IJMMS 2004:36, 1937–1942
PII. S0161171204309051
http://ijmms.hindawi.com
© Hindawi Publishing Corp.
COEFFICIENT ESTIMATES FOR RUSCHEWEYH DERIVATIVES
MASLINA DARUS and AJAB AKBARALLY
Received 5 September 2003
We consider functions f , analytic in the unit disc and of the normalized form f (z) =
n
z+ ∞
n=2 an z . For functions f ∈ R̄δ (β), the class of functions involving the Ruscheweyh
derivatives operator, we give sharp upper bounds for the Fekete-Szegö functional |a3 −µa22 |.
2000 Mathematics Subject Classification: 30C45.
1. Introduction. Let S denote the class of normalized analytic univalent functions f
defined by
f (z) = z +
∞
an zn
(1.1)
n=2
in the unit disc D = {z : |z| < 1}. Suppose that
zf (z)
> 0, z ∈ D ,
S ∗ = f ∈ S : Re
f (z)
zf (z) < βπ , z ∈ D
S ∗ (β) = f ∈ S : arg
f (z) 2
(1.2)
are classes of starlike and strongly starlike functions of order β (0 < β ≤ 1), respectively. Note that S ∗ (β) ⊂ S ∗ for 0 < β < 1 and S ∗ (1) = S ∗ [5]. Kanas [2] introduced the
subclass R̄δ (β) of function f ∈ S as the following.
Definition 1.1. For δ ≥ 0, β ∈ (0, 1], a function f normalized by (1.1) belongs to
R̄δ (β) if, for z ∈ D − {0} and D δ f (z) ≠ 0, the following holds:
δ
arg z D f (z) ≤ βπ ,
δ
D f (z) 2
(1.3)
where D δ f denotes the generalized Ruscheweyh derivative which was originally defined
as the following.
Definition 1.2 [6]. Let D n f and f be defined by (1.1). Then for n ∈ N ∪ {0},
D n f (z) =
z
∗ f (z),
(1 − z)n+1
(1.4)
where ∗ denotes the Hadamard product of two analytic functions and N is a set of
natural numbers.
1938
M. DARUS AND A. AKBARALLY
Later in [1], Al-Amiri generalized the Ruscheweyh derivative D δ for real numbers
δ ≥ −1 as a Hadamard product of the functions f and z/(1 − z)δ+1 .
Note that R̄0 (β) = S ∗ (β) for each β ∈ (0, 1] and R̄0 (1) = S ∗ . In this note, we obtain
sharp estimates for |a2 |, |a3 | and the Fekete-Szegö functional for the class R̄δ (β). For
the subclass S ∗ , sharp upper bounds for the functional |a3 − µa22 | have been obtained
for all real µ [3, 4].
2. Preliminary results. In proving our results, we will need the following lemmas.
However, we first denote P to be the class of analytic functions with positive real part
in D.
∞
Lemma 2.1. Let p ∈ P and let it be of the form p(z) = 1+ n=1 cn zn with Rep(z) > 0.
Then
(i) |cn | ≤ 2 for n ≥ 1,
(ii) |c2 − c12 /2| ≤ 2 − |c1 |2 /2.
Lemma 2.2. Let δ ≥ 0 and β ∈ (0, 1]. If f ∈ R̄δ (β) and is given by (1.1), then
a2 ≤ 2β ,
δ+1

2β


if β ≤

 (δ + 2)(δ + 1)
a3 ≤


6β2


if β ≥
(δ + 2)(δ + 1)
1
,
3
(2.1)
1
.
3
Proof. Let F (z) = D δ f (z) = z + A2 z2 + A3 z3 + · · · . Since f ∈ R̄δ (β) and D δ f (z) ∈
S (β), then
∗
zF (z)
= p β (z)
F (z)
(2.2)
β
z 1 + 2A2 z + 3A3 z2 + · · ·
= 1 + c1 z + c2 z 2 + · · · ,
2
3
z + A2 z + A3 z + · · ·
(2.3)
and so
which implies that
β(β − 1) 2
c1 + βA2 c1 + A3 z3 + · · · .
z + 2A2 z2 + 3A3 z3 + · · · = z + βc1 + A2 z2 + βc2 +
2
(2.4)
Equating the coefficients, we have
A2 = βc1 ,
c12
3
β
A3 =
c2 −
+ β2 c12 .
2
2
4
(2.5)
(2.6)
COEFFICIENT ESTIMATES FOR RUSCHEWEYH DERIVATIVES
1939
Now, for δ ≥ −1, D δ f has the Taylor expansion
D δ f (z) = z +
∞
Γ (n + δ)
an zn ,
(n
−
1)!Γ (1 + δ)
n=2
z ∈ D,
(2.7)
where Γ (n + δ) denotes Euler’s functions with
Γ (n + δ) = δ(δ + 1) · · · (δ + n − 1)Γ (δ).
(2.8)
Then
z + A2 z 2 + A 3 z 3 + · · · = z +
Γ (2 + δ)
Γ (3 + δ)
a2 z2 +
a3 z3 + · · · .
Γ (1 + δ)
2Γ (1 + δ)
(2.9)
Equating the coefficients in (2.9), we have
a2
Γ (2 + δ)
= a2 (δ + 1) = A2 .
Γ (1 + δ)
(2.10)
Then, from (2.5), we obtain
a2 =
βc1
.
δ+1
(2.11)
It follows that from Lemma 2.1(i)
a2 ≤ 2β ,
δ+1
(2.12)
whereas the coefficient of z3 in (2.9) is
a3
Γ (3 + δ)
(δ + 1)(δ + 2)
= a3
= A3 .
2Γ (1 + δ)
2
(2.13)
From (2.6), we obtain
c12
β
2
3 2 2
a3 =
c2 −
+ β c1 .
(δ + 1)(δ + 2) 2
2
4
(2.14)
It follows from Lemma 2.1(ii) that
a3 ≤
2 c 1 β
3 2
2
2−
+ β2 c1 ,
(δ + 1)(δ + 2) 2
2
4
(2.15)
that is,
a3 ≤









2β
(δ + 2)(δ + 1)
if β ≤
1
,
3
6β2
(δ + 2)(δ + 1)
if β ≥
1
.
3
(2.16)
1940
M. DARUS AND A. AKBARALLY
3. Results. We first consider the functional |a3 − µa22 | for complex µ.
Theorem 3.1. Let f ∈ R̄δ (β) and β ∈ (0, 1]. Then for µ complex,
a3 − µa2 ≤
2
β 3(δ + 1) − 2µ(δ + 2) 2β
max 1,
.
(δ + 1)(δ + 2)
(δ + 1)
(3.1)
For each µ there is a function in R̄δ (β) such that equality holds.
Proof. From (2.11) and (2.14), we write
c2
βc1 2
β
3
2
c2 − 1 + β2 c12 − µ
,
(δ + 1)(δ + 2) 2
2
4
δ+1
c2
1
β2 3(δ + 1) − 2µ(δ + 2) 2
=
β c2 − 1
+
c1 .
(δ + 1)(δ + 2)
2
2(δ + 1)2 (δ + 2)
a3 − µa22 =
(3.2)
It follows from (3.2) and Lemma 2.1(ii) that
2 2 c 1 β 3(δ + 1) − 2µ(δ + 2) 2
β
c1 ,
2−
+
(δ + 1)(δ + 2)
2
2(δ + 1)2 (δ + 2)
2
β 3(δ + 1) − 2µ(δ + 2) − β(δ + 1) 2
2β
c 1 ,
=
+
(δ + 1)(δ + 2)
2(δ + 1)2 (δ + 2)
a3 − µa2 ≤
2
(3.3)
which on using Lemma 2.1(i), that is, |c1 | ≤ 2, gives
a3 − µa2 ≤
2





2β
(δ + 1)(δ + 2)
2
β 6(δ + 1) − 4µ(δ + 2) 



(δ + 1)2 (δ + 2)
if κ(δ) ≤ β(δ + 1),
(3.4)
if κ(δ) ≥ β(δ + 1),
where κ(δ) = |β2 (3(δ + 1) − 2µ(δ + 2))|.
Equality is attained for functions in R̄δ (β) given by
z D δ f (z)
1 + z2 β
,
=
D δ f (z)
1 − z2
z D δ f (z)
1+z β
,
=
D δ f (z)
1−z
respectively.
We next consider the cases where µ is real and prove the following.
(3.5)
COEFFICIENT ESTIMATES FOR RUSCHEWEYH DERIVATIVES
1941
Theorem 3.2. Let f ∈ R̄δ (β) and β ∈ (0, 1]. Then for µ real,
a3 − µa2 ≤
2
 2
β 6(δ + 1) − 4µ(δ + 2)





(δ + 1)2 (δ + 2)





2β

(δ + 1)(δ + 2)





 β2 4µ(δ + 2) − 6(δ + 1)



(δ + 1)2 (δ + 2)
if µ ≤
if
(6β − 2)(δ + 1)
,
4β(δ + 2)
(6β − 2)(δ + 1)
(2 + 6β)(δ + 1)
≤µ≤
,
4β(δ + 2)
4β(δ + 2)
if µ ≥
(2 + 6β)(δ + 1)
.
4β(δ + 2)
(3.6)
For each µ, there is a function in R̄δ (β) such that equality holds.
Proof. Here we consider two cases.
Case (i): µ ≤ 3(δ + 1)/2(δ + 2).
In this case, (3.2) and Lemma 2.1(ii) give
2 c1 β2 6(δ + 1) − 4µ(δ + 2) β
c 1 2 ,
2−
+
(δ + 1)(δ + 2)
2
4(δ + 1)2 (δ + 2)
2β
β2 6(δ + 1) − 4µ(δ + 2) − 2β(δ + 1) c1 2 ,
=
+
2
(δ + 1)(δ + 2)
4(δ + 1) (δ + 2)
a3 − µa2 ≤
2
(3.7)
and so, using the fact that |c1 | ≤ 2, we obtain
a3 − µa2 ≤
2
 2
β 6(δ + 1) − 4µ(δ + 2)





(δ + 1)2 (δ + 2)





2β
(δ + 1)(δ + 2)
if µ ≤
(6β − 2)(δ + 1)
,
4β(δ + 2)
(6β − 2)(δ + 1)
3(δ + 1)
if
≤µ≤
.
4β(δ + 2)
2(δ + 2)
(3.8)
Equality is attained on choosing c1 = c2 = 2 and c1 = 0, c2 = 2, respectively, in (3.2).
Case (ii): µ ≥ 3(δ + 1)/2(δ + 2).
It follows from (3.2) and Lemma 2.1(ii) that
2 c1 β2 4µ(δ + 2) − 6(δ + 1) β
c 1 2 ,
2−
+
2
(δ + 1)(δ + 2)
2
4(δ + 1) (δ + 2)
β2 4µ(δ + 2) − 6(δ + 1) − 2β(δ + 1) 2β
c1 2 ,
=
+
2
(δ + 1)(δ + 2)
4(δ + 1) (δ + 2)
a3 − µa2 ≤
2
(3.9)
and so, using the fact that |c1 | ≤ 2, we obtain
a3 − µa2 ≤
2





2β
(δ + 1)(δ + 2)


β2 4µ(δ + 2) − 6(δ + 1)


(δ + 1)2 (δ + 2)
if
3(δ + 1)
(6β + 2)(δ + 1)
≤µ≤
,
2(δ + 2)
4β(δ + 2)
if µ ≤
(6β + 2)(δ + 1)
.
4β(δ + 2)
(3.10)
Equality is attained on choosing c1 = 0, c2 = 2 and c1 = 2i, c2 = −2, respectively, in
(3.2). Thus the proof is complete.
1942
M. DARUS AND A. AKBARALLY
Theorem 3.3. Let f ∈ R̄δ (β) and let it be given by (1.1). Then
a3 − a2 ≤
2β
(δ + 1)(δ + 2)
if β ≤
3(δ + 1)
.
5δ + 1
(3.11)
Proof. Write
a3 − a2 ≤ a3 − 2 a2 + 2 a2 2 − a2 .
2
3
3
(3.12)
Then since (6β − 2)(δ + 1)/4β(δ + 2) ≤ 2/3 for β ≤ 3(δ + 1)/(5δ + 1), it follows from
Theorem 3.2 that
a3 − a2 ≤
2 2 2β
+ a2 − a2 = λ(x),
(δ + 1)(δ + 2) 3
(3.13)
where x = |a2 | ∈ [0, 2β/(δ + 1)]. Since λ(x) attains its maximum value at x = 0, the
theorem follows. This is sharp.
Acknowledgments. The work presented here was supported by IRPA Grant 09-0202-0080-EA208. The author would like to thank all the referees for their suggestions
regarding the contents of the note.
References
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[2]
[3]
[4]
[5]
[6]
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Maslina Darus: School of Mathematical Sciences, Faculty of Science and Technology, Universiti
Kebangsaan Malaysia, 43600 Bangi, Selangor Darul Ehsan, Malaysia
E-mail address: maslina@pkrisc.cc.ukm.my
Ajab Akbarally: School of Mathematical Sciences, Faculty of Science and Technology, Universiti
Kebangsaan Malaysia, 43600 Bangi, Selangor Darul Ehsan, Malaysia