Abstract. Alexander’s projective capacity for the polydisk and the ellipsoid in C

POLONICI MATHEMATICI LXII.3 (1995)

Alexander’s projective capacity for polydisks and ellipsoids in _C

^N

by Mieczys law Je ¸drzejowski (Krak´ ow)

Abstract. Alexander’s projective capacity for the polydisk and the ellipsoid in C

^N

is computed. Sharper versions of two inequalities concerning this capacity and some other capacities in C

^N

are given. A sequence of orthogonal polynomials with respect to an appropriately defined measure supported on a compact subset K in C

^N

is proved to have an asymptotic behaviour in C

^N

similar to that of the Siciak homogeneous extremal function associated with K.

1. Introduction. Let S be the unit sphere in C

^N

. Let σ denote the Lebesgue surface area measure on S. Let

s

N

:= R

S

dσ.

Let H

n

= H

n

(C

^N

) denote the set of all homogeneous polynomials of degree n (with complex coefficients) of N complex variables.

Let K be a compact subset of C

^N

. Let

kf k

_K

:= sup{|f (z)| : z ∈ K}, where f : K → C is a continuous function.

Definition 1.1 (see [1], [6]). Alexander’s projective capacity γ(K) is γ(K) := lim

n→∞

γ

n

(K)

^1/n

= inf

n

γ

n

(K)

^1/n

, where

γ

n

(K) := inf{kQk

K

},

the infimum being taken over all homogeneous polynomials Q ∈ H

n

, nor-

1991 Mathematics Subject Classification: Primary 31C15.

Key words and phrases: ellipsoid, projective capacity, extremal function.

This research was supported by KBN Grant No. 2 1077 91 01.

[245]

malized so that

1 s

N

R

S

log(|Q(z)|

^1/n

) dσ(z) = κ

N

, where

κ

N

:= 1 s

N

R

S

log |z

N

| dσ(z).

It is known that

κ

N

= − 1 2

N −1

X

j=1

1 j .

Let a

j

> 0 and R

j

> 0 (j = 1, . . . , N ). In this paper Alexander’s pro- jective capacity for the polydisk {(z

1

, . . . , z

N

) ∈ C

^N

: |z

1

| ≤ R

₁

, . . . , |z

N

| ≤ R

N

} and the ellipsoid {(z

₁

, . . . , z

N

) ∈ C

^N

: a

1

|z

₁

|

²

+ . . . + a

N

|z

_N

|

²

≤ 1} is computed.

2. Preliminaries. Let K be a compact subset of C

^N

.

Definition 2.1 (see [4]–[6]). The Siciak homogeneous extremal function Ψ

K

is

Ψ (z) = Ψ

K

(z) := lim

n→∞

Ψ

n

(z)

^1/n

, z ∈ C

^N

, where

Ψ

n

(z) := sup{|Q(z)|},

the supremum being taken over all Q ∈ H

n

normalized so that kQk

K

= 1.

Proposition 2.2 (see [4], p. 304). If R

j

> 0 (j = 1, . . . , N ) and P is the polydisk

P := {(z

1

, . . . , z

N

) ∈ C

^N

: |z

1

| ≤ R

₁

, . . . , |z

N

| ≤ R

_N

}, then

Ψ

P

(z

1

, . . . , z

N

) = max{|z

1

|/R

1

, |z

2

|/R

2

, . . . , |z

N

|/R

N

}.

Proposition 2.3 (see [5], p. 342). If a

j

> 0 (j = 1, . . . , N ) and E is the ellipsoid

E := {(z

1

, . . . , z

N

) ∈ C

^N

: a

1

|z

1

|

²

+ . . . + a

N

|z

N

|

²

≤ 1}, then

Ψ

E

(z

1

, . . . , z

N

) = (a

1

|z

₁

|

²

+ . . . + a

N

|z

_N

|

²

)

^1/2

. Definition 2.4 (see [6], p. 53). The constant τ (K) is

τ (K) := exp



− 1 s

N

R

S

log Ψ

K

(z) dσ(z)

 . Theorem 2.5 (see [2]). If K is a compact subset of C

^N

then

γ(K) = e

^κN

τ (K).

3. Main result

Theorem 3.1. If a

j

> 0 (j = 1, . . . , N ) and E is the ellipsoid E := {(z

1

, . . . , z

N

) ∈ C

^N

: a

1

|z

1

|

²

+ . . . + a

N

|z

N

|

²

≤ 1}, then

γ(E) = exp



− 1 2 · 1

2πi

R

C

z

^{N −1}

Log z dz (z − a

1

) . . . (z − a

N

)

 ,

where C is any contour in the right half-plane {z ∈ C : Re z > 0} enclosing all the points a

1

, . . . , a

N

and Log z is the principal branch of the logarithm.

In particular , if a

j

6= a

_k

for j 6= k, then γ(E) = exp



− 1 2

N

X

j=1

a

^{N −1}_j

log a

j

Q

N

k=1,k6=j

(a

j

− a

k

)

 .

If a

1

= . . . = a

N

= 1/R

²

(R > 0), then γ(E) = Re

^κN

= R exp



− 1 2

N −1

X

j=1

1 j

 .

Theorem 3.2. If R

j

> 0 (j = 1, . . . , N ) and P is the polydisk P := {(z

1

, . . . , z

N

) ∈ C

^N

: |z

1

| ≤ R

₁

, . . . , |z

N

| ≤ R

_N

} then

γ(P ) =

 Y

^N

k=1

M

⁽⁻¹⁾

k−1

k



1/2

, where

M

k

:= Y

1≤i1<...<ik≤N

 X

^k

j=1

R

²_ij

 . We first prove some lemmas. Let

D = D

^{N −1}

:=

n

(r

1

, . . . , r

N

) ∈ R

^N

: r

1

≥ 0, . . . , r

N

≥ 0,

N

X

j=1

r

²_j

= 1 o

,

Σ = Σ

^{N −1}

:= n

(θ

1

, . . . , θ

N

) ∈ R

^N

: θ

1

≥ 0, . . . , θ

_N

≥ 0,

N

X

j=1

θ

j

= 1 o . Lemma 3.3. If f : D → R is a continuous function then

1 s

N

R

S

f (|z

1

|, . . . , |z

_N

|) dσ(z) = 1 vol(Σ)

R

Σ

f ( p

θ

1

, . . . , p

θ

N

) dω(θ), where ω is the Lebesgue surface area measure on the hyperplane {(θ

1

, . . . , θ

N

)

∈ R

^N

: P

N

j=1

θ

j

= 1} and vol(Σ) := R

Σ

dω(θ).

P r o o f. Our proof starts with the observation that (3.1) 1

s

N

R

S

f (|z

1

|, |z

₂

|, . . . , |z

_N

|) dσ(z)

= 1 e s

N

R

D

f (r

1

, . . . , r

N

)r

1

. . . r

N

d e σ(r), where σ is the Lebesgue surface area measure on the unit sphere in R e

^N

and

e s

N

:= R

D

r

1

. . . r

N

d e σ(r).

Indeed, consider the following coordinates in C

^N

(z

j

= x

j

+ iy

j

for j = 1, . . . , N ):

x

j

= r

j

cos ψ

j

, y

j

= r

j

sin ψ

j

(r

j

≥ 0, 0 ≤ ψ

_j

≤ 2π, j = 1, . . . , N ), where

r

1

= r cos φ

1

, r

2

= r sin φ

1

cos φ

2

, r

3

= r sin φ

1

sin φ

2

cos φ

3

,

.. .

r

N −1

= r sin φ

1

sin φ

2

. . . sin φ

N −2

cos φ

N −1

, r

N

= r sin φ

1

sin φ

2

. . . sin φ

N −2

sin φ

N −1

(r ≥ 0, 0 ≤ φ

j

≤ π/2, j = 1, . . . , N − 1).

It is easy to see that

dx

1

. . . dx

N

dy

₁

. . . dy

_N

= r

1

. . . r

N

dr

1

. . . dr

N

dψ

₁

. . . dψ

_N

and

dr

1

. . . dr

N

= r

^{N −1}

sin

^{N −2}

φ

1

sin

^{N −3}

φ

2

. . . sin φ

N −2

dr dφ

₁

. . . dφ

_{N −1}

. Therefore

dx

1

. . . dx

N

dy

₁

. . . dy

_N

= r

^{2N −1}

cos φ

1

sin

^{2N −3}

φ

1

cos φ

2

sin

^{2N −5}

φ

2

. . . cos φ

N −1

sin φ

N −1

dr dφ

₁

. . . dφ

_{N −1}

dψ

₁

. . . dψ

_N

, where

r ≥ 0, 0 ≤ φ

j

≤ π/2 for j = 1, . . . , N − 1, 0 ≤ ψ

j

≤ 2π for j = 1, . . . , N.

Hence

dσ = cos φ

1

sin

^{2N −3}

φ

1

cos φ

2

sin

^{2N −5}

φ

2

. . . cos φ

N −1

sin φ

N −1

dφ

₁

. . . dφ

_{N −1}

dψ

₁

. . . dψ

_N

. In the same manner we can see that

d e σ = sin

^{N −2}

φ

1

sin

^{N −3}

φ

2

. . . sin φ

N −2

dφ

₁

. . . dφ

_{N −1}

, where 0 ≤ φ

j

≤ π/2, j = 1, . . . , N − 1.

Now it is easy to see that (3.1) is true.

It remains to prove that (3.2) 1

s e

N

R

D

f (r

1

, . . . , r

N

)r

1

. . . r

N

d σ(r) e

= 1

vol(Σ)

R

Σ

f ( p

θ

1

, . . . , p

θ

N

) dω(θ), which is equivalent to the equality

(3.3) A e

1

= e A

2

,

where A e

1

:= 1

s

^∗_N

R

D∗

f



r

1

, . . . , r

N −1

,

 1 −

N −1

X

j=1

r

²_j



1/2



r

1

. . . r

N −1

dr

1

. . . dr

N −1

,

A e

2

:= 1 vol(Σ

∗

)

R

Σ∗

f p

θ

1

, . . . , p θ

N −1

,

 1 −

N −1

X

j=1

θ

j



1/2



dθ

1

. . . dθ

N −1

,

D

∗

:= n

(r

1

, . . . , r

N −1

) ∈ R

^{N −1}

: r

1

≥ 0, . . . , r

_{N −1}

≥ 0,

N −1

X

j=1

r

²_j

≤ 1 o ,

Σ

∗

:=

n

(θ

1

, . . . , θ

N −1

) ∈ R

^{N −1}

: θ

1

≥ 0, . . . , θ

N −1

≥ 0,

N −1

X

j=1

θ

j

≤ 1 o ,

s

^∗_N

:= R

D_∗

r

1

. . . r

N −1

dr

1

. . . dr

N −1

, and vol(Σ

∗

) := R

Σ_∗

dθ

1

. . . dθ

N −1

. Indeed, consider the following parametri- zation of D:

r

j

= r

j

for j = 1, . . . , N − 1, r

N

=

 1 −

N −1

X

j=1

r

²_j



1/2

, and the parametrization of the simplex Σ:

θ

j

= θ

j

for j = 1, . . . , N − 1, θ

N

= 1 −

N −1

X

j=1

θ

j

.

Obviously, d e σ =

 1 +

N −1

X

j=1

 ∂r

N

∂r

j



2



1/2

dr

1

. . . dr

N −1

= dr

1

. . . dr

N −1

q

1 − r

₁²

− . . . − r

²_{N −1}

and

dω =

 1 +

N −1

X

j=1

 ∂θ

_N

∂θ

j



2



1/2

dθ

1

. . . dθ

N −1

=

√

N dθ

1

. . . dθ

N −1

. Now it follows easily that (3.2) is equivalent to (3.3). Therefore it suffices to prove (3.3). We change the variables:

r

j

= pθ

j

for j = 1, . . . , N − 1.

Clearly,

dr

1

. . . dr

N −1

= dθ

1

. . . dθ

N −1

2

^{N −1}

pθ

1

. . . θ

N −1

. We see at once that (3.3) is true, which completes the proof.

Lemma 3.4 (see [3], Lemma 6.3). If a

1

, . . . , a

N

∈ R and a

j

6= a

_k

for j 6= k, then

N

X

j=1

1 Q

N

k=1, k6=j

(a

j

− a

_k

) = 0, (3.4)

N

X

j=1

a

^{N −1}_j

Q

N

k=1, k6=j

(a

j

− a

k

) = 1.

(3.5)

P r o o f. Let W (x) := −1 + P

N

j=1

W

j

(x) and V (x) := −x

^{N −1}

+

N

X

j=1

a

^{N −1}_j

W

j

(x), where W

j

(x) :=

N

Y

k=1 k6=j

x − a

k

a

j

− a

_k

. It is clear that W (x) = P

N −1

i=0

B

i

x

ⁱ

and V (x) = P

N −1

j=0

C

j

x

^j

(where B

i

, C

j

∈ R). We see that W

j

(a

j

) = 1 and W

j

(a

k

) = 0 for k 6= j. Hence W (a

j

) = 0 for j = 1, . . . , N and V (a

j

) = 0 for j = 1, . . . , N , which implies W ≡ 0 and V ≡ 0. Therefore B

N −1

= 0 and C

N −1

= 0, which is the desired conclusion.

Lemma 3.5 (see [3], Lemma 6.2). If f : Σ

^{N −1}

→ R is a continuous function, then

1 vol(Σ

^{N −1}

)

R

Σ^{N −1}

f (θ

1

, . . . , θ

N

) dω(θ) = (N − 1)

1

R

0

x

^{N −2}

L(x) dx,

where

L(x) := 1

vol(Σ

^{N −2}

)

R

Σ^{N −2}

f (ξ

1

x, . . . , ξ

N −1

x, 1 − x) d ω(ξ), e

ω is the Lebesgue surface area measure on the simplex e Σ

^{N −2}

= n

(ξ

1

, . . . , ξ

N −1

) ∈ R

^{N −1}

: ξ

1

≥ 0, . . . , ξ

_{N −1}

≥ 0,

N −1

X

j=1

ξ

j

= 1 o and vol(Σ

^{N −2}

) = R

Σ^{N −2}

d ω(ξ). e

P r o o f. As in the proof of Lemma 3.3, we obtain dω(θ) =

√

N dθ

1

. . . dθ

N −1

. Therefore

1 vol(Σ

^{N −1}

)

R

Σ^{N −1}

f (θ

1

, . . . , θ

N

) dω(θ)

= 1

vol(Σ

∗^{N −1}

)

R

Σ^{N −1}_∗

f 

θ

1

, . . . , θ

N −1

, 1 −

N −1

X

j=1

θ

j



dθ

1

. . . dθ

N −1

,

where vol(Σ

_∗^{N −1}

) = R

Σ^{N −1}_∗

dθ

1

. . . dθ

N −1

. We change the variables:

θ

j

= ξ

j

x for j = 1, . . . , N − 2, θ

N −1

=

 1 −

N −2

X

j=1

ξ

j

 x,

where 0 ≤ x ≤ 1 and (ξ

1

, . . . , ξ

N −2

) ∈ Σ

_∗^{N −2}

. It is easy to check that dθ

1

. . . dθ

N −1

= x

^{N −2}

dxdξ

₁

. . . dξ

_{N −2}

. Therefore

1 vol(Σ

∗^{N −1}

)

R

Σ^{N −1}_∗

f 

θ

1

, . . . , θ

N −1

, 1 −

N −1

X

j=1

θ

j



dθ

1

. . . dθ

N −1

= 1

vol(Σ

_∗^{N −1}

)

1

R

0

x

^{N −2}

× R

Σ_∗^{N −2}

f 

ξ

1

x, . . . , ξ

N −2

x,  1 −

N −2

X

j=1

ξ

j



x, 1 − x 

dξ

₁

. . . dξ

_{N −2}

dx

= vol(Σ

_∗^{N −2}

) vol(Σ

∗^{N −1}

)

1

R

0

x

^{N −2}

 1

vol(Σ

∗^{N −2}

)

× R

Σ_∗^{N −2}

f



ξ

1

x, . . . , ξ

N −2

x,

 1 −

N −2

X

j=1

ξ

j



x, 1 − x



dξ

₁

. . . dξ

_{N −2}



dx

= ((N − 2)!)

⁻¹

((N − 1)!)

⁻¹

1

R

0

x

^{N −2}

×

 1

vol(Σ

^{N −2}

)

R

Σ^{N −2}

f (ξ

1

x, . . . , ξ

N −2

x, ξ

N −1

x, 1 − x) d ω(ξ) e

 dx

= (N − 1)

1

R

0

x

^{N −2}

L(x) dx, which completes the proof.

Lemma 3.6. If a

^j

> 0 for j = 1, . . . , N , then

(3.6) 1

vol(Σ

^{N −1}

)

R

Σ^{N −1}

log(a

1

θ

1

+ . . . + a

N

θ

N

) dω(θ)

= −

N −1

X

j=1

1 j + 1

2πi

R

C

z

^{N −1}

Log z dz (z − a

1

) . . . (z − a

N

) , where C is any contour in the right half-plane enclosing all the points a

1

, . . . , a

N

and Log z is the principal branch of the logarithm. In particu- lar , if a

j

6= a

_k

for j 6= k, then

(3.7) 1

vol(Σ

^{N −1}

)

R

Σ^{N −1}

log(a

1

θ

1

+ . . . + a

N

θ

N

) dω(θ)

= −

N −1

X

j=1

1 j +

N

X

j=1

a

^{N −1}_j

log a

j

Q

N

k=1, k6=j

(a

j

− a

_k

) . If a

1

= . . . = a

N

= 1/R

²

(R > 0), then the above integral is equal to

−2 log R.

P r o o f. A trivial verification shows that the lemma is true if a

1

= a

2

= . . . = a

N

= 1/R

²

(R > 0). This result can also be obtained from (3.6) (by the residue theorem), because g

^{(N −1)}

(z) = (N − 1)!(Log z + P

N −1

j=1

1/j) if g(z) := z

^{N −1}

Log z.

Let us prove (3.6). Both its sides are continuous functions of the param- eters a

j

. Therefore it suffices to prove (3.7). The proof is by induction on N . It is easy to check the case N = 2:

1 vol(Σ

¹

)

R

Σ¹

log(a

1

θ

1

+ a

2

θ

2

) dω(θ) = −1 + a

1

log a

1

a

1

− a

₂

+ a

2

log a

2

a

2

− a

₁

.

Assuming (3.7) to hold for N − 1 (N ≥ 3), we will prove it for N , applying

Lemma 3.5. We first compute (with f (θ) := log( P

N

j=1

a

j

θ

j

)):

vol(Σ

^{N −2}

)L(x) = R

Σ^{N −2}

f (ξ

1

x, . . . , ξ

N −1

x, 1 − x) d e ω(ξ)

= R

Σ^{N −2}

log 

^{N −1}

X

j=1

a

j

ξ

j

x + a

N

(1 − x)  d e ω(ξ)

We have a

N

(1 − x) ≡ a

N

(1 − x) P

N −1

j=1

ξ

j

on Σ

^{N −2}

. Therefore vol(Σ

^{N −2}

)L(x) = R

Σ^{N −2}

log 

^{N −1}

X

j=1

A

j

ξ

j

 d ω(ξ), e

where A

j

= A

j

(x) := a

N

+ (a

j

− a

_N

)x for j = 1, . . . , N − 1. By assumption,

L(x) = 1

vol(Σ

^{N −2}

)

R

Σ^{N −2}

log 

^{N −1}

X

j=1

A

j

ξ

j

 d ω(ξ) e

= −

N −2

X

j=1

1 j +

N −1

X

j=1

(A

j

(x))

^{N −2}

log A

j

(x) Q

N −1

k=1, k6=j

(A

j

(x) − A

k

(x))

= −

N −2

X

j=1

1 j +

N −1

X

j=1

(a

N

+ (a

j

− a

_N

)x)

^{N −2}

log(a

N

+ (a

j

− a

_N

)x) Q

N −1

k=1, k6=j

(a

j

− a

_k

)x Applying Lemma 3.5 we obtain

1 vol(Σ

^{N −1}

)

R

Σ^{N −1}

log

 X

^N

j=1

a

j

θ

j



dω(θ) = (N − 1)

1

R

0

x

^{N −2}

L(x)dx = G

1

+ G

2

,

where

G

1

:= (N − 1)

1

R

0

x

^{N −2}



−

N −2

X

j=1

1 j



dx = −

N −2

X

j=1

1 j , G

2

:= (N − 1)

N −1

X

j=1

R

1

0

(a

N

+ (a

j

− a

N

)x)

^{N −2}

log(a

N

+ (a

j

− a

N

)x) dx Q

N −1

k=1, k6=j

(a

j

− a

_k

) . Integrating by parts the integral in G

2

, we obtain G

2

= Z

1

+ Z

2

, where

Z

1

:= − 1 N − 1

N −1

X

j=1

a

^{N −1}_j

− a

^{N −1}_N

Q

N

k=1, k6=j

(a

j

− a

_k

) , Z

2

:=

N −1

X

j=1

a

^{N −1}_j

log a

j

− a

^{N −1}_N

log a

N

Q

N

k=1, k6=j

(a

j

− a

k

) .

Applying (3.4) and (3.5) (see Lemma 3.4) we get Z

1

= − 1

N − 1



N −1

X

j=1

a

^{N −1}_j

Q

N

k=1, k6=j

(a

j

− a

_k

) − a

^{N −1}_N

N −1

X

j=1

1 Q

N

k=1, k6=j

(a

j

− a

_k

)



= − 1 N − 1



^{N −1}

X

j=1

a

^{N −1}_j

Q

N

k=1, k6=j

(a

j

− a

k

) + a

^{N −1}_N

Q

N −1

k=1

(a

N

− a

k

)



= − 1 N − 1

N

X

j=1

a

^{N −1}_j

Q

N

k=1, k6=j

(a

j

− a

_k

) = − 1 N − 1 . By (3.4),

Z

2

=

N −1

X

j=1

a

^{N −1}_j

log a

j

Q

N

k=1, k6=j

(a

j

− a

_k

) − a

^{N −1}_N

log a

N N −1

X

j=1

1 Q

N

k=1, k6=j

(a

j

− a

_k

)

=

N −1

X

j=1

a

^{N −1}_j

log a

j

Q

N

k=1, k6=j

(a

j

− a

_k

) + a

^{N −1}_N

log a

N

Q

N −1

k=1

(a

N

− a

_k

)

=

N

X

j=1

a

^{N −1}_j

log a

j

Q

N

k=1, k6=j

(a

j

− a

_k

) .

Thus G

1

+ Z

1

+ Z

2

is equal to the right-hand side of (3.7), which proves the lemma.

Lemma 3.7. Let {g

1

, . . . , g

l

} and {h

₁

, . . . , h

l

} be two sets of integers.

Let b

hj

> 0 for j = 1, . . . , l. Let S((g), (h)) be the set of all bijections {g

1

, . . . , g

l

} → {h

1

, . . . , h

l

}. Then

(3.8) X

%∈S((g),(h))

Q

l

j=1

b

_%(gj)

Q

l

j=1

(b

_%(g1)

+ b

_%(g2)

+ . . . + b

_%(gj)

) = 1.

P r o o f. We proceed by induction on l. The lemma is true for l = 1.

Assuming (3.8) to hold for l − 1 (l ≥ 2), we will prove it for l. Let k ∈ {1, . . . , l}. Let

S

k

((g), (h)) := {% ∈ S((g), (h)) : %(g

l

) = h

k

}.

Let

T

k

:= X

%∈Sk((g),(h))

Q

l

j=1

b

%(gj)

Q

l

j=1

(b

%(g1)

+ b

%(g2)

+ . . . + b

%(gj)

) . It is easy to check that

T

k

= b

hk

b

h1

+ . . . + b

hl

X

µ

Q

l−1 j=1

b

_µ(gj)

Q

l−1

j=1

(b

_µ(g1)

+ . . . + b

_µ(gj)

) ,

where the summation extends over all bijections µ : {g

1

, . . . , g

l−1

} → {h

₁

, . . . , h

l

} \ {h

_k

}. By the induction hypothesis,

T

k

= b

hk

b

h1

+ . . . + b

hl

. Obviously, S((g), (h)) = S

l

k=1

S

k

((g), (h)) and

S

i

((g), (h)) ∩ S

j

((g), (h)) = ∅ for i 6= j.

Therefore the left-hand side of (3.8) is equal to P

l

k=1

T

k

= 1, which com- pletes the proof.

P r o o f o f T h e o r e m 3.1. Applying Theorem 2.5, Proposition 2.3, Lemma 3.3 and Lemma 3.6 gives

log γ(E) = κ

N

+ log τ (E) = κ

N

− 1 s

N

R

S

log Ψ

E

(z) dσ(z)

= κ

N

− 1 2 · 1

s

N

R

S

log(a

1

|z

₁

|

²

+ . . . + a

N

|z

_N

|

²

) dσ(z)

= κ

N

− 1 2 · 1

vol(Σ)

R

Σ

log(a

1

θ

1

+ . . . + a

N

θ

N

) dω(θ)

= − 1 2

N −1

X

j=1

1 j − 1

2



−

N −1

X

j=1

1 j + 1

2πi

R

C

z

^{N −1}

Log z dz (z − a

1

) . . . (z − a

N

)



= − 1 2 · 1

2πi

R

C

z

^{N −1}

Log z dz (z − a

1

) . . . (z − a

N

) .

The particular case a

j

6= a

k

for j 6= k is an easy consequence of the above formula (it suffices to apply the residue theorem).

P r o o f o f T h e o r e m 3.2. We have to show that (3.9) log γ(P ) = 1

2

N

X

j=1

X

1≤k1<...<kj≤N

(−1)

^j−1

log(b

k1

+ . . . + b

kj

),

where b

j

:= R

²_j

for j = 1, . . . , N. We first observe that (3.10) log γ(P )

= 1 2



−

N −1

X

j=1

1

j − 1

vol(Σ)

R

Σ

max

 log θ

1

b

1

, . . . , log θ

N

b

N

 dω(θ)



.

Indeed, combining Theorem 2.5 with Proposition 2.2 and Lemma 3.3 gives

log γ(P ) = κ

N

+ log τ (P ) = κ

N

− 1 s

N

R

S

log Ψ

P

(z) dσ(z)

= κ

N

− 1 s

N

R

S

log



max  |z

₁

| R

1

, . . . , |z

_N

| R

N



dσ(z)

= κ

N

− 1 s

N

R

S

max



log |z

₁

| R

1

, . . . , log |z

_N

| R

N

 dσ(z)

= − 1 2

N −1

X

j=1

1 j − 1

2 · 1 s

N

R

S

max



log |z

1

|

²

R

²₁

, . . . , log |z

N

|

²

R

²_N

 dσ(z)

= 1 2



−

N −1

X

j=1

1

j − 1

vol(Σ)

R

Σ

max

 log θ

1

b

1

, . . . , log θ

N

b

N

 dω(θ)

 . Let S(N ) denote the set of all permutations of {1, . . . , N }. For % ∈ S(N ) define

Λ

%

:=



(θ

1

, . . . , θ

N

) ∈ Σ

^{N −1}

: θ

_%(1)

b

%(1)

≥ . . . ≥ θ

_{%(N )}

b

%(N )

 , t(%(1), . . . , %(N )) := 1

vol(Σ)

R

Λ%

max

 log θ

1

b

1

, . . . , log θ

N

b

N

 dω(θ)

= 1

vol(Σ)

R

Λ%

log θ

%(1)

b

_%(1)

dω(θ).

Obviously,

(3.11) 1

vol(Σ)

R

Σ

max

 log θ

1

b

1

, . . . , log θ

N

b

N

 dω(θ)

= X

%∈S(N )

t(%(1), . . . , %(N )).

We next show that (3.12) t(%(1), . . . , %(N ))

=

Q

N i=1

b

%(i)

Q

N

i=1

(b

%(1)

+ . . . + b

%(i)

)



−

N −1

X

j=1

1 j



+

N

X

j=1

(−1)

^j

( Q

N

i=1

b

_%(i)

) log(b

_%(1)

+ . . . + b

_%(j)

) ( Q

j

i=1

(b

%(i)

+ . . . + b

%(j)

))( Q

N

i=j+1

(b

%(j+1)

+ . . . + b

%(i)

)) , where Q

N

i=j+1

(b

_%(j+1)

+ . . . + b

_%(i)

) := 1 for j = N .

Without loss of generality we can assume that % = %

0

, where %

0

(j) := j for j = 1, . . . , N . Clearly,

(3.13) t(1, . . . , N ) = 1 vol(Σ)

R

Λ

log θ

1

b

1

dω(θ), where Λ := Λ

%0

.

We first prove that

(3.14) 1

vol(Σ)

R

Λ

log θ

1

b

1

dω(θ)

=

Q

N i=1

b

i

Q

N

i=1

(b

1

+ . . . + b

i

) 1 vol(Σ)

R

Σ

log



^N

X

j=1

η

j

b

1

+ . . . + b

j

 dω(η).

Let

Σ

(∗)

:=

n

(η

2

, . . . , η

N

) ∈ R

^{N −1}

: η

2

≥ 0, . . . , η

N

≥ 0,

N

X

j=2

η

j

≤ 1 o ,

Λ

(∗)

:=



(θ

2

, . . . , θ

N

) ∈ R

^{N −1}

: 1 − P

N j=2

θ

j

b

1

≥ θ

2

b

2

≥ . . . ≥ θ

N

b

N

≥ 0

 . Analysis similar to that in the proof of Lemma 3.3 shows that (3.14) is equivalent to

(3.15) 1

vol(Σ

_(∗)

)

R

Λ(∗)

log 1 − P

N j=2

θ

j

b

1

dθ

2

. . . dθ

N

=

Q

N i=1

b

i

Q

N

i=1

(b

1

+ . . . + b

i

) · 1 vol(Σ

_(∗)

)

× R

Σ_(∗)

log



N

X

j=1

η

j

b

1

+ b

2

+ . . . + b

j



dη

2

. . . dη

N

,

where η

1

:= 1 − P

N j=2

η

j

.

The proof of (3.15) is immediate. We change the variables:

θ

k

=

N

X

j=k

b

k

b

1

+ . . . + b

j

η

j

, for k = 2, . . . , N, where (θ

2

, . . . , θ

N

) ∈ Λ

(∗)

and (η

2

, . . . , η

N

) ∈ Σ

(∗)

. Obviously,

1 − P

N j=2

θ

j

b

1

=

N

X

j=1

η

j

b

1

+ . . . + b

j

and

dθ

2

. . . dθ

N

=

Q

N i=1

b

i

Q

N

i=1

(b

1

+ . . . + b

i

) dη

₂

. . . dη

_N

. This establishes the formula (3.15). Hence (3.14) is also true.

Let us observe that the simplex Λ has the vertices A

k

=

 b

1

b

1

+ . . . + b

k

, b

2

b

1

+ . . . + b

k

, . . . , b

k

b

1

+ . . . + b

k

, 0, . . . , 0



(k = 1, . . . , N ) and η

1

, . . . , η

N

are the barycentric coordinates on Λ.

Let c

j

:= b

1

+ . . . + b

j

and a

j

:= 1/c

j

(for j = 1, . . . , N ). Applying (3.13), (3.14) and (3.7) (see Lemma 3.6) gives

t(1, . . . , N ) =

Q

N i=1

b

i

Q

N

i=1

(b

1

+ . . . + b

i

) (3.16)

×



−

N −1

X

j=1

1 j +

N

X

j=1

a

^{N −1}_j

log a

j

Q

N

k=1, k6=j

(a

j

− a

_k

)

 . We have, for j = 1, . . . , N ,

a

^{N −1}_j

· log a

_j

Q

N

k=1, k6=j

(a

j

− a

k

)

= (1/c

j

)

^{N −1}

log(1/c

j

) Q

N

k=1, k6=j

(1/c

j

− 1/c

_k

)

= − c

^{N −1}_j

Q

N i=1

c

i

c

^{N −1}_j

· Q

N i=1

c

i

· (1/c

j

)

^{N −1}

log c

j

Q

N

k=1, k6=j

(1/c

j

− 1/c

_k

)

= − ( Q

N

i=1

c

i

) log c

j

c

j

· Q

N

i=1, i6=j

(c

i

− c

j

)

= − ( Q

N

i=1

(b

1

+ . . . + b

i

)) log(b

1

+ . . . + b

j

) (−1)

^j−1

( Q

j

i=1

(b

i

+ . . . + b

j

))( Q

N

i=j+1

(b

j+1

+ . . . + b

i

)) , where Q

N

i=j+1

(b

j+1

+ . . . + b

i

) := 1 for j = N . On substituting the above expression into (3.16) we obtain

t(1, . . . , N ) =

Q

N i=1

b

i

Q

N

i=1

(b

1

+ . . . + b

i

)



−

N −1

X

j=1

1 j



+

N

X

j=1

(−1)

^j

( Q

N

i=1

b

i

) log(b

1

+ . . . + b

j

) ( Q

j

i=1

(b

i

+ . . . + b

j

))( Q

N

i=j+1

(b

j+1

+ . . . + b

i

)) ,

and (3.12) is proved for % = %

0

. In the same manner we can see that (3.12) is true for each % ∈ S(N ).

By Lemma 3.7,

(3.17) X

%∈S(N )

Q

N i=1

b

_%(i)

Q

N

i=1

(b

%(1)

+ . . . + b

%(i)

) = 1.

Combining (3.10) with (3.11), (3.12) and (3.17) gives log γ(P ) = − 1

2



^{N −1}

X

j=1

1

j + X

%∈S(N )

t(%(1), . . . , %(N ))



= − 1 2



^{N −1}

X

j=1

1 j



1 − X

%∈S(N )

Q

N i=1

b

_%(i)

Q

N

i=1

(b

%(1)

+ . . . + b

%(i)

)



+ 1 2

N

X

j=1

X

%∈S(N )

(−1)

^j−1

× ( Q

N

i=1

b

_%(i)

) log(b

_%(1)

+ . . . + b

_%(j)

) ( Q

j

i=1

(b

%(i)

+ . . . + b

%(j)

))( Q

N

i=j+1

(b

%(j+1)

+ . . . + b

%(i)

))

= 1 2

N

X

j=1

X

%∈S(N )

(−1)

^j−1

× ( Q

N

i=1

b

%(i)

) log(b

%(1)

+ . . . + b

%(j)

) ( Q

j

i=1

(b

_%(i)

+ . . . + b

_%(j)

))( Q

N

i=j+1

(b

_%(j+1)

+ . . . + b

_%(i)

))

= 1 2

N

X

j=1

X

1≤k1<...<kj≤N

(−1)

^j−1

B(k

1

, . . . , k

j

) log(b

k1

+ . . . + b

kj

), where

B(k

1

, . . . , k

j

) := X

%

Q

N i=1

b

%(i)

( Q

j

i=1

(b

%(i)

+ . . . + b

%(j)

))( Q

N

i=j+1

(b

%(j+1)

+ . . . + b

%(i)

)) and the summation runs over all permutations % ∈ S(N ) such that %({1, . . . . . . , j}) = {k

1

, . . . , k

j

}. Every such permutation is a combination of a bijec- tion

µ : {1, . . . , j} → {k

1

, . . . , k

j

} and a bijection

ν : {j + 1, . . . , N } → {1, . . . , N } \ {k

1

, . . . , k

j

}.

Therefore B(k

1

, . . . , k

j

)

=

 X

µ

Q

j i=1

b

µ(i)

Q

j

i=1

(b

_µ(i)

+ . . . + b

_µ(j)

)



X

ν

Q

N

i=j+1

b

ν(i)

Q

N

i=j+1

(b

_ν(j+1)

+ . . . + b

_ν(i)

)

 , where the summation extends over all bijections µ, ν as above. By Lemma 3.7, both the factors in the above product are equal to 1. Hence B(k

1

, . . . , k

j

)

= 1. This establishes the formula (3.9), and the proof of Theorem 3.2 is complete.

4. The capacities D and %. Let K be a compact subset of C

^N

. Let kf k

K

denote the supremum norm of a function f : K → C. For a nonnega- tive integer s put

h

s

:= s + N − 1 N − 1



and m

s

:= s + N N

 .

Let e

1

(z), e

2

(z), . . . be all monomials z

^α

:= z

₁^α1

. . . z

_N^αN

ordered so that the degrees of the e

j

(z) are nondecreasing and the monomials of a fixed degree are ordered lexicographically. For an integer j ≥ 1 let α(j) := (α

1

, . . . , α

N

), where z

₁^α1

. . . z

_N^αN

= e

j

(z). Put |α(j)| := α

1

+ . . . + α

N

. For an integer k (1 ≤ k ≤ h

s

) let β(s, k) := α(m

s−1

+ k), where m

−1

:= 0, and put e

s,k

(z) := e

j

(z), where j = m

s−1

+ k. It is easy to check that e

s,1

, . . . , e

s,hs

are all monomials z

^α

of degree s ordered lexicographically. Put M

s,k

:= inf{kqk

K

: q ∈ H

s,k

},

where

H

s,k

:= n

e

s,k

(z) + X

j<k

c

j

e

s,j

(z) : c

j

∈ C o . Let τ

s,k

:= (M

s,k

)

^1/s

. Put

Σ = Σ

^{N −1}

:= n

θ = (θ

1

, . . . , θ

N

) ∈ R

^N

: θ

j

≥ 0 for j = 1, . . . , N ;

N

X

j=1

θ

j

= 1 o , Σ

0

= Σ

₀^{N −1}

:= {θ ∈ Σ

^{N −1}

: θ

j

> 0 for j = 1, . . . , N }.

For θ ∈ Σ let us define the directional Chebyshev constants:

τ (K, θ) := lim sup{τ e

s,k

: s → ∞, β(s, k)/s → θ}, τ e

−

(K, θ) := lim inf{τ

s,k

: s → ∞, β(s, k)/s → θ}.

It is known (see [3]) that τ (K, θ) = e e τ

₋

(K, θ) for each θ ∈ Σ

0

and that

log τ (K, θ) is a convex function on Σ e

0

.

Definition 4.1 (see [3]). The homogeneous transfinite diameter D(K) is

D(K) := exp

 1

vol Σ

R

Σ

log τ (K, θ) dω(θ) e

 , where vol Σ := R

Σ

dω(θ) and ω denotes the Lebesgue surface area measure on the hyperplane {θ = (θ

1

, . . . , θ

N

) ∈ R

^N

: P

N

j=1

θ

j

= 1}.

Definition 4.2 (see [6]). The Chebyshev constant %(K) is

%(K) := lim

s→∞

(%

s

(K))

^1/s

,

where %

s

(K) := inf{kQk

K

}, the infimum being taken over all polynomials Q ∈ H

s

normalized so that kQk

S

= 1.

Definition 4.3. Let µ be a positive Radon measure supported on K.

The pair (K, µ) is said to have the Bernstein–Markov property if for every λ > 1 there exists an M > 0 such that for all polynomials p

kpk

_K

≤ M λ

^{deg p}

kpk

₂

, where kpk

₂

:=  R

K

|p|

²

dµ 

1/2

.

A few examples of pairs with the Bernstein–Markov property can be found e.g. in [7] and [8].

Definition 4.4. K is called unisolvent with respect to homogeneous poly- nomials if no nonzero homogeneous polynomial vanishes identically on K.

5. Orthogonal polynomials associated with K. Suppose that the pair (K, µ) has the Bernstein–Markov property. For two integers s, k (s ≥ 0, 1 ≤ k ≤ h

s

) put

C

s,k

:= inf{kpk

2

: p ∈ H

s,k

}.

It is easy to check that there exists at least one polynomial A

s,k

(z) ∈ H

s,k

attaining the infimum. For each s, {A

s,k

(z)}

k=1,...,hs

is obviously a sequence of orthogonal polynomials in the space L

²

(K, µ). If kA

s,k

k

₂

> 0 then put

B

s,k

(z) := A

s,k

(z)/kA

s,k

k

₂

, z ∈ C

^N

.

For an integer j ≥ 1, let A

j

:= A

s,k

and B

j

:= B

s,k

, where the integers s, k (s ≥ 0, 1 ≤ k ≤ h

s

) are chosen so that j = m

s−1

+ k.

Proposition 5.1. For each θ ∈ Σ

0

,

τ (K, θ) = lim{(kA e

j

k

2

)

^1/|α(j)|

: j → ∞, α(j)/|α(j)| → θ}.

P r o o f. This is an easy consequence of the Bernstein–Markov property and the equality e τ (K, θ) = e τ

−

(K, θ) for θ ∈ Σ

0

([3], Lemma 4.1).

Theorem 5.2. Let K be unisolvent with respect to homogeneous polyno-

mials. Then:

(a) kA

j

k

₂

> 0 for each j, (b) lim

j→∞

(kB

j

k

_K

)

^1/|α(j)|

= 1,

(c) Ψ

K

(z) = lim sup

_j→∞

|B

j

(z)|

^1/|α(j)|

, z ∈ C

^N

.

P r o o f. The theorem can be proved in the same manner as Theorem 1 of [8] (it suffices to replace e

j

(z) + P

i<j

c

i

e

i

(z) by e

s,k

(z) + P

i<k

b

i

e

s,i

(z), where e

s,k

= e

j

). Z´ eriahi’s result is true for z ∈ C

^N

\ b K, where b K is the polynomially convex hull of K, because for z ∈ b K one cannot prove that the sequence (j

d

) in the proof of Theorem 1 of [8] is not bounded. In our proof the sequence (j

d

) must be unbounded for each z ∈ C

^N

, because we deal with the homogeneous polynomials and their degrees must increase.

6. Comparison of the capacities D, % and γ. Let K be a compact subset of the unit ball B := {z ∈ C

^N

: P

N

j=1

|z

_j

|

²

≤ 1}. It is known that

%(K)/ √

N ≤ D(K) ≤ %(K)

^1/N

and

γ(K)/

√

N ≤ D(K) ≤ exp(−κ

N

/N )γ(K)

^1/N

(see [3], Theorem 5.7 and Corollary 5.8). It turns out that the exponents in the estimates above cannot be improved. This is an easy consequence of (6.1), (6.3), (6.4) and (6.6) in the following theorem:

Theorem 6.1. If ε > 0 and C > 0, then there exist compact subsets K

1

, K

2

, K

4

, K

5

of the unit sphere S and compact subsets K

3

, K

6

of the unit ball B such that

D(K

1

) > C%(K

1

)

^{1/N +ε}

, (6.1)

D(K

2

) < C%(K

2

)

^{1−1/N −ε}

, (6.2)

D(K

3

) < C%(K

3

)

^1−ε

, (6.3)

D(K

4

) > Cγ(K

4

)

^{1/N +ε}

, (6.4)

D(K

5

) < Cγ(K

5

)

^{1−1/N −ε}

, (6.5)

D(K

6

) < Cγ(K

6

)

^1−ε

. (6.6)

P r o o f. It is known that for every compact subset K of C

^N

, γ(K) ≤ %(K) ≤ γ(K) exp(−κ

N

)

(see [6], Proposition 12.1). Therefore it suffices to prove (6.1)–(6.3). Let P = P (r

1

, . . . , r

N

) := {z ∈ C

^N

: |z

1

| = r

₁

, . . . , |z

N

| = r

_N

}, where r

1

, . . . , r

N

are real positive numbers. It is known that

Ψ

P

(z) = max(|z

1

|/r

₁

, . . . , |z

N

|/r

_N

), z ∈ C

^N

(see [4], p. 304). For every compact set K we have %(K)

⁻¹

= sup{Ψ

K

(z) :

z ∈ B} (see [6], Theorem 8.2). Hence %(P ) = min(r

1

, . . . , r

N

).

It is also known that

D(P ) =

 Y

^N

j=1

r

j



1/N

(see e.g. [3], Corollary 6.4). Therefore we can take K

1

:= P (r

1

, . . . , r

N

)

(where r

1

= . . . = r

N −1

= (1/(N − 1) − δ)

^1/2

and r

N

= ((N − 1)δ)

^1/2

, 0 < δ < 1/(N (N − 1))),

K

2

:= P (R

1

, . . . , R

N

)

(where R

1

= . . . = R

N −1

= δ

^1/2

and R

N

= (1 − (N − 1)δ)

^1/2

, 0 < δ < 1/N ) and

K

3

:= P (s

1

, . . . , s

N

)

(where s

1

= . . . = s

N −1

= s

N

= δ

^1/2

, 0 < δ < 1/N ). It is easy to check that (6.1)–(6.3) are true if δ > 0 is sufficiently small. This proves the theorem.

It is known that D(K) = γ(K)

^1/2

for every compact set K such that K ⊂ {(z

1

, z

2

) ∈ C

²

: |z

1

|

²

+ |z

2

|

²

= 1}

(see [6], Theorem 12.3). We can ask whether the following conjecture is true for the unit sphere S ⊂ C

^N

(N ≥ 3):

There exist positive numbers δ, m and M such that for every compact subset K of S,

mγ(K)

^δ

≤ D(K) ≤ M γ(K)

^δ

.

Corollary 6.2. The above conjecture is false for each N ≥ 3.

P r o o f. Suppose that this conjecture is true for some N . Then δ ≥ 1 − 1/N by (6.5) and δ ≤ 1/N by (6.4). This is impossible for N ≥ 3.

References

[1] H. A l e x a n d e r, Projective capacity , in: Conference on Several Complex Variables, Ann. of Math. Stud. 100, Princeton Univ. Press, 1981, 3–27.

[2] U. C e g r e l l and S. K o l o d z i e j, An identity between two capacities, Univ. Iagel. Acta Math. 30 (1993), 155–157.

[3] M. J ¸ e d r z e j o w s k i, The homogeneous transfinite diameter of a compact subset of C

^N

, Ann. Polon. Math. 55 (1991), 191–205.

[4] J. S i c i a k, On an extremal function and domains of convergence of series of homo- geneous polynomials, ibid. 10 (1961), 297–307.

[5] —, On some extremal functions and their applications in the theory of analytic func- tions of several complex variables, Trans. Amer. Math. Soc. 105 (1962), 322–357.

[6] —, Extremal Plurisubharmonic Functions and Capacities in C

ⁿ

, Sophia Kokyuroku

in Math. 14, Sophia University, Tokyo, 1982.

[7] J. S i c i a k, Families of polynomials and determining measures, Ann. Fac. Sci. Tou- louse 9 (1988), 193–211.

[8] A. Z ´ e r i a h i, Capacit´ e, constante de ˇ Cebyˇ sev et polynˆ omes orthogonaux associ´ es ` a un compact de C

ⁿ

, Bull. Sci. Math. (2) 109 (1985), 325–335.

INSTITUTE OF MATHEMATICS JAGIELLONIAN UNIVERSITY REYMONTA 4

30-059 KRAK ´OW, POLAND

Re¸ cu par la R´ edaction le 6.5.1994