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

Jordan Bell [email protected] Department of Mathematics, University of Toronto April 3, 2014

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.

1 Poisson kernel

Let T = R/2πZ. Let  Z 2π 1/p 1 p kfkLp = |f(t)| dt . 2π 0

If f ∈ L1(T), let 1 Z 2π fˆ(k) = f(t)e−iktdt. 2π 0 Define X 1 − r2 P (t) = r|n|eint = , 0 ≤ r < 1, t ∈ . r 1 − 2r cos t + r2 T n∈Z

One checks that Pr is an approximation to the identity, which implies that for 1 − 1 f ∈ L (T), for almost all θ ∈ T we have (f ∗ Pr)(θ) → f(θ) as r → 1 . For f ∈ L1(T), for any θ we have

X |n| in(θ−t) 1 + r f(t) r e ≤ kfk 1 , 1 − r L |n|≤N L1 1 − If f is continuous, then f ∗ Pr converges to f uniformly on T as r → 1 . This is proved for example in Lang’s Complex Analysis, fourth ed., chapter VIII, §5.

1 hence by the dominated convergence theorem we have

X Z 2π Z 2π X lim r|n| f(t)ein(θ−t)dt = lim f(t) r|n|ein(θ−t)dt N→∞ N→∞ |n|≤N 0 0 |n|≤N Z 2π X = f(t) r|n|ein(θ−t)dt, 0 n∈Z and so 1 Z 2π (f ∗ Pr)(θ) = f(t)Pr(θ − t)dt 2π 0 1 Z 2π X = f(t) r|n|ein(θ−t)dt 2π 0 n∈Z X 1 Z 2π = r|n| f(t)ein(θ−t)dt 2π 0 n∈Z X = r|n|einθfˆ(n). n∈Z 2 Harmonic functions

1 For f ∈ L (T), define uf on |z| < 1 by

iθ uf (re ) = (f ∗ Pr)(θ).

In polar coordinates, the Laplacian is

∂2 1 ∂ 1 ∂2 ∆ = + + . ∂r2 r ∂r r2 ∂θ2 Then   iθ X −n inθ ˆ X n inθ ˆ (∆uf )(re ) = ∆  r e f(n) + r e f(n) n<0 n≥0 X X = fˆ(n)∆ r−neinθ
+ fˆ(n)∆ rneinθ
n<0 n≥0 X X = fˆ(n) · 0 + fˆ(n) · 0 n<0 n≥0 = 0.

Hence uf is harmonic on the open unit disc.

2 3 Fatou’s theorem

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

X n F (z) = anz , an ∈ C. n≥0

iθ By the Cauchy integral formula, for n ≥ 0 and for any 0 < r < 1, γr(θ) = re , we have Z (n) n! F (ζ) F (0) = n+1 dζ 2πi γr ζ Z 2π iθ n! F (re ) iθ = iθ n+1 rie dθ 2πi 0 (re ) Z 2π n! iθ −inθ = n F (re )e dθ. 2πr 0 Hence, for n ≥ 0 and for 0 < r < 1,

Z 2π 1 iθ −inθ n F (re )e dθ = anr . 2π 0 On the other hand, for n < 0, we have

Z 2π Z n! iθ −inθ n! F (ζ) n F (re )e dθ = n+1 dζ, 2πr 0 2πi γr ζ

F (ζ) and because 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, 1 Z 2π F (reiθ)e−inθdθ = 0. 2π 0 Let F : D → C be holomorphic, and suppose there is some M such that iθ |F (z)| ≤ M for all z ∈ D. For 0 < r < 1, define fr : T → C by fr(θ) = F (re ). From our above work, we have

( n anr n ≥ 0, fbr(n) = 0 n < 0.

For 0 < r < 1, note that kfrkL2 ≤ kfrkL∞ ≤ M, so, by Parseval’s identity,

X 2 2 |fbr(n)| ≤ M . n∈Z On the other hand, X 2 X 2 2n |fbr(n)| = |an| r . n∈Z n≥0

3 It follows that X 2 2 |an| ≤ M . n≥0 Define f ∈ L2(T) by ( a n ≥ 0, fˆ(n) = n 0 n < 0; this defines an element of L2( ) if and only if P |fˆ(n)|2 < ∞, and indeed T n∈Z X |fˆ(n)|2 ≤ M 2. n∈Z

As f ∈ L2(T), f ∈ L1(T). Then by our work in §1, for almost all θ ∈ T we have X lim r|n|einθfˆ(n) = f(θ), r→1− n∈Z which means here that for almost all θ ∈ T,

X n inθ lim anr e = f(θ). r→1− n≥0

Thus, for almost all θ ∈ T, lim F (reiθ) = f(θ). r→1− 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.

4 Bergman spaces

This section somewhat follows Problem 24 of Halmos. Let µ be Lebesgue mea- dz∧dz sure on D. dµ(z) = dx ∧ dy = −2i . If U is a nonempty bounded open subset of C and 1 ≤ p < ∞, let Ap(U) denote the set of functions f : U → C that are holomorphic and that satisfy Z 1/p p kfkAp(U) = |f(z)| dµ(z) < ∞, U and let A∞(U) denote the set of functions f : U → C that are holomorphic and that satisfy

kfkA∞(U) = sup |f(z)| < ∞. z∈U It is apparent that Ap(U) is a vector space over C. By Minkowski’s inequality, p k·kAp(U) is a norm, and thus A (U) is a normed space. If p ≤ q then by Jensen’s inequality we have 1 1 p − q kfkAp(U) ≤ µ(D) kfkAq (U) ,

4 and so Aq(U) ⊆ Ap(U). 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. 1 Lemma 1. If z0 ∈ C, R > 0, and f ∈ A (D(z0,R)), then 1 Z f(z0) = 2 f(z)dµ(z). πR D(z0,R)

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

kgkr = sup |g(z)|. |z−z0|≤r

We have kFn − fkr → 0 as n → ∞. Then,

Z Z Z

f(z)dµ(z) − Fn(z)dµ(z) = f(z) − Fn(z)dµ(z) D(z0,r) D(z0,r) D(z0,r) Z ≤ |f(z) − Fn(z)|dµ(z) D(z0,r) Z ≤ kf − Fnkr dµ(z) D(z0,r) 2 = kf − Fnkr · πr , which tends to 0 as n → ∞. Thus Z Z f(z)dµ(z) = lim Fn(z)dµ(z) n→∞ D(z0,r) D(z0,r) Z n X k = lim ak(z − z0) dµ(z) n→∞ D(z0,r) k=0 n Z X k = lim ak (z − z0) dµ(z) n→∞ k=0 D(z0,r) n Z X k = lim ak z dµ(z). n→∞ k=0 D(0,r) For k ≥ 1, using polar coordinates we have Z Z r Z 2π zkdµ(z) = (ρeiθ)kρdθdρ D(0,r) 0 0 Z r Z 2π = ρk+1eikθdθdρ 0 0 Z r 0 = ρk+1 · dρ 0 k = 0.

5 Therefore Z 2 f(z)dµ(z) = lim a0 · πr n→∞ D(z0,r) 2 = a0 · πr .

That is, for each 0 < r < R we have 1 Z f(z0) = 2 f(z)dµ(z). (1) πr D(z0,r)

1 Because f ∈ L (D(z0,R)), Z Z lim f(z)dµ(z) = f(z)dµ(z). r→R D(z0,r) D(z0,R)

Thus, taking the limit as r → R of (1), we obtain

1 Z f(z0) = 2 f(z)dµ(z). πR D(z0,R)

If z0 ∈ C and S ⊆ C, denote

d(z0,S) = inf |z0 − z|, z∈S and for z0 ∈ U, let r(z0) = d(z0, ∂U).

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) < ∞.

p Lemma 2. If 1 ≤ p ≤ ∞, z0 ∈ U, and f ∈ A (U), then

 1 1/p |f(z0)| ≤ 2 kfkAp(U) . πr(z0)

p p 1 Proof. As f ∈ A (U) we have f ∈ A (D(z0, r(z0))) ⊆ A (D(z0, r(z0))). Using

6 1 1 Lemma 1 and H¨older’sinequality, we get, with p + q = 1 (q is infinite if p = 1),

1 Z |f(z )| = f(z)dµ(z) 0 πr(z )2 0 D(z0,r(z0)) 1 Z ≤ 2 |f(z)|dµ(z) πr(z0) D(z0,r(z0))

1 1/q ≤ µ(D(z , r(z ))) kfk p 2 0 0 A (D(z0,r(z0))) πr(z0)

1 2 1/q = (πr(z ) ) kfk p 2 0 A (D(z0,r(z0))) πr(z0)

1 2 1/q ≤ 2 (πr(z0) ) kfkAp(U) πr(z0) 1 1 2 1− p = 2 (πr(z0) ) kfkAp(U) πr(z0)  1 1/p = 2 kfkAp(U) . πr(z0)

Now we prove that Ap(U) is a complete metric space, showing that it is a Banach space. Theorem 3. If 1 ≤ p ≤ ∞, then Ap(U) is a Banach space.

p Proof. Suppose that fn ∈ A (U) is a Cauchy sequence. We have to show that p p there is some f ∈ A (U) such that fn → f in A (U). The space H(U) of holomorphic functions on U is a Fr´echet space: there is an increasing sequence of compact sets Ki ⊂ U 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 p both compact sets. If z0 ∈ Ki then r(z0) ≥ ri. Thus if z0 ∈ Ki and g ∈ A (U), using Lemma 2 we get

 1 1/p  1 1/p |g(z0)| ≤ 2 kgkAp(U) ≤ 2 kgkAp(U) . πr(z0) πri

From this and the fact that kfn − fmkAp(U) → 0 as m, n → ∞, we get that

pKi (fn − fm) → 0, m, n → ∞.

That is, fn is a Cauchy sequence in each of the seminorms pKi , and as H(U) is a Fr´echet space it follows that there is some f ∈ H(U) such that fn → f in H(U). In particular, for all z0 ∈ U 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).

7 On the other hand, Lp(U) is a Banach space, and hence there is some g ∈ p L (U) such that kfn − gkLp(U) → 0 as n → ∞. This implies that there is some subsequence fa(n) such that for almost all z0 ∈ U, fa(n)(z0) → g(z0). Thus, for p almost all z0 ∈ U we have f(z0) = g(z0). Therefore, in L (U) we have f = g and so

kfn − fkAp(U) = kfn − fkLp(U) → 0, n → ∞.

5 The Hilbert space A2(D)

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 Z Z hf, gi = fg∗dµ = f(z)g(z)dµ(z). D D

2 2 As hf, fi = kfkA2(D), it follows that A (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 n ≥ 0 and z ∈ D, define en : D → C by

rn + 1 e (z) = · zn. n π

2 Then en are an orthonormal basis for A (D). Proof. If E is a subset of a Hilbert space and v ∈ H, we write v ⊥ E if hv, ei = 0 for all e ∈ E. If E is an orthonormal set in H, E is an orthonormal basis if and only if v ⊥ E implies that v = 0. This is proved in John B. Conway, A Course in Functional Analysis, second ed., p. 16, Theorem 4.13. For n 6= m, Z r r n + 1 n m + 1 m hen, emi = z z dµ(z) D π π p(n + 1)(m + 1) Z = znzmdµ(z) π D p(n + 1)(m + 1) Z 1 Z 2π = (reiθ)n(re−iθ)mrdθdr π 0 0 p(n + 1)(m + 1) Z 1 Z 2π = rn+m+1eiθ(n−m)dθdr π 0 0 p(n + 1)(m + 1) Z 1 Z 2π = rn+m+1eiθ(n−m)dθdr π 0 0 = 0,

8 while Z 1 Z 2π n + 1 2n+1 hen, eni = r dθdr π 0 0 Z 1 = 2(n + 1) r2n+1dr 0 = 1.

Therefore en is an orthonormal set. Hence, to show that it is an orthonormal 2 basis for A (D) we have to show that if hf, eni = 0 for all n ≥ 0 then f = 0. For 0 < r < 1, let Dr be the open disc centered at 0 of radius r, and let P∞ n kgkr = sup|z|≤r |g(z)|. Let f(z) = n=0 anz , and for each 0 < r < 1 this power series converges uniformly in Dr. Then Z Z ∞ ∗ X n m femdµ = anz z dµ(z) Dr Dr n=0 ∞ Z X n m = an z z dµ(z) n=0 Dr ∞ Z r Z 2π X n+m+1 iθ(n−m) = an ρ e dθdρ n=0 0 0 ∞ Z r X n+m+1 = an ρ · 2π · δn,mdρ n=0 0 Z r 2m+1 = 2πam ρ dρ 0 r2m+2 = 2πa m 2m + 2

∗ 1 One checks that fem ∈ A (D), and hence Z Z ∗ ∗ lim femdµ(z) = femdµ(z). r→1 Dr D Therefore 1 hf, e i = πa . m m m + 1

As hf, emi = 0 for each m, this gives us that am = 0 for all m and hence f = 0. 2 This shows that en is an orthonormal basis for A (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 C, not necessarily bounded.

9 6 Hardy spaces W In a Hilbert space H, if Sα, α ∈ I are subsets of H, let Sα denote the S α∈I closure in H of Sα. Thus, to say that a set {vα} is an orthonormal basis α∈I W for a Hilbert space H is to say that {vα} is orthonormal and that α∈I {vα} = H. Let S1 = {z ∈ C : |z| = 1}, and let µ be normalized arc length, so that 1 1 n µ(S ) = 1. Define en : S → C by en(z) = z , for n ∈ Z. It is a fact that 2 1 en, n ∈ Z are an orthonormal basis for the Hilbert space L (S ), with inner product Z hf, gi = fg∗dµ. S1 2 1 W We define the Hardy space H (S ) to be n≥0{en}. As it is a closed subspace of the Hilbert space L2(S1), it is itself a Hilbert space. For f ∈ L2(S1), we denote f ∗(z) = f(z). The following is Problem 26 of Halmos. Note f ∗(z) = f(z). Theorem 5. If f ∈ H2(S1) and f ∗ = f, then f is constant.

2 1 2 1 Proof. If gn ∈ L (S ) and gn → g ∈ L (S ), then

∗ ∗ kgn − g k = kgn − gk → 0.

Thus g 7→ g∗ is continuous L2(S1) → L2(S1). 2 1 2 1 If g ∈ L (S ), then, as en, n ∈ Z is an orthonormal basis for L (S ), we have P ∗ g = limN→∞ |n|≤N hg, eni en, and so, as en = e−n,

∗ X ∗ X X g = lim (hg, eni en) = lim hg, enie−n = lim hg, e−nien. N→∞ N→∞ N→∞ |n|≤N |n|≤N |n|≤N

Therefore if n ∈ Z then ∗ hg , eni = hg, e−ni. (2) For n > 0, ∗ hf, eni = hf , eni = hf, e−ni = 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 f ∈ H2(S1). It follows that f ∈ span{e0}, and thus that f is constant. If g ∈ L2(S1), define Re g ∈ L2(S1) by

g + g∗ Re g = 2 and Re g ∈ L2(S1) by g − g∗ Im g = . 2i

10 P ∗ P ∗ P g = hg, eni en and, by (2), g = hg , eni en = hg, e−nien, so n∈Z n∈Z n∈Z ! 1 X X 1 X   Re g = hg, e i e + hg, e ie∗ = hg, e i + hg, e i e , 2 n n n n 2 n −n n n∈Z n∈Z n∈Z and ! 1 X X 1 X   Im g = hg, e i e − hg, e ie = hg, e i − hg, e i e . 2i n n −n n 2i n −n n n∈Z n∈Z n∈Z (3) g = Re g + iIm g, and we have (Re g)∗ = Re g and (Im g)∗ = Im g; that is, both Re g and Im g 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 R2, 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 z0 ∈ Ω there is some open neighborhood of z on which g has a primitive f 1 (g might not have a primitive defined on Ω, e.g. g(z) = z on Ω = C \{0}), and there is a constant c such that u(x, y) = Re f(x + iy) + c for all (x, y) in this neighborhood. u and v(x, y) = Im f(x + iy) + c are called harmonic conjugates. Theorem 6. If u ∈ L2(S1) and u∗ = u, then there is a unique v ∈ L2(S1) such ∗ 2 1 that v = v, hv, e0i = 0, and u + iv ∈ H (S ). Proof. Define D : {u ∈ L2(S1): u∗ = u} → H2(D) by  hu, e i n = 0,  0 hDu, eni = hu, eni + hu, e−ni n > 0, 0 n < 0.

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

X 2 2 X 2 | hDu, eni | = | hu, e0i | + | hu, eni + hu, e−ni| n≥0 n>0 2 X 2 2 ≤ | hu, e0i | + 2 | hu, eni | + |hu, e−ni| n>0 2 X 2 = | hu, e0i | + 2 | hu, eni | n6=0 ≤ 2 kuk2 .

This is finite, hence Du ∈ H2(S1).

11 2 1 ∗ For any g ∈ L (S ) and n ∈ Z, by (2) we have hg , eni = hg, e−ni. As ∗ u = u, if n ∈ Z then hu, eni = hu, e−ni. Using this, we check that Re Du = u. Put v = Im Du, hence Du = u + iv. hu, e0i = hu, e0i gives hDu, e0i = hDu, e0i, and applying this and (3) we get hv, e0i = 0. Thus v satisfies the ∗ 2 1 conditions v = v, hv, e0i = 0, and u + iv ∈ H (S ). We are not obliged to do so, but let’s write out the Fourier coefficients of v. If n ∈ Z then, using hu, eni = hu, e−ni,

hv, eni = hIm Du, eni 1   = hDu, e i − hDu, e i 2i n −n 0 n = 0   1   = 2i hu, eni + hu, e−ni n > 0     1 − 2i hu, e−ni + hu, eni n < 0 0 n = 0   1   = 2i hu, eni + hu, e−ni n > 0  1   − 2i hu, eni + hu, e−ni n < 0  0 n = 0  1 = i hu, eni n > 0  1 − i hu, eni n < 0.

Thus hv, eni = −isgn(n) hu, eni. 2 1 hf,eni+hf,e−ni If f ∈ H (S ), then, as hRe f, eni = 2 ,   hf,e0i+hf,e0i n = 0  2 hf,e i+hf,e i hf,e i+hf,e i hDRe f, eni = n −n + −n n n > 0  2 2 0 n < 0.  hf,e0i+hf,e0i n = 0  2 = hf, eni n > 0  0 n < 0 ( hf,e0i+hf,e0i n = 0 = 2 hf, eni n 6= 0 Thus

( hf,e0i−hf,e0i 2 n = 0 hf − DRe f, eni = 0 n 6= 0 ( i · hIm f, e i n = 0 = 0 0 n 6= 0

12 13