Journal of Inequalities in Pure and
Applied Mathematics
SOME APPLICATIONS OF A FIRST ORDER DIFFERENTIAL
SUBORDINATION
volume 5, issue 3, article 78,
2004.
SUKHJIT SINGH AND SUSHMA GUPTA
Department of Mathematics,
Sant Longowal Institute of Engineering & Technology
Longowal-148 106(Punjab)-India.
Received 28 January, 2004;
accepted 19 July, 2004.
Communicated by: A. Lupaş
EMail: sukhjit_d@yahoo.com
EMail: sushmagupta1@yahoo.com
Abstract
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Abstract
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
zp0(z) δ
zq0(z) δ
α
(p(z)) p (z) +
≺ (q(z)) q(z) +
, z ∈ E,
βp(z) + γ
βq(z) + γ
α
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
and as applications, find several sufficient conditions for starlikeness and univalence of functions analytic in the unit disc E.
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2000 Mathematics Subject Classification: Primary 30C45, Secondary 30C50.
Key words: Univalent function, Starlike function, Convex function, Differential subordination.
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1
2
3
4
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Main Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Applications to Univalent Functions . . . . . . . . . . . . . . . . . . . . .
4.1
Conditions for Strongly Starlikeness . . . . . . . . . . . . . .
4.2
Conditions for Starlikeness . . . . . . . . . . . . . . . . . . . . . .
4.3
Conditions for Univalence . . . . . . . . . . . . . . . . . . . . . .
References
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1.
Introduction
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),
z ∈ E, ψ(p(0), 0) = h(0)
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
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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the type
(1.2)
p(z) +
zp0 (z)
≺ h(z),
βp(z) + γ
z ∈ E,
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) +
βp(z) + γ
βq(z) + γ
= h(z), z ∈ E,
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)
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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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)
+α 1+ 0
> 0, z ∈ E
Mα = f ∈ A : Re (1 − α)
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)
Re
1+ 0
> 0, z ∈ E.
f (z)
f (z)
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.
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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2.
Preliminaries
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
t→∞
said to be a subordination 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.
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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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
1+z
zf 0 (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
3a
zf 0 (z)
≺1+
z.
f (z)
2−a
Lemma 2.3 and Lemma 2.4 are due to Ponnusamy and Singh [26] and Ponnusamy [25], respectively.
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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3.
Main Results
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
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.
Theorem 3.1. For a function q ∈ R(α, β, γ, δ), analytic and univalent in E, set
zq 0 (z)
= Q(z). Suppose that q satisfies the following conditions:
βq(z)+γ
(i) Re(βq(z) + γ) > 0, z ∈ E, and
i
h 0
0 (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) +
βp(z) + γ
βq(z) + γ
= h(z),
h(0) = 1,
Some Applications of a First
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Sukhjit Singh and Sushma Gupta
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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
z0 p0(z0 )
mζ0 q 0 (ζ0 )
α
α
(3.2) (p(z0 )) p (z0 ) +
= (q(ζ0 )) q(ζ0 ) +
.
βp(z0 ) + γ
βq(ζ0 ) + γ
Consider the function
zq 0 (z)
L(z, t) = (q(z))α q(z) + (1 + t)
βq(z) + γ
α
= (q(z)) (q(z) + (1 + t)Q(z))δ
= 1 + a1 (t)z + · · · .
δ
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
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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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→∞
δ
zq 0 (z)
α
(q(z))
q(z)
+
(1
+
t)
βq(z)+γ
L(z, t)
h
i
.
=
1+t
a1 (t)
q 0 (0) α + δ 1 + β+γ
Taking limits on both sides
zq 0 (z)
δ
(q(z))α q(z)
+ ( 1t + 1) βq(z)+γ
t
L(z, t)
i .
h
lim
= lim
1
+1
t→∞ a1 (t)
t→∞
t
q 0 (0)t1−δ αt + δ 1t + β+γ
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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When 0 < δ < 1,
L(z, t)
= 0.
t→∞ a1 (t)
lim
Thus, there exists a positive constant such that
|L(z, t)| < |a1 (t)|, t ≥ 0.
When δ = 1, we have
L(z, t)
(q(z))α Q(z)
lim
= 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
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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
|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
0
(3.3)
0
0
z∂L/∂z
α zq (z) (q(z) + (1 + t)Q(z)) z(q (z) + (1 + t)Q (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,
∂L/∂t
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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z ∈ E,
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
/ h(E) for |ζ0 | = 1 and t ≥ 0. Moreover, h is univalent in E so
get L(ζ0 , t) ∈
that the subordination (3.1) is well-defined. In view of (3.2), we can write
δ
z0 p0(z0 )
α
= L(ζ0 , m − 1) ∈
/ h(E),
(p(z0 )) p (z0 ) +
βp(z0 ) + γ
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which is a contradiction to (3.1). Hence p ≺ q in E. This completes the proof.
Remark 3.1. 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 3.2. 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
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
(ii) log(βq(z) + γ) is convex in E.
If an analytic function p ∈ R(0, β, γ, 1) satisfies the differential subordination
zp0 (z)
zq 0 (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 3.3. 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),
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z ∈ E,
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then p ≺ q in E.
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Writing α = −1, β = 1, γ = 0 and δ = 1 in Theorem 3.1, we obtain the
following result of Ravichandran and Darus [28]:
Corollary 3.4. Let q ∈ A0 be univalent in E such that
p ∈ A0 satisfies the differential subordination
zp0 (z)
zq 0 (z)
≺
,
p2 (z)
q 2 (z)
then p(z) ≺ q(z) in E.
zq 0 (z)
q 2 (z)
is starlike in E. If
z ∈ E,
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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4.
Applications to Univalent Functions
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 +0 δ ≥ 0.
(z)
Let λ, Re λ > 0, be a complex number. For f and g ∈ A, let G(z) = zgg(z)
be a
univalent function which satisfies
(i) Re λ1 G(z) > 0 in E; and
h
0 (z)
(ii) Re αδ − 1 zG
+1+
G(z)
Some Applications of a First
order Differential Subordination
zG00 (z)
G0 (z)
i
> 0, z ∈ E.
Sukhjit Singh and Sushma Gupta
Then, if f ∈ A satisfies the differential subordination
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δ
α zf 0 (z)
zf 00 (z)
zf 0 (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)
Contents
z ∈ E,
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in E.
Proof. The proof of the theorem follows by writing p(z) =
β = 1/λ and γ = 0 in Theorem 3.1.
Close
zf 0 (z)
,
f (z)
q(z) =
zg 0 (z)
,
g(z)
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.
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Theorem 04.2. Let δ, 0 < δ ≤ 1, be fixed and let f and g ∈ A. Assume that
(z)
G(z) = zgg(z)
is univalent in E which satisfies the following conditions:
(i) Re G(z) > 0 i.e. g is starlike in E; and
h
i
0 (z)
zG00 (z)
(ii) Re 1δ − 2 zG
+
1
+
> 0, z ∈ E.
0
G(z)
G (z)
Then the differential subordination
(4.1)
zf 0 (z)
f (z)
1−δ zf 00 (z)
1+ 0
f (z)
Sukhjit Singh and Sushma Gupta
≺
implies that
zf 0 (z)
f (z)
≺
zg 0 (z)
g(z)
Some Applications of a First
order Differential Subordination
δ
0
zg (z)
g(z)
1−δ 00
1+
zg (z)
g 0 (z)
δ
,
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) =
zg 0 (z)
be univalent in E and satisfy the following conditions:
g(z)
(i) Re λ1 G(z) > 0, z ∈ E; and
(ii)
zG0 (z)
G(z)
is starlike in E.
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If an analytic function f ∈ A satisfies the differential subordination
zf 0 (z)
zf 00 (z)
(4.2) (1 − λ)
+λ 1+ 0
f (z)
f (z)
zg 0 (z)
zg 00 (z)
≺ (1 − λ)
+λ 1+ 0
, z ∈ E,
g(z)
g (z)
then
zf 0 (z)
f (z)
≺
zg 0 (z)
g(z)
in E.
Remark 4.1. 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 4.4. 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]):
Some Applications of a First
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0
(z)
Corollary 4.5. 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)
zg 00 (z)
zf 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:
0
(z)
Corollary 4.6. Let g ∈ A be starlike in E. Set zgg(z)
= G(z). Assume that G is
convex univalent in E. If f ∈ A satisfies the differential subordination
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
Title Page
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,
JJ
J
in E.
Setting p(z) = f 0 (z), q(z) = g 0 (z) and β =
we obtain
1
λ
and γ = 0 in Theorem 3.1,
Theorem 4.7. 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 λ1 g 0 (z) > 0, z ∈ E; and
i
h
00 (z)
000 (z)
(ii) Re αδ − 1 zgg0 (z)
+ 1 + zgg00 (z)
> 0 in E.
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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 4.7, we obtain the following result:
Corollary 4.8. Let λ ∈ C with Re λ > 0. Let g ∈ A be such that
(i) Re
1 0
g (z)
λ
> 0, z ∈ E and
Some Applications of a First
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Sukhjit Singh and Sushma Gupta
(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),
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z ∈ E,
then f 0 (z) ≺ g 0 (z) in E.
Taking δ = λ = 1, α = −1 in Theorem 4.7, we obtain the following result:
00
(z)
Corollary 4.9. Let g ∈ A be such that zg
is starlike in E. If an analytic funcg 02 (z)
tion
f,
f (0) = 0, satisfies the differential subordination
zf 00 (z)
zg 00 (z)
≺
,
f 02 (z)
g 02 (z)
z ∈ E,
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then f 0 (z) ≺ g 0 (z) in E.
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Setting α = 0 and δ = 1 in Theorem 4.7, we get:
Corollary 4.10. 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)
Some Applications of a First
order Differential Subordination
z ∈ E,
then f 0 (z) ≺ g 0 (z) in E.
4.1.
Sukhjit Singh and Sushma Gupta
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Conditions for Strongly Starlikeness
A function f ∈ A is said to be strongly starlike of order α, 0 < α ≤ 1, if
arg
απ
zf 0 (z)
<
,
f (z)
2
z ∈ E.
Let the class of all such functions be denoted by S̃(α).
1+z µ
Consider G(z) = ( 1−z
) , 0 < µ < 1. Let α, δ and λ be fixed real numbers
such that 0 < δ ≤ 1, α + δ ≥ 0 and λ > 0. Then
(i) Re λ1 G(z) > 0 in E; and
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(ii) Re
h
α
δ
0 (z)
− 1 zG
+1+
G(z)
zG00 (z)
G0 (z)
i
= Re
h
α 2µz
δ 1−z 2
+
1+z 2
1−z 2
i
> 0, z ∈ E.
Thus, all the conditions of Theorem 4.1 are satisfied. Therefore, we obtain
the following result:
Theorem 4.11. 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
zf 0 (z)
f (z)
α δ
zf 0 (z)
zf 00 (z)
(1 − λ)
+λ 1+ 0
f (z)
f (z)
αµ µ
δ
1+z
1+z
2λµz
≺
+
,
1−z
1−z
1 − z2
then f ∈ S̃(µ).
Some Applications of a First
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Sukhjit Singh and Sushma Gupta
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Contents
Remark 4.2. Theorem 4.11 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
δ
0 α zf 0 (z)
zf 00 (z)
νπ
zf (z)
arg
(1 − λ)
+λ 1+ 0
<
,
f (z)
f (z)
f (z)
2
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then f ∈ S̃(µ), where
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ν = αµ +
2
µπ
λµ
δ arctan tan
+
(1−µ)/2
π
2
(1 − µ)
(1 + µ)(1+µ)/2 cos µπ
2
.
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Deductions.
Writing λ = 1 and α = 1 − δ in Theorem 4.11, 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
δ ν
0 1−δ zf (z)
zf 00 (z)
1+z
≺
, z ∈ E,
1+ 0
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 4.11, 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 00 (z)
1+z
2λµz
zf 0 (z)
+λ 1+ 0
≺
+
, z ∈ E,
(1 − λ)
f (z)
f (z)
1−z
1 − z2
then f ∈ S̃(µ).
For λ = 0, α = −1 and δ = 1, Theorem 4.11 gives the following result of
Ravichandran and Darus [28]:
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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(iii). For µ, 0 < µ < 1, a fixed real number, let f ∈ A satisfy
1 + zf 00 (z)/f 0 (z)
2µz(1 − z)µ−1
≺
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 4.11, 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
zf 00 (z)
1+z
1+ 0
≺
,
f (z)
1−z
for all z ∈ E. Then f ∈ S̃(µ), where
2
µπ
µ
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 4.11, we obtain the following result of
Padmanabhan [23]:
Some Applications of a First
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(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).
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4.2.
Conditions for Starlikeness
Consider G(z) =
we have
1+az
, −1
1−z
< a ≤ 1. Then, for |α| ≤ δ, 0 < δ ≤ 1 and λ > 0,
(i) Re λ1 G(z) > λ1 ( 1−a
) > 0 in E; and
2
h
i
0 (z)
zG00 (z)
(ii) Re αδ − 1 zG
+
1
+
> Re (1 − αδ )
G(z)
G0 (z)
E.
1
1+az
−
1
2
> 0, z ∈
Thus G satisfies all the conditions of Theorems 4.1. Therefore, we obtain:
Theorem 4.12. Let |α| ≤ δ, 0 < δ ≤ 1 be fixed and λ be a real number, λ > 0.
If f ∈ A satisfies the differential subordination
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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α δ
zf (z)
zf 0 (z)
zf 00 (z)
(1 − λ)
+λ 1+ 0
f (z)
f (z)
f (z)
α δ
1 + (λ + a)z
λaz
1 + az
≺
+
,
1−z
1−z
1 + az
0
where −1 < a ≤ 1, then f ∈ St( 1−a
).
2
Deductions.
Taking λ = 1, α = 1 − δ and a = 0 in Theorem 4.12, we obtain the following
result:
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(i). Let f ∈ A satisfy
0 1−δ δ
(1 + z)δ 1
zf (z)
zf 00 (z)
1+ 0
≺
, ≤ δ ≤ 1.
f (z)
f (z)
1−z 2
Then f ∈ St(1/2).
Remark 4.3. For δ = 1, Deduction (i) yields the well-known result: K ⊂
St(1/2).
Writing α = 0 and δ = 1 in Theorem 4.12, 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 00 (z)
1 + (a + λ)z
λaz
zf 0 (z)
(1 − λ)
+λ 1+ 0
≺
+
,
f (z)
f (z)
1−z
1 + az
0
where −1 < a ≤ 1 and z ∈ E, then zf (z)/f (z) ≺ (1 + az)/(1 − z).
Taking λ = δ = 1 and α = −1 in Theorem 4.12, we have the following
result:
(iii). If f ∈ A satisfies
1 + zf 00 (z)/f 0 (z)
(1 + a)z
≺1+
, −1 < a ≤ 1,
0
zf (z)/f (z)
(1 + az)2
Some Applications of a First
order Differential Subordination
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z ∈ E,
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then zf 0 (z)/f (z) ≺ (1 + az)/(1 − z).
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Remark 4.4. 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 4.12:
(iv). Let f ∈ A satisfy
1+
zf 00 (z)
1 + (a + 1)z
az
≺
+
, −1 < a ≤ 1,
f 0 (z)
1−z
1 + az
z ∈ E.
Some Applications of a First
order Differential Subordination
Then zf 0 (z)/f (z) ≺ (1 + az)/(1 − z).
Taking α = δ = λ = 1 in Theorem 4.12, we obtain the following result:
Sukhjit Singh and Sushma Gupta
(v). If f ∈ A satisfies
zf 0 (z)
f (z)
zf 00 (z)
1+ 0
f (z)
≺
1 + az
1−z
2
(1 + a)z
+
,
(1 − z)2
for all z in E and −1 < a ≤ 1, then zf 0 (z)/f (z) ≺ (1 + az)/(1 − z).
Remark 4.5. Theorem 3 of Padmanabhan [23] corresponds to a = 0 in deduction (v).
Writing a = 1 and α = 1 − δ in Theorem 4.12, we obtain:
Theorem 4.13. 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
where t is a real number with |t| ≥ δ δ λ 2δ−1
. Then f ∈ St.
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Proof. As the proof runs on the same lines as in [1], so we omit it.
Deductions.
For λ = 1 in Theorem 4.13, 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
δ
0 1−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.
Writing δ = 1 in Theorem 4.13, 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 0 (z)
zf 00 (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 4.13, we obtain the following result of Mocanu [13]:
(iii). If f ∈ A satisfies
zf 00 (z)
1+ 0
6= it, z ∈ E,
f (z)
√
where t is a real number and |t| ≥ 3, then f ∈ St.
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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4.3.
Conditions for Univalence
Consider g 0 (z) = 1+az
, −1 < a ≤ 1. It can be easily checked that all the
1−z
conditions of Theorem 4.7 are satisfied. Thus, we obtain the following result:
Theorem 4.14. Let |α| ≤ δ, 0 < δ ≤ 1, be fixed real numbers. Let λ be a
positive real number. If f ∈ A satisfies the differential subordination
(f 0 (z))α−δ [(f 0 (z))2 + λzf 00 (z)]δ
#δ
α−δ "
2
1 + az
(1 + a)z
1 + az
+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 4.14 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.
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
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Taking α = −1 and δ = λ = 1 in Theorem 4.14, we obtain:
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(ii). If f ∈ A satisfies
(1 + a)z
zf 00 (z)
≺
,
0
2
(f (z))
(1 + az)2
z ∈ E,
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,
(f 0 (z))2
Some Applications of a First
order Differential Subordination
z ∈ E,
Sukhjit Singh and Sushma Gupta
then Re f 0 (z) > 1/2 for all z in E.
Writing α = 0 and δ = 1 in Theorem 4.14, we get the following result:
Contents
(iv). Let λ be a positive real number. If f ∈ A satisfies
f 0 (z) + λ
zf 00 (z)
1 + az
λ(1 + a)z
≺
+
,
f 0 (z)
1−z
(1 + az)(1 − z)
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]):
(v). Let λ be a positive real number.If f ∈ A satisfies
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.
f 0 (z) + λ
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Writing g 0 (z) = 1 + az, 0 < a ≤ 1, in Theorem 4.7, then, for |α| ≤ δ,
0 < δ ≤ 1 and λ > 0, we obtain:
(i) Re λ1 g 0 (z) > 0 in E; and
h
00 (z)
(ii) Re αδ − 1 zgg0 (z)
+1+
zg 000 (z)
g 00 (z)
i
1
= Re (1 − αδ ) 1+az
+ αδ > 0, z ∈ E.
Thus, all the conditions of Theorem 4.7 are satisfied. Therefore, we obtain:
Theorem 4.15. 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)
0
α−δ
(f (z))
0
2
00
Some Applications of a First
order Differential Subordination
Sukhjit Singh and Sushma Gupta
δ
α−δ
[(f (z)) + λzf (z)] ≺ (1 + az)
2
δ
[(1 + az) + λaz] ,
where 0 < a ≤ 1, then |f 0 (z) − 1| < a.
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Remark 4.6. 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)
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Remark 4.7. 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
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Deductions.
For α = δ = 1, Theorem 4.15 gives the following result:
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(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 4.15, we obtain:
(ii). If f ∈ A satisfies
Some Applications of a First
order Differential Subordination
az
zf 00 (z)
≺
, 0 < a ≤ 1,
0
2
(f (z))
(1 + az)2
z ∈ E,
then |f 0 (z) − 1| < a in E.
Sukhjit Singh and Sushma Gupta
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Taking α = 0 and δ = 1 in Theorem 4.15, we obtain:
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(iii). Let f ∈ A satisfy
f 0 (z) + λ
λaz
zf 00 (z)
≺ (1 + az) +
,
0
f (z)
1 + az
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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 4.6 and 4.7 can
also be derived in the case of deductions (i), (ii) and (iii).
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References
[1] N.E. CHO AND J.A. KIM, On a sufficient condition and an angular estimation for φ-like functions, Taiwanese J. Math., 2(4) (1998), 397–403.
[2] M. DARUS AND D.K. THOMAS, α−logarithmically convex functions,
Indian J. Pure Appl. Math., 29(10) (1998), 1049–1059.
[3] P. EENIGENBURG, S.S. MILLER, P.T. MOCANU AND M.O. READE,
On a Briot-Bouquet differential subordination, General Inequalities 3,
Birkhäuser, Basel, (1983), 339–348.
Some Applications of a First
order Differential Subordination
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Some Applications of a First
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Some Applications of a First
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Some Applications of a First
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