Volume 8 (2007), Issue 1, Article 12, 11 pp.
ON CERTAIN SUBCLASS OF BAZILEVIČ FUNCTIONS
DONG GUO AND MING-SHENG LIU
S CHOOL OF M ATHEMATICAL S CIENCES ,
S OUTH C HINA N ORMAL U NIVERSITY,
G UANGZHOU 510631, G UANGDONG ,
P. R. C HINA
liumsh@scnu.edu.cn
Received 22 September, 2006; accepted 26 February, 2007
Communicated by H.M. Srivastava
P∞
A BSTRACT. Let H be the class of functions f (z) of the form f (z) = z + n=2 an z n , which
are analytic in the unit disk U = {z : |z| < 1}. In this paper, the authors introduce a subclass M (α, λ, ρ) of H and study its some properties. The subordination relationships, inclusion
relationships, coefficient estimates, the integral operator and covering theorem are proven here
for each of the function classes. Furthermore, some interesting Fekete-Szegö inequalities are
obtained. Some of the results, presented in this paper, generalize the corresponding results of
earlier authors.
Key words and phrases: Starlike function, Bazilevič Function, Subordination relationships, Inclusion relationship, Coefficient
estimates, Integral operator, Covering theorem, Fekete-Szegö inequalities.
2000 Mathematics Subject Classification. Primary 30C45; Secondary 26A33, 33C05.
1. I NTRODUCTION
Let H denote the class of functions f of the form
∞
X
(1.1)
f (z) = z +
an z n ,
n=2
which are analytic in the open unit disk U = {z : |z| < 1}, and let S denote the class of all
functions in H which are univalent in the disk U . Suppose also that S ∗ , K and α − K denote the
familiar subclasses of H consisting of functions which are, respectively, starlike in U, convex
in U and α− convex in U. Thus we have
0 zf (z)
∗
S = f : f ∈ H and R
> 0, z ∈ U ,
f (z)
zf 00 (z)
K = f : f ∈ H and R 1 + 0
> 0, z ∈ U
f (z)
This research is partly supported by the Doctoral Foundation of the Education committee of China (No. 20050574002).
242-06
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D ONG G UO AND M ING -S HENG L IU
and
zf 00 (z)
zf 0 (z)
α − K = f : f ∈ H and R α 1 + 0
+ (1 − α)
> 0,
f (z)
f (z)
z∈U
.
Let f (z) and F (z) be analytic in U . Then we say that the function f (z) is subordinate to F (z)
in U , if there exists an analytic function ω(z) in U such that |ω(z)| ≤ |z| and f (z) = F (ω(z)),
denoted f ≺ F or f (z) ≺ F (z). If F (z) is univalent in U , then the subordination is equivalent
to f (0) = F (0) and f (U ) ⊂ F (U ) (see [18]).
Assuming that α > 0, λ ≥ 0, ρ < 1, a function p(z) = 1 + p1 z + p2 z 2 + · · · is said to be in
the class Pρ if and only if p(z) is analytic in the unit disk U and Rp(z) > ρ, z ∈ U . A function
f (z) ∈ H is said to be in the class B(λ, α, ρ) if and only if it satisfies
α
α 0
f (z)
zf (z) f (z)
(1.2)
R(1 − λ)
+λ
> ρ, z ∈ U,
z
f (z)
z
α
α f (z)
f (z)
where we choose the branch of the power z
such that
= 1. It is obvious that
z
z=0
the subclass B(1, α, 0) is the subclass of Bazilevič functions, which is the subclass of univalent
functions S, we set B(α, ρ) ≡ B(1, α, ρ). The function class B(λ, α, ρ) was introduced and
studied by Liu [10]. Some special cases of the function class B(λ, α, ρ) had been studied by
Bazilevič [1], Chichra [2], Ding, Ling and Bao [3], Liu [9] and Singh [19], respectively.
Liu [11] introduced the following class B(λ, α, A, B, g(z)) of analytic functions, and studied
its some properties.
B(λ, α, A, B, g(z))
α
α
0
zg 0 (z)
f (z)
zf (z) f (z)
1 + Az
= f ∈H : 1−λ
+λ
≺
,
g(z)
g(z)
f (z)
g(z)
1 + Bz
where α > 0, λ ≥ 0, −1 ≤ B < A ≤ 1, g(z) ∈ S ∗ .
Fekete and Szegö [4] showed that for f ∈ S given by (1.1),

3 − 4u,
if µ ≤ 0,



|a3 − µa22 | ≤
1 + 2e−2/(1−µ) , if 0 ≤ µ < 1,



4 − 3µ,
if µ ≥ 1.
As a result, many authors studied similar problems for some subclasses of H or S (see [6, 7,
8, 13, 14, 15, 20]), which is popularly referred to as the Fekete-Szegö inequality or the FeketeSzegö problem. Li and Liu [12] obtained the Fekete-Szegö inequality for the function class
B(λ, α, ρ).
Recently, Patel [17] introduced the following subclass Mp (λ, µ, A, B) of p−valent Bazilevič
functions, and studied some of its properties.
P
n
An analytic function f (z) = z p + ∞
the class Mp (λ, µ, A, B) if
n=p+1 an z is said to be in
P
n
and only if there exists a p−valent starlike function g(z) = z p + ∞
n=p+1 bn z such that
µ
0
zf 0 (z) f (z)
zf 00 (z) zf 0 (z)
zf (z) zg 0 (z)
1 + Az
+λ 1+ 0
−
+µ
−
≺p
,
f (z)
g(z)
f (z)
f (z)
f (z)
g(z)
1 + Bz
where µ ≥ 0, λ > 0, −1 ≤ B < A ≤ 1.
In the present paper, we introduce the following subclass of analytic functions, and obtain
some interesting results.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
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BAZILEVI Č F UNCTIONS
3
Definition 1.1. Assume that α ≥ 0, λ ≥ 0, 0 ≤ ρ < 1, f ∈ H. We say that f (z) ∈ M (α, λ, ρ)
if and only if f (z) satisfies the following inequality:
0
α
0
zf 00 (z) zf 0 (z)
zf (z)
zf (z) f (z)
R
+λ 1+ 0
−
+α
−1
> ρ, z ∈ U.
f (z)
z
f (z)
f (z)
f (z)
It is evident that M (α, 0, ρ) = B(α, ρ)(α ≥ 0) and M (0, α, 0) = α − K(α ≥ 0).
2. P RELIMINARIES
To derive our main results, we shall require the following lemmas.
Lemma 2.1 ([16]). If −1 ≤ B < A ≤ 1, β > 0 and the complex number γ satisfies R(λ) ≥
−β(1−A)
, then the differential equation
1−B
q(z) +
zq 0 (z)
1 + Az
=
,
βq(z) + γ
1 + Bz
z ∈ U,
has a univalent solution in U given by

z β+γ (1 + Bz)β(A−B)/B


Rz
− γ, B =
6 0,


 β 0 tβ+γ−1 (1 + Bt)β(A−B)/B dt β
(2.1)
q(z) =


z β+γ exp(βAz)


 R z β+γ−1
− γ,
B = 0.
β 0 t
exp(βAt)dt β
If φ(z) = 1 + c1 z + c2 z 2 + · · · is analytic in U and satisfies
(2.2)
φ(z) +
zφ0 (z)
1 + Az
≺
,
βφ(z) + γ
1 + Bz
(z ∈ U),
then
φ(z) ≺ q(z) ≺
1 + Az
,
1 + Bz
(z ∈ U),
and q(z) is the best dominant of (2.2).
Lemma 2.2 ([11]). Suppose that F (z) is analytic and convex in U, and 0 ≤ λ ≤ 1, f (z) ∈ H,
g(z) ∈ H. If f (z) ≺ F (z) and g(z) ≺ F (z). Then
λf (z) + (1 − λ)g(z) ≺ F (z).
P
n
Lemma 2.3 ([18]). Let p(z) = 1 + ∞
n=1 pn z ∈ P0 . Then
1
2
p2 − p1 ≤ 2 − 1 |p21 |
2 2
and |pn | ≤ 2 for all n ∈ N+ .
Lemma 2.4 ([1]). Let α ≥ 0, f ∈ H and for |z| < R ≤ 1,
0
α zf (z) f (z)
R
> 0,
f (z)
z
then f (z) is univalent in |z| < R.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
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4
D ONG G UO AND M ING -S HENG L IU
3. M AIN R ESULTS AND T HEIR P ROOFS
Theorem 3.1. Let α ≥ 0 and λ > 0. If f (z) ∈ M (α, λ, ρ). Then
α
1 + (1 − 2ρ)z
zf 0 (z) f (z)
≺ q(z) ≺
(3.1)
, (z ∈ U),
f (z)
z
1−z
where
q(z) = R z
0
λz 1/λ (1 − z)−2(1−ρ)/λ
,
t(1−λ)/λ (1 − t)−2(1−ρ)/λ dt
and q(z) is the best dominant of (3.1).
Proof. By applying the method of the proof of Theorem 3.1 in [17] mutatis mutandis, we can
prove this theorem.
With the aid of Lemma 2.4, from Theorem 3.1, we have the following inclusion relation.
Corollary 3.2. Let α ≥ 0, 0 ≤ ρ < 1 and λ ≥ 0, then
M (α, λ, ρ) ⊂ M (α, 0, ρ) ⊂ M (α, 0, 0) ⊂ S.
Theorem 3.3. Let α ≥ 0 and λ2 > λ1 ≥ 0, 1 > ρ2 ≥ ρ1 ≥ 0, then
M (α, λ2 , ρ2 ) ⊂ M (α, λ1 , ρ1 ).
Proof. Suppose that f (z) ∈ M (α, λ2 , ρ2 ). Then, by the definition of M (α, λ2 , ρ2 ), we have
0
α
0
zf (z) f (z)
zf 00 (z) zf 0 (z)
zf (z)
(3.2) R
+ λ2 1 + 0
−
+α
−1
f (z)
z
f (z)
f (z)
f (z)
> ρ2
(z ∈ U).
Since α ≥ 0 and λ2 > λ1 ≥ 0, by Theorem 3.1, we obtain
0
α zf (z) f (z)
(3.3)
R
> ρ2 (z ∈ U).
f (z)
z
Setting λ = λλ12 , so that 0 ≤ λ < 1, we find from (3.2) and (3.3) that
0
α
0
zf (z) f (z)
zf 00 (z) zf 0 (z)
zf (z)
R
+ λ1 1 + 0
−
+α
−1
f (z)
z
f (z)
f (z)
f (z)
0
α
0
zf (z) f (z)
zf 00 (z) zf 0 (z)
zf (z)
= λR
+ λ2 [1 + 0
−
+α
−1
f (z)
z
f (z)
f (z)
f (z)
α
zf 0 (z) f (z)
+ (1 − λ)R
> ρ2 ≥ ρ1 (z ∈ U),
f (z)
z
that is, f (z) ∈ M (α, λ1 , ρ1 ). Hence, we have M (α, λ2 , ρ2 ) ⊂ M (α, λ1 , ρ1 ), and the proof of
Theorem 3.3 is complete.
Remark 3.4. Theorem 3.3 obviously provides a refinement of Corollary 3.2. Setting α =
0, ρ2 = ρ1 = 0 in Theorem 3.3, we get Theorem 9.4 of [5].
With the aid of Lemma 2.2, by using the method of our proof of Theorem 3.3, we can prove
the following inclusion relation.
Theorem 3.5. Let µ ≥ 0, −1 ≤ B < A ≤ 1 and λ2 > λ1 ≥ 0, then
Mp (λ2 , µ, A, B) ⊂ Mp (λ1 , µ, A, B).
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
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BAZILEVI Č F UNCTIONS
5
By applying the method of the proof of Theorem 3.13, Theorem 3.6 and Theorem 3.11 in
[17] mutatis mutandis, we can prove the following three results.
Theorem 3.6. Let α ≥ 0, λ > 0 and γ > 0. If f (z) ∈ H satisfies
0
α 0
zf (z)
zf (z) f (z)
zf 00 (z) zf 0 (z)
γ
+λ 1+ 0
−
+α
−1
6= it,
f (z)
z
f (z)
f (z)
f (z)
p
where t is a real number satisfying |t| > λ(λ + 2γ), then
α 0
zf (z) f (z)
> 0, (z ∈ U).
R
f (z)
z
(z ∈ U),
Theorem 3.7. Suppose that α > 0 and 0 ≤ ρ < 1. If f (z) ∈ H satisfies
0
α zf (z) f (z)
Re
> ρ, (z ∈ U),
f (z)
z
then f (z) ∈ M (α, λ, ρ) for |z| < R(λ, ρ), where λ > 0, and
√

 (1+λ−ρ)− (1+λ−ρ)2 −(1−2ρ) , ρ 6= 1 ,
1−2ρ
2
R(λ, ρ) =
 1
,
ρ = 21 .
1+2λ
The bound R(λ, ρ) is the best possible.
For a function f ∈ H, we define the integral operator Fα,δ as follows:
α1
Z
α + δ z δ−1
α
(3.4)
Fα,δ (f ) = Fα,δ (f )(z) =
t f (t) dt
(z ∈ U),
zδ
0
where α and δ are real numbers with α > 0, δ > −α.
Theorem 3.8. Let α and δ be real numbers with α > 0, 0 ≤ ρ < 1, δ > max{−α, −αρ} and
let f (z) ∈ H. If
0
α
π
zf
(z)
f
(z)
arg
≤ β (0 ≤ ρ < 1; 0 < β ≤ 1),
−
ρ
2
f (z)
z
then
0
π
zFα,δ (f ) Fα,δ (f ) α
arg
≤ β,
−
ρ
2
Fα,δ (f )
z
where Fα,δ (f ) is the operator given by (3.4).
Now we derive the Fekete-Szegö inequality for the function class M (α, λ, ρ).
P
n
Theorem 3.9. Suppose that f (z) = z + ∞
n=2 an z ∈ M (α, λ, ρ). Then
|a2 | ≤
2(1 − ρ)
,
(1 + λ)(1 + α)
and for each µ ∈ C, the following bound is sharp
|a3 − µa22 | ≤
2(1 − ρ)
(1 + 2λ)(2 + α)
(1 − ρ)[2λ(3 + α) − (2 + α)(α − 1 + 2µ + 4µλ)] × max 1, 1 +
.
(1 + λ)2 (1 + α)2
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
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6
D ONG G UO AND M ING -S HENG L IU
P
k
Proof. Since f (z) ∈ M (α, λ, ρ), by Definition 1.1, there exists a function p(z) = 1+ +∞
k=1 pk z ∈
P0 , such that
α
0
zf 00 (z) zf 0 (z)
zf (z)
zf 0 (z) f (z)
+λ 1+ 0
−
+α
−1
f (z)
z
f (z)
f (z)
f (z)
= (1 − ρ)p(z) + ρ, z ∈ U.
Equating coefficients, we obtain
1−ρ
p1 ,
(1 + λ)(1 + α)
i
h
(α+2)(α−1)
2
(1
−
ρ)
λ(3
+
α)
−
2
1−ρ
p2 .
a3 =
p2 +
2
2
(1 + λ) (1 + α) (1 + 2λ)(2 + α) 1
(1 + 2λ)(2 + α)
a2 =
Thus, we have
1−ρ
1 2
p2 − p1
a3 −
=
(1 + 2λ)(2 + α)
2
2
(1 − ρ) [2λ(3 + α) − (2 + α)(α − 1) − 2µ(1 + 2λ)(2 + α)] + (1 − ρ)(1 + λ)2 (1 + α)2 2
p1 .
+
2(1 + λ)2 (1 + α)2 (1 + 2λ)(2 + α)
µa22
By Lemma 2.3, we obtain that |a2 | =
1−ρ
|p |
(1+λ)(1+α) 1
≤
2(1−ρ)
,
(1+λ)(1+α)
|a3 − µa22 | ≤ H(x) = A +
and
ABx2
,
4
where x = |p1 | ≤ 2,
A=
2(1 − ρ)
,
(1 + 2λ)(2 + α)
B=
|C| − (1 + λ)2 (1 + α)2
,
(1 + λ)2 (1 + α)2
and
C = (1 + λ)2 (1 + α)2 + (1 − ρ)[2λ(3 + α) − (2 + α)(α − 1 + 2µ + 4µλ)].
So, we have

|c| ≤ (1 + λ)2 (1 + α)2 ,
 H(0) = A,
|a3 − µa22 | ≤
A|C|
 H(2) =
, |c| ≥ (1 + λ)2 (1 + α)2 .
(1+λ)2 (1+α)2
Here equality is attained for the function given by
α
zf 0 (z) f (z)
(3.5)
f (z)
z

λz 1/λ (1 − z 2 )(ρ−1)/λ


R
,

z (1−λ)/λ


t
(1 − t2 )(ρ−1)/λ dt

0




1 + (1 − 2ρ)z 2



,

1 − z2
=


λz 1/λ (1 − z)2(ρ−1)/λ


Rz
,


t(1−λ)/λ (1 − t)2(ρ−1)/λ dt


0





 1 + (1 − 2ρ)z ,
1−z
λ > 0, |c| ≤ (1 + λ)2 (1 + α)2 ,
λ = 0, |c| ≤ (1 + λ)2 (1 + α)2
λ > 0, |c| ≥ (1 + λ)2 (1 + α)2 ,
λ = 0, |c| ≥ (1 + λ)2 (1 + α)2 .
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
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BAZILEVI Č F UNCTIONS
7
Setting λ = 0 in Theorem 3.9, we have the following corollary.
Corollary 3.10. If f (z) ∈ B(α, ρ) given by (1.1), then
2(1 − ρ)
,
1+α
and for each µ ∈ C, the following bound is sharp
2(1 − ρ)
(1 − ρ)(2 + α)(1 − 2µ − α) 2
|a3 − µa2 | ≤
max 1, 1 +
.
2+α
(1 + α)2
|a2 | ≤
Notice that M (0, α, 0) ≡ α − K, and from Theorem 3.9, we have the following corollary.
Corollary 3.11. Let α ≥ 0. If f (z) ∈ α − K given by (1.1). Then
2
|a2 | ≤
,
1+α
and for each µ ∈ C, the following bound is sharp
1
6α
+
2
−
4µ
−
8µα
.
|a3 − µa22 | ≤
max 1, 1 +
1 + 2α
(1 + α)2
Theorem 3.12 (Covering Theorem). Let α ≥ 0, λ ≥ 0 and f (z) ∈ M (α, λ, ρ), then the unit
disk U is mapped by f (z) on a domain that contains the disk |ω| < r1 , where
r1 =
(1 + α)(1 + λ)
.
2(1 + α)(1 + λ) + 2(1 − ρ)
Proof. Let ω0 be any complex number such that f (z) 6= ω0 (z ∈ U), then ω0 6= 0 and (by
Corollary 3.2) the function
1
ω0 f (z)
= z + a2 +
z2 + · · · ,
ω0 − f (z)
ω0
is univalent in U, so that
1
a2 + ≤ 2,
ω0 Therefore, according to Theorem 3.9, we obtain
|ω0 | ≥
(1 + α)(1 + λ)
= r1 .
2(1 + α)(1 + λ) + 2(1 − ρ)
Thus we have completed the proof of Theorem 3.12.
Remark 3.13. Setting α = λ = ρ = 0 in Theorem 3.12, we get the well-known 14 − covering
theorem for the familiar class S ∗ of starlike functions.
If 0 ≤ µ ≤ µ1 and µ is a real number, Theorem 3.9 can be improved as follows.
P
n
Theorem 3.14. Suppose that f (z) = z + ∞
n=2 an z ∈ M (λ, α, ρ) and µ ∈ R. Then
(3.6) |a3 − µa22 | + µ|a2 |2
2(1 − ρ)
≤
(1 + 2λ)(2 + α)
(3.7)
(1 − ρ)[2λ(3 + α) − (2 + α)(α − 1)]
1+
(1 + λ)2 (1 + α)2
|a3 − µa22 | + (µ1 − µ)|a2 |2 ≤
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
2(1 − ρ)
,
(1 + 2λ)(2 + α)
,
0 ≤ µ ≤ µ0 ,
µ0 ≤ µ ≤ µ1 ,
http://jipam.vu.edu.au/
8
D ONG G UO AND M ING -S HENG L IU
and these inequalities are sharp, where
µ0 =
1
2λ − α(2 + α)
(1 + λ)2 (1 + α)2
+
+
,
2 2(1 + 2λ)(2 + α) 2(1 + 2λ)(2 + α)(1 − ρ)
µ1 =
1
2λ − α(2 + α)
(1 + λ)2 (1 + α)2
+
+
.
2 2(1 + 2λ)(2 + α) (1 + 2λ)(2 + α)(1 − ρ)
Proof. From Theorem 3.9, we get
2(1 − ρ)
2(1 − ρ)
(3.8) |a3 − µa22 | ≤
+
(1 + 2λ)(2 + α) (1 + 2λ)(2 + α)
|(1 − ρ)[2λ(3 + α) − (2 + α) [(α − 1) + 2µ(1 + 2λ)]] + (1 + λ)2 (1 + α)2 |
1
·
−
|p1 |2 .
4(1 + λ)2 (1 + α)2
4
Using (3.8) and a2 =
1−ρ
p,
(1+λ)(1+α) 1
if 0 ≤ µ ≤ µ0 , we obtain
2(1 − ρ)
2(1 − ρ)
+
(1 + 2λ)(2 + α) (1 + 2λ)(2 + α)
(1 − ρ)[2λ(3 + α) − (2 + α)(α − 1) − 2µ(1 + 2λ)(2 + α)]
×
|p1 |2
4(1 + λ)2 (1 + α)2
2(1 − ρ)2 [2λ(3 + α) − (2 + α)(α − 1)]
2(1 − ρ)
+
|p1 |2 − µ|a2 |2 .
=
2
2
(1 + 2λ)(2 + α)
4(1 + 2λ)(2 + α)(1 + λ) (1 + α)
|a3 − µa22 | ≤
Hence
|a3 − µa22 | + µ|a2 |2
2(1 − ρ)2 [2λ(3 + α) − (2 + α)(α − 1)]
2(1 − ρ)
+
|p1 |2
2
2
(1 + 2λ)(2 + α)
4(1 + 2λ)(2 + α)(1 + λ) (1 + α)
2(1 − ρ)
(1 − ρ)[2λ(3 + α) − (2 + α)(α − 1)]
1+
,
≤
(1 + 2λ)(2 + α)
(1 + λ)2 (1 + α)2
≤
0 ≤ µ ≤ µ0 .
If µ0 ≤ µ ≤ µ1 , from (3.8), we obtain
|a3 − µa22 |
2(1 − ρ)
2(1 − ρ)
+
(1 + 2λ)(2 + α) (1 + 2λ)(2 + α)
−2(1 + λ)2 (1 + α)2 − (1 − ρ)[2λ(3 + α) − (2 + α)(α − 1 + 2µ + 4µλ)]
×
|p1 |2
4(1 + λ)2 (1 + α)2
2(1 − ρ)
=
− (µ1 − µ)|a2 |2 .
(1 + 2λ)(2 + α)
≤
Therefore
|a3 − µa22 | + (µ1 − µ)|a2 |2 ≤
2(1 − ρ)
,
(1 + 2λ)(2 + α)
µ0 ≤ µ ≤ µ1 .
Here equality is attained for the function given by (3.5), and the proof of Theorem 3.14 is
complete.
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
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BAZILEVI Č F UNCTIONS
9
Theorem 3.15. Let f (z) ∈ H, α ≥ 0, λ ≥ 0 and 0 < k ≤ 1. If
0
zf (z) f (z) α
(3.9) f (z)
z
0
zf (z)
zf 00 (z) zf 0 (z)
+λ 1 + 0
−
+α
−1
− 1 < k,
f (z)
f (z)
f (z)
z ∈ U,
then
k
,
(1 + λ)(1 + α)
and for each µ ∈ C, the following bound is sharp
|a2 | ≤
|a3 − µa22 |

α

k(1 + 2λ)(2 + α) 1 − 2µ − 1+2λ
+
k
≤
max 1,

(1 + 2λ)(2 + α)
2(1 + λ)2 (1 + α)2


2λ
(1+2λ)(2+α) .

Proof. By (3.9), there exists a function p(z) ∈ P0 such that for all z ∈ U
α
0
zf 00 (z) zf 0 (z)
zf (z)
zf 0 (z) f (z)
+λ 1+ 0
−
+α
−1
f (z)
z
f (z)
f (z)
f (z)
2k
+ 1 − k.
=
1 + p(z)
Equating the coefficients, we obtain
k
p1 ,
2(1 + λ)(1 + α)
h
i
k 2 λ(3 + α) − (2+α)(α−1)
2
k
1
(1 + 2λ)(2 + α)a3 = −
p2 − p21 +
p21 .
2
2
2
2
4(1 + λ) (1 + α)
Thus, we have
k
1 2
2
a3 − µa2 = −
p2 − p1
2(1 + 2λ)(2 + α)
2
i
h
k 2 λ(3 + α) − (2+α)(α−1)
−
µ(1
+
2λ)(2
+
α)
2
+
p21 ,
2
4(1 + λ) (1 + α)2 (1 + 2λ)(2 + α)
a2 = −
so that, by Lemma 2.3, we get that |a2 | =
|a3 −
µa22 |
k
|p |
2(1+λ)(1+α) 1
≤
k
,
(1+λ)(1+α)
and
Bx2
≤ H(x) = A +
,
4
where x = |p1 | ≤ 2,
A=
k
,
(1 + 2λ)(2 + α)
B=
k 2 |C|
k
−
2
2
[(1 + λ) (1 + α) ] [(1 + 2λ)(2 + α)]
and
C=
1 − 2µ
α
λ
−
+
.
2
2(1 + 2λ) (1 + 2λ)(2 + α)
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
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10
D ONG G UO AND M ING -S HENG L IU
Therefore
|a3 − µa22 | ≤

 H(0) = A,
 H(2) = Ak(1+2λ)(2+α)|C| ,
(1+λ)2 (1+α)2
Here equality is attained for the function given by

2
λz 1/λ exp(−kz )/(2λ)

Rz

,
2 )/(2λ)
(1−λ)/λ
(−kt

exp
dt

0 t



α 
 1 − kz 2 ,
0
zf (z) f (z)
=
1/λ exp−kz/λ

f (z)
z

R z λz

(1−λ)/λ exp−kt/λ dt ,

t
0




 1 − kz,
|c| ≤
(1+λ)2 (1+α)2
,
k(1+2λ)(2+α)
|c| ≥
(1+λ)2 (1+α)2
.
k(1+2λ)(2+α)
λ > 0, |c| ≤
(1+λ)2 (1+α)2
,
k(1+2λ)(2+α)
λ = 0, |c| ≤
(1+λ)2 (1+α)2
,
k(1+2λ)(2+α)
λ > 0, |c| ≥
(1+λ)2 (1+α)2
,
k(1+2λ)(2+α)
λ = 0, |c| ≥
(1+λ)2 (1+α)2
.
k(1+2λ)(2+α)
This completes the proof of Theorem 3.15.
Setting λ = 0, we get the following corollary.
Corollary 3.16. Let f (z) ∈ H, α ≥ 0 and 0 < k ≤ 1. If
0
zf (z) f (z) α
< k, z ∈ U,
−
1
f (z)
z
then
k
,
(1 + α)
and for each µ ∈ C, the following bound is sharp
k
k(2 + α)
2
|a3 − µa2 | ≤
max 1,
|1 − 2µ − α| .
2+α
2(1 + α)2
|a2 | ≤
Corollary 3.17. Let f (z) ∈ H, α ≥ 0 and 0 < k ≤ 1. If
0
00
zf
(z)
zf
(z)
(1 − α)
+α 1+ 0
− 1 < k, z ∈ U,
f (z)
f (z)
then
k
,
1+α
and for each µ ∈ C the following bound is sharp
(
1 − 2µ +
k(1
+
2α)
k
|a3 − µa22 | ≤
max 1,
2(1 + 2α)
(1 + α)2
|a2 | ≤
)
α 1+2α
.
Setting α = 1 in Corollary 3.17, we have the following corollary.
Corollary 3.18. Let f (z) ∈ H and 0 < k ≤ 1. If
00 zf (z) f 0 (z) < k, z ∈ U,
then
|a2 | ≤
k
,
2
and for each µ ∈ C the following bound is sharp
k
k|4 − 6µ|
2
|a3 − µa2 | ≤ max 1,
.
6
4
J. Inequal. Pure and Appl. Math., 8(1) (2007), Art. 12, 11 pp.
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BAZILEVI Č F UNCTIONS
11
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