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Journal of Function Spaces and Applications
VolumeΒ 2012, Article IDΒ 905650, 15 pages
http://dx.doi.org/10.1155/2012/905650
Research Article

A Note on Some Uniform Algebra Generated by Smooth Functions in the Plane

1DΓ©partement de MathΓ©matiques, LMAM, UMR 7122, UniversitΓ© Paul Verlaine, Ile du Saulcy, 57045 Metz, France
2FakultÀt Allgemeinwissenschaften, Georg-Simon-Ohm-Hochschule Nürnberg, Keßlerplatz 12, 90489 Nürnberg, Germany

Received 13 March 2010; Accepted 19 October 2010

Academic Editor: N.Β Kalton

Copyright Β© 2012 Raymond Mortini and Rudolf Rupp. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We determine, via classroom proofs, the maximal ideal space, the Bass stable rank as well as the topological and dense stable rank of the uniform closure of all complex-valued functions continuously differentiable on neighborhoods of a compact planar set 𝐾 and holomorphic in the interior 𝐾∘ of 𝐾. In this spirit, we also give elementary approaches to the calculation of these stable ranks for some classical function algebras on 𝐾.

1. Introduction

Let 𝐾 be a compact subset of the complex plane β„‚. We denote by 𝐴1(𝐾) the uniform closure on 𝐾 of all complex-valued functions continuously differentiable on neighborhoods of 𝐾 and holomorphic in the interior 𝐾∘ of 𝐾. As usual, let 𝑃(𝐾) be the uniform closure on 𝐾 of the set of polynomials βˆ‘π‘›π‘—=0π‘Žπ‘—π‘§π‘—,𝑅(𝐾) the uniform closure on 𝐾 of those rational functions that have no poles in 𝐾, and 𝐴(𝐾) the uniform algebra of all continuous functions on 𝐾 that are holomorphic in 𝐾∘. Finally, 𝐢(𝐾) is the algebra of all complex-valued continuous functions on 𝐾. We obviously have that 𝑃(𝐾)βŠ†π‘…(𝐾)βŠ†π΄1(𝐾)βŠ†π΄(𝐾)βŠ†πΆ(𝐾). Whereas 𝑃(𝐾),𝑅(𝐾),𝐴(𝐾), and 𝐢(𝐾) are classical objects that are well understood (see, e.g., the books by Browder [1], and Gamelin [9]), we do not know a paper or textbook dealing with this intermediate algebra 𝐴1(𝐾).

In the present paper we are mainly interested in solving the BΓ©zout equation βˆ‘π‘›π‘—=1π‘₯𝑗𝑓𝑗=1 in these algebras, with particular emphasis on 𝐴1(𝐾). By giving classroom proofs of our results, we hope that we will make the notions of stable ranks appearing below accessible to a larger group of analysts, especially students willing to work in function theory.

So let 𝐴 be any of the algebras above. The leading questions will be the following.(1)Suppose that (𝑓,𝑔) is a pair of functions in 𝐴 such that 𝑓 and 𝑔 have no common zeros on 𝐾. Does there exist a solution (π‘₯,𝑦)∈𝐴2 of the BΓ©zout equation π‘₯𝑓+𝑦𝑔=1 such that π‘₯ is invertible?(2)Let (𝑓,𝑔)∈𝐴2 be arbitrary. Is it possible to uniformly approximate (𝑓,𝑔) by a pair (𝑒,𝑣) of functions in 𝐴2 such that 𝑒 and 𝑣 have no common zeros on 𝐾?

These questions originally stem from algebraic 𝐾-theory. In the abstract setting of commutative rings or Banach algebras, they run under the heading β€œBass stable rank” and β€œtopological stable rank,” and go back to Bass and Rieffel.

The intention of our paper is now twofold. We first present an elementary approach to the calculation of the stable ranks for the classical algebras 𝑃(𝐾),𝑅(𝐾), and 𝐢(𝐾) without using sophisticated methods or notions from algebraic topology and without using deep results as the Arens-Taylor-Novodvorsky theory (see the introduction to [5], for details of this theory). These results were known.

Then, we apply our techniques to the new algebra 𝐴1(𝐾). So, in Section 4 we determine the maximal ideal space of 𝐴1(𝐾), and the 𝐴1(𝐾)-convex sets. These results will be used in Section 5 to determine the Bass, topological, and dense stable rank of 𝐴1(𝐾), which represent the main results of our paper.

2. The Central Definitions

Here we recall those definitions necessary to understand this paper. Let 𝑋 be a compact Hausdorff space, and let 𝐢(𝑋) be the algebra of all complex-valued continuous functions on 𝑋 endowed with the supremum norm. A uniform algebra 𝐴 on 𝑋 is a uniformly closed subalgebra of 𝐢(𝑋) separating the points of 𝑋 and having as unit the constant function 1. Its maximal ideal space, or spectrum, 𝑀(𝐴), is the set of all nonzero, multiplicative linear functionals on 𝐴. As usual we will identify a function π‘“βˆˆπ΄ with its Gelfand transform 𝑓 defined on 𝑀(𝐴) by 𝑓(π‘š)=π‘š(𝑓).

In the sequel, let β€–π‘“β€–πΈβˆΆ=supπ‘§βˆˆπΈ|𝑓(𝑧)|. A closed subset 𝐸 of the spectrum 𝑀(𝐴) of 𝐴 is called 𝐴-convex, if 𝐸 coincides with its 𝐴-convex hull:||||𝐸=π‘šβˆˆπ‘€(𝐴)βˆΆπ‘“(π‘š)≀max𝐸||𝑓||ξ‚Όβˆ€π‘“βˆˆπ΄.(2.1)

It is well known that 𝐸 can be identified with the spectrum of the algebra 𝐴|𝐸 (see [9, page 39]). Let us also note that 𝐾𝑀(𝑃(𝐾))= [9, page 27]. In this setting||||𝐾=π‘§βˆˆβ„‚βˆΆπ‘“(𝑧)≀max𝐾||𝑓||ξ‚Όβˆ€π‘“βˆˆπ‘ƒ(𝐾)(2.2) and 𝐾 is called the polynomial convex hull of 𝐾. Note that 𝐾 is the union of 𝐾 with all the bounded components of ℂ⧡𝐾, which we will call holes.

An 𝑛-tuple (𝑓1,…,𝑓𝑛)βˆˆπ΄π‘› is said to be invertible (or unimodular) if there exists (π‘₯1,…,π‘₯𝑛)βˆˆπ΄π‘› such that βˆ‘π‘›π‘—=1π‘₯𝑗𝑓𝑗=1. The set of all invertible 𝑛-tuples is denoted by π‘ˆπ‘›(𝐴). An (𝑛+1)-tuple (𝑓1,…,𝑓𝑛,𝑔)βˆˆπ‘ˆπ‘›+1(𝐴) is called reducible if there exists (π‘Ž1,…,π‘Žπ‘›)βˆˆπ΄π‘› such that (𝑓1+π‘Ž1𝑔,…,𝑓𝑛+π‘Žπ‘›π‘”)βˆˆπ‘ˆπ‘›(𝐴).

The Bass stable rank of 𝐴, denoted by bsr(𝐴), is the smallest integer 𝑛 such that every element in π‘ˆπ‘›+1(𝐴) is reducible. If no such 𝑛 exists, then bsr(𝐴)=∞.

The topological stable rank, tsr(𝐴), of 𝐴 is the least integer 𝑛 for which π‘ˆπ‘›(𝐴) is dense in 𝐴𝑛, or infinite if no such 𝑛 exists. This notion goes back to Rieffel [19].

The following concept was introduced by Corach and SuΓ‘rez [7, page 542]. A version from [18] reads as follows.

The dense stable rank, dsr(𝐴), of 𝐴 is the least integer 𝑛 such that for every 𝐴-convex set 𝐸 in the character space of 𝐴 the Gelfand transform of any 𝑛-tuple (𝑓1,…,𝑓𝑛)βˆˆπ΄π‘› satisfying βˆ‘π‘›π‘—=1|𝑓𝑗|β‰₯𝛿>0 on 𝐸 can be uniformly approximated on 𝐸 by the Gelfand transform of 𝑛-tuples (π‘Ž1,…,π‘Žπ‘›) that are invertible in 𝐴.

It is well known that bsr(𝐴)≀dsr(𝐴)≀tsr(𝐴) (see [7, 17]).

Stable ranks of various real or complex function algebras have mainly been determined by Corach and SuΓ‘rez [4, 6, 7], Rupp [20–22], Rupp and Sasane [24, 25], Mikkola and Sasane [14], and Mortini and Wick [16, 17].

3. The Stable Ranks of 𝑃(𝐾),𝑅(𝐾),and𝐢(𝐾): A Classroom Approach

As a major tool, we use the following lemma, due to Corach and SuΓ‘rez [4, 6].

Lemma 3.1 (see [4, page 636] and [6, page 608]). Let 𝐴 be a commutative unital Banach algebra. Then, for π‘”βˆˆπ΄, the set 𝑅𝑛𝑓(𝑔)=ξ€½ξ€·1,…,π‘“π‘›ξ€Έβˆˆπ΄π‘›βˆΆξ€·π‘“1,…,𝑓𝑛,π‘”π‘–π‘ π‘Ÿπ‘’π‘‘π‘’π‘π‘–π‘π‘™π‘’(3.1) is open-closed inside 𝐼𝑛𝑓(𝑔)=ξ€½ξ€·1,…,π‘“π‘›ξ€Έβˆˆπ΄π‘›βˆΆξ€·π‘“1,…,𝑓𝑛,π‘”βˆˆπ‘ˆπ‘›+1ξ€Ύ(𝐴).(3.2)
In particular, for 𝑛=1, if πœ™βˆΆ[0,1]→𝐼1(𝑔) is a continuous curve and (πœ™(0),𝑔) is reducible, then (πœ™(1),𝑔) is reducible.

In the case where 𝑛=1 a pretty proof goes as follows (see [17]).

Let (π‘₯,𝑦)∈𝐴2 be so that 1=π‘₯𝑓+𝑦𝑔. Suppose that 𝑓𝑛→𝑓 and (𝑓𝑛,𝑔) is reducible, say 𝑓𝑛+π‘’π‘›π‘”βˆˆπ΄βˆ’1. Write𝑓𝑛+𝑒𝑛𝑔=𝑓π‘₯𝑓𝑛+𝑦𝑓+π‘¦π‘”π‘›ξ€Έβˆ’π‘“+𝑒𝑛𝑔.(3.3)

Since β€–π‘“π‘›βˆ’π‘“β€–π΄β†’0, we may choose 𝑛0 so big that for all 𝑛β‰₯𝑛0 the elements π‘₯𝑓𝑛+𝑦𝑔 are invertible in 𝐴. Hence 𝑓+β„Žπ‘›π‘” is invertible for some β„Žπ‘›. Thus, 𝑅1(𝑔) is closed within 𝐼1(𝑔). Now if 𝑒=𝑓+β„Žπ‘” is invertible, then every small perturbation of 𝑒 is invertible, too. Thus, 𝑅1(𝑔) is open.

3.1. The Algebras  𝑅(𝐾) and 𝑃(𝐾)

Corach and SuΓ‘rez showed in [4] that bsr(𝑃(𝐾))=bsr(𝑅(𝐾))=1. In our opinion, their proof on page 638 contains a gap, since, in general, the boundary of a component of 𝐾 is not accessible via a path staying outside 𝐾. For example, let 𝐾 be the union of the closed disk 𝔻 and a spiral outside 𝔻 that clusters at 𝕋=πœ•π”».

So, for the reader’s convenience, we present a short proof here (that is very close to the original one, though). We begin with a Lemma, used in [4], without a proof. We think that this needs a proof though, since a priori it is not clear that a domain admits accessible boundary points in the sense described below.

Lemma 3.2. Let πΎβŠ†β„‚    be compact, π‘ŽβˆˆπΎβˆ˜. If π‘”βˆˆπ΄(𝐾) does not vanish at π‘Ž, then there exist π‘βˆˆπœ•πΎ and a piecewise 𝐢1-path outside the zero set of 𝑔 joining π‘Ž and 𝑏 within 𝐾∘βˆͺ{𝑏}.

Proof. Let 𝐺 be the connected component of π‘Ž in 𝐾∘. Since 𝑔(π‘Ž)β‰ 0, the maximum principle implies that 𝑔 does not vanish identically on the boundary of 𝐺, a set that is contained in 𝐾. Let π‘€βˆˆπœ•πΊ with 𝑔(𝑀)β‰ 0. Choose a disk π‘ˆ centered at 𝑀 so that 𝑔≠0 on π‘ˆβˆ©πΎ. Let π‘€ξ…žβˆˆπ‘ˆβˆ©πΊ. Since, as an open set, 𝐺 is pathconnected, there is a path 𝛾1 in 𝐺 joining 𝑀′ with π‘Ž. Since at most finitely many zeros of 𝑔 could belong to such a path, we can easily avoid these zeros by perturbing a little bit the original path 𝛾1. Next, on the segment joining 𝑀′ with 𝑀, there is a first boundary point of 𝐺, say 𝑏. The combined path now joins π‘Ž with 𝑏 and stays outside the zero set of 𝑔.

Theorem 3.3 (Corach-SuΓ‘rez). Let πΎβŠ†β„‚β€‰β€‰be compact. Then,(1)bsr(𝑅(𝐾))=1; (2)tsr(𝑅(𝐾))=1   if and only if 𝐾∘=βˆ…;(3)tsr(𝑅(𝐾))=2   if and only if πΎβˆ˜β‰ βˆ….

Proof. (1) Let (𝑓,𝑔) be an invertible pair in 𝑅(𝐾), and let (π‘Ÿπ‘›) be a sequence of rational functions with poles outside 𝐾 uniformly approximating 𝑓 on 𝐾. By Lemma 3.1, in order to show the reducibility of (𝑓,𝑔), it suffices to show the reducibility of (π‘Ÿπ‘›,𝑔) for 𝑛 large. Now π‘Ÿπ‘›=𝑝/π‘ž for some polynomials 𝑝 and π‘ž. We may assume that 𝑝 and π‘ž have no common zeros. Obviously, π‘ž is invertible in 𝑅(𝐾). Also, since the reducibility of (𝑓1,𝑔) and (𝑓2,𝑔) implies that of (𝑓1𝑓2,𝑔), it suffices to show that invertible pairs of the form (π‘§βˆ’π‘Ž,𝑔) are reducible. We have to deal with three cases.Case 1. Let π‘Žβˆˆβ„‚β§΅πΎ. Then, π‘§βˆ’π‘Ž is invertible in 𝑅(𝐾) and so (π‘§βˆ’π‘Ž,𝑔) is reducible.Case 2. Let π‘Žβˆˆπœ•πΎ satisfy 𝑔(π‘Ž)β‰ 0. Choose a sequence, (π‘Žπ‘›), of points outside 𝐾 converging to π‘Ž. Then, the pairs (π‘§βˆ’π‘Žπ‘›,𝑔) are reducible and hence, by Lemma 3.1, (π‘§βˆ’π‘Ž,𝑔), too.Case 3. Now let π‘ŽβˆˆπΎ0 and 𝑔(π‘Ž)β‰ 0. Choose according to Lemma 3.2 a curve 𝛾 in 𝐾, joining π‘Ž with a boundary point 𝑏 of 𝐾 such that 𝛾 stays outside the zero set of 𝑔. The curve πœ™βˆΆ[0,1]→𝐼1(𝑔), given by πœ™(𝑑)=π‘§βˆ’π›Ύ(𝑑), is a continuous curve, and (πœ™(1),𝑔)=(π‘§βˆ’π‘,𝑔) is reducible. Thus, by Lemma 3.1, (πœ™(0),𝑔)=(π‘§βˆ’π‘Ž,𝑔) is reducible. (2), (3) We first note that tsr(𝑅(𝐾))≀2. In fact, let (𝑓,𝑔) be a pair of functions in 𝑅(𝐾). Then, for any πœ€>0, there exist two rational functions π‘Ÿπ‘“,π‘Ÿπ‘” with poles off 𝐾 such that β€–β€–π‘“βˆ’π‘Ÿπ‘“β€–β€–πΎ+β€–β€–π‘”βˆ’π‘Ÿπ‘”β€–β€–πΎ<πœ€.(3.4) By slightly perturbing, if necessary, the zeros of π‘Ÿπ‘“, we may assume that π‘Ÿπ‘“ and π‘Ÿπ‘” have no zeros in common. Thus, (π‘Ÿπ‘“,π‘Ÿπ‘”) is an invertible pair and so tsr(𝑅(𝐾))≀2.
Now if 𝐾 has no interior points, then any π‘“βˆˆπ‘…(𝐾) can be uniformly approximated on 𝐾 by rational functions without poles and zeros on 𝐾. Hence, tsr(𝑅(𝐾))=1 if 𝐾∘=βˆ…. If πΎβˆ˜β‰ βˆ…, then, by Rouché’s theorem, the function π‘§βˆ’π‘Ž with π‘ŽβˆˆπΎβˆ˜ cannot be uniformly approximated by holomorphic functions invertible in a neighborhood of π‘Ž. Thus, tsr(𝑅(𝐾))β‰₯2. Keeping in mind that tsr(𝑅(𝐾))≀2, we deduce that tsr(𝑅(𝐾))=2.

A similar result holds of course for the algebras 𝑃(𝐾), too. Since we were unable to find the assertions on the topological stable rank in the literature, we provide them for the reader’s convenience. Recall that for compact sets in ℂ𝑛 the Bass stable rank for 𝑃(𝐾) was determined by Corach and SuΓ‘rez in [7].

Theorem 3.4. Let πΎβŠ†β„‚ be a compact set. Then,(1)bsr(𝑃(𝐾))=1; (2)tsr(𝑃(𝐾))=1 if and only if 𝐾∘=βˆ… and 𝐾 has no holes;(3)tsr(𝑃(𝐾))=2 if and only if πΎβˆ˜β‰ βˆ… or 𝐾 has holes.

Proof. All three assertions follow as above, by noticing that 𝑀(𝑃(𝐾)) coincides with the polynomial convex hull, 𝐾, of 𝐾 (i.e., the union of 𝐾 with all its holes) and that 𝑅(𝐾)=𝑃(𝐾)=‖⋅‖𝑃(𝐾).

Our original intention for the present paper was to give such an elementary proof for the algebra 𝐴(𝐾). Our method would have been to approximate π‘“βˆˆπ΄(𝐾) by functions in 𝐴(𝐾) that have no zeros on πœ•πΎ. A β€œproof” of that claim that appeared in [13, page 268, Theorem  9] does not seem to be correct, though. In fact, even for functions 𝑓 in the disk algebra, one can have that 0 is an interior point of 𝑓(πœ•π”») [26], the image of the unit circle under 𝑓. Hence, contrary to the claim in that paper, there does not exist πœ‚ sufficiently small with πœ‚βˆ‰π‘“(πœ•π”») (note also the misprint in that paper, where the function 𝑔 in the assertion πœ‚βˆ‰π‘”(πœ•π”») should have been 𝑓). Thus, we were led to consider functions continuously differentiable in a neighborhood of 𝐾, where such a Peano-curve phenomenon does not occur.

We will need the following classical result based on Sard’s Lemma [12, page 81] that tells us that the image of the set of critical points of a 𝐢1-map 𝑓 in ℝ𝑛 has Lebesgue measure zero. As it is less time consuming to present a proof here than to browse through monographs not readily available, we add these few lines.

Theorem 3.5. Let π‘ˆβŠ†β„π‘›   be open and π‘“βˆΆβ„π‘›β†’β„π‘› a 𝐢1-map on π‘ˆ. Suppose that πΎβŠ†π‘ˆβ€‰β€‰is compact and nowhere dense. Then, 𝑓(𝐾)is nowhere dense, too.

Proof. Let 𝑀 be in the target space. We have to show that 𝑀 is not an interior point of 𝑓(𝐾). If π‘€βˆ‰π‘“(𝐾), then we are done. So let π‘€βˆˆπ‘“(𝐾). Consider the image 𝑓(𝐢) of the set 𝐢 of critical points of 𝑓 within a fixed compact neighborhood π‘Š of 𝐾. Then, 𝑓(𝐢) is a compact set having Lebesgue measure zero by Sard’s Lemma. Thus, there exist closed disks 𝐷𝑛 with centers 𝑀𝑛 and radii πœ€π‘› such that 𝑀𝑛→𝑀,πœ€π‘›β†’0, and π·π‘›βˆ©π‘“(𝐢)=βˆ…. It suffices to show that each of these disks contains points that do not belong to 𝑓(𝐾). Let 𝐸𝑛=πΎβˆ©π‘“βˆ’1(𝐷𝑛). Note that each point in 𝐸𝑛 is a regular point for 𝑓. Thus, we may cover 𝐸𝑛 by a finite number of closed disks π‘‰βŠ†π‘Š such that 𝑓 is a diffeomorphism of 𝑉 onto 𝑓(𝑉). By our hypothesis, πΈπ‘›βˆ©π‘‰ is nowhere dense in 𝑉. Thus, 𝑓(πΈπ‘›βˆ©π‘‰) is nowhere dense in 𝐷𝑛. Therefore, 𝑓(𝐸𝑛⋃)=𝑉𝑓(πΈπ‘›βˆ©π‘‰) is a finite union of nowhere dense, compact sets. Hence 𝑓(𝐸𝑛) is nowhere dense, too. Note that 𝑓(𝐸𝑛)βŠ†π·π‘› and so  𝑓(𝐸𝑛)=𝑓(𝐾)βˆ©π·π‘›. Thus, 𝑓(𝐾)βˆ©π·π‘› is nowhere dense. Since the disks 𝐷𝑛 are eventually arbitrarily close to 𝑀, we conclude that 𝑀 is not an interior point of 𝑓(𝐾).

3.2. The Algebra 𝐢(𝐾)

For arbitrary compact Hausdorff spaces 𝑋, Vasershtein [27] and Rieffel [19] gave the following formula for the Bass and topological stable ranks of 𝐢(𝑋,β„‚) and 𝐢(𝑋,ℝ) (see also [15], for a self-contained, easy proof).

Theorem 3.6. Let 𝑋 be a compact Hausdorff space. Then, tsr(𝐢(𝑋,β„‚))=bsr(𝐢(𝑋,β„‚))=dim𝑋2ξ‚„+1,tsr(𝐢(𝑋,ℝ))=bsr(𝐢(𝑋,ℝ))=dim𝑋+1.(3.5)

Using the additional fact from dimension theory that the covering dimension (or Čech-Lebesgue dimension) of πΎβŠ†β„‚ is less than or equal to one if and only if 𝐾 has no interior points, and two otherwise, it follows that the Bass stable rank of 𝐢(𝐾) is one whenever 𝐾∘=βˆ… and two if πΎβˆ˜β‰ βˆ…. The proofs of these facts on the covering dimension are rather technical. We would like to present two elementary approaches, independent of dimension theory. The first one determines directly the Bass stable rank of 𝐢(𝐾). The second approach determines the topological stable rank and then deduces the Bass stable rank. But first we present an elementary proof of Rieffel’s result [19] that the Bass and topological stable ranks of 𝐢(𝑋) coincide.

Theorem 3.7. Let 𝑋 be a compact Hausdorff space. Then, bsr(𝐢(𝑋))=tsr(𝐢(𝑋)).

Proof. Let 𝑁=bsr(𝐢(𝑋))<∞, and let 𝐹∢=(𝑓1,…,𝑓𝑁) be an 𝑁-tuple in 𝐢(𝑋). If 𝐹 is invertible, we are done. So we may assume that the 𝑓𝑗 have at least one zero in common. Consider the sets 𝐸𝑛=ξƒ―π‘₯βˆˆπ‘‹βˆΆπ‘ξ“π‘—=1||𝑓𝑗||(π‘₯)2β‰₯ξ‚€1𝑛2ξƒ°.(3.6) Choose by Urysohn’s Lemma a function π‘”π‘›βˆˆπΆ(𝑋) with 0≀𝑔𝑛≀1 such that 𝑔𝑛 vanishes identically on 𝐸𝑛 and is constant one on ⋂𝑁𝑗=1𝑍(𝑓𝑗). Then, the (𝑁+1)-tuple (𝑓1,…,𝑓𝑁,𝑔𝑛) is invertible in 𝐢(𝑋). Since bsr(𝐢(𝑋))=𝑁, this tuple is reducible. Hence, there exist β„Žπ‘—βˆˆπΆ(𝑋) so that πΉπ‘›ξ€·π‘“βˆΆ=1+β„Ž1𝑔𝑛,…,𝑓𝑁+β„Žπ‘π‘”π‘›ξ€Έ(3.7) is invertible in 𝐢(𝑋). Now on 𝑋||||ξ‚€||𝐹||+1πΉβˆ’π‘›ξ‚πΉπ‘›||𝐹𝑛||||||≀3𝑛,(3.8) where |𝑄|=βˆ‘π‘π‘—=1|𝑐𝑗|2 whenever 𝑄=(𝑐1,…,𝑐𝑁) (this can easily be seen by considering the cases π‘₯βˆˆπΈπ‘›,π‘₯βˆ‰πΈπ‘›, and |𝐹|(π‘₯)=0). Thus, tsr(𝐢(𝑋))≀bsr(𝐢(𝑋)).
The reverse inequality holds for any Banach algebra 𝐴, (see [19]). In fact, let π‘š=tsr(𝐴)<∞, and suppose that (𝑓1,…,π‘“π‘š,𝑔)βˆˆπ‘ˆπ‘š+1(𝐴). Choose (π‘₯1,…,π‘₯π‘š,π‘₯)βˆˆπ΄π‘š+1 so that βˆ‘π‘šπ‘—=1π‘₯𝑗𝑓𝑗+π‘₯𝑔=1. Approximating (π‘₯1,…,π‘₯π‘š) by an invertible π‘š-tuple (𝑦1,…,π‘¦π‘š) yields that βˆ‘πΉβˆΆ=π‘šπ‘—=1𝑦𝑗𝑓𝑗+π‘₯𝑔≠0 on 𝑀(𝐴). Moreover, there is (β„Ž1,…,β„Žπ‘š)βˆˆπ΄π‘š with βˆ‘π‘₯=π‘šπ‘—=1β„Žπ‘—π‘¦π‘—. Hence, 𝐹=π‘šξ“π‘—=1𝑦𝑗𝑓𝑗+β„Žπ‘—π‘”ξ€Έβ‰ 0on𝑀(𝐴).(3.9) This shows that (𝑓1+β„Ž1𝑔,…,π‘“π‘š+β„Žπ‘šπ‘”)βˆˆπ‘ˆπ‘š(𝐴)  and so bsr(𝐴)β‰€π‘š=tsr(𝐴).
The cases where tsr(𝐴)=∞ or bsr(𝐴)=∞ immediately follow.

Theorem 3.8 (see [27]). Let πΎβŠ†β„‚   be compact and suppose that 𝐾∘=βˆ…. Then, bsr(𝐢(𝐾))=1.

Proof. Let (𝑓,𝑔) be an invertible pair in 𝐢(𝐾). Hence, 𝑓≠0 on 𝑍(𝑔)={π‘₯βˆˆπΎβˆΆπ‘”(π‘₯)=0}. By [2, Theorem 4.29], there exists a rational function π‘Ÿ without poles and zeros in 𝑍(𝑔) such that on 𝑍(𝑔) we have 𝑓=π‘Ÿπ‘’β„Ž for some continuous function β„Ž. Since 𝐾 has no interior points, we may shift the poles and zeros of π‘Ÿ so that the new rational function, 𝑠, has no zeros and poles in 𝐾 and satisfies β€–π‘Ÿβˆ’π‘ β€–π‘(𝑔)<12min𝑍(𝑔)|π‘Ÿ|.(3.10) Thus, ‖𝑠/π‘Ÿβˆ’1‖𝑍(𝑔)<1 and, hence, by a standard reasoning with series in Banach algebras, 𝑠=π‘’π‘žπ‘Ÿ on 𝑍(𝑔). So ξ‚€π‘Ÿπ‘“=π‘ π‘ π‘’β„Žξ‚=π‘ π‘’β„Žβˆ’π‘žon𝑍(𝑔).(3.11) We may assume that π‘ž and β„Ž are continuously extended to β„‚. Thus, 𝐹=π‘ π‘’β„Žβˆ’π‘ž is a continuous zero-free extension of 𝑓|𝑍(𝑔) to 𝐾.
Next we use a simple version of a technique in [23]. Let the closed neighborhood π‘ˆ of 𝑍(𝑔) be chosen so that β€–πΉβˆ’π‘“β€–πΎβˆ©π‘ˆ<12min𝐾||𝐹||.(3.12) Hence, again, ‖𝑓/πΉβˆ’1β€–πΎβˆ©π‘ˆ<1 and so, on πΎβˆ©π‘ˆ,𝑓=𝑒𝐻𝐹 for some function 𝐻∈𝐢(𝐾).
Now let ξƒ―1β„Ž=π‘”ξ€·πΉβˆ’π‘’βˆ’π»π‘“ξ€ΈonπΎβ§΅π‘ˆ,0onπΎβˆ©π‘ˆ.(3.13) Then, β„ŽβˆˆπΆ(𝐾) and 𝑓+π‘’π»β„Žπ‘”=𝑒𝐻𝐹 on 𝐾. Therefore, the pair (𝑓,𝑔) is reducible.

We will now determine with our classroom proofs the stable ranks of 𝐢(𝐾) for arbitrary compacta in β„‚.

Theorem 3.9. Let πΎβŠ†β„‚   be a compact set. Then,(1)bsr(𝐢(𝐾))=tsr(𝐢(𝐾))=1   if and only if 𝐾∘=βˆ…;(2)bsr(𝐢(𝐾))=tsr(𝐢(𝐾))=2   if and only if πΎβˆ˜β‰ βˆ….

Proof. In view of Theorem 3.7, it suffices to show the assertions for the topological stable rank.
(1) Suppose that 𝐾∘=βˆ…, and let π‘“βˆˆπΆ(𝐾). By Weierstrass’ approximation theorem, there exists for each πœ€>0 a polynomial 𝑝 in the real variables π‘₯ and 𝑦 such that β€–π‘“βˆ’π‘β€–πΎ<πœ€. Since 𝐾∘=βˆ…, we may use Theorem 3.5 to conclude that 𝑝(𝐾) is nowhere dense in β„‚. In particular, there exists πœ‚βˆˆβ„‚ with arbitrarily small modulus, say |πœ‚|<πœ€, such that π‘βˆ’πœ‚ has no zeros on 𝐾. Thus, β€–π‘“βˆ’(π‘βˆ’πœ‚)‖𝐾||πœ‚||β‰€πœ€+<2πœ€.(3.14) Hence, 𝑓 has been approximated by invertible functions on 𝐾 and so tsr(𝐢(𝐾))=1.
Now if πΎβˆ˜β‰ βˆ…, then, after a suitable affine transformation, we may assume that π”»βŠ†πΎβˆ˜. But the identity function 𝑧 cannot be uniformly approximated on 𝔻 by invertible functions 𝑒𝑛 in 𝐢(𝐾). Otherwise |π‘§βˆ’π‘’π‘›(𝑧)|<πœ€ for all π‘§βˆˆπ”»; hence, by Brouwer’s fixed-point theorem, [2, page 108], π‘§βˆ’π‘’π‘›(𝑧) has a fixed point in 𝔻 and so 𝑒𝑛 has a zero, a contradiction. Thus, tsr(𝐢(𝐾))=1 implies that 𝐾∘=βˆ….
(2) We show that tsr(𝐢(𝐾))≀2. Having done this, (2) is the logical negation of (1).
So let (𝑓,𝑔)∈𝐢(𝐾)2. As above, by Weierstrass’ approximation theorem, there exist for each πœ€>0 two polynomials 𝑝 and π‘ž in the real variables π‘₯ and 𝑦 such that β€–π‘“βˆ’π‘β€–πΎ+β€–π‘”βˆ’π‘žβ€–πΎ<πœ€. Now we consider the set 𝐸∢=𝐾×{0} as a subset in β„‚2=ℝ4 and look upon the map (𝑝,π‘ž) as a map from ℝ4 to ℝ4. Since 𝐸 has no interior points, we may conclude with Theorem 3.5 as in (1) that there exist (πœ‚1,πœ‚2)βˆˆβ„‚2 with |πœ‚1|+|πœ‚2|<πœ€ such that (π‘βˆ’πœ‚1,π‘žβˆ’πœ‚2) is an invertible pair in 𝐢(𝐸). Hence, (𝑓,𝑔) has been approximated by invertible pairs in 𝐢(𝐾).

4. The Algebra 𝐴1(𝐾): Its Spectrum, Its 𝐴1(𝐾)-Convex Sets

Suppose that πΎβŠ†β„‚ is given. Does the algebra 𝐴1(𝐾) resemble 𝐴(𝐾) or 𝑅(𝐾)? For which 𝐾 the algebra 𝐴1(𝐾) is strictly contained in 𝐴(𝐾)? We refer the reader to the books by Gaier [8] and Gamelin [9] for a thorough study of the algebras 𝑅(𝐾) and 𝐴(𝐾). In particular, they give descriptions of those compacta for which both coincide (Vitushkin’s theorem).

Concerning our algebra 𝐴1(𝐾), since every continuous function can be uniformly approximated on compact planar sets by polynomials in the real variables π‘₯ and 𝑦, it would be natural to guess that 𝐴1(𝐾) coincides with 𝐴(𝐾). For instance, if 𝐾∘=βˆ…, then, by the Stone-Weierstrass theorem just mentioned, 𝐴1(𝐾)=𝐢(𝐾) (note that 𝑧,π‘§βˆˆπ΄1(𝐾)). It will follow from our representation of 𝐴1(𝐾) (Theorem 4.2) that this is not true in general. Indeed, 𝐴1(𝐾) will be β€œcloser” to 𝑅(𝐾) than to 𝐴(𝐾), and this is due to the following theorem that will be the main tool for our study of 𝐴1(𝐾).

Theorem 4.1 (see [1, page 160] or [9, page 26]). Let π‘“βˆˆπΆ1(π‘ˆ) for some neighborhood π‘ˆ of 𝐾, and suppose that πœ•π‘“=0 on 𝐾. Then, 𝑓|πΎβˆˆπ‘…(𝐾).

We start now with a representation of 𝐴1(𝐾) in terms of the algebras 𝑅(𝑆1) and 𝐢(𝑆2) for some specified compact sets 𝑆1 and 𝑆2. Let 𝐻(Ξ©) be the set of all holomorphic functions in an open set Ξ©βŠ†β„‚.

Theorem 4.2. If πΎβŠ†β„‚β€‰β€‰is compact, then 𝐴1(𝐾)=𝐢(𝐾)βˆ©π‘…(𝐾∘).

Proof. (1) Let π‘“βˆˆπΆ1(π‘ˆ) for a neighborhood π‘ˆ of 𝐾, and suppose that 𝑓 is holomorphic in 𝐾∘. Then, πœ•π‘“=0 on 𝐾∘, and so, by Theorem 4.1, π‘“βˆˆπ‘…(𝐾∘). Since 𝐢(𝐾)βˆ©π‘…(𝐾∘) is uniformly closed, we obtain that 𝐴1(𝐾)βŠ†πΆ(𝐾)βˆ©π‘…(𝐾∘).
(2) To show the reverse inclusion, let π‘“βˆˆπΆ(𝐾)βˆ©π‘…(𝐾∘). We may think of 𝑓 as being continuous on the whole plane (Tietze). Let πœ€>0. Choose a rational function π‘Ÿ without poles in 𝐾∘ such that β€–π‘Ÿβˆ’π‘“β€–πΎβˆ˜<πœ€. Note that possible poles of π‘Ÿ belong to πœ•πΎ or to ℂ⧡𝐾. By shifting the boundary poles, if necessary, we may assume that π‘Ÿ has no poles on 𝐾. Let 𝑉 be a neighborhood of 𝐾∘, so that β€–π‘Ÿβˆ’π‘“β€–π‘‰<πœ€, too. By the Stone-Weierstrass theorem, let 𝑝=𝑝(π‘₯,𝑦) be a polynomial of the two real variables π‘₯ and 𝑦 such that β€–π‘βˆ’π‘“β€–πΎ<πœ€. Now choose π‘žβˆˆπΆβˆž(β„‚) such that π‘ž=1 on 𝐾∘,π‘ž=0 on 𝐾⧡𝑉, and 0β‰€π‘žβ‰€1. Finally let 𝑔=π‘Ÿπ‘ž+(1βˆ’π‘ž)𝑝.(4.1) Then, π‘”βˆˆπΆ1(π‘ˆ) for some open set π‘ˆ with πΎβŠ†π‘ˆ. Also, 𝑔=π‘Ÿ on 𝐾∘; hence, 𝑔 is holomorphic in 𝐾∘, and so π‘”βˆˆπ΄1(𝐾). Moreover, on 𝐾⧡𝑉,|π‘”βˆ’π‘“|=|π‘βˆ’π‘“|<πœ€ and on πΎβˆ©π‘‰ we have ||||=||||≀||||+||||≀||||+||||ξ€Έ+||||π‘”βˆ’π‘“π‘ž(π‘Ÿβˆ’π‘)+(π‘βˆ’π‘“)π‘Ÿβˆ’π‘π‘βˆ’π‘“π‘Ÿβˆ’π‘“π‘“βˆ’π‘π‘βˆ’π‘“β‰€πœ€+2πœ€=3πœ€.(4.2) Thus, we have uniformly approximated each π‘“βˆˆπΆ(𝐾)βˆ©π‘…(𝐾∘) by functions in 𝐴1(𝐾). Thus, 𝐢(𝐾)βˆ©π‘…(𝐾∘)βŠ†π΄1(𝐾).

The proof shows that a set of generators of 𝐴1(𝐾) is given by𝒒=π‘“βˆˆπΆ(𝐾)βˆΆπ‘“|πΎβˆ˜ξ€Ύrationalwithoutpolesin𝐾.(4.3)

Corollary 4.3. The following assertions hold.(a) If 𝐾=𝐾∘, then 𝐴1(𝐾)=𝑅(𝐾),   and if  𝐾∘=βˆ…, then 𝐴1(𝐾)=𝐢(𝐾).(b) There exist compact sets πΎβŠ†β„‚β€‰β€‰β€‰withβ€‰β€‰πΎβˆ˜=βˆ… for which 𝑅(𝐾)⫋𝐴1(𝐾).(c) There exist compact sets πΎβŠ†β„‚   with  𝑅(𝐾)⫋𝐴1(𝐾)⫋𝐴(𝐾)⫋𝐢(𝐾).(d) There exist compact sets πΎβŠ†β„‚   with  𝑅(𝐾)⫋𝐴1(𝐾)=𝐴(𝐾)⫋𝐢(𝐾).

Proof. (a) This immediately follows from Theorem 4.2.
(b) A Swiss cheese 𝑆 with π‘†βˆ˜=βˆ… for which 𝑅(𝑆)≠𝐢(𝑆) appears in [8, page 103]. Hence, 𝑅(𝑆)⫋𝐴1(𝑆)=𝐢(𝑆).
(c) Let 𝑆 be the Swiss cheese above (with π‘†βˆ˜=βˆ…), and let π‘†ξ…ž be a Swiss cheese satisfying (𝑆′)∘=𝑆′, 𝑅(π‘†ξ…ž)≠𝐴(π‘†ξ…ž) (see [8, page 104]). We may assume that π‘†βˆ©π‘†β€²=βˆ…. Then, 𝐾=𝑆βˆͺ𝑆′ satisfies 𝑅(𝐾)⫋𝐴1(𝐾)⫋𝐴(𝐾)⫋𝐢(𝐾).
(d) Let 𝑆 be as above, and choose any π‘†ξ…žξ…ž with (π‘†ξ…žξ…ž)βˆ˜β‰ βˆ… such that 𝐴(π‘†ξ…žξ…ž)=𝑅(π‘†ξ…žξ…ž). If π‘†βˆ©π‘†ξ…žξ…ž=βˆ…, then 𝐾=𝑆βˆͺπ‘†ξ…žξ…ž satisfies 𝑅(𝐾)⫋𝐴1(𝐾)=𝐴(𝐾)⫋𝐢(𝐾).

Theorem 4.4. The maximal ideal space of 𝐴1(𝐾) can be identified with 𝐾 via point evaluations. Also, the Shilov boundary of 𝐴1(𝐾) coincides with πœ•πΎ, the topological boundary of 𝐾.

Proof. We use Theorem 4.2 and the fact that all functions defined on 𝐾, respectively, 𝐾∘, and appearing here admit continuous extensions to the whole plane. Let (𝑓1,…,𝑓𝑛) satisfy βˆ‘π‘›π‘—=1|𝑓𝑗|β‰₯𝛿>0 on 𝐾. Since 𝑀(𝑅(𝐸))=𝐸 for every compact set πΈβŠ†β„‚ (see [9, page 27]), there exists (π‘Ÿ1,…,π‘Ÿπ‘›)βˆˆπ‘…(𝐾∘)𝑛 such that 𝑛𝑗=1π‘Ÿπ‘—π‘“π‘—=1on𝐾∘.(4.4) Since 𝑀(𝐢(𝐾))=𝐾, there is (π‘ž1,…,π‘žπ‘›)∈𝐢(𝐾)𝑛 such that 𝑛𝑗=1π‘žπ‘—π‘“π‘—=1on𝐾.(4.5) Let 𝑉 be a neighborhood of 𝐾∘ so that |βˆ‘π‘›π‘—=1π‘Ÿπ‘—π‘“π‘—βˆ’1|<1/2 on 𝑉. Choose a function π‘žβˆˆπΆβˆžβˆ˜(β„‚), β€–π‘žβ€–βˆžβ‰€1, with π‘ž=1 on 𝐾∘ and π‘ž=0 on 𝐾⧡𝑉. Let 𝑠𝑗=π‘Ÿπ‘—π‘ž+(1βˆ’π‘ž)π‘žπ‘—. Then, π‘ π‘—βˆˆπΆ(𝐾)βˆ©π‘…(𝐾∘)=𝐴1(𝐾). Moreover, βˆ‘π‘›π‘—=1𝑠𝑗𝑓𝑗≠0 on 𝐾. In fact, on 𝐾⧡𝑉 we have 𝑛𝑗=1𝑠𝑗𝑓𝑗=𝑛𝑗=1π‘Ÿπ‘—π‘“π‘—ξƒͺπ‘ž+(1βˆ’π‘ž)𝑛𝑗=1π‘žπ‘—π‘“π‘—=𝑛𝑗=1π‘žπ‘—π‘“π‘—=1(4.6) and on πΎβˆ©π‘‰ we have |||||𝑛𝑗=1𝑠𝑗𝑓𝑗|||||=|||||𝑛𝑗=1π‘Ÿπ‘—π‘“π‘—ξƒͺ|||||=|||||π‘žξƒ©π‘ž+(1βˆ’π‘ž)β‹…1𝑛𝑗=1π‘Ÿπ‘—π‘“π‘—ξƒͺ||||||||||βˆ’1+1β‰₯1βˆ’π‘›ξ“π‘—=1π‘Ÿπ‘—π‘“π‘—|||||β‰₯1βˆ’12.(4.7) Since zero-free functions in 𝐴1(𝐾) are obviously invertible, we obtain that βˆ‘π‘›π‘—=1𝑠𝑗𝑓𝑗 is invertible, too. Hence, the ideal 𝐼 generated by (𝑓1,…,𝑓𝑛) coincides with 𝐴1(𝐾) and so 𝑀(𝐴1(𝐾))=𝐾.
That the Shilov boundary of 𝐴1(𝐾) is πœ•πΎ works in the same manner as for 𝑅(𝐾) (see [9, page 27]).

Next we determine the 𝐴-convex subsets of 𝑀(𝐴) for a large class of function algebras. It applies in particular to 𝑅(𝐾),𝐴1(𝐾), and 𝐴(𝐾). We do not claim originality of this result, but we could not find a reference.

Recall that a hole 𝐻 of a compact set πΎβŠ†β„‚ is a bounded component of ℂ⧡𝐾.

Theorem 4.5. Let πΎβŠ†β„‚β€‰β€‰be compact, and let 𝐴 be a uniform algebra with 𝑃(𝐾)βŠ†π΄βŠ†π΄(𝐾) and such that 𝑀(𝐴)=𝐾 (via point evaluation). Suppose that 𝑆 is a closed subset of 𝐾. Then 𝑆 is 𝐴-convex if and only if each hole of 𝑆 contains a hole of 𝐾.

Proof. Let π‘†βŠ†πΎ   be 𝐴-convex. Suppose that there exists a hole 𝐻 of 𝑆 with π»βŠ†πΎβˆ˜. Note that πœ•π»βŠ†π‘†. By virtue of the maximum principle for holomorphic functions, we see that for any π‘§βˆˆπ» and π‘“βˆˆπ΄||||𝑓(𝑧)≀maxπœ•π»||𝑓||≀max𝑆||𝑓||.(4.8) Thus, 𝐻 is contained in the 𝐴-convex hull of 𝑆, which coincides, by our hypothesis, with 𝑆; a contradiction. Hence, if 𝑆 is 𝐴-convex, then each hole of 𝑆 contains a hole of 𝐾.
To prove the converse, let 𝑆 satisfy the condition above. We first note that our hypothesis on the spectrum actually implies that 𝑅(𝐾)βŠ†π΄. In fact, for π‘Žβˆˆβ„‚β§΅πΎ, the function π‘§βˆ’π‘Ž has no zeros on the spectrum 𝐾 of the algebra, so is invertible. Thus, each rational function with poles outside 𝐾 belongs to 𝐴. So actually 𝑅(𝐾)βŠ†π΄.
Choose any point π‘ŽβˆˆπΎβ§΅π‘†, and let 𝑓≑0 in a neighborhood of 𝑆 and 𝑓≑1 in a neighborhood of π‘Ž. By Runge’s theorem, there exists a rational function π‘Ÿ with poles off 𝑆βˆͺ{π‘Ž} such that |π‘“βˆ’π‘Ÿ|<1/4 on 𝑆βˆͺ{π‘Ž}. Since each hole of 𝑆 contains a hole of 𝐾 and the unbounded component of ℂ⧡𝑆 contains the unbounded component of ℂ⧡𝐾, we may assume, by Runge again, that the poles of π‘Ÿ are outside 𝐾. So π‘Ÿβˆˆπ‘…(𝐾) and ||||β‰₯3π‘Ÿ(π‘Ž)4>14β‰₯max𝑆|π‘Ÿ|.(4.9) Thus, π‘Ž does not belong to the 𝐴-convex hull of 𝑆. Since π‘ŽβˆˆπΎβ§΅π‘† was arbitrary, 𝑆 is 𝐴-convex.

Theorem 4.6. Let πΎβŠ†β„‚β€‰β€‰β€‰be compact and 𝑆 an 𝐴1(𝐾)-convex subset of 𝐾. Then, 𝐴1(ξ€Έ|𝐾)𝑆=π‘…π‘†βˆ©πΎβˆ˜ξ‚βˆ©πΆ(π‘†βˆ©πœ•πΎ).(4.10)

Proof. Let π‘“βˆˆπ΄1(𝐾). To show the inclusion βŠ†, we may assume, by Theorem 4.2, that π‘Ÿ=𝑓|𝐾∘ is a rational function without poles in 𝐾. Thus, π‘Ÿβˆ£π‘†βˆ©πΎβˆ˜βˆˆπ‘…(π‘†βˆ©πΎβˆ˜). So π‘“βˆ£π‘†ξ‚€βˆˆπ‘…π‘†βˆ©πΎβˆ˜ξ‚βˆ©πΆ(π‘†βˆ©πœ•πΎ)=βˆΆπ‘….(4.11)
To show the reverse inclusion, let π‘“βˆˆπ‘…. Note that π‘“βˆˆπΆ(𝑆). We may extend 𝑓 continuously to β„‚. Let π‘Ÿ be a rational function without poles in π‘†βˆ©πΎβˆ˜ so that |π‘Ÿβˆ’π‘“|<πœ€ on π‘†βˆ©πΎβˆ˜. By moving possible poles on πœ•πΎ a little bit, we may assume that π‘Ÿ has no poles in 𝑆. Note that 𝑆=(π‘†βˆ©πΎβˆ˜)βˆͺ(π‘†βˆ©πœ•πΎ). Since 𝑆 is 𝐴1(𝐾)-convex, by Theorem 4.5, each hole of 𝑆 contains a hole of 𝐾. Also, the unbounded component of ℂ⧡𝑆 contains the unbounded component of ℂ⧡𝐾. Thus, we may assume, using Runge’s theorem, that π‘Ÿ has no poles in 𝐾. Let 𝑉 be a neighborhood of π‘†βˆ©πΎβˆ˜ so that |π‘Ÿβˆ’π‘“|<πœ€ on 𝑉.
Also, let β„Ž be a polynomial of the real variables π‘₯ and 𝑦 with |π‘“βˆ’β„Ž|<πœ€ on 𝐾. In particular, |π‘“βˆ’β„Ž|<πœ€ on π‘†βˆ©πœ•πΎ. Choose π‘žβˆˆπΆβˆž(β„‚), 0β‰€π‘žβ‰€1, so that π‘ž=1 on 𝐾∘ and π‘ž=0 on 𝐾⧡𝑉. Then, 𝐹∢=π‘Ÿπ‘ž+(1βˆ’π‘ž)β„Ž is in 𝐢1(π‘ˆ) for some neighborhood π‘ˆ of 𝐾. Moreover, 𝐹 is holomorphic in 𝐾∘. Thus, 𝐹∈𝐴1(𝐾). Now, on 𝑆, we have that |πΉβˆ’π‘“|<3πœ€ (see the estimate in Theorem 4.2). Thus, π‘“βˆˆ(𝐴1(𝐾))|𝑆.

Remark 4.7. Let us mention the following interesting and related result by Izzo [11, Theorem 2.2 and Corollary 2.5]. Let 𝐸 be a closed subset of 𝐾. Then, 𝐴(𝐾)|𝐸 is dense in 𝐢(𝐸) if and only if 𝐸 has no interior points and each component of ℂ⧡𝐸 contains a component of ℂ⧡𝐾. Note that the second condition is, in view of Theorem 4.5, equivalent to 𝐸 being 𝐴(𝐾)-convex.

5. The Stable Ranks of 𝐴1(𝐾)

Let us call an algebra 𝐴 of holomorphic functions on a planar open set Ξ©  stable if (𝑓(𝑧)βˆ’π‘“(π‘Ž))/(π‘§βˆ’π‘Ž)∈𝐴 whenever π‘“βˆˆπ΄ and π‘ŽβˆˆΞ©. It is clear that the algebras 𝑃(𝐾),𝑅(𝐾),𝐴1(𝐾), and 𝐴(𝐾) are all stable, where for Ξ© we take 𝐾∘.

Lemma 5.1. For all compact subsets πΎβŠ†β„‚,  one has tsr(𝐴1(𝐾))≀2.

Proof. Let (𝑓,𝑔) be a pair of functions in 𝐴1(𝐾). By definition, for πœ€>0, there is a neighborhood π‘ˆ of 𝐾 and functions 𝐹 and 𝐺 in 𝐢1(π‘ˆ)∩𝐻(𝐾∘) so that β€–πΉβˆ’π‘“β€–πΎ+β€–πΊβˆ’π‘”β€–πΎ<πœ€. Since πœ•πΎ has no interior points, we conclude from Theorem 3.5 that the images 𝐹(πœ•πΎ) and 𝐺(πœ•πΎ) have no interior points. In particular, the value zero is not an interior point of 𝐹(πœ•πΎ) and 𝐺(πœ•πΎ). Thus, we may choose πœ‚1 and πœ‚2 in β„‚ with |πœ‚π‘—|<πœ€/2 so that πœ‚1βˆ‰πΉ(πœ•πΎ) and πœ‚2βˆ‰πΊ(πœ•πΎ). Therefore, πΉβˆ’πœ‚1 and πΊβˆ’πœ‚2 have no zeros on πœ•πΎ. Since both functions are holomorphic in 𝐾∘, they have only finitely many zeros at all. Hence, there exist (holomorphic) polynomials 𝑝 and π‘ž and functions 𝑒 and 𝑣 zero-free on 𝐾 such that πΉβˆ’πœ‚1=𝑝𝑒 and πΊβˆ’πœ‚2=π‘žπ‘£. Since 𝐴1(𝐾) is a stable algebra, 𝑒 and 𝑣 actually belong to 𝐴1(𝐾). By moving, if necessary, the zeros of 𝑝 a little bit, we may assume that 𝑝 and π‘ž have no common zeros. By construction we have β€–πΉβˆ’π‘π‘’β€–πΎ+β€–πΊβˆ’π‘žπ‘£β€–πΎβ‰€||πœ‚1||+||πœ‚2||β‰€πœ€,(5.1) and so β€–π‘“βˆ’π‘π‘’β€–πΎ+β€–π‘”βˆ’π‘žπ‘£β€–πΎβ‰€2πœ€.(5.2) Since the functions 𝑝𝑒 and π‘žπ‘£ have no common zeros on 𝐾, they form, by Theorem 4.4, an invertible pair in 𝐴1(𝐾).

We are now able to prove the main results of this paper.

Theorem 5.2. Let πΎβŠ†β„‚   be a compact set. Then,(1)bsr(𝐴1(𝐾))=1; (2)tsr(𝐴1(𝐾))=1  if and only if 𝐾∘=βˆ…;(3)tsr(𝐴1(𝐾))=2  if and only if πΎβˆ˜β‰ βˆ….

Proof. (1) follows from [21, Corollary 1.3] (that was based on the Arens-Taylor-Novodvorsky theory) and Theorem 4.4. It can also be seen in the following way, using the simpler results that bsr(𝑅(𝐸))=1 for every compact set πΈβŠ†β„‚ and bsr(𝐢(𝑆))=1 whenever 𝑆 is a compact set in β„‚ with empty interior (see Theorem 3.8). So let (𝑓,𝑔) be an invertible pair in 𝐴1(𝐾). By Theorems 4.2 and 3.3, there exists π‘Ÿβˆˆπ‘…(𝐾∘) so that 𝑓+π‘Ÿπ‘”β‰ 0on𝐾∘.(5.3) We may assume that π‘Ÿ is a rational function without poles in 𝐾∘. By moving the poles a little bit, if necessary, we may also assume that π‘Ÿ has no poles on 𝐾.
Let π‘žβˆˆπΆβˆž(β„‚) be taken so that π‘ž=0 on 𝐾∘ and that 𝑍(π‘ž)βˆ©π‘(𝑓+π‘Ÿπ‘”)=βˆ…,(5.4) where 𝑍(β„Ž)={π‘§βˆˆπΎβˆΆβ„Ž(𝑧)=0}. Then, (𝑓+π‘Ÿπ‘”,π‘žπ‘”) is an invertible pair in 𝐢(𝐾). Since bsr(𝐢(πœ•πΎ))=1, there is β„ŽβˆˆπΆ(𝐾) so that 𝐻∢=(𝑓+π‘Ÿπ‘”)+β„Ž(π‘žπ‘”)β‰ 0onπœ•πΎ.(5.5) Now by Theorem 4.2, β„Žπ‘žβˆˆπ΄1(𝐾) and π‘Ÿ+β„Žπ‘žβˆˆπ΄1(𝐾) (note that π‘ž vanishes identically on 𝐾∘). Thus, 𝐻∈𝐴1(𝐾). Moreover, by (5.3) and (5.5), we obtain that 𝑓+(π‘Ÿ+β„Žπ‘ž)𝑔≠0 on 𝐾. Since, by Theorem 4.4, 𝑀(𝐴1(𝐾))=𝐾, we conclude that bsr(𝐴1(𝐾))=1.
(2), (3) By Lemma 5.1 we have that tsr(𝐴1(𝐾))≀2. Thus, it suffices to show (2). So assume that 𝐾∘=βˆ…. Then, by Corollary 4.3, 𝐴1(𝐾)=𝐢(𝐾). Hence, by Theorem 3.9, tsr(𝐴1(𝐾))=tsr(𝐢(𝐾))=1. If π‘ŽβˆˆπΎβˆ˜β‰ βˆ…, then, by Rouché’s theorem, the function π‘§βˆ’π‘Ž cannot be uniformly approximated by functions holomorphic and invertible in a neighborhood of π‘Ž. Hence, tsr(𝐴1(𝐾))β‰₯2.

Our final theorem determines the dense stable rank of a large class of function algebras whose spectrum is, via point evaluation, a compact set in β„‚. In particular it applies to 𝑅(𝐾),𝐴1(𝐾),𝐴(𝐾), and 𝑃(𝐾). For compact sets in ℂ𝑛 the dense stable rank for 𝑃(𝐾) has been determined by Corach and SuΓ‘rez in [7].

Theorem 5.3. Let πΎβŠ†β„‚β€‰β€‰be a compact set. Suppose that 𝐴 is a uniform algebra with 𝑃(𝐾)βŠ†π΄βŠ†π΄(𝐾) and that 𝑀(𝐴)=𝐾. Then, dsr(𝐴)=1.

Proof. As in the proof of Theorem 4.5, we observe that the hypothesis implies that 𝑅(𝐾)βŠ†π΄. Let 𝐸 be an 𝐴-convex set, and suppose that π‘“βˆˆπ΄ does not vanish on 𝐸. Since 𝐸 is 𝐴-convex, by Theorem 4.5, each hole of 𝐸 contains a hole of 𝐾. Moreover, the unbounded component of ℂ⧡𝐸 contains the unbounded component of ℂ⧡𝐾. Hence, by [2, Corollary 4.30], there exists a rational function π‘Ÿ with zeros and poles off 𝐾 and some continuous function β„ŽβˆˆπΆ(𝐸) so that 𝑓|𝐸=π‘Ÿπ‘’β„Ž. Now π‘Ÿ is invertible in 𝐴 and so the function (𝑓/π‘Ÿ)|𝐸 belongs to 𝐴|𝐸 and admits a continuous logarithm on 𝐸. Note that 𝐸=𝑀(𝐴|𝐸), because 𝐸 is 𝐴-convex. Thus, by [9, Corollary 6.2, page 88], (𝑓/π‘Ÿ)|𝐸 actually admits a logarithm in 𝐴|𝐸; that is, 𝑓|𝐸=π‘Ÿπ‘’π‘” for some π‘”βˆˆπ΄|𝐸. Uniformly approximating 𝑔 on 𝐸 by functions β„Žπ‘› in 𝐴 now shows that, on 𝐸, 𝑓 is the uniform limit of functions of the form π‘Ÿπ‘’β„Žπ‘› that are invertible in 𝐴. Hence, dsr(𝐴)=1.

We remark that Theorem 5.3 yields another proof that bsr(𝐴)=1 for these algebras, since the Bass stable rank is always dominated by the dense stable rank (see [7, 17]). In particular, we therefore have a rather short proof for bsr(𝐴(𝐾))=1 (see [5, 25]). We do not know, though, whether tsr(𝐴(𝐾))≀2 for every compact set πΎβŠ†β„‚.

Acknowledgments

The authors thank Joel Feinstein for some comments concerning Corollary 4.3 and, in particular, for reminding them of Theorem 4.1. They also thank the referee for his remarks that improved the exposition of our paper.

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