Volume 10 (2009), Issue 3, Article 77, 6 pp.
STARLIKENESS CONDITIONS FOR AN INTEGRAL OPERATOR
PRAVATI SAHOO AND SAUMYA SINGH
D EPARTMENT OF M ATHEMATICS
BANARAS H INDU U NIVERSITY
BANARAS 221 005, I NDIA
pravatis@yahoo.co.in
bhu.saumya@gmail.com
Received 30 June, 2009; accepted 26 August, 2009
Communicated by R.N. Mohapatra
A BSTRACT. Let for fixed n ∈ N, Σn denotes the class of function of the following form
∞
1 X
f (z) = +
ak z k ,
z
k=n
∗
which are analytic in the punctured open unit disk ∆ = {z ∈ C : 0 < |z| < 1}. In the present
paper we defined and studied an operator in
µ
µ1
Z c + 1 − µ z f (t)
c+µ
F (z) =
t
dt ,
for f ∈ Σn and c + 1 − µ > 0.
z c+1
t
0
Key words and phrases: Meromorphic functions; Differential subordination; Starlike functions, Convex functions.
2000 Mathematics Subject Classification. 30C45, 30D30.
1. I NTRODUCTION
Let H(∆) = H denote the class of analytic functions in ∆, where ∆ = {z ∈ C : |z| < 1}.
For a fixed positive integer n and a ∈ C, let
H[a, n] = {f (z) ∈ H : f (z) = a + an z n + an+1 z n+1 + · · · },
with H0 = H[0, 1]. Let An be the class of analytic functions defined on the unit disc with the
normalized conditions f (0) = 0 = f 0 (0) − 1, that is f ∈ An has the form
(1.1)
f (z) = z +
∞
X
ak z k ,
(z ∈ ∆ and n ∈ N).
k=n+1
Let A1 = A and let S be the class of all functions f ∈ A which are univalent in ∆.
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P RAVATI S AHOO
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A function f ∈ A is said to be in S ∗ iff f (∆) is a starlike domain with respect to the origin.
Let for 0 ≤ α < 1,
zf 0 (z)
∗
S (α) = f ∈ A : Re
> α, z ∈ ∆
f (z)
be the Tclass of all starlike functions of order α. So S ∗ (0) ≡ S ∗ . We denote Sn∗ (α) ≡
S ∗ (α) An for n ∈ N.
A function f ∈ A is said to be in C iff f (∆) is a convex domain. Let for 0 ≤ α < 1,
zf 00 (z)
C(α) = f ∈ A : Re 1 +
> α, z ∈ ∆
f (z)
be the class of convex functions of order α. So C(0) ≡ C.
Let for fixed n ∈ N, Σn denote the class of meromorphic functions of the following form
∞
1 X
ak z k ,
(1.2)
f (z) = +
z k=n
which are analytic in the punctured open unit disk ∆∗ = {z : z ∈ C and 0 < |z| < 1} =
∆ − {0}. Let Σ0 = Σ.
A function f ∈ Σ is said to be meromorphically starlike of order α in ∆∗ if it satisfies the
condition
0 zf (z)
− Re
> α, (0 ≤ α < 1; z ∈ ∆∗ ).
f (z)
We denote by Σ∗ (α), the subclass of
T Σ consisting of all meromorphically starlike functions of
order α in ∆∗ and Σ∗n (α) ≡ Σ∗ (α) Σn for n ∈ N.
We say that f (z) is subordinate to g(z) and f ≺ g in ∆ or f (z) ≺ g(z) (z ∈ ∆) if there
exists a Schwarz function w(z), which (by definition) is analytic in ∆ with w(0) = 0 and
|w(z)| < 1, such that f (z) = g(w(z)), z ∈ ∆. Furthermore, if the function g is univalent in ∆,
f (z) ≺ g(z) (z ∈ ∆) ⇔ f (0) = g(0) andf (∆) ⊂ g(∆).
In the present paper, for f (z) ∈ Σn , we define and study a generalized operator I[f ]
µ
µ1
Z c + 1 − µ z f (t)
c+µ
(1.3)
I[f ] = F (z) =
t dt , (c + 1 − µ > 0, z ∈ ∆∗ ),
z c+1
t
0
which is similar to the Alexander transform when c = µ = 1 and is similar to Bernardi transformation when µ = 1 and c > 0.
2. M AIN R ESULTS
For our main results we need the following lemmas.
T
Lemma 2.1 (Goluzin [5]). If f ∈ An S ∗ , then
n
f (z) 2
1
Re
> .
z
2
This inequality is sharp with extremal function f (z) =
z
2
(1−z n ) n
.
Lemma 2.2 ([9]). Let u and v denote complex variables, u = α + iρ, v = σ + iδ and let Ψ(u, v)
be a complex valued function that satisfies the following conditions:
(i) Ψ(u, v) is continuous in a domain Ω ⊂ C2 ;
(ii) (1, 0) ∈ Ω and Re(Ψ(1, 0)) > 0;
2
(iii) Re(Ψ(iρ, σ)) ≤ 0 whenever (iρ, σ) ∈ Ω, σ ≤ − 1+ρ
and ρ, σ are real.
2
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I NTEGRAL O PERATOR
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If p(z) ∈ H[a, n] is a function that is analytic in ∆, such that (p(z), zp0 (z)) ∈ Ω and
Re(Ψ(p(z), zp0 (z))) > 0 hold for all z ∈ ∆, then Re p(z) > 0, when z ∈ ∆.
Lemma 2.3 ([9, p. 34], [8]). Let p ∈ H[a, n]
(i) If Ψ ∈ Ψn [Ω, M, a], then
Ψ(p(z), zp02 p00 (z); z) ∈ Ω ⇒ |p(z)| < M.
(ii) If Ψ ∈ Ψn [M, a], then
|Ψ(p(z), zp02 p00 (z); z)| < M ⇒ |p(z)| < M.
Lemma 2.4 ([6]). Let h(z) be an analytic and convex univalent function in ∆, with h(0) = a,
c 6= 0 and Re c ≥ 0. If p ∈ H[a, n] and
p(z) +
zp0 (z)
≺ h(z),
c
then
p(z) ≺ q(z) ≺ h(z),
where
Z z
c
c
q(z) ≺
t n −1 f (t)dt,
c
nz n 0
The function q is convex and the best dominant.
z ∈ ∆.
Theorem 2.5. Let c > 0 and 0 < µ < 1. If f ∈ Σ∗n (α) for 0 < α < 1, then I(f ) = F (z) ∈ Σ∗n (β),
where
(2.1)
β = β(α, c, µ)
1h
=
2c + 2αµ + n + 2
4µ
i
p
2
− [4(c − αµ)] + (n + 2)(n + 2 + 4c + 4µα) − 16α − 8µn .
Proof. Here we have the conditions
(2.2)
0 < α < 1, 0 < µ < 1
and
c > 0,
which will imply that β < 1.
Let f (z) ∈ Σ∗n (α). We first show that F (z) defined by (1.3) will become nonzero for z ∈ ∆∗ .
Again since f ∈ Σ∗n (α), we have f (z) 6= 0, for z ∈ ∆∗ .
1
∗
Let g(z) = (f (z))
µ , then a simple computation shows that g(z) ∈ Sn (αµ).
If we define
1
g(z) { 1−αµ }
Ig =
,
z
then I(g) ∈ Sn∗ and by Goluzin’s subordination result (by Lemma 2.1), we obtain
n2
Ig
1
≺
.
z
1+z
From the relation between Ig , g and f we get that
2
g(z)
≺ (1 + z) n (αµ−1) ,
z
which implies
2
z(f (z))µ ≺ (1 + z) n (1−αµ)
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 77, 6 pp.
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P RAVATI S AHOO
AND
S AUMYA S INGH
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and since 0 < αµ < 1, we have z(f (z))µ ≺ (1 + z) n . Combining this with
2
min Re(1 + z) n = 0,
|z|=1
we deduce that
Re[z(f (z))µ ] > 0.
(2.3)
By differentiating (1.3), we obtain
(c + 1)(F (z))µ + z
(2.4)
d
(F (z))µ = (c + 1 − µ)(f (z))µ .
dz
If we let
P (z)
= (F (z))µ ,
z
(2.5)
then (2.4) becomes
1
c+1−µ
z(f (z))µ .
P (z) + zP 0 (z) =
c
c
Hence from (2.3) we have
zP 0 (z)
,
Re Ψ(P (z), zP (z)) = Re P (z) +
c
0
(2.6)
where Ψ(r, s) = r + sc . To show that Re P (z) > 0, condition (iii) of Lemma 2.2 must be
satisfied. Since c > 0, (2.6) implies that
σ
n(1 + ρ2 )
Re Ψ(iρ, σ) = Re iρ +
≤−
≤ 0,
c
2c
2
)
when σ ≤ − n(1+ρ
, for all ρ ∈ R. Hence from (2.6) we deduce that Re P (z) > 0, which
2
implies that F (z) 6= 0 for z ∈ ∆∗ .
We next determine β such that F ∈ Σ∗n (β). Let us define p(z) ∈ H[1, n] by
(2.7)
−
zF 0 (z)
= (1 − β)p(z) + β.
F (z)
By applying (part iii) of Lemma 2.2 again with different Ψ we finish the proof of the theorem.
Since f ∈ Σ∗n (α), by differentiating (2.4) we easily get
Re Ψ(p(z), zp0 (z)) > 0,
where
(1 − β)σ
− α.
c + 1 − µβ − µ(1 − β)p(z)
For β ≤ β(α, c, µ), where β(α, c, µ) is given by (2.1), a simple calculation shows that the
admissibility condition (iii) of Lemma 2.2 is satisfied. Hence by Lemma 2.2, we get Re p(z) >
0. Using this result in (2.7) together with β < 1 shows that F (z) ∈ Σ∗n (β).
Ψ(r, s) = (1 − β)r + β +
Theorem 2.6. Let 0 < c + 1 − µ < 1. If, for 0 < α < 1, f ∈ Σ∗ (α), then I(f ) ∈ Σ∗ (β), where
i
p
1 h
(2.8) β = β(α, µ, c) =
2c + 2αµ + 3 − [2(c − αµ)]2 + 3(3 + 4c) − 4µ(2 + α) .
2µ
The proof is very similar to that of Theorem 2.5.
In the special case when the meromorphic function given in (1.2) has a coefficient a0 = 0, it
is possible to obtain a stronger result than (2.8).
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 77, 6 pp.
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I NTEGRAL O PERATOR
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Theorem 2.7. Let c > 0, 0 < µ < 1, 0 < α < 1, f ∈ Σ∗1 (α), then I(f ) ∈ Σ∗1 (β), where
i
p
1 h
c + αµ + 1 − (c − αµ)2 + 4(c + 1 − µ) .
(2.9)
β = β(α, µ, c) =
2µ
The proof is similar to that of Theorem 2.5.
Corollary 2.8. Let n ≥ 1, c + n + 1 > 0 and g(z) ∈ H[0, n]. If | ((g(z))µ )0 | ≤ λ and
µ1
Z z
1
µ c
(g(t)) t dt ,
(2.10)
F(z) = c+1
z
0
then
λ
| ((F(z))µ )0 | ≤
.
c+n+1
Proof. From (2.10) we deduce (c + 1)(F(z))µ + z ((F(z))µ )0 = g µ (z). If we set z ((F(z))µ )0 =
P (z), then P ∈ H[0, n] and
(c + 1)P (z) + zP 0 (z) = z(g µ (z))0 ≺ λz.
From part(i) of Lemma 2.3, it follows that this differential subordination has the best dominant
λz
P (z) ≺ Q(z) =
.
c+n+1
Hence we have
λ
| ((F(z))µ )0 | ≤
.
c+n+1
Corollary 2.9. Let c + n + 1 > 0 and f ∈ Σn be given as
1
f (z) = + g(z),
z
where n ≥ 1 and g(z) ∈ H[0, n]. Let F be defined by
µ1
Z z
1
1
1
µ c
(g(t)) t dt .
(2.11)
F(z) ≡ + G(z) = + c+1
z
z
z
0
Then
n(c + n + 1)
.
| ((g(z))µ )0 | ≤ √
n2 + 1
Proof. From Corollary 2.8 we obtain
n
,
| ((G(z))µ )0 | ≤ √
2
n +1
since from (2.11), we have
|z 2 ((F(z))µ )0 + 1| = |G0 (z)|.
Hence from [2], we conclude that F ∈ Σ∗n .
Corollary 2.10. Let n be a fixed positive integer and c > 0. Let q be a convex function in ∆,
with q(0) = 1 and let h be defined by
n+1 0
(2.12)
h(z) = q(z) +
zq (z).
c
If f ∈ Σn and F (z) is given by (1.3), then
c+1−µ 2
−
z ((f (z))µ )0 ≺ h(z) ⇒ −z 2 ((F (z))µ )0 ≺ q(z),
c
and this result is sharp.
J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 77, 6 pp.
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P RAVATI S AHOO
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S AUMYA S INGH
Proof. From the definition of h(z), it is a convex function. If we obtain
p(z) = −z 2 (F µ (z))0 ,
then p ∈ H[1, n + 1] and from (2.3), we get
c+1−µ 2
1
p(z) + zp0 (z) = −
z ((f (z))µ )0 ≺ h(z).
c
c
The conclusion of the corollary follows by Lemma 2.4.
Corollary 2.11. Let n ≥ 1 and c > 0. Let f ∈ Σn and let F (z) given by (1.3). If λ > 0, then
λc
|z 2 ((f (z))µ )0 + 1| < λ ⇒ |z 2 ((F (z))µ )0 + 1| <
.
c+n+1
In particular,
c+n+1
|z 2 ((f (z))µ )0 + 1| <
⇒ |z 2 ((F (z))µ )0 + 1| < 1.
c
Hence (F (z))µ is univalent.
Proof. If we take
q(z) = 1 +
then (2.12) becomes
λcz
,
c+n+1
h(z) = 1 + λz.
The conclusion of the corollary follows by Corollary 2.10.
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J. Inequal. Pure and Appl. Math., 10(3) (2009), Art. 77, 6 pp.
http://jipam.vu.edu.au/