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)