selected problems in classical function ... - crm.umontreal.ca

Introduction: Spaces of Analytic Functions Analytic Functions with Positive Boundary Values Analytic Content: Approximating z̄ in Bergman and Smirnov Norms Putnam’s Inequality for Toeplitz Operators in Bergman Spaces Final Remarks

Selected Problems in Classical Function Theory

Dmitry [email protected]

http://shell.cas.usf.edu/ dkhavins/

University of South Florida

August 17, 2013


Outline

1 Introduction: Spaces of Analytic Functions

2 Analytic Functions with Positive Boundary Values

3 Analytic Content: Approximating z̄ in Bergman and SmirnovNorms

4 Putnam’s Inequality for Toeplitz Operators in Bergman Spaces

5 Final Remarks


Place of Action

open disk

not a disk!, a f. c. domain G


Place of Action

open disk

not a disk!,

a f. c. domain G


Place of Action

open disk

not a disk!, a f. c. domain G


Bergman spaces

Definition

For 0 < p <∞, define the Bergman space as

Ap =

{f analytic in D :

(∫D|f (z)|pdA(z)

) 1p

=: ‖f ‖p <∞

},

where dA = 1πdxdy denotes normalized area measure in the unit

disk D.

Ap(G ) are defined accordingly

Stefan Bergman (b. 1895 d.1977)


Bergman spaces

Definition

For 0 < p <∞, define the Bergman space as

Ap =

{f analytic in D :

(∫D|f (z)|pdA(z)

) 1p

=: ‖f ‖p <∞

},

where dA = 1πdxdy denotes normalized area measure in the unit

disk D. Ap(G ) are defined accordingly

Stefan Bergman (b. 1895 d.1977)


Hardy spaces

Definition

For 0 < p <∞, define the Hardy space as

Hp :=

{f analytic in D : sup

0<r≤1

1

2π

∫ 2π

0|f (re it)|pdt =: ‖f ‖pHp <∞

}.

(H∞ = {f analytic in D : ‖f ‖∞ := sup {|f (z)|, z ∈ D} <∞}).

Godfrey Harold Hardy (b. 1877 d. 1947)


Hardy classes in multiply connected domains

Definition

An analytic function f (z) in G belongs the the Hardy class Hp(G )for p : 0 < p <∞ if the subharmonic function |f |p has a harmonicmajorant in G .

Remark: Hardy classes (spaces) are conformally equivalent, i.e., ifϕ : K → G is the conformal mapping of an n-connected circulardomain K onto G , then f ∈ Hp(G ) if and only if f ◦ ϕ ∈ Hp(K ) .


V. I. Smirnov classes

Let G be an n-connected domain in thecomplex plane bounded by Jordan rectifiable curves γ1,...,γn andlet Γ =

⋃ni=1 γi .

Definition

An analytic function f (z) in G is said to belong to the Smirnovclass Ep(G ) for p : 0 < p <∞ if there exists a sequence ofrectifiable curves {Γi} in G converging to Γ such that

lim supi→∞

∫Γi

|f (z)|p|dz | <∞.


V. I. Smirnov (1887 - 1974)

Remark (i) The critical difference between Smirnov classes andHardy classes is that the former are NOT conformally equivalent.

(ii) For p ≥ 1 Hardy functions are represented by Poisson integralsof their boundary values, while Smirnov functions are representedby Cauchy integrals, i.e., in the former case the kernel is positive,

in the latter - complex.


J. Neuwirth - D. J. Newman (1930 - 2007) Theorem, 1967

Theorem

Let f ∈ H1/2(D) be ≥ 0 a.e. on the unit circle T. Then f isconstant. (The function z

(1+z)2 implies that 1/2 is sharp.)

Remarks (i) If “a.e.” is meant wrt the harmonic measure thetheorem automatically extends to arbitrary domains.(ii)

Theorem

(L. De Castro - DK, 2012) Let G be a simply connected Smirnovdomain with rectifiable boundary Γ. Let p0 ≥ 1 be defined as thesmallest p ≥ 1 such that f ∈ Ep(G ) and f has real boundaryvalues a.e. on Γ imply that f is a constant. Then all f ∈ Ep0/2

such that f ≥ 0 a.e. on Γ are constants.

(iii) The theorem is false for non-Smirnov domains (DK -1982).



Theorem



Remarks (i) If “a.e.” is meant wrt the harmonic measure thetheorem automatically extends to arbitrary domains.

(ii)

Theorem






Theorem



Remarks (i) If “a.e.” is meant wrt the harmonic measure thetheorem automatically extends to arbitrary domains.(ii)

Theorem





Neuwirth - Newman Theorem in Multiply ConnectedDomains

S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains.

f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function.

But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e.,

soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.

Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.

Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D.

f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ.

The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G .

Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



S.C. Smirnov Domains. f (z) = B(z)S(z)F 2(z), where B(z) is atransplanted Blaschke product, S(z) is a bounded singular innerfunction, and F (z) ∈ Ep0 is an outer function. But, on Γ,B(z)S(z)F 2(z) = |f (z)| = F (z)F (z) a.e., soF (z) = B(z)S(z)F (z) ∈ Ep0(G ) which implies thatF (z) + F (z) ∈ Ep0(G ) and is real-valued, hence a constant.Thus,f (z) = const · B(z)S(z) is a bounded function with non-negativeboundary values, hence a constant as well.Difficulty for M.C.D. f = QBSF 2, where B is the generalizedBlaschke product, S is a singular inner function, F 2 is an outerfactor, F ∈ Ep0 , Q is an invertible bounded analytic function, and|Q| is a local constant on Γ. The problem is that |B| and |S | arelocal constants a.e. on the boundary of G . Hence, f ≥ 0 a.e. on Γonly yields on Γ that f = QBSF 2 = |QBS |F F , i.e., F coincideswith different analytic functions on different boundary components.



Conjecture

(L. De Castro - DK, 2012) Let G be a f.c. Smirnov domain. Ifp0 ≥ 1 is the smallest index for which all functions in Ep0 with realboundary values are constants, then all Ep0/2 - functions withpositive boundary values are constants.

Example L. De Castro - DK, 2012.

In an n - connected domain G of “cardioid type”with m interiorcusps on ∂G all E 1-functions with positive boundary values areconstants. (For such domains p0 = 2, cf. LDC - DK, 2012)).



Conjecture



In an n - connected domain G of “cardioid type”with m interiorcusps on ∂G all E 1-functions with positive boundary values areconstants.

(For such domains p0 = 2, cf. LDC - DK, 2012)).



Conjecture



In an n - connected domain G of “cardioid type”with m interiorcusps on ∂G all E 1-functions with positive boundary values areconstants. (For such domains p0 = 2, cf. LDC - DK, 2012)).


Analytic Content

Let G be a finitely connected region in C bounded by n simpleclosed analytic curves γj , j = 1, . . . , n.

Definition

The analytic content of a domain G is

λ(G ) := infφ∈H∞(G)

‖ z̄ − φ ‖H∞(G)


Theorem

(H. Alexander - ’73, DK -’82) Let A and P be the area andperimeter of G , respectively. Then

2A

P≤ λ(G ) ≤

√A

π,

so P2 ≥ 4πA. Moreover, λ(G ) =√

Aπ ⇔ G is a disk.

Question: For which G does the equality 2AP = λ(G ) hold?


Theorem


2A

P≤ λ(G ) ≤

√A

π,

so P2 ≥ 4πA.

Moreover, λ(G ) =√




Theorem


2A

P≤ λ(G ) ≤

√A

π,

so P2 ≥ 4πA. Moreover, λ(G ) =√




Equivalent form of achievement of lower bound

Theorem

TFAE:

(i) λ = 2AP ;

(ii)There is φ ∈ H∞(G ) such that z̄(s)− iλ ˙̄z(s) = φ(z(s)) on Γ,where s is the arc-length parameter;(iii) 1

A

∫G fdA = 1

P

∫Γ fds for all f ∈ H∞(G ).

Remarks:

(iii) holds for annuli G = {r < |z | < R}.(ii) implies that if Γ contains a circular arc, G is a disk or anannulus.

(DK-’82) Are these the only two possibilities? YES!(A.Abanov, CB, DK, R. Teodorescu (2012 - 2014))

In the simply connected case the extremal domain is a disk(DK -’86).



Theorem

TFAE:(i) λ = 2A

P ;

(ii)There is φ ∈ H∞(G ) such that z̄(s)− iλ ˙̄z(s) = φ(z(s)) on Γ,where s is the arc-length parameter;(iii) 1

A

∫G fdA = 1

P


Remarks:






Theorem

TFAE:(i) λ = 2A

P ;(ii)There is φ ∈ H∞(G ) such that z̄(s)− iλ ˙̄z(s) = φ(z(s)) on Γ,where s is the arc-length parameter;

(iii) 1A

∫G fdA = 1

P


Remarks:






Theorem

TFAE:(i) λ = 2A

P ;(ii)There is φ ∈ H∞(G ) such that z̄(s)− iλ ˙̄z(s) = φ(z(s)) on Γ,where s is the arc-length parameter;(iii) 1

A

∫G fdA = 1

P


Remarks:






Theorem

TFAE:(i) λ = 2A


A

∫G fdA = 1

P


Remarks:

(iii) holds for annuli G = {r < |z | < R}.

(ii) implies that if Γ contains a circular arc, G is a disk or anannulus.





Theorem

TFAE:(i) λ = 2A


A

∫G fdA = 1

P


Remarks:






Theorem

TFAE:(i) λ = 2A


A

∫G fdA = 1

P


Remarks:


(DK-’82) Are these the only two possibilities?

YES!(A.Abanov, CB, DK, R. Teodorescu (2012 - 2014))




Theorem

TFAE:(i) λ = 2A


A

∫G fdA = 1

P


Remarks:





E p Analytic Content

Definition

Forp ≥ 1, λEp := inf

φ∈Ep(G)‖ z̄ − φ ‖Ep(G) .

Theorem

(Z. Guadarrama - DK, 2007) Let A,P denote the area andperimeter of the f.c. domain G. For p ≥ 1, q = p

p−1 , we have

2Aq√

P≤ λEp ≤

√A

πP

1p .

Questions. (i) Are disks and annuli the only extremal domains forall λEp ,p≥1, as well? (ii) Do the extremal functions for λEp

characterize the domain G ?



Definition


φ∈Ep(G)‖ z̄ − φ ‖Ep(G) .

Theorem


p−1 , we have

2Aq√

P≤ λEp ≤

√A

πP

1p .

Questions. (i) Are disks and annuli the only extremal domains forall λEp ,p≥1, as well?

(ii) Do the extremal functions for λEp




Definition


φ∈Ep(G)‖ z̄ − φ ‖Ep(G) .

Theorem


p−1 , we have

2Aq√

P≤ λEp ≤

√A

πP

1p .

Questions. (i) Are disks and annuli the only extremal domains forall λEp ,p≥1, as well? (ii) Do the extremal functions for λEp



Constant Best Approximations Characterize Disks

Theorem

(Z. Guadarrama - DK, 2007) Let Γ := ∂G be real analytic andp ≥ 1. If the best approximation (BA) to z̄ in Ep is a constant,then G is a disk.

Starting Point: WLOG BA is zero. Duality yieldsf ∗z̄dz = const|z |pds on Γ, where f ∗ is the extremal for the dual

problem. Dividing by z , f ∗(z)z dz = const|z |p−2ds. For p = 1, the

argument principle ⇒ f ∗ is a unimodular constant, and hence G isa disk. p > 1 is more complicated, p ∈ N has to be treatedseparately. If G is s.c., the regularity hypothesis can be relaxedsignificantly to assume merely that G is a Smirnov domain.Problem. Extend the above theorem to f.c. Smirnov domains.



Theorem


Starting Point:

WLOG BA is zero. Duality yieldsf ∗z̄dz = const|z |pds on Γ, where f ∗ is the extremal for the dual





Theorem


Starting Point: WLOG BA is zero.

Duality yieldsf ∗z̄dz = const|z |pds on Γ, where f ∗ is the extremal for the dual





Theorem



problem.

Dividing by z , f ∗(z)z dz = const|z |p−2ds. For p = 1, the




Theorem



problem. Dividing by z ,

f ∗(z)z dz = const|z |p−2ds. For p = 1, the




Theorem



problem. Dividing by z , f ∗(z)z dz = const|z |p−2ds.

For p = 1, theargument principle ⇒ f ∗ is a unimodular constant, and hence G isa disk. p > 1 is more complicated, p ∈ N has to be treatedseparately. If G is s.c., the regularity hypothesis can be relaxedsignificantly to assume merely that G is a Smirnov domain.Problem. Extend the above theorem to f.c. Smirnov domains.



Theorem




argument principle ⇒ f ∗ is a unimodular constant

, and hence G isa disk. p > 1 is more complicated, p ∈ N has to be treatedseparately. If G is s.c., the regularity hypothesis can be relaxedsignificantly to assume merely that G is a Smirnov domain.Problem. Extend the above theorem to f.c. Smirnov domains.



Theorem




argument principle ⇒ f ∗ is a unimodular constant, and hence G isa disk.

p > 1 is more complicated, p ∈ N has to be treatedseparately. If G is s.c., the regularity hypothesis can be relaxedsignificantly to assume merely that G is a Smirnov domain.Problem. Extend the above theorem to f.c. Smirnov domains.



Theorem




argument principle ⇒ f ∗ is a unimodular constant, and hence G isa disk. p > 1 is more complicated,

p ∈ N has to be treatedseparately. If G is s.c., the regularity hypothesis can be relaxedsignificantly to assume merely that G is a Smirnov domain.Problem. Extend the above theorem to f.c. Smirnov domains.



Theorem




argument principle ⇒ f ∗ is a unimodular constant, and hence G isa disk. p > 1 is more complicated, p ∈ N has to be treatedseparately.

If G is s.c., the regularity hypothesis can be relaxedsignificantly to assume merely that G is a Smirnov domain.Problem. Extend the above theorem to f.c. Smirnov domains.



Theorem




argument principle ⇒ f ∗ is a unimodular constant, and hence G isa disk. p > 1 is more complicated, p ∈ N has to be treatedseparately. If G is s.c., the regularity hypothesis can be relaxedsignificantly to assume merely that G is a Smirnov domain.

Problem. Extend the above theorem to f.c. Smirnov domains.



Theorem






Do BA to z̄ Characterize Annuli

Theorem

(ZG - DK, 2007) Let Γ := ∂G be real analytic and p = 1(!). If thebest approximation (BA) to z̄ in E 1 is a rational functiong(z) = c

z−a , then G is an annulus centered at a.

Conjecture

The theorem holds for all p > 1 and all f.c. Smirnov domains.

Unknown Territory: Study domains where BA to z̄ in Ep are,say, rational functions. It is easy to see that it leads to a largerclass than (by now celebrated) quadrature domains.



Theorem



Conjecture


Unknown Territory: Study domains where BA to z̄ in Ep are,say, rational functions.

It is easy to see that it leads to a largerclass than (by now celebrated) quadrature domains.



Theorem



Conjecture


Unknown Territory: Study domains where BA to z̄ in Ep are,say, rational functions. It is easy to see that it leads to a largerclass than (by now celebrated) quadrature domains.


Ap Analytic Content

Definition

Forp ≥ 1, λAp := inf

φ∈Ap(G)‖ z̄ − φ ‖Ap(G) .

Theorem

(ZG - DK, 2007) (i) Let G be a Smirnov domain and p ≥ 1. If BAto z̄ in Ap is a constant, then G is a disk.

(ii)If BA to z̄ in Ap is g(z) = cz−a , then G is an annulus centered

at a.

The Bergman spaces are easier!


Here is why

((i), p > 1, BA= 0):The duality yields

uz̄ = const|z |p

z̄, u ∈W 1,q

0 is a solution of the dual problem.

Integrating wrt z̄ gives (in G !):

u = const|z |p + h, where h ∈ H∞(G ).

Hence h|Γ is real-valued, so |z | = const on Γ.

What are the sharp bounds for λAp? Nothing is known aboutthe domains with other rational BA (in Ap).


Here is why ((i), p > 1, BA= 0):

The duality yields

uz̄ = const|z |p

z̄, u ∈W 1,q







Here is why ((i), p > 1, BA= 0):The duality yields

uz̄ = const|z |p

z̄, u ∈W 1,q







Putnam’s Inequality (1970) on E 2 and the IsoperimetricInequality

Let φ be analytic in a neighborhood of the f.c. domainG , ∂G := Γ,A := A(G ) = area of G ,P := P(Γ) = perimeter of G .Let T := Tφ : E 2 → E 2,Tf = φf , [T ∗,T ] = T ∗T − TT ∗.Putnam’s inequality then states

A(φ(G ))

π≥ ‖[T ∗,T ]‖ ≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).

φ = z =⇒ P2 ≥ 4πA, the isoperimetric inequality. Thus, Putnam’sinequality in E 2 context is sharp.Question: What are the bounds in the Bergman space context?



Let φ be analytic in a neighborhood of the f.c. domainG , ∂G := Γ,A := A(G ) = area of G ,P := P(Γ) = perimeter of G .

Let T := Tφ : E 2 → E 2,Tf = φf , [T ∗,T ] = T ∗T − TT ∗.Putnam’s inequality then states

A(φ(G ))

π≥ ‖[T ∗,T ]‖ ≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).




Let φ be analytic in a neighborhood of the f.c. domainG , ∂G := Γ,A := A(G ) = area of G ,P := P(Γ) = perimeter of G .Let T := Tφ : E 2 → E 2,Tf = φf , [T ∗,T ] = T ∗T − TT ∗.

Putnam’s inequality then states

A(φ(G ))

π≥ ‖[T ∗,T ]‖ ≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).





A(φ(G ))

π≥ ‖[T ∗,T ]‖ ≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).





A(φ(G ))

π≥ ‖[T ∗,T ]‖

≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).





A(φ(G ))

π≥ ‖[T ∗,T ]‖ ≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).





A(φ(G ))

π≥ ‖[T ∗,T ]‖ ≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).

φ = z =⇒ P2 ≥ 4πA, the isoperimetric inequality.

Thus, Putnam’sinequality in E 2 context is sharp.Question: What are the bounds in the Bergman space context?




A(φ(G ))

π≥ ‖[T ∗,T ]‖ ≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).

φ = z =⇒ P2 ≥ 4πA, the isoperimetric inequality. Thus, Putnam’sinequality in E 2 context is sharp.

Question: What are the bounds in the Bergman space context?




A(φ(G ))

π≥ ‖[T ∗,T ]‖ ≥ 4(A(φ(G )))2

‖φ′‖2E2(G)

· P(DK - 1985).



Putnam’s inequality in A2

Observation: (M. Fleeman -DK (2012), folowing S. Axler - J.Shapiro (1985)). For a subnormal operator T the equality inPutnam’s inequality occurs only if the spectrum sp(T ) is a diskand the spectral measure “sits” on the circumference.The crux isthe equality case in the upper bound for the analytic content:

λ(G ) ≤√

Aπ . Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic

‖z̄g−f ‖2} ≤ infhanalytic

‖z̄−h‖∞ = λ.

A calculation (straightforward but tedious!) reveals that inA2(D), ‖[T ∗z ,Tz ]‖ = 1/2, i.e., the upper bound is two timessmaller than in the general Putnam inequality.Question: Should Putnam’s inequality in the context of theBergman space be corrected by a factor 1/2?



Observation: (M. Fleeman -DK (2012), folowing S. Axler - J.Shapiro (1985)).

For a subnormal operator T the equality inPutnam’s inequality occurs only if the spectrum sp(T ) is a diskand the spectral measure “sits” on the circumference.The crux isthe equality case in the upper bound for the analytic content:

λ(G ) ≤√

Aπ . Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic


‖z̄−h‖∞ = λ.




Observation: (M. Fleeman -DK (2012), folowing S. Axler - J.Shapiro (1985)). For a subnormal operator T the equality inPutnam’s inequality occurs only if the spectrum sp(T ) is a diskand the spectral measure “sits” on the circumference.

The crux isthe equality case in the upper bound for the analytic content:

λ(G ) ≤√

Aπ . Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic


‖z̄−h‖∞ = λ.





λ(G ) ≤√

Aπ . Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic


‖z̄−h‖∞ = λ.





λ(G ) ≤√

Aπ .

Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic


‖z̄−h‖∞ = λ.





λ(G ) ≤√

Aπ . Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic

‖z̄g−f ‖2}

≤ infhanalytic

‖z̄−h‖∞ = λ.





λ(G ) ≤√

Aπ . Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic


‖z̄−h‖∞ = λ.





λ(G ) ≤√

Aπ . Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic


‖z̄−h‖∞ = λ.

A calculation (straightforward but tedious!) reveals that inA2(D), ‖[T ∗z ,Tz ]‖ = 1/2, i.e., the upper bound is two timessmaller than in the general Putnam inequality.

Question: Should Putnam’s inequality in the context of theBergman space be corrected by a factor 1/2?




λ(G ) ≤√

Aπ . Indeed,

‖[T ∗z ,Tz ]‖ = sup‖g‖2=1

{ inff analytic


‖z̄−h‖∞ = λ.



Toeplitz Operators on the Bergman Space andSaint-Venant Inequality

The torsional rigidity ρ of a domain G measures the resilience ofthe beam of cross section G to twisting. In terms of the Rayleightype quotient :

ρ := supψ∈C∞0

(2‖ψ‖1

‖∇ψ‖2

)2

.

Theorem

(S. Bell - T. Ferguson - E. Lundberg, 2012)

ρ

Area(φ(G ))≤ ‖[T ∗φ ,Tφ]‖ (Putnam’s) ≤ A(φ(G ))

π.

Hence, taking φ = z, we obtain from the “isoperimetric sandwich”

ρ ≤ (Area(G ))2

π.



The torsional rigidity ρ of a domain G measures the resilience ofthe beam of cross section G to twisting.

In terms of the Rayleightype quotient :

ρ := supψ∈C∞0

(2‖ψ‖1

‖∇ψ‖2

)2

.

Theorem


ρ


π.


ρ ≤ (Area(G ))2

π.




ρ := supψ∈C∞0

(2‖ψ‖1

‖∇ψ‖2

)2

.

Theorem


ρ


π.


ρ ≤ (Area(G ))2

π.




ρ := supψ∈C∞0

(2‖ψ‖1

‖∇ψ‖2

)2

.

Theorem


ρ

Area(φ(G ))≤ ‖[T ∗φ ,Tφ]‖

(Putnam’s) ≤ A(φ(G ))

π.


ρ ≤ (Area(G ))2

π.




ρ := supψ∈C∞0

(2‖ψ‖1

‖∇ψ‖2

)2

.

Theorem


ρ

Area(φ(G ))≤ ‖[T ∗φ ,Tφ]‖ (Putnam’s)

≤ A(φ(G ))

π.


ρ ≤ (Area(G ))2

π.




ρ := supψ∈C∞0

(2‖ψ‖1

‖∇ψ‖2

)2

.

Theorem


ρ


π.


ρ ≤ (Area(G ))2

π.




ρ := supψ∈C∞0

(2‖ψ‖1

‖∇ψ‖2

)2

.

Theorem


ρ


π.

Hence, taking φ = z,

we obtain from the “isoperimetric sandwich”

ρ ≤ (Area(G ))2

π.




ρ := supψ∈C∞0

(2‖ψ‖1

‖∇ψ‖2

)2

.

Theorem


ρ


π.


ρ ≤ (Area(G ))2

π.


J.-F. Olsen - M.C. Reguera Theorem, 2013

Bell - Ferguson - Lundberg estimate ρ ≤ (Area(G))2

π

was missing bya factor of 2 the celebrated Saint-Venant inequality conjectured in1856. It was first proved by G. Polya in 1948.

Conjecture

(Bell - Ferguson - Lundberg, 2012) For the Bergman space

‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .

For simply connected domains G it is now a theorem! (J.-F. Olsen- M.C. Reguera, May 2013). Hence,

Saint-Venant’s Inequality, a new proof: ρ ≤ (Area(G))2

2π .The proof is tour de force calculation with power series. This iswhy the statement is restricted to s.c. domains.M. Fleeman -DK noted that refining Olsen - Reguera’s proofimplies that the equality for the self-commutator upper bound ins.c. domains holds only for disks. This yields an alternative proofthat St.-Venant’s inequality becomes equality only for disks.




π was missing bya factor of 2 the celebrated Saint-Venant inequality conjectured in1856.

It was first proved by G. Polya in 1948.

Conjecture


‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .







π was missing bya factor of 2 the celebrated Saint-Venant inequality conjectured in1856. It was first proved by G. Polya in 1948.

Conjecture


‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .








Conjecture


‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .

For simply connected domains G it is now a theorem! (J.-F. Olsen- M.C. Reguera, May 2013).

Hence,







Conjecture


‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .



2π .

The proof is tour de force calculation with power series. This iswhy the statement is restricted to s.c. domains.M. Fleeman -DK noted that refining Olsen - Reguera’s proofimplies that the equality for the self-commutator upper bound ins.c. domains holds only for disks. This yields an alternative proofthat St.-Venant’s inequality becomes equality only for disks.





Conjecture


‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .



2π .The proof is tour de force calculation with power series.

This iswhy the statement is restricted to s.c. domains.M. Fleeman -DK noted that refining Olsen - Reguera’s proofimplies that the equality for the self-commutator upper bound ins.c. domains holds only for disks. This yields an alternative proofthat St.-Venant’s inequality becomes equality only for disks.





Conjecture


‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .



2π .The proof is tour de force calculation with power series. This iswhy the statement is restricted to s.c. domains.

M. Fleeman -DK noted that refining Olsen - Reguera’s proofimplies that the equality for the self-commutator upper bound ins.c. domains holds only for disks. This yields an alternative proofthat St.-Venant’s inequality becomes equality only for disks.





Conjecture


‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .



2π .The proof is tour de force calculation with power series. This iswhy the statement is restricted to s.c. domains.M. Fleeman -DK noted that refining Olsen - Reguera’s proofimplies that the equality for the self-commutator upper bound ins.c. domains holds only for disks.

This yields an alternative proofthat St.-Venant’s inequality becomes equality only for disks.





Conjecture


‖[T ∗z ,Tz ]‖ ≤ Area(G)2π .





Further Research

Find the “book” proof of Olsen - Reguera theorem, freeing itfrom the power series calculation and extending the result toarbitrary domains.

Is the sharp upper bound for the A2-content equal Area(G)2π ,

similarly to the situation with the analytic content?

What is the sharp lower bound for the A2 - content expressedin terms of “simple” geometric characteristics (e.g., area,perimeter, principal frequency) of the domain?

Refine the “isoperimetric sandwiches” to include theconnectivity of the domain. This is virtually unknownterritory. For the celebrated Carleman’s inequality boundingA2(G ) norm in terms of E 1(G ) norm, some initial steps weremade in S. Jacobs’ thesis in 1973 (unpublished). For theanalytic content, there is a result of DK - D. Luecking from1984. Nothing else has been done since.


Further Research





Refine the “isoperimetric sandwiches” to include theconnectivity of the domain.

This is virtually unknownterritory. For the celebrated Carleman’s inequality boundingA2(G ) norm in terms of E 1(G ) norm, some initial steps weremade in S. Jacobs’ thesis in 1973 (unpublished). For theanalytic content, there is a result of DK - D. Luecking from1984. Nothing else has been done since.


Further Research





Refine the “isoperimetric sandwiches” to include theconnectivity of the domain. This is virtually unknownterritory.

For the celebrated Carleman’s inequality boundingA2(G ) norm in terms of E 1(G ) norm, some initial steps weremade in S. Jacobs’ thesis in 1973 (unpublished). For theanalytic content, there is a result of DK - D. Luecking from1984. Nothing else has been done since.


Further Research





Refine the “isoperimetric sandwiches” to include theconnectivity of the domain. This is virtually unknownterritory. For the celebrated Carleman’s inequality boundingA2(G ) norm in terms of E 1(G ) norm, some initial steps weremade in S. Jacobs’ thesis in 1973 (unpublished).

For theanalytic content, there is a result of DK - D. Luecking from1984. Nothing else has been done since.


Further Research





Refine the “isoperimetric sandwiches” to include theconnectivity of the domain. This is virtually unknownterritory. For the celebrated Carleman’s inequality boundingA2(G ) norm in terms of E 1(G ) norm, some initial steps weremade in S. Jacobs’ thesis in 1973 (unpublished). For theanalytic content, there is a result of DK - D. Luecking from1984.

Nothing else has been done since.


Further Research





Refine the “isoperimetric sandwiches” to include theconnectivity of the domain. This is virtually unknownterritory. For the celebrated Carleman’s inequality boundingA2(G ) norm in terms of E 1(G ) norm, some initial steps weremade in S. Jacobs’ thesis in 1973 (unpublished). For theanalytic content, there is a result of DK - D. Luecking from1984. Nothing else has been done since.


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selected problems in classical function ... - crm.umontreal.ca

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