Harmonic and Real Analysis Herbert Koch Universit¨Atbonn Wintersemester 2014-2015

Notes for Harmonic and real analysis Herbert Koch Universit¨atBonn Wintersemester 2014-2015

Recommended literature: [10, 7, 14, 13, 15]

Contents

Chapter 1. Introduction 5 1. An example 5 Chapter 2. Fourier series 9 1. Deﬁnitions 9 2. The convolution and Young’s inequality 10 3. Lp convergence of partial sums 15 4. Regularity and Fourier series 16 5. Complex Interpolation 17 6. The Hardy-Littlewood maximal function and real interpolation 21 7. Higher dimensions 30 Chapter 3. Harmonic functions and the Poisson kernel 31 1. Basic properties and the Poisson kernel 31 2. Boundary behaviour of harmonic functions 36 3. Almost everywhere convergence 37 4. Subharmonic functions 38 5. The theorems of Riesz 40 6. Conjugate functions 41 7. The Hilbert transform 43

n Chapter 4. The Fourier transform on R 47 1. Definition and first properties 47 2. The Fourier transform of Schwartz functions 48 3. The Fourier transform on tempered distributions 51 4. Oscillatory integrals and stationary phase 58 5. Quantization 64 6. Cotlar’s Lemma and the Theorem of Calderón-Vaillancourt 70 Chapter 5. Singular integrals of Calderón-Zygmund type 75 1. The setting 75 2. The Calderón-Zygmund theorem 77 3. Examples 81 Chapter 6. Hardy and BMO 87 1. More general maximal functions and the Hardy space Hp 87 2. The atomic decomposition 91 3. Duality and BMO 95 4. Relation to harmonic functions 98 5. Div-curl type results 100 6. Application to elliptic PDEs 101 7. Pointwise estimates and perturbations of elliptic equations 107

3 4 CONTENTS

Chapter 7. Littlewood-Paley theory and square functions 111 1. The range 1 < p < ∞ 111 2. Square functions, tents and Carleson measures 114 Bibliography 117 CHAPTER 1

Introduction

Harmonic analysis is concerned with describing, decomposing and ana- lyzing functions and operators with some ’structure’ coming from the structure of the Euclidean space. Its relevance comes from the insight that the same structures are relevant in different areas of mathematics like partial differential equations, signal processing, Fourier analysis and mathematical physics. Key notions are Fourier transform, maximal functions, square function, BMO and Hardy spaces, Calderón-Zygmund operators and their relation to partial differential equations.

1. An example The series ∞ X 1 (1.1) zn n n=1 converges absolute for |z| < 1, and uniformly on every ball of radius < 1 since m m X 1 X 1 − |z|m+1 1 f(z) = |z|n ≤ |z|n = ≤ . n 1 − |z| 1 − |z| n=1 n=0 I claim that, given ε > 0, it converges uniformly in {z : |z| ≤ 1, |z − 1| > ε}. This is proven by an argument going back to Abel and mimics an integration by parts: m−1 m−1 m−1 j ! X zj 1 X X X 1 1 = zj + zk − j m j j + 1 j=n j=n j=n k=n m−1 m−1 1 X X zn − zj+1 1 1 = zj + − . m 1 − z j j + 1 j=n j=n To verify the formula, compare the coeﬃcients of zj, n ≤ j ≤ m − 1: m−1 1 1 X 1 1 = + − . j m l l + 1 l=j Then

m j X z 4 ≤ j nε j=n 5 6 1. INTRODUCTION and the partial sums are a Cauchy sequence. The limit is continuous, as a uniform limit of continuous functions. It is not hard to identify it: 1 f 0(x) = 1 − x which is the derivative of − ln(1 − x) and a check at x = 0 shows that f(x) = − ln(1 − x). Hence Im(1 − z) f(z) = − ln (1 − z) = − ln |1 − z| − i arctan C Re(1 − z) Im z = − ln |1 − z| + i arctan 1 − Re z for |z| ≤ 1, z 6= 1. Deﬁne ∞ ∞ X (eix)n X einx 1 sin(x) g(x) := = = − ln |2 − 2 cos x| + i arctan n n 2 1 − cos(x) n=1 n=1 since p |1 − eix| = (1 − cos(x))2 + sin(x)2 = p2 − 2 cos(x). By elementary geometry of the inscribed angle or the addition theorem sin(x) 2 sin(x/2) cos(x/2) = 1 − cos(x) 1 − cos2(x/2) + sin2(x/2) π−x cos(x/2) sin( 2 ) π − x = = π−x = tan( ) sin(x/2) cos( 2 ) 2 and we obtain for the imaginary part ∞ X sin(nx) π − x (1.2) h (x) := = . 0 n 2 n=1 for 0 < x < 2π. Deﬁne the absolute convergent series ∞ X cos(nx) h (x) = . 1 n2 n=1 and for 0 < x < 2π n Z x Z x X sin(jx) h0(t)dt = lim dx n→∞ j π π j=1  n n  X cos(jx) X cos(jπ) = − lim  +  n→∞ j2 j2 j=1 j=1

= − h1(x) + h1(π) and −h1 is a primitive of h0. Thus there exists a ∈ R such that x2 πx h (x) = − + a 1 4 2 1. AN EXAMPLE 7

R 2π 1 and we want to determine a. Since 0 cos(nt)dt = n (sin(2πn)+sin(0)) = 0 we get Z 2π 3 2π 3 0 = h1(t)dt = − π + 2aπ 0 3 π2 and a = 6 . The evaluation at t = 0 gives the value of the Riemann ζ function at the value 2. Lemma 1.1. ∞ X 1 π2 (1.3) ζ(2) = = . n2 6 n=1 Similarly ∞ X 1 π4 (1.4) ζ(4) = = . n4 90 n=1

We have seen diﬀerent notions of convergence. (1) Point-wise convergence (2) Uniform convergence (3) Absolute convergence (4) Convergence in L2 in connection with Fourier series. The notions are important since we want to diﬀerentiate resp. integrate term by term.

CHAPTER 2

Fourier series

1. Deﬁnitions

We denote the one dimensional torus by T = R\Z. Functions on the torus are identiﬁed with periodic functions on R with period 1. We denote the space of Radon measures ( complex Borel measures which are ﬁnite on compact sets) on a metric space X by M - these are objects which can be written as

µ+ + µ− + iµi− − iµi− with non negative Radon measures µ+, µ−, µi+ and µi−.

Definition 2.1. Let µ ∈ T . We deﬁne the Fourier coeﬃcients

Z 1 −2inπx µˆ(n) =µ ˆn = e µ 0 and we write formally

∞ X µ ∼ µˆ(n)e2inπx n=−∞ for the relation between the series and the measure µ. If f is an integrable function we write

Z 1 −2inπx fˆ(n) = fˆn = e f(x)dx 0 and ∞ X f ∼ fˆ(n)e2inπx. n=−∞

2inπx The functions (e )n are orthonormal in the sense that

Z 1 if n = m e\2iπnx(m) e2inπxe2imπxdx = 0 if n 6= m

The convergence questions of the Fourier series are interesting, important, and a prototype for similar question in all areas of analysis.

9 10 2. FOURIER SERIES

We deﬁne the Dirichlet kernel through N X ˆ 2inπx SN f(x) = f(n)e n=−N N Z X 0 = e−2inπx f(x0)dx0e2inπx n=−N T Z X 0 = e2inπ(x−x )f(x0)dx0 T n=−N N Z 0 0 0 = DN (x − x )f(x )dx T =DN ∗ f(x) where N X sin[(2N + 1)πx] D (x) = einx = . N sin πx n=−N We may formulate the convergence question as: When and in which sense does DN ∗ f(x) converge to f(x)?

2. The convolution and Young’s inequality Motivated by the occurrence of the convolution we take a closer look n at its properties. For X = T, R or R we deﬁne the convolution of two continuous functions with compact support by Z f ∗ g(x) = f(x − y)g(y)dy.

We will need the convolution in much more general context. For that we consider Z If,g,h = f(x)g(x − y)h(y)dxdy Rn×Rn Lemma 2.2. Suppose that 1 ≤ p, q, r ≤ ∞ and 1 1 1 + + = 2. p q r Then the integral defining I is integrable for all f ∈ Lp, g ∈ Lq and h ∈ Lr and |If,g,h| ≤ kfkLp kgkLq khkLr Proof. The case when one of the exponents is ∞ and the others are 1 is simple. Hence we may assume that 1 ≤ p, q, r < ∞. Let p/r0 q/r0 F1(x, y) =|f(x)| |g(x − y)| , q/p0 r/p0 F2(x, y) =|g(x − y)| |h(y)| , p/q0 r/q0 F3(x, y) =|f(x)| |h(y)| where 0 denotes the Hölderdual exponent 1 1 1 1 1 1 + = + = + = 1, p p0 q q0 r r0 2. THE CONVOLUTION AND YOUNG’S INEQUALITY 11 with the obvious interpretation of the exponent ∞. Then, applying Hölder’s 1 1 1 inequality twice, since p0 + q0 + r0 = 1 Z Ifgh ≤ F1F2F3dxdy Z 1/q q ≤ (F1F2) dxdy kF3kLq0

≤kF1kLr0 kF2kLp0 kF3kLq0 p q q q p r r0 r0 p0 p0 q0 q0 =kfkLp kgkLq kgkLq khkLr kfkLp khkLr =kfkLp kgkLq khkLr The lemma is the ’dual’ statement to Young’s inequality. Proposition 2.3 (Young’s inequality). Suppose that 1 ≤ p, q, r ≤ ∞ and 1 1 1 + = 1 + . p q r If f ∈ Lp and g ∈ Lq then for almost all x the integrand of Z f(x − y)g(y)dy

r n is integrable and it deﬁnes a function in L (R ). Moreover

p q kf ∗ gkLr(Rn) ≤ kfkL kgkL . If r = ∞ then the integrand is integrable for all x and f ∗ g is a bounded continuous function.

0 Proof. Let h ∈ Lr . By Lemma 2.2 Z

h(x)f(x − y)g(y)dxdy ≤ kfkLp kgkLq khk r0 . L It is an exercise to work out the details, including the last statement. The convolution has nice algebraic properties. (1) f ∗ g(x) = g ∗ f(x) (2)( f ∗ g) ∗ h(x) = f ∗ (g ∗ h)(x) (3) f ∗ (g + h)(x) = f ∗ g(x) + f ∗ h(x) ˆ (4) f[∗ gn = fngˆn for integrable functions on the torus. The Dirichlet kernel satisﬁes 1 1 |D (x)| ≤ c min{|N|, max{ , }} N x 1 − x 1 1 for 0 ≤ x ≤ 1. If 100N ≤ |x| ≤ 2 this follows from 1 |D (x)| ≤ N | sin πx| 1 and for |x| ≤ 100N it follows from the Taylor expansion of sin Nπx. We claim that there exists a constant c > 0 so that

ln(1 + N)/c ≤ kDN kL1 ≤ c ln(1 + N). 12 2. FOURIER SERIES

The upper estimate follows by integrating the upper bound. For the lower bound we restrict the integration to 1 1 {x ∈ [0, 1] : d(Nx − , ) ≤ }. 2 Z 100 For those values x −1 |DN (x)| ≥ c min{N, (sin πx) } and integration gives the lower bound.

Definition 2.4. Let 0 < s < 1, X a metric space. We say that : X → C is H¨oldercontinuous with exponent s if |f(x) − f(y)| ≤ cd(x, y)s. The best constant is the H¨oldersemi-norm.

Lemma 2.5. If f : T → C is H¨oldercontinuous with exponent s then

DN ∗ f → f uniformly as N → ∞. Proof. A term-wise integration shows that R D (x)dx = 1. Thus T N Z 1

|DN ∗ f(x) − f(x)| = (f(x − y) − f(x))DN (y)dy 0 Z δ ≤ |DN (y)||(f(x − y) − f(x)|dy −δ Z 1−δ

+ DN (y)(f(x − y) − f(x))dy δ Z δ −1 s ≤2c |y| |y| dykfkC˙ s −δ Z 1−δ f(x − y) − f(x) + sin((2N + 1)πy)dy δ sin(πy) Z 1−δ 4c s ≤ δ kfkC˙ s + hx(y) sin((2N + 1)πy)dy s δ It remains to estimate the second term on the right hand side. We denote 1 it by B and use that sin((2N + 1)πy) = − sin((2N + 1)π(y + 2N+1 )). Thus Z 1−δ 1 B ≤ (hx(y) − hx(y − ) sin((2N + 1)πy)dy δ 2N + 1 1−δ+ 1 Z Z 2N+1 + |hx(y)|dy + hx(y)dy 1 δ− 2N+1 1−δ −s −1 −1 ≤(2N + 1) δ kfkC˙ s + (2N + 1) δ−2kfksup. − s The estimate holds for all δ. We choose δ = N 3 and obtain 2 − s − 2s −1+ 2s |DN f(x) − f(x)| ≤ c N 3 + N 3 + N 3 .

2. THE CONVOLUTION AND YOUNG’S INEQUALITY 13

We can improve the convergence by using Ces´aromeans resp. the Fej´er kernel.

N−1 1 X σ f = D ∗ f(x) N N n n=1 KN ∗ f(x) where N−1 2 X 1 sin Nπx K = D (x) = . N n N sin πx n=0 It satisfies |n| KˆN (n) = 1 − N + which is seen by checking the Fourier transform C (2.1) 0 ≤ K (x) ≤ min{N 2, | sin(πx)|−2} N N by definition (left hand side) and as for DN (right hand side), Z 1 Z 1 Kn(x) = |KN (x)| = 1 0 0 which we can read off the Fourier coefficients and

0 2 −2 (2.2) |KN (x)| ≤ C min{N , | sin(πx)| }. 1 1 which is a straight forward calculation for 100N ≤ x ≤ 1 − 100N , and which follows from Taylor expansion in the remaining interval. We call a function approximate identity.

n Definition 2.6. The family Φn of functions on T, R or R is called approximate identity if R 1 (1) 0 Φn(x)dx = 1 R 1 (2) supN 0 |Φn(x)|dx < ∞ R 1−δ (3) For all δ > 0 one has δ |Φn(x)|dx → 0 as n → ∞

Proposition 2.7. Let Φn be an approximate identity. If f ∈ C(T) then

Φn ∗ f → f

p uniformly as n → ∞. If f ∈ L (T), 1 ≤ p < ∞, then

kΦn ∗ f − fkLp(T) → 0 as n → ∞. If µ ∈ M(T) then

Φn ∗ µ → µ in the sense of measures. 14 2. FOURIER SERIES

Proof. The proof uses some constructions which we will need often. Let f : T → R be continuous, and Φn and approximate identity. Then Z 1 f(x) − f ∗ Φn(x) = (f(x) − f(x − y))Φn(y)dy 0 Z δ = (f(x) − f(x − y))Φn(y)dy −δ Z 1−δ + (f(x) − f(x − y))Φn(y)dy δ We deﬁne the modulus of continuity

ωf (t) = sup |f(x) − f(y))|. |x−y|

Then limt→0 ωf (t) = 0 since f is uniformly continuous. Then Z 1−δ |f(x) − f ∗ Φn(x)| ≤ 2ω(δ)kΦnkL1 + 2kfksup |Φn(y)|dy −δ and

lim sup sup |f(x) − f ∗ Φn(x)| ≤ 2ω(δ) sup kΦnkL1 n→∞ x n This holds for all δ, hence the lim sup is zero. Assume now that f ∈ Lp for some 1 ≤ p < ∞. Continuous functions are p dense in L (T). Given ε there exists a continuous function fε such that

kf − fεkLp(T) ≤ ε. Then

p p I = kf − f ∗ ΦnkL ≤ kfε − fε ∗ ΦnkL + kf − fεkLp(T) + k(f − fε) ∗ ΦnkLp(T). By Young’s inequality with p = r and q = 1

p k(f − fε) ∗ ΦnkLp(T) ≤ kΦnkL1 kf − fεkL Then

In ≤ kfε − fε ∗ Φnksup + ε(1 + kΦnkL1 ) and

lim sup In ≤ lim sup kfε − fε ∗ Φnksup + ε(1 + kΦnkL1 ) n→∞ n→∞ ≤ε(1 + kΦnkL1 ) This holds for all ε and hence the lim sup is 0. The convolution of a measure with an L1 function is a integrable function. The extension of the dual Young’s inequality is easy. Let µ be a measure and h a continuous function. If (Φn) is an approximate identity then the same is true for Φ˜ n(t) = Φn(−t) and Z Z Z µ ∗ Φnh(x)dx = Φ˜ n ∗ h(x)µ → hµ

This is the deﬁnition of the convergence of measures. 3. Lp CONVERGENCE OF PARTIAL SUMS 15

Corollary 2.8. The trigonometric polynomials are dense in Lp for 2 1 ≤ p < ∞ and in C(T). For f ∈ L the identity of Plancherel ∞ 2 X ˆ 2 (2.3) kfkL2 = |fn| n=−∞ 2πinx 2 holds. The complex exponentials (e )n are an orthonormal basis of L (T) and for f, g ∈ L2 one has Parsevals identity Z X fgdx¯ = fˆngˆn.

2 The inner product of L (T) is given by the formula in the corollary. The 2 space is a Hilbert space i.e. kfkL2 = hf, fi is a norm, and the space if complete with this norm.

Proof. (Kn) is an approximate identity. By Proposition 2.7 we have p p f ∗ Kn → f in L for any f ∈ L if p < ∞. Now Kˆn is a trigonometric polynomial, and hence f ∗Kn is a trigonometric polynomial. This implies the density. The identity of Plancherel is trivial for trigonometric polynomials: When we expand them and integrate all nondiagonal terms will give 0. Now n ∞ 2 2 X ˆ 2 j 2 X ˆ 2 kfk 2 = lim kf ∗ Knk 2 = lim |fj| (1 − ) = |fj| L n→∞ L n→∞ n j=−n j=−∞ 2 by monotone convergence. Let f, g ∈ L (T). Then, using Plancherels formula 1 Z 1 1 Z 1 fg¯ + gfdt¯ = |f + g|2 − |f − g|2dx 2 0 4 0 ∞ 1 X = |fˆ +g ˆ |2 − |fˆ − gˆ |2 4 n n n n j=−∞ ∞ 1 X = fˆ gˆ +g ˆ fˆ 2 n n n n j=−∞ which is Parsevals identity for the real part for the formula. We replace f by if to obtain imaginary part of the formula. The complex exponentials e2πinx are clearly orthonormal. if R fe2πinxdx = n for all n then f = 0 by Plancherels identity. 3. Lp convergence of partial sums Proposition 2.9. Let 1 ≤ p < ∞. Then the following is equivalent: p (1) For all f ∈ L we have kDN ∗ f − fkLp → 0 (2) sup kDN ∗fkLp < ∞. n kfkLp Proof. (1) =⇒ (2) is a consequence of the uniform boundedness principle. The reverse implication follows from the density of trigonometric polynomials and Young’s inequality. 1 Corollary 2.10. There exists f ∈ L so that DN ∗ f does not converge 1 to f in L . There exists f ∈ C(T) so that DN ∗f does not converge uniformly to f. 16 2. FOURIER SERIES

Proof. We have seen that

kDN kL1 ≥ c ln(1 + N). −1 1 Let fδ(x) = (2δ) χ|x|≤δ. Then kfkL1 = 1 and, if δ < 4N , kDN ∗ fkL1 ≥ c ln(1 + N). This contradicts (2) and hence there exists f ∈ L1 so that 1 DN (−y) DN ∗ f does not converge to f in L . If f(y) = if DN (y) 6= 0 and |DN (−y)| 0 otherwise then

|DN ∗ f(0)| ≥ c ln(1 + N). By dominated convergence

DN lim DN ∗ (0) → DN ∗ f(0) ε→0 ε + |DN | and hence

sup sup kDN ∗ fksup = ∞. N≥1 f∈C(T),|f|≤1 Arguing as in the proposition with the uniform boundedness principle this implies that there is f ∈ C(T) such that DN ∗f does not converge uniformly to f. The variant of the argument gives a stronger statement: There exist 1 f ∈ L (resp. f ∈ C(T) so that lim supN→∞ kDN ∗ fkL1 → ∞ (resp. lim supN→∞ kDN ∗ fksup = ∞). Exercise: Find such functions.

4. Regularity and Fourier series Proposition 2.11 (Bernstein’s inequality). Let f be a trigonometric polynomial with fˆ(k) = 0 for |k| > n. Then 0 kf kLp ≤ CnkfkLp Proof. We deﬁne the de la Vall´eePoussin kernel 2N k−1 1 X X V (x) = e2πijx N N k=N+1 j=1−k

=2K2N (x) − KN (x) 1 sin2 2Nπx − sin2 Nπx = N sin2 πx Then, by Young’s inequality (and a tedious bound of the L1 norm) 0 0 kf kLp =k(VN ∗ f) kLp 0 =kVN ∗ fkLp 0 ≤kVN kL1 kfkLp ≤CNkfkLp .

We deﬁne Sobolev spaces on the torus. 5. COMPLEX INTERPOLATION 17

s Definition 2.12. Let s ∈ R. We deﬁne the Hilbert space H (T) as the completion of the trigonometric polynomials in the norm 1/2  ∞  X 2 s ˆ 2 kfkHs :  (1 + n ) |f(n)|  . j=−∞

s 0 2 1 0 2 If s ≥ 0 then H (T) ⊂ H (T) = L (T). If f ∈ H then f ∈ L and 0 2 2 2 kf kL2 + kfkL2 = kfkH1 . This holds for trigonometric polynomials, and we use it to deﬁne the derivative of functions in H1. Theorem 2.13. Let 0 < s ≤ 1. Then Cs ⊂ Hσ for all σ < s. Proof. We claim that X 2 −2js 2 (2.4) |f(n)| ≤ c2 kfk s . C (T) 2j ≤|n|≤2j+1 Then

∞ ∞ ∞ X 2σ ˆ 2 X 2jσ X 2 X 2j(σ−s) |n| |f(n)| ≤ 2 2 |f(n)| ≤ C 2 kfkCs n=−∞ j=0 2j ≤|n|≤2j+1 j=0 To prove the claim (2.4) we observe that it follows from

kKn ∗ f − fkL2 ≤ sup |Kn ∗ f(x) − f(x)| x Z 1/2 s ≤2 |Kn(y)||y| dykfkCs 0 −s ≤cn kfkCs (see (2.1)) with n = 2j−1 since X 2 |f(n)| ≤ kK2j−1 ∗ f − fkL2 . 2j ≤|n|≤2j+1 5. Complex Interpolation Holomorphic functions satisfy a maximum principle.

Lemma 2.14. Let U ⊂ C be open, f : U → C holomorphic. If z0 ∈ U and |f(z0)| = max{|f(z)| : z ∈ U} then f is constant.

Proof. 1) There is nothing to show if f(z0) = 0. Otherwise we divide by f(z0) and may assume that f(z0) = 1 The Taylor series ∞ X j f(z) = 1 + aj(z − z0) j=1 18 2. FOURIER SERIES converges in a neighborhood of z0. There is a ﬁrst non-vanishing coeﬃcient m am with m ≥ 1. There exists b with b = am. We consider g(z) = f(z/b). It has the form ∞ m X n g(z) = 1 + z + anz . n=m+1 Then d |g(1 + t)| = m dt t=0 and we obtain a contradiction. 2) By the Cauchy integral formula, resp. the mean value property of harmonic functions, for r > 0 and small Z 2π 1 it f(z0) = f(z0 + re )dt 2π 0 This implies that the maximum is assumed at the boundary. If it is assumed in an interior point then |f(z)| is constant and the holomorphic function z → lnC f(z) has constant real part. The Cauchy-Riemann equations imply now that f is constant. Lemma 2.15 (Three lines theorem). Let f : {z ∈ C : 0 ≤ Re z ≤ 1} → C be bounded, continuous and holomorphic in the interior. Then 1−x x |f(x + iy)| ≤ sup |f(y)| sup |f(1 + iy)| . y y Proof. Let z−1 −z εz2 fε(z) = e f(z) sup |f(iy)| sup |f(1 + iy)| . y y Then ε(1−y2) |fε| ≤ ce → 0 as y → 0.

Thus there exists z = x + iy with 0 ≤ x ≤ 1 where |fε| is maximal. Then either fε is constant and hence identically 0, or the maximum is assumed at the boundary. But then

sup |fε(z)| ≤ max{sup |fε(iy)|, sup |fε(1 + iy)|} y y and hence 1−x x |f(z)| ≤ eε sup |f(y)| sup |f(1 + iy)| y y Theorem 2.16 (Riesz-Thorin interpolation theorem). Let µ and ν be measures on X resp. Y , Tz maps characteristic functions of measurable sets of bounded µ measure to functions which are integrable over sets of ﬁnite measure, for 0 ≤ Re z ≤ 1 such that for all measurable sets A ⊂ X and B ⊂ Y of ﬁnite measure the map Z z → (TzχA(y))χB(y)dy 5. COMPLEX INTERPOLATION 19 is bounded and continuous and holomorphic in the interior of the strip. Sup- pose that 1 ≤ p0, p1, q0, q1 ≤ ∞ and, 1 Re z 1 − Re z = + pz p1 p0 1 Re z 1 − Re z = + . qz q1 q0 Suppose that Z

(Tiy0 f)gν ≤ C0kfkLp0 kgk q0 L 0 Z

(T1+iy0 f)gν ≤ C1kfkLp1 kgk q0 . L 1 Then Tλ has a unique extension to a continuous linear operator

pz qz Tλ : L (µ) → L (ν) 1−Re z Re z with norm C0 C1 . Proof. By duality and density it suffices to prove Z 1−Re z Re z (Tzf)gν ≤ C C kfkLp kgk q0 0 1 L for 0 < t < 1, for finite sums of characteristic functions of sets of finite measure. We fix z0, p = pz0 , q = qz0 , f and g a finite sum of characteristic functions of sets of finite measure on X resp Y ,

q0−q0 t−z p−p t−z 0 0 0 t p t q fz(x) = |f(x)| 0 f(x), gλ(y) = |g(x)| 0 g(y)

ft(x) = f(x) gt(y) = g(y) 0 0 p0 p q q |fiy0 (x)| = |f(x)| , |giy0 (y)| 0 = |g(y)| p p (Re z0 − 1) − 1 = Re z0 − 1 p0 p1 0 0 p1 p q q |f1+iy0 (x)| = |f(x)| , |g1+iy0 (y)| 1 = |g(y)| . We taking the diﬀerence quotients and using dominated convergence d Z Z d Z d (T f )(y)g (y)ν(y) = ( T )f (y)g (y)ν(y) + (T f )g ν(y) dz z z z dz z z z λ dz z z Z d + (T f ) g ν(y) z z dz z R and hence Tzfzgzν is holomorphic in z. Boundedness and continuity are immediate. By H¨older’sinequality and the construction Z

(Tiy0 fiy0 )giy0 ν(y) ≤kTiy0 fiy0 kLq0 kgiy0 k q0 L 0

≤c0kf 0 kLp0 kg 0 k q0 iy iy L 0 q0/q0 ≤c kfkp/p0 kgk 0 . 0 Lp Lq0 20 2. FOURIER SERIES and Z 0 0 p/p1 q /q1 (T1+iy0 f1+iy0 )g1+iy0 dy ≤ C1kfk p kgk 0 . L Lq By the three lines theorem

Z 0 1−Re z0 0 0 Re z0 p/p0 q/q0 p/p1 q /q1 Tz0 fg ≤ C0kfkLp kgk q0 C1kfkLp kgk q0 L 0 L (1−Re z ) Re z 1−Re z0 Re z0 q0 0 + 0 p + q0 q0 =C1−Re z0 CRe z0 kfk p0 p1 kgk 0 1 0 1 Lp Lq0 1−Re z0 Re z0 =C0 C1 kfkLp kgkLq0 . The lemma of Schur is a consequence.

Lemma 2.17 (Schur). Let K : X × Y → C be µ × ν measurable and suppose that Z Z sup |K(x, y)|ν(y) ≤ C1, sup |K(x, y)|µ(x) ≤ C∞ x y Let 1 ≤ p ≤ ∞ The linear map defined on characteristic functions by Z T (χA)(y) = K(x, y)dµ(x) A has a unique extension to a continuous linear map T : Lp(X, µ) → Lp(Y, ν) which satisfies 1 1 p 1− p kT fkLp(Y,ν) ≤ C1 C∞ kfkLp(X,µ). ∞ 1 Proof. If f ∈ L (X, µ) then |T f(y)| ≤ C∞kfkL∞ . If f ∈ L then Z Z Z | T f(y)|ν(y) ≤ |K(x, y)||f(x)|µ(x)ν(y) Z ≤ sup |K(x, y)|dν(y)kfkL1(X,µ) x ≤C1kfkL1 . This holds first for simple functions. There is a unique extension to T : 1 1 ∞ L (X, µ) → L (Y, ν) bounded by C1 and a unique extension to L (X, µ) → ∞ L (Y, ν) bounded by C∞. The Riesz-Thorin theorem implies the full statement. Alternatively we may estimate as in Young’s inequality:

f(x)K(x, y)g(y)dµ(x)dν(y)

1/p 1/p0 ≤kf(x)|K(x, y)| kLp(µ×ν)k|K(x − y)| g(y)kLp0 (µ×ν) 1/p 1/p0 ≤kfkLp(µ)kgkLp0 (ν) sup k|K(x, .)| kLp(ν) sup k|K(., y()| kLp0 (µ). x < This second proof gives a stronger statement: Existence of the deﬁning integral for almost all y. 6. THE HARDY-LITTLEWOOD MAXIMAL FUNCTION AND REAL INTERPOLATION21

We could derive the bound in Young’s inequality via the Theorem of Riesz-Thorin from the extremal cases p = 1, q = ∞ and p = 1, q = 1. With this type of argument we would however loose the existence of the integral for almost all y. Up to that Young’s inequality with p = 1 is a special case of Schur’s lemma.

p Lemma 2.18 (The Hausdorﬀ-Young inequality). Let f ∈ L (T, C), 1 ≤ p ≤ 2. Then ˆ kfklp0 ≤ kfkLp Proof. The case p = 2 is the Plancherel identity. The case p = 1 is trivial. If 1 1 − λ = λ + p 2 then 1 1 − λ = p0 2 The assertion follows from Theorem 2.16 with µ the Lebesgue measure and ν the counting measure.

6. The Hardy-Littlewood maximal function and real interpolation 1 n Definition 2.19. Let f ∈ Lloc(R ). We define the (uncentered) Hardy- Littlewood maximal function as Z −1 Mf(x) = sup |BR(y)| |f(z)|dz x3BR(y) BR(y) Theorem 2.20 (Hardy-Littlewood maximal function). Let 1 ≤ p ≤ ∞ p n and f ∈ L (R ). The Hardy-Littlewood maximal function Mf is measurable and finite almost everywhere. It satisfies p kMfk p n ≤ c(n) kfk p n . L (R ) p − 1 L (R ) and n n 3 m ({x : Mf(x) > λ}) ≤ kMfk 1 λ L 1 n Definition 2.21. Let f ∈ Lloc. We call x ∈ R Lebesgue point if Z Z −1 −1 lim sup (BR(x1)) f(y)dy − (Br(x2)) f(y)dy = 0 R→0 BR(x1)∩Br(x2)3x,r≤R We may (and usually do ) assume that at a Lebesgue point f is equal to the limit of the averages as R → 0.

1 Theorem 2.22. Let f ∈ Lloc. Then there is a set A of measure 0 so n that all point in R \A are Lebesgue points. Proof. We may assume that f is compactly supported and integrable. Given ε > 0 there exists a continuous compactly supported function g with 22 2. FOURIER SERIES kf − gkL1 < ε. Since for continuous functions all points are Lebesgue points it remains to consider h = f − g. Then

At ={x : lim sup |fBR(x1) − fBr(x2)| > t} r,R

⊂ {x : lim sup |hBR(x1) − hBr(x2)| > t} r,R ⊂ {x : Mh(x) > t/2} cε and its measure is bounded by t . We let ε → 0 to see that n m (At) = 0. Denote the set of Lebesgue points by L. Then

n [ R \L = A1/j j is a countable union of sets of measure 0, hence its union has measure 0. Proof of Theorem 2.20. The proof consists of two steps, the ﬁrst being the proof of the weak type inequality for p = 1, and the second being an interpolation argument. Lemma 2.23 (Covering argument of Vitali). Let (X, d) be a metric space and (Bi)1≤i≤N = (Bri (xi))1≤i≤N by a ﬁnite set of balls. Then there exists a pairwise disjoint subset (Bij )1≤j≤M so that

N M [ [ B ⊂ B (x ). i 3rij ij i=1 j=1 Proof. We choose the balls recursively. Suppose we have chosen disjoint balls Bij for j < m. Let Bijm be the ball of the largest radius which is disjoint to the previous ones. If there is no such ball we are done. This process ends at some point. We have to verify the covering statement.

Let Bri (xi) be one of the balls. If it has been chosen it is certainly contained in the union to the right. If not there is a largest index m so that rim ≥ ri. Since we did not choose Bri (xi) in this step it has a nonempty intersection with one of the balls B with j ≤ m. But then r ≤ r and rij i ij B (x ) ⊂ B (x ). ri i 3rij ij

1 n Let f ∈ L (R ) and t > 0. The set U = {x : Mf > t} is open. For each point x ∈ U there is a ball Br(y) 3 x with Z n |f(y)|dy > λm (Br(y)). Br(y) Let K ⊂ U be a compact set. Since K is compact, and covered by these balls there exists a ﬁnite number of such balls (Bri (xi))1≤i≤N which cover K. By Lemma 2.23 there is a subset of disjoint balls (B (x )) rij ij

N M [ [ K ⊂ B (x ) ⊂ B (x ) ri i 3rij ij i=1 j=1 6. THE HARDY-LITTLEWOOD MAXIMAL FUNCTION AND REAL INTERPOLATION23

Thus  M  [ mn(K) ≤mn(K) B (x )  3rij ij  j=1

 M  [ ≤3nmn B (x )  rij ij  j=1 Z ≤3nt−1 |f| Mf>t 3n ≤ kfk 1 t L and n n n 3 m ({x : Mf(x) > t}) = sup m (K) ≤ kfkL1 . K t

Clearly Mf(t) ≤ kfkL∞ . Let 1 ≤ p < ∞. There is the bath tube representation Z Z ∞ (2.5) |f|dx = mn+1({(x, t) : 0 ≤ t < |f(x)|}) = mn({|f| > t})dt 0 and Z Z ∞ Z ∞ (2.6) |f|pdx = mn({|f| > s1/p})ds = p tp−1mn({|f| > t})dt. 0 0 Given t > 0 we deﬁne ( f if |f| ≤ t ft = f t |f| if |f| > t and t f = f − ft. The maximal function is sublinear, i.e. t Mf(x) ≤ Mft(x) + Mf (x). Thus Z ∞ p p p−1 n kMfkLp =2 p t m ({Mf > 2t})dt 0 Z ∞ p p−1 n n t ≤2 p t m ({Mft > t}) + m ({Mf > t}) dt 0 Z ∞ ≤2pp tp−1mn({Mf t > t})dt 0 Z ∞ Z ≤2p3n tp−2 |f|dxdt 0 |f|>t Z Z |f(x)| =2pp3n |f(x)| tp−2dt Rn 0 2p3n = kfkp p − 1 Lp 24 2. FOURIER SERIES and 23n 1/p kMfk p ≤ kfk p . L p − 1 L By Tschebychevs inequality −p p µ({|f| > λ}) ≤ λ kfkLp . We deﬁne the weak norm by 1/p (2.7) kfk p = sup λµ({|f| > λ}) Lw(µ) λ This is an abuse of notation, since the ’norm’ does in general not satisfy the triangle inequality but only (2.8) kf + gk p ≤ C kfk p + kgk p . Lw Lw Lw ∞ ∞ If p = ∞ we set Lw = L .

Theorem 2.24 (Marcinkiewicz). Let 1 ≤ p0 < p1 ≤ ∞, 1 ≤ q0 6= q1 ≤ ∞. Suppose that T is a sublinear operator mapping the span of characteristic function of µ measurable sets of ﬁnite to functions which are ν integrable over ν measurable sets of ﬁnite ν measure, such that,

(1) For all finite collections of measurable sets Aj ⊂ X and B ⊂ Y of finite measure the map Z N X R 3 (a) → T ajχAj ν B is continuous. (2) T (tf)(y) = tT (f)(y) for t > 0 and |T (f +g)(x)| ≤ |T f(x)|+|T g(x)| almost everywhere. R 1/p 1/q0 1/p 1/q0 (3) T (χA)χBdy ≤ min{c0µ(A) 0 µ 0 (B), c1µ(A) 1 µ 1 (B)} If 0 < λ < 1 1 1 − λ λ 1 1 − λ λ = + , = + p p0 p1 q q0 q1 p q then T defines a unique continuous sublinear operator from Lw to Lw.

kT fk q ≤ c(p, q, λ)kfk p Lw Lw and Z 1/p T (tχ )g(y)dν ≤ c(p, q, λ)|t|µ(A) kgk 0 A Lq Y w If p ≤ q then T : Lq → Lp and

kT fkLq ≤ c(p, q, λ)kfkLp . We apply the Marcinkiewicz interpolation theorem to an important version of Young’s inequality before we prove it. Lemma 2.25 (Weak Young’s inequality). Let 1 1 1 1 < p, q, r < ∞, + = 1 + p q r 6. THE HARDY-LITTLEWOOD MAXIMAL FUNCTION AND REAL INTERPOLATION25

p p and f ∈ L and g ∈ Lw. Then the integral Z f ∗ g(x) = f(x − y)g(y)dy exists for almost all x and

kf ∗ gk r ≤ ckfk p kgk q . L L Lw 1 Proof. Truncating g = g1 + g as above it is not hard to see that the integral exists for almost all x. By Young’s inequality

kf ∗ gkLr ≤ kfkLp kgkLq for all such triple. We ﬁx p and f ∈ Lp and deﬁne T g = f ∗ g, T : Lq → Lr for all admissible triple. Thus, by duality, Z

T ghdx ≤ kgkLq khk r0 L and we can apply the interpolation theorem of Markinciewicz for all admissible exponent beside q = 1 or r = ∞:

kf ∗ gk r ≤ ckfk p kgk q . Lw L Lw q Now we ﬁx g ∈ Lw and deﬁne p r T f = f ∗ g, T : Lw → Lw. This implies Z ˜ ˜ 1/p 1/r0 T fχAdx kfkL∞ |A| |B| , B and, since p ≤ r, kT fk r ≤ ckfk p kgk q L L Lw for triples which we obtain by interpolation, which are those of the lemma. The assumption of the theorem implies for all simple functions

Z 0 1/p0 1/q (2.9) T fχAgν ≤ ckfkL∞(µ)kgkL∞(ν)µ(A) ν(B) 0 B with a similar estimate for the index 1. Allowing for a factor 4 in the constant it suﬃces to verify this for nonnegative functions f and g. We can then write X X f = fjχAj , g = gjχBj j

P ∞ P ∞ with 0 ≤ fj, gj, fj = kfkL , gj = |gkL , Aj ⊂ Aj1 and Bj ⊂ Bj−1 for all j. We expand the sum and use the sublinearity to arrive at (2.9). p Lemma 2.26. If 1 < p < ∞ and f ∈ Lw the inequalities Z 1/p 1 −1/p0 1 kfk p ≤ sup µ(A) fµ ≤ kfk p Lw Lw 2 µ(A)>0 A p − 1 p hold. As a consequence there exists an equivalent norm so that Lw is a Banach space. 26 2. FOURIER SERIES

Proof. Set A = {|f(x)| > t}. Then Z Z ∞ tp|{|f(x)| > t}| ≤ tp−1|f|dµ ≤tp−1 µ({|f| > s})ds A t Z ∞ p−1 p −p ≤t kfkLp s ds w t 1 p ≤ kfk p p − 1 Lw Thus, for f > 0 Z p p−1 1/p0 −1/p0 kfkLp ≤ sup t µ(|f| > t) sup(µ(A)) fdµ w A A p p Z − 0 0 p−1 p0 p −1/p ≤t t kfkLp sup(µ(A)) fdµ w A A Z p−1 −1/p0 =kfkLp sup(µ(A)) fdµ w A A

p−1 which yields the left inequality in the lemma after dividing by kfk p . The Lw inequality on the right hand side follows from the second inequality above. The quantity in the middle satisﬁes the triangle inequality. Completeness is easy.

Let f be a simple function with kfk p = 1. Proof of Theorem 2.24. Lw We claim that

(2.10) kT fk q ≤ ckfk p Lw Lw This follows from Z 1/q0 T fχBdy ≤ ckfk p ν(B) Lw by Lemma 2.26

1 Z − 0 0 kT fk q ≤ c sup ν(B) q T fν ≤ c kfk p . Lw Lw B B

X f = fjχAj

j j+1 with fj(x) = f(x) if 2 ≤ |f| ≤ 2 . We decompose fj into the positive and negative part, and argue similar for both. Let fj ≥ 0. We can write it as X fjχAj = tlχCl with tl ≥ 0 and Cl+1 ⊂ Cl ⊂ Aj for all l. Then Z 1/p 1/q0 ∞ 0 0 T (fjχAj )ν ≤ ckfkL µ(Aj) µ(B) . B P since tj ≤ kfkL∞ . We obtain the same estimate with the index 1. 6. THE HARDY-LITTLEWOOD MAXIMAL FUNCTION AND REAL INTERPOLATION27

Let kfk p = 1 and s = ν(B). Then Lw Z Z X |T f|ν ≤c |T fχA |ν j B j B 0 0 X j 1/p0 1/q 1/p1 1/q ≤ 2 min{µ (Aj)ν 0 (B), µ (Aj)ν 1 (B)} j X j−jp/p 1/q0 j−jp/p 1/q0 ≤c min{2 0 s 0 , 2 1 s 1 }   X j−jp/p 1/q0 X j−jp/p 1/q0 ≤c  2 1 s 1 + 2 0 s 0  j≤J j>J p p J(1− ) 1/q0 J(1− ) 1/q0 ≤C(2 p1 s 1 + 2 p0 s 0 ) 0 ≤2Cs1/q if we choose J appropriately so that both summands have the same size: Jp 1 1 1 1 − = 0 − 0 ln2 s p0 p1 q0 q1 then (check if p = p0 and p = p1 , plus linearity) ln s p 2 1 J(1− p )+ 0 2 1 q1 = s q0 . The same argument yields Z 1/p (2.11) fχAg ≤ cµ(A) kfkL∞ kgk q0 . Lw We read (2.10) and (2.11) as L∞ resp L1 estimates. We deﬁne the Lorentz spaces rsp. norms Z q dt (2.12) kfkq := q tµ({x : |f(x)| > t})1/p . Lpq t Then kfk p = kfk p∞ ≤ kfk pq Lw L L since Z t tµ({|f(x)| > t})1/p ≤ qsq−1dsµ({|f(x)| > t})1/p 0 Z ∞ ds ≤q (sµ({|f| > s})1/p)q . 0 s and, if q1 < q2

q2 q1 q2−q1 q2 kfkLpq2 ≤ kfkLpq1 kfkLp∞ ≤ kLpq1 hence

(2.13) kfkLpq1 ≤ kfkLpq1 . Lemma 2.27. Let 1 < p < ∞ and 1 ≤ q ≤ ∞. Then the generalized H¨olderinequality Z

(2.14) fgµ ≤ ckfkLpq kgk p0q0 L 28 2. FOURIER SERIES holds and Z (2.15) kfkLpq ≤ c sup fgµ. kgk 0 0 ≤1 Lp q Proof. To prove (2.14) for functions f and g we deﬁne monotonically decreasing functions f ∗, g∗ : (0, ∞) → [0, ∞) so that |{f ∗ > s}| = µ({|f| > s}) for all s > 0. With this notation the Hardy-Littlewood inequality holds: Z Z ∞ (2.16) fgµ ≤ f ∗(t)g∗(t)dt. 0 We have Z Z Z ∞ Z ∞ |f(x)g(x)|µ = χ|f(x)|≥sχ|g(x)|≥tdsdt X 0 0 Z Z ∞ Z ∞ = χ{|f(x)|>s,|g(x)|>t}dsdt X 0 0 Z ∞ Z ∞ = µ({|f(x)| > s, g(x) > t})ds dt 0 0 Z ∞ Z ∞ ≤ m1({f ∗ > s} ∩ {|g∗ > t})ds dt 0 0 Z ∞ Z ∞ Z ∞ = χf ∗>s(u)χg∗>t(u)du dt ds 0 0 0 Z ∞ = f ∗(u)g∗(u)du. 0 Inequality (2.14) follows by H¨older’sinequality from this estimate: Z Z ∞ fgdµ ≤ f ∗g∗du 0 ∞ 1/q ∞ 1/q0 Z dt Z 0 0 0 dt ≤ tq/p(f ∗(t))q tq /p (g∗(t))q 0 t 0 t and, with the Riemann Stieltjes integral (or less transparently, by substitu- tion with s = f ∗(t)) q Z ∞ dt Z ∞ tq/p(f ∗(t))q = (f ∗(t))qdtq/p p 0 t 0 Z ∞ = tq/pd(f ∗)q 0 Z ∞ ds =q (sµ({|f| ≥ s})1/p)q . 0 s This completes the proof of (2.14). p p/p0 f Let q < ∞ and f ∈ L (µ). Then, with g = |f| |f| Z p p−1 fgµ = kfkLp , kgkLp0 = kfkLp . 6. THE HARDY-LITTLEWOOD MAXIMAL FUNCTION AND REAL INTERPOLATION29

pq P Let f ∈ L . We write it as f = fjχAj as above. The ﬁrst Ansatz is

X q−p 0 f g = 2jµ(A )1/p 2j(p/p ) χ . j |f| Aj j Then Z fgµ ∼ kfkLpq(µ). PN Recall that this sum is ﬁnite, j=−N . For κ > 0 we set

X q−p 0 f gκ = 2j(1+pκ)µ(A )1/p 2j(p/p ) χ . j |f| Bj j −κ(j+N) with Bj ⊂ Aj and µ(Bj) = 2 µ(Aj). A tedious calculation shows κ that kg kLpq(µ) is bounded independent of the the exponents provided κ is suﬃciently large. The case q = ∞ follows from Lemma 2.26 and

p q kfkLpq(µ) ∼ kkfjχAj kL kl . We read (2.11) as 1/q 1/p kT fkLq1 = ktν({y : |T f(y)| > t}) kL1 ≤ cktµ({x : |f(x)| > t) kL1 Now the real interpolation argument of for the maximal function with a small modiﬁcation in the deﬁnition of ft which yields

(2.17) kT fkLqr ≤ kfkLpr . for 1 ≤ r ≤ ∞. If p ≤ q we set r = p and conclude by

kT fkLq ≤ kT fkLqp ≤ ckfkLp . The transition from simple functions to general functions is an approximation argument. It suffices to prove norm continuity on simple functions. Let p p q0 p ≤ q and f, g ∈ L (µ). The argument is similar for Lw. Let h ∈ L . It suffices to show Z | (T f − T g))hν| ≤ ckhkLq0 kf − gkLp . Let Z φ(f) = | T fhν|. It satisfies for t > 0 φ(tf) = tφ(f), 0 ≤ φ(f + g) ≤ φ(f) + φ(g) and

φ(f) ≤ ckfkLp(µ)khkLq0 (ν). The continuity follows from

|φ(f) − φ(g)| ≤ ckhkLq0 (ν)kf − gkLp(ν). Continuity on simple functions of Z T fhdµ 30 2. FOURIER SERIES follows from the continuity assumption of the theorem. We consider Z η(t) = T (f + t(g − f))hdµ for 0 ≤ t ≤ 1. If η(t) = 0 for some t then Z Z

T ghµ + T fhµ ≤ ckhk q0 kf − gkLp . L Otherwise it does not change sign, and the assertion follows again from the continuity of φ. 7. Higher dimensions The definition of the convolution and the Fourier transform carries over to several different situations. n 7.1. Fourier transform on T . We define the n dimensional torus by n n n n R \T . Again we identify functions on T by 1 periodic functions on R . n We define the Fourier coefficients for m ∈ Z by Z fˆ(m) = f(x)e−2πim·xdx. Tn Many but not all of the statements have analogues in this higher dimensional 2πim·x setting. In particular the functions (e )m are an orthonormal basis. n 7.2. The case of a general lattice. Let vi ∈ R , 1 ≤ i ≤ n be a basis n in R . Its linear combinations with integer coefficients define a lattice in n R . We denote it by L. The dual lattice consists of all vectors in the dual n space (which we identify with R ) which map the elements of the lattice L ∗ n to Z. We denote it by L . Again we identify functions on the torus R /L with L periodic functions. The Fourier coefficients are defined for m ∈ L∗ by Z fˆ(m) = e−2πim·xf(x)dx. Rn/L The functions −1/2 2πim·x | det(v1, . . . , vn)| e are an orthonormal basis. n 7.3. The case of R . We will later look at the Fourier transform which n for integrable functions in R is defined by Z fˆ(ξ) = e2πix·ξf(x)dx. CHAPTER 3

Harmonic functions and the Poisson kernel

1. Basic properties and the Poisson kernel n Definition 3.1. Let U ⊂ R be open. A two times diﬀerentiable function u is called harmonic if

n X 2 ∆u = ∂xj u = 0 j=0

(1) The real part of holomorphic functions is harmonic. (2) Harmonic function satisfy the mean value property. A function u is harmonic if and only if

1 Z u(x) = u(x + ry)dHn−1(y) n|B1(0)| ∂B1(0)

whenever Br(x) ⊂ U. This follows by computing Z Z d n−1 n−1 u(x + ry)dH (y) = y · ∇y(x + ry)dH (y) dr ∂B1(0) ∂B1(0) Z n = ∆y(x + ry)dm (y) B1(0) =0

and a slightly reﬁned argument gives the converse. (3) They satisfy a maximum principle: If the maximum is assumed in the interior then a harmonic function is constant on the connected part of its domain of deﬁnition. (4) There is the fundamental solution

− 1 ln |x| if n = 2 g(x) = 2π 1 |x|2−n if n ≥ 3 n(n−2)|B1(0)|

and −∆g ∗ f = f

s n whenever f ∈ C (R ) with compact support: First ∆g(x) = 0 if x 6= 0, and then, formally

1 xj ∂jg ∗ f = − n ∗ f n|B1(0)| |x| 31 32 3. HARMONIC FUNCTIONS AND THE POISSON KERNEL

and

Z Z 2 xj − yj n(xj − yj)(xl − yl) − δjl|x − y| ∂l f(y) n = n+2 f(y)dy n |x − y| n |x − y| R \Br(x) R \Br(x) Z 1−n yjyl − r 2 f(x − y)dσ. ∂Br(0) |y|

The ﬁrst term vanishes for the Laplacian ∆ (if we sum over j = l. The second converges to −n|B1(0)|f(x) as r → 0. See (Evans, PDE) for details. The H¨olderregularity is needed to show existence of second derivatives.

1.1. The Greens function in the half space and the Poisson 0 n kernel. We denote x for the ﬁrst n − 1 components of x ∈ R and by en the nth vector for the standard basis. We deﬁne the Greens function in the half space by

0 gH (x, y) = g(x, y) − g(x, (y , −yn) = gH (y, x).

In physical terms g is the potential of an electric field with the charge in y, with a conducting boundary at xn = 0, normalized so that g vanishes in this hyperplane. This effect is obtained by putting a particle with the opposite charge at the reflected point. If f is Höldercontinuous then

Z u(x) = gH (x, y)f(y)dy yn>0 satisﬁes the inhomogeneous Dirichlet problem

−∆u = f u(x) = 0

Let u be a bounded continuous function on xm ≥ 0, harmonic in xn > 0, satisfying

|u(x)| ≤ c|x|−1, |Du(x)| ≤ c|x|−2. 1. BASIC PROPERTIES AND THE POISSON KERNEL 33

Then, for ε > 0 and xn > ε, since the second part of the Poisson kernel is harmonic in H, by several applications of the divergence theorem, (3.1) Z 0 = u(y)∆ygH (x, y)dy yn>0,|x−y|>ε Z Z 0 0 n−1 0 1 = − u(y )∂yn gH (x, y )m (y ) − u(x + εy)dσ(y) Rn−1 n|B1(0)| |y|=1 Z − εn−1 u(x + εy)y · ∇g(x + εy − x˜)dσ(y) |y|=1 Z − ∇yu∇ygH (x, y)dy yn>0,|x−y|>ε Z Z 0 0 n−1 0 1 = − u(y )∂yn gH (x, y )m (y ) − u(x + εy)dσ Rn−1 n|B1(0)| |y|=1 Z − εn−1 u(x + εy)y · ∇g(x + εy − x˜)dσ |y|=1 ε Z + y · ∇u(x + εy)dσ (n − 2)n|B1(0)| |y|=1 Z + εn−1 g(x + εy − x˜)y · ∇u(x + εy)dσ |y|=1 Z + ∆ugH (x, y)dy yn>0,|x−y|>ε Z → − u(x) + u(y0)P (x, y0)dmn−1(y0) Rn−1 where we deﬁne the Poisson kernel by

0 0 1 2xn P (x, y ) := −∂yn gH (x, y ) = 0 n . n|B1| |x − y | There is a small modiﬁcation by an additional logarithmic term if n = 2. We claim that 1 Z 2x (3.2) n mn−1(y0) = 1. 2 0 0 2 n/2 n|B1| Rn−1 (xn + |x − y | ) To see this we apply the identity above to n−1 ut(x) = t ∂xn g(x + ten).

0 0 and ut ∗ f → f almost everywhere where ut(x ) = P (x + ten, 0) and Z 0 0 0 0 ut ∗ f = P (x + ten, y )f(y )dy

p n−1 if f ∈ L (R ). n−1 Lemma 3.2. Suppose that g ∈ C(R ) has compact support. Let Z u(x) = P (x, y0)f(y0)dy0. Rn−1 Then u is bounded and harmonic on xn > 0 and continuous up to xn = 0. Moreover u(x0, 0) = g(x0). 1.2. Green’s function and Poisson kernel on the unit ball. We define the reflection at the unit sphere by x n\{0} 3 x → =x ˜ R |x|2 and we define gB(x, y) = g(y − x) − g(|y|(x − y˜)) If |x| = 1 and y 6= 0 2y · x 1 |y|2|x − y˜|2 =|y|2(|x|2 − + ) |y|2 |y|2 = |y|2 + 2y · x + 1 = |x − y|2 and thus gB(x, y) = 0 if |x| = 1. Also gB(x, y) = gB(y, x). This has the same interpretation as for the half space. In particular, if f ∈ C(B1(0)) and Z u(x) = gB(x, y)f(y)dy B1(0) then u it two times continuous differentiable in B1(0), −∆u = f and u(x) = 0 if |x| = 1. We define the Poisson kernel for |x| < 1 and |y| = 1 by X P (x, y) = − yi∂igB(x, y) i 1 1 − |x|2 = n . n|B1| |x − y| Let g ∈ C(∂B) and Z u(x) = P (x, y)g(y)dσ(y) ∂B1(0) is harmonic and continuous up to the boundary. Moreover u(x) = g(x) if |x| = 1. The first statement is obvious. The second statement follows as for the half space, with modifications for the approximate identity on the sphere, which we do not discuss here. 1. BASIC PROPERTIES AND THE POISSON KERNEL 35

Lemma 3.3. Suppose that f ∈ C(∂B1(0)) and Z u(x) = P (x, y)f(y)dσ ∂B1(0) then u is continuous on the closure, twice continuously differentiable and it satisfies ∆u = 0 and u = f on ∂B1(0). 1.2.1. The Poisson integral in the two dimensional case. We define the Poisson integral in polar coordinates for 0 ≤ r < 1 ∞ X |n| 2πinx Pr(x) = r e n=−∞ 1 − r2 (3.3) = 1 − 2r cos 2πx + r2 1 + re2πix = Re 1 − ri2πix where the second identity is an evaluation of the two geometric series, and the third is an algebraic manipulation. Comparison with the kernel above shows that 2πix 2πiy P (re , e ) = Pr(x − y). s There is an alternative derivation in this case. Let f ∈ C (T). We expand it into a Fourier series ∞ X Z f(x) = e−2πinyf(y)dy e2πinx n=−∞ If f is real then Z Z e−2πinyf(y)dy = e2πinyf(y)dy and ∞ X Z u(re2πix) = r|n| e2πinx−yf(y)dy n=−∞ T Z = Pr(x − y)f(y)dy T Let 1 + re2πix 2r sin(2πx) (3.4) Q (x) = Im = . r 1 − re2πix 1 − 2r cos(2πx) + r2 Then

Pr(x) + iQr(x) 2πix is a holomorphic function if z = re . We denote the unit disc by D. 36 3. HARMONIC FUNCTIONS AND THE POISSON KERNEL

2. Boundary behaviour of harmonic functions

Given a real measure µ ∈ M(T) we deﬁne the harmonic function Z 2πix u(re ) = Pr(x − y)µ(y).

We recall that 1 + re2πix P (x) = Re r 1 − re2πix satisﬁes

(1)0 ≤ Pr(x) for 0 ≤ r < 1 R (2) Pr(x)dx = 1 for 0 ≤ r < 1 R 1−δ (3) δ Pr(x)dx → 0 as r → ∞ and hence it is an approximate identity. Theorem 3.4. The function u satisﬁes Z 1 Z 1 2πix 2πix sup |u(re )|dx = lim |u(re )|dx = kµkM. 0≤r≤1 0 r→1 0 Any such function determines a measure uniquely. (1) µ is absolutely continuous with respect to the Lebesgue measure if and only if u(re2πi.) converges in L1 to the density with respect to the Lebesgue measure. (2) Let 1 < p ≤ ∞ and µ = fm1. The following assertions are equivalent: p • f ∈ L (T). 2πi. • sup0

Lemma 3.5. (1) Suppose that F ∈ C(D) is harmonic in the open disc. Then Fr = Pr ∗ F1 for 0 ≤ r < 1. (2) If F is harmonic in the interior then Fsr = Pr ∗Fs for 0 ≤ r, s, < 1. (3) If F is harmonic in the interior and 1 ≤ p ≤ ∞ then r → kFrkLp(T) is non-decreasing.

Proof. Let F be as in (1). Then Pr ∗ F1 deﬁnes a continuous harmonic function (since Pr is an approximate identity as r → 1). Thus 2πix v = F (re ) − Pr ∗ F1(x) is harmonic and zero at the boundary. By the maximum principle v = 0. Point (2) follows by rescaling. By (2) and

Young’s inequality kFrskLp(T) ≤ kFrkLp(T). Proof of Theorem 3.4. Let F be harmonic in D with

(3.5) sup kFrkL1(T) = C < ∞. r The Fr deﬁnes a uniformly bounded family of real measures on T. The space of measures M is dual to the separable space C(T) of continuous functions. By the theorem of Banach Alaoglou the closed unit ball in M(T) is weak* 3. ALMOST EVERYWHERE CONVERGENCE 37

compact, and there exists a weakly converging subsequence Frj → µ as rj → 1. Then, for 0 < r < 1

Pr ∗ µ = lim Pr ∗ Fr = lim Fr r = Fr j→∞ j j→∞ j and, for f ∈ C(T), Z Z Z Z fFrdx = fPr ∗ µdx = Pr ∗ fµ → fµ as r → 1. Thus Fr → µ weak* in M(T) and any harmonic function which satisﬁes (3.5) determines a measure on T. Moreover

kµkM( ) ≤ lim inf kFrkL1( ). T r→1 T We have seen that Young’s inequality implies the converse statement and 1 hence we have equality. If f ∈ L (T) then Proposition 2.7 implies conver- 1 1 gence in L (T). Conversely, if Tr → f in L then µ = fdx. The similar statements of the next point follow by the same type of argument. Finally, if f is continuous then Fr = Pr ∗ f deﬁnes by Proposition 2.7 a continuous function. The converse statement is obvious. 3. Almost everywhere convergence

Definition 3.6. Let (Φn) by an approximate identity. We say it is radially bounded if there exists functions Ψn so that |Φn| ≤ Ψn, Ψn is radial 1 and radially nonincreasing ( ψn(x) ≥ ψn(y) if 0 ≤ |x| ≤ |y| ≤ 2 ) and supn kΨnkL1 < ∞. 1 Lemma 3.7. Let (Φn) and Ψn be as in the deﬁnition. Then for f ∈ L (T)

sup |Φn ∗ f(x)| ≤ sup kψnkL1(T)Mf(x). n n Proof. Let K be even and radially nonincreasing. We claim that

|K ∗ f(x)| ≤ kKkL1 Mf(x). Assume that this is true. Then the statement follows from

|Φn ∗ f(x)| ≤ ψn ∗ |f|(x) ≤ kψnkL1 Mf(x). The claim holds for even characteristic functions, and for radially nonincreasing simple functions, and by a further approximation it holds in the generality of the lemma.

Theorem 3.8. Let (Φn) be a radially bounded approximate identity. Let 1 f ∈ L (T). Then Φn ∗ f → f almost everywhere as n → ∞.

Proof. Pick ε > 0 and g ∈ C(T) with kf − gkL1(T) < ε/2. Then 1/2 |{x ∈ T : lim sup |Φn ∗ f(x) − f(x)| > ε }| n→∞ 1/2 ≤|{x ∈ T : lim sup |Φn ∗ (f − g)(x) − (f − g)(x)| > ε }| n→∞ −1/2 ≤3 sup kψnkL1 ε kf − gkL1 n 1/2 ≤3 sup kψnkL1 ε n 38 3. HARMONIC FUNCTIONS AND THE POISSON KERNEL

As a consequence

|{x ∈ T : lim sup |Φn ∗ f(x) − f(x)| > 0}| = 0. n→∞

4. Subharmonic functions n Definition 3.9. Let U ⊂ R be open. A function f : U → [−∞, ∞ is called subharmonic if it is continuous (with the obvious meaning for the value −∞) and if for all x ∈ U there exists r > 0 so that 1 Z f(x) ≤ f(x + sy)dσ(y). n|B1(0)| ∂B1(0) for all 0 ≤ s < r.

Lemma 3.10. (1) If f and f are subharmonic then max{f, g} is subharmonic. (2) A function f ∈ C2(U) is subharmonic if and only if ∆f ≥ 0. (3) If F is holomorphic then ln |F | and |F |α for α > 0 are subharmonic. (4) If f is subharmonic and φ is increasing and convex then φ ◦ f is subharmonic.

Proof. The first claim follows immediately from the definition. Subhar- monicity follows from ∆f ≥ 0 by the argument for the mean value property: d 1 Z Z f(x − sy)dσ(y) = ∆f(x + y)dy. dr n|B1(0)| ∂B1(0) B1(0) It also implies the converse implication. Let x be a point where ∆f(x) < 0. Then for small radii the mean value inequality is violated. We turn to the fourth point. By monotonicity and the mean value inequality resp. definition and Jensen’s inequality ! 1 Z φ(f(x)) ≤φ f(x + ry)dσ(y) n|B1(0)| ∂B1(0) 1 Z ≤ φ ◦ f(x + ry)dσ(y) n|B1(0)| ∂B1(0) If F is holomorphic then ln |F | is continuous, with values in [−∞, ∞). If F (z0) 6= 0 then Re ln F (z) = ln |F (z)| is harmonic in a neighborhood and it satisfies the mean value identity near z0. If F (z0) = −∞ there is nothing to prove. Since |F |α = exp(α ln |F |) the last statement follows from the previous points. n Lemma 3.11. Let Ω ⊂ R be a bounded domain, u, v : Ω → R continuous and subharmonic resp harmonic. If u ≤ v on ∂Ω then u ≤ v in Ω.

Proof. The function u − v is subharmonic, ≤ 0 at ∂Ω. The deﬁnition of subharmonic function implies the maximum principle, and the maximum is assumed at the boundary. 4. SUBHARMONIC FUNCTIONS 39

n Lemma 3.12. Let Ω ⊂ R be open, f :Ω → [−∞, ∞) be subharmonic, n Br(x) ⊂ R . Then 1 Z f(x) ≤ f(x + ry)dσ(y) n|B1(0)| ∂B1(0) Proof. Let fn(x) = max{f(x), −n}. It is subharmonic and it suﬃces to verify the assertion for fn, or, by an abuse of notation, we assume that f is bounded. Let u be the harmonic function on Br(x) which coincides with f at the boundary. By the previous lemma f ≤ u on Br(x), and hence 1 Z 1 Z f(x) ≤ u(x) ≤ u(x + ry)dσ = f(x + ry)dσ n|B1(0)| ∂B1(0) n|B1(0)| ∂B1(0) as claimed. The prove implies also Corollary 3.13. Suppose that f is subharmonic on the open set U ⊂ n R and Br(x0) ⊂ U. Then, if x ∈ Br(x0) Z x − x f(x) ≤ P ( 0 , y)f(x + ry)dσ(y). r 0 Proposition 3.14. Suppose that g is a nonnegative subharmonic func- 2 tion on B1(0) ⊂ R which satisﬁes 2πix |kgk|1 := sup kg(re )kL1 < ∞ 0 t}| ≤ t |kgk|1 (2) If 1 < p ≤ ∞ and 2πix |kgk|p := sup kg(re )kLp(T) < ∞ r then ∗ kg kLp ≤ c|kgk|p

Proof. Suppose that grj → µ as j → ∞ with rj → 1. Existence of such a sequence and µ follows from the weak* compactness argument. Then, as in the proof of Theorem 3.4 combined with Lemma 3.13 2πix g(re ) ≤ Pr ∗ µ(x) and lim kgrkL1 = kµkM = |kgk|1. r→1 But now 2πix 0 ≤ g(re ) ≤ Pr ∗ µ(x) ≤ Mµ(x) and hence g∗(x) ≤ Mµ(x) 40 3. HARMONIC FUNCTIONS AND THE POISSON KERNEL

This implies the weak inequality. The estimate in Lp follows in the same fashion. 5. The theorems of Riesz Definition 3.15 (complex Hardy space). Let 1 ≤ p ≤ ∞. We denote p the unit disc by D. The space H (D) is the space of holomorphic functions F on D such that

kF kHp(D) = sup kFrkLp(T) < ∞. 0≤r<1 1 ∗ 1 Theorem 3.16. If F ∈ H (D) then F ∈ L (T). 1 1 Proof. The function |F | 2 is subharmonic and |Fr| 2 is bounded in 2 1/2 ∗ 2 L (T), uniformly in r. By Proposition 3.14 (|F | ) ∈ L . This implies the statement of the theorem.

By Theorem 3.4 Fr = Pr ∗µ for a complex measure µ on T. By Theorem ∗ 3.16 there is the integrable majorant |F | for the functions Fr. The functions Fr converge to the measure µ in the sense of measures. Because of the integrable majorant µ is absolutely continuous with respect the Lebesgue- 1 1 measure and can be written as µ = fm with f ∈ L . Moreover Fr = Pr ∗f. But then, due to the pointwise convergence of Theorem 3.8 Fr → f as r → 1 almost everywhere. We obtain the second version of the Riesz theorem. 1 Theorem 3.17. If F ∈ H (D) and f = limr→1 Fr(x) almost everywhere 1 then f ∈ L (T) and Fr(x) = Pr ∗ f(x). Theorem 3.18. Suppose that µ ∈ M(T) satisﬁes µˆ(n) = 0 for n < 0. Then µ is absolutely continuous with respect to the Lebesgue measure.

Proof. Let F be defined by Pr ∗ µ. Then the real and imaginary part of F are harmonic. Harmonic functions are smooth and even analytic (this can be seen from the representation by the Poisson formula) and near z = 0 we can expand F in a Taylor expansion (by expanding the Poisson kernel) which converges in D X n m F = amnz z¯ . m,n≥0 1 We apply the Laplace operator ∆ = 4∂z∂z¯ with ∂z = 2 (∂x − i∂y) and 1 ∂z¯ = 2 (∂x + i∂y). Then X n m 0 = ∆F = amn∂z∂z¯(z z¯ ) n,m≥0 n m n−1 m−1 with ∂z∂z¯(z z¯ ) = nmz z¯ . Thus we can write ∞ X n n F (z) = anz + bnz¯ . n=0 Now we apply the assumption: For k < 0 Z e−2πikxµ(x) = 0 6. CONJUGATE FUNCTIONS 41 and hence Z Z Z −2πikx −2πikx |k| −2πikx e Pr ∗ µdx = Pre µ = r e µ = 0 since z−k is holomorphic and hence real and imaginary part are harmonic. 1 Thus F ∈ H (D) and the assertion follows from Theorem 3.17. 1 Theorem 3.19. Let F ∈ H (D). Suppose that F is non identically 0 and let f be as in Theorem 3.17. Then ln |f| ∈ L1. In particular is does not vanish on a set of positive measure. Proof. We first assume that F (0) 6= 0. Since ln |F | is subharmonic Z ln |F (0)| ≤ ln |Fr|dx. T We want to take the limit as r → 1. We know that Fr → f almost everywhere and ln |F | ≤ |F |∗ ∗ is an integrable majorant for the positive part. Let gr = |F | − ln |Fr| ≥ 0. ∗ It converges almost everywhere to g1 = |F | − ln |f| and by the Lemma of fatou Z Z Z ∗ g1dr ≤ lim inf g1 ≤ |F | − ln |F (0)|. r→1 We have to remove the assumption that F (0) 6= 0. There exists a point z0 ∈ B1/2(0) with F (z0) 6= 0. We apply a Möbiustransform to transport z0 to 0. The Möbiustransform maps circles to circles with possibly different radii and centers. For large circle the L1 norm is bounded by |F |∗ before the transform. This is considerable sharpening of a statement from complex analysis.

6. Conjugate functions

Definition 3.20. Let u be harmonic and real valued on D. We define the conjugate function u˜ to be the unique harmonic function with u˜(0) = 0 such that u + iu˜ is holomorphic. Lemma 3.21. If u is constant then u˜ = 0. If f = u + iv is holomorphic and v(0) = 0 then u˜ = v. If u is harmonic on D then u˜br(n) = −i sign(n)ubr(n) where   −1 if n < 0 sign(n) = 0 if n = 0  1 if n > 0 Proof. The function u + iv − (u + iu˜) is purely imaginary and holomorphic, hence constant. We expand X 2πikx ur(x) = ak(r)e k∈Z 42 3. HARMONIC FUNCTIONS AND THE POISSON KERNEL where Z 2πix −2πikx ak = u(re )e dx and ak = a−k since u is real valued. Then 2 X 2 00 0 2 2πikx r ∆u = (r ak(r) + rak − k ak(r))e k∈Z |k| and hence ak(r) = akr and X |k| 2πikx ur(x) = akr e k∈Z with ak = a−k. Then ∞ X k u + iv = u(0) + 2 akz k=1 is holomorphic with real part u. Checking the Fourier coefficients gives the claimed relation of the Fourier coefficients. We begin with an important weak-L1 bound due to Besicovitch and Kolmogorov.

Theorem 3.22. Let u be harmonic in D and satisfy

|kuk|1 = sup kurkL1(T) < ∞. 0 t}| ≤ |kuk| 1 . t L

Proof. By Theorem 3.4 ut = Pt ∗µ for a measure µ. It suﬃces to verify the assertion for nonnegative measures µ. Let ∗ Es = {x :u ˜ (x) > s}. We deﬁne the holomorphic function F = −u˜ + iu. Let 1 Z y ωs(x, y) := 2 2 dt π (−∞,−s)∪(s,∞) (x − t) + y which is harmonic for y > 0 and nonnegative. Moreover 1 • ωs(x, y) ≥ 2 if |x| > s, 2y • ωs(0, y) ≤ πs since Z ∞ y Z ∞ 1 dx 2 2 dx = 2 0 x + y 0 (x/y) + 1 y Z ∞ 1 = 2 dz 0 1 + z = arctan(∞) − arctan(0) π = 2 7. THE HILBERT TRANSFORM 43 and 1 Z y 2 Z ∞ dt 2y ωs(0, y) = 2 2 dt ≤ 2 ≤ . π (−∞,−s)∪(s,∞) t + y π s/y 1 + t πs

1 The composition ω ◦ F is harmonic and, if x ∈ Es then ω(F (x)) ≥ 2 for some 0 < r < 1. By Proposition 3.14 1 |E | ≤ |{y :(ω ◦ F )∗ ≥ }| ≤ 6|kω F k| . s s 2 s 1

Since ωs ◦ F ≥ 0 the mean value property implies 2 u(0) 2 |kuk| |kω ◦ F k| = ω (F (0)) = ω (iu(0)) ≤ = 1 s 1 s s π s π s which gives 12 |kuk| |E | ≤ 1 . s π s Lemma 3.23. Let u be harmonic on D. Then 2 2 2 ku˜rkL2 + u(0) = kurkL2 Proof. This is seen by Plancherel and the relation of the Fourier coef- ﬁcients. 2 Corollary 3.24. If f ∈ L (T) and ur = Pr ∗ f then the limit g(x) = 2 limr→1 u˜r(x) exists almost everywhere and in L .

Proof. By the previous lemma |ku˜k|2 ≤ |ku˜k|2. By Theorem 3.4 ur and 2 2 u˜r converge in L to some functions f and f˜ in L . Since |ur(x)| ≤ Mf(x) ˜ and |u˜r(x)| ≤ Mf(x) we obtain pointwise convergence.

7. The Hilbert transform

We formally introduce the Hilbert transform H of a measure µ ∈ M(T) as

Hµ =µ ˜ = lim P\r ∗ µ r→1 It is not clear whether and in which sense the limit exists. It does for µ = fm1 for a square integrable f.

Corollary 3.25. If u is harmonic and |kuk|1 < ∞ then

g = lim u˜r r→1 exists almost everywhere. It satisﬁes the weak bound c |{x : |g(x)| > t}| ≤ kfk 1 t L 1 2 Proof. By Theorem 3.4 ur = Pr ∗ µ. If µ = fm with f ∈ L then the almost everywhere convergence follows from Corollary 3.24. Next we 44 3. HARMONIC FUNCTIONS AND THE POISSON KERNEL

1 1 2 consider µ = fm with f ∈ L (T). Given ε > 0 there exists g ∈ L with kf − gkL2 ≤ ε. Let vr = Pr ∗ g. Then, by Theorem 3.22, 2πix 2πix 1/2 |Eε| =|{| lim sup |(u + iu˜)(re ) − (u + iu˜)(se )| > ε }| s,r→1

= {| lim sup |(u − v + i(˜u − v˜))(re2πix) s,r→1

2πix 1/2 − (u − v + i(˜u − v˜))(se )| > ε } c ≤ kf − gk 1 ε1/2 L ≤cε1/2. Thus 2πix 2πix |{| lim sup |(u + iv)(re ) − (u + iv)(se )| > 0}| = lim |Eε| = 0. s,r→1 ε→0 Let µ be a general measure on T. We decompose into the absolutely 1 continuous part and a singular part: There exists f ∈ L (T) and a set B of measure 0 so that Z µ(A) = fdx + µ(A ∩ B). A The first part we have dealt with above, and we consider a Borel measure µ such that there exists B of measure 0 so that µ(A) = µ(A ∩ B). Since µ(B) = sup µ(K) K⊂B,compact for every µ measurable set, given ε > 0 there exists K so that µ(A) < µ(K) + ε. We define µK (A) = µ(A ∩ K). K If I is an open interval disjoint from K and if ut = Pt ∗ µ and w = u + iu˜ then u has a continous extension to the image of I on the arc of the unit circle defined by I and the real part vanishes. By the Schwarz reflection principle w has a holomorphic extension across this arc, and hence all limits exist in the complement of the compact set K of measure 0. As in the first step we derive the existence of the limit almost everywhere. The bound is a consequence of Theorem 3.22. Theorem 3.26. If 1 < p < ∞ then π tan( 2p ) if 1 < p ≤ 2 kHfkLp ≤ π kfkLp cot( 2p ) if p ≥ 2 Proof. We have seen that

kHfkL2 ≤ kfkL2 By the interpolation theorem of Markinciewicz this implies together with the weak estimate that kHfkLp ≤ cpkfkLp for 1 < p ≤ 2. Since Z Z Hfgdx = − fHgdx 7. THE HILBERT TRANSFORM 45 the estimate (up to the sharp constant) for p ≥ 2 follows by duality. We provide a second proof following Grafakos’ variant of Pichorides’ proof, which yields a sharp constant. Lemma 3.27. For a, b real and 1 < p ≤ 2 π |b|p ≤ tan( )|a|p − B Re[(|a| + ib)p] 2p b with sinp−1( π ) B = 2p b (p−1)π sin( 2p ) We set a = f(x) and b = Hf(x) and integrate. Then Z 1 Z 1 Z 1 p p π p p |Hf(x)| dx ≤ tan ( ) |f(x)| dx − Bp Re[(|f(x)| + iHf(x)) ]dx 0 2p 0 0 and the theorem follows once we show that Z Re[(|f(x)| + iHf(x))p]dx ≥ 0.

Let u + iv be the harmonic (holomorphic) extensions of f + iHf. Lemma 3.28. The function g(x, y) = Re((|x| + iy))p) is subharmonic and g(u, v) is also subharmonic. Then Z 1 0 ≤ g(u(0), v(0)) ≤ Re[(|f(x)| + iHf(x))p]dx 0 which concludes the proof up to a duality argument. Sharpness follows by taking 1 + z s f = Re 1 − z for suitable s. ∞ The Hilbert transform is not bounded in L : The real part of i lnC (1 − 2πix 1 2πix e ) is π(x − 2 ) for 0 < x < 2π and the imaginary part is ln |1 − e |, which is not bounded. We recall that 1 + e2πix cos πx Im = = cot(πx). 1 − e2πix sin πx 1 Proposition 3.29. If f ∈ C (T) then Z f(x) = lim cot(π(x − y))f(y)dy ε→0 |x−y|>ε Proof. By symmetry Z 1 Z cot(π(x − y)f(y)dy = cot(π(y)(f(x − y) − f(x + y))dy |x−y|>ε 2 |x−y|>ε and the second integrand is bounded. 46 3. HARMONIC FUNCTIONS AND THE POISSON KERNEL

Proposition 3.30. Suppose that kfkL∞ ≤ 1. Then Z 1 2 eα|Hf(x)|dx ≤ 0 cos α π for 0 ≤ α < 2 . Proof. Let f be as in the proposition. Let u be the harmonic extension of f and F =u ˜−iu. Then by the maximum principle |u| ≤ 1 and cos(αu) ≥ cos α. Hence Re eαf = Re eαu˜e−iαu = cos(αu)eαu˜ ≥ cos αeαu˜ and with the mean value property Z Re eαFr(x)dx = Re e−iα(u(0)) = cos αu(0) ≤ 1 T and therefore Z 1 eαu˜r(x)dx ≤ . T cos α The estimate follows now by the Lemma of Fatou, and the same argument for −f.

Theorem 3.31. Let SN be the partial sum of Fourier series and 1 < p < ∞. Then

sup sup kSN fkLp(T) < ∞ N kfkLp(T ) p and for all f ∈ L (T) p SN f → f in L as N → ∞. Proof. We write (which we check by checking the eﬀect on the Fourier coeﬃcients) 1 − iH 1 + iH S f =e−2πiNx e2πi(2N)x e−2πiNxf N 2 2 1 + (a e2πiNx + a e2πiNx) 2 N −N and obtain by the boundedness of the Hilbert transform

sup kSN fkLp(T) ≤ cpkfkLp(T). N This implies the assertion by Proposition 2.9. Pointwise convergence is true but it is considerably harder, and much more recent. ßß CHAPTER 4

The Fourier transform on Rn

1. Deﬁnition and ﬁrst properties Definition 4.1. The Fourier transform of a complex valued function 1 n f ∈ L (R ) is Z fˆ(ξ) = e−2πiξ·xf(x)dx.

Properties: (1) fˆ is a bounded continuous function which converges to 0 as |ξ| → ∞. Moreover ˆ |f(ξ)| ≤ kfkL1(Rn). 1 n n (2) If f ∈ L (R ), h, η ∈ R and fh(x) = f(x − h) then ˆ −2πih·y ˆ fh(ξ) = e f(ξ) and e−\2πiηxf = fˆ(ξ + η) 1 n (3) If f, g ∈ L (R ) then f[∗ g = fˆ(ξ)ˆg(ξ) and by Fubini Z Z fgˆ(ξ)dξ = fˆ(x)g(x)dx.

n n (4) If A : R → R is an invertible linear map then f[◦ A(ξ) = | det A|−1fˆ(A−tξ). Lemma 4.2. exp(\−π|x|2)(ξ) = exp(−π|ξ|2) Proof. We calulate more than we need. Z Z 1 1/2 −π|x|2 n − ln(t) In = e dm = |{x : |x| ≤ }|dt 0 π Z 1 −n/2 n/2 =|B1(0)|π | ln(t)| dt 0 Z ∞ −n/2 n −s =|B1(0)|π s 2 e ds 0 n + 2 =|B (0)|π−n/2Γ( ). 1 2 In particular I2 = Γ(2) = 1 Alternatively, by Fubini In+m = InIm

47 n 48 4. THE FOURIER TRANSFORM ON R and hence In = 1. As a consequence n/2 n π (4.1) m (B1(0)) = n+2 Γ( 2 ) and 1 1 Γ( ) = Γ(3/2) = π1/2. 2 2 Let n = 1, f(x) = e−π|x|2 and

Z 2 2 Z 2 e−πx −2πiξxdx = e−π|ξ| e−π(x−iξ) dx.

We denote Z 2 J(t) = e−π(x−it) dx. Then d Z 2 J(t) = e−π(x−it) (2πi)(x − it)dx dt Z d 2 = − i e−π(x−it) dx dx =0 This implies the formula in the one dimensional case. The higher dimensional case follows by Fubini. An example:

Z 2πixξ 1 −2π|ξ| (4.2) e 2 dx = πe R 1 + x by the residue theorem since 1 1 1 1 = − . x2 + 1 2i x + i x − i

2. The Fourier transform of Schwartz functions n n Definition 4.3. The space of Schwartz functions S(R ) on R is the space of all smooth functions for which the seminorms sup |xα∂βf(x)| are bounded for all multiindices. Properties : (1) If f ∈ S and g ∈ C∞ with derivatives bounded by polynomials then fg ∈ S. n 2πiη·x (2) If f ∈ S and h, η ∈ R then fh(x) = f(x − h) ∈ S and e f ∈ S. 1 n (3) S ⊂ L (R ) and the Fourier transform is deﬁned for functions in S. ˆ (4) (−\2πi)xjf = ∂ξj f ˆ (5) ∂dxj f = 2πiξjf 2. THE FOURIER TRANSFORM OF SCHWARTZ FUNCTIONS 49

(6) It follows that, if f ∈ S, α β ˆ |β|−|α| |α|+|β| \α β ξ ∂ξ f = (2π) (−i) ∂x x f is a bounded continuous function since it is the Fourier transform of a Schwartz function. The inversion theorem is a consequence: n Theorem 4.4. If f, g ∈ S(R ) then Z (4.3) f(x) = e2πix·ξfˆ(ξ)dξ Rn Z Z (4.4) fgdx = fˆgdξ.ˆ Rn Rn 2 In particular the Fourier transform deﬁnes a unitary map on L (R). We denote the inverse transform by fˇ. Proof. The family of functions n −π|tx|2 φt = t e is an approximate identity as t → ∞. Its Fourier transform is 2 −π |ξ| φˆt = e t2 Thus, by the dominated convergence theorem and Fubini

Z Z |ξ|2 2πix·ξ 2πixξ −π e fˆ(ξ)dξ = lim e e t2 fˆ(ξ)dξ t→∞

ZZ |ξ|2 2πi(x−y)ξ −π = lim e e t2 dξf(y)dy t→∞

Z 2 2 = lim tne−πt |x−y| f(y)dy t→∞ =f(x). Now, since R fgˆ = R fgˆ and gˆ = gˇ, Z Z fgdξ = fgdξˆˇ Z = fgdxˆ Z = fˆgdξˆ

We obtain n Proposition 4.5. The Fourier transform maps S(R ) to itself. It is invertible and it satisﬁes (4.5) f[∗ g = fˆgˆ

(4.6) fgc = fˆ∗ gˆ n 50 4. THE FOURIER TRANSFORM ON R

Proof. Only the last statement has to be shown. We apply the inverse Fourier transform to the right hand side of the second equality. The statement then follows from the ﬁrst equality.

2.1. The Fourier transform of complex Gaussians. Let A = Ar + iAi be a symmetric invertible complex matrix with Ar positive deﬁnite. The function f(x) = exp(−πxtAx) is a Schwartz function. We want to compute its Fourier transform. We t 2 t write Ar = B B and y = Bx. Then, with g(y) = exp(−π|y| −iπy Cy) with −t C = B AiB we have fˆ(ξ) = | det B|−1gˆ(B−tξ).

If Ai = 0 we know the Fourier transform of g and obtain t −1/2 t −1 (4.7) exp(−\πx Arx) = (det Ar) exp(−πξ Ar ξ). Lemma 4.6. The Fourier transform of exp(−πxtAx) is, by an abuse of notation (4.8) (det A)−1/2 exp(−πξtA−1ξ). where (det A)−1/2 denotes the product of the square roots of the eigenvalues of A. The eigenvalues have positive real part, and the square root is uniquely defined. This is not necessarely true for the determinant. It is both remark- able and important that both f and the formula of its Fourier transform are well defined for invertible complex matrices with nonnegative real part. Proof. By the considerations above it suffices to prove the statement for A = 1 + iC. By the Schur decomposition we can also diagonalise A and it suffices to verify the formula in one dimension. Let

−1 2 Z 2 h(τ, ξ) = (1 + iτ)1/2eπ(1+iτ) ξ e−2πiξx−π(1+iτ)x dx

We diﬀerentiate with respect to ξ:

dh −1 2 Z 2πiξ 2 =(1 + iτ)1/2eπ(1+iτ) ξ [− − 2πix]e−2πiξx−π(1+iτ)x dx dξ 1 + iτ

−1 2 Z d 2 = − (1 + iτ)−1/2eπ(1+iτ) ξ e−2πiξx−π(1+iτ)x dx dx =0 hence h(τ, ξ) = h(τ, 0) and Z dh(τ, 0) 1 1 2 −π(1+iτ)x2 =i(1 + iτ) 2 [ − πx ]e dx dτ 2(1 + iτ) Z i − 3 d −π(1+iτ)x2 = (1 + iτ) 2 e dx 4π dx2 =0 This completes the proof since h(0, 0) = 1. 3. THE FOURIER TRANSFORM ON TEMPERED DISTRIBUTIONS 51

3. The Fourier transform on tempered distributions 0 n The space of tempered distributions S (R ) is the topological dual of S, or, in diﬀerent words, the set of all continuous linear maps from S to C. Definition 4.7. T : S → C is a tempered distribution if (1) T is linear (2) T is continuous, i.e. there exist N and C so that |T (f)| ≤ C sup |xα∂βf| |α|≤N,|β|≤N,x Let µ be a measure. It deﬁnes a distribution by Z Tµ(f) = fdµ.

Let 1 ≤ p ≤ ∞. Then g ∈ Lp deﬁnes a tempered distribution by Z Tg(f) = fgdx

We often identity f and Tf . It is important that the formulas do not lead to ambiguities when several interpretations are possible. Let φ ∈ C∞ with (polynomially) bounded derivatives. We deﬁne (φT )(f) = T (φf)

(∂xj T )(f) = −T (∂xj f) and (φ ∗ T )(f) = T (φ˜ ∗ f) where φ˜(x) = φ(−x). Then φ ∗ T is the smooth (C∞) function φ ∗ T (x) = T (φ(x − .)) since this expression is continuously diﬀerentiable, and Z Z φ ∗ T (f) = T (φ˜ ∗ f) = T ( φ(x − y)f(x)dx) = T (φ(x − .))f(x)dx.

The last inequality would be obvious for sums, and it is veriﬁed by writing the integral as a limit of sums. Every Schwartz function deﬁnes a tempered distribution. Test functions are dense in the space of tempered distribution in the sense that, given a ∞ tempered distribution T there exists a sequence of functions φj ∈ C0 such that Z φjfdx → T (f) ∞ for all φ ∈ S: Let η ∈ C0 have integral 1. Then Z T (t−nφ(t.) ∗ φ(x/t)f) = T (t−nφ(t(x − .)))φ(x/t)f(x)dx → T (f).

Definition 4.8 (Support of a distribution). The support of the distribution T is deﬁned as complement of the union of all open sets U with the property that T (φ) = 0 if φ is supported in U. Lemma 4.9. Let T be a distribution with compact support. The convolution with T deﬁnes a continuous linear map on S. n 52 4. THE FOURIER TRANSFORM ON R

We will not go much into the notion of continuity here. One way of formulating it: For all N > 0 there exists M and C such that α β α β sup |x ∂x T ∗ φ| ≤ C sup |x ∂ φ|. |α|+|β|≤N |α|+|β|≤M Proof. We assume that the support of T is contained in the ball of radius R. Then there exists N so that α |T (φ)| ≤ CN sup |∂ φ| |α|≤N,|x|≤R and T ∗ φ(x) = T (φ(x − .)) satisfies α −M |T ∗ φ(x)| ≤ CN sup |∂ φ(y)| ≤ C(1 + (|x| − R)+) |α|≤N,|x−y|≤R for all M. Moreover ∂xi (T ∗ φ) = T ∗ (∂xi φ). Let T be a tempered distribution and S a compactly supported distribution. We define S˜(φ) = S(φ˜) and S ∗ T (φ) = T ∗ S(φ) = T (S˜ ∗ φ) = T (S(φ(.−))). Definition 4.10. The Fourier transform of the tempered distribution T is defined by Tˆ(φ) = T (φˆ) Properties: • Since we identify functions with tempered distributions it is important that this does not introduce ambiguities: Z Z ˆ ˆ ˆ (4.9) Tˆf (φ) = Tf (φ) = fφdx = fφdx

• S\∗ T = SˆTˆ whenever both sides satisfy the conditions for convo- lutions resp. products. • Similarly STc = Sb ∗ Tb. Definition 4.11. A distribution is called homogeneous of degree m ∈ C if T (φ) = λm+nT (φ(λx)) for λ > 0. An homogeneous distribution is tempered (exercise) Lemma 4.12. The Fourier transform of a homogeneous distribution of degree −m is a homogeneous distribution of degree −m − n. Proof. Tˆ(φ) = T (φˆ) = T (λ−m−nφˆ(./λ)) = Tˆ(λ−mφ(λ(x)).

3. THE FOURIER TRANSFORM ON TEMPERED DISTRIBUTIONS 53

Examples:

Z (4.10) 1(ˆ φ) = φˆ(ξ)dξ = φ(0) = δ0(φ)

Z Z −2πixξ (4.11) δˆ0(φ) = δ0(φˆ) = φˆ(0) = φdx = δ0(e )φ(ξ)dξ

The same proof as for Fourier series shows

Theorem 4.13 (Hausdorﬀ-Young inequality). Let 1 ≤ p ≤ 2 and 1/p + 1/p0 = 1. Then ˆ kfkLp0 ≤ kfkLp and

Lemma 4.14 (Riemann-Lebesgue). Let f ∈ L1 . Then fˆ and fˇ are continuous and lim fˆ(ξ) = 0 ξ→∞

∞ n Proof. Let fj ∈ C0 (R ) with kf − fjkL1 < 1/j.Then

kfˆ− fˆjksup ≤ 1/j and fj → f uniformly. The assertion holds for fj since fj ∈ S, and hence for f by uniform convergence.

3.1. The Fourier transform of 1/x. Let f(x) = 1 if x > 0 and 0 otherwise. Then formally Z ∞ 1 e−2πixξdx = . 0 2πiξ How do we deﬁne the right hand side? There are at least four possibilities

1 1 Z 1 (4.12) (φ) = (φ(x) − φ(−x))dx 2πix 2 2πix

1 1 (4.13) = lim 2πix + i0 t→0,t>0 2πix + t

1 1 (4.14) = lim 2πiξ − i0 t→0,t<0 2πiξ − t and as principle value 1 Z 1 (4.15) (φ) = lim (φ(ξ) − φ(−ξ))dξ 2πiξ ε→0 |ξ|>ε 2πiξ It is an exercise to compare the diﬀerent deﬁnitions and to choose the correct one. n 54 4. THE FOURIER TRANSFORM ON R

3.2. Gaussians, heat and Schr¨odingerequation. We consider the heat equation ut − ∆u = 0

We make the Ansatz that u and ut are tempered distributions in x uniformly in time. A Fourier transform in x leads to 2 2 uˆt + (2π) |ξ| uˆ = 0 and 2 2 uˆ(t, ξ) = e−4π t|ξ| uˆ(0, ξ) for t > 0. The inverse Fourier transform of e−4π2t|ξ|2 is 2 1 − |x| g (x) = e 4t t (4πt)n/2 and u(t, x) = gt ∗ u(0,.)(x) for t > 0. Similarly we deal with the Schr¨odingerequation

i∂tu + ∆u = 0 and obtain with −n 2 1/2 |x| gt(x) = (4πit) e 4it formally u(t) = gt ∗ u0.

We denote the map u0 → u(t) by S(t)u0. It is a unitary group: (1) S(t + s) = S(t)S(s) for s, t ∈ R, S(0) = 1. 2 2 (2) For all u0 ∈ L the map t → S(t)u0 ∈ L is continuous. Lemma 4.15. Let 1 ≤ p ≤ 2. Then 1 1 −n( p − 2 ) (4.16) kS(t)u0kLp0 ≤ (4π|t|) ku0kLp p 2 for all u0 ∈ L . Suppose that u0 ∈ L and u(t, x) = S(t)u0(x). Then the support of the space time Fourier transform of u is the ’characteristic set’ {(τ, ξ): τ = −2π|ξ|2} and Z uˆ(ψ) = ψˆ(−2π|ξ|2, ξ)dξ Rn The estimate follows as for Hausdorﬀ Young. The remaining part is an exercise. Lemma 4.16 (Strichartz estimates for the Schr¨odingerequation). Let 0 2(n+2) p n 2 p = n . If f ∈ L (R × R ) and u0 ∈ L then the solution to

i∂tu + ∆u = f, u(0, x) = u0(x) given by Duhamel’s formula satisﬁes 0 sup ku(t)kL2(Rn) + kukLp ( × n) ≤ c ku0kL2 + kfkLp(R×Rn) . t R R 3. THE FOURIER TRANSFORM ON TEMPERED DISTRIBUTIONS 55

Proof. Formally Z Z t −n/2 − n k S(t − s)f(s)dsk 0 ≤ (4π) |t − s| n+2 kf(s)k p n Lp (Rn) L (R ) s

kS(t − s)f(s)k 0 ≤ ckfk p n . Lp (R×Rn) L (R×R ) ∗ p n 2 n Let T be the map from L ((−∞, 0) × R ) to L (R ) which maps Z 0 f → S(−s)f(s)ds. −∞ On the Fourier side we see that Z 0 TT ∗f(t) = i S(t − s)f(−s)ds −∞ and 2 ∗ 2 kT k 0 =kT k p n 2 L2→Lp ((0,∞)×Rn) L ((0,∞)×R )→L ∗ =kTT k 0 p((0,∞)×Rn)→Lp ((0,∞)×Rn) which is bounded by the ﬁrst step. Remark: Restriction theorem. 3.3. The Poisson summation formula. We denote the Dirac measure at the point x by δx. Let X T = δk k∈Zn resp X T (f) = f(k). k∈Zn Theorem 4.17. The Fourier transform of T is T . This statement is by the deﬁnition of the Fourier transform of distributions equivalent to the Poisson summation formula X X f(k) = fˆ(k) k∈Zn k∈Zn n for all f ∈ S(R ). Proof. If f ∈ S then X F (x) = f(x − k) k∈Zn n is a smooth periodic function, hence, if m ∈ Z , X Z Fˆ(m) = f(x − k)e−2πimxdx [0,1)n k∈Zn Z = f(x)e−2πimxdx

=fˆ(m) n 56 4. THE FOURIER TRANSFORM ON R and X F (x) = fˆ(m)e2πimx m∈Zn which is the Poisson summation formula.

3.4. Homogeneous distributions and the Laplace operator. We call T radial if for every f ∈ S and any orthogonal matrix O the identity T (f) = T (f ◦ O) holds. The Fourier transform of a radial distribution is radial.

Lemma 4.18. let 0 < Re m < n. Then the Fourier transform of πm/2 π(n−m)/2 |x|m−n is |ξ|−m Γ(m/2) Γ((n − m)/2) Proof. We evaluate using polar coordinates and that the n − 1 dimen- 2πn/2 sional Hausdorﬀ measure of the unit sphere is n|B1| = Γ(n/2) for m > −n

n −1 ∞ Z 2 π 2 Z 2 |x|me−π|x| dx = rm+n−2e−πr 2πrdr Γ(n/2) 0 Z ∞ 1 m+n −1 −s = s 2 e ds m/2 π Γ(n/2) 0 Γ((n + m)/2) = πm/2Γ(n/2)

Let T be homogeneous radial distribution of homogeneity m ∈ C. We claim that there exists a radial function t of homogeneity m so that Z T (φ) = tφdx

∞ whenever φ ∈ C0 (R\{0}). An incorrect proof is given by choosen a smooth cutoﬀ function η supported in the interval B1/2(0) with integral 1 and calculate for the n-th unit vector

t(en) =T (δen ) Z m+n = |en − y| φ(y)T (δen−y)dy B1/2(0) Z m+n =T ( |en − y| φ(y)δen−ydy) B1/2(0) m+n =T (φ(en + .)|.| ). To do it correctly one has to apply this argument to an approximate identity. It is clear that t is radial and of homogeneity m. If Re m > −n then this function is locally integrable and the identity holds for all functions in S, which is seen by a smooth truncation argument and rescaling. Now |x|−m for m < n deﬁnes a homogeneous radial distribution of degree −m. Its Fourier transform is a radial distribution of degree m − n. By the 3. THE FOURIER TRANSFORM ON TEMPERED DISTRIBUTIONS 57 considerations above it is given by a radial function of degree m − n, hence m−n cm|x| . To determine the constant we use the ﬁrst step:

Z 2 Γ((n − m)/2)πm/2 =Γ(n/2) |x|−me−π|x| dx Z m−n −π|ξ|2 =Γ(n/2) cm|ξ| e

(n−m)/2 =cmΓ(m/2)π

We apply these considerations to the Laplace operator for n ≥ 3. Then, if u and f are tempered distributions

−∆u = f implies 4π2|ξ|2uˆ = fˆ and formally fˆ uˆ = (4π2)−1 |ξ|2 The function 1 |ξ|2 is locally integrable and deﬁnes a homogeneous distribution of degree −2. Its inverse Fourier transform is π2−n/2 |x|2−n Γ((n − 2)/2) and hence u = g ∗ f with Γ((n + 2)/2) g(x) = |x|2−n. n(n − 2)πn/2 Moreover the inverse Fourier transform of

iξj 4π2|ξ|2 is πn/2−1 x − j 2nΓ((n + 2)/2) |x|n which remains true for n = 2. n 58 4. THE FOURIER TRANSFORM ON R

4. Oscillatory integrals and stationary phase We begin with a study of complex Gaussian integrals. The ﬁrst lemma contains the algebraic part. n Proposition 4.19. Let p(x) be a polynomial in R . and let A = Ar + iAi be symmetric with A invertible and inverse (aij) and Ar positive semi deﬁnite. Then  k Z ∞ n −πxtAx − 1 X 1 X 2 (4.17) p(x)e dx = (det A) 2 a ∂ p(x) (4π)kk!  ij ij k=0 i,j=1 x=0

If Ar is not positive deﬁnite we understand both sides as the limit as ε → 0 for A + ε1.

Proof. It suffices to consider Ar positive definite since both sides are continuous in A as long as A is invertible. By a change of coordinates we reduce the assertion to diagonal matrices A. It suffices to prove it for monomials, since polynomials are sums of monomials. An application of the theorem of Fubini reduces it to the one dimensional case. (Check carefully. There are binomial coefficients on the right hand side). So we want to evaluate for Re µ > 0 ∞ Z 2 xme−πµ|x| dx −∞ Both sides vanish if m is odd, and it remains to prove Z ∞ 2k −πµ|x|2 −1/2−k 1 x e dx = µ k (2k)! −∞ (2π) k! which in turn follows by induction on k ∞ ∞ Z 2 1 Z d 2 x2(k+1)e−πµ|x| dx = − x2k+1 e−πµ|x| dx −∞ 2πµ −∞ dx ∞ 2k + 1 Z 2 = x2ke−πµ|x| 2πµ −∞ − 1 −k 2k + 1 2k 2k = µ 2 ∂ x 2(4π)k+1k! x

− 1 −k 1 2(k+1) 2(k+1) = µ 2 ∂ x (4π)k+1(k + 1)! x We consider integrals of the type Z I(τ) = eτφ(x)a(x)dx Rn where a is smooth and compactly supported, φ is smooth with nonnegative real part and τ is supposed to be large. This integral is clearly bounded and we are interested in the behavior for large τ. Proposition 4.20. Suppose that there exists κ > 0 so that |Dφ| − Re φ ≥ κ > 0 4. OSCILLATORY INTEGRALS AND STATIONARY PHASE 59 in the support of φ and that Re φ ≤ 0. Then, if N > 0 there exists cN so that −N (4.18) |I(τ)| ≤ cN τ Proof. There exists a smooth function η so that |∇Φ| ≥ κ/3 if x ∈ supp η ∩ supp a and Re Φ < −κ/3 for x ∈ supp (1 − η) ∩ supp a. We split the two cases by writing a = aη + a(1 − η). For the second term we get pointwise bounds of the integrand by e−τκ/3, which gives the estimate for this term. So we may assume that |Dφ| ≥ κ in the support of a. We calculate Z Z a∇φ eτφ(x)a(x)dx =τ −1 ∇eτφdx |∇φ|2 Z X a∂jφ =τ −1 eτφ ∂ dx. j |∇φ|2 j where we used the divergence theorem to integrate by parts. We obtain the claim by induction. The next situation with a fairly complete understanding is the one dimensional situation. Lemma 4.21 (Van der Corput). We consider Z d I = eih(x)a(x)dx c for real functions h under the assumption that a is compactly supported and that for one j ≥ 1 h(j) ≥ κ > 0. If j = 1 we assume in addition that h0 is monotone. Then

1 − j 0 j |I| ≤ κ (ka kL1 + 23 kaksup) if a is compactly supported. Proof. We begin with the case j = 1 and proceed by induction. Z d Z ih(x) ih 0 e a(x)dx =i e ∂x(a/h )dx c Z Z ih 0 ih 0 =i e ax/h dx + i e a∂x(1/h )dx

+ ieih(d)a(d)/h0(d) − ieih(c)a(c)/h0(c) n 60 4. THE FOURIER TRANSFORM ON R for the ﬁrst term we obtain the estimate with L1 and by monotony the second term is bounded by Z d Z d ih 0 0 e a∂x(1/h )dx ≤kaksup ∂x(1/h )dx c c 0 −1 0 −1 =kaksup|h (c) − h (d) | −1 ≤kaksupκ We assume that we have proven the estimate for j − 1 ≥ 1 and we want to prove it for j. So h(j) ≥ κ and h(j−1) has at most one zero, which we assume to be 0. Let us assume that it is 0. We choose δ > 0. If |x| ≥ δ then |h(j−1)(x)| ≥ κδ. We apply the previous argument on both sides and obtain −1/(j−1) −1/(j−1) 0 j−1 |I| ≤ 2δ sup +δ κ )(ka kL1 + 43 kaksup) |x|≤δ|a(x)| We choose δ = κ−1/j. This requires c < −δ and d ≥ δ, the modiﬁcations for the other cases are trivial. In higher dimensions the only large contributions can come from stationary points, i.e. points x in the support of a where Re φ(x) = 0 and ∇φ(x) = 0. We always assume that these points are nondegenerate in the sense that the Hesse matrix D2φ is invertible. We assume that Re φ ≥ 0. Theorem 4.22. Under the assumptions above we assume that 0 is the only stationary point of φ in the support of a where Re φ(0) = 0, and that it is nondegenerate. Let Z I(τ) = e−τπφadx Rn We write φ = φ(0) + xT Ax + ψ(x). Given N there exists M and C so that (4.19) k M  n  −τφ(0) − n −k − 1 X 1 X 2 −τψ(x) I − e τ 2 (det A) 2 a ∂ (a(x)e ) (4π)kk!  ij ij k=0 i,j=1 x=0 ≤ Cτ −N . There is a suggestive notation for the estimate:

−τπφ(0) −n/2 −1/2 1 X 2 −τψ(x) I − e τ det(A) exp aij∂ij (a(x)e ) 4πτ x=0 = O(τ −∞)

Proof. Again it suﬃces to consider Ar positive deﬁnite. Let η ∈ ∞ C0 (B1(0)), identically 1 on B1/2. We may assume that a is supported in a small ball around x = 0 and that ∂αψ(0) = 0 4. OSCILLATORY INTEGRALS AND STATIONARY PHASE 61

1 1 for |α| ≤ 3. We claim that for 3 < s < 2 Z s −τπφ I1(τ) = (1 − η(xτ ))e adx and N ∈ N there exists CN such that −N |I1(τ)| ≤ cN τ We observe that D|Dφ(x)|−1 ≤ c|x|−2 near x = 0 in the support of a and each integration by parts brings a gain of τ −1. After many integrations by parts we end up with Z Cτ −k |x|−2kdx ≤ Cτ −kτ 2ks. 2|x|≥τ −s 1 Since s < 2 this implies the assertion. Let pM be the Taylor expansion of ae−τπψ at 0 of order M. Then −τπψ M+1 M+1 |pM − ae | ≤ CM |x| τ 3 Z 1/4 −πτxT Ax −τπψ (M+1)( 1 −s) |I2(τ)| = η(xτ )e (pM − ae )dx ≤ cτ 3

We obtain the statement by chosen M large. Finally we consider - under the assuming that Re A is positive definite - Z s −τπxT Ax I3(τ) = (1 − η(xτ ))e pM dx in the same fashion as the first term, using also the Gaussian decay, which, however, does not enter in quantitative fashion since the integration by parts leads to factors |x|−N , which ensures integrability at ∞. This shows that Z s −τπφ(0) −τπxT Ax −N I(τ) − (1 − η(xτ ))e e pM dx ≤ cN τ if M is sufficiently large. The second term has been evaluated in Proposition 4.19. 4.1. Example 1: Korteweg-de Vries equation and the Airy function. The Korteweg-de Vries

ut + uxxx + 6uux = 0 is among the most fascinating partial differential equation. It describes one dimensional water waves, for which it describes a second order approximation: To first order there is a linear transport resp. wave equation, but waves are in general slower than the wave speed of the wave equation. It it strongly related to Schrödingeroperators

ψ → ψxx + uψ. There is a complicated but amazingly explicite way of getting formulas for solutions via the so called inverse scattering theory. Here we will look at the linear part

ut + uxxx = 0 n 62 4. THE FOURIER TRANSFORM ON R

By a Fourier transform it(2π)3ξ3 uˆ(ξ) = e uˆ0(ξ) and with Z 3 3 g(x) = ei[(2π) ξ +2πxξ]dξ

−1/3 −1/3 u(t) = t g(.t ) ∗ uˆ0 The Airy function is deﬁned by

Z 3 Ai(x) = (2π)−1 ei(ξ /3+ixξ)dξ which, as usual, is an abuse of notation and has to be as limit of the integral with the additional factor e−εξ2 or as inverse Fourier transform. Then g(x) = 31/3(2π)−2 Ai(x(3t)−1/3). By the Lemma of van der Corput Ai(x) is uniformly bounded, and since we can differentiate under the integral and apply Proposition 4.20 it is a smooth function (the noncompact interval does not change this fact, but one has to pay attention to what happens at infinity). The phase function is φ(ξ) = ξ3/3 + xξ with derivative ξ2 + x. We consider first the√ case when x is negative and there are two stationary points ξ = ± −x. Let x < −10 and choose a cutoff function ∞ η ∈ C0 ((−2, 2)), identically 1 in (−1, 1) and write

Z √ Z 3 Ai(x) =(2π)−1 η(ξ − −x) ei(ξ /3+ixξ)dξ

Z √ Z 3 + (2π)−1 η(ξ + −x) ei(ξ /3+ixξ)dξ

Z √ √ Z 3 + (2π)−1 1 − η(ξ + −x) − η(ξ − −x) ei(ξ /3+ixξ)dξ

The last term decays faster than any negative power of |x| by Proposition 4.20. The second is the complex conjugate of the first one. We apply Theorem 4.22 to the first term and obtain the first part of Lemma 4.23 (Asymptotics of Airy function).

−1/2 − 1 2 3/2 π − 1 Ai(x) = π (−x) 4 sin( (−x) + ) + O(r 2 3 4 if x < 0 and N −1 − 1 2 3/2 X k 1 − 3k Ai(x) − (2π) x 4 exp(− x ) (−9) Γ(3k + )/(2k)!x 2 3 2 k=0 2 3/2 − 1 − 3k ≤c exp(− x )x 4 2 . N 3 It remains to prove the asymptotics for x > 0. We ﬁrst verify for x > 1

Z 3 Ai(x) = (2π)−1 ei(ξ+iη) +ix(ξ+iη)dξ 4. OSCILLATORY INTEGRALS AND STATIONARY PHASE 63 for 0 ≤ η ≤ ξ, by diﬀerentiating with respect to η, and checking continuity at η = 0. Then

Z √ 3 √ Ai(x) =(2π)−1 ei(ξ+i x) /3+ix(ξ+i x)dξ

Z √ −1 − 2 x3/2 − xξ2+iξ3/3 =(2π) e 3 e dξ.

The proof of Proposition 4.22 consists of taking a Taylor expansion. Here we take a Taylor expansion of exp iξ3/3 and evaluate the integrals. Mutliplication by |ξ|s on the Fourier side roughly corresponds to taking s derivatives. Looking at the expansion of the Airy function suggests that about half a derivative of the Airy function is bounded. A precise version of that is the contents of the next lemma. Proposition 4.24. There exists C so that Z 1 +is i(ξ3/3+xξ) |ξ| 2 e dξ ≤ C(1 + |s|) for s ∈ R. Proof. We decompose the integral into an integral over (−2, 2) which is always bounded, and integrals over (−∞, −1) and (1, ∞) using η as above. Then estimate follows if we prove the estimate Z ∞ 1 +is i(ξ3/3+xξ) (1 − η(ξ))|ξ| 2 e dξ ≤ C(1 + |s|). 0 If x > −1/2 we deﬁne t = ξ3/3 + xξ and get Z ∞ 1 +is i(ξ3/3+xξ) (1 − η(ξ))|ξ| 2 e dξ 0 ∞ Z 1 1 d 3 2 +is i(ξ /3+xξ) = (1 − η(ξ(t)))|ξ(t)| 2 2 e 0 (ξ(t) + x) dξ we integrate once by parts to obtain√ the desired√ estimate for x ≥ −1/2. If x < 0 we decompose into [± −x − 2, ± x + 2] and the complement. On the complement we argue as for x > −1/2. In these intervals we apply the Lemma of von der Corput with j = 2. The second derivative is bounded from below by (−x)1/2 and

d 1/2+is 1 ξ − 1 +is |ξ| = ( + is) |ξ| 2 dξ 2 |ξ| which gives the estimate. Lemma 4.25. Let S(t) be the unitary group deﬁned by the Airy equation. Then 1 − 1 −( 1 − 1 ) k|D| p 2 S(t)u k 0 ≤ c|t| p 2 ku k p 0 Lp (R) 0 L (R) Here |D|su = F −1(|ξ|suˆ). The analogue of the Strichartz estimate is an immediate consequence. 1 Proof. Let for 0 ≤ Re z ≤ 2 z2 −1 iξ3/3 Tzu0 = e F (e uˆ0) n 64 4. THE FOURIER TRANSFORM ON R

If z = is then by Plancherel

kTisu0kL∞ ≤ ku0kL2

1 and if z = + is T 1 is the convolution by a bounded function according 2 2 +is to the previous Proposition by

1 −s2 ce 4 (1 + |s|) which is uniformly bounded, hence the operator is bounded from L1 to L∞. By the Theorem of Riesz-Thorin on complex interpolation

kTtu0kLp0 ≤ cku0kLp

1 1 1 1 if 1 ≤ p ≤ 2, 0 ≤ t ≤ 2 , p + p0 = 1, p0 = t. This implies the assertion for t = 1. For general t −1/3 1/3 gt(t, x) = t g1(x/t ) and 1 −1/2 k|D| 2 gtkL∞ ≤ c|t| .

The same argument yields the claimed estimate.

4.2. Example 2: The half wave equation.

5. Quantization n n n Quantization means a map from functions on R ×R to operators on R . We choose a so called semiclassical calculus with a semi classical parameter h > 0, basically Planck’s constant in physics. n We begin with several preliminary observations. Denote (x, y) ∈ R × m n m n+m R . If u ∈ S(R ) and v ∈ S(R ) then (x, y) → u(x)v(y) ∈ S(R ). If 0 m n+m n T ∈ S (R ) and u ∈ S(R ) then x → T (u(x, .)) ∈ S(R ) and for the partial Fourier transform we have

n+m (Fyu)(x, η) ∈ S(R ).

5.1. Semiclassical Fourier transform. We deﬁne Z − i hx,ξi ˆ Fhφ(ξ) = e h φ(x)dx = φ(ξ/(2πh)) Rn with inverse 1 Z i −1 − h hx,ξi Fh φ(x) = n e φ(ξ)dξ. (2πh) Rn 1 Our Fourier transform is the one with h = 2π and we could restrict to this h in the sequel. n n We denote point in R × R by (x, ξ) and we call x position and ξ momentum. 5. QUANTIZATION 65

5.2. Deﬁnition of quantization. n n Definition 4.26. Let a ∈ S(R ×R ). We deﬁne the Weyl quantization to be the operator

1 Z Z i x + y w h hx−y,ξi (4.20) (a (x, hD)u)(x) := n e a( , ξ)u(y)dydξ, (2πh) Rn Rn 2 the standard quantization

1 Z Z i h hx−y,ξi (4.21) (a(x, hD)u)(x) := n e a(x, ξ)u(y)dydξ, (2πh) Rn Rn and, for 0 ≤ t ≤ 1, the t quantization Z Z 1 i hx−y,ξi (4.22) (Opt(a)u)(x) := e h a(tx + (1 − t)y, ξ)u(y)dydξ. 2πh Rn Rn We denote by 1 Dj = ∂ i xj and Dα for multiindices α with the obvious meaning. α α (1) Opt(ξ ) = (hD) u P α (2) If a(x, ξ) = |α|≤N aα(x)ξ then X α a(x, hD)u = aα(x)(hD) u |α|≤N w h h (3) hx, hDi u = 2 hD, xui + 2 hx, Dui n n Lemma 4.27. Let a ∈ S(R × R ) and 0 ≤ t ≤ 1. Then Opt(a) defines 0 ∗ a continuous linear operator from S to S. The formal adjoint is Opt(a) = Op1−t(¯a). Proof. We have Z Opt(a)u(x) = Kt(x, y)u(y)dx Rn with 1 Z i h hx−y,ξi Kt(x, y) = n e a(tx + (1 − t)y, ξ)dξ (2πh) Rn −1 = Fh (a(tx + (1 − t)y, .) (x − y) The partial Fourier transform g(z, w) = (F −1a(z, .))(w) is a Schwartz function by the considerations above. The equations tx + (1 − t)y = z, x − y = w 0 n define linear maps, and hence Kt is a Schwartz function. Let T ∈ S (R ). Then n x → T (Kt(x, .)) ∈ S(R ) which proves the first assertion. The statement about the formal adjoint follows from the definition. In particular, if a is real then aw(x, hD) is formally self adjoint. 0 n n Lemma 4.28. Let a ∈ S (R × R ) and 0 ≤ t ≤ 1. Then Opt(a) defines n 0 n a continuous linear operator from S(R ) to S (R ). n 66 4. THE FOURIER TRANSFORM ON R

Proof. By the same argument as above the partial inverse Fourier transform defines a distribution, and the contructions above of the kernel n n n defines a tempered distribution T on R × R . If u, v ∈ S(R ) then n u(x)v(y) ∈ S(R ). n and, given v ∈ S(R ) the map u → T (u(x)v(y)) defines a unique tempered distribution which coincides with the quantization whenever the previous definition applies. 5.3. Simple examples. Lemma 4.29. If a(x, ξ) = b(x) then

Opt(a)u = b(x)u Proof. The claim follows from Fourier inversion if t = 0. It suffices to n proof it for a ∈ S(R ). Let u ∈ S and compute ∂t Opt u Z Z n 1 i X h hx−y,ξi = n e (∂jb)(tx + (1 − t)y)(xj − yj)u(y)dydξ (2πh) n n R R j=1 Z n Z 1 X i h hx−y,ξi = n ∂ξj · e (∂jb)(tx + (1 − t)y)u(y)dy dξ ih(2πh) n n R j=1 R = 0 by the divergence theorem (applied on balls with radii tending to infinity, and using the decay of the Schwartz functions), and since the inner integral defines a Schwartz function. Lemma 4.30. Let l(x, ξ) = hx∗, xi + hξ∗, ξi be a linear function (x∗, ξ∗ ∈ n R ). Then ∗ ∗ Opt(l)u = hx , xiu + hξ , hDui. Pn If c(x, ξ) = J=1 cj(x)ξj then n h X cw(x, hD)u = (Dj(c u) + c Dju) 2 j j j=1 Proof. The first statement is a consequence of the previous considerations. For the second we use the definition Z Z 1 X x + y i w h hx−y,ξi c (x, hD)u = n cj( )ξje u(y)dξdy (2πh) Rn Rn 2 Z Z 1 X x + y i h hx−y,ξi = − n cj( )hDyj e u(y)dξdy (2πh) Rn Rn 2 Z Z 1 X h x + y i h hx−y,ξi = n ∂xj cj( )e u(y)dξdy (2πh) i Rn Rn 2 Z Z 1 X x + y i h hx−y,ξi + n cj( )e hDyj u(y)dξdy (2πh) Rn Rn 2 h X = (∂ c )u + h c (x)D u 2i xj j j xj This implies the second statement. 5. QUANTIZATION 67

Lemma 4.31. The following identities hold: w w (Dxj a) = [Dxj , a ] and w w h(Dξj a) = −[xj, a ] There is symplectic quadratic form (4.23) σ(z, w) = ξ · y − x · η, z = (x, ξ), w = (y, η) 2n on R . We can write it via the Euclidean inner product 0 1 σ(z, w) = hJz, wi J = . −1 0 Lemma 4.32. w i l i l(x,D) i hx∗,xi+ i hx∗,ξ∗i ∗ e h u = e h u := e h 2h u(x + ξ ).

2n If l, m ∈ R then i l(x,hD) i m(x,hD) i σ(l,m) i (l+m)(x,hD) e h e h u = e 2h e h Consider the PDE

ih∂sv + l(x, hD)v = 0 Its solution is - independent of t

i hx∗,xi+ i hx∗,ξ∗i ∗ v(1, x) = e h 2h u(x + ξ ). which is the meaning of the not rigorously deﬁned second equality.

Proof.

i 1 Z Z i i ∗ ∗ x+y h l w h hx−y,ξi h (hξ ,ξi+hx , 2 i) (e ) u = n e e u(y)dydξ (2πh) Rn Rn i hx∗,xi e h Z Z i ∗ i ∗ h hx−y+ξ ,ξi 2h hx ,yi = n e (e u(y))dydξ (2πh) Rn Rn i hx∗,xi e h Z Z i i ∗ ∗ h hx−y,ξi 2h (hx ,y+ξ i) ∗ = n e (e u(y + ξ )) dy dξ (2πh) Rn Rn i hx∗,xi+ i hx∗,ξ∗i ∗ =e h 2h u(x + ξ ). This implies the ﬁrst assertion. The second assertion follows from an application of the ﬁrst formula: Let l = (x∗, ξ∗) and m = (y+; η∗). Then

i l w i w i l w h i hy∗,ξ∗i i hy∗,.i ∗ i (e h ) (e h ) u(x) =(e h ) e 2h e h u(. + η )

i (hx∗,ξ∗i+hy∗,η∗i) i hy∗,ξ∗i i hx∗+y∗,xi ∗ ∗ =e 2h e h e h u(x + η + ξ ) i (hy∗,ξ∗i−hx∗,η∗i) i (l+m) w =e 2h (e h ) u(x).

If symbols only depend on ξ and not on x then all quantization yield the same result. n 68 4. THE FOURIER TRANSFORM ON R

5.4. Composition of semiclassical pseudodiﬀerential operators. Theorem 4.33. Let Q be a nonsingular selfadjoint n × n matrix. Then − 1 Z ih hQD,Di | det Q| 2 iπ sign Q − i hQ−1y,yi e 2 u(x) = e 4 e 2h u(x + y)dy n/2 (2πh) Rn 1 Z Z i ihσ(Dz,Dw) − h σ(z1,w1) e u(z, w) = 2n e u(z + z1, w + w1)dz1dw1 (2πh) R2n R2n Here sign Q is the signature, the number of positive eigenvalues minus the number of negative eigenvalues. Proof. The second statement is a special case of the ﬁrst one with the quadratic form given by the matrix 0 −J . J 0 We compute i hQD,Di i hQhD,hDi e h u(x) =e 2h u(x)

1 Z Z i i h hx−y,ξi 2h hQξ,ξi = n e e u(y)dydξ (2πh) Rn Rn − 1 Z | det Q| 2 iπ sign Q − i hQ−1(x−y),x−yi = e 4 e 2h u(y)dy (2πh)n/2 − 1 Z | det Q| 2 iπ sign Q − i hQ−1(y),yi = e 4 e 2h u(x + y)dy (2πh)n/2 where we exchanged the order of integration (ﬁrst multiplying by e−ε|ξ|2 and then we passed to the limit as ε → 0. The ξ integration yields the inverse Fourier transform of the complex Gaussian. The last integral follows by the translation invariance of the integration. With Proposition 4.22 we get an asymptotic series for this integral in powers of h. Theorem 4.34. F −1aw(x, hD)F = aw(hD, −x) n n Proof. We prove the claim for a ∈ S(R × R ). The integral kernel −1 w Kh(x, y) of Fh a Fh is 1 Z Z Z i 0 0 0 0 h (hx ,xi+hx −y ,ζi−hy ,yi) 0 0 0 0 2n e a((x + y )/2, ζ)dy dx dζ (2πh) Rn Rn Rn 1 Z Z Z i 0 x+y −n h 2(hx ,ζ+ 2 i−hz,y+ζi) 0 = 2n 2 e a(z, ζ)dx dzdζ (2πh) Rn Rn Rn 1 Z i x + y h hx−y,zi = n e a(z, − )dz (2πh) Rn 2 where we used in the last step that

−1 2i hx0,ζ+ x+y n x + y F (e h 2 ) = 2 δ (ζ + ). h 0 2 5. QUANTIZATION 69

Interpretation: The map (x, ξ) = J(x, ξ) is a symplectic map. The semiclassical Fourier transform quantizes the symplectic map J. We deﬁne

A(D) = σ((Dx,Dξ); (Dy,Dη)) Theorem 4.35 (Composition for Weyl quantization). With

a#b(x, ξ) := eihA(D)(a(x, ξ)b(y, η)) y=x,ξ=η we have (a#b)wu = awbwu and with i σ(h(D ,D ,D ,Dη) c(x, ξ) = e h x ξ y (a(x, ξ)b(y, η) x=y,ξ=η a(x, hD)b(x, hD)u = c(x, hD)u. Proof. Let Z − i hl,(x,ξ)i aˆ(l) = e h a(x, ξ)dxdξ R2n By the inversion formula

1 Z i w h l(x,hD) a (x, hD) = n aˆ(l)e dl (2πh) R2n We apply this formula for a and b to get

1 Z Z i i w w ˆ h l(x,hD) h m(x,hD) a (x, hD)b (x, hD) = 4n aˆ(l)b(m)e e dmdl (2πh) R2n R2n 1 Z i h r(x,hD) = 2n cˆ(r)e dr (2πh) R2n where, by Lemma 4.32

1 Z iσ(l,r−l) ˆ 2h cˆ(r) = 2n aˆ(l)b(r − l)e dl (2πh) R2n 1 Z Z 1 Z i i h hl+m−r,zi 2h σ(l,m) ˆ = 2n 2n e dz e aˆ(l)b(m)dldm (2πh) R2n R2n (2πh) R2n Hence, since

i σ(hD ,hD ) i (hl,zi+hm,wi) i σ(l,m) i (hl,zi+hm,wi) e 2h z w e h = e 2h e h

Z Z 1 i σ(hD ,hD ) i (hl,zi+hm,wi) 2h z w h ˆ c(z) = 4n e e aˆ(l)b(m)dldm 2n 2n z=w (2πh) R R The statement follows now by Theorem 4.33. The functions are called symbols. The proposition allows to compute the symbol of the composition from the symbols of the operators. The method of stationary phase allows to obtain an expansion in terms of powers of h. This expansion is given by formally expanding the exponential, and then applying the operators to the functions. n 70 4. THE FOURIER TRANSFORM ON R

Since there is exactly one stationary point we can apply this expansion even if we have unbounded symbols. The first term is always the product. The second term is n X A(D)a(x, ξ)b(y, η)|x=y,ξ=η = h ∂xj a∂ξj b − ∂ξj a∂xj b. j=1 This is the first term in the expansion of the operator. The expression on the RHS is h times the Poisson bracket of the symbols a and b, h{a, b}. Here we already see an important property of the semiclassical limit h → 0: The composition becomes closer and closer to a multiplication. This suggest that as first approximation to composition of operators we may consider the multiplication. In particular we expect some approximate inverse to be given by taking the inverse of the symbol. So if we consider the elliptic equation with smooth coefficients ij 2 −a ∂j u = f ij then the operator is given by the classical symbol a (x)ξiξj. One can proof that

ij −1,w ij 2 D[(1 + a (x)ξiξj) a ∂ij − 1] is bounded operator on Lp. To carry that program out we need (1) Criteria for Lp boundedness (2) A calculus for pseudodiﬀerential operators, i.e. the operators obtained by quantizing symbols.

6. Cotlar’s Lemma and the Theorem of Calderón-Vaillancourt In this section we will prove L2 boundedness of semiclassical operators under fairly weak assumptions on the symbol a. As a start we consider Schwartz functions. n n Lemma 4.36. Let a ∈ S(R × R ). Then α β k Opt(a)fkL2(Rn) ≤ c(n) sup sup |∂x ∂ξ a(x, ξ)|kfkL2 |α|,|β|≤n+1 x.ξ It suffices to assume that the quantity on the right hand side is bounded. −1 Proof. The partial Fourier transforma ˇ = Fh za(w, .)(z) is a Schwartz function. It satisfies β −1 β −|β| γ β |z aˇ| = |F ((hD)ξ a)| ≤ c(n) sup h |ξ ∂ξ a(x, ξ)| |γ|≤n+1 We obtain for the integral kernel −n Kt(x, y) = h aˇ(tx + (1 − t)y, (x − y)) and −n −n−1 α β |Kt(x, y)| ≤ c(n)h (1 + |x − y|/h) sup sup |ξ ∂ξ a|. |α|,|β|≤n+1 w,ξ 6. COTLAR’S LEMMA AND THE THEOREM OF CALDERON-VAILLANCOURT´ 71

Then Z Z α β sup |Kt(x, y)|dy + sup |Kt(x, y)|dx ≤ c2(n) sup sup |ξ ∂ξ a| x y |α|,|β|≤n+1 w,ξ and the assertion follows from Schur’s lemma 2.17. The product of two functions with disjoint support is zero. This fails to be true for the composition of pseudodiﬀerential operators, but it remains approximately true. We focus on h = 1 in the sequel and recover the assertion for 0 < h by a standard rescaling.

1/2 1 1 Lemma 4.37. Let h > 0, u˜(x) = u(h x) and ah(x, ξ) = a(h 2 x, h 2 ξ). Then w 1 1 w a (h 2 x, hD)u(h 2 x) = ah (x, D)˜u. Proof. aw(h1/2x, hD)u(h1/2x) 1/2 1 Z Z h x + y i 1/2 h hh x−y,ξi = n a( , ξ)e u(y)dydξ (2πh) Rn Rn 2 Z Z 1 x + y ihx−y,ξi = n ah( , ξ)e u˜(y)dydξ (2π) Rn Rn 2 ∞ n n Lemma 4.38. Let a, b ∈ C (R × R ) with d = d(supp a, supp b) > 0 and α β α β n |∂x ∂ξ a| + |∂x ∂ξ a| ≤ B for x, ξ ∈ R , |α|, |β| ≤ 6n + N + 2 Then w w −N 2 ka (x, D)b (x, D)fkL2 ≤ c(n, N)d B kfkL2 . Proof. We recall that the symbol of the composition is

1 Z Z i − 2 σ(w1,w2) c(z) = 2n e a(z + w1)b(z + w2)dw1dw2 (2π) R2n R2n The point 0 is√ the unique critical point, and the integrand vanishes for |(w1, w2)| ≤ d/ 2. The integrations by parts in the stationary phase argument lead to bound (since the same type of argument holds for derivatives) α 2 −N |∂ c| ≤ cN B d . This implies the statement of the Theorem of Calderón-Vaillancourt below. At this point we prove the statement under the additional assumption that b is supported in a unit ball B1(x0, ξ0) Then α −N −M |∂z c(z)| ≤ cd (1 + |z − (x0, ξ0)|) . This implies that the symbol c is a shifted Schwartz function. Since w w [c (x, D)u](x + x0) = c (x + x0,D)u(. + x0) and −ihξ0,xi w ihξ0,yi w e c (x, D)(e u(y))(x) = c (x, D − ξ0)u n 72 4. THE FOURIER TRANSFORM ON R and neither translation nor multiplication by the complex exponential changes the L2 the boundedness assertion follows from Lemma 4.36. It remain to count the number of derivatives we need. For the application of (4.36) we need n+1 derivatives and a decay of power n+1. Each derivative in the argument above gives one inverse power of d resp. one inverse power −1 of |z − (x0, ξ0)| or one of |w| . The integration is over a 4n dimensional space and it uses up 4n inverse powers of w. Thus we need 4n+(2n+2)+N derivatives for this argument. It is clearly not optimal. This special case will suffice to proof the Theorem of CalderónVail- lancourt 4.40, which in turn implies the full statement of the proposition here.

Lemma 4.39 (Cotlar’s lemma). Let H be a Hilbert space and Tj : H → H + a family of operators, and γ : Z → R such that ∗ 2 ∗ 2 kTj TkkL(H) ≤ γ (j − k), kTjTk kL(H) ≤ γ (j − k) for all j, k. Suppose that ∞ X γ(l) =: A < ∞ l=−∞ Then N X k TjkL(H) ≤ A j=1 for all N. PN Proof. We have with T = j=1 Tj N ∗ n X ∗ ∗ (T T ) = Tj1 Tk1 ...Tjn Tkn j1,...jn,k1,...,kn=1 Now ∗ ∗ Y ∗ kTj1 Tk1 ...Tjn Tkn k ≤ kTj1 kkTkn k kTki Tji+1 k Y kT ∗ T ...T ∗ T k ≤ kT +T ∗ k j1 k1 jn kn ji ki We multiply and take square roots Y Y kT ∗ T ...T ∗ T k ≤ (kT kkT )1/2k kT T ∗ k1/2 kT +T ∗ k1/2 j1 k1 jn kn j1 kn ki ji+1 ji ki and thus with B = sup kTjk N ∗ n X Y 2n−1 k(T T ) k ≤ B γ(ji − ki)γ(ki − ji+1) ≤ NBA

j1,...jn,k1,...,kn=1 Since T ∗T is selfadjoint 1/n 1 2 ∗ ∗ n 1/n 2− n kT kL(H) = kT T kL(H) = k(T T ) kL(H) ≤ (NB) A We let n tend to inﬁnity to obtain the assertion. 8n+4 n n Let a ∈ C (R × R ) satisfy β α (4.24) sup |x ∂x a(x, ξ)| ≤ B x,ξ,|α|+|β|≤c(n) 6. COTLAR’S LEMMA AND THE THEOREM OF CALDERON-VAILLANCOURT´ 73

Lemma 4.40 (Calderón-Vaillancourt). , The operators T = aw(x, D) satisfies kT fkL2 ≤ c(n)BkfkL2 . Proof. We choose a function η supported in [−1, 1]n so that X η(x − k) = 1. k∈Zn ∞ 2n 2n (Chooseη ˜ ∈ C0 ((−1, 1) ) nonnegative and identially 1 on [−1/2, 1/2] . Then set η˜(z) η(z) = P k η˜(z + k) 2n Given k ∈ Z we define ak(z) = a(z)η(z + k) w Tkf = ak f We claim that ∗ ∗ −2n−1 (4.25) kTk TlkL(L2) + kTk TlkL(L2) ≤ C(1 + |k − l|) Since the square roots of the bounds on the RHS are summable this implies X k TklfkL2 ≤ CkfkL2 . |k|≤N It is not difficult to let N tend to infinity. Both terms on the left hand side have the same structure and it suffices to deal with one of them. Now ∗ w w Tk Tl = ak (x, D)al (x, D) We apply Proposition 4.36 twice to see that ∗ 2 kTk TlkL(L2) ≤ CB and, if |k − l| ≥ 2n, by Proposition 4.38 ∗ 2 −2n−1 kTk TlkL(L2) ≤ cB |k − l| . ß

CHAPTER 5

Singular integrals of Calder´on-Zygmund type

1. The setting The basic setting of this chaper are spaces of homogeneous type (X, d, µ) where (X, d) is a complete metric space and µ is a Borel measure, ﬁnite on bounded sets, which satisﬁes a doubling condition: There exists b so that

µ(B3r(x)) ≤ bµ(Br(x)) Lemma 5.1. Let A be a closed bounded set. Then A is compact. Proof. We prove that A is totally bounded. Closed totally bounded sets in a complete metric space are compact. Since A is bounded there exists a ball A ⊂ BR(x0). Let R > r > 0. There is at most a ﬁnite number of disjoint balls with centers in A with radii = r/3 by the doubling condition: there exists k so that R ≤ 3k−3r.

Let Br/3(xj) a ball which intersects A. Then

A ⊂ BR ⊂ B3k−1(r/3)(xj) and by the doubling condition k−1 µ(A) ≤ µ(BR) ≤ b µ(Br/3(xj).

Let Bj be N such disjoint balls. The measure of there union is at least 1−k [ Nb µ(B) ≤ µ( Bj) ≤ bµ(B) and hence there are at most bk such balls. Take a maximal such sequence Br/3(xi). Then as for the Lemma of Vitali the union of Br(xi) covers A. Theorem 5.2. The measure µ is σ ﬁnite and regular, i.e. (5.1) µ(A) = sup µ(K) = inf µ(U) K⊂A A⊂U for every Borel set.

Proof. Let xo ∈ X. We deﬁne the Borel measure

µR(A) = µ(BR(x0) ∩ A).

It is ﬁnite and µ = limn→∞ µn, and hence it is σ ﬁnite. Let A be the set of Borel set for (5.1) holds. We will show that A contains all compact sets, with A it contains its complement, and it closed under countable unions. So it is a σ algebra containing all open sets. The Borel σ algebra is the smallest such σ algebra, and hence A consists of all Borel sets.

75 76 5. SINGULAR INTEGRALS OF CALDERON-ZYGMUND´ TYPE

For compact sets K inner regularity (approximation by compacts) is trivial. Since K = ∩Uj with Uj(d(x, K) < 1/j) is countable intersection µ(K) = limj µ(Uj) and outer regularity follows. A is an inner regular Borel set if and only if X\A is outer regular - this obvious for compact metric spaces, and requires use of σ ﬁniteness in general. S Let A = Aj be a countable union of inner regular sets, and let ε > 0. Then there exists compact sets Kj and open sets Uj with

Kj ⊂ Aj ⊂ Uj and −1−j µ(Uj) ≤ µ(Kj) + 2 ε. SN Then j=1 Kj is compact and

N ∞ [ [ µ( Kj) → µ( Kj). j=1 j=1

Moreover [ µ(A) ≤ µ( Kj) + ε and inner regularity follows. Outer regularity is similar but simpler.

The most important such space is the Euclidean space with the Lebesgue measure.

Definition 5.3. We call a linear operator T Calder´on-Zygmundopera- tor if p p (1) There exists 1 < p0 ≤ ∞ such that T : L 0 (µ) → L 0 (µ)and

kT fkLp0 (µ) ≤ AkfkLp0 (µ).

(2) There exists a continuous kernel function K : X × X\{(x, x): x ∈ X} which satisﬁes Z T f(x) = K(x, y)f(y)dµ

whenever f is compactly supported and x is not contained in the support of f. Moreover Z |K(x, y) − K(x, z)|dµ(x) ≤ A d(x,y)≥2d(y,z) for all y, z ∈ X.

The Hilbert transform on R or T is the most important example. It is bounded by the constant 1 as linear operator on L2, and has the kernel −1 c(x − y) for R. 2. THE CALDERON-ZYGMUND´ THEOREM 77

2. The Calderón-Zygmund theorem We denote the doubling constant by b in this section. Theorem 5.4. Let T be a Calderón-Zygmundoperator. Then the weak type estimate holds for all f ∈ Lp0 ∩ L1. A Z µ({x : |T f| > t}) ≤ (2p0 + b4) |f|dµ t X p Let 1 < p ≤ p0. Then T defines a unique bounded operator from L (µ) to Lp(µ) which satisfies

4 p kT fk p ≤ c(p )Ab kfk p L 0 p − 1 L The second statement follows from the weak type inequality by the interpolation theorem of Marcinkiewicz 2.24 or even from the simpler version in the proof of the bounds for the maximal function Theorem 2.20. The uniform constant as p → p0 involves a second application of complex interpolation Theorem 2.16. n We begin the proof by the Whitney covering lemma, first in R . A dyadic cube is a cube of the form j n Qs,k = 2 ([0, 1) + k) n n with j ∈ Z and k ∈ Z . A dyadic cube has 2 children and 1 parent. Given n s the cubes cover R . n n If U ⊂ R is open, and not equal to R we can cover it by dyadic disjoint cubes of size 2s between the distance of the cube to the boundary, and 2n times the distance to the boundary. This is called a Whitney covering. We formalize this for spaces of homogeneous type. Lemma 5.5 (Whitney). Let U be open, U 6= X. Then there exists a countable sequence of balls Bj so that 1 (1) The balls 3 Bj are disjoint. (2) The balls Bj cover U (3) The balls 2Bj are contained in U. (4) The balls 3Bj are not contained in U. Proof. Given x ∈ U we define r(x) = d(x, X\U)/3. The balls Br(x)/3(x) cover U. Let K ⊂ U be compact. There is a finite number of the balls covering K. By the lemma of Vitali there is a disjoint subset Bj = Br(xj )/3(xj) so that the balls Br(xj )(xj) this cover K. Fix x0 ∈ X and let

Kl = {x ∈ U : d(x, x0) ≤ l, d(x, X\U) ≤ 1/l} S Then Kl is compact, monoton (Kl ⊂ Kl+1) and U = Kl. We proceed recursively, and always keep the balls we have already chosen. In the sequel we denote the doubling constant by b. Lemma 5.6 (Calder´on-Zygmund decomposition). Let f ∈ L1(µ) and 1 R t > µ(X) |f|dµ . Then there exists a decomposition f = g + b 78 5. SINGULAR INTEGRALS OF CALDERON-ZYGMUND´ TYPE and X b = bk k so that (1) |g(x)| ≤ b2t, R |g|dµ ≤ R |f|dµ. (2) bk is supported in balls Bk. They satisfy Z Z |bk|dµ ≤ 2tbµ(Bk), bkdµ = 0

P b2 R (3) k µ(Bk) ≤ t fdµ Proof. Let U = {x : Mf(x) > t} and let Bj be the balls of the Whitney decomposition. By Theorem 2.20 b µ(U) ≤ kfk 1 t L We deﬁne recursively j−1 [ Qj = Bj ∩ {X\ Ql}. l=1 Then 1 (5.2) B ⊂ Q ⊂ B , 3 j j j the Qj are disjoint, and their union is U. We deﬁne ! 1 Z bj = χQj f − fdx µ(Qj) Qj and f(x) if x∈ / U g(x) = 1 R f if x ∈ Qj Qj Qj We verify the properties. It is immediate that R |g|dµ ≤ inf |f|dµ.By construction X f = g + b with b = bj Since Z Z |f|dµ ≤ |f|dµ ≤ tbµ(B) Bj 3Bj −1 and µ(Qj) ≥ b µ(Bj) we have |g| ≤ b2t in U. Outside U |g(x)| ≤ Mf(x) ≤ t almost everywhere. 2. THE CALDERON-ZYGMUND´ THEOREM 79

Similarly

Z Z Z −1 |bj|dµ = f − µ(Qj) fdµ dµ Bj Qj Qj Z ≤2 |f|dµ 3Bj ≤2btµ(Bj) since 3Bj ∩ X\U 6= {}, using the deﬁnition of the maximal function. Now, by the weak type estimate of the maximal function X X µ(Bj) ≤b µ(Qj) =bµ(U) b2 ≤ kfk 1 . t L We turn to the proof of weak type estimate of Theorem 5.4.

Proof. We may choose A by multiplying T by a constant. There is nothing to show if t ≤ µ(X)−1 R |f|dµ. If t ≤ µ(X)−1 R |f|dµ then Rn Z µ(X) ≤ |f|dµ/t and the weak type estimate is trivial. Hence we assume t > µ(X)−1 R |f|dµ. We decompose f = g + b as in the Calder´on-Zygmund decomposition. Then

kgkL1 ≤ kfkL1 by the Calder´on-Zygmund decomposition. Thus

−p0 p0 µ(|T (g)(x)| > t/2) ≤(2/t) kT gkLp0 2Ap0 Z ≤ |g|p0 dµ t (5.3) (2A)p0 2p0−2 ≤ b kgk 1 t L (2A)p0 2p0−2 ≤ b kfk 1 . t L Since X 2 t µ(3Bj) ≤ b kfkL1 it suﬃces to bound (together with Tschebycheﬀ’s inequality) Z X Z |T b|dµ ≤ |T bj|dµ. X\U j X\2Bj

We observe that for x ∈ X\2Bj and zj the center Z Z T bj(x) = K(x, y)bjdµ = (K(x, y) − K(x, zj))bj(y)dµ(y) Bj 80 5. SINGULAR INTEGRALS OF CALDERON-ZYGMUND´ TYPE R since the bj = 0. Thus Z ZZ |T bj|dµ(x) ≤ |K(x, y) − K(x, zj)|dµ(x)|bj|dµ(y) X\2Bj d(x,zj )≥2rj Z ≤A |bj|dµ

2 ≤2Atb µ(Qj). and hence 4 2 2 X 4Ab µ({T b > t/2} ∩ (X\U)} ≤ kT bk 1 ≤ 4Ab µ(B ) ≤ kfk 1 t L (X\U) j t L and hence A p0 p0−1 2(p0−1) −1 2 4 µ({T f > t}) ≤ 2 A b + A b + 4b kfk 1 . t L −2 We choose A = b . Remark 5.7. (1) There are obvious vector valued versions. We consider weakly continuous functions, i.e. f : U → E is weakly measurable if for every element of the dual space e∗ e∗ ◦ f is measurable. Definition and estimate for the maximal function work without change for weakly measuable functions which assume values in Banach spaces. The interpolation theorems of Marcinkiewicz and Riesz-Thorin hold for operators from Lp(µ; E) to Lq(ν, F ) for real resp. complex Banach spaces E and F . In that case the kernel function has values in the Banach space of continuous linear operators from E to F . Such a map to linear operators is measuable if for all e ∈ E and f ∗ ∈ F ∗ the map x → f ∗(K(x, y)e is measurable. The integral for weakly continuous integrable functions with values in Banach spaces is defined through the application of a linear form. p p p0 p0 (2) If T : L 0 (µ) → L 0 (µ) then the adjoint maps L 0 to L 0 . If K(x, y) is the kernel function of T then K(y, x) is the kernel function of T ∗. Thus, if Z |K(x, y) − K(z, y)|dµ(x) ≤ A d(y,x)≥2d(y,z) ∗ p 0 then T is a bounded operator on L (µ) for 1 < p ≤ p0. So under this condition p2 kT fk p ≤ c(b, p ) kfk p . L 0 p − 1 L Lemma 5.8. Suppose that there exists ε > 0 such that d(x, x˜) + d(y, y˜)ε min{µ(B (x)), µ(B (˜x)}|K(x, y)−K(˜x, y˜)| ≤ c d(x,y) d(˜x,y˜) d(x, y) + d(˜x, y˜) for all x, y, y,˜ y˜ then the kernel condition is satisfied for T and T ∗. In particular, if T : Lp0 → Lp0 has an integral kernel which satisfies the condition above then it defines a unique bounded operator from Lp to Lp. 3. EXAMPLES 81

In particular this holds for the Euclidean space with the Lebesgue measure if −n−1 (5.4) |DxK(x, y)| + |DyK(x, y)| ≤ c|x − y| where one of the conditions suﬃces for either p ≥ p0 or p ≤ p0. It is not hard to see that if T is a Calder´on-Zygmund operator with integral kernal k then

(5.5) µ(Bd(x,y)(x))|K(x, y)| ≤ c(b)A

3. Examples 3.1. Fourier multipliers. n n Lemma 5.9 (Mihlin-Hörmander). Let a ∈ C (R \{0}) and assume that sup ||ξ|αDαa(ξ)| ≤ A |α|≤n+2 The operator T : L2 3 f → F −1afˆ ∈ L2 has a convolution kernel K(x, y) = k(x − y) with sup |z|n|k(z)| + |z|n+1|Dk(z)| ≤ c(n)A z It is a Calderon-Zgymund operator. ∞ 1 Proof. Let φ ∈ C0 ({x : 2 ≤ |x| ≤ 4}). We claim that k(x) = φ(x)T is a differentiable function which satisfies (5.6) |k(x)| + |Dk(x)| ≤ c Since the integral kernel of the Fourier multiplier a(λ(x)) is λ−nk(x/n) and since the conditions are invariant under rescaling this implies the full desired estimate. The Fourier multiplier of the convolution by φ is the convolution of the Fourier transform of k with a. Then kˆ ∈ S and Z kξˆ αdξ = 0

Thus, with pn+1 the Taylor polynomial of degree n + 1 at ξ Z φˆ ∗ a(ξ) = φˆ(η)(a(ξ − η) − pn(η))dη and by the Taylor formula −n−2 n+2 |(a(ξ − η) − pn+1(η))| ≤ c(n)|ξ| |η| for |η| ≤ |ξ|/2 and hence

Z ˆ |φ ∗ a(ξ)| ≤ φ(η)(a(ξ − η) − pn+1(η))dη |η|≤|ξ|/2 Z + c |φˆ(η)||η|n+2dξ |η|≥|ξ|/2 ≤c(n)(1 + |ξ|−n−2) This implies the assertion. 82 5. SINGULAR INTEGRALS OF CALDERON-ZYGMUND´ TYPE

The Riesz transforms deﬁned by a Fourier multiplier are important examples: −1 a(ξ) = ξj|ξ| with a convolution kernel (for n ≥ 2)

2Γ((n + 1)/2 xj i n−1 n+1 π 2 |x| since −1 Γ((n − 1)/2 1−n F|x| = n−1 |x| π 2 Then 2Γ((n + 1)/2 xj − yj K(x, y) = i n−1 n+1 π 2 |x − y| It satisfies the conditions on the derivatives above. For n = 1 we obtain again the Hilbert transform. We denote the Riesz transforms by Rj. Similarly, if n ≥ 2, the second order Riesz transforms defined by the Fourier multiplier ξiξj |ξ|2 then the kernel function is 1 xixj − δij c n . n |x|2+n Again this follows for n ≥ 3 from taking derivatives of the inverse Fourier transform of |ξ|−2, and we obtain a formula for the constant c. The case n = 1 is trivial - the second order Riesz transform is the identity. For n = 2 we may take a detour via the Cauchy kernel and calculate 1 F −1 ξ1 + iξ2 and then take the real part. We denote the second order Riesz transforms Rij. If ∆u = f for u ∈ S then 2 ∂xixj u = Rijf and we obtain from the CalderónZygmund estimate 2 2 p k∂ uk p ≤ c(n) kfk p . ij L p − 1 L 3.2. The heat equation. The heat equation leads to a Calderón- Zygmund operator on a space of homogeneous type. Consider

ut − ∆u = f If u ∈ S then iτ uˆ = fˆ t iτ + |ξ|2 and hence kutkL2 ≤ kfkL2 3. EXAMPLES 83 and similarly 2 kDxukL2 ≤ ckfkL2 The convolution kernel is of T : f → ut is

2 2 2 1 − |x| −1 |x| 1 − |x| ∂ e 4t = t (2nπ + ) e 4t t (4πt)n/2 4t (4πt)n/2 for t > 0 and 0 otherwise. We deﬁne the metric d((x, t), (y, s)) = max{|x − y|, |t − s|1/2}. It is easy to verify the triangle inequality. Let µ be the Lebesgue measure. The doubling condition with b = 3n+2 is immediate. Then |g(x, t)| ≤ c(|x| + p|t|)−n−2 and p −n−3 |∇xg| ≤ c(|x| + |t|) p −n−4 |∂tg| ≤ c(|x| + |t|) . This implies the kernel conditions of Lemma 5.8. The L2 boundedness follows from the Fourier transform.

0 3.3. Weyl quantization of symbols in S1,0. m n Definition 5.10. Let m ∈ R, ρ ≥ δ ∈ [0, 1]. We say that a ∈ Sρ,δ(R ) ∞ n n if a ∈ C (R × R ) and if for all N ∈ N there exists cN such that α β m−ρ|β|+δ|α| |∂x ∂ξ a(x, ξ)| ≤ c|α|+|β|(1 + |ξ|) . −1 Let k(w, z) = Fξ a(w, z). The kernel of the standard quatization is K(x, y) = k(x, x − y) and the kernel of Opt(a) is K(x, y) = k(tx + (1 − t)y, x − y) Let g(w, z) = k(x − (1 − t)z, z) and let b be the Fourier transform of g with respect to z. Then b(x, D) = Opt a. Equivalently Z b(x, ξ) = (2π)−2n eihη−ξ,zia(x − (1 − t)z, η)dηdz

m Lemma 5.11. Suppose that 0 ≤ t ≤ 1 and ρ ≥ δ ≥ 0 and a ∈ Sρ,δ. Then m m m b ∈ Sρ,δ. Similarly, if b ∈ Sρ,δ then a ∈ Sρ,δ, We only prove one direction. The reverse direction is similar.

∞ Proof. Let ρ ∈ C (B2(0)), identically 1 on B1(0). Then, for all α, β and N Z α β −2n ihη−ξ,zi −N ∂ ∂ (2π) e ρ(η)a(x − (1 − t)z, η)dηdz ≤ c(1 + |ξ|) x ξ 84 5. SINGULAR INTEGRALS OF CALDERON-ZYGMUND´ TYPE by stationary phase. Now let R > 1 and ρR(η) = ρ(η/(2R))−ρ(η/R). Then Z α β −2n ihη−ξ,zi ∂ ∂ (2π) e ρR(η)a(x − (1 − t)z, η)dηdz x ξ m+δ|α|−ρ|β| −N ≤ cN,α,βR (1 + |ξ|/R + R/|ξ|) . To see this we apply the derivatives - the x derivatives fall directly on a, the ξ derivatives after an integration by parts. This yields the factor Rm+δα−ρβ. Now we change coordinates tox ˜ = Rδx and ξ˜ = R−δξ. In the new variables we use stationary phase. The only stationary point in the phase function is (ξ, 0), and with each integration by parts we gain a power of the distance to the support. 0 Theorem 5.12. The kernel K(x, y) of Opta(x, D) with a ∈ S1,0 satisfies −n −n−1 |K(x, y)| ≤ C|x − y| , |∇x,yK(x, y)| ≤ c|x − y| 0 In particular Opt a for a ∈ S1,0 defines a Calderón-Zygmund operator 2 with p0 = 2. The boundedness on L is a consequence of the Theorem of Calderón-Vaillancourt, and the kernel estimates (which have to be verified) imply that the assumption on the kernel of Theorem 5.4 is satisfied. It suffices to prove the bound for K ∇yK for t = 1 by the previous lemma. ∞ n Proof. Let φ ∈ C (R ) be nonnegative, radial, supported in B2(0), indentically 1 on B1(0). Let η0(x) = φ and j+1 j ηj(x) = φ(2 x) − φ(2 x) fpr j ≥ 1. Then X ηj(x) = 1 j for x 6= 0. We define aj(x, ξ) = ηj(ξ)a(x, ξ) and −1 kj(x, z) = Fξ aj (x, z) Then (this follows by several integrations by parts) α β jn+|β| j −M (5.7) |∂x ∂z kj(x, z)| ≤ c(M, α, β)2 (1 + 2 |z|) if j ≥ 1 and, since a0 is a Schwartz function with respect to ξ α β −M (5.8) |∂x ∂z k0(x, z)| ≤ c(M, α, β)(1 + |z|) . The case |z| ≥ 2−j−2 is the first inequality is see by scaling ξ = 2jξ˜, x = 2−jx˜. The case |z| ≤ 2−j−2 follows by integration. Finally X k(x, z) = kj(x, z) j satisfies α β −n−|α| |∂x ∂z k(x, z)| ≤ c|z| 3. EXAMPLES 85

We claim that K(x, y) = k(x, x − y) satisﬁes the Calder´onZygmund condition, as well as K(y, x). Then −n−1 |∇yK(x, y)| ≤ C|x − y|

CHAPTER 6

Hardy and BMO

This section follows to a large extend the work of Feﬀerman and Stein [6] and Stein [13].

1. More general maximal functions and the Hardy space Hp We fix a measurable function φ for which there is a radial and radially decreasing majorant φ∗, |φ| ≤ φ∗. Then we have seen that |φ ∗ f(x)| ≤ cMf(x). −n We define φt(x) = t φ(x/t). For N ∈ N we define the norm α β kfkN = sup sup |x ∂ f| |α|+|β|≤N x and the set of functions

FN = {φ : kΦkN ≤ 1}. Definition 6.1. We deﬁne

(6.3) MN f(x) = sup sup |f ∗ φt(x)|. φ:kφkN ≤1 t It is important in the following theorem that we allow p ≤ 1. Theorem 6.2. Let f be a tempered distribution and 0 < p ≤ ∞. Then the following conditions are equivalent n R p (1) There exists φ ∈ S(R ) with φ = 1 so that Mφf ∈ L . p (2) There exist seminorms N so that MN φ ∈ L (3) M ∗ f ∈ Lp. e−π|x|2 We deﬁne the real Hardy space Hp as the set of all functions for which the equivalent conditions of the Theorem hold. If p > 1 then any maximal function of f majorizes a multiple of f. The second and the third are bounded by the standard Hardy-Littlewood maximal function and hence Hp = Lp in that case. For p = 1 the same argument shows that H1 ⊂ L1. The spaces Lp for p < 1 are deﬁned in the obvious fashion. They are not Banach spaces, and they do not imbed into the space of distributions.

87 88 6. HARDY AND BMO

There are typical elements of Hp called atoms. Definition 6.3 (Atoms). Let 0 < p ≤ 1 A p atom is a bounded function a for which there is a ball B = Br(x0) so that supp a ⊂ B |a| ≤ |B|−1/p Z xαadx = 0 for all multiindices α with 1 |α| ≤ n( − 1) p n Lemma 6.4. Let φ ∈ S(R ) and let a be a atom with the ball Br(x0). Then there exist ε and c so that |x − x0| − n −ε(n,p) M a ≤ C|B (x )|−1/p(1 + ) p φ r 0 r In particular Z p |Mφa| dx ≤ c(n, p).

Proof. Exercise We introduce a modified nontangential maximal function for a ≥ 1 a M f(x) = sup sup |f ∗ φt(x − y)| t |y|≤at Lemma 6.5. a n/p kM fkLp ≤ ca kfkLp Proof. The claim follows from a n ∗ (6.4) |{x : M f > λ}| ≤ ca |{x : Mφf > λ}| by integration. ∗ a Let O = {x : Mφf > λ}. Suppose that M f(x) > λ. Then there exist ˜ (˜x, t) with f ∗ φt˜(˜x) > λ and |x − x˜| ≤ at. Then Bt(˜x) ⊂ O and hence |O ∩ B (x)| at > a−n. |Bat(x)| n Let A = R \O and |O ∩ B (x)| A∗ = {x ∈ A : r < a−n for some r} |Br(x)| Then n ∗ n n (6.5) |R \A | ≤ (3a) |R \A| implies (6.4). To prove (6.5) we turn to an argument in measure theory. Suppose that n −n ∗ A ⊂ R is a closed set and let 0 < γ < 1 (γ = 1 − a ) . Let A ⊂ A be the set of all points x so that |A ∩ B| ≤ γ |B| 1. MORE GENERAL MAXIMAL FUNCTIONS AND THE HARDY SPACE Hp 89 for some ball B containing x. Then n ∗ R \A = x : M(χRn\A) > 1 − γ and by the estimate for the maximal function 3n | n\A∗| ≤ | n\A| R 1 − γ R For the proof of Theorem 6.2 we have to study the effect of changing the function is the definition of the maximal function. This is the easier part of the proof.

Lemma 6.6. Let φ, ψ ∈ S with R φ = 1. For all M > 0 There exists a sequence η(k) ∈ S such that for all N

(k) −kM kη kN ≤ cN,M 2 and X (k) ψ = η ∗ Φ2−k

∞ Proof. We ﬁx ρ ∈ C0 (B2(0)), identically 1 on B1 and deﬁne −k 1−k ρk(ξ) = ρ(2 ξ) − ρ(2 ξ) for k ≥ 1 and ρ0 = ρ. Then ∞ ˆ X ˆ ψ = ρkψ. k=0 R ˆ ˆ 1 Now 1 = φdx = φ(0). Without loss of generality we assume |φ(ξ)| ≥ 2 for |ξ| ≤ 2. Then

∞ X ρk(ξ) ψˆ(ξ) = Ψ(ˆ ξ)φˆ(2−kξ) =η ˆ(k)φˆ(2−kξ) ˆ −k k=0 φ(2 ξ) ˆ Now Ψ is a Schwartz function which leads to the claimed decay. The proof gives actually a stronger statement: Given M there exists N so that the claim holds for kψkN < ∞. We turn to the proof of the theorem.

Proof. Let Φ ∈ S with R Φ = 1. We claim that there is are constants c and N so that

∗ (6.6) kMN fkLp ≤ ckMΦfkLp and

∗ (6.7) kMΦfkLp ≤ ckMΦfkLp These two estimates imply all assertions of the theorem. 90 6. HARDY AND BMO

To proof the ﬁrst inequality we choose ψ ∈ S. The expansion gives ∞ X (k) Mψf(x) = sup |f ∗ ψt(X)| ≤ sup |f ∗ Φ2−kt ∗ ηt (x)| t t>0 k=0 Z −n X (k) ≤ sup t |f ∗ Φ2−kt(x − y)||η (y/t)|dy t k Z X |y| −N −n |y| N (k) ≤ sup sup |f ∗ Φ2−kt(x − y)|(1 + −k ) t (1 + −k ) |η (y/t)|dy t y 2 t 2 t k with N > n/p and Z |y| t−n (1 + )N |η(k)(y/t)|dy ≤ c2−k 2−kt if (which is ensured by Lemma 6.6 ) (k) −k(N+1) kη kN+n ≤ c2 . We claim that

|y| −N ∗ (6.8) k sup sup |f ∗ Φt(x − y)(1 + ) kLp ≤ ckMφf(x)kLp . t y t Then

|y| −N −jN 2j sup sup |f ∗ Φt(x − y)(1 + ) ≤ sup 2 M f(x) t y t j=0,1... and the assertion follows Lemma 6.5. This implies (6.6).

To complete the proof we will prove ∗ (6.9) kMφfkLp ≤ ckfkLp . Let ∗ Fλ = {x : MN f(x) ≤ λMφf(x)}. Since Z Z Z ∗ p −p p p −p ∗ p |MΦf| dx ≤ λ |MN f| dx ≤ c λ |MΦf| dx Rn\F Rn\F we obtain Z Z ∗ p ∗ p |MΦf| dx ≤ 2 |MΦf| dx Rn F provided we choose λp ≥ 2cp. We claim that on F and any q > 0 ∗ q 1/q (6.10) MΦf(x) ≤ c[M|MΦf| (x)] . This implies the desired estimate via Z Z Z Z ∗ p ∗ p q p/q p |MΦf| dx ≤ 2 |MΦf| dx ≤ c1 [M|MΦf| ] dx ≤ c2 |MΦf| dx Rn F by the estimate for the Hardy-Littlewood maximal function. It remains to prove (6.10). Let ∗ ∗ f(x, t) = f ∗ Φt(x), f (x) = MΦf(x) 2. THE ATOMIC DECOMPOSITION 91

1 1 Proposition 6.7. Let f ∈ Lloc with Me−2π|x|2 f ∈ L and λ > 0. Then there is a decomposition X f = g + b, b = bk and a collection of dyadic cubes Qk so that (1) |g| ≤ c(n)λ ∗ R (2) supp bk ⊂ Qk and bkdx = 0 (3) The Qk are disjoint and [ Qk = {x : MN f > λ}

∞ n ∗ n n Proof. We ﬁx ζ ∈ C0 ((0, 1) ) ), identically one on [0, 1] . For k ∈ Z and l ∈ Z we deﬁne −l ζkl = ζ(2 x − k) ∗ ˜ which is supported in Qkl and identically 1 in Qkl.

Let O = {x : MN f(x) > λ}, let Qkj ,lj be a Whitney decomposition (with disjoint cubes, and edge lengths comparable to the distance to the complement), and

ζkj ,lj ηj = P i ζki,li a partition of unity. Then, if lj is the edge length,

α −lj |α| |∂ ηj| ≤ c2 . We deﬁne R fηjdx bj = (f − cj)ηj, cj = R ηj n Then, by the deﬁnition of Qj there exist r and x ∈ R \O such that

Qj ⊂ Br(x0) 1/n and r ≤ c(n)|Qj| . But then x − x kη ( 0 )k ≤ c(n) j r N and Z n n ηjfdx ≤ c(n)r MN f(x0) ≤ c(n)r λ.

Thus |cj| ≤ c(n)λ. Now |g| ≤ cMN f(x) ≤ cλ for X/∈ O. Together this gives the bound on g. Theorem 6.8. Suppose that 0 < p ≤ 1 and f ∈ Hp. Then there exists a sequence of p atoms aj and a summable sequence λj so that X f = λjaj and X p p |λj| ≤ c(n, p)kMe−π|x|2 fkLp . 2. THE ATOMIC DECOMPOSITION 93

Remark 6.9. Since X p X p |Mφ λjaj| ≤ |λj| |Mφaj| j j

X p = |λj||Mφaj| j any such sum is bounded in Hp. The sum X aj

λj converges in the space of tempered distributions. Proof. We only consider p = 1. We have seen that H1 ⊂ L1. Let f ∈ H1. It is integrable. For each integer l we apply the Calder´on-Zygmund l l l l P l decomposition at level 2 and we write f = g + b , b = j bj. We claim that gl → f 1 l in H for l → ∞, or, equvalently, kb kH1 → 0 as l → ∞. This follows from Z l j kb kH1 ∼ Me−π|x|2 b dx Z X j ≤ Me−π|x|2 bkdx j Z ≤ (MN f)dx S l Qj Z = MN fdx → 0 l MN f>2 Since |gl| ≤ c2l we have gl → 0 as l → −∞ in the sense of distributions. Hence X X f = gl+1 − gl = bl − bl+1 l l in the sense of distributions. The diﬀerence gl+1 − gl is supported in l k O = {x : Mnf > 2 } and l+1 l l l+1 X l l X l+1 l+1 X l g − g = b − b = (f − cj)ηj − (f − cj ηj ) = Aj j j j with l l l X l+1 l+1 l X l+1 Aj = (f − cj)ηj − (f − cm )ηm ηj + cj,mηm m m with R (f − cl+1ηl )ηl+1dx c = m j m , j,m R l+1 ηm dx 94 6. HARDY AND BMO

l since ηj is a partition of unity and hence X cj,m = 0. j Then Z l Ajdx = 0

l ˜l,∗ supp Aj ⊂ Qj l l |Aj| ≤ c2 by construction. We set l −1 −l l −1 l aj = c 2 |Qj| Aj and l l l λj = c2 |Qj|. l Then the aj are atoms, and Z X l X l l X l l λj = c 2 |Qj| = c 2 |{MN f > 2 }| ≤ c MN fdx. l It is not hard to see that H1 is a Banach space. We can use the atomic decomposition to define a norm: X X (6.12) kfkH1 = inf{ |λk| : there exists atoms with f = λkak}. It is a consequence that the span of atoms is dense in H1. Corollary 6.10. Let T be a Calderón-Zygmundoperator. Then T defines a unique continuous operator from the Hardy space H1 to L1. If T is a convolution operator satisfying the assumptions of the Mihlin-Hörmander theorem then T defines a unique continuous operator on H1. Proof. We only prove the first part. The second part is an exercise. Let a be an atom. We want to prove that

kT akL1 ≤ c(n). By translation invariance we may assume that the corresponding ball has center 0, and be rescalling we may assume that the radius is 1. Then

kT akLp0 ≤ cka|Lp0 ≤ c(n) where p0 is the exponent of the Calderón-Zygmund operator. We use this bound on B2(0). Outside we argue as for the proof of the boundedness of Calderón-Zygmund operators. Now let f ∈ H1. By the atomic decomposition X f = λjaj We define X T f = λjT aj 3. DUALITY AND BMO 95 where the right hand side converges in L1. There is no other choice for the definition, but wellposedness has to be proven. Suppose X X f = λjaj = µjbj with atoms (aj), (bj) and summable sequences λj and µj. We have to show that for ε > 0 there exists N0 so that for all t > 0 and N > N0 N N X X (6.13) |{x : | λjT aj − µjT bj| > t}| < ε/t. j=1 j=1 Then ∞ ∞ X X λjT aj = µjT bj j=1 j=1 follows. Inequality (6.13) follows from two properties: (1) The weak type inequality for CalderónZygmund operators c |{x : |T g(x)| > t}| ≤ kgk 1 t L (2) The convergence of the partial sums in L1.

By the convergence there exists forε ˜ > 0 and N0 so that ∞ N ∞ N X X X X k λjajdx − λjajkL1 + k µjbjdx − µjbjkL1 < ε˜ j=1 j=1 j=1 j=1 and then N N X X 0 |{x : | λjT aj − µjT bj| > t}| < cε /t. j=1 j=1

3. Duality and BMO 1 n Definition 6.11. Let f ∈ Lloc(R ). The sharp maximal function is deﬁned by

Z Z ] −n −1 f (x) = sup |Br(y)| f(w) − |Br(y)| f(z)dz dw ∈ [0, ∞] Br(y)3x Br(y) Br(y) Properties (1)( f + g)](x) ≤ f ](x) + g](x) ] ] p (2) f (x) ≤ 2Mf(x). Hence kf kLp ≤ cn p−1 kfkLp Definition 6.12. We deﬁne BMO as the space of all function for which the (semi) norm ] kfkBMO = kf ksup is ﬁnite. Certainly L∞ ⊂ BMO. Moreover ln(|x|) ∈ BMO. 96 6. HARDY AND BMO

1 Theorem 6.13. Let L : H → R be a continous linear map. Then there exists f ∈ BMO such that for every atom Z (6.14) L(a) = afdx,

kfkBMO = kLk(H1)∗ . Vice versa: let f ∈ BMO. Then (6.14) defines a continuous linear functional on H1. Proof. We prove the second part first. If a is an atom with ball B then Z Z ] fadx = (f − fB)adx ≤ kf − fBkL1(B9)kakL∞ ≤ f (x0). B ∞ 1 1 thus for f ∈ L and g ∈ H or f ∈ BMO and g ∈ Ha Z 1 fgdx ≤ kfkBMOkgkH This implies the second statement. Now let L be a linear function on H1 of norm at most 1. let B be a ball. Then 2 L (B) 3 f → L(f − fB) defines a linear functional on L2 which is represented by a function gB so that Z B L(f − fB) = g fdx

R b and in particular B g = 0. We search for a function g so that B g − gB = g for all balls B. Let B ⊂ B0 be two balls and gB resp. gB0 the functions constructed above. For f ∈ L2(B) with R fdx = 0 we have

Z Z 0 L(f) = fgBdx = fgB dx B B0 thus for such f Z 0 (gB − gB )fdx = 0. B Thus gB − gB0 is constant on B. We deﬁne g = gB1(0) if |x| < 1. Choose BR(0) cR = g − g for x ∈ B1(0) and deﬁne

g(x) = gBR(0)(x) − cR for |x| < R and R ≥ 1. This gives a consistent choice, and by the considerations of the ﬁrst part g ∈ BMO. The following theorem has been proven by diﬀerent methods by John and Nirenberg [8]. 3. DUALITY AND BMO 97

Theorem 6.14 (John-Nirenberg inequality). There exists δ > 0 such that for all balls B and all functions f ∈ BMO

Z δ |f(x)−fB | e kfkBMO dx ≤ c|B| B Proof. This is a consequence of the previous proof and the bound for the maximal function. We assume that kgkBMO ≤ 1. If p > 1 and g ∈ p0 R L (B) with B g = 0 then

0 1 1 p 1− 0 kgk 1 ≤ c |B| p kgk 0 = p|B| p kgk 0 H 1 − p0 Lp Lp Then −1 p p p |B| kf − fBkLp ≤ cp kgkBMO and by Tschebychev (cp)p |{x ∈ B : |f − f | > λ}| ≤ |B|. B λp 1 If λ > 2c we choose p = λ/(2c) and get λ 1 2c |{x ∈ B : |f − f | > λ}| ≤ |B| ≤ e−δλ|B|. B 2 The assertion follows by integration with respect to λ. p n p n Theorem 6.15. Let 1 < p0 < ∞, T : L 0 (R ) → L 0 (R ) linear and continuous with an integral kernel K(x, y) which satisﬁes Z |K(x, y) − K(˜x, y)|dy ≤ A n R \B2|y−y˜|(y) Then kT fkBMO ≤ ckfkL∞

Proof. We ﬁx a ball B = Br(x0) and ω = |B|. We decompose f = n f1 + f2 = χB2r(x0)f + χR \B2r(x0)f and use Jensen’s inequality to control the ﬁrst term. 1 Z Z Z p0 −1 −1 −1 p0 ω T f1 − ω T f1 ≤ ω |T f1|

−1/p0 ≤Aω kf1kLp0

≤kfkL∞

Since for x, x˜ ∈ B1

Z |T f2(x) − T f2(˜x)| = (K(x, y) − K(˜x, y))f(y)dy n R \BR(x0) Z ≤ |K(x, y) − K(˜x, y)|dykfkL∞ n R \B2R(x0) ≤AkfkL∞ . 98 6. HARDY AND BMO

Theorem 6.16. The inequality Z Z ] fgdx ≤ c f MN gdx holds whenever g ∈ H1 and f is bounded. Proof. We apply the atomic decomposition, together with its proof. We have X l g = λj,laj Then Z Z X j fgdx = λj,l fakdx ˜l Qj Z X j = (f − fBl )akdx ˜l j j,l Qj Z X λj,l ] ≤ l f (x)dx |Q | l j,l j Qj X Z =c 2k f ](x)dx l l MN G>2 Z ] ≤c f Mngdx

Corollary 6.17. Suppose that 1 < p < ∞. Then ] kfkLp ≤ cpkf kLp Proof. Suppose that f ∈ Lp ∩ L∞. Then Z kfkLp = sup fgdx kgk 0 ≤1 Lp Z = sup fgdx g∈H1,kgk 0 ≤1 Lp Z ] ≤cn sup f Mgdx g∈H1,kgk 0 ≤1 Lp ] ≤cn sup kf kLp kMgkLp0 g∈H1,kgk 0 ≤1 Lp ] ≤cnpkf kLp . 4. Relation to harmonic functions

We consider harmonic functions u on the upper halfplane {xn+1 > 0} in n+1 R . We denote the coornate xn+1 = t. If u is such a function we deﬁne u∗(x) = sup |u(x, t)| t>0 n with x ∈ R . 4. RELATION TO HARMONIC FUNCTIONS 99

Theorem 6.18. The following is equivalent. (1) u∗ ∈ Lp (2) There exists f ∈ Hp so that

u(t, x) = Pt ∗ f where t Pt(x) = cn n+1 (|x|2 + t2) 2 is the Poisson kernel. n Proof. We restrict the proof to p > n+1 . We observe that

−n 1 −n Pt(x) = t n+1 =: t φ(x/t) (1 + (|x|/t)2) 2 satisﬁes α β |x ∂x φ| ≤ cαβ and hence, with u(t, x) = Pt ∗ f ∗ p p ku kL ≤ ckMe−π|x|2 fkL n A closer check shows that the condition p > n+1 is not needed. ∗ p Now suppose that u ∈ L . We claim that with fε = u(ε, x)

Pt ∗ fε = u(ε + t, x) n Both functions are bounded and harmonic and they coincide on R . By Liouville’s theorem they are equal. Thus the family fε is uniformly bounded p in H . Hence there is a sequence fεj converging to some f in the sense of distributions. But then u(t, x) = Pt ∗ f p and f ∈ H . Proposition 6.19 (Analog of Cauchy-Riemann). Suppose that

F = (u0, u1, . . . un) are harmonic functions which satisfy the Cauchy-Riemann type equations n ε X ∂uj ∂uj ∂ui ∂ uε + = 0, = . t 0 ∂ ∂x ∂x j=1 xj i j Then k sup |u0(x, t)|kL1 ≤ c sup kF (., t)kL1 . t>0 t Proof. The key fact is that |F |q n−1 is subharmonic for q > n as in Chapter 3 for n = 2. The claim follows as in Theorem 3.16 based on Proposition (3.14). 1 1 1 Theorem 6.20. Let f ∈ L . Then f ∈ H iﬀ Rjf ∈ L for all Riesz transforms. Remark 6.21. The Riesz transforms deﬁne a distribution, which we assume to be in L1. 100 6. HARDY AND BMO

Proof. We deﬁne ε ε ε Fε = (u0, u1, . . . un) where ε ε u0(t, x) = f ∗ Pt ε ε j uj(t, x) = f ∗ Qt with j cnxj Q1 = n+1 . (1 + |x|2) 2 Then n ε X ∂uj ∂uj ∂ui ∂ uε + = 0, = t 0 ∂ ∂x ∂x j=1 xj i j and |Fε(x, t)| ≤ |Fε(., 0)| ∗ Pt(x) Since Fε(x, 0) = (f ∗ φε,Rjf ∗ φε) we get k|Fε|kL1 ≤ c By Fatou Z sup |F (x, t)|dx ≤ c t>0 Rn

5. Div-curl type results Here we follow Coifman, Lions, Meyer and Semmes [3, 4]. 1 1 Corollary 6.22 (div-curl lemma 1). Let n ≥ 2, p + q = 1, 1 < p, q < p n n ∞, f, g ∈ L (R ; R ) div f ∈ Lp, curl g ∈ Lq. Then hf, gi ∈ H1 ∞ n Proof. We ﬁx φ ∈ C0 (R ) supported in B2(0), idntically 1 on B1/2(0) with R φdx = 1. Then g = ∇G for some function g and hf, gihf, ∇Gi = ∇ · (Gf) and Z −1 ∇ · (Gf)φt(x) = − t Gf(x − y)(∇φ)t(y)dy

−1−n ≤t kGk r kfk r0 L (B2t(x)) L (B2t(x)) −1−n r˜ r ≤t kgkL (B2t(x))kfkL (B2t(x)) 0 0 ≤2n(M|g|r˜(x))1/r˜(M|f|r (x))1/r hence n r˜ 1/r˜ r0 1/r0 M hf, gi(x) ≤ 2 kM|g| k kM|f| k ≤ ckgk p kfk q φ Lp/r˜ Lq/r0 L L 6. APPLICATION TO ELLIPTIC PDES 101 provided 1 1 1 1 < r˜ < p, 1 < r0 < q, + = 1 + r˜ r0 n Corollary 6.23. Let f and g as above, n = 2 and ∆u = hf, gi. Then u is continuous. n i n Corollary 6.24. Let u : R → R satisfy ∂ju ∈ L . Then det(Du) ∈ H1 Proof. We expand the determinant with respect to the first row. Then det Du = h∇u, F i where F is given by subdeterminants. For smooth function i ∂iF = 0 (see Evans, Partial Differential Equations, Theorem 2 in Section 8.1) and hence this is true for the functions at hand. The statement follows now from the previous assertion. 1 Remark 6.25. If f ∈ H is nonnegative on B1(0) then Z f| ln f|dx ≤ c B1/2(0) Hence the determinant has some higher integrability if it is nonnegative. Let (u, p) be a solution to the Navier-Stokes equations. Then n X i j ∆p = − (∂ju )(∂iu ) i,j=1 Pn i and i=1 ∂iu = 0. We fix i. Then i curl ∇u = 0 and ∇ · ∂iu = 0 and hence ∆p ∈ H1 Then, if n = 2 2 kpksup ≤ ck∇ukL2 6. Application to elliptic PDEs n 1 Let U ⊂ R be open. We denote by H (U) the set of all functions in L2(U) with distributional derivatives in L2, with norm 2 2 2 kukH1 = kukL2 + k|Du|kL2 . It is a closed subspace of the Hilbert space (L2)n+1 and hence a Hilbert space with inner product Z X hu, vi = uv¯ + ∂ju∂jvdx j 102 6. HARDY AND BMO

1 ∞ We denote by H0 (U) the closure of C0 (U). It encodes the boundary value 0. We assume n ≥ 3 and we will rely on two properties for balls B = Br(x0), the Sobolev inequality 1 (6.15) kuk 2n ≤ ck|Du|kL2(B) for u ∈ H0 (B1) L n−2 (B) and the composition with Lipschitz functions g with g(0) = 0,

(6.16) kg ◦ ukH1 ≤ kgkLipschitzkukH1 . ij ∞ Let (a )1≤i,j≤n be real functions in L (U) such that there exists κ > 0 with X ij 2 a ξiξj ≥ κ|ξ| i,j n for almost every x and every ξ ∈ R . Let f, F j ∈ L2(U). We call u a weak solution to

X ij X i ∂ia ∂ju = ∂if + g i,j i if this equation holds in the sense of distributions. In that case it holds 1 ∗ 1 ∞ in (H0 ) - i.e. we may test it with functions in H0 instead of C0 . If U is bounded then the Lemma of Lax-Milgram and the Poincar´einequality 1 imply existence of a unique weak solution u ∈ H0 (U). Theorem 6.26. There exists C such that the following is true. Let 1 u ∈ H (B1(0)) be a nonnegative weak solution to n X ij ∂ia ∂ju = 0 i,j=1 in a ball B. Then the Harnack inequality sup u ≤ C inf u. 1 x∈ 1 B x∈ 2 B 2 holds. This has been proven by diﬀerent methods by De Giorgi [5] and Nash [12, 11]. We follow the proof of Moser [9].

Proof. We begin with the preliminary L∞ estimate with a technique known as Moser iteration.

ij Lemma 6.27. There exists c depending only on n, ka k∞/κ such that the following is true. Suppose that u is a solution on B1(0). Then

kuk ∞ ≤ ckuk 2 L (B1/2(0)) L (B1(0))

1 Proof. Given 2 ≤ r < R ≤ 1 let  1 if |x| ≤ r  η(x) = 0 if |x| ≥ R  R−|x| R−r if r ≤ |x| ≤ R 6. APPLICATION TO ELLIPTIC PDES 103

Then, neglecting that f(u) = |u|2(p−1)u is not Lipschitz (this can be ﬁxed by truncating at height H, and letting H → ∞, Z ij 2 2(p−1) 0 = a ∂iu∂j(η |u| u)dx B1(0) Z 2p − 1 ij p−1 p−1 = 2 a ∂i(η|u| u)∂j(η|u| u)dx p B1(0) 2p − 1 2 Z − (2 − ) aij(∂ η)∂ (η|u|p−1u)|u|p−1udx p2 p i j 2 2p − 1 Z − ( − ) (aij∂ η∂ η)|u|2pdx p p2 i j 2p − 1 Z = aij∂ v∂ vdx p2 i j 2(p − 1) Z − aij∂ v(∂ η)|u|p−1udx p2 i j 1 Z − aij(∂ η)(∂ η)|u|2pdx p2 i j and hence, with v = η|u|p−1u,

2p 2 c 2p (6.17) kuk 2pn ≤ ckvk 1 ≤ kuk 2p H 2 L (BR(0)) L n−2 (Br(0)) (R − r) k n 1 For k ∈ N we write pk = n−2 , and for 1 ≤ j ≤ k we set rj = 1 − 2j . Then

21/pk kukL2pk (B ) ≤ c(k − 1) ku| 2pk−1 rk L (Brk−1 ) k Y 2 ( n−2 )j n 2 ≤ (cj ) kukL (B1). j=1 Since k j ! Y 2 ( n−2 )j X 1 i (cj ) n = exp (ln c + 2 ln i)( n ) j=1 i=1 n−2 is uniformly bounded we obtain the statement of the Lemma. Now let u be a positive solution. The we obtain (6.17) for p > 0 as long as we avoid p = 1/2. In this case, if 1 > p0 > 0

(6.18) kuk ∞ ≤ ckuk p L (B1/4(0)) L 0 (B1/2(0)) and, with negative exponents (how we have to add ε to get a Lipschitz function) −1 −1 (6.19) ku k ∞ ≤ cku k p . L (B1/4(0)) L 0 (B1/2(0))

Suppose we knew that there exists p0 so that −1 (6.20) kuk p ku k p ≤ C L 0 (B1/2(0)) L 0 (B1/2(0)) 104 6. HARDY AND BMO

Then Z p0 1/p0 sup u(x)dx ≤c1 u dx) dx x∈B1/4 B1/2 Z 1 −p0 − p ≤c2( u dx) 0 B1/2 ≤c3 inf u(x) x∈B1/4 which implies the claim by a covering argument. We claim that v = η(ln u − ln uB) ∈ BMO and

kvkBMO ≤ c(n) Then, by the inequality of John-Nirenberg, Theorem 6.14, Z eδ|v−vB |dx ≤ c(n) B and hence, Z Z Z Z uδdx u−δdx = eδ(v−vB )dx e−δ(v−vB )dx ≤ c(n)2. B B B and this implies the previous statement. We repeat the calculation above for p = 0, andη ˜ as above, but related to a ball Br(x0) ⊂ B1(0), Z ij −1 0 = a ∂iu∂j(˜ηu )dx B1(0) Z Z ij ij −2 = a ∂iu(∂jη˜)udx − a η∂˜ iu∂juu dx B1(0) Z Z ij ij = a ∂i(ln u)∂jηdx˜ − ηa˜ ∂i(ln u)∂j(ln u)dx B1(0)

k∇ ln(u)k 1 ≤ c H (B5/6(0)) and hence, by Poincar´e’sinequality Z

| ln(u) − ln(u)B3/4 |dx ≤ c. B3/4

We obtain for all balls contained in B3/4 −2 2 k∇ ln ukL (Br(x)) ≤ cr hence (with Lemma 6.31 and more arguments, sorry, I’ll complete that later)

kη(ln u − (ln u)B)kBMO ≤ c Theorem 6.28. p > n and q > n/2. Then there exists s > 0 so that the following is true. Suppose that F i ∈ Lp(B), f ∈ Lq(B), u ∈ H1, n n 1 − = 2 − = s p q n X ij i ∂ia ∂ju = ∂iF + g in B i,j=1 6. APPLICATION TO ELLIPTIC PDES 105

Then

|u(x) − u(y)| h − n −s 1− n −s 2− n −s i 2 p p q q sup s ≤ c R kukL2(B) +R kF kL +R kfkL 1 |x − y| x,y∈ 2 B,x6=y

Proof. We prove this statement ﬁrst for F = g = 0 and B = B1(0). Let ωk = sup u − inf u. B −k B2−k 2 We claim that there exists γ < 1 depending only on the quantities of the Harnack inequality so that

(6.21) ωk+1 < γωk. Indeed, by the Harnack inequality applied to

vk = u − inf u, wk = sup u − u B −k 2 B2−k sup u ≤ inf u + C( inf u − inf u) B −k B −k−1 B −k B2−k−1 2 2 2 and inf u ≥ sup u − C( sup u − sup u) B −k−1 2 B2−k B2−k−1 B2−k hence (1 + C)ωk+1 ≤ (C − 1)ωk, C−1 −k−1 −k which implies (6.21) with γ = C+1 < 1. Then, if 2 |x| ≤ 2 k − ln |x| 2 |u(0) − u(x)| ≤ ωk ≤ γ ω0 ≤ γ kukL (B10) and γ− ln |x| = e− ln γ ln |x| = |x|− ln |γ|. n Now we consider U = R . n Theorem 6.29. Let n ≥ 3 There is a Green’s function on R which satisﬁes |x − x˜|s |g(x, y)| ≤ c|x − y|2−n, |g(x, y) − g(˜x, y)| ≤ (|x − y| + |x˜ − y|)2+s−n

Proof. Let F and f be supported in B1(0). By Lax Milgram (with 2n H the space of functions in L n−2 with derivatives in L2, equipped with the R 2 1 2 norm |Du| dx) there is a unique solution u ∈ Hloc with Du ∈ L and 2n u ∈ L n−2 and h i 2n 2 2 2n kuk + kDukL ≤ c kF kL (B1(0)) + kgk . L n−2 L n+2 (B1(0))

By the previous H¨olderestimate, if |x0| = 3, h i s 2 2n kukC (B1(x0)) ≤ c kF kL (B1(0)) + kgk L n+2 (B1(0)) We ﬁx x and consider 2n x L n+2 (B1(0)) 3 f → u . 106 6. HARDY AND BMO

By duality there exists a unique gx(y) with x sup kg (y)k 2n ≤ C x L n−2 and Z u(x) = gx(y)f(y)dy.

Clearly ij ji h∂ia ∂ju, vi = h∂ia ∂jv, ui Let T : f → u. Then hT f, hi = hf, T hi This operator is self adjoint. Hence gx(y) = gy(x) and, repeating the previous argument we obtain the assertion. Now we complete the prove of Theorem 6.28 by using the kernel estimates. Let gx(y) be the Green’s function. We claim that Z x x˜ s | (g (y) − g (y))f(y)dy| ≤ c|x − x˜| kfkLp(Rn) B1 By rescaling and translating we may assume thatx ˜ = 0 and |x| = 1. We decompose f = f1 + f2 with f1 = χB2(0)f. Then Z 0 0 g (y)f(y)dy ≤ kfkLp kg (.)k p0 L (B2(0)) B2(0) and the same holds for x and Z x 0 x 0 (g (y) − g (y))f(y)dy ≤ kfkLp kg − g k p0 n . L (R \B2(0)) B2(0)

n 0 n n−2 Since p < n−2 and by Theorem 6.29 g(x, .) ∈ Lw 0 kg (.)k p0 ≤ c(n). L (B2(0)) Similarly, again by Theorem 6.29 0 x kg − g k r n ≤ c(n) Lw(R \B2(0)) for 1 n − 2 s = − . r n n This allows to bound the second term, provided we choose s in the theorem smaller than for the estimate of the solution to the homogeneous problem. We proceed similarly with the term ∇F , for which we need 0 k∇g k q0 ≤ c L (B2(0)) and 0 x k∇(g − g )k q0 n ≤ c. L (R \B2(0)) 7. POINTWISE ESTIMATES AND PERTURBATIONS OF ELLIPTIC EQUATIONS 107

This follows by Cacciopoli’s inequality n − n 0 q0 2 0 k∇g k q0 ≤R k∇g k 2 L (B2R\BR) L (B2R\BR) n n 0 − −1 ≤cR q 2 kg k 2 0 L (B4R\BR/2) n −(n−1) ≤cR q0 1− n ≤cR q which is summable over dyadic radii provided q > n. Similarly, if R > 3, n 0 x 1−s− q k∇(g − g )k q0 ≤ cR L (B2R\BR) which is clearly summable provided q is suﬃciently large.

7. Pointwise estimates and perturbations of elliptic equations We obtain an alternative argument for the boundedness of Calder´on zygmund operators with kernels satisfying −n −n−1 |K(x, y)| ≤ c|x − y| , |Dx,yK(x, y)| ≤ c|x − y| Theorem 6.30. Let r > 1. Then T f ](x) ≤ c(M|f|r)1/r(x).

Proof. We ﬁx a ball B1(0) and decompose f = f1 + f2 with f1 =

χB3(0)f. Then r 1/r kT f1kLr ≤ ckf1kLr ≤ c(M|f| (x)) and, for x, x˜ ∈ B1(0),

|T f2(x) − T f2(˜x)| ≤ cMf2(x). Thus for p > r ] r 1/r kT fkLp ≤ k(T f) kLp ≤ ck(M|f| ) kLp ≤ ckfkLp . This proof uses the Calder´on-Zygmund estimate. Instead we could set r = p0, get bounded for p > p0, and apply duality. Lemma 6.31. Let f ∈ BMO. Then

|x1 − x2| + r1 + r2 |fBr (x1) − fBr (x2)| ≤ c(| ln r1/r2| + ln )kfkBMO 1 2 r1 + r2 Proof. We prove this statement in several steps. We have

Z |f − f | ≤ |B (x)|−1 f − f dx BR(y) Br(x) r BR(y) Br(x) Z n −1 n ] ≤(R/r) |BR(y)| |f − fBR(y)|dx ≤ (R/r) f (y) if Br(x) ⊂ BR(x). If max{|x1 − x2|, r1} < r2 < 2r1 then we obtain immediately the assertion. Moreover, by applying the argument k times |f − f | ≤ 2nkf ](x) B2kr(x) Br(x) 108 6. HARDY AND BMO

We choose

k = max{| ln r2/r1|, ln |x1 − x2|/(r1 + r2)} + 2 k and R = 2 min{r1, r2}. Then |f − f | ≤|f − f | + |f − f | Br1 (x1) Br2 (x2) Br1 (x1) BR(x1) Br2 (x2) BR(x2)

+ |fBR(x1) − fBR(x2)| n ] ≤(2k + 1)2 f (x1) ij ∞ n Let a ∈ L (R ) and ij 2 a ξiξj ≥ κ|ξ| n for all ξ ∈ R and almost every x. We consider u with second derivatives in Lp and and n X ij 2 a ∂iju = f. i,j=1

Theorem 6.32. Let p ∈ (1, ∞), aij ∈ L∞ and κ > 0. There exists ε > 0 such that if ij ka kBMO < ε then 2 kD ukLp ≤ ckfkLp . Remark 6.33. This theorem provides an important and strong theme in linear and nonlinear partial differential equations: Often not pointwise conditions on coefficients are important, but uniform conditions on every scale. This type of result goes back to Caffarelli and Xabre [1] in 1995 by different techniques. Proof. We claim that with r > 1 2 ] r 1/r 2 r 1/r (6.22) D u ≤ c(M|f| ) + ckakBMO(M|D u| ) . Then 2 ij 2 kD ukLp ≤ ckfkLp + cka kBMOkD ukLp . ij 1 The claim follows if cka kBMO < 2 . As for the proof of Theorem 6.30 we fix a ball B = B1(0). To simplify ij ij the notation we assume that aB = δ . We rewrite the equation as ∆u = f + (δij − aij)∂iju and, redoing the argument above ij ij k∂ u − ∂ uBkL1(B) ≤ckfkLr(2B) + Mf(x) ij ij ij + ck(δ − a )∂ uk 1+r L 2 (2B) Z + c |y|−n−1|(δij − aij)∂iju|dy. |y|>2 7. POINTWISE ESTIMATES AND PERTURBATIONS OF ELLIPTIC EQUATIONS 109

1 2 1 Then,by H¨olderinequality, with q = 1+r − r and the triangle inequality ij ij ij h ij ij ij ij i 2 r 1/r k(δ − a )∂ uk 1+r ≤ ka − a )kLq + |a − a | (M|D u| (0)) L 2 (2B) 2B 2B B 2 r 1/r ≤ckakBMO(M|D u| (0)) where we also used Lemma 6.31 and the proof of Theorem 6.14. Similarly

Z −n−1 ij ij 2 |y| |a − δ ||D u|dy B2k (0)\B2k−1(0) − nk ij ij ij −k r0 ij ≤2 2 ka − aB kLr0 (B (0)) + |a − aB| 2k 2k B2k (0) − nk 2 × 2 r kD uk r L (B2k ) −k 2 r 1/r ≤2 (1 + k)kakBMO(M|D u| (0)) . This implies (6.22).

CHAPTER 7

Littlewood-Paley theory and square functions

1. The range 1 < p < ∞ We ﬁx a function φ ∈ S with R φdx = 0. This conditions can be relaxed in the sequel. For f ∈ Lp we deﬁne

2 n (T f)(t, x) = f ∗ φt(x) =: F (t, x) ∈ L (R × R+, dxdt/t) Then the formal adjoint operator is Z dt T ∗F (x) = φ (y − x)F (t, y) dy. t t Then T ∗T is the Fourier multiplier given by Z ∞ dt m(ξ) = φ(tξ)φ(tξ) 0 t which is homogeneous of degree 0 and smooth. 2 n 2 n 2 We consider T as an operator from L (R ) to L (R ,L (dt/t)). Its kernel is given by the convolution kernel K(x − y) = (t → t−nφ((x − y)/t)) ∈ L2(dt/t) and Z ∞ 1/2 |K(z)| = t−1−2n|φ(z/t)|2dt 0 z =|z|−nω( ) |z| where Z ∞ 1/2 ω(ν) = t−1−2n|φ(ν/t)|2dt 0 is bounded since |φ(ν/t)| ≤ c min{tn+ε, 1} Similarly Z ∞ 1/2 −3−2n 2 |∂xi K(z)| = t |(∂iφ)(z/t)| dt 0 z =|z|−n−1ω ( ) i |z| The same type of estimate holds for T ∗. The Calder´on-Zygmund Theorem 5.4 applies to vector valued functions, see the Remark 5.7 and we obtain

111 112 7. LITTLEWOOD-PALEY THEORY AND SQUARE FUNCTIONS

Proposition 7.1. Let 1 < p < ∞,A > 0 and ε > 0. We assume that n φ ∈ C(R ) satisﬁes (7.1) Z |φ(x)| ≤ A(1 + |x|)−n−ε, |Dφ(x)| ≤ A(1 + |x|−n−1−ε, φdx = 0.

Then there exists c depending on p, n and ε so that p2 ∞ kT fkLp( n,L2( dt )) ≤ c (A + kωkL )kfkLp(Rn) R t p − 1 and 2 ∗ p ∞ kT F kLp(Rn ≤ c (A + kωkL )kF kLp( n;L2( dt )) p − 1 R t Proof. We have seen that the operator satisﬁes the L2 bound under the assumption. The kernel estimates also hold. Hence the assertion follows ∗ for 1 < p ≤ 2 for T and T . Duality gives the full statement. Examples are (1) The derivatives of the heat kernel, evaluated at t = 1:

2 2 −n/2 2 n − |x| −n/2 − |x| π (|x| − )e 2 , π x e 2 2 j (2) The derivatives of the Poisson kernel 2 |x| − (n + 1) xj cn n+3 , cn n+2 (1 + |x|2) 2 (1 + |x|2) 2 Given φ we search ψ so that ∗ TψTφf = f where use the index φ and ψ with the obvious meaning. For that we need more angular regularity. Let i = (i1, i2) be a pair of (nonequal) indices between 1 and n, and we deﬁne the angular derivatives

Di = xi1 ∂i2 − xi2 ∂i1 . Let α be the analogue of the multiindices for pairs of indices, and we denote - by an abuse of notation Dα for the obvious product of angular derivatives. The angular derivative of the Fourier transform is the Fourier transform of angular derivatives. Lemma 7.2. Suppose that φ satisﬁes the conditions (7.1) and α −n−ε (7.2) |D φ| ≤ cα|x| for all such multiindices and assume and φˆ does not vanish identically on any set {λν : λ > 0} with |ν| = 1. Then there exists ψ ∈ S satisfying R ψ = 0 and hence (7.1) and Z ∞ dt φˆ(tξ)ψˆ(tξ) = 1 0 t for all ξ 6= 0. 1. THE RANGE 1 < p < ∞ 113

Remark 7.3. In particular ∗ ∗ TψTφf = TψTφf = f for all f ∈ L2 and hence for all f in Lp. Proof. Let |ν| = 1. Then there exist ε > 0, r and R so that Z R |φˆ(ν/t)| > ε r by continuity and compactness we can choose ε, r and R independent of ν. Given ν we ﬁx a smooth compactely supported function ψν on [r/2, 2R] with Z dt φˆ(tν)ψ (t) = 1 ν t Then Z dt ξ → φˆ(tξ/|ξ|)ψ (t) = ρ (ξ) ν t ν is smooth due to condition (7.2). Locally we can divide by ρν. We use a n−1 homogeneous partition ηk of unity on S to construct ψ by ˆ X ηk(ξ) ψ(ξ) = ψνk (|ξ|) ∈ S. ρνk (ξ) supported in B2R\Br/2. Definition 7.4 (Square function). Let φ satisfy (7.1) . We deﬁne Z dt1/2 s (x) = |f ∗ φ (x)|2 f t t Z Z !1/2 ∗ −n−1 2 sf (x) = t |f ∗ φt(x + y)| dydt t |y|

ksf kLp ≤ ckfkLp , ∗ ksf kLp ≤ ckfkLp and kgf kLp ≤ ckfkLp . If φ satisﬁes the assumption of Lemma 7.2 then

kfkLp ≤ cksf kLp , ∗ ∗ kf kLp ≤ cksf kLp . If φ satisﬁes the assumptions of Lemma 7.2 with:

For all ξ 6= 0 there exists k so that φˆ(2kξ) 6= 0. replacing the integral condition then 114 7. LITTLEWOOD-PALEY THEORY AND SQUARE FUNCTIONS

kfkLp ≤ ckgf kLp .

Proof. The claims on sf follow from Proposition 7.1 and Lemma 7.2. For s∗(f) we observe that Z Z ∞ Z ∗ 2 −n−1 2 ksf kL2 = t |f ∗ φt(x − y)| dydtdx Rn 0 |y|≤t Z Z Z t 2 dt = |f ∗ (φ(. + y))t)(x)| dxdy |y|≤1 Rn 0 t Z Z ∞ 2 2 dt =|B1| |fˆ| |φˆ(tξ)| dξ Rn 0 t and we obtain the L2 bound. It also follows that with ψ as in Lemma 7.2 - assuming that the assumptions are satisfies - ψ defines a left inverse to the vector values operator. Similarly we obtain the kernel bound for Tφ(.+y) for all |y| ≤ 1, and hence also when we integrate y with respect to the unit balls. Replacing the t integration by a summation for gf does not require sub- stantial changes. Only the analogue of Lemma 7.2 needs some consideration. We want to find ψ so that ∞ X φˆ(2kξ)ψˆ(2kξ) = 1 k=−∞ for ξ 6= 0. This is done as in Lemma 7.2. 2. Square functions, tents and Carleson measures Theorem 6.10 applies here and gives Z t 2 dt k |φt ∗ f| kL1(Rn) ≤ ckfkH1 . 0 t We also have Lemma 7.6. Let φ satisfy (7.1). Then Z Z R −1 2 dt 2 sup |BR| |f ∗ φt| dy ≤ ckukBMO x,R BR(x) 0 t

Proof. As usual is suﬃces to prove the bound for B = B1(0). We may assume that f2B = 0, since we may add a constant. We write

f = f1 + f2 2 2 n dt with f1 = fχ2B. Then f1 ∈ L and Tφf1 ∈ L (R ×(0, ∞), dx t ). It remains to consider f2. We claim that

|T φf2(t, x)| = |f2 ∗ φt(x)| ≤ ctkfkBMO if |x| + t ≤ 1. This follows from Lemma 6.31, Z |y|1|φ(y)|dy ≤ cT |y|≥1 and scaling. The inequality is a consequence of the bounds on φ. 2. SQUARE FUNCTIONS, TENTS AND CARLESON MEASURES 115

2 We call |F | a Carleson measure if CF is bounded. The following duality statement holds. Proposition 7.7. Z Z ∞ Z dt ∗ F (t, x)G(t, x) dx ≤ c sF CGdx. Rn 0 t Proof. We deﬁne τ(x) for a large constant A by Z τ Z n −n−1 2 o τ(x) = sup τ > 0 : t G (t, x − y)dtdy ≤ ACG(x) . 0 |y|

We claim that there exists A0 and c > 0 so that if A ≥ A0 then for all balls B = Br(x)

(7.3) |{x ∈ B : τ(x) ≥ r}| ≥ c|Br(x)|. Then, for any nonnegative function H Z Z ∞ Z Z τ(x) Z H(y, t)tndydt ≤ c−1 H(y, t)dydtdx. Rn 0 Rn 0 |y|≤t We take H = F Gt−n−1 and get Z Z ∞ dt Z F (t, x)G(t, x) dx ≤c−1 s∗(F )(x)× Rn 0 t Z τ(x) Z t−n−1G2(t, x − y)dydt 0 |y|≤τ Z A ∗ ≤ sF CGdx c Rn by the deﬁnition of τ(x). It remains to prove (7.3). Let B = Br(x0) be a ball, y ∈ B, and T = {(t, x): |x − y| + t < 3r} the tent over the ball of 3 times the radius around y. Thus, by an application of Fubini, similar to the treatment of s∗, Z Z r Z Z 2 2 dt G (t, x − y)dydt ≤ |B1(0)| |G(t, y)| dy B 0 |y|

This implies (7.3) if we choose A0 > cn. 116 7. LITTLEWOOD-PALEY THEORY AND SQUARE FUNCTIONS

Theorem 7.8. Suppose that φ satisﬁes the assumptions of Lemma 7.2. Then kfkBMO ≤ ckCT f kL∞ and ∗ kfkH1 ≤ cksT f kL1 Proof. Let ψ be as in Lemma (7.2). Then Z Z dt fgdx = T fT g dx φ ψ t By duality and Proposition 7.7 Z kfkBMO ≤C sup fgdx : kgkH1 ≤ 1 Z dt =C sup T fT g dx : kgk 1 ≤ 1 t H Z ∗ =C sup CT f sT gdx : kgkH1 ≤ 1 ∗ ≤CkCT f kL∞ sup{ksT gkL1 : kgkH1 ≤ 1} kfkBMO ≤ ckCT f kL∞ and similarly kgkH1 ≤ ckST gkL1 . Bibliography

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