Weighted-Hardy functions with Hadamard gaps on the unit ball

Applied Mathematics and Computation 212 (2009) 229–233

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Weighted-Hardy functions with Hadamard gaps on the unit ball Songxiao Li a, Stevo Stevic´ b,* a b

Department of Mathematics, JiaYing University, 514015 Meizhou, GuangDong, China Mathematical Institute of the Serbian Academy of Science, Knez Mihailova 36/III, 11000 Beograd, Serbia

a r t i c l e

i n f o

a b s t r a c t We prove that an analytic function f on the unit ball B with Hadamard gaps, that is, P (the homogeneous polynomial expansion of f) satisfying f ðzÞ ¼ 1 n k¼1 P nk ðzÞ nkþ1 =nk P k > 1 for all k 2 N, belongs to the space Bap ðBÞ ¼ f j sup0 0 if and only if lim supk!1 kP nk kp n1 k a . Also we prove that the following asymptotic relation holds kf kBap  supk2N kP nk kp n1 k a ¼ 0. These results confirm two limr!1 ð1  r2 Þa kRfr kp ¼ 0 if and only if limk!1 kPnk kp n1 k conjectures from the following recent paper [S. Stevic´, On Bloch-type functions with Hadamard gaps, Abstr. Appl. Anal. 2007 (2007) 8 pages (Article ID 39176)]. Ó 2009 Elsevier Inc. All rights reserved.

Keywords: Holomorphic function a-Bloch space Weighted-Hardy space Hadamard gaps Unit ball

1. Introduction Let B ¼ Bn be the open unit ball of Cn ; S ¼ oB be its boundary, D ¼ B1 the unit disk in C; dr the normalized rotation invariant Lebesgue measure on S, i.e., rðoBÞ ¼ 1 and HðBÞ the class of all holomorphic functions on B. P For f 2 HðBÞ with the Taylor expansion f ðzÞ ¼ jbjP0 ab zb , let

Rf ðzÞ ¼

X

jbjab zb

jbjP0

be the radial derivative of f, where b ¼ ðb1 ; b2 ; . . . ; bn Þ is a multi-index and zb ¼ zb11    zbnn . It is easy to see that

Rf ðzÞ ¼

1 X

kP k ðzÞ

k¼0

P if f ðzÞ ¼ 1 k¼0 P k ðzÞ (the homogeneous polynomial expansion of f). As usual, we write

kfr kp ¼

Z

jf ðrfÞjp drðfÞ

1=p

S

if p 2 ð0; 1Þ, and where fr ðfÞ ¼ f ðrfÞ; 0 6 r < 1. If p ¼ 1, then kf k1 ¼ supz2B jf ðzÞj. The a-Bloch space Ba ðBÞ ¼ Ba ; a > 0, is the space of all holomorphic functions f on B such that

ba ðf Þ ¼ supð1  jzj2 Þa jRf ðzÞj < 1: z2B

Ba is a normed space with the norm

kf kBa ¼ jf ð0Þj þ ba ðf Þ: * Corresponding author. E-mail addresses: [email protected], [email protected] (S. Li), [email protected] (S. Stevic´). 0096-3003/$ - see front matter Ó 2009 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2009.02.019

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230

Let Ba0 ðBÞ ¼ Ba0 denote the subspace of Ba consisting of those f such that

lim ð1  jzj2 Þa jRf ðzÞj ¼ 0:

jzj!1

This space is called the little a-Bloch space. For a ¼ 1 the a-Bloch space and the little a-Bloch space become the Bloch space B and the little Bloch space B0 . Some information on these spaces can be found, for example, in the following papers [4,7– 9,13,15,17,21,22] (on some related spaces of holomorphic functions on the unit ball, see also [5,18]). The weighted Hardy space Bap ðBÞ ¼ Bap consists of all f 2 HðBÞ such that

kf kBap :¼ sup ð1  r 2 Þa kRfr kp < 1: 0
The little weighted Hardy space is defined as follows:

Bap;0 ðBÞ ¼

  f j limð1  r2 Þa kRfr kp ¼ 0; f 2 HðBÞ : r!1

We say that an analytic function f on the unit disk D has Hadamard gaps if f ðzÞ ¼ k 2 N. In [19] Yamashita proved the following result:

P1

k¼1 ak z

nk

where nkþ1 =nk P k > 1 for all

Theorem A. Assume that f is an analytic function on D with Hadamard gaps. Then for a > 0, the following two statements hold. a < 1. (a) f 2 Ba ðDÞ if and only if lim supk!1 jak jn1 k a 1a (b) f 2 B0 ðDÞ if and only if limk!1 jak jnk ¼ 0.

P An analytic function on B with the homogeneous expansion f ðzÞ ¼ 1 k¼1 P nk ðzÞ (here, P nk is a homogeneous polynomial of degree nk ) is said to have Hadamard gaps if nkþ1 =nk P k > 1 for all k 2 N. Motivated by Theorem A, in [16] the second author of this paper proved the following result. P Theorem B. Assume a > 0; p ¼ 1; 2; 1, and f ðzÞ ¼ 1 k¼1 P nk ðzÞ is an analytic function on B with Hadamard gaps. Then the following statements hold. a < 1. (a) f 2 Bap ðBÞ if and only if lim supk!1 kPnk kp n1 k a ¼ 0. (b) f 2 Bap;0 ðBÞ if and only if limk!1 kP nk kp n1 k

For related results see, e.g., [1–3,6,10,11,14,20,22] and references therein. In [16] S. Stevic´ conjectured that Theorem B also holds for the case p P 1. Our aim here is to confirm the conjecture. Moreover we give an asymptotic relationship for the norm of the functions belonging to Bap . The main result in this paper is the following. P Theorem 1. Assume that a > 0; p > 0, and f ðzÞ ¼ 1 k¼1 P nk ðzÞ is an analytic function on B with Hadamard gaps. Then the following statements hold true: a < 1. Moreover, the following asymptotic relation holds: (a) f 2 Bap if and only if lim supk!1 kP nk kp n1 k

a kf kBap  sup kPnk kp n1 : k

ð1Þ

k2N

a ¼ 0. (b) f 2 Bap;0 if and only if limk!1 kPnk kp n1 k

Throughout this paper, constants are denoted by C, they are positive and may differ from one occurrence to the other. The notation A  B means that there is a positive constant C such that B=C 6 A 6 CB. 2. Proof of the main result Before proving Theorem 1 we quote an auxiliary result (see [23]). Lemma 1. Assume that p 2 ð0; 1Þ. If ðnk Þ is an increasing sequence of positive integers satisfying nkþ1 =nk P k > 1 for all k, then there is a positive constant A depending only on p and k such that 1 1 X jak j2 A k¼1

!1=2 6

1 2p

for any sequence ðak Þk 2 N.

p !1=p !1=2 Z 2p X  1 1 X  2 ink h  a e dh 6 A ja j   k   k¼1 k 0 k¼1

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Proof of Theorem 1. (a) The proof combines ideas from [14] and [16]. Let f 2 Bap . By Proposition 1.4.7 in [12], the definition of the radial derivative and Lemma 1, we have

Z

jRf ðrfÞjp drðfÞ ¼

S

Z Z 2p S



jRf ðrfeih Þjp

0

Z

1 X

S

dh drðfÞ ¼ 2p !p=2

n2k jPnk ðfÞj2 r2nk

p Z Z 2p X  dh 1   nk Pnk ðrfeih Þ drðfÞ   2p  k¼1 S 0

drðfÞ P npk r pnk

k¼1

Z S

jP nk ðfÞjp drðfÞ;

ð2Þ

for each k 2 N. Hence

ð1  rÞa nk r nk kPnk kp 6 ð1  r 2 Þa kRfr kp 6 kf kBap :

ð3Þ 

 1 nþ1

Choosing r ¼ 1  n1 in (3), and using the well-known inequality 1 þ n k

6 4; n 2 N, we obtain

a sup kPnk kp n1 6 Ckf kBap k

ð4Þ

k2N

as desired. a < 1. If p 2 ð0; 2, from (2) and by using the known inequality Now assume that lim supk!1 kPnk kp n1 k 1 X

!q

ak

6

1 X

k¼1

aqk ;

k¼1

where ak P 0; k 2 N; q 2 ð0; 1, we have that

kf kpBap  sup ð1  r 2 Þpa

Z

0
1 X

S

6 C sup ð1  r 2 Þpa 0
6 C sup ð1  r p Þpa 0
!p=2

n2k jP nk ðfÞj2 r 2nk

1 X k¼1 1 X

6 C sup ð1  r Þ 0
ð5Þ

r pnk npk kPnk kpp r pnk npk kPnk kpp

k¼1 p paþ1

drðfÞ

k¼1

1 X X n¼1

! npk kPnk kpp

rpn

ð6Þ

nk 6n

!

 p 1 X X pa a C sup ð1  r p Þpaþ1 nk ðrp Þn 6 sup kPnk kp n1 k k2N

0
n¼1

0
n¼1

nk 6n

 p 1 X a 6 C sup kPnk kp n1 sup ð1  r p Þpaþ1 npa ðr p Þn k k2N

 p a 6 C sup kPnk kp n1 ; k

ð7Þ

k2N

where we have used the fact that there is a positive constant C independent of n such that

X

npka 6 Cnpa

nk 6n

(here is used the assumption that nkþ1 =nk P k > 1 and the formula for the sum of a geometric progression) and the following well-known estimate: 1 X

nb qn 6 Cð1  qÞðbþ1Þ ;

n¼1

b > 0; q 2 ½0; 1Þ, see, for example, [23]. From (4) and (7) the asymptotic relation in (1) follows in this case. If p P 2, then by Jensen’s inequality from (5) we get

kf kpBa p

2 pa

6 C sup ð1  r Þ 0
1 X

!p=2 n2k r 2nk kPnk k2p

ð8Þ

k¼1

and consequently

 2 1 X a kf k2Bap 6 C sup kPnk kp n1 sup ð1  r 2 Þ2a n2k a r 2nk k k2N

0
k¼1

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232

from which the implication follows as in the case p 2 ð0; 2. (b) Let f 2 Bap;0 , then for every e > 0 there is a d > 0 such that

ð1  r2 Þa

Z

1=p jRf ðrfÞjp drðfÞ < e;

ð9Þ

S

whenever d < r < 1. From the first inequality in (3), and (9), we have

ð1  rÞa nk r nk kPnk kp < e 1 for every k 2 N and r 2 ðd; 1Þ. Choosing r ¼ 1  n1 ; where nk > 1d ; we obtain k

a n1 kPnk kp < 4e: k

From this and since

lim

k!1

e is an arbitrary positive number it follows that

a kP nk kp n1 k

¼ 0:

a ¼ 0. Then for every Now assume that limk!1 kP nk kp n1 k

a1

kPnk kp 6 enk ;

e > 0 there is a k0 2 N such that

for k > k0 :

From this, (2) and by the proof of (a), when p 2 ð0; 2, we have that

ð1  r2 Þpa kRfr kpp 6 Cð1  r 2 Þpa

1 X k¼1

r pnk npk kP nk kpp 6 Cð1  r2 Þpa

k0 X

rpnk npk kPnk kpp þ C ep ð1  r2 Þpa

1 X k¼k0 þ1

k¼1

6 Cð1  r 2 Þpa Pnk

! 1 X X pa p p p paþ1 ðr Þ þ C e ð1  r Þ nk r pn

6 Cð1  r 2 Þpa Pnk

1 X ðr p Þ þ C ep ð1  r p Þpaþ1 npa rpn 6 Cð1  r 2 Þpa Pnk ðr p Þ þ C ep ;

0

0

n¼1

r pnk npka

nk 6n

n¼1

0

ð10Þ

where P nk is a polynomial of degree nk0 . 0 Letting r ! 1  0 in (10), and using the fact that

lim ð1  r 2 Þpa Pnk ðrÞ ¼ 0

r!10

0

it follows that:

lim supð1  r 2 Þa kRfr kp 6 C 1=p e: r!10

Since e is an arbitrary positive number the implication follows in this case. The implication for case p P 2 follows similarly, from the following inequality:

ð1  r2 Þpa kRfr kpp 6 Cð1  r 2 Þpa

1 X

!p=2 n2k r 2nk kPnk k2p

;

k¼1

which easily follows from (2) (see (8)). Hence, we omit this part of the proof.  Acknowledgement The first author of this paper is supported by NSF of Guangdong Province (No. 7300614). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]

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