A note on convergence in Bergman spaces over bounded symmetric domains in C
N(N > 1)
Maher M. H. Marzuq
Abstract. In this paper we prove a result that generalizes the result of [7] on the unit disk to bounded star-shaped circular domains.
1. Introduction
Let D be a bounded symmetric domain in C
N(N > 1), and O ∈ D. D is circular and star-shaped with respect to the origin, i. e. tz ∈ D when z ∈ D and t ∈ C with |t| < 1, [4]. We denote by H(D) the space of holomorphic functions on D.
For p > 0 the Bergman space A
pis defined on D by
A
p= A
p(D) = n
f : f ∈ H(D) and ¯¯ f ¯¯
Ap
= µ 1
V Z
D
¯¯ f (z) ¯¯
pdv
z¶
1p
< ∞ o ,
or equivalently [5],
A
′p= n
f : f ∈ H(D) and ¯¯ f ¯¯
A′p
= sup
0≤r<1
µ 1 V
Z
D
¯¯ f (rz ) ¯¯
pdv
z¶
1p
< ∞ o . (1)
Where V is the Euclidean volume of D and dv
zis the Euclidean element of volume at z ∈ D.
It is well known that a complete orthonormal system (CONS) of homoge- nous polynomials { Ψ
kν} ν = 1, . . . ; m
k= ¡
N+k−1k
¢ ; k = 0, 1, . . . exist on a bounded star-shaped domain [2].
2000
Mathematics Subject Classification.Primary 32A36-42B05.
87
We will have the following lemmas which will be used in the proof of Theorem 2.1.
Lemma 1.1. Let D be a bounded star-shaped circular domain. Then any holomorphic function on D has a Fourier series expansion
f (z) = X
∞ k=0mk
X
ν=1
c
kν(f)ψ
kν(z); c
kν(f ) = lim
r→1
Z
D
f (rz) ¯ ψ
kνdv
z, (2)
where the series converges absolutely and uniformly on compact subsets of D.
Proof. The proof of Lemma 1.1 follows the method of the proof of the Lemma in [3] and the proof of Theorem [1].
Lemma 1.2.
A
p(D) = A
′p(D) and ∥ f ∥
Ap= ∥ f ∥
A′p,
[5] where D is bounded star-shaped circular domain in C
N(N > 1).
2. The following Theorem will extend a special case of a result of [7].
Theorem 2.1. Let D be a bounded star-shaped circular domain in C
N(N > 1) and f ∈ A
p(D) (0 < p < ∞ ), then ∥ f
r− f ∥
Ap→ 0 as r → 1, where f
r(z) = f (rz).
Furthermore, the set of polynomials in z is dense in A
p(D); also A
p(D) is separable.
Proof. Since rz ∈ D ¯ for fixed r, 0 ≤ r < 1, f
r(z) = f(rz) is holomorphic on ¯ D and hence bounded on ¯ D. Thus f
r∈ A
pand hence to L
p. Also for z ∈ D, lim
r→1f
r(z) = f (z), by continuity of f .
Finally by (1) and Lemma 1.2, we have ∥ f
r∥
Ap→ ∥ f ∥
Apas r → 1.
Thus the hypothesis of [6] are satisfied, so that
∥ f
r− f ∥
Ap→ 0 as r → 1 .
Let f ∈ A
p(0 < p < ∞ ). Given ϵ > 0, there exists r
0(r
0< 1), such that
°° f − f
r0
°°
Ap
< ϵ
2 . (3)
Now let S
n,r0
(z) denote the n
thpartial sum of the Fourier series (2) of f
r0
(z). Since S
n,r0
→ f
r0
uniformly on ¯ D by Lemma 1.1,
°° S
n,r0
− f
r0
°°
Ap
< ϵ
2 , (4)
for n sufficiently large, thus by (3) and (4)
°° S
n,r0
− f °°
Ap
< °° S
n,r0
− f
r0
°°
Ap
+ °° f − f
r0
°°
Ap
< ϵ 2 + ϵ
2 = ϵ . Hence, the linear combination of { Ψ
kν} is dense in A
p, but as in [3],
m
P
kν=1
c
kνψ
kν(z) =
mk
P
ν=1
A
kνZ
kν, where Z
kνdenote the monomial z
ν11. . . z
νNn¡ k = ν
1+ · · · + ν
N; k = 0, 1, . . . ; ν = 1, . . . ; m
k= ¡
N+k−1k
¢¢ . Thus the polynomials are dense in A
p.
3. We have the following Corollaries:
Corollary 1. Let f, g ∈ A
2(D), then (f, g) = lim
r→1
µ 1 V
Z
D
f (rz )g(rz)dv
z¶ .
Proof.
¯¯ (f, g) − (f
r, g
r) ¯¯ ≤ 1 V
Z
D
¯¯ f(z)g(z) − f
r(z)g
r(z) ¯¯ dv
z= 1 V
Z
D
¯¯ f (z) ¡
g(z) − g
r(z) ¢ + ¡
f (z) − f
r(z) ¢
g
r(z) ¯¯ dv
z≤ 1 V
Z
D
¯¯ f (z) ¯¯¯¯ g(z) − g
r(z) ¯¯ dv
z+ 1 V
Z
D
¯¯ f(z) − f
r(z) ¯¯¯¯ g
r(z) ¯¯ dv
z.
By Schwarz inequality,
| (f, g) − (f
r, g
r) | < 1 V
µZ
D
| f (z) |
2dv
z¶
12
µZ
D
| g(z) ¯ − g(rz) |
2dv
z¶
12
+ 1 V
µZ
D
| f (z) − f (rz) |
2dv
z¶
12
µZ
D
| g(rz) |
2dv
z¶
12
,
or
| (f, g) − (f
r, g
r) | ≤ ∥ f ∥
A2∥ g − g
r∥
A2+ ∥ f − f
r∥
A2∥ g
r∥
A2. (5) Now ∥ g
r∥
A2→ ∥ g ∥
A2as r → 1 by Lemma 1.2, so the right side of (5) tends to zero by Theorem 2.1.
Therefore
(f, g) = lim
r→1
(f
r, g
r) = lim
r→1
µ 1 V
Z
D
f (rz)g(rz) dv
z¶ .
Corollary 2. For f ∈ A
p(1 ≤ p < ∞ ), c
kν(f ) =
Z
D
f (z) ¯ ψ
kνdv
z, (6)
where c
kνis given by (2).
Proof. By Holder’s inequality for 1 < p < ∞ ,
¯¯ ¯¯ Z
D
¡ f
r(z) ¯ ψ
kν(z) − f (z) ¯ ψ
kν(z) ¢ dv
z¯¯
¯¯
≤ µZ
D
¯¯ f (rz) − f (z) ¯¯
pdv
z¶
1p
µZ
D
¯¯ ψ
kν(z) ¯¯
qdv
z¶
1q
,
(7)
where
1p+
1q= 1. The right side of (7) equals ∥ f
r− f ∥
Ap∥ ψ
1ν∥
Aq. But ψ
kνis homogeneous polynomial on C
N(N > 1), so it is bounded on compact set ¯ D and hence is in A
q. By Theorem 2.1 ∥ f
r− f ∥
Ap→ 0 as r → 1. Thus formula (6) follows. If p=1
¯¯ ¯¯ Z
D
µ
f
r(z) ¯ ψ
kν− f (z) ¯ ψ
kν¶ dv
z¯¯
¯¯ ≤ µZ
D
| f(rz) − f (z) | dv
z¶ sup
z∈D
¯¯ ψ
kν(z) ¯¯ (8)
= °° f
r− f °°
A1
sup
z∈D
¯¯ ψ
kν(z) ¯¯ . But sup
z∈D
