Invariant distances and metrics in complex analysis

Invariant Distances and Metrics in Complex Analysis — revisited Marek Jarnicki Peter Pflug Jagiellonian University, Inst...

Author: Jarnicki M. | Pflug P.

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Invariant Distances and Metrics in Complex Analysis — revisited Marek Jarnicki Peter Pflug Jagiellonian University, Institute of Mathematics, Reymonta 4, 30-059 ´ w, Poland Krako E-mail address: [email protected] ¨t Oldenburg, Institut fu ¨r Mathematik, Carl von Ossietzky Universita Postfach 2503, D-26111 Oldenburg, Germany E-mail address: [email protected]

2000 Mathematics Subject Classification: 32F45, 32A07.

Contents Preface

5

Chapter 1. Holomorphically invariant objects 1.1. Holomorphically contractible families of functions 1.2. Holomorphically contractible families of pseudometrics 1.3. Effective formulas for elementary Reinhardt domains 1.4. The converse to the Lempert theorem 1.5. Generalized holomorphically contractible families 1.6. Properties of the generalized Möbius and Green functions 1.7. Examples 1.8. Properties of dmin and dmax G G 1.9. Relative extremal function 1.10. Analytic discs method 1.11. Coman conjecture 1.12. Product property

7 7 9 27 30 43 46 50 70 72 74 90 93

Chapter 2. Hyperbolicity and completeness 2.1. ci –hyperbolicity versus c–hyperbolicity 2.2. Hyperbolicity for Reinhardt domains 2.3. Hyperbolicities for balanced domains 2.4. Hyperbolicities for Hartogs type domains 2.5. c–completeness for Reinhardt domains 2.6. c–completeness for complete circular domains R (k) 2.7. γG –completeness for Zalcman domains 2.8. Kobayashi completeness and smoothly bounded pseudoconvex domains 2.9. Kobayashi completeness and unbounded domains

105 105 106 112 114 116 123 123 129 130

Chapter 3. Bergman metric 3.1. The Bergman kernel 3.2. The Lu Qi-Keng problem 3.3. Bergman exhaustiveness 3.4. L2h –domains of holomorphy 3.5. Bergman completeness

137 137 146 153 168 172

Symbols

187

Open problems

191 3

4

Contents

References

193

Index

201

Preface In 1993, the authors published their book “Invariant Distances and Metrics in Complex Analysis”, in which they discussed the state of affairs in the domain covered by the title of that book. In the meantime, some open questions mentioned in the book have been solved, more explicit formulas for different invariant functions on certain concrete domains were found (see also [Kob 1998]). Moreover, the classical Green function became important in studying Bergman completeness which finally led to the result that any hyperconvex bounded domain is Bergman complete. Simultaneously, a new development started, namely the study of the Green functions with multipoles. This led to the creation of a lot of new objects. Recently, there was the surprising example of the symmetrized bidisc which initiated a lot of new activities. The symmetrized bidisc is not biholomorphically equivalent to a convex domain but, nevertheless, its Carathéodory distance and its Lempert function coincide. Hence, it becomes again an interesting question, for which domains these two objects are equal. The main idea of this work is to describe what happened during the last 10 years in the area. The main source is our old book; in particular, if we quote a result before 1993 we send the reader to our book and not to the original source. This kind of quotation seems to be easier for the authors as well as for the reader. As always, the authors have to apologize for their choice of the material they present. Of course, it reflects their personal taste. At the end of the writing process a lot of our colleagues helped us to find typographical and mathematical errors in the manuscript and to essentially improve our presentation. We like to thank them all, in particular, thanks are due to our colleagues Z. Blocki, A. Edigarian, P. Jucha, N. Nikolov, H. Youssfi, P. Zapalowski, and W. Zwonek. Nevertheless, the authors are responsible for all mistakes which remained. Finally, we are deeply indebted to the following institutions: Committee of Scientific Research (KBN), Warsaw (grant no. 5 P03A 033 21), Niedersächsisches Ministerium f¨ ur Wissenschaft und Kunst (Az. 15.3-50 113(55) PL), Volkswagenstiftung (RiP-program at Oberwolfach) and the German Research Foundation (DFG), Bonn (Az. PF 227/8-1). Without their financial support this work would have never been possible. Last but not least we like to thank our universities for support during the time of preparing this survey. Krak´ ow — Oldenburg, February 2004

Marek Jarnicki Peter Pflug

5

CHAPTER 1

Holomorphically invariant objects 1.1. Holomorphically contractible families of functions Let us begin with the following definition of a holomorphically contractible family (cf. [J-P 1993], § 4.1). Definition 1.1.1. A family (dG )G of functions dG : G × G −→ R+ 1 , where G runs over all domains G ⊂ Cn (with arbitrary n ∈ N), is said to be holomorphically contractible if the following two conditions are satisfied: (A) for the unit disc E ⊂ C we have z−a dE (a, z) = mE (a, z) := , a, z ∈ E 1 − az (the function mE : E × E −→ [0, 1) is called the M¨ obius distance), (B) for any domains G ⊂ Cn , D ⊂ Cm , every holomorphic mapping F : G −→ D is a contraction with respect to dG and dD , i.e. dD (F (a), F (z)) ≤ dG (a, z),

a, z ∈ G.

(1.1.1)

Notice that there is also another version of the definition of the holomorphically invariant family in which the normalization condition (A) is replaced by the condition (A0 ) dE = pE , where 1 1 + mE pE := log 2 1 − mE is the Poincaré distance. Both definitions are obviously equivalent in the sense that (dG )G fulfills (A,B) iff the family (tanh−1 dG )G satisfies (A0 ,B). In our opinion the normalization condition (A) is more handy in calculations. The following contractible families seem to be the most important. • M¨ obius pseudodistance: c∗G (a, z) : = sup{mE (f (a), f (z)) : f ∈ O(G, E)} = sup{|f (z)| : f ∈ O(G, E), f (a) = 0},

(a, z) ∈ G × G;

the function cG := tanh−1 c∗G is called the Carathéodory pseudodistance. 1

n n A+ := {x ∈ A : x ≥ 0} (A ⊂ R), An + := (A+ ) , e.g. R+ = [0, +∞), Z+ = {0, 1, 2, . . . }, R+ ,

Zn +. 7

8

1. Holomorphically invariant objects

• Higher order M¨ obius function: (k)

mG (a, z) := sup{|f (z)|1/k : f ∈ O(G, E), orda f ≥ k},

(a, z) ∈ G × G, k ∈ N,

where orda f denotes the order of zero of f at a. • Pluricomplex Green function: gG (a, z) := sup{u(z) : u : G −→ [0, 1), log u ∈ PSH(G), ∃C=C(u,a)>0 ∀w∈G : u(w) ≤ Ckw − ak},

(a, z) ∈ G × G,

where PSH(G) denotes the family of all functions plurisubharmonic on G (and k k is the Euclidean norm in Cn ) 2 . • Lempert function: ∗ e kG (a, z) := inf{mE (λ, µ) : λ, µ ∈ E : ∃ϕ∈O(E,G) : ϕ(λ) = a, ϕ(µ) = z}

= inf{µ ∈ [0, 1) : ∃ϕ∈O(E,G) : ϕ(0) = a, ϕ(µ) = z},

(a, z) ∈ G × G.

It is well known that (1)

(k)

∗ c∗G = mG ≤ mG ≤ gG ≤ e kG ,

and for any holomorphically contractible family (dG )G we have ∗ c∗G ≤ dG ≤ e kG ,

(1.1.2)

i.e. the Möbius family is minimal and the Lempert family is maximal. ∗ Put e kG := tanh−1 e . The pseudodistance kG kG := sup{d :

d : G × G −→ R+ is a pseudodistance with d ≤ e kG }

is called the Kobayashi pseudodistance — cf. [J-P 1993], Ch. 3. Observe that (kG )G satisfies (A0 , B). Notice that one can consider conditions weaker than (B), for instance: (B0 ) Condition (1.1.1) holds for every injective holomorphic mapping F : G −→ D; (B00 ) Condition (1.1.1) holds for every biholomorphic mapping F : G −→ D. For example: ∗ )G of Hahn functions • The family (HG ∗ HG (a, z) := inf{mE (λ, µ) : ∃ϕ∈O(E,G) : ϕ is injective, ϕ(λ) = a, ϕ(µ) = z}

= inf{|µ| : ∃ϕ∈O(E,G) : ϕ is injective, ϕ(0) = a, ϕ(µ) = z},

(a, z) ∈ G × G,

∗ ∗ satisfies (A,B0 ). Obviously, e kG ≤ HG . • The family (bG )G of Bergman pseudodistances (see § 3.5) satisfies (A,B00 ). 2

For relations between the pluricomplex and classical Green functions in the unit ball see [Car 1997]. For a different pluricomplex Green function see [Ceg 1995], [Edi-Zwo 1998a].

1.2. Holomorphically contractible families of pseudometrics

9

Remark 1.1.2. The notion of the holomorphically contractible family (dG )G (Definition 1.1.1) may be extended to the case where G runs through all connected complex manifolds, complex analytic sets, or even complex spaces. In particular, one can define the ∗ Möbius pseudodistance c∗M , the Lempert function e kM (defined as 1 for pairs of points for which there is no analytic disc passing through them), and the Kobayashi pseudodistance kM for arbitrary connected complex analytic set M . The following elementary example points out some new problems appearing in this case. Let M := {(z, w) ∈ E 2 : z 2 = w3 } be the Neil parabola. M is a connected one– dimensional analytic subset of E 2 with Reg M = M∗ = M \ {(0, 0)} 3 4 . The set M has a global bijective holomorphic parametrization p

E 3 λ −→ (λ3 , λ2 ) ∈ M. • The mapping q := p−1 is holomorphic on M∗ and continuous on M . Note that q(z, w) = z/w, (z, w) ∈ M∗ , q(0, 0) = 0. • The mapping q|M∗ : M∗ −→ E∗ is biholomorphic. Thus c∗M∗ ((a, b), (z, w)) = mE∗ (q(a, b), q(z, w)) = mE (q(a, b), q(z, w)), ∗ ∗ e kM ((a, b), (z, w)) = e kE (q(a, b), q(z, w)), ∗ ∗

(a, b), (z, w) ∈ M∗ .

• For any ϕ ∈ O(E, M ) there exists a ψ ∈ O(E, E) such that ϕ = p ◦ ψ. Hence ∗ e kM ((a, b), (z, w)) = mE (q(a, b), q(z, w)),

(a, b), (z, w) ∈ M.

• For any f ∈ O(M, E) the holomorphic function h := f ◦ p : E −→ E satisfies h0 (0) = 0. Conversely, for any h ∈ O(E, E) with h0 (0) = 0 the function f := h ◦ q is holomorphic on M . Hence c∗M ((a, b), (z, w)) = sup{|h(q(z, w))| : h ∈ O(E, E), h(q(a, b)) = 0, h0 (0) = 0}, (a, b), (z, w) ∈ M. It is a little bit surprising that, despite the elementary description, ? an effective formula for c∗M is not known ? One can prove that for any λ0 ∈ E∗ , we have sup{|h| : h ∈ O(E, E), h(λ0 ) = 0, h0 (0) = 0} = sup{|B| : B is a Blaschke product of order ≤ 3, B(λ0 ) = 0, B 0 (0) = 0}. 1.2. Holomorphically contractible families of pseudometrics Parallel to the category of holomorphically contractible families of functions (in the sense of Definition 1.1.1) one studies holomorphically contractible families of pseudometrics (cf. [J-P 1993], § 4.1). Definition 1.2.1. A family (δG )G of C–pseudometrics δG : G × Cn −→ R+ , G ⊂ Cn , δG (a; λX) = |λ|δG (a; X), 3

a ∈ G, X ∈ Cn , λ ∈ C,

Reg M denotes the set of all regular points of M . 4 A := A \ {0} (A ⊂ Cn ), An := (A )n , e.g. E , C , (Zn ) , Cn . ∗ ∗ ∗ ∗ ∗ ∗ + ∗

10


where G runs over all domains G ⊂ Cn , is said to be holomorphically contractible if the following two conditions are satisfied: |X| (A) δE (a; X) = γE (a; X) := 1−|a| a ∈ E, X ∈ C, 2, n (B) for any domains G ⊂ C , D ⊂ Cm and for every holomorphic mapping F : G −→ D we have

δD (F (a); F 0 (a)(X)) ≤ δG (a; X),

(a, X) ∈ G × Cn .

(1.2.3)

The following holomorphically contractible families of pseudometrics correspond to the holomorphically contractible families of functions from § 1.1. • Carathéodory–Reiffen pseudometric: γG (a; X) := sup{|f 0 (a)(X)| : f ∈ O(G, E), f (a) = 0},

(a, X) ∈ G × Cn ;

we have γG (a; X) =

1 ∗ cG (a, a + λX) = C∗ 3λ→0 |λ| lim

lim 0 00

z ,z →a

c∗G (z 0 , z 00 ) , kz 0 − z 00 k

z 0 −z 00 →X kz 0 −z 00 k

(a, X) ∈ G × Cn , kXk = 1. • Higher order Reiffen pseudometric: 1/k n 1 o (k) γG (a; X) := sup f (k) (a)(X) : f ∈ O(G, E), orda f ≥ k , k! (a, X) ∈ G × Cn , k ∈ N; we have 1 (k) m (a, a + λX), |λ| G and if G is biholomorphic to a bounded domain, then (k)

γG (a; X) =

lim

C∗ 3λ→0

(a, X) ∈ G × Cn ,

(k)

(k)

γG (a; X) =

lim 0 00

z ,z →a z 0 −z 00 →X kz 0 −z 00 k

mG (z 0 , z 00 ) , kz 0 − z 00 k

(a, X) ∈ G × Cn , kXk = 1.

• Azukawa pseudometric: 1 gG (a, a + λX), |λ| C∗ 3λ→0

AG (a; X) := lim sup

(a, X) ∈ G × Cn ;

if G is a bounded hyperconvex domain, then AG (a; X) =

lim 0 00

z ,z →a z 0 −z 00 →X kz 0 −z 00 k

gG (z 0 , z 00 ) , kz 0 − z 00 k

(a, X) ∈ G × Cn , kXk = 1,

cf. [Zwo 2000c], Corollary 4.4. • Kobayashi–Royden pseudometric: κG (a; X) := inf{α ≥ 0 : ∃ϕ∈O(E,G) : ϕ(0) = a, αϕ0 (0) = X},

(a, X) ∈ G × Cn ;


11

if G is taut, then κG (a; X) =

lim

C∗ 3λ→0

1 e kG (a, a + λX), |λ|

(a, X) ∈ G × Cn ,

cf. [Pan 1994]. (1) (k) It is well known that γG = γG ≤ γG ≤ AG ≤ κG . Moreover, for any holomorphically contractible family of pseudometrics (δG )G we have γG ≤ δG ≤ κG for any G. Notice that (cf. [J-P 1993], [Jar-Pfl 1995b]): • γG is Lipschitz continuous; (k) (k) • γG is upper semicontinuous; if γG (a; X) > 0, (a, X) ∈ G × (Cn )∗ , then γG is (k) continuous (cf. [Nik 2000]); in particular, if G is bounded, then γG is continuous; • AG is upper semicontinuous; • κG is upper semicontinuous; if G is taut, then κG is continuous. Similarly as in the case of contractible functions, one can consider conditions weaker than (B), for example: (B0 ) Condition (1.2.3) holds for every injective holomorphic mapping F : G −→ D; (B00 ) Condition (1.2.3) holds for every biholomorphic mapping F : G −→ D. For example: • The family (hG )G of Hahn pseudometrics hG (a; X) := inf{α ≥ 0 : ∃ϕ∈O(E,G) : ϕ is injective, ϕ(0) = a, αϕ0 (0) = X}, (a, X) ∈ G × Cn , fulfills (A,B0 ). Obviously, κG ≤ hG . • The families of Wu and Bergman pseudometrics satisfy (A,B00 ) (see §§ 1.2.6, 3.5). ∗ ∗ ≡ HG iff κG ≡ hG iff G is simply connected. Remark 1.2.2. (a) If G ⊂ C, then e kG ∗ ∗ (b) e kG ≡ HG and κG ≡ hG for any domain G ⊂ Cn with n ≥ 3 (cf. [Ove 1995]). (c) Let D1 , D2 ⊂ C be domains. Then (cf. [JarW 2000], [JarW 2001]): • if at least one of the domains D1 , D2 is simply connected or biholomorphic to C∗ , ∗ ∗ then hD1 ×D2 ≡ κD1 ×D2 and HD ≡e kD ; 1 ×D2 1 ×D2 ∗ ∗ • otherwise hD1 ×D2 6≡ κD1 ×D2 and HD 6≡ e kD ; see also [Choi 1998]. 1 ×D2 1 ×D2

1.2.1. Inner pseudodistances. For a domain G ⊂ Cn let D(G) be the family of all pseudodistances ρ : G × G → R+ such that ∀a∈G ∃M,r>0 : ρ(z, w) ≤ M kz − wk,

z, w ∈ B(a, r) ⊂ G,

where B(a, r) denotes the Euclidean ball with center at a and radius r. Notice that for any holomorphically contractible family of pseudodistances (dG )G (with the normalization condition (A) or (A0 )), we have dG ∈ D(G) for any G. Let F be one of the following three families of curves in Cn : • Fin := the family of all curves, • Fi := the family of all rectifiable curves (in the Euclidean sense), • Fic := the family of all piecewise C 1 –curves.

12


For any ρ ∈ D(G) we define the inner pseudodistance for ρ w.r.t. the family F: ρF (a, z) := inf{Lρ (α) : α : [0, 1] −→ G, α(0) = a, α(1) = z, α ∈ F},

(a, z) ∈ G × G,

where Lρ (α) is the ρ–length of α: Lρ (α) := sup

N nX

o ρ(α(tj−1 ), α(tj )) : 0 = t0 < t1 < · · · < tN = 1, N arbitrary .

j=1

Note that: • ρF ∈ D(G), • ρF ≥ ρ, • LρF (α) = Lρ (α) for any α ∈ F, • (ρG )F = ρF for F ⊂ G, • (ρF )F = ρF , • ρF = (tanh ρ)F . We put: • ρin := ρFin (cf. [Rin 1961]), • ρi := ρFi (cf. [J-P 1993]), • ρic := ρFic (cf. [Ven 1989]). Note that ρ ≤ ρin ≤ ρi ≤ ρic . We say that ρ is inner if ρ = ρic (in particular, ρ = ρin = ρi = ρic ); see also [Bar 1995]. In particular, we introduce the inner Carathéodory pseudodistance (ciG )G . It is known that: • ciG = cic G for any G; 1 ic i • cin G = cG = cG if G is biholomorphic to a bounded domain or G ⊂ C — notice in i that ? in the general case the equality cG = cG remains still open ? ; • ci 6= c — for instance cA 6≡ ciA if A ⊂ C is an annulus (cf. [J-P 1993], Example 2.5.7, see also [Jar-Pfl 1993a]); • miE = pE = piE . in On the other hand, the Kobayashi pseudodistance is obviously inner, i.e. kG = kG = i ic kG = kG for any G — cf. [J-P 1993], Proposition 3.3.1. If (dG )G is a holomorphically contractible family of pseudodistances (with the nori ic malization condition (A) or (A0 )), then the families (din G )G , (dG )G , (dG )G are holomor0 phically contractible with the normalization condition (A ). 1.2.2. Integrated forms. The idea of inner pseudodistances is strictly connected with the idea of integrated forms from differential geometry. More precisely, for a domain G ⊂ Cn , let M(G, K) (K ∈ {R, C}) denote the space of all K–pseudometrics 5 η : G × Cn −→ R+ , η(a; tX) = |t|η(a; X), (a, X) ∈ G × Cn , t ∈ K, such that ∀a∈G ∃M,r>0 : η(z; X) ≤ M kXk, 5

z ∈ B(a, r) ⊂ G, X ∈ Cn .

Notice that so far we have used only C–pseudometrics (cf. Definition 1.2.1).


13

(k)

If η ∈ M(G, K) is Borel measurable (e.g. η ∈ {γG , γG , AG , κG }), then we define the integrated form of η: R ( η)(a, z) := inf{Lη (α) : α : [0, 1] −→ G, α(0) = a, α(1) = z, α ∈ Fic }, a, z ∈ G, where Lη (α) is the η–length of α Z Lη (α) :=

1

η(α(t); α0 (t))dt.

0

One can easily prove that

R

R R R η ∈ D(G) and ( η)ic = η, i.e. η is always inner.

1.2.3. Buseman pseudometric. Let h : Cn −→ R+ be such that: • h(λX) = |λ|h(X), X ∈ Cn , λ ∈ C, • there exists a constant M > 0 for which h(X) ≤ M kXk, X ∈ Cn . Define the Buseman seminorm for h b h := sup{q : q is a C-seminorm, q ≤ h}; note that b h is a C–seminorm (in particular, b h is continuous) and b h ≤ h; cf. [J-P 1993], § 4.3). If η ∈ M(G, C) (cf. § 1.2.2), then we define the Buseman pseudometric associated to η, ηb(a; X) := (η(a; ·))b (X), (a, X) ∈ G × Cn ; cf. [J-P 1993], § 4.3. In particular, we define the Kobayashi–Buseman pseudometric κ bG . Recall that: • if η is upper semicontinuous, then so is ηb; • if (δG )G is a holomorphically contractible family of pseudometrics, then so is (δbG )G . Remark 1.2.3. (a) One can easily prove that if η is a continuous metric (η(a; X) > 0, (a, X) ∈ G × (Cn )∗ ), then so is ηb (cf. the proof of Proposition 1.2.13(a)). (b) The following example (due to W. Jarnicki) shows that if η is a continuous pseudometric, then ηb need not be continuous. Let η : C2 × C2 −→ R+ , η(z; X) := max{0, |X2 | − kzk|X1 |}. Then η is a continuous pseudometric. Observe that η(0; X) = |X2 | = ηb(0; X) and ηb(z; ·) ≡ 0 for z 6= 0 (in particular, ηb is not continuous). Indeed, for z 6= 0 we have ηb(z; (X1 , 0)) ≤ η(z; (X1 , 0)) = 0 and ηb(z; (0, X2 )) ≤ ηb(z; (0, 2X2 )) ≤ ηb(z; (X2 /kzk, X2 )) + ηb(z; (−X2 /kzk, X2 )) ≤ η(z; (X2 /kzk, X2 )) + η(z; (−X2 /kzk, X2 )) = 0. 1.2.4. Derivatives. It is natural to conjecture that (*) for any ρ ∈ D(G) (cf. § 1.2.1) R there exists a Borel measurable pseudometric η = η(ρ) ∈ M(G, K) such that ρic = η and if (dG )G is a holomorphically contractible family of pseudodistances, then (η(dG ))G is a contractible family of K–pseudometrics.

14


R R Remark 1.2.4. (a) Recall that kG = κG = κ b G , where κ b G is the Kobayashi–Buseman R pseudometric (§ 1.2.3); cf. [Ven 1996] for a generalization of the formula kG = κG to the case of analytic spaces. Thus, we can take η(kG ) := κG or η(kG ) := κ bG . (b) Notice that in general κG is not determined by kG ; there exists a pseudoconvex Hartogs domain G ⊂ C2 such that kG ≡ 0 and κG 6≡ 0 (cf. [J-P 1993], Example 3.5.10). (c) It is known that the problem (*) has a positive solution in the category of so– called C 1 –pseudodistances, i.e. those pseudodistances ρ ∈ D(G) for which the limit 1 (Dρ)(a; X) = lim ρ(z, z + λY ) C∗ 3λ→0 |λ| z→a Y →X

exists for all (a, X) ∈ G × Cn and the function G × Cn 3 (a, X) −→ (Dρ)(a; X) is continuous. If ρ is a C 1 –pseudodistance, then (Dρ)(a; X) =

0

lim 00

z ,z →a z 0 −z 00 →X kz 0 −z 00 k

ρ(z 0 , z 00 ) , kz 0 − z 00 k

(a, X) ∈ G × Cn , kXk = 1;

R ρi = ρic = (Dρ), ρi is a C 1 –pseudodistance, and Dρi = Dρ (cf. [J-P 1993], Proposition 4.3.9). In particular, since cG is a C 1 –pseudodistance, we have η(cG ) = η(ciG ) = γG . (d) M. Kobayashi proved in [KobM 2000] that if G is taut, then kG is a C 1 –pseudodistance. In the case K = C the problem (*) seems to be open (cf. [J-P 1993], the remark after Theorem 4.3.10). Surprisingly, in the case K = R (*) has the following complete solution. For ρ ∈ D(G) define 1 (Dρ)(a; X) := lim sup ρ(a, a + tX), (a, X) ∈ G × Cn ; R∗ 3t→0 |t| cf. [Ven 1989]. We say that Dρ is the weak derivative of ρ. One can prove that: • Dρ ∈ M(G, R), Dρ is Borel measurable. • 1 (Dρ)(a; X) = lim sup ρ(a, a + tY ), (a, X) ∈ G × Cn , R∗ 3t→0 |t| Y →X

(Dρ)(a; X) = lim sup

z→a z−a →X kz−ak

ρ(a, z) , ka − zk

(a, X) ∈ G × Cn , kXk = 1.

• Lρ (α)R= LDρ (α) for any piecewise C 1 –curve α : [0, 1] −→ G. In particular, ρic = (Dρ). • If (dG )G is a holomorphically contractible family of pseudodistances, then (Dd R G )G is a holomorphically contractible family of R–pseudometrics. • (DkG ) = kG .


15

1.2.5. Complex geodesics. Recall that a holomorphic mapping ϕ : E −→ G (G is a domain in Cn ) is called a complex geodesic if c∗G (ϕ(λ0 ), ϕ(λ00 )) = mE (λ0 , λ00 ) for any λ0 , λ00 ∈ E. Let (dG )G be a holomorphically contractible family of functions. Fix a domain G ⊂ Cn and let z00 , z000 ∈ G, z00 6= z000 . We say that ϕ ∈ O(E, G) is a dG –geodesic for (z00 , z000 ) if there exist λ00 , λ000 ∈ E such that z00 = ϕ(λ00 ), z000 = ϕ(λ000 ), and dG (z00 , z000 ) = mE (λ00 , λ000 ). ∗ If ϕ is a dG –geodesic for (z00 , z000 ), then dG (z00 , z000 ) = e kG (z00 , z000 ). Obviously, any complex ∗ 0 00 0 geodesic is a cG –geodesic for (z0 , z0 ) with arbitrary z0 , z000 ∈ ϕ(E), z00 6= z000 . Let (δG )G be a holomorphically contractible family of pseudometrics. Let z0 ∈ G, X0 ∈ Cn∗ . We say that ϕ ∈ O(E, G) is a δG –geodesic for (z0 , X0 ) if there exist λ0 ∈ E, α0 ∈ C such that z0 = ϕ(λ0 ), X0 = α0 ϕ0 (λ0 ), and δG (z0 ; X0 ) = γE (λ0 ; α0 ). If ϕ is a δG –geodesic for (z0 , X0 ), then δG (z0 ; X0 ) = κG (z0 ; X0 ). Proposition 1.2.5 ([J-P 1993], Proposition 8.1.3). For a mapping ϕ ∈ O(E, G) the following conditions are equivalent: (i) ∃λ00 ,λ000 ∈E : c∗G (ϕ(λ00 ), ϕ(λ000 )) = mE (λ00 , λ000 ), i.e. ϕ is a complex c∗G –geodesic for λ00 6=λ00 0

(ϕ(λ00 ), ϕ(λ000 )); (ii) ∀λ0 ,λ00 ∈E : c∗G (ϕ(λ0 ), ϕ(λ00 )) = mE (λ0 , λ00 ), i.e. ϕ is a complex geodesic; (iii) ∀λ∈E : γG (ϕ(λ); ϕ0 (λ)) = γE (λ; 1), i.e. ϕ is a complex γG –geodesic for any pair (ϕ(λ), ϕ0 (λ)); (iv) ∃λ0 ∈E : γG (ϕ(λ0 ); ϕ0 (λ0 )) = γE (λ0 ; 1), i.e. ϕ is a complex γG –geodesic for (ϕ(λ0 ), ϕ0 (λ0 )). Consequently, any complex c∗G – or γG –geodesic ϕ is a complex geodesic. Moreover, ϕ is injective, proper, and regular. In particular, ϕ(E) is a 1-dimensional complex submanifold of G. Proposition 1.2.6 ([J-P 1993], Proposition 8.1.5). Let G ⊂ Cn be a taut domain. Then the following conditions are equivalent: ∗ and γG = κG 6 ; kG (i) c∗G = e ∗ (ii) c∗G = e kG ; (iii) for any z00 , z000 ∈ G, z00 6= z000 , there exist ϕ ∈ O(E, G) and f ∈ O(G, E) such that z00 , z000 ∈ ϕ(E) and f ◦ ϕ = idE ; (iv) for any z00 , z000 ∈ G there exist a holomorphic embedding ϕ : E → G and a holomorphic retraction r : G → ϕ(E) such that z00 , z000 ∈ ϕ(E). Moreover, any holomorphic mapping ϕ E → G satisfying (iii) or (iv) is a complex geodesic. Conversely, for any complex geodesic ϕ there exists f (resp. r) such that (iii) (resp. (iv)) is fulfilled. Recently Proposition 1.2.5 was generalized in the following way in [EHHM 2003], Corollary 9. Proposition 1.2.7. A holomorphic mapping ϕ : E −→ G (G is a domain in Cn ) is a complex geodesic iff there exist λ00 , λ000 ∈ E, λ00 6= λ000 , such that ciG (ϕ(λ00 ), ϕ(λ000 )) = pE (λ00 , λ000 ). 6

For example, G is a convex domain — cf. Lempert Theorem 8.2.1 in [J-P 1993].

16


Proof. (Here we present a direct proof independent of [EHHM 2003].) Using a suitable automorphism of E, we may assume that λ00 = 0 and λ000 =: t0 ∈ (0, 1). Recall that piE = pE . Hence, for any t ∈ [0, t0 ], we have pE (0, t0 ) = pE (0, t) + pE (t, t0 ) ≥ ciG (ϕ(0), ϕ(t)) + ciG (ϕ(t), ϕ(t0 )) ≥ ciG (ϕ(0), ϕ(t0 )) = pE (0, t0 ). Consequently, ciG (ϕ(0), ϕ(t)) = pE (0, t) for any t ∈ [0, t0 ]. Let tk & 0 be such that ϕ(tk ) − ϕ(0) −→ X0 ∈ ∂Bn kϕ(tk ) − ϕ(0)k (observe that X0 = ϕ0 (0)/kϕ0 (0)k if ϕ0 (0) 6= 0). Recall that DciG = γG . We get pE (0, tk ) ci (ϕ(0), ϕ(tk )) = lim G = γG (ϕ(0); X0 )kϕ0 (0)k. k→∞ k→∞ tk tk

1 = γE (0; 1) = lim

Hence ϕ0 (0) 6= 0 and 1 = γG (ϕ(0); ϕ0 (0)) which, by Proposition 1.2.5, implies that ϕ is a complex geodesic. Remark 1.2.8. (a) Complex geodesics were recently studied by many authors. For instance: • in [Jar-Pfl 1995a] for convex complex ellipsoids Ep := {(z1 , . . . , zn ) ∈ Cn :

n X

|zj |2pj < 1},

j=1

p = (p1 , . . . , pn ), pj ≥ 1/2, j = 1, . . . , n (n ≥ 2);

7

• in [Zwo 1997] to prove the following result showing that in the category of complex ellipsoids the symmetry of the pluripolar Green function gEp is a very rare phenomenon: Theorem. For a complex ellipsoid Ep the following conditions are equivalent: (i) kEp (λ1 b, λ2 b) = pE (λ1 , λ2 ), b ∈ ∂Ep , λ1 , λ2 ∈ E; (ii) gEp (λb, 0) = gEp (0, λb), b ∈ ∂Ep , λ ∈ E; (iii) gEp is symmetric; (iv) Ep is convex; • in [Vis 1999a], [Vis 1999b], [Vis 1999c] for some classes of convex Reinhardt domains; • in [Pfl-You 2003] for the so–called minimal ball Mn := {z = (z1 , . . . , zn ) ∈ Cn : kzk2 + |z12 + · · · + zn2 | < 1} (which will be studied in § 3.1). 7

Observe that E(1,...,1) coincides with the open unit Euclidean ball Bn . Moreover, Ep is convex if and only if pj ≥ 1/2, j = 1, . . . , n (cf. [J-P 1993], § 8.4).


17

(b) Consider the following general problem: Given convex balanced do a 9bounded main G ⊂ Cn (n ≥ 2) with Minkowski function hG 8 , find conditions on a, b ∈ G and r, R ∈ (0, 1), under which the Carathéodory ball Bc∗G (a, r) := {z ∈ G : c∗G (a, z) < r} coincides with the norm ball BhG (b, R) := {z ∈ Cn : hG (z − b) < R} 10 . Since c∗G (0, ·) = hG (·), we always have Bc∗G (0, r) = BhG (0, r),

r ∈ (0, 1).

In the case where G = Ep,α := {(z1 , . . . , zn ) ∈ Cn : 2α|z1 |p1 |z2 |p2 +

n X

|zj |2pj < 1},

j=1

p = (p1 , . . . , pn ) ∈ Rn>0 , α ≥ 0,

11

the problem was studied in: [Sch 1993] (the case n = 2, α = 0, p1 = p2 = 1), [Sre 1995], [Zwo 1995] (the case α = 0, p1 = · · · = pn = 1), [Sch-Sre 1996] (the case α = 0, 1 < p1 = · · · = pn ∈ / N), [Zwo 1996], [Zwo 2000b] (the case α = 0), [Vis 1999a] (the general case). The methods introduced by W. Zwonek and developed by B. Visintin are based on complex geodesic. The most general result is the following theorem from [Vis 1999a]. Theorem. Assume that α ≥ 0 and p ∈ Rn>0 are such that Ep,α is convex. Then Bc∗Ep,α (a, r) = BhEp,α (b, R) for some a, b ∈ Ep,α , a 6= 0, r, R ∈ (0, 1) iff α = 0, {j ∈ {1, . . . , n} : aj 6= 0} = {j0 }, pj0 = 1, and pj = 1/2 for all j 6= j0 . Remark 1.2.9. Let G ⊂ Cn be a domain and let (z0 , X0 ) ∈ G × Cn . Recall that a mapping ϕ ∈ O(E, G) is called a κG –geodesic for (z0 , X0 ) if there are a λ0 ∈ E and an α0 ∈ C such that ϕ(λ0 ) = z0 and κG (z0 ; X0 ) = γE (λ0 ; α0 ) (see Chapter VIII in [J-P 1993]). If G is taut, then for any pair (z0 , X0 ) ∈ G × Cn there is a κG –geodesic for (z0 , X0 ). If G is even convex, then any κG –geodesic is a complex geodesic. Now let G = Ep , where p = (p1 , . . . , pn ) with pj > 0. Observe that Ep is not necessarily convex. Fix a pair (z0 , X0 ) ∈ Ep × (Cn )∗ . Then any κEp –geodesic ϕ for (z0 , X0 ), 8

Observe the difference between the Hahn pseudometric hG : G × Cn −→ R+ and the Minkowski function hG : Cn −→ R+ ; since the Hahn pseudometric will be not used in the sequel, there will be no confusion for the reader. 9 Notice that under our assumptions h G is a complex norm. 10 Recall that in the case of the unit disc we have

a(1 − r 2 ) r(1 − |a|2 ) BmE (a, r) = B , , 1 − r 2 |a|2 1 − r2 |a|2 11

a ∈ E, r ∈ (0, 1).

n n A>0 := {x ∈ A : x > 0} (A ⊂ R), An >0 := (A>0 ) , e.g. R>0 , R>0 . Observe that Ep,0 = Ep .

18


where ϕj 6≡ 0, j = 1, . . . , n, is necessarily of the following form (see [Pfl-Zwo 1996]) 1 − α λ 1 pj j ϕj (λ) = Bj (λ) aj , j = 1, . . . , n, (1.2.4) 1 − α0 λ where Bj is a Blaschke product and the complex numbers aj , αj fulfill the following conditions • aj ∈ C∗ , αP j ∈ E, j = 1, . . . , n, α0 ∈ E, n • 1 + |α0 |2 = j=1 |aj |2 (1 + |αj |2 ), Pn • α0 = j=1 |aj |2 αj . Moreover, if pj ≥ 12 , then Bj ≡ 1 or Bj (λ) =

λ−αj 1−αj λ

with |αj | < 1. λ−α

Additionally, if αj ∈ E for all j = 1, . . . , n, then either Bj ≡ 1 or Bj (λ) = 1−αjjλ for all j = 1, . . . , n. Using this result, the Kobayashi metric for the non convex domain E(1,m) , 0 < m < 12 , is obtained. First observe that the following mappings z − a eiθ (1 − |a|2 )1/(2m) z 1 2 E(1,m) 3 z −→ , ∈ E(1,m) , a ∈ E, θ ∈ R, 1 − az1 (1 − az1 )1/m are automorphisms. Therefore, to know κE(1,m) , it suffices to calculate κE(1,m) (0, b); · , b ≥ 0. The simple part is given by the following formulas • κE(1,m) (0, 0); X) = hE(1,m) (X), where hE(1,m) denotes the Minkowski function of E(1,m) , X ∈ C2 ; |X2 | b > 0, X1 = 0; • κE(1,m) (0, b); X) = 1−b 2, • κE(1,m) (0, b); X) =

|X1 | , (1−b2m )1/2

b > 0, X2 = 0.

To discuss the remaining case (b > 0 and (without loss of generality) X = (X1 , 1) ∈ C2∗ ), we put b|X | 2 1 ν := ν(m, b, X) := . m 1 Moreover, in the case ν ≤ 4m(1−m) set t := t(m, b, X) :=

2m2 ν p . 1 + 2m(m − 1)ν + 1 + 4m(m − 1)ν

Observe that then the following function ξ 2m − tξ 2m−2 − (1 − t)b2m ,

ξ ∈ R,

has exactly one zero x = x(m, b, X) in the interval (0, 1). Now we are able to give the remaining formulas. Theorem. Let m ∈ (0, 21 ), b > 0, X = (X1 , 1) ∈ C2 , ν = ν(b, m, X), and x = x(b, m, X) 1 . Then: if ν ≤ 4m(1−m) • if ν ≤ 1, then κE(1,m) (0, b); X) =

m x2m−1 =: κ1 (ν); b (1 − m)x2m + mx2m−2 − b2m


• if ν ≥

1 4m(1−m) ,

19

then

m κE(1,m) (0, b); X) = b 1 • if 1 < ν < 4m(1−m) , then

p

(1 − b2m )ν + b2m =: κ2 (ν); 1 − b2m

κE(1,m) (0, b); X) = max{κ1 (ν), κ2 (ν)}. The minimum in the last formula is equal to κ1 (ν) for ν ≤ ν0 and equal to κ2 (ν) for ν > ν0 , where 2m t0 x2m 0 −b ν0 := 2 , t0 := 2m−2 xo − b2m t0 (1 − m) + m and x0 is the only solution in the interval (0, 1) of the following equation ξ 4m−2 − 1 − 2m + 2m2 + b2m + ξ 2m 1 + (1 − 2m)b2m + ξ 2m−2 1 + (2m − 1)b2m − (1 − m)2 ξ 4m − m2 ξ 4m−2 − b2m = 0 It turns out that there is a mapping ϕ ∈ O(E, E(1,m) ) of the form (1.2.4) which is not a κE(1,m) –geodesic for (ϕ(0), ϕ0 (0)). Moreover, for a b > 0 such that (0, b) ∈ G the function κE(1,m) ((0, b); (·, 1)) is not differentiable on C. 1.2.6. Wu pseudometric. The Wu pseudometric has been introduced by H. Wu in [Wu 1993] (and [Wu]). Various properties of the Wu pseudometric have been studied in [Che-Kim 1996], [Che-Kim 1997], [Kim 1998], [Che-Kim 2003], [Jar-Pfl 2003a], [Juc 2002]. Following [Jar-Pfl 2003a], let us formulate the definition of the Wu pseudometric in an abstract setting. Let h : Cn −→ R+ be a C–seminorm. Put: I = I(h) := {X ∈ Cn : h(X) < 1} (I is convex), V = V (h) := {X ∈ Cn : h(X) = 0} ⊂ I (V is a vector subspace of Cn ), U = U (h) := the orthogonal complement of V with respect to the standard Hermitian Pn scalar product hz, wi := j=1 zj wj in Cn , I0 := I ∩ U , h0 := h|U (h0 is a norm on U , I = I0 + V ). For any pseudo–Hermitian scalar product s : Cn × Cn −→ C 12 , let p qs (X) := s(X, X), X ∈ Cn , E(s) := {X ∈ Cn : qs (X) < 1}. Consider the family F of all pseudo–Hermitian scalar products s : Cn × Cn −→ C such that I ⊂ E(s), equivalently, qs ≤ h. In particular, V ⊂ I = I0 + V ⊂ E(s) = E(s0 ) + V, 12

That is, • s(·, w) : Cn −→ C is C–linear for any w ∈ Cn , • s(z, w) = s(w, z) for any z, w ∈ Cn , • s(z, z) ≥ 0 for any z ∈ Cn (if s(z, z) > 0 for any z ∈ (Cn )∗ , then s is a Hermitian scalar product).

20


where s0 := s|U ×U (note that E(s0 ) = E(s) ∩ U ). Let Vol(s0 ) denote the volume of E(s0 ) with respect to the Lebesgue measure of U . Since I0 is bounded, there exists an s ∈ F with Vol(s0 ) < +∞. Observe that for any basis e = (e1 , . . . , em ) of U (m := dimC U ) we have C(e) Vol(s0 ) = , det S where C(e) > 0 is a constant (independent of s) and S = S(s0 ) denotes the matrix representation of s0 in the basis e, i.e. Sj,k := s(ej , ek ), j, k = 1, . . . , m. In particular, if U = Cm × {0}n−m and e = (e1 , . . . , em ) is the canonical basis, then C(e) = Λ2m (Bm ), where Λ2m denotes the Lebesgue measure in Cm . We are interested in finding an s ∈ F, for which Vol(s0 ) is minimal, equivalently, det S(s0 ) is maximal. Observe that, if s has the above property with respect to h (i.e. the volume of E(s0 ) is minimal), then, for any C–linear isomorphism L : Cn −→ Cn , the scalar product L(s)

Cn × Cn 3 (X, Y ) −→ s(L(X), L(Y )) ∈ C has the analogous property with respect to h ◦ L. In particular, this permits us to reduce the situation to the case where U = Cm × {0}n−m and next to assume that m = n (by restricting all the above objects to Cm ' Cm × {0}n−m ). Lemma 1.2.10. There exists exactly one element sh ∈ F such that Vol(sh0 ) = min{Vol(s0 ) : s ∈ F} < +∞. Proof. ([Wu], [Wu 1993]) We may assume U (h) = Cn . First we prove that the set F is compact. It is clear that F is closed. To prove that F is bounded, observe that q |s(ej , ek )| ≤ s(ej , ej )s(ek , ek ) = qs (ej )qs (ek ) ≤ h(ej )h(ek ), s ∈ F, j, k = 1, . . . , n, where e1 , . . . , en is the canonical basis in Cn . Consequently, the entries of the matrix S(s) are bounded (by a constant independent of s). Recall that Λ2n (Bn ) , Vol(s) = det S(s) Now, using compactness of F, we see that there exists an sh ∈ F such that Vol(sh ) = min{Vol(s) : s ∈ F} < +∞. It remains to show that sh is uniquely determined. Suppose that s0 , s00 ∈ F, s0 6= s00 , are both minimal and let S 0 , S 00 denote the matrix representation of s0 , s00 , respectively. We know that µ := det S 0 = det S 00 is maximal (with respect to any basis (e1 , . . . , en )) in the class F. Take a basis e1 , . . . , en such that the matrix A := S 00 (S 0 )−1 is diagonal and let d1 , . . . , dn be the diagonal elements. Note that 1 = det A = d1 · · · dn and that for at least one j ∈ {1, . . . , n} we have dj 6= 1. Put s := 12 (s0 + s00 ). Then s ∈ F. Let S = S(s) be the matrix representation of s. We have det S =

1 1 det(S 0 + S 00 ) = n det(In + A) det S 0 2n 2 p 1 + d1 1 + dn = ··· µ > d1 · · · dn µ = µ; 2 2


21

contradiction (In denotes the unit matrix).

p n h Put sbh := m · √ sh (m := dim U (h)), √ Wh := qsbh (Wh(X) = ms (X, X), X ∈ C ). Obviously, Wh ≤ mh and Wh ≡ √mh iff h = qs for some pseudo–Hermitian scalar product s. For instance, Wk k = nk k, where k k is the Euclidean norm in Cn . √ Moreover, W(Wh) ≡ mWh. Remark 1.2.11. Assume that U (h) = Cn . Let L : Cn −→ Cn be a C–linear isomorphism such that | det L| = 1 and h ◦ L = h. Then Vol(sh ) = Vol(L(sh )) and hence sh = L(sh ), i.e. sh (X, Y ) = sh (L(X), L(Y )), X, Y ∈ Cn . √ Proposition 1.2.12. (a) h ≤ Wh ≤ mh. (b) If h(X) := max{h1 (X1 ), h2 (X2 )}, X = (X1 , X2 ) ∈ Cn1 × Cn2 , then sbh (X, Y ) = sbh1 (X1 , Y1 ) + sbh2 (X2 , Y2 ),

X = (X1 , X2 ), Y = (Y1 , Y2 ) ∈ Cn1 × Cn2 .

In particular, 1/2 Wh(X) = (Wh1 (X1 ))2 + (Wh2 (X2 ))2 ,

X = (X1 , X2 ) ∈ Cn1 × Cn2 .

Proof. ([Wu], [Wu 1993]) (a) Using a suitable C–linear isomorphism we may reduce the situation to the case where: • U = Cn , • sh (X, Y ) = hX, Y i, X, Y ∈ Cn , • min{kXk : h(X) = 1} = kX∗ k = a > 0, X∗ = (0, . . . , 0, a) ∈ ∂I; in particular, since I is a balanced convex domain, I ⊂ {(X 0 , Xm ) ∈ Cm−1 × C : |Xm | < a}. √ √ We only need to show that a ≥ 1/ n. Suppose that a < 1/ n and let 0 < b < 1 be such that a2 + b2 = 1. Put c := a/b. Note that (n − 1)c2 < 1. Let L : Cn −→ Cn be the C–linear isomorphism √ L(X) := (c n − 1X 0 , Xn ), X = (X 0 , Xn ) ∈ Cn−1 × C. −1

Obviously, sh◦L

= L−1 (sh ), so

√ −1 Vol(sh◦L ) = Λ2n (Bn )| det L|2 = Λ2n (Bn )(c n − 1)2(n−1) . √ On the other hand, L(I) ⊂ B(a n) ⊂ Cn . Indeed, for X = (X 0 , Xn ) we have kL(X)k2 = (n − 1)c2 kX 0 k2 + |Xn |2 = (n − 1)c2 kXk2 + (1 − (n − 1)c2 )|Xn |2 < (n − 1)c2 + (1 − (n − 1)c2 )a2 = a2 + (1 − a2 )(n − 1)(a2 /b2 ) = na2 < 1. √ −1 Consequently, Vol(sh◦L ) ≤ Λ2n (Bn )(a n)2n . Thus, using the above inequality, we get

√ √ (a n − 1/b)2(n−1) ≤ (a n)2n .

Put f (t) := t(1 − t)n−1 , 0 ≤ t ≤ 1. Then 1 1 n−1 1− = f (1/n); n n contradiction (because f is strictly increasing in the interval [0, 1/n] and a2 < 1/n). f (a2 ) = a2 (1 − a2 )n−1 ≥

22


(b) We may assume that U (hj ) = Cnj , j = 1, 2. Put s∗ (X, Y ) :=

n1 n2 sh2 (X2 , Y2 ), sh1 (X1 , Y1 ) + n1 + n2 n1 + n2 X = (X1 , X2 ), Y = (Y1 , Y2 ) ∈ Cn1 × Cn2 .

We only need to prove that det S(sh ) = det S(s∗ ) (all matrix representations are taken in the canonical bases of Cn1 and Cn2 , respectively). Let s := sh . Since I(h) = I(h1 ) × I(h2 ) ⊂ E(s∗ ), we get det S(s) ≥ det S(s∗ ). Let L : Cn1 × Cn2 −→ Cn1 × Cn2 be the isomorphism of the form L(X1 , X2 ) := (X1 , −X2 ). Then h ◦ L = h and, consequently, s = L(s) (Remark 1.2.11), i.e. s(X, Y ) = s(L(X), L(Y )),

X, Y ∈ Cn1 × Cn2 .

Hence s((X1 , X2 ), (Y1 , Y2 )) = 0 if (X2 = 0 and Y1 = 0) or (X1 = 0 and Y2 = 0). Indeed, s((X1 , 0), (0, Y2 )) = s(L(X1 , 0), L(0, Y2 )) = s((X1 , 0), (0, −Y2 )) = s((X1 , 0), −(0, Y2 )) = −s((X1 , 0), (0, Y2 )). Consequently, s(X, Y ) = s1 (X1 , Y1 ) + s2 (X2 , Y2 ),

X = (X1 , X2 ), Y = (Y1 , Y2 ) ∈ Cn1 × Cn2 ,

where sj is a Hermitian scalar product in Cnj , j = 1, 2. It is clear that I(hj ) ⊂ E(sj ), j = 1, 2. Let cj ≤ 1 be the minimal number such that I(hj ) ⊂ E(c−2 j sj ), j = 1, 2. Assume that Xj0 ∈ ∂I(hj ) is such that sj (Xj0 , Xj0 ) = c2j , j = 1, 2. In particular, qs (X10 , X20 ) ≤ 1, so c21 + c22 ≤ 1. We have −2nj

det S(shj ) ≥ cj

det S(sj ),

j = 1, 2,

and, therefore, 1 2n2 det S(s) = det S(s1 ) det S(s2 ) ≤ c2n det S(sh1 ) det S(sh2 ) 1 c2

2 n2 1 ≤ c2n det S(sh1 ) det S(sh2 ) 1 (1 − c1 ) n1 n n2 n 2 1 ≤ det S(sh1 ) det S(sh2 ) = det S(s∗ ), n1 + n2 n1 + n2

since the maximum of the function f (t) = tn1 (1 − t)n2 , 0 ≤ t ≤ 1, is attained at t = n1 /(n1 + n2 ). For a domain G ⊂ Cn and η ∈ M(G, C) (cf. § 1.2.1), we define the Wu pseudometric (Wη)(a; X) := (Wb η (a; ·))(X),

(a, X) ∈ G × Cn ,

where ηb is the Buseman pseudometric associated to η (cf. § 1.2.3). Observe that Wη ∈ M(G, C). R Recall that a Borel measurable metric η ∈ M(G, C) is said to be complete if any η–Cauchy sequence is convergent to a point from G (cf. [J-P 1993], § 7.3).


23

Proposition 1.2.13. (a) If η ∈ M(G, C) is a continuous metric, then so is Wη (cf. Example 1.2.15). (b) If η ∈ M(G, C) is a continuous complete metric, then so is Wη. (c) If (δG )G is a holomorphically contractible family of pseudometrics, then: • for any biholomorphic mapping F : G −→ D, G, D ⊂ Cn , we have (WδD )(F (z); F 0 (z)(X)) = (WδG )(z; X),

(z, X) ∈ G × Cn ;

• for any holomorphic mapping F : G −→ D, G ⊂ Cn1 , D ⊂ Cn2 , we have √ (WδD )(F (z); F 0 (z)(X)) ≤ n2 (WδG )(z; X), (z, X) ∈ G × Cn1 , but, for example, the family (WκG )G is not holomorphically contractible (cf. Example 1.2.14). In the case η = κG , the above properties (a) — (c) were formulated (without proof) in [Wu], [Wu 1993]. Proof. (a) Fix a point z0 ∈ G ⊂ Cn . Let sz := sηb(z;·) , z ∈ G. We are going to show that sz −→ sz0 when z −→ z0 . By our assumptions, there exist r > 0, c > 0 such that η(z; X) ≥ ckXk,

z ∈ B(z0 , r) ⊂ G, X ∈ Cn .

In particular, the sets Iz := {X ∈ Cn : ηb(z; X) < 1},

z ∈ B(z0 , r),

are contained in the ball B(C) with C := 1/c. Moreover, |b η (z; X) − ηb(z0 ; X)| ≤ ϕ(z)kXk,

X ∈ Cn ,

where ϕ(z) −→ 0 when z −→ z0 . Hence (1 + Cϕ(z))−1 Iz ⊂ Iz0 ⊂ (1 + Cϕ(z))Iz ,

z ∈ B(z0 , r),

and, consequently, Iz0 ⊂ (1 + Cϕ(z))E(sz ) = E((1 + Cϕ(z))−2 sz ), Iz ⊂ (1 + Cϕ(z))E(sz0 ) = E((1 + Cϕ(z))−2 sz0 ),

(1.2.5) z ∈ B(z0 , r).

Hence, Vol(sz0 ) ≤ Vol((1 + Cϕ(z))−2 sz ) = (1 + Cϕ(z))2n Vol(sz ), Vol(sz ) ≤ Vol((1 + Cϕ(z))−2 sz0 ) = (1 + Cϕ(z))2n Vol(sz0 ),

z ∈ B(z0 , r).

Thus Vol(sz ) −→ Vol(sz0 ) when z −→ z0 . Take a sequence zν −→ z0 . Since |szν (ej , ek )| ≤ η(zν ; ej )η(zν ; ek ),

j, k = 1, . . . , n, ν ∈ N,

we may assume that szν −→ s∗ , where s∗ is a pseudo–Hermitian scalar product. We already know that Vol(s∗ ) = Vol(sz0 ). Moreover, by (1.2.5), Iz0 ⊂ E(s∗ ). Consequently, the uniqueness of sz0 implies that s∗ = sz0 .

24


R R (b) Recall that η = ηb — cf. [J-P 1993],RProposition 4.3.5(b). By (a), Wη is a continuous metric. In particular, the distance (Wη) is well defined. By Proposition 1.2.12(a) we get R R ηb ≤ (Wη), which directly implies the required result. (c) Recall that the family (δbG )G is holomorphically contractible (cf. § 1.2.3). If F is biholomorphic, then the result is obvious because for any z ∈ G, the mapping F 0 (z) is a C–linear isomorphism and δbD (F (z); F 0 (z)(X)) = δbG (z; X), X ∈ Cn . In the general case, using Proposition 1.2.12(a), we get √ (WδD )(F (z); F 0 (z)(X)) ≤ n2 δbD (F (z); F 0 (z)(X)) √ √ ≤ n2 δbG (z; X) ≤ n2 (WδG )(z; X), (z, X) ∈ G × Cn1 . √ Example 1.2.14. Let Gε := {(z1 , z2 ) ∈ B2 : |z1 | < ε}, 0 < ε < 1/ 2. Recall that κB2 (0; X) = kXk and κGε (0; X) = max{kXk, |X1 |/ε}, X = (X1 , X2 ). Then r |X1 |2 |X2 |2 (WκGε )(0; (X1 , X2 )) = + , X = (X1 , X2 ) ∈ C2 . 2 ε 1 − ε2 In particular, √ 1 = (WκGε )(0; (0, 1)). (WκB2 )(0; (0, 1)) = 2 > √ 1 − ε2 Consequently, the family (WκG )G is not contractible with respect to inclusions. We point out that Proposition 1.2.13(a) gives us the continuity of Wη only in the case where η is a continuous metric. The following Example 1.2.15 shows that if η is only upper semicontinuous, then Wη need not be upper semicontinuous. We do not know whether Wη is upper semicontinuous in the case where η is a continuous pseudometric. Observe that the upper semicontinuity (or at least Borel measurability) of Wη appears in a natural R way when one defines (Wη). In the case where η = κG , the upper semicontinuity of WκG is claimed for instance in [Wu 1993] (Theorem 1), [Che-Kim 1996] (Proposition 2), [Juc 2002] (Theorem 0), but so far there is no proof. (k) ? Let η ∈ {γG , AG , κG } (cf. § 1.2). Is Wη upper semicontinuous ? Example 1.2.15. There is an upper semicontinuous metric η such that Wη is not upper semicontinuous. Indeed, let η : B2 × C2 −→ R+ , η(z; X) := kXk for z 6= 0, and η(0;√ X) := 2 max{kXk, |X1 |/ε}, X = (X1 , X2 ) ∈ C (ε > 0 small). Then (Wη)(z; √ X) = 2kXk for z 6= 0, and (by Example 1.2.14) {X ∈ C2 : (Wη)(0; X) < 1} 6⊂ B(1/ 2), so Wη is not upper semicontinuous. Example 1.2.16. There exists a bounded domain G ⊂ C2 such that WκG is not continuous (see Proposition 2 in [Che-Kim 1996], where such a continuity is claimed). Indeed, let D ⊂ C2 be a domain such that (cf. [J-P 1993], Example 3.5.10): • there exists a dense subset M ⊂ C such that (M × C) ∪ (C × {0}) ⊂ D, • κD (z; (0, 1)) = 0, z ∈ A := M × C,


25

• there exists a point z 0 ∈ D \ A such that κD (z 0 ; X) ≥ ckXk, X ∈ C2 , where c > 0 is a constant. For R > 0 let DR := {z = (z1 , z2 ) ∈ D : |zj − zj0 | < R, j = 1, 2}. It is known that κDR & κD when R % +∞. Observe that z 0 ∈ DR and κDR (z 0 ; X) ≥ κD (z 0 ; X) ≥ ckXk,

X ∈ C2 .

Hence, by Proposition 1.2.12(a), (WκDR )(z 0 ; X) ≥ ckXk, X ∈ C2 . In particular, (WκDR )(z 0 ; (0, 1)) ≥ c. Fix a sequence M 3 zk −→ z10 . Note that {zk } × (z20 + RE) ⊂ DR , which implies that κDR ((zk , z20 ); (0, 1)) ≤ 1/R, k = 1, 2, . . . . In particular, √ √ (WκDR )((zk , z20 ); (0, 1)) ≤ 2κDR ((zk , z20 ); (0, 1)) ≤ 2/R, k = 1, 2, . . . . √

Now it clear that if R >

2 c ,

then

lim sup(WκDR )((zk , z20 ); (0, 1)) ≤

√

2/R < c ≤ (WκDR )(z 0 ; (0, 1)),

k→+∞

which shows that for G := DR the pseudometric WκG is not continuous. √ Remark 1.2.17. We point out the role played in the definition of W by the factor m. Put f Wh := qsh , f Wη(a; X) = (f Wb η (a; ·))(X), (a, X) ∈ G × Cn . Let D ⊂ C2 and D 3 zk −→ z0 ∈ D be such that: • κD (zk ; ·) is not a metric (in particular, m(k) := dim U (b κD (zk ; ·)) ≤ 1, k ∈ N), • κD (z0 ; ·) is a metric (take, for instance, the domain D from Example 1.2.16). Put G := D × E ⊂ C3 . Then 1 1 (f WκG )2 ((zk , 0); (0, 1)) = sκG ((zk ,0);·) ((0, 1), (0, 1)) = ≥ , k ∈ N, m(zk ) + 1 2 1 1 f G )2 ((z0 , 0); (0, 1)) = sκG ((z0 ,0);·) ((0, 1), (0, 1)) = = , (Wκ m(z0 ) + 1 3 and, therefore, f WκG is not upper semicontinuous at ((z0 , 0), (0, 1)) (the example is due to W. Jarnicki). Remark 1.2.18. The Wu metric in complex ellipsoids E(1,m) was studied in [Che-Kim 1996] (m ≥ 21 ) and [Che-Kim 1997] (0 < m < 12 ). In a recent paper [Che-Kim 2003] the same authors proved the following two results. Let G := Bn ∩ U , where U is open in Cn . Then there exists a neighborhood V of ∂G ∩ ∂Bn such that WκG = WκBn in V ∩ G. Let p = (p1 , . . . , pn ) ∈ Nn , pj ≥ 2, j = 1, . . . , n. Then any strongly pseudoconvex point a ∈ ∂Ep has a neighborhood V such that WκEp is a K¨ ahler metric with constant negative curvature in V ∩ Ep . 1.2.7. Regularity of contractible pseudodistances and pseudometrics. Let us mention a few new results related to different regularity properties of contractible objects. T∞ n • Let (Gj )∞ j=1 be a sequence of domains in C such that Gj+1 ⊂⊂ Gj and j=1 Gj = G, where G is a domain in Cn . It is an open question to find conditions under which

26


cGj −→ cG or kGj −→ kG . M. Kobayashi in [KobM 2002] proved the following two results: (a) If G is strongly pseudoconvex, then cGj −→ cG locally uniformly. (b) If G is a bounded domain such that every point b ∈ ∂G admits a weak peak function (i.e. a function f holomorphic in a neighborhood of G such that f (b) = 1 and |f | < 1 on G), then kGj −→ kG locally uniformly. • The behavior of the Bergman, Carathéodory, and Kobayashi metrics on a smooth bounded pseudoconvex domain G ⊂ Cn near a boundary point of finite type, where the Levi form of ∂G has at least n − 2 positive eigenvalues, was studied in [Cho 1995]. The behavior of the Kobayashi metric near boundary points of exponentially-flat infinite type in bounded domains in C2 was studied in [Lee 2001]. Lower and upper nontangential bounds for the Carathéodory metric of a smooth bounded pseudoconvex domain G ⊂ Cn near an h–extendible boundary point (a boundary point is said to be h–extendible if its Catlin multitype coincides with its D’Angelo type) were proved in [Nik 1997] and [Nik 1999]. Some localization theorems for contractible functions and metrics were proved in [Nik 2002]. • Let G be a strongly pseudoconvex balanced domain with C ∞ (resp. real analytic) boundary. Then there is an open neighborhood U = U (0) ⊂ G such that κG is C ∞ (resp. real analytic) on U × (Cn )∗ ; cf. [Pan 1993]. • Let Dm := E(1,m) × (C2 )∗ ⊂ C4 , m > 0. It was proved in [Ma 1995] that: (a) κE(1,m) ∈ C 2 (Dm ) for m ≥ 1, (b) κE(1,m) is piecewise C 3 on Dm and κE(1,m) ∈ / C 3 (Dm ) for m ≥ 32 . • Let G, D ⊂ Cn be domains and let a ∈ G, b ∈ D. We say that a holomorphic mapping F : G −→ D with F (a) = b is Carathéodory extremal if | det F 0 (a)| = sup{| det Φ0 (a)| : Φ ∈ O(G, D), Φ(a) = b}. In the cases: G = Bn , D = Em,p , a = b = 0, G = Em,p , D = Bn , a = b = 0, where k n o X kzj k2pj < 1 , Em,p := z ∈ Cm1 × · · · × Cmk : j=1

m = (m1 , . . . , mk ) ∈ Nk , m1 + · · · + mk = n, p = (p1 , . . . , pk ) ∈ Rn>0 , the Carathéodory extremal mappings are characterized in [Ma 1997].

1.3. Effective formulas for elementary Reinhardt domains

27

1.3. Effective formulas for elementary Reinhardt domains For α = (α1 , . . . , αn ) ∈ Rn∗ and c ∈ R put Dα,c := {z ∈ Cn : |z1 |α1 · · · |zn |αn < ec (∀j∈{1,...,n} : αj < 0 =⇒ zj 6= 0)},

Dα := Dα,0 ;

Dα,c is called an elementary Reinhardt domain. We say that Dα,c is of rational type if α ∈ R · Zn . The domain Dα,c is of irrational type if it is not of rational type. Without loss of generality we may assume that α1 , . . . , αk < 0 and αk+1 , . . . , αn > 0 for some k ∈ {0, . . . , n}. If k < n, then we put tk := min{αk+1 , . . . , αn }. Let V0 := {(z1 , . . . , zn ) ∈ Cn : z1 · · · zn = 0}. For α ∈ Zn and r ∈ N, put Φ(z) := z α , Φ(r) (a)(X) :=

X β∈Zn + , |β|=r

1 β D Φ(a)X β , β!

a ∈ Dα , X ∈ Cn .

To simplify notation, for z ∈ Dα , write |z α | := |z1 |α1 · · · |zn |αn (observe that this notation agrees with the standard one if α ∈ Zn ). The following effective formulas for holomorphically contractible functions and pseudometrics on Dα are known. Theorem 1.3.1 ([J-P 1993] (§ 4.4), [Pfl-Zwo 1998], [Zwo 1999a], [Zwo 2000a]). Let a = (a1 , . . . , an ) ∈ Dα . Assume that a1 · · · as 6= 0, as+1 = · · · = an = 0 for some s ∈ {k + 1, . . . , n}. Put r := orda (z α − aα ). For z ∈ Dα and X ∈ Cn consider the following four cases. (1) k < n, Dα is of rational type (we may assume that α ∈ Zn and α1 , . . . , αn are relatively prime). Then: c∗Dα (a, z) = mE (aα , z α ), ∗ gD (a, z) = (mE (aα , z α ))1/r , α

kDα (a, z) = min{pE (ζ1 , ζ2 ) : ζ1 , ζ2 ∈ E, aα = ζ1tk , z α = ζ2tk }, ( min{pE (ζ1 , ζ2 ) : ζ1 , ζ2 ∈ E, aα = ζ1tk , z α = ζ2tk }, e kDα (a, z) = pE (0, |z α |1/r ),

s = n, z ∈ / V0 , s 0 and q F ≡ 0 if

48


Proof. It suffices to consider only the case n = 1. Take a disc B(a, r) b Ω, ε > 0, and a continuous function w ∈ C(∂B(a, r)) such that w ≥ v on ∂B(a, r). We want to R 2π 1 show that v(a) ≤ 2π w(a + reiθ )dθ + ε. For any point b ∈ ∂B(a, r) there exists an 0 i = i(b) ∈ A such that vi (b) < w(b)+ε. Hence there exists an open arc I = I(b) ⊂ ∂B(a, r) with b ∈ I such that vi (λ) < w(λ) + ε, λ ∈ I. By a compactness argument, we find SN b1 , . . . , bN ∈ ∂B(a, r) such that ∂B(a, r) = j=1 I(bj ). By assumption, there exists an i0 ∈ A such that vi0 ≤ min{vi(b1 ) , . . . , vi(bN ) }. Then Z 2π Z 2π 1 1 iθ v(a) ≤ vi0 (a) ≤ vi0 (a + re )dθ ≤ w(a + reiθ )dθ + ε. 2π 0 2π 0 Proposition 1.6.4. For any function p : G −→ Z+ we get mG (p, ·) = inf{mG (q, ·) : q : G −→ Z+ , q ≤ p, #|q| < +∞}. Proof. The case where |p| is finite is trivial. The case where the set |p| is countable follows from Remark 1.6.1(h). In the general case let Ak := {a ∈ G : p(a) = k} and let S∞ Bk be a countable (or finite) dense subset of Ak , k ∈ Z+ . Put B := k=0 Bk , p0 := p·χB . Then p0 ≤ p, the set |p0 | is at most countable, and mG (p, ·) ≡ mG (p0 , ·). Consequently, the problem reduces to the countable case. Proposition 1.6.5. (a) mG (p, ·) ≥

Y

[mG (a, ·)]p(a) ,

gG (p, ·) ≥

a∈G

Y

[gG (a, ·)]p(a) .

a∈G

(b) If G ⊂ C, then gG (p, z) =

Y

[gG (a, z)]p(a) ,

z ∈ G.

a∈G

In particular, dmin E (p, z) = mE (p, z) = gE (p, z) =

Y

[mE (a, z)]p(a) ,

z ∈ E.

a∈E

Notice that the formula in (b) is not true for G ⊂ Cn , n ≥ 2; cf. Example 1.7.17. Proof. (a) Use Remark 1.6.1(c) and Propositions 1.6.2, 1.6.4. (b) By Proposition 1.6.2 we may assume that the set |p| is finite. Let Y u := [gG (a, ·)]p(a) . a∈|p|

By (a) we only need to show that gG (p, ·) ≤ u. Now, by Remark 1.6.1(h), we may assume that G b C is regular with respect to the Dirichlet problem. Then the function log u is subharmonic on G and harmonic on G \ |p|. The function v := log gG (p, ·) − log u is locally bounded from above in G and lim supz→ζ v(z) ≤ 0, ζ ∈ ∂G. Consequently, v extends to a subharmonic function on G and, by the maximum principle, v ≤ 0 on G, i.e. gG (p, ·) ≤ u on G.

1.6. Properties of the generalized M¨ obius and Green functions

49

Proposition 1.6.6 ([Edi-Zwo 1998b], [Lár-Sig 1998b]). Let G, D ⊂ Cn be domains and let F : G −→ D be a proper holomorphic mapping. (a) Let q : D −→ R+ . Assume that det F 0 (a) 6= 0, a ∈ F −1 (|q|). Then gD (q, F (z)) = gG (q F , z) = gG (q ◦ F, z), 0

In particular, if B ⊂ D is such that det F (a) 6= 0, a ∈ F gD (B, F (z)) = gG (F

−1

(B), z),

−1

z ∈ G. (B), then

z ∈ G.

(b) Assume that D is convex. Then for any point b ∈ D such that det F 0 (a) 6= 0, a ∈ F −1 (b), we have mD (b, F (z)) = mG (F −1 (b), z),

z ∈ G.

Notice that (a) may be false if det F 0 (a) = 0 for some a ∈ F −1 (|q|) — cf. Example 1.7.4. Moreover, (b) need not be true if D is not convex — cf. Example 1.7.7. For the behavior of the pluricomplex Green function under coverings see [Azu 1995], [Azu 1996]. Proof. (a) We only need to show gD (q, F (z)) ≥ gG (q ◦ F, z), z ∈ G; cf. Remark 1.6.1(e). Put S := {z ∈ G : det F 0 (z) = 0}, Σ := F (S). It is well-known that F |G\F −1 (Σ) : G \ F −1 (Σ) −→ D \ Σ is a holomorphic covering. Let N denote its multiplicity. Let u : G −→ [0, 1) be a logarithmically plurisubharmonic function such that u(z) ≤ C(a)kz − akq(F (a)) ,

a, z ∈ G.

Define v(w) := max{u(z) : z ∈ F −1 (w)},

w ∈ D.

Since F is proper, log v ∈ PSH(D) (cf. [Kli 1991], Proposition 2.9.26). Take a b ∈ D with q(b) > 0 (recall that b ∈ / Σ) and let F −1 (b) = {a1 , . . . , aN } (aj 6= ak for j 6= k). There exist open neighborhoods U1 , . . . , UN , V of a1 , . . . , aN , b, respectively, such that F |Uj : Uj −→ V is biholomorphic, j = 1, . . . , N . Let gj := (F |Uj )−1 , j = 1, . . . , N . Shrinking the neighborhoods, if necessary, we may assume that there is a constant M > 0 such that kgj (w) − aj k ≤ M kw − bk, w ∈ V . Then, for w ∈ V , we get v(w) = max{u ◦ gj (w) : j = 1, . . . , N } ≤ max{C(aj )kgj (w) − aj kq(b) : j = 1, . . . , N } ≤ max{C(aj ) : j = 1, . . . , N }M q(b) kw − bkq(b) . Consequently, gD (q, ·) ≥ v and, therefore, gD (q, F (z)) ≥ v(F (z)) ≥ u(z), z ∈ G, which gives the required inequality. (b) By Remark 1.6.1(e) we only need to check the inequality “≥”. Since D is convex, the Lempert theorem implies that mD (b, ·) = gD (b, ·) (cf. [J-P 1993], Theorem 8.2.1). Hence, by (a) we get mD (b, F (z)) = gD (b, F (z)) = gG (F −1 (b), z) ≥ mG (F −1 (b), z),

z ∈ G.

50


1.7. Examples Example 1.7.1 ([Car-Ceg-Wik 1999]). Let T := {(z1 , z2 ) ∈ E∗ × E : |z2 | < |z1 |} be the Hartogs triangle. Let p : T −→ R+ . Consider the biholomorphism F

E∗ × E 3 (z1 , z2 ) −→ (z1 , z1 z2 ) ∈ T. The set E 2 \ (E∗ × E) is pluripolar. Hence, by Remark 1.6.1(e,i), gT (p, F (z)) = gE∗ ×E (p ◦ F, z) = gE 2 (p0 , z), 0

z ∈ E∗ × E,

0

where p := p ◦ F on E∗ × E and p := 0 on {0} × E. In particular, gT (a, z) = max{mE (a1 , z1 ), mE (a2 /a1 , z2 /z1 )},

a = (a1 , a2 ), z = (z1 , z2 ) ∈ T.

Example 1.7.2. For any non-empty sets A1 , . . . , An ⊂ E we have mE n (A1 × · · · × An , z) = gE n (A1 × · · · × An , z) = max{mE (A1 , z1 ), . . . , mE (An , zn )} n Y o = max mE (aj , zj ) : j = 1, . . . , n , z = (z1 , . . . , zn ) ∈ E n . aj ∈Aj

In particular, for any non-empty set A ⊂ E we have mE n (A × {0}n−1 , z) = gE n (A × {0}n−1 , z) = max{mE (A, z1 ), |z2 |, . . . , |zn |}, z = (z1 , . . . , zn ) ∈ E n ; cf. Example 1.7.17. Indeed, by Propositions 1.6.2, 1.6.4 we may assume that A1 , . . . , An are finite. Let Y λ−a Fj (λ) := , λ ∈ E, j = 1, . . . , n, 1 − aλ a∈Aj

be the corresponding Blaschke products. The mapping F

E n 3 (z1 , . . . , zn ) −→ (F1 (z1 ), . . . , Fn (zn )) ∈ E n is proper. Moreover, det F 0 (z) = F10 (z1 ) · · · Fn0 (zn ) 6= 0 for z ∈ A1 × · · · × An . Consequently, by Proposition 1.6.6, mE n (A1 × · · · × An , z) = gE n (A1 × · · · × An , z) = gE n (0, F (z)) = max{|Fj (zj )| : j = 1, . . . , n} = max{mE (A1 , z1 ), . . . , mE (An , zn )},

z = (z1 , . . . , zn ) ∈ E n .

Example 1.7.3. Recall that for p = (p1 , . . . , pn ) ∈ Rn>0 (n ≥ 2), we put Ep := {(z1 , . . . , zn ) ∈ Cn :

n X j=1

|zj |2pj < 1}.

1.7. Examples

51

Fix (ν1 , . . . , νn ) ∈ Nn . The mapping F

Bn 3 (z1 , . . . , zn ) −→ (z1ν1 , . . . , znνn ) ∈ E(1/ν1 ,...,1/νn ) ν −1

is proper. Let (a1 , . . . , an ) ∈ Bn be such that aj j

6= 0, j = 1, . . . , n, and let √ ν A := F −1 (F (a)) = {(ε1 a1 , . . . , εn an ) : εj ∈ j 1, j = 1, . . . , n}.

Then, by Proposition 1.6.6, gBn (A, z) = gE(1/ν1 ,...,1/νn ) (F (a), F (z)),

z ∈ Bn ;

roughly speaking, the multi-pole pluricomplex Green function for the Euclidean ball is expressed by the standard one-pole pluricomplex Green function for an ellipsoid. Notice that for some special cases the function gE(1/ν1 ,...,1/νn ) (F (a), F (·)) may be effectively calculated. For example, let n = 2, ν1 = 1, ν2 = 2, a = (0, s) (s ∈ (0, 1)). Then A = {(0, −s), (0, s)} and gB2 ({(0, −s), (0, s)}, (z1 , z2 )) = gE(1,1/2) ((0, s2 ), (z1 , z22 ))  1/2 (1−s2 )(1−|z1 |2 −|z2 |2 )   1 − if s|z1 | ≥ |z2 − s| 2  |1−sz | 2   1/2 2 2 2 1 | −|z2 | ) , = 1 − (1−s )(1−|z if s|z1 | ≥ |z2 + s| |1+sz2 |2   √  1/2 2  |z1 |2 −z22 |2 + ∆  2(1−s2 Re z22 )|z1 |2 +|s2 −s if s|z1 | < min{|z2 − s|, |z2 + s|} 2|1−s2 z 2 |2 2

where 2 4 ∆ := −4|z1 |4 (s2 Im z22 )2 + 4|z1 |2 (1 − s2 Re z22 ) s2 − s2 |z1 |2 − z22 + s2 − s2 |z1 |2 − z22 ; cf. [Edi-Zwo 1998b] (see also [Com 2000] for a different approach). We would like to point out that even ? for the case |p| = {a1 , a2 }, p(a1 ) 6= p(a2 ), a formula for gBn (p, ·) is not known ? F

Example 1.7.4. Let B2 3 (z1 , z2 ) −→ (z1 , z22 ) ∈ E(1,1/2) , a := (0, 0). Then det F 0 (0) = 0 and gB2 (0, ·) 6≡ [gE(1,1/2) (0, F (·))]s for any s > 0. Indeed, p gB2 ((0, 0), (z1 , z2 )) = hB2 (z1 , z2 ) = |z1 |2 + |z2 |2 , p |z2 | + 4|z1 |2 + |z2 |2 gE(1,1/2) ((0, 0), (z1 , z2 )) = hE(1,1/2) (z1 , z2 ) = , 2 where hD is the Minkowski function. In particular, for small t > 0, we get √ gB2 ((0, 0), (t, t)) = t 2, gE(1,1/2) ((0, 0), (t, t2 )) ≈ t, which implies the required result. Example 1.7.5 ([JarW 2003]). Let p = (p1 , . . . , pn ) ∈ Rn>0 , E := Ep . Put A = AE,k := {z ∈ E : z1 · · · zk = 0},

k = 1, . . . , n.

52


Our aim is to find effective formulae for mE (AE,k , z) and gE (AE,k , z), where z = (z1 , . . . , zn ) ∈ E. It is clear that we may assume that p1 |z1 |2p1 ≤ · · · ≤ pk |zk |2pk . 29 Put qs :=

s X 1 , 2p j j=1

rs (z) := 1 −

n X

|zj |2pj ,

cs (z) := rs (z)/qs ,

s = 1, . . . , k

n o d = d(z) := max s ∈ {1, . . . , k} : 2ps |zs |2ps ≤ cs (z) , RE (A, z) :=

(rn = 1),

j=s+1 30

d Y

d 2p 2p1 Y 1 |z1 · · · zd | j j qd . |zj | = qdqd (2pj )2pj P n cd (z) 1− |z |2pj j=1 j=1 j=d+1

j

Then: (a) (b) (c) (d)

gE (A, z) = RE (A, z); mE (A, z) = gE (A, z) = RE (A, z) if pj ≥ 1/2, j = d + 1, . . . , n; mE (A, z) = gE (A, z) = RE (A, z) for k = 1, n = 2, p2 ≥ 1/2; mE (A, z) 6= gE (A, z) if there exists a j0 ∈ {k + 1, . . . , n} with pj0 < 1/2, |z` | = 6 0 small enough, ` = 1, . . . , k, j0 , z` = 0, ` = k + 1, . . . , j0 − 1, j0 + 1, . . . , n; (e) mE (A, z) = gE (A, z) = RE (A, z) for k = n = 2, p1 ≤ p2 , and either p2 ≥ 1/2 or 8p1 + 4p2 (1 − p2 ) > 1.

? It is an open question whether mE (A, z) = gE (A, z) = RE (A, z) if pj ≥ 1/2, j = k + 1, . . . , n (with arbitrary n and k) ? Proof of (a). Step 1. mE n (AE n,k , ζ) = gE n (AE n,k , ζ) = |ζ1 · · · ζk |, ζ ∈ E n , where AE n,k := {ζ ∈ E n : ζ1 · · · ζk = 0}. Indeed, it is clear that |ζ1 · · · ζk | ≤ mE n (AE n,k , ζ) ≤ gE n (AE n,k , ζ). It remains to prove that u(ζ) := gE n (AE n,k , ζ) ≤ |ζ1 · · · ζk |, ζ ∈ E n . We proceed by induction on k (with arbitrary n and logarithmically plurisubharmonic function u : E n −→ [0, 1) such that u(ζ) ≤ C(a)kζ − ak, a ∈ AE n,k , ζ ∈ E n ). For k = 1 the inequality follows from the Schwarz type lemma for logarithmically subharmonic functions u(·, ζ2 , . . . , ζn ). For k > 1 we first apply the case k = 1 and get u(ζ1 , . . . , ζn ) ≤ |ζ1 |, ζ ∈ E n . Next we apply the inductive assumption to the functions u(ζ1 , ·)/|ζ1 |, ζ1 ∈ E∗ . Step 2. Consider the mapping c (z) 2p1 c (z) 2p1 ιz d d 1 d E d 3 (ζ1 , . . . , ζd ) −→ ζ1 , . . . , ζd , zd+1 , . . . , zn ∈ E. 2p1 2pd Using the holomorphic contractivity and Step 1, we get mE (A, z) ≤ gE (A, z) ≤ RE (A, z). It remains to prove that gE (A, z) ≥ RE (A, z). 29

In particular, if p1 = · · · = pk , then the condition simply means that |z1 | ≤ · · · ≤ |zk |. 30 Observe that z d+1 · · · zk 6= 0.

1.7. Examples

53

Step 3. gE (A, z) ≥ RE (A, z). We may assume that z1 · · · zd 6= 0. 1 Qn First consider the case d = k = n. Put f (ζ) := qnqn j=1 ζj (2pj )2pj , ζ ∈ E. Pn |ζ |2pj qn j j=1 < 1, ζ ∈ E 31 . Observe that |f (z)| = RE (A, z) and |f (ζ)| ≤ qnqn qd Thus gE (A, z) ≥ mE (A, z) ≥ RE (A, z). Now assume that d < n. Put E0 := E(pd+1 ,...,pn ) . Observe that we only need to find a logarithmically plurisubharmonic function v : E0 −→ [0, 1), v 6≡ 0, such that • v(ζ 0 ) ≤ |ζj |, ζ 0 = (ζd+1 , . . . , ζn ) ∈ E0 , j = d + 1, . . . , k 32 , • the mapping E0 3 ζ0 −→ v(ζ 0 )rdqd (ζ 0 ) ∈ R+ attains its maximum for ζ 0 = (zd+1 , . . . , zn ) 33 . Indeed, suppose that such a v is already constructed and let M be the maximal value of the function E0 3 ζ 0 −→ v(ζ 0 )rdqd (ζ 0 ). Put d

u(ζ) :=

1 qdqd Y |ζj |(2pj ) 2pj v(ζ 0 ), M j=1

ζ = (ζ1 , . . . , ζn ) = (ζ1 , . . . , ζd , ζ 0 ).

Then log u ∈ PSH(E) and u(ζ) ≤ C(a)|ζj | ≤ C(a)kζ − ak for any ζ ∈ E and a ∈ A with aj = 0, where j ∈ {1, . . . , k}. Moreover, for ζ ∈ E we have: Pd Pd 2p qdqd j=1 |ζj |2pj qd 1 j=1 |ζj | j qd 0 u(ζ) ≤ v(ζ ) = v(ζ 0 )rdqd (ζ 0 ) < 1. M qd M rd (ζ 0 ) Consequently, u : E −→ [0, 1) and, therefore, gE (A, z) ≥ u(z) =

1 RE (A, z)v(z 0 )rdqd (z 0 ) = RE (A, z). M

Step 4. Construction of the function v. We may assume that zd+1 , . . . , zn ≥ 0. For α = (αd+1 , . . . , αn ) ∈ Rn−d define + vα (ζ 0 ) :=

k Y

|ζj |1+αj

n Y

j=d+1

|ζj |αj .

j=k+1

Obviously v : E0 −→ [0, 1), log v ∈ PSH(E0 ), and v(ζ 0 ) ≤ |ζj |, ζ 0 ∈ E0 , j = d + 1, . . . , k. ϕα It is enough to find an α such that the function E0 ∩ Rn−d 3 t0 −→ vα (t0 )rdqd (t0 ) attains + 31

We have used the following elementary inequality d Y j=1

32

w aj j

≤

Pd

j=1

Pd

Pd

w j aj

j=1 wj

j=1

wj

,

Notice that this Pn condition2pis empty if d = k. 33 r (ζ 0 ) = 1 − j. d j=d+1 |ζj |

a1 , . . . , ad ≥ 0, w1 , . . . , wd > 0.

54


its maximum at t0 = z 0 . In particular,

∂ϕα 0 ∂tj (z )

= 0, j = d + 1, . . . , n. Hence

2p

zj j 0 = 1+αj − 2pj qd , rd (z 0 )

j = d + 1, . . . , k,

2p

zj j 0 =αj − 2pj qd , rd (z 0 )

j = k + 1, . . . , n,

which gives formulas for αd+1 , . . . , αn . To prove that there are no other points like this, rewrite the above equations in the form 2p

rd (z 0 ) =

2pj qd zj j , 1 + αj

rd (z 0 ) =

2pj qd zj αj

j = d + 1, . . . , k,

2pj

,

j = k + 1, . . . , n.

The left side is decreasing in any of the variables zd+1 , . . . , zn , while the right sides are increasing. Thus, at most one common zero is allowed. It remains to check whether αj ≥ 0, j = d + 1, . . . , n. Obviously, αj ≥ 0, j = k + 1, . . . , n. In the remaining cases, using the definition of the number d, we have: 2p

αj =

2pj qd zj j − rd (z 0 ) ≥ 0, rd (z 0 )

j = d + 1, . . . , k.

Proof of (b). By the proof of (a), we only have to check whether mE (A, z) ≥ RE (a, z) in the case where d < n. First observe that it is sufficient to find a function h ∈ O(E0 ), h 6≡ 0, such that: • h(ζ 0 ) = 0 if ζd+1 · · · ζk = 0, • the function E0 3 ζ 0 −→ |h(ζ 0 )|rdqd (ζ 0 ) ∈ R+ attains its maximum for ζ 0 = (zd+1 , . . . , zn ). Indeed, suppose that such an h is already constructed and let M be the maximal value of the function E0 3 ζ 0 −→ |h(ζ 0 )|rdqd (ζ 0 ). Put d

f (ζ) :=

1 qdqd Y ζj (2pj ) 2pj h(ζ 0 ), M j=1

ζ ∈ E.

Obviously f (ζ) = 0 for ζ ∈ A. Similarly as in (a) we prove that |f | < 1 and |f (z)| = RE (A, z). Thus mE (A, z) ≥ |f (z)| = RE (A, z). To construct h assume that zd+1 , . . . , zn ≥ 0 and define hα (ζ 0 ) :=

k Y j=d+1

ζj eαj ζj

n Y

eαj ζj ,

j=k+1

where α = (αd+1 , . . . , αn ) ∈ Rn−d + . It is enough to find an α such that the function 0 0 qd E0 ∩ Rn−d 3 t −→ h (t )r (t) attains its maximum at t0 = (zd+1 , . . . , zn ). Considering α + d

1.7. Examples

55

the partial derivatives results in the following equations: 2pj −1

0=

zj 1 +αj − 2pj qd , zj rd (z 0 )

j = d + 1, . . . , k,

2p −1

0 =αj − 2pj qd

zj j , rd (z 0 )

j = k + 1, . . . , n.

We continue as in the proof of (a).

Proof of (c). Assertion (c) follows directly from (b).

Proof of (d). Step 1. Suppose that mE (A, z) = RE (A, z). Let f ∈ O(E, E) be such that f |A ≡ 0 and |f (z)| = RE (A, z) (cf. [Jar-Jar-Pfl 2003], Property 2.5). Put h(ζ 0 ) :=

∂df (0, ζ 0 ), ∂z1 . . . ∂zd

ζ 0 ∈ E0 .

We have h(ζ 0 ) = 0 if ζd+1 · · · ζk = 0. For ζ 0 ∈ E0 consider the mapping c (ζ 0 ) 2p1 c (ζ 0 ) 2p1 ιζ 0 d d 1 d E d 3 (ξ1 , . . . , ξd ) −→ ξ1 , . . . , ξd , ζ 0 ∈ E. 2p1 2pd Applying the Schwarz lemma to the mapping f ◦ ιζ 0 , ζ 0 ∈ E0 , we get |h(ζ 0 )|rdqd (ζ 0 ) ≤qdqd

d Y

1 (2pj ) 2pj ,

ζ 0 ∈ E0 ,

j=1

|h(z 0 )|rdqd (z 0 ) =qdqd

d Y

1 (2pj ) 2pj .

j=1

Thus we have constructed a mapping h as in the proof of (b) (consequently, the equality mE (A, z) = RE (A, z) is equivalent to the existence of the mapping h). Step 2. For any p ∈ (0, 1) and q > 0 there exists c = c(p, q) ∈ (0, 1) such that for any function f ∈ O(E) if the function E 3 λ −→ |f (λ)|(1 − |λ|p )q attains its maximum at λ0 6= 0, then |λ0 | ≥ c. Let 1 , t ∈ [0, 1). ϕ(t) := (1 − tp )q Observe that there exists a b ∈ (0, 1) such that ϕ is strictly concave on [0, b). Moreover, ϕ(t) − ϕ(0) = +∞. t−→0+ t lim

Consequently, there exists a c ∈ (0, b) such that b ϕ(0) + (ϕ(c) − ϕ(0)) > ϕ(b) + 2. c

56


Suppose that f ∈ O(E) is such that the function E 3 λ −→ |f (λ)|/ϕ(|λ|) attains its maximum at λ0 6= 0 with |λ0 | < c. We may assume that |f (λ0 )| = ϕ(|λ0 |). Consider the function t ψ |f (λ0 ) − f (0)|. [0, b] 3 t −→ |f (0)| + |λ0 | From ψ(0) = |f (0)| ≤ ϕ(0) = 1, ψ(|λ0 |) ≥ ϕ(|λ0 |), and the convexity condition we get: b b |f (λ0 ) − f (0)| ≥ ϕ(0) + |ϕ(|λ0 |) − ϕ(0)| |λ0 | |λ0 | b ≥ ϕ(0) + |ϕ(c) − ϕ(0)| > ϕ(b) + 2. c The Schwarz lemma and the maximum principle imply that there exists a λ∗ ∈ E with |λ∗ | = b and |f (λ∗ ) − f (0)| |f (λ0 ) − f (0)| ≥ . |λ∗ | |λ0 | This means that ψ(b) = |f (0)| +

|f (λ∗ )| ≥|f (λ∗ ) − f (0)| − |f (0)| = |f (0)| + |f (λ∗ ) − f (0)| − 2|f (0)| ≥ψ(b) − 2|f (0)| > ϕ(b) + 2 − 2|f (0)| ≥ ϕ(b) = ϕ(|λ∗ |); contradiction. Step 3. We may assume that pk+1 < 1/2. Assume that 0 < |zj | < ε, j = 1, . . . , k + 1, zj = 0, j = k + 2, . . . , n, with 0 < ε < c(2pk+1 , qk ). Observe that d(z) = k provided ε is small enough. Let h be as in Step 1. Then the mapping E 3 λ −→ |h(λ, 0, . . . , 0)|(1 − |λ|2pk+1 )qk attains its maximum at λ = zk+1 , which contradicts Step 2.

Proof of (e). See [JarW 2003].

Example 1.7.6. Let P = P (R) := {z ∈ C : 1/R < |z| < R} (R > 1). Put q := 1/R2 and let Q∞ (1 − az q 2ν )(1 − az q 2ν ) , Π(a, z) = ΠR (a, z) := Q∞ ν=1 2ν−1 )(1 − 1 q 2ν−1 ) ν=1 (1 − azq az z f (a, z) = fR (a, z) := 1 − Π(a, z), 1/R < a < R, z ∈ P. a Using the same methods as in the proof of Proposition 5.5 in [J-P 1993], one can prove that for any function p : P −→ Z+ such that |p| = {a1 , . . . , aN } is finite, if aj = |aj |eiϕj , |aj | = R1−2sj , sj ∈ (0, 1), j = 1, . . . , N , then we get mP (p, z) =

N f (b, −|z|) Y |f (|aj |, e−iϕj z)|kj , |Rz|` j=1

where • ` = `(p) := ds1 + · · · + sN e, • b = b(p) := R1−2(`−(s1 +···+sN )) ,

z ∈ P,

1.7. Examples

57

• f (R, ·) :≡ 1. Example 1.7.7. If D is not convex, then Proposition 1.6.6(b) need not be true. Indeed, let P (R), ΠR , and fR be as Example 1.7.6. Consider F : P (R) −→ P (R2 ), F (z) := z 2 , and suppose that mP (R2 ) (1, z 2 ) = mP (R) ({−1, +1}, z), z ∈ P (R), R > 1. Then, using Example 1.7.6, we get fR2 (1, −|z|2 ) 1 |fR2 (1, z 2 )| = |fR (1, z)fR (1, −z)|, 2 2 R |z| R|z|

z ∈ P (R).

Consequently, 1 (1 + |z|2 )ΠR2 (1, −|z|2 )|(1 − z 2 )ΠR2 (1, z 2 )| = |(1 − z)ΠR (1, z)(1 + z)ΠR (1, −z)|, R|z| and hence 1 (1 + |z|2 )ΠR2 (1, −|z|2 )|ΠR2 (1, z 2 )| = |ΠR (1, z)ΠR (1, −z)|, R|z|

z ∈ P (R);

contradiction (at least for big R) (take z = 1 and then let R −→ +∞). Remark 1.7.8. Let G ⊂ Cn and assume that p : G −→ R+ , |p| = {a1 , . . . , aN }. Directly from the definition of the function dmin we get the following estimate: G s nY dmin [mE (µj , z)]max p(Bj ) : s ∈ N, µ1 , . . . , µs ∈ E, µj 6= µk (j 6= k), G (p, z) = sup j=1

o B1 ∪ · · · ∪ Bs = |p|, Bj ∩ Bk = ∅ (j 6= k), ∃f ∈O(G,E) : f |Bj ≡ µj , j = 1, . . . , s ≤ sup

s nY

o [mG (Bj , z)]max p(Bj ) : s ∈ N, B1 ∪ · · · ∪ Bs = |p|, Bj ∩ Bk = ∅ (j 6= k)

j=1

=: d0G (p, z),

z ∈ G.

0 Recall that in the case where p = χA we have dmin G (χA , ·) ≡ mG (A, ·) ≡ dG (χA , ·) (cf. Proposition 1.5.4). In particular, if N = 2, |p| = {a, b}, α = p(a) ≥ p(b) = β, then we get n o α β α dmin (αχ + βχ , z) ≤ max [m (a, z)] [m (b, z)] , [m ({a, b}, z)] G G G {a} {b} G

= d0G (αχ{a} + βχ{b} , z), dmin G (αχ{a}

z ∈ G.

d0G (αχ{a}

+ βχ{b} , ·) 6≡ + βχ{b} , ·). Notice that in general In fact, let G = P be an annulus as in Example 1.7.6. Take 1/R < a, b < R, a 6= b, ab 6= 1, p := 2χ{a} + χ{b} . We are going to show that there exists a z ∈ P such that [mP (a, z)]2 mP (b, z) > [mP ({a, b}, z)]2 , 2 dmin P (p, z) < [mP (a, z)] mP (b, z) n o = max [mP (a, z)]2 mP (b, z), [mP ({a, b}, z)]2 = d0P (p, z).

58


First observe that there are points z (near b) such that [mP (a, z)]2 mP (b, z) > [mP ({a, b}, z)]2 . For, using the effective formula from Example 1.7.6, we have: f (1/a, −|z|) 2 f (1/b, −|z|) [mP (a, z)]2 mP (b, z) = |f (a, z)| |f (b, z)|, R|z| R|z| f (c, −|z|) 2 [mP ({a, b}, z)]2 = |f (a, z)f (b, z)| , ` (R|z|) with c = c(a, b) and ` = `(a, b) as in Example 1.7.6. Consequently, we only need to find a z ∈ P \ {a, b} such that f (1/a, −|z|) 2 f (1/b, −|z|) f (c, −|z|) 2 > |f (b, z)|. R|z| R|z| (R|z|)` Observe that at z = b the right hand side of the above formula is zero while the left hand side is strictly positive. Thus, by continuity, we can easily find the required z, say z0 . Let ϕ ∈ O(P, E) be an extremal function for dmin P (p, z0 ) (Proposition 1.8.4). We may assume that ϕ(a) = 0. There are two cases: 2 2 0 (a) ϕ(b) = 0: Then dmin P (p, z0 ) = |ϕ(z0 )| ≤ [mP ({a, b}, z0 )] < dP (p, z0 ). min 2 (b) ϕ(b) 6= 0: Then dP (p, z0 ) = |ϕ(z0 )| mE (ϕ(b), ϕ(z0 )) ≤ [mP (a, z0 )]2 mP (b, z0 ) = 0 0 dP (p, z0 ). The equality dmin P (p, z0 ) = dP (p, z0 ) would imply that ϕ is simultaneously extremal for mP (a, z0 ) and mP (b, z0 ). Using Robinson’s lemma ([J-P 1993], Lemma 5.6), we know that such extremal functions are uniquely determined up to rotations. Hence f (1/a, −e−iϕ0 z) f (1/b, −e−iϕ0 z) ϕ(z) = eiθa f (a, z) = eiθb f (b, z); Rz Rz contradiction (both sides have different zeros). Example 1.7.9 ([Jar-Pfl 1999a]). Let F be a primitive polynomial of n–complex variables i.e. F ∈ P(Cn ) is a polynomial which cannot be represented in the form F = f ◦ Q, where f is a polynomial of one complex variable of degree ≥ 2 and Q ∈ P(Cn ). Notice that a monomial z α (α = (α1 , . . . , αn ) ∈ Nn ) is primitive iff the numbers α1 , . . . , αn are relatively prime. One can prove (cf. [Cyg 1992]) that there exists a finite set S ⊂ C such that for any b ∈ C \ S: • the fiber F −1 (b) is connected, • F 0 (a) 6= 0 for a ∈ F −1 (b). Thus, if b ∈ / S, then the fiber F −1 (b) is a connected (n − 1)–dimensional algebraic manifold. In particular, for any b ∈ / S, the fiber F −1 (b) has the plurisubharmonic Liouville property, i.e. any plurisubharmonic function u : F −1 (b) −→ [−∞, 0) is constant (cf. [Jar-Pfl 1999a], Proposition 6). Put r(a) := orda (F − F (a)). Let D ⊂ C be a domain. Put G := F −1 (D). Then G is a domain. Indeed, since the set F −1 (S) is thin, it suffices to prove that the set G0 := F −1 (D \ S) is connected. Suppose G0 = U1 ∪ U2 , where U1 , U2 are non-empty disjoint open sets. Put Bj := {w ∈ D \ S : F −1 (w) ⊂ Uj }, j = 1, 2. Since the fibers over

1.7. Examples

59

points from D \ S are connected, we conclude that B1 , B2 are disjoint and Bj = F (Uj ), j = 1, 2. In particular, Bj is open, non-empty, and D \ S = B1 ∪ B2 ; contradiction. (a) Let p : G −→ R+ be such that |p| is finite. Then gG (p, z) = gD (pF , F (z)),

z ∈ G,

(1.7.7)

where pF (b) := max

n p(a) r(a)

o : a ∈ |p| ∩ F −1 (b) ,

b ∈ D.

In particular, gG (a, z) = [gD (F (a), F (z))]1/r(a) ,

a, z ∈ G.

(b) Let p : G −→ Z+ be such that |p| is finite. Then mG (p, z) = mD (p0 , F (z)),

z ∈ G,

(1.7.8)

where p0 (b) := max

nl p(a) m r(a)

o : a ∈ |p| ∩ F −1 (b) ,

b ∈ D.

In particular, mG (kχ{a} , z) = mD (k 0 χ{F (a)} , F (z)), l m k where k 0 := r(a) .

a, z ∈ G, k ∈ N,

Indeed, in both cases the inequalities “≥” follow from Remark 1.6.1(e). To prove the opposite inequality in (a), take a logarithmically plurisubharmonic function u : G −→ [0, 1) such that u(z) ≤ C(a)kz − akp(a) ,

a, z ∈ G.

For any b ∈ D \ S, the function u|F −1 (b) is constant. Consequently, there exists a logarithmically subharmonic function u e : D \ S −→ [0, 1) such that u = u e ◦ F on G \ F −1 (S). The function u e extends to a logarithmically subharmonic function on D (the extended function will be denoted by the same letter u e). By the identity principle for plurisubharmonic functions we get u = u e ◦ F in G. We want to show that F

u e(w) ≤ const(b)|w − b|p F

(b)

,

b, w ∈ D;

(1.7.9)

F

then u e ≤ gD (p , ·) and hence u = u e ◦F ≤ gD (p , F ), which gives the required inequality. Fix a b ∈ D with pF (b) > 0, and let a ∈ F −1 (b) be such that p(a) > 0. Observe that there exist ε > 0, δ > 0, and M > 0 such that B(b, ε) ⊂ D and ∀w∈B(b,ε) ∃z(w)∈B(a,δ) : F (z(w)) = w, kz(w) − akr(a) ≤ M |w − b|. Indeed, let X ∈ Cn , kXk = 1, be such that ord0 (ϕ−b) = r(a), where ϕ(λ) := F (a+λX). Then |ϕ(λ) − b| ≥ (1/M )|λ|r(a) for |λ| < δ (where M > 0, δ > 0) and ϕ is open. Consequently, ϕ(B(δ)) ⊃ B(b, ε) for some ε > 0. Thus for any w ∈ B(b, ε) there exists a λ(w) ∈ B(δ) such that z(w) := a + λ(w)X satisfies all the required conditions.

60


Consequently, u e(w) = u(z(w)) ≤ C(a)kz(w) − akp(a) ≤ C(a)M p(a)/r(a) |w − b|p(a)/r(a) , Since the set of all a ∈ F

−1

w ∈ B(b, ε).

(b) with p(a) > 0 is finite, we get (1.7.9).

In the situation of (b) let f ∈ O(G, E) be such that orda f ≥ p(a), a ∈ G. For any b ∈ D \ S the function f |F −1 (b) must be constant. Hence there exists a function fe ∈ O(D \ S, E) such that f = fe◦ F . Using the Riemann removable singularity theorem, we extend holomorphically fe to the whole D. Take b ∈ D and a ∈ F −1 (a) such that p(a) > 0. Then, using the same argument as in (a), we get |fe(w)| = |f (z(w))| ≤ const kz(w) − akorda f ≤ const kz(w) − akp(a) ≤ |w − b|p(a)/r(a) for w in a neighborhood of b. Consequently, ordb fe ≥ p0 (b), b ∈ G, and, therefore, |f (z)| = |fe(F (z))| ≤ mD (p0 , F (z)), z ∈ G. The case where F (z) = z α and D = E was studied in [J-P 1993], § 4.4 (one pole), and [Edi-Zwo 1999] (many poles). ? Are formulas (1.7.7), (1.7.8) true for arbitrary p ? In the special case where F (z) = z α we get the following example. Example 1.7.10. Let α = (α1 , . . . , αn ) ∈ Nn , G := {z ∈ Cn : |z α | < 1}. Assume that α1 , . . . , αn are relatively prime. Then for any function p : G −→ R+ (resp. p : G −→ Z+ ) such that |p| is finite, the following formulae are true: α

gG (p, z) = mE (pz , z α ),

mG (p, z) = mE (p0 , z α ),

z ∈ G,

where α

pz (b) : = sup

n p(a)

o : a ∈ G, aα = b ,

r(a) o nl p(a) m p0 (b) : = sup : a ∈ G, aα = b , r(a)

34

b ∈ E.

Example 1.7.9 may be extended in the following way. Example 1.7.11 ([Jar-Pfl 1999a], Theorem 1). Let αj = (αj,1 , . . . , αj,n ) ∈ (Zn+ )∗ ,

j = 1, . . . , m ≤ n − 1, m ≥ 2, be such that rank A = m, where A := [αj,k ]. Assume that AZn = Zm 35 . Let F = (F1 , . . . , Fm ) : Cn −→ Cm 34

Observe that r(a) = 1 if a1 . . . an 6= 0, and r(a) = the sum of those αi for which ai = 0 if a1 · · · an= 0. 35 One can prove that AZn = Zm iff the greatest common divisor of all determinants of m × m submatrices of A equals 1.

1.7. Examples

61

be given by the formula Fj (z) := z αj , j = 1, . . . , m. Define G := F −1 (E m ) = {z ∈ Cn : |z αj | < 1, j = 1, . . . , m}. Let p : G −→ R+ (resp. p : G −→ Z+ ) be such that |p| is finite and for any a ∈ |p| we have rank F 0 (a) = m (in particular, r(a) = 1). Then 36 (1.7.10) gG (p, z) = gE m (pF , F (z)), mG (p, z) = mE m (pF , F (z)) z ∈ G, where

n o pF (b) := max p(a) : a ∈ |p| ∩ F −1 (b) ,

b ∈ Em.

In particular, if p = χ{a} , then gG (a, z) = gE m (F (a), F (z)),

mG (a, z) = mE m (F (a), F (z)),

z ∈ G.

Indeed, put V0 := {w = (w1 , . . . , wm ) ∈ Cm : w1 · · · wm = 0} and observe that for any w ∈ E m \ V0 the fiber Vw := F −1 (w) is connected 37 . For (the proof is due to W. Zwonek), let w = (u1 e2πiθ1 , . . . , um e2πiθm ). Take arbitrary two points a, b ∈ F −1 (w), a = (r1 e2πiϕ1 , . . . , rn e2πiϕn ), b = (s1 e2πiψ1 , . . . , sn e2πiψn ). We have F (r) = F (s) = u, Aϕ = θ mod Zm , Aψ = θ mod Zm . We have to find a curve γ : [0, 1] −→ F −1 (w) such that γ(0) = a, γ(1) = b. Write γ(t) = (R1 (t)e2πi(ϕ1 +σ1 (t)) , . . . , Rn (t)e2πi(ϕn +σn (t)) ), where R : [0, 1] −→ Rn+ is continuous, σ : [0, 1] −→ Rn is such that the mapping t −→ (e2πiσ1 (t) , . . . , e2πiσn (t) ) is continuous, F (R(t)) = u, Aσ(t) = 0 mod Zm , t ∈ [0, 1], R(0) = r, R(1) = s, σ(0) = 0 mod Zn , σ(1) = ψ − ϕ mod Zn . Note that the set {x ∈ Rn+ : F (x) = u} is connected. Hence we can easily find an R with the required properties. To find a σ it would be sufficient to know that the set T := {x ∈ Rn : Ax ∈ Zm }/mod Zn is connected. Since AZn = Zm , we get T = {x ∈ Rn : Ax = 0}/mod Zn , which directly implies that T is connected (because A−1 (0) is connected). The inequalities “≥” in (1.7.10) follow from Remark 1.6.1(e). For the proof of “≤” in the case of a generalized Green function, let u : G −→ [0, 1) be such that log u ∈ PSH(G) and u(z) ≤ C(a)kz − akp(a) for any a, z ∈ G. For any w ∈ E m \ V0 , since Vw is a connected algebraic set, the function u|Vw is constant. Hence there exists a logarithmically plurisubharmonic function v : E m \ V0 −→ [0, 1) such that u = v ◦ F on G \ F −1 (V0 ). By the Riemann type extension theorem for plurisubharmonic functions, v extends to a logarithmically plurisubharmonic function on E m . By the identity principle for plurisubharmonic functions we get u = v ◦ F in G. Fix a b ∈ E m with pF (b) > 0 and let a ∈ F −1 (b) be such that p(a) > 0. By our assumption (rank F 0 (a) = m) there exists an m–dimensional vector subspace L ⊂ Cn g such that the mapping L 3 z −→ F (a + z) is biholomorphic in a neighborhood of 0 ∈ L. Then kg(z) − bk ≥ (1/M )kzk, z ∈ L ∩ B(δ) (for some M, δ > 0) and g(L ∩ B(δ)) ⊃ B(b, ε) 0 Notice that the formula (1.7.10) may be not true if rank F (a) ≤ m − 1 — cf. Example 1.7.13. 37 In fact, one can prove that V is connected iff AZn = Zm . w

36

62


(cf. Example 1.7.9). Hence, for any w ∈ B(b, ε) there exists a z(w) ∈ L ∩ B(δ) such that F (a + z(w)) = w. Finally, v(w) = v(F (a + z(w))) = u(a + z(w)) ≤ C(a)kz(w)kp(a) ≤ C(a)M kg(z(w)) − bkp(a) = C(a)M kw − bkp(a) , F e and, consequently, v(w) ≤ C(b)kw − bkp (b) , b, w ∈ E m . Hence

u(z) = v(F (z)) ≤ gE m (pF , F (z)),

z ∈ G,

F

which implies that gG (p, z) ≤ gE m (p , F (z)), z ∈ G. In the case of the generalized Möbius function we use an analogous argument (as in Example 1.7.9). Remark 1.7.12. Example 1.7.11 may be extended to more general mappings F and domains G. The condition rank F 0 (a) = m, a ∈ |p|, may be also weakened; cf. [Jar-Pfl 1999a]. The cases which are not covered by Example 1.7.11 are in general much more difficult (even in the case where p = χ{a} ). Example 1.7.13 ([Jar-Pfl 1999a], Proposition 3). Let αj = (αj,1 , . . . , αj,n ) ∈ (Zn+ )∗ , j = 1, . . . , m, m ≥ 2, A := [αj,k ], G := {z ∈ Cn : |z αj | < 1, j = 1, . . . , m}. 38 Fix an a ∈ G with aαj = 0, j = 1, . . . , m. Assume that a = (a1 , . . . , as , 0, . . . , 0) with a1 · · · as 6= 0 and 1 ≤ s ≤ n − 1. Put   β1  e := [αj,k ] j=1,...,m =  ..  A . . k=s+1,...,n

Notice that rj := orda z

αj

βm

= |βj | > 0, j = 1, . . . , m.

Then the following conditions are equivalent: e (i) rank A = rank A; (ii) gG (a, z) = max{|z αj |1/rj : j = 1, . . . , m}, z ∈ G; 39 (iii) gG (a, z) = sup{|z α |1/r : α ∈ (Z+ )n , |z α | < 1 in G, r = orda z α > 0}, z ∈ G; (iv) gG (a, (z 0 , λz 00 )) = |λ|gG (a, z), z = (z 0 , z 00 ) ∈ G ⊂ Cs × Cn−s , λ ∈ E; (r+1) (v) lim supθ→0+ θ1 mG (a, (z 0 , θz 00 )) < +∞, (z 0 , z 00 ) ∈ G ⊂ Cs × Cn−s . Indeed, to prove (i) =⇒ (ii), let L(z) := gG (a, z),

R(z) := max{|z αj |1/rj : j = 1, . . . , m},

z ∈ G.

The inequality L ≥ R follows from the definition of gG . To prove that L ≤ R it suffices to show that L(z) ≤ R(z) for any z ∈ G0 := G ∩ ((C∗ )s × Cn−s ). 38

We do not assume that m ≤ n − 1, rank A = m, AZn = Zm . 39 Note that (ii) gives an effective formula for g (a, ·). G

1.7. Examples

63

By virtue of (i), for any k = 1, . . . , s, the system of equations αj,s+1 xs+1 + · · · + αj,n xn = −αj,k ,

j = 1, . . . , m,

has a rational solution (Qs+1,k /µk , . . . , Qn,k /µk ) with Qs+1,k , . . . , Qn,k ∈ Z, µk ∈ N. Put Qk,k := µk and Qj,k := 0, j, k = 1, . . . , s, j 6= k. Then αj,1 Q1,k + · · · + αj,n Qn,k = 0,

j = 1, . . . , m, k = 1, . . . , s.

(1.7.11)

Let Qj := (Qj,1 , . . . , Qj,s ) ∈ Zs , j = 1, . . . , n. Define Φ : (C∗ )s × Cn−s −→ (C∗ )s × Cn−s , Φ(ξ, η) := (ξ Q1 , . . . , ξ Qs , ξ Qs+1 η1 , . . . , ξ Qn ηn−s ) = (ξ1µ1 , . . . , ξsµs , ξ Qs+1 η1 , . . . , ξ Qn ηn−s ), (ξ, η) = (ξ1 , . . . , ξs , η1 , . . . , ηn−s ) ∈ (C∗ )s × Cn−s . Observe that Φ is surjective. Indeed, for z = (z1 , . . . , zn ) ∈ (C∗ )s × Cn−s , take an arbitrary ξj ∈ (zj )1/µj , j = 1, . . . , s, and define ηj := zs+j /ξ Qs+j , j = 1, . . . , n − s. Moreover, if z = Φ(ξ, η), then by (1.7.11) we get z αj = ξ αj,1 Q1 +···+αj,n Qn η βj = η βj ,

j = 1, . . . , m.

(1.7.12)

Let D := {η ∈ Cn−s : |η βj | < 1, j = 1, . . . , m}. Using (1.7.12) we get the equality Φ((C∗ )s × D) = G0 . Fix a ξ0 ∈ (C∗ )s such that a = Φ(ξ0 , 0). Then, for any z = Φ(ξ, η) ∈ G0 , we have gG (a, z) = gG (Φ(ξ0 , 0), Φ(ξ, η)) ≤ g(C∗ )s ×D ((ξ0 , 0), (ξ, η)) = gD (0, η) = hD (η) = max{|η βj |1/rj : j = 1, . . . , m} = max{|z αj |1/rj : j = 1, . . . , m}. The implications (ii) =⇒ (iii) =⇒ (iv) =⇒ (v) are trivial. e < rank A. We may assume that (v) =⇒ (i). Suppose that rank A     β1 α1  ..   ..  2 ≤ t := rank A = rank  .  , rank  .  < t. βt αt Then there exist c1 , . . . , ct ∈ Z such that c1 β1 + · · · + ct βt = 0 and |c1 | + · · · + |ct | > 0. To simplify notation, assume that c1 , . . . , cu ≥ 0, cu+1 , . . . , ct < 0 for some 1 ≤ u ≤ t − 1. Let d := ac1 α1 +···+ct αt , f (z) :=

r := c1 r1 + · · · + cu ru = −(cu+1 ru+1 + · · · + ct rt ),

z c1 α1 +···+cu αu − dz −(cu+1 αu+1 +···+ct αt ) , 1 + |d|

z ∈ G.

Observe that f ∈ O(G, E), orda f ≥ r + 1, and f 6≡ 0 (because α1 , . . . , αt are linearly independent). Fix a b = (b0 , b00 ) ∈ G ⊂ Cs × Cn−s with f (b) 6= 0. Observe that f (b0 , θb00 ) = θr f (b), 0 ≤ θ ≤ 1. Thus we get 1 (r+1) 1 m (a, (b0 , θb00 )) ≥ |f (b0 , θb00 )|1/(r+1) = θ−1/(r+1) |f (b)|1/(r+1) −→ +∞; θ→0+ θ G θ contradiction.

64


Example 1.7.14. Let n = 3, m = 2, α1 := (1, 1, 0), α2 := (1, 0, 1), F (z) = (z1 z2 , z1 z3 ), G = {(z1 , z2 , z3 ) ∈ C3 : |z1 z2 | < 1, |z1 z3 | < 1}. Observe that: • rank A = 2, AZ3 = Z2 , • r(a) = 1 iff a 6= 0, • r(0) = 2, and • rank F 0 (a) = 2 iff a1 6= 0. (a) It is well known that gG (0, z) = hG (z) = max{|z1 z2 |1/2 , |z1 z3 |1/2 },

z ∈ G.

Moreover, one can prove (see (c)) that (2p)

mG (0, z) = gG (0, z) = max{|z1 z2 |1/2 , |z1 z3 |1/2 }, (2p+1) mG (0, z)

= max{|z1 z2 |

p+1 2p+1

, |z1 z3 |

p+1 2p+1

},

p ∈ N,

p ∈ Z+ , z ∈ G.

(b) By Example 1.7.11, if a1 6= 0, then (k)

mG (a, z) = gG (a, z) = mE 2 (F (a), F (z)) z ∈ G, k ∈ N.

= max{mE (a1 a2 , z1 z2 ), mE (a1 a3 , z1 z3 )}, (c) By Example 1.7.13, if a3 6= 0, then (2p)

mG ((0, 0, a3 ), z) = gG ((0, 0, a3 ), z) = max{|z1 z2 |1/2 , |z1 z3 |},

z ∈ G, p ∈ N.

Moreover, (2p+1)

mG

p+1

1

(0, 0, a3 ), z) = max{|z1 z2 | 2p+1 , |(z1 z2 )p z1 z3 | 2p+1 , |z1 z3 |},

z ∈ G, p ∈ Z+ .

Indeed, the inequality “≥” is obvious. Thus we have to show that (2p+1)

L(z) := mG

p+1

1

((0, 0, a3 ), z) ≤ max{|z1 z2 | 2p+1 , |(z1 z2 )p z1 z3 | 2p+1 , |z1 z3 |} =: R(z). (1.7.13)

The inequality is obviously true if z1 = 0 (because {0} × C2 ⊂ G). Take z = (z1 , z2 , z3 ) ∈ G with z1 6= 0. Then  p+1  if |z2 | ≥ |z3 | |z1 z2 | 2p+1 1 R(z) = |(z1 z2 )p z1 z3 | 2p+1 if |z1 z32 | ≤ |z2 | ≤ |z3 | .   |z1 z3 | if |z2 | ≤ |z1 z32 | Using standard argument, we reduce the proof of (1.7.13) to the cases where |z2 | = |z1 z32 | or |z2 | = |z3 |. If |z2 | = |z1 z32 |, then we have L(z) ≤ gG (a, z) = |z1 z3 | = R(z). If |z2 | = |z3 |, then L(z) ≤ gG (a, z) = |z1 z2 |1/2 .

(1.7.14)

1.7. Examples

65

Take an arbitrary f ∈ O(G, E) with orda f ≥ 2p + 1. We know that f (z) = fe(z1 z2 , z1 z3 ), where fe ∈ O(E 2 , E) (cf. Example 1.7.11). Inequality (1.7.14) shows that |fe(λ, eiθ λ)| ≤ |λ|p+1/2 ,

λ ∈ E, θ ∈ R.

Hence |fe(λ, eiθ λ)| ≤ |λ|p+1 , and, therefore, if |z2 | = |z3 |, then

λ ∈ E, θ ∈ R,

p+1

L(z) ≤ |z1 z2 | 2p+1 = R(z). (d) Similar formulae hold at points (0, a2 , 0) with a2 6= 0. (e) In the case a2 a3 6= 0, by Example 1.7.13, we already know that gG ((0, a2 , a3 ), z) ≥ max{|z1 z2 |, |z1 z3 |},

z ∈ G.

6≡

One can prove that a z z − a z z dk/2e/k o n 3 1 2 2 1 3 (k) gG ((0, a2 , a3 ), z) ≥ mG (a, z) ≥ max |z1 z2 |, |z1 z3 |, , |a2 | + |a3 | 6≡

z ∈ G.

(k)

? It seems that effective formulae for mG ((0, a2 , a3 ), ·) and gG ((0, a2 , a3 ), ·) are not known ? Let us mention that it must be (k)

(k)

mG ((0, a2 , a3 ), (eiϕ z1 , eiψ z2 , eiψ z3 )) = mG ((0, a2 , a3 ), z), gG ((0, a2 , a3 ), (eiϕ z1 , eiψ z2 , eiψ z3 )) = gG ((0, a2 , a3 ), z),

z ∈ G, ϕ, ψ ∈ R,

and (k)

mG (a, z) = gG ((0, a2 , a3 ), z) = max{|z1 z2 |, |z1 z3 |},

z ∈ G ∩ {a3 z2 − a2 z3 = 0}.

Similarly as in the one-pole case (cf. [J-P 1993], Proposition 4.2.7(h)), the generalized Green function may be characterized in terms of the Monge–Ampère operator (ddc · )n . Theorem 1.7.15 ([Lel 1989]). Let G ⊂ Cn be a bounded hyperconvex domain. Assume that the set |p| is finite. Then the function u := gG (p, ·) is a unique solution of the following problem:  u ∈ C(G, [0, 1]), log u ∈ PSH(G),    u = 1 on ∂G,  ∀a∈|p| ∃C(a)>0 ∀z∈G : u(z) ≤ C(a)kz − akp(a) ,    c (dd log u)n = 0 in G \ |p|. 40 The proof is beyond the scope of this article. 40

Recall that for a locally bounded function v ∈ PSH(D) (D ⊂ Cn ) we have (ddc v)n = 0 iff v is maximal, i.e. for any domain D0 b D and for any function v0 ∈ PSH(D0 ) upper semicontinuous on D0 , if v0 ≤ v on ∂D0 , then v0 ≤ v in D0 ; cf. [J-P 1993], Appendix, § MA.

66


Remark 1.7.16. (a) Recall that even in the case of the single pole Green, the Green function gG is not symmetric. Thus one can for instance ask whether for a bounded hyperconvex domain G ⊂ Cn we have lima→b gG (a, z) = 1 for arbitrary b ∈ ∂G and z ∈ G. The question is also interesting from the point of view of the boundary behavior of the Bergman function. D. Coman in [Com 1998] proved that if G is a bounded domain with a plurisubharmonic peak function ρ at a point b ∈ ∂G (i.e. ρ ∈ PSH(G) ∩ C(G), ρ(b) = 0, and ρ(z) < 0, z ∈ G \ {b}) such that ρ is Hölder continuous at b, and n log |ρ(z)| o 1 max : z ∈ G, r ≤ |z − b| ≤ 1/2 = O log log , r −→ 0, log |z − b| r then lima→b inf z∈K gG (a, z) = 1 for any compact K b G \ {b}. In particular, the result is true in the case where G is a pseudoconvex domain with smooth boundary and b is of finite type. G. Herbort in [Her 2000] proved that if G is a bounded hyperconvex with a Hölder continuous bounded plurisubharmonic exhaustion function, then lima→b inf z∈K g(a, z) = 1 for any K b G and b ∈ ∂G. In particular, the result holds if G is a bounded pseudoconvex domain with C 2 boundary. (b) Let G be a bounded strictly hyperconvex domain, i.e. there exist a domain U ⊂ Cn , G b U , and a function ρ ∈ PSH(U )∩C(U ) such that G = {z ∈ U : ρ(z) < 0}. S. Nivoche proved [Niv 1994], [Niv 1995], [Niv 2000] that in this case, for every a ∈ G, we have (k)

(k)

gG (a, z) = lim mG (a, z) = sup mG (a, z), k→+∞

z ∈ G,

k∈N

(k)

(k)

AG (a; X) = lim γG (a; X) = sup γG (a; X), k→+∞

X ∈ Cn \ P,

k∈N

where P ⊂ Cn is pluripolar; in fact, P = ∅ as it was shown by N. Nikolov in [Nik 2000]. (k) Observe that for an elementary Reinhardt domain Dα of irrational type all the mDα ’s (k)

vanish and mDα 6→ gDα ; cf. Theorem 1.3.1. (c) In the case n = 1 the above result was generalized by N. Nikolov and W. Zwonek in [Nik-Zwo 2003b], Theorem 2. They proved that if G ⊂ C is a domain for which the set of one-point connected components of C \ G is polar, then (k)

gG = sup mG , k∈N

(k)

AG = sup γG . k∈N

Moreover, they gave an example of a hyperconvex domain G ⊂ C for which the above equalities do not hold. (d) Recently E. A. Poletsky [Pol 2002] proved the following important theorem. Let G ⊂ Cn be a bounded strictly hyperconvex domain and let u be a negative plurisubharmonic function on G with zero boundary values, i.e. lim inf z→ζ u(z) = 0, ζ ∈ ∂G. Then there exist functions pk : G −→ R+ , |pk | finite, k = 1, 2, . . . , such that log gG (pk , ·) −→ u in L1 (G). Moreover, if u is continuous and ψ ∈ C0 ((−∞, 0]), then Z Z ψ(u(z))(ddc log gG (pk , ·))n −→ ψ(u(z))(ddc u)n . G

G

1.7. Examples

67

Example 1.7.17 ([Car-Wie 2003]). Let p : E n −→ R+ be such that |p| = {a1 , . . . , aN } ⊂ E × {0}n−1 . Put aj = (cj , 0, . . . , 0), kj := p(aj ), j = 1, . . . , N , and assume that k1 ≥ · · · ≥ kN . Then g

En

(p, z) =

N Y

k −kj+1

uj j

(z),

z ∈ En,

j=1

where kN +1 := 0 and uj (z) : = max{mE (c1 , z1 ) · · · mE (cj , z1 ), |z2 |, . . . , |zn |} = max{mE ({c1 , . . . , cj }, z1 ), |z2 |, . . . , |zn |} = mE n ({a1 , . . . , aj }, z),

j = 1, . . . , N.

Moreover, if k1 , . . . , kN ∈ N, then mE n (p, ·) = gE n (p, ·). The results extends easily to the case where |p| = {a1 , . . . , aN } ⊂ E × {c}n−1 ⊂ E × E n−1 . Observe that if k1 = · · · = kN = 1, then the above formula coincides with that from Example 1.7.2. QN n k −k Indeed, let u := j=1 uj j j+1 . Notice that u is continuous on E , log u is plurisubn harmonic, and u = 1 on ∂(E ). Take 1 ≤ s ≤ N and z = (z1 , . . . , zn ) in a small neighborhood of as . Then for j = s, . . . , N , we get uj (z) ≤ max{const |z1 − cs |, |z2 |, . . . , |zn |} ≤ const kz − as k. Consequently, u(z) ≤ const

N Y j=s

k −kj+1

uj j

(z) ≤ const

N Y

kz − as kkj −kj+1 = const kz − as kks .

j=s

Thus gE n (p, ·) ≥ u. To prove the opposite inequality we consider first the case n = 2. By virtue of Theorem 1.7.15, we only need to verify that the function log u is maximal on E 2 \ {a1 , . . . , aN }. Fix a point b = (b1 , b2 ) ∈ E 2 \ {a1 , . . . , aN }. Observe that the functions log uj (z), j = 1, . . . , N , are maximal on E 2 \ {a1 , . . . , aN } (cf. [Kli 1991], Example 3.1.2). It is clear that there exists at most one j0 ∈ {1, . . . , N } such that m(c1 , b1 ) · · · m(cj0 , b1 ) = |b2 |. Consequently, all the functions log uj with j 6= j0 are pluriharmonic near b. Since PN log u = j=1 (kj − kk+1 ) log uj , we easily conclude that log u is maximal near b. Now, consider the general case n ≥ 3. Take a point b = (b1 , . . . , bn ) ∈ E n \ {a1 , . . . , aN }

68


and let max{|b2 |, . . . , |bn |} = |bs0 |. If bs0 = 0 (i.e. b2 = · · · = bn = 0), then consider the F mapping E 3 λ −→ (λ, 0, . . . , 0) ∈ E n and use Remark 1.6.1(e): gE n (p, b) = gE n (p, F (b1 )) ≤ gE (p ◦ F, b1 ) =

N Y

[mE (cj , b1 )]kj −kj+1 = u(b).

j=1

If bs0 6= 0, then let qs := bs /bs0 ∈ E, s = 2, . . . , n. Consider the mapping F

E 2 3 (λ, ξ) −→ (λ, q1 ξ, . . . , qn ξ) ∈ E n . Using Remark 1.6.1(e) and the case n = 2, we get gE n (p, b) = gE n (p, F (b1 , bs0 )) ≤ gE 2 (p ◦ F, (b1 , bs0 )) =

N Y

[max{mE (c1 , b1 ) · · · m(cj , b1 ), |bs0 |}]kj −kj+1 = u(b).

j=1

Example 1.7.18. Using Proposition 1.6.6 and Theorem 1.4.1 we get gE 2 ({(a, b), (b, a)}, (z, w)) = c∗G2 ((a + b, ab), (z + w, zw)) n 2αab − (a + b) 2αzw − (z + w) o = max mE , : α ∈ ∂E , 2 − (a + b) 2 − (z + w) (a, b), (z, w) ∈ E 2 , a 6= b. (1.7.15) Notice that the above case is not covered by Example 1.7.17. For any points (a1 , b1 ), (a2 , b2 ) ∈ E 2 with mE (a1 , a2 ) = mE (b1 , b2 ) > 0 there exists an h ∈ Aut(E) such that h(a1 ) = b2 , h(a2 ) = b1 ), and consequently, formula (1.7.15) may be easily extended to such pairs of points. ? In the case where 0 < mE (a1 , a2 ) 6= mE (b1 , b2 ) > 0, an effective formula for gE 2 ({(a1 , b1 ), (a2 , b2 )}, ·) is still unknown ? ∗ Recall that by the Lempert theorem, if G ⊂ Cn is convex, then c∗G = e kG and, consequently, all holomorphically contractible families coincide on G. The following example shows that this is not true in the category of generalized holomorphically contractible families.

Example 1.7.19 (Due to W. Zwonek). Let D := {(z, w) ∈ C2 : |z| + |w| < 1},

√ √ At := {(t, t), (t, − t)},

0 < t 1.

Then mD (At , (0, 0)) < gD (At , (0, 0)) < dmax D (At , (0, 0)) for small t. p Indeed, let G := {(z, w) ∈ C2 : |z| + |w| < 1} and let F : D −→ G, F (z, w) := (z, w2 ). Note that F is proper and locally biholomorphic in a neighborhood of At . Moreover, At = F −1 (t, t). Using Proposition 1.6.6, we conclude that gD (At , (0, 0)) = gG ((t, t), (0, 0)).

1.7. Examples

69

Observe that mD (At , (0, 0)) = mG ((t, t), (0, 0)). In fact, the inequality ‘≥” follows from (H) (applied to F ). The opposite inequality may be proved as follows. Let f ∈ O(D, E) be such that f |At = 0. Define √ √ fe(z, w) := 21 (f (z, w) + f (z, − w)), (z, w) ∈ G. Note that fe is well defined, |fe| < 1, fe(t, t) = 0, fe is continuous, and fe is holomorphic on D ∩ {w 6= 0}. In particular, fe is holomorphic on D. Consequently, |f (0, 0)| = |fe(0, 0)| ≤ mG ((t, t), (0, 0)). Suppose that mD (Atk , (0, 0)) = gD (Atk , (0, 0)) for a sequence tk & 0. Then gG ((tk , tk ), (0, 0)) = gD (Atk , (0, 0)) = mD (Atk , (0, 0)) = mG ((tk , tk ), (0, 0)) ≤ gG ((tk , tk ), (0, 0)),

k = 1, 2, . . . .

Thus mG ((tk , tk ), (0, 0)) = gG ((tk , tk ), (0, 0)), k = 1, 2, . . . . Consequently, using [J-P 1993], § 2.5, and [Zwo 2000c], Corollary 4.4 (cf. § 1.2), we conclude that γG ((0, 0); (1, 1)) = AG ((0, 0); (1, 1)), where γG (resp. AG ) denotes the Carathéodory–Reiffen (resp. Azukawa) metric of G (cf. § 1.2). Hence, by Propositions 4.2.7 and 2.2.1(d) from [J-P 1993], using the fact that D is the convex envelope of G, we get 41 2 = hD (1, 1) = γG ((0, 0); (1, 1)) = AG ((0, 0); (1, 1)) = hG (1, 1) = 3−2√5 ; contradiction. To see the inequality gD (At , (0, 0)) < dmax D (A t , (0, 0)), we may argue as follows. We know (cf. [Zwo 2000c], Corollary 4.5 42 ) that gD (At , (0, 0)) = gG ((t, t), (0, 0)) ≈ gG ((0, 0), (t, t)) = hG (t, t) =

2t √ 3− 5

for small t > 0. On the other hand √ √ e∗ e∗ dmax D (At , (0, 0)) = min{kD ((t, − t), (0, 0)), kD ((t, t), (0, 0))} √ √ √ = min{hD (t, − t), hD (t, t)} = t + t. √ It remains to observe that 3−2t√5 < t + t for small t > 0. Example 1.7.20. Let G := E 2 , a− := (− 12 , 0), a+ := ( 12 , 0), b := (0, 13 ), p := 2χa− +χa+ . 0 0 Then dmin E 2 (p, b) ≤ dE 2 (p, b) < mE 2 (p, b), where dE 2 (p, ·) is defined in Remark 1.7.8. min 2 Recall that dE 2 (A, ·) ≡ mE 2 (A, ·) (A ⊂ E ) — Proposition 1.5.4. 41 42

Recall thatn hD (resp. hG ) denotes the Minkowski function for D (resp. G). Let G ⊂ C be a bounded hyperconvex domain. Then 0

lim 00

z ,z →a z 0 6=z 00

gG (z 0 , z 00 ) = 1, gG (z 00 , z 0 )

a ∈ G.

70


Indeed, by Example 1.7.17, mE 2 (p, b) = u1 (b)u2 (b) = max{ 12 ,

1 1 3 } max{ 2

· 12 ,

1 3}

=

1 2

·

1 3

= 16 .

On the other side: d0E 2 (p, b) = max{[mE 2 (a− , b)]2 mE 2 (a+ , b), [mE 2 ({a− , a+ }, b)]2 } 1 2 1 1 3 }] max{ 2 , 3 }, [max{ 12 · 12 , 13 }]2 } = 18 .

= max{[max{ 12 , = max{ 18 ,

[mE 2 ({− 12 , 12 } × {0}, b)]2 }

1.8. Properties of dmin and dmax G G Remark 1.8.1. If D ⊂ Cm is a Liouville domain (i.e. O(D, E) ' E), then min 0 dmin G×D (p, (z, w)) = dG (p , z),

(z, w) ∈ G × D,

where p0 (z) := sup{p(z, w) : w ∈ D}, z ∈ G. max Proposition 1.8.2. (a) The functions dmin G (p, ·) and dG (p, ·) are upper semicontinuous. (b) If p : G −→ Z+ , then dmin G (p, ·) ∈ C(G).

Proof. (a) The case of dmax G (p, ·) is obvious. To prove the upper semicontinuity of min dmin (p, ·), fix a z ∈ G and suppose that dmin 0 G G (p, zk ) −→ α > β > dG (p, z0 ) for a sequence zk −→ z0 . Take functions fk ∈ O(G, E), k ∈ N, such that fk (zk ) = 0 and Q sup p(fk−1 (µ)) −→ α. By a Montel argument we may assume that fk −→ f0 µ∈fk (G) |µ| Q −1 locally uniformly in G with f0 ∈ O(G, E), f0 (z0 ) = 0. Since µ∈f0 (G) |µ|sup p(f0 (µ)) < β, Q we can find a finite set A ⊂ G such that f0 |A is injective and a∈A |f0 (a)|p(a) < Q p(a) β. Consequently, < β and fk |A is injective for k 1. Finally, a∈A |fk (a)| Q −1 sup p(fk (µ)) |µ| < β, k 1; contradiction. µ∈fk (G) (b) In view of (a), it suffices to prove that for every f ∈ O(G, E) the function Q −1 uf (z) := µ∈f (G) [mE (µ, f (z))]sup p(f (µ)) , z ∈ G, is continuous on G. Observe that Q uf (z) = inf M { µ∈M [mE (µ, f (z))]kf (µ) }, where M runs over all finite sets M ⊂ f (|p|) such that kf (µ) := sup p(f −1 (µ)) < +∞, µ ∈ M . Thus uf = inf M {|hM |}, where hM ∈ O(G, E). Consequently, since the family (hM )M is equicontinuous, the function uf is continuous on G. Example 1.8.3. Let p : E × C −→ R+ , p( k1 , k) := k12 , k = 2, 3, . . . , and p(z, w) := 0 otherwise. Notice that |p| is discrete. Then by Remark 1.8.1, min 0 dmin E×C (p, (z, w)) = dE (p , z) =

∞ Y

2

[mE (1/k, z)]1/k ,

(z, w) ∈ E × C.

k=2

In particular, dmin E×C (p, ·) is discontinuous at (0, w) ∈ E × C \ |p|.

1.8. Properties of dmin and dmax G G

71

Proposition 1.8.4. If #|p| < +∞, then for any z0 ∈ G there exists an extremal function for dmin G (p, z0 ), i.e. a function fz0 ∈ O(G, E) with fz0 (z0 ) = 0 and Y −1 |µ|sup p(fz0 (µ)) = dmin G (p, z0 ). µ∈fz0 (G)

Proof. Fix a z0 ∈ G and let fk ∈ O(G, E), fk (z0 ) = 0 be such that Y −1 αk := |µ|sup p(fk (µ)) −→ α := dmin G (p, z0 ). µ∈fk (G)

Let Ak ⊂ |p| be such that fk |Ak is injective, fk (Ak ) = fk (|p|), and p(a) = sup p(fk−1 (fk (a))),

a ∈ Ak .

Thus αk = a∈Ak |fk (a)|p(a) . We may assume that Ak = B is independent of k and for any a ∈ B the fiber Ba := fk−1 (fk (a)) ∩ |p| is also independent of k. Moreover, we may that fk −→ f0 locally uniformly in G. Then f0 ∈ O(G, E), f0 (z0 ) = 0, and Q assume p(a) |f (a)| = α. Observe that f0 (B) = f0 (|p|). Let B0 ⊂ B be such that f0 |B0 is 0 a∈B injective and f0 (B0 ) = f0 (B). We have Y Y −1 −1 α≥ |µ|sup p(f0 (µ)) = |µ|sup p(f0 (µ)) Q

µ∈f0 (|p|)

=

Y

µ∈f0 (B0 ) max{p(b):b∈B, f0 (b)=f0 (a)}

|f0 (a)|

a∈B0

≥

Y

|f0 (a)|p(a) = α.

a∈B

Proposition 1.8.5. log dmin G (p, ·) ∈ PSH(G). Proof. By virtue of Proposition 1.8.2(a), we only need to show that for any f ∈ O(G, E) Q −1 the function uf (z) := µ∈f (G) [mE (µ, f (z))]sup p(f (µ)) , z ∈ G, is log–plurisubharmonic on G. The proof of Proposition 1.8.2 shows that uf = inf M vM , where vM is a log–pluriQ subharmonic function given by the formula vM (z) := µ∈M [mE (µ, f (z))]kf (µ) and M runs over a family of finite sets. Observe that vM1 ∪M2 ≤ min{vM1 , vM2 }. It remains to apply Lemma 1.6.3. Proposition 1.8.6. If Gk % G and pk % p, then min dmin Gk (pk , z) & dG (p, z),

max dmax Gk (pk , z) & dG (p, z),

z ∈ G.

Proof. By (H) and (M) (Definition 1.5.3) the sequence is monotone and for the limit max function u we have u ≥ dmin G (p, ·) (resp. u ≥ dG (p, ·)). Fix a z0 ∈ G. In the case of the minimal family suppose that u(z0 ) > α > dmin G (G, z0 ). Let fk ∈ Q −1 O(Gk , E) be such that fk (z0 ) = 0 and µ∈fk (Gk ) |µ|sup pk (fk (µ)) −→ u(z0 ). By a Montel argument we may assume that fk −→ f0 locally uniformly in G with f0 ∈ O(G, E), Q −1 f0 (z0 ) = 0. Since µ∈f0 (G) |µ|sup p(f0 (µ)) < α, we can find a finite set A ⊂ G such that Q Q f |A is injective and a∈A |f0 (a)|p(a) < α. Consequently, a∈A |fk (a)|pk (a) < α and fk |A Q −1 in injective for k 1. Finally, µ∈fk (Gk ) |µ|sup pk (fk (µ)) < α, k 1; contradiction.

72


In the case of the maximal family for any a ∈ G and ε > 0 there exists a k(a, ε) ∈ N ∗ ∗ such that z0 , a ∈ Gk , e kG (a, z0 ) ≤ e kG (a, z0 ) + ε, and pk (a) ≥ p(a) − ε for k ≥ k(a, ε). k Hence inf dmax Gk (pk , z0 ) =

k∈N

inf

k∈N, a∈Gk

∗ [e kG (a, z0 )]pk (a) k

∗ ≤ inf inf{[e kG (a, z0 ) + ε]pk (a) : 0 < ε 1, k ≥ k(a, ε)} a∈G

∗ ≤ inf inf{[e kG (a, z0 ) + ε]p(a)−ε : 0 < ε 1} = dmax G (p, z0 ). a∈G

Example 1.8.7. Let G := {z ∈ Cn : |z α | < 1}, where α = (α1 , . . . , αn ) ∈ Nn is such that α1 , . . . , αn are relatively prime. Then Y 0 min 0 α dmin [mE (µ, z α )]p (µ) , z ∈ G, G (p, z) = dE (p , z ) = µ∈E

where p0 (λ) = sup{p(a) : aα = λ}, λ ∈ E. Indeed, it is known (cf. Example 1.7.9) that any function f ∈ O(G, E) has the form f = fe ◦ F , where F (z) := z α and fe ∈ O(E, E). Thus Y −1 e−1 dmin [mE (µ, fe(F (z)))]sup p(F (f (µ))) : fe ∈ O(E, E)} G (p, z) = sup{ µ∈fe(F (G))

= sup{

Y

0 e−1 0 [mE (µ, fe(F (z)))]sup p (f (µ)) : fe ∈ O(E, E)} = dmin E (p , F (z)).

µ∈fe(E)

1.9. Relative extremal function Definition 1.9.1. Let G ⊂ Cn be a domain. For A ⊂ G the relative extremal function of A in G is given by the formula (cf. [Kli 1991], § 4.5) ωA,G := sup{u ∈ PSH(G), u ≤ 0, u|A ≤ −1}. Let

∗ ωA,G

denote the upper semicontinuous regularization of ωA,G .

Remark 1.9.2. Let F : G −→ D be a holomorphic mapping and let A ⊂ G, B ⊂ D be such that F (A) ⊂ B. Then ωB,D (F (z)) ≤ ωA,G (z),

z ∈ G.

Theorem 1.9.3 ([Edi 2001]). Let G ⊂ Cn be a domain, let p : G −→ R+ be such that the set #|p| is finite. Fix an R > 0 so small that • B(a, R1/p(a) ) b G for any a ∈ |p|, • B(a, R1/p(a) ) ∩ B(b, R1/p(b) ) = ∅ for any a, b ∈ |p|, a 6= b. S Let Ar := a∈|p| B(a, r1/p(a) ), 0 < r < R. Then R log ωAr ,G & log gG (p, ·) when r & 0. r

1.9. Relative extremal function

73

Proof. Let vr :=

log

R ωAr ,G , r

0 < r < R.

Step 1. vr1 ≤ vr2 for 0 < r1 < r2 . Indeed, fix 0 < r1 < r2 < R and define u :=

vr1 = log rR2

R r1

log

ωAr1 ,G

R r2

log

.

Then u ∈ PSH(G) and u ≤ 0. It suffices to show that u ≤ −1 on Ar2 . Fix an a ∈ |p|. 1/k Let k := p(a). Take a z ∈ B(a, r2 ). Then (cf. [Kli 1991], Lemma 4.5.8): log rR1 ωB(a,r1/k ), B(a,R1/k ) (z) log rR1 log+ kz−ak 1 R1/k − 1 ≤ −1. u(z) ≤ = 1/k R R log r2 log r2 log R1/k r1

Let v := lim vr = lim r→0+

r→0+

log

R ωAr ,G . r

Note that v ∈ PSH(G). Step 2. vr ≥ log gG (p, ·), 0 < r < R. In particular, v ≥ log gG (p, ·). Indeed, fix 0 < r < R and let ur :=

log gG (p, ·) . log Rr

Then ur ∈ PSH(G) and ur ≤ 0. Fix a ∈ |p| and z ∈ B(a, r1/k ) (k := p(a)). Then ur (z) ≤

k log gB(a,R1/k ) (a, z) log

R r

=

kz−ak R1/k log Rr

k log

≤ −1.

Thus ur ≤ ωAr ,G . Step 3. v ≤ log gG (p, ·). Indeed, it suffices to check the growth of v near every point a ∈ |p|. Fix an a ∈ |p| and let z ∈ B(a, R1/k ), z 6= a (k := p(a)). Let 0 < r < kz − akk . Then R v(z) − k log kz − ak ≤ log ωAr ,G (z) − k log kz − ak r R ≤ log ωB(a,r1/k ), B(a,R1/k ) (z) − k log kz − ak r + kz−ak R log R1/k − 1 − k log kz − ak ≤ − log R. = log 1/k r log R1/k r

74


1.10. Analytic discs method From some general point of view the invariant objects we have studied so far may be divided into three groups: (a) objects related to certain extremal problems concerning holomorphic mappings (k) f : G −→ E, e.g. c∗G (a, z), mG (a, z), γG (a; X), γG (a; X), mG (p, z), dmin G (p, z); (b) objects related to certain extremal problems concerning logarithmically plurisubharmonic functions u : G −→ [0, 1), e.g. gG (a, z), AG (a; X), gG (p, z); (c) objects related to certain extremal problems concerning analytic discs ∗ ∗ kG ϕ : E −→ G, e.g. e kG (a, z), H ∗ (a, z), κG (a; X), hG (a; X), e (p, z). At the end of the eighties E. A. Poletsky invented and partially developed a general method which reduces in some sense problems of type (b) to (c). This method found various important applications, due mainly to A. Edigarian (cf. [Edi 2002] and the references given there) and E. A. Poletsky (cf. [Pol 1991], [Pol 1993], [Edi-Pol 1997]) — see for instance § 1.12. In the present section we are mainly inspired by the exposition of the analytic disc theory presented in [Lár-Sig 1998b] and [Edi 2002]. Definition 1.10.1. Let G ⊂ Cn be a domain. By a disc functional (on G) we mean any function Ξ : O(E, G) −→ R. The envelope of a disc functional Ξ : O(E, G) −→ R is the function EΞ : G −→ R defined by the formula EΞ (z) := inf{Ξ(ϕ) : ϕ ∈ O(E, G), ϕ(0) = z},

z ∈ G.

Definition 1.10.2. The following four types of disc functionals play important role in complex analysis: • Poisson functional: Z 2π 1 p ΞPoi (ϕ) := p(ϕ(eiθ ))dθ, ϕ ∈ O(E, G), 2π 0 where p : G −→ [−∞, ∞) is an upper semicontinuous function 43 . • Green functional: X p ΞGre (ϕ) := p(ϕ(λ)) log |λ|,

ϕ ∈ O(E, G), p : G −→ R+ .

44

λ∈E∗

• Lelong functional: X p ΞLel (ϕ) := p(ϕ(λ)) ordλ (ϕ − ϕ(λ)) log |λ|,

ϕ ∈ O(E, G), p : G −→ R+ .

λ∈E∗

43 44

The P Poisson functionalPmay be defined for more general functions p — see [Edi 2002]. f (λ) := inf f (λ) (f : A −→ [−∞, 0]). λ∈A

B⊂A λ∈B #B 1 and Φ ∈ C ∞ (Ur × ∂E, G), where Ur := B(r) ⊂ C, such that: (i) Φ(·, ξ) ∈ O(E, G), ξ ∈ ∂E, (ii) Φ(0, ξ) = ϕ0 (ξ), ξ ∈ ∂E, (iii) Z 2π Z p ΞPoi (Φ(·, eit ))dt ≤ 0

2π

u0 (ϕ0 (eit ))dt + ε.

(1.10.17)

0

Proof. Since u0 is upper semicontinuous (Lemma 1.10.6), there exists a v ∈ C(G, R) with v ≥ u0 such that Z 2π Z 2π it v(ϕ0 (e ))dt ≤ u0 (ϕ0 (eit ))dt + 2ε . 0

0

For any ξ0 ∈ ∂E there exist ϕ ∈ O(E, G), 0 < δ < dist(ϕ(E), ∂G), an open arc I ⊂ ∂E, and r > 1 such that: • ξ0 ∈ I, ϕ(0) = ϕ0 (ξ0 ), p • ΞPoi (ϕ + z − ϕ0 (ξ0 )) < v(z) + ε/4, z ∈ B(ϕ0 (ξ0 ), δ), • ϕ0 (ξ) ∈ B(ϕ0 (ξ0 ), δ), ξ ∈ I, • Φ0 (Ur × I) b G, where Φ0 (λ, ξ) := ϕ(λ) + ϕ0 (ξ) − ϕ0 (ξ0 ). SN0 By a compactness argument we find a finite covering ∂E = ν=1 Iν , r > 1, and functions Φν ∈ C ∞ (Ur × Iν , G), ν = 1, . . . , N0 , such that: • Φν (·, ξ) ∈ O(E, G), ξ ∈ Iν , • Φν (0, ξ) = ϕ0 (ξ), ξ ∈ Iν , • Φν (Ur × Iν ) b G, p • ΞPoi (Φν (·, ξ)) < v(ϕ0 (ξ)) + ε/4, ξ ∈ Iν , ν = 1, . . . , N0 . SN0 Let K be the closure of the set ϕ0 (∂E)∪ ν=1 Φν (Ur ×Iν ) and let C > 0 be such that C > max{v(z) : z ∈ K}. There exist disjoint closed arcs Jν ⊂ Iν , ν ∈ A ⊂ {1, . . . , N0 }, such that [ ε Λ(∂E \ Jν ) < 8C . ν∈A

78


We may assume that A = {1, . . . , N } for some N ≤ N0 . Fix open disjoint arcs Kν with SN Jν ⊂ Kν ⊂ Iν , ν = 1, . . . , N , and let ρ ∈ C ∞ (∂E, [0, 1]) be such that ρ = 1 on ν=1 Jν SN and supp ρ ⊂ ν=1 Kν . Now we define Φ : Ur × ∂E −→ G by the formula ( Φν (ρ(ξ)λ, ξ), (λ, ξ) ∈ Ur × Kν Φ(λ, ξ) := . SN ϕ0 (ξ), (λ, ξ) ∈ Ur × (∂E \ ν=1 Kν ) It is clear that Φ is well defined, Φ ∈ C ∞ (Ur × ∂E, G), Φ(Ur × ∂E) ⊂ K, and Φ satisfies (i) and (ii). It remains to check (iii). Let Jeν := {t ∈ [0, 2π) : eit ∈ Jν }, ν = 1, . . . , N . We have: Z 2π N Z X p p ΞPoi (Φ(·, eit ))dt ≤ ΞPoi (Φν (·, eit ))dt + 8ε 0

≤

N Z X ν=1

ν=1 it

v(ϕ0 (e ))dt +

Jeν

3ε 8

Z

Jeν

2π

≤

it

v(ϕ0 (e ))dt + 0

ε 2

Z ≤

2π

u0 (ϕ0 (eit ))dt + ε.

0

Lemma 1.10.9. There exists 1 < s < r such that for any j ≥ 1 there exist an open annulus Aj ⊃ ∂E and Φj ∈ O(Us × Aj , G) with: (i) Φj −→ Φ uniformly on Us × ∂E, (ii) there exist 1 < sj < s and kj ∈ N, kj ≥ j, such that the mapping (λ, ξ) −→ Φj (λξ kj , ξ) extends to a mapping Ψj ∈ O(Usj × Usj , G), (iii) Ψj (0, ξ) = ϕ0 (ξ), ξ ∈ Usj . Proof. Let Φj (λ, ξ) := ϕ0 (ξ) +

Z 2π j X 1 (Φ(λ, eiθ ) − ϕ0 (eiθ ))e−ikθ dθ ξ k , 2π 0

k=−j

(λ, ξ) ∈ Ur × (Ur )∗ ; observe that the second term is the j–th partial sum of the Fourier series of the function ξ −→ Φ(λ, ξ) − ϕ0 (ξ); Φj is holomorphic and Φj (0, ξ) = ϕ0 (ξ), ξ ∈ (Ur )∗ . Moreover, for any 1 < t < r, Φj −→ Φ uniformly on Ut × ∂E. Indeed, it follows directly from Fourier series theory that Φj (λ, ·) −→ Φ(λ, ·) uniformly on ∂E for any λ ∈ Ur . Thus we only need to show that the series Z 2π ∞ X 1 (Φ(λ, eiθ ) − ϕ0 (eiθ ))e−ikθ dθ ξ k 2π 0 k=−∞

converges uniformly on Ut × ∂E. Using integration by parts, we obtain Z 2π (Φ(λ, eiθ ) − ϕ0 (eiθ ))e−ikθ dθ ξ k 0

≤

1 k2

∂2 2 (Φ(λ, eiθ ) − ϕ0 (eiθ )) , λ∈Ut , θ∈[0,2π) ∂θ sup

which implies the required convergence.

k ∈ Z∗ , (λ, ξ) ∈ Ut × ∂E,

1.10. Analytic discs method

79

Fix 1 < t < r. It follows that Φj (Ut × ∂E) b G for j ≥ j0 . Hence, one can find an open annulus Aj ⊃ ∂E such that Φj (Ut × Aj ) ⊂ G, j ≥ j0 . For any ξ ∈ (Ur )∗ the mapping Φj (·, ξ) − ϕ0 (ξ) has a zero at λ = 0. For any λ ∈ Ur the mapping Φj (λ, ·) − ϕ0 has a pole of order ≤ j at ξ = 0. Consequently, for any k ≥ j the mapping (λ, ξ) −→ Φj (λξ k , ξ) extends holomorphically to E × E. It remains to check (ii). Recall that Φj (0, ·) = ϕ0 . Hence there exists δj > 0 such that Φj (Uδj × E) ⊂ G. Since Φj (Ut × Aj ) ⊂ G, j ≥ j0 , we can find 0 < ρj < 1 such that Φj (E × (E \ Uρj )) ⊂ G, k j ≥ j0 . Now, let kj ≥ j be so big that ρj j < δj . Then Ψj (λ, ξ) := Φj (λξ kj , ξ) ∈ G, (λ, ξ) ∈ E × E, j ≥ j0 . Lemma 1.10.10. There exist 1 < s < r and Ψ ∈ O(Us × Us , G) such that: (i) Ψ (0, ξ) = ϕ0 (ξ), ξ ∈ Us , (ii) Z 2π Z 2π p p ΞPoi (Ψ (·, eit ))dt ≤ ΞPoi (Φ(·, eit ))dt + ε. 0

(1.10.18)

0

Proof. Let Φj , Ψj be as in Lemma 1.10.9. Then, for j ≥ j(ε) we have Z 2π Z 2π Z 2π 1 p it p(Φj (ei(θ+kj t) , eit ))dθ dt ΞPoi (Ψj (·, e ))dt = 2π 0 0 0 Z 2π Z 2π Z 2π Z 2π 1 1 = p(Φj (eiθ , eit ))dθdt ≤ p(Φ(eiθ , eit ))dθdt + ε 2π 0 2π 0 0 0 Z 2π p = ΞPoi (Φ(·, eit ))dt + ε.

0

Lemma 1.10.11. There exists a θ0 ∈ R such that if we put ϕ(λ) e = ϕθ0 (λ) := Ψ (eiθ0 λ, λ),

λ ∈ Us ,

then p ΞPoi (ϕ) e ≤

Proof. We have Z 2π Z 0

0

1 2π

Z

2π p ΞPoi (Ψ (·, eit ))dt.

0

2π

p(Ψ (eiθ , eit ))dθdt =

Z 0

2π

Z

(1.10.19)

2π

p(Ψ (eiθ eit , eit ))dtdθ.

0

Consequently, there exists a θ0 ∈ R such that Z 2π Z 2π Z 2π 1 1 p iθ0 it it p(Ψ (e e , e ))dt ≤ p(Ψ (eiθ eit , eit ))dtdθ ΞPoi (ϕθ0 ) = 2π 0 (2π)2 0 0 Z 2π Z 2π Z 2π 1 1 p iθ it = p(Ψ (e , e ))dθdt = ΞPoi (Ψ (·, eit ))dt. (2π)2 0 2π 0 0 Now, using (1.10.19), (1.10.18), and (1.10.17) gives (1.10.16).

The following result is a direct corollary of the definition of the function ωU,G and Theorem 1.10.7.

80


Proposition 1.10.12. Let F : G −→ D be a holomorphic covering, let V ⊂ D be open, and let U := F −1 (V ). Then ωV,D ◦ F = ωU,G . Proof. The inequality “≤” follows from Remark 1.9.2. The opposite inequality follows from Theorem 1.10.7 and Remark 1.10.3(c,d): ωU,G = EΞ −χU = EΞ −χV ◦F = EΞ −χV ◦F = EΞ −χV ◦ F = ωV,D ◦ F. Poi

Poi

Poi

Poi

1.10.2. Green, Lelong, and Lempert functionals. For any function p : G −→ R+ let Gp (G) := {u ∈ PSH(G) : u ≤ 0, ∀a∈G ∃C(u,a)∈R ∀z∈G : u(z) ≤ p(a) log kz − ak + C(a)}. Observe that log gG (p, z) = sup{u(z) : u ∈ Gp (G)}, z ∈ G (cf. Definition 1.5.1). Proposition 1.10.13. log gG (p, ·) ≤ EpLel . Consequently, p p p • for Ξ ∈ {ΞGre , ΞLel , ΞLem } if EΞ ∈ PSH(G), then EΞ ∈ Gp (G) and log gG (p, ·) ≡ EΞ , • if E

b k

G ΞPoi

(p,·)

45

∈ PSH(G) E

b k

, then E

G ΞPoi

(p,·)

b k

G ΞPoi

(p,·)

∈ Gp (G) and

≤ log gG (p, ·) ≤ EpLel .

Proof. Take ϕ ∈ O(E, G), u ∈PGp (G), and a finite set B ⊂ E∗ ∩ ϕ−1 (|p|). We are going to prove that u(ϕ(0)) ≤ λ∈B p(ϕ(λ)) ordλ (ϕ − ϕ(λ)) log |λ|, which implies that p u(ϕ(0)) ≤ ΞLel (ϕ) and, consequently, log gG (p, ·) ≤ EpLel . Let X v(ξ) := u(ϕ(ξ)) − p(ϕ(λ)) ordλ (ϕ − ϕ(λ)) log mE (λ, ξ). λ∈B

Then v ∈ SH(Ur \ B) for some r > 1 and v = u ◦ ϕ ≤ 0 on ∂E. Moreover, one can easily check that v is locally bounded from above in Ur . Hence v extends subharmonically to Ur and, by the maximum principle, v ≤ 0 on E. In particular, v(0) ≤ 0, which gives the required inequality. Proposition 1.10.14. EpGre = EpLel . Proof. We have to prove that n X o L(z) := inf p(ϕ(λ)) ordλ (ϕ − ϕ(λ)) log |λ| : ϕ ∈ O(E, G), ϕ(0) = z λ∈E∗

= inf

n X

o p(ϕ(λ)) log |λ| : ϕ ∈ O(E, G), ϕ(0) = z =: R(z),

z ∈ G.

λ∈E∗

The inequality “L ≤ R” is obvious. Fix a z ∈ G and an arbitrary constant C > L(z). We want to show that C ≥ R(z). 45

∗ (p, ·). Recall that b kG (p, ·) = log e kG


81

Since C > L(z), there exist ϕ ∈ O(E, G), ϕ(0) = z, and a finite set B ⊂ E∗ ∩ϕ−1 (|p|) such that X p(ϕ(λ)) ordλ (ϕ − ϕ(λ)) log |λ| < C. λ∈B

Write B = {b1 , . . . , bN }, aj := ϕ(bj ), r(j) := ordbj (ϕ − aj ), j = 1, . . . , N . Consider the family of all systems c of pairwise different points cj,k ∈ E, j = 1, . . . , N , k = 1, . . . , r(j), r(j) such that cj,1 · · · cj,r(j) = bj , j = 1, . . . , N . Define polynomials Y Qc,µ,ν (λ) := (λ − cj,k ), ν = 1, . . . , r(µ), j=1,...,N k=1,...,r(j) (j,k)6=(µ,ν) r(µ)

Pc,µ (λ) :=

X Qc,µ,ν (λ) , Qc,µ,ν (cµ,ν ) ν=1

µ = 1, . . . , N, λ ∈ C.

Observe that • deg Pc,j ≤ r(1) + · · · + r(N ) − 1, • Pc,j (cµ,ν ) = 0 if µ 6= j and Pc,j (cj,ν ) = 1, • Pc,1 + · · · + Pc,N ≡ 1. Define ϕc (λ) :=

N X

ϕ(λ) − a µ Pc,µ (λ) (λ − c ) · · · (λ − c ) + a . µ,1 µ µ,r(µ) (λ − bµ )r(µ) µ=1

Observe that ϕc ∈ O(E, Cn ), ϕc (0) = ϕ(0) = z, and ϕc (cj,k ) = aj for all j = 1, . . . , N , k = 1, . . . , r(j). Moreover, r(j) N X X

p(aj ) log |cj,k | =

j=1 k=1

N X

p(aj )r(j) log |bj | < C.

j=1

It remains to observe that ϕc (E) ⊂ G provided that cj,k ≈ bj , j = 1, . . . , N , k = 1, . . . , r(j). Theorem 1.10.15. If |p| is finite, then E

b k

G ΞPoi

(p,·)

= EpLel . Consequently, by Theorem

1.10.7, EpLel ∈ PSH(G) and, therefore, by Propositions 1.10.13 and 1.10.14, E

b k

G ΞPoi

(p,·)

= log gG (p, ·) = EpLel = EpGre .

Moreover, by Proposition 1.6.2, for arbitrary p : G −→ R+ we get the following Poletsky formula log gG (p, ·) = EpLel = EpGre . The Poletsky formula and the main ideas of the proof are due to E. A. Poletsky, cf. [Pol-Sha 1989], [Pol 1991], [Pol 1993]. The first complete proof was given by A. Edigarian in [Edi 1997b]. We follow the exposition of A. Edigarian.

82


Proof. We may assume that p 6≡ 0. By Proposition 1.10.13 we only need to show that EpLel ≤ E kbG (p,·) . Fix a ϕ0 ∈ O(E, G) and ε > 0. It suffices to find a ϕ e ∈ O(E, G) such ΞPoi

that ϕ(0) e = ϕ0 (0) and p ΞLel (ϕ) e ≤

1 2π

Z

2π

b kG (p, ϕ0 (eit ))dt + ε.

0

The existence of ϕ will be a consequence of a sequence of lemmas (Lemmas 1.10.16 — 1.10.20): Lemma 1.10.16. There exist: • 1 < s < r, • Φ ∈ C ∞ (Ur × ∂E, G), • N ∈ N, • a1 , . . . , aN ∈ |p|, • σ1 , . . . , σN ∈ C ∞ (∂E, C∗ ), • disjoint closed arcs J1 , . . . , JN ⊂ ∂E such that: (i) Φ(·, ξ) ∈ O(E, G), Φ(0, ξ) = ϕ0 (ξ), ξ ∈ ∂E, (ii) if |σν (ξ)| < s, then |σµ (ξ)| > s, µ 6= ν, and Φ(σν (ξ), ξ) = aν , SN (iii) |σν (ξ)| < 1, ξ ∈ Jν , ν = 1, . . . , N , Λ(∂E \ ν=1 Jν ) < ε, (iv) σν (ξ) 6= σµ (ξ), ξ ∈ ∂E, ν 6= µ, (v) 2πN max {p(aν ) max∂E log |σν |} < ε/2, ν=1,...,N R 2π R 2π PN (vi) ν=1 p(aν ) 0 log |σν (eit )|dt ≤ 0 b kG (p, ϕ0 (eit ))dt + ε. Proof. Let u0 := b kG (p, ·). Since u0 is upper semicontinuous, there exists a v ∈ C(G, R) with v ≥ u0 such that Z 2π Z 2π v(ϕ0 (eit ))dt ≤ u0 (ϕ0 (eit ))dt + 2ε . 0

0

For any ξ0 ∈ ∂E there exist ϕ ∈ O(E, G), λ0 ∈ E∗ , δ > 0, an open arc I ⊂ ∂E, and r > 1 such that: • ξ0 ∈ I, ϕ(0) = ϕ0 (ξ0 ), ϕ(λ0 ) =: a ∈ |p|, • p(a) log |λ0 | < v(z) + ε/8, z ∈ B(ϕ0 (ξ0 ), δ) ⊂ G, • ϕ(λ) + (1 − λ/λ0 )(z − ϕ0 (ξ0 )) ∈ G, (λ, z) ∈ Ur × B(ϕ0 (ξ0 ), δ), • ϕ0 (ξ) ∈ B(ϕ0 (ξ0 ), δ), ξ ∈ I, • Φ0 (Ur × I) b G, where Φ0 (λ, ξ) := ϕ(λ) + (1 − λ/λ0 )(ϕ0 (ξ) − ϕ0 (ξ0 )). SN0 By a compactness argument we find a covering ∂E = ν=1 Iν , r > 1, functions Φν ∈ C ∞ (Ur × Iν , G), ν = 1, . . . , N0 , and points λ1 , . . . , λN0 ∈ E∗ such that: • Φν (·, ξ) ∈ O(E, G), ξ ∈ Iν , • Φν (0, ξ) = ϕ0 (ξ), ξ ∈ Iν , • Φν (λν , ξ) =: aν ∈ |p|, ξ ∈ Iν , • Φν (Ur × Iν ) b G, • p(aν ) log |λν | < v(ϕ0 (ξ)) + ε/8, ξ ∈ Iν , ν = 1, . . . , N0 . Replacing Φν by the function (λ, ξ) −→ Φν (eiθν λ, ξ) with suitable θν ≈ 0, we may assume that the points λ1 , . . . , λN0 have different arguments.


Fix 1 < s < s0 < r with 2πN0

max

ν=1,...,N0

83

p(aν ) log s0 < ε/8. Let K be the closure of

the set N0 [

ϕ0 (∂E) ∪

Φν (Ur × Iν )

ν=1

and let C > 0 be such that C > 2πN0

max

ν=1,...,N0

p(aν )| log |λν || + max{v(z) : z ∈ K}.

There exist disjoint closed arcs Jν ⊂ Iν , ν ∈ A ⊂ {1, . . . , N0 }, such that [ ε Λ(∂E \ Jν ) < 8C . ν∈A

We may assume that A = {1, . . . , N } for some N ≤ N0 . Fix open disjoint arcs Kν with SN Jν ⊂ Kν ⊂ Iν , ν = 1, . . . , N , and let ρ ∈ C ∞ (∂E, [0, 1]) be such that ρ = 1 on ν=1 Jν SN and supp ρ ⊂ ν=1 Kν . We define Φ : Ur × ∂E −→ G by the formula ( Φν (ρ(ξ)λ, ξ), (λ, ξ) ∈ Ur × Kν Φ(λ, ξ) := . SN ϕ0 (ξ), (λ, ξ) ∈ Ur × (∂E \ ν=1 Kν ) It is clear that Φ is well defined, Φ ∈ C ∞ (Ur × ∂E, G), and Φ satisfies (i). Let Kν = {eiθ : θ ∈ (αν , βν )}, Jν = {eiθ : θ ∈ [γν , δν ]} with αν < γν < δν < βν . We may assume that ρ increases on (αν , γν ) and decreases on (δν , βν ). Then the set Jν0 := {ξ ∈ Kν : |λν |/ρ(ξ) ≤ s} is a closed arc with Jν ⊂ Jν0 ⊂ Kν . Take a σν ∈ C ∞ (∂E, R>0 λν ) with • σν (ξ) = λν /ρ(ξ), ξ ∈ Jν0 , • s < |σν (ξ)| < s0 , ξ ∈ Kν \ Jν0 , • |σν (ξ)| = s0 , ξ ∈ ∂E \ Kν . Then (ii), (iii), (iv), and (v) are satisfied. It remains to check (vi). Let Jeν := {θ ∈ [0, 2π) : eiθ ∈ Jν }, ν = 1, . . . , N . We have: N X ν=1

Z p(aν )

2π

log |σν (eit )|dt ≤

0

N X

Z Jeν

ν=1

Z ≤ 0

log |λν |dt +

p(aν )

2π

v(ϕ0 (eit ))dt +

ε 2

ε 8

≤

N Z X ν=1

Z ≤

Jeν

v(ϕ0 (eit ))dt +

ε 4

2π

b kG (p, ϕ0 (eit ))dt + ε.

0

Lemma 1.10.17. There exists a j0 ∈ N such that for any j ≥ j0 there exist 1 < sj < s, Ψj ∈ O(Usj × Usj , G), and τν,j ∈ O(Usj \ U 1/sj ), ν = 1, . . . , N , such that: (i) Ψj (0, ξ) = ϕ0 (ξ), ξ ∈ Usj , (ii) |τν,j | −→ |σν | uniformly on ∂E, (iii) Ψj (τν,j (ξ), ξ) = aν , ξ ∈ Usj \ U 1/sj with |τν,j (ξ)| < sj , (iv) |τν,j (ξ)| < 1, ξ ∈ Jν , ν = 1, . . . , N .

84


Proof. Recall that for any ξ ∈ ∂E the numbers 0, σ1 (ξ), . . . , σN (ξ) are pairwise different. Let P : C × ∂E −→ C be defined by the formula P (λ, ξ) := ϕ0 (ξ)

N N N Y λ − σ` (ξ) X λaµ Y λ − σ` (ξ) + ; −σ` (ξ) σ (ξ) σµ (ξ) − σ` (ξ) µ=1 µ

`=1

`=1 `6=µ

observe that P (·, ξ) is the Lagrange interpolation polynomial with P (0, ξ) = ϕ0 (ξ), P (σν (ξ), ξ) = aν , ν = 1, . . . , N . We will prove that there exists a function Φ0 ∈ C ∞ (Us × ∂U ) such that Φ(λ, ξ) = P (λ, ξ) + (λ − σ1 (ξ)) · · · (λ − σN (ξ))Φ0 (λ, ξ),

(λ, ξ) ∈ Us × ∂E.

Indeed, the only problem is to check that Φ0 is C ∞ near a point (σν (ξ0 ), ξ0 ) with |σν (ξ0 )| < s. Then |σµ (ξ0 )| > s for µ 6= ν, and there exists a neighborhood V of ξ0 such that |σν (ξ)| < s, |σµ (ξ)| > s, µ 6= ν, for ξ ∈ V . Observe that b ξ), Φ(λ, ξ) = aν + (λ − σν (ξ))Φ(λ, P (λ, ξ) = aν + (λ − σν (ξ))Pb(λ, ξ),

(λ, ξ) ∈ Us × V,

b and Pb are C ∞ mappings. Hence where Φ b ξ) − Pb(λ, ξ)) Φ0 (λ, ξ) = (Φ(λ,

Y µ6=ν

1 , λ − σµ (ξ)

(λ, ξ) ∈ Us × V,

and, consequently, Φ0 ∈ C ∞ (Us × V ). Notice that Φ0 (0, ·) ≡ 0. Let Φ0,j and σν,j be the j–th partial sums of the Fourier series of Φ0 and σν , respectively, i.e. Z 2π j X 1 Φ0 (λ, eit )e−ikt dt ξ k , Φ0,j (λ, ξ) : = 2π 0 k=−j

σν,j (ξ) : =

Z 2π j X 1 σν (eit )e−ikt dt ξ k , 2π 0

(λ, ξ) ∈ Us × C∗ ;

k=−j

cf. the proof of Lemma 1.10.9. One can easily show that Φ0,j −→ Φ0 and σν,j −→ σν uniformly on Ut × ∂E for any 1 < t < s. Define Pj (λ, ξ) := ϕ0 (ξ)

N N N Y λ − σ`,j (ξ) X λaµ Y λ − σ`,j (ξ) + , −σ`,j (ξ) σ (ξ) σ (ξ) − σ`,j (ξ) µ,j µ,j µ=1

`=1

`=1 `6=µ

Φj (λ, ξ) := Pj (λ, ξ) + (λ − σ1,j (ξ)) · · · (λ − σN,j (ξ))Φ0,j (λ, ξ). Then • • • • •

Φ0,j ∈ O(Us × C∗ ), for any λ ∈ Us the function Φ0,j (λ, ·) has a pole of order ≤ j at ξ = 0, for any ξ ∈ C∗ the function Φ0,j (·, ξ) has a zero at λ = 0, Φj −→ Φ uniformly on Ut × ∂E, 1 < t < s, Φj is holomorphic on Ut × ∂E, 1 < t < s, j 1,


85

• for any λ ∈ Us the function Φj (λ, ·) has a pole of order ≤ j at ξ = 0, j 1. Suppose that j 1 is such that σν,j (ξ) 6= 0, ξ ∈ ∂E. In particular, the set Zν,j := SN −1 E∗ ∩ σν,j (0) is finite. Observe that Φj ∈ O(Us × (E ∗ \ Zj )), where Zj := ν=1 Zν,j . Put Bj := B1,j · · · BN,j , where Bν,j denotes the Blaschke product for Zν,j with the zeros counted with multiplicities 46 . For every ξ ∈ C∗ \ Zj with |σν,j (ξ)| < s, we get Φj (σν,j (ξ), ξ) = aν . For any k ≥ j: • the mapping Φj,k (λ, ξ) := Φj (λξ k Bj (ξ), ξ) is holomorphic on E × E and • the mapping σν,j,k (ξ) := σν,j (ξ)/(ξ k Bj (ξ)) is meromorphic in C∗ and zero-free holomorphic in E∗ . Moreover, Φj,k (σν,j,k (ξ), ξ) = aν for all ξ ∈ (Us )∗ such that |σν,j,k (ξ)ξ k Bj (ξ)| < s. Using the same method as in the proof of Lemma 1.10.9, we get the required result with Ψj (λ, ξ) := Φj,kj (λ, ξ) = Φj (λξ kj Bj (ξ), ξ), τν,j (ξ) := σν,j,kj (ξ) = σν,j (ξ)/(ξ kj Bj (ξ)), where kj ≥ j is sufficiently big (and 1 < sj < s, sj ≈ 1).

Taking in Lemma 1.10.17 a j 1 gives the following result. Lemma 1.10.18. There exist 1 < t < s, Ψ ∈ O(Ut × Ut , G), τν ∈ O(Ut \ U 1/t , C∗ ), ν = 1, . . . , N , such that: (i) Ψ (0, ξ) = ϕ0 (ξ), ξ ∈ Ut , (ii) |τν (ξ)| < 1, ξ ∈ Jν , (iii) Ψ (τν (ξ), ξ) = aν , ξ ∈ (Ut )∗ with |τν (ξ)| < t, ν = 1, . . . , N , (iv) 2πN max {p(aν ) max∂E log |τν |} < ε/2, ν=1,...,N R 2π R 2π PN PN (v) ν=1 p(aν ) 0 log |τν (eit )|dt ≤ ν=1 p(aν ) 0 log |σν (eit )|dt + ε. Lemma 1.10.19. There exist η0 ∈ ∂E, k, c > 0, and 0 < ρ < 1 such that the functions f (ξ) := Ψ (η0 ξ k , ξ),

F (λ, η) := η

ρλ + e−c/k 1 + e−c/k ρλ

satisfy Z 0

2π p ΞLel (f (F (·, eit )))dt ≤

N X ν=1

Z p(aν )

2π

log |τν (eit )|dt + ε.

0

Proof. Since τν (ξ) 6= 0, ξ ∈ Ut \ U 1/t , there exists c 1 such that ηe−c − τ (ξ) ν log < log |τν (ξ)| + ε/(2M ), η ∈ E, ξ ∈ ∂E, ν = 1, . . . , N, (1.10.20) −c 1 − τν (ξ)ηe PN where M := ν=1 p(aν ). Let λ − 1 ψ(λ) := exp c , λ ∈ C \ {−1}; λ+1 observe that ψ(E) = E∗ , ψ(∂E \ {−1}) = ∂E. 46

That is, the function σν,j /Bν,j extends to a zero-free holomorphic function on E∗ .

86


Define ϕν (λ; η, ξ) :=

ηψ(λ) − τν (ξ) 1 − τν (ξ)ηψ(λ)

,

(λ, η, ξ) ∈ (C \ {−1}) × ∂E × Jν ;

we have |ϕν (λ; η, ξ)| = 1, (λ, η, ξ) ∈ (∂E \ {−1}) × ∂E × Jν . Moreover, ϕν (t; η, ξ) −→ −τν (ξ) when t −→ −1− . Thus ϕν (·; η, ξ) is an inner function with non-zero radial limits. Consequently, by ..., ϕν (·; η, ξ) is a Blaschke product. Moreover, since ψ 0 (λ) 6= 0, the zeros of ϕν (·; η, ξ) are simple and, by the implicit mapping theorem, for any point (λ0 , η0 , ξ0 ) with ϕν (λ0 ; η0 , ξ0 ) = 0, there exists a holomorphic function h = hλ0 ,η0 ,ξ0 defined in neighborhood V0 of (η0 , ξ0 ) such that h(η0 , ξ0 ) = λ0 and ϕν (h(η, ξ); η, ξ) = 0, (η, ξ) ∈ V0 . Observe that Z ληψ 0 (λ) 1 dλ, h(η, ξ) = 2πi ∂B(λ0 ,r) ηψ(λ) − τν (ξ) where B(λ0 , r) is so small that λ = λ0 is the only zero of ϕν (·; η0 , ξ0 ) in B(λ0 , r). Let (λν,` )∞ `=1 be the zeros of ϕν (·; η0 , ξ0 ) in E∗ . Since ϕν (·; η0 , ξ0 ) is a Blaschke product, we get ∞ ηe−c − τ (ξ) Y ν |ϕν (0; η0 , ξ0 )| = = |λν,` |. 1 − τν (ξ)ηe−c `=1 Hence, using (1.10.20), we conclude that there exist L ∈ N and ρ > 1 such that L X

log(|λν,` |/ρ) < log |τν (ξ0 )| + ε/(2M ).

`=1

Consequently, L X

log(|hλν,` ,η0 ,ξ0 (η, ξ)|/ρ) < log |τν (ξ)| + ε/(2M )

`=1

for (η, ξ) in a neighborhood of (η0 , ξ0 ). Using a compactness argument we see that there exist L ∈ N and ρ > 1 such that for any point (η, ξ) ∈ ∂E × Jν there exist λν,1 (η, ξ), . . . , λν,L (η, ξ) such that ϕν (λν,` (η, ξ); η, ξ) = 0,

` = 1, . . . , L,

and L X

log(|λν,` (η, ξ)|/ρ) < log |τν (ξ)| + ε/(2M ).

`=1

Let λ + e−c/k λ−1 = 1 + (1 − e−c/k ) , λ ∈ C \ {−ec/k }. −c/k 1+e λ 1 + e−c/k λ Observe that ψk −→ 1 locally uniformly in E and λ−1 47 k Log ψk (λ) −→ c λ+1 k locally uniformly in E. Consequently, ψk −→ ψ locally uniformly in E. ψk (λ) :=

47

limk→+∞ k Log ψk (λ) = limk→+∞ k(1 − e−c/k )

λ−1 1+e−c/k λ

= c λ−1 . λ+1


87

Fix a 1 < t0 < 1/ρ and let Vν be a neighborhood of Jν such that |τν (ξ)| < 1, ξ ∈ Vν . Let k0 ∈ N be such that ξψk (ρλ) ∈ Vν , (λ, ξ) ∈ Ut0 × Jν , k ≥ k0 . Hence, by (iii) of Lemma 1.10.18, we get Ψ (τν (ξψk (ρλ)), ξψk (ρλ)) = aν ,

(λ, ξ) ∈ Ut0 × Jν , k ≥ k0 .

Recall that ηψkk (ρλ) − τν (ξψk (ρλ)) −→ ηψ(ρλ) − τν (ξ) uniformly with respect to (λ, η, ξ) ∈ Ut0 × ∂E × Jν . Hence, by the Hurwitz theorem, for k 1, there are zeros λν,`,k (η, ξ) of the function λ −→ ηψkk (ρλ) − τν (ξψk (ρλ)) which are so close to λν,` (η, ξ) that L X

log |λν,`,k (η, ξ)| < log |τν (ξ)| + ε/(2M ),

(η, ξ) ∈ ∂E × Jν .

`=1

Observe that Ψ (ηψkk (ρλν,`,k (η, ξ)), ξψk (ρλν,`,k (η, ξ))) = aν ,

(η, ξ) ∈ ∂E × Jν , k 1.

Hence p (λ −→ Ψ (ηψkk (ρλ), ξψk (ρλ))) < ΞLel

N X

p(aν ) log |τν (ξ)| + ε/2,

ν=1

(η, ξ) ∈ Q :=

N [

(∂E × Jν ).

ν=1

Consider the diffeomorphism H : (∂E)2 −→ (∂E)2 given by H(η, ξ) := (ηξ −k , ξ). Let S := H(Q). Then Λ(S) = Λ(Q) ≥ 2π(2π − ε) (because the modulus of the Jacobian of H is equal to 1). Consequently, there exists an η0 ∈ ∂E such that Λ(R) ≥ 2π − ε, where R := {ξ ∈ ∂E : (η0 , ξ) ∈ S}. We have p ΞLel (λ −→ Ψ (η0 (ξψk (ρλ))k , ξψk (ρλ))) ≤

N X

p(aν ) log |τν (ξ)| + ε/2,

ξ ∈ R.

ν=1

Finally, by Lemma 1.10.18, we conclude that Z 2π Z N X p ΞLel (λ −→ Ψ (η0 (eit ψk (ρλ))k , eit ψk (ρλ)))dt ≤ p(aν ) 0

ν=1

which implies directly the required result.

2π

log |τν (eit )|dt + ε,

0

Lemma 1.10.20. There exists a θ0 ∈ R such that the mapping ϕ(ξ) := f (F (eiθ0 ξ, ξ)) satisfies p ΞLel (ϕ) ≤

1 2π

Z 0

2π p ΞLel (f (F (·, eit )))dt.

(1.10.21)

88


Proof. First we will prove that for any ϕ ∈ O(E, G) we have Z p ΞLel (ϕ) = (log |λ|)∆vϕ (λ)dΛ2 (λ),

(1.10.22)

E

where vϕ (λ) :=

1 X p(ϕ(b)) ordb (ϕ − ϕ(b)) log mE (b, λ), 2π

λ ∈ Ur ,

b∈Bϕ

Bϕ := {b ∈ E∗ : p(ϕ(b)) > 0} = E∗ ∩ ϕ−1 (|p|) (for some r > 1). Observe that Bϕ is discrete and vϕ ∈ SH(Ur ). To prove (1.10.22) we use the Riesz representation formula: Z

Z

2π

vϕ (eiθ )dθ

(log |λ|)∆vϕ (λ)dΛ2 (λ) = 2πvϕ (0) − E

0

X

=

p p(ϕ(b)) ordb (ϕ − ϕ(b)) log |b| = ΞLel (ϕ).

b∈Bϕ

Next we are going to show that for any function h ∈ O(E) with h(E) ⊂ E we have ∆vϕ◦h = ∆(vϕ ◦ h) in E,

(1.10.23)

with ∆−∞ := 0. If ϕ ≡ const or h ≡ const or h(E) ∩ Bϕ = ∅, then (1.10.23) is obviously true. Assume that ϕ 6≡ const and h 6≡ const and h(E) ∩ Bϕ 6= ∅. It is clear that vϕ◦h and vϕ ◦ h are harmonic on E \ h−1 (Bϕ ) and, consequently, ∆vϕ◦h = ∆(vϕ ◦ h) = 0 on E \ h−1 (Bϕ ). Take b ∈ Bϕ and c ∈ h−1 (b). Write h(λ) = (λ − c)m g(λ), where g ∈ O(E) and g(c) 6= 0. Then vϕ ◦ h(λ) =

1 p(ϕ(b)) ordb (ϕ − ϕ(b))m log |λ − c| + u(λ), 2π

where u is harmonic near c. Thus ∆(vϕ ◦ h) = p(ϕ(b)) ordb (ϕ − ϕ(b))mδc = p(ϕ(h(c))) ordc (ϕ ◦ h − ϕ ◦ h(c))δc = ∆vϕ◦h in a neighborhood of c. Applying (1.10.23) to ϕ := f and h := F (·, ξ), we get Z p ΞLel (f (F (·, ξ)) = (log |λ|)∆vf ◦F (·,ξ) (λ)dΛ2 (λ) E Z = (log |λ|)∆λ (vf ◦ F (λ, ξ))dΛ2 (λ). (1.10.24) E

Now we need the following auxiliary result.


89

Lemma 1.10.21. Let w ∈ PSH(Ur × Ur ) (r > 1) and let wθ (ξ) := w(eiθ ξ, ξ). Then there exists a θ0 ∈ R such that Z 2π Z Z 1 (log |λ|)∆λ w(λ, eiθ )dΛ2 (λ) dθ. (log |λ|)∆wθ0 (λ)dΛ2 (λ) ≤ 2π 0 E E Proof. The Riesz representation formula gives: Z Z 2π 1 1 (log |λ|)∆wθ (λ)dΛ2 (λ) + w(ei(θ+t) , eit )dt. w(0, 0) = wθ (0) = 2π E 2π 0 Hence 1 2π

2πw(0, 0) =

Z

2π

0

Z

(log |λ|)∆wθ (λ)dΛ2 (λ) dθ

E

1 + 2π

Z

2π

Z

0

2π

w(ei(θ+t) , eit )dt dθ.

0

On the other hand, using the Riesz representation formula for the functions w(0, ·) and w(·, eiθ ), we get Z Z 2π Z 2π 2πw(0, 0) = (log |λ|)∆λ w(0, λ)dΛ2 (λ) + w(0, eiθ )dθ ≤ w(0, eiθ )dθ E 2π

Z =

0

Consequently, Z 2π Z 0

0

0

Z 2π 1 Z 1 (log |λ|)∆λ w(λ, eiθ )dΛ2 (λ) + w(eit , eiθ )dt dθ. 2π E 2π 0

Z (log |λ|)∆wθ (λ)dΛ2 (λ) dθ ≤

E

0

2π

Z

(log |λ|)∆λ w(0, λ)dΛ2 (λ) dθ,

E

which implies the required result.

Applying Lemma 1.10.21 to (1.10.24) gives (1.10.21).

Remark 1.10.22. (a) There is a counterpart of the Poletsky formula from Theorem 1.10.15 for the Azukawa pseudometric AG (cf. § 1.2). Recently, N. Nikolov and W. Zwonek in [Nik-Zwo 2003b], Theorem 1, proved that for any domain G ⊂ Cn we have AD = ΓG = ΓG1 , where o n L (a) ϕ : ϕ ∈ O(E, G), ϕ(0) = a, ϕ(k) (0) = k!tX, k := ord0 (ϕ − a) , ΓG (a; X) := inf |t| o n L (a) ϕ ΓG1 (a; X) := inf : ϕ ∈ O(E, G), ϕ(0) = a, ϕ0 (0) = tX, ord0 (ϕ − a) = 1 , |t| Y χ{a} 48 Lϕ (a) := |λ|ordλ (ϕ−a) = exp(ΞLel (ϕ)), (a, X) ∈ G × Cn . λ∈ϕ−1 (a)∩E∗

48

Q

λ∈∅

χ{a}

... := 1. See Remark 1.10.3(f) for the definition of ΞLel (ϕ).

90


(b) Let G ⊂ Cn be a domain and let a, z0 ∈ G, z0 6= a, X0 ∈ Cn∗ . Following [Nik-Zwo 2003b], we say that a mapping ϕ ∈ O(E, G) is gG –extremal for (a, z0 ) χ{a} (resp. AG –extremal for (a, X0 )) if a, z0 ∈ ϕ(E), ϕ(0) = z0 , and log gG (a, z0 ) = ΞLel (ϕ) Lϕ (a) (k) (resp. ϕ(0) = a, ϕ (0) = tk!X0 , k := ord0 (ϕ − a), and AG (a; X0 ) = |t| ) up to an automorphism of E (i.e. we are allowed to substitute ϕ by ϕ ◦ h, where h ∈ Aut(E)). It follows from Proposition 3 in [Nik-Zwo 2003b] that for ϕ ∈ O(E, G), ϕ 6≡ const, a ∈ ϕ(E), the following conditions are equivalent: (i) ϕ is gG –extremal for a pair (a, z0 ) with a 6= z0 ∈ ϕ(E); (ii) ϕ is gG –extremal for any pair (a, z) with a 6= z ∈ ϕ(E); (iii) ϕ is AG –extremal for any pair (a, ϕ(k) (λ)) with λ ∈ ϕ−1 (a), k := ordλ (ϕ − a). Moreover, if G ⊂ C is such that ∂G is not polar, then a mapping ϕ ∈ O(E, G), a ∈ ϕ(E), ϕ 6≡ const, is a gG –extremal for some (a, z0 ) (a 6= z0 ∈ ϕ(E)) iff ϕ = π ◦ ψ, ψ−λ where π : E −→ G is a universal covering, ψ ∈ O(E, E), and the function hλ ◦ ψ = 1−λψ is a Blaschke product for any λ ∈ π −1 (a).

1.11. Coman conjecture Definition 1.11.1. Let G be a domain in Cn and let p : G −→ R+ . Define the Coman function

δG (p, z) := inf

n Y

|µa |p(a) : (µa )a∈|p| ⊂ E,

a∈|p|

o ∃ϕ∈O(E,G) : ϕ(0) = z, ϕ(µa ) = a, a ∈ |p| ,

z ∈ G;

we put δG (p, z) := 1 if the defining family is empty. We put δG (A, ·) := δG (χA , ·) (A ⊂ G), δG (a, ·) := δG ({a}, ·) (a ∈ G). Remark 1.11.2. (a) Directly from Proposition 1.10.13 it follows that gG (p, ·) ≤ δG (p, ·). ∗ (b) Obviously, δG (a, ·) = e kG (a, ·) (a ∈ G). Q p(a) (c) a∈|p| [mE (a, ·)] = gE (p, ·) = δE (p, ·) (for any p : E −→ R+ ). Q Indeed, we only need to prove that δE (p, ·) ≤ a∈|p| [mE (a, ·)]p(a) . Fix a z0 ∈ E z−a (a, z ∈ E). Let µa := ϕ−1 (a), a ∈ |p|. and let ϕ := h−z0 , where ha (z) := 1−az Q Q Then ϕ(0) = z0 , ϕ(µa ) = a, a ∈ |p|, and a∈|p| |µa |p(a) = a∈|p| [mE (µa , 0)]p(a) = Q p(a) . a∈|p| [mE (a, z0 )] (d) Let F : G −→ D be a holomorphic mapping and let q : D −→ R+ be such that #F −1 (b) = 1 for any b ∈ |q| (e.g. F is bijective). Then δD (q, F (z)) ≤ δG (q ◦ F, z),

z ∈ G.

1.11. Coman conjecture

91

Indeed, δD (q, F (z)) = inf

n Y

o |µb |q(b) : ∃ψ∈O(E,D) : ψ(0) = F (z), ψ(µb ) = b, b ∈ |q|

b∈|q|

≤ inf

n

Y

o |µa |q(F (a)) : ∃ϕ∈O(E,G) : ϕ(0) = z, ϕ(µa ) = a, a ∈ F −1 (|q|) .

a∈F −1 (|q|)

The Coman conjecture says that gG (p, ·) ≡ δG (p, ·) for any convex bounded domain G and function p with #|p| < +∞ (cf. [Com 2000]). The conjecture was motivated by the Lempert theorem and Remark 1.11.2(b). D. Coman proved that his conjecture is true in the case where G = B2 is the unit ball in C2 , |p| = {a1 , a2 }, and p(a1 ) = p(a2 ) (cf. [Com 2000]). Example 1.11.3. The first counterexample was given by M. Carlehed and J. Wiegerinck in [Car-Wie 2003]: G = E 2 ⊂ C2 , |p| = {a1 , a2 } ⊂ E × {0}, p(a1 ) 6= p(a2 ). Let c1 , c2 , d ∈ E∗ , c1 6= c2 , |c1 c2 | < |d| < |c1 |, p2,1 := 2χ(c1 ,0) + χ(c2 ,0) . Then gE 2 (p2,1 , (0, d)) < δE 2 (p2,1 , (0, d)). Indeed, by Example 1.7.17, gE 2 (p2,1 , z) = [max{mE (c1 , z1 ), |z2 |}][max{mE (c1 , z1 )mE (c2 , z1 ), |z2 |}]. Hence, by Example 1.7.2, if p1,1 := χ(c1 ,0) + χ(c2 ,0) , then gE 2 (p2,1 , z) = [gE 2 ((c1 , 0), z)][gE 2 (p1,1 , z)] ≤ [δE 2 ((c1 , 0), z)][δE 2 (p1,1 , z)] = [inf{|λ| : ∃ϕ∈O(E,E 2 ) : ϕ(0) = z, ϕ(λ) = (c1 , 0)}]× × [inf{|λ1 λ2 | : ∃ϕ∈O(E,E 2 ) : ϕ(0) = z, ϕ(λ1 ) = (c1 , 0), ϕ(λ2 ) = (c2 , 0)}] ≤ inf{|λ21 λ2 | : ∃ϕ∈O(E,E 2 ) : ϕ(0) = z, ϕ(λ1 ) = (c1 , 0), ϕ(λ2 ) = (c2 , 0)} = δE 2 (p2,1 , z). Suppose that gE 2 (p2,1 , (0, d)) = δE 2 (p2,1 , (0, d)). Then there exist ϕν ∈ O(E, E 2 ) and λν,1 , λν,2 ∈ E∗ such that ϕν (0) = (0, d), ϕν (λν,1 ) = (c1 , 0), ϕν (λν,2 ) = (c2 , 0), and |λ2ν,1 λν,2 | −→ [max{mE (c1 , 0), |d|}][max{mE (c1 , 0)mE (c2 , 0), |d|}] = |c1 d|. Using Montel argument we easily conclude that there are following three situations: (a) There exist ψ1 , ψ2 ∈ O(E, E) and ζ1 , ζ2 ∈ E such that ψ1 (0) = 0, ψ2 (0) = d, ψ1 (ζ1 ) = c1 , ψ2 (ζ1 ) = 0, ψ1 (ζ2 ) = c2 , ψ2 (ζ2 ) = 0, and |ζ12 ζ2 | = |c1 d|. Then, by the Schwarz lemma, λ−ζ λ − ζ2 1 · |ψ1 (λ)| ≤ |λ|, |ψ2 (λ)| ≤ , λ ∈ E. 1 − ζ 1λ 1 − ζ 2λ Hence |c1 | ≤ |ζ1 |, |d| ≤ |ζ1 ζ2 | and, consequently, |c1 | = |ζ1 | and |d| = |ζ1 ζ2 |. Thus |ψ1 (λ)| ≡ |λ|. It follows that |c2 | = |ζ2 | and |d| = |ζ1 ζ2 | = |c1 c2 |; contradiction. (b) There exist ψ1 , ψ2 ∈ O(E, E) and ζ1 ∈ E such that ψ1 (0) = 0, ψ2 (0) = d, ψ1 (ζ1 ) = c1 , ψ2 (ζ1 ) = 0, and |ζ12 | = |c1 d|.

92


Then, by the Schwarz lemma, |ψ1 (λ)| ≤ |λ|,

λ−ζ 1 |ψ2 (λ)| ≤ , 1 − ζ 1λ

λ ∈ E.

Hence |c1 | ≤ |ζ1 |, |d| ≤ |ζ1 | and, consequently, |c1 | = |ζ1 | = |d|; contradiction. (c) There exist ψ1 , ψ2 ∈ O(E, E) and ζ2 ∈ E such that ψ1 (0) = 0, ψ2 (0) = d, ψ1 (ζ2 ) = c2 , ψ2 (ζ2 ) = 0, and |ζ2 | = |c1 d|. Then, by the Schwarz lemma, λ−ζ 2 |ψ2 (λ)| ≤ , λ ∈ E. 1 − ζ 2λ Hence |c1 d| = |ζ2 | = |ψ2 (0)| ≥ |d|; contradiction. Example 1.11.4. Recently, P. J. Thomas and N. V. Trao [Tho-Tra 2003] (see also [Die-Tra 2003]) found a counterexample with G = E 2 , p = χB×C , #B = #C = 2. Let a ∈ (0, 1), a3/2 < γ < a, Aε := {−a, a} × {−ε, ε}, ε ∈ (0, 1). Then there exists a small ε > 0 such that gE 2 (Aε , (0, γ)) < δE 2 (Aε , (0, γ)). Indeed, first recall that γ + ε γ − ε o n gE 2 (Aε , (0, γ)) = max a2 , 1 + εγ 1 − εγ (cf. Example 1.7.2). Suppose that there exists a sequence εk & 0 such that gE 2 (Aεk , (0, γ)) = δE 2 (Aεk , (0, γ)),

k = 1, 2, . . . . (k)

We may assume that gE 2 (Aεk , (0, γ)) = a2 , k ∈ N. Let ϕk : E −→ E 2 and ξσ,τ ∈ E (σ, τ ∈ {−1, +1}) be such that Y (k) (k) ϕk (0) = (0, γ), ϕk (ξσ,τ ) = (σa, τ εk ), |ξσ,τ | −→ a2 . σ,τ ∈{−1,+1}

By Montel argument we may assume that ϕk −→ ϕ ∈ O(E, E 2 ) locally uniformly in E (k) and ξσ,τ −→ ξσ,τ ∈ E with ϕ(0) = (0, γ) and ϕ(ξσ,τ ) = (σa, 0) for (σ, τ ) ∈ J, where J := {(σ, τ ) ∈ {−1, +1} : ξσ,τ ∈ E}. Observe that Y (σ,τ )∈J

|ξσ,τ | =

Y

|ξσ,τ | = a2 ,

(σ,τ )∈{−1,+1}

in particular, J 6= ∅. It is clear that ξσ,τ 6= ξσ0 ,τ 0 for (σ, τ ), (σ 0 , τ 0 ) ∈ J with σ 6= σ 0 . Put T := {ξσ,τ : (σ, τ ) ∈ J}. Q Let ϕk =: (fk , gk ), ϕ =: (f, g). We have |g(z)| ≤ ξ∈T mE (ξ, z), z ∈ E. In particular, Q γ = |g(0)| ≤ ξ∈T |ξ|. Consequently, if #T = #J, then γ ≤ a2 < a3/2 ; contradiction. From now on assume that #T < #J. It suffices to consider the following four cases: (a) #T = 1, #J = 2: J = {(−1, −1), (−1, 1)}, ξ−1,−1 = ξ−1,1 =: ξ−1 . (b) #T = 2, #J = 3: J = {(−1, −1), (−1, 1), (1, −1)}, ξ−1,1 = ξ−1,1 = ξ−1 .

1.12. Product property

93

(c) #T = 3, #J = 4: J = {(−1, −1), (−1, 1), (1, −1), (1, 1)}, ξ−1,−1 = ξ−1,1 =: ξ−1 , ξ1,−1 6= ξ1,1 . (d) #T = 2, #J = 4: J = {(−1, −1), (−1, 1), (1, −1), (1, 1)}, ξ−1,−1 = ξ−1,1 =: ξ−1 , ξ1,−1 = ξ1,1 =: ξ1 . f −σa Put fσ := hσa ◦ f = 1−σaf . If (σ, τ ) ∈ J, then fσ (ξσ,τ ) = 0. Hence |fσ (z)| ≤ mE (ξσ,τ , z), z ∈ E. In particular, a = |fσ (0)| ≤ |ξσ,τ |. If (σ, −1), (σ, 1) ∈ J and ξσ,−1 6= ξσ,1 , then |fσ (z)| ≤ mE (ξσ,−1 , z)mE (ξσ,1 , z), z ∈ E. In particular, a = |fσ (0)| ≤ |ξσ,−1 ξσ,1 |. If (σ, −1), (σ, 1) ∈ J and ξσ,−1 = ξσ,1 =: ξσ , then f 0 (ξσ ) = 0 (if f 0 (ξσ ) 6= 0, then by the Hurwitz theorem, for big k, the equation fk (z) = σa has exactly one solution in a (k) (k) (k) (k) neighborhood of ξσ , which is false since fk (ξσ,τ ) = σa, ξσ,−1 6= ξσ,1 , and ξσ,τ −→ ξσ ). We have fσ (ξσ ) = fσ0 (ξσ ) = 0. Hence |fσ (z)| ≤ [mE (ξσ , z)]2 , z ∈ E. In particular, a = |fσ (0)| ≤ |ξσ |2 . Consequently: In the case (a) we get a2 = |ξ−1 |2 ≥ a — contradiction. 2 2 In the case (b) we get a2 = |ξ−1 ξ1,−1 | ≥ a · a. Hence |ξ−1 | = |ξ1,−1 | = a. Since |g(z)| ≤ mE (ξ−1 , z)mE (ξ1,−1 , z), z ∈ E, we have γ = |g(0)| ≤ |ξ−1 ξ1,−1 | = a1/2 · a — contradiction. 2 2 ξ1,−1 ξ1,1 | ≥ a · a. Hence |ξ−1 | = |ξ1,−1 ξ1,1 | = a. Since In the case (c) we get a2 = |ξ−1 |g(z)| ≤ mE (ξ−1 , z)mE (ξ1,−1 , z)mE (ξ1,1 , z), z ∈ E, we have γ = |g(0)| ≤ |ξ−1 ξ1,−1 ξ1,1 | = a1/2 · a — contradiction. 2 2 | = |ξ12 | = a and, conseξ12 | ≥ a · a. Hence |ξ−1 In the case (d) we get a2 = |ξ−1 2 quently, by the Schwarz lemma, hσa ◦ f = fσ = ασ hξσ , where |ασ | = 1, σ ∈ {−1, +1}, which implies that f = ha (α−1 h2ξ−1 ) = h−a (α1 h2ξ1 ). In particular, −a = f (ξ−1 ) = 2a h−a (α1 h2ξ1 (ξ−1 )) and a = f (ξ1 ) = ha (α−1 h2ξ−1 (ξ1 ). Then 1+a = −α1 h2ξ1 (ξ−1 ) = 2 1 2 2 2 2 α−1 hξ−1 (ξ1 ). Recall that −σa = fσ (0) = ασ ξσ . Hence 1+a2 = ξ2 hξ1 (ξ−1 ) = ξ21 h2ξ−1 (ξ1 ). 1

Put t := ξ−1 /ξ1 . Note that |t| = 1 and t 6= 1. We have contradiction.

2 1+a2

−1

1/t−1 2 t−1 2 = ( 1−at ) = ( 1−a/t ) —

Example 1.11.5. Let D, At be as in Example 1.7.19. Taking ϕ(λ) := (λ2 /4, λ/2), we √ max easily see that δD (At , (0, 0)) ≤ 4t < t + t = dD (At , (0, 0)), 0 < t 1. ? We do not know whether gD (At , (0, 0)) < δD (At , (0, 0)) for small t > 0 ? 1.12. Product property 1.12.1. Product property for relative extremal function. Theorem 1.12.1 ([Edi-Pol 1997], [Edi 2002]). Let Gj ⊂ Cnj be a domain, Aj ⊂ Gj , j = 1, 2. Assume that A1 , A2 are open or A1 , A2 are compact. Then ωA1 ×A2 ,G1 ×G2 (z1 , z2 ) = max{ωA1 ,G1 (z1 ), ωA2 ,G2 (z2 )},

(z1 , z2 ) ∈ G1 × G2 .

94


Moreover, if G1 , G2 are bounded, then for arbitrary subsets A1 ⊂ G1 , A2 ⊂ G2 we have ∗ ∗ ∗ ωA (z1 , z2 ) = max{ωA (z1 ), ωA (z2 )}, 1 ×A2 ,G1 ×G2 1 ,G1 2 ,G2

(z1 , z2 ) ∈ G1 × G2 .

We need a few auxiliary results. Proposition 1.12.2 ([Nos 1960], Chapter III). Let ϕ ∈ O(E, E) be an inner function, ϕ 6≡ const. Assume that ϕ is not a Blaschke product. Then there exists a ζ ∈ ∂E such that ϕ∗ (ζ) = 0 49 . Proposition 1.12.3 ([Nos 1960], Chapter II). Let ϕ ∈ H∞ (E) and let A ⊂ C be a compact polar set. Assume that there exists a set I ⊂ ∂E of positive measure such that ϕ∗ (ζ) ∈ A, ζ ∈ I. Then ϕ ≡ const. Lemma 1.12.4. Let A ⊂ E be a compact polar set and let π : E −→ E \ A be a universal covering. Then π is an inner function. Moreover, if 0 ∈ / A, then π is a Blaschke product. Proof. Obviously π ∗ (ζ) ∈ A ∪ ∂E for each ζ ∈ ∂E such that π ∗ (ζ) exists. Hence, by Proposition 1.12.3, we conclude that π ∗ (ζ) ∈ ∂E for almost all ζ ∈ ∂E and, consequently, π is an inner function. Now, if 0 6∈ A, then Proposition 1.12.2 implies that π is a Blaschke product. Remark 1.12.5. Let B be a finite Blaschke product and let ϕ ∈ O(E, E). Then ϕ is an inner function iff B ◦ ϕ is inner. Lemma 1.12.6 (L¨ owner theorem, [Edi 2002]). Let ϕ ∈ O(E, E) be an inner function such that ϕ(0) = 0. Then for any open set I ⊂ ∂E we have Λ((ϕ∗ )−1 (I)) = Λ(I) 50 . Proof. We may assume that I is an arc. Put J := (ϕ∗ )−1 (I) (observe that J is measurable). Consider the following holomorphic functions: Z 2π 1 uI (z) : = P (z, θ)χI (eiθ )dθ, 2π 0 Z 2π 1 uJ (z) : = P (z, θ)χJ (eiθ )dθ, z ∈ E, 2π 0 u : = uI ◦ ϕ − uJ , where P (z, θ) denotes the Poisson kernel. Let A denote the set of all ζ ∈ ∂E such that: • u∗J (ζ) does not exist or • u∗J (ζ) exists but u∗J (ζ) 6= χI (ζ) or • ϕ∗ (ζ) does not exist or • ϕ∗ (ζ) exists and ϕ∗ (ζ) ∈ ∂I (here ∂I denotes the boundary of I in ∂E). Note that A is of zero measure (use Proposition 1.12.3). Observe that u∗ (ζ) = 0 on J \ A. Moreover, u∗ (ζ) ≤ 0 on (∂E \ J) \ A. Thus u∗ ≤ 0 almost everywhere on ∂E and hence u ≤ 0. In particular, u(0) = (1/2π)(Λ(I) − Λ(J)) ≤ 0. Using the same argument to the arc ∂E \ I shows that Λ(∂E \ I) ≤ Λ(∂E \ J), which finishes the proof. ∗ ϕ (ζ) := limr→1 ϕ(rζ). 50 Recall that Λ denotes the Lebesgue measure on ∂E. 49


95

Lemma 1.12.7 ([Edi 2002]). Let (Ij )kj=1 ⊂ ∂E be a family of disjoint open arcs, let Sk I := j=1 Ij , and let α := Λ(I). Then for every ε > 0 there exists a finite Blaschke product B such that: • B(0) = 0, • B 0 (z) 6= 0 for z ∈ B −1 (0), and • B −1 (Jε ) ⊂ I, where Jε = {eiθ : 0 < θ < α − ε}. Proof. We may assume that α < 2π. Let Ij = {eiθ : θ1,j < θ < θ2,j }, j = 1, . . . , k, J0 := {eiθ : 0 < θ < α}. Define Qk Qk iθ2,j ) − eiα j=1 (z − eiθ1,j ) j=1 (z − e B0 (z) = Qk . Qk iθ2,j ) − iθ1,j ) j=1 (z − e j=1 (z − e One can prove ([Edi 2002], the proof of Lemma 4.8) that B0 is a finite Blaschke product with B0 (I) = J0 , B0 (∂E \ I) ⊂ ∂E \ J0 , and B0 (∂E \ I) = ∂E \ J 0 . Suppose that N Y z − aj mj B0 (z) = eiτ . 1 − aj z j=1 Take a closed arc Je0 ⊂ J0 such that Λ(Je0 ) ≥ α−ε. Then for different points aj,1 , . . . , aj,mj , sufficiently close to aj , such that aj ∈ {aj,1 , . . . , aj,mj }, if e0 (z) = eiτ B

mj N Y Y z − aj,` , 1 − aj,` z j=1 `=1

e0 (∂E \ I) ⊂ ∂E \ Je0 . then B e iθ z) (with suitable θ). Finally, we put B(z) := B(e

n

Proposition 1.12.8 (cf. [Lev-Pol 1999]). Let G ⊂ C be a domain and let A ⊂ G. Then ωA,G = sup{ωU,G : A ⊂ U ⊂ G, U open}. In particular, if A is compact, then for any neighborhood basis (Uj )∞ j=1 of A with G ⊃ Uj+1 ⊂ Uj , we have ωA,G = lim ωUj ,G . j→∞

Proof. Let u ∈ PSH(G), u ≤ 0, u ≤ −1 on A. Fix 0 < ε < 1 and define Uε := {z ∈ G : u < −1 + ε}. Then

u 1−ε

≤ ωUε ,G . Consequently, u ≤ (1 − ε) sup{ωU,X : A ⊂ U, U open}.

Taking ε −→ 0, we get the required result.

n

Proposition 1.12.9 (cf. [Blo 2000]). Let G ⊂ C be a bounded domain and let A ⊂ G. ∗ Put Aε := {z ∈ G : ωA,G (z) < −1 + ε}, 0 < ε < 1. Then ∗ ωA,G ∗ ≤ ωAε ,G ≤ ωA,G . 1−ε

96


∗ Consequently, ωAε ,G % ωA,G as ε & 0. ∗ Proof. Put N := {z ∈ G : ωA,G (z) < ωA,G (z)} and Q := A \ N . It is well-known (see ∗ ∗ e.g. Theorem 4.7.6 in [Kli 1991]) that N is pluripolar and ωQ,G = ωA,G (cf. [Jar-Pfl 2000], ∗ ∗ ∗ Lemma 3.5.3). We have ωQ,G = ωA,G = ωA,G = −1 on Q. Hence Q ⊂ Aε and ωA,G = ∗ ωQ,G ≥ ωAε ,Ω . Put u := Hence, u ≤ ωAε ,G .

∗ ωA,G 1−ε .

Note that u ∈ PSH(G), u ≤ 0, and u ≤ −1 on Aε .

Proof of Theorem 1.12.1. For the proof of the inequality “≥” it suffices to consider projections prj : G1 × G2 −→ Gj , j = 1, 2, and use Remark 1.6.1(e). We move to the opposite inequality. First assume that A1 , A2 are open. Put u1 = −χA1 and u2 = −χA2 . Let (z1 , z2 ) ∈ G1 × G2 be fixed and let β ∈ R be such that max{ωA1 ,G1 (z1 ), ωA2 ,G2 (z2 )} < β. By Theorem 1.10.7 there are ϕj ∈ O(E, Gj ), j = 1, 2, such that ϕ1 (0) = z1 , ϕ2 (0) = z2 , and Z 2π 1 uj (ϕj (eiθ ))dθ < β, j = 1, 2. 2π 0 Note that ϕ−1 disjoint 1 (A1 ) ∩ ∂E is an open set in ∂E. So, we may choose a finite set ofS m 1 1 1 1 open arcs I11 , . . . , Im ⊂ ϕ−1 (A ) ∩ ∂E such that Λ(I ) > −2πβ where I = 1 1 j=1 Ij . S k Similarly we choose I12 , . . . , Ik2 with I 2 = j=1 Ij2 . By Lemma 1.12.7 we may find Blaschke products B1 , B2 and a closed arc I ⊂ ∂E with Λ(I) > −2πβ such that B1−1 (I) ⊂ I 1 and B2−1 (I) ⊂ I 2 . Let A be the union of sets of critical values of B1 and B2 . Note that 0 is not in A. Let π be a holomorphic universal covering of E\A by E with π(0) = 0. Observe that π is inner e = Λ(I). There are (Lemma 1.12.4). If Ie = π −1 (I), then according to Lemma 1.12.6, Λ(I) liftings ψ1 and ψ2 of E into E such that π = B1 ◦ ψ1 = B2 ◦ ψ2 and ψ1 (0) = ψ2 (0) = 0. By Remark 1.12.5, ψ1 , ψ2 are inner. Also non-tangential boundary values of ψ1 and ψ2 on Ie belong to I 1 and I 2 , respectively. Put ϕ e1 = ϕ1 ◦ ψ1 and ϕ e2 = ϕ2 ◦ ψ2 . Then Z 2π e Λ(I) 1 max{u1 (ϕ e1 (eiθ )), u2 (ϕ e2 (eiθ ))}dθ ≤ − < β. 2π 0 2π By Fatou’s theorem the same inequality holds if we replace ϕ ej (z), j = 1, 2, with ϕ ej (rz), where r < 1 is sufficiently close to 1. Hence, ωA1 ×A2 ,G1 ×G2 (z1 , z2 ) < β. Since β was arbitrary, we get the proof. The case where A1 , A2 are compact follows from Proposition 1.12.8. We move to the second part of the theorem. First note that for any (z1 , z2 ) ∈ G1 ×G2 we have max{ωA1 ,G1 (z1 ), ωA2 ,G2 (z2 )} ≤ ωA1 ×A2 ,G1 ×G2 (z1 , z2 ) ∗ ∗ ≤ −ωA (z1 )ωA (z2 ). (1.12.25) 1 ,G1 2 ,G2


97

Indeed, we only need to prove the second inequality. Let u ∈ PSH(G1 × G2 ), u ≤ 0, u ≤ −1 on A1 × A2 . Then u(·, z2 ) ≤ −ωA2 ,G2 (z2 )ωA1 ,G1 (·),

z2 ∈ A2 ,

u(z1 , ·) ≤ −ωA1 ,G1 (z1 )ωA2 ,G2 (·),

z1 ∈ A1 .

Take a z1 ∈ G1 . If ωA1 ,G1 (z1 ) = 0, then u(z1 , ·) ≤ 0 = −ωA1 ,G1 (z1 )ωA2 ,G2 (·). If ωA1 ,G1 (z1 ) 6= 0, then let v := u(z1 , ·)/(−ωA1 ,G1 (z1 )). Then v ∈ PSH(G2 ), v ≤ 0, and v ≤ −1 on A2 . Hence v ≤ ωA2 ,G2 . Fix an ε > 0. Then by (1.12.25) ωA1 ×A2 ,G1 ×G2 (z1 , z2 ) ≤ −(1 − ε)2

on (A1 )ε × (A2 )ε .

∗ ωA (z1 , z2 ) ≤ −(1 − ε)2 1 ×A2 ,G1 ×G2 It follows that on G1 × G2

on (A1 )ε × (A2 )ε .

Hence

∗ ∗ ∗ (1 − ε)2 ωA ≤ ω(A ≤ ωA . 1 ×A2 ,G1 ×G2 1 ×A2 ,G1 ×G2 1 )ε ×(A2 )ε ,G1 ×G2

Thus, using the first part of the theorem and Proposition 1.12.9, we get ∗ ωA (z1 , z2 ) = lim ω(A1 )ε ×(A2 )ε ,G1 ×G2 (z1 , z2 ) 1 ×A2 ,G1 ×G2 ε→0

= lim max{ω(A1 )ε ,G1 (z1 ), ω(A2 )ε ,G2 (z2 )} ε→0

∗ ∗ = max{ωA (z1 ), ωA (z2 )}, 1 ,G1 2 ,G2

(z1 , z2 ) ∈ G1 × G2 .

Remark 1.12.10. Using analytic discs method F. Lárusson, P. Lassere, and R. Sigurdsson proved in [L´ ar-Las-Sig 1998a] the following result. Theorem. Let G ⊂ Cn be a convex domain and let A ⊂ G be an open or compact convex set. Then for any α ∈ [−1, 0) the level set {z ∈ G : ωA,G (z) < α} is convex. 1.12.2. Product property for the generalized Green function. Proposition 1.6.2 and Theorems 1.10.15, 1.12.1 imply the following product property for the generalized Green function (cf. [Edi 2001]). Theorem 1.12.11. For any domains G1 ⊂ Cn1 , G2 ⊂ Gn2 and for any sets A1 ⊂ G1 , A2 ⊂ G2 , the pluricomplex Green function with many poles has the product property: gG1 ×G2 (A1 × A2 , (z1 , z2 )) = max{gG1 (A1 , z1 ), gG2 (A2 , z2 )},

(z1 , z2 ) ∈ G1 × G2 .

In particular, gG1 ×G2 ((a1 , a2 ), (z1 , z2 )) = max{gG1 (a1 , z1 ), gG2 (a2 , z2 )}, (a1 , a2 ), (z1 , z2 ) ∈ G1 × G2 . A different proof, using Poletsky’s methods, was given by A. Edigarian in [Edi 1999]. The case of the pluricomplex Green function with one pole has been solved in [Edi 1997a]; some particular cases have been previously solved (using different methods based on the Monge–Ampère operator):

98


the case where both G1 and G2 are domains of holomorphy — in [J-P 1993], Theorem 9.8, the case where at least one of the domains G1 , G2 is a domain of holomorphy — in [Jar-Pfl 1995b]. Remark 1.12.12. One could try to generalize the above product property to arbitrary pole functions pj : Gj −→ R+ with maxGj pj = 1, j = 1, 2. For instance, one could conjecture that gG1 ×G2 (p, (z1 , z2 )) = max{gG1 (p1 , z1 ), gG2 (p2 , z2 )},

(z1 , z2 ) ∈ G1 × G2 ,

where p(a1 , a2 ) := min{p1 (a1 ), p2 (a2 )}. Unfortunately, such a formula is false. Take for instance G1 = G2 = E, p1 := χ{0} + 21 χ{c} , p2 := χ{0} , where 0 < c < 1. Observe that p = χ{(0,0)} + 12 χ{(c,0)} . Hence, by Example 1.7.17, we get 1/2 gE 2 (p, (z1 , z2 )) = max{|z1 |, |z2 |} max{|z1 |m(z1 , c), |z2 |} . In particular, if z1 = c, z2 = c2 , then gE 2 (p, (c, c2 )) = c3/2 . On the other hand, max{gE (p1 , c), gE (p2 , c2 )} = c2 . Remark 1.12.13. Example 1.11.4 shows that, in general, the Coman function does not satisfy the product property. Indeed, let B, C, and γ be as in the example. Then, by Remark 1.11.2(c), we have δE 2 (B × C, (0, γ)) > gE 2 (B × C, (0, γ)) = max{gE (B, 0), gE (C, γ)} = max{δE (B, 0), δE (C, γ)}. Proposition 1.12.14 ([Die-Tra 2003]). For any domains G ⊂ Cn , D ⊂ Cm , the following conditions are equivalent: (i) for any finite set A ⊂ G, and for any point b ∈ D we have δG×D (A × {b}, (z, w)) = max{δG (A, z), δD (b, w)},

(z, w) ∈ G × D;

(ii) δD (b, w) = gD (b, w), b, w ∈ D. Proof. (i) =⇒ (ii): By Theorem 1.10.15 we have (N )

gD (b, w) = inf δD (b, w), N ∈N

w ∈ D,

where (N )

δD (b, w) := inf

N nY

|µj | : µ1 , . . . , µN ∈ E, µj 6= µk ,

j=1

o ∃ψ∈O(E,D) : ψ(µj ) = b, j = 1, . . . , N, ψ(0) = w , (N )

w ∈ D.

By Remark 1.11.2(a), it suffices to show that δD (b, ·) ≤ δD (b, ·) (for every N ).


99

Fix N ∈ N, w0 ∈ D, and ε > 0. Let µ1 , . . . , µN ∈ E and ψ ∈ O(E, D) be such QN (N ) that µj 6= µk , ψ(µj ) = b, j = 1, . . . , N , ψ(0) = w0 , and j=1 |µj | ≤ δD (b, w0 ) + ε. Take an arbitrary ϕ ∈ O(E, G) such that ϕ(µj ) 6= ϕ(µk ). Put A := {ϕ(µ1 ), . . . , ϕ(µN )}, z0 := ϕ(0). Then δD (b, w0 ) ≤ max{δG (A, z0 ), δD (b, w0 )} = δG×D (A × {b}, (z0 , w0 )) ≤

N Y

(N )

|µj | ≤ δD (b, w0 ) + ε.

j=1

Letting ε −→ 0 we conclude the proof. (ii) =⇒ (i): Directly from the definition we get the inequality δG (A, z) ≤ δG×D (A × {b}, (z, w)),

(z, w) ∈ G × D.

Moreover, by Remark 1.11.2(a) and Theorem 1.12.11, δD (b, w) = gD (b, w) ≤ max{gG (A, z), gD (b, w)} = gG×D (A × {b}, (z, w)) ≤ δG×D (A × {b}, (z, w)),

(z, w) ∈ G × D.

Thus δG×D (A × {b}, (z, w)) ≥ max{δG (A, z), δD (b, w)},

(z, w) ∈ G × D.

Let A = {a1 , . . . , aN }. Fix (z0 , w0 ) ∈ G × D and ε > 0. To prove the inequality δG×D (A × {b}, (z0 , w0 )) ≤ max{δG (A, z0 ), δD (b, w0 )}, we may assume that max{δG (A, z0 ), δD (b, w0 )} + ε < 1. Consider the following two cases: (a) δD (b, w0 ) ≤ δG (A, z0 ). Take µ1 , . . . , µN ∈ E and ϕ ∈ O(E, G) such that ϕ(0) = z0 , ϕ(µj ) = aj , j = QN 1, . . . , N , and j=1 |µj | < δG (A, z0 ) + ε. We may assume that µj 6= 0, j = 1, . . . , N . Indeed, suppose that µ1 · · · µN −1 6= 0, and µN = 0. Then we may substitute ϕ z−ε by ϕ e := ϕ ◦ B, where B(z) := z 1−εz , z ∈ E. Observe that B(E) = E, so there exist µ e1 , . . . , µ eN −1 ∈ E∗ with B(e µj ) = µj , j = 1, . . . , N − 1. Hence ϕ(e e µj ) = aj , j = 1, . . . , N − 1, ϕ(ε) e = ϕ(0) e = z0 , and |µ1 · · · µN −1 ε| < ε. QN We may also assume that δD (b, w0 ) < j=1 |µj |. QN Indeed, if δD (b, w0 ) = j=1 |µj |, then we may substitute ϕ by ϕ(z) e := ϕ(tz), z ∈ E, and µj by µj /t, j = 1, . . . , N , with suitable 0 < t < 1, t ≈ 1. QN Take η ∈ E and ψ ∈ O(E, D) such that ψ(0) = w0 , ψ(η) = b, and |η| < j=1 |µj |. QN QN e Define α := (−µj ) ∈ E, t := −η/α ∈ E, ψ(z) := ψ(tz), z ∈ E, B := hµ , j=1

j=1

j

χ : E −→ G × D, χ := (ϕ, ψe ◦ hα ◦ B). e α (B(0)))) = (z0 , ψ(h e α (α))) = (z0 , ψ(0)) e We have χ(0) = (ϕ(0), ψ(h = (z0 , ψ(0)) = e α (B(µj )))) = (aj , ψ(h e α (0))) = (aj , ψ(−α)) e (z0 , w0 ). Moreover, χ(µj ) = (ϕ(µj ), ψ(h =

100


(aj , ψ(η)) = (aj , b), j = 1, . . . , N . Hence δG×D (A × {b}, (z0 , w0 )) ≤

N Y

|µj | < δG (A, z0 ) + ε.

j=1

(b) δG (A, z0 ) < δD (b, w0 ). Take η ∈ E and ψ ∈ O(E, D) such that ψ(0) = w0 , ψ(η) = b, and |η| < δD (b, w0 ) + ε/2. Take µ1 , . . . , µN ∈ E and ϕ ∈ O(E, G) such that ϕ(0) = z0 , ϕ(µj ) = aj , j = QN 1, . . . , N , and j=1 |µj | < δD (b, w0 ) + ε/2. Using the same argument as in (a), we may assume that µj 6= 0, j = 1, . . . , N . QN 1 Lemma 1.12.15. Let µ1 , . . . , µN ∈ E∗ , α := j=1 |µj |. Assume that α < β < α k+1 z−t k with k ∈ N. For t ∈ [0, 1], put ft (z) := z( 1−tz ) , z ∈ E. Then there exist t ∈ [0, 1] and QN µ e1 , . . . , µ eN ∈ E such that ft (e µj ) = µj , j = 1, . . . , N , and j=1 |e µj | = β. Proof. (Due to W. Zwonek 51 ) Observe that ft (E) = E. Hence, for any j ∈ {1, . . . , N } there exists a µ ej ∈ E such that ft (e µj ) = µj . Let Φj (t) := min{|e µj | : ft (e µj ) = µj }, 1 Φ := Φ1 · · · ΦN . We have Φ(0) = α k+1 > β, Φ(1) = α < β. We only need to show that each function Φj is continuous. Fix a j ∈ {1, . . . , N } and let Φj (t) = |e µj (t)|, t ∈ [0, 1]. Suppose that [0, 1] 3 ts −→ t0 and |e µj (ts )| ≤ m < Φj (t0 ). Then, without loss of generality, µ ej (ts ) −→ µ ej ∈ E. Therefore, ft0 (e µj ) = µj , i.e. Φj (t0 ) ≤ m; contradiction. So, Φj is lower semicontinuous. On the other hand, by the Hurwitz theorem, for any ε > 0, the equation fts (z) = µj must have a solution in the disc B(e µj (t0 ), ε), provided s 1. Hence Φj (ts ) ≤ Φj (t0 ) + ε, s 1, and finally lims→+∞ Φj (ts ) ≤ Φj (t0 ). Using Lemma 1.12.15 with β := δD (b, w0 ) + ε/2, we may modify ϕ and µ1 , . . . , µN QN in such a way that |η| < j=1 |µj | < δD (b, w0 ) + ε. Now we continue as in (a). Remark 1.12.16. Lemma 1.12.15 was improved by N. Nikolov, namely: QN z−t Let µ1 , . . . , µN ∈ E∗ , α := j=1 |µj |, α < β < 1. For t ∈ [0, 1], put ft (z) := z 1−tz , z ∈ E. Then there exist t ∈ [0, 1] and µ e1 , . . . , µ eN ∈ E such that ft (e µj ) = µj , j = 1, . . . , N , QN and j=1 |e µj | = β. Indeed, the case where β < α1/2 reduces to the proof of Lemma 1.12.15 (with k = 1). If β ≥ α1/2 , then put Ψj (t) := max{|e µj | : ft (e µj ) = µj }, Ψ := Ψ1 · · · ΨN . Observe that Ψ (0) = α1/2 . Similarly as in Lemma 1.12.15 we prove that Ψj is continuous on [0, 1). Moreover, Ψj (t) −→ 1 when t −→ 1, j = 1, . . . , N . Remark 1.12.17. Recently, Theorem 1.12.14 has been extended in the following way in [Nik-Zwo 2004]. Theorem. Let D ⊂ Cn and G ⊂ Cm be domains and let z ∈ D, w, b ∈ G, A ⊂ D. Then #A max{δD (A, z), lG (b, w)} ≤ δD×G (A × {b}, (z, w)) ≤ max{δD (A, z), δG (b, w)}, 51

The original proof in [Die-Tra 2003] contains an essential gap.


101

where N nY

N lG (b, w) := inf

|λj | : (λj )N j=1 ⊂ E, ∃ϕ∈O(E,G) :

j=1

o ϕ(0) = w, ϕ(λj ) = b, #{k : λk = λj } ≤ ordλj (ϕ − b), j = 1, . . . , N , ∞ lG (b, w) := inf

∞ nY

N ∈ N,

|λj | : (λj )∞ j=1 ⊂ E, ∃ϕ∈O(E,G) :

j=1

o ϕ(0) = w, ϕ(λj ) = b, #{k : λk = λj } ≤ ordλj (ϕ − b), j = 1, 2, . . . , . Moreover, for any N ∈ N ∪ {∞} the equality δD×G (A × {b}, (z, w)) = max{δD (A, z), δG (b, w)} N holds for any A ⊂ D with #A = N if and only if δG (b, w) = lG (b, w).

Moreover, they proved the following result. Theorem. Let A, B ⊂ E, #A = #B = 2, and z, w ∈ E be such that δE (A, z) = δE (B, w). Then δE (A, z) = min{δE 2 (C, (z, w)) : C ⊂ A × B} if and only if there is an h ∈ Aut(E) with h(z) = w and h(A) = B. Consequently, if ζ ∈ E \ A, then there exist uncountably many ξ ∈ E for which δE (A, ζ) = δE (B, ξ) < min{δE 2 (C, (ζ, ξ)) : C ⊂ A × B} ≤ δE 2 (A × B, (ζ, ξ)) and, therefore, gE 2 (A × B, (ζ, ξ)) < δE 2 (A × B, (ζ, ξ)), cf. Example 1.11.4. 1.12.3. Product property for dmin and dmax G G . The case of generalized Green function suggests that the product property might hold for other generalized holomorphically contractible families (dG )G , i.e. dG×D (A × B, (z, w)) = max{dG (A, z), dD (B, w)}, n

(z, w) ∈ G × D,

(P)

m

for any domains G ⊂ C , D ⊂ C and for any sets ∅ 6= A ⊂ G, ∅ 6= B ⊂ D. Notice that the inequality “≥” follows from (H) applied to the projections G×D −→ G, G×D −→ D. The definition applies to the standard holomorphically contractible families and means that dG×D ((a, b), (z, w)) = max{dG (a, z), dD (b, w)},

(a, b), (z, w) ∈ G × D.

∗ )G , (c∗G )G , (gG )G have the product Recall that the standard (non generalized) families (e kG property; cf. [J-P 1993], Ch. 9; see also [Mey 1997] (for a proof of the product property for the M¨ obius functions based on functional analysis methods) and [Jar-Pfl 1999b] (for the case of complex spaces). Moreover, it is known that the higher order Möbius functions (k) (mG )G with k ≥ 2 have no product property; cf. [J-P 1993], Ch. 9.

102


Proposition 1.12.18. The system (dmax G )G has the product property. ∗ Proof. Fix (z0 , w0 ) ∈ G × D and ε > 0. Let (a, b) ∈ A × B be such that e kG (a, z0 ) ≤ max ∗ max ∗ e dG (A, z0 )+ε, kD (b, w0 ) ≤ dG (B, w0 )+ε. Then using the product property for (e kG )G , we get dmax (A × B, (z0 , w0 )) ≤ e k∗ ((a, b), (z0 , w0 )) G×D

G×D

∗ ∗ = max{e kG (a, z0 ), e kD (b, w0 )} max ≤ max{dmax G (A, z0 ), dD (B, w0 )} + ε.

? We do not know whether the system (dmin G )G has the product property ? So far we were able to manage ([Jar-Jar-Pfl 2003]) only the case where #B = 1 — see Proposition 1.12.20 (see also [Die-Tra 2003]). Recall that dmin G (A, ·) = mG (A, ·) — Proposition 1.5.4. Proposition 1.12.19. Assume that for any n ∈ N, the system (mG )G has the following special product property: |Ψ (z, w)| ≤ (max |Ψ |) max{mG (A, z), mD (B, w)}, G×D

(z, w) ∈ G × D,

(P0 )

where G, D ⊂ Cn are balls with Pn respect to arbitrary C–norms, A ⊂ D, B ⊂ G are finite and non-empty, Ψ (z, w) := j=1 zj wj , and Ψ |A×B = 0. Then the system (mG )G has the product property (P) in the full generality. Moreover, if (P0 ) holds with #B = 1, then (P) holds with #B = 1. Proof. (Cf. [J-P 1993], the proof of Theorem 9.5.) Fix arbitrary domains G ⊂ Cn , D ⊂ Cm , non-empty sets A ⊂ G, B ⊂ G, and (z0 , w0 ) ∈ G × D. We have to prove that for any F ∈ O(G × D, E) with F |A×B = 0 the following inequality is true: |F (z0 , w0 )| ≤ max{mG (A, z0 ), mD (B, w0 )}. By Remark 1.6.1(h), we may assume that A, B are finite. ∞ Let (Gν )∞ ν=1 , (Dν )ν=1 be sequences of relatively compact subdomains of G and D, respectively, such that A ∪ {z0 } ⊂ Gν % G, B ∪ {w0 } ⊂ Dν % D. By Remark 1.6.1(h), it suffices to show that |F (z0 , w0 )| ≤ max{mGν (A, z0 ), mDν (B, w0 )}, 0

ν ≥ 1.

0

Fix a ν0 ∈ N and let G := Gν0 , D := Dν0 . It is well known that F may be approximated locally uniformly in G×D by functions of the form Ns X fs,µ (z)gs,µ (w), (z, w) ∈ G × D, (1.12.26) Fs (z, w) = µ=1

where fs,µ ∈ O(G), gs,µ ∈ O(D), s ≥ 1, µ = 1, . . . , Ns . Notice that Fs −→ 0 uniformly on A×B. Using Lagrange interpolation formula, we find polynomials Ps : Cn ×Cm −→ C such that Ps |A×B = Fs |A×B and Ps −→ 0 locally uniformly in Cn × Cm . The functions Fbs := Fs − Ps , s ≥ 1, also have form (1.12.26) and Fbs −→ F locally uniformly in G × D. Hence, without loss of generality, we may assume that Fs |A×B = 0, s ≥ 1. Let


103

ms := max{1, kFs kG0 ×D0 } and Fes := Fs /ms , s ≥ 1. Note that ms −→ 1 and, therefore, Fes −→ F uniformly on G0 × D0 . Consequently, we may assume that Fs (G0 × D0 ) b E, s ≥ 1. It is enough to prove that |Fs (z0 , w0 )| ≤ max{mG0 (A, z0 ), mD0 (B, w0 )},

s ≥ 1.

Fix an s = s0 ∈ N and let N := Ns0 , fµ := fs0 ,µ , gµ := gs0 ,µ , µ = 1, . . . , N . Let f := (f1 , . . . , fN ) : G −→ CN and g := (g1 , . . . , gN ) : D −→ CN . Put K := {ξ = (ξ1 , . . . , ξN ) ∈ CN : |ξµ | ≤ kfµ kG0 , µ = 1, . . . , N, |Ψ (ξ, g(w))| ≤ 1, w ∈ D0 }. It is clear that K is an absolutely convex compact subset of CN with f (G0 ) ⊂ K. Let L := {η = (η1 , . . . , ηN ) ∈ CN : |ηµ | ≤ kgµ kD0 , µ = 1, . . . , N, |Ψ (ξ, η)| ≤ 1, ξ ∈ K}. Then again L is an absolutely convex compact subset of CN , and moreover, g(D0 ) ⊂ L. ∞ Let (Wσ )∞ σ=1 (resp. (Vσ )σ=1 ) be a sequence of absolutely convex bounded domains N in C such that Wσ+1 b Wσ and Wσ & K (resp. Vσ+1 b Vσ and Vσ & L). Put Mσ := kΨ kWσ ×Vσ , σ ∈ N. By (P0 ) and by the holomorphic contractibility applied to the mappings f : G0 −→ Wσ , g : D0 −→ Vσ , we have |Fs0 (z0 , w0 )| = |Ψ (f (z0 ), g(w0 ))| ≤ Mσ max{mWσ (f (A), f (z0 )), mV σ (g(B), g(w0 ))} ≤ Mσ max{mG0 (f −1 (f (A)), z0 ), mD0 (g −1 (g(B)), w0 )} ≤ Mσ max{mG0 (A, z0 ), mD0 (B, w0 )}. Letting σ −→ +∞ we get the required result.

Proposition 1.12.20. The system (mG )G has the product property (P) whenever #B = 1, i.e. for any domains G ⊂ Cn , D ⊂ Cm , for any set A ⊂ G, and for any point b ∈ D we have mG×D (A × {b}, (z, w)) = max{mG (A, z), mD (b, w)},

(z, w) ∈ G × D.

Proof. By Proposition 1.12.19, we only need to check (P) in the case, where D is a bounded convex domain, A is finite, and B = {b}. Fix (z0 , w0 ) ∈ G×D. Let ϕ : E −→ D be a holomorphic mapping such that ϕ(0) = b and ϕ(mD (b, w0 )) = w0 (cf. [J-P 1993], Ch. 8). Consider the mapping F : G × E −→ G × D, F (z, λ) := (z, ϕ(λ)). Then mG×D (A × {b}, (z0 , w0 )) ≤ mG×E (A × {0}, (z0 , mG (b, w0 ))). Consequently, it suffices to show that mG×E (A × {0}, (z0 , λ)) ≤ max{mG (A, z0 ), |λ|},

λ ∈ E.

(1.12.27)

The case where mG (A, z0 ) = 0 is elementary: for an f ∈ O(G × E, E) with f |A×{0} = 0 we have f (z0 , 0) = 0 and hence |f (z0 , λ)| ≤ |λ|, λ ∈ E (by the Schwarz lemma). Thus, we may assume that r := mG (A, z0 ) > 0. First observe that it suffices to prove (1.12.27) only on the circle |λ| = r. Indeed, if the inequality holds on that circle, then by the maximum principle for subharmonic functions (applied to the function mG×E (A × {0}, (z0 , ·))) it

104


holds for all |λ| ≤ r. In the annulus {r < |λ| < 1} we apply the maximum principle to 1 the subharmonic function λ −→ |λ| mG×E (A × {0}, (z0 , λ)). Now fix a λ0 ∈ E with |λ0 | = r. Let f be an extremal function for mG (A, z0 ) with f |A = 0 and f (z0 ) = λ0 . Consider F : G −→ G × E, F (z) := (z, f (z)). Then mG (A × {0}, (z0 , λ0 )) ≤ mG (A, z0 ) = max{mG (A, z0 ), |λ0 |}, which completes the proof.

CHAPTER 2

Hyperbolicity and completeness 2.1. ci –hyperbolicity versus c–hyperbolicity Recall that a domain G ⊂ Cn is called ciG –hyperbolic (or shortly ci –hyperbolic), respectively cG –hyperbolic (shortly c–hyperbolic), if ciG , respectively cG , is a true distance on G. In virtue of the inequality cG ≤ ciG , if G is cG –hyperbolic, then it is ciG –hyperbolic. If G is bounded, then cG is a distance. In the general case, the following result due to J.-P. Vigué (cf. [Vig 1996]) gives a characterization of ciG –hyperbolicity. Theorem 2.1.1. Let G ⊂ Cn be a domain. Then the following properties are equivalent: (i) G is ciG –hyperbolic; (ii) there is no non-constant C 1 –curve α : [0, 1] −→ G such that γG (α; α0 ) ≡ 0; (iii) for any point a ∈ G there exists a neighborhood U = U (a) ⊂ G such that cG (a, z) 6= 0, z ∈ U \ {a}. Proof. (i) =⇒ (ii): Suppose the contrary, namely, that there exists a C 1 –curve α : [0, 1] −→ G such that γG (α; α0 ) ≡ 0,

α0 (t0 ) 6= 0 for a t0 ∈ [0, 1].

Obviously, then for any 0 ≤ t0 < t00 ≤ 1 we have ciG (α(t0 ), α(t00 )) = 0. In virtue of α0 (t0 ) 6= 0 there are two different points α(t0 ), α(t00 ) showing that G is not ci –hyperbolic. Contradiction. (ii) =⇒ (iii): We proceed by assuming the contrary. So let a ∈ G be such a point that there exists a sequence of points (z j )j∈N ⊂ G \ {a}, z j −→ a, such that cG (a, z j ) = 0, j ∈ N. We have to find a C 1 –curve which does fulfill the property stated in (ii). Observe that A := {z ∈ G : cG (a, z) = 0} = {z ∈ G : f (a) = f (z), f ∈ O(G, E)} is an analytic subset of G. In virtue of the existence of the points z j ∈ A \ {a} tending to a, the dimension of the analytic set A in a is at least 1. Therefore, there is a C 1 –curve α : [0, 1] −→ Reg A such that α0 6≡ 0. On the other side, since this curve lies in A, we have γG (α; α0 ) ≡ 0; contradiction. (iii) =⇒ (i): Fix a, b ∈ G, a 6= b, and choose a neighborhood U = U (a) ⊂ G according to (iii). Moreover, let V = V (a) b U , b ∈ / V . Obviously, 0 < cG (a, z), z ∈ U \ {a}. Applying the continuity of cG there is a C > 0 such that cG (a, ·)|∂V ≥ C. Thus for any C 1 –curve α : [0, 1] −→ G, α(0) = a, α(1) = b, there is a t0 ∈ (0, 1) with α(t0 ) ∈ ∂V ; therefore, Lc (α) ≥ cG (a, α(t0 )) + cG (α(t0 ), b) ≥ C > 0. Hence, ciG (a, b) ≥ C > 0. Moreover, there is the following general relation between γG –hyperbolicity and local c–hyperbolicity. 105

106

2. Hyperbolicity and completeness

Proposition 2.1.2. Any domain G ⊂ Cn that is γG –hyperbolic (i.e. γG (z; X) > 0, z ∈ G, X ∈ Cn \ {0}) is locally cG –hyperbolic (i.e. for any a ∈ G there exists a neighborhood U = U (a) ⊂ G such that cG is a distance on U ). In particular, G is ci –hyperbolic. Proof. Fix an a ∈ G and suppose that z j , wj −→ a, z j 6= wj , cG (z j , wj ) = 0, j = 1, 2, . . . . j −wj We may assume that kzzj −w j k −→ X0 ∈ ∂Bn . Then cG (z j , wj ) = 0; j→∞ kz j − w j k

γG (a; X0 ) = lim contradiction (cf. §1.2).

1

Observe that the result is true for any C –pseudodistance (cf. §1.2.4). Remark 2.1.3. ? It seems to be unknown whether c–hyperbolicity implies γ–hyperbolicity ? Example 2.1.4. There is a domain G ⊂ C3 which is not cG –hyperbolic and not γG – hyperbolic, but nevertheless ciG –hyperbolic (see [Vig 1996]). This G is constructed via an example of a 1–dimensional complex space and then applying the Remmert embedding theorem. We omit here details. Remark 2.1.5. Notice that the Example 2.1.4 is not explicitly given. So ? it is interesting to find an effective example of that type; moreover, the question whether such an example is possible in C2 is still an open one ? 2.2. Hyperbolicity for Reinhardt domains Before we shall discuss the different notions of hyperbolicity in the case of pseudoconvex Reinhardt domains, recall the effective formulas for the Kobayashi pseudodistance on elementary Reinhardt domains (cf. Theorem 1.3.1). Let Vj := {z ∈ Cn : zj = 0}, (Ajk )j=1,...,n, k=1,...,n

Moreover, for a matrix A = row. Put ΦA : Cn∗ → Cn∗ ,

j = 1, . . . , n. ∈ Z(n × n), we denote by Aj its j–th 1

n

Φ(z) := (z A , . . . , z A ).

Theorem 2.2.1 ([Zwo 1999a]). Let G be a pseudoconvex Reinhardt domain in Cn . Then the following properties are equivalent: (i) G is cG –hyperbolic; (ii) G is e kG –hyperbolic; (iii) G is Brody–hyperbolic (i.e. O(C, G) = C); (iii’) log G 1 contains no affine lines, and either Vj ∩G = ∅ or Vj ∩G is c–hyperbolic as a domain in Cn−1 , j = 1, . . . , n; (iv) there exist A = (Ajk )j=1,...,n, k=1,...,n ∈ Z(n × n), rank A = n, and a vector C = (C1 , . . . , Cn ) ∈ Rn such that 1

log G := {x ∈ Rn : (ex1 , . . . , exn ) ∈ G}.

2.2. Hyperbolicity for Reinhardt domains

107

• G ⊂ G(A, C) := DA1 ,C1 ∩ · · · ∩ DAn ,Cn , • either Vj ∩ G = ∅ or Vj ∩ G is c–hyperbolic as a domain in Cn−1 , j = 1, . . . , n; (iv’) there exist A ∈ Z(n × n), | det A| = 1, and a vector C ∈ Rn such that • G ⊂ G(A, C) (cf. (iv)), • either Vj ∩ G = ∅ or Vj ∩ G is c–hyperbolic as a domain in Cn−1 , j = 1, . . . , n; (v) G is algebraically equivalent to a bounded domain (i.e. there is a matrix A ∈ Z(n × n) such that ΦA is defined on G and gives a biholomorphic mapping from G to the bounded domain ΦA (G)); (vi) G is kG –complete. In the sequel a domain of the type G(A, C) (cf. (iv) in Theorem 2.2.1) will be shortly called a quasi-elementary Reinhardt domain. To prove Theorem 2.2.1 we need the following lemmas. Lemma 2.2.2 ([Zwo 1999a]). Let G(A, C) be as in Theorem 2.2.1. Then: e ∈ Z(n × n), | det A| e = 1, and a vector C e ∈ Rn such that (a) there is a matrix A e C); e G(A, C) ⊂ G(A, (b) cG(A,C) (z, w) > 0 for any points z, w ∈ G(A, C) ∩ Cn∗ , z 6= w. Proof. Fix a matrix A and a vector C as in Lemma 2.2.2. Step 1. To prove (a) it suffices to construct a sequence of quasi-elementary Reinhardt domains G0 := G(A, C) ⊂ · · · ⊂ GN such that | det Gj | < | det Gj−1 |, where det G(A, C) := det A. Assume that Gj has been already constructed. Let Gj = G(B, D) with a matrix B ∈ Z(n × n), | det B| ≥ 1, and a vector D ∈ Rn . In case when | det B| > 1 we describe how to get Gj+1 . Put • S(Gj ) := {α ∈ Zn : z α ∈ H∞ (Gj )}, • B(Gj ) := S(Gj ) \ (S(Gj ) + S(Gj )). It is known (cf. [J-P 1993], Lemma 2.7.6) that S := S(G(B, D)) = Zn ∩ (Q+ B 1 + · · · + Q+ B n ), B := B(G(B, D)) ⊂ Zn ∩ (Q ∩ [0, 1)B 1 + · · · + Q ∩ [0, 1)B n ) ∪ {B 1 , . . . , B n }. Claim: B 6⊂ {B 1 , . . . , B n }. Assume the contrary, i.e. B ⊂ {B 1 , . . . , B n }. Define r(B) := min{r ∈ N : if x ∈ Qn , xB ∈ Zn , then rx ∈ Zn }. Observe that B −1 B ∈ Z(n×n), i.e. all the rows of B −1 are special vectors in the definition of the number r(B). So r(B)B −1 ∈ Z(n × n), from which r(B)n = det(r(B)B −1 B) = det(r(B)B −1 ) det(B) follows. Therefore, if r(B) = 1 then | det B| = 1, which gives the contradiction. So it remains to prove that r(B) = 1. Take an arbitrary x ∈ Qn with xB ∈ Zn . We have to show that x ∈ Zn . In fact: we write xB = uB + νB, where u = (u1 , . . . , un ), uj := xj − [xj ] ≥ 0 and ν = (ν1 , . . . , νn ), νj := [xj ] ∈ Z, j = 1, . . . , n (here [x] denotes the largest integer smaller or equal x). Obviously, uB ∈ Zn . Applying the above description of S, it follows that

108


uB = u1 B 1 + · · · + un B n ∈ S (recall that B j is the j–th row of B). In virtue of the assumption, uB is an entire linear combination of the vectors B 1 , . . . , B n ; in particular, (recall that the B j ’s are linear independent) u ∈ Zn . Hence, x = u+ν ∈ Zn , i.e. r(B) = 1. So the claim is verified. Therefore, there is a β ∈ B \ {B 1 , . . . , B n } such that β = t1 B 1 + · · · + tn B n with tj ∈ [0, 1) and one of the tj ’s is positive. We may assume that t1 > 0. We denote by e that matrix whose rows B e j are given by B e 1 := β, B e j := B j , j = 2, . . . , n. Moreover, B Pn e e with C1 := j=1 tj Cj and Cj := Cj , j = 2, . . . , n, we put e C), e where C e := (C e1 , . . . , C en ). Gj+1 := G(B, e = t1 | det B| < | det B| and Gj ⊂ Gj+1 . Hence, (a) is verified. Then | det B| Step 2. Recall that for a matrix A ∈ Z(n × n) the mapping ΦA : Cn∗ −→ Cn∗ ,

1

n

ΦA (z) := (z A , . . . , z A ), z ∈ Cn∗ ,

is proper iff det A 6= 0, and that in this case its multiplicity is given by | det A|. In e of (a), is a biholomorphic mapping from Cn∗ to itself. particular, the mapping ΦAe, A Now fix two different points z, w ∈ G(A, C) ∩ Cn∗ . Then cG(A,C) (z, w) ≥ cG(A, n (z, w) = cE n (Ψ (z), Ψ (w)) > 0, e C) e (z, w) = cG(A, e C)∩C e ∗

e1 C

where Ψ (z) := (Φ1 (z)/e

en C

, . . . , Φn (z)/e

) with ΦAe =: (Φ1 , . . . , Φn ).

Lemma 2.2.3. Let Ω ⊂ Rn be a convex domain containing no straight lines. Then there are linearly independent vectors A1 , . . . , An ∈ Zn and a C ∈ Rn such that Ω ⊂ {x ∈ Rn : hx, Aj i < Cj , j = 1, . . . , n}. Proof. Cf. [Vla 1993].

Proof of Theorem 2.2.1. First, observe that the implications (i) =⇒ (ii) =⇒ (iii) are obvious and that (iv) =⇒ (iv’) is true due to Lemma 2.2.2. The remaining proof uses induction on the dimension n. Obviously, the theorem is true in the case n = 1. Now, let n ≥ 2. (iii) =⇒ (iii’): The first condition is an obvious consequence of (iii). The second one follows from the induction process. (iii’) =⇒ (iv): Note that the second condition in (iv) follows from applying the theorem in the case n − 1. From (iii) we see that log G does not contain straight lines. Therefore, we immediately get (iv) from Lemma 2.2.3. (iv’) =⇒ (i): Take z, w ∈ G, z 6= w. Case 1: If both points belong to Cn∗ , then, in virtue of Lemma 2.2.2, we have cG (z, w) ≥ cG(A,C) (z, w) > 0. w ∈ / Cn∗ . Without loss of generality we may assume that w = Case 2: Let z ∈ (w1 , . . . , wk , 0, . . . , 0) with w1 · · · wk 6= 0. Then k < n and Ajs ≥ 0, j = 1, . . . , n, s = k + 1, . . . , n. Since rank A = n we find a j ∈ {1, . . . , n} and an r ∈ {k + 1, . . . , n} such j j that Ajr > 0. Thus wA = 0 6= z A . Therefore, Cn∗ ,

j

j

cG (z, w) ≥ cG(Aj ,Cj ) (z, w) ≥ cD (z A , wA ) > 0, where D := eCj E.


109

Case 3: Letz, w ∈ / Cn∗ . We may assume that z1 = 0 and z2 6= w2 . Consequently, 2 π2,...,n (G) is c–hyperbolic and π2,...,n (z) 6= π2,...,n (w). Therefore, cG (z, w) ≥ cπ2,...,n (G) (π2,...,n (z), π2,...,n (w)) > 0. Hence G is c–hyperbolic. (iv’) =⇒ (v): By (iv’) we know that there is a matrix A ∈ Z(n × n), | det A| = 1, and a vector C ∈ Rn with G ⊂ G(A, C). Moreover, the mapping ΦA : Cn∗ −→ Cn∗ , 1 n ΦA (z) := (z A , . . . , z A ) is biholomorphic. Therefore, if the domain G is contained in Cn∗ , then ΦA : G → ΦA (G) is a biholomorphic mapping and ΦA (G) is bounded. The remaining case is done by induction: Obviously, the case n = 1 is clear. So we may assume that n ≥ 2 and, without loss of generality, that Vn ∩ G 6= ∅. Claim: It suffices to prove (v) under the additional assumption that Vn ∩ G 6= ∅ and πj (G) is bounded, j = 1 . . . , n − 1.

(2.2.1)

e := G ∩ Vn . By assumption, G e is a c-hyperbolic pseudoconvex Reinhardt In fact, put G e ∈ Z((n−1)×(n−1)) domain in Cn−1 . By the induction hypothesis there exists a matrix A such that e Φ e(G) e is bounded, and Φ e : G e −→ Φ e(G) e is biholomorphic. Φ e is defined on G, A

A

A

A

Put e 0 A B := ∈ Z(n × n). 0 1 Then ΦB satisfies condition (2.2.1), and so the claim has been verified. For the remaining part of the proof of (v) we may now assume that (2.2.1) is fulfilled. Without loss of generality assume that Vj ∩ G 6= ∅, j = 1, . . . , k,

Vj ∩ G = ∅, j = k + 1, . . . , n − 1.

e := V1 ∩ · · · ∩ Vk ∩ G. Then G e is a (non empty) c–hyperbolic pseudoconvex Put G e αn 6= 0. The fact Reinhardt domain. Then there is α = (0, . . . , 0, αk+1 , . . . , αn ) ∈ S(G), e that G ∩ Vn 6= ∅ implies αn > 0. Moreover, in virtue of (2.2.1), it is clear that ej := e (the number 1 in ej is at the j–th place), j = k +1, . . . , n−1. (0, . . . , 0, 1, 0, . . . , 0) ∈ S(G) Thus n−1 X h αj i αj 1 e ⊂ S(G). α e := α+ +1− ej ∈ S(G) αn αn αn j=k+1

Define

A :=

0 ...

In−1 0

α ek+1

...

α en−1

0 . 1

Then A fulfills all the required properties. Hence condition (v) is proved. (v) =⇒ (vi): By assumption we may assume that G is a bounded pseudoconvex Reinhardt domain. Fix a point w ∈ G. To verify that G is k–complete we only have to 2

πi1 ,...,ik (z1 , . . . , zn ) := (zi1 , . . . , zik ).

110


disprove the existence of a sequence (z j )j∈N ⊂ G such that (kG (w, z j ))j∈N is bounded, but z j −→ z 0 ∈ ∂G. j→∞

Case z 0 ∈ Cn∗ : We may assume that z 0 = (1, . . . , 1). It is clear that there is an α ∈ Rn , α 6= 0, such that G ⊂ Dα , where Dα denotes the elementary Reinhardt domain for α. Moreover, we may assume that αj 6= 0, j = 1, . . . , k, and αk+1 = . . . αn = 0, where k ≥ 1. So we get j kG (w, z j ) ≥ kDα (w, z j ) = max{kDαe (w, e zej ), kCn−k ((wk+1 , . . . , wn ), (zk+1 , . . . , znj ))}

= kDαe (w, e zej ), (2.2.2) where α e := (α1 , . . . , αk ), w e := (w1 , . . . , wk ), zej := (z1j , . . . , zkj ). Observe that the sequence (e z j )j∈N converges to the boundary point ze0 of Dαe . Then, applying Theorem 1.3.1, we see that the sequence in (2.2.2) tends to infinity. / Cn∗ : Let us assume that zj0 6= 0 for j = 1, . . . , k with a suitable k, Case z 0 ∈ 0 0 ≤ k < n, and zk+1 = · · · = zn0 = 0. We have to discuss two subcases: (a) There is an s ∈ {k + 1, . . . , n} such that G ∩ Vs = ∅. Then kG (w, z j ) ≥ kπs (G) (ws , zsj ). Here, πs (G) is a plane Reinhardt domain not containing the origin, but 0 ∈ ∂πs (G). Therefore, the right side tends to infinity. (b) All the intersections G ∩ Vj , j = k + 1, . . . , n, are non empty. Obviously, k > 0, otherwise z 0 = 0 ∈ G; contradiction. Then kG (w, z j ) ≥ kGe ((w1 , . . . , wk ), (z1j , . . . , zkj )), e := π1,...k (G). Since (z 0 , . . . , z 0 ) ∈ ∂ G e and G e is a Reinhardt domain of the first where G 1 k case, the right side again tends to infinity. Hence, the Kobayashi completeness of G has been verified. What remains is to mention is that (vi) trivially implies (iv). Remark 2.2.4. Observe that Theorem 2.2.1 shows that all notions of hyperbolicity coincide in the class of pseudoconvex Reinhardt domains. That’s why we will often speak only of hyperbolic pseudoconvex Reinhardt domains. Moreover, in that class “hyperbolic” and “Kobayashi-complete” are the same notions. Remark 2.2.5. The following pseudoconvex Reinhardt domain D := {z ∈ C3 : max{|z1 z2 |, |z1 z3 |, |z2 |, |z3 |} < 1} is not k–hyperbolic since C × {0} × {0} ⊂ D; in particular, D is not c–hyperbolic. e := D \ (C × {0} × {0}). Then D e is c–hyperbolic (the functions z1 z2 , z1 z3 , z2 , Let D e and z3 separate the points of D). Observe that D is the envelope of holomorphy of e e D, i.e. D = H(D). Hence, in general, c–hyperbolicity of a Reinhardt domain and its envelope of holomorphy may be different. But in the two-dimensional case, there is the following positive result [Die-Hai 2003]. Theorem 2.2.6. Let G ⊂ C2 be a c–hyperbolic Reinhardt domain. Then its envelope of holomorphy H(G) is c–hyperbolic.


111

Proof. Recall that the envelope of holomorphy H(D) of a Reinhardt domain D ⊂ Cn∗ satisfies the following properties • H(D) ⊂ Cn∗ , • log H(D) = conv(log D). Put G∗ := G ∩ C2∗ . G∗ is a Reinhardt domain. Assume that log H(G∗ ) contains an affine line `. Fix a point x0 ∈ log G \ `. Denote by `0 the line passing through x0 which is parallel to `. Then `0 ⊂ log H(G∗ ). Let `0 = {(a1 t + b1 , a2 t + b2 ) : t ∈ R}, where a1 , a2 , b1 , b2 ∈ R and a21 + a22 6= 0. Hence A := {(ea1 λ+b1 , ea2 λ+b2 ) : λ ∈ C} ⊂ H(G∗ ). Using Liouville’s theorem and the fact that G is c–hyperbolic, we get A ∩ G = ∅ or `0 ∩ log G = ∅; a contradiction. Assume now that log H(G) contains an affine line. As in the previous step, this leads to a non trivial entire map ϕ : C → H(G) ∩ Cn∗ . Recall that H(G∗ ) = H(G) ∩ Cn∗ (see Theorem 2.5.9 in [Jar-Pfl 2000]). Hence, H(G∗ ) contains an affine line; a contradiction. Without loss of generality, assume finally that H(G) ∩ V2 6= ∅. Denote this intersection by G0 ⊂ C. Suppose that G0 is not c–hyperbolic. Then either G0 = C or G0 = C∗ . Therefore, either A1 := C × {0} ⊂ H(G) or A2 := C∗ × {0} ⊂ H(G). In virtue of the c–hyperbolicity of G, we conclude that A1 ∩ G = ∅ or that A2 ∩ G = ∅. Therefore, G ∩ V2 = ∅; a contradiction. Thus Theorem 2.2.1 implies that H(G) is c–hyperbolic. We conclude this section with the following result which will be useful later. Proposition 2.2.7 ([Zwo 2000a]). Let G ⊂ Cn be a hyperbolic pseudoconvex Reinhardt domain. Then the following conditions are equivalent: (i) G is algebraically equivalent to an unbounded Reinhardt domain; (ii) G is algebraically equivalent to a bounded Reinhardt domain D, for which there is a j0 , 1 ≤ j0 ≤ n, such that D ∩ Vj0 6= ∅, but D ∩ Vj0 = ∅. Proof. (i) =⇒ (ii): We may assume that G is an unbounded hyperbolic pseudoconvex Reinhardt domain. In virtue of Theorem 2.2.1, there are a bounded Reinhardt domain D and a biholomorphic mapping ΦA : D −→ G (here we use the notation from Theorem 2.2.1). Suppose that D satisfies the following property: if D ∩ Vj 6= ∅ then D ∩ Vj 6= ∅,

j = 1, . . . , n.

Without loss of generality, we may assume that there is a k ∈ {0, 1, . . . , n} such that D ∩ Vj 6= ∅, j = 1, . . . , k, (Arj )r=1,...,n, j=1,...,n

D ∩ Vj = ∅, j = k + 1, . . . , n.

(2.2.3)

Arj

Now, let A = ∈ Z(n × n). Then ≥ 0, j = 1, . . . , k, r = 1, . . . , n. Moreover, using (2.2.3) and that D is bounded, gives a positive M such that |zj | ≥ M, Hence, sup{|z diction.

Ar

z ∈ D, k + 1 ≤ j ≤ n.

| : z ∈ D} < ∞, r = 1, . . . , n, which implies that G is bounded; contra-

112


(ii) =⇒ (i): Observe that the mapping D 3 z −→ (z1 , . . . , zj0 −1 ,

1 , zj +1 , . . . , zn ) zj0 0

e Thus maps D biholomorphically onto an unbounded pseudoconvex Reinhardt domain D. e and so is G. D is algebraically equivalent to D 2.3. Hyperbolicities for balanced domains Recall that any balanced domain is k–hyperbolic if and only if it is bounded (cf. Theorem 7.1.2 in [Jar-Pfl 1999a]). Example 2.3.1 ([Azu 1983], see also [J-P 1993], Example 7.1.4). Observe that there is an unbounded pseudoconvex balanced domain G ⊂ C2 , that is Brody–hyperbolic. To be more concrete G is defined as ( z1 |z2 |eϕ( z2 ) if z2 6= 0 2 G := {z ∈ C : h(z) < 1}, where h(z) := , |z1 | if z2 = 0 and

∞ n X 1 1 o ϕ(λ) := max log |λ|, log − λ , k2 k j=2

λ ∈ C.

Recently, S.-H. Park [Par 2003] has shown that G is almost e k–hyperbolic, i.e. e kG (z, w) > 0, whenever z1 6= w1 or (z1 = w1 6= 0 and z2 6= w2 ). ? It is still unclear what happens to e kG ((0, z2 ), (0, w2 )) ? Nevertheless, there is the following result (cf. [Par 2003]). Proposition 2.3.2. For any n ≥ 3 there exists a pseudoconvex balanced domain G ⊂ Cn such that • G is Brody–hyperbolic, • G is not e kG –hyperbolic. Proof. Obviously, it suffices to construct such an example G in C3 . Then, in the general case, G × E n−3 will do the job in Cn . √ So let n = 3. Put rj := ej , sj := 1/(rj2 + rj ), tj := j/sj , εj := 2−j−1 , and ηj := tj sj , j ∈ N. Then ∞ X j=1

εj = 1/2,

∞ X j=1

∞

εj log

X 1 1 ≥ εj log > −∞. ηj t j j=1

For j ∈ N define Qj (z) := z1 z2 − sj (z3 − z2 )(z3 − 2z2 ),

z = (z1 , z2 , z3 ) ∈ C3 .

Put G := {z ∈ C3 : h(z) < 1} with h(z) := max{|z1 |, |z2 |/2, h0 (z)},

2.3. Hyperbolicities for balanced domains

where h0 (z) :=

113

∞ ∞ X Y |Qj (z)| Qj (z) εj εj log . = exp ηj ηj j=1 j=1

We claim that G is a pseudoconvex balanced domain that is Brody–hyperbolic, but not e k–hyperbolic. Step 1. h is absolutely homogeneous and positive definite. It suffices to discuss h0 . Fix z ∈ C3 and λ ∈ C. Then: ∞ ∞ ∞ X X |Qj (z)| |Qj (z)| |Qj (λz)| X 2 = εj log(|λ| ) + εj log = log |λ| + εj log . εj log η η ηj j j j=1 j=1 j=1 j=1

∞ X

Hence, h0 (λz) = |λ|h0 (z). Assume now that h(z) = 0. Then z1 = z2 = 0 = h0 (z) which implies that −∞ =

∞ X

∞

εj log

j=1

|Qj (0, 0, z3 )| X 1 1 = εj log + log(|z3 |2 ), ηj tj 2 j=1

from which we obtain that z3 = 0. Hence, h is positively defined. Step 2. h0 ∈ PSH(C3 ) (in particular, G is pseudoconvex). Fix a positive R and let z ∈ (RE)3 . Then |Qj (z)| ≤ (1+6)R2 . Recall that ηj −→ ∞. Therefore, there is a jR such that |Qj (z)|/ηj < 1,

z ∈ (RE)3 , j ≥ jR .

So it follows that h0 ∈ PSH((RE)3 ) for arbitrary R. Hence, h0 ∈ PSH(C3 ). Step 3. G is not e k–hyperbolic. Let ϕj ∈ O(C, C3 ), ϕj (λ) := (sj λ(λ − 1), 1, λ + 1), j ∈ N. Observe that Qj ◦ ϕj = 0 on C, j ∈ N. Therefore, ϕj (λ) ∈ G if |λ| < rj . In particular, e kG (0, 1, 1), (0, 1, 2) = e kG ϕj (0), ϕj (1) ≤ kE (0, 1/rj ) −→ 0, j→∞

meaning that G is not e k–hyperbolic. Step 4. G is Brody–hyperbolic. Let f = (f1 , f2 , f3 ) ∈ O(C, G). In virtue of the form of G, fj is bounded and so fj ≡: aj , j = 1, 2. Suppose that f3 is not constant. Then, in virtue of Picard’s theorem, we have C \ {w} ⊂ f3 (C) for a suitable w ∈ C. Hence, h(a1 , a2 , ·) < 1 on C \ {w}. Using Liouville’s theorem for subharmonic functions, we conclude that h0 (a1 , a2 , ·) ≡ const. Note that h0 (a1 , a2 , λ) = 0 if Qj (a1 , a2 , λ) = 0 for at least one j. Therefore, h0 (a1 , a2 , ·) ≡ 0. To get a contradiction we discuss different cases of a1 , a2 . Case a2 = 0: Then Qj (a1 , 0, λ) = −sj λ2 , j ∈ N. Therefore, log h0 (a1 , 0, 1) =

∞ X j=1

contradiction.

∞

εj log

|Qj (a1 , 0, 1)| X 1 = εj log > −∞; ηj t j j=1

114


Case a2 6= 0, a1 = 0: Then Qj (0, a2 , 0) = −2sj a22 , j ∈ N. Therefore, log h0 (0, a2 , 0) =

∞ X

εj log

j=1

∞ X 1 + log(2|a2 |2 ) εj > −∞; tj j=1

contradiction. Case a1 a2 6= 0: Then log h0 (a1 , a2 , a2 ) =

∞ X

εj log

j=1

|a1 a2 | > −∞; ηj

contradiction. Hence, G is Brody–hyperbolic.

Remark 2.3.3. ? It remains an open question whether such an example does exist in C2 ? 2.4. Hyperbolicities for Hartogs type domains Let G ⊂ Cn be an arbitrary domain. A domain D = D(G) ⊂ G × Cm is called a Hartogs domain over G with m–dimensional balanced fibers if for any z ∈ G the fiber Dz := {w ∈ Cm : (z, w) ∈ D} is a non empty balanced domain in Cm . Recall that for such a D there exists exactly one upper semicontinuous function H : G × Cm −→ [0, ∞), H(z, λw) = |λ|H(z, w), z ∈ G, w ∈ Cm , λ ∈ C, such that DH = D = {(z, w) ∈ G × Cm : H(z, w) < 1}. Conversely, any such H leads to a Hartogs domain over G with m–dimensional balanced fibers. Recall that D = DH is pseudoconvex iff G is pseudoconvex and log H ∈ PSH(G × Cm ). Then we have the following hyperbolicity criterion (cf. [DDT-Tho 1998], see also [DDT-PVD 2000]). Theorem 2.4.1. Let D = DH ⊂ G × Cm be a Hartogs domain over G ⊂ Cn with m–dimensional balanced fibers. If D is k–hyperbolic, then G is k–hyperbolic and, for any compact set K ⊂ G, the function log H is bounded from below on K × ∂Bm . Proof. If D is k–hyperbolic then kG (z 0 , z 00 ) ≥ kD ((z 0 , 0), (z 00 , 0)) > 0 for all z 0 , z 00 ∈ G, z 0 6= z 00 . Hence G is k–hyperbolic. Assume now that there are two sequences (z j )j∈N ⊂ G, lim z j =: z 0 ∈ G, (wj )j∈N ⊂ ∂Bm , lim wj = w0 ∈ ∂Bm such that limj→∞ H(z j , wj ) = 0. We may assume that (z j , wj ) ∈ D, j ∈ N. Then ϕj ∈ O(C, G × Cm ), ϕj (λ) := (z j , λwj ), maps Rj E into D for a suitable sequence (Rj )j∈N with Rj −→ ∞. Therefore, j→∞

j

j

j

kD ((z , 0), (z , w )) = kD (ϕj (0), ϕj (1)) ≤ kE (0, 1/Rj ) −→ 0; hence, kD ((z 0 , 0), (z 0 , w0 )) = 0; contradiction.

2.4. Hyperbolicities for Hartogs type domains

115

Remark 2.4.2. ? It seems to be not known whether the converse of Theorem 2.4.1 also holds ? Nevertheless, the following special case is true (cf. [DDT-Tho 1998]). Proposition 2.4.3. Let G := E and u : E −→ [−∞, ∞) upper semicontinuous. Put H(z, w) := |w|eu(z) and D := DH . Assume that u is locally bounded from below. Then D is k–hyperbolic. Proof. In virtue of Theorem 7.2.2 in [J-P 1993], it suffices to show that the Kobayashi– Royden pseudometric is locally positive definite, i.e. for any point p0 = (z0 , w0 ) ∈ D there exist a neighborhood U = U (p0 ) ⊂ D and a positive number C such that κD (p; X) ≥ CkXk, p ∈ U , X ∈ C2 . First, observe that g(r) := − inf{u(λ) : λ ∈ E, |λ| ≤ r} < ∞,

r ∈ (0, 1).

Now, let s ∈ (0, 1) and fix (z0 , w0 ) ∈ D, |z0 | < s, and X ∈ C2 \ {0}. Let f ∈ O(E, D) with f (0) = (z0 , w0 ) and αf 0 (0) = X for α ∈ C∗ . In virtue of the Schwarz Lemma, we see that |f 0 (0)| ≤ 1 − |z0 |2 ≤ 1. Put r0 := 1+2s 2+s . Applying the Schwarz Lemma, it follows that, if |f1 (λ)| ≥ r0 , then z − f (λ) |f1 (λ)| − |z0 | r0 − |z0 | 1 0 1 |λ| ≥ ≥ ≥ . ≥ 1 − z 0 f1 (λ) 1 − |f1 (λ)||z0 | 1 − r0 |z0 | 2 Put Ω := {λ ∈ E : |f1 (λ)| < r0 }. Then sup kf2 | ≤ eg(r0 ) and B1 (0, 1/2) ⊂ Ω. Thus, Ω

|f20 (0)| ≤ 2eg(r0 ) . Then n |X2 | o 1 1 o √ ≥ min 1, kXk. 2eg(r0 ) 2eg(r0 ) 2 Since f was arbitrarily chosen we get n 1 1 o κD ((z, w); X) ≥ √ min 1, g(r ) kXk, (z, w) ∈ D, |z| < s. 2e 0 2 n |α| ≥ max |X1 |,

Hence, D is k–hyperbolic.

Remark 2.4.4. In Remark 2.2.5 we mentioned that, if a Reinhardt domain in C2 is c– hyperbolic, then its envelope of holomorphy is also c–hyperbolic. In the class of Hartogs domains and the case of k–hyperbolicity, such a conclusion is false even in dimension 2 (cf. [Die-Hai 2003]). Let u : [0, 1) −→ (−∞, 0) be continuous function satisfying limt%1 ϕ(t) = −∞. Put u(z1 ) := ϕ(|z1 |). Then the domain D := {z ∈ E × C : |z2 | < e−u(z1 ) } is k–hyperbolic (see Proposition 2.4.3). Recall that H(D) = {z ∈ E × C : |z2 | < e−bu(z1 ) }, where u b is the largest subharmonic minorant of u. In virtue of the maximum principle for subharmonic function, it is clear that u b ≡ −∞. Therefore, H(D) = E × C which is not k–hyperbolic.

116


Remark 2.4.5. So far, we discussed hyperbolicity. We close this part by a remark on the opposite situation. There is the following result due to E. Fornæss and N. Sibony [For-Sib 1981]: Let D ⊂ C2 be a domain which can be monotonically exhausted by domains Dj , where each of the Dj is biholomorphically equivalent to B2 . If κD 6≡ 0, then D is biholomorphically equivalent either to B2 or to E × C. Observe that B2 (0, j) % C2 , κC2 ≡ 0, but, obviously, C2 is neither biholomorphic to B2 nor to E × C. It turned out that there is a domain D ⊂ Cn , n ≥ 2, Dj % D, each Dj is biholomorphically equivalent to Bn , such that • κD ≡ 0, • ∃u∈PSH(Cn ) : D = {z ∈ Cn : u(z) < 0} and u|D 6≡ const; in particular, D is not biholomorphic to Cn . Domains of that type are called short Cn ’s (see [For 2001]). The domain D is obtained in the following way: Let d ∈ N, d ≥ 2, and η > 0. Denote by Autd,η the set of all polynomial automorphisms Φ of Cn of the form Φ(z) = Φ(z1 , . . . , zn ) = (z1d + P1 (z), P2 (z), . . . , Pn (z)), where deg Pj ≤ d − 1, j = 1, . . . , n − 1, and where each coefficient of the polynomials Pj has a modulus at most η. Choosing sufficiently good sequences aj & 0, aj ∈ (0, 1), and j Fj ∈ Autd,ηj where ηj := adj , j ∈ N, gives D as D = {z ∈ Cn : lim Fk ◦ · · · ◦ F1 (z) = 0}. k→∞

2.5. c–completeness for Reinhardt domains In this chapter Carathéodory completeness for Reinhardt domains will we discussed. Recall that a domain G ⊂ Cn is called to be cG –complete (shortly, c–complete) (respectively, cG –finitely compact (shortly, c–finitely compact)) if cG is a distance and if any cG –Cauchy sequence does converge to a point in G (in the standard topology) (respectively, if cG is a distance and if any cG –ball with a finite radius is a relatively compact subset of G). Moreover, recall that any cG –complete domain G is pseudoconvex. Theorem 2.5.1. Let G ⊂ Cn be a pseudoconvex Reinhardt domain. Then the following conditions are equivalent: (i) G is cG –finitely compact; (ii) G is cG –complete; P∞ (iii) there is no sequence (zν )ν∈N ⊂ G with ν=1 gG (zν , zν+1 ) < ∞; (iv) G is bounded and fulfills the following so called Fu–condition: if G ∩ Vj 6= ∅, then G ∩ Vj 6= ∅,

(2.5.4)

where Vj := {z ∈ Cn : zj = 0}. This result is due W. Zwonek ([Zwo 2000a], see also [Zwo 2000b]); earlier partial results can be found in [Pfl 1984] (see also [J-P 1993]) and [Fu 1994]. For the proof of Theorem 2.5.1 we shall need the following three lemmas. Lemma 2.5.2 ([Zwo 2000a]). Let G ⊂ Cn∗ be a pseudoconvex Reinhardt domain. Then kG = e kG . In particular, the Lempert function e kG is continuous on G × G.

2.5. c–completeness for Reinhardt domains

117

Proof. Observe that T := log G is a convex domain in Rn and that the mapping Φ

T + iRn 3 z −→ (ez1 , . . . , ezn ) ∈ G is a holomorphic covering. Therefore, for z, w ∈ G we have e kG (z, w) = inf{e kT +iRn (e z , w) e : ze, w e ∈ T + iRn with Φ(e z ) = z, Φ(w) e = w} = inf{kT +iRn (e z , w) e : ze, w e ∈ T + iRn with Φ(e z ) = z, Φ(w) e = w} = kG (z, w). Here we have used the theorem of Lempert.

Lemma 2.5.3. Let Ω ⊂ Rn be a unbounded convex domain which is contained in n X (−∞, R) for a certain number R. Then, for any point a ∈ Ω there exist a vector

j=1

v ∈ Rn− \ {0} and a neighborhood V = V (a) ⊂ Ω such that V + R+ v ⊂ Ω. Proof. Take w.l.o.g. the point a = 0. Then the continuity of the Minkowski function h of Ω and the assumptions on Ω lead to a vector v on the unit sphere with h(v) = 0. Obviously, v ∈ Rn− \ {0} and R+ v ⊂ Ω. Finally, using the convexity of Ω, we see that for any open ball V ⊂ Ω with center a the following inclusion holds: V + R+ v ⊂ Ω. Lemma 2.5.4 ([Hay-Ken 1976]). Let H := {λ ∈ C : Re λ < 0}, b < 0, and M < 0. Moreover, let u ∈ SH(H), u < 0, and u(λ) ≤ M for all λ with Re λ = b. Then u ≤ M on {λ ∈ C Re λ ≤ b}. Now we are in the position to proceed with the proof of the above theorem. Proof of Theorem 2.5.1. Observe that the following two implications (i) =⇒ (ii) =⇒ (iii) are obvious. Moreover, the proof of (iv) =⇒ (i) follows along the same lines as the one of Theorem 7.4.6 in [J-P 1993]. Therefore, we shall prove only (iii) =⇒ (iv): Suppose that this implication is false. Then, in virtue of Proposition 2.2.7, we may assume that G is bounded and doesn’t fulfill the Fu–condition (2.5.4). Moreover, without loss of generality we only have to deal with the following situation: G ∩ Vj 6= ∅, but G ∩ Vj = ∅, j = 1, . . . , k, G ∩ Vj = ∅, j = k + 1, . . . , n,

1 ≤ k ≤ n.

In fact, if G ∩ Vj 6= ∅, then one can go to the intersection of G with those coordinate axes. Hence G ⊂ Cn∗ . We may also assume that (1, . . . , 1) ∈ G. Observe that log G is convex, bounded into all positive directions, unbounded into the first k negative directions, and bounded in the remaining negative directions. Thus, in virtue of Lemma 2.5.3 we find a small ball V = V (0) ⊂ log G with center 0 and a vector v ∈ Rn− \{0} such that V +R+ v ⊂ log G. It is clear that vj = 0, j = k +1, . . . , n. Without loss of generality, we may assume that vj < 0, j = 1, . . . , `, where ` ≤ k, v1 = −1, and v`+1 = . . . vn = 0. Hence, (ex1 e−t , ex2 etv2 , . . . , exn etvn ) ∈ G,

t > 0, x ∈ V.

118


Then, with α := −v, we have an ε > 0 such that (eλ , µ2 eλα2 , . . . , µ` eλα` , 1, . . . , 1) ∈ G,

λ ∈ H, e−ε < |µj | < eε , j = 2, . . . , `.

Put A := {(µ2 , . . . , µ` ) ∈ C`−1 : e−ε < |µj | < eε , j = 2, . . . , `}, HR := {λ ∈ C : Re λ < R}, R ≥ 0, Φ : C × A −→ C` ,

Φ(λ, µ) := (eλ , µ2 eλα2 , . . . , µ` eλα` ).

It is clear that Φ(HR × A) =: DR ⊂ C`∗ , R ≥ 0, is a pseudoconvex Reinhardt domain with DR % D∞ := Φ(C × A) ⊂ C`∗ . Therefore, R→∞

e kD∞ (Φ(−1, 1 . . . , 1), Φ(λ, 1, . . . , 1)) ≤ e kC (−1, λ) = 0, λ ∈ C. In virtue of Lemma 2.5.3, e kD∞ (Φ(−1, 1, . . . , 1), z) = 0 for all z ∈ D∞ ∩ M , where M := Φ(C × {(1, . . . , 1)}). Observe that D0 ×{(1,P . . . , 1)} ⊂ G, but (0, . . . , 0, 1, . . . , 1) ∈ / G. Now choose positive ∞ numbers aj , j ∈ N, with j=1 aj < ∞. It suffices to find points z j ∈ D0 , j ∈ N, with limj→∞ z1j = 0 such that gG ((z j , 1, . . . , 1), (z j+1 , 1, . . . , 1)) ≤ gD0 (z j , z j+1 ) ≤ aj , j ∈ N. Applying that e kDR is continuous on GR × GR , the theorem of Dini, and that lim e kDR (Φ(−1, 1, . . . , 1), z) = e kD∞ (Φ(−1, 1, . . . , 1), z) = 0,

R→∞

z ∈ D∞ ∩ Φ(C × {(1, . . . , 1)}), e−2 < |z1 | < e−1 , we conclude that this convergence is a uniform one. Hence we have a sequence (Rj )j∈N , limj→∞ Rj = ∞, such that ∗ e kD (Φ(−1, 1, . . . , 1), Φ(λ, 1, . . . , 1)) < aj , R j

−2 ≤ Re λ ≤ −1.

Observe that the mapping ψR : D0 −→ DR , ψ(z) := (eR z1 , z2 eα2 R , . . . , z` eα` R ), is biholomorphic. Therefore, ∗ e kD (Φ(−1 − Rj , 1 . . . , 1), Φ(λ, 1, . . . , 1)) < aj , 0

−2 − Rj ≤ Re λ ≤ −1 − Rj .

Define uj (λ) := log gD0 (Φ(−1 − Rj , 1, . . . , 1), Φ(λ, 1, . . . , 1)),

λ ∈ H0 .

Observe that u ∈ SH(H0 ). In virtue of Lemma 2.5.4 it follows that uj (λ) < log aj whenever Re λ ≤ −1 − Rj . Therefore, we may take z j := Φ(−1 − Rj , 1, . . . , 1) as the desired point-sequence. Remark 2.5.5. Obviously, any cG –finitely compact domain G is cG –complete. We point out that the converse (due to N. Sibony and M.A. Selby) is also known for domains in the plane (see [J-P 1993], Theorem 7.4.7). ? Whether the two notions for the c– completeness do coincide for all bounded domains is still unknown ? We only mention


119

that there is a one-dimensional complex space X that is cX –complete but not cX –finitely compact (see [Jar-Pfl-Vig 1993]). Remark 2.5.6. Let G ⊂ C2 be a bounded pseudoconvex Reinhardt domain, a ∈ G, and z 0 ∈ ∂G ∩ C2∗ . Then cG (a, z) −→0 ∞ (see [Zwo 2000b]). So that part of ∂G not lying on z→z

an coordinate axis is cG –infinitely far away from any point of G. We point out that this phenomenon remains not true in higher dimensions. Example 2.5.7 (cf. [Zwo 2000b]). Let α > 0 be an irrational number. Put G := {z ∈ C3 : |z1 ||z2 |α |z3 |α+1 < 1, |z2 ||z3 | < 1, |z3 | < 1}. Then G is a pseudoconvex Reinhardt domain. Fixing points z 0 ∈ G ∩ C3∗ and w ∈ C3∗ with |w1 ||w2 |α |z3 |α+1 = 1, |w2 |−1 |w3 |2 < 1, and |w3 | < 1, we get lim sup cG (z 0 , z) < ∞. G3z→w

Moreover, the biholomorphic map Φ : G ∩ C3∗ −→ C3∗ ,

[α]+1 [α]+3 z3 , z2 z32 , z3 ), ∗

Φ(z) := (z1 z2

has as its image a bounded pseudoconvex Reinhardt G domain contained in {z ∈ C3∗ : |z2 | < 1, |z3 | < 1, |z1 ||z2 |α−[α]−1 |z3 |[α]−α < 1}. In the class of Reinhardt domains we have the following characterization of hyperconvexity (cf. [Zwo 2000a] and [Car-Ceg-Wik 1999]). Theorem 2.5.8. Let G ⊂ Cn be a pseudoconvex Reinhardt domain. Then the following conditions are equivalent: (i) G is hyperconvex; (ii) G is bounded and fulfills the Fu–condition. Proof. The direction (ii) =⇒ (i) follows directly from Theorem 2.5.1(i). To prove the converse, suppose that G doesn’t fulfill the conditions in (ii). According to Proposition 2.2.7, we may assume that G is bounded and doesn’t fulfill the Fu condition. Hence, without loss of generality, we may assume that G = D0 (compare the proof of Theorem 2.5.1), i.e. G := {(ζ, µ2 ζ α2 , . . . , µn ζ αn ) ∈ Cn : ζ ∈ E∗ , µj ∈ C, e−ε0 < |µj | < eε0 , j = 2, . . . , n}, where ε0 > 0, αj > 0, j = 2, . . . , n. Let u ∈ PSH(G) ∩ C(G), u < 0, be such that {z ∈ G : u(z) < −ε} b G for any ε > 0. Define v(z) := sup{u(z1 eiθ1 , . . . , zn eiθn ) : θj ∈ R}. Obviously, v is an exhausting function of G with v(z) = v(|z1 |, . . . , |zn |). Therefore, the function E∗ 3 λ −→ v(|λ|, |λ|α2 , . . . , |λ|αn ) is subharmonic and bounded from above by 0. Hence it can be continued as a function v ∗ ∈ SH(E). Then, in virtue of the hyperconvexity of G, it follows that v ∗ (0) = 0 implying that v = 0 — contradiction.

120


Recall that the Carathéodory distance is not inner. So, in general, we have cG ≤ ciG , R R (1) R (k) cG 6= ciG (cf. §1.2.1). Moreover, it is known that ciG = γG = γG ≤ γG , k ∈ N. Thus, R (k) cG –complete =⇒ ciG –complete =⇒ γG –complete, k ∈ N. (A domain G is called δG –complete (shortly, δ–complete) if δG is a distance and any R δG –Cauchy sequence in G does converge to a point in G, δ ∈ {ci , γ (k) , k ∈ N}. See [J-P 1993] for more details.) In fact there is the following result (see [Zwo 2001b], [Zap 2003]). Theorem 2.5.9. Let G ⊂ Cn be a bounded pseudoconvex Reinhardt domain. Then the following properties are equivalent: (i) G is cG –complete; (ii) G is ciG –complete; R (k) (iii) G is γG –complete, k ∈ N; R (k) (iv) there is a k ∈ N such that G is γG –complete. In order to be able to prove Theorem 2.5.9, we first recall a fact on multi-dimensional Vandermonde’s determinants (for example, see [Sic 1962]), namely: Let Xs := (s, . . . , s) ∈ Cn 3 , s ∈ N, and Nk := #{α ∈ Zn+ : |α| ≤ k}, k ∈ N. Then det (Xsα )1≤s≤Nk , |α|≤k 6= 0. (2.5.5) Using this information we get the following P Lemma 2.5.10. Let P (z) = 1≤|β|≤k bβ z β , z ∈ Cn , be a polynomial in Cn . Then there are numbers (Nj )1≤j≤k ⊂ N, (cj,s )1≤j≤k, 1≤s≤Nj ⊂ C, and vectors (Xj,s )1≤j≤k, 1≤s≤Nj ⊂ Cn such that Nj k X X X j! β pβ (z)Xj,s , z ∈ Cn , P (z) = cj,s β! j=1 s=1 |β|=j

where pβ (z) :=

n Y

pβ,j (z),

pβ,j (z) := zj (zj − 1) · · · (zj − βj + 1),

z = (z1 , . . . , zn ) ∈ Cn .

j=1

Proof. The proof is by induction on k ∈ N. Obviously, the case k = 1 is true. So we may Lemma 2.5.10 holds for a k ∈ N. Now take a polynomial P (z) = P assume that β n 1≤|β|≤k+1 bβ z , z ∈ C , and write X X X P (z) = bβ z β + bβ (z β − pβ (z)) + bβ pβ (z), z ∈ Cn . 1≤|β|≤k

|β|=k+1

|β|=k+1

Observe that the first two terms are of degree less than or equal to k. The third one may be written as X X (k + 1)! β!bβ bβ pβ (z) = pβ (z) . β! (k + 1)! |β|=k+1

3

|β|=k+1

Notice that here Xs is a vector and not the s-th coordinate of a vector.


121

Using (2.5.5), we find (cs )1≤s≤Nk+1 ⊂ C and (Xs )1≤s≤Nk+1 ⊂ Cn such that Nk+1 X β!bβ = cs Xsβ , (k + 1)! s=1

|β| = k + 1.

Hence, Lemma 2.5.10 has been proved.

Proof of Theorem 2.5.9. It remains to prove (iv) =⇒ (i). We may assume that n ≥ 2. Suppose that G doesn’t fulfill the Fu–condition, but R (k) is γG –complete for a suitable k. According to the proof of Theorem 2.5.1, we may assume that G ⊂ Cn∗ and that log G = {0} × (log δ, − log δ)n−1 + R>0 v, where δ ∈ (0, 1), v ∈ (−∞, 0)n , and v1 = −1. Put γ := −v. Observe that a monomial z α (α ∈ Zn ) is bounded on G, if and only if hα, γi ≥ 0. Put χ : (0, 1) −→ G, χ(t) := (tγ1 , . . . , tγn ). (k)

For a fixed t ∈ (0, 1), we are going to estimate γG (χ(t); χ0 (t)). Fix an f ∈ O(G, E), ordχ(t) f ≥ k. Then, using Laurent expansion, we get Z X 1 f (ζ)dζ1 . . . dζn α f (z) = aα z , where aα = (2πi)n |ζ1 |=r1 ,...,|ζn |=rn ζ α+1 is independent of r = (r1 , . . . , rn ) ∈ G ∩ Rn>0 . Note that 1 for any r ∈ G. rα From (2.5.6) it follows that aα = 0 if hα, γi < 0. Therefore, X f (z) = aα z α , z ∈ G. |aα | ≤

(2.5.6)

α∈Zn : hα,γi≥0

Taking r1 < 1 in (2.5.6) arbitrarily large and rj arbitrarily close to δ (or to δ −1 ), j = 2, . . . , n, then |aα | ≤ δ |α2 |+···+|αn | . Taking derivatives we have k k 1 (s) f (χ(t))(X1 tγ1 − s , . . . , Xn tγn − s ) s! X X 1 = aα pβ (α)X β thα,γi−k , β!

hα,γi≥0

|β|=s

Since ordχ(t) f ≥ k, it follows that X X 1 aα thα,γi−k pβ (α)X β = 0, β! hα,γi≥0

s ∈ N, X = (X1 , . . . , Xn ) ∈ Cn .

|β|=s

0 ≤ s < k, X ∈ Cn .

(2.5.7)

122


Moreover, 1 (k) f (χ(t))(χ0 (t)) k! 1 X 1 = aα hα, γik thα,γi−k + k! k! hα,γi≥0

X

aα thα,γi−k

hα,γi≥0

X k! (pβ (α) − αβ )γ β . (2.5.8) β!

|β|=k

Applying Lemma 2.5.10 and (2.5.7) shows that the second term in (2.5.8) vanishes. Hence, we have 1 (k) 1 X f (χ(t))(χ0 (t)) = aα hα, γik thα,γi−k . k! k! hα,γi≥0

In virtue of the above estimate, we obtain the following inequality X 1 (k) δ (|α2 |+···+|αn |)/k hα, γit(hα,γi/k)−1 , γG (χ(t)); χ0 (t)) ≤ √ k k! α∈Zn :hα,γi>0

t ∈ (0, 1).

What remains to show is that Lγ (k) (χ|(0,1/2] ) is finite, which would give the desired G contradiction. So we have the following estimate: Z Lγ (k) (χ|(0,1/2] ) ≤ G

0

1/2

(k)

γG (χ(t); χ0 (t))dt

Z 1/2 X 1 (|α2 |+···+|αn |)/k δ hα, γit(hα,γi/k)−1 dt ≤ √ k k! hα,γi≥0 0 X 1 k = √ δ (|α2 |+···+|αn |)/k hα,γi/k k 2 k! hα,γi>0 X k = √ δ (|α2 |+···+|αn |)/k k k! α2 ,...,αn ∈Z α

X

0 0 1 ∈Z:α1 >−hα ,γ i

1 2hα,γi/k

k 1 ≤ √ δ (|α2 |+···+|αn |)/k (hα0 ,γ 0 i+[−hα0 ,γ 0 i])/k , k 2 k! α2 ,...,αn ∈Z X

where α0 := (α2 , . . . , αn ), γ 0 := (γ2 , . . . , γn ). Obviously, the last number is finite, which finishes the proof. Remark 2.5.11. Observe that in the case γ ∈ Qn the above proof may be essentially simplified. Namely, then the punctured unit disc can be embedded into G. So the R (k) non– γG –completeness of G follows immediately from the one of E∗ . Remark 2.5.12. Let G ⊂ Cn be an arbitrary domain and A ⊂ G finite. In generalization of the notion of cG –finite compactness we say that G is mG (A, ·)–finitely compact if for any R > 0 the set {z ∈ G : mG (A, z) < R} is relatively compact in G. Obviously, any

R (k) 2.7. γG –completeness for Zalcman domains

123

mG (A, ·)–finitely compact domain is cG –finitely compact. ? Is there a geometrical characterization for mG (A, ·)–finite compactness in the class of all pseudoconvex Reinhardt domains as in Theorem 2.5.1 ? 2.6. c–completeness for complete circular domains Let G ⊂ Cn be a bounded pseudoconvex balanced (:= complete circular) domain. Then there is an h = hG ∈ PSH(Cn ) with h(λz) = |λ|h(z) (λ ∈ C, z ∈ Cn ) such that G = Gh = {z ∈ Cn : h(z) < 1} and, since G is pseudoconvex, log h ∈ PSH(Cn ) (see Chapter 1). It is known (due to T. Barth) that h is continuous, whenever G is kG –complete. In dimensions larger than 2 the converse statement becomes false; in fact there is a counterexample due to Jarnicki–Pflug ([J-P 1993], Theorem 7.5.7). In particular, this example is not c–complete. ? It is still open which conditions on h may imply that G = Gh is c–complete. Moreover, in dimension 2, so far it is not known whether the continuity of h implies the Kobayashi completeness or even the Carathéodory completeness ? 2.7.

R

(k)

γG –completeness for Zalcman domains

First, we introduce the class of domains we like to study. Let (aj )j∈N and (rj )j∈N be sequences of positive real numbers such that: • 2rj < aj , j ∈ N, • aj & 0, j→∞

• B(aj , rj ) ⊂ E, B(aj , rj ) ∩ B(ak , rk ) = ∅, j 6= k. S∞ Then G := E∗ \ j=1 B(aj , rj ) is called a Zalcman type domain. The main result here is the following one due to P. Zapalowski (see [Zap 2002] and [Zap 2004]). R (`) Theorem 2.7.1. For any k ∈ N there exists a Zalcman type domain G which is γG – R (m) complete, but not γG –complete, whenever m ≤ k < `. Remark 2.7.2. ? It seems to be an open problem whether for different k, l ∈ N, k < l, R (k) R (l) there exists a Zalcman type domain G, which is γG –complete, but not γG –complete R (k) R (sk) ? Note that for l = sk, s ∈ N, it is impossible because of γG ≤ γG . Before giving the proof of TheoremR2.7.1 we mention the following sufficient condition for a Zalcman type domain to be not γ (k) –complete. Proposition 2.7.3. Let G ⊂ C be a Zalcman type domain (as above) and let k ∈ N, α ∈ (0, 1), and c > 0. Assume that (k)

γG (t; 1) ≤ c|t|−α ,

t ∈ (−1, 0).

(2.7.9)

124


Then G is not

R

(`)

γG –complete for any ` ≥ k.

Proof. Fix ` ∈ N, ` ≥ k, and a point t ∈ (−1, 0). Take an f ∈ O(G, E), f (t) = f 0 (t) = (`) · · · = f (`−1) (t) = 0, with (γG (t; 1))` = `!1 |f (`) (t)|. We define ( f (z) if z 6= t `−k g(z) = (z−t) . 0, if z = t Then g is holomorphic with g (m) (t) = 0,

m = 0, . . . , k − 1, and g (k) (t) =

k! (`) f (t). `!

Moreover, in virtue of the maximum principle, we have kgkG ≤ dist(t, ∂G)−(`−k) . Therefore, h := g dist(t, ∂G)`−k ∈ O(G, E) and we obtain

(k)

γG (t; 1)

k

≥

` 1 (k) dist(t, ∂G)`−k (`) (`) |h (t)| = |f (t)| = dist(t, ∂G)(`−k) γG (t; 1) . k! `! (2.7.10)

Finally, from the assumed inequality (2.7.9) the following estimate follows (`)

γG (t; 1) ≤

0 ck/` |t|−αk/` = c0 |t|−α , (`−k)/` |t|

where c0 := ck/` and α0 := (αk + (` − k))/` < 1. Then integrating along the segment R (`) (−1/2, 0) shows that G is not γG –complete. Consequently, to find examples as claimed in Theorem 2.7.1 we should try to deal (k) with situations where the boundary behavior of γG is of the following type (k)

γG (·; 1) ≤ c dist(·, ∂G)−1 | log dist(·, ∂G)|−α with some α > 1, c > 0. Lemma 2.7.4. Let G ⊂ C be a Zalcman type domain and k ∈ N. Then there exists a C > 0 such that ∞ X rj |f (k) (z)| ≤ C 1 + , z ∈ (− 12 , 0), f ∈ O(G, E). k+1 (a − z) j j=1 Proof. Choose numbers e aj ∈ (0, aj ) and rej ∈ (e aj , 1) such that B(as , rs ) ⊂ B(e aj , rej ), s > j, and B(e aj , rej ) ∩ B(aj , rj ) = ∅. Put j [ Gj := E \ B(e aj , rej ) ∪ B(as , rs ) . s=1


125

Gj is a (j + 2)–connected domain with Gj ⊂ G, j ∈ N. Then, for a sufficiently small positive εj (we may assume that εj −→ 0), we have j→∞

j [ aj , rej + εj ) ∪ B(as , rs + εj ) b Gj . Gj,εj := (1 − εj )E \ B(e s=1

In virtue of the Cauchy integral, we see that Z Z f (ζ) k! f (ζ) k! dζ − dζ f (k) (z) = 2πi |ζ|=1−εj (ζ − z)k+1 2πi |ζ−eaj |=erj +εj (ζ − z)k+1 Z j X k! f (ζ) − dζ, z ∈ Gj,εj , f ∈ O(G, E). 2πi (ζ − z)k+1 |ζ−as |=rs +εj s=1 p Let z ∈ (−1/2, 0). Then z ∈ (−1/2, e aj − rej − εj − 2(k+1) rej + εj ) and z < −εj for all sufficiently large j. Hence we obtain Z Z k! 2π 1 − εj k! 2π rej + εj (k) |f (z)| ≤ dt + dt 2π 0 |(1 − εj )eit − z|k+1 2π 0 |(e rj + εj )eit + e aj − z|k+1 Z j X rs + ε j k! 2π dt + it + a − z|k+1 2π |(r + ε )e s j s 0 s=1 j

≤ k!

X rej + εj rs + ε j 1 − εj p + + (1/2 − εj )k+1 ( 2(k+1) rej + εj )k+1 s=1 (1/2(as − z − εj ))k+1

Observe that here the assumption 2rj < aj , j ∈ N, is used to estimate the third term. Since εj −→ 0, we finally receive the following inequality j X p |f (k) (z)| ≤ k! 2k+1 + rej + 2k+1 s=1

rs . (as − z)k+1

Recall that rej −→ 0. Therefore, letting j −→ ∞, we obtain j→∞

∞ X |f (k) (z)| ≤ k!2k+1 1 + s=1

rs . k+1 (as − z)

Lemma 2.7.5. For every k ∈ N there are a e k ∈ N and a Zalcman type domain G such that R (m) e (a) lim sup ( γG )(−1/2k−1 , z) < ∞, 1 ≤ m ≤ k, (−1,0)3z→0

(b)

R (`) lim ( γG )(w, z)

G3z→0

= ∞,

w ∈ G, k < `.

Observe that Lemma 2.7.5 implies immediately Theorem 2.7.1.

126


Proof of Theorem 2.7.1. Fix a k ∈ N and take the Zalcman type domain from Lemma 2.7.5. Let m ∈ N, 1 ≤ m ≤ k. As a direct consequence of (a) and the fact that the R (m) R (m) γ –completeness is equivalent to the γG –finite compactness we see that G is not R G (m) γG –complete. R (`) It remains to see that G is γG –complete whenever ` > k. So let us fix such an `, a R (`) point w ∈ G, and a boundary point z 0 ∈ ∂G. We have to show that lim0 ( γG )(w, z) = z→z ∞. Case 1: If z 0 = 0, then using (b) we are done. Case 2: If |z 0 | = 1, it follows that R (`) R (`) lim0 ( γG )(w, z) ≥ lim0 ( γE )(w, z) = lim0 cE (w, z) = ∞. z→z

z→z

z→z

0

Case 3: If z ∈ ∂B(aj , rj ) for some j, then R (`) lim0 ( γG )(w, z) ≥ lim0 cG (w, z) ≥ lim0 cC\B(aj ,rj ) (w, z) z→z z→z z→z r r rj rj j j = lim0 cE∗ , = lim0 cE , = ∞, w − aj z − aj w − aj z − aj z→z z→z since |rj /(z − aj )| −→ 1 as z → z 0 .

What remains is the proof of Lemma 2.7.5. Proof of Lemma 2.7.5. Let k ∈ N, aj := 2−j , and rk,j := 2−j j −k−1 , j ∈ N. Since 2 1 2 1 s 1 s √ √ lims→∞ s−1 = √ < 1, we may choose a e k ∈ N such that s−1 < 1, s ≥ e k. k k k 2 2 2 Put [ Gk := E∗ \ B(aj , rk,j ). j≥e k

Obviously, Gk is a Zalcman type domain. To prove (a) it suffices to verify the following inequality: c (m) 1 ∃c=c(k)>0 : γGk (z; 1) ≤ z ∈ [− k−1 , 0), m ≤ k. k+1 , 2e −z(− log(−z)) m

(2.7.11)

1 , 0). Then there exist a unique N ∈ N, N ≥ e k, and a b ∈ (1, 2] In fact, let z ∈ [− k−1 2e N such that z = −b/2 . Therefore, N X j=e k

N N ∞ X X rk,j rk,j 2jm 2N m X j 1 2N m ≤ = ≤ δ ≤ , m+1 m+1 k+1 k+1 (aj − z) j N 1 − δ N k+1 aj j=0 j=e k

j=e k

(2.7.12) ∞ X j=N

rk,j ≤ (aj − z)m+1

∞ X j=N

rk,j = (−z)m+1

∞ X j=N

2N (m+1) 2j j k+1 bm+1

≤

∞ N (m+1) X

2 2N N k+1

j=0

1 2N m+1 ≤ . j 2 N k+1 (2.7.13)

(The second inequality in (2.7.12) follows easily from the observation that there is a 2(s−1)m 2sm e positive δ < 1 such that (s−1) k+1 ≤ δ sk+1 , s ≥ k, m ≤ k.)


We put b c := ∞ X j=e k

1 1−δ .

127

Using (2.7.12) and (2.7.13) we get

rk,j (b c + 2)2k (log 2)k+1 2N m C1 ≤ =: . (aj − z)m+1 bm (log(2m /b))k+1 (−z)m (− log(−z))k+1

In virtue of Lemma 2.7.4, we obtain 2CC1 C1 |f (m) (z)| ≤ C 1 + ≤ , m k+1 m (−z) (− log(−z)) (−z) (− log(−z))k+1

f ∈ O(G, E),

which finally proves (2.7.11). (Note that we may take C = k!2k+1 ≥ m!2m+1 , m ≤ k; thus the constant C = C(k) from Lemma 2.7.4 works for all m, m ≤ k.) To prove (b) we claim that (`)

∀`>k ∃c=c(k,`)>0 : γGk (z; 1) ≥

c , |z| log(1/|z|)

|z|
k, and a w ∈ Gk . Take 1 , and a C 1 –curve α : [0, 1] −→ Gk connecting z with w. Then we a z ∈ Gk , |z| < k−2 2e have Z 1 Z tα Z tα d |α0 (t)|dt (`) 0 dt |α(t)|dt γGk (α(t); α (t))dt ≥ c ≥c |α(t)| log(1/|α(t)|) |α(t)| log(1/|α(t)|) 0 0 0 Z tα d 1 e ≥c − log log(1/|α(t)|) dt = c(log log − log log 2k−2 ), dt |z| 0 1 where tα := sup{t ∈ [0, 1] : |α(τ )| < k−2 , 0 ≤ τ ≤ t}. 2e Since the curve α was an arbitrary one connecting z and w in Gk , it follows that

R

(`)

γGk (w, z) ≥ c(log log

1 e − log log 2k−2 ) −→ ∞, z→0 |z|

Hence, (b) is verified. 1 What remains is the proof of (2.7.14). Fix a z ∈ Gk ∩ B(0, k−2 ). Then we have to 2e find an f ∈ O(Gk , E) satisfying the following conditions: • f (z) = f 0 (z) = . . . f (`−1) (z) = 0, c • |f (`) (z)| ≥ , where c is independent of z. (|z| log(1/|z|))` Again we write z as z = beiθ /2N with N ∈ N, b ∈ (1, 2], and θ ∈ [0, 2π). Observe that N ≥ e k − 1. Put f (λ) :=

`−1 X

αb,θ,j (2−N −j−1 − λ)−1 + 2N +1 βb,θ ,

λ ∈ Gk ,

(2.7.15)

j=0

where αb,θ,0 := 1 and αb,θ,1 , . . . , αb,θ,`−1 , βb,θ ∈ C depend only on b and θ such that (obviously, f ∈ O(Gk )) f (z) = · · · = f (`−1) (z) = 0.

128


We proceed under the assumption that we already have chosen f as in (2.7.15). Then it follows that `−1 X 1 beiθ |f (`) (z)| = αb,θ,j `!( N +j+1 − N )−`−1 2 2 j=0 `−1 X ≥ αb,θ,j ( j=0

2N +j+1 `+1 ) = 2(N +1)(`+1) |B`,b,θ |, 1 − 2j+1 beiθ

j

P`−1

2 where B`,b,θ := j=0 αb,θ,j ( 1−2j+1 )`+1 . beiθ Moreover, let us assume that |B`,b,θ | ≥ B` > 0, where B` is independent of b and θ. Then

kf kGk ≤

`−1 X `+1 |αb,θ,j | + 2N +1 |βb,θ | ≤ α r r k,N +` j=0 k,N +j+1

= α(` + 1)2N +` (N + `)k+1 ≤ c2N (N − 1)k+1 ≤ c2N (N − 1)` , where α := max{|αb,θ,j |, |βb,θ |} and c depends only on k and `. Put g := f /kf kGk ∈ O(Gk , E). Then the following estimate is true: ` 2`+1 B` 2N ` 2N log 2 e c |g (`) (z)| ≥ ≥ e c ≥ , 1 ` N c(N − 1) b(log 2 − log b) (|z| log(1/|z|))` where e c1 and e c are constants that only depend on k. In order to finish the proof of Lemma 2.7.4 we need the following lemma. Lemma 2.7.6. For an ` ∈ N there are positive numbers α and B` such that for every z = beiθ /2N , where b ∈ [1, 2), θ ∈ [0, 2π], and N ≥ e k − 1, there exist complex numbers αb,θ,j , j = 1, . . . , ` − 1, and βb,θ such that • max{|αb,θ,j |, j = 1, . . . , ` − 1, |βb,θ |, b and θ as above }| ≤ α, • min{|B`,b,θ | : b ∈ [1, 2], θ ∈ [0, 2π]} ≥ B` , • f (z) = f 0 (z) = . . . f (`−1) (z) = 0 (for f see (2.7.15)). Proof. Let f be a function as in (2.7.14) with unknown numbers αb,θ,j . Then the condition f 0 (z) = · · · = f (`−1) (z) = 0 gives the following system of ` − 1 equations `−1 X s! j=0

2N +j+1 s+1 αb,θ,j = 0, 1 − 2j+1 beiθ

s = 1, . . . , ` − 1,

which is equivalent to `−1 X s! j=1

s+1 s+1 2j 1 α = − , b,θ,j 1 − 2j+1 beiθ 1 − 2beiθ

s = 1, . . . , ` − 1.

To simplify further discussions we put Ab,θ,j :=

2j , 1 − 2j+1 beiθ

j = 0, . . . , ` − 1.

(2.7.16)

2.8. Kobayashi completeness and smoothly bounded pseudoconvex domains

129

Observe that |Ab,θ,j | ∈ [1/8, 1] and that Ab,θ,µ 6= Ab,θ,ν for µ 6= ν. Now we can rewrite the system of equations (2.7.16) in the following form `−1 X

s+1 As+1 b,θ,j αb,θ,j = −Ab,θ,0 ,

s = 1, . . . , ` − 1.

j=1

From here we conclude that i h det As+1 b,θ,j

j,s=1,...,`−1

2 `−1 Y Ab,θ,j = j=1

Y

|Ab,θ,µ − Ab,θ,ν | ≥ ε > 0,

1≤µ 0 such that all the |αb,θ,j | ≤ α e. Finally, the lower estimate remains. Since |B`,b,θ | is continuous with respect to (b, θ) it suffices to show that B`,b,θ 6= 0 or equivalently, `−1 X

`+1 A`+1 b,θ,j αb,θ,j 6= −Ab,θ,0 .

j=1

Suppose that this is false. Then the αb,θ,j ’s fulfill the following ` equations `−1 X

s+1 As+1 b,θ,j αb,θ,j = −Ab,θ,0 ,

s = 1, . . . , `,

j=1

implying that Ab,θ,0 /Ab,θ,j = 1, j = 1, . . . , ` − 1. But this is impossible. Thus also the lower estimate has been proved. 2.8. Kobayashi completeness and smoothly bounded pseudoconvex domains It is well known that there is a bounded pseudoconvex domain G (due to N. Sibony) with a C ∞ –boundary except of one point that is not kG –complete (see [J-P 1993], Theorem 7.5.9). ? On the other hand, for a smoothly bounded pseudoconvex domain G it is still an open question whether it is kG –complete ?

130


Let D ⊂ Cn be a smoothly bounded pseudoconvex domain and let z0 ∈ ∂D. Then there is a neighborhood U = U (z0 ) and a function r ∈ C ∞ (U, R) such that D ∩ U = {z ∈ U : r(z) < 0} and grad r(z) 6= 0, z ∈ U . Moreover, let V(z0 ) be the set of all germs of non–constant holomorphic mappings ψ : C0 −→ Cn with ψ(0) = z0 . According to d’Angelo, the domain D is said to be of finite type at z0 if n ord (r ◦ ψ) o 0 τ (D, z0 ) := sup : ψ ∈ V(z0 ) < ∞. ord0 ψ The domain D is said to be of finite type if D is of finite type at all of its boundary points. Then there is the following result due to [Bed-For 1978] (see also [For-Sib 1989], [For-McN 1994]). Theorem 2.8.1. Let D ⊂ C2 be a bounded pseudoconvex domain. Assume that D is of finite type. Then any boundary point a ∈ ∂D is a peak point with respect to C(D) ∩ O(D), i.e. there exists an f ∈ C(D) ∩ O(D) such that f (a) = 1 and |f (z)| < 1, z ∈ D \ {a}. In particular, we have Corollary 2.8.2. Any bounded pseudoconvex domain with a smooth boundary, which is of finite type, is c–complete. Observe, that to conclude that a domain is c–complete, a less strong condition is already sufficient; namely we have (see [J-P 1993]): Let D ⊂ Cn be a c–hyperbolic domain. Then the following two conditions are equivalent : (i) D is c–finitely compact; (ii) for any z0 ∈ D and for any sequence (zj )j ⊂ D without accumulation points in D, there is an f ∈ O(D, E) with f (z0 ) = 0 and sup{|f (zj )| : j ∈ N} = 1. ? Up to now it is an open problem whether all bounded pseudoconvex domains of finite type are k–complete or even c–complete ? 2.9. Kobayashi completeness and unbounded domains Let D ⊂ Cn be an arbitrary domain and let a ∈ ∂D. The point a is called to be a local holomorphic peak point of D, if there is a neighborhood U = U (a) such that a is a peak point with respect to C(U ∩ D) ∩ O(U ∩ D). When D is unbounded, we say that D has a local holomorphic peak point at infinity if there is an r > 0 and an f ∈ C(D \ B(0, r), E) ∩ O(D \ B(0, r), E) such that limz→∞ f (z) = 1. Recall that a bounded domain is locally k–complete iff it is k–complete (see [J-P 1993], Theorem 7.5.5). For an unbounded domain we have the following result (see [Gau 1999]). Theorem 2.9.1. Let D ⊂ Cn be an unbounded domain. Assume that D has a local holomorphic peak point at any point of ∂D ∪ {∞}. Then D is k–complete.

2.9. Kobayashi completeness and unbounded domains

131

Example 2.9.2. Put D := {z ∈ C2 : u(z) := |z1 |2 (1 + |z2 |2 ) < 1}. Obviously, {0} × C ⊂ D. Thus, D is not k–hyperbolic. On the other hand, since u is strongly psh, any a ∈ ∂D is a local holomorphic peak point. So this example shows that the condition at infinity in Theorem 2.9.1 is in some sense necessary. The proof of Theorem 2.9.1 is based on the following lemma. Lemma 2.9.3. Let D ⊂ Cn be an arbitrary domain and let a ∈ Cn ∪ {∞} be a boundary point of D. Assume that a is a local holomorphic peak point of D. Then: for any neighborhood U = U (a) there exists a neighborhood V = V (a) ⊂ U such that for any ϕ ∈ O(E, D), ϕ(0) ∈ V , one has ϕ(λ) ∈ U , |λ| < 1/2. Proof. We give the proof only for a = ∞ (the finite case is similar). Without loss of generality, let U = U (∞) := Cn \ B(0, ρ). By assumption, there is an r > 0 and an f ∈ C(D \ B(0, r), E) ∩ O(D \ B(0, r), E) such that limD3z→∞ f (z) = 1. We may assume that r = ρ. Put u(z) := log |f (z)|, z ∈ D \ B(0, r). Then u ∈ C(D \ B(0, r), [−∞, 0)) ∩ PSH(D \ B(0, r)) and

lim

u(z) = 0.

D3z→∞

Fix numbers r0 , r00 , r < r0 < r00 , such that sup{u(z) : z ∈ D ∩ ∂B(0, r0 )} =: c0 < 0, inf{u(z) : z ∈ D ∩ ∂B(0, r00 )} =: c00 > c0 , and f (z) 6= 0, kzk ≥ r0 . We define u b : D −→ (−∞, 0),   u(z), u b(z) := max{u(z),   c0 +c00 2 ,

0

00

c +c 2

},

if kzk ≥ r00 if r0 < kzk < r00 . if kzk ≤ r0

Obviously, u b is a global negative psh peak function at ∞, i.e. u b is a negative continuous function on D, psh on D, such that limD3z→∞ u b(z) = 0. Fix an ψ ∈ C(E, D) ∩ O(E, D). Observe that u b ◦ ψ ∈ C(E) ∩ SH(E). Let α < 0. Put E(ψ, α) := {θ ∈ [0, 2π] : u b ◦ ψ(eiθ ) ≥ 2α}. Assume that α ≤ u b ◦ ψ(0). Then Z 2π 1 α ≤b u ◦ ψ(0) ≤ u b ◦ ψ(eiθ )dθ 2π 0 Z 1 α ≤ u b ◦ ψ(eiθ )dθ ≤ 2π − Λ1 (E(ψ, α)) . 2π [0,2π]\E(ψ,α) π Hence, Λ1 (E(ψ, α)) ≥ π.

(2.9.17)

132


Put v(z) := log an ε > 0 such that

|f (z)−1| , 2

z ∈ D. Then v ∈ C(D \ B(0, r)) ∩ PSH(D \ B(0, r)). Choose

sup{(u + εv)(z) : z ∈ D ∩ ∂B(0, r0 )} =: c01 < 0, inf{(u + εv)(z) : z ∈ D ∩ ∂B(0, r00 )} =: c001 > c01 . Define vb : D −→ (−∞, 0) as   (u + εv)(z), vb(z) := max{(u + εv)(z),   c01 +c001 2 ,

if kzk ≥ r00 c01 +c00 1 2

},

if r0 < kzk < r00 . if kzk ≤ r0

Then vb ∈ C(D) ∩ PSH(D), vb < 0, and limD3z→∞ vb(z) = −∞. (Such a function is sometimes called a global psh antipeak function at ∞.) Let ψ be as above. Applying the Poisson integral representation, we get Z 2π Z 2π 1 1 − |λ|2 1 1 vb ◦ ψ(λ) ≤ vb ◦ ψ(eiθ ) iθ dθ ≤ vb ◦ ψ(eiθ )dθ, |λ| ≤ . (2.9.18) 2π 0 |e − λ|2 6π 0 2 Now, choose L > 0 such that L U 0 := {z ∈ D : vb(z) < − } ⊂ Cn \ B(0, r). 6 Then there is an α0 > 0 such that V := {z ∈ D : u b(z) > −α0 } ⊂ {z ∈ D : u b(z) ≥ −2α0 } ⊂ {z ∈ D : vb(z) < −L}. Now let ϕ ∈ O(E, D) such that ϕ(0) ∈ V . Obviously, we may also assume that ϕ ∈ C(E, D). Applying (2.9.17) and (2.9.18) we have for λ ∈ E, |λ| ≤ 1/2: Z 2π Z 1 1 LΛ1 (E(ϕ, α0 )) L iθ vb ◦ ϕ(λ) ≤ vb ◦ ϕ(e )dθ ≤ vb ◦ ϕ(eiθ )dθ ≤ − ≤− , 6π 0 6π E(ϕ,α0 ) 6π 6 i.e. ϕ(λ) ∈ U 0 .

Corollary 2.9.4. Let D ⊂ Cn and a be as in Lemma 2.9.3. Let U = U (a) be any neighborhood of a. Then there exists a neighborhood V = V (a) ⊂ U such that for any connected component V 0 of D ∩ V the following inequality is true: 2κD (z; X) ≥ κV 0 (z; X),

z ∈ V 0 , X ∈ Cn .

Remark 2.9.5. Observe that in Lemma 2.9.3 only the existence of a local psh peak function and a local psh antipeak function was needed. Other localization results for unbounded domains may be found in [Nik 2002]. Proof of Theorem 2.9.1. Step 1. We prove that D is k–hyperbolic if D has a local holomorphic peak point at infinity. Assume this is not the case. Then there exist z0 ∈ D, (zj )j ⊂ D with zj −→ z0 , and Xj ∈ Cn with kXj k = 1 such that κD (zj ; Xj ) < 1/j, j ∈ N (see Theorem 7.2.2 in [J-P 1993]). Therefore, we find functions ϕj ∈ O(E, D) such that ϕj (0) = zj and


133

kϕ0j (0)k > j, j ∈ N. In virtue of the Cauchy inequalities, we may further assume that there is a sequence (λj )j ∈ 12 E, λj −→ 0, such that kϕj (λj )k −→ ∞. Put ϕ ej := ϕj ◦ (−hλj ). Then ϕ ej ∈ O(E, D) with ϕ ej (λj ) = zj and kϕ ej (0)k −→ ∞. Put R := 2kz0 k + 1. Then there is an R0 > R for which Lemma 2.9.3 can be used. Since kϕ ej (0)k > R0 for large j, we get for these j that kϕ ej (λj )k > R, which contradicts the fact that zj −→ z0 . Step 2. Here we prove that D is k–complete. Assume the contrary. Then there are a point z0 ∈ D and a sequence (zj )j ⊂ D such that A := sup{kD (z0 , zj ) : j ∈ N} < ∞ and either zj −→ z ∗ ∈ ∂D or zj −→ ∞. Again we discuss only the second case. The first is similar. Let f ∈ C(D \ B(0, r), E) ∩ O(D \ B(0, r), E) be the local holomorphic peak function at infinity. Choose C 1 –curves αj : [0, 1] → D with Z 1 αj (0) = z0 , αj (1) = zj , κD (αj (t); αj0 (t))dt < A + 1. 0

According to Corollary 2.9.4, we find an R > max{kz0 k, r} such that for any connected component U of D ∩ (Cn \ B(0, R)) the following is true: 2κD (z; X) ≥ κU (z; X), z ∈ U , X ∈ Cn . Moreover, observe that |f | ≤ C < 1 on D ∩ ∂B(0, R). We may assume that all kzj k > R. Fix an j, put tj := sup{t ∈ [0, 1] : kαj (t)k ≤ R}, and let Uj denote that connected component of (Cn \ B(0, R)) ∩ D containing αj ((tj , 1]). Then Z 1 Z 1 2 κD (αj (t); αj0 (t))dt ≥ κUj (αj (t); αj0 (t))dt 0

tj

Z

1

≥

κE (f ◦ αj (t); (f ◦ αj )0 (t))dt ≥ min{kE (f (zj ), λ) : |λ| ≤ C} −→ ∞; j→∞

tj

contradiction.

Remark 2.9.6. The following domain (see [Par 2003]) D := {(z, w) ∈ C3 × C : |z1 z2 z3 | < 1, 0 < |w| < e− max{|zj |:j=1,2,3} } is k–complete, but there is no local psh peak function at infinity; in particular, there is no local holomorphic peak function at infinity. Indeed, D is a pseudoconvex Reinhardt domain which is Brody–hyperbolic. Hence, it is k–complete (see Theorem 2.2.1). Assume now that there exists a local psh peak function at ∞. Hence there is an R > 1 and a ϕ ∈ C(D\B(0, R))∩PSH(D\B(0, R)), ϕ < 0, such that limD3(z,w)→∞ ϕ(z, w) = 0. Fix an a ∈ C with |a| = 2R and define Da := {z ∈ C2 : 2R|z1 z2 | < 1} and ua (z) := max{|z1 |, |z2 |, |a|}, Moreover, put Ω := {(z, λ) ∈ Da × C : |λ| < e−ua (z) }.

z ∈ Da .

134


Finally, set ϕa : Ω −→ [−∞, ∞) as ϕa (z, λ) := ϕ(z, a, λ). Then ϕa ∈ PSH(Ω) and 1 ϕa (·, 0) ∈ PSH(Da ). In virtue of the Liouville theorem, there is a ψ ∈ SH( 2R E) such that ϕa (z, 0) = ψ(z1 z2 ), z ∈ Da . So, applying the Oka theorem, we get 1 1 ψ(0) = lim sup ψ = lim sup ϕa t, , 0 =: C ≤ 0. 2 2Rt 2Rt R3t→∞ R3t→∞ 1 In the case when C=0, the maximum principle would imply that ψ ≡ 0 on 2R E and thus ϕ(·, a, 0) ≡ 0 on Da which contradicts the assumption ϕ < 0. Hence, C < 0. Now choose a t0 > 2|a| such that for all t > t0 the following inequality is true: 3 1 1 lim ϕ t, , a, λ = ϕa t, , 0 < C. 2 2 06=λ→0 2Rt 2Rt 4

So, for t > t0 , there is an εt > t such that 1 1 ϕ t, , a, λ < C, 2Rt2 2

0 < |λ| < e−εt .

1 Now observe that max{t, 2Rt 2 , |a|} = t, t > t0 , and εt −→ ∞ if t −→ ∞. Therefore, we 1 may choose a sequence (tj , 2Rt ⊂ D ∩ (R2 × {a} × C) such that t0 < tj −→ ∞ 2 , a, λj ) j j

and |λj | < e−εj . Hence, we have 0 = lim ϕ(tj , j→∞

1 C , a, λj ) ≤ ; 2Rt2j 4

a contradiction. The above example shows that the conditions in Theorem 2.9.1 are too strong. Observe that any bounded boundary point z0 is obviously a local psh antipeak point (take 0k for a large R). simply log kz−z R ? Does Theorem 2.9.1 remain true if one only assumes that any boundary point admits a local psh peak and antipeak function ? In this context observe that there exists a smoothly bounded pseudoconvex domain D ⊂ C3 such that each boundary point of D is a global psh peak point, but some boundary point is not a local holomorphic peak point (see [Yu 1997]). Example 2.9.7. Theorem 2.9.1 has been used in [Gau 1999] to prove the following results. (a) Let P be a real valued subharmonic polynomial on C without harmonic terms. Then D := {(z, w) ∈ C × C : Re w + P (z) < 0} is k–complete. (b) Let P be a real valued convex polynomial on Cn , P (0) = grad P (0) = 0, without harmonic terms, such that the set {z ∈ Cn : P (z) = 0} does not contain a nontrivial analytic set. Then D := {(z, w) ∈ Cn × C : Re w + P (z) < 0}


is a convex k–complete domain.

4

4

135

See also Theorem 7.1.8 in [J-P 1993] for the following characterization of k–complete convex domains: A convex domain G is k–complete iff G contains no complex lines iff G is biholomorphic to a bounded convex domain.

CHAPTER 3

Bergman metric 3.1. The Bergman kernel In this chapter we will discuss a metric on domains which is invariant under biholomorphic mappings, namely the Bergman metric. To do so we have to recall first the Bergman kernel function and the Bergman kernel. Let G ⊂ Cn be a domain. We denote by L2h (G) the Hilbert space of all square integrable functions on G which are holomorphic; it is a closed subspace of L2 (G). The key tool in this chapter is the following extension theorem due to K. Ohsawa and K. Takegoshi [Ohs-Tak 1987]. Theorem 3.1.1. Let D be a bounded pseudoconvex domain in Cn and H an affine subspace of Cn . Then there is a positive constant C, which depends only on the diameter of D and n, such that for any f ∈ L2h (D∩H) there is an F ∈ L2h (D) such that F |D∩H = f and kF kL2h (D) ≤ Ckf kL2h (D∩H) . Moreover, we recall the following one-dimensional result (see [Lin 1977], [Che 2000]) which will be used in the sequel. Theorem 3.1.2. Let D ⊂ C be a bounded domain, z0 ∈ ∂D, and f ∈ L2h (D). Then for any ε > 0 there exist a neighborhood U = U (z0 ) and a function g ∈ L2h (D ∪ U ) such that kf − gkL2h (D) ≤ ε. In particular, the subspace of all functions in L2h (D), bounded near z0 , is dense in L2h (D). In [Che 2000], complete Kaehler metrics were used to solve a corresponding ∂– problem in order to find g. Here we give a proof which is based on Berndtsson’s solution of a ∂–problem (see [Pfl 2000]). Proof. We may assume that z0 = 0 ∈ ∂D and that D ⊂ E. Fix f ∈ L2h (D) and a sufficiently small ε ∈ (0, 1/2). Put ψ(z) := − log(log(1/|z|)), z 6= 0. Observe that ∂2ψ 2 2 −2 ψ ∈ C ∞ (C∗ ) ∩ SH(C∗ ) and | ∂ψ |z|−2 > 0. ∂z | = ∂z∂z = (log |z| ) ∞ Moreover, let χ ∈ C (R, [0, 1]), ( 1, if t ≤ 1 − log 2 χ(t) := , 0, if t > 1 be such that |χ0 | ≤ 3. Finally, we define ρε (z) := χ(−ψ(z) − log(log(1/ε)) + 1), z ∈ C∗ . Observe that √ ρε (z) = 0 if 0 < |z| < ε, and ρε (z) = 1 if |z| > ε. 137

138

3. Bergman metric

Then α := ∂(ρε f ) is a C ∞ ∂–closed (0, 1)–form on Dε := D ∪ B(0, ε). Now we like to apply the following theorem of Berndtsson. Theorem ([Ber 1996]). Let Ω ⊂ Cn be a bounded pseudoconvex domain. Let ϕ, ψ ∈ PSH(Ω), ψ strongly psh, be such that for any X ∈ Cn the following inequality n n X 2 X ∂2ψ ∂ψ (z)Xj X k ≥ (z)Xj ∂zj ∂z k ∂z j j=1 j,k=1

Pn holds on Ω. Let δ ∈ (0, 1) and α = j=1 αj dz j a ∂–closed (0, 1)–form. Then there exists a solution u ∈ L2loc (Ω) of ∂u = α such that Z Z X n 4 |u|2 e−ϕ+δψ dΛ2n (z) ≤ ψ j,k αj αk e−ϕ+δψ dΛ2n (z), 2 δ(1 − δ) Ω Ω j,k=1

2

ψ ). where (ψ jk ) denotes the inverse matrix of ( ∂z∂j ∂z k

Take ϕ := (1/2)ψ and δ := 1/2. Then there exists a function uε , ∂uε = α on Dε \ {0} such that Z Z 4 |α|2 −ϕ+δψ e dΛ2 (z) |u|2 e−ϕ+δψ dΛ2 (z) ≤ δ(1 − δ)2 Dε \{0} | ∂ 2 ψ | Dε \{0} ∂z∂z Z |χ0 |2 · |f |2 dΛ2 (z). = 16 √ z∈D, ε≤|z|≤ ε

Then the function fε := uε − ρε f belongs to L2h (Dε \ {0}) and kf − fε kL2h (D) ≤ k(1 − ρε )f kL2h (D) + 160kf kL2h (D∩B(0,√ε)) ≤ Ckf kL2h (D∩B(0,√ε)) −→ 0, ε→0

where C is a general positive constant. It remains to note that fε ∈ O(Dε ) (use Laurent series), which finishes the proof. We note that under some proper assumptions this result can be generalized to higher dimensions. Observe that the point evaluation functional L2h (G) 3 f −→ f (w) (w ∈ G) is continuous. Therefore, there is a uniquely defined function KG (·, w) ∈ L2h (G) such that Z f (w) = f (z)KG (z, w)dΛ(z), f ∈ L2h (g), w ∈ G. G

The function KG is the Bergman kernel function for G. Recall that KG can be given with the help of a complete orthonormal system (ϕj )j∈N ⊂ L2h , where N ⊂ N; namely X KG (z, w) = ϕj (z)ϕj (w), z, w ∈ G. j∈N

Remark 3.1.3. Recall that there are domains Gk ⊂ C2 for which dim L2h (Gk ) = k [Wig 1984]. ? It is unknown whether dim L2h (G) = ∞, if G ⊂ Cn , n > 1, is a pseudoconvex domain with L2h (G) 6= {0} ?

3.1. The Bergman kernel

139

The function KG is holomorphic in z and antiholomorphic in w; moreover, we have KG (z, w) = KG (w, z), z, w ∈ G. If Φ : G −→ D is a biholomorphic mapping between the domains D and G, then KD (Φ(z), Φ(w)) det Φ0 (z)det Φ0 (w) = KG (z, w),

z, w ∈ G.

Moreover, there is a transformation law even for proper holomorphic mappings due to S. Bell (see [J-P 1993], Theorem 6.1.8). Theorem 3.1.4. Let F : G −→ D be a proper holomorphic mapping of order m between the bounded domains G, D ⊂ Cn . Let u := det F 0 and denote by Φ1 , . . . , Φm the local inverses of F defined on D0 := D \ {F (z) : z ∈ G, u(z) = 0}. Put Uk := det Φ0k . Then m X

KG (z, Φk (w))Uk (w) = u(z)KD (F (z), w),

z ∈ G, w ∈ D0 .

k=1

The function kG (z) := KG (z, z) 1 , z ∈ G, is called the Bergman kernel of G. In the case when L2h (G) 6= {0}, then kG is also given as n |f (z)|2 o 2 kG (z) = sup : f ∈ L (G) \ {0} . h kf k2L2 h

Observe that kD |G ≤ kG whenever G ⊂ D. For the Bergman kernel there is the following localization result (see [J-P 1993], Theorem 6.3.5). Theorem 3.1.5. Let Dj ⊂ Cn , j = 1, 2, be two bounded pseudoconvex domains and let z0 ∈ ∂D1 . Assume that there is a neighborhood U = U (z0 ) of z0 such that D1 ∩ U = D2 ∩ U . Then there exist positive numbers m, M and a neighborhood V = V (z0 ) such that mkD1 (z) ≤ kD2 (z) ≤ M kD1 (z), z ∈ V ∩ D1 . In general, it is not easy to find explicit formulas for the Bergman kernel function. In most of the known examples the formulas are obtained using an explicit complete orthonormal system (ϕj )j ∈ L2h (G). Example 3.1.6 (see Examples 6.1.5 and 6.1.6 in [J-P 1993]). (a) For the Euclidean ball Bn we have −(n+1) n! , z, w ∈ Bn . KBn (z, w) = n 1 − hz, wi π (b) Let E n be the n–dimensional polydisc. Then KE n (z, w) =

1

n −2 1 Y 1 − z w , j j π n j=1

z, w ∈ E n .

Observe that the symbol kD (·) is a function on D, while the Kobayashi pseudodistance kD ( · , · ) is defined on D × D; we hope there will be no confusion for the reader.

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3. Bergman metric

(c) Put Dp := {z ∈ C2 : |z1 |2 + |z2 |2/p < 1}, p > 0. Then KDp (z, w) =

p 1 p−2 (p + 1)(1 − z1 w 1 ) + (p − 1)z2 w 2 (1 − z w ) , 1 1 π2 ((1 − z1 w1 )p − z2 w2 )3

z, w ∈ Dp .

Observe (by a simple calculation) that KD2 has no zeros on D2 × D2 . (d) Recently, using Theorem 3.1.4, the following formula for the Bergman kernel function of Gn has been found in [Edi-Zwo 2004] (see Remark 1.4.17 for a definition of Gn ). h i det (1−λ1j µ )2 k Fn (z, w) 1≤j,k≤n KGn (πn (λ), πn (µ)) = KGn (z, w) = = n Qn , 2 n 0 0 π π det πn (λ)det πn (µ) j,k=1 (1 − λj µk ) λ, µ ∈ E n \ {ζ ∈ E n : det πn0 (ζ) = 0} = E n \ {ζ ∈ E n : ζj = ζk for some j 6= k}, where πn : Cn −→ Cn , πn (λ1 , . . . , λn ) :=

X

λj1 · · · λjk

1≤j1 0 such that w+rE n b M\V . In view of Remark 6.1.4 in [J-P 1993], we may find a C ∞ –function u : Cn −→ [0, ∞), supp u ⊂ w + rE n , such that Z f (w) =

f (z)u(z)dΛ2n (z),

f ∈ O(M).

M

Therefore, KM (·, w) = PM (u). Applying Lemma 3.1.12, it follows that zn+1 KM (π(z), w) = zn+1 PM u (π(z)) = PK χ · u ◦ π (z) Z α(ζ) ∧ α(ζ) = ζn+1 u ◦ π(ζ)KK (z, ζ) n(n+1) K (−1) 2 (2i)n Z K (z, ϕ(η)) K (z, ψ(η)) K K = (n + 1)2 u(η) + dΛ2n (η) ϕn+1 (η) ψn+1 (η) M\V K (z, ϕ(w)) K (z, ψ(w)) K K = (n + 1)2 + , z ∈ K. ϕn+1 (η) ψn+1 (η) Hence the lemma is proved.

3.1. The Bergman kernel

145

Now we are in the position to finish the proof of Theorem 3.1.9. According to Lemma 3.1.11, we have KK (z, w) = Ch(z • w), where n(n+1) 2

2(2i)n (−1) ω(∂K)

C=

and h(t) :=

2n n+1 − , t ∈ C. (1 − t)n+1 (1 − t)n

Hence, KK (ϕ(z), ϕ(w)) = Ch(x),

KK (ϕ(z), ψ(w)) = Ch(y),

(3.1.6)

where x := hz, wi + t, t := ϕn+1 (z)ϕn+1 (w), and y := hz, wi − t. In virtue of Lemma 3.1.13, we get KM (z, w) = C(n + 1)2

h(x) − h(y) t

.

Using the abbreviation r := 1 − hz, wi, the last expression can be written as Q :=

h(x) − h(y) (r + t)n+1 − (r − t)n+1 (r + t)n − (r − t)n = 2n − (n + 1) . 2 2 n+1 t t(r − t ) t(r2 − t2 )n

Then [n [ n−1 2] 2 ] X X 2n n + 1 n−2k 2k n+1 n Q= 2 2 r t − 2 2 rn−2k−1 t2k . (r − t2 )n+1 2k + 1 (r − t2 )n 2k + 1 k=0 k=0 n−2k n+1 n Since 2k+1 = 2k+1 2k+1 , we proceed with our calculations and get [n 2] X 2 n + 1 n−1−2k 2k 2 2 Q= 2 r t 2nr − (n − 2k)(r − t ) , (r − t2 )n+1 2k + 1 k=0

which immediately leads to the formula in Theorem 3.1.9.

Example 3.1.14 (See Example 6.1.9 in [J-P 1993]). Applying Theorem 3.1.9, the biholomorphic mapping 1 M2 3 (z1 , z2 ) −→ √ (z1 + iz2 , z1 − iz2 ) ∈ G2 := {z ∈ C2 : |z1 | + |z2 | < 1} 2 leads to the following formula of the Bergman kernel function of the domain G2 : KG2 (z, w) =

2 3(1 − hz, wi)2 (1 + hz, wi) + 4z1 z2 w1 w2 (5 − 3hz, wi) · , 3 π2 (1 − hz, wi)2 − 4z1 z2 w1 w2

z, w ∈ G2 .

Observe that this formula may be also derived from the one in Example 3.1.6(c) using the proper holomorphic mapping D2 3 z −→ (z12 , z2 ) ∈ G2 and Bell’s transformation law. p p Fix points z, w ∈ G2 and write, for abbreviation, ξj := zj wj . Then |ξ1 |+ |ξ2 | < 1, and so 4|ξ1 ξ2 | < (1 − |ξ1 | − |ξ2 |)2 . Therefore, the numerator in the formula above allows

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3. Bergman metric

the following estimate 3(1−hz, wi)2 (1+hz, wi)+4z1 z2 w1 w2 (5−3hz, wi) = 3(1−ξ1 −ξ2 )(1−(ξ1 −ξ2 )2 )+8ξ1 ξ2 ≥ 3(1 − |ξ1 | − |ξ2 |)(1 − |ξ1 − ξ2 |2 ) − 2(1 − |ξ1 | − |ξ2 |)2 ≥ 3(1 − |ξ1 | − |ξ2 |)2 (1 + |ξ1 − ξ2 |) − 2(1 − |ξ1 | − |ξ2 |)2 > (1 − |ξ1 | − |ξ1 |)2 > 0. Hence, the Bergman kernel function KG2 has no zeros on G2 × G2 . Remark 3.1.15. In [You 2002] an explicit formula for the Bergman kernel function is given even for a more general domain Ω which could be thought as some interpolation between the minimal balls and the Euclidean balls. Here, we only describe Ω. Fix d ∈ N and two d–tuples m = (m1 , . . . , md ) ∈ Nd and n = (n1 , . . . , nd ) ∈ Nd . Moreover, let a = (a1 , . . . , ad ) ∈ [1, ∞)d . Then the domain Ω = Ωd,m,n,a is given as d n o X 2a Ω := Z = (Z(1), . . . , Z(d)) ∈ Mm1 ,n1 (C) × · · · × Mmd ,nd (C) : kZ(j)k∗ j < 1 , j=1

where Mp,q (C) denotes the space of all p × q–matrices with complex entities, and where p X q q X 1/2 X 2 kM k∗ := ( |zjk |2 + | zjk |) , M = (zjk )j=1,...,p, k=1,...,q ∈ Mp,q (C). j=1 k=1

k=1

Observe that for d = 1 = a = m, n1 = n the domain Ωd,m,n,a is just the minimal ball M ⊂ Cn . 3.2. The Lu Qi-Keng problem For a while it was a question (posed by Lu Qi-Keng [LQK 1966]) whether the Bergman kernel function of a simply connected domain G ⊂ Cn , n ≥ 2, has no zeros. Such a domain is called a Lu Qi-Keng domain. A first example of a simply connected domain of holomorphy which is not a Lu Qi-Keng domain was given by H.P. Boas [Boa 1986] (see also [Skw 1980]). In fact, it turned out that the set of domains of holomorphy not being Lu Qi-Keng form a nowhere dense set in a suitable topology. For a more detailed discussion of this topic see [Boa 1996] (see also [Boa 2000]). Example 3.2.1. Let D = Dp = {z ∈ C2 : |z1 |2 + |z2 |2/p < 1}, p a positive integer, be the third example of 3.1.6. Then there is the proper holomorphic mapping F : Dp −→ Gp := {z ∈ C2 : |z1 | + |z2 |2/p < 1},

F (z1 , z2 ) := (z12 , z2 ).

Using Bell’s transformation law (see Theorem 3.1.4), we obtain 1 √ √ KGp ((z12 , 0), (w1 , 0))2z1 = KD ((z1 , 0), ( w1 , 0)) − KD ((z1 , 0), (− w1 , 0)) √ , 2 w1 whenever z1 ∈ E, w1 ∈ E \ {0}. Now, applying Example 3.1.6(c), it follows that p+1 −p−2 −p−2 KGp ((z12 , 0), (w12 , 0))2z1 = (1 − z w ) − (1 + z w ) . 1 1 1 1 2w1 π 2

3.2. The Lu Qi-Keng problem

147

Then, if z1 6= 0, the kernel function KGp ((z12 , 0)(w12 , 0)) has a zero iff (1 + x)p+2 = 1+λ (1 − x)p+2 , where x := z1 w1 . Observe that λ −→ 1−λ maps E biholomorphically to the 1+λ p+2 right half-plane. Hence, ( 1−λ ) = 1 has a non-zero solution iff p > 2. We point out that also KGp ((0, z2 ), (0, w2 )) has zeros. Example 3.2.2. Next, we study domains of the following type n m o n X X Ωn,m := (z, w) ∈ Cn × Cm : |zj | + |wk |2 < 1 , j=1

k=1

where n ∈ N and m ∈ N0 . In a first step, let n = 1 and m ∈ N0 . Then, using Bell’s transformation law for the proper holomorphic mapping F : Bk −→ Ω1,m , F (z) := (z12 , z2 , . . . , zk ), where k := m+1, we get KΩ1,m ((z12 , z2 , . . . , zk ), (w12 , w2 , . . . , wk )) k! 1 1 = k − , π 4z1 w1 (1 − hz, wi)k+1 (1 + z1 w1 − he z , wi) e k+1

z, w ∈ Bk , z1 w1 6= 0,

where ze := (z2 , . . . , zk ) and w e := (w2 , . . . , wk ). In the case m + 2 > 4, a similar reasoning as above gives z1 , w1 ∈ E∗ such that KΩ1,m ((z12 , 0, . . . , 0), (w12 , 0, . . . , 0)) = 0. If m+2 ≤ 4, an easy calculation shows that KΩ1,m has no zeros on Ω1,m × Ω1,m . Hence, the Bergman kernel function of Ω1,m has a zero iff m + 2 > 4. Finally, using the above result for n = 1, induction over n, and the deflation method, we are led to the following result: The Bergman kernel function of Ωn,m has zeros iff 2n + m > 4 3 . In particular, the convex domain Ωn,0 , n ≥ 3, is not Lu Qi-Keng. ? So far it is not known whether there is a convex domain in C2 which is not a Lu Qi-Keng domain ? Let n, k ∈ N, m ∈ N0 , and a ∈ (0, 1]. Put 1 2k X Na,k (z) := αε2k1 ,...,εn+1 (z) + a2k α2k (z) ,

z ∈ C n × Cm ,

ε1 ,...,εn+1 ∈{+1,−1}

where αε1 ,...,εn+1 (z) := over, put

Pn

j=1 εj |zj |

+ εn+1

Pm

j=1

|zn+j |2 and α(z) :=

Pn+m j=1

|zj |2 . More-

Ωa,k,n,m := {z ∈ Cn+m : Na,k (z) < 1}. The following result is due to Nguyên Viêt Anh [Viê 2000]. Theorem 3.2.3. The domain Ωa,k,n,m is strongly convex, algebraic 4 , complete Reinhardt. Moreover, if 2n − m > 4, then there is a positive integer M = M (a, n, m) such that for all k ≥ M the domain Ωa,k,n,m is not a Lu Qi-Keng domain. In particular, for m = 0 there are strongly convex algebraic complete Reinhardt domains in Cn , n ≥ 3, which are not Lu Qi-Keng. 3

Recall Example 3.1.14. 4 Here ”algebraic” means that the domain is given as the sublevel set of a real polynomial.

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3. Bergman metric

? What are effective values for the number M (a, n, m) ? To prove Theorem 3.2.3, we need the following Lemma. Lemma 3.2.4. Suppose fj : Rq −→ R+ is a convex function, j = 1, . . . , p. Then for k ∈ N the following function X ρ(z) := (ε1 f1 (z) + · · · + εp fp (z))2k , z ∈ Rq , ε1 ,...,εp ∈{−1,+1}

is also a convex one. Proof. Fix z, w ∈ Rq . Then X ρ(z) + ρ(w) ≥ 2

ε1

ε1 ,...,εp ∈{−1,+1}

f1 (z) + f1 (w) fp (z) + fp (w) 2k + · · · + εp . 2 2

Now recall the following formula 2k

X

(ε1 b1 + · · · εp bp )2k = 2p

ε1 ,...,εp ∈{−1,+1}

X k1 +···kp =k

p 1 (2k)!b2k 1 · · · bp . (2k1 )! · · · (2kp )!

Plugging it into the first expression we get f (z) + f (w) 2k1 f (z) + f (w) 2kp X ρ(z) + ρ(w) 2p (2k)! 1 1 p p ≥ ··· . 2 (2k1 )! · · · (2kp )! 2 2 k1 +···kp =k

In virtue of the positivity and convexity of the functions fj the last inequality gives (ρ(z) + ρ(w))/2 ≥ ρ((z + w)/2), i.e. ρ is a convex function. Proof of Theorem 3.2.3. Put ρ(z) :=

X

αε2k1 ,...,εn+1 (z) + a2k α2k (z) − 1.

ε1 ,...,εn+1 ∈{−1,+1}

Then ρ is the defining function of the domain Ω = Ωa,k,n,m . Using the above Pmexpansion, we see that ρ is a polynomial with positive coefficients in |z1 |2 , . . . , |zn |2 and j=1 |zn+j |2 . Hence, Ω is an algebraic complete Reinhardt domain with a smooth boundary. Moreover, in virtue of the Lemma 3.2.4, we may see that Ω is strongly convex. Observe that Ω ⊂ Ωn,m , where Ωn,m is the domain from Example 3.2.2, and that Na,k ≤ Na,l when l ≤ k. Moreover, lim Na,k (z) =

k→∞

n X j=1

|zj | +

m X

|zn+k |2 ,

z ∈ Ωn,m .

k=1

What remains is to apply Ramadanov’s theorem (see [J-P 1993], Theorem 6.1.15), Example 3.2.2, and the Hurwitz theorem. Example 3.2.5. We also mention that the minimal ball M ⊂ Cn , n ≥ 4, is non Lu Qi-Keng [Pfl-You 1998]. This result is proved exploiting the explicit formula given in Theorem 3.1.9. In fact, let first n ≥ 5:


149

Put f : R → R, f (t) := −(n + 1) arctan

2(n2 − 1)t 2t + 2π − arctan . 1 − t2 (n − 1)2 − (n + 1)2 t2

Observe that f (0) = 2π and f (1/2) < 0 (here we need that n ≥ 5); so f (t0 ) = 0 for a certain t0 ∈ (0, 1/2). Therefore 1 − it n+1 n − 1 + it0 (n + 1) 0 = . 1 + it0 n − 1 − it0 (n + 1) √ √ Put z0 := it0 (1, 0, . . . , 0), w0 := −it0 (0, 1, 0, . . . , 0) ∈ Cn . A simple calculation gives that N (z0 ) = N (w0 ) = t0 < 1/2; thus, z0 , w0 ∈ M. Then, in virtue of Theorem 3.1.9, it follows that P[ n2 ] n+1 2j 2 1 j=0 2j+1 (it0 ) (n + 2j + (n − 2j)(it0 ) ) . KM (z0 , w0 ) = (1 − (it0 )2 )n+1 n(n + 1)Λ2n (M) Computing the binomial expression leads to KM (z0 , w0 ) =

(n − 1 + (n + 1)it0 )(1 + it0 )n+1 − (n − 1 − (n + 1)it0 )(1 − it0 )n+1 = 0. n(n + 1)Λ2n (M)2it0 (1 − (it0 )2 )n+1

It remains the case n = 4: Consider the function g : R −→ R, g(s) := −28s4 + 50s3 − 10s2 − 15s + 5. Then g(0) = 5 and g(2/5) < 0. Therefore, there exists a s0 ∈ (0, 2/5) with g(s0 ) = 0. Put p p √ √ √ √ s0 (1 − i) s0 (i − 1) (i − i, −i − i, 0, 0). (i + i, −i + i, 0, 0), w0 := z0 := 2 2 Then N (z0 ) = N (w0 ) < 1/2, i.e. z0 , w0 ∈ M. A little calculation gives from the formula in Theorem 3.1.9 that g(s0 ) KM (z0 , w0 ) = = 0. 5Λ2n (M)((1 − s0 )2 + s20 )5 Hence, the Bergman kernel function vanishes at the point (z0 , w0 ). ? It is an open question whether the three dimensional minimal ball is a Lu Qi-Keng domain ? For further open problems see also [Boa 2000]. Other examples of domains, that are not Lu Qi-Keng, may be found in [Die-Her 1999], [Eng 2000], and [Che 2002]. We close this section discussing consequences of the following result. Theorem 3.2.6 ([Eng 1997], [Eng 2000], [Che 2002]). Let D ⊂ Cn be a bounded pseudoconvex domain, ϕ > 0 a positive function on D, − log ϕ ∈ PSH(D), such that 0 1/ϕ ∈ L∞ loc (D) fails to have a sesqui-holomorphic extension near a point z ∈ D (i.e. there 0 is no function f : V × V → C, V ⊂ D a neighborhood of z , satisfying: f is holomorphic in the first coordinates and antiholomorphic in the latter, f (z, z) = 1/ϕ(z) for all z ∈ V ).

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3. Bergman metric

Let U = U (z 0 ) ⊂ D be a neighborhood. Then there is an mU ∈ N such that the Bergman kernel function KΩm (( · , 0), ( · , 0)) of Ωm := {(z, w) ∈ D × Cm : kwk2 < ϕ(z)} has a zero in U × U , m ≥ mU . Proof. The proof of Theorem 3.2.6 is based on an extension theorem for L2h –functions (see [Ohs 2001]) and a description of the Bergman kernel with weights due to E. Ligocka (see [Lig 1989]). Applying the first result we are led 5 to the following formula lim KΦk (z, z)1/k =

k→∞

1 , Φ(z)

z ∈ D,

(3.2.7)

where KΦk denotes the reproducing kernel function of the Hilbert space L2h (D, Φk ). (Observe that in the case Φ ≡ 1 this is just the classical Bergman kernel function.) Then there is a mU such that KΦm has zeros on U × U , m ≥ mU . Otherwise, we may assume that U is simply connected and that all the functions KΦk have no zeros 1/k on U × U . Next we choose a sesqui–holomorphic branch of KΦk on U × U . Since the function 1/Φ is locally bounded, using (3.2.7) we see that the sequence KΦk (z, z)1/k is locally bounded on U . Therefore, applying |KΦk (z, w)|2 ≤ KΦk (z, z)KΦk (w, w), z, w ∈ U , shows that (KΦk ) is locally bounded on U × U . Hence, it converges locally uniformly to a sesqui–holomorphic function L on U × U with L(z, z) = 1/Φ(z), z ∈ U ; a contradiction. It remains to recall Ligocka’s formula ∞ X (j + m)! KΦm+j (z, w)ht, sij , KΩm (z, t), (w, s) = m j!π j=0

(z, t), (w, s) ∈ Ωm .

Thus we have m! KΩm (z, 0), (w, 0) = m KΦm (z, w), z, w ∈ D. π Therefore, in virtue of the above claim, it follows that there is an mU such that for any m ≥ mU the function KΩm (·, 0), (·, 0) has zeros on U × U . We should mention that the original formulation in [Che 2002] is much stronger as the one given here. Applying Theorem 3.2.6 for certain complex ellipsoids we obtain the following consequences. Corollary 3.2.7 ([Che 2002]). (a) For any k ≥ 1, not an even integer, there exists an m = m(k) ∈ N such that Ω := Ωk := {(z, w) ∈ E × Cm : |z|k + kwk2 < 1} is not Lu Qi-Keng. (b) For any k ∈ N there exists a natural number m = m(k) such that, if Ω := Ωk := {(z, w) ∈ E × Cm : |z| 5

We omit that proof.

2k+1 2

+ kwk2 < 1},

(z 0 , w0 ) := (0, 0, . . . , −1) ∈ ∂Ω,


151

then Ω is convex with a C k –boundary and there are sequences ((zj0 , wj0 ))j , ((zj00 , wj00 ))j ⊂ Ω,

lim (zj0 , wj0 ) = lim (zj00 , wj00 ) = (z 0 , w0 ),

j→∞

j→∞

KΩ ((zj0 , wj0 ), (zj00 , wj00 )

such that = 0, j ∈ N. In particular, the set {(z, w) ∈ Ω × Ω : KΩ (z, w) = 0} accumulates at ((z 0 , w0 ), (z 0 , w0 )). Proof. (a) Take D = E and ϕ(z) := 1 − |z|k , z ∈ E. Then − log ϕ ∈ SH(E) and 1/ϕ is not real analytic at 0. So it cannot be extended to a sesqui-holomorphic function near 0. Hence, in virtue of Theorem 3.2.6, there is a neighborhood U = U (0) and an m = m(k) ∈ N such that KΩk (( · , 0), ( · , 0)) has at least one zero in U × U . (b) Fix a k. In virtue of part (a), there is an m = m(k) such that KΩ has a zero at a point ((z 0 , w0 ), (z 00 , w00 )) ∈ Ω × Ω. Put m o n X 2k+1 m+1 2 + |ζj |2 + Re ζm+1 < 0 . D := ζ ∈ C : |ζ1 | j=2

Observe that 2

Φ(ζ) :=

4 2k+1 4

,

2ζ2 2ζm ζm+1 + 1 ,..., , ζm+1 − 1 ζm+1 − 1 ζm+1 − 1

(ζm+1 − 1) 2k+1 defines a biholomorphic map from D to Ω. Moreover, for any positive ε, √ √ 2 Fε (ζ) := (ε 2k+1 ζ1 , εζ2 , . . . , εζm , εζm+1 ) is a biholomorphic mapping from D to D. Therefore, KΩ (Φ ◦ Fε ◦ Φ−1 (z 0 , w0 ), Φ ◦ Fε ◦ Φ−1 (z 00 , w00 )) = 0,

ε > 0.

It remains to mention that limε→0 Φ ◦ Fε ◦ Φ−1 (z 0 , w0 ) = limε→0 Φ ◦ Fε ◦ Φ−1 (z 00 , w00 ) = (z 0 , w0 ). ? It would be interesting to find in the situation of Corollary 3.2.7 concrete numbers m = m(k) ? So far, we saw that some of the domains Ep ⊂ Cn are not Lu Qi-Keng, some of them are. ? Describe all the vectors p = (p1 , . . . , pn ) for which the Bergman kernel function of Ep is zero–free ? Remark 3.2.8. Recall that in the situation of Corollary 3.2.7 (b) we have lim kΩ (z) = lim0 KΩ (z, z) = ∞

z→z 0

z→z

(apply Theorem 6.1.17 in [J-P 1993]). Therefore, Corollary 3.2.7 (b) shows that KΩ does not continuously extend as a map Ω × Ω −→ C. ? It is unknown whether this negative phenomenon does also occur for C ∞ –smooth convex domains ? In addition to Remark 3.2.8 we recall that the Bergman kernel function KD , D ⊂ Cn a smooth bounded strictly pseudoconvex domain, can be smoothly extended to D × D \ ∇(∂D), where ∇(∂D) := {(z, z) : z ∈ ∂D} (see [Ker 1972]). This result was generalized by Bell and Boas (see [Bel 1986], [Boa 1987]) to the following statements:

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3. Bergman metric

(a) Let D ⊂ Cn be a smoothly bounded pseudoconvex domain. Let Γ1 , Γ2 ⊂ ∂D be two open disjoint subsets of the boundary consisting of points of finite type (in the sense of d’Angelo). Then KD extends smoothly to (D ∪ Γ1 ) × (D ∪ Γ2 ). (b) Let D be as in (a) and assume that D satisfies condition (R) 6 . If Γ1 , Γ2 are disjoint open subsets of ∂D and Γ1 consists of points of finite type, then KD extends smoothly to (D ∪ Γ1 ) × (D ∪ Γ2 ). There was the question whether a similar extension phenomenon might be probable for any smoothly bounded pseudoconvex domain. That this is not true is shown by So-Chin Chen [Chen 1996]. Theorem 3.2.9. Let D ⊂ Cn be a smoothly bounded pseudoconvex domain, n ≥ 2. Suppose that its boundary contains a non–trivial complex variety V . Then KD cannot continuously extended to D × D \ ∇(∂D). Proof. Take a regular point z 0 ∈ V and denote by n the outward unit normal at z 0 . Then the smoothness assumption gives an ε0 > 0 such that w − εn ∈ D,

ε ∈ (0, ε0 ), w ∈ ∂D ∩ B(z 0 , ε0 ).

Moreover, we choose a holomorphic disc in V , i.e. a holomorphic embedding ϕ : E −→ V , with ϕ(0) = z 0 and ϕ(E) ⊂ V ∩ B(z 0 , ε0 ). Now assume that KD ∈ C(D × D \ ∇(∂D)). Then sup |KD (z 0 , ϕ(λ))| < ∞. |λ|=1/2

Applying Theorem 6.1.17 in [J-P 1993] and the maximum principle leads to sup |KD (z 0 , ϕ(λ))| = lim sup |KD (z 0 − εn, ϕ(λ) − εn)| |λ|=1/2

ε→0 |λ|=1/2

≥ lim KD (z 0 − εn, z 0 − εn) = ∞; ε→0

a contradiction.

Example 3.2.10 ([Chen 1996]). Fix a smooth real valued function r : R −→ R with the following properties: (i) r(t) = 0 if t ≤ 0, (ii) r(t) > 1 if t > 1, (iii) r00 (t) ≥ 100r0 (t) for all t, (iv) r00 (t) > 0 if t > 0, (v) r0 (t) > 100, if r(t) > 1/2. For s > 1 put Ω := Ωs := {z ∈ C2 : ρ(z) < 0}, where ρ(z) := ρs (z) := |z1 |2 − 1 + r(|z2 |2 − s2 ). Then Ωs is a smoothly bounded pseudoconvex domain in C2 , it is convex and satisfies condition (R), and it is strictly pseudoconvex everywhere except on the set {z ∈ C2 : |z1 | = 1, 0 ≤ |z2 | ≤ s} ⊂ ∂Ω. 6

A bounded domain is said to satisfy condition (R) if the Bergman projection L2 (D) −→ L2h (D) sends C ∞ (D) ∩ L2 (D) to C ∞ (D) ∩ O(D).

3.3. Bergman exhaustiveness

153

Obviously, this set contains non–trivial analytic varieties. So Ω is an example for a domain treated in Theorem 3.2.9. 3.3. Bergman exhaustiveness In the study of the Bergman kernel it is important to know its boundary behavior. We define Definition 3.3.1. Let D ⊂ Cn be a domain and z 0 ∈ ∂D. We say that D is Bergman exhaustive at z 0 (for short, b–exhaustive) if limD3z→z0 kD (z) = ∞. Moreover, if D is b–exhaustive at any of its boundary points, then D is called b–exhaustive. Obviously, any b–exhaustive domain is pseudoconvex. There are a lot of general results giving sufficient condition for a pseudoconvex domain to be b–exhaustive at a boundary point. Besides Theorem 6.1.17 in [J-P 1993] the most general is the following one that relates b–exhaustiveness to the boundary behavior of certain level sets of the Green function. For an arbitrary domain D ⊂ Cn and a point z ∈ D we define Az := Az (D) := {w ∈ D : log gD (z, w) ≤ −1}. Then: Theorem 3.3.2. Let D be a bounded pseudoconvex domain in Cn and z0 ∈ ∂D. Assume that lim Λ2n (Az (D)) = 0. z→z0

Then D is b–exhaustive at z0 . Theorem 3.3.2 is a simple consequence of the following result ([Che 1999], [Her 1999]). Theorem 3.3.3. For any n ∈ N there exists a positive number Cn such that for every bounded pseudoconvex domain D ⊂ Cn the following is true: Z |f (z)|2 ≤ Cn |f (w)|2 dΛ2n (w), f ∈ L2h (D), z ∈ D. kD (z) Az Proof. Let D be a bounded pseudoconvex domain in Cn , z0 ∈ D, and fix an f ∈ L2h (D), f 6= 0. Put Dt := {z ∈ D : dist(z, ∂D) > t}, ∞

0 < t < 1 sufficiently small.

n

Moreover, let ψ1 ∈ C (C , R) be a non negative polyradial symmetric function with R 1 z n ψ(z)dΛ 2n (z) = 1 and supp ψ1 ⊂ Bn (0, 1); put ψt (z) := t2n ψ1 ( t ), z ∈ C , t > 0. Cn On Dt we define ϕt (z) := 2nVt (z) + exp(Vt (z)) + tkzk2 ,

ϕ(z) := 2n log gD (z0 , ·) + gD (z0 , ·),

where Vt := log gD (z0 , ·) ∗ ψt . Finally, we choose a χ ∈ C ∞ (R, [0, 1]) with χ(t) = 1 if t ≤ −2, χ(t) = 0 if t ≥ −1, and |χ0 | ≤ 2. We define the following ∂–closed (0, 1)–form αt on Dt , αt := ∂(χ ◦ Vt · f ) = χ0 (Vt )f ∂Vt .

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3. Bergman metric

Observe that αt is a smooth form whose support R is contained in the set {−2 ≤ Vt ≤ −1}. Moreover, ϕt ≥ −4n on supp αt . Therefore, Dt |αt |2 e−ϕt dΛ2n < ∞. For the Levi form of ϕt we have the following estimate Lϕt (z; X) ≥ eVt (z) |Vt0 (z)X|2 ≥ e−2 |Vt0 (z)X|2 ,

z ∈ supp αt , X ∈ Cn .

Let Q denote the inverse matrix of the coefficient matrix of L. Then, if z ∈ supp αt , we have n X Qj,k (z)αtj (z)αtk (z) exp(−ϕt (z)) ≤ e2 |χ0 (Vt (z))|2 |f (z)|2 e−ϕt (z) ≤ 4e4n+2 |f (z)|2 . j,k=1

Therefore, in virtue of Lemma 4.4.1 in [Hör 1979], there exists a solution ut ∈ C ∞ (Dt ) of the equation ∂ut = αt with the following estimates Z Z |ut |2 e−ϕt dΛ2n ≤ 4e4n+2 |f |2 dΛ2n . Dt

supp αt

Put ( ut e−ϕt /2 vt := 0

on Dt . on D \ Dt

Then the family (vt )t belongs to L2 (D) and satisfies the following uniform estimate Z Z |vt |2 dΛ2n ≤ 4e4n+2 |f |2 dΛ2n D

Az0

(observe that supp αt ⊂ {−2 ≤ Vt ≤ −1} ⊂ Az0 ). In virtue of the Alaoglu–Bourbaki theorem, we may find a function v ∈ L2 (D) satisfying Z Z |v|2 dΛ2n ≤ 4e4n+2

D

|f |2 dΛ2n .

Az0

Put u := veϕ/2 . Then Z Z Z |u|2 dΛ2n ≤ e |v|2 dΛ2n ≤ 4e4n+3 D

|f |2 dΛ2n .

(3.3.8)

Az0

Using distributional derivatives, we find an fe ∈ O(D) such that fe = χ ◦ log gD (z0 , ·) − u almost everywhere on D. Moreover, take a neighborhood U ⊂ D of z0 , log gD (z0 , ·) · f ≤ −3 on U . Then f − fe = u almost everywhere on U . In virtue of (3.3.8), it follows that Z |f − fe|2 e−ϕ dΛ2n < ∞. U ϕ

Observe that e is not locally integrable near z0 ; hence fe(z0 ) = f (z0 ). Summarizing, we have found an fe ∈ L2h (D) with f (z0 ) = fe(z0 ) and Z kfekL2h (D) ≤ (1 + 4e4n+3 ) |f |2 dΛ2n . Az0


155

Consequently, |f (z0 )|2 ≤ kfek2L2 (D) ≤ (1 + 4e4n+3 ) h kD (z0 )

Z

|f |2 dΛ2n ,

Az0

which finishes the proof.

Proof of Theorem 3.3.2. In virtue of Theorem 3.3.3 we know that there is a constant Cn > 0 such that Z 1 ≤ Cn dΛ2n (w) ≤ Cn Λ2n (Az (D)) −→ 0; kD (z) Az Therefore, kD (z) −→ ∞.

z→z0

Moreover, combining Theorem 3.3.2 and a result due to Blocki we have the following (see also [Ohs 1993]). Theorem 3.3.4. For a bounded hyperconvex domain D ⊂ Cn (i.e. there is a negative u ∈ PSH(D) such that the sublevel sets {z ∈ D : u(z) < −ε}, ε > 0, are relatively compact in D), the following is true: Λ2n (Az (D)) −→ 0. In particular, any hyperconvex z→∂D

domain is b–exhaustive. Proof. According to [Blo 1996], there is a function u ∈ C(D) ∩ PSH(D) satisfying the following properties: u|∂D = 0 and (ddc u)n ≥ Λ2n . Applying [Blo 1993], we get for a point z0 ∈ ∂D: Z Z n (− log gD (z, w)) dΛ2n (w) ≤ lim (− max{log gD (z, ·), −k})n (ddc u)n D

k→∞

D

≤ n!kukn−1 L∞ (D) |u(z)| −→ 0, z→z0

where the last inequality is due to Demailly (see [Dem 1987]). Finally, in virtue of Theorem 3.3.3, we get Z Z 1 ≤ Cn dΛ2n (w) ≤ Cn (− log gD (z, w))n dΛ2n (w) −→ 0. z→z0 kD (z) Az (D) D Since z0 is arbitrary, it follows that kD (z) −→ ∞, i.e. D is b–exhaustive. z→∂D

Example 3.3.5. (1) There is a large class of bounded pseudoconvex domains which are hyperconvex, namely Theorem 3.3.6 ([Ker-Ros 1981], [Dem 1987]). Any bounded pseudoconvex domain D ⊂ Cn with a Lipschitz boundary is hyperconvex. In particular, if D has a C 1 –boundary, then it is hyperconvex. (2) Hyperconvexity is even a local property. Theorem 3.3.7 ([Ker-Ros 1981]). Suppose that D is a bounded domain in Cn such that every z0 ∈ ∂D has a neighborhood U = U (z0 ) for which D ∩ U is hyperconvex. Then D itself is hyperconvex.

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3. Bergman metric

(3) Put D := {z ∈ C2 : |z1 | < |z2 | < 1}. Then D is b–exhaustive but not hyperconvex. (For other examples of this type see also Theorems 3.3.8 and 3.3.9 and Example 3.3.23.) For D even more is true. Namely, there is a sequence (zk )k ⊂ D tending to 0 such that Λ2n (Azk (D))k 6→ 0. For Reinhardt domains in C2 we have (see [Zwo 2001a]) the following general result for the pole boundary behavior of the Green function. Theorem 3.3.8. Let D ⊂ C2 be a bounded pseudoconvex Reinhardt domain such that D ∩ (C∗ × {0}) = E∗ × {0}. Moreover, suppose that for a z 0 ∈ D: {v ∈ R2 : (log |z10 |, log |z20 |) + R+ v ⊂ log D} = R+ (0, −1). Then gD (z, w) −→ 0, D3z→0

w ∈ D ∩ C∗ ,

and, therefore, Λ2n (Az (D)) −→ 0. D3z→0

In particular, D is b–exhaustive at the origin but not hyperconvex. Proof. We may assume that D = {z ∈ E 2 : |z2 | < ρ(|z1 |)}, where ρ : [0, 1) −→ [0, 1], ρ(r) = 0 iff r = 0 and ∀A>0 ∃B∈R : log ρ(et ) ≤ At + B,

t ∈ −R+ .

Take a z ∈ D close to 0. Then for w ∈ D with |w1 | = 2|z1 | we have gD (z, w) ≥ gE (z1 , w1 ) ≥

|z1 | |z1 − w1 | ≥ . |1 − z1 w1 | 2

Now we claim that log gD (z, w) ≥ log

|z1 | log |w2 | , 2 log ρ(2|z1 |)

w ∈ D, |w1 | ≥ 2|z1 |.

(3.3.9)

If |w1 | = 2|z1 | then (3.3.9) is true since the second factor is larger or equal 1. Moreover, in virtue of Remark 2.5.6, we know that cD (z, w) −→ ∞ whenever w −→ w∗ , w∗ ∈ ∂D ∩C2∗ . In particular, gD (z, w) −→∗ 1. Therefore, the maximality of the Green function implies w→w

inequality (3.3.9). It remains to show that the right side of (3.3.9) tends to 0 if z tends to 0. In fact, the right side can be written with t = log |z1 | < 0 and v(t) := log ρ(et ) as t − log 2 t − log 2 t + log 2 log |w2 | =: f (t). log |w2 | = v(t + log 2) t + log 2 v(t + log 2) According to our assumption, we know that for any A > 0 we have lim inf t→−∞

particular, since A is arbitrary, limt→−∞

v(t) t

= ∞. Therefore,

finishes the proof.

v(t) t

≥ A. In

lim f (t) = 0, which

t→−∞

A Reinhardt domain satisfying the conditions of Theorem 3.3.8 is given, for example, by D := {z ∈ E∗ × E : |z2 | < e−1/|z1 | }. Hence, D is b–exhaustive but not hyperconvex. For circular domains we have the following result.


157

Theorem 3.3.9 ([Jar-Pfl-Zwo 2000]). Any bounded pseudoconvex balanced domain is b–exhaustive. Proof. Let D = Dh = {z ∈ Cn : h(z) < 1} be a bounded pseudoconvex balanced domain. Fix a boundary point z0 and let M be an arbitrary positive number. Put H := Cz0 . Then, in virtue of the theorem of Ohsawa (see Theorem 3.1.1), we have kD∩H (z) ≤ CkD (z), z ∈ D ∩ H, where C is a suitable positive number. Since D ∩ H is a plane disc, there is an s ∈ (0, 1) such that M < kD∩H (sz0 ). Using the continuity of kD leads to an open neighborhood U = U (z0 ) ⊂ D \ {0} such that kD (z) > M , z ∈ U . 1 Now fix a z ∈ U and define uz : h(z) E −→ R, uz (λ) := kD (λz). This function is 1 subharmonic and radial, so u|[0, h(z) is an increasing function. Therefore, M < uz (1) ≤ ) uz (λ) = kD (λz), 1 ≤ |λ|
1} is an open neighborhood of z0 . Since M is arbitrary, we have lim inf kD (z) = ∞ proving D3z→z0

the theorem.

In the case of a bounded pseudoconvex balanced domain with a continuous Minkowski function, Theorem 3.3.9 was proved in [Jar-Pfl 1989] (see Theorem 7.6.7 in [J-P 1993]). Observe that any bounded hyperconvex balanced domain is taut and therefore its Minkowski function h is continuous. Obviously, there are a lot of bounded balanced pseudoconvex domains with a non-continuous Minkowski function. Moreover, we mention that there exists a bounded pseudoconvex balanced domain D which is not fat (i.e. int D 6= D); see Example 3.1.12 in [J-P 1993]. ? Describe all bounded pseudoconvex circular domain D (i.e. ∀z∈D, θ∈R : eiθ z ∈ D) which are b–exhaustive ? Example 3.3.10. Let D ⊂ Cn be a bounded domain, H : D × Cm −→ [0, ∞) such that log H ∈ PSH(D × Cm ), H(z, λw) = |λ|H(z, w), (z, w) ∈ D × Cm and λ ∈ C. Put GD := {(z, w) ∈ D × Cm : H(z, w) < 1}. GD is a Hartogs domain with m–dimensional fibers. Assume that GD is bounded and pseudoconvex. Then we have the following result. Theorem 3.3.11. Let GD be bounded pseudoconvex as above and let (z0 , w0 ) ∈ ∂GD . Assume that one of the following conditions is satisfied: (a) z0 ∈ D, (b) z0 ∈ ∂D and lim kD (z) = ∞, D3z→z0

(c) there is a neighborhood U = U ((z0 , w0 )) such that U ∩ GD ⊂ {(z, w) ∈ Cn × Cm : kwk < kz − z0 kδ } Then

for some δ > 0. lim kGD ((z, w)) = ∞. In particular, if D is b–exhaustive, then so is

GD 3(z,w)→(z0 ,w0 )

GD . For a proof see [Jar-Pfl-Zwo 2000].

158

3. Bergman metric

Example 3.3.12. The following example shows that Theorem 3.3.11 is far away from being optimal. Fix sequences (aj )j∈N ⊂ (0, 1) and (nj )j∈N ⊂ N with limj→∞ aj = 0 Pk aj and nj ≥ j. Put on Ek := E \ {aj : j = 1, . . . , k}, uk (λ) := j=1 ( 2|λ−a )nj . Observe j| that uk (0) < 0. Define E∞ := E \ ({0} ∪ {aj : j ∈ N}). Then the sequence (uk )k is locally bounded from above on E∞ and globally bounded from below; moreover, it is an increasing sequence of subharmonic functions. It turns out that u := limk→∞ uk ∈ SH(E∞ ) and lim u(x) ≤ 0. Finally, we define the following bounded pseudoconvex (−1,0)3x→0

Hartogs domain with one-dimensional fibers GE∞ := {(z, w) ∈ E∞ × C : |w| < e−u(z) }. Obviously, the point (0, 0) ∈ ∂GE∞ does not satisfy any of the conditions in Theorem 3.3.11. Nevertheless, a correct choice of the nj ’s may show that GE∞ satisfies the cone condition of Theorem in [J-P 1993] at (0, 0). Therefore, kGE∞ ((z, w))

−→

GE∞ 3(z,w)→(0,0)

∞.

The discussion of the other boundary points with the help of Theorem 3.3.11 and Theorem 6.1.17 in [J-P 1993] even proves that GE∞ is b–exhaustive. ? Try to give a complete description of those bounded pseudoconvex Hartogs domains with m–dimensional fibers that are b–exhaustive ? In the complex plane there is even a full characterization of bounded domains being b–exhaustive in terms of the potential theory (see [Zwo 2002]). To be able to present this result we recall a few facts from the classical plane potential theory. 3.3.1. A short course in plane potential theory. (See [Ran 1995]) Let K ⊂ C be compact and P(K) := {µ : µ a probabilistic measure of K}. For µ ∈ P(K), Z pµ (λ) := log |λ − ζ|dµ(ζ), λ ∈ C, K

is the logarithmic potential of µ. Recall that pµ ∈ SH(C) and that pµ |C\K is a harmonic function. To any such a µ one associates its energy Z Z Z I(µ) := pµ (λ)dµ(λ) = log |λ − ζ|dµ(λ)dµ(ζ). K

K

K

A probabilistic Borel measure ν ∈ P(K) is called the equilibrium measure of K if I(ν) = supµ∈P(K) I(µ). It is known that the equilibrium measure exists and is unique if K is not a polar set; then we write νK . Moreover, the logarithmical capacity of any set M ⊂ C is given by cap(M ) := exp(sup{I(µ) : K ⊂ M compact , µ ∈ P(K)}). In the case when M = K compact and not polar then cap(K) = eI(νK ) . Moreover, if M is any Borel set then: M is polar iff cap(M ) = 0. For further applications we collect a few well known properties of the logarithmic capacity: (1) if M1 ⊂ M2 then cap(M1 ) ≤ cap(M2 );


159

S∞ (2) if M1 ⊂ M2 ⊂ M3 ⊂ . . . are Borel sets then cap( j=1 Mj ) = limj→∞ cap(Mj ); T∞ (3) if K1 ⊃ K2 ⊃ K3 . . . are compact sets, then cap Kk −→ cap( k=1 Kk ); k→∞ SN (4) if M = j=1 Mj , Mj Borel sets with diam M ≤ d, N ∈ N ∪ {∞}, then N

X 1 1 ≤ ; log d − log cap M log d − log cap Mj j=1 (4’) if M = then

SN

j=1

Mj , Mj Borel sets with dist(Mj , Mk ) ≥ d > 0, k 6= j, N ∈ N∪{∞}, N

X 1 1 ≥ ; + d log ( cap M ) j=1 log+ capdMj (5) Theorem of Frostman. Let K ⊂ C be a non polar compact subset and νK its equilibrium measure. Then pνK ≥ log cap K on C and pνK = log cap K on K \ F , F ⊂ ∂K a suitable polar Fσ -set. Moreover, pνK (z) = log cap K for z ∈ ∂K, whenever z is regular for the Dirichlet problem for the unbounded component of C \ K. (6) cap B(z, r) = cap(∂B(z, r)) = r and cap K = cap(∂K) ≤ diam K for any compact set K ⊂ C. For a compact set in the complex plane we introduce its Cauchy transform. Definition 3.3.13. Let K ⊂ C be compact. The function fK : C \ K −→ C, (R dνK (ζ) , if K is not polar K z−ζ , fK (z) := 0, if K is polar is called the Cauchy transform of K. (Recall that νK is the equilibrium measure of K.) Obviously, fK ∈ O(C \ K) and fK |D ∈ L2h (D) for any bounded domain D ⊂ C \ K. Then: Lemma 3.3.14 ([Zwo 2002]). For a ρ ∈ (0, 12 ) there exist positive numbers C1 , C2 such that for any pair of disjoint compact sets K, L ⊂ ρE and any domain D ⊂ ρE \ (K ∪ L) the following inequalities hold: |hfK , fL iL2h (D) | ≤ C2 − C1 log dist(K, L), kfK k2L2 (D) h

≤ C2 − C1 log(cap K).

(3.3.10) (3.3.11)

Proof. Obviously, both inequalities are true for any constants Cj when K or L is a polar set. So we may assume that both sets are not polar. Applying the Fubini theorem, we get the following inequality Z Z dν (ζ) Z dν (η) K L |hfK , fL iL2h (D) | = dΛ2 (z) D K z−ζ L z−η Z Z Z 1 ≤ dΛ2 (z)dνL (η)dνK (ζ). |z − ζ||z − η| K L ρE Now we discuss the interior integral.

160

3. Bergman metric

Take ζ, η ∈ ρE, ζ 6= η. Then Z dΛ2 (z) dΛ2 (z) ≤ |z − ζ||z − η| |z||z − (ζ − η)| ρE E Z Z Z dΛ2 (z) dΛ2 (z) dΛ2 (z) = = + . 1 1 1 1 |z||z − 1| E |z||z − 1| E\ 2ρ E |z||z − 1| 2ρ E |ζ−η| |ζ−η|

Z

Observe that the first term in the last expression is finite and independent of η and ζ. For the second summand we proceed as follows: Z

1 Z |ζ−η| Z 2π dΛ2 (z) drdθ = 1 1 1 |z||z − 1| |1 − reiθ | E\ 2ρ E 0 2ρ |ζ−η| 1 1 Z |ζ−η| Z 2π drdθ Z |ζ−η| e2iθ C1 dr eiθ ≤ ≤ −C1 log |ζ − η|, = + 1 + iθ e 1 1 r r r2 (1 − r ) r 0 2ρ 2ρ

where C1 is independent of the discussed ζ, η. Consequently, Z dΛ2 (z) ≤ C2 − C1 log |ζ − η|, |z − ζ||z − η| ρE

ζ, η ∈ ρE, ζ 6= η,

where C1 , C2 are positive constants. Coming back to the beginning, we obtain Z Z |hfK , fL iL2h (D) | ≤ C2 − C1 log |ζ − η|dνK dνL , K

L

which ends the proof.

The main notion here will be the following potential theoretic function. Definition 3.3.15. Let D ⊂ C be a bounded domain. Put αD : D −→ (−∞, ∞], Z 1/2 Z 1/2 dr dr αD (z) := = . 3 3 −r log(cap(B(z, r) \ D)) −r log(cap(B(z, r) \ D)) 0 0 Remark 3.3.16. We denote by Ak (z) the annulus with center z and radii 1/2k+1 , 1/2k , i.e. Ak (z) := {w ∈ C : 1/2k+1 ≤ |w − z| ≤ 1/2k }. Then for a bounded domain D ⊂ C there is an alternative description of αD , namely: ∞

∞

k=2

k=1

X 1X 22k 22k ≤ αD (z) ≤ 8 , 8 − log cap(Ak (z) \ D) − log cap(Ak (z) \ D)

z ∈ D.

To get the lower estimate one only has to use the monotonicity of cap, whereas the upper estimate is based on property (4) of cap. Moreover, αD is semicontinuous from below on D and continuous on D; here use properties (4) and (6) of cap and Fatou’s lemma, respectively the Lebesgue theorem.


161

Remark 3.3.17. For a point z0 = x0 + iy0 ∈ C we define the annuli with respect to the ek (z0 ) := {z = x + iy ∈ C : 1/2k+1 ≤ max{|x − x0 |, |y − y0 |} ≤ maximum norm, i.e. A k e r) := {z = x + iy ∈ C : max{|x − Re a|, |y − 1/2 }, where k ∈ N. Moreover, let B(a, Im a|} < r}, where a ∈ C and r > 0. Then we may define a similar notion to αD , namely Z α eD (z) :=

1/2

0

dr , e r) \ D) −r3 log cap(B(z,

z ∈ D.

We only note that both functions αD and α eD are comparable and that for the new functions inequalities like the ones in Remark 3.3.16 hold. It turns out that, in general, the function αD is not continuous on D (see the next Example 3.3.18). Example 3.3.18 ([Zwo 2002]). Now fix n ∈ N and put en (0) ∩ Mn := A

j

n

2n 2n3

3

+i

k 2n 2n3

o 3 : |j|, |k| = 0, . . . , 2n − 1 .

3

Then Mn has ln := (21+n − 1)2 − (2n − 1)2 elements. We denote them by zn,k , k = 1, . . . , ln . Then we define the following plane domain e 1/4) \ D := B(0,

ln ∞ [ [

e n,k , rn ) ∪ {0} . B(z

n=2 k=1

e rn )) = n2 22n(1+n2 ) , n ≥ 2. Here the radii rn > 0 are chosen such that − log(cap(B(0, Observe that the distance between two different zn,k ’s is equal to dn := 2n 12n3 . Therefore, 3

e rn ) ≥ dn − 2e−n2 22n+2n dn − 2rn ≥ dn − 2 cap B(0, 1 2 1 1 ≥ dn − 2 2 2n+2n3 = n n3 (1 − 2 n+n3 ) =: bn ≥ n+1 n3 > 0. n 2 2 2 n 2 2 2 Hence two of the different “balls” have a distance which is at least bn . In a next step we are going to estimate α eD (0), namely: α eD (0) ≤ C1 = C1

∞ X n=1 ∞ X

∞

n=1 k=1

2n 2+2n3

∞ X

2 2 1 < ∞. 3 = 4C1 2 2n 2n n2 n 2 2 n=1 n=1

Therefore, αD (0) < ∞.

l

n XX 22n 22n ≤ C1 en (0) \ D) e − log cap(A − log cap(B(z

n,k , rn ))

162

3. Bergman metric

en (0). Then To see that αD is not continuous at 0 take an arbitrary point z ∈ A α eD (z) ≥ C2

3 nX −1

j=1

= C2

22(n+j) en+j (z) \ D) − log cap(A

3 nX −1

j=1

22(n+j) log

21+n+n3

1 en+j (z)\D) cap(A

+ log 21+n+n3

.

To continue with the estimate we note that there exists a C3 > 0 such that e n,k , rn ) ⊂ A en+j (z)} ≥ C3 22(n3 −j) , #{k = 1, . . . , ln : B(z

j = 1, . . . , n3 − 1.

Moreover, applying properties (4) and (4’) of cap we obtain l

3

n X 22+2n 1 1 4 ≤ ≤ 2 2n+2n3 = 2 2n en+j (z) \ D) n 2 n 2 e − log cap(A k=1 − log cap(B(zn,k , rn ))

and 1 log+

2n+1 2n3

≥

1 en+j \D) cap(A

C3 22(n log+

3

−j)

,

1

j = 1 . . . , n3 − 1.

e 2n+1 2n3 cap(B(0,r n ))

3

e rn )), cap(A en+j (z)\D)} < 1, if n ≥ n0 Now, observing that 22(1+n+n ) max{cap(B(0, for a suitable n0 ∈ N, leads for n ≥ n0 to α eD (z) ≥ C2

3 nX −1

j=1

22(n+j) 2 log

1 en+j (z)\D) 21+n+n3 cap(A

3

3 n −1 n3 − 1 C2 C3 X 22(n+j) 22(n −j) ≥ = C −→ ∞. 4 2 3 2 j=1 n2 22n +2n n2 n→∞

Hence limD3z→0 αD (z) = limD3z→0 α eD (z) = ∞. Finally, we formulate the main result. Theorem 3.3.19 ([Zwo 2002]). Let D ⊂ C be a bounded domain, z0 ∈ ∂D. Then the following properties are equivalent: (i) D is b–exhaustive at z0 (i.e. limD3z→z0 kD (z) = ∞); (ii) limD3z→z0 αD (z) = ∞. Proof. For the whole proof we may assume that D ⊂ 21 E and z0 = 0 ∈ ∂D. (ii) =⇒ (i): Assume that the statement in (i) is not true. Then there exists a sequence (zk )k∈N ⊂ D ∩ B(0, 1/8) with limk→∞ zk = 0 and supk∈N kD (zk ) =: M < ∞. Put Knk := An (zk ) \ D, n ≥ 2, k ∈ N. Since zk ∈ D there is an Nk ∈ N such that k Kn = ∅ for all k > Nk . Observe that necessarily Nk −→ ∞. k→∞

By assumption (see Remark 3.3.16) we know that Nk X 1 22n αD (zk ) ≤ Sk := −→ ∞. 8 − log cap Knk k→∞ n=2


163

Put k Kn,j

:= Knk ∩{zk +reiθ : r > 0, −π/3+(j −1)2π/3 ≤ θ ≤ π/3+(j −1)2π/3},

j = 1, 2, 3.

In virtue of property (4) of the function cap, we have 3

X 1 1 ≤ . k k − log cap Kn − log cap Kn,j j=1 k k e k := K k Choose j(n, k) such that cap Kn,j ≤ cap Kn,j(n,k) and put K n n,j(n,k) . Then Nk X 1 22n Sk ≤ −→ ∞. e k k→∞ 3 n=2 − log cap Kn

Define

(R fn,k (z) :=

dνn,k (ζ) e k z−eiθn,k ζ K n

if cap Knk 6= 0 if cap Knk = 0

0

,

e k, z ∈C\K n

e nk . where νn,k := νKe k and θn,k such that arg(zk − eiθn,k ζ) ∈ [−π/3, π/3] for all ζ ∈ K n Then Z Z dνn,k (ζ) dνn,k (ζ) e3 C ≥ C3 2n , |fn,k (zk )| ≥ Re ≥ iθ n,k ζ e k |zk − ζ| e k zk − e K K n n e3 , C3 are fixed positive constants. where C There are two cases to be discussed. The first one: assume that there are a subsequence of (zk ), denoted again by (zk ) and a sequence (nk )k ⊂ N with nk ≤ Nk , k ∈ N, such that 22nk lim = ∞. (3.3.12) e nk k→∞ − log cap K k

Put fk := fnk ,k . Then, in virtue of Lemma 3.3.14, we have ek . kfk k2L2 (D) ≤ C2 − C1 log cap K nk h

Therefore, taking (3.3.12) into account, it follows that limk→∞ kD (zk ) = ∞; a contradiction. The second case: we have a certain positive constant C4 such that 22 n e nk − log cap K

≤ C4 ,

k ∈ N, n = 2, 3, . . . , Nk .

e nk , fk,n := f e k . We are going to choose complex numbers ak,n with Put ck,n := cap K Kn ak,n fk,n (zk ) ≥ 0 such that if Nk X fk := ak,n fk,n , n=2

then (|fk (zk )|)k is unbounded whereas (kfk kL2h (D) )k remains bounded by a positive constant C. In that situation we have |fk (zk )|2 1 M ≥ kD (zk ) ≥ ≥ |fk (zk )|2 ; 2 kfk kL2 (D) C h

164

3. Bergman metric

a contradiction. In a first step observe the following inequalities (see Lemma 3.3.14): |2 Rehfk,m , fk,n iL2h (D) | ≤ kfk,m k2L2 (D) + kfk,n k2L2 (D) ≤ 2C2 − C1 log(ck,m ck,n ), h

h

when |n − m| ≤ 1, and |2 Rehfk,m , fk,n iL2h (D) | ≤2C2 + 2C1 max | log(

1 1 1 1 − m+2 )|, | log( n+1 − n+2 )| 2m+1 2 2 2

≤2C2 + C5 mn, when |n − m| ≥ 2. e k,n = 0. Then Put ak,n := 0 if cap K kfk k2L2 (D) ≤ C6

Nk X

h

n=1

≤ C7 Let |ak,n | :=

Nk X

2 − log ck,n bk,n ,

|ak,n ||ak,m |nm

n,m=2, |n−m|≥2

|ak,n |2 (− log ck,n ) +

n=2 n

Nk X

|ak,n |2 (− log ck,n ) + C6 Nk X

2 n|ak,n | .

n=2

where the numbers bk,n ≥ 0 will be fixed later. Then

|fk (zk )| =

Nk X

ak,n fk,n (zk ) ≥ C3

n=2

Nk X

2n bk,n 2n . − log c k,n n=2

So we have to look for numbers bk,n such that |fk (zk )| −→ ∞, but k→∞

kfk k2L2 (D) ≤ C7 h

Nk X

Nk X 2 n 22n 22n (bk,n )2 + bk,n n − log ck,n 2 − log ck,n n=2 n=2

(3.3.13)

remains bounded. PNk 22n Put νk,n := − log n=2 νk,n −→ ∞, νk,n ≤ C4 , k ∈ N, ck,n , k ∈ N. Recall that Sk = k→∞

k and Nk −→ ∞. So we may find sequences (nk,j )qj=0 , where nk,0 = 1, nk,qk = Nk , and

k→∞

qk −→ ∞ such that k→∞

νk,nk,j +1 + · · · + νk,nk,j+1 > 1, and

l 1 < , 2l j+1

j = 0, . . . , qk − 1, l > nk,j+1 .

Now we take bk,nk,j +1 = · · · = bk,nj+1 :=

1 , (j + 1)(νk,nj +1 + · · · + νk,nk,j+1 )

j = 0, . . . , qk − 1.

With this setting we finally obtain that |fk (zk )| −→ ∞ and that (kfk kL2h (D) )k remains k→∞

bounded (compare (3.3.13)). So this part of the proof is complete.


165

(i) =⇒ (ii): Suppose that there is a sequence (zk )k ⊂ D, zk −→ 0, such that, for a k→∞

suitable positive number M , αD (zk ) ≤ M for all k. Then ∞ X

22n ≤ 8M. − log cap(An (zk ) \ D) n=2 2n

In particular, if ck,n := cap(An (zk ) \ D) then log ck,n ≤ −2 8M , k, n ∈ N, n ≥ 2, and therefore we may find an n0 ∈ N such that log ck,n + 1 < −(n + 1) log 2 − 1, n > n0 , k ∈ N. Let z ∈ An (zk ), 1 ≤ n < n0 , then 1 1 1 1 1 + ≥ |z − zk | + |zk | ≥ |z| ≥ |z − zk | − |zk | ≥ n0 − n0 +1 = n0 +1 , 2 2n0 +1 2 2 2 when |zk | < 2n01+1 , i.e. for any k ≥ k0 , k0 suitably chosen. Choose a domain D0 ⊃ D such 1 1 ) = D0 ∩ B(0, 22n ), such that An (zk ) \ D0 = ∅, 1 ≤ n < n0 , k ≥ k0 . that D0 ∩ B(0, 22n 0 0 Applying the localization result (cf. Theorem 3.1.5) for the Bergman kernel, we still know that lim kD0 (zk ) = ∞. k→∞

Now, fix a k ≥ k0 . Recall that there is an n1 > 2n0 such that B(zk , 2n11 ) ⊂ D0 . We exhaust D0 by a sequence of domains Dj0 b D0 with real analytic boundaries such that P∞ 22n 0 0 e n=2 − log cap(An (zk )\Dj0 ) < 8M , ∂(A2 (zk ) \ Dj ) = ∂B(zk , 1/4), Kn := AN (zK ) \ Dj is e n , if K e n 6= ∅, is a regular point either empty or non polar, and any boundary point of K with respect to the unbounded component of its complement. So Frostman’s theorem (see Theorem 3.3.4 in [Ran 1995]) together with the continuity principal for logarithmic potentials (see Theorem 3.1.3 in [Ran 1995]) implies the logarithmic potential pn := pµK , fn e n 6= ∅, is continuous on C. For an n ≥ 3 such that K e n = ∅ put pn := −∞. if n ≥ 3 and K ∞ e For n ≥ 3 choose χn ∈ C (R, [0, 1]) such that χn = 0, if Kn = ∅, or ( e n + 1/2 1, if t ≤ log cap K χn (t) := , 0, if t ≥ −(n + 1) log 2 − 1/2 and |χ0n (t)| ≤

2 en −M1 log cap K

where M1 is a suitable positive number. e n 6= ∅ then pn (z) ≥ For n ≥ 3 define fn := fKe n and ϕn := χn ◦ pn . Note that if K e n ). So ϕn is −(n + 1) log 2, z ∈ / An−1 (zk ) ∪ An (zk ) ∪ An+1 (zk ), and that pn ∈ C ∞ (C \ K a smooth function with support in An−1 (zk ) ∪ An (zk ) ∪ An+1 (zk ) such that ϕn |Ke n = 1 1 n en. and ∂p /K ∂z (z) = 2 f n (z), z ∈ For n = 2 we put p2 (z) := log |z| and take a χ2 ∈ C ∞ (R, [0, 1]) such that ( 0, if t ≤ − log 8 or t ≥ − log 2 χ2 (t) = , 1, if t is near − log 4 and |χ02 | ≤ Then

2 log 4 .

Again, let ϕ2 := χ2 ◦ p2 and put f2 := 1. ∂ϕ |fn (z)| n , (z) ≤ en ∂z −M2 log cap K

z∈

1 e n , n ≥ 2. E\K 2

166

3. Bergman metric

Finally, we define ϕ := sup{ϕn : n ≥ 2}. Note that the supremum is taken over at most three functions. ϕ is a Lipschitz function satisfying ϕ|∂D = 1 and ϕ = 0 in a neighborhood of zk . Now let f ∈ L2h (D0 ). Then the Cauchy formula and the Green formula lead to the following equations: Z Z Z f (λ)dλ 1 (f ϕ)(λ)dλ 1 f (λ) ∂ϕ 1 dΛ2 (λ) . |f (zk )| = = = 2π ∂Dj λ − zk 2π ∂Dj λ − zk π Dj λ ∂λ Applying various versions of the Schwarz inequalities and Lemma 3.3.14 finally gives the following inequalities |f (zk )| ≤ Z ∞ X 2n

|fn−1 (λ)| |f (λ)| |fn+1 (λ)| + + dΛ2 (λ) ≤ e n−1 en e n+1 en − log cap K − log cap K − log cap K An (zk )\K n=2 ∞ 1/2 X kfn−1 k2D0 kfn k2D0 kfn+1 k2D0 M4 kf kL2 (An (zk )\Ke n ) + + 2 h e n−1 )2 e n+1 )2 (− log cap Kn ) (− log cap K (− log cap K n=2 ∞ ∞ X 1/2 X 1/2 p 22n fkf kD0 , ≤ M5 kf k2L2 (A (z )\Ke ) ≤ M n k n e h n=2 n=2 − log cap Kn |f (λ)|

where the constant on the right side is independent of k. This estimate is true for all sufficiently large k. Therefore, (kD0 (zk ))k is bounded; a contradiction. Remark 3.3.20. There are similar considerations as in Theorem 3.3.19 for the so called point evaluation. To be more precise, let z0 ∈ ∂D, where D ⊂ C is a bounded domain. Recall that V := {f ∈ L2h (D) : f is holomorphic in D ∪ {z0 }} is dense in L2h (D) (cf. Theorem 3.1.2). Therefore, we may define the evaluation functional on V , i.e. Φz0 : V −→ C, Φz0 (f ) := f (z0 ). The point z0 is called to be a bounded evaluation point for L2h (D) if Φ extends to a continuous functional on L2h (D). There is the following description of such points [Hed 1972]. Theorem. Let D and z0 be as above. Then αD (z0 ) = ∞ iff z0 is not a bounded evaluation point for L2h (D). Observe, if z0 is not a bounded evaluation point then D is b–exhaustive at z0 . Nevertheless, the converse statement is false (see Example 3.3.18). Remark 3.3.21. For a bounded domain D ⊂ C there are analogous notions like the Bergman kernel taking derivatives into account, namely the n–th Bergman kernel (n)

kD (z) := sup{|f (n) (z)|2 : f ∈ L2h (D) \ {0}, kf kL2h (D) = 1}, (0)

n ∈ N0 , z ∈ D.

Observe that kD = kD . Moreover, one has the following potential theoretic function Z 1/2 dr (n) αD (z) := , z ∈ D, n ∈ N0 . 2n+3 r (− log cap(B(z, r) \ D) 0


167

(0)

Observe that αD = αD . There is the following relation between these notions (see [Pfl-Zwo 2003a]) Theorem. Let n ∈ N0 and d > 1. Then there is a C > 0 such that • for any domain D ⊂ C with diam D < d (n)

(n)

CαD (z) ≤ kD (z), •

for any domain D ⊂ C with

1 d

z ∈ D;

< diam D < d

(n)

(n)

(n)

kD (z) ≤ C max{1, αD (z)(log αD (z))2 },

z ∈ D. (n)

? Let D ⊂ C be a domain and z0 ∈ ∂D. Is it true that limD3z→z0 kD (z) = ∞ (n) implies that limD3z→z0 αD (z) = ∞ ? With the help of the above theorem there is a complete description of those Zalcman domains which are b–exhaustive at all of its boundary points. Corollary 3.3.22 ([Juc 2003]). Let ∞ [

D := E \

B(xk , rk ) ∪ {0}

k=1 7

be a Zalcman domain , where xk > xk+1 > 0, limk→∞ xk = 0, rk > 0 with B(xk , rk ) ⊂ E, B(xk , rk ) ∩ B(xj , rj ) = ∅, k, j ≥ 1, k 6= j. Assume that xk+1 ≤ Θ2 , k ∈ N. ∃Θ1 ∈(0,1) ∃Θ2 ∈(Θ1 ,1) : Θ1 ≤ xk P∞ −1 Then D is b–exhaustive iff D is b–exhaustive at 0 iff k=1 x2 log rk = ∞ iff αD (0) = ∞. k

Observe that special cases were treated also in [Ohs 1993] and [Che 1999]. Moreover, we mention that the domains D in Corollary 3.3.22 are fat domains, but not all of them are b–exhaustive (for another example see [Jar-Pfl-Zwo 2000]). Proof. First, observe that for every boundary point z0 except the origin we have lim

D3z→z0

kD (z) = ∞

(use Theorem 6.1.17 in [J-P 1993]). Obviously, B(xk+1 − rk+1 /2, rk+1 ) ⊂ B(0, δ) \ D, δ ∈ (xk+1 , xk ). Then ∞ Z xk X dr αD (0) ≥ 3 −r log cap(B(0, r) \ D) k=1 xk+1 Z ∞ ∞ ∞ x k X X X −1 −1 dr ≥ C , ≥ (x − x ) ≥ k k+1 rk+1 2 3 x x3k log rk+1 xk+1 −r log 2 k+1 log rk+1 2 k=k0

k=k0

k=k0

where C is a constant. Observe that for the last inequality the assumption on the centers xk was used. 7

Observe that we use here a slightly more general notion than the one of a Zalcman type domain in Section 2.7.

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3. Bergman metric

Now, the divergence of the series in the corollary implies that αD (0) = ∞. In virtue of the lower semicontinuity of the function αD it follows that limD3z→0 αD (z) = ∞. On the other hand we have ∞ Z xk Z 1/2 X dr αD (0) = + 3 log cap(B((0, r) \ D) −r x1 k=1 xk+1 ≤C1 +

j ∞ ∞ ∞ X X xk − xk+1 X −1 −1 X 1 ≤ C + C 1 2 x3k+1 log rj log rj x2k j=1 j=k

k=1

≤C1 +

∞ X

k=1

j X

2(j−k) Θ2 −1 C2 log rj x2j j=1 k=1

≤ C1 + C3

∞ X j=1

−1 , x2j log rj

where C1 ≥ 0 and C2 , C3 > 0 are suitable numbers. Observe that the last three inequalities follow from the assumptions on the centers xk . If the series in the corollary does converge, then αD (0) < ∞. Moreover, directly from the definition we see that αD restricted to the interval (−1/4, 0] is monotonically increasing. Hence lim sup0>x→0 αD (x) ≤ αD (0) < ∞. So, the corollary is proved. Example 3.3.23. We discuss the particular case of a Zalcman domain, namely xk := (1/2)k and rk := (1/2)kN (k) , where Nk ∈ N, k ≥ 2. Then we have D is b–exhaustive

∞ X

iff

k=2

22k = ∞. kN (k) log 2

On the other hand, following Ohsawa [Ohs 1993] we have D is hyperconvex

iff

∞ X

1/N (k) = ∞.

k=2

So we see that there are plenty of Zalcman domains which are not hyperconvex but, nevertheless, they are b–exhaustive. 3.4. L2h –domains of holomorphy The boundary behavior of the Bergman kernel may be used to give a complete description of L2h –domains of holomorphy 8 . The precise result is the following one. Theorem 3.4.1 ([Pfl-Zwo 2002]). For a bounded domain D ⊂ Cn the following conditions are equivalent: (i) D is an L2h –domain of holomorphy; (ii) lim supD3z→z0 kD (z) = ∞ for every boundary point z0 ∈ ∂D. 8

Recall that a domain D ⊂ Cn is an L2h –domain of holomorphy if for any pair of open sets U1 , U2 ⊂ Cn with ∅ 6= U1 ⊂ D ∩ U2 6= U2 , U2 connected, there is an f ∈ L2h (G) such that for any F ∈ O(U2 ): f |U1 6= F |U1 .

3.4. L2h –domains of holomorphy

169

Remark 3.4.2. There is also the following more geometric condition which is equivalent to (i) of Theorem 3.4.1: (iii) for any boundary point z0 ∈ ∂D and for any open neighborhood U = U (z0 ) the set U \ D is not pluripolar 9 . Proof of Theorem 3.4.1. The case n = 1 may be found in [Con 1995]. So we will always assume that n ≥ 2. (ii) =⇒ (i): Suppose that G is not an L2h –domain of holomorphy. Then there are concentric polydiscs P b Pe satisfying P ⊂ D, ∂P ∩ ∂D 6= ∅, and Pe 6⊂ D such that for any function g ∈ L2h (D) there exists an gb ∈ H∞ (Pe) with gb|P = g|P . Let a be the center of P and let L be an arbitrary complex line through a. Then L ∩ Pe \ D =: K is a polar set (in L). Indeed, suppose that K is not polar. Fix a compact non–polar subset K 0 ⊂ K. Then, according to Theorem 9.5 in [Con 1995], there is a non–trivial function f ∈ L2h (L \ K 0 ) which has no holomorphic extension to L. Since K 0 ∩ D = ∅, Theorem 3.1.1 guarantees the existence of a function F ∈ L2h (D) with F |L∩D = f |L∩D . Hence, we find Fb ∈ O(Pe) such that Fb|P = F |P . In particular, Fb|L∩Pe extends f to the whole of L; a contradiction. So L ∩ Pe ∩ D is connected 10 . Since L is arbitrary, D ∩ Pe is connected. Therefore, for any function g ∈ L2h (D) there exists a unique holomorphic extension gb ∈ H∞ (Pe) with gb|D∩Pe = g|D∩Pe . Consider the linear space A := {(g, gb) : g ∈ L2h (D)} ⊂ L2h (D) × H∞ (Pe) equipped with the norm k(g, gb)k := kgkL2h (D) + kb g kH∞ (Pe) . Then A is a Banach space. Observe that the mapping A 3 (g, gb) −→ g ∈ L2h (D) is a one-to-one, surjective, continuous, linear mapping. Hence, in view of the Banach open mapping theorem, its inverse map is also continuous, i.e. there is a C > 0 such that k(g, gb)k ≤ CkgkL2h (D) ,

g ∈ L2h (D).

In particular, kb g kH∞ (Pe) ≤ CkgkL2h (D) . So we are led to the following estimate o n |g(z)|2 e, 0 6≡ g ∈ L2h (D) ≤ C 2 . : z ∈ D ∩ P sup{kD (z) : z ∈ D ∩ Pe} = sup kgk2L2 (D) h

In particular, lim supz→w kD (z) ≤ C 2 for a point w ∈ ∂P ∩ ∂D 6= ∅ (recall that such a point exists); a contradiction. Before we are able to start the proof of (i) =⇒ (ii) we need some auxiliary results. Lemma 3.4.3. Let G ⊂ C be a bounded domain and let a ∈ ∂G. Assume that lim supG3z→a kG (z) < ∞. Then there is a neighborhood U = U (a) such that U \ G is polar. 9

Recall that a set P ⊂ Cn is called to be pluripolar if there is a u ∈ PSH(Cn ), u 6≡ −∞, such −1 that P ⊂ u (−∞). 10 Recall that for a plane domain G and a relatively closed polar subset M ⊂ G the open set G \ M is connected.

170

3. Bergman metric

Proof. Suppose that Lemma 3.4.3 is not true. First we claim that for any r > 0 the intersection B1 (a, r) ∩ ∂G is not polar. Otherwise, there is an r0 > 0 such that B1 (a, r0 )∩∂G is polar. Observe that B1 (a, r0 /4)\G is not polar. Therefore, there exists a b0 ∈ B1 (a, r0 /4) \ G. Choose a point b ∈ B1 (a, r0 /4) ∩ G. Since B1 (a, r0 ) ∩ ∂G is polar, there exists an s ∈ (0, r0 /2) such that ∂B1 (b0 , s) ∩ ∂G = ∅, ∂B1 (b0 , s) ∩ G 6= ∅, and ∂B1 (b0 , s) ⊂ B1 (a, r0 ). Hence, ∂B1 (b0 , s) ⊂ G. Therefore, for any z ∈ ∂B1 (b0 , s), one has [b0 , z] ∩ ∂G 6= ∅. Then, in virtue of Theorem 5.1.7 in [Arm-Gar 2001], it follows that ∂B1 (b0 , s) is a polar set; a contradiction. Hence, there is a sequence (aj )j ⊂ ∂G, aj −→ a, where all the points aj are regular boundary points of G. Suppose for a moment that we knew that limG3z→b kG (z) = ∞ for a regular boundary point b of G. Then lim supG3z→a kG (z) = ∞ which obviously contradicts the assumption of Lemma 3.4.3. It remains to prove the following claim: Let b ∈ ∂G be a regular point. Then kG (z) −→ ∞ when G 3 z −→ b. Here we will use the following relation of the Bergman kernel and the Azukawa metric given in [Ohs 1995], namely, there is a C > 0 such that p kG (z) ≥ CAG (z; 1), z ∈ G. Define Gp := {z ∈ G : log gG (p, z) < −1}, p ∈ G, and r(p) := diam Gp . Then, using [Zwo 2000c], we get e AG (p, 1) = eAGp (p, 1) ≥ eAB1 (p,r(p)) (p, 1) = −→ ∞ as p −→ b. r(p) So it remains to show that r(p) −→ 0 when p −→ b. Suppose this is not true. Then we find an ε > 0, sequences G 3 pj −→ b and G 3 zj −→ z ∗ ∈ G such that |pj − zj | ≥ ε and log gG (pj , zj ) < −1, j ∈ N. Choose a small disc V around z ∗ with b ∈ / V . Then we have e := G ∪ V . Now, observe that g e (pj , ·) = g e (·, pj ) and gG (pj , zj ) ≥ gGe (pj , zj ), where G G G that log gGe (pj , ·) −→ 0 pointwise. Applying that log gGe (pj , ·) are harmonic functions, the Vitali theorem implies that log gGe (pj , ·) tends uniformly to 0 on some small neighborhood of z ∗ contradicting that log gG (pj , zj ) < −1 for all j. Lemma 3.4.4. Let D ⊂ Cn , n ≥ 2, be a domain and let 0 < r < t. For any z 0 ∈ Cn−1 define Dz0 := {zn ∈ tE : (z 0 , zn ) ∈ D} =: tE \ K(z 0 ). 0 Assume that K(0 ) is polar and that there is a neighborhood V of 00 such that for almost all z 0 ∈ V the set K(z 0 ) is also polar. Then there is a neighborhood V 0 ⊂ V of 00 such that for any f ∈ L2h (D) there exists an fb ∈ O(V 0 × rE) with f = fb on D ∩ (V 0 × rE). Proof. Since K(00 ) is a polar set, there is an s with r < s < t such that K(00 )∩∂B1 (0, s) = ∅. Therefore, we find a neighborhood V 0 = V 0 (00 ) ⊂ V such that K(z 0 ) ∩ ∂B1 (0, s) = ∅, z 0 ∈ V 0 . Then we may define Z 1 f (z 0 , λ) 0 b f (z , zn ) := dλ, (z 0 , zn ) ∈ V 0 × B1 (0, s). 2π ∂B1 (0,s) λ − zn Obviously, fb ∈ O(V 0 × B1 (0, s)).

3.4. L2h –domains of holomorphy

171

On the other hand, using that f ∈ L2h (D), the Fubini theorem and the assumptions made in Lemma 3.4.4 give that for almost all z 0 ∈ V 0 the function f (z 0 , ·) ∈ L2h (B1 (0, t) \ 0 K(z 0 )) and K(z 0 ) is polar. 1 (0, t) Hence, f (z , ·) extends to a holomorphic function on B 0 0 11 for almost all z ∈ V . Applying the Cauchy integral formula, we obtain f (z 0 , zn ) = fb(z 0 , zn ), (z 0 , zn ) ∈ V 0 × B1 (0, s), for almost all z 0 ∈ V 0 . Since this set is dense in (V 0 × B1 (0, s)) ∩ D, we have reached the claim in Lemma 3.4.4. Now we are able to complete the proof of Theorem 3.4.1. (i) =⇒ (ii): Fix a boundary point w ∈ ∂D. First we discuss the case when w ∈ / / D, j ∈ N. By int(D). Then there is a sequence (zj )j ⊂ Cn such that zj −→ w and zj ∈ rj we denote the largest radius such that Bj := Bn (zj , rj ) does not intersect D. Select wj ∈ ∂Bj ∩ ∂D. Then, wj −→ w. Observe that the domain D satisfies the general outer cone condition at wj (see Theorem 6.1.17 in [J-P 1993]). Therefore, limD3z→wj kD (z) = ∞. Hence, (ii) follows. From now on we assume that w ∈ int(D). Suppose that (ii) is not true for w. Then there are a polydisc P b D with the center at w and a constant C > 0 such that kD (z) ≤ C,

z ∈ D ∩ P.

Now, let L be a complex line through P . Then (L ∩ P ) \ D is a polar set (in L) or it is empty. Indeed, otherwise we apply Lemma 3.4.3. Therefore, sup{kD∩L (z) : z ∈ L ∩ P ∩ D} = ∞. Then, in virtue of Theorem 3.1.1, it follows that sup{kD (z) : z ∈ L ∩ D ∩ P } = ∞; a contradiction. Observe that there is a complex line L∗ passing through w and P ∩D. We may assume that w = 0 and, after a linear change of coordinates, that P = E n and L∗ = {(0, . . . , 0)}× C. So the assumptions of Lemma 3.4.4 are fulfilled with respect to some neighborhood V ⊂ E n−1 of 00 ∈ Cn−1 . Therefore, there is a neighborhood V 0 = V 0 (00 ) ⊂ V such that for any f ∈ L2h (D) there is an fb ∈ O V 0 ×B1 (0, 1/2) with f = fb on D ∩ V 0 ×B1 (0, 1/2) contradicting the assumption in (i). Remark 3.4.5. In [Irg 2003], the following generalization of Theorem 3.4.1 may be found. Theorem. Let (X, π) be a Riemann domain over Cn such that π(X) is bounded. Let b π e π (X, b) be the envelope of holomorphy and (X, e) the L2h (X)–envelope of holomorphy of b e (X, π). Then (X, π b) embeds into (X, π e) and the difference of these two sets is a pluripolar e subset of X. Applying Theorem 3.3.9, Theorem 3.4.1 may be used to get the following result. Corollary 3.4.6 ([Jar-Pfl 1996]). Any bounded balanced domain of holomorphy is an L2h –domain of holomorphy. 11

Recall that a relatively closed polar subset of a plane domain is a removable set of singularities for square–integrable holomorphic functions.

172

3. Bergman metric

? It is an open problem to characterize those unbounded domains of holomorphy that are L2h –domains of holomorphy. Even more, so far there is no description of such unbounded domains that carry a non trivial L2h –function ? 3.5. Bergman completeness

kG

Let G ⊂ Cn be a domain such that kG (z) > 0, z ∈ G 12 . Then the Bergman kernel is a logarithmically psh function on G. So n X 1/2 ∂2 βG (z; X) := log kG (z)Xj Xk , z ∈ G, X ∈ Cn , ∂zj ∂zk j,k=1

gives a hermitian pseudometric on G. It is the Bergman pseudometric. Recall that there is another description of the Bergman pseudometric. Let G ⊂ Cn be as above. We define MG (z; X) = sup{|f 0 (z)X| : f ∈ L2h (G), kf kL2h (G) = 1, f (z) = 0}, z ∈ G, X ∈ Cn . Then, MG (z; X) βG (z; X) = p . kG (z)

(3.5.14)

Observe that βG is a metric if ∀z∈G and ∀X∈Cn , X6=0 ∃g,f ∈L2h (g) : g(z) 6= 0 and f (z) = 0, f 0 (z)X 6= 0.

(3.5.15)

The Bergman pseudodistance on G is given by nZ 1 o bG (z, w) := inf βG (γ(t); γ 0 (t))dt : γ ∈ C 1 ([0, 1], G) : γ(0) = z, γ(1) = w , z, w ∈ G. 0

Under the condition (3.5.15), the function bG is in fact a distance. One of the main questions here is to decide which domain in Cn is bG –complete. Definition 3.5.1. A domain G ⊂ Cn satisfying kG (z) > 0 for all z ∈ G is called Bergman-complete (for short b–complete or bG –complete) if bG is a distance and if for any bG –Cauchy sequence (zj )j ⊂ G there is a point a ∈ G such that limj→∞ zj = a. Obviously, for any bounded domain D ⊂ Cn , βD is a metric and bD is a distance on D. For a not necessarily bounded domain D, we have the following sufficient criterion (see [Che-Zha 2002]). Theorem 3.5.2. Let D ⊂ Cn be a pseudoconvex domain (not necessarily bounded). Assume for any point w ∈ D there is an r > 0 such that Aw (D; r) := {z ∈ D : log gD (w, z) < −r} ⊂⊂ D. 13 Then βD is a metric on D and bD is a distance. 12

f (z) 6= 0. 13

Observe that this condition is fulfilled if for any z ∈ G there exists an f ∈ L2h (G) such that Observe that this condition is always true for a bounded domain.

3.5. Bergman completeness

173

Proof. The construction of functions verifying condition (3.5.15) is done by solving a ∂–problem. For more details, the reader may consult [Che-Zha 2002]. From the above theorem we immediately get the following one–dimensional result (see [Che-Zha 2002]). Corollary 3.5.3. Any hyperbolic Riemann surface has a Bergman metric and distance. In particular, any plane domain D ⊂ C, such that C\D is not a polar set, has a Bergman distance. Remark 3.5.4. Moreover, any domain D ⊂ Cn which carries either a bounded continuous strictly psh function or a negative function u ∈ PSH(D) such that {z ∈ D : u(z) < −r} b D, r > 0, satisfies (3.5.15), i.e. D allows a Bergman distance. For more details see [Che-Zha 2002]. Moreover, the following result due to N. Nikolov (private communication) is also a consequence of Theorem 3.5.2. Corollary 3.5.5. Let D ⊂ Cn be an unbounded domain. Assume that there are R > 0 and ψ ∈ PSH(D \ B(R)) such that: • ψ < 0 on D \ B(R), • lim ψ(z) = 0, z→∞

• lim sup ψ(z) < 0, a ∈ (∂D) \ B(R). z→a

Then D has the Bergman metric. Proof. Fix a z0 ∈ D and choose positive numbers R3 > R2 > R1 > R such that kz0 k < R1 and 2 inf ψ ≥ sup ψ =: c < 0. D\B(R2 )

D∩∂B(R1 )

Moreover, put d :=

inf w∈D∩∂B(R1 )

log gB(R3 ) (z0 , w) > −∞,

u(w) := 2ψ(w)(d/c) − d,

w ∈ D \ B(R).

Observe that u(w) ≤ d ≤ log gB(R3 ) (z0 , w),

w ∈ D ∩ ∂B(R1 ),

u(w) ≥ 0 ≥ log gB(R3 ) (z0 , w),

w ∈ D ∩ ∂B(R2 ).

Hence, the following function   w ∈ D ∩ B(R1 ) log gB(R3 ) (z0 , w), v(w) := max{log gB(R3 ) (z0 , w), u(w)}, w ∈ D ∩ (B(R2 ) \ B(R1 ))   u(w), w ∈ D \ B(R2 ) is psh on D with logarithmic pole at z0 . Therefore, v + d ≤ log gD (z0 , ·) on D. Since v = u ≥ 0 on D \ B(R2 ), we have logD (z0 , ·) ≥ d on D \ B(R2 ). And if w ∈ D ∩ B(R2 ), then log gD (z0 , w) ≥ log gB(R3 ) (z0 , w) + d ≥ log gB(z0 ,R3 +kz0 k) (z0 , w) + d.

174

3. Bergman metric

Let B(z0 , s) b D ∩ B(R2 ), then there is a d1 < 0 such that log gD (z0 , w) ≥ d + d1 , w ∈ D ∩ B(R2 ) \ B(z0 , s). Therefore, Az0 (D; d + d1 ) b D and, since z0 was arbitrarily chosen, Theorem 3.5.2 finishes the proof. It is an old result due to Bremermann that a bounded b–complete domain in Cn is pseudoconvex. On the other side there is the following sufficient conditions for a bounded domain to be b–complete due to Kobayashi. Theorem 3.5.6. Let D ⊂ Cn be a bounded pseudoconvex domain. (a) Assume that |f (z)| lim sup p < kf k2L2 (D) , h kD (z) z→∂D

f ∈ L2h (D) \ {0}.

Then D is b–complete. (a’) Let H ⊂ L2h (D) be a dense subspace. Moreover, assume that for any sequence (zj )j ∈ D, zj −→ z0 ∈ ∂D, and any g ∈ H, there is a subsequence (zjk )k such that |g(zjk )| lim p = 0. kD (zjk )

k→∞

Then D is b–complete. (b) For all z, w ∈ D we have bD (z, w) ≥ arccos √

|KD (z,w)| kD (z)

√

kD (w)

.

The statements (a) and (b) are due to Z. Blocki (see [Blo 2002], [Blo 2003]). Observe that (b) explains the connection between the Bergman distance and the Skwarczy´ nski distance (see for more details [J-P 1993]). Proof. (a) Suppose D is not b–complete. Then, in virtue of the proof of Lemma 7.6.4 in [J-P 1993], we may find an f ∈ L2h (D), kf kL2h (D) = 1, and real number θj such that eiθj KD (·, zj ) p −→ f kD (zj ) j→∞

in L2h (D).

|f (z )| Therefore, taking the scalar product with f , we get √ j −→ kf k2 ; a contradiction. kD (zj )

(a’) Suppose again that D is not b–complete and choose f and θj as above. Moreover, take a g ∈ H with kg − f kL2h (D) < 1/2. Then, in virtue of our assumption, there is a subsequence (zjk )k such that |f (zjk )| |g(zjk )| 1 1 ←− p ≤ kf − gkL2h (D) + p −→ kf − gkL2h (D) < ; k→∞ 2 kD (zjk ) kD (zjk ) k→∞ a contradiction.

Remark 3.5.7. Most of the results on b–completeness will be based on Theorem 3.5.6. In order to verify b–completeness one could also try to find good quantitative estimates for the Bergman distance or the Bergman metric near the boundary. For example, there are the following two positive results.


175

Theorem ([Die-Ohs 1995]). Let D ⊂ Cn be a C 2 –smooth bounded pseudoconvex domain and let z0 ∈ D. Then there exist positive constants C1 and C2 such that bD (z0 , z) ≥ C1 log | log(C2 dist(z, ∂D))| − 1,

z ∈ D.

(3.5.16)

Theorem ([Blo 2002]). Let D be as above and let z0 ∈ D. Then there is a positive constant C such that 1 log dist(z,∂D) , z ∈ D, z sufficiently close to ∂D. (3.5.17) bD (z0 , z) ≥ C 1 log log dist(z,∂D) In fact, both estimates (3.5.16) and (3.5.17) remain true in a more general situation, namely, for bounded pseudoconvex domains, not necessarily smooth, which allow a good bounded psh exhaustion function. On the other side, the following example shows that there are certain obstacles, even for smooth domains, to allow a good boundary behavior of the Bergman metric. Theorem ([Die-Her 2000]). Let a ∈ (0, 1). Then there exists a bounded pseudoconvex domain D ⊂ C2 given as D = {z ∈ C2 : r(z) < 0} with a smooth boundary, 0 ∈ ∂D, where the defining function r is of the form r(z) = Re z1 + b|z1 |2 + ρ(z2 ) for suitable ρ ∈ PSH(C), ρ(0) = 0, and b > 0 such that there are no positive constant C and no neighborhood U = U (0) ⊂ C2 such that Pn ∂r j=1 ∂zj (z)Xj βD (z; X) ≥ C 1 1+2a , z ∈ D ∩ U. 1 |r(z)| log |r(z)| As a consequence of Theorem 3.5.6 we get (see [Blo-Pfl 1998], [Her 1999]) Theorem 3.5.8. Let D ⊂ Cn be a bounded pseudoconvex domain. Assume that lim

D3z→∂D

Λ2n (Az (D)) = 0.

Then D is b–complete. In particular, any hyperconvex bounded domain is b–complete. Proof. Fix an f ∈ L2h (D) \ {0}. Using Theorem 3.3.3 we have Z |f (z)|2 ≤ Cn |f (w)|2 dΛ2n (w), z ∈ D. kD (z) Az (D) Then the assumption and the Theorem 3.5.6(a’) immediately gives the proof. Finally, it suffices to recall that a hyperconvex domain fulfills the condition on the level sets of the Green function. Remark 3.5.9. In [Che 2003], a similar result is announced even for arbitrary domains, namely: Let D ⊂ Cn be a (not necessarily bounded) domain. Assume that there is a strictly psh function u : D −→ [−1, 0) such that all sublevel sets {z ∈ D : u(z) < c}, c ∈ (−1, 0), are relatively compact subsets of D. Then D is b–complete. For weaker results see also [Che-Zha 2002].

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3. Bergman metric

A direct consequence of Theorem 3.5.6 is the following sufficient criterion for b– completeness. Corollary 3.5.10. Let D ⊂ Cn be a bounded b–exhaustive domain. Assume that there is a dense subspace H ⊂ L2h (D) such that any f ∈ H is bounded near z0 ,

z0 ∈ ∂D.

Then D is b–complete. Applying this result together with Theorem 3.1.2 leads to the following sufficient criterion for a plane domain to be b–complete (see [Che 2000]). Corollary 3.5.11. Any bounded b–exhaustive domain D ⊂ C is b–complete. Remark 3.5.12. For a bounded pseudoconvex domain, a localization result for the Bergman metric is well known ([J-P 1993]; for a sharper version see also [Her 2003]) This implies that a bounded pseudoconvex domain in Cn is b–complete iff D is locally b–complete, i.e. for any a ∈ ∂D there is an open neighborhood U = U (a) such that any connected component V of D ∩ U is b–complete. Due to N. Nikolov [Nik 2003a], there is an analogous result in the plane case for the unbounded situation, namely: Theorem 3.5.13. Let D ⊂ C be a domain such that C \ D is not a polar set. Assume that D is locally b–complete 14 . Then D is b–complete. The proof of Theorem 3.5.13 is based on the following lemma. Lemma 3.5.14. Let D ⊂ C be a domain such that C \ D is not polar. Moreover, let a ∈ ∂D and U = U (a) be an open neighborhood of a. Then there exists a neighborhood V = V (a) ⊂ U and a constant C > 0 such that CβUb (z; 1) ≤ βD (z; 1),

z ∈ V ∩ D,

b denotes that connected component of D ∩ U with z ∈ U b. where U Proof. Since C \ D is not polar, there is an r0 > 0 suchthat C \ D ∪ B(a, r0 ) is not polar. Hence, log gD∪B(a,r0 ) is harmonic on D ∪ B(a, r0 ) \ {a}. Fix an r1 ∈ (0, r0 ) and define D1 := D ∪ B(a, r1 ). Applying that gD1 (a, z) ≥ gD∪B(a,r0 ) (a, z), z ∈ D1 , we have inf{log gD1 (a, z) − |z − a|2 : z ∈ ∂B(a, r1 ) ∩ D} =: m > −∞. Put

( max{|z − a|2 + m, log gD1 (a, z)}, u(z) := log gD1 (a, z),

if z ∈ D ∩ B(a, r1 ) . if z ∈ D \ B(a, r1 )

Observe that |z − a|2 + m ≤ gD1 (a, z), z ∈ D ∩ B(a, r1 ). Therefore, 0 ≥ u ∈ SH(D1 ) and u(z) = |z − a|2 + m, z ∈ B(a, r2 ), for a sufficiently small r2 < r1 . Choose numbers 0 < r4 < r3 < r2 and a C ∞ cut–off function χ such that χ ≡ 1 on B(a, r4 ) and χ ≡ 0 outside of B(a, r3 ). 14

Observe that here the point ∞ is counted as a boundary point of D; so we assume also that there is a compact set K ⊂ C such that any (non empty) connected component of D \ K is b–complete.


177

e be the connected component of D ∩ U with Fix a point z0 ∈ D ∩ B(a, r4 ) and let U 2 e e z0 ∈ U . Take an f ∈ Lh (U ) with f (z0 ) = 0. Put ( e ∂(χf )(z), if z ∈ U . α(z) := e 0, if z ∈ D \ U ∞ Then α is a ∂–closed C(0,1) –form on D satisfying the following inequality Z Z e |α(z)|2 e−6 log gD (z0 ,z)−u(z) dΛ2 (z) ≤ C |f (z)|2 dΛ2 (z) < ∞, D

e U

e > 0 is independent of f and z0 . Observe that the subharmonic weight function where C is strictly subharmonic near the support of α. Therefore, using Hörmander’s L2 –theory 15 , we get a function h ∈ C ∞ (D) with ∂h = α on D such that Z 2 khkL2 (D) ≤ |h(z)|2 e−6 log gD (z0 ,z)−u(z) dΛ2 (z) ≤ C 0 kf k2L2 (Ue ) , D

h

0

where C is a positive number which is independent of f and z0 . Moreover, since the second integral is finite, it follows that h(z0 ) = h0 (z0 ) = 0. Hence, the function ( e (χf )(z) − h(z), if z ∈ U b f (z) := e −h(z), if z ∈ D \ U is holomorphic on D satisfying fb(z0 ) = 0, fb0 (z0 ) = f 0 (z0 ), and kfbkL2h (D) ≤ Ckf kL2 (Ue ) , h where C > 0 is independent of f and z0 . Therefore, in virtue of (3.5.14), we get e D (z0 ; 1) ≥ β e (z0 ; 1). Since z0 was arbitrary, the lemma is proved. Cβ U Now, we turn to the proof of Theorem 3.5.13. Proof of Theorem 3.5.13. First of all, let us mention that, in virtue of Corollary 3.5.3, D has a Bergman metric. Suppose now that D is not b–complete. Then there is a b–Cauchy sequence (zj )j ⊂ D with zj −→ a ∈ ∂D or zj −→ ∞. The second case can be reduced to the first one using 1 the biholomorphic transformation z −→ z−c , where c ∈ / D. So we only have to deal with the first case. By the assumption, there is an open neighborhood U = U (a) such that any connected component of U ∩ D is b–complete. Fix a positive r1 such that B(a, r1 ) b U . Applying Lemma 3.5.14, we may find positive numbers r2 < r1 and C such that and b and U e denote the conCβD (z; 1) ≤ min{βUb (z; 1), βUe (z; 1)}, z ∈ D ∩ B(a, r2 ), where U b ∩U e . Choose an nected components of D ∩ U and D ∩ B(a, r1 ), respectively, with z ∈ U 15

Here we use the following form of H¨ ormander’s result.

∞ (D). Assume that Theorem. Let D ⊂ Cn be a pseudoconvex domain, ϕ ∈ PSH(D), and α ∈ C(0,1)

∂α = 0 and that on an open set U ⊂ D, supp α ⊂ U , the function ϕ can be written as ϕ = ψ + χ, ψ, χ ∈ PSH(U ), such that Lψ(z; X) ≥ RCkXk2 , z ∈ U , X ∈ Cn . Then there exists an h ∈ C ∞ (D), R ∂h = α, such that D |h|2 e−ϕ dΛ2n ≤ C 0 D |α|2 e−ϕ dΛ2n , where C 0 > 0 depends only on C.

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3. Bergman metric

r3 ∈ (0, r2 ). Put e, U e d := inf{bUe (z, w) : z ∈ ∂B(a, r3 ) ∩ D, w ∈ ∂B(a, r2 ) ∩ D, z, w ∈ U a connected component of D ∩ B(a, r1 )}. In virtue of the inequality c ≤ b, it follows that d > 0. So we may take an index k0 ∈ N such that bD (zk , z` ) < Cd 2 and zk ∈ B(a, r3 ), k, ` ≥ k0 . Fix such k, ` with zk 6= z` . Then there is a C 1 –curve αk,` : [0, 1] −→ D such that Z 1 0 2bD (zk , z` ) > βD αk,` (t); αk,` (t) dt. 0

Suppose this curve is not lying in B(a, r1 ). Then there are numbers 0 < s1 < s2 < 1 such that αk,` (s1 ) ∈ ∂B(a, r3 ), αk,` (s2 ) ∈ ∂B(a, r2 ), and αk,` ([s1 , s2 ]) ⊂ B(a, r1 ). Hence, Z ss 0 2bD (zk , z` ) > βD αk,` (t); αk,` (t) dt ≥ CbUe (zk , z` ) ≥ dc, s1

e is the connected component of D ∩ B(a, r1 ) containing this part of the curve; a where U contradiction. ek,` denotes Hence, we obtain for k, ` ≥ k0 that CbUek,` (zk , z` ) ≤ bD (zk , z` ), where U the connected component of U ∩ D which contains the curve αk,` . Hence, (zj )j is even a bUk,` –Cauchy sequence; a contradiction. Remark 3.5.15. Let D ⊂ C be an unbounded b–complete domain. Due to N. Nikolov (private communication) the following inverse statement to that of Theorem 3.5.13 is true: For any open disc U ⊂ C, any connected component of U ∩ D (resp. D \ U ) is also b–complete. Moreover, there is the following general result for balanced domains. Theorem 3.5.16. Let D ⊂ Cn be a bounded pseudoconvex balanced domain. Then D is b–complete. P∞ Proof. Recall that any f ∈ O(D) can be written as a series k=1 Qk , where Qk are homogeneous polynomials, and that this convergence is an L2h –convergence. Therefore, the bounded holomorphic functions on D are dense in L2h (D). Then Theorem 3.3.9 and Corollary 3.5.10 finishes the proof. ? Characterize the b–complete bounded circular pseudoconvex domains ? Remark 3.5.17. There are also sufficient conditions for Hartogs domains with m–dimensional fibers to be b–complete (see [Jar-Pfl-Zwo 2000]). Theorem. Let D ⊂ Cn be a domain and let GD be bounded pseudoconvex Hartogs domain with m–dimensional balanced fibers. (a) Assume that D is b–exhaustive, that H∞ (D) is dense in L2h (D), and that there is an ε > 0 such that D × P (0, ε) ⊂ GD . Then GD is b–complete. (b) Assume that D is ci –complete, then GD is b–complete.


179

For a proof see [Jar-Pfl-Zwo 2000]. For further results on b–complete Hartogs domains see also [Che 2001b]. ? Is there a complete characterization of such domains which are b–complete ? Moreover, the following result may be found in [Che 2001a]. Theorem. Let u ∈ PSH(Cn ), u 6≡ −∞, and h ∈ O(Cm ), h 6≡ 0, be such that u ∈ C(Cn \ u−1 (−∞)). Let r > 0 and assume that 1

Ω := {(z 0 , z 00 ) ∈ Bn (r) × Bm (r) ⊂ Cn × Cm : u(z 0 ) + e |h(z00 )| < 1} is a domain. Then Ω is b–complete. If, in addition, there is a point (z00 , z000 ) ∈ Bn (r)×Bm (r) with u(z00 ) = −∞, h(z000 ) = 0, then Ω is not hyperconvex. Observe that the boundary behavior of the level sets of the Green function implies both b–exhaustiveness and b–completeness. We already saw that there exist b–exhaustive domains not being b–complete. It was a long standing question whether any b–complete domain is automatically b–exhaustive. The first counterexample to that question was given by W. Zwonek [Zwo 2001a] (see also [Zwo 2002]). The following Theorem 3.5.18 (see [Juc 2003]) gives even a large variety of domains that are b–complete but not b– exhaustive. Theorem 3.5.18. Let D ⊂ C be a Zalcman domain as in Corollary 3.3.22. Then: D is b-complete

iff

∞ X k=1

xk

√

1 = ∞. − log rk

Proof. The proof “=⇒” is similar to the one of Theorem 3.3.19; so it is omitted here. Proof of “⇐=”: Suppose that D is not b–complete. Then D is not b–exhaustive. Therefore, in view of Corollary 3.3.22, we have ∞ X

1 √ = ∞ and x − log rk k k=1

lim

j→∞

1 = 0. −x2j log rj

(3.5.18)

Moreover, there is a bD –Cauchy–sequence (zk )k ⊂ D with limk→∞ zk = 0. We may even assume that bD (zk , zk+1 ) < 21k . So there exist C 1 –curves γk : [0, 1] −→ D such that LβD (γk ) < 1/2k . Gluing all these curves together we obtain a piecewise C 1 –curve γ : [0, 1) → D with a finite βD –length. We claim that the Bergman kernel remains bounded along γ. In fact, if not then there is a sequence (wk )k ⊂ γ([0, 1)) such that lim kD (wk ) = ∞,

k→∞

lim wk = 0.

k→∞

Obviously, the sequence (wk )k is again a bD –Cauchy-sequence. As in the proof of Theorem 3.5.6, there exist an f ∈ L2h (D) and a subsequence (wkj )j such that |f (wkj |2 = 1. j→∞ kD (wkj ) lim

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3. Bergman metric

Applying Theorem 3.1.2, we find a g ∈ L2h (D), locally bounded near 0, such that kf − gkL2h (D) < 1/2. Therefore, |f (wkj )| |f (wkj )| |g(wkj )| 1 1 0 ←− p ≥p − kf − gkL2h (D) ≥ p − −→ ; j→∞ kD (wkj ) kD (wkj ) kD (wkj ) 2 j→∞ 2 a contradiction. Hence, there is a positive C such that kD (γ(t)) ≤ C, t ∈ [0, 1). To be able to continue we need the following lemma. Lemma 3.5.19. Let D be a domain as above satisfying (3.5.18) and let γ : [0, 1) −→ D be a piecewise C 1 –curve with limt→1 γ(t) = 0. Then Z τp lim MD (γ(t); γ 0 (t))dt = ∞. τ →1

0

p √ Proof. We may assume that |γ(0)| > x1 and that x1 − log r1 < xj − log rj , j ≥ j0 for a suitable j0 (use (3.5.18)). Now, fix an N ∈ N, N ≥ j0 , and let zN ∈ D be an arbitrary point with xN +2 ≤ |zN | ≤ xN +1 . We define xN − zN f , f := fB(x1 ,r1 ) − x1 − zN B(xN ,rN ) where fK denotes the Cauchy transform of K. Or more explicit, we have 1 xN − zN 1 f (z) = − , z ∈ D. x1 − z x1 − zN xN − z 1 Therefore, we see that f (zN ) = 0 and f 0 (zN ) = (x1 −zxNN)2−x (xN −zN ) . What remains is to estimate the L2h (D)–norm of the function f . Applying the relation between xn and xn+1 , we get |xN − zN | p |xN − zN | kfB(xN ,rN ) kL2h (D) ≤ C2 − log rN , kf kL2h (D) ≤ kfB(x1 ,r1 ) kL2h (D) + |x1 − zN | |x1 − zN |

where C1 , C2 are positive constants, independent of N and zN . Therefore, if xN +2 ≤ |z| ≤ xN +1 then: p |x1 − xN | C3 √ MD (z; X) ≥ |X| ≥ |X| 2 √ , C2 |x1 − γ(t)||xn − γ(t)|2 − log rN xN − log rN where C3 > 0 is a constant (use again that xk+1 ≤ Θ2 xk for all k). Finally, we obtain Z τp ∞ X xn+1 − xN +2 lim MD (γ(t); γ 0 (t))dt ≥ C3 2 √ τ →1 0 x N − log rN N =j 0

≥

∞ X N =j0

C3

∞ X Θ1 xN − Θ22 xN 1 √ √ ≥ C = ∞, 4 2 xN − log rN xN − log rN N =j 0

where C4 > 0. Hence, the proof of this lemma is complete. Now, applying Lemma 3.5.19 leads to the following contradiction: Z τ Z 1 τp ∞ > lim βD (γ(t); γ 0 (t))dt ≥ MD (γ(t); γ 0 (t))dt = ∞. τ →1 0 C 0


181

The boundary behavior of the Bergman metric on a Zalcman domain is partially described in the following result whose proof is based on methods of the proof of Theorem 3.5.18. TheoremP 3.5.20 ([Juc 2003]). Let D be a domain as in Theorem 3.5.18. ∞ (a) If k=1 x2 √−1log r < ∞, then lim sup βD (t; 1) < ∞. k

k

(−1,0)3t→0

(b) If lim sup βD (t; 1) < ∞, then lim supk→∞ (−1,0)3t→0

x2k

√ 1 − log rk

< ∞.

? It seems to be open how to characterize those Zalcman domains that are β– exhaustive, i.e. limz→∂D βD (z; 1) = ∞ ? The b–completeness means heuristically that boundary points are infinitely far away from inner points. So one might think that for a b–complete domain the Bergman metric βD becomes infinite at the boundary. The following example shows that this is not true. Example 3.5.21. There exists a b–complete bounded domain D in the plane which is not βD –exhaustive, i.e. there is a boundary sequence (wk )k ⊂ D such that (βD (wk ; 1))k∈N is bounded [Pfl-Zwo 2003a]). To be more precise: Put 1 1 2πj xn := n+1 + n+2 , zn,j := exp(i 4n ), n ∈ N, j = 0, . . . , 24n − 1. 2 2 2 9n Moreover, let rn := exp(−C1 2 ), n ∈ N, where C1 > 0 is chosen such that • the discs B(zn,j , rn ) ⊂ C, n ∈ N, j = 0, . . . 24n − 1, are pairwise disjoint, • B(zn,j , rn ) ⊂ An (0), n ∈ N, j = 0, . . . , 24n − 1. Then there is a sequence (nk )k ⊂ N such that the domain D := E \

∞ [ k=1

k −1 24n [

B(znk , j, rn )

j=0

is a domain satisfying the above desired properties. 3.5.1. Reinhardt domains and b–completeness. In the class of pseudoconvex Reinhardt domains there is a complete geometric characterization of b–complete domains (see [Zwo 1999b], [Zwo 2000b]). Let D ⊂ Cn be a pseudoconvex Reinhardt domain. Then Ω := ΩD := log D is a convex domain in Rn . Let us fix a point a ∈ Ω. Put C(Ω, a) := {v ∈ Rn : a + R+ v ⊂ Ω}. It is easy to see that C(Ω, a) is a closed convex cone with vertex at 0, i.e. tx ∈ C(Ω, a) for all x ∈ C(Ω, a) and t ∈ R+ . Moreover, this cone is independent of the point a, i.e. C(Ω, a) = C(Ω, b), b ∈ Ω. So we will write shortly C(Ω) := C(Ω, a). Observe that C(Ω) = {0} iff Ω ⊂⊂ Rn . We define now e e C(D) := {v ∈ C(ΩD ) : exp(a + R+ v) ⊂ D}, C0 (D) := C(ΩD ) \ C(D). e Observe that the definition of C(D) and C0 (D) is independent of the point a.

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3. Bergman metric

With the help of this geometric notions, there is the following complete description of those bounded Reinhardt domains which are Bergman complete. Theorem 3.5.22 ([Zwo 1999b]). Let D ⊂ Cn be a bounded pseudoconvex Reinhardt domain. Then the following conditions are equivalent: (i) D is b–complete; (ii) C0 (D) ∩ Qn = ∅. Example 3.5.23. Put D1 := {z ∈ C2 : |z1 |2 /2 < |z2 | < 2|z1 |2 , |z1 | < 2}. Obviously, D1 is a bounded pseudoconvex domain which contains the point (1, 1). Then it turns out that C0 (D1 ) = R>0 (−1, −2); so it contains the rational vector (−1, −2). Using the map Φ : C2∗ −→ C2∗ , Φ(z) := (z13 z2−1 , z1−1 z2 ),

z = (z1 , z2 ),

we see that D1 is biholomorphic to e 1 := {z ∈ C2 : 1/2 < |z2 | < 2, |z1 z2 | < 2}. D ∗ e 1 and, therefore also D1 , is not b–complete. It may be directly seen that D On the other hand, let √ √ 1 D2 := {z ∈ C2 : |z1 | 2 < |z2 | < 2|z1 | 2 , |z1 | < 2}. 2 Again, D2 is a bounded pseudoconvex Reinhardt domain; now a simple calculation gives √ C0 (D2 ) = R>0 (−1, − 2), i.e. C0 (D2 ) does not contain any rational vector. Hence the Theorem 3.5.22 tells us the D2 is b–complete. Recall that D2 is not hyperconvex. The next example can be found in [Her 1999]. Let D := {z ∈ C2 : |z2 |2 < exp(−1/|z1 |2 ), |z1 | < 1}. e Again, D is a bounded pseudoconvex Reinhardt domain. Here we have C(D) = C(D) = {0} × R− and C0 (D) = ∅. So Theorem 3.5.22 gives that D is b–complete (in [Her 1999], a direct proof of this fact is presented). Again, observe that D is not hyperconvex. For the proof of Theorem 3.5.22 we need the following lemma. Lemma 3.5.24. Let C ⊂ Rn be a convex closed cone with C ∩ Qn = {0}. Assume that C contains no straight lines. Then for any positive δ and any vector v ∈ C \ {0} there is a β ∈ Zn such that hβ, vi > 0

and

hβ, wi < δ, w ∈ C, kwk = 1.

Since this lemma is based on the geometric number theory, we will omit its proof. For more details, we refer to [Zwo 1999b]. Proof of Theorem 3.5.22. In a first step we are going to verify (i) =⇒ (ii): Suppose that (ii) does not hold, i.e. there is a non trivial vector v ∈ C0 (D) ∩ Qn . We may assume that 0 ∈ log D, v = (v1 , . . . , vn ) ∈ Zn− , and that v1 , . . . , vn are relaγ tively prime. It suffices to see that the Bergman length LβD of the curve (0, 1) −→ −v1 −vn (t , . . . , t ) ∈ D is finite.


183

In fact, put ϕ(λ) := (λ−v1 , . . . , λ−vn ), λ ∈ E∗ . Then ϕ ∈ O(E∗ , D). Now let u(λ) := kD (ϕ(λ)), λ ∈ E∗ . To continue we need a part of the following lemma (see [Zwo 2000b]). Lemma 3.5.25. Let D ⊂ Cn be a pseudoconvex Reinhardt domain, α ∈ Zn , and p ∈ (0, ∞). Then the following properties hold: • the monomial z α belongs to Lph (D) iff h p2 α + 1, vi < 0 for any v ∈ C(D) \ {0}; • if hα, vi < 0 for any v ∈ C(D) \ {0}, then z α ∈ H∞ (D); • if z α ∈ H∞ (D), then hα, vi ≤ 0 for any v ∈ C(D). In virtue of Lemma 3.5.25 (p = 2) it follows that ∞ X X aα |λ|−2hα,vi = bj |λ|2j , u(λ) = α∈Zn : hα+1,vi 0 such that D ⊂ Bs (0, R) × π es (D), where π es := πs+1,...,n denotes the projection of Cn onto Cn−s if s ≥ 1 or the identity if s = 0. Then π es (D) is a bounded pseudoconvex Reinhardt domain with π es (z 0 ) ∈ ∂e πs (D), where all coordinates of π es (z 0 ) are different from zero. Hence, π es (D) satisfies the general outer cone condition at π es (z 0 ). In virtue 00 of Theorem 6.1.17 in [J-P 1993], it follows that limz00 →eπs (z0 ) kπes (D) (z ) = ∞. Using the monotonicity and the product formula of the Bergman kernel, we finally get kD (z) ≥ kBs (0,R) (z 0 )kπes (D) (z 00 )

−→

D3z=(z 0 ,z 00 )→z 0

∞.

In the remaining part of the proof we assume that there is at least one j such that zj0 = 0, but Vj ∩ D = ∅. 16

Observe here that log |λ|2j0 is harmonic on E∗ .

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3. Bergman metric

We may assume that D ∩ Vj 6= ∅, j = 1, . . . , k, D ∩ Vj = ∅, j = k + 1, . . . , n 17 , 0 zk+1 = 0, and 1 ∈ D. e Let v ∈ Qn ∩C(D) \{0}. Then, by assumption, we know that v ∈ C(D). Therefore, limt→∞ exp(tv) = w ∈ D. So, if vj < 0 then wj = 0, and if vj = 0 then wj = 1. In particular, if there is a v ∈ C(D) ∩ Qn , vj < 0, then j ≤ k. Observe that Rk− × {0}n−k ⊂ C(D). Now, we claim that for any v ∈ C(D) \ Rk × n−k {0} we have that v ∈ / Rk × Qn−k . In fact, suppose that v ∈ Rk × Qn−k . Then vj < 0 for a some j > k. Then we may choose a suitable vector w ∈ Rk− × {0}n−k ⊂ C(D) such that ve := v + w ∈ C(D) ∩ Qn and vej < 0. Hence, j ≤ k; a contradiction. Put π : Rn −→ Rn , π(x) := (0, . . . , 0, xk+1 , . . . , xn ), where x = (x1 , . . . , xn ). Then π C(D) is a closed convex cone in {0}k × Rk− . In virtue of the above property, we conclude that π C(D) ∩ {0}k × Qn−k = {0}. 0 Recall that zk+1 = 0. Now, let z j ∈ D ∩ Cn∗ be a sequence tending to z 0 18 . Put xj := (log |z1j |, . . . , log |znj |) ∈ Rn . Obviously, kxj k −→ ∞. Moreover, without loss of generality, we may assume that the sequence (xj /kxj k)j converges to a vector ve ∈ C(D). Fix an α ∈ Zn such that z α ∈ L2h (D). Then, using Lemma 3.5.25, we conclude that inf{−hα + 1, wi : w ∈ C(D), kwk = 1} =: δ0 > 0. Two cases have to be discussed. Case 1: vej < 0 for some j > k. Applying Lemma 3.5.24 for C = π(C(D)), v = π(e v ), and δ0 , we get the existence of a β ∈ {0}k × Zn−k such that hβ, vei = hβ, π(e v )i > 0,

hβ, wi = kπ(w)khβ,

π(w) i < δ, w ∈ C(D), π(w) 6= 0. kπ(w)k

Observe that hβ, wi = 0 if π(w) = 0. Then z α+β ∈ L2h (D) (use Lemma 3.5.25) and |(z j )α | |(z j )α | p ≤ kz α+β kL2h (D) j α+β = kz α+β kL2h (D) |(z j )−β | −→ 0. j→∞ |(z ) | kD (z j ) Hence, the assumption of Theorem 3.5.6 is fulfilled. Case 2. vek+1 = · · · = ven = 0. Recall that kπ(xj )k −→ ∞. So we may assume that π(xj ) −→ w e = (0, . . . , 0, w ek+1 , . . . , w en ). kπ(xj )k If w e ∈ π(C(D)), then, in virtue of Lemma 3.5.24, there is a β ∈ {0}k × Zn−k such that hβ, wi e > 0 and hβ, wi < δ0 , w ∈ C(D) \ {0}. 17

Then necessarily, k < n. 18 Observe that it suffices to prove (3.5.19) for sequences in Cn . ∗


185

e be the smallest convex closed cone containing π(C(D)) and If w e∈ / π(C(D)), let C k n−k e e Therefore, −w. e Then C ⊂ {0} × R and w e∈ / C. e ui < 0, u ∈ C e \ {0}} {βe ∈ {0}k × Rn−k : hβ, is a non–empty convex open cone (see [Vla 1993], §25). So it contains a β ∈ {0}k × Zn−k . π(w) Thus, hβ, −wi e < 0 and hβ, wi = kπ(w)khβ, kπ(w)k i < 0 < δ0 , w ∈ C(D), π(w) 6= 0. Now we are able to complete the proof as in case 1 using the β we just constructed. Namely, we conclude that z α+β ∈ L2h (D) and n Y |(z j )α | p ≤ kz α+β kL2h (D) |(z j )−β | = kz α+β kL2h (D) | |(zνj )−βν | −→ 0. j→∞ kD (z j ) ν=k+1

Hence, Theorem 3.5.6 may be applied.

Finally, we will prove that part of Lemma 3.5.25 used during the proof of Theorem 3.5.22. Proof of Lemma 3.5.25. We restrict ourselves to prove only the following statement (the other ones in Lemma 3.5.25 may be taken as an exercise!): if D is as in Theorem 3.5.22 (in particular, D is bounded) and if hα + 1, vi < 0, v ∈ C(D) \ {0}, then z α ∈ L2h (D).

(†)

Assume that 1 ∈ D. In the case when C(D) = {0}, then (†) is obvious. So let us assume that C(D) 6= {0}. Then there is a δ0 < 0 such that hα + 1, vi < δ0 , v ∈ C(D), kvk = 1. We claim that for any ε > 0 there is a cone T such that log D \ T is bounded and kw − vk < ε, v ∈ T , w ∈ C(D), kvk = kwk = 1. Indeed, fix an ε > 0 and let h be the Minkowski function of the convex set log D. Observe that h is continuous and h−1 (0) = C(D). Therefore, there is a δ > 0 such that {w ∈ Rn : h(w) ≤ δ, kwk = 1} ⊂ {w ∈ Rn : kwk = 1, ∃v∈C(D) : kvk = 1, kv − wk < ε}. Set T as the smallest cone containing {w ∈ Rn : h(w) ≤ δ, kwk = 1}. Then log D \ T is bounded; otherwise there would exist an unbounded sequence xj ∈ log D \ T such that j h(xj ) < 1. Therefore, h( kxxj k ) < kx1j k , i.e. xj ∈ T for large j; a contradiction. Z

Now observe that hα + 1, vi ≤ δ20 kvk, v ∈ T , and Z Z |z α |2 dΛ2n (z) < ∞ iff e2hα+1,xi dΛn (x) < ∞ iff e2hα+1,xi dΛn (x) < ∞.

D

log D

T

So it remains to estimate the last integral. We get Z Z Z e2hα+1,xi dΛn (x) ≤ eδ0 kxk dΛn (x) < T

Hence, the monomial z α ∈ L2h (D).

T

eδ0 kxk dΛn (x) < ∞.

Rn

Remark 3.5.26. ? Up to our knowledge, so far there is no complete description for b–complete unbounded Reinhardt domains ?

Symbols Chapter 1 n n n A+ := {x ∈ A : x ≥ 0} (A ⊂ R), An + := (A+ ) , e.g. R+ = [0, +∞), Z+ = {0, 1, 2, . . . }, R+ , Z+

.

7

. . . . . . . . . . . . . . . . . . . . . . . .

7

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

z−a mE (a, z) := | 1−az | = the M¨ obius distance . . . . . . . . . . . . . . . . . . . . . . . .

7

N = {1, 2, . . . } the set of natural numbers E = the unit disc

pE :=

1 2

log

1+mE 1−mE

= the Poincar´ e distance

c∗G = the M¨ obius pseudodistance

. . . . . . . . . . . . . . . . . . . . . . .

7

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

O(G, D) = the family of all holomorphic mappings G −→ D cG = the Carath´ eodory pseudodistance (k)

mG = k–th M¨ obius function

. . . . . . . . . . . . . . . .

7

. . . . . . . . . . . . . . . . . . . . . . . . .

7

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

orda f = the order of zero of f at a

. . . . . . . . . . . . . . . . . . . . . . . . . . .

gG = the pluricomplex Green function

. . . . . . . . . . . . . . . . . . . . . . . . .

PSH(G) = the family of all functions plurisubharmonic on G

8 8

. . . . . . . . . . . . . . .

8

k k = the Euclidean norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ∗ = the Lempert function e kG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8 8

∗ e kG := tanh−1 e kG

8

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

kG = the Kobayashi pseudodistance ∗ = the Hahn function HG

. . . . . . . . . . . . . . . . . . . . . . . . . .

8

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

8

Reg M = the set of regular points A∗ := A \ {0} (A ⊂

Cn ),

An ∗

:= (A∗

. . . . . . . . . . . . . . . . . . . . . . . . . . . )n ,

e.g.

E∗ , C∗ , (Zn + )∗ ,

(k)

9

. . . . . . . . . . . . . . .

9

. . . . . . . . . . . . . . . . . . . . . .

10

. . . . . . . . . . . . . . . . . . . . . . . . . .

10

γG = the Carath´ eodory–Reiffen pseudometric γG = the k-th Reiffen pseudometric

Cn ∗

AG = the Azukawa pseudometric . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10

κG = the Kobayashi–Royden pseudometric

. . . . . . . . . . . . . . . . . . . . . . .

10

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

D(G) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

hG = the Hahn pseudometric

B(a, r) := {z ∈ Cn : kz − ak < r}, B(r) := B(0, r), Bn := B(1)

. . . . . . . . . . . . . . .

11

Lρ (α) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

ρF

12

ρin ,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ρi ,

ρic

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

M(G, K) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . R η = the integrated form of η . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

12

187

13

188

Symbols

Lη (α) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b h = the Buseman seminorm

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ηb = the Buseman pseudometric

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

κ b G = the Kobayashi–Busemann pseudometric Dρ = the weak derivative of ρ Pn

Ep := {(z1 , . . . , zn ) ∈ Cn :

j=1

13

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

|zj |2pj < 1} = complex ellipsoid . . . . . . . . . . . . . .

17

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20

Vol(s0 )

Λk = the Lebesgue measure in Rk

. . . . . . . . . . . . . . . . . . . . . . . . . . .

20

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20

In = the unit matrix sbh

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

21

Wη = the Wu pseudometric

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Dα,c := {z ∈ Cn : if αj < 0, then zj 6= 0, |z1 |α1 · · · |zn |αn < ec }, Dα := Dα,0 |z α |

16

. . . . . . . . . . . . . .

I(h) := {X ∈ Cn : h(X) < 1}

sh

13 13

. . . . . . . . . . . . . . . . . . . . . .

n n A>0 := {x ∈ A : x > 0} (A ⊂ R), An >0 := (A>0 ) , e.g. R>0 , R>0

U (h)

13

:= |z1

|α 1

· · · |zn

|α n

(α ∈

G2 = the symmetrized bidisc

Rn )

. . . . . . . .

22 27

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

27

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

σ2 := {(λ1 + λ2 , λ1 λ2 ) : λ1 , λ2 ∈ ∂E} . . . . . . . . . . . . . . . . . . . . . . . . . .

31

Σ2 := {(2λ, λ2 ) : λ ∈ E} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

λ−a . ha (λ) := 1−aλ 2ap−s Fa (s, p) := 2−as

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

A(m × n) = the set of all (m × n)–matrices with entries in A gG (p, ·) = the generalized Green function |p| := {z ∈ G : p(z) > 0} gG (A, ·) := gG (χA , ·)

. . . . . . . . . . . . . . .

42

. . . . . . . . . . . . . . . . . . . . . . . .

43

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

mG (p, z) = the generalized M¨ obius function

. . . . . . . . . . . . . . . . . . . . . . .

43 43

mG (A, ·) := mG (χA , ·)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

mG (a, ·) := mG ({a}, ·)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

RG + := {p : G −→ R+ }

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

dG (A, ·) := dG (χA , ·)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

dG (a, ·) := dG ({a}, ·)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

dmin G (p, ·)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

dmax G (p, ·)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

∗ (p, ·) e kG

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

q F (a) := q(F (a)) orda (F − F (a)) . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

AG,k := {z ∈ G : z1 · · · zk = 0} . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

52

P(Cn )

= the space of all complex polynomials of n–complex variables . . . . . . . . . . . .

58

ωA,G = the relative extremal function . . . . . . . . . . . . . . . . . . . . . . . . . .

72

EΞ

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

74

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

74

p ΞPoi

Symbols p ΞGre

189

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

p . . . . . . ΞLel p ΞLem . . . . . . Ep := EΞ p , E p Gre Lel Lel

74

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

74

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

p p , E p := EΞ p , Ep Poi := EΞ Lem := EΞ

. . . . . . . . . . . . . . .

75

b kG (p, z) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

Λ = the Lebesgue measure on ∂E

75

Pp (G)

Gre

Poi

Lem

. . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ω bG (p, ·) = the generalized relative extremal function Gp (G)

76

. . . . . . . . . . . . . . . . . . .

76

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

80

δG (p, ·) = the Coman function

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

90

δG (A, ·) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

90

δG (a, ·)

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

H∞ (G) = the space of all bounded holomorphic functions on G

90

. . . . . . . . . . . . . .

94

. . . . . . . . . . . . . . . . . . . . . . . . . .

106

Chapter 2 Vj := {z ∈

Cn

: zj = 0}, j = 1, . . . , n

log G := {(x1 , . . . , xn ) ∈ Rn : (ex1 , . . . , exn ) ∈ G} (G ⊂ Cn )

. . . . . . . . . . . . . . . .

G(A, C) = quasi-elementary Reinhardt domain, where A ∈ Z(n × n), C ∈ Rn .

106

. . . . . . . .

107

πi1 ,...,ik (z1 , . . . , zn ) := (zi1 , . . . , zik ), z ∈ Cn . . . . . . . . . . . . . . . . . . . . . . .

109

H(G) = the envelope of holomorphy of a domain G . . . . . . . . . . . . . . . . . . . .

110

Chapter 3 L2h (G) = the space of square integrable holomorphic functions on G

. . . . . . . . . . . .

137

. . . . . . . . . . . . . . . . . . . . . .

138

kG (·) = the Bergman kernel of G . . . . . . . . . . . . . . . . . . . . . . . . . . . .

139

Mn = the minimal ball

142

KG (·, ·) = the Bergman kernel function on G

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Az = Az (D) := {w ∈ D : log gD (z, w) ≤ −1} . . . . . . . . . . . . . . . . . . . . . . .

153

νK = the equilibrium measure of a non polar compact set K . . . . . . . . . . . . . . . .

158

cap M = the logarithmic capacity of M

. . . . . . . . . . . . . . . . . . . . . . . . .

158

fK = the Cauchy transform of a compact set K . . . . . . . . . . . . . . . . . . . . . .

159

αD (·) = the potential theoretic function of D . . . . . . . . . . . . . . . . . . . . . . .

160

(n) kD (·)

= the n–th Bergman kernel

. . . . . . . . . . . . . . . . . . . . . . . . . . .

166

MG (z; X) =

:f ∈

L2h (G),

. . . . . . . . . . . . . . . . . . . . . . . . . kf kL2 (G) = 1, f (z) = 0}, z ∈ G ⊂

Aw (D; r) := {z ∈ D : log gD (w, z) < −r} . Rn

2k+1

,

1 2k

160

βG (z; X) = the Bergman pseudometric sup{|f 0 (z)X|

1

. . . . . . . . . . . . . . . . . .

Ak (z) = the annulus with center z and radii

Cn ,

X∈

Cn

172

. . . .

172

. . . . . . . . . . . . . . . . . . . . . . . .

172

h

Rn

C(Ω, a) = C(Ω) := {v ∈ : a + R+ v ∈ Ω}, where Ω ⊂ is a convex domain . . . . . . . . 181 e C(D) := {v ∈ C(ΩD ) : exp(a + R+ v) ⊂ D}, where ΩD := log D and D is a pseudoconvex Reinhardt domain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 0 e C (D) := C(ΩD ) \ C(D) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 1 := (1, . . . , 1) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

183

Open problems Chapter 1 1.1. Find a formula for 1.2. Is

cin G

1.3. Let η 1.4.

c∗M ,

where M := {(z, w) ∈ E 2 : z 2 = w3 } . . . . . . .

= ciG ? . . . . (k) ∈ {γG , AG , κG }.

. . . . . . . . . . . . . . . . . . . . . . . Is Wη upper semicontinuous ?

9 12

. . . . . . . . .

24

∗ Decide whether: c∗Gn ≡ e kG and Gn cannot be exhausted by domains n biholomorphic to convex domains . . . . . . . . . . . . . . . .

42

1.5. Find a formula for gBn (p, ·) in the case where |p| = {a1 , a2 }, p(a1 ) 6= p(a2 )

51

1.6. Decide whether mE (A, z) = gE (A, z) = RE (A, z) if pj ≥ 1/2, j = k + 1, . . . , n (with arbitrary n and k) . . . . . . . . . . . . . . . . . . .

52

1.7. 1.8.

0

F

Are formulas gG (p, ·) = gD (p , F (·)), mG (p, ·) = mD (p , F (·)) true for arbitrary p ? . . . . . . . . . . . . . . . . . . . . . . . .

60

= . . .

65

1.9. Find an effective formula for gE 2 ({(a1 , b1 ), (a2 , b2 )}, (z, w)) with arbitrary (a1 , b1 ), (a2 , b2 ) ∈ E 2 . . . . . . . . . . . . . . . . . . . .

68

1.10.

Find formulas for {(z1 , z2 , z3 )

(k) mG ((0, a2 , a3 ), ·) and gG ((0, a2 , a3 ), ·), where G ∈ C3 : |z1 z2 | < 1, |z1 z3 | < 1} . . . . . . . . .

Decide whether gD (At , (0, 0)) < δD (At , (0, 0)) for √small t > √ 0, where D := {(z, w) ∈ C2 : |z| + |w| < 1}, At := {(t, t), (t, − t)} . . . .

1.11. Decide whether the system

(dmin G )G

has the product property . . . . . .

93 102

Chapter 2 2.1. Is any c–hyperbolic domain γ–hyperbolic ? . . . . . . . . . . . . . . .

106

3

2.2. Give an effective example of a domain in C which is not c–hyperbolic and not γ–hyperbolic, but ci –hyperbolic. Does such an example exist in C2 ? . 106 2.3. Is the Azukawa domain e k–hyperbolic ? . . . . . . . . . . . . . . . . 112 2.4. Is there a pseudoconvex balanced domain in C2 which is Brody–hyperbolic but not e k–hyperbolic ? . . . . . . . . . . . . . . . . . . . . . . 114 2.5. Is the converse of Theorem 2.4.1 true ? . . . . . . . . . . . . . . . .

115

2.6. Is any c–complete domain c–finitely compact ?

118

. . . . . . . . . . . .

2.7. Give a characterization of all mG (A, ·)–finitely compact Reinhardt domains 191

123

192

Open problems

2.8. What are conditions for the Minkowski function h of a pseudoconvex balanced domain G = Gh implying that G is c– or k–complete ? What about the 2–dimensional situation ? . . . . . . . . . . . . . . . . . 123 R (k) 2.9. Let k, l ∈ N, k < R l.(l)Is there a Zalcman type domain that is γ –complete, but not γ –complete ? . . . . . . . . . . . . . . . . . . . 123 R 2.10. Characterize the γ (k) –complete Zalcman type domains . . . . . . . . . 123 2.11. Is any bounded pseudoconvex domain G ⊂ Cn with a smooth boundary kG –complete ? . . . . . . . . . . . . . . . . . . . . . . . 129 2.12. Is any bounded pseudoconvex domain of finite type k– or c–complete ? . 130 2.13. Does Theorem 2.9.1 remain true if one only assumes that any boundary point, including ∞, admits a local psh peak and antipeak function ? . . 134 Chapter 3 3.1. Is there a pseudoconvex domain D ⊂ Cn , n > 1, such that dim L2h (D) < ∞ ? 138 3.2. Has KGn (with n ≥ 3) zeros ? . . . . . . . . . . . . . . . . . . . . 140 3.3. Is any bounded convex domain in C2 a Lu Qi-Keng domain ? . . . . . . 147 3.4. What are the effective values of M (a, n, m) ? . . . . . . . . . . . . . 148 3.5. Is the three-dimensional minimal ball a Lu Qi-Keng domain ? . . . . . . 149 3.6. Find the concrete values for m(k) . . . . . . . . . . . . . . . . . . 151 3.7. Describe all p = (p1 , . . . , pn ) for which Ep is a Lu Qi-Keng domain . . . . 151 3.8. Does the Bergman kernel function of a C ∞ –smooth convex domain Ω extend continuously to Ω × Ω ? . . . . . . . . . . . . . . . . . . . 151 3.9. Describe all bounded circular pseudoconvex domain which are b–exhaustive 157 3.10. Try to give a complete description of those bounded pseudoconvex Hartogs domains with m–dimensional fibers that are b–exhaustive . . . . 158 (n) (n) 3.11. Does limD3z→z0 kD (z) = ∞ imply that limD3z→z0 αD (z) = ∞, z0 ∈ ∂D ? 167 3.12. Characterize those unbounded domains of holomorphy that carry a non trivial L2h –function or that are L2h –domains of holomorphy . . . . . . . 172 3.13. Characterize the b–complete bounded circular pseudoconvex domains . . 178 3.14. Give characterization of b–complete Hartogs domains with m–dimensional fibers . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 3.15. Give a complete description of those Zalcman domains which are β–exhaustive 181 3.16. Give a full description of all unbounded pseudoconvex Reinhardt domains which are b–complete . . . . . . . . . . . . . . . . . . . . 185

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Index

dG –geodesic, 15 disc functional, 74

Azukawa pseudometric, 10 b–exhaustive, 153 Bergman complete, 172 exhaustive, 153 kernel, 139 kernel function, 138 pseudodistance, 8, 172 pseudometric, 11, 172 β–exhaustive, 181 bG –complete, 172 Brody–hyperbolic, 106, 112 Buseman pseudometric, 13 seminorm, 13

elementary Reinhardt domain, 27 envelope of a disc functional, 74 equilibrium measure, 158 extremal function, 47 mapping, 90 finite type, 130 Fu–condition, 116 γ–hyperbolic (γG –hyperbolic), 105 generalized holomorphically contractible family, 44 with integer-valued weights, 44 M¨ obius function, 43 pluricomplex Green function, 43 relative extremal function, 76 geodesic, 15 Green function, 8, 43 functional, 74

c–complete (cG –complete), 116 c–finitely compact (cG –finitely compact), 116 c–hyperbolic (cG –hyperbolic), 105 C–pseudometric, 9 C 1 –pseudodistance, 14 Carath´ eodory extremal mapping, 26 pseudodistance, 7 –Reiffen pseudometric, 10 Cauchy transform, 159 ci –complete (ciG –complete), 120 ci –hyperbolic (ciG –hyperbolic), 105 Coman conjecture, 91 function, 90 complete pseudometric, 22 complex ellipsoid, 16 geodesic, 15 contraction, 7

Hahn function, 8 pseudometric, 11 Hartogs domain over G with m–dimensional fibers, 114 triangle, 50 Hermitian scalar product, 19 holomorphically contractible family of functions, 7 pseudometrics, 10 hyperbolicity Brody, 112 c, 105 ci , 105

deflation identity, 141 δG –geodesic, 15 201

202 hyperconvex, 119, 155 inner Carath´ eodory pseudodistance, 12 pseudodistance, 12 integrated form, 13 irrational type, 27 e k–hyperbolic (e kG –hyperbolic), 112 K–pseudometric, 12 Kobayashi –Busemann pseudometric, 13 pseudodistance, 8, 9 –Royden pseudometric, 10 k–th M¨ obius function, 8 k–th Reiffen pseudometric, 10 L2h –domain of holomorphy, 168 Lelong functional, 74 Lempert function, 8, 9 functional, 75 length of a curve, 12, 13 Liouville property, 58 local holomorphic peak point, 130 holomorphic peak point at infinity, 130 local/global psh peak function at ∞, 131, 132 locally b–complete, 176 logarithmic capacity, 158 potential, 158 Lu Qi-Keng domain, 146 maximal function, 65 mG (A, ·)–finitely compact, 122 minimal ball, 16, 142 Minkowski function, 17 M¨ obius distance, 7 function, 43 function of higher order, 8 pseudodistance, 7, 9 Neil parabola, 9 n–th Bergman kernel, 166 order of zero, 8 peak point, 130 pluricomplex Green function, 8, 43 pluripolar, 169 Poincar´ e distance, 7 Poisson functional, 74 Poletsky formula, 81 primitive polynomial, 58

Index product property, 93, 97 for the generalized Green function, 97 for the relative extremal function, 93 pseudo–Hermitian scalar product, 19 pseudometric, 9 quasi–elementary Reinhardt domain, 107 rational type, 27 Reiffen pseudometric of higher order, 10 relative extremal function, 72 short Cn ’s, 116 strictly hyperconvex domain, 66 symmetrized bidisc, 42 symmetrized bidisc, 31 theorem of Frostman, 159 type irrational, 27 rational, 27 weak derivative, 14 peak function, 26 Wu pseudometric, 11, 19, 22 Zalcman domain, 167 type domain, 123

Invariant distances and metrics in complex analysis

Invariant distances and metrics in complex analysis

Invariant Distances and Metrics in Complex Analysis