The Bergman kernel: explicit formulas, deflation, Lu Qi-Keng problem and Jacobi Polynomials

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by  Tomasz Beberok

THE BERGMAN KERNEL: EXPLICIT FORMULAS, DEFLATION,
LU QI-KENG PROBLEM AND JACOBI POLYNOMIALS
Tomasz Beberok
In this paper we investigate the Bergman kernel function forintersection of two
complex ellipsoids
6 (z,w ,w ) Cn+2 : z 2+ + z 2+ w q < 1, z 2+ + z 2+ w r < 1 .
1 { 1 2 ∈ | 1| ··· | n| | 1| | 1| ··· | n| | 2| }
0
2
n Keyword: Lu Qi-Keng problem, Bergman kernel, Routh-Hurwitz theorem,
a Jacobi polynomials
J

2
AMS Subject Classifications: 32A25; 33D70
1

]
V
1 Introduction
C
.
h Let D be a bounded domain in Cn. The Bergman space L2(D) is the space of all
t a
a square integrable holomorphic functions onD. Then the Bergmankernel K (z,w)
m D
is defined [2] by
[

1 ∞
v K (z,w) = Φ (z)Φ (w), (z,w) D D,
D j j
9 ∈ ×
8 Xj=0
9
2 where Φ ( ): j = 0,1,2,... is a complete orthonormal basis for L2(D). It is
0 { j · } a
defined for arbitrary domains, but it is hard to obtain concrete representations
.
1
for the Bergman kernel except for special cases like a Hermitian ball or polydisk.
0
6 Refer to [6] for more on this topic.
1 In 2015 [1] the author of this paper computed the Bergman kernel for Dq,r =
: 1
v z = (z ,w ,w ) C3: z 2+ w q < 1, z 2+ w r < 1 explicitly. The goal of
1 1 2 1 1 1 2
i { ∈ | | | | | | | | }
X thispaperistoextendtheresultin[1]tohigherdimensionalcase, namelytoDq,r =
n
r (z,w ,w ) Cn+2 : z 2+ + z 2+ w q < 1, z 2+ + z 2+ w r < 1 .
a { 1 2 ∈ | 1| ··· | n| | 1| | 1| ··· | n| | 2| }
This paper will be organized as follows. In Section 2, we will compute ex-
plicit formula of the Bergman kernel for Dq,r and we show deflation identity
n
between domains Dq,r and Dq,r = z = (z ,w ,w ) C3: z 2/n + w q <
n 1/n { 1 1 2 ∈ | 1| | 1|
1, z 2/n + w r < 1 . In Section 3, we show some relation between Bergman
1 2
| | | | }
kernel for D2,2 and Jacobi polynomials. In Section 4, we investigate the Lu Qi-
n
Keng problem for D2,2. In the final section, we consider the Bergman kernel for
n
Ωr := (z,w) C Cn: z 2 + w r < 1,..., z 2 + w r < 1 .
n { ∈ × | | | 1| | | | n| }
1
2 Bergman Kernel
Let ζ = (z,w ,w ) Cn+2. Put Φ (ζ) = ζκ = zαwγ1wγ2 for each multi-index
1 2 ∈ κ 1 2
κ = (α,γ ,γ ). Since Dq,r is complete Reinhardt domain, the collection of Φ
1 2 n { κ}
such that each α 0 and γ 0 is a complete orthogonal set for L2(Dq,r).
i ≥ j ≥ n
Proposition 2.1 Let α Z for i = 1,...,n and γ 0, γ 0. Then, we have
i + 1 2
∈ ≥ ≥
πn+2Γ(2γ1+2 + 2γ2+2 +1) n Γ(α +1)
zαwγ1wγ2 2 = q r i=1 i ,
k 1 2 kL2(Dnq,r) (γ +1)(γ +1)Γ(2γ1+2 + 2γ2+2 + α +n+1)
2 2 q Qr | |
where α = α + +α .
1 n
| | ···
Proof.
zαwγ2wγ3 2 = z 2α w 2γ2 w 2γ3 dV(z)dV(w)
k 1 2 kL2(Dnq,r) | | | 1| | 2|
Z
Dnq,r
We introduce polar coordinate in each variable by putting z = reiθ, w = s eiλ1,
1 1
w = s eiλ2. After doing so, and integrating out the angular variables we have
2 2
(2π)n+2 r2α+1s2γ1+1s2γ2+1dV(r)dV(s),
1 2
ZRe(Dnq,r)
where Re(Dq,r) = (r,s) Rn R2 : r 2 +sq < 1, r 2 +sr < 1 . Next we
n ∈ + × + k k 1 k k 2
use spherical coordinates in the r variable to obtain
(cid:8) (cid:9)
(2π)n+2 ρ2|α|+2n−1ω2α+1s2γ1+1s2γ2+1dσ(ω)dV(s)dρ,
1 2
ZS∗Re(Dnq,r)ZSn+1−1
where S Re(Dq,r) = (ρ,s) R R2 : ρ2 +sq < 1, ρ2 +sr < 1 . Using (6)
∗ n ∈ + × + 1 2
form Lemma 1 in [4] we obtain
(cid:8) (cid:9)
β(α+1) 1 (1−ρ2)1/q (1−ρ2)1/r
(2π)n+2 ρ2|α|+2n−1s2γ1+1s2γ2+1ds ds dρ,
2n−1 1 2 1 2
Z0 Z0 Z0
where β(α) = Qni=1Γ(αi) . Integrating out of s and s variables, we have
Γ(α1+···+αn) 1 2
(2π)n+2β(α+1) 1ρ2|α|+2n−1(1 ρ2)2γ1q+2+2γ2r+2 dρ
2n+1(γ +1)(γ +1) −
1 2 Z0
After a little calculation using well known fact
1 Γ((a+1)/p)Γ(b+1)
xa(1 xp)bdx = ,
− pΓ((a+1)/p+b+1)
Z0
we obtain desired result.
Now we discuss the Bergman kernel for Dq,r.
n
2
Theorem 2.1 The Bergman kernel for Dq,r is given by
n
Lq,r(a,b)
KDnq,r((z,w1,w2),(η,ξ1,ξ2)) = πn+2(1−z1η1n−···−znηn)2q+r2+1,
where
w ξ w ξ
a = 1 1 , b = 2 2
2 2
(1−z1η1 −···−znηn)q (1−z1η1 −···−znηn)r
2 ∂ 2 ∂
Lq,r (x,y) = nLq,r(x,y)+ xLq,r(x,y)+ yLq,r(x,y) and
n+1 n q∂x n r∂y n
2(q(1 x)(y +1)+r(x+1)(1 y))
Lq,r(x,y) = − −
1 qr(1 x)3(1 y)3
− −
Proof. By Proposition 2.1, we have
K ((z,w ,w ),(η,ξ ,ξ ))
Dnq,r 1 2 1 2
= 1 ∞ ∞ (γ1 +1)(γ2 +1)Γ(2γ1q+2 + 2γ2r+2 +|α|+n+1)µανγ1νγ2,
πn+2 Γ(2γ1+2 + 2γ2+2 +1) n Γ(α +1) 1 2
Xα=0γ1X,γ2=0 q r i=1 i
Q
where µα = (z η )α1 (z η )αn and ν = w ξ , ν = w ξ . Sum out of each α ,
1 1 ····· n n 1 1 1 2 2 2 i
we have
1 ∞ (γ1 +1)(γ2 +1)Γ(2γ1q+2 + 2γ2r+2 +n+1)νγ1νγ2,
πn+2(1−τ)q2+2r+1 γ1X,γ2=0 Γ(2γ1q+2 + 2γ2r+2 +1)(1−τ)2γq1+2γr2 1 2
where τ = z η + + z η . Now we will consider sequence of functions Lq,r
1 1 ··· n n n
defined as follows
∞ (γ +1)(γ +1)Γ(2γ1+2 + 2γ2+2 +n+1)
Lq,r(x,y) = 1 2 q r xγ1yγ2
n Γ(2γ1+2 + 2γ2+2 +1)
γ1X,γ2=0 q r
Using the identity Γ(t+1) = tΓ(t) we easily obtain recursion formula
2 ∂ 2 ∂
Lq,r (x,y) = nLq,r(x,y)+ xLq,r(x,y)+ yLq,r(x,y)
n+1 n q∂x n r∂y n
Moreover
∞
2γ +2 2γ +2
Lq,r(x,y) = (γ +1)(γ +1) 1 + 2 +1 xγ1yγ2
1 1 2 q r
γ1X,γ2=0 (cid:18) (cid:19)
Hence
2(q(1 x)(y +1)+r(x+1)(1 y))
Lq,r(x,y) = − −
1 qr(1 x)3(1 y)3
− −
which completes the proof.
3
2.1 Deflation
Now we will consider following domains
Dq,r = (z,w ,w ) C3: z 2/n + w q < 1, z 2/n + w r < 1
1/n { 1 2 ∈ | | | 1| | | | 2| }
Similarly as in Sect. 2, we have
Proposition 2.2 Let α 0, γ 0 and γ 0. Then, we have
1 2
≥ ≥ ≥
nπ3Γ(2γ1+2 + 2γ2+2 +1)Γ(nα+n)
zαwγ1wγ2 2 = q r .
k 1 2 kL2(D1q/,rn) (γ2 +1)(γ2 +1)Γ(2γ1q+2 + 2γ2r+2 +nα+n+1)
Therefore
K ((0,w ,w ),(0,ξ ,ξ ))
Dq,r 1 2 1 2
1/n
∞ (γ +1)(γ +1)Γ(2γ1+2 + 2γ2+2 +n+1)
= 1 2 q r (w ξ )γ1(w ξ )γ2
π3n!Γ(2γ1+2 + 2γ2+2 +1) 1 1 2 2
γ1X,γ2=0 q r
Hance
1
KD1q/,rn((0,w1,w2),(0,ξ1,ξ2)) = π3n!Lqn,r(w1ξ1,w2ξ2)
Comparing formulas for KD1q/,rn and KDnq,r, we obtain following deflation identity
Proposition 2.3 For every n N and every positive numbers q and r, we have
∈
n!
πn−1KD1q/,rn((0,w1,w2),(0,ξ1,ξ2)) = KDnq,r((0,...,0,w1,w2),(0,...,0,ξ1,ξ2))
Note that, we have some kind of deflation result similar to that obtained in [3].
3 Some representations of Bergman kernel for D2,2
n
Jacobi polynomials are a class of classical orthogonal polynomials. They are or-
thogonal with respect to the weight (1 x)k(1 + x)l on the interval [ 1,1]. For
− −
k, l > 1 the Jacobi polynomials are given by the formula
−
( 1)d ∂
P(k,l)(z) = − (1 z)−k(1+z)−l (1 z)k(1+z)l(1 z2)d .
d 2dd! − ∂zd − −
(cid:8) (cid:9)
The Jacobi polynomials are defined via the hypergeometric function as follows
(k +1) 1 z
(k,l) d
P (z) = F d,1+k +l+d;k +1, − , (1)
d d! 2 1 − 2
(cid:18) (cid:19)
where (k+1) is Pochhammer’s symbol and F is Gaussian or ordinary hyperge-
d 2 1
ometric function defined for z < 1 by the power series
| |
∞
(a) (b)
F (a,b;c;z) = n nzn
2 1
n!(c)
n
n=0
X
4
Appell series F defined by
1
∞ ∞ (a) (b) (b′)
F (a;b,b′;c,x,y) = m+n n mxnym, ( x < 1, y < 1)
1
n!m!(c) | | | |
n+m
n=0m=0
XX
is one of a natural two-variable extension of hypergeometric series F .
2 1
The following reduction formulas can be proved (see, for details [5], p. 238-239)
F (a;b,b′;b+b′;x,y) = (1 y)−a F (a,b;b+b′;(x y)/(1 y)) (2)
1 2 1
− − −
F (a;b,b′;c;x,x) = F (a,b+b′;c;x) (3)
1 2 1
Comparing functions F and L2,2 it is easy to see
1 n
Γ(3+n)
F (3+n;2,2;3;x,y) = L2,2(x,y)
2 1 n
The following is the main theorem of this section.
Theorem 3.1 The Bergman kernel for D2,2 can be expressed in the following ways
n
(i) K ((0,w ,w ),(0,ξ ,ξ )) =
Dn2,2 1 2 1 2
Γ(3+n) n n−i n n 1 (2) (2) (w ξ )i(w ξ )k
− i k 1 1 2 2
2πn+2 i k (3) (1 w ξ )2+i(1 w ξ )2+k
i=0 k=0(cid:18) (cid:19)(cid:18) (cid:19) i+k − 1 1 − 2 2
XX
(ii) For every m N 0 , we have
∈ ∪{ }
C (2+m)(2 ν ν )P(32,21)(x′+1)
K ((0,w ,w ),(0,ξ ,ξ )) = m − 1 − 2 m 1−x′
D22m,2+1 1 2 1 2 π2m+3(1 ν1 ν2 ν1ν2)m+3
− − −
C (2m+1)P(32,−12)(x′+1)
m m 1−x′ ,
− π2m+3(1 ν ν ν ν )m+2
1 2 1 2
− − −
2
where ν = w ξ , ν = w ξ , x′ = ν1−ν2 and C = Γ(4+2m)m! .
1 1 1 2 2 2 2−ν1−ν2 m 6(52)m
(cid:16) (cid:17)
(iii) For every m N, we have K ((0,w ,w ),(0,ξ ,ξ )) =
∈ D22m,2 1 2 1 2
2P(32,−21)(x′+1) (2 ν ν )P(23,21)(x′+1)
Γ(3+2m)m! m 1−x′ − − 1 − 2 m−1 1−x′ ,
π2m+26(5) (1 ν ν ν ν )m+2
2 m−1 − 1 − 2 − 1 2
(iv) If w ξ = w ξ then
1 1 2 2
Γ(3+n)(3+nw ξ )
K ((0,w ,w ),(0,ξ ,ξ )) = 1 1
Dn2,2 1 2 1 2 6πn+2(1 w ξ )4+n
− 1 1
5
Proof. For the proof of (i), if we apply the recursion formula (see [8])
n n−i n n 1 (b) (b′)
F (a+n;b,b′;c;x,y) = − i k
1
i k (c)
i=0 k=0(cid:18) (cid:19)(cid:18) (cid:19) i+k
XX
xiykF (a+i+k;b+i,b′ +k;c+i+k;x,y),
1
×
then using formula
1
F (3+i+k;2+i,2+k;3+i+k;x,y) =
1 (1 x)2+i(1 y)2+k
− −
we obtain (i).
In order to prove the second statements we need the following well-known contigu-
ous relation
cF (a;b,b′;c;x,y) (c a)F (a;b,b′;c+1;x,y)
1 1
− −
aF (a+1;b,b′;c+1;x,y) = 0
1
−
It follows that
2m+1
F (4+2m;2,2;3;x,y) = F (2m+4;2,2;4;x,y)
1 1
− 3
4+2m
+ F (2m+5;2,2;4;x,y)
1
3
By (2), we have
2m+1 x y
F (4+2m;2,2;3;x,y) = F 2m+4,2;4; −
1 − 3(1 y)4+2m2 1 1 y
− (cid:18) − (cid:19)
4+2m x y
+ F 2m+5,2;4; −
3(1 y)5+2m2 1 1 y
− (cid:18) − (cid:19)
Then by (see, for details [5], p. 66)
z −a a a+1 1
F (a,b;2b;z) = 1 F , ;b+ ;[z/(2 z)]2 ,
2 1 2 1
− 2 2 2 2 −
(cid:18) (cid:19)
(cid:16) (cid:17)
we have
(2m+1)24+2m 5 5
F (4+2m;2,2;3;x,y) = F m+2,m+ ; ;(x′)2
1 − 3(2 x y)4+2m2 1 2 2
− − (cid:18) (cid:19)
(4+2m)25+2m 5 5
+ F m+ ,m+3; ;(x′)2 ,
3(2 x y)5+2m2 1 2 2
− − (cid:18) (cid:19)
2
where x′ = x−y . Next by (see, for details [5], p. 64)
2−x−y
(cid:16) (cid:17)
F (a,b;c;z) =(1 z)−a F (a,c b;c;z/(z 1))
2 1 2 1
− − −
=(1 z)−b F (c a,b;c;z/(z 1)),
2 1
− − −
6
we can easily get F (4+2m;2,2;3;x,y) =
1
2m+1 5 x′ 2
F m+2, m; ;
− 3(1 x y xy)m+22 1 − 2 x′ 1
!
− − − (cid:18) − (cid:19)
(2+m)(2 x y) 5 5 x′ 2
+ − − F m+ ,m+3; ; ,
3(1 x y xy)m+32 1 2 2 x′ 1
!
− − − (cid:18) − (cid:19)
Finally, by (1) we obtain (ii). The formula (iii) can be obtained by the same
method and we omit the details.
Nowwewillprove(iv). This isaformalexercise butweinclude itforcompleteness:
From (3)
F (3+n;2,2;3;x,y) = F (3+n,4;3;x)
1 2 1
Now it is relatively easy to compute the following result
3+nx
F (3+n,4;3;x) =
2 1 3(1 x)4+n
−
Thus we prove (iv).
4 Lu Qi-Keng problem
The explicit formula of the Bergman kernel function for the domains Dq,r enables
n
us to investigate whether the Bergman kernel has zeros in Dq,r Dq,r or not. We
n × n
will call this kind of problem a Lu Qi-Keng problem. If the Bergman kernel for a
bounded domain does not have zeros, then the domain will be called a Lu Qi-Keng
domain.
By Theorem 1.2 form [1] if n = 1, then Dq,r is a Lu Qi-Keng domain for all
n
positive real numbers q and r. Combining deflation identity (Proposition 2.3) and
Proposition 4.4 from [1], if n = 2 and q = r, then Dr,r is not a Lu Qi-Keng domain.
2
Using the same method as in [1] we will prove that it is also true for n = 3.
Denote by
G(x,y) =3r3(1 x)3(1 y)3+22r2(1 x)2(1 y)2(1 xy)
− − − − −
+24r(1 x)(1 y) x(2x+1)y2+(x 8)xy +x+y +2
− − −
+8(1 xy) x2y(4y+7)+x2 +y2 +xy(7y 38)+7x+7y +4 .
(cid:0) (cid:1)
− −
(cid:0) (cid:1)
Since
r3(1 x)5(1 y)5
G(x,y) = − − Lr,r(x,y)
2 3
then the Bergman kernel KD3r,r has zero inside D2r,r ×D2r,r if polynomial G(ǫx,ǫy)
does not satisfy the stability property (see [1], for details) for some 0 < ǫ < 1. By
following Šiljak and Stipanović [7] we consider polynomial
z3G(eiη,1/z) = d(z) =d z3 +d z2 +d z +d ,
3 2 1 0
7
where
d = 3r3( 1+t)3 22r2( 1+t)2t+24rt( 1 t+2t2) 8t(1+7t+4t2),
0
− − − − − −
d = 8 9r3( 1+t)3 +336t2 56t3 +22r2( 1+t)2(1+2t)
1
− − − −
24r(1 10t+8t2 +t3),
− −
d = 9r3( 1+t)3 22r2(2 3t+t3) 8( 7+42t+t3)
2
− − − − −
24r(1+8t 10t2 +t3),
− −
d = 22r2( 1+t)2 3r3( 1+t)3 24r( 2+t+t2)+8(4+7t+t2)
3
− − − − −
and t = eiη.
With the polynomial d(z) we associate the Schur-Cohn 3 3 matrix
×
d d d
11 12 13
M(eiη) = d d d ,
21 22 23
 
d d d
31 32 33
 
d = j (d d d d ), where 1 j k. The matrix M(eiη) is
jk l=1 m−j+l m−k+l − j−l k−l ≤ ≤
defined when j > k to become Hermitian. After some calculation (with the help
P
of a computer program Maple or Mathematica), we have
3
detM(eiη) = 130459631616r3sin12(η/2) g (r)cos(nη) ,
n
−
" #
n=0
X
where
g (r) = 26624 24672r2+15724r4 2430r6
0
− −
g (r) = 12288+22496r2 20822r4+3645r6
1
−
g (r) = 512+2208r2+5012r4 1458r6
2
− −
g (r) = r2( 32+86r2 +243r4)
3
−
Since
3
g (r) = 38400
n
n=0
X
Therefore it is easy to see that for every r > 0 there exist η such that detM(eiη) <
0. Hence there exist 1 > ǫ > 0, such that polynomial G(ǫx,ǫy) does not satisfy
the stability property (see [7]). As a consequence of above consideration, we have
following corollary
Corollary 4.1 For any r > 0, domains Dr,r and Dr,r are not Lu Qi-Keng.
3 1/3
Now we will study Lu Qi-Keng problem for D2,2 in the case when n > 3. By
n
representation (iv) from Theorem 3.1
Γ(3+n)(3+nw ξ )
K ((0,w ,w ),(0,ξ ,ξ )) = 1 1 .
Dn2,2 1 1 1 1 6πn+2(1 w ξ )4+n
− 1 1
8
Hence
K (0,i 3/n,i 3/n),(0, i 3/n, i 3/n) = 0.
Dn2,2 − −
(cid:16) p p p p (cid:17)
For brevity, we shall summarize these last statements by saying that
Proposition 4.1 Domain D2,2 is Lu Qi-Keng if and only if n = 1.
n
At the end of this section we would like to present following relations between
zeros of the Bergman kernel for domains Dq,r and Dq,r .
n 1/n
Proposition 4.2 For any positive real numbers q and r
(i) If KDnq,r has zeros, then KD1q/,rn also has zeros,
(ii) If Dq,r is Lu Qi-Keng domain, then Dq,r is Lu Qi-Keng domain.
1/n n
Proof. Note that the zero set is a bi-holomorphic invariant object. Since any
point (z,w ,w ) Dq,r can be mapped equivalently onto the form (0,w ,w ) by
1 2 ∈ n 1 2
following automorphism of the Dq,r
n
f f
(1 a 2)1/q (1 a 2)1/r
Dq,r (z,w ,w ) Ψ (z), −k k w , −k k w Cn+2,
n ∋ 1 2 7→ a (1 z,a )2/q 1 (1 z,a )2/r 2 ∈
(cid:18) −h i −h i (cid:19)
where
hz,ai 1 a+z 1 a 2
1+√1−kak2 − −k k
Ψ (z) = (cid:18) (cid:19) .
a 1 z,a p
−h i
Therefore, we need only consider the zeroes restricted to 0 D D, where
{ } × ×
D := z C : z < 1 . The results now follows from Proposition 2.3.
{ ∈ | | }
5 Additional Results
Our purpose in this section is to consider domains Ωr defined for every positive
n
real number r by
Ωr := (z,w) C Cn: z 2 + w r < 1,..., z 2 + w r < 1
n ∈ × | | | 1| | | | n|
(cid:8) (cid:9)
The reader can see that following proposition is completely analogous to the
results presented earlier.
Proposition 5.1 For j = 1,2,...,n let β 0. Then for α 0, we have
j
≥ ≥
πn+1Γ 2( n β +n)+1 Γ(α+1)
zαwβ 2 = r j=1 j .
L2(Ωn) Γ 2( n(cid:16) βP+n)+α+2 (cid:17)n (β +1)
(cid:13) (cid:13) r j=1 j j=1 j
(cid:13) (cid:13) (cid:16) (cid:17)
P Q
Now we will give an explicit formula for the kernel K of Ωr.
Ωrn n
9
Theorem 5.1 If w,ζ Dn, then the Bergman kernel for Ωr is
∈ n
n n
2(1+ν )
πn+1K ((0,w),(0,ζ))= (1 ν )−2 + k ,
Ωrn − k r(1 ν )2 (1 ν )2(1 ν )
1 n k
k=1 k=1 − ··· − −
Y X
where ν = w ζ for k = 1,...,n. Moreover if ν = ν = = ν , then
k k k 1 2 ··· n
(2n r)ν +2n+r
πn+1K ((0,w),(0,ζ))= − 1 .
Ωr
n r(1 ν )2n+1
1
−
Proof. As a consequence of Proposition 5.1, we have
∞ Γ 2( n β +n)+2 n (β +1)
r j=1 j j=1 j
πn+1K ((0,w),(0,ζ)) = νβ,
Ωrn (cid:16) PΓ 2( n β +(cid:17)nQ)+1
β1,.X..,βn≥0 r j=1 j
(cid:16) (cid:17)
P
where νβ = νβ1 νβn. Then, using the fact that aΓ(a) = Γ(a+1),
1 ····· n
∞ n
2 r
πn+1K ((0,w),(0,ζ)) = β + +β +n+ (β +1)νβ.
Ωr 1 n j
n r ··· 2
β1,.X..,βn≥0 (cid:16) (cid:17)Yj=1
Hence, the kernel is
n ∞ n
2
(β +1) (β +1)νβ,
rπn+1 k j
Xk=0β1,.X..,βn≥0 Yj=1
where β = r/2 1. After some calculation using formulas
0
−
∞ ∞
1+x
(m+1)xm = (1 x)−2 and (m+1)2xm = ,
− (1 x)3
m=0 m=0 −
X X
we obtain desired formula.
r
5.1 Zeros of Bergman kernels on Ω
n
In this section, we will prove that the Bergman kernel function of Ωr for any
n
natural number n and positive real number r is zero-free.
Proposition 5.2 For any positive integer n, the domain Ωr is a Lu Qi-Keng
n
domain.
Proof. It is obvious that the Bergman kernel for Ωr has no zeros. The Bergman
1
kernel function of Ωr has zeros iff
n
n n
r 1+ν r 2ν
k k
+ = n+ + = 0,
2 1 ν 2 1 ν
k k
k=1 − k=1 −
X X
10