A GENERALIZATION OF A RESULT OF CHOA ON ANALYTIC FUNCTIONS WITH HADAMARD GAPS
Stevo Stevi´ c
Abstract. In this paper we obtain a sufficient and necessary con- dition for an analytic function f on the unit ball B with Hadamard gaps, that is, for f (z) =P∞
k=1Pnk(z) (the homogeneous polyno- mial expansion of f ) satisfying nk+1/nk≥ λ > 1 for all k ∈ N, to belong to the weighted Bergman space
Apα(B) =© f |
Z
B
|f (z)|p(1 − |z|2)αdV (z) < ∞, f ∈ H(B)ª . We find a growth estimate for the integral mean
µZ
∂B
|f (rζ)|pdσ(ζ)
¶1/p ,
and an estimate for the point evaluations in this class of functions.
Similar results on the mixed norm space Hp,q,α(B) and weighted Bergman space on polydisc Ap~α(Un) are also given.
1. Introduction and preliminaries
Let z = (z
1, . . . , z
n) and w = (w
1, . . . , w
n) be points in the complex vector space C
n. By hz, ωi = P
nk=1
z
kw ¯
kwe denote the complex inner product of z and w, and |z| = p
hz, zi. Let U denote the unit disc in the complex plane, dm(z) = rdrdθ/π the normalized Lebesgue area measure on U, B the unit ball of C
n, B(a, r) the open unit ball centered at a of radius r, dV the normalized Lebesgue measure on B, dσ the normalized surface measure on the boundary S of B, and P (0, r
1, r
2) = {w | r
1<
|w| < r
1}. By H(B) we denote the class of all functions analytic in B.
Received February 7, 2005.
2000 Mathematics Subject Classification: Primary 32A35.
Key words and phrases: analytic functions, Hadamard gap, Bergman space, unit ball.
For an f ∈ H(B), the radial derivative Rf of f is defined by Rf (z) =
X
n k=1z
j∂f
∂z
j(z) = X
∞ k=0kP
k(z) if f (z) = P
∞k=0
P
k(z) is the homogeneous polynomial expansion of f.
As usual, we write
||f ||
p= µZ
S
|f (ζ)|
pdσ(ζ)
¶
1/pif p ∈ (0, ∞).
The expression A ³ B means that there are finite positive constants C and C
0such that
CA ≤ B ≤ C
0A.
In [6] J. Miao investigates analytic functions f with Hadamard gaps on the unit disk U, that is, those f such that f (z) = P
∞k=1
a
kz
nkwhere n
k+1/n
k≥ λ > 1 for all k ∈ N, which belong to the space B
pdefined as follows
B
p= {f ∈ H(U ) | ||f ||
Bp< ∞}, where
||f ||
Bp= sup
a∈U
µZ
U
|f
0(z)|
p(1 − |z|
2)
p−2(1 − |ϕ
a(z)|
2)dm(z)
¶
1/p, ϕ
a(z) = (a − z)/(1 − ¯az), or to its subspace B
0pconsisting of those f such that
|a|→1−0
lim Z
U
|f
0(z)|
p(1 − |z|
2)
p−2(1 − |ϕ
a(z)|
2)dm(z) = 0.
He proves the following result.
Theorem A. Let p ∈ (0, ∞). If f (z) = P
∞k=1
a
kz
nkis an analytic function on U such that n
k+1/n
k≥ λ > 1 for all k ∈ N, then the following statements are equivalent:
(i) f ∈ B
p; (ii) f ∈ B
p0; (iii) P
∞k=1
|a
k|
p< ∞.
Some other classical results of this type can be found, for example, in [4, 5, 13, 14].
An analytic function on B with the homogeneous expansion f (z) = P
∞k=1
P
nk(z) (here, P
nkis a homogeneous polynomial of degree n
k) is
said to have Hadamard gaps if n
k+1/n
k≥ λ > 1 for all k ∈ N. In [3],
among others, J. S. Choa generalizes Theorem A proving the following result.
Theorem B. Let p ∈ (0, ∞) and f (z) = P
∞k=1
P
nk(z) be an analytic function on B with Hadamard gaps. Then the following statements are equivalent:
(i) ||f ||
Xp= ¡R
B
|Rf (z)|
p(1 − |z|
2)
p−1dV (z) ¢
1/p< ∞;
(ii) P
∞k=1
||P
nk||
pp< ∞.
The weighted Bergman space A
pα= A
pα(B), α > −1, p > 0, is the space of all analytic functions f on B for which
||f ||
Apα= µZ
B
|f (z)|
p(1 − |z|
2)
αdV (z)
¶
1/p< ∞.
The weighted Bergman space on the unit disk, polydisc or on the unit ball has been investigated recently a great deal, see, for example, [1, 2, 7, 8, 9, 10, 11, 12] and the references in there.
Motivated by Theorems A and B, in this paper, we investigate an- alytic functions with Hadamard gaps, which belong to the weighted Bergman space A
pα(B), the mixed norm space (see, Section 3) and on the weighted Bergman space on U
n(Section 4). One of the main results is the following theorem.
Theorem 1. Let p ∈ (0, ∞), α > −1 and f (z) = P
∞k=1
P
nk(z) be an analytic function on B with Hadamard gaps. Then the following statements are equivalent:
(i) f ∈ A
pα; (ii) P
∞k=0
||Pnk||pp
nα+1k
< ∞.
Now we gather auxiliary results which are used in the proof of the main result. The following two lemmas can be found, for example, in [6].
Lemma 1. [14] Let p ∈ (0, ∞). If (n
k) is an increasing sequence of positive integers satisfying n
k+1/n
k≥ λ > 1 for all k, then there is a positive constant A depending only on p and λ such that
1 A
Ã
∞X
k=1
|a
k|
2!
1/2≤ Ã
1 2π
Z
2π0
¯ ¯
¯ ¯
¯ X
∞ k=1a
ke
inkθ¯ ¯
¯ ¯
¯
p
dθ
!
1/p≤ A Ã
∞X
k=1
|a
k|
2!
1/2for any number a
k, k ∈ N.
Lemma 2. Let α > 0, p > 0, n ≥ 0, a
n≥ 0, I
n= {k | 2
n≤ k <
2
n+1, k ∈ N}, t
n= P
k∈In
a
kand f (x) = P
∞n=1
a
nx
n. Then there is a positive constant K depending only on p and α such that
1 K
X
∞ n=0t
pn2
nα≤
Z
10
(1 − x)
α−1f
p(x)dx ≤ K X
∞ n=0t
pn2
nα. The following lemma is well known.
Lemma 3. Let a
n≥ 0 and n
0∈ N. Then for p ∈ (0, 1], 1
n
1−p0Ã
nX
0n=1
a
pn!
≤ Ã
nX
0n=1
a
n!
p≤ Ã
nX
0n=1
a
pn!
and for p ≥ 1, Ã
nX
0n=1
a
pn!
≤ Ã
nX
0n=1
a
n!
p≤ n
0p−1Ã
nX
0n=1
a
pn! .
2. Proofs of the main results
In this section first we prove Theorem 1.
Proof of Theorem 1. As in the proof of Proposition 2 in [3], first we use polar coordinates and Proposition 1.4.7 of [7]. We have
||f ||
pAp α= 2n Z
10
Z
S
|f (rζ)|
pdσ(ζ)(1 − r
2)
αr
2n−1dr
= 2n Z
10
Z
S
1 2π
Z
2π0
|f (re
iθζ)|
pdθdσ(ζ)(1 − r
2)
αr
2n−1dr
= 2n Z
10
Z
S
1 2π
Z
2π0
¯ ¯
¯ ¯
¯ X
∞ k=1P
nk(ζ)r
nke
inkθ¯ ¯
¯ ¯
¯
p
dθdσ(ζ)(1 − r
2)
αr
2n−1dr.
By the second inequality in Lemma 1 and the change ρ = r
2, we get
||f ||
pApα
≤ 2nA
pZ
10
Z
S
Ã
∞X
k=1
|P
nk(ζ)|
2r
2nk!
p/2dσ(ζ)(1 − r
2)
αr
2n−1dr
= nA
pZ
10
Z
S
Ã
∞X
k=1
|P
nk(ζ)|
2ρ
nk!
p/2dσ(ζ)(1 − ρ)
αρ
n−1dρ
≤ nA
pZ
10
Z
S
à X
∞k=1
|P
nk(ζ)|
2ρ
nk!
p/2dσ(ζ)(1 − ρ)
αdρ.
By the first inequality in Lemma 1 and the change ρ = r
2n, it follows that
||f ||
pAp α≥ 2n
A
pZ
10
Z
S
Ã
∞X
k=1
|P
nk(ζ)|
2r
2nk!
p/2dσ(ζ)(1 − r
2)
αr
2n−1dr
= 1 A
pZ
10
Z
S
Ã
∞X
k=1
|P
nk(ζ)|
2ρ
nk/n!
p/2dσ(ζ)(1 − ρ
1/n)
αdρ
≥ C(α) A
pZ
10
Z
S
Ã
∞X
k=1
|P
nk(ζ)|
2ρ
nk!
p/2dσ(ζ)(1 − ρ)
αdρ,
since ρ
1/n≥ ρ for ρ ∈ [0, 1], which implies X
∞k=1
|P
nk(ζ)|
2ρ
nk/n≥ X
∞ k=1|P
nk(ζ)|
2ρ
nkand since
(1 − ρ
1/n)
α= (1 − ρ)
α(1 + ρ
1/n+ ρ
2/n+ · · · + ρ
(n−1)/n)
α≥ C(α)(1 − ρ)
α, where C(α) = 1/n
αwhen α ≥ 0 and C(α) = 1 when α ∈ (−1, 0).
Thus
||f ||
pAp α³
Z
S
Z
10
Ã
∞X
k=1
|P
nk(ζ)|
2ρ
nk!
p/2(1 − ρ)
αdρdσ(ζ).
By Lemma 2 applied to the integral Z
10
à X
∞k=1
|P
nk(ζ)|
2ρ
nk!
p/2(1 − ρ)
αdρ, we obtain
||f ||
pAp α³
Z
S
X
∞ k=01 2
(α+1)k
X
nm∈Ik
|P
nm(ζ)|
2
p/2
dσ(ζ).
Because n
k+1/n
k≥ λ > 1 for all k ∈ N, the number of P
nmwhen
n
m∈ I
kis at most [log
λ2] + 1. Using this fact and Lemma 3 it follows
that
||f ||
pAp α³
Z
S
X
∞ k=01 2
(α+1)kX
nm∈Ik
|P
nm(ζ)|
p
dσ(ζ)
³ X
∞ k=01 2
(α+1)kX
nm∈Ik
||P
nm||
pp.
Form this and since n
m³ 2
kwhen n
m∈ I
k, we get
||f ||
pAp α³
X
∞ k=0||P
nk||
ppn
α+1kas desired.
Corollary 1. Let p ∈ (0, ∞), α > −1 and f (z) = P
∞k=1
P
nk(z) be an analytic function on B with Hadamard gaps. Then the following statements are equivalent:
(i) R
(l)f ∈ A
pα; (ii) P
∞k=0
||Pnk||pp
nα+1−lpk
< ∞.
Proof. Since f has Hadamard gaps and R
(l)f (z) = P
∞k=1
n
lkP
nk(z) it follows that R
(l)f has Hadamard gaps too. Applying Theorem 1 to the function R
(l)f we obtain that R
(l)f ∈ A
pαif and only if
X
∞ k=0||n
lkP
nk||
ppn
α+1k=
X
∞ k=0||P
nk||
ppn
α+1−lpk< ∞, finishing the proof.
Remark 1. Setting l = 1 and α = p − 1 in Corollary 1, we obtain Theorem B.
Since for n = 1, ||P
nk||
pp= |a
k|
p, k ∈ N, from Theorem 1 it follows the following corollary:
Corollary 2. Let p ∈ (0, ∞), α > −1 and let f (z) = P
∞k=1
a
kz
nkbe an analytic function on U with Hadamard gaps. Then the following statements are equivalent:
(i) f ∈ A
pα(U );
(ii) P
∞k=0 |ak|p nα+1k
< ∞.
Remark 2. Theorem 1 gives an estimation of the growth of the sequence ||P
nk||
p. Since the series
X
∞ k=0||P
nk||
ppn
α+1kis convergent, we have
||P
nk||
p= o ¡
n
(α+1)/pk¢ .
Theorem 2. Let p ∈ (0, ∞), α > −1 and f (z) = P
∞k=1
P
nk(z) be an analytic function on B with Hadamard gaps belonging to A
pα. Then
||f (z)||
p= o
µ 1
(1 − |z|)
(α+1)/p¶
as |z| → 1 − 0.
Proof. Let z = rζ = |z|ζ and I
n= {k | 2
n≤ k < 2
n+1, k ∈ N}. Let first p ≥ 1. From Remark 2 we have that for every ε > 0 there is an k
0∈ N such that
(1) ||P
nk||
p≤ εn
(α+1)/pkfor k ≥ k
0.
Without loss of generality we may assume that k
0= 1, since for every l ∈ N,
|z|→1−0
lim (1 − |z|)
(α+1)/pX
l k=1||P
nk(z)||
p= 0.
We have
||f (z)||
p≤ X
∞ k=1||P
nk(z)||
p= X
∞ k=1||P
nk(ζ)||
p|z|
nk≤ ε X
∞ k=1n
(α+1)/pk|z|
nk≤ ε X
∞ k=0X
nm∈Ik
n
(α+1)/pm|z|
nm(2)
≤ ε X
∞ k=02
(k+1)(α+1)/pX
nm∈Ik
|z|
nm= ε X
∞ k=02
(k+1)[(α+1)/p−1]2
k+1X
nm∈Ik
|z|
nm≤ 4ε
C
1|z| + X
∞ k=12
(k+1)[(α+1)/p−1]X
nm∈Ik
2
k−1|z|
2k
≤ 4εC
1|z| + C
2X
∞ k=12
(k−1)[(α+1)/p−1]X
m∈Ik−1
|z|
m≤ 4εC
1|z| + C
2X
∞ k=1k
[(α+1)/p−1]|z|
k,
where C
1= 2
(α+1)/p−2χ
I0(n
1), χ
I0(·) the characteristic function of the set I
0, and C
2= 2
2(α+1)/pε([log
λ2] + 1). From (2) it follows that there is a positive constant C such that
||f (z)||
p≤ εC X
∞ k=0(k + 1)
[(α+1)/p−1]|z|
k, z ∈ B.
It is well known [14, p.77] that (3)
X
∞ k=0(k + 1)
[(α+1)/p−1]|z|
k³ (1 − |z|)
−(α+1)/p, from which the result follows in this case.
Assume now that p ∈ (0, 1). As in the first case we may assume that (1) holds for every k ∈ N. Then it holds
||f (z)||
pp≤ X
∞ k=1||P
nk(z)||
pp≤ X
∞ k=1||P
nk(ζ)||
pp|z|
pnk≤ ε X
∞ k=1n
(α+1)k|z|
pnk. As in the first case we obtain that there is a constant M
1> 0 such that
||f (z)||
pp≤ εM
1X
∞ k=0(k + 1)
α|z|
kp.
Using (3) and the inequality (1−x)
p≤ 1−px, for x ∈ [0, 1] and p ∈ [0, 1], we obtain
||f (z)||
pp≤ εM
2(1 − |z|
p)
α+1≤ εM
2( p(1 − |z|) )
α+1, as desired.
Theorem 3. Let p ∈ (0, ∞), α > −1 and f (z) = P
∞k=1
P
nk(z) be an analytic function on B with Hadamard gaps belonging to A
pα. Then (4) |f (z)| = o
µ 1
(1 − |z|)
(n+α)/p¶
as |z| → 1 − 0.
Proof. By subharmonicity of |f |
p, p > 0, we obtain
(5) |f (z)|
p≤ 2
n(1 − |z|)
nZ
B(z,(1−|z|)/2)
|f (w)|
pdV (w).
For w ∈ B(z, (1 − |z|)/2), we have
(6) 1
2 (1 − |z|) < (1 − |w|) < 3
2 (1 − |z|).
By Theorem 2 it follows that for every ε > 0 there is a δ > 1/3 such that
(7) ||f (z)||
pp≤ ε
(1 − |z|)
α+1, |z| > δ.
From (5)–(7) and by polar coordinates, we have
|f (z)|
p≤ C (1 − |z|)
n+αZ
B(z,(1−|z|)/2)
|f (w)|
p(1 − |w|)
αdV (w)
≤ C
(1 − |z|)
n+αZ
P (0,3|z|−12 ,1+|z|2 )
|f (w)|
p(1 − |w|)
αdV (w)
= 2nC
(1 − |z|)
n+αZ
1+|z|2 3|z|−1
2
||f (rζ)||
pp(1 − r)
αr
2n−1dr
≤ 2nCε
(1 − |z|)
n+αZ
1+|z|2 3|z|−1
2
dr 1 − r
= 2nCε ln 3 (1 − |z|)
n+α, from which the result follows.
Remark 3. The basic estimate for the point evaluations on the weighted Bergman space is the following (see, Corollary 3.5 in [1])
(8) |f (z)| ≤ ||f ||
Apα(1 − |z|)
(n+α)/p,
the equality holds at some z ∈ B if and only if
f (w) = λ(1−hz, wi)−2(n+α)/pfor some λ ∈ C and every w ∈ B.
Thus, our Theorem 3 can be considered as an improvement of es-
timate (8) in the subclass of A
pαconsisting of the functions having
Hadamard gaps.
3. The case of mixed norm space
In this section we consider analytic functions with Hadamard gaps on the mixed norm space. The mixed norm space H
p,q,α(B), p, q > 0 and α ∈ (−1, ∞), consists of all f ∈ H(B) such that
||f ||
p,q,α= µZ
10
||f (rζ)||
qp(1 − r)
αdr
¶
1/p< ∞.
Note that when p = q, H
p,q,α(B) is just weighted Bergman space. For f ∈ H
p,q,α(B), the following result holds:
Theorem 4. Let f ∈ H
p,q,αand f (z) = P
∞k=1
P
nk(z) be an analytic function on B with Hadamard gaps. Then the following statements holds:
(i) If p ∈ (0, 2], then P
∞k=0
||Pnk||qp
nα+1k
< ∞ implies f ∈ H
p,q,α(B);
(ii) If p ≥ 2, then f ∈ H
p,q,α(B) implies P
∞k=0
||Pnk||qp
nα+1k
< ∞.
Proof. (i) Similarly to the proof of Theorem 1 we have
||f ||
pHp,q,α= Z
10
ÃZ
S
1 2π
Z
2π0
¯ ¯
¯ ¯
¯ X
∞ k=1P
nk(ζ)r
nke
inkθ¯ ¯
¯ ¯
¯
p
dθdσ(ζ)
!
q/p(1 − r)
αdr
≤ A
pZ
10
Z
S
Ã
∞X
k=1
|P
nk(ζ)|
2r
2nk!
p/2dσ(ζ)
q/p
(1 − r)
αdr
≤ A
pZ
10
ÃZ
S
X
∞ k=1|P
nk(ζ)|
pr
pnkdσ(ζ)
!
q/p(1 − r)
αdr
= A
pZ
10
Ã
∞X
k=1
||P
nk||
ppr
pnk!
q/p(1 − r)
αdr
³ Z
10
Ã
∞X
k=1
||P
nk||
ppρ
nk!
q/p(1 − ρ)
αdρ
³ X
∞ k=01 2
(α+1)k
X
nm∈Ik
||P
nm||
pp
q/p
³ X
∞ k=0||P
nk||
qpn
α+1k,
where in the second inequality we used the condition p/2 ≤ 1 and in the first asymptotic relation “³” the change ρ = r
p.
(ii) Since p ≥ 2, in the above sequence of relations the reverse in- equalities hold, from which the result follows.
Theorem 4, by a similar argument to the proof of Theorem 2, gives the next corollary.
Corollary 3. Let p ≥ 2, α > −1 and f (z) = P
∞k=1
P
nk(z) be an analytic function on B with Hadamard gaps belonging to H
p,q,α(B).
Then
||f (z)||
p= o
µ 1
(1 − |z|)
(α+1)/q¶
as |z| → 1 − 0.
4. A version of Theorem 1 on the polydisc
Let L
p~α(U
n) denote the class of all measurable functions defined on the unit polydisc U
n= {z ∈ C
n| |z
i| < 1, i = 1, . . . , n} such that
||f ||
pLp~ α(Un)
=
Z
Un
|f (z)|
pY
n j=1(1 − |z
j|
2)
αjdm(z
j) < ∞,
where ~α = (α
1, . . . , α
n), α
j> −1, j = 1, . . . , n. The weighted Bergman space on U
nis defined as A
p~α(U
n) = L
p~α(U
n) ∩ H(U
n).
Theorem 5. Let p ∈ (0, ∞), α
j> −1, j = 1, . . . , n, and f (z) = X
k
a
kz
mk= X
k1,...,kn≥1
a
k1,...,knz
1mk1· · · z
nmknbe an analytic function on U
n, with Hadamard gaps in each variable, that is, there are λ
j> 1, j = 1, . . . , n, such that m
kj+1/m
kj≥ λ
jfor every k
j∈ N. Then the following statements are equivalent:
(i) f ∈ A
pα~(U
n);
(ii) P
∞k1=1
· · · P
∞kn=1
|ak1,...,kn|p Qn
j=1mαj +1kj
< ∞.
Proof. In order to avoid too much calculations we may assume that n = 2. It means that the function f can be written as follows
(9) f (z
1, z
2) = X
k1,k2≥1
a
k1,k2z
m1k1z
m2 k2= X
∞ k1=1P
k1(z
2)z
1mk1,
where
(10) P
k1(z
2) =
X
∞ k2=1a
k1,k2z
2mk2.
Let z
j= r
je
iθj, j = 1, 2. Then from (9) and by Fubini’s theorem we have
||f ||
pAp~ α
= 1
π
2× Z
10
Z
2π0
µ Z
10
Z
2π0
¯ ¯
¯ ¯ X
∞ k1=1P
k1(z
2)z
1mk1¯ ¯
¯ ¯
p
(11)
dθ
1(1 − r
21)
α1r
1dr
1¶
dθ
2(1 − r
22)
α2r
2dr
2. Fix z
2for a moment. By Corollary 2 applied to the function
g
z2(z
1) = X
∞ k1=1P
k1(z
2)z
1mk1and to the integral I(z
2) =
Z
10
Z
2π0
¯ ¯
¯ ¯ X
∞ k1=1P
k1(z
2)z
1mk1¯ ¯
¯ ¯
p
dθ
1(1 − r
12)
α1r
1dr
1we obtain
(12) I(z
2) ³
X
∞ k1=1|P
k1(z
2)|
pm
αk11+1.
Substituting (12) in (10) and applying Corollary 2 to the functions in (11), it follows that
||f ||
pAp~ α
³ X
∞ k1=11 m
αk11+1Z
10
Z
2π0
|P
k1(z
2)|
pdθ
2(1 − r
22)
α2r
2dr
2³ X
∞ k1=11 m
αk1+11
Z
10
Z
2π0
¯ ¯
¯ ¯ X
∞ k2=1a
k1,k2r
m2 k2e
iθ2mk2¯ ¯
¯ ¯
p
dθ
2(1 − r
22)
α2r
2dr
2³ X
∞ k1=11 m
αk1+11
X
∞ k2=1|a
k1,k2|
pm
αk2+12
= X
∞ k1=1X
∞ k2=1|a
k1,k2|
pm
αk1+11
m
αk2+12
,
as desired.
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