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ACTA UNIVERSITATIS APULENSIS

No 16/2008

DIFFERENTIAL SUBORDINATIONS AND
SUPERORDINATIONS FOR ANALYTIC FUNCTIONS
˘ AGEAN
˘
DEFINED BY THE GENERALIZED SAL
DERIVATIVE

Veronica Oana Nechita

Abstract. Let q1 and q2 be univalent in the unit disk U , with q1 (0) =
q2 (0) = 1. We give applications of first order differential subordination
and superordination to obtain sufficient conditions for normalized analytic
Dλm f (z)
≺ q2 (z)
functions f ∈ A to satisfy one of the conditions q1 (z) ≺ m+1
Dλ f (z)
Dm+1 f (z)

≺ q2 (z), where Dλm f (z) is the generalized S˘al˘agean
or q1 (z) ≺ z λm
{Dλ f (z)}2
differential operator.
2000 Mathematics Subject Classification:30C45
Keywords and phrases: Differential subordinations, Differential superordinations, Generalized S˘al˘agean derivative.
1. Introduction
Let H = H (U ) denote the class of functions analytic in U = {z ∈ C : |z| < 1}.
For n a positive integer and a ∈ C, let
H [a, n] = {f ∈ H : f (z) = a + an z n + ...} .
We also consider the class


A = f ∈ H : f (z) = z + a2 z 2 + ... .
143

V.O. Nechita - Differential subordinations and superordinations for...
We denote by Q the set of functions f that are analytic and injective on
U \ E (f ), where



E (f ) = ζ ∈ ∂U : lim f (z) = ∞ ,
z→ζ

and are such that f ′ (ζ) 6= 0 for ζ ∈ ∂U \ E (f ).
Since we use the terms of subordination and superordination, we review
here those definitions. Let f, F ∈ H. The function f is said to be subordinate
to F, or F is said to be superordinate to f , if there exists a function w analytic
in U , with w (0) = 0 and |w (z)| < 1, and such that f (z) = F (w (z)). In
such a case we write f ≺ F or f (z) ≺ F (z). If F is univalent, then f ≺ F
if and only if f (0) = F (0) and f (U ) ⊂ F (U ).
Since most of the functions considered in this paper and conditions on
them are defined uniformly in the unit disk U, we shall omit the requirement
”z ∈ U ”.
Let ψ : C3 ×U → C, let h be univalent in U and q ∈ Q. In [3], the authors
considered the problem of determining conditions on admissible functions ψ
such that


ψ p (z) , zp′ (z) , z 2 p′′ (z) ; z ≺ h (z)
(1)

implies p (z) ≺ q (z), for all functions p ∈ H [a, n] that satisfy the differential
subordination (1). Moreover, they found conditions so that the function q is
the ”smallest” function with this property, called the best dominant of the
subordination (1).
Let ϕ : C3 × U → C, let h ∈ H and q ∈ H [a, n]. Recently, in [4], the
authors studied the dual problem and determined conditions on ϕ such that


h (z) ≺ ϕ p (z) , zp′ (z) , z 2 p′′ (z) ; z
(2)

implies q (z) ≺ p (z), for all functions p ∈ Q that satisfy the above differential
superordination. Moreover, they found conditions so that the function q is
the ”largest” function with this property, called the best subordinant of the
superodination (2).
∞
P
P
n = 2∞ an z n and g (z) = z +
bn z n ,

For two functions f (z) = z +
n=2

the Hadamard product (or convolution) of f and g is defined by
(f ∗ g) (z) := z +

∞
X
n=2

144

an b n z n .

V.O. Nechita - Differential subordinations and superordinations for...
Let λ > 0. The generalized S˘al˘agean derivative of a function f is defined
in [1] by


Dλo f (z) = f (z) , Dλ1 f (z) = (1 − λ) f (z)+λzf ′ (z) , Dλm f (z) = Dλ1 Dλm−1 f (z) .
If f (z) = z +

∞
P

an z n , we write the generalized S˘al˘agean derivative as a

n=2

Hadamard product
(
Dλm f

(z) = f (z)∗ z +

∞
X

m

[1 + (n − 1) λ] z

n=2

n

)

∞
X
= z+
[1 + (n − 1) λ]m an z n .
n=2

When λ = 1, we get the classic S˘al˘agean derivative [7], denoted by
D f (z).
In this paper we will determine some properties on admissible functions
defined with the generalized S˘al˘agean derivative.
m

2. Preliminaries

In our present investigation we shall need the folllowing results.
Theorem 1 [[3], Theorem 3.4h., p.132] Let q be univalent in U and let θ
and φ be analytic in a domain D containing q (U ), with φ (w) 6= 0, when
w ∈ q (U ). Set Q (z) = zq ′ (z) · φ [q (z)], h (z) = θ [q (z)] + Q (z) and suppose
that either
(i) h is convex or
(ii) Q is starlike.
In addition, assume that
(iii) Re

zh′ (z)
> 0.
Q (z)

If p is analytic in U , with p (0) = q (0), p (U ) ⊂ D and
θ [p (z)] + zp′ (z) · φ [p (z)] ≺ θ [q (z)] + zp′ (z) · φ [q (z)] = h (z) ,
then p ≺ q, and q is the best dominant.
145

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By taking θ (w) := w and φ (w) := γ in Theorem 1, we get
Corollary 2 Let q be univalent in U , γ ∈ C∗ and suppose




zq ′′ (z)
1
Re 1 + ′
> max 0, −Re
.
q (z)
γ
If p is analytic in U, with p (0) = q (0) and
p (z) + γzp′ (z) ≺ q (z) + γzq ′ (z) ,
then p ≺ q, and q is the best dominant.
Theorem 3 [5] Let θ and φ be analytic in a domain D and let q be univalent
in U , with q (0) = a, q (U ) ⊂ D. Set Q (z) = zq ′ (z) · φ [q (z)], h (z) =
θ [q (z)] + Q (z) and suppose that


 ′
θ [q (z)]
> 0 and
(i) Re
φ [q (z)]
(ii) Q (z) is starlike.
If p ∈ H [a, 1] ∩ Q, p (U ) ⊂ D and θ [p (z)] + zp′ (z) · φ [p (z)] is univalent
in U , then
θ [q (z)] + zp′ (z) · φ [q (z)] ≺ θ [p (z)] + zp′ (z) · φ [p (z)] ⇒ q ≺ p
and q is the best subordinant.
By taking θ (w) := w and φ (w) := γ in Theorem 3, we get
Corollary 4 [2] Let q be convex in U , q (0) = a and γ ∈ C, Reγ > 0. If
p ∈ H [a, 1] ∩ Q and p (z) + γzp′ (z) is univalent in U , then
q (z) + γzq ′ (z) ≺ p (z) + γzp′ (z) ⇒ q ≺ p
and q is the best subordinant.

3. Main results
146

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Theorem 5 Let q be univalent in U with q (0) = 1, γ ∈ C∗ and suppose




1
zq ′′ (z)
> max 0, −Re
.
Re 1 + ′
q (z)
γ
If f ∈ A and
Dλm f (z)
γ
+
m+1
Dλ f (z) λ

(

then

Dm+2 f (z) · Dλm f (z)
1 − λ  m+1
2
Dλ f (z)

)

≺ q (z) + γzq ′ (z)

(3)

Dλm f (z)
≺ q (z)
Dλm+1 f (z)

and q is the best dominant.
Proof. Let
p (z) :=

Dλm f (z)
.
Dλm+1 f (z)

By a simple computation we get
′

z {Dλm f (z)}′ z Dλm+1 f (z)
zp′ (z)
=
−
.
p (z)
Dλm f (z)
Dλm+1 f (z)

(4)

By using the identity
z

{Dλm f



1
1 m+1
Dλm f (z) ,
(z)} = Dλ f (z) + 1 −
λ
λ
′

(5)

we obtain from (4) that


1
Dλm+2 f (z)
1
zp′ (z)
=
−
p (z)
λ p (z) Dλm+1 f (z)
and
Dλm f (z)
γ
p (z) + γzp′ (z) = m+1
+
Dλ f (z) λ

(

Dm+2 f (z) · Dλm f (z)
1 − λ  m+1
2
Dλ f (z)

)

The subordination (3) from hypothesis becomes

p (z) + γzp′ (z) ≺ q (z) + γzq ′ (z) .
We obtain the conclusion of our theorem by simply applying Corrolary 2.
For m = 0, we have the following result.
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Corollary 6 Let q be univalent in U with q (0) = 1, γ ∈ C∗ and suppose




1
zq ′′ (z)
> max 0, −Re
.
Re 1 + ′
q (z)
γ
If f ∈ A and
i
γ
f (z)
(2 − λ)
+
λ
(1 − λ) f (z) + λzf ′ (z)


λ2 z 2 f ′′ (z) f (z) − (1 − λ) f 2 (z)
γ
≺ q (z) + γzq ′ (z)
1−
λ
[(1 − λ) f (z) + λzf ′ (z)]2
h

then

1−

f (z)
≺ q (z)
(1 − λ) f (z) + λzf ′ (z)

and q is the best dominant.
For λ = 1, the generalized S˘al˘agean derivative becomes the classic S˘al˘agean
derivative, denoted by Dm f (z). In this case we have the following consequence of Theorem 5.
Corollary 7 [6] Let q be univalent in U with q (0) = 1, γ ∈ C∗ and suppose




1
zq ′′ (z)
> max 0, −Re
.
Re 1 + ′
q (z)
γ
If f ∈ A and


Dm f (z)
Dm+2 f (z) · Dm f (z)
≺ q (z) + γzq ′ (z)
+γ 1−
Dm+1 f (z)
[Dm+1 f (z)]2
then

Dm f (z)
≺ q (z)
Dm+1 f (z)

and q is the best dominant.
We next consider the case when λ = 1 and m = 0.

148

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Corollary 8 [6] Let q be univalent in U with q (0) = 1, γ ∈ C∗ and suppose




zq ′′ (z)
1
Re 1 + ′
> max 0, −Re
.
q (z)
γ
If f ∈ A and


f ′′ (z) · f (z)
f (z)
≺ q (z) + γzq ′ (z)
+γ 1−
(1 − γ) ′
2
′
zf (z)
[f (z)]
then

f (z)
≺ q (z)
zf ′ (z)

and q is the best dominant.
For our next application, we select in Theorem 5 a particular dominant
q.
Corollary 9 Let A, B, γ ∈ C, A 6= B such that |B| ≤ 1 and Reγ > 0. If
f ∈ A satisfies the subordination
)
(
1 + Az
γ
(A − B) z
Dλm+2 f (z) · Dλm f (z)
Dλm f (z)
,
+γ
1−
≺
+
2
 m+1
m+1
1 + Bz
Dλ f (z) λ
(1 + Bz)2
Dλ f (z)
then

and q (z) =

1 + Az
Dλm+1 f (z)
≺
m
Dλ f (z)
1 + Bz

1 + Az
is the best dominant.
1 + Bz

We apply Corrolary 4 to obtain the following result.
Theorem 10 Let q be convex in U , q (0) = 1(and γ ∈ C, Reγ > 0. If f ∈
) A,
m+2
m
m
m
Dλ f (z)
f (z) · Dλ f (z)
D
Dλ f (z)
γ
is
1 − λ  m+1
∈ H [1, 1] ∩ Q, m+1
+
2
m+1
λ
Dλ f (z)
Dλ f (z)
Dλ f (z)
univalent in U and
(
)
m+2
m
m
D
f
(z)
γ
f
(z)
·
D
D
f
(z)
λ
λ
q (z) + γzq ′ (z) ≺ m+1
+
1 − λ  m+1
,
2
Dλ f (z) λ
Dλ f (z)
149

V.O. Nechita - Differential subordinations and superordinations for...
then
q (z) ≺

Dλm f (z)
Dλm+1 f (z)

and q is the best subordinant.
Corollary 11 Let q be convex in U , q (0) = 1 and γ ∈ C, Reγ > 0. If f ∈ A,
h
i
γ
f (z)
f (z)
∈ H [1, 1]∩Q, 1 − (2 − λ)
+
′
(1 − λ)
(z)
λ
(1 − λ) f (z) + λzf ′ (z)

 f (z) 2+ 2λzf
λ z f ′′ (z) f (z) − (1 − λ) f 2 (z)
γ
is univalent in U and
1−
λ
[(1 − λ) f (z) + λzf ′ (z)]2
h
i
γ
f (z)
q (z) + γzq ′ (z) ≺ 1 − (2 − λ)
+
λ
(1 − λ) f (z) + λzf ′ (z)


λ2 z 2 f ′′ (z) f (z) − (1 − λ) f 2 (z)
γ
,
1−
λ
[(1 − λ) f (z) + λzf ′ (z)]2
then
f (z)
q (z) ≺
(1 − λ) f (z) + λzf ′ (z)
and q is the best subordinant.
Proof. The conclusion follows from Theorem 10 for m = 0.
If we take λ = 1 in Theorem 10, we have the next corollary.
Corollary 12 [6] Let q be convex in U , q (0) =1 and γ ∈ C, Reγ > 0. 
If
m
m
m+2
m
D f (z)
D f (z)
D
f (z) · D f (z)
f ∈ A, m+1
∈ H [1, 1]∩Q, m+1
+γ 1 −
D
f (z)
D
f (z)
[Dm+1 f (z)]2
is univalent in U and


Dm+2 f (z) · Dm f (z)
Dm f (z)
′
,
+γ 1−
q (z) + γzq (z) ≺ m+1
D
f (z)
[Dm+1 f (z)]2
then
q (z) ≺

Dλm f (z)
Dλm+1 f (z)

and q is the best subordinant.
We combine the results of Theorem 5 and Theorem 10 to obtain the
following ”sandwich theorem”.
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Corollary 13 Let q1 , q2 be convex in U , q1 (0) = q(
2 (0) = 1, γ ∈ C, Reγ > 0. )
m
m
Dλ f (z) γ
Dλ f (z)
Dm+2 f (z) · Dλm f (z)
∈ H [1, 1]∩Q, m+1
+
If f ∈ A, m+1
1 − λ  m+1
2
Dλ f (z)
Dλ f (z) λ
Dλ f (z)
is univalent in U and
(
)
m+2
m
m
D
f (z) · Dλ f (z)
Dλ f (z) γ
1 − λ  m+1
q1 (z)+γzq1′ (z) ≺ m+1
+
≺ q2 (z)+γzq2′ (z) ,
2
Dλ f (z) λ
Dλ f (z)

then

q1 (z) ≺

Dλm f (z)
≺ q2 (z)
Dλm+1 f (z)

and the functions q1 and q2 are respectively the best subordinant and the best
dominant.
Theorem 14 Let q be univalent in U with q (0) = 1, γ ∈ C∗ and suppose




1
zq ′′ (z)
> max 0, −Re
.
Re 1 + ′
q (z)
γ
If f ∈ A and
2

γ  zDλm+1 f (z) γ zDλm+2 f (z) 2γ z Dλm+1 f (z)
+
−
≺ q (z) + γzq ′ (z)
1+
λ {Dλm f (z)}2 λ {Dλm f (z)}2
λ {Dλm f (z)}3
(6)
then
Dm+1 f (z)
≺ q (z)
z λm
{Dλ f (z)}2


and q is the best dominant.
Proof. Let

Dλm+1 f (z)
p (z) := z m
.
{Dλ f (z)}2

By calculating the logarithmic derivative of p, we get
′

z Dλm+1 f (z)
zp′ (z)
z {Dλm f (z)}′
=1+
.
−2
p (z)
Dλm f (z)
Dλm+1 f (z)
151

(7)

V.O. Nechita - Differential subordinations and superordinations for...
We use the identity (5) in (7) to obtain


Dm+1 f (z)
zp′ (z)
1
Dλm+2 f (z)
=
1 + m+1
−2 m
p (z)
λ
D f (z)
Dλ f (z)
and
 zDm+1 f (z) γ zDm+2 f (z) 2γ z Dm+1 f (z) 2
γ
λ
λ
λ
+
−
.
p (z)+γzp′ (z) = 1 +
λ {Dλm f (z)}2 λ {Dλm f (z)}2 λ {Dλm f (z)}3


The subordination (6) becomes
p (z) + γzp′ (z) ≺ q (z) + γzq ′ (z) .
We obtain the conclusion of our theorem by simply applying Corrolary 2.
In Theorem 14 we let m = 0 and obtain the following result.
Corollary 15 Let q be univalent in U with q (0) = 1, γ ∈ C∗ and suppose




zq ′′ (z)
1
> max 0, −Re
.
Re 1 + ′
q (z)
γ
If f ∈ A and
(1 + γ)

z 3 f ′′ (z)
z 3 [f ′ (z)]2
(1 − λ) z
z 2 f ′ (z)
+
γλ
−
2γλ
+ [λ + (3λ − 1) γ]
f (z)
[f (z)]2
[f (z)]2
[f (z)]3
≺ q (z) + γzq ′ (z)

then
(1 − λ)

z 2 f ′ (z)
z
+λ 2
≺ q (z)
f (z)
f (z)

and q is the best dominant.
We consider λ = 1 in Theorem 14 to get the following corrolary.
Corollary 16 [6] Let q be univalent in U with q (0) = 1, γ ∈ C∗ and suppose




1
zq ′′ (z)
> max 0, −Re
.
Re 1 + ′
q (z)
γ
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If f ∈ A and
2

(1 + γ)

zDm+1 f (z)
zDm+2 f (z)
z {Dm+1 f (z)}
+
γ
−
2γ
≺ q (z) + γzq ′ (z)
2
2
3
m
m
m
{D f (z)}
{D f (z)}
{D f (z)}

then

Dm+1 f (z)
z
≺ q (z)
{Dm f (z)}2

and q is the best dominant.
For m = 0 and λ = 1 in Theorem 10, we obtain the following result.
Corollary 17 [6] Let q be univalent in U with q (0) = 1, γ ∈ C∗ and suppose




1
zq ′′ (z)
> max 0, −Re
.
Re 1 + ′
q (z)
γ
If f ∈ A and
zf ′ (z)
2
2 − γz
{f (z)}
then



z
f (z)

′′

≺ q (z) + γzq ′ (z)

z 2 f ′ (z)
≺ q (z)
[f (z)]2

and q is the best dominant.
We take a particular dominant q in Theorem 10 to get the next corolarry.
Corollary 18 Let A, B, γ ∈ C, A 6= B such that |B| ≤ 1 and Reγ > 0. If
f ∈ A satisfies the subordination

2

γ  zDλm+1 f (z) γ zDλm+2 f (z) 2γ z Dλm+1 f (z)
1+
+
−
λ {Dλm f (z)}2 λ {Dλm f (z)}2
λ {Dλm f (z)}3
≺

1 + Az
(A − B) z
,
+γ
1 + Bz
(1 + Bz)2

z

1 + Az
Dλm+1 f (z)
2 ≺
m
1 + Bz
{Dλ f (z)}

then

and q (z) =

1 + Az
is the best dominant.
1 + Bz
153

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Next, applying Corrolary 4 we have the following theorem.
Theorem 19 Let q be convex in U , q (0) = 1 and γ ∈ C, Reγ > 0. If

Dm+1 f (z)
γ  zDλm+1 f (z) γ zDλm+2 f (z)
f ∈ A, z λm
∈
H
[1,
1]
∩
Q,
1
+
+
−
λ {Dλm f (z)}2 λ {Dλm f (z)}2
{Dλ f (z)}2
2

2γ z Dλm+1 f (z)
is univalent in U and
λ {Dλm f (z)}3
2

γ  zDλm+1 f (z) γ zDλm+2 f (z) 2γ z Dλm+1 f (z)
q (z)+γzq (z) ≺ 1 +
+
−
,
λ {Dλm f (z)}2 λ {Dλm f (z)}2 λ {Dλm f (z)}3
′



then

Dλm+1 f (z)
{Dλm f (z)}2

q (z) ≺ z
and q is the best subordinant.

Corollary 20 Let q be convex in U , q (0) = 1 and γ ∈ C, Reγ > 0. If f ∈ A,
z 2 f ′ (z)
(1 − λ) z
z 2 f ′ (z)
z
+
+λ 2
∈ H [1, 1]∩Q, (1 + γ)
+[λ + (3λ − 1) γ]
(1 − λ)
f (z)
f (z)
f (z)
[f (z)]2
z 3 [f ′ (z)]2
z 3 f ′′ (z)
− 2γλ
is univalent in U and
γλ
[f (z)]2
[f (z)]3
q (z) + γzq ′ (z) ≺ (1 + γ)

(1 − λ) z
z 2 f ′ (z)
+ [λ + (3λ − 1) γ]
f (z)
[f (z)]2

z 3 f ′′ (z)
z 3 [f ′ (z)]2
+γλ
− 2γλ
,
[f (z)]2
[f (z)]3
then
q (z) ≺ (1 − λ)

z 2 f ′ (z)
z
+λ 2
f (z)
f (z)

and q is the best subordinant.
Proof. The conclusion follows from Theorem 14 by taking m = 0.
We write Theorem 14 and Theorem 19 together and obtain the following
”sandwich theorem”.

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Corollary 21 Let q1 , q2 be convex in U , q1 (0) = q2 (0) = 1, γ ∈ C, Reγ > 0.

Dm+1 f (z)
γ  zDλm+1 f (z) γ zDλm+2 f (z)
If f ∈ A, z λm
∈
H
[1,
1]∩Q,
1
+
+
−
λ {Dλm f (z)}2 λ {Dλm f (z)}2
{Dλ f (z)}2
2

2γ z Dλm+1 f (z)
is univalent in U and
λ {Dλm f (z)}3
q1 (z) +

then

γzq1′

2

γ  zDλm+1 f (z) γ zDλm+2 f (z) 2γ z Dλm+1 f (z)
+
−
(z) ≺ 1 +
λ {Dλm f (z)}2 λ {Dλm f (z)}2
λ {Dλm f (z)}3
≺ q2 (z) + γzq2′ (z) ,


Dλm+1 f (z)
q1 (z) ≺ z m
≺ q2 (z)
{Dλ f (z)}2

and the functions q1 and q2 are respectively the best subordinant and the best
dominant.

References
[1] F.M. Al-Oboudi, On univalent functions defined by a generalized S˘al˘agean
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V.O. Nechita - Differential subordinations and superordinations for...
Author:
Veronica Oana Nechita
Department of Mathematics,
Babe¸s-Bolyai University,
Cluj-Napoca,
Romania
email: vnechita@math.ubbcluj.ro

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