introduction to bloch spaces - uma · 2016. 6. 27. · introduction to bloch spaces vicent asensio...

Introduction to Bloch spaces

Vicent Asensio LopezCarmen Molina LaluezaAntonio Zarauz Moreno

Workshop on Complex Analysis and Operator TheoryAdvisor: Oscar Blasco

24 de junio de 2016

V. Asensio, C. Molina, A. Zarauz 24 de junio de 2016 1 / 30

Outline

1 Bloch and little Bloch spaces

2 Hyperbolic distances

3 Bergman projection and dualityBergman projectionDuality

4 References

Bloch and little Bloch spaces V. Asensio, C. Molina, A. Zarauz 24 de junio de 2016 2 / 30

The Bloch space

We define the set:

B = {f ∈ H(D) : sup|z|<1{(1− |z |2)|f ′(z)| <∞}}

The seminorm||f ||B = sup

|z|<1{(1− |z |2)|f ′(z)|}

And the norm||f ||Bloch = |f (0)|+ ||f ||B

The Bloch space is the set B with the norm || · ||Bloch.

B is Banach

The Bloch space is Banach.

Proof Let {fn}n ⊂ B be a Cauchy sequence in B: ∀ε > 0 ∃n0 s.t. ∀n,m ≥n0, ||fn − fm||Bloch < ε. Hence,

|fn(0)− fm(0)| < ε

(1− |z |2)|f ′n(z)− f ′m(z)| < ε when |z | < 1

Then, {fn(0)}n is Cauchy on C (complete), so we can consider a0 =limn→∞|fn(0)|. In addition, {f ′n(z)}n is Cauchy on C, we call g(z) = limn→∞f ′n(z).Since

|f ′n(z)− f ′m(z)| < ε

1− |z |2

f ′n(z)→ g uniformly on compact sets of D, then g ∈ H(D).

B is Banach

|fn(0)− fm(0)| < ε

(1− |z |2)|f ′n(z)− f ′m(z)| < ε when |z | < 1

|f ′n(z)− f ′m(z)| < ε

1− |z |2

B is Banach

|fn(0)− fm(0)| < ε

(1− |z |2)|f ′n(z)− f ′m(z)| < ε when |z | < 1

|f ′n(z)− f ′m(z)| < ε

1− |z |2

B is Banach

|fn(0)− fm(0)| < ε

(1− |z |2)|f ′n(z)− f ′m(z)| < ε when |z | < 1

|f ′n(z)− f ′m(z)| < ε

1− |z |2

B is Banach

|fn(0)− fm(0)| < ε

(1− |z |2)|f ′n(z)− f ′m(z)| < ε when |z | < 1

|f ′n(z)− f ′m(z)| < ε

1− |z |2

Since D is simply connected there exists primitive for g. Take f such thatf ′(z) = g(z) ∀|z | < 1 and f (0) = a0.Thus, ∀ε > 0 ∃n0 s.t. ∀n,m ≥ n0, (1−|z |2)|f ′n(z)−f ′m(z)| < ε when |z | < 1.When n tends to infinity,

(1− |z |2)|f ′(z)− f ′m(z)| < ε when |z | < 1

Then, ||fn − f ||Bloch → 0.Plus,

(1− |z |2)|f ′(z)| ≤ (1− |z |2)|f ′(z)− f ′m(z)|+ (1− |z |2)|f ′n(z)| << ε+ (1− |z |2)|f ′n(z)| <∞

and taking supremes we have f ∈ B.

(1− |z |2)|f ′(z)− f ′m(z)| < ε when |z | < 1

The little Bloch space

The little Bloch space, denoted by B0, is the closed subspace of Bconsisting of functions f ∈ B with

lim|z|→1(1− |z |2)|f ′(z)| = 0

Examples:

The set of polynomials P is contained in B0. (the derivate is boundedin D)

H∞ ⊂ B. We use the Poisson Kernel Pw (e it) = 1−|w |2(1−we it)2 with

|w | < 1, we have∫ 2π

0 Pw (e it) dt2π = 1. Thus, for f ∈ H∞

|f ′(z)| =1

∣∣∣∣ ∫Γr

(w − z)2dw

∣∣∣∣ ≤≤ ||f ||∞

r(1− |z|2

∫ 2π

(1− |z|2

|1− zr e

it |2dt

r ||f ||∞r2 − |z |2

Taking r → 1,

|f ′(z)| ≤ ||f ||∞1− |z |2

f (z) = Log(1− z) ∈ B − B0.

(1−|z |2)|f ′(z)| = (1−|z |2)1

|1− z |≤ (1−|z |2)

(1− |z |)= 1+ |z | ≤ 2

So f ∈ B, but, taking the limit, for example, in the positive real axis,

lim|z|→1(1− |z |2)|f ′(z)| = 2

Then, f 6∈ B0.

The following are equivalent:

1. f ∈ B0

2. limr→1− ||fr − f ||B = 0 where r ∈ (0, 1) and fr (z) = f (rz).

3. limn→∞||f − pn||Bloch = 0 with pn ∈ P

To prove this theorem, we will use the following result:

Given f ∈ H(D), we denote by fr the function fr (z) = f (rz). Then,fr → f uniformly on compact sets of D, when r → 1−.

Then, we also have f ′r → f ′ uniformly on compact sets, when r → 1−.

The following are equivalent:

1. f ∈ B0

2. limr→1− ||fr − f ||B = 0 where r ∈ (0, 1) and fr (z) = f (rz).

3. limn→∞||f − pn||Bloch = 0 with pn ∈ P

To prove this theorem, we will use the following result:

Given f ∈ H(D), we denote by fr the function fr (z) = f (rz). Then,fr → f uniformly on compact sets of D, when r → 1−.

Then, we also have f ′r → f ′ uniformly on compact sets, when r → 1−.

Proof (1)⇒(2). Assume f ∈ B0. For any ε > 0 there exists δ ∈ (0, 1) s.t.

(1− |z |2)|f ′(z)| < ε, δ2 < |z | < 1

Consider

||fr − f ||B = sup{(1− |z |2)|rf ′(rz)− f ′(z)| : |z | < 1} ≤≤ sup{(1− |z |2)|rf ′(rz)− f ′(z)| : δ < |z | < 1}+

+ sup{(1− |z |2)|rf ′(rz)− f ′(z)| : |z | ≤ δ}

By the previous lemma, rf ′(rz) = f ′r (z)→ f ′(z) when |z | ≤ δ ⇒ the secondsupreme is zero.If δ < r < 1, and δ < |z | < 1, then δ2 < r |z | < 1. Thus,

(1− |z |2)|rf (rz)| ≤ (1− r2|z |2)|f ′(rz)| < ε

Hencesup{(1− |z |2)|rf ′(rz)− f ′(z)| : δ < |z | < 1} ≤ 2ε

for δ < r < 1, and we have for any ε > 0

limr→1− ||fr − f ||B ≤ 2ε

(1− |z |2)|f ′(z)| < ε, δ2 < |z | < 1

Consider

+ sup{(1− |z |2)|rf ′(rz)− f ′(z)| : |z | ≤ δ}

(1− |z |2)|rf (rz)| ≤ (1− r2|z |2)|f ′(rz)| < ε

limr→1− ||fr − f ||B ≤ 2ε

(1− |z |2)|f ′(z)| < ε, δ2 < |z | < 1

Consider

+ sup{(1− |z |2)|rf ′(rz)− f ′(z)| : |z | ≤ δ}

(1− |z |2)|rf (rz)| ≤ (1− r2|z |2)|f ′(rz)| < ε

limr→1− ||fr − f ||B ≤ 2ε

(1− |z |2)|f ′(z)| < ε, δ2 < |z | < 1

Consider

+ sup{(1− |z |2)|rf ′(rz)− f ′(z)| : |z | ≤ δ}

(1− |z |2)|rf (rz)| ≤ (1− r2|z |2)|f ′(rz)| < ε

limr→1− ||fr − f ||B ≤ 2ε

(1− |z |2)|f ′(z)| < ε, δ2 < |z | < 1

Consider

+ sup{(1− |z |2)|rf ′(rz)− f ′(z)| : |z | ≤ δ}

(1− |z |2)|rf (rz)| ≤ (1− r2|z |2)|f ′(rz)| < ε

limr→1− ||fr − f ||B ≤ 2ε

(2)⇒(3) Since fr can be approximated by polynomials in H∞ and || ||B ≤|| ||H∞ , we may approximate by polynomials in B. Thus, for any fr and forany ε > 0, ∃{pn} ∈ P and n0 ≥ 0 s.t. ∀n ≥ n0,

||fr − pn||B < ε

Then, using (2), ∃δ > 0 s.t. ∀δ < r < 1

||f − pn||B ≤ ||f − fr ||B + ||fr − pn||B < 2ε

Without loss of generality, we take pn(0) = f (0). Then,

limn→∞||f − pn||Bloch = 0

(3)⇒(1) Straightforward because P ⊂ B0 and B0 is closed.

||fr − pn||B < ε

Then, using (2), ∃δ > 0 s.t. ∀δ < r < 1

||f − pn||B ≤ ||f − fr ||B + ||fr − pn||B < 2ε

||fr − pn||B < ε

Then, using (2), ∃δ > 0 s.t. ∀δ < r < 1

||f − pn||B ≤ ||f − fr ||B + ||fr − pn||B < 2ε

||fr − pn||B < ε

Then, using (2), ∃δ > 0 s.t. ∀δ < r < 1

||f − pn||B ≤ ||f − fr ||B + ||fr − pn||B < 2ε

||fr − pn||B < ε

Then, using (2), ∃δ > 0 s.t. ∀δ < r < 1

||f − pn||B ≤ ||f − fr ||B + ||fr − pn||B < 2ε

Outline

4 References

Hyperbolic distances V. Asensio, C. Molina, A. Zarauz 24 de junio de 2016 12 / 30

Pseudo-hyperbolic distance

We remind that the Mobius transformation is ϕa(z) = a−z1−az . Then the

pseudo-hyperbolic distance in D is defined to be

ρ(z , ω) = |ϕz(ω)| ∀z , ω ∈ D.

Note that ρ(z , 0) = |z |.We show that is indeed a distance. Clearly ρ(z , ω) ≥ 0 and ρ(z , ω) = 0 iffz = ω. To see the symmetric property, we note that

ρ(z , ω) =

∣∣∣∣ z − ω1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− ωz

∣∣∣∣ = ρ(ω, z).

ρ(z , ω) = |ϕz(ω)| ∀z , ω ∈ D.

ρ(z , ω) =

∣∣∣∣ z − ω1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− ωz

∣∣∣∣ = ρ(ω, z).

ρ(z , ω) = |ϕz(ω)| ∀z , ω ∈ D.

ρ(z , ω) =

∣∣∣∣ z − ω1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− ωz

∣∣∣∣ = ρ(ω, z).

ρ(z , ω) = |ϕz(ω)| ∀z , ω ∈ D.

ρ(z , ω) =

∣∣∣∣ z − ω1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− ωz

∣∣∣∣ = ρ(ω, z).

ρ(z , ω) = |ϕz(ω)| ∀z , ω ∈ D.

ρ(z , ω) =

∣∣∣∣ z − ω1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− ωz

∣∣∣∣ = ρ(ω, z).

ρ(z , ω) = |ϕz(ω)| ∀z , ω ∈ D.

ρ(z , ω) =

∣∣∣∣ z − ω1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− ωz

∣∣∣∣ = ρ(ω, z).

ρ(z , ω) = |ϕz(ω)| ∀z , ω ∈ D.

ρ(z , ω) =

∣∣∣∣ z − ω1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− ωz

∣∣∣∣ = ρ(ω, z).

ρ(z , ω) = |ϕz(ω)| ∀z , ω ∈ D.

ρ(z , ω) =

∣∣∣∣ z − ω1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− zω

∣∣∣∣ =

∣∣∣∣ ω − z

1− ωz

∣∣∣∣ = ρ(ω, z).

For the triangle inequality, we first prove that ρ(z , ω) ≤ ρ(z , 0) + ρ(0, ω).

That is,∣∣∣ z−ω1−zω

∣∣∣ ≤ |z |+ |ω|. Let |z | = a, |ω| = b and θ be the angle between

z and ω. Therefore

a2 + b2 − 2ab cos θ

1 + a2b2 − 2ab cos θ≤ (a + b)2.

Estimating the first factor, we have

1 + a2b2 − 2ab cos θ= 1− 1− a2 − b2 + a2b2

1 + a2b2 − 2ab cos θ

≤ 1− (1− a2)(1− b2)

1 + a2b2 + 2ab

=(a + b)2

(1 + ab)2.

z and ω. Therefore

1 + a2b2 − 2ab cos θ≤ (a + b)2.

1 + a2b2 − 2ab cos θ= 1− 1− a2 − b2 + a2b2

≤ 1− (1− a2)(1− b2)

1 + a2b2 + 2ab

=(a + b)2

(1 + ab)2.

z and ω. Therefore

1 + a2b2 − 2ab cos θ≤ (a + b)2.

1 + a2b2 − 2ab cos θ= 1− 1− a2 − b2 + a2b2

≤ 1− (1− a2)(1− b2)

1 + a2b2 + 2ab

=(a + b)2

(1 + ab)2.

z and ω. Therefore

1 + a2b2 − 2ab cos θ≤ (a + b)2.

1 + a2b2 − 2ab cos θ= 1− 1− a2 − b2 + a2b2

≤ 1− (1− a2)(1− b2)

1 + a2b2 + 2ab

=(a + b)2

(1 + ab)2.

z and ω. Therefore

1 + a2b2 − 2ab cos θ≤ (a + b)2.

1 + a2b2 − 2ab cos θ= 1− 1− a2 − b2 + a2b2

≤ 1− (1− a2)(1− b2)

1 + a2b2 + 2ab

=(a + b)2

(1 + ab)2.

z and ω. Therefore

1 + a2b2 − 2ab cos θ≤ (a + b)2.

1 + a2b2 − 2ab cos θ= 1− 1− a2 − b2 + a2b2

≤ 1− (1− a2)(1− b2)

1 + a2b2 + 2ab

=(a + b)2

(1 + ab)2.

z and ω. Therefore

1 + a2b2 − 2ab cos θ≤ (a + b)2.

1 + a2b2 − 2ab cos θ= 1− 1− a2 − b2 + a2b2

≤ 1− (1− a2)(1− b2)

1 + a2b2 + 2ab

=(a + b)2

(1 + ab)2.

To extend the previous argument, we now claim that for any f ∈ H(D),

ρ(f (z), f (ω)) ≤ ρ(z , ω) ∀z , ω ∈ D.

We select the holomorphic function ϕf (z) ◦ f ◦ ϕz , which fixes 0:

ϕz(0) = z

f (ϕz(0)) = f (z)

ϕf (z)(f (ϕz(0))) = ϕf (z)(f (z)) = 0

so by Schwarz’s Lemma we get that

|ϕf (z) ◦ f ◦ ϕz(ω)| ≤ |ω|.

Denoting ω′ = ϕz(ω) and using the fact that ϕ−1z = ϕz , we obtain

|ϕf (z)(f (ω′))| ≤ |ϕz(ω′)|.

ρ(f (z), f (ω)) ≤ ρ(z , ω) ∀z , ω ∈ D.

ϕz(0) = z

f (ϕz(0)) = f (z)

ϕf (z)(f (ϕz(0))) = ϕf (z)(f (z)) = 0

|ϕf (z) ◦ f ◦ ϕz(ω)| ≤ |ω|.

|ϕf (z)(f (ω′))| ≤ |ϕz(ω′)|.

ρ(f (z), f (ω)) ≤ ρ(z , ω) ∀z , ω ∈ D.

ϕz(0) = z

f (ϕz(0)) = f (z)

ϕf (z)(f (ϕz(0))) = ϕf (z)(f (z)) = 0

|ϕf (z) ◦ f ◦ ϕz(ω)| ≤ |ω|.

|ϕf (z)(f (ω′))| ≤ |ϕz(ω′)|.

ρ(f (z), f (ω)) ≤ ρ(z , ω) ∀z , ω ∈ D.

ϕz(0) = z

f (ϕz(0)) = f (z)

ϕf (z)(f (ϕz(0))) = ϕf (z)(f (z)) = 0

|ϕf (z) ◦ f ◦ ϕz(ω)| ≤ |ω|.

|ϕf (z)(f (ω′))| ≤ |ϕz(ω′)|.

ρ(f (z), f (ω)) ≤ ρ(z , ω) ∀z , ω ∈ D.

ϕz(0) = z

f (ϕz(0)) = f (z)

ϕf (z)(f (ϕz(0))) = ϕf (z)(f (z)) = 0

|ϕf (z) ◦ f ◦ ϕz(ω)| ≤ |ω|.

|ϕf (z)(f (ω′))| ≤ |ϕz(ω′)|.

ρ(f (z), f (ω)) ≤ ρ(z , ω) ∀z , ω ∈ D.

ϕz(0) = z

f (ϕz(0)) = f (z)

ϕf (z)(f (ϕz(0))) = ϕf (z)(f (z)) = 0

|ϕf (z) ◦ f ◦ ϕz(ω)| ≤ |ω|.

|ϕf (z)(f (ω′))| ≤ |ϕz(ω′)|.

ρ(f (z), f (ω)) ≤ ρ(z , ω) ∀z , ω ∈ D.

ϕz(0) = z

f (ϕz(0)) = f (z)

ϕf (z)(f (ϕz(0))) = ϕf (z)(f (z)) = 0

|ϕf (z) ◦ f ◦ ϕz(ω)| ≤ |ω|.

|ϕf (z)(f (ω′))| ≤ |ϕz(ω′)|.

Hyperbolic distance

β(z , ω) =1

1 + ρ(z , ω)

1− ρ(z , ω)

is the hyperbolic distance. It can also be seen as∫ ρ(z,ω)

1− t2=

∣∣∣∣∣∫

[0,ϕz (ω)]

1− |ξ|2

∣∣∣∣∣ =

∫γa,b

1− |η|2

where we are given a Mobius transformation ϕa, and we used the change ofvariables ξ = ϕa(η), for an arbitrary a. Note that such measure is invariant.From these reformulation, it can be proved that the triangle inequality issatisfied. The other ones follow from the fact that ρ is already a distance.Moreover, β is invariant under Mobius transformations because ρ so is.

Hyperbolic distance

β(z , ω) =1

1 + ρ(z , ω)

1− ρ(z , ω)

1− t2=

∣∣∣∣∣∫

[0,ϕz (ω)]

1− |ξ|2

∣∣∣∣∣ =

∫γa,b

1− |η|2

Hyperbolic distance

β(z , ω) =1

1 + ρ(z , ω)

1− ρ(z , ω)

1− t2=

∣∣∣∣∣∫

[0,ϕz (ω)]

1− |ξ|2

∣∣∣∣∣ =

∫γa,b

1− |η|2

Hyperbolic distance

β(z , ω) =1

1 + ρ(z , ω)

1− ρ(z , ω)

1− t2=

∣∣∣∣∣∫

[0,ϕz (ω)]

1− |ξ|2

∣∣∣∣∣ =

∫γa,b

1− |η|2

Hyperbolic distance

β(z , ω) =1

1 + ρ(z , ω)

1− ρ(z , ω)

1− t2=

∣∣∣∣∣∫

[0,ϕz (ω)]

1− |ξ|2

∣∣∣∣∣ =

∫γa,b

1− |η|2

Hyperbolic distance

β(z , ω) =1

1 + ρ(z , ω)

1− ρ(z , ω)

1− t2=

∣∣∣∣∣∫

[0,ϕz (ω)]

1− |ξ|2

∣∣∣∣∣ =

∫γa,b

1− |η|2

Hyperbolic distance

β(z , ω) =1

1 + ρ(z , ω)

1− ρ(z , ω)

1− t2=

∣∣∣∣∣∫

[0,ϕz (ω)]

1− |ξ|2

∣∣∣∣∣ =

∫γa,b

1− |η|2

Hyperbolic distance

β(z , ω) =1

1 + ρ(z , ω)

1− ρ(z , ω)

1− t2=

∣∣∣∣∣∫

[0,ϕz (ω)]

1− |ξ|2

∣∣∣∣∣ =

∫γa,b

1− |η|2

For any Bloch function f , |f (z)− f (ω)| ≤ ‖f ‖Bβ(z , ω).

‖·‖B is invariant under Mobius transformations.

We shall rewrite ‖f ‖B by sup|z|<1

{|(f ◦ ϕz)′(0)|

}because

(f ◦ ϕz)′(0) = f ′(ϕz(0))ϕ′z(0) = f ′(z)ϕ′z(0) = f ′(z)(1− |z |2),

where ϕ′z(ω) = 1−|z|2(1−zω)2 , and the result follows from

ϕa ◦ ϕb(0) = ϕϕa(b)(0).

{|(f ◦ ϕz)′(0)|

}because

ϕa ◦ ϕb(0) = ϕϕa(b)(0).

{|(f ◦ ϕz)′(0)|

}because

ϕa ◦ ϕb(0) = ϕϕa(b)(0).

{|(f ◦ ϕz)′(0)|

}because

ϕa ◦ ϕb(0) = ϕϕa(b)(0).

{|(f ◦ ϕz)′(0)|

}because

ϕa ◦ ϕb(0) = ϕϕa(b)(0).

{|(f ◦ ϕz)′(0)|

}because

ϕa ◦ ϕb(0) = ϕϕa(b)(0).

{|(f ◦ ϕz)′(0)|

}because

ϕa ◦ ϕb(0) = ϕϕa(b)(0).

Indeed, we are going to demonstrate that

‖f ‖B = sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

It is enough to take ω = 0. Since f ∈ H(D), by the fundamental theoremof algebra,

f (z)− f (0) =

∫[0,z]

f ′(ξ)dξ = z

0f ′(zt)dt.

‖f ‖B = sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

f (z)− f (0) =

∫[0,z]

f ′(ξ)dξ = z

0f ′(zt)dt.

‖f ‖B = sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

f (z)− f (0) =

∫[0,z]

f ′(ξ)dξ = z

0f ′(zt)dt.

According to f is Bloch, we then estimate

|f (z)− f (0)| ≤ |z |‖f ‖B∫ 1

1− |z |2t2

= ‖f ‖B∫ |z|

1− s2

2‖f ‖B

∫ |z|0

1− s+

1 + sds

2‖f ‖B (− log(1− |z |) + log(1 + |z |))

= ‖f ‖B1

(1 + |z |1− |z |

)= ‖f ‖Bβ(z , 0).

By a Mobius transformation, we may extend the argument for every ω.

|f (z)− f (0)| ≤ |z |‖f ‖B∫ 1

1− |z |2t2

= ‖f ‖B∫ |z|

1− s2

2‖f ‖B

∫ |z|0

1− s+

1 + sds

2‖f ‖B (− log(1− |z |) + log(1 + |z |))

= ‖f ‖B1

(1 + |z |1− |z |

)= ‖f ‖Bβ(z , 0).

|f (z)− f (0)| ≤ |z |‖f ‖B∫ 1

1− |z |2t2

= ‖f ‖B∫ |z|

1− s2

2‖f ‖B

∫ |z|0

1− s+

1 + sds

2‖f ‖B (− log(1− |z |) + log(1 + |z |))

= ‖f ‖B1

(1 + |z |1− |z |

)= ‖f ‖Bβ(z , 0).

|f (z)− f (0)| ≤ |z |‖f ‖B∫ 1

1− |z |2t2

= ‖f ‖B∫ |z|

1− s2

2‖f ‖B

∫ |z|0

1− s+

1 + sds

2‖f ‖B (− log(1− |z |) + log(1 + |z |))

= ‖f ‖B1

(1 + |z |1− |z |

)= ‖f ‖Bβ(z , 0).

|f (z)− f (0)| ≤ |z |‖f ‖B∫ 1

1− |z |2t2

= ‖f ‖B∫ |z|

1− s2

2‖f ‖B

∫ |z|0

1− s+

1 + sds

2‖f ‖B (− log(1− |z |) + log(1 + |z |))

= ‖f ‖B1

(1 + |z |1− |z |

)= ‖f ‖Bβ(z , 0).

|f (z)− f (0)| ≤ |z |‖f ‖B∫ 1

1− |z |2t2

= ‖f ‖B∫ |z|

1− s2

2‖f ‖B

∫ |z|0

1− s+

1 + sds

2‖f ‖B (− log(1− |z |) + log(1 + |z |))

= ‖f ‖B1

(1 + |z |1− |z |

)= ‖f ‖Bβ(z , 0).

|f (z)− f (0)| ≤ |z |‖f ‖B∫ 1

1− |z |2t2

= ‖f ‖B∫ |z|

1− s2

2‖f ‖B

∫ |z|0

1− s+

1 + sds

2‖f ‖B (− log(1− |z |) + log(1 + |z |))

= ‖f ‖B1

(1 + |z |1− |z |

)= ‖f ‖Bβ(z , 0).

To see the other inequality, we compute lımω→z

|f (ω)− f (z)|β(ω, z)

lımω→z

|f (ω)− f (z)||ω − z |

|ω − z |β(ω, z)

The first term goes to |f ′(z)| since modulus is a continuous function, so it

suffices to show that lımω→z

|ω − z |β(ω, z)

= (1− |z |2).

Note that β(z , ω) can be rewritten by1

(|1− ωz |+ |ω − z ||1− ωz | − |ω − z |

|f (ω)− f (z)|β(ω, z)

lımω→z

|f (ω)− f (z)||ω − z |

|ω − z |β(ω, z)

= (1− |z |2).

(|1− ωz |+ |ω − z ||1− ωz | − |ω − z |

|f (ω)− f (z)|β(ω, z)

lımω→z

|f (ω)− f (z)||ω − z |

|ω − z |β(ω, z)

= (1− |z |2).

(|1− ωz |+ |ω − z ||1− ωz | − |ω − z |

|f (ω)− f (z)|β(ω, z)

lımω→z

|f (ω)− f (z)||ω − z |

|ω − z |β(ω, z)

= (1− |z |2).

(|1− ωz |+ |ω − z ||1− ωz | − |ω − z |

In real analysis, for a = a(x) > 0,

lımx→0

12 log a(x)+x

a(x)−x

x= lım

(a(x)+xa(x)−x − 1

= lımx→0

a(x) + x − (a(x)− x)

a(x)− x

= lımx→0

a(x)− x.

So for our case, lımω→z

|1− ωz | − |ω − z |=

1− |z |2and hence

|f ′(z)|(1− |z |2) ≤ sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

}as wanted.

Any analytic function f is Bloch iff there exists C > 0 such that

|f (z)− f (ω)| ≤ Cβ(z , ω) ∀z , ω ∈ D.Hyperbolic distances V. Asensio, C. Molina, A. Zarauz 24 de junio de 2016 21 / 30

lımx→0

12 log a(x)+x

a(x)−x

x= lım

(a(x)+xa(x)−x − 1

= lımx→0

a(x) + x − (a(x)− x)

a(x)− x

= lımx→0

a(x)− x.

|1− ωz | − |ω − z |=

1− |z |2and hence

|f ′(z)|(1− |z |2) ≤ sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

}as wanted.

lımx→0

12 log a(x)+x

a(x)−x

x= lım

(a(x)+xa(x)−x − 1

= lımx→0

a(x) + x − (a(x)− x)

a(x)− x

= lımx→0

a(x)− x.

|1− ωz | − |ω − z |=

1− |z |2and hence

|f ′(z)|(1− |z |2) ≤ sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

}as wanted.

lımx→0

12 log a(x)+x

a(x)−x

x= lım

(a(x)+xa(x)−x − 1

= lımx→0

a(x) + x − (a(x)− x)

a(x)− x

= lımx→0

a(x)− x.

|1− ωz | − |ω − z |=

1− |z |2and hence

|f ′(z)|(1− |z |2) ≤ sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

}as wanted.

lımx→0

12 log a(x)+x

a(x)−x

x= lım

(a(x)+xa(x)−x − 1

= lımx→0

a(x) + x − (a(x)− x)

a(x)− x

= lımx→0

a(x)− x.

|1− ωz | − |ω − z |=

1− |z |2and hence

|f ′(z)|(1− |z |2) ≤ sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

}as wanted.

lımx→0

12 log a(x)+x

a(x)−x

x= lım

(a(x)+xa(x)−x − 1

= lımx→0

a(x) + x − (a(x)− x)

a(x)− x

= lımx→0

a(x)− x.

|1− ωz | − |ω − z |=

1− |z |2and hence

|f ′(z)|(1− |z |2) ≤ sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

}as wanted.

lımx→0

12 log a(x)+x

a(x)−x

x= lım

(a(x)+xa(x)−x − 1

= lımx→0

a(x) + x − (a(x)− x)

a(x)− x

= lımx→0

a(x)− x.

|1− ωz | − |ω − z |=

1− |z |2and hence

|f ′(z)|(1− |z |2) ≤ sup

{|f (z)− f (ω)|β(z , ω)

, z , ω ∈ D, z 6= ω

}as wanted.

β(z , ω) = sup {|f (z)− f (ω)| : ‖f ‖B ≤ 1}.

We have to prove that β(z , 0) = sup{|f (z)− f (0)| : ‖f ‖B ≤ 1} since β and‖·‖B are Mobius invariant.We already know that |f (z)− f (0)| ≤ ‖f ‖Bβ(z , 0).Thus, sup{|f (z)− f (0)| : ‖f ‖B ≤ 1} ≤ β(z , 0).For the other inequality, we are going to construct a function g whose normis 1 and satisfies that |g(z) − g(0)| = β(z , 0). For z 6= 0, z = |z |e−iθ, wethen define

g(ω) =1

1 + ωe iθ

1− ωe iθ, ω ∈ D.

β(z , ω) = sup {|f (z)− f (ω)| : ‖f ‖B ≤ 1}.

g(ω) =1

1 + ωe iθ

1− ωe iθ, ω ∈ D.

β(z , ω) = sup {|f (z)− f (ω)| : ‖f ‖B ≤ 1}.

g(ω) =1

1 + ωe iθ

1− ωe iθ, ω ∈ D.

β(z , ω) = sup {|f (z)− f (ω)| : ‖f ‖B ≤ 1}.

g(ω) =1

1 + ωe iθ

1− ωe iθ, ω ∈ D.

β(z , ω) = sup {|f (z)− f (ω)| : ‖f ‖B ≤ 1}.

g(ω) =1

1 + ωe iθ

1− ωe iθ, ω ∈ D.

β(z , ω) = sup {|f (z)− f (ω)| : ‖f ‖B ≤ 1}.

g(ω) =1

1 + ωe iθ

1− ωe iθ, ω ∈ D.

Since g ′(ω) =e iθ

1− e2iθω2,

‖g‖B = sup

{1− |ω|2 1

|1− e2iθω2|: ω ∈ D

}≥ 1− |z |2

|1− e2iθz2|= 1.

On the other hand, |1− e2iθω2| ≥ 1− |ω|2, then

‖g‖B = sup

{1− |ω|2 1

|1− e i2θω2|: ω ∈ D

}≤ 1.

1− e2iθω2,

‖g‖B = sup

{1− |ω|2 1

|1− e2iθω2|: ω ∈ D

}≥ 1− |z |2

|1− e2iθz2|= 1.

‖g‖B = sup

{1− |ω|2 1

|1− e i2θω2|: ω ∈ D

}≤ 1.

1− e2iθω2,

‖g‖B = sup

{1− |ω|2 1

|1− e2iθω2|: ω ∈ D

}≥ 1− |z |2

|1− e2iθz2|= 1.

‖g‖B = sup

{1− |ω|2 1

|1− e i2θω2|: ω ∈ D

}≤ 1.

1− e2iθω2,

‖g‖B = sup

{1− |ω|2 1

|1− e2iθω2|: ω ∈ D

}≥ 1− |z |2

|1− e2iθz2|= 1.

‖g‖B = sup

{1− |ω|2 1

|1− e i2θω2|: ω ∈ D

}≤ 1.

1− e2iθω2,

‖g‖B = sup

{1− |ω|2 1

|1− e2iθω2|: ω ∈ D

}≥ 1− |z |2

|1− e2iθz2|= 1.

‖g‖B = sup

{1− |ω|2 1

|1− e i2θω2|: ω ∈ D

}≤ 1.

1− e2iθω2,

‖g‖B = sup

{1− |ω|2 1

|1− e2iθω2|: ω ∈ D

}≥ 1− |z |2

|1− e2iθz2|= 1.

‖g‖B = sup

{1− |ω|2 1

|1− e i2θω2|: ω ∈ D

}≤ 1.

1− e2iθω2,

‖g‖B = sup

{1− |ω|2 1

|1− e2iθω2|: ω ∈ D

}≥ 1− |z |2

|1− e2iθz2|= 1.

‖g‖B = sup

{1− |ω|2 1

|1− e i2θω2|: ω ∈ D

}≤ 1.

Outline

4 References

Bergman projection and duality V. Asensio, C. Molina, A. Zarauz 24 de junio de 2016 24 / 30

Bergman Kernel

The Bergman kernel is defined by:

K : D× D −→ L1(dA)

(z , ω) 7−→ hz(ω) = 1(1−zω)2

Bergman projection

Consider f ∈ L1(dA). The Bergman projection from L1(D) ontoL1(dA) is defined as follows

Pf (z) =

∫DK (z , ω)f (ω)dA(ω), ∀z ∈ D. (1)

Bergman Kernel

The Bergman kernel is defined by:

K : D× D −→ L1(dA)

(z , ω) 7−→ hz(ω) = 1(1−zω)2

Bergman projection

Consider f ∈ L1(dA). The Bergman projection from L1(D) ontoL1(dA) is defined as follows

Pf (z) =

∫DK (z , ω)f (ω)dA(ω), ∀z ∈ D. (1)

P(L1(D)) ⊆ H(D)

Pf (z) =

∫Df (ω)

[+∞∑n=0

(n + 1)ωnzn

]dA(ω)

+∞∑n=0

(∫Df (ω)ωndA(ω)

)︸︷︷︸

An∈C

By Cauchy-Hadamard theorem, the radius of convergence of theprevious power series is bounded by 1 in view of

R−1 = limn→∞

(n + 1)An ≤ limn→∞

√(n + 1)‖f ‖L1(D) = 1⇒ R ≥ 1,

hence Pf ∈ H(D).

P(L∞(D)) ⊆ BGiven f ∈ L∞(D), we have:

Pf ′(z) = 2

f (ω)ω

(1− ωz)3dA(ω);

|Pf ′(z)| ≤ 2‖f ‖∞∫D

|ω|dA(ω)

|1− ωz |3

= 2‖f ‖∞∫ 1

∫ 2π

|1− re−itz |3dt

TIL/EF=

4‖f ‖∞π

[∫ π

(|1− r |z |e it |2)32

]r2 dr

P(L∞(D)) ⊆ BThe number A = |1− r |z |e it |2 lies in(

[(1− r |z |2) + r |z |t2

], (1− r |z |2) + r |z |t2

|Pf ′(z)| =4

π‖f ‖∞

)3∫ 1

[∫ π

((1− r |z |2) + r |z |t2)32

]r2 dr

2‖f ‖∞

(1− r |z |)3

∫ π

dt[1 +

(√r |z|

1−r |z| t

P(L∞(D)) ⊆ B

(cont) =π2

2‖f ‖∞

r2√r |z |(1− r |z |)2

∫ √r|z|

1−r|z|

(1 + s2)32

︸︷︷︸√

r|z|1−r|z|<+∞⇒Int≤1

≤ π2

2‖f ‖∞

|z |2r2√r |z |(1− r |z |)2

2√|z |‖f ‖∞

|z |52

∫ |z|0

(1− s)2ds

)≤ π2

2√|z |‖f ‖∞

1− |z |− 1

2√|z |‖f ‖∞

1− |z |

≤ π2

2√|z |‖f ‖∞

1− |z |2

P(L∞(D)) ⊆ BThus,

(1− |z |2)|Pf ′(z)| ≤ π2

2√|z |

Given ε ∈ (0, 1), by Cauchy’s formula we can fathom

‖Pf ′‖B = max

{sup|z|>ε { π2

2√|z|} , 1

(1−ε)2

}Define ϕ : [0, 1) → C, ϕ(ε) =

(1−ε)2 . It is a continuous function

satisfying [0,+∞) ⊂ Im(ϕ), hence there exists ε′ ∈ [0,+∞) such

that ϕ(ε′) = π2

Eventually, ‖P‖L∞→B ≤ π2

P(L∞(D)) ⊇ BConversely, let g(z) = g(0) + zg ′(0) + g1(z) be a Bloch functionand

f (z) = g(0) + zg ′(0) +(1− |z |2)

zg ′1(z).

We have that f ∈ L∞(D) in light of the inequality

‖f ‖∞ ≤ |g(0)|+ |g ′(0)|+ ‖g‖B ≤ |g(0)|+ 2‖g‖B ≤ 2‖g‖Bloch

It is readily seen that Pf = g using the following result.

LemmaFor every u ∈ P,

u(z) =

(1− |ω|2)

u′(ω)

(1− ωz)2dA(ω).

Duality

The main result of this section states that (A1(D))∗ = B. For thatpurpose, we require a previous lemma.

For every f ∈ P, g ∈ H(D), the following results hold:

1.∫D f (z)g(z) dA(z) =

∫D(1− |z |2)f (z)[zg ′(z) + 2g(z)] dA(z).

2.∣∣∣∫D f (z)g(z) dA(z)

∣∣∣ ≤ C‖g‖B‖f ‖A1 .

Duality

The main result of this section states that (A1(D))∗ = B. For thatpurpose, we require a previous lemma.

For every f ∈ P, g ∈ H(D), the following results hold:

1.∫D f (z)g(z) dA(z) =

∫D(1− |z |2)f (z)[zg ′(z) + 2g(z)] dA(z).

2.∣∣∣∫D f (z)g(z) dA(z)

∣∣∣ ≤ C‖g‖B‖f ‖A1 .

(A1(D))∗ ⊇ BLet g ∈ B; using the dual pair we can define a linear functional inA1(D) given by

Tg f =

∫Df (z)g(z) dA(z), ∀f ∈ A1(D). (2)

This functional is uniformly bounded thanks to the previous lemma:

|Tg (f )| =

∣∣∣∣∫Df (z)g(z) dA(z)

∣∣∣∣ ≤ C‖g‖B‖f ‖A1 ;

‖Tg‖ = sup {|Tg (f )| : f ∈ P, ‖f ‖ ≤ 1} ≤ C‖g‖B.

(A1(D))∗ ⊆ BLet Φ ∈ (A1(D))∗; Hahn-Banach theorem states that there exists acontinuous linear functional Φ : L1(D) → C satisfying ‖Φ‖A1(D) =

‖Φ‖L1(D).

!!A1(D)

Φ // C

Since (L1(D))∗ = L∞(D), we can find g ∈ L∞(D) such that:

Φ(f ) =

∫Df (z)g(z) dA(z), ∀f ∈ L1(D).

(A1(D))∗ ⊆ BChoosing g = P(g) ∈ B, we conclude that Φ is represented by g :

Φ(f ) = Φ(Pf ) = Φ(Pf ) =

∫DPf (z)g dA(z)

∫Df (z)Pg dA(z) =

∫Df (z)g(z) dA(z).

In order to estimate the norm ‖g‖B in terms of ‖g‖L∞(D) and‖Φ‖L1(D), we compute:

‖Φ‖A1(D) = ‖Φ‖L1(D) = ‖g‖L∞(D)

‖g‖B = ‖Pg‖B ≤ ‖P‖L∞→B‖g‖L∞(D) ≤ π2

2 α‖g‖L∞(D)

Outline

4 References

References V. Asensio, C. Molina, A. Zarauz 24 de junio de 2016 29 / 30

K. Zhu. Operator theory in function spaces. Springer, New York. 2ndEdition, 2007.

K. Zhu. Spaces of Holomorphic Functions in the Unit Ball. Springer,New York. 1st Edition, 2005.

References V. Asensio, C. Molina, A. Zarauz 24 de junio de 2016 30 / 30

introduction to bloch spaces - uma · 2016. 6. 27. · introduction to bloch spaces vicent asensio...

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