Calder on-Zygmund theory Updated May 23, 2020 Plan 2 Outline: - PowerPoint PPT Presentation

Calder´ on-Zygmund theory Updated May 23, 2020

Plan 2 Outline: Statement and motivation Proof via Marcinkiewicz and duality Applications to Hilbert and inverse-Fourier transform General Calder´ on-Zygmund kernels

Motivations 3 Q: What happens with Riesz transform ż 1 T α f p x q : “ | x ´ y | α f p y q d y when α “ d ? Or with Hilbert transform? A: Integral not defined even for nice f due to singularity at x “ y , but could truncate to | x ´ y | ě ǫ , perhaps. Singularity as | x | Ñ 8 bad too; kernel 1 K p x q : “ | x | d 1 | x |ě ǫ obeys K P L 1,w , but Schur’s test requires L r ,w with r ą 1

Calder´ on-Zygmund theorem 4 Theorem (Calder´ on-Zygmund) Consider the measure space p R d , L p R d q , λ q for d ě 1 . For all A , B ą 0 , all M ą 1 and all p P p 1, 8q there is C p P p 0, 8q such that for all measurable kernels K : R d Ñ R satisfying K P L 2 with Fourier transform p K obeying } p K } 8 ď A and ż ˇ ˇ ˇ K p x ´ z q ´ K p x q ˇ d x ď B , sup | x |ą M | z | z P R d � t 0 u the convolution operator T K f : “ K ‹ f is well defined by the integral expression for all f P L 1 and extends continuously to a map L p Ñ L p for each p P p 1, 8q with } T K } L p Ñ L p ď C p @ p P p 1, 8q :

Remarks 5 1st condition ensures K is locally integrable and K ‹ f meaningful for f P L 1 (by Young convolution inequality) 2nd condition often stated as K P C 1 p R d � t 0 uq with ˇ ˇ ˜ B ˇ ∇ K p x q ˇ ď @ x P R d � t 0 u : | x | d ` 1 which (along with 1st condition) gives ˜ B { d | K p x q| ď | x | d and so we cannot hope for more than K P L 1,w (and so Schur’s test is still out). Upshot: Trading local regularity against integrability

Strategy of proof 6 1st condition implies T K is strong type p 2, 2 q with 2nd condition this implies T K is weak type p 1, 1 q . This is the key novelty; requires so called Calderon-Zygmund decomposition of R d into sets where f is bounded and sets where f has bounded integral Marcinkiewicz interpolation gives T K is strong type p p , p q for all p P p 1, 2 s Duality: true also for p P r 2, 8q

Improved Young convolution inequality 7 Recall: By Young convolution inequality f P L 2 and g P L 1 implies integral f ‹ g converges absolutely a.e. and } f ‹ g } 2 ď } f } 2 } g } 1 Need a slight improvement: Lemma @ f P L 1 @ g P L 2 : } f ‹ g } 2 ď } f } 2 } p g } 8 Proof: Let f P L 1 and g P L 2 . Fourier transform isometry so } y f ‹ g } 2 ď } f } 1 } g } 2 Hence g ÞÑ y f ‹ g continuous. If g P L 1 , then y f ‹ g “ p f p g so true for g P L 2 as well. Hence } f ‹ g } 2 “ } y f ‹ g } 2 “ } p g } 2 ď } p f p f } 2 } p g } 8 “ } f } 2 } p g } 8

Strong type p 2, 2 q 8 Corollary The operator T K is strong type p 2, 2 q with } T K } L 2 Ñ L 2 ď A Proof: For f P L 1 , T K f well defined via K ‹ f and obeys } T K f } 2 ď A } f } 2 by above lemma. So T K extends to L 2 with } T K } L 2 Ñ L 2 ď A

Calder´ on-Zygmund decomposition 9 Dyadic cube is any cube of the form 2 n x ` r 0, 2 n q d for x P Z d and n P Z . Lemma (Calder´ on-Zygmund decomposition) Let f P L 1 and t ą 0 . Then there exist disjoint dyadic cubes t Q i u i P I such that ż | f | d λ ď 2 d t λ p Q i q @ i P I : t λ p Q i q ă Q i and ď λ -a.e. on R d � | f | ď t Q i i P I

Proof of Calder´ on-Zygmund decomposition 10 Call a dyadic cube Q good if ż 1 | f | d λ ď t λ p Q q Q ş | f | d λ ď t 2 n , all and call it bad otherwise. For n P Z such that dyadic cubes of side-length 2 n good. Let t Q i u i P I enumerate the set of all bad dyadic cubes Q such that the (unique) dyadic cube Q 1 containing Q and having side length twice as that of Q is good. Then ż ż Q 1 | f | d λ ď t λ p Q 1 q “ 2 d t λ p Q q t λ p Q q ă | f | d λ ď Q because Q is bad and Q 1 is good. If x lies only in good cubes, Lebesgue differentiation shows | f p x q| ď t a.e. (Need a version for dyadic cubes; proved when discussed martingale convergence.)

Weak type p 1, 1 q 11 Proposition T K is weak type p 1, 1 q . Explicitly, ` ˘ ď c D c P p 0, 8q @ f P L 1 @ t ą 0: λ | T K f | ą t t } f } 1 where c depends only d and the constants A and B

Decomposition of f 12 Pick f P L 1 and t ą 0 and let t Q i u i P I be as above. Set ď F : “ R d � Q i i P I define g : R d Ñ R by # ş 1 Q i f d λ if x P Q i for some i P I λ p Q i q g p x q : “ f p x q if x P F and abbreviate h p x q : “ f p x q ´ g p x q Note that ż h “ 0 on F ^ @ i P I : h d λ “ 0 Q i Union bound + additivity: ` ˘ ` ˘ ` ˘ | T K f | ą t ď λ | T K g | ą t { 2 ` λ | T K h | ą t { 2 λ Now estimate each term separately . . .

Tails of T K g 13 Will use that T K maps L 2 Ñ L 2 with norm ď A (proved above). Need to estimate ż ż ÿ } g } 2 g 2 d λ ` g 2 d λ 2 “ F Q i i P I ż ż ´ ¯ 2 ÿ 1 ď t | f | d λ ` f d λ λ p Q i q λ p Q i q F Q i i P I ż ÿ p 2 d t q 2 λ p Q i q ď t | f | d λ ` F i P I ż ż ÿ | f | d λ ` 4 d t ď t | f | d λ F Q i i P I ż ż F c | f | d λ “ p 4 d ` 1 q t } f } 1 | f | d λ ` 4 d t “ t F Hence 2 ď 4 A 2 p 4 d ` 1 q ` ˘ 2 ď 4 A 2 ď 4 t 2 } T K g } 2 t 2 } g } 2 λ | T K g | ą t { 2 } f } 1 t

Tails of T K h 14 ş Consider h i : “ h 1 Q i . Let y i : “ the center of Q i . As Q i h d λ “ 0, ż ż ` ˘ T K h i p x q “ K p x ´ y q h i p y q d y “ K p x ´ y q ´ K p x ´ y i q h i p y q d y Q i Q i ? Let Q 1 i : “ the cube of M d -times the side length of Q i centered at y i . By Tonelli and 2nd condition: ż | T K h i | d λ R d � Q 1 i ż ´ż ¯ˇ ˇ ˇ ˇ ˇ K p x ´ y i ` y ´ y i q ´ K p x ´ y i q ˇ d x ˇ h i p y q ˇ d y ď R d � Q 1 Q i ż i ż ď B | h | d λ ď 2 B | f | d λ Q i Q i which uses | x ´ y i | ą M | y ´ y i | for all x R Q 1 i and all y P Q i . Then . . .

Tails of T K h continued ... 15 . . . abbreviating F 1 : “ R d � Ť i ě 1 Q 1 i we thus get ż ˇ ˇ T K h | d λ ď 2 B } f } 1 F 1 On the other hand, ÿ ? d q d ÿ λ p R d � F 1 q ď λ p Q 1 i q “ p M λ p Q i q i P I i P I ? ? ż ÿ d q d d q d ď p M | f | d λ ď p M } f } 1 t t Q i i P I and so ? ż d q d ` 4 B ` ˘ ˇ ď λ p R d � F 1 q ` 2 ˇ T K h | d λ ď p M | T K h | ą t { 2 } f } 1 λ t t F 1 So claim holds with ? c : “ 4 A 2 p 4 d ` 1 q ` p M d q d ` 4 B

Proof of Calder´ on-Zygmund theorem 16 Marcinkiwicz: T K strong type p p , p q for p P p 1, 2 s . Now let q P p 2, 8q and let p be such that p ´ 1 ` q ´ 1 “ 1. Then duality between L p and L q gives ż ˇ ˇ ˇ ˇ @ f P L 1 X L p @ g P L q : g p K ‹ f q d λ ˇ ď } T K } L p Ñ L p } f } p } g } q ˇ For f P L 1 integral K ‹ f converges absolutely. So by Fubini-Tonelli: ż ˇ ˇ ˇ ˇ @ f P L 1 X L p @ g P L q X L 1 : ˇ p T K g q f d λ ˇ ď } T K } L p Ñ L p } f } p } g } q Density of L p X L 1 in L p gives @ g P L q X L 1 : } T K g } q ď } T K } L p Ñ L p } g } q . which implies that T K extends continuously to a map L q Ñ L q with } T } L q Ñ L q ď } T K } L p Ñ L p (Equality holds by duality.)

Application to Hilbert transform 17 Recall: Hf defined as the ǫ Ó 0 limit of convolution-type operator H ǫ f : “ K ǫ ‹ f where K ǫ p x q : “ 1 π x 1 p ǫ ,1 { ǫ q p| x |q Convergence pointwise for f P C 1 p R q X L 1 and in L 2 for f P L 2 Theorem (Hilbert transform in L p ) We have } H ǫ } L p Ñ L p ă 8 . @ p P p 1, 8q : sup 0 ă ǫ ă 1 In particular, for all p P p 1, 8q , there exists a continuous linear operator H : L p Ñ L p such that @ f P L p : ǫ Ó 0 Hf in L p . H ǫ f Ý Ñ

Proof of Theorem 18 For strong type p 2, 2 q , use Fourier calculation to get ż K ǫ p z q “ ´ 2i sin p 2 π zt q p d t π t ǫ ă t ă 1 { ǫ Hence, A : “ sup 0 ă ǫ ă 1 } p K ǫ } 8 ă 8 . For weak type p 1, 1 q , compute ˇ ˇ ˇ K ǫ p x ´ z q ´ K ǫ p x q ˇ ˇ ˇ ˇ ˇ x ´ z ´ 1 1 ˇ ` 1 ˇ ˇ ˇ 1 p ǫ ,1 { ǫ q p| x ´ z |q ´ 1 p ǫ ,1 { ǫ q p| x |q ˇ ď ˇ x | x | ď 2 | z | | x | 2 ` 1 | x | 1 tp 1 { ǫ ´| z | ,1 { ǫ `| z |q p| x |q ` 1 | x | 1 tp ǫ ´| z | , ǫ `| z |q p| x |q Need to integrate this over | x | ą 2 | z | . First term easy. For the other two terms we note that . . .

Proof of Theorem continued ... 19 . . . for any a , b ą 0 with max t 2 b , a ´ b u ă a ` b , ż a ` b ´ ¯ d x a ` b x “ log max t 2 b , a ´ b u max t 2 b , a ´ b u Examining a ´ b ă 2 b and a ´ b ą 2 b separately, RHS ď log p 2 q . Now use this with a : “ ǫ , 1 { ǫ and b “ | z | to get ż ˇ ˇ ˇ K ǫ p x ´ z q ´ K ǫ p x q ˇ d x ă 8 B : “ sup sup | x |ą 2 | z | 0 ă ǫ ă 1 z P R � t 0 u So t H ǫ u 0 ă ǫ ă 1 obey conditions of Calder´ on-Zygmund theorem with uniform A and B (and M : “ 2). So we get } H ǫ } L p Ñ L p ă 8 . sup 0 ă ǫ ă 1 It remains to address convergence H ǫ f Ñ Hf . . .

Proof of Theorem continued ... 20 . . . which we already know in L 2 by Fourier techniques. We will use interpolation for L p -norms. Given p P p 1, 8q , choose ˜ p P p 1, p q when p ă 2 or ˜ p P p p , 8q when p ą 2. Then 1 p “ p 1 ´ θ q 1 p ` θ 1 2 for some θ P p 0, 1 q and so ˜ p X L 2 : @ f P L ˜ 2 } H ǫ f ´ H δ f } 1 ´ θ } H ǫ f ´ H δ f } p ď } H ǫ f ´ H δ f } θ . p ˜ Now } H ǫ f ´ H δ f } 2 Ñ 0 as ǫ , δ Ó 0 by the claim in L 2 while ´ ¯ } H ǫ f ´ H δ f } ˜ p ď 2 sup } H ǫ } L ˜ } f } ˜ p . p Ñ L ˜ p 0 ă ǫ 1 ă 1 Completeness of L p shows H ǫ f Ñ Hf for each f P L ˜ p X L 2 . As p X L 2 dense in L p , true for all f P L p . L ˜