ISSN1842-6298 (electronic), 1843-7265 (print) Volume5(2010), 285 – 295
A CLASS OF ANALYTIC FUNCTIONS BASED ON
CONVOLUTION
K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
Abstract. We introduce a classT Spg(α) of analytic functions with negative coefficients defined
by convolution with a fixed analytic functiong(z) =z+P∞n=2bnz
n,b
n>0,|z|<1. We obtain the
coefficient inequality, coefficient estimate, distortion theorem, a convolution result, extreme points and integral representation for functions in the classT Spg(α).
1
Introduction
LetS denote the class of functions
f(z) =z+
∞
X
n=2
anzn (1.1)
that are analytic and univalent in the unit disk ∆ ={z:|z|<1}. LetS∗
(α) denote the subfamily ofS consisting of functions starlike of orderα, 0≤α <1.
LetT be the class of all analytic functions with negative coefficients of the form
f(z) =z−
∞
X
n=2
anzn (an≥0, z∈∆) (1.2)
The subclass of T consisting of starlike functions of order α denoted by T∗
(α) = T∩S∗
(α) was studied by Silverman [7]. In [8], the subclass T Sp(α) of functions of
the form (1.2) for which
Re
zf′
(z) f(z) −α
≥
zf′
(z) f(z) −1
, −1≤α <1
has been studied. We need the following theorem proved in [8].
Theorem 1. A necessary and sufficient condition forf of the form
f(z) =z−P∞
n=2anzn, an≥0 to be in T Sp(α), −1≤α <1 is that
P∞
n=2[2n−(α+ 1)]an≤1−α.
In [1], Ahuja has studied the class Tλ(α), consisting of functions f(z) ∈ T
sat-wheref ∗g is the Hadamard product of analytic functions. Recall that for any two functionsf(z) and g(z) given by
the Hadamard product (or convolution) off and g is defined by
(f∗g)(z) :=z+
∞
X
n=2
anbnzn=: (g∗f)(z).
Classes of analytic functions defined by convolution with given analytic functions have been investigated for their properties (see for example [2, 3, 4, 9] to mention a few). In this paper, motivated by [1], we introduce and study a class T Spg(α),
We obtain coefficient inequality, coefficient estimate, distortion theorem, convo-lution result, extreme points and integral representation for functions in the class T Spg(α).Unless mentioned otherwise, the function g is taken as in definition 2.
2
The class
T S
gp
(
α
)
We begin with a necessary and sufficient condition to be in the classT Spg(α).
Theorem 3. A functionf(z) =z−P∞
n=2anzn∈T Spg(α) if and only if
P∞
Proof. From the definitions of the classes T Spg(α) and T Sp(α) it follows on using
Theorem1 that
f ∈T Spg(α)⇔f∗g∈T Sp(α)⇔ ∞
X
n=2
[2n−(α+ 1)]anbn≤1−α.
Remark 4. For −1≤α <1 and g(z) = 1−zz, we have T Spg(α) =T Sp(α) [8]
Theorem 5. If f ∈T Spg(α), then
an≤
1−α [2n−(α+ 1)]bn
with equality only for the functions of the form
fn(z) =z−
1−α [2n−(α+ 1)]bn
zn.
Proof. Iff ∈T Spg(α), then
[2n−(α+ 1)]anbn ≤ ∞
X
n=2
[2n−(α+ 1)]anbn
≤ 1−α
implies an≤ [2n−1(α−+1)]α bn.
We have equality for
fn(z) =z−
(1−α)zn [2n−(α+ 1)]bn
∈T Spg(α).
Theorem 6. If f ∈T Spg(α), then
r− 1−α
min[(2n−1−α)bn]
r2 ≤ |f(z)| ≤r+ 1−α
min[(2n−1−α)bn]
r2,
|z|=r <1. The result is sharp for f(z) =z− 1−α min[(2n−1−α)bn]z
Proof. Let|z|=r <1.For f(z) =z−P∞
n=2anzn, we have
|f(z)| ≤ |z|+
∞
X
n=2
|an||z|n
≤ r+
∞
X
n=2
anrn≤r+ ∞
X
n=2
anr2
= r+r2
∞
X
n=2
an (2.1)
Since min[(2n−1−α)bn]≤[2n−(α+ 1)]bn, in view of Theorem3,
min[(2n−1−α)bn] ∞
X
n=2
an ≤ ∞
X
n=2
[2n−(α+ 1)]anbn
≤ 1−α
Hence
|f(z)| ≤r+r2 1−α
min[(2n−1−α)bn]
.
Similarly we can show that |f(z)| ≥r−r2 1−α min[(2n−1−α)bn].
We now prove a convolution result for the familyT Spg(α).
Definition 7. Let
f1(z) =z−
∞
X
n=2
anzn, an≥0
and
f2(z) =z−
∞
X
n=2
bnzn, bn>0.
Then the Quasi Hadamard product (f1∗f2)(z) is defined by
(f1∗f2)(z) =f1(z)∗f2(z) =z−
∞
X
n=2
Theorem 8. Let f ∈T Sk1
Now, we will show that
In order to prove (2.4), it suffices to show that
[2n−(γ+ 1)]
1−γ fngnhn≤
[2n−(α+ 1)][2n−(β+ 1)]
(1−α)(1−β) fngnb1nb2n 1/2
or equivalently
p
fngn≤
1−γ [2n−(γ+ 1)]hn
[2n−(α+ 1)][2n−(β+ 1)] (1−α)(1−β) b1nb2n
1/2
(2.6)
From (2.5), we have
p
fngn≤
s
(1−α)(1−β)
[2n−(α+ 1)][2n−(β+ 1)]b1nb2n
(2.7)
In view of (2.6) and (2.7), it is enough to prove that
s
(1−α)(1−β)
[2n−(α+ 1)][2n−(β+ 1)]b1nb2n
≤ 1−γ
[2n−(γ+ 1)]hn
[2n−(α+ 1)][2n−(β+ 1)] (1−α)(1−β) b1nb2n
1/2
⇔ 2n−(γ+ 1)
1−γ ≤ 1 hn
s
[2n−(α+ 1)][2n−(β+ 1)] (1−α)(1−β) b1nb2n
×
s
[2n−(α+ 1)][2n−(β+ 1)] (1−α)(1−β) b1nb2n
⇔ 2n−(γ+ 1)
1−γ ≤ 1 hn
[2n−(α+ 1)][2n−(β+ 1)] (1−α)(1−β) b1nb2n
⇔ 2n−(γ+ 1)≤B−Bγ (2.8)
where
B = 1 hn
[2n−(α+ 1)][2n−(β+ 1)] (1−α)(1−β) b1nb2n.
Then B≥2n−1>1 and (2.8) is equivalent to
γ ≤ B−(2n−1)
B−1 (2.9)
= 1−
2n−1 B
1−B1 (2.10)
= 1−
(2n−1)hn(1−α)(1−β)
[2n−(α+1)][2n−(β+1)]b1nb2n
1− hn(1−α)(1−β)
[2n−(α+1)][2n−(β+1)]b1nb2n
=G(n). (2.11)
Theorem 9. If
The last inequality will be satisfied if
Solving for β, we have
β≤ [2n−(α+ 1)]
2−2(2n−1)(1−α)2
[2n−(α+ 1)]2−2(1−α)2 .
Hence the result follows.
3
Extreme points and Integral representation
Theorem 10. Let
f1(z) =z, fn(z) =z−
1−α [2n−(α+ 1)]bn
zn,
n= 2,3, . . . thenf ∈T Spg(α) if and only if it can be expressed in the form
f(z) =
∞
X
n=2
µnfn(z);
where µn≥0 andP ∞
n=1µn= 1. In particular the extreme points of T Spg(α) are the functions f1(z) =z and
fn(z) =z− 1−
α [2n−(α+ 1)]bn
zn,
n= 2,3, . . ..
Proof. First let f be expressed as in the above theorem. This means that we can write
f(z) =
∞
X
n=1
µnfn(z) =z− ∞
X
n=2
(1−α)µn
[2n−(α+ 1)]bn
zn
= z−
∞
X
n=2
tnzn
Thereforef ∈T Spg(α), sinceP∞n=2 2n −(α+1)
1−α tnbn=
P∞
n=2µn= 1−µ1 <1
Conversely, If f ∈T Spg(α), by Theorem 3, we have
an≤
1−α [2n−(α+ 1)]bn
so we may setµn= [2n
−(α+1)]anbn
1−α and µ1= 1−
Then
f(z) = z−
∞
X
n=2
anzn=z− ∞
X
n=2
1−α [2n−(α+ 1)]bn
µnzn
= z−
∞
X
n=2
µn[z−fn(z)]
= 1−
∞
X
n=2
µn
! z+
∞
X
n=2
µnfn(z) = ∞
X
n=1
µnfn(z).
Remark 11.
1. For g(z) = 1−zz, we obtain corollary 1 in [8].
2. For g(z) = 1−zz and replacingα by 1+2α, in T∗
(α), we obtain the corresponding result in [7].
Theorem 12. If f is in T Spg(α), then
(f ∗g)(z) =exp Z z
0
1 +αρ(t) t(1−ρ(t))dt
for some ρ(z), |ρ(z)|<1, z∈∆ where
g(z) =z+
∞
X
n=2
bnzn,
bn>0 for n= 2,3, . . .. Also
(f∗g)(z) =z exp Z
x
ln(1−xz)−(1+α)
dµ(x)
where µ(x) is probability measure on X={x:|x|= 1}.
Proof. Letf ∈T Spg(α) and
ω= z(f ∗g)
′
(z) (f ∗g)(z) .
We have Re ω−α >|ω−1|. Therefore
and
ω−1
ω−α =ρ(z), where|ρ(z)|<1, z∈∆.
This gives
(f∗g)′
(z) (f∗g)(z) =
1−αρ(z) z(1−ρ(z)) and therefore
(f∗g)(z) =exp Z z
0
1−αρ(t) t(1−ρ(t))dt
.
Now set X={x:|x|= 1}. Then we have ω−1
ω−α =xz and hence
(f ∗g)′
(z) (f ∗g)(z) =
1−αxz z(1−xz) =
1 z +
(1−α)x 1−xz .
Integrating we obtain,
ln(f ∗g)(z)
z = (−1 +α)ln(1−xz).
This proves the second representation
(f ∗g)(z) =z exp Z
x
ln(1−xz)−(1+α)
dµ(x)
.
Remark 13. Takingg(z) = (1−zz)λ+1,λ >−1,T S
g
p(α)reduces to the classD(α, β, λ) of [6], so that Theorems 2.1, 2.4, 2.7 in [6] are consequences of our Theorems 3,10, 12.
Acknowledgment. The authors thank the referees for very useful comments which helped to correct an error in Theorem9and to improve the paper.
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K.G. Subramanian T.V. Sudharsan
School of Mathematical sciences, Department of Mathematics, Universiti Sains Malaysia, SIVET College,
11800 Penang, Malaysia. Chennai 601 302 India.
e-mail: [email protected] e-mail: [email protected]
R. Thirumalaisamy H. Silverman
