SOME SUBORDINATION CRITERIA CONCERNING THE S ˇ AL ˇ AGEAN OPERATOR

advertisement
SOME SUBORDINATION CRITERIA CONCERNING
THE SǍLǍGEAN OPERATOR
Subordination Criteria
KAZUO KUROKI AND SHIGEYOSHI OWA
Department of Mathematics
Kinki University
Higashi-Osaka, Osaka 577-8502, Japan
EMail: freedom@sakai.zaq.ne.jp
owa@math.kindai.ac.jp
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Title Page
Contents
Received:
14 September, 2008
Accepted:
11 January, 2009
Communicated by:
H.M. Srivastava
2000 AMS Sub. Class.:
26D15.
Key words:
Janowski function, Starlike, Convex, Univalent, Subordination.
Abstract:
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.
JJ
II
J
I
Page 1 of 23
Go Back
Full Screen
Close
Contents
1
Introduction, Definition and Preliminaries
3
2
Subordinations for the Class Defined by the Sǎlǎgean Operator
11
3
Subordination Criteria for Other Analytic Functions
17
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Title Page
Contents
JJ
II
J
I
Page 2 of 23
Go Back
Full Screen
Close
1.
Introduction, Definition and Preliminaries
Let A denote the class of functions f (z) of the form:
(1.1)
f (z) = z +
∞
X
an z n
n=2
Subordination Criteria
which are analytic in the open unit disk
Kazuo Kuroki and Shigeyoshi Owa
U = {z : z ∈ C
|z| < 1}.
and
vol. 10, iss. 2, art. 36, 2009
Also, let P denote the class of functions p(z) of the form:
(1.2)
p(z) = 1 +
∞
X
Title Page
pn z n
Contents
n=1
which are analytic in U. If p(z) ∈ P satisfies Re p(z) > 0 (z ∈ U), then we say
that p(z) is the Carathéodory function (cf. [1]).
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),
JJ
II
J
I
Page 3 of 23
Go Back
Full Screen
Close
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)
and f (U) ⊂ g(U).
For the function p(z) ∈ P, we introduce the following function
p(z) =
(1.3)
1 + Az
1 + Bz
(−1 5 B < A 5 1)
Subordination Criteria
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
p(z) =
1 + Az
1 + Bz
(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
vol. 10, iss. 2, art. 36, 2009
Title Page
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|

Kazuo Kuroki and Shigeyoshi Owa
(z ∈ U).
Contents
JJ
II
J
I
Page 4 of 23
Go Back
Full Screen
Close
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
(1.6)
1 − |A|2
Re p(z) >
= 0.
2(1 − AB)
Hence, we see that the generalized Janowski function maps the open unit disk U
onto some domain which is on the right half-plane.
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
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) =
Title Page
Contents
p(z) 6= 0 namely,
1 + Az 6= 0
(z ∈ U).
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.
JJ
II
J
I
Page 5 of 23
Go Back
Full Screen
Close
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.
D0 f (z) = f (z) = z +
(1.10)
an z n ,
n=2
(1.11)
1
0
D f (z) = Df (z) = zf (z) = z +
∞
X
Title Page
Contents
JJ
II
J
I
Page 6 of 23
nan z n ,
n=2
(1.12)
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
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
Subordination Criteria
∞
X
j
j−1
D f (z) = D D f (z) = z +
nj an z n .
n=2
Also, we consider the following differential operator
Z z
∞
X
f (ζ)
−1
dζ = z +
n−1 an z n ,
(1.13)
D f (z) =
ζ
0
n=2
Go Back
Full Screen
Close
(1.14)
∞
X
D−j f (z) = D−1 D−(j−1) 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
(1.15)
Dj f (z) = z +
∞
X
Subordination Criteria
n j an z n
(j = 0, ±1, ±2, . . . ).
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
n=2
We consider the subclass Sjk (α) as follows:
k
D f (z)
k
>α
Sj (α) = f (z) ∈ A : Re
Dj f (z)
Title Page
(z ∈ U ; 0 5 α < 1) .
Sjj+1 (α)
In particular, putting k = j + 1, we also define
j+1
D f (z)
j+1
Sj (α) = f (z) ∈ A : Re
>α
Dj f (z)
by
(z ∈ U ; 0 5 α < 1) .
II
J
I
Page 7 of 23
Full Screen
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),
JJ
Go Back
Remark 2. Noting
D1 f (z)
zf 0 (z)
=
,
D0 f (z)
f (z)
Contents
S12 (α) ≡ K(α)
(0 5 α < 1).
Close
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 :
≺
Dj f (z)
1 + Bz
(z ∈ U; A 6= B, |B| 5 1) .
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
Remark 3. Noting
vol. 10, iss. 2, art. 36, 2009
Dk f (z)
1 + (1 − 2α)z
≺
Dj f (z)
1−z
⇐⇒
Re
Dk f (z)
Dj f (z)
>α
(z ∈ U; 0 5 α < 1),
Title Page
Contents
we see that
P01 (1 − 2α, −1) ≡ S ∗ (α),
P12 (1 − 2α, −1) ≡ K(α)
(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
(i)
Q(z) is starlike and univalent in U;
JJ
II
J
I
Page 8 of 23
Go Back
Full Screen
Close
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,
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
and
vol. 10, iss. 2, art. 36, 2009
φ p(z) + zp0 (z)ψ p(z) ≺ φ q(z) + zq 0 (z)ψ q(z) =: h(z)
(z ∈ U),
Title Page
then
p(z) ≺ q(z)
(z ∈ U)
Contents
and q(z) is the best dominant of this subordination.
JJ
II
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.
J
I
Lemma 1.2. Let the function f (z) ∈ A be chosen so that f (z)
6= 0 (z ∈ U).
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|
Page 9 of 23
Go Back
Full Screen
Close
(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),
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
where
h(z) =
Subordination Criteria
1 + Az
1 + Bz
β−1 1 + Az α(1 + Az)2 + α(A − B)z
(1 − α)
+
,
1 + Bz
(1 + Bz)2
Title Page
Contents
then
zf 0 (z)
1 + Az
≺
f (z)
1 + Bz
(z ∈ U).
JJ
II
J
I
Page 10 of 23
Go Back
Full Screen
Close
2.
Subordinations for the Class Defined by the Sǎlǎgean
Operator
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
1−β
1+β
β(1 − α) (1 + β) Re 1 − AB − |A − B|
+
+
−1 = 0,
+
1 − |B|2
1 + |A| 1 + |B|
α
(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
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Title Page
Contents
JJ
II
J
I
Page 11 of 23
Go Back
Full Screen
k
β k
D f (z)
D f (z) Dk+1 f (z) Dj+1 f (z)
(2.1)
(1−α)+α
+ k
− j
≺ h(z),
Dj f (z)
Dj f (z)
D f (z)
D 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
Close
then
Dk f (z)
1 + Az
≺
Dj f (z)
1 + Bz
(z ∈ U).
Proof. If we define the function p(z) by
Dk f (z)
p(z) = j
D f (z)
(z ∈ U),
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
then p(z) is analytic in U with p(0) = 1. Further, since
k
k+1
D f (z) Dj+1 f (z)
D 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)
vol. 10, iss. 2, art. 36, 2009
Title Page
Contents
(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
JJ
II
J
I
Page 12 of 23
Go Back
Full Screen
Close
we obtain
1−β
1+β
zQ0 (z)
=
+
− 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
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
and
zh0 (z)
β(1 − α)
zQ0 (z)
=
+ (1 + β)q(z) +
.
Q(z)
α
Q(z)
(2.3)
Title Page
Contents
Hence,
(i) For the complex numbers A and B such that
JJ
II
|B| < 1, A 6= B,
J
I
and
Re(1 − AB) = |A − B|,
Page 13 of 23
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|
Go Back
Full Screen
and
Close
Re
zh0 (z)
Q(z)
β(1 − α) (1 + β) Re 1 − AB − |A − B|
>
+
α
1 − |B|2
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|)
Subordination Criteria
and
0 zh (z)
β(1 − α) (1 + β)(1 − |A|2 ) (1 − β)(1 − |A|)
Re
>
+
+
=0
Q(z)
α
2(1 + |A|)
2(1 − AB)
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
(z ∈ U).
Since all the conditions of Lemma 1.1 are satisfied, we conclude that
Dk f (z)
1 + Az
≺
j
D f (z)
1 + Bz
Title Page
Contents
(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
1−β
1+β
β(1 − α) (1 + β) Re 1 − AB − |A − B|
+
+
− 1 = 0,
+
2
1 − |B|
1 + |A| 1 + |B|
α
JJ
II
J
I
Page 14 of 23
Go Back
Full Screen
Close
(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
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
h(z) =
then
Subordination Criteria
β−1 1 + Az α(1 + Az)2 + α(A − B)z
1 + Az
(1 − α)
+
,
1 + Bz
1 + Bz
(1 + Bz)2
Dj+1 f (z)
1 + Az
≺
j
D f (z)
1 + Bz
Contents
(z ∈ U).
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
α = 1, β = A = 0, and B =
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
Dj+2 f (z)
1 + µ − (1 − µ)eiθ z
≺
Dj+1 f (z)
1 + µ + (1 − µ)eiθ z
Title Page
(z ∈ U; 0 5 θ < 2π)
JJ
II
J
I
Page 15 of 23
Go Back
Full Screen
Close
for some µ (0 5 µ < 1), then
Dj+1 f (z)
1+µ
≺
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
> µ =⇒ Re
>
Re
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
> µ =⇒ Re
>
Re 1 + 0
f (z)
f (z)
2
(z ∈ U; 0 5 µ < 1).
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
(z ∈ U; 0 5 µ < 1),
Title Page
Contents
namely
f (z) ∈ K(µ)
=⇒
f (z) ∈ S
∗
1+µ
2
(z ∈ U; 0 5 µ < 1).
And, taking µ = 0, we find that every convex function is starlike of order 21 . This
fact is well-known as the Marx-Strohhäcker theorem in Univalent Function Theory
(cf. [4, 9]).
JJ
II
J
I
Page 16 of 23
Go Back
Full Screen
Close
3.
Subordination Criteria for Other Analytic Functions
In this section, by making use of Lemma 1.1, we consider some subordination critej
ria concerning 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
Subordination Criteria
(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|
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Title Page
(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 j
Dj+1 f (z)
D f (z)
1−α+α j
≺ h(z),
(3.1)
z
D f (z)
Contents
JJ
II
J
I
Page 17 of 23
Go Back
Full Screen
where
h(z) =
then
1 + Az
1 + Bz
β
+
α(A − B)z(1 + Az)β−1
,
(1 + Bz)β+1
Dj f (z)
1 + Az
≺
z
1 + Bz
(z ∈ U).
Close
Proof. If we define the function p(z) by
p(z) =
Dj f (z)
z
(z ∈ U),
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).
We also set
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
1 + Az
q(z) =
,
1 + Bz
vol. 10, iss. 2, art. 36, 2009
β
φ(z) = z ,
and
ψ(z) = αz
β−1
for z ∈ U. Then, the function q(z) is analytic and univalent in U and satisfies
Re q(z) > 0
(z ∈ U)
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
Title Page
Contents
JJ
II
J
I
Page 18 of 23
Go Back
Full Screen
1−β
1+β
zQ0 (z)
=
+
− 1 and
Q(z)
1 + Az 1 + Bz
zh0 (z)
β zQ0 (z)
= +
.
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
Q(z)
Q(z)
(z ∈ U)
Close
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|
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Title Page
(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
(3.2)
then
f (z)
z
β−1 Contents
JJ
II
J
I
Page 19 of 23
f (z)
0
(1 − α)
+ αf (z)
z
β
1 + Az
α(A − B)z(1 + Az)β−1
≺
+
,
1 + Bz
(1 + Bz)β+1
f (z)
1 + Az
≺
z
1 + Bz
(z ∈ U).
Go Back
Full Screen
Close
Also, taking
α = 1, β = A = 0,
1 − ν iθ
e
and B =
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
Subordination Criteria
(z ∈ U; 0 5 θ < 2π)
f (z)
ν
≺
z
ν + (1 − ν)eiθ z
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Title Page
(z ∈ U).
Contents
Further, making
α = β = 1, A = 0,
and B =
1 − ν iθ
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 21 5 ν < 1 , then
ν
f (z)
≺
z
ν + (1 − ν)eiθ z
JJ
II
J
I
Page 20 of 23
Go Back
Full Screen
(z ∈ U ; 0 5 θ < 2π)
(z ∈ U).
Close
The above corollaries give:
0 zf (z)
(3.3) Re
> ν =⇒
f (z)
Re
f (z)
z
>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
1
,
2
Here, taking ν =
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.
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Title Page
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
1−β
1+β
β
+
+
− 1 = 0,
α 1 + |A| 1 + |B|
Contents
JJ
II
J
I
Page 21 of 23
Go Back
(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
Full Screen
Close
then
f 0 (z) ≺
1 + Az
1 + Bz
(z ∈ U).
Here, making
α = 1, β = A = 0,
1 − ν iθ
and B =
e
ν
1
5 ν < 1, 0 5 θ < 2π
2
in Theorem 3.5, we have:
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Corollary 3.6. If f (z) ∈ A satisfies
zf 00 (z)
ν
1+ 0
≺
f (z)
ν + (1 − ν)eiθ z
for some ν 12 5 ν < 1 , then
f 0 (z) ≺
ν
ν + (1 − ν)eiθ z
(z ∈ U ; 0 5 θ < 2π)
Title Page
Contents
(z ∈ U).
JJ
II
J
I
Page 22 of 23
Go Back
Also, from Corollary 3.6 we have:
zf 00 (z)
(3.6) Re 1 + 0
> ν =⇒ Re (f 0 (z)) > ν
f (z)
Full Screen
1
z ∈ U; 5 ν < 1 .
2
Close
References
[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.
Subordination Criteria
Kazuo Kuroki and Shigeyoshi Owa
vol. 10, iss. 2, art. 36, 2009
Title Page
[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.
Contents
JJ
II
J
I
Page 23 of 23
Go Back
Full Screen
Close
Download