MATEMATIQKI VESNIK
originalni nauqni rad
research paper
65, 2 (2013), 178–186
June 2013
SANDWICH-TYPE RESULTS FOR A CLASS OF FUNCTIONS
DEFINED BY A GENERALIZED DIFFERENTIAL OPERATOR
Priyabrat Gochhayat
Abstract. By making use of a generalized differential operator a new class of non-Bazilevič
functions is introduced. Differential sandwich-type theorem for the above class is investigated.
Relevant connections of the results, which are presented in this paper, with various other known
results are also pointed out.
1. Introduction
Let H be the class of functions analytic in the open unit disk
U := {z : z ∈ C and |z| < 1}
and H[a, n] (n ∈ N := {1, 2, 3, · · · }) be the subclass of H consisting of functions of
the form
f (z) = a + an z n + an+1 z n+1 + · · · .
Let A(⊂ H) be the class of all analytic functions given by the power series
f (z) = z +
∞
P
n=2
an z n
(z ∈ U ).
Recalling the principle of subordination between analytic functions, we say
that f is subordinate to g, written as f ≺ g in U or f (z) ≺ g(z) (z ∈ U ), if there
exists a function ω, analytic in U satisfying the conditions of the Schwarz lemma
( i.e. ω(0) = 0 and |ω(z)| < 1) such that f (z) = g(ω(z)) (z ∈ U). It follows that
f (z) ≺ g(z) (z ∈ U) =⇒ f (0) = g(0) and
f (U ) ⊂ g(U).
In particular, if g is univalent in U, then the reverse implication also holds (cf. [13]).
2010 AMS Subject Classification: 30C45, 30C80
Keywords and phrases: Univalent function; derivative operator; differential subordination;
differential superordination.
The present investigation is supported under the Fast Track Research Project for Young
Scientist, Department of Science and Technology, New Delhi, Government of India. Sanction
Letter No. 100/IFD/12100/2010-11.
178
179
Sandwich-type theorem . . .
Furthermore, f is said to be subordinate to g in the disk Ur := {z : z ∈
C and |z| < r}, if the function fr (z) = f (rz) is subordinate to gr (z) = g(rz) in U.
Definition 1.1. Let p, h ∈ H and let ϕ(r, s, t; z) : C3 × U → C. If p(z)
and ϕ(p(z), zp0 (z), z 2 p00 (z); z) are univalent and if p(z) satisfies the second order
superordination
h(z) ≺ ϕ(p(z), zp0 (z), z 2 p00 (z); z),
(z ∈ U )
(1.1)
then p(z) is a solution of the differential superordination (1.1). (Note that, if f is
subordinate to F , then F is superordinate to f ).
An analytic function q is called a subordinant of the differential superordination, or more precisely a subordinant if q ≺ p, for all p satisfying (1.1). A univalent
subordinant qe that satisfies q ≺ qe, for all subordinants q of (1.1) is said to be the
best subordinant. Note that the best subordinant is unique upto a rotation of U.
Recently in [10], Miller and Mocanu obtained conditions on h, q and ϕ for which
the following implication holds:
h(z) ≺ ϕ(p(z), zp0 (z), z 2 p00 (z); z) ⇒ q(z) ≺ p(z) (z ∈ U ).
Using the results due to Miller and Mocanu [10], Bulboaca [6] considered certain classes of first order differential superordination as well as superordinationpreserving integral operators [5]. More recently using the result of Bulboaca [6],
Ali et al. [1] obtained some sufficient conditions for functions to satisfy
q1 (z) ≺ zf 0 (z)/f (z) ≺ q2 (z),
(z ∈ U)
where q1 , q2 are univalent in U with q1 (0) = 1 = q2 (0).
by
k,α
We now introduce the generalized differential operator Dλ,δ
: A → A defined
k,α
Dλ,δ
f (z) = z +
∞
P
µ
¶
n+δ−1
an z n
δ
(z ∈ U ; λ, δ ≥ 0; k, α ∈ N0 := N ∪ {0}).
[nα + (n − 1)nα λ]k
n=2
(1.2)
k,α
Note that the differential operator Dλ,δ
unifies many operators of A.
In particular:
k,0
k,1
(= D1,1
) ≡ Sălăgean operator (cf. [12]),
D0,0
0,α
Dλ,δ
≡ Ruscheweyh differential operator of order δ (cf. [11]),
k,0
Dλ,0
≡ Al-Oboudi operator(cf. [2]),
k,0
k,1
D1,δ
(= D0,δ
) ≡ differential operator studied by Al-shaqsi and Darus (cf. [4]).
k,α
By using the operator Dλ,δ
, we now define a new subclass of analytic functions
as follows:
180
P. Gochhayat
Definition 1.2. The function f ∈ A is said to be in the class GDk,α
λ,δ (φ)
(λ, δ ≥ 0; k, α ∈ N0 ) if and only if it satisfies the condition
Ã
!1+µ
z
k,α 0
(Dλ,δ f ) (z)
≺ φ(z)
(z ∈ U ; 0 ≤ µ ≤ 1),
(1.3)
k,α
(Dλ,δ
f )(z)
where φ(z) is an analytic function with positive real part on U with φ(0) =
1, φ0 (0) > 0 which maps the unit disc U onto a region which is symmetric with
respect to the real axis.
Note that for k = 0, δ = 0 and φ = (1 + z)/(1 − z) the class reduces to the
class of functions of non-Bazilevič type which is recently introduced and studied by
Obradović [3] as follows:
µ
¶
µ
¶1+µ
1+z
z
1+z
α
0
f ∈ GDλ
⇔ f (z)
≺
(z ∈ U; 0 ≤ µ ≤ 1).
1−z
f (z)
1−z
In the present article we investigate certain subordination and superordination
results of the class GDk,α
λ,δ (φ), together with differential sandwich type theorem as
an interesting consequence of the results.
2. Preliminaries
To establish our main results, we need the following:
Definition 2.1. ([10, Definition 2, p. 817]; see also [9, Definition 2.2b, p. 21])
Let Q be the set of all functions f that are analytic and injective on U \ E(f ),
where
n
o
E(f ) := ζ : ζ ∈ ∂U and lim f (z) = ∞
z→ζ
0
and are such that f (ζ) 6= 0 for ζ ∈ ∂U \ E(f ).
Lemma 2.2. [9, Theorem 3.4h, p. 132] Let q be univalent in the open unit disk
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)), h(z) = θ(q(z)) + Q(z) and suppose that
1. Q is starlike in U , and
³ 0 ´
(z)
2. < zh
> 0 for z ∈ U .
Q(z)
If p is analytic in U, 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 ≺ q and q is the best dominant.
Lemma 2.3. [6] Let q be univalent in the open unit disk U and ϑ and ϕ be
analytic in a domain D containing q(U ). Suppose that
³ 0
´
(q(z))
1. < zϑ
> 0 for z ∈ U, and
ϕ(q(z))
181
Sandwich-type theorem . . .
2. zq 0 (z)ϕ(q(z)) is starlike univalent in U .
If p ∈ H[q(0), 1] ∩ Q, with p(U ) ⊆ D, and ϑ(p(z)) + zp0 (z)ϕ(p(z)) is univalent
in U and
ϑ(q(z)) + zq 0 (z)ϕ(q(z)) ≺ ϑ(p(z)) + zp0 (z)ϕ(p(z)),
(z ∈ U)
then q ≺ p and q is the best subordinant.
Lemma 2.4. [8] If −1 ≤ B < A ≤ 1, β > 0 and the complex number γ is
constrained by
β(1 − A)
<(γ) ≥ −
,
(1 − B)
then the following differential equation
q(z) +
1 + Az
zq 0 (z)
=
βq(z) + γ
1 + Bz
(z ∈ U )
has a univalent solution in U given by

z β+γ (1+Bz)β(A−B)/B
γ

 β R z tβ+γ−1 (1+Bt)β(A−B)/B dt − β ;
0
q(z) =
β+γ
exp(βAz)

 R zz β+γ−1
− βγ ;
β
0
t
exp(βAt)dt
B 6= 0
(2.1)
B = 0.
If the function φ(z) = 1 + c1 z + c2 z 2 + · · · is analytic in U and satisfies
φ(z) +
zφ0 (z)
1 + Az
≺
βφ(z) + γ
1 + Bz
then
φ(z) ≺ q(z) ≺
1 + Az
1 + Bz
(z ∈ U ),
(z ∈ U)
(2.2)
and q(z) is the best dominant of (2.2).
3. Main results
We have the following subordination and superordination results:
Theorem 3.1. Let the function q be analytic in U such that q(z) 6= 0 (z ∈ U).
0
(z)
is univalent starlike in U. Let
Suppose that zqq(z)
½
¾
q(z) zq 00 (z) zq 0 (z)
< 1+
+ 0
−
> 0 (β ∈ C; β 6= 0)
(3.1)
β
q (z)
q(z)
and
³
Ω(α, k, λ, δ; f )(z) :=
α,k
Dλ,δ
f
"
+β
´0
Ã
(z)
z
α,k
(Dλ,δ
f )(z)
α,k
z(Dλ,δ
f )00 (z)
α,k
(Dλ,δ
f )0 (z)
!1+µ
Ã
+ (1 + µ) 1 −
α,k
z(Dλ,δ
f )0 (z)
α,k
(Dλ,δ
f )(z)
!#
.
(3.2)
182
P. Gochhayat
If q satisfies
Ω(α, k, λ, δ; f )(z) ≺ q(z) + β
then
³
α,k
Dλ,δ
f
´0
Ã
(z)
zq 0 (z)
q(z)
(β ∈ C; β 6= 0),
!1+µ
z
α,k
(Dλ,δ
f )(z)
≺ q(z)
(0 ≤ µ ≤ 1)
(3.3)
and q is the best dominant.
Proof. Let the function p be defined by
Ã
!1+µ
³
´0
z
α,k
p(z) := Dλ,δ f (z)
α,k
(Dλ,δ
f )(z)
(z ∈ U ; z 6= 0; f ∈ A).
Logarithmic differentiation yields
Ã
!
α,k
α,k
z(Dλ,δ
f )00 (z)
z(Dλ,δ
f )0 (z)
zp0 (z)
=
+ (1 + µ) 1 −
.
α,k
α,k
p(z)
(Dλ,δ
f )0 (z)
(Dλ,δ
f )(z)
0
(z)
Let θ(w) := w and φ(w) := β/w; by letting Q(z) = zq 0 (z)φ(q(z)) = β zqq(z)
and
0
(z)
h(z) = θ(q(z))+Q(z) = q(z)+β zqq(z)
, we observe that Q(z) is univalent starlike in U
³ 0 ´
zh (z)
and < Q(z) > 0. Thus assertion (3.3) of Theorem 3.1 follows by an applications
of Lemma 2.2. This completes the proof.
³
´τ
1+Az
1+z
By taking q(z) = 1+Bz
, −1 ≤ B < A ≤ 1 and q(z) = 1−z
, 0 < τ ≤ 1, in
Theorem 3.1, we get the following:
Corollary 3.2. Assume that (3.1) holds. If f ∈ A and
Ω(α, k, λ, δ; f )(z) ≺
1 + Az
(A − B)z
+β
1 + Bz
(1 + Az)(1 + Bz)
(β ∈ C; β 6= 0),
where Ω(α, k, λ, δ; f ) is defined in (3.2), then
³
and
1+Az
1+Bz
α,k
Dλ,δ
f
´0
Ã
(z)
z
α,k
(Dλ,δ
f )(z)
!1+µ
≺
1 + Az
1 + Bz
(0 ≤ µ ≤ 1)
is the best dominant.
Corollary 3.3. Assume that (3.1) holds. If f ∈ A and
µ
¶τ
1+z
2βτ z
Ω(α, k, λ, δ; f )(z) ≺
(β ∈ C; β 6= 0),
+
1−z
(1 − z 2 )
183
Sandwich-type theorem . . .
where Ω(α, k, λ, δ; f ) is defined in (3.2), then
Ã
!1+µ µ
¶τ
³
´0
z
1+z
α,k
Dλ,δ f (z)
≺
α,k
1−z
(Dλ,δ
f )(z)
³
´τ
1+z
is the best dominant.
and 1−z
(0 ≤ µ ≤ 1)
Theorem 3.4. Let the function q be analytic in U such that q(z) 6= 0 (z ∈ U).
0
(z)
Suppose that zqq(z)
is univalent starlike in U. Furthermore assume that
½
¾
q(z)
<
> 0 (β ∈ C; β 6= 0).
(3.4)
β
If f ∈ A, with
³
0 6=
α,k
Dλ,δ
f
´0
Ã
(z)
!1+µ
z
∈ H[q(0), 1] ∩ Q,
α,k
(Dλ,δ
f )(z)
and Ω(α, k, λ, δ; f ) is univalent in U, then
q(z) + β
zq 0 (z)
≺ Ω(α, k, λ, δ; f )(z)
q(z)
implies
³
q(z) ≺
α,k
Dλ,δ
f
´0
Ã
(z)
z
α,k
(Dλ,δ
f )(z)
(β ∈ C; β 6= 0),
!1+µ
(0 ≤ µ ≤ 1)
(3.5)
where q is the best subordinant and Ω(α, k, λ, δ; f ) is given by (3.2).
Proof. By setting ϑ(w) := w and ϕ := β/w, we observe that ϑ and ϕ are
analytic in C and C \ {0} respectively. ϕ(w) 6= 0 (w ∈ C \ {0}) and q is univalent
convex yields
µ 0
¶
µ
¶
ϑ (q(z))
q(z)
<
=<
> 0 (β ∈ C; β 6= 0).
ϕ(q(z))
β
Application of Lemma 2.3 gives the assertion (3.5) of Theorem 3,4. This completes
the proof.
Now by combining Theorems 3.1 and 3.4, we get the following differential
Sandwich-type theorem:
Theorem 3.5. Let the function q1 and q2 be univalent in U such that q1 (z) 6= 0
zq 0 (z)
zq 0 (z)
and q2 (z) 6= 0 (z ∈ U) with q11(z) and q22(z) is univalent starlike. Furthermore
assume that q1 and q2 satisfies (3.1) and (3.4), respectively. If f ∈ A, with
Ã
!1+µ
³
´0
z
α,k
0 6= Dλ,δ f (z)
∈ H[q(0), 1] ∩ Q, and Ω(α, k, λ, δ; f )
α,k
(Dλ,δ
f )(z)
184
P. Gochhayat
is univalent in U, then
q1 (z) + β
zq10 (z)
zq 0 (z)
≺ Ω(α, k, λ, δ; f )(z) ≺ q2 (z) + β 2
q1 (z)
q2 (z)
(β ∈ C; β 6= 0),
implies
³
α,k
q1 (z) ≺ Dλ,δ
f
´0
Ã
(z)
z
α,k
(Dλ,δ
f )(z)
!1+µ
≺ q2 (z)
(0 ≤ µ ≤ 1)
(3.6)
where q1 and q2 are respectively the best subordinant and the dominant.
Taking k = 0, δ = 0; µ → 0 and 1 we have the following:
Corollary 3.6. For k = 0, δ = 0; µ → 0, let the functions q1 and q2 be
zq 0 (z)
zq 0 (z)
univalent in U such that q1 (z) 6= 0 and q2 (z) 6= 0 (z ∈ U ) with q11(z) and q22(z)
is univalent starlike. Furthermore assume that q1 and q2 satisfy (3.1) and (3.4),
respectively. If f ∈ A, with
zf 0 (z)
∈ H[q(0), 1] ∩ Q,
f (z)
and
· 00
µ
¶¸
zf 0 (z)
zf (z)
zf 0 (z)
+β
+
1
−
f (z)
f 0 (z)
f (z)
is univalent in U, then
q1 (z) + β
zq10 (z)
zf 0 (z)
zq 0 (z)
≺
≺ q2 (z) + β 2
q1 (z)
f (z)
q2 (z)
(β ∈ C; β 6= 0),
implies
zf 0 (z)
≺ q2 (z)
f (z)
where q1 and q2 are respectively the best subordinant and the dominant.
q1 (z) ≺
Corollary 3.7. [7] For k = 0, δ = 0; µ → 1, let the functions q1 and q2 be
zq 0 (z)
zq 0 (z)
univalent in U such that q1 (z) 6= 0 and q2 (z) 6= 0 (z ∈ U ) with q11(z) and q22(z)
is univalent starlike. Furthermore assume that q1 and q2 satisfies (3.1) and (3.4),
respectively. If f ∈ A, with
z 2 f 0 (z)
∈ H[q(0), 1] ∩ Q,
(f (z))2
and
· 00
µ
¶¸
z 2 f 0 (z)
zf (z)
zf 0 (z)
+β
+2 1−
(f (z))2
f 0 (z)
f (z)
is univalent in U, then
q1 (z) + β
z 2 f 0 (z)
zq 0 (z)
zq10 (z)
≺
≺ q2 (z) + β 2
2
q1 (z)
(f (z))
q2 (z)
(β ∈ C; β 6= 0),
185
Sandwich-type theorem . . .
implies
z 2 f 0 (z)
≺ q2 (z)
(f (z))2
q1 (z) ≺
where q1 and q2 are respectively the best subordinant and the dominant.
Theorem 3.8. For −1 ≤ B < A ≤ 1 and 0 ≤ µ ≤ 1, if
Ã
!1+µ
³
´0
z
α,k
Dλ,δ f (z)
α,k
(Dλ,δ
f )(z)
Ã
!#
"
α,k
α,k
z(Dλ,δ
f )0 (z)
z(Dλ,δ f )00 (z)
1 + Az
+ (1 + µ) 1 −
≺
+β
α,k
α,k
0
1
+ Bz
(Dλ,δ f ) (z)
(Dλ,δ f )(z)
then
³
α,k
Dλ,δ
f
´0
Ã
(z)
where
q(z) =
!1+µ
z
≺ q(z) ≺
α,k
(Dλ,δ
f )(z)
1 + Az
1 + Bz
(β > 0)
(z ∈ U ),

βz 1/β (1+Bz)(A−B)/βB

 R z t(1−β)/β (1+Bt)(A−B)/βB dt ;
(B 6= 0)

 Rz
(B = 0),
0
0
βz 1/β exp(Az/β)
;
t(1−β)/β exp(At/β)dt
(3.7)
(3.8)
(3.9)
and q is the best dominant of (3.8).
Proof. Write
³
´0
α,k
φ(z) = Dλ,δ
f (z)
Ã
!1+µ
z
α,k
(Dλ,δ
f )(z)
(z ∈ U ; z 6= 0; f ∈ A).
Therefore, we observe that φ(z) = 1 + c1 z + c2 z 2 + · · · is analytic in U. Taking
logarithmic differentiation and using (3.7) gives
φ(z) + β
1 + Az
zφ0 (z)
≺
φ(z)
1 + Bz
Application of Lemma 2.4 yields
φ(z) ≺ q(z) ≺
1 + Az
1 + Bz
(z ∈ U ),
where q(z) is defined by (3.3) is the best dominant. This completes the proof.
Taking k = 0, δ = 0; µ → 0 and 1, we get the following corollaries, respectively:
Corollary 3.9. For −1 ≤ B < A ≤ 1 and µ → 0, if β ∈ C
· 00
µ
¶¸
zf (z)
zf 0 (z)
1 + Az
zf 0 (z)
+β
+
1
−
≺
(β > 0)
0
f (z)
f (z)
f (z)
1 + Bz
186
P. Gochhayat
then
zf 0 (z)
1 + Az
≺ q(z) ≺
f (z)
1 + Bz
(z ∈ U),
where q is the best dominant.
Corollary 3.10. For −1 ≤ B < A ≤ 1 and µ → 1, if β ∈ C
¶¸
· 00
µ
z 2 f 0 (z)
zf (z)
zf 0 (z)
1 + Az
+β
+2 1−
≺
(β > 0)
2
0
(f (z))
f (z)
f (z)
1 + Bz
then
1 + Az
z 2 f 0 (z)
≺ q(z) ≺
(f (z))2
1 + Bz
(z ∈ U ),
where q is the best dominant.
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(received 16.06.2011; in revised form 17.01.2012; available online 15.03.2012)
Department of Mathematics, Sambalpur University, Jyoti Vihar, Burla, Sambalpur, 768019,
Odisha, India
E-mail: pb gochhayat@yahoo.com