Paper-18-Acta24.2010. 125KB Jun 04 2011 12:03:11 AM

Acta Universitatis Apulensis
ISSN: 1582-5329

No. 24/2010
pp. 201-209

A NOTE ON DIFFERENTIAL SUPERORDINATIONS USING
SĂLĂGEAN AND RUSCHEWEYH OPERATORS
Alina Alb Lupaş
Abstract. In the present paper we define a new operator using the Sălăgean
and Ruscheweyh operators. Denote by SRm the Hadamard product of the Sălăgean
operator S m and the Ruscheweyh operator Rm , given by SRm : An → An , SRm f (z)
= (S m ∗ 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. We study some differential superordinations
regarding the operator SRm .
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 }
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),

for z ∈ U,

(1)

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).
201

A. Alb Lupas - A note on differential superordinations using Sălăgean...

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. (Sălăgean [4]) For f ∈ An , n ∈ N, m ∈ N ∪ {0}, the operator S m
is defined by S m : An → An ,
S 0 f (z) = f (z)
S 1 f (z) = zf ′ (z)
...
S

m+1

f (z) = z (S m f (z))′ ,

z ∈ U.

P

P∞
j
m
m
j
Remark 1. If f ∈ An , f (z) = z+ ∞
j=n+1 aj z , then S f (z) = z+ j=n+1 j aj z ,
z ∈ U.
Definition 2. (Ruscheweyh [3]) For f ∈ An , n ∈ N, m ∈ N ∪ {0}, the operator
Rm is defined by Rm : An → An ,
R0 f (z) = f (z)
R1 f (z) = zf ′ (z)
...
(m + 1) R

m+1

f (z) = z (Rm f (z))′ + mRm f (z) ,

z ∈ U.

P
P∞
j
m
m
j
Remark 2. If f ∈ An , f (z) = z+ ∞
j=n+1 aj z , then R f (z) = z+ j=n+1 Cm+j−1 aj z ,
z ∈ U.
Definition 3. [2] 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
z→ζ

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 [2]) 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
202

for z ∈ U,

A. Alb Lupas - A note on differential superordinations using Sălăgean...

then
q(z) ≺ p(z),
where q(z) = nzγγ/n
best subordinant.

Rz
0

for z ∈ U,

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

Lemma 2. (Miller and Mocanu [2]) 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. [1] Let m ∈ N. Denote by SRm the operator given by the
Hadamard product (the convolution product) of the Sălăgean operator S m and the
Ruscheweyh operator Rm , SRm : An → An ,
SRm f (z) = (S m ∗ Rm ) f (z) .

Remark 3. If f ∈ An , f (z) = z +
P∞
m
m 2 j
j=n+1 Cm+j−1 j aj z .

P∞

j=n+1 aj z

j,

then SRm f (z) = z +

Theorem 1. Let h be a convex function, h(0) = 1. Let m ∈ N ∪ {0}, f ∈ An
m
and suppose that z1 SRm+1 f (z) + m+1
z (SRm f (z))′′ is univalent and (SRm f (z))′ ∈
H [1, n] ∩ Q. If
h(z) ≺

m
1
SRm+1 f (z) +
z (SRm f (z))′′ ,

z
m+1

for z ∈ U,

(2)

then
q(z) ≺ (SRm f (z))′ ,
Rz

1
−1
n

for z ∈ U,

dt. The function q is convex and it is the best subordiwhere q(z) = 1 1 0 h(t)t
nz n
nant.

P
m
m+1 a2 z j−1
Proof. With notation p (z) = (SRm f (z))′ = 1 + ∞
j
j=n+1 Cm+j−1 j
203

A. Alb Lupas - A note on differential superordinations using Sălăgean...
P
j
and p (0) = 1, we obtain for f (z) = z + ∞
j=n+1 aj z ,
1
m
′
m+1
m
p (z) + zp (z) = z SR
f (z) + z m+1 (SR f (z))′′ . Evidently p ∈ H[1, n].

Then (2) becomes
h(z) ≺ p(z) + zp′ (z),

for z ∈ U.

By using Lemma 1, we have
q(z) ≺ p(z),
where q(z) =
dinant.

1
1
nz n

Rz
0

for z ∈ U,

i.e. q(z) ≺ (SRm f (z))′ ,

for z ∈ U,

1

h(t)t n −1 dt. The function q is convex and it is the best subor-

Corollary No. 1 Let h(z) = 1+(2β−1)z
be a convex function in U , where 0 ≤
1+z
m
β < 1. Let m ∈ N∪{0}, f ∈ An and suppose that z1 SRm+1 f (z)+ m+1
z (SRm f (z))′′
′
is univalent and (SRm f (z)) ∈ H [1, n] ∩ Q. If
h(z) ≺

m
1
SRm+1 f (z) +
z (SRm f (z))′′ ,
z
m+1

for z ∈ U,

(3)

then
q(z) ≺ (SRm f (z))′ ,
for z ∈ U,
R z t n1 −1
where q is given by q(z) = 2β − 1 + 2(1−β)
1
0 1+t dt, for z ∈ U. The function q is
nz n
convex and it is the best subordinant.
Proof. Following the same steps as in the proof of Theorem 1 and considering
p(z) = (SRm f (z))′ , the differential superordination (3) becomes
h(z) =

1 + (2β − 1)z
≺ p(z) + zp′ (z),
1+z

for z ∈ U.

By using Lemma 1 for γ = 1, we have q(z) ≺ p(z), i.e.,
Z z
Z z
1
1
1
1 + (2β − 1) t
1
−1
n
dt
dt
=
q(z) =
h
(t)
t
t n −1
1
1
1+t
nz n 0
nz n 0
1
Z
2(1 − β) z t n −1
= 2β − 1 +
dt ≺ (SRm f (z))′ , for z ∈ U.
1
1
+
t
0
nz n
The function q is convex and it is the best subordinant.

204

A. Alb Lupas - A note on differential superordinations using Sălăgean...

Theorem No. 2 Let q be convex in U and let h be defined by h (z) = q (z) +
m
zq ′ (z) . If m ∈ N ∪ {0}, f ∈ An , suppose that z1 SRm+1 f (z) + m+1
z (SRm f (z))′′ is
′
univalent, (SRm f (z)) ∈ H [1, n] ∩ Q and satisfies the differential superordination
h(z) = q (z) + zq ′ (z) ≺

1
m
SRm+1 f (z) +
z (SRm f (z))′′ ,
z
m+1

for z ∈ U,

(4)

then
q(z) ≺ (SRm f (z))′ ,
1
1
nz n

Rz

for z ∈ U,

1

h(t)t n −1 dt. The function q is the best subordinant.
P
m
m+1 a2 z j−1 .
Proof. Let p (z) = (SRm f (z))′ = 1 + ∞
j
j=n+1 Cm+j−1 j

where q(z) =

0

m
(SRm f (z))′′ , for
Differentiating, we obtain p(z) + zp′ (z) = z1 SRm+1 f (z) + z m+1
z ∈ U and (4) becomes

q(z) + zq ′ (z) ≺ p(z) + zp′ (z) ,

for z ∈ U.

Using Lemma 2, we have
q(z) ≺ p(z), for z ∈ U, i.e. q(z) =

1
nz

1
n

Z

z

1

h(t)t n −1 dt ≺ (SRm f (z))′ , for z ∈ U,

0

and q is the best subordinant.
Theorem 3. Let h be a convex function, h(0) = 1. Let m ∈ N, f ∈ An and
m
suppose that (SRm f (z))′ is univalent and SR zf (z) ∈ H [1, n] ∩ Q. If
h(z) ≺ (SRm f (z))′ ,
then
q(z) ≺

SRm f (z)
,
z

for z ∈ U,

(5)

for z ∈ U,

Rz
1
where q(z) = 1 1 0 h(t)t n −1 dt. The function q is convex and it is the best subordinz n
nant.
P
m
m 2 j
m
z+ ∞
j=n+1 Cm+j−1 j aj z
=
Proof. Consider p (z) = SR zf (z) =
z
P
m
m 2 j−1 . Evidently p ∈ H[1, n].
1+ ∞
j=n+1 Cm+j−1 j aj z
Differentiating, we obtain p (z) + zp′ (z) = (SRm f (z))′ .
Then (5) becomes
h(z) ≺ p(z) + zp′ (z),
205

for z ∈ U.

A. Alb Lupas - A note on differential superordinations using Sălăgean...

By using Lemma 1, we have
q(z) ≺ p(z),
where q(z) =
dinant.

1
1
nz n

Rz
0

for z ∈ U,

i.e. q(z) ≺

SRm f (z)
,
z

for z ∈ U,

1

h(t)t n −1 dt. The function q is convex and it is the best subor-

Corollary 2. Let h(z) =

1+(2β−1)z
1+z

be a convex function in U , where 0 ≤ β < 1.

Let m ∈ N ∪ {0}, f ∈ An and suppose that (SRm f (z))′ is univalent and
H [1, n] ∩ Q. If
h(z) ≺ (SRm f (z))′ ,
for z ∈ U,
then
q(z) ≺

SRm f (z)
,
z

SRm f (z)
z

∈

(6)

for z ∈ U,

R z t n1 −1
where q is given by q(z) = 2β − 1 + 2(1−β)
1
0 1+t dt, for z ∈ U. The function q is
nz n
convex and it is the best subordinant.
Proof. Following the same steps as in the proof of Theorem 3 and considering
p(z) =

SRm f (z)
,
z

the differential superordination (6) becomes
h(z) =

1 + (2β − 1)z
≺ p(z) + zp′ (z),
1+z

for z ∈ U.

By using Lemma 1 for γ = 1, we have q(z) ≺ p(z), i.e.,
Z z
Z z
1
1
1
1
1 + (2β − 1) t
−1
q(z) =
h (t) t n dt =
t n −1
dt
1
1
1+t
nz n 0
nz n 0
1
Z
2(1 − β) z t n −1
SRm f (z)
= 2β − 1 +
dt
≺
, for z ∈ U.
1
z
0 1+t
nz n
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) .
m
If m ∈ N∪{0}, f ∈ An , suppose that (SRm f (z))′ is univalent, SR zf (z) ∈ H [1, n]∩Q
and satisfies the differential superordination
h(z) = q (z) + zq ′ (z) ≺ (SRm f (z))′ ,
then
q(z) ≺

SRm f (z)
,
z
206

for z ∈ U,

for z ∈ U,

(7)

A. Alb Lupas - A note on differential superordinations using Sălăgean...

where q(z) =

1
1
nz n

Rz
0

1

h(t)t n −1 dt. The function q is the best subordinant.

Proof. Let p (z) =

SRm f (z)
z

=

z+

P∞

j=n+1

m
Cm+j−1
j m a2j z j
z

=

P
m
m 2 j−1 . Evidently p ∈ H[1, n].
1+ ∞
j=n+1 Cm+j−1 j aj z
Differentiating, we obtain p(z)+zp′ (z) = (SRm f (z))′ , for z ∈ U and (7) becomes
q(z) + zq ′ (z) ≺ p(z) + zp′ (z) ,

for z ∈ U.

Using Lemma 2, we have
q(z) ≺ p(z), for z ∈ U,

i.e. q(z) =

1
1

nz n

Z

z

SRm f (z)
, for z ∈ U,
z

1

h(t)t n −1 dt ≺

0

and q is the best subordinant.
Theorem 5. Let h be a 
convex function, h(0) = 1. Let m ∈ N ∪ {0}, f ∈ An
′
m+1 f (z)
zSRm+1 f (z)
is univalent and SR
and suppose that
SRm f (z)
SRm f (z) ∈ H [1, n] ∩ Q. If
h(z) ≺



zSRm+1 f (z)
SRm f (z)

then

′

SRm+1 f (z)
,
SRm f (z)

q(z) ≺

for z ∈ U,

,

(8)

for z ∈ U,

Rz
1
where q(z) = 1 1 0 h(t)t n −1 dt. The function q is convex and it is the best subordin
nz
nant.
P
m+1 f (z)
C m+1 j m+1 a2j z j
z+ ∞
Pj=n+1 m+j
=
=
Proof. Consider p (z) = SR
SRm f (z)
z+ ∞
Cm
j m a2 z j
j=n+1

P
1+ ∞
C m+1 j m+1 a2j z j−1
Pj=n+1 m+j
m
m 2 j−1 .
1+ ∞
j=n+1 Cm+j−1 j aj z

m+j−1

j

Evidently p ∈ H[1, n].
′

m f (z))′
(SRm+1 f (z))
′
We have p′ (z) = SRm f (z) −p (z)· (SR
SRm f (z) and p (z)+zp (z) =
Then (8) becomes

h(z) ≺ p(z) + zp′ (z),



zSRm+1 f (z)
SRm f (z)

for z ∈ U.

By using Lemma 1, we have
q(z) ≺ p(z),

for z ∈ U,

i.e. q(z) ≺

207

SRm+1 f (z)
,
SRm f (z)

for z ∈ U,

′

.

A. Alb Lupas - A note on differential superordinations using Sălăgean...

where q(z) =
dinant.

1
1
nz n

Rz
0

1

h(t)t n −1 dt. The function q is convex and it is the best subor-

Corollary 3. Let h(z) =

1+(2β−1)z
1+z

be a convex function in U , where 0 ≤ β < 1.


m+1 f (z) ′
m+1 f (z)
is univalent, SR
Let m ∈ N ∪ {0}, f ∈ An and suppose that zSR
m
SR f (z)
SRm f (z) ∈
H [1, n] ∩ Q. If
′

zSRm+1 f (z)
,
for z ∈ U,
(9)
h(z) ≺
SRm f (z)
then

SRm+1 f (z)
,
for z ∈ U,
SRm f (z)
R z t n1 −1
where q is given by q(z) = 2β − 1 + 2(1−β)
1
0 1+t dt, for z ∈ U. The function q is
nz n
convex and it is the best subordinant.
Proof. Following the same steps as in the proof of Theorem 5 and considering
q(z) ≺

p(z) =

SRm f (z)
,
z

the differential superordination (9) becomes
h(z) =

1 + (2β − 1)z
≺ p(z) + zp′ (z),
1+z

for z ∈ U.

By using Lemma 1 for γ = 1, we have q(z) ≺ p(z), i.e.,
Z z
Z z
1
1
1
1
1 + (2β − 1) t
−1
h (t) t n dt =
t n −1
q(z) =
dt
1
1
1+t
nz n 0
nz n 0
1
Z
SRm+1 f (z)
2(1 − β) z t n −1
dt
≺
, for z ∈ U.
= 2β − 1 +
1
SRm f (z)
0 1+t
nz n
The function q is convex and it is the best subordinant.
Theorem 6. Let q be convex in U andlet h be defined
+ zq ′ (z) .
′ by h (z) = q (z)m+1
m+1
f (z)
f (z)
If m ∈ N ∪ {0}, f ∈ An , suppose that zSR
is univalent, SR
SRm f (z)
SRm f (z) ∈
H [1, n] ∩ Q and satisfies the differential superordination
h(z) = q (z) + zq ′ (z) ≺
then
q(z) ≺



zSRm+1 f (z)
SRm f (z)

SRm+1 f (z)
,
SRm f (z)
208

′

,

for z ∈ U,

for z ∈ U,

(10)

A. Alb Lupas - A note on differential superordinations using Sălăgean...

where q(z) =

1
1
nz n

Rz
0

1

h(t)t n −1 dt. The function q is the best subordinant.

Proof. Let p (z) = p (z) =

SRm+1 f (z)
SRm f (z)

=

P
m+1 m+1 2 j
z+ ∞
Cm+j
j
aj z
Pj=n+1
∞
m
z+ j=n+1 Cm+j−1
j m a2j z j

=

P
m+1 m+1 2 j−1
1+ ∞
Cm+j
j
aj z
Pj=n+1
.
∞
m
1+ j=n+1 Cm+j−1
j m a2j z j−1

Evidently p ∈ H[1, n].


m+1 f (z) ′
Differentiating, we obtain p(z) + zp′ (z) = zSR
, for z ∈ U and (10)
SRm f (z)
becomes
q(z) + zq ′ (z) ≺ p(z) + zp′ (z) , for z ∈ U.
Using Lemma 2, we have
q(z) ≺ p(z), for z ∈ U, i.e. q(z) =

1
1

nz n

Z

z

1

h(t)t n −1 dt ≺

0

SRm+1 f (z)
, for z ∈ U,
SRm f (z)

and q is the best subordinant.

References
[1] Alina Alb Lupaş, A note on differential subordinations using Sălăgean and
Ruscheweyh operators, Romai Journal, Proceedings of CAIM 2009, (to appear).
[2] S.S. Miller, P.T. Mocanu, Subordinants of Differential Superordinations, Complex Variables, vol. 48, no. 10, 2003, 815-826.
[3] St. Ruscheweyh, New criteria for univalent functions, Proc. Amer. Math.
Soc., 49 (1975), 109-115.
[4] 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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