Harmonic Analysis

Home , Hilbert transform, Plancherel theorem

Ralph Chill

Harmonic analysis

– in Banach spaces –

January 15, 2014

c by R. Chill

Contents

1 Bochner-Lebesgue and Bochner-Sobolev spaces ...... 1 1.1 The Bochner integral ...... 1 1.2 Bochner-Lebesgue spaces ...... 5 1.3 The convolution ...... 7 1.4 Bochner-Sobolev spaces ...... 11

2 The Fourier transform ...... 15 2.1 The Fourier transform in L1 ...... 15 2.2 The Fourier transform on L2 ...... 21 2.3 The Fourier transform on ...... 22 S 2.4 The Fourier transform on 0 ...... 24 2.5 Elliptic and parabolic equationsS in RN ...... 29 2.6 The Marcinkiewicz multiplier theorem ...... 29

3 Singular integrals ...... 31 3.1 The Marcinkiewicz interpolation theorem ...... 31 3.2 The Hardy-Littlewood maximal operator ...... 34 3.3 The Rubio de Francia extrapolation theorem ...... 44 3.4 Calderon-Zygmund operators ...... 51 3.5 The Hilbert transform ...... 58 3.6 Covering and decomposition ...... 63

References ...... 65

Index ...... 67

Chapter 1 Bochner-Lebesgue and Bochner-Sobolev spaces

1.1 The Bochner integral

Let X and Y be Banach spaces, and let (Ω, ,µ) be a measure space. A A function f : Ω X is called step function, if there exists a sequence (An) → ⊆ of mutually disjoint measurable sets and a sequence (xn) X such that A P ⊆ f = 1A xn. A function f : Ω X is called mesurable, if there exists a n n → sequence ( fn) of step functions fn : Ω X such that fn f pointwise µ- almost everywhere. → → Remark 1.1. Note that there may be a difference to the definition of mesura- bility of scalar valued functions. Measurability of a function is here depending on the measure µ. However, if the measure space (Ω, ,µ) is complete in the sense that µ(A) = 0 and B A implies B , then theA above definition of measurability and the classical⊆ definition of∈ measurability A coincide. Note that one may always consider complete measure spaces. Lemma 1.2. If f : Ω X is measurable, then f : Ω R is measurable. More generally, if f : Ω X→ is measurable and if g :kXk Y→ is continuous, then g f : Ω Y is measurable.→ → ◦ → Proof. This is an easy consequence of the definition of measurability and the continuity of g. Note that in particular the norm : X R is continuous. k · k → Lemma 1.3. If f : Ω X and g : Ω K are measurable, then f g : Ω X is → → → measurable. Similarly, if f : Ω X and g : Ω X0 are measurable, then g, f X0,X : Ω K is measurable. → → h i → Proof. For the proof it suffices to use the definition of measurability and to show that the (duality) product of two step functions is again a step function. This is, however, straightforward.

Theorem 1.4 (Pettis). A function f : Ω X is measurable if and only if x0, f is measurable for every x X (we say that→ f is weakly measurable) andh if therei 0 ∈ 0

1 2 1 Bochner-Lebesgue and Bochner-Sobolev spaces exists a µ-null set N such that f (Ω N) is separable (we say that f is almost separably valued). ∈ A \

For the following proof of Pettis’ theorem, see Hille &Phillips [Hille and Phillips (1957)].

Proof. Suﬃciency. Assume that f is measurable. Then f is weakly measurable by Lemma 1.2. Moreover, by deﬁnition, there exists a sequence ( fn) of test functions and a µ-null set N such that ∈ A

fn(t) f (t) for all t Ω N. → ∈ \ Hence, [ f (Ω N) fn(Ω). \ ⊆ n

Since for every step function fn the range is countable, the set on the right- hand side of this inclusion is separable, and hence f is almost separably valued. Necessity. Assume that f is weakly measurable and almost separably valued. We first show that f is measurable. By assumption, there exists a k k µ-null set and a sequence (xn) in X such that D := xn : n N is dense in f (Ω N). By the Hahn-Banach theorem, there exists{ a sequence∈ } (x ) in X \ n0 ∗ such that xn0 = 1 and xn0 ,xn = xn . Since f is weakly measurable, xn0 , f is measurablek k for everyhn. Asi a consequence,k k sup x , f is measurable.|h Buti| n |h n0 i| supn xn0 , f = f on Ω N by the choice of the sequence (xn0 ) and the density of D |hin thei| f (Ωk k N). Since\ our measure space (Ω, ,µ) is supposed to be complete, we obtain\ that f is measurable. In a similarA way, one shows that k k f x is measurable for every x X, and in particular for x = xn. k −Nowk fix m N and define ∈ ∈

Am1 := f x1 inf f xk , {k − k ≤ 1 k m k − k} ≤ ≤ Am2 := f x2 inf f xk Am1, {k − k ≤ 1 k m k − k} \ ≤ ≤ Am3 := f x3 inf f xk (Am1 Am2), {k − k ≤ 1 k m k − k} \ ∪ ≤ ≤ . . . . m 1 [− Amm := f xm inf f xk ( Amk). {k − k ≤ 1 k m k − k} \ ≤ ≤ k=1

Then (Amn)1 n m is a family of measurable, mutually disjoint sets such that Sm ≤ ≤ 1 n=1 Amn = Ω. Deﬁne

1 We are grateful to Anton Claußnitzer for the deﬁnition of the sets Amn and the functions fm. 1.1 The Bochner integral 3

Xm fm := 1Amn xn. n=1

Then ( fm) is a sequence of step functions, ( fm f )m is decreasing pointwise everywhere, and since D is dense in f (Ω kN),− k \

lim fm(t) f (t) = 0 for every t Ω N. m →∞ k − k ∈ \ that is, fm f µ-almost everywhere. As a consequence, f is measurable. → Corollary 1.5. If ( fn) is a sequence of measurable functions Ω X such that fn f pointwise µ-almost everywhere, then f is measurable. → →

Proof. We assume that this corollary is known in the scalar case, that is, when X = K. By Pettis’s theorem (Theorem 1.4), for all n there exists a µ-null set Nn ∈ A such that fn(Ω Nn) is separable. Moreover there exists a µ-null set N0 Ω \ S ∈ such that fn(t) f (t) for all t Ω N0. Let N := n 0 Nn; as a countable union of µ-null sets, N→is a µ-null set.∈ \ ≥ Then f (restricted to Ω N) is the pointwise limit everywhere of the se- \ quence ( fn). In particular f is weakly measurable. Moreover, f (Ω N) is separable since \ [ f (Ω N) fn(Ω N), \ ⊆ n \ and since fn(Ω N) is separable. The claim follows from Pettis’ theorem. \ R A measurable function f : Ω X is called integrable if f dµ < . → Ω k k ∞ P Lemma 1.6. For every integrable step function f : Ω X, f = 1A xn the series P → n n n xnµ(An) converges absolutely and its limit is independent of the representation of f . P Proof. Let f = 1A xn be an integrable step function. The sets (An) are n n ⊆ A mutually disjoint and (xn) X. Then ⊆ X Z xn µ(An) = f dµ < . n k k Ω k k ∞

P Let f : Ω X be an integrable step function, f = n 1An xn. We deﬁne the Bochner integral→ (for integrable step functions) by Z X f dµ := xn µ(An). Ω n

Lemma 1.7. (a) For every integrable function f : Ω X there exists a sequence → ( fn) of integrable step functions Ω X such that fn f and fn f pointwise → k k ≤ k k → 4 1 Bochner-Lebesgue and Bochner-Sobolev spaces

µ-almost everywhere. (b) Let f : Ω X be integrable. Let ( fn) be a sequence of integrable step functions → such that fn f and fn f pointwise µ-almost everywhere. Then k k ≤ k k → Z x := lim fn dµ exists n →∞ Ω and Z x f dµ. k k ≤ Ω k k Proof. (a) Let f : Ω X be integrable. Then f : Ω R is integrable. → k k → Therefore there exists a sequence (gn) of integrable step functions such that 0 gn f and gn f pointwise µ-almost everywhere. ≤ ≤ k k → k k Since f is measurable, there exists a sequence ( fñ) of step functions Ω X → such that fñ f pointwise µ-almost everywhere. Put → fñ gn f := . n 1 fñ + k k n (b) For every integrable step function g : Ω X one has → Z Z g dµ g dµ. Ω ≤ Ω k k Hence, for every n, m Z Z

fn fm dµ fn fm dµ, Ω − ≤ Ω k − k R and by Lebesgue’s dominated convergence theorem the sequence ( fn dµ) R Ω is a Cauchy sequence. When we put x = limn fn dµ then →∞ Ω Z Z x liminf fn dµ = f dµ. n k k ≤ →∞ Ω k k Ω k k Let f : Ω X be integrable. We define the Bochner integral → Z Z f dµ := lim fn dµ, n Ω →∞ Ω where ( fn) is a sequence of step functions Ω X such that fn f and → k k ≤ k k fn f pointwise µ-almost everywhere. The definition of the Bochner integral → for integrable functions is independent of the choice of the sequence ( fn) of step functions, by Lemma 1.7. Moreover, if f is a step function, then this definition of the Bochner integral and the previous definition coincide. Finally, by Lemma 1.7 (b), 1.2 Bochner-Lebesgue spaces 5 Z Z f dµ f dµ (triangle inequality). (1.1) k Ω k ≤ Ω k k Remark 1.8. We will also use the following notation for the Bochner integral: Z Z f oder f (t) dµ(t), Ω Ω and if Ω = (a,b) is an interval in R:

Z b Z b f oder f (t) dµ(t). a a If µ = λ is the Lebesgue measure then we also write

Z b f (t)dt. a Lemma 1.9. Let f : Ω X be integrable and T (X,Y). Then T f : Ω Y is integrable and → ∈ L → Z Z T f dµ = T f dµ. Ω Ω Proof. Exercise.

Theorem 1.10 (Lebesgue, dominated convergence). Let ( fn) be a sequence of integrable functions. Suppose there exists an integrable function g : Ω R and → an (integrable) measurable function f : Ω X such that fn g and fn f pointwise µ-almost everywhere. Then → k k ≤ → Z Z f dµ = lim fn dµ. n Ω →∞ Ω Proof. By the triangle inequality and the classical Lebesgue dominated convergence theorem, Z Z Z f dµ fn dµ f fn dµ 0 as n . k Ω − Ω k ≤ Ω k − k → → ∞

1.2 Bochner-Lebesgue spaces

Deﬁnition 1.11( p spaces). For every 1 p < we deﬁne L ≤ ∞ Z p(Ω;X):= f : Ω X measurable : f p dµ < . L { → Ω k k ∞} 6 1 Bochner-Lebesgue and Bochner-Sobolev spaces

We also deﬁne

∞(Ω;X):= f : Ω X measurable : C 0 such that µ( f C ) = 0 . L { → ∃ ≥ {k k ≥ } } Lemma 1.12. For every 1 p < we put ≤ ∞ Z p 1/p f p := f dµ . k k Ω k k We also put f := inf C 0 : µ( f C ) = 0 . k k∞ { ≥ {k k ≥ } } p Then p is a seminorm on (Ω;X) (1 p ). k · k L ≤ ≤ ∞ Remark 1.13. A function : X R+ on a real or complex vector space is called a seminorm if k · k → (i) x = 0 x = 0, ⇒ k k (ii) λx = λ x for every λ K and all x X, k k | |k k ∈ ∈ (iii) x + y x + y for all x, y X. k k ≤ k k k k ∈ Deﬁnition 1.14( Lp spaces). For every 1 p we put ≤ ≤ ∞ p Np := f (Ω;X): f p = 0 { ∈ L k k } = f p(Ω;X): f = 0µ-almost everywhere . { ∈ L } We deﬁne the quotient space

p p L (Ω;X):= (Ω;X)/Np, L which is the space of all equivalence classes

p [ f ]:= f + Np, f (Ω;X). ∈ L Lemma 1.15. For every [ f ] Lp(Ω;X) ( f p(Ω;X)) the value ∈ ∈ L

[ f ] p := f p k k k k is well deﬁned, i.e. independent of the representant f . The function p is a norm on Lp(Ω;X). The space Lp(Ω;X) is a Banach space when equipped withk · k this norm.

Remark 1.16. As in the scalar case we will in the following identify functions f p(Ω;X) with their equivalence classes [ f ] Lp(Ω;X), and we say that Lp is∈ a Lfunction space although we should be aware∈ that it is only a space of equivalence classes of functions.

Remark 1.17. For Ω = (a,b) an interval in R and for µ = λ the Lebesgue measure we simply write 1.3 The convolution 7

Lp(a,b;X):= Lp((a,b);X).

We can do so since the spaces Lp([a,b];X) and Lp((a,b);X) coincide since the end points a and b have Lebesgue measure zero and there is no danger of confusion. { } { }

Lemma 1.18. Let Ω Rd be open and bounded. Then C(Ω¯ ;X) Lp(Ω;X) for every 1 p . ⊂ ⊆ ≤ ≤ ∞ Proof. Actually, for ﬁnite measure spaces, we have the more general inclusions p q 1 L∞(Ω;X) L (Ω;X) L (Ω;X) L (Ω;X) ⊆ ⊆ ⊆ if 1 q p . ≤ ≤ ≤ ∞ Lemma 1.19. Let the measure space (Ω, ,µ) be such that Lp(Ω) is separable for 1 p < (e.g. Ω Rd be an open set withA the Lebesgue measure). Let X be separable. Then≤ Lp∞(Ω;X) is⊂ separable for 1 p < . ≤ ∞ p Proof. By assumption the spaces L (Ω) and X are separable. Let (hn) p ⊆ L (Ω;X) and (xn) X be two dense sequences. Then the set ⊆

:= f : Ω X : f = hn xm F { → } is countable. It suﬃces to shows that Lp(Ω;X) is total, i.e. span is dense in Lp(Ω;X). This is an exercise. F ⊆ F

Theorem 1.20. Let Ω be as in lemma 1.19. Let 1 < p < and assume that X is reﬂexive. Then the space Lp(Ω;X) is reﬂexive and ∞

p p L (Ω;X)0 L 0 (Ω;X0).

Proof. Without proof.

1.3 The convolution

Theorem 1.21 (Young’s inequality). Let T L1(RN; (X,Y)) and f Lp(RN;X) (1 p ). Then for almost every x RN the∈ integral L ∈ ≤ ≤ ∞ ∈ Z T f (x):= T(x y) f (y) dy ∗ RN − converges absolutely, and for the function T f thus deﬁned one has ∗ T f Lp(RN;Y) and ∗ ∈ T f p T 1 f p . k ∗ kL ≤ k kL k kL 8 1 Bochner-Lebesgue and Bochner-Sobolev spaces

Proof. The case p = is almost trivial. Actually, the strong continuity of the shift semigroup on∞ L1 yields continuity (and thus measurability) of T f while the boundedness of T f and Young’s inequality are immediate from∗ the triangle inequality. ∗ Assume now that p = 1. By Tonnelli’s theorem, we have Z Z T(x y) f (y) dy dx RN RN k − kk k Z Z = T(x y) f (y) dx dy RN RN k − kk k

= T 1 f 1 , k kL k kL and from this equality follows the claim. Assume now 1 < p < . From the previous case we deduce that for almost all x RN ∞ ∈ T(x ) f ( ) p L1(RN), k − · kk · k ∈ and thus 1 T(x ) p f ( ) Lp(RN). k − · k k · k ∈ 1 p N N On the other hand, T(x ) p0 L 0 (R ) for every x R . By Holder’s¨ inequality, for almost everyk −x · k RN∈, ∈ ∈ T(x ) f ( ) L1(RN), k − · kk · k ∈ and Z Z !p T(x y) f (y) dy dx RN RN k − kk k Z Z ! p Z p0 T(x y) dy T(x y) f (y) p dy dx ≤ RN RN k − k RN k − kk k Z Z p 1 p = T −1 T(x y) f (y) dx dy k kL RN RN k − kk k p p = T f k kL1 k kLp < . ∞ For every T L1(RN; (X,Y)) and every f L1(RN;X) we call the function T f Lp(RN;Y∈) the convolutionL of T and∈f . It is a fundamental tool in harmonic∗ ∈ analysis and the theory of partial differential equations. One first property is the following regularizing effect of the convolution. We recall that NN we adopt multi-index notation. For example, for every multi-index α 0 we define ∈ 1.3 The convolution 9

XN α := α , | | k k=1 YN α!:= αk!, and k=1 YN xα := xαk (x CN). k ∈ k=1

Moreover, we denote by ∂k the partial derivative operator with respect to the k-th variable, and deﬁne the α-th partial derivative

α α1 αN ∂ := ∂1 ...∂N .

Let Ω RN be an open set. For every function f C(Ω;X) we define the support ⊆ ∈ supp f := x Ω : f (x) , 0 , { ∈ } where the closure has to be taken in Ω! We then define for k N0 ∈ ∪ {∞} Ck(Ω;X):= f Ck(Ω;X) : supp f is compact , c { ∈ } the space of compactly supported Ck-functions. In the special case X = K we define (Ω):= C∞(Ω). D c Elements of (Ω) are called test functions. D 1 N N Lemma 1.22 (Regularization). For every f L (R ;X) and every ϕ Cc∞(R ) one has f ϕ C (RN;X) and ∈ ∈ ∗ ∈ ∞ ∂α( f ϕ) = f ∂αϕ. ∗ ∗ Lemma 1.23 (Strong continuity of the shift-group). For every x RN and every 1 p we define the shift operator S(x) (Lp(RN;X)) by ∈ ≤ ≤ ∞ ∈ L (S(x) f )(y):= f (x + y)( f Lp(RN;X), y RN). ∈ ∈ Then S(x) is an isometric isomorphism and, if p < , ∞ p N lim S(x) f f Lp = 0 for every f L (R ;X). x 0 k − k ∈ → Proof. The first statement about S(x) being an isometric isomorphism is easy 1 (with S(x)− = S( x)). Next, for every simple step function f = 1Q x with a cube Q RN, the− second statement follows easily from Lebesgue’s⊗ dominated convergence⊆ theorem. By linearity, the second statement holds for every f in the dense subspace 10 1 Bochner-Lebesgue and Bochner-Sobolev spaces

N D := span 1Q x : Q R a cube, x X . { ⊗ ⊆ ∈ } p N Now ﬁx f L (R ;X) and let ε > 0. Then there exists g D such that f g Lp < ∈ ∈ k − k N ε. Moreover, there exists δ > 0 such that S(x)g g Lp < ε for every x R with x < δ. Hence, for every x RN withk x <− δ k ∈ k k ∈ k k S(x) f f p S(x) f S(x)g p + S(x)g g p + g f p k − kL ≤ k − kL k − kL k − kL 2 g f p + S(x)g g p ≤ k − kL k − kL < 3ε. R 1 RN If ϕ L ( ) is such that RN ϕ = 1, then we call the sequence (ϕn)n given by ∈ N N ϕn(x):= n ϕ(nx)(x R , n N) ∈ ∈ an approximate identity or an approximate unit. The reason for this notion follows from the following lemma.

Lemma 1.24 (Property of an approximate identity). Let f Lp(RN;X) (1 ∈ ≤ p < ) and let (ϕn)n be an approximate identity. Then ∞ p N lim f ϕn = f in L (R ;X). n →∞ ∗ Proof. By Tonnelli’s theorem, the Holder¨ inequality, by the strong continuity of the shift-group and by Lebesgue’s dominated convergence theorem we have Z Z p p f ϕn f Lp = f (x y)ϕn(y) dy f (x) dx k ∗ − k RN RN − − Z Z !p f (x y) f (x) ϕn(y) dy dx ≤ RN RN k − − k| | Z Z p p 1 f (x y) f (x) ϕn(y) dy ϕn −1 dx ≤ RN RN k − − k | | k kL Z Z p 1 = ϕn −1 f (x y) f (x) dxϕn(y) dy k kL RN RN k − − k Z Z p 1 y = ϕn −1 f (x ) f (x) dxϕ(y) dy k kL RN RN k − n − k 0 (n ). → → ∞ Corollary 1.25. For every 1 p < the space C (RN;X) is dense in Lp(RN;X). ≤ ∞ c∞ Proof (by regularization and truncation). Let f Lp(RN;X). In the ﬁrst step, the ∈ regularization step, we choose an approximate identity (ϕn) starting with a N p N test function ϕ Cc∞(R ). By Young’s inequality, f ϕn L (R ;X), by Lemma ∈ N ∗ ∈ 1.22, f ϕn C (R ), and by Lemma 1.24, ∗ ∈ ∞ 1.4 Bochner-Sobolev spaces 11

lim f ϕn f Lp = 0. n →∞ k ∗ − k

In the second step, the truncation step, we choose a sequence (ψm)m of test functions satisfying 0 ψm 1 and ψm = 1 on the ball B(0,m) (such ≤ ≤ functions can be obtained by convolving characteristic functions χB(0,2m) with appropriate positive test functions, relying on Lemma 1.22). It is clear from Lebesgue’s dominated convergence theorem, that for every g Lp(RN;X) one has ∈ lim gψm g Lp = 0. m →∞ k − k Combining the preceding two equalities, we ﬁnd a sequence (mn)n in N such that lim ( f ϕn)ψm f Lp = 0, n n →∞ k ∗ − k N and since ( f ϕn)ψm C (R ;X), the claim is proved. ∗ n ∈ c∞ Corollary 1.26. Let f Lp(RN;X) be such that ∈ Z f ϕ = 0 for every ϕ (RN). RN ∈ D Then f = 0.

Proof. The assumption implies that Z f ϕ(x) = f (y)ϕ(x y) dy = 0 for every x RN, ϕ (RN), ∗ RN − ∈ ∈ D which just means that

f ϕ = 0 for every ϕ (RN). ∗ ∈ D

The claim now follows upon choosing an approximate identity (ϕn) out of a test function ϕ and by applying Lemma 1.24.

1.4 Bochner-Sobolev spaces

Let Ω RN be an open set, 1 p and k N. We deﬁne the Bochner- Sobolev⊆ space ≤ ≤ ∞ ∈

k,p p N p W (Ω;X):= u L (Ω;X): α N0 vα L (Ω;X) ϕ (Ω) { ∈ Z∀ ∈ ∃ ∈ Z ∀ ∈ D α α f ∂ ϕ = ( 1)| | vαϕ Ω − Ω } 12 1 Bochner-Lebesgue and Bochner-Sobolev spaces

k,p The functions vα in this deﬁnition of the space W (Ω;X) are uniquely deter- α α mined. We write vα =: ∂ u and we call the function ∂ u the weak α-th partial derivative of u. The space Wk,p(Ω;X) becomes a Banach space for the norm X α u k,p := ∂ u p . k kW k kL α NN ∈ 0 α k | |≤ Similarly as in the case of the Lp-spaces we write Wk,p(a,b;X) instead of Wk,p((a,b);X). In the special case when p = 2 and X = H is a Hilbert space, we also write Hk(Ω;H):= Wk,2(Ω;H). This space is a Hilbert space for the inner product X α α u,v k := ∂ u,∂ v 2 . h iH h iL α NN ∈ 0 α k | |≤

The resulting norm Hk is equivalent to the norm Wk,2 deﬁned above. The main resultsk about · k Sobolev spaces of scalar-valuedk · k functions remain true for Sobolev spaces of Banach space valued functions if interpreted prop- erly. In particular, the Sobolev embedding theorem, a version of the product rule, the integration by parts formula and Poincare’s´ inequality remain true. Even a version of the Rellich-Kondrachev theorem remains true.

Lemma 1.27. For every < a < b < and every 1 p one has W1,p(a,b;X) −∞ ∞ ≤ ≤ ∞ ⊆ Cb((a,b);X). For every u W1,p(a,b;X) and every s, t (a,b) one has ∈ ∈ Z t u(t) u(s) = u0(r) dr. − s Lemma 1.28. Assume that the embedding V , H is continuous and let u W1,2(0,T;H) L (0,T;V). Then u is weakly continuous→ with values in V, that∈ ∩ ∞ is, for every v V the function t v,u(t) V V is continuous on [0,T]. ∈ 0 7→ h i 0, Proof. Since every function u W1,2(0,T;H) is continuous (and hence weakly continuous) with values in H,∈ the claim follows from [Temam (1984), Lemma 1.4, page 263] .

Lemma 1.29. Assume that the embedding V , H is continuous and let (un) be a sequence such that →

1,2 un * u in W (0,T;H) and w ∗ un u in L∞(0,T;V). →

Then there exists a subsequence of (un) (which we denote again by (un)) such that 1.4 Bochner-Sobolev spaces 13

un(t) * u(t) in V for every t [0,T]. ∈ Proof. Using the fact that the point evaluation in t [0,T] from W1,2(0,T;H) into H is bounded and linear, and maps weakly convergent∈ sequences into weakly convergent sequences, the assumption implies that for every t [0,T] ∈

un(t) * u(t) in H.

Let now w H and t [0,T]. Then one has ∈ 0 ∈

w,un(t) u(t) V V = w,un(t) u(t) H H 0. h − i 0, h − i 0, −→

Using the fact that H0 is dense in V0 and that the sequence (un(t)) is bounded in V, the claim follows from Lemma ??.

Chapter 2 The Fourier transform

2.1 The Fourier transform in L1

1 N Let X be a Banach space with norm := X. For every f L (R ;X) we deﬁne the Fourier transform f and| the · | adjoint| · | Fourier transform∈ ¯ f by F F Z ixy f (x):= e− f (y) dy and F RN Z ¯ f (x):= eixy f (y) dy (x RN). F RN ∈ The integrals are absolutely convergent, and we have the trivial estimates

N f (x) , ¯ f (x) f 1 for every x R . |F | |F | ≤ k kL ∈ In particular, the functions f and ¯ f are bounded. F F Theorem 2.1 (Riemann-Lebesgue). For every f L1(RN;X) one has f , ¯ f RN ∈ F F ∈ C0( ;X). Proof. The fact that the Fourier transform f is continuous follows easily from Lebesgue’s dominated convergence theorem.F Next, for every x RN, ∈ x , 0,

Z x x 1 iπ · ixy ixy x 2 f (x) = (e− e− e | | ) f (y) dy F 2 RN − Z 1 πx eixy( f (y) f (y )) dy = + 2 . 2 RN − x | | Since the shift group on L1(RN;X) is strongly continuous, we thus obtain Z 1 πx f x f y f y y x ( ) ( ) ( + 2 ) d 0 as . kF k ≤ 2 RN k − x k → | | → ∞ | |

15 16 2 The Fourier transform

The arguments for the adjoint Fourier transform are similar.

Corollary 2.2. The Fourier transform and the adjoint Fourier transform are 1 RN F RN bounded, linear operators from L ( ;X) into C0( ;X). Weneed the following basic lemma in order to prove the inversion formula for the Fourier transform.

Lemma 2.3 (F´ejerkernel). One has, for a > 0, Z sin2 ax x a 2 d = π. R x Proof. We deﬁne

Z 2 ∞ sin ax f e λx x (λ):= − 2 d (λ (0, )). 0 x ∈ ∞ Then f C ((0, )) and ∈ ∞ ∞ Z 2 Z 2 ∞ sin ax 1 sin ax lim f (λ) = dx = dx, and λ 0+ x2 2 x2 → 0 R lim f (λ) = 0. λ →∞ A simple computation shows

Z 2 ∞ λx sin ax f 0(λ) = e− dx, and − 0 x Z ∞ λx 2 f 00(λ) = e− sin ax dx 0 Z iax iax !2 ∞ λx e e− = e− − dx 0 2i 1 1 2 1 = + −4 λ 2ia − λ λ + 2ia 1 2 − 2λ = . 4 λ − λ2 + 4a2 As a consequence, 1 λ2 f 0(λ) = log . 4 λ2 + 4a2 In order to integrate this function, we make the ansatz ! 1 λ2 f (λ) = λ log + g(λ) , 4 λ2 + 4a2 2.1 The Fourier transform in L1 17 which leads to the equation

8a2 g0(λ) = , −λ2 + 4a2 that is, λ g(λ) = 4aarctan + C − 2a

Together with the condition limλ f (λ) = 0 we thus ﬁnd →∞ ! 1 λ2 π λ f (λ) = λ log + 4a( arctan ) . 4 λ2 + 4a2 2 − 2a

This yields π lim f (λ) = a , λ 0 2 → which implies the claim.

N Before stating the following theorem we deﬁne for every r R with rk 0 the set ∈ ≥ N Qr := [ r ,r ]. − k k k=1 Theorem 2.4 (Inversion formula for the Fourier transform I). Let f L1(RN;X). For every R > 0 we put ∈ Z Z 1 ixy N gR(x):= e f (y) dy dr (x R ). N N (2πR) [0,R] Qr F ∈

1 N Then gR L (R ;X) and ∈ lim gR f L1 = 0. R k − k →∞ Proof. For every R > 0 and every x RN we compute, using Fubini’s theorem, ∈ 18 2 The Fourier transform Z Z 1 eixy f (y) dy dr N N (2πR) [0,R] Qr F Z Z Z 1 iy(x z) = e − dy f (z) dz dr N N N (2πR) [0,R] R Qr Z Z N 1 Y sin(r (x z )) k k k r f z z = N − d ( ) d (πR) RN N xk zk [0,R] k=1 − Z N Y 1 R (x z )2 f (z) dz = 2 R π k k RN 2 − k=1 sin ( 2 (xk zk)) Z − = kR(x z) f (z) dz R − = kR f (x) ∗ where YN 2 R sin ( 2 xk) N kR(x):= (x R ) R 2 ∈ k=1 2 πxk is the F´ejerkernel. Note that

1 N kR L (R ), ∈ kR 0, ≥ N N kR(x) = R k2(Rx) for every x R and Z ∈ kR(x) dx = 1 (Lemma 2.3) RN for every R > 0. Hence, (kR)R is an approximate identity, and the claim follows from Young’s inequality%∞ and Lemma 1.24.

Corollary 2.5 (Inversion formula for the Fourier transform II). Let f L1(RN;X) be such that f L1(RN;X). Then ¯ f L1(RN;X) and ∈ F ∈ F ∈ 1 f = ¯ ( f ) and (2π)N F F 1 f = ( ¯ f ). (2π)N F F

Proof. Since Z ¯ f (x) = eixy f (y) dy = f ( x) for every x RN, F RN F − ∈ we immediately obtain ¯ f L1(RN;X). F ∈ 2.1 The Fourier transform in L1 19

Now let gR be deﬁned as in the preceding theorem. For every R > 0 and every x RN we then have ∈ 1 gR(x) ¯ ( f )(x) = − (2π)N F F " Z Z Z # 1 1 = eixy f (y) dy dr eixy f (y) udy N N N N (2π) R [0,R] Qr F − R F Z Z 1 1 = eixy f (y) dy dr, N N N c (2π) R [0,R] Qr F and hence, for every L > 0

1 g x ¯ f x limsup R( ) N ( )( ) R − (2π) F F ≤ →∞ Z Z 1 h 1 limsup eixy f (y) dy dr N N N c ≤ (2π) R R [L,R] Qr F →∞Z Z 1 i eixy f y y r + limsup N ( ) d d R R [0,R]N [L,R]N Qc F →∞ \ r N Z N 1 1 h (R L) NLR − i f y y f 1 N limsup −N ( ) d + limsup N L ≤ (2π) R R ([ L,L]N)c |F | R R kF k Z →∞ − →∞ 1 f y y N ( ) d ≤ (2π) ([ L,L]N)c |F | − Since L > 0 was arbitrary, and since Z lim f (y) dy = 0, L ([ L,L]N)c |F | →∞ − we thus obtain

1 N lim gR(x) = ¯ ( f )(x) for every x R . R (2π)N F F ∈ →∞ Combining this with the ﬁrst inversion formula, we obtain the ﬁrst identity. The second identity is proved similarly.

¯ 1 RN RN Corollary 2.6. The Fourier transforms , : L ( ;X) C0( ;X) are injective. F F →

Remark 2.7. The Fourier transform on L1 is not surjective onto C . F 0 Lemma 2.8 (Fourier transform and convolution). For every T L1(RN; (X,Y)) and every f L1(RN;X) one has ∈ L ∈ 20 2 The Fourier transform

(T f ) = T f and F ∗ F F ¯ (T f ) = ¯ T ¯ f. F ∗ F F Proof. For every x RN we compute, using Fubini’s theorem, ∈ Z ixy (T f )(x) = e− T f (y) dy F ∗ RN ∗ Z Z ixy = e− T(y z) f (z) dy dz RN RN − Z Z ix(y+z) = e− T(y) dy f (z) dz RN RN Z Z ixy ixz = e− T(y) dy e− f (z) dz RN RN = T(x) f (x). F F The second identity is proved similarly. Lemma 2.9 (Fourier transforms of partial derivatives). For every f C (RN;X), every multi-index α NN and every x RN one has ∈ c∞ ∈ 0 ∈ (∂α f )(x) = (ix)α f (x). F F Proof. For every k 1,...,N we obtain, using integration by parts, ∈ { } Z ixy (∂k f )(x) = e− ∂k f (y) dy F RN Z ixy = (∂ke− ) f (y) dy − RN Z ixy = ixk e− f (y) dy RN = ix f (x). kF The general formula for higher derivatives follows by induction. Corollary 2.10 (Fourier transforms of vector-valued test functions). For every f C (RN;X) we have f L1(RN;X). ∈ c∞ F ∈ Proof. Let p : CN C be any polynomial. The preceding lemma implies → (p(∂) f )(x) = p(ix) f (x) for every x RN. F F ∈ By the Lemma of Riemann-Lebesgue, the left-hand side of this equality is uniformly bounded in x RN. Hence, for every polynomial p : CN C we have ∈ → sup p(ix) f (x) < . x RN | F | ∞ ∈ 2.2 The Fourier transform on L2 21

Choosing p such that p(ix) = 1 + x k for some k N large enough, we obtain the claim. | | ∈

2.2 The Fourier transform on L2

Theorem 2.11 (Parseval’s identity). a) For every T L1(RN; (X,Y)) and every f L1(RN;X) one has ∈ L ∈ Z Z T(x) f (x) dx = T(x) f (x) dx. RN F RN F

b) For every f , g L1(RN) one has ∈ Z Z f (x)g(¯x) dx = f (x) ¯ g(x) dx. RN F RN F

c) For every f , g L1(RN) such that f , g L1(RN) one has ∈ F F ∈ Z Z 1 f x g ¯x x f x g x x ( ) ( ) d = N ( ) ( ) d . RN (2π) RN F F

Similar identities hold if we replace everywhere by ¯ and vice versa. F F Proof. (a) We calculate, using Fubini’s theorem, Z Z Z ixy T(x) f (x) dx = e− T(y) dy f (x) dx RN F RN RN Z Z ixy = T(y) e− f (x) dx dy RN RN Z = T(y) f (y) dy. RN F (b) is proved in a similar way and (c) follows from (b) by using the Inversion Formula II (Corollary 2.5). ¯ N 2 N Theorem 2.12 (Plancherel). The Fourier transforms , :Cc∞(R ) L (R ) extend uniquely to bounded, linear operators on L2(RFN).F The operators→ 1 , √2πN F 1 ¯ : L2(RN) L2(RN) are unitary and √2πN F → 1 1 ( )∗ = ¯ . √2πN F √2πN F Proof. From Parseval’s identity (Theorem 2.11 (c)) we obtain, that for every f , g C (RN) ∈ c∞ 22 2 The Fourier transform

N f, g 2 = (2π) f, g 2 , hF F iL h iL and in particular, f 2 = (2π)N f 2 . kF kL2 k kL2 N 2 N As a consequence, since Cc∞(R ) is dense in L (R ), extends in a unique way to a bounded, linear operator on L2(RN). Moreover,F we see from the above equality that 1 is isometric. As a consequence, this operator is √2πN F injective and has closed range. However, from the inversion formula we see N 1 that C∞(R ) is contained in the range. Hence, is surjective, and thus c √2πN F unitary. The arguments for ¯ are similar. F Theorem 2.13 (Plancherel in Hilbert space). Let H be a Hilbert space. Then ¯ N 2 N the Fourier transforms , :Cc∞(R ;H) L (R ;H) extend to bounded, linear operators on L2(RN;H)F. TheF operators 1 → , 1 ¯ : L2(RN;H) L2(RN;H) √2πN F √2πN F → are unitary and 1 1 ( )∗ = ¯ . √2πN F √2πN F Proof. The proof is very similar to the previous proof, once one has proved the following variant of Parseval’s identity (Theorem 2.11 (c)) for every f , g L1(RN;H) such that f , g L1(RN;H): ∈ F F ∈ Z Z 1 f x g x x f x g x x ( ), ( ) H d = N ( ), ( ) H d . RN h i (2π) RN hF F i Remark 2.14. Kwapien has shown the following result: if the Fourier trans- N 2 N from extends from Cc∞(R ;X) to a bounded, linear operator on L (R ;X) (X being a general Banach space), then X is already isomorphic to a Hilbert space, that is, there exists an inner product on X which induces an equivalent norm. We will not prove this result here.

2.3 The Fourier transform on S We deﬁne the space Z N N N β α 2 (R ;X):= f C∞(R ;X): α, β N0 : x ∂ f (x) dx < . S { ∈ ∀ ∈ RN k k ∞}

Elements of (RN) (that is, X = C) are called the rapidly decreasing functions or SchwartzS (test) functions. Clearly, the space of (classical) test functions N N N x2 Cc∞(R ) = (R ) is a subspace of (R ), but the function f (x) = e− is an example ofD a Schwartz test functionS which does not have compact support. 2.3 The Fourier transform on 23 S It is an exercise to show that Z N N N β α (R ;X) = f C∞(R ;X): α, β N0 : x ∂ f (x) dx < S { ∈ ∀ ∈ RN k k ∞} N N β α = f C∞(R ;X): α, β N0 : sup x ∂ f (x) < . { ∈ ∀ ∈ x RN k k ∞} ∈ The space (RN;X) is equiped with the topology induced by the countable S family of seminorms ( α,β)α,β NN , where k · k ∈ 0

1 Z ! 2 β α f α,β := x ∂ f (x) dx . k k RN k k

This countable family of seminorms induces in a natural way a metric d given by X f g α,β d( f, g):= cα,β k − k , 1 + f g α,β α,β NN k − k ∈ 0 P where the coefficients cα,β > 0 are fixed such that α,β NN cα,β < . We have ∈ 0 ∞ N N fn f in (R ;X) α, β N : fn f 0 → S ⇔ ∀ ∈ 0 k − kα,β → d( fn, f ) 0, ⇔ → and the space (RN;X) is complete. In other words, the countable family of seminorms turnsS (RN;X) into a Frechet´ space. From the definitionS of the space (RN;X) we immediately obtain the following lemma which is, however, worthS of being stated separately.

Lemma 2.15. For every f (RN;X) and every polynomial p : CN C the product p f and the (sum of) partial∈ S derivative p(∂) f belong again to (R→N;X). In other words, the mappings S

f p f and 7→ f p(∂) f 7→ leave the space (RN) invariant. S Lemma 2.16. For every f (RN;X) and every polynomial p : CN C one has f , ¯ f C (RN;X), and∈ S → F F ∈ ∞ (p(∂) f ) = p(i ) f, F · F (p( i ) f ) = p(∂) f, F − · F ¯ (p(∂) f ) = p( i ) ¯ f, and F − · F ¯ (p(i ) f ) = p(∂) ¯ f. F · F 24 2 The Fourier transform

Proof. Let f (RN;X) and k 1,...,N . Then ∈ S ∈ { } Z ixy ( i k f )(x) = e− iyk f (y) dy F − · − RN Z ∂ ixy = (e− ) f (y) dy RN ∂xk Z ixy = ∂k e− f (y) dy RN = ∂ f (x). kF Moreover, by an integration by parts, Z ixy ∂ (∂k f )(x) = e− f (y) dy F RN ∂yk Z ∂ ixy = (e− ) f (y) dy − RN ∂yk Z ixy = ixk e− f (y) dy − RN = ix f (x). − kF The ﬁrst two equalities follow from these two identities and by induction. The proofs for the adjoint Fourier transform ¯ are similar. F Theorem 2.17. For every f (RN;X) one has f , ¯ f (RN;X) and the Fourier transforms , ¯ : (R∈N S;X) (RN;X) areF linearF isomorphisms.∈ S F F S → S Proof.

2.4 The Fourier transform on S0 We call

N N 0(R ;X):= ( (R );X) S L S := T : (RN) X : T is linear and continuous { S → } the space of (vector-valued) tempered distributions. It is equiped with the following “topology”: a sequence (Tn) of tempered distributions converges in (RN;X) to a tempered distribution T if S0 N lim Tn,ϕ = T,ϕ for every ϕ (R ). n →∞h i h i ∈ S Many classical function spaces are included in the space of tempered distributions. For example, the weighted spaces L1(RN, 1 dx;X)(k N ) 1+ x k 0 | | ∈ 2.4 The Fourier transform on 25 S0 are in a natural way contained in the space of tempered distributions via the mapping

1 N 1 N L (R , dx;X) 0(R ;X), 1 + x k → S | | f T , 7→ f where Z N T f ϕ = f (x)ϕ(x) dx (ϕ (R )). RN ∈ S Note that the integral is absolutely convergent.

Lemma 2.18. Let f , g L1(RN, 1 dx;X) be such that T = T . Then f = g. 1+ x k f g ∈ | | Proof. By linearity, it suﬃces to show that T = 0 implies f = 0. So let f f ∈ L1(RN, 1 dx;X) be such that T = 0. Then 1+ x k f | | Z f (x)ϕ(x) dx = 0 for every ϕ (RN), RN ∈ S which implies Z f (x y)ϕ(y) dy = 0 for every ϕ (RN), x RN. RN − ∈ S ∈ Hence f ϕ = 0 for every ϕ (RN). ∗ ∈ S N Choosing an approximate identity (ϕn) out of a test function ϕ (R ), we obtain f = 0. ∈ S

Note that the classical space Lp(RN;X) (1 p ) is a subspace of 1 N 1 ≤ ≤ ∞ p L (R , dx;X) for some k N0 large enough. Hence, L functions are 1+ x k ∈ tempered| distributions| via the above embedding. N Conversely we say that a distribution T 0(R ;X) belongs to Lp(RN, 1 dx;X) (1 p , k N ) if there exists∈ Sf Lp(RN, 1 dx;X) 1+ x k 0 1+ x k | | ≤ ≤ ∞ ∈ ∈ | | such that T = T f . By the preceding lemma, the function f , if it exists, is uniquely determined. We simply write T Lp(RN, 1 dx;X) if the tempered ∈ 1+ x k distribution T belongs to this space. | | For every tempered distribution T (RN;X) and every multi-index α NN we deﬁne the partial derivative ∂∈αT S (RN;X) by ∈ 0 ∈ S α α α N ∂ T,ϕ := ( 1)| | T,∂ ϕ (ϕ (R )). h i − h i ∈ S Moreover, we deﬁne the Fourier transforms T, ¯ T (RN;X) by F F ∈ S0 26 2 The Fourier transform

T,ϕ := T, ϕ and hF i h F i ¯ T,ϕ := T, ¯ ϕ . hF i h F i Finally, for every polynomial p : CN C we deﬁne the product pT by → pT,ϕ := T,pϕ . h i h i Lemma 2.19. For every f (RN;X), every multi-index α NN and every poly- ∈ S ∈ 0 nomial p : CN C one has → α ∂ T f = T∂α f ,

T f = T f , F F ¯ T f = T ¯ f , and F F pT f = Tp f , that is, the distributional partial derivatives, Fourier transforms and products are consistent with the corresponding classical operators on (RN;X) ( (RN;X)). S ⊆ S0 RN NN Proof. For every ϕ ( ) and every α 0 one has, by deﬁnition of the distributional derivative∈ S and by integration∈ by parts,

α α α ∂ T f ,ϕ = ( 1)| | T f ,∂ ϕ h i − Zh i α α = ( 1)| | f (x)∂ ϕ(x) dx − RN Z = ∂α f (x)ϕ(x) dx RN = T α ,ϕ . h ∂ f i This proves the ﬁrst equality. Using Parseval’s identity, we obtain

T f ,ϕ = T f , ϕ hF i Zh F i = f (x) ϕ(x) dx RN F Z = f (x)ϕ(x) dx RN F

= T f ,ϕ , h F i and this proves the second equality. The third one is proved similarly. The fourth equality uses only the associativity of the product: 2.4 The Fourier transform on 27 S0

pT f ,ϕ = T f ,pϕ h i Zh i = f (x)p(x)ϕ(x) dx RN = T ,ϕ . h p f i Remark 2.20. In the above lemma, the function f can be replaced by any other function for which the formula of integration by parts (ﬁrst equality) or Parseval’s identity (second and third equality) still holds. For example, in the second and third equality, one may take f L1(RN;X) or, if X is a Hilbert space, f L2(RN;X). ∈ ∈ N N Theorem 2.21. The Fourier transforms , ¯ : 0(R ;X) 0(R ;X) are linear, bijective and continuous. F F S → S

Proof. This follows immediately from Theorem 2.17.

From Theorem 2.21, but also from the Riemann-Lebesgue Lemma (The- orem 2.1), Plancherel’s Theorem (Theorem 2.12), the Hausdorﬀ-Young The- orem and Theorem 2.17 we obtain the following picture for the Fourier transform. In the following diagram, a (horizontal) double arrow means that the Fourier transform is an isomorphism between the spaces in the same line. Vertical arrows mean inclusion / natural embeddings.

N N 0(R ;X) ks F +3 0(R ;X) 6 S O Z 5 S O Z

1 N F RN L (R ;X) / C0( ;X) U V

Lp(RN;X) F / Lp0 (RN;X) O X Fourier type p O

L2(RN;X) ks F +3 L2(RN;X) ? X Hilbert ?

(RN;X) ks F +3 (RN;X) S S From Lemma 2.16 we immediately obtain the following analogon for tempered distributions.

Lemma 2.22. For every T (RN;X) and every polynomial p : CN C one has ∈ S0 → 28 2 The Fourier transform

(p(∂)T) = p(i ) T, F · F (p( i )T) = p(∂) T, F − · F ¯ (p(∂)T) = p( i ) ¯ T, and F − · F ¯ (p(i )T) = p(∂) ¯ T. F · F Theorem 2.23 (Fourier characterization of Sobolev spaces). Let H be a Hilbert space and k N. Then the Fourier transforms , ¯ map the Sobolev space Hk(RN;H) isomorphically∈ onto the weighted space LF2(RFN,(1 + x 2)kdx;H). | | Proof. Let ﬁrst k = 1. Then we have

f H1(RN;H) ∈ 2 N f, ∂ f, ..., ∂N f L (R ;H) ⇔ 1 ∈ 2 N (by Plancherel) f, (∂ f ),..., (∂N f ) L (R ;H) ⇔F F 1 F ∈ 2 N (by Lemma 2.22) f, ix1 f,..., ixN f L (R ;H) ⇔FZ F F ∈ 2 2 (1 + x1 + + xN) f (x) dx < ⇔ RN ··· F ∞ f L2(RN,(1 + x 2) dx;H). ⇔F ∈ | | The case k 2 is proved by induction and the assertion for ¯ is proved similarly. ≥ F

For the Sobolev spaces we thus have the following picture, in which again vertical arrows stand for inclusions and all (horizontal) double arrows mean that the Fourier transform is an isomorphism.

L2(RN;H) ks F +3 L2(RN;H) O O

H1(RN;H) ks F +3 L2(RN,(1 + x 2) dx;H) O O | |

H2(RN;H) ks F +3 L2(RN,(1 + x 2)2 dx;H) | |

Hk(RN;H) ks F +3 L2(RN,(1 + x 2)k dx;H) | |

ks F +3

Remark 2.24. This definition justifies to define the fractional Sobolev space Hs(RN;H) for s R (thus including Sobolev spaces of negative order) by ∈ 2.6 The Marcinkiewicz multiplier theorem 29

s N 1 2 N 2 s H (R ;H):= − L (R ,(1 + x ) dx;H). F | | Note that for negative s R the weighted space on the right-hand side of ∈ N 2 N this deﬁnition is only included in 0(R ;H), but not in L (R ;H). As a consequence, if s R is negative, thenS Hs(RN;H) is a subspace of the space of tempered distributions∈ which actually includes L2(RN;X).

2.5 Elliptic and parabolic equations in RN

2.6 The Marcinkiewicz multiplier theorem

Chapter 3 Singular integrals

3.1 The Marcinkiewicz interpolation theorem

Let (Ω,µ) be a measure space and (X, X) be a Banach space. Given a measurable function f : Ω X and a parameter| · | λ > 0, we shortly write → f X > λ := t Ω : f (t) X > λ , and we deﬁne the distribution function {|m | : (0, }) {[0∈, ] by| | } f ∞ → ∞

m (λ):= µ( f X > λ )(λ > 0). f {| | } Lemma 3.1. Let Φ : [0, ) [0, ) be diﬀerentiable, increasing and such that Φ(0) = 0. Let f : Ω X∞ be a→ measurable∞ function. Then → Z Z ∞ Φ( f X) dµ = Φ0(λ)m f (λ) dλ. Ω | | 0 Proof. This follows from a simple application of Tonnelli’s theorem:

Z Z Z f (t) | |X Φ( f X) dµ = Φ0(λ) dλ dµ(t) Ω | | Ω 0 Z Z ∞ = Φ0(λ) dµ dλ 0 f X>λ Z {| | } ∞ = Φ0(λ)m f (λ) dλ. 0 Example 3.2. For Φ(λ) = λp (p 1) we obtain ≥ Z Z p ∞ p dλ f X dµ = p λ m f (λ) , Ω | | 0 λ and in particular, f Lp(Ω;X) if and only if ∈

31 32 3 Singular integrals

1 p p λ λm f (λ) L (0, ), 7→ ∈ ∗ ∞ where p dλ L (0, ):= Lp((0, ); ). ∗ ∞ ∞ λ Note that f Lp(Ω;X) thus implies that ∈ p supλ m f (λ) < . λ>0 ∞

Denote by M(Ω;X) be the space of all measurable functions Ω X. Let → (Ω1,µ1) and (Ω2,µ2) be two measure spaces, and let (X, X) and (Y, Y) be two Banach spaces. We say that a (not necessarily linear)| · | operator| ·T | : p L (Ω1,X) M(Ω2,Y) satisﬁes a weak-(p,q) estimate, or we say that T is → p weak-(p,q) if there exists a constant C 0 such that, for every f L (Ω1;X) and every λ > 0 ≥ ∈ !q C f Lp m (λ) k k (if q < ), T f ≤ λ ∞ or T f L C f p (if q = ). k k ∞ ≤ k kL ∞ We say that an operator T on a subspace of M(Ω1;X) with values in M(Ω2,Y) subadditive if for every f , g in the domain of T

T( f + g) Y T f Y + Tg Y almost everywhere, | | ≤ | | | | and we say that it is homogeneous if for every α K and every f in the domain of T ∈ T(α f ) Y = α T f Y almost everywhere. | | | || | Finally, T is said to be sublinear if it is both subadditive and homogeneous.

Theorem 3.3 (Marcinkiewicz interpolation). Let (Ω1,µ1), (Ω2,µ2) be two p measure spaces, X, Y be two Banach spaces, 1 p0 < p1 , and let T : L 0 + p ≤ ≤ ∞ L 1 (Ω ;X) M(Ω2,Y) be a subadditive operator which is both weak-(p0,p0) and 1 → weak-(p ,p ). Then, for every p0 0. For each λ > 0 we write f = f0 + f with ∈ 1 1

f0 = f 1 f >cλ , {| |X } f1 = f 1 f cλ . {| |X≤ } p p Then f0 L 0 (Ω ;X) and f L 1 (Ω ;X), and, by the subadditivity of T, ∈ 1 1 ∈ 1

T f (t) Y T f0(t) Y + T f (t) Y for every t Ω2. | | ≤ | | | 1 | ∈ 3.1 The Marcinkiewicz interpolation theorem 33

As a consequence, for every λ0 > 0,

λ0 λ0 m (λ0) m ( ) + m ( ). T f ≤ T f0 2 T f1 2

By assumption, there exist constants C0, C1 0 independent of f , f0, f1, λ, λ0 such that ≥

!p0 λ0 2C0 f0 Lp0 mT f0 ( ) k k , and 2 ≤ λ0 !p1 λ0 2C1 f1 Lp1 mT f1 ( ) k k (p1 < ), 2 ≤ λ0 ∞ T f L C f L (p = ). k 1k ∞ ≤ 1 k k ∞ 1 ∞ The case p = . Choose c := (2C ) 1 and λ = λ. Then m ( λ ) = 0. Hence 1 ∞ 1 − 0 T f1 2 Z p ∞ p 1 T f Lp = p λ − mT f (λ) dλ k k 0 Z ∞ p 1 λ p λ − mT f0 ( ) dλ ≤ 0 2 Z Z ∞ p 1 p0 p0 p0 p λ − − (2C0) f (t) X dµ(t) dλ ≤ 0 f >cλ | | {| |X } Z Z f (t) /c | |X p0 p 1 p0 = p(2C0) f (t) X λ − − dλ dµ(t) Ω1 | | 0 p p0 p p0 p = (2C0) (2C1) − f Lp . p p0 k k − The case p < . Similarly as above we obtain, for every c > 0, 1 ∞ Z p ∞ p 1 T f Lp = p λ − mT f (λ) dλ k k 0 Z ∞ p 1 λ p λ − mT f0 ( ) dλ ≤ 0 2 Z ∞ p 1 λ + p λ − mT f1 ( ) dλ 0 2 Z Z ∞ p 1 p0 p0 p0 p λ − − (2C0) f (t) X dµ(t) dλ ≤ 0 f X>cλ | | Z Z{| | } ∞ p 1 p1 p1 p1 + p λ − − (2C1) f (t) X dµ(t) dλ 0 f cλ | | {| |X≤ } p p 1 1 (2C0) 0 (2C0) 0 p p( + )( p p + p p ) f Lp . ≤ p p0 p p c 0 c 0 k k − 1 − − − 34 3 Singular integrals

This completes the proof.

Remark 3.4. By minimizing over c > 0, we obtain that the constant C 0 in Theorem 3.3 can be chosen as ≥

1 1 p 1 1 p θ 1 θ C = 2p ( + ) C0 C1− , p p0 p p − 1 − where the C0, C1 0 are the constants from the weak-(pi,pi) estimates, and θ (0,1) is chosen≥ such that ∈ 1 θ 1 θ = + − . p p0 p1

3.2 The Hardy-Littlewood maximal operator

N Let (X, X) be a Banach space and N N. For every f M(R ;X) we deﬁne the maximal| · | function M f : RN [0, ∈ ] by ∈ → ∞

N M f (x):= sup f (y) X dy (x R ), Q x Q | | ∈ 3 ? where Q is any cube with sides parallel to the axes, that is, ball with respect to the norm, and where for every measurable set B RN we have set | · |∞ ⊆ Z 1 = , B B | | B B being the Lebesgue measure? of B. By continuity, the deﬁnition does not change| | if one considers only the supremum over all cubes with rational cen- ters and rational radii, so that one sees that M f is measurable. The operator M : M(RN;X) M(RN) is called the Hardy-Littlewood maximal operator. It is easily seen→ that M is sublinear.

Lemma 3.5 (Covering lemma in R). Let K R be a compact set, and let (Iα)α A S ⊆ ∈ be a family of intervals such that K = α A Iα. Then there exists a ﬁnite subfamily ∈ (Iα )1 j n such that j ≤ ≤ [n K Iα and ⊆ j j=1 Xn 1I (x) 2 for every x R. αj ≤ ∈ j=1 3.2 The Hardy-Littlewood maximal operator 35

Theorem 3.6. The Hardy-Littlewood maximal operator M is weak (1,1) and strong (p,p) for every 1 < p . ≤ ∞ Proof (for the case N = 1). The estimate

M f L f L k k ∞ ≤ k k ∞ follows immediately from the deﬁnition. Hence, M is strong ( , ) = weak ( , ). By the Marcinkiewicz interpolation theorem it suﬃces∞ to∞ show that M∞is∞ weak (1,1). Now let f L1(RN;X). Let λ > 0 and let K M f > λ be any compact ∈ ⊆ { } subset. For every x K there exists an interval Ix containing x such that ∈

f X > λ. Ix | | S ? Clearly, K x K Ix, so that, by Lemma 3.5, there exists a ﬁnite subset ⊆ ∈ x ,...,xn K such that { 1 } ⊆ [n K Ix and ⊆ j j=1 Xn 1I (x) 2 for every x R. xj ≤ ∈ j=1

Hence,

Xn K Ix | | ≤ | j | j=1 n Z X 1 f X ≤ λ I | | j=1 xj Z n 1 X 1I f X ≤ λ xj | | R j=1 2 f 1 . ≤ λ k kL Since this inequality holds for every compact subset K M f > λ , the inner regularity of the Lebesgue measure yields ⊆ { }

2 m (λ) f 1 . M f ≤ λ k kL For the general case N > 1, we need the following covering lemma, which is a variant of Lemma 3.5. 36 3 Singular integrals

N Lemma 3.7 (Vitali). Let K R be a compact set, and let (Qα)α A be a family of N ⊆ ∈ cubes in R such that [ K Qα. ⊆ α A ∈ Then there exist α , ..., αn A such that 1 ∈ [n K Q 5 and ⊆ αj, j=1 Xn 1Q 1. αj ≤ j=1

Proof.

Proof (of Theorem 3.6 for the general case N 1). The beginning of the proof is the same as in the case N = 1. We only need≥ to prove a weak (1,1) estimate. Let f L1(RN;X). Let λ > 0 and let K M f > λ be any compact subset. For ∈ ⊆ { } every x K there exists a cube Qx containing x such that ∈

f X > λ. Qx | | S ? Clearly, K x K Qx, so that, by Lemma 3.7, there exists a ﬁnite subset ⊆ ∈ x ,...,xn K such that { 1 } ⊆ [n K Qx 5 and ⊆ j, j=1 Xn N 1Q (x) 1 for every x R . xj ≤ ∈ j=1

Hence, 3.2 The Hardy-Littlewood maximal operator 37

Xn K Qx 5 | | ≤ | j, | j=1 Xn N = 5 Qx | j | j=1 Xn Z N 1 5 f X ≤ λ Q | | j=1 xj Z n 5N X 1Qx f X ≤ λ N j | | R j=1 5N f 1 . ≤ λ k kL Since this inequality holds for every compact subset K M f > λ , the inner regularity of the Lebesgue measure yields ⊆ { }

5N m (λ) f 1 . M f ≤ λ k kL

Lemma 3.8. If f L1(RN;X) is not identically 0, then M f < L1(RN). ∈ Proof. If f L1(RN;X) is not identically 0, then there exist r > 0 and ε > 0 such ∈ that Z f (y) X dy ε. Qr(0) | | ≥ Now, for any x RN with x r one has ∈ | | ≥

A weight w : RN R is a measurable function which is strictly positive almost everywhere.→ Given a weight w, we denote also by w the weighted Lebesgue measure w(x)dλ(x), so that, for a measurable set B RN ⊆ Z w(B):= 1B(x)w(x) dλ(x). RN

If f : RN X is in addition a measurable function, the we denote its distri- → bution function with respect to the weighted Lebesgue measure by w f , that is, 38 3 Singular integrals

w (λ):= w( f X > λ ). f {| | } p N p p N We denote by Lw(R ;X) the weighted L space L (R ,w(x) dλ(x);X). We say that a locally integrable weight w satisﬁes the Muckenhoupt Ap condition or that w is an Ap weight (1 p < ), and we write w Ap, if there exists a constant C 0 such that ≤ ∞ ∈ ≥ Mw(x) Cw(x) for every x RN, if p = 1, ≤ ∈ ! !p 1 1 p − N w w − 0 C for every cube Q R , if p > 1. Q Q ≤ ⊆ ? ? The smallest possible constant C 0 for which the above inequality (for p = 1 ≥ or for p > 1) holds is called the Ap-constant of the weight; it is denoted by

[w]Ap . Theorem 3.9. For every 1 p < the weak (p,p) estimate ≤ ∞ Z C p wM f (λ) p f X w (3.1) ≤ λ RN | | holds if and only if w Ap. ∈ Proof. Necessity. Assume that there exists a constant C 0 such that for every p N ≥ R f L (Ω;X) the inequality (3.1) holds. Let Q R be a cube such that f X > ∈ ⊆ Q | | 0, and let λ > 0 be such that f X > λ. Then Q | |

Q M( f 1Q) > λ > ⊆ { } and therefore w(Q) w (λ). ≤ M( f 1Q) From inequality (3.1) follows Z C p w(Q) p f X w, ≤ λ Q | | and since this inequality holds for every λ > 0 such that f X > λ, we obtain Q | | R f p w w(Q) Q X > C | | . (3.2) p R p Q ≤ ( f X) | | Q | |

The case p = 1. We choose any measurable subset B Q and f = 1B in the above inequality. Then this inequality becomes ⊆

w(Q) w(B) C . Q ≤ B | | | | 3.2 The Hardy-Littlewood maximal operator 39

This inequality implies for almost every x Q ∈ w(Q) Cw(x), Q ≤ | | and maximizing over all cubes Q containing x, we obtain

Mw(x) Cw(x), ≤ that is, w A1. ∈ 1 p The case 1 < p < . We choose f = w − 0 1Q in inequality (3.2) which then becomes ∞ Z !p Z 1 1 p 1 p w(Q) w − 0 C w − 0 , Q ≤ | | Q Q N or, since Q R was arbitrary, w Ap. ⊆ p N ∈ N Suﬃciency. Let f Lw(R ;X). For every cube Q R one has, by Holder’s¨ inequality, ∈ ⊆

!p Z !p 1 1 1 p p f X = f X w w− | | Q | | Q | | Q Z ! Z !p 1 ? 1 1 − p 1 p0 f Xw w − ≤ Q Q | | Q Q | | Z | | ! 1 p Q [w]A f w | | . (3.3) ≤ p Q | |X w(Q) | | Q Now let λ > 0 and let K M f > λ be any compact subset. For every x K ⊆ { } ∈ there exists a cube Qx containing x such that

f X > λ. Qx | | S ? Clearly, K x K Qx. We consider⊆ noew∈ the case N = 1. By the covering lemma (3.5, there exists a ﬁnite subset x ,...,xn K such that { 1 } ⊆ [n K Qx and ⊆ j j=1 Xn 1Q (x) 2 for every x R. xj ≤ ∈ j=1

Hence, if we combine this with inequality (3.3), 40 3 Singular integrals

Xn w(K) w(Qx ) ≤ j j=1 n Z X [w]Ap p p f Xw ≤ λ Q | | j=1 xj Z n [w]Ap X p 1Q f w ≤ λp xj | |X R j=1

2[w]Ap p p f p . ≤ λ k kLw Since this inequality holds for every compact subset K M f > λ , the inner regularity of the weighted Lebesgue measure yields ⊆ { }

2[w]Ap p wM f (λ) p f p . ≤ λ k kLw Remark 3.10. The proof of Theorem 3.9 shows in the necessity part that if C 0 is a constant such that weak (p,p)-inequality (3.1) holds, then ≥

[w]A C. p ≤ On the other hand, the suﬃciency part shows that in the case N = 1

2[w]Ap p wM f (λ) p f p , ≤ λ k kLw that is, C can be chosen equal to 2[w]Ap in (3.1), if N = 1. In the general case N N 1 we obtain that C = 5 [w]A is a possible constant in (3.1). ≥ p We list a few properties of Muckenhoupt weights. The ﬁrst few properties are mainly simple consequences of the Holder¨ inequality.

Lemma 3.11. a) For 1 p q < one has Ap Aq, and for every w Ap ≤ ≤ ∞ ⊆ ∈

[w]A [w]A . q ≤ p 1 p b) For 1 < p < one has w Ap if and only if w 0 Ap , and then ∞ ∈ − ∈ 0 1 1 p0 p 1 [w − ]A = [w] −p . p0 A

1 p c) If w, v A and 1 < p < , then wv Ap and ∈ 1 ∞ − ∈ 1 p 1 p [wv − ]A [w]A [v] − . p ≤ 1 A1 3.2 The Hardy-Littlewood maximal operator 41

1 q 1 q p q 1 d) If 1 p < q, w Aq and v A , then u := (w u ) − Ap and ≤ ∈ ∈ 1 − − ∈ 1 p 1 q p p 1 [u]A ([w] − [v] − ) − . p ≤ Aq A1 Proof. (a) For p = 1 and q > 1 one has

!q 1 ! 1 − − 1 q0 1 1 w − supw(x)− = (inf w(x))− [w]A1 w . Q ≤ x Q x Q ≤ Q ∈ ∈ ? ? For p > 1, the inclusion follows immediately from Holder’s¨ inequality. 1 p0 (b) The Ap0 condition for w − is

! !p0 1 1 p (1 p )(1 p) − sup w − 0 w − 0 − < , Q Q Q ∞ ? ? 1 p 1 but since (p0 1)(p 1) = 1, the left-hand side is equal to [w] − . − − Ap (c) We compute, using a similar inequality as in the proof of (a),

! !p 1 1 p 1 p (1 p)(1 p ) − wv − w − 0 v − − 0 Q Q ! !p 1 ? ? − p 1 1 p p0 1 1 p − − 0 w[v]A ( v) − v[w]A ( w) − ≤ Q 1 Q Q 1 Q p 1 = [w?]A [v] − . ? ? ? 1 A1

Theorem 3.12 (Reverse H¨olderinequality). Let w Ap for some 1 0,C 0 depending only on [w]A such that ≥ p 1 ! 1+ε w1+ε C w for every cube Q RN. Q ≤ Q ⊆ ? ? Corollary 3.13. a) For every w Ap (1 0 depending ∈ ∞ only on [w]Ap such that w Ap ε. In other words, ∈ − [ Ap = Aq. 1 q

0 such that w Ap. ∈ ∞ ∈ c) If w Ap for some 1 p < , then there exists δ > 0,C 0 such that ∈ ≤ ∞ ≥ 42 3 Singular integrals

!δ w(S) S C | | for every cube Q RN and every S Q. w(Q) ≤ Q ⊆ ⊆ | |

1 p0 Proof. (a) Let w Ap for some 1 0 and∈ C 0 such that ≥

1 ! 1+ε (1 p )(1+ε) 1 p N w − 0 C w − 0 for every cube Q R . Q ≤ Q ⊆ ? ? Fix q (1,p) such that 1 q0 = (1 p0)(1 ε). Then the preceding inequality gives∈ − − − !q 1 !p 1 1 q − p 1 1 p − w − 0 C − w − 0 , Q ≤ Q which ﬁnally yields w? Aq. ? ∈ 1 p0 (b) Since w Ap and w − Ap0 , we can choose, by the reverse Holder¨ inequality and∈ by Lemma 3.11∈ (a), a common ε > 0 such

1 ! 1+ε w1+ε C w and Q ≤ Q 1 ? ! 1+ε ? (1+ε)(1 p ) 1 p N w − 0 C w − 0 for every cube Q R . Q ≤ Q ⊆

? ? 1+ε From both inequalities together follows w Ap. ∈ (c) Fix a cube Q RN and S Q. Let ε > 0 be such that w satisfies the reverse Holder¨ inequality⊆ with exponent⊆ 1 + ε. Then, by Holder’s¨ inequality and by the reverse Holder¨ inequality, Z w(S) = 1S w Q 1 Z ! ε 1+ε ε w S 1+ε ≤ Q | | ! ε S 1+ε Cw(Q) | | . ≤ Q | | We say that a weight satisfies the Muckenhoupt A condition or that w is an A weight, and we write w A , if w satisfies the∞ property in Corollary 3.13 (c).∞ By Corollary 3.13 (c), one∈ has∞ [ Ap A , ⊆ ∞ 1 p< ≤ ∞ 3.2 The Hardy-Littlewood maximal operator 43 and one can show that one actually has equality (compare with Corollary 3.13 (a)).

Theorem 3.14 (Muckenhoupt). For every 1 < p < and every Muckenhoupt ∞ weight w Ap there exists a constant C 0 (depending only on [w]A ) such that ∈ ≥ p p M f p C f p for every f Lw(Ω;X). k kLw ≤ k kLw ∈ In other words, the Hardy-Littlewood maximal operator M is strong (p,p) on the p weighted Lebesgue space Lw.

Proof. Let w be a Muckenhoupt Ap-weight for some 1 < p < . By the corollary to the reverse Holder¨ inequality (Corollary 3.13 (a)), there∞ exists q < p such that w Aq. By Theorem 3.9, the Hardy-Littlewood maximal operator M is ∈ q weak (q,q) on L . On the other hand, M satisﬁes the strong = weak ( , ) w ∞ ∞ estimate on Lw∞ = Lw∞. By Marcinkiewicz’ interpolation theorem (Theorem p 3.3), M satisﬁes therefore a strong (p,p)-estimate on Lw. We conclude this section by noting that in some cases it is also useful to consider the centered maximal operator

N Mc f (x):= sup f X (x R ), r>0 Qr(x) | | ∈ ? in which the supremum is only taken over all cubes centered at x, and the dyadic maximal operator

N Md f (x):= sup f X (x R ) k Z Q k (x) | | ∈ ∈ 2 ? in which the supremum is only taken over all dyadic cubes centered at x, that is, cubes with radius = 2k for some k Z. Clearly, for every f M(RN;X), ∈ ∈ Md f Mc f M f pointwise everywhere. These trivial inequalities show that the centered≤ ≤ and the dyadic maximal operators also satisfy strong (p,p) p estimates on Lw whenever 1 < p < and w Ap. However, one actually has a sort of equivalence between the∞ maximal∈ operators in the sense that, for every f M(RN;X), ∈ N N M f 2 Mc f 4 M f pointwise everywhere. ≤ ≤ d We may exploit this fact later on. We also deﬁne the sharp maximal operator

] N M f (x):= sup f fQ X (x R ), Q x Q | − | ∈ 3 ? where the supremum is taken over all cubes containing x, and where fQ :=

Q f is the mean of f over Q. We say that a function f has bounded mean

> 44 3 Singular integrals

] N oscillation if M f L∞(R ), and we consider the space of all functions of bounded mean oscillation∈

N N ] N BMO(R ;X):= f M(R ;X): M f L∞(R ) , { ∈ ∈ } which is equipped with the seminorm

] f BMO := M f L . k k k k ∞ Note that constant functions have mean oscillation equal to 0, and one would have to take the quotient of BMO with respect to the constant functions in order to obtain a Banach space.

3.3 The Rubio de Francia extrapolation theorem

by Sebastian Krol´ The aim of the section is the proof of a variant of the following boundedness principle by Rubio de Francia [?]: The boundedness properties of a linear operator depend only on the weighted L2 inequalities that it satisﬁes.

This is a final version of the extrapolation theorem Muckenhoupt’s Ap weights. The first variant of the principle is was given by Rubio de Francia in 1982 [Rubio de Francia (1982)]. Below, we shall present the proof of the extrapolation theorem of Rubio de Francia in its first formulation. The underlying philosophy of Rubio de Francia’s extrapolation theory has been summarized by A. Cordoba [Cordoba´ (1988)]:

There are no Lp spaces only weighted L2. This was the basic idea in the original (non constructive) proof of the extrapolation theorem for Muckenhoupt’s Ap weights. Although originally given for operators, it was realized that the operators do not play any role and all the statements can be given in therms of families of nonnegative measurable functions. It is a setting observed by Cruz-Uribe and Perez [Cruz-Uribe and Perez´ (2000)]. Below we follow a presentation given by Duoandikoetxea [Duoandikoetxea (2011)] - a version of the extrapolation theorem with sharp bounds - it is of main interest in studying the sharp dependence of the norms of operators in therms of the Ap-constant of the weights.

Subsequently, given p [1, ), a weight w, and a family of pairs of nonnegative, measurable functions,∈ ∞ we put F 3.3 The Rubio de Francia extrapolation theorem 45 Z p p,w = ( f, g) : g wdx < . F { ∈ F Rn ∞}

Theorem 3.15. Let p0 (1, ) and be a family of pairs of nonnegative, measurable functions. Assume that∈ there∞ existsF an increasing function N : (0, ) (0, ) such ∞ → ∞ that for every Muckenhoupt weight w Ap we have: ∈ 0 !1/p !1/p Z 0 Z 0 p0 p0 g wdx N([w]p0 ) f wdx ( f, g) p0,w . (3.4) Rn ≤ Rn ∈ F

Then for every p (1, ) and every Muckenhoupt weight w Ap we have ∈ ∞ ∈ Z !1/p Z !1/p p p g wdx CNp([w]p) f wdx ( f, g) p,w , Rn ≤ Rn ∈ F where C does not depend on w, and Np([w]Ap ) is given by  p p0  N [w]Ap (2 M Lp(w)) − , if p p0,  k k ≤  p p   p0 1 − 0  Np([w]Ap ):=  − p 1   p 1 −   N [w] − 2 M 1 p , if p > p0.   Ap L (w 0 )    k k p0 − 

The proof of the Theorem 3.15 is based on the following results: the factorization of Muckenhoupt’s Ap weights and the construction of A1 weights via Rubio de Francia’s iteration algorithm.

Lemma 3.16 (Factorization). p p a) Let 1 p < p0 < . If w Ap and u A1, then v := wu − 0 Ap and [v]A ≤ ∞ ∈ ∈ ∈ 0 p0 ≤ p0 p [w]A [u] − . p A1 1 p0 1 p p0 p 1 b) Let 1 < p0 < p < . If w Ap and u A1, then v := (w − u − ) − Ap0 p∞1 p p∈ ∈ ∈ 0− − 0 p 1 p 1 and [v]A [w] − [u] − . p0 ≤ Ap A1 Proof. The statements of Lemma 3.16 follow directly from the deﬁnition of Muckenhoupt’s Ap classes and Holder’s¨ inequality. Here, we provide only the proof of the statement (b).

n p 1 p 1 p 1 Fix a cube Q R . Note that − > 1, and ( − )0 = − . By Holder’s¨ ⊆ p0 1 p0 1 p0 p inequality, we easily get − − −

p 1 p p 0− − 0 Z Z p 1 p p Z ! p 1 Z ! p 1 1 1 0− − 0 1 1 vdx = w p 1 u p 1 dx wdx − udx − . (3.5) Q Q − − ≤ Q Q | | Q | | Q | | Q | | Q 46 3 Singular integrals R u x 1 u 1 udx 1 x Q On the other hand, since ( )− [ ]A1 ( Q Q )− for almost every , ≤ | | ∈ and (p0 1)(p 1) = 1, we have − 0 − p p !p 1 !p 1 p p ! − 0 Z 0 Z 0 − 0 Z p 1 1 1 p − 1 1 p − p 1 1 − − v − 00 dx w − 0 dx [u] − udx . (3.6) Q ≤ Q A1 Q | | Q | | Q | | Q Therefore, combining (3.5) and (3.6) we obtain the desired conclusion.

Recall that, by Muckenhoupt’s theorem (Theorem 3.14, the Hardy- p Littlewood maximal operator M is bounded on L for every p (1, ) and w ∈ ∞ every Muckenhoupt weight w Ap. For every p (1, ) and every Muck- ∈ ∈ ∞ enhoupt weight w Ap we deﬁne therefore deﬁne the Rubio de Francia ∈ operator R = Rp,w as follows

k X∞ M f p R f := ( f Lw), (2 M p )k ∈ k=0 k kLw k where M denotes the k-th iterate of M, and M p is the operator norm of Lw p k k M on L , that is, M p = sup M f p . w Lw f p 1 Lw k k k kLw ≤ k k Lemma 3.17 (Rubio de Francia’s iteration algorithm). Let R be the Rubio de p Francia extrapolation operator, 1 < p < , w Ap, and h L . Then: ∞ ∈ ∈ w a) One has

h Rh pointwise almost everywhere, and | | ≤ Rh p 2 h p . k kLw ≤ k kLw b) If h 0, then ≥ M(Rh) 2 M p Rh pointwise almost everywhere, ≤ k kLw that is, Rh A and ∈ 1 [Rh]A 2 M p . 1 ≤ k kLw Proof. The proof is immediate. p Proof (of Theorem 3.15). The case p < p0. Fix w Ap and ( f, g) p,w with f L ∈ ∈ F ∈ w and g , 0. Let p f L h := f + k k w g. g p k kLw Note that h p 2 f p . By Lemma 3.17, k kLw ≤ k kLw

Rh A1 with [Rh]A 2 M p . ∈ 1 ≤ k kLw 3.3 The Rubio de Francia extrapolation theorem 47

Consequently, Lemma 3.16 yields

p p0 p0 p v := w(Rh) − Ap with [v]A [w]A [Rh] − . ∈ 0 p0 ≤ p A1

g p g p k kLw k kLw p0 Furthermore, since g f p h f p Rh (see Lemma 3.17), note that g Lv , ≤ k kLw ≤ k kLw ∈ that is, ( f, g) p0,v. Thus, combining∈ F Holder’s¨ inequality (with respect to the weighted mea- p p p sure wdx - note also 0 > 1 and ( 0 ) = 0 ,), our assumption (3.4), and p p 0 p p0 Lemma 3.17, we easily get: −

Z Z p p p p (p p0) (p0 p) g w dx = g w(Rh) p0 − (Rh) p0 − dx Rn Rn p p Z ! p Z !1 p p p p 0 p − 0 g 0 w(Rh) − 0 dx (Rh) w dx ≤ Rn Rn p p Z ! Z !1 p p0 p0 p0 p p p p p − − 0 0 N [w]Ap [Rh]A f w(Rh) − dx (Rh) w dx ≤ 1 Rn Rn Z p pp p p 0 N [w]Ap (2 M L ) − (Rh) w dx ≤ k k w Rn Z p pp p p p 0 N [w]Ap (2 M L ) − 2 h w dx ≤ k k w Rn Z p p pp p 2 p 0 2 N [w]Ap (2 M L ) − f w dx. ≤ k k w Rn

p The case p > p0. Fix w Ap and ( f, g) p,w with f Lw and g , 0. By a duality argument we can write∈ ∈ F ∈

p0 Z ! p (Z ) gpwdx := sup gp0 ϕwdx : ϕ L(p/p0)0 (w), ϕ = 1,ϕ 0 . Rn Rn ∈ k k ≥

Fix such a function ϕ. Let h stand for the function given by

p   p p0  p p0  p p f − g − − 0 p0 1 p0       h w − = ϕ +   +    w.   f p   g p   k kLw k kLw p 0 1 p0 Note that h L 1 p , and w − Ap0 . Therefore, by Rubio de Francia’s iteration ∈ w − 0 ∈ algorithm (Lemma 3.17), we obtain that the operator R = R 1 p is bounded p0,w − 0 p0 on L 1 p , with norm less than or equal to 2, h Rh, and Rh A1 with [Rh]A1 w − 0 ≤ ∈ ≤ 1 p0 1 p p0 p 1 2 M p . Lemma 3.16 shows that v := w − (Rh) − − Ap . Moreover, L 0 ) 0 k k 1 p ∈ w − 0 48 3 Singular integrals since p0 p p p0 p p p p 0 0 p p 0 0 p0 0 p 1 p 1 0 p 1 p 1 g g p− h − w− − g p− (Rh) − w− − , ≤ k kLw ≤ k kLw note that ( f, g) p0,v. ∈ F p p p 1 p p p 1 − 0 0− − 0 0− Finally, since ϕw h p 1 w p 1 (Rh) p 1 w p 1 = v, we obtain ≤ − − ≤ − − Z Z Z p0 p0 p0 p0 g ϕw dx g vdx N([v]Ap ) f v dx Rn ≤ Rn ≤ 0 Rn Z p p p 1 − 0 0− p0 p0 p 1 p 1 1 = N([v]Ap ) f (Rh) − w − w− w dx 0 Rn p p Z ! 0 Z !1 0 p − p p0 p p0 1 p0 N([v]Ap ) f w dx (Rh) w − dx ≤ 0 Rn Rn p0 p0 p p Z ! p Z !1 p − 0 − p 1 p0 p p0 1 p0 2 − N([v]Ap ) f w dx h w − dx ≤ 0 Rn Rn p0 p p Z ! p − 0 p 1 p0 p 2 − 3N([v]Ap ) f w dx , ≤ 0 Rn where we applied Holder’s¨ inequality. Therefore, the proof is complete.

The example considered below well illustrates the underlying ideas of Rubio de Francia’s extrapolation theory, which are not so transparent in the setting of pairs of functions presented above. This theory was summarized by A. Cordoba as follows: There are no Lp spaces only L2; compare with (3.7) below.

Example 3.18. Let X be a Banach space. Let T be a sublinear operator, which 2 is bounded on Lw(X) for every Muckenhoupt weight w A2. Assume that, for every C > 0, ∈

sup T 2 : w A2, [w]A C < . {k kLw(X) ∈ 2 ≤ } ∞ How can we apply Rubio de Francia’s extrapolation principle to show that T p extends to a bounded operator on L (X) for every p (1, ) and every w ∈ ∞ Muckenhoupt weight w Ap? ∈ Set [ 2 := ( f X, T f X): f L (X) , F { | | | | ∈ w } w A ∈ 2 and N(t):= sup T 2 : w A2, [w]A t (t > 0). {k kLw(X) ∈ 2 ≤ } 3.3 The Rubio de Francia extrapolation theorem 49

Then, 2 2 w ( f X, T f X): f L (X) . F , ⊇ { | | | | ∈ w } By Rubio de Francia’s extrapolation theorem (Theorem 3.15), for every p (1, ) and every Muckenhoupt weight w Ap there exists a constant Cp,w such∈ that∞ ∈ Z Z [ p p 2 p T f Xw dx Cp,w f Xw dx for every f Lv(X) with T f Lw(X). N | | ≤ N | | ∈ ∈ R R v A ∈ 2

Therefore, if [ 2 p p,w := f L (X): T f L (X) D { ∈ v ∈ w } v A ∈ 2 p is dense in Lw(X), then T admits a unique extension to a bounded operator p on L (X), and T p Cp,w. w Lw(X) k k ≤ p In fact, we show that p,w L (X). Note ﬁrst that D ⊇ w [ [ p 2 Lv(X) = Lv(X). (3.7) p (1, ) v A ∈ ∞ 2 v Ap ∈ ∈

2 Indeed, let 0 , f Lw(X). In the case p < 2, following the lines of the corre- ∈ 2 sponding part of the proof of Theorem 3.15, it is easy to check that f Lv(X) p 2 2 ∈ with v := wu − A2, where u := Rh for h := f X. For p > 2, f Lv(X) with 1 ∈ | | p∈ p 2 p 1 p0 1 p0 v := (wu − ) − A2, where u := Rh for h given by h w − = f X w. ∈ p | | In particular, T f is well-deﬁned for all f Lw(X). Therefore, it is suﬃcient p p ∈ to show that T f X L for all f L (X). | | ∈ w ∈ w For this purpose, consider the (formal) adjoint operator M0 to the Hardy- p Littlewood maximal operator M (as an operator on Lw), that is,

M(hw) p0 M0h := ( f L ). w ∈ w

p0 1 p Note that M is bounded on L . Indeed, we ﬁrst note that w 0 Ap with 0 w − ∈ 0 1 p p0 1 [w − 0 ]A [w] − - it follows immediately from the deﬁnition of the Ap p0 ≤ Ap 0 p0 p0 class. Therefore, since f w L if and only if f Lw , and f p = f p , 1 p0 L 0 L 0 ∈ w − ∈ k k 1 p k k w w − 0 we have Z ! 1 Z ! 1 p0 p0 p0 p0 1 p0 (M0 f ) w dx = (M( f w)) w − dx M p f p . L 0 L 0 RN RN ≤ k k 1 p k k w w − 0 Consequently, M0 p M p . L 0 L 0 k k w ≤ k k 1 p w − 0 50 3 Singular integrals

Consider also the (formal) adjoint operator R0 to R (see Lemma 3.17) given by: X∞ M0(h) p0 R0h := ( f L ). k ∈ w (2 M0 p0 ) k=0 k kLw

The counterpart of Lemma 3.17 holds for the operator R0, namely: p a) For h L 0 one has ∈ w

h R0h pointwise almost everywhere, and | | ≤ R0h p0 2 h p0 . k kLw ≤ k kLw b) If, in addition, h 0, then ≥

M(wR0h) 2 M p wR0h pointwise almost everywhere, ≤ k kLw that is, wR h A and 0 ∈ 1

[wR0h]A1 2 M p0 . ≤ k kLw

p p0 1 Let 0 , f L (X) and h L with h 0. Set v := R( f X) R hw. Note that, ∈ w ∈ w ≥ | | − 0 by Lemma 3.16 (a), v A2 with [v]A2 [R( f X)]A1 [R0h]A1 . Then, by Holder’s¨ inequality, we obtain∈ ≤ k k

2 p T L (X) R( f X) L R0h p0 k k v k k k k w k kLw Z T L2(v;X) R( f X)R0(h)w dx ≥ k k RN k k 1 1 Z ! 2 Z ! 2 2 1 T L2(X) f XR( f X)− R0(h)w dx R( f X)R0(h)w dx ≥ k k v RN | | | | RN | | 1 1 Z ! 2 Z ! 2 2 1 T f X R( f X)− R0(h)w dx R( f X)R0(h)w dx ≥ RN | | | | RN | | Z T f X R0(h)w dx ≥ RN | | Z T f X hw dx. ≥ RN | |

p0 p This implies that T f X (L ) = L . Therefore, our claim holds. | | ∈ w ∗ w 3.4 Calderon-Zygmund operators 51 3.4 Calderon-Zygmund operators

Let X and Y be two Banach spaces. We say that a measurable kernel K : RN RN (X,Y) satisﬁes the standard conditions if there exist constants C ×0 and→δ >L0 such that ≥ C K(x, y) (X,Y) if x y > 0, (S1) | |L ≤ x y N | − | | − | δ y y0 1 K(x, y) K(x, y0) (X,Y) C | − | if 0 < y y0 x y , (S2) | − |L ≤ x y N+δ | − | ≤ 2| − | | − | δ x x0 1 K(x, y) K(x0, y) (X,Y) C | − | if 0 < x x0 x y . (S3) | − |L ≤ x y N+δ | − | ≤ 2| − | | − | Moreover, we call a bounded, linear operator T : Lp(RN;X) Lp(RN;Y) a (generalized) Calderon-Zygmund operator (1 p ﬁxed)→ if there exists a kernel satisfying the standard conditions such≤ that≤ ∞ Z T f (x) = K(x, y) f (y) dy RN (3.8) N for every f Cc(R ;X) and almost every x < supp f. ∈ Theorem 3.19 (Weak (1,1) estimate for Calderon-Zygmund operators). Ev- ery Calderon-Zygmund operator T : Lp(RN;X) Lp(RN;Y) (1 0 and every f Lp L1(RN;X) one has ≥ ∈ ∩ Z N C mT f (λ) = m( x R : T f (x) Y > λ ) f (x) X dx. { ∈ | | } ≤ λ RN | |

Proof. Fix λ > 0 and f Lp L1(RN;X). Applying the corollary of [Stein (1970), ∈ ∩ Theorem 4, Chapter I.3.4] to the function f ( ) X, we obtain a decomposition Rn = F Ω, F Ω = , such that | · | ∪ ∩ ∅

f (x) X λ for almost every x F, | |[≤ ∈ Ω = Qj for cubes Qj such that m(Qj Q ) = 0 for j , k, ∩ k j

Here Qj,2 denotes, similarly as before, the double cube which has the same center as Qj but whose sides are twice as long as those of Q. We set

 f x x F  ( ) for , g(x):=  R ∈  1  Q Q f (x) dx for x Qj, | j| j ∈ and we set b(x) = f (x) g(x). Then − b(x) = 0 for x F, and Z ∈ b(x) dx = 0 for each cube Qj. Qj

1 N g L L R X g 1 f 1 g C Moreover, ∞( ; ), L (X) L (X) and L∞(X) λ. Since T f =∈Tg +∩Tb, it followsk k that ≤ k k k k ≤

n m( x R : T f (x) Y > λ ) { ∈ | | } ≤ n λ n λ m( x R : Tg(x) Y > ) + m( x R : Tb(x) Y > ), ≤ { ∈ | | 2 } { ∈ | | 2 } and it suﬃces to estimate both terms on the right hand side separately. First, we estimate Tg. First of all, g Lp(X) and ∈ Z p p p 1 p 1 p 1 p 1 g Lp(X) = g(x) X dx C − λ − g L1(X) C − λ − f L1(X) k k Rn | | ≤ k k ≤ k k By using in addition the assumption of boundedness of T,

p p p p 1 Tg C g Cλ − f 1 , k kLp(Y) ≤ k kLp(X) ≤ k kL (X) and this implies

n λ C m( x R : Tg(x) Y > ) f 1 . { ∈ | | 2 } ≤ λ k kL (X) P Second, we estimate Tb. Let bj = bχQj . Then b = j bj and it suffices to estimate Tb . j R Fix x F and fix j. Since b = 0, we have ∈ Qj Z Tbj(x) = K(x, y) K(x,xj) b(y) dy, Qj − where xj is the center of the cube Qj. In particular, 3.4 Calderon-Zygmund operators 53 Z Z Z Tbj(x) Y dx K(x, y) K(x,xj) (X,Y) b(y) X dy dx L F | | ≤ F Qj | − | | | Z Z = K(x, y) K(x,xj) (X,Y) dx b(y) X dy L Qj F | − | | | Z Z K(x, y) K(x,xj) (X,Y) dx b(y) X dy ≤ Q Qc | − |L | | j j,2 Z CK b(y) X dy, ≤ Qj | | where we have used the fact that F is a subset of the complement of the double cube Qj,2. Of course, we also used that K satisfies the second standard condition (S2). From the preceding estimate we obtain Z X Z Tb(x) Y dx Tbj(x) Y dx | | ≤ | | F j F Z CK b(y) X dy ≤ Ω | | 2CK f 1 . ≤ k kL (X) This estimate implies

λ C m( x F : Tb(x) Y > ) f 1 . { ∈ | | 2 } ≤ λ k kL (X) On the other hand, λ C m( x Ω : Tb(x) Y > ) m(Ω) f 1 . { ∈ | | 2 } ≤ ≤ λ k kL (X) The preceding two estimates together give the estimate for Tb.

Lemma 3.20 (Good-λ inequality). Fix a weight w A . Then there exist C 0 and δ > 0 such that for every λ > 0, and every γ > 0 small∈ ∞ enough the inequality≥

n δ n w( x R : T¯ ∗ f (x) > 2λ and M f (x) γλ ) Cγ w( x R : T¯ ∗ f (x) > λ ) (3.9) { ∈ ≤ } ≤ { ∈ } holds.

n Proof. First, we can assume that w( x R : T¯ ∗ f (x) > λ ) , 0, for otherwise the above inequality is clearly satisﬁed.{ ∈ Since the measure} w(x) dx is outer regular, there exists an open set Uλ such that

n x R : T¯ ∗ f (x) > λ Uλ and { ∈ n} ⊆ (3.10) w(U ) 2w( x R : T¯ ∗ f (x) > λ ). λ ≤ { ∈ } 54 3 Singular integrals

By Whitney’s lemma (Lemma 3.30), there exists a sequence (Qk) of mutually S ¯ disjoint cubes with sides parallel to the coordinate axes such that Uλ = k Qk and such that Qk,4 (the cube which has the same center as the cube Qk c and satisﬁes diamQk,4 = 4diamQk) has nonempty intersection with Uλ. The cubes Qk are of the form B(x¯k,rk) for some center x¯k and some radius rk > 0, and hence Qk,4 = B(x¯k,4rk). We prove that there exists a constant C 0 which is independent of f such that for every γ > 0 small enough and every≥ k,

m( x Q : T¯ ∗ f (x) > 2λ and M f (x) γλ ) Cγm(Q ). (3.11) { ∈ k ≤ } ≤ k We may in fact assume that γ is small, since the above inequality is trivial for γ C 1. ≥ − Fix k. We may assume that there exists ξk Qk such that M f (ξk) γλ, because otherwise the inequality (3.11) is obviously∈ satisﬁed. Moreover,≤ since Q Uc is nonempty, there exists x Q such that k,4 ∩ λ k ∈ k,4

T¯ ∗ f (x ) λ. k ≤

Now let Q˜ k := B(xk,16rk) be the cube centered at xk and satisfying diamQ˜ k = 16diamQk. Deﬁne f1 = f χQ˜ and f2 = f χQ˜ c , so that f = f1 + f2. By subadditivity k k of the maximal operator T¯ ∗, we have

m( x Q : T¯ ∗ f (x) > 2λ and M f (x) γλ ) { ∈ k ≤ } ≤ λ m( x Q : T¯ ∗ f (x) > and M f (x) γλ )+ (3.12) ≤ { ∈ k 1 2 ≤ } 3λ + m( x Q : T¯ ∗ f2(x) > and M f (x) γλ ), { ∈ k 2 ≤ } and it suﬃces to estimate the two terms on the right-hand side of this inequality. Since ξ Q Q˜ , it follows that k ∈ k ⊆ k Z Z 1 1 n n f1(y) X dy = f (y) X dy 16 M f (ξk) 16 γλ, m Q n m Q ˜ ( k) R | | ( k) Qk | | ≤ ≤ so that the weak (1,1) estimates from Theorem 3.19 and Corollary ?? yield Z n λ 2C m( x R : T¯ ∗ f1 > ) f1(y) X dy Cγm(Qk) (3.13) { ∈ 2 } ≤ λ Rn | | ≤ for some constant C 0 which is independent of f , γ, k and λ. Next, we shall estimate, for small γ≥ > 0, the second term on the right-hand side of (3.12). Fix x Q = B(x¯ ,r ) and ε > 0. Then ∈ k k k 3.4 Calderon-Zygmund operators 55 Z T f x K x y f y dy ε 2( ) Y = ( , ) 2( ) Y | | B(x,ε)c Z

K(xk, y) f2(y) dy c Y ≤ B(xk,ε) Z + K(xk, y) K(x, y) (X,Y) f2(y) X dy c L B(xk,ε) | − | | | Z + K(x, y) (X,Y) f2(y) X dy L B(x,ε)MB(xk,ε) | | | | =: I1 + I2 + I3, where B(x,ε) M B(xk,ε) denotes the symmetric diﬀerence of B(x,ε) and B(xk,ε). We have Z I K x y f y dy T¯ f x 1 = ( k, ) ( ) Y ∗ ( k) λ (B(x ,ε) B(x ,8r ))c ≤ ≤ k ∪ k k by the choice of x , since B(x ,ε) B(x ,8r ) = B(x ,sup ε,8r ) is a cube cen- k k ∪ k k k { k} tered at xk, and by the deﬁnition of T¯ ∗. Furthermore, by using ξk Qk Q˜ k again, and by proceeding similarly as in the estimates (??), one obtains∈ ⊆

I2 CM f (ξ ) Cγλ. ≤ k ≤

Note that I3 = 0 whenever ε 16rk. On the other hand, for ε 16rk one has B(x,ε) M B(x ,ε) B(x,2ε) B(≤x,ε/2), and therefore, ≥ k ⊆ \ Z CK I3 n f (y) X dy ≤ B(x,2ε) B(x,ε/2) x y | | \ | − | n Z CK2 n f (y) X dy ≤ ε B(x,2ε) | | CM f (ξ ) Cγλ. ≤ k ≤ We note that the preceding estimates can also be made for ε = 0 if one interpretes T0 = T. Taking all the above estimates together, and taking the supremum over ε 0, we ﬁnd that there exists a constant C 0 which is independent of f , λ≥, k, γ such that for every x Q , ≥ ∈ k

T¯ ∗ f2(x) Y Cγλ + λ. | | ≤ 1 Taking γ > 0 so small that Cγ < 2 , it follows that 3λ m( x Q : T¯ ∗ f2(x) > and M f (x) γλ ) = 0. { ∈ k 2 ≤ } As a consequence, we have proved (3.11). Now, since the weight w belongs to A , there exist constants δ > 0 such that ∞ 56 3 Singular integrals

w(E) m(E) δ C for every cube Q and every measurable E Q. w(Q) ≤ m(Q) ⊆

Hence, the estimate (3.11) implies that, for every k,

δ w( x Q : T¯ ∗ f (x) > 2λ and M f (x) γλ ) Cγ w(Q ). { ∈ k ≤ } ≤ k Summing up in k and recalling the inequality (3.10) yields the estimate (3.9).

Theorem 3.21 (Coifman-Feﬀerman). Let T : Lp(RN;X) Lp(RN;Y) be a → Calderon-Zygmund operator. Then, for every Muckenhoupt weight w Ap the p N p ∈ N operator T extends to a bounded, linear operator from Lw(R ;X) into Lw(R ;Y). Proof. By the good-λ inequality (3.9) from Lemma 3.20, for every f L1 Lp(RN;X), ∈ ∩ Z p T¯ ∗ f (x) w(x) dx = Rn Z ∞ p 1 = C λ − w( T¯ ∗ f > 2λ ) dλ 0 { } Z Z ∞ p 1 δ ∞ p 1 C λ − w( M f > γλ ) dλ + Cγ λ − w( T¯ ∗ f > λ ) dλ ≤ 0 { } 0 { } Z Z p δ p = C M f (x) w(x) dx + Cγ T¯ ∗ f (x) w(x) dx. Rn Rn

Taking γ > 0 so small that Cγδ 1 , we obtain the claim. ≤ 2 From the preceding theorem and the Rubio de Francia extrapolation theorem (Theorem 3.15, we immediately obtain the following main result of this section.

Theorem 3.22. Let T : Lp(RN;X) Lp(RN;Y) be a Calderon-Zygmund operator. → Then, for every 1 < q < and every Muckenhoupt weight w Aq the operator T ∞ q N q ∈ N extends to a bounded, linear operator from Lw(R ;X) into Lw(R ;Y). Lemma 3.23. If T is a Calderon-Zygmund operator, then, for each s > 1,

] s 1 p N N M (T f )(x) Cs M( f )(x) s ( f L (R ;X), x R ), ≤ | |X ∈ ∈ where M] is the sharp maximal operator.

Proof. Fix s > 1. Let f Lp(RN;X). Given a cube Q RN and given x Q, we ∈ ⊆ ∈ decompose f = f1 + f2, where f1 = f 12Q and f2 = f 1(2Q)c , and we let a := T f2(x). Then 3.4 Calderon-Zygmund operators 57

T f (y) a Y Q | − | ? T f1 Y + T f2 T f2(x) Y. ≤ Q | | Q | − | ? ? Since T is bounded on Lp, the ﬁrst term on the right-hand side of this inequality can be estimated by

T f2 T f2(x) Y Q | − | Z ? (K(y,z) K(x,z)) f (z) dz dy ≤ Q RN 2Q − \ Y ? Z y x δ C | − | f (z) dz dy N+δ X ≤ Q RN 2Q x z | | \ Z| − | ? X∞ f (z) X CLδ | | dz dy ≤ k k+1 x z N+δ Q k= 1 2 L< x z <2 L − | − | | − | ?X Z δ ∞ 1 CL f (z) X dz ≤ (2kL)N+δ k+1 | | k= 1 x z <2 L − | − | X∞ 2N C f (z) dz kδ X ≤ 2 x z <2k+1L | | k=1 | − | 2N C ? M f (x) ≤ 1 2δ − 1 CM˜ ( f s )(x) s ≤ | |X N Lemma 3.24. Fix 1 p < and w Ap. If f M(R ;X) is such that Md f p ≤ ∞ ∈ ∈ ∈ L (RN), then w Z Z p ] p Md f w C M f w. RN | | ≤ RN | | Proof. 58 3 Singular integrals 3.5 The Hilbert transform

Let X be a Banach space. For every f (R;X) we define the Hilbert transform H f : R X by ∈ S → Z 1 H f (x):= P.V. f (x y) dy R y − Z 1 = lim f (x y) dy (x R). ε 0 y ε y − ∈ → | |≥ The limit actually exists for every Schwartz test functions f and every x R, since ∈ Z Z 1 ε f (x y) dy = ln y f (x y) ε∞ + ln y f (x y) − + ln y f 0(x y) dy y ε y − | | − | | − −∞ y ε | | − | |≥ Z | |≥ = lnε( f (x + ε) f (x ε)) + ln y f 0(x y) dy − − y ε | | − Z | |≥ ln y f 0(x y) dy as ε 0. → R | | − → In particular, if we take x = 0 and X = C, we see that the principal value Z 1 P.V. f (y) dy R y is well defined as tempered distribution. Theorem 3.25. If X is a Hilbert space, then the Hilbert transform extends to a bounded, linear operator on L2(R;X). We give two different proofs for this fact. Proof (First proof of Theorem 3.25). Our first proof uses the Fourier transform. We prove that for every f (R;H) one has ∈ S 1 H f = − Msgn f, F F where Msgn is the multiplication operator associated with the sign function, that is, Msgn g := sgn g for appropriate functions g. If the above equality is true, then ·

Lemma 3.26 (Cotlar). Let (Tn)n Z be a sequence of bounded, linear operators on a Hilbert space H. Assume that ∈

TnTm∗ (H), Tn∗ Tm (H) an m, | |L | |L ≤ − where 3.5 The Hilbert transform 59

X 1 a 2 =: A < . n ∞ n Z ∈ Then: a) For every finite I Z ⊆ X Tn (H) A. | |L ≤ n I ∈ b) For every finite family (Iα)1 α m of finite, mutually disjoint Iα Z one has ≤ ≤ ⊆ Xm 2 TI∗ TIα (H) A , | α |L ≤ α=1

where X TIα := Tn. n Iα ∈ c) The series X∞ Tn n= −∞ is strongly convergent. Proof. (a) Recall that for every operator T (H) and every k N one has ∈ L ∈ 2 2k k T (H) = TT∗ (H) and T (H) = (T∗T) (H). | |L | |L | |L | |L In fact, using the Cauchy-Schwarz inequality in its full form (including the equality), we obtain

T∗T (H) = sup T∗Tx H | |L x 1 | | | |H≤ = sup T∗Tx, y H x , y 1h i | |H | |H≤ = sup Tx,Ty H x , y 1h i | |H | |H≤ 2 = sup Tx H x 1 | | | |H≤ 2 = T (H), | |L which is the above equality for k = 1. Iterating this equality, and using that (T∗T)∗ = T∗T, we obtain ﬁrst

4 2 2 T (H) = T∗T (H) = (T∗T) (H) | |L | |L | |L and then the above equality for all powers k = 2m (m N). The full claim follows by writing k N in its binary extension. Having proved∈ the above equal- ∈ 60 3 Singular integrals ity, one sees that assertion (a) follows from (b), namely when one chooses m = 1. (b) Take now a ﬁnite family (Iα)1 α m of ﬁnite, mutually disjoint Iα Z. Let ≤ ≤ ⊆ [m ∆ := Iα Iα Z Z, × ⊆ × α=1 the union being disjoint. Then

Xm X S := T T = T T I∗α Iα i∗ j α=1 (i,j) ∆ ∈ is selfadjoint, nonnegative and, for every k N one has ∈ k k S (H) = S (H). | |L | |L However, X k S = T∗ Tj ...T∗ Tj i1 1 ik k (i ,j ),...,(i ,j ) ∆ 1 1 k k ∈ We have the two estimates

Ti∗ Tj1 ...Ti∗ Tjk (H) a(i1 j1) a(ik jk) | 1 k |L ≤ − ····· − and 1 1 2 2 Ti∗ Tj1 ...Ti∗ Tjk (H) a(0) a(j1 i2) a(jk 1 ik) a(0) . | 1 k |L ≤ · − ····· − − · Hence

X 1 1 1 1 k 2 2 2 2 S H a(0) a(i1 j1) a(j1 i2) a(jk 1 ik) a(ik jk) | | ( ) ≤ − · − ····· − − · − L (i ,j ),...,(i ,j ) ∆ 1 1 k k ∈  2k 1 X  −  1  a(0) I  a(j) 2  , ≤ | |   j Z ∈ Sm where I = α=1. Taking the k-th root and letting k tend to inﬁnity, we obtain the claim. P (c) Assume that there exists x H such that the series n∞= Tnx does not ∈ −∞ converge. Then there exists ε > 0 and a sequence (Iα)α N of mutually disjoint, ∈ ﬁnite subsets Iα Z such that, using the notation from (b), ⊆

TI u H ε. | α | ≥ Then we can choose m N large enough so that ∈ 3.5 The Hilbert transform 61

Proof (Second proof of Theorem 3.25). For every n Z we deﬁne ∈ n n+1 ∆n := x R : 2 x 2 , { ∈ ≤ | | ≤ } 1 1 kn L (R) by kn(x):= 1 (x), and ∈ x ∆n 2 Tn (L (R;H)) by Tn f := kn f. ∈ L ∗

From the equality T = Tn, and from Young’s inequality 1.21 we obtain n∗ −

TnTm∗ (H) = Tn∗ Tm (H) = TnTm (H) kn km L1 . | |L | |L | |L ≤ k ∗ k Assume that n m, and let x R be positive. Then we can estimate ≤ ∈ Z 1 1

kn km(x) = 1∆m (x y) dy | ∗ | 2n y 2n+1 y x y − ≤| |≤ −  m n+1 m+1 n+1 0 if x < [2 2 ,2 + 2 ],  −  4 2 m if x [2m 2n+1,2m + 2n+1],  · − ∈ −  ≤  n 2m m n+1 m+1 n+1 2 2 2− if x [2 + 2 ,2 2 ],  · · ∈ −  4 2 m if x [2m+1 2n+1,2m+1 + 2n+1]. · − ∈ − While the second and fourth estimate follow directly from crude estimates R in the integral, the third estimate uses the equality R kn = 0, which allows one to write Z 1 1 kn km(x) dy ∗ 2n y 2n+1 y x y Z≤| |≤ − 1 1 1 = ( ) dy 2n y 2n+1 y x y − x Z ≤| |≤ − 1 1 = dy. 2n y 2n+1 x y x ≤| |≤ − 62 3 Singular integrals

As a consequence,

n m n m kn km 1 2 8 2 − + 2 2 − k ∗ kL ≤ · · · n m 36 2−| − |. ≤ ·

The sequence (Tn)n Z thus satisﬁes the hypotheses of Cotlar’s lemma, so that, for every f L2∈(R;H) the limit ∈ N 1  N 1  X−  X−  lim Tn f = lim  kn f N N   ∗ →∞ n= N →∞ n= N − − exists in L2(R;H). Since for every f (R;H) and every x R ∈ S ∈  N 1  Z  X−    1 lim  kn f (x) = f (x y) dy N   ∗ n n y − →∞ n= N 2− y 2 − ≤| |≤ = H f (x), we have proved the boundedness of the Hilbert transform in L2(R;H).

Corollary 3.27. If X is a Hilbert space, then for every 1 < p < and every Muck- ∞ enhoupt weight w Ap the Hilbert transform extends to a bounded, linear operator p ∈ on Lw(R;X). We say that a Banach space X has the Hilbert transform property if, for some 1 < p < , the Hilbert transform extends to a bounded, linear operator on Lp(R;X). By∞ Theorem 3.25, Hilbert spaces have the Hilbert transform property. From Theorem 3.22, we have the following general result.

Corollary 3.28. A Banach space X has the Hilbert transform property if and only if for every 1 < p < and every Muckenhoupt weight w Ap the Hilbert transform ∞ p ∈ extends to a bounded, linear operator on Lw(R;X). Corollary 3.29. a) Hilbert spaces have the Hilbert transform property. b) If X has the Hilbert transform property and if 1 < p < , then Lp(Ω;X) has the Hilbert transform property ((Ω, ,µ) being any measure∞ space). A c) The spaces Lp(Ω) have the Hilbert transform property if 1 < p < . ∞ Proof. (a) We already remarked that Hilbert spaces have the Hilbert transform property by Theorem 3.25. (b) This follows from the boundedness of the Hilbert transform on the scalar-valued Lp(R) (Corollary 3.27) and Tonnelli’s theorem. In fact, for every f Lp(R;Lp(Ω)) (this space can be identiﬁed with Lp(R Ω) and Lp(Ω;Lp(R))), ∈ × 3.6 Covering and decomposition 63 Z H f p H f x p udx Lp(R;Lp(Ω)) = ( ) Lp(Ω) k k R k k Z Z = H f (x,ω) p dω dx R Ω | | Z Z = H f (x,ω) p dx dω Ω R | | Z Z C f (x,ω) p dx dω ≤ Ω R | | Z Z = C f (x,ω) p dω dx R Ω | | p = C f , k kLp(R;Lp(Ω)) where C is the operator norm of the Hilbert transform on Lp(R).

3.6 Covering and decomposition

This section contains several covering and decomposition results which where useful in this chapter.

Lemma 3.30 (Whitney decomposition of open sets). Let U (RN be an open, proper subset. Then there exists a sequence (Qk) of mutually disjoint open (dyadic) cubes such that [ U = Q¯ k and k diamQ dist(Q ,Uc) 4diamQ . k ≤ k ≤ k Proof. Let Q := (0,1)N be a reference cube and consider for each k Z the collection ∈ := 2k(x + Q): x ZN Dk { ∈ } of dyadic cubes of side length equal to 2k, and [ := D Dk k Z ∈ the collection of all dyadic cubes. Let now, for every k Z, ∈ U := x U : 2k < dist(x,Uc) 2k+1 , k { ∈ ≤ } 64 3 Singular integrals so that [˙ U = Uk. k Z ∈ We now put := Q : Q¯ U , Qk { ∈ Dk ∩ k ∅} and [ := . Q Qk k Z ∈ One easily observes, by deﬁnition of the , that for every Q Qk ∈ Q diamQ dist(Q ,Uc) 4 √N diamQ . k ≤ k ≤ k Moreover, [ U = Q¯ , Q ∈Q but it may happen that the cubes in the collection are not mutually disjoint. In order to obtain a mutually disjoint union, weQ choose for every Q the ∈ Q unique cube Qmax which contains Q and which has maximal diameter (such a cube exists∈ and Q is uniquely determined due to the fact that U is a proper subset of RN, that is, the complement is nonempty). So if we take the collection 0 := Qmax : Q , then this collection satisﬁes all required properties. Q { ∈ Q} References

[Cordoba´ (1988)]C ordoba,´ A. : Jos´eLuis Rubio de Francia (1949–88): a sketch of his life and works. Rev. Mat. Iberoamericana 4 (1), 1988, 1–10. [Cruz-Uribe and Perez´ (2000)] Cruz-Uribe, D., Perez,´ C. : Two weight extrapolation via the maximal operator. J. Funct. Anal. 174 (1), 2000, 1–17. URL http://dx.doi.org/10.1006/jfan.2000.3570 [Duoandikoetxea (2001)] Duoandikoetxea, J. : Fourier analysis. Vol. 29 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2001. [Duoandikoetxea (2011)] Duoandikoetxea, J. : Extrapolation of weights revisited: new proofs and sharp bounds. J. Funct. Anal. 260 (6), 2011, 1886–1901. URL http://dx.doi.org/10.1016/j.jfa.2010.12.015 [Hille and Phillips (1957)] Hille, E., Phillips, R. S. : Functional Analysis and Semi-Groups. Amer. Math. Soc., Providence, R.I., 1957. [Rubio de Francia (1982)] Rubio de Francia, J. L. : Factorization and extrapolation of weights. Bull. Amer. Math. Soc. (N.S.) 7 (2), 1982, 393–395. URL http://dx.doi.org/10.1090/S0273-0979-1982-15047-9 [Stein (1970)] Stein, E. M. : Singular Integrals and Diﬀerentiability Properties of Functions. Princeton University Press, Princeton, N.J., 1970. [Temam (1984)] Temam, R. : Navier-Stokes Equations. Vol. 2 of Studies in Mathematics and its Applications. Elsevier Science Publishers, 1984.

Index

p approximate identity, 10 boundedness in Lw, 62 approximate unit, 10 Hilbert transform property, 62

Bochner integral, 3, 4 inequality weak type (p,p), 38 Calderon-Zygmund operator, 51 Young, 7 centered maximal operator, 43 integrable function, 3 convolution, 8 integral, 3, 4 inversion formula, 17, 18 distribution product, 26 Lebesgue’s theorem, 5 tempered distribution, 24 lemma dyadic maximal operator, 43 Cotlar, 58 covering lemma, 34 Fejer kernel, 16 Vitali, 36 Fourier transform dimension N = 1, 34 adjoint, 15 Whitney, 63 inversion formula, 17, 18 on , 24 maximal operator S on 0, 25 centered, 43 on LS1, 15 dyadic, 43 on L2, 21 Hardy-Littlewood, 34 function sharp, 43, 56 almost separably valued, 2 measurable function, 1 integrable, 3 Muckenhoupt weight maximal function, 34 A , 42 ∞ measurable, 1 Ap, 38 rapidly decreasing, 22 multi-index, 8 Schwartz test function, 22 step function, 1 operator test function, 9 Calderon-Zygmund, 51 weakly measurable, 1 maximal operator, 34 weight, 37 Rubio de Francia, 46 subadditive, 32 Hardy-Littlewood maximal operator, 34 sublinear, 32 Hilbert transform, 58 boundedness in L2, 58 Parseval’s identity, 21

67 68 Index partial derivative, 9 step function, 1 for distributions, 25 support, 9 weak partial derivative, 12 Pettis’ theorem, 1 tempered distribution, 24 Plancherel’s theorem, 21, 22 test function, 9 principal value, 58 theorem Coifman-Feﬀerman, 56 regularization, 9 Lebesgue, 5 reverse Holder¨ inequality, 41 Marcinkiewicz interpolation, 32 Riemann-Lebesgue theorem, 15 Muckenhoupt, 43 Rubio de Francia Parseval, 21 extrapolation theorem, 45 Pettis, 1 iteration, 46 Plancherel, 21 operator, 46 in Hilbert spaces, 22 Schwartz test function, 22 reverse Holder¨ inequality, 41 sharp maximal operator, 43, 56 Riemann-Lebesgue, 15 shift-group, 9 Rubio de Francia extrapolation, 45 Sobolev space, 11 Young, 7 fractional, 28 space weakly measurable, 1 Lebesgue, 6 weight, 37 Sobolev Muckenhoupt fractional, 28 A , 42 ∞ Sobolev space, 11 Ap, 38 standard conditions, 51 Whitney decomposition, 63