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Subordination properties for a certain class of analytic functions defined by the Salagean operator
Applied Mathematics Letters 22 (2009) 1581–1585
Contents lists available at ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
Subordination properties for a certain class of analytic functions defined by the Salagean operator M.K. Aouf Faculty of Science, Mansoura University, Mansoura 35516, Egypt
article
abstract
info
Article history: Received 6 May 2008 Received in revised form 9 May 2009 Accepted 12 May 2009
In this paper we derive several subordination results for a certain class of analytic functions defined by the Salagean operator. © 2009 Elsevier Ltd. All rights reserved.
Keywords: Analytic Subordinating factor sequence Salagean operator Hadamard product
1. Introduction Let A denote the class of functions f (z ) of the form: f (z ) = z +
∞ X
ak z k
(1.1)
k=2
which are analytic in the open unit disc U = {z : z ∈ C and |z | < 1}. We also denote by K the class of functions f (z ) ∈ A that are convex in U. Salagean [1] introduced the following operator which is popularly known as the Salagean derivative operator: D0 f (z ) = f (z ), 0
D1 f (z ) = Df (z ) = zf (z ) and, in general, Dn f (z ) = D(Dn−1 f (z ))
(n ∈ N = {1, 2, . . .}).
We can easily find from (1.1) that Dn f (z ) = z +
∞ X
kn ak z k
(f ∈ A; n ∈ N0 = N ∪ {0}).
(1.2)
k=2
Let Gn (λ, b) denote the subclass of A consisting of functions f (z ) which satisfy:
Re 1 +
1 b
(1 − λ)
Dn f (z ) z
0 + λ(Dn f (z )) − 1 > 0
E-mail address: [email protected]. 0893-9659/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2009.05.005
(1.3)
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M.K. Aouf / Applied Mathematics Letters 22 (2009) 1581–1585
or which satisfy the following inequality:
(1 − λ) Dn f (z ) + λ(Dn f (z ))0 − 1 z < 1 (z ∈ U ; λ ≥ 0; n ∈ N0 ; b ∈ C \ {0}). (1 − λ) Dn f (z ) + λ(Dn f (z ))0 − 1 + 2b
(1.4)
z
We note that:
f (z ) 1 0 (1 − λ) + λf (z ) − 1 > 0; z ∈ U ; (i) G0 (λ, b) = G(λ, b) = f ∈ A : Re 1 +
(1.5)
1 Dn f (z ) (ii) Gn (0, b) = Gn (b) = f (z ) ∈ A : Re 1 + −1 > 0; z ∈ U ;
(1.6)
b
z
b
(iii) Gn (1, b) = Rn (b) = f (z ) ∈ A : Re 1 +
1 b
z
0
(D f (z )) − 1 n
> 0; z ∈ U ;
(1.7)
1 f (z ) −1 > 0; z ∈ U ; (iv) G0 (0, b) = G(b) = f (z ) ∈ A : Re 1 +
(1.8)
1 0 (v) G0 (1, b) = R(b) = f (z ) ∈ A : Re 1 + f (z ) − 1 > 0; z ∈ U ;
(1.9)
b
z
b
f (z ) (vi) G0 (0, 1 − α) = Gα = f (z ) ∈ A : Re > α; 0 ≤ α < 1; z ∈ U ;
(1.10)
o n 0 (vii) G0 (1, 1 − α) = Rα = f (z ) ∈ A : Re f (z ) > α; 0 ≤ α < 1; z ∈ U .
(1.11)
z
The class R(b) was studied by Halim [2], the class Gα was studied by Chen [3,4] and the class Rα was studied by Ezrohi [5]. Definition 1 (Hadamard Product or Convolution). Given two functions f and g in the class A, where f (z ) is given by (1.1) and g (z ) is given by g (z ) = z +
∞ X
bk z k ;
(1.12)
k=2
the Hadamard product (or convolution) f ∗ g is defined (as usual) by
(f ∗ g )(z ) = z +
∞ X
ak bk z k = (g ∗ f )(z )
(z ∈ U ).
k=2
Definition 2 (Subordination Principal). For two functions f and g, analytic in U, we say that the function f (z ) is subordinate to g (z ) in U, and write f (z ) ≺ g (z ) (z ∈ U ), if there exists a Schwarz function w(z ), which (by definition) is analytic in U with w(0) = 0 and |w(z )| < 1, such that f (z ) = g (w(z )) (z ∈ U ). Indeed it is known that f (z ) ≺ g (z )(z ∈ U ) ⇒ f (0) = g (0) and
f (U ) ⊂ g (U ).
Furthermore, if the function g is univalent in U, then we have the following equivalence [6, p. 4]: f (z ) ≺ g (z )(z ∈ U ) ⇔ f (0) = g (0) and
f (U ) ⊂ g (U ).
Definition 3 (Subordinating Factor Sequence). A sequence {bk }∞ k=1 of complex numbers is said to be a subordinating factor sequence if, whenever f (z ) of the form (1.1) is analytic, univalent and convex in U, we have the subordination given by ∞ X
ak bk z k ≺ f (z )
(z ∈ U ; a1 = 1).
k=1
2. Main result To prove our main result we need the following lemmas. Lemma 1 ([7]). The sequence {bk }∞ k=1 is a subordinating factor sequence if and only if
( Re 1 + 2
∞ X k=1
) bk z
k
> 0,
(z ∈ U ).
(1.13)
M.K. Aouf / Applied Mathematics Letters 22 (2009) 1581–1585
1583
Now, we prove the following lemma which gives a sufficient condition for functions belonging to the class Gn (λ, b). Lemma 2. Let the function f (z ) which is defined by (1.1) satisfy the following condition: ∞ X [1 + λ(k − 1)]kn |ak | ≤ |b| ,
(λ ≥ 0; n ∈ N0 ; b ∈ C \ {0});
(2.1)
k=2
then f (z ) ∈ Gn (λ, b). Proof. Suppose that the inequality (2.1) holds. Then we have for z ∈ U,
n n (1 − λ) D f (z ) + λ(Dn f (z ))0 − 1 − (1 − λ) D f (z ) + λ(Dn f (z ))0 + 2b − 1 z z ∞ ∞ X X [1 + λ(k − 1)] kn ak z k−1 − 2b + [1 + λ(k − 1)] kn ak z k−1 = k =2 k=2 ( ) ∞ ∞ X X n k−1 n k−1 [1 + λ(k − 1)] k |ak | |z | [1 + λ(k − 1)] k |ak | |z | ≤ − 2 |b| − k =2
( ≤2
k=2
∞ X
) [1 + λ(k − 1)] kn |ak | − |b|
≤ 0,
k=2
which shows that f (z ) belongs to the class Gn (λ, b).
Let G∗n (λ, b) denote the class of functions f (z ) ∈ A whose coefficients satisfy the condition (2.1). We note that G∗n (λ, b) ⊆ Gn (λ, b). Employing the technique used earlier by Attiya [8] and Srivastava and Attiya [9], we prove: Theorem 1. Let f (z ) ∈ G∗n (λ, b). Then
(1 + λ)2n (f ∗ g )(z ) ≺ g (z ), 2 [(1 + λ)2n + |b|]
(z ∈ U )
(2.2)
for every function g in K , and Re (f (z )) > −
The constant factor
[(1 + λ)2n + |b|]
(1 + λ)2n
(1+λ)2n 2[(1+λ)2n +|b|]
,
(z ∈ U ).
(2.3)
in the subordination result (2.2) cannot be replaced by a larger one.
P∞ Proof. Let f (z ) ∈ G∗n (λ, b) and let g (z ) = z + k=2 ck z k ∈ K . Then we have ! ∞ X (1 + λ)2n (1 + λ)2n k (f ∗ g )(z ) = z+ ak ck z . 2 [(1 + λ)2n + |b|] 2 [(1 + λ)2n + |b|] k=2
(2.4)
Thus, by Definition 3, the subordination result (2.2) will hold true if the sequence
(1 + λ)2n ak 2 [(1 + λ)2n + |b|]
∞ (2.5) k=1
is a subordinating factor sequence, with a1 = 1. In view of Lemma 1, this is equivalent to the following inequality:
( Re 1 +
∞ X k=1
(1 + λ)2n ak z k (1 + λ)2n + |b|
Now, since
Ψ (k) = [1 + λ(k − 1)] kn
) > 0,
(z ∈ U ).
(2.6)
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M.K. Aouf / Applied Mathematics Letters 22 (2009) 1581–1585
is an increasing function of k(k ≥ 2), we have
( Re 1 +
∞ X k=1
(1 + λ)2n ak z k (1 + λ)2n + |b|
)
(
∞ X 1 (1 + λ)2n = Re 1 + z + (1 + λ)2n ak z k (1 + λ)2n + |b| (1 + λ)2n + |b| k=2
)
≥1−
∞ X 1 (1 + λ)2n [1 + λ(k − 1)] kn |ak | r k r− n n (1 + λ)2 + |b| (1 + λ)2 + |b| k=2
>1−
|b| (1 + λ)2n r− r =1−r >0 (1 + λ)2n + |b| (1 + λ)2n + |b|
(|z | = r < 1) ,
where we have also made use of assertion (2.1) of Lemma 2. Thus (2.6) holds true in P U. This proves the inequality (2.2). The ∞ z k inequality (2.3) follows from (2.2) by taking the convex function g (z ) = 1− = z + k=2 z . To prove the sharpness of the z constant
(1+λ)2n , 2[(1+λ)2n +|b|]
f0 ( z ) = z −
we consider the function f0 (z ) ∈ G∗n (λ, b) given by
| b| z2. (1 + λ)2n
(2.7)
Thus from (2.2), we have z (1 + λ)2n f0 ( z ) ≺ , 2 [(1 + λ)2n + |b|] 1−z
(z ∈ U ).
(2.8)
Moreover, it can easily be verified for the function f0 (z ) given by (2.7) that
min Re |z |≤r
1 (1 + λ)2n f ( z ) =− . 0 2 [(1 + λ)2n + |b|] 2
This shows that the constant
(1+λ)2n 2[(1+λ)2n +|b|]
is the best possible.
(2.9)
Putting n = 0 in Theorem 1, we have: Corollary 1. Let the function f (z ) defined by (1.1) be in the class G(λ, b) and suppose that g (z ) ∈ K . Then
(1 + λ) 2 (1 + λ + |b|)
(f ∗ g )(z ) ≺ g (z ) (z ∈ U )
(2.10)
and Re (f (z )) > −
1 + λ + |b| 1+λ
(z ∈ U ).
(2.11)
1+λ The constant factor 2(1+λ+| in the subordination result (2.10) cannot be replaced by a larger one. b|)
Putting n = λ = 0 in Theorem 1, we have: Corollary 2. Let the function f (z ) defined by (1.1) be in the class G(b) and suppose that g (z ) ∈ K . Then 1 2 (1 + |b|)
(f ∗ g )(z ) ≺ g (z ) (z ∈ U )
(2.12)
and Re (f (z )) > − (1 + |b|)
(z ∈ U ).
(2.13)
1
The constant factor 2(1+|b|) in the subordination result (2.12) cannot be replaced by a larger one. Putting n = λ = 0 and b = 1 − α, 0 ≤ α < 1, in Theorem 1, we have: Corollary 3. Let the function f (z ) defined by (1.1) be in the class Gα and suppose that g (z ) ∈ K . Then 1 2 (2 − α)
(f ∗ g )(z ) ≺ g (z ) (z ∈ U )
(2.14)
M.K. Aouf / Applied Mathematics Letters 22 (2009) 1581–1585
1585
and Re (f (z )) > − (2 − α) The constant factor
1 2(2−α)
(z ∈ U ).
(2.15)
in the subordination result (2.14) cannot be replaced by a larger one.
Putting n = 0, λ = 1 and b = 1 − α, 0 ≤ α < 1, in Theorem 1, we have: Corollary 4. Let the function f (z ) defined by (1.1) be in the class Rα and suppose that g (z ) ∈ K . Then 2 2 (3 − α)
(f ∗ g )(z ) ≺ g (z ) (z ∈ U )
(2.16)
and Re (f (z )) > −
3−α 2
(z ∈ U ).
(2.17)
1 The constant factor 3−α in the subordination result (2.16) cannot be replaced by a larger one.
Acknowledgement The author is grateful to the referees for their comments and suggestions. References [1] G.S. Salagean, Subclasses of univalent functions, in: Lecture Notes in Mathematics, 1013, Springer-Verlag, 1983, pp. 362–372. [2] S.A. Halim, On a class of functions of complex n o order, Tamkang J. Math. 30 (2) (1999) 147–153. f (z ) z
> α , Tamkang J. Math. 5 (1974) 231–234. n o f (z ) > α , Bull. Inst. Math. Acad. Sinica 3 (1) (1975) 65–70. [4] M.-P. Chen, On the regular functions satisfying Re z [5] T.G. Ezrohi, Certain estimates in special classes of univalent functions regular in the circle |z | < 1, Dopovidi Akademiji Nauk Ukrajins Koji RSR (1965) [3] M.-P. Chen, On functions satisfying Re
984–988. [6] S.S. Miller, P.T. Mocanu, Differential Subordinations: Theory and Applications, in: Series on Monographs and Textbooks in Pure and Appl. Math., vol. 255, Marcel Dekker, Inc, New York, 2000. [7] H.S. Wilf, Subordinating factor sequence for convex maps of the unit circle, Proc. Amer. Math. Soc. 12 (1961) 689–693. [8] A.A. Attiya, On some application of a subordination theorems, J. Math. Anal. Appl. 311 (2005) 489–494. [9] H.M. Srivastava, A.A. Attiya, Some subordination results associated with certain subclass of analytic functions, J. Inequal. Pure Appl. Math. 5 (2004) 1–6. Art. 82.
Contents lists available at ScienceDirect
Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml
Subordination properties for a certain class of analytic functions defined by the Salagean operator M.K. Aouf Faculty of Science, Mansoura University, Mansoura 35516, Egypt
article
abstract
info
Article history: Received 6 May 2008 Received in revised form 9 May 2009 Accepted 12 May 2009
In this paper we derive several subordination results for a certain class of analytic functions defined by the Salagean operator. © 2009 Elsevier Ltd. All rights reserved.
Keywords: Analytic Subordinating factor sequence Salagean operator Hadamard product
1. Introduction Let A denote the class of functions f (z ) of the form: f (z ) = z +
∞ X
ak z k
(1.1)
k=2
which are analytic in the open unit disc U = {z : z ∈ C and |z | < 1}. We also denote by K the class of functions f (z ) ∈ A that are convex in U. Salagean [1] introduced the following operator which is popularly known as the Salagean derivative operator: D0 f (z ) = f (z ), 0
D1 f (z ) = Df (z ) = zf (z ) and, in general, Dn f (z ) = D(Dn−1 f (z ))
(n ∈ N = {1, 2, . . .}).
We can easily find from (1.1) that Dn f (z ) = z +
∞ X
kn ak z k
(f ∈ A; n ∈ N0 = N ∪ {0}).
(1.2)
k=2
Let Gn (λ, b) denote the subclass of A consisting of functions f (z ) which satisfy:
Re 1 +
1 b
(1 − λ)
Dn f (z ) z
0 + λ(Dn f (z )) − 1 > 0
E-mail address: [email protected]. 0893-9659/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2009.05.005
(1.3)
1582
M.K. Aouf / Applied Mathematics Letters 22 (2009) 1581–1585
or which satisfy the following inequality:
(1 − λ) Dn f (z ) + λ(Dn f (z ))0 − 1 z < 1 (z ∈ U ; λ ≥ 0; n ∈ N0 ; b ∈ C \ {0}). (1 − λ) Dn f (z ) + λ(Dn f (z ))0 − 1 + 2b
(1.4)
z
We note that:
f (z ) 1 0 (1 − λ) + λf (z ) − 1 > 0; z ∈ U ; (i) G0 (λ, b) = G(λ, b) = f ∈ A : Re 1 +
(1.5)
1 Dn f (z ) (ii) Gn (0, b) = Gn (b) = f (z ) ∈ A : Re 1 + −1 > 0; z ∈ U ;
(1.6)
b
z
b
(iii) Gn (1, b) = Rn (b) = f (z ) ∈ A : Re 1 +
1 b
z
0
(D f (z )) − 1 n
> 0; z ∈ U ;
(1.7)
1 f (z ) −1 > 0; z ∈ U ; (iv) G0 (0, b) = G(b) = f (z ) ∈ A : Re 1 +
(1.8)
1 0 (v) G0 (1, b) = R(b) = f (z ) ∈ A : Re 1 + f (z ) − 1 > 0; z ∈ U ;
(1.9)
b
z
b
f (z ) (vi) G0 (0, 1 − α) = Gα = f (z ) ∈ A : Re > α; 0 ≤ α < 1; z ∈ U ;
(1.10)
o n 0 (vii) G0 (1, 1 − α) = Rα = f (z ) ∈ A : Re f (z ) > α; 0 ≤ α < 1; z ∈ U .
(1.11)
z
The class R(b) was studied by Halim [2], the class Gα was studied by Chen [3,4] and the class Rα was studied by Ezrohi [5]. Definition 1 (Hadamard Product or Convolution). Given two functions f and g in the class A, where f (z ) is given by (1.1) and g (z ) is given by g (z ) = z +
∞ X
bk z k ;
(1.12)
k=2
the Hadamard product (or convolution) f ∗ g is defined (as usual) by
(f ∗ g )(z ) = z +
∞ X
ak bk z k = (g ∗ f )(z )
(z ∈ U ).
k=2
Definition 2 (Subordination Principal). For two functions f and g, analytic in U, we say that the function f (z ) is subordinate to g (z ) in U, and write f (z ) ≺ g (z ) (z ∈ U ), if there exists a Schwarz function w(z ), which (by definition) is analytic in U with w(0) = 0 and |w(z )| < 1, such that f (z ) = g (w(z )) (z ∈ U ). Indeed it is known that f (z ) ≺ g (z )(z ∈ U ) ⇒ f (0) = g (0) and
f (U ) ⊂ g (U ).
Furthermore, if the function g is univalent in U, then we have the following equivalence [6, p. 4]: f (z ) ≺ g (z )(z ∈ U ) ⇔ f (0) = g (0) and
f (U ) ⊂ g (U ).
Definition 3 (Subordinating Factor Sequence). A sequence {bk }∞ k=1 of complex numbers is said to be a subordinating factor sequence if, whenever f (z ) of the form (1.1) is analytic, univalent and convex in U, we have the subordination given by ∞ X
ak bk z k ≺ f (z )
(z ∈ U ; a1 = 1).
k=1
2. Main result To prove our main result we need the following lemmas. Lemma 1 ([7]). The sequence {bk }∞ k=1 is a subordinating factor sequence if and only if
( Re 1 + 2
∞ X k=1
) bk z
k
> 0,
(z ∈ U ).
(1.13)
M.K. Aouf / Applied Mathematics Letters 22 (2009) 1581–1585
1583
Now, we prove the following lemma which gives a sufficient condition for functions belonging to the class Gn (λ, b). Lemma 2. Let the function f (z ) which is defined by (1.1) satisfy the following condition: ∞ X [1 + λ(k − 1)]kn |ak | ≤ |b| ,
(λ ≥ 0; n ∈ N0 ; b ∈ C \ {0});
(2.1)
k=2
then f (z ) ∈ Gn (λ, b). Proof. Suppose that the inequality (2.1) holds. Then we have for z ∈ U,
n n (1 − λ) D f (z ) + λ(Dn f (z ))0 − 1 − (1 − λ) D f (z ) + λ(Dn f (z ))0 + 2b − 1 z z ∞ ∞ X X [1 + λ(k − 1)] kn ak z k−1 − 2b + [1 + λ(k − 1)] kn ak z k−1 = k =2 k=2 ( ) ∞ ∞ X X n k−1 n k−1 [1 + λ(k − 1)] k |ak | |z | [1 + λ(k − 1)] k |ak | |z | ≤ − 2 |b| − k =2
( ≤2
k=2
∞ X
) [1 + λ(k − 1)] kn |ak | − |b|
≤ 0,
k=2
which shows that f (z ) belongs to the class Gn (λ, b).
Let G∗n (λ, b) denote the class of functions f (z ) ∈ A whose coefficients satisfy the condition (2.1). We note that G∗n (λ, b) ⊆ Gn (λ, b). Employing the technique used earlier by Attiya [8] and Srivastava and Attiya [9], we prove: Theorem 1. Let f (z ) ∈ G∗n (λ, b). Then
(1 + λ)2n (f ∗ g )(z ) ≺ g (z ), 2 [(1 + λ)2n + |b|]
(z ∈ U )
(2.2)
for every function g in K , and Re (f (z )) > −
The constant factor
[(1 + λ)2n + |b|]
(1 + λ)2n
(1+λ)2n 2[(1+λ)2n +|b|]
,
(z ∈ U ).
(2.3)
in the subordination result (2.2) cannot be replaced by a larger one.
P∞ Proof. Let f (z ) ∈ G∗n (λ, b) and let g (z ) = z + k=2 ck z k ∈ K . Then we have ! ∞ X (1 + λ)2n (1 + λ)2n k (f ∗ g )(z ) = z+ ak ck z . 2 [(1 + λ)2n + |b|] 2 [(1 + λ)2n + |b|] k=2
(2.4)
Thus, by Definition 3, the subordination result (2.2) will hold true if the sequence
(1 + λ)2n ak 2 [(1 + λ)2n + |b|]
∞ (2.5) k=1
is a subordinating factor sequence, with a1 = 1. In view of Lemma 1, this is equivalent to the following inequality:
( Re 1 +
∞ X k=1
(1 + λ)2n ak z k (1 + λ)2n + |b|
Now, since
Ψ (k) = [1 + λ(k − 1)] kn
) > 0,
(z ∈ U ).
(2.6)
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M.K. Aouf / Applied Mathematics Letters 22 (2009) 1581–1585
is an increasing function of k(k ≥ 2), we have
( Re 1 +
∞ X k=1
(1 + λ)2n ak z k (1 + λ)2n + |b|
)
(
∞ X 1 (1 + λ)2n = Re 1 + z + (1 + λ)2n ak z k (1 + λ)2n + |b| (1 + λ)2n + |b| k=2
)
≥1−
∞ X 1 (1 + λ)2n [1 + λ(k − 1)] kn |ak | r k r− n n (1 + λ)2 + |b| (1 + λ)2 + |b| k=2
>1−
|b| (1 + λ)2n r− r =1−r >0 (1 + λ)2n + |b| (1 + λ)2n + |b|
(|z | = r < 1) ,
where we have also made use of assertion (2.1) of Lemma 2. Thus (2.6) holds true in P U. This proves the inequality (2.2). The ∞ z k inequality (2.3) follows from (2.2) by taking the convex function g (z ) = 1− = z + k=2 z . To prove the sharpness of the z constant
(1+λ)2n , 2[(1+λ)2n +|b|]
f0 ( z ) = z −
we consider the function f0 (z ) ∈ G∗n (λ, b) given by
| b| z2. (1 + λ)2n
(2.7)
Thus from (2.2), we have z (1 + λ)2n f0 ( z ) ≺ , 2 [(1 + λ)2n + |b|] 1−z
(z ∈ U ).
(2.8)
Moreover, it can easily be verified for the function f0 (z ) given by (2.7) that
min Re |z |≤r
1 (1 + λ)2n f ( z ) =− . 0 2 [(1 + λ)2n + |b|] 2
This shows that the constant
(1+λ)2n 2[(1+λ)2n +|b|]
is the best possible.
(2.9)
Putting n = 0 in Theorem 1, we have: Corollary 1. Let the function f (z ) defined by (1.1) be in the class G(λ, b) and suppose that g (z ) ∈ K . Then
(1 + λ) 2 (1 + λ + |b|)
(f ∗ g )(z ) ≺ g (z ) (z ∈ U )
(2.10)
and Re (f (z )) > −
1 + λ + |b| 1+λ
(z ∈ U ).
(2.11)
1+λ The constant factor 2(1+λ+| in the subordination result (2.10) cannot be replaced by a larger one. b|)
Putting n = λ = 0 in Theorem 1, we have: Corollary 2. Let the function f (z ) defined by (1.1) be in the class G(b) and suppose that g (z ) ∈ K . Then 1 2 (1 + |b|)
(f ∗ g )(z ) ≺ g (z ) (z ∈ U )
(2.12)
and Re (f (z )) > − (1 + |b|)
(z ∈ U ).
(2.13)
1
The constant factor 2(1+|b|) in the subordination result (2.12) cannot be replaced by a larger one. Putting n = λ = 0 and b = 1 − α, 0 ≤ α < 1, in Theorem 1, we have: Corollary 3. Let the function f (z ) defined by (1.1) be in the class Gα and suppose that g (z ) ∈ K . Then 1 2 (2 − α)
(f ∗ g )(z ) ≺ g (z ) (z ∈ U )
(2.14)
M.K. Aouf / Applied Mathematics Letters 22 (2009) 1581–1585
1585
and Re (f (z )) > − (2 − α) The constant factor
1 2(2−α)
(z ∈ U ).
(2.15)
in the subordination result (2.14) cannot be replaced by a larger one.
Putting n = 0, λ = 1 and b = 1 − α, 0 ≤ α < 1, in Theorem 1, we have: Corollary 4. Let the function f (z ) defined by (1.1) be in the class Rα and suppose that g (z ) ∈ K . Then 2 2 (3 − α)
(f ∗ g )(z ) ≺ g (z ) (z ∈ U )
(2.16)
and Re (f (z )) > −
3−α 2
(z ∈ U ).
(2.17)
1 The constant factor 3−α in the subordination result (2.16) cannot be replaced by a larger one.
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