Calderon-Zygmund theory

Updated May 23, 2020

Plan 2

Outline:Statement and motivationProof via Marcinkiewicz and dualityApplications to Hilbert and inverse-Fourier transformGeneral Calderon-Zygmund kernels

Motivations 3

Q: What happens with Riesz transform

Tαf pxq :“ż

1|x´ y|α

f pyqdy

when α “ d? Or with Hilbert transform?A: Integral not defined even for nice f due to singularity atx “ y, but could truncate to |x´ y| ě ε, perhaps.Singularity as |x| Ñ 8 bad too; kernel

Kpxq :“1|x|d

1|x|ěε

obeys K P L1,w, but Schur’s test requires Lr,w with r ą 1

Calderon-Zygmund theorem 4

Theorem (Calderon-Zygmund)

Consider the measure space pRd,LpRdq, λq for d ě 1. For allA, B ą 0, all M ą 1 and all p P p1,8q there is Cp P p0,8q such thatfor all measurable kernels K : Rd Ñ R satisfying

K P L2 with Fourier transform pK obeying }pK}8 ď A

andsup

zPRdrt0u

ż

|x|ąM|z|

ˇ

ˇKpx´ zq ´Kpxqˇ

ˇdx ď B,

the convolution operator TKf :“ K ‹ f is well defined by the integralexpression for all f P L1 and extends continuously to a map Lp Ñ Lp

for each p P p1,8q with

@p P p1,8q : }TK}LpÑLp ď Cp

Remarks 5

1st condition ensures K is locally integrable and K ‹ fmeaningful for f P L1 (by Young convolution inequality)

2nd condition often stated as K P C1pRd r t0uqwith

@x P Rd r t0u :ˇ

ˇ∇Kpxqˇ

ˇ ďB

|x|d`1

which (along with 1st condition) gives

|Kpxq| ďB{d|x|d

and so we cannot hope for more than K P L1,w (and so Schur’stest is still out).

Upshot: Trading local regularity against integrability

Strategy of proof 6

1st condition implies TK is strong type p2, 2qwith 2nd condition this implies TK is weak type p1, 1q. Thisis the key novelty; requires so called Calderon-Zygmunddecomposition of Rd into sets where f is bounded and setswhere f has bounded integralMarcinkiewicz interpolation gives

TK is strong type pp, pq for all p P p1, 2sDuality: true also for p P r2,8q

Improved Young convolution inequality 7

Recall: By Young convolution inequality f P L2 and g P L1

implies integral f ‹ g converges absolutely a.e. and

}f ‹ g}2 ď }f }2}g}1

Need a slight improvement:

Lemma

@f P L1 @g P L2 : }f ‹ g}2 ď }f }2 }pg}8

Proof: Let f P L1 and g P L2. Fourier transform isometry so

}yf ‹ g}2 ď }f }1}g}2

Hence g ÞÑ yf ‹ g continuous. If g P L1, then yf ‹ g “ pfpg so truefor g P L2 as well. Hence

}f ‹ g}2 “ }yf ‹ g}2 “ }pfpg}2 ď }pf }2}pg}8 “ }f }2}pg}8

Strong type p2, 2q 8

Corollary

The operator TK is strong type p2, 2q with }TK}L2ÑL2 ď A

Proof: For f P L1, TKf well defined via K ‹ f and obeys

}TKf }2 ď A}f }2

by above lemma. So TK extends to L2 with }TK}L2ÑL2 ď A

Calderon-Zygmund decomposition 9

Dyadic cube is any cube of the form

2nx` r0, 2nqd

for x P Zd and n P Z.

Lemma (Calderon-Zygmund decomposition)

Let f P L1 and t ą 0. Then there exist disjoint dyadic cubes tQiuiPIsuch that

@i P I : tλpQiq ă

ż

Qi

|f |dλ ď 2dtλpQiq

and|f | ď t λ-a.e. on Rd r

ď

iPI

Qi

Proof of Calderon-Zygmund decomposition 10

Call a dyadic cube Q good if

1λpQq

ż

Q|f |dλ ď t

and call it bad otherwise. For n P Z such thatş

|f |dλ ď t2n, alldyadic cubes of side-length 2n good. Let tQiuiPI enumerate theset of all bad dyadic cubes Q such that the (unique) dyadiccube Q1 containing Q and having side length twice as that of Qis good. Then

tλpQq ăż

Q|f |dλ ď

ż

Q1|f |dλ ď tλpQ1q “ 2dtλpQq

because Q is bad and Q1 is good.If x lies only in good cubes, Lebesgue differentiation shows|f pxq| ď t a.e. (Need a version for dyadic cubes; proved whendiscussed martingale convergence.)

Weak type p1, 1q 11

Proposition

TK is weak type p1, 1q. Explicitly,

Dc P p0,8q @f P L1 @t ą 0 : λ`

|TKf | ą t˘

ďct}f }1

where c depends only d and the constants A and B

Decomposition of f 12

Pick f P L1 and t ą 0 and let tQiuiPI be as above. Set

F :“ Rd rď

iPI

Qi

define g : Rd Ñ R by

gpxq :“

#

1λpQiq

ş

Qif dλ if x P Qi for some i P I

f pxq if x P F

and abbreviatehpxq :“ f pxq ´ gpxq

Note thath “ 0 on F ^ @i P I :

ż

Qi

hdλ “ 0

Union bound + additivity:

λ`

|TKf | ą t˘

ď λ`

|TKg| ą t{2˘

` λ`

|TKh| ą t{2˘

Now estimate each term separately . . .

Tails of TKg 13

Will use that TK maps L2 Ñ L2 with norm ď A (proved above).Need to estimate

}g}22 “ż

Fg2dλ`

ÿ

iPI

ż

Qi

g2dλ

ď

ż

Ft|f |dλ`

ÿ

iPI

´ 1λpQiq

ż

Qi

f dλ¯2

λpQiq

ď

ż

Ft|f |dλ`

ÿ

iPI

p2dtq2λpQiq

ď tż

F|f |dλ` 4dt

ÿ

iPI

ż

Qi

|f |dλ

“ tż

F|f |dλ` 4dt

ż

Fc|f |dλ “ p4d ` 1qt}f }1

Hence

λ`

|TKg| ą t{2˘

ď4t2 }TKg}22 ď

4A2

t2 }g}22 ď

4A2p4d ` 1qt

}f }1

Tails of TKh 14

Consider hi :“ h1Qi . Let yi :“ the center of Qi. Asş

Qihdλ “ 0,

TKhipxq “ż

Qi

Kpx´ yqhipyqdy “ż

Qi

`

Kpx´ yq ´Kpx´ yiq˘

hipyqdy

Let Q1i :“ the cube of M?

d-times the side length of Qi centeredat yi. By Tonelli and 2nd condition:ż

RdrQ1i

|TKhi|dλ

ď

ż

Qi

´

ż

RdrQ1i

ˇ

ˇKpx´ yi ` y´ yiq ´Kpx´ yiqˇ

ˇdx¯

ˇ

ˇhipyqˇ

ˇdy

ď Bż

Qi

|h|dλ ď 2Bż

Qi

|f |dλ

which uses |x´ yi| ą M|y´ yi| for all x R Q1i and all y P Qi. Then. . .

Tails of TKh continued . . . 15

. . . abbreviating F1 :“ Rd rŤ

iě1 Q1i we thus getż

F1

ˇ

ˇTKh|dλ ď 2B}f }1

On the other hand,

λpRd r F1q ďÿ

iPI

λpQ1iq “ pM?

dqdÿ

iPI

λpQiq

ďpM?

dqd

t

ÿ

iPI

ż

Qi

|f |dλ ďpM?

dqd

t}f }1

and so

λ`

|TKh| ą t{2˘

ď λpRd r F1q`2t

ż

F1

ˇ

ˇTKh|dλ ďpM?

dqd ` 4Bt

}f }1

So claim holds with

c :“ 4A2p4d ` 1q ` pM?

dqd ` 4B

Proof of Calderon-Zygmund theorem 16

Marcinkiwicz: TK strong type pp, pq for p P p1, 2s.Now let q P p2,8q and let p be such that p´1 ` q´1 “ 1. Thenduality between Lp and Lq gives

@f P L1 X Lp @g P Lq :ˇ

ˇ

ˇ

ż

gpK ‹ f qdλˇ

ˇ

ˇď }TK}LpÑLp}f }p}g}q

For f P L1 integral K ‹ f converges absolutely. So byFubini-Tonelli:

@f P L1 X Lp @g P Lq X L1 :ˇ

ˇ

ˇ

ż

pTKgqf dλˇ

ˇ

ˇď }TK}LpÑLp}f }p}g}q

Density of Lp X L1 in Lp gives

@g P Lq X L1 : }TKg}q ď }TK}LpÑLp}g}q.

which implies that TK extends continuously to a map Lq Ñ Lq

with}T}LqÑLq ď }TK}LpÑLp

(Equality holds by duality.)

Application to Hilbert transform 17

Recall: Hf defined as the ε Ó 0 limit of convolution-typeoperator Hεf :“ Kε ‹ f where

Kεpxq :“1

πx1pε,1{εqp|x|q

Convergence pointwise for f P C1pRq X L1 and in L2 for f P L2

Theorem (Hilbert transform in Lp)

We have@p P p1,8q : sup

0ăεă1}Hε}LpÑLp ă 8.

In particular, for all p P p1,8q, there exists a continuous linearoperator H : Lp Ñ Lp such that

@f P Lp : Hεf ÝÑεÓ0

Hf in Lp.

Proof of Theorem 18

For strong type p2, 2q, use Fourier calculation to get

pKεpzq “ ´2iπ

ż

εătă1{ε

sinp2πztqt

dt

Hence, A :“ sup0ăεă1 }pKε}8 ă 8.

For weak type p1, 1q, computeˇ

ˇKεpx´ zq ´Kεpxqˇ

ˇ

ď

ˇ

ˇ

ˇ

1x´ z

´1x

ˇ

ˇ

ˇ`

1|x|

ˇ

ˇ1pε,1{εqp|x´ z|q ´ 1pε,1{εqp|x|qˇ

ˇ

ď2|z||x|2

`1|x|

1tp1{ε´|z|,1{ε`|z|qp|x|q `1|x|

1tpε´|z|,ε`|z|qp|x|q

Need to integrate this over |x| ą 2|z|. First term easy. For theother two terms we note that . . .

Proof of Theorem continued . . . 19

. . . for any a, b ą 0 with maxt2b, a´ bu ă a` b,

ż a`b

maxt2b,a´bu

dxx“ log

´ a` bmaxt2b, a´ bu

¯

Examining a´ b ă 2b and a´ b ą 2b separately, RHS ď logp2q.Now use this with a :“ ε, 1{ε and b “ |z| to get

B :“ sup0ăεă1

supzPRrt0u

ż

|x|ą2|z|

ˇ

ˇKεpx´ zq ´Kεpxqˇ

ˇdx ă 8

So tHεu0ăεă1 obey conditions of Calderon-Zygmund theoremwith uniform A and B (and M :“ 2). So we get

sup0ăεă1

}Hε}LpÑLp ă 8.

It remains to address convergence Hεf Ñ Hf . . .

Proof of Theorem continued . . . 20

. . . which we already know in L2 by Fourier techniques. We willuse interpolation for Lp-norms.

Given p P p1,8q, choose p P p1, pqwhen p ă 2 or p P pp,8qwhen p ą 2. Then 1

p “ p1´ θq 1p ` θ 1

2 for some θ P p0, 1q and so

@f P Lp X L2 : }Hεf ´Hδf }p ď }Hεf ´Hδf }θ2 }Hεf ´Hδf }1´θ

p .

Now }Hεf ´Hδf }2 Ñ 0 as ε, δ Ó 0 by the claim in L2 while

}Hεf ´Hδf }p ď´

2 sup0ăε1ă1

}Hε}LpÑLp

¯

}f }p.

Completeness of Lp shows Hεf Ñ Hf for each f P Lp X L2. AsLp X L2 dense in Lp, true for all f P Lp.

Uniform convergence? 21

Q: Is Hεf Ñ Hf uniform in f P Lp with }f }p ď 1?

A: Not in L2 (and by duality in interpolation, not in Lp) because

}Hε ´H}L2ÑL2 “ }pKε ´ pK}8

and RHS does not tend to zero because Kε is continuous andpKpzq :“ p´iqsgnpzq is not.

Strong vs norm convergence 22

Definition (Strong and norm convergence of operators)

A sequence tTnuně1 of linear operators on a normed linearspace V is said to converge strongly to a linear operator T if

@f P V : limnÑ8

}Tnf ´ Tf } “ 0

The sequence tTnuně1 converges to T in (operator) norm if

limnÑ8

}Tn ´ T} “ 0

So, on Lp with p P p1,8q, we get Hε Ñ H strongly but not inoperator norm.

Cotlar’s approach 23

Cotlar’s identity:

pHf q2 “ f 2 `H`

f pHf q˘

By induction: For n ě 1 and p :“ 2n,

}Hf }2p2p ď p}f }2p

2p ` p›

›Hpf pHf qq}pp

ď p}f }2p2p ` pp}H}LpÑLpqp}f }p2p}Hf }p2p

Proves }H}LpÑLp ă 8 for p P t2n : n ě 1u. Interpolation +duality gives this for all p P p1,8q.

Partial Fourier inversions 24

For f P L1, define

Tnf pxq :“ż n

´n

pf pkqe´2πik¨xdk,

where pf :“ Fourier transform of f . Know that, if pf P L1, thenTnf Ñ f pointwise. Q: Convergence in Lp?

TheoremLet p P p1,8q. Then, for each n ě 1, the operator Tn extendscontinuously to a map Lp Ñ Lp and

@f P Lp : Tnf ÝÑnÑ8

f in Lp

Proof: homework

A.e. convergence 25

Lp convergence gives a.e. convergence along a subsequence.

Need for subsequences removed by L. Carleson (1966) for L2

for L2 and by R. Hunt (1968) for Lp (1 ă p ă 8). Key idea:Carleson operator

T‹f pxq :“ supně1

ˇ

ˇ

ˇ

ż n

´n

pf pkqe´2πik¨xdkˇ

ˇ

ˇ,

is weak type p2, 2q.

Hard proof (ą 100 pages). M. Lacey and C. Thiele in “A proofof boundedness of the Carleson operator” (Math. Res. Lett. 7(2000), no. 4, 361–370) give a proof in under 20 pages.

No a.e. convergence for L1 functions (A.N. Kolmogorov’scounterexample)

Calderon-Zygmund theory, general kernels 26

Definition (Calderon-Zygmund type)

Given A, B ą 0 and M ą 1, a linear operator T : CcpRdq Ñ L0 we

say that T is Calderon-Zygmund type with parameters pA, B, Mq if

@f P CcpRdq : }Tf }2 ď A}f }2

and there is a measurable kernel K : Rd ˆRd Ñ R such that

supyPRd

supzPRdrt0u

ż

|x´y|ąM|z|

ˇ

ˇKpx, y` zq ´Kpx, yqˇ

ˇdx ď B

for which T admits the integral representation

@f P CcpRdq : Tf p¨q “

ż

Kp¨, yqf pyqdy λ-a.e.

with integral absolutely convergent λ-a.e.

Strong type p2, 2q 27

Note: Assuming strong type p2, 2q! Sufficient conditions exist:

Lemma

Let pX,F , µq and pY,G, νq be σ-finite measure spaces and letK : XˆY Ñ R be F b G-measurable and such that K P L2pµb νq.Then for each f P L2, the integral in

Tf pxq :“ż

Kpx, yqf pyqνpdyq

converges absolutely for ν-a.e. x P X and defines a continuous linearoperator L2pνq Ñ L2pµq. Moreover,

}T}L2pνqÑL2pµq ď }K}L2pµbνq

These are usually too weak to be used here. Strong type p2, 2qproperty usually verified by “Hilbert space” techniques.

Main theorem 28

Theorem (Calderon-Zygmund, general form)

For all A, B ą 0, M ą 1, d ě 1 and p P r1, 2s there is Cp P p0,8qsuch that for every Calderon-Zygmund-type operator T withparameters pA, B, Mq, we have:(1) T is weak type p1, 1q with (its extension to L1 satisfying)

@t ą 0@f P L1 : λ`

|Tf | ą t˘

ďC1

t}f }1

(2) For each p P p1, 2s, T is strong type pp, pq with (its extensionto Lp satisfying)

@f P Lp : }T}p ď Cp}f }p

If K‹px, yq :“ Kpy, xq is also C.Z.-type with the same A, B, M, then Tis also strong type pp, pq for every p P r2,8q with Cp :“ C p

p´1

Proof: main changes 29

The proof for p P p1, 2s is taken nearly verbatim. Dualityargument requires some work. First some functional analysis:

LemmaLet V be a normed linear space and let T : DompTq Ñ V be a linearoperator on V with dense linear DompTq. For each φ P V‹,

@f P DompTq : pT‹φqpf q :“ φpTf q

defines a linear functional T‹φ on DompTq. The map φ ÞÑ T‹φ islinear and so T‹ is a linear operator called the adjoint of T.If T is bounded, then T‹φ P V‹ and T‹ extends to a continuous linearoperator T‹ : V‹ Ñ V‹ with

}T‹} ď }T}.

(Equality holds by the Hahn-Banach theorem.)

Proof of Lemma 30

Linearity of T‹ clear. For T bounded,

@f P DompTq @φ P V‹ :ˇ

ˇpT‹φqpf qˇ

ˇ ď }φ} }Tf } ď }T} }φ} }f }

As DompTq is dense in V , T‹φ extends continuously to V with

}T‹φ} ď }T}}φ}

Hence }T‹} ď }T}.

Adjoint of integral operators 31

Lemma

Let p, q P p1,8q be such that p´1 ` q´1 “ 1 and let T : CcpRq Ñ L0

be a linear operator such that

@f P CcpRdq : Tf p¨q “

ż

Kp¨, yqf pyqdy λ-a.e.

with the integral convergent λ-a.e. If T is continuous as amap Lp Ñ Lp, its adjoint T‹ admits the integral representation

@f P CcpRdq : T‹f p¨q “

ż

Kpy, ¨qf pyqdy λ-a.e.

where the integral converges λ-a.e.

Proof of Lemma 32

For f , g P CcpRdq, Fubini-Tonelli gives

ż

gpTf qdλ “

ż

gpxq´

ż

|x´y|ąεKpx, yqf pyqdy

¯

dx

“

ż

f pyq´

ż

Kpx, yqgpxqdx¯

dy “ż

f prTgqdλ

where rTg :“ş

Kpy, ¨qf pyqdy converges absolutely.Riesz representation:

φgpTf q “ φrTgpf q

Using that g ÞÑ φg is bijective isometry of pLpq‹ Ñ Lq, we nowidentify T‹φg with rTg.

Proof of Calderon-Zygmund theorem, p P r2, 8q 33

Pick p P p2,8q and let q be Holder dual. Let rT be definedusing K‹-kernel (which is C.Z.-type). Then Lemma says

rT‹ “ T (1)

and so}T}LpÑLp ď }rT}LqÑLq ď Cq (2)