MERGELYAN TYPE THEOREMS FOR SOME FUNCTION SPACES

Let F be a relatively closed subset of the unit disc D. If A is any of the Hardy spaces Hp(D), 0 < p < ∞, A|F denotes the functions on F being uniform limits of elements from Hp(D). Let F̃ consist of all z ∈ D such that |f(z)| ≤ sup{|f(z)|z ∈ F} for any bounded analytic function in D. It is proved that A|F consist of all functions f that can be decomposed as f = u + v, where u belongs to Hp(D) and v is a uniformly continuous function on the set F̃ , analytic at interior points of F̃ . Let A be a linear space of analytic functions and F a subset of the complex plane C such that each f ∈ A is defined on F . We denote by A|F the functions being uniformly approximable on F by sequences from A. The aim with this paper is to give a partial solution to problem 8.5 no. 2 in [7]. If A is any of the classical Hardy spaces H(D), 0 < p <∞, our main result is that A|F coincides (modulo the approximating space) with a well defined algebra of uniformly continuous analytic functions on F . Before giving a precise formulation of the main result, we need some definitions. Let Cua(F ) denote the functions on F being analytic in its interior F 0 and admitting continuous extension to the extended complex plane C∪{∞}. If F is a compact subset of C and P consists of the polynomials, a famous theorem of S. N. Mergelyan [7] can be formulated as

Let A be a linear space of analytic functions and F a subset of the complex plane C such that each f ∈ A is defined on F .We denote by A| F the functions being uniformly approximable on F by sequences from A. The aim with this paper is to give a partial solution to problem 8.5 no. 2 in [7].If A is any of the classical Hardy spaces H p (D), 0 < p < ∞, our main result is that A| F coincides (modulo the approximating space) with a well defined algebra of uniformly continuous analytic functions on F .
Before giving a precise formulation of the main result, we need some definitions.
Let C ua (F ) denote the functions on F being analytic in its interior F 0 and admitting continuous extension to the extended complex plane C ∪{∞}.If F is a compact subset of C and P consists of the polynomials, a famous theorem of S. N. Mergelyan [7] can be formulated as where F is the union of F and the bounded components of C/F .

A. Stray
Suppose now we replace P by the set H(C) consisting of all entire functions.Also allow F to be a closed but possibly unbounded subset of C. Then it can be proved that (I): where F is the union of F and certain components of C/F .A component V is to be included in F if and only if V ∪ {∞} is not arcwise connected in C ∪ {∞}.For details see [8], [9] and [10].
In general A may contain unbounded functions.For this reason it is natural to look for an identity like (I) if we seek to describe A|F in terms of uniformly continuous analytic functions.Let us use the notation g B = sup{|g(x)| : z ∈ B} if g is a function defined on the set B. We also define the hull of F with respect to A: We look for spaces A satisfying the following: ( * ): Our main result is the ( * ) is valid for the classical Hardy sapces H P (D) in the unit disc D, 0 < p < ∞, when F is any relatively closed subset of D. Also note that the two introductory examples are special cases of ( * ).We refer to [3] or [5] for the basic theory of Our main result is There are two special classes of sets F where a short proof of the decomposition of g can be found.It may be instructive to consider these cases prior to the general proof.
Let us firs assume that F is a Farrell set for H p (D). (See [8] for definition and various properties of these sets).Then we can find polynomials p n , n = 1, 2 such that as claimed.
In our second example, we assume that F can be written as a Blaschke sequence S = {ζ ν }, meaning that (1) Then it is well known that the Blashcke product and again we have a decomposition The geometric properties of the set F are quite different in the two cases just discussed.To clarify this, let F nt denote the non tangential closure of F on the unit circle where C may depend only on z.We also define It is well known that F is a Farrell set for H p (D) if and only if the linear measure |F t | of F t is zero ( [8]).On the other hand, the condition (1) is easily seen to imply that |F nt | = 0.
We have thus obtained the decomposition of H p (D)| F in two rather different situations.The general proof will be divided into parts reflecting the "geometry" of the cases considered above.We shall argue as in the proof where F was a Farrell set, but the polynomials p n will be replaced by functions from H p (D) having a uniformly continuous restriction to F .
The key part of the proof is an approximation argument related to the set F nt : Lemma 1.Given f ∈ H p (D), 0 < p < ∞, and > 0, there is an open set V and f 1 ∈ H p (D) with the following properties: For the moment we take Lemma 1 for granted.Consider the "tangential" part F t of F ∩ T .Let K be a compact subset of F t .We assume there is a number δ = δ(K) such that By a construction due to J. Detraz ([2, Prop.3.1]), we can find an outer function Moreover, G K extends to be continuous and non zero at any e iθ ∈ T \K.
We can find an increasing sequence of such sets K n ⊂ F t \V with corresponding outer functions G n , such that |F t \V \K n | → 0 and such that G = n G n has the following properties Consider finally the set It is evident that the linear meausre |L| of L is zero.By a general version of the Rudin-Carleson theorem ( [2]) there is H ∈ H ∞ (D) with continuous extension to L ∪ (T \L) such that H = 0 in D and H = 0 on L.
For n = 1, 2, . . .we consider the functions U n in H p (D) given by where f 1 satisfies the conclusions of Lemma 1.
It follows from the construction of f 1 , G and H, that U n | F is uniformly continuous.This implies that U n | F ∈ C ua ( F ), by the maximum principle.To be a little bit more specific, suppose z 0 ∈ F ∩ T and that U n (z) → 0 as z → z 0 and z ∈ F .Then |U n | < in F ∩ ∆(z 0 ), for some disc centered at z 0 .Choose a polynomial p peaking at z 0 such that |U n p| < on F .Then if p(z 0 ) = 1, we have lim sup z→z0 z∈ We turn to the proof of Theorem 1.If f ∈ H p (D) and > 0 is given, we have shown (modulo proving Lemma 1) that there is a function The proof of Theorem 1 now follows the introductory argument we gave in the special case where F is a Farrell set.
Let us finally prove Lemma 1.We may assume that f is bounded in D.
Since f ∈ H p (D), the radial limits f (e iθ ), 0 ≤ θ < 2π, belong to L p (dθ).We assume that K is included in the Lebesgue set for f and that uniformly in e iθ ∈ K as r → 0. Such a set K can be found with |F nt \K| as small we please.Fix δ > 0 so small that |f (e iθ ) − f (z)| < if e iθ ∈ K, |z| > 1 − δ and z ∈ T (θ, α).We are now in a convenient position for applying Vitushkin's scheme for approximation (see [13] or [4]).Let {∆ j } N j=1 be a finite collection of open discs with centers z j ∈ K and a common radius r < δ.Following Vitushkin's scheme, let As a preliminary approximation to f we define r j is a finite sum which we shall explain in some detail.
We assume f is defined outside of D by f (z −1 ) = f (z).For general properties of the T ϕ -operator we refer to [11] or [4, page 30].Here we only note that We assume ∂ϕj ∂z ≤ A r , where A is a numerical constant.Since f ∈ H p (D), we have in particular that f ∈ L p (dx dy) locally.Therefore the convolution term V j is continuous as a function of ς.If α is close to π, Hölders inequality gives that Note also that V j is analytic outside ∆ j .According to Vitushkin's scheme, the functions r j should be analytic outside a compact subset of ∆ j \D and with the property that (V j − r j )(ς) has a zero of order 3 at ∞.In addition we should require (4): where A 1 is a numerical constant.In our simple situation, the existence of {r j } is rather evident ([4, page 210-214]).From the individual bounds (3), it is part of Vitushkin's scheme that (5): for some numerical constant A 2 .We have not claimed that {∆ 1 j } N j=1 cover all of K.In fact we shall assume that ∆ j ∩ ∆ k = φ if j = k.In addition we assume that for some numerical constant A 3 , where ∆ and inspecting these four terms separately.
Note that the (Fatou) boundary values From ( 3) and ( 5) we also get if r is sufficiently small.The function f K satisfies the conditions for f 1 in Lemma 1 except that f K is only analytic (and hence continuous) near a subset P K of K.But since |P K | ≥ A 3 |K|, Lemma 1 follows by repeating our construction countably many times.The main reason why repetition works, is that the T ϕ-operator preserves continuity and analyticity ([4, page 30]).
It remains to show that This means that R(X)| F ⊂ B, where R(X) is the uniform closure on X by the rational functions with poles off X.
This solves completely problem 8.5 no. 2 in [7] for the space H p (D), 0 < p < ∞.For p = ∞ the problem is still open.
For p = ∞, some information about H ∞ | F can be obtained from the work by Carl Sundberg in [12].If f ∈ BMOA and f | F is bounded, Sundberg shows that f ∈ H ∞ | F .On the other hand, our proof above shows that any f ∈ H ∞ | F can be written as f = u + v with u ∈ H ∞ and v ∈ ∩ p>0 H p (D).Several questions arises from this.Here we only mention the following: Let f ∈ BMOA be bounded on a relatively closed set F ⊂ D.
Is there g ∈ H ∞ such that the restriction (f − g)| F is uniformly continuous on F ?