• Nie Znaleziono Wyników

Effective formulas for invariant functions

N/A
N/A
Protected

Academic year: 2021

Share "Effective formulas for invariant functions"

Copied!
22
0
0

Pełen tekst

(1)

POLONICI MATHEMATICI LXIX.2 (1998)

Effective formulas for invariant functions

—case of elementary Reinhardt domains

by Peter Pflug (Oldenburg) and W lodzimierz Zwonek (Krak´ow)

Abstract. We find effective formulas for the invariant functions, appearing in the theory of several complex variables, of the elementary Reinhardt domains. This gives us the first example of a large family of domains for which the functions are calculated explicitly.

0. Introduction. Holomorphically invariant functions and pseudomet- rics have proved to be very useful in the theory of several complex variables.

Nevertheless, the problem of finding effective formulas for them has turned out to be very difficult. So far there have been very few examples of domains for which explicit formulas for these functions are known.

Among many different invariant functions and pseudometrics let us men- tion the Lempert and Green functions, the Kobayashi and Carath´ eodory pseudodistances as well as their infinitesimal versions, i.e. the Kobayashi–

Royden, Carath´ eodory and Azukawa pseudometrics.

Due to Lempert’s theorem (see [L1,2]) all holomorphically invariant func- tions and pseudometrics coincide in the class of convex domains. But even in the convex case it is difficult to find explicit formulas. Among the few ex- isting results in this direction let us mention here the special case of convex (see [BFKKMP], [JP2]) and non-convex (see [PZ]) ellipsoids. Another class of non-convex domains for which some of the functions were calculated is the class of elementary Reinhardt domains (see [JP1,2]). In our paper we extend the results obtained in the latter domains to all invariant functions and pseudometrics mentioned above. The formulas obtained enable us to understand better the mutual relations between the invariant objects and give surprising solutions to some problems.

1991 Mathematics Subject Classification: Primary 32H15.

The research started while the second author was visiting Universit¨ at Oldenburg; the stay was enabled by Volkswagen Stiftung Az. I/71 062. The second author was also supported by KBN Grant No 2 PO3A 060 08.

[175]

(2)

1. Definitions, notations and main results. By E we will always denote the unit disc in C. We put

m(λ

1

, λ

2

) := |λ

1

− λ

2

|

|1 − λ

1

λ

2

| , λ

1

, λ

2

∈ E, γ

E

(λ; α) := |α|

1 − |λ|

2

, λ ∈ E, α ∈ C.

Let D be a domain in C

n

. Following [L1], [Ko1], [Kl1,2], [C], [A], [R]

and [JP2] for (w, z) ∈ D × D and (w, X) ∈ D × C

n

we define the following functions:

e k

D

(w, z) := inf{m(λ

1

, λ

2

) : ∃ϕ ∈ O(E, D), ϕ(λ

1

) = w, ϕ(λ

2

) = z}, k

D

(w, z) := tanh k

D

(w, z),

where k

D

is the largest pseudodistance smaller than or equal to e k

D

:=

tanh

−1

e k

D

,

g

D

(w, z) := sup{u(z) : log u ∈ PSH(D, [−∞, 0)),

∃M, R > 0 : u(ζ) ≤ M kζ − wk for ζ ∈ D, kζ − wk < R}, c

D

(w, z) := sup{m(ϕ(w), ϕ(z)) : ϕ ∈ O(D, E)};

and also their infinitesimal versions:

κ

D

(w; X) := inf{γ

E

(λ; α) : ∃ϕ ∈ O(E, D), ϕ(λ) = w, αϕ

0

(λ) = X}, A

D

(w; X) := lim sup

λ→0, λ6=0

g

D

(w, w + λX)

|λ| ,

γ

D

(w; X) := sup{γ

E

(ϕ(w); ϕ

0

(w)X) : ϕ ∈ O(D, E)}.

The function e k

D

(resp. g

D

, k

D

, c

D

) is called the Lempert function (resp. the Green function, the Kobayashi and Carath´ eodory pseudodistance). The func- tion κ

D

(resp. A

D

and γ

D

) is called the Kobayashi–Royden (resp. Azukawa and Carath´ eodory–Reiffen) pseudometric.

Note that the functions e k

D

, k

D

and c

D

are always symmetric, whereas g

D

need not have this property. For the basic properties of the functions defined we refer the interested reader to [JP2]. Let us mention here only some basic relations:

e k

D

≥ k

D

≥ c

D

, e k

D

≥ g

D

≥ c

D

, κ

D

≥ A

D

≥ γ

D

.

A mapping ϕ ∈ O(E, D) is called a e k

D

-geodesic for (w, z), w 6= z, if ϕ(λ

1

) = w, ϕ(λ

2

) = z and m(λ

1

, λ

2

) = e k

D

(w, z) for some λ

1

, λ

2

∈ E.

The class of domains we are interested in is defined below.

For α = (α

1

, . . . , α

n

) ∈ R

n+

, n > 1, (R

+

:= (0, ∞)) define

D

α

:= {z ∈ C

n

: |z

1

|

α1

. . . |z

n

|

αn

< 1}.

(3)

We say that α is of rational type if there are t > 0 and β = (β

1

, . . . , β

n

) ∈ N

n

such that α = tβ; otherwise α is of irrational type. Note that if α is of rational type we may without loss of generality assume that all α

j

’s are relatively prime natural numbers. We also define

D e

α

:= {z ∈ D

α

: z

1

. . . z

n

6= 0}.

For α ∈ N

n

, we set

z

α

:= z

1α1

. . . z

αnn

, F

α

(z) := z

α

, F

(r)α

(z)X := X

β1+...+βn=r

1 β

1

! . . . β

n

!

β1+...+βn

F

α

(z)

∂z

β11

. . . ∂z

βnn

X

β

, z, X ∈ C

n

. Note that the domain D

α

is always unbounded, Reinhardt, complete, and pseudoconvex but not convex.

As mentioned in the introduction, some of the invariant functions for the domains D

α

are explicitly known. We gather the results known so far in the following theorem.

Theorem 1 (see [JP2]). If α ∈ N

n

, where α

j

’s are relatively prime, then c

Dα

(w, z) = m(w

α

, z

α

),

g

Dα

(w, z) = (m(w

α

, z

α

))

1/r

, γ

Dα

(w; X) = γ

E

(w

α

; (F

α

)

0

(w)X),

A

Dα

(w; X) = (γ

E

(w

α

; F

(r)α

(w)X))

1/r

, w, z ∈ D

α

, X ∈ C

n

, where r is the order of the zero of F

α

(·) − F

α

(w) at w.

If α is of irrational type, then c

Dα

(w, z) = 0,

γ

Dα

(w; X) = 0, w, z ∈ D

α

, X ∈ C

n

.

We extend the results of Theorem 1 to other invariant functions and pseudometrics and we find the remaining formulas for the Green function (and the Azukawa pseudometric) in the irrational case. The results are pre- sented in two theorems: for rational and irrational α. In both theorems the formulas for the Lempert function may seem incomplete (not all the cases are covered); nevertheless, because of the symmetry of both functions one easily obtains the formulas in the remaining cases.

Theorem 2. Assume that α ∈ N

n

with α

j

’s relatively prime. Let (w, z) ∈

D

α

× D

α

and (w, X) ∈ D

α

× C

n

. Set J := {j ∈ {1, . . . , n} : w

j

= 0} =

{j

1

, . . . , j

k

}. Then

(4)

e k

Dα

(w, z) =

 

 

min{m(λ

1

, λ

2

) : λ

1

, λ

2

∈ E,

λ

min{α1 j}

= w

α

, λ

min{α2 j}

= z

α

} if w, z ∈ e D

α

,

|z

α

|

1/(αj1+...+αjk)

if J 6= ∅, k

Dα

(w, z) = min{m((w

α

)

1/min{αj}

, (z

α

)

1/min{αj}

)},

where the minimum is taken over all possible roots; and in the infinitesimal case we have

κ

Dα

(w; X)

=

 

  γ

E



(w

α

)

1/min{αk}

; (w

α

)

1/min{αk}

1 min{α

k

}

n

X

j=1

α

j

X

j

w

j



if J = ∅, (|w

1

|

α1

. . . |X

j1

|

αj1

. . . |X

jk

|

αjk

. . . |w

n

|

αn

)

1/(αj1+...+αjk)

if J 6= ∅.

Observe that if min{α

j

} = 1, then e k

Dα

(w, z) = g

Dα

(w, z) for w, z ∈ e D

α

; otherwise, if w

α

6= z

α

, then the Green function is strictly less than the Lempert function.

As opposed to the rational case, in the irrational case not only the Lem- pert function, Kobayashi pseudodistance and Kobayashi–Royden pseudo- metric, but also the Green function and the Azukawa pseudometric have not been calculated so far.

Theorem 3. Assume that α is of irrational type. Let (w, z) ∈ D

α

× D

α

and (w, X) ∈ D

α

× C

n

. Set J := {j ∈ {1, . . . , n} : w

j

= 0} = {j

1

, . . . , j

k

}.

Then

e k

Dα

(w, z) =

m((|w

1

|

α1

. . . |w

n

|

αn

)

1/min{αj}

,

(|z

1

|

α1

. . . |z

n

|

αn

)

1/min{αj}

) if w, z ∈ e D

α

, (|z

1

|

α1

. . . |z

n

|

αn

)

1/(αj1+...+αjk)

if J 6= ∅, k

Dα

(w, z) = m

 Y

n

j=1

|w

j

|

αj



1/ min{αj}

,

 Y

n

j=1

|z

j

|

αj



1/ min{αj}

 ,

g

Dα

(w, z) =  0 if J = ∅,

(|z

1

|

α1

. . . |z

n

|

αn

)

1/(αj1+...+αjk)

if J 6= ∅;

and in the infinitesimal case we have κ

Dα

(w; X)

=

 

 

 

 

 

 

 γ

E

  Y

n

j=1

|w

j

|

αj



1/min{αk}

;

 Y

n

j=1

|w

j

|

αj



1/min{αk}

1 min{α

k

}

n

X

j=1

α

j

X

j

w

j



if J = ∅,

(|w

1

|

α1

. . . |X

j1

|

αj1

. . . |X

jk

|

αjk

. . . |w

n

|

αn

)

1/(αj1+...+αjk)

if J 6= ∅,

(5)

A

Dα

(w; X)

=  0 if J = ∅,

(|w

1

|

α1

. . . |X

j1

|

αj1

. . . |X

jk

|

αjk

. . . |w

n

|

αn

)

1/(αj1+...+αjk)

if J 6= ∅.

Observe that for an arbitrary balanced pseudoconvex domain D we al- ways have e k

D

(0, z) = h

D

(z) for z ∈ D, where h

D

denotes the Minkowski function for D. In the above formula, k

Dα

(0, z) < h

Dα

(z) for 0 6= z ∈ D

α

. It would be interesting to find the general form of k

D

(0, ·) for an arbitrary balanced pseudoconvex domain D.

2. Auxiliary results. For z ∈ C

n

put

T

z

:= {(e

1

z

1

, . . . , e

n

z

n

) : θ

j

∈ R}.

Note that T

z

is a group with multiplication defined as follows:

(e

1

z

1

, . . . , e

n

z

n

) ◦ (e

i eθ1

z

1

, . . . , e

i eθn

z

n

) := (e

i(θ1+ eθ1)

z

1

, . . . , e

i(θn+ eθn)

z

n

).

Define T

z,α

to be the subgroup of T

z

generated by the set

{(e

iαjα112k1π

z

1

, . . . , e

iαjnαn2knπ

z

n

) : j

1

, . . . , j

n

∈ {1, . . . , n}, k

1

, . . . , k

n

∈ Z}.

Note that if α is of rational type, then T

z,α

is finite; more precisely, if we assume that α ∈ N

n

and α

j

’s are relatively prime, then

T

z,α

= {(ε

1

z

1

, . . . , ε

n

z

n

) : ε

αjj

= 1 for all j}.

However, if α is of irrational type, then a well-known theorem of Kronecker (see [HW], Theorem 439) gives

(1) T

z,α

= T

z

.

For µ ∈ E

we define

Φ

µ

: C

n−1

3 (λ

1

, . . . , λ

n−1

) →

(e

αnλ1

, . . . , e

αnλn−1

, µe

−α1λ1

. . . e

−αn−1λn−1

) ∈ D

α

. Put

V

µ

:= Φ

µ

(C

n−1

), µ ∈ E

, V

0

:= {z ∈ C

n

: z

1

. . . z

n

= 0}.

Note that

[

µ∈E

V

µ

= D

α

.

Remark 4. Let µ ∈ E

. Assume that w, z ∈ V

µ

, and X ∈ C

n

satisfies P

n

j=1

α

j

X

j

/w

j

= 0. Then

e k

Dα

(w, z) = 0, κ

Dα

(w; X) = 0.

In fact, w = Φ

µ

(λ) and z = Φ

µ

(γ) for some λ, γ ∈ C

n−1

, so

e k

Dα

(w, z) = e k

Dα

µ

(λ), Φ

µ

(γ)) ≤ e k

Cn−1

(λ, γ) = 0.

(6)

For the second equality note that assuming Φ

µ

(λ) = w we have Φ

0µ

(λ)(Y ) =



α

n

w

1

Y

1

, . . . , α

n

w

n−1

Y

n−1

, −

n−1

X

j=1

α

j

w

n

Y

j



, Y ∈ C

n−1

. One may easily verify that

Φ

0µ

(λ)(C

n−1

) = n

X ∈ C

n

:

n

X

j=1

α

j

X

j

/w

j

= 0 o . Note that

0 = κ

Cn−1

(λ; Y ) ≥ κ

Dα

µ

(λ); Φ

0µ

(λ)Y ), Y ∈ C

n−1

, which finishes the proof.

In the proof of Lemma 5 below we shall replace E in the definition of the Lempert function with H := {x + iy : 1 > x > −1}.

Lemma 5. Fix w, z ∈ D

α

. Take any e z ∈ T

z,α

. Then for any ϕ ∈ O(E, D

α

) such that ϕ(λ

1

) = w, ϕ(λ

2

) = z for some λ

1

6= λ

2

, there is ϕ ∈ O(E, D e

α

) such that ϕ(λ e

1

) = w and ϕ(λ e

2

) = z. Consequently, e

e k

Dα

(w, z) = e k

Dα

(w, e z) for any z ∈ T e

z,α

.

P r o o f. Take any mapping ϕ ∈ O(H, D

α

) with ϕ(0) = w and ϕ(it) = z for some t > 0. Define (for k

n

∈ Z fixed)

ϕ : H 3 λ → e

1

(λ), . . . , ϕ

n−2

(λ), e

−2knπλ/t

ϕ

n−1

(λ), e

αn−12knπλ/(αnt)

ϕ

n

(λ)) ∈ D

α

. We have

ϕ(0) = w, e ϕ(it) = (z e

1

, . . . , z

n−1

, e

i(αn−1n)2knπ

z

n

).

Note that we may replace α

n−1

above with any other α

j

, and z

n

with e

i(αjn)2knπ

z

n

, and also we may continue the procedure with the next com- ponents z

j

varying, which finishes the proof.

Remark 6. From the proof of Lemma 5 we also have the following property:

Fix α ∈ N

n

with α

j

’s relatively prime and 0 < δ

1

≤ m(λ

1

, λ

2

) ≤ δ

2

< 1.

Take any ψ ∈ O(E, C

n

) with ψ(E) b (C

)

n

and choose z ∈ C

n

such that z

jαj

= ψ

αjj

2

) for j = 1, . . . , n. Then there is a mapping e ψ ∈ O(E, C

n

) such that e ψ(E) b (C

)

n

, ψ(λ

1

) = e ψ(λ

1

), e ψ(λ

2

) = z and

ψ

α11

(λ) . . . ψ

αnn

(λ) = e ψ

1α1

(λ) . . . e ψ

nαn

(λ), λ ∈ E,

mkψ

j

k

E

≤ k e ψ

j

k

E

≤ M kψ

j

k

E

, j = 1, . . . , n,

where m, M > 0 depend only on δ

1

and α.

(7)

Lemma 7. Fix L

11

, L

21

b E, L

2

b C

and α ∈ (R

+

)

n

. Assume that there is δ > 0 such that for any λ

1

∈ L

11

and λ

2

∈ L

21

we have m(λ

1

, λ

2

) ≥ δ. Then there is L

2

⊂ K b C

such that for any z

1

, z

2

∈ L

2

and for any λ

1

∈ L

11

and λ

2

∈ L

21

there is ψ ∈ O(E, C

) with ψ(λ

j

) = z

j

for j = 1, 2, and ψ(E) ⊂ K.

Moreover , there is e K b C

such that for any numbers z

1

, . . . , z

n

∈ L

2

and w

1

, . . . , w

k

∈ L

2

, k < n, with

|z

1

|

α1

. . . |z

n

|

αn

= 1

there are functions ψ

j

∈ O(E, C

) with ψ

j

(E) ⊂ e K for j = 1, . . . , n and ψ

1α1

(λ) . . . ψ

αnn

(λ) = e

, λ ∈ E,

ψ

j

1

) = z

j

, j = 1, . . . , n, ψ

j

2

) = w

j

, j = 1, . . . , k.

P r o o f. For the first part it is sufficient to consider L

11

= {λ

1

} and L

21

= {λ

2

} with m(λ

1

, λ

2

) = δ. The general case is then obtained by composing the functions with automorphisms of E and the dilatation Rλ, where 0 ≤ R < 1, since the images of new functions are contained in that of the original one.

Define

L := exp

−1

(L

2

) ∩ (R × [0, 2π)) b C.

Now put

K := {exp(h(λ)) : λ ∈ E, and h is of the type

h(λ) = aλ + b, a, b ∈ C, h(λ

1

) = e z

1

, h(λ

2

) = z e

2

, e z

1

, z e

2

∈ L}.

Observe that K b C

. The mappings we are looking for are of the form exp ◦h, where h is one of the functions appearing in the definition of K.

For the proof of the second part of the lemma we set w

j

for j = k + 1, . . . , n − 1 to be any number from L

2

and we take mappings ψ

1

, . . . , ψ

n−1

as in the first part of the lemma. Define

ψ

n

(λ) := e

i eθ

α11

(λ) . . . ψ

n−1αn−1

(λ))

1/αn

,

where the branches of powers are chosen arbitrarily and e θ ∈ R is chosen so that ψ

n

1

) = z

n

.

Lemma 8. Let L

11

, L

21

, L

2

, δ be as in Lemma 7. Fix α ∈ N

n

, where α

j

’s are relatively prime. Then there is K b C

such that for any mappings ψ

j

∈ O(E, C

), j = 1, . . . , n, with

ψ

1α1

. . . ψ

αnn

= 1, λ ∈ E,

(8)

and ψ

j

1

), ψ

j

2

) ∈ L

2

, where λ

1

∈ L

11

and λ

2

∈ L

21

, there are functions ψ e

j

∈ O(E, C

) such that

ψ e

1α1

. . . e ψ

nαn

= 1 on E,

ψ e

j

1

) = ψ

j

1

), ψ e

j

2

) = ψ

j

2

), ψ e

j

(E) ⊂ K, j = 1, . . . , n.

P r o o f. Put z

j

:= ψ

j

1

) and w

j

:= ψ

j

2

), j = 1, . . . , n. From Lemma 7 there are e ψ

j

, j = 1, . . . , n − 1, as desired. Put

ψ e

n

(λ) := 1

( e ψ

α11

(λ) . . . e ψ

n−1αn−1

(λ))

1/αn

.

We choose the branch of the power 1/α

n

so that e ψ

n

1

) = z

n

; note also that ψ e

nαn

2

) = w

αnn

. From Remark 6 we may change e ψ := ( e ψ

1

, . . . , e ψ

n

) so that all the desired properties are preserved and, additionally, e ψ

n

2

) = w

n

.

Now we present a lemma which is a weaker infinitesimal version of Lemma 7.

Lemma 9. Let w ∈ C

, X ∈ C and λ

1

∈ E. Then there is a mapping ψ ∈ O(E, C

) such that

ψ(λ

1

) = w, ψ

0

1

) = X.

Moreover , for given numbers w

1

, . . . , w

n

∈ C

, X

1

, . . . , X

k

∈ C (k < n) and α ∈ (R

+

)

n

, where |w

1

|

α1

. . . |w

n

|

αn

= 1, there are mappings ψ

j

∈ O(E, C

), j = 1, . . . , n, such that

ψ

j

1

) = w

j

, j = 1, . . . , n, ψ

0j

1

) = X

j

, j = 1, . . . , k, ψ

1α1

(λ) . . . ψ

αnn

(λ) = e

, λ ∈ E.

P r o o f. The first part goes as in the proof of Lemma 7 (note that we do not need to specify more, since we do not demand so much about the mapping ψ as in Lemma 7). The mapping we are looking for is of the form exp(aλ + b).

For the second part of the lemma let X

j

(j = k + 1, . . . , n − 1) be any complex number. Take ψ

j

as given in the first part of the lemma (for j = 1, . . . , n − 1) with w replaced with w

j

and X replaced with X

j

. Put

ψ

n

(λ) := e

i eθ

α11

(λ) . . . ψ

n−1αn−1

(λ))

1/αn

,

where the branches of powers are chosen arbitrarily and e θ ∈ R is chosen so that ψ

n

1

) = w

n

.

Now we are able to give formulas for the Lempert function and the

Kobayashi–Royden metric for special points.

(9)

Lemma 10. Fix w ∈ V

0

. Let z ∈ D

α

and X ∈ C

n

. Then e k

Dα

(w, z) = (|z

1

|

α1

. . . |z

n

|

αn

)

1/(αj1+...+αjk)

,

κ

Dα

(w; X) = (|w

1

|

α1

. . . |X

j1

|

αj1

. . . |X

jk

|

αjk

. . . |w

n

|

αn

)

1/(αj1+...+αjk)

, where J := {j ∈ {1, . . . , n} : w

j

= 0} = {j

1

, . . . , j

k

}.

P r o o f. Without loss of generality we may assume that w

1

= . . . = w

k

= 0, w

k+1

, . . . , w

n

6= 0, n ≥ k ≥ 1. We prove both equalities simultaneously.

First we consider the case z ∈ e D

α

(resp. X

j

6= 0 for all j = 1, . . . , k).

Take any ϕ ∈ O(E, D

α

) such that ϕ(0) = w, ϕ(t) = z (resp. ϕ(0) = w, tϕ

0

(0) = X) for some t > 0. We have

ϕ(λ) = (λψ

1

(λ), . . . , λψ

k

(λ), ψ

k+1

(λ), . . . , ψ

n

(λ)),

ψ

j

∈ O(E, C), j = 1, . . . , n.

Put

u(λ) :=

n

Y

j=1

j

(λ)|

αj

.

We know that log u ∈ SH(E) and u ≤ 1 on ∂E, so the maximum principle for subharmonic functions implies that u ≤ 1 on E. In particular, u(t) ≤ 1 (resp. u(0) ≤ 1), so

Q

n

j=1

|z

j

|

αj

t

α1+...+αk

≤ 1 (resp.

Q

k

j=1

|X

j

|

αj

Q

n

j=k+1

|w

j

|

αj

t

α1+...+αk

≤ 1), which gives

t ≥

 Y

n

j=1

|z

j

|

αj



1/(α1+...+αk)

(resp. t ≥  Y

k

j=1

|X

j

|

αj

n

Y

j=k+1

|w

j

|

αj



1/(α1+...+αk)

).

Therefore,

e k

Dα

(w, z) ≥

 Y

n

j=1

|z

j

|

αj



1/(α1+...+αk)

resp. κ

Dα

(w; X) ≥  Y

k

j=1

|X

j

|

αj

n

Y

j=k+1

|w

j

|

αj



1/(α1+...+αk)

).

To get equality put

t :=  Y

n

j=1

|z

j

|

αj



1/(α1+...+αk)

(10)

(resp. t :=

 Y

k

j=1

|X

j

|

αj

n

Y

j=k+1

|w

j

|

αj



1/(α1+...+αk)

) and consider the mapping

ϕ(λ) := (λψ

1

(λ), . . . , λψ

k

(λ), ψ

k+1

(λ), . . . , ψ

n

(λ)), λ ∈ E, where ψ

j

∈ O(E, C

), j = 1, . . . , n, Q

n

j=1

ψ

jαj

(λ) = e

on E and ψ

j

(t) = z

j

/t, j = 1, . . . , k, ψ

j

(t) = z

j

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

ψ

j

(0) = w

j

, j = k + 1, . . . , n (see Lemma 7)

(resp. ψ

j

(0) = X

j

/t, j = 1, . . . , k, ψ

j

(0) = w

j

, j = k + 1, . . . , n, ψ

j0

(0) = X

j

/t, j = k + 1, . . . , n; see Lemma 9).

Then ϕ ∈ O(E, D

α

), ϕ(0) = w, ϕ(t) = z (resp. tϕ

0

(0) = X), which finishes that case.

We are left with the case z ∈ V

0

(resp. X

j

= 0 for some 1 ≤ j ≤ k). If there is j such that w

j

= z

j

= 0 (resp. w

j

= X

j

= 0), then the mapping

C

n−1

3 (z

1

, . . . , ˇ z

j

, . . . , z

n

) → (z

1

, . . . , 0, . . . , z

n

) ∈ D

α

gives

0 = e k

Cn−1

((w

1

, . . . , ˇ w

j

, . . . , w

n

), (z

1

, . . . , ˇ z

j

, . . . , z

n

)) ≥ e k

Dα

(w, z), (resp.

0 = κ

Cn−1

((w

1

, . . . , ˇ w

j

, . . . , w

n

); (X

1

, . . . , ˇ X

j

, . . . , X

n

)) ≥ κ

Dα

(w; X)).

Now, there only remains the Lempert function and then we may assume that for all j we have |w

j

| + |z

j

| > 0.

For fixed 1 > β > 0 define the mapping ϕ := (ϕ

1

, . . . , ϕ

n

) as follows:

ϕ

j

(λ) :=

 

 

 

 

 λ − β

1 − βλ ψ

j

(λ) if w

j

= 0, λ + β

1 + βλ ψ

j

(λ) if z

j

= 0, ψ

j

(λ) if w

j

z

j

6= 0, where ψ

j

∈ O(E, C

), Q

n

j=1

ψ

αjj

(λ) = e

on E and ϕ(β) = w, ϕ(−β) = z (the values of ψ

j

(β) and ψ

j

(−β) are prescribed if only w

j

z

j

6= 0; for w

j

z

j

= 0 only one of them is prescribed; more precisely, take j

1

such that z

j1

= 0, and define ψ

j1

(−β) so that |ψ

1

(−β)|

α1

. . . |ψ

n

(−β)|

αn

= 1; note also that there is j

2

such that w

j2

= 0, so ψ

j2

(β) has no fixed value—it is the reason why we are allowed to use Lemma 7). Note also that ϕ ∈ O(E, D

α

). As β > 0 may be chosen arbitrarily small this completes the proof of the lemma.

In the next step we prove a formula for the Lempert function in the spe-

cial case of the domain D

(1,...,1)

. Following (to some extent) the ideas from

[JPZ] and [PZ] we extend the formulas to the general case using a technique

(11)

which could be called transport of geodesics. Roughly speaking, the idea is to transport the formulas from simpler domains to more complex ones with the help of “good” mappings. In [JPZ] and [PZ] it was the Euclidean ball that was a model domain. In the present paper it is the domain D

(1,...,1)

.

Lemma 11. If w, z ∈ V

0

, then

e k

D(1,...,1)

(w, z) = 0.

Assume that w ∈ e D

(1,...,1)

. Then

e k

D(1,...,1)

(w, z) = (m(w

1

. . . w

n

, z

1

. . . z

n

))

1/k

, where

k := max{#{j : z

j

= 0}, 1}.

P r o o f. The first part is a consequence of Lemma 10. Moreover, also the case z ∈ V

0

is a consequence of Lemma 10.

Consider now the case w, z ∈ e D

(1,...,1)

. We may assume that w

1

. . . w

n

6=

z

1

. . . z

n

(the other case is covered by Remark 4).

Consider the following mapping (see Lemma 7):

ϕ(λ) := (ψ

1

(λ), . . . , ψ

n−1

(λ), e

−iθ

λψ

n

(λ)), where

λ

1

:= w

1

. . . w

n

, λ

2

:= z

1

. . . z

n

,

ψ

j

∈ O(E, C

), j = 1, . . . , n, ψ

1

(λ) . . . ψ

n

(λ) = e

, λ ∈ E, ψ

j

1

) = w

j

, ψ

j

2

) = z

j

, j = 1, . . . , n − 1

(using Lemma 7 we may even assume that ψ

j

(E) ⊂ K b C

, j = 1, . . . , n;

compare Remark 12 below).

Note that ϕ ∈ O(E, D

(1,...,1)

), ϕ(λ

1

) = w, ϕ(λ

2

) = z. Therefore, com- bining this with the contractivity property of the Lempert function we have

m(w

1

. . . w

n

, z

1

. . . z

n

) ≥ e k

D(1,...,1)

(w, z) = m(w

1

. . . w

n

, z

1

. . . z

n

).

This completes the proof.

Remark 12. From the proof of Lemma 11 we see that for any w, z ∈ D e

(1,...,1)

with w

1

. . . w

n

6= z

1

. . . z

n

there is a e k

D(1,...,1)

-geodesic for (w, z) of the form



ψ

1

(λ), . . . , ψ

n−1

(λ), e

λ − β 1 − βλ ψ

n

(λ)

 with ψ

1

(λ) . . . ψ

n

(λ) = 1 and ψ

j

(E) b C

.

The domains D

α

, although very regular, do not have a property which

is crucial in the theory of holomorphically invariant functions: they are not

(12)

taut. Therefore, we have no guarantee that they admit e k

Dα

-geodesics. How- ever, as Lemma 13 will show, they do admit them at least in the rational case and for points which are “separated” by the Lempert function. The existence of the geodesics will play a great role in the proof of the formula for the Lempert function in the rational case.

Lemma 13. Assume that α ∈ N

n

and α

j

’s are relatively prime. Let w, z ∈ D e

α

with w

α

6= z

α

. Then there is a bounded e k

Dα

-geodesic ϕ ∈ O(E, D

α

) for (w, z).

P r o o f. We know that (use contractivity of the Lempert function) t := e k

Dα

(w, z) ≥ m(w

α

, z

α

) > 0;

consequently, there are mappings ϕ

(k)

= (ϕ

(k)1

, . . . , ϕ

(k)n

), k = 1, 2, . . . , such that ϕ

(k)

∈ O(E, D

α

), ϕ

(k)

(0) = w and ϕ

(k)

(t

k

) = z, where t

k

≥ t

k+1

→ t > 0. We have

ϕ

(k)j

= B

j(k)

ψ

j(k)

, j = 1, . . . , n, where B

j(k)

is a Blaschke product and ψ

j(k)

∈ O(E, C

).

Put ψ

(k)

:= (ψ

j(k)

)

nj=1

. There are two possibilities (due to the maximum principle for subharmonic functions—remember the pseudoconvexity of D

α

):

ψ

(k)

(E) ⊂ D

α

, (2)

ψ

(k)

(E) ⊂ ∂D

α

. (3)

Below we prove that without loss of generality we may restrict our attention to a case which is some kind of generalization of (3).

Take any k such that (2) is satisfied. First, notice that the mapping ψ e

(k)

:= ((ψ

(k)1

)

α1/(α1...αn)

, . . . , (ψ

n(k)

)

αn/(α1...αn)

)

is in O(E, D

(1,...,1)

). From Remark 12 there is a e k

D(1,...,1)

-geodesic for ( e ψ

(k)

(0), e ψ

(k)

(t

k

)) of the form

µ

(k)

:=



ψ b

(k)1

, . . . , b ψ

(k)n−1

, e

k

λ − β

k

1 − β

k

λ ψ b

(k)n

 ,

where b ψ

1(k)

. . . b ψ

n(k)

= 1 on E, such that µ

(k)

(0) = e ψ

(k)

(0) and µ

(k)

(R

k

t

k

) = ψ e

(k)

(t

k

), β

k

∈ E, R

k

≤ 1.

Coming back to the domain D

α

we see that instead of considering ϕ

(k)

with the property (2) we may consider the mapping (note that (α

1

. . . α

n

)/α

j

∈ N)

ϕ e

(k)

(λ) := (B

j(k)

(λ)(µ

(k)j

)

1...αn)/αj

(R

k

λ))

nj=1

,

because ϕ e

(k)

∈ O(E, D

α

), ϕ e

(k)

(0) = w and ϕ e

(k)

(t

k

) = z.

(13)

Therefore we may assume that (irrespective of which case we consider, (2) or (3))

ϕ

(k)j

= B

j(k)

ψ

j(k)

, j = 1, . . . , n,

where (ψ

1(k)

)

α1

. . . (ψ

(k)n

)

αn

= 1 and |B

j(k)

| ≤ 1 (although B

j(k)

’s need not longer be Blaschke products).

Choosing a subsequence if necessary, we may assume that for all j = 1, . . . , n, {B

j(k)

}

k=1

converges locally uniformly on E. Keeping in mind that ϕ

(k)

(0) = w and ϕ

(k)

(t

k

) = z we see that in view of Lemma 8 there is K b C

such that we may assume that ψ

j(k)

(E) ⊂ K for any j, k (we may apply Lemma 8 because L

2

:= {ψ

(k)j

(t

k

), ψ

j(k)

(0)}

j,k

b C

, which follows from the convergence and boundedness of {B

j(k)

}

k=1

, the fact that w

j

z

j

6= 0 for j = 1, . . . , n, and the equality (ψ

(k)1

)

α1

. . . (ψ

n(k)

)

αn

= 1), and then choosing a subsequence if necessary we deduce that ϕ

(k)

converges to a mapping ϕ ∈ O(E, D

α

) with ϕ(E) b (C

)

n

such that ϕ(0) = w and ϕ(t) = z.

The maximum principle for subharmonic functions implies, however, that ϕ(E) ⊂ D

α

. This completes the proof of the lemma.

3. The rational case—proof of Theorem 2. In the present section we provide the proof of Theorem 2. Since the theorem consists of a number of formulas, we prove them below one by one. We start with the Lempert function, which is basic in the calculation of other functions.

We begin with a formula for the M¨ obius function which seems to be known, but we have not been able to find any references in the literature.

Its proof is elementary but it needs tedious calculations, so we skip it.

Lemma 14. Fix 0 < s ≤ 1. Then for any λ

1

∈ (0, 1) and λ

2

∈ E we have m(λ

s1

, λ

s2

) ≥ m(λ

1

, λ

2

),

where λ

s1

∈ (0, 1) and the power λ

s2

is chosen so that the left-hand side of the formula is smallest possible.

Proof of the formula for e k

Dα

in the rational case. The case w

1

. . . w

n

= 0 is a consequence of Lemma 10. The case w, z ∈ e D

α

, w

α

= z

α

follows from Remark 4. We are left with the case w, z ∈ e D

α

, w

α

6= z

α

. By Lemma 13 there is a bounded e k

Dα

-geodesic ϕ ∈ O(E, D

α

) for (w, z) = (ϕ(λ

1

), ϕ(λ

2

)).

Proceeding as in the proof of Lemma 13 we may assume that ϕ

j

= B

j

ψ

j

, j = 1, . . . , n,

where B

j

is a Blaschke product (up to a constant |c

j

| = 1), ψ

j

(E) ⊂ K b C

and ψ

1α1

. . . ψ

αnn

= 1. In fact, consider the decomposition of ϕ

j

as above with

(14)

Blaschke product B

j

. Put

ψ := (ψ e

jαj/(α1...αn)

)

nj=1

.

Consider two cases. If ψ

α11

. . . ψ

nαn

is not constant on E, then e ψ ∈ O(E, D

(1,...,1)

) and it is a e k

D(1,...,1)

-geodesic for ( e ψ(λ

1

), e ψ(λ

2

)): otherwise, there would be b ψ ∈ O(E, D

(1,...,1)

) such that b ψ(λ

1

) = e ψ(λ

1

), b ψ(λ

2

) = e ψ(λ

2

) and b ψ(E) b D

(1,...,1)

and taking ϕ(λ) := (B b

j

(λ) b ψ

j 1...αn)/αj

(λ))

nj=1

we get a mapping such that ϕ(λ b

1

) = ϕ(λ

1

), ϕ(λ b

2

) = ϕ(λ

2

) and ϕ(E) b b D

α

— contradiction. By Remark 12 there is a e k

D(1,...,1)

-geodesic µ for ( e ψ(λ

1

), e ψ(λ

2

))

= (µ(λ

1

), µ(λ

2

)), where b ψ

1

. . . b ψ

n

= 1 and b ψ

j

(E)’s are relatively compact in C

. Taking now (B

j

(λ)µ

j

(λ)

1...αn)/αj

)

nj=1

instead of ϕ we get the desired property.

In case ψ

α11

. . . ψ

nαn

= e

, we may assume that ψ

j

(E) ⊂ K b C

for some K because of Lemma 8 (and then without loss of generality we may assume that e

= 1).

Therefore, ϕ(E) is contained in some polydisk. Consequently, ϕ(E) is contained in some smooth bounded pseudoconvex complete Reinhardt do- main G ⊂ D

α

, which arises from the domain D

α

by “cutting the ends”

and “smoothing the corners”. Therefore, ϕ is a e k

G

-geodesic for (w, z). Us- ing the results of [E], [Pa] we find that there are mappings h

j

∈ H

(E, C), j = 1, . . . , n, and % : ∂E → (0, ∞) such that

1

λ h

j

(λ)ϕ

j

(λ) = %(λ)α

j

|(ϕ

(λ))

α

|, j = 1, . . . , n, for almost all λ ∈ ∂E (we easily exclude the case (ϕ

(λ))

α

= 0 for λ from some subset of ∂E with non-zero Lebesgue measure). Using the result of Gentili (see [Ge]) we deduce that for some b

j

∈ C

, j = 1, . . . , n, β ∈ E,

ϕ

j

(λ)h

j

(λ) = b

j

(1 − βλ)(λ − β), j = 1, . . . , n, λ ∈ E, where b

j

j

= b

k

k

, j, k = 1, . . . , n. Consequently, we may take

B

j

(λ) = c

j

 λ − β 1 − βλ



rj

, |c

j

| = 1,

where r

j

∈ {0, 1} and not all r

j

’s are 0. Without loss of generality we may assume that β = 0 (we then change only λ

1

and λ

2

).

Now we come back to the domain D

α

. We may assume that r

1

= . . . =

r

k

= 1 and r

k+1

= . . . = r

n

= 0 (1 ≤ k ≤ n). We want to have for some

λ

1

, λ

2

∈ E (without loss of generality we may assume that c

j

= 1—if neces-

sary we change w and z with the help of rotations of suitable components,

(15)

so the Lempert function does not change)

λ

1

ψ

j

1

) = w

j

, j = 1, . . . , k, ψ

j

1

) = w

j

, j = k + 1, . . . , n, λ

2

ψ

j

2

) = z

j

, j = 1, . . . , k, ψ

j

2

) = z

j

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

Taking the α

j

th power and multiplying the equalities we get λ

α11+...+αk

= w

α

, λ

α21+...+αk

= z

α

.

The formulas above describe all possibilities which may yield candidates for the realization of the Lempert function. Now for all possible λ

1

, λ

2

as above we find mappings which map λ

1

and λ

2

to w and z. Note that there are mappings ψ

j

∈ O(E, C

), j = 2, . . . , n, such that (see Lemma 7)

ψ

j

1

) = w

j

(w

α

)

1/(α1+...+αk)

= w

j

λ

1

, j = 2, . . . , k, ψ

j

2

) = z

j

(z

α

)

1/(α1+...+αk)

= z

j

λ

2

, j = 2, . . . , k, ψ

j

1

) = w

j

, ψ

j

2

) = z

j

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

ψ

1

(λ) := 1

α22

(λ) . . . ψ

nαn

(λ))

1/α1

, λ ∈ E.

Put

ϕ(λ) := (λψ

1

(λ), . . . , λψ

k

(λ), ψ

k+1

(λ), . . . , ψ

n

(λ)).

The α

1

st root in the definition of ψ

1

is chosen so that ϕ

1

1

) = w

1

, and we know that ϕ

α11

2

) = z

1α1

. One may also easily verify that ϕ(λ

1

) = w and ϕ

j

2

) = z

j

for j = 2, . . . , n, which, however, in view of Lemma 5 shows that there is also a mapping ϕ ∈ O(E, D e

α

) such that ϕ(λ e

1

) = w and ϕ(λ e

2

) = z.

Therefore we have proved that

e k

Dα

(w, z) = min{m(λ

1

, λ

2

) : λ

1

, λ

2

∈ E,

λ

α1j1+...+αjk

= w

α

, λ

α2j1+...+αjk

= z

α

}, where the minimum is taken over all possible subsets {j

1

, . . . , j

k

} ⊂ {1, . . . , n}. Now Lemma 14 finishes the proof (remark that without loss of generality we may assume that w

j

> 0 for j = 1, . . . , n).

Proof of the formula for k

Dα

in the rational case. Note that tanh

−1

of the desired formula is equal to tanh

−1

of the Lempert function off the axis, satisfies the triangle inequality and is continuous. The definition of the Kobayashi pseudodistance and its continuity (see [JP2]) finish the proof.

To finish the proof it remains to compute the Kobayashi–Royden pseu-

dometric κ

Dα

. We do that by defining an operator which connects κ

Dα

to

the Kobayashi pseudodistance.

(16)

Following M. Jarnicki and P. Pflug (see [JP2]), for a domain D ⊂ C

n

we define

Dk

D

(w; X) := lim sup

λ→0, λ6=0

k

Dα

(w, w + λX)

|λ| , w ∈ D, X ∈ C

n

. This function differs from that in [JP2], but it is no larger, so the inequality below, which is crucial for our considerations, remains true:

(4) Dk

D

(w; X) ≤ κ

D

(w; X), w ∈ D, X ∈ C

n

. Lemma 15. Let α ∈ N

n

, where α

j

’s are relatively prime. Then Dk

Dα

(w; X)

= γ

E



n

Y

j=1

|w

j

|

αj



1/min{αk}

;



n

Y

j=1

|w

j

|

αj



1/min{αk}

1 min{α

k

}

n

X

j=1

α

j

X

j

w

j



for w ∈ e D

α

and X ∈ C

n

.

P r o o f. Without loss of generality we may assume that w

j

> 0 for j = 1, . . . , n, and α

n

= min{α

k

}. Using the formula for k

D

α

we get (5) Dk

Dα

(w; X) = lim sup

λ→0, λ6=0

| Q

n

j=1

(w

j

+ λX

j

)

αjn

− Q

n

j=1

w

αjjn

|

|1 − Q

n

j=1

(w

j

+ λX

j

)

αjn

Q w

αjjn

| · |λ| . Applying the Taylor formula we get, for λ close to 0,

(w

j

+ λX

j

)

αjn

= w

αjjn

+ α

j

α

n

w

jαjn

λX

j

w

j

+ ε

j

(λ), j = 1, . . . , n, where ε

j

(λ)/λ → 0 as λ → 0. Substituting the last equalities in (5) we get

Dk

Dα

(w; X) = lim sup

λ→0, λ6=0

( Q

n

j=1

|w

αjj

|

1/αn

)|λ|

P

n

j=1 αjXj

αnwj

(1 − Q

n

j=1

|w

j

|

jn

)|λ| , which equals the desired value.

Proof of the formula for κ

Dα

in the rational case. If J 6= ∅, then in view of Lemma 10 we are done. The case P

n

j=1

α

j

X

j

/w

j

= 0 follows from Remark 4.

Take w ∈ e D

α

. Without loss of generality we may assume that w

j

∈ R

+

for j = 1, . . . , n, and α

n

= min{α

j

}. Below, for X ∈ C

n

with P

n

j=1

α

j

X

j

/w

j

6= 0 we shall construct a mapping ϕ ∈ O(E, D

α

) such that ϕ(λ

1

) = w, tϕ

0

1

) = X,

where

λ

1

:= (w

α11

. . . w

nαn

)

1/αn

> 0, t := (w

α11

. . . w

nαn

)

1/αn

n

X

j=1

α

j

X

j

α

n

w

j

.

The existence of such a ϕ finishes the proof by Lemma 15 and (4).

(17)

Define ϕ(λ) :=



ψ

1

(λ), . . . , ψ

n−1

(λ), λ

1α1

(λ) . . . ψ

n−1αn−1

(λ))

1/αn

 , where (see Lemma 9)

ψ

j

1

) = w

j

, tψ

j0

1

) = X

j

, j = 1, . . . , n − 1.

We choose the (1/α

n

)th power so that ϕ

n

1

) = w

n

. After some elementary transformation we get tϕ

0n

1

) = X

n

, which finishes the proof.

4. The irrational case—proof of Theorem 3. As in the rational case we start with the proof of the formula for the Lempert function. First, we make use of the special properties of the domains of irrational type to get:

Lemma 16. Let α be of irrational type. Then for any w, z ∈ D

α

, e k

Dα

(w, z) = e k

Dα

( w, e z), e w ∈ T e

w

, z ∈ T e

z

.

P r o o f. Certainly it is enough to prove that

e k

Dα

(w, z) = e k

Dα

(w, z) e whenever z ∈ T e

z

. Assume that

(6) e k

Dα

(w, e z

1

) < e k

Dα

(w, e z

2

) =: ε for some z e

1

, e z

2

∈ T

z

. Then in view of Lemma 5,

(7) e k

Dα

(w, e z) = ε

for all z ∈ T e

ez2

. Because of (1) we have z e

1

∈ T

z

= T

ze2

= T

ze2

. To- gether with (6) and (7), the last statement contradicts, however, the upper- semicontinuity of the Lempert function.

As an immediate corollary of Lemma 16 we get

Corollary 17. Let α be of irrational type. Then for any z ∈ D

α

, e k

Dα

(z, e z) = 0 for any z ∈ T e

z

.

Proof of the formula for e k

Dα

in the irrational case. The case J 6= ∅ is covered by Lemma 10. Consider now the remaining case. In view of Lemma 16 we have

e k

Dα

(w, z) = e k

Dα

((|w

1

|, . . . , |w

n

|), (|z

1

|, . . . , |z

n

|)).

Choose a sequence {α

(k)

}

k=1

⊂ (Q

+

)

n

such that α

(k)

→ α. First notice that in view of Theorem 2, if x, y ∈ (R

+

)

n

∩ D

α(k)

, then

(8) e k

D

α(k)

(x, y)

= m((x

α

(k) 1

1

. . . x

α

(k)

nn

)

1/min{α(k)j }

, (y

α

(k) 1

1

. . . y

α

(k)

nn

)

1/min{α(k)j }

).

We may assume that min{α

j

} = α

n

and min{α

(k)j

} = α

(k)n

.

(18)

First we prove that

e k

Dα

(w, z) ≥ m((|w

1

|

α1

. . . |w

n

|

αn

)

1/αn

, (|z

1

|

α1

. . . |z

n

|

αn

)

1/αn

).

Indeed, otherwise there is a mapping ϕ ∈ O(E, D

α

) such that ϕ(λ

1

) = (|w

1

|, . . . , |w

n

|), ϕ(λ

2

) = (|z

1

|, . . . , |z

n

|) and

m(λ

1

, λ

2

) < m((|w

1

|

α1

. . . |w

n

|

αn

)

1/αn

, (|z

1

|

α1

. . . |z

n

|

αn

)

1/αn

).

Then we may choose k so large that ϕ(E) ⊂ D

α(k)

and

m(λ

1

, λ

2

) < m((|w

1

|

α(k)1

. . . |w

n

|

α(k)n

)

1/α(k)n

, (|z

1

|

α(k)1

. . . |z

n

|

α(k)n

)

1/α(k)n

), which, however, contradicts (8).

To get equality consider the mapping ϕ(λ) := (ψ

1

(λ), . . . , ψ

n−1

(λ), λψ

n

(λ)), where (see Lemma 7) ϕ

j

∈ O(E, C

) for j = 1, . . . , n − 1,

λ

1

:= (|w

1

|

α1

· . . . · |w

n

|

αn

)

1/αn

> 0, λ

2

:= (|z

1

|

α1

. . . |z

n

|

αn

)

1/αn

> 0;

ψ

j

1

) = |w

j

|, ψ

j

2

) = |z

j

|, j = 1, . . . , n − 1.

Define also

ψ

n

(λ) := 1

α11

(λ) . . . ψ

αn−1n−1

(λ))

1/αn

, λ ∈ E.

The α

n

th root is chosen so that ϕ

n

1

) = |w

n

|. One may also easily check from the form of ψ

j

’s in the proof of Lemma 7 that then ϕ

n

2

) > 0, so ϕ

n

2

) = |z

n

|. This completes the proof.

Just as in the rational case we have:

Proof of the formula for k

Dα

in the irrational case. Note that tanh

−1

of the desired formula satisfies the triangle inequality and coincides with the tanh

−1

of the Lempert function off the axis. The definition of the Kobayashi pseudodistance and its continuity finish the proof.

Having the formula for the Lempert function we get Proof of the formula for g

Dα

in the irrational case.

Case I: J = ∅. Corollary 17 implies that g

Dα

(w, z) = 0 for any z ∈ T

w

. The maximum principle for plurisubharmonic functions (applied to g

Dα

(w, ·)) implies that g

Dα

(w, z) = 0 for any z with |z

j

| ≤ |w

j

|, which means that g

Dα

(w, ·) vanishes on a set with non-empty interior (remember that w

1

. . . w

n

6= 0); but g

Dα

(w, ·) is logarithmically plurisubharmonic, so it must vanish on D

α

.

Case II: J 6= ∅. This case is a simple consequence of Lemma 10, the

inequality g ≤ e k

, the definition of the Green function and the fact that the

function (|z

1

|

α1

. . . |z

n

|

αn

)

1/(αj1+...+αjk)

is logarithmically plurisubharmonic

on D

α

.

(19)

Proof of the formula for A

Dα

in the irrational case. The result follows from the formula for the Green function and the definition of the Azukawa pseudometric.

Now, as in the rational case, we finish up the proof by showing the formula for κ

Dα

.

Lemma 18. Let α be of irrational type. Then Dk

Dα

(w; X)

= γ

E

  Y

n

j=1

|w

j

|

αj



1/min{αk}

;

 Y

n

j=1

|w

j

|

αj



1/min{αk}

1 min{α

k

}

n

X

j=1

α

j

X

j

w

j



for w ∈ e D

α

and X ∈ C

n

.

P r o o f. Without loss of generality we may assume that α

n

= min{α

k

}.

The formula for the Kobayashi pseudodistance gives us (9) Dk

Dα

(w; X) = lim sup

λ→0, λ6=0

| Q

n

j=1

|w

j

+ λX

j

|

αjn

− Q

n

j=1

|w

j

|

αjn

|

|1 − Q

n

j=1

|w

j

+ λX

j

|

αjn

Q |w

j

|

αjn

| · |λ| . Note that α

j

n

≥ 1. Therefore, applying the Taylor formula we get, for λ close to 0,

|w

j

+ λX

j

|

αjn

= |w

j

|

αjn

+ α

j

α

n

|w

j

|

αjn



Re  λX

j

w

j



+ ε

j

(λ),

j = 1, . . . , n, where ε

j

(λ)/λ → 0 as λ → 0. Substituting the last equalities in (9) we get

Dk

Dα

(w; X) = lim sup

λ→0, λ6=0

Q

n

j=1

(|w

j

|

αj

)

1/αn

Re λ P

n j=1

αjXj

αnwj



1 − Q

n

j=1

|w

j

|

jn

|λ| , which equals the desired value.

Proof of the formula for κ

Dα

in the irrational case. If J 6= ∅, then in view of Lemma 10 we are done. Also the case P

n

j=1

α

j

X

j

/w

j

= 0 follows from Remark 4. Below we deal with the remaining cases.

Take w ∈ e D

α

. Without loss of generality we may assume that w

j

∈R

+

for j = 1, . . . , n, and α

n

= min{α

j

}. Below, for X ∈ C

n

with P

n

j=1

α

j

X

j

/w

j

6= 0 we construct a mapping ϕ ∈ O(E, D

α

) such that ϕ(λ

1

) = w and tϕ

0

1

)

= X, where

λ

1

:= (w

α11

. . . w

nαn

)

1/αn

> 0, t := (w

α11

. . . w

αnn

)

1/αn

n

X

j=1

α

j

X

j

α

n

w

j

.

The existence of such a ϕ finishes the proof by Lemma 18 and (4).

Cytaty

Powiązane dokumenty

The purpose of this article is to present the selection of performance measures for effective implementation of the city strategy on the example of the Balanced Scorecard with

State its domain

0. It is well known that holomorphically contractible families of functions or pseudometrics give very useful and powerful tools in complex analysis. invariant pseudometric);

We use a method from harmonic analysis to reduce the computation of the 2-dimensional case to the problem of finding the kernel function of a weighted space of entire functions in

“Effective formulas for invariant functions—case of elementary Reinhardt

and [PT], namely: Let D be an oriented diagram of n components, and let cr(D) denote the number of crossings in D. , b n ) be base points of D, one point for each component of D,

V., A theorem of regularity in universal linearly invariant families of functions, Proceedings of the International Conference of Constructed Theory of Functions Varna 1984,

Analogously as in the definition of the adjoint systems, the boundary of the set is called an adjoint type hypersurface of the order q. X ^(nn); rffj* being the extremal system