Rational approximation on compact, nowhere dense subsets of the complex plane

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Rational approximation on compact,

nowhere dense subsets of the complex plane

Alan Groot

July 17, 2014

Bachelorscriptie

Supervisors: prof. dr. Tom Koornwinder, P.J.I.M. de Paepe, prof. dr. Jan

Wiegerinck

Korteweg-de Vries Instituut voor Wiskunde

Faculteit der Natuurwetenschappen, Wiskunde en Informatica Universiteit van Amsterdam

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Abstract

Let C denote the complex plane. Let X ⊂ C be compact and nowhere dense. We investigate the denseness of R(X), the linear space of rational functions on C with poles off X, in C(X) and in Lp_{(X) (1 ≤ p < ∞) in the uniform norm and p-norm on X}

respectively and we study the role of the underlying compact, nowhere dense set X. Let R(X) and Rp(X) denote the closures of R(X) in C(X) and Lp(X) in the respective norms. We prove equality Rp_{(X) = L}p_{(X) holds when 1 ≤ p < 2 and we prove the}

Hartogs-Rosenthal theorem. We give an example of a compact, nowhere dense set X such that R2(X) 6= L2(X) using the Bergman kernel and bounded point evaluations.

The question of whether R(X) = C(X) or Rp_{(X) = L}p_{(X) for 1 ≤ p < ∞ can be}

formulated entirely in terms of capacities: the analytic capacity for the first and the q-capacity for the second question (1/p + 1/q = 1). Vitushkin’s theorem connects the question of denseness of R(X) in C(X) to the analytic capacity, while Hedberg’s theorem connects the question of denseness of R(X) in Lp_{(X) to q-capacity. We use the criteria}

obtained by Vitushkin and Hedberg to answer the following questions: does there exist a compact, nowhere dense set X ⊂ C such that

(I) Rp_{(X) = L}p_{(X) for all 1 ≤ p < ∞, but R(X) 6= C(X) and}

(II) Rp(X) = Lp(X) for all 1 ≤ p < p∗ for a certain 2 ≤ p∗ < ∞, but Rp(X) 6= Lp(X) for all p∗ ≤ p < ∞?

Furthermore, we investigate the role played by bounded point evaluations in approxima-tion in the p-norm. Again, the existence of a bounded point evaluaapproxima-tion can be expressed in terms of q-capacity (due to Hedberg) and we use this criterion to construct a compact, nowhere dense set X ⊂ C such that

1. R2_{(X) has no bounded point evaluations, but}

2. R2_{(X) 6= L}2_(X).

Title: Rational approximation on compact, nowhere dense subsets of the complex plane Authors: Alan Groot, alan.groot@student.uva.nl, 10282831

Supervisors: prof. dr. Tom Koornwinder, P.J.I.M. de Paepe, prof. dr. Jan Wiegerinck Second grader: prof. dr. Jan Wiegerinck

Date: July 17, 2014

Korteweg-de Vries Instituut voor Wiskunde Universiteit van Amsterdam

Science Park 904, 1098 XH Amsterdam http://www.science.uva.nl/math

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1. Introduction

In January of 2014, I went to Jan Wiegerinck to talk about my bachelor’s thesis. He taught the second-year course Functietheorie and it was a course I enjoyed, so it seemed logical to me that I would ask him to be my supervisor. Because I did not want to approach him empty-handed, I looked at his site and found a page on which he suggested topics for a bachelor’s thesis. In these topics, I saw the topic of the Kakeya needle problem. Around that time, I was interested in the moving sofa problem and the Kakeya needle problem seemed to focus on a similar type of question. We talked about it and he gave me the article [10] by Fefferman and asked me to read it as an introduction. The next time we would meet, I would have read the article and tell him what I thought of the article. The article was an interesting read. However, it was only seven pages long and consisted of quite elementary arguments, making me wonder whether it was difficult enough to challenge other readers and myself and most importantly, whether I could turn this into a thesis of 30 pages.

When I expressed my doubts, Jan suggested me to look at approximation problems instead and made me look into Gaier’s book [16], especially at the parts concerning Runge’s and Mergelyan’s theorem. This resonated better with me. This time, however, it would be easy to make the 30 pages, but there would not be a lot (if anything) for me to clarify about Gaier’s texts. Then Jan sent me an e-mail telling me that I could look at the 2013 article Approximation by rational functions on compact nowhere dense subsets of the complex plane by J. E. Brennan and C. N. Mattingly ([5]). This article combined the challenging part with the clarification part and it formed the basis for my thesis. What I liked about the article was the construction of sets and using these constructions to prove statements about function spaces.

In this thesis, we will investigate various approximation problems. For a compact, nowhere dense set X ⊂ C, let R(X) denote the linear space of rational functions with poles off X, let C(X) denote the space of continuous functions on X endowed with the uniform norm and let Lp(X) (for 1 ≤ p < ∞) denote the space of all Lebesgue-measurable functions on X that are p-integrable with respect to area or two-dimensional Lebesgue measure (dA) endowed with the p-norm on X. Let R(X) and Rp_{(X) denote the closures}

of R(X) in C(X) and Lp(X) respectively. We want to determine whether R(X) = C(X) and whether Rp_{(X) = L}p_{(X) for this set X and for which values of 1 ≤ p < ∞ this}

equality holds. Also, we investigate the influence of the underlying compact, nowhere dense set X.

The approximation problem R(X) = C(X) is a problem on the subject of uniform algebras. For more information on uniform algebras, we refer to Gamelin’s book [17]. Swiss cheeses are discussed in more depth in [11]. The problem Rp(X) = Lp(X) is a problem in the theory of function spaces and in functional analysis. This problem has a connection with the invariant subspace problem for subnormal operators on a Hilbert space (cf. [5, p. 202]). For more information, see [2, 4]. Also, this problem can be reformulated as a ‘stability problem’ in Sobolev spaces, see [1, Chapter 11].

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the general setting for the approximation problems we will discuss in this thesis (cf. [5, Section 2]). We establish the importance of the assumptions that X is compact and nowhere dense and formulate the main questions (I) and (II) we will investigate in this thesis. We identify the dual spaces of C(X) and Lp_{(X) (1 ≤ p < ∞) and we use duality}

to answer the question whether R(X) is dense in C(X) or Lp_{(X) partly: we will find}

sufficient conditions on the Cauchy transform of a complex regular Borel measure µ or of a q-integrable function k ∈ Lq_{(X) (1/p + 1/q = 1). These conditions will allow us}

to conclude that Rp_{(X) = L}p_{(X) holds for 1 ≤ p < 2. We will use bounded point}

evaluations for Rp(X) to formulate a necessary condition for R(X) to be dense in Lp(X) (1 ≤ p < ∞) and use this condition to construct a compact, nowhere dense set X such that R2_{(X) 6= L}2_{(X), solving a special case of (I).}

In Chapter 3, we will develop a partial theory of capacities (cf. [5, Section 3]). We are able to formulate the question whether R(X) = C(X) or Rp(X) = Lp(X) (1 ≤ p < ∞) entirely in terms of capacities of associated sets. For R(X) = C(X), Vitushkin’s theorem formulates this problem entirely in terms of analytic capacity γ. In the same way, for Rp(X) = Lp(X) (1 ≤ p < ∞), Hedberg’s theorem formulates this problem entirely in terms of q-capacity (Bq and Cq). Special attention is needed for the case q = 2, in which

Cq cannot be used. We establish properties of analytic capacity and q-capacity and we

establish bounds on the analytic capacities and q-capacities of certain sets.

In Chapter 4, we will construct compact, nowhere dense sets to answer our main questions (I) and (II) (cf. [5, Section 4]). To do this, the construction of the corner quarters Cantor set is needed. We will use the criteria obtained in Chapter 3 in terms of capacities to show that these sets indeed answer our questions.

In Chapter 5, we will investigate the role played by bounded point evaluations in approximation in p-norm further (cf. [5, Section 5]). Again, the question whether a point z0 ∈ X is a bounded point evaluation for Rp(X), can be expressed entirely in

terms of capacities of associated sets. We ask ourselves whether the entire absence of bounded point evaluations for Rp(X) on X ensures us that Rp(X) = Lp(X) (for a given 2 ≤ p < ∞). We answer this question partially by showing a construction of a compact, nowhere dense set X, due to Fernstr¨om, that has no bounded point evaluations for R2_(X),

but nevertheless satisfies R2(X) 6= L2(X).

Chapter 6 will discuss the goals we originally wanted to achieve.

Chapter 7 will be a summary in Dutch about what we have done in this thesis. Its intended target audience consists of students in the last year of their pre-university sec-ondary education (vwo) and first-year students of mathematics.

In the appendix, we establish the denseness of smooth, compactly supported functions f : C → C in various function spaces and we give an elementary proof of Lemma 4 in [14].

We originally set out to discuss the entire article [5] and to turn it into something that would be readable for a third-year student of mathematics. The article ends with a question about a compact, nowhere dense set X that satisfied certain properties. Since it was unresolved when Jan gave the article (and to my knowledge, it is unresolved as of yet), Jan asked me to see if I could develop some necessary conditions on the set X for these properties to hold. Also, he asked me not to end up with a lot more than 30 pages. In the end, I want to thank my supervisors Tom Koornwinder, Peter de Paepe and Jan Wiegerinck for helpful discussions, comments and insights.

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2. Preliminaries

2.1. Definitions and examples

In this thesis, we will study various approximation problems. Let X be a (non-empty) compact, nowhere dense1 _{subset of the complex plane C (with the usual Euclidean} topol-ogy). Let R(X) be the linear space of rational functions with poles off X (that is, the poles of these rational functions lie in the complement Xc_{= C \ X of X). We will denote} the planar Lebesgue measure (dA) of a measurable set E ⊂ C by |E| (this is not to be confused with the Euclidean norm |z| =p<(z)2_{+ =(z)}2 ₌ √_{zz of a complex number}2_,

for which we will use the same notation). We define C(X) as the space of functions continuous on X endowed with the supremum or uniform norm k · k (the compactness of X turns this into a Banach space). For 1 ≤ p < ∞, we define Lp_{(X) as the space of all}

Lebesgue-measurable functions that are p-integrable with respect to area measure. That is (with some abuse of notation),

Lp(X) = f : Z X |f |p_{dA < ∞} ,

where we consider two functions that agree almost everywhere on X (with respect to two-dimensional Lebesgue measure) equal. We denote the p-norm on Lp_{(X) by k · k}

p, so for f ∈ Lp(X) we have kf kp = Z X |f |pdA 1_p .

Furthermore, we will denote the closure of R(X) in C(X) with respect to the supremum norm by R(X) and for 1 ≤ p < ∞, we will denote the closure of R(X) in Lp_{(X) with}

respect to the p-norm by Rp(X).

We will denote open disks of radius r with centre z, closed disks of radius r with centre z and circles of radius r with centre z by B(z, r) := {w ∈ C : |z − w| < r}, B(z, r) := {w ∈ C : |z − w| ≤ r} and C(z, r) := {w ∈ C : |z − w| = r} respectively. For a non-empty set A ⊂ C, we will define d(z, A) = inf{|z − a| : a ∈ A} for all z ∈ C. For two non-empty sets A, B ⊂ C, we define d(A, B) = inf{|a − b| : a ∈ A, b ∈ B}.

Let us emphasize the importance of the underlying set X for general approximation problems. We will give two examples.

Example 2.1.1. Let X = C(0, 1) be the unit circle. Then the continuous function f given by f (z) = z cannot be approximated uniformly by polynomials. Suppose (pn) is

a sequence of polynomials converging uniformly to f . Then by part 2 of Theorem 5.1.2

1

For E ⊂ C we will denote its closure by E, its interior by E◦ and its boundary E \ E◦ by ∂E. A set X ⊂ C is called nowhere dense if the closure X of X has an empty interior: in symbols X◦ = ∅. Because a compact set X ⊂ C is closed (by for example Heine-Borel or using that C is a Hausdorff space), a compact set X is nowhere dense if and only if it has an empty interior.

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from [9, p. 55], if we would integrate over the unit circle by running around once in counterclockwise direction, we would have

0 = Z X pn(z)dz → Z X zdz = 2πi 6= 0, a contradiction.

Example 2.1.2. Let us suppose that X is not nowhere dense. Suppose that X equals B(0, 1), the closed unit disk, and we are interested in whether R(X) = C(X). Suppose (fn) is a sequence in R(X) converging uniformly to some function f ∈ C(X). Note that

each fn is holomorphic on the interior B(0, 1) of X. By Weierstrass’ theorem (Theorem

7.2.1 of [9, p. 82]), the limit function f would be holomorphic on B(0, 1) as well. Now not every continuous function f on X is holomorphic on B(0, 1): for example, the continuous function f (z) = z is not holomorphic outside 0. This means that R(X) 6= C(X).

The last example shows that every compact set X that is not nowhere dense contains a continuous function that cannot be approximated uniformly by rational functions with poles off X (i.e. R(X) 6= C(X)), namely the continuous function f : z 7→ z. This follows from the fact that every compact set X with non-empty interior contains an open disk and therefore a point a 6= 0, in which f is not holomorphic. This shows the importance of the assumption that the compact set X is nowhere dense.

Let’s investigate the question whether R(X) equals C(X) further. In the last example, we considered a compact set with positive measure (without further notice, we will assume that we mean two-dimensional Lebesgue measure whenever we say measure). This is because X had a non-empty interior. We could ask ourselves whether compact, nowhere dense sets of positive measure even exist and if so, would the absence of an interior ensure us that R(X) = C(X)? The Swiss mathematician Alice Roth constructed a compact, nowhere dense set X called the Swiss cheese of positive measure for which R(X) 6= C(X). The next example shows how such a set is constructed. We do not intend to give all the details.

Example 2.1.3 (Construction of a Swiss cheese). This construction follows [16, p. 110– 112]. Let D = B(0, 1) be the open unit disk and let {∆j}

∞

j=1 be a sequence of open disks

∆j of radius rj such that the closures of all ∆j are contained in D, the ∆j have mutually

disjoint closures, the radii satisfy P∞

j=1rj < 1 and most importantly, such that

X := D \

∞

[

j=1

∆j

has no interior. By Heine-Borel, X is compact. By takingP∞ j=1rj < 1, we ensure that |X| = |D| − ∞ [ j=1 ∆j = π − π ∞ X j=1 r_j2 > π − π ∞ X j=1 rj > 0,

so indeed we have a compact, nowhere dense set X of positive measure. For R ∈ R(X), we have Z ∂D R(z)dz = ∞ X j=1 Z ∂∆j R(z)dz (2.1)

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Figure 2.1.: Construction of the Swiss cheese

by Cauchy’s integral theorem and the residue theorem.3 _{Now if f ∈ R(X), this means}

we have Z ∂D f (z)dz − ∞ X j=1 Z ∂∆j f (z)dz = Z ∂D [f (z) − R(z)] dz − ∞ X j=1 Z ∂∆j [f (z) − R(z)] dz (2.2) ≤ 2πkf − Rk + 2πkf − Rk ∞ X j=1 rj < 4πkf − Rk (2.3)

by the M-L inequality. Since f ∈ R(X), the right-hand term in (2.3) can be made arbitrarily small. This means that

Z ∂D f (z)dz = ∞ X j=1 Z ∂∆j f (z)dz (2.4)

3_{This is because for each R ∈ R(X), there are only finitely many ∆}

j that contain poles of R, say

∆j1, . . . , ∆jn. Because the other ∆j do not contain poles of R, for those ∆jwe have

R

∂∆jR(z)dz = 0

by Cauchy’s integral theorem. By the residue theorem (Theorem 5.4.11 in [9, p. 65]) applied to D, we have Z ∂D R(z)dz = n X i=1 Z ∂∆_ji R(z)dz = n X i=1 Z ∂∆_ji R(z)dz + 0 = n X i=1 Z ∂∆_ji R(z)dz + X other ∆j Z ∂∆j R(z)dz = ∞ X j=1 Z ∂∆j R(z)dz.

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for all f ∈ R(X) as well. Now suppose we have constructed X in such a way that 0 /∈ X. Then the function f defined by f (z) = |z|/z is continuous. However,

Z

∂D

f (z)dz = 2πi, while by the M-L inequality

∞ X j=1 Z ∂∆j f (z)dz ≤ ∞ X j=1 Z ∂∆j f (z)dz ≤ ∞ X j=1 2πrj < 2π,

so the equality in (2.1) does not hold for this f ∈ C(X). The fact that (2.1) holds for all f ∈ R(X), but not for all f ∈ C(X), proves that R(X) 6= C(X).

This construction is an example of the proofs that will follow: we will argue by duality and look at bounded linear functionals (such as f 7→ R_∂Df (z)dz −P∞

j=1

R

∂∆jf (z)dz in

the last example) to find out whether R(X) is dense in C(X) or Lp(X) (for 1 ≤ p < ∞). We will specify this more precisely further on.

Thus far, we have seen two examples of sets X for which R(X) 6= C(X). We could ask ourselves whether equality is even possible, but by taking X = {a} for some a ∈ C, it is easy to see that equality is indeed possible. However, there is a certain limitation to the questions we can ask ourselves.

First of all, when equality R(X) = C(X) holds for a certain compact, nowhere dense set X, we also have Rp_{(X) = L}p_{(X) for all 1 ≤ p < ∞. This follows from the fact that}

for 1 ≤ p < ∞, the continuous functions on X are dense in Lp_{(X) with respect to the}

p-norm (Theorem 3.14 in [24, p. 71]) and that if fn→ g uniformly on X, then fn → g in

p-norm as well (all fn and g continuous), since X is compact and therefore has a finite

measure. This means that for any g ∈ Lp_{(X) and > 0, there is an f ∈ C(X) such that}

kg − f kp < ₂ and for this f ∈ C(X), there is an R ∈ R(X) such that kf − Rkp < ₂,

proving that Rp_{(X) = L}p_{(X) for all 1 ≤ p < ∞.}

Secondly, we will prove that Rp_{(X) is always equal to L}p_{(X) for 1 ≤ p < 2 and that if}

Rp∗(X) 6= Lp∗(X) for some 2 ≤ p∗ < ∞, then Rp(X) 6= Lp(X) for all p ≥ p∗. Therefore, a few questions pop up: does there exist a compact, nowhere dense set X ⊂ C such that

(I) Rp_{(X) = L}p_{(X) for all 1 ≤ p < ∞, but R(X) 6= C(X) and}

(II) Rp(X) = Lp(X) for all 1 ≤ p < p∗ for a certain 2 ≤ p∗ < ∞, but Rp(X) 6= Lp(X) for all p∗ ≤ p < ∞?

Also, we ask ourselves how we can determine these equalities. We will devote the rest of this section to proving that for a compact, nowhere dense set X ⊂ C, the equality Rp(X) = Lp(X) holds for all 1 ≤ p < 2 and to proving that there exists a compact, nowhere dense set X for which R2_{(X) 6= L}2_{(X) (answering the second question positively}

for p∗ = 2). The Hartogs-Rosenthal theorem, stating that R(X) = C(X) whenever the compact, nowhere dense set X satisfies |X| = 0, will be proven as a corollary.

2.2. The Cauchy transform, bounded point

evalutions and the Bergman kernel

2.2.1. Uniform approximation

As we have remarked before, we would like to argue by duality. For a normed vector space V , we will denote its dual by V∗. We will denote the operator norm on V∗ by k · k.

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We will define the annihilator of a subspace M ⊂ V as follows: for a subspace M ⊂ V , let

M⊥ = {Φ ∈ V∗ : Φ(m) = 0 for all m ∈ M }.

If Φ ∈ M⊥ _{we will say that Φ annihilates M . It is known that for a compact set X ⊂ C,} the space C(X) endowed with the supremum norm k · k is a Banach space. Now if we want to show that R(X) = C(X), then by Theorem 4.7 from [25, p. 96] we are done when we can show that every bounded linear functional Φ that annihilates R(X), also annihilates C(X). To make this easier, we want to identify the dual space C(X)∗ of C(X). This space turns out to be the space of complex regular Borel measures on X by Theorem 6.19 of [24, p. 139].

Let us explain first what we mean by complex measures. For a set V , let Σ be a sigma-algebra on V . A countable collection {Ei}∞i=1 of sets in Σ is called a partition of E if the

Ei are pairwise disjoint and E =S ∞

i=1Ei. A complex measure µ on Σ is a complex-valued

function on Σ such that

µ(E) =

∞

X

i=1

µ(Ei)

for every partition {Ei} of E. Given such a measure µ, we define a positive4 set function

|µ| on Σ by |µ|(E) = sup ∞ X i=1 |µ(Ei)|

where the supremum is taken over all partitions {Ei} of E. It turns out this set function

is in fact a measure, the total variation measure of µ (cf. Theorem 6.2 of [24, p. 125]). Furthermore, if µ is a complex measure on V , then even |µ|(V ) < ∞ (by Theorem 6.4 of [24, p. 126]). The number |µ|(V ) will be called the total variation of µ and is not to be confused with the total variation measure.

Now if we return to a compact set X ⊂ C, then a complex Borel measure µ is a measure defined on the Borel sigma-algebra B on X. The Borel sigma-algebra B on X is defined as the smallest sigma-algebra that contains all the open sets in X (open with respect to the subspace topology). Such a measure µ is called regular if the corresponding total variation measure |µ| satisfies

|µ|(E) = inf{|µ|(V ) : E ⊂ V, V open} and |µ|(E) = sup{|µ|(K) : K ⊂ E, K compact} for all E ∈ B5_{. This leads to the following theorem, a less-known version of the Riesz}

representation theorem that can be found in e.g. [24, p. 139].

Theorem 2.2.1 (Riesz representation theorem). Let X be a compact subset of the com-plex plane. For every bounded linear functional on C(X) there exists a unique comcom-plex regular Borel measure µ such that

Φ(f ) = Z

X

f dµ (2.5)

for all f ∈ C(X). Moreover, if Φ and µ are related as in (2.5), then kΦk = |µ|(X).

4_{A set function will be called positive whenever it assigns non-negative values to sets.}

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The above theorem now tells us that if for every complex regular Borel measure µ that annihilates R(X) (in the sense that R_Xf dµ = 0 for all f ∈ R(X)) we are able to conclude that µ = 0, then R(X) = C(X). To investigate whether µ = 0 for a given complex regular Borel measure that annihilates R(X), we will look at the corresponding Cauchy transform and the Newtonian potential of this measure.

Lemma 2.2.2. Let µ be any complex regular Borel measure on X. For z ∈ C, define6 ˆ µ(z) = Z X dµζ ζ − z and ˜µ(z) = Z X d|µ|ζ |ζ − z|. Then both are finite dA-almost everywhere in C.

Proof. We will use the fact that if a measureable function has a finite integral when integrated over a set, that function is finite almost everywhere on that set. It is sufficient to prove that the latter is finite dA-almost everywhere in C, because |ˆµ(z)| ≤ ˜µ(z) for all z ∈ C. Because X is compact, we can find R > 1 large enough so that X ⊂ B(0, R − 1). Because the following integrand is non-negative, we can apply Fubini to get

Z B(0,R) Z X d|µ|ζ |ζ − z|dAz = Z X Z B(0,R) 1 |ζ − z|dAzd|µ|ζ.

Note that for all ζ ∈ X, we have B(0, R) ⊂ B(ζ, 2R) by the triangle inequality, so we can estimate the inner integral:

Z B(0,R) 1 |ζ − z|dAz ≤ Z B(ζ,2R) 1 |ζ − z|dAz = Z 2π 0 Z 2R 0 1 rrdrdθ = 4πR. This means that

Z B(0,R) Z X d|µ|ζ |ζ − z|dAz = Z X Z B(0,R) 1 |ζ − z|dAzd|µ|ζ ≤ Z X 4πRd|µ|ζ = 4πR|µ|(X) < ∞,

because all complex regular Borel measures µ on X have finite total variation, so ˜µ is finite dA-almost everywhere in B(0, R). For z /∈ B(0, R), we have |ζ − z| > 1 for all ζ ∈ X, because X ⊂ B(0, R − 1), so that ˜ µ(z) = Z X d|µ|ζ |ζ − z| ≤ Z X 1 1d|µ|ζ = |µ|(X) < ∞,

so ˜µ is finite outside B(0, R). Combining this gives that ˜µ is finite dA-a.e.

The above lemma tells us that these for a complex regular Borel measure µ on X, the functions ˆµ and ˜_{µ are well-defined for dA-almost every z ∈ C. This motivates the} following definition.

Definition 2.2.3. For a complex regular Borel measure µ on X, the corresponding functions ˆµ and ˜µ as defined in Lemma 2.2.2 will be called the Cauchy transform and the Newtonian potential respectively.

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Why the notion of the Cauchy transform of a measure is convenient, will be clear from the following lemma. Also, note that if a complex regular Borel measure µ annihilates R(X), then ˆ_{µ(z) = 0 for z ∈ C \ X, because ζ 7→} 1

ζ−z ∈ R(X) whenever z ∈ C \ X.

Therefore, it is very natural to look at the Cauchy transform.

Lemma 2.2.4. Let µ be a complex regular Borel measure on X. If ˆ_{µ = 0 a.e.-dA in C,} then µ = 0 as a measure.

To be able to prove this lemma, we will need the Cauchy-Pompeiu formula for com-pactly supported, continuously differentiable functions. Recall that the complex partial derivatives are defined as (z = x + iy)

∂ ∂z := 1 2 ∂ ∂x − i ∂ ∂y , ∂ ∂z := 1 2 ∂ ∂x + i ∂ ∂y .

The following version of the Cauchy-Pompeiu formula is proved in [24, p. 422].

Theorem 2.2.5 (Cauchy-Pompeiu). Suppose that f : C → C is a compactly supported, continuously differentiable function. Then

f (z) = −1 π Z C ∂f ∂z(ζ) · 1 ζ − zdAζ. We are now able to prove Lemma 2.2.4.

Proof of Lemma 2.2.4. We will follow Zalcman’s proof from [28, p. 118]. Let f : C → C be a compactly supported, continuously differentiable function. Then f restricts to a continuous function on X. By Fubini’s theorem and Cauchy-Pompeiu,

Z X f (z)dµz = − 1 π Z X Z C ∂f ∂z(ζ) · 1 ζ − zdAζdµz = −1 π Z C Z X ∂f ∂z(ζ) · 1 ζ − zdµzdAζ = −1 π Z C ∂f ∂z(ζ) Z X 1 ζ − zdµzdAζ = −1 π Z C ∂f ∂z(ζ) · ˆµ(ζ)dAζ = 0, since ˆ_{µ = 0 almost everywhere-dA in C.}

Now let f ∈ C(X). Then by Tietze’s extension theorem (Theorem 20.4 in [24, p. 422]) there exists a continuous, compactly supported function F : C → C such that F (z) = f (z) for all z ∈ X. By Theorem A.1.2, there exists a sequence (fn) of smooth,

compactly supported functions fn : C → C that converges to F uniformly on C, so by

restricting the fn to functions fn : X → C, we have fn → f as n → ∞ uniformly on X

as well. For these restrictions of fn we have just seen that

R

Xfn(z)dµz = 0 for all n ≥ 1.

This means that Z X f (z)dµz = Z X f (z)dµz− Z X fn(z)dµz ≤ kf − fnk · |µ|(X) → 0

as n → ∞ (since |µ|(X) < ∞). As f ∈ C(X) was arbitrary, this means thatR_Xf (z)dµz =

0 for all f ∈ C(X). By the uniqueness part of the Riesz representation theorem, we have µ = 0.

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Corollary 2.2.6 (Hartogs-Rosenthal). If X is compact, nowhere dense set in C with |X| = 0, then R(X) = C(X).

Proof. Let µ be a complex regular Borel measure on X that annihilates R(X). This means that for all z ∈ C \ X, we have

ˆ µ(z) = Z X 1 ζ − zdµζ = 0.

Since |X| = 0, this means that ˆ_{µ(z) = 0 for dA-almost every z ∈ C. By Lemma 2.2.4,} µ = 0. This means that every complex regular Borel measure µ on X that annihilates R(X) also annihilates C(X), so R(X) = C(X).

2.2.2. Approximation in p-norm

In the case we want to determine whether Rp_{(X) = L}p_{(X) for 1 ≤ p < ∞, we would like}

to argue by duality as well. As we have seen, this is a trivial question when |X| = 0, so in some instances we will assume that |X| > 0. To argue by duality, we need to identify the dual space Lp_(X)∗ _{of the Banach space L}p_{(X) for 1 ≤ p < ∞. This space turns}

out to be Lq_{(X), where q satisfies 1/p + 1/q = 1 (q = +∞ when p = 1). From now on,}

whenever p and q occur with the same decoration (e.g. p∗ and q∗, pn and qn or simply p

and q), it is tacitly assumed that p and q are so-called dual or conjugate exponents, that is 1/p + 1/q = 1. We have the following theorem, Theorem 6.16 from [24, p. 136]. Theorem 2.2.7. Let X be a compact set and let 1 ≤ p < ∞. For every bounded linear functional Φ on Lp_{(X) there exists a unique g ∈ L}q_{(X) such that}

Φ(f ) = Z

X

f gdA (2.6)

for all f ∈ Lp(X). Moreover, if Φ and g are related as in (2.6), then kΦk = kgkq7.

Again, to determine whether Rp_{(X) = L}p_{(X), we need to determine whether each k ∈}

Lq(X) that annihilates R(X) (that is, R_Xf kdA = 0 for all f ∈ R(X)), also annihilates Lp(X). Again, we will look at the Cauchy transform ˆk(z) of a k ∈ Lq(X) defined by

ˆ k(z) = Z X k(ζ) ζ − zdAζ

for z ∈ C. This will also be well-defined for dA-almost every z ∈ C. Again, it is natural to look at the Cauchy transform ˆk of k ∈ Lq_{(X). If} R

Xf kdA = 0 for every f ∈ R(X),

then ˆ_{k(z) = 0 if z ∈ C \ X, since ζ 7→ 1/(ζ − z) is a rational function with poles off X if} z ∈ C \ X. Therefore, if k ∈ Lq(X) annihilates R(X), then ˆk = 0 in points outside X.

We will show that every k ∈ Lq_{(X) yields a Cauchy transform that is well-defined}

dA-almost everywhere on C after the auxiliary lemmas below. We will use these lemmas in various instances.

Lemma 2.2.8. Suppose X ⊂ C is compact. If 1 ≤ a < b ≤ ∞, then

7_{For q = ∞, the norm k · k}

∞on the space L∞(X) of all measurable, essentially bounded functions f is

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1. Lb_{(X) ⊂ L}a_(X),

2. Rb_{(X) ⊂ R}a_{(X) if b < ∞.}

Proof. First, suppose b < ∞. We will prove a more general statement, namely that for 0 < r < s < ∞ we have kf kr ≤ kf ks|X|

1 r−

1

s. Let p = s/r > 1 and let q = p/(p − 1).

Applying H¨older’s inequality to fr _yields

Z X |f |r_{dA ≤} Z X (|f |r)pdA 1_p Z X 1qdA 1_q = Z X |f |s_dA r_s |X|1−r_s

and from this, kf kr ≤ kf ks|X|

1 r−

1

s follows by taking the rth root. Because X is compact,

|X| is finite. This and the inequality above immediately prove the first statement when b < ∞.

If f ∈ L∞(X), then for any 1 ≤ a < ∞ we have Z

X

|f |a_{dA ≤ ess sup(|f |)}a_{· |X| < ∞}

since X is compact and hence has a finite measure, so f ∈ La_{(X) for all 1 ≤ a < ∞.}

Now we prove the second statement. Because X is compact and each f ∈ R(X) is continuous on X, we have R(X) ⊂ Lr_{(X) for all 0 < r < ∞. Now suppose that}

g ∈ Rb_{(X). Then there exists a sequence (f}

n) in R(X) such that kg − fnkb → 0. But

then, by the above inequality,

kg − fnka≤ kg − fnkb|X| 1 a− 1 b → 0 as well, so that g ∈ Ra_(X).

Corollary 2.2.9. Let 1 ≤ a < ∞ be given. If X ⊂ C is compact and Rb(X) = Lb(X) for some b > a, where 1 ≤ b < ∞, then Ra_{(X) = L}a_(X).

Proof. Suppose X is compact and Rb_{(X) = L}b_{(X) for some b > a. Then by the second}

statement of the preceding lemma, we know that Rb(X) ⊂ Ra(X), so Lb(X) = Rb(X) ⊂ Ra(X) ⊂ La(X).

Now by taking closures with respect to the a-norm on La(X), we get

La(X) = C(X) ⊂ Lb_{(X) ⊂ R}a_{(X) = R}a_{(X) ⊂ L}a_{(X) = L}a_(X),

as the continuous functions are dense in La_{(X) with respect to the a-norm (Theorem}

3.14 in [24, p. 71]) and both Ra(X) and La(X) are closed with respect to the a-norm, proving this corollary.

Lemma 2.2.10. Suppose X ⊂ C is compact. Suppose k ∈ Lq_{(X). Then}

ˆ k(z) = Z X k(ζ) ζ − zdAζ is finite for dA-almost every z ∈ C.

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Proof. The proof will be similar to the proof of Lemma 2.2.2. Since X is compact, there exists a R > 1 such that X ⊂ B(0, R − 1). First we will show that ˆk is finite a.e.-dA in B(0, R). Integrating over B(0, R) gives

Z B(0,R) ˆ k(z)dAz ≤ Z B(0,R) Z X |k(ζ)| |ζ − z|dAζdAz = Z X Z B(0,R) |k(ζ)| |ζ − z|dAzdAζ

by Fubini’s theorem. Since B(0, R) ⊂ B(ζ, 2R) for all ζ ∈ X, we can estimate the inner integral by Z B(0,R) |k(ζ)| |ζ − z|dAz ≤ Z B(ζ,2R) |k(ζ)| |ζ − z|dAz = Z 2π 0 Z 2R 0 |k(ζ)|drdθ = 4πR · |k(ζ)|. This means that

Z B(0,R) ˆ k(z)dAz ≤ Z X Z B(0,R) |k(ζ)| |ζ − z|dAzdAζ ≤ 4πR Z X |k(ζ)|dAζ. (2.7)

By Lemma 2.2.8, we know k ∈ L1(X) since k ∈ Lq(X) for q > 1. This means that in (2.7), the right hand side is finite and therefore the left hand side as well. This proves that ˆk is finite dA-almost everywhere on B(0, R).

For z ∈ C with |z| ≥ R, we know that |ζ − z| > 1 for all ζ ∈ X by our choice of R, so that ˆ k(z) ≤ Z X |k(ζ)| |ζ − z|dAζ ≤ Z X |k(ζ)|dAζ.

Again the right hand side is finite since k is integrable, which means that ˆk(z) is finite for all z ∈ C with |z| ≥ R. Combining this with the finiteness for dA-almost every z ∈ C with |z| < R, we conclude that ˆ_{k is finite dA-almost everywhere in C.}

Again, if the Cauchy transform ˆ_{k is equal to 0 for dA-almost every z ∈ C, then k(z) = 0} for dA-almost every z ∈ X.

Lemma 2.2.11. Suppose X ⊂ C is compact and that k ∈ Lq_{(X). If ˆ}_{k = 0 for dA-almost}

every z ∈ C, then k = 0 for dA-almost every z ∈ X.

Proof. The proof will be very similar to the one given to prove Lemma 2.2.4. First, let f : C → C be a continuously differentiable, compactly supported function. Then f restricts to a continuous function on X. By Cauchy-Pompeiu and the same calculations as in Lemma 2.2.4, we get Z X f (z)k(z)dAz = − 1 π Z C ∂f ∂z(ζ) Z X k(z) z − ζdAz dAζ = − 1 π Z C ∂f ∂z(ζ) · ˆk(ζ)dAζ, = 0 because ˆ_{k = 0 dA-almost everywhere in C by assumption.}

Now let f ∈ C(X) be a continuous function. By Tietze’s extension theorem (Theorem 20.4 in [24, p. 422]), there exists a compactly supported, continuous function F : C → C such that F (x) = f (x) for all x ∈ X. By Theorem A.1.2, there exists a sequence (fn) of

compactly supported, smooth functions fn : C → C such that

Z

C

|f (z) − fn(z)|pdAz

1/p → 0

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as n → ∞. This also means that Z X |f (z) − fn(z)|pdAz 1/p → 0

as n → ∞, so that if k · kp denotes the p-norm on X, then kf − fnkp → 0 as n → ∞ as

well. By restricting these functions fn to X, we have just seen that

R

Xfn(z)k(z)dAz = 0.

By H¨older’s inequality, this means that Z X f (z)k(z)dAz = Z X f (z)k(z)dAz− Z X fn(z)k(z)dAz = Z X [f (z) − fn(z)]k(z)dAz ≤ Z X |f (z) − fn(z)||k(z)|dAz ≤ kf − fnkpkkkq → 0

as n → ∞, since kkkq < ∞. This proves that

R

Xf (z)k(z)dAz = 0 for all f ∈ C(X).

By Theorem 3.14 in [24], C(X) is dense in Lp_{(X) with respect to the p-norm. Since}

f 7→ R_Xf (z)k(z)dAz is a bounded linear functional on Lp(X) (by H¨older’s inequality)

that vanishes on the dense subspace C(X) of Lp(X), it vanishes on the whole space Lp(X) and is therefore equal to the zero functional. By the uniqueness part of Theorem 2.2.7, this means that k = 0 in Lq_{(X). Hence k = 0 dA-almost everywhere on X.}

As it turns out, the Cauchy transform ˆk of a k ∈ Lq_{(X) is even continuous. A proof}

of the following lemma can be found in [5, p. 205].

Lemma 2.2.12. Suppose X ⊂ C is compact. If k ∈ Lq_{(X) for q > 2, then} ˆ k(z1) − ˆk(z2) ≤ C|z1 − z2|1−2/q for all z1, z2 ∈ C for a constant C only depending on q (independent of

the chosen points z1 and z2).

Now suppose that we have a compact, nowhere dense set X ⊂ C. Suppose that 1 ≤ p < 2 and suppose that k ∈ Lq_{(X) annihilates R(X). Then we have seen before}

that ˆ_{k(z) = 0 for z ∈ C \ X. Since q > 2 when 1 ≤ p < 2, the Cauchy transform ˆk of} k is continuous by Lemma 2.2.12. Because X is nowhere dense in C, the complement C \ X is dense in C, and the continuous functionˆk is equal to 0 on this dense subset of C. Therefore, ˆk = 0 everywhere on the complex plane. By Lemma 2.2.11, this means that k = 0. This proves that every k ∈ Lq_{(X) that annihilates R(X) also annihilates L}p_(X).

We conclude that Rp_{(X) = L}p_{(X). Since 1 ≤ p < 2 was arbitrary, we have just proved}

the following theorem.

Theorem 2.2.13. Suppose X ⊂ C is compact and nowhere dense. Then Rp_{(X) = L}p_(X)

for all 1 ≤ p < 2.

Whenever 2 ≤ p < ∞, the situation can sometimes change quite heavily, since we are not automatically guaranteed that Rp_{(X) = L}p_{(X) as well. To illustrate this, we}

will end this section by constructing a compact, nowhere dense set X ⊂ C such that R2(X) 6= L2(X) (note that by Corollary 2.2.9, this means that Rp(X) 6= Lp(X) for all 2 ≤ p < ∞ and that R(X) 6= C(X) by the remarks at the end of Section 2.1). In this construction we will use bounded point evaluations and the Bergman kernel.

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2.2.3. Bounded point evaluations and the Bergman kernel

We are going to investigate the role bounded point evaluations play in approximating in p-norm. Let X be a compact, nowhere dense subset of the complex plane with |X| > 0 and let 1 ≤ p < ∞. Suppose there exist z0 ∈ X and C > 0 such that

|f (z0)| ≤ Ckf kp

for all f ∈ R(X), where C is independent of f . This map f 7→ f (z0) extends from R(X)

to a unique bounded linear functional on its closure Rp_{(X) by Theorem 4.19 of [26, p.}

99]. A point z0 ∈ X is called a bounded point evaluation for Rp(X) if the map f 7→ f (z0),

defined for f ∈ R(X), can be extended to a bounded linear functional on Rp_{(X) (if}

z0 ∈ X is a bounded point evaluation for Rp(X), we will also say that X has a bounded

point evalution for Rp(X)). Since Lp(X) is a normed space, by Hahn-Banach there exists a bounded linear functional Φ on Lp_{(X) that is equal to f 7→ f (z}

0) when restricted to

the subspace Rp_{(X) (e.g. Theorem 5.19 in [26, p. 133]). From Theorem 2.2.7, we know}

that there exists a function k ∈ Lq(X) such that f (z0) =

Z

X

f kdA (2.8)

for all f ∈ Rp_{(X). Now if we take the function g : z 7→ (z − z}

0)k(z), then it is easy to

see that indeed g ∈ Lq(X). For this g ∈ Lq(X), we see that Z

X

f (z)(z − z0)k(z)dA = f (z)(z − z0)|z=z0 = 0,

for all f ∈ Rp_{(X), since z 7→ f (z)(z − z}

0) ∈ Rp(X) for all f ∈ Rp(X) (the vertical

bar denotes evaluation in the point z = z0). Again, it is an easy check to see that

z 7→ f (z)(z − z0) is indeed in Rp(X). This means that the function g ∈ Lq(X) annihilates

Rp_{(X). However, g does not annihilate L}p_{(X) (here, the assumption that |X| > 0 comes}

into play: if |X| = 0, then g would annihilate Lp(X) as well): from (2.8) it follows that R

X kdA = 1, by evaluating the rational function z 7→ 1 ∈ R(X). Therefore, k 6= 0 in

Lp_{(X) (that is, k 6= 0 on a subset of X of positive measure). This means that g 6= 0}

in Lp(X) as well, so indeed g does not annihilate Lp(X) (the only function in Lq(X) that annilates Lp_{(X) is the zero function by Theorem 2.2.7). From this we conclude that}

Rp_{(X) 6= L}p_{(X) whenever there exists a bounded point evaluation for R}p_(X).

Theorem 2.2.14. Suppose X ⊂ C is a compact, nowhere dense set of positive measure. If X has a bounded point evaluation for some 1 ≤ p < ∞, then Rp_{(X) 6= L}p_(X).

By Theorem 2.2.13, this means that for 1 ≤ p < 2, X has no bounded point evaluations for Rp(X). However, for p ≥ 2, it might happen that X has a bounded point evaluation. We will use this to prove the following theorem [5, p. 207, Theorem 2.6].

Theorem 2.2.15. There exists a compact set X having no interior such that R2_{(X) 6=}

L2(X).

To prove this theorem, we need to explain a little bit about Bergman kernels. We will start by discussing the Bergman kernel of an open set G ⊂ C. Let G ⊂ C be a bounded,

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open set. The Bergman space A2_{(G) consists of all complex functions f holomorphic on}

G that are square integrable with respect to area measure, that is Z G |f (z)|2_dA z 1/2 < ∞.

If H(G) denotes the set of all holomorphic functions on G, then we can write A2_{(G) =}

L2_{(G) ∩ H(G). The Bergman space is an inner product space when endowed with the}

inner product hf, gi = R_Gf (z)g(z)dAz and induced norm k · k2. In fact, it is even a

Hilbert space. To prove this claim, we can show that for any compact K ⊂ G we have the estimate

sup

z∈K

|f (z)| ≤ CKkf k2 (2.9)

for all f ∈ A2(G), for some constant CK > 0 only depending on K (note that the norm

k · k2 is not restricted to K). This is a consequence of the following two lemmas, that can

be found in Conway [8, p. 6].

Lemma 2.2.16. If f is analytic in a neighbourhood of B(a, r), then f (a) = 1

πr2

Z

B(a,r)

f (z)dAz.

Corollary 2.2.17. If f ∈ A2_{(G), a ∈ G and 0 < r < d(a, ∂G), then}

|f (a)| ≤ 1

r√πkf k2.

It is not hard to verify that for every compact K ⊂ G, we have d(a, ∂G) > > 0 for all a ∈ K, for some > 0 independent of a ∈ K (we could take = d(K, ∂G), for instance: the boundedness of G implies that this number is finite). By Corollary 2.2.17, this means |f (a)| ≤ CKkf k2 for all f ∈ A2(G) and a ∈ K, where CK = 1/(

√

π), proving the estimate (2.9).

Because L2_{(G) is a Hilbert space and A}2_{(G) is a linear subspace, it is sufficient to show}

that A2(G) is closed. Now suppose that (fn) is a sequence in A2(G) converging to an

f ∈ L2_{(G). Then the sequence (f}

n) is a Cauchy sequence in L2(G). From (2.9) applied

to the compact set {z}, we see that (fn(z)) is a Cauchy sequence, so the pointwise limit

g(z) = limn→∞fn(z) exists for all z ∈ G. It is not hard to deduce from (2.9) that fn→ g

uniformly on compact subsets K ⊂ G. Let > 0 and K ⊂ G compact. Then since (fn)

is a Cauchy sequence, there exists an N ≥ 1 so that kfn− fmk2 < _2C_K for all m, n ≥ N .

From (2.9), we get sup a∈K |fn(a) − fm(a)| < 2 for all m, n ≥ N . In particular,

sup

a∈K

|fN(a) − fm(a)| <

2

for all m ≥ N . Now pick a ∈ K. Then |fN(a) − fm(a)| < ₂ for all m ≥ N , so by taking

m → ∞, we get |fN(a) − g(a)| ≤ ₂. Since a ∈ K was arbitrary, this means that

sup

a∈K

|fN(a) − g(a)| ≤

2.

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By the triangle inequality, these inequalities imply that sup

a∈K

|fn(a) − g(a)| ≤ sup a∈K

|fn(a) − fN(a)| + sup a∈K |fN(a) − g(a)| < 2 + 2 =

for all n ≥ N . This proves that fn → g uniformly on compact subsets K ⊂ G. By

Theorem 7.2.1 from [9, p. 82], this means that g is holomorphic on G. Furthermore, by Theorem 3.12 from [24, p. 70], there exists a subsequence (fnk) of (fn) such that fnk → f

pointwise dA-almost everywhere on G. Since fn → g pointwise everywhere on G, this

means that f = g dA-almost everywhere on G, so that f ∈ A2_{(G). Therefore, A}2_{(G) is}

a closed subspace of the Hilbert space L2_{(G) and hence a Hilbert space itself.}

Now for any ζ ∈ G, the set {ζ} is compact. From (2.9) we obtain |f (ζ)| ≤ Cζkf k2

for all f ∈ A2_{(G) for a constant C}

ζ independent of f ∈ A2(G). This means the linear

functional f 7→ f (ζ) is continuous on A2(G). By the standard Riesz representation theorem for Hilbert spaces, for each ζ ∈ G, there exists a unique ηζ ∈ A2(G) such that

f (ζ) = hf, ηζi =

Z

G

f (z)ηζ(z)dAz

for all f ∈ A2_{(G). We now define the Bergman kernel K of G by}

K(z, ζ) = ηζ(z).

Note that for a fixed ζ ∈ G, we have z 7→ K(z, ζ) ∈ A2(G) and that f (ζ) =

Z

G

f (z)K(z, ζ)dAz

for all f ∈ A2_{(G). Plugging in f = K(·, ζ) gives}

K(ζ, ζ) = Z

G

K(z, ζ)K(z, ζ)dAz = kK(·, ζ)k22 < ∞,

since z 7→ K(z, ζ) is square-integrable. In particular, K(ζ, ζ) is a real number for all ζ ∈ G. Also, note that for any ζ ∈ G, we have

|f (ζ)| = Z G f (z)K(z, ζ)dAz ≤ Z G f (z)K(z, ζ) dAz ≤ kf k2·kK(·, ζ)k2 = p K(ζ, ζ)kf k2

for all f ∈ A2_{(G) by Cauchy-Schwarz.}

For a compact set Ω ⊂ C, we can use the same arguments to prove that when ζ ∈ Ω◦, ζ ∈ Ω is bounded point evaluation for R2_{(X). Since R}2_{(X) is a closed subspace of the}

Hilbert space L2_{(X), it is a Hilbert space itself and the Riesz representation theorem}

gives us a unique function K(·, ζ) ∈ R2(X) such that f (ζ) =

Z

Ω

f (z)K(z, ζ)dAz

for all f ∈ R2_{(X) when ζ ∈ Ω}◦_{. Again, we will call K the Bergman kernel for Ω. Note}

however, that this time, K(·, ζ) ∈ R2(X) for ζ ∈ Ω◦. This is the definition of the Bergman kernel that we will use in the remainder of this section (cf. [3, p. 311]).

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Lemma 2.2.18. Suppose that G1 and G2 are compact sets and that G1 ⊂ G2. Let K1

and K2 be the Bergman kernels of G1 and G2 respectively. Then for any ζ ∈ G◦1, we have

K2(ζ, ζ) ≤ K1(ζ, ζ).

Proof. Fix ζ ∈ G◦₁. Clearly, every f ∈ R2_(G

2) is an element of R2(G1) when restricted to

G1. In particular, K2(·, ζ) restricts to R2(G1). This gives (cf. [3, p. 312])

0 ≤ Z G1 |K1(z, ζ) − K2(z, ζ)|2dAz = Z G1 h K1(z, ζ)K1(z, ζ) − K1(z, ζ)K2(z, ζ) − K2(z, ζ)K1(z, ζ) + K2(z, ζ)K2(z, ζ) i dAz = K1(ζ, ζ) − K2(ζ, ζ) − K2(ζ, ζ) + Z G1 K2(z, ζ)K2(z, ζ)dAz ≤ K1(ζ, ζ) − 2K2(ζ, ζ) + Z G2 K2(z, ζ)K2(z, ζ)dAz ≤ K1(ζ, ζ) − K2(ζ, ζ).

(Note that K2(ζ, ζ) = K2(ζ, ζ) and that the integral increases by enlarging the domain

of integration.) Rearrangement gives K2(ζ, ζ) ≤ K1(ζ, ζ).

We are now able to prove Theorem 2.2.15. The proof is due to Brennan and Mattingly [5, p. 207].

Proof of Theorem 2.2.15. Let D denote the open unit disk B(0, 1). Choose a countable set S = {α1, α2, . . .} ⊂ D\{0} that is dense in D. Let a1 = α1. Remove from D an open

disk D1 = B(a1, r1) of radius r1 < |a1| so that 0 /∈ B(a1, r1). We define Ω1 = D\D1.

Note that 0 ∈ Ω1 and that even 0 ∈ Ω◦1.

Let K1 denote the Bergman kernel of Ω1. Because 0 ∈ Ω◦1, for any rational function f

with poles off Ω1 we now have

|f (0)|2 ≤ K1(0, 0)

Z

Ω1

|f (z)|2dAz.

Next, let a2 = αj2 be the first point of S not contained in D1. Remove a second disk

D2 = B(a2, r2) in such a way that

1. D1 and D2 have disjoint closures,

2. 0 ∈ Ω2 := D \ (D1∪ D2) and 0 ∈ Ω◦2,

3. K2(0, 0) − K1(0, 0) < 1₂, where Kj denotes the Bergman kernel of Ωj for j = 1, 2.

Note that 0 ≤ K2(0, 0) − K1(0, 0) by Lemma 2.2.18. All these conditions can be satisfied

simultaneously by taking the radius r2 sufficiently small. For the last condition, this

follows from the following lemma from [3, p. 312].

Lemma 2.2.19. Let Ω be a closed region bounded by the unit circle and a finite number of disjoint circles lying inside the open unit disk. Fix a ∈ Ω◦, assume that B(a, r) ⊂ Ω◦ and, put

Ωr= Ω \ B(a, r).

Let K(z, ζ) and Kr(z, ζ) be the Bergman kernels for Ω and Ωr, respectively. Then

Kr(z, ζ) → K(z, ζ) uniformly as r → 0 for z and ζ in any fixed compact subset of

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It suffices to take {0} as the fixed compact subset of Ω◦₁ \ {a2}. We continue this

procedure and obtain a sequence of disks Dn, n = 1, 2, . . . and a sequence of closed sets

Ωn= D \

Sn

i=1Di, n = 1, 2, . . ., such that

1. the disks Dn have disjoint closures,

2. T∞

i=1Ωi has no interior, contains the point z = 0 and for any n ≥ 1, we have 0 ∈ Ω◦n,

3. 0 ≤ Kn+1(0, 0) − Kn(0, 0) < ₂1n, where Kj denotes the Bergman kernel of Ωj for

j = 1, 2, . . .. Define X =T∞

i=1Ωi. Then X has no interior, since S ∩ X = ∅ and X has a positive

measure whenever P∞

i=1r 2

i < 1 (this is something we can achieve by taking each ri

sufficiently small in each step). We will now show that R2_{(X) 6= L}2_{(X) by showing that}

X has a bounded point evaluation for R2_{(X) at 0. Suppose that f is a rational function}

without poles on X. Because f has only finitely many poles, for some n0, f will not have

any poles in Ωn0 and therefore no poles in Ωn for all n ≥ n0. This means that

|f (0)|2 ≤ Kn(0, 0)

Z

Ωn

|f (z)|2dAz

for all n ≥ n0. Because the sequence8 (|f |2· 1Ωn)

∞

n=n0 of integrable functions on Ωn0

converges pointwise to the function |f |2 _{· 1}

X on Ωn0 and is dominated by the integrable

function |f |2· 1Ω_n0 on Ωn0, the dominated convergence theorem tells us that

Z Ωn |f (z)|2_dA z = Z Ω_n0 |f (z)|2_{· 1} ΩndAz → Z Ω_n0 |f (z)|2_{· 1} XdAz = Z X |f (z)|2_dA z

as n → ∞. Also, because for any n ≥ 1 Kn(0, 0) = K1(0, 0) + n X i=2 Ki(0, 0) − Ki−1(0, 0) ≤ K1(0, 0) + ∞ X i=2 Ki(0, 0) − Ki−1(0, 0) ≤ K1(0, 0) + ∞ X i=2 1 2i−1 = K1(0, 0) + 1, we have |f (0)|2 _{≤ (K} 1(0, 0) + 1) Z X |f (z)|2_dA z.

By taking square roots, we get

|f (0)| ≤pK1(0, 0) + 1kf k2

This inequality holds for all rational functions f with no poles on X. This means that 0 is a bounded point evaluation for R(X) (since K1(0, 0) < ∞, we saw this before), so by

extending this bounded linear functional on R(X) to a bounded linear functional on its closure R2_{(X), 0 is a bounded point evalution for R}2_{(X), so R}2_{(X) 6= L}2_{(X) by Theorem}

2.2.14 (since X has a positive measure).

8_{The function 1}

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3. Capacities and properties of

capacities

As we have seen, there are ways to determine whether R(X) = C(X) or Rp_{(X) =}

Lp(X) for 1 ≤ p < ∞ for a compact, nowhere dense set X. For example, we can look at complex regular Borel measures µ on X and check if the corresponding Cauchy transform ˆµ vanishes almost everywhere on the complex plane, or we can look at the Cauchy transforms ˆk of functions k ∈ Lq(X) that annihilate Rp(X) to see if Rp(X) = Lp_{(X). In Lemma 2.2.12, we saw that for q > 2, the Cauchy transform ˆ}_{k of a k ∈ L}q_{(X) is}

in fact continuous. However, when 1 < q ≤ 2 or equivalently 2 ≤ p < ∞, it is not possible to use the same lemma to check if ˆk is continuous. However, when k ∈ Lq(X) annihilates Rp_{(X), then ˆ}_{k = 0 on C \ X. We want to see whether ˆk is sufficiently continuous to}

conclude that ˆk = 0 dA-almost everywhere on the complex plane. This can be done using capacities, see for example [5, Subsection 3.5]. The same remark applies to the Cauchy transfrom ˆµ of a complex regular Borel measure µ on X, again see [5, Subsection 3.5]. Capacities and their uses are related to function spaces. More information about capacities and their connection to function spaces can be found in [1]. For our purposes, it is sufficient to say that a capacity C is a set function defined on a certain collection F of subsets of C containing at least the Borel sigma-algebra on C such that

1. C(∅) = 0,

2. if E, F ∈ F and E ⊂ F , then C(E) ≤ C(F ).

We will need two kinds of capacities, corresponding to the two different problems of determining whether R(X) = C(X) or Rp_{(X) = L}p_{(X) for 1 ≤ p < ∞. The first}

problem can be expressed entirely in terms of analytic capacity, while the second one can be expressed entirely in terms of q-capacity.

3.1. Uniform approximation

In order to solve the problem whether R(X) = C(X) for a compact, nowhere dense set X, we will consider the analytic capacity. Let ˆ_{C = C ∪ {∞} denote the Riemann sphere.} For a compact set K ⊂ C, let Ω(K) denote the unbounded component of ˆC \ K. For a function f that is defined and analytic on Ω(K), we define f0(∞) = limz→∞zf (z) (cf.

[17, p. 196]).

Definition 3.1.1. The analytic capacity γ(K) of a compact set K is defined by γ(K) = sup |f0(∞)|,

where the supremum runs over all functions f that are defined and analytic on Ω(K) and satisfy f (∞) = 0 and sup_Ω(K)_{|f | < ∞. This definition is extended to all sets E ⊂ C by} setting

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Note that the analytic capacity γ(∅) of the empty set ∅ is zero, since all bounded entire functions f are constant by Liouville’s theorem and hence f = 0 by the assumption f (∞) = 0. We will only need the following properties of analytic capacity. All proofs of these properties can be found in [17, Chapter VIII], except for the last two. The fourth property was proved by Tolsa in [27, Theorem 1.4, p. 108].

Lemma 3.1.2 (Monotonicity). If E ⊂ F ⊂ C, then γ(E) ≤ γ(F ).

Lemma 3.1.3 (Translation-invariance and scaling). If E ⊂ C and z0, a ∈ C, then γ(z0+

aE) = |a|γ(E).

Lemma 3.1.4 (Outer regularity for compact sets). If K ⊂ C is compact, then γ(K) = inf{γ(U ) : K ⊂ U, U open}.

Lemma 3.1.5 (Tolsa). Let E ⊂ C be compact. Let Ei for i ≥ 1 be Borel sets such that

E =S∞ i=1Ei. Then γ(E) ≤ C ∞ X i=1 γ (Ei) ,

where C > 0 is a constant independent of the sets chosen above.

Corollary 3.1.6 (Semiadditivity for Borel sets). Let Ei for i ≥ 1 be Borel sets. Then

for the same constant C as in Tolsa’s theorem,

γ ∞ [ i=1 Ei ! ≤ C ∞ X i=1 γ(Ei).

Proof. Let Ei for i ≥ 1 be Borel sets. Let E =S ∞

i=1Ei. Let K ⊂ E be compact. Then

K =S∞

i=1Fi where all Fi = Ei∩ K are Borel sets (since K is a Borel set), so by Tolsa’s

theorem and monotonicity

γ(K) ≤ C ∞ X i=1 γ(Fi) ≤ C ∞ X i=1 γ(Ei).

Now taking the supremum over all compact sets K ⊂ E yields

γ(E) ≤ C

∞

X

i=1

γ(Ei).

The advantage of having the analytic capacity is the following theorem by Vitushkin. A proof can be found in [17, p. 207].

Theorem 3.1.7 (Vitushkin). For a compact, nowhere dense set X ⊂ C, the following are equivalent:

(a) R(X) = C(X), (b) lim sup

r↓0

γ(B(x, r) \ X)

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In order to determine whether the last statement holds for dA-almost every x ∈ X, we have the aid of another theorem by Vitushkin, about the instability of analytic capacity (cf. [17, p. 207]).

Theorem 3.1.8 (Instability of analytic capacity). If X ⊂ C is compact, then for dA-almost every z ∈ C, either

(a) lim r↓0 γ(B(z, r) \ X) r = 1 or (b) lim r↓0 γ(B(z, r) \ X) r2 = 0.

Remark. It may be trivial, but it is very important to note that the instability of analytic capacity does not state that exactly one of the following statement holds:

(a) lim

r↓0

γ(B(z, r) \ X)

r = 1 for dA-almost every z ∈ C, or

(b) lim

r↓0

γ(B(z, r) \ X)

r2 = 0 for dA-almost every z ∈ C.

That this is not true, can be seen by considering Roth’s Swiss cheese X of positive measure from Chapter 2. Since R(X) 6= C(X) for this set X, we have that

lim sup

r↓0

γ(B(x, r) \ X)

r = 0

for dA-almost every x ∈ X by Vitushkin’s theorem. Hence, statement (a) does not hold, since X has a positive measure.

Now for every x ∈ C \ X, there exists a δ > 0 such that B(x, r) ⊂ C \ X for all 0 < r < δ, since C \ X is open. It can be proved that the analytic capacity of an open disk is equal to its radius (Corollary 1.5 in [17, p. 196]), so that

lim r↓0 γ(B(x, r) \ X) r = limr↓0 γ(B(x, r)) r = limr↓0 r r = 1

for all x ∈ C \ X, since for all x ∈ C \ X we have B(x, r) ⊂ C \ X for all r small enough. Hence

lim

r↓0

γ(B(x, r) \ X)

r2 = +∞

for all x ∈ C \ X. Since C \ X is not a null set, this means that (b) is also violated.

3.2. Approximation in p-norm

We will define various kinds of q-capacities, all of which turn out to be equivalent for 1 < q < 2. When q = 2, this might not be the case. Therefore, in the following, we will be careful in stating for which values of q the results hold.

To give our definitions, we will need the Bessel kernel, for our purposes defined by [23, p. 279]

G(z) = 1 2π|z|

−1

e−|z|.

for all z ∈ C unequal to 0. More properties of this function can be found in Sections 1.2.3, 1.2.4 and 1.2.5 of [1].

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Definition 3.2.1. For 1 < q < 2, the Riesz q-capacity Cq(E) of an arbitrary set E ⊂ C is defined by Cq(E) = inf Z C |f (z)|qdAz

where the infimum is taken over all functions f ∈ Lq_{(C) such that f is a non-negative}

real-valued function and

Z

C

f (w)

|z − w|dAw ≥ 1

for all z ∈ E. It can be shown (Theorem 2.2 in [13, p. 242]) that for Borel sets E ⊂ C Cq(E) = (sup µ(E))q,

where the supremum runs over all positive finite (i.e. µ(C) < ∞) Borel measures µ on the plane concentrated on E for which the corresponding Newtonian potential satisfies k˜µkp ≤ 1, where k · kp denotes the p-norm on Lp(C).

Definition 3.2.2. For 1 < q ≤ 2, the Bessel q-capacity Bq(E) of an arbitrary set E ⊂ C

is defined by Bq(E) = inf Z C |f (z)|q_dA z

where the infimum is taken over all functions f ∈ Lq_{(C) such that f is a non-negative} real-valued function and1

Z

C

G(z − w)f (w)dAw ≥ 1

for all z ∈ E.

The following properties can be found in [23] and [1].

Lemma 3.2.3 (Monotonicity of Riesz and Bessel q-capacity). Let 1 < q < 2 (q = 2 for the Bessel q-capacity as well). Then by Proposition 2.3.4 from [1],

Bq(E) ≤ Bq(F ), Cq(E) ≤ Cq(F )

for all E ⊂ F ⊂ C.

Lemma 3.2.4 (Translation-invariance of Riesz and Bessel q-capacity). Let 1 < q < 2 (q = 2 for the Bessel q-capacity as well). Then,

Bq(E + z0) ≤ Bq(E), Cq(E + z0) ≤ Cq(E)

for all E ⊂ C and z0 ∈ C.

Proof. For Bessel q-capacity, this is Lemma 6 [23, p. 282]. For Riesz q-capacity, the proof by change of variable is also obvious.

1_{Note that the exact constant in front of the Bessel kernel G is not that important: if instead of G,}

the function cG was used (for some real-valued c > 0) in the definition of the Bessel q-capacity (1 < q ≤ 2), then the resulting capacities differ by a constant of cq _{and are therefore equivalent. See}

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Lemma 3.2.5 (Subadditivity of Riesz and Bessel q-capacity). Let 1 < q < 2 (q = 2 for the Bessel q-capacity as well) and let Ei ⊂ C for i ≥ 1. Then by Proposition 2.3.6 from

[1] Bq ∞ [ i=1 Ei ! ≤ ∞ X i=1 Bq(Ei), Cq ∞ [ i=1 Ei ! ≤ ∞ X i=1 Bq(Ei).

Lemma 3.2.6 (Ascending chain property of Riesz and Bessel q-capacity). Let 1 < q < 2 (q = 2 for the Bessel q-capacity as well) and let E1 ⊂ E2 ⊂ · · · ⊂ C be arbitrary sets with

union E. Then by Proposition 2.3.12 from [1] Bq(E) = lim

i→∞Bq(Ei), Cq(E) = limi→∞Cq(Ei).

Another important property of these capacities is that certain sets, in particular Borel sets, are capacitable. We will define this in the following lemma. A proof can be found in [1, p. 28, Theorem 2.3.11] and in the remarks directly following the statement of the theorem.

Lemma 3.2.7 (Capacitability). Let 1 < q < 2 (q = 2 for the Bessel q-capacity as well). Then, in particular, all Borel sets are capacitable for the Riesz and Bessel q-capacities. This means that for all Borel sets E ⊂ C

Cq(E) = sup{Cq(K) : K ⊂ E, K compact} = inf{Cq(G) : E ⊂ G, G open}

Bq(E) = sup{Bq(K) : K ⊂ E, K compact} = inf{Bq(G) : E ⊂ G, G open}

for 1 < q < 2 (q = 2 for the Bessel q-capacity as well).

Note that by capacitability, it is sufficient to define Cq (1 < q < 2) and Bq (1 < q ≤ 2)

for compact sets. Also note that for both q-capacities, the q-capacity of the empty set is indeed zero. Clearly both q-capacities are positive and in both definitions, the zero function is admissible in the definition when E is the empty set.

The equivalence of the two different types of q-capacity is proven in [22, p. 313, Lemma 3] for 1 < q < 2. Note the omission of the case q = 2. In [22, p. 313], our Bq is Hedberg’s

cq and our Cq is the qth power of his γq. Using this, we reformulate this lemma in our

capacities. Using our definitions, Hedberg’s Lemma 3 reads Cq(E)1/q ≤ ABq(E)1/q ≤ ACq(E)1/q

for all Borel sets E ⊂ C, for a constant A > 0 only depending on q. Rewriting this gives the following lemma.

Lemma 3.2.8 (Equivalence of Riesz and Bessel q-capacities). Let 1 < q < 2. Then there is constant A > 0 independent of E (but depending on q) such that

A−qCq(E) ≤ Bq(E) ≤ Cq(E)

for all Borel sets E ⊂ C.

These capacities have the following estimates for disks and line segments. When we write F ≈ G for two functions F and G, for example in the following lemma, we mean that there exist positive constants A1 and A2 such that

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for all x under consideration (x may be a set, a positive real number etc.) and where the constants A1 and A2 do not depend on x. Such functions will be called equivalent. The

equivalence of the Riesz and Bessel q-capacity for 1 < q < 2 can be restated as follows: for fixed 1 < q < 2, we have Bq ≈ Cq for all Borel sets in C. Proofs of the following two

lemmas can be found in [23, Lemma 7, Lemma 8].

Lemma 3.2.9. If 1 < q < 2, then for any x ∈ C and 0 < r ≤ 1, we have Bq(B(x, r)) ≈

r2−q. By Bq ≈ Cq for fixed 1 < q < 2 for all Borel sets, the same statement holds for Cq.

The following statement only holds for the Bessel 2-capacity. The Riesz 2-capacity is the zero capacity, by the remark directly following Proposition 5.1.2 in [1, p. 131]. Lemma 3.2.10. If q = 2 and 0 < δ < 1 is fixed, then for all x ∈ C and 0 < r ≤ δ < 1 we have B2(B(x, r)) ≈ log 1 r −1 .

Lemma 3.2.11. Let 1 < q < 2. If Lr is any straight line segment (that may or may not

contain the endpoints) of length 0 < r < 1, then Cq(Lr) ≈ r2−q. The same statement

holds for Bq for 1 < q < 2.

Proof. Fix 1 < q < 2. Let x be one of the endpoints of the straight line segment Lr.

Then for any > 0, we know Lr ⊂ B(x, r + ). By monotonicity, this means that

Cq(Lr) ≤ Cq(B(x, r + )) for all > 0. For < 1 − r we can apply Lemma 3.2.9 to get

Cq(Lr) ≤ Cq(B(x, r + )) ≤ A2(r + )2−q for some A2 > 0 independent of r. Letting ↓ 0

yields Cq(Lr) ≤ A2r2−q.

To find A1, we need something totally different. We follow the idea from [4, p. 411–

412]. Let ` be the line through the endpoints of Lr. For a Borel set B, define the positive

Borel measure ν on C by ν(B) = m(B ∩ Lr), where m denotes the one-dimensional

Lebesgue measure on the line `. The measure ν has total mass ν(C) = m(Lr) = r, so

that the Borel measure µ = r−1_{ν is a Borel measure of total mass 1 (i.e., µ(C) = 1).} This enables us to apply Lemma 2(i) from [4], which states that for a constant K1 > 0

independent of r, k˜µkp_p = Z C Z Lr dµ(ζ) |ζ − z| p dA(z) ≤ K1 sup z∈C Z Lr dµ(ζ) |ζ − z|2−q p−1 . Now applying the reverse triangle inequality yields

K1 sup z∈C Z Lr dµ(ζ) |ζ − z|2−q p−1 ≤ K1 sup z∈C Z Lr dµ(ζ) ||ζ| − |z||2−q p−1 = K1 1 rsup_z∈C Z Lr dm(ζ) ||ζ| − |z||2−q p−1 . We will now estimate the integral R_L

r

dm(ζ)

||ζ|−|z||2−q. Since Lr is a line segment of length r,

we know that for all ζ ∈ Lr, |ζ| ∈ [c, c + r] for some constant c ∈ R. For z ∈ C, there are

three possibilities (draw some pictures!).

1. |z| ∈ [c, c + r]: In this case, we can estimate R_L

r dm(ζ) ||ζ|−|z||2−q by 2 · Rr 0 1 x2−qdx. Namely, the integralR|z| c 1 |z|−x2−qdx can be estimated by Rr 0 1

x2−qdx in itself (the shapes of the

functions are the same, however the latter integral is taken over a larger interval) and a similar reasoning holds for R_|z|c+r _|z|−x12−qdx.

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2. |z| < c: In this case, the entire integral R Lr dm(ζ) ||ζ|−|z||2−q = Rc+r c 1 (x−|z|)2−qdx can be estimated byR₀r _x2−q1 dx.

3. |z| > c + r: This case is similar to the second. In conclusion, in all cases, it suffices to estimateR_L

r dm(ζ) ||ζ|−|z||2−q by 2· Rr 0 1 x2−qdx. Returning

to our previous inequalities, this means that K1 1 rsupz∈C Z Lr dm(ζ) ||ζ| − |z||2−q p−1 ≤ K1 1 r · 2 · Z r 0 1 x2−qdx p−1 = K1 1 r · 2 · 1 q − 1r q−1 p−1 .

By introducing the new constant K2 = K1· (2 ·_q−11 )p−1 and combining all above

inequal-ities, we get Z C Z Lr dµ(ζ) |ζ − z| p dA(z) ≤ K2r(q−2)(p−1) = K2r(q−2)p/q.

If we now define the Borel measure σ by σ = K₂−1r(2−q)/qµ, then k˜σkp = kK2−1r (2−q)/q_µk_˜ p = K2−1r (2−q)/q_k˜_µk p ≤ K2−1r (2−q)/q _K 2r(q−2)p/q 1/p = 1. This means that the measure σ is admissible in the definition of q-capacity, so that

Cq(Lr) ≥ (σ(Lr))q = (K2−1r (2−q)/q_µ(L r))q = K −q 2 r 2−q_.

Note that the constant K₂−q does not depend on r. Taking A1 = K2−q and A2 the same

constant as before proves the lemma.

The q-capacities we introduced here, will help us in determining whether Rp_{(X) =}

Lp_{(X) for 2 ≤ p < ∞. The theorem that makes this clear, is due to Hedberg [22, p. 316,}

Corollary 1]. Hedberg formulates this theorem using Bessel capacities and we will do this as well.2

Theorem 3.2.12 (Hedberg #1). Let X ⊂ C be a compact, nowhere dense set of positive measure and let 2 ≤ p < ∞. Then the following are equivalent:

(a) Rp_{(X) = L}p_(X),

(b) Bq(G \ X) = Bq(G) for all open sets G ⊂ C,

(c) Bq(B(x, r) \ X) = Bq(B(x, r)) for all x ∈ C and r > 0,

(d) lim sup

r↓0

Bq(B(x, r) \ X)

r2 > 0 for dA-almost all x ∈ C.

Note that by the equivalence of Bq and Cq for 1 < q < 2 (or equivalently 2 < p < ∞),

the last statement in Hedberg’s theorem holds for Bq if and only if it holds for Cq for

1 < q < 2. This leads to the following corollary.

2_{It is not clearly stated in [22] whether Corollary 1 in [22, p. 316] only holds for compact, nowhere}

dense sets of positive measure, but as we have seen, it trivially holds that R(X) = C(X) (by Hartogs-Rosenthal) and that Rp(X) = Lp(X) = {0} whenever X is a compact, nowhere dense set of measure 0. Therefore, in the constructions in the following chapters, we will only use Hedberg’s theorems whenever a compact, nowhere dense set X has a positive measure. Since the sets we will construct need to have a positive measure, this is no restriction.

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Corollary 3.2.13 (Hedberg #2). Let X ⊂ C be a compact, nowhere dense set of positive measure and let 2 < p < ∞. Then the following are equivalent:

(a) Rp_{(X) = L}p_(X),

(b) lim sup

r↓0

Bq(B(x, r) \ X)

r2 > 0 for dA-almost all x ∈ C,

(c) lim sup

r↓0

Cq(B(x, r) \ X)

We can prove even more, namely that for a compact, nowhere dense set X ⊂ C, it is sufficient to look at the behaviour, as r tends to 0, of the q-capacity of sets B(x, r) \ X for which x is an element of X.

Corollary 3.2.14 (Hedberg #3). Let X ⊂ C be a compact, nowhere dense set of positive measure and let 2 < p < ∞. Then the following are equivalent:

(a) Rp_{(X) = L}p_(X),

(b) lim sup

r↓0

Bq(B(x, r) \ X)

r2 > 0 for dA-almost all x ∈ X,

(c) lim sup

r↓0

Cq(B(x, r) \ X)

r2 > 0 for dA-almost all x ∈ X,

(d) lim sup

r↓0

Bq(B(x, r) \ X)

Proof. The equivalence of (a) and (d) follows from Hedberg #2. The equivalence of (b) and (c) is clear from the equivalence of Bq and Cq for 2 < p < ∞. We now prove the

equivalence of (b) and (d). That (d) implies (b) is clear, so we will prove (b) implies (d). Suppose (b) holds. For every x ∈ C \ X, there exists a δ with 0 < δ < 1 such that B(x, r) ⊂ C \ X for all r with 0 < r ≤ δ, since C \ X is open. Then by Lemma 3.2.9 (because 1 < q < 2) and using that r2 _{≤ r}2−q _{whenever 0 < r < 1, we have}

A1r2 ≤ A1r2−q ≤ Bq(B(x, r)) ≤ A2r2−q

for all 0 < r < δ, for positive constants A1 and A2 independent of r. Therefore,

Bq(B(x, r) \ X) r2 = Bq(B(x, r)) r2 ≥ A1r2 r2 = A1

for all 0 < r < δ, since B(x, r) \ X = B(x, r) for these values of r. Hence lim sup

r↓0

Bq(B(x, r) \ X)

r2 ≥ A1 > 0.

Since x ∈ C \ X was arbitrary, we have lim sup

r↓0

Bq(B(x, r) \ X)

r2 > 0

for all x ∈ C \ X. By assumption, we have lim sup

r↓0

Bq(B(x, r) \ X)

r2 > 0

for dA-almost every x ∈ X. Therefore, the above inequality holds for every x ∈ C \ X and dA-almost every x ∈ X, which means for dA-almost every x ∈ C. So (d) holds.

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What helps in determining whether any one of these equivalent statement in Hedberg #3 holds, is that the Riesz q-capacity for 1 < q < 2 also displays a certain instability. This was proved by Fernstr¨om in [13, p. 245, Theorem 4.1].

Theorem 3.2.15 (Instability of Riesz q-capacity). Let 1 < q < 2 and let E ⊂ C be a Borel set. Then for dA-almost every z ∈ C, either

(a) lim r↓0 Cq(E ∩ B(z, r)) Cq(B(z, r)) = 1 or (b) lim r↓0 Cq(E ∩ B(z, r)) r2 = 0.

Remark. A similar remark as the one we made after the instability of analytic capacity applies here.

We will use the criteria we obtained above in terms of capacities in order to determine whether R(X) is dense in C(X) with respect to the uniform norm or R(X) is dense in Lp_{(X) with respect to the p-norm extensively in the following chapters.}

Rational approximation on compact, nowhere dense subsets of the complex plane