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1
On a Subclass of Multivalent Functions
Defined by Differential Operator
by
Zain AL-abideen Abbas Nassir
Department of Mathematics, College of Computer Science
and Mathematics, University of AL-Qadisiya, Diwaniya, Iraq.
E-mail:[email protected]
Abstract: In this paper, we introduce a class of multivalent functions defined by
differential operator. We obtain some geometric properties, like, coefficient
inequalities, distortion and growth theorem, closure theorem, extreme points and
Hadamard product (or convolution).
Keyword: Multivalent functions, Differential operator, Extreme points and
Hadamard product.
2010 Mathematics Subject classification: 30C45, 30C50.
2
1. Introduction
Let π(π) denote the class of functions of the form:
π π§ = π§π β πππ§π
β
π+1
, (ππ β₯ 0, π β₯ π + 1, π β β = 1,2, β¦ , (1.1)
which are π- valent in the unit disk π = {π§ β β: π§ < 1}.
For a function π(π§) in the class π(π) we define
π·π0π π§ = π π§ ,
π·π1π π§ = π·ππ π§ =
π§
ππ β² π§
π·π2π π§ = π· π·ππ π§ ,
π·πππ π§ = π· π·π
πβ1π π§ = π§π β
π
π
π
πππ§π
β
π=π+1
, π β β. 1.2
For π = 1, the differential operator π·π was introduced by salagean [4].
Let π β π (π§) denote the Hadamard product of the functionsπ(π§) and π(π§),
that is, if π(π§) is given by (1.1) and π(π§) is given by
π π§ = π§π β πππ§π
β
π=π+1
, ππ β₯ 0 , (1.3)
then
π β π π§ = π§π β πππππ§π
β
π=π+1
. (1.4)
For a function π(π§) of the form (1.1), we define the class πππ
(π½, πΌ, π’, π£) as
follows:
3
Definition (1.1): Let π(π§) β π(π) be given by (1.1). Then the class
πππ
(π½, πΌ, π’, π£) is
πππ π½, πΌ, π’, π£
= π β π π : π§ π·π
ππ π§
β²β²+ 2 β π π·π
ππ π§
β²
π’ + π£ + 2π½ π·πππ π§
β²+ π½π§ π·π
ππ π§
β²β² < πΌ, 0
β€ π½ < 1,0 < π’ β€ 1,0 β€ π£ < 1, π, π β β and 0 < πΌ < 1 1.5
The same properties have been found for other classes in [2], [3] and [1].
2. Coefficient inequality
The following theorem has given a necessary and sufficient condition for a
function π to be in the class πππ
(π½, πΌ, π’, π£).
Theorem (2.1): Let a function π(π§) β π π . Then π(π§) β πππ
(π½, πΌ, π’, π£) if
and only if
π[π + 1 + πΌ π½ π + 1 + π’ + π£ β π] π
π
π
ππ
β
π=π+1
β€ π 1 + πΌ π½ π + 1 + π’ + 1 , (2.1)
where 0 β€ π½ < 1,0 < π’ β€ 1, π, π β β and 0 < πΌ < 1.
Proof: Suppose that the inequality (2.1) holds true and π§ = 1. Then, we have
π§ π·πππ π§
β²β²+ 2 β π π·π
ππ π§
β²
β πΌ π’ + π£ + 2π½ π·πππ π§
β²+ π½π§ π·π
ππ π§
β²β²
4
= π π β π + 1 π
π
π
πππ§πβ1
β
π=π+1
β ππ§πβ1
β πΌπ π½ π + 1 + π’ + π£ π§πβ1
β ππΌ π½ π + 1 + π’ + π£ π
π
π
πππ§πβ1
β
π=π+1
β€ π π β π + 1 π
π
π
ππ π§ πβ1
β
π=π+1
β ππ§πβ1
β πΌπ π½ π + 1 + π’ + π£ π§ πβ1
+ ππΌ π½ π + 1 + π’ + π£ π
π
π
ππ π§ πβ1
β
π=π+1
= π π + 1 + πΌ π½ π + 1 + π’ + π£ β π π
π
π
ππ
β
π=π+1
β π 1 + πΌ π½ π + 1 + π’ + π£ β€ 0,
by hypothesis.
Hence, by maximum modulus principle, π β πππ
(π½, πΌ, π’, π£). Then from (1.5),
we have
π§ π·π
ππ π§
β²β²+ (2 β π) π·π
ππ π§
β²
(π’ + π£ + 2π½) π·πππ π§
β²+ π½π§ π·π
ππ π§
β²β²
= π π β π + 1
π
π π
πππ§πβ1βπ=π+1 β ππ§πβ1
π π½ π + 1 + π’ + π£ π§πβ1 β π π½ π + 1 + π’ + π£ π
π π
πππ§πβ1βπ=π+1
.
Since π π(π§) β€ π§ for all π§ π§ β π , we get
5
π π π πβπ+1
π
π πππ π§πβ1β
π=π+1 βππ§πβ1
π π½ π+1 +π’+π£ π§πβ1β π π½ π+1 +π’+π£ π
π πππ π§πβ1β
π=π+1
< πΌ.
Letting π§ β 1β , through real values, we obtain the inequality (2.1), so the
proof is complete.
Corollary (2.1): Let π β πππ
(π½, πΌ, π’, π£). Then
ππ β€π 1 + πΌ π½ π + 1 + π’ + π£
π
π π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π , π = π + 1, π + 2, β¦ 2.2
3. Distortion and Growth Theorem
Theorem (3.1): Let the function π(π§) defined by (1.2) be in the class
πππ
(π½, πΌ, π’, π£). Then for π§ = π < 1 and π β₯ π + 1, we have
π·πππ π§ β₯ 1 +
π 1 + πΌ π½ π + 1 + π’ + π£
π + 1 πΌ π½ π + 2 + π’ + π£ + 2 π ππ , 3.1
and
π·πππ π§ β€ 1 β
π 1 + πΌ π½ π + 1 + π’ + π£
π + 1 πΌ π½ π + 2 + π’ + π£ + 2 π ππ . 3.2
The result is sharp for the function π π§ is given by
π π§ = π§π
βπ[1 + πΌ π½ π + 2 + π’ + π£ ]
π+1
π π
π + 1 [πΌ π½ π + 2 + π’ + π£ + 2]π§π+1 , π§
β π . (3.3)
Proof: Let π β πππ
(π½, πΌ, π’, π£). Then by Theorem (2.1), we get
6
π + 1
π
π
π + 1 πΌ π½ π + 2 + π’ + π£ + 2 ππ
β
π=π+1
β€ π π + 1 + πΌ π½ π + 2 + π’ + π£ β π π
π
πβ
π=π+1
β€ π 1 + πΌ π½ π + 2 + π’ + π£
or
ππ
β
π=π+1
β€π 1 + πΌ π½ π + 2 + π’ + π£
π+1
π π
π + 1 πΌ π½ π + 2 + π’ + π£ + 2 . 3.4
Hence,
π·πππ π§ β€ π§ π +
π
π
π
ππ π§ πβ
π=π+1
β€ π§ π + π + 1
π
π
π§ π+1 ππ
β
π=π+1
= ππ + π + 1
π
π
ππ+1 ππ
β
π=π+1
β€ ππ +π 1 + πΌ π½ π + 1 + π’ + π£
π + 1 πΌ π½ π + 1 + π’ + π£ + 2 ππ+1
= 1 +π 1 + πΌ π½ π + 1 + π’ + π£
π + 1 πΌ π½ π + 1 + π’ + π£ + 2 π ππ . 3.5
Similarly,
π·πππ π§ β₯ π§ π β
π
π
π
ππ π§ πβ
π=π+1
β₯ π§ π β π + 1
π
π
π§ π+1 ππ
β
π=π+1
= ππ β π + 1
π
π
ππ+1 ππ
β
π=π+1
β₯ ππ βπ 1 + πΌ π½ π + 1 + π’ + π£
π + 1 πΌ π½ π + 1 + π’ + π£ + 2 ππ+1
= 1 βπ 1 + πΌ π½ π + 1 + π’ + π£
π + 1 πΌ π½ π + 1 + π’ + π£ + 2 π ππ . 3.6
From (3.5) and (3.6), we get (3.1) and (3.2) and the proof is complete.
7
Theorem (3.2): Let the function π(π§) defined by (1.2) be in the class
πππ
(π½, πΌ, π’, π£). Then for π§ = π < 1 and π β₯ π + 1, we have
π·πππ π§
β² β₯
π
π+
π 1 + πΌ π½ π + 1 + π’ + π£
πΌ π½ π + 1 + π’ + π£ + 2 ππ , 3.7
and
π·πππ π§
β² β€
π
πβ
π 1 + πΌ π½ π + 1 + π’ + π£
πΌ π½ π + 1 + π’ + π£ + 2 ππ , 3.8
The result is sharp for the function π is given by (3.3).
Proof: The proof is similar to that of Theorem (3.1)
4. Closure Theorem
Let the functionππ π§ π = 1,2, β¦ , π be defined by
ππ π§ = π§π β ππ ,ππ§π
β
π=π+1
, ππ ,π β₯ 0 . 4.1
We shall prove the following results for the closure functions in the class
πππ π½, πΌ, π’, π£ .
Theorem (4.1): Let the functions ππ π§ π = 1,2, β¦ , π be defined by (4.1) be
in the class πππ π½, πΌ, π’, π£ . Then the function π(π§) defined by
π π§ = ππππ π§
π
π=1
, ππ β₯ 0 ,
is also in theclass πππ π½, πΌ, π’, π£ , where
ππ = 1
π
π=1
.
Proof: According to the definition of π(π§), it can be written as
8
π π§ = ππ π§π β ππ ,ππ§π
β
π=π+1
π
π=1
= πππ§π β ππππ ,ππ§
π
β
π=π+1
π
π=1
π
π=1
= π§π β ππππ ,ππ§π
π
π=1
β
π=π+1
.
Furthermore, since the functions ππ π§ (π = 1,2, β¦ , π) are in the class
πππ
(π½, πΌ, π’, π£), then
π[π + 1 + πΌ π½ π + 1 + π’ + π£ β π] π
π
π
ππ ,π
β
π=π+1
β€ π 1 + πΌ π½ π + 1 + π’ + π£ .
Hence
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π π
π
π
ππππ ,π
π
π=1
β
π=π+1
= ππ π π + 1 + πΌ π½ π + 1 + π’ + π£ β π π
π
πβ
π=π+1
ππ ,π
π
π=1
β€ π 1 + πΌ π½ π + 1 + π’ + π£ ,
which implies that π(π§) is in the class πππ
(π½, πΌ, π’, π£).
Corollary (4.1): Let the function ππ π§ (π = 1,2) defined by (4.1) be in the
classπππ
(π½, πΌ, π’, π£).
Then the function π(π§) defined by
π π§ = 1 β πΎ π1 π§ + πΎπ2 π§ , 0 β€ πΎ < 1 , is also in the class
πππ
(π½, πΌ, π’, π£).
5. Extreme points:
9
Theorem (5.1): Let ππ = π§π and
ππ = π§π βπ 1 + πΌ π½ π + 1 + π’ + π£
π
π
ππ π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π§π , π β₯ π + 1; π β β . 5.1
Then the function π(π§) is in the class πππ
(π½, πΌ, π’, π£) if and only if it can be
expressed in the form
π π§ = ππππ π§
β
π=π+1
,
where
ππ β₯ 0 and ππ
β
π=π
= 1. (5.2)
Proof: Suppose that π(π§) is expressed in the form
π π§ ππππ π§
β
π=π
= πππ§π + ππ π§π βπ 1 + πΌ π½ π + 1 + π’ + π£
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π π§π
β
π=π+1
= π§π ππ + ππ
β
π=π+1
β π 1 + πΌ π½ π + 1 + π’ + π£
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π πππ§π
β
π=π+1
= π§π β πππ§π
β
π=π+1
,
where
ππ =π 1 + πΌ π½ π + 1 + π’ + π£
π
π
ππ π + 1 + πΌ π½ π + 1 + π’ + π£ β π
ππ .
Therefore π β πππ(π½, πΌ, π’, π£), since
10
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£ π
π
β
π=π+1
= ππ
β
π=π+1
= 1 β ππ β€ 1.
Conversely, assume that π β πππ(π½, πΌ, π’, π£), then by (2.1) we may set
ππ =
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£ ππ , π β₯ π + 1
and
1 β ππ
β
π=π+1
= ππ .
Then
π π§ = π§π β πππ§π
β
π=π+1
= π§π β π 1 + πΌ π½ π + 1 + π’ + π£
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π π§π
β
π=π+1
,
= π§π β ππ(
β
π=π+1
π§π β ππ(π§)) = π§π(1 β ππ
β
π=π+1
) + ππππ π§
β
π=π+1
= π§πππ + ππππ π§
β
π=π+1
= ππππ π§
β
π=π
.
This completes the proof.
11
6. Hadamard product:
Let the function ππ π§ (π = 1,2) defined by (4.1). The Hadamard product of
the functions π1(π§) and π2(π§) is defined by
π1 β π2 π§ = π§π + ππ,1ππ,2π§π
β
π=π+1
. 6.1
Theorem (6.1): Let the functions ππ π§ π = 1,2 defined by (4.1) be in the
class πππ
(π½, πΌ, π’, π£) and π β₯ π + 1 . Then π1 β π2 π§ β πππ
(π½, πΏ, π’, π£) ,
where
πΏ β€
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π 2 β π + 1 β π π 1 + πΌ π½ π + 1 + π’ + π£ 2
π½ π + 1 + π’ + π£ π 1 + πΌ π½ π + 1 + π’ + π£ 2 β π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π 2 .
The result is sharp for the functions ππ(π§) given by
ππ π§ =π 1 + πΌ π½ π + 1 + π’ + π£
π+1
π π
π + 1 πΌ π½ π + 2 + π’ + π£ + 2 π§π+1 , π = 1,2 . 6.2
Proof: Since the functions ππ π§ (π = 1,2) belong to the class πππ
(π½, πΌ, π’, π£),
then from Theorem (2.1), we have
π
π
ππ π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£ ππ ,1
β
π=π+1
β€ 1. (6.3)
Employing the technique used earlier by Schild and Silverman [5], we need to
find the largest πΏ such that
π
π π
π π + 1 + πΏ π½ π + 1 + π’ + π£ β π
π 1 + πΏ π½ π + 1 + π’ + π£ ππ ,1ππ ,2
β
π=π+1
β€ 1. 6.4
By Cauchy β Schwarz inequality, we get
12
π
π π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£ ππ ,1ππ ,2
β
π=π+1
β€ 1. 6.5
Thus, it is sufficient to show that
π
π π
π π + 1 + πΏ π½ π + 1 + π’ + π£ β π
π 1 + πΏ π½ π + 1 + π’ + π£ ππ ,1ππ ,2 β€
π
π π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£ ππ ,1ππ ,2
that is, that
ππ ,1ππ ,2 β€ 1 + πΏ π½ π + 1 + π’ + π£ π + 1 + πΌ π½ π + 1 + π’ + π£ β π
1 + πΌ π½ π + 1 + π’ + π£ π + 1 + πΏ π½ π + 1 + π’ + π£ β π . 6.6
But from (6.5), we have
ππ ,1ππ ,2 β€π 1 + πΌ π½ π + 1 + π’ + π£
π
π π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π . 6.7
Thus it is enough to show that
π 1 + πΌ π½ π + 1 + π’ + π£
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π
β€ 1 + πΏ π½ π + 1 + π’ + π£ π + 1 + πΌ π½ π + 1 + π’ + π£ β π
1 + πΌ π½ π + 1 + π’ + π£ π + 1 + πΏ π½ π + 1 + π’ + π£ β π ,
which implies
πΏ β€
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π 2 β π + 1 β π π 1 + πΌ π½ π + 1 + π’ + π£ 2
π½ π + 1 + π’ + π£ π 1 + πΌ π½ π + 1 + π’ + π£ 2 β π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π 2 .
This completes the proof.
Theorem (6.2): Let the functions ππ π§ (π = 1,2) defined by (4.1) be in the
classπππ
(π½, πΌ, π’, π£). Then the function
π π§ = π§π β ππ ,12 + ππ ,2
2 π§π
β
π=π+1
, (6.8)
is in the class πππ
(π½, πΏ, π’, π£), where
13
πΏ β€ π + 1 β π 1 + πΌ π½ π + 1 + π’ + π£ β π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π½ π + 1 + π’ + π£ 1 + πΌ π½ π + 1 + π’ + π£ β π½ π + 1 + π’ + π£ π + 1 + πΌ π½ π + 1 + π’ + π£ β π .
The result is sharp for the functions ππ π§ (π = 1,2) given by (6.2).
Proof: From Theorem (2.1), we have
π
π
ππ π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£
2
ππ ,π2
β
π=π+1
β€
π
π
ππ π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£
β
π=π+1
ππ ,π
2
β€ 1,
it follows that
1
2
π
π
ππ π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£
2
ππ ,12 + ππ ,2
2 β€ 1
β
π=π+1
. 6.9
But π β πππ(π½, πΏ, π’, π£) if and only if
π
π
ππ π + 1 + πΏ π½ π + 1 + π’ + π£ β π
π 1 + πΏ π½ π + 1 + π’ + π£ (ππ ,1
2 + ππ ,22 )
β
π=π+1
β€ 1, 6.10
the inequality (6.10) will satisfies if
π
π
π
π π + 1 + πΏ π½ π + 1 + π’ + π£ β π
π 1 + πΏ π½ π + 1 + π’ + π£ β€
π
π
π
π π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π 1 + πΌ π½ π + 1 + π’ + π£ , π β₯ π + 1
so that
πΏ β€ π + 1 β π 1 + πΌ π½ π + 1 + π’ + π£ β π + 1 + πΌ π½ π + 1 + π’ + π£ β π
π½ π + 1 + π’ + π£ 1 + πΌ π½ π + 1 + π’ + π£ β π½ π + 1 + π’ + π£ π + 1 + πΌ π½ π + 1 + π’ + π£ β π .
This completes the proof.
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defined by using a differential operator, Bulletin of Institute of Mathematics, 5(2) (2010),
181-200.
14
[2] M. K. Aout A. O. Msotafa, Certain class of p-valent functions defined by convolution,
General Math., 20(1) (2012), 85-98.
[3] W. G. Atshan, S. R. Kulkarni, same application of generalized Ruscheweyh
derivatives involving a general fractional derivative operator to a class of analytic
functions with negative coefficient, surveys in Mathematics and its Applications, 5(2010),
35-47.
[4] G. S .SαΊ·lαΊ·gean, Subclass of univalent functions, Lecture Notes in Math.,Springer-
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