Volume 10 (2009), Issue 2, Article 36, 11 pp.
SOME SUBORDINATION CRITERIA CONCERNING THE SǍLǍGEAN
OPERATOR
KAZUO KUROKI AND SHIGEYOSHI OWA
D EPARTMENT OF M ATHEMATICS
K INKI U NIVERSITY
H IGASHI -O SAKA , O SAKA 577-8502
JAPAN
freedom@sakai.zaq.ne.jp
owa@math.kindai.ac.jp
Received 14 September, 2008; accepted 11 January, 2009
Communicated by H.M. Srivastava
A BSTRACT. Applying Sǎlǎgean operator, for the class A of analytic functions f (z) in the open
unit disk U which are normalized by f (0) = f 0 (0) − 1 = 0, the generalization of an analytic
function to discuss the starlikeness is considered. Furthermore, from the subordination criteria
for Janowski functions generalized by some complex parameters, some interesting subordination
criteria for f (z) ∈ A are given.
Key words and phrases: Janowski function, Starlike, Convex, Univalent, Subordination.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION , D EFINITION AND P RELIMINARIES
Let A denote the class of functions f (z) of the form:
(1.1)
f (z) = z +
∞
X
an z n
n=2
which are analytic in the open unit disk
U = {z : z ∈ C
and |z| < 1}.
Also, let P denote the class of functions p(z) of the form:
(1.2)
p(z) = 1 +
∞
X
pn z n
n=1
which are analytic in U. If p(z) ∈ P satisfies Re p(z) > 0
the Carathéodory function (cf. [1]).
253-08
(z ∈ U), then we say that p(z) is
2
K AZUO K UROKI AND S HIGEYOSHI OWA
By the familiar principle of differential subordination between analytic functions f (z) and
g(z) in U, we say that f (z) is subordinate to g(z) in U if there exists an analytic function w(z)
satisfying the following conditions:
w(0) = 0 and |w(z)| < 1
(z ∈ U),
such that
f (z) = g w(z)
(z ∈ U).
We denote this subordination by
f (z) ≺ g(z)
(z ∈ U).
In particular, if g(z) is univalent in U, then it is known that
f (z) ≺ g(z)
(z ∈ U)
⇐⇒
f (0) = g(0)
f (U) ⊂ g(U).
and
For the function p(z) ∈ P, we introduce the following function
1 + Az
(−1 5 B < A 5 1)
1 + Bz
which has been investigated by Janowski [3]. Thus, the function p(z) given by (1.3) is said to
be the Janowski function. And, as a generalization of the Janowski function, Kuroki, Owa and
Srivastava [2] have discussed the function
1 + Az
p(z) =
1 + Bz
for some complex parameters A and B which satisfy one of following conditions

 (i) |A| 5 1, |B| < 1, A 6= B, and Re 1 − AB = |A − B|
p(z) =
(1.3)

(ii) |A| 5 1, |B| = 1, A 6= B, and 1 − AB > 0.
Here, for some complex numbers A and B which satisfy condition (i), the function p(z) is
analytic and univalent in U and p(z) maps the open unit disk U onto the open disk given by
1 − AB |A − B|
(1.4)
p(z) − 1 − |B|2 < 1 − |B|2 .
Thus, it is clear that
(1.5)
Re(1 − AB) − |A − B|
Re p(z) >
=0
1 − |B|2
(z ∈ U).
Also, for some complex numbers A and B which satisfy condition (ii), the function p(z) is
analytic and univalent in U and the domain p(U) is the right half-plane satisfying
1 − |A|2
Re p(z) >
= 0.
2(1 − AB)
(1.6)
Hence, we see that the generalized Janowski function maps the open unit disk U onto some
domain which is on the right half-plane.
Remark 1. For the function
1 + Az
1 + Bz
defined with the condition (i), the inequalities (1.4) and (1.5) give us that
p(z) =
p(z) 6= 0 namely,
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
1 + Az 6= 0
(z ∈ U).
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S OME S UBORDINATION C RITERIA C ONCERNING THE S ǍL ǍGEAN O PERATOR
3
Since, after a simple calculation, we see the condition |A| 5 1, we can omit the condition
|A| 5 1 in (i).
Hence, the condition (i) is newly defined by the following conditions
(1.7)
|B| < 1, A 6= B, and Re 1 − AB = |A − B|.
A function f (z) ∈ A is said to be starlike of order α in U if it satisfies
0 zf (z)
(1.8)
Re
>α
(z ∈ U)
f (z)
for some α (0 5 α < 1). We denote by S ∗ (α) the subclass of A consisting of all functions f (z)
which are starlike of order α in U.
Similarly, if f (z) ∈ A satisfies the following inequality
zf 00 (z)
>α
(z ∈ U)
(1.9)
Re 1 + 0
f (z)
for some α (0 5 α < 1), then f (z) is said to be convex of order α in U. We denote by K(α)
the subclass of A consisting of all functions f (z) which are convex of order α in U.
As usual, in the present investigation, we write
S ∗ (0) ≡ S ∗
and K(0) ≡ K.
The classes S ∗ (α) and K(α) were introduced by Robertson [7].
We define the following differential operator due to Sǎlǎgean [8].
For a function f (z) and j = 1, 2, 3, . . . ,
∞
X
(1.10)
D0 f (z) = f (z) = z +
an z n ,
n=2
(1.11)
0
1
D f (z) = Df (z) = zf (z) = z +
∞
X
nan z n ,
n=2
(1.12)
j
D f (z) = D D
j−1
∞
X
f (z) = z +
n j an z n .
n=2
Also, we consider the following differential operator
Z z
∞
X
f (ζ)
−1
(1.13)
D f (z) =
dζ = z +
n−1 an z n ,
ζ
0
n=2
(1.14)
−j
D f (z) = D
−1
D
−(j−1)
∞
X
f (z) = z +
n−j an z n
n=2
for any negative integers.
Then, for f (z) ∈ A given by (1.1), we know that
∞
X
j
(1.15)
D f (z) = z +
nj an z n (j = 0, ±1, ±2, . . . ).
n=2
We consider the subclass Sjk (α) as follows:
k
D f (z)
k
Sj (α) = f (z) ∈ A : Re
>α
Dj f (z)
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
(z ∈ U ; 0 5 α < 1) .
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4
K AZUO K UROKI AND S HIGEYOSHI OWA
In particular, putting k = j + 1, we also define Sjj+1 (α) by
j+1
D f (z)
j+1
Sj (α) = f (z) ∈ A : Re
> α (z ∈ U ; 0 5 α < 1) .
Dj f (z)
Remark 2. Noting
D1 f (z)
zf 0 (z)
=
,
D0 f (z)
f (z)
0
z zf 0 (z)
D2 f (z)
zf 00 (z)
=
=
1
+
,
D1 f (z)
zf 0 (z)
f 0 (z)
we see that
S01 (α) ≡ S ∗ (α), S12 (α) ≡ K(α)
(0 5 α < 1).
Furthermore, by applying subordination, we consider the following subclass
Dk f (z)
1 + Az
k
Pj (A, B) = f (z) ∈ A : j
≺
(z ∈ U ; A 6= B, |B| 5 1) .
D f (z)
1 + Bz
In particular, putting k = j + 1, we also define
Dj+1 f (z)
1 + Az
j+1
Pj (A, B) = f (z) ∈ A :
≺
(z ∈ U; A 6= B, |B| 5 1) .
Dj f (z)
1 + Bz
Remark 3. Noting
1 + (1 − 2α)z
Dk f (z)
≺
Dj f (z)
1−z
we see that
⇐⇒
P01 (1 − 2α, −1) ≡ S ∗ (α),
Re
Dk f (z)
Dj f (z)
>α
P12 (1 − 2α, −1) ≡ K(α)
(z ∈ U; 0 5 α < 1),
(0 5 α < 1).
In our investigation here, we need the following lemma concerning the differential subordination given by Miller and Mocanu [5] (see also [6, p. 132]).
Lemma 1.1. Let the function q(z) be analytic and univalent in U. Also let φ(ω) and ψ(ω) be
analytic in a domain C containing q(U), with
ψ(ω) 6= 0
ω ∈ q(U) ⊂ C .
Set
Q(z) = zq 0 (z)ψ q(z)
and
h(z) = φ q(z) + Q(z),
and suppose that
Q(z) is starlike and univalent in U;
(i)
and
(ii)
Re
zh0 (z)
Q(z)
= Re
!
φ0 q(z)
zQ0 (z)
+
>0
Q(z)
ψ q(z)
(z ∈ U).
If p(z) is analytic in U, with
p(0) = q(0)
and
p(U) ⊂ C,
and
φ p(z) + zp0 (z)ψ p(z) ≺ φ q(z) + zq 0 (z)ψ q(z) =: h(z)
(z ∈ U),
then
p(z) ≺ q(z)
(z ∈ U)
and q(z) is the best dominant of this subordination.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
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S OME S UBORDINATION C RITERIA C ONCERNING THE S ǍL ǍGEAN O PERATOR
5
By making use of Lemma 1.1, Kuroki, Owa and Srivastava [2] have investigated some subordination criteria for the generalized Janowski functions and deduced the following lemma.
6= 0 (z ∈ U).
Lemma 1.2. Let the function f (z) ∈ A be chosen so that f (z)
z
Also, let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy one of
following conditions:
(i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be such that
β(1 − α) (1 + β) Re 1 − AB − |A − B|
1−β
1+β
+
+
+
− 1 = 0,
2
α
1 − |B|
1 + |A| 1 + |B|
(ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be such that
β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|)
+
+
= 0.
α
2(1 + |A|)
2(1 − AB)
If
(1.16)
zf 0 (z)
f (z)
β zf 00 (z)
1+α 0
f (z)
≺ h(z)
(z ∈ U),
where
h(z) =
1 + Az
1 + Bz
then
β−1 1 + Az α(1 + Az)2 + α(A − B)z
(1 − α)
+
,
1 + Bz
(1 + Bz)2
1 + Az
zf 0 (z)
≺
f (z)
1 + Bz
(z ∈ U).
2. S UBORDINATIONS FOR THE C LASS D EFINED BY THE S ǍL ǍGEAN O PERATOR
First of all, by applying the Sǎlǎgean operator for f (z) ∈ A, we consider the following
subordination criterion in the class Pjk (A, B) for some complex parameters A and B.
j
Theorem 2.1. Let the function f (z) ∈ A be chosen so that D fz (z) 6= 0 (z ∈ U).
Also, let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy one of
following conditions:
(i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that
β(1 − α) (1 + β) Re 1 − AB − |A − B|
1−β
1+β
+
+
+
− 1 = 0,
2
α
1 − |B|
1 + |A| 1 + |B|
(ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that
β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|)
+
+
= 0.
α
2(1 + |A|)
2(1 − AB)
If
(2.1)
Dk f (z)
Dj f (z)
β k
D f (z) Dk+1 f (z) Dj+1 f (z)
(1 − α) + α
+
−
≺ h(z),
Dj f (z)
Dk f (z)
Dj f (z)
where
h(z) =
1 + Az
1 + Bz
β−1 1 + Az α(1 + Az)2 + α(A − B)z
(1 − α)
+
,
1 + Bz
(1 + Bz)2
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
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6
K AZUO K UROKI AND S HIGEYOSHI OWA
then
Dk f (z)
1 + Az
≺
j
D f (z)
1 + Bz
(z ∈ U).
Proof. If we define the function p(z) by
p(z) =
Dk f (z)
Dj f (z)
(z ∈ U),
then p(z) is analytic in U with p(0) = 1. Further, since
k
k+1
D f (z)
D f (z) Dj+1 f (z)
0
zp (z) =
−
,
Dj f (z)
Dk f (z)
Dj f (z)
the condition (2.1) can be written as follows:
β β−1
p(z)
(1 − α) + αp(z) + αzp0 (z) p(z)
≺ h(z)
(z ∈ U).
We also set
q(z) =
1 + Az
,
1 + Bz
φ(z) = z β (1 − α + αz),
and ψ(z) = αz β−1
for z ∈ U. Then, it is clear that the function q(z) is analytic and univalent in U and has a positive
real part in U for the conditions (i) and (ii).
Therefore, φ and ψ are analytic in a domain C containing q(U), with
ψ(ω) 6= 0
ω ∈ q(U) ⊂ C .
Also, for the function Q(z) given by
α(A − B)z(1 + Az)β−1
Q(z) = zq 0 (z)ψ q(z) =
,
(1 + Bz)β+1
we obtain
zQ0 (z)
1−β
1+β
=
+
− 1.
Q(z)
1 + Az 1 + Bz
(2.2)
Furthermore, we have
h(z) = φ q(z) + Q(z)
β 1 + Az
1 + Az
α(A − B)z(1 + Az)β−1
=
1−α+α
+
1 + Bz
1 + Bz
(1 + Bz)β+1
and
(2.3)
zh0 (z)
β(1 − α)
zQ0 (z)
=
+ (1 + β)q(z) +
.
Q(z)
α
Q(z)
Hence,
(i) For the complex numbers A and B such that
|B| < 1, A 6= B,
and
Re(1 − AB) = |A − B|,
it follows from (2.2) and (2.3) that
0 zQ (z)
1−β
1+β
Re
>
+
− 1 = 0,
Q(z)
1 + |A| 1 + |B|
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
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S OME S UBORDINATION C RITERIA C ONCERNING THE S ǍL ǍGEAN O PERATOR
7
and
Re
zh0 (z)
Q(z)
β(1 − α) (1 + β) Re 1 − AB − |A − B|
>
+
1 − |B|2
α
1−β
1+β
+
+
−1=0
1 + |A| 1 + |B|
(z ∈ U).
(ii) For the complex numbers A and B such that
|B| = 1, |A| 5 1, A 6= B,
and
1 − AB > 0,
from (2.2) and (2.3), we get
0 zQ (z)
1−β
1
(1 − β)(1 − |A|)
Re
>
+ (1 + β) − 1 =
= 0,
Q(z)
1 + |A| 2
2(1 + |A|)
and
Re
zh0 (z)
Q(z)
>
β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|)
+
=0
+
α
2(1 + |A|)
2(1 − AB)
(z ∈ U).
Since all the conditions of Lemma 1.1 are satisfied, we conclude that
1 + Az
Dk f (z)
≺
j
D f (z)
1 + Bz
(z ∈ U),
which completes the proof of Theorem 2.1.
Remark 4. We know that a function f (z) satisfying the conditions in Theorem 2.1 belongs to
the class Pjk (A, B).
Letting k = j + 1 in Theorem 2.1, we obtain the following theorem.
j
Theorem 2.2. Let the function f (z) ∈ A be chosen so that D fz (z) 6= 0 (z ∈ U).
Also, let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy one of
following conditions
(i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that
β(1 − α) (1 + β) Re 1 − AB − |A − B|
1−β
1+β
+
+
+
− 1 = 0,
2
α
1 − |B|
1 + |A| 1 + |B|
(ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that
β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|)
+
+
= 0.
α
2(1 + |A|)
2(1 − AB)
If
(2.4)
Dj+1 f (z)
Dj f (z)
β Dj+2 f (z)
1 − α + α j+1
D f (z)
≺ h(z),
where
h(z) =
then
1 + Az
1 + Bz
β−1 1 + Az α(1 + Az)2 + α(A − B)z
(1 − α)
+
,
1 + Bz
(1 + Bz)2
Dj+1 f (z)
1 + Az
≺
j
D f (z)
1 + Bz
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
(z ∈ U).
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8
K AZUO K UROKI AND S HIGEYOSHI OWA
Remark 5. A function f (z) satisfying the conditions in Theorem 2.2 belongs to the class
Pjj+1 (A, B). Setting j = 0 in Theorem 2.2, we obtain Lemma 1.2 proven by Kuroki, Owa
and Srivastava [2].
Also, if we assume that
1 − µ iθ
e
(0 5 µ < 1, 0 5 θ < 2π),
1+µ
Theorem 2.2 becomes the following corollary.
j
Corollary 2.3. If f (z) ∈ A D fz (z) 6= 0 in U satisfies
α = 1, β = A = 0,
and
B=
Dj+2 f (z)
1 + µ − (1 − µ)eiθ z
≺
Dj+1 f (z)
1 + µ + (1 − µ)eiθ z
(z ∈ U; 0 5 θ < 2π)
for some µ (0 5 µ < 1), then
1+µ
Dj+1 f (z)
≺
j
D f (z)
1 + µ + (1 − µ)eiθ z
(z ∈ U).
From the above corollary, we have
j+2
j+1
D f (z)
D f (z)
1+µ
Re
> µ =⇒ Re
>
(z ∈ U; 0 5 µ < 1).
j+1
j
D f (z)
D f (z)
2
In particular, making j = 0, we get
0 zf 00 (z)
zf (z)
1+µ
Re 1 + 0
> µ =⇒ Re
>
(z ∈ U; 0 5 µ < 1),
f (z)
f (z)
2
namely
1+µ
∗
(z ∈ U; 0 5 µ < 1).
f (z) ∈ K(µ) =⇒ f (z) ∈ S
2
And, taking µ = 0, we find that every convex function is starlike of order 12 . This fact is
well-known as the Marx-Strohhäcker theorem in Univalent Function Theory (cf. [4, 9]).
3. S UBORDINATION C RITERIA FOR OTHER A NALYTIC F UNCTIONS
In this section, by making use of Lemma 1.1, we consider some subordination criteria conj
cerning the analytic function D fz (z) for f (z) ∈ A.
Theorem 3.1. Let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B which
satisfy one of following conditions
(i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that
β
1−β
1+β
+
+
− 1 = 0,
α 1 + |A| 1 + |B|
(ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that
β (1 − β)(1 − |A|)
+
= 0.
α
2(1 + |A|)
(3.1)
If f (z) ∈ A satisfies
j
β D f (z)
Dj+1 f (z)
1−α+α j
≺ h(z),
z
D f (z)
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
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S OME S UBORDINATION C RITERIA C ONCERNING THE S ǍL ǍGEAN O PERATOR
where
h(z) =
then
1 + Az
1 + Bz
β
+
9
α(A − B)z(1 + Az)β−1
,
(1 + Bz)β+1
1 + Az
Dj f (z)
≺
z
1 + Bz
(z ∈ U).
Proof. If we define the function p(z) by
Dj f (z)
(z ∈ U),
z
then p(z) is analytic in U with p(0) = 1 and the condition (3.1) can be written as follows:
β
β−1
p(z) + αzp0 (z) p(z)
≺ h(z)
(z ∈ U).
p(z) =
We also set
1 + Az
, φ(z) = z β , and ψ(z) = αz β−1
1 + Bz
for z ∈ U. Then, the function q(z) is analytic and univalent in U and satisfies
Re q(z) > 0
(z ∈ U)
q(z) =
for the condition (i) and (ii).
Thus, the functions φ and ψ satisfy the conditions required by Lemma 1.1.
Further, for the functions Q(z) and h(z) given by
Q(z) = zq 0 (z)ψ q(z) and h(z) = φ q(z) + Q(z),
we have
zQ0 (z)
1−β
1+β
zh0 (z)
β zQ0 (z)
=
+
− 1 and
= +
.
Q(z)
1 + Az 1 + Bz
Q(z)
α
Q(z)
Then, similarly to the proof of Theorem 2.1, we see that
0 0 zQ (z)
zh (z)
Re
> 0 and Re
>0
(z ∈ U)
Q(z)
Q(z)
for the conditions (i) and (ii).
Thus, by applying Lemma 1.1, we conclude that p(z) ≺ q(z)
The proof of the theorem is completed.
(z ∈ U).
Letting j = 0 in Theorem 3.1, we obtain the following theorem.
Theorem 3.2. Let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy
one of following conditions:
(i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that
β
1−β
1+β
+
+
− 1 = 0,
α 1 + |A| 1 + |B|
(ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that
β (1 − β)(1 − |A|)
+
= 0.
α
2(1 + |A|)
(3.2)
If f (z) ∈ A satisfies
β−1 β
f (z)
f (z)
1 + Az
α(A − B)z(1 + Az)β−1
0
(1 − α)
+ αf (z) ≺
+
,
z
z
1 + Bz
(1 + Bz)β+1
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
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10
K AZUO K UROKI AND S HIGEYOSHI OWA
then
f (z)
1 + Az
≺
z
1 + Bz
(z ∈ U).
Also, taking
α = 1, β = A = 0,
1 − ν iθ
and B =
e
ν
1
5 ν < 1, 0 5 θ < 2π
2
in Theorem 3.2, we have
Corollary 3.3. If f (z) ∈ A satisfies
for some ν
1
2
zf 0 (z)
ν
≺
f (z)
ν + (1 − ν)eiθ z
5 ν < 1 , then
(z ∈ U; 0 5 θ < 2π)
f (z)
ν
≺
z
ν + (1 − ν)eiθ z
(z ∈ U).
Further, making
α = β = 1, A = 0,
1 − ν iθ
and B =
e
ν
1
5 ν < 1, 0 5 θ < 2π
2
in Theorem 3.2, we get
Corollary 3.4. If f (z) ∈ A satisfies
2
ν
0
f (z) ≺
ν + (1 − ν)eiθ z
for some ν 12 5 ν < 1 , then
(z ∈ U ; 0 5 θ < 2π)
f (z)
ν
≺
z
ν + (1 − ν)eiθ z
The above corollaries give:
0 zf (z)
(3.3)
Re
> ν =⇒
f (z)
Re
f (z)
z
(z ∈ U).
>ν
1
z ∈ U; 5 ν < 1 ,
2
and
(3.4)
Re
p
f 0 (z)
>ν
=⇒
Re
f (z)
z
>ν
1
z ∈ U; 5 ν < 1 .
2
Here, taking ν = 12 , we find some results that are known as the Marx-Strohhäcker theorem in
Univalent Function Theory (cf. [4], [9]).
Setting j = 1 in Theorem 3.1, we obtain the following theorem.
Theorem 3.5. Let α (α 6= 0), β (−1 5 β 5 1), and some complex parameters A and B satisfy
one of following conditions
(i) |B| < 1, A 6= B, and Re(1 − AB) = |A − B| be so that
β
1−β
1+β
+
+
− 1 = 0,
α 1 + |A| 1 + |B|
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
http://jipam.vu.edu.au/
S OME S UBORDINATION C RITERIA C ONCERNING THE S ǍL ǍGEAN O PERATOR
11
(ii) |B| = 1, |A| 5 1, A 6= B, and 1 − AB > 0 be so that
β (1 − β)(1 − |A|)
+
= 0.
α
2(1 + |A|)
If f (z) ∈ A satisfies
β
β
1 + Az
α(A − B)z(1 + Az)β−1
zf 00 (z)
0
+
(3.5)
f (z)
1+α 0
≺
,
f (z)
1 + Bz
(1 + Bz)β+1
then
1 + Az
f 0 (z) ≺
(z ∈ U).
1 + Bz
Here, making
1 − ν iθ
1
α = 1, β = A = 0, and B =
e
5 ν < 1, 0 5 θ < 2π
ν
2
in Theorem 3.5, we have:
Corollary 3.6. If f (z) ∈ A satisfies
zf 00 (z)
ν
≺
(z ∈ U ; 0 5 θ < 2π)
1+ 0
f (z)
ν + (1 − ν)eiθ z
for some ν 12 5 ν < 1 , then
ν
(z ∈ U).
f 0 (z) ≺
ν + (1 − ν)eiθ z
Also, from Corollary 3.6 we have:
zf 00 (z)
(3.6)
Re 1 + 0
> ν =⇒
f (z)
0
Re (f (z)) > ν
1
z ∈ U; 5 ν < 1 .
2
R EFERENCES
[1] P.L. DUREM, Univalent Functions, Springer-Verlag, New York, Berlin, Heidelberg, Tokyo, 1983.
[2] K. KUROKI, S. OWA AND H.M. SRIVASTAVA, Some subordination criteria for analytic functions,
Bull. Soc. Sci. Lett. Lodz, Vol., 52 (2007), 27–36.
[3] W. JANOWSKI, Extremal problem for a family of functions with positive real part and for some
related families, Ann. Polon. Math., 23 (1970), 159–177.
[4] A. MARX, Untersuchungen uber schlichte Abbildungen, Math. Ann., 107 (1932/33), 40–67.
[5] S.S. MILLER AND P.T. MOCANU, On some classes of first-order differential subordinations, Michigan Math. J., 32 (1985), 185–195.
[6] S.S. MILLER AND P.T. MOCANU, Differential Subordinations, Pure and Applied Mathematics
225, Marcel Dekker, 2000.
[7] M.S. ROBERTSON, On the theory of univalent functions, Ann. Math., 37 (1936), 374–408.
[8] G.S. SǍLǍGEAN, Subclass of univalent functions, Complex Analysis-Fifth Romanian-Finnish
Seminar, Part 1(Bucharest, 1981), Lecture Notes in Math., vol. 1013, Springer, Berlin, 1983, pp.
362–372.
[9] E. STROHHÄCKER, Beitrage zur Theorie der schlichten Funktionen, Math. Z., 37 (1933), 356–380.
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 36, 11 pp.
http://jipam.vu.edu.au/