# Normed Domains of Holomorphy

Authors: Steven G. Krantz
Journal: International Journal of Mathematics and Mathematical Sciences
 (2010)

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## Abstract

We treat the classical concept of domain of holomorphy inℂn when the holomorphic functions considered are restricted to lie in some Banach space. Positive and negative results are presented. A new view of the case n=1 is considered.


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## Body

## 1. Introduction

In this paper adomainΩ⊆ℂn is a connected open set. We let 𝒪(Ω) denote the algebra of holomorphic functions on Ω.We will use the following notation:D denotes the unit disc in the complex plane. We let D2=D×D denote the bidisc, and Dn=D×D×⋯×D the polydisc in ℂn. The symbol B=Bn is the unit ball in ℂn.A domainΩ⊆ℂn is said to be Runge if any holomorphic f on Ω is the limit, uniformly on compact subsets of Ω, of polynomials.In the classical function theory of several complex variables there are two fundamental concepts: domain of holomorphy and pseudoconvex domain. The Levi problem, which was solved comprehensively in the 1940s and 1950s, asserts that these two concepts are equivalent: a domainΩ⊆ℂn is a domain of holomorphy if and only if it is pseudoconvex. These matters are discussed in some detail in [1].Roughly speaking, ifΩ is a domain of holomorphy, then there is a holomorphic function f on Ω such that f cannot be analytically continued to any larger domain. Generally speaking one cannot say much about the nature of this f—whether it is bounded, or satisfies some other growth condition.In the paper [2], Sibony presents the following remarkable example.Example 1.1.
There is a bounded pseudoconvex Runge domainΩS⊆ℂ2, with ΩS being a proper subset of the bidisc D2=D×D, such that any bounded holomorphic function φ on Ω analytically continues to all of D2.Of course this result of Sibony can be extended toℂn in a variety of ways. For one thing, one may take the product of the Sibony domain with the polydisc in ℂn-2 to obtain an example in ℂn. Alternatively, one may replace the first (or z) variable in the Sibony construction with a tuple in ℂn-1 to obtain a counterexample in ℂn. The Sibony example and its implications are studied extensively in [3]. See also [4].It is interesting to note that in some sense, the Sibony example is generic. In fact we have the following proposition.Proposition 1.2.
The collection of domainsΩ⊆D2, with Ω≠D2, such that any bounded holomorphic function on Ω analytically continues to all of D2 (as in the Sibony example above) is uncountable.Proof.
We very quickly review the key steps of the Sibony construction.
Letpj be a countable collection of points in the unit disc D, with no interior accumulation point, so that every boundary point of D is an accumulation point of the set {pj}. Now define
(1.1)φ(ζ)=∑jλjlog|ζ-pj2|.
Here {λj} is a summable sequence of positive, real numbers. Notice that the function φ—being the sum of subharmonic functions—is subharmonic. Define
(1.2)V0(ζ)=exp(φ(ζ)).
Then V0 has the properties:  (i)
V0 is subharmonic;(ii)
0<V0(ζ)≤1 for all ζ∈D;(iii)
the functionV0 is continuous.The last property holds because the sequence {pj} is discrete and V0 takes the value 0 only at the pj.
Now define the domain(1.3)M(D,V0)={(z,w)∈ℂ2:z∈D,w∈ℂ,|w|<exp(-V0(z))}.
Since V0 is positive, we see that this definition makes sense and that M(D,V0) is a proper subset of D2.
The remainder of Sibony's argument shows that any bounded, holomorphic function onM(D,V0) analytically continues to a bounded, holomorphic function on D2. We will not repeat it but refer the reader to [2, 5].
The key fact in the Sibony construction is that the points{pj} form a discrete set that accumulates at every boundary point of D. Apart from this property, there is complete freedom in choosing the pj. We begin by showing how to construct two biholomorphically distinct instances of Sibony domains and then consider at the end how to produce uncountably many biholomorphically distinct domains.
Define, forℓ=1,2,…,
(1.4)Sℓ={z∈D:|z|=1-2-ℓ-1}.
Set  {pj1}= (the sequence consisting of 4 equally spaced points on S1, 8 equally spaced points on S2, 16 equally spaced points on S3, etc.),{pj2}= (the sequence consisting of 8 equally spaced points on S1, 16 equally spaced points on S2, 32 equally spaced points on S3, etc.).Define a domain Ω1 using the Sibony construction, as above, with the sequence {pj1} and define a domain Ω2 using the Sibony construction with the sequence {pj2}. We claim that Ω1 and Ω2 are biholomorphically inequivalent.
To see this, suppose the contrary. So there is a biholomorphic mappingΦ:Ω1→Ω2. By the usual classical arguments (see the proof of Proposition 11.1.2 in [1]), we see that Φ must commute with rotations in the w variable. It follows that any disc in Ω1 of the form
(1.5)dz={(z,w):|w|<exp(-V0(z))},
for z∈D fixed, must be mapped to a similar disc in Ω2.
Further observe that each of the discsdpj⊆Ω1 is a totally geodesic submanifold in the Kobayashi metric. This assertion follows immediately from the existence of the maps
(1.6)D→iΩ1→πD,
where i is the injection
(1.7)i(w)=(pj,w),
and π is the projection
(1.8)π(pj,w)=w.
Observe that π∘i=id. Similar reasoning shows that the disc d*={(z,0)}⊆Ω1 is totally geodisc. Of course similar remarks apply to the corresponding discs in Ω2.
Now it is essential to notice that, inΩ1, the vertical discs of the form dz for z not one of the pj are not totally geodesic. This follows because, at the point (z,0), the Kobayashi extremal disc in the vertical direction 〈0,1〉 will not be the rigid disc ζ↦(0,ζ) but rather a disc that curves into one of the spikes above a nearby pj.
A final observation that we need to make is this. The vertical totally geodesic discs will be mapped to each other by the biholomorphic mappingΦ. But more is true. Because d* is totally geodesic, in fact the totally geodesic discs located at the points pj that lie on S1 in Ω1 must be mapped to the totally geodesic discs located at the points pj that lie on S1 in Ω2 (because both circles consist of points that have the same Kobayashi distance from the origin). This is impossible because S1 in Ω1 has four such points while S1 in Ω2 has eight such points. That is the required contradiction.
Now it is clear how to construct uncountably many inequivalent domains of Sibony type. For if𝒜={αj} is any exponentially increasing sequence of positive integers, then, we may associate to it a domain Ω̃ of Sibony-type with α1 points on S1, α2 points on S2, and so forth. The preceding argument shows that different choices of 𝒜 result in biholomorphically inequivalent domains. And there are clearly uncountably many such sequences. That completes the argument.Recall that ifΩ⊆ℂn is a domain and ℱ is a family of functions on Ω, then we say that Ω is convex with respect to ℱ if, whenever K⊆Ω is relatively compact, then
(1.9)K̂≡{z∈Ω:|f(z)|≤supw∈K|f(w)|}
is also relatively compact in Ω. It is well known that any domain of holomorphy (i.e., any pseudoconvex domain) is convex with respect to the family of holomorphic functions on the domain.It is natural in our discussion to consider a domainΩ and the family 𝒢 of all bounded holomorphic functions on Ω.Proposition 1.3.
The Sibony domainΩS is convex with respect to the family 𝒢.Proof.
Let𝒢 be as above for the domain ΩS and let ℱ be the usual family of all holomorphic functions on ΩS. Let K be a compact subset of ΩS and K̂ℱ the hull with respect to ℱ. Then, since ΩS is a domain of holomorphy, K̂ℱ is still compact in ΩS. Let z∈ΩS be a point that does not lie in K̂ℱ. Then there is a holomorphic function f on ΩS so that |f(z)|>supw∈K̂ℱ|f(w)|. Let ϵ=|f(z)|-supw∈K̂ℱ|f(w)|>0. Since ΩS is Runge, there is a polynomial p so that |p(m)-f(m)|<ϵ/3 on K∪{z}. But then p is bounded on ΩS and, by the triangle inequality, |p(z)|>supw∈K̂ℱ|f(w)|. Thus K̂ℱ=K̂𝒢. This shows that ΩS is convex with respect to the family of bounded holomorphic functions on ΩS.In fact the argument just presented (for which I thank Erik Løw) shows that any bounded, pseudoconvex, Runge domain is convex with respect to the family of bounded, holomorphic functions.The Sibony result has an interesting and important interpretation in terms of the corona problem. We have the following proposition.Proposition 1.4.
LetΩ⊆ℂn be a bounded domain. Suppose that X is a Banach space of holomorphic functions on Ω that contains H∞(Ω). Let Ω′ be a strictly larger domain that contains Ω. Assume that any element of X analytically continues to a holomorphic function on Ω′ (one often assumes that the extended function satisfies a similar norm estimate to that specified by the norm on X, but that is not necessary and one does not impose that condition at this time). Then the corona problem cannot be solved in the space X. That is to say, if f1,f2,…,fk are holomorphic functions in X with no common zero, then there do not  exists elements g1,g2,…,gk∈X such that
(1.10)f1g1+f2g2+⋯fkgk≡1
on Ω.Proof.
Assume to the contrary that suchg1,g2,…,gk exist. Of course each gj analytically continues to Ω′.
LetP=(p1,p2,…,pn) be a point of Ω′∖Ω. Set fj(z)=zj-pj. Then the fj have no common zero in Ω. So, by hypothesis, the gj exist. And these functions extend analytically to Ω′. But then
(1.11)f1g1+f1g2+⋯fngn≡1
on Ω′. Since the fj all vanish at P, we see that, at P, the left-hand side of this last equation vanishes. That is clearly a contradiction. Hence the gj do not exist.Of course this last proposition means in particular that the point evaluations onΩ are not weak-* dense in the maximal ideal space of H∞(Ω); see [6] for more on these matters.By contrast to Sibony's result, Catlin [7] has shown that any smoothly bounded, pseudoconvex domain in ℂn supports a bounded holomorphic function that cannot be analytically continued to any larger domain. In fact he has proved something sharper.Theorem 1.5.
LetΩ⊆ℂn be a smoothly bounded pseudoconvex domain. Then there is a function in C∞(Ω¯), holomorphic on the interior, which cannot be analytically continued to any larger domain.Hakim and Sibony [8] have proved something even more decisive.Theorem 1.6.
LetΩ⊆ℂn be a smoothly bounded pseudoconvex domain. Then the maximal ideal space (or spectrum) of the algebra C∞(Ω¯)∩𝒪(Ω) is in fact Ω¯.It should be stressed that the proofs of the last two results use an algebraic formalism of Hörmander [9] which entails the loss of some derivatives; so it is essential to be working with functions that are C∞ on Ω¯. Attempts to adapt the arguments to other function spaces are doomed to failure.Pflug and Zwonek [10] have shown that the situation for L2 holomorphic functions is very neat and elegant.Theorem 1.7.
LetΩ⊆ℂn be any  pseudoconvex domain. There is an L2 holomorphic function on Ω that cannot be continued to any larger domain if and only if, for all w∈∂U and all neighborhoods U of w, U∖Ω is not pluripolar.There is also a characterization in terms of geometric regularity of the boundary, expressed terms of external balls (see [11]).Theorem 1.8.
Suppose thatΩ in ℂn is a domain of holomorphy and that, for each z0∈∂Ω, there is a sequence wν∈cΩ¯ such that wν→z0 and there are 0<r≤1 and α>0 such that B(wν,r|z0-wν|α)∩Ω¯=∅. Then there is an L2 holomorphic function on Ω that cannot be analytically continued to any larger domain.It is natural to ask for a characterization of those domainsΩ which are domains of holomorphy in the traditional sense but not domains of holomorphy for bounded holomorphic functions. One would also like to know whether there are analogous results for Lp holomorphic functions, 1≤p<∞.The purpose of the present paper is to consider these matters. While we cannot provide a full answer to the questions just posed, we can certainly give some useful partial results, and point in some new directions. The work in [3] contains a detailed consideration of questions of this kind for the case of H∞.We mention in passing that the paper [12] contains some results that bear on the questions posed here. The arguments presented in [12] appear to be incomplete.


## 2. Some Notation

Let us say that a domainΩ⊆ℂn is of type HLp, 1≤p≤∞, if there is a holomorphic function f on Ω, f∈Lp(Ω), which cannot be analytically continued to any larger domain. We instead say that Ω is of type ELp if there is a strictly larger domain Ω̂ so that every holomorphic Lp function on Ω analytically continues to Ω̂. Obviously HLp and ELp are disjoint.We are interested in giving an extrinsic description of those domains which are of typeHLp and those which are of type ELp. It is not the case in higher dimensions, for instance, that HL∞ domains are the same as domains that are convex with respect to 𝒢=H∞ (see [3]). So we seek other characterizations.


## 3. The Situation in the Complex Plane

Matters in one complex variable are fairly well understood.First of all, we should note the example of(3.1)Ω=D∖{0}.
Of course, by the Riemann removable singularities theorem, any bounded holomorphic function on Ω analytically continues to all of D. So Ω is not a domain of type HL∞. It is a domain of type EL∞.In fact it may be noted (for the domainΩ in the last paragraph) that, if p≥2, then any holomorphic function that is Lp(Ω) will analytically continue to all of D (see [13]). So this Ω is a domain of type ELp. By contrast, if p<2, then the function f(ζ)=1/ζ is holomorphic on Ω and in Lp(Ω). But of course this fdoes not analytically continue to the full disc D. So, for p<2, the domain is of type HLp.The treatment in [13] of the matter just discussed is rather abstract, and it is worthwhile to have a traditional function-theoretic treatment of these matters. We provide one now. We thank Richard Rochberg for a helpful conversation about this topic. So let f be holomorphic on D∖{0} and assume that f∈L2(D) (the case p>2 follows immediately from this one). We write f(ζ)=∑j=-∞∞ajζj. For 0<a<b<1 and k a negative integer, consider the expression
(3.2)A=∫abr∫02πf(ζ)e-ikθdθdr,where it is understood thatζ=reiθ.On the one hand,(3.3)|A|≤∫a≤|ζ|≤b|f(ζ)|dA(ζ)≤∥f∥L2·|{ζ:a≤|ζ|≤b}|1/2=∥f∥L2·[π(b2-a2)]1/2.
On the other hand,
(3.4)|A|=|∫abrakrkdr|=|ak·[rk+2k+2]ab|=|ak(bk+2k+2-ak+2k+2)|.
If k<-2 and a=b/2, this gives a contradiction as b→0+. Of course the cases k=-2 and k=-1 can be handled separately because ζ-2 and ζ-1 are certainly not in L2.Remark 3.1.
We note that the proof goes through forp<2 up until the very end. One must note that ζ-1 in fact does lie in Lp for p<2. So there is no removable singularities theorem for this range of p.Remark 3.2.
A standard result coming from potential theory is that, ifΩ⊆ℂ, P∈Ω, and f is holomorphic on Ω∖{P}, then |f(z)|=o(log[1/|z-P|]) (where we are using Landau's notation) implies that f continues analytically to all of Ω. The philosophy here is that a function f satisfying this growth hypothesis has a singularity at P that is milder than the singularity of the Green's function. This point of view is particularly useful in the study of removable singularities for harmonic functions. This result is not of any particular interest for us because it is not formulated in the language of Lebesgue spaces. In any event, it is weaker than the result presented above for L2 because the logarithm function is certainly square integrable. It is a pleasure to thank Al Baernstein and David Minda for helpful remarks about these ideas.The enemy in the results discussed at the beginning of this section is thatΩ is not equal to the interior of its closure. In fact we have the following proposition.Proposition 3.3.
Suppose that the bounded domainΩ⊆ℂ is the interior of its closure. Then Ω is a domain of type HLp for 1≤p≤∞.Proof.
The proof that we now present is an adaptation and simplification of an argument from [4].
Let{pj} be a countable, dense subset of cΩ¯. For each j, the function φj(ζ)=1/(ζ-pj) is holomorphic and bounded on Ω and does not analytically continue past pj.
Now, for eachj, let dj be an open disc centered at pj which has nontrivial intersection with Ω. Consider the linear mapping
(3.5)Ij:𝒪(Ω∪dj)∩Lp(Ω∪dj)→𝒪(Ω)∩Lp(Ω)
given by restriction. Of course each of the indicated spaces is equipped with the Lp norm, and is therefore a Banach space. We note that the example above of φj shows that Ij is not surjective. As a result, the open mapping principle tells us that the image ℳj of Ij is of first category in 𝒪(Ω)∩Lp(Ω). Therefore, by the Baire category theorem,
(3.6)ℳ≡⋃jℳj
is of first category in 𝒪(Ω)∩Lp(Ω). But this just says that the set of Lp holomorphic functions on Ω that can be analytically continued to some pj is of first category. Therefore the set of Lp holomorphic functions that cannot be analytically continued across the boundary is dense in 𝒪(Ω)∩Lp(Ω). That completes the proof.The key point of the proof just presented is that, for each point not in the closure of the given domain, there is a function holomorphic on the domain (and in the given function space) that does not analytically continue past the point. Such functions are trivial to construct in one complex variable, not so in higher dimensions.We note in passing that whenΩ⊆ℂ is the unit disc D then it is easy to construct a bounded holomorphic function that does not analytically continue to a larger domain. Let {pj} be a discrete set in D that accumulates at every boundary point and so that
(3.7)∑j1-|pj|<∞.
For example, takep1, p2, p3, p4 to be equally spaced points at distance 1/4 from ∂D,p5,p6,…,p12 to be equally spaced points at distance 1/8 from ∂D,p13,p6,…,p28 to be equally spaced points at distance 1/16 from ∂D,and so forth. Then the Blaschke product with zeros at thepj will do the job. If Ω is a simply connected domain having a Jordan curve as its boundary, then conformal mapping together with Carathéodory's theorem about continuous boundary extension will give a bounded, holomorphic, non-continuable function on this Ω.We close this section by noting that, ifΩ is a domain of holomorphy in ℂn and if V={z∈Ω:f(z)=0} for some holomorphic f on Ω (we call V a variety), then Ω′=Ω∖V is also a domain of holomorphy (if φ is a holomorphic function on Ω that does not analytically continue to a larger domain then φ/f is a holomorphic function on Ω′ that does not analytically continue to any larger domain. And it is easy to see that Ω′ is an EL∞ domain; see [14, page 19] for the details).


## 4. Complications in Dimensionn

As we have indicated, the example of Sibony exhibits a domain which isnot of type HL∞ (instead it is of type EL∞). The theorem of Catlin shows that all smoothly bounded, pseudoconvex domains are of type HL∞.It of course makes sense to focus this discussion on pseudoconvex domains. If a domainΩ is not pseudoconvex, then there will perforce be a larger domain Ω′ to which all holomorphic functions (regardless of growth) on Ω analytically continue. So this situation is not interesting.Thus we see that the domains of interest for us will be pseudoconvex domains that donot have smooth boundary. Our first result is as follows.Proposition 4.1.
LetD1,D2,…,Dn be bounded domains in ℂ, each of which is equal to the interior of its closure. Define
(4.1)Ω=D1×D2×⋯×Dn.
Then Ω is a domain of type HLp for any 1≤p≤∞.Proof.
Fixp as indicated. Then, by Proposition 3.3, there is a holomorphic function ψj on Dj, for 1≤j≤n, such that ψj is holomorphic and Lp on Dj and does not analytically continue to any larger domain.
But then
(4.2)ψ(z1,…,zn)=ψ1(z1)·ψ2(z2)⋯ψn(zn)
is holomorphic and Lp on Ω and does not analytically continue to any larger domain.Proposition 4.2.
LetΩ⊆ℂn be bounded and convex. Let 1≤p≤∞. Then Ω is a domain of type HLp.Proof.
Just imitate the proof of Proposition3.3. The main point to note is that if q∉Ω¯ and ν is a unit normal vector from ∂Ω out through q, then the function
(4.3)ψ(z)=1(z-q)·ν
is holomorphic and bounded on Ω and is singular at q. So the rest of the proof goes through as before.In fact more is true.Proposition 4.3.
LetΩ⊆ℂn be bounded and strongly pseudoconvex with C2 boundary. Let 1≤p≤∞. Then Ω is a domain of type HLp.Proof.
Of course we again endeavor to apply the argument of the proof of  Proposition3.3. It is enough to restrict attention to points q in cΩ which are sufficiently close to ∂Ω. If q is such a point, then there is a larger strongly pseudoconvex domain Ω′ with C2 boundary such that Ω¯⊆Ω′ and q∈∂Ω′. Now let Lq(z) be the Levi polynomial (see [1]) for Ω′ at q. Then there is a neighborhood U of q so that
(4.4){z∈U:Lq(z)=0}∩Ω′¯={q}.
Thusfq(z)=1/Lq(z) is holomorphic on Ω′∩U and singular at q. Let φ be a Cc∞ function that is compactly supported in U and is identically equal to 1 in a small neighborhood of q. We wish to choose a bounded function h so that
(4.5)g(z)=φ(z)Lq(z)+h
is holomorphic on Ω′. This entails solving the ∂¯-problem
(4.6)∂¯h=-∂¯φ(z)Lq(z).
Of course the data on the right-hand side of this equation is ∂¯-closed with bounded coefficients. By work in [15] or [16] we see that a bounded solution h exists.
This gives us a functiong that is (i) holomorphic on Ω′ and (ii) singular at q. This is just what we need, for points q in cΩ that are close to ∂Ω, in order to imitate the proof of  Proposition 3.3. That completes our argument. See also [4, Theorem  3.6] for a similar result with a somewhat different proof in the case p=∞.For finite type domains we can prove the following result. LetΩ⊆ℂ2 be given by Ω={z∈ℂ2:ρ(z)<0}. Recall that a point q in the boundary of a domain Ω is said to be of finite geometric typem in the sense of Kohn if there is a nonsingular, one-dimensional analytic variety φ:D→ℂ2 with φ(0)=q and(4.7)|ρ(φ(ζ))|≤C·|φ(ζ)-q|m,
and so that there is no other nonsingular, one-dimensional analytic variety satisfying a similar inequality with m replaced by m+1. These ideas are discussed in detail in [1, Chapter 10].It is known that the geometric definition of finite type given in the last paragraph is equivalent to a more analytic one in terms of commutators of vector fields. Namely, let(4.8)L=∂ρ∂z1∂∂z2-∂ρ∂z2∂∂z1
be a complex tangential vector field to ∂Ω and L¯ its conjugate. A first-order commutator is a Lie bracket of the form [L,L¯]. A second-order commutator is a Lie bracket of the form [L,M] or [L¯,M], where M is a first-order commutator, and so forth. We say that a point q∈∂Ω is of analytic typem if all the commutators ℒ up to and including order m-1 have the property that
(4.9)ℒ(ρ)[q]=0,
but there is a commutator ℒ′ of order m such that
(4.10)ℒ′(ρ)[q]≠0.It is a result of Kohn [17] and Bloom and Graham [18] that, when Ω⊆ℂ2, a point q∈∂Ω is of geometric finite type if and only if it is of analytic finite type. Details of these matters may be found in [1].Now it is easy to see that the notion of analytic finite type varies semi-continuously with smooth variation ofρ. In particular, if each point of ∂Ω is of some finite type, then the type of the point will vary semi-continously. So there is an upper bound M for all types of points in ∂Ω. In this circumstance we say that Ω is a domain of finite type (at most) M.As a result of these considerations, one has the following lemma.Lemma 4.4.
LetΩ={z∈ℂ2:ρ(z)<0} be a domain of finite type M. Then there are domains Ω′ of finite type so that Ω′⊃Ω¯. In particular, if ψ is a smooth, negative function with ∥ψ∥CM+1 sufficiently small and ρ′=ρ+ψ then Ω′≡{z∈ℂ2:ρ'(z)<0} will contain Ω¯ and be of finite type.Now we have the following proposition.Proposition 4.5.
LetΩ⊆ℂ2 be smoothly bounded and of finite type m. Let 1≤p≤∞. Then Ω is a domain of type HLp.Proof.
The argument is similar to that for the last few propositions. Ifq∉Ω¯ and is sufficiently close to ∂Ω, then we may use the last lemma and the discussion preceding that to construct a finite type domain Ω′⊃Ω¯ and with q∈∂Ω′. Now the theorem of Bedford and Fornæss [19] gives us a peaking function fq for the point q on the domain Ω′. That is to say,  (i)
fq is continuous on Ω′¯;(ii)
fq is holomorphic on Ω′;(iii)
|fq(z)|≤1 for all z∈Ω′¯;(iv)
fq(q)=1;(v)
|fq(z)|<1 for all z∈Ω′¯∖{q}.Then the function gq(z)=1/[1-fq(z)] is holomorphic on Ω and singular at q.
The rest of the argument is completed as in the proof of the last proposition.We note that the Kohn-Nirenberg domain [20] shows that, even on a finite type domain in ℂ2, we cannot hope for a holomorphic separating function like Lq in the strongly pseudoconvex case. But the peak function of Bedford-Fornæss suffices for our purposes.Proposition 4.6.
LetΩ⊆ℂn be a bounded analytic polyhedron. Certainly Ω is then a domain of holomorphy. We have that Ω is a domain of type HLp for 1≤p≤∞.Proof.
We know by the standard definition (see [1]) that
(4.11)Ω={z∈ℂn:|f1(z)|<1,|f2(z)|<1,…,|fk(z)|<1}
for some holomorphic functions fj. Now if q∉Ω¯, then there is some complex constant λ with |λ|>1 and some j so that fj(q)=λ. That being the case, the function
(4.12)ψ(z)=1λ-fj(z)
is a function that is bounded and holomorphic on Ω but singular at q. Now the proof can be completed as in the previous propositions.Proposition 4.7.
LetΩ⊆ℂn be a complete circular domain. Assume that Ω is pseudoconvex. Then Ω is a domain of type HLp, 1≤p≤∞.Proof.
Letq be a point that does not lie in Ω¯. Let q* be the nearest point to q in the boundary of Ω, and let ν be the unit outward normal vector at q*. Set
(4.13)fq(z)=(z-q)·ν.
Then fq is holomorphic, and we claim that the zero set 𝒵q of fq does not intersect Ω¯. Suppose to the contrary that it does.
Letx be a point that lies in both Ω¯ and in 𝒵q. Of course any point that can be obtained by rotating the coordinates of x will also lie in Ω¯. One such choice of rotations will give a point that lies on the ray from the origin out to q. But that rotated point will be further from the origin than q itself (by the Pythagorean theorem). Since it lies in Ω¯, then so does q (because the domain is complete circular). That is a contradiction. Therefore x does not exist and Ω¯ and the zero set of fq are disjoint.
As a result, the functiongq≡1/fq is holomorphic and bounded on Ω and singular at q. The proof may now be completed as in the preceding propositions.The next result points in the general direction that any reasonable pseudoconvex domain will be of typeHLp for 1≤p≤∞.Proposition 4.8.
LetΩ be a bounded, pseudoconvex domain with a Stein neighborhood basis. (Here a Stein neighborhood basis  for Ω is a decreasing collection of pseudoconvex domains Ωj such that ∩jΩj=Ω¯; see [21] for further details in this matter.) Then Ω is a domain of type HLp for 1≤p≤∞.Remark 4.9.
Of course a domain with Stein neighborhood basis can have rough boundary. So this proposition says something new and with content.Proof.
Letϵ>0. By definition of Stein neighborhood basis, there is a pseudoconvex domain Ω̃ so that Ω̃⊇Ω¯. Therefore (see [1, Chapter 3]) there is a smoothly bounded, strongly pseudoconvex domain Ω̃̃ so that Ω̃⊇Ω̃̃⊃Ω¯. Let P∈∂Ω̃̃. Then we may imitate the construction in the proof of  Proposition 3.3 to find a function that is holomorphic and bounded on Ω, extends past the boundary of Ω, but is singular at P. Now the rest of the argument—elementary functional analysis—is just as in the proof of  Proposition 3.3.The interest of Propositions4.5, 4.6, and 4.7 is that the domains constructed there have only Lipschitz boundary. We know for certain (thanks to Catlin and Hakim/Sibony) that pseudoconvex domains with smooth boundary are of type HL∞. And there are domains with rough boundary, such as the Sibony domain, that are of type EL∞. So the last two propositions give examples of domains with rough boundary which are of type HL∞.


## 5. Other Properties ofHLp and ELp Domains

In [4] an example is given which shows that the increasing union of HL∞ domains need not be HL∞. Indeed, it is well known (see [22]) that any domain of holomorphy is the increasing union of analytic polyhedra (see Proposition 3.3). Of course an analytic polyhedron is HLp for 1≤p≤∞, but the Sibony domain (which is certainly the union of analytic polyhedra) described above is pseudoconvex and not HL∞. Berg in addition shows (see his Theorem 1.15) that the decreasing intersection of HL∞ domains is HL∞.Now we describe some other related examples. Again see [4] for cognate ideas.Example 5.1.
There is a decreasing sequenceΩ1⊇Ω2⊇⋯ of EL∞ domains such that the intersection domain Ω0≡∩jΩj is not EL∞.
To see this, we follow the construction of [2, page 206]. Let {aj} be a sequence in the unit disc D with no interior accumulation point and such that every boundary point of D is the nontangential limit of some subsequence. Let λj be a summable sequence of positive real numbers. Define, for ϵ>0 and z∈D,
(5.1)φϵ(z)=∑jϵλjlog|z-aj2|.
Then certainly φϵ is subharmonic and negative on D. Further note that the functions φϵ increase pointwise to the identically 0 function as ϵ→0+. Now set
(5.2)V0ϵ(z)=exp(φϵ(z)).
Then V0ϵ is also subharmonic, 0≤V0ϵ<1. The function takes the value 0 only at the points aj.
Finally define the domains(5.3)Mϵ(D,V0ϵ)={(z,w)∈ℂ2:z∈D,w∈ℂ,|w|<exp(-V0ϵ(z))}.
Each Mϵ(D,V0ϵ) is pseudoconvex. And the argument of Sibony shows that it is a domain of type EL∞. But notice that the function exp(-V0ϵ(z)) decreases pointwise to the function that is identically equal to 1/e as ϵ→0+. Hence the domains Mϵ(D,V0ϵ) decrease to the bidisc {(z,w):z∈D,|w|<1/e}. And the latter is a domain of type HL∞.
So we have produced a decreasing sequence ofEL∞ domains whose intersection is HL∞.We now give a separate proof, which has independent interest, of the contrapositive of Proposition4.8.Proposition 5.2.
IfΩ is a bounded domain of type ELp, 1≤p≤∞ then Ω does not have a Stein neighborhood basis.Proof.
Suppose that every holomorphicLp function on Ω analytically continues to a larger domain Ω̂. Seeking a contradiction, we assume that Ω has a Stein neighborhood basis. Choose a pseudoconvex domain U⊇Ω¯ so that Ω̂∖U is nonempty.
Now there is some holomorphic functiong on U that does not analytically continue to any larger open domain. Therefore the restriction of g to Ω is a holomorphic Lp function g̃ on Ω that analytically continues to U but no further. This contradicts the fact that g̃ must analytically continue to Ω̂. We conclude that Ω cannot have a Stein neighborhood basis.We close with the following useful property ofEL∞ domains.Proposition 5.3.
LetΩ be a bounded, EL∞ domain in ℂn, so that any bounded, holomorphic function f on Ω analytically continues to some bounded, holomorphic function f̂ on some Ω̂. Let f be a bounded, holomorphic function on Ω so that |f| is bounded from 0 by some η>0. Then f̂ will be nonvanishing.Proof.
Of courseg=1/f makes sense on Ω and is holomorphic and bounded; so it analytically continues to some bounded, holomorphic function ĝ on Ω̂. But of course 1≡f·g analytically continues to the identically 1 function on Ω̂. So we see that f̂·ĝ≡1 on Ω̂. We conclude then that f̂ cannot vanish.


## 6. Relationship with the∂¯-Problem

In the paper [5], Sibony exhibits a smoothly bounded, pseudoconvex domain on which the equation
(6.1)∂¯u=f,
for f a ∂¯-closed (0,1) form with bounded coefficents, has no bounded solution u. This is important information for function theory, and also for the theory of partial differential equations.It is natural to speculate that there is some relation between those domains on which the∂¯-equation satisfies uniform estimates and those domains which are of type HL∞. In that vein, we offer the following result.Proposition 6.1.
LetΩ⊆ℂn be a bounded domain which is of finite type m and so that the ∂¯-equation ∂¯u=f satisfies uniform estimates on Ω. That is to say, there is a universal constant C>0 so that, given a ∂¯-closed (0,1) form f with bounded coefficients, there is a solution u to the equation ∂¯u=f with
(6.2)∥u∥L∞≤C·∥f∥L∞.
Then Ω is a domain of type HL∞.Remark 6.2.
It is important to notice in this last proposition that the domainΩ need not have C∞ boundary. For type 2, it suffices for the boundary to be C2. For type m≥2, it suffices for the boundary to be Cm.
It is known that strongly pseudoconvex domains [15], finite type domains in ℂ2[23], and the polydisc [24] all satisfy uniform estimates for the ∂¯-problem.Proof.
It is known (see, e.g., [25, 26] or [27, 28]) that the strongly pseudoconvex points in ∂Ω form an open, dense set. Let q∈∂Ω be such a point and let ϵ>0. Let ν be the unit outward normal vector to ∂Ω at q and set q′=q+ϵν. If ϵ is small then there is a “bumped domain’’ Ω′ with these properties.  (i)
There is a small neighborhoodU of q so that U∩∂Ω consists only of strongly pseudoconvex points.(ii)
∂Ω∖U=∂Ω′∖U.(iii)
∂Ω′∩U is strongly pseudoconvex and lies outside Ω.(iv)
distEuclid(q,∂Ω′)>0.(v)
q′∈∂Ω′.We exhibit the situation in Figure 1.
Now letLq′ be the Levi polynomial for ∂Ω′ at q′. Let φ∈Cc∞(U) be identically equal to 1 in a small neighborhood of q′.
We do not knowa priori that the ∂¯-problem satisfies uniform estimates on the domain Ω'. But we may apply the construction of Beatrous and Range [29] to see that this is in fact the case (we thank Frank Beatrous and R. Michael Range for helpful remarks regarding this device). In detail, suppose that f is a ∂¯ closed form on Ω′. Solve ∂¯u=f on Ω with uniform estimates. Let χ be a cutoff function which is 0 on U and identically 1 in the complement of a slightly larger strongly pseudoconvex neighborhood of q. Let u0=χu, extended as zero across the perturbed part of the boundary. Let f0=f-∂¯u0, which is defined and bounded on Ω′ and vanishes in a neighborhood of ∂Ω∖U. We can therefore solve ∂¯v=f0 in Ω′, with uniform estimates, by [29], Theorem 1.1. The solution in Ω′ to the original equation is then u0+v. And that solution is bounded.
Now we use this last result to solve the equation(6.3)∂¯u=(∂¯φ)·1Lq′
on Ω′. The data on the righthand side is ∂¯-closed and has bounded coefficients. So there is a bounded solution u by our hypothesis.
Set(6.4)h(z)=φ(z)·1Lq′(z)-u.
Then h is holomorphic and bounded on Ω and does not analytically continue past q′. So we may complete the argument just as in the proofs of  Proposition 3.3.Figure 1
The domainsΩ and Ω′.Corollary of the proof
IfΩ is a smoothly bounded domain on which uniform estimates for the ∂¯-equation hold, and if Ω′ is a domain obtained from Ω by perturbing the strongly pseudoconvex points (so that the perturbed points are also strongly pseudoconvex), then the ∂¯-problem on Ω′ also satisfies uniform estimates.We conclude this section by noting that in fact the proof of Theorem1.1 in [29] goes through verbatim if “strongly pseudconvex’’ is replaced by “finite type’’ in ℂ2. As a result, in view of the discussion above, we have the following proposition.Proposition 6.3.
IfΩ is a smoothly bounded domain in ℂ2 on which uniform estimates for the ∂¯-equation hold, and if Ω′ is a domain obtained from Ω by perturbing the finite type points (so that the perturbed points are also finite type), then the ∂¯-problem on Ω' also satisfies uniform estimates.


## 7. Peak Points

We have seen peak points and peaking functions put to good use in the proof of Proposition4.5. Now we will see them in a more general context.LetΩ be a domain of type EL∞. So Ω is pseudoconvex, and there is a strictly larger domain Ω̂ so that every bounded holomorphic function on Ω analytically continues to a bounded holomorphic function f̂ on Ω̂. Of course the operator T:f↦f̂ is linear. It is one-to-one and onto. It follows from the closed graph theorem that T continuous. Now we have a lemma.Lemma 7.1.
The operatorT has norm 1.Proof.
Of course the norm ofT is at least 1. Suppose that it is actually greater than 1. Then there is an H∞ function f on Ω so that f has norm 1, and its extension f̂ has norm greater than 1. For k being a positive integer consider gk=fk. Then the extension of gk to Ω̂ is ĝk=(f̂)k. As k→+∞, the norm of ĝk tends to +∞ while the norm of gk remains 1. That is a contradiction.Proposition 7.2.
LetΩ⊆ℂn be a domain and let q∈∂Ω be a peak point (see the proof of Proposition 4.5). Let fq be the peaking function. Then there cannot be a domain Ω̂ which properly contains Ω so that (i) any bounded holomorphic function on Ω analytically continues to Ω̂ and (ii)q lies in the interior of Ω̂.Proof.
Suppose to the contrary that there is such a domainΩ̂. Then the holomorphic function fq analytically continues to a function f̂q on Ω̂. Of course fq has H∞ norm 1. Thus the extended function f̂q will also have norm 1. But f̂q(q)=1. This contradicts the maximum modulus principle unless fq≡1. But that is impossible by the definition of peak function.Remark 7.3.
In fact one does not need the full force ofq being a peak point in order for this last result to hold. It is sufficient, for instance, for the nontangential limit of f at q to be 1, and the values of f at other points of Ω have modulus smaller than 1.
It may also be noted that, by a result of Basener [30], the set of peak points for a domain is contained in the closure of the strongly pseudoconvex points. This observation is helpful in applying the last proposition.


## 8. Concluding Remarks

It would have been best if we could have given a characterization ofHLp domains or ELp domains. Unfortunately such a result is beyond our reach at this time.We hope that the information gathered here will help to inform the situation and lead, in future work, to increased understanding of this fascinating problem. It is clear that there is a spectrum of domains of holomorphy, and it is in our best interest to understand the elements of this spectrum.



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