)

:
Similarly,for
x
2
I

using(3.8),wegetif
"
I
=

1
1
j
I
j
w

f
x
2
I

:
X
J

I;J
2
D
"
J
(
';h
J
)
h
J
(
x
)
>
g
=
1
j
I
j
w

f
x
2
I

:
X
J

I

;J
2
D
"
J
(
';h
J
)
h
J
(
x
)
>


g
=
1
2
j
I

j
w

f
x
2
I

:
X
J

I

;J
2
D
"
J
(
'

;h
J
)
h
J
(
x
)
>P

;
3
g
1
2
B
(
P

)

:
Combiningthetwolefthandsidesweobtainfor
"
I
=

1
1
j
I
j
!
f
x
2
I
+
:
X
J

I;J
2
D
"
J
(
';h
J
)
h
J
(
x
)
>
g
1
2
(
B
(
P
+
)+
B
(
P

))

2
:
Letususenowthesimpleinformation(3.9):ifwetakethesupremuminthelefthandside
overallfunctions
'
,suchthat
hj
'
j
w
i
I
=
F;
h
'
i
I
=
f;
h
!
i
=
w
,andweights
!
:
h
!
i
=
w
,in
dyadic
A
1
with
A
1
-normatmost
Q
,andsupremumoverall
"
J
=

1(only
"
I
=

1stays
wegetaquantitysmallerorequalthantheone,wherewehavethesupremumover
allfunctions
'
,suchthat
hj
'
j
!
i
=
F;
h
'
i
I
=
f;
h
!
i
=
w
,andweights
!
:
h
!
i
=
w
,indyadic
A
1
with
A
1
-normatmost
Q
,andanunrestrictedsupremumoverall
"
J
=

1including
83
"
I
=

1.Thelatterquantityisofcourse
B
(
F;w;m;f;
).Soweproved(3.5).
Toprove(3.6)werepeatverbatimthesamereasoning,onlykeepingnow
"
I
=1.Weare
done.
Remark.
Thistheoremisasortof\fancy"concavityproperty,theattentivereaderwould
seethat(3.5),(3.6)representbi-concavitynotunlikedemonstratedbythecelebratedBurk-
holder'sfunction.Wewillusetheconsequenceofbi-concavityencompassedby(3.7).There
isstillanotherconcavityifweallowtohave
j
"
J
j
1.
Theorem3.2.2.
Intheof
B
weallownowtotakesupremumoverall
j
"
j
j
1
.
Let
P;P
+
;P

2

;P
=(
F;w;m;f;
)
,
P
+
=(
F
+
;w
+
;m;f
+
;
)
,
P

=(
F

;w

;m;f

;
)
.Then
B
(
P
)

1
2
(
B
(
P
+
)+
B
(
P

))

0
:
(3.10)
Proof.
Werepeattheproofof(3.5)butwith
"
I
=0.
Theorem3.2.3.
Ford
F;w;f;
function
B
isdecreasingin
m
.
Proof.
Let
m
=min(
m

;m
+
)=
m

.Andlet
m
+
>m
.Then(3.5)becomes
B
(
F;w;m;f;
)
B
(
F;w;m
+
;f;
)

0
:
Thisiswhatwewant.
84
3.3Theunweightedestimate:theexactBellmanfunction
Wedealwiththecasewhenthereisnoweight,i.e.withthecasewhen
w
=1a.e.We
noticethatthisistheboundaryofourdomain
w
=
m
.
Introduceafunction
B
0
(
f;F
)=sup






8
<
:
x
:
X
I
ˆ
I
0
;I
2
D
"
I
(
';h
I
)
h
I
(
x
)
>

9
=
;






;
wherethesupremumistakenoverallfamilies
f
"
I
g
suchthat
j
"
I
j
=1,andallfunctions
'
with
hj
'
ji
I
0
=
F
,
h
'
i
I
0
=
f
.
Let
0
=
f
(
f;F
):
F
>
j
f
jg
bethedomainof
B
0
.
Denote
B
0
(
f;F
)=
8
>
>
>
<
>
>
>
:
1
;
6
F
1

(


F
)
2

2

f
2
;
>
F;
(
F;f;
)
2

0
:
Ourmaintheoremisthefollowing.
Theorem3.3.1.
Forany
(
f;F
)
2

0
itholdsthat
B
0
(
F;f;
)=
B
0
(
F;f;
)
.
Firstly,itwillbemoreconvenienttoworkwithaslightlymofunction.Weneeda

21.
Afunction
 
iscalledamartingaletransformofafunction
'
,ifforsome
family
f
"
I
g
,with
j
"
I
j
=1,
 
(
x
)=
h
 
i
I
0
+
X
I
ˆ
I
0
;I
2
D
"
I
(
';h
I
)
h
I
(
x
)
;x
2
I
0
:
85
Denote
B
(
g;f;F
)=sup
jf
x
:
 
(
x
)
>
0
gj
;
wherethesupremumistakenoverallfunctions
'
with
hj
'
ji
I
0
=
F
,
h
'
i
I
0
=
f
,andall
martingaletransforms
 
of
'
with
h
 
i
I
0
=
g
.Itiseasytoseethat
B
0
(
f;F
)=
B
(

g;f;F
)
:
Denote=
f
(
g;f;F
):
F
>
j
f
jg
and
B
(
g;f;F
)=
8
>
>
>
<
>
>
>
:
1
;

g
6
F
1

(
g
+
F
)
2
g
2

f
2
;

g
>
F;
(
g;f;F
)
2

:
Thenourmaintheoremisequivalenttothefollowingone.
Theorem3.3.2.
Forany
(
g;f;F
)
2

itholds
B
(
g;f;F
)=
B
(
g;f;F
)
.
Corollary3.3.3.
Foranyfunction
'
2
L
1
,anynumber

>
0
andanyfamily
f
"
I
g
with
j
"
I
j
=1
itholds






8
<
:
x
:
X
I
ˆ
I
0
;I
2
D
"
I
(
';h
I
)
h
I
(
x
)
>

9
=
;






6
2
k
'
k
1

Proof.
Itiseasytoverifythat
sup

B
0
(
f;F
)


F

=2
:
Thus,






8
<
:
x
:
X
I
ˆ
I
0
;I
2
D
"
I
(
';h
I
)
h
I
(
x
)
>

9
=
;






6
2
F

=2
k
'
k
1

:
86
Corollary3.3.4.
Foranyfunction
'
2
L
1
,anynumber

>
0
andanyfamily
f
"
I
g
with
j
"
I
j
=1
itholds






8
<
:
x
:
X
I
ˆ
I
0
;I
2
D
"
I
(
';h
I
)
h
I
(
x
)
>

9
=
;






6
4
k
'
k
1

Proof.






8
<
:
x
:
X
I
ˆ
I
0
;I
2
D
j
"
I
(
';h
I
)
h
I
(
x
)
j
>

9
=
;






=






8
<
:
x
:
X
I
ˆ
I
0
;I
2
D
"
I
(
';h
I
)
h
I
(
x
)
>

9
=
;






+






8
<
:
x
:
X
I
ˆ
I
0
;I
2
D
"
I
(

';h
I
)
h
I
(
x
)
>

9
=
;






6
4
k
'
k
1

(3.11)
Westarttoproveourmaintheorem.
3.3.1
B
>
B
Weneedatechnicallemma.
Lemma3.3.5.
Let
x

betwopointsin

suchtat
j
f
+

f

j
=
j
g
+

g

j
and
x
=
1
2
(
x
+
+
x

)
.
Then
B
(
x
)

B
(
x
+
)+
B
(
x

)
2

0
:
(3.12)
Giventhelemma,weprovethefollowingtheorem.
Theorem3.3.6.
Foranypoint
x
2

itholds
B
(
x
)
>
B
(
x
)
.
87
Proof.
Letusapoint
x
2
andapairofadmissiblefunctions
'
,
 
on
I
0
corresponding
to
x
.Forany
I
2
D
bythesymbol
x
I
wedenotethepoint(
h
 
i
I
;
h
'
i
I
;
hj
'
ji
I
;
).Wenotice
thatsince
 
isamartingaletransformof
'
,wealwayshave
j
f
I
+

f
I

j
=
j
g
I
+

g
I

j
;
and
x
I
=
x
I
+
+
x
I

2
:
Usingconsequentlymaininequalityforthefunction
B
wecanwritedownthefollowingchain
ofinequalities
B
(
x
)

1
2

B
(
x
I
+
0
)+
B
(
x
I

0
)


X
I
2
D;
j
I
j
=2

n
1
j
I
j
B
(
x
I
)=
Z
1
0
B
(
x
(
n
)
(
t
))
dt;
where
x
(
n
)
(
t
)=
x
I
,if
t
2
I
,
j
I
j
=2

n
.
Notethat
x
(
n
)
(
t
)
!
(
 
(
t
)
;'
(
t
)
;
j
'
(
t
)
j
)almosteverywhere(atanyLebesguepoint
t
),and
therefore,since
B
iscontinuousandbounded,wecanpasstothelimitintheintegral.So,
wecometotheinequality
B
(
x
)

Z
1
0
B
(
 
(
t
)
;'
(
t
)
;
j
'
(
t
)
j
)
dt

Z
f
t
:
 
(
t
)

0
g
=


f
t
2
I
0
:
 
(
t
)

0
g


(3.13)
wherewehaveusedtheproperty
B
(
g;f;
j
f
j
)=1for
g

0.Now,takingsupremumin(3.13)
overalladmissiblepairs
'
,
 
,wegettherequiredestimate
B
(
x
)
B
(
x
).
88
3.3.2
B
(
g;f;F
)
6
B
(
g;f;F
)
Thissectionisdevotedtothefollowingtheorem.
Theorem3.3.7.
Foranypoint
x
2

itholds
B
(
x
)
6
B
(
x
)
.
Toprovethetheoremweneedtopresenttwosequencesoffunctions
f
'
n
g
,
f
 
n
g
,such
that

Forevery
n
thefunction
 
n
isamartingaletransformof
'
n
;

Forevery
n
:
hj
'
n
ji
I
0
=
F
,
h
'
n
i
I
0
=
f
,
h
 
n
i
I
0
=
g
;

Itholdsthat
B
(
g;f;F
)=lim
n
!1
jf
x
:
 
n
(
x
)
>
0
gj
.
Weneedthefollowing
22.
Wecallapair(
'; 
)admissibleforthepoint(
g;f;F
)if
 
isamartingale
transformof
'
,and
hj
'
ji
I
0
=
F
,
h
'
i
I
0
=
f
,
h
 
i
I
0
=
g
.
23.
Wecallapair(
'; 
)an
"
-extremizerforapoint(
g;f;F
),ifthispairis
admissibleforthispointand
jf
x
:
 
(
x
)
>
0
gj
>
B
(
g;f;F
)

"
.
Thefollowinglemmaisalmostobvious.
Lemma3.3.8.
1.Forapositivenumber
s
:
B
(
sg;sf;sF
)=
B
(
g;f;F
)
.Moreover,ifa
pair
(
'; 
)
isadmissibleforapoint
(
g;f;F
)
then
(
s';s 
)
isadmissiblefor
(
sg;sf;sF
)
.
Ifapair
(
'; 
)
isan
"
-extremizerforapoint
(
g;f;F
)
then
(
s';s 
)
isan
"
-extremizer
for
(
sg;sf;sF
)
.
2.
B
(
g;f;F
)=
B
(
g;

f;F
)
.Moreover,ifapair
(
'; 
)
isadmissibleforapoint
(
g;f;F
)
then
(

'; 
)
isadmissiblefor
(
g;

f;F
)
.Ifapair
(
'; 
)
isan
"
-extremizerforapoint
(
g;f;F
)
then
(

'; 
)
isan
"
-extremizerfor
(
g;

f;F
)
.
89
Thenextlemmaisakeytoour\splitting"technique.
Lemma3.3.9.
Supposetwopairs
(
'

; 

)
areadmissibleforpoints
(
g

;f

;F

)
corre-
spondingly.Supposealsothat
F
=
F
+
+
F

2
;f
=
f
+
+
f

2
;;g
=
g
+
+
g

2
;
j
f
+

f

j
=
j
g
+

g

j
:
Thenapair
(
'; 
)
isadmissibleforthepoint
(
g;f;F
)
,where
'
(
x
)=
8
>
>
>
<
>
>
>
:
'

(2
x
)
;x
2
[0
;
1
2
)
'
+
(2
x

1)
;x
2
[
1
2
;
1]
;
 
(
x
)=
8
>
>
>
<
>
>
>
:
 

(2
x
)
;x
2
[0
;
1
2
)
 
+
(2
x

1)
;x
2
[
1
2
;
1]
:
Proof.
Itisclearthat
h
'
i
I
0
=
f
,
h
 
i
I
0
=
g
,and
hj
'
ji
I
0
=
F
.Allweneedtoproveisthat
foranyinterval
I
itistruethat
j
(
 ;h
I
)
j
=
j
(
';h
I
)
j
:
Foranyinterval
I
6
=
I
0
itisobvious,sincepairs(
'

; 

)areadmissibleforcorresponding
points.Thus,weneedtoshowthat
j
(
';h
I
0
)
j
=
j
(
 ;h
I
0
)
j
:
But
(
';h
I
0
)=
h
'
i
[
1
2
;
1]
h
'
i
[0
;
1
2
]
=
h
'
+
i
[0
;
1]
h
'

i
[0
;
1]
=
f
+

f

;
(
 ;h
I
0
)=
h
 
i
[
1
2
;
1]
h
 
i
[0
;
1
2
]
=
h
 
+
i
[0
;
1]
h
 

i
[0
;
1]
=
g
+

g

;
90
whichourproof.
Wegeneralizethislemmaalittle.
Lemma3.3.10.
Supposetwopairs
(
'

; 

)
areadmissibleforpoints
(
g

;f

;F

)
corre-
spondingly.Supposealsothat
F
=
F
+
+
F

2
;f
=
f
+
+
f

2
;;g
=
g
+
+
g

2
;
j
f
+

f

j
=
j
g
+

g

j
:
Suppose
I
isadyadicintervalwith\sons"
I

.Supposethatapair

;

isadmissiblefor
somepoint
(
g
0
;f
0
;F
0
)
.Supposethat
8
x
2
I

x
)=
'
I
(
x
)
;

x
)=
 
I
(
x
)
;
wherethepair
(
'; 
)
isadmissibleforthepoint
(
g;f;F
)
.Thenthepair

1
;

1
)
,d
below,isadmissibleforthepoint
(
g
0
;f
0
;F
0
)
:

1
(
x
)=
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:

x
)
;x
62
I
'
I
+
+
(
x
)
;x
2
I
+
'
I


(
x
)
;x
2
I

;

1
(
x
)=
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:

x
)
;x
62
I
 
I
+
+
(
x
)
;x
2
I
+
 
I


(
x
)
;x
2
I

Essentiallythislemmasaysthatifwehavepairs(
'

; 

),andandapair(
'; 
)
intheLemma3.3.9,thenwecansplitthispairinto(
'

; 

),nedon
I

correspondingly.
TheproofoftheLemma3.3.10isessentiallythesameastheproofoftheLemma3.3.9.
91
3.3.2.1Changeofvariables
Itwillbemoreconvenientforustoworkinvariables
y
1
=
f

g
2
;y
2
=

f

g
2
;F:
We
M
(
y
1
;y
2
;F
)=
B
(
g;f;F
).Thenallpropertiesof
B
areeasilytranslatedto
propertiesof
M
.Moreover,the\splitting"lemmas3.3.9,3.3.10remaintruefor
y
1
or

y
2
.
Ifwehaveapoint(
y
1
;y
2
;F
)thenby(
'
(
y
1
;y
2
;F
)
; 
(
y
1
;y
2
;F
)
)wedenoteanadmissible
pairforthispoint.Anindividualfunction
'
(
y
1
;y
2
;F
)
isalwayssuchthatthereisafunction
 
(
y
1
;y
2
;F
)
,suchthatthepair(
'
(
y
1
;y
2
;F
)
; 
(
y
1
;y
2
;F
)
)isadmissiblefor(
y
1
;y
2
;F
).
3.3.2.2Theproofof
B
>
B
Wewillworkinthe
y
-variables.Inthesevariablesitistruethatthefunction
M
isconcave
when
y
1
or
y
2
isThisisprovedintheTheorem3.3.4.Analogouslytotheprevious
we
M
(
y
1
;y
2
;F
)=
B
(
g;f;F
)
:
Weprovethat
M
(1
;
1
;F
)
>
M
(1
;
1
;F
)
:
92
Fixalargeinteger
r
andset

=
1
2
r
.Wenoticethefollowingchainofinequalities:
M
(1
;
1
;F
)
>
1
2
(
M
(1
;
1

;F
+

(1

F
))+
M
(1
;
1+
;F


(1

F
)))=
=
1
2
(
M
(1
;
1

;F
+

(1

F
))+
M
(1+
;
1
;F


(1

F
)))
:
(3.14)
Applyingthesameconcavityweseethat
M
(1
;
1

;F
+

(1

F
))
>

M
(1
;
0
;
1)+(1


)
M
(1
;
1
;F
)=

+(1


)
M
(1
;
1
;F
)
:
Moreover,bytheconcavity
M
(1+
;
1
;F


(1

F
))
>
(



2
)
M
(1+
;
0
;
1+

)+(1


)
M
(1+
;
1+
;
(1+

)(
F


(2

F
)))+

2
M
(1+
;
1
;F


(1

F
))
>



2
+(1


)
M
(1
;
1
;F


(2

F
))(3.15)
Therefore,weget
M
(1
;
1
;F
)
>
1
2


+(1


)
M
(1
;
1
;F
)+



2
+(1


)
M
(1
;
1
;F


(2

F
))

;
or
M
(1
;
1
;F
)
>
2



2
1+

+
1


1+

M
(1
;
1
;F


(2

F
))
:
Noticethatitistrueforany
F
.Wenowdenote
F
k
=2

(2

F
)(1+

)
k
:
93
Then,clearly,
F
0
=
F
,and
F
k
+1
=
F
k


(2

F
k
).Withthisnotationweget
M
(1
;
1
;F
)
>
2



2
1+

K
X
k
=0

1


1+


k
+

1


1+


K
+1
M
(1
;
1
;F
K
+1
)
:
3.3.2.3Thecase
F
>
2
Inthiscasewehave
F
k
+1
>
F
k
,andthereforethepoint(1
;
1+
;F
k


(1

F
k
))always
liesinThus,wecantake
K
ashugeaswewant.Therefore,
M
(1
;
1
;F
)
>
2



2
1+

1
X
k
=0

1


1+


k
=
2



2
2

:
Thisistrueforarbitrarysmall

,andthus
M
(1
;
1
;F
)
>
1.
3.3.2.4Thecase
F
6
2
Inthiscasetoassurethat(1
;
1+
;F
k


(1

F
k
))
2
weneed
F
k


(1

F
k
)
>

,which
implies
(1+

)
K
+1
6
2
2

F
:
Take
K
2
[
log
2
2

F
log(1+

)

10
;
log
2
2

F
log(1+

)
+10],suchthatthisinequalityholds.Thenweget
M
(1
;
1
;F
)
>
2



2
1+

K
X
k
=0

1


1+


k
=
2



2
2

 
1


1


1+


K
+1
!
:
Itisonlylefttonoticethatwithourchoiseof
K
wehave

1


1+


K
+1
!
(2

F
)
2
4
;
!
0
;
94
andtherefore
M
(1
;
1
;F
)
>
1

(2

F
)
2
4
=
M
(1
;
1
;F
)
:
Weleavetheproofofthegeneralinequality
M
(
y
1
;y
2
;F
)
>
M
(
y
1
;y
2
;F
)tothereader.
Infact,itisasimpleuseoftheconcavityof
M
alongthelinethatconnects(
y
1
;
0
;y
1
)with
(
y
1
;y
2
;F
).
3.3.3Buildingtheextremalsequenseforpoints(1
;
1
;F
)
TheaimofthisSectionistoprovethat
B
(
g;f;F
)
6
B
(
g;f;F
)byaconstructionofan
extremalsequenseofpairs(
'
n
; 
n
).Forthesakeofsimplicity,wedoitonlyforthecase
f

g
=2.
Duetothehomogeneityandsymmetryofthefunction
B
itisenoughtoprovethat
B
(
g;f;F
)
6
B
(
g;f;F
)
for
f
>
0,
f

g
=2.Inthenewvariablesitmeansthatweconsiderthecase
y
1
=1,and
y
2
6
y
1
=1.Aswehaveseen,for
f
>

g
wehave
B
(
g;f;F
)=
B
(
g;f;F
)=1,andsowe
needtoconsiderthecase
f
6

g
,i.e.
y
2
>
0.Webuildthe
"
-extremizerforthepoint
(
F;
1
;
1).
Fixalargeinteger
r
andlet

=2

r
.Asbefore,denote
I
0
=[0
;
1].Alsodenote
J
i
=[2

i
;
2

i
+1
),Denote
m
i
(
x
)=2
i
x

1|thelinearfunctionfrom
J
k
onto
I
0
.
Weneedthefollowinglemma.
Lemma3.3.11.
Suppose

=2

r
issmallenough.Also,asmallnumber
">
0
.Suppose
F
1
=
F


(2

F
)
,andthepair
(
'
(1
;
1
;F
1
)
; 
(1
;
1
;F
1
)
)
isadmissible.Thenthereexistsan
95
admissiblepair
(
'
(1
;
1
;F
)
; 
(1
;
1
;F
)
)
suchthat
jf
x
:
 
(1
;
1
;F
)
>
0
gj
>
2



2
1+

+
1


1+

jf
x
:
 
(1
;
1
;F
1
)
>
0
gj
":
(3.16)
Proof.
First,weexplainourstrategy.Inwhatfollows,wealwaysassumethatfunctionson
theright-handsidearealreadyWespecifytheirlater;however,weclearly
indicatepointstowhichthefunctionsareadmissible.
We
'
(1
;
1
;F
)
(
x
)=
8
>
>
>
<
>
>
>
:
'
(1
;
1

;F
+

(1

F
))
(
m
1
(
x
))
;x
2
J
1
'
(1
;
1+
;F


(1

F
))
(2
x
)
;x
2
[0
;
1
2
)
:
 
(1
;
1
;F
)
(
x
)=
8
>
>
>
<
>
>
>
:
 
(1
;
1

;F
+

(1

F
))
(
m
1
(
x
))
;x
2
J
1
 
(1
;
1+
;F


(1

F
))
(2
x
)
;x
2
[0
;
1
2
)
:
BytheLemma3.3.10weseethat
 
(1
;
1
;F
)
isamartingaletransformof
'
(1
;
1
;F
)
.We
96
next
'
(1
;
1
;F
)
(
x
)=
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
'
(1
;
0
;
1)
(
m
1
(
x
)

))
;x
2
m

1
1
(
I
0
)
'
(1
;
1
;F
)
(
m
k
(
m
1
(
x
)))
;x
2
m

1
1
m

1
k
(
I
0
)
;k
=1
:::r

'
(1+
;
1
;F


(1

F
))
(2
x
)
;x
2
[0
;
1
2
)
:
'
(1
;
1
;F
)
(
x
)=
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
 
(1
;
0
;
1)
(
m
1
(
x
)

))
;x
2
m

1
1
(
I
0
)
 
(1
;
1
;F
)
(
m
k
(
m
1
(
x
)))
;x
2
m

1
1
m

1
k
(
I
0
)
;k
=1
:::r
 
(1+
;
1
;F


(1

F
))
(2
x
)
;x
2
[0
;
1
2
)
:
(3.17)
BytheLemma3.3.8andamultipleapplicationoftheLemma3.3.10,westillgetan
admissiblepairforthepoint(1
;
1
;F
).
Finally,
'
(1+
;
1
;F


(1

F
))
(
x
)=
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
'
(1+
;
0
;
1+

)
(
m
k
(
x

))
;x
2


J
k
;k
=1
:::r
'
(1+
;
1
;F


(1

F
))
(
x

2
)
;x
2
[0
;
2
)
(1+

)
'
(1
;
1
;F


(2

F
))
(
m
k
(
x
))
;x
2
J
k
;k
=1
:::r
 
(1+
;
1
;F


(1

F
))
(
x
)=
8
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
:
 
(1+
;
0
;
1+

)
(
m
k
(
x

))
;x
2


J
k
;k
=1
:::r
 
(1+
;
1
;F


(1

F
))
(
x

2
)
;x
2
[0
;
2
)
(1+

)
 
(1
;
1
;F


(2

F
))
(
m
k
(
x
))
;x
2
J
k
;k
=1
:::r
(3.18)
Again,theLemma3.3.8andtheLemma3.3.10assurethatthepairisadmissible.
97
Bringingeverythingtogether,weget
'
(1
;
1
;F
)
(
x
)=
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
'
(1
;
0
;
1)
(
m
1
(
x
)

))
;x
2
m

1
1
(
I
0
)
'
(1
;
1
;F
)
(
m
k
(
m
1
(
x
)))
;x
2
m

1
1
m

1
k
(
I
0
)
;k
=1
:::r

'
(1+
;
0
;
1+

)
(
m
k
(
2
x

))
;x
2

2
m

1
k
(
I
0
)
;k
=1
:::r

'
(1+
;
1
;F


(1

F
))
(
2
x

2
)
;x
2
[0
;

2
2
)

(1+

)
'
(1
;
1
;F


(2

F
))
(
m
k
(2
x
))
;x
2
1
2
m

1
k
(
I
0
)
;k
=1
:::r:
 
(1
;
1
;F
)
(
x
)=
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
 
(1
;
0
;
1)
(
m
1
(
x
)

))
;x
2
m

1
1
(
I
0
)
 
(1
;
1
;F
)
(
m
k
(
m
1
(
x
)))
;x
2
m

1
1
m

1
k
(
I
0
)
;k
=1
:::r
 
(1+
;
0
;
1+

)
(
m
k
(
2
x

))
;x
2

2
m

1
k
(
I
0
)
;k
=1
:::r
 
(1+
;
1
;F


(1

F
))
(
2
x

2
)
;x
2
[0
;

2
2
)
(1+

)
 
(1
;
1
;F


(2

F
))
(
m
k
(2
x
))
;x
2
1
2
m

1
k
(
I
0
)
;k
=1
:::r:
(3.19)
Wenowspecifysoffunctionsontheright-handside.Thepair(
'
(1
;
0
;
1)
; 
(1
;
0
;
1)
)
isa
"
2
-extremizerforthepoint(1
;
0
;
1).Thepair(
'
(1+
;
0
;
1+

)
; 
(1+
;
0
;
1+

)
)isa
"



2
-
extremizerforthepoint(1+
;
0
;
1+

).
Thepair(
'
(1
;
1
;F


(2

F
))
; 
(1
;
1
;F


(2

F
))
)isgiveninthelemma.Asforthepair
(
'
(1+
;
1
;F


(1

F
))
; 
(1+
;
1
;F


(1

F
))
)|wetakeanyadmissiblepairforthispoint.
Itisaneasycalculationthatthefunction
 
(1
;
1
;F
)
theinequality(3.16).More-
over,itiseasytoseethatfor
any
pair,by(3.19)wehave
h
'
(1
;
1
;F
)
i
I
0
h
 
(1
;
1
;F
)
i
I
0
=
2.Thus,whatweneedtoshowisthatthereexistsanadmissiblepair(
'
(1
;
1
;F
)
; 
(1
;
1
;F
)
)
98
thattheself-similaritycondition(3.19)
Todothat,wetakeanyadmissiblepair(~
'
(1
;
1
;F
)
;
~
 
(1
;
1
;F
)
)and
'
0
(1
;
1
;F
)
(
x
)=
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
'
(1
;
0
;
1)
(
m
1
(
x
)

))
;x
2
m

1
1
(
I
0
)
~
'
(1
;
1
;F
)
(
m
k
(
m
1
(
x
)))
;x
2
m

1
1
m

1
k
(
I
0
)
;k
=1
:::r

'
(1+
;
0
;
1+

)
(
m
k
(
2
x

))
;x
2

2
m

1
k
(
I
0
)
;k
=1
:::r

'
(1+
;
1
;F


(1

F
))
(
2
x

2
)
;x
2
[0
;

2
2
)

(1+

)
'
(1
;
1
;F


(2

F
))
(
m
k
(2
x
))
;x
2
1
2
m

1
k
(
I
0
)
;k
=1
:::r:
 
0
(1
;
1
;F
)
(
x
)=
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
 
(1
;
0
;
1)
(
m
1
(
x
)

))
;x
2
m

1
1
(
I
0
)
~
 
(1
;
1
;F
)
(
m
k
(
m
1
(
x
)))
;x
2
m

1
1
m

1
k
(
I
0
)
;k
=1
:::r
 
(1+
;
0
;
1+

)
(
m
k
(
2
x

))
;x
2

2
m

1
k
(
I
0
)
;k
=1
:::r
 
(1+
;
1
;F


(1

F
))
(
2
x

2
)
;x
2
[0
;

2
2
)
(1+

)
 
(1
;
1
;F


(2

F
))
(
m
k
(2
x
))
;x
2
1
2
m

1
k
(
I
0
)
;k
=1
:::r:
(3.20)
Thenthepair(
'
0
(1
;
1
;F
)
; 
0
(1
;
1
;F
)
)isadmissibletopoint(1
;
1
;F
).ItistruebytheLemma
3.3.10,andbyaneasycalculationthatshowsthatallaveragesareasweneed.Wenow
99
inductively
'
n
+1
(1
;
1
;F
)
(
x
)=
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
'
(1
;
0
;
1)
(
m
1
(
x
)

))
;x
2
m

1
1
(
I
0
)
'
n
(1
;
1
;F
)
(
m
k
(
m
1
(
x
)))
;x
2
m

1
1
m

1
k
(
I
0
)
;k
=1
:::r

'
(1+
;
0
;
1+

)
(
m
k
(
2
x

))
;x
2

2
m

1
k
(
I
0
)
;k
=1
:::r

'
(1+
;
1
;F


(1

F
))
(
2
x

2
)
;x
2
[0
;

2
2
)

(1+

)
'
(1
;
1
;F


(2

F
))
(
m
k
(2
x
))
;x
2
1
2
m

1
k
(
I
0
)
;k
=1
:::r:
 
n
+1
(1
;
1
;F
)
(
x
)=
8
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
<
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>
:
 
(1
;
0
;
1)
(
m
1
(
x
)

))
;x
2
m

1
1
(
I
0
)
 
n
(1
;
1
;F
)
(
m
k
(
m
1
(
x
)))
;x
2
m

1
1
m

1
k
(
I
0
)
;k
=1
:::r
 
(1+
;
0
;
1+

)
(
m
k
(
2
x

))
;x
2

2
m

1
k
(
I
0
)
;k
=1
:::r
 
(1+
;
1
;F


(1

F
))
(
2
x

2
)
;x
2
[0
;

2
2
)
(1+

)
 
(1
;
1
;F


(2

F
))
(
m
k
(2
x
))
;x
2
1
2
m

1
k
(
I
0
)
;k
=1
:::r:
(3.21)
Thenforany
n
wegetanadmissiblepairtothepoint(1
;
1
;F
).
Weneedtonoticethat
Z
I
0
j
'
n
+1
(1
;
1
;F
)

'
n
(1
;
1
;F
)
j
2
dx
=
X
k
j
J
k
j
2
Z
I
0
j
'
n
(1
;
1
;F
)

'
n

1
(1
;
1
;F
)
j
2
dx
=
=
1


2
Z
I
0
j
'
n
(1
;
1
;F
)

'
n

1
(1
;
1
;F
)
j
2
dx
=
=(
1


2
)
n
Z
I
0
j
'
1
(1
;
1
;F
)

'
0
(1
;
1
;F
)
j
2
dx:
(3.22)
100
Thus,wecantake
'
(1
;
1
;F
)
=lim
'
n
+1
(1
;
1
;F
)
in
L
2
(
I
0
)
:
Similarly
 
(1
;
1
;F
)
=lim
 
n
+1
(1
;
1
;F
)
in
L
2
(
I
0
)
:
Itisclearthatthepair(
'
(1
;
1
;F
)
; 
(1
;
1
;F
)
)theself-similarityconditions(3.19).More-
over,sincethelimitin
L
2
impliestheliminin
L
1
,wegetthatalltheaveragesareasneeded.
Moreover,foreveryinterval
I
:
j
(
'
(1
;
1
;F
)
;h
I
)
j
=lim
j
(
'
n
(1
;
1
;F
)
;h
I
)
j
=
j
(
 
n
(1
;
1
;F
)
;h
I
)
j
=
j
(
 
(1
;
1
;F
)
;h
I
)
j
;
andthuswegetanadmissiblepair.Theproofofthelemmais
Wearenowreadytothewholeconstruction.Weconsiderasequence
F
k
=2

(2

F
)(1+

)
k
:
Thenitisclearthe
F
0
=
F
and
F
k
+1
=
F
k


(2

F
k
).
3.3.3.1Thecase
F
>
2
Wetakeahugenumber
N
andasmallnumber
"
.Forapoint(1
;
1
;F
N
)wetakeanyadmis-
siblepair(
'
(1
;
1
;F
N
)
; 
(1
;
1
;F
N
)
).UsingtheLemma3.3.11
N
timeswebuildanadmissible
pair(
'
(1
;
1
;F
)
; 
(1
;
1
;N
)
).Moreover,weget
jf
x
:
 
(1
;
1
;F
)
(
x
)
>
0
gj
>
2



2
1+

N
X
k
=0

1


1+


k

N":
101
Wenowspecifythechoiseof

,
N
and
"
.Weasmall

,sothat
2



2
2

=1

˙
.
Thenahugenumber
N
,suchthat
N
P
k
=0

1


1+


k
>
1+

2


˙
1+

2



2
.Finally,averysmall
number
"
,suchthat
N"<˙
.Thenweget
jf
x
:
 
(1
;
1
;F
)
(
x
)
>
0
gj
>
2



2
1+


1+

2


˙
1+

2



2


˙
=1

3
˙:
where
˙
isanarbitrarysmallnumber.
3.3.3.2Thecase
F<
2
Weremindthatourverysteprequiresthatthepoint(1
;
1+
;F


(1

F
))tobeinour
domain.Thus,theonthe
N
-thiterationweneedthatthepoint(1
;
1+
;F
N


(1

F
N
))
isinthedomain=
f
(
y
1
;y
2
;F
):
F
>
j
y
1

y
2
jg
.Thisyieldstotheinequality
(1+

)
N
+1
<
2
2

F
:
Thus,weshouldstopatthe
K
-thstepwith
(1+

)
N
+1
ˇ
2
2

F
:
Herethesign\
ˇ
"meansthat
N
2
[
log
2
2

F
log(1+

)

10
;
log
2
2

F
log(1+

)
+10]
:
102
WeagainapplytheLemma3.3.11
N
timesandget
jf
x
:
 
(1
;
1
;F
)
(
x
)
>
0
gj
>
2



2
1+

N
X
k
=0

1


1+


k

N"
=
2



2
2

 
1


1


1+


N
+1
!

N"
Finally,since
N
2
[
log
2
2

F
log(1+

)

10
;
log
2
2

F
log(1+

)
+10]
wegetthat

!
0implies1


1


1+


N
+1
!
1

(2

F
)
2
4
,whichourproof.
3.3.4HowtotheBellmanfunction
B
Inthissectionweexplainhowdidwesearchforthefunction
B
andit.Westartwith
thefollowinglemma.Let
x

betwopointsinsuchtat
j
f
+

f

j
=
j
g
+

g

j
and
x
=
1
2
(
x
+
+
x

).Then
B
(
x
)

B
(
x
+
)+
B
(
x

)
2

0
:
(3.23)
Proof.
Fix
x

2
andlet(
'

; 

)betwopairsoffunctionsgivingthesupremumfor
B
(
x
+
),
B
(
x

)respectivelyuptoasmallnumber
>
0.Write
'

=
f

+
X
I

I
0
;I
2
D
(
';h
I
)
h
I
; 

=
g

+
X
I

I
0
;I
2
D
"
I
(
';h
I
)
h
I
;
Consider
'
(
t
):=
8
>
>
>
<
>
>
>
:
'
+
(2
t

1)
;
if
t
2
[
1
2
;
1]
'

(2
t
)
;
if
t
2
[0
;
1
2
)
:
103
and
 
(
t
):=
8
>
>
>
<
>
>
>
:
 
+
(2
t

1)
;
if
t
2
[
1
2
;
1]
 

(2
t
)
;
if
t
2
[0
;
1
2
)
Since
j
x
+
1

x

1
j
=
j
x
+
2

x

2
j
,thefunction
 
isamartingaletransformof
'
,andthepair
(
'; 
)isanadmissiblepairofthetestfunctionscorrespondingtothepoint
x
.Therefore,
B
(
x
)

1
j
I
0
j


f
t
2
I
0
:
 
(
t
)

0
g


=
1
2
j
I
+
0
j


f
t
2
[
1
2
;
1]:
 
(
t
)

0
g


+
1
2
j
I

0
j


f
t
2
[0
;
1
2
):
 
(
t
)

0
g



1
2
B
(
x
+
)+
1
2
B
(
x

)

2
:
Sincethisinequalityholdsforanarbitrarysmall

,wecanpasstothelimit

!
0,what
givesustherequiredassertion.
Corollary3.3.12.
Thelemmameansthatifwechangevariables
f
=
y
1

y
2
,
g
=

y
1

y
2
,andintroduceafunction
M
(
y
1
;y
2
;F
):=
B
(
g;f;F
)
dinthedomain
G
:=
f
y
=
(
y
1
;y
2
;F
)
2
R
3
:
j
y
1

y
2
j
F
g
,thenwegetthatforeachd
y
2
,
M
(
F;y
1
;

)
isconcave
andforeachd
y
1
,
M
(
F;

;y
2
)
isconcave.
3.3.4.1Theboundary
F
=
y
1

y
2
Westartwithconsideringaboundarycase
F
=
f
or,inthe
y
variables,
F
=
y
1

y
2
.It
meansthatweconsideronlynon-negativefunctions
'
.Bythehomogeneityofthefunction
M
weneedtoafunction
S
ofvariable
s
=
y
1
y
2
,suchthat

S
(
y
1
y
2
)

00
y
1
y
1
6
0
;
and

S
(
y
1
y
2
)

00
y
2
y
2
6
0
:
(3.24)
104
Wenoticethatwhen
g
!
0wehave
s
!
1andwemusthave
S
!
0.Thus,wegeta
condition
S
(
s
)
!
0
;
as
s
!
1
:
(3.25)
Moreover,wehaveseenthatif
f
>

g
then
B
(
g;f;F
)=1.Inparticular,itholdswhen
f
=

g
.Therefore,wehave
M
(
y
1
;

y
1
;
0)=1.Thisimpliesthat
S
(
s
)
!
1
;
as
s
!
:
Frominequalities(3.24)wegetthat
S
00
(
s
)
6
0
;;s
2
S
00
(
s
)+2
S
0
(
s
)
6
0
;s
2
(

;

1]
:
Makethesecondinequalityanequation(wearelookingforthe
best
nontrivial
S
).Weget
S
(
s
)=
c
1
+
c
2
s
:
Theboundaryconditionsimplythat
S
(
s
)=1

1
s
;
andtherefore
M
(
y
1
;y
2
;y
1

y
2
)=1

y
2
y
1
=
y
1

y
2
y
1
;
or
B
(
g;f;f
)=
2
f
f

g
:
105
Thus,wegetananswer
M
(
y
1
;y
2
;y
1

y
2
)=
8
>
>
>
<
>
>
>
:
1
;y
2
6
0
y
1

y
2
y
1
;y
2
>
0
;
(3.26)
or
B
(
g;f;f
)=
8
>
>
>
<
>
>
>
:
1
;f
>

g
2
f
f

g
;f
6

g:
3.3.4.2Thedomain
Weremindthereaderthatfora
y
1
thefunction
M
isconcaveinvariables(
F;y
2
).We
alsoremindthesymmetrycondition,i.e.
M
(
y
1
;y
2
;F
)=
M
(
y
2
;y
1
;F
)
:
Letustiatethisequationin
y
2
andset
y
2
=
y
1
.Thenwegetanequation:
M
y
1
(
y
1
;y
1
;F
)=
M
y
2
(
y
1
;y
1
;F
)
:
Moreover,duetothesymmetryitisenoughto
M
for
y
2
6
y
1
.Asbefore,wesawthat
for
f
>

g
wehave
B
(
g;f;F
)=1,i.e.
for
y
2
6
0,wehave
M
(
y
1
;y
2
;F
)=1.(3.27)
106
Thus,itisenoughtoconsiderthecase0
6
y
2
6
y
1
.Denote
y
1
=
f
(
y
2
;F
):
F
>
j
y
2

y
1
jg
|thesectionoffor
y
1
.Wewantto
M
satisfyingconcavityinthishyperplane{we
aregoingtolookfor
M
(andwewillchecklaterthatitisconcave)thatsolvesMonge{Ampere
(MA)equationin
y
1
withboundaryconditions(3.26)and(3.27).In
y
1
,thereisapoint
P
:=(0
;y
1
;y
1
).Letusmakeaguessthatthecharacteristics(andweknowbyPogorelov's
theoremthattheyformthefoliationof
y
1
bystraightlines)ofourMAequationin
y
1
formthefanoflineswithcommonpoint
P
=(
y
1
;y
1
;
0).ByPogorelov'stheoremwealso
knowthatthereexistsfunctions
t
1
;t
2
;t
constantoncharacteristicssuchthat
M
=
t
1
F
+
t
2
y
2
+
t;
(3.28)
suchthat
t
1
=
t
1
(
t
;
y
1
)
;t
2
=
t
2
(
t
;
y
1
)(wethinkthat
y
1
isaparameter),that
0=(
t
1
)
0
t
F
+(
t
2
)
0
t
y
2
+1
;
(3.29)
that
t
1
=
@M
(

;y
2
;F
)
@F
;t
1
=
@M
(

;y
2
;F
)
@y
2
:
(3.30)
Letuscallcharacteristics
L
t
.Extendoneofthemfrom
P
till
y
2
=
y
1
.Werecallanother
boundarycondition:
If
y
2
=
y
1
)
@M
@y
2
=
@M
@y
1
:
(3.31)
Orifwedenotetheintersectionof
L
t
with
y
2
=
y
1
by(
y
1
;y
1
;F
(
t
))weget
t
2
(
t
;
y
1
)=
@M
@y
1
(
y
1
;y
1
;F
(
t
))
:
(3.32)
107
Wewanttoprovenowthat
Onthewhole
L
t
wehave
F
(
t
)
t
1
+2
y
1
t
2
=0
:
(3.33)
Infact,our
M
is0homogeneous.Soeverywhere
FM
0
F
+
y
1
M
0
y
1
+
y
2
M
0
y
2
=0.Applythis
topoint(
y
1
;y
1
;F
(
t
)),wherewecanuse(3.32)andget
F
(
t
)
t
1
+
t
2
y
1
+
t
2
y
1
=0,whichis
(3.33)inonepoint.Butthenallentriesareconstantson
L
t
,therefore,(3.33)follows.
Nowuseourguessthat
L
t
fanfrom
P
=(
y
1
;y
1
;
0).Plugthiscoordinatesinto0=
(
t
1
)
0
t
F
+(
t
2
)
0
t
y
2
+1,whichis(3.29).Thenwegetthecrucial(andtrivial)ODE
t
0
1
(
t
)=

1
y
1
)
t
1
(
t
)=

1
y
1
t
+
C
1
(
y
1
)
:
(3.34)
Letboundaryline
F
=
y
1

u
correspondsto
t
=
t
0
.Thenweuse(3.28)and(3.26):
(

1
y
1
t
0
+
C
1
(
y
1
))(
y
1

u
)+
t
2
u
+
t
0
=1

u
y
1
:
Using(3.33)wecanplug
t
2
expressedvia
F
(
t
).Butbyion
F
(
t
0
)=0.Soweget
(

1
y
1
t
0
+
C
1
(
y
1
))(
y
1

u
)+
t
0
=1

u
y
1
:
Or
C
1
(
y
1
)
y
1

(
t
0
+
C
1
(
y
1
)
y
1
)
u
y
1
=1

u
y
1
:
Varying
u
weget
C
1
(
y
1
)=
1
y
1
,
t
0
=0.Nowfrom(3.34)weget
t
1
(
t
)=
1
y
1
(1

t
)
:
(3.35)
108
Afterthat(3.29)and(3.33)becomethesystemoftwolinear\ODE"son
F
(
t
)and
t
2
(
t
):
8
>
>
>
<
>
>
>
:

1
y
1
F
(
t
)+
y
1
t
0
2
(
t
)+1=0
2
y
1
t
2
(
t
)+
F
(
t
)
1
y
1
(1

t
)=0
:
(3.36)
We
t
2
=

1
y
1
(1

t
)
t
.Wethearbitraryconstantfor
t
2
bynoticingthatthe
secondequationof(3.36)at
t
0
=0impliesthat
t
2
(0)=0as
F
(
t
0
)=
F
(0)=0by
Hence(3.29)becomes

1
y
1
F
+
1
y
1
(2
t

1)
y
2
+1=0
:
(3.37)
Given(
y
1
;y
2
;F
)
2

y
1
\f
0

y
2

y
1
g
,we
t
from(3.37)andplugitinto(3.28),in
whichweknowalready
t
(
t
)and
t
2
(
t
).Namely,weknowthat
M
(
y
1
;y
2
;F
)=
1
y
1
F

1
y
1
t
(1

t
)
y
2
+
t:
(3.38)
Plugging
t
=
1
2
F

(
y
1

y
2
)
y
2
from(3.37)intothisequationweobtain
M
(
y
1
;y
2
;F
)=1

(
F

y
1

y
2
)
2
4
y
1
y
2
:
(3.39)
Wenoticethatontheline
F
=
y
2
+
y
1
weget
M
=1.Thus,wegetthefollowinganswer
for
M
:
M
(
y
1
;y
2
;F
)=
8
>
>
>
<
>
>
>
:
1

(
F

y
1

y
2
)
2
4
y
1
y
2
;F
6
y
1
+
y
2
1
;F
>
y
1
+
y
2
:
(3.40)
109
Inourinitialcoordinatesweget
B
(
g;f;F
)=
8
>
>
>
<
>
>
>
:
1

(
F
+
g
)
2
g
2

f
2
;F
6

g
1
;F
>

g:
3.4Theweightedestimate
3.4.1tialpropertiesof
B
translatedtoerentialpropertiesof
B
Itisconvenienttointroduceanauxiliaryfunctionsof4and3variables:
e
B
(
x;y;f;
):=
B
(
x

;y;
f

)
:
Ofcourse
B
(
F;w;m;f;
)=
m
e
B
(
F
m
;
w
m
;f;
)=
mB
(
F

;
w
m
;
f

)
:
(3.41)
Lemma3.4.1.
Function
B
increasesintheandinthesecondvariable.
Proof.
WeknowthatbytheRHSof(3.41)isgettingbiggerif

isgettingsmaller.
Soletusconsider

1
>
2
;
1
=

2
+

,andvariables
F;w;m;f
andchoose
˚
1
(and
aweight
!
),
h
˚
1
i
=
f
+
";
hj
˚
1
j
!
i
=
F
,whichalmostrealizesthesupremum
B
(
F;w;m;f
+
";
1
).Consider
˚
2
suchthat
˚
2
=
˚
1

h
.Function
h
willbechosenlater,howeverwe
saynowthat
h
isequaltoacertainconstant
a
onasmalldyadicinterval
`
andiszero
otherwise.Constant
a
andinterval
`
wewillchoselater.But
"
:=
h
h
i
willbechosenvery
soon.Function
˚
2
competesforsupremizing
B
at(
hj
˚
2
j
!
i
;w;m;f;
2
).Wechoose
"
insuch
110
awaythat
h
˚
1
i

1
=
f
+
"

1
=
f

1


=
h
˚
2
i

2
:
(3.42)
Letusprovethat(3.42)impliesthat
hj
˚
1
j
!
i

1

hj
˚
2
j
!
i

2
:
(3.43)
By(3.42)thisisthesameas
hj
˚
2
+
h
j
!
i
hj
˚
2
j
!
i

h
˚
1
i
h
˚
2
i
=
h
˚
2
i
+
"
h
˚
2
i
:
Thepreviousinequalitybecomes
hj
˚
2
+
h
j
!
i
hj
˚
2
j
!
i

1+
h
h
i
h
˚
2
i
:
Bytriangleinequalitythelatterinequalitywouldfollowfromthefollowingone
hj
˚
2
j
!
ih
˚
2
i
hj
h
j
!
i
h
h
i
:
Wecanthinkthattheminimum
m
of
!
isattainedonawholetinydyadicinterval
`
(we
aretalkingabout
almost
supremums).Put
h
tobeacertain
a>
0onthisintervalandzero
otherwise.Ofcoursewechoose
a
tohave
h
h
i
=
"
,where
"
waschosenbefore.Nowthe
previousdisplayinequalitybecomes
hj
˚
2
j
!
ih
˚
2
i
m;
111
whichisobvious.
Noticethat
B
(
hj
˚
2
ji
;w;m;f;
2
)asasupremumislargerthanthe
!
-measureofthelevel
set
>
2
ofthemartingaletransformof
˚
2
.Butthisisalsothemartingaletransformof
˚
1
.
The

1
-levelsetforanymartingaletransformof
˚
1
issmaller,as

1
>
2
.Butrecallthatwe
alreadysaidthat
˚
1
(andweight
!
)almostrealizesitsownsupremum
B
(
F;w;m;f
+
";
1
)=
B
(
hj
˚
1
ji
;w;m;
h
˚
1
i
;
1
)So
B
(
hj
˚
1
ji
;w;m;
h
˚
1
i
;
1
)
B
(
hj
˚
2
ji
;w;m;
h
˚
2
i
;
2
)
:
Inothernotationsweget
B
(
hj
˚
1
ji

1
;
w
m
;
h
˚
1
i

1
)

B
(
hj
˚
2
ji

2
;
w
m
;
h
˚
2
i

2
)
:
LetusdenotetheargumentontheLHSas(
x
1
;y
1
;z
1
),andontheRHSas(
x
2
;y
2
;z
2
).Notice
that
y
1
=
y
2
=:
y
triviallyand
z
1
=
z
2
=:
z
by(3.42).Noticealsothat
x
1
<x
2
by(3.43).
Moreoverbychoosing

verysmallwecanrealizeany
x
1
<x
2
ascloseto
x
2
aswewant.
Thenthelastdisplayinequalityreadsas
B
(
x
1
;y;z
)

B
(
x
2
;y;z
)
:
Soweprovedthatfunction
B
increasesinthevariable.
Theincreaseinthesecondvariableiseasy.Chooseadyadicinterval
I
onwhichinf
I
!>
m
,but
h
!
i
I
=
inf
I
!<Q
=:[
!
]
A
1
.Fornon-constant
!
thisisalwayspossible,justtakea
smallintervalcontainingapoint
x
0
,where
!
(
x
0
)
>m
.Thenaugment
!
on
I
slightlyto
get
!
1
with
h
!
1
i
=
w
+
"
.Itiseasytoseethatasaresultwehavethenewweightwith
112
the
A
1
normatmost
Q
,thesameglobalum
m
butalargerglobalaverage
h
!
i
.The
!
1
measureofthelevelsetofthemartingaletransformwillbebiggerthan
!
measureof
thesamelevelsetofthesamemartingaletransform,and
w=m
alsogrowsto(
w
+
"
)
=m
.
Allothervariablesstaythesame.Soiftheoriginal
!
(andsome
˚
)were(almost)realizing
supremum,wewouldget
B
(
x;y
1
;z
)

B
(
x;y
2
;z
)
for
y
1
=
w=m;y
2
=(
w
+
"
)
=m
.
Theorem3.4.2.
Function
B
from
(3.3)

t
!
t

1
B
(
t;t;
)
isincreasingfor
j

j


t

Q

:
(3.44)
B
isconcave
:
(3.45)
B
(
x

;y;
f

)

1
4

B
(
x

dx



;y

dy;
f





)+
B
(
x

dx



;y

dy;
f
+




)+
B
(
x
+
dx

+

;y
+
dy;
f



+

)+
B
(
x
+
dx

+

;y
+
dy;
f
+


+

)


0
:
(3.46)
Proof.
TheserelationsfollowfromTheorem3.2.3,Theorem3.2.2,andTheorem3.2.1(actu-
allyfrom(3.7))correspondingly.
Wecanchooseextremelysmall
"
0
andinsidethedomainwecanmollify
B
bya
convolutionofitwith
"
0
-bellfunction
 
supportedinaballofradius
"
0
=
10.
Multiplicativeconvolutioncanbeviewedastheintegrationwith
1

5
 
(
x

x
0

),where

=
"
0
=
10.Here
x
0
isapointinsidethedomainofforfunction
B
.
Thisnewfunctionwecall
B
again.Itisexactlyastheinitialfunction
B
,anditobviously
allthesamerelationships,inparticularitTheorems3.2.1,3.2.2,3.2.3.Only
113
itsdomainof
"
0
issmaller(slightly)thanTheadvantagehoweveristhatthe
new
B
issmooth.Webuild
B
bythisnew
B
.Anewfunction
B
bythenew
B
as
in(3.41)willbesmooth.Actuallythenew
B
shouldbedenoted
B

0
,wheresuperscript
denotesouroperationofbutwedropthesuperscriptforthesakeofbrevity.In
fact,alltheseareforthesakeofconvenience,thenewfunctionssatisfytheold
inequalitiesintheuniformway,independentlyof
"
0
.Property(3.46)canbenowrewritten
bytheuseofTaylor'sformula:
Theorem3.4.3.


2
B


dx
x




2


2
B

 
dy
y

2

(1+

2
)
B

 



2


2
B


dx
x




dy
y
+2
B

dy
y


+2
B


dx
x






+
+2
B


dx
x







2
B





2

0
:
Proof.
ThisisjustTaylor'sformulaappliedto(3.46).
Denoting
˘
=
dx
x
=
dy
y
;
=


weobtainthefollowingquadraticforminequality
Theorem3.4.4.

˘
2
[

2
B

+

2
B

+2
B

]


2
[

2
B

+(1+

2
)
B

+2
B

+2
B

+2
B

]+
+2
˘
[

2
B

+
B

+
B

+
B

+
B

]

0
:
114
NowletuscombineTheorem3.4.4andTheorem3.2.2.Infact,Theorem3.2.2implies

2
B


2


2
B


2

B


2
:
Weplugitintothesecondtermabove.AlsoTheorem3.2.2implies
2
B

˘


2
B

˘
2

B


2
;
2
B

˘


2
B

˘
2

B


2
;
Wewillplugitintothethirdtermabove.Thenusingthenotation
 
(
;;
):=


2
B


2
B



2
B

(whichisnon-negativebytheconcavityof
B
initstwovariablesbytheway)weintroduce
thenotations
K
:=
 
(
;;
)+(


2
B



2
B

)
;
L
:=

 
(
;;
)+(

2
B

)



2
B

;
N
:=

(1+3

+

2
)
B


2
B


(

2
B

)



2
B

:
Andwegetthatthefollowingquadraticformisnon-negative:
˘
2
K
+
˘L
+

2
N
:=
˘
2
[
 
(
;;
)+(


2
B



2
B

)

]+
115
˘
[

 
(
;;
)+(

2
B

)



2
B

]+

2
[

(1+3

+

2
)
B


2
B


(

2
B

)



2
B


]

0
:
Therefore,
K
ispositive,and
N

L
2
4
K
:
(3.47)
Nowwewillestimate
L
frombelow,
K
fromaboveandasaresultwewillobtainthe
estimateof
N
frombelow,whichwillbringusourproof.
Butweneedsomeaprioriestimates,andforthatwewillneedtomollify
B
=
B

0
invariables
;
.Againwemakeamultiplicativeconvolutionwithabell-typefunction.Let
usexplainwhyweneedit.Let
^
Q
:=sup
G
B:
Wewanttoprovethat
^
Q=Q
!1
:
(3.48)
Firstweneedtonoticethat
Z
1
1
=
2
 
(
t;t;
)
dt

C
(
^

+
^
Q
Q

)
; 
(
;;
):=


2
B


2
B



2
B

:
(3.49)
Infact,consider

2
[
Q=
4
;Q=
2],
b
(
t
):=
B
(
t;t;
)ontheinterval
j

j

=:
t
0

t

1.
Let
`
(
t
)=
b
(1)
t

^

.Wesawthat
b
(
t
)
=t
isincreasingand
b
isconcave,and
b
isunder
`
,andsobyelementarypictureofconcavefunctionhavingproperty
b
(

)
=

increasingand
b
(

)concaveontheinterval[
t
0
0
;
1]wegetthatthemaximumof
`
(

)

b
(

)isattainedonthe
leftend-point.Theleftend-point
t
0
0
isthemaximumof
t
0
=
j

j

and1

whichis
c=Q
.
116
Therefore,
`
(
t
)

b
(
t
)
j
(
t
=(max(


;
c
Q
))

`
(max(


;
c
Q
))

C
^

max(


;
1
Q
)

^

+
^
Q
Q
;
andtheabovevalueismaximumof
g
(
t
):=
`
(
t
)

b
(
t
)on[
t
0
0
;
1].Bythesamepropertythat
b
(
t
)
=t
isincreasingwegetthat
g
0
(1)=
`
0
(1)

b
0
(1)=
b
(1)

b
0
(1)

0
:
CombiningthiswithTaylor'sformulaon[
t
0
;
1]wegetfor
g
:=
`

b
(gisconvexofcourse):

(1

t
0
)
g
0
(1)+
Z
1
t
0
dt
Z
1
t
g
00
(
s
)
ds
=positive+
Z
1
t
0
(
s

t
0
)
g
00
(
s
)
ds

sup
g

^

+
^
Q
Q
:
(3.50)
Thisimplies(3.49)because
g
00
(
t
)=
1
t
2
 
(
t;t;
)
;t
2
[1
=
2
;
1].
Considernowfunction
a
(
t
):=
B
(
t;;
)Wealsohavethesametypeofconsideration
appliedtoconvexfunction
^


a
(
t
)bringingus
Z
1
1
=
2


2
B

(
t;;
)
dt

C
^
:
(3.51)
Similarly,
Z
1
1
=
2


2
B

(
;t;
)
dt

C
^
:
(3.52)
Weusedherethat
B


0
;B


0,whichisnottosee.
Forthefutureestimateswewant(3.49),(3.51),(3.52)toholdnotinaveragebutpoint-
wise.
117
Toachievethereplacementof\in-average"estimates(3.49),(3.51),(3.52)bytheirpoint-
wiseanalogsletusconsideryetanothernowitisof
B
:
B
new
(
;;
):=2
Z
1
1
=
2
B
(
t;t;
)
dt:
Thedomainofof
B
new
isonlyintinywiththedomainof
of
B
.Infact,thelatteris
f
(
;;
):
j

j
;
1



Q
g
,andtheformerisjust
G
:=
f
(
;;
):
j

j
1
2
;
2



Q
g
.
Ifwereplace(
;;
)by(
t;t;
)
;
1
=
2

t

1
;
everywhereintheinequalityofTheorem
3.4.4,andthenintegratetheinequalitywith2
R
1
1
=
2
:::dt
,wewillgetTheorem3.4.4butfor
B
new
.
Itisnottoseethat(3.49)becomesapointwiseestimatefor
B
new
(just
tiatetheformulafor
B
new
in
;;
andmultiplyby
;;
appropriately):


2
(
B
new
)


2

(
B
new
)



2
(
B
new
)


C
(
^

+
^
Q
Q

)
:
(3.53)
Thispointwiseestimateautomaticallyimplynew\average"estimate:
2
Z
1
1
=
2



2
s
2
(
B
new
)

(
s;;
)

2

(
B
new
)

(
:;;:
)


2
(
B
new
)



C
(
^

+
^
Q
Q

)
:
Thismeansexactlythatthefunction
~
B
:=(
B
new
)
new
:=2
Z
1
1
=
2
B
(
s;;
)
ds
still(3.53).ItalsoclearlytheinequalityofTheorem3.4.4because(aswe
118
noticedabove)
B
new
thisinequality.Toseethisfactjustreplaceall

'sinthe
inequalityofTheorem3.4.4appliedto
B
new
by
s
andintegrate2
R
1
1
=
2
:::ds
.
Nowletusseethat
~
B
=(
B
new
)
new
alsoapointwiseanalogof(3.51),namely,
that


2
~
B

(
;;
)

C
^
:
(3.54)
Toshow(3.54)wejustrepeatwhathasbeendoneabove.Let~
g
(
t
):=
^


B
new
(
t;;
).
Thenwehave:1)0

~
g

^

on[
t
0
;
1],2)~
g
0
(1)

0(wesawthat
B
,andhence
B
new
,are
increasingintheargument),3)~
g
isconvex.Thenwesawin(3.50)that
Z
1
1
=
2
s
2
~
g
00
(
s
)
ds

Z
1
1
=
2
~
g
00
(
s
)
ds

C
^
:
Butthisisexactly(3.54).
Sofarweconstructedafunction
~
B
=(
B
new
)
new
thatpointwiseinequalities
(3.53),(3.54)andtheinequalityofTheorem3.4.4.Wearelefttoseethatbyintroducing
^
B
:=2
Z
1
1
=
2
~
B
(
;s;
)
ds
wekeep(3.53),(3.54)andtheinequalityofTheorem3.4.4validandalsoensure


2
^
B

(
;;
)

C
^
:
(3.55)
Walreadyjustsawthat(3.53),(3.54)andtheinequalityofTheorem3.4.4arevalidfor
^
B
justbyaveragingthesameinequalitiesfor
~
B
.Wecanseethat(3.55)holdsbytherepetition
ofwhathasbeenjustdone.Namely,consider^
g
(
t
):=
^


~
B
(
;t;
).Thenwehave:1)
119
0

^
g

^

on[
t
0
;
1],2)^
g
0
(1)

0(wesawthat
B
,andhence
B
new
,
~
B
areincreasingin
theargument),3)^
g
isconvex.Using(3.50)againinexactlythesamemanneraswedid
withproving(3.54)weget
Z
1
1
=
2
s
2
^
g
00
(
s
)
ds

Z
1
1
=
2
^
g
00
(
s
)
ds

C
^
:
Butthisisexactly(3.55).
Wedrop\hat",andfromnowon
^
B
isjustdenotedby
B
.Wecansummarizeitsproperties
asfollows.
0

 
(
;;
)

C
(
^

+
^
Q
Q

)
:
(3.56)
0


2
B

(
;;
)

C
^
:
(3.57)
0


2
B

(
;;
)

C
^
:
(3.58)
Recallthat(nowwiththis
B
):
˘
2
K
+
˘L
+

2
N
:=
˘
2
[
 
(
;;
)+(


2
B



2
B

)

]
˘
[

 
(
;;
)+(

2
B

)



2
B

]

2
[

(1+3

+

2
)
B


2
B


(

2
B

)



2
B


]

0
:
Wewillchoosesoonappropriate

0
;
1

1
100

0
and


˝
0
withsomesmall
˝
.Letus
120
introduce
k
:=
Z

0

1
K
=
Z

0

1
[
 
(
;;
)+(


2
B



2
B

)

]
;
n
:=
Z

0

1
N
=
Z

0

1
[

(1+3

+

2
)
B


2
B


(

2
B

)



2
B


]
;
`
:=
Z

0

1
[

 
(
;;
)+(

2
B

)



2
B

]
:
Estimateof
k
fromabove.
Theintegrandof
k
isobviouslypositiveand
 
termdominates
otherterms(by(3.56),(3.57),(3.58)andthesmallnessof

).Therefore,
0

k

C
1
(
^

0
+
C
^
Q
Q

2
0
)+
C
2
^

2
0

C
(
^

0
+
C
^
Q
Q

2
0
)
;
(3.59)
if
Q
isverylarge.Wechoose(wearesorryforastrangewayofwriting

0
,whywedothat
willbeseeninthenextsection)

0
=
c

Q
^
Q

ˆ
;ˆ
=1
;
1
=
1
100
s
Q
^
Q

0
:
(3.60)
Here
c
isasmallpositiveconstant.Wealsochoosetohave

runningonlyonthefollowing
interval

2
[0
;
0
]
;
0
:=
˝

Q
^
Q

ˆ

0
;ˆ
=1
;
(3.61)
where
˝
isasmallpositiveconstant.
Estimateof
`
frombelow.
Estimatingfrombelowwecanskipthenon-negativeterm


2
B

.Also
Z

0

1

 
(
;;
)

C
^

0

C
^
Q
Q

2
0
:
121
Ontheotherhand,
Z

0

1
(

2
B

)




2
0
B

(

0
;;
)


2
1
^
Q;
asgivesapointwiseestimate
B


C
^
Q:
(3.62)
Recallthat

2
[
Q=
4
;Q=
2].Wealsowillprovesoontheobstaclecondition(3.74),which
saysthat
B
(1
;;
)


8
:
(3.63)
If
B

(

0
;;
)wouldbesmallerthan
Q=
40(andthen
B

(
s;;
)

Q=
40forall
s
2
[

0
;
1]byconcavityof
B
initsrstvariable)wewouldnotbeabletoreachatleast
Q
4

8
.In
fact,byourchoiceof

0
in(3.60)wehave
B
(

0
;;
)

^

0

cQ:
(3.64)
If
B

(

0
;;
)

Q
40
,andsothisderivative
B

(
s;;
)

Q
40
on
s
2
[

0
;
1](concavity),we
cannotreach
Q=
(4

8)for
s
=1ifwestartwithvalueof
B
in(3.64)at
s
=

0
.Butthefact
thatwecannotreach
Q=
(4

8)contradictsto(3.63).Therefore,
B

(

0
;;
)

Q
40
;
(3.65)
and
`


2
0
40
Q


2
1
^
Q

C
^

0

C
^
Q
Q

2
0
:
(3.66)
122
As

1
=
1
100

0
r
Q
^
Q
(see(3.60)),thesecondtermisdominatedbythethethirdterm
isdominatedbythebecauseofthechoiceof

0
in(3.61),thefourthtermisdominated
bytheonebecause
Q
2
>>
^
Q
,see[P]foramuchbetterestimate.
Finally,
`


2
0
80
Q

c
2
0
Q:
(3.67)
And
k
is
0

k

C
(
^

0
+
C
^
Q
Q

2
0
)=

0
^
Q
(

+
1
Q

0
)
:
Wegot
n

`
2
4
k

c

4
0
Q
2

0
^
Q
(

+
1
Q

0
)
:
(3.68)
Estimateof
n
fromabove.
By(3.65),(3.62)and(3.57)weget
Z

0

1

(

2
B

)




Z

0

1

2
B




2
0
+
C
^

2
1
+
c
^

2
0


0
:
Negativityisbythechoiceof

1
in(3.60)andbythefactthat


c
s
Q
^
Q
;
(3.69)
whichismuchoverdonein(3.61).
Therefore,weget,combiningwith(3.68)(here
>
isanabsoluteconstantanditisat
leastthemaximumofallour3

+

2
)
c

3
0
Q
2
^
Q
(

+
1
Q

0
)

n

(1+

)
Z

0

1
(
e
1
1+


2
B

)

;
123
or
Z

0

1
(

e
1
1+


2
B

)



C

3
0
Q
3
^
Q
(

+

0
)
:
(3.70)
Function
B
issmooth,concavein

andsymmetricin

(thelatterisbyIn
particular
B

(
;;
0)=0.Soafterintegratingin

on[0
;
]
;<
0
weget
Z

0

1
(

B

)


C
3
0
Q
2
^
Q
[log(

0
+

)

log

0
]=
C
3
0
Q
2
^
Q
log(1+
Q

0

)
:
(3.71)
Integrateagainin

on[0
;
0
].Wegettheintegralover[

1
;
0
]oftheoscillationof
B
,
whichis
Z

0

1
[
B
(
;;
0)

B
(
;;
0
)]


C
3
0
Q
2
^
Q


0
Q
(1+
Q

0

0
)log(1+
Q

0

0
)
:
Butthisoscillationissmallerthan
C
^

2
0
.Wegettheinequality
C
4
0
Q
^
Q
(1+
Q

0

0
)log(1+
Q

0

0
)


2
0
^
Q:
(3.72)
Noticethat

0
,

0
,

0

0
areallpowersof
Q
^
Q
,whichweexpecttobeasortof
1
(log
Q
)
p
.
Thenwegettheestimateintermsof
powers
of
Q
^
Q
:
C
2
0
Q
2
^
Q
2

0

0
log(1+
Q

0

0
)

1
:
(3.73)
Letuscountthepowersof
Q
^
Q
:

2
0
bringspower2|by(3.60),

0

0
bringspower1by(3.61),
sototallywehave
1
(log
Q
)
5
p
log
Q
(log
Q
)
:::
inthelefthandside.
Wecanseethatif
^
Q

Q
log
p
Q
with
p<
1
5
,then(3.73)leadstoacontradiction.Sowe
124
proved
Theorem3.4.5.
Theweightedweaknormofthemartingaletransformforweights
w
2
A
dyadic
1
canreach
c
[
w
]
A
1
log
p
[
w
]
A
1
foranypositive
p<
1
=
5
.
3.4.2Obstacleconditionsfor
B
.
Nowwewanttoshowthefollowingobstacleconditionfor
B
,whichwealreadyused:
if
j

j
<
1
4
;
then
B
(1
;;
)


8
:
(3.74)
Let
I
:=[0
;
1].Givennumbers
j
f
j
<
4
;
F
m
=

itisenoughtoconstructfunctions
'; ;w
on
I
suchthat
Put
'
=

a
on
I

,=
b
on
I
++
,zerootherwise.And
w
=1on
I

[
I
++
,and
w
=
Q
otherwise.Thenput
 
:=(
';h
I

)
h
I


(
';h
I
+
)
h
I
+
:
Let0
<a<b
and
a
iscloseto
b
.Put

=(
a
+
b
)
=
4.Thenaverageof
'
issmallwithrespect
to

andwecanprescribeit.
F
=(
a
+
b
)
=
4
;m
=1.Ontheotherhand,function
 
(which
isamartingaletransformof
'
h
'
i
)isatleast

(
';h
I
+
)
h
I
+

1
2
b


on
I
+

,whose
w
-measureismorethan
1
3
w
(
I
).So
B
(1
;;
)

1
3
;
(3.75)
forallsmall

and


Q
.Thisiswhatwewantedtoprove.
125
Chapter4
Oneweightestimatesinmetricspaces
4.1Mainresults
Herewegiveaproofofthe
A
2
conjectureingeometricallydoublingmetricspaces(GDMS),
i.e.ametricspacewhereonecannotmorethanaamountofdisjointballsofradius
r
inaballofradius2
r
.
Ourmainresultisthefollowing.
Theorem4.1.1
(
A
2
theoremforageometricallydoublingmetricspace)
.
Let
X
beageo-
metricallydoublingmetricspace,

and
T
asabove,
w
2
A
2

.Inadditionweassumethat

isadoublingmeasure.Then
k
T
k
L
2
(
w
)
!
L
2
(
w
)
6
C
(
T
)[
w
]
2

:
(4.1)
Theproofisorganizedasfollows:
1.Aconstructionofaprobabilityspaceofrandom\dyadic"latticeinametricspaceis
giveninSection4.2;
2.Averagingtrickofonen[HPTV](butwethinkweit)isgiveninSection
4.4.3;
3.Alinearestimateofweighteddyadicshiftonmetricspacefrom[NV],whichuses
126
Bellmanfunctiontechnique,isgiveninSections4.4.1and4.4.2.Foranotherproof
ofthelinearestimateforweighteddyadicshifts,whichcanbeeasilyadjustedtothe
metriccase,wereferto[T].
4.2Themainconstruction
4.2.1FirstStep
Consideracompactdoublingmetricspace
X
withmetric
d
anddoublingconstant
A
.Instead
of
d
(
x;y
)wewrite
j
xy
j
.Precisely,theisthefollowing.
24.
Suppose(
X;
j
:
j
)isametricspace.Wecallitgeometricallydoublingwith
constant
A
,ifforany
x
2
X
and
r>
0wecannomorethan
A
disjointballsofradius2
intheball
B
(
x;r
).
Asauthorsof[HM],weessentiallyusetheideaofMichaelChrist[Chr],butrandomize
hisconstructioninatway.Therefore,wewanttoguardthereaderthateventhough
onthesurfacetheproofbelowisveryclosetotheprooffrom[HM],however,ourconstruction
isessentiallyt,andsotheproofoftheassertioninourmainlemma,whichwasnot
hardin[HM],becomesmuchmoresubtlehere.
Wenowproceedtotheconstruction.
Foranumber
k>
0wesaythataset
G
isa
k
-gridif
G
ismaximal(withrespectto
inclusion)set,suchthatforany
x;y
2
G
wehave
d
(
x;y
)
>k
.
Letfromnowondiam
X
=1.Takeasmallpositivenumber

˝
1dependingonthe
doublingconstantof
X
andalargenaturalnumber
N
,andforevery
M
>
N

G
M
=
f
z

M
g
,
acertain

M
-gridof
X
.Nowtake
G
N
andrandomlychoosea
G
N

1
=

N

1
-gridin
G
N
.
127
Thentake
G
N

1
andrandomlychoosea
G
N

2
=

N

2
-gridin
G
N

1
.Dothis
N
times.
Noticethat
G
0
consistsofjustonerandompointof
G
N
.
Weexplainwhatis\randomly".Since
X
isacompactmetricspace,all
G
k
'sare
Therefore,therearemany(
N

1)-gridsin
G
N
.Wechooseoneofthemwitha
probability
1
numberof(
N

1)-gridsin
G
N
:
Ourlemmaisthefollowing.
Lemma4.2.1.
For
k
=0
;:::;N
[
y
2
G
N

k
B
(
y;
3

N

k
)=
X:
Remark11.
For
N
+
k;k

0
;
insteadof
N

k
thisisobvious.
Proof.
Take
x
2
X
.Then,since
G
N
ismaximal,thereexistsapoint
y
0
2
G
N
,such
that
j
xy
0
j
6

N
.Since
G
N

1
ismaximalin
G
N
,thereisapoint
y
1
2
G
N

1
,suchthat
j
y
0
y
1
j
6

N

1
.Similarlyweget
y
2
;:::;y
k
andthen
j
xy
k
j
6
j
xy
0
j
+
:::
+
j
xy
k
j
6

N
+
:::
+

N

k
=

N

k
(1+

+
:::
+

k
)
6

N

k
1


6
2

N

k
:
Oncewehavealloursets
G
N
,weintroducearelationship
˚
betweenpoints.Wefollow
[HM]and[Chr].
Takeapoint
y
k
+1
2
G
k
+1
.Thereexistsatmostone
y
k
2
G
k
,suchthat
j
y
k
+1
y
k
j
6

k
4
.
128
Thisistruesinceiftherearetwosuchpoints
y
1
k
;y
2
k
,then
j
y
1
k
y
2
k
j
6

k
2
;
whichisacontradiction,since
G
k
wasa

k
-gridin
G
k
+1
.
Alsothereexistsatleastone
z
k
2
G
k
suchthat
j
y
k
+1
z
k
j
6
3

k
.Thisistruebythe
lemma.
Now,ifthereexistsan
y
k
asabove,weset
y
k
+1
˚
y
k
.Ifno,thenwepickoneof
z
k
as
aboveandset
y
k
+1
˚
z
k
.Forallother
x
2
G
k
weset
y
k
+1
6˚
x
.Thenextendbytransitivity.
Wealsoassumethat
y
k
˚
y
k
.Thisisif
y
k
onthelefthappenedtobelongalreadyto
G
k
+1
.
Wedothisprocedurerandomlyandindependently,andtreatsamefamiliesof
G
k
'swith
t
˚
-lawastfamilies.
Takenowapoint
y
k
2
G
k
and
Q
y
k
=
[
z
˚
y
k
;z
2
G
`
B
(
z;

`
100
)
:
Lemma4.2.2.
Forevery
k
wehave
X
=
[
y
k
2
G
k
clos(
Q
y
k
)
Remark12.
Thereisonlyonepointin
G
0
,and
clos
(
Q
y
)
;y
2
G
0
;
isjust
X
.Butforsmall

,
X
=
S
y
1
2
G
1
clos(
Q
y
1
)isagenuine(andrandom)splittingof
X
.
Proof.
Takeany
x
2
X
.Bythepreviouslemma,forevery
m>k
thereexistsapoint
129
x
m
2
G
m
,suchthat
j
xx
m
j
6
3

m
.Inparticular,
x
m
!
x
.Fixforamoment
x
m
.Then
therearepoints
y
m

1
2
G
m

1
;:::;y
k
2
G
k
,suchthat
x
m
˚
y
m

1
˚
:::
˚
y
k
.In
particular,
x
m
2
Q
y
k
,where
y
k
dependson
x
m
.Then
j
y
k
x
j
6
j
y
k
x
m
j
+
j
x
m
x
j
6
j
y
k
x
m
j
+3

m
6
j
y
k
x
m
j
+3

k
:
Moreover,bythechainof
˚
's,weknowthat
j
y
k
x
m
j
6
10

k
.Therefore,
j
y
k
x
j
6
15

k
:
Weclaimthattheset
f
y
k
g
=
f
y
k
(
x
m
)
g
m
>
k
isindependentlyon
k
.Thisistruesince
all
y
k
'sareseparatedfromeachotherandbythedoublingofourspace(weare
theball
B
(
x;
15

k
)withballs
B
(
y
k
;
k
)).
So,takeansubsequence
x
m
thatcorrespondstoonepoint
y
k
2
G
k
.Thenweget
x
m
2
Q
y
k
,
x
m
!
x
,so
x
2
clos
Q
y
k
,andwearedone.
Remark13.
Sincethespace
X
iscompact,ourrandomprocedureconsistsofmany
steps.Therefore,ourprobabilityspaceisdiscreet.Wesuggesttothinkaboutallprobabilities
justasnumberofgoodeventsdividedbynumberofallevents.
However,allourestimateswillnotdependonnumberofsteps(and,therefore,diameter
of
X
),whichisessential.
Remark14.
WenoticethatintheEuclidianspace,say,
R
,thisproceduredoesnotgivea
standarddyadiclattice.
130
4.2.2Secondstep:technicallemmata

~
Q
y
k
=
X
n
[
z
k
6
=
y
k
;z
k
2
G
k
clos
Q
z
k
:
Inparticular,
Q
y
k
ˆ
~
Q
y
k
ˆ
clos(
Q
y
k
)
:
Lemma4.2.3
(Lemma4.5in[HM])
.
Let
m
beanaturalnumber,
">
0
,and

m
>
100
"
.
Suppose
x
2
clos
Q
y
k
and
dist
(
x;X
n
~
Q
y
k
)
<
k
.Thenforanychain
z
k
+
m
˚
z
k
+
m

1
˚
:::
˚
z
k
+1
˚
z
k
;
suchthat
x
2
clos
Q
z
k
+
m
,thefollowingrelationshipshold
j
z
i
z
j
j
>

j
100
;k
6
j<i
6
k
+
m:
Proof.
Suppose
j
z
i
z
j
j
<

j
100
.Weconsideracasewhen
z
k
=
y
k
.Since
z
j
˚
z
k
=
y
k
,we
have
B
(
z
j
;

j
200
)
ˆ
Q
y
k
ˆ
~
Q
y
k
.Therefore,

j
200
6
dist
(
z
j
;X
n
~
Q
y
k
)
6
dist
(
x;X
n
~
Q
y
k
)+
dist
(
x;z
i
)+
dist
(
z
i
;z
j
)
<
k
+5

i
+

j
100
If

islessthan,say,
1
1000
,thenwegetacontradiction.
Theonlynotobviousestimateisthat
dist
(
x;z
i
)
<
5

i
.Itistruesince
x
2
clos
Q
z
k
+
m
.
Wehaveprovedthelemmawithassumptionthat
z
k
=
y
k
.Letusgetridofthisassump-
131
tion.Weknowthat
x
2
clos
Q
z
k
+
m
ˆ
clos
Q
z
k
:
Alsowehave
x
2
clos
Q
y
k
,so,since
~
Q
z
k
=
X
n
[
u
k
6
=
z
k
clos
Q
u
k
ˆ
X
n
clos
Q
y
k
;
weget
x
2
X
n
~
Q
z
k
.Inparticular,
dist
(
x;X
n
~
Q
z
k
)=0
<
k
,andweareinthesituation
ofthepart.Thisourproof.
Lemma4.2.4.
Fix
x
k
2
G
k
.Then
P
(
9
x
k

1
2
G
k

1
:
j
x
k
x
k

1
j
<

k

1
1000
)
>
a
(4.2)
forsome
a
2
(0
;
1)
.
Proof.
Weremindthatweareinacompactmetricsituation.Byrescalingwecanthinkthat
weworkwith
G
1
andchoose
G
0
.Wecaneventhinkthatthemetricspaceconsistsof
manypoints,itis
X
:=
G
2
.Theset
G
1
ˆ
X
consistsofpointshavingthefollowing
properties:
1.
8
x;y
2
G
1
wehave
j
xy
j

;
2.if
z
2
X
n
G
1
then
9
x
2
G
1
suchthat
j
zx
j
<
.
Thesetwopropertiesareequivalenttosayingthatthesubset
G
1
of
X
consistsofpoints
suchthat
8
x;y
2
G
1
wehave
j
xy
j

andwecannotaddanypointfrom
X
to
G
1
without
violatingthatproperty.Inotherwords:
G
1
isa
maximal
setwithproperty1.
Recallthatheretheword\maximal"meansmaximalwithrespecttoinclusion,not
132
maximalinthesenseofthenumberofelements.
Nowweconsiderthenewmetricspace
Y
=
G
1
and
G
0
isanymaximalsubsetsuchthat
8
x;y
2
G
0
;
j
xy
j
1
:
(4.3)
Inotherwords,wehave1.
8
x;y
2
G
0
wehave
j
xy
j
1;
2.if
z
2
Y
n
G
0
then
9
x
2
G
0
suchthat
j
zx
j
<
1.
Therearelymanysuchmaximalsubsets
G
0
of
Y
.Weprescribeforeachchoicethe
sameprobability.Nowwewanttoprovetheclaimthatisevenstrongerthan(4.2).Namely,
wearegoingtoprovethatgiven
y
2
Y
P
(
9
x
0
2
G
0
:
x
0
=
y
)
>
a;
(4.4)
where
a
dependsonlyon

andtheconstantsofgeometricdoublingofourcompactmetric
space.
Let
Y
beanymetricspacewithitelymanyelements.Wewillcolorthepointsof
Y
intoredandgreencolors.Thecoloringiscalledproperif
1.everyredpointdoesnothaveanyotherredpointatdistance
<
1;
2.everygreenpointhasatleastoneredpointatdistance
<
1.
Given
apropercoloring
of
Y
thecollectionofredpointsiscalled1-
lattice
.Itisamaximal
(byinclusion)collectionofpointsatdistance

1fromeachother.
Whatweneedtotheproofis
Lemma4.2.5.
Let
Y
beametricspaceasabove.Assume
Y
hasthefollowingproperty:
Ineveryballofradiuslessthan
1
thereareatmost
d
elements
:
(4.5)
133
Let
L
beacollectionof
1
-latticesin
Y
.Elementsof
L
arecalled
L
.Let
v
2
Y
.Then
thenumberof1-latticesLsuchthatvbelongstoL
thetotalnumberof1-latticesL

a>
0
;
where
a
dependsonlyon
d
.
Proof.
Given
v
2
Y
considerallsubsetsof
B
(
v;
1)
n
v
,thiscollectioniscalled
S
.Let
S
2S
.
Wecall
W
S
thecollectionofallpropercoloringssuchthat
v
isgreen,allelementsof
S
are
red,andallelementsof
B
(
v;
1)
n
S
aregreen.Wecall
~
S
allpointsin
Y
,whicharenotin
B
(
v;
1),butatdistance
<
1fromsomepointin
S
.
Allpropercoloringsof
Y
suchthat
v
isredarecalled
B
.Letusshowthat
card
W
S

card
B:
(4.6)
Noticethatif(4.6)wereproved,wewouldbedonewithLemma4.2.5,
a

2

d
+1
,and,
consequently,theproofofthemainlemmawouldbe
a

2



D
,where
D
isa
geometricdoublingconstant.
Toprove(4.6)letusshowthatwecanrecoloranypropercoloringfrom
W
S
intotheone
from
B
,andthatthismapisinjective.Let
L
2
W
S
.We
1.Color
v
intored;
2.Color
S
intogreen;
3.Elementsof
~
S
wereallgreenbefore.Weleavethemgreen,butweamongthemall
those
y
thatnowintheopenball
B
(
y;
1)in
Y
allelementsaregreen.Wecallthemyellow
(temporarily)anddenotethem
Z
;
4.Weenumerate
Z
inanyway(non-uniquenessishere,butwedonotcare);
134
5.Intheorderofenumerationcoloryellowpointstored,ensuringthatweskiprecoloring
ofapointin
Z
ifitisat
<
1distancetoanypreviouslycoloredyellow-to-redpointfrom
Z
.
Afterseveralstepsallgreenandyellowelementsof
~
S
willhavethepropertythatatdistance
<
1thereisaredpoint;
6.Colortherestofyellow(ifany)intogreenandstop.
Weresultinapropercoloring(itiseasytocheck),whichisobviously
B
.Suppose
L
1
;L
2
aretwotpropercoloringin
W
S
.Noticethatthecolorsof
v;S;B
(
v;
1)
n
S
,
~
S
arethe
sameforthem.Sotheyersomewhereelse.Butourproceduredoesnottouch\somewhere
else".Sothemocolorings
L
0
1
;L
0
2
thatweobtainafterthealgorithm1-6willas
wellmaybeevenmore).Soourmap
W
S
!
B
(beingnotuniquelyd)ishowever
injective.Weproved(4.6).
Thus,theproofoftheLemma4.2.4isished.
Remark.
WearegratefultoMichaelShapiroandDapengZhanwhohelpedustoprove
Lemma4.2.4.
4.2.3Mainandtheorem
Fixanumber

,0
<<
1.Laterthechoiceof

willbedictatedbytheon-Zygmund
propertiesoftheoperator
T
.Alsoatlybig
r
.Thecoiceof
r
willbemadeinthis
section.
25
(Badcubes)
.
Takea\cube"
Q
=
Q
x
k
.Wesaythat
Q
isgoodifthereexists
135
acube
Q
1
=
Q
x
n
,suchthatif

k
6

r

n
(
k
>
n
+
r
)
theneither
dist
(
Q;Q
1
)
>

k

n
(1


)
or
dist
(
Q;X
n
Q
1
)
>

k

n
(1


)
:
Remark15.
Noticethat

k
=
`
(
Q
)justby
If
Q
isnotgoodwecallitbad.
Theorem4.2.6.
Fixacube
Q
x
k
.Then
P
(
Q
x
k
isbad
)
6
1
2
:
Remark16
(Discussion)
.
Thistheoremmakessensebecausewhenweacube
Q
k
,say,
k
>
N
,sothegrid
G
k
isnotevenrandom,wecanmakebigcubesrandom.Andweclaim
thatforbigquantityofchoices,ourbigcubeswillhave
Q
k
either\inthemiddle"orfar
away,butnotclosetotheboundary.
26.
For
Q
=
Q
x
k


Q
(
"
)=

Q
=
f
x
:
dist
(
x;Q
)
6

k
and
dist
(
x;X
n
Q
)
6

k
g
Lemma4.2.7.
Letusstartwithlevel
N
bya

N
-grid(non-random),andlet
k<N
,
136
x
k
denotingthepointsofthe(random)grid
G
k
.Fixapoint
x
2
X
.
P
(
9
x
k
2
G
k
:
x
2

Q
x
k
)
6
"

forsome
>
0
.
Proofofthetheorem.
Takethecube
Q
x
k
.Thereisaunique(random!)point
x
k

s
such
that
x
k
2
Q
x
k

s
.Then
dist
(
Q
x
k
;X
n
Q
x
k

s
)
>
dist
(
x
k
;X
n
Q
x
k

s
)

diam
(
Q
x
k
)
>
dist
(
x
k
;X
n
Q
x
k

s
)

C
k
:
Assumethat
dist
(
x
k
;X
n
Q
x
k

s
)
>
2

k

(
k

s
)(1


)
andthat
s
>
r
(thisassumptionis
obvious,otherwise
Q
x
k

s
doesnotgoodnessof
Q
x
k
).
Then,if
r
isbigenough(

r
(1


)
<
1
C
)weget
dist
(
Q
x
k
;X
n
Q
x
k

s
)
>

k

(
k

s
)(1


)
;
andso
Q
x
k
isgood.Therefore,
P
(
Q
x
k
isbad)
6
C
X
s
>
r
P
(
x
k
2

Q
k

s
(
"
=2


))
6
C
X
s
>
r

s
6
100
C
r
:
Bythechoiceof

,fortlylarge
r
thisislessthan
1
2
.
Proofofthelemma.
Let
x
k
besuchthat
x
2
clos
Q
x
k
(seeLemma4.2.2).Wewillestimate
P
(
dist
(
x;X
n
~
Q
k
)
<
k
)
j
x
2
clos
Q
x
k
).Fixthelargest
m
suchthat500
"
6

m
.Choosea
137
point
x
k
+
m
suchthat
x
2
clos
Q
x
k
+
m
.Thenbythemainlemma
P
(
9
x
k
+
m

1
2
G
k
+
m

1
:
j
x
k
+
m
x
k
+
m

1
j
<

k
+
m

1
1000
)
>
a:
Therefore,
P
(
8
x
k
+
m

1
2
G
k
+
m

1
:
j
x
k
+
m
x
k
+
m

1
j
>

k
+
m

1
1000
)
6
1

a:
Letnow
x
k
+
m
˚
x
k
+
m

1
:
Then
P
(
8
x
k
+
m

2
2
G
k
+
m

2
:
j
x
k
+
m

1
x
k
+
m

2
j
>

k
+
m

2
1000
)
6
1

a:
SobyLemma4.2.3
P
(
dist
(
x;X
n
~
Q
k
)
<
k
)
6
P
(
j
x
k
+
j
x
k
+
j

1
j
>

k
+
j

1
1000
8
j
=1
;:::;m
)
6
(1

a
)
m
6
C"

for

=
log(1

a
)
log(

)
:
4.2.4Probabilitytobe\good"isthesameforeverycube
Wemakethelaststeptomaketheprobabilitytobe\good"notjustboundedawayfrom
zero,butthesameforallcubes.Weusetheideafrom[M].
Takeacube
Q
(
!
).Takearandomvariable
˘
Q
(
!
0
),whichisequallydistributedon[0
;
1].
138
Weknowthat
P
(
Q
isgood)=
p
Q
>a>
0
:
Wecall
Q
\reallygood"if
˘
Q
2
[0
;
a
p
Q
]
:
Otherwise
Q
joinsbadcubes.Then
P
(
Q
isreallygood)=
a;
andwearedone.
4.3TheHaarshiftdecomposition
Taketwostepfunctions,
f
and
g
.Wean
N
-grid
G
N
in
X
,and\cubes"onlevel
N
,suchthat
f
and
g
areconstantsoneverysuchcube.Thenwestartourrandomization
process.
Aswementioned,thisprocessconsistsofmanysteps,soallprobabilistictermi-
nologybecomestrivial:wehaveaprobabilityspace.
Startingfrom
G
N
,wego\up"andoneachlevelgetdyadiccubes(randomChrist'scubes).
Theyhavetheusualstructureofbeingeitherdisjointoronecontainingtheother.Foreach
dyadiccube
Q
wehaveseveraldyadicsons,theyaredenotedby
s
i
(
Q
),
i
=1
;:::;M
(
Q
)

M
.
Thenumber
M
hereisuniversalanddependsonlyongeometricdoublingconstantsofthe
space
X
.
27.
By
E
k
wedenotesetofalldyadic\cubes"ofgeneration
k
.Wecall
Q
i
k
ˆ
Q
j
k

1
,
Q
i
k
2E
k
sons
of
Q
j
k

1
.
139
Witheverycube
Q
=
Q
x
k
weassociate
Haarfunctions
h
j
Q
,
j
=1
;:::;M

1,with
followingproperties:
1.
h
j
Q
issupportedon
Q
;
2.
h
j
Q
takesconstantvaluesoneach\son"of
Q
;
3.Foranytwocubes
Q
and
R
,wehave(
h
j
Q
;h
i
R
)=0,and(
h
j
Q
;
1)=0;
4.
k
h
j
Q
k
1
6
C
p

(
Q
)
.
Wenoticethatthelastpropertyimpliesthat
k
h
j
Q
k
2
6
C
.
Weuseangularbracketstodenotetheaverage:
h
f
i

:=
1

(
Q
)
R
Q
f
.Whenweaverage
overthewholespace
X
,wedroptheindexandwrite
h
f
i
=
1

(
X
)
R
X
f
.
Ourmain\tool"isgoingtobethefamous\dyadicshifts".Precisely,wecallby
S
m;n
the
operatorgivenbythekernel
f
!
X
L
2D
Z
L
a
L
(
x;y
)
f
(
y
)
dy;
where
a
L
(
x;y
)=
X
I
ˆ
L;J
ˆ
L
g
(
I
)=
g
(
L
)+
m;g
(
J
)=
g
(
L
)+
n
c
L;I;J
h
j
J
(
x
)
h
i
I
(
y
)
;
where
h
i
I
;h
j
J
areHaarfunctionsnormalizedin
L
2
(

)andsatisfying(iv),and
j
c
L;I;J
j
p

(
I
)
p

(
J
)

(
L
)
.Oftenwewillskipsuperscripts
i;j
.
28.
Wecallthenumber
m
+
n
+1the
complexity
ofashift
S
m;n
.
Ournextaimistodecomposethebilinearformoftheoperator
T
intobilinearforms
140
ofdyadicshifts,whichareestimatedintheSection4.4.2.Therestwillbetheso-called
\paraproducts",estimatedintheSection4.4.1.
Functions
f
˜
X
g[f
h
j
Q
g
formanorthogonalbasisinthespace
L
2
(
X;
).Therefore,we
canwrite
f
=
h
f
i
˜
X
+
X
Q
X
j
(
f;h
j
Q
)
h
j
Q
;g
=
h
g
i
˜
X
+
X
R
X
i
(
g;h
i
R
)
h
i
R
:
First,westateandproofthetheorem,thatsaysthatessentialpartofbilinearformof
T
canbeexpressedintermsofpairofcubes,wherethesmallestoneisgood.Wefollowthe
ideaofonen[H].Infact,thework[H]improvedon\good-bad"decompositionof[NTV],
[NTV2],[NTV3]byreplacinginequalitiesbyanequality.
Theorem4.3.1.
Let
T
beanylinearoperator.Thenthefollowingequalityholds:
ˇ
good
E
X
Q;R;i;j
`
(
Q
)
>
`
(
R
)
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)=
E
X
Q;R;i;j
`
(
Q
)
>
`
(
R
)
;R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
:
Thesameistrueifwereplace
>
by
>
.
Proof.
Wedenote
˙
1
(
T
)=
X
`
(
Q
)
>
`
(
R
)
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
:
˙
1
(
T
)=
X
`
(
Q
)
>
`
(
R
)
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
:
Wewouldliketogetarelationshipbetween
E
˙
1
(
T
)and
E
˙
1
(
T
).
141
We
R
andwrite(using
g
good
:=
P
R
isgood
(
g;h
i
R
)
h
i
R
)
X
Q
X
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)=
0
@
T
(
f
h
f
i
˜
X
)
;
X
R
isgood
(
g;h
i
R
)
h
i
R
1
A
=

T
(
f
h
f
i
˜
X
)
;g
good

:
Takingexpectations,weobtain
E
X
Q;R
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
1
R
isgood
=
E
(
T
(
f
h
f
i
˜
X
)
;g
good
)=(
T
(
f
h
f
i
˜
X
)
;
E
g
good
)=
ˇ
good
(
T
(
f
h
f
i
˜
X
)
;g
)=
ˇ
good
E
X
Q;R
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
:
(4.7)
Next,suppose
`
(
Q
)
<`
(
R
).Thengoodnessof
R
doesnotdependon
Q
,andso
ˇ
good
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)=
E

(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
1
R
isgood
j
Q;R

:
Letusexplainthisequality.Therighthandsideisconditioned:meaningthatthelefthand
sideinvolvesthefractionofthenumberofalllatticescontaining
Q;R
inthislatticeandsuch
that
R
(thelargerone)isgoodtothenumberoflatticescontaining
Q;R
init.Thisfraction
isexactly
ˇ
good
.Nowweapairof
Q;R
,
`
(
Q
)
<`
(
R
),andmultiplybothsidesbythe
probabilitythatthispairisinthesamedyadiclatticefromourfamily.Thisprobabilityis
justtheratioofthenumberofdyadiclatticesinourfamilycontainingelements
Q
and
R
tothenumberofalldyadiclatticesinourfamily.Aftermultiplicationbythisratioandthe
142
summationofalltermswith
`
(
Q
)
<`
(
R
)weget,
ˇ
good
E
X
`
(
Q
)
<`
(
R
)
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)=
E
X
`
(
Q
)
<`
(
R
)
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
1
R
isgood
:
(4.8)
Nowweuse(4.7)andthen(4.8):
ˇ
good
E
X
Q;R
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)=
E
X
Q;R
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
1
R
isgood
=
=
E
X
`
(
Q
)
<`
(
R
)
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
1
R
isgood
+
E
X
`
(
Q
)
>
`
(
R
)
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
1
R
isgood
=
=
ˇ
good
E
X
`
(
Q
)
<`
(
R
)
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)+
E
X
`
(
Q
)
>
`
(
R
)
;R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
;
(4.9)
andtherefore
E
X
`
(
Q
)
>
`
(
R
)
;R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)=
ˇ
good
E
X
`
(
Q
)
>
`
(
R
)
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
:
(4.10)
Thisisthemaintrick.Tohavethewholesumexpressedasthemultipleofthesum,
wherethe
smaller
insizecubeisgood,is
veryuseful
aswewillsee.Itgivesextra
decayonmatrixcots(
Th
j
Q
;h
i
R
)andallowsustorepresentouroperatoras\convex
combinationofdyadicshifts".
143
So,wehaveobtainedthat
E
˙
1
(
T
)=
ˇ

1
good

E
˙
1
(
T
)
:
Thus,toestimate
E
˙
1
(
T
)itisenoughtoestimate
E
˙
1
(
T
).Absolutelythesamesym-
metricallyholdsfor
˙
2
(
T
).
4.3.1Paraproducts
Inthissubsectionwetakecareoftheterms
h
f
i
˜
X
and
h
g
i
˜
X
.Thesetermswillleadtoso
calledparaproducts.Infact,letusintroducethreeauxiliaryoperators:
ˇ
(
f
):=
ˇ
T˜
X
(
f
):=
X
Q;j
h
f
i
Q
(
T˜
X
;h
j
Q
)
h
j
Q
;(4.11)
ˇ

(
f
):=
X
Q;j
(
f;h
j
Q
)(
T

˜
X
;h
j
Q
)
˜
Q

(
Q
)
=(
ˇ
T

˜
X
)

(
f
);(4.12)
o
(
f
):=
h
f
ih
T˜
X
i
˜
X
:
(4.13)
Recallthat
h
'
i
denotes
1

(
X
)
R
X
'
.Theseoperatorsdependonthedyadicgridwe
chose.Weshallneedthefollowingtechnicallemma.
Lemma4.3.2.
(
ˇ
(
f
)
;g
)=
h
f
i
(
T˜
X
;g
h
g
i
˜
X
)+
X
(
ˇh
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
;
(
ˇ

(
f
)
;g
)=
h
g
i
(
T

˜
X
;f
h
f
i
˜
X
)+
X
(
ˇ

h
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
:
144
Proof.
Thesecondequalityfollowsfromtheoneandtheof
ˇ

.Weprovethe
equality.Wewillnotwritesuperscripts
i
and
j
inHaarfunctions.
Wewrite
ˇ
(
f
)=
h
f
i
ˇ
(
˜
X
)+
X
(
f;h
i
Q
)
ˇ
(
h
i
Q
)
:
Noticethat
ˇ
(
˜
X
)=
X
(
T˜
X
;h
i
Q
)
h
i
Q
=
T˜
X
h
T˜
X
i
;
andthat
ˇ
(
f
)isorthogonalto
˜
X
.Thus,
(
ˇ
(
f
)
;g
)=(
ˇ
(
f
)
;g
h
g
i
˜
X
)=
h
f
i
(
ˇ
(
˜
X
)
;
X
(
g;h
j
R
)
h
j
R
)+
X
(
ˇh
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)=
=
h
f
i
(
T˜
X
;g
h
g
i
˜
X
)+
X
(
ˇh
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)
;
asdesired.Thelastequalityistruebecause
h
T˜
X
i
isorthogonalto
g
h
g
i
˜
X
.
Noticethat
ˇ;ˇ

dependontherandomdyadicgrid.Weintroducearandomoperator
~
T
=
Tf

ˇ
(
f
)

ˇ

(
f
)
:
Nowwestatethefollowingveryusefullemma.
Lemma4.3.3.
(
Tf;g
)=
ˇ

1
good
E
X
Q;R
smallerisgood
(
~
Th
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
E
(
ˇ
(
f
)
;g
)+
E
(
ˇ

(
f
)
;g
)+
h
f
ih
g
i
(
T˜
X
;˜
X
)
:
145
Proof.
First,wewrite
(
Tf;g
)=
X
(
Th
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
h
f
i
(
T˜
X
;g
)+
h
g
i
(
T

˜
X
;f
h
f
i
˜
X
)
:
Wetakeexpectationsnow.Noticethatonlythetermintheright-handsidedepends
onadyadicgrid.Therefore,
(
Tf;g
)=
E
X
(
Th
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
h
f
i
(
T˜
X
;g
)+
h
g
i
(
T

˜
X
;f
h
f
i
˜
X
)
:
Wefocusontheterm.BytheTheorem4.3.1,weknowthat
E
X
(
Th
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)=
ˇ

1
good
E
X
smallerisgood
(
Th
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)=
=
ˇ

1
good
E
X
smallerisgood
(
~
Th
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
+
ˇ

1
good
E
X
smallerisgood
(
ˇh
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
ˇ

1
good
E
X
smallerisgood
(
ˇ

h
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)
:
(4.14)
Thetermisoneofthosethatwewanttogetintheright-handside.
Ontheotherhand,wewanttogetaresultforparaproducts,similartotheTheorem
4.3.1.Indeed,itisclearthat
(
ˇh
i
Q
;h
j
R
)=
h
h
i
Q
i
R
(
T˜
X
;h
j
R
)
;
146
whichisnon-zeroonlyif
R
ˆ
Q
,and
R
6
=
Q
.So,
E
X
smallerisgood
(
ˇh
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)=
E
X
R
ˆ
Q
h
h
i
Q
i
R
(
T˜
X
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)
1
R
isgood
=
=
E
X
R
(
T˜
X
;h
j
R
)(
g;h
j
R
)
1
R
isgood
X
Q
:
R
(
Q
(
f;h
i
Q
)
h
h
i
Q
i
R
:
(4.15)
Wenowseethatsince
f
=
h
f
i
˜
X
+
P
Q
(
f;h
i
Q
)
h
i
Q
,wehave
h
f
i
R
h
f
i
=(
f;
(
R
)

1
˜
R
)
h
f
i
=
X
Q
:
R
(
Q
(
f;h
i
Q
)
h
h
i
Q
i
R
=
X
Q
(
f;h
i
Q
)
h
h
i
Q
i
R
:
Therefore,
E
X
R
(
T˜
X
;h
j
R
)(
g;h
j
R
)
1
R
isgood
X
Q
(
f;h
i
Q
)
h
h
i
Q
i
R
=
E
X
R
(
T˜
X
;h
j
R
)(
g;h
j
R
)
1
R
isgood
(
h
f
i
R
h
f
i
)
:
(4.16)
Nowitisclearthatwecantaketheexpectationinside(wehaveno
Q
anymore,whichwas
preventingusfromdoingthat),andsoweget
E
X
smallerisgood
(
ˇh
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)=
ˇ
good
E
X
R
(
T˜
X
;h
j
R
)(
g;h
j
R
)(
h
f
i
R
h
f
i
)
:
Makingallabovestepsbackwards,weget
E
X
smallerisgood
(
ˇh
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)=
ˇ
good
E
X
(
ˇh
i
Q
;h
j
R
)(
f;h
j
Q
)(
g;h
j
R
)
147
Therefore,
ˇ

1
good
E
X
smallerisgood
(
ˇh
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
ˇ

1
good
E
X
smallerisgood
(
ˇ

h
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)=
=
E
X
(
ˇh
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
E
X
(
ˇ

h
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)=
=
E
(
ˇ
(
f
)
;g
)+
E
(
ˇ

(
f
)
;g
)

E
[
h
f
i
(
T˜
X
;g
h
g
i
˜
X
)]

E
[
h
g
i
(
T

˜
X
;f
h
f
i
˜
X
)]
:
(4.17)
Wenowusethatlasttwotermsdonotdependonthedyadicgrid,andsowedropexpecta-
tions.Finally,
(
Tf;g
)=
E
X
smallerisgood
(
~
Th
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
E
(
ˇ
(
f
)
;g
)+
E
(
ˇ

(
f
)
;g
)

h
f
i
(
T˜
X
;g
h
g
i
˜
X
)
h
g
i
(
T

˜
X
;f
h
f
i
˜
X
)+
h
f
i
(
T˜
X
;g
)+
h
g
i
(
T

˜
X
;f
h
f
i
˜
X
)=
=
E
X
smallerisgood
(
~
Th
i
Q
;h
j
R
)(
f;h
i
Q
)(
g;h
j
R
)+
E
(
ˇ
(
f
)
;g
)+
E
(
ˇ

(
f
)
;g
)+
h
f
ih
g
i
(
T˜
X
;˜
X
)
:
(4.18)
Thisiswhatwewanttoprove.
Thefollowinglemma,whichwillbeprovedlater,takescareofparaproducts.
Lemma4.3.4.
Theoperators
ˇ
,
ˇ

areboundedon
L
2
(
X;w
)
,and
k
ˇ
k
2
;w
6
C

[
w
]
2

:
Thesameistruefor
ˇ

.
148
Wepostponetheproofofthislemma.Wealsonoticethattheoperator
o
(
f
)=
h
f
ih
T˜
X
i
˜
X
isclearlyboundedwithdesiredconstant.Infact,as
T
isboundedintheunweighted
L
2
,we
have
h
T˜
X
i
2
k
T
k
2
L
2
=:
C
0
k
o
(
f
)
k
2
2
;w
=
h
f
i
2
h
T˜
X
i
2
w
(
X
)
6
C
0
h
f
2
w
ih
w

1
i
w
(
X
)
6
C
0
[
w
]
2
k
f
k
2
2
;w
:
We,therefore,shouldtakecareonlyoftheterm,with
~
T
.Wenowerasethetilde,and
write
T
insteadof
~
T
.Eventhough
T
isnotaCalderon-Zygmundoperatoranymore,all
furtherestimatesaretruefor
T
(i.e.,foraCZOminusparaproducts),see,forexample,[HM]
or[HPTV].
149
4.3.2Estimatesof
˙
1
Ournextstepistodecompose
˙
1
intorandomdyadicshifts.Wewrite
˙
1
(
T
)=
X
`
(
Q
)
>
`
(
R
)
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)=
=
E
X
`
(
Q
)
>


r
0
`
(
R
)
;
R
ˆ
Q;
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)+
+
E
X
`
(
R
)
6
`
(
Q
)


r
0
`
(
R
)
;
R
ˆ
Q;
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)+
+
E
X
`
(
R
)
6
`
(
Q
)
;
R
\
Q
=
;
;
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
:
(4.19)
Essentially,wewillprovethatthenormofeveryexpectationisboundedby
C
(
T
)

E
X
n


"
(
T
)

n
k
S
n
k
:
First,westateourchoicefor

,whichwehaveseenintheofgoodcubes.
29.
Put

=
"
2

(
"
+log
2
(
C
))
;
where
C
isthedoublingconstantofthefunction

.
Remark17.
Weremarkthatthischoiceof

makeLemmata4.3.5and4.3.6true.
150
Theestimateofthesecondsumiseasy.Infact,
E
X
`
(
R
)
6
`
(
Q
)


r
0
`
(
R
)
;
R
ˆ
Q;
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
6
Cr
0
[
w
]
2
k
f
kk
g
k
:
Thisisboundedbyatmost
r
0
expressionsforshiftsofboundedcomplexity,sojustsee[NV].
Formoredetails,see[HPTV]
Wedenote

in
=
E
X
`
(
Q
)
>


r
0
`
(
R
)
;
R
ˆ
Q;
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
;

out
=
E
X
`
(
R
)
6
`
(
Q
)
;
R
\
Q
=
;
;
R
isgood
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
:
4.3.3Estimateof
in
.
Weusethefollowinglemma.
Lemma4.3.5.
Let
T
beasbefore;suppose
`
(
Q
)
>


r
0
`
(
R
)
and
R
ˆ
Q
.Let
Q
1
betheson
of
Q
thatcontains
R
.Then
j
(
Th
j
Q
;h
i
R
)
j
.
`
(
R
)
"
2
`
(
Q
)
"
2


(
R
)

(
Q
1
)

1
2
:
Wenoticethat

(
Q
1
)


(
Q
).
151
Wewrite

in
=
X
n
>
r
0
X
`
(
Q
)=


n
`
(
R
)
;R
isgood
;R
ˆ
Q
(
Th
j
Q
;h
i
R
)(
f;h
j
Q
)(
g;h
i
R
)
;
j

in
j
6
X
n
>
r
0
X
`
(
Q
)=


n
`
(
R
)
;
R
isgood
;
R
ˆ
Q
j
(
Th
j
Q
;h
i
R
)
jj
(
f;h
j
Q
)
jj
(
g;h
i
R
)
j
6
6
C
X
n
>
r
0
X
`
(
Q
)=


n
`
(
R
)
;
R
isgood
;
R
ˆ
Q
`
(
R
)
"
2
`
(
Q
)
"
2


(
R
)

(
Q
)

1
2
j
(
f;h
j
Q
)
jj
(
g;h
i
R
)
j
=
=
C
X
n
>
r
0

n"
2
X
`
(
Q
)=


n
`
(
R
)
;
R
isgood
;
R
ˆ
Q


(
R
)

(
Q
)

1
2
j
(
f;h
j
Q
)
jj
(
g;h
i
R
)
j
:
(4.20)
We

functions
f
and
g
and
S
n
asanoperatorwiththefollowingquadraticform:
(
S
n
u;v
)=
X
`
(
Q
)=


n
`
(
R
)
;
R
isgood
;
R
ˆ
Q



(
R
)

(
Q
)

1
2
(
u;h
j
Q
)(
v;h
i
R
)
;
where

ischosenso
j
(
f;h
j
Q
)
jj
(
g;h
i
R
)
j
=

(
f;h
j
Q
)(
g;h
i
R
).Thenclearly
S
n
isadyadicshift
ofcomplexity
n
,andso,seeSection4.4.2,
j
(
S
n
f;g
)
j
6
Cn
a
[
w
]
2
k
f
k
w
k
g
k
w

1
:
152
Therefore,
j

in
j
6
X
n
Cn
a

n"
2
[
w
]
2
k
f
k
w
k
g
k
w

1
6
C
[
w
]
2
k
f
k
w
k
g
k
w

1
:
4.3.4Estimatesfor
out
Weusethefollowinglemmafrom[HM].
Lemma4.3.6.
Let
T
beasbefore,
`
(
R
)
6
`
(
Q
)
and
R
\
Q
=
;
.Thenthefollowingholds
j
(
Th
j
Q
;h
i
R
)
j
.
`
(
Q
)
"
2
`
(
R
)
"
2
D
(
Q;R
)
"
sup
z
2
R

(
z;D
(
Q;R
))

(
Q
)
1
2

(
R
)
1
2
;
where
D
(
Q;R
)=
`
(
Q
)+
`
(
R
)+dist(
Q;R
)
.
Remark18.
Weshouldclarifyonethinghere.If
T
wasaCalderon-Zygmundoperator,this
estimatewouldbestandard,see[NTV],[NTV2]or,formetricspaces,[HM].We,however,
subtractedfrom
T
twooperators:paraproductandadjointtoparaproduct.However,an
easyargument(see[HPTV])showsthatif
R
\
Q
=
;
,then(
Th
j
Q
;h
i
R
)=(
~
Th
j
Q
;h
i
Q
)(forthe
of
~
T
seeLemma4.3.4andthereon).
Supposenowthat
D
(
Q;R
)
˘


s
`
(
Q
).Weaskthequestion:whatistheprobability
P
(
R
ˆ
Q
(
s
+
s
0
+10)
j
Q;R
2
D
!
)
;
where
s
0
isatlybignumber.WeusetheLemma4.2.7.Supposethat
R
\
Q
(
s
+
s
0
+10)
=
;
:
153
Supposealso
R
=
R
x
(so
x
isthe\center"of
R
).Then
dist
(
x;Q
(
s
+
s
0
+10)
)
6
dist
(
x;Q
)
6
dist
(
Q;R
)
6
C

s
`
(
Q
)=
=
C

s

s
+
s
0
+10
`
(
Q
(
s
+
s
0
+10)
)=
C
s
0
+10
`
(
Q
(
s
+
s
0
+10)
)
:
(4.21)
So
x
2

Q
(
s
+
s
0
+10)
(

s
0
+10
)),andtheprobabilityofthisisestimatedby


(
s
0
+10)
<
1
2
for
tlybig
s
0
(weremindthat

=log

(1

a
)).Therefore,
P
(
R
ˆ
Q
(
s
+
s
0
+10)
j
Q;R
2
D
!
)
>
1
2
:
154
So
j

out
j
6
2
E
X
t;s
X
`
(
Q
)=


t
`
(
R
)
;
D
(
Q;R
)
˘


s
`
(
Q
)
;
R
\
Q
=
;
j
(
Th
j
Q
;h
i
R
)
jj
(
f;h
j
Q
)
jj
(
g;h
i
R
)
j
1
R
isgood
1
R
ˆ
Q
(
s
+
s
0
+10)
6
2
E
X
t;s
X
`
(
Q
)=


t
`
(
R
)
;
D
(
Q;R
)
˘


s
`
(
Q
)
;
R
\
Q
=
;
R;Q
ˆ
Q
s
+
s
0
+10
`
(
Q
)
"
2
`
(
R
)
"
2
D
(
Q;R
)
"
sup
z
2
R

(
z;D
(
Q;R
))

(
Q
)
1
2

(
R
)
1
2

j
(
f;h
j
Q
)
jj
(
g;h
i
R
)
j
1
R
isgood
6
6
2
E
X
t;s
X
`
(
Q
)=


t
`
(
R
)
;
D
(
Q;R
)
˘


s
`
(
Q
)
;
R
\
Q
=
;
;
R;Q
ˆ
Q
s
+
s
0
+10

t"
2

`
(
Q
)
D
(
Q;R
)

"

(
Q
)
1
2

(
R
)
1
2
sup
z
2
R

(
z;D
(
Q;R
))

j
(
f;h
j
Q
)
jj
(
g;h
i
R
)
j
1
R
isgood
6
6
C
2
E
X
t;s

t"
2

s"
X
`
(
Q
)=


t
`
(
R
)
;
D
(
Q;R
)
˘


s
`
(
Q
)
;
R
\
Q
=
;
;
R;Q
ˆ
Q
s
+
s
0
+10

(
Q
)
1
2

(
R
)
1
2
sup
z
2
R

(
z;D
(
Q;R
))
j
(
f;h
j
Q
)
jj
(
g;h
i
R
)
j
1
R
isgood
:
(4.22)
Wenow
S
n
aswedidbefore:
(
S
n
u;v
)=
X
`
(
Q
)=


t
`
(
R
)
;
D
(
Q;R
)
˘


s
`
(
Q
)
;
R
\
Q
=
;
;
R;Q
ˆ
Q
s
+
s
0
+10


(
Q
)
1
2

(
R
)
1
2
sup
z
2
R

(
z;D
(
Q;R
))
(
u;h
j
Q
)(
v;h
i
R
)
1
R
isgood
:
155
Weneedtoestimatethecot.Wewrite

(
z;D
(
Q;R
))
˘

(
z;

s
`
(
Q
))
˘

(
z;

s

s
0

10
`
(
Q
))
˘
˘

(
z;`
(
Q
(
s
+
s
0
+10)
))
˘

(
z;diam
(
Q
(
s
+
s
0
+10)
))
>

(
B
(
z;diam
(
Q
(
s
+
s
0
+20)
)))
>
>

(
Q
(
s
+
s
0
+10)
)
;
(4.23)
andtherefore
j

(
Q
)
1
2

(
R
)
1
2
sup
z
2
R

(
z;D
(
Q;R
))
j
6
C

(
Q
)
1
2

(
R
)
1
2

(
Q
s
+
s
0
+10
)
:
Wenoticethat
C
doesnotdependon
s
sinceweusedthedoublingpropertyof

onlyfor
transmissionfrom


s
`
(
Q
)to


s

s
0

10
`
(
Q
).
Weconcludethat
S
n
isadyadicshiftofcomplexityatmost
C
(
s
+
t
).Therefore,see
Section4.4.2,
j

out
j
6
2
C
E
X
t;s

t"
2

s"
(
s
+
t
)
a
[
w
]
2
k
f
k
w
k
g
k
w

1
6
C
[
w
]
2
k
f
k
w
k
g
k
w

1
;
andourproofiscompleted.
4.4Therestoftheproof
4.4.1ParaproductsandBellmanfunction
NowwewillprovetheLemma4.3.4.
156
Weremindthatthequadraticformofourparaproduct
ˇ
isthefollowing:
(
ˇ
(
f
)
;g
):=
X
R
X
i
h
f
i

(
T˜
X
;h
i
R
)(
g;h
i
R
)
:
Operator
T
isboundedin
L
2
(

)and

isdoubling.Therefore,itiswellknownthat
cots
b
R
:=
b
i
R
:=(
T˜
X
;h
i
R
)satisfyCarlesonconditionforanyofourlatticesof
Christ'sdyadiccubes:
8
Q
2D
X
R
2D
;R
ˆ
Q
j
b
R
j
2

B
(
Q
)
:
(4.24)
Thebestconstant
B
hereiscalledtheCarlesonconstantanditisdenotedby
k
b
k
C
.It
isknownthatforour
b
R
:=(
T˜
X
;h
i
R
)theCarlesonconstantisboundedby
B
T
:=
C
k
T
k
L
2
(

)
!
L
2
(

)
.
IfwewouldbeonthelinewithLebesguemeasure

and
w
wouldbeausualweightin
A
2
,thenthesumwouldfollowtheestimateofO.Beznosova[B]:
j
ˇ
T˜
X
(
f;g
)
j
C
p
B
T
[
w
]
A
2
:
(4.25)
Butthesameistrueinoursituation.Toprovethat,oneshouldanalyzetheproofin
[B]andseethatitusedalwaysconditionson
w
and
b
separately.Theywerealwayssplitby
Cauchy{Schwarzinequality.Theonlyinequality,where
w
and
b
meetwasofthetype:let
Q
beaChrist'scubeofacertainlattice,then
X
R
ˆ
Q;R
2D
h
w
i

b
2
R

[
w
]
A
1
k
b
k
C
Z
Q
w;
(4.26)
157
where
[
w
]
A
1
=sup
1

(
B
)
Z
B
w

exp
0
@

1

(
B
)
Z
B
w
1
A
:
Letusexplainthelastinequality.Wewrite
h
w
i

6
[
w
]
A
1

exp

h
w
i


=[
w
]
A
1

exp

2
h
w
1
2
i


6
[
w
]
A
1
h
w
1
2
i
2

6
[
w
]
A
1
inf
x
2
R
M
(
w
1
2
˜
R
)
2
:
Finally,wenoticethat
f
b
2
R
g
isaCarlesonsequence,andourexplanationwiththe
followingwellknowntheorem.
Theorem4.4.1.
Suppose
f

K
g
isaCarlesonsequence.Thenforanypositivefunction
F
thefollowinginequalityholds:
X
K

K
inf
K
F
(
x
)
6
Z
F
(
x
)

(
x
)
:
Inallotherestimatesin[B]thesumswith
Q
w
(seethebeforeLemma3.2
of[NV])andthesumswith
b
arealwaysestimatedseparately.Thesumswheretheterms
containtheproductof
Q
w
and
b
Q
nevergotestimatedbyBellmantechnique:theygot
splitThen(4.25)followsinourmetricsituationaswell.
4.4.2WeightedestimatesfordyadicshiftsviaBellmanfunction
Thissectionisherejustforthesakeofcompleteness.Infact,itjustrepeatsthearticleof
Nazarov{Volberg[NV].Inthissectionweprovethefollowingtheorem.
158
Theorem4.4.2.
Let
S
m;n
beadyadicshiftofcomplexity
m
+
n
+1
.Then
k
S
m;n
k
w
6
C
(
m
+
n
+1)
a
[
w
]
2

:
Remark19.
Wenoticethatthebestknown
a
isequaltoone.Itcanbegottenusing
thetechniquefrom[HLM+]orfrom[T].However,fortheapplicationwemadeinthe
previoussections,namely,thelinear
A
2
boundforanarbitraryon{Zygmundoperator
ongeometricallydoublingmetricspace,theactualvalueof
a
isnotimportant.
Wedenote
˙
=
w

1
.Webeginwiththefollowingfamouslemma.
Lemma4.4.3.
h
j
I
=

j
I
h
w;j
I
+

j
I
˜
I
;
where
1)
j

j
I
j
q
h
w
i

,
2)
j

j
I
j
j
(
h
w;j
I
;w
)

j
w
(
I
)
,where
w
(
I
):=
R
I
w
,
3)
f
h
w;j
I
g
I
issupportedon
I
,orthogonaltoconstantsin
L
2
(
w
)
,
4)
h
w;j
I
assumesoneachson
s
i
(
I
)
aconstantvalue,
5)
k
h
w;j
I
k
L
2
(
w
)
=1
.

Let

I
w
:=
X
sonsof
I
jh
w
i

(
I
)
h
w
i

j
:
Itisaeasytoseethatthedoublingpropertyofmeasure

implies
j
(
h
w;j
I
;w
)

j
C

I
w
)

(
I
)
1
=
2
:
(4.27)
159
Therefore,theproperty2)abovecanberewrittenas
2')
j

j
I
j
C
j

I
w
j
h
w
i

1

(
I
)
1
=
2
.
Fix
˚
2
L
2
(
w
)
; 
2
L
2
(
˙
).Weneedtoprove
j
(
S
m;n
˚w; ˙
)
j
C
(
n
+
m
+1)
a
k
˚
k
w
k
 
k
˙
:
(4.28)
Remark20.
Innextcalculationswedropthesuperscript
i
and
j
inHaarfunctions
h
i
I
and
h
w;i
I
.Thereadershouldalwaysassumethatwesumupoverall
i
's.
Weestimate(
S
m;n
˚w; ˙
)as
j
X
L
X
I;J
c
L;I;J
(
˚w;h
I
)

(
 ˙;h
J
)

j
X
L
X
I;J
j
c
L;I;J
(
˚w;h
w
I
)

q
h
w
i

(
 ˙;h
˙
J
)

j
q
h
˙
i

j
+
X
L
X
I;J
j
c
L;I;J
h
˚w
i


I
w
h
w
i

(
 ˙;h
˙
J
)

q
h
˙
i

p
I
j
+
X
L
X
I;J
j
c
L;I;J
h
 ˙
i


J
˙
h
˙
i

(
˚w;h
w
I
)

q
h
w
i

p
J
j
+
X
L
X
I;J
j
c
L;I;J
h
˚w
i

h
 ˙
i


I
w
h
w
i


J
˙
h
˙
i

p
I
p
J
j
=:
I
+
II
+
III
+
IV:
Wecannoticethatbecause
j
c
L;I;J
j
p

(
I
)
p

(
J
)

(
L
)
eachsuminside
L
canbeestimated
byaperfectproductof
S
and
R
terms,where
R
L
(
˚w
):=
X
I
ˆ
L:::
h
˚w
i

j

I
w
j
h
w
i


(
I
)
p

(
L
)
160
S
L
(
˚w
):=
X
I
ˆ
L:::
(
˚w;h
w
I
)

q
h
w
i

p

(
I
)
p

(
L
)
andthecorrespondingtermsfor
 ˙
.Sowehave
I

X
L
S
L
(
˚w
)
S
L
(
 ˙
)
;II

X
L
S
L
(
˚w
)
R
L
(
 ˙
)
;
III

X
L
R
L
(
˚w
)
S
L
(
 ˙
)
;IV

X
L
R
L
(
˚w
)
R
L
(
 ˙
)
:
Now
S
L
(
˚w
)

s
X
I
ˆ
L:::
j
(
˚w;h
w
I
)

j
2
q
h
w
i

;S
L
(
 ˙
)

s
X
J
ˆ
L:::
j
(
 ˙;h
˙
J
)
j
2
q
h
˙
i

(4.29)
Therefore,
I
6
C
[
w
]
1
=
2
A
2
k
˚
k
w
k
 
k
˙
:
(4.30)
Terms
II;III
aresymmetric,soconsider
III
.UsingBellmanfunction(
xy
)

onecan
provenow
Lemma4.4.4.
Let
Q
:=[
w
]
A
2
and

2
(0
;
1
=
2)
.
Thesequence
˝
I
:=
h
w
i


h
˙
i



j

I
w
j
2
h
w
i
2

+
j

I
˙
j
2
h
˙
i
2



(
I
)
formaCarlesonmeasurewithCarlesonconstantatmost
c

Q

.
Proof.
Weneedaverysimple
Sublemma
.Let
Q>
1,0
<<
1
2
.Indomain
Q
:=
f
(
x;y
):
X>o;y>
0
;
1
<xy

Q
161
thefunction
B
Q
(
x;y
):=
x

y

thefollowingestimateofitsHessianmatrix:

d
2
B
Q
(
x;y
)


(1

2

)
x

y


(
dx
)
2
x
2
+
(
dy
)
2
y
2

:
Theform

d
2
B
Q
(
x;y
)

0everywherein
x>
0
;y>
0.Alsoobviously0

B
Q
(
x;y
)

Q

in
Q
.
Proof.
Directcalculation.
FixnowaChrist'scube
I
andlet
s
i
(
I
)
;i
=1
;:::;M
,beallitssons.Denote
a
=
(
h
w
i

;
h
˙
i

),
b
i
=(
h
w
i

i
(
I
)
;
h
˙
i

i
(
I
)
),
i
=1
;:::;M
,bepoints{obviously{in
Q
,where
Q
temporarilymeans[
w
]
A
2
.Consider
c
i
(
t
)=
a
(1

t
)+
b
i
t;
0

t

1and
q
i
(
t
):=
B
Q
(
c
i
(
t
)).
WewanttouseTaylor'sformula
q
i
(0)

q
i
(1)=

q
0
i
(0)

Z
1
0
dx
Z
x
0
q
00
i
(
t
)
dt:
(4.31)
Noticetwothings:Sublemmashowsthat

q
00
i
(
t
)

0everywhere.Moreover,itshowsthat
if
t
2
[0
;
1
=
2],thenthefollowingqualitativeestimateholds

q
00
i
(
t
)

c
(
h
w
i

h
˙
i

)


(
h
w
i

i
(
I
)
h
w
i

)
2
h
w
i
2

+
(
h
˙
i

i
(
I
)
h
˙
i

)
2
h
˙
i
2


(4.32)
Thisrequiresasmallexplanation.Ifweareonthesegment[
a;b
i
],thenthecoordinate
ofsuchapointcannotbelargerthan
C
h
w
i

,where
C
dependsonlyondoublingof

(not
w
).Thisisobvious.Thesameistrueforthesecondcoordinatewiththeobviouschangeof
w
to
˙
.Butthereisnosuchtypeofestimatefrombelowonthissegment:thecoordinate
cannotbesmallerthan
k
h
w
i

,but
k
may(andwill)dependonthedoublingof
w
(so
162
ultimatelyonits[
w
]
A
2
norm.Infact,atthe\right"endpointof[
a;b
i
].Thecoordinate
is
h
w
i

i
(
I
)

R
I
w
(
s
i
(
I
))

C
R
I
w
(
I
))=
C
h
w
i

,with
C
onlydependingon
thedoublingof

.Buttheestimatefrombelowwillinvolvethedoublingof
w
,whichwe
mustavoid.Butif
t
2
[0
;
1
=
2],andweareonthe\lefthalf"ofinterval[
a;b
i
]thenobviously
thecoordinateis

1
2
h
w
i

andthesecondcoordinateis

1
2
h
˙
i

.
Wedonotneedtointegrate

q
00
i
(
t
)forall
t
2
[0
;
1]in(4.31).Wecanonlyuseintegration
over[0
;
1
=
2]noticingthat

q
00
i
(
t
)

0otherwise.Thenthechainrule
q
00
i
(
t
)=(
B
Q
(
c
i
(
t
))
00
=(
d
2
B
Q
(
c
i
(
t
)(
b
i

a
)
;b
i

a
)
immediatelygivesus(4.32)withconstant
c
dependingonthedoublingof

but
independent
ofthedoublingof
w
.
Nextstepistoaddall(4.31),withconvexcots

(
s
i
(
I
))

(
I
)
,andtonoticethat
P
M
i
=1

(
s
i
(
I
))

(
I
)
q
0
i
(0)=
r
B
Q
(
a
)
P
M
i
=1

(
a

b
i
)

(
s
i
(
I
))

(
I
)
=0,becauseby
a
=
M
X
i
=1

(
s
i
(
I
))

(
I
)
b
i
:
Noticethattheadditionofall(4.31),withconvexcots

(
s
i
(
I
))

(
I
)
givesusnow(wetake
intoaccount(4.32)andpositivityof

q
00
i
(
t
))
B
Q
(
a
)

M
X
i
=1

(
s
i
(
I
))

(
I
)
B
Q
(
b
i
)

cc
1
(
h
w
i

h
˙
i

)

M
X
i
=1

(
h
w
i

i
(
I
)
h
w
i

)
2
h
w
i
2

+
(
h
˙
i

i
(
I
)
h
˙
i

)
2
h
˙
i
2


:
163
Weusedherethedoublingof

again,bynoticingthat

(
s
i
(
I
))

(
I
)

c
1
(recallthat
s
i
(
I
)and
I
arealmostballsofcomparableradii).Werewritethepreviousinequalityusingour
of
I
w;

I
˙
listedaboveasfollows

(
I
)
B
Q
(
a
)

M
X
i
=1

(
s
i
(
I
))
B
Q
(
b
i
)

cc
1
(
h
w
i

h
˙
i

)



I
w
)
2
h
w
i
2

+

I
˙
)
2
h
˙
i
2



(
I
)
:
Noticethat
B
Q
(
a
)=
h

w
i

h
˙
i



.Nowweiteratetheaboveinequalityandgetforany
ofChrist'sdyadic
I
's:
X
J
ˆ
I;J
2D
(
h
w
i

h
˙
i

)



J
w
)
2
h
w
i
2

+

J
˙
)
2
h
˙
i
2



(
J
)

CQ


(
I
)
:
ThisisexactlytheCarlesonpropertyofthemeasure
f
˝
I
g
indicatedinourLemma4.4.4,
withCarlesonconstant
CQ

.Theproofshowedthat
C
dependedonlyon

2
(0
;
1
=
2)and
onthedoublingconstantofmeasure

.
Now,usingthislemma,westarttoestimateour
S
L
'sand
R
L
's.For
S
L
(
 ˙
)wealready
hadestimate(4.29).
Toestimate
R
L
(
˚w
)letusdenoteby
P
L
maximalstoppingintervals
K
2D
;K
ˆ
L
,
wherethestoppingcriteriaare1)either
j

K
w
j
h
w
i


1
m
+
n
+1
,or
j

K
˙
j
h
˙
i


1
m
+
n
+1
,or2)
g
(
K
)=
g
(
L
)+
m
.
164
Lemma4.4.5.
If
K
isanystoppingintervalthen
X
I
ˆ
K;`
(
I
)=2

m
`
(
L
)
jh
˚w
i

j
j

I
w
j
h
w
i


(
I
)
p

(
L
)

2
e

(
m
+
n
+1)
hj
˚
j
w
i

p

(
K
)
p

(
L
)
p
˝
K
h
w
i

=
2

h
˙
i

=
2

:
(4.33)
Proof.
Ifwestopbythecriterion,then
X
I
ˆ
K;`
(
I
)=2

m
`
(
L
)
jh
˚w
i

j
j

I
w
j
h
w
i


(
I
)
p

(
L
)

2
X
I
ˆ
K;`
(
I
)=2

m
`
(
L
)
jh
˚w
i

j

(
I
)
1

(
K
)

(
K
)
p

(
L
)


2
hj
˚
j
w
i


(
K
)
p

(
L
)

2(
m
+
n
+1)
hj
˚
j
w
i


j

K
w
j
h
w
i

+
j

K
˙
j
h
˙
i



(
K
)
p

(
L
)


2(
m
+
n
+1)
hj
˚
j
w
i

p

(
K
)
p

(
L
)
p
˝
K
h
w
i

=
2

h
˙
i

=
2

:
Nowreplacing
h
w
i

=
2

h
˙
i

=
2

by
h
w
i

=
2

h
˙
i

=
2

doesnotgrowtheestimatebymore
than
e

asallpairsofson/fatherintervalslargerthan
K
andsmallerthan
L
willhavethere
averagescomparedbyconstantatmost1

1
m
+
n
+1
.Andthereareatmost
m
suchintervals
between
K
and
L
.
Ifwestopbythesecondcriterion,then
K
isoneof
I
's,
g
(
I
)=
g
(
L
)+
m
,and
jh
˚w
i

j
j

I
w
j
h
w
i


(
I
)
p

(
L
)
jh
˚w
i

j

(
K
)
p

(
L
)
j

K
w
j
h
w
i


hj
˚
j
w
i

p

(
K
)
p

(
L
)
p
˝
K
h
w
i

=
2

h
˙
i

=
2

:
Nowwereplace
h
w
i

=
2

h
˙
i

=
2

by
h
w
i

=
2

h
˙
i

=
2

asbefore.
165
Now
R
L
(
˚w
)

C
(
m
+
n
+1)
h
w
i

=
2

h
˙
i

=
2

X
K
2P
L
hj
˚
j
w
i

p

(
K
)
p

(
L
)
p
˝
K

C
(
m
+
n
+1)
h
w
i

=
2

h
˙
i

=
2


X
K
2P
L
hj
˚
j
w
i
2


(
K
)

(
L
)

1
=
2
(
e
˝
L
)
1
=
2
;
where
e
˝
L
=
X
K
2P
L
˝
K
:
Noticethatthesequence
f
e
˝
L
g
L
2D
formaCarlesonsequence(measure)withconstantat
most
C
(
m
+1)
Q

.
Nowwemakeatrick!Wewillestimatetherighthandsideas
R
L
(
˚w
)

C
(
m
+
n
+1)
h
w
i

=
2

h
˙
i

=
2


X
K
2P
L
hj
˚
j
w
i
p


(
K
)

(
L
)

1
=p
(
e
˝
L
)
1
=
2
;
where
p
=2

1
m
+
n
+1
.Infact,

X
K
ˆ
L;Kismaximal
hj
˚
j
w
i
2


(
K
)

(
L
)

p=
2

X
K
2P
L
hj
˚
j
w
i
p



(
K
)

(
L
)

p=
2
:
Butifif0

j

m
,then(
C

j
)

1
m
+
n
+1

C
,andthereforeintheformulaabove


(
K
)

(
L
)

1

1
2(
m
+
n
+1)

C

(
K
)

(
L
)
,and
C
dependsonlyonthedoublingconstantof

.So
166
thetrickisTherefore,usingCauchyinequality,onegets
R
L
(
˚w
)

C
(
m
+
n
+1)
h
w
i

=
2

h
˙
i

=
2


X
K
2P
L
hj
˚
j
p
w
i

h
w
i
p

1


(
K
)

(
L
)

1
=p
(
e
˝
L
)
1
=
2
:
Wecanreplaceall
h
w
i
p

1

by
h
w
i
p

1

payingthepricebyconstant.Thisisagainbecauseall
intervalslargerthan
K
andsmallerthan
L
willhavethereaveragescomparedbyconstant
atmost1

1
m
+
n
+1
.Andthereareatmost
m
suchintervalsbetween
K
and
L
.Finally,
R
L
(
˚w
)

C
(
m
+
n
+1)
h
w
i

=
2

h
˙
i

=
2


X
K
2P
L
hj
˚
j
p
w
i


(
K
)

(
L
)

1
=p
h
w
i
1

1
p

(
e
˝
L
)
1
=
2
(4.34)
Weneedthestandardnotations:if

isanarbitrarypositivemeasurewedenote
M

f
(
x
):=sup
r>
0
1

(
B
(
x;r
))
Z
B
(
x;r
)
j
f
(
x
)
j

(
x
)
:
Inparticular
M
w
willstandforthismaximalfunctionwith

=
w
(
x
)

.
From(4.34)weget
R
L
(
˚w
)

C
(
m
+
n
+1)
h
w
i
1

=
2

h
˙
i

=
2

inf
L
M
w
(
j
˚
j
p
)
1
=p
(
e
˝
L
)
1
=
2
(4.35)
Now
S
L
(
 ˙
)
R
L
(
˚w
)

C
(
m
+
n
+1)
h
w
i
1

=
2

h
˙
i
1

=
2

inf
L
M
w
(
j
˚
j
p
)
1
=p
h
˙
i
1
=
2


(
e
˝
L
)
1
=
2
s
X
J
ˆ
L:::
j
(
 ˙;h
˙
J
)
j
2
;
(4.36)
167
R
L
(
 ˙
)
R
L
(
˚w
)

C
(
m
+
n
+1)
h
w
i
1



h
˙
i
1



inf
L
M
w
(
j
˚
j
p
)
1
=p
inf
L
M
˙
(
j
 
j
p
)
1
=p
e
˝
L
:
(4.37)
NowweusetheCarlesonpropertyof
f
e
˝
L
g
L
2D
.WeneedasimplefolkloreLemma.
Lemma4.4.6.
Let
f

L
g
L
2D
Carlesonmeasurewithintensity
B
relatedtodyadic
lattice
D
onmetricspace
X
.Let
F
beapositivefunctionon
X
.Then
X
L
(inf
L
F
)

L

2
B
Z
X
F:
(4.38)
X
L
inf
L
F
h
˙
i


L

CB
Z
X
F
˙
:
(4.39)
Nowuse(4.36).Thentheestimateof
III

P
L
S
L
(
 ˙
)
R
L
(
˚w
)willbereducedto
estimating
(
m
+
n
+1)
Q
1

=
2

X
L
inf
L
M
w
(
j
˚
j
p
)
2
=p
h
˙
i

e
˝
L

1
=
2

(
m
+
n
+1)
2
Q

Z
R
(
M
w
(
j
˚
j
p
))
2
=p
w

1
=
2

(
1
2

p
)
1
=p
(
m
+
n
+1)
2
Q

Z
R
˚
2
w

1
=
2

(
m
+
n
+1)
3
Q

Z
R
˚
2
w

1
=
2
:
Hereweused(4.39)andtheusualestimatesofmaximalfunction
M

in
L
q
(

)when
q
ˇ
1.
Ofcoursefor
II
weusethesymmetricreasoning.
Now
IV
:weuse(4.37)
X
L
R
L
(
 ˙
)
R
L
(
˚w
)

(
m
+
n
+1)
Q
1


X
L
inf
L
M
w
(
j
˚
j
p
)
1
=p
inf
L
M
˙
(
j
 
j
p
)
1
=p
e
˝
L

C
(
m
+
n
+1)
2
Q
Z
R
(
M
w
(
j
˚
j
p
))
1
=p
(
M
˙
(
j
 
j
p
))
1
=p
w
1
=
2
˙
1
=
2

168

C
(
m
+
n
+1)
2
Q

Z
R
(
M
w
(
j
˚
j
p
))
2
=p
w

1
=
2

Z
R
(
M
˙
(
j
 
j
p
))
2
=p
˙

1
=
2

C
(
m
+
n
+1)
4
Q

Z
R
˚
2
w

1
=
2

Z
R
 
2
˙

1
=
2
:
Hereweused(4.38)andtheusualestimatesofmaximalfunction
M

in
L
2
=p
(

)when
p
ˇ
2
;p<
2.
169
BIBLIOGRAPHY
170
BIBLIOGRAPHY
[R1]Reznikov,Alexander
SharpconstantsinthePaneyah-Logvinenko-Sereda
theorem
C.R.Math.Acad.Sci.Paris348(2010),no.3-4,141{144.
[NR1]Nazarov,A.I.;Reznikov,A.B.
Ontheexistenceofanextremalfunction
incriticalSobolevtraceembeddingtheorem
J.Funct.Anal.258(2010),
no.11,3906{3921.
[NR2]NazarovA.,ReznikovA.,
AttainabilityofinthecriticalSobolev
traceembeddingtheoremonmanifolds
|AmericanMathematicalSoci-
etyTranslations{Series2AdvancesintheMathematicalSciences2010;252pp;
hardcoverVolume:229.
[R2]Reznikov,Alexander
Sharpweaktypeestimatesforweightsintheclass
A
p
1
;p
2
,Rev.Mat.Iberoam.
29
(2013),no.2,433{478;doi10.4171/rmi/726;
arXiv:1105.4848v1.
[RVV]ReznikovA.,VasyuninV.,VolbergA.
Anobservation:oftheweight
w
doesnotincreasethe
A
p
1
;p
2
-\norm"of
w
arXiv:1008.3635
[BR1]BeznosovaO.,ReznikovA.
EquivalentofdyadicMucken-
houptandReverseolderclassesintermsofCarlesonsequences,
weakclasses,andcomparabilityofdyadic
L
log
L
and
A
1
constants
,
arXiv:1201.0520;acceptedtoRevistaMatematicaIberoamericana.
[BR2]BeznosovaO.,ReznikovA.
Sharpestimatesinvolving
A
1
and
L
log
L
con-
stants,andtheirapplicationstoPDE
,arXiv:1107.1885;acceptedtotheSt.
PetersburgMathJournal.
[RTV]ReznikovA.,TreilS.,VolbergA.
Asharpestimateofweighteddyadicshifts
ofcomplexity0and1
,arXiv:1104.5347.
[RV2]ReznikovA.,VolbergA.
Random\dyadic"latticeingeometricallydou-
blingmetricspaceand
A
2
conjecture
,arXiv:1103.5246;
171
[NRV]NazarovF.,ReznikovA.,VolbergA.
Theproofof
A
2
conjectureinageo-
metricallydoublingmetricspace
,arXiv:1106.1342;acceptedtotheIndiana
UniversityMathJournal.
[NRVV]NazarovF.,ReznikovA.,VasyuninV.,VolbergV.
A
1
conjectureandBellmanfunctions
,2010.Availableat
http://sashavolbordpress.com/2010/11/a11
7loghilb11
21
2010.pdf
[CURV]Cruz-UribeD.,ReznikovA.,VolbergA.
Logarithmicbumpconditions
andthetwoweightboundednessofCaldern-Zygmundoperators
,
arXiv:1112.0676;submittedtotheAdvancesinMathematics.
[NRTV1]Nazarov,F.,Reznikov,A.,Treil,S.,Volberg,A.
ABellmanfunctionproofof
the
L
2
bumpconjecture
,arXiv:1202.2406;acceptedtotheJournald'Analyse
Mathematique.
[NRTV2]Nazarov,F.,Reznikov,A.,Treil,S.,Volberg,A.
Carleson{Buckleymeasures
beyondthescopeof
A
1
andtheirapplications
,arXiv:1202.2931.
[NRV1]NazarovF.,ReznikovA.,VolbergA.,
Bellmanapproachtotheone-sided
bumpingforweightedestimatesofon{Zygmundoperators
,
arXiv:1306.2653.
[RV]Reznikov,A.,Volberg,A.
Cauchyindependentmeasuresandsuper-
additivityofanalyticcapacity
,arXiv:1211.2675;submittedtotheMathRe-
searchLetters.
[ACUM]AndersonT.,Cruz-UribeD.,MoenK.,
Logarithmicbumpconditions
foron-ZygmundOperatorsonspacesofhomogeneoustype
,
arXiv:1308.2026.
[B]
O.Beznosova
,LinearboundforthedyadicparaproductonweightedLebesgue
spaceL2(w).J.Funct.Anal.255(2008),no.4,994{1007
[BMP]BeznosovaO.,MoraesJ.C.,PereyraM.C.,
SharpboundsforT-Haarmulti-
plierson
L
2
.
ToappearinProceedingsElEscorial2012
[Chr]ChristM.,
A
T
(
b
)
theoremwithremarksonanalyticcapacityandthe
Cauchyintegral
,Colloq.Math.
60/61
(1990),no.2,601-628;
172
[CUMP]Cruz-UribeD.,MartellJ.M.,PerezC.,
Sharpweightedestimatesforclas-
sicaloperators
,Adv.inMath.,229(2012),408{441.
[1]
D.Cruz-Uribe,J.M.Martell,C.P

erez,
Weights,Extrapolationandthe
TheoryofRubiodeFrancia
,OperatorTheory:AdvancesandApplications,215,
Birkhauser,Basel,(2011).
[CWW]ChangA.,WilsonM.,WT.,
Someweightednorminequalitiescon-
cerningtheScodingeroperators
,Comment.Math.Helvetici,60(1985),
217{286.
[CUMP1]Cruz-UribeD.,MartellJ.M.,PerezC.,
Weights,Extrapolationandthe
TheoryofRubiodeFrancia
,OperatorTheory:AdvancesandApplications,
215,Birkhauser,Basel,(2011)
[F]FermanC.,
Theuncertaintyprinciple
,Bull.Amer.Math.Soc.,9(1983),
129{206
[HavJ]Havin,V.,oricke,B.
Theuncertaintyprincipleinharmonicanalysis.
ErgebnissederMathematikundihrerGrenzgebiete(3)[ResultsinMathematics
andRelatedAreas(3)],28.Springer-Verlag,Berlin,1994.xii+543pp.ISBN:
3-540-56991-X
[H]onenT.,
Thesharpweightedboundforgeneralon-Zygmund
operators
,arXiv:1007.4330v1;
[HyLa]onenT.,LaceyM.,
The
A
p
-
A
1
inequalityforgeneralon{
Zygmundoperators
,arXiv:1106.4797;
[HLM+]onen,T.,Lacey,M.,Martikainen,H.,Orponen,T.,Reguera,M.,Sawyer,E.,
Uriarte-Tuero,I.,
Weakandstrongtypeestimatesformaximaltrunca-
tionsofon-Zygmundoperatorson
A
p
weightedspaces.
J.Anal.
Math.118(2012),no.1,177{220;
[HM]onenT.,MartikainenH.,
Non-homogeneous
Tb
theoremandrandom
dyadiccubesonmetricmeasurespaces
,arXiv:0911.4387;
[HP]onen,T.Perez,C.
Sharpweightedboundsinvolving
A
1
.Anal.PDE6
(2013),no.4,777{818.
[HPTV]onenT.,PerezC.,TreilS.,VolbergA.,
Sharpweightedestimatesfor
dyadicshiftsandthe
A
2
conjecture
,arXiv:1010.0755v2;
173
[KV]KonyaginS.,VolbergA.,
Onmeasureswiththedoublingcondition.
(Rus-
sian)Izv.Akad.NaukSSSRSer.Mat.
51
(1987),no.3,666{675;translationin
Math.USSR-Izv.
30
(1988),no.3,629{638;
[Lac1]Lacey,M.
TheTwoWeightInequalityfortheHilbertTransform:A
Primer
,arXiv:1304.5004;
[Lac2]Lacey,M.
TwoWeightInequalityfortheHilbertTransform:AReal
VariableCharacterization,II
,arXiv:1301.4663;
[Lac3]Lacey,M.
Onthe
A
2
inequalityforon{ZygmundOperators
,
arXiv:1106.4802;
[LPR]LaceyM.,PetermichlS.,RegueraM.,
Sharp
A
2
inequalityforHaarshift
operators
,arXiv:0906.1941;
[LSUT]Lacey,M.,Sawyer,S.,Uriarte-Tuero,
I.ATwoWeightInequalityforthe
HilberttransformAssuminganEnergyHypothesis
,arXiv:1001.4043;
[LSSUT]Lacey,M.,Sawyer,S.,Shen,C.Y.,Uriarte-Tuero,I.
TwoWeightInequal-
ityfortheHilbertTransform:ARealVariableCharacterization,I
,
arXiv:1201.4319;
[L]LernerA.,
AsimpleproofofA2conjecture
,Intern.Math.Res.Notices
IMRN,2012
[L1]LernerA.,
OnanestimateofCaldern-Zygmundoperatorsbydyadic
positiveoperators
,arXiv:1202.1860.
[M]
H.Martikainen
,Vector-valuednon-homogeneous
Tb
theoremonmetricmea-
surespaces,arXiv:1004.3176;
[MP]
J.MoraesandM.C.Pereyra
,
Weightedestimatesfordyadicparaproducts
and
t
-Haarmultiplierswithcomplexity
(
m;n
),ArXiv:1108.3109.
[NV]Nazarov,F.,Volberg,A.
TheBellmanfunction,thetwo-weightHilbert
transform,andembeddingsofthemodelspaces
K

.Dedicatedtothe
memoryofThomasH.WJ.Anal.Math.87(2002),385414.
[NV1]NazarovF.,VolbergA.,
Asimplesharpweightedestimateofthedyadic
shiftsonmetricspacewithgeometricdoubling
,arxiv1104.4893;
174
[NV2]NazarovF.,VolbergA.,
Onanalyticcapacityofportionsofcontinuum
andaquestionofT.Murai
,Proc.Sympos.PureMath.,79,Amer.Math.
Soc.,Providence,RI,2008;
[NTV]NazarovF.,TreilS.,VolbergA.,
The
Tb
-theoremonnon-homogeneous
spaces
,ActaMath.,190(2003),151-239;
[NTV2]Nazarov,F.,Treil,S.,Volberg,A.
TheBellmanfunctionsandtwo-weight
inequalitiesforHaarmultipliers
,J.Amer.Math.Soc.12(1999),no.4,909{
928
[NTV3]NazarovF.,TreilS.,VolbergA.,
Twoweightinequalitiesforindividual
Haarmultipliersandotherwelllocalizedoperators
,Math.Res.Lett.
15
(2008),no.4,583-597;
[NTV4]NazarovF.,TreilS.,VolbergA.,
TwoweightestimatefortheHilberttrans-
formandcoronadecompositionfornon-doublingmeasures
,Preprint
2005,arXiv:1003.1596;
[Pan]Panejah,B.
SometheoremsofPaley-Wienertype.
Dokl.Akad.NaukSSSR
13847{50(Russian);translatedasSovietMath.Dokl.21961533536.
[P]
C.P

erez,
Weightednorminequalitiesforsingularintegraloperators
.J.London
Math.Soc.(2)
49
(1994),no.2,296{308
[Pe]
C.P

erez
OnconditionsfortheboundednessoftheHardy-Littlewood
maximaloperatorbetweenweighted
L
p
-spaceswithentweights.
Proc.Lon-
donMath.Soc.
71
(1995),No.3,135{157.
[PTV]PerezC.,TreilS.,VolbergA.,
On
A
2
conjectureandcoronadecomposition
ofweights
,arXiv:1006.2630;
[Pet1]Petermichl,S.
ThesharpboundfortheHilberttransformonweighted
Lebesguespacesintermsoftheclassical
A
p
characteristic.
Amer.J.
Math.129(2007),no.5,1355{1375.
[Pet2]Petermichl,S.
ThesharpweightedboundfortheRiesztransforms
.Proc.
Amer.Math.Soc.136(2008),no.4,1237{1249.
[RegS]Reguera,M.,Scurry,J.,
Onjointestimatesformaximalfunctionsand
singularintegralsonweightedspaces
.Proc.Amer.Math.Soc.141(2013),
no.5,1705{1717.
175
[RegTh]Reguera,M.,Thiele,C.
TheHilberttransformdoesnotmap
L
1
(
Mw
)
to
L
1
;
1
(
w
)
.
Math.Res.Lett.19(2012),no.1,1{7;
[SlVa]SlavinL.,VasyuninV.,
Sharpresultsintheintegral-formJohn-Nirenberg
inequality.
Trans.Amer.Math.Soc.363(2011),no.8,4135{4169
[St]SteinE.,
Noteontheclass
L
log
L
.
StudiaMath.,31:305-310,1969.
[Tol]TolsaX.,
Analyticcapacity,theCauchytransform,andnon-
homogeneouson-Zygmundtheory.
Toappear(2012).
[T]TreilS.,
Sharp
A
2
estimatesofHaarshiftsviaBellmanfunction
,
arXiv:1105.2252v1;
[Va1]VasyuninV.,
Theexactconstantintheinverseolderinequalityfor
Muckenhouptweights
.St.PetersburgMathematicalJournal,15:49{79,2004.
[Va2]VasyuninV.,
Mutualestimatesof
L
p
-normsandtheBellmanfunc-
tion
Journalofmathematicalscience,Volume156,Number5,766{798,DOI:
10.1007/s10958-009-9288-3;
[VaVo]
V.Vasyunin,A.Volberg
,
TheBellmanfunctionforthesimplesttwo-weight
inequality:aninvestigationofaparticularcase
,St.PetersburgMath.J.18
(2007),no.2,201{222.
[V]VolbergV.,
on-Zygmundcapacitiesandoperatorsonnonhomo-
geneousspaces.
CBMSRegionalConf.Ser.inMath.100,Amer.Math.Soc.,
Providence,2003.
[Wil]WilsonM.,
WeightedLittlewood-PaleyTheoryandExponential-Square
Integrability
LectureNotesinMathematics,1924.Springer,Berlin,2008.
xiv+224pp.ISBN:978-3-540-74582-2
[W]Wittwer,J.
Asharpestimateonthenormofthemartingaletransform
.
Math.Res.Lett.7(2000),no.1,1{12.
176