TITLE MARCH 2012 - Pakistan Academy of Sciences

M.K. Aouf et al 27
∞
∑ k=2(1 − λ)(k − 1) b k |a k |
and convex in U, we have
<
2(β − 1) − ∑
∞
k=2
|k − (2β − 1)[1 + λ(k − 1)]| b k |a k | < 1.
∞
d k a k z k This completes the proof of Lemma 2.
≺ f(z) (a 1 = 1; z ∈ U ).
k=2
Corollary 1. Let the function f(z) defined by
(1.1) be in the class T(g; λ, β), then
2. MAIN RESULTS
2(β − 1)
|a k | ≤
.
{(1 − λ)(k − 1) + |k − (2β − 1)[1 + λ(k − 1)]|}b
Unless otherwise mentioned, we assume
k
(2.3)
throughout this paper that 0 ≤ λ < 1, β > 1, z ∈
U and g(z) is given by (1.7) with b k+1 ≥ The result is sharp for the function
b k (k ≥ 2).
2(β − 1)
f(z) = z +
.
{(1 − λ)(k − 1) + |k − (2β − 1)[1 + λ(k − 1)]|}b
To prove our main result we need the following
k
lemmas.
(2.4)
∞
Lemma 1. [15]. The sequence {d k } k=1 is a
Let T ∗ (g; λ, β) denote the subclass of functions
subordinating factor sequence if and only if
f(z) ∈ A whose coefficients satisfy the condition
∞
(2.2). We note that T ∗ (g; λ, β) ⊆ T(g, λ, β).
Re 1 + 2 d k z k > 0. (2.1)
Thereom 1. Let f(z) ∈ T ∗ (g; λ, β). Then
k=1
[1 − λ + |3 − 2β − λ(2β − 1)|]b 2
Now, we prove the following lemma which gives
(f ∗ h)(z) ≺ h(z),
2{2(β − 1) + [1 − λ + |3 − 2β − λ(2β − 1)|]b 2 }
a sufficient condition for functions belonging to
(2.5)
the class T(g, λ, β):
for every function h ∈ K, and
Lemma 2. A function f(z) of the form (1.1) is Re{f(z)} > − {2(β − 1) + [1 − λ + |3 − 2β − λ(2β − 1)|]b 2 }
.
[1 − λ + |3 − 2β − λ(2β − 1)|]b
said to be in the class T(g, λ, β) if
2
(2.6)
(1 − λ)(k − 1)
[1−λ+|3−2β−λ(2β−1)|]b
∞
∑k=2 k − (2β − 1)
+
[1 + λ(k − 1)] b k |a k | ≤ 2(β − 1). (2.2)
The constant factor
2
2{2(β−1)+[1−λ+|3−2β−λ(2β−1)|]b 2 }
in the subordination result (2.5) is the best
estimate.
Proof. Assume that the inequality (2.2) holds true.
Then it suffices to show that
Proof. Let f(z) ∈ T ∗ (g; λ, β) and suppose that
∞
z(f ∗ g)′(z)
(1 − λ)(f ∗ g)(z) + λz(f ∗ g)′(z) − 1
h(z) = z + ∑k=2 h k z k ∈ K, then
< 1.
[1 − λ + |3 − 2β − λ(2β − 1)|]b
z(f ∗ g)′(z)
(1 − λ)(f ∗ g)(z) + λz(f ∗ g)′(z) − (2β − 1) 2
(f ∗ h)(z)
2{2(β − 1) + [1 − λ + |3 − 2β − λ(2β − 1)|]b 2 }
∞
[1 − λ + |3 − 2β − λ(2β − 1)|]b 2
We have
=
2{2(β − 1) + [1 − λ + |3 − 2β − λ(2β − 1)|]b 2 } z + h k a k z k .
k=2
(2.7)
z(f ∗ g)′(z)
(1 − λ)(f ∗ g)(z) + λz(f ∗ g)′(z) − 1 Thus, by using Definition 3, the subordination
result holds true if
z(f ∗ g)′(z)
(1 − λ)(f ∗ g)(z) + λz(f ∗ g)′(z) − (2β − 1) ∞
[1 − λ + |3 − 2β − λ(2β − 1)|]b 2
2{2(β − 1) + [1 − λ + |3 − 2β − λ(2β − 1)|]b
∞
∑ k=2(1 − λ)(k − 1) b k |a k ||z| k−1
2 } a k
k=1
≤
2(β − 1) − ∑
∞
k=2
|k − (2β − 1)[1 + λ(k − 1)]| b k |a k ||z| k−1