CLASSICAL SPACES OF HOLOMORPHIC FUNCTIONS

Transcription

1 CLASSICAL SPACES OF HOLOMORPHIC FUNCTIONS MARCO M. PELOSO Contents. Hardy Spaces on the Unit Disc.. Review from complex analysis.2. Hardy spaces 5.3. Harmonic Hardy classes 7.4. Fatou s theorem 9.5. The zero sets of functions in H p 3.6. Boundary behaviour of functions in Hardy spaces 6.7. The Cauchy Szegö projection 9 2. Bergman spaces on the unit disc Function spaces with reproducing kernel The Bergman spaces Biholomorphic invariance L p -boundedness of a family of integral operators The Bergman kernel and projection on the unit disc The Paley Weiner and Bernstein spaces The Fourier transform The Paley Wiener theorems The Paley Wiener spaces The Bernstein spaces Function theory on the upper half plane Hardy spaces on the upper-half plane Factorization and boundary behaviour of functions in H p (U) H 2 and the Paley Wiener theorem revisited H, the atomic decomposition and the space of bounded mean oscillations Weighted Bergman spaces on the upper-half plane The Paley Wiener theorem for Bergman spaces Further topics Hardy spaces on tube domains in C n Weighted Bergman spaces on the unit ball in C n The Cauchy integral along Lipschitz curves Real variable Hardy spaces Hankel and Toeplitz operators 56 References 56 Appunti per il corso Argomenti Avanzati di Analisi Complessa per i Corsi di Laurea in Matematica dell Università di Milano, a.a. 2/2. November 2, 2.

2 SPACES OF HOLOMORPHIC FUNCTIONS In these notes we consider spaces of holomorphic functions defined on domains of the complex plane or, toward the end of the notes, on domains of the n-dimensional complex space C n. We begin with the Hardy spaces on the unit disc.. Hardy Spaces on the Unit Disc.. 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 < }. Here and in what follows, we will denote by D(z, r) the disc centered at z C and radius r > ; we will sometimes abbreviate D r to denote the disc D(, r). If f is holomorphic on D then it can be written as sum of a converging power series f(z) = k= a k z k. From Cauchy s formula we know that, if γ is a simple closed curve contained in D and z lies in the interior of γ (which is a well-defined domain by the Jordan curve theorem), then f(z) = f(ζ) i ζ z dζ. If the function f is actually holomorphic in a slightly larger disc D R with R >, then we may take the unit circonference D as curve of integration and obtain f(z) = f(ζ) i ζ z dζ, for all z D. One of the main themes of this course is trying to recover the holomorphic function f (in a given class) from its boundary values, once these values are suitably defined. Setting z = re iη, we rewrite the above identity as f(re iη ) = = = using the periodicity of the exponential. Notice that γ D e iθ f(e iθ ) e iθ re iη dθ f(e iθ ) dθ rei(η θ) f(e i(η θ) ) dθ, (.) reiθ + re iθ = r n e inθ n= =: C r (e iθ ), (.2) where C r (e iθ ) is the Cauchy kernel (thought of as a convolution kernel, see (.4) below).

3 2 M. M. PELOSO Therefore, using the uniform convergence on compact subsets of D of the power series, we have that (.) gives the identity f(re iη ) = = n= n= r n f(e iθ )e in(η θ) dθ r n e inη f(e iθ )e inθ dθ. (.3) We recall that the unit circle D = {ζ C : ζ = e iθ } inherits the multiplicative structure from C and it is homeomorphic (as a topological group) to the interval [ π, π] (or any interval of length ) once we identify the two end points of the interval. This group T = R/Z is called the torus. Of course, the identification between T and D is given by the mapping t e it. Given f, g L (T) (where, we recall, T is identified with the unit circle) the convolution on T is defined as f g(e iη ) = f(e iθ )g(e i(η θ) ) dθ. (.4) Therefore, the identity (.) can be expressed as convolution on the group T as f(re iη ) = (f C r )(e iη ). (.5) Moreover, in (.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) ˆf(n) = f(e iθ )e inθ dθ. It is clear that in order to define the Fourier coefficients is sufficient to assume that f is an L -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 u(re iη ) = u(e iθ r 2 ) re i(η θ) 2 dθ = (Pu)(re iη ), (.6) where u is a function harmonic in a domain containing D and < r <. 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 Dirichlet problem 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 L (T) and that Pg is a harmonic function 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

4 SPACES OF HOLOMORPHIC FUNCTIONS 3 other words, a classical question is whether the Dirichlet problem (for the Laplacian on the unit disc) { f = on D (.7) f = g on D, where g L p ( D) is an assigned function, 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 (.5) we may think of the Poisson integral as a convolution integral operator on T by writing P r (e iθ ) = r2 re iθ 2, (.8) so that the Poisson integral of a function g L (T) is defined by the formula (Pg)(re iη ) = g(e iθ r 2 ) re i(η θ) 2 dθ = (P r g)(e iη ). (.9) In complex analysis we were used to think of the Poisson kernel, that we denote temporarely as P z (e iθ ) as a function of the variables z = re iη D and e iθ D, where P z (e iθ ) = = r 2 e iθ re iη 2 = r 2 re i(η θ) 2 (.) r 2 2r cos(θ η) + r 2. (.) In the current setting it is more convenient to think of the Poisson kernel as a family of functions {P r } defined on the torus T, hence depending on the variable e iη, and of the Poisson integral of a function g as the convolution of g with P r, as in (.9). According to this point of view, the collection of functions {P r } <r< on T is a summability kernel (or a approximation of the identity), A family of functions {Φ t } t> is called a summability kernel if they satisfy the following properties: () Φ t (e iθ ) dθ = for t > ; (2) Φ t L C with C independent of t; (3) for every δ >, lim Φ t (e iθ ) dθ =. t + δ θ π We recall that if {Φ t } t> is a summability kernel on T and g L p (T), p <, then Φ t g g in the L p -norm, and that if g C(T), then Φ t g g in the sup-norm. We collect here the basic properties of the Poisson kernel and integral. Proposition.. The function P (re iη ) = P r (e iη ) is harmonic in D as function of z = re iη and P r (e iη ) dη = for all < r <. Moreover, as collection {P r } of functions on T we have (i) for p <, f P r L p f L p; (ii) for p <, lim r f P r f L p = ;

5 4 M. M. PELOSO (iii) if g C(T), then lim r g P r g L =. Finally, for f L p (T), p, P(f) is harmonic in D. Proof. It is clear that it suffices to show {P r } is a summability kernel and that P (re iη ) is harmonic in the variable z = re iη D. The latter fact is well known, as well as that P r (e iη ) dη =. In order to prove that {P r } 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. Lemma.2. For < r < we have that and the convergence is uniform on T. P r (e iη ) = k= r k e ikη, Proof. Recall that we always identify the unit circle D with T. We can write r 2 re iη 2 = re iη + (.2) re iη = = k= k= r k e ikη + r k e ikη. k= r k e ikη The statement about the uniform convergence is clear, by the Weierstrass M-test. Notice that by (.2) we obtain a relation between the Poisson and Cauchy kernels, that is, P r = 2Re C r. (.3) Lemma.3. Let g L ( D). Then for every < r < we have that ( Pr g ) (e iη ) = k= ĝ(k)r k e ikη. Proof. Using the uniform convergence of the series development of P r we see that ( Pr g ) (e iη ) = and the convergence is uniform on T. = = k= k= g(e iθ ) k= r k e ik(η θ) dθ ( g(e iθ )e ikθ dθ ĝ(k)r k e ikη, ) r k e ikη

6 SPACES OF HOLOMORPHIC FUNCTIONS 5.2. 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.4. For < p < and < r <, for a function f defined on D we set ( /p M p (f, r) = f(re iθ ) dθ) p. We define the Hardy space H p = H p (D) as { } H p = f H(D) : sup <r< M p (f, r) <, and for f H p we set f H p = sup M p (f, r). <r< Furthermore, we define H as the space of holomorphic functions that are bounded on the unit disc, endowed with the sup-norm. We mention in passing that, also when < p < we set ( f L p = f(e iθ ) p dθ and, with an abuse of language, we call it the L p -norm. However, setting d(f, g) = f g p L p L p becomes a complete metric space. Notice that f + g L p 2 ( p)/p ( f L p + g L p). Remark.5. Aside from the case p =, also the space H 2 can be described at once. For, if f is holomorphic on D, then it admits power series expansion f(z) = + n= a nz n. Using the uniform convergence on compact subsets of D we have Therefore, M 2 (f, r) 2 = = = n= f(re it ) 2 dt n,m= r 2n a n 2. sup M 2 (f, r) = <r< ) /p a n a m r n+m e it(n m) dt ( + a n 2) /2, that is, f H 2 if and only if + n= a n 2 is finite. In particular, it follows that H 2 is a Hilbert space. Observe that, in particular, a function f in H 2 can be extended to the boundary D, having as boundary values the function f L 2 ( D) given by Exercise I.2. f(e it ) = n= n= a n e int.

7 6 M. M. PELOSO Moreover, by Lemma.3, P( f)(r ) = P r f = + n= a nr n e in( ) = f(r ). We can call f boundary values since f(r ) = P r f f in L 2 (T) as r. This fundamental relation is a first indication of the deep connection between the theories of Fourier series and of Hardy spaces. Remark.6. For < p < q we have {} H q H p. For, notice that for < p < q we clearly have H q H p. Moreover, all holomorphic polynomials are bounded, hence in all the H p spaces and these are non-trivial. Moreover, using Lemma 2.5 (that we will see later on) one can show that f(z) = ( z) a Hp if and only if p < a. In these notes we will essentially restrict ourselves to the case p, although the Hardy spaces turn out to be of great interest also and, from a certain point of view, especially, in the case < p. We point out that the Hardy space H can be considered as a limit case for both theories < p and < p < 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 H p -norm is actually a lim. Remark.7. Given f holomorphic in D, the function M p (f, r) is increasing in r, < r <. This statement holds true for the full range < p, but its proof is elementary only in the case p and we will restrict to this case. For < ϱ < and a function f defined in D we write f ϱ := f(ϱ ) to denote a function on the unit circle. For < r, ϱ <, given a function f holomorphic in D, notice that f r is holomorphic in a ngbh of D so that f(rϱe iη ) = (f r P ϱ )(e iη ). Hence, we have that is, M p (f, r) is increasing in r and M p (f, rϱ) = f r P ϱ L p (T) f r L p (T) P ϱ L (T) = M p (f, r), sup M p (f, r) = lim M p(f, r). <r< r Proposition.8. For p H p is a Banach space. Proof. It is clear that H p is a norm, so we only need to prove that it is complete. Let < r, ϱ < and notice that for f holomorphic on D we have This shows that f(rϱe it ) = f r P ϱ (e it ) f r L p (T) P ϱ L p (T) C ϱ f H p. sup f(z) C ϱ f H p, z ϱ

8 SPACES OF HOLOMORPHIC FUNCTIONS 7 and therefore the convergence in the H p -norm implies the uniform convergence on compact subsets. Thus, let {f n } be a Cauchy sequence in the H p -norm, and let f be the function uniform limit on compact subsets of D. Then f is holomorphic and for < r < fixed, using the uniform convergence, for p < we have M p (f n f, r) p = = lim m + f n (re it ) f(re it ) p dt f n (re it ) f m (re it ) p dt = lim m + M p(f n f m, r) p lim f n f m p m + H. p Therefore, f n f H p lim f n f m H p < ε, m + for n sufficiently large. The case p = is similar and we leave the details to the reader Harmonic Hardy classes. We now analyze the action of the Poisson integral on the spaces L p (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 H p -growth condition, < p <, that is, sup M p (f, r) <, (.4) <r< and in an obvious way when p =. Proposition.9. Let f be harmonic in D and satisfy (.4), < p. Then there exists f L p ( D) such that f = P( f), that is, f(re it ) = ( f P r )(e it ) = Moreover, if < p <, f r f in the L p -norm. f(e iθ r 2 ) re i(t θ) 2 dθ. We observe that this statement is false when p =. For instance, f(re iη ) = P r (e iη ) satisfies the hypotheses with p =, but there exists no L -function g on D such that P r g and P r g = P r. We leave the details as an exercise. 3 Proof. Let f r = f(r ), < r <. Let {r n } be a strictly increasing sequence in (, ), r n. Then {f rn } is a bounded set in L p (T). By the Banach-Alaoglu theorem (see [Ru]), there exists a subsequence {f rnj } converging in the weak- topology to a funtion f L p (T), < p, that is, for every g L p (T) we have that 2 Exercise I.4. 3 Exercise I.5. f rnj (e iθ )g(e iθ ) dθ f(e iθ )g(e iθ ) dθ as j +.

9 8 M. M. PELOSO (Here we need the restriction p >, since the Banach-Alaoglu theorem requires that the space containing {f r } to be the dual space of another Banach space and L (T) is not.) Let < r < and let j be such r nj > r for j j. Then, f rnj is continuous on D and harmonic in D, hence it satisfies the Poisson reproducing formula f(re it ) = f rnj ( (r/rnj )e it) = (f rnj P r/rnj )(e it ) = = f rnj (e iθ )P r/rnj (e i(t θ) ) dθ f rnj (e iθ )P r (e i(t θ) ) dθ + [ ] f rnj (e iθ ) P r/rnj (e i(t θ) ) P r (e i(t θ) ) dθ. (.5) Since P r C( D) L p ( D) for all < r <, the first term on the right hand side above converges to f(e iθ )P r (e i(t θ) ) dθ. Now we show that the second term on the right hand side of (.5) tends to as j +. For, g j := P r/rnj P r in the L p -norm, since g j C(T) and tend to uniformly. Therefore, f rnj (e iθ )g j (e i(t θ) ) dθ f rnj L p g j L p C g j L p as j +. Notice that we have used the assumption that f satisfies (.4). From (.5) it follows that f(re it ) = f(e iθ )P r (e i(t θ) ) dθ, as we wished to show. The last part of the statement is clear: f r = f P r as functions on D and since {P r } is a summability kernel, f r = f P r f in the L p -norm, for < p <. Remark.. Notice that, in particular Prop..9 applies to H p -functions. Given f H p, we have constructed the boundary value function f as weak- limit of a sequence {f nj }, where n j. It is easy to see that this construction is independent of the choice of the sequence {f nj }. (This is obvious in particular when < p < since f is the limit in the L p -norm of the whole family {f r }. ) Moreveor, if f r g in L p ( D), it also easy to see that g = f. In particular, the boundary value function of an H 2 -function identified in Remark.5 coincides with the one constructed in Prop..9. We now prove a partial converse of Prop..9. Proposition.. Let g L p ( D), p <. Then the Poisson integral of g, P(g) is harmonic in D, satisfies the H p -growth condition (.4), and (in the notation of Prop..9) P(g) = g.

10 SPACES OF HOLOMORPHIC FUNCTIONS 9 Proof. Notice that, as functions on D, f r = f(r ) := g P r satisfy the norm estimate f r L p g L p P r L = g L p. Therefore, sup M p (f, r) g L p. <r< This shows that the harmonic function P(g) satisfies the H p -growth condition (.4). Next, if < p <, by the previous proposition, f := P(g) admits a boundary value function f such that P( f) = f, that is, P( f g) =. Therefore, P r ( f g) = in L p (T) for all < r <, since {P r } is a summability kernel, = P r ( f g) f g in L p (T). (Notice that here we need to restrict to the case p <.) Finally, in the case p =, we simply notice that f r := P r g g in L ( D) as r, so that we may define again the boundary values of f as the limit of the family {f r }, that is, f = g. This concludes the proof. Remark.2. Recall the Dirichlet problem (.7). If the boundary datum g C( D) we know that the solution f = P(g) is harmonic, continuous in the closure of the unit disc and f restricted to D coincides with g (e.g. see [CA-notes]). Prop.. can be restated by saying that if g L p ( D), p <, then f = P(g) is harmonic in D and can be extdended to a boundary value function f that coincides with g. Then, we say that f = P(g) is (the) solution of the Dirichlet problem (.7) with boundary datum in L p. We conclude this part with the following consequence of the last two propositions. Corollary.3. Define Then, for < p <, is an isometric isomorphism. h p (D) = { f harmonic on D : f h p := sup M p (f, r) < }. <r< P : L p ( D) h p (D).4. Fatou s theorem. Perhaps, the main result about the Hardy spaces is that if f H p with < p, then f, as in the case p = 2, admits boundary values, that is, lim r f(reit ) = f(e it ) exists for a.e. t [, ), and f L p ( D). Actually, more is true: the function f converges to f when z approaches a point e it D, for a.e. t [, ), within the non-tangential approach regions. Definition.4. We define the subset of the unit disc Γ α (e it ) = { z D : z e it < α( z ) }. (.6) The set Γ α (e it ) is also called the Stoltz region, with vertex in e it and aperture α >. Remark.5. We need to make a few comments on the definition of Γ α (e it ). (i) It is easy to see that the regions Γ α (e it ) are increasing with α and that they would be empty for α.

11 M. M. PELOSO (ii) The region Γ α (e it ) is also called a non-tangential approach region, since outside a fixed compact subset of D, it is contained in a cone with vertex in e it and aperture C α α, where C α > is a suitable constant. For, assume that ε α r <, where c α is a (small) positive constant, that we are going to fix momentarily, and let z = re iη Γ α (e it ). Then so that re iη e it = e iη e it ( r)e iη e iη e it ( r), e iη e it (α + )( r). If c α is chosen sufficientily small, then we must have η t < π/4 (that is, z Γ α (e it ) and r c α implies η t < π/4). Next, e iη e it = 2 sin ( η t /2 ) 2 2 π η t, as long as η t π/4, which is the case. Therefore, z = re iη Γ α (e it ) and r c α implies η t π 2 2 (α + )( r) = C αα( r), (.7) as claimed. In particular, we may take C α = 3. Theorem.6. (Fatou) Let f H p, < p. Then lim f(z) = f(e it ), Γ α(e it ) z e it where f is the function as in Prop..9, and Γ α (e it ) is as in (.6). 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 Mf(e it ) = sup f(e i(t θ) ) dθ. <δ π 2δ We recall that M is weak-type (, ), that is, it satisfies the following estimate: there exists a constant C > such that for every λ >, {t : Mf(e it ) > λ} C λ f L (T). The next result shows the reason for our interest in the Hardy Littlewood maximal function. Proposition.7. Let g L ( D), α > a fixed parameter. Then there exists C α > such that for every e it D sup P(g)(z) C α Mg(e it ). z Γ α(e it ) Proof. The proof is divided in a few steps. (Step.) Assume first that z = re iη varies in a fixed compact subset of D, namely assume that r c α, with c α as in Remark.5 (ii). Notice that for all θ P r (e iθ ) = θ <δ r2 re iθ 2 r2 ( r) 2 2 r. (.8)

12 SPACES OF HOLOMORPHIC FUNCTIONS Then, for z = re iη as above, P(g)(z) = (g P r )(e iη ) g(e i(η θ) ) P r (e iθ ) dθ 2 g(e i(η θ) ) dθ c α = 2 c α C α Mg(e it ), g(e i(t θ) ) dθ where we have used the translation invariance of the integral on T. (Step 2.) Now we assume that z = re iη Γ α (e it ) with r < c α. By Remark.5 (ii) we know that We break the integral on T as follows: where η t 3α( r). P(g)(z) = (g P r )(e iη ) g(e i(η θ) ) P r (e iθ ) dθ N+ g(e i(η θ) ) P r (e iθ ) dθ, E j j= E = { θ T : θ 3α( r) } ; for j =,..., N, E j = { θ T : 2 j 3α( r) θ < 2 j 3α( r) } and N is chosen so that ; 2 N 3α( r) π/4 < 2 N 3α( r); E N+ = { θ T : π/4 < θ π }. Notice that, for θ E j (which implies in particular that θ π/4), arguing as in Remark.5 (ii) we have that Therefore, for θ E j we have that re iθ 2 j α( r). P r (e iθ ) = r2 re iθ 2 r 2 2 2j α 2 ( r) j α 2 ( r). Finally, for θ E N+, re iθ / 2 so that P r (e iθ ) 2( r 2 ) 2. (Step 3.) Having set up the above bounds, we finally come to the estimate in the statement.

13 2 M. M. PELOSO Using the inequalities in Step 2 we have that g(e i(η θ) ) P r (e iθ ) dθ 2 E r θ 3α( r) C α 6α( r) C α Mg(e it ), using the change of variables θ θ t + η. Next, using again the change of variables θ θ t + η, E N+ g(e i(η θ) ) P r (e iθ ) dθ 2 = 2 g(e i(η θ) ) dθ θ 3α( r) θ π θ π cmg(e it ). Finally, we estimate the central part of the integral: N j= E j g(e i(η θ) ) P r (e iθ ) dθ C α C α N j= N j= N j= g(e i(t θ) ) dθ g(e i(η θ) ) dθ g(e i(t θ) ) dθ 2 2j α 2 g(e i(η θ) ) dθ ( r) E j 2 2j α 2 g(e i(η θ) ) dθ ( r) θ 2 j 3α( r) 2 2j α 2 g(e i(t θ) ) dθ ( r) θ 2 j+ 3α( r) N j= N j= C α Mg(e it ). 2 j 2 j+ g(e i(t θ) ) dθ 3α( r) θ 2 j+ 3α( r) 2 j Mg(e it ) This concludes the proof. We are now ready to prove Thm..6. Proof of Thm..6. Since H H p for all p <, it suffices to deal with the case p <. Let f L p ( D) be as in Prop..9 (so that P( f) = f) and for a given γ > let g C( D) be such that f g L p γ; notice that here we need p <. By the solution of the Dirichlet problem we know that P(g)(z) g(e it ) as D z e it (even without the restriction z Γ α (e it )).

14 SPACES OF HOLOMORPHIC FUNCTIONS 3 We use Prop..7 and argue as in the proof of the Lebesgue differentiation theorem, { e it D : lim sup f(z) f(e it ) ε } Γ α(e it ) z e it { e it D : lim sup f(z) P(g)(z) ε/3 } Γ α(e it ) z e it + { e it D : lim sup P(g)(z) g(e it ) ε/3 } + { e it D : g(e it ) f(e it ) ε/3 } Γ α(e it ) z e it { e it D : C α M( f g)(e it ) ε/3 } ( 3 g(e it ) f(e it ) ) p + dt C ε g f L + C ε p g f p L C p Cε, ε p γp {e it D: g(e it ) f(e it ) ε/3} if we choose γ sufficiently small. This shows that { e it D : lim sup f(z) f(e it ) ε } Cε, Γ α(e it ) z e it which implies that a.e., and the statement follows. lim f(z) = f(e it ), Γ α(e it ) z e it ε.5. The zero sets of functions in H p. In this section we describe the zero sets of functions in the Hardy spaces, for the full range < 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 H p -functions that will have important consequences. We begin by recalling a result from complex analysis. Proposition.8. (Jensen s formula) Let r > and let f be holomorphic in a ngbh of D(, r). Let ζ,..., ζ N be the zeros of f in D(, r), counting multiplicity. Assume that f does not vanish at the origin and on the circle of radius r. Then log f() + log N j= r ζ j = log f(re iθ) dθ. 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.9. Let f H(D), f() and {ζ, ζ 2,... } its zeros counting multiplicity. Then, + log f() + log j= where, for t >, log + t = max(log t, ). ζ j sup <r< log + f(re iθ ) dθ,

15 4 M. M. PELOSO Proof. Clearly, for every < r <, N r log f() + log ζ j log + f(re iθ ) dθ j= Letting r we obtain the result. sup <r< log + f(re iθ ) dθ. Corollary.2. Let f H p, < p, f, and let {ζ, ζ 2,... } its zeros counting multiplicity. Then ( ζ j ) <. (.9) j= Proof. If f vanishes of order k at the origin, we consider f(z)/z k, which is still in H p and it is zero-free at the origin. Notice that Then we applying Cor..9 to obtain that log + f(re iθ ) f(re iθ ) p. + < log ζ j <, j= which implies that + j= ζ j converges; hence + j= ζ j converges. But this is equivalent to + j= ( ζ j ) <. We recall that for ζ D the function ϕ ζ (z) = z ζ ζz is a biholomorphic mapping of D onto itself. In this setting ϕ ζ is also called a Blaschke factor. Proposition.2. Let {ζ, ζ 2,... } be points in D satisfying condition (.9). Then the infinite product + j= converges uniformly on compact subsets of D. ζ j ζ j ϕ ζ j Proof. It suffices to show that, for r < be fixed and z r, the series + ζ j ζ j ϕ ζ j (z) < j= converges uniformly. A simple calculation now shows that + ζ j ζ j ϕ ζ j ( ζ j ) + ζ j z( ζ j ) ζ j (z) = ζ j ζ j z + r r ( ζ j ).

16 SPACES OF HOLOMORPHIC FUNCTIONS 5 By the Weierstrass M-test the convergence is uniform in z r, and the conclusion follows. Definition.22. Let < p, f H p and let {ζ, ζ 2,... } its zeros different from the origin, counting multiplicity. We set + B(z) = z k ζ j ζ j ϕ ζ j (z), j= where k 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 = F B is called the canonical factorization of f H p. Proposition.23. Let < p, f H p and let f = F B be its canonical factorization. Then F H p and F H p = f H p. Proof. Notice that B = z k + j= ζ j ζ j ϕ ζ j and every factor satisfies ζ j ζ j ϕ ζ j, so that B. Therefore, f(z) F (z) for all z D, so that f H p F H p. We set B N = N j= ζ j ζ j ϕ ζ j and F N = f/b N. We recall that ϕ ζj is defined and holomorphic in the disc D(, / ζ j ) which is strictly larger than the unit disc D and that ϕ ζj (e iη ) = for all η. Therefore, B N converge uniformly on D to a function B N as r, with B N = on D. Hence, Moreover, F N p H = lim M p(f p N, r) p r = lim r lim r inf θ B N (re iθ ) = f p H. p (f/b N )(re iθ ) p dθ (f)(re iθ ) p dθ M p (F, r) p = lim M p(f N, r) p N + lim F N p N + H = f p p H. p Therefore, F H p f H p, and the equality follows. Corollary.24. Let {ζ, ζ 2,... } be points in D satisfying the condition + j= ( ζ j ) < and let B = z k + j= ζ j ζ j ϕ ζ j be the corresponding Blaschke product. Then B H and B = a.e. on D. Proof. In the proof of Prop..23 we have already noticed that B H and that its boundary value function B is such that B. Since B H, its canonical factorization is B = B. By Prop..23 again it follows that = 2 H 2 = B 2 H 2 = B(e iθ ) 2 dθ.

17 6 M. M. PELOSO Together with the estimate B, it implies that B = a.e. on D. We conclude this part with the extension of Thm..6 to the range < p. Theorem.25. (Fatou) Let f H p, < p. Then lim f(z) =: f(e it ), Γ α(e it ) z e it where Γ α (e it ) is as in Def..6, exists a.e. The function f L p ( D) and f L p = f H p. 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..23, f admits the canonical factorization f = F B, and the function F H p and is zero-free. Therefore, F p/2 is well defined, holomorphic, and in H 2. By Thm..6, F p/2 admits boundary values F p/2 a.e. Therefore, also F admits non-tangential boundary values a.e. By Cor..24 B admits boundary values B and B = a.e. Hence f admits non-tangential boundary values f = ( F p/2 ) 2/p B. Finally, f p L p = = ( F p/2 ) 2/p B(e iθ ) p dθ F p/2 (e iθ ) 2 dθ = F p/2 2 H 2 = sup F (re iθ ) p dθ <r< = F p H = f p H p, by Prop..23 again. This proves the theorem..6. Boundary behaviour of functions in Hardy spaces. The goal of this section is to describe the boundary behaviour of functions in H p and to characterize them in terms of the Fourier coefficients of their boundary values. In the previous Section.3 we studied the classes of functions that are Poisson integrals of L p -functions on D. In this case, we had to restrict to the case p >. In the section we study the boundary behaviour of functions in H p, and we will be able to remove the restriction p >. In this analysis, we will consider the action of the Cauchy kernel, that is more closely related to H p -functions, since, in particular, it produces holomorphic functions, unlike the Poisson integral. In the previous section we learned that for < p, if f H p, then f converges pointwise non-tangentially a.e. to a boundary function that, with a momentary abuse of notation, we denote by f. We also know that f r f in L p ( D) when < p <. We now show that this holds true also in the case < p. Proposition.26. Let < p, f H p, and f be its boundary value function, as in Thm..25. Then f r f in L p as r.

18 SPACES OF HOLOMORPHIC FUNCTIONS 7 Proof. We first prove the case p =. Let f H and let f = BF its canonical factorization. Set g = F /2, h = BF /2. (.2) Then both g and h are in H 2 and g r g, h r h in L 2 ( D) as r. Then, f r f L = g r h r g h L (g r g)h r L + g(h r h) L g r g L 2 h r L 2 + g L 2 h r h L 2 as r. This proves the case p =. When < p < we proceed similarly, but we have to use an iteration process. Let f H p and define g and h as before. Then, g, h H 2p. If the results holds true in H 2p, that is, g r g, h r h in L 2p ( D) then the previous arguments applies (with a constant appearing in the triangular inequality) and therefore the results holds true in H p. Thus, after k log 2 /p steps we can prove the results for H p. Recall that we denote by C r (e iθ ) 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 (.3) (using the uniform convergence of the power series expansion of C r ) we have that C(g)(re iη ) = (C r g)(e iη ) = n= ĝ(n)r n e inη. Hence, we see that if g C( D) and ĝ(n) = when n <, then C(g) = P(g). More generally we have Lemma.27. Let g L p ( D), p. Then P(g) = C(g) if and only if ĝ(n) for all n <. Proof. It suffices to assume that g L ( D). We use the uniform convergence of the power series expansion of P r and C r. We have P(g)(re iη ) = (P r g)(e iη ) = g(e iθ ) r n e i(η θ) dθ Analogously, = n= ĝ(n)r n e inη. C(g)(re iη ) = (C r g)(e iη ) = = n= ĝ(n)r n e inη. n= g(e iθ ) It follows that P(g) = C(g) if and only if ĝ(n) for all n <. n= r n e i(η θ) dθ

19 8 M. M. PELOSO Theorem.28. Let f H p, p and let f be its boundary value function. ˆ f(n) = for n < and Then P( f) = C( f) = f. On the other hand, if p and g L p ( D) is such that ĝ(n) =, then f := P(g) = C(g) is in H p and f = g. Proof. Let f(z) = + n= a nz n be the power series expansion of f in D, f H p, p. Notice that it suffices to consider the case p =. Then f r (e iη ) = n= a n r n e inη and for all < r <, f r (n) = a n r n, by the uniform convergence. Since by Prop..26 f r f in L ( D), f r (n) f(n) as r, so that { a n for n f(n) = for n <. This clearly implies that f(re iη ) = n= ( = (P r f)(re iη ), ) f(e iθ )e inθ dθ r n e inη that is, f = P( f). The fact that C( f) = P( f) follows from the previous Lemma.27. When p <, he second part of the statement follows immediately from Lemma.27, Prop.. 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 state it for sake of completeness. Corollary.29. For p we denote by H p ( D) the subspace of L p ( D) of the functions that are boundary values of functions in H p. Then, H p ( D) = { g L p ( D) : ĝ(n) = for n < }. 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, endowed with the sup-norm. A(D) = H(D) C(D), Corollary.3. For p < the polynomials are dense in H p, while the closure of P in the sup-norm is A(D) and A(D) H. Furthermore, if f A(D), then f r f uniformly on D.

20 SPACES OF HOLOMORPHIC FUNCTIONS 9 Proof. We denote by σ N (g) the Fejér trigonometric polynomial of an integrable function g on T = D, that is, N ( σ N (g)(e iη ) = k ) ĝ(k)e ikη. N + k= N It is well known (see [Ka] or your Real Analysis lecture notes) that σ N (g) g in L p ( D), as N +, when p <. Now, let p <, f H p, f its boundary function. Then, σ N ( f)(e N ( iη ) = k ) ˆ f(k)e ikη, N + k= by Prop..28. Clearly σ N ( f) is the restriction to the boundary of a polynomial p N and it holds that p N = C(σ N ( f)). Then, we obtain a sequence of polynomials in D such that f p N H p = f p N L p ( D) = f σ N ( f) L p ( D) as N +. Thus, P is dense in H p, 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, sup f(z) p N (z) = sup f(e iη ) p N (e iη ) z D e iη D = sup e iη D f(e iη ) σ N (f)(e iη ) < ε choosing N large enough. Moreover, if f A(D), then f C( D), f = P( f) and therefore, f r = P r 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..22. By Cor..24 we know that B = a.e. on D. Let {ζ j } 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..7. The Cauchy Szegö projection. Definition.3. We define the Cauchy Szegö projection on L 2 ( D) as the Hilbert space orthogonal projection as ( + S(g)(e iη ) = S ĝ(n)e in( )) (e iη ) = ĝ(n)e inη. n= It is clear that S is a projection (that is, S 2 = S) and that it is self-adjoint (that is, S = S); hence an orthogonal projection. The next result follows from the previous lemma. Proposition.32. For g L 2 ( D) we have that S(g) = C(g). n=

21 2 M. M. PELOSO In particular we have obtained a characterization of S as (boundary limit of an) integral operator. Namely, S(g)(e iη ) = lim g(e i(η θ) ) dθ. (.2) r reiθ With an abuse of notation, we will simply write S(g)(e iη ) = g(e i(η θ) ) dθ. (.22) eiθ Notice however that the integral is not absolutely convergent since g L 2 and the function /( e iθ ) has a non-integrable singularity in θ =. The integral in (.22) has to be interpreted as in (.2). We wish to extend the action of S to all the spaces L p ( D) and prove its L p -boundedness, for the appropriate range of p s. The following result, of which we will not provide a complete proof, answers the question. Theorem.33. The operator S, initially defined on the dense subset L 2 L p, extends to an operator S : L p ( D) L p ( D) that is of weak-type (, ) and is bounded for < p <. Sketch of the proof. Recall the definition S(g) = lim r C r g and the relation between the Cauchy and Poisson kernels (.3). From the properties of the Poisson kernel we see that S (g) = lim r Re C r g extends to a bounded linear operator for p <. Thus, in order to prove the theorem, it suffices to prove the statement for (twice) the imaginary part of C r, where 2Im C r (e iη ) = [ i re iη ] re iη 2r sin η = re iη 2 = 2r sin η 2r cos η + r 2 (.23) = i n= sgn(n)r n e inη := Q r (e iη ), (.24) (where the last equality follows from the power series expansions of the right hand side of the first line in display). The kernel Q r is called the conjugate Poisson kernel, the reason being that for every < r <, Q r is the harmonic function conjugate to the real harmonic function P r on the unit disc. The family of functions on T {Q r } does not form a summability kernel, and the properties of the mapping g lim r Q r g are substantially different from the ones we obtain when Q r is replaced by the Poisson kernel. We give a name to the operator g lim r Q r g, we write Q(g) = lim r Q r g, (.25)

22 SPACES OF HOLOMORPHIC FUNCTIONS 2 and call Q the Hilbert transform 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. Using the fact that the (convolution) kernels Q r are odd and therefore they have built in some cancellation, it is indeed possible to show that Q is L p -bounded for < p <. Notice that the pointwise limit of the kernels lim Q r(e iη ) = sin η r cos η = cot(η/2) := Q(eiη ) and that Q L ( D). It turns out (and it is not hard to see) that, Q(g)(e iη ) = p.v.(q g)(e iη ) =: lim g(e i(η θ) )Q(θ) dθ. ε + ε< θ π We will not prove that Q is weak-type (, ) and bounded on L p, < p <, but refer to [Ka], Ch. 3, e.g., or also [Gr]. It can be shown that the Cauchy Szegö projection is not bounded from L ( D) to itself. 4 Thm..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 =, i.e. we set v() =. In fact, it is not difficult to see that the boundedness of the Cauchy Szegö projection on L p ( D) is equivalent to the boundedness of the harmonic conjugate operator T. Corollary.34. Let u be a real harmonic function satisfying the H p -growth condition (.4), < p <, and let v be its harmonic conjugate. Then, also v satisfies the H p -growth condition (.4) and the mapping T : ũ ṽ is bounded on L p ( D), < p <. Proof. By Prop..9 u = P r ũ, with ũ L p ( D). Now, v r = Q r ũ and ṽ = Q(ũ). By the previous theorem ṽ L p = Q(ũ) L p C ũ L p, and we are done. 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 Exercise.

23 22 M. M. PELOSO 2. Bergman spaces on the unit disc 2.. 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 C z > such that for all f H we have f(z) C z f H. (2.) This implies that the linear functional L z (f) = f(z), f H, is bounded. By the Riesz Fisher theorem, there exists k z H such that, for all f H we have f, k z H = f(z). We define a kernel function K : Ω Ω C by setting Such kernel K is called the reproducing kernel for H. Proposition 2.. The kernel K satisfies the following properties: (i) for all f H and z Ω we have f(z) = f, K(, z) H ; (ii) for all z, w Ω, K(w, z) = K(z, w). K(w, z) = k z (w). (2.2) Proof. Property (i) is obvious by construction. For (ii), using the reproducing property of K we have and we are done. K(w, z) = k z (w) = k z, K(, w) H = K(, w), k z H = K(z, w) The following result 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) f, H(, z) H = 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 as we wished to show. H(z, w) = H(, w), K(, z) H = K(, z), H(, w) H = K(w, z) = K(z, w), Notice that (i) and (ii) 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.. In this notes we will be concerned only with spaces H for which (2.) holds and that consist of holomorphic functions.

24 SPACES OF HOLOMORPHIC FUNCTIONS 23 It will also always be the case that, for each compact subset E of Ω there exists C = C E > such that for all f H we have sup f(z) C f H. (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 {ϕ j } be an orthonormal basis for H. Then the series j= ϕ j (z)ϕ j (w) 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 l 2, for z E we have ( + ϕ j (z) 2) /2 {ϕj = (z)} = sup a l 2 j ϕ j (z) Therefore, j= sup z E = sup f(z) f H = sup f H = z E C E. sup f(z) {a j } l 2 = ( + ϕ j (z) 2) /2 CE, j= so that the Cauchy Schwarz inequality gives that j= ϕ j (z)ϕ j (w) CE 2, and the convergence is uniform on E E. Next, for z Ω fixed, the previous argument also shows that {ϕ j (z)} l 2, so that + j= ϕ j(z)ϕ j converges in H to a function h z. By the theory of Fourier series in H we have that, for all f H f, h z H = j= f, ϕ j H ϕ j (z) = f(z). Therefore, the function H(w, z) = h z (w) = + j= ϕ j(z)ϕ j (w) satisfies the conditions of Lemma 2.2 and hence coincides with the reproducing kernel K(w, z). Notice that the argument above also shows that the series + j= ϕ j(z)ϕ j converges in the H-norm, for all z Ω fixed. An interesting consequence of this proposition is the following result that shows that the reproducing kernel K satisfies an extremal property. j=

25 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, f H = Proof. Let {ϕ k } be an orthonormal basis for H. We have as claimed. K(z, z) = k ( = ϕ k (z) 2 sup {a k } l 2, {a k } l 2 = = sup f(z) 2, f H, f H = k a k ϕ k (z) ) 2 Notice that the reproducing kernel K satisfies the condition K(z, z) and in fact K(z, z) > unless f(z) = 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 H 2 (D) is a Hilbert space with reproducing kernel, with inner product f, g H 2 = f, g L 2 ( D). Its reproducing kernel is the Cauchy Szegö kernel K(z, w) = wz. Proof. We showed that f H 2 = f L 2 ( D). Therefore, H 2 is a Hilbert space with inner product f, g H 2 = f, g L 2 ( D), for f, g H 2. (We also showed that we can describe the inner product in H 2 by setting f, g H 2 = {a n }, {b n } l 2, where f(z) = + n= a nz n and g = + n= b nz n. However, we will not use this characterization at this stage.) We now show that the point evalutions are bounded functionals on H 2. Let z D and let z < r <. By the Cauchy formula f(z) = f(ζ) i ζ =r ζ z dζ = f(re iθ ) re iθ z reiθ dθ, (2.4) and by the Cauchy Schwartz inequality we have ( f(z) r r z f H 2. ) /2 ( f(re iθ ) 2 dθ This shows that H 2 is a Hilbert space with reproducing kernel. r 2 re iθ z 2 dθ ) /2

26 SPACES OF HOLOMORPHIC FUNCTIONS 25 Consider now the kernel K(z, w) = wz. It is immediate to see that K(, w) H2 for every w D fixed. Moreover, f, K(, z) H 2 = f, K(, z) L 2 ( D) = f(e iθ ) e iθ z dθ = lim r r = f(z), f(re iθ ) re iθ z reiθ dθ by (2.4). The conclusion now follows from Lemma The Bergman spaces. Let Ω be a domain in C. We denote by da the Lebesgue measure on C, which of course is identified with real euclidean space R 2, and we write E to denote the Lebesgue measure of a measurable set E. For < p < we denote by L p (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 < p we define the Bergman space A p as A p (Ω) = L p (Ω) H(Ω). Proposition 2.7. Let Ω C be a domain and let < p. Then the space A p (Ω) is a closed subspace of L p (Ω). When p = 2, A 2 is a Hilbert space with reproducing kernel. Proof. We show that the functions in the Bergman space A p satisfy (2.3). Let E be a compact subset of Ω and let δ be half the distance of E from Ω, i.e. δ = 2dist(E, Ω). For every z E, D(z, δ) Ω. From the mean value property f(z) = f(z + re iθ ) dθ, (valid for every r > such that D(z, r) Ω) multiplying by r both sides and integrating in r [, δ] we obtain the identity 5 f(z) = f(w) da(w). D(z, δ) D(z,δ) Therefore, for every z E, using Hölder s inequality, for p, f(z) p f(w) p da(w) (2.5) D(z, δ) D(z,δ) D(z, δ) f p L = p πδ 2 f p L. p We remark that the above inequality (2.5) holds also for < p <, 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) C E f L p. (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.

27 26 M. M. PELOSO This inequality show that if {f n } is a sequence of holomorphic functions converging in the L p -norm to a function f, then {f n } converges to f also uniformly on compact subsets. This proves that A p is closed < p <. The case p = is obvious. When p = 2 (2.6) is exactly (2.3) and this concludes the proof. Notice that the constant C E in inequality (2.6), depends on the distance of the set E from the boundary of Ω, and, more precisely C E δ 2/p, as δ = 2dist(E, Ω). Definition 2.8. Given a domain Ω C, let K = K Ω be its Bergman kernel, and for z Ω write k z = K(, z). We define the Bergman projection on Ω the mapping B defined on L 2 (Ω) as (Bf)(z) = f, k z = f(w)k z (w) da(w) Ω = f(w)k(z, w) da(w). Ω Proposition 2.9. The Bergman projection is the Hilbert space orthogonal projection of L 2 (Ω) onto its closed subspace A 2 (Ω). Proof. First of all, notice that Bf is well defined for all f L 2, since for all z Ω, k z A 2 L 2, and the L 2 -pairing f, k z makes sense (and converges). Next, we use the characterization of the Bergman kernel given by Prop. 2.3, that is, k z = j ϕ j(z)ϕ j, where {ϕ j } is an orthonormal basis for A 2, where the convergence is in the L 2 -norm, for all z Ω fixed. Therefore, (Bf)(z) = f, k z = f, ϕ j (z)ϕ j j = j ϕ j (z) f, ϕ j = j a j ϕ j (z), where a j = f, ϕ j is the j-th Fourier coefficient of f, w.r.t. the orthonormal system {ϕ j }, (which is not complete in L 2 ). Clearly, {a j } l 2 and j a j 2 f 2 L 2, by Bessel s inequality. Hence, the series j f, ϕ j ϕ j converges in L 2 and therefore Bf L 2 f L 2. It is also clear that Bf A 2 since A 2 is closed, so that B : L 2 A 2 and it is bounded. It remains to show that B is the orthogonal projection, that is, that B 2 = B and B = B. The former identity follows by the reproducing property of the Bergman kernel when acting on functions in A 2, while the latter one follows from the hermitian symmetry property: for f, g L 2 Bf, g = f, ϕ j ϕ j, g = f, ϕ j ϕ j, g j j = f, j g, ϕ j ϕ j = f, Bg. This shows that B = B and we are done.

28 SPACES OF HOLOMORPHIC FUNCTIONS 27 Remark 2.. We remark that, up to this point, we have not discussed whether, on a given domain Ω C, the Bergman space A 2 (Ω) is trivial, i.e. reduces to {}, 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 A 2 (C) = {}, 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 approximation theorem 6 shows that, if Ω is a bounded domain, the space of holomorphic polynomials is (contained and) dense in A 2 (Ω). We conclude this remark by observing that, as a consequence of the extremal property in Cor. 2.4 we have that, it Ω Ω are domains in C and K, K denote their Bergman kernels resp., then for all z Ω. K (z, z) K(z, z) Before proceeding any further, we present an example of Bergman space and kernel. Proposition 2.. Let D be the unit disc. Then, the set of functions {ϕ k }, where ϕ k = (k + )/πz k, k =,, 2,..., is a complete orthonormal system in A 2 (D) and the Bergman kernel has the expression K(z, w) = π ( zw) 2. Proof. It is clear that {ϕ k } is an orthonormal system: (j + )(k + ) ϕ j, ϕ k = π = δ jk 2(k + ) = δ jk, r 2k+ dr e i(j k)θ dθ r j+k+ dr where δ jk 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 A 2 (D). For, let f A 2 (D) be orthogonal to all the z k s and let f(z) = n a nz n be its power series expansion in D. By integrating over the closed disc z r for some fixed r < and interchanging the summation and integration orders, we obtain that a n = for all n; hence f =. Thus, {ϕ k } is a complete orthonormal system for A 2 (D) and, by Prop. 2.3 the Bergman kernel is given by the sum k ϕ k(z)ϕ k (w), that we easily compute: k k + π zk w k = π = π (k + )(zw) k k ( zw) 2, 6 True? Recall? Reference?

29 28 M. M. PELOSO as we wished to show. 7 Notice the Bergman kernel for the unit disc K(z, w) satisfies the condition K(z, w) and K(z, z) > for all z, w D Biholomorphic invariance. One of the main features of the Bergman kernel is its invariance under conformal mappings. This invariance is the content of the next proposition. Proposition 2.2. Let Ω, Ω 2 C be domains, Φ : Ω Ω 2 be a biholomorphic mapping, and let K Ω, B and K Ω2, B 2 be the Bergman kernels and projections of Ω and Ω 2, resp. Then we have and, for f A 2 (Ω 2 ) we K Ω (z, w) = Φ (z)k Ω2 ( Φ(z), Φ(w) ) Φ (w), (2.7) ) B (Φ (f Φ) = Φ ( ) (B 2 f) Φ. (2.8) Proof. We begin by observing that if Φ : Ω Ω 2 is a biholomorphic mapping, then the determinant jacobian of Φ thought as a mapping between domains in R 2 equals Φ (z) 2. For, if we write Φ = u + iv, u, v real functions, then Φ is identified with the mapping Then so that, using the Cauchy-Riemann equations, Ω z = x + iy (x, y) ( u(x, y), v(x, y) ). ( ) x u Jac Φ = y u, x v y v det(jac Φ) = ( x u) 2 + ( x v) 2 = x Φ 2 = Φ 2. This identity implies that the mapping f Φ (f Φ) is an isometric isomorphism of A 2 (Ω 2 ) onto A 2 (Ω ). Notice also the analogous statements hold true with Φ in place of Φ and with the roles of Ω and Ω 2 switched. For z, w Ω we set H(z, w) = Φ (z)k Ω2 ( Φ(z), Φ(w) ) Φ (w), and easily see that H satisfies the conditions in Lemma 2.2, thus implying that H coincides with the Bergman kernel on Ω. 8 7 Exercise. Using the residue theorem, show that for every f A 2 (D) and z D we have the identity f(z) = ZZ f(w) π ( zw) da(w). 2 D Once you have completed your argument, explain where you use the assumption f A 2 (D). Can this assumption be relaxed? 8 Exercise.