Integral operators preserving subordination and superordination for multivalent functions
UDC 517.9 We obtain subordination, superordination and sandwich-preserving new theorems for certain integral operators defined on multivalent functions. The sandwich-type theorem for these integral operators is also derived and our results extend some earlier ones. Combining these new theorems with...
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2021
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507020607619072 |
|---|---|
| author | Aouf , M. K. Bulboacă, T. Seoudy, T. Aouf , M. K. Bulboacă, T. Seoudy, T. Aouf , M. K. Bulboacă, T. Seoudy, T. |
| author_facet | Aouf , M. K. Bulboacă, T. Seoudy, T. Aouf , M. K. Bulboacă, T. Seoudy, T. Aouf , M. K. Bulboacă, T. Seoudy, T. |
| author_sort | Aouf , M. K. |
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| collection | OJS |
| datestamp_date | 2022-03-26T11:03:03Z |
| description | UDC 517.9
We obtain subordination, superordination and sandwich-preserving new theorems for certain integral operators defined on multivalent functions. The sandwich-type theorem for these integral operators is also derived and our results extend some earlier ones. Combining these new theorems with some previous related results, we give interesting subordination and superordination consequences for a wide class of analytic integral operators. |
| doi_str_mv | 10.37863/umzh.v73i6.437 |
| first_indexed | 2026-03-24T02:02:40Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v73i6.437
UDC 517.5
M. K. Aouf (Dep. Math., Mansoura Univ., Egypt),
T. Bulboacă (Babeş-Bolyai Univ., Cluj-Napoca, Romania),
T. M. Seoudy (Dep. Math., Fayoum Univ., Egypt and Jamoum Univ. College, Umm Al-Qura Univ.,
Makkah, Saudi Arabia)
INTEGRAL OPERATORS PRESERVING SUBORDINATION
AND SUPERORDINATION FOR MULTIVALENT FUNCTIONS
IНТЕГРАЛЬНI ОПЕРАТОРИ, ЯКI ЗБЕРIГАЮТЬ СУБОРДИНАЦIЮ
ТА СУПЕРОРДИНАЦIЮ ДЛЯ БАГАТОЗНАЧНИХ ФУНКЦIЙ
We obtain subordination, superordination and sandwich-preserving new theorems for certain integral operators defined on
multivalent functions. The sandwich-type theorem for these integral operators is also derived and our results extend some
earlier ones. Combining these new theorems with some previous related results, we give interesting subordination and
superordination consequences for a wide class of analytic integral operators.
Отримано новi теореми щодо субординацiї, суперординацiї та збереження порядку для деяких iнтегральних опе-
раторiв на багатозначних функцiях. Також доведено теореми типу стискання для iнтегральних операторiв, якi уза-
гальнюють деякi вiдомi результати. Комбiнуючи цi новi теореми з деякими вiдомими вiдповiдними результатами,
ми отримуємо цiкавi наслiдки щодо субординацiї та суперординацiї для широкого класу аналiтичних iнтегральних
операторiв.
1. Introduction. Let H(\mathrm{U}) be the class of functions analytic in the open unit disk \mathrm{U} = \{ z \in \BbbC :
| z| < 1\} , and let denote by H[a, n] the subclass of H(\mathrm{U}) consisting of functions of the form
f(z) = a+ anz
n + an+1z
n+1 + . . . , a \in \BbbC , n \in \BbbN = \{ 1, 2, . . .\} .
Also, let \scrA (p) denote the subclass of H(\mathrm{U}) of the form
f(z) = zp +
\infty \sum
k=p+1
akz
k, z \in \mathrm{U}, p \in \BbbN ,
and denote \scrA := \scrA (1).
If f and F are members of H(\mathrm{U}), then the function f is said to be subordinate to F, or F is
said to be superordinate to f, if there exists a function w \in H(\mathrm{U}) with w(0) = 0 and | w(z)| < 1
for z \in \mathrm{U}, such that f(z) = F (w(z)) for all z \in \mathrm{U}, and in such a case we write f(z) \prec F (z).
If F is univalent, then f(z) \prec F (z) if and only if f(0) = F (0) and f(\mathrm{U}) \subset F (\mathrm{U}) (see [12, 13]).
Let \Psi : \BbbC 2\times \mathrm{U} \rightarrow \BbbC and h be univalent in \mathrm{U}. If p \in H(\mathrm{U}) and satisfies the first order differential
subordination
\Psi
\bigl(
p(z), zp\prime (z); z
\bigr)
\prec h(z), (1.1)
then p is a solution of the differential subordination (1.1). The univalent function q is called a
dominant of the solutions of the differential subordination (1.1) if p(z) \prec q(z) for all p satisfying
(1.1). A univalent dominant \~q that satisfies \~q(z) \prec q(z) for all the dominants of (1.1) is called the
best dominant.
c\bigcirc M. K. AOUF, T. BULBOACĂ, T. M. SEOUDY, 2021
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 6 749
750 M. K. AOUF, T. BULBOACĂ, T. M. SEOUDY
Similarly, if \Phi : \BbbC 2 \times \mathrm{U} \rightarrow \BbbC such that p and \Phi
\bigl(
p(z), zp\prime (z); z
\bigr)
are univalent in \mathrm{U} and if p
satisfies first order differential superordination
h(z) \prec \Phi
\bigl(
p(z), zp\prime (z); z
\bigr)
, (1.2)
then p is called a solution of the differential superordination (1.2). An analytic function q is called a
subordinant of the solutions of the differential superordination (1.2) if q(z) \prec p(z) for all p satisfying
(1.2). A univalent subordinant \~q that satisfies q(z) \prec \~q(z) for all the subordinants of (1.2) is called
the best subordinant (see [12, 13]).
For the parameters \alpha , \beta , \gamma \in \BbbC with \beta \not = 0 and p \in \BbbN , we introduce the integral operators \mathrm{I}p\alpha ,\beta ,\gamma :
\scrK p
\alpha ,\gamma \rightarrow \scrA (p) with \scrK p
\alpha ,\gamma \subset \scrA (p) defined by
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z) =
\left[ \alpha p+ \gamma
z(\alpha - \beta )p+\gamma
z\int
0
f\alpha (t)t\gamma - 1dt
\right] 1
\beta
, (1.3)
where all powers are principal ones.
Remark 1.1. For p = 1 and \alpha = \beta we obtain
\mathrm{I}\beta ,\gamma [f ](z) =
\left[ \beta + \gamma
z\gamma
z\int
0
f\beta (t)t\gamma - 1dt
\right] 1
\beta
, (1.4)
where \mathrm{I}\beta ,\gamma is the integral operator introduced by Miller and Mocanu [12], and studied in [1 – 3] and
more other articles (see [4 – 6]).
In the present paper we obtain sufficient conditions on the functions g1, g2 and on the parameters
\alpha , \beta , \gamma such that the following sandwich-type result holds:
z
\biggl(
g1(z)
zp
\biggr) \alpha
\prec z
\biggl(
f(z)
zp
\biggr) \alpha
\prec z
\biggl(
g2(z)
zp
\biggr) \alpha
implies
z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g1](z)
zp
\Biggr) \beta
\prec z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
zp
\Biggr) \beta
\prec z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g2](z)
zp
\Biggr) \beta
.
Moreover, our result is sharp, i.e., the functions z
\biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g1](z)
zp
\biggr) \beta
and z
\biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g2](z)
zp
\biggr) \beta
are,
respectively, the best subordinant and the best dominant.
Combining these new theorems with some previous related results, we give subordination and
superordination consequences for a wide class of analytic integral operators.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 6
INTEGRAL OPERATORS PRESERVING SUBORDINATION AND SUPERORDINATION . . . 751
2. Preliminaries. The following definitions and lemmas will be required in our present investi-
gation.
Definition 2.1 [12]. Denote by \scrQ the set of all functions q that are analytic and injective on
\mathrm{U} \setminus E(q), where
E(q) =
\Bigl\{
\zeta \in \partial \mathrm{U} : \mathrm{l}\mathrm{i}\mathrm{m}
z\rightarrow \zeta
q(z) = \infty
\Bigr\}
,
and are such that q\prime (\zeta ) \not = 0 for \zeta \in \partial \mathrm{U} \setminus E(q). Further, let denote by \scrQ (a) the subclass of the
functions q \in \scrQ for which q(0) = a.
Definition 2.2. A function L(z; t) : \mathrm{U}\times [0,+\infty ) \rightarrow \BbbC is called a subordination (or a Loewner)
chain if L(\cdot ; t) is analytic and univalent in \mathrm{U} for all t \geq 0 and L(z; s) \prec L(z; t) when 0 \leq s \leq t.
The next known lemma gives a sufficient condition so that the L(z; t) function will be a subor-
dination chain.
Lemma 2.1 [14, p. 159]. Let L(z; t) = a1(t)z + a2(t)z
2 + . . . with a1(t) \not = 0 for all t \geq 0 and
\mathrm{l}\mathrm{i}\mathrm{m}t\rightarrow +\infty | a1(t)| = +\infty . Suppose that L(\cdot ; t) is analytic in \mathrm{U} for all t \geq 0, L(z; \cdot ) is continuously
differentiable on [0,+\infty ) for all z \in \mathrm{U}. If L(z; t) satisfies
\mathrm{R}\mathrm{e}
\biggl[
z
\partial L(z; t)/\partial z
\partial L(z; t)/\partial t
\biggr]
> 0, z \in \mathrm{U}, t \geq 0,
and
| L(z; t)| \leq K0 | a1(t)| , | z| < r0 < 1, t \geq 0,
for some positive constants K0 and r0, then L(z; t) is a subordination chain.
Lemma 2.2 [8]. Suppose that the function H : \BbbC 2 \rightarrow \BbbC satisfies the condition
\mathrm{R}\mathrm{e}H(is, t) \leq 0, s \in \BbbR , t \leq -
n
\bigl(
1 + s2
\bigr)
2
,
where n \in \BbbN . If the function p(z) = 1 + pnz
n + pn+1z
n+1 + . . . is analytic in \mathrm{U} and
\mathrm{R}\mathrm{e}H
\bigl(
p(z), zp\prime (z)
\bigr)
> 0, z \in \mathrm{U},
then \mathrm{R}\mathrm{e} p(z) > 0 for z \in \mathrm{U}.
The next result deals with the solutions of the Briot – Bouquet differential equation (2.1), and
more general forms of the following lemma may be found in [9] (Theorem 1).
Lemma 2.3 [9]. Let \lambda , \mu \in \BbbC with \lambda \not = 0 and k \in H(\mathrm{U}) with k(0) = c. If \mathrm{R}\mathrm{e}
\bigl[
\lambda k(z)+\mu
\bigr]
> 0,
z \in \mathrm{U}, then the solution of the differential equation
q(z) +
zq\prime (z)
\lambda q(z) + \mu
= k(z) (2.1)
with q(0) = c is analytic in \mathrm{U} and satisfies \mathrm{R}\mathrm{e}
\bigl[
\lambda q(z) + \mu
\bigr]
> 0, z \in \mathrm{U}.
Lemma 2.4 [12]. Let p \in \scrQ (a) and q(z) = a+ anz
n + an\gamma +1z
n+1 + . . . be analytic in \mathrm{U} with
q(z) \not \equiv a and n \geq 1. If q is not subordinate to p, then there exist two points z0 = r0e
i\theta \in \mathrm{U} and
\zeta 0 \in \partial \mathrm{U} \setminus E(q), and a number m \geq n such that
q(\mathrm{U}r0) \subset p(\mathrm{U}), q(z0) = p(\zeta 0), and z0p
\prime (z0) = m\zeta 0p(\zeta 0),
where \mathrm{U}r0 =
\bigl\{
z \in \BbbC : | z| < r0
\bigr\}
.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 6
752 M. K. AOUF, T. BULBOACĂ, T. M. SEOUDY
Lemma 2.5 [14]. Let \Phi : \BbbC 2 \rightarrow \BbbC , q \in H[a, 1] and set \Phi
\bigl(
q(z), zq\prime (z)
\bigr)
= h(z). If L(z; t) =
= \Phi
\bigl(
q(z), tzq\prime (z)
\bigr)
is a subordination chain and q \in H[a, 1] \cap \scrQ (a), then
h(z) \prec \Phi
\bigl(
p(z), zp\prime (z)
\bigr)
implies that q(z) \prec p(z). Furthermore, if \Phi
\bigl(
q(z), zq\prime (z)
\bigr)
= h(z) has a univalent solution q \in \scrQ (a),
then q is the best subordinant.
Let c \in \BbbC with \mathrm{R}\mathrm{e} c > 0, n \in \BbbN and
Cn = Cn(c) =
n
\mathrm{R}\mathrm{e} c
\biggl[
| c|
\sqrt{}
1 + 2\mathrm{R}\mathrm{e}
\Bigl( c
n
\Bigr)
+ \mathrm{I}\mathrm{m} c
\biggr]
.
If R is the univalent function R(z) =
2Cnz
1 - z2
, then the open door function Rc,n is defined by
Rc,n(z) = R
\biggl(
z + b
1 + bz
\biggr)
, z \in \mathrm{U},
where b = R - 1(c).
Remark that Rc,n is univalent in \mathrm{U}, Rc,n(0) = c and Rc,n(\mathrm{U}) = R(\mathrm{U}) is the complex plane slit
along the half-lines \mathrm{R}\mathrm{e}w = 0, \mathrm{I}\mathrm{m}w \geq Cn and \mathrm{R}\mathrm{e}w = 0, \mathrm{I}\mathrm{m}w \leq - Cn. Moreover, if c > 0, then
Cn+1 > Cn and \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty Cn = \infty , hence Rc,n(z) \prec Rc,n+1(z) and \mathrm{l}\mathrm{i}\mathrm{m}n\rightarrow \infty Rc,n(\mathrm{U}) = \BbbC . In this
paper we will use the notation Rc := Rc,1.
3. Main results. Unless otherwise mentioned, we assume throughout this section that \alpha , \beta , \gamma \in
\in \BbbC with \beta \not = 0, \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) \geq 1 and p \in \BbbN .
First we will determine the subset \scrK p
\alpha ,\gamma \subset \scrA (p) such that the integral operator \mathrm{I}p\alpha ,\beta ,\gamma given by
(1.3) will be well-defined.
If we denote by \scrA n
p the class of functions
\scrA n
p =
\bigl\{
f \in H(\mathrm{U}) : f(z) = zp + ap+nz
p+n + . . .
\bigr\}
, n \in \BbbN ,
then \scrA (p) = \scrA 1
p.
Lemma 3.1. Let \phi ,\Phi \in H[1, n] with \phi (z) \cdot \Phi (z) \not = 0 for z \in \mathrm{U}. Let \alpha , \beta , \gamma , \delta \in \BbbC with \beta \not = 0,
\alpha p+ \delta = \beta p+ \gamma and \mathrm{R}\mathrm{e}(\alpha p+ \delta ) > 0. If the function f belongs to \scrA n
p and satisfies
\alpha
zf \prime (z)
f(z)
+
z\phi \prime (z)
\phi (z)
+ \delta \prec R\alpha p+\delta ,n(z), (3.1)
then
F (z) =
\left[ \beta p+ \gamma
z\gamma \Phi (z)
z\int
0
f\alpha (t)\phi (t)t\delta - 1dt
\right] 1
\beta
= zp +Ap+nz
p+n + . . . \in \scrA n
p ,
F (z)
zp
\not = 0, z \in \mathrm{U}, and
\mathrm{R}\mathrm{e}
\biggl[
\beta
zF \prime (z)
F (z)
+
z\Phi \prime (z)
\Phi (z)
+ \gamma
\biggr]
> 0, z \in \mathrm{U}.
(All powers are principal ones.)
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 6
INTEGRAL OPERATORS PRESERVING SUBORDINATION AND SUPERORDINATION . . . 753
Proof. The idea of the proof is similar with those of Theorem 2.5c of [12]. Thus, the subor-
dination (3.1) implies that
f(z)
zp
\not = 0 for all z \in \mathrm{U}. Since \mathrm{R}\mathrm{e}(\alpha p + \delta ) > 0, like in the proof of
Lemma 1.2c of [12] we can easily show that the function
p(z) =
1
z\alpha p+\delta \varphi (z)
\biggl[
f(z)
zp
\biggr] - \alpha
z\int
0
\biggl[
f(t)
tp
\biggr] \alpha
t\alpha p+\delta - 1\varphi (t)dt =
=
1
z\delta f\alpha (z)\varphi (z)
z\int
0
f\alpha (t)t\delta - 1\varphi (t)dt =
1
\alpha p+ \delta
+ pnz
n + . . .
is analytic in \mathrm{U} and p \in H[1/(\alpha p + \delta ), n]. Differentiating the above definition formula of p, it is
easy to show that the function p satisfies the differential equation
zp\prime (z) + P (z)p(z) = 1
with
P (z) = \alpha
zf \prime (z)
f(z)
+
z\phi \prime (z)
\phi (z)
+ \delta .
Starting from this point the proof is similar with those of Theorem 2.5c of [12], hence it will omitted.
Lemma 3.1 is proved.
Note that for the special case p = 1 the above lemma represents the Integral Existence Theorem
[12] (Theorem 2.5c) (see also [10, 11]).
Lemma 3.2. Let \alpha , \beta , \gamma , \delta \in \BbbC with \beta \not = 0 and \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) > 0. If f belongs to \scrK p
\alpha ,\gamma , where
\scrK p
\alpha ,\gamma =
\biggl\{
f \in \scrA (p) : \alpha
zf \prime (z)
f(z)
+ \gamma \prec R\alpha p+\gamma (z)
\biggr\}
,
then \mathrm{I}p\alpha ,\beta ,\gamma [f ] \in \scrA (p),
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
zp
\not = 0 for all z \in \mathrm{U}, and
\mathrm{R}\mathrm{e}
\left[ \beta z
\Bigl(
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
\Bigr) \prime
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
+ \gamma
\right] > 0, z \in \mathrm{U},
where \mathrm{I}p\alpha ,\beta ,\gamma is the integral operator defined by (1.3).
Proof. Taking in the Lemma 3.1 the values n := 1, \alpha := \alpha , \beta := \beta , \gamma := (\alpha - \beta )p+\gamma , \delta := \gamma
and \varphi (z) = \Phi (z) \equiv 1, it follows that the assumption of this lemma holds. Also, the subordination
condition (3.1) becomes
\alpha
zf \prime (z)
f(z)
+ \gamma \prec R\alpha p+\gamma ,1(z),
while F (z) = \mathrm{I}p\alpha ,\beta ,\gamma [f ](z).
Lemma 3.2 is proved.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 6
754 M. K. AOUF, T. BULBOACĂ, T. M. SEOUDY
Theorem 3.1. Let \alpha , \beta , \gamma \in \BbbC with \beta \not = 0 such that \mathrm{R}\mathrm{e}(\alpha p+\gamma ) \geq 1. For f, g \in \scrK p
\alpha ,\gamma , suppose
that the function \phi , defined by
\phi (z) = z
\biggl(
g(z)
zp
\biggr) \alpha
, (3.2)
satisfies the inequality
\mathrm{R}\mathrm{e}
\biggl[
1 +
z\phi \prime \prime (z)
\phi \prime (z)
\biggr]
> - \delta 0, z \in \mathrm{U}, (3.3)
where \delta 0 is given by
\delta 0 =
\left\{
1 + | \alpha p+ \gamma - 1| 2 -
\bigm| \bigm| 1 - (\alpha p+ \gamma - 1)2
\bigm| \bigm|
4\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1)
, if \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) > 1,
0, if \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) = 1.
(3.4)
Then the subordination condition
z
\biggl(
f(z)
zp
\biggr) \alpha
\prec z
\biggl(
g(z)
zp
\biggr) \alpha
(3.5)
implies that
z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
zp
\Biggr) \beta
\prec z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g](z)
zp
\Biggr) \beta
,
and the function z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g](z)
zp
\Biggr) \beta
is the best dominant. (All powers are principal ones.)
Proof. Let define the functions F and G by
F (z) := z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
zp
\Biggr) \beta
, G(z) := z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g](z)
zp
\Biggr) \beta
, z \in \mathrm{U}, (3.6)
respectively, where all powers are principal ones. From Lemma 3.2 it follows that F, G belong to
\scrA . First we will show that if
q(z) = 1 +
zG\prime \prime (z)
G\prime (z)
, z \in \mathrm{U}, (3.7)
then
\mathrm{R}\mathrm{e} q(z) > 0, z \in \mathrm{U}. (3.8)
From (1.3) and the definitions of the functions G and \phi , we obtain
\phi (z) =
\biggl(
1 - 1
\alpha p+ \gamma
\biggr)
G(z) +
1
\alpha p+ \gamma
zG\prime (z). (3.9)
Differentiating both side of (3.9) with respect to z, we have
\phi \prime (z) = G\prime (z) +
zG\prime \prime (z)
\alpha p+ \gamma
, (3.10)
and combining (3.7) and (3.10) we easily get
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 6
INTEGRAL OPERATORS PRESERVING SUBORDINATION AND SUPERORDINATION . . . 755
k(z) := 1 +
z\phi \prime \prime (z)
\phi \prime (z)
= q(z) +
zq\prime (z)
q(z) + \alpha p+ \gamma - 1
. (3.11)
According to (3.11), from (3.3) it follows that
\mathrm{R}\mathrm{e}
\bigl[
k(z) + \alpha p+ \gamma - 1
\bigr]
> - \delta 0 +\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) \geq 0, z \in \mathrm{U},
whenever
\delta 0 \leq \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1). (3.12)
Supposing that the inequality (3.12) holds, according to Lemma 2.3 we conclude that the diffe-
rential equation (3.11) has a solution q \in H(\mathrm{U}) with k(0) = q(0) = 1.
If we let
H(u, v) = u+
v
u+ \alpha p+ \gamma - 1
+ \delta ,
from (3.11) and (3.4) we obtain
\mathrm{R}\mathrm{e}H
\bigl(
q(z), zq\prime (z)
\bigr)
> 0, z \in \mathrm{U}.
To verify the condition
\mathrm{R}\mathrm{e}H(is, t) \leq 0 for s \in \BbbR , t \leq - 1 + s2
2
, (3.13)
first we see that
\mathrm{R}\mathrm{e}H(is, t) =
t\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1)
| \alpha p+ \gamma - 1 + is| 2
+ \delta = \delta \leq \delta 0 = 0
for \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) = 0. If \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) > 0, then
\mathrm{R}\mathrm{e}H(is, t) = \mathrm{R}\mathrm{e}
\biggl[
is+
t
is+ \alpha p+ \gamma - 1
+ \delta 0
\biggr]
=
=
t\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1)
| \alpha p+ \gamma - 1 + is| 2
+ \delta \leq -
Kp;\alpha ,\gamma ;\delta (s)
2 | \alpha p+ \gamma - 1 + is| 2
for t \leq - 1 + s2
2
,
where
Kp;\alpha ,\gamma ;\delta (s) :=
\bigl[
\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) - 2\delta
\bigr]
s2 - 4\delta \mathrm{I}\mathrm{m}(\alpha p+ \gamma - 1)s -
- 2\delta | \alpha p+ \gamma - 1| 2 +\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1).
We need to determine the value
\delta 0 = \mathrm{s}\mathrm{u}\mathrm{p}
\bigl\{
\delta : Kp;\alpha ,\gamma ;\delta (s) \geq 0, s \in \BbbR , \delta \leq \delta 0
\bigr\}
.
(i) If \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) - 2\delta = 0, then
Kp;\alpha ,\gamma ;\delta (s) = \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1)
\bigl[
- 2 \mathrm{I}\mathrm{m}(\alpha p+ \gamma - 1) s+ 1 - | \alpha p+ \gamma - 1| 2
\bigr]
\geq 0
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756 M. K. AOUF, T. BULBOACĂ, T. M. SEOUDY
for all s \in \BbbR if and only if (\alpha p+ \gamma - 1) \in (0, 1], and in this case \delta 0 = (\alpha p+ \gamma - 1)/2. Thus, it is
easy to see that the definition relation (3.4) could be used for this special case.
(ii) If \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) - 2\delta \not = 0, then Kp;\alpha ,\gamma ;\delta (s) \geq 0 for all s \in \BbbR if and only if
\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) - 2\delta \geq 0 (3.14)
and
4\delta 2 \mathrm{I}\mathrm{m}2(\alpha p+ \gamma - 1) -
-
\bigl[
\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) - 2\delta
\bigr] \bigl[
\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) - 2\delta | \alpha p+ \gamma - 1| 2
\bigr]
\leq 0. (3.15)
Using the fact that the inequality (3.15) is equivalent to
\chi (\delta ) := - 4\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) \delta 2+
+2
\bigl(
1 + | \alpha p+ \gamma - 1| 2
\bigr)
\delta - \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) \leq 0,
a simple computation shows that the function \chi has the positive zeros 0 < \delta 0 \leq \delta 1, where \delta 0 is
given by (3.4). Since \chi (\delta ) \leq 0 for all \delta \leq \delta 0 and
\chi
\biggl(
\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1)
2
\biggr)
= \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) \mathrm{I}\mathrm{m}2(\alpha p+ \gamma - 1) \geq 0,
it follows that \delta 0 \leq
\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1)
2
, i.e., the condition (3.14) holds for \delta = \delta 0.
Moreover, because
\delta 0 \leq
\mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1)
2
< \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) if \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) > 0,
we conclude that the inequality (3.12) holds whenever \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) > 0. Obviously, it holds also
for \mathrm{R}\mathrm{e}(\alpha p+ \gamma - 1) = 0, since in this case \delta 0 = 0.
In conclusion, for the assumed value of \delta 0 given by (3.4), we proved that Kp;\alpha ,\gamma ;\delta (s) \geq 0 for all
s \in \BbbR which implies that (3.13) holds.
By using Lemma 2.2, we conclude that the inequality (3.8) holds, and from the definition rela-
tion (3.7) it follows that G is convex. Hence, G is a univalent function in \mathrm{U}.
Next, we will prove that the subordination condition (3.5) implies that F (z) \prec G(z), where the
functions F and G are defined by (3.6). For this purpose, let define the function L(z; t) by
L(z; t) =
\biggl(
1 - 1
\alpha p+ \gamma
\biggr)
G(z) +
1 + t
\alpha p+ \gamma
zG\prime (z), z \in \mathrm{U}, t \geq 0. (3.16)
If we denote L(z; t) = a1(t)z + . . . , then
a1(t) =
\partial L(0; t)
\partial z
=
\biggl(
1 +
t
\alpha p+ \gamma
\biggr)
G\prime (0) = 1 +
t
\alpha p+ \gamma
,
hence \mathrm{l}\mathrm{i}\mathrm{m}t\rightarrow +\infty | a1(t)| = +\infty . By using the fact that \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) > 0, we obtain that a1(t) \not = 0 for
all t \geq 0.
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INTEGRAL OPERATORS PRESERVING SUBORDINATION AND SUPERORDINATION . . . 757
Since \mathrm{R}\mathrm{e} q(z) > 0, z \in \mathrm{U}, and \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) \geq 1, we deduce that
\mathrm{R}\mathrm{e}
\biggl[
z
\partial L(z; t)/\partial z
\partial L(z; t)/\partial t
\biggr]
= \mathrm{R}\mathrm{e}
\bigl[
\alpha p+ \gamma - 1 + (1 + t)q(z)
\bigr]
> 0, z \in \mathrm{U}, t \geq 0.
From the definition (3.16), since \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) \geq 1, for all t \geq 0, we have that
| L(z; t)|
| a1(t)|
=
\bigm| \bigm| (\alpha p+ \gamma - 1)G(z) + (1 + t)zG\prime (z)
\bigm| \bigm|
| \alpha p+ \gamma + t|
\leq
\leq | \alpha p+ \gamma - 1| | G(z)| + (1 + t)| zG\prime (z)|
| \alpha p+ \gamma + t|
. (3.17)
Since G is convex, the following known growth and distortion sharp inequalities are true (see [7]):
r
1 + r
\leq | G(z)| \leq r
1 - r
, if | z| \leq r,
1
(1 + r)2
\leq | G\prime (z)| \leq 1
(1 - r)2
, if | z| \leq r.
(3.18)
By using the right-hand sides of these inequalities in (3.17), we obtain
| L(z; t)|
| a1(t)|
\leq r
(1 - r)2
t+ 1 + | \alpha p+ \gamma - 1| (1 - r)
| \alpha p+ \gamma + t|
, | z| \leq r, t \geq 0. (3.19)
The assumption \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) \geq 1 implies
| t+ \alpha p+ \gamma | \geq | \alpha p+ \gamma | , | t+ \alpha p+ \gamma | \geq | t+ 1| , t \geq 0,
and from (3.19) we conclude that
| L(z; t)|
| a1(t)|
\leq r
(1 - r)2
\biggl[
1 +
| \alpha p+ \gamma - 1| (1 - r)
| \alpha p+ \gamma |
\biggr]
, | z| \leq r, t \geq 0.
Thus, the second assumption of Lemma 2.1 holds, and according to this lemma we obtain that
the function L(z; t) is a subordination chain.
By using Lemma 2.4, we will show that F (z) \prec G(z). Without loss of generality, we can assume
that \phi and G are analytic and univalent in \mathrm{U} and G\prime (\zeta ) \not = 0 for | \zeta | = 1. If not, then we could replace
\phi with \phi \rho (z) = \phi (\rho z) and G with G\rho (z) = G(\rho z), where \rho \in (0, 1). These new functions have the
desired properties on \mathrm{U} and we can use them in our proof. Therefore, the results would follow by
letting \rho \rightarrow 1.
From the definition of the subordination chain it follows
\phi (z) =
\biggl(
1 - 1
\alpha p+ \gamma
\biggr)
G(z) +
1
\alpha p+ \gamma
zG\prime (z) = L(z; 0)
and
L(z; 0) \prec L(z; t), t \geq 0,
which implies
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758 M. K. AOUF, T. BULBOACĂ, T. M. SEOUDY
L(\zeta ; t) /\in L(\mathrm{U}; 0) = \phi (\mathrm{U}), \zeta \in \partial \mathrm{U}, t \geq 0. (3.20)
According to Lemma 2.4, if F (z) \not \prec G(z) it follows that there exist two points z0 \in \mathrm{U}, \zeta 0 \in \partial \mathrm{U}
and a number m = 1 + t0 \geq 1 such that
F (z0) = G(\zeta 0) and z0F
\prime (z0) = (1 + t0) \zeta 0G
\prime (\zeta 0), t0 \geq 0. (3.21)
Hence, by virtue of (3.21), we have
L(\zeta 0; t0) =
\biggl(
1 - 1
\alpha p+ \gamma
\biggr)
G(\zeta 0) +
1 + t0
\alpha p+ \gamma
\zeta 0G
\prime (\zeta 0) =
=
\biggl(
1 - 1
\alpha p+ \gamma
\biggr)
F (z0) +
1
\alpha p+ \gamma
z0F
\prime (z0) \in \phi (\mathrm{U}),
which contradicts the above remark (3.20), i.e., L(\zeta 0; t0) /\in \phi (\mathrm{U}). Consequently, the subordination
condition (3.5) implies that F (z) \prec G(z), and considering F = G we conclude that the function G
is the best dominant.
Theorem 3.1 is proved.
Remark 3.1. (i) Taking p = 1 and \alpha = \beta in Theorem 3.1, we obtain a subordination results for
the class integral operators studied in [1, 2].
(ii) Note that in [1] (Theorem 1) the author supposed that 0 < \beta + \gamma \leq 1, in [2] (Theorem 3.1)
the assumption was extended to 0 < \beta + \gamma \leq 2, while the above theorem extends the range of these
parameters to \mathrm{R}\mathrm{e}(\beta + \gamma ) \geq 1.
According to this last remark, for the special case p = 1 and \beta + \gamma > 0, combining Theorem 3.1
with Theorem 3.1 in [2], we obtain the following result.
Corollary 3.1. Let \beta , \gamma \in \BbbC with \beta \not = 0 such that \beta + \gamma > 0. For f, g \in \scrK 1
\alpha ,\gamma , suppose that the
function \phi , defined by
\phi (z) = z
\biggl[
g(z)
z
\biggr] \beta
,
satisfies the inequality
\mathrm{R}\mathrm{e}
\biggl[
1 +
z\phi \prime \prime (z)
\phi \prime (z)
\biggr]
> \widetilde \delta , z \in \mathrm{U},
where \widetilde \delta is given by
\widetilde \delta =
\left\{
1 - (\beta + \gamma ), if 0 < \beta + \gamma \leq 1,
1 - (\beta + \gamma )
2
, if 1 \leq \beta + \gamma \leq 2,
- 1
2(\beta + \gamma - 1)
, if \beta + \gamma \geq 2.
Then the subordination condition
z
\biggl[
f(z)
z
\biggr] \beta
\prec z
\biggl[
g(z)
z
\biggr] \beta
implies z
\biggl[
\mathrm{I}\beta ,\gamma [f ](z)
z
\biggr] \beta
\prec z
\biggl[
\mathrm{I}\beta ,\gamma [g](z)
z
\biggr] \beta
,
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INTEGRAL OPERATORS PRESERVING SUBORDINATION AND SUPERORDINATION . . . 759
where the integral operator \mathrm{I}\beta ,\gamma is given by (1.4). Moreover, the function z
\biggl[
\mathrm{I}\beta ,\gamma [g](z)
z
\biggr] \beta
is the best
dominant.
We now derive the following superordination result.
Theorem 3.2. Let \alpha , \beta , \gamma \in \BbbC with \beta \not = 0 such that \mathrm{R}\mathrm{e}(\alpha p+\gamma ) > 1. For f, g \in \scrK p
\alpha ,\gamma , suppose
that the function \phi defined by (3.2) satisfies the condition (3.3), where \delta 0 is given by (3.4).
If the function z
\biggl(
f(z)
zp
\biggr) \alpha
is univalent in \mathrm{U} and z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
zp
\Biggr) \beta
\in \scrQ (0), then the superordi-
nation condition
z
\biggl(
g(z)
zp
\biggr) \alpha
\prec z
\biggl(
f(z)
zp
\biggr) \alpha
(3.22)
implies that
z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g](z)
zp
\Biggr) \beta
\prec z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
zp
\Biggr) \beta
,
and the function z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g](z)
zp
\Biggr) \beta
is the best subordinant.
Proof. Like in the proof of Theorem 3.1, suppose that the functions F, G and q are defined by
(3.6) and (3.7), respectively. Applying a similar method as in the proof of Theorem 3.1, we get that
the inequality (3.8) holds, and from the definition (3.7) it follows that G is convex. Hence, G is a
univalent function in \mathrm{U}.
Next, we will prove that the superordination condition (3.22) implies that G(z) \prec F (z). For this,
we define the function L(z; t) by
L(z; t) =
\biggl(
1 - 1
\alpha p+ \gamma
\biggr)
G(z) +
t
\alpha p+ \gamma
zG\prime (z), z \in \mathrm{U}, t \geq 0. (3.23)
If we denote L(z; t) = a1(t)z + . . . , then
a1(t) =
\partial L(0; t)
\partial z
=
\biggl(
1 +
t - 1
\alpha p+ \gamma
\biggr)
G\prime (0) = 1 +
t - 1
\alpha p+ \gamma
.
Hence, \mathrm{l}\mathrm{i}\mathrm{m}t\rightarrow +\infty | a1(t)| = +\infty , and, using the assumption \mathrm{R}\mathrm{e}(\alpha p + \gamma ) > 1, we obtain a1(t) \not = 0
for all t \geq 0.
Using the facts that \mathrm{R}\mathrm{e} q(z) > 0, z \in \mathrm{U}, and \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) > 1, we get
\mathrm{R}\mathrm{e}
\biggl[
z
\partial L(z; t)/\partial z
\partial L(z; t)/\partial t
\biggr]
= \mathrm{R}\mathrm{e}
\bigl[
\alpha p+ \gamma - 1 + tq(z)
\bigr]
> 0, z \in \mathrm{U}, t \geq 0.
From the definition (3.23), since \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) > 1, for all t \geq 0, we have
| L(z; t)|
| a1(t)|
=
| (\alpha p+ \gamma - 1)G(z) + tzG\prime (z)|
| \alpha p+ \gamma + t - 1|
\leq
\leq | \alpha p+ \gamma - 1| | G(z)| + t| zG\prime (z)|
| \alpha p+ \gamma + t - 1|
.
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760 M. K. AOUF, T. BULBOACĂ, T. M. SEOUDY
Since G is convex, using in the above relation the right-hand sides of the inequalities (3.18), we
obtain
| L(z; t)|
| a1(t)|
\leq r
(1 - r)2
t+ | \alpha p+ \gamma - 1| (1 - r)
| \alpha p+ \gamma + t - 1|
, | z| \leq r, t \geq 0. (3.24)
The assumption \mathrm{R}\mathrm{e}(\alpha p+ \gamma ) > 1 implies
| t - 1 + \alpha p+ \gamma | \geq | \alpha p+ \gamma - 1| , | t - 1 + \alpha p+ \gamma | > | t| , t \geq 0,
and, from (3.24), we conclude that
| L(z; t)|
| a1(t)|
<
r(2 - r)
(1 - r)2
, | z| \leq r, t \geq 0.
Hence, all the assumptions of Lemma 2.1 hold and we conclude that the function L(z; t) is a
subordination chain.
According to Lemma 2.5, the supeordination condition (3.22) implies that G(z) \prec F (z), and
since the differential equation
\phi (z) =
\biggl(
1 - 1
\alpha p+ \gamma
\biggr)
G(z) +
1
\alpha p+ \gamma
zG\prime (z) = \Phi
\bigl(
G(z), zG\prime (z)
\bigr)
has a univalent solution G, the function G is the best subordinant.
Theorem 3.2 is proved.
Remark 3.2. Taking p = 1 and \alpha = \beta in Theorem 3.2, we obtain a superordination result
that generalizes the result from Theorem 3.1 in [3], where a similar implication was obtained for
1 < \beta + \gamma \leq 2. In the present paper this result was extended by assuming that \mathrm{R}\mathrm{e}(\beta + \gamma ) > 1.
Combining the above-mentioned subordination and superordination results involving the operator
\mathrm{I}p\alpha ,\beta ,\gamma , we have the following sandwich-type result.
Theorem 3.3. Let \alpha , \beta , \gamma \in \BbbC with \beta \not = 0 such that \mathrm{R}\mathrm{e}(\alpha p + \gamma ) > 1. For fk, gk \in \scrK p
\alpha ,\gamma ,
k = 1, 2, suppose that the functions \phi k, defined by
\phi k(z) = z
\biggl(
gk(z)
zp
\biggr) \alpha
,
satisfy the inequalities
\mathrm{R}\mathrm{e}
\biggl[
1 +
z\phi \prime \prime
k(z)
\phi \prime
k(z)
\biggr]
> - \delta 0, z \in \mathrm{U}, k = 1, 2,
where \delta 0 is given by (3.4).
If the function z
\biggl(
f(z)
zp
\biggr) \alpha
is univalent in \mathrm{U} and z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
zp
\Biggr) \beta
\in \scrQ (0), then the condition
z
\biggl(
g1(z)
zp
\biggr) \alpha
\prec z
\biggl(
f(z)
zp
\biggr) \alpha
\prec z
\biggl(
g2(z)
zp
\biggr) \alpha
implies that
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INTEGRAL OPERATORS PRESERVING SUBORDINATION AND SUPERORDINATION . . . 761
z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g1](z)
zp
\Biggr) \beta
\prec z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [f ](z)
zp
\Biggr) \beta
\prec z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g2](z)
zp
\Biggr) \beta
,
and the functions z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g1](z)
zp
\Biggr) \beta
and z
\Biggl(
\mathrm{I}p\alpha ,\beta ,\gamma [g2](z)
zp
\Biggr) \beta
are, respectively, the best subordinant
and the best dominant.
Remark 3.3. (i) Taking p = 1 and \alpha = \beta in Theorem 3.3, we obtain the sandwich superordi-
nation result that generalizes the result from Theorem 3.2 in [3].
(ii) While in this previously mentioned article the assumption for the parameters \beta , \gamma \in \BbbC was
1 < \beta + \gamma \leq 2, we proved now that the implication holds for \mathrm{R}\mathrm{e}(\beta + \gamma ) > 1.
Thus, for the special case p = 1 and \beta + \gamma > 1 we deduce the following sandwich-type result.
Corollary 3.2. Let \beta , \gamma \in \BbbC with \beta \not = 0 such that \beta + \gamma > 1. For f, g1, g2 \in \scrK 1
\alpha ,\gamma , suppose that
the functions \phi k, k = 1, 2, defined by
\phi k(z) = z
\biggl[
gk(z)
z
\biggr] \beta
, k = 1, 2,
satisfy the inequality
\mathrm{R}\mathrm{e}
\biggl[
1 +
z\phi \prime \prime
k(z)
\phi \prime
k(z)
\biggr]
> \widehat \delta , z \in \mathrm{U}, k = 1, 2,
where \widehat \delta is given by
\widehat \delta =
\left\{
1 - (\beta + \gamma )
2
, if 1 < \beta + \gamma \leq 2,
- 1
2(\beta + \gamma - 1)
, if \beta + \gamma \geq 2.
If the function z
\biggl[
f(z)
z
\biggr] \beta
is univalent in \mathrm{U} and z
\biggl[
\mathrm{I}\beta ,\gamma [f ](z)
z
\biggr] \beta
\in \scrQ (0), then the condition
z
\biggl[
g1(z)
z
\biggr] \beta
\prec z
\biggl[
f(z)
z
\biggr] \beta
\prec z
\biggl[
g2(z)
z
\biggr] \beta
implies that
z
\biggl[
\mathrm{I}\beta ,\gamma [g1](z)
z
\biggr] \beta
z
\biggl[
\mathrm{I}\beta ,\gamma [f ](z)
z
\biggr] \beta
\prec z
\biggl[
\mathrm{I}\beta ,\gamma [g2](z)
z
\biggr] \beta
,
where the integral operator \mathrm{I}\beta ,\gamma is given by (1.4). Moreover, the functions z
\biggl[
\mathrm{I}\beta ,\gamma [g1](z)
z
\biggr] \beta
and
z
\biggl[
\mathrm{I}\beta ,\gamma [g2](z)
z
\biggr] \beta
are, respectively, the best subordinant and the best dominant.
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 6
762 M. K. AOUF, T. BULBOACĂ, T. M. SEOUDY
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Received 17.10.18
ISSN 1027-3190. Укр. мат. журн., 2021, т. 73, № 6
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| id | umjimathkievua-article-437 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:02:40Z |
| publishDate | 2021 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/3b/3da023f1d4f2cacfd83744c6cca38a3b.pdf |
| spelling | umjimathkievua-article-4372022-03-26T11:03:03Z Integral operators preserving subordination and superordination for multivalent functions Integral operators preserving subordination and superordination for multivalent functions Integral operators preserving subordination and superordination for multivalent functions Aouf , M. K. Bulboacă, T. Seoudy, T. Aouf , M. K. Bulboacă, T. Seoudy, T. Aouf , M. K. Bulboacă, T. Seoudy, T. Analytic function convex function differential subordination and superordination subordination chain integral operator Analytic function convex function differential subordination and superordination subordination chain integral operator UDC 517.9 We obtain subordination, superordination and sandwich-preserving new theorems for certain integral operators defined on multivalent functions. The sandwich-type theorem for these integral operators is also derived and our results extend some earlier ones. Combining these new theorems with some previous related results, we give interesting subordination and superordination consequences for a wide class of analytic integral operators. UDC 517.9 Iнтегральнi оператори, якi зберiгають субординацiю та суперординацiю для багатозначних функцiй Отримано нові теореми щодо субординації, суперординації та збереження порядку для деяких інтегральних операторів на багатозначних функціях.&nbsp;Також доведено теореми типу стискання для інтегральних операторів, які узагальнюють деякі відомі результати.&nbsp;Комбінуючи ці нові теореми з деякими відомими відповідними результатами, ми отримуємо цікаві наслідки щодо субординації та суперординації для широкого класу аналітичних інтегральних операторів. Institute of Mathematics, NAS of Ukraine 2021-06-18 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/437 10.37863/umzh.v73i6.437 Ukrains’kyi Matematychnyi Zhurnal; Vol. 73 No. 6 (2021); 749 - 762 Український математичний журнал; Том 73 № 6 (2021); 749 - 762 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/437/9024 Copyright (c) 2021 M. K. Aouf , T. Bulboacaă, T. Seoudy |
| spellingShingle | Aouf , M. K. Bulboacă, T. Seoudy, T. Aouf , M. K. Bulboacă, T. Seoudy, T. Aouf , M. K. Bulboacă, T. Seoudy, T. Integral operators preserving subordination and superordination for multivalent functions |
| title | Integral operators preserving subordination and superordination for multivalent functions |
| title_alt | Integral operators preserving subordination and superordination for multivalent functions Integral operators preserving subordination and superordination for multivalent functions |
| title_full | Integral operators preserving subordination and superordination for multivalent functions |
| title_fullStr | Integral operators preserving subordination and superordination for multivalent functions |
| title_full_unstemmed | Integral operators preserving subordination and superordination for multivalent functions |
| title_short | Integral operators preserving subordination and superordination for multivalent functions |
| title_sort | integral operators preserving subordination and superordination for multivalent functions |
| topic_facet | Analytic function convex function differential subordination and superordination subordination chain integral operator Analytic function convex function differential subordination and superordination subordination chain integral operator |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/437 |
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