#### Abstract

We treat the classical concept of domain of holomorphy in 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 is considered.

#### 1. Introduction

In this paper a *domain * is a connected open set. We let denote the algebra of holomorphic functions on .

We will use the following notation: denotes the unit disc in the complex plane. We let denote the bidisc, and the polydisc in . The symbol is the unit ball in .

A domain is said to be *Runge* if any holomorphic on is the limit, uniformly on compact subsets of , 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 domain is a domain of holomorphy if and only if it is pseudoconvex. These matters are discussed in some detail in [1].

Roughly speaking, if is a domain of holomorphy, then there is a holomorphic function on such that cannot be analytically continued to any larger domain. Generally speaking one cannot say much about the nature of this —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 domain , with being a proper subset of the bidisc , such that any *bounded* holomorphic function on analytically continues to all of .

Of course this result of Sibony can be extended to in a variety of ways. For one thing, one may take the product of the Sibony domain with the polydisc in to obtain an example in . Alternatively, one may replace the first (or ) variable in the Sibony construction with a tuple in to obtain a counterexample in . 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 domains , with , such that any bounded holomorphic function on analytically continues to all of (as in the Sibony example above) is uncountable.*

*Proof. *We very quickly review the key steps of the Sibony construction.

Let be a countable collection of points in the unit disc , with no interior accumulation point, so that every boundary point of is an accumulation point of the set . Now define
Here is a summable sequence of positive, real numbers. Notice that the function —being the sum of subharmonic functions—is subharmonic. Define
Then has the properties: (i) is subharmonic; (ii) for all ; (iii)the function is continuous. The last property holds because the sequence is discrete and takes the value 0 only at the .

Now define the domain
Since is positive, we see that this definition makes sense and that is a proper subset of .

The remainder of Sibony's argument shows that any bounded, holomorphic function on analytically continues to a bounded, holomorphic function on . We will not repeat it but refer the reader to [2, 5].

The key fact in the Sibony construction is that the points form a discrete set that accumulates at every boundary point of . Apart from this property, there is complete freedom in choosing the . 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, for ,
Set (the sequence consisting of 4 equally spaced points on , 8 equally spaced points on , 16 equally spaced points on , etc.), (the sequence consisting of 8 equally spaced points on , 16 equally spaced points on , 32 equally spaced points on , etc.). Define a domain using the Sibony construction, as above, with the sequence and define a domain using the Sibony construction with the sequence . We claim that and are biholomorphically inequivalent.

To see this, suppose the contrary. So there is a biholomorphic mapping . By the usual classical arguments (see the proof of Proposition in [1]), we see that must commute with rotations in the variable. It follows that any disc in of the form
for fixed, must be mapped to a similar disc in .

Further observe that each of the discs is a totally geodesic submanifold in the Kobayashi metric. This assertion follows immediately from the existence of the maps
where is the injection
and is the projection
Observe that . Similar reasoning shows that the disc is totally geodisc. Of course similar remarks apply to the corresponding discs in .

Now it is essential to notice that, in , the vertical discs of the form for not one of the are *not* totally geodesic. This follows because, at the point , the Kobayashi extremal disc in the vertical direction will not be the rigid disc but rather a disc that curves into one of the spikes above a nearby .

A final observation that we need to make is this. The vertical totally geodesic discs will be mapped to each other by the biholomorphic mapping . But more is true. Because is totally geodesic, in fact the totally geodesic discs located at the points that lie on in must be mapped to the totally geodesic discs located at the points that lie on in (because both circles consist of points that have the same Kobayashi distance from the origin). This is impossible because in has four such points while in has eight such points. That is the required contradiction.

Now it is clear how to construct uncountably many inequivalent domains of Sibony type. For if is any exponentially increasing sequence of positive integers, then, we may associate to it a domain of Sibony-type with points on , points on , and so forth. The preceding argument shows that different choices of result in biholomorphically inequivalent domains. And there are clearly uncountably many such sequences. That completes the argument.

Recall that if is a domain and is a family of functions on , then we say that is convex with respect to if, whenever is relatively compact, then is also relatively compact in . 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 domain and the family of all *bounded holomorphic functions* on .

Proposition 1.3. *The Sibony domain is convex with respect to the family .*

*Proof. *Let be as above for the domain and let be the usual family of *all* holomorphic functions on . Let be a compact subset of and the hull with respect to . Then, since is a domain of holomorphy, is still compact in . Let be a point that does not lie in . Then there is a holomorphic function on so that . Let . Since is Runge, there is a polynomial so that on . But then is bounded on and, by the triangle inequality, . Thus . This shows that is convex with respect to the family of bounded holomorphic functions on .

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. *Let be a bounded domain. Suppose that is a Banach space of holomorphic functions on that contains . Let be a strictly larger domain that contains . Assume that any element of analytically continues to a holomorphic function on (one often assumes that the extended function satisfies a similar norm estimate to that specified by the norm on , 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 . That is to say, if are holomorphic functions in with no common zero, then there do not exists elements such that
**
on .*

*Proof. *Assume to the contrary that such exist. Of course each analytically continues to .

Let be a point of . Set . Then the have no common zero in . So, by hypothesis, the exist. And these functions extend analytically to . But then
on . Since the all vanish at , we see that, at , the left-hand side of this last equation vanishes. That is clearly a contradiction. Hence the do not exist.

Of course this last proposition means in particular that the point evaluations on are not weak- dense in the maximal ideal space of ; see [6] for more on these matters.

By contrast to Sibony's result, Catlin [7] has shown that any smoothly bounded, pseudoconvex domain in supports a bounded holomorphic function that cannot be analytically continued to any larger domain. In fact he has proved something sharper.

Theorem 1.5. *Let be a smoothly bounded pseudoconvex domain. Then there is a function in , 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. *Let be a smoothly bounded pseudoconvex domain. Then the maximal ideal space (or spectrum) of the algebra is in fact .*

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 on . Attempts to adapt the arguments to other function spaces are doomed to failure.

Pflug and Zwonek [10] have shown that the situation for holomorphic functions is very neat and elegant.

Theorem 1.7. *Let be any pseudoconvex domain. There is an holomorphic function on that cannot be continued to any larger domain if and only if, for all and all neighborhoods of , 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 that in is a domain of holomorphy and that, for each , there is a sequence such that and there are and such that . Then there is an holomorphic function on that cannot be analytically continued to any larger domain.*

It is natural to ask for a characterization of those domains 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 holomorphic functions, .

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 .

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 domain is of type , , if there is a holomorphic function on , , which cannot be analytically continued to any larger domain. We instead say that is of type if there is a strictly larger domain so that every holomorphic function on analytically continues to . Obviously and are disjoint.

We are interested in giving an extrinsic description of those domains which are of type and those which are of type . It is not the case in higher dimensions, for instance, that domains are the same as domains that are convex with respect to (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
Of course, by the Riemann removable singularities theorem, any bounded holomorphic function on analytically continues to all of . So is *not* a domain of type . It *is* a domain of type .

In fact it may be noted (for the domain in the last paragraph) that, if , then any holomorphic function that is will analytically continue to all of (see [13]). So this is a domain of type . By contrast, if , then the function is holomorphic on and in . But of course this * does not* analytically continue to the full disc . So, for , the domain is of type .

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 be holomorphic on and assume that (the case follows immediately from this one). We write . For and a negative integer, consider the expression

where it is understood that .

On the one hand,
On the other hand,
If and , this gives a contradiction as . Of course the cases and can be handled separately because and are certainly *not* in .

*Remark 3.1. *We note that the proof goes through for up until the very end. One must note that in fact *does* lie in for . So there is no removable singularities theorem for this range of .

*Remark 3.2. *A standard result coming from potential theory is that, if , , and is holomorphic on , then (where we are using Landau's notation) implies that continues analytically to all of . The philosophy here is that a function satisfying this growth hypothesis has a singularity at 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 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 that is not equal to the interior of its closure. In fact we have the following proposition.

Proposition 3.3. *Suppose that the bounded domain is the interior of its closure. Then is a domain of type for .*

*Proof. *The proof that we now present is an adaptation and simplification of an argument from [4].

Let be a countable, dense subset of . For each , the function is holomorphic and bounded on and does not analytically continue past .

Now, for each , let be an open disc centered at which has nontrivial intersection with . Consider the linear mapping
given by restriction. Of course each of the indicated spaces is equipped with the norm, and is therefore a Banach space. We note that the example above of shows that is not surjective. As a result, the open mapping principle tells us that the image of is of first category in . Therefore, by the Baire category theorem,
is of first category in . But this just says that the set of holomorphic functions on that can be analytically continued to some is of first category. Therefore the set of holomorphic functions that *cannot* be analytically continued across the boundary is dense in . 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 when is the unit disc then it is easy to construct a bounded holomorphic function that does not analytically continue to a larger domain. Let be a discrete set in that accumulates at every boundary point and so that For example, take

, , , to be equally spaced points at distance 1/4 from , to be equally spaced points at distance 1/8 from , to be equally spaced points at distance 1/16 from ,and so forth. Then the Blaschke product with zeros at the will do the job. If 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 .

We close this section by noting that, if is a domain of holomorphy in and if for some holomorphic on (we call a *variety*), then is also a domain of holomorphy (if is a holomorphic function on that does not analytically continue to a larger domain then is a holomorphic function on that does not analytically continue to any larger domain. And it is easy to see that is an domain; see [14, page 19] for the details).

#### 4. Complications in Dimension

As we have indicated, the example of Sibony exhibits a domain which is *not* of type (instead it is of type ). The theorem of Catlin shows that all smoothly bounded, pseudoconvex domains are of type .

It of course makes sense to focus this discussion on pseudoconvex domains. If a domain is *not* pseudoconvex, then there will perforce be a larger domain to which all holomorphic functions (regardless of growth) on analytically continue. So this situation is not interesting.

Thus we see that the domains of interest for us will be pseudoconvex domains that do *not* have smooth boundary. Our first result is as follows.

Proposition 4.1. *Let be bounded domains in , each of which is equal to the interior of its closure. Define
**
Then is a domain of type for any .*

*Proof. *Fix as indicated. Then, by Proposition 3.3, there is a holomorphic function on , for , such that is holomorphic and on and does not analytically continue to any larger domain.

But then

is holomorphic and on and does not analytically continue to any larger domain.

Proposition 4.2. *Let be bounded and convex. Let . Then is a domain of type .*

*Proof. *Just imitate the proof of Proposition 3.3. The main point to note is that if and is a unit normal vector from out through , then the function
is holomorphic and bounded on and is singular at . So the rest of the proof goes through as before.

In fact more is true.

Proposition 4.3. *Let be bounded and strongly pseudoconvex with boundary. Let . Then is a domain of type .*

*Proof. *Of course we again endeavor to apply the argument of the proof of Proposition 3.3. It is enough to restrict attention to points in which are sufficiently close to . If is such a point, then there is a larger strongly pseudoconvex domain with boundary such that and . Now let be the Levi polynomial (see [1]) for at . Then there is a neighborhood of so that

Thus is holomorphic on and singular at . Let be a function that is compactly supported in and is identically equal to 1 in a small neighborhood of . We wish to choose a bounded function so that
is holomorphic on . This entails solving the -problem
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 exists.

This gives us a function that is (i) holomorphic on and (ii) singular at . This is just what we need, for points in that are close to , 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 .

For finite type domains we can prove the following result. Let be given by . Recall that a point in the boundary of a domain is said to be of *finite geometric type * in the sense of Kohn if there is a nonsingular, one-dimensional analytic variety with and

and so that there is no other nonsingular, one-dimensional analytic variety satisfying a similar inequality with replaced by . 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
be a complex tangential vector field to and its conjugate. A *first-order commutator* is a Lie bracket of the form . A *second-order commutator* is a Lie bracket of the form or , where is a first-order commutator, and so forth. We say that a point is of *analytic type * if all the commutators up to and including order have the property that
but there is a commutator of order such that

It is a result of Kohn [17] and Bloom and Graham [18] that, when , a point 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 of . In particular, if each point of is of some finite type, then the type of the point will vary semi-continously. So there is an upper bound for all types of points in . In this circumstance we say that is a domain of finite type (at most) .

As a result of these considerations, one has the following lemma.

Lemma 4.4. *Let be a domain of finite type . Then there are domains of finite type so that . In particular, if is a smooth, negative function with sufficiently small and then will contain and be of finite type.*

Now we have the following proposition.

Proposition 4.5. *Let be smoothly bounded and of finite type . Let . Then is a domain of type .*

*Proof. *The argument is similar to that for the last few propositions. If and is sufficiently close to , then we may use the last lemma and the discussion preceding that to construct a finite type domain and with . Now the theorem of Bedford and Fornæss [19] gives us a peaking function for the point on the domain . That is to say, (i) is continuous on ; (ii) is holomorphic on ; (iii) for all ; (iv); (v) for all . Then the function is holomorphic on and singular at .

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 , we cannot hope for a holomorphic separating function like in the strongly pseudoconvex case. But the peak function of Bedford-Fornæss suffices for our purposes.

Proposition 4.6. *Let be a bounded analytic polyhedron. Certainly is then a domain of holomorphy. We have that is a domain of type for .*

*Proof. *We know by the standard definition (see [1]) that
for some holomorphic functions . Now if , then there is some complex constant with and some so that . That being the case, the function
is a function that is bounded and holomorphic on but singular at . Now the proof can be completed as in the previous propositions.

Proposition 4.7. *Let be a complete circular domain. Assume that is pseudoconvex. Then is a domain of type , .*

*Proof. *Let be a point that does not lie in . Let be the nearest point to in the boundary of , and let be the unit outward normal vector at . Set
Then is holomorphic, and we claim that the zero set of does not intersect . Suppose to the contrary that it does.

Let be a point that lies in both and in . Of course any point that can be obtained by rotating the coordinates of will also lie in . One such choice of rotations will give a point that lies on the ray from the origin out to . But that rotated point will be further from the origin than itself (by the Pythagorean theorem). Since it lies in , then so does (because the domain is complete circular). That is a contradiction. Therefore does not exist and and the zero set of are disjoint.

As a result, the function is holomorphic and bounded on and singular at . 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 type for .

Proposition 4.8. *Let be a bounded, pseudoconvex domain with a Stein neighborhood basis. (Here a Stein neighborhood basis for is a decreasing collection of pseudoconvex domains such that ; see [21] for further details in this matter.) Then is a domain of type for .*

*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. *Let . By definition of Stein neighborhood basis, there is a pseudoconvex domain so that . Therefore (see [1, Chapter 3]) there is a smoothly bounded, strongly pseudoconvex domain so that . Let . Then we may imitate the construction in the proof of Proposition 3.3 to find a function that is holomorphic and bounded on , extends past the boundary of , but is singular at . Now the rest of the argument—elementary functional analysis—is just as in the proof of Proposition 3.3.

The interest of Propositions 4.5, 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 . And there are domains with rough boundary, such as the Sibony domain, that are of type . So the last two propositions give examples of domains with rough boundary which are of type .

#### 5. Other Properties of and Domains

In [4] an example is given which shows that the increasing union of domains need not be . 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 for , but the Sibony domain (which is certainly the union of analytic polyhedra) described above is pseudoconvex and not . Berg in addition shows (see his Theorem ) that the decreasing intersection of domains is .

Now we describe some other related examples. Again see [4] for cognate ideas.

*Example 5.1. *There is a decreasing sequence of domains such that the intersection domain is not .

To see this, we follow the construction of [2, page 206]. Let be a sequence in the unit disc with no interior accumulation point and such that every boundary point of is the nontangential limit of some subsequence. Let be a summable sequence of positive real numbers. Define, for and ,
Then certainly is subharmonic and negative on . Further note that the functions increase pointwise to the identically 0 function as . Now set
Then is also subharmonic, . The function takes the value 0 only at the points .

Finally define the domains
Each is pseudoconvex. And the argument of Sibony shows that it is a domain of type . But notice that the function decreases pointwise to the function that is identically equal to as . Hence the domains decrease to the bidisc . And the latter is a domain of type .

So we have produced a decreasing sequence of domains whose intersection is .

We now give a separate proof, which has independent interest, of the contrapositive of Proposition 4.8.

Proposition 5.2. *If is a bounded domain of type , then does not have a Stein neighborhood basis.*

*Proof. *Suppose that every holomorphic function on analytically continues to a larger domain . Seeking a contradiction, we assume that has a Stein neighborhood basis. Choose a pseudoconvex domain so that is nonempty.

Now there is some holomorphic function on that does not analytically continue to any larger open domain. Therefore the restriction of to is a holomorphic function on that analytically continues to but no further. This contradicts the fact that must analytically continue to . We conclude that cannot have a Stein neighborhood basis.

We close with the following useful property of domains.

Proposition 5.3. *Let be a bounded, domain in , so that any bounded, holomorphic function on analytically continues to some bounded, holomorphic function on some . Let be a bounded, holomorphic function on so that is bounded from by some . Then will be nonvanishing.*

*Proof. *Of course makes sense on and is holomorphic and bounded; so it analytically continues to some bounded, holomorphic function on . But of course analytically continues to the identically 1 function on . So we see that on . We conclude then that cannot vanish.

#### 6. Relationship with the -Problem

In the paper [5], Sibony exhibits a smoothly bounded, pseudoconvex domain on which the equation for a -closed form with bounded coefficents, has no bounded solution . 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 . In that vein, we offer the following result.

Proposition 6.1. *Let be a bounded domain which is of finite type and so that the -equation satisfies uniform estimates on . That is to say, there is a universal constant so that, given a -closed form with bounded coefficients, there is a solution to the equation with
**
Then is a domain of type .*

*Remark 6.2. *It is important to notice in this last proposition that the domain need not have boundary. For type 2, it suffices for the boundary to be . For type , it suffices for the boundary to be .

It is known that strongly pseudoconvex domains [15], finite type domains in [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 form an open, dense set. Let be such a point and let . Let be the unit outward normal vector to at and set . If is small then there is a “bumped domain’’ with these properties. (i)There is a small neighborhood of so that consists only of strongly pseudoconvex points. (ii). (iii) is strongly pseudoconvex and lies outside . (iv). (v). We exhibit the situation in Figure 1.

Now let be the Levi polynomial for at . Let be identically equal to 1 in a small neighborhood of .

We do not know *a priori* that the -problem satisfies uniform estimates on the domain . 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 is a closed form on . Solve on with uniform estimates. Let be a cutoff function which is 0 on and identically 1 in the complement of a slightly larger strongly pseudoconvex neighborhood of . Let , extended as zero across the perturbed part of the boundary. Let , which is defined and bounded on and vanishes in a neighborhood of . We can therefore solve in , with uniform estimates, by [29], Theorem . The solution in to the original equation is then . And that solution is bounded.

Now we use this last result to solve the equation
on . The data on the righthand side is -closed and has bounded coefficients. So there is a bounded solution by our hypothesis.

Set
Then is holomorphic and bounded on and does not analytically continue past . So we may complete the argument just as in the proofs of Proposition 3.3.

*Corollary of the proof*

If is a smoothly bounded domain on which uniform estimates for the -equation hold, and if is a domain obtained from by perturbing the strongly pseudoconvex points (so that the perturbed points are also strongly pseudoconvex), then the -problem on also satisfies uniform estimates.

We conclude this section by noting that in fact the proof of Theorem in [29] goes through verbatim if “strongly pseudconvex’’ is replaced by “finite type’’ in . As a result, in view of the discussion above, we have the following proposition.

Proposition 6.3. *If is a smoothly bounded domain in on which uniform estimates for the -equation hold, and if is a domain obtained from by perturbing the finite type points (so that the perturbed points are also finite type), then the -problem on also satisfies uniform estimates.*

#### 7. Peak Points

We have seen peak points and peaking functions put to good use in the proof of Proposition 4.5. Now we will see them in a more general context.

Let be a domain of type . So is pseudoconvex, and there is a strictly larger domain so that every bounded holomorphic function on analytically continues to a bounded holomorphic function on . Of course the operator is linear. It is one-to-one and onto. It follows from the closed graph theorem that continuous. Now we have a lemma.

Lemma 7.1. *The operator has norm 1.*

*Proof. *Of course the norm of is at least 1. Suppose that it is actually greater than 1. Then there is an function on so that has norm 1, and its extension has norm greater than 1. For being a positive integer consider . Then the extension of to is . As , the norm of tends to while the norm of remains 1. That is a contradiction.

Proposition 7.2. *Let be a domain and let be a peak point (see the proof of Proposition 4.5). Let be the peaking function. Then there cannot be a domain which properly contains so that (i) any bounded holomorphic function on analytically continues to and (ii) lies in the interior of .*

*Proof. *Suppose to the contrary that there is such a domain . Then the holomorphic function analytically continues to a function on . Of course has norm 1. Thus the extended function will also have norm 1. But . This contradicts the maximum modulus principle unless . But that is impossible by the definition of peak function.

*Remark 7.3. *In fact one does not need the full force of being a peak point in order for this last result to hold. It is sufficient, for instance, for the nontangential limit of at to be 1, and the values of at other points of 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 of domains or 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.

#### Acknowledgments

The author is supported in part by the National Science Foundation and by the Dean of the Graduate School at Washington University. The author thanks Abtin Daghighi for helpful comments and corrections throughout the paper.