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J. Nonlinear Sci. Appl., 10 (2017), 954–963
Research Article
Journal Homepage: www.tjnsa.com - www.isr-publications.com/jnsa
On second-order differential subordinations for a class of analytic functions
defined by convolution
Arzu Akgül
Kocaeli University, Faculty of Arts and Sciences, Department of Mathematics, Umuttepe Campus, İzmit-Kocaeli, Turkey.
Communicated by E. Savaş
Abstract
Making use of the convolution operator we introduce a new class of analytic functions in the open unit disk and investigate
c
some subordination results. 2017
All rights reserved.
Keywords: Analytic functions, univalent function, differential subordination, convex function, Hadamard product, best
dominant.
2010 MSC: 30C45.
1. Introduction
Let C be complex plane and U = {z ∈ C : |z| < 1} = U\ {0}, open unit disc in C. Let H(U) be the
class of analytic functions in U. For p ∈ N+ = {1, 2, 3, · · · } and a ∈ C, let H[a, k] be the subclass of H(U)
consisting of the functions of the form
f(z) = a + ak zk + ak+1 zk+1 + · · ·
with H0 ≡ H[0, 1] and H ≡ H[1, 1]. Let Ap be the class of all analytic functions of the form
∞
X
p
f(z) = z +
ak z k
(1.1)
k=p+1
in the open unit disk U with A1 = A. For functions f ∈ Ap given by equation (1.1) and g ∈ Ap defined
by
∞
X
p
g(z) = z +
b k zk ,
k=p+1
their Hadamard product (or convolution) [7] of f and g is defined by
p
(f ∗ g)(z) := z +
∞
X
ak bk zk .
k=p+1
A function f ∈ H(U) is univalent if it is one to one in U. Let S denote the subclass of A consisting of
Email address: [email protected] (Arzu Akgül)
doi:10.22436/jnsa.010.03.08
Received 2016-08-13
A. Akgül, J. Nonlinear Sci. Appl., 10 (2017), 954–963
955
functions univalent in U. If a function f ∈ A maps U onto a convex domain and f is univalent, then f is
called a convex function. Let
00
zf (z)
K = f ∈ A : < 1+ 0
> 0, z ∈ U
f (z)
denote the class of all convex functions defined in U and normalized by f(0) = 0, f0 (0) = 1. Let f and F
be members of H(U). The function f is said to be subordinate to ϕ, if there exists a Schwarz function w
analytic in U with
w(0) = 0 and |w(z)| < 1, (z ∈ U),
such that
f(z) = ϕ(w(z)).
We denote this subordination by
f(z) ≺ ϕ(z) or f ≺ ϕ.
Furthermore, if the function ϕ is univalent in U, then we have the following equivalence [5, 13]
f(z) ≺ ϕ(z) ⇐⇒ f(0) = ϕ(0) and f(U) ⊂ ϕ(U).
The method of differential subordinations (also known as the admissible functions method) was first
introduced by Miller and Mocanu in 1978 [11] and the theory started to develop in 1981 [12]. All the
details were captured in a book by Miller and Mocanu in 2000 [13]. Let Ψ : C3 × U −→ C and h be
univalent in U. If p is analytic in U and satisfies the second-order differential subordination
0
00
Ψ p(z), zp (z), zp (z); z ≺ h(z),
(1.2)
then p is called a solution of the differential subordination. The univalent function q is called a dominant
of the solution of the differential subordination or more simply dominant, if p ≺ q for all p satisfying
(1.2). A dominant q1 satisfying q1 ≺ q for all dominants (1.2) is said to be the best dominant of (1.2).
For functions f, g ∈ Ap , the linear operator Qm
λ,p : Ap −→ Ap (λ > 0, m ∈ N ∪ {0}) is defined by:
Q0λ,p (f ∗ g)(z) = (f ∗ g)(z),
Q1λ,p (f ∗ g)(z) = Qλ,p ((f ∗ g)(z))
= (1 − λ)(f ∗ g)(z) +
= zp +
0
λz
((f ∗ g)(z))
p
∞
X
p + λ(k − p)
ak bk zk ,
p
k=p+1
Q2λ,p (f ∗ g)(z)
= Qλ,p [Qλ , p(f ∗ g)(z)] .
Thus, we get
Qm
λ,p (f ∗ g)(z)
∞
X
p + λ(k − p) m
m−1
p
= Qλ,p Qλ,p (f ∗ g)(z) = z +
ak bk zk ,
p
(λ > 0) .
(1.3)
k=p+1
From (1.3) it can be easily seen that
0
λz
m+1
m
Qm
λ,p (f ∗ g)(z) = Qλ,p (f ∗ g)(z) − (1 − λ) Qλ,p (f ∗ g)(z), (λ > 0) .
p
The operator Qm
λ,p (f ∗ g) was introduced and studied by Selveraj and Selvakumaran [19], Aouf and
Mostafa [4], and for p = 1 was introduced by Aouf and Mostafa [3]. Recent years, Özkan [16], Özkan
A. Akgül, J. Nonlinear Sci. Appl., 10 (2017), 954–963
956
and Altntaş [17], Lupaş [9], and Lupaş [10] (also see [1, 2]) investigated some applications and results of
subordinations of analytic functions given by convolution. Also Bulut [6] used the same techniques by
using Komatu integral operator. In some of this study, the results given by Lupaş [10] and Lupaş [9] were
generalized. In order to prove our main results we need the following lemmas.
Lemma 1.1 ([8]). Let h be convex function with h(0) = a and let γ ∈ C∗ := C\{0} be a complex number with
<{γ} > 0. If p ∈ H[a, k] and
1 0
p(z) + zp (z) ≺ h(z),
(1.4)
γ
then
p(z) ≺ q(z) ≺ h(z),
where
γ
q(z) =
nzγ/n
Zz
t(γ/n)−1 h(t)dt.
0
The function q is convex and is the best dominant of the subordination (1.4).
Lemma 1.2 ([15]). Let <{µ} > 0, n ∈ N, and let
n2 + |µ|2 − n2 − µ2 w=
.
4n<{µ}
Let h be an analytic function in U with h(0) = 1 and suppose that
00
zh (z)
< 1+ 0
> −w.
h (z)
If
p(z) = 1 + pn zn + pn+1 zn+1 + · · ·
is analytic in U and
p(z) +
1 0
zp (z) ≺ h(z),
µ
(1.5)
then
p(z) ≺ q(z),
where q is a solution of the differential equation
q(z) +
n 0
zq (z) = h(z),
µ
given by
µ
q(z) =
nzσ/n
Zz
t
(µ/n )−1
q(0) = 1,
h(t)dt
(z ∈ U).
0
Moreover q is the best dominant of the subordination (1.5).
Lemma 1.3 ([14]). Let r be a convex function in U and let
0
h(z) = r(z) + nβzr (z), (z ∈ U),
where β > 0 and n ∈ N. If
p(z) = r(0) + pn zn + pn+1 zn+1 + · · · ,
(z ∈ U)
A. Akgül, J. Nonlinear Sci. Appl., 10 (2017), 954–963
is holomorphic in U and
957
0
p(z) + βzp (z) ≺ h(z),
then
p(z) ≺ r(z),
and this result is sharp.
In the present paper, making use of the subordination results of [13] and [18] we will prove our main
results.
Definition 1.4. Let <λ,m (β) be the class of functions f ∈ A satisfying
0
> β,
< (Qm
λ (f ∗ g)(z))
where z ∈ U, 0 6 β < 1.
2. Main results
Theorem 2.1. The set <λ,m (β) is convex.
Proof. Let
∞
X
fj ∗ gj (z) = z +
ak,j bk,j zk
(z ∈ U; j = 1, ....m)
k=2
be in the class of <λ,m (β). Then, by Definition 1.4, we get
∞
X
0
< Qm
= < 1+
ak,j bk,j kzk−1 > β.
λ (fj ∗ gj )(z)
k=2
For any positive numbers λ1 , λ2 , ..., λ` such that
X̀
λj = 1,
j=1
we have to show that the function
h(z) =
X̀
λ fj ∗ gj (z)
j=1
is member of <λ,m (β), that is,
0
< (Qm
h(z))
> β.
λ
(2.1)
Thus, we have
Qm
λ h(z) = z +
∞
X

(1 + λ (k − 1))m 
k=2
X̀

λj ak,j bk,j  zk .
j=1
If we differentiate (2.2) with respect to z, then we obtain
0
(Qm
λ h(z)) = 1 +
∞
X


X̀
(1 + λ (k − 1))m 
λj ak,j bk,j  kzk−1
k=2
j=1
(2.2)
A. Akgül, J. Nonlinear Sci. Appl., 10 (2017), 954–963
= 1+
X̀
λj
∞
X
958
(1 + λ (k − 1))m ak,j bk,j kzk−1 .
k=2
j=1
Thus, we have
<
0
(Qm
λ h(z))
= 1+
X̀
λj <
X̀
(1 + λ (k − 1))
m
ak,j bk,j kz
k−1
k=2
j=1
> 1+
∞
X
λj (β − 1)
j=1
= β.
Thus, the inequality (2.1) holds and we obtain desired result.
Theorem 2.2. Let q be convex function in U with q(0) = 1 and let
h(z) = q(z) +
0
1
zq (z)
γ+1
(z ∈ U),
where γ is a complex number with <{γ} > −1. If f ∈ <σ,θ (β) and F = Υγ (f ∗ g), where
Zz
γ + 1 γ−1
F(z) = Υγ (f ∗ g)(z) = γ
t
(f ∗ g)(t)dt,
z
(2.3)
0
then,
0
(Qm
λ (f ∗ g)(z)) ≺ h(z)
(2.4)
implies
0
(Qm
λ F(z)) ≺ q(z),
and this result is sharp.
Proof. From the equality (2.3) we can write
Zz
γ
z F(z) = (γ + 1) tγ−1 (f ∗ g)(t)dt,
(2.5)
0
by differentiating the equality (2.5) with respect to z, we obtain
0
(γ) F(z) + zF (z) = (γ + 1) (f ∗ g)(z).
If we apply the operator Qm
λ to the last equation, then we get
0
m
m
(γ) Qm
λ F(z) + z (Qλ F(z)) = (γ + 1) Qλ (f ∗ g)(z).
(2.6)
If we differentiate (2.6) with respect to z, we can obtain
0
(Qm
λ F(z)) +
00
0
1
m
z (Qm
λ F(z)) = (Qλ f(z)) .
γ+1
(2.7)
By using the differential subordination given by (2.4) in the equality (2.7), we have
0
(Qm
λ F(z)) +
00
1
z (Qm
λ F(z)) ≺ h(z).
γ+1
(2.8)
A. Akgül, J. Nonlinear Sci. Appl., 10 (2017), 954–963
959
Now, we define
0
p(z) = (Qm
λ F(z)) .
(2.9)
Then by a simple computation we get
"
p(z) = z +
∞
X
(1 + λ (k − 1))m
k=2
γ+1
ak bk zk
γ+k
#0
= 1 + p1 z + p2 z + · · · ,
(p ∈ H[1, 1]) .
Using the equation (2.9) in the subordination (2.8), we obtain
p(z) +
0
0
1
1
zp (z) ≺ h(z) = q(z) +
zq (z).
γ+1
γ+1
If we use Lemma 1.2, then we get
p(z) ≺ q(z).
So we obtain the desired result and q is the best dominant.
Example 2.3. If we choose in Theorem 2.2
γ = i + 1,
q(z) =
1
,
1−z
thus we get
h(z) =
(i + 2) − (i + 1)z
.
(i + 2)(1 − z)2
If (f ∗ g) ∈ <λ,m (β) and F is given by
Zz
i+2 i
F(z) = Υi (f ∗ g)(z) = i+1 t (f ∗ g)(t)dt,
z
0
then by Theorem 2.2, we obtain
0
(Qm
λ f(z)) ≺ h(z) =
0
(i + 2) − (i + 1)z
1
=⇒ (Qm
.
λ F(z)) ≺ q(z)) =
2
(i + 2)(1 − z)
1−z
Theorem 2.4. Let <{γ} > −1 and let
1 + |γ + 1|2 − γ2 + 2γ
w=
.
4<{γ + 1}
Let h be an analytic function in U with h(0) = 1 and assume that
00
zh (z)
< 1+ 0
> −w.
h (z)
If f ∗ g ∈ <λ,m (β) and F = Υδγ (f ∗ g), where F is defined by equation (2.3), then
0
(Qm
λ (f ∗ g)(z)) ≺ h(z)
implies
0
(Qm
λ F(z)) ≺ q(z),
(2.10)
A. Akgül, J. Nonlinear Sci. Appl., 10 (2017), 954–963
960
where q is the solution of the differential equation
h(z) = q(z) +
given by
0
1
zq (z),
γ+1
q(0) = 1,
Zz
γ+1 γ
q(z) = γ+1 t (f ∗ g)(t)dt.
z
0
Moreover q is the best dominant of the subordination (2.10).
Proof. If we choose n = 1 and µ = γ + 1 in Lemma 1.2, then the proof is obtained by means of the proof
of Theorem 2.4.
Letting
1 + (2β − 1)z
,
1+z
in Theorem 2.4, we obtain the following result.
h(z) =
06β<1
Corollary 2.5. If 0 6 β < 1, 0 6 ξ < 1, λ > 0, <{γ} > −1, and F = Υγ (f ∗ g) is defined by the equation (2.3),
then
Υγ (<λ,m (β)) ⊂ <λ,m (ξ),
where
ξ = min <{q(z)} = ξ(γ, β)
|z|=1
and this result is sharp. Also,
ξ = ξ(γ, β) = (2β − 1) + 2(γ + 1)(1 − β)τ(γ),
where
Z1
tγ
dt.
1+t
τ(γ) =
0
(2.11)
(2.12)
Proof. Let f ∈ <λ,m (β). Then from Definition 1.4 it is known that
0
< (Qm
(f
∗
g)(z))
> β,
λ
which is equivalent to
0
(Qm
λ (f ∗ g)(z)) ≺ h(z).
By using Theorem 2.2, we have
0
(Qm
λ F(z)) ≺ q(z).
If we take
1 + (2β − 1)z
, 0 6 β < 1,
1+z
then h is convex and by Theorem 2.4, we obtain
Z
Z
0
(1 − β)(γ + 1) z tγ
γ + 1 z γ 1 + (2β − 1)t
m
(Qλ F(z)) ≺ q(z) = γ+1
t
dt = (2β − 1) + 2
dt.
z
1+t
zγ+1
0
0 1+t
h(z) =
On the other hand if <{γ} > −1, then from the convexity of q and the fact that q(U) is symmetric with
respect to the real axis, we get
0
< (Qm
F(z))
> min <{q(z)} = <{q(1)} = ξ(γ, β) = 2β − 1 + 2(1 − β)(γ + 1)τ(γ),
(2.13)
λ
|z|=1
A. Akgül, J. Nonlinear Sci. Appl., 10 (2017), 954–963
961
where τ(γ) is given by equation (2.12). From inequality (2.13), we get
Υγ (<λ,m (β)) ⊂ <λ,m (ξ),
where ξ is given by (2.11).
Theorem 2.6. Let q be a convex function with q(0) = 1 and h a function such that
0
(z ∈ U).
h(z) = q(z) + zq (z),
If f ∈ A, then the following subordination
0
0
(Qm
λ (f ∗ g)(z)) ≺ h(z) = q(z) + zq (z)
(2.14)
implies that
Qm
λ (f ∗ g)(z)
≺ q(z),
z
and the result is sharp.
Proof. Let
Qm
λ (f ∗ g)(z)
p(z) =
.
z
(2.15)
Differentiating (2.15), we have
0
0
(Qm
λ (f ∗ g)(z)) = p(z) + zp (z),
and thus (2.14) becomes
(z ∈ U)
0
0
p(z) + zp (z) ≺ h(z) = q(z) + zq (z).
Hence by applying Lemma1.3, we conclude that
p(z) ≺ q(z),
that is,
Qm
λ (f ∗ g)(z)
≺ q(z),
z
and this result is sharp.
Theorem 2.7. Let q be a convex function with q(0) = 1 and h the function
0
h(z) = q(z) + zq (z) (z ∈ U).
If m ∈ N, f ∈ A and verifies the differential subordination
Qm+1
(f ∗ g)(z)
λ
m
Qλ (f ∗ g)(z)
then
!0
≺ h(z),
(2.16)
Qm+1
(f ∗ g)(z)
λ
≺ q(z),
m
Qλ (f ∗ g)(z)
and this result is sharp.
Proof. For the function f ∈ A, given by the equation (1.1), we have
Qm
λ (f ∗ g)(z) = z +
∞
X
k=2
(1 + λ (k − 1))m
γ+1
ak b k z k ,
k+γ
(z ∈ U).
A. Akgül, J. Nonlinear Sci. Appl., 10 (2017), 954–963
962
Let us consider
Qm+1 (f ∗ g)(z)
p(z) = λm
=
Qλ (f ∗ g)(z)
z+
∞
P
(1 + λ (k − 1))m+1
k=2
∞
P
z+
(1 + λ (k − 1))m
k=2
1+
=
∞
P
γ+1
k
k+γ ak bk z
(1 + λ (k − 1))m+1
k=2
∞
P
1+
(1 + λ (k − 1))m
k=2
γ+1
k
k+γ ak bk z
γ+1
k−1
k+γ ak bk z
.
γ+1
k−1
k+γ ak bk z
We get
0
(Qm+1
(f ∗ g)(z))
(Qm
λ
λ (f ∗ g)(z))
(p(z)) =
−
p(z)
m
m
Qλ (f ∗ g)(z)
Qλ (f ∗ g)(z))
0
0
and
0
p(z) + zp (z) =
zQm+1
(f ∗ g)(z)
λ
m
Qλ (f ∗ g)(z)
!0
(z ∈ U).
Thus, the relation (2.16) becomes
0
0
p(z) + zp (z) ≺ h(z) = q(z) + zq (z),
(z ∈ U),
and by using Lemma 1.3, we obtain
p(z) ≺ q(z),
that is,
Qm
λ (f ∗ g)(z)
≺ q(z).
z
Acknowledgment
The author would like to thank the anonymous referees for their valuable suggestions which improved
the presentation of the paper.
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