DOI: 10.1515/auom-2015-0003
An. Şt. Univ. Ovidius Constanţa
Vol. 23(1),2015, 37–51
On certain differential sandwich theorems
involving a generalized Sălăgean operator and
Ruscheweyh operator
Andrei Loriana
Abstract
In the present paper we introduce sufficient conditions for subordination and superordination involving the operator DRλm,n and also we
obtain sandwich-type results.
1
Introduction
Let H (U ) be the class of analytic function in the open unit disc of the complex
plane U = {z ∈ C : |z| < 1}. Let H (a, n) be the subclass of H (U ) consisting
of functions of the form f (z) = a + an z n + an+1 z n+1 + . . . .
n+1
Let An = {f ∈ H(U
+ . . . , oz ∈ U } and A = A1 .
n ) : f (z) = z + an+1 z
00
(z)
Denote by K = f ∈ A : Re zff 0 (z)
+ 1 > 0, z ∈ U , the class of normalized convex functions in U .
Let the functions f and g be analytic in U . We say that the function f is
subordinate to g, written f ≺ g, if there exists a Schwarz function w, analytic
in U , with w(0) = 0 and |w(z)| < 1, for all z ∈ U, such that f (z) = g(w(z)),
for all z ∈ U . In particular, if the function g is univalent in U , the above
subordination is equivalent to f (0) = g(0) and f (U ) ⊂ g(U ).
Key Words: analytic functions, differential operator, differential subordination, differential superordination.
2010 Mathematics Subject Classification: Primary 30C45.
Received: 5 May, 2014.
Revised: 20 June, 2014.
Accepted: 29 June, 2014.
37
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
38
Let ψ : C3 × U → C and h be an univalent function in U . If p is analytic
in U and satisfies the second order differential subordination
ψ(p(z), zp0 (z), z 2 p00 (z); z) ≺ h(z),
for z ∈ U,
(1)
then p is called a solution of the differential subordination. The univalent
function q is called a dominant of the solutions of the differential subordination,
or more simply a dominant, if p ≺ q for all p satisfying (1). A dominant qe
that satisfies qe ≺ q for all dominants q of (1) is said to be the best dominant
of (1). The best dominant is unique up to a rotation of U .
Let ψ : C2 ×U → C and h analytic in U . If p and ψ p (z) , zp0 (z) , z 2 p00 (z) ; z
are univalent and if p satisfies the second order differential superordination
h(z) ≺ ψ(p(z), zp0 (z), z 2 p00 (z) ; z),
z ∈ U,
(2)
then p is a solution of the differential superordination (2) (if f is subordinate to
F , then F is called to be superordinate to f ). An analytic function q is called
a subordinant if q ≺ p for all p satisfying (2). An univalent subordinant qe that
satisfies q ≺ qe for all subordinants q of (2) is said to be the best subordinant.
Miller and Mocanu [17] obtained conditions h, q and ψ for which the following implication holds
h(z) ≺ ψ(p(z), zp0 (z), z 2 p00 (z) ; z) ⇒ q (z) ≺ p (z) .
P∞
P∞
For two functions f (z) = z + j=2 aj z j and g(z) = z + j=2 bj z j analytic
in the open unit disc U , the Hadamard product (or convolution product) of
f (z) and g (z), written as (f ∗ g) (z), is defined by
f (z) ∗ g (z) = (f ∗ g) (z) = z +
∞
X
aj bj z j .
j=2
Definition 1. (Al Oboudi [7]) For f ∈ A, λ ≥ 0 and n ∈ N, the operator Dλm
is defined by Dλm : A → A,
Dλ0 f (z)
=
f (z)
Dλ1 f
=
(1 − λ) f (z) + λzf 0 (z) = Dλ f (z)
(z)
...
Dλm f (z)
=
(1 − λ) Dλm−1 f (z) + λz (Dλm f (z))0 = Dλ Dλm−1 f (z) , for z ∈ U.
Remark 1. If f ∈ A and f (z) = z +
P∞
m
j
j=2 [1 + (j − 1) λ] aj z , for z ∈ U .
P∞
j=2
aj z j , then Dλm f (z) = z +
Remark 2. For λ = 1 in the above definition we obtain the Sălăgean differential operator [20].
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
39
Definition 2. (Ruscheweyh [19]) For f ∈ A and n ∈ N, the operator Rn is
defined by Rn : A → A,
R0 f (z)
= f (z)
R1 f (z)
= zf 0 (z)
...
0
= z (Rn f (z)) + nRn f (z) ,
P∞
Remark 3. If f ∈ A, f (z) = z + j=2 aj z j , then
(n + 1) R
n+1
f (z)
Rn f (z) = z +
∞
X
(n + j − 1)!
j=2
n! (j − 1)!
z ∈ U.
aj z j for z ∈ U
.
Using the results of Miller and Mocanu [17], [18], Bulboaca [14] considered
certain classes of first order differential superordinations. Recently, many authors [15], [22], [21], [16], [23], have used the results of Bulboaca and obtain
certain sufficient conditions applying first order differential subordinations and
superordinations. In order to prove our subordination and superordination results, we make use of the following:
Definition 3. [18] Denote by Q the set of all functions f that are analytic
and injective on U \E (f ), where E (f ) = {ζ ∈ ∂U : lim f (z) = ∞}, and are
z→ζ
such that f 0 (ζ) 6= 0 for ζ ∈ ∂U \E (f ).
Lemma 1. [18] Let the function q be univalent in the unit disc U and 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 0 (z) φ (q (z)) and h (z) = θ (q (z)) + Q (z). Suppose that
1. Q is
starlike
univalent in U and
zh0 (z)
2. Re Q(z) > 0 for z ∈ U .
If p is analytic with p (0) = q (0), p (U ) ⊆ D and
θ (p (z)) + zp0 (z) φ (p (z)) ≺ θ (q (z)) + zq 0 (z) φ (q (z)) ,
then p (z) ≺ q (z) and q is the best dominant.
Lemma 2. [14] Let the function q be convex univalent in the open unit disc
U and ν and
0 φ beanalytic in a domain D containing q (U ). Suppose that
(q(z))
1. Re νφ(q(z))
> 0 for z ∈ U and
2. ψ (z) = zq 0 (z) φ (q (z)) is starlike univalent in U .
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
40
If p (z) ∈ H [q (0) , 1] ∩ Q, with p (U ) ⊆ D and ν (p (z)) + zp0 (z) φ (p (z)) is
univalent in U and
ν (q (z)) + zq 0 (z) φ (q (z)) ≺ ν (p (z)) + zp0 (z) φ (p (z)) ,
then q (z) ≺ p (z) and q is the best subordinant.
2
Main results
Definition 4. Let λ ≥ 0 and n, m ∈ N. Denote by DRλm,n : A → A the
operator given by the Hadamard product of the generalized Sălăgean operator
Dλm and the Ruscheweyh operator Rn ,
DRλm,n f (z) = (Dλm ∗ Rn ) f (z) ,
for any z ∈ U and each nonnegative integers m, n.
P∞
Remark 4. If f ∈ A and f (z) = z + j=2 aj z j , then
P∞
m
2 j
DRλm,n f (z) = z + j=2 [1 + (j − 1) λ] (n+j−1)!
n!(j−1)! aj z , for z ∈ U .
This operator was studied in [12] and [13].
Remark 5. For λ = 1, m = n, we obtain the Hadamard product SRn [1] of
the Sălăgean operator S n and Ruscheweyh derivative Rn , which was studied
in [2], [3].
Remark 6. For m = n we obtain the Hadamard product DRλn [4] of the
generalized Sălăgean operator Dλn and Ruscheweyh derivative Rn , which was
studied in [5], [6], [8], [9], [10], [11].
Using simple computation one obtains the next result.
Proposition 1. [12].For m, n ∈ N and λ ≥ 0 we have
DRλm+1,n f (z) = (1 − λ) DRλm,n f (z) + λz (DRλm,n f (z))
and
0
0
z (DRλm,n f (z)) = (n + 1) DRλm,n+1 f (z) − nDRλm,n f (z) .
(3)
(4)
First, our purpose is to find sufficient conditions for certain normalized
analytic functions f such that
m,n+1
z δ DRλ
f (z)
z ∈ U, 0 < δ ≤ 1
1+δ ≺ q2 (z) ,
(DRλm,n f (z))
where q1 and q2 are given univalent functions.
q1 (z) ≺
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
41
z δ DRm,n+1 f (z)
λ
1+δ ∈ H (U ) , z ∈ U , f ∈ A, m, n ∈ N, λ ≥ 0,
(DRλm,n f (z))
0 < δ ≤ 1 and let the function q (z) be convex and univalent in U such that
q (0) = 1. Assume that
zq 0 (z) zq 00 (z)
α
+ 0
> 0, z ∈ U,
(5)
Re 1 + q (z) −
β
q (z)
q (z)
Theorem 1. Let
for α, β ∈ C,β 6= 0, z ∈ U, and
ψλm,n (α, β, δ; z) := α
z δ DRλm,n+1 f (z)
(DRλm,n f (z))
1+δ
+
(6)
#
DRλm,n+1 f (z)
− (1 + δ) (n + 1)
.
+β δ (n + 1) − 1 + (n + 2)
DRλm,n f (z)
DRλm,n+1 f (z)
"
DRλm,n+2 f (z)
If q satisfies the following subordination
ψλm,n (α, β, δ; z) ≺ αq (z) +
βzq 0 (z)
,
q (z)
(7)
z ∈ U,
(8)
for α, β ∈ C, β 6= 0 then
z δ DRλm,n+1 f (z)
(DRλm,n f (z))
1+δ
≺ q (z) ,
and q is the best dominant.
m,n+1
z δ DRλ
f (z)
1+δ , z ∈ U , z 6= 0,
(DRλm,n f (z))
0 < δ ≤ 1, f ∈ A. The function p is analytic in U and p (0) = 1.
Differentiating this function,
with respect to z,we get
0
0
m,n+1
m,n
m,n+1
z (DRλ
z (DRλ
f (z))
f (z))
z δ DRλ
f (z)
0
.
zp (z) =
δ
+
−
(1
+
δ)
m,n
m,n+1
1+δ
DRλ f (z)
DRλ
f (z)
(DRλm,n f (z))
By using the identity (4), we obtain
Proof. Let the function p be defined by p (z) :=
DRλm,n+2 f (z)
DRλm,n+1 f (z)
zp0 (z)
= δ (n + 1)−1+(n + 2)
−(1
+
δ)
(n
+
1)
.
p (z)
DRλm,n f (z)
DRλm,n+1 f (z)
(9)
β
By setting θ (w) := αw and φ (w) := w
, α, β ∈ C, β 6= 0 it can be easily
verified that θ is analytic in C, φ is analytic in C\{0} and that φ (w) 6= 0,
w ∈ C\{0}.
0
(z)
Also, by letting Q (z) = zq 0 (z) φ (q (z)) = βzq
q(z) ,we find that Q (z) is
starlike univalent in U.
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
42
0
(z)
Let h (z) = θ (q (z)) + Q (z) = αq (z) + βzq
q(z) , z ∈ U.
If wederivethe function
Q, with respect to z, perform calculations, we
0
zq 0 (z)
zq 00 (z)
(z)
α
=
Re
1
+
q
> 0.
have Re zh
Q(z)
β (z) − q(z) + q 0 (z)
z δ DRm,n+1 f (z)
0
(z)
λ
By using (9), we obtain αp (z) + β zpp(z)
=α
1+δ +
m,n
DRλ
f (z))
(
h
i
m,n+2
DR
f (z)
DRm,n+1 f (z)
+β δ (n + 1) − 1 + (n + 2) DRλm,n+1 f (z) − (1 + δ) (n + 1) DRλm,n f (z) .
λ
λ
0
0
(z)
βzq (z)
By using (7), we have αp (z) + βzp
p(z) ≺ αq (z) + q(z) .
Therefore, the conditions of Lemma 1 are met, so we have p (z) ≺ q (z),
z ∈ U, i.e.
m,n+1
z δ DRλ
f (z)
(DRλm,n f (z))
1+δ
≺ q (z), z ∈ U, and q is the best dominant.
1+Az
Corollary 1. Let q (z) = 1+Bz
, −1 ≤ B < A ≤ 1, m, n ∈ N, λ ≥ 0, z ∈ U.
Assume that (5) holds. If f ∈ A and
ψλm,n (α, β, δ; z) ≺ α
(A − B) z
1 + Az
+β
,
1 + Bz
(1 + Az) (1 + Bz)
for α, β ∈ C, β 6= 0, −1 ≤ B < A ≤ 1, 0 < δ ≤ 1 where ψλm,n is defined in
(6), then
z δ DRλm,n+1 f (z)
1 + Az
≺
1+δ
m,n
1
+ Bz
(DR f (z))
λ
and
1+Az
1+Bz
is the best dominant.
Proof. For q (z) =
1+Az
1+Bz ,
−1 ≤ B < A ≤ 1, in Theorem 1 we get the corollary.
Theorem 2. Let q be convex and univalent in U, such that q (0) = 1, m, n ∈ N,
λ ≥ 0. Assume that
α
q (z) > 0, for α, β ∈ C, β 6= 0, z ∈ U.
(10)
Re
β
m,n+1
z δ DRλ
f (z)
m,n
(α, β, δ; z) is
1+δ ∈ H [q (0) , 1] ∩ Q and ψλ
(DRλm,n f (z))
univalent in U , where ψλm,n (α, β, δ; z) is as defined in (6), then
If f ∈ A,0 < δ ≤ 1,
αq (z) +
βzq 0 (z)
≺ ψλm,n (α, β, δ; z) ,
q (z)
z ∈ U,
(11)
implies
q (z) ≺
z δ DRλm,n+1 f (z)
(DRλm,n f (z))
and q is the best subordinant.
1+δ
,
z ∈ U,
(12)
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
Proof. Let the function p be defined by p (z) :=
m,n+1
z δ DRλ
f (z)
1+δ
(DRλm,n f (z))
43
, z ∈ U , z 6= 0,
0 < δ ≤ 1, f ∈ A.
β
By setting ν (w) := αw and φ (w) := w
it can be easily verified that ν is
analytic in C, φ is analytic in C\{0} and that φ (w) 6= 0, w ∈ C\{0}.
Since q is convex and univalent function, it follows that Re
Re α
β q (z) > 0, for α, β ∈ C, β 6= 0.
ν 0 (q(z))
φ(q(z))
=
By using (11) we obtain
αq (z) + β
zq 0 (z)
zp0 (z)
≺ αp (z) + β
.
q (z)
p (z)
Using Lemma 2, we have
q (z) ≺ p (z) =
z δ DRλm,n+1 f (z)
(DRλm,n f (z))
1+δ
,
z ∈ U,
and q is the best subordinant.
Corollary 2. Let q (z) =
that (10) holds. If f ∈ A,
α
1+Az
1+Bz , −1 ≤ B
m,n+1
z δ DRλ
f (z)
1+δ
(DRλm,n f (z))
< A ≤ 1, m, n ∈ N, λ ≥ 0. Assume
∈ H [q (0) , 1] ∩ Q and
(A − B) z
1 + Az
+β
≺ ψλm,n (α, β, δ; z) ,
1 + Bz
(1 + Az) (1 + Bz)
for α, β ∈ C, β 6= 0, −1 ≤ B < A ≤ 1, where ψλm,n is defined in (6), then
z δ DRλm,n+1 f (z)
1 + Az
≺
1+δ
1 + Bz
(DRλm,n f (z))
and
1+Az
1+Bz
is the best subordinant.
Proof. For q (z) =
1+Az
1+Bz ,
−1 ≤ B < A ≤ 1 in Theorem 2 we get the corollary.
Combining Theorem 1 and Theorem 2, we state the following sandwich
theorem.
Theorem 3. Let q1 and q2 be analytic and univalent in U such that q1 (z) 6= 0
and q2 (z) 6= 0, for all z ∈ U , with zq10 (z) and zq20 (z) being starlike univalent.
Suppose that q1 satisfies (5) and q2 satisfies (10). If f ∈ A,
m,n+1
z δ DRλ
f (z)
(DRλm,n f (z))
1+δ
∈
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
44
H [q (0) , 1] ∩ Q, 0 < δ ≤ 1 and ψλm,n (α, β, δ; z) is as defined in (6) univalent
in U , then
αq1 (z) +
βzq10 (z)
βzq20 (z)
≺ ψλm,n (α, β, δ; z) ≺ αq2 (z) +
,
q1 (z)
q2 (z)
for α, β ∈ C, β 6= 0, implies
q1 (z) ≺
z δ DRλm,n+1 f (z)
(DRλm,n f (z))
1+δ
≺ q2 (z) ,
δ ∈ C, δ 6= 0,
and q1 and q2 are respectively the best subordinant and the best dominant.
1+A1 z
, q2 (z) =
For q1 (z) = 1+B
1z
we have the following corollary.
1+A2 z
1+B2 z ,
where −1 ≤ B2 < B1 < A1 < A2 ≤ 1,
Corollary 3. Let m, n ∈ N, λ ≥ 0. Assume that (5) and (10) hold for q1 (z) =
1+A1 z
1+B1 z
and q2 (z) =
1+A2 z
1+B2 z ,
respectively. If f ∈ A, 0 < δ ≤ 1,
H [q (0) , 1] ∩ Q and
α
m,n+1
z δ DRλ
f (z)
1+δ
(DRλm,n f (z))
∈
(A1 − B1 ) z
1 + A1 z
+β
≺ ψλm,n (α, β, δ; z)
1 + B1 z
(1 + A1 z) (1 + B1 z)
≺α
(A2 − B2 ) z
1 + A2 z
+β
,
1 + B2 z
(1 + A2 z) (1 + B2 z)
for α, β ∈ C, β 6= 0, −1 ≤ B2 ≤ B1 < A1 ≤ A2 ≤ 1, where ψλm,n is defined in
(6), then
z δ DRλm,n+1 f (z)
1 + A1 z
1 + A2 z
≺
≺
,
1+δ
m,n
1 + B1 z
1
+ B2 z
(DR f (z))
λ
1+A1 z
hence 1+B
and
1z
spectively.
1+A2 z
1+B2 z
are the best subordinant and the best dominant, re-
Next, our purpose is to find sufficient conditions for certain normalized
analytic functions f such that
δ
m+1,n
m,n
aDRλ
f (z)+bDRλ
f (z)
q1 (z) ≺
≺ q2 (z) , z ∈ U
(a+b)z
where q1 and q2 are given univalent functions
δ
m+1,n
m,n
aDRλ
f (z)+bDRλ
f (z)
Theorem 4. Let
∈ H (U ) , f ∈ A, z ∈ U ,
(a+b)z
δ, a, b ∈ C, δ 6= 0, a + b 6= 0, m, n ∈ N, λ ≥ 0 and let the function q (z) be
convex and univalent in U such that q (0) = 1, z ∈ U . Assume that
α
zq 0 (z) zq 00 (z)
Re 1 + q (z) −
+ 0
> 0,
(13)
β
q (z)
q (z)
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
45
for α, β ∈ C, β 6= 0, z ∈ U, and
ψλm,n
(a, b, α, β, δ; z) := α
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ
+
h
i
βδ aDRλm+2,n f (z) + (b − a) DRλm+1,n f (z) − bDRλm,n f (z)
.
λ aDRλm+1,n f (z) + bDRλm,n f (z)
(14)
If q satisfies the following subordination
ψλm,n (a, b, α, β, δ; z) ≺ αq (z) +
βzq 0 (z)
,
q (z)
(15)
z ∈ U, δ ∈ C, δ 6= 0,
(16)
for α, β ∈ C, β 6= 0, z ∈ U, then
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ
≺ q (z) ,
and q is the best dominant.
δ
m+1,n
m,n
aDRλ
f (z)+bDRλ
f (z)
Proof. Let the function p be defined by p (z) :=
,
(a+b)z
z ∈ U , z 6= 0, δ, a, b ∈ C, δ 6= 0, a + b 6= 0, f ∈ A. The function p is analytic in
U and p (0) = 1
Differentiating this function, with respect to z, we get
0
zp (z) = δ
1
·
a+b
"
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ−1
·
#
0
a DRλm+1,n f (z) + b (DRλm,n f (z))0
aDRλm+1,n f (z) + bDRλm,n f (z)
−
.
z
z2
We have
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ
1
·
aDRλm+1,n f (z) + bDRλm,n f (z)
h
i
0
az DRλm+1,n f (z) + bz (DRλm,n f (z))0 − aDRλm+1,n f (z) − bDRλm,n f (z) (17)
zp0 (z) = δ
By using the identity (3) we obtain
h
i
m+2,n
m+1,n
m,n
δ
aDR
f
(z)
+
(b
−
a)
DR
f
(z)
−
bDR
f
(z)
λ
λ
λ
zp (z)
=
m+1,n
m,n
p (z)
λ aDRλ
f (z) + bDRλ f (z)
0
(18)
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
46
β
By setting θ (w) := αw and φ (w) := w
, α, β ∈ C, β 6= 0 it can be easily
verified that θ is analytic in C, φ is analytic in C\{0} and that φ (w) 6= 0,
w ∈ C\{0}.
0
(z)
Also, by letting Q (z) = zq 0 (z) φ (q (z)) = βzq
q(z) ,we find that Q (z) is
starlike univalent in U.
0
(z)
Let h (z) = θ (q (z)) + Q (z) = αq (z) + βzq
q(z) , z ∈ U
If wederivethe function
Q, with respect to z, perform calculations, we
0
00
zh0 (z)
(z)
(z)
α
have Re Q(z) = Re 1 + β q (z) − zqq(z)
> 0.
+ zqq0 (z)
By using (9), we obtain
!δ
aDRλm+1,n f (z) + bDRλm,n f (z)
zp0 (z)
αp (z) + β
+
=α
p (z)
(a + b) z
h
i
βδ aDRλm+2,n f (z) + (b − a) DRλm+1,n f (z) − bDRλm,n f (z)
+
.
λ aDRλm+1,n f (z) + bDRλm,n f (z)
0
0
(z)
(z)
By using (15), we have αp (z) + β zpp(z)
≺ αq (z) + β zqp(z)
.
From Lemma 1, we have p (z) ≺ q (z), z ∈ U, i.e.,
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ
≺ q (z) ,
z ∈ U, δ ∈ C, δ 6= 0, and q is the best dominant.
1+Az
Corollary 4. Let q (z) = 1+Bz
, z ∈ U, −1 ≤ B < A ≤ 1, m, n ∈ N, λ ≥ 0.
Assume that (13) holds. If f ∈ A and
ψλm,n (a, b, α, β, δ; z) ≺ α
(A − B) z
1 + Az
+β
,
1 + Bz
(1 + Az) (1 + Bz)
for α, β ∈ C, β 6= 0, −1 ≤ B < A ≤ 1, where ψλm,n is defined in (14), then
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
and
1+Az
1+Bz
!δ
≺
1 + Az
,
1 + Bz
δ ∈ C, δ 6= 0,
is the best dominant.
Proof. For q (z) =
1+Az
1+Bz ,
−1 ≤ B < A ≤ 1, in Theorem 4 we get the corollary.
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
47
Theorem 5. Let q be convex and univalent in U such that q (0) = 1. Assume
that
α
Re
q (z) > 0, for α, β ∈ C, β 6= 0.
(19)
β
δ
m+1,n
m,n
aDRλ
f (z)+bDRλ
f (z)
If f ∈ A, δ, a, b ∈ C, δ 6= 0, a + b 6= 0,
∈
(a+b)z
H [q (0) , 1] ∩ Q and
ψλm,n (a, b, α, β, δ; z) is univalent in U , where ψλm,n (a, b, α, β, δ; z) is as defined
in (14), then
βzq 0 (z)
αq (z) +
≺ ψλm,n (a, b, α, β, δ; z)
(20)
q (z)
implies
q (z) ≺
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ
,
δ ∈ C, δ 6= 0, z ∈ U,
(21)
and q is the best subordinant.
δ
m+1,n
m,n
aDRλ
f (z)+bDRλ
f (z)
,
Proof. Let the function p be defined by p (z) :=
(a+b)z
z ∈ U , z 6= 0, a, b ∈ C, a + b 6= 0, δ ∈ C, δ 6= 0, f ∈ A. The function p is
analytic in U and p (0) = 1.
β
By setting ν (w) := αw and φ (w) := w
it can be easily verified that ν is
analytic in C, φ is analytic in C\{0} and that φ (w) 6= 0, w ∈ C\{0}.
Since q is convex and univalent function, it follows that Re
Re α
q
(z)
> 0, for α, β ∈ C, β 6= 0.
β
ν 0 (q(z))
φ(q(z))
=
By using (11) we obtain
αq (z) + β
zp0 (z)
zq 0 (z)
≺ αp (z) + β
.
q (z)
p (z)
From Lemma 2, we have
q (z) ≺ p (z) =
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ
,
z ∈ U, δ ∈ C, δ 6= 0,
and q is the best subordinant.
1+Az
1+Bz ,
−1 ≤ B < A ≤ 1, z ∈ U, m, n ∈ N,
δ
m+1,n
m,n
aDRλ
f (z)+bDRλ
f (z)
λ ≥ 0. Assume that (19) holds. If f ∈ A,
∈
(a+b)z
Corollary 5. Let q (z) =
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
48
H [q (0) , 1] ∩ Q, δ ∈ C, δ 6= 0 and
α
1 + Az
(A − B) z
+β
≺ ψλm,n (a, b, α, β, δ; z) ,
1 + Bz
(1 + Az) (1 + Bz)
for α, β ∈ C, β 6= 0, −1 ≤ B < A ≤ 1, where ψλm,n is defined in (14), then
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
1 + Az
≺
1 + Bz
and
1+Az
1+Bz
!δ
, δ ∈ C, δ 6= 0, a, b ∈ C, a + b 6= 0
is the best subordinant.
Proof. For q (z) =
1+Az
1+Bz ,
−1 ≤ B < A ≤ 1, in Theorem 5 we get the corollary.
Combining Theorem 4 and Theorem 5, we state the following sandwich
theorem.
Theorem 6. Let q1 and q2 be convex and univalent in U such that q1 (z) 6= 0
and q2 (z) 6= 0, for all z ∈ U . Suppose that q1 satisfies (13) and q2 satisfies
(19).
δ
m+1,n
m,n
aDRλ
f (z)+bDRλ
f (z)
∈ H [q (0) , 1] ∩ Q , δ ∈ C, δ 6= 0 ,
If f ∈ A,
(a+b)z
a, b ∈ C, a + b 6= 0 and ψλm,n (a, b, α, β, δ; z) is as defined in (14) univalent in
U , then
αq1 (z) +
βzq10 (z)
βzq20 (z)
≺ ψλm,n (a, b, α, β, δ; z) ≺ αq2 (z) +
,
q1 (z)
q2 (z)
for α, β ∈ C, β 6= 0, implies
q1 (z) ≺
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ
≺ q2 (z) ,
z ∈ U, δ ∈ C, δ 6= 0,
and q1 and q2 are respectively the best subordinant and the best dominant.
1+A1 z
For q1 (z) = 1+B
, q2 (z) =
1z
we have the following corollary.
1+A2 z
1+B2 z ,
where −1 ≤ B2 < B1 < A1 < A2 ≤ 1,
Corollary 6. Let m, n ∈ N, λ ≥ 0. Assume that (13) and (19) hold for
δ
m+1,n
DRλ
f (z)
1+A1 z
1+A2 z
q1 (z) = 1+B
and q2 (z) = 1+B
, respectively. If f ∈ A,
∈
DRm,n f (z)
1z
2z
H [q (0) , 1] ∩ Q and
α
λ
1 + A1 z
(A1 − B1 ) z
+β
≺ ψλm,n (a, b, α, β, δ; z)
1 + B1 z
(1 + A1 z) (1 + B1 z)
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
≺α
49
1 + A2 z
(A2 − B2 ) z
+β
,
1 + B2 z
(1 + A2 z) (1 + B2 z)
for α, β ∈ C, β 6= 0, −1 ≤ B2 ≤ B1 < A1 ≤ A2 ≤ 1, where ψλm,n is defined in
(14), then
1 + A1 z
≺
1 + B1 z
aDRλm+1,n f (z) + bDRλm,n f (z)
(a + b) z
!δ
≺
1 + A2 z
,
1 + B2 z
z ∈ U, δ ∈ C, δ 6= 0, a, b ∈ C, a + b 6= 0
1+A1 z
and
hence 1+B
1z
spectively.
1+A2 z
1+B2 z
are the best subordinant and the best dominant, re-
Acknowledgement. This work was partially suported by the strategic
project POSDRU/159/1.5/S/138963 - ”Perform”.
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Loriana ANDREI,
Department of Mathematics and Computer Science,
University of Oradea,
1 Universitatii Street, 410087, Oradea, Romania.
Department of Mathematics,
University of Pitesti,
Targul din Vale Street, No. 1, 110040, Pitesti, Romania
Email: lori andrei@yahoo.com
ON CERTAIN DIFFERENTIAL SANDWICH THEOREMS INVOLVING A
GENERALIZED SALAGEAN OPERATOR AND RUSCHEWEYH OPERATOR
52