Singular integral operators and the Riesz transform

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1 Singular integral operators and the Riesz transform Jordan Bell Department of Mathematics, University of Toronto November 17, Calderón-Zygmund kernels Let ω n 1 be the measure of S n 1. It is ω n 1 πn{ Γpn{q. Let v n be the measure of the unit ball in R n. It is For k, N ě 0 and φ P C 8 pr n q let v n ω n 1 n πn{ nγpn{q. p k,n pφq max α ďk sup xpr n p1 ` x q N pb α φqpxq. A Borel measurable function K : R n zt0u Ñ C is called a Calderón- Zygmund kernel if there is some B such that 1. Kpxq ď B x n, x 0. ş Kpx yq Kpxq dx ď B, y 0 x ě y 3. ş R 1ă x ăr Kpxqdx 0, 0 ă R 1 ă R ă 8. The following lemma gives a tractable condition under which Condition is satisfied. 1 Lemma 1. If p Kqpxq ď C x n 1 for all x 0 then for y 0, Kpx yq Kpxq dx ď v n n C. x ě y 1 Camil Muscalu and Wilhelm Schlag, Classical and Multilinear Harmonic Analysis, volume I, p. 167, Lemma 7.. 1

2 Proof. For x ą y ą 0, if 0 ď t ď 1 then x ty ě x t y ě x y ą x x x. Write fptq Kpx tyq, for which f 1 ptq p Kqpx tyq y. By the fundamental theorem of calculus, thus Kpx yq Kpxq fp1q fp0q Kpx yq Kpxq ď Then using x ty ą x, 1 Kpx yq Kpxq ď C y f 1 ptqdt 1 p Kqpx tyq y dt ď C y 0 p Kqpx tyq ydt, 1 0 x ty n 1 dt. n 1 ˆ x n`1 C y x n 1. For y ą 0, using spherical coordinates, Kpx yq Kpxq dx ď n`1 C y x n 1 dx x ě y x ě y n`1 C y 8 v n n`1 C y v n n`1 C y v n n C. ˆ rγ n 1 dσpγq y S n 1 8 r dr y 1 y r n 1 dr For a Calderón-Zygmund kernel K, for f P S pr n q, for x P R n, and for ɛ ą 0, using Condition 3 with R 1 ɛ and R 1, 3 Kpx yqfpyqdy x y ěɛ ɛď x y ď1 Kpx yqpfpyq fpxqqdy ` Kpx yqfpyqdy. x y ě1 By Condition 1 there is some B such that Kpxq ď B x n, and combining this with fpyq fpxq ď } f} 8 y x, Kpx yqpfpyq fpxqq ď B } f} 8 y x n`1, See p. 115, Lemma 6.15.

3 which is integrable on t x y ď 1u. Then by the dominated convergence theorem, Kpx yqpfpyq fpxqqdy Kpx yqpfpyq fpxqqdy. ɛď x y ď1 x y ď1 Lemma. For a Calderón-Zygmund kernel K, for f P S pr n q, and for x P R n, the it Kpx yqfpyqdy exists. x y ěɛ Singular integral operators For a Calderón-Zygmund kernel K on R n, for f P S pr n q, and for x P R n, let pt fqpxq Kpx yqfpyqdy. x y ěɛ We call T a singular integral operator. By Lemma this makes sense. We prove that singular integral operators are L Ñ L bounded. 4 Theorem 3. There is some C n such that for any Calderón-Zygmund kernel K and any f P S pr n q, }T f} ď C n B }f}. Proof. For 0 ă r ă s ă 8 and for ξ P R n define m r,s pξq e πix ξ 1 ră x ăs pxqkpxqdx. R n Take r ă ξ 1 ă s, for which m r,s pξq e πix ξ Kpxqdx ` ră x ă ξ 1 e πix ξ Kpxqdx. ξ 1 ă x ăs For the first integral, using Condition 3 with R 1 r and R ξ 1 and then 4 Camil 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., Theorem ; Talbut.pdf 3

4 using Condition 1, ˇ ră x ă ξ 1 e πix ξ Kpxqdx For the second integral, let z ξ, and e πix ξ Kpxqdx ξ 1 ă x ăs ξ ˇ pe πix ξ 1qKpxqdx ˇ ˇ ră x ă ξ 1 ď e πix ξ 1 Kpxq dx x ă ξ 1 ď π x ξ Kpxq dx x ă ξ 1 ď π ξ π ξ v n ξ 1. x ă ξ 1 B x n`1 dx e πipx`zq ξ Kpxqdx ξ 1 ă x ăs e πix ξ Kpx zqdx. ξ 1 ă x z ăs Let R e πix ξ Kpx zqdx e πix ξ Kpx zqdx ξ 1 ă x ăs e πix ξ Kpxqdx ` ξ 1 ă x z ăs Kpxqdx, ξ 1 ă x`z ăs ξ 1 ă x ăs with which ξ 1 ă x ăs e πix ξ Kpxqdx 1 e πix ξ pkpxq Kpx zqqdx ` R ξ 1 ă x ăs. On the one hand, applying Condition, as z 1 ξ, e πix ξ pkpxq Kpx zqqdx ˇ ξ 1 ă x ăs ˇ ď Kpxq Kpx zq dx x ą ξ 1 Kpxq Kpx zq dx On the other hand, let ď B. x ą z D D 1 D tx : ξ 1 ă x ` z ă su tx : ξ 1 ă x ă su. For x P D 1 we have x ě x ` z z ą 1 ξ 1 ξ 1 ξ, 4

5 and for x P D we have x ą 1 ξ, so for x P D, Applying Condition 1, x ą 1 ξ. Kpxq ď B x n ă n B ξ n. Furthermore, for x P D 1 zd we have x ď ξ 1, and for x P D zd 1 we have x ď x ` z ` z x ` z ` 1 ξ ď 1 ξ ` 1 ξ 3 ξ. Hence so Therefore and then D Ă λpdq ď R ď n B ξ n ˇˇˇˇ ˇ ξ 1 ă x ăs " 1 x : ξ ă x ď 3 *, ξ ˆ n 3 v n ξ e πix ξ Kpxqdx n ˆ3 ξ n v n. n ˆ3 ξ n v n 3 n v n B ˇ ď 1 B ` 1 3n v n B, and finally m r,spξq ď πv n ` 1 B ` 1 3n v n B C n B. Define pt r,s fqpxq 1 ră y ăs pyqkpyqfpx yqdy, R n x P R n. Then ˆ zt r,s fpξq e πix ξ 1 ră y ăs pyqkpyqfpx yqdy dx R n R n ˆ 1 ră y ăs pyqkpyq e πix ξ fpx yqdx dy R n R n 1 pyqkpyqe πiy ξ ră y ăs fpξqdy p R n m r,s pξq p fpξq, 5 The way I organize the argument, I want to use }m r,s} 8 ď C nb, while we have only obtained this bound for r ă ξ 1 ă 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

6 and so T z r,s f T z r,s fpξq dξ m r,s pξq p fpξq dξ ď }m r,s } 8 p f, R n R n by Plancherel s theorem and the inequality we got for m r,s pξq, }T r,s f} ď }m r,s} 8 }f} ď pc nbq }f}. For each x P R n, pt r,s fqpxq Ñ pt fqpxq as r Ñ 0 and s Ñ 8, and thus using Fatou s lemma, pt fqpxq dx ď inf pt r,s fqpxq dx pc n Bq }f} R n rñ0,sñ8. R n That is, }T f} ď C n B }f}. 3 The Riesz transform Let For 1 ď j ď n, let c n 1 Γ ` n`1 πv n 1 π. pn`1q{ x j K j pxq c n x n`1. This is a Calderón-Zygmund kernel. For φ P S pr n q define pr j φqpxq K j px yqφpyqdy K j pyqφpx yqdy. y x ěɛ y ěɛ We call each R j, 1 ď j ď n, a Riesz transform. For 1 ď j ď n define W j : S pr n q Ñ C by xw j, φy Γ ` n`1 K j pyqφpyqdy. (1) For ɛ ą 0, ˇ ɛď y ď1 K j pyqφpyqdy ˇ ˇ ď ɛď y ď1 ɛď y ď1 y ěɛ c n } φ} 8 ω n 1 1 K j pyqpφpyq φp0qqdy ˇ c n y n } φ} 8 y dy c n } φ} 8 ω n 1 p1 ɛq. ɛ r n`1 r n 1 dr 6

7 For y ě 1, y ě1 K j pyqφpyq dy ď c y n n p1 ` y q 1{ p 0,1 pφqdy y ě1 8 c n ω r n n p1 ` rq 1 r n 1 dr 1 c n ω n log. It then follows from the dominated convergence theorem that the it K j pyqφpyqdy y ěɛ exists, which shows that the definition (1) makes sense. It is apparent that W j is linear. Then prove that if φ k Ñ φ in S pr n q then xw j, φ k y Ñ xw j, φy. This being true means that W j P S 1 pr n q, namely that each W j is a tempered distribution. For a function f : R n Ñ C, write rfpxq fp xq, pτ y fqpxq fpx yq. For u P S 1 pr n q and h P S pr n q, define A xh u, φy u, r E h φ, φ P S pr n q. It is a fact that h u P S 1 pr E n q, and this tempered distribution is induced by the C 8 function x ÞÑ Au, τ x r h. 6 The Fourier transform of a tempered distribution u is defined by A xpu, φy u, φ p E, φ P S pr n q, where pφpξq e πix ξ φpxqdx, ξ P R n. R n It is a fact that pu is itself a tempered distribution. Finally, for a tempered distribution u and a Schwartz function h, we define xhu, φy xu, hφy, φ P S pr n q. It is a fact that hu is itself a tempered distribution. It is proved that 7 zφ u p φpu. The left-hand side is the Fourier transform of the tempered distribution φ u, and the right-hand side is the product of the Schwartz function p φ and the tempered distribution pu. 6 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 116, Theorem Loukas Grafakos, Classical Fourier Analysis, second ed., p. 10, Proposition.3.. 7

8 Lemma 4. For 1 ď j ď n, for φ P S pr n q, and for x P R n, pr j φqpxq pφ W j qpxq. We will use the following identity for integrals over S n 1. 8 Lemma 5. For ξ 0 and for 1 ď j ď n, S n 1 sgn pξ θqθ j dσpθq ω n n 1 ξ j ξ. Proof. It is a fact that S n 1 sgn pθ k qθ j dσpθq # 0 k j ş S n 1 θ j dσpθq k j. () It suffices to prove the claim when ξ P S n 1. A j pa i,k q i,k P SO n prq such that 9 For 1 ď j ď n there is A j e j ξ, for which a i,j ξ i. Using that A T j A 1 j and that σ is invariant under Opnq we calculate sgn pξ θqθ j dσpθq sgn pa j e j θqpaa 1 θq j dσpθq S n 1 S n 1 sgn pe j A 1 j θqpaa 1 θq j dσpθq S n 1 sgn pe j θqpaθq dpa 1 j j σqpθq S n 1 sgn pe j θqpaθq j dσpθq S n 1 nÿ sgn pθ j q a j,k θ k dθ. S n 1 Applying Lemma and a j,j ξ j, this becomes sgn pξ θqθ j dσpθq ξ j θ j dσpθq ξ j θ j dσpθq. S n 1 S ξ n 1 S n 1 Hence for each 1 ď j ď n, S n 1 sgn pξ θqθ j dσpθq ξ j ξ k 1 S n 1 θ 1 dσpθq. 8 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 61, Lemma

9 It is a fact that 10 RS n 1 fpθqdσpθq R R Rds? fps, φqdφ? R s S n R s. Using this with fpθq fpθ 1,..., θ n q θ 1 and using that the measure of RS n is R n ω n 1, we calculate S n 1 θ 1 dσpθq ds? s dφ? 1 s S n 1 s p1 s q n 1 ωn s dφ 1 ω n p1 s q n 3 sds ω n 1 0 ω n n 1. 0 u n 3 du We now calculate the Fourier transform of the W j. We show that the Fourier transform of the tempered distribution W j is induced by the function ξ ÞÑ i ξj ξ.11 Theorem 6. For 1 ď j ď n and for φ P S pr n q, A xwj, φe iφpxq x j R x dx. n Proof. We calculate E Γ ` n`1 A W j, p φ Γ ` n`1 Γ ` n`1 Γ ` n`1 ξ ěɛ ɛď ξ ď1{ɛ ɛď ξ ď1{ɛ R n φpxq K j pξq p φpξqdξ K j pξq p φpξqdξ ˆ R n e πix ξ φpxqdx e πix ξ ɛď ξ ď1{ɛ ξ ξj ξ ξ j dξ n`1 n`1 dξ 10 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 441, Appendix D.. 11 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 60, Proposition dx. 9

10 For the inside integral, because θ ÞÑ cosp πrx j θ j qθ j is an odd function, ˆ e πix ξ ξ j dξ e πix prθq rθ j dσpθq r n 1 dr ɛď ξ ď1{ɛ ξ n`1 ɛďrď1{ɛ S rn`1 ˆ n 1 e πirx θ θ j dσpθq r 1 dr ɛďrď1{ɛ S ˆ n 1 i sinp πrx θqθ j dσpθq r 1 dr ɛďrď1{ɛ S ˆ n 1 i sinpπrx θqθ j dσpθq r 1 dr ɛďrď1{ɛ S n 1 i sinpπrx θqr 1 dr θ j dσpθq. S n 1 ɛďrď1{ɛ Call the whole last expression f ɛ pxq. It is a fact that for 0 ă a ă b ă 8, b sin t dt ˇ t ˇ ď 4, thus for x 0, As 1 a f ɛ pxq ď 4ω n 1. f ɛpxq i sgn px θq π S θ jdσpθq, n 1 applying the dominated convergence theorem yields φpxq e πix ξ ξ j dξ dx R n ɛď ξ ď1{ɛ ξ n`1 ˆ φpxq i sgn px θq π R n S θ jdσpθq dx n 1 i π ˆ φpxq sgn px θqθ j dσpθq dx. R n S n 1 Then using Lemma 5 and putting the above together we get A W j, φ p E Γ ` n`1 i π ˆ φpxq sgn px θqθ j dσpθq dx R n S n 1 i π Γ ` n`1 φpxq ω n x j R n 1 x dx. n We work out that π Γ ` n`1 ω n n 1 1, 1 Loukas Grafakos, Classical Fourier Analysis, second ed., p. 63, Exercise

11 and therefore completing the proof. A W j, φ p E i φpxq x j R x dx, n Because R j h h W j, A yrj h, φe A E A ph Wj x, φ xwj, p E hφ Theorem 7. For 1 ď j ď n and for h P S pr n q, yr j hpξq i ξ j ξ p hpξq, ξ P R n. R n i p hpξqφpξq ξ j ξ dξ. In other words, the multiplier of the Riesz transform R j is m j pξq i ξj ξ. 4 Properties of the Riesz transform Theorem 8. where Iphq h for h P S pr n q. nÿ I Rj, Proof. For h P S pr n q, hence yr j hpξq i ξ j ξ y R j hpξq i ξ j ξ i ξ j ξ p hpξq ξ j ξ p hpξq, nÿ Taking the inverse Fourier transform, yr j h ph. nÿ Rj h h, i.e. nÿ Rj I. 11

12 For a tempered distribution u, for 1 ď j ď n, we define xb j u, φy p 1q xu, B j φy, φ P S pr n q. It is a fact that B j u is itself a tempered distribution. One proves that yb j u pπiξ j qpu. Each side of the above equation is a tempered distribution. Then nÿ n x u yb j u ÿ ÿ n pπiξ j q pu 4π ξj pu 4π ξ pu. Suppose that f is a Schwartz function and that u is a tempered distribution satisfying u f, called Poisson s equation. Then For 1 ď j, k ď n, Using Theorem 7, Therefore 4π ξ pu p f. B j B k u F 1 pf pb j B k uqq F 1 ppπiξ j qpπiξ k qpuq F 1 4π ξ j ξ k ˆξj F 1 ξ k ξ p f. pf 4π ξ R j R k f F 1 F pr j R k fq ˆ F 1 i ξ j R ξ y k f ˆ F 1 i ξ j ξ i ξ k F 1 ˆ ξjξ k ξ p f B j B k u R j R k f.. f ξ p 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, B j B k u R j R k f. 1