Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 5, Issue 3, Article 78, 2004
SOME APPLICATIONS OF A FIRST ORDER DIFFERENTIAL SUBORDINATION
SUKHJIT SINGH AND SUSHMA GUPTA
D EPARTMENT OF M ATHEMATICS ,
S ANT L ONGOWAL I NSTITUTE OF E NGINEERING & T ECHNOLOGY
L ONGOWAL -148 106(P UNJAB )-I NDIA .
sukhjit_d@yahoo.com
sushmagupta1@yahoo.com
Received 28 January, 2004; accepted 19 July, 2004
Communicated by A. Lupaş
A BSTRACT. Let p and q be analytic functions in the unit disc E = {z : |z| < 1}, with p(0) =
q(0) = 1. Assume that α and δ are real numbers such that 0 < δ ≤ 1, α + δ ≥ 0. Let β and γ be
complex numbers with β 6= 0. In the present paper, we investigate the differential subordination
δ
δ
zp0(z)
zq0(z)
α
α
(p(z)) p (z) +
≺ (q(z)) q(z) +
, z ∈ E,
βp(z) + γ
βq(z) + γ
and as applications, find several sufficient conditions for starlikeness and univalence of functions
analytic in the unit disc E.
Key words and phrases: Univalent function, Starlike function, Convex function, Differential subordination.
2000 Mathematics Subject Classification. Primary 30C45, Secondary 30C50.
1. I NTRODUCTION
Let A be the class of functions f , analytic in E, having the normalization f (0) = f 0 (0) − 1 =
0. Denote by A0 , the class of functions f which are analytic in E and satisfy f (0) = 1 and
f (z) 6= 0, z ∈ E. Furthermore, let St(α), St and K denote the usual subclasses of A consisting
of functions which are starlike of order α, 0 ≤ α < 1, starlike (with respect to the origin) and
convex in E, respectively.
For the analytic functions f and g, we say that f is subordinate to g in E, written as f (z) ≺
g(z) in E (or simply f ≺ g), if g is univalent in E, f (0) = g(0), and f (E) ⊂ g(E).
Let ψ : D → C (C is the complex plane) be an analytic function defined on a domain
D ⊂ C2 . Further, let p be a function analytic in E with (p(z), zp0 (z)) ∈ D for z ∈ E, and let h
be a univalent function in E. Then p is said to satisfy first order differential subordination if
(1.1)
ψ(p(z), zp0 (z)) ≺ h(z),
ISSN (electronic): 1443-5756
c 2004 Victoria University. All rights reserved.
023-04
z ∈ E, ψ(p(0), 0) = h(0)
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A univalent function q is said to be the dominant of the differential subordination (1.1), if
p(0) = q(0) and p ≺ q for all p satisfying (1.1). A dominant q̃ of (1.1) that satisfies q̃ ≺ q for
all dominants q of (1.1) is said to be the best dominant of (1.1).
The fascinating theory of differential subordination was put on sound footing by Miller and
Mocanu [6] in 1981. Later, they used it (e.g. see [3], [7] and [10]) to obtain the best dominant
for the Briot-Bouquet differential subordination of the type
zp0 (z)
(1.2)
p(z) +
≺ h(z), z ∈ E,
βp(z) + γ
which was first studied by Ruscheweyh and Singh [29]. The observation that the differential
subordination (1.2) had several interesting applications to univalent functions led to its generalization in many ways (see [8], [16]).
In the present paper, we consider a generalization of the form
δ
δ
zp0 (z)
zq 0 (z)
α
α
(1.3)
(p(z)) p(z) +
≺ (q(z)) q(z) +
= h(z), z ∈ E,
βp(z) + γ
βq(z) + γ
where α, δ are suitably chosen real numbers, β, γ ∈ C, β 6= 0. Using the characterization of a
subordination chain and a beautiful lemma of Miller and Mocanu [6], we determine conditions
on q under which it becomes best dominant of differential subordination (1.3), Section 3. In
Section 4, we apply our results to find several new sufficient conditions for starlikeness, strongly
starlikeness and univalence of function f ∈ A. Large number of known results also follow as
particular cases from our results.
The motivation to study the differential subordination (1.3) was provided by the class H(γ), γ ≥
0, of γ-starlike functions defined by Lewandowski et al. [4], defined as under:
(
)
0 1−γ γ
zf (z)
zf 00 (z)
H(γ) = f ∈ A : Re
1+ 0
> 0, z ∈ E .
f (z)
f (z)
Recently, Darus and Thomas [2], too, investigated the class H(γ) and proved that the functions
in this class are starlike.
Finally, we define few classes of analytic functions, which will be required by us in the
present paper.
Miller, Mocanu and others studied extensively (e.g. see [11], [12], [14], [15] and [17]) the
class of α-convex functions, Mα , α real, defined as under
zf 00 (z)
zf 0 (z)
Mα = f ∈ A : Re (1 − α)
+α 1+ 0
> 0, z ∈ E
f (z)
f (z)
0
where f (z)fz (z) 6= 0 in E. They proved that Mα ⊂ Mβ ⊂ St for 0 ≤ α/β ≤ 1 and Mα ⊂ M1 =
K for α ≥ 1.
In [30], Silverman defined the following subclass of the class of starlike functions:
1 + zf 00 (z)/f 0 (z)
Gb = f ∈ A : − 1 < b, z ∈ E .
zf 0 (z)/f (z)
N. Tuneski ([22], [31], [32]) dedicated lot of his work to the study of this class and obtained
some interesting conditions of starlikeness.
Another class which is of considerable interest and has been investigated in many articles
([23], [27]), consists of functions f ∈ A which satisfy
0
zf (z)
zf 00 (z)
> 0, z ∈ E.
Re
1+ 0
f (z)
f (z)
J. Inequal. Pure and Appl. Math., 5(3) Art. 78, 2004
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S OME A PPLICATIONS OF A F IRST ORDER D IFFERENTIAL S UBORDINATION
3
It is known that the functions in this class are starlike, too. We observe that all the classes mentioned above consist of algebraic expressions involving the analytic representations of convex
and starlike functions.
2. P RELIMINARIES
We shall need the following definition and lemmas.
Definition 2.1. A function L(z, t), z ∈ E and t ≥ 0 is said to be a subordination chain if L(·, t)
is analytic and univalent in E for all t ≥ 0, L(z, ·) is continuously differentiable on [0, ∞) for
all z ∈ E and L(z, t1 ) ≺ L(z, t2 ) for all 0 ≤ t1 ≤ t2 .
Lemma 2.1 ([24, p. 159]). The function L(z, t) : E × [0, ∞) → C of the form L(z, t) =
a1 (t)z + · · · with a1 (t) 6= 0 for all t ≥ 0, and lim |a1 (t)| = ∞, is said to be a subordination
t→∞
chain if and only if there exist constants r ∈ (0, 1] and M > 0 such that
(i) L(z, t) is analytic in |z| < r for each t ≥ 0, locally absolutely continuous in t ≥ 0 for
each |z| < r, and satisfies
|L(z, t)| ≤ M |a1 (t)|, for |z| < r, t ≥ 0,
(ii) there exists a function p(z, t) analytic in E for all t ∈ [0, ∞) and measurable in [0, ∞)
for each z ∈ E, such that Re p(z, t) > 0 for z ∈ E, t ≥ 0 and
∂L(z, t)
∂L(z, t)
=z
p(z, t),
∂t
∂z
in |z| < r, and for almost all t ∈ [0, ∞).
Lemma 2.2. Let F be analytic in E and let G be analytic and univalent in E except for points
ζ0 such that limz→ζ0 F (z) = ∞, with F (0) = G(0). If F 6≺ G in E, then there is a point
z0 ∈ E and ζ0 ∈ ∂E (boundary of E) such that F (|z| < |z0 |) ⊂ G(E), F (z0 ) = G(ζ0 ) and
z0 F 0 (z0 ) = mζ0 G0 (ζ0 ) for m ≥ 1.
Lemma 2.2 is due to Miller and Mocanu [6].
Lemma 2.3. Let f ∈ A be such that f 0 (z) ≺ (1 + az) in E, where 0 < a ≤ 1. Then
µ
zf 0 (z)
1+z
≺
, z ∈ E,
f (z)
1−z
where 0 < a ≤ √ 2 sin(πµ/2)
5+4 cos(πµ/2)
.
Lemma 2.4. Let f ∈ A be such that f 0 (z) ≺ (1 + az) in E, where 0 < a ≤ 21 . Then we have
zf 0 (z)
3a
≺1+
z.
f (z)
2−a
Lemma 2.3 and Lemma 2.4 are due to Ponnusamy and Singh [26] and Ponnusamy [25],
respectively.
3. M AIN R ESULTS
Definition 3.1. Let α and δ be fixed real numbers, with 0 < δ ≤ 1, α + δ ≥ 0. Further, let
β and γ be complex numbers such that β 6= 0. Then, by R(α, β, γ, δ), we denote the class of
functions p ∈ A0 , such that the function
δ
zp0 (z)
α
P (z) = (p(z)) p(z) +
, P (0) = p(0), z ∈ E,
βp(z) + γ
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is well defined in E.
For fixed α and δ, where α + δ ≥ 0, 0 < δ ≤ 1 and for given β, γ ∈ C with β 6= 0, the class
R(α, β, γ, δ) is non-empty, as the function p(z) = 1 + pn z n , where z ∈ E, pn ∈ C and n ∈ N,
belongs to R(α, β, γ, δ).
We are now in a position to state and prove our main theorem.
0
zq (z)
Theorem 3.1. For a function q ∈ R(α, β, γ, δ), analytic and univalent in E, set βq(z)+γ
= Q(z).
Suppose that q satisfies the following conditions:
(i) Re(βq(z)
h 0 + γ) >0 0,i z ∈ E, and
(z)
(z)
(ii) Re αδ zqq(z)
+ zQ
> 0, z ∈ E.
Q(z)
If p ∈ R(α, β, γ, δ) satisfies the first order differential subordination
δ
δ
zp0 (z)
zq 0 (z)
α
α
(3.1) (p(z)) p(z) +
≺ (q(z)) q(z) +
= h(z), h(0) = 1,
βp(z) + γ
βq(z) + γ
for all z ∈ E. Then p ≺ q and q is the best dominant of differential subordination (3.1). All the
powers chosen here are the principal ones.
Proof. Without any loss of generality, we assume that q is univalent on Ē. If not, then we can
replace p, q, and h by pr (z) = p(rz), qr (z) = q(rz) and hr (z) = h(rz), respectively, where
0 < r < 1. These new functions satisfy the conditions of the theorem on Ē. We would then
prove that pr ≺ qr , and by letting r → 1− , we obtain p ≺ q.
We need to prove that p is subordinate to q in E. If possible, suppose that p 6≺ q in E.
Then by Lemma 2.2, there exist points z0 ∈ E and ζ0 ∈ ∂E such that p(|z| < |z0 |) ⊂ q(E),
p(z0 ) = q(ζ0 ) and z0 p0 (z0 ) = mζ0 q 0 (ζ0 ) for m ≥ 1. Now
δ
δ
mζ0 q 0 (ζ0 )
z0 p0(z0 )
α
α
= (q(ζ0 )) q(ζ0 ) +
.
(3.2)
(p(z0 )) p (z0 ) +
βp(z0 ) + γ
βq(ζ0 ) + γ
Consider the function
α
L(z, t) = (q(z))
zq 0 (z)
q(z) + (1 + t)
βq(z) + γ
δ
= (q(z))α (q(z) + (1 + t)Q(z))δ
= 1 + a1 (t)z + · · · .
As q ∈ R(α, β, γ, δ), so L(z, t) is analytic in E for all t ≥ 0 and is continuously differentiable
on [0, ∞) for all z ∈ E.
∂L(z, t)
1+t
0
a1 (t) =
= q (0) α + δ 1 +
.
∂z
β+γ
z=0
Since q is univalent in E, q 0 (0) 6= 0 and in view of condition (i), β + γ 6= 0. Thus, a1 (t) 6= 0
and lim |a1 (t)| = ∞. Moreover,
t→∞
α
zq 0 (z)
t) βq(z)+γ
(q(z)) q(z) + (1 +
L(z, t)
h
=
a1 (t)
0
q (0) α + δ 1 +
J. Inequal. Pure and Appl. Math., 5(3) Art. 78, 2004
1+t
β+γ
i
δ
.
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S OME A PPLICATIONS OF A F IRST ORDER D IFFERENTIAL S UBORDINATION
5
Taking limits on both sides
lim
t→∞
L(z, t)
= lim
t→∞
a1 (t)
δ
zq 0 (z)
+ ( 1t + 1) βq(z)+γ
i .
h
1
+1
1
α
t
0
1−δ
q (0)t
+ δ t + β+γ
t
(q(z))α
q(z)
t
When 0 < δ < 1,
L(z, t)
= 0.
t→∞ a1 (t)
Thus, there exists a positive constant such that
lim
|L(z, t)| < |a1 (t)|, t ≥ 0.
When δ = 1, we have
(q(z))α Q(z)
L(z, t)
= 0
= φ(z), (say).
t→∞ a1 (t)
q (0)/β + γ
In view of condition (ii), it is obvious that φ is a normalized starlike (with respect to the origin),
and hence, univalent function. By the Growth theorem for univalent functions [18, p. 217], the
function φ is bounded for all z, |z| < r, where r ∈ (0, 1]. Thus, we conclude that for 0 < δ ≤ 1,
there exist constants K0 > 0 and r0 ∈ (0, 1] such that
lim
|L(z, t)| ≤ K0 |a1 (t)|, |z| < r0 , t ∈ [0, ∞).
Now
∂L(z, t)
= (q(z))α δ(q(z) + (1 + t)Q(z))δ−1 Q(z).
∂t
z∂L/∂z
α zq 0 (z) (q(z) + (1 + t)Q(z)) z(q 0 (z) + (1 + t)Q0 (z))
=
+
∂L/∂t
δ q(z)
Q(z)
Q(z)
α
0
α zq (z) zQ0 (z)
=
+ 1 (βq(z) + γ) + (1 + t)
+
.
δ
δ q(z)
Q(z)
In view of conditions (i)) and (ii), we obtain
z∂L/∂z
Re
> 0, z ∈ E,
∂L/∂t
(3.3)
for α +δ ≥ 0 and 0 < δ ≤ 1. Thus, all the conditions of Lemma 2.1 are satisfied. Therefore, we
deduce that L(z, t) is a subordination chain. In view of Definition 2.1, L(z, t1 ) ≺ L(z, t2 ) for
0 ≤ t1 ≤ t2 . Since L(z, 0) = h(z), we get L(ζ0 , t) ∈
/ h(E) for |ζ0 | = 1 and t ≥ 0. Moreover, h
is univalent in E so that the subordination (3.1) is well-defined. In view of (3.2), we can write
δ
z0 p0(z0 )
α
(p(z0 )) p (z0 ) +
= L(ζ0 , m − 1) ∈
/ h(E),
βp(z0 ) + γ
which is a contradiction to (3.1). Hence p ≺ q in E. This completes the proof.
Remark 3.2. From (3.3), we observe that the condition (i) in Theorem 3.1 can be omitted in
case α + δ = 0.
Setting α = 0 and δ = 1 in Theorem 3.1, we obtain the following result on Briot-Bouquet
differential subordination (also see [7]):
Corollary 3.3. Let β, γ ∈ C with β 6= 0. Let q ∈ R(0, β, γ, 1), be a univalent function which
satisfies the following conditions:
(i) Re(βq(z) + γ) > 0, z ∈ E and
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(ii) log(βq(z) + γ) is convex in E.
If an analytic function p ∈ R(0, β, γ, 1) satisfies the differential subordination
zq 0 (z)
zp0 (z)
p(z) +
≺ q(z) +
, z ∈ E,
βp(z) + γ
βq(z) + γ
then p ≺ q in E.
Taking α = δ = 1 and γ = 0, Theorem 3.1 provides the following result (also see [5] and
[21]):
Corollary 3.4. Let β be a complex number, β 6= 0. Let q ∈ R(1, β, 0, 1), be convex univalent
in E for which Re βq(z) > 0, z ∈ E. If an analytic function p ∈ R(1, β, 0, 1) satisfies the
differential subordination
βp2 (z) + zp0 (z) ≺ βq 2 (z) + zq 0 (z),
z ∈ E,
then p ≺ q in E.
Writing α = −1, β = 1, γ = 0 and δ = 1 in Theorem 3.1, we obtain the following result of
Ravichandran and Darus [28]:
0
(z)
is starlike in E. If p ∈ A0 satisfies
Corollary 3.5. Let q ∈ A0 be univalent in E such that zq
q 2 (z)
the differential subordination
zp0 (z)
zq 0 (z)
≺
, z ∈ E,
p2 (z)
q 2 (z)
then p(z) ≺ q(z) in E.
4. A PPLICATIONS TO U NIVALENT F UNCTIONS
In this section, we obtain sufficient conditions for a function to be starlike, strongly starlike
and univalent in E.
Theorem 4.1. Let α and δ be fixed real numbers, with 0 < δ ≤ 1, α + δ ≥ 0. Let λ, Re λ > 0,
0 (z)
be a univalent function which satisfies
be a complex number. For f and g ∈ A, let G(z) = zgg(z)
(i) Re hλ1 G(z) > 0 in E; and
0 (z)
(ii) Re αδ − 1 zG
+1+
G(z)
zG00 (z)
G0 (z)
i
> 0, z ∈ E.
Then, if f ∈ A satisfies the differential subordination
0 α δ
zf (z)
zf 0 (z)
zf 00 (z)
(1 − λ)
+λ 1+ 0
f (z)
f (z)
f (z)
0 α δ
zg (z)
zg 0 (z)
zg 00 (z)
≺
· (1 − λ)
+λ 1+ 0
,
g(z)
g(z)
g (z)
then,
zf 0 (z)
f (z)
≺
zg 0 (z)
g(z)
z ∈ E,
in E.
Proof. The proof of the theorem follows by writing p(z) =
γ = 0 in Theorem 3.1.
zf 0 (z)
f (z)
, q(z) =
zg 0 (z)
g(z)
, β = 1/λ and
Letting λ = 1 and α = 1 − δ in Theorem 4.1, we obtain the following general subordination
theorem for the class of δ-starlike functions.
Theorem 4.2. Let δ, 0 < δ ≤ 1, be fixed and let f and g ∈ A. Assume that G(z) =
univalent in E which satisfies the following conditions:
J. Inequal. Pure and Appl. Math., 5(3) Art. 78, 2004
zg 0 (z)
g(z)
is
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S OME A PPLICATIONS OF A F IRST ORDER D IFFERENTIAL S UBORDINATION
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(i) Re G(z)
> 0 i.e. g is starlike in iE; and
h
0 (z)
zG00 (z)
1
(ii) Re δ − 2 zG
+
1
+
> 0, z ∈ E.
G(z)
G0 (z)
Then the differential subordination
0 1−δ δ 0 1−δ δ
zf (z)
zf 00 (z)
zg (z)
zg 00 (z)
(4.1)
1+ 0
≺
1+ 0
,
f (z)
f (z)
g(z)
g (z)
implies that
zf 0 (z)
f (z)
≺
zg 0 (z)
g(z)
z ∈ E,
in E.
For δ = 1, α = 0, Theorem 4.1 gives the following subordination result.
Theorem 4.3. Let λ, Re λ > 0, be a complex number. For g ∈ A, let G(z) =
in E and satisfy the following conditions:
(i) Re λ1 G(z) > 0, z ∈ E; and
0 (z)
(ii) zG
is starlike in E.
G(z)
zg 0 (z)
g(z)
If an analytic function f ∈ A satisfies the differential subordination
zf 0 (z)
zf 00 (z)
zg 0 (z)
zg 00 (z)
(4.2) (1 − λ)
+λ 1+ 0
≺ (1 − λ)
+λ 1+ 0
,
f (z)
f (z)
g(z)
g (z)
then
zf 0 (z)
f (z)
≺
zg 0 (z)
g(z)
be univalent
z ∈ E,
in E.
Remark 4.4. We observe that when λ is a positive real number, Theorem 4.3 provides a general
subordination result for the class Mλ .
Putting δ = λ = 1, α = −1 in Theorem 4.1, we obtain the following general subordination
result for the class Gb :
0
(z)
Theorem 4.5. For a function g ∈ A, set zgg(z)
= G(z). Suppose that G(z) is a univalent
function which satisfies
2zG0 (z) zG00 (z)
Re 1 −
+
> 0, z ∈ E.
G(z)
G(z)
If an analytic function f ∈ A satisfies the differential subordination
1 + zf 00 (z)/f 0 (z)
1 + zg 00 (z)/g 0 (z)
≺
,
zf 0 (z)/f (z)
zg 0 (z)/g(z)
then
zf 0 (z)
f (z)
≺
zg 0 (z)
g(z)
z ∈ E,
in E.
Taking δ = 1, α = 0 and λ = 1 in Theorem 4.1, we obtain the following Marx-Strohhäcker
differential subordination theorem of first type (also see [9]):
0
(z)
Corollary 4.6. Let g ∈ A be starlike in E. Suppose that zgg(z)
= G(z) is univalent and log G(z)
is convex in E. If an analytic function f ∈ A satisfies the differential subordination
1+
then
zf 0 (z)
f (z)
≺
zg 0 (z)
g(z)
zf 00 (z)
zg 00 (z)
≺
1
+
,
f 0 (z)
g 0 (z)
z ∈ E,
in E.
Writing α = δ = λ = 1 in Theorem 4.1, we have the following subordination result:
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S UKHJIT S INGH
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0
(z)
= G(z). Assume that G is convex univalent
Corollary 4.7. Let g ∈ A be starlike in E. Set zgg(z)
in E. If f ∈ A satisfies the differential subordination
zf 0 (z) z 2 f 00 (z)
zg 0 (z) z 2 g 00 (z)
+
≺
+
,
f (z)
f (z)
g(z)
g(z)
then
zf 0 (z)
f (z)
≺
zg 0 (z)
g(z)
z ∈ E,
in E.
Setting p(z) = f 0 (z), q(z) = g 0 (z) and β =
1
λ
and γ = 0 in Theorem 3.1, we obtain
Theorem 4.8. Let α and δ be fixed real numbers, with 0 < δ ≤ 1 and α + δ ≥ 0. Let λ ∈ C
with Re λ > 0. Assume that g ∈ A satisfies the following conditions:
(i) Re hλ1 g 0 (z) > 0, z ∈ E; and
i
zg00 (z)
zg 000 (z)
α
(ii) Re δ − 1 g0 (z) + 1 + g00 (z) > 0 in E.
If f ∈ A satisfies the differential subordination
(f 0 (z))α−δ [(f 0 (z))2 + λzf 00 (z)]δ ≺ (g 0 (z))α−δ [(g 0 (z))2 + λzg 00 (z)]δ ,
for all z ∈ E, then f 0 (z) ≺ g 0 (z) in E.
Writing α = δ = 1 in Theorem 4.8, we obtain the following result:
Corollary 4.9. Let λ ∈ C with Re λ > 0. Let g ∈ A be such that
(i) Re λ1 g 0 (z) > 0, z ∈ E and
(ii) g 0 is convex in E.
If f ∈ A satisfies the differential subordination
f 02 (z) + λzf 00 (z) ≺ g 02 (z) + λzg 00 (z),
z ∈ E,
then f 0 (z) ≺ g 0 (z) in E.
Taking δ = λ = 1, α = −1 in Theorem 4.8, we obtain the following result:
00
(z)
is starlike in E. If an analytic function f,
Corollary 4.10. Let g ∈ A be such that zg
g 02 (z)
f (0) = 0, satisfies the differential subordination
zf 00 (z)
zg 00 (z)
≺
,
f 02 (z)
g 02 (z)
z ∈ E,
then f 0 (z) ≺ g 0 (z) in E.
Setting α = 0 and δ = 1 in Theorem 4.8, we get:
Corollary 4.11. Let λ be a complex number, Re λ > 0. Let g ∈ A be such that
(i) Re λ1 g 0 (z) > 0, z ∈ E, and
(ii) log g 0 (z) is convex in E.
If f ∈ A satisfies the differential subordination
f 0 (z) + λ
zf 00 (z)
zg 00 (z)
0
≺
g
(z)
+
λ
,
f 0 (z)
g 0 (z)
z ∈ E,
then f 0 (z) ≺ g 0 (z) in E.
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S OME A PPLICATIONS OF A F IRST ORDER D IFFERENTIAL S UBORDINATION
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4.1. Conditions for Strongly Starlikeness. A function f ∈ A is said to be strongly starlike of
order α, 0 < α ≤ 1, if
0
arg zf (z) < απ , z ∈ E.
f (z) 2
Let the class of all such functions be denoted by S̃(α).
Consider G(z) = ( 1+z
)µ , 0 < µ < 1. Let α, δ and λ be fixed real numbers such that
1−z
0 < δ ≤ 1, α + δ ≥ 0 and λ > 0. Then
(i) Re hλ1 G(z) > 0 in E; and
i
h
i
0 (z)
zG00 (z)
α 2µz
1+z 2
(ii) Re αδ − 1 zG
=
Re
+
> 0, z ∈ E.
+
1
+
G(z)
G0 (z)
δ 1−z 2
1−z 2
Thus, all the conditions of Theorem 4.1 are satisfied. Therefore, we obtain the following
result:
Theorem 4.12. Let α, δ and λ be fixed real numbers such that 0 < δ ≤ 1, α + δ ≥ 0 and
λ > 0. For a real number µ, 0 < µ < 1, if f ∈ A satisfies the differential subordination
0 α δ αµ µ
δ
zf (z)
zf 0 (z)
zf 00 (z)
1+z
1+z
2λµz
(1 − λ)
+λ 1+ 0
≺
+
,
f (z)
f (z)
f (z)
1−z
1−z
1 − z2
then f ∈ S̃(µ).
Remark 4.13. Theorem 4.12 can also be written as:
Let α, δ, λ and µ be fixed real numbers such that 0 < δ ≤ 1, α + δ ≥ 0, λ > 0 and 0 < µ < 1.
If an analytic function f in A satisfies
0 α δ 0
00
zf (z)
zf (z)
zf (z)
νπ
(1 − λ)
+λ 1+ 0
,
arg
<
f (z)
f (z)
f (z)
2
then f ∈ S̃(µ), where
2
µπ
λµ
ν = αµ + δ arctan tan
+
.
π
2
(1 − µ)(1−µ)/2 (1 + µ)(1+µ)/2 cos µπ
2
Deductions.
Writing λ = 1 and α = 1 − δ in Theorem 4.12, we obtain the following result, essentially
proved by Darus and Thomas [2]:
(i). Let δ, 0 < δ ≤ 1, and µ, 0 < µ < 1, be fixed real numbers. Let f ∈ A satisfy
0 1−δ δ ν
zf (z)
zf 00 (z)
1+z
1+ 0
≺
, z ∈ E,
f (z)
f (z)
1−z
then f ∈ S̃(µ), where
2δ
µπ
µ
ν = (1 − δ)µ +
arctan tan
+
.
π
2
(1 − µ)(1−µ)/2 (1 + µ)(1+µ)/2 cos µπ
2
Letting α = 0 and δ = 1 in Theorem 4.12, we get the following result of Mocanu [15]:
(ii). Let λ, λ > 0 and µ, 0 < µ < 1, be fixed real numbers. If f ∈ A satisfies the differential
subordination
µ
zf 0 (z)
zf 00 (z)
1+z
2λµz
(1 − λ)
+λ 1+ 0
≺
+
, z ∈ E,
f (z)
f (z)
1−z
1 − z2
then f ∈ S̃(µ).
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S UKHJIT S INGH AND S USHMA G UPTA
For λ = 0, α = −1 and δ = 1, Theorem 4.12 gives the following result of Ravichandran and
Darus [28]:
(iii). For µ, 0 < µ < 1, a fixed real number, let f ∈ A satisfy
2µz(1 − z)µ−1
1 + zf 00 (z)/f 0 (z)
≺1+
zf 0 (z)/f (z)
(1 + z)µ+1
for all z ∈ E. Then f ∈ S̃(µ).
Writing α = 0 and δ = λ = 1 in Theorem 4.12, we obtain the following result of Nunokawa
and Thomas [20]:
(iv). For µ, 0 < µ < 1, a fixed real number, let f ∈ A satisfy
ν
1+z
zf 00 (z)
,
1+ 0
≺
f (z)
1−z
for all z ∈ E. Then f ∈ S̃(µ), where
µπ
µ
2
ν = arctan tan
+
.
π
2
(1 − µ)(1−µ)/2 (1 + µ)(1+µ)/2 cos µπ
2
Taking α = δ = λ = 1 in Theorem 4.12, we obtain the following result of Padmanabhan
[23]:
(v). For µ, 0 < µ < 1, a fixed real number, let f ∈ A satisfy
µ µ
zf 0 (z)
zf 00 (z)
1+z
1+z
2µz
1+ 0
≺
+
,
f (z)
f (z)
1−z
1−z
1 − z2
for all z ∈ E. Then f ∈ S̃(µ).
4.2. Conditions for Starlikeness. Consider G(z) = 1+az
, −1 < a ≤ 1. Then, for |α| ≤
1−z
δ, 0 < δ ≤ 1 and λ > 0, we have
(i) Re hλ1 G(z) > λ1 ( 1−a
) > 0 in E; and
2
i
0 (z)
00 (z)
zG
α
1
1
(ii) Re αδ − 1 G(z) + 1 + zG
>
Re
(1
−
)
−
> 0, z ∈ E.
0
G (z)
δ
1+az
2
Thus G satisfies all the conditions of Theorems 4.1. Therefore, we obtain:
Theorem 4.14. Let |α| ≤ δ, 0 < δ ≤ 1 be fixed and λ be a real number, λ > 0. If f ∈ A
satisfies the differential subordination
0 α δ
zf (z)
zf 0 (z)
zf 00 (z)
(1 − λ)
+λ 1+ 0
f (z)
f (z)
f (z)
α δ
1 + az
1 + (λ + a)z
λaz
≺
+
,
1−z
1−z
1 + az
where −1 < a ≤ 1, then f ∈ St( 1−a
).
2
Deductions.
Taking λ = 1, α = 1 − δ and a = 0 in Theorem 4.14, we obtain the following result:
(i). Let f ∈ A satisfy
0 1−δ δ
zf (z)
zf 00 (z)
(1 + z)δ 1
1+ 0
≺
, ≤ δ ≤ 1.
f (z)
f (z)
1−z 2
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S OME A PPLICATIONS OF A F IRST ORDER D IFFERENTIAL S UBORDINATION
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Then f ∈ St(1/2).
Remark 4.15. For δ = 1, Deduction (i) yields the well-known result: K ⊂ St(1/2).
Writing α = 0 and δ = 1 in Theorem 4.14, we obtain the following form of the result of
Mocanu [14]:
(ii). Let λ be a real number, such that λ > 0. If f ∈ A satisfies the differential subordination
zf 0 (z)
zf 00 (z)
1 + (a + λ)z
λaz
(1 − λ)
+λ 1+ 0
≺
+
,
f (z)
f (z)
1−z
1 + az
where −1 < a ≤ 1 and z ∈ E, then zf 0 (z)/f (z) ≺ (1 + az)/(1 − z).
Taking λ = δ = 1 and α = −1 in Theorem 4.14, we have the following result:
(iii). If f ∈ A satisfies
(1 + a)z
1 + zf 00 (z)/f 0 (z)
≺1+
, −1 < a ≤ 1, z ∈ E,
0
zf (z)/f (z)
(1 + az)2
then zf 0 (z)/f (z) ≺ (1 + az)/(1 − z).
Remark 4.16. For a = 1, Deduction (iii) gives Theorem 3 in [22] and Corollary 8 in [28].
Theorem 3 (in case γ = 0) in [31] is improved by writing a = 1 − 2α in above deduction.
For λ = δ = 1 and α = 0, we obtain from Theorem 4.14:
(iv). Let f ∈ A satisfy
zf 00 (z)
1 + (a + 1)z
az
1+ 0
≺
+
, −1 < a ≤ 1, z ∈ E.
f (z)
1−z
1 + az
Then zf 0 (z)/f (z) ≺ (1 + az)/(1 − z).
Taking α = δ = λ = 1 in Theorem 4.14, we obtain the following result:
(v). If f ∈ A satisfies
2
zf 0 (z)
zf 00 (z)
1 + az
(1 + a)z
1+ 0
≺
+
,
f (z)
f (z)
1−z
(1 − z)2
for all z in E and −1 < a ≤ 1, then zf 0 (z)/f (z) ≺ (1 + az)/(1 − z).
Remark 4.17. Theorem 3 of Padmanabhan [23] corresponds to a = 0 in deduction (v).
Writing a = 1 and α = 1 − δ in Theorem 4.14, we obtain:
Theorem 4.18. Let δ, 1/2 < δ ≤ 1, be fixed. For λ, λ > 0, let f ∈ A satisfy
0 1−δ δ
zf (z)
zf 0 (z)
zf 00 (z)
(1 − λ)
+λ 1+ 0
6= it,
f (z)
f (z)
f (z)
√ 2+λ δ−1/2
. Then f ∈ St.
where t is a real number with |t| ≥ δ δ λ 2δ−1
Proof. As the proof runs on the same lines as in [1], so we omit it.
Deductions.
For λ = 1 in Theorem 4.18, we obtain the following result which improves the Theorem 2 of
Darus and Thomas [2]:
(i). Let δ, 1/2 < δ ≤ 1, be fixed. Let f ∈ A satisfy
0 1−δ δ
zf (z)
zf 00 (z)
1+ 0
6= it,
f (z)
f (z)
δ−1/2
3
where t is a real number with |t| ≥ δ δ 2δ−1
. Then f ∈ St.
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12
S UKHJIT S INGH AND S USHMA G UPTA
Writing δ = 1 in Theorem 4.18, we get the the following result obtained by Nunokawa [19]:
(ii). Let λ, λ > 0 be a real number. If f ∈ A satisfies the differential subordination
zf 00 (z)
zf 0 (z)
(1 − λ)
+λ 1+ 0
6= it,
f (z)
f (z)
p
where t is a real number and |t| ≥ λ(λ + 2). Then f is starlike in E.
Taking λ = δ = 1 in Theorem 4.18, we obtain the following result of Mocanu [13]:
(iii). If f ∈ A satisfies
zf 00 (z)
6= it, z ∈ E,
1+ 0
f (z)
√
where t is a real number and |t| ≥ 3, then f ∈ St.
4.3. Conditions for Univalence. Consider g 0 (z) = 1+az
, −1 < a ≤ 1. It can be easily
1−z
checked that all the conditions of Theorem 4.8 are satisfied. Thus, we obtain the following
result:
Theorem 4.19. Let |α| ≤ δ, 0 < δ ≤ 1, be fixed real numbers. Let λ be a positive real number.
If f ∈ A satisfies the differential subordination
#δ
α−δ "
2
1
+
az
(1
+
a)z
1
+
az
(f 0 (z))α−δ [(f 0 (z))2 + λzf 00 (z)]δ ≺
+λ
,
1−z
1−z
(1 − z)2
for all z ∈ E and −1 < a ≤ 1, then f 0 (z) ≺ (1 + az)/(1 − z) in E.
Deductions.
For α = δ = 1, Theorem 4.19 gives the following result:
(i). Let λ be a positive real number. If f ∈ A satisfies the differential subordination
2
1 + az
(1 + a)z
0
2
00
(f (z)) + λzf (z) ≺
+λ
, z ∈ E,
1−z
(1 − z)2
for all z ∈ E and −1 < a ≤ 1, then f 0 (z) ≺ (1 + az)/(1 − z) in E.
Taking α = −1 and δ = λ = 1 in Theorem 4.19, we obtain:
(ii). If f ∈ A satisfies
zf 00 (z)
(1 + a)z
≺
,
z ∈ E,
(f 0 (z))2
(1 + az)2
then f 0 (z) ≺ (1 + az)/(1 − z).
In particular, a = 0 gives the following interesting result:
(iii). If f ∈ A satisfies
zf 00 (z)
≺ z, z ∈ E,
(f 0 (z))2
then Re f 0 (z) > 1/2 for all z in E.
Writing α = 0 and δ = 1 in Theorem 4.19, we get the following result:
(iv). Let λ be a positive real number. If f ∈ A satisfies
zf 00 (z)
1 + az
λ(1 + a)z
f (z) + λ 0
≺
+
,
f (z)
1−z
(1 + az)(1 − z)
0
z ∈ E,
where −1 < a ≤ 1, then f 0 (z) ≺ (1 + az)/(1 − z). For a = 1, this result can be
expressed as (also see [1]):
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S OME A PPLICATIONS OF A F IRST ORDER D IFFERENTIAL S UBORDINATION
13
(v). Let λ be a positive real number.If f ∈ A satisfies
f 0 (z) + λ
zf 00 (z)
6= it,
f 0 (z)
p
where t is a real number and |t| ≥ λ(λ + 2), then Re f 0 (z) > 0 in E.
Writing g 0 (z) = 1 + az, 0 < a ≤ 1, in Theorem 4.8, then, for |α| ≤ δ, 0 < δ ≤ 1 and λ > 0,
we obtain:
(i) Re hλ1 g 0 (z) > 0 in E; and
i
00 (z)
000 (z)
1
(ii) Re αδ − 1 zgg0 (z)
+ 1 + zgg00 (z)
= Re (1 − αδ ) 1+az
+ αδ > 0, z ∈ E.
Thus, all the conditions of Theorem 4.8 are satisfied. Therefore, we obtain:
Theorem 4.20. Let |α| ≤ δ, 0 < δ ≤ 1, be fixed real numbers. Let λ be a positive real number.
If f ∈ A satisfies the differential subordination
(4.3)
(f 0 (z))α−δ [(f 0 (z))2 + λzf 00 (z)]δ ≺ (1 + az)α−δ [(1 + az)2 + λaz]δ ,
where 0 < a ≤ 1, then |f 0 (z) − 1| < a.
Remark 4.21. Let f ∈ A satisfy (4.3). Then by Lemma 2.3, f ∈ S̃(µ), where µ is given by
2 sin(πµ/2)
0<a≤ p
5 + 4 cos(πµ/2)
.
Remark 4.22. For 0 < a ≤ 1/2, let f ∈ A satisfy (4.3). Then by Lemma 2.4,
zf 0 (z)
3a
≺1+
z, z ∈ E.
f (z)
2−a
Deductions.
For α = δ = 1, Theorem 4.20 gives the following result:
(i). For λ, λ > 0, if f ∈ A satisfies
(f 0 (z))2 + λzf 00 (z) ≺ (1 + az)2 + λaz, 0 < a ≤ 1,
z ∈ E,
then |f 0 (z) − 1| < a in E.
Writing α = −1 and δ = λ = 1 in Theorem 4.20, we obtain:
(ii). If f ∈ A satisfies
zf 00 (z)
az
≺
, 0 < a ≤ 1,
0
2
(f (z))
(1 + az)2
z ∈ E,
then |f 0 (z) − 1| < a in E.
Taking α = 0 and δ = 1 in Theorem 4.20, we obtain:
(iii). Let f ∈ A satisfy
f 0 (z) + λ
zf 00 (z)
λaz
≺ (1 + az) +
,
0
f (z)
1 + az
z ∈ E,
where 0 < a ≤ 1 and λ is a positive real number. Then |f 0 (z) − 1| < a in E.
We note that results similar to the ones proved in Remarks 4.21 and 4.22 can also be derived
in the case of deductions (i), (ii) and (iii).
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S UKHJIT S INGH AND S USHMA G UPTA
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