Rational Approximation on theComplex Plane

A Thesis submitted for the degree of Master of Scienceat the University of Amsterdam

Supervisor: Prof. dr. J.J.O.O. Wiegerinck

by

Gang Huang

Korteweg-de Vries Institute for Mathematics

Amsterdam, The Netherlands

August, 2011

Acknowledgements

I would like to thank my supervisor Prof. dr. Jan Wiegerinck for his kindlyguidance, for much input on this thesis. It is such a privilege to learn doingresearch from him. I am also grateful to Prof. dr. Ale Jan Homburg forhis excellent advice on choosing courses and practical matters. Mukul Tyagihelped me a lot on typesetting. Finally, I am deeply indebted to all lecturerswhose courses I attended.

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Contents

Introduction 5

1 The Fine Topology 71.1 Logarithmic capacity . . . . . . . . . . . . . . . . . . . . . . . 71.2 Definition and some properties of the fine topology . . . . . . 91.3 The connection between the fine topology and thinness . . . . 10

2 Rational Approximation Theory 112.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Analytic capacity and Melnikov’s estimate of the Cauchy in-

tegral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Vitushkin’s criteria for R(X) = A(X) . . . . . . . . . . . . . 132.4 Peak points and bounded point derivations . . . . . . . . . . 15

3 Finely Holomorphic Functions 183.1 Definition and some properties of finely holomorphic functions 183.2 Rational approximation of finely holomorphic functions . . . 20

4 Rational Approximation On A Special Set 244.1 An example of fine domains . . . . . . . . . . . . . . . . . . . 244.2 An illustrating example . . . . . . . . . . . . . . . . . . . . . 254.3 Rational approximation on the illustrating example . . . . . . 264.4 Discussion and open problems . . . . . . . . . . . . . . . . . . 32

Bibliography 34

4

Introduction

In this thesis, rational approximation of complex-valued functions definedon subsets of the complex plane is studied. Let f(z) be a function defined ona subset X of the complex plane C. By rational approximation of f(z) wemean to find a sequence of rational functions fn(z) with poles off X whichconverges uniformly on X to f(z). A rational function is a quotient of twopolynomial functions.

The domains of functions are crucial to rational approximation, and theyhave been characterized in terms of geometric features, topological proper-ties, continuous analytic capacity, peak points, etc. The famous Runge’stheorem states that any function which is holomorphic on a neighbourhoodof a compact set admits rational approximation. Let f be continuous on acompact set X, holomorphic on the interior of X. Mergelyan constructeda class of compact sets with empty interiors on which f does not allow ra-tional approximation. In terms of continuous analytic capacity, Vitushkin([9]) obtained criteria of X on which f allows rational approximation. Wecan see that the classical maximal domain of rational approximation is acompact set satisfying Vitushkin’s criteria.

To meet the need of potential theory, mathematicians have attempted tocarry over certain analytic properties into more general domains since thebeginning of the 20th Century. A coherent theory of finely holomorphicfunctions defined on a finely open set has been developed. Bent Fuglede([4]) shows that a finely holomorphic function and its finely derivatives ofall orders admit rational approximation on finely open sets.

It is natural to ask the following question: are finely holomorphic func-tions the best extension of the notion of holomorphic functions? Have wegot some alternatives from the rational approximation’s point of view? Ouridea is that it is possible to find some new domains which are different fromfinely open sets and compact sets. We aim to investigate rational approx-imation of certain classes of functions on new domains. We hope that oureffort can serve as a tiny inspiration for clarifying directions of extendingholomorphic functions beyond classical domains.

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In this thesis, we have constructed a set which is neither open nor finelyopen, and have defined a new topology on it. We have obtained a partialresult of rational approximation on basis elements of this set. We have topoint out that our result is only valid for this special set. Generalising thisto a system of theorems is anything to us but intuitive.

The fine topology, rational approximation of holomorphic functions on com-pact sets and rational approximation of finely holomorphic functions on finedomains are reviewed in the first three chapters respectively. Our examplecomes in Chapter 4.

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Chapter 1

The Fine Topology

1.1 Logarithmic capacity

We start to introduce some classical potential theory in the Complex Planeserved as background information on logarithmic capacity. The Definition1.1.1-1.1.3 in this section are adapted from Ransford’s book. [1]

Definition 1.1.1 Let µ be a finite Borel measure on C with compact sup-port.(a) The logarithmic potential of µ is defined as a function pµ : C 7→ [−∞,∞)by

pµ(z) :=

∫C

log |z − w|dµ(w).

(b) The energy of µ is defined by

I(µ) :=

∫Cpµ(z)dµ(z).

Definition 1.1.2 Let E be a subset of C, and define B(E):= µ|µ is a finiteBorel measure, supp µ is a compact subset of E, µ 6= 0 . E is said to bepolar, if the energy I(µ) = −∞ for every µ ∈ B(E).

Definition 1.1.3 Let K be a compact subset of C, and define P(K) :=µ|µis a Borel probability measure on K. If there exists ν ∈ P(K) s.t.

I(ν) = supµ∈P(K)

I(µ),

then ν is called an equilibrium measure for K.

The property that every compact non-polar subset K of C has a uniqueequilibrium measure guarantees the following definition of the logarithmic

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capacity of a compact set is well defined. (see [1,Thm 3.3.2, Thm 3.7.6])

Definition 1.1.4 The logarithmic capacity of any compact subset K ofC is given by

c(K) =

exp(I(ν)) if K is non-polar,

0 if K is polar,

where I(ν) is the energy of the equilibrium measure ν for K.

Definition 1.1.5 For an arbitrary subset E of C, the inner logarithmiccapacity of E is defined by

c∗(E) = supc(K) : K is a compact subset of E

and the outer logarithmic capacity of E is defined by

c∗(E) = infc∗(ω) : ω is an open set containing E.

It is clear from the definitions that (outer/inner) logarithmic capacity ismonotone as a set function and c∗(E) ≤ c∗(E). But it is not subadditive,i.e. there exist some compact subsets A, B of C, such that c(A

⋃B) >

c(A) + c(B) (see [6]). This makes it difficult to compute logarithmic capac-ity of general sets except some sets with relatively simple geometric features,so we have to rely heavily on estimates.

Theorem 1.1.6(a) The inner and outer logarithmic capacities of any bounded Borel set Ecoincide. In that case, we write c(E) = c∗(E) = c∗(E).(b) If B1 ⊂ B2 ⊂ · · · are Borel subsets of C and B = ∪∞n=1Bn, thenc(B) = limn→∞ c(Bn).Proof: For (a), see[7,p170]. For (b), see[1, Thm5.1.3(b)].

Example 1.1.7 Given a ∈ C, and r > 0, we write D(a, r) = z ∈ C :|z − a| < r as an open disc in the complex plane, and D(a, r) = z ∈ C :|z − a| ≤ r as a closed disc. It is well known that the logarithmic capacityof a closed disc equals its radius, i.e., c(D(a, r)) = r. This was proved by arelation between logarithmic capacity and Green’s functions. (See [1],[2])

For an open disc, we also have c∗(D(a, r)) = c∗(D(a, r)) = c(D(a, r)) = r.Since D(a, r) =

⋃∞n=1 D(a, r− 1

n+1) and D(a, r− 1n+1) ⊂ D(a, r− 1

n+2), ∀n ∈ N,

Theorem 1.1.6(b) implies that c(D(a, r)) = limn→∞ c(D(a, r − 1n+1) ≤

limn→∞ c(D(a, r− 1n+1) = r. On the other hand, D(a, r− 1

n+1) ⊂ D(a, r), ∀n.

By monotonicity, c(D(a, r− 1n+1) ≤ c(D(a, r)),∀n. This implies that c(D(a, r) ≥

limn→∞ c(D(a, r − 1n+1) = r.

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1.2 Definition and some properties of the fine topol-ogy

Let τ be the Euclidean topology of the complex plane C, let τ ′ be the topol-ogy on [−∞,∞) defined by declaring the following sets to be open: (a, b),[−∞, a) and any union of them. By a subharmonic function on C, we meana function f : (C, τ) → ([−∞,∞), τ ′) which is upper semicontinuous withrespect to τ and satisfies the local sub-mean inequality. The following ex-ample shows that there exist discontinuous subharmonic functions on C.u(z) =

∑∞n=1 2−n log |z − 2−n| is subharmonic on C. The sequence 2−n

converges to 0 as n → ∞ and u(2−n) = −∞ for all n, but u(0) > −∞, sou(z) is discontinuous at 0. Hence, topologies which make all subharmonicfunctions on C continuous are finer than the Euclidean topology. The in-tersection of all these topologies, which is the coarsest one, is finer thanthe Euclidean one as well. This suggests the following definition of a newtopology on C.

Definition 1.2.1 The fine topology on C is defined as the coarsest topologymaking all the subharmonic functions on C continuous.

It is equivalent to define fine topology as the coarsest topology making allthe superharmonic functions continuous. In the sequel, topological notionsrelative to the fine topology will be distinguished by the prefix “fine(ly)”,and any other ones will refer to the Euclidean topology. For example, a finedomain is understood as a finely connected, finely open set in C.

Theorem 1.2.2(a) A finely open set U ⊂ C is finely connected if and only if U is connected.[14, Corollary 9.8](b) The complex plane is locally finely connected, i.e. let x ∈ C, for everyfinely open set U containing x, there exists a finely connected, finely openset V such that x ∈ V ⊂ U . [14, Corollary 9.11]

Theorem 1.2.3 X is a finely compact set if and only if X is finite. [2,Lemma 7.1.2 (3)]Proof : The “if” part is obvious. Conversely, let X be a finely compactset. Since every open covering of X is a finely open covering of X, X iscompact and hence bounded. Suppose X is infinite. Since it is bounded,there exist x ∈ X and a sequence xn ∈ X \ x such that xn → x. Bytranslation and dilation of X, we can assume, without loss of generality,that x = 0 and xn are in the unit disc for all n. Define the functionUy on C as Uy(x) = − log |x − y|. Define a superharmonic function u asu(x) =

∑∞n=1 2−nUxn(x)/Uxn(0), x ∈ C. Let V0 = x : u(x) < 2 and let

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Vnn>1 be a sequence of disjoint open discs centred at the points xn. LetV−1 := C\(xn : n > 1∪0). Then Vn : n > −1 is a finely open cover ofX, but it does not have a finite subcover. This contradicts the assumption,so X is finite.

1.3 The connection between the fine topology andthinness

Definition 1.3.1 [8] Let ζ ∈ C, a subset E of C is said to be thin atζ if: (a) ζ is not a limit point of E in the Euclidean topology , or (b)there exists a superharmonic function u on a neighbourhood of ζ such thatu(ζ) < lim infx→ζ,x∈E u(x).

Theorem 1.3.2 (Wiener’s criterion)Let ζ ∈ C, define An = z ∈ C : 2−(n+1) ≤ |z − ζ| ≤ 2−n. A subset E of Cis thin at ζ if and only if∑

n≥1

n

log(1/c∗(An ∩ E))<∞,

where c∗ denotes the outer logarithmic capacity.

The following Theorem of H. Cartan shows the connection between thin-ness and the fine topology. It can be read as an alternative definition of thinsets, and it can be used to check whether a set is finely open.

Theorem 1.3.3 The following are equivalent:(a) A subset E of C is thin at ζ ∈ C;(b) ζ is not a fine limit point of E;(c) C \ (E \ ζ) is a fine neighbourhood of ζ.

Theorem 1.3.4 If a set E in C is thin at x, then there are arbitrarilysmall circles centred at y which do not intersect E. [2, Theorem 7.3.9]

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Chapter 2

Rational ApproximationTheory

2.1 Introduction

Let f(z) be a function defined on a subset X of the complex plane C. Toapproximate f(z) we mean to find a sequence of functions fn(z) which con-verges uniformly to f(z) on X. Approximation problems of a function ofone complex variable then fall into three directions: to describe the set X;to classify the sequence fn(z); to classify the function f(z) admitting anapproximation.

In this chapter, we set X to be a compact subset of C in the Euclideantopology. We will discuss approximation problems on finely open sets andmore general domains in the next chapter.

By Runge’s theorem, any function which is holomorphic on a neighbour-hood of X is the uniform limit of a sequence of rational functions with polesoutside X on X. If a function can be approximated by holomorphic func-tions, it can be approximated by rational functions automatically. Hence,the sequence fn(z) is chosen to be polynomials or rational functions. Byrational approximation we mean approximation of a function by rationalfunctions.

By Weierstrass’s theorem, the uniform limit of continuous functions is con-tinuous. Then a necessary condition for function f(z) to be approximatedby polynomials or rational functions with poles off X is that f(z) is continu-ous. We focus on holomorphic functions f(z) in this chapter, and investigatefinely harmonic and finely holomorphic functions in the next chapter.

Another way to state approximation problems is by uniform algebras. Let

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C(X) denote the algebra of all continuous complex-valued functions f onX with the supremum norm ||f ||∞ = supz∈X |f(z)|. Then (C(X), || · ||∞)becomes a Banach algebra. Let A(X) be the subalgebra of C(X) consistingof functions that are holomorphic on the interior X of X. Let R(X) bethe algebra of rational functions with poles outside X. Let R(X) be theclosure of R(X) in the metric induced by supremum norm. That is, R(X)contains all functions on X which are uniform limits of rational functionswith poles outside X. Let P (X) denote the algebra of all functions that areuniform limits of polynomials on X.

Evidently, we have P (X) ⊂ R(X) ⊂ A(X) ⊂ C(X), and they are all com-plete. Approximation problems can be written as: under what conditionsdo two algebras coincide? For example, if R(X) = C(X), then any functionf ∈ C(X) is the uniform limit of a sequence of rational functions with polesoutside X.

Mergelyan proved that P (X) = A(X) if the complement of X is connected,and he constructed an example which is the so-called Swiss cheese to pointout that a simple geometric criterion for rational approximation is impossi-ble. To describe compact sets on which rational approximation is possible,one approach is by peak points and bounded point derivation, and anotherone is by continuous analytic capacity. Vitushkin ([9]) obtained two criteriafor R(X) = A(X) in terms of continuous analytic capacity and derived thetheorem of peak points as a consequence of his criteria.

2.2 Analytic capacity and Melnikov’s estimate ofthe Cauchy integral

The study of analytic capacity was initialised by Ahlfors, and the continuousanalytic capacity was a natural extension by Erokin to fit into the approx-imation problems. We shall use Vitushkin’s formulation of them ([9]) sincethey will be consistent with the notions in Vitushkin’s criteria used later.

Definition 2.2.1 Let G be a bounded subset of C. Denote by F (G) thecollection of functions defined on the whole complex plane, that are boundedin modulus by 1, vanish at ∞, and holomorphic outside some closed subsetof G. Let FC(G) be the collection of functions belonging to F (G) that arecontinuous on the whole complex plane.(a) The analytic capacity of G is defined by

γ(G) = supf∈F (G)

| limz→∞

zf(z)|

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(b) The continuous analytic capacity of G is defined by

α(G) = supf∈FC(G)

| limz→∞

zf(z)|

γ(G) and α(G) are monotonic set functions. Also, both the analytic ca-pacity and the continuous analytic capacity of an open disc are equal to itsradius.

In aid of the continuous analytic capacity, Melnikov obtained an upperbound of the Cauchy integral which is the key to the proofs of Vitushkin’scriteria and the theorem of bounded point derivation.

Theorem 2.2.2 Let G be a closed subset of D(0, 1) = z ∈ C : |z| 6 1.Let f(z) be a function continuous on D(0, 1) and holomorphic on D(0, 1)\G,then

|∫|z|=1

f(z)dz| 6 C max|z|61|f(z)|α(G)

where C is an absolute constant and α(G) is the continuous analytic capacityof G. [13]

2.3 Vitushkin’s criteria for R(X) = A(X)

We state the criteria for R(X) = A(X) in terms of the continuous analyticcapacity. In this section, let K(z, δ) denote the open disc centred at z withradius δ.

Definition 2.3.1 Let X ∈ C be a compact set. The set ∪i∂Ki(X), whereKi(X) are all components of the complement of X, is called the outerboundary of X. We denote the outer boundary of X by ∂′X. The set∂′′X = ∂X \ ∂′X is called the inner boundary of X.

Theorem 2.3.2 Let X ∈ C be a closed set and f(z) a function contin-uous on the whole complex plane and holomorphic outside a compact setX∗. If f ∈ R(X), then f ∈ R(X∪Xδ) for any δ > 0; here Xδ is the set of allpoints of the plane whose distance from X∗ is not less than δ. [9, Chapter5, §1, Theorem 1]

Lemma 2.3.3 Let E ∈ C be a compact set and let E denote the inte-rior of E. let ρ > 1, δ > 0 and m > 0 be numbers such that, for any pointz and for any δ 6 δ,

α(K(z, δ) \ E) 6 mα(K(z, ρδ) \ E),

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then A(E) = R(E). [9, Chapter 5, §3, lemma 2]

Theorem 2.3.4 Let E ∈ C be a compact set. The following conditionsare equivalent:(1) A(E) = R(E);(2) α(G \ E) = α(G \ E) for every bounded open set G;

(3) for each z ∈ E \ E, lim supδ→0α(K(z,δ)\E)α(K(z,δ)\E) < ∞. [9, Chapter 5, §3,

Theorem 1]Proof :(1) → (2): We fix the bounded open set G. Let C(G \ E, 1) be the set offunctions that are continuous on the whole complex plane, bounded in mod-ulus by 1, and vanish at infinity, such that each function from C(G \E, 1)is holomorphic outside a closed bounded subset of G \E. Given a functionf ∈ C(G \ E, 1), f is analytic outside some closed subset G′ ⊂ G \ E.Since G is open, G′ has a positive distance from the boundary of G. Sincef ∈ C(G \ E, 1) ⊂ A(E) and A(E) = R(E), f ∈ R(E). By Theorem2.3.2, f ∈ R(E∪E′), where E′ is some neighbourhood of Gc. Hence, ∀ε > 0,there exists a rational function rε(z) such that supz∈E∪E′ |rε(z)− f(z)| < ε.Since the poles of rε(z) can lie only in the complement of E ∪ E′ andsupz∈E∪E′ |rε| 6 1 + ε, the function fε(z) := rε(z) − rε(∞) belongs toC(G \ E, 1 + 2ε). Since ε is arbitrarily small, we have the following ex-pressions as close as we please:

γ(G \ E, f(z)) =1

2πi

∫∂(G\E)

f(z)dz;

γ(G \ E, fε(z)) =1

2πi

∫∂(G\E)

fε(z)dz.

Hence, α(G \E) > supf∈C(G\E,1)|γ(G \E, f)| = α(G \E). But, G \E ⊂G \E implies α(G \E) 6 α(G \E). Consequently, α(G \E) = α(G \E).

(2)→ (3): Obviously.

(3) → (1): We claim that if for some r > 1, lim supδ→0α(K(z,δ)\E)α(K(z,rδ)\E) < ∞,

then A(E) = R(E). Assume A(E) 6= R(E), then it follows from Lemma2.3.3 that there exists a K(z1, δ1) such that δ1 < 1

2 and α(K(z1, δ1) \E) > 1 · α(K(z1, 4δ1) \ E). It follows from this result and (1) → (2)that for the set E1 = E ∩ K(z1, δ1), A(E1) 6= R(E1). Repeating this ar-guments we obtained a sequence of discs K(zn, δn), n = 1, 2, · · · such thatδn <

12δn−1(δ0 = 1) and α(K(zn, δn) \ En−1

) > n · α(K(zn, 4nδn) \ En−1),where En = En−1 ∩ K(zn, δn) and En−1

denotes the interiors of the setEn−1(E0 = E).

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Suppose K(zn, δn) \K(zn−1, δn−1) 6= ∅, then K(zn, 4nδn) \En−1 contains awhole disc K ′(z, δn) as a subset, therefore:

α(K(zn, δn)) > α(K(zn, δn)\En−1 ) > n·α(K(zn, 4nδn)\En−1) > nα(K ′(z, δn)) = nδn.

Then α(K(zn, δn)) > δn, which is contradicted to the fact that the continu-ous analytic capacity equals to its radius. Hence, we have:

K(z1, δ1) ⊃ K(z2, δ2) ⊃ · · · ⊃ K(zn, δn) ⊃ · · · .

Consequently, En = E ∩ K(zn, δn) for any n, and for any r > 1 and suffi-ciently large n we have:

K(zn, δn) ⊂ K(z, 2δn) ⊂ K(z, 2rδn) ⊂ K(zn, 4nδn),

where z = limn→∞ zn. From this we have:

α(K(z, 2δn) \ E)>α(K(zn, δn) \ E)=α(K(zn, δn) \ En−1

)

>nα(K(zn, 4nδn) \ En−1)

>nα(K(zn, 4nδn) \ E)

>nα(K(z, 2rδn) \ E),

So, α(K(z, 2rδn) \ E) < δn and α(K(z, 2δn) \ E) > 0 when n > 1. If zis an interior point of E, then α(K(z, 2δn) \ E) = α(0) = 0 for large n. Ifz is a point of the complement of E, then α(K(z, 2rδn) \ E) = 2rδn > δnfor large n. Hence, z is neither an interior point of E nor a point in thecomplement of E. So z is a boundary point of E and

lim supδ→0

α(K(z, δ) \ E)α(K(z, rδ) \ E)

> lim supn→∞

α(K(z, 2δn) \ E)α(K(z, 2rδn) \ E)

=∞.

This is contradicted to given condition. Hence, we have A(E) = R(E). Since

lim supδ→0α(K(z,δ)\E)α(K(z,δ)\E) <∞ implies lim supδ→0

α(K(z,δ)\E)α(K(z,rδ)\E) <∞, (3)→ (1)

is proved.

2.4 Peak points and bounded point derivations

Definition 2.4.1 Let X be a compact subset of C. The point ζ ∈ X is saidto a peak point of the algebra R(X) if there exists a function f ∈ R(X) suchthat |f(ζ)| = 1 and |f(z)| < 1,∀z ∈ X, z 6= ζ. [9, p195]

Definition 2.4.2 Let X be a compact subset of C. If A is a function

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algebra on X, a (bounded) point derivation D on A at x ∈ X is a nonzero(bounded) linear functional on A satisfying Dfg = f(x)Dg+ g(x)Df for allf and g in A. [10]

Recall that R(X) is the algebra of rational functions with poles outside Xand R(X) is the closure of R(X) in the metric induced by supremum norm.

Theorem 2.4.3 Let X be a compact subset of C and x ∈ X. There isa bounded point derivation on R(X) at x if and only if there exists a con-stant k such that |f ′(x)| 6 k · ‖f‖, for all f ∈ R(X).Proof :We fix a x ∈ X, suppose there exists a constant k such that |f ′(x)| 6 k ·‖f‖,for all f ∈ R(X). Then there exists a bounded point derivation D ofR(X) on x by Df = Cf

′(x), where C is a nonzero constant. For any

f ∈ R(X) \ R(X), there exists a sequence fn ∈ R(X) converging uni-formly to it on X, and hence fn is a uniform Cauchy sequence. Then∀ε > 0, there exists an positive integer Nε s.t. whenever n, n′ > Nε, we have|f ′n(x) − f ′n′(x)| 6 k · ‖fn − fn′‖ < ε. So the numerical sequence f ′n(x) isconvergent, i.e. limn→∞ f

′n(x) is well-defined. We can extend D to D on

R(X) as:

Df =

Cf ′(x) if f ∈ R(X),

C limn→∞ f′n(x) if f ∈ R(X) \R,

where C is a nonzero constant. It is easy to see D is a point derivation.Also, |Df | = C limn→∞ |f

′n(x)| 6 C limn→∞ k‖fn‖ = Ck‖f‖. Hence D is a

bounded point derivation on R(X).

Conversely, suppose D is a bounded point derivation on R(X) at x. D1 =1D1 + 1D1 implies D1 = 0. D is linear implies Da = 0, for any com-plex number a. We set Dx = K, where K is a complex number. ThenDx2 = xDx + xDx = C2x. By iteration, we have Dxn = Cnxn−1

. Let1g ∈ R(X), 0 = D(gg ) = g(x)D(1

g ) + 1g(x)

Dg implies D(1g ) = − Dg

g(x)2. Let

fg ∈ R(X), D(fg ) = 1

g(x)Df + f(x)D(1

g ) = g(x)Df−f(x)Dgg(x)2

. We conclude

that D on R(X) at x is of the form Df = Cf′(x), where C is a nonzero

constant. D is bounded implies |Df | 6 C ′ · ‖f‖, for all f ∈ R(X), whereC ′ is a nonzero constant. Hence, |f ′(x)| 6 C′

C · ‖f‖, for all f ∈ R(X).

Inspired by the above theorem, Hallstrom ([10]) extended the notion ofbounded point derivation to higher orders as : for a positive integer t, thereis a bounded point derivation of order t on R(X) at x if there is a con-stant k such that |f (t)(x)| ≤ k · ‖f‖ for all f ∈ R(X). With the aid ofMelnikov’s estimate, he obtained the following characterization for boundedpoint derivation of higher orders in terms of analytic capacity.

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Theorem 2.4.4 Let An = z ∈ C : 2−(n+1) ≤ |z − ζ| ≤ 2−n. Thereis a bounded point derivation of order t on R(X) at x ∈ X if and only if∑∞

n=0(2t+1)nγ(An(x) \X) <∞. [10, Theorem 1’]

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Chapter 3

Finely HolomorphicFunctions

3.1 Definition and some properties of finely holo-morphic functions

Definition 3.1.1 Let U be a finely open subset of C. Let f : U → C.(a) f is said to be finely continuous (fine-to-Euclidean) if for every opensubset V of C, the set f−1(V ) is a finely open subset of U .(b) Let z is a fine limit point of U , we say L is the fine limit of f at z if forevery open neighborhood N of L, there exists a finely open neighborhoodV of z such that f((V ∩ U) \ z) ⊂ N . We write:

F- limω→z

f(ω) = L.

(c) Let z ∈ U , if the fine limit of f(ω)−f(z)ω−z in the direction ω → z exists,

then the fine derivative of f at z, denoted by F-f′(z), is defined as:

F-f′(z) = F- lim

ω→z

f(ω)− f(z)

ω − z.

In this case, f is said to be finely differentiable at z.

Regarding the above definitions, a function which is finely differentiableon U is finely continuous on U . All subharmonic functions on C are finelycontinuous. The following theorem which is called the Brelot property showsthat a finely continuous function is locally continuous .

Theorem 3.1.2 Let U be a finely open subset of C. Consider a count-able family of finely continuous functions fn : U → C. Every point of Uhas a fine neighbourhood V ⊂ U (V a compact set if we like) such that therestriction of each fn to V is continuous (Euclidean-to-Euclidean). [5]

18

The theory of holomorphic functions defined on domains (open and con-nected subsets) in C dominates the classical complex analysis. After thegeneralisation of domains to fine domains, it is natural to attempt to definea class of functions on fine domains which are analogous to holomorphicfunctions. The following definition of finely homomorphic functions fromthe analytic point of view inherits the original idea of defining functionswhich are complex differentiable as holomorphic functions.

Definition 3.1.3 Let U be a finely open subset of C. A function f : U → Cis said to be finely holomorphic if f is finely differentiable on U and itsfine derivative is finely continuous on U .[15]

The following theorem provides a local characterization of finely holomor-phic functions and can be used as alternative definitions.

Theorem 3.1.4 Let U be a finely open subset of C. The following state-ments are equivalent:(a) A function f : U → C is finely holomorphic.(b) Every point of U has a compact fine neighbourhood V ⊂ U , such thatthe restriction f |V belongs to R(V ).(c) Every point of U has a fine neighbourhood V ⊂ U on which f coincideswith the Cauchy-Pompeiu transform of some function ϕ ∈ L2

c(C) such thatϕ = 0 a.e. in V :

f(z) =

∫C

1

z − ζϕ(ζ)dλ(ζ), z ∈ V

where λ denotes the Lebesgue measure on C, L2c(C) denotes the square

(Lebesgue) integrable complex-valued functions on C with compact sup-port. [4]

Theorem 3.1.5 Let U be a finely open subset of C. Any finely holomorphicfunction f : U → C is finely infinitely differentiable, and the fine derivativesF-f (n) are finely holomorphic on U . [4, Proposition 9]

Theorem 3.1.6 (Cartan) If x is a fine limit point of the set E and g isan extended real-valued function having a fine limit λ at x, then there is afine neighbourhood V f of x such that lim y→x

y∈E\x∩V fg(y) = λ. [16, Theorem

10.15]

19

3.2 Rational approximation of finely holomorphicfunctions

Lemma 3.2.1 Let U ⊂ C be finely open and K ⊂ C be compact. For anyreal number α, the function hα defined by

hα(z) =

∫K\U|z − ζ|−αdλ(ζ), z ∈ U,

(a) is finite and finely continuous everywhere in U .(b) Every point of U therefore has a compact fine neighbourhood V in U inwhich each hα is bounded (even continuous), i.e.

Cα = supz∈V

hα(z) <∞.

[4, lemma 7].

In the following theorem, we make a convention to let f (0) = f .

Theorem 3.2.2 Let U be a finely open subset of C. If f : U → C isfinely holomorphic, then every point of U has a compact fine neighbour-hood V ⊂ U satisfying: there exists a sequence of rational functions gj

with poles off V such that, for each integer n ≥ 0, the n’th derivative g(n)j

converges uniformly on V to the n’th fine derivative F-f (n) as j → ∞. [4,Theorem 11]Proof : By Theorem 3.1.4 (c), for any point z ∈ U , there exists a fineneighbourhood Uz ⊂ U such that:

f(z) =

∫C

1

z − ζϕ(ζ)dλ(ζ) =

∫S\Uz

1

z − ζϕ(ζ)dλ(ζ), z ∈ Uz ,

where ϕ ∈ L2c(C) such that ϕ = 0 a.e. in Uz and with compact support S.

Let f (n), n = 0, 1, · · · , denote the functions obtained by differentiating theabove integral under the integral sign, i.e.,

f (n)(z) := (−1)nn!

∫S\Uz

1

(z − ζ)n+1ϕ(ζ)dλ(ζ), z ∈ Uz .

By Lemma 3.2.1, z has a compact fine neighbourhood V ⊂ Uz such that:for every integer n > 0,

Cn+1 = supz∈V

hn+1(z) = supz∈V

∫S\Uz

|z − ζ|−(n+1)dλ(ζ) <∞.

20

(1) Firstly, we prove that the fine derivatives F-f (n) equals to f (n) on V foreach n. We fix n = 0, 1, 2, · · · , consider function g(ω) = ω−n−1, ω ∈ C \ 0.The Taylor expansion of g(ω) at any z ∈ C \ 0 is given by :

g(ω) =∞∑k=0

1

k!(ω − z)kg(k)(z).

We define:

R(z, ω) := ω−n−1 −1∑

k=0

1

k!(ω − z)kg(k)(z)

= ω−n−1z−n−2[(n+ 1)ωn+2 − (n+ 2)zωn+1 + zn+2]

= ω−n−1z−n−2n+2∑i=0

Pi(z)(ω − z)i,

where Pi(z) are polynomials in z. By Taylor’s theorem, limω→zR(z,ω)ω−z = 0.

But R(z,ω)ω−z = ω−n−1z−n−2

∑n+2i=0 Pi(z)(ω − z)i−1. It yields that Pi = 0, i =

0, 1. Therefore, ∀w 6= z,

R(z, ω)

(ω − z)2= ω−n−1z−n−2

n∑j=0

Pj+2(z)(ω − z)j

= ω−n−1z−n−2∑i,j>0,j=n,i=n+2

aijziωj

=

r=n+2,s=n+1∑r,s>0

br,sz−rω−s.

where br,s are constant coefficients. Choose ω, z ∈ V such that ω 6= z,

1

(ω − z)2[f (n)(ω)− f (n)(z)− (ω − z)f (n+1)(z)]

=(−1)nn!

∫S\Uz

(ω − ζ)−n−1 − (z − ζ)−n−1 + (ω − z)(n+ 1)(z − ζ)−n−2

(ω − z)2ϕ(ζ)dλ(ζ)

=(−1)nn!

∫S\Uz

R(ω − ζ, z − ζ)

[(ω − ζ)− (z − ζ)]2ϕ(ζ)dλ(ζ)

=(−1)nn!

r=n+2,s=n+1∑r,s>0

br,s

∫S\Uz

(z − ζ)−r(ω − ζ)−sϕ(ζ)dλ(ζ).

21

By Lemma 3.2.1 and Holder’s inequality, we have:

|∫S\Uz

(z − ζ)−r(ω − ζ)−sϕ(ζ)dλ(ζ)|

6‖ϕ‖L2(

∫S\Uz

|z − ζ|−2r|ω − ζ|−2sdλ(ζ))1/2

6‖ϕ‖L2 [h4r(z)]1/4[h4s]

1/4

6‖ϕ‖L2C1/44r C

1/44s .

Then by triangle inequality, we have:

|(−1)nn!

r=n+2,s=n+1∑r,s>0

br,s

∫S\Uz

(z−ζ)−r(ω−ζ)−sϕ(ζ)dλ(ζ)| 6 n!‖ϕ‖L2

r=n+2,s=n+1∑r,s>0

|br,s|C1/44r C

1/44s .

Since the above sum has finite numbers of terms, there exists a positiveconstant A such that:

|f(n)(ω)− f (n)(z)

ω − z− f (n+1)(z)| 6 A‖ϕ‖L2 |ω − z|.

When n = 0, by Definition 3.1.1, ∀z ∈ V, F-f ′(z) = f (1)(z). By induc-tion, Ff (n)(z) = f (n)(z), ∀z ∈ V,∀n = 0, 1, · · · .

(2) Secondly, we prove that there exists a sequence of holomorphic functionsfj such that the fine-derivatives F-f (n) can be uniformly approximated by

f(n)j , ∀n = 0, 1, 2, · · · .

Choose a decreasing sequence of open sets ωj , j = 1, 2, · · · such that V ⊂ ωjand

⋂j=1 ωj = V . Define ϕj = 1C\ωj · ϕ on C, where 1 denotes the

indicator function.

‖ϕj − ϕ‖L2 = (

∫C|ϕj − ϕ|2dλ)1/2 = (

∫ωj\Uz

ϕ2dλ)1/2

‖ϕj − ϕ‖L2 → 0 as j →∞, since ωj is decreasing and⋂j=1 ωj = V ⊂ Uz .

Define a sequence of functions fj:

fj(z) =

∫C

1

z − ζϕj(ζ)dλ(ζ) =

∫S\ωj

1

z − ζϕ(ζ)dλ(ζ), z ∈ ωj .

Let a ∈ ωj , we can write 1z−ζ = 1

a−ζ1

1− z−aζ−a

= 1a−ζ

∑∞n=0( z−aζ−a)n. Then, fj(z)

can be written as a convergent power series:

fj(z) = −∞∑n=0

(z − a)n∫S\ωj

1

(ζ − a)n+1ϕ(ζ)dλ(ζ), z ∈ ωj .

22

Hence, fj is holomorphic on ωj , and infinitely differentiable on ωj . Let f(n)j

denote the n-th derivatives of fj , then

f(n)j (z) = (−1)nn!

∫C

1

(z − ζ)−(n+1)ϕj(ζ)dλ(ζ), n = 0, 1, 2, · · ·

Hence, by Holder’s inequality and Lemma 3.2.1, for every fixed n,

supz∈V|f (n)j (z)−F-f (n)(z)|

=supz∈V|f (n)j (z)− f (n)(z)|

=supz∈V

n!|∫S\Uz

(z − ζ)−(n+1)(ϕj(ζ)− ϕ(ζ))dλ(ζ)|

≤n! supz∈V

∫S\Uz

|z − ζ|−(n+1)|ϕj(ζ)− ϕ(ζ)|dλ(ζ)

≤n! supz∈V

(

∫S\Uz

|z − ζ|−2(n+1)dλ(ζ))1/2(

∫S\Uz

|ϕj(ζ)− ϕ(ζ)|2dλ(ζ))1/2

≤n!(C2(n+1))1/2‖ϕj − ϕ‖L2

C2(n+1) is finite and ‖ϕj − ϕ‖L2 → 0 as j → ∞ implies supz∈V |f(n)j (z) −

Ff (n)(z)| → 0 as j →∞.

Hence, we have a sequence of holomorphic functions fj defined on openneighbourhoods of compact set V such that, for each integer n ≥ 0, the n’th

derivative f(n)j converges uniformly on V to the n’th fine derivative F-f (n)

as j →∞.

Following the discussion in the third paragraph of Section 2.2.1, we canfind a sequence of desired rational functions.

Theorem 3.2.3A finely holomorphic function f , defined on a fine domain U , is uniquelydetermined by the sequence of its fine derivatives F-f (n)(z0), n ≥ 0, at anypoint z0 ∈ U . [4]

23

Chapter 4

Rational Approximation OnA Special Set

4.1 An example of fine domains

We compare two examples to serve as a starting point of investigating do-mains different from fine domains. Example 4.1.2 is a modification of [17,Example 2.4.2]. In this section we adhere to the notations in Example1.1.7 and Theorem 1.3.2.

Theorem 4.1.1 If a1 > a2 > · · · > 0, then the numerical series∑∞

n=1 an <∞ if and only if

∑∞k=1 2ka2k <∞.

Example 4.1.2 Let an = 12(2−n + 2−n−1), rn = 2−(n+1)3 , n = 1, 2, · · · .

We define U := D(0, 1) \⋃∞n>1 D(an, rn). We have 0 ∈ U since an − rn =

3/2−2−n3−3n2−2n

2n+1 > 0, ∀n.

U is a fine neighbourhood of 0 (finely open set containing 0). To showthis, we apply Wiener’s test (Theorem 1.3.2) on the set C \ U :

∑n>1

n

log(1/c∗(An ∩ (C \ U)))

=∑n>1

n

log(1/c∗(D(an, rn)))

=∑n>1

n

log(1/rn)

=∑n>1

n

(n+ 1)3 log 2

24

By ratio test,∑

k>1 2k 2k

(2k+1)3 log 2<∞, since

limk→∞

22k+2

(2k+1 + 1)3

(2k + 1)3

22k= lim

k→∞4(

1 + 2−k

2 + 2−k)3 =

1

2< 1.

Then,∑

n>1n

(n+1)3 log 2< ∞ by Theorem 4.1.1. It implies that C \ U is

thin at 0. Hence U is a fine neighbourhood of 0 by Theorem 1.3.3(c).

Example 4.1.3 We define U := D(0, 1) \⋃∞n>1 D(an, rn), where rn =

2−(n+1)(15/8) . We also have 0 ∈ U , but in contrast, U is not finely open.Again, we apply Wiener’s test on C \ U :∑

n>1

n

log(1/c∗(An ∩ (C \ U)))=

∑n>1

n

(n+ 1)(15/8) log 2.

By ratio test,∑

k>1 2k 2k

(2k+1)(15/8) log 2=∞, since

limk→∞

22k+2

(2k+1 + 1)(15/8)

(2k + 1)(15/8)

22k= lim

k→∞4(

1 + 2−k

2 + 2−k)15/8 =

4

215/8> 1.

Then,∑

n>1n

(n+1)(15/8) log 2= ∞ by Theorem 4.1.1. It implies that C \ U

is not thin at 0. Hence U is not finely open by Theorem 1.3.3(c).

4.2 An illustrating example

In this section, we first provide a concrete example of sets which are neitherin the fine topology nor in the Euclidean topology.

Example 4.2.1 We define G := D(0, 1) \⋃∞n>1 D(an, rn), where rn =

2−(n+1)(15/8) . We have 0 ∈ G since

an − rn =

32 −

1

2(n+1)[(n+1)7/8−1]

2n+1> 0,∀n.

G is not open since 0 is a boundary point, G is not closed neither. G isnot finely open by the argument in Example 4.1.3, and G is not finelyclosed by Wiener’s criterion. We also notice that the fine interior and theEuclidean interior of G coincide.

Next, we define a basis B and the topology τ generated by B on G whichwill be called G-basis and G-topology respectively. Although G-topology isnot finer than the fine topology and G-holomorphic functions which will be

25

defined in the next section are not more general than holomorphic functions,we hope our attempt serves as a starting point to investigate more generalcases beyond holomorphic functions. Hence, G stands for general in thischapter. It is easy to check our following definition that G-basis is indeed abasis for a topology on G.

Definition 4.2.2 The G-basis for G is the collection B of subsets of Gsuch that:(1) For each x ∈ G, x 6= 0, all the discs D(x, r) ⊂ G belong to B.(2) For 0, sets D(0, R) \ ∪∞n>1D(an, r

′n) belong to B if

(2.1) 0 < R < 1;(2.2) ∀n, 0 /∈ D(an, r

′n);

(2.3) there exists a δ > 0 such that rn 6 r′n 6 2−(n+1)1+δ .

Definition 4.2.3 We define G-topology τ as the topology generated by theG-basis, that is: a subset U of G is in τ if for each x ∈ U , there is a basiselement B ∈ B such that x ∈ B and B ⊂ U .

G becomes a topological space with the G-topology τ . A set in τ will becalled a G-open set. Let τ ′ denote the fine topology on C. We have thesubspace fine topology τ on G as τ = G ∩ U |U ∈ τ ′. The basis elementsof 0 are not finely open, so the G-topology τ and the subspace fine topologyτ do not coincide. On the other hand, the subspace fine topology is notincluded in τ neither. Similar reasoning shows τ is also different with theEuclidean topology.

Definition 4.2.4 A function f : G → C is said to have a G-derivativeat z ∈ G, which is denoted by G-f ′(z), if for any ε > 0, there exists aG-neighbourbood V ⊂ G of z such that:

|f(z)− f(z)

z − z− Gf ′(z)| < ε,∀z ∈ V.

An interesting question arises from here. Can we define a class of functionson G that is in a sense analogous to finely holomorphic functions on finedomains? Can we then obtain an approximation theorem similar to the the-orem of rational approximation of finely holomorphic functions (Theorem3.2.2)? This will be discussed in the next section.

4.3 Rational approximation on the illustrating ex-ample

We recall thatG := D(0, 1)\⋃∞n>1 D(an, rn) and U := D(0, 1)\

⋃∞n>1 D(an, rn),

where rn = 2−(n+1)(15/8) . Since our main concern is rational approximation,

26

we have the following conjecture.

Conjecture 1: Let f : G → C be in a certain class of functions (Wewill speak of G-holomorphic functions), then every point of G has a G-neighbourhood V ⊂ G satisfying: there exists a sequence of rational func-tions gj with poles off V such that, for each integer n > 0, the n′th derivative

g(n)j converges uniformly to the n′th G-derivative Gf (n) as j → ∞ on the

closure of V (with respect to the Euclidean topology), denoted by V .

We notice that if f is a G-holomorphic function, it should at least have G-derivatives of all orders at every point z ∈ G relative to a G-neighbourhoodcontaining z. We discuss three classes of functions which are likely to fitinto this conjecture.

Case 1 A function f defined on G is said to be a G-holomorphic func-tion if f ∈ R(U), where R(U) is the uniform closure of the space of rationalfunctions defined on U with poles off U .

Case 2 f is said to be a G-holomorphic function, if every point of G hasa Gneighbourhood V ( G such that f |V ∈ R(V ), where R(V ) is the uni-form closure of the space of rational functions defined on V with poles off V .

Case 3 f is said to be a G-holomorphic function, if f has the G-derivativeat every point in G and its G-derivative is continuous on G relative to theG-topology.

Discussion on Case 1Let A(U) denotes the functions which are continuous on U and holomorphicon the interior of U , denoted by U. Gamelin showed that A(U) = R(U)([18,page 52]). Let f ∈ R(U) and fn be the sequence of rational functions con-verging uniformly to f on U . We know f is holomorphic on the interiorof U , but the behaviour of f on the boundary of U is unknown to us. Inparticular, f may not have the derivative at 0 relative to G.

We need to investigate the existence of G-derivatives of f on G. For anyz ∈ G but z 6= 0, f has the derivatives of all orders, and hence has G-derivatives of all orders. Therefore we only need to check the existence ofthe G-derivative of f at 0.

Lemma 4.3.1 There exist bounded point derivations of all orders of R(U)at 0.

27

Proof : We have

∞∑n=0

(2t+1)nγ(An(0) \ U)

=

∞∑n=1

(2t+1)nγ(D(an, rn))

=

∞∑n=1

(2t+1)nrn

=

∞∑n=1

(2t+1)n2−(n+1)(15/8)

By Cauchy’s root test,∑∞

n=1(2t+1)n2−(n+1)(15/8) <∞,∀t = 1, 2, · · · , since

limn→∞

[2n(t+1)2−(n+1)15/8 ]1/n = limn→∞

2(t+1)2−(n+1n

)15/8n7/8= 0.

By Theorem 2.4.4,∑∞

n=0(2t+1)nγ(An(0) \ U) < ∞, t = 1, 2, · · · impliesthat there exists a bounded point derivation of order t of R(U) at 0, t =1, 2, · · · .

LetD be the bounded point derivation onR(U) at 0. ThenDf = limn→∞ f′n(0)

by the proof of Theorem 2.4.3. For a base element B ( G in the G-basiscontaining 0, we define the function:

F (z) =

f ′(z) if z ∈ B \ 0,Df = limn→∞ f

′n(0) if z = 0,

where D is the bounded point derivation of R(U) at 0.

We therefore make two conjectures.Conjecture 2 F (z) is the G-derivative of f relative to B.Conjecture 3 F ∈ R(B).

If we prove these two, then we can prove our Conjecture 1. The fol-lowing proposition is the proof of Conjecture 2.

Proposition 4.3.1: There exists a base element B ( G in the G-basiscontaining 0 such that f has the G-derivative at 0 relative to B, whichequals to Df .Proof : Put P = ∪∞j>1∂D(aj , rj) ∪ z : |z| = 1. We derive the integralrepresentation of f(z) on G.

28

For z = 0, we know f is continuous at 0. Since fn(z) is a rational func-tion with poles off G, by Cauchy’s formula, we have

fn(0) =1

2πi

∫P

fn(ζ)

ζdζ, ∀n.

Since fn(z) converges uniformly on U to f , we have ∀ε > 0, there exists aN > 0 such that whenever n > N , supζ∈U |fn(ζ)−f(ζ)| < ε. Hence we have

|∫P

fn(ζ)

ζdζ −

∫P

f(ζ)

ζdζ|

6∫P

|fn(ζ)− f(ζ)||ζ|

d|ζ|

<ε(∞∑j=1

∫∂D(aj ,rj)

1

|ζ|d|ζ|+

∫z:|z|=1

1

|ζ|d|ζ|)

6ε(∞∑j=1

2πrjaj − rj

+ 2π)

We perform the ratio test:

limj→∞

rj+1

aj+1 − rj+1

aj − rjrj

= limj→∞

2−(j+2)15/8

32− 1

2(j+2)[(j+2)7/8−1]

2j+2

32− 1

2(j+1)[(j+1)7/8−1]

2j+1

2−(j+1)15/8

= limj→∞

2−(j+2)15/82j+2

32 −

1

2(j+2)[(j+2)7/8−1]

(32 −

1

2(j+1)[(j+1)7/8−1])2(j+1)15/8

2j+1

= limj→∞

2−(j+2)15/8+(j+1)15/8+1

32 −

1

2(j+1)[(j+1)7/8−1]

32 −

1

2(j+2)[(j+2)7/8−1]

=0.

So,∞∑j=1

2πrjaj − rj

<∞.

Hence, we have ∀ε > 0, there exists a N > 0 such that whenever n > N ,

|∫P

fn(ζ)

ζdζ −

∫P

f(ζ)

ζdζ| < Cε,

where C is a positive constant. Hence, the sequence 12πi

∫Pfn(ζ)ζ dζ converges

to 12πi

∫Pf(ζ)ζ dζ.

29

Since fn(z) converges uniformly on U to f , we have,

f(0) = limn→∞

fn(0) = limn→∞

1

2πi

∫P

fn(ζ)

ζdζ =

1

2πi

∫P

f(ζ)

ζdζ.

For z ∈ G, but z 6= 0, by the Cauchy’s formula we have

f(z)=1

2πi

∫P

f(ζ)

ζ − zdζ.

For each rational function fn, again by Cauchy’s formula we have:

f′n(0) =

1

2πi

∫P

fn(ζ)

ζ2dζ.

Since fn(z) converges uniformly on U to f , by the same argument on lastpage we have:

limn→∞

f′n(0) = lim

n→∞

1

2πi

∫P

fn(ζ)

ζ2dζ =

1

2πi

∫P

f(ζ)

ζ2dζ.

Next, we consider:

|f(z)− f(0)

z − 0− limn→∞

f′n(0)|

=1

2π|∫P

f(ζ)

z(ζ − z)dζ −

∫P

f(ζ)

zζdζ −

∫P

f(ζ)

ζ2dζ|

=1

2π|∫P

f(ζ)z

ζ2(ζ − z)dζ|

=|z|2π|∫P

f(ζ)

ζ2(ζ − z)dζ|

6|z|2π

maxζ∈U|f(ζ)|

∫P

1

|ζ2(ζ − z)|d|ζ|

=|z|2π

maxζ∈U|f(ζ)|(

∞∑j=1

∫∂D(aj ,rj)

1

|ζ2(ζ − z)|d|ζ|+

∫∂D(0,1)

1

|ζ2(ζ − z)|d|ζ|).

Let ds denotes arc length, by Holder’s inequality we have:∫∂D(aj ,rj)

1

|ζ2(ζ − z)|d|ζ|

=

∫|ζ−aj |=rj

ds

|ζ2(ζ − z)|

6(

∫|ζ−aj |=rj

ds

|ζ|4)1/2(

∫|ζ−aj |=rj

ds

|ζ − z|2)1/2

30

It is not difficult to compute that:∫|ζ−aj |=rj

ds

|ζ − z|2=

2πrj|z − aj |2 − (rj)2

,

∫|ζ−aj |=rj

ds

|ζ|46

2πrj|aj − rj |4

.

Then we have:

|f(z)− f(0)

z − 0− limn→∞

f′n(0)|

6|z|maxζ∈U|f(ζ)|(

∞∑j=1

rj|aj − rj |2

(1

|z − aj |2 − r2j

)1/2 +

∫∂D(0,1)

1

|ζ2(ζ − z)|d|ζ|).

Let B = D(0, R) \ ∪∞n>1D(aj , r′j) be a Gbasis element of 0, which is small

enough. All z in B are close to 0, so∫∂D(0,1)

1

|ζ2(ζ − z)|d|ζ| < C ′,

where C ′ is a positive constant. We also have

∞∑j=1

rj|aj − rj |2

(1

|z − aj |2 − r2j

)1/2 6∞∑j=1

rj|aj − rj |2

(1

r′2j − r2

j

)1/2 <∞,

since the ratio test shows

limj→∞

rj+1

rj(

aj − rjaj+1 − rj+1

)2[r′2j − r2

j

r′2j+1 − r2

j+1

]1/2

= limj→∞

2−(j+2)15/8+(j+1)15/8+2(

32 −

1

2(j+1)[(j+1)7/8−1]

32 −

1

2(j+2)[(j+2)7/8−1]

)2(2−2(j+1)η − 2−2(j+1)15/8

2−2(j+2)η − 2−2(j+2)15/8)1/2

= limj→∞

2−(j+2)15/8+(j+1)15/8+2(

32 −

1

2(j+1)[(j+1)7/8−1]

32 −

1

2(j+2)[(j+2)7/8−1]

)22−(j+1)η+(j+2)η(1− 2−2(j+1)15/8+2(j+1)η

1− 2−2(j+2)15/8+2(j+2)η)1/2

= limj→∞

2−(j+2)15/8+(j+1)15/8−(j+1)η+(j+2)η+2(

32 −

1

2(j+1)[(j+1)7/8−1]

32 −

1

2(j+2)[(j+2)7/8−1]

)2(1− 2−2(j+1)15/8+2(j+1)η

1− 2−2(j+2)15/8+2(j+2)η)1/2

=0,

where 1 < η < 158 is a constant.

Since f is continuous on U , it is bounded on U . Let K denote a positiveconstant, we have

|f(z)− f(0)

z − 0− limn→∞

f′n(0)| 6 K|z|.

31

Hence, ∀ε > 0, there exists a G-neighbourhood G′ ⊂ B with |z| < ε/K,∀z ∈G′ such that

|f(z)− f(0)

z − 0− limn→∞

f′n(0)| < ε.

Conjecture 3: There exists a base element B ( G in the G-basis con-taining 0 such that F ∈ R(B).

Unfortunately, we cannot prove it at the present due to limit of time. Butwe have found a promising way to continue our proof. We can extend f(z)into C \G. Then f is continuous over C and equal to 0 on a neighbourhoodof z = ∞. Recall that ∂

∂z = 12( ∂∂x −

1i∂∂y ). We also recall that Green’s

formula is ∫Ω

∂ϕ(z)

∂zdzdz =

∫∂Ωϕ(z)dz,

where Ω is a domain bounded by smooth arcs and ϕ(z) is a continuouslydifferentiable function. Then f has the integral representation:

f(z) =1

2πi

∫C

∂f

∂ζ

1

ζ − zdλ(ζ).

We set

−∂f∂ζ

= ϕ,

then we get

f(z) =1

2πi

∫Cϕ(ζ)

1

z − ζdλ(ζ).

We hope we can recover Lemma 3.2.1 on our situation and follow the wayin the proof of Theorem 3.2.2 to prove it later.

4.4 Discussion and open problems

As we can see, the last two cases of G-holomorphic functions resemble to theequivalent definitions of finely holomorphic functions. We then can ask thefollowing questions:

Problem 1 Will Case 1, Case 2 and Case 3 are equivalent?

We can easily see Case 1 implies Case 2 and Case 3, but other directionsneed some effort. Suppose the answer to Problem 1 is positive, then we canuse it to define a class of functions which can be called G-holomorphic func-tion. Theorem 3.2.3 shows that a finely holomorphic function is uniquely

32

determined by the sequence of its fine derivatives at any point of the finedomain where it is defined. Also, Borel proved that a monogenic functionis uniquely determined by the sequence of its derivatives at any point of areduced domain. Then we can ask:

Problem 2: Is a G-holomorphic function defined on G uniquely determinedby the sequence of its G-derivatives at any point of G? For example, let f, gbe two G-holomorphic functions such that G-fn(0) = G-gn(0), ∀n. Does thisimply f = g on G.

Remark 1 We have taken a specific radius in our example, but as we cansee, our Definition 4.2.2, 4.2.3 and Proposition 4.3.2 work for a rangeof radii as well.

Remark 2 The G-basis for our example is defined in order to obtain Propo-sition 4.3.2. It is possible to define some more general basis.

Remark 3 The boundaries of our example, which are circles, is crucialto obtaining the estimate in the proof of Proposition 4.3.2. But line inte-grals are hard to estimate in general. Melnikov’s estimate is the only resultknown to us of estimating Cauchy’s integral. In order to generalise our ex-ample, we expect to use area integrals and analytic capacity instead of lineintegrals.

33

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[3] E. Kaniuth, A Course in Commutative Banach Algebras, Springer, NewYork, 2009

[4] B. Fuglede, Sur les fonctions finement holomorphes, Ann. Inst. Fourier.31(1981), no.4, 57-88

[5] B. Fuglede, Finely holomorphic functions. A survey, Rev. RoumaineMath. Pures Appl. 33 (1988), no. 4, 283-295.

[6] P. Pyrih, Logarithmic capacity is not subadditive – a fine topology ap-proach,Commentationes Mathematicae Universitatis Carolinae, Vol. 33(1992), No. 1, 67-72

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