Fatou’s theorem, Bergman spaces, and Hardy spaces on the circle

Jordan Bell
April 3, 2014

1 Introduction

In this note I am writing out proofs of some facts about Fourier series, Bergman spaces, and Hardy spaces. §§1–3 follow the presentation in Stein and Shakarchi’s Real Analysis and Fourier Analysis. The questions in Halmos’s Hilbert Space Problem Book that deal with Hardy spaces are: §§24–35, 67, 116–117, 124–125, 127, 193–199, and I present solutions to some of these in §§4–5, on Bergman spaces, and §6, on Hardy spaces.

2 Poisson kernel

Let 𝕋=/2π. Let


If fL1(𝕋), let




One checks that Pr is an approximation to the identity, which implies that for fL1(𝕋), for almost all θ𝕋 we have (f*Pr)(θ)f(θ) as r1-.11 1 If f is continuous, then f*Pr converges to f uniformly on 𝕋 as r1-. This is proved for example in Lang’s Complex Analysis, fourth ed., chapter VIII, §5.

For fL1(𝕋), for any θ we have


hence by the dominated convergence theorem we have

limN|n|Nr|n|02πf(t)ein(θ-t)𝑑t = limN02πf(t)|n|Nr|n|ein(θ-t)dt
= 02πf(t)nr|n|ein(θ-t)dt,

and so

(f*Pr)(θ) = 12π02πf(t)Pr(θ-t)𝑑t
= 12π02πf(t)nr|n|ein(θ-t)dt
= nr|n|12π02πf(t)ein(θ-t)𝑑t
= nr|n|einθf^(n).

3 Harmonic functions

For fL1(𝕋), define uf on |z|<1 by


In polar coordinates, the Laplacian is



(Δuf)(reiθ) = Δ(n<0r-neinθf^(n)+n0rneinθf^(n))
= n<0f^(n)Δ(r-neinθ)+n0f^(n)Δ(rneinθ)
= n<0f^(n)0+n0f^(n)0
= 0.

Hence uf is harmonic on the open unit disc.

4 Fatou’s theorem

Let D={z:|z|<1}. If F:D is holomorphic, let it have the power series


By the Cauchy integral formula, for n0 and for any 0<r<1, γr(θ)=reiθ, we have

F(n)(0) = n!2πiγrF(ζ)ζn+1𝑑ζ
= n!2πi02πF(reiθ)(reiθ)n+1rieiθ𝑑θ
= n!2πrn02πF(reiθ)e-inθ𝑑θ.

Hence, for n0 and for 0<r<1,


On the other hand, for n<0, we have


and because F(ζ)zn+1 is holomorphic on D for n<0, by the residue theorem the right-hand side of the above equation is equal to 0. Hence, for n<0 and for 0<r<1,


Let F:D be holomorphic, and suppose there is some M such that |F(z)|M for all zD. For 0<r<1, define fr:𝕋 by fr(θ)=F(reiθ). From our above work, we have


For 0<r<1, note that frL2frLM, so, by Parseval’s identity,


On the other hand,


It follows that


Define fL2(𝕋) by


this defines an element of L2(𝕋) if and only if n|f^(n)|2<, and indeed


As fL2(𝕋), fL1(𝕋). Then by our work in §2, for almost all θ𝕋 we have


which means here that for almost all θ𝕋,


Thus, for almost all θ𝕋,


In words, we have proved that if F is a bounded holomorphic function on the unit disc, then it has radial limits at almost every angle. This is Fatou’s theorem.

5 Bergman spaces

This section somewhat follows Problem 24 of Halmos. Let μ be Lebesgue measure on D. dμ(z)=dxdy=dzdz¯-2i.

If U is a nonempty bounded open subset of and 1p<, let Ap(U) denote the set of functions f:U that are holomorphic and that satisfy


and let A(U) denote the set of functions f:U that are holomorphic and that satisfy


It is apparent that Ap(U) is a vector space over . By Minkowski’s inequality, Ap(U) is a norm, and thus Ap(U) is a normed space. If pq then by Jensen’s inequality we have


and so


Ap(U) is called a Bergman space. It is not apparent that it is a complete metric space. We show this using the following lemmas. We use the following lemma to prove the lemma after it, and use that lemma to prove the theorem.

Lemma 1.

If z0, R>0, and fA1(D(z0,R)), then


Put Fn(z)=k=0nak(z-z0)k, with ak=f(k)(z0)k!. For 0<r<R, define


We have Fn-fr0 as n. Then,

|D(z0,r)f(z)𝑑μ(z)-D(z0,r)Fn(z)𝑑μ(z)| = |D(z0,r)f(z)-Fn(z)dμ(z)|
= f-Fnrπr2,

which tends to 0 as n. Thus

D(z0,r)f(z)𝑑μ(z) = limnD(z0,r)Fn(z)𝑑μ(z)
= limnD(z0,r)k=0nak(z-z0)kdμ(z)
= limnk=0nakD(z0,r)(z-z0)k𝑑μ(z)
= limnk=0nakD(0,r)zk𝑑μ(z).

For k1, using polar coordinates we have

D(0,r)zk𝑑μ(z) = 0r02π(ρeiθ)kρ𝑑θ𝑑ρ
= 0r02πρk+1eikθ𝑑θ𝑑ρ
= 0rρk+10k𝑑ρ
= 0.


D(z0,r)f(z)𝑑μ(z) = limna0πr2
= a0πr2.

That is, for each 0<r<R we have

f(z0)=1πr2D(z0,r)f(z)𝑑μ(z). (1)

Because fL1(D(z0,R)),


Thus, taking the limit as rR of (1), we obtain


If z0 and S, denote


and for z0U, let


This is the radius of the largest open disc centered at z0 that is contained in U (it is equal to the union of all open discs centered at z0 that are contained in U, and thus makes sense). As U is open, r(z0)>0, and as U is bounded, r(z0)<.

Lemma 2.

If 1p, z0U, and fAp(U), then


As fAp(U) we have fAp(D(z0,r(z0)))A1(D(z0,r(z0))). Using Lemma 1 and Hölder’s inequality, we get, with 1p+1q=1 (q is infinite if p=1),

|f(z0)| = |1πr(z0)2D(z0,r(z0))f(z)𝑑μ(z)|
= 1πr(z0)2(πr(z0)2)1/qfAp(D(z0,r(z0)))
= 1πr(z0)2(πr(z0)2)1-1pfAp(U)
= (1πr(z0)2)1/pfAp(U).

Now we prove that Ap(U) is a complete metric space, showing that it is a Banach space.

Theorem 3.

If 1p, then Ap(U) is a Banach space.


Suppose that fnAp(U) is a Cauchy sequence. We have to show that there is some fAp(U) such that fnf in Ap(U). The space H(U) of holomorphic functions on U is a Fréchet space: there is an increasing sequence of compact sets KiU whose union is U, and the pKi seminorms on H(U) are the supremum of a function on Ki. (See Henri Cartan, Elementary Theory of Analytic Functions of One or Several Complex Variables, §V.1.3.) For each of these compact sets Ki, let ri be the distance between Ki and U, which are both compact sets. If z0Ki then r(z0)ri. Thus if z0Ki and gAp(U), using Lemma 2 we get


From this and the fact that fn-fmAp(U)0 as m,n, we get that


That is, fn is a Cauchy sequence in each of the seminorms pKi, and as H(U) is a Fréchet space it follows that there is some fH(U) such that fnf in H(U). In particular, for all z0U we have fn(z0)f(z0) as n (because each z0 is included in one of the compact sets Ki, on which the fn converge uniformly to f and hence pointwise to f).

On the other hand, Lp(U) is a Banach space, and hence there is some gLp(U) such that fn-gLp(U)0 as n. This implies that there is some subsequence fa(n) such that for almost all z0U, fa(n)(z0)g(z0). Thus, for almost all z0U we have f(z0)=g(z0). Therefore, in Lp(U) we have f=g and so


6 Inner products

In this section we follow Problem 25 of Halmos. In this section we restrict our attention to the Bergman space A2(D), where D is the open unit disc, on which we define the inner product


As f,f=fA2(D)2, it follows that A2(D) is a Hilbert space with this inner product. If we have a Hilbert space we would like to find an explicit orthonormal basis.

Theorem 4.

If n0 and zD, define en:D by


Then en are an orthonormal basis for A2(D).


If E is a subset of a Hilbert space and vH, we write vE if v,e=0 for all eE. If E is an orthonormal set in H, E is an orthonormal basis if and only if vE implies that v=0. This is proved in John B. Conway, A Course in Functional Analysis, second ed., p. 16, Theorem 4.13. For nm,

en,em = Dn+1πznm+1πz¯m𝑑μ(z)
= (n+1)(m+1)πDznz¯m𝑑μ(z)
= (n+1)(m+1)π0102π(reiθ)n(re-iθ)mr𝑑θ𝑑r
= (n+1)(m+1)π0102πrn+m+1eiθ(n-m)𝑑θ𝑑r
= (n+1)(m+1)π0102πrn+m+1eiθ(n-m)𝑑θ𝑑r
= 0,


en,en = n+1π0102πr2n+1𝑑θ𝑑r
= 2(n+1)01r2n+1𝑑r
= 1.

Therefore en is an orthonormal set. Hence, to show that it is an orthonormal basis for A2(D) we have to show that if f,en=0 for all n0 then f=0.

For 0<r<1, let Dr be the open disc centered at 0 of radius r, and let gr=sup|z|r|g(z)|. Let f(z)=n=0anzn, and for each 0<r<1 this power series converges uniformly in Dr. Then

Drfem*𝑑μ = Drn=0anznz¯mdμ(z)
= n=0anDrznz¯m𝑑μ(z)
= n=0an0r02πρn+m+1eiθ(n-m)𝑑θ𝑑ρ
= n=0an0rρn+m+12πδn,m𝑑ρ
= 2πam0rρ2m+1𝑑ρ
= 2πamr2m+22m+2

One checks that fem*A1(D), and hence




As f,em=0 for each m, this gives us that am=0 for all m and hence f=0. This shows that en is an orthonormal basis for A2(D). ∎

Steven G. Krantz, Geometric Function Theory: Explorations in Complex Analysis, p. 9, §1.2, writes about the Bergman space A2(Ω), where Ω is a connected open subset of , not necessarily bounded.

7 Hardy spaces

In a Hilbert space H, if Sα,αI are subsets of H, let αISα denote the closure in H of αISα. Thus, to say that a set {vα} is an orthonormal basis for a Hilbert space H is to say that {vα} is orthonormal and that αI{vα}=H.

Let S1={z:|z|=1}, and let μ be normalized arc length, so that μ(S1)=1. Define en:S1 by en(z)=zn, for n. It is a fact that en,n are an orthonormal basis for the Hilbert space L2(S1), with inner product


We define the Hardy space H2(S1) to be n0{en}. As it is a closed subspace of the Hilbert space L2(S1), it is itself a Hilbert space. For fL2(S1), we denote f*(z)=f(z)¯.

The following is Problem 26 of Halmos. Note f*(z)=f(z)¯.

Theorem 5.

If fH2(S1) and f*=f, then f is constant.


If gnL2(S1) and gngL2(S1), then


Thus gg* is continuous L2(S1)L2(S1).

If gL2(S1), then, as en,n is an orthonormal basis for L2(S1), we have g=limN|n|Ng,enen, and so, as en*=e-n,


Therefore if n then

g*,en=g,e-n¯. (2)

For n>0,


the first equality is because f*=f, the second equality is by what we showed for any element of L2(S1), and the third equality is because fH2(S1). It follows that fspan{e0}, and thus that f is constant. ∎

If gL2(S1), define RegL2(S1) by


and RegL2(S1) by


g=ng,enen and, by (2), g*=ng*,enen=ng,e-n¯en, so



Img=12i(ng,enen-ng,e-n¯en)=12in(g,en-g,e-n¯)en. (3)

g=Reg+iImg, and we have (Reg)*=Reg and (Img)*=Img; that is, both Reg and Img are real valued, like how the real and imaginary parts of a complex number are both real numbers.

The following is Problem 35 of Halmos. In words, it states that a real valued L2 function u has a corresponding real valued L2 function v (made unique by demanding that v have 0 constant term) such that the sum u+iv is an element of the Hardy space H2. This v is called the Hilbert transform of u. This is analogous to how if u is harmonic on an open subset Ω of 2, then g(x+iy)=ux(x,y)-iuy(x,y) satisfies the Cauchy-Riemann equations at every point in Ω and hence is holomorphic on Ω. Since g is holomorphic on Ω, for every z0Ω there is some open neighborhood of z on which g has a primitive f (g might not have a primitive defined on Ω, e.g. g(z)=1z on Ω={0}), and there is a constant c such that u(x,y)=Ref(x+iy)+c for all (x,y) in this neighborhood. u and v(x,y)=Imf(x+iy)+c are called harmonic conjugates.

Theorem 6.

If uL2(S1) and u*=u, then there is a unique vL2(S1) such that v*=v, v,e0=0, and u+ivH2(S1).


Define D:{uL2(S1):u*=u}H2(D) by


As |a+b|22|a|2+2|b|2, and using Parseval’s identity,

n0|Du,en|2 = |u,e0|2+n>0|u,en+u,e-n¯|2
= |u,e0|2+2n0|u,en|2

This is finite, hence DuH2(S1).

For any gL2(S1) and n, by (2) we have g*,en=g,e-n¯. As u*=u, if n then u,en=u,e-n¯. Using this, we check that ReDu=u.

Put v=ImDu, hence Du=u+iv. u,e0=u,e0¯ gives Du,e0=Du,e0¯, and applying this and (3) we get v,e0=0. Thus v satisfies the conditions v*=v, v,e0=0, and u+ivH2(S1). We are not obliged to do so, but let’s write out the Fourier coefficients of v. If n then, using u,en=u,e-n¯,

v,en = ImDu,en
= 12i(Du,en-Du,e-n¯)
= {0n=012i(u,en+u,e-n¯)n>0-12i(u,e-n+u,en¯)¯n<0
= {0n=012i(u,en+u,e-n¯)n>0-12i(u,en+u,e-n¯)n<0
= {0n=01iu,enn>0-1iu,enn<0.

Thus v,en=-isgn(n)u,en.

If fH2(S1), then, as Ref,en=f,en+f,e-n¯2,

DRef,en = {f,e0+f,e0¯2n=0f,en+f,e-n¯2+f,e-n¯+f,en2n>00n<0.
= {f,e0+f,e0¯2n=0f,enn>00n<0
= {f,e0+f,e0¯2n=0f,enn0


f-DRef,en = {f,e0-f,e0¯2n=00n0
= {iImf,e0n=00n0