Abstract. We compute the constant sup

POLONICI MATHEMATICI LVI.2 (1992)

A constant in pluripotential theory

by Zbigniew B locki (Krak´ow)

Abstract. We compute the constant sup

n ₁

deg P



max

S log |P | − R

S

log |P | dσ



: P a polynomial in C ⁿ

o

,

where S denotes the euclidean unit sphere in C ⁿ and σ its unitary surface measure.

Let S be the euclidean unit sphere in C ⁿ and σ its unitary surface mea- sure. Setting

L := {u ∈ PSH(C ⁿ ) : sup{u(z) − log ₊ |z| : z ∈ C ⁿ } < ∞}

we define

(1) c n := sup n

max S u − R

S

u dσ : u ∈ L o .

The aim of this note is to find the exact value of the constant c n (for its applications see [2]). We start with the following

Lemma 1. For a ∈ C we have

R

S

log |z n + a| dσ(z) =



 

 

− 1 2

n−1

X

j=1

1

j (1 − |a| ² ) ^j if |a| ≤ 1,

log |a| if |a| ≥ 1.

P r o o f. The function z 7→ log |z n + a| is harmonic in {|z| < |a|}, thus for

|a| ≥ 1 the statement is obvious. Now let |a| < 1. Without loss of generality we may assume a ≥ 0. We have

R

S

log |z n + a| dσ(z) = n − 1 π

1

R

0

r(1 − r ² ) ⁿ⁻²

2π

R

0

log |re ^it + a| dt dr

1991 Mathematics Subject Classification: Primary 31C10, 32F05.

(see [3], p. 15). Using the equality 1

2π

2π

R

0

log |re ^it + a| dt = max{log r, log a}

we see that

R

S

log |z n + a| dσ(z) = 2(n − 1) log a

a

R

0

r(1 − r ² ) ⁿ⁻² dr

+ 2(n − 1)

1

R

a

r(1 − r ² ) ⁿ⁻² log r dr

= − log a[(1 − r ² ) ⁿ⁻¹ ] ^a ₀ +



− (1−r ² ) ⁿ⁻¹ log r + log r + 1 2

n−1

X

j=1

1

j (1−r ² ) ^j

 1 a

= − 1 2

n−1

X

j=1

1

j (1 − a ² ) ^j .

Lemma 2. For every n ≥ 2 there exists a unique % n ∈ (0, 1) such that (1 + % n )(1 − % ² _n ) ⁿ⁻¹ = 1. Moreover , {% n } decreases to 0.

P r o o f. For % ∈ [0, 1] let

f n (%) := (1 + %)(1 − % ² ) ⁿ⁻¹ − 1 . We have

f _n ⁰ (%) = −(2n − 1)(1 + %)(1 − % ² ) ⁿ⁻²



% − 1 2n − 1

 .

Hence f n (0) = 0, f n increases in (0, 1/(2n − 1)), decreases in (1/(2n − 1), 1) and f n (1) = −1. Thus the first statement is clear. Now since f n+1 (% n ) <

f n (% n ) = 0 we see that the sequence {% n } is decreasing. Let then ε := lim % _n . Since

0 = lim f n (% n ) ≤ lim 2(1 − ε ² ) ⁿ⁻¹ − 1 we conclude that ε must be equal to 0.

Theorem. The supremum in (1) is attained for the function z 7→

log |z n + % n | where % ₁ = 1 and for n ≥ 2, % n is given by Lemma 2. Therefore by Lemma 1 we have

c 1 = log 2 ,

c n = log(1 + % n ) + 1 2

n−1

X

j=1

1

j (1 − % ² _n ) ^j for n ≥ 2 .

P r o o f. Let u ∈ L. First we want to show that it is enough to take functions in the supremum which depend on one variable. For n ≥ 2 we have

R

S

u dσ = R

D

R

Γ

u(z ⁰ , z n ) dσ z

(z ⁰ ) A(|z n |) dλ(z _n ) ,

where D is the unit disc in C, λ its Lebesgue measure, z ⁰ = (z 1 , . . . , z n−1 ) ∈ C ⁿ⁻¹ ,

Γ z

:= {(z ⁰ , z n ) ∈ C ⁿ : |z ⁰ | ² = 1 − |z n | ² } ⊂ S , σ z

the unitary surface measure on Γ z

and

A(r) := n − 1

π (1 − r ² ) ⁿ⁻² . By subharmonicity of the function z ⁰ 7→ u(z ⁰ , z n ) we have

R

Γ

u(z ⁰ , z n ) dσ z

(z ⁰ ) ≥ u(0, z n ) and therefore

R

S

u dσ ≥ R

S

u(0, z n ) dσ(z) .

Since σ is invariant under unitary transformations we may assume that max S u = u(0, . . . , 0, 1), hence

max S u − R

S

u dσ ≤ max

|z

|=1 u(0, z n ) − R

S

u(0, z n ) dσ(z) .

We may thus consider only functions from L depending on the one variable z n . Define

c ⁰ _n := sup  1 d



max S log |P | − R

S

log |P | dσ



: P ∈ P d (C ⁿ )



= sup  1 d



|z max

|=1 log |P (z n )| − R

S

log |P (z n )| dσ(z) 

: P ∈ P d (C)



where P d (C ⁿ ) denotes the set of all polynomials in C ⁿ of degree ≤ d. If P (ζ) = α(ζ + a 1 ) . . . (ζ + a d ) then

1 d



|z max

|=1 log |P (z n )| − R

S

log |P (z n )| dσ(z)



≤ 1 d

d

X

k=1



|z max

|=1 log |z n + a k | − R

S

log |z n + a k | dσ(z) 

≤ max

k=1,...,d



|z max

|=1 log |z n + a k | − R

S

log |z n + a k | dσ(z) 

.

Therefore c ⁰ _n = sup n

|z max

|=1 log |z n + %| − R

S

log |z n + %| dσ(z) : % ∈ [0, 1] o (since for % > 1 one has log(1 + %) − log % < log 2, and by Lemma 1, it is enough to take only % ∈ [0, 1] in the supremum). For % ∈ [0, 1] put

(2) g n (%) := max

|z

|=1 log |z n + %| − R

S

log |z n + %| dσ(z) . By Lemma 1

g n (%) = log(1 + %) + 1 2

n−1

X

j=1

1

j (1 − % ² ) ^j . Therefore g ₁ ⁰ (%) = 1/(1 + %) and for n ≥ 2

g ⁰ _n (0) = 1 ,

g ⁰ _n (%) = (1 + %)(1 − % ² ) ⁿ⁻¹ − 1

(1 + %)% for % > 0 , g ⁰ _n (1) = −1/2 .

Hence, by Lemma 2 it is enough to prove that c ⁰ _n = c n . Of course c ⁰ _n ≤ c _n , let us then take any u ∈ L. By the approximation property of L (see [4]) there exists a sequence of polynomials {P k } such that P _k ∈ P _k and

u =



lim sup

k→∞

1

k log |P k |

 ∗

(v ^∗ denotes the upper regularization of v). Now Fatou’s Lemma gives max S u − R

S

u dσ ≤ lim sup

k→∞

1 k



max S log |P k | − R

S

log |P k | dσ  . The proof is complete.

Put

H := {v ∈ PSH(C ⁿ ) : v(az) = v(z) + log |a| for all a ∈ C , z ∈ C ⁿ } and define

κ n := sup n

max S v − R

S

v dσ : v ∈ H o . Then H ⊂ L and

κ n = − R

S

log |z n | dσ(z) = 1 2

n−1

X

j=1

1

j

(see [4] and [1]).

Corollary. c 1 = log 2, κ n < c n < κ n + log(1 + % n ) for n ≥ 2, lim n→∞ (c n − κ _n ) = 0 and {c n } is increasing.

P r o o f. Everything but the last statement is clear. Taking g n defined by (2) we see that g n ≤ g _n+1 , hence

c n = max

[0,1] g n < max

[0,1] g n+1 = c n+1 .

References

[1] H. A l e x a n d e r, Projective capacity , in: Ann. of Math. Stud. 100, Princeton Univ.

Press, 1981, 3–27.

[2] J.-P. D e m a i l l y, Potential theory in several complex variables, preprint, 1989.

[3] W. R u d i n, Function Theory in the Unit Ball of C ⁿ , Springer, 1980.

[4] J. S i c i a k, Extremal plurisubharmonic functions and capacities in C ⁿ , Sophia Kok- yuroku in Math. 14 (1982).

INSTITUTE OF MATHEMATICS JAGIELLONIAN UNIVERSITY REYMONTA 4

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Re¸ cu par la R´ edaction le 18.3.1991