Directory UMM :Journals:Journal_of_mathematics:AUA:
ACTA UNIVERSITATIS APULENSIS
No 19/2009
CERTAIN CLASSES OF ANALYTIC FUNCTIONS OF COMPLEX
ORDER
C.Selvaraj and C.Santhosh Moni
Abstract.By making use of the Hadamard product, we define a new class of
analytic functions of complex order. Coefficient inequalities, sufficient condition and
an interesting subordination result are obtained.
2000 Mathematics Subject Classification: 30C45, 30C50.
Keywords and Phrases: analytic functions, Hadamard product, generalized hypergeometric functions, coefficient inequalities, subordinating sequence.
1. Introduction, Definitions And Preliminaries
Let A denote the class of functions of the form
f (z) = z +
∞
X
ak z k
ak ≥ 0,
(1)
k=2
which are analytic in the open disc U = {z ∈ C \ | z |< 1} and S be the class of
function f ∈ A which are univalent in U.
P∞
k
The
Hadamard
product
of
two
functions
f
(z)
=
z
+
k=2 ak z and g(z) =
P∞
z + k=2 bk z k is given by
∞
X
ak bk z k .
(2)
(bk ≥ 0 for k ≥ 2)
(3)
(f ∗ g)(z) = z +
k=2
For a fixed function g ∈ A defined by
g(z) = z +
∞
X
bk z k
k=2
m f : A −→ A by
We now define the following linear operator Dλ,
g
0
Dλ,
g f (z) = (f ∗ g)(z),
55
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
′
1
Dλ,
g f (z) = (1 − λ)(f (z) ∗ g(z)) + λ z(f (z) ∗ g(z)) ,
m−1
m
1
Dλ,
g f (z) = Dλ, g (Dλ, g f (z)).
(4)
(5)
If f ∈ A, then from (4) and (5) we may easily deduce that
m
Dλ,
g f (z) = z +
∞
X
m
1 + (k − 1)λ ak bk
k=2
zk
,
(k − 1)!
(6)
where m ∈ N0 = N ∪ {0} and λ ≥ 0.
Remark 1. It is interesting to note that several integral and differential operator
m f (z), here we list few of them.
follows as a special case of Dλ,
g
m f (z) reduces to an operator introduced recently
1. When g(z) = z/(1 − z), Dλ,
g
by F. Al-Oboudi [1].
2. Let the coefficients bk be of the form
bk =
(α1 )k−1 . . . (αq )k−1
(β1 )k−1 . . . (βs )k−1 (k − 1)!
(7)
(α1 , . . . αq , β1 , . . . βs ∈ C and βj 6= 0, −1, −2, . . . for j={1,2,. . . ,s})
m f (z) reduces to the well-known Dziok-Srivastava operaand if m = 0, then Dλ,
g
tor. It is well known that Dziok-Srivastava operator includes as its special cases
various other linear operator which were introduced and studied by Hohlov,
Carlson and Shaffer and Ruscheweyh. For details we refer to [6, 7, 8]
m f (z) generalizes the well-known operators like
Apart from these, the operator Dλ,
g
S˘
al˘agean operator [14] and Bernardi-Libera-Livingston operator.
m f (z), we define Hm (g, δ; A, B) to be the class of functions
Using the operator Dλ,
g
λ
f ∈ A satisfying the inequality
m+1
1 Dλ, g f (z)
1 + Az
1+
−1 ≺
,
m
δ Dλ, g f (z)
1 + Bz
z ∈ U,
(8)
where δ ∈ C \ {0}, A and B are arbitrary fixed numbers, −1 ≤ B < A ≤ 1, m ∈ N0 .
Remark 2. Several interesting well-known and new subclasses of analytic functions can be obtained by specializing the parameters in Hλm (g, δ; A, B). Here we list
a few of them.
56
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
z
, λ = 1 in (8), then the class Hλm (g, δ; A, B) reduces
1−z
to the well- known class
1 + Az
1 Dm+1 f (z)
m
−1 ≺
H (δ; A, B) := f : 1 +
δ Dm f (z)
1 + Bz
1. If we let g(z) =
where Dm f is the well- known S˘al˘
agean operator. The class Hm (δ; A, B) was
introduced and studied by Attiya [4].
∞
X
(α1 )k−1 . . . (αq )k−1
ak z k in (8), then we have
(β1 )k−1 . . . (βs )k−1 (k − 1)!
n=2
following class of functions
1 + Az
1 Dm+1 (α1 , β1 )f (z)
m
−1 ≺
Hλ (α1 , β1 ; A, B) := f : 1 +
δ Dm (α1 , β1 )f (z)
1 + Bz
2. If we let g(z) = z +
introduced and studied by C. Selvaraj and K.R.Karthikeyan in [17].
z
, λ = 1, m = 0 in (8), then the class Hλm (g, δ; A, B) reduces
1−z
to the well- known class of starlike functions of complex order [12].
3. If we let g(z) =
Further, we note that several subclass of analytic functions can be obtained by
specializing the parameters in Hλm (g, δ; A, B) (see for example [3, 11, 12]).
We use Ω to denote the class of bounded analytic functions w(z) in U which
satisfy the conditions w(0) = 1 and | w(z) |< 1 for z ∈ U.
2.Coefficient estimates
Theorem 1. Let the function f (z) defined by (1) be in the class Hλm (g, δ; A, B).
(a) If (A − B)2 | δ |2 > (k − 1) 2B(A − B)λ Re{δ} + (1 − B 2 )λ2 (k − 1) , let
G=
(A − B)2 | δ |2
,
(k − 1) 2B(A − B)λ Re{δ} + (1 − B 2 )λ2 (k − 1)
k = 2, 3, . . . , m − 1,
M = [G] (Gauss symbol) and [G] is the greatest integer not greater than G.
Then, for j = 2, 3, . . . , M + 2
| aj |≤
1
[1 + (j − 1)λ]m λj−1 (j − 1)!bj
57
j
Y
k=2
| (A − B)δ − (k − 2)B |
(9)
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
and for
| aj |≤
j >M +2
1
[1 + (j − 1)λ]m λj−1 (j − 1) (M + 1)! bj
M
+3
Y
| (A − B)δ − (k − 2)B | .
k=2
(10)
(b) If (A − B)2 | δ |2 ≤ (k − 1) 2B(A − B)λ Re{δ} + (1 − B 2 )λ2 (k − 1) , then
| aj |≤
(A − B) | δ |
λ (j − 1) [1 + (j − 1)λ]m bj
j ≥ 2.
(11)
The bounds in (9) and (11) are sharp for all admissible A, B, δ ∈ C \ {0} and for
each j.
Proof. Since f (z) ∈ Hλm (g, δ; A, B), the inequality (8) gives
m+1
m+1
m
m
| Dλ,
f
(z)
−
D
f
(z)
|=
[(A
−
B)δ
+
B]D
f
(z)
−
BD
f
(z)
w(z).
λ,
g
λ,
g
g
λ, g
(12)
Equation (12) may be written as
∞
X
[1 + (k − 1)λ]m λ(k − 1) bk ak z k
(13)
k=2
=
(A − B)δz +
m
[(A − B)δ − B(k − 1)λ][1 + (k − 1)λ] bk ak z
k=2
Or equivalently
j
X
∞
X
[1 + (k − 1)λ]m λ(k − 1) bk ak z k +
k=2
∞
X
k
w(z).
ck z k
k=j+1
=
j−1
X
(A−B)δz+ [(A−B)δ−B(k−1)λ][1+(k−1)λ]m bk ak z k w(z), (14)
k=2
for certain coefficients ck . Explicitly ck = [1 + (k − 1)λ]m λ(k − 1)bk ak − [(A − B)δ −
B(k − 2)λ][1 + (k − 2)λ]m bk−1 ak−1 z −1 . Since | w(z) |< 1, we have
X
∞
X
j
m
k
k
[1 + (k − 1)λ] λ(k − 1) bk ak z +
ck z
k=2
k=j+1
58
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C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
j−1
X
m
k
≤ (A − B)δz +
[(A − B)δ − B(k − 1)λ][1 + (k − 1)λ] bk ak z .
k=2
Let z = reiθ , r < 1, applying the Parseval’s formula (see [9] p.138) on both sides of
the above inequality and after simple computation , we get
j
X
[1 + (k − 1)λ]2m λ2 (k − 1)2 b2k | ak |2 r2k +
k=2
∞
X
| ck |2 r2k
k=j+1
2
2
2
≤ (A − B) | δ | r +
j−1
X
| (A − B)δ − B(k − 1)λ |2 [1 + (k − 1)λ]2m b2k | ak |2 r2k .
k=2
Let r −→ 1− , then on some simplification we obtain
[1 + (j − 1)λ]2m λ2 (j − 1)2 b2j | aj |2 ≤ (A − B)2 | δ |2
+
j−1
X
k=2
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k | ak |2
j ≥ 2.
(16)
Now the following two cases arises:
(a) Let (A − B)2 | δ |2 > (k − 1)λ 2B(A − B)Re(δ) + (1 − B 2 )λ(k − 1) , suppose
that j ≤ M + 2, then for j = 2, (16) gives
| a2 |≤
(A − B) | δ |
,
(1 + λ)m λ b2
which gives (9) for j = 2. We establish (9) for j ≤ M + 2, from (16), by
mathematical induction. Suppose (9) is valid for j = 2, 3, . . . , (k − 1). Then
it follows from (16)
[1 + (j − 1)λ]2m λ2 (j − 1)2 b2j | aj |2
≤ (A − B)2 | δ |2
+
j−1
X
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k | ak |2
j−1
X
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k
k=2
≤ (A − B)2 | δ |2
+
k=2
59
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
×
k
Y
1
2
| (A − B)δ − (n − 2)B |
[1 + (k − 1)λ]2m b2k {λk−1 (k − 1)!}2 n=2
2
2
= (A − B) | δ | +
j−1
X
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2
k=2
×
k
Y
1
2
| (A − B)δ − (n − 2)B |
{λk−1 (k − 1)!}2
n=2
= (A − B)2 | δ |2 +
| (A − B)δ − Bλ |2 −λ2
1
λ2 (1!)2
| (A − B)δ − 2Bλ |2 −4λ2
(A − B)2 | δ |2 +
1
λ4 (2!)2
(A − B)2 | δ |2 | (A − B)δ − Bλ |2
+ . . . up to (k = j − 1)
=
j
Y
1
| (A−B)δ−(k−2)B |2 .
{λj−2 (j − 2)!}2
k=2
Thus, we get
| aj |≤
1
[1 + (j − 1)λ]m λj−1 (j − 1)!bj
j
Y
| (A − B)δ − (k − 2)B |,
k=2
which completes the proof of (9).
Next, we suppose j > M + 2. Then (16) gives
[1 + (j − 1)λ]2m λ2 (j − 1)2 b2j | aj |2 ≤ (A − B)2 | δ |2
+
M
−2
X
k=2
+
j−1
X
k=M +3
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k | ak |2
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k | ak |2 .
On substituting upper estimates for a2 , a3 , . . . , aM +2 obtained above and simplifying, we obtain (10).
60
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
(b) Let (A − B)2 | δ |2 ≤ (k − 1)λ 2B(A − B)Re(δ) + (1 − B 2 )λ(k − 1) , then it
follows from (16)
[1 + (j − 1)λ]2m λ2 (j − 1)2 b2j | aj |2 ≤ (A − B)2 | δ |2
(j ≥ 2),
which proves (11).
The bounds in (9) are sharp for the functions f (z) given by
(
(A−B)δ
if B 6= 0,
z(1 + Bz) B
m
Dλ, g f (z) =
z exp(Aδz)
if B = 0.
Also, the bounds in (11) are sharp for the functions fk (z) given by
(A−B)δ
z(1 + Bz) Bλ(k−1)
if B 6= 0,
m
Dλ,
g fk (z) =
Aδ
z k−1
if B = 0.
z exp λ(k−1)
Remark 2. If we let λ = 1, g(z) =
due to Attiya [4].
z
in Theorem 1, we get the result
1−z
3. A sufficient condition for a function to be in Hλm (g, δ; A, B)
Theorem 2. Let the function f (z) defined by (1) and let
∞
X
k=2
[1 + (k − 1)λ]m (k − 1)+ | (A − B)δ − B(k − 1) | λ bk | ak |≤ (A − B) | δ | (17)
holds, then f (z) belongs to Hλm (g, δ; A, B).
Proof. Suppose that the inequality holds. Then we have for z ∈ U
m+1
m
m
| Dλ,
g f (z) − Dλ, g f (z) | − (A − B)δDλ, g f (z)−
m+1
m
B Dλ,
f
(z)
−
D
f
(z)
λ,
g
g
∞
∞
X
X
m
k
m
k
[1 + (k − 1)λ] bk ak z −
[1 + (k − 1)λ] λ(k − 1) bk ak z − (A − B)δ z +
=
k=2
k=2
B
∞
X
k=2
61
[1 + (k − 1)λ] λ(k − 1)bk ak z
m
k
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
≤
∞
X
k=2
[1 + (k − 1)λ]m (k − 1)λ+ | (A − B)δ − B(k − 1)λ | bk | ak | rk − (A − B) | δ | r.
Letting r −→ 1− , then we have
m+1
m
m
| Dλ,
g f (z) − Dλ, g f (z) | − (A − B)δDλ, g f (z)−
≤
∞
X
k=2
m+1
m
B Dλ,
g f (z) − Dλ, g f (z)
[1 + (k − 1)λ]m (k − 1)λ+ | (A − B)δ − B(k − 1)λ | bk | ak | −(A − B) | δ |≤ 0.
Hence it follows that
Letting
m+1
Dλ, g f (z)
m
Dλ, g f (z) − 1
< 1,
Dm+1 f (z)
B Dλ,mg f (z) − 1 − (A − B)δ
λ, g
w(z) =
B
m+1
Dλ,
g f (z)
m f (z)
Dλ,
g
m+1
Dλ,
g f (z)
m f (z)
Dλ,
g
z ∈ U.
−1
,
− 1 − (A − B)δ
then w(0) = 0, w(z) is analytic in | z |< 1 and | w(z) |< 1. Hence we have
m+1
Dλ,
g f (z)
m f (z)
Dλ,
g
=
1 + [B + δ(A − B)]w(z)
1 + Bw(z)
which shows that f belongs to Hλm (g, δ; A, B).
3. Subordination Results For the Class Hλm (g, δ; A, B)
Definition 1. A sequence {bk }∞
k=1 of complex numbers is called a subordinating
factor sequence if, whenever f (z) is analytic, univalent and convex in U, we have
the subordination given by
∞
X
bk ak z k ≺ f (z)
k=1
62
(z ∈ U, a1 = 1).
(18)
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
Lemma 1. The sequence {bk }∞
k=1 is a subordinating factor sequence if and only
if
∞
X
k
>0
Re 1 + 2
bk z
(z ∈ U).
(19)
k=1
For convenience, we shall henceforth
σk (δ, λ, m, A, B)
(20)
= [1 + (k − 1)λ]m λ (k − 1)+ | (A − B)δ − B(k − 1) | bk
to be real.
e m (g, δ; A, B) denote the class of functions f ∈ A whose coefficients satisfy
Let H
λ
e m (g, δ; A, B) ⊆ Hm (g, δ; A, B).
the conditions (17). We note that H
λ
λ
e m (g, δ; A, B)
Theorem 3. Let the function f (z) defined by (1) be in the class H
λ
where −1 ≤ B < A ≤ 1. Also let C denote the familiar class of functions f ∈ A
which are also univalent and convex in U. Then
and
σ2 (δ, λ, m, A, B)
(f ∗ g)(z) ≺ g(z)
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
Re(f (z)) > −
(A − B)|δ| + σ2 (δ, λ, m, A, B)
σ2 (δ, λ, m, A, B)
(z ∈ U; g ∈ C),
(z ∈ U).
(21)
(22)
σ2 (δ, λ, m, A, B)
is the best estimate.
The constant
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
e m (δ; A, B) and let g(z) = z + P∞ bk z k ∈ C. Then
Proof. Let f (z) ∈ H
k=2
λ
σ2 (δ, λ, m, A, B)
(f ∗ g)(z)
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
∞
X
σ2 (δ, λ, m, A, B)
k
z+
ak bk z .
=
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
k=2
Thus, by Definition 1, the assertion of the theorem will hold if the sequence
∞
σ2 (δ, λ, m, A, B)
ak
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
k=1
is a subordinating factor sequence, with a1 = 1. In view of Lemma 1, this will be
true if and only if
∞
X
σ2 (δ, λ, m, A, B)
k
ak z
>0
(z ∈ U).
(23)
Re 1 + 2
2
(A
−
B)|δ|
+
σ
(δ,
λ,
m,
A,
B)
2
k=1
63
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
Now
Re 1+
∞
X
σ2 (δ, λ, m, A, B)
ak z k
(A − B)|δ| + σ2 (δ, λ, m, A, B)
k=1
= Re 1+
σ2 (δ, λ, m, A, B)
a1 z
(A − B)|δ| + σ2 (δ, λ, m, A, B)
∞
X
1
σ2 (δ, λ, m, A, B)ak z k
(A − B)|δ| + σ2 (δ, λ, m, A, B)
k=2
σ2 (δ, λ, m, A, B)
r
≥ 1−
(A − B)|δ| + σ2 (δ, λ, m, A, B)
+
∞
X
1
k
+
σk (δ, λ, m, A, B)|ak |r .
|(A − B)|δ| + σ2 (δ, λ, m, A, B)|
k=2
Since σk (δ, λ, m, A, B) is a real increasing function of k (k ≥ 2)
σ2 (δ, λ, m, A, B)
r+
1−
(A − B)|δ| + σ2 (δ, λ, m, A, B)
∞
X
1
σk (δ, λ, m, A, B)|ak |rk
|(A − B)|δ| + σ2 (δ, λ, m, A, B)|
k=2
(A − B)|δ|
σ2 (δ, λ, m, A, B)
r+
r
>1−
(A − B)|δ| + σ2 (δ, λ, m, A, B)
(A − B)|δ| + σ2 (δ, λ, m, A, B)
= 1 − r > 0.
Thus (23) holds true in U. This proves the inequality (21).PThe inequality (22)
z
k
= z+ ∞
follows by taking the convex function g(z) = 1−z
k=2 z in (21). To
σ2 (δ, λ, m, A,B)
, we consider f0 (z) ∈
prove the sharpness of the constant
2 (A−B)|δ|+σ2 (δ, λ, m, A,B)
e m (g, δ;
H
λ
A, B) given by
f0 (z) = z −
(A − B)|δ|
z2
σ2 (δ, λ, m, A, B)
(−1 ≤ B < A ≤ 1).
Thus from (21), we have
z
σ2 (δ, λ, m, A, B)
f0 (z) ≺
.
1−z
2 (A − B)|δ| + σ2 (b, λ, m, A, B)
64
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C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
It can be easily verified that
1
σ2 (δ, λ, m, A, B)
f0 (z)
=−
min Re
2
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
This shows that the constant
σ2 (δ, λ, m, A,B)
2 (A−B)|δ|+σ2 (δ, λ, m, A,B)
(z ∈ U).
is best possible.
Remark 3.By specializing the parameters, the above result reduces to various
other results obtained by several authors.
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[12.] M. A. Nasr and M. K. Aouf, Starlike function of complex order, J. Natur.
Sci. Math. 25 (1985), no. 1, 1–12.
[13.] S. Ruscheweyh, New criteria for univalent functions, Proc. Amer. Math.
Soc., 49(1975), 109-115.
65
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
[14.] G. S
¸ . S˘
al˘agean, Subclasses of univalent functions, in Complex analysis—
fifth Romanian-Finnish seminar, Part 1 (Bucharest, 1981), 362–372, Lecture Notes
in Math., 1013, Springer, Berlin.
[15.] N. S. Sohi and L. P. Singh, A class of bounded starlike functions of complex
order, Indian J. Math. 33 (1991), no. 1, 29–35.
[16.] C.Selvaraj and K.R.Karthikeyan, Differential sandwich theorems for certain subclasses of analytic functions, Math. Commun. 12 (2008), no. 2.
[17.] C.Selvaraj and K.R.Karthikeyan, Certain classes of analytic functions
of complex order involving a family of generalized differential operators, J. Math.
Inequ. Appl. 2 (2008), no. 4, 449-458.
[18.] H. S. Wilf, Subordinating factor sequences for convex maps of the unit
circle, Proc. Amer. Math. Soc. 12 (1961), 689–693.
C.Selvaraj
Department of Mathematics,
Presidency College,
Chennai-600 005,
Tamilnadu, India
email: [email protected]
C. Santhosh Moni
R.M.K.Engineering College
R.S.M.Nagar,
Kavaraipettai-601206,
Thiruvallur District, Gummidipoondi Taluk,
Tamilnadu,India
email: [email protected].
66
No 19/2009
CERTAIN CLASSES OF ANALYTIC FUNCTIONS OF COMPLEX
ORDER
C.Selvaraj and C.Santhosh Moni
Abstract.By making use of the Hadamard product, we define a new class of
analytic functions of complex order. Coefficient inequalities, sufficient condition and
an interesting subordination result are obtained.
2000 Mathematics Subject Classification: 30C45, 30C50.
Keywords and Phrases: analytic functions, Hadamard product, generalized hypergeometric functions, coefficient inequalities, subordinating sequence.
1. Introduction, Definitions And Preliminaries
Let A denote the class of functions of the form
f (z) = z +
∞
X
ak z k
ak ≥ 0,
(1)
k=2
which are analytic in the open disc U = {z ∈ C \ | z |< 1} and S be the class of
function f ∈ A which are univalent in U.
P∞
k
The
Hadamard
product
of
two
functions
f
(z)
=
z
+
k=2 ak z and g(z) =
P∞
z + k=2 bk z k is given by
∞
X
ak bk z k .
(2)
(bk ≥ 0 for k ≥ 2)
(3)
(f ∗ g)(z) = z +
k=2
For a fixed function g ∈ A defined by
g(z) = z +
∞
X
bk z k
k=2
m f : A −→ A by
We now define the following linear operator Dλ,
g
0
Dλ,
g f (z) = (f ∗ g)(z),
55
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
′
1
Dλ,
g f (z) = (1 − λ)(f (z) ∗ g(z)) + λ z(f (z) ∗ g(z)) ,
m−1
m
1
Dλ,
g f (z) = Dλ, g (Dλ, g f (z)).
(4)
(5)
If f ∈ A, then from (4) and (5) we may easily deduce that
m
Dλ,
g f (z) = z +
∞
X
m
1 + (k − 1)λ ak bk
k=2
zk
,
(k − 1)!
(6)
where m ∈ N0 = N ∪ {0} and λ ≥ 0.
Remark 1. It is interesting to note that several integral and differential operator
m f (z), here we list few of them.
follows as a special case of Dλ,
g
m f (z) reduces to an operator introduced recently
1. When g(z) = z/(1 − z), Dλ,
g
by F. Al-Oboudi [1].
2. Let the coefficients bk be of the form
bk =
(α1 )k−1 . . . (αq )k−1
(β1 )k−1 . . . (βs )k−1 (k − 1)!
(7)
(α1 , . . . αq , β1 , . . . βs ∈ C and βj 6= 0, −1, −2, . . . for j={1,2,. . . ,s})
m f (z) reduces to the well-known Dziok-Srivastava operaand if m = 0, then Dλ,
g
tor. It is well known that Dziok-Srivastava operator includes as its special cases
various other linear operator which were introduced and studied by Hohlov,
Carlson and Shaffer and Ruscheweyh. For details we refer to [6, 7, 8]
m f (z) generalizes the well-known operators like
Apart from these, the operator Dλ,
g
S˘
al˘agean operator [14] and Bernardi-Libera-Livingston operator.
m f (z), we define Hm (g, δ; A, B) to be the class of functions
Using the operator Dλ,
g
λ
f ∈ A satisfying the inequality
m+1
1 Dλ, g f (z)
1 + Az
1+
−1 ≺
,
m
δ Dλ, g f (z)
1 + Bz
z ∈ U,
(8)
where δ ∈ C \ {0}, A and B are arbitrary fixed numbers, −1 ≤ B < A ≤ 1, m ∈ N0 .
Remark 2. Several interesting well-known and new subclasses of analytic functions can be obtained by specializing the parameters in Hλm (g, δ; A, B). Here we list
a few of them.
56
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
z
, λ = 1 in (8), then the class Hλm (g, δ; A, B) reduces
1−z
to the well- known class
1 + Az
1 Dm+1 f (z)
m
−1 ≺
H (δ; A, B) := f : 1 +
δ Dm f (z)
1 + Bz
1. If we let g(z) =
where Dm f is the well- known S˘al˘
agean operator. The class Hm (δ; A, B) was
introduced and studied by Attiya [4].
∞
X
(α1 )k−1 . . . (αq )k−1
ak z k in (8), then we have
(β1 )k−1 . . . (βs )k−1 (k − 1)!
n=2
following class of functions
1 + Az
1 Dm+1 (α1 , β1 )f (z)
m
−1 ≺
Hλ (α1 , β1 ; A, B) := f : 1 +
δ Dm (α1 , β1 )f (z)
1 + Bz
2. If we let g(z) = z +
introduced and studied by C. Selvaraj and K.R.Karthikeyan in [17].
z
, λ = 1, m = 0 in (8), then the class Hλm (g, δ; A, B) reduces
1−z
to the well- known class of starlike functions of complex order [12].
3. If we let g(z) =
Further, we note that several subclass of analytic functions can be obtained by
specializing the parameters in Hλm (g, δ; A, B) (see for example [3, 11, 12]).
We use Ω to denote the class of bounded analytic functions w(z) in U which
satisfy the conditions w(0) = 1 and | w(z) |< 1 for z ∈ U.
2.Coefficient estimates
Theorem 1. Let the function f (z) defined by (1) be in the class Hλm (g, δ; A, B).
(a) If (A − B)2 | δ |2 > (k − 1) 2B(A − B)λ Re{δ} + (1 − B 2 )λ2 (k − 1) , let
G=
(A − B)2 | δ |2
,
(k − 1) 2B(A − B)λ Re{δ} + (1 − B 2 )λ2 (k − 1)
k = 2, 3, . . . , m − 1,
M = [G] (Gauss symbol) and [G] is the greatest integer not greater than G.
Then, for j = 2, 3, . . . , M + 2
| aj |≤
1
[1 + (j − 1)λ]m λj−1 (j − 1)!bj
57
j
Y
k=2
| (A − B)δ − (k − 2)B |
(9)
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
and for
| aj |≤
j >M +2
1
[1 + (j − 1)λ]m λj−1 (j − 1) (M + 1)! bj
M
+3
Y
| (A − B)δ − (k − 2)B | .
k=2
(10)
(b) If (A − B)2 | δ |2 ≤ (k − 1) 2B(A − B)λ Re{δ} + (1 − B 2 )λ2 (k − 1) , then
| aj |≤
(A − B) | δ |
λ (j − 1) [1 + (j − 1)λ]m bj
j ≥ 2.
(11)
The bounds in (9) and (11) are sharp for all admissible A, B, δ ∈ C \ {0} and for
each j.
Proof. Since f (z) ∈ Hλm (g, δ; A, B), the inequality (8) gives
m+1
m+1
m
m
| Dλ,
f
(z)
−
D
f
(z)
|=
[(A
−
B)δ
+
B]D
f
(z)
−
BD
f
(z)
w(z).
λ,
g
λ,
g
g
λ, g
(12)
Equation (12) may be written as
∞
X
[1 + (k − 1)λ]m λ(k − 1) bk ak z k
(13)
k=2
=
(A − B)δz +
m
[(A − B)δ − B(k − 1)λ][1 + (k − 1)λ] bk ak z
k=2
Or equivalently
j
X
∞
X
[1 + (k − 1)λ]m λ(k − 1) bk ak z k +
k=2
∞
X
k
w(z).
ck z k
k=j+1
=
j−1
X
(A−B)δz+ [(A−B)δ−B(k−1)λ][1+(k−1)λ]m bk ak z k w(z), (14)
k=2
for certain coefficients ck . Explicitly ck = [1 + (k − 1)λ]m λ(k − 1)bk ak − [(A − B)δ −
B(k − 2)λ][1 + (k − 2)λ]m bk−1 ak−1 z −1 . Since | w(z) |< 1, we have
X
∞
X
j
m
k
k
[1 + (k − 1)λ] λ(k − 1) bk ak z +
ck z
k=2
k=j+1
58
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C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
j−1
X
m
k
≤ (A − B)δz +
[(A − B)δ − B(k − 1)λ][1 + (k − 1)λ] bk ak z .
k=2
Let z = reiθ , r < 1, applying the Parseval’s formula (see [9] p.138) on both sides of
the above inequality and after simple computation , we get
j
X
[1 + (k − 1)λ]2m λ2 (k − 1)2 b2k | ak |2 r2k +
k=2
∞
X
| ck |2 r2k
k=j+1
2
2
2
≤ (A − B) | δ | r +
j−1
X
| (A − B)δ − B(k − 1)λ |2 [1 + (k − 1)λ]2m b2k | ak |2 r2k .
k=2
Let r −→ 1− , then on some simplification we obtain
[1 + (j − 1)λ]2m λ2 (j − 1)2 b2j | aj |2 ≤ (A − B)2 | δ |2
+
j−1
X
k=2
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k | ak |2
j ≥ 2.
(16)
Now the following two cases arises:
(a) Let (A − B)2 | δ |2 > (k − 1)λ 2B(A − B)Re(δ) + (1 − B 2 )λ(k − 1) , suppose
that j ≤ M + 2, then for j = 2, (16) gives
| a2 |≤
(A − B) | δ |
,
(1 + λ)m λ b2
which gives (9) for j = 2. We establish (9) for j ≤ M + 2, from (16), by
mathematical induction. Suppose (9) is valid for j = 2, 3, . . . , (k − 1). Then
it follows from (16)
[1 + (j − 1)λ]2m λ2 (j − 1)2 b2j | aj |2
≤ (A − B)2 | δ |2
+
j−1
X
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k | ak |2
j−1
X
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k
k=2
≤ (A − B)2 | δ |2
+
k=2
59
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
×
k
Y
1
2
| (A − B)δ − (n − 2)B |
[1 + (k − 1)λ]2m b2k {λk−1 (k − 1)!}2 n=2
2
2
= (A − B) | δ | +
j−1
X
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2
k=2
×
k
Y
1
2
| (A − B)δ − (n − 2)B |
{λk−1 (k − 1)!}2
n=2
= (A − B)2 | δ |2 +
| (A − B)δ − Bλ |2 −λ2
1
λ2 (1!)2
| (A − B)δ − 2Bλ |2 −4λ2
(A − B)2 | δ |2 +
1
λ4 (2!)2
(A − B)2 | δ |2 | (A − B)δ − Bλ |2
+ . . . up to (k = j − 1)
=
j
Y
1
| (A−B)δ−(k−2)B |2 .
{λj−2 (j − 2)!}2
k=2
Thus, we get
| aj |≤
1
[1 + (j − 1)λ]m λj−1 (j − 1)!bj
j
Y
| (A − B)δ − (k − 2)B |,
k=2
which completes the proof of (9).
Next, we suppose j > M + 2. Then (16) gives
[1 + (j − 1)λ]2m λ2 (j − 1)2 b2j | aj |2 ≤ (A − B)2 | δ |2
+
M
−2
X
k=2
+
j−1
X
k=M +3
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k | ak |2
| (A − B)δ − B(k − 1)λ |2 −(k − 1)2 λ2 [1 + (k − 1)λ]2m b2k | ak |2 .
On substituting upper estimates for a2 , a3 , . . . , aM +2 obtained above and simplifying, we obtain (10).
60
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
(b) Let (A − B)2 | δ |2 ≤ (k − 1)λ 2B(A − B)Re(δ) + (1 − B 2 )λ(k − 1) , then it
follows from (16)
[1 + (j − 1)λ]2m λ2 (j − 1)2 b2j | aj |2 ≤ (A − B)2 | δ |2
(j ≥ 2),
which proves (11).
The bounds in (9) are sharp for the functions f (z) given by
(
(A−B)δ
if B 6= 0,
z(1 + Bz) B
m
Dλ, g f (z) =
z exp(Aδz)
if B = 0.
Also, the bounds in (11) are sharp for the functions fk (z) given by
(A−B)δ
z(1 + Bz) Bλ(k−1)
if B 6= 0,
m
Dλ,
g fk (z) =
Aδ
z k−1
if B = 0.
z exp λ(k−1)
Remark 2. If we let λ = 1, g(z) =
due to Attiya [4].
z
in Theorem 1, we get the result
1−z
3. A sufficient condition for a function to be in Hλm (g, δ; A, B)
Theorem 2. Let the function f (z) defined by (1) and let
∞
X
k=2
[1 + (k − 1)λ]m (k − 1)+ | (A − B)δ − B(k − 1) | λ bk | ak |≤ (A − B) | δ | (17)
holds, then f (z) belongs to Hλm (g, δ; A, B).
Proof. Suppose that the inequality holds. Then we have for z ∈ U
m+1
m
m
| Dλ,
g f (z) − Dλ, g f (z) | − (A − B)δDλ, g f (z)−
m+1
m
B Dλ,
f
(z)
−
D
f
(z)
λ,
g
g
∞
∞
X
X
m
k
m
k
[1 + (k − 1)λ] bk ak z −
[1 + (k − 1)λ] λ(k − 1) bk ak z − (A − B)δ z +
=
k=2
k=2
B
∞
X
k=2
61
[1 + (k − 1)λ] λ(k − 1)bk ak z
m
k
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
≤
∞
X
k=2
[1 + (k − 1)λ]m (k − 1)λ+ | (A − B)δ − B(k − 1)λ | bk | ak | rk − (A − B) | δ | r.
Letting r −→ 1− , then we have
m+1
m
m
| Dλ,
g f (z) − Dλ, g f (z) | − (A − B)δDλ, g f (z)−
≤
∞
X
k=2
m+1
m
B Dλ,
g f (z) − Dλ, g f (z)
[1 + (k − 1)λ]m (k − 1)λ+ | (A − B)δ − B(k − 1)λ | bk | ak | −(A − B) | δ |≤ 0.
Hence it follows that
Letting
m+1
Dλ, g f (z)
m
Dλ, g f (z) − 1
< 1,
Dm+1 f (z)
B Dλ,mg f (z) − 1 − (A − B)δ
λ, g
w(z) =
B
m+1
Dλ,
g f (z)
m f (z)
Dλ,
g
m+1
Dλ,
g f (z)
m f (z)
Dλ,
g
z ∈ U.
−1
,
− 1 − (A − B)δ
then w(0) = 0, w(z) is analytic in | z |< 1 and | w(z) |< 1. Hence we have
m+1
Dλ,
g f (z)
m f (z)
Dλ,
g
=
1 + [B + δ(A − B)]w(z)
1 + Bw(z)
which shows that f belongs to Hλm (g, δ; A, B).
3. Subordination Results For the Class Hλm (g, δ; A, B)
Definition 1. A sequence {bk }∞
k=1 of complex numbers is called a subordinating
factor sequence if, whenever f (z) is analytic, univalent and convex in U, we have
the subordination given by
∞
X
bk ak z k ≺ f (z)
k=1
62
(z ∈ U, a1 = 1).
(18)
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
Lemma 1. The sequence {bk }∞
k=1 is a subordinating factor sequence if and only
if
∞
X
k
>0
Re 1 + 2
bk z
(z ∈ U).
(19)
k=1
For convenience, we shall henceforth
σk (δ, λ, m, A, B)
(20)
= [1 + (k − 1)λ]m λ (k − 1)+ | (A − B)δ − B(k − 1) | bk
to be real.
e m (g, δ; A, B) denote the class of functions f ∈ A whose coefficients satisfy
Let H
λ
e m (g, δ; A, B) ⊆ Hm (g, δ; A, B).
the conditions (17). We note that H
λ
λ
e m (g, δ; A, B)
Theorem 3. Let the function f (z) defined by (1) be in the class H
λ
where −1 ≤ B < A ≤ 1. Also let C denote the familiar class of functions f ∈ A
which are also univalent and convex in U. Then
and
σ2 (δ, λ, m, A, B)
(f ∗ g)(z) ≺ g(z)
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
Re(f (z)) > −
(A − B)|δ| + σ2 (δ, λ, m, A, B)
σ2 (δ, λ, m, A, B)
(z ∈ U; g ∈ C),
(z ∈ U).
(21)
(22)
σ2 (δ, λ, m, A, B)
is the best estimate.
The constant
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
e m (δ; A, B) and let g(z) = z + P∞ bk z k ∈ C. Then
Proof. Let f (z) ∈ H
k=2
λ
σ2 (δ, λ, m, A, B)
(f ∗ g)(z)
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
∞
X
σ2 (δ, λ, m, A, B)
k
z+
ak bk z .
=
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
k=2
Thus, by Definition 1, the assertion of the theorem will hold if the sequence
∞
σ2 (δ, λ, m, A, B)
ak
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
k=1
is a subordinating factor sequence, with a1 = 1. In view of Lemma 1, this will be
true if and only if
∞
X
σ2 (δ, λ, m, A, B)
k
ak z
>0
(z ∈ U).
(23)
Re 1 + 2
2
(A
−
B)|δ|
+
σ
(δ,
λ,
m,
A,
B)
2
k=1
63
C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
Now
Re 1+
∞
X
σ2 (δ, λ, m, A, B)
ak z k
(A − B)|δ| + σ2 (δ, λ, m, A, B)
k=1
= Re 1+
σ2 (δ, λ, m, A, B)
a1 z
(A − B)|δ| + σ2 (δ, λ, m, A, B)
∞
X
1
σ2 (δ, λ, m, A, B)ak z k
(A − B)|δ| + σ2 (δ, λ, m, A, B)
k=2
σ2 (δ, λ, m, A, B)
r
≥ 1−
(A − B)|δ| + σ2 (δ, λ, m, A, B)
+
∞
X
1
k
+
σk (δ, λ, m, A, B)|ak |r .
|(A − B)|δ| + σ2 (δ, λ, m, A, B)|
k=2
Since σk (δ, λ, m, A, B) is a real increasing function of k (k ≥ 2)
σ2 (δ, λ, m, A, B)
r+
1−
(A − B)|δ| + σ2 (δ, λ, m, A, B)
∞
X
1
σk (δ, λ, m, A, B)|ak |rk
|(A − B)|δ| + σ2 (δ, λ, m, A, B)|
k=2
(A − B)|δ|
σ2 (δ, λ, m, A, B)
r+
r
>1−
(A − B)|δ| + σ2 (δ, λ, m, A, B)
(A − B)|δ| + σ2 (δ, λ, m, A, B)
= 1 − r > 0.
Thus (23) holds true in U. This proves the inequality (21).PThe inequality (22)
z
k
= z+ ∞
follows by taking the convex function g(z) = 1−z
k=2 z in (21). To
σ2 (δ, λ, m, A,B)
, we consider f0 (z) ∈
prove the sharpness of the constant
2 (A−B)|δ|+σ2 (δ, λ, m, A,B)
e m (g, δ;
H
λ
A, B) given by
f0 (z) = z −
(A − B)|δ|
z2
σ2 (δ, λ, m, A, B)
(−1 ≤ B < A ≤ 1).
Thus from (21), we have
z
σ2 (δ, λ, m, A, B)
f0 (z) ≺
.
1−z
2 (A − B)|δ| + σ2 (b, λ, m, A, B)
64
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C.Selvaraj, C.Santhosh Moni - Certain Classes of Analytic Functions Of...
It can be easily verified that
1
σ2 (δ, λ, m, A, B)
f0 (z)
=−
min Re
2
2 (A − B)|δ| + σ2 (δ, λ, m, A, B)
This shows that the constant
σ2 (δ, λ, m, A,B)
2 (A−B)|δ|+σ2 (δ, λ, m, A,B)
(z ∈ U).
is best possible.
Remark 3.By specializing the parameters, the above result reduces to various
other results obtained by several authors.
References
[1] F. M. Al-Oboudi, On univalent functions defined by a generalized S˘
al˘
agean
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C.Selvaraj
Department of Mathematics,
Presidency College,
Chennai-600 005,
Tamilnadu, India
email: [email protected]
C. Santhosh Moni
R.M.K.Engineering College
R.S.M.Nagar,
Kavaraipettai-601206,
Thiruvallur District, Gummidipoondi Taluk,
Tamilnadu,India
email: [email protected].
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