Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator
UDC 517.5The purpose of the present paper is to investigate some subordination, other properties and inclusion relations for functions in certain subclasses of univalent functions in the open unit disc which are defined by $k$-Ruscheweyh derivative operator.  
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| author | Seoudy, T. M. Seoudy , T. M. |
| author_facet | Seoudy, T. M. Seoudy , T. M. |
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| description | UDC 517.5The purpose of the present paper is to investigate some subordination, other properties and inclusion relations for functions in certain subclasses of univalent functions in the open unit disc which are defined by $k$-Ruscheweyh derivative operator.
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| doi_str_mv | 10.37863/umzh.v74i1.2337 |
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DOI: 10.37863/umzh.v74i1.2337
UDC 517.5
T. M. Seoudy (Fayoum Univ., Egypt and Jamoum Univ. College, Umm Al-Qura Univ., Makkah, Saudi Arabia)
SOME SUBCLASSES OF UNIVALENT FUNCTIONS
ASSOCIATED WITH \bfitk -RUSCHEWEYH DERIVATIVE OPERATOR
ДЕЯКI ПIДКЛАСИ УНIВАЛЕНТНИХ ФУНКЦIЙ,
АСОЦIЙОВАНИХ З ОПЕРАТОРОМ \bfitk -ПОХIДНОЇ РУШЕВЕЯ
The purpose of the present paper is to investigate some subordination, other properties and inclusion relations for functions
in certain subclasses of univalent functions in the open unit disc which are defined by k-Ruscheweyh derivative operator.
Мета цiєї роботи — дослiдження деякого пiдпорядкування та iнших властивостей, а також спiввiдношень включення
для функцiй деяких пiдкласiв унiвалентних функцiй у вiдкритому одиничному диску, якi визначаються оператором
k-похiдної Рушевея.
1. Introduction. Denote \scrA by the class of analytic functions in the open unit disc \BbbU =
\bigl\{
z \in \BbbC :
| z| < 1
\bigr\}
of the form
f(z) = z +
\infty \sum
n=1
an+1z
n+1, z \in \BbbU . (1.1)
For function g \in \scrA given by
g(z) = z +
\infty \sum
n=1
bn+1z
n+1, z \in \BbbU ,
the Hadamard (or convolution) product of f and g is defined by
(f \ast g) (z) = z +
\infty \sum
n=1
an+1bn+1z
n+1 = (g \ast f)(z), z \in \BbbU .
For the functions f and g analytic in \BbbU , we say that f is subordinate to g, written f(z) \prec g(z) if
there exists a Schwarz function w (which is analytic in \BbbU , with w(0) = 0, and | w(z)| < 1 (z \in \BbbU )
such that f(z) = g
\bigl(
w(z)
\bigr)
for all z \in \BbbU . Furthermore, if g is univalent in \BbbU , then we have the
following equivalence (see [1, 6, 8]):
f(z) \prec g(z) \leftrightarrow f(0) = g(0) and f(\BbbU ) \subset g(U).
Consider the first-order differential subordination
\scrH
\bigl(
g(z), zg\prime (z)
\bigr)
\prec h(z).
A univalent function q is called dominant, if g(z) \prec q(z) for all analytic functions g that satisfies
this differential subordination. A dominant \widetilde q is called the best dominant, if \widetilde q(z) \prec q(z) for all
dominants q (see [1, 8]).
For v \in \BbbC , k \in \BbbR and n \in \BbbN = \{ 1, 2, 3, . . .\} , the Pochhammer k-symbol (v)n,k is given by
(see [3])
c\bigcirc T. M. SEOUDY, 2022
122 ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
SOME SUBCLASSES OF UNIVALENT FUNCTIONS ASSOCIATED WITH k-RUSCHEWEYH DERIVATIVE . . . 123
(v)n,k = v(v + k)(v + 2k) . . .
\bigl(
v + (n - 1)k
\bigr)
=
n\prod
i=1
\bigl(
v + (i - 1)k
\bigr)
, (1.2)
we define the function F\delta ,k(z) by
F\delta ,k(z) =
z
(1 - z)
\delta +k
k
, \delta > - k, k > 0, z \in \BbbU . (1.3)
Corresponding to the function F\delta ,k(z), we consider a linear operator \scrR \delta
k : \scrA \rightarrow \scrA (\delta > - k, k > 0),
which is defined by means of the following Hadamard product (or convolution):
\scrR \delta
kf(z) = F\delta ,k(z) \ast f(z) = z +
\infty \sum
n=1
(\delta + k)n,k
(k)n,k
an+1z
n+1, z \in \BbbU . (1.4)
It is easily verified from (1.4) that
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
=
\biggl(
\delta + k
k
\biggr)
\scrR \delta +k
k f(z) - \delta
k
\scrR \delta
kf(z), k > 0. (1.5)
Moreover, for f \in \scrA , we observe that:
(1) \scrR \delta
1f(z) = \scrR \delta f(z) (\delta > - 1) , where \scrR \delta denotes the Ruscheweyh derivative of order \delta
(see [10]);
(2) \scrR 0
kf(z) = f(z) and \scrR k
kf(z) = zf \prime (z).
By using k-Ruscheweyh derivative operator \scrR \delta
k, we introduce the following subclass of univalent
functions.
Definition 1. For fixed parameters A, B with - 1 \leq B < A \leq 1 and 0 \leq \lambda < 1, we say that a
function f \in \scrA is in the class \scrS \delta
k(\lambda ;A,B) if it satisfies the subordination condition
1
1 - \lambda
\Biggl(
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
- \lambda
\Biggr)
\prec 1 +Az
1 +Bz
,
which is equivalent to\bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm|
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
- 1
B
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
- [B + (A - B)(1 - \lambda )]
\bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm|
< 1, z \in \BbbU .
We note that
(1) \scrS \delta
k(\lambda ; 1, - 1) = \scrS \delta
k(\lambda ) =
\Biggl\{
f \in \scrA : \Re
\Biggl\{
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
\Biggr\}
> \lambda
\Biggr\}
;
(2) \scrS 0
k(\lambda ;A,B) = \scrS \ast (\lambda ;A,B) =
\biggl\{
f \in \scrA :
1
1 - \lambda
\biggl\{
zf \prime (z)
f(z)
- \lambda
\biggr\}
\prec 1 +Az
1 +Bz
\biggr\}
and
\scrS 0
k(\lambda ; 1, - 1) = \scrS \ast (\lambda ) =
\biggl\{
f \in \scrA : \Re
\biggl\{
zf \prime (z)
f(z)
\biggr\}
> \lambda
\biggr\}
;
(3) \scrS k
k (\lambda ;A,B) = \scrC (\lambda ;A,B) =
\biggl\{
f \in \scrA :
1
1 - \lambda
\biggl\{
1 +
zf \prime \prime (z)
f \prime (z)
- \lambda
\biggr\}
\prec 1 +Az
1 +Bz
\biggr\}
and
\scrS k
k (\lambda ; 1, - 1) = \scrC (\lambda ) =
\biggl\{
f \in \scrA : \Re
\biggl\{
1 +
zf \prime \prime (z)
f \prime (z)
\biggr\}
> \lambda
\biggr\}
.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
124 T. M. SEOUDY
To prove main results, we need the following lemmas.
Lemma 1 [4]. Let h be a convex function in \BbbU with h(0) = 1. Suppose also that the function
g of the form g(z) = 1 + cnz
n + cn+1z
n+1 + . . . is analytic in \BbbU . Then
g(z) +
zg\prime (z)
\sigma
\prec h(z), \Re \{ \sigma \} \geq 0, \sigma \not = 0, (1.6)
implies
g(z) \prec Q(z) =
\sigma
n
z -
\sigma
n
z\int
0
t
\sigma
n
- 1h(t) dt \prec h(z),
and Q is the best dominant of (1.6).
Lemma 2 [12] (see also [8]). Let \nu be a positive measure on the unit interval [0, 1]. Let h(z, t)
be a complex-valued function defined on \BbbU \times [0, 1] such that h(\cdot , t) is analytic in \BbbU for each t \in [0, 1],
and h(z, \cdot ) is \nu -integrable on [0, 1] for all z \in \BbbU . In addition, suppose that \Re
\bigl\{
h(z, t)
\bigr\}
> 0, h( - r, t)
is real, and
\Re
\biggl\{
1
h(z, t)
\biggr\}
\geq 1
h( - r, t)
, | z| \leq r < 1, t \in [0, 1].
If the function H is defined by
H(z) =
1\int
0
h(z, t) d\nu (t),
then
\Re
\biggl\{
1
H(z)
\biggr\}
\geq 1
H( - r)
, | z| \leq r < 1.
Lemma 3 [5]. Suppose \lambda \not = 0 be a real number,
\gamma
\lambda
> 0, \eta \in [0, 1), and let g be an analytic
function in \BbbU of the form g(z) = 1 + anz
n + an+1z
n+1 + . . . , z \in \BbbU , with
P (z) \prec 1 +Rz, R =
\gamma M
n\lambda + \gamma
,
where
M = Mn(\lambda , \gamma , \eta ) =
(1 - \eta )| \lambda |
\biggl(
1 +
n\lambda
\gamma
\biggr)
\bigm| \bigm| 1 - \lambda + \lambda \eta
\bigm| \bigm| +\sqrt{} 1 +
\biggl(
1 +
n\lambda
\gamma
\biggr) 2
.
If p is an analytic function in \BbbU of the form p(z) = 1+ bnz
n+ bn+1z
n+1+ . . . , z \in \BbbU , and satisfies
the subordination
g(z)
\bigl[
1 - \lambda + \lambda ((1 - \eta )p(z) + \eta )
\bigr]
\prec 1 +Mz,
then \Re
\bigl\{
p(z)
\bigr\}
> 0 for all z \in \BbbU .
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
SOME SUBCLASSES OF UNIVALENT FUNCTIONS ASSOCIATED WITH k-RUSCHEWEYH DERIVATIVE . . . 125
Lemma 4 ([7], Corollary 3.2). If - 1 \leq B < A \leq 1, \eta > 0, and the complex number \zeta satisfies
\Re \{ \zeta \} \geq - \eta (1 - A)
1 - B
,
then the following differential equation:
q(z) +
zq\prime (z)
\eta q(z) + \zeta
=
1 +Az
1 +Bz
with q(0) = 1
has a univalent solution in \BbbU given by
q(z) =
\left\{
z\eta +\zeta (1 +Bz)\eta (A - B)/B
\eta
\int z
0
t\eta +\zeta - 1(1 +Bt)\eta (A - B)/B dt
- \zeta
\eta
, if B \not = 0,
z\eta +\zeta \mathrm{e}\mathrm{x}\mathrm{p}(\eta Az)
\eta
\int z
0
t\eta +\zeta - 1 \mathrm{e}\mathrm{x}\mathrm{p}(\eta At) dt
- \zeta
\eta
, if B = 0.
Moreover, if the function g is analytic in \BbbU and satisfies the following subordination:
g(z) +
zg\prime (z)
\eta g(z) + \zeta
\prec 1 +Az
1 +Bz
, (1.7)
then
g(z) \prec q(z) \prec 1 +Az
1 +Bz
,
and q is the best dominant of (1.7).
For real or complex numbers \alpha 1, \alpha 2 and \beta 1 with \beta 1 /\in \BbbZ -
0 = \{ 0, - 1, - 2, . . . \} , the Gauss
hypergeometric function 2F1 is defined by
2F1(\alpha 1, \alpha 2;\beta 1; z) = 1 +
\alpha 1\alpha 2
\beta 1
z
1!
+
\alpha 1 (\alpha 1 + 1)\alpha 2 (\alpha 2 + 1)
\beta 1 (\beta 1 + 1)
z2
2!
+ . . . .
We note that the above series converges absolutely for z \in \BbbU and hence represents an analytic
function in \BbbU (see, for details, [11], Chapter 14).
Each of the identities (asserted by Lemma 5 below) is well-known (cf., e.g., [11], Chapter 14).
Lemma 5 ([11], Chapter 14). For real or complex numbers \alpha 1, \alpha 2 and \beta 1 with \beta 1 /\in \{ 0, - 1,
- 2, . . . \} , we have
1\int
0
t\alpha 2 - 1(1 - t)\beta 1 - \alpha 2 - 1(1 - tz) - \alpha 1 dt =
\Gamma (\alpha 2)\Gamma (\beta 1 - \alpha 2)
\Gamma (\beta 1)
2F1(\alpha 1, \alpha 2;\beta 1; z), (1.8)
\Re \{ \beta 1\} > \Re \{ \alpha 2\} > 0,
2F1(\alpha 1, \alpha 2;\beta 1; z) = (1 - z) - \alpha 1
2F1
\biggl(
\alpha 1, \beta 1 - \alpha 2;\beta 1;
z
z - 1
\biggr)
, (1.9)
and
2F1(\alpha 1, \alpha 2;\beta 1; z) = 2F1(\alpha 2, \alpha 1;\beta 1; z). (1.10)
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
126 T. M. SEOUDY
2. Properties involving the operator \bfscrR \bfitdelta
\bfitk . Unless otherwise mentioned, we assume throughout
this paper that - 1 \leq B < A \leq 1, \delta > - k, k > 0, \theta > 0, 0 \leq \lambda < 1 and all the powers are
considered the principal ones.
Theorem 1. If the function f \in \scrA satisfies the subordination condition
(1 - \theta )
\scrR \delta
kf(z)
z
+ \theta
\scrR \delta +k
k f(z)
z
\prec 1 +Az
1 +Bz
, (2.1)
then
\scrR \delta
kf(z)
z
\prec Q(z) \prec 1 +Az
1 +Bz
, (2.2)
where the function Q given by
Q(z) =
\left\{
A
B
+
\biggl(
1 - A
B
\biggr)
(1 +Bz) - 1
2F1
\biggl(
1, 1;
\delta + (1 + \theta )k
\theta k
;
Bz
1 +Bz
\biggr)
, B \not = 0,
1 +
\delta + k
\delta + (1 + \theta )k
Az, B = 0,
is best dominant of (2.1). Furthermore,
\Re
\biggl\{
\scrR \delta
kf(z)
z
\biggr\}
> M, z \in \BbbU , (2.3)
where
M =
\left\{
A
B
+
\biggl(
1 - A
B
\biggr)
(1 - B) - 1
2F1
\biggl(
1, 1;
\delta + (1 + \theta )k
\theta k
;
B
B - 1
\biggr)
, B \not = 0,
1 - \delta + k
\delta + (1 + \theta )k
A, B = 0.
The estimate in (2.3) is the best possible.
Proof. Letting
g(z) =
\scrR \delta
kf(z)
z
, z \in \BbbU , (2.4)
then g is analytic in \BbbU . Differentiating (2.4) with respect to z and using identity (1.5) in the resulting
relation, we get
(1 - \theta )
\scrR \delta
kf(z)
z
+ \theta
\scrR \delta +k
k f(z)
z
= g(z) +
\theta k
\delta + k
zg\prime (z) \prec 1 +Az
1 +Bz
= h(z).
Now, by using Lemma 1 for \sigma =
\delta + k
\theta k
, and making a change of variables followed by the use of
identities (1.8), (1.9), and (1.10) with \alpha 1 = 1, \alpha 2 =
\delta + k
\theta k
, and \beta 1 - \alpha 2 = 1, we deduce that
\scrR \delta
kf(z)
z
\prec Q(z) =
\delta + k
\theta k
z -
\delta +k
\theta k
z\int
0
t
\delta +k
\theta k
- 1 1 +At
1 +Bt
dt =
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
SOME SUBCLASSES OF UNIVALENT FUNCTIONS ASSOCIATED WITH k-RUSCHEWEYH DERIVATIVE . . . 127
=
\left\{
A
B
+
\biggl(
1 - A
B
\biggr)
(1 +Bz) - 1
2F1
\biggl(
1, 1;
\delta + (1 + \theta )k
\theta k
;
Bz
1 +Bz
\biggr)
, B \not = 0,
1 +
\delta + k
\delta + (1 + \theta )k
Az, B = 0,
which proves assertion (2.2). From here, to prove inequality (2.3) it is sufficient to show that
\mathrm{i}\mathrm{n}\mathrm{f}
\bigl\{
\Re
\bigl\{
Q(z)
\bigr\}
: | z| < 1
\bigr\}
= Q( - 1).
Indeed, we have
\Re
\biggl\{
1 +Az
1 +Bz
\biggr\}
\geq 1 - Ar
1 - Br
, | z| \leq r < 1.
Setting
h (z, s) =
1 +Asz
1 +Bsz
and d\nu (s) =
\delta + k
\theta k
s
\delta +k
\theta k
- 1 ds, 0 \leq s \leq 1,
which is a positive measure on the closed interval [0, 1], we get
Q(z) =
1\int
0
h(z, s) d\nu (s)
and
\Re
\bigl\{
Q(z)
\bigr\}
\geq
1\int
0
1 - Asr
1 - Bsr
d\nu (s) = Q( - r), | z| \leq r < 1.
Letting r \rightarrow 1 - in the above inequality we obtain (2.3). Finally, the estimate (2.3) is the best possible
as the function Q is the best dominant of (2.1).
Theorem 1 is proved.
For a function f \in \scrA the generalized Bernardi – Libera – Livingston integral operator J\mu : \scrA \rightarrow
\rightarrow \scrA is defined by (see [2])
J\mu f(z) =
\mu + 1
z\mu
z\int
0
t\mu - 1f(t) dt, \mu > - 1. (2.5)
It is easy to verify that, for all f \in \scrA , we have
z
\Bigl(
\scrR \delta
kJ\mu f(z)
\Bigr) \prime
= (1 + \mu )\scrR \delta
kf(z) - \mu \scrR \delta
kJ\mu f(z). (2.6)
Theorem 2. If the function f \in \scrA satisfies the subordination condition
\scrR \delta
kf(z)
z
\prec 1 +Az
1 +Bz
(2.7)
and J\mu is the integral operator defined by (2.5), then
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
128 T. M. SEOUDY
\scrR \delta
kJ\mu f(z)
z
\prec K (z) \prec 1 +Az
1 +Bz
,
where the function K given by
K(z) =
\left\{
A
B
+
\biggl(
1 - A
B
\biggr)
(1 +Bz) - 1
2F1
\biggl(
1, 1;\mu + 2;
Bz
1 +Bz
\biggr)
, B \not = 0,
1 +
\mu + 1
\mu + 2
Az, B = 0,
is the best dominant of (2.7). Furthermore,
\Re
\biggl\{
\scrR \delta
kJ\mu f(z)
z
\biggr\}
> L, z \in \BbbU , (2.8)
where
L =
\left\{
A
B
+
\biggl(
1 - A
B
\biggr)
(1 - B) - 1
2F1
\biggl(
1, 1;\mu + 2;
B
B - 1
\biggr)
, B \not = 0,
1 - \mu + 1
\mu + 2
A, B = 0.
The estimate in (2.8) is the best possible.
Proof. Let
\varphi (z) =
\scrR \delta
kJ\mu f(z)
z
, z \in \BbbU . (2.9)
Then \varphi is analytic in \BbbU . Differentiating (2.9) with respect to z and using identity (2.6) in the
resulting relation, we get
\scrR \delta
kf(z)
z
= \varphi (z) +
z\varphi \prime (z)
\mu + 1
\prec 1 +Az
1 +Bz
.
Employing the same technique that we used in the proof of Theorem 1, the remaining part of the
theorem can be proved similarly.
Theorem 3. If the function f \in \scrA satisfies the condition\bigm| \bigm| \bigm| \bigm| \bigm| (1 - \theta )
\scrR \delta
kf(z)
z
+ \theta
\scrR \delta +k
k f(z)
z
- 1
\bigm| \bigm| \bigm| \bigm| \bigm| < M1 =
\delta + (1 + \theta )k
\delta + k
N1, z \in \BbbU , (2.10)
with
\delta + (1 + \theta )k
\delta + k
N1 \leq 1, (2.11)
where
N1 = \mathrm{m}\mathrm{i}\mathrm{n}
\bigl\{
x \in (0, 1) : \Phi (x) = 0
\bigr\}
(2.12)
and
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
SOME SUBCLASSES OF UNIVALENT FUNCTIONS ASSOCIATED WITH k-RUSCHEWEYH DERIVATIVE . . . 129
\Phi (x) =
\Biggl[ \biggl(
\theta k(1 - \lambda )
\delta + k
\biggr) 2
- 2
\theta k(1 - \lambda )
\delta + k
-
\biggl(
\delta + (1 + \theta ) k
\delta + k
\biggr) 2
\Biggr]
x2 -
- 2
\theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| x+
\biggl(
\theta k(1 - \lambda )
\delta + k
\biggr) 2
, (2.13)
then f \in \scrS \delta
k(\lambda ).
Proof. Let
g(z) =
\scrR \delta
kf(z)
z
, z \in \BbbU .
Then g is analytic in \BbbU . From assumption (2.11), according to Theorem 1 for the special case
A = M1 and B = 0, we obtain that assumption (2.10) implies
g(z) \prec 1 +
\delta + k
\delta + (1 + \theta )k
M1z = 1 +N1z,
which is equivalent to
| g(z) - 1| < N1, z \in \BbbU . (2.14)
Setting
p(z) =
1
1 - \lambda
\Biggl(
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
- \lambda
\Biggr)
, (2.15)
assumption (2.10) could be written as\bigm| \bigm| \bigm| \bigm| \biggl( 1 - \theta k(1 - \lambda )
\delta + k
\biggr)
g(z) +
\theta k(1 - \lambda )
\delta + k
p(z)g(z) - 1
\bigm| \bigm| \bigm| \bigm| < \delta + (1 + \theta ) k
\delta + k
N1, z \in \BbbU . (2.16)
Now, we will show that (2.16) implies \Re
\bigl\{
p(z)
\bigr\}
> 0 for all z \in \BbbU , that is, f \in \scrS \delta
k(\lambda ).
Supposing that this last inequality is false, since p(0) = 1, there exist z0 \in \BbbU and a number
x \in \BbbR such that p(z0) = ix. Therefore, in order to show that (2.16) implies \Re
\bigl\{
p(z)
\bigr\}
> 0 for all
z \in \BbbU , it is sufficient to obtain a contradiction with (2.16), for example,
E =
\bigm| \bigm| \bigm| \bigm| \biggl( 1 - \theta k(1 - \lambda )
\delta + k
\biggr)
g(z0) +
\theta k(1 - \lambda )
\delta + k
p(z0)g(z0) - 1
\bigm| \bigm| \bigm| \bigm| \geq \delta + (1 + \theta )k
\delta + k
N1. (2.17)
Thus, if we put g(z0) = u+ iv, then
E2 =
\bigm| \bigm| \bigm| \bigm| \biggl( 1 - \theta k(1 - \lambda )
\delta + k
\biggr)
\varphi (z0) +
\theta k(1 - \lambda )
\delta + k
P (z0)\varphi (z0) - 1
\bigm| \bigm| \bigm| \bigm| 2 =
=
\bigl(
u2 + v2
\bigr) \biggl( \theta k(1 - \lambda )
\delta + k
\biggr) 2
x2 + 2xv
\theta k(1 - \lambda )
\delta + k
+
\bigm| \bigm| \bigm| \bigm| \biggl( 1 - \theta k(1 - \lambda )
\delta + k
\biggr)
\varphi (z0) - 1
\bigm| \bigm| \bigm| \bigm| 2 . (2.18)
By using (2.14), we have
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
130 T. M. SEOUDY\bigm| \bigm| \bigm| \bigm| \biggl( 1 - \theta k(1 - \lambda )
\delta + k
\biggr)
\varphi (z0) - 1
\bigm| \bigm| \bigm| \bigm| = \bigm| \bigm| \bigm| \bigm| \biggl( 1 - \theta k(1 - \lambda )
\delta + k
\biggr)
(\varphi (z0) - 1) - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| \geq
\geq \theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1. (2.19)
Now, we will prove that under our assumptions the next inequality holds:
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1 \geq 0. (2.20)
Thus, if we denote
l =
\theta k(1 - \lambda )
\delta + k
> 0, m =
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
> 0,
then the function \Phi given by (2.13) becomes
\Phi (x) = (l2 - 2l - m)x2 - 2l| 1 - l| x+ l2,
and
\Phi (0) = l2 > 0, \Phi (1) = - 2(\delta + k)(| 1 - l| + 1 - l) - m < 0.
If l =
\theta k(1 - \lambda )
\delta + k
= 1, it is obvious that (2.20) holds for any number N1 . If l \not = 1, since
\Phi
\biggl(
l
| 1 - l|
\biggr)
= - l2(1 +m)
| 1 - l| 2
< 0,
we deduce that if N1 is given by (2.12), then inequality (2.20) is also true.
Hence, from (2.18), (2.19) and (2.20), we obtain that
E2 - M2
1 \geq (u2 + v2)
\biggl(
\theta k (1 - \lambda )
\delta + k
\biggr) 2
x2 + 2xv
\theta k (1 - \lambda )
\delta + k
+
+
\biggl(
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
-
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
N2
1 .
Denoting
F (x) = (u2 + v2)
\biggl(
\theta k(1 - \lambda )
\delta + k
\biggr) 2
x2 + 2xv
\theta k(1 - \lambda )
\delta + k
+
+
\biggl(
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
-
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
N2
1 ,
since (u2 + v2)
\biggl(
\theta k(1 - \lambda )
\delta + k
\biggr) 2
> 0, it follows that the inequality F (x) \geq 0 holds for all x \in \BbbR , if
and only if the discriminant \Delta \leq 0, that is,
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SOME SUBCLASSES OF UNIVALENT FUNCTIONS ASSOCIATED WITH k-RUSCHEWEYH DERIVATIVE . . . 131
\Delta = 4
\Biggl\{
v2
\biggl(
\theta k(1 - \lambda )
\delta + k
\biggr) 2
- (u2 + v2)
\biggl(
\theta k (1 - \lambda )
\delta + k
\biggr) 2
\times
\times
\Biggl[ \biggl(
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
-
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
N2
1
\Biggr] \Biggr\}
\leq 0,
which is equivalent to
v2
\Biggl[
1 -
\biggl(
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
+
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
N2
1
\Biggr]
\leq
\leq u2
\Biggl[ \biggl(
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
-
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
N2
1
\Biggr]
. (2.21)
Putting \varphi (z0) = 1 + \rho ei\varepsilon for some \varepsilon \in \BbbR , it is easy to show that
v2
u2
=
\rho 2 \mathrm{s}\mathrm{i}\mathrm{n}2 \varepsilon
(1 + \rho \mathrm{c}\mathrm{o}\mathrm{s} \varepsilon )2
\leq \rho 2
1 - \rho 2
, \varepsilon \in \BbbR .
From (2.14) we have \rho \leq N1 < 1 and, by using the above inequality, we obtain
v2
u2
\leq \rho 2
1 - \rho 2
\leq N2
1
1 - N2
1
. (2.22)
Since the function
T (\rho ) =
\rho 2
1 - \rho 2
, \rho \in [0, 1),
is a strictly increasing function on [0, 1), we need to determine the maximum value of N1 \in [0, 1)
(this condition follows from the previous comments) such that
N2
1 \leq
\biggl(
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
-
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
N2
1 .
A simple computation shows that this value is given by (2.11), where \Phi has the form (2.13). Ac-
cording to the above reasons, from (2.22) it follows that
v2
u2
\leq \rho 2
1 - \rho 2
\leq N2
1
1 - N2
1
\leq
\biggl(
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
-
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
N2
1
1 -
\biggl(
\theta k (1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
+
\biggl(
\delta + (1 + \theta ) k
\delta + k
\biggr) 2
N2
1
,
hence
v2
u2
\leq
\biggl(
\theta k(1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
-
\biggl(
\delta + (1 + \theta )k
\delta + k
\biggr) 2
N2
1
1 -
\biggl(
\theta k (1 - \lambda )
\delta + k
-
\bigm| \bigm| \bigm| \bigm| 1 - \theta k(1 - \lambda )
\delta + k
\bigm| \bigm| \bigm| \bigm| N1
\biggr) 2
+
\biggl(
\delta + (1 + \theta ) k
\delta + k
\biggr) 2
N2
1
,
which is equivalent to (2.21), that is, \Delta \leq 0. Therefore, (2.17) holds.
Theorem 3 is proved.
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132 T. M. SEOUDY
Theorem 4. If the function f \in \scrA such that
\scrR \delta
kf(z)
z
\not = 0 for all z \in \BbbU and satisfies the
subordination condition
(1 - \theta )
\biggl(
\scrR \delta
kf(z)
z
\biggr) \mu
+ \theta
\scrR \delta +k
k f(z)
\scrR \delta
kf(z)
\biggl(
\scrR \delta
kf(z)
z
\biggr) \mu
\prec 1 +M2z, (2.23)
where
M2 =
\left\{
(1 - \lambda )\theta k
\delta + k
\biggl(
1 +
\theta k
\mu (\delta + k)
\biggr)
\bigm| \bigm| \bigm| \bigm| 1 - \theta
(1 - \lambda )k
\delta + k
\bigm| \bigm| \bigm| \bigm| +
\sqrt{}
1 +
\biggl(
1 +
\theta k
\mu (\delta + k)
\biggr) 2
, \mu > 0,
(1 - \lambda )\theta k
\delta + k
, \mu = 0,
then f \in \scrS \delta
k(\lambda ).
Proof. If \mu = 0, then assumption (2.23) is equivalent to
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
- \lambda \prec (1 - \lambda )(1 + z),
which implies that f \in \scrS \delta
k(\lambda ).
If we consider \mu > 0, let define the function
g(z) =
\biggl(
\scrR \delta
kf(z)
z
\biggr) \mu
, z \in \BbbU , (2.24)
where we choose the principal value of the power function. Then g is analytic in \BbbU with g(0) = 1,
and differentiating (2.24) with respect to z, we obtain
(1 - \theta )
\biggl(
\scrR \delta
kf(z)
z
\biggr) \mu
+ \theta
\scrR \delta +k
k f(z)
\scrR \delta
kf(z)
\biggl(
\scrR \delta
kf(z)
z
\biggr) \mu
= g(z) +
\theta k
\mu (\delta + k)
zg\prime (z),
which in view of Lemma 1 yields
g(z) \prec 1 +
\mu (\delta + k)
\mu (\delta + k) + \theta k
M2z.
Also, the subordination assumption (2.23) can be written as\biggl[
(1 - \theta ) + \theta
\biggl(
(1 - \lambda )k
\delta + k
p(z) +
\lambda k + \delta
\delta + k
\biggr) \biggr]
g(z) \prec 1 +M2z,
where p(z) is given by (2.15). Therefore, from Lemma 3 we deduce that \Re
\bigl\{
p(z)
\bigr\}
> 0, z \in \BbbU .
Theorem 4 is proved.
Remarks. 1. Putting \delta = 0 in Theorem 4, we obtain the result of Liu [5] (Theorem 2.2).
2. Taking \delta = 0 and \theta = 1 in Theorem 4, we have the result of Liu [5] (Corollary 2.1).
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SOME SUBCLASSES OF UNIVALENT FUNCTIONS ASSOCIATED WITH k-RUSCHEWEYH DERIVATIVE . . . 133
3. Taking \delta = 0 and \theta = \mu = 1 in Theorem 4, we get the result of Mocanu and Oros [9]
(Corollary 2.2, with n = 1).
4. Putting \delta = 0 and \theta =
1
1 - \lambda
(0 \leq \lambda < 1) in Theorem 4, we obtain the result of Liu [5]
(Corollary 2.2).
5. Taking \delta = 0, \theta =
1
1 - \lambda
(0 \leq \lambda < 1) and \mu = 1 in Theorem 4, we have the result of
Mocanu and Oros [9] (Corollary 2.4).
3. Inclusion relationships for the class \bfscrS \bfitdelta
\bfitk (\bfitlambda ;\bfitA ,\bfitB ).
Theorem 5. If f \in \scrS \delta +k
k (\lambda ;A,B) such that \scrR \delta
kf(z) \not = 0 for all z \in \BbbU \ast = \BbbU \setminus \{ 0\} and\biggl(
\delta
k
+ \lambda
\biggr)
(1 - B) + (1 - \lambda )(1 - A) \geq 0, (3.1)
then
1
1 - \lambda
\Biggl(
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
- \lambda
\Biggr)
\prec q1(z) \prec
1 +Az
1 +Bz
, (3.2)
where
q1(z) =
1
1 - \lambda
\biggl(
1
Q1(z)
- \delta
k
- \lambda
\biggr)
(3.3)
and
Q1(z) =
\left\{
\int 1
0
s
\delta
k
\biggl(
1 +Bzs
1 +Bz
\biggr) (1 - \lambda )(A - B)
B
ds, B \not = 0,
\int 1
0
s
\delta
k \mathrm{e}\mathrm{x}\mathrm{p}
\bigl[
(1 - \lambda )(s - 1)Az
\bigr]
ds, B = 0,
and q1 is the best dominant of (3.2). If , in addition to (3.1),
A \leq -
\biggl(
\delta
k
+ \lambda + 1
\biggr)
B
1 - \lambda
, - 1 \leq B < 0, (3.4)
then f \in \scrS \delta
k(\rho 1), where
\rho 1 =
\delta + k
k
\biggl[
2F1
\biggl(
1,
(1 - \lambda )(B - A)
B
;
\delta + 2k
k
;
B
B - 1
\biggr) \biggr] - 1
- \delta
k
.
The result is the best possible.
Proof. Let
\phi (z) =
1
1 - \lambda
\Biggl(
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
- \lambda
\Biggr)
, z \in \BbbU . (3.5)
Then \phi is analytic in \BbbU with \phi (0) = 1. Using identity (1.5) in (3.5), we get
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
134 T. M. SEOUDY
\delta + k
k
\scrR \delta +k
k f(z)
\scrR \delta
kf(z)
= (1 - \lambda )\phi (z) +
\delta
k
+ \lambda , (3.6)
and differentiating (3.6) with respect to z, we obtain
1
1 - \lambda
\left( z
\Bigl(
\scrR \delta +k
k f(z)
\Bigr) \prime
\scrR \delta +k
k f(z)
- \lambda
\right) = \phi (z) +
z\phi \prime (z)
(1 - \lambda )\phi (z) +
\delta
k
+ \lambda
\prec 1 +Az
1 +Bz
. (3.7)
From the assumption (3.1), by using Lemma 4, we get
\phi (z) \prec q1(z) \prec
1 +Az
1 +Bz
,
where q1 is given (3.3) is the best dominant of (3.7), and this proves (3.2).
Now, we will show that
\mathrm{i}\mathrm{n}\mathrm{f}
\bigl\{
\Re
\bigl\{
q1(z)
\bigr\}
: | z| < 1
\bigr\}
= q1( - 1),
or, equivalently,
\mathrm{i}\mathrm{n}\mathrm{f}
\biggl\{
\Re
\biggl\{
1
Q1(z)
\biggr\}
: | z| < 1
\biggr\}
=
1
Q1( - 1)
. (3.8)
Denoting \alpha 1 =
(1 - \lambda )(B - A)
B
, \alpha 2 =
\delta + k
k
and \beta 1 =
\delta + 2k
k
, since \beta 1 > \alpha 2 > 0, from (1.8),
(1.9), and (1.10), we deduce that
Q1(z) = (1 +Bz)\alpha 1
1\int
0
s\alpha 2 - 1(1 +Bzs) - \alpha 1 ds =
\Gamma (\alpha 2)
\Gamma (\beta 1)
2F1
\biggl(
1, \alpha 1;\beta 1;
Bz
Bz + 1
\biggr)
=
k
\delta + k
H(z),
where
H(z) = 2F1
\biggl(
1, \alpha 1;\beta 1;
Bz
Bz + 1
\biggr)
, (3.9)
whenever B \not = 0. From (3.4), excepting the case of equality, we have \beta 1 > \alpha 1 > 0, hence from
(1.9) we get
H(z) =
1\int
0
h(z, s)d\nu (s),
where
h(z, s) =
1 +Bz
1 + (1 - s)Bz
and
d\nu (s) =
\Gamma (\beta 1)
\Gamma (\alpha 1)\Gamma (\beta 1 - \alpha 1)
s\alpha 1 - 1(1 - s)\beta 1 - \alpha 1 - 1 ds, 0 \leq s \leq 1,
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SOME SUBCLASSES OF UNIVALENT FUNCTIONS ASSOCIATED WITH k-RUSCHEWEYH DERIVATIVE . . . 135
which is a positive measure on [0, 1]. Using the fact - 1 \leq B < 0, it is easy to check that
\Re
\biggl\{
1
H(z)
\biggr\}
> 0, z \in \BbbU ,
h ( - r, s) \in \BbbR , 0 \leq r < 1; s \in [0, 1],
\Re
\biggl\{
1
h(z, s)
\biggr\}
\geq 1 - (1 - s)Br
1 - Br
=
1
h( - r, s)
, | z| \leq r < 1, s \in [0, 1].
Therefore, according to Lemma 2, we deduce that
\Re
\biggl\{
1
H(z)
\biggr\}
\geq 1
H( - r)
, | z| \leq r < 1,
and by letting r \rightarrow 1 - , taking into the account the relation (3.9), we obtain inequality (3.8).
Further, by taking A \uparrow - (\delta /k + \lambda + 1)B
1 - \lambda
for the case A = - (\delta /k + \lambda + 1)B
1 - \lambda
, we conclude
that (3.8) holds whenever the inequality (3.4) is satisfied, which prove f \in \scrS \delta
k(\rho 1). The result is the
best possible as the function q1 is the best dominant of (3.7).
Theorem 5 is proved.
Theorem 6. If f \in \scrS \delta
k(\lambda ;A,B) such that \scrR \delta
kJ\mu f(z) \not = 0 for all z \in \BbbU \ast and
(\lambda + \mu )(1 - B) + (1 - \lambda )(1 - A) \geq 0, (3.10)
then J\mu f(z) \in \scrS \delta
k(\lambda ;A,B), where the operator J\mu is defined by (2.5). Furthermore, if f \in
\in \scrS \delta
k(\lambda ;A,B), then
1
1 - \lambda
\Biggl(
z
\bigl(
\scrR \delta
kJ\mu f(z)
\bigr) \prime
\scrR \delta
kJ\mu f(z)
- \lambda
\Biggr)
\prec q3(z) \prec
1 +Az
1 +Bz
, (3.11)
where
q3(z) =
1
1 - \lambda
\biggl(
1
Q3(z)
- \lambda - \mu
\biggr)
and
Q3(z) =
\left\{
\int 1
0
s\mu
\biggl(
1 +Bzs
1 +Bz
\biggr) (1 - \lambda )(A - B)
B
ds, B \not = 0,\int 1
0
s\mu \mathrm{e}\mathrm{x}\mathrm{p}
\bigl[
(1 - \lambda )(s - 1)Az
\bigr]
ds, B = 0,
and q3 is the best dominant of (3.11). If , in addition to (3.10),
A \leq - (\lambda + \mu + 1)B
1 - \lambda
with - 1 \leq B < 0,
then J\mu f(z) \in \scrS \delta
k(\rho 3), where
\rho 3 = (1 + \mu )
\biggl[
2F1
\biggl(
1,
(1 - \lambda )(B - A)
B
; \mu + 2;
B
B - 1
\biggr) \biggr] - 1
- \mu .
The result is best possible.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
136 T. M. SEOUDY
Proof. Let
\phi (z) =
1
1 - \lambda
\Biggl(
z
\bigl(
\scrR \delta
kJ\mu f(z)
\bigr) \prime
\scrR \delta
kJ\mu f(z)
- \lambda
\Biggr)
, z \in \BbbU . (3.12)
Then \phi is analytic in \BbbU with \phi (0) = 1. Using the identity (2.6) in (3.12), we get
(1 + \mu )
\scrR \delta
kf(z)
\scrR \delta
kJ\mu f(z)
= (1 - \lambda )\phi (z) + \lambda + \mu . (3.13)
Differentiating (3.13) with respect to z, we obtain
1
1 - \lambda
\Biggl(
z
\bigl(
\scrR \delta
kf(z)
\bigr) \prime
\scrR \delta
kf(z)
- \lambda
\Biggr)
= \phi (z) +
z\phi \prime (z)
(1 - \lambda )\phi (z) + \lambda + \mu
\prec 1 +Az
1 +Bz
,
and employing the same technique that used in the proof of Theorem 5, the remaining part of the
theorem can be proved similarly.
References
1. T. Bulboacă, Differential subordinations and superordinations. Recent results, House Sci. Book Publ., Cluj-Napoca
(2005).
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Received 27.01.20,
after revision — 04.05.20
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 1
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| id | umjimathkievua-article-2337 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:22:19Z |
| publishDate | 2022 |
| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/6b/a2986582c2edfbf4488ac33e6d09336b.pdf |
| spelling | umjimathkievua-article-23372022-03-27T15:39:11Z Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator Seoudy, T. M. Seoudy , T. M. Univalent functions, Pochhammer ksymbol, Differential subordination, Hadamard product, kRuscheweyh derivative operator UDC 517.5The purpose of the present paper is to investigate some subordination, other properties and inclusion relations for functions in certain subclasses of univalent functions in the open unit disc which are defined by $k$-Ruscheweyh derivative operator. &nbsp; УДК 517.5 Деякі підкласи унівалентних функцій, асоційованих з оператором $k$-похідної Рушевея Мета цієї роботи – дослідження деякого підпорядкування та інших властивостей, а також співвідношень включення для функцій деяких підкласів унівалентних функцій у відкритому одиничному диску, які визначаються оператором $k$-похідної Рушевея. Institute of Mathematics, NAS of Ukraine 2022-01-24 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2337 10.37863/umzh.v74i1.2337 Ukrains’kyi Matematychnyi Zhurnal; Vol. 74 No. 1 (2022); 122 - 136 Український математичний журнал; Том 74 № 1 (2022); 122 - 136 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2337/9185 Copyright (c) 2022 Tamer Seoudy |
| spellingShingle | Seoudy, T. M. Seoudy , T. M. Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator |
| title | Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator |
| title_alt | Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator |
| title_full | Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator |
| title_fullStr | Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator |
| title_full_unstemmed | Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator |
| title_short | Some subclasses of univalent functions associated with $k$-Ruscheweyh derivative operator |
| title_sort | some subclasses of univalent functions associated with $k$-ruscheweyh derivative operator |
| topic_facet | Univalent functions Pochhammer ksymbol Differential subordination Hadamard product kRuscheweyh derivative operator |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2337 |
| work_keys_str_mv | AT seoudytm somesubclassesofunivalentfunctionsassociatedwithkruscheweyhderivativeoperator AT seoudytm somesubclassesofunivalentfunctionsassociatedwithkruscheweyhderivativeoperator |