Singular Integral Operators and the Riesz Transform

Singular integral operators and the Riesz transform

Jordan Bell [email protected] Department of Mathematics, University of Toronto November 17, 2017

1 Calder´on-Zygmund kernels

n´1 Let ωn´1 be the measure of S . It is 2πn{2 ωn 1 “ . ´ Γpn{2q

n Let vn be the measure of the unit ball in R . It is n{2 ωn´1 2π vn “ “ . n nΓpn{2q

For k, N ě 0 and φ P C8pRnq let N α pk,N pφq “ max sup p1 ` |x|q |pB φqpxq|. α k n | |ď xPR A Borel measurable function K : Rnzt0u Ñ C is called a Calder´on- Zygmund kernel if there is some B such that 1. |Kpxq| ď B|x|´n, x ‰ 0

2. |x|ě2|y| |Kpx ´ yq ´ Kpxq|dx ď B, y ‰ 0

3. ş Kpxqdx “ 0, 0 ă R1 ă R2 ă 8. R1ă|x|ăR2 Theş following lemma gives a tractable condition under which Condition 2 is satisﬁed.1 Lemma 1. If |p∇Kqpxq| ď C|x|´n´1 for all x ‰ 0 then for y ‰ 0,

n |Kpx ´ yq ´ Kpxq|dx ď vn2 C. ż|x|ě2|y| 1Camil Muscalu and Wilhelm Schlag, Classical and Multilinear Harmonic Analysis, volume I, p. 167, Lemma 7.2.

1 Proof. For |x| ą 2|y| ą 0, if 0 ď t ď 1 then |x| |x| |x ´ ty| ě |x| ´ t|y| ě |x| ´ |y| ą |x| ´ “ . 2 2 Write fptq “ Kpx´tyq, for which f 1ptq “ ´p∇Kqpx´tyq¨y. By the fundamental theorem of calculus,

1 1 Kpx ´ yq ´ Kpxq “ fp1q ´ fp0q “ f 1ptqdt “ ´ p∇Kqpx ´ tyq ¨ ydt, ż0 ż0 thus 1 1 |Kpx ´ yq ´ Kpxq| ď |p∇Kqpx ´ tyq||y|dt ď C|y| |x ´ ty|ń´1dt. ż0 ż0 |x| Then using |x ´ ty| ą 2 , |x| ń´1 |Kpx ´ yq ´ Kpxq| ď C|y| “ 2n`1C|y||x|ń´1. 2 ˆ ˙ For |y| ą 0, using spherical coordinates,2

|Kpx ´ yq ´ Kpxq|dx ď 2n`1C|y||x|´n´1dx ż|x|ě2|y| ż|x|ě2|y| 8 “ 2n`1C|y| |rγ|´n´1dσpγq rn´1dr n´1 ż2|y| ˆżS ˙ 8 n`1 ´2 “ vn2 C|y| r dr ż2|y| n`1 1 “ vn2 C|y| ¨ 2|y| n “ vn2 C.

For a Calder´on-Zygmund kernel K, for f P S pRnq, for x P Rn, and for ą 0, 3 using Condition 3 with R1 “ and R2 “ 1,

Kpx ´ yqfpyqdy ż|xý|ě “ Kpx ´ yqpfpyq ´ fpxqqdy ` Kpx ´ yqfpyqdy. żď|xý|ď1 ż|xý|ě1 By Condition 1 there is some B such that |Kpxq| ď B|x|ń, and combining this with |fpyq ´ fpxq| ď }∇f}8 |y ´ x|, ń`1 |Kpx ´ yqpfpyq ´ fpxqq| ď B }∇f}8 |y ´ x| , 2See http://individual.utoronto.ca/jordanbell/notes/sphericalmeasure.pdf 3https://math.aalto.fi/~parissi1/notes/harmonic.pdf, p. 115, Lemma 6.15.

2 which is integrable on t|x ´ y| ď 1u. Then by the dominated convergence theorem,

lim Kpx ´ yqpfpyq ´ fpxqqdy “ Kpx ´ yqpfpyq ´ fpxqqdy. Ñ0 żď|xý|ď1 ż|xý|ď1 Lemma 2. For a Calderón-Zygmundkernel K, for f P S pRnq, and for x P Rn, the limit lim Kpx ´ yqfpyqdy Ñ0 ż|xý|ě exists.

2 Singular integral operators

For a Calder´on-Zygmund kernel K on Rn, for f P S pRnq, and for x P Rn, let

pT fqpxq “ lim Kpx ´ yqfpyqdy. Ñ0 ż|x´y|ě We call T a singular integral operator. By Lemma 2 this makes sense. We prove that singular integral operators are L2 Ñ L2 bounded.4

Theorem 3. There is some Cn such that for any Calder´on-Zygmundkernel K and any f P S pRnq, }T f}2 ď CnB }f}2 . Proof. For 0 ă r ă s ă 8 and for ξ P Rn deﬁne

´2πix¨ξ mr,spξq “ e 1ră|x|ăspxqKpxqdx. n żR Take r ă |ξ|´1 ă s, for which

´2πix¨ξ ´2πix¨ξ mr,spξq “ e Kpxqdx ` e Kpxqdx. ´1 ´1 żră|x|ă|ξ| ż|ξ| ă|x|ăs ´1 For the ﬁrst integral, using Condition 3 with R1 “ r and R2 “ |ξ| and then

4Camil Muscalu and Wilhelm Schlag, Classical and Multilinear Harmonic Analysis, volume I, p. 168, Proposition 7.3; Elias M. Stein, Singular Integrals and Diﬀerentiability Properties of Functions, p. 35, §3.2, Theorem 2; http://math.uchicago.edu/~may/REU2013/REUPapers/ Talbut.pdf

3 using Condition 1,

e´2πix¨ξKpxqdx “ pe´2πix¨ξ ´ 1qKpxqdx ˇ ră|x|ă|ξ|´1 ˇ ˇ ră|x|ă|ξ|´1 ˇ ˇż ˇ ˇż ˇ ˇ ˇ ˇ ˇ ˇ ˇ ď ˇ |e´2πix¨ξ ´ 1||Kpxq|dx ˇ ˇ ˇ ˇ ´1 ˇ ż|x|ă|ξ| ď 2π|x||ξ||Kpxq|dx ´1 ż|x|ă|ξ| ď 2π|ξ| B|x|´n`1dx ´1 ż|x|ă|ξ| ´1 “ 2π|ξ| ¨ vn|ξ| .

ξ For the second integral, let z “ 2|ξ|2 , and

e´2πix¨ξKpxqdx “ ´ e´2πipx`zq¨ξKpxqdx ´1 ´1 ż|ξ| ă|x|ăs ż|ξ| ă|x|ăs “ ´ e´2πix¨ξKpx ´ zqdx. ´1 ż|ξ| ă|x´z|ăs Let

R “ e´2πix¨ξKpx ´ zqdx ´ e´2πix¨ξKpx ´ zqdx ´1 ´1 ż|ξ| ă|x|ăs ż|ξ| ă|x´z|ăs “ ´ e´2πix¨ξKpxqdx ` e´2πix¨ξKpxqdx, ´1 ´1 ż|ξ| ă|x`z|ăs ż|ξ| ă|x|ăs with which 1 R e´2πix¨ξKpxqdx “ e´2πix¨ξpKpxq ´ Kpx ´ zqqdx ` . ´1 2 ´1 2 ż|ξ| ă|x|ăs ż|ξ| ă|x|ăs 1 On the one hand, applying Condition 2, as |z| “ 2|ξ| ,

e´2πix¨ξpKpxq ´ Kpx ´ zqqdx ď |Kpxq ´ Kpx ´ zq|dx ˇ |ξ|´1ă|x|ăs ˇ |x|ą|ξ|´1 ˇż ˇ ż ˇ ˇ ˇ ˇ ˇ ˇ “ |Kpxq ´ Kpx ´ zq|dx ż|x|ą2|z| ď B. On the other hand, let

´1 ´1 D “ D14D2 “ tx : |ξ| ă |x ` z| ă su4tx : |ξ| ă |x| ă su.

For x P D1 we have 1 1 1 |x| ě |x ` z| ´ |z| ą ´ “ , |ξ| 2|ξ| 2|ξ|

4 1 and for x P D2 we have |x| ą |ξ| , so for x P D, 1 |x| ą . 2|ξ| Applying Condition 1,

|Kpxq| ď B|x|´n ă 2nB|ξ|n.

´1 Furthermore, for x P D1zD2 we have |x| ď |ξ| , and for x P D2zD1 we have 1 1 1 3 |x| ď |x ` z| ` |z| “ |x ` z| ` ď ` “ . 2|ξ| |ξ| 2|ξ| 2|ξ| Hence 1 3 D Ă x : ă |x| ď , 2|ξ| 2|ξ| " * so n n 3 3 ´n λpDq ď vn “ |ξ| vn. 2|ξ| 2 ˆ ˙ ˆ ˙ Therefore 3 n |R| ď 2nB|ξ|n ¨ |ξ|´nv “ 3nv B 2 n n ˆ ˙ and then

´2πix¨ξ 1 1 n e Kpxqdx ď B ` ¨ 3 vnB, ˇ |ξ|´1ă|x|ăs ˇ 2 2 ˇż ˇ ˇ ˇ and ﬁnally5 ˇ ˇ ˇ 1 ˇ1 |m pξq| ď 2πv ` B ` ¨ 3nv B “ C B. r,s n 2 2 n n Deﬁne

n pTr,sfqpxq “ 1ră|y|ăspyqKpyqfpx ´ yqdy, x P R . n żR Then

´2πix¨ξ Tr,sfpξq “ e 1ră|y|ăspyqKpyqfpx ´ yqdy dx n n żR ˆżR ˙ z ´2πix¨ξ “ 1ră|y|ăspyqKpyq e fpx ´ yqdx dy n n żR ˆżR ˙ ´2πiy¨ξ “ 1ră|y|ăspyqKpyqe fpξqdy n żR “ mr,spξqfpξq, p

5 The way I organize the argument, I want to use }mr,s}8 ď CnB, while we have only obtained this bound for r ă |ξ|´1pă s. To make the argument correct I may need to do things in a diﬀerent order, e.g. apply Fatou’s lemma and then use an inequality instead of using an inequality and then apply Fatou’s lemma.

5 and so 2 2 2 2 2 Tr,sf “ |Tr,sfpξq| dξ “ |mr,spξqfpξq| dξ ď }mr,s}8 f , 2 n n 2 › › żR żR › › › › › › by Plancherel’s› z › theoremz and the inequality we gotp for |mr,spξq|, › p› 2 2 2 2 2 }Tr,sf}2 ď }mr,s}8 }f}2 ď pCnBq }f}2 . n For each x P R , pTr,sfqpxq Ñ pT fqpxq as r Ñ 0 and s Ñ 8, and thus using Fatou’s lemma,

2 2 2 2 |pT fqpxq| dx ď lim inf |pTr,sfqpxq| dx “ pCnBq }f}2 . n rÑ0,sÑ8 n żR żR That is,

}T f}2 ď CnB }f}2 .

3 The Riesz transform

Let 1 Γ n`1 c “ “ 2 . n πv πpn`1q{2 n´1 ` ˘ For 1 ď j ď n, let xj Kjpxq “ cn . |x|n`1 This is a Calder´on-Zygmund kernel. For φ P S pRnq deﬁne

pRjφqpxq “ lim Kjpx ´ yqφpyqdy “ lim Kjpyqφpx ´ yqdy. Ñ0 Ñ0 ż|y´x|ě ż|y|ě We call each Rj, 1 ď j ď n, a Riesz transform. n For 1 ď j ď n deﬁne Wj : S pR q Ñ C by n`1 Γ 2 xWj, φy “ n`1 lim Kjpyqφpyqdy. (1) π 2 Ñ0 y ` ˘ ż| |ě For ą 0,

Kjpyqφpyqdy “ Kjpyqpφpyq ´ φp0qqdy ˇ ď|y|ď1 ˇ ˇ ď|y|ď1 ˇ ˇż ˇ ˇż ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ´n ˇ ˇ ˇ ď ˇ cn|y| ¨ }∇φ}8 |y|dy ˇ żď|y|ď1 1 ´n`1 n´1 “ cn }∇φ}8 ωn´1 r ¨ r dr ż “ cn }∇φ}8 ωn´1p1 ´ q.

6 For |y| ě 1,

´n 2 ´1{2 |Kjpyqφpyq|dy ď cn |y| p1 ` |y| q p0,1pφqdy ż|y|ě1 ż|y|ě1 8 ´n ´1 n´1 “ cnωn r p1 ` rq ¨ r dr ż1 “ cnωn log 2.

It then follows from the dominated convergence theorem that the limit

lim Kjpyqφpyqdy Ñ0 ż|y|ě exists, which shows that the deﬁnition (1) makes sense. It is apparent that n Wj is linear. Then prove that if φk Ñ φ in S pR q then xWj, φky Ñ xWj, φy. 1 n This being true means that Wj P S pR q, namely that each Wj is a tempered distribution. For a function f : Rn Ñ C, write

fpxq “ fp´xq, pτyfqpxq “ fpx ´ yq.

1 n n For u P S pR q andrh P S pR q, deﬁne

n xh ˚ u, φy “ u, h ˚ φ , φ P S pR q. A E It is a fact that h˚u P S 1pRnq, andr this tempered distribution is induced by the 8 6 C function x ÞÑ u, τxh . The Fourier transform of a tempered distribution u is deﬁned by A E r n xu, φy “ u, φ , φ P S pR q, where A E p p ´2πix¨ξ n φpξq “ e φpxqdx, ξ P R . n żR It is a fact that u isp itself a tempered distribution. Finally, for a tempered distribution u and a Schwartz function h, we deﬁne

p n xhu, φy “ xu, hφy , φ P S pR q. It is a fact that hu is itself a tempered distribution. It is proved that7

φ ˚ u “ φu.

The left-hand side is the Fourier transform of the tempered distribution φ˚u, and z pp the right-hand side is the product of the Schwartz function φ and the tempered distribution u. 6Loukas Grafakos, Classical Fourier Analysis, second ed., p. 116, Theoremp 2.3.20. 7Loukas Grafakos,p Classical Fourier Analysis, second ed., p. 120, Proposition 2.3.22.

7 Lemma 4. For 1 ď j ď n, for φ P S pRnq, and for x P Rn,

pRjφqpxq “ pφ ˚ Wjqpxq.

We will use the following identity for integrals over Sn´1.8

Lemma 5. For ξ ‰ 0 and for 1 ď j ď n,

2ωn´2 ξj sgn pξ ¨ θqθjdσpθq “ . n´1 n ´ 1 |ξ| żS

Proof. It is a fact that

0 k ‰ j sgn pθkqθjdσpθq “ (2) n´1 |θ |dσpθq k “ j. żS # Sn´1 j It suﬃces to prove the claim when ξş P Sn´1. For 1 ď j ď n there is 9 Aj “ pai,kqi,k P SOnpRq such that

Ajej “ ξ,

T ´1 for which ai,j “ ξi. Using that Aj “ Aj and that σ is invariant under Opnq we calculate

´1 sgn pξ ¨ θqθjdσpθq “ sgn pAjej ¨ θqpAA θqjdσpθq n´1 n´1 żS żS ´1 ´1 “ sgn pej ¨ Aj θqpAA θqjdσpθq n´1 żS ´1 “ sgn pej ¨ θqpAθqjdpAj σqpθq n´1 ˚ żS “ sgn pej ¨ θqpAθqjdσpθq Sn´1 ż n “ sgn pθjq aj,kθkdθ. n´1 S k“1 ż ÿ Applying Lemma 2 and aj,j “ ξj, this becomes

ξj sgn pξ ¨ θqθjdσpθq “ ξj|θj|dσpθq “ |θj|dσpθq. n´1 n´1 |ξ| n´1 żS żS żS Hence for each 1 ď j ď n,

ξj sgn pξ ¨ θqθjdσpθq “ |θ1|dσpθq. n´1 |ξ| n´1 żS żS 8Loukas Grafakos, Classical Fourier Analysis, second ed., p. 261, Lemma 4.1.15. 9http://www.math.umn.edu/~garrett/m/mfms/notes/08_homogeneous.pdf

8 It is a fact that10

R Rds fpθqdσpθq “ fps, φqdφ? . ? 2 2 n´1 2 2 n´2 R ´ s żRS ż´R ż R ´s S n´2 Using this with fpθq “ fpθ1, . . . , θnq “ |θ1| and using that the measure of RS n´2 is R ωn´1, we calculate

1 ds |θ1|dσpθq “ |s|dφ? ? 2 n´1 2 n´2 1 ´ s żS ż´1 ż 1´s S 1 2 n´2 ´ 1 “ p1 ´ s q 2 2 ωn´2|s|dφ ż´1 1 2 n´3 “ 2ωn´2 p1 ´ s q 2 sds 0 ż1 n´3 “ ωn´2 u 2 du ż0 2ω “ n´2 . n ´ 1

We now calculate the Fourier transform of the Wj. We show that the Fourier transform of the tempered distribution Wj is induced by the function ξ ÞÑ ξj 11 ´i |ξ| .

Theorem 6. For 1 ď j ď n and for φ P S pRnq,

xj Wj, φ “ ´iφpxq dx. n |x| A E żR x Proof. We calculate

n`1 Γ 2 Wj, φ “ n`1 lim Kjpξqφpξqdξ π 2 Ñ0 ξ ` ˘ ż| |ě A E n`1 p Γ 2 p “ n`1 lim Kjpξqφpξqdξ π 2 Ñ0 ξ 1 ` ˘ ż ď| |ď { n`1 Γ p ξj 2 lim e´2πix¨ξφ x dx dξ “ n`1 p q n`1 π 2 Ñ0 ξ 1 n |ξ| ` ˘ ż ď| |ď { ˆżR ˙ Γ n`1 ξ 2 lim φ x e´2πix¨ξ j dξ dx. “ n`1 p q n`1 π 2 Ñ0 n ξ 1 |ξ| ` ˘ żR ˜ż ď| |ď { ¸ 10Loukas Grafakos, Classical Fourier Analysis, second ed., p. 441, Appendix D.2. 11Loukas Grafakos, Classical Fourier Analysis, second ed., p. 260, Proposition 4.1.14.

9 For the inside integral, because θ ÞÑ cosp´2πrxjθjqθj is an odd function, ξ rθ e´2πix¨ξ j dξ e´2πix¨prθq j dσ θ rn´1dr n`1 “ n`1 p q |ξ| n´1 r żď|ξ|ď1{ żďrď1{ ˆżS ˙ ´2πirx¨θ ´1 “ e θjdσpθq r dr n´1 żďrď1{ ˆżS ˙ ´1 “ i sinp´2πrx ¨ θqθjdσpθq r dr n´1 żďrď1{ ˆżS ˙ ´1 “ ´i sinp2πrx ¨ θqθjdσpθq r dr n´1 żďrď1{ ˆżS ˙ ´1 “ ´i sinp2πrx ¨ θqr dr θjdσpθq. n´1 żS ˜żďrď1{ ¸

Call the whole last expression fpxq. It is a fact that for 0 ă a ă b ă 8,

b sin t dt ď 4, ˇ a t ˇ ˇż ˇ ˇ ˇ thus for x 0, ˇ ˇ ‰ ˇ ˇ |fpxq| ď 4ωn´1. As12 π lim fpxq “ ´i sgn px ¨ θq θjdσpθq, Ñ0 n´1 2 żS applying the dominated convergence theorem yields

ξ lim φ x e´2πix¨ξ j dξ dx p q n`1 Ñ0 n |ξ| żR ˜żď|ξ|ď1{ ¸ π “ φpxq ´i sgn px ¨ θq θjdσpθq dx n n´1 2 żR ˆ żS ˙ π “ ´ i φpxq sgn px ¨ θqθjdσpθq dx. 2 n n´1 żR ˆżS ˙ Then using Lemma 5 and putting the above together we get

n`1 Γ 2 π Wj, φ “ n`1 ¨ ´i φpxq sgn px ¨ θqθjdσpθq dx π 2 2 n Sn´1 ` ˘ żR ˆż ˙ A E n`1 p π Γ 2 2ωn´2 xj “ ´i n`1 φpxq dx. 2 π 2 n n ´ 1 |x| ` ˘ żR We work out that n`1 π Γ 2 2ωn´2 n`1 ¨ “ 1, 2 2 n 1 π` ˘ ´ 12Loukas Grafakos, Classical Fourier Analysis, second ed., p. 263, Exercise 4.1.1.

10 and therefore xj Wj, φ “ ´i φpxq dx, n |x| A E żR completing the proof. p

Because Rjh “ h ˚ Wj,

ξj Rjh, φ “ hWj, φ “ Wj, hφ “ ´ihpξqφpξq dξ. n |ξ| A E A E A E żR Theorem 7.yFor 1 ď jpďxn and for xh P pS pRnq, p

ξj n Rjhpξq “ ´i hpξq, ξ P R . |ξ| y p

ξj In other words, the multiplier of the Riesz transform Rj is mjpξq “ ´i |ξ| .

4 Properties of the Riesz transform

Theorem 8. n 2 Í “ Rj , j“1 ÿ where Iphq “ h for h P S pRnq. Proof. For h P S pRnq, ξ2 2 ξj ξj ξj j R hpξq “ í Rjhpξq “ í ¨ í hpξq “ ´ hpξq, j |ξ| |ξ| |ξ| |ξ|2 hence y y p p n 2 Rj h “ ´h. j“1 ÿ Taking the inverse Fourier transform,y p

n 2 Rj h “ ´h, j“1 ÿ i.e. n 2 Rj “ ´I. j“1 ÿ

11 For a tempered distribution u, for 1 ď j ď n, we deﬁne n xBju, φy “ p´1q xu, Bjφy , φ P S pR q.

It is a fact that Bju is itself a tempered distribution. One proves that

Bju “ p2πiξjqu. Each side of the above equation is a tempered distribution. Then y p n n n 2 2 2 2 2 2 ∆u “ Bj u “ p2πiξjq u “ ´4π ξj u “ ´4π |ξ| u. j“1 j“1 j“1 ÿ ÿ ÿ Supposex that f isy a Schwartz functionp and that u isp a temperedp distribution satisfying ∆u “ f, called Poisson’s equation. Then ´4π2|ξ|2u “ f. For 1 ď j, k ď n, p p ´1 BjBku “ F pF pBjBkuqq ´1 “ F pp2πiξjqp2πiξkquq

´1 2 f “ F ´4π ξjξk ¨ p 2 2 ˜ ´4π |ξ| ¸ p ξjξk “ F ´1 f . |ξ|2 ˆ ˙ Using Theorem 7, p

´1 RjRkf “ F F pRjRkfq

´1 ξj “ F í Rkf |ξ| ˆ ˙ ξj ξk “ F ´1 í y¨ í f |ξ| |ξ| ˆ ˙ ξjξk “ F ´1 ´ f . p |ξ|2 ˆ ˙ Therefore p BjBku “ ´RjRkf. Theorem 9. If f is a Schwartz function and u is a tempered distribution satisfying ∆u “ f, then for 1 ď j, k ď n, BjBku “ ´RjRkf.