Paper2-Acta25-2011. 133KB Jun 04 2011 12:03:16 AM

Acta Universitatis Apulensis
ISSN: 1582-5329

No. 25/2011
pp. 31-40

CERTAIN DIFFERENTIAL SUPERORDINATIONS USING A
GENERALIZED SĂLĂGEAN AND RUSCHEWEYH OPERATORS

Alb Lupaş Alina
Abstract. In the present paper we define a new operator using the generalized
Sălăgean and Ruscheweyh operators. Denote by DRλm the Hadamard product of
the generalized Sălăgean operator Dλm and the Ruscheweyh operator Rm , given
by DRλm : A → A, DRλm f (z) = (Dλm ∗ Rm ) f (z) and An = {f ∈ H(U ), f (z) =
z +an+1 z n+1 +. . . , z ∈ U } is the class of normalized analytic functions with A1 = A.
We study some differential superordinations regarding the operator DRλm .
2000 Mathematics Subject Classification: 30C45, 30A20, 34A40.
Keywords: Differential superordination, convex function, best subordinant, differential operator.
1. Introduction and definitions
Denote by U the unit disc of the complex plane U = {z ∈ C : |z| < 1} and H(U )
the space of holomorphic functions in U .

Let
An = {f ∈ H(U ), f (z) = z + an+1 z n+1 + . . . , z ∈ U }
with A1 = A and
H[a, n] = {f ∈ H(U ), f (z) = a + an z n + an+1 z n+1 + . . . , z ∈ U }
for a ∈ C and n ∈ N.
If f and g are analytic functions in U , we say that f is superordinate to g, written
g ≺ f , if there is a function w analytic in U , with w(0) = 0, |w(z)| < 1, for all z ∈ U
such that g(z) = f (w(z)) for all z ∈ U . If f is univalent, then g ≺ f if and only if
f (0) = g(0) and g(U ) ⊆ f (U ).
Let ψ : C2 × U → C and h analytic in U . If p and ψ (p (z) , zp′ (z) ; z) are
univalent in U and satisfies the (first-order) differential superordination
h(z) ≺ ψ(p(z), zp′ (z); z),

31

for z ∈ U,

(1)

A. Alb Lupaş - Certain differential superordinations using a generalized...

then p is called a solution of the differential superordination. The analytic function
q is called a subordinant of the solutions of the differential superordination, or more
simply a subordinant, if q ≺ p for all p satisfying (1). An univalent subordinant qe
that satisfies q ≺ qe for all subordinants q of (1) is said to be the best subordinant of
(1). The best subordinant is unique up to a rotation of U .
Definition 1 (Al Oboudi [4]) For f ∈ A, λ ≥ 0 and m ∈ N, the operator Dλm is
defined by Dλm : A → A,
Dλ0 f (z) = f (z)
Dλ1 f (z) = (1 − λ) f (z) + λzf ′ (z) = Dλ f (z)
...



= (1 − λ) Dλm−1 f (z) + λz (Dλm f (z))′ = Dλ Dλm−1 f (z) , f or z ∈ U.
P
j
m
Remark 1 If f ∈ A and f (z) = z + ∞
j=2 aj z , then Dλ f (z) = z +
P∞

m
j
j=2 [1 + (j − 1) λ] aj z , for z ∈ U .
Remark 2 For λ = 1 in the above definition we obtain the Sălăgean differential
operator [7].
Definition 2 (Ruscheweyh [6]) For f ∈ A, m ∈ N, the operator Rm is defined by
Rm : A → A,
Dλm f (z)

R0 f (z) = f (z)
R1 f (z) = zf ′ (z)
...
f (z) = z (Rm f (z))′ + mRm f (z) , z ∈ U.
P
P∞
m
j
j
m
Remark 3 If f ∈ A, f (z) = z + ∞

j=2 aj z , then R f (z) = z +
j=2 Cm+j−1 aj z ,
z ∈ U.
Definition 3 ([5]) We denote by Q the set of functions that are analytic and injective
on U \E (f ), where E (f ) = {ζ ∈ ∂U : lim f (z) = ∞}, and f ′ (ζ) 6= 0 for ζ ∈
(m + 1) R

m+1

z→ζ

∂U \E (f ). The subclass of Q for which f (0) = a is denoted by Q (a).
We will use the following lemmas.
Lemma 1 (Miller and Mocanu [5]) Let h be a convex function with h(0) = a, and
let γ ∈ C\{0} be a complex number with Re γ ≥ 0. If p ∈ H[a, n] ∩ Q, p(z) + γ1 zp′ (z)
is univalent in U and
h(z) ≺ p(z) +

1 ′
zp (z),

γ

for z ∈ U,

then
q(z) ≺ p(z),

for z ∈ U,
32

A. Alb Lupaş - Certain differential superordinations using a generalized...
Rz
where q(z) = nzγγ/n 0 h(t)tγ/n−1 dt, for z ∈ U. The function q is convex and is the
best subordinant.
Lemma 2 (Miller and Mocanu [5]) Let q be a convex function in U and let h(z) =
q(z) + γ1 zq ′ (z), for z ∈ U, where Re γ ≥ 0.
If p ∈ H [a, n] ∩ Q, p(z) + γ1 zp′ (z) is univalent in U and
q(z) +

1

1 ′
zq (z) ≺ p(z) + zp′ (z) ,
γ
γ

for z ∈ U,

then
q(z) ≺ p(z),
where q(z) =

γ
nz γ/n

Rz
0

for z ∈ U,

h(t)tγ/n−1 dt, for z ∈ U. The function q is the best subordinant.

2. Main results

Definition 4 ([2]) Let λ ≥ 0 and m ∈ N. Denote by DRλm the operator given by
the Hadamard product (the convolution product) of the generalized Sălăgean operator
Dλm and the Ruscheweyh operator Rm , DRλm : A → A,
DRλm f (z) = (Dλm ∗ Rm ) f (z) .
P
j
Remark 4 If f ∈ A and f (z) = z + ∞
j=2 aj z , then
P
∞
m
m
DRλm f (z) = z + j=2 Cm+j−1
[1 + (j − 1) λ] a2j z j , for z ∈ U .
Remark 5 For λ = 1 we obtain the Hadamard product SRn [1] of the Sălăgean
operator S n and Ruscheweyh operator Rn .
Theorem 1 Let h be a convex function, h(0) = 1. Let λ ≥ 0, m ∈ N, f ∈ A and supm(1−λ)
m+1

DRλm+1 f (z) − (mλ+1)z
DRλm f (z) is univalent and (DRλm f (z))′ ∈
pose that (mλ+1)z
H [1, 1] ∩ Q. If
h(z) ≺

m+1
m (1 − λ)
DRλm+1 f (z) −
DRλm f (z) ,
(mλ + 1) z
(mλ + 1) z

for z ∈ U,

(2)

then
q(z) ≺ (DRλm f (z))′ ,
where q(z) =

1
m+ λ
1
z m+ λ

Rz
0

for z ∈ U,

1

h (t) tm−1+ λ dt. The function q is convex and it is the best

subordinant.
P
m
m
Proof. With notation p (z) = (DRλm f (z))′ = 1+ ∞

j=2 Cm+j−1 [1 + (j − 1) λ] ·

ja2j z j−1 and p (0) = 1, we obtain for f (z) = z +
33

P∞

j=2 aj z

j,

A. Alb Lupaş - Certain differential superordinations using a generalized...


m(1−λ)
′
m+1
1
m
m
DR

f
(z)
−
m
−
1
+
p (z) + zp′ (z) = m+1
λ
λz
λ (DRλ f (z)) −
λz DRλ f (z)
m(1−λ)
λ
m+1
and p (z) + mλ+1
zp′ (z) = (mλ+1)z
DRλm+1 f (z) − (mλ+1)z DRλm f (z) .
Evidently p ∈ H[1, 1].
Then (2) becomes
h(z) ≺ p (z) +

λ
zp′ (z) ,
mλ + 1

By using Lemma 1 for γ = m +
q(z) ≺ p(z),
1
z m+ λ

0

and n = 1, we have

i.e. q(z) ≺ (DRλm f (z))′ ,

for z ∈ U,
Rz

1
m+ λ

where q(z) =

1
λ

for z ∈ U.

for z ∈ U,

1

h (t) tm−1+ λ dt. The function q is convex and it is the best

subordinant.
Corollary 1 ([3]) Let h be a convex function, h(0) = 1. Let n ∈ N, f ∈ A
n
z (SRn f (z))′′ is univalent and (SRn f (z))′ ∈
and suppose that z1 SRn+1 f (z) + n+1
H [1, 1] ∩ Q. If
h(z) ≺

1
n
SRn+1 f (z) +
z (SRn f (z))′′ ,
z
n+1

for z ∈ U,

(3)

then
q(z) ≺ (SRn f (z))′ ,

for z ∈ U,

R
1 z

where q(z) = z 0 h(t)dt. The function q is convex and it is the best subordinant.
λ
Theorem 2 Let q be convex in U and let h be defined by h (z) = q (z) + mλ+1
zq ′ (z),

m(1−λ)
m+1
DRλm+1 f (z) − (mλ+1)z
·
λ ≥ 0, m ∈ N. If f ∈ A, suppose that (mλ+1)z
′
m
m
DRλ f (z) is univalent, (DRλ f (z)) ∈ H [1, 1] ∩ Q and satisfies the differential superordination

h(z) = q (z)+

m+1
m (1 − λ)
λ
zq ′ (z) ≺
DRλm+1 f (z)−
DRλm f (z) , (4)
mλ + 1
(mλ + 1) z
(mλ + 1) z

for z ∈ U, then
q(z) ≺ (DRλm f (z))′ ,
1
m+ λ

Rz

for z ∈ U,

1

h (t) tm−1+ λ dt. The function q is the best subordinant.
P
m
m
2 j−1 .
Proof. Let p (z) = (DRλm f (z))′ = 1 + ∞
j=2 Cm+j−1 [1 + (j − 1) λ] jaj z

where q(z) =

1
z m+ λ

0



m+1
Differentiating, we obtain p (z) + zp′ (z) = m+1
f (z) − m − 1 + λ1 ·
λz DRλ
m+1
m+1
λ
n
′
f (z) −
(DRλm f (z))′ − m(1−λ)
λz DRλ f (z) and p (z) + mλ+1 zp (z) = (mλ+1)z DRλ
34

A. Alb Lupaş - Certain differential superordinations using a generalized...
m(1−λ)
m
(mλ+1)z DRλ f

(z) , for z ∈ U and (4) becomes

q(z) +

λ
λ
zq ′ (z) ≺ p(z) +
zp′ (z) ,
mλ + 1
mλ + 1

Using Lemma 2 for γ = m +
q(z) ≺ p(z), for z ∈ U,

1
λ

for z ∈ U.

and n = 1, we have

i.e. q(z) =

m+
z

1
λ

1
m+ λ

Z

z

0

1

h (t) tm−1+ λ dt ≺ (DRλm f (z))′ ,

for z ∈ U, and q is the best subordinant.
Corollary 2 ([3]) Let q be convex in U and let h be defined by h (z) = q (z) +
n
zq ′ (z) . If n ∈ N, f ∈ A, suppose that z1 SRn+1 f (z) + n+1
z (SRn f (z))′′ is univalent,
(SRn f (z))′ ∈ H [1, 1] ∩ Q and satisfies the differential superordination
h(z) = q (z) + zq ′ (z) ≺

1
n
SRn+1 f (z) +
z (SRn f (z))′′ ,
z
n+1

for z ∈ U,

(5)

then
q(z) ≺ (SRn f (z))′ ,

for z ∈ U,

R
1 z

where q(z) = z 0 h(t)dt. The function q is the best subordinant.
Theorem 3 Let h be a convex function, h(0) = 1. Let λ ≥ 0, m ∈ N, f ∈ A and
m f (z)
DRλ
suppose that (DRλm f (z))′ is univalent and
∈ H [1, 1] ∩ Q. If
z
h(z) ≺ (DRλm f (z))′ ,
then
q(z) ≺
where q(z) =

1
z

Rz
0

(6)

for z ∈ U,

h(t)dt. The function q is convex and it is the best subordinant.

Proof. Consider p (z) =
1+

DRλm f (z)
,
z

for z ∈ U,

m f (z)
DRλ
z

=

z+

P∞

j=2

m
Cm+j−1
[1+(j−1)λ]m a2j z j
z

=

P∞

+ (j − 1) λ]m a2j z j−1 . Evidently p ∈ H[1, 1].
We have p (z) + zp′ (z) = (DRλm f (z))′ , for z ∈ U .
Then (6) becomes
m
j=2 Cm+j−1 [1

h(z) ≺ p(z) + zp′ (z),

for z ∈ U.

By using Lemma 1 for γ = 1 and n = 1, we have
q(z) ≺ p(z),

for z ∈ U,

i.e. q(z) ≺
35

DRλm f (z)
,
z

for z ∈ U,

A. Alb Lupaş - Certain differential superordinations using a generalized...
Rz
where q(z) = z1 0 h(t)dt. The function q is convex and it is the best subordinant.
Corollary 3 ([3]) Let h be a convex function, h(0) = 1. Let n ∈ N, f ∈ A and
n
suppose that (SRn f (z))′ is univalent and SR zf (z) ∈ H [1, 1] ∩ Q. If
h(z) ≺ (SRn f (z))′ ,
then
q(z) ≺

SRn f (z)
,
z

for z ∈ U,

(7)

for z ∈ U,

Rz
where q(z) = z1 0 h(t)dt. The function q is convex and it is the best subordinant.
Theorem 4 Let q be convex in U and let h be defined by h (z) = q (z) + zq ′ (z) . If
m f (z)
DRλ
∈ H [1, 1] ∩ Q
λ ≥ 0, m ∈ N, f ∈ A, suppose that (DRλm f (z))′ is univalent,
z
and satisfies the differential superordination
h(z) = q (z) + zq ′ (z) ≺ (DRλm f (z))′ ,
then
q(z) ≺
where q(z) =

1
z

Rz
0

DRλm f (z)
,
z

for z ∈ U,

(8)

for z ∈ U,

h(t)dt. The function q is the best subordinant.

Proof. Let p (z) =

m f (z)
DRλ
z

=

z+

P∞

j=2

m
Cm+j−1
[1+(j−1)λ]m a2j z j
z

=

P∞

+ (j − 1) λ]m a2j z j−1 . Evidently p ∈ H[1, 1].
Differentiating, we obtain p(z) + zp′ (z) = (DRλm f (z))′ , for z ∈ U and (8) becomes
q(z) + zq ′ (z) ≺ p(z) + zp′ (z) , for z ∈ U.

1+

m
j=2 Cm+j−1 [1

Using Lemma 2 for γ = 1 and n = 1, we have
Z
DRλm f (z)
1 z
h(t)dt ≺
, for z ∈ U,
q(z) ≺ p(z), for z ∈ U, i.e. q(z) =
z 0
z
and q is the best subordinant.
Corollary 4 ([3]) Let q be convex in U and let h be defined by h (z) = q (z)+zq ′ (z) .
n
If n ∈ N, f ∈ A, suppose that (SRn f (z))′ is univalent, SR zf (z) ∈ H [1, 1] ∩ Q and
satisfies the differential superordination
h(z) = q (z) + zq ′ (z) ≺ (SRn f (z))′ ,
then
q(z) ≺

SRn f (z)
,
z
36

for z ∈ U,

for z ∈ U,

(9)

A. Alb Lupaş - Certain differential superordinations using a generalized...
Rz
where q(z) = z1 0 h(t)dt. The function q is the best subordinant.
Theorem 5 Let h(z) = 1+(2β−1)z
be a convex function in U , where 0 ≤ β < 1.
1+z
Let λ ≥ 0, m ∈ N, f ∈ A and suppose that (DRλm f (z))′ is univalent and
H [1, 1] ∩ Q. If
h(z) ≺ (DRλm f (z))′ ,
for z ∈ U,
then
q(z) ≺

DRλm f (z)
,
z

m f (z)
DRλ
z

∈

(10)

for z ∈ U,

where q is given by q(z) = 2β − 1 + 2(1 − β) ln(1+z)
, for z ∈ U. The function q is
z
convex and it is the best subordinant.
Proof. Following the same steps as in the proof of Theorem and considering
p(z) =

m f (z)
DRλ
,
z

the differential superordination (10) becomes
h(z) =

1 + (2β − 1)z
≺ p(z) + zp′ (z),
1+z

for z ∈ U.

By using Lemma 1 for γ = 1 and n = 1, we have q(z) ≺ p(z), i.e.,
Z
Z
DRλm f (z)
1 z
1 z 1 + (2β − 1)t
1
q(z) =
h(t)dt =
dt = 2β−1+2(1−β) ln(z+1) ≺
,
z 0
z 0
1+t
z
z
for z ∈ U.
The function q is convex and it is the best subordinant.
Theorem 6 
Let h be a convex
function, h(0) = 1. Let λ ≥ 0, m ∈ N, f ∈ A and

m+1
zDRλ
f (z)
m f (z)
DRλ

suppose that

′

h(z) ≺

is univalent and
zDRλm+1 f (z)
DRλm f (z)

then
q(z) ≺
where q(z) =

1
z

Rz
0

m+1
DRλ
f (z)
m f (z)
DRλ

!′

DRλm+1 f (z)
,
DRλm f (z)

,

∈ H [1, 1] ∩ Q. If

for z ∈ U,

(11)

for z ∈ U,

h(t)dt. The function q is convex and it is the best subordinant.

Proof. Consider p (z) =

m+1
DRλ
f (z)
m f (z)
DRλ

P
m+1
1+ ∞
Cm+j
[1+(j−1)λ]m+1 a2j z j−1
Pj=2
.
∞
m
1+ j=2 Cm+j−1 [1+(j−1)λ]m a2j z j−1

=

P
m+1
z+ ∞
Cm+j
[1+(j−1)λ]m+1 a2j z j
Pj=2
∞
m
z+ j=2 Cm+j−1 [1+(j−1)λ]m a2j z j

Evidently p ∈ H[1, 1].
37

=

A. Alb Lupaş - Certain differential superordinations using a generalized...
′

′

(DRλm f (z))
(DRλm+1 f (z))
−
p
(z)
·
m
m f (z) .
DR f (z)
DRλ
′
 λ
m+1
zDRλ f (z)
.
Then p (z) + zp′ (z) =
DRm f (z)
We have p′ (z) =

λ

Then (11) becomes

h(z) ≺ p(z) + zp′ (z),

for z ∈ U.

By using Lemma 1 for γ = 1 and n = 1, we have
q(z) ≺ p(z),

for z ∈ U,

DRλm+1 f (z)
,
DRλm f (z)

i.e. q(z) ≺

for z ∈ U,

Rz
where q(z) = z1 0 h(t)dt. The function q is convex and it is the best subordinant.
Corollary 5 ([3]) Let h
 be a convex function, h(0) = 1. Let n ∈ N, f ∈ A and
suppose that

zSRn+1 f (z)
SRn f (z)

′

SRn+1 f (z)
SRn f (z)

is univalent and

h(z) ≺



zSRn+1 f (z)
SRn f (z)

then
q(z) ≺

′

SRn+1 f (z)
,
SRn f (z)

∈ H [1, 1] ∩ Q. If

for z ∈ U,

,

(12)

for z ∈ U,

Rz
where q(z) = z1 0 h(t)dt. The function q is convex and it is the best subordinant.
′
Theorem 7 Let q be convex in U and let
 h be defined
′ by h (z) = q (z) + zq (z) .
m+1
m+1
zDRλ
f (z)
DRλ
f (z)
If λ ≥ 0, m ∈ N, f ∈ A, suppose that
is
univalent,
m
DR f (z)
DRm f (z) ∈
λ

H [1, 1] ∩ Q and satisfies the differential superordination
!′
zDRλm+1 f (z)
′
,
h(z) = q (z) + zq (z) ≺
DRλm f (z)
then
q(z) ≺

where q(z) =

1
z

Rz
0

DRλm+1 f (z)
,
DRλm f (z)

λ

for z ∈ U,

for z ∈ U,

h(t)dt. The function q is the best subordinant.

Proof. Let p (z) =

m+1
DRλ
f (z)
m f (z)
DRλ

P
m+1
Cm+j
[1+(j−1)λ]m+1 a2j z j−1
1+ ∞
Pj=2
.
∞
m
1+ j=2 Cm+j−1 [1+(j−1)λ]m a2j z j−1

=

P
m+1
Cm+j
[1+(j−1)λ]m+1 a2j z j
z+ ∞
Pj=2
∞
m
z+ j=2 Cm+j−1 [1+(j−1)λ]m a2j z j

Evidently p ∈ H[1, 1].
38

=

(13)

A. Alb Lupaş - Certain differential superordinations using a generalized...

Differentiating, we obtain p(z) +

zp′ (z)

=

becomes



m+1
zDRλ
f (z)
m f (z)
DRλ

q(z) + zq ′ (z) ≺ p(z) + zp′ (z) ,

′

, for z ∈ U and (13)

for z ∈ U.

Using Lemma 2 for γ = 1 and n = 1, we have
q(z) ≺ p(z), for z ∈ U,

1
i.e. q(z) =
z

Z

DRλm+1 f (z)
, for z ∈ U,
DRλm f (z)

z

h(t)dt ≺

0

and q is the best subordinant.
Corollary 6 ([3]) Let q be convex
U andlet h be defined by h (z) = q (z)+zq ′ (z) .
 in n+1
′
n+1 f (z)
f (z)
is univalent, SR
If n ∈ N, f ∈ A, suppose that zSR
SRn f (z)
SRn f (z) ∈ H [1, 1] ∩ Q
and satisfies the differential superordination
h(z) = q (z) + zq ′ (z) ≺
then
q(z) ≺



zSRn+1 f (z)
SRn f (z)

SRn+1 f (z)
,
SRn f (z)

′

for z ∈ U,

,

(14)

for z ∈ U,

Rz
where q(z) = z1 0 h(t)dt. The function q is the best subordinant.
Theorem 8 Let h(z) = 1+(2β−1)z
be a convex function in U , where 0 ≤ β < 1. Let
1+z
′

m+1
m+1
DRλ
f (z)
zDRλ
f (z)
is
univalent,
λ ≥ 0, m ∈ N, f ∈ A and suppose that
m
DR f (z)
DRm f (z) ∈
λ

H [1, 1] ∩ Q. If

h(z) ≺

zDRλm+1 f (z)
DRλm f (z)

then
q(z) ≺

!′

DRλm+1 f (z)
,
DRλm f (z)

,

λ

for z ∈ U,

(15)

for z ∈ U,

, for z ∈ U. The function q is
where q is given by q(z) = 2β − 1 + 2(1 − β) ln(1+z)
z
convex and it is the best subordinant.
Proof. Following the same steps as in the proof of Theorem and considering
p(z) =

m+1
DRλ
f (z)
m f (z) ,
DRλ

the differential superordination (15) becomes

h(z) =

1 + (2β − 1)z
≺ p(z) + zp′ (z),
1+z
39

for z ∈ U.

A. Alb Lupaş - Certain differential superordinations using a generalized...

By using Lemma 1 for γ = 1 and n = 1, we have q(z) ≺ p(z), i.e.,
1
q(z) =
z

Z

0

z

1
h(t)dt =
z

Z

0

z

DRλm+1 f (z)
1 + (2β − 1)t
1
dt = 2β−1+2(1−β) ln(z+1) ≺
,
1+t
z
DRλm f (z)

for z ∈ U.
The function q is convex and it is the best subordinant.
References

[1] Alina Alb Lupaş, Adriana Cătaş, Certain differential subordinations using
Sălăgean and Ruscheweyh operators, Proceedings of International Conference on
Complex Analysis and Related Topics, The 11th Romanian-Finish Seminar, Alba
Iulia, 2008, (to appear).
[2] Alina Alb Lupaş, Certain differential subordinations using a generalized Sălăgean
operator and Ruscheweyh operator, submitted 2009.
[3] Alina Alb Lupaş, Certain differential superordinations using Sălăgean and
Ruscheweyh operators, Analele Universităţii din Oradea, Fascicola Matematica, Tom
XVII, Issue no. 2, 2010, 201-208.
[4] F.M. Al-Oboudi, On univalent functions defined by a generalized Sălăgean
operator, Ind. J. Math. Math. Sci., 2004, no.25-28, 1429-1436.
[5] S.S. Miller, P.T. Mocanu, Subordinants of Differential Superordinations, Complex Variables, vol. 48, no. 10, 2003, 815-826.
[6] St. Ruscheweyh, New criteria for univalent functions, Proc. Amer. Math.
Soc., 49 (1975), 109-115.
[7] G.St. Sălăgean, Subclasses of univalent functions, Lecture Notes in Math.,
Springer Verlag, Berlin, 1013 (1983), 362-372.
Alb Lupaş Alina
Department of Mathematics
University of Oradea
str. Universităţii nr. 1, 410087, Oradea, Romania
email: dalb@uoradea.ro

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