CLASSICAL SPACES OF HOLOMORPHIC FUNCTIONS

MARCO M. PELOSO

Contents 1. Hardy Spaces on the Unit Disc 1 1.1. Review from 1 1.2. Hardy spaces 5 1.3. Harmonic Hardy classes 7 1.4. Fatou’s theorem 9 1.5. The zero sets of functions in Hp 13 1.6. Boundary behaviour of functions in Hardy spaces 16 1.7. The Cauchy–Szeg¨oprojection 19 2. Bergman spaces on the unit disc 22 2.1. Function spaces with reproducing kernel 22 2.2. The Bergman spaces 25 2.3. Biholomorphic invariance 28 2.4. Lp-boundedness of a family of integral operators 29 2.5. The Bergman kernel and projection on the unit disc 33 3. The Paley–Weiner and Bernstein spaces 35 3.1. The 35 3.2. The Paley–Wiener theorems 40 3.3. The Paley–Wiener spaces 45 3.4. The Bernstein spaces 49 4. Function theory on the upper half plane 51 4.1. Hardy spaces on the upper-half plane 51 4.2. Factorization and boundary behaviour of functions in Hp(U). 54 4.3. H2 and the Paley–Wiener theorem revisited 55 4.4. H1, the atomic decomposition and the of bounded mean oscillations 55 4.5. Weighted Bergman spaces on the upper-half plane 55 4.6. The Paley–Wiener theorem for Bergman spaces 55 5. Further topics 56 5.1. Hardy spaces on tube domains in Cn 56 5.2. Weighted Bergman spaces on the unit ball in Cn 56 5.3. The Cauchy integral along Lipschitz curves 56 5.4. Real variable Hardy spaces 56 5.5. Hankel and Toeplitz operators 56 References 56

Appunti per il corso Argomenti Avanzati di Analisi Complessa per i Corsi di Laurea in Matematica dell’Universit`adi Milano, a.a. 2011/12. – November 12, 2011. SPACES OF HOLOMORPHIC FUNCTIONS 1

In these notes we consider spaces of holomorphic functions defined on domains of the or, toward the end of the notes, on domains of the n-dimensional complex space Cn. We begin with the Hardy spaces on the unit disc.

1. Hardy Spaces on the Unit Disc

1.1. Review from complex analysis. Let C denote the field of complex numbers z = x + iy, where x, y ∈ R are the real and imaginary part, respectively, of z. We will denote by D the unit disc, that is, D = z ∈ C : |z| < 1 .

Here and in what follows, we will denote by D(z0, r) the disc centered at z0 ∈ C and radius r > 0; we will sometimes abbreviate Dr to denote the disc D(0, r). If f is holomorphic on D then it can be written as sum of a converging power series +∞ X k f(z) = akz . k=0 From Cauchy’s formula we know that, if γ is a simple closed curve contained in D and z lies in the of γ (which is a well-defined domain by the Jordan curve theorem), then 1 Z f(ζ) f(z) = dζ . 2πi γ ζ − z

If the function f is actually holomorphic in a slightly larger disc DR with R > 1, then we may take the unit circonference ∂D as curve of integration and obtain 1 Z f(ζ) f(z) = dζ , 2πi ∂D ζ − z for all z ∈ D. One of the main themes of this course is trying to recover the f (in a given class) from its boundary values, once these values are suitably defined. Setting z = reiη, we rewrite the above identity as Z 2π iθ iη 1 iθ e f(re ) = f(e ) iθ iη dθ 2π 0 e − re 1 Z 2π 1 = f(eiθ) dθ i(η−θ) 2π 0 1 − re Z 2π 1 i(η−θ) 1 = f(e ) iθ dθ , (1.1) 2π 0 1 − re using the periodicity of the exponential. Notice that +∞ 1 X = rneinθ 1 − reiθ n=0 iθ =: Cr(e ) , (1.2) iθ where Cr(e ) is the Cauchy kernel (thought of as a convolution kernel, see (1.4) below). 2 M. M. PELOSO

Therefore, using the uniform convergence on compact subsets of D of the power series, we have that (1.1) gives the identity

+∞ X 1 Z 2π f(reiη) = rn f(eiθ)ein(η−θ) dθ 2π n=0 0 +∞ X 1 Z 2π = rneinη f(eiθ)e−inθ dθ . (1.3) 2π n=0 0 We recall that the unit circle ∂D = {ζ ∈ C : ζ = eiθ} inherits the multiplicative structure from C and it is homeomorphic (as a topological group) to the interval [−π, π] (or any interval of length 2π) once we identify the two end points of the interval. This group T = R/2πZ is called the torus. Of course, the identification between T and ∂D is given by the mapping t 7→ eit. Given f, g ∈ L1(T) (where, we recall, T is identified with the unit circle) the convolution on T is defined as 1 Z 2π f ∗ g(eiη) = f(eiθ)g(ei(η−θ)) dθ . (1.4) 2π 0 Therefore, the identity (1.1) can be expressed as convolution on the group T as iη iη f(re ) = (f ∗ Cr)(e ) . (1.5) Moreover, in (1.3) we recognize the n-th Fourier coefficient of the restriction of f on the unit circle ∂D (function that, at this stage, is assumed to be continuous) 1 Z 2π fˆ(n) = f(eiθ)e−inθ dθ . 2π 0 It is clear that in order to define the Fourier coefficients is sufficient to assume that f is an L1-function on the unit circle. Another integral formula, with its relative integral kernel, that will play a key role in these notes is the Poisson reproducing formula for the unit disc D 1 Z 2π 1 − r2 u(reiη) = u(eiθ) dθ i(η−θ) 2 2π 0 |1 − re | = (Pu)(reiη) , (1.6) where u is a function harmonic in a domain containing D and 0 < r < 1. The operator P is called the Poisson operator and Pu the Poisson integral of the function u defined on the unit circle. More generally, the solution of the on the unit D (see [CA-notes]) shows that if we start with a continuous function g on the unit circle ∂D, the function u := Pg is harmonic in D, continuous on the closure of the unit disc and coincides with g on the unit circle.

It is worth to observe that Pg is well defined for all g ∈ L1(T) and that Pg is a in D. One of the main questions that we are going to address is if, and in which sense, the function Pg admits boundary values and, in this case, if such values coincide with the function g. In SPACES OF HOLOMORPHIC FUNCTIONS 3 other words, a classical question is whether the Dirichlet problem (for the Laplacian on the unit disc) ( ∆f = 0 on D (1.7) f = g on ∂D , where g ∈ Lp(∂D) is an assigned function, 1 ≤ p < ∞, admits a solution and in what sense the solution f admits g as values on the boundary. As in the case of the reproducing formula (1.5) we may think of the Poisson integral as a convolution integral operator on T by writing 1 − r2 P (eiθ) = , (1.8) r |1 − reiθ|2 so that the Poisson integral of a function g ∈ L1(T) is defined by the formula 1 Z 2π 1 − r2 (Pg)(reiη) = g(eiθ) dθ = (P ∗ g)(eiη) . (1.9) i(η−θ) 2 r 2π 0 |1 − re | In complex analysis we were used to think of the , that we denote temporarely iθ iη iθ as P˜z(e ) as a function of the variables z = re ∈ D and e ∈ ∂D, where 1 − r2 1 − r2 P˜ (eiθ) = = (1.10) z |eiθ − reiη|2 |1 − rei(η−θ)|2 1 − r2 = . (1.11) 1 − 2r cos(θ − η) + r2 In the current setting it is more convenient to think of the Poisson kernel as a family of iη functions {Pr} defined on the torus T, hence depending on the variable e , and of the Poisson integral of a function g as the convolution of g with Pr, as in (1.9). According to this point of view, the collection of functions {Pr}00 is called a summability kernel if they satisfy the following properties: Z 2π 1 iθ (1) Φt(e ) dθ = 1 for t > 0; 2π 0 (2) kΦtkL1 ≤ C with C independent of t; Z iθ (3) for every δ > 0, lim |Φt(e )| dθ = 0. t→+∞ δ≤|θ|≤π p We recall that if {Φt}t>0 is a summability kernel on T and g ∈ L (T), 1 ≤ p < ∞, then p Φt ∗ g → g in the L -norm, and that if g ∈ C(T), then Φt ∗ g → g in the sup-. We collect here the basic properties of the Poisson kernel and integral. iη iη iη Proposition 1.1. The function P (re ) = Pr(e ) is harmonic in D as function of z = re and Z 2π 1 iη Pr(e ) dη = 1 2π 0 for all 0 < r < 1. Moreover, as collection {Pr} of functions on T we have (i) for 1 ≤ p < ∞, kf ∗ PrkLp ≤ kfkLp ; (ii) for 1 ≤ p < ∞, limr→1− kf ∗ Pr − fkLp = 0; 4 M. M. PELOSO

(iii) if g ∈ C(T), then limr→1− kg ∗ Pr − gkL∞ = 0. Finally, for f ∈ Lp(T), 1 ≤ p ≤ ∞, P(f) is harmonic in D. iη Proof. It is clear that it suffices to show {Pr} is a summability kernel and that P (re ) is harmonic in the variable z = reiη ∈ D. 1 R 2π iη The latter fact is well known, as well as that 2π 0 Pr(e ) dη = 1. In order to prove that {Pr} is a summability kernel, it remains to show that condition (3) is satisfied. This is easily verified and we leave the details to the reader. 2 Lemma 1.2. For 0 < r < 1 we have that +∞ iη X |k| ikη Pr(e ) = r e , k=−∞ and the convergence is uniform on T. Proof. Recall that we always identify the unit circle ∂D with T. We can write 1 − r2 1 1 = + − 1 (1.12) |1 − reiη|2 1 − reiη 1 − re−iη +∞ +∞ X X = rkeikη + rke−ikη − 1 k=0 k=0 +∞ X = r|k|eikη . k=−∞ The statement about the uniform convergence is clear, by the Weierstrass M-test. Notice that by (1.12) we obtain a relation between the Poisson and Cauchy kernels,2 that is,

Pr = 2Re Cr − 1 . (1.13) Lemma 1.3. Let g ∈ L1(∂D). Then for every 0 < r < 1 we have that +∞  iη X |k| ikη Pr ∗ g (e ) = gˆ(k)r e . k=−∞

Proof. Using the uniform convergence of the series development of Pr we see that +∞ 1 Z 2π X P ∗ g(eiη) = g(eiθ) r|k|eik(η−θ) dθ r 2π 0 k=−∞ +∞ X  1 Z 2π  = g(eiθ)e−ikθ dθ r|k|eikη 2π k=−∞ 0 +∞ X = gˆ(k)r|k|eikη , k=−∞ and the convergence is uniform on T. 2 SPACES OF HOLOMORPHIC FUNCTIONS 5

1.2. Hardy spaces. We are finally ready to define the classical Hardy spaces. We will denote by H(Ω) the space of holomorphic functions on a given domain Ω. Definition 1.4. For 0 < p < ∞ and 0 < r < 1, for a function f defined on D we set Z 2π 1/p  1 iθ p  Mp(f, r) = |f(re )| dθ . 2π 0 We define the Hp = Hp(D) as p n o H = f ∈ H(D) : sup Mp(f, r) < ∞ , 0

We mention in passing that, also when 0 < p < 1 we set Z 2π 1/p  1 iθ p  kfkLp = |f(e )| dθ 2π 0 p p and, with an abuse of language, we call it the L -norm. However, setting d(f, g) = kf − gkLp p (1−p)/p L becomes a . Notice that kf + gkLp ≤ 2 (kfkLp + kgkLp ).

Remark 1.5. Aside from the case p = ∞, also the space H2 can be described at once. For, P+∞ n if f is holomorphic on D, then it admits power series expansion f(z) = n=0 anz . Using the uniform convergence on compact subsets of D we have Z 2π 2 1 it 2 M2(f, r) = |f(re )| dt 2π 0 +∞ 1 Z 2π X = a a rn+meit(n−m) dt 2π n m 0 n,m=0 +∞ X 2n 2 = r |an| . n=0 Therefore, +∞ 1/2  X 2 sup M2(f, r) = |an| , 0

1Exercise I.2. 6 M. M. PELOSO

˜ ˜ P+∞ n in(·) ˜ Moreover, by Lemma 1.3, P(f)(r·) = Pr ∗f = n=0 anr e = f(r·). We can call f “boundary 2 − values” since f(r·) = Pr ∗ f˜ → f˜ in L (T) as r → 1 . This fundamental relation is a first indication of the deep connection between the theories of and of Hardy spaces.

q p Remark 1.6. For 0 < p < q ≤ ∞ we have {0} ( H ( H . For, notice that for 0 < p < q ≤ ∞ we clearly have Hq ⊆ Hp. Moreover, all holomorphic polynomials are bounded, hence in all the Hp spaces and these are non-trivial. Moreover, using Lemma 2.15 (that we will see later on) one can show that 1 1 f(z) = ∈ Hp if and only if p < . (1 − z)a a

In these notes we will essentially restrict ourselves to the case 1 ≤ p ≤ ∞, although the Hardy spaces turn out to be of great interest also and, from a certain point of view, especially, in the case 0 < p ≤ 1. We point out that the Hardy space H1 can be considered as a limit case for both theories 1 < p ≤ ∞ and 0 < p < 1 and we will say more about it in Section 4.4 when we deal with the Hardy spaces on the upper half-plane. The first observation we make is that the “sup” in the definition of the Hp-norm is actually a “lim”.

Remark 1.7. Given f holomorphic in D, the function Mp(f, r) is increasing in r, 0 < r < 1. This statement holds true for the full range 0 < p ≤ ∞, but its proof is elementary only in the case p ≥ 1 and we will restrict to this case. For 0 < % < 1 and a function f defined in D we write f% := f(%·) to denote a function on the unit circle. For 0 < r, % < 1, given a function f holomorphic in D, notice that fr is holomorphic in a ngbh of D so that iη iη f(r%e ) = (fr ∗ P%)(e ) . Hence, we have

Mp(f, r%) = kfr ∗ P%kLp(T)

≤ kfrkLp(T) kP%kL1(T) = Mp(f, r) , that is, Mp(f, r) is increasing in r and

sup Mp(f, r) = lim Mp(f, r) . 0

Proof. It is clear that k · kHp is a norm, so we only need to prove that it is complete. Let 0 < r, % < 1 and notice that for f holomorphic on D we have it it |f(r%e )| = fr ∗ P%(e ) ≤ kfrkLp(T)kP%kLp0 (T)

≤ C%kfkHp . This shows that sup |f(z)| ≤ C%kfkHp , |z|≤% SPACES OF HOLOMORPHIC FUNCTIONS 7 and therefore the convergence in the Hp-norm implies the uniform convergence on compact subsets. p Thus, let {fn} be a Cauchy in the H -norm, and let f be the function uniform limit on compact subsets of D. Then f is holomorphic and for 0 < r < 1 fixed, using the uniform convergence, for p < ∞ we have Z 2π p 1 it it p Mp(fn − f, r) = |fn(re ) − f(re )| dt 2π 0 Z 2π 1 it it p = lim |fn(re ) − fm(re )| dt m→+∞ 2π 0 p = lim Mp(fn − fm, r) m→+∞ p ≤ lim kfn − fmk p . m→+∞ H Therefore, kfn − fkHp ≤ lim kfn − fmkHp < ε , m→+∞ for n sufficiently large. The case p = ∞ is similar and we leave the details to the reader.2 2 1.3. Harmonic Hardy classes. We now analyze the action of the Poisson integral on the spaces Lp(T). This analysis has interest on its own right, but also will serve for studying the boundary behaviour of functions in (the holomorphic) Hardy spaces. We consider now harmonic functions on D that satisfy the Hp-growth condition, 0 < p < ∞, that is, sup Mp(f, r) < ∞ , (1.14) 0

Proposition 1.9. Let f be harmonic in D and satisfy (1.14), 1 < p ≤ ∞. Then there exists f˜ ∈ Lp(∂D) such that f = P(f˜), that is, 1 Z 2π 1 − r2 f(reit) = (f˜∗ P )(eit) = f˜(eiθ) dθ . r i(t−θ) 2 2π 0 |1 − re | p Moreover, if 1 < p < ∞, fr → f˜ in the L -norm. iη iη We observe that this statement is false when p = 1. For instance, f(re ) = Pr(e ) satisfies 1 the hypotheses with p = 1, but there exists no L -function g on ∂D such that Pr → g and 3 Pr ∗ g = Pr. We leave the details as an exercise. − Proof. Let fr = f(r·), 0 < r < 1. Let {rn} be a strictly increasing sequence in (0, 1), rn → 1 . p Then {frn } is a bounded set in L (T). By the Banach-Alaoglu theorem (see [Ru]), there exists a subsequence {f } converging in the weak-∗ topology to a funtion f˜ ∈ Lp(T), 1 < p ≤ ∞, rnj that is, for every g ∈ Lp0 (T) we have that Z 2π Z 2π f (eiθ)g(eiθ) dθ → f˜(eiθ)g(eiθ) dθ as j → +∞ . rnj 0 0

2Exercise I.4. 3Exercise I.5. 8 M. M. PELOSO

(Here we need the restriction p > 1, since the Banach-Alaoglu theorem requires that the space 1 containing {fr} to be the of another Banach space– and L (T) is not.) Let 0 < r < 1 and let j be such r > r for j ≥ j . Then, f is continuous on D and 0 nj 0 rnj harmonic in D, hence it satisfies the Poisson reproducing formula it it it f(re ) = fr (r/rn )e = (fr ∗ P )(e ) nj j nj r/rnj Z 2π 1 iθ i(t−θ) = fr (e )P (e ) dθ nj r/rnj 2π 0 Z 2π Z 2π 1 iθ i(t−θ) 1 iθ h i(t−θ) i(t−θ) i = fr (e )Pr(e ) dθ + fr (e ) P (e ) − Pr(e ) dθ . nj nj r/rnj 2π 0 2π 0 (1.15)

p0 Since Pr ∈ C(∂D) ⊂ L (∂D) for all 0 < r < 1, the first term on the right hand side above converges to Z 2π 1 iθ i(t−θ) f˜(e )Pr(e ) dθ . 2π 0 Now we show that the second term on the right hand side of (1.15) tends to 0 as j → +∞. For, p0 gj := P − Pr → 0 in the L -norm, since gj ∈ C(T) and tend to 0 uniformly. Therefore, r/rnj Z 2π iθ i(t−θ) fr (e )gj(e ) dθ ≤ kfr kLp kgjk p0 ≤ Ckgjk p0 → 0 nj nj L L 0 as j → +∞. Notice that we have used the assumption that f satisfies (1.14). From (1.15) it follows that Z 2π it 1 iθ i(t−θ) f(re ) = f˜(e )Pr(e ) dθ , 2π 0 as we wished to show. The last part of the statement is clear: fr = f˜∗ Pr as functions on ∂D and since {Pr} is a p summability kernel, fr = f˜∗ Pr → f˜ in the L -norm, for 1 < p < ∞. 2 Remark 1.10. Notice that, in particular Prop. 1.9 applies to Hp-functions. Given f ∈ Hp, we have constructed the boundary value function f˜ as weak-∗ limit of a − sequence {fnj }, where nj → 1 . It is easy to see that this construction is independent of the ˜ choice of the sequence {fnj }. (This is obvious in particular when 1 < p < ∞ since f is the limit p in the L -norm of the whole family {fr}.) p Moreveor, if fr → g in L (∂D), it also easy to see that g = f˜. In particular, the boundary value function of an H2-function identified in Remark 1.5 coincides with the one constructed in Prop. 1.9. We now prove a partial converse of Prop. 1.9. Proposition 1.11. Let g ∈ Lp(∂D), 1 ≤ p < ∞. Then the Poisson integral of g, P(g) is harmonic in D, satisfies the Hp-growth condition (1.14), and (in the notation of Prop. 1.9)

P](g) = g . SPACES OF HOLOMORPHIC FUNCTIONS 9

Proof. Notice that, as functions on ∂D, fr = f(r·) := g ∗ Pr satisfy the norm estimate kfrkLp ≤ kgkLp kPrkL1 = kgkLp . Therefore,

sup Mp(f, r) ≤ kgkLp . 0

We conclude this part with the following consequence of the last two propositions. Corollary 1.13. Define p  h (D) = f harmonic on D : kfkhp := sup Mp(f, r) < ∞ . 0

1.4. Fatou’s theorem. Perhaps, the main result about the Hardy spaces is that if f ∈ Hp with 0 < p ≤ ∞, then f, as in the case p = 2, admits “boundary values”, that is, lim f(reit) = f˜(eit) exists for a.e. t ∈ [0, 2π) , r→1− and f˜ ∈ Lp(∂D). Actually, more is true: the function f converges to f˜ when z approaches a point eit ∈ ∂D, for a.e. t ∈ [0, 2π), within the non-tangential approach regions. Definition 1.14. We define the subset of the unit disc it  it Γα(e ) = z ∈ D : |z − e | < α(1 − |z|) . (1.16) it it The set Γα(e ) is also called the Stoltz region, with vertex in e and aperture α > 1. it Remark 1.15. We need to make a few comments on the definition of Γα(e ). it (i) It is easy to see that the regions Γα(e ) are increasing with α and that they would be empty for α ≤ 1. 10 M. M. PELOSO

it (ii) The region Γα(e ) is also called a non-tangential approach region, since outside a fixed it compact subset of D, it is contained in a cone with vertex in e and aperture Cαα, where Cα > 1 is a suitable constant. For, assume that 1 − εα ≤ r < 1, where cα is a (small) positive constant, that we are going iη it to fix momentarily, and let z = re ∈ Γα(e ). Then iη it iη it iη iη it |re − e | = e − e − (1 − r)e ≥ |e − e | − (1 − r) , so that |eiη − eit| ≤ (α + 1)(1 − r) . it If cα is chosen sufficientily small, then we must have |η − t| < π/4 (that is, z ∈ Γα(e ) and r ≥ 1 − cα implies |η − t| < π/4). √ iη it  2 2 Next, |e − e | = 2 sin |η − t|/2 ≥ π |η − t|, as long as |η − t| ≤ π/4, which is the case. iη it Therefore, z = re ∈ Γα(e ) and r ≥ 1 − cα implies π |η − t| ≤ √ (α + 1)(1 − r) = Cαα(1 − r) , (1.17) 2 2 as claimed. In particular, we may take Cα = 3.

Theorem 1.16. (Fatou) Let f ∈ Hp, 1 < p ≤ ∞. Then lim f(z) = f˜(eit) , it it Γα(e )3z→e

it where f˜ is the function as in Prop. 1.9, and Γα(e ) is as in (1.16). In order to prove Fatou’s theorem, we need to prove a result concerning the Hardy–Littlewood maximal function, defined (in the case of the torus T) as 1 Z Mf(eit) = sup |f(ei(t−θ))| dθ . 0<δ≤π 2δ |θ|<δ We recall that M is weak-type (1, 1), that is, it satisfies the following estimate: there exists a constant C > 0 such that for every λ > 0,

it C {t : Mf(e ) > λ} ≤ kfk 1 . λ L (T) The next result shows the reason for our interest in the Hardy–Littlewood maximal function. 1 Proposition 1.17. Let g ∈ L (∂D), α > 1 a fixed parameter. Then there exists Cα > 0 such that for every eit ∈ ∂D it sup |P(g)(z)| ≤ CαMg(e ) . it z∈Γα(e ) Proof. The proof is divided in a few steps. (Step 1.) Assume first that z = reiη varies in a fixed compact subset of D, namely assume that 1 − r ≥ cα, with cα as in Remark 1.15 (ii). Notice that for all θ 1 − r2 1 − r2 2 P (eiθ) = ≤ ≤ . (1.18) r |1 − reiθ|2 (1 − r)2 1 − r SPACES OF HOLOMORPHIC FUNCTIONS 11

Then, for z = reiη as above,

Z 2π iη 1 i(η−θ) iθ |P(g)(z)| = |(g ∗ Pr)(e )| ≤ |g(e )|Pr(e ) dθ 2π 0 2 1 Z 2π ≤ |g(ei(η−θ))| dθ cα 2π 0 2 1 Z 2π = |g(ei(t−θ))| dθ cα 2π 0 it ≤ CαMg(e ) , where we have used the translation invariance of the integral on T.

iη it (Step 2.) Now we assume that z = re ∈ Γα(e ) with 1 − r < cα. By Remark 1.15 (ii) we know that |η − t| ≤ 3α(1 − r) . We break the integral on T as follows:

Z 2π iη 1 i(η−θ) iθ |P(g)(z)| = |(g ∗ Pr)(e )| ≤ |g(e )|Pr(e ) dθ 2π 0 N+1 Z X i(η−θ) iθ ≤ |g(e )|Pr(e ) dθ , j=0 Ej where  ◦ E0 = θ ∈ T : |θ| ≤ 3α(1 − r) ;  j−1 j ◦ for j = 1,...,N, Ej = θ ∈ T : 2 3α(1 − r) ≤ |θ| < 2 3α(1 − r) and N is chosen so that ; 2N−13α(1 − r) ≤ π/4 < 2N 3α(1 − r);  ◦ EN+1 = θ ∈ T : π/4 < |θ| ≤ π .

Notice that, for θ ∈ Ej (which implies in particular that |θ| ≤ π/4), arguing as in Remark 1.15 (ii) we have that |1 − reiθ| ≥ 2jα(1 − r) .

Therefore, for θ ∈ Ej we have that

1 − r2 1 − r2 P (eiθ) = ≤ r |1 − reiθ|2 22jα2(1 − r)2 1 ≤ 2 . 22jα2(1 − r) √ iθ Finally, for θ ∈ EN+1, |1 − re | ≥ 1/ 2 so that

iθ 2 Pr(e ) ≤ 2(1 − r ) ≤ 2 .

(Step 3.) Having set up the above bounds, we finally come to the estimate in the statement. 12 M. M. PELOSO

Using the inequalities in Step 2 we have that Z Z i(η−θ) iθ 2 i(η−θ) |g(e )|Pr(e ) dθ ≤ |g(e )| dθ E0 1 − r |θ|≤3α(1−r) Z 1 i(t−θ) ≤ Cα |g(e )| dθ 6α(1 − r) |θ|≤3α(1−r) it ≤ CαMg(e ) , using the change of variables θ 7→ θ − t + η. Next, using again the change of variables θ 7→ θ − t + η, Z Z i(η−θ) iθ 1 i(η−θ) |g(e )|Pr(e ) dθ ≤ 2 |g(e )| dθ EN+1 2π |θ|≤π 1 Z = 2 |g(ei(t−θ))| dθ 2π |θ|≤π ≤ cMg(eit) .

Finally, we estimate the central part of the integral:

N N X Z X 1 Z |g(ei(η−θ))|P (eiθ) dθ ≤ 2 |g(ei(η−θ))| dθ r 22jα2(1 − r) j=1 Ej j=1 Ej N Z X 1 i(η−θ) ≤ 2 2j 2 |g(e )| dθ 2 α (1 − r) j j=1 |θ|≤2 3α(1−r) N Z X 1 i(t−θ) ≤ 2 2j 2 |g(e )| dθ 2 α (1 − r) j+1 j=1 |θ|≤2 3α(1−r) N Z X −j 1 i(t−θ) ≤ Cα 2 j+1 |g(e )| dθ 2 3α(1 − r) j+1 j=1 |θ|≤2 3α(1−r) N X −j it ≤ Cα 2 Mg(e ) j=1 it ≤ CαMg(e ) .

This concludes the proof. 2 We are now ready to prove Thm. 1.16.

Proof of Thm. 1.16. Since H∞ ⊂ Hp for all p < ∞, it suffices to deal with the case p < ∞. Let f˜ ∈ Lp(∂D) be as in Prop. 1.9 (so that P(f˜) = f) and for a given γ > 0 let g ∈ C(∂D) be ˜ such that kf−gkLp ≤ γ; notice that here we need p < ∞. By the solution of the Dirichlet problem it it it we know that P(g)(z) → g(e ) as D 3 z → e (even without the restriction z ∈ Γα(e )). SPACES OF HOLOMORPHIC FUNCTIONS 13

We use Prop. 1.17 and argue as in the proof of the Lebesgue differentiation theorem,

eit ∈ ∂D : lim sup |f(z) − f˜(eit)| ≥ ε it it Γα(e )3z→e

≤ eit ∈ ∂D : lim sup |f(z) − P(g)(z)| ≥ ε/3 it it Γα(e )3z→e

+ eit ∈ ∂D : lim sup |P(g)(z) − g(eit)| ≥ ε/3 + eit ∈ ∂D : |g(eit) − f˜(eit)| ≥ ε/3 it it Γα(e )3z→e Z 3|g(eit) − f˜(eit)|p  it ˜ it ≤ e ∈ ∂D : CαM(f − g)(e )| ≥ ε/3 + dt {eit∈∂D:|g(eit)−f˜(eit)|≥ε/3} ε 0 00 C C p C p ≤ kg − f˜k 1 + kg − f˜k ≤ γ ε L εp Lp εp ≤ Cε , if we choose γ sufficiently small. This shows that

eit ∈ ∂D : lim sup |f(z) − f˜(eit)| ≥ ε ≤ Cε , it it Γα(e )3z→e which implies that lim f(z) = f˜(eit) , it it Γα(e )3z→e a.e., and the statement follows. 2 1.5. The zero sets of functions in Hp. In this section we describe the zero sets of functions in the Hardy spaces, for the full range 0 < p ≤ ∞. Surprisingly, it turns out that these are the same for all values of p. The description of such sets will allow us to prove a factorization theorem for Hp-functions that will have important consequences. We begin by recalling a result from complex analysis. Proposition 1.18. (Jensen’s formula) Let r > 0 and let f be holomorphic in a ngbh of D(0, r). Let ζ1, . . . , ζN be the zeros of f in D(0, r), counting multiplicity. Assume that f does not vanish at the origin and on the circle of radius r. Then N Y r 1 Z 2π log |f(0)| + log = log |f(reiθ)| dθ . |ζ | 2π j=1 j 0 For a proof we refer the reader to any classical text in complex analysis, as well as for the theory of infinite products. A consequence of this result is the following corollary.

Corollary 1.19. Let f ∈ H(D), f(0) 6= 0 and {ζ1, ζ2,... } its zeros counting multiplicity. Then, +∞ Z 2π Y 1 1 iθ log |f(0)| + log ≤ sup log+ |f(re )| dθ , |ζ | 0 0, log+ t = max(log t, 0). 14 M. M. PELOSO

Proof. Clearly, for every 0 < r < 1, N Y r 1 Z 2π log |f(0)| + log ≤ log |f(reiθ)| dθ |ζ | 2π + j=1 j 0 Z 2π 1 iθ ≤ sup log+ |f(re )| dθ . 0

p 2 Corollary 1.20. Let f ∈ H , 0 < p ≤ ∞, f 6≡ 0, and let {ζ1, ζ2,... } its zeros counting multiplicity. Then +∞ X (1 − |ζj|) < ∞ . (1.19) j=1 Proof. If f vanishes of order k at the origin, we consider f(z)/zk, which is still in Hp and it is zero-free at the origin. Notice that iθ iθ p 0 ≤ log+ |f(re )| ≤ |f(re )| . Then we applying Cor. 1.19 to obtain that +∞ Y 1 0 < log < ∞ , |ζ | j=1 j Q+∞ 1 Q+∞ which implies that converges; hence |ζj| converges. But this is equivalent to j=1 |ζj | j=1 P+∞ j=1(1 − |ζj|) < ∞. We recall that for ζ 2∈ D the function z − ζ ϕζ (z) = 1 − ζz is a biholomorphic mapping of D onto itself. In this setting ϕζ is also called a Blaschke factor.

Proposition 1.21. Let {ζ1, ζ2,... } be points in D satisfying condition (1.19). Then the infinite product +∞ Y ζj − ϕ |ζ | ζj j=1 j converges uniformly on compact subsets of D. Proof. It suffices to show that, for r < 1 be fixed and |z| ≤ r, the series +∞ X ζj 1 + ϕ (z) < ∞ |ζ | ζj j=1 j converges uniformly. A simple calculation now shows that

ζ |ζj|(1 − |ζj|) + ζ z(1 − |ζj|) 1 + j ϕ (z) = j ζj |ζj| |ζj||1 − ζjz| 1 + r ≤ (1 − |ζ |) . 1 − r j SPACES OF HOLOMORPHIC FUNCTIONS 15

By the Weierstrass M-test the convergence is uniform in |z| ≤ r, and the conclusion follows.

p 2 Definition 1.22. Let 0 < p ≤ ∞, f ∈ H and let {ζ1, ζ2,... } its zeros different from the origin, counting multiplicity. We set +∞ Y ζj B(z) = zk − ϕ (z) , |ζ | ζj j=1 j where k ≥ 0 is the order of zero of f at the origin. Then, B is a well-defined holomorphic function on D, and there exists a holomorphic, zero-free function F on D such that f/B = F . The function B is called a Blaschke product and f = FB is called the canonical factorization of f ∈ Hp. Proposition 1.23. Let 0 < p ≤ ∞, f ∈ Hp and let f = FB be its canonical factorization. p Then F ∈ H and kF kHp = kfkHp . ζ ζ Proof. Notice that B = zk Q+∞ − j ϕ and every factor satisfies j ϕ ≤ 1, so that |B| ≤ 1. j=1 |ζj | ζj |ζj | ζj Therefore, |f(z)| ≤ |F (z)| for all z ∈ D, so that kfkHp ≤ kF kHp . QN ζj We set BN = − ϕ and FN = f/BN . We recall that ϕ is defined and holomorphic j=1 |ζj | ζj ζj iη in the disc D(0, 1/|ζj|) which is strictly larger than the unit disc D and that |ϕζj (e )| = 1 for − all η. Therefore, BN converge uniformly on D to a function B˜N as r → 1 , with |B˜N | = 1 on ∂D. Hence, p p kFN kHp = lim Mp(FN , r) r→1− Z 2π 1 iθ p = lim |(f/BN )(re )| dθ r→1− 2π 0 Z 2π 1 1 iθ p ≤ lim iθ |(f)(re )| dθ r→1− infθ |BN (re )| 2π 0 p = kfkHp . Moreover, p p Mp(F, r) = lim Mp(FN , r) N→+∞ p p ≤ lim kFN k p = kfk p . N→+∞ H H

Therefore, kF kHp ≤ kfkHp , and the equality follows. 2 P+∞ Corollary 1.24. Let {ζ1, ζ2,... } be points in D satisfying the condition j=1(1 − |ζj|) < ∞ ζ and let B = zk Q+∞ − j ϕ be the corresponding Blaschke product. Then B ∈ H∞ and |B˜| = 1 j=1 |ζj | ζj a.e. on ∂D. Proof. In the proof of Prop. 1.23 we have already noticed that B ∈ H∞ and that its boundary value function B˜ is such that |B˜| ≤ 1. Since B ∈ H∞, its canonical factorization is B = 1 · B. By Prop. 1.23 again it follows that Z 2π 2 2 1 ˜ iθ 2 1 = k1kH2 = kBkH2 = |B(e )| dθ . 2π 0 16 M. M. PELOSO

Together with the estimate |B˜| ≤ 1, it implies that |B˜| = 1 a.e. on ∂D. We conclude this part with the extension of Thm. 1.16 to the range 0 <2 p ≤ 1. Theorem 1.25. (Fatou) Let f ∈ Hp, 0 < p ≤ 1. Then lim f(z) =: f˜(eit) , it it Γα(e )3z→e

it p where Γα(e ) is as in Def. 1.16, exists a.e. The function f˜ ∈ L (∂D) and ˜ kfkLp = kfkHp . Proof. Notice that, for this range of p’s, the existence of the boundary values f˜ of f is part of the statement, as well as the equality of the norms of f and f˜. By Prop. 1.23, f admits the canonical factorization f = FB, and the function F ∈ Hp and is zero-free. Therefore, F p/2 is well defined, holomorphic, and in H2. By Thm. 1.16, F p/2 admits boundary values Fgp/2 a.e. Therefore, also F admits non-tangential boundary values a.e. By Cor. 1.24 B admits boundary values B˜ and |B˜| = 1 a.e. Hence f admits non-tangential boundary values f˜ = Fgp/22/pB˜. Finally, 1 Z 2π ˜ p p/22/p ˜ iθ p kfkLp = Fg B(e )| dθ 2π 0 Z 2π 1 p/2 iθ 2 p/2 2 = Fg (e ) dθ = kF kH2 2π 0 1 Z 2π = sup |F (reiθ)|p dθ 0 1. In the section we study the boundary behaviour of functions in Hp, and we will be able to remove the restriction p > 1. In this analysis, we will consider the action of the Cauchy kernel, that is more closely related to Hp-functions, since, in particular, it produces holomorphic functions, unlike the Poisson integral. In the previous section we learned that for 0 < p ≤ ∞, if f ∈ Hp, then f converges pointwise non-tangentially a.e. to a boundary function that, with a momentary abuse of notation, we p denote by f˜. We also know that fr → f˜ in L (∂D) when 1 < p < ∞. We now show that this holds true also in the case 0 < p ≤ 1. Proposition 1.26. Let 0 < p ≤ 1, f ∈ Hp, and f˜ be its boundary value function, as in Thm. p − 1.25. Then fr → f˜ in L as r → 1 . SPACES OF HOLOMORPHIC FUNCTIONS 17

Proof. We first prove the case p = 1. Let f ∈ H1 and let f = BF its canonical factorization. Set g = F 1/2 , h = BF 1/2 . (1.20)

2 2 − Then both g and h are in H and gr → g˜, hr → h˜ in L (∂D) as r → 1 . Then, ˜ ˜ ˜ kfr − fkL1 = kgrhr − g˜hkL1 ≤ k(gr − g˜)hrkL1 + kg˜(hr − h)kL1 ˜ ≤ kgr − g˜kL2 khrkL2 + kg˜kL2 khr − hkL2 → 0 as r → 1−. This proves the case p = 1. When 0 < p < 1 we proceed similarly, but we have to use an iteration process. Let f ∈ Hp 2p 2p and define g and h as before. Then, g, h ∈ H . If the results holds true in H , that is, gr → g˜, 2p hr → h˜ in L (∂D) then the previous arguments applies (with a constant appearing in the p triangular inequality) and therefore the results holds true in H . Thus, after k ≥ log2 1/p steps we can prove the results for Hp.

iθ2 Recall that we denote by Cr(e ) the Cauchy kernel. We will denote by C(g) the Cauchy integral of a given function g on ∂D. Recall that, for g ∈ C(∂D), C(g) is a holomorphic in D and arguing as in (1.3) (using the uniform convergence of the power series expansion of Cr) we have that +∞ iη iη X n inη C(g)(re ) = (Cr ∗ g)(e ) = gˆ(n)r e . n=0 Hence, we see that if g ∈ C(∂D) andg ˆ(n) = 0 when n < 0, then C(g) = P(g). More generally we have Lemma 1.27. Let g ∈ Lp(∂D), 1 ≤ p ≤ ∞. Then P(g) = C(g) if and only if gˆ(n) for all n < 0. Proof. It suffices to assume that g ∈ L1(∂D). We use the uniform convergence of the power series expansion of Pr and Cr. We have +∞ 1 Z 2π X P(g)(reiη) = (P ∗ g)(eiη) = g(eiθ) rnei(η−θ) dθ r 2π 0 n=−∞ +∞ X = gˆ(n)rneinη . n=−∞ Analogously,

+∞ 1 Z 2π X C(g)(reiη) = (C ∗ g)(eiη) = g(eiθ) rnei(η−θ) dθ r 2π 0 n=0 +∞ X = gˆ(n)rneinη . n=0 It follows that P(g) = C(g) if and only ifg ˆ(n) for all n < 0. 2 18 M. M. PELOSO

Theorem 1.28. Let f ∈ Hp, 1 ≤ p ≤ ∞ and let f˜ be its boundary value function. Then ˆ f˜(n) = 0 for n < 0 and P(f˜) = C(f˜) = f . On the other hand, if 1 ≤ p ≤ ∞ and g ∈ Lp(∂D) is such that gˆ(n) = 0, then f := P(g) = C(g) is in Hp and f˜ = g. P+∞ n p Proof. Let f(z) = n=0 anz be the power series expansion of f in D, f ∈ H , 1 ≤ p ≤ ∞. Notice that it suffices to consider the case p = 1. Then +∞ iη X n inη fr(e ) = anr e n=0 n ˜ and for all 0 < r < 1, fbr(n) = anr , by the uniform convergence. Since by Prop. 1.26 fr → f 1 ˜ − in L (∂D), fbr(n) → fb(n) as r → 1 , so that ( a for n ≥ 0 fb˜(n) = n 0 for n < 0 . This clearly implies that +∞ X  1 Z 2π  f(reiη) = f˜(eiθ)e−inθ dθ rneinη 2π n=0 0 iη = (Pr ∗ f˜)(re ) , that is, f = P(f˜). The fact that C(f˜) = P(f˜) follows from the previous Lemma 1.27. When 1 ≤ p < ∞, he second part of the statement follows immediately from Lemma 1.27, Prop. 1.11 and the holomorphicity of the Cauchy integrals. If p = ∞, we obtain C(g) = P(g) as before, and clearly f = P(g) = C(g) ∈ H∞ and the equality f˜ = g follows from the case p < ∞. The next result is now obvious, and we state2 it for sake of completeness. Corollary 1.29. For 1 ≤ p ≤ ∞ we denote by Hp(∂D) the subspace of Lp(∂D) of the functions that are boundary values of functions in Hp. Then, Hp(∂D) = g ∈ Lp(∂D) :g ˆ(n) = 0 for n < 0 .

We denote by P the set of the (complex) polynomials in z, and by A(D) the space of holomorphic functions on D that are continuous up to the boundary, that is, A(D) = H(D) ∩ C(D) , endowed with the sup-norm.

Corollary 1.30. For 1 ≤ p < ∞ the polynomials are dense in Hp, while the closure of P in ∞ ˜ the sup-norm is A(D) and A(D) ( H . Furthermore, if f ∈ A(D), then fr → f uniformly on ∂D. SPACES OF HOLOMORPHIC FUNCTIONS 19

Proof. We denote by σN (g) the Fej´ertrigonometric polynomial of an integrable function g on T = ∂D, that is, N X  |k|  σ (g)(eiη) = 1 − gˆ(k)eikη . N N + 1 k=−N p It is well known (see [Ka] or your lecture notes) that σN (g) → g in L (∂D), as N → +∞, when 1 ≤ p < ∞. Now, let 1 ≤ p < ∞, f ∈ Hp, f˜ its boundary function. Then, N X  |k|  ˆ σ (f˜)(eiη) = 1 − f˜(k)eikη , N N + 1 k=0 ˜ by Prop. 1.28. Clearly σN (f) is the restriction to the boundary of a polynomial pN and it holds ˜ that pN = C(σN (f)). Then, we obtain a sequence of polynomials in D such that ˜ ˜ ˜ kf − pN kHp = kf − pfN kLp(∂D) = kf − σN (f)kLp(∂D) → 0 as N → +∞. Thus, P is dense in Hp, 1 ≤ p < ∞. Next, it is clear that A(D) ⊂ H∞ and that the closure of P in the sup-norm is contained in A(D) = H(D) ∩ C(D). The same proof as before shows that P is dense in A(D). For, iη iη sup |f(z) − pN (z)| = sup |f(e ) − pN (e )| z∈D eiη∈∂D iη iη = sup |f(e ) − σN (f)(e )| eiη∈∂D < ε choosing N large enough. Moreover, if f ∈ A(D), then f˜ ∈ C(∂D), f = P(f˜) and therefore, fr = Pr ∗ f˜ → f˜ uniformly on ∂D. It only remains to show that A(D) is strictly contained in H∞. It suffices to consider an infinite Blaschke product B, as defined in Def. 1.22. By Cor. 1.24 we know that |B˜| = 1 a.e. on ∂D. Let {ζj} be the (infinite) sequence of the zeros of B in D, then they must accumulate at boundary points. Clearly B cannot be continuous at those points. 2 1.7. The Cauchy–Szeg¨oprojection. Definition 1.31. We define the Cauchy–Szeg¨oprojection on L2(∂D) as the Hilbert space orthogonal projection as +∞ +∞  X  X S(g)(eiη) = S gˆ(n)ein(·) (eiη) = gˆ(n)einη . n=−∞ n=0 It is clear that S is a projection (that is, S2 = S) and that it is self-adjoint (that is, S∗ = S); hence an orthogonal projection. The next result follows from the previous lemma. Proposition 1.32. For g ∈ L2(∂D) we have that S(g) = Cg(g) . 20 M. M. PELOSO

In particular we have obtained a characterization of S as (boundary limit of an) integral operator. Namely, Z 2π iη 1 i(η−θ) 1 S(g)(e ) = lim g(e ) iθ dθ . (1.21) r→1− 2π 0 1 − re With an abuse of notation, we will simply write Z 2π iη 1 i(η−θ) 1 S(g)(e ) = g(e ) iθ dθ . (1.22) 2π 0 1 − e Notice however that the integral is not absolutely convergent since g ∈ L2 and the function 1/(1 − eiθ) has a non-integrable singularity in θ = 0. The integral in (1.22) has to be interpreted as in (1.21). We wish to extend the action of S to all the spaces Lp(∂D) and prove its Lp-boundedness, for the appropriate range of p’s. The following result, of which we will not provide a complete proof, answers the question. Theorem 1.33. The operator S, initially defined on the dense subset L2 ∩ Lp, extends to an operator S : Lp(∂D) → Lp(∂D) that is of weak-type (1, 1) and is bounded for 1 < p < ∞.

Sketch of the proof. Recall the definition S(g) = limr→1− Cr ∗ g and the relation between the Cauchy and Poisson kernels (1.13). From the properties of the Poisson kernel we see that

S1(g) = lim Re Cr ∗ g r→1− extends to a bounded linear operator for 1 ≤ p < ∞. Thus, in order to prove the theorem, it suffices to prove the statement for (twice) the imaginary part of Cr, where 1h 1 1 i 2Im C (eiη) = − r i 1 − reiη 1 − re−iη 2r sin η 2r sin η = = (1.23) |1 − reiη|2 1 − 2r cos η + r2 +∞ X |n| inη iη = −i sgn(n)r e := Qr(e ) , (1.24) n=−∞ (where the last equality follows from the power series expansions of the right hand side of the first line in display).

The kernel Qr is called the conjugate Poisson kernel, the reason being that for every 0 < r < 1, Qr is the harmonic function conjugate to the real harmonic function Pr on the unit disc.

The family of functions on T {Qr} does not form a summability kernel, and the properties of the mapping g 7→ limr→1− Qr ∗ g are substantially different from the ones we obtain when Qr is replaced by the Poisson kernel. We give a name to the operator g 7→ limr→1− Qr ∗ g, we write

Q(g) = lim Qr ∗ g , (1.25) r→1− SPACES OF HOLOMORPHIC FUNCTIONS 21 and call Q the on T. (In fact, this operator is the closely related to the classical Hilbert transform on the real line. We will return to the latter operator in Section 4.3). The operator Q belongs to a fundamental class of operators, called singular integrals. Us- ing the fact that the (convolution) kernels Qr are odd and therefore they have built in some cancellation, it is indeed possible to show that Q is Lp-bounded for 1 < p < ∞. Notice that the pointwise limit of the kernels

iη 1 sin η iη lim Qr(e ) = = cot(η/2) := Q(e ) r→1− 1 − cos η and that Q 6∈ L1(∂D). It turns out (and it is not hard to see) that, Z Q(g)(eiη) = p.v.(Q ∗ g)(eiη) =: lim g(ei(η−θ))Q(θ) dθ . ε→0+ ε<|θ|≤π We will not prove that Q is weak-type (1, 1) and bounded on Lp, 1 < p < ∞, but refer to [Ka], Ch. 3, e.g., or also [Gr]. 2 It can be shown that the Cauchy–Szeg¨oprojection is not bounded from L1(∂D) to itself.4 Thm. 1.33 also gives an answer to the problem of the harmonic conjugate. Suppose u is a real harmonic function on the unit disc. Then we know that u admits a harmonic conjugate v, that is, a real harmonic function such that u + iv is holomorphic in D. The function v is uniquely determined modulo constants, so we require v to vanish at the origin z = 0, i.e. we set v(0) = 0. In fact, it is not difficult to see that the boundedness of the Cauchy– Szeg¨oprojection on Lp(∂D) is equivalent to the boundedness of the harmonic conjugate operator T . Corollary 1.34. Let u be a real harmonic function satisfying the Hp-growth condition (1.14), 1 < p < ∞, and let v be its harmonic conjugate. Then, also v satisfies the Hp-growth condition (1.14) and the mapping T :u ˜ 7→ v˜ is bounded on Lp(∂D), 1 < p < ∞. p Proof. By Prop. 1.9 u = Pr ∗ u˜, withu ˜ ∈ L (∂D). Now, vr = Qr ∗ u˜ andv ˜ = Q(˜u). By the previous theorem kv˜kLp = kQ(˜u)kLp ≤ Cku˜kLp , and we are done. 2 There exists an anologous theory for the Hardy on the upper half-plane. The two theory have many common aspects, but they also present significant differencies, since the upper half-plane is unbounded, while D is not. Moreover, the symmetry properties of the two domains are different (although analogous) and certain formulas appear simpler in one setting or another. We will present some further aspects of this theory in the setting of the upper half-plane in Sections 4.1 - 4.4.

4Exercise. 22 M. M. PELOSO

2. Bergman spaces on the unit disc

2.1. Function spaces with reproducing kernel. In this section we briefly describe the theory of reproducing kernels. Let H be a Hilbert space of functions defined on a set Ω and suppose that the point evaluations are bounded (linear) functionals on H, that is, for each z ∈ Ω, there exists a constant Cz > 0 such that for all f ∈ H we have |f(z)| ≤ CzkfkH . (2.1) This implies that the linear functional

Lz(f) = f(z) , f ∈ H , is bounded. By the Riesz–Fisher theorem, there exists kz ∈ H such that, for all f ∈ H we have

hf, kziH = f(z) . We define a kernel function K :Ω × Ω → C by setting

K(w, z) = kz(w) . (2.2) Such kernel K is called the reproducing kernel for H. Proposition 2.1. The kernel K satisfies the following properties:

(i) for all f ∈ H and z ∈ Ω we have f(z) = hf, K(·, z)iH; (ii) for all z, w ∈ Ω, K(w, z) = K(z, w). Proof. Property (i) is obvious by construction. For (ii), using the reproducing property of K we have

K(w, z) = kz(w) = hkz,K(·, w)iH = hK(·, w), kziH = K(z, w) and we are done. The following result2 establishes conditions that characterize the reproducing kernel of H. Lemma 2.2. Let H(z, w) be a function on Ω × Ω such that (i) H(·, w) ∈ H for all w ∈ Ω fixed; (ii) hf, H(·, z)iH = f(z) for all f ∈ H and z ∈ Ω. Then, H(z, w) coincides with the reproducing kernel K(z, w) of H. Proof. This is simple: using the reproducing property of H(·, z) and K(·, z), for every w ∈ Ω fixed we have

H(z, w) = hH(·, w),K(·, z)iH = hK(·, z),H(·, w)iH = K(w, z) = K(z, w) , as we wished to show. Notice that (i) and (ii)2 together imply the hermitian-symmetric property H(w, z) = H(z, w) for all z, w ∈ Ω, as in the proof of (ii) in Prop. 2.1.

In this notes we will be concerned only with spaces H for which (2.1) holds and that consist of holomorphic functions. SPACES OF HOLOMORPHIC FUNCTIONS 23

It will also always be the case that, for each compact subset E of Ω there exists C = CE > 0 such that for all f ∈ H we have sup |f(z)| ≤ CkfkH . (2.3) z∈E If (2.3) holds, then clearly the convergence in H implies the uniform convergence on compact subsets of Ω. Proposition 2.3. Let H be a Hilbert space of holomorphic functions on Ω for which condition (2.3) holds. Let {ϕj} be an orthonormal basis for H. Then the series +∞ X ϕj(z)ϕj(w) j=1 converges uniformly on compact subsets of Ω × Ω to the reproducing kernel K(z, w) of H. Proof. It suffices to consider the case of subsets of the form E × E, where E is a compact subset of Ω. By the Riesz-Fisher theorem for `2, for z ∈ E we have +∞ +∞ 1/2  X 2 X |ϕj(z)| = {ϕj(z)} = sup ajϕj(z) `2 j=1 k{aj }k`2 =1 j=1 = sup |f(z)| kfkH=1 ≤ sup sup |f(z)| kfkH=1 z∈E ≤ CE . Therefore, +∞ 1/2  X 2 sup |ϕj(z)| ≤ CE , z∈E j=1 so that the Cauchy–Schwarz inequality gives that +∞ X 2 ϕj(z)ϕj(w) ≤ CE , j=1 and the convergence is uniform on E × E. 2 P+∞ Next, for z ∈ Ω fixed, the previous argument also shows that {ϕj(z)} ∈ ` , so that j=1 ϕj(z)ϕj converges in H to a function hz. By the theory of Fourier series in H we have that, for all f ∈ H +∞ X hf, hziH = hf, ϕjiHϕj(z) = f(z) . j=1 P+∞ Therefore, the function H(w, z) = hz(w) = j=1 ϕj(z)ϕj(w) satisfies the conditions of Lemma 2.2 and hence coincides with the reproducing kernel K(w, z). P+∞ Notice that the argument above also shows that the series j=1 ϕj(z)ϕj converges in the H-norm, for all z ∈ Ω fixed. 2 An interesting consequence of this proposition is the following result that shows that the reproducing kernel K satisfies an extremal property. 24 M. M. PELOSO

Corollary 2.4. Let H be a space of holomorphic functions on Ω for which condition (2.3) holds and let K(z, w) be its reproducing kernel. Then K(z, z) = sup |f(z)|2 . f∈H,kfkH=1

Proof. Let {ϕk} be an orthonormal basis for H. We have

X 2 K(z, z) = |ϕk(z)| k  X 2 = sup akϕk(z) 2 {ak}∈` , k{ak}k`2 =1 k = sup |f(z)|2 , f∈H,kfkH=1 as claimed. Notice that2 the reproducing kernel K satisfies the condition K(z, z) ≥ 0 and in fact K(z, z) > 0 unless f(z) = 0 for all f ∈ H. We conclude this introduction on Hilbert spaces with reproducing kernels with a concrete example, that we have in fact already encountered. Proposition 2.5. The Hardy space H2(D) is a Hilbert space with reproducing kernel, with inner ˜ product hf, giH2 = hf, g˜iL2(∂D). Its reproducing kernel is the Cauchy–Szeg¨okernel 1 K(z, w) = . 1 − wz ˜ 2 Proof. We showed that kfkH2 = kfkL2(∂D). Therefore, H is a Hilbert space with inner product ˜ 2 hf, giH2 = hf, g˜iL2(∂D), for f, g ∈ H . (We also showed that we can describe the inner product in H2 by setting

hf, giH2 = h{an}, {bn}i`2 , P+∞ n P+∞ n where f(z) = n=0 anz and g = n=0 bnz . However, we will not use this characterization at this stage.) We now show that the point evalutions are bounded functionals on H2. Let z ∈ D and let |z| < r < 1. By the Cauchy formula Z Z 2π iθ 1 f(ζ) 1 f(re ) iθ f(z) = dζ = iθ re dθ , (2.4) 2πi |ζ|=r ζ − z 2π 0 re − z and by the Cauchy–Schwartz inequality we have

Z 2π 1/2 Z 2π 2 1/2  1 iθ 2   1 r  |f(z)| ≤ |f(re )| dθ iθ 2 dθ 2π 0 2π 0 |re − z| r ≤ kfk 2 . r − |z| H This shows that H2 is a Hilbert space with reproducing kernel. SPACES OF HOLOMORPHIC FUNCTIONS 25

1 2 Consider now the kernel K(z, w) = 1−wz . It is immediate to see that K(·, w) ∈ H for every w ∈ D fixed. Moreover, Z 2π ˜ iθ ˜ ˜ 1 f(e ) hf, K(·, z)iH2 = hf, K(·, z)iL2(∂D) = −iθ dθ 2π 0 1 − e z Z 2π iθ 1 f(re ) iθ = lim iθ re dθ r→r1− 2π 0 re − z = f(z) , by (2.4). The conclusion now follows from Lemma 2.2. 2 2.2. The Bergman spaces. Let Ω be a domain in C. We denote by dA the Lebesgue on C, which of course is identified with real euclidean space R2, and we write |E| to denote the Lebesgue measure of a measurable set E. For 0 < p < ∞ we denote by Lp(Ω) the space of the Lebesgue measurable functions that are p-integrable with respect to dA, and by L∞(Ω) the space of measurable functions that are essentially bounded. Definition 2.6. For 0 < p ≤ ∞ we define the Bergman space Ap as Ap(Ω) = Lp(Ω) ∩ H(Ω). Proposition 2.7. Let Ω ⊂ C be a domain and let 0 < p ≤ ∞. Then the space Ap(Ω) is a closed subspace of Lp(Ω). When p = 2, A2 is a Hilbert space with reproducing kernel. Proof. We show that the functions in the Ap satisfy (2.3). Let E be a compact 1 subset of Ω and let δ be half the distance of E from ∂Ω, i.e. δ = 2 dist(E, ∂Ω). For every z ∈ E, D(z, δ) ⊂ Ω. From the mean value property 1 Z 2π f(z) = f(z + reiθ) dθ , 2π 0 (valid for every r > 0 such that D(z, r) ⊂ Ω) multiplying by r both sides and integrating in r ∈ [0, δ] we obtain the identity5 1 ZZ f(z) = f(w) dA(w) . |D(z, δ)| D(z,δ) Therefore, for every z ∈ E, using H¨older’s inequality, for 1 ≤ p ≤ ∞, 1 ZZ |f(z)|p ≤ |f(w)|p dA(w) (2.5) |D(z, δ)| D(z,δ) 1 1 ≤ kfkp = kfkp . |D(z, δ)| Lp πδ2 Lp We remark that the above inequality (2.5) holds also for 0 < p < 1, but this is a consequence of the fact that |f|p is subharmonic when f is holomorphic, and to avoid introducing another notion, we will not prove this fact here; see [Kr], e.g. Since E can be covered by a finite number of such discs

sup |f(z)| ≤ CEkfkLp . (2.6) z∈E

5 This identity is called the mean value property and it is valid for all holomorphic functions f and discs D(z, δ) contained in its region of holomorphicity. 26 M. M. PELOSO

This inequality show that if {fn} is a sequence of holomorphic functions converging in the p L -norm to a function f, then {fn} converges to f also uniformly on compact subsets. This proves that Ap is closed 1 < p < ∞. The case p = ∞ is obvious. When p = 2 (2.6) is exactly (2.3) and this concludes the proof. Notice that the constant CE in inequality (2.6), depends on the distance of the set E from −2/p 1 the boundary of Ω, and, more precisely CE ∼ δ , as δ = 2 dist(E, ∂Ω) → 0. 2 Definition 2.8. Given a domain Ω ⊂ C, let K = KΩ be its Bergman kernel, and for z ∈ Ω 2 write kz = K(·, z). We define the Bergman projection on Ω the mapping B defined on L (Ω) as Z (Bf)(z) = hf, kzi = f(w)kz(w) dA(w) Ω Z = f(w)K(z, w) dA(w) . Ω Proposition 2.9. The Bergman projection is the Hilbert space orthogonal projection of L2(Ω) onto its closed subspace A2(Ω).

2 2 2 Proof. First of all, notice that Bf is well defined for all f ∈ L , since for all z ∈ Ω, kz ∈ A ⊂ L , 2 and the L -pairing hf, kzi makes sense (and converges). Next, we use the characterization of the Bergman kernel given by Prop. 2.3, that is, kz = P 2 2 j ϕj(z)ϕj, where {ϕj} is an orthonormal basis for A , where the convergence is in the L -norm, for all z ∈ Ω fixed. Therefore, X (Bf)(z) = hf, kzi = f, ϕj(z)ϕj j X X = ϕj(z)hf, ϕji = ajϕj(z) , j j where aj = hf, ϕji is the j-th Fourier coefficient of f, w.r.t. the orthonormal system {ϕj}, 2 2 P 2 2 (which is not complete in L ). Clearly, {aj} ∈ ` and j |aj| ≤ kfkL2 , by Bessel’s inequality. P 2 Hence, the series jhf, ϕjiϕj converges in L and therefore

kBfkL2 ≤ kfkL2 . It is also clear that Bf ∈ A2 since A2 is closed, so that B : L2 → A2 and it is bounded. It remains to show that B is the orthogonal projection, that is, that B2 = B and B∗ = B. The former identity follows by the reproducing property of the Bergman kernel when acting on functions in A2, while the latter one follows from the hermitian symmetry property: for f, g ∈ L2 X X hBf, gi = hf, ϕjiϕj, g = hf, ϕjihϕj, gi j j X = f, hg, ϕjiϕj = hf, Bgi . j This shows that B∗ = B and we are done. 2 SPACES OF HOLOMORPHIC FUNCTIONS 27

Remark 2.10. We remark that, up to this point, we have not discussed whether, on a given domain Ω ⊂ C, the Bergman space A2(Ω) is trivial, i.e. reduces to {0}, is infinite dimensional, or even if it is not trivial, but finite dimensional. We mention that all these cases can occur. It is easy to see that A2(C) = {0}, and the finite dimensional examples are also fairly elementary; we refer the reader to [Wi]. However, it is important to point out that, an application of the Weierstrass approxima- tion theorem6 shows that, if Ω is a bounded domain, the space of holomorphic polynomials is (contained and) dense in A2(Ω). We conclude this remark by observing that, as a consequence of the extremal property in Cor. 2.4 we have that, it Ω ⊆ Ω0 are domains in C and K,K0 denote their Bergman kernels resp., then K0(z, z) ≤ K(z, z) for all z ∈ Ω. Before proceeding any further, we present an example of Bergman space and kernel.

Proposition 2.11. Let D be the unit disc. Then, the set of functions {ϕk}, where ϕk = p(k + 1)/πzk, k = 0, 1, 2,... , is a complete orthonormal system in A2(D) and the Bergman kernel has the expression 1 1 K(z, w) = . π (1 − zw)2

Proof. It is clear that {ϕk} is an orthonormal system: p Z 1 Z 2π (j + 1)(k + 1) i(j−k)θ j+k+1 hϕj, ϕki = e dθ r dr π 0 0 Z 1 2k+1 = δjk2(k + 1) r dr 0 = δjk , where δjk denotes the Kronecker’s delta. By the uniqueness of the Taylor expansion, it is also clear that the holomorphic polinomials (i.e. the polynomials in z) are dense in A2(D). For, let f ∈ A2(D) be orthogonal to all the zk’s P n and let f(z) = n anz be its power series expansion in D. By integrating over the closed disc |z| ≤ r for some fixed r < 1 and interchanging the summation and integration orders, we obtain that an = 0 for all n; hence f = 0. 2 Thus, {ϕk} is a complete orthonormal system for A (D) and, by Prop. 2.3 the Bergman P kernel is given by the sum k ϕk(z)ϕk(w), that we easily compute: X k + 1 1 X zkwk = (k + 1)(zw)k π π k k 1 1 = , π (1 − zw)2

6True? Recall? Reference? 28 M. M. PELOSO as we wished to show.7 Notice the Bergman kernel2 for the unit disc K(z, w) satisfies the condition K(z, w) 6= 0 and K(z, z) > 0 for all z, w ∈ D.

2.3. Biholomorphic invariance. One of the main features of the Bergman kernel is its invari- ance under conformal mappings. This invariance is the content of the next proposition.

Proposition 2.12. Let Ω1, Ω2 ⊂ C be domains, Φ:Ω1 → Ω2 be a biholomorphic mapping, and let KΩ1 , B1 and KΩ2 , B2 be the Bergman kernels and projections of Ω1 and Ω2, resp. Then we have 0  0 KΩ1 (z, w) = Φ (z)KΩ2 Φ(z), Φ(w) Φ (w) , (2.7) 2 and, for f ∈ A (Ω2) we  0  0  B1 Φ (f ◦ Φ) = Φ (B2f) ◦ Φ . (2.8)

Proof. We begin by observing that if Φ : Ω1 → Ω2 is a biholomorphic mapping, then the determinant jacobian of Φ thought as a mapping between domains in R2 equals |Φ0(z)|2. For, if we write Φ = u + iv, u, v real functions, then Φ is identified with the mapping Ω 3 z = x + iy ≡ (x, y) 7→ u(x, y), v(x, y) . Then ∂ u ∂ u Jac Φ = x y , ∂xv ∂yv so that, using the Cauchy-Riemann equations,

2 2 2 0 2 det(Jac Φ) = (∂xu) + (∂xv) = |∂xΦ| = |Φ | . 0 2 This identity implies that the mapping f 7→ Φ (f ◦ Φ) is an isometric of A (Ω2) 2 onto A (Ω1). Notice also the analogous statements hold true with Φ−1 in place of Φ and with the roles of Ω1 and Ω2 switched.

For z, w ∈ Ω1 we set 0  0 H(z, w) = Φ (z)KΩ2 Φ(z), Φ(w) Φ (w) , and easily see that H satisfies the conditions in Lemma 2.2, thus implying that H coincides with 8 the Bergman kernel on Ω1.

7Exercise. Using the residue theorem, show that for every f ∈ A2(D) and z ∈ D we have the identity 1 ZZ f(w) f(z) = 2 dA(w) . π D (1 − zw) Once you have completed your argument, explain where you use the assumption f ∈ A2(D). Can this assumption be relaxed? 8Exercise. SPACES OF HOLOMORPHIC FUNCTIONS 29

Finally, (2.8) follows easily from (2.7). Let z ∈ Ω1 Z  0  0 B1 Φ (f ◦ Φ) (z) = Φ (ζ)(f ◦ Φ)(ζ)KΩ1 (z, ζ) dA(ζ) Ω1 Z 0   0 2 = Φ (z) f Φ(ζ) KΩ2 Φ(z), Φ(ζ) |Φ (ζ)| dA(ζ) Ω1 Z 0  = Φ (z) f(ω)KΩ2 Φ(z), ω dA(ω) Ω2 0   = Φ (z) (B2f) ◦ Φ (z) , as we wanted to prove. 2

2.4. Lp-boundedness of a family of integral operators. In the next section we will prove the boundedness of the Bergman projection B when acting on Lp, 1 < p < ∞ as a consequence of a more general result concerning the boundedness of a family of operators on weighted Lp- spaces. This greater generality does not introduce extra arguments in the proof and at the same time provides a variety of possible applications. We begin by recalling a general result concerning the boundedness of integral operators with positive kernels.

Proposition 2.13. (Schur’s test) Let (X , dµX ), (Y, dµY ) be measure spaces. Let T be the integral operator given by Z T f(x) = K(x, y)f(y) dµY (y) , Y where K is a measurable positive kernel on X × Y. Let 1 < p, p0 < ∞ be conjugate exponents. Suppose there exist positive functions ψ : Y → (0, +∞), ϕ : X → (0, +∞), such that Z p0 p0 (i) K(x, y)ψ (y) dµY (y) ≤ Cϕ(x) ; ZY p p (ii) K(x, y)ϕ (x) dµX (x) ≤ Cψ(y) . X Then T : Lp(Y) → Lp(X ) is bounded. The proof is a fairly simple application of H¨older’sinequality. We leave the details as an exercise. For ν > −1 we consider the weights (1 − |w|2)ν and the kernels 1 1 K (z, w) = . ν π |1 − zw|2+ν We are interested in the boundedness of the integral operators T : Lp(1 − |w|2)νdA → Lp(1 − |w|2)νdA , where Z 2 ν T f(z) = f(w)Kν(z, w) (1 − |w| ) dA(w) . D 30 M. M. PELOSO

We begin with the following integral estimate. Given two positive quantities A and B depend- 1 ing on a variable x, we write A ≈ B as x → x0 if there exists c > 0 such that c ≤ A(x)/B(x) ≤ c as x → x0. Lemma 2.14. Let γ > −1, t ∈ R and for z ∈ D set Z (1 − |w|2)γ It(z) = 2+γ+t dA(w) . D |1 − zw| Then,  1 if t < 0    It(z) ≈ log 1/(1 − |z|) if t = 0 (1 − |z|2)−t if t > 0 , for |z| → 1−.

Notice that the function It is bounded on every compact subsets of D. Therefore, the be- haviour of It(z) is non-trivial only for z approaching the boundary. The proof of the lemma is based on another integral estimate. Lemma 2.15. Let τ ∈ R and for % ∈ (0, 1) set Z 2π 1 Jτ (%) = iθ 1+τ dθ . 0 |1 − %e | Then, as % → 1−,  1 if τ < 0    Jτ (%) ≈ log 1/(1 − %) if τ = 0 (1 − %)−τ if τ > 0 .

Proof. Notice that, for all θ ∈ [0, 2π), (1 − %) ≤ |1 − %eiθ| ≤ 1 + %, so that the region of integration [0, 2π) is contained in the union of the disjoint sets Ej, j = 0,...,N, where  j iθ j+1 Cj = θ : 2 (1 − %) ≤ |1 − %e | < 2 (1 − %)   9 and N = blog2 (1 + %)/(1 − %) c. We also observe that, if we set  iθ j Ej = θ : |1 − %e | < 2 (1 − %) , then Cj = Ej+1 \ Ej, j = 0,...,N. Moreover, j |Ej| ≤ c2 (1 − %) for some constant c > 0 (and independent of j and %).10

9We denote by btc the integral part of the real number t. 10For, |1 − %eiθ|2 < 22(j+1)(1 − %)2 if and only if 1 − A ≤ cos θ, where A = (22(j+1) − 1)(1 − %)2/2%. But this is equivalent to |θ| ≤ cos−1(1 − A) ≈ (1 − %), as % → 1−. SPACES OF HOLOMORPHIC FUNCTIONS 31

Then, if τ > 0 N X Z 1 J (%) ≤ dθ τ |1 − %eiθ|1+τ j=0 Cj N X 1 ≤ |Ej| j 1+τ j=0 2 (1 − %) −τ ≤ Cτ (1 − %) . When τ = 0 we need to refine the above estimate. We obtain N X 1 Jτ (%) ≤ |Cj| 2j(1 − %) j=0 N X 1 ≤ |Ej| ≤ cN 2j(1 − %) j=0 ≤ c log(1/(1 − %)) .

− iθ −1−τ When τ < 0 is clear that Jτ is bounded as % → 1 since |1 − e | is integrable on [0, 2π). Finally, the reverse inequalities are easy to prove and we leave the details to the reader.11 2 Proof of Lemma 2.14. Integrating in polar coordinates w = reiθ and using the invariance in the integral over the circle we see that Z 1 Z 2π 1 I (z) = dθ(1 − r)γr dr . t 2+γ+t 0 0 1 − |z|reiθ If t > 0, then τ := 1 + γ + t is also > 0 and Lemma 2.15 gives that Z 1 (1 − r)γ Z 1 (1 − r)γ I (z) ≈ r dr = r dr . t 1+γ+t 1+γ+t 0 1 − |z|r 0 (1 − r) + (1 − |z|)r

We split the latter integral into the regions G1 = {r : (1 − r) ≤ (1 − |z|)r} and G2 = {r : (1 − r) > (1 − |z|)r}, that is, 0, 1/(2 − |z|) and 1/(2 − |z|), 1. Then, Z 1/(2−|z|) 1 Z 1 (1 − r)γ I (z) ≤ r dr + r dr t 1+t 1+γ+t 0 (1 − r) 1/(2−|z|) (1 − |z|)r Z 1/(2−|z|) Z 1 1 C γ ≤ 1+t dr + 1+γ+t (1 − r) dr 0 (1 − r) (1 − |z|) 1/(2−|z|) C ≤ γ,t . (1 − |z|)t This proves the estimate from above in the case t > 0. It is not hard to see that the estimate from below follows from the same argument. The case t < 0 is easy, so is the case t = 0, whose details are left to the reader. 2 11Exercise. 32 M. M. PELOSO

We finally come to the main result of this section. For ν > −1 define the measure on the unit 2 ν disc dAν(w) = (1 − |w| ) dA(w).

Theorem 2.16. For α, ν > −1 consider the integral operator Tν defined as ZZ (1 − |w|2)ν Tνf(z) = f(w) 2+ν dA(w) . D |1 − wz| p p Then, if α + 1 < p(ν + 1), Tν : L (D, dAα) → L (D, dAα) is bounded. Proof. We are going to use Schur’s test Prop. 2.13 and the integral estimates in Lemma 2.14. Notice that in this case the pairs (Y, dµY ) and (X , dµX ) are both equal to (D, dAα) and that we write the operator Tν as ZZ (1 − |w|2)ν−α Tν(f)(z) = f(w) 2+ν dAα(w) , D |1 − wz| (1−|w|2)ν−α that is, as having integral kernel H(z, w) = |1−wz|2+ν w.r.t. the measure dAα. Let a > 0 to be chosen later and define ψ(w) = (1 − |w|2)−a. When ν − ap0 > −1, we may apply Lemma 2.14 with γ = ν − ap0 and t = ap0 and obtain ZZ p0 p0 Tν(ψ )(z) = ψ (w)H(z, w) dAα(w) D ZZ (1 − |w|2)ν−ap0 = 2+ν dA(w) D |1 − wz| 0 0 ≤ C(1 − |z|2)−ap = Cψp (z) . Notice that this argument is possible if the parameter a satisfies ν + 1 0 < a < . (2.9) p0 On the other hand, we apply Lemma 2.14 again with γ = α−ap > −1 and t = ν−(α−ap) > 0, and obtain ZZ p p Tν(ψ )(w) = ψ (z)H(z, w) dAα(z) D ZZ 2 α−ap 2 ν−α (1 − |z| ) = (1 − |w| ) 2+ν dA(z) D |1 − wz| ≤ C(1 − |w|2)−ap = Cψp(w) . Again, this argument is possible if the parameter a satisfies α − ν α − ν < a < . (2.10) p p Therefore, we can apply Prop. 2.13 if we can find a satisfying both (2.9) and (2.10), that is, ν+1 α−ν α−ν if the intervals (A, B) = (0, p0 ) and (C,D) = ( p , p ) have non-empty intersection. Since C > 0, this happens if and only if C < B, that is, α − ν ν + 1 < . p p0 Recalling that 1/p0 = 1 − 1/p, the statement follows. 2 SPACES OF HOLOMORPHIC FUNCTIONS 33

2.5. The Bergman kernel and projection on the unit disc. In this section we consider the Bergman projection B on the unit disc D. Since C(D) ⊂ L∞(D) ⊂ Lp(D) for all 0 < p < ∞, we have that B is well defined on a dense subset of Lp if 0 < p ≤ 2, and on all Lp if 2 < p ≤ ∞. Main goal of this section is proving the follwing result. Theorem 2.17. Let 1 ≤ p ≤ ∞. Then, the Bergman projection B is bounded B : Lp(D) → Ap(D) if and only if 1 < p < ∞. Proof. The sufficiency follows easily from Thm. 2.16. Observe that the bounedness of the operator T with kernel |K(z, w)| implies the boundedness of P . Thus, it suffices to apply Thm. 2.16 with ν = α = 0. To prove the necessity, observe that if B : L1(D, dA) → A1(D, dA) were bounded, B : L∞(D) → A∞(D) would also be bounded. Thus, it suffices to show that B is not bounded ∞ 2 2+ν −(2+ν) on L (D). Consider the family of functions fz(w) = (1 − |w| ) (1 − zw) . Then, kfzkL∞ = 1 for all z ∈ D. But, for all ν > −1, ZZ 2 2+2ν (1 − |w| ) kT f k ∞ ≥ |T f (z)| = dA(w) ν z L ν z 2(2+ν) D |1 − wz| ≥ C log(1 − |z|)−1 , that clearly is not bounded. Therefore, B cannot be bounded on L∞ and on L1 and we are done. 2 We now discuss some consequences of the boundedness of the Bergman projection. The first one is the question of the description of the dual space of Ap(D). It is clear that A2(D) is a Hilbert space and can be identified with its dual space, that is, A2(D)0 = A2(D) , where the identification is w.r.t. to its own inner product. By H¨older’sinequality we also immediately have taht any g ∈ Ap0 (D) defines a linear func- tional on Ap(D), again via the L2(D)-inner product ZZ Lg(f) = f(z)g(z) dA(z) . D However, it is not apriori clear that every g ∈ Ap0 (D), g 6= 0, gives rise to a non-trivial functional on Ap(D). Theorem 2.18. Let 1 < p < ∞, then the dual space of Ap(D) can be identified with Ap0 (D), where p0 is the conjugate exponent. Proof. Let L be a continuous linear functional on Ap(D), that is, L ∈ Ap(D)0. By the Hahn– Banach theorem, L extends to a linear functional on Lp(D), so that there exists g ∈ Lp0 (D) RR p p such that L(f) = D f(z)g(z) dA(z) for all f ∈ A (D). Using the fact that B is bounded on L and the density of L2 ∩ Lq in Lq for q = p, p0, we see that L(f) = hf, gi = hB(f), gi = hf, B(g)i =: hf, hi , with h ∈ Ap0 (D).Hence, Ap(D)0 ⊆ Ap0 (D). It only remains to see that a non-zero element of Ap0 (D) cannot be identically zero on Ap(D). By symmetry we may assume that p < 2 < p0. Then Ap0 ⊂ A2 ⊂ Ap. If h ∈ Ap0, h 6= 0, and 34 M. M. PELOSO hf, hi = 0 for all f ∈ Ap, then h ∈ A2 and in particular hf, hi = 0 for all f ∈ A2. Hence h = 0, a contradiction. 2 We show that the problem of the conjugate function has a simpler solution w.r.t. the case of the Hardy spaces. Given a real harmonic function u in D we denote by v = T u the harmonic conjugate in D such that v(0) = 0. Hence, f = u + iv is holomorphic in D and Im f(0) = 0. Theorem 2.19. For 1 < p < ∞, the mapping u 7→ v is bounded in the Lp-norm, that is, there exists a constant C > 0 such that

kvkLp(D) ≤ CkukLp(D) for all real harmonic functions u on D.

We remark that, using the weighted Bergman projection Bν with ν ≥ 1, it is possible to prove that the the harmonic conjugation mapping u 7→ v is also bounded in the L1(dA)-norm.12

Proof. By Thm. 2.17 we know that there exist c > 0 such that kB(g)kLp ≤ ckgkLp , for all g ∈ Lp(D). Next, if f ∈ Ap, by the mean value property applied to the function f(w)(1 − zw¯ )−2 which is holomorphic for w ∈ D, we have ZZ ¯ 1 f(w) B(f)(z) = 2 dA(w) π D (1 − zw¯) 1  ZZ f(w)  = 2 dA(w) π D (1 − zw¯ ) = f(0) . Therefore, if f ∈ Ap, f = u + iv and Im f(0) = 0 we have |v| ≤ |u + iv| = |f + f(0)| so that

kIm fkLp ≤ kf + f(0)kLp = kB(f + f)kLp = 2kB(Re f)kLp

≤ 2ckRe fkLp . This proves the statement under the initial a priori assumption that f ∈ Ap, that is, both u and v are in Lp. p Suppose now that u is real harmonic in L (D). For r < 1 consider ur. Then ur is harmonic in p the disc |z| < 1/r, so it is continuous in the closure of D; hence ur ∈ L (D) and kurkLp → kukLp as r → 1− monotonically from below. It is easy to check that for each r < 1, the function harmonic conjugate of ur is exactly vr, where v is the harmonic conjugate of u. Therefore, vr are also harmonic in the disc |z| < 1/r and in Lp and they converge pointwise to v. Hence, kvrkLp ≤ ckurkLp ≤ ckukLp . The conclusion now follows from the dominated convergence theorem. 2

12Exercise. SPACES OF HOLOMORPHIC FUNCTIONS 35

3. The Paley–Weiner and Bernstein spaces The growth of the modulus of an entire function can be described in terms of the decay of the Fourier transform of its restriction to the real line. With this approach in mind, we introduce (or recall) the main properties of the Fourier transform. The reader who is already familiar with this subject, can skip the next two sections.

3.1. The Fourier transform. In this section we briefly review a few basic facts about the Fourier transform and the space of rapidly decreasing functions, the . For sake of simplicity, we will limit ourselves to the real line R, although the extension to several variables, that is, to Rn, is straightforward. There are many excellent references that serve as introduction to the Fourier transform– for simplicity we mention only [Ka] or [Gr] and the references therein. For f ∈ L1(R) we define the Fourier transform of f as Z fˆ(ξ) = f(x)e−ixξ dx .

We will also denote it by Ff to emphasyse the mapping F : L1(R) → C(R) ∩ L∞(R). It is ˆ 1 ˆ ˆ clear that f is well defined for f ∈ L and that kfkL∞ ≤ kfkL1 . Moreover, f is continuous, as an immediate application of the dominated convergence theorem. We now see the first elementary properties of the Fourier transform. Given a function f on R we write

τyf(x) = f(x − y) .

Moreover, we write Dx to denote the differentiation operator d/dx. Proposition 3.1. Let f, g ∈ L1. Then, the following hold. −iyξ ixη (i) For any y, η ∈ R, F(τyf)(ξ) = e fˆ(ξ) and τηfˆ(ξ) = F(e f)(ξ). (ii) F(f ∗ g) = fˆgˆ. j 1 ˆ k j ˆ j  (iii) If x f ∈ L for 0 ≤ j ≤ k, then f ∈ C and Dξf(ξ) = F (−ix) f (ξ). k j 1 j j (iv) Suppose f ∈ C , Dxf ∈ L for 0 ≤ j ≤ k, then F(Dxf)(ξ) = (iξ) fˆ(ξ). Proof. (i) is elementary: Z Z −ixξ −i(x+y)ξ (τyf)ˆ(ξ) = f(x − y)e dx = f(x)e dx

= e−iyξfˆ(ξ) , and analogously for the other part of the statement. (ii) is also elementary. By Fubini’s theorem, ZZ (f ∗ g)ˆ(ξ) = f(x − y)g(y) dye−ixξ dx ZZ = f(x − y)e−i(x−y)ξ dxg(y)e−iyξ dy

= fˆ(ξ)ˆg(ξ) . (iii) is also elementary. It suffices to pass the differentiation under the integral sign. 36 M. M. PELOSO

j In order to prove (iv), we first assume that f is such that Dxf ∈ C(0), where we denote by 13 C(0) the space of continuous functions that vanish at ∞ (that is lim|x|→+∞ g(x) = 0). Under this assumption, assume that j = 1 first. Then, since f 0 vanishes at ∞, by integration by parts we have Z Z f 0(x)e−ixξ dx = − f(x)(−iξ)e−ixξ dx

= iξfˆ(ξ) . The case when j > 1 follows by induction. The general case now follows by approximating 1 f ∈ L with f0 ∈ C0 and using Lemma 3.2.

21 Lemma 3.2. The Fourier transform maps L continuously into C(0)(R): 1 F : L (R) → C(0)(R) . 1 ˆ Proof. We only need to show that for f ∈ L , f vanishes at ∞ (that is lim|x|→+∞ f(x) = 0). 1 1 For f ∈ L ∩ C0 we have that Z −ixξ |ξ||fˆ(ξ)| ≤ ξ f(x)e dx

≤ C0 , ˆ ˆ 1 1 1 i.e. |ξ|f is bounded, that is f ∈ C(0). Finally, the result follows by the density of L ∩ C0 in L .

2 We now compute a fundamental Fourier transform. Lemma 3.3. Let a > 0 and f(x) = e−ax2 . Then r π 2 fˆ(ξ) = e−ξ /4a . a Proof. By the previous proposition, we can differentiate under the integral sign and get

2  i   2  (fˆ)0(ξ) = F−ixe−aπx (ξ) = F e−aπx 0 (ξ) 2πa  i  = (iξ)fˆ(ξ) 2πa ξ = − fˆ(ξ) . 2πa Therefore, the function g = eξ2/4πafˆ(ξ) satisfies the differential equation d g = 0 , dξ R −t2 √i.e. it is a constant. In order to compute the constant, we select ξ = 0 and recall that e dt = π, so that r Z 2 π fˆ(0) = e−ax dx = . a

13The one we are adopting is not the standard notation for the space of continuous functions that vanish at ∞. Typically, this latter space is denoted by C0. We have decided to keep the subscript “0” to denote functions with compact support. SPACES OF HOLOMORPHIC FUNCTIONS 37

Remark 3.4. We will consider two types of dilations, that in fact are dual to each other in the integral pairing. Given a function f on R and t > 0 we define −1 t ft(x) = t f(x/t) and f (x) = f(tx) . (3.1) 1 R R −1 R Notice that, if f ∈ L then ft(x) dx = t f(x/t) dx = f(y) dy. Moreover, we have that t ˆ ˆ t −1 ˆ ˆ fbt(ξ) = f(tξ) = (f) (ξ) and fb (ξ) = t f(ξ/t) = (f)t(ξ) , (3.2) (We leave the details as an Exercise.) Finally, notice that, if the integral is well defined (for instance if f, g ∈ L2) Z Z t ft(x)g(x) dx = f(x)g (x) dx , as a simple change of variables shows.

We now prove Theorem 3.5. (Inversion Theorem) Let f ∈ L1 be such that fˆ ∈ L1. Then 1 Z f(x) = eiξxfˆ(ξ) dξ . 2π Proof. Let t > 0 and set 2 ψ(ξ) = eixξ−tξ . Then, r 2 π 2 (Fψ)(y) = τ (Fe−tξ )(y) = τ e−y /4t = ϕ√ (x − y) , x t x t √ where ϕ(x) = πe−x2/4. Next notice that, if f, g ∈ L1, then Z ZZ Z f(x)ˆg(x) dx = f(x)g(ξ)e−ixξ dxdξ = fˆ(ξ)g(ξ) dξ . (3.3)

Therefore,

Z 2 Z eixξ−tξ fˆ(ξ) dξ = ψ(ξ)fˆ(ξ) dξ Z √ √ = ϕ t(x − y)f(y) dy = f ∗ ϕ t(x) . √ R R −x2/4 √ 1 Since ϕ(x) dx = π e dx = 2π, we have that f ∗ ϕ t → 2πf in the L norm as t → 0. On the left hand side of the equalities above we can use the dominated convergence theorem to show that R eixξ−tξ2 fˆ(ξ) dξ → R eixξfˆ(ξ) dξ as t → 0. As an immediate consequence of the Inversion Theorem2 we have the following. Corollary 3.6. If f ∈ L1 and fˆ = 0, then f = 0. We now extend the definition of the Fourier transform to functions in L2. 38 M. M. PELOSO

1 2 ˆ 2 Theorem 3.7. () Let f ∈ L ∩ L . Then f ∈ L and F|L1∩L2 extends uniquely to a unitary isomorphism of L2 onto L2(dξ/2π); in other words, for all f ∈ L2(R), 1 kfk 2 = kfˆk 2 . L 2π L Proof. Consider the space X = {f ∈ L1 : fˆ ∈ L1}. We use the facts that f ∈ L1 implies fˆ ∈ L∞ 1 ∞ 2 ∞ 2 ∞ ∞ and that L ∩L ⊂ L to see that C0 ⊂ X ⊂ L . (We denote by C0 the space of C functions ∞ 2 having compact support.) Since C0 is dense in L , also X is. Notice that, for g ∈ X , by the inversion theorem, setting h¯ =g/ ˆ 2π 1 Z 1 Z hˆ(ξ) = e−ixξgˆ(x) dx = eixξgˆ(x) dx = g(ξ) . 2π 2π Therefore, Z Z Z 1 Z fg¯ dx = fh,ˆ dx = fhˆ dξ = fˆgˆ dξ . 2π Thus, F restricted to X preserves the L2-scalar products, so that, by taking f = g we see that 1 ˆ 2 kfk2 = 2π kfk2. Since F(X ) = X and X is dense in L , F extends by continuity to a unique unitary isomorphism of L2 onto L2(dξ/2π). Finally, one has to show that such an extension coincides with F on L1 ∩ L2. Suppose 1 2 −π|x|2 1 2 f ∈ L ∩ L and let ϕ(x) = e . Consider f ∗ ϕt. Then f ∗ ϕt ∈ L ∩ L , F(f ∗ ϕt) = −πt|·|2 1 1 2 fˆF(e ) ∈ L (since fˆ is bounded). Therefore, f ∗ ϕt ∈ X and f ∗ ϕt → f both in L and L 2 norms, and F(f ∗ ϕt) → fˆ both uniformly and in L norm. 2 We concluded this section with a brief introduction to a class of functions that are smooth and rapidly decreasing. This class turns out to be of fundamental importance, it is called the Schwartz space and it is denoted by S(R) (or simply as S). We define the %(j,k) as j k %(j,k)(f) = sup |x Dxf(x)| x∈R and define ∞ S = {f ∈ C (R): %(j,k)(f) < ∞, for all integers j, k} .

p ∞ It is cleat that S is contained in all the L -spaces, and that S contains C0 . j It is also clear that if ϕ ∈ S, p is a polinomial and j a positive integer, then Dx pϕ) ∈ S. We will soon see that this correspondence is continuous on S in the topology given by the above, see Prop. 3.9. The family of seminorms {%(j,k)} defines a topology on S as follows. A ngbh basis at a point f ∈ S is given by the sets U(j1,k1),...,(jN ,kN );ε  U(j1,k1),...,(jN ,kN );ε = g ∈ S : %(jh,kh)(g − f) < ε , for N a positive integer and ε > 0. A consequence of Prop. 3.1 and the Inversion Theorem is the following. Proposition 3.8. The Fourier transform F : S → S is a bijection. SPACES OF HOLOMORPHIC FUNCTIONS 39

We remark that in fact F is in fact a continuous bijection.14 We will prove the continuity in Prop. 3.9 We remark that, as a consequence of these facts, we obtain a simple proof that the convolution of two Schwartz functions is still a Schwartz function. For, ϕ ∗ ψ = F −1(ϕ ˆψˆ), whereϕ ˆψˆ ∈ S.

Proposition 3.9. The following are continuous mappings from S into itself: j  (i) for any polinomial p and j ∈ N, T : S 3 ϕ 7→ Dx pϕ ∈ S; (ii) the Fourier transform F : S → S. Proof. Exercise. (Notice that, in order to prove the continuity of F, and therefore the one of −1 F , we need to show that for every semi-norm %(j,k), there exist integers j1, k1, . . . , jN , kN and C > 0 such that for every F ∈ S,  X %(j,k) F(f) ≤ C %(jj ,kj )(f) . j=1 But,  j k ˆ %(j,k) F(f) = sup |ξ Dξ (f)| ξ∈R and then we use Prop. 3.1.) 2 Proposition 3.10. The space S is a complete metric space (i.e. a Fr´echet space) in the topology ∞ defined by the seminorms %(j,k). Moreover, C0 is dense in S. Proof. We only need to prove that S is complete in the given topology. Let {fn} be a Cauchy sequence in S, then %(j,k)(fm − fn) → 0 as m, n → +∞ for all (j, k). k In particular, {Dxfn} converges uniformly to a function gk for all k. Notice that

g0(x + t) − g0(x) = lim fn(x + t) − fn(x) k→+∞ Z t 0 = lim fn(x + s) ds k→+∞ 0 Z t = g1(x + s) ds . 0 k Therefore, Dxg0 = g1, and by induction on k we obtain that gk = Dxg0. Finally, j k k j k k |x ||D fn(x) − D g0(x)| = lim |x ||D fn(x) − D fm(x)| x x m→+∞ x x

≤ lim %(j,k)(fn − fm) m→+∞ ≤ ε , for j, k ≥ k0. Then %(j,k)(fn − g0) → 0 as n → +∞.

14We remark that, in ordet to prove the continuity of a linear mapping T : S → S we need to show that for every semi-norm %(j,k), there exist pair of integers (j1, k1),..., (jN , kN ) and C > 0 such that for every ϕ ∈ S, ` ´ X %(j,k) T (ϕ) ≤ C %(jj ,kj )(ϕ) . j=1 40 M. M. PELOSO

∞ ∞ In order to show that C0 is dense in S, observe that, if η ∈ C0 with η(0) = 1 and ϕ ∈ S, ε ∞ ε ε then, for every ε > 0, η ϕ is in C0 , where η (x) = η(εx). Now it is easy to check that η ϕ → ϕ ε in the S-topology, as ε → 0 that is, %(j,k)(η ϕ − ϕ) → 0, as ε → 0. 2

3.2. The Paley–Wiener theorems. We now study the connection between the Fourier trans- form and the growth of (the maximum modulus of) certain holomorphic functions. We identify the real line R with the subset of the complex plane {z : Im z = 0}. Our presentation of the Paley–Wiener theorem begins with the simple observation that if f is in L1(R) and has compact support, then its Fourier transform fˆ can be extended to an entire function. For, by definition Z b fˆ(ξ) = f(x)e−iξx dx a and, if we set Z b G(ζ) = f(x)e−iζx dx a we immediately see that: ◦ G is holomorphic in C, i.e. G is entire, since we can differentiate under the integral sign and in particular obtain that Z b G0(ζ) = −ixf(x)e−iζx dx ; a ◦ G extends fˆ, that is, G = fˆ; R ◦ finally, we can also estimate |G| by Z b −iζx M|η| |G(ζ)| ≤ sup |e | |f(x)| dx = e kfkL1 , x∈[a,b] a where M = max(|a|, |b|) and η = Im ζ; hence, G is bounded in every horizontal strip S = {z = x + iy : a ≤ y ≤ b}. In few words, we will see that a function on the real line whose Fourier transform has some support property is the restriction of a function holomorphic on a certain domain and satisfying some growth condition. Since we are more interested in the Hilbert space theory, we will concentrate on the case of functions that are L2 on the real line. Theorem 3.11. (Paley–Wiener Thm. for the strip) Let f ∈ L2(R) and a > 0. Then the following are equivalent:

(i) f is the restriction to the real line of a function F holomorphic in the strip Sa = {z = x + iy : |y| < a} and Z sup |F (x + iy)|2 dx < ∞ ; (3.4) |y|

Proof. (ii) ⇒ (i). Assume that (ii) holds and for z ∈ Sa define 1 Z F (z) = fˆ(ξ)eizξ dξ . (3.5) 2π R −yξ ˆ a|ξ| ˆ 2 Since |e f(ξ)| ≤ e f(ξ) ∈ L (R), F is well defined and F|R = f by the inversion theorem. Moreover, izξ a|ξ| |Dz(fˆ(ξ)e )| ≤ C|fˆ(ξ)|e for z ∈ Sa and some constant C > 0. Then, it is easy to see that we pass the differentiation 0 under the integral sign to show that F (z) exists in Sa; hence, F is holomorphic in Sa. −y(·) Notice that, writing Fy = F (· + iy), Fy is the inverse Fourier transform of e fˆ, so that by Plancherel’s theorem we see that Z 1 Z |F (x + iy)|2 dx = |fˆ(ξ)|2e−2ξy dξ R 2π R 1 ˆ a|·| 2 ≤ fe 2 . 2π L (R) −y(·) ˆ (i) ⇒ (ii). Let Fy = F (· + iy). We wish to show that Fcy = e f, where f = F|R ≡ F0, since by Plancherel’s theorem we would then have a|·| ˆ 2 −y(·) ˆ 2 ke fkL2(R) = sup ke fkL2(R) |y| 0 consider  G(x + iy) = ϕε ∗ F (· + iy) (x) .

Then G it is easy to see that holomorphic in the strip Sa and for |y| < a

sup kG(· + iy)kL2(R) ≤ sup kϕεkL1 kF (· + iy)kL2(R) < ∞ , |y| 0 (3.6) the upper half-plane. The domain U is conformally equivalent to the unit disc D, unbounded and its boundary is the real line. We will concentrate most of our upcoming efforts on the analysis of function spaces on such domain. We recall also that, if F is a function defined on U, we denote by Fy the function R 3 x 7→ F (x + iy). Theorem 3.12. (Paley–Wiener Thm. for the upper half-plane) The following properties hold true. (I) Let F be a function holomorphic on the upper half-plane U such that Z sup |F (x + iy)|2 dx < ∞ . (3.7) y>0 R 2 2 + Then, there exists f ∈ L (R) such that Fy → f in L , as y → 0 , i.e. Z lim |F (x + iy) − f(x)|2 dx = 0 ; (3.8) y→0+ R and fˆ(ξ) = 0 for ξ < 0 . (3.9) (II) Let f ∈ L2(R) be such that (3.9) holds. For z ∈ U we define F (z) by 1 Z F (z) = fˆ(ξ)eizξ dξ . (3.10) 2π R Then F in holomorphic in U and satisfies (3.7) and (3.8). The carefull reader has certainly realized the similarity of (3.7) with the H2-growth condition. In fact, we anticipate the definition of H2(U): we define 2  H (U) = F ∈ H(U) : sup kFykL2(R) < ∞ , y>0 and set kF kH2(U) = sup kFykL2(R) . y>0 We will come back to this theorem in Section 4.3 where we study the Hardy space H2(U).

Proof of Thm. 3.15. (II) Define F as in (3.10) for z ∈ U, i.e. for y > 0. If is clear that F is holomorphic in U and that, writing Fy = F (· + iy), Fy is the inverse Fourier transform of e−y(·)fˆ. By Plancherel’s theorem we have Z 2 1 ˆ 2 −2yξ 2 kFykL2(R) = |f(ξ)| e dξ ≤ kfkL2 . 2π R

15Exercise: Find and prove the Paley–Wiener theorem in the hypothesis that fˆ ∈ L2 has support in [a, b]. SPACES OF HOLOMORPHIC FUNCTIONS 43

This shows that (3.7) holds. Moreover, Z Z 2 1 ˆ −yξ  2 |Fy(x) − f(x)| dx = f(ξ) e − 1 dξ → 0 R 2π R as y → 0+, by the dominated convergence theorem; that is, (3.8) holds too.

(I) Consider F1 = F (· + i) and set f1 = F1|R. Then F1 satisfies the hypotheses in Thm. 3.11 (i) with a = 1, since Z Z Z 2 2 0 2 sup |F1(x + iy)| dx = sup |F (x + i(1 + y))| dx ≤ sup |F (x + iy )| dx < ∞ . (3.11) |y|<1 R |y|<1 R y0>0 R  Therefore, arguing as in the proof of the implication (i) ⇒ (ii) of Thm. 3.11, F F1(· + iy) = −y(·) e fˆ1 for |y| < 1, that is,  −y(·) F F (· + i(1 + y)) = e fˆ1 , ξ 2 2 and morevoer, e fˆ1 ∈ L . We define f ∈ L by setting ξ   fˆ = e fˆ1 = lim F F (· + i(1 + y)) = lim F F (· + iy) , (3.12) y→−1− y→0+ where the limits above are in the L2-norm. Hence, (3.8) holds true. On the other hand, from (3.11) it follows that Z Z 1 ˆ −yξ 2 2 2 |f1(ξ)e | dξ = kF1(· + iy)kL2(R) = |F (x + i(1 + y))| dx < ∞ 2π R R uniformly for y > −1. Letting y → +∞, since the left hand side above remain bounded, we 16 −ξ obtain that fˆ1(ξ) = 0 for ξ < 0. By (3.12) it follows that e fˆ(ξ) = 0 for ξ < 0, that is, fˆ(ξ) = 0 for ξ < 0. The following corollary2 is essentially a restatement of the previous theorem. We state for sake of completeness and leave the details to the reader. Corollary 3.13. Let f ∈ L2(R). Then the following are equivalent. (i) There exists a function F holomorphic on the upper half-plane U such that Z sup |F (x + iy)|2 dx < ∞ y>0 R and Z lim |F (x + iy) − f(x)|2 dx = 0 ; y→0+ R (ii) fˆ(ξ) = 0 for ξ < 0 .

Before presenting the last version of the Paley–Wiener theorem (for the whole complex plane), we need to recall a result from classical complex analysis, a consequence of the maximum modulus principle. For its proof see e.g. [CA-notes].17

16Make sure you agree with this assertion. 17Not yet though. 44 M. M. PELOSO

Theorem 3.14. (Phragmen–Lindel¨of) Let F be holomorphic in U and continuous on its closure. Suppose that |F (x)| ≤ M for x ∈ R and that for every ε > 0 |F (z)| ≤ e(a+ε)|z| , for z ∈ U and |z| sufficiently large, where a ≥ 0 is given. Then |F (z)| ≤ Meay for all z ∈ U. Theorem 3.15. (Paley–Wiener, third version) Let F be an entire function and a > 0. Then the following are equivalent. 2 (i) F|R ∈ L (R) and |F (z)| = o(ea|z|) as |z| → +∞ ; (3.13) (ii) there exists f ∈ L2(R) with supp fˆ ⊆ [−a, a] such that (3.10) holds, i.e., 1 Z F (z) = fˆ(ξ)eizξ dξ . 2π R Proof. (ii) ⇒ (i). Assume that (ii) holds, then 1 Z a 1 Z a ˆ izξ ˆ −yξ |F (z)| = f(ξ)e dξ ≤ |f(ξ)|e dξ 2π −a 2π −a  1 Z a 1/2 1 Z a 1/2 ≤ |fˆ(ξ)|2 dξ e−2yξ dξ 2π −a 2π −a  1 1/2 = kfk 2 sinh(2ya) . L (R) 2πy

Notice that, for y ∈ R, the function sinh(2ya)/y is C∞, even and positive, and that sinh(2ya)/y ≤ e2a|y|/|y|, as |y| → +∞. Therefore, as |y| → +∞, ea|y| |F (z)| ≤ kfkL2(R) , p2π|y| 2 which is stronger than (3.13). Finally, by the Fourier inversion theorem, F|R = f ∈ L (R). (i) then follows.

(i) ⇒ (ii). Assume that (i) holds, and suppose first that F|R is bounded. Condition (3.13) implies that the hypoheses of the Phragmen–Lindel¨oftheorem are satisfied so that |F (z)| ≤ Meay in U. Set G(z) = eiazF (z). Then G is entire and |G(z)| ≤ M in U. We now claim that G satisfies (3.7), the H2-growth condition on U. Assuming the claim for now we finish the proof. By Thm. 3.15 we obtain that suppg ˆ ⊂ ˆ ˆ [0, +∞), where g = G|R. Since f(ξ) =g ˆ(ξ + a), supp f ⊂ [−a, +∞), where f = F|R. We may repeat the previous argument for G(z) = e−iazF (z) in the lower half-space and obtain that supp fˆ ⊂ (−∞, a], so that supp fˆ ⊂ [−a, a] . (3.14) Now set 1 Z H(z) = fˆ(ξ)eizξ dξ . 2π R SPACES OF HOLOMORPHIC FUNCTIONS 45

Then H is an entire function (by the first part of the proof) and, by the inversion theorem, coincides with F on the real line, hence on all of C. This prove the assertion under the assumption F|R bounded (and modulo the claim). If F is not bounded, consider Z Fϕ(z) = F (z − u)ϕ(u) du , R where ϕ be an arbitrary continuous function with compact support in R. The function Fϕ satisfies condition (3.13) and Fϕ|R is bounded, as it is easy to check. It follows then that  supp F Fϕ|R ⊂ [−a, a] .  ˆ But F Fϕ|R = fϕˆ, and by the arbitrariness of ϕ we obtain (3.14). It only remains to prove the claim. Let ϕ be a continuous function with compact support in R, kϕk 2 ≤ 1. We define L Z Gϕ(z) = G(z − u)ϕ(u) du . R Using the fact that the ϕ has compact support, by differentiating under the integral sign it is easy to see that Gϕ is holomorphic on the closure of U. Moreover, Gϕ is bounded, since

|Gϕ(z)| ≤ MkϕkL1 , and it holds that

|Gϕ|R(x)| ≤ kG|RkL2 kϕkL2 ≤ kG|RkL2 . Then we can apply the Phragmen–Lindel¨oftheorem (with a = 0) to obtain that

|Gϕ(z)| ≤ kG|RkL2 for all z ∈ U. This in particular implies that Z

G(x + iy)ϕ(−x) dx ≤ kG|RkL2 R for y > 0. Therefore, sup kGykL2 ≤ kG|RkL2 y>0 This proves the claim and therefore the theorem. 2 3.3. The Paley–Wiener spaces. Our starting point is Thm. 3.15 where we dealt with entire functions with a prescribed order of growth. Definition 3.16. Let F be an entire function. We say that F is an entire function of exponential type if there exists constants A, B > 0 such that, for all z ∈ C |F (z)| ≤ AeB|z| . If F is entire function of exponential type, we call a the type of F where log M(r) a = lim sup , r→+∞ r and where M(r) = sup|z|=r |F (z)|. We denote by Ea the space of functions of exponential type at most a. 46 M. M. PELOSO

Remark 3.17. Consider the following conditions on an entire function F :

(i) for every ε > 0 there exists Cε > 0 such that (a+ε)|z| |F (z)| ≤ Cεe ; (ii) there exists C > 0 such that |F (z)| ≤ Cea|z| ; (iii) as |z| → +∞ |F (z)| = o(ea|z|) . Then clearly, (iii) ⇒ (ii) ⇒ (i) ⇒ “F is of exponential type at most a”. We recall that an entire function F is said to be of finite order if there exists % > 0 such that

% |F (z)| = O(e|z| ) as |z| → +∞ . The infimum of the values % for which such relation holds is called the order of F . Thus, an entire function of exponential type is of finite order 1, although the converse is not true. p Definition 3.18. Let a > 0, 1 ≤ p ≤ ∞. We define the Paley–Wiener space PWa as Z a p n −ixy p o PWa = f : f(x) = g(y)e dy , where g ∈ L (−a, a) , −a and we set

kfk p = 2πkgk p . PWa L p p In other words, PWa is the image via the Fourier transform of the L -functions that are supported in [−a, a]. Remark 3.19. In these notes we will essentially first concentrate on the case p = 2, in which 2 case we simply write PWa to denote the Paley–Wiener space PWa . Notice that, it follows from the Plancherel formula that

kfk 2 = kgˆk 2 = 2πkgk 2 = kgˆk 2 = kfk 2 . PWa PWa L L L

Hence, by polarization, for f, ϕ ∈ PWa,

hf, ϕiPWa = hf, ϕiL2 . (3.15) Forthermore, we remark that, from Thm. 3.15 we obtain the following characterization of PWa: 2 “A function f is in PWa if and only if f ∈ L (R) and f = F|R (that is, f is the restriction to the real line of a function F ), where F is an entire function of exponential type such that |F (z)| = o(ea|z|) for |z| → +∞.” It follows that PWa coincides with the subspace of Ea of the functions whose restriction to 2 2 the real line are in L . For, if F ∈ Ea and F|R ∈ L then F satisfies (3.13) in (i) of Thm. 3.15 for 0 0 0 0 any a > a. By Thm. 3.15 itself it then follows that supp F(F|R) ⊆ [−a , a ], for every a > a. a|z| Hence supp F(F|R) ⊆ [−a, a] and by Thm. 3.15 again |F (z)| = o(e ) for |z| → +∞. The remaining implication is trivial. SPACES OF HOLOMORPHIC FUNCTIONS 47

Theorem 3.20. The Paley–Wiener space PWa is a Hilbert space with reproducing kernel w.r.t. the inner product given by (3.15). Its reproducing kernel is the function a K(x, y) = sinc(a(x − y)) , π 18 where sinc t = sin t/t. Hence, for every f ∈ PWa and x ∈ R a Z f(x) = f(y) sinc(a(x − y)) dy . (3.16) π R 2 Proof. Let f ∈ PWa. It is clear that f ∈ L (R), f = F|R for an entire function F as in Remark 3.19, and that f =g ˆ, for some g ∈ L2(R) with supp g ⊆ [−a, a], so that 1 Z a f(x) = fˆ(ξ)eixξ dξ . (3.17) 2π −a Notice that fˆ ∈ L1 so that the integral converges absolutely. Then,

1 1 1/2 1/2 |f(x)| ≤ kfˆk 1 ≤ (2a) kfˆk 2 = (2a) kfk 2 , 2π L (−a,a) 2π L L 2 and it is immediate to check that PWa is closed. This shows that PWa, as a subspace of L (R), is a Hilbert space with reproducing kernel. Next, notice that from (3.17), if we could switch the integration order (e.g. if f ∈ L1(R)), we would obtain 1 Z a Z f(x) = f(y)e−iyξ dyeixξ dξ 2π −a R Z 1 Z a = f(y) ei(x−y)ξ dξ dy . (3.18) R 2π −a

Therefore, the candidate for the reproducing kernel of PWa is the function 1 Z a K(x, y) = ei(x−y)ξ dξ . 2π −a ε ∞ However, we may switch the integration order if we approximate f with fϕ where ϕ ∈ C0 , ϕ identically 1 in a ngbh of the origin (so that ϕε(x) = ϕ(εx) → 1 as ε → 0). −1 Since K(·, y) = F (χ(−a,a))(· − y), K(·, y) ∈ PWa for every y ∈ R. Since K satisfies the conditions of Lemma 2.2, K(x, y) is the reproducing kernel for PWa. Notice also that the resulting identity (3.18) gives an absolutely convergent integral since K(x, ·) ∈ L2(R), for every x ∈ R. Finally, since 1 Z a 1 sin at a eitξ dξ = = sinc at , 2π −a π t π we have a K(x, y) = sinc(a(x − y)) . π This proves the theorem. 2 18This function is called the (unnormalized) “cardinal sine”, or “sinc” function for short. It is clear from the formulas above that the case a = π is particularly simple. The function sinc(π·) is called the “normalized” sinc function. 48 M. M. PELOSO

Theorem 3.21. The following properties hold. a 1/2 π  (1) The set {( π ) sinc a(x − n a ) }n∈Z is an orthonormal basis for PWa. (2) For every f ∈ PWa we have the orthogonal expansion

X π π f(x) = fn  sinc a(x − n ) . (3.19) a a n∈Z

The series converges absolutely, it holds also when x ∈ C and it converges uniformly on compact subsets of C. (3) For every f ∈ PWa we have the Parseval relation

π X π kfk2 = |fn |2 . (3.20) L2 a a n∈Z

Proof. Recall that √1 F is a unitary isomorphism of L2 onto itself. Therefore, √1 F maps an 2π 2π 2 orthonormal basis of L (−a, a) onto an orthonormal basis of PWa. Let

1 in π t ϕn(t) = √ e a n ∈ Z . 2a

Then setting ψ = √1 ϕˆ , {ψ } becomes an orthonormal basis for PW , where n 2π n n n∈Z a

Z a 1 it(n π −ξ) ψn(ξ) = √ e a dt 2 aπ −a 1 1 π  = √ π sin a(n − ξ) aπ n a − ξ a r a π = sinc a(ξ − n ) . π a

This proves (1).

(2) By the theory of abstract Fourier series, from (1) it follows that, for every f ∈ PWa,

X f = hf, ψniψn , n∈Z where the series converges in the Hilbert space norm. Then, applying (3.16) we have

a X  Z π  π f(x) = f(y) sinc a(n − y) dy sinc a(x − n ) π a a n∈Z R X π π = fn  sinc a(x − n ) , a a n∈Z SPACES OF HOLOMORPHIC FUNCTIONS 49 where the convergence is the L2-norm. In order to prove the absolute convergence we first prove (3). The convergence is the L2-norm allows to switch the summation and integration orders Z X π π π π kfk2 = fm fn  sinc a(x − m ) sinc a(x − n ) dx L2 a a a a R m,n∈Z X π π π = fm fn  δ a a a m,n m,n∈Z π X π = |fn |2 . a a n∈Z This proves (3). Now, using the Cauchy–Schwarz inequality, it easy to see that the series in (2) converges absolutely since r X π  π  π  X π  21/2 f n sinc a(x − n ) ≤ kfk 2 sinc a(x − n ) a a a L a n∈Z n∈Z  = kfkL2 sinc a(x − ·) L2 rπ = kfk 2 . a L

Finally, the functions on the two sides of (3.19) are restrictions of functions in Ea and they coincides on R, hence they equal when x is let vary on the whole complex plane. 2 3.4. The Bernstein spaces. We now define a scale of spaces closely related to the Paley– Wiener spaces. p Definition 3.22. Let a > 0, 1 ≤ p ≤ ∞. We define the Bernstein space Ba as p n p o Ba = f : f ∈ L (−a, a) , f = F|R for some F ∈ Ea , and we set kfk p = kfk p . Ba L p p In other words, Ba consists of the subspace of L (R) given of the restriction of entire function of exponential type at most a. 2 By the Paley–Wiener theorem, we clearly have that Ba = PWa. In order to prove they are complete we need the following important result proved by Plancherel and P´olya. We remark that it holds true also when 0 < p < 1, but we prove only in the case 1 ≤ p < ∞ to keep our presentation within reasonable space. p Theorem 3.23. (Plancherel–P´olya) Let 1 < p < ∞, a > 0. Then, for every f ∈ Ba we have Z Z |f(x + iy)|p dx ≤ eap|y| |f(x)|p dx . (3.21) R R The proof is based on an application of the Phragmen–Lindel¨oftheorem. However, it is somewhat technical and we refer to [Yo], p. 80. p Corollary 3.24. For 1 < p < ∞ and a > 0, Ba is a Banach space. 50 M. M. PELOSO

p p Proof. We only need to prove that Ba is closed in L . We first observe that, for every compact set E ⊂ C there exists a constant C = CE > 0 such p that for every f ∈ Ba we have sup |f(z)| ≤ CEkfkLp . (3.22) z∈E For, we may assume that E = Q(z0, δ) is a closed square,  Q(z0, δ) = z = x + iy : |x − x0| , |y − y0| ≤ δ , for some δ > 0. Notice that, for every z ∈ Q(z0, δ), D(z, δ) ⊂ Q(z, δ) ⊂ Q(z0, 2δ). By the mean-value inequality (2.5), for z ∈ we obtain ZZ p 1 p |f(z)| ≤ 2 |f(w)| dA(w) πδ Q(z,δ) ZZ 1 p ≤ 2 |f(w)| dA(w) πδ Q(z0,2δ) Z x0+2δ Z p ≤ Cδ |f(u + iv)| du dv x0−2δ R Z x0+2δ Z ap|v| p ≤ Cδ e |f(u)| du dv x0−2δ R p ≤ CEkfkLp . p This clearly implies that the convergence in Ba forces the uniform convergence on compact subsets in C. p p Therefore, if fn ∈ Ba, n = 1, 2,... , and fn → f in L , then fn → f uniformly on compact p subsets of C. Hence, f is entire, of exponential type at most a, that is, f ∈ Ba.

2 2 2 We now consider the orthogonal projection Pa : L (R) → Ba which is given by the formula a P (f) = f ∗ sinc(a·) , f ∈ L2 . a π We do not have the necessary tools to prove the next result, and we refer to any classical text in Fourier Analysis, or the notes [HA-notes]. The argument essentially relies on the boundedness of the Hilbert transform on R. Theorem 3.25. Let a > 0, 1 < p < ∞. Then, the operator P (f) = f ∗ sinc, initially defined 2 p p p on the dense subset L ∩ L , extends to Pa : L → Ba. p Proof. It is obvious that Pa(L ) ⊂ Ea. We notice that Pa = cTχ(−a,a) , where Tm denotes the

Fourier multiplier operator, with multiplier m = χ(−a,a). The fact that Tχ(−a,a) is a bounded operator follows as a simple consequence of the Lp-boundedness of Hilbert transform. 2 Dipartimento di Matematica, Universita` degli Studi di Milano, Via C. Saldini 50, 20133 Milano, Italy E-mail address: [email protected] URL: http://wwww.mat.unimi.it/~peloso/