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(1)

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

n

k=1

z

k

w ¯

k

we 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.

(2)

For an f ∈ H(B), the radial derivative Rf of f is defined by Rf (z) =

X

n k=1

z

j

∂f

∂z

j

(z) = X

k=0

kP

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)|

p

dσ(ζ)

1/p

if p ∈ (0, ∞).

The expression A ³ B means that there are finite positive constants C and C

0

such that

CA ≤ B ≤ C

0

A.

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

k

z

nk

where n

k+1

/n

k

≥ λ > 1 for all k ∈ N, which belong to the space B

p

defined 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

0p

consisting 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

k

z

nk

is 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

nk

is 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],

(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−1

dV (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

)

α

dV (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

Z

0

¯ ¯

¯ ¯

¯ X

k=1

a

k

e

inkθ

¯ ¯

¯ ¯

¯

p

!

1/p

≤ A Ã

X

k=1

|a

k

|

2

!

1/2

for any number a

k

, k ∈ N.

(4)

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

k

and f (x) = P

n=1

a

n

x

n

. Then there is a positive constant K depending only on p and α such that

1 K

X

n=0

t

pn

2

Z

1

0

(1 − x)

α−1

f

p

(x)dx ≤ K X

n=0

t

pn

2

. The following lemma is well known.

Lemma 3. Let a

n

≥ 0 and n

0

∈ N. Then for p ∈ (0, 1], 1

n

1−p0

Ã

n

X

0

n=1

a

pn

!

Ã

n

X

0

n=1

a

n

!

p

Ã

n

X

0

n=1

a

pn

!

and for p ≥ 1, Ã

n

X

0

n=1

a

pn

!

Ã

n

X

0

n=1

a

n

!

p

≤ n

0p−1

Ã

n

X

0

n=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

1

0

Z

S

|f (rζ)|

p

dσ(ζ)(1 − r

2

)

α

r

2n−1

dr

= 2n Z

1

0

Z

S

1

Z

0

|f (re

ζ)|

p

dθdσ(ζ)(1 − r

2

)

α

r

2n−1

dr

= 2n Z

1

0

Z

S

1

Z

0

¯ ¯

¯ ¯

¯ X

k=1

P

nk

(ζ)r

nk

e

inkθ

¯ ¯

¯ ¯

¯

p

dθdσ(ζ)(1 − r

2

)

α

r

2n−1

dr.

By the second inequality in Lemma 1 and the change ρ = r

2

, we get

||f ||

pAp

α

≤ 2nA

p

Z

1

0

Z

S

Ã

X

k=1

|P

nk

(ζ)|

2

r

2nk

!

p/2

dσ(ζ)(1 − r

2

)

α

r

2n−1

dr

= nA

p

Z

1

0

Z

S

Ã

X

k=1

|P

nk

(ζ)|

2

ρ

nk

!

p/2

dσ(ζ)(1 − ρ)

α

ρ

n−1

(5)

≤ nA

p

Z

1

0

Z

S

à X

k=1

|P

nk

(ζ)|

2

ρ

nk

!

p/2

dσ(ζ)(1 − ρ)

α

dρ.

By the first inequality in Lemma 1 and the change ρ = r

2n

, it follows that

||f ||

pAp α

2n

A

p

Z

1

0

Z

S

Ã

X

k=1

|P

nk

(ζ)|

2

r

2nk

!

p/2

dσ(ζ)(1 − r

2

)

α

r

2n−1

dr

= 1 A

p

Z

1

0

Z

S

Ã

X

k=1

|P

nk

(ζ)|

2

ρ

nk/n

!

p/2

dσ(ζ)(1 − ρ

1/n

)

α

C(α) A

p

Z

1

0

Z

S

Ã

X

k=1

|P

nk

(ζ)|

2

ρ

nk

!

p/2

dσ(ζ)(1 − ρ)

α

dρ,

since ρ

1/n

≥ ρ for ρ ∈ [0, 1], which implies X

k=1

|P

nk

(ζ)|

2

ρ

nk/n

X

k=1

|P

nk

(ζ)|

2

ρ

nk

and 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

α

when α ≥ 0 and C(α) = 1 when α ∈ (−1, 0).

Thus

||f ||

pAp α

³

Z

S

Z

1

0

Ã

X

k=1

|P

nk

(ζ)|

2

ρ

nk

!

p/2

(1 − ρ)

α

dρdσ(ζ).

By Lemma 2 applied to the integral Z

1

0

à X

k=1

|P

nk

(ζ)|

2

ρ

nk

!

p/2

(1 − ρ)

α

dρ, we obtain

||f ||

pAp α

³

Z

S

  X

k=0

1 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

nm

when

n

m

∈ I

k

is at most [log

λ

2] + 1. Using this fact and Lemma 3 it follows

(6)

that

||f ||

pAp α

³

Z

S

 X

k=0

1 2

(α+1)k

X

nm∈Ik

|P

nm

(ζ)|

p

 dσ(ζ)

³ X

k=0

1 2

(α+1)k

X

nm∈Ik

||P

nm

||

pp

.

Form this and since n

m

³ 2

k

when n

m

∈ I

k

, we get

||f ||

pAp α

³

X

k=0

||P

nk

||

pp

n

α+1k

as 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

lk

P

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

pα

if and only if

X

k=0

||n

lk

P

nk

||

pp

n

α+1k

=

X

k=0

||P

nk

||

pp

n

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

k

z

nk

be 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

< ∞.

(7)

Remark 2. Theorem 1 gives an estimation of the growth of the sequence ||P

nk

||

p

. Since the series

X

k=0

||P

nk

||

pp

n

α+1k

is 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)/pk

for 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)/p

X

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=1

n

(α+1)/pk

|z|

nk

≤ ε X

k=0

X

nm∈Ik

n

(α+1)/pm

|z|

nm

(2)

≤ ε X

k=0

2

(k+1)(α+1)/p

X

nm∈Ik

|z|

nm

= ε X

k=0

2

(k+1)[(α+1)/p−1]

2

k+1

X

nm∈Ik

|z|

nm

(8)

≤ 4ε

C

1

|z| + X

k=1

2

(k+1)[(α+1)/p−1]

X

nm∈Ik

2

k−1

|z|

2k

≤ 4εC

1

|z| + C

2

X

k=1

2

(k−1)[(α+1)/p−1]

X

m∈Ik−1

|z|

m

≤ 4εC

1

|z| + C

2

X

k=1

k

[(α+1)/p−1]

|z|

k

,

where C

1

= 2

(α+1)/p−2

χ

I0

(n

1

), χ

I0

(·) the characteristic function of the set I

0

, and C

2

= 2

2(α+1)/p

ε([log

λ

2] + 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|

k

³ (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=1

n

(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

1

X

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.

(9)

Proof. By subharmonicity of |f |

p

, p > 0, we obtain

(5) |f (z)|

p

2

n

(1 − |z|)

n

Z

B(z,(1−|z|)/2)

|f (w)|

p

dV (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+α

Z

B(z,(1−|z|)/2)

|f (w)|

p

(1 − |w|)

α

dV (w)

C

(1 − |z|)

n+α

Z

P (0,3|z|−12 ,1+|z|2 )

|f (w)|

p

(1 − |w|)

α

dV (w)

= 2nC

(1 − |z|)

n+α

Z

1+|z|

2 3|z|−1

2

||f (rζ)||

pp

(1 − r)

α

r

2n−1

dr

2nCε

(1 − |z|)

n+α

Z

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)/p

for 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

pα

consisting of the functions having

Hadamard gaps.

(10)

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

1

0

||f (rζ)||

qp

(1 − r)

α

dr

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,α

and 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

1

0

ÃZ

S

1

Z

0

¯ ¯

¯ ¯

¯ X

k=1

P

nk

(ζ)r

nk

e

inkθ

¯ ¯

¯ ¯

¯

p

dθdσ(ζ)

!

q/p

(1 − r)

α

dr

≤ A

p

Z

1

0

 Z

S

Ã

X

k=1

|P

nk

(ζ)|

2

r

2nk

!

p/2

dσ(ζ)

q/p

(1 − r)

α

dr

≤ A

p

Z

1

0

ÃZ

S

X

k=1

|P

nk

(ζ)|

p

r

pnk

dσ(ζ)

!

q/p

(1 − r)

α

dr

= A

p

Z

1

0

Ã

X

k=1

||P

nk

||

pp

r

pnk

!

q/p

(1 − r)

α

dr

³ Z

1

0

Ã

X

k=1

||P

nk

||

pp

ρ

nk

!

q/p

(1 − ρ)

α

³ X

k=0

1 2

(α+1)k

 X

nm∈Ik

||P

nm

||

pp

q/p

³ X

k=0

||P

nk

||

qp

n

α+1k

,

(11)

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)|

p

Y

n j=1

(1 − |z

j

|

2

)

αj

dm(z

j

) < ∞,

where ~α = (α

1

, . . . , α

n

), α

j

> −1, j = 1, . . . , n. The weighted Bergman space on U

n

is 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

k

z

mk

= X

k1,...,kn≥1

a

k1,...,kn

z

1mk1

· · · z

nmkn

be 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

j

for 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,k2

z

m1k1

z

m2 k2

= X

k1=1

P

k1

(z

2

)z

1mk1

,

(12)

where

(10) P

k1

(z

2

) =

X

k2=1

a

k1,k2

z

2mk2

.

Let z

j

= r

j

e

j

, j = 1, 2. Then from (9) and by Fubini’s theorem we have

||f ||

pAp

~ α

= 1

π

2

× Z

1

0

Z

0

µ Z

1

0

Z

0

¯ ¯

¯ ¯ X

k1=1

P

k1

(z

2

)z

1mk1

¯ ¯

¯ ¯

p

(11)

1

(1 − r

21

)

α1

r

1

dr

1

2

(1 − r

22

)

α2

r

2

dr

2

. Fix z

2

for a moment. By Corollary 2 applied to the function

g

z2

(z

1

) = X

k1=1

P

k1

(z

2

)z

1mk1

and to the integral I(z

2

) =

Z

1

0

Z

0

¯ ¯

¯ ¯ X

k1=1

P

k1

(z

2

)z

1mk1

¯ ¯

¯ ¯

p

1

(1 − r

12

)

α1

r

1

dr

1

we obtain

(12) I(z

2

) ³

X

k1=1

|P

k1

(z

2

)|

p

m

αk11+1

.

Substituting (12) in (10) and applying Corollary 2 to the functions in (11), it follows that

||f ||

pAp

~ α

³ X

k1=1

1 m

αk11+1

Z

1

0

Z

0

|P

k1

(z

2

)|

p

2

(1 − r

22

)

α2

r

2

dr

2

³ X

k1=1

1 m

αk1+1

1

Z

1

0

Z

0

¯ ¯

¯ ¯ X

k2=1

a

k1,k2

r

m2 k2

e

2mk2

¯ ¯

¯ ¯

p

2

(1 − r

22

)

α2

r

2

dr

2

³ X

k1=1

1 m

αk1+1

1

X

k2=1

|a

k1,k2

|

p

m

αk2+1

2

= X

k1=1

X

k2=1

|a

k1,k2

|

p

m

αk1+1

1

m

αk2+1

2

,

as desired.

(13)

References

[1] F. Beatrous and J. Burbea, Holomorphic Sobolev spaces on the ball, Dissertationes Math. 276 (1989), 1–57.

[2] G. Benke and D. C. Chang, A note on weighted Bergman spaces and the Ces´aro operator, Nagoya Math. J. 159 (2000), 25–43.

[3] J. S. Choa, Some properties of analytic functions on the unit ball with Hadamard gaps, Complex Variables Theory Appl. 29 (1996), no. 3, 277–285.

[4] P. Duren, Theory of Hpspaces, Pure and Applied Mathematics Vol. 38, Academic Press, New York, 1970.

[5] J. H. Mathews, Coefficients of uniformly normal-Bloch functions, Yokohama Math. J. 21 (1973), 29–31.

[6] J. Miao, A property of analytic functions with Hadamard gaps, Bull. Austral.

Math. Soc. 45 (1992), no. 1, 105–112.

[7] W. Rudin, Function theory in the unit ball of Cn, Grundlehren der Mathematis- chen Wissenschaften, 241, Springer-Verlag, New York-Berlin, 1980.

[8] J. -H. Shi, Inequalities for the integral means of holomorphic functions and their derivatives in the unit ball of Cn, Trans. Amer. Math. Soc. 328 (1991), no. 2, 619–637.

[9] A. Siskakis, Weighted integrals of analytic functions, Acta Sci. Math. 66 (2000), 651–664.

[10] S. Stevi´c, On an area inequality and weighted integrals of analytic functions, Result Math. 41 (2002), no. 3-4, 386–393.

[11] , Weighted integrals and conjugate functions in the unit disk, Acta Sci.

Math. 69 (2003), no. 1-2, 109–119.

[12] , Weighted integrals of holomorphic functions on the polydisc, Z. Anal.

Anwendungen 23 (2004), no. 3, 577–587.

[13] S. Yamashita, Gap series and α-Bloch functions, Yokohama Math. J. 28 (1980), no. 1-2, 31–36.

[14] A. Zygmund, Trigonometric series, Cambridge University Press, London, 1959.

Mathematical Institute of Serbian Academy of Science Knez Mihailova 35/I, 11000 Beograd, Serbia

E-mail: [email protected]; [email protected]

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