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On the coefficients of Bazilevi c ˘ functions and circularly symmetric functions
Applied Mathematics Letters 24 (2011) 991–995
Contents lists available at ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
On the coefficients of Bazilevi˘c functions and circularly symmetric functions Qin Deng School of Science, Hangzhou Dianzi University, Hangzhou 310018, PR China
article
abstract
info
Article history: Received 27 August 2010 Accepted 7 January 2011
In this work, we study the coefficients of Bazilevi˘c functions and circularly symmetric functions, and obtain exact estimates. © 2011 Elsevier Ltd. All rights reserved.
Keywords: Analytic functions Univalent functions Bazilevi˘c functions Circularly symmetric functions
1. Introduction Throughout this work, A denotes the class of analytic functions f (z ) in the unit disk U = {z ∈ C : |z | < 1} normalized such that f (0) = 0 and f ′ (0) = 1. Let S , K , S ∗ , Sc and Bα denote the subclasses of A of functions that are univalent, convex, starlike, close-to-convex and Bazilevi˘c, respectively. We also denote by P the class of analytic functions p with p(0) = 1 and Re{p(z )} > 0 in U. Note that P is known as the Carathéodory class. Also, it is well known that the inclusion relations K ⊂ S ∗ ⊂ Sc ⊂ Bα ⊂ S
are valid. See [1–3] for further information. Following [3,4], for f (z ) ∈ A , we see that f (z ) ∈ Bα if and only if zf ′ (z )
f (z )
f (z ) g (z )
α
∈ P,
z∈U
(1.1)
for some g (z ) ∈ S ∗ , α ≥ 0. Let D be a domain in C with 0 ∈ D. We shall say that D is circularly symmetric if, for every R with 0 < R < +∞, D ∩{|z | = R}, is either empty, the whole circle |z | = R, or a single arc on |z | = R which contains z = R and is symmetric with respect to the real axis. Following [5,6], we shall denote by Y the class of those functions f (z ) in S which map U onto a circularly symmetric domain. The elements of Y will be called circularly symmetric functions. Associated with each f (z ) in S is a well defined logarithmic function log
f (z ) z
=2
∞ −
γn z n ,
z ∈ U.
(1.2)
n =1
The numbers γn are called the logarithmic coefficients of f . Thus the Koebe function k(z ) = z (1 − z )−2 has logarithmic coefficients γn = 1n , n ∈ N = {1, 2, 3, . . .}. E-mail address: [email protected]. 0893-9659/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2011.01.012
992
Q. Deng / Applied Mathematics Letters 24 (2011) 991–995
The inequality |γn | ≤ 1n holds for functions f (z ) in S ∗ , but is false for the full class S, even in order of magnitude. Indeed (see Theorem 8.4 on p. 242 of [1]) there exists a bounded function f (z ) ∈ S with logarithmic coefficients γn ̸= O(n−0.83 ). In the paper [5], it is shown that the inequality |γn | ≤ 1n is false for functions f (z ) in Sc . f (z ) λ z
Let f (z ) ∈ S , ψ(z ) = [
∑∞
] = 1+
n =1
Dn (λ)z n , 0 < λ < 1. For the estimation of coefficients |Dn (λ)|, when λ > − 12
1 , 4
Milin (see [7]) gives a sharp estimate. But when 0 < λ ≤ 14 , we only obtain |Dn (λ)| = O(n log n). Estimating the sharp order of |Dn (λ)| is a difficult problem which is still unresolved. In this work, for when f (z ) is a Bazilevi˘c function Bα , we study the coefficients |Dn | and obtain an exact estimate for the order of |Dn (λ)|. Furthermore, by using the Milin–Lebejev method, we investigate the coefficients |Dn | and obtain a sharp estimate for the order of |Dn (λ)| when f (z ) ∈ Y . Throughout the work, let A denote the absolute constant, whose value varies in different places. 2. On the coefficients of Bazilevi˘c functions We use the symbols {f }n to denote the coefficients of z n in the Taylor expansion of a function f (z ). In order to obtain Theorem 1, we need the following lemmas. Lemma 1 ([2]). Let f (z ) ∈ S. Then, for z = reiθ ,
∫ 0
1 2
Lemma 2 ([8]). Let p(z ) ∈ P . Then, for z = reiθ ,
∫
≤ r < 1,
2 2π ′ f (z ) dθ ≤ A(1 − r )−1 log 1 . f (z ) 1−r
(2.1) 1 2
≤ r < 1,
2π ′ p (z ) dθ ≤ 4 log 1 + r . p(z ) 1−r 0
(2.2)
Lemma 3 ([2]). Let f (z ) ∈ S. Then, for z = reiθ , 0 < r < 1,
|f (z )| ≤
r
(1 − r )2
.
(2.3)
Theorem 1. Let f (z ) ∈ Bα , ψ(z ) = [
f (z ) λ z
] =1+
∑∞
n =1
Dn (λ)z n , 0 < λ < 1. Then
3
|Dn | ≤ An2λ−1 (log n) 2 .
(2.4)
The exponent 2λ − 1 is the best possible. Proof. Assume without loss of generality that |f (r )| = max|z |=r |f (z )|; otherwise we consider the function e−iθ0 f (reiθ0 ). It is clear that
nDn =
z
f (z )
λ ′
z
= {λ[f (z )]λ−1 f ′ (z )z 1−λ − λ[f (z )]λ z −λ }n n
= {λh1 (z )}n − {λh2 (z )}n , λ−1 ′
(2.5) λ −λ
where h1 (z ) = [f (z )] f (z )z , h2 (z ) = [f (z )] z By means of Lemma 3, we get that, for r < 1,
|{h2 (z )}n | ≤
r −n
2π
∫
2π
1−λ
|f (z )|λ |z |−λ dθ ≤
0
.
r −n
1
2π (1 − r )2λ
· 2π = r −n (1 − r )−2λ .
(2.6)
Let r = 1 − 1n ; we obtain from (2.6) that, for n = 2, 3, . . . ,
|{h2 (z )}n | ≤ An2λ .
(2.7)
A short calculation shows that
[
zh1 (z ) = h1 (z ) (λ − 1) ′
zf ′ (z ) f (z )
+
zf ′′ (z ) f ′ (z )
]
+ (1 − λ) .
zf ′ (z ) f (z ) Because f (z ) ∈ Bα , there exists a starlike function g (z ) ∈ S ∗ such that Re{ f (z ) ( g (z ) )α } > 0, α ≥ 0. Write p(z ) = then
zf ′′ (z ) f ′ (z )
=
zp′ (z ) p(z )
− 1 − (α − 1)
zf ′ (z ) f (z )
+α
zg ′ (z ) g (z )
.
(2.8) zf ′ (z ) f (z )
( gf ((zz)) )α ; (2.9)
Q. Deng / Applied Mathematics Letters 24 (2011) 991–995
993
By applying (2.9), from (2.8) we obtain zh′1 (z ) = (λ − α)
zf ′ (z ) f (z )
zp′ (z )
h1 (z ) +
p(z )
h1 ( z ) + α
zg ′ (z ) g (z )
h1 (z ) − λh1 (z ).
(2.10)
By means of Lemmas 1 and 3, we obtain n −
|{h1 (z )}m |2 ≤ r −2n
m=0
∞ −
|{h1 (z )}m |2 r 2m ≤
m=0
=
r −2n+2−2λ
2π
∫
2π
r −2n
∫
2π
2π
|h1 (z )|2 dθ
0
′ 2 f (z ) dθ ≤ Ar −2n+2 (1 − r )−4λ−1 log 1 . |f (z )|2λ f (z ) 1−r
0
(2.11)
Let r = 1 − 1n , we obtain from (2.11) that, for n = 2, 3, . . . , n −
|{h1 (z )}m |2 ≤ An4λ+1 log n.
(2.12)
m=0
From (2.12) we obtain
|{h1 (z )}n | ≤
n −
|{h1 (z )}m | ≤ n
1 2
m=0
n −
12 1
≤ An2λ+1 (log n) 2 .
|{h1 (z )}m |2
(2.13)
m=0
By means of Lemma 2, we obtain
′ ∫ zp (z ) r −n 2π zp′ (z ) 1 −n +1 ≤ log . p(z ) dθ ≤ Ar p(z ) 2 π 1 − r 0 n
(2.14)
Let r = 1 − 1n ; we obtain from (2.14) that, for n = 2, 3, . . . ,
′ zp (z ) p(z ) ≤ A log n. n
(2.15)
We obtain from (2.12) and (2.15) that
′ ′ n−1 n − − zp (z ) zp (z ) |{h1 (z )}m | ≤ A(log n) p(z ) h1 (z ) = {h1 (z )}m p(z ) n n −m m=0 m=0 ≤ A(log n)n
1 2
n −
12 |{h1 (z )}m |
2
3
≤ An2λ+1 (log n) 2 .
(2.16)
m=0
By means of Lemmas 1 and 3 we obtain that
−n ∫ 2π ′ ≤ r zf (z ) ||h1 (z ) dθ f (z ) 2 π 0 n ∫ 2π r −n f ′ (z ) 2 1 = |z |2−λ | | |f (z )|λ dθ ≤ Ar −n+2 (1 − r )−2λ−1 log . 2π 0 f (z ) 1−r
′ zf (z ) f ( z ) h1 ( z )
(2.17)
Let r = 1 − 1n ; we obtain from (2.17) that, for n = 2, 3, . . . ,
′ zf (z ) h ( z ) 1 f (z )
n
≤ An2λ+1 log n.
(2.18)
Because g (z ) ∈ S ∗ ⊂ S, we get from (2.18) that
′ zg (z ) h ( z ) 1 g (z )
n
≤ An2λ+1 log n.
(2.19)
Combining (2.13), (2.16), (2.18) and (2.19), we obtain from (2.10) that, for n = 2, 3, . . . , 3
n|{h1 (z )}n | = |{zh′1 (z )}n | ≤ An2λ+1 (log n) 2 .
(2.20)
994
Q. Deng / Applied Mathematics Letters 24 (2011) 991–995
Then 3
|{h1 (z )}n | ≤ An2λ (log n) 2 .
(2.21)
Combining (2.7) and (2.21), we obtain from (2.5) that, for n = 2, 3, . . . , 3
n|Dn | ≤ An2λ (log n) 2 ,
(2.22)
which implies that, for n = 2, 3, . . . , 3
|Dn | ≤ An2λ−1 (log n) 2 .
(2.23)
The Koebe function k(z ) = z (1 − z )
−2
implies that the exponent 2λ − 1 is the best possible.
Since Sc ⊂ Bα , we obtain the following corollary: Corollary 1. Let f (z ) ∈ Sc , ψ(z ) = [
f (z ) λ z
] =1+
∑∞
n =1
Dn (λ)z n , 0 < λ < 1. Then
3
|Dn | ≤ An2λ−1 (log n) 2 .
(2.24)
The exponent 2λ − 1 is the best possible. 3. On the coefficients of circularly symmetric functions In this section, by using the Milin–Lebejev method, we shall study the coefficients |Dn | when f (z ) ∈ Y . In order to obtain Theorem 2, we need the following lemma. Lemma 4 ([9]). Let f (z ) ∈ Y . Then, for n ≥ 2,
|γn | ≤ An−1 log n,
(3.1)
where the exponent −1 is the best possible. Theorem 2. Let f (z ) ∈ Y , ψ(z ) = [
f (z ) λ z
] =1+
∑∞
n=1
Dn (λ)z n , 0 < λ < 1. Then
|Dn | ≤ An2λ−1 log n.
(3.2)
The exponent 2λ − 1 is the best possible. Proof. We define dk (2λ) by the expansion 1
(1 − z )
2λ
=
∞ −
dk (2λ)z k ,
|z | < 1.
(3.3)
k=0
In the paper [7], it is shown that n −1 −
dk (2λ) = dn (2λ + 1) ≤ An2λ .
(3.4)
k=0
It is clear that z ψ ′ (z ) =
z ψ ′ (z )
ψ(z )
ψ(z ).
(3.5)
So ∞ − k=1
kDk z k =
∞ − k=1
2λkγk z k
∞ −
Dk z k ,
D0 = 1.
(3.6)
k=0
Comparing the coefficients of the same powers of z in (3.6) and applying Lemma 4, we obtain
n n −1 − − |nDn | = 2λ kγk Dn−k ≤ A log n |Dk |. k=1 k=0
(3.7)
Q. Deng / Applied Mathematics Letters 24 (2011) 991–995
995
Applying Schwarz’s inequality, we obtain from (3.7) that
n −1 n −1 − |Dk |2 − |Dn | ≤ 2 (log n) · dk (2λ) . n d (2λ) k=0 k=0 k 2
A
2
(3.8)
It is well known that (see [7])
n −1 n − − 2λ |Dk |2 ≤ dn (2λ + 1) exp dn−l (2λ)1k , d (2λ) dn (2λ + 1) l=1 k=0 k
(3.9)
where
1k =
k −
l|γl |2 −
l =1
k − 1 l =1
l
< 0.312.
(3.10)
Combining (3.4) and (3.10), we obtain from (3.9) that n −1 − |Dk |2 ≤ Adn (2λ + 1) ≤ An2λ . d (2λ) k=0 k
(3.11)
Combining (3.4) and (3.11), we obtain from (3.8) that
|Dn |2 ≤ An4λ−2 (log n)2 ,
(3.12)
which implies that
|Dn | ≤ An2λ−1 log n.
(3.13)
The Koebe function k(z ) = z (1 − z )
−2
implies that the exponent 2λ − 1 is the best possible.
The inequality |γn | ≤ holds for functions f (z ) in S . Similarly, by using the Milin–Lebejev method we obtain: 1 n
∗
∑∞ f (z ) Corollary 2. Let f (z ) ∈ S ∗ , ψ(z ) = [ z ]λ = 1 + n=1 Dn (λ)z n , 0 < λ < 1. Then |Dn | ≤ An2λ−1 .
(3.14)
The exponent 2λ − 1 is the best possible. Since K ⊂ S ∗ , we have: Corollary 3. Let f (z ) ∈ K , ψ(z ) = [
f (z ) λ z
] =1+
∑∞
n =1
Dn (λ)z n , 0 < λ < 1. Then
|Dn | ≤ An2λ−1 .
(3.15)
The exponent 2λ − 1 is the best possible. Acknowledgements Project 10926058 and 61003194 were supported by the National Natural Science Foundation of China. References [1] P.L. Duren, Univalent Functions, Springer-Verlag, New York, 1983. [2] Ch. Pommerenke, Univalent Functions, Vandenkoeck and Ruprecht, Göttingen, 1975. [3] I.E. Bazilevi˘c, On a case of integrability in quadratures of the Löwner–Kufarev equation, Matematicheskii Sbornik (N.S.) 37 (3) (1955) 471–476 (in Russian). [4] Y.C. Kim, A note on growth theorem of Bazilevi˘c functions, Appl. Math. Comput. 208 (2009) 542–546. [5] D. Girela, Logarithmic coefficients of univalent functions, Ann. Acad. Sci. Fenn. Math. 25 (2000) 337–350. [6] J.A. Jenkins, On circularly symmetric functions, Proc. Amer. Math. Soc. 6 (1955) 620–624. [7] I.M. Milin, Univalent Functions and Orthonormal Systems, Moscow, 1971 (in Russian). [8] Z. Ye, On the successive coefficients of close-to-convex functions, J. Math. Anal. Appl. 283 (2003) 689–695. [9] Q. Deng, On circularly symmetric functions, Appl. Math. Lett. 23 (12) (2010) 1483–1488.
Contents lists available at ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
On the coefficients of Bazilevi˘c functions and circularly symmetric functions Qin Deng School of Science, Hangzhou Dianzi University, Hangzhou 310018, PR China
article
abstract
info
Article history: Received 27 August 2010 Accepted 7 January 2011
In this work, we study the coefficients of Bazilevi˘c functions and circularly symmetric functions, and obtain exact estimates. © 2011 Elsevier Ltd. All rights reserved.
Keywords: Analytic functions Univalent functions Bazilevi˘c functions Circularly symmetric functions
1. Introduction Throughout this work, A denotes the class of analytic functions f (z ) in the unit disk U = {z ∈ C : |z | < 1} normalized such that f (0) = 0 and f ′ (0) = 1. Let S , K , S ∗ , Sc and Bα denote the subclasses of A of functions that are univalent, convex, starlike, close-to-convex and Bazilevi˘c, respectively. We also denote by P the class of analytic functions p with p(0) = 1 and Re{p(z )} > 0 in U. Note that P is known as the Carathéodory class. Also, it is well known that the inclusion relations K ⊂ S ∗ ⊂ Sc ⊂ Bα ⊂ S
are valid. See [1–3] for further information. Following [3,4], for f (z ) ∈ A , we see that f (z ) ∈ Bα if and only if zf ′ (z )
f (z )
f (z ) g (z )
α
∈ P,
z∈U
(1.1)
for some g (z ) ∈ S ∗ , α ≥ 0. Let D be a domain in C with 0 ∈ D. We shall say that D is circularly symmetric if, for every R with 0 < R < +∞, D ∩{|z | = R}, is either empty, the whole circle |z | = R, or a single arc on |z | = R which contains z = R and is symmetric with respect to the real axis. Following [5,6], we shall denote by Y the class of those functions f (z ) in S which map U onto a circularly symmetric domain. The elements of Y will be called circularly symmetric functions. Associated with each f (z ) in S is a well defined logarithmic function log
f (z ) z
=2
∞ −
γn z n ,
z ∈ U.
(1.2)
n =1
The numbers γn are called the logarithmic coefficients of f . Thus the Koebe function k(z ) = z (1 − z )−2 has logarithmic coefficients γn = 1n , n ∈ N = {1, 2, 3, . . .}. E-mail address: [email protected]. 0893-9659/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2011.01.012
992
Q. Deng / Applied Mathematics Letters 24 (2011) 991–995
The inequality |γn | ≤ 1n holds for functions f (z ) in S ∗ , but is false for the full class S, even in order of magnitude. Indeed (see Theorem 8.4 on p. 242 of [1]) there exists a bounded function f (z ) ∈ S with logarithmic coefficients γn ̸= O(n−0.83 ). In the paper [5], it is shown that the inequality |γn | ≤ 1n is false for functions f (z ) in Sc . f (z ) λ z
Let f (z ) ∈ S , ψ(z ) = [
∑∞
] = 1+
n =1
Dn (λ)z n , 0 < λ < 1. For the estimation of coefficients |Dn (λ)|, when λ > − 12
1 , 4
Milin (see [7]) gives a sharp estimate. But when 0 < λ ≤ 14 , we only obtain |Dn (λ)| = O(n log n). Estimating the sharp order of |Dn (λ)| is a difficult problem which is still unresolved. In this work, for when f (z ) is a Bazilevi˘c function Bα , we study the coefficients |Dn | and obtain an exact estimate for the order of |Dn (λ)|. Furthermore, by using the Milin–Lebejev method, we investigate the coefficients |Dn | and obtain a sharp estimate for the order of |Dn (λ)| when f (z ) ∈ Y . Throughout the work, let A denote the absolute constant, whose value varies in different places. 2. On the coefficients of Bazilevi˘c functions We use the symbols {f }n to denote the coefficients of z n in the Taylor expansion of a function f (z ). In order to obtain Theorem 1, we need the following lemmas. Lemma 1 ([2]). Let f (z ) ∈ S. Then, for z = reiθ ,
∫ 0
1 2
Lemma 2 ([8]). Let p(z ) ∈ P . Then, for z = reiθ ,
∫
≤ r < 1,
2 2π ′ f (z ) dθ ≤ A(1 − r )−1 log 1 . f (z ) 1−r
(2.1) 1 2
≤ r < 1,
2π ′ p (z ) dθ ≤ 4 log 1 + r . p(z ) 1−r 0
(2.2)
Lemma 3 ([2]). Let f (z ) ∈ S. Then, for z = reiθ , 0 < r < 1,
|f (z )| ≤
r
(1 − r )2
.
(2.3)
Theorem 1. Let f (z ) ∈ Bα , ψ(z ) = [
f (z ) λ z
] =1+
∑∞
n =1
Dn (λ)z n , 0 < λ < 1. Then
3
|Dn | ≤ An2λ−1 (log n) 2 .
(2.4)
The exponent 2λ − 1 is the best possible. Proof. Assume without loss of generality that |f (r )| = max|z |=r |f (z )|; otherwise we consider the function e−iθ0 f (reiθ0 ). It is clear that
nDn =
z
f (z )
λ ′
z
= {λ[f (z )]λ−1 f ′ (z )z 1−λ − λ[f (z )]λ z −λ }n n
= {λh1 (z )}n − {λh2 (z )}n , λ−1 ′
(2.5) λ −λ
where h1 (z ) = [f (z )] f (z )z , h2 (z ) = [f (z )] z By means of Lemma 3, we get that, for r < 1,
|{h2 (z )}n | ≤
r −n
2π
∫
2π
1−λ
|f (z )|λ |z |−λ dθ ≤
0
.
r −n
1
2π (1 − r )2λ
· 2π = r −n (1 − r )−2λ .
(2.6)
Let r = 1 − 1n ; we obtain from (2.6) that, for n = 2, 3, . . . ,
|{h2 (z )}n | ≤ An2λ .
(2.7)
A short calculation shows that
[
zh1 (z ) = h1 (z ) (λ − 1) ′
zf ′ (z ) f (z )
+
zf ′′ (z ) f ′ (z )
]
+ (1 − λ) .
zf ′ (z ) f (z ) Because f (z ) ∈ Bα , there exists a starlike function g (z ) ∈ S ∗ such that Re{ f (z ) ( g (z ) )α } > 0, α ≥ 0. Write p(z ) = then
zf ′′ (z ) f ′ (z )
=
zp′ (z ) p(z )
− 1 − (α − 1)
zf ′ (z ) f (z )
+α
zg ′ (z ) g (z )
.
(2.8) zf ′ (z ) f (z )
( gf ((zz)) )α ; (2.9)
Q. Deng / Applied Mathematics Letters 24 (2011) 991–995
993
By applying (2.9), from (2.8) we obtain zh′1 (z ) = (λ − α)
zf ′ (z ) f (z )
zp′ (z )
h1 (z ) +
p(z )
h1 ( z ) + α
zg ′ (z ) g (z )
h1 (z ) − λh1 (z ).
(2.10)
By means of Lemmas 1 and 3, we obtain n −
|{h1 (z )}m |2 ≤ r −2n
m=0
∞ −
|{h1 (z )}m |2 r 2m ≤
m=0
=
r −2n+2−2λ
2π
∫
2π
r −2n
∫
2π
2π
|h1 (z )|2 dθ
0
′ 2 f (z ) dθ ≤ Ar −2n+2 (1 − r )−4λ−1 log 1 . |f (z )|2λ f (z ) 1−r
0
(2.11)
Let r = 1 − 1n , we obtain from (2.11) that, for n = 2, 3, . . . , n −
|{h1 (z )}m |2 ≤ An4λ+1 log n.
(2.12)
m=0
From (2.12) we obtain
|{h1 (z )}n | ≤
n −
|{h1 (z )}m | ≤ n
1 2
m=0
n −
12 1
≤ An2λ+1 (log n) 2 .
|{h1 (z )}m |2
(2.13)
m=0
By means of Lemma 2, we obtain
′ ∫ zp (z ) r −n 2π zp′ (z ) 1 −n +1 ≤ log . p(z ) dθ ≤ Ar p(z ) 2 π 1 − r 0 n
(2.14)
Let r = 1 − 1n ; we obtain from (2.14) that, for n = 2, 3, . . . ,
′ zp (z ) p(z ) ≤ A log n. n
(2.15)
We obtain from (2.12) and (2.15) that
′ ′ n−1 n − − zp (z ) zp (z ) |{h1 (z )}m | ≤ A(log n) p(z ) h1 (z ) = {h1 (z )}m p(z ) n n −m m=0 m=0 ≤ A(log n)n
1 2
n −
12 |{h1 (z )}m |
2
3
≤ An2λ+1 (log n) 2 .
(2.16)
m=0
By means of Lemmas 1 and 3 we obtain that
−n ∫ 2π ′ ≤ r zf (z ) ||h1 (z ) dθ f (z ) 2 π 0 n ∫ 2π r −n f ′ (z ) 2 1 = |z |2−λ | | |f (z )|λ dθ ≤ Ar −n+2 (1 − r )−2λ−1 log . 2π 0 f (z ) 1−r
′ zf (z ) f ( z ) h1 ( z )
(2.17)
Let r = 1 − 1n ; we obtain from (2.17) that, for n = 2, 3, . . . ,
′ zf (z ) h ( z ) 1 f (z )
n
≤ An2λ+1 log n.
(2.18)
Because g (z ) ∈ S ∗ ⊂ S, we get from (2.18) that
′ zg (z ) h ( z ) 1 g (z )
n
≤ An2λ+1 log n.
(2.19)
Combining (2.13), (2.16), (2.18) and (2.19), we obtain from (2.10) that, for n = 2, 3, . . . , 3
n|{h1 (z )}n | = |{zh′1 (z )}n | ≤ An2λ+1 (log n) 2 .
(2.20)
994
Q. Deng / Applied Mathematics Letters 24 (2011) 991–995
Then 3
|{h1 (z )}n | ≤ An2λ (log n) 2 .
(2.21)
Combining (2.7) and (2.21), we obtain from (2.5) that, for n = 2, 3, . . . , 3
n|Dn | ≤ An2λ (log n) 2 ,
(2.22)
which implies that, for n = 2, 3, . . . , 3
|Dn | ≤ An2λ−1 (log n) 2 .
(2.23)
The Koebe function k(z ) = z (1 − z )
−2
implies that the exponent 2λ − 1 is the best possible.
Since Sc ⊂ Bα , we obtain the following corollary: Corollary 1. Let f (z ) ∈ Sc , ψ(z ) = [
f (z ) λ z
] =1+
∑∞
n =1
Dn (λ)z n , 0 < λ < 1. Then
3
|Dn | ≤ An2λ−1 (log n) 2 .
(2.24)
The exponent 2λ − 1 is the best possible. 3. On the coefficients of circularly symmetric functions In this section, by using the Milin–Lebejev method, we shall study the coefficients |Dn | when f (z ) ∈ Y . In order to obtain Theorem 2, we need the following lemma. Lemma 4 ([9]). Let f (z ) ∈ Y . Then, for n ≥ 2,
|γn | ≤ An−1 log n,
(3.1)
where the exponent −1 is the best possible. Theorem 2. Let f (z ) ∈ Y , ψ(z ) = [
f (z ) λ z
] =1+
∑∞
n=1
Dn (λ)z n , 0 < λ < 1. Then
|Dn | ≤ An2λ−1 log n.
(3.2)
The exponent 2λ − 1 is the best possible. Proof. We define dk (2λ) by the expansion 1
(1 − z )
2λ
=
∞ −
dk (2λ)z k ,
|z | < 1.
(3.3)
k=0
In the paper [7], it is shown that n −1 −
dk (2λ) = dn (2λ + 1) ≤ An2λ .
(3.4)
k=0
It is clear that z ψ ′ (z ) =
z ψ ′ (z )
ψ(z )
ψ(z ).
(3.5)
So ∞ − k=1
kDk z k =
∞ − k=1
2λkγk z k
∞ −
Dk z k ,
D0 = 1.
(3.6)
k=0
Comparing the coefficients of the same powers of z in (3.6) and applying Lemma 4, we obtain
n n −1 − − |nDn | = 2λ kγk Dn−k ≤ A log n |Dk |. k=1 k=0
(3.7)
Q. Deng / Applied Mathematics Letters 24 (2011) 991–995
995
Applying Schwarz’s inequality, we obtain from (3.7) that
n −1 n −1 − |Dk |2 − |Dn | ≤ 2 (log n) · dk (2λ) . n d (2λ) k=0 k=0 k 2
A
2
(3.8)
It is well known that (see [7])
n −1 n − − 2λ |Dk |2 ≤ dn (2λ + 1) exp dn−l (2λ)1k , d (2λ) dn (2λ + 1) l=1 k=0 k
(3.9)
where
1k =
k −
l|γl |2 −
l =1
k − 1 l =1
l
< 0.312.
(3.10)
Combining (3.4) and (3.10), we obtain from (3.9) that n −1 − |Dk |2 ≤ Adn (2λ + 1) ≤ An2λ . d (2λ) k=0 k
(3.11)
Combining (3.4) and (3.11), we obtain from (3.8) that
|Dn |2 ≤ An4λ−2 (log n)2 ,
(3.12)
which implies that
|Dn | ≤ An2λ−1 log n.
(3.13)
The Koebe function k(z ) = z (1 − z )
−2
implies that the exponent 2λ − 1 is the best possible.
The inequality |γn | ≤ holds for functions f (z ) in S . Similarly, by using the Milin–Lebejev method we obtain: 1 n
∗
∑∞ f (z ) Corollary 2. Let f (z ) ∈ S ∗ , ψ(z ) = [ z ]λ = 1 + n=1 Dn (λ)z n , 0 < λ < 1. Then |Dn | ≤ An2λ−1 .
(3.14)
The exponent 2λ − 1 is the best possible. Since K ⊂ S ∗ , we have: Corollary 3. Let f (z ) ∈ K , ψ(z ) = [
f (z ) λ z
] =1+
∑∞
n =1
Dn (λ)z n , 0 < λ < 1. Then
|Dn | ≤ An2λ−1 .
(3.15)
The exponent 2λ − 1 is the best possible. Acknowledgements Project 10926058 and 61003194 were supported by the National Natural Science Foundation of China. References [1] P.L. Duren, Univalent Functions, Springer-Verlag, New York, 1983. [2] Ch. Pommerenke, Univalent Functions, Vandenkoeck and Ruprecht, Göttingen, 1975. [3] I.E. Bazilevi˘c, On a case of integrability in quadratures of the Löwner–Kufarev equation, Matematicheskii Sbornik (N.S.) 37 (3) (1955) 471–476 (in Russian). [4] Y.C. Kim, A note on growth theorem of Bazilevi˘c functions, Appl. Math. Comput. 208 (2009) 542–546. [5] D. Girela, Logarithmic coefficients of univalent functions, Ann. Acad. Sci. Fenn. Math. 25 (2000) 337–350. [6] J.A. Jenkins, On circularly symmetric functions, Proc. Amer. Math. Soc. 6 (1955) 620–624. [7] I.M. Milin, Univalent Functions and Orthonormal Systems, Moscow, 1971 (in Russian). [8] Z. Ye, On the successive coefficients of close-to-convex functions, J. Math. Anal. Appl. 283 (2003) 689–695. [9] Q. Deng, On circularly symmetric functions, Appl. Math. Lett. 23 (12) (2010) 1483–1488.