*Corresponding author
E-mail address: [email protected] Received October 29, 2018
1
Available online at http://scik.org Eng. Math. Lett. 2019, 2019:5
https://doi.org/10.28919/eml/3940 ISSN: 2049-9337
STUDY OF A NEW CLASS FOR HIGHER-ORDER DERIVATIVES OF
HARMONIC MULTIVALENT FUNCTIONS
ABBAS KAREEM WANAS1,* AND FIRAS HUSSEAN MAGHOOL2 1Department of Mathematics, College of Science, University of Al-Qadisiyah, Iraq
2 Department of Mathematics, College of Computer Science and Information Technology, University of Al-Qadisiyah, Iraq
Copyright Β© 2019 the authors. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract: In this paper, we define and investigate a new class for higher-order derivatives of harmonic multivalent functions. We obtain coefficient inequalities, distortion bounds, extreme points, convex combination. Our results extend corresponding previously known results.
Keywords: harmonic multivalent functions; extreme points; convex combination; higher-order derivatives.
2010 AMS Subject Classification: 30C45, 30C50.
1.
I
NTRODUCTIONA continuous function π = π’ + ππ£ is a complex valued harmonic function in a complex
domain β , if both π’ and π£ are real harmonic in β. In any simply connected domain π· β β, we
analytic part of π . A necessary and sufficient condition for π to be locally univalent and sense-preserving in π· is that |ββ²(π§)| > |πβ²(π§)| in π· (see Clunie and Sheil-Small [3]).
Denote by πΎβ(π) the class of functions π = β + π that are harmonic univalent and sense-
preserving in the open unit disk π = {π§ β β: |π§| < 1}. For π = β + π β πΎβ(π) , we may express
the analytic functions β and π as
β(π§) = π§π+ β π ππ§π
β
π=π+1
, π(π§) = β πππ§π
β
π=π
, |ππ| < 1 . (1.1)
Also denote by πβ(π) the subclass of πΎβ(π) containing of all functions π = β + π , where β
and πare given by
β(π§) = π§πβ β π ππ§π
β
π=π+1
, π(π§) = β β πππ§π
β
π=π
, (ππ β₯ 0, ππ β₯ 0, |ππ| < 1 ) . (1.2)
We denote by π΄πΎβ(π, π, πΎ, πΌ, π½) the class of all functions of the form (1.1) that satisfy the
condition:
π π {π§π
(π+2)(π§) + π(π+1)(π§)
π(π+1)(π§) + πΎπ§π(π+2)(π§)} > π½ |
π§π(π+2)(π§) + π(π+1)(π§)
π(π+1)(π§) + πΎπ§π(π+2)(π§)β 1| + πΌ, (1.3)
where π β π = {1,2, β¦ }, π β ππ= π βͺ {0} , π > π , 0 β€ πΌ < π β π , π½ β₯ 0 , 0 β€ πΎ < 1 and
for each π = β + πΜ β πΎπ»(π), we have
π(π)(π§) = πΏ(π, π)π§πβπ+ β πΏ(π, π)π π β
π=π+1
π§πβπ+ β πΏ(π, π)π π (π§Μ β
π=π
)πβπ,
πΏ(π, π) = π!
(π β π)!= {
1 (π = 0)
π(π β 1) β¦ (π β π + 1) (π β 0) .
Let π΄πβ(π, π, πΎ, πΌ, π½) be the subclass of π΄πΎβ(π, π, πΎ, πΌ, π½), where
π΄πβ(π, π, πΎ, πΌ, π½) = πβ(π) β© π΄πΎβ(π, π, πΎ, πΌ, π½).
Remark 1.1.
(1) If π = 1 and π = 0, we have π΄πβ(1,0, πΎ, πΌ, π½) = πΎπβ(πΎ, πΌ, π½) which was studied
by Atshan and Wanas [2].
(2) If π = 1and π = π½ = 0, we have π΄πβ(1,0, πΎ, πΌ, 0) = πΆ(πΎ, πΌ) which was studied by
STUDY OF A NEW CLASS FOR HIGHER-ORDER DERIVATIVES OF HARMONIC MULTIVALENT β¦
(3) If π = 1 and π = πΎ = π½ = 0, we have π΄πβ(1,0,0, πΌ, 0) = πΆ(πΌ) which was studied by
Silverman [5] for analytic part.
Lemma 1.1. (Aqlan, [1]) Let π€ = π’ + ππ£ and π be a real number. Then π π(π€) β₯ π if and
only if
|π€ + (1 β π)| β₯ |π€ β (1 + π)|.
Lemma 1.2. (Aqlan, [1]) Let π€ = π’ + ππ£ and π, π‘ be a real number. Then π π(π€) β₯
π|π€ β 1| + π‘ if and only if
π π{π€(1 + ππππ) β ππππ} β₯ π‘ (βπ β€ π β€ π).
2.
M
AIN RESULTIn our first theorem, we introduce a sufficient coefficient bound for harmonic function in
π΄πΎβ(π, π, πΎ, πΌ, π½).
Theorem 2.1. Let π = β + π with β and π are given by (1.1). If
β π!
(π β π β 1)
β
π=π+1
β€ π!
(π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) β πΌ](1 β |ππ|), (2.1)
where π β π, π β π0 , π > π, 0 β€ πΌ < π β π , π½ β₯ 0 , 0 β€ πΎ < 1, then π is harmonic
multivalent sense preserving in π and π β π΄πΎβ(π, π, πΎ, πΌ, π½).
Proof. For proving π β π΄πΎβ(π, π, πΎ, πΌ, π½), we must show that (1.3) holds true. by using Lemma
1.2 , it is sufficient to show that
π π {π§π(π+2)(π§) + π(π+1)(π§)
π(π+1)(π§) + πΎπ§π(π+2)(π§)(1 + π½πππ) β π½πππ} > πΌ (βπ β€ π β€ π) ,
or equivalently
π π {(1 + π½π
ππ)(π§π(π+2)(π§) + π(π+1)(π§)) β π½πππ(π(π+1)(π§) + πΎπ§π(π+2)(π§))
π(π+1)(π§) + πΎπ§π(π+2)(π§) } > πΌ.
If we put
and
π(π§) = π(π+1)(π§) + πΎπ§π(π+2)(π§).
In view of Lemma 1.1, we only need to prove that
|πΏ(π§) + (1 β πΌ)π(π§)| β |πΏ(π§) β (1 + πΌ)π(π§)| β₯ 0.
Now,
|πΏ(π§) + (1 β πΌ)π(π§)|
= |(1 + π½πππ) ( π!
(π β π β 2)! π§πβπβ1+ β
π!
(π β π β 2)!πππ§πβπβ1
β
π=π+1
+ β(π β π β 2)!π! ππ(π§)πβπβ1
β
π=π
+ π!
(π β π β 1)!π§πβπβ1+ β
π!
(π β π β 1)!πππ§πβπβ1
β
π=π+1
+ β π!
(π β π β 1)!ππ(π§)πβπβ1
β
π=π
)
βπ½πππ( π!
(π β π β 1)!π§πβπβ1+ β
π!
(π β π β 1)!πππ§πβπβ1
β
π=π+1
+ β π!
(π β π β 1)!ππ(π§)πβπβ1
β
π=π
+ πΎπ!
(π β π β 2)!π§πβπβ1+ β
πΎπ!
(π β π β 2)!πππ§πβπβ1
β
π=π+1
+ β πΎπ!
(π β π β 2)!ππ(π§)πβπβ1
β
π=π
)
+(1 β πΌ) ( π!
(π β π β 1)!π§πβπβ1+ β
π!
(π β π β 1)!πππ§πβπβ1
β
π=π+1
+ β π!
(π β π β 1)!ππ(π§)πβπβ1
β
π=π
+ πΎπ!
(π β π β 2)!π§πβπβ1+ β
πΎπ!
(π β π β 2)!πππ§πβπβ1+
β
π=π+1
β πΎπ!
(π β π β 2)!ππ(π§)πβπβ1
β
π=π
)|
= | π!
(π β π β 1)![((π β π β 1)(π½πππ(1 β πΎ) + (1 β πΌ)πΎ) + π β π + 1 β πΌ)] π§πβπβ1
+ β π!
(π β π β 1)![(π β π β 1)(π½πππ(1 β πΎ) + (1 β πΌ)πΎ) + π β π + 1 β πΌ]πππ§πβπβ1
β
π=π+1
+ β π!
(π β π β 1)![(π β π β 1)(π½πππ(1 β πΎ) + (1 β πΌ)πΎ) + π β π + 1 β πΌ]ππ(π§)πβπβ1
β
π=π
|
β₯ π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) + (1 β πΌ)πΎ) + π β π + 1 β πΌ]|π§|πβπβ1
β β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) + (1 β πΌ)πΎ) + π β π + 1 β πΌ]|ππ||π§|πβπβ1
β
STUDY OF A NEW CLASS FOR HIGHER-ORDER DERIVATIVES OF HARMONIC MULTIVALENT β¦
+ β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) + (1 β πΌ)πΎ) + π β π + 1 β πΌ]|ππ||π§|πβπβ1
β
π=π
.
Similarly
|πΏ(π§) β (1 + πΌ)π(π§)|
= | βπ!
(π β π β 1)![(π β π β 1)(π½πππ(πΎ β 1) + (1 + πΌ)πΎ) β π + π + 1 + πΌ]π§πβπβ1
+ β π!
(π β π β 1)![(π β π β 1)(π½πππ(1 β πΎ) + (1 + πΌ)πΎ) + π β π β 1 β πΌ]πππ§πβπβ1
β
π=π+1
+ β π!
(π β π β 1)![(π β π β 1)(π½πππ(1 β πΎ) β (1 + πΌ)πΎ) + π β π β 1 β πΌ]ππ(π§)πβπβ1
β
π=π
|
β€ π!
(π β π β 1)![(π β π β 1)(π½(πΎ β 1) + (1 + πΌ)πΎ) β π + π + 1 + πΌ]|π§|πβπβ1
+ β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β (1 + πΌ)πΎ) + π β π β 1 β πΌ] |ππ||π§|πβπβ1
β
π=π+1
+ β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β (1 + πΌ)πΎ) + π β π + 1 β πΌ]|ππ||π§|πβπβ1
β
π=π
.
Then
|πΏ(π§) + (1 β πΌ)π(π§)| β |πΏ(π§) β (1 + πΌ)π(π§)|
β₯ 2π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β β 2π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]|ππ|
β
π=π+1
β β 2π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]|ππ| β₯
β
π=π
0.
The harmonic multivalent function
π(π§) = π§π+ β (π β π β 1)! π₯π
π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]π§π
β
π=π+1
+ β (π β π β 1)! π¦π
π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ](π§)π
β
π=π
where
ο₯
ο₯
+ =p 1
n
|π₯π| +
ο₯
ο₯
=p n
|π¦π| =(πβπβ1)!π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ], show that
the coefficient bound given by (2.1) is sharp. The functions of the form (2.2) are in the class
π΄πΎβ(π, π, πΎ, πΌ, π½), because
β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π=π+1
Γ
Γ (π β π β 1)! |π₯π|
π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
+ β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π=π
Γ
Γ (π β π β 1)! |π¦π|
π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
= β |π₯π|
β
π=π+1
+ β|π¦π|
β
π=π
= π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ].
The restriction placed in Theorem 2.1 on the moduli of the coefficients of π = β + π enables
us to conclude for arbitrary rotation of the coefficients of π that the resulting functions would
still be harmonic multivalent and π β π΄πΎβ(π, π, πΎ, πΌ, π½).
In the following theorem, it is shown that the condition (2.1) is also necessary for functions in
π΄πβ(π, π, πΎ, πΌ, π½).
Theorem 2.2. Let π = β + π with β and π are given by (1.2) . Then π β π΄πβ(π, π, πΎ, πΌ, π½) if and
only if
β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]ππ
β
π=π+1
+ β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]ππ
β
π=π
β€π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
(π β π β 1)! , (2.3)
STUDY OF A NEW CLASS FOR HIGHER-ORDER DERIVATIVES OF HARMONIC MULTIVALENT β¦
Proof. Since π΄πβ(π, π, πΎ, πΌ, π½) β π΄πΎβ(π, π, πΎ, πΌ, π½), we only need to prove the "only if " part of
the theorem. Assume that π β π΄πβ(π, π, πΎ, πΌ, π½). Then by (1.3), we have
π π {π§π(π+2)(π§) + π(π+1)(π§)
π(π+1)(π§) + πΎπ§π(π+2)(π§)(1 + π½πππ) β π½πππ} > πΌ,
This is equivalent to
π π {(1 + π½π
ππ) (π§π(π+2)(π§) + π(π+1)(π§)) β π½πππ(π(π+1)(π§) + πΎπ§π(π+2)(π§))
π(π+1)(π§) + πΎπ§π(π+2)(π§) }
= π π {
π!
(π β π β 1)! [(π β π β 1)(π½πππ(1 β πΎ) β πΌπΎ) + π β π β πΌ]π§πβπβ1 π!
(π β π β 1)! (1 + πΎ(π β π β 1))π§πβπβ1β ββπ=π+1(π β π β 1)! (1 + πΎ(π β π β 1))ππ! ππ§πβπβ1
βββπ=π+1(π β π β 1)! [(π β π β 1)(π½ππ! ππ(1 β πΎ) β πΌπΎ) + π β π β πΌ]πππ§πβπβ1
βββ (π β π β 1)! (1 + πΎ(π β π β 1))ππ! π(πΜ )πβπβ1 π=π
β β (π β π β 1)! [(π β π β 1)(π½ππ! ππ(1 β πΎ) β πΌπΎ) + π β π β πΌ]π
π(πΜ )πβπβ1
β
π=π
} β₯ 0. (2.4)
The above required condition (2.4) must hold for all values of π§ in π. Upon choosing the values
of π§ on the positive real axis where 0 β€ π§ = π < 1, we must have
π π { π!
(π β π β 1)! [π β π β πΌ β πΌπΎ(π β π β 1)]ππβπβ1β ββπ=π+1(π β π β 1)! [π β π β πΌ β πΌπΎ(π β π β 1)]ππ! π ππβπβ1
π!
(π β π β 1)! (1 + πΎ(π β π β 1))ππβπβ1β ββπ=π+1(π β π β 1)! (1 + πΎ(π β π β 1))ππ! π ππβπβ1
β ββπ=π(π β π β 1)! [π β π β πΌ β πΌπΎ(π β π β 1)]ππ! π ππβπβ1βπ½πππ[(π β π β 1)! (π β π β 1)(πΎ β 1)ππ! πβπβ1
β ββπ=π(π β π β 1)! (1 + πΎ(π β π β 1))ππ! π ππβπβ1
β βπ=π+1β (π β π β 1)! (π β π β 1)(1 β πΎ)ππ! π ππβπβ1+ ββπ=π(π β π β 1)! (π β π β 1)(1 β πΎ)ππ! π ππβπβ1]
} β₯ 0.
Since π π(-πππ) β₯ β|πππ| = β1, and letting π βΆ 1β, the above inequality reduces to
π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ] β ββπ=π+1(π β π β 1)!π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]ππ
π!
β ββπ=π+1(π β π β 1)!π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]ππ
β₯ 0.
This gives (2.3), and the poof is complete.
Next, we establish the distortion bounds for the function in π΄πβ(π, π, πΎ, πΌ, π½) which yields a
covering result for this class.
Theorem 2.3. Let π β π΄πβ(π, π, πΎ, πΌ, π½). Then for |π§| = π < 1, we have
|π(π§)| β₯ (1 β ππ)ππβ (π β π)[(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ](1 β ππ)
(π + 1)[(π β π)(π½(1 β πΎ) + 1 β πΌπΎ) + 1 β πΌ] (2.5)
and
|π(π§)| β€ (1 + ππ)ππ+
(π β π)[(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ](1 β ππ)
(π + 1)[(π β π)(π½(1 β πΎ) + 1 β πΌπΎ) + 1 β πΌ] . (2.6)
Proof. Assume that π β π΄πβ(π, π, πΎ, πΌ, π½). Then by (2.3), we get
|π(π§)| = |π§πβ β π ππ§π
β
π=π+1
β β ππ(π§)π
β
π=π
| β₯ (1 β ππ)ππβ β (ππ + ππ)ππ
β
π=π+1
β₯ (1 β ππ)ππβ β (ππ+ ππ)ππ+1
β
π=π+1
= (1 β ππ)ππβ (π β π)!
(π + 1)! [(π β π)(π½(1 β πΎ) + 1 β πΌπΎ) + 1 β πΌ]Γ
Γ β (π + 1)! [(π β π)(π½(1 β πΎ) + 1 β πΌπΎ) + 1 β πΌ] (π β π)!
β
π=π+1
(ππ+ ππ)ππ+1
β₯ (1 β ππ)ππβ (π β π)!
(π + 1)! [(π β π)(π½(1 β πΎ) + 1 β πΌπΎ) + 1 β πΌ]
β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π=π+1
(ππ+ ππ)ππ+1
β₯ (1 β ππ)ππβ(π β π)[(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ](1 β ππ)
(π + 1)[(π β π)(π½(1 β πΎ) + 1 β πΌπΎ) + 1 β πΌ] .
Relation (2.6) can be proved by using the similar statements. So the proof is completes.
The following covering result follows from the inequality (2.5) of theorem 2.3.
STUDY OF A NEW CLASS FOR HIGHER-ORDER DERIVATIVES OF HARMONIC MULTIVALENT β¦ {π€: |π€| < (1 β ππ) β
(π β π)[(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ](1 β ππ)
(π + 1)[(π β π)(π½(1 β πΎ) + 1 β πΌπΎ) + 1 β πΌ] } β π(π)
In the next result, we discuss extreme points of π΄πβ(π, π, πΎ, πΌ, π½).
Theorem2.4. Let π be given by (1.2). Then π β π΄πβ(π, π, πΎ, πΌ, π½) if and only if π can be
expressed as
π(π§) = β(ππβπ(π§) + ππππ(π§)) β
π=π
, (π§ β π),
where βπ(π§) = π§π,
βπ(π§) = π§πβ π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]π§π, π = π + 1, π + 2, β¦
and
ππ(π§) = π§πβ
π! (π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
π! (π β π β 1)π, π = π, π + 1, β¦ ,
β(ππ + ππ) β
π=π
= 1, (ππ β₯ 0, ππ β₯ 0).
Inparticular , the extreme points of π΄πβ(π, π, πΎ, πΌ, π½) are {βπ} and {ππ}.
Proof: Assume that π can be expressed by (2.7). Then, we have
π(π§) = β(ππβπ(π§) + ππππ(π§)) β
π=π
= β(ππ+ ππ)π§π β
π=π
β β π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]
β
π=π+1
πππ§π
β β π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ] π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]
β
π=π
ππ(π§Μ )π
= π§πβ β π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]
β
π=π+1
πππ§π
β β π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ] π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]
β
π=π
ππ(π§Μ )π.
β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π =π+1
Γ
Γ π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ] π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]ππ
+ β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π=π
Γ
Γ π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ] π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]ππ
=π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
(π β π β 1)! (β(ππ+ ππ)
β
π=π
β ππ)
= π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
(π β π β 1)! (1 β ππ)
β€ π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
(π β π β 1)! ,
and so π β π΄πβ(π, π, πΎ, πΌ, π½).
Conversely, let π β π΄πβ(π, π, πΎ, πΌ, π½). putting
ππ=π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ] ππ, (π = π + 1, π + 2, β¦ )
and
ππ =
π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]
π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ] ππ, (π = π, π + 1, β¦ ).
We define ππ = 1 β ββπ=π+1ππβ ββπ=πππ. There fore
π(π§) = π§πβ β π
ππ§πβ β ππ β
π=π
(π§Μ )π
β
π=π+1
= π§πβ β π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]πππ§π
β
π =π+1
β β π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
π! (π β π β 1)! [(π β π β 1)(π½(1 β πΎ) + 1 β πΌπΎ) + π β π β πΌ]ππ(π§Μ )π
β
π=π
= π§πβ β (π§πβ β
π(π§))ππ β β (π§πβ ππ(π§))ππ β
π =π β
STUDY OF A NEW CLASS FOR HIGHER-ORDER DERIVATIVES OF HARMONIC MULTIVALENT β¦ = (1 β β ππβ β ππ
β π =π β π =π+1 ) π§π+ β π πβπ(π§) β β ππ ππ(π§) β π =π β π =π+1 = ππβπ(π§) + β ππβπ(π§) β β ππ ππ(π§) β π =π β π =π+1 = β (ππβπ(π§) + ππ ππ(π§)) β π =π ,
and this completes the proof of theorem 2.4.
Theorem 2.5. The class π΄πβ(π, π, πΎ, πΌ, π½) is closed under convex combinations.
Proof. For π = 1,2,3, β¦, let ππ β π΄πβ(π, π, πΎ, πΌ, π½) where ππ is given by
ππ(π§) = π§πβ β π π,ππ§πβ β ππ,π β π=π (π§Μ )π β π=π+1 .
Then by (2.3), we have
β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π =π+1
ππ,π
+ β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π=π
ππ,π
β€ π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
(π β π β 1)! (2.8)
For ββπ=1 ππ = 1 , 0 β€ ππ β€ 1, the convex combination of ππ may be written as
β ππ β π=1 ππ(π§) = π§πβ β (β ππ β π=1 ππ,π) ππβ β (β ππ β π=1 ππ,π) β π=π (π§Μ )π β π=π+1 .
Then by (2.8), we have
β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β π =π+1 (β ππ β π=1 ππ,π) + β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β π=π (β ππ β π=1 ππ,π) = β ππ β π=1 { β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π =π+1
+ β π!
(π β π β 1)![(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
β
π=π
ππ,π}
β€ β ππ
β
π=1
π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
(π β π β 1)!
= π! [(π β π β 1)(π½(1 β πΎ) β πΌπΎ) + π β π β πΌ]
(π β π β 1)! .
Therefore
β ππ β
π=1
ππ(π§) β π΄πβ(π, π, πΎ, πΌ, π½).
This completes the proof.
Conflict of Interests
The authors declare that there is no conflict of interests.
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