Journal of Inequalities in Pure and Applied Mathematics

Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 4, Article 134, 2006
SUBORDINATION RESULTS FOR A CLASS OF ANALYTIC FUNCTIONS
DEFINED BY A LINEAR OPERATOR
B.A. FRASIN
D EPARTMENT OF M ATHEMATICS
A L AL -BAYT U NIVERSITY
P.O. B OX : 130095
M AFRAQ , J ORDAN
bafrasin@yahoo.com
Received 21 June, 2005; accepted 01 June, 2006
Communicated by A. Sofo
A BSTRACT. In this paper, we derive several interesting subordination results for certain class of
analytic functions defined by the linear operator L(a, c)f (z) which introduced and studied by
Carlson and Shaffer [2].
Key words and phrases: Analytic functions, Hadamard product, Subordinating factor sequence.
2000 Mathematics Subject Classification. Primary 30C45; Secondary 30A10, 30C80.
1. I NTRODUCTION AND D EFINITIONS
Let A denote the class of functions of the form:
f (z) = z +
(1.1)
∞
X
an z n
n=2
which are analytic in the open unit disc ∆ = {z : |z| < 1} . For two functions f (z) and g(z)
given by
(1.2)
f (z) = z +
∞
X
an z
n
and g(z) = z +
n=2
n=2
their Hadamard product (or convolution) is defined by
(1.3)
(f ∗ g)(z) := z +
∞
X
n=2
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
189-05
∞
X
an c n z n .
cn z n
2
B.A. F RASIN
Define the function φ(a, c; z) by
(1.4)
φ(a, c; z) :=
∞
X
(a)n
n=0
(c)n
(c ∈
/ Z−
0 := {0, −1, −2, . . .}, z ∈ ∆),
z n+1
where (λ)n is the Pochhammer symbol given, in terms of Gamma functions,
(1.5)
(λ)n :=
=
Γ(λ + n)
Γ(λ)
(
1,
n = 0,
λ(λ + 1)(λ + 2) . . . (λ + n − 1), n ∈ N : {1, 2, . . .}.
Corresponding to the function φ(a, c; z), Carlson and Shaffer [2] introduced a linear operator
L(a, c) : A → A by
L(a, c)f (z) := φ(a, c; z) ∗ f (z),
(1.6)
or, equivalently, by
L(a, c)f (z) := z +
∞
X
(a)n
n=1
(c)n
an+1 z n+1
(z ∈ ∆).
Note that L(1, 1)f (z) = f (z), L(2, 1)f (z) = zf 0 (z) and L(3, 1)f (z) = zf 0 (z) + 12 z 2 f 00 (z).
For −1 ≤ α < 1, β ≥ 0, we let L(a, c; α, β) consist of functions f in A satisfying the
condition
aL(a + 1, c)f (z)
aL(a + 1, c)f (z)
(1.7)
Re
− (a − 1) > β − a + α, (z ∈ ∆)
L(a, c)f (z)
L(a, c)f (z)
The family L(a, c; α, β) is of special interest for it contains many well-known as well as
many new classes of analytic univalent functions. For L(1, 1; α, 0), we obtain the family of
starlike functions of order α (0 ≤ α < 1) and L(2, 1; α, 0) is the family of convex functions
of order α (0 ≤ α < 1). For L(1, 1; 0, β) and L(2, 1; 0, β), we obtain the class of uniformly
β- starlike functions and uniformly β- convex functions, respectively, introduced by Kanas
and Winsiowska ([3],[4]) (see also the work of Kanas and Srivastava [5], Goodman ([7],[8]),
Rønning ([10],[11]), Ma and Minda [9] and Gangadharan et al. [6]).
Before we state and prove our main result we need the following definitions and lemmas.
Definition 1.1 (Subordination Principle). Let g(z) be analytic and univalent in ∆. If f (z) is
analytic in ∆, f (0) = g(0), and f (∆) ⊂ g(∆), then we see that the function f (z) is subordinate
to g(z) in ∆, and we write f (z) ≺ g(z).
Definition 1.2 (Subordinating Factor Sequence). A sequence {bn }∞
n=1 of complex numbers is
called a subordinating factor sequence if, whenever f (z) is analytic , univalent and convex in
∆, we have the subordination given by
(1.8)
∞
X
bn an z n ≺ f (z)
(z ∈ ∆, a1 = 1).
n=2
Lemma 1.1 ([14]). The sequence {bn }∞
n=1 is a subordinating factor sequence if and only if
(
)
∞
X
(1.9)
Re 1 + 2
bn z n > 0
(z ∈ ∆).
n=1
J. Inequal. Pure and Appl. Math., 7(4) Art. 134, 2006
http://jipam.vu.edu.au/
S UBORDINATION R ESULTS FOR A C LASS OF A NALYTIC F UNCTIONS
3
Lemma 1.2. If
∞
X
(1.10)
σn (a, c; α, β) |an | ≤ 1 − α
n=2
where, for convenience,
(1.11)
(1 + β)(a)n + [1 − α − a(1 + β)](a)n−1
(c)n−1
(−1 ≤ α < 1; β ≥ 0, n ≥ 2),
σn (a, c; α, β) :=
then f (z) ∈ L(a, c; α, β).
Proof. It suffices to show that
aL(a + 1, c)f (z)
aL(a + 1, c)f (z)
β
− a − Re
− a ≤ 1 − α.
L(a, c)f (z)
L(a, c)f (z)
We have
aL(a + 1, c)f (z)
aL(a + 1, c)f (z)
β
− a − Re
−a
L(a, c)f (z)
L(a, c)f (z)
aL(a + 1, c)f (z)
− a
≤ (1 + β) L(a, c)f (z)
P∞ a(a+1)n−1 −a(a)n−1 (1 + β) n=2
|an | |z|n−1
(c)n−1
≤
P
n−1
(a)n−1
1− ∞
n=2 (c)n−1 |an | |z|
P
(a)n −a(a)n−1
(1 + β) ∞
|an |
n=2
(c)n−1
.
≤
P
(a)n−1
1− ∞
n=2 (c)n−1 |an |
The last expression is bounded above by 1 − α if
∞
X
(1 + β)(a)n + [1 − α − a(1 + β)](a)n−1
(c)n−1
n=2
|an | ≤ 1 − α
and the proof is complete.
Let L? (a, c; α, β) denote the class of functions f (z) ∈ A whose coefficients satisfy the condition (1.10). We note that L? (a, c; α, β) ⊆ L(a, c; α, β).
2. M AIN T HEOREM
Employing the techniques used earlier by Srivastava and Attiya [13], Attiya [1] and Singh
[12], we state and prove the following theorem.
Theorem 2.1. Let the function f (z) defined by (1.1) be in the class L? (a, c; α, β) where −1 ≤
α < 1 ; β ≥ 0; a > 0; c > 0. Also let K denote the familiar class of functions f (z) ∈ A which
are also univalent and convex in ∆. Then
σ2 (a, c; α, β)
(2.1)
(f ∗ g)(z) ≺ g(z)
( z ∈ ∆; g ∈ K),
2[1 − α + σ2 (a, c; α, β)]
and
1 − α + σ2 (a, c; α, β)
(2.2)
Re(f (z)) > −
,
(z ∈ ∆).
σ2 (a, c; α, β)
The constant
σ2 (a,c;α,β)
2[1−α+σ2 (a,c;α,β)]
is the best estimate.
J. Inequal. Pure and Appl. Math., 7(4) Art. 134, 2006
http://jipam.vu.edu.au/
4
B.A. F RASIN
Proof. Let f (z) ∈ L? (a, c; α, β) and let g(z) = z +
(2.3)
P∞
n=2 cn z
n
∈ K. Then
σ2 (a, c; α, β)
σ2 (a, c; α, β)
(f ∗ g)(z) =
2[1 − α + σ2 (a, c; α, β)]
2[1 − α + σ2 (a, c; α, β)]
z+
∞
X
!
an c n z n .
n=2
Thus, by Definition 1.2, the assertion of our theorem will hold if the sequence
∞
σ2 (a, c; α, β)
(2.4)
an
2[1 − α + σ2 (a, c; α, β)]
n=1
is a subordinating factor sequence, with a1 = 1. In view of Lemma 1.1, this will be the case if
and only if
(
)
∞
X
σ2 (a, c; α, β)
(2.5)
Re 1 + 2
an z n > 0
(z ∈ ∆).
2[1
−
α
+
σ
(a,
c;
α,
β)]
2
n=1
Now
)
∞
X
σ2 (a, c; α, β)
Re 1 +
an z n
1 − α + σ2 (a, c; α, β) n=1
σ2 (a, c; α, β)
= Re 1 +
z
1 − α + σ2 (a, c; α, β)
)
∞
X
1
σ2 (a, c; α, β)an z n
+
1 − α + σ2 (a, c; α, β) n=1
σ2 (a, c; α, β)
≥1−
r
1 − α + σ2 (a, c; α, β)
)
∞
X
1
n
−
σn (a, c; α, β)an r .
1 − α + σ2 (a, c; α, β) n=1
(
Since σn (a, c; α, β) is an increasing function of n (n ≥ 2)
σ2 (a, c; α, β)
1−
r
1 − α + σ2 (a, c; α, β)
)
∞
X
1
−
σn (a, c; α, β)an rn
1 − α + σ2 (a, c; α, β) n=1
>1−
σ2 (a, c; α, β)
1−α
r−
r
1 − α + σ2 (a, c; α, β)
1 − α + σ2 (a, c; α, β)
(|z| = r)
> 0.
Thus (2.5) holds true in ∆. This proves the inequality
P∞ n (2.1). The inequality (2.2) follows by
z
taking the convex function g(z) = 1−z = z + n=2 z in (2.1). To prove the sharpness of the
σ2 (a,c;α,β)
constant 2[1−α+σ
, we consider the function f0 (z) ∈ L? (a, c; α, β) given by
2 (a,c;α,β)]
(2.6)
f0 (z) = z −
1−α
z2
σ2 (a, c; α, β)
(−1 ≤ α < 1; β ≥ 0).
Thus from (2.1), we have
(2.7)
σ2 (a, c; α, β)
z
f0 (z) ≺
.
2[1 − α + σ2 (a, c; α, β)]
1−z
J. Inequal. Pure and Appl. Math., 7(4) Art. 134, 2006
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S UBORDINATION R ESULTS FOR A C LASS OF A NALYTIC F UNCTIONS
It can easily verified that
(2.8)
min Re
σ2 (a, c; α, β)
1
f0 (z)
=−
2[1 − α + σ2 (a, c; α, β)]
2
This shows that the constant
σ2 (a,c;α,β)
2[1−α+σ2 (a,c;α,β)]
is best possible.
5
(z ∈ ∆),
Corollary 2.2. Let the function f (z) defined by (1.1) be in the class L? (1, 1; α, β) and satisfy
the condition
∞
X
(2.9)
[n(1 + β) − (α + β)] |an | ≤ 1 − α
n=2
then
β+2−α
(f ∗ g)(z) ≺ g(z)
2(β + 3 − 2α)
(−1 ≤ α < 1; β ≥ 0; z ∈ ∆; g ∈ K)
(2.10)
and
Re(f (z)) > −
(2.11)
The constant
β+2−α
2(β+3−2α)
β + 3 − 2α
,
β+2−α
(z ∈ ∆).
is the best estimate.
Corollary 2.3. Let the function f (z) defined by (1.1) be in the class L? (1, 1; α, 0) and satisfy
the condition
∞
X
(2.12)
(n − α) |an | ≤ 1 − α,
n=2
then
2−α
(f ∗ g)(z) ≺ g(z)
6 − 4α
(2.13)
(z ∈ ∆; g ∈ K)
and
Re(f (z)) > −
(2.14)
The constant
2−α
6−4α
3 − 2α
,
2−α
(z ∈ ∆).
is the best estimate.
Putting α = 0 in Corollary 2.3, we obtain
Corollary 2.4 ([12]). Let the function f (z) defined by (1.1) be in the class L? (1, 1; 0, 0) and
satisfy the condition
∞
X
(2.15)
n |an | ≤ 1
n=2
then
(2.16)
1
(f ∗ g)(z) ≺ g(z)
3
(z ∈ ∆; g ∈ K)
and
3
Re(f (z)) > − ,
2
The constant 1/3 is the best estimate.
(2.17)
J. Inequal. Pure and Appl. Math., 7(4) Art. 134, 2006
(z ∈ ∆).
http://jipam.vu.edu.au/
6
B.A. F RASIN
Corollary 2.5. Let the function f (z) defined by (1.1) be in the class L? (2, 1; α, β) and satisfy
the condition
∞
X
(2.18)
n[n(1 + β) − (α + β)] |an | ≤ 1 − α,
n=2
then
β+2−α
(f ∗ g)(z) ≺ g(z)
2β + 5 − 3α
(−1 ≤ α < 1; β ≥ 0; z ∈ ∆; g ∈ K)
(2.19)
and
Re(f (z)) > −
(2.20)
The constant
β+2−α
2β+5−3α
2β + 5 − 3α
,
2(β + 2 − α)
(z ∈ ∆).
is the best estimate.
Corollary 2.6. Let the function f (z) defined by (1.1) be in the class L? (2, 1; α, 0) and satisfy
the condition
∞
X
(2.21)
n(n − α) |an | ≤ 1 − α,
n=2
then
2−α
(f ∗ g)(z) ≺ g(z)
5 − 3α
(2.22)
(z ∈ ∆; g ∈ K)
and
Re(f (z)) > −
(2.23)
The constant
2−α
5−3α
5 − 3α
,
2(2 − α)
(z ∈ ∆).
is the best estimate.
Putting α = 0 in Corollary 2.6, we obtain
Corollary 2.7. Let the function f (z) defined by (1.1) be in the class L? (2, 1; 0, 0) and satisfy
the condition
∞
X
(2.24)
n2 |an | ≤ 1
n=2
then
(2.25)
2
(f ∗ g)(z) ≺ g(z)
5
(z ∈ ∆; g ∈ K)
and
(2.26)
Re(f (z)) >
−5
,
4
(z ∈ ∆).
The constant 2/5 is the best estimate.
J. Inequal. Pure and Appl. Math., 7(4) Art. 134, 2006
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S UBORDINATION R ESULTS FOR A C LASS OF A NALYTIC F UNCTIONS
7
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J. Inequal. Pure and Appl. Math., 7(4) Art. 134, 2006
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