Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2009, Article ID 807943, 12 pages
doi:10.1155/2009/807943
Research Article
Univalence of Certain Linear Operators Defined by
Hypergeometric Function
R. Aghalary and A. Ebadian
Department of Mathematics, Faculty of Science, Urmia University, Urmia, Iran
Correspondence should be addressed to A. Ebadian, a.ebadian@urmia.ac.ir
Received 11 January 2009; Accepted 22 April 2009
Recommended by Vijay Gupta
The main object of the present paper is to investigate univalence and starlikeness of certain integral
operators, which are defined here by means of hypergeometric functions. Relevant connections of
the results presented here with those obtained in earlier works are also pointed out.
Copyright q 2009 R. Aghalary and A. Ebadian. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction and Preliminaries
Let H denote the class of all analytic functions f in the unit disk D {z ∈ C : |z| < 1}. For
n ≥ 0, a positive integer, let
An ∞
nk
,
f ∈ H : fz z ank z
1.1
k1
with A1 : A, where A is referred to as the normalized analytic functions in the unit disc. A
function f ∈ A is called starlike in D if fD is starlike with respect to the origin. The class of
all starlike functions is denoted by S∗ : S∗ 0. For α < 1, we define
S∗ α zf z
> α, z ∈ D ,
f ∈ A : Re
fz
1.2
and it is called the class of all starlike functions of order α. Clearly, S∗ α ⊆ S∗ for 0 < α < 1.
For functions fj z, given by
fj z ∞
ak,j zk ,
k0
j 1, 2 ,
1.3
2
Journal of Inequalities and Applications
we define the Hadamard product or convolution of f1 z and f2 z by
∞
ak,1 ak,2 zk : f2 ∗ f1 z.
f1 ∗ f2 z :
1.4
k0
An interesting subclass of S the class of all analytic univalent functions is denoted by
Uα, μ, λ and is defined by
U α, μ, λ μ
μ1
z
z
α
f z − 1 < λ, z ∈ D ,
f ∈ A : 1 − α
fz
fz
1.5
where 0 < α ≤ 1, 0 ≤ μ < αn, and λ > 0.
The special case of this class has been studied by Ponnusamy and Vasundhra 1 and
Obradović et al. 2.
For a,b,c ∈ C and c / 0,-1,-2,. . ., the Gussian hypergeometric series Fa,b;c;z is defined
as
Fa, b; c; z ∞
an bn zn
cn
n0
n!
,
z ∈ D,
1.6
where an aa 1a 2 · · · a n − 1 and a0 1. It is well-known that Fa, b; c; z is
analytic in D. As a special case of the Euler integral representation for the hypergeometric
function, we have
F1, b; c; z Γc
ΓbΓc − b
1
1 b−1
t 1 − tc−b−1 dt,
0 1 − tz
z ∈ D, Re c > Re b > 0.
1.7
Now by letting
φa; c; z : F1, a; c; z,
1.8
zφa; c 1; z cφa; c; z − cφa; c 1; z.
1.9
it is easily seen that
For f ∈ A, Owa and Srivastava 3 introduced the operator Ωλ : A → A defined by
Ωλ fz Γ2 − λ λ d
z
Γ1 − λ dz
z
ft
0 z
− tλ
dt,
2, 3, 4 . . .,
λ /
1.10
Journal of Inequalities and Applications
3
which is extensions involving fractional derivatives and fractional integrals. Using definition
of φa; c; z : F1, a; c; z we may write
Ωλ fz zφ2; 2 − λ; z ∗ fz.
1.11
This operator has been studied by Srivastava et al. 4 and Srivastava and Mishra 5.
k
Also for λ < 1, Re α > 0, and fz z ∞
k2 ak z , let us define the function F by
Fz : λz 1−λ
α
1
t1/α−2 ftzdt
0
1.12
∞
z 1 − λ
ak
zk .
−
1α 1
k
k2
This operator has been investigated by many authors such as Trimble 6, and
Obradović et al. 7.
If we take
∞
ψ m, γ, z 1 1 − m
1
zk ,
−
1γ
1
k
k2
1.13
then we can rewrite operator F defined by 1.11 as
fz
Fz z ψλ, α, z ∗
.
z
1.14
From the definition of ψm, γ, z it is easy to check that
1
1 z
zψ m, γ, z ψ m, γ, z 1 1 − m
.
γ
γ
1−z
1.15
For f ∈ Uα, μ, λ with z/fzμ ∗φa; c1; z / 0 for all z ∈ D we define the transform
G by
Gz z
z/fz
μ
1/μ
1
,
∗ φa; c 1; z
1.16
where a, c ∈ C and c /
0, −1, −2, . . . .
Also for f ∈ Uα, μ, λ with z/fzμ ∗ ψm, γ, z /
0 for all z ∈ D we define the
transform H by
Hz z
where m < 1 and γ /
0; Re γ ≥ 0.
z/fz
μ
1
∗ ψm, γ, z
1/μ
,
1.17
4
Journal of Inequalities and Applications
In this investigation we aim to find conditions on α, μ, λ such that f ∈ Uα, μ, λ
implies that the function f to be starlike. Also we find conditions on α, μ, λ, m, γ, a, c for each
f ∈ Uα, μ, λ; the transforms G and H belong to Uα, μ, λ and S∗ .
For proving our results we need the following lemmas.
Lemma 1.1 cf. Hallenbeck and Ruscheweyh 8. Let hz be analytic and convex univalent in
the unit disk D with h0 1. Also let
gz 1 b1 z b2 z2 · · ·
1.18
be analytic in D. If
gz zg z
≺ hz
c
0,
z ∈ U; c /
1.19
then
gz ≺ ψz c
zc
z
tc−1 htdt ≺ hz
z ∈ D; Re c ≥ 0; c / 0.
1.20
0
and ψz is the best dominant of 1.20.
Lemma 1.2 cf. Ruscheweyh and Stankiewicz 8. If f andg are analytic and F and G are
convex functions such that f ≺ F, g ≺ G, then f ∗ g ≺ F ∗ G.
Lemma 1.3 cf. Ruscheweyh and Sheil-Small 9. Let F and G be univalent convex functions in
D. Then the Hadamard product F ∗ G is also univalent convex in D.
2. Main Results
We follow the method of proof adopted in 1, 10.
Theorem 2.1. Let n be positive integer with n ≥ 2. Also let n1/2n < α ≤ 1 and n1−α < μ < αn.
If fz z an1 zn1 · · · belongs to Uα, μ, λ, Then f ∈ S∗ γ whenever 0 < λ ≤ λα, μ, n, γ,
where
⎧
⎪
αn − μ 2α 1 − γ − 1
⎪
⎪
⎪
⎪
,
⎪ ⎨ αn − μ
2 μ2 2α1 − γ − 1
λ α, μ, n, γ :
⎪
⎪
⎪ αn − μ
1 − γ ⎪
⎪
⎪
,
⎩
n μγ − μ
0≤γ ≤
μ − n1 − α
,
μ1 n
2.1
μ − n1 − α
< γ < 1.
μ1 n
Proof. Let us define
pz z
fz
μ
.
2.2
Journal of Inequalities and Applications
5
Since f ∈ Uα, μ, λ, we have
1 − α
z
fz
μ
μ1
z
α
α
f z pz − zf z
fz
μ
1 αn − μ an1 zn · · ·
2.3
1 λωz,
where ωz is an analytic function with |ωz| < 1 and ω0 ω 0 · · · ωn−1 0 0. By
Schwarz lemma, we have |ωz| ≤ |z|n . By 2.3, it is easy to check that
pz 1 −
μλ
α
1
ωtz
dt,
μ/α1
0t
2.4
zf z
1 λωz
1 − α α
1
.
fz
1 − μλ/α 0 ωtz/ tμ/α1 dt
Therefore
1
1−γ
zf z
−γ
fz
1
α − μλ 0 ωtz/tμ/α1 dt α/ 1 − γ 1 λωz
α − 1 − αγ / 1 − γ
.
1
α α − μλ 0 ωtz/tμ/α1 dt
2.5
We need to show that f ∈ S∗ γ. To do this, according to a well-known result 9 and 2.5 it
suffices to show that
1
α− 1−αγ / 1− γ α − μλ 0 ωtz/tμ/α1 dt α/ 1−γ 1λωz
− iT,
/
1
α α−μλ 0 ωtz/tμ/α1 dt
T ∈ R,
2.6
which is equivalent to
⎡
1 ⎤
ωz μ αγ 1 − α /α − i 1 − γ T 0 ωtz/tμ/α1 dt
⎦/
λ⎣
− 1,
α 1 − γ 1 iT T ∈ R.
2.7
Suppose that Bn denote the class of all Schwarz functions ω such that ω0 ω 0 · · · ωn−1 0 0, and let
1 ωz μ αγ 1 − α /α − i 1 − γ T 0 ωtz/tμ/α1 dt ,
M
sup
α 1 − γ 1 iT z∈D,ω∈Bn ,T ∈R
2.8
6
Journal of Inequalities and Applications
then, f ∈ S∗ γ if λM ≤ 1. This observation shows that it suffices to find M. First we notice
that
⎫
⎧
2 ⎪
2
⎪
⎨ 1 μ/ n − μ /α
αγ 1 − α /α2 1 − γ T 2 ⎬
.
M ≤ sup
⎪
α 1 − γ 1 T 2 ⎭
⎩
T ∈R ⎪
2.9
Define φ : 0, ∞ → R by
2
2 αn − μ μ αγ 1 − α 1 − γ α2 x
φx .
√
αn − μ α 1 − γ 1 x
2.10
Differentiating φ with respect to x, we get
2
3
3 √
2 1
x/2
μ
αn
−
μ
α
1
−
γ
αγ 1 − α 1 − γ α2 x
φ x 2 2
αn − μ α2 1 − γ 1 x
√
2
2 αn − μ α 1 − γ
αn − μ μ αγ 1 − α 1 − γ α2 x /2 1 x
−
.
2 2
αn − μ α2 1 − γ 1 x
2.11
Case 1. Let 0 < γ < μ − n1 − α/μ1 n. Then we see that φ has its only critical point in the
positive real line at
x0 1
2
1 − γ α2
#
$
2
μ2 2α1 − γ − 1
2
.
− αγ 1 − γ
2
αn − μ
2.12
Furthermore, we can see that φ x > 0 for 0 ≤ x < x0 and φ x < 0 for x > x0 . Hence φx
attains its maximum value at x0 and
2
αn − μ μ2 2α 1 − γ − 1
φx ≤ φx0 2
2 .
αn − μ
2α 1 − γ − 1 αn − μ μ2 2α1 − γ − 1
2.13
Case 2. Let γ > μ − n1 − α/μ1 n, then it is easy to see that φ x < 0, and so φx attains
its maximum value at 0 and
n μγ − μ
,
αn − μ 1 − γ
φx ≤ φ0 ∀x ≥ 0.
Now the required conclusion follows from 2.13 and 2.14.
2.14
Journal of Inequalities and Applications
7
By putting γ 0 in Theorem 2.1 we obtain the following result.
Corollary 2.2. Let n be the positive integer with n ≥ 2. Also let n 1/2n < α ≤ 1 and n1 − α <
n1
∗
μ < αn. If fz
z an1 z · · · belongs to Uα, μ, λ, then f ∈ S whenever 0 < λ ≤ αn −
√
μ 2α − 1/ αn − μ2 μ2 2α − 1.
Remark 2.3. Taking α 1, μ 1 in Theorem 2.1 and Corollary 2.2 we get results of 10.
We follow the method ofproof adopted in 11.
Theorem 2.4. Let n ≥ 2, a /
0, c ∈ C with Re c ≥ 0 /
c and the function ϕz 1 b1 z b2 z2 · · ·
0 be univalent convex in D. If fz z an1 zn1 · · · ∈ Uα, μ, λ and φa; c; z defined
with bn /
by 1.8 satisfy the conditions
z
fz
μ
∗ φa; c 1; z /
0 ∀z ∈ D,
2.15
φa; c; z ≺ ϕz,
then the transform G defined by 1.16 has the following:
1 G ∈ Uα, μ, λ|bn ||c|/|c n|,
2 G ∈ S∗ whenever
√
|c n| αn − μ 2α − 1
0<λ≤
.
2
2
|bn ||c| αn − μ μ 2α − 1
2.16
Proof. From the definition of G we obtain
z
Gz
μ
z
fz
μ
∗ φa; c 1; z.
2.17
Differentiating z/Gzμ shows that
μ μ
μ1
z
z
z
z
μ
−μ
G z.
Gz
Gz
Gz
2.18
μ μ
z
z
∗ φa; c 1; z z
∗ φa; c 1; z .
z
fz
fz
2.19
It is easy to see that
From 1.9 and 2.19 we deduce that
μ
μ
μ
z
z
z
z
∗ φa; c 1; z c
∗ φa; c; z − c
∗ φa; c 1; z,
fz
fz
fz
2.20
8
Journal of Inequalities and Applications
or
μ μ
μ
z
z
z
c
c
∗ φa; c; z.
z
Gz
Gz
fz
2.21
Let us define
z
pz 1 − α
Gz
μ
μ1
z
α
G z : 1 dn zn · · · ,
Gz
2.22
then pz is analytic in D, with p0 1 and p 0 · · · pn−1 0 0. Combining 2.18 with
2.21, one can obtain
pz αc
1
μ
z
Gz
μ
αc
−
μ
z
fz
μ
∗ φa; c; z.
2.23
Differentiating pz yields
zp z μ μ z
z
αc
αc
z
z
−
∗ φa; c; z.
1
μ
Gz
μ
fz
2.24
In view of 2.21, 2.23, and 2.24, we obtain
μ
μ
z
z
αc2
αc
cpz zp z c 1 −
∗ φa; c; z
μ
Gz
μ fz
μ μ αc
z
z
αc
1
z
z
−
∗ φa; c; z
μ
Gz
μ
fz
μ
αc
z
c 1
∗ φa; c; z
μ
fz
2.25
μ
μ αc2
αc
z
z
−
∗ φa; c; z −
∗ φa; c; z
μ fz
μ
fz
$
#
μ
μ μ1
z
z
z
c
∗ φa; c; z − cα
−
f z ∗ φa; c; z
fz
fz
fz
#
$
μ
μ1
z
z
c 1 − α
α
f z ∗ φa; c; z.
fz
fz
Hence
$
#
μ
μ1
1 z
z
pz zp z 1 − α
α
f z ∗ φa; c; z.
c
fz
fz
2.26
Journal of Inequalities and Applications
9
Since 1 λzn and ϕz 1 b1 z b2 z2 · · · are convex and
1 − α
z
fz
μ
μ1
z
α
f z ≺ 1 λzn ,
fz
φa; c; z ≺ ϕz,
2.27
by using Lemmas 1.2 and 1.3, from 2.26 we deduce that
1
pz zp z ≺ 1 bn λzn .
c
2.28
It now follows from Lemma 1.1 that
pz ≺ ψz c
zc
z
tc−1 1 bn λzn dt.
2.29
λbn c n
z ,
cn
2.30
0
Therefore
pz ≺ 1 and the result follows from the last subordination and Corollary 2.2.
It is well-known that see, 12 if c, a > 0 and c ≥ max{2, a}, then φa; c; z is univalent
convex function in D. So if we take ϕz φa; c; z in the Theorem 2.4, we obtain the
following.
Corollary 2.5. For n ≥ 2, c, a > 0, and c ≥ max{2, a}, let the function fz z an zn1 · · · ∈
Uα, μ, λ and φa; c; z defined by 1.8 satisfy the condition
z
fz
μ
∗ φa; c 1; z /
0 ∀z ∈ D.
2.31
Then the transform G defined by 1.16 has the following:
1 G ∈ Uα, μ, λ|an |c/|cn |c n;
2 G ∈ S∗ whenever
√
c n|cn | αn − μ 2α − 1
0<λ≤
.
2
|an |c αn − μ μ2 2α − 1
2.32
By putting a c on the 1.8, we get φc; c; z 1/1 − z which is evidently convex.
So by taking ϕz 1/1 − z on Theorem 2.4 we have the following.
10
Journal of Inequalities and Applications
Corollary 2.6. For n ≥ 2, c ∈ C with Re c ≥ 0 /
c, let the function fz zan zn1 · · · ∈ Uα, μ, λ
and φa; c; z defined by 1.8 satisfy the condition
z
fz
μ
∗ φa; c 1; z /
∀z ∈ D.
2.33
√
|c n| αn − μ 2α − 1
0 < λ ≤ .
2
2
|c| αn − μ μ 2α − 1
2.34
Then the transform G defined by 1.16 has the following:
1 G ∈ Uα, μ, λ|c|/|c n|;
2 G ∈ S∗ whenever
Remark 2.7. Taking α 1 and μ 1 on Corollary 2.6, we get a result of 11.
By putting c 1 − M and a 2 on Theorem 2.10 we obtain the following.
k
Corollary 2.8. Let n ≥ 2 and ϕz 1 ∞
/ 0 be univalent convex function in D.
k1 bk z with bn Also let M ∈ C with Re M < 1 and fz z an1 zn1 · · · ∈ Uα, μ, λ, satisfy
Ω
M
z
fz
μ 0
/
z ∈ D,
2.35
and let G be the function which is defined by
Gz z
1
μ ΩM z/fz
1/μ
.
2.36
If
φ2; 1 − M; z ≺ ϕz,
2.37
then we have the following:
1 G ∈ Uα, μ, λ|bn ||1 − M|/|n 1 − M|;
2 G ∈ S∗ whenever
√
|1 − M n| αn − μ 2α − 1
0<λ≤
.
2
|bn ||1 − M| αn − μ μ2 2α − 1
2.38
Remark 2.9. We note that if M < −1, then φ2; 1 − M; z is convex function, and so we can
replace ϕz with φ2; 1 − M; z in Corollary 2.8 to get other new results.
Journal of Inequalities and Applications
11
In 13, Pannusamy and Sahoo have also considered the class Uα, μ, λ for the case
α 1 with μ n.
Theorem 2.10. For m < 1, γ /
0; Re γ > 0, n ≥ 2, let fz z an1 zn1 · · · ∈ Uα, μ, λ and
ψm, γ, z defined by 1.13 satisfy the condition
z
fz
μ
∗ ψ m, γ, z /
0 ∀z ∈ D.
2.39
Then the transform H defined by 1.17 has the following:
1 H ∈ Uα, μ, λ1 − m/|1 nγ|;
2 H ∈ S∗ whenever
√
1 nγ αn − μ 2α − 1
.
0<λ≤
2
1 − m αn − μ μ2 2α − 1
2.40
Proof. Let us define
z
pz 1 − α
Hz
μ
α
z
Hz
μ1
H z,
2.41
then pz is analytic in D, with p0 1 and p 0 · · · pn−1 0 0. Using the same method
as on Theorem 2.4 we get
#
z
pz γzp z 1 − α
fz
μ
$ μ1
z
z
α
f z ∗ 1 1 − m
.
fz
1−z
2.42
Since 1 λzn and hz 1 1 − mz/1 − z are convex,
μ
μ1
z
z
α
f z ≺ 1 λzn .
1 − α
fz
fz
2.43
Using Lemmas 1.2 and 1.3, from 2.42 it yields
pz γzp z ≺ 1 − mλzn .
2.44
It now follows from Lemma 1.1 that
pz ≺
1
γz1/γ
z
0
t1/γ −1 1 1 − mλtn dt.
2.45
12
Journal of Inequalities and Applications
Therefore
− m n
pz − 1 ≤ λ1
|z| ,
1 nγ 2.46
and the result follows from 2.46 and Corollary 2.2.
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