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ACTA UNIVERSITATIS APULENSIS

No 19/2009

NEIGHBORHOOD AND PARTIAL SUM PROPERTY FOR
UNIVALENT HOLOMORPHIC FUNCTIONS IN TERMS OF
KOMATU OPERATOR

Shahram Najafzadeh and Ali Ebadian

Abstract. In the present paper, we investigate some important properties of
a new class of univalent holomorphic functions by using Komatu operator. For
example coefficient estimates, extreme points, neighborhoods and partial sums.
2000 Mathematics Subject Classification: 30C45, 30C80.
1. Introduction and preliminaries
Let S denote the class of functions f that are analytic in the open unit disk
D = {z ∈ C : |z| < 1} of the form
f (z) = z +

∞
X

ak z k ,

(1)

k=2

and consider the subclass T consisting of functions f which are univalent in D and
are of the form
∞
f (z) = z −

X

|ak |z k .

(2)

k=2

For α ≥ 0, 0 ≤ β < 1, c ≥ −1 and δ ≥ 0, we let Scδ (α, β) be the subclass of S
consisting of functions of the form (1) and satisfying the condition
Re

(

Kcδ (f )
z[Kcδ (f )]′

)



 Kδ (f )



c
> α
−

1
 + β.
 z[Kcδ (f )]′


The operator Kcδ (f ) is the Komato operator [3] defined by
Kcδ (f ) =

Z

0

1

1
(c + 1)δ c
t log
Γ(δ)
t


81

δ−1

f (tz)
dt.
t

(3)

Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...

We also let
T Scδ (α, β) = Scδ (α, β) ∩ T .

(4)

By applying a simple calculation for f ∈ Scδ (α, β) we get
Kcδ (f )

=z−


∞ 
X
c+1 δ

k=2

c+k

|ak |z k .

(5)

The family T Scδ (α, β) is of special interest for contains many well-known as well as
new classes of analytic functions. For example T Sc0 (0, β) is the family of starlike
functions of order at most β1 .
In our present investigation, we need the following elementary Lemmas.

Lemma 1.1. If α ≥ 0, 0 ≤ β < 1 and γ ∈ R then Re ω > α|ω − 1| + β if and
only if Re [ω(1 + αeiγ ) − αeiγ ] > β where ω be any complex number.
Lemma 1.2. With the same condition in Lemma 1.1, Re ω > α if and only if
|ω − (1 + α)| < |ω + (1 − α)|.
The main aim of this paper is to verify coefficient bounds and extreme points
of the general class T Scδ (α, β). Furthermore, we obtain neighborhood property for
functions in T Scδ (α, β). Also partial sums of functions in the class Scδ (α, β) are
obtained.
2.Coefficient bounds for Scδ (α, β)
In this section we find a necessary and sufficient condition and extreme points
for functions in the class T Scδ (α, β).
Theorem 2.1. If


∞
X
[(1 + α) − k(α + β)] c + 1 δ

k=2

(1 − β)

c+k

ak < 1,

(6)

then f (z) ∈ Scδ (α, β).
Proof. Let (6) hold, we will show that (3) is satisfied and so f (z) ∈ Scδ (α, β). By
Lemma 1.2, it is enough to show that
|ω − (1 + α|ω − 1| + β)| < |ω + (1 − α|ω − 1| − β)|,
where ω =
R

Kcδ (f )
.
z[Kcδ (f )]′

By letting B =

z[Kcδ (f )]′
|z[Kcδ (f )]′ |

and by using (5) we may write

= |ω + 1 − β − α|ω − 1||
∞

X
1
c+1

=
2z
−
βz
−
[1 + (1 − β)k + α − αak ]


|z[Kcδ (f )]′ | 
c
+k
k=2





82

δ



ak z  .

k

Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...

This implies that
∞
X
|z|
c+1
R>
2
−
β
−
[k + (1 + α) − k(α + β)]
|z[Kcδ (f )]′ |
c
+k
k=2

"



δ

#

ak .

Similarly, if L = |ω − 1 − α|ω − 1| − β| we get
∞
X
|z|
c+1
β
+
L<
[−k + (1 + α) − k(α + β)]
δ
′
|z[Kc (f )] |
c+k
k=2

"



δ

#

ak .

It is easy to verify that R − L > 0 if (6) holds and so the proof is complete.
Theorem 2.2. Let f ∈ T . Then f is in T Kcδ (α, β) if and only if


∞
X
[(1 + α) − k(α + β)] c + 1 δ

1−β

k=2

c+k

ak < 1.

Proof. Since T is the subclass of S and T Scδ (α, β) = Scδ (α, β) ∩ T , and using
Theorem 2.1, we need only to prove the necessity of theorem. Suppose that f ∈
Kcδ (f )
T Kcδ (α, β). By Lemma 1.1, and letting ω = z[K
δ (f )]′ in (3) we obtain
c

Re (ω(1 + αeiγ ) − αeiγ ) > β
or



P∞  c+1 δ

zk

δ

zk



z − k=2 c+k ak


 (1 + αeiγ ) − αeiγ − β 

Re 
δ


 > 0,
P∞
c+1
ak z k−1
z 1 − k=2 k c+k

then


1 − β −

Re 


P∞

k=2 (1

− βk)




c+1
c+k

1−

ak

P∞



δ

rk

k=2 k

−

αeiγ

c+1
c+k

δ

P∞

k=2 (1

ak z k−1

− k)





c+1
c+k

δ

ak

z k−1




 > 0.


The above inequality must hold for all z in D. Letting z = re−iθ where 0 ≤ r < 1
we obtain


1 − β −

Re 


P∞

k=2 (1

− βk)




c+1
c+k

1−

P∞

ak

k=2 k



−

αeiγ

c+1
c+k

83

δ

P∞

k=2 (1

ak rk−1



− k)



c+1
c+k

δ

ak

rk−1




 > 0.


Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...
By letting r → 1 through half line z = re−iθ (0 ≤ r < 1) and the mean value
theorem we have
Re

"

Therefore

∞
X

c+1
[(1 − βk) − α(1 − k)]
(1 − β) −
c
+k
k=2


∞
X

c+1
[(1 − βk) + α(1 − k)]
c
+k
k=2

This implies that



δ

(1 − β)

#

ak ] > 0 .

ak < 1 − β.



∞
X
(1 + α) − k(α + β) c + 1 δ

k=2

δ

c+k

ak < 1,

and the proof is complete.
Theorem 2.3. Let f1 (z) = z and
fk (z) = z −

1−β
[(1 + α) − k(α + β)]



c+k
c+1

δ

z k , k ≥ 2.

(7)

Then f ∈ T Scδ (α, β) if and only if it can be expressed in the form f (z) = ∞
k=1 ηk fk (z)
P
δ (α, β) are
where ηk ≥ 0 and ∞
η
=
1.
In
particular,
the
extreme
points
of
T
S
k
c
k=1
the functions defined by (7).
Proof. Let f be expressed as in the above theorem. This means that we can
write
P

f (z)

=

∞
X

ηk fk (z) = η1 f1 (z) +

k=1
∞
X

ηk z −

k=2
∞
X

ηk −

∞
X

∞
X

(1 − β)
[(1 + α) − k(α + β)]
k=2

(1−β)
[(1+α)−k(α+β)]

c+k
c+1



c+k
c+1

1−β

δ

ηk z k

ηk . Therefore f ∈ T Scδ (α, β) since
tk



c+1
c+k

δ

=

∞
X

ηk = 1 − η1 < 1.

k=2

Conversely, suppose that f ∈ T Scδ (α, β). Then by (6), we have
ak <

δ

tk z k ,

∞
X
[(1 + α) − k(α + β)]

k=2



k=2

k=1

where tk =

ηk fk (z)

k=2

= η1 z +
=z

∞
X

1−β
[(1 + α) − k(α + β)]
84



c+k
c+1

δ

, k ≥ 2.

Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...

So
f (z)

∞
X

1−β
=z−
[(1
+
α)
− k(α + β)]
k=2
=z−
=



c+k
c+1

δ

ηk z k

∞
X

ηk (z − fk (z))
k=2
!
∞
∞
X
X

1−

ηk z −

= η1 z −

k=2
∞
X

ηk fk (z)

k=2

ηk fk (z) =

k=2

∞
X

ηk fk (z).

k=1

This completes the proof.
3.Neighborhood Property
In this section we study neighborhood property for functions in the class T Scδ (α, β).
This concept was introduced by Goodman [2] and Ruscheweyh [6]. See also [1], [4],
[5], and [7].
Definition 3.1. For functions f belong f to S of the form (1) and γ ≥ 0, we
define η − γ-neighborhood of f by
Nγη (f ) = {g(z) ∈ S : g(z) = z +

∞
X

∞
X

bk z k ,

k=2

k η+1 |ak − bk | ≤ γ},

k=2

where η is a fixed positive integer.
By using the following lemmas we will investigate the η − γ-neighborhood of
functions in T Scδ (α, β).
P
k
Lemma 3.2. Let m ≥ 0 and −1 ≤ θ < 1. If g(z) = z + ∞
k=2 bk z satisfies
∞
X

k ρ+1 |bk | ≤

k=2

1−θ
,
1+α

then g(z) ∈ Scρ (α, θ).
Proof. By using of Theorem 2.1, it is sufficient to show that
(1 + α) − k(α + θ)
1−θ
But

(1 + α) − k(α + θ)
1−θ





ρ+1
ρ+k

ρ+1
ρ+k
85

δ

δ

≤

=

k ρ+1
(1 + α).
1−θ

1+α
1−θ



ρ+1
ρ+k

δ

.

Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...

Therefore it is enough to prove that
Q(k, ρ) =



ρ+1 δ
ρ+k
k ρ+1



≤ 1.

The result follows because the last inequality holds for all k ≥ 2.
P
k
Lemma 3.3. Let f (z) = z − ∞
k=2 ak z ∈ T , − 1 ≤ β < 1, α ≥ 0 and ǫ ≥ 0. If
f (z)+ǫz
∈ T Scδ (α, β) then
1+ǫ
∞
X

k ρ+1 ak ≤

k=2

2ρ+1 (1 − β)(1 + ǫ)
(1 − α − 2β)



c+2
c+1

δ

where either ρ = 0 and c ≥ o or ρ = 1 and 1 ≤ c ≤ 2. The result is sharp with the
extremal function
f (z) = z −
Proof. Letting g(z) =

(1 − β)(1 + ǫ)
(1 − α − 2β)

f (z)+ǫz
1+ǫ

c+2
c+1

δ

z 2 , (z ∈ D).

we have

g(z) = z −
In view of theorem 2.3, g(z) =



∞
X
ak

1+ǫ
k=2

P∞

k=1 ηk gk (z)

z k , (z ∈ D).
P∞

where ηk ≥ 0,

(1 − β)(1 + ǫ)
g1 (z) = z and gk (z) = z −
[(1 + α) − k(α + β)]



= 1,

k=1 ηk

c+k
c+1

δ

z k (k ≥ 2).

So we obtain
= η1 z +

g(z)

=z−

∞
X

ηk

k=2
∞
X

ηk

k=2

Since ηk ≥ 0 and
∞
X

k=2

P∞

k=2 ηk

k

ρ+1

"

(1 − β)(1 + ǫ)
z−
[(1 + α) − k(α + β)]

(1 − β)(1 + ǫ)
[(1 + α) − k(α + β)]





c+k
c+1

δ

δ !

zk .

c+k
c+1

δ #

c+k
c+1

z

k

#

≤ 1, it follows that

ak ≤ sup k
k≥2

ρ+1

"

(1 − β)(1 + ǫ)
[(1 + α) − k(α + β)]



.

Since whenever ρ = 0 and c ≥ 0 or ρ = 1 and 1 ≤ c ≤ 2 we conclude
W (k, ρ, α, β, ǫ, c, δ) = k ρ+1

"

(1 − β)(1 + ǫ)
[(1 + α) − k(α + β)]
86



c+k
c+1

δ #

,

Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...

is a decreasing function of k, the result will follow. So the proof is complete.
Theorem 3.4. Let either ρ = 0 and c ≥ 0 or ρ = 1 and 1 ≤ c ≤ 2. Suppose
−1 ≤ β < 1, and
−1 ≤ θ <

(1 − α − 2β)(c + 1)δ − 2ρ+1 (1 − β)(1 + ǫ)(c + 2)δ (1 + α)
,
(1 − α − 2β)(c + 1)δ (1 + α)

∈ T Scδ (α, β). Then the η − γ-neighborhood of f is the subset
f (z) ∈ T and f (z)+ǫz
1+ǫ
of Scη (α, θ), where
γ=

(1 − θ)(1 − α − 2β)(c + 1)δ − 2η+1 (1 − β)(1 + ǫ)(c + 2)δ (1 + α)
.
(1 − α − 2β)(c + 1)δ (1 + α)

The result is sharp.
P
P∞
k
k
η
Proof. For f (z) = z − ∞
k=2 |ak |z , let g(z) = z +
k=2 bk z be in Nγ (f ). So by
Lemma 3.3, we have
∞
X

k η+1 |bk |

k=2

=

∞
X

k η+1 |ak − bk − ak |

k=2

2η+1 (1 − β)(1 + ǫ)
≤γ+
(1 − α − 2β)



c+2
c+1

δ

.

By using Lemma 3.2, g(z) ∈ Scη (α, β) if
γ+

2η+1 (1 − β)(1 + ǫ)
(1 − α − 2β)



c+2
c+1

δ

≤

1−θ
.
1+α

That is,
γ≤

(1 − θ)(1 − α − 2β)(c + 1)δ − 2η+1 (1 − β)(1 + ǫ)(c + 2)δ (1 + α)
(1 − α − 2β)(c + 1)δ (1 + α)

and the proof is complete.
4.Partial Sums
In this section we verify some properties of partial sums of functions in the class
(see [5] and [8])
Theorem 4.1. Let f (z) ∈ Scδ (α, β), and define the partial sums f1 (z) and fn (z)

Scδ (α, β).
by

f1 (z) = z and fn (z) = z +

n
X

k=2

87

ak z k

(n ∈ N , n > 1).

(8)

Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...

If

∞
X

ck |ak | ≤ 1

(9)

k=2

where

[(1 + α) − k(α + β)]
ck =
1−β

Then fk (z) ∈ Scδ (α, β). Moreover
Re



f (z)
fn (z)



Re

>1−


1
cn+1

fn (z)
f (z)



c+1
c+k

δ

(10)

.

(z ∈ D, n ∈ N )

,

>



cn+1
.
1 + cn+1

(11)
(12)

Proof. It is easy to show that f1 (z) = z ∈ Scδ (α, β). So by Theorem 3.3, and
condition (9), we have N1η (z) ⊂ Scδ (α, β), so fk ∈ Scδ (α, β). Next, for the coefficient
ck it is easy to show that
ck+1 > ck > 1.
Therefore by using (9) we obtain
n
X

|ak | + cn+1

k=2

∞
X

|ak | ≤

k=n+1

∞
X

ck |ak | ≤ 1.

(13)

k=2

By putting
h1 (z)

1
f (z)
− 1−
fn (z)
cn+1


f (z)
= 1 + cn+1
−1
fn (z)

= cn+1





k
z+ ∞
k=2 ak z
Pn
−1
z + k=2 ak z k

!

P

= 1 + cn+1
= 1 + cn+1



"

1+

P∞

= 1 + cn+1

k−1
1+ ∞
k=2 ak z
Pn
−1
1 + k=2 ak z k−1

− 1 − nk=2 ak z k−1
P∞
1 + k=2 ak z k−1

k=2 ak z

k−1

P

P

!

#

ak z k−1
cn+1 ∞
Pk=n+1
=1+
,
n
1 + k=2 ak z k−1
P

and using (13), for all z ∈ D we have

P


 cn+1 ∞

ak z k−1
P


 
Pk=n+1
n
 h1 (z) − 1  

cn+1 ∞
1+ k=2 ak z k−1
k=n+1 |ak |


=
P
P
P
≤
≤ 1,
∞
n
 h (z) + 1  

k−1
ak z
2 − 2 k=2 |ak | − cn+1 ∞
|ak |
1
 2 + cn+1 Pk=n+1

k=n+1
n


1+
ak z k−1
k=2

88

Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...

which proves (11). Similarly, if we put
h2 (z)

and using (13) we obtain

cn+1
fn (z)
−
(1 + cn+1 )
f (z)
1 + cn+1
P
(1 + cn+1 ) ∞
ak z k−1
P∞ k=n+1k−1
,
=1−
1 + k=2 ak z

=







 h2 (z) − 1 


 h (z) + 1  ≤ 1, (z ∈ D),
2

which yields the condition (12). So the proof is complete.

References
[1] Altintas and S. Owa, Neighborhoods of certain analytic functions with negative
coefficients, Int. J. Math. and Math. Sci. 19 (1996), 797-800.
[2] A. W. Goodman, Univalent functions with analytic curves, Proc. Amer.
Math. Soc. 8 (1957), 598-601.
[3] Y. Komato, On analytic prolongation of a family of operators, Mathematica
(Cluj). 39 (55) ( 1990), 141-145.
[4] Shahram Najafzadeh and S. R. Kulkarni, Study of certain properties of univalent functions based on Ruscheweyh derivative, Southeast Asian Bulletin of Mathematics 32 (2008), 478-488.
[5] Thomas Rosy, K. G. Subramanian and G. Murugusundaramoorthy, Neighborhoods and partial sums of starlike functions based on Ruscheweyh derivative, J.
Ineq. pure and appl. Math. 4 (4) (2003), art. 64.
[6] S. Ruscheweyh, Neighborhoods of univalent functions, Proc. Amer. Math.
Soc. 81 (4) (1981), 521-527.
[7] H. Silverman, Neighbohoods of classes of analytic functions, Fareast. J. Math.
Sci. 3 (2) (1995), 165-169.
[8] H. Silverman, Partial sums of starlike and convex functions, J. Math. Anal.
and Appl., 209 (1997), 221-227.
Shahram Najafzadeh
Department of Mathematics
University of Maragheh
Maragheh, Iran
email:najafzadeh1234@yahoo.ie
89

Sh. Najafzadeh, A. Ebadian - Neighborhood and partial sum property ...

Ali Ebadian
Department of Mathematics
Urmia University
Faculty of Science
Urmia, Iran
email:a.ebadian@urmia.ac.ir

90