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Conference for Tony O’Farrell, June 2013

SWISS CHEESE

&

CHAMPAGNE

BUBBLES

Ivan Netuka

Mathematical Institute, Faculty of Mathematics and Physics, Charles University in Prague

IvanNetuka (CharlesUniversity in Prague) CHEESE & CHAMPAGNE 1 / 20

SWISS CHEESE


SWISS CHEESE

A Swiss cheese is a nowhere dense

compact set K ⊂ C formed byremoving from the closure B of theopen unit disc B a collection of opendiscs Bn := B(xn, rn) such thatBn ⊂ B , Bn are pairwise disjoint, and∑∞

n=1 rn < ∞.


SWISS CHEESE

A Swiss cheese is a nowhere dense

compact set K ⊂ C formed byremoving from the closure B of theopen unit disc B a collection of opendiscs Bn := B(xn, rn) such thatBn ⊂ B , Bn are pairwise disjoint, and∑∞

n=1 rn < ∞.

For a compact set X in C, R(X ) is the uniform closure in C (X ) of thealgebra R0(X ) of rational functions with poles outside X .


Theorem (F. Hartogs and A. Rosenthal).

If X is a compact subset of C and λ2(X ) = 0, then R(X ) = C (X ).




[An extension of the Weierstrass Approximation Theorem.]




[An extension of the Weierstrass Approximation Theorem.]

Proof.

For a complex Radon measure ν on X , the Cauchy transform is defined by

ν̂(ζ) =

∫

dν(z)

z − ζ, ζ ∈ C.

Facts:

ν̂ = 0 on X c if and only if ν⊥R(X )

ν̂ = 0 λ2 - a.e. implies ν = 0


For a Swiss cheese K , R(K ) is called the Swiss cheese algebra.



Theorem (A. Roth; S. N. Mergelyan).

Every Swiss cheese algebra is a proper subalgebra of C(K ).



Theorem (A. Roth; S. N. Mergelyan).

Every Swiss cheese algebra is a proper subalgebra of C(K ).

Proof.

Let µ be the measure on ∂B ∪⋃∞

n=1 ∂Bn which coincides with dz on ∂B

and −dz on⋃∞

n=1 ∂Bn. Since∑∞

n=1 rn < ∞, µ is a finite measure, andobviously µ 6= 0. Cauchy’s theorem gives

∫

K

r dµ = 0 , r ∈ R0(K ).

Thus∫

K

f dµ = 0 , f ∈ R(K ).

Hence R(K ) 6= C (K ) by the Hahn-Banach Theorem.


Summary

R(X ) = C (X ) for every compact set X with λ2(X ) = 0

R(K ) 6= C (K ) for every Swiss cheese K


Corollary.

λ2(K ) > 0 for every Swiss cheese K.


Corollary.

λ2(K ) > 0 for every Swiss cheese K.

A direct proof (O. Wesler).

x 0 S

B

CHxL

Take x ∈ S , the chord C (x), and denoteZ1 :=

{

x ∈ S : λ1(C (x) ∩ K ) > 0}

.Suppose that λ2(K ) = 0. Fubini’s theo-

rem: λ1(Z1) = 0. Put Z2 :={

x ∈ S\Z1 :C (x) meets only finitely many discs Bn

}

.For x ∈ Z2, C (x) must pass, in this finitestring of discs, through points of tangen-

cy. There is countably many points oftangency among B and the Bn. Hence Z2

is countable, thus λ1(Z1 ∪ Z2) = 0.Let In be the projection of Bn onto S . Then In is an open interval of length2 rn. Almost all points of S belong to infinitely many In. Borel-Cantelli

lemma (the easy half):∑∞

n=1 rn = ∞, a contradiction.


Swiss cheese origin remark. It is often stated that Swiss cheese isbasically due to Mergelyan overlooking the work of Swiss mathematicianAlice Roth (Commentarii Mathematici Helvetici, 1938).


Swiss cheese origin remark. It is often stated that Swiss cheese isbasically due to Mergelyan overlooking the work of Swiss mathematicianAlice Roth (Commentarii Mathematici Helvetici, 1938).

In the article Alice in Switzerland: the life and mathematics of Alice Roth

(Mathematical Intelligencer, 2005), the authors say:. . . her Swiss cheese has been modified (to an entire variety of cheeses).


TONY’S FINE CHEESE



A point derivation on R(X ) at x ∈ X is a linear functional D on R(X )such that

D(fg) = f (x)Dg + g(x)Df , f , g ∈ R(X ).

D is called a bounded point derivation at x , if D is continuous.

Equivalent to: there exists α ∈ R such that

|r ′(x)| ≤ α ||r ||X , r ∈ R0(X ).



A point derivation on R(X ) at x ∈ X is a linear functional D on R(X )such that

D(fg) = f (x)Dg + g(x)Df , f , g ∈ R(X ).

D is called a bounded point derivation at x , if D is continuous.

Equivalent to: there exists α ∈ R such that

|r ′(x)| ≤ α ||r ||X , r ∈ R0(X ).

Facts:

There exists a Swiss cheese K such that R(K ) admits no boundedpoint derivations. (J. Wermer)


There exists a Swiss cheese K (Tony’s fine cheese) with the propertythat R(K ) admits a bounded point derivation at exactly one point.


There exists a Swiss cheese K (Tony’s fine cheese) with the propertythat R(K ) admits a bounded point derivation at exactly one point.

This result stands in marked contrast to the following result.

For every compact X ⊂ C, the set of points of X at which R(X )admits a (not necessarily bounded) point derivation has no isolated

points. (A. Browder)


ANOTHER CHEESE


ANOTHER CHEESE

Theorem.

There exists a sequence {xn} of distinct points in B := B(0, 1) ⊂ R2 and

a sequence {an} ∈ l1 such that, for every bounded harmonic function h

on B,

h(0) =∞

∑

n=1

an h(xn).


ANOTHER CHEESE

Theorem.

There exists a sequence {xn} of distinct points in B := B(0, 1) ⊂ R2 and

a sequence {an} ∈ l1 such that, for every bounded harmonic function h

on B,

h(0) =∞

∑

n=1

an h(xn).

Proof.

Choose pairwise disjoint closed discs Bn := B(xn, rn) in B such thatλ2

(

B\⋃∞

n=1 Bn

)

= 0 (the Vitali Covering Theorem). Then, for an := r2n ,

h(0) = (1/π)

∫

B

h dλ2 =

∞∑

n=1

an · (1/πr2n ) ·

∫

Bn

h dλ2 =

∞∑

n=1

an h(xn).


Remark

a quadrature formula∫

Bh dλ2 =

∑∞n=1 πan h(xn)

an obvious modification for Rd , d > 2


UNAVOIDABLE SETS

A relatively closed subset A of a connected open set U 6= ∅ in Rd , d ≥ 2, is

unavoidable, if Brownian motion (Bt), starting in U\A and killed when

leaving U, hits A almost surely. Equivalently, denoting µU\Ay the harmonic

measure at y with respect to U\A, µU\Ay (A) = 1 for every y ∈ U\A.

y

U\A

U


CHAMPAGNE BUBBLES

Let X be a countable subset of U having no accumulation point in U, andlet rx > 0, x ∈ X , be such that the closed balls B(x , rx), the bubbles, arepairwise disjoint and

supx∈X

(

rx/dist(x , ∂U))

< 1.

Then the union A of all B(x , rx) isrelatively closed in U.

The (connected) open set U\A is calledchampagne subdomain of U.

The set of centers of all bubbles forming A

will be denoted by XA.

Rosé champagne


UNAVOIDABLE SETS OF BUBBLES



Taking TF (ω) := inf{t ≥ 0 : Bt(ω) ∈ F}, we have: A :=⋃

x∈X B(x , rx)unavoidable if

Py [TA < TUc ] = 1 for some (all) y ∈ U\A,

µU\Ay (A) = 1 for some (all) y ∈ U\A.







Still unavoidable, if we omit finitely many of the bubles!







Still unavoidable, if we omit finitely many of the bubles!

Question

How small may an unavoidable set A of bubbles be?


Investigated by J. R. Akeroyd (2002); J. Ortega-Cerdà and K. Seip (2004);T. Carroll and J. Ortega-Cerdà (2007); S. J. Gardiner and M. Ghergu(2010), J. O’Donovan (2010), J. Pres (2012). Also W. Hansen andN. Nadirashvili (1994); T. Lundh (2001); T. Carroll, J. O’Donovan andJ. Ortega-Cerdà (2012).



Let us define the (capacity) function ϕ(t) := 1/ log 1t, d = 2,

ϕ(t) := td−2, d > 2, t ∈ (0, 1).



Let us define the (capacity) function ϕ(t) := 1/ log 1t, d = 2,

ϕ(t) := td−2, d > 2, t ∈ (0, 1).

Theorem (S. J. Gardiner and M. Ghergu; J. Pres).

If d ≥ 2, then, for all ε > 0 and δ > 0, there exists a champagne

subdomain B(0, 1)\A such that A is unavoidable and

∑

x∈XA

ϕ(rx)1+ε < δ.


Now we fix an arbitrary function h : (0, 1) → (0, 1) such thatlim inft→0 h(t) = 0.

Examples

h(t) = ϕε(t), t ∈ (0, 1), ε > 0; h(t) =(

log log . . . log (1/ϕ(t)))−ε

,t small


Now we fix an arbitrary function h : (0, 1) → (0, 1) such thatlim inft→0 h(t) = 0.

Examples

h(t) = ϕε(t), t ∈ (0, 1), ε > 0; h(t) =(

log log . . . log (1/ϕ(t)))−ε

,t small

Theorem (W. Hansen and I. Netuka).

Let U 6= ∅ be a connected open set in Rd , d ≥ 2. Then, for every δ > 0,

there exists a champagne subdomain U\A such that A is unavoidable and

∑

x∈XA

ϕ(rx)h(rx) < δ.


Remark (by Gardiner/Ghergu and Pres)

The result is optimal. Indeed, if A is an unavoidable set of bubbles, then∑

x∈XAϕ(rx) = ∞, hence the factor h(rx) cannot be omitted.


THE KEY RESULT FOR B(0, 1)


THE KEY RESULT FOR B(0, 1)

Let d ≥ 2, δ > 0, and (Rk) be a sequence in (0, 1) which is strictlyincreasing to 1. Then there exist finite sets Xk in ∂B(0, Rk) and rk > 0such that,

∑∞k=1 #Xk ϕ(rk) h(rk) < δ and, taking

A :=⋃

x∈Xk , k∈N

B(x , rk),

the set B(0, 1)\A is a champagnesubdomain, A is unavoidable and

∑

x∈XA

ϕ(rx) h(rx) < δ.


A GENERAL CRITERION FOR UNAVOIDABLE SETS


A GENERAL CRITERION FOR UNAVOIDABLE SETS

Let A ⊂ U be relatively closed and let (Vn) be an exhaustion of U bybounded open sets (Vn ր U, V n ⊂ Vn+1). Suppose that αn ≥ 0 and, forall n ∈ N and z ∈ ∂Vn, Pz [TA < TV c

n+1] ≥ αn.

Then, for all m > n and z ∈ Vn, Pz [TA < TUc ] ≥ 1 −∏

n≤j<m(1 − αj).

In particular, A is unavoidable provided∑∞

n=1 αn = ∞.


THANK YOU!


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