Extension Properties of Pluricomplex Green Functions

Transcription

1 Syracuse University SURFACE Mathematics - Dissertations Mathematics 2013 Extension Properties of Pluricomplex Green Functions Sidika Zeynep Ozal Follow this and additional works at: Part of the Mathematics Commons Recommended Citation Ozal, Sidika Zeynep, "Extension Properties of Pluricomplex Green Functions" (2013). Mathematics - Dissertations. Paper 72. This Dissertation is brought to you for free and open access by the Mathematics at SURFACE. It has been accepted for inclusion in Mathematics - Dissertations by an authorized administrator of SURFACE. For more information, please contact surface@syr.edu.

2 ABSTRACT In 1985, Klimek introduced an extremal plurisubharmonic function on bounded domains in C n that generalizes the Green s function of one variable. This function is called the pluricomplex Green function of Ω with logarithmic pole at a and is denoted by g Ω (, a). The aim of this thesis was to investigate the extension properties of g Ω (, a). Let Ω 0 be a bounded domain of C n and E be a compact subset of Ω 0 such that Ω := Ω 0 \E is connected. In general, g Ω (, a) cannot be extended as a pluricomplex Green function to any subdomain of Ω 0 that is strictly larger that Ω. In this thesis it was proved that if Ω 0 is a pseudoconvex, bounded complete Reinhardt domain in C n and E 0 is a strictly logarithmically convex, Reinhardt compact subset of Ω 0 with E {z 1 z n = 0} =, there exists a subdomain Ω Ω of Ω 0 such that g Ω (z, 0) = g Ω(z, 0) for any z Ω. It was also shown that in C 2, one can omit the condition E {z 1 z 2 = 0} =. The methods required to prove the results heavily use the relation between the plurisubharmonicity of poyradial functions on Reinhardt domains and convexity of related functions. Special classes of convex functions were introduced and discussed for this purpose. These methods were also used to discuss the extension properties of the pluricomplex Green functions when Ω 0 is equal to unit the bidisk in C 2 and in that case a complete solution of the problem was given.

3 EXTENSION PROPERTIES OF PLURICOMPLEX GREEN FUNCTIONS by Sidika Zeynep Ozal B.S., Bilkent University, 2003 M.S., Sabanci University, 2006 Dissertation Submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in Mathematics Syracuse University June 2013

4 c 2013 Sidika Zeynep Ozal All Rights Reserved

5 To Ayşe Safiye Özal and Ahmet Tayfun Özal

6 Acknowledgments I am greatly indebted to my advisor Dan Coman. This thesis would not have been written without his patience, guidance and support. He has shown me how a mathematician should be and it is my dream to be like him. I would like to thank all of my professors at the Department of Mathematics, who supported me throughout this process. Especially, I am grateful to all of the members of my defense committee, Dan Coman, Leonid Kovalev, Liviu Movileanu, Eugene Poletsky, Gregory Verchota and Andrew Vogel. I also would like to express my gratitude to the Özal and the Kurşungöz families, and my friends for their love and support. Last but not least, I thank Kağan Kurşungöz, for his understanding and encouragement. v

7 Contents Acknowledgments v 1 Introduction 1 2 Preliminaries Pluricomplex Green Function Plurisubharmonic Functions on Reinhardt Domains Convex Functions Polyradial Plurisubharmonic Functions Pluricomplex Green Functions on Reinhardt Domains Extension for general Reinhardt domains Statement of Theorem Special Classes of Convex Functions Proof of Theorem Proof of Theorem Reinhardt subdomains of the unit bidisk 37 vi

8 4.1 Introduction The case E {z 1 z 2 = 0} = Basic Subdomains Ω B Construction of Ω The case E {z 1 z 2 = 0} Basic Subdomains Ω B Construction of Ω Bibliography 64 vii

9 List of Figures 3.1 The case Ω C 2 and E {z 1 z 2 = 0} = Logarithmic image of the case Ω C 2 and E {z 1 z 2 = 0} = K and K E constructed in the case Ω C 2 and E {z 1 z 2 = 0} = The case Ω C 2 and E {z 1 z 2 = 0} Logarithmic image of the case Ω C 2 and E {z 1 z 2 = 0} ω in the case Ω C 2 and E {z 1 z 2 = 0} viii

10 Chapter 1 Introduction A function in C n is said to be plurisubharmonic if it is an upper semicontinuous function whose restriction to any complex line is a subharmonic function of one variable. These functions were defined by Oka [O] and Lelong [LE1] in They are useful in studying holomorphic functions and domains of holomorphy and they are invariant under holomorphic mappings. Pluripotential theory deals with the study of such functions. [B] and [KI] provide in depth surveys about pluripotential theory. For smooth plurisubharmonic functions, the Monge-Ampère operator is defined as the nth exterior power of dd c, where d = + is the exterior derivative and d c = i( ). Its definition can be extended to certain classes of non-smooth plurisubharmonic functions ([B-T1], [B-T2], [D1], [D2], [F-S], [S]). Let Ω be a domain in C n and let P SH(Ω) denote the set of plurisubharmonic functions on Ω. For S = {a 1,, a k } Ω and W = {ν 1,, ν k } 1

11 CHAPTER 1. INTRODUCTION 2 R +, we consider the Dirichlet problem u P SH(Ω) L loc (Ω \ S), u(z) = ν j log z a j + O(1) as z a j, j = 1,, k (dd c u) n = k j=1 νn j δ aj, (1.1) where δ aj denotes the Dirac mass at a j. Functions that satisfy (1.1) are called pluricomplex Green functions on Ω with logarithmic poles in S of weights in W. On bounded domains in C n, the pluricomplex Green functions, denoted by g Ω (, S, W ), can be obtained by taking the supremum of negative plurisubharmonic functions that have logarithmic poles at a j with corresponding weights ν j, j = 1,, k. Klimek [K2] used this method to construct g Ω (, a) in the case of one pole and D ly [D1] showed that for hyperconvex domains, this function is continuous and is the unique solution to (1.1). Moreover, if Ω is convex, Lempert [LM1, LM2] used extremal disks for Kobayashi metric to construct g Ω (, a). Lelong [LE2] showed that g Ω (, S, W ) is continuous and is the unique solution to (1.1) for hyperconvex domains in the case of multiple poles. It should be noted that g Ω (, S, W ) is a negative plurisubharmonic function on Ω that satisfies (1.1) and is also invariant under biholomorphic mappings. Poletsky [PO1] and Nivoche [N1] used this latter property of pluricomplex Green functions to give two other methods of constructing g Ω (, a). A non-constant holomorphic mapping from the unit disk C to Ω is called an analytic disk. Let z Ω and f be an analytic disk such that f(0) = z. Poletsky [PO1] showed that if f 1 (a) = {t 1, t 2, } is the set of all preimages of the pole a, counted with multiplicities

12 CHAPTER 1. INTRODUCTION 3 then g Ω (z, a) is the infimum of the sum of log t j, t j f 1 (a), where the infimum is taken over all analytic disks f : Ω that map 0 to z Ω. Edigarian [E] used similar constructions. This method is applicable to the case of pluricomplex Green functions with multiple poles of arbitrary weights. Nivoche [N1] showed that if Ω is strictly hyperconvex and h m (z) = sup ( 1 m log f(z) ), where the supremum is taken over all holomorphic functions f : Ω that vanish at least to order to m at the pole a then g Ω (z, a) = lim m h m (z) for all z Ω. The complex Monge-Ampère operator is nonlinear. That nonlinearity creates difficulties in studying pluricomplex Green functions with multiple poles. Lempert [LM1, LM2] showed that if the boundary of Ω is C -smooth or real-analytic, then g Ω (, a) is C or real-analytic on Ω \ {a}, respectively. These regularity results do not hold for multiple poles. Coman [CO1, CO2] obtained the formula for the pluricomplex Green function of the unit ball B n C n with two poles that have the same weight and showed that g B n(, {a 1, a 2 }) is only of class C 1,1 (B n \ {a 1, a 2 }) and not of class C 2. In the case of one pole, the pluricomplex Green function is a higher dimensional analogue of the Green function in C. But unlike that function, it is not symmetric in general, i.e. g Ω (z, a) g Ω (a, z). Cegrell [C1, C2] introduced a symmetric Green function, denoted by W Ω, obtained by taking the supremum over a set of symmetric plurisubharmonic functions with logarithmic pole. If g Ω is symmetric, it is equal to W Ω but in general, W Ω g Ω. Edigarian and Zwonek [E-Z] provided a new approach to W Ω based on a result by Poletsky [PO2]. Domains of holomorphy are another difference between analysis in C and C n, n 2. A domain of holomorphy Ω is the maximal domain of existence for holomorphic functions, which means there exists a holomorphic function defined on Ω that cannot be extended

13 CHAPTER 1. INTRODUCTION 4 to a larger open set as a holomorphic function. These domains are important in several complex variables as many problems can only be solved on such domains. In C, every open set is a domain of holomorphy. This does not hold in C n, n 2. In 1906, F. Hartogs showed that there exists a domain H in C 2 such that every holomorphic function on H extends holomorphically to a strictly larger domain that contains H. In this thesis, we mainly work with Reinhardt domains. Identifying whether a Reinhardt domain is a domain of holomorphy is simpler. A Reinhardt domain Ω is called logarithmically convex if the set {(log z 1,, log z n ) : (z 1,, z n ) Ω} R n is convex. Then Ω is a domain of holomorphy if and only if it is logarithmically convex. Let Ω 0 be a domain in C n, n 2 and let E be a compact subset of Ω 0 such that Ω 0 \ E is connected. Hartogs extension theorem states that every holomorphic function f : Ω 0 \E C has a unique holomorphic extension to Ω 0. A natural question then is whether pluricomplex Green functions of certain domains can be extended as plurisubharmonic functions on larger domains, and in particular, as pluricomplex Green functions of larger domains. In the above setting, in general the pluricomplex Green function of Ω 0 \ E cannot be extended to Ω 0, neither as a plurisubharmonic function nor as a pluricomplex Green function. However, in some cases we can extend the pluricomplex Green function of Ω 0 \ E to a strictly larger subdomain of Ω 0. In Chapter 3, we show that for some Reinhardt subdomains Ω of a pseudoconvex complete Reinhardt domain Ω 0 in C n, the pluricomplex Green function g Ω (, 0) can be extended as a pluricomplex Green function of a strictly larger Reinhardt subdomain of Ω 0. Theorem Let Ω 0 be a pseudoconvex, bounded complete Reinhardt domain in C n, and

14 CHAPTER 1. INTRODUCTION 5 let Ω = Ω 0 \ E, where E 0 is a Reinhardt compact subset of Ω 0 that satisfies the following properties: E is strictly logarithmically convex, i.e. l(e) is strictly convex, E {z 1... z n = 0} =, where l(z 1,, z n ) = (log z 1,, log z n ) for (z 1,, z n ) C n. Then there exists a Reinhardt domain Ω such that Ω Ω Ω 0 and g Ω (z, 0) = g Ω(z, 0), z Ω. In the case of C 2, we have a more general result. Theorem Let Ω 0 be a pseudoconvex, bounded complete Reinhardt domain in C 2, and let Ω = Ω 0 \ E, where E 0 is a Reinhardt compact, strictly logarithmically convex subset of Ω 0. Then there exists a Reinhardt domain Ω such that Ω Ω Ω 0 and g Ω (z, 0) = g Ω(z, 0), z Ω. The plurisubharmonicity of polyradial functions on Reinhardt domains can be discussed by related convex functions. Using some results on a special class of convex functions and a method introduced by Klimek [K3], we will be able to find Ω. In Chapter 4, we will discuss this problem in the unit bidisk 2 C 2. In this case, we can say more about Ω. When E is strictly logarithmically convex, we show that there exists a unique largest Ω. If E is logarithmically convex, but not strictly logarithmically convex, we

15 CHAPTER 1. INTRODUCTION 6 show that in some cases there are infinitely many maximal subdomains of 2 with respect to inclusion that can be taken as Ω.

16 Chapter 2 Preliminaries In this thesis, we will work with pluricomplex Green functions on Reinhardt domains. This preliminary chapter starts with a brief summary of pluricomplex Green functions. After providing some basic results about convex functions defined on domains that are not necessarily convex, we will investigate the relation between these convex functions and plurisubharmonic functions on Reinhardt domains. We will close this chapter with some results about pluricomplex Green functions on Reinhardt domains. 2.1 Pluricomplex Green Function Plurisubharmonic functions are generalizations of subharmonic functions to C n. These functions are extremely useful as they can be used to understand some features of holomorphic functions, and in describing pseudoconvex domains and domains of holomorphy. Let Ω be an open set in C n and let f : Ω [, ) be an upper semicontinuous function which is not identically equal to on any connected component of Ω. We say 7

17 CHAPTER 2. PRELIMINARIES 8 that f is plurisubharmonic if for each a Ω and b C n, the function λ f(a + λb) is subharmonic or identically equal to on every component of the set {λ C : a+λb Ω}. The class of plurisubharmonic functions on Ω is denoted by P SH(Ω). Some of the simplest examples of plurisubharmonic functions are log h, h p, for p 0, where h is a holomorphic function. It should be noted that the composition f h of a plurisubharmonic function f with a holomorphic mapping h is either plurisubharmonic or identically equal to on each connected component of its domain of definition. Now let Ω be a bounded domain in C n. If f is a plurisubharmonic function in a neighborhood of a Ω, we say that f has a logarithmic pole at a if there exists a constant C such that f(z) log z a + C for all z near a. Klimek [K2] introduced the following extremal function: g Ω (z, a) = sup {f(z) : f PSH(Ω, [, 0)) and f has a logarithmic pole at a}. The function g Ω (, a) is called the pluricomplex Green function of Ω with pole at a. It is a generalization to higher dimensions of the Green function for the Laplace operator in C. g Ω (, a) is negative and plurisubharmonic in Ω, it has a logarithmic pole at a, and it is maximal in Ω \ {a}. It is also decreasing under composition with holomorphic mappings, which implies that it is invariant under biholomorphic mappings. Basic examples of pluricomplex Green functions are g B n(z, 0) = log z and g n(z, 0) = max 1 i n log z i, where B n and n denote the unit ball and the unit polydisk in C n, respectively, and z = (z 1,, z n ) C n. Note that g B n(, 0) is differentiable but g 2(, 0) is only continuous. This shows that the unit ball and the unit polydisk in C n are not biholomorphic to each other.

18 CHAPTER 2. PRELIMINARIES 9 For smooth plurisubharmonic functions, the Monge-Ampère operator is defined by (dd c v) n := dd c v dd c v }{{} n times where d = + is the exterior derivative, d c = i( ), and so dd c = 2i. The definition of the Monge-Ampère operator can be extended to certain classes of non-smooth plurisubharmonic functions ([B-T1], [B-T2], [D1], [D2], [F-S], [S]), for example, it can be defined for locally bounded plurisubharmonic functions ([B-T1], [B-T2]) or plurisubharmonic functions that are locally bounded away from compact subsets of pseudoconvex domains ([D1], [D2], [S]). Recall that a bounded domain Ω in C n is called hyperconvex if there exists a continuous plurisubharmonic function ρ : Ω (, 0) such that {z Ω : ρ(z) < c} Ω for each c < 0. It is proved by D ly [D1] that if Ω is hyperconvex then g Ω (, a) is the unique solution to the Dirichlet problem u P SH(Ω) C( Ω \ {a}), (dd c u) n = (2π) n δ a in Ω, u(z) log z a = O(1) as z a, u(z) 0 as z Ω, where δ a is the Dirac mass at a.

19 CHAPTER 2. PRELIMINARIES Plurisubharmonic Functions on Reinhardt Domains Let Ω be a domain in C n. If (z 1,, z n ) Ω implies that (e iθ 1 z 1,, e iθn z n ) Ω for every (e iθ 1,, e iθn ) ( ) n, Ω is called a Reinhardt domain. Suppose that g is a function defined on a Reinhardt domain Ω that satisfies g(z 1,, z n ) = g( z 1,, z n ) for all (z 1,, z n ) Ω. Then g is called a polyradial function. The plurisubharmonicity of polyradial functions can be studied in terms of related convex functions. We first recall some properties of convex functions defined on domains that are not necessarily convex and then investigate the connection between polyradial plurisubharmonic functions on Reinhardt domains and the related convex functions Convex Functions For any two points x 1, x 2 R n, we will denote the line segment between x 1 and x 2 by [x 1, x 2 ]. Let E be any set in R n and x 1, x 2 E. A function u : E R is called convex if for any x 1, x 2 E such that [x 1, x 2 ] E, we have u(λx 1 + (1 λ)x 2 ) λu(x 1 ) + (1 λ)u(x 2 ), 0 λ 1. (2.1) For D an open set in R n, a function u : D R n is called locally convex if for any x D, there exists a ball B r (x) = {y R n : x y < r} D on which u is convex. We refer to [Y, Section 2] for a discussion of several notions of convexity of functions defined on arbitrary sets. In particular, we note that our notion of convexity here corresponds to the notion of interval convexity in [Y]. In the paper of Bertin [BE], some interesting

20 CHAPTER 2. PRELIMINARIES 11 properties of convex functions and their extension properties are discussed in great generality. We are going to work with convex functions defined on domains that are not necessarily convex. In this case, proving local convexity will be enough, as is shown by the following well-known theorem (see e.g. [Y, Section 2]): Theorem Let D be any open set in R n and u be a real-valued function defined on D. Then, u is convex if and only if u is locally convex. The following lemma will be used several times in the subsequent chapters. Lemma If f : (, a] (, 0) is a convex function then f(x) f(a), for any x (, a). Proof. Let x (, a). Hence for any T < x, 0 < x T a T < 1, and x can be written as a convex combination of T and a as x = ( ) ( x T a T a + a x ) a T T. Then, using the convexity of f, f(x) = f (( ) x T a + a T ( ) ) a x T x T a T a T f(a) + a x a T f(t ) < x T a T f(a). The result follows by taking the limit as T. We conclude this section with a simple fact about closed, unbounded convex sets. It will be used in proving some results about pluricomplex Green functions in C 2. Lemma Let E be a closed, convex, unbounded subset of R 2 such that E {(x 1, x 2 ) R 2 : x 1 < 0, m < x 2 < 0}. (2.2) Then, for any p = (p 1, p 2 ) E, we have l p E, where l p = {(p 1 + t, p 2 ) : t 0}.

21 CHAPTER 2. PRELIMINARIES 12 Proof. Let p be an arbitrary point in E. Since E satisfies equation (2.2) and is unbounded, there exists a sequence of points {x j } j 1 = {(x j 1, x j 2)} j 1 E with x j 1 as j. E is convex, therefore the line segment [p, x j ] lies in E for any j. As m < x j 2 < 0, the angle between the line segment [p, x j ] and the half-line l p approaches 0 as j goes to. Hence, any point in l p is a limit point of the set A = j 1 [p, x j ]. Obviously, A Ā E as E is closed and the result follows Polyradial Plurisubharmonic Functions As mentioned before, the plurisubharmonicity of a polyradial function depends on the convexity of a related function. Before we discuss this, we need to give some definitions. Define functions l(z 1,, z n ) = (log z 1,, log z n ), (z 1,, z n ) C n, and e(x 1,, x n ) = (e x 1,, e xn ), (x 1,, x n ) [, ) n, where log 0 = and e = 0. For a Reinhardt domain Ω C n, denote its logarithmic image by ω = l(ω) R n. For any ω R n open define ê(ω) = int {(z 1,, z n ) C n : ( z 1,, z n ) e(ω)}. (2.3) The following proposition shows the relation between polyradial plurisubharmonic functions on a Reinhardt domain Ω and the related convex functions when Ω {z C n : z 1 z n 0}(see e.g. [J-P1, Proposition (a)], [V, Section 11.4]).

22 CHAPTER 2. PRELIMINARIES 13 Proposition Let Ω {z C n : z 1 z n 0} be a Reinhardt domain and g be a real valued polyradial function. Then the function g is plurisubharmonic on Ω if and only if g e is convex on ω = l(ω) R n. The plurisubharmonic functions that we will consider will have a logarithmic pole at 0, hence the origin will be an element of Ω. However, they will be bounded from above, specifically, they will be negative functions. We now show that in this case Proposition can still be applied. Proposition Let Ω C n be a Reinhardt domain containing the origin and let g : Ω [, 0) be an upper semicontinuous polyradial function such that g(z) > for z 0. Then, g is plurisubharmonic on Ω if and only if g e is convex on ω = l(ω) R n. Proof. Let A = {z 1... z n = 0} C n and Ω A = Ω \ A. Then Ω A is a Reinhardt domain and ω = l(ω) R n = l(ω A ). Since g is upper bounded and upper semicontinuous it follows that g is plurisubharmonic on Ω if and only if its restriction g ΩA is plurisubharmonic on Ω A. By Proposition 2.2.4, g ΩA is plurisubharmonic on Ω A if and only if g e is convex on ω. 2.3 Pluricomplex Green Functions on Reinhardt Domains Recall that pluricomplex Green functions are invariant under biholomorphic transformations. We see that if Ω 0 is a Reinhardt domain in C n then g Ω (, 0) is a polyradial function since it is invariant under transformations of the form (z 1, z n ) (e iθ 1 z 1,, e iθn z n ), where (e iθ 1,, e iθn ) ( ) n.

23 CHAPTER 2. PRELIMINARIES 14 Now we can define the convex function related to the pluricomplex Green function. Recall that ω = l(ω) R n and let 1 = (1,, 1) R n. A function u : ω R is said to have normalized growth at if, for any a = (a 1,, a n ) ω so that a + t 1 ω for all t 0, there exists a constant C a such that u(a + t 1) = u(a 1 + t,, a n + t) t + C a, t 0. The convex envelope of ω with normalized growth at is defined by u ω (x) = sup{u(x) : u : ω (, 0) convex with normalized growth at }. It is easily seen that if ω 1 ω 2, then u ω1 u ω2. The following lemma shows the relation between polyradial functions with logarithmic pole at 0 and the convex functions with normalized growth at. Lemma ([K3]). Let Ω be a Reinhardt domain containing the origin and let g be a polyradial plurisubharmonic function on Ω. Then g has a logarithmic pole at 0 if and only if g e has normalized growth at. Proof. The polyradial plurisubharmonic function g has a logarithmic pole at 0 if and only if for every z Ω near 0 there exists a constant M = M(z) such that g(r z ) log rz + M for 0 < r 1. Writing e xj+t = r z j for j = 1,..., n, where r 1 and t = log r, we see that this is equivalent to g(e(x + t 1)) t + M + log e x, t 0, i.e. g e has normalized growth at. The following result is stated by Klimek [K3]. We include a proof for the convenience of

24 CHAPTER 2. PRELIMINARIES 15 the reader. Theorem ([K3]). If Ω C n is a bounded Reinhardt domain, then u ω (x) = g Ω (e(x), 0), x ω. Proof. By Proposition and Lemma 2.3.1, the function g Ω (e( ), 0) is convex and negative on ω, with normalized growth at. Hence g Ω (e(x), 0) u ω (x) for x ω. Next let u be an element in the defining family of u ω. By Proposition and Lemma the function u l gives rise to a negative polyradial plurisubharmonic function on Ω with logarithmic pole at 0, so u g Ω (e( ), 0) on ω. This implies that u ω g Ω (e( ), 0) on ω. Since the pluricomplex Green functions are invariant under biholomorphic mappings, it is now easily seen by Theorem that u l( n (0,r))(x) = max 1 j n {x j} log r, where n (0, r) := {w C n : w j < r for all j = 1,, n}. One has the following results for the function u ω. Proposition ([K3]). Let Ω 0 0 be a bounded Reinhardt domain in C n, Ω 0 be a Reinhardt subdomain of Ω 0, ω 0 = l(ω 0 ) R n, and ω = l(ω) R n. Then the following hold: (i) If x 0 ω and L = {x 0 + tb : t 0} is a ray to with L \ {x 0 } ω for some b (0, + ) n, then u ω (x) max 1 j n {x j x 0 j} + u ω(x 0 ), x L \ {x 0 }, (2.4)

25 CHAPTER 2. PRELIMINARIES 16 where u ω(x 0 ) = lim sup x x 0, x ω u ω (x). (ii) If, in addition, Ω 0 is pseudoconvex, then for any x 0 ω 0 ω lim x x 0 x ω u ω (x) = 0. (2.5) Proof. (i) Let x L \ {x 0 } and fix x 1 [x, x 0 ], x x 1 x 0. Define ψ(t) = (ψ 1 (t),..., ψ n (t)) = x 1 + t(x 1 x), t 0. Then ψ(0) = x 1, ψ( 1) = x and ψ((, 0]) L \ {x 0 }. By Theorem the function u ω is convex on ω, so the function v := u ω ψ is convex on (, 0]. Hence for t < 1, ( u ω (x) = v( 1) = v 1 t t + t + 1 ) 0 t 1 t v(t) + t + 1 v(0). t Fix a polydisk n (0, r) Ω. Then u ω (y) u l( n (0,r))(y) = max 1 j n {y j} log r, y l( n (0, r)). Hence for t small enough, ψ(t) l( n (0, r)) and we obtain u ω (x) 1 t = max 1 j n = max 1 j n ( ) max {ψ j(t)} log r + t + 1 u ω (x 1 ) 1 j n t { ψ } j(t) + log r + u ω(x 1 ) + u ω (x 1 ) t t { (x j x 1 j) 1 } t x1 j + log r + u ω(x 1 ) + u ω (x 1 ). t

26 CHAPTER 2. PRELIMINARIES 17 Letting t we get u ω (x) max 1 j n {x j x 1 j} + u ω (x 1 ). Finally, we let x 1 x 0 and obtain (2.4). (ii) Let x 0 ω 0 ω. Since there exists a polydisk n (0, r) Ω 0 we have that (, log r) n ω 0. Moreover, ω 0 is convex since Ω 0 is pseudoconvex. It follows that there exists an affine mapping u : R n R, u(x) = u(x 1,..., x n ) = n j=1 a jx j + b, where (a 1,..., a n ) [0, ) n, b R, such that u(x 0 ) = 0 and u < 0 on ω 0. If c := (a a n ) 1 > 0 then the function c u has normalized growth at, so it is an element of the defining family of u ω 0. Hence c u u ω 0 u ω < 0 on ω, which implies (2.5). We now let Ω 0 be a complete Reinhardt domain, i.e. (z 1,, z n ) Ω 0 implies that (r 1 e iθ 1 z 1,, r n e iθn z n ) Ω 0 for all 0 r j 1 and (e iθ 1,, e iθn ) ( ) n. Then Ω 0 = {z C n : h Ω 0(z) < 1}, where h Ω 0 is the Minkowski functional of Ω 0, defined by h Ω 0(z) := inf { t > 0 : z t Ω 0}, z C n. Note that h Ω 0(λ z) = λ h Ω 0(z), for λ C and z C n. The pluricomplex Green functions of certain domains can be found using the Minkowski functional (see [J-P1, Proposition ]) as follows: Proposition Let Ω 0 be a bounded pseudoconvex complete Reinhardt domain with Minkowski functional h Ω 0. Then g Ω 0(z, 0) = log h Ω 0(z), for all z Ω 0. Proposition will form a base for the discussions in the next two chapters. We will

27 CHAPTER 2. PRELIMINARIES 18 work on a bounded, pseudoconvex complete Reinhardt domain Ω 0 and on its Reinhardt subdomains. Using the pluricomplex Green function obtained from Proposition for Ω 0 and Theorem 2.3.2, we will find the pluricomplex Green functions of certain subdomains.

28 Chapter 3 Extension for general Reinhardt domains 3.1 Statement of Theorem A domain D C n is called a domain of existence for the class of plurisubharmonic functions if there exists a plurisubharmonic function on D that cannot be extended as a plurisubharmonic function to any domain strictly larger than D. For example, pseudoconvex domains are domains of existence for plurisubharmonic functions. Indeed, let D be a pseudoconvex domain, i.e. D has a continuous plurisubharmonic exhaustion function ρ, that is, for any r R, the set {z D : ρ(z) < r} is relatively compact in D. Therefore, lim z z 0 D ρ(z) =. Hence ρ cannot be extended as a plurisubharmonic function beyond the boundary of D and D is a domain of existence for plurisubharmonic functions. Bremermann [BR] and Cegrell [C2] showed independently that domains of existence for plurisubharmonic functions are not necessarily pseudoconvex. When D has C 2 boundary, 19

29 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 20 Bedford and Burns [B-B] used the Levi form of D to give a sufficient condition for D to be a domain of existence for plurisubharmonic functions. Cegrell [C1] generalized this result to domains with general boundary by giving a sufficient geometric condition on a dense subset of the boundary. A natural question is whether pluricomplex Green functions of certain domains can be extended as plurisubharmonic functions on larger domains and, in particular, as pluricomplex Green functions of larger domains. More specifically, let Ω 0 C n be a bounded, pseudoconvex domain with 0 Ω 0. Let E 0 be a compact subset of Ω 0 such that Ω := Ω 0 \ E is connected. We consider the following two questions: 1. For what kind of sets E does there exist a domain Ω with Ω Ω Ω 0 such that g Ω (z, 0) = g Ω(z, 0), z Ω? 2. Is there a largest domain Ω with the above property? These questions can be viewed in analogy to the classical Hartogs extension phenomenon for holomorphic functions on Ω recalled in the Introduction. In C n, we prove the following theorem, which relates to question (1). Theorem Let Ω 0 be a pseudoconvex, bounded complete Reinhardt domain in C n, and let Ω = Ω 0 \ E, where E 0 is a Reinhardt compact subset of Ω 0 that satisfies the following properties: E is strictly logarithmically convex, i.e. l(e) is strictly convex, E {z 1... z n = 0} =.

30 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 21 Then there exists a Reinhardt domain Ω such that Ω Ω Ω 0 and g Ω (z, 0) = g Ω(z, 0), z Ω. In C 2, one can omit the condition E {z 1 z 2 = 0} = in Theorem Theorem Let Ω 0 be a pseudoconvex, bounded complete Reinhardt domain in C 2, and let Ω = Ω 0 \ E, where E 0 is a Reinhardt compact, strictly logarithmically convex subset of Ω 0. Then there exists a Reinhardt domain Ω such that Ω Ω Ω 0 and g Ω (z, 0) = g Ω(z, 0), z Ω. These theorems do not address question (2), about the maximality of the extension domain Ω. We give a complete answer to this question in Chapter 4, in the special case when Ω 0 = 2 is the unit bidisk in C Special Classes of Convex Functions The proof of Theorem relies on analyzing the envelopes of certain classes of convex functions defined on solid hypercylinders in R n. These are introduced and studied in this section. Let β R n 1 be a compact, convex set, and q = (q 1,..., q n ) R n with q n > 0. Consider the set K = {(x, 0) + tq : x β, t R},

31 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 22 which is a solid hypercylinder in R n. The boundary of K is K = {(x, 0)+tq : x β, t R}, where β is the boundary of β R n 1. Given a hyperplane H that is not parallel to q, we let D H := K H. Note that D H is a compact convex set. Definition Let ω R n. A function f : ω R is called q-linear if f(x+tq) = f(x)+t holds for any x ω and t R such that x + tq ω. Let now f : K R be a given convex, q-linear function, such that f is bounded above on D H, for some hyperplane H not parallel to q. We let V = V ( ; K, f) : K R be the convex envelope of f defined by V (x; K, f) = sup {w(x) : w : K R is convex and w(y) f(y) for y K}, x K. (3.1) Lemma The function V is convex and V (x) = f(x) for x K. Proof. Since f is q-linear and bounded above on D H, it follows that f is bounded above on each compact subset of K. Let x K and H x be the hyperplane parallel to H that contains x. Then f M on D Hx = H x K, for some constant M. If w is an element of the defining family of V then the restriction of w to the convex set D Hx is convex and w f M on the boundary of this set regarded as a subset of H x. We conclude that w(x) M. It follows that the functions in the defining family of V are locally uniformly upper bounded on K, hence V is convex fuction on K. Obviously, f is an element of the defining family of V and therefore f V on K by

32 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 23 definition. On the other hand, any element of the defining family of V is dominated by f on K, hence so is V. Therefore V = f on K. In order to prove further properties of V we need an alternate description using q-linear extensions of convex envelopes on slices D H = H K, where H is a hyperplane not parallel to q. We define v(x; D H, f) = sup{w(x) : w : D H R is convex and w(y) f(y) on D H }, x D H. Here D H denotes the boundary of D H seen as a subset of H. We denote by v H = v H ( ; K, f) the q-linear extension of v( ; D H, f) to K defined by v H (x) = v(y; D H, f) + t, where x = y + tq, y D H, t R, x K. This function is clearly q-linear. Lemma If H is a hyperplane not parallel to q then V = v H on K. In particular, the function V is q-linear. Proof. Note that v( ; D H, f) is a convex function on D H, as it is the supremum of a family of uniformly upper bounded convex functions. If H = H + tq is a hyperplane parallel to H, for some fixed t R, then D H = D H + tq. For any function w on D H we can define a function w on D H by w (x + tq) = w(x) + t, x D H. Then, a simple calculation shows that w is convex on D H if and only if w is convex on D H.

33 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 24 Moreover, since f is q-linear, we have that w(x) f(x) if and only if w (x + tq) f(x + tq), where x D H. Hence v(x + tq; D H, f) = v(x; D H, f) + t, x D H. (3.2) Formula (3.2) implies that v H = v H if H is a hyperplane parallel to H. Also, as f is a convex function in the defining family of v( ; D H, f), we have that v(x; D H, f) = f(x) for x D H and the q-linearity of the functions v H and f implies that v H = f on K. Next we will show that v H is also a convex function on K. Let x, x 1, x 2 be points in K such that x = µx 1 + (1 µ)x 2 for 0 µ 1. Now write x = x 0 + λq, x j = x j,0 + λ j q where x 0, x j,0 H and λ, λ j R, j = 1, 2. We claim that x 0 = µx 1,0 + (1 µ)x 2,0 and λ = µλ 1 + (1 µ)λ 2. Indeed, we have that x 0 µx 1,0 (1 µ)x 2,0 + (λ µλ 1 (1 µ)λ 2 )q = 0 If x 1,0 = x 2,0 = x 0 then our claim follows. Otherwise, the vectors p := x 2,0 x 1,0 0 and q are linearly independent, since p is parallel to H and q is not parallel to H. As x 0 µx 1,0 (1 µ)x 2,0 = sp for some s R, we conclude that s = λ µλ 1 (1 µ)λ 2 = 0, which implies our claim. Using the definition of v H and the fact that the function v( ; D H, f)

34 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 25 is convex on D H, we obtain v H (x) = v(x 0 ; D H, f) + λ µ v(x 1,0 ; D H, f) + (1 µ)v(x 2,0 ; D H, f) + µλ 1 + (1 µ)λ 2 = µ ( v(x 1,0 ; D H, f) + λ 1) + (1 µ) ( v(x 2,0 ; D H, f) + λ 2) = µ v H (x 1 ) + (1 µ)v H (x 2 ), hence v H is a convex function on K. Since v H is convex on K and v H = f on K, v H is an element of the defining family of V, so v H V. On the other hand, take any x K. Then, there exists a hyperplane H x that is parallel to H. Since V DH is an element of the defining family of v( ; D H, f), we see that V (x) v(x; D H, f) = v H (x) = v H (x), which concludes the proof. For the proof of Theorem we actually need to work with envelopes of convex functions on certain subsets of K, which we now introduce. For a fixed hyperplane H that is not parallel to q, let D := K H, C := K H, and define K := {x + tq : x D, t < 0}, K := {x + tq : x C, t < 0}. Note that D is a compact convex set and K = K D is the boundary of K in R n. Definition If E is a compact convex subset of K so that D E, we let K E = K \ E and K E = K E K. The convex function that we will need can now be introduced. Let f : K R be a given

35 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 26 convex, q-linear function, such that f is bounded above on D. For each subset K E of K, we let u( ; K E, f) : K E R be the convex envelope of f defined by u(x; K E, f) := sup {w(x) : w : K E R is convex and w(y) f(y) for y K E }. (3.3) We can now state the main result of this section. Theorem For any subset K E of K as above, we have that u( ; K E, f) = V KE, where V = V ( ; K, f) is the function defined in (3.1). In particular, u(x; K E, f) = u(x; K, f), x K E. Proof. Note that K = K E if we take E = D. Therefore, the second conclusion of the theorem follows at once from the first one. A similar argument to that in the proof of Lemma shows that the function u( ; K E, f) is convex on K E and u(x; K E, f) = f(x) for x K E. By Lemma 3.2.2, V is a function in the defining family of u( ; K E, f), so V u( ; K E, f) on K E. On the other hand, let x K E. As E is convex, there exists a hyperplane H x such that H E =, so H is not parallel to q and D H K E. Since u( ; K E, f) DH is an element of the defining family of v( ; D H, f) we have by Lemma that u(x; K E, f) v(x; D H, f) = v H (x) = V (x). Hence u(x; K E, f) = V (x) for x K E, and the theorem is proved.

36 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS Proof of Theorem Let Ω 0 be a pseudoconvex, bounded complete Reinhardt domain in C n. Let Ω = Ω 0 \ E, where E 0 is a Reinhardt compact subset of Ω 0 that satisfies the following properties: E is strictly logarithmically convex, i.e. l(e) is strictly convex, E {z 1... z n = 0} =. For z Ω \ {0}, let L z = {ζz : ζ C}. Ω will be partitioned into the following sets: Ω 1 = {z Ω \ {0} : L z E = }, which is an open set, Ω 2 = the connected component of 0 in Ω \ Ω 1, Ω 3 = Ω \ (Ω 1 Ω 2 ). Figure 3.1: The case Ω C 2 and E {z 1 z 2 = 0} = Recall that l(z 1,, z n ) = (log z 1,, log z n ), (z 1,, z n ) C n,

37 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 28 and e(x) = (e x 1,, e xn ), (x 1,, x n ) [, + ) n, with log 0 = and e = 0. Set l(ω) R n = ω, l(ω 0 ) R n = ω 0, l(ω i ) R n = ω i, for i = 1, 2, 3. Figure 3.2: Logarithmic image of the case Ω C 2 and E {z 1 z 2 = 0} = We study g Ω by working with the associated convex function u ω in logarithmic coordinates. As E {z 1,..., z n = 0} =, l(e) is a compact set in R n and we define a solid hypercylinder K by K = {x + t 1 : t R} x l(e) where 1 = (1,, 1) R n. Let H be a fixed hyperplane that is not parallel to 1, and define D := K H, C := K H, K := {x + t 1 : x D, t < 0}, K := {x + t 1 : x C, t < 0}.

38 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 29 We assume that H is chosen so that K l(e). Note that K does not necessarily lie in ω 0. Let E = {x + t 1 : t 0} K. x l(e) Then E is a compact, convex set and we set K E := K \ E. Note that K E = ω 2. Figure 3.3: K and K E constructed in the case Ω C 2 and E {z 1 z 2 = 0} = Lemma We have that g Ω (z, 0) = g Ω 0(z, 0) for z Ω 1, and u ω (x) = u ω 0(x) for x ω 1. Proof. Since Ω Ω 0, g Ω 0(z, 0) g Ω (z, 0) for all z Ω. Fix z Ω 1. Observe that for ζ C, ζz Ω if and only if ζ < R := 1 h Ω 0 (z), where h Ω 0 is the Minkowski functional of Ω0. Since z Ω 0 we have h Ω 0(z) < 1, so R > 1. Consider the holomorphic function f z : (R) Ω, f z (ζ) = ζz. Since the pluricomplex Green function is decreasing under holomorphic maps we obtain g Ω (ζz, 0) g (R) (ζ, 0) = log ζ R. Hence, letting ζ = 1, g Ω(z, 0) log h Ω 0(z) = g Ω 0(z, 0). Thus g Ω (z, 0) = g Ω 0(z, 0) for z Ω 1. Since u ω (x) = g Ω (e(x), 0), this implies that u ω (x) = u ω 0(x) for x ω 1. Lemma The function f := log h Ω 0 e is a 1-linear convex function on R n. Moreover, f = u ω 0 on ω 0.

39 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 30 Proof. Since the Minkowski functional is defined for all C n, f is defined on R n. As h Ω 0(ζz) = ζ h Ω 0(z), for ζ C, we have that ( f(x + t 1) = log h Ω 0 (e(x + t 1)) = log h Ω 0 e t (e x 1,, e xn ) ) = t + log h Ω 0 (e(x)) = t + f(x), so f is a 1-linear function. Now f = log h Ω 0 e = g Ω 0(, 0) e = u ω 0 on ω 0, which shows that f is convex on ω 0. Since f is 1-linear, it follows that f is convex on R n, using a similar argument to that in the proof of Lemma Lemma If f = log h Ω 0 e then u ω (x) = u(x; K E, f) for x ω 2 = K E. Proof. Recall that the function u( ; K E, f) is defined in (3.3). By Lemmas and 3.3.2, u ω = u ω 0 = f on ω 1, so u ω = u ω 0 = f on K E since these functions are continuous. As u ω KE is a convex function that is equal to f on K E, we have u ω u( ; K E, f) on K E. For the opposite inequality, we consider the function u ω (x), if x ω 1 ω 3, u(x) = u(x; K E, f), if x ω 2 = K E. (3.4) We will show that u is an element of the defining family of u ω, hence u u ω on ω, and in particular u( ; K E, f) u ω on ω 2. By Theorem 3.2.5, Lemma and Lemma 3.2.3, the function u( ; K E, f) is 1-linear convex on K E and u( ; K E, f) = f on K E. We have u = u ω < 0 on ω 1 ω 3. Moreover, since f < 0 on K E it follows that u( ; K E, f) < 0 on K E, so u is a negative function. It has normalized growth at on ω 1 as u ω has normalized growth at on ω. For x K E

40 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 31 and t R with x + t 1 K E, the 1-linearity of u( ; K E, f) shows that u(x + t 1; K E, f) = t + u(x; K E, f), thus u( ; K E, f) has normalized growth at on ω 2 as well. It remains to show that u is a (locally) convex function. Note that it suffices to show that u is convex in a small neighborhood of each point of K E. Since u ω is convex on ω and u( ; K E, f) is convex on K E, this amounts to proving the convexity inequality for points x, x 1 ω 1, x 2 ω 2, such that the segment [x 1, x 2 ] ω and x [x 1, x 2 ]. We write x = sx 1 + (1 s)x 2, 0 < s < 1. Recall that u = u( ; K E, f) = f = u ω on K E ω 2 and u ω u = u( ; K E, f) on K E. If x ω 1 K E then u(x) = u ω (x) su ω (x 1 ) + (1 s)u ω (x 2 ) su(x 1 ) + (1 s)u(x 2 ), since u ω (x 2 ) u(x 2 ; K E, f) = u(x 2 ). We assume next that x ω 2, and we let {y} = [x 1, x 2 ] K E, so y = tx 1 + (1 t)x 2 with s t < 1. Then x = s y + ( ) 1 s t t x 2. As y K E, it follows by above that u(y) tu(x 1 ) + (1 t)u(x 2 ). Since u = u( ; K E, f) is convex on ω 2 = K E we obtain u(x) s ( t u(y) + 1 s ) u(x 2 ) t s t ( tu(x 1 ) + (1 t)u(x 2 ) ) + ( 1 s ) u(x 2 ) = su(x 1 ) + (1 s)u(x 2 ). t This shows that u is (locally) convex on ω, hence an element of the defining family of u ω.

41 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 32 The proof of the lemma is complete. Since the function u defined in (3.4) is equal to u ω on ω 2 by Lemma 3.3.3, it should be noted that in fact we have u ω = u on ω. Theorem can now be proved using these lemmas. Proof of Theorem It suffices to show that there exists a domain ω such that ω ω ω 0 and u ω = u ω ω. In the above setting, let C = l(e) K, and let conv(c) denote the convex hull of C. Define Ẽ = {x + t 1 : t 0} K. x conv(c) Then Ẽ is a compact, convex set and we let K Ẽ := K \ Ẽ. Since l(e) is convex, conv(c) l(e), so Ẽ E and K Ẽ K E. Now we show that KẼ K E. We prove in fact that K E \ K KẼ. Let x K E \ K. Then x ( l(e) \ C) K E. Since l(e) is strictly convex, there exists a hyperplane H such that H l(e) = {x}, hence H C =. Since C is compact, if ɛ > 0 is small enough the hyperplane H ɛ = H + ɛ 1 does not intersect C. It follows that x and conv(c) lie on opposite sides of H ɛ, for some ɛ > 0, so x Ẽ. Let ω = ω 1 ω 3 ω 2, where ω 2 = KẼ. Since l(e) K = conv(c) K = C we have K E = KẼ, so ω is a domain contained in ω 0 and ω ω. We have u ω u ω on ω. For the opposite inequality we define the function ũ on ω, u ω (x), if x ω 1 ω 3, ũ(x) = u(x; KẼ, f), if x ω 2 = KẼ,

42 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 33 where f = log h Ω 0 e is as in Lemma Since K E = KẼ and, by Theorem 3.2.5, u(x; K E, f) = u(x; KẼ, f) for x K E, Lemma and its proof (see (3.4)) imply that the function ũ is an element of the defining family of u ω. Thus ũ u ω on ω. As u( ; K E, f) = u( ; KẼ, f) KE, it follows from Lemma that ũ = u ω on ω. Therefore u ω u ω on ω, hence u ω = u ω ω. This concludes the proof of Theorem Proof of Theorem If E {z 1 z 2 = 0} =, the existence of Ω follows from Theorem So, let E {z 1 z 2 = 0}. This implies that E intersects {z 1 = 0} or {z 2 = 0}. Since Ω is connected and Reinhardt, it can intersect only one of them. Without loss of generality, suppose that E {z 1 = 0}. Following the notation in the proof of Theorem 3.1.1, we consider the partition of Ω, Ω 1 = {z Ω \ {0} : L z E = }, which is an open set, Ω 2 = the connected component of 0 in Ω \ Ω 1, Ω 3 = Ω \ (Ω 1 Ω 2 ). Figure 3.4: The case Ω C 2 and E {z 1 z 2 = 0}

43 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 34 If Φ(x 1, x 2 ) = x 2 x 1, the function Φ l(e) attains its minimum at a unique point a = (a 1, a 2 ) l(e), as l(e) is closed and strictly convex. Let L = {x ω : x = a + t 1, t 0}. Then the regions ω j = l(ω j ) R 2, j = 1, 2, 3, are as in Figure 3.5. Figure 3.5: Logarithmic image of the case Ω C 2 and E {z 1 z 2 = 0} Now let ω 2 = {y + (s, 0) : y L \ {a}, s 0}, ω = ω 1 ω 2 ω 3. By the construction of a and since l(e) is strictly convex, it follows easily that ω is a domain such that ω ω ω 0. Theorem will follow if we prove that u ω = u ω ω. Since ω ω we have u ω u ω on ω. To complete the proof, we need to show that u ω u ω on ω. Lemma shows that u ω = u ω = u ω 0 = f on ω 1 L, where the function f = log h Ω 0 e is as in Lemma Let us introduce the function w(x 1, x 2 ) = x 2 a 2 + f(a), x = (x 1, x 2 ) ω 2.

44 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 35 Figure 3.6: ω in the case Ω C 2 and E {z 1 z 2 = 0} We claim that w = f on L. Indeed, L has equation x 2 = x 1 + a 2 a 1, x 1 a 1, and by the 1-linearity of f, w(x 1, x 1 + a 2 a 1 ) = x 1 a 1 + f(a 1, a 2 ) = f(x 1, x 1 + a 2 a 1 ). Consider now the function u defined on ω by u ω (x), if x ω 1 ω 3, u(x) = w(x), if x ω 2. Clearly u < 0 and u has normalized growth at. To show that u is (locally) convex we proceed as in the proof of Lemma Let (x 1, x 2 ) ω 2 and x 1 be so that (x 1, x 2 ) L. Then f(, x 2 ) is a convex function on (, x 1], so by Lemma 2.2.2, f(x 1, x 2 ) f(x 1, x 2 ) = w(x 1, x 2 ) = w(x 1, x 2 ). Therefore we can apply the same argument as the one used in the proof of the convexity of the function defined in (3.4). We conclude that u is an element of the defining family of the function u ω, so u u ω on ω. We will prove that u ω w on ω 2. This implies that u ω u u ω on ω, which finishes the proof. To this end, we define m = inf{x 2 : x = (x 1, x 2 ) l(e)} >,

45 CHAPTER 3. EXTENSION FOR GENERAL REINHARDT DOMAINS 36 as 0 E. Since l(e) is strictly convex, Lemma shows that if (x 1, x 2) L and x 2 m then (x 1, x 2) ω 2 for all x 1 < x 1. We partition ω 2 as follows: ω 2 = {(x 1, x 2 ) ω 2 : x 2 m}, ω 2 = ω 2 \ ω 2. Let v be any element of the defining family of u ω. If (x 1, x 2 ) ω 2 and x 1 is so that (x 1, x 2 ) L then Lemma applied to the convex function v(, x 2 ) on (, x 1] implies that v(x 1, x 2 ) v(x 1, x 2 ) u ω (x 1, x 2 ) = f(x 1, x 2 ) = w(x 1, x 2 ) = w(x 1, x 2 ). If x ω 2, then, since l(e) is convex, we can write x = tx 1 + (1 t)x 2, 0 < t < 1, with points x 1 ω 2 and x 2 L. Since v u ω = f = w on L and w is an affine function, it follows that v(x) tv(x 1 ) + (1 t)v(x 2 ) tw(x 1 ) + (1 t)w(x 2 ) = w(x). So v w, and hence u ω w, on ω 2. This concludes the proof of Theorem

46 Chapter 4 Reinhardt subdomains of the unit bidisk 4.1 Introduction In Chapter 3, the following questions were introduced: Let Ω 0 C n be a bounded, pseudoconvex domain with 0 Ω 0. Let E 0 be a compact subset of Ω 0 such that Ω := Ω 0 \ E is connected. 1. For what kind of sets E does there exist a domain Ω with Ω Ω Ω 0 such that g Ω (z, 0) = g Ω(z, 0), z Ω? (4.1) 2. Is there a largest domain Ω with the above property? We proved two theorems related to question (1) on complete Reinhardt domains Ω 0 when E 0 is a Reinhardt compact, strictly logarithmically convex subset of Ω 0. Here a complete 37

47 CHAPTER 4. REINHARDT SUBDOMAINS OF THE UNIT BIDISK 38 answer to both questions is obtained when Ω 0 = 2 and E 0 is a Reinhardt compact, logarithmically convex subset of 2. It should be noted that if the pluricomplex Green function of a given subdomain Ω 2 is identically equal to that of 2, the largest domain Ω that satisfies the extension property given by equation (4.1) is 2 itself. We have the following theorem: Theorem Let Ω = 2 \E where E 0 is a Reinhardt compact, logarithmically convex subset of 2 such that int E {(z 1, z 2 ) C 2 : z 1 = z 2 } =. Then g Ω (z, 0) = g 2(z, 0), for all z Ω. Proof. As in Chapter 3, we set ω = l(ω) R 2, ω 0 = l( 2 ) R 2 = {(x 1, x 2 ) : x 1 < 0, x 2 < 0}. Without loss of generality, assume that E {(z 1, z 2 ) C 2 : z 1 z 2 }. Then, l(e) lies in some strip S = {(x 1, x 2 ) R 2 : m x 2 M, x 1 x 2 }, where m = inf{x 2 : (x 1, x 2 ) l(e)} and M = sup{x 2 : (x 1, x 2 ) l(e)}. Note that < m M < 0. Obviously, u ω (x 1, x 2 ) u ω 0(x 1, x 2 ) = max{x 1, x 2 }. Lemma shows that for any (x 1, x 2 ) ω with x 1 > x 2, u ω (x 1, x 2 ) = x 1. Also, for any (x 1, x 2 ) ω with x 1 = x 2, u ω (x 1, x 2 ) = u ω (x 2, x 2 ) = lim u ω (y 1, y 2 ) = x 2. (4.2) (y 1,y 2 ) (x 1,x 2 ) (y 1,y 2 ) ω, y 1 >y 2 Now, let (x 1, x 2 ) ω \ S with x 1 x 2. Then, {(x 1, x 2 ) : < x 1 x 2 } ω and by

48 CHAPTER 4. REINHARDT SUBDOMAINS OF THE UNIT BIDISK 39 Lemma 2.2.2, u ω (x 1, x 2 ) u ω (x 2, x 2 ) = x 2. Hence we have shown that u ω (x 1, x 2 ) = u ω 0(x 1, x 2 ), (x 1, x 2 ) (ω \ S) {(x 1, x 2 ) ω : x 1 = x 2 }. Note that for any point in {(x 1, x 2 ) ω : x 2 = m} or {(x 1, x 2 ) ω : x 2 = M}, u ω (x 1, x 2 ) = lim u ω (y 1, y 2 ) = x 2 (4.3) (y 1,y 2 ) (x 1,x 2 ) (y 1,y 2 ) ω, y 2 <m and u ω (x 1, x 2 ) = lim u ω (y 1, y 2 ) = x 2. (4.4) (y 1,y 2 ) (x 1,x 2 ) (y 1,y 2 ) ω, y 2 >M If E {(z 1, z 2 ) C 2 : z 1 z 2 = 0} =, then l(e) is unbounded. Any point in S \ l(e) lies on a segment [P 1, P 2 ] ω, where the points P 1, P 2 {(x 1, x 2 ) ω : x 2 = m} {(x 1, x 2 ) ω : x 2 = M} {(x 1, x 2 ) ω : x 1 = x 2 }. Since u ω is (locally) convex, equations (4.2), (4.3) and (4.4) yield that u ω (x 1, x 2 ) x 2. If E {(z 1, z 2 ) C 2 : z 1 z 2 = 0} =, then l(e) is bounded and S \ l(e) has an unbounded connected component. For any point in the bounded component(s) of S \ l(e), the previous argument shows that u ω (x 1, x 2 ) x 2. Any point in the unbounded component of S \l(e) lies on a segment [Q 1, Q 2 ] ω, where the points Q 1, Q 2 {(x 1, x 2 ) ω : x 2 = m} {(x 1, x 2 ) ω : x 2 = M}. Equations (4.3) and (4.4) show that u ω (x 1, x 2 ) x 2. Hence, we conclude that u ω = u ω 0 on ω and the theorem is proved. So, if Ω 2 is a domain that satisfies the requirements of Theorem 4.1.1, Ω = 2. Hence, we will focus on the domains that do not fall into this category.