Approximation of Holomorphic Functions in the Complex Plane

Home , Holomorphic function

Auðunn Skúta Snæbjarnarson

FacultyFaculty of of Physical Physical Sciences Sciences UniversityUniversity of of Iceland Iceland 20142014

APPROXIMATION OF HOLOMORPHIC FUNCTIONS IN THE COMPLEX PLANE

Auðunn Skúta Snæbjarnarson

60 ECTS thesis submitted in partial fulllment of a Magister Scientiarum degree in Mathematics

Advisor Ragnar Sigurðsson

Faculty Representative Reynir Axelsson

M.Sc. committee Ragnar Sigurðsson, Jón Ingólfur Magnússon

Faculty of Physical Sciences School of Engineering and Natural Sciences University of Iceland Reykjavik, October 2014 Approximation of Holomorphic Functions in the Complex Plane Approximation of Holomorphic Functions 60 ECTS thesis submitted in partial fulllment of a M.Sc. degree in Mathematics

Faculty of Physical Sciences School of Engineering and Natural Sciences University of Iceland Dunhagi 5 IS-107, Reykjavik, Reykjavik Iceland

Telephone: 525 4000

Bibliographic information: Auðunn Skúta Snæbjarnarson, 2014, Approximation of Holomorphic Functions in the Complex Plane, M.Sc. thesis, Faculty of Physical Sciences, University of Iceland.

Printing: Háskólaprent, Fálkagata 2, 107 Reykjavík Reykjavik, Iceland, October 2014 .. Abstract

This thesis is a study of approximation of holomorphic functions, by polynomials, rational functions or entire functions. Hörmander's L2 existence theorem for the Cauchy-Riemann operator is proved and used to prove a generalization of the Bernstein-Walsh theorem, which describes the equivalence between possible holomorphic continuation of a function f in a neighbourhood of a compact set K in terms of the decay of the sequence (dn(f, K)) of best approximations of f by polynomials of degree less than or equal to n. The generalization uses the best approximation of f by rational functions with poles in a prescribed set instead of polynomials. A theorem of Vitushkin is proved. It characterizes in terms of analytic capacity the compact sets K having the property that every function f, continuous on K and holomorphic in the interior of K, can be approximated uniformly on K by rational functions. Finally, Mergelyan's theorem is used to prove a generalization of Arake- lian's theorem, which describes uniform approximation of holomorphic functions on possibly unbounded sets by entire functions.

Útdráttur

Þessi ritgerð fjallar um nálganir á fáguðum föllum, með margliðum, ræðum föllum eða heilum föllum. L2-tilvistarsetning Hörmanders fyrir Cauchy-Riemann virkjann er sönnuð og hún er notuð til þess að sanna alhængu á setningu Bernstein-Walsh, sem lýsir jafngildi milli mögulegrar fágaðar framlengingar á falli f á opinni grennd við þjappað hlutmengi K og runu bestu nálgana (dn(f, K)) á f með margliðum af stigi minna eða jöfnu n. Alhængin notar bestu nálganir á f með ræðum föllum með skaut í gefnu mengi. Setning Vitushkins er sönnuð, en hún lýsir hvernig fáguð rýmd mengis er notuð til þess að auðkenna þjöppuð mengi K með þann eiginleika að sérhvert fall f, samfellt á K og fágað á innmengi K, megi nálga í jöfnu mæli á K með ræðum föllum. Að lokum er setning Mergelyans beitt til þess að sanna alhængu á setningu Arakelians, sem lýsir nálgun í jöfnum mæli á fáguðum föllum á ótakmörkuðum mengjum með heilum föllum.

Contents

Acknowledgments ix

1 Introduction 1

2 Preliminaries 7 2.1 Holomorphic functions ...... 7 2.2 Harmonic functions ...... 10 2.3 Subharmonic functions ...... 11 2.4 The Dirichlet problem ...... 12 2.4.1 The Perron method ...... 13 2.4.2 Criteria for regularity ...... 14 2.5 Green's functions ...... 15 2.6 Distributions ...... 16 2.6.1 Denitions ...... 16 2.6.2 Dierential equations in the sense of distributions ...... 17 2.7 Results from functional analysis ...... 18

3 Hörmander's L2 theorem 19 3.1 Hörmander ...... 19 3.2 An application: Brunn-Minkowski type inequalities ...... 24

4 A generalized version of the Bernstein-Walsh theorem 29 4.1 Denitions ...... 29 4.2 The estimation procedure ...... 30 4.2.1 The construction of F ...... 31 4.2.2 Showing that F ∈ R(a, n) ...... 32 4.2.3 Estimation of kF − fkK ...... 34 4.3 Discussion on the estimate ...... 36

5 Vitushkin's Theorem 41 5.1 Introduction ...... 41 5.2 Theorems and estimates of analytic capacity ...... 44 5.2.1 Green's functions and analytic capacity ...... 50 5.3 Rational approximation ...... 51 5.3.1 Partition of f ...... 51 5.3.2 Proof of Vitushkin's Theorem ...... 57

vii Contents

6 Arakelian's Theorem 67 6.1 Arakelian ...... 67

Bibliography 71

viii Acknowledgements

First I would like to thank my supervisor Ragnar Sigurðsson for his excellent guid- ance and inspiration. I would also like to thank the other members of the complex analysis research group at the university, Benedikt Steinar Magnússon, Jón Ingólfur Magnússon and Reynir Axelsson for all their help and advice. I thank my father for making mathematics the obvious choice for me and I thank my mother for her encouragement and support.

1 Introduction

When studying polynomial approximation of functions dened on Rn the Weierstraÿ approximation theorem gives a very strong result.

Theorem 1.0.1 (Weierstraÿ). Let K be a compact subset of Rn and f ∈ C(K). For every > 0 there exists a polynomial P such that sup |P (x) − f(x)| < . x∈K When we look at functions dened on C, the results become more complicated. Since all polynomials in the variable z are holomorphic the following theorem implies that if a function f is to be approximated uniformly by polynomials on an open set it must be holomorphic. Theorem 1.0.2 (Cor. 1.2.6. of [6]). If is an open set, for all Ω ⊂ C fn ∈ O(Ω) and uniformly on compact subsets of then . n ∈ N fn → f Ω f ∈ O(Ω) Here O(Ω) is the set of functions holomorphic on Ω. The following question now arises:

Can we nd some necessary and/or sucent conditions such that a function f on a compact set K can be approximated by polynomials? From Theorem 1.0.2 it follows that the following conditions are necessary on f • f is continuous on K. • f is holomorphic on int(K). We denote the set of functions which satisfy these conditions by A(K). But can we nd some sucient or necessary conditions on the set K so that every function in A(K) can be approximated in K? Suppose its compliment, Kc, has a bounded component U and assume we can approximate every function g ∈ A(K) uniformly in K by polynomials. We choose some ζ ∈ U and consider the function on K given by −1. By assumption we can nd a sequence of polynomials g(z) = (z − ζ) (gn)n∈N such that gn → g uniformly on K. By the maximum principle we have

sup |gn(z) − gm(z)| ≤ sup |gn(z) − gm(z)| ≤ sup |gn(z) − gm(z)| z∈U z∈∂U z∈K so gn converges uniformly in U to some limit G ∈ A(U). We notice that G(z)(z−ζ) = 1 on ∂U, so G(z)(z − ζ) = 1 on all of U. This gives a contradiction when z = ζ. Our conclusion is:

1 1 Introduction

• If every function f ∈ A(K) can be approximated by polynomials on K, then Kc is connected. To continue this discussion we need to dene a new set of functions.

Denition 1.0.3. Let E ⊂ C be an arbitrary set and f be a function on E. We say that f ∈ O(E) if there exists some open set Ω ⊂ C and a holomorphic function, g ∈ O(Ω) such that E ⊆ Ω and g|E = f. In 1885, Runge proved a considerably general result concerning the question stated above: Theorem 1.0.4 (Runge, Cor. 1.3.2 of [6]). Let K be a compact set. Every function in O(K) can be uniformly approximated by polynomials on K if and only if the set Kc is connected. Quite obviously we have O(K) ⊆ A(K). Therefore it is rather natural to wonder if Runge's theorem can be extended to the set A(K). The answer to that question is quite satisfyingly yes, and was proved by Mergelyan in 1952. It is startling that a period of sixty-seven years elapsed between the appearance of Runge's theorem and that of Mergelyan's theorem. Especially if one examines the large number of papers written on this subject during those years. The explanation is perhaps that people thought that Mergelyan's theorem was too good to be true. Theorem 1.0.5 (Mergelyan, Th. 12.2.1 of [4]). Let K be a compact set. Every function in A(K) can be uniformly approximated on K by polynomials if and only if the set Kc is connected.

Let K ⊂ C be compact and f : K → C be any function. For n ≥ 1, we dene

dn(f, K) := inf{kf − pkK ; p is a polynomial of degree at most n}.

By the preceding discussion we see that limn→∞ dn(f, K) = 0 if and only if f ∈ A(K). Quite simple result, but for f ∈ A(K) it can be of great interest to study the decay of the sequence because it helps us giving answer to an important (dn(f, K))n∈N question: What is the largest open set to which f can be extended holomorphically? To describe the equivalence between the decay of and the possible (dn(f, K))n∈N analytic extension of , the so called Green's function , for the set is f gK C \ K introduced, which will be discussed in Section 2.5. For the Green's function to be dened, K must be regular, which means that the Dirichlet problem has a solution on . For a real number we dene for C \ K R > 1 KR = {z ∈ C; gK (z) < log(R)} the following result.

Theorem 1.0.6 (Bernstein-Walsh, Th. BW1 in [8]). Let K ⊂ C be regular and f be a function on K. Let gK and KR be as above, then

1/n 1 lim sup(dn(f, K)) < n→∞ R if and only if f is the restriction to K of a function holomorphic on KR.

2 In Chapter 4 we will give a proof of a generalized version of the theorem of Bernstein-Walsh, considering compact sets K without assuming Kc to be connected. Of course, without assuming Kc to be connected we cannot hope to approximate functions in O(K) with polynomials. Therefore we will rather consider approximation by rational functions. We immediadely run into some complications, the most obvious one being: How should one dene a similar quantity to dn(f, K) for rational functions? We will dene the notation Rd(f, K, a, n), where a is a vector, describing the position of poles of the rational functions and n describes the maximum degree of each pole. We will have to dene a Green's function for each of the components of Kc, but another complication then arises. Previously we needed just one real number R to describe the holomorphic extension of f, but now we need many. It is simple enough to use a vector R rather than a real number, but how is one to relate those numbers to the decay of Rd(f, K, a, n)? Since Rd(f, K, a, n) doesn't even dene a sequence, as dn(f, K) does in the variable n, one will have to ask: what is the best way to measure the decay of Rd(f, K, a, n)? Those questions will be answered in Chapter 4. In the proof of the generalization we rely heavily on Hörmander's L2 theorem, which gives an estimate on the solution of the inhomogeneous ∂¯-equation. In Chapter 3 we will discuss and prove this theorem. Similar methods are considered in Claveras [2]. In light of the discussion it should be clear that Runge's theorem extends to an arbritrary compact K by considering approximation by rational functions rather than polynomials.

Theorem 1.0.7 (Runge, Th. 12.1.1 of [4]). Let K be a compact set and A be a set which contains at least one point from each bounded component of Kc. Then every function in O(K) can be uniformly approximated on K by rational functions with poles in A. Mergelyan's theorem only extends to a certain degree though.

Theorem 1.0.8 (Mergelyan, Th. 12.2.7 of [4]). Let K be a compact such that Kc has only nitely many components. Let A be a set which contains at least one point from every bounded component of Kc. Then every function in A(K) can be uniformly approximated on K by rational functions with poles in A.

Notice the extra condition in Mergelyan's case, that Kc may only have nitely many components. This extra condition can denitely not be abolished completely as shown by the following counter-example, sometimes called the Swiss-cheese example, by Alice Roth [13]. We denote the open disc with center z and radius r with ∆(z, r). Example. Consider the open unit disc ∆ = ∆(0, 1). Let ∆j = ∆(aj, rj) be a sequence of closed discs satisfying the following conditions:

1. They are pairwise disjoint.

3 1 Introduction

2. The union S ∆ is dense in ∆. j∈N j 3. P r < 1 j∈N j

Assuming for the moment such sequences (an)n∈ and (rn)n∈ exist, we let K = N N ∆ \ S ∆ . Notice that A(K) = C(K) since K has no interior. Consider the j∈N j measure µ on K given by Z Z X Z fdµ = f(z)dz − f(z)dz, f ∈ C(K). K ∂∆ ∂∆ j∈N j We see that if is a rational function with poles in then R since by R C \ K K Rdµ = 0 the residue theorem all the residues cancel out. Considering the function f(z) =z ¯, we get

Z Z X Z X X fdµ = zdz¯ − zdz¯ = 2πi − rj2πi = 2πi(1 − rj) 6= 0. K ∂∆ ∂∆ j∈N j j∈N j∈N Therefore f(z) =z ¯ can not be approximated by rational functions on K. But f ∈ A(K), so not every function in A(K) can be approximated by rational functions. Lemma 1.0.9. Sequences and satisfying 1.-3. in the example above (an)n∈N (rn)n∈N exist.

Proof. For z ∈ C and E ⊂ C, we denote by d(z, E) the distance between z and E. Let (s ) be any sequence of positive numbers such that P s < 1. Let n n∈N n∈N n K0 = K, and dene by induction Kn+1 = Kn \∆(an, rn) where an is equal to a point where the function c gains its maximum (considered as a function of ) and d(z, Kn) z c . Note that since is compact, exists. rn = min{sn, d(an,Kn)/2} Kn an It is obvious that (an)n∈ and (rn)n∈ satisfy 1. and 3. above. Suppose they do N N not satisfy 2. Then there exists an open disc ∆(b, ) ⊂ ∆ \ S ∆ . By the way j∈N j is chosen this means that for all , . Since ¯ is an |an − am| ≥ n, m ∈ N n 6= m ∆ compact this is a contradiction.

Even though the condition that Kc must have only nitely many components can't be discarded altogether, it can be lightened. There does in fact exist a compact set K such that the set Kc has innitely many components and every function in A(K) can be approximated by rational functions. In 1967 Vitushkin [15] gave necessary and sucient conditions on K for that to be the case. So far the conditions on K that we have considered have been purely topological. Vitushkin's conditions are on the other hand not topological but depend rather on a concept of analytic capacity. We will devote Chapter 5 to a discussion of Vitushkin's theorem. Finally, consider a closed set E ⊂ C, possibly unbounded, and assume that Ec is connected. Since non-constant polynomials are unbounded we cannot expext to be able to approximate every f ∈ A(E) uniformly on E by polynomials, therefore we rather consider approximation by entire functions.

4 Theorem 1.0.10 (Arakelian, [12]). Let E ⊂ C be a closed set satisfying the following condition: Ec is connected and for every closed disc ∆¯ ⊂ C the union of the bounded components of the set C \ (E ∪ ∆)¯ is a bounded set. Then every f ∈ A(E) can be approximated uniformly on E by entire functions. In Chapter 6 we use Mergelyan's theorem to give a generalization of this theorem for sets E for which Ec isn't required to be connected. For that to be possible we need to consider approximations by functions in O(C \ A), where A is a closed set containing at least one element from every bounded component of Ec.

2 Preliminaries

In this chapter we collect some basic facts necessary for the rest of the thesis. Most of the theorems are be stated without proof, but sometimes we give a short reason- ing instead. We start by looking at some fundemental results concerning complex analysis. More explicitly we discuss holomorphic functions, harmonic and subharmonic functions, the Dirichlet problem and Green's functions. We also discuss a few basic results from distribution theory and functional analysis.

2.1 Holomorphic functions

Throughout this discussion we identify the complex plane, C with R2 in the usual way. Let Ω be an open set in the complex plane and let f be a complex valued function in C1(Ω). If the real coordinates are denoted by x and y, then we set z = x + iy and z¯ = x − iy. Also we have z +z ¯ z − z¯ x = and y = . 2 2i We dene partial dierential operators in the following way

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

Now, the dierential of f can be expressed as a linear combination of dz and dz¯: ∂f ∂f ∂f ∂f df = dx + dy = dz + dz.¯ ∂x ∂y ∂z ∂z¯

Denition 2.1.1. A function f ∈ C1(Ω) is said to be holomorphic (analytic) if ∂f/∂z¯ = 0 in Ω, or equivalently if df is proportional to dz. If the function f is holomorphic we write f 0 rather then ∂f/∂z. We denote the set of all holomorphic functions on Ω by O(Ω).

Recall the classic Green-formula for functions dened in R2. Z Z ∂Q ∂P P dx + Qdy = − dx ∧ dy. ∂Ω Ω ∂x ∂y By applying it with P = f and Q = if we get the following result:

7 2 Preliminaries

Theorem 2.1.2 (Green's theorem). Let Ω be a bounded open set in C such that the boundary consists of a nite number of regular C1 curves. If f ∈ C1(U) for some open set U such that Ω ⊂ U, then Z Z ∂f Z ∂f fdz = 2i dx ∧ dy = − dz ∧ dz.¯ ∂Ω Ω ∂z¯ Ω ∂z¯ Z Corollary 2.1.3. If f ∈ C1(U) is holomorphic and Ω is as before, then fdz = 0. ∂Ω It is rather spectacular to see how one can manipulate Green's theorem for complex functions to derive countless statements in the theory of complex analysis, as we will see. Let f, Ω and U be as before. By xing some point ζ ∈ Ω and applying Theorem 2.1.2 to the function f(z)/(z − ζ), noting that the function 1/(z − ζ) is holomorphic on the set U \ ∆(ζ, /2), we get the following Z ∂f/∂z¯ Z f(z) Z f(z) Z 2π − dz∧dz¯ = dz = dz− if(ζ+eiθ)dθ. Ω\∆(ζ,) (z − ζ) ∂(Ω\∆(ζ,)) (z − ζ) ∂Ω (z − ζ) 0 Since (z − ζ)−1 is integrable over Ω and f is continuous at ζ we can let tend to zero to get the following result:

Theorem 2.1.4 (Cauchy-Pompeiu's formula, Th. 1.2.1 of [6]). If f ∈ C1(U), and Ω is as before, then 1 Z f(z) Z ∂f/∂z¯ f(ζ) = dz + dz ∧ dz¯ , ζ ∈ Ω. 2πi ∂Ω (z − ζ) Ω z − ζ Corollary 2.1.5. Let U and Ω be as before. 1 Z f(z) (i) If f ∈ O(U), then f(ζ) = dz for all ζ ∈ Ω. 2πi ∂Ω (z − ζ) 1 Z ∂f/∂z¯ (ii) If 1 , then for all . f ∈ Cc (Ω) f(ζ) = dz ∧ dz¯ ζ ∈ C 2πi Ω z − ζ Now, consider some function 1 and dene g ∈ Cc (C) 1 Z g(z) −1 Z g(ζ − z) u(ζ) := dz ∧ dz¯ = dz ∧ dz.¯ 2πi C (z − ζ) 2πi C z Since z−1 is integrable over every compact set, it is legitimate to dierentiate under the sign of integration to get

∂u −1 Z (∂g/∂ζ¯)(ζ − z) Z ∂g/∂z¯ ¯(ζ) = dz ∧ dz¯ = dz ∧ dz¯ = g(ζ) ∂ζ 2πi C z C z − ζ Hence u is the solution to the partial dierential equation ∂u/∂z¯ = g. A more general result is stated in the following theorem:

8 2.1 Holomorphic functions

Theorem 2.1.6 (Th. 1.2.2 of [6]). If µ is a complex measure with compact support in C, the integral Z dµ(z) u(ζ) = C z − ζ denes a holomorphic C∞ function outside the support of µ. If dµ = (2πi)−1gdz∧dz¯ for some k , , we have k and . g ∈ Cc (C) k ≥ 1 u ∈ C (C) ∂u/∂z¯ = g By applying Theorem 2.1.4 and Theorem 2.1.6 we get the following corollary:

Corollary 2.1.7 (Cor. 1.2.3 of [6]). Every u ∈ O(Ω) is in C∞(Ω). Hence u0 ∈ O(Ω) if u ∈ O(Ω).

It now makes sense to dene by induction the notation u(1) = u0 and u(n+1) = (u(n))0. Consider some function u ∈ O(Ω) and let K ⊂ Ω be compact. Choose some ∞ such that in some neighbourhood of . Since is holomorphic ψ ∈ Cc (Ω) ψ = 1 K u we notice that ∂(ψu)/∂z¯ = u∂ψ/∂z¯ and by Theorem 2.1.4, we get for every ζ ∈ K that 1 Z u(z)∂ψ/∂z¯ ψ(ζ)u(ζ) = dz ∧ dz.¯ 2πi Ω z − ζ Since ∂ψ/∂z¯ = 0 in K, the integrand is bounded so dierentiation under the integral sign is allowed, which yields the following result:

Theorem 2.1.8 (Th. 1.2.4 of [6]). For every compact set K ⊂ Ω and every open neighbourhood of there are constants , such that ω ⊂ Ω K Cj j ∈ N

(j) sup |u (z)| ≤ CjkukL1(ω), u ∈ O(Ω). z∈K Consider a sequence, , of holomorphic functions on , which is uniformly (un)n∈N Ω bounded on some compact set K ⊂ Ω. The preceding theorem implies that the sequence 0 is also uniformly bounded on . By applying Ascoli's theorem we (un)n∈N K come to the following conclusion:

Theorem 2.1.9 (Montel, Th. in Ch. 7.1.1 of [11]). If un ∈ O(Ω) and the sequence is uniformly bounded on every compact subset of , there is a subsequence |un| Ω unj converging uniformly on every compact subset of Ω to a limit u ∈ O(Ω). Finally, we state the Riemann mapping theorem, which is applied frequently in the thesis.

Theorem 2.1.10 (Riemann, Th. 6.4.2 of [4]). If U ⊂ C is open , simply connected and U 6= C, then there is an bijective holomorphic map from U to the open unit disc.

9 2 Preliminaries 2.2 Harmonic functions

We start with a formal denition.

Denition 2.2.1. Let U be an open subset of C. A function h : U → R is said to be harmonic if h ∈ C2(Ω) and

∂2 ∂2 ∆h := + h = 0. ∂x2 ∂y2 The following basic result allows us to construct numerous examples of harmonic functions. It also provides a useful tool for deriving their elementary properties from those of analytic functions.

Theorem 2.2.2 (Th. 1.1.2 of [10]). Let D be a domain in C. (i) If f is holomorphic on D and h = Re f, then h is harmonic on D.

(ii) If h is harmonic on D, and if D is simply connected, then h = Re f for some holomorphic f on D. Moreover f is unique up to adding a constant.

Notice that the condition that D should be simply connected is crucial to the second part of the theorem. To see that we can consider the function h(z) = log |z| which is clearly harmonic on the set C\{0}. It cannot be the real part of an analytic function g on C \{0} because if it were then | exp(g(z))/z| = 1 on C \{0} so by the maximum principle the analytic function exp(g(z))/z is equal to a constant C. That would mean that g0(z) = 1/z which in turn would give Z Z 1 0 = g0(z)dz = dz = 2πi. |z|=1 |z|=1 z However, since discs are simply connected, every harmonic function is locally the real part of an analytic function which leads to the following corollary.

Corollary 2.2.3 (Cor. 1.1.4 of [10]). If h is a harmonic function on an open subset U of C, then h ∈ C∞(U). A simple consequence of Theorem 2.2.2 and Cauchy's integral formula is the following theorem.

Theorem 2.2.4 (Mean-value property, Th. 1.1.6 of [10]). Let h be a function harmonic on an open neighbourhood of the disc ∆(z, ρ), then

1 Z 2π h(z) = h(z + ρeiθ)dθ. 2π 0 This section ends with two further ways in which harmonic functions behave like analytic ones, an identity principle and a maximum principle.

10 2.3 Subharmonic functions

Theorem 2.2.5 (Identity principle, Th. 1.1.7 of [10]). Let h and k be harmonic functions on a domain D in C. If h = k on a non-empty open subset U of D, then h = k on D. Theorem 2.2.6 (Maximum principle, Th. 1.1.8 of [10]). Let h be a harmonic function on a domain D in C. (i) If h attains a local maximum on D then h is constant. (ii) If h extends continuously to D and h ≤ 0 on ∂D, then h ≤ 0 on D.

2.3 Subharmonic functions

The analogue of the Laplacian for functions of one real variable is the dierential operator d2/dx2. Thus the analogue of harmonic functions on R is the null space of this operator, which consists of the ane functions. A convex function, f : R → R is a function whose graph satises the following property: If l is a line in the plane which cuts the graph of f in the points (a, f(a)) and (b, f(b)) (a < b) then the graph of f on the interval [a, b] lies below l. Subharmonic functions satisfy the complex analogue of this property. We start by dening semicontinuous functions. Denition 2.3.1. Let X be a topological space. We say that a function u : X → [−∞, ∞[ is upper semicontinuous if the set {x ∈ X : u(x) < α} is open in X for each α ∈ R. We say that u : X →] − ∞, ∞] is lower semicontinuous if −u is upper semicontinuous.

Let f ∈ C2(Ω), for some open set Ω ⊂ C. Then f is subharmonic if ∆f ≥ 0. This could seem like a suitable denition for subharmonicity, but it turns out that it is way to strict. One of the great virtues of subharmonic functions is their exibility, and this would be lost if we were to assume they were smooth.

Denition 2.3.2. Let U be an open subset of C. A function u : U → [−∞, ∞[ is called subharmonic if it satises the following properties: (i) It is upper semicontinuous.

(ii) For every z ∈ U and r > 0 such that ∆(z, r) ⊂ U, and for every continuous function h on ∆(z, r) which is harmonic on ∆(z, r) and satises h ≥ u on ∂∆(z, r), it holds that h ≥ u on ∆(z, r). Almost immediately from denition and the mean-value property of harmonic function we can derive the following theorem, which some textbooks prefer as the original denition: Theorem 2.3.3 (Submean inequality, special case of Th. 3.2.3(ii) of [7]). A function u : U → [−∞, ∞[ is subharmonic if and only if it is upper semicontinuous and for every z ∈ U and every r > 0 such that ∆(z, r) ⊂ U we have 1 Z 2π u(z) ≤ u(z + reit)dt. 2π 0

11 2 Preliminaries

Even though subharmonic functions need not be smooth, they can never the less be approximated by others which are.

Theorem 2.3.4 (Th. 2.7.2 of [10]). Let u : U → [−∞, ∞[ be subharmonic. There ∞ exist open sets U1 ⊂ U2 ⊂ U3... increasing to U and functions un ∈ C (Un) such that un is subharmonic on Un for all n and un & u. We gain numerous examples of subharmonic functions by noticing that for a holomorphic function f, the function log |f| is subharmonic. Further examples can be generated from the following result:

Theorem. 2.3.5 (Th. 3.2.2 and Cor. 2.4.3 of [7]). (i) If u, v are functions subharmonic on U and α, β ≥ 0, then αu + βv is subharmonic on U. (ii) If is a family of subharmonic functions on and the function (u)i∈I U supi∈I ui is upper semicontinuous on and takes values in , then is U [−∞, ∞[ supi∈I ui subharmonic on U. (iii) If is a conformal mapping between open subsets and of , f : U1 → U2 U1 U2 C and if u is subharmonic on U2, then u ◦ f is subharmonic on U1. Using (iii) in the preceding theorem, we can extend the denition of subharmonicity to the Riemann sphere. Given a function u dened on a neighbourhood U of ∞, we say u is subharmonic on U if u ◦ g−1 is subharmonic on g(U), where g is a conformal mapping of U onto an open subset of C. It can be seen that it does not matter which map g is chosen.

Theorem 2.3.6 (Maximum principle, Th. 2.3.1 of [10]). Let u be a subharmonic function on a domain D in C. (i) If u attains a global maximum in D, then u is constant. (ii) If is bounded and for all , then on . D lim supz→ζ u(z) ≤ 0 ζ ∈ ∂D u ≤ 0 D

2.4 The Dirichlet problem

Let U ⊂ C be an open set, U 6= C. Let f be a given continuous function on ∂U. Does there exist a continuous function u on U such that u = f on ∂U and u is harmonic on U? If u exists, is it unique? These two questions taken together are called the Dirichlet problem for the domain U. It has many motivations from physics, for instance the distribution of heat in a heat-conducting, thin lm in thermal equilibrium is known to be a harmonic function. Suppose one knew the temperature at the edge of such a lm, then by solving the Dirichlet problem one could calculate the heat at every point of that lm.

12 2.4 The Dirichlet problem

The uniqueness part of the question is easily answered. Suppose u1 and u2 were two functions solving the Dirichlet problem discussed above. Then u1 − u2 would be harmonic on U and equal to 0 on ∂U. By the maximum principle we get u1 = u2. We thus see that if such a function exists it is unique, it is much harder to determine whether such a function exists.

2.4.1 The Perron method The key idea of solving the Dirichlet problem will be enshrined in the following denition. This method of gaining the solution is often called the Perron method.

Denition 2.4.1. Let D be a proper subdomain of the Riemann sphere, Cb, and let be a bounded function. The associated Perron function φ : ∂D → R HDφ : D → R is dened by

HDφ(z) = sup u(z) z ∈ D u∈U Where U denotes the family of all subharmonic functions u on D which satisfy for each . lim supz→z0 u(z) ≤ φ(z0) z0 ∈ ∂D The motivation for this denition is that if the Dirichlet problem has a solution at all, then HDφ is the solution. Indeed, if h is such a solution then h ∈ U and so h ≤ HDφ. On the other hand, by the maximum principle, if u ∈ U, then u ≤ h on D, and so HDφ ≤ h. Therefore h = HDφ. The only question left to answer is: when does a solution exist? Consider for a moment the domain D = {z ∈ C : 0 < |z| < 1} and the continuous function φ : ∂D → R given by 0 if |z| = 1, φ(z) = 1 if |z| = 0. If as in the notation above, then by dening it can u ∈ U u(0) =: lim supz→0 u(z) be seen that u is subharmonic on all ∆ := ∆(0, 1). Hence the Perron function HDφ is be the same as the one if we were considering the Dirichlet problem on ∆ with constant boundary function 0. But the solution to that is the zero function hence

HDφ = 0, which is obviously does not solve the Dirichlet problem on D with boundary function φ because of the discontinuity at 0. We see that the Dirichlet problem for D = {z ∈ C : 0 < |z| < 1} is unsolvable. The reason for that is that the isolated boundary point 0 lacks sucient inuence on the subharmonic functions in U. Denition 2.4.2. Let be a proper subdomain of and let . A barrier at D Cb ζ0 ∈ ∂D ζ0 is a subharmonic function b defnied on D ∩ N where N is an open neighbourhood of ζ0 satisfying b < 0 on D ∩ N

13 2 Preliminaries and lim b(z) = 0. z→ζ0 A boundary point at which a barrier exists is called regular, otherwise irregular. If every ζ ∈ ∂D is regular, then D is called a regular domain. Now for the main result:

Theorem 2.4.3 (Cor. 4.1.8 of [10]). Let D be a regular domain, and let φ : ∂D → R be a continuous function. Then there exists a unique harmonic function h on D such that limz→ζ h(z) = φ(ζ) for all ζ ∈ ∂D. There is also a converse to the preceding theorem, which means that regularity is not only sucient to guarantee the existence of a solution but also necessary:

Theorem 2.4.4 (Discussed after Cor. 4.1.8 of [10]). Let D be a domain which is not regular. Then there exists a continuous function φ : ∂D → R such that no function h ∈ C(D) which is harmonic on D is equal to φ on the boundary.

2.4.2 Criteria for regularity Although the results of the previous section seem to solve the Dirichlet problem completely, it leaves a certain question unanswered. How can we tell if a boundary point of a set D is regular?

Theorem 2.4.5. If D is a simply connected domain such that Cb \ D contains at least two points, then D is a regular domain.

Proof. We need to show that every boundary point is regular. Given z0 ∈ ∂D, pick z1 ∈ ∂D \{z0}. Applying a conformal mapping of the sphere if necessery, we can assume that and . Then is a simply connected subdomain of z0 = 0 z1 = ∞ D C\{0} and therefore there exists a branch of the logarithm on D, which we call log. Put N = ∆(0, 1) and dene a function b on N ∩ D by

1 b(z) = Re . log(z)

Then clearly b satises the conditions to be a barrier at 0. Since regularity is clearly a local property, but not a global one, it is easy to generalize this theorem.

Corollary 2.4.6. Let be a subdomain of , let and be the component D Cb z0 ∈ ∂D C of ∂D which contains z0. If C 6= {z0} then z0 is a regular boundary point of D.

14 2.5 Green's functions 2.5 Green's functions

In essence, a Green's function for D is a family of fundamental solutions of the Lapla- cian. In Chapter 4 they become very useful in constructing holomorphic functions with a prescribed growth rate.

Denition 2.5.1. Let D be a proper regular subdomain of Cb. A Green's function for D is a map gD : D × D →] − ∞, ∞] such that for each w ∈ D:

(i) gD(·, w) is harmonic on D \{w} and bounded outside each neighbourhood of w.

(ii) limz→w gD(z, w) = ∞, more explicitly

log |z| + O(1) if w = ∞, g (z, w) = D − log |z − w| + O(1) if w 6= ∞.

(iii) limz→ζ gD(z, w) = 0 for every ζ ∈ ∂D. For example, a Green's function for the unit disc ∆ = ∆(0, 1) is

1 − zw¯ g∆(z, w) = log . z − w Having solved the Dirichlet problem, the construction of a Green's function becomes very easy. For a domain D and w ∈ D, w 6= ∞ we solve the Dirichlet problem on D with a boundary function φ : ∂D → R given by φ(z) = log |z − w|. Say the solution is h then we have gD(z, w) = h(z) − log |z − w|. If w = ∞ the Green's function is constructed in a similar way. From the construction and the uniqueness of the solution to the Dirichlet problem it should be clear that the Green's function for a given domain is unique. Therefore we usually talk about the Green's function with a denite article, rather than a Green's function. Some of the most basic properties of Green's functions are the following.

Theorem 2.5.2 (Theorems 4.4.3, 4.4.5 and 4.4.8 of [10]). Let D1,D2 be a regular domains in such that . Cb D1 ⊂ D2 (i) gD1 (z, w) > 0 z, w ∈ D1. (ii) gD1 (z, w) = gD1 (w, z) z, w ∈ D1. (iii) gD1 (z, w) ≤ gD2 (z, w) z, w ∈ D1.

15 2 Preliminaries 2.6 Distributions

Various diculties in the theory of partial dierential equations and Fourier analysis are some of the main reasons to extend the space of continuous functions and explore the theory of distributions.

2.6.1 Denitions

Let be an open set in n.A distribution in is a linear form on ∞ such X R u X Cc (X) that for every compact set K ⊂ X there exist constants C and k such that

X α ∞ |u(φ)| ≤ C sup |∂ φ|, φ ∈ Cc (K). |α|≤k

The set of distributions in X is denoted by D 0(X). If the same integer k can be used in the estimate above for every compact K we say that u is of order ≤ k and the set of such distributions is denoted by D 0k(X). The union

0 [ 0k DF (X) = D (X) k∈N is the space of distributions of nite order. Some people choose to dene distributions in a dierent but equivalent way. We state the equivalence of the two denitions as a theorem.

Theorem 2.6.1 (Th. 2.1.4 of [5]). A linear form on ∞ is a distribution if and u Cc only if when for every sequence ∞ converging to in u(φj) → 0 j → ∞ φj ∈ Cc (X) 0 the sense that there exists some compact K ⊂ X such that supp φj ⊂ K for every j α and ∂ φj → 0 uniformly for every α.

1 Now, for each f ∈ Lloc(X) we can dene a distribution in X with Z ∞ φ → f(x)φ(x)dx, φ ∈ Cc (X). X

1 It is easily seen that if f, g ∈ Lloc(X) give the same distribution in that way then 1 f and g are equal almost everywhere. Therefore we can identify the space Lloc(X) as a subspace of D 0(X), if it is considered as a space of equvialence classes, and we write f ∈ D 0(X) meaning Z f(φ) = f(x)φ(x)dx. X

1 If f ∈ Lloc(X) then f has order 0 since Z ∞ |f(φ)| ≤ |f(x)|dx · sup |φ|, φ ∈ Cc (X). K

16 2.6 Distributions

2.6.2 Dierential equations in the sense of distributions

If f is a continuous function on X such that ∂kf is dened everywhere on X and is continuous, integration by parts gives Z Z ∂kf(x)φ(x)dx = − f(x)∂kφ(x)dx. X X

In the sense of distributions, this means (∂kf)(φ) = −f(∂kφ). It is therefore reasonable to dene dierentiation on the set of distributions in a similar way. For u ∈ D 0(X) we dene

∞ (∂ku)(φ) := −u(∂kφ), φ ∈ Cc (X). For similar reasons we can dene the product of distributions and innitely dierentiable functions. If u ∈ D 0(X) and f ∈ C∞(X) we dene

∞ (fu)(φ) := u(fφ), φ ∈ Cc (X). It is not quite obvious that the classical rules of dierentiation will carry over to the space of distributions.

Theorem 2.6.2 (Discussed in Ch. 3.1 of [5]). Let u ∈ D 0(X) and f ∈ C∞(X). Then

∂k(∂ju) = ∂j(∂ku).

∂k(fu) = (∂kf)u + f(∂ku). Extending the denition of dierentiation to distributions in such a way can sometimes be useful when solving classical partial dierential equations.

Theorem 2.6.3 (Cor. 3.1.6 of [5]). If X ⊂ R is open, u ∈ D 0(X) and if

(m) (m−1) u + am−1u + ... + a0u = f,

∞ m where f ∈ C(X) and the coecients aj ∈ C (X), then u ∈ C (X) so the equation is fullled in the classical sense.

The function 1 is a fundemental solution of the ¯-operator, which means E(z) = πz ∂ that ∂E/∂z¯ = δ0 in the sense of distributions. The singular support of E, i.e. the complement of the open set where E is C∞, is the set singsupp(E) = {0}. The following theorem shows that every solution of ∂u/∂z¯ = f with f a C∞ function is in fact a C∞ function and a solution in the classical sense. Theorem 2.6.4 (Th. 4.4.1 of [5]). If P has a fundamental solution E, with singsupp(E) = {0} and X is any open set in Rn, then

singsupp(u) = singsupp(P u), u ∈ D 0(X).

17 2 Preliminaries 2.7 Results from functional analysis

Let X be some vector space over a scalar eld F. A functional f on X is a linear map from X to F. In other words, a scalar valued function on X such that

f(αx + βy) = αf(x) + βf(y), α, β ∈ F, x, y ∈ X. If the vector space X is normed it is easily seen that the following statements are equivalent for a functional f:

1. f is continuous.

2. f is continuous at 0.

3. f(B) is bounded, where B is the unit ball in X.

4. There exists a real number C ≥ 0 such that |f(x)| ≤ Ckxk for all x ∈ X.

A functional f is said to be bounded if it satises one of the preceding statements. In light of this it makes sense to dene a norm on the set of bounded functionals by

kfk := sup{|f(x)| ; kxk ≤ 1}.

We now state two fundamental results from functional analysis.

Theorem 2.7.1 (Hahn-Banach, Th. 8.9 of [9]). Let f0 be a bounded functional dened on a linear subspace M of a real or complex normed vector space X. Then f0 has an extension to a bounded functional f on X such that kfk = kf0k. Theorem 2.7.2 (Riesz, Th. 12.10 of [9]). Let H be an Hilbert space with inner product h·, ·i and let f be a bounded functional on H. Then there exists an a ∈ H such that f(x) = hx, ai for all x ∈ H, further kak = kfk.

18 3 Hörmander's L2 theorem

The goal of this chapter is to prove Hörmander's theorem for the solution of the ∂¯-equation with L2-estimates in the case of one complex variable. As an application we prove the Brunn-Minkowski inequality. The proof will be based on the one given in notes by Bo Berndtsson [1]. For the rest of the thesis we will let dλ denote the usual Lebesgue-measure on C.

3.1 Hörmander

Hörmander's theorem is as follows:

Theorem 3.1.1 (Hörmander's Theorem). Let Ω be a domain in C and φ ∈ C2(Ω) be a real function such that ∂2 . Let 2 , then there is a ∆φ = 4 ∂z∂z¯φ > 0 f ∈ Lloc(Ω) 2 function u ∈ Lloc(Ω) such that ∂u = f ∂z¯ in the sense of distributions and

Z Z |f|2 |u|2e−φdλ ≤ 4 e−φdλ. Ω Ω ∆φ The proof of this theorem needs some preparation. Since we are estimating the norm of considered as an element of the Hilbert space 2 2 u Lφ(Ω) = {g ∈ Lloc(Ω) : R 2 −φ it will be useful to know the adjoint of the -operator in this Ω |g| e dλ < ∞} ∂ space:

Lemma 3.1.2. The adjoint of the ∂¯-operator with respect to the inner product Z −φ hf, giφ = fge¯ dλ Ω is given by the equation ∂ ∂¯∗α = −eφ (e−φα). φ ∂z Proof. The set ∞ is dense in 2 . Therefore it is sucient to show that Cc (Ω) Lφ(Ω) ∂f hf, ∂¯∗αi = h , αi φ φ ∂z¯ φ

19 3 Hörmander's L2 theorem for every ∞ : α ∈ Cc (Ω) Z Z ∂ Z ∂ ¯∗ ¯∗ −φ −φ φ −φ −φ hf, ∂φαiφ = f∂φαe dλ = fe (−e (e α))dλ = − f (e α¯)dλ Ω Ω ∂z Ω ∂z¯ Z ∂ −φ ∂f = ( f)¯αe dλ = h , αiφ. Ω ∂z¯ ∂z¯

The following proposition is the key idea of the proof of Hörmander's theorem. It reduces the proof of an existence statement to the proof of an inequality.

Proposition 3.1.3. Let f and φ be as in Theorem 3.1.1. The following two statements are equivalent:

(i) There exists a function which satises the equation ∂u and the estimate u ∂z¯ = f Z |u|2e−φdλ ≤ C. Ω

(ii) The inequality Z Z −φ 2 ¯∗ 2 −φ | fαe¯ dλ| ≤ C |∂φα| e dλ Ω Ω holds true for all 2 α ∈ Cc (Ω). Proof. Let's begin assuming that (i) is true. Interpreted in the sense of distributions the equation ∂u implies ∂z¯ = f Z ∂α Z 2 (1) − u dλ = fαdλ, α ∈ Cc (Ω). Ω ∂z¯ Ω If we interchange the function α for the function αe¯ −φ we get Z Z ¯∗ −φ −φ 2 (2) u∂φαe dλ = fαe¯ dλ, α ∈ Cc (Ω). Ω Ω Now by combining equations of statement (i) and equation (2) and applying the Cauchy-Schwarz inequality, we get Z Z Z Z −φ 2 ¯∗ −φ 2 2 −φ ¯∗ 2 −φ | fαe¯ dλ| = | u∂φαe dλ| ≤ ( |u| e dλ)( |∂φα| e dλ) Ω Ω Ω Ω Z ¯∗ 2 −φ ≤ C |∂φα| e dλ Ω for all 2 . Hence is true. α ∈ Cc (Ω) (ii) Now, let's assume that (ii) is true. Let

¯∗ 2 E := {∂φα; α ∈ Cc (Ω)},

20 3.1 Hörmander and consider E as a subspace of Z 2 2 2 −φ Lφ(Ω) = {g ∈ Lloc(Ω); |g| e dλ < ∞}. Ω We dene an antilinear functional on E by Z ¯∗ −φ H(∂φα) = fαe¯ dλ. Ω Notice that the operator ¯∗ is one-to-one on the space of 2-functions with compact ∂φ C support so H is well dened. Also, according to assumption, the functional H is continuous and of norm not exceeding C. The Hahn-Banach extension theorem now guarantees an antilinear extension of to 2 with the same norm. The Riesz H Lφ(Ω) representation theorem then implies that there is some element 2 with u ∈ Lφ(Ω) norm not exceeding C such that Z H(g) = uge¯ −φdλ Ω for all 2 . If we set ¯∗ then g ∈ Lφ(Ω) g = ∂φα Z Z −φ ¯∗ −φ ue ∂φαdλ = fαe¯ dλ, Ω Ω which according to equation implies that ∂u in the sense of distributions. (2) ∂z¯ = f

1 Corollary 3.1.4. Let µ > 0 be in Lloc(Ω). The following two statements are equivalent.

(i) Statement (ii) of Proposition 3.1.3 is fullled for every f which satises the estimate Z |f|2 e−φdλ ≤ C. Ω µ (ii) The inequality Z Z 2 −φ ¯∗ 2 −φ µ|α| e dλ ≤ |∂φα| e dλ Ω Ω holds true for all 2 α ∈ Cc (Ω).

Proof. Assume (ii) is true. Then the Cauchy-Schwarz inequality gives

Z Z Z 2 Z Z −φ 2 f √ −φ 2 |f| −φ 2 −φ ¯∗ 2 −φ | fαe¯ dλ| = | √ µαe dλ| ≤ e dλ µ|α| e dλ ≤ C |∂φα| e dλ, Ω Ω µ Ω µ Ω Ω hence (i) holds.

21 3 Hörmander's L2 theorem

Now assume (i) is true. For any function h ∈ L2 (Ω) with norm less than C we √ φ can set f = h µ and then we get by assumption that Z Z √ −φ 2 ¯∗ 2 −φ | hα¯ µe dλ| ≤ C |∂φα| e dλ. Ω Ω By the converse Hölder's inequality, we get (ii). To complete the proof of Hörmander's theorem it is enough to prove an inequality as in statement (ii) of Proposition 3.1.3. This will be accomplished by the following integral identity.

Proposition 3.1.5. Let be a domain in , 2 and 2 , then Ω C φ ∈ C (Ω) α ∈ Cc (Ω) Z Z 2 Z ∆φ 2 −φ ∂α −φ ¯∗ 2 −φ |α| e dλ + e dλ = |∂φα| e dλ. Ω 4 Ω ∂z¯ Ω Proof. Applying Lemma 3.1.2, we get

Z Z Z ∂ ¯∗ 2 −φ ¯∗ ¯∗ −φ ¯∗ −φ |∂φα| e dλ = (∂φα)(∂φα)e dλ = ( ∂φα)¯αe dλ. Ω Ω Ω ∂z¯ also ∂ ∂φ ∂α ∂α ∂φ ∂¯∗α = −eφ (e−φα) = −eφ(− e−φα + e−φ ) = − + α. φ ∂z ∂z ∂z ∂z ∂z so ∂ ∆α ∂φ ∂α ∆φ ∂α ∆φ ∂¯∗α = − + · + α = ∂¯∗ + α. ∂z¯ φ 4 ∂z ∂z¯ 4 φ ∂z¯ 4 Therefore we get Z Z Z Z ¯∗ 2 −φ ∂ ¯∗ −φ ¯∗ ∂α −φ ∆φ 2 −φ |∂φα| e dλ = ( ∂φα)¯αe dλ = ∂φ αe¯ dλ + |α| e dλ Ω Ω ∂z¯ Ω ∂z¯ Ω 4 Z ∂α ∂α Z ∆φ = e−φdλ + |α|2e−φdλ Ω ∂z¯ ∂z¯ Ω 4 Z 2 Z ∂α −φ ∆φ 2 −φ = e dλ + |α| e dλ. Ω ∂z¯ Ω 4

From the last two propositions the proof of Hörmander's theorem follows almost immediately. Proof (Hörmander's Theorem, 3.1.1). From Proposition 3.1.5 we get Z Z ∆φ 2 −φ ¯∗ 2 −φ 2 |α| e dλ ≤ |∂φα| e dλ, α ∈ Cc (Ω). Ω 4 Ω

22 3.1 Hörmander

According to Corollary 3.1.4 Z Z −φ 2 ¯∗ 2 −φ | fαe¯ dλ| ≤ C |∂φα| e dλ. Ω Ω

Where R |f|2 −φ From Corollary 3.1.3 we see that there exists a function C = 4 Ω ∆φ e dλ. such that ∂ and u ∂z¯u = f Z Z |f|2 |u|2e−φdλ ≤ 4 e−φdλ. Ω Ω ∆φ

Corollary 3.1.6. Suppose φ ∈ C2(Ω) satises ∆φ > 0. Let u be a C1 function in a domain such that Ω Z uhe¯ −φdλ = 0 Ω for every holomorphic function in 2 . Then h Lφ(Ω)

Z Z |∂u/∂z|2 |u|2e−φdλ ≤ 4 e−φdλ. (3) Ω Ω ∆φ Proof. Hörmander's theorem states that the equation

∂v ∂u = ∂z ∂z has some solution v satisfying (3) with u replaced by v. But, the condition that u is orthogonal to all holomorphic functions means that u is the solution to this equation which has minimal norm in 2 . Hence satises the estimate as well which is Lφ(Ω) u what the corollary claims.

Now, lets consider the weight function φ in Hörmander's theorem for a moment. We notice that the Laplacian of the weight function has to maintain a strict inequality on all of Ω. That can be very inconvenient since we might want to use some subharmonic function as a weight function for which the Laplacian is denitely not guaranteed to be strictly positive. Therefore we shall state and prove a second version of Hörmander's theorem:

Theorem 3.1.7 (Hörmander's Theorem (Second Version)). Let 2 . If f ∈ Lloc(Ω) 2 2 a > 0 and φ ∈ C (Ω) is subharmonic, then there is a function u ∈ Lloc(Ω) such that ∂u in the sense of distributions and ∂z = f Z Z a |u(z)|2e−φ(z)(1 + |z|2)−adλ(z) ≤ |f(z)|2e−φ(z)(1 + |z|2)2−adλ(z). (4) Ω Ω

23 3 Hörmander's L2 theorem

Proof. Let ψ(z) = φ(z) + a log(1 + |z|2), let r = |z| and get ∂ ∂ 4a ∆ψ(z) ≥ a∆ log(1 + |z|2) = ar−1 (r log(1 + r2)) = > 0. ∂r ∂r (1 + r2)2 Hörmanders Theorem therefore applies to so there exists some such that ∂u ψ u ∂z = f and Z Z |f|2 |u|2e−ψdλ ≤ 4 e−ψdλ. Ω Ω ∆ψ Now we have e−ψ = e−φ(1 + |z|2)−a and 4e−ψ 4e−φ(1 + |z|2)−a e−φ(1 + |z|2)2−a ≤ = , ∆ψ 4a/(1 + |z|2)2 a which implies (4). Still there are some inconvenient restrictions for the weight function φ, it is required to be in C2(Ω). In fact we can do better: Theorem 3.1.8. The preceding theorem holds true for any subharmonic weight function φ which is not constantly equal to −∞ in any component of Ω.

Proof. According to Theorem 2.3.4 there are open sets, Y1 ⊂ Y2 ⊂ Y3 ⊂ ... increasing ∞ to Ω and a sequence of subharmonic functions φj ∈ C (Yj) with φj+1 ≤ φj in Yj for all j and limj→∞ φj = φ in X. According to Theorem 3.1.7 there exists a function , such that ∂uj in and uj ∂z = f Yj Z Z 2 −φj (z) 2 −a 2 −φ(z) 2 2−a a |uj(z)| e (1 + |z| ) dλ(z) ≤ |f(z)| e (1 + |z| ) dλ(z) (5) Yj Ω where on the right-hand side we have extended the domain of integration and in- creased the weight. We see that the sequence (un)n≥j is bounded in Yj for every , if it is considered as a sequence in the space 2 . Since the unit ball is com- j Lφ(Yj) pact in the weak topology, we can therefore nd a subsequence which is weakly unk convergent in 2 for every to some limit . Since dierential operators are Lφ(Yj) j u continuous in the weak topology of D 0(X), we have ∂u/∂z¯ = f. Also, by monotone convergence, letting j → ∞ we see that equation (4) holds true with uj replaced with u and Yj replaced with X.

3.2 An application: Brunn-Minkowski type inequalities

In this section we give an application of Corollary 3.1.6. It will be applied in a proof of a functional form of the Brunn-Minkowski inequality, which can in fact be considered a generalization of it.

24 3.2 An application: Brunn-Minkowski type inequalities

Proposition 3.2.1. Let φ ∈ C2(R) be a convex function with φ00 > 0. Let u ∈ C1(R) be a function such that Z u(x)e−φ(x)dx = 0. R Then Z Z 0 2 −φ(x) |u (x)| |u(x)| e dx ≤ 00 dx. R R φ (x) This statement is very similar to the one of Corollary 3.1.6. In a sense just a real version of it. We have replaced subharmonic φ by a convex φ and we require u to be orthogonal to constants, which is the kernel of the regular dierential operator, rather than to be orthogonal to the space of holomorphic functions, which is the kernel of the ∂/∂z¯ operator. We omit a formal proof which is very similar to the proof of the complex version. A generalized version of the Brunn-Minkowski theorem is as follows:

Theorem 3.2.2. Let φ(t, x) ∈ C2(Rm × Rn) be a convex function, (this is to be interpreted as t ∈ Rm and x ∈ Rn), and assume that

m n ∆xφ(t, x) > 0 t ∈ R , x ∈ R , where is the Laplacian of when considered as a function of n with ∆xφ φ x ∈ R t constant. Dene a function φ˜ on Rm by Z −φ(t,x) m φ˜(t) = − log e dx , t ∈ R . Rn Then φ˜ is a convex function. Proof. We start by making some reductions. By Fubini's theorem we may assume that n = 1, then the case n > 1 follows by a simple induction argument. Since convexity means convexity on any line, we may also assume that m = 1. In the following argument we write φt := ∂φ/∂t and φx := ∂φ/∂x. We have Z φ˜(t) = − log e−φ(t,x)dx . R Dierentiating with respect to t we get

R φ (t, x)e−φ(t,x)dx φ˜0(t) = t , R e−φ(t,x)dx and dierentiating once more gives

R (φ − (φ )2)e−φdx R e−φdx + R φ e−φdx2 φ˜00 = tt t t . R e−φdx2

25 3 Hörmander's L2 theorem

We simplify:

R (φ − (φ )2)e−φdx R φ e−φdx2 R (φ − (φ )2)e−φdx φ˜00 = tt t + t = tt t + (φ˜0)2 R e−φdx R e−φdx R e−φdx R (φ − (φ )2)e−φdx (−(φ˜0)2 + 2(φ˜0)2) R e−φdx = tt t + R e−φdx R e−φdx R (φ − (φ )2)e−φdx − R (φ˜0)2e−φdx + 2φ˜0 R φ e−φdx = tt t t R e−φdx R (φ − (φ )2 − (φ˜0)2 + 2φ˜0φ )e−φdx R (φ − (φ − φ˜0)2)e−φdx = tt t t = tt t . R e−φdx R e−φdx

˜0 Now, dene the function u(t, x) := φt(t, x) − φ (t). We notice that Z −φ(t,x) u(t, x)e dx = 0 t ∈ R.

Since by assumption φxx > 0 we can use Proposition 3.2.1. Note that ux = φtx. Z Z 2 ˜0 2 −φ(t,x) (φtx(t, x)) (φt(t, x) − φ (t)) e dx ≤ dx. φxx(t, x) Thus we get

R (φ − (φ − φ˜0)2)e−φdx R (φ − (φ )2/φ )e−φdx φ˜00 = tt t ≥ tt tx xx . R e−φdx R e−φdx Since the function φ is convex, the determinant of its Hessian matrix is non-negative and we have 2 2 (φtx) φttφxx − (φtx) φtt − = ≥ 0, φxx φxx and thus φ˜00 ≥ 0.

The preceding theorem is a functional form of the Brunn-Minkowski inequality, which can be stated as follows.

Theorem 3.2.3. Let be a convex open set in m n. For any m, let D R × R t ∈ R Dt be a slice of D n Dt := {x ∈ R :(t, x) ∈ D}.

Let |Dt| be the Lebesgue measure of Dt. Then the function

t −→ − log |Dt| is convex.

26 3.2 An application: Brunn-Minkowski type inequalities

Proof. Take φ to be the function that equals 0 in D, and ∞ outside D. This function can be written as an increasing limit of smooth convex functions which satisfy the criteria in the theorem above. Therefore the function Z ˜ −φ(t,x) φ(t) = − log e dx = − log |Dt| Rn is convex. There is another, perhaps more common, way of stating the Brunn-Minkowski inequality:

Theorem 3.2.4 (Brunn-Minkowski). Let and be open convex sets in n. D0 D1 R For any t ∈ [0, 1] we have

1 1 1 |tD1 + (1 − t)D0| n ≥ t|D1| n + (1 − t)|D0| n .

Proof. We dene a set D ∈ R × Rn with

D := {(t, x); x ∈ tD1 + (1 − t)D0, t ∈ [0, 1]}. Then

Dt = tD1 + (1 − t)D0, t ∈ [0, 1], where as before, were the slices of given by n Dt D Dt := {x ∈ R :(t, x) ∈ D}. It can be seen that D is convex. Theorem 3.2.3 then implies that |Dt| ≥ min{|D0|, |D1|}. To see that assume |Dt| < min{|D0|, |D1|}. Then − log |Dt| > max{− log |D0|, − log |D1|} which is a contradiction because − log |Dt| is a convex function of t. Now, let A and B be convex sets in Rn and t ∈]0, 1[.

1 1 1 1 tA (1 − t)B n |A| n |B| n |A + B| n = + ≥ min{ , }. t 1 − t t (1 − t) Choosing 1 |A| n t = 1 1 |A| n + |B| n we get 1 1 1 |A + B| n ≥ |A| n + |B| n . By substituting B and A we obtain D0 := 1−t D1 := t

1 1 1 |tD1 + (1 − t)D0| n ≥ t|D1| n + (1 − t)|D0| n , which is the desired result.

4 A generalized version of the Bernstein-Walsh theorem

From now on, for a given set E ⊂ C, any bounded component of the set Ec will be refered to as a hole of E. The polynomial hull of a compact set K is the set

Kˆ := {z ∈ C; |f(z)| ≤ sup |f(z)| for all f ∈ O(C)} z∈K and it is known that Kˆ is the union of K and all its holes. Runge's theorem states that for a compact K ⊂ C, a function f ∈ O(K), and > 0 there exists a rational function R = p/q with poles outside of K such that kf −RkK < . In practice, it could be benecial to know the least possible degree of the polynomials p and q for such an estimate to be possible. The goal of this chapter is to estimate this least possible degree of p and q with respect to and prove a theorem similar to the Bernstein-Walsh theorem. We start by some denitions.

4.1 Denitions

Let m and m+1. We let a = (a1, a2, ..., am) ∈ C n = (n0, n1, n2, ..., nm) ∈ N R(a, n) denote the set of all rational functions p which satisfy the following conditions: r = q

• If a is a zero of the polynomial q, then a = ak for some k ∈ {1, 2, ..., m}.

• If ak is a zero of q its multiplicity is at most nk. • The degree of the polynomial p is at most

m X m + 2 + nk. k=0

Now let be a compact subset of and assume that for all . K C ak 6∈ K k ∈ {1, 2, ..., m} Then we set

Rd(f, K, a, n) = inf{kf − rkK : r ∈ R(a, n)}. The goal of this chapter is to estimate Rd(f, K, a, n) as well as possible. Giving precise information on the degree of each pole of the optimal approximating rational function. Now suppose that K ⊂ C is compact and f ∈ O(K). Let U be a neighbourhood of K on which f is holomorphic. The unbounded component of Kc will be denoted

29 4 A generalized version of the Bernstein-Walsh theorem

by Ω0 and the holes of K not coverd by U will be denoted by Ω1, Ω2,...,Ωm. There are only nitely many such holes, as stated in the following lemma.

Lemma 4.1.1. Let K be a compact set and U be a neighbourhood of K. Let (Ωi)i∈I be the family of holes in K. Then Ωi ⊆ U for all but nitely many i ∈ I. ˆ Proof. Note that since K is compact, K is as well. The family (Ωi)i∈I along with the open set U is an open cover of Kˆ . Since Kˆ is compact there is a nite subcover. Continuing our discussion, we assume that the vector a contains precisely one point from each of the holes Ω1,...,Ωm. If a didn't contain a point from some hole then we could not expect to approximate f with functions from ∪nR(a, n). On the other hand, if some hole of K would contain more than one point of a one could just pick one and omit the others. By redening the holes we can assume that ak ∈ Ωk for all k ∈ {1, 2, ..., m}. We assume that K is regular, since if it was not we could nd a slightly bigger K0 which is regular and K ⊂ K0 ⊂ U. We dene if gΩ0 (z, ∞) z ∈ Ω0, G0(z) := if 0 z ∈ C \ Ω0 and for k ∈ {1, 2, ..., m} if gΩk (z, ak) z ∈ Ωk, Gk(z) := if 0 z ∈ C \ Ωk.

Where gD is the Green's function for the regular domain D. Notice that by the denition of the Green's function, for all . Gk ∈ C(C \{ak}) k At last we dene level sets of the set K, with respect to those Green's functions with the following notation. For a vector m+1 we set r = (r0, r1, r2, ..., rm) ∈ R for Kr = {z ∈ C : Gk(z) ≤ log(rk) k ∈ {0, 1, ..., m}}.

We see that if rk < 1 for some k then Kr = ∅ because the Green's functions are positive. We let m+1 be a vector such that for R = (R0,R1,R2, ...Rm) ∈ R Rk > 1 all k and f ∈ A(KR). Since f is holomorphic on a whole neighbourhood of K we know such an R exists.

4.2 The estimation procedure

In this section we use Hörmander's L2-estimates to construct a holomorphic function F ∈ R(a, n). Later we see that this function estimates the function f properly.

30 4.2 The estimation procedure

4.2.1 The construction of F First we state a lemma:

Lemma 4.2.1. The function as constructed above is subharmonic on . The G0 C function as constructed above is subharmonic on for . Gk C \{ak} k ∈ {1, 2, ..., m}

Proof. Let k 6= 0. By denition Gk is harmonic on Ωk \{ak}, it is also harmonic on since there it is constant. In particular it is subharmonic on those sets. We C \ Ωk are only left to show that it is subharmonic on ∂Ωk. For any z0 ∈ ∂Ωk and > 0 we have Z 2π 1 iθ Gk(z0 + e )dθ ≥ 0 = Gk(z0). 2π 0

Therefore, by the sub-mean inequality theorem, Gk is subharmonic in z0. The proof for G0 is similar. Now to the actual construction. For all k ∈ {0, 1, 2, ..., m} we choose some and dene . Let 1 be functions which satisfy rk ∈]1,Rk[ r = (r0, r1, ..., rm) βk ∈ C (R) the following

for all • 0 ≤ βk(x) ≤ 1 x ∈ R.

• βk(x) = 0 if x ≥ log(Rk).

• βk(x) = 1 if x ≤ log(rk). • 0 3 3 kβkkR ≤ = . 2(log(Rk) − log(rk)) 2 log(Rk/rk)

We set χk(z) := βk(Gk(z)). Then we have:

• 0 ≤ χk ≤ 1,

• χk(z) = 1 if Gk(z) ≤ log(rk),

• χk(z) = 0 if Gk(z) ≥ log(Rk).

For any z such that log(rk) ≤ Gk(z) ≤ log(Rk) we then get

∂ 0 ∂ 3Mk | χk(z)| = |βk(G(z))|| Gk(z)| ≤ . ∂z¯ ∂z¯ 2 log(Rk/rk)

Where we dene Mk by ∂ Mk := sup | Gk(z)|. log(rk)≤Gk(z)≤log(Rk) ∂z¯

31 4 A generalized version of the Bernstein-Walsh theorem

Finally we dene the global cuto function with

1 if z ∈ K, χ(z) := χk(z) if z ∈ Ωk. Take particular notice that χ ∈ C1(C). This cuto function is designed such that the product fχ is dened and dierentiable on all C, holomorphic in a neighbourhood of K and has a compact support. Now dene

m X G := 2nkGk. k=0 Since each of the is subharmonic on , we see that is subharmonic on Gk C \{ak} G . Therefore we can use it as a weight function in Theorem 3.1.8. C \{a1, a2, ..., am} By Hörmander's theorem we can nd a function 2 such u ∈ Lloc(C \{a1, ..., am}) that ∂ ∂ ∂ u = (fχ) = f χ (1) ∂z¯ ∂z¯ ∂z¯ in the sense of distributions and Z Z ∂ |u(z)|2(1+|z|2)−2e−G(z)dλ(z) ≤ |f(z) χ(z)|2e−G(z)dλ(z). ∂z¯ C\{a1,a2,...,am} C\{a1,a2,...,am} By Theorem 2.6.4 we see that u ∈ C∞ and (1) holds in the classical sense. Since the set has measure zero, we can consider to be the domain of {a1, a2, ..., am} C integration, instead of in the following calculations. We now C \{a1, a2, ..., am} dene our approximation function with

F := fχ − u.

Then is holomorphic on since by the denition of F C \{a1, a2, ..., am} u ∂ ∂ ∂ ∂ F = (fχ − u) = (fχ) − (fχ) = 0. ∂z¯ ∂z¯ ∂z¯ ∂z¯ Next we are going to show that F is in fact a member of the set R(a, n), and after that we show that F approximates f properly on K.

4.2.2 Showing that F ∈ R(a, n) Notice that since the function has compact support ∂ does as well. There- fχ ∂z¯(fχ) fore the function u is holomorphic near ∞, and we can use the mean value theorem for holomorphic functions. Now we estimate the growth rate of F at ∞ and each of the points of . We start at . For a large enough we get {a1, a2, ...am} ∞ z ∈ C 1 Z |F (z)| = |u(z)| ≤ |u(ζ)|dλ(ζ). π ∆(z,1)

32 4.2 The estimation procedure

Using the inequality of Cauchy-Schwarz and estimating one of the integrals by integrating over C rather than ∆(z, 1) we get

1 1 1 Z 2 Z 2 |F (z)| ≤ (1 + |ζ|2)2eG(ζ)dλ(ζ) |u(ζ)|2(1 + |ζ|2)−2e−G(ζ)dλ(ζ) π ∆(z,1) C Notice that the second integral in the product is independent of the variable z. In the next subsection we see that it is actually nite. We can therefore estimate it by some constant . Since is large is equal to . We C1 z G(z) 2n0G0(z) = 2n0gΩ0 (z, ∞) get the following

1 1 ! 2 Z 2 C1 2 2 2n g (ζ,∞) C1 2 2 2n g (ζ,∞) |F (z)| ≤ (1 + |ζ| ) e 0 Ω0 dλ(ζ) ≤ π sup (1 + |ζ| ) e 0 Ω0 . π ∆(z,1) π ζ∈∆(z,1) By denition of the Green's function we have as gΩ0 (z, ∞) = log |z| + O(1) z → ∞.

In other words, there is a constant C2 such that for large enough z we get

|gΩ0 (z, ∞)| ≤ log |z| + C2. Therefore we have

1 C1 C1 |F (z)| ≤ (1 + |z + 1|2)2e2n0(log |z+1|+C2) 2 = eC2n0 (1 + |z + 1|2)|z + 1|n0 . π1/2 π1/2

This last function has growth rate equal to a polynomial of degree n0 + 2. In other words, we have shown that there is a constant C such that

|F (z)| ≤ C|z|n0+2, for z large enough. That is F (z) = O(zn0+2) as z → ∞. The estimation for the growth rate at each of the points in {a1, a2, ..., am} is similar. For z close enough to ak, F = u on ∆(z, |ak − z|/2), also u is holomorphic there. Let |ak−z| and get = 2 (i) 4|a − z|−2 Z |F (z)| = |u(z)| ≤ k |u(ζ)|dλ(ζ) π ∆(z,) 1 (ii) Z 2 −2 2 2 G(ζ) ≤ C1|ak − z| (1 + |ζ| ) e dλ(ζ) ∆(z,) 1 ! 2 (iii) |a − z|2 −2 k 2 2 −2nkGk(ζ) ≤ C1|ak − z| sup (1 + |ζ| ) e 4 ζ∈B(z,) (iv) |ak−z| 1 −1 −2nk(log( )+C2) ≤ C3|ak − z| (e 2 ) 2 −nk−1 ≤ C|ak − z| .

33 4 A generalized version of the Bernstein-Walsh theorem

In (i) we used the mean value theorem, in (ii) we used Cauchy-Schwarz and estimated one integral by a constant as before, in (iii) we estimated the integral by it's supremum and in (iv) we used the denition of the Green's function. This implies −ni−1 F (z) = O(|z − ak| ) as z → ak.

In other words, F has a pole of degree at most nk + 1 at ak. Therefore there is function P holomorphic in C such that

−(n1+1) −(n2+1) −(nk+1) F (z) = P (z)(z − a1) (z − a2) ··· (z − am) .

Now, since the growth rate of F at innity is O(zn0+2) the growth rate of P at P innity is O(zm+2+ nk ). Therefore by Liouville's theorem, P is a polynomial of P degree, at most m + 2 + nk and so F is a member of the set R(a, n). The only thing left to show now is that F approximates f properly on K.

4.2.3 Estimation of kF − fkK Let be such that and . r ∈ R 1 < r < min{r0, r1, ..., rm} n = max{n0, n1, ..., nm} Let s be a constant such that Gk(z) < log(r) for all z ∈ K + ∆(0, s) and all k ∈ {0, 1, .., m}. Then for any z ∈ K, u is holomorphic on ∆(z, s), because there we have χ = 1 and so we get ∂u/∂z¯ = ∂(fχ)/∂z¯ = ∂f/∂z¯ = 0. Therefore we can use the mean value theorem on u and so we have 1 Z |F (z) − f(z)| = |u(z)| ≤ 2 |u(ζ)|dλ. s π ∆(z,s)

Using Cauchy-Schwarz, and then integrating over C rather than ∆(z, s) in the second integral of the product we have

1 Z 1 Z 1 2 2 G(ζ) 2 2 2 −2 −G(ζ) 2 |F (z)−f(z)| ≤ 2 ( (1+|ζ| ) e dλ(ζ)) ( |u(ζ)| (1+|ζ| ) e dλ(ζ)) . s π ∆(z,s) C Now, u was constructed by Hörmander's theorem so that the second integral of this product could be estimated. The rst integral of the product is estimated by the supremum of the integrand,

1 ! 2 1 1 Z ∂χ 2 |F (z)−f(z)| ≤ s2π sup (1 + |ζ|2)2eG(ζ) |f(ζ) (ζ)|2e−G(ζ)dλ(ζ) . 2 ¯ s π ζ∈∆(z,s) C ∂ζ

By the choice of s and the denition of G, the supremum of eG in the rst factor can be estimated by r2n. The integral in the second factor can be taken over the support of ∂χ/∂z¯ rather than C.

1 rn Z ∂χ 2 |F (z) − f(z)| ≤ sup (1 + |ζ|2) |f(ζ) (ζ)|2e−G(ζ)dλ(ζ) . 1 ¯ sπ 2 ζ∈K+∆(0,s) supp(∂χ/∂ζ¯) ∂ζ

34 4.2 The estimation procedure

Now, by the denition of χ we see that the sets supp(∂χk/∂z¯) form a partion of supp(∂χ/∂z¯). Therefore we can write the integral as a sum of smaller integrals

1 m ! 2 rn Z ∂χ 2 X k 2 −2nkGk(ζ) |F (z)−f(z)| ≤ 1 sup (1+|ζ| ) |f(ζ) (ζ)| e dλ(ζ) . 2 supp ¯ ∂ζ¯ sπ ζ∈K+∆(0,s) k=0 (∂χk/∂ζ)

We have also changed G to 2nkGk because supp(∂χk/∂z¯) ⊂ Ωk and G = 2nkGk on Ωk. By dention of χk we have that Gk(z) ≤ log(rk) on supp(∂χk/∂z¯), hence we can −2n estimate −2nkGk in the integral by k . Above we have already estimated that e rk

2 ∂χk 3Mk ≤ . ∂z¯ 2 C 2(log (Rk/rk))

Where Mk was dened by ∂ Mk = sup | Gk(z)|. log(rk)

1 n m 2 Z ! 2 r 2 X 3Mk −2nk 2 |F (z)−f(z)| ≤ 1 sup (1+|ζ| ) rk |f(ζ)| dλ(ζ) . 2 2(log(Rk/rk)) sπ ζ∈K+∆(0,s) k=0 KR

Here we have also estimated the integral of |f|2 over supp(∂χ/∂z¯) by integrating over a larger set KR. Simplifying we get

1 m 2 2n ! 2 kfkL ,K X 3Mk r |F (z) − f(z)| = 2 R sup (1 + |ζ|2) . 1 2nk 2 2(log(Rk/rk)) sπ ζ∈K+∆(0,s) k=0 rk

Now, remembering that the real numbers rk were chosen arbitrarily from the set ]1,Rk[, we could choose them to minimize this expression. By dierentiating

−nk rk log(Rk/rk)

−nk with respect to rk we nd that the minimal value is at rk = Rke . Putting those values in the above expression gives us the estimate

1 m 2n ! 2 3ekfkL2,KR 2 X 2 r kF − fkK ≤ sup (1 + |ζ| ) (Mknk) . 1 2nk 2 2sπ ζ∈K+∆(0,s) k=0 Rk But since we have already shown that F ∈ R(a, n) we see that Rd(f, K, a, n) can be estimated by the same expression.

35 4 A generalized version of the Bernstein-Walsh theorem

At last it can be easily seen that Rd(f, K, a, n) is invariant under shifting in the plane. That is to say, if for any we set , w ∈ C Kw = K + w = {z + w; z ∈ K} fw(z) = f(z − w) and aw = (a1 + w, a2 + w, ..., am + w) then

Rd(f, K, a, n) = Rd(fw,Kw, aw, n).

Thus we can estimate the expression sup(1 + |ζ|2) by the diameter of K, getting the following result

1 m 2n ! 2 3ekfkL2,KR 2 X 2 r Rd(f, K, a, n) ≤ (1 + (diam(K) + s) ) (Mknk) . 1 2nk 2 2sπ k=0 Rk 4.3 Discussion on the estimate

The estimate on R(f, K, a, n) given above includes a lot of constants which are very hard to calculate explicitly, but the expression still gives us some information about the approximation of f by rational functions. The rst thing to check would be whether this estimate guaranties that f can be approximated arbitrary well by rational functions. It does indeed! Remember that n = max{n0, n1, ..., nm}, and that by denition r < Rk for all k. Therefore by choosing n = n0 = n1 = ... = nm we see that m X r2n (M n)2 −→ 0 as n → ∞, k R2n k=0 k because each of the terms of the sum tends to zero. Thus we can make Rd(f, K, a, n) arbitrarily small by letting n be large enough. At rst we explore what happens if we let n tend to innity in the most simple way.

Theorem 4.3.1. Let K, m, f, a be as before. For any n ∈ N we dene n∗ := m+1. If m+1 is a vector such that (n, n, n, ..., n) ∈ N R = (R0,R1, ..., Rm) ∈ R f ∈ O(int(KR)), then

∗ 1/n −1 lim sup(Rd(f, K, a, n )) ≤ (min{R0,R1, ..., Rm}) . n→∞ We skip the proof of this theorem momenteraly since it can be considered as a corollary to the next one. By letting the vector n tend to innity in a more complicated manner we get dierent results. For any x ∈ R we denote the smallest integer larger than or equal to x by dxe. Theorem 4.3.2. Let be as before and let m+1 be K, m, f, a l = (l0, l1, ..., lm) ∈ R+ an arbitrary vector. For any dene ∗ . Let n ∈ N nl = (dl0ne, dl1ne, ..., dlm, ne) m+1 be a vector such that . Then R = (R0,R1, ..., Rm) ∈ R f ∈ A(KR)

1 l l ∗ n 0 1 lm −1 lim sup(Rd(f, K, a, nl )) ≤ (min{R0 ,R1 , ..., Rm }) . n→∞

36 4.3 Discussion on the estimate

Proof. Let l = max{l0, l1, ..., lm}. From the calculations above we have seen that there are constants A, Mk and r < min{R0,R1, ..., Rm} such that

1 m 2dnle ! 2 ∗ X 2 r Rd(f, K, a, n ) ≤ A (Mkdnlke) . l 2dnlke k=0 Rk Therefore we see that

1  2n 2n m l ! 1 1 X r ∗ n n 2 lim sup(Rd(f, K, a, nl )) ≤ lim sup A  (Mknlk) l  n→∞ n→∞ k k=0 Rk ! rl = max k∈{0,1,...,m} lk Rk

l l0 l1 lm −1 = r (min{R0 ,R1 , ..., Rm }) .

Now, we see in the calculations above that r could have been chosen arbitrarily close to 1. Thus

1 l l ∗ n 0 1 lm −1 lim sup(Rd(f, K, a, nl )) ≤ (min{R0 ,R1 , ..., Rm }) . n→∞

We can generalize the last theorem a little bit more:

Theorem 4.3.3. Theorem 4.3.2 holds true for any vector R = (R0,R1, ..., Rm) ∈ m+1 such that R f ∈ O(int(KR))

Proof. If f ∈ O(int(KR)) then f ∈ A(Kr) for any r = (r0, r1, ..., rm) such that rk < Rk for all K. Then

1 l l ∗ n 0 1 lm −1 lim sup(Rd(f, K, a, nl )) ≤ (min{r0 , r1 , ..., rm }) . n→∞ Letting r tend to R in an obvious manner, we get the result. Notice that by setting l = (1, 1, ..., 1) in the above theorems we get Theorem 4.3.1. We can also prove a sort of converse to the preceding theorem.

Theorem 4.3.4. Let K, m and a be as before and f ∈ C(K). Let R = (R0,R1, ..., Rm) ∈ ]1, ∞[m+1 be a vector such that

1 l l ∗ n 0 1 lm −1 lim sup(Rd(f, K, a, nl )) ≤ (min{R0 ,R1 , ..., Rm }) n→∞ for every m+1. Then is the restriction to of a holomorphic l = (l0, l1, ..., lm) ∈ R+ f K function on int(KR).

37 4 A generalized version of the Bernstein-Walsh theorem

Proof. Note that int . For an (KR) = {z ∈ C; Gk(z) < log(Rk), k = 0, 1, ..., m} arbitrary r = p/q ∈ ∪nR(a, n) we dene m m |r(z)| X X VK,r(z) := log − (n0 − nk)G0(z) − nkGk(z), krkK k=1 k=1 where n0 = deg(p) and nk is the degree of ak as a root of q. Notice that VK,r is subharmonic and bounded on . Since on , by the C \{a1, a2, ..., am} VK,r ≤ 0 ∂K maximum principle, on . Obviously we have on , thus we VK,r ≤ 0 C \ K VK,r ≤ 0 K have seen that m m X X |r(z)| (n0 − nk)G0(z) + nkGk(z) ≥ log , z ∈ C. krkK k=1 k=1 Taking the exponential on both sides gives

m m X X |r(z)| ≤ krkK exp((n0 − nk)G0(z) + nkGk(z)), z ∈ C. k=1 k=1 Let be the -th hole of as before, and the unbounded component of . Ωk k K Ω0 C \ K It is sucient to show that f is the restriction to K of a function holomorphic on (Ωk ∩ int(KR)) ∪ U, for some neighbourhood U of K, for every k. Let k ≥ 1 be xed and be such that l0 l1 lm lk . By assumption we l = (l1, l2, ..., ln) min{R0 ,R1 , ..., Rm } = Rk have ∗ 1 1 lim sup(Rd(f, K, a, n )) n ≤ l lk n→∞ Rk so there is a constant M such that M Rd(f, K, a, n∗) < , n ∈ . l nlk N Rk

Now, pick ∗ such that nlk for all . We claim rn ∈ R(a, nl ) krn − fkK < M/R n that the series P∞ converges uniformly on compact subsets of r0 + k=1(rk − rk−1) (Ωk ∩int(KR))∪U, where U is some neighbourhood of K, to a holomorphic function which agrees with on . By the denition of the sequence , this series F f K (rn)n∈N converges on some neighbourhood U of K. Let 1 < ρ < Rk and dene Bρ := int . Note that if then on . Ωk ∩ {z ∈ C; Gk(z) ≤ ρ} ⊂ Ωk ∩ (KR) j 6= k Gj = 0 Bρ Therefore

sup |rn(z) − rn−1(z)| z∈Bρ m ! X ≤ sup krn − rn−1kK exp((dnl0e + 2 + m)G0(z) + dnlkeGk(z)) z∈B ρ k=1

≤ sup ((krn − fkK + kf − rn−1kK ) exp(dnlkeGk(z))) z∈Bρ n 2M ρlk ≤ ρdnlke ≤ 2MRρ . R(n−1)lk Rlk

38 4.3 Discussion on the estimate

So the series converges. The case when k = 0 is very similar.

The preceding theorems are very similar to the Bernstein-Walsh theorem, which can actually be considered as a corollary to Theorems 4.3.1 and 4.3.4.

Denition 4.3.5. Let K be a compact subset of the complex plane and f be a function on K. Then we dene

dn(f, K) := inf{kf − pkK : p is a polynomial}.

By noticing that dn(f, K) = Rd(f, K, ∅, (n)), the following corollary is just a special case of the above theorem.

Corollary 4.3.6 (Bernstein-Walsh). Let K be a compact subset of C which is regular and has no holes. Let be the Green's function for with the pole at innity gK C \ K and let . Let . Then KR = {z ∈ C; gK (z) < log(R)} f ∈ C(K)

1/n 1 lim sup(dn(f, K)) ≤ n→∞ R if and only if f is the restriction to K of a holomorphic function on KR. For a given function f holomorphic on a neighbourhood U of a compact set K and some > 0, it can sometimes be useful to know some upper limit to the degree of poles of a rational function r necessary so that kr − fkK < . We shall state the problem more explicitly. Let and ∗ be as before. Given some we want to nd K, m, a, f, R, l nl > 0 N ∈ N such that ∗ Rd(f, K, a, nl ) < , n ≥ N.

We have already seen that there are constants A, Mk and r ∈]1, min{R0, ..., Rm}[ such that 1 m 2dnle ! 2 ∗ X 2 r Rd(f, K, a, n ) ≤ A (Mkdnlke) . l 2dnlke k=0 Rk We put this expression less than and solve for a lower bound on n,

m 2dnle 2 X 2 r (Mkdnlke) < . 2dnlke A k=0 Rk We demand every term in this sum to be small enough,

2dnle 2 2 r −1 (Mkdnlke) < (m + 1) , k ∈ {0, 1, ..., m}. 2dnlke A Rk

39 4 A generalized version of the Bernstein-Walsh theorem

The expression on the left is decreasing with n so we can estimate dnle with nl. !n rl n < , k ∈ {0, 1, ..., m}. lk A(m + 1)1/2M l Rk k k Remembering that r could have been chosen arbitarily close to 1, we can assume that l lk . By taking both sides to the power , and then taking the (r /Rk ) < 1 −1 logarithm yields

1 A(m + 1)1/2M l n + B log > B log k k , k ∈ {0, 1, ..., m}. k n k

For some constants dependent on l lk . It is clear that as grows, the term Bk r /Rk n Bk log(1/n) will be neglectable compared to n. Thus by changing the constant Bk to some other constant 0 , it suces to choose Bk

A(m + 1)1/2M l n > B0 log k k , k ∈ {0, 1, ..., m}. k

By taking 0 0 and 0 1/2 B = max{B0, ..., Bm} A = max{A(m + 1) Mklk; k = 0, 1, ..., m} we get the following theorem.

Theorem 4.3.7. Let and ∗ be as before. There are constants and K, m, a, f, l nl A B such that if n > B log(A/) then

∗ Rd(f, K, a, nl ) < .

40 5 Vitushkin's Theorem

In this chapter we prove Vitushkin's theorem, which characterizes the compact sets K for which A(K) = R(K). Before we can even state this theorem we must dene some new concepts, the ones of analytic capacity and continuous analytic capacity. We discuss some properties of those concepts and introduce a third, the analytic diameter, before moving on to the proof of the main theorem. (The proof given here follows the steps of [3].)

5.1 Introduction

For an arbitrary subset E of the complex plane, we say that a function f is holomorphic o E if there is a compact set K ⊆ E such that f is holomorphic on C \ K. Let K be a compact set in the complex plane and f be a function dened on some subset of the Riemann sphere Cb. We say that the function f is admissible for K if f is holomorphic on Cb \ K and if f satises kfk ≤ 1 and f(∞) = 0. Cb\K For an arbitrary bounded subset E of the complex plane we say that the function f is admissible for E if it is admissible for some compact subset of E. The set of functions which are admissible for E is denoted by A (E). The set of functions which are admissible for E and are dened and continuous on the whole Riemann-sphere is denoted by CA (E). In other words

CA (E) = C(Cb) ∩ A (E). The analytic capacity of a set E is dened by γ(E) := sup{|f 0(∞)| : f ∈ A (E)} and the continuous analytic capacity of a set E is dened by α(E) := sup{|f 0(∞)| : f ∈ CA (E)}, where the derivative of f at innity is dened by f 0(∞) = lim z(f(z) − f(∞)). z→∞ Observe that f 0(∞) is equal to the derivative of the function g(z) := f(1/z), g(0) = f(∞) at the origin. With those concepts at hand, we can now state the main theorem of this chapter.

41 5 Vitushkin's Theorem

Theorem 5.1.1 (Vitushkin's theorem). Let K be a compact set. The following statemens are equivalent.

(i) R(K) = A(K)

(ii) For every bounded open set D, we have α(D \ K) = α(D \ int(K)).

(iii) For every z ∈ C, r > 1 and δ > 0, we have α(∆(z, δ) \ int(K)) ≤ α(∆(z, rδ) \ K). (iv) There exist , and such that for every and , we r ≥ 1 c > 0 δ0 > 0 z ∈ C δ < δ0 have α(∆(z, δ) \ int(K)) ≤ cα(∆(z, rδ) \ K).

(v) For every z ∈ ∂K, there exists r ≥ 1 such that

α(∆(z, δ) \ int(K)) lim sup < ∞. δ→0 α(∆(z, rδ) \ K)

Before we can prove the theorem we must explore some properties of analytic capacity and also introduce analytic diameter. Any function f which is holomorphic at ∞ can be represented near ∞ as a Laurent series

∞ a1 a2 X −k f(z) = a0 + + 2 + ... = ak(z − z0) , z − z0 (z − z0) k=0 where z0 is any given complex number. It is easy to see that the rst two coecients, a0 and a1, are independent of the choice of z0. In fact

0 a0 = f(∞) and a1 = f (∞).

The third coecient, a2, depends on z0 though and we set

β(f, z0) := a2.

It is easy to see that if R > 0 is large, then 1 Z β(f, z0) = f(z)(z − z0)dz. 2πi |z|=R

Since the analytic capacity of a set E was dened as the supremum of the second coecient of the Laurent series of an admissible function on E, it is reasonable to do something similar for the third coecient. For each bounded set E we dene

( sup |β(f,z)| f∈A (E) if γ(E) > 0, β(E, z) = γ(E) 0 if γ(E) = 0

42 5.1 Introduction and we dene the analytic diameter of E by

β(E) = inf β(E, z). z∈C

Any point w0 such that β(E, w0) = β(E) is called the analytic centre of E. In fact, each bounded set has an analytic centre. To show that we need only to show that β(E, z) is a continuous function in the variable z and if β(E, z) is not equal to the constant 0 then limz→∞ β(E, z) = ∞. Lemma 5.1.2. The function z → β(E, z) is continuous for every bounded set E. Proof. Simple calculations show that

0 β(f, z1) = β(f, z0) + (z0 − z1)f (∞). (1)

From this it follows that for an admissible function f

|β(f, z1)| ≤ β(E, z0)γ(E) + |z1 − z0|γ(E). Taking the supremum over admissible functions we get

β(E, z1) − β(E, z0) ≤ |z0 − z1|.

Since z0 and z1 are interchangeable

|β(E, z1) − β(E, z0)| ≤ |z0 − z1|.

In particular β(E, z) is a Lipschitz-continuous function of z.

Lemma 5.1.3. If γ(E) > 0 then limz→∞ β(E, z) = +∞. Proof. Suppose γ(E) > 0. Let f be an admissible function for E such that f 0(∞) > 0. From the equation (1), we obtain

0 γ(E)β(E, z1) ≥ |β(f, z1)| ≥ |z0 − z1||f (∞)| − |β(f, z0)|.

By xing z0 and letting z1 tend to innity we see that

lim β(E, z) = +∞. z→∞

43 5 Vitushkin's Theorem 5.2 Theorems and estimates of analytic capacity

Some important theorems concerning analytic capacity and diameter follow almost instantly from denitions:

Theorem 5.2.1.

(i) For every set E we have α(E) ≤ γ(E) ≤ β(E).

(ii) If U is an open set then γ(U) = α(U).

(iii) If E ⊂ F , then γ(E) ≤ γ(F ) and α(E) ≤ α(F ).

(iv) If a and z0 are complex, then γ(z0 + aE) = |a|γ(E) and α(z0 + aE) = |a|α(E).

(v) The analytic capacity of a compact set K depends only on the boundary of Kb. That is γ(K) = γ(∂K) = γ(Kb) = γ(∂Kb). Proof. (i) Since CA (E) ⊂ A (E), we have α(E) ≤ γ(E). For > 0, choose an admissible function f for E such that f(∞) > γ(E) − . Then β(f 2, z) > (γ(E) − )2 for all z. Consequently

β(E, z) = sup |β(f, z)|/γ(E) ≥ γ(E) f∈A (E) for all z, and β(E) ≥ γ(E). (ii) From (i) we have α(U) ≤ γ(U). If f is an admissible function for U there is a compact set K ⊂ U such that f is admissible for K. Choose some compact J ⊂ U such that K ⊂ int(J). Now, by Tietze's extension theorem, we can nd some function f ∗ ∈ CA (U) such that f ∗ = f on Cb \ J. In particular those functions agree at innity so α(U) ≥ γ(U). Hence α(U) = γ(U). (iii) If E ⊂ F then A (E) ⊂ A (F ) and CA (E) ⊂ CA (F ). (iv) If f is an admissible function for E, then g(z) := f(z0 + az) is an admissible function for z0 + aE. (v) If f is an admissible function for Kb, then the function g dened with g(z) = f(z) on Cb \ Kb and g(z) = 0 on Kb is admissible for any of the other sets. Now let K be a compact connected subset of the plane. It turns out that the problem of calculating the analytic capacity of such a set can be reduced to nding the Riemann-mapping from Cb \ Kb to the open unit disc. Theorem 5.2.2. Let K be a compact connected subset of the complex plane which contains more than one point. Let g be the conformal map of Cb \Kb onto the interior of the unit disc, satisfying g(∞) = 0 and g0(∞) ∈ R, g0(∞) > 0. Then γ(K) = g0(∞).

44 5.2 Theorems and estimates of analytic capacity

Proof. Since γ(K) = γ(Kb) we can assume that K = Kb. Since g is admissible for K, g0(∞) ≤ γ(K). If h is also admissible, then h ◦ g−1 is a holomorphic function on ∆(0, 1) satisfying (h ◦ g−1)(0) = 0 and |h ◦ g−1| ≤ 1. By Schwarz's lemma |h ◦ g−1(w)| ≤ |w|, which gives |h(z)| ≤ |g(z)| for all z ∈ Cb \ K. Consequently 0 0 0 |h (∞)| = limz→∞ |zh(z)| ≤ limz→∞ |zg(z)| = g (∞) and γ(K) ≤ g (∞). This theorem allows us to calculate the analytic capacity of any set for which this Riemann-mapping is known.

Theorem 5.2.3. The analytic capacity of a disc is equal to its radius. The analytic capacity of a line segment of length L is L/4. Proof. The function r g(z) = z − z0 is a conformal mapping of ∆(z0, r) onto the unit disc which satises g(∞) = 0. We 0 obtain γ(∆(z0, r)) = g (∞) = r. For the second assertion, we consider the Joukowsky map

L 1 h(z) = z + , 4 z which maps the unit disc conformally onto Cb \ [−L/2, L/2], with h(0) = ∞. We obtain

L 1 γ([−L/2, L/2]) = (h−1)0(∞) = lim zh−1(z) = lim h−1(z) + h−1(z) z→∞ z→∞ 4 h−1(z)

L L = lim ((h−1(z))2 + 1) = . z→∞ 4 4

Calculations of capacity of further sets are not necessary for the upcoming proof of Vitushkin's theorem. We include a list of examples though, for the interested reader (Table 5.1. of [10])1.

1The table in the book by Ransford refers to logarithmic capacity but not analytic capacity. For a connected compact set this is the same thing. That can by seen by comparing Theorem 5.2.9 of this thesis with Theorem 5.2.1. of [10].

45 5 Vitushkin's Theorem

K γ(K) disc, radius r r

L line segment, length L 4 a + b ellipse, semi axes a, b 2 31/2Γ3(1/3) equilateral triangle, side h · h 8π2 33/4Γ2(1/4) isosceles right triangle, short side h · h 27/2π3/2 Γ2(1/4) square, side h · h 4π3/2 Γ(1/n) regular n-gon, side h · h 21+2/nπ1/2Γ(1/2 + 1/n) circular arc, radius r, angle α r sin(α/4)

4 half-disc, radius r · r 33/2

For a disconnected compact set K a conformal mapping from Cb \ K to the unit disc of course does not exist, but there is still a function for which the supremum is gained.

Theorem 5.2.4. Let K ⊂ C be compact and holomorphically convex. Then there is a unique admissible function g for K satisfying g0(∞) = γ(K).

Proof. The existence of such a function is always assured by the fact that A (K) is a normal family of holomorphic functions on Cb \ K. Suppose g0, g1 ∈ A (K) both have derivatives at ∞ equal to γ(K). Then the 0 function g = (g0 + g1)/2 ∈ A (K) also satises g (∞) = γ(K). Let h = (g − g0)/2. Then g1 = g + h and g0 = g − h. It suces to show that h = 0. Since by denition we have |g0|, |g1| ≤ 1, we get 1 ≥ |g ± h|2 = |g|2 + |h|2 ± 2Re(gh¯) and obtain |g|2 + |h|2 ≤ 1. Let k = h2/2. Then 1 − |g|2 (1 − |g|)(1 + |g|) (1 − |g|)(1 + 1) |k| ≤ = ≤ = 1 − |g|, 2 2 2

46 5.2 Theorems and estimates of analytic capacity so |g| + |k| ≤ 1. If k is not zero, we can write its Laurent series near ∞ as a a k(z) = n + n+1 + ... zn zn+1

0 0 where an 6= 0. Since h(∞) = h (∞) = 0, also k(∞) = k (∞) = 0 and n > 1. n−1 Consequently for > 0 small enough, |an||z| ≤ 1 in a neighbourhood of K. Let

n−1 f = g + a¯nz k.

Then f(∞) = 0. In a neighbourhood of K we get

n−1 |f| ≤ |g| + |an||z| |k| ≤ |g| + |k| ≤ 1.

So by the maximum principle we have |f| ≤ 1 on Cb \ K. Therefore, f is admissible for K. However 0 0 2 f (∞) = g (∞) + |an| > γ(K). This contradiction shows that k = 0. Hence also h = 0, and the theorem is estab- lished.

The function g mentioned in the theorem above is called the Ahlfors function for K. Notice the condition K = Kb in the theorem. It is quite easy to look past this condition since for an arbitrary compact K we lose the uniqueness property only because an admissible function can change in the bounded components of Kc without eecting the behavior in the unbounded one. We are only interested in its behaviour in the unbounded part. Therefore, for an arbitrary K we say that the Ahlfors function for Kb is also the Ahlfors function for K. Now we turn to estimates for admissible functions in terms of analytic capacity.

Theorem 5.2.5. Let K be compact and let f be admissible for K. γ(K) (i) If z ∈ b \ Kb, then |f(z)| ≤ . C d(z, K) 4γ(K) (ii) If z ∈ b \ Kb, then |f 0(z)| ≤ . C d(z, K)2

∞ X an (iii) If K ⊂ ∆(0,R) and f(z) = represents f near ∞, then |a | ≤ enRn−1γ(K). zn n n=1

Proof. (i) For z ∈ Cb \ Kb xed and a < d(z, K), let ! a f(ζ) − f(z) h(ζ) := . ζ − z 1 − f(z)f(ζ)

47 5 Vitushkin's Theorem

ζa f(ζ) − f(z) 0 γ(K) ≥ |h (∞)| = lim |ζh(ζ)| = lim = a|f(z)|. ζ→∞ ζ→∞ ζ − z 1 − f(z)f(ζ) Hence |f(z)| ≤ γ(K)/a. Since a is an arbitrary positive number greater than d(z, K) we have the desired result. (ii) Let d = d(z, K). Again, xing z ∈ Cb \ Kb, for any ζ such that |ζ − z| = d/2 by (i) we have |f(ζ)| ≤ 2γ(K)/d(z, K). Hence Z 0 1 f(ζ) 4γ(K) |f (z)| = 2 dζ ≤ 2 . 2πi |ζ−z|=d/2 (ζ − z) d(z, K)

(iii) Fix R0 so that R0 > R. By (i) we have |f(z)| ≤ γ(K)/(R0 − R) if |z| = R0. Hence Z n 1 n−1 γ(K)R0 |an| = f(z)z dz ≤ . 2πi |z|=R0 R0 − R n−1 Setting R0 = nR/(n − 1), and noting that (n/(n − 1)) ≤ e, we get the desired result. Even though direct calculations of analytic capacity are usually extremely dicult we have some estimates which sometimes are sucient: Theorem 5.2.6. (i) If K is compact and connected then γ(K) ≤ diam(K) ≤ 4γ(K).

(ii) If E is a bounded measurable subset of the complex plane, then Area(E) ≤ 4πα(E)2 ≤ 4πγ(E)2.

(iii) If E is bounded, then β(E) ≤ 6 diam(E).

(iv) If z0 ∈ E and w0 is the analytic centre of E then |z0 − w0| ≤ 24 diam(E). Before we prove this theorem we must mention a known result from complex analysis, which we state without proof.

Theorem 5.2.7 (Koebe's 1/4-Theorem, Th. 14.14 of [14]). If f ∈ O(∆(0, 1)) is injective, f(0) = 0 and f 0(0) = 1 then ∆(0, 1/4) is contained in the image of f. Proof of Theorem 5.2.6. (i) Any bounded set E is contained in a disc ∆ with radius diam(E). So the inequality γ(E) ≤ γ(∆) = diam(E) holds for an arbitrary E. For the second inequality, rst assume that K contains only one point. Any admissible function for K has a removable singularity at the point of K. Since it is bounded on the Riemann sphere it is equal to a constant and its derivative at innity is 0. We obtain γ(K) = 0.

48 5.2 Theorems and estimates of analytic capacity

Now assume that K contains at least two points. Let g be the Ahlfors function of −1 K. Let z1 ∈ K be xed. Dene f(z) := γ(K)/(g (z) − z1). Then f is an injective 0 function on the unit disc, f(0) = 0 and f (0) = 1. If z2 ∈ K is not equal to z1 then γ(E)/(z2 − z1) does not belong to the range of f. By Theorem 5.2.7 we have γ(K)/|z2 − z1| ≥ 1/4 or |z1 − z2| ≤ 4γ(K). Since z1, z2 ∈ K were arbitrary, we obtain diam(K) ≤ 4γ(K). (ii) It suces to prove the theorem for compact sets K. We consider the convo- lution ZZ dxdy f(ζ) = , K z − ζ of the locally integrable function 1/z with the characteristic function of K. The function f is continuous on the Riemann sphere, and f is holomorphic o K. Also f(∞) = 0 and ZZ ζ ZZ f 0(∞) = lim ζf(ζ) = lim dxdy = 1dxdy = Area(K). ζ→∞ K ζ→∞ z − ζ K

1 Let a = (Area(K)/π) 2 . For ζ xed, the disc ∆(ζ, a) has the same area as K. Since 1/|z − ζ| is a decreasing function of |z − ζ|, we obtain ZZ dxdy ZZ dxdy Z 2π Z a rdrdθ |f(ζ)| ≤ ≤ = = 2πa. K |z − ζ| ∆(ζ,a) |z − ζ| 0 0 r

It follows that f/(2πa) is an admissible function for K, continuous on Cb, and f 0(∞) (Area(K))1/2 α(K) ≥ = √ . 2πa 2 π This is the desired inequality.

(iii) Let z0 ∈ E, then E ⊂ ∆(z0, diam(E)). By a simple shifting in the plane and Theorem 5.2.5 (iii) we have |β(f, z0)| ≤ 6 · diam(E)γ(E) for any function f which is admissible for E. Thus we have β(E, z0) ≤ 6 · diam(E) for all z0 ∈ E and we get β(E) = infz β(E, z) ≤ 6 · diam(E). (iv) As in the proof of Lemma 5.1.3, for any and any admissible function z1 ∈ C f we have 0 γ(E)β(E, z1) ≥ |z0 − z1||f (∞)| − |β(f, z0)|. 0 By substituting z1 = w0 and letting f be an admissible function such that |f (∞)| ≥ γ(E)/2 we get γ(E) γ(E)β(E) ≥ |z − w | − |β(f, z )|. 0 0 2 0

By (iii) we have β(E) ≤ 6 · diam(E) and as in the proof of (iii) we have |β(f, z0)| ≤ 6 · diam(E)γ(E) so γ(E) 6γ(E)diam(E) ≥ |z − w | − 6γ(E)diam(E). 0 0 2 Simplifying, we get the desired result.

49 5 Vitushkin's Theorem

The next theorem is an existence theorem, guaranteeing the existence of a function f satisfying certain capacity estimates. This theorem will be extremely important in the upcoming proof of Vitushkin's Theorem since it will be used in the actual approximating step.

Theorem 5.2.8. Let E be bounded and let w0 be the analytic centre of E. If a and b are complex numbers such that |a| ≤ γ(E) and |b| ≤ β(E)γ(E), then there is a 0 function f such that f/20 is admissible for E, f (∞) = a and β(f, w0) = b. Proof. Let h be an admissible function for E, such that h0(∞) = γ(E)/2. Let z0 = w0 + 2β(h, w0)/γ(E). By equation (1) in the proof of Lemma 5.1.2 we have

0 β(h, z0) = β(h, w0) + (w0 − z0)h (∞) γ(E) = β(h, w ) + (w − w − 2β(h, w )/γ(E)) = 0. 0 0 0 0 2

Also |w0 − z0| ≤ 2β(E). Let f0 be an admissible function for E such that β(f0, z0) = γ(E)β(E)/2. Let 0 . Then 0 , f1 = 2f0 − 4f0(∞)h/γ(E) f1(∞) = f1(∞) = 0

0 0 β(f1, w0) = β(f1, z0) + (z0 − w0)f1(∞) = β((2f0 − 4f0(∞)h/γ(E)), z0) 0 = 2β(f0, z0) − 4f (∞)β(h, z0)/γ(E) = γ(E)β(E) and |f1| ≤ 2|f0| + 4|h| ≤ 6. Set . Then , 0 , f2 = 2h + (w0 − z0)f1/β(E) f2(∞) = 0 f2(∞) = γ(E) β(f , w ) β(f , w ) = 2β(h, w ) + (w − z ) 1 0 2 0 0 0 0 β(E) 0 = −2(w0 − z0)h (∞) + (w0 − z0)γ(E) = 0 and |f2| ≤ 2|h| + |f1||w0 − z0|/β(E) ≤ 2 + 6 · 2 = 14. Now f f f := a 1 + b 2 γ(E) β(E)γ(E) is the desired function.

5.2.1 Green's functions and analytic capacity

Green's functions can be helpful in the computation of analytic capacity. Let K be a connected compact set of the plane, and suppose f is the Ahlfors function of K. (Since K is connected f is essentially just the Riemann mapping of Cb \Kb to the unit disc such that f(∞) = 0, f 0(∞) > 0). We notice that the function g := − log |f| is harmonic on C\Kb. In fact we see that g satises every condition a Green's function with pole at ∞ should have. Therefore by the uniqueness property of the Green's function we see that g (z, ∞) = − log |f|. Cb\Kb

50 5.3 Rational approximation

∞ X ak Now, f can be expressed as a Laurent series near innity as f(z) = . Therefore zk k=1 we see that

∞ X ak g (z, ∞) = − log | | = log |z| − log |a | + o(1) as z → ∞. Cb\K zk 1 k=1

Now remembering that a1 = γ(K) we see the following Theorem 5.2.9. Let K be a connected compact set and D be the component of Cb \ K which contains ∞. Then

gD(z, ∞) = log |z| − log |γ(K)| + o(1) as z → ∞.

5.3 Rational approximation

Let K ⊂ C be compact and suppose f ∈ A(K). Now we develope an approximation procedure, to approximate f with some function which extends analytically beyond the interior of K. Since f is continuous to the boundary of K we can extend it to be continuous on the whole complex plane and with compact support. That is to say, we can assume that f ∈ Cc(C).

5.3.1 Partition of f Our rst step is to write f as a nite sum of functions, each one easier to approximate than the original function f. First we nd a suitable partition of unity:

Theorem 5.3.1. Let δ > 0. There exist discs ∆k,δ = ∆(zk, δ) which cover the complex plane, and continuously dierentiable functions gk,δ such that

(i) gk,δ is supported on ∆k,δ. (ii) P g = 1. k∈N k,δ

∂gk,δ 4 (iii) ≤ . ∂z ∞ δ

(iv) No point z is contained in more than 25 of the disks ∆k,δ. Proof. Let 1 be a function such that , RR and g ∈ Cc (∆(0, 1/2)) g ≥ 0 g(z)dxdy = 1 . Let be a partition of the plane into squares whose sides have |∂g/∂z¯| ≤ 16 {Ek}k∈N length , and set , where is the characteristic function for the set 1/2 gk,1 = g ∗χEk χEk Ek. Then each gk,1 is supported on the disc ∆k,1 = ∆(zk, 1) where zk is the centre of Ek. Now, let gk,δ(z) = gk,1(z/δ) and ∆k,δ = ∆(δzk, δ).

51 5 Vitushkin's Theorem

Now we let δ > 0 and gk,δ be a partition of unity as mentioned in the theorem above, each supported in ∆k,δ. We dene for each k an new function

1 ZZ f(z) − f(ζ) ∂g 1 ZZ f(z) ∂g G (ζ) := k,δ dxdy = f(ζ)g (ζ) + k,δ dxdy. k,δ π z − ζ ∂z k,δ π z − ζ ∂z These functions have many good properties as stated in the next theorem.

Theorem 5.3.2. Let g be a continuously dierentiable function supported on the disc . Let be holomorphic on some open set and ∆(z0, δ) f ∈ Cc(C) U 1 ZZ f(z) − f(ζ) ∂g G(ζ) := dxdy, ζ ∈ . π z − ζ ∂z C

Then G ∈ C(Cb) and:

(i) G is holomorphic on U and o the disc ∆(z0, δ). (ii) G(∞) = 0.

(iii) 0 1 RR ∂g . G (∞) = − π f(z) ∂z dxdy (iv) 1 RR ∂g . β(G, w) = − π f(z)(z − w) ∂z dxdy (v) ∂g . kGk∞ ≤ 2δω(f; 2δ) ∂z ∞ (vi) 0 ∂g . |G (∞)| ≤ 2δω(f; 2δ) ∂z ∞ α(∆(z0, δ) \ U) (vii) 1 RR ∂g 2 ∂g . π f(z)(z − z0) ∂z dxdy ≤ 4δ ω(f; 2δ) ∂z ∞ α(∆(z0, δ) \ U) where ω(f; ·) is the modulus of continuity of f.

Proof. (i) Since ∂g/∂z¯ has support in ∆(z0, δ) we see that G is holomorphic o ∆(z0, δ). If f is holomorphic in a neighbourhood of ζ we get ∂G ∂ 1 ZZ f(z) ∂g (ζ) = f(ζ)g(ζ) + dxdy ∂ζ¯ ∂ζ¯ π z − ζ ∂z ∂g ∂g = f(ζ) (ζ) − f(ζ) (ζ) = 0. ∂ζ¯ ∂ζ¯ So G is holomorphic wherever f is and o the support of g. (ii)

G(∞) = lim G(z) ζ→∞ 1 ZZ f(z) − f(ζ) ∂g 1 ZZ = lim dxdy = 0dxdy = 0. π ζ→∞ z − ζ ∂z π

52 5.3 Rational approximation

(iii)

G0(∞) = lim ζG(ζ) ζ→∞ 1 ZZ ζ(f(z) − f(ζ)) ∂g 1 ZZ ∂g = lim dxdy = − f(z) dxdy. π ζ→∞ z − ζ ∂z π ∂z (iv) We have

G0(∞) 1 ZZ f(z) − f(ζ) ∂g 1 ZZ f(z) ∂g G(ζ) − = dxdy + dxdy ζ − w π z − ζ ∂z π (ζ − w) ∂z 1 ZZ f(z)(z − w) + f(ζ)(w − ζ) ∂g = dxdy, π (ζ − w)(z − ζ) ∂z¯ so, remembering that f has compact support, we get G0(∞) β(G, w) = lim (G(ζ) − )(ζ − w)2 ζ→∞ ζ − w 1 ZZ (ζ − w)(f(z)(z − w) + f(ζ)(w − ζ)) ∂g = lim dxdy π ζ→∞ (z − ζ) ∂z¯ 1 ZZ ∂g = − f(z)(z − w) dxdy. π ∂z¯

∂g kGk∞ ≤ 2δω(f; 2δ) . ∂z¯ ∞ (vi) is continuous on and holomorphic on and o . By the de- G Cb U ∆(z0, δ) 0 nition of continuous analytic capacity we have |G (∞)| ≤ kGk∞α(∆(z0, δ) \ U). (vii) Very similar to the proof of (v) and (vi) with f(z) replaced by f(z) = f(z)(z − z0). If ζ ∈ ∆(z0, δ), we can evaluate |f(z)(z − z0) − f(ζ)(ζ − z0)| as in the integral in the proof of (v) by 2δω(f; 2δ) rather than ω(f; 2δ). Now, let the largest open set on which f is holomorphic be called U. Since f c has compact support U is bounded. Therefore only nitely many of the discs ∆k,δ intersect U c. Theorem 5.3.2 gives that the function Gk,δ is holomorphic on U and o ∆k,δ, it also states that Gk,δ(∞) = 0. Consequently from Liouville's theorem we see that if

53 5 Vitushkin's Theorem

c ∆k,δ ∩ U = ∅, then Gk,δ = 0. Therefore the function Gk,δ is equal to the constant 0 for all but nitely many k. At last we notice that for every ζ ∈ C ZZ X X 1 f(z) ∂gk,δ f(ζ) − G (ζ) = f(ζ) − f(ζ)g (ζ) + dxdy k,δ k,δ π z − ζ ∂z X 1 ZZ f(z) ∂ X = f(ζ) − f(ζ) g (ζ) − g dxdy k,δ π z − ζ ∂z k,δ = 0.

The conclusion is X f = Gk,δ where the sum is only over the nitely many non-zero terms. Observe that the number of terms depends on δ. Now, if we only approximate each of the Gk,δ suitably well then f will be approximated.

Theorem 5.3.3. Let K be compact, and let f ∈ A(K). For each δ > 0 let {gk,δ} be a partition of unity as discussed above, and let Gk,δ be as above. Suppose there is a constant independent of , a function and functions such r ≥ 1 δ a(δ) Hk,δ ∈ C(Cb) that

(i) limδ→0 a(δ) = 0,

(ii) kHk,δk ≤ a(δ),

(iii) Hk,δ is holomorphic o ∆(zk, rδ) \ K,

(iv) Hk,δ − Gk,δ has a triple zero at ∞. Then f ∈ R(K).

Proof. Let δ be xed. As before, we can assume that f has compact support, and that P is a nite sum. By assumption is holomorphic in a f = k Gk,δ (iii) Hk,δ neighbourhood of K. Therefore Runge's Theorem states that Hk,δ ∈ R(K) and it is sucient to show that X k Gk,δ − Hk,δkK ≤ b(δ), k where b is some function such that b(δ) → 0 as δ → 0. By Theorems 5.3.1 (iii) and 5.3.2 (v) we have kGk,δ − Hk,δk ≤ kGk,δk + kHk,δk ≤ 8ω(f; 2δ) + a(δ). 3 3 3 By assumption (iii) and (iv) the function (z − zk) (Gk,δ − Hk,δ)/(r δ ) is holomorphic in a neighbourhood of . On the boundary of the disc Cb \ ∆(zk, rδ) ∆(zk, rδ) we get

|z − z |3|G − H | k k,δ k,δ ≤ |G − H | ≤ |H | + |G | ≤ a(δ) + 8ω(f; 2δ). (r3δ3) k,δ k,δ k,δ k,δ

54 5.3 Rational approximation

By the maximum principle this inequality then holds when |z − zk| ≥ rδ. This inequality holds trivially when |z − zk| ≤ rδ. Consequently r3δ3(a(δ) + 8ω(f; 2δ)) |Gk,δ(z) − Hk,δ(z)| ≤ 3 , z ∈ C. |z − zk|

Now x some complex number z0 ∈ K. Each disc ∆k,δ = ∆(zk, δ) passes through at least one, and at most two of the circles , . Let Cn := {|z − z0| = nδ} n ∈ N N(n) be the number of discs which meet Cn. If n ≥ 2 and the disc ∆k,δ meets Cn it is contained in the annulus An = {(n − 2)δ ≤ |z − z0| ≤ (n + 2)δ}. If ∆k,δ meets C1, 2 it is contained in A1 = {|z − z0| ≤ 3δ}. The area of An is 8nδ π if n ≥ 2 and the 2 area of A1 is 9δ π. Since, by Theorem 5.3.1, each point is contained in at most 25 2 discs the sum of the areas of the discs meeting Cn is less than 9nδ π · 25. The area of each disc is δ2π so we see that 9nδ2π · 25 N(n) ≤ = 225n. δ2π

If n ≥ 2 and ∆k,δ meets Cn, then |z0 − zk| ≥ (n − 1)δ so r3(a(δ) + 8ω(f; 2δ)) |G (z ) − H (z )| ≤ . k,δ 0 k,δ 0 (n − 1)3

Using the estimate |Gk,δ(z0) − Hk,δ(z0)| ≤ a(δ) + 8ω(f; 2δ) when ∆k,δ meets C1, we obtain

∞ X X N(n)r3(a(δ) + 8ω(f; 2δ)) |G (z ) − H (z )| ≤ N(1)(a(δ) + 8ω(f; 2δ)) + k,δ 0 k,δ 0 (n − 1)3 k n=2 ∞ ! X n ≤ 225(a(δ) + 8ω(f; 2δ)) 1 + r3 . (n − 1)3 n=2 Hence P , where k k(Gk,δ − Hk,δ)k ≤ b(δ)

∞ ! X n b(δ) = 225(a(δ) + 8ω(f; 2δ)) 1 + r3 . (n − 1)3 n=2 We see that b(δ) → 0 as δ → 0.

The existance of functions Hk satisfying the conditions in Theorem 5.3.3 is highly dependant on analytic capacity.

Theorem 5.3.4. Suppose K, f, δ, gk,δ and Gk,δ are as above. Suppose there is a function c(δ) > 0 and a constant r ≥ 1 independent of δ such that

(i) limδ→0 c(δ) = 0, (ii) 0 , |Gk,δ(∞)| ≤ c(δ)γ(∆(zk, rδ) \ K)

55 5 Vitushkin's Theorem

(iii) |β(Gk,δ, wk)| ≤ c(δ)γ(∆(zk, rδ) \ K)β(∆(zk, rδ) \ K), where wk is the analytic centre of ∆(zk, rδ) \ K. Then f ∈ R(K).

Proof. By Theorem 5.2.8 there is a function Hk,δ holomorphic o ∆(zk, rδ) \ K such that , 0 0 , and Hk,δ(∞) = 0 Hk,δ(∞) = Gk,δ(∞) β(Hk,δ, wk) = β(Gk,δ, wk) kHk,δk∞ ≤ . Since is open we can arrange so that . Now 20c(δ) ∆(zk, rδ) \ K Hk,δ ∈ C(Cb) Hk,δ − Gk,δ has a triple zero at ∞ so we can apply Theorem 5.3.3. Now, using the theory which we have discussed the proof of Mergelyan's classical result is easy.

Theorem 5.3.5. Suppose K is compact and the set Kc has only nitely many components. Then every function f ∈ A(K) can be approximated uniformly on K by rational functions.

Proof. It is sucient to approximate f with a function which has an analytic extension to a neighbourhood of K. Extend f continuously to Cb, so that f vanishes near ∞ and construct gk,δ and Gk,δ as before. The indices can be arranged so that ∆k,δ meets ∂K if and only if 1 ≤ k ≤ m. Since Gk,δ is holomorphic in a neighbourhood of K for k > m it is sucient to nd functions H1,δ,H2,δ, ..., Hm,δ, holomorphic in a neighbourhood of K such that

m X (Gk,δ − Hk,δ) ≤ b(δ). k=1

Where b is some function such that limδ→0 b(δ) = 0. The desired approximating function is then

m m ∞ X X X fδ = f + (Hk,δ − Gk,δ) = Hk,δ + Gk,δ, k=1 k=1 k=m+1 where of course the last sum can be considered as nite since Gk = 0 for all but nitely many k. Suppose that 1 ≤ k ≤ m and that δ is suitably small. More exactly, suppose δ is smaller than half the diameter of every bounded component of Kc. Let c E = ∆(zk, 3δ) \ K. Then E contains an arc in K whose diameter exceeds δ. By Theorems 5.2.6 and 5.2.1 we get

δ β(E) ≥ γ(E) ≥ . 4

∂gk,δ 4 Estimating α(∆(zk, δ) \ K) ≤ δ and ≤ as in Theorem 5.3.1, from ∂z δ Theorem 5.3.2 we get ∞

0 and |Gk,δ(∞)| ≤ 8δω(f; 2δ) ≤ 32ω(f; 2δ)γ(E)

56 5.3 Rational approximation

2 |β(Gk,δ, zk)| ≤ 16δ ω(f; 2δ) ≤ 256ω(f; 2δ)γ(E)β(E).

If wk is the analytic centre of E = ∆(zk, 3δ) \ K then by Theorem 5.2.6 0 0 4 |wk − zk||Gk,δ(∞)| ≤ 72δ|Gk,δ(∞)| ≤ 10 ω(f; 2δ)γ(E)β(E).

From the equality β(Gk,δ, wk) = β(Gk,δ, zk) + (wk − zk)Gk,δ(∞) discussed late in section 5.1 we get 5 |β(Gk,δ, wk)| ≤ 10 ω(f; 2δ)γ(E)β(E).

Now we have veried the estimates of Theorem 5.3.4 for the functions G1,δ,G2,δ, ..., Gm,δ with c(δ) = 105ω(f; 2δ) and r = 3.

5.3.2 Proof of Vitushkin's Theorem

According to Vitushkin's theorem the question whether A(K) equals R(K) or not depends only on some estimates of continuous analytic capacity, but not on analytic diameter. The nal steps of Vitushkin's proof are essentially methods to obtain the estimate of as in Theorem 5.3.4 from the estimate of 0 in the |β(Gk,δ, wk)| |Gk,δ(∞)| same theorem. First we need two lemmas.

Lemma 5.3.6. Let p > 0 be an integer. Let {cn,k : 1 ≤ k ≤ pn, 1 ≤ n < ∞} be a double sequence such that 0 ≤ cn,k ≤ 1 for all n, k. Then !2 X cn,k X ≤ 2p c , n n,k n,k n,k assuming that the left sum converges. Proof. Let Ppn . Then and Ppn . By an = k=1 cn,k/n an ≥ 0 an ≤ k=1 1/n = pn/n = p induction it is easily seen that for all N we have

N !2 N ! X X an ≤ 2p nan . n=1 n=1 Letting N → ∞ we get the desired result.

Lemma 5.3.7. Let p > 0 be an integer and let E be a bounded set. Let {Ej} be a family of subsets of E such that every disc ∆(z, γ(E)) meets at most p of the sets Ej. Then X γ(Ej) ≤ 200pγ(E). j

If fj is an admissible function for Ej for all j, then X |fj(z)| ≤ 100p, j where the series is being summed over those indices of j for which fj is dened at z.

57 5 Vitushkin's Theorem

Proof. Let z0 be xed and let N(n) be the number of the sets Ej which meet ∆(z0, (n + 1)γ(E)) but not ∆(z0, nγ(E)). We renumber the sets with double indices En,1,En,2...En,N(n), n = 1, 2, 3, ...N, such that En,k meets ∆(z0, (n + 1)γ(E)) but not ∆(z0, nγ(E)) for all n, k. We take for granted that for n ≥ 1 the annulus An := {nγ(E) ≤ |z −z0| ≤ (n+1)γ(E)} can be covered by 20n discs of radius γ(E). Therefore N(n) ≤ 20pn if n ≥ 1. Also we have d(z ,E ) n ≤ 0 n,k ≤ n + 1. γ(E)

Let fn,k be an admissible function for En,k for all n, k. By Theorem 5.2.5 (i) and the estimate above we have

γ(En,k) γ(En,k) |fn,k(z0)| ≤ ≤ , n ≥ 1. d(z0,En,k) nγ(E)

Using Lemma 5.3.6 and estimating |f0,k(z0)| ≤ 1, we obtain

1 ! 2 X X γ(En,k) X γ(En,k) |f (z )| ≤ p + ≤ p + 40p . n,k 0 nγ(E) γ(E) n,k n6=0,k n,k P Let a = γ(Ej)/γ(E). The estimate becomes X p |fj(z0)| ≤ p + 40pa. j If is an admissible function for such that 0 , then the function hj Ej hj(∞) = γ(Ej)/2 P hj h := √j p + 40pa is admissible for E. So P γ(Ej) γ(E) ≥ h0(∞) = j √ , 2(p + 40pa) which implies a ≤ 2(p + p40pa). √ Solving for a we get a ≤ (82 + 8 105)p ≤ 200p, which proves the rst inequality. Also, X p p 2 |fj(z0)| ≤ p + 40pa ≤ p + 8000p ≤ 100p. j

Before we prove Vitushkins theorem, which gives a description of the compact sets K such that R(K) = A(K), we completely describe those functions f ∈ C(K) that satisfy f ∈ R(K).

58 5.3 Rational approximation

Theorem 5.3.8. Let K be compact. The following statements are equivalent for f ∈ C(Cb). (i) f ∈ R(K). (ii) If 1 , then g ∈ Cc (∆(z0, δ)) ZZ ∂g ∂g f(z) dxdy ≤ 2πδω(f; 2δ) γ(∆(z0, δ) \ K). ∂z¯ ∂z¯ ∞

(iii) There exists r ≥ 1 and a function a(δ) which tends to zero as δ → 0, such that ZZ ∂g ∂g ∂g f(z) dxdy ≤ δa(δ) + γ(∆(z0, rδ) \ K) ∂z¯ ∂x ∞ ∂y ∞ whenever 1 . g ∈ Cc (∆(z0, δ)) Proof. . Let 1 . First we shall assume that (i) ⇒ (ii) g ∈ Cc (∆(z0, δ)) f ∈ R(K) is holomorphic in a neighbourhood of K. Then, according to Theorem 5.3.2, the function 1 ZZ f(ζ) − f(z) ∂g G(ζ) = dxdy π ζ − z ∂z¯ is holomorphic o a compact subset of ∆(z0, δ)\K and is bounded by 2δω(f, 2δ)k∂g/∂z¯k∞. Since 1 ZZ ∂g G0(∞) = − f(z) dxdy, π ∂z¯ we obtain the estimate in (ii). Now let's assume that f ∈ R(K) is arbitrary. Let q be a rational function such that and let . Now let kf − qkK < /2 U = {z ∈ C; |f(z) − q(z)| < } ∞ be a cuto function such that in a neighbourhood of , and ψ ∈ Cc (U) ψ = 1 K dene h := ψq + (1 − ψ)f. Then h ∈ C( ) ∩ O(K) and kf − hk < . We have Cb Cb ZZ ZZ ZZ ∂g ∂g ∂g f(z) dxdy ≤ (f(z) − h(z)) dxdy + h(z) dxdy ∂z¯ ∂z¯ ∂z¯ ZZ ∂g ∂g ≤ dxdy + 2πδω(h; 2δ) γ(∆(z0, δ) \ K). ∂z¯ ∂z¯ ∞ Letting → 0 and noting that ω(h; 2δ) ≤ ω(f; 2δ) + 2 we obtain the estimate in (ii). (ii) ⇒ (iii).This is obvious. (iii) ⇒ (i). Suppose f satises the estimate in (iii). We verify the estimates of Theorem 5.3.4, for an appropriate function c(δ). Fix δ > 0 and k, and set E = ∆(zk, (r + 2)δ) \ K. By Theorems 5.3.2 and 5.3.4, it suces to show that ZZ ∂gg,δ f(z) dxdy ≤ c(δ)γ(E) ∂z¯

59 5 Vitushkin's Theorem and ZZ ∂gk,δ f(z)(z − wk) dxdy ≤ c(δ)γ(E)β(E). ∂z¯ The former estimate follows easily from (iii) since

∂gk,δ ∂gk,δ ∂gk,δ 16 + ≤ 4 ≤ ∂x ∞ ∂y ∞ ∂z¯ ∞ δ and γ(∆(z0, rδ) \ K) ≤ γ(E), providing only c(δ) ≥ 16a(δ). Obtaining the second estimate is much harder.

Set β0 := min{δ, β(E)}. Note that γ(∆(zk, (r + 2)δ)) = (r + 2)δ and by Theorem 5.2.1 (i) we have γ(E) ≤ β(E) ≤ (r + 2)β(E). Therefore

γ(E) ≤ (r + 2)β0. Let be a partition of unity like the one in Theorem 5.3.1 associated with hj = gj,β0 the discs ∆(tj, β0). From the properties of gk,δ and hj, we obtian

∂ ∂gk,δ ∂hj 8 8 16 (gk,δhj) ≤ + ≤ + ≤ , ∂x ∞ ∂x ∞ ∂x ∞ δ β0 β0 and

∂ ∂ ((z − tj)gkhj) ≤ kgkhjk∞ + β0 (gk,δhj) ≤ 17. ∂x ∞ ∂x ∞ There are similar estimates for the derivative with respect to y. From (iii) we now obtain ZZ ∂ f(z) (gk,δhj)dxdy ∂z¯ ∂ ∂ ≤ β0a(β0) (gk,δhj) + (gk,δhj) γ(∆(tj, rβ0) \ K) ∂x ∞ ∂y ∞ 16 16 ≤ β0a(β0) + γ(∆(tj, rβ0) \ K) = 32a(β0)γ(∆(tj, rβ0) \ K) β0 β0 and similarly ZZ ∂ f(z)(z − tj) (gk,δhj)dxdy ∂z¯ ZZ ∂ = f(z) ((z − tj)gk,δhj)dxdy ∂z¯

≤ 34β0a(β0)γ(∆(tj, rβ0) \ K).

60 5.3 Rational approximation

Now let J be the set of those indices j such that ∆(tj, β0) meets ∆(zk, δ). Note that any such disc is contained in ∆(zk, (r + 2)δ). From the preceding estimates we get: ZZ ∂gk,δ f(z)(z − wk) dxdy ∂z¯ ZZ X ∂ ≤ f(z)(z − wk) (gk,δhj)dxdy ∂z¯ j∈J ZZ ZZ X ∂ X ∂ ≤ f(z)(z − tj) (gk,δhj)dxdy + |tj − wk| f(z) (gk,δhj)dxdy ∂z¯ ∂z¯ j∈J j∈J X X ≤34β0a(β0) γ(∆(tj, rβ0) \ K) + 32a(β0) |tj − wk|γ(∆(tj, rβ0) \ K) j∈J j∈J ≤A + B. We estimate A and B separately. By construction, each z is contained in at most 25 of the discs ∆(tj, β0) (Theorem 5.3.1 (iv)). It follows that every disc of radius (r + 2)β0 meets at most M(r) of the discs ∆(tj, rβ0),where M is some function of r independent of δ and β0. Since γ(E) ≤ (r + 2)β0, every disc of radius γ(E) meets at most M(r) of the sets ∆(tj, rβ0) \ K. By Lemma 5.3.7 we have

A ≤ 34β0a(β0)(200M(r)γ(E)) ≤ 6800M(r)a(β0)γ(E)β(E).

Note that β0 ≤ δ so that takes care of A. B is harder. Choose an admissible function fj for ∆(tj, rβ0) \ K such that |t − w | (t − w )f 0(∞) = j k γ(∆(t , rβ ) \ K). j k j 2 j 0 By Lemma 5.3.7, the function

X fj F := 100M(r) j∈J is admissible for E. For R large we get

100M(r)γ(E)β(E) = 100M(r)γ(E)β(E, wk) ≥ 100M(r)|β(F, wk)| Z 100M(r) = F (z)(z − wk)dz 2π |z|=R

1 X Z X Z = (t − w ) f (z)dz + (z − t )f (z)dz 2π j k j j j j∈J |z|=R j∈J |z|=R Z 1 X |tj − wk| X = γ(∆(t , rβ ) \ K) + (z − t )f (z)dz 2π 2 j 0 j j j∈J j∈J |z|=R Z 1 X X 1 ≥ |tj − wk|γ(∆(tj, rβ0) \ K) − (z − tj)fj(z)dz . 4π 2π j∈J j∈J |z|=R

61 5 Vitushkin's Theorem

From the estimate of a2 in Theorem 5.2.5 (iii), we obtain Z X 1 X (z − tj)fj(z)dz ≤ 2erβ0 γ(∆(tj, rβ0) \ K) 2π j∈J |z|=R j∈J ≤ 400erM(r)γ(E)β(E). Where the second inequality comes from Lemma 5.3.7 as before. Consequently

1 X |t − w |γ(∆(t , rβ ) \ K) 4π j k j 0 j∈J ≤100M(r)γ(E)β(E) + 400erM(r)γ(E)β(E). Which leads to

X 6 B = 32a(β0) |tj − wk|γ(∆(tj, rβ0) \ K) ≤ 10 rM(r)a(β0)γ(E)β(E). j∈J Combining A and B we obtain the required estimate: ZZ ∂gk,δ f(z)(z − wk) dxdy ≤ c(δ)γ(E)β(E), ∂z¯ with c(δ) = 107rM(r) sup a(x). x≤δ

In his paper, Vitushkin gives other criteria necessary and sucient for a function f ∈ C(Cb) to belong to R(K). Since they are not necessary for the proof of Vitushkin's main theorem we state them without proof. We denote the square with centre z0 and side length δ with S(z0, δ). Theorem 5.3.9 (Th. IV.2.2. of [15]). The following statements are equivalent to the ones given in the preceding theorem.

(iv) There is a constant c such that Z

f(z)dz ≤ cω(f, 2δ)γ(S(z0, δ) \ K), ∂S(z0,δ)

for all squares S(z0, δ). (v) There exists r ≥ 1 and a function c(δ) which tends to zero as δ → 0, such that Z

f(z)dz ≤ c(δ)γ(S(z0, rδ) \ K), ∂S(z0,δ)

for all squares S(z0, δ).

62 5.3 Rational approximation

We are nally ready to prove Vitushkin's theorem

Proof of Theorem 5.1.1. (iv) ⇒ (i). Suppose (iv) is true and let f ∈ C(Cb) ∩ A(K). If 1 , then g ∈ Cc (∆(z0, δ)) 1 ZZ f(ζ) − f(z) ∂g G(ζ) = dxdy π ζ − z ∂z¯ is holomorphic o ∆(z0, δ) and on int(K). G is bounded by 2δω(f; 2δ)k∂g/∂z¯k∞. Hence ZZ ∂g 0 ∂g f(z) dxdy = π|G (∞)| ≤ 2πδω(f; 2δ) α(∆(z0, δ) \ int(K)) ∂z¯ ∂z¯ ∞

∂g ≤ 2πcδω(f; 2δ) γ(∆(z0, rδ) \ K). ∂z¯ ∞ This shows that statement (iii) in Theorem 5.3.8 holds with c(δ) = 2πcω(f; 2δ). Therefore f ∈ R(K) and A(K) = R(K). (i) ⇒ (ii). Suppose R(K) = A(K) and let D be an open set. Since α is monotone it is sucient to show that α(D\K) ≥ α(D\int(K)). Let > 0 and choose f ∈ C(Cb) which is admissible for D \ int(K) and |f 0(∞)| ≥ α(D \ int(K)) − . Then there is some compact set L ⊂ D \ int(K) such that f is holomorphic on Cb \ L. Let ∞ be a cuto function such that and in a neighbourhood χ ∈ Cc (D) kχk∞ ≤ 1 χ = 1 of L. Let M ≥ 1 be a constant such that ZZ ZZ |∂χ/∂w¯| M ≥ |∂χ/∂w¯|dλ(w) and M ≥ dλ(w) |z − w| for all z. Since f ∈ A(K) = R(K) we can nd Q ∈ R(K) such that kf − Qk < . K M

Dene 2 . Then is an open neighbourhood of . U := {z ∈ Cb; |f(z) − Q(z)| < M } U K Dene 1 ZZ f(w) − Q(w) ∂χ r(z) := (w)dλ(w). π U z − w ∂w¯ Since ∂χ/∂z¯ is supported on D \ L, r is holomorphic o that set. Also

2 ZZ |∂χ/∂w¯| |r(z)| ≤ dλ(w) ≤ 2, M C |z − w| for all z so krk∞ ≤ 2. Let g := χQ + (1 − χ)f + r. Then g is holomorphic o D since f and r are and supp(χ) ⊂ D. Also, g is holomorphic on U because there we have ∂g ∂χ ∂r = (Q − f) + = 0. ∂z¯ ∂z¯ ∂z¯

63 5 Vitushkin's Theorem

If z ∈ Cb \ supp(χ) then |g(z)| ≤ |f(z)| + |r(z)| ≤ 1 + 2. If z ∈ U, then |g(z)| ≤ |f(z)| + |χ(z)||Q(z) − f(z)| + |r(z)| ≤ 1 + 3.

We have seen that g < 1 + 3 o D and on U. Since it is also holomorphic there we see that g/(1 + 3) is admissible for D \ K and therefore

(1 + 3)α(D \ K) ≥ |g0(∞)| = |f 0(∞) + r0(∞)|

≥ α(D \ int(K)) − − lim zr(z) z→∞ ZZ 1 ∂χ ≥ α(D \ int(K)) − − (f(w) − Q(w)) dλ(w) π U ∂w¯ 2 ZZ ≥ α(D \ int(K)) − − |∂χ/∂w¯|dλ(w) M ≥ α(D \ int(K)) − 3.

By letting → 0 we see that α(D \ K) ≥ α(D \ int(K)). (ii) ⇒ (iii) Assuming (ii) and using the monotonicity of α we get

α(∆(z0, δ) \ int(K)) = α(∆(z0, δ) \ K) ≤ α(∆(z0, rδ) \ K). (iii) ⇒ (iv) and (iv) ⇒ (v) are obvious. We have seen that conditions (i) − (iv) are equivalent and that they imply (v). In the last step we show that if the equivalent statements (i) − (iv) fail, then (v) also fails. Suppose . We construct three sequences, ∞ , ∞ and R(K) 6= A(K) (Kn)n=0 (δn)n=0 ∞ of compact sets, positive real numbers and complex numbers respectively. (zn)n=1 Set K0 = K and δ0 = 1. For n ≥ 1, if A(Kn−1) 6= R(Kn−1) it follows from (iv) (by taking , and δn−1 ) that there exist and δn−1 such that c = n r = 4n δ0 = 2 zn δn < 2

α(∆(zn, δn) \ int(Kn−1)) > nα(∆(zn, 4nδn) \ Kn−1).

Let Kn = Kn−1 ∩ ∆(zn, δn). Then ∆(zn, δn) \ int(Kn−1) = ∆(zn, δn) \ int(Kn), ∆(zn, δn) \ Kn−1 = ∆(zn, δn) \ Kn and since α is monotone we have

α(∆(zn, δn) \ int(Kn)) = α(∆(zn, δn) \ int(Kn−1))

> nα(∆(zn, 4nδn) \ Kn−1) ≥ α(∆(zn, δn) \ Kn−1)

= α(∆(zn, δn) \ Kn).

Therefore (ii) fails for Kn and A(Kn) 6= R(Kn). By induction we see that this holds true for all n. Next we show that ∆(zn, δn) ⊂ ∆(zn−1, δn−1). Suppose on the contrary that ∆(zn, δn) \ ∆(zn−1, δn−1) 6= ∅. Then the set ∆(zn, 4nδn) \ Kn−1 contains as a subset a disc (for some ) and ∆(w, δn) w ∈ C

δn = α(∆(zn, δn)) ≥ α(∆(zn, δn) \ int(Kn−1))

> nα(∆(zn, 4nδn) \ Kn−1) ≥ α(∆(w, δn)) = δn.

64 5.3 Rational approximation

Therefore

∆(zn, δn) ⊂ ∆(zn−1, δn−1) n ≥ 1. Consequently, the sequence ∞ has a limit , also for all . (zn)n=1 z Kn = K ∩ ∆(zn, δn) n For any r ≥ 1 and n suciently large we have

∆(zn, δn) ⊂ δ(z, 2δn) ⊂ ∆(z, 2rδn) ⊂ ∆(z, 4nδn). Therefore, for n large enough, we obtain

α(∆(z, 2δn) \ int(K)) ≥ α(∆(zn, δn) \ int(K))

= α(∆(zn, δn) \ int(K ∩ ∆(zn−1, δn−1)))

= α(∆(zn, δn) \ int(Kn−1))

> nα(∆(zn, 4nδn) \ Kn−1)

≥ nα(∆(zn, 4nδn) \ K)

≥ nα(∆(z, 2rδn) \ K). and α(∆(z, δ) \ int(K)) α(∆(z, 2δ ) \ int(K)) lim sup ≥ lim n δ→0 α(∆(z, rδ) \ K) n→∞ α(∆(z, 2rδn) \ K) ≥ lim n = ∞. n→∞ Thus (v) fails at z. Evidently z ∈ ∂K.

For an arbitrary compact set K it can be very hard to tell if it satises the above criteria and determine if A(K) = R(K). The following few corollaries give some descriptions on K which are easier to visualize. Setting r = 1 in Theorem 5.1.1 (v), reversing the fraction and using the estimate α(∆(z, δ) \ int(K)) ≤ δ, we get the following corollary

Corollary 5.3.10. If K is compact, and γ(∆(z, δ) \ K) lim inf > 0 δ→0 δ for all z ∈ ∂K, then R(K) = A(K). Here we have interchanged α for γ since it is the same on open sets. The next corollary includes Mergelyan's theorem.

Corollary 5.3.11. If every point on the boundary of the compact set K belongs to the boundary of some component of Kc, then

R(K) = A(K) .

65 5 Vitushkin's Theorem

Proof. Let z ∈ ∂K. For suciently small δ > 0, the interior of the set ∆(z, δ) \ K contains an arc whose diameter exceeds δ/2. By Theorem 5.2.6 γ(∆(z, δ)\K) ≥ δ/8. By Corollary 5.3.10 R(K) = A(K).

For some set E, and z ∈ C we dene the lower Lebesgue density at the point z to be Area(∆(z, δ) ∩ E) lim inf . δ→0 πδ2 In fact it is not very hard to visualize what the lower Lebesgue density tells you about a set locally. For example, every interior point of E has obviusly lower Lebesgue density equal to one.

Corollary 5.3.12. If K is compact and Kc has positive lower Lebesgue density at every point of ∂K, then A(K) = R(K).

Proof. The lower Lebesgue density of K at the point z ∈ ∂K is given with

Area(∆(z, δ) \ K) lim inf . δ→0 πδ2 By Theorem 5.1.1 and estimating by Theorem 5.2.6, we obtain the conclusion.

66 6 Arakelian's Theorem

6.1 Arakelian

Arakelian's theorem concerns uniform approximation by entire functions on possibly unbounded closed subsets E of the complex plane C. To motivate the following denition, notice that if E is a closed set with no holes and ∆ is a closed disc in C the intersection ∆ ∩ E obviously has no holes, but the union ∆ ∪ E may well have some holes, even innitely many.

Denition 6.1.1. A set E ⊂ C is called an Arakelian set if it is closed, Ec is conneced, and for every closed disc ∆ ⊂ C the union of all holes of the set E ∪ ∆ is a bounded set.

The classical Arakelian theorem is as follows

Theorem 6.1.2 (Arakelian [12]). Let E ⊂ C be an Arakelian set and f be a continuous function on E which is holomorphic on the interior of E. Then for every > 0 there is an entire function h such that

|f(z) − h(z)| < , z ∈ E.

In Chapter 1 we discussed how easily approximation theorems can be generalized to a bigger variety of sets by approximating by rational functions rather than polynomials. In this section we prove a slight generalization of the Arakelian theorem concerning sets with holes. This proof is based on the one given by Jean-Pierre Rosay and Walter Rudin [12], changed a little because of the generalization. We dene a new kind of an Arakelian set.

Denition 6.1.3. A closed set E ⊂ C is called an Arakelian set with holes if Eb is an Arakelian set and if for any closed disc ∆ the intersection ∆ ∩ E has nitely many holes.

Our generalized version of the theorem is as follows:

Theorem 6.1.4. Let E be an Arakelian set with holes and f be a continuous function on E which is holomorphic on the interior of E. Let A be a closed set which contains at least one point from every hole of E. Then there exists a function h ∈ O(C \ A) such that |f(z) − h(z)| < , z ∈ E.

67 6 Arakelian's Theorem

Proof. For convenience we can assume that A contains exactly one point from every hole of E. We denote the holes of E by Ω1, Ω2, .... We let m denote the number of holes, note that . The point in (which is a singleton by m ∈ N ∪ {∞} Ωk ∩ A assumption) is denoted by ak. For every we let (k) be a sequence such that (k) and k ∈ N (rn ) ∆(ak, r1 ) ⊂ Ωk (k) rn & 0 as n → ∞. Let [ (k) Pn := ∆(ak, rn ). 1≤k≤m

Let Rn be a sequence such that Rn % ∞ and ∆(0,Rn)∪Hn ⊂ ∆(0,Rn+1) where Hn is dened to be the union of all holes of ∆(0,Rn) ∪ Eb. This sequence exists because E is an Arakelian set with holes. Now dene

E0 = E and En = (Eb ∪ ∆(0,Rn) ∪ Hn) \ Pn.

Notice that En ⊂ En+1 for all n and [ En = C \ A. n∈N

We dene now a sequence of holomorphic functions hn such that hn ∈ A(En) for all n and uniformly on compact subsets of , for some function . hn → h C\A h ∈ O(C\A) Let and assume that . Let ∞ be such that: h0 = f hn−1 ∈ A(En−1) ψn ∈ Cc (C) for all • 0 ≤ ψn(z) ≤ 1 z ∈ C.

• ψn = 1 in some neighbourhood Un of (∆(0,Rn) ∪ Hn) \ Pn.

• supp(ψn) ⊂ ∆(0,Rn+1) \ Pn+1. Since the function 1 R ∂ψn/∂w¯ is continuous and goes to zero at it is z → π z−w dλ(w) ∞ bounded by some constant Mn ≥ 1. Since E is an Arakelian set with holes, the set En−1 ∩ ∆(0,Rn+1) has nitely many holes and Mergelyan's theorem states that there is a rational function Qn with poles in A such that |hn−1(z) − Qn(z)| ≤ n+1 , z ∈ En−1 ∩ ∆(0,Rn+1). 2 Mn Dene Z 1 ∂ψn −1 rn(z) := (hn−1(w) − Qn(w)) (w)(z − w) dλ(w). π En−1 ∂z¯ Note that since supp ∂ψn we see that is holomorphic on . ( ∂z¯ ) ⊂ ∆(0,Rn+1) \ Un rn Un Finally we let

hn := ψnQn + (1 − ψn)hn−1 + rn.

Then hn is dened and continuous on Un ∪ En−1. Since ψ(z) = 1 on Un we have hn(z) = Qn(z) + rn(z) on Un. Hence hn is holomorphic on Un. On the interior of En−1 the Cauchy-Pompeiu's formula gives ∂h ∂ψ ∂ψ ∂r ∂ψ ∂ψ n = Q n − h n + n = (Q − h ) n + (h − Q ) n = 0. ∂z¯ n ∂z¯ n−1 ∂z¯ ∂z¯ n n−1 ∂z¯ n−1 n ∂z¯

68 6.1 Arakelian

Since En ⊂ En−1 ∪ Un we see that hn ∈ A(En). We have constructed a sequence of functions such that for all . Next we note that (hn)n∈N hn ∈ A(En) n ∈ N Z ¯ 1 ∂ψn(w) |rn(z)| ≤ sup |hn−1(ζ) − Qn(ζ)| dλ(w) ≤ n+1 , ζ∈En−1 π En−1 z − w 2 so for z ∈ En−1 we have |h (z) − h (z)| ≤ |r (z)| + ψ (z)|Q (z) − h (z)| < . n n−1 n n n n−1 2n

This shows that the sequence (hm)m≥(n−1) converges uniformly to a holomorphic function h on En−1. This is true for all n so we see that the sequence (hn) converges uniformly to a holomorphic function h on compact subsets of C \ A. Therefore h ∈ O(C \ A). On E we have ∞ X h = h0 + (hn − hn−1), n=1 and since h0 = f by denition, we get ∞ ∞ X X |f(z) − h(z)| ≤ |h (z) − h (z)| ≤ = , z ∈ E. n n−1 2n n=1 n=1

On sets with no interior a considerably stronger version of the theorem can be derived without much extra eort.

Corollary 6.1.5. Let E and A be as in the previous theorem and assume further that the interior of is empty. Given a continuous function and E ω : E → R+ f ∈ C(E), there is a function h ∈ O(C \ A) such that |h(z) − f(z)| ≤ ω(z), z ∈ E. Proof. From the previous theorem there is a function such that g1 ∈ O(C \ A)

|g1(z) − log(ω(z))| < 1 z ∈ E.

Let g2(z) = g1(z) − 1, we have

Re(g2(z)) = Re(g1(z)) − 1 < log(ω(z)) z ∈ E. By the same theorem we can nd , such that g3 ∈ O(C \ A)

|g3(z) − f(z) exp(−g2(z))| < 1, z ∈ E. Hence

|g3(z) exp(g2(z)) − f(z)| < | exp(g2(z))| < ω(z), which concludes the proof of the corollary.

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