Quasicircles, quasiconformal extensions and the Corona … › ~pietro › Seattle19-Slides › MariaJose-Gonzalez.pdfCarleson’s Theorem Carleson, L., Interpolation by bounded analytic

Quasicircles, quasiconformal extensions andthe Corona Theorem

María José GonzálezUniversidad de Cádiz

Celebrating J. Garnett and D. MarshallSeattle 2019

María José González Universidad de CádizQuasicircles, quasiconformal extensions and the Corona Theorem

Corona Problem

Theorem (Carleson). Let f1(z), ..., fn(z) be given functions inH∞(D) with ||fk ||∞ ≤ 1, k = 1,2, ...,n, and verifying that forsome δ > 0,

|f1(z)|+ |f2(z)|+ ...|fn(z)| ≥ δ > 0,

Then, there exist gknk=0 ∈ H∞(D), so that:

n∑k=1

fkgk = 1

and ||gk ||∞ ≤ C(n, δ).

The functions fk and gk are called corona data and coronasolutions respectively.

OPEN PROBLEM: Is Corona true for any domain in the plane?


Corona Problem


|f1(z)|+ |f2(z)|+ ...|fn(z)| ≥ δ > 0,


n∑k=1

fkgk = 1

and ||gk ||∞ ≤ C(n, δ).




Corona Problem


|f1(z)|+ |f2(z)|+ ...|fn(z)| ≥ δ > 0,


n∑k=1

fkgk = 1

and ||gk ||∞ ≤ C(n, δ).




Carleson’s Theorem

Carleson, L., Interpolation by bounded analytic functions andthe corona problem, Ann. of Math. 76 (1962), 547-559.

Carleson measures: µ a positive measure in D , andf ∈ Hp, p ≥ 1∫

D|f |p dµ ≤ ‖f‖pp iff µ(Q) ≤ c l(Q)

for any Carleson cube Q ⊂ D.

Carleson Contour: System of curves Γ where the analyticfunction is not too big, not too small, i.e. ε < |f | < εk ; k < 1,and arc-lengh Γ is a Carleson measure.

Corona decomposition


















Interpolation

Let B(z) and C(z) be Blaschke products with zeros (bn) and(cn) respectively, and such that |B(z)|+ |C(z)| ≥ δ > 0

The following two problems are equivalent

1 Solve Corona problem with corona data B(z) and C(z).

2 Construct f ∈ H∞(D), with ‖f‖ ≤ c(δ) such that f (bn) = 0and f (cn) = 1

f = B h; h ∈ H∞(D)

1− f = C g g ∈ H∞(D)

Then B h + C g = 1


Interpolation





f = B h; h ∈ H∞(D)

1− f = C g g ∈ H∞(D)

Then B h + C g = 1


Interpolation





f = B h; h ∈ H∞(D)

1− f = C g g ∈ H∞(D)

Then B h + C g = 1


Solving interpolation

By duality, if wn is the sequence of 0’s and 1’s

‖f‖∞ = sup

∣∣∣∣∑ G(wn) wn

(B C)′(wn)

∣∣∣∣ ; G ∈ H1, ‖G‖1 ≤ 1

Let Γ be the Carleson contour for C(z) with εk < δ/2.Recall that |B(z)|+ |C(z)| > δ.

‖f‖∞ = sup

∣∣∣∣∑ G(cn)

(B C)′(cn)

∣∣∣∣ = sup

∣∣∣∣ 12πi

∫Γ

G(z)

B(z) C(z)dz∣∣∣∣

< c(δ)

∫Γ|G(z)|ds < C(δ)‖G‖1 ≤ c(δ).


Solving interpolation

By duality, if wn is the sequence of 0’s and 1’s

‖f‖∞ = sup

∣∣∣∣∑ G(wn) wn

(B C)′(wn)

∣∣∣∣ ; G ∈ H1, ‖G‖1 ≤ 1

Let Γ be the Carleson contour for C(z) with εk < δ/2.Recall that |B(z)|+ |C(z)| > δ.

‖f‖∞ = sup

∣∣∣∣∑ G(cn)

(B C)′(cn)

∣∣∣∣ = sup

∣∣∣∣ 12πi

∫Γ

G(z)

B(z) C(z)dz∣∣∣∣

< c(δ)

∫Γ|G(z)|ds < C(δ)‖G‖1 ≤ c(δ).


∂- problem

Wolff,T. Published by Gamelin, T. W. (1980), Wolff’s proof of thecorona theorem, Israel Journal of Mathematics, 37, 113-119

∂- problem: Corona problem for two functions: f1 and f2. Firstfind solutions ϕ1 and ϕ2 NOT necessarily analytic. Set

g1 = ϕ1 + b f2

g2 = ϕ2 − b f1

Want b such that∂ϕ1 + ∂b f2 = 0

∂ϕ2 − ∂b f1 = 0

Therefore, to solve corona it is enough to solve

∂b = ϕ2∂ϕ1 − ϕ1∂ϕ2

Wolff‘s choice: ϕj = fj/∑|fj |2


∂- problem

Wolff,T. Published by Gamelin, T. W. (1980), Wolff’s proof of thecorona theorem, Israel Journal of Mathematics, 37, 113-119

∂- problem: Corona problem for two functions: f1 and f2. Firstfind solutions ϕ1 and ϕ2 NOT necessarily analytic. Set

g1 = ϕ1 + b f2

g2 = ϕ2 − b f1

Want b such that∂ϕ1 + ∂b f2 = 0

∂ϕ2 − ∂b f1 = 0

Therefore, to solve corona it is enough to solve

∂b = ϕ2∂ϕ1 − ϕ1∂ϕ2

Wolff‘s choice: ϕj = fj/∑|fj |2


Solving ∂-problem: particular case

Let µ be a Carleson measure in D.

Want to solve∂b = µ

First try:

b(z) = F (z) =1π

∫C

dµ(w)

w − z

BUT F(z) might NOT be bounded. By duality

‖b‖∞ = supG∈H1,‖G‖1≤1

∣∣∣∣ 12πi

∫∂D

F (z)G(z) dz∣∣∣∣

' sup

∣∣∣∣∫D∂ (F G) dx dy

∣∣∣∣ ≤ sup

∫D|G| dµ ≤ c ‖G‖1 ≤ c

where the constant c depends on the Carleson constant of µ.


Solving ∂-problem: particular case

Let µ be a Carleson measure in D.

Want to solve∂b = µ

First try:

b(z) = F (z) =1π

∫C

dµ(w)

w − z

BUT F(z) might NOT be bounded. By duality

‖b‖∞ = supG∈H1,‖G‖1≤1

∣∣∣∣ 12πi

∫∂D

F (z)G(z) dz∣∣∣∣

' sup

∣∣∣∣∫D∂ (F G) dx dy

∣∣∣∣ ≤ sup

∫D|G| dµ ≤ c ‖G‖1 ≤ c

where the constant c depends on the Carleson constant of µ.


Multiply connected domains

Carleson, L., On H∞ in multiply connected domains.Conference on Harmonic Analysis in Honor of AntoniZygmund, 1983, pp. 349-372.

Denjoy domains with thick boundary: Domains havingboundary E ⊂ R satisfying, for some ε > 0,

|(x − r , x + r) ∩ E | ≥ εrfor every x ⊂ E and r > 0.

Jones, P. W. and Marshall, D. E., Critical points of Green’sfunction, harmonic measure, and the corona problem, Ark.för Mat. 23 (1985), no. 2, 281-314.

They extend the result to thick Cantor sets and they showthat it is enough to solve the corona problem at the criticalpoints of the Green function.

Idea: Construct an explicit projection from H∞(D) onto H∞Γ (Ω).María José González Universidad de Cádiz

Quasicircles, quasiconformal extensions and the Corona Theorem

















No thickness

Garnett, J.B., Jones, P.W.: The corona theorem for Denjoydomains. Acta Math. 155, 27-40 (1985).

NO assumption on thickness1 Divide the domain Ω into two almost disjoint regions: Ω1

where ω(z,E) > ε and Ω2 = Ω \ Ω1(cε).2 Solve Corona in each of the components of Ω1, because

they are simply connected (Max. principle).3 Find explicit solutions in Ω2 using that f (z) is analytic.4 Glue them together by solving a ∂-problem: First find a

Carleson contour Γ in Ω1 ∩ Ω2, define the region (diamondshape) D = z; dist(z, Γ) < α y. Then there is Ψsupported on D, which is 0 or 1 on each side of D, andsuch that |y∇ψ| is a Carleson mesaure. Set

ϕj = Gj,1(1−Ψ) + Gj,2Ψ

Solve the corresponding ∂-problem ( No duality, by hand!!!)María José González Universidad de Cádiz


More results

Moore, C.N.: The corona Theorem for domains whoseboundary lies in a smooth curve. P. Am. Math. Soc. 100,266-270 (1987).Handy, J.: The Corona Theorem on the Complement ofCertain Square Cantor Sets. J. Anal. Math. 108, 1-18(2009).Newdelman, B. M.: Homogeneous subsets of a lipschitzgraph and the corona theorem. Publ. Mat. 55, 93-121(2011).

Open: Corona for the 1/3-Cantor set or Corona fordomains with bounday contained on a chord- arc curve (nothickness).


More results

Moore, C.N.: The corona Theorem for domains whoseboundary lies in a smooth curve. P. Am. Math. Soc. 100,266-270 (1987).Handy, J.: The Corona Theorem on the Complement ofCertain Square Cantor Sets. J. Anal. Math. 108, 1-18(2009).Newdelman, B. M.: Homogeneous subsets of a lipschitzgraph and the corona theorem. Publ. Mat. 55, 93-121(2011).

Open: Corona for the 1/3-Cantor set or Corona fordomains with bounday contained on a chord- arc curve (nothickness).


Different approach ( joint work with J.M. Salamanca)

Let ρ be a quasiconformal mapping on the plane withcomplex dilatation µ, ‖µ‖∞ < 1..

∂ρ− µ∂ρ = 0

A Jordan curve Γ passing through∞ is a quasicircle if it isthe image of the real line R under a quasiconformalmapping on the plane.

Geometric properties of Γ←→ Conditions on µ.


Cauchy Integral

Given a function f on a rectifiable curve Γ, define its Cauchyintegral F (z) = CΓf (z) off Γ by

F (z) =1

2π

∫Γ

f (w)

w − zdw , z /∈ Γ.

If F+ and F− are the restrictions of F to Ω+ and Ω−, and if f+and f− denote their boundary values, then the classical Plemeljformula states that

f±(z) = ±12

f (z) +1

2πP.V.

∫Γ

f (w)

w − zdw , z ∈ Γ.

The singular integral is also called the Cauchy integral.

In particular the jump of F is f . Write j(F ) = f .


Cauchy Integral

Given a function f on a rectifiable curve Γ, define its Cauchyintegral F (z) = CΓf (z) off Γ by

F (z) =1

2π

∫Γ

f (w)

w − zdw , z /∈ Γ.

If F+ and F− are the restrictions of F to Ω+ and Ω−, and if f+and f− denote their boundary values, then the classical Plemeljformula states that

f±(z) = ±12

f (z) +1

2πP.V.

∫Γ

f (w)

w − zdw , z ∈ Γ.

The singular integral is also called the Cauchy integral.

In particular the jump of F is f . Write j(F ) = f .


Passing to a Denjoy domain

Consider a domain of the form

Ω = C \ E

where E ⊂ Γ is a compact set with positive length contained ina rectifiable quasicircle Γ = ρ (∂R).

Set Ω0 = ρ−1(Ω) and E0 = ρ−1(E). Note that Ω0 is a Denjoydomain.

Define the space:

H∞(Ω0, µ) = g = f ρ : f ∈ H∞(Ω)

Then,∂f = 0 on Ω⇔ ∂g − µ∂g = 0 on Ω0

The jump of g across E0 is given by j(g) = j(f ) ρ.


Passing to a Denjoy domain

Consider a domain of the form

Ω = C \ E

where E ⊂ Γ is a compact set with positive length contained ina rectifiable quasicircle Γ = ρ (∂R).

Set Ω0 = ρ−1(Ω) and E0 = ρ−1(E). Note that Ω0 is a Denjoydomain.

Define the space:

H∞(Ω0, µ) = g = f ρ : f ∈ H∞(Ω)

Then,∂f = 0 on Ω⇔ ∂g − µ∂g = 0 on Ω0

The jump of g across E0 is given by j(g) = j(f ) ρ.


Dictionary

Idea: Translate Corona data in Ω to Corona data in the Denjoydomain Ω0.

Let f ∈ H∞(Ω) and g = f ρ ∈ H∞(Ω0, µ).

Set g = CR(j(g)), where j(g) is the jump of g. Then g isanalytic in Ω0, but NOT necessarily bounded.Define

H = g − g

Then∂H = µ ∂g

Since H has no jump across E0, we can consider that thisequation holds on all C.

REMARK: H ∈ L∞(C) iff g ∈ H∞(Ω0).


Dictionary




H = g − g

Then∂H = µ ∂g




Dictionary




H = g − g

Then∂H = µ ∂g




Solving ∂

Recall that f ∈ L∞(Γ) and g = f ρ. Want to solve ∂H = µ∂g

H(z0) =1π

∫C

∂Hz − z0

dxdy =1π

∫C

µ∂gz − z0

dxdy

for all z0 ∈ C.

PROBLEM: Find conditions on µ so that H ∈ L∞(C).

Theorem: If the quasicircle is C1+α, then the conformalmapping can be entended to a quasiconformal map with|µ|2/y1+ε Carleson measure. The converse is also true.

Theorem: Under such conditions, H is bounded in C , andCorona holds. (We recover Moore‘s result).

Idea: Show that the dictionary provides corona data on Ω0, useGarnett-Jones to find Corona solutions on Ω0, use thedictionary to get analytic functions in Ω. Modify them to get thecorona solutions ( localization argument by T. Gamelin)


Solving ∂


H(z0) =1π

∫C

∂Hz − z0

dxdy =1π

∫C

µ∂gz − z0

dxdy

for all z0 ∈ C.






Solving ∂


H(z0) =1π

∫C

∂Hz − z0

dxdy =1π

∫C

µ∂gz − z0

dxdy

for all z0 ∈ C.






Solving ∂


H(z0) =1π

∫C

∂Hz − z0

dxdy =1π

∫C

µ∂gz − z0

dxdy

for all z0 ∈ C.






More conditions

Theorem: If µ satisfies any of the following two conditions, thenCorona holds

1 ∫0

µ∗(t)|t |

log(1/|t |)dt <∞

where µ∗(t) = sup|µ(z)| : 0 < |Im(z)| < |t |

2 ∫R

σ(y)

|y |3/2 dy <∞

where σ(y) = (∫R |µ(x + iy)|2 dx)1/2, a.a. y ∈ R, and

|µ(z0)| .∫|z−z0|<C|y0|

|µ(z)|dxdy , z0 ∈ C \ R


Smooth curves

Remark : In both cases, for all a ∈ R,∫C

|µ(z)||z − a|

dxdy|y |

< M

and therefore the quasicircle is smooth (Gutlyanskii and Martio)

Example: Let h be the conformal map taking D onto the ballB(9/10,1/10),h(z) = (9 + z)/10. Consider:

g(z) = 2z +1− z

log(1− z)

and set f = g h.

Then f is conformal in D and, Γ = f (∂D) is smooth. It has aquasiconformal extension satisfying condition (2), BUT it isNOT Dini smooth.


Smooth curves

Remark : In both cases, for all a ∈ R,∫C

|µ(z)||z − a|

dxdy|y |

< M

and therefore the quasicircle is smooth (Gutlyanskii and Martio)

Example: Let h be the conformal map taking D onto the ballB(9/10,1/10),h(z) = (9 + z)/10. Consider:

g(z) = 2z +1− z

log(1− z)

and set f = g h.

Then f is conformal in D and, Γ = f (∂D) is smooth. It has aquasiconformal extension satisfying condition (2), BUT it isNOT Dini smooth.


T HANK YOU DON AND JOHN

for your Math and for your Kindness.


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Quasicircles, quasiconformal extensions and the Corona … › ~pietro › Seattle19-Slides › MariaJose-Gonzalez.pdfCarleson’s Theorem Carleson, L., Interpolation by bounded analytic