Directory UMM :Journals:Journal_of_mathematics:EJQTDE:
Surveys in Mathematics and its Applications
ISSN 1842-6298 (electronic), 1843-7265 (print)
Volume 5 (2010), 285 – 295
A CLASS OF ANALYTIC FUNCTIONS BASED ON
CONVOLUTION
K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
P
Abstract. We introduce a class T Spg (α) of analytic functions with negative coefficients defined
by convolution with a fixed analytic function g(z) = z +
∞
n=2
bn z n , bn > 0, |z| < 1. We obtain the
coefficient inequality, coefficient estimate, distortion theorem, a convolution result, extreme points
and integral representation for functions in the class T Spg (α).
1
Introduction
Let S denote the class of functions
f (z) = z +
∞
X
an z n
(1.1)
n=2
that are analytic and univalent in the unit disk ∆ = {z : |z| < 1}. Let S ∗ (α) denote
the subfamily of S consisting of functions starlike of order α, 0 ≤ α < 1.
Let T be the class of all analytic functions with negative coefficients of the form
f (z) = z −
∞
X
an z n
(an ≥ 0, z ∈ ∆)
(1.2)
n=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
′
′
zf (z)
zf (z)
Re
−α ≥
− 1 , −1 ≤ α < 1
f (z)
f (z)
has been studied. We need the following theorem proved in [8].
Theorem 1.
and sufficient condition for f of the form
P A necessary
n , a ≥ 0 to be in T S (α), −1 ≤ α < 1 is that
fP(z) = z − ∞
a
z
n
p
n=2 n
∞
[2n
−
(α
+
1)]a
≤
1
−
α.
n
n=2
2010 Mathematics Subject Classification: 30C45; 30C50.
Keywords: Analytic functions; Starlike functions; Convolution.
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286
K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
In [1], Ahuja has studied the class Tλ (α), consisting of functions f (z) ∈ T satλ f (z))′
isfying the condition Re z(D
> α, z ∈ ∆, λ > −1, α < 1, where the operator
Dλ f (z)
Dλ f is the Ruscheweyh derivative [5] of f defined by
z
Dλ f (z) = f (z) ∗ (1−z)
λ+1 . In fact we can write
z(f ∗ g)′ (z)
z
Tλ (α) = f ∈ T : Re
> α, g(z) =
,
(f ∗ g)(z)
(1 − z)λ+1
where f ∗ g is the Hadamard product of analytic functions.
Recall that for any two functions f (z) and g(z) given by
f (z) = z +
∞
X
an z n ,
n=2
g(z) = z +
∞
X
bn z n ,
n=2
the Hadamard product (or convolution) of f and g is defined by
(f ∗ g)(z) := z +
∞
X
an bn z n =: (g ∗ f )(z).
n=2
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
we introduce and study a class T Spg (α),
P∞ by [1],
n
−1 ≤ α < 1 where g(z) = z + n=2 bn z , bn > 0, z ∈ ∆ is a fixed analytic function.
P
n
Definition 2. Let g(z) = z + ∞
n=2 bn z be a fixed analytic function in ∆ = {z :
|z| < 1} and for n ≥P
2, bn > 0,. The class T Spg (α) consists of analytic functions of
n
the form f (z) = z − ∞
n=2 an z , where z ∈ ∆ and for n ≥ 2, an ≥ 0, satisfying the
inequality,
z(f ∗ g)′ (z)
z(f ∗ g)′ (z)
Re
−α ≥
− 1 , −1 ≤ α < 1
(f ∗ g)(z)
(f ∗ g)(z)
We obtain coefficient inequality, coefficient estimate, distortion theorem, convolution 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 Spg (α)
We begin with a necessary and sufficient condition to be in the class T Spg (α).
P∞
g
n
Theorem
3.
A
function
f
(z)
=
z
−
n=2 an z ∈ T Sp (α) if and only if
P∞
n=2 [2n − (α + 1)]an bn ≤ 1 − α.
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
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287
Analytic functions based on convolution
Proof. From the definitions of the classes T Spg (α) and T Sp (α) it follows on using
Theorem 1 that
f ∈ T Spg (α) ⇔ f ∗ g ∈ T Sp (α) ⇔
∞
X
[2n − (α + 1)]an bn ≤ 1 − α.
n=2
Remark 4. For −1 ≤ α < 1 and g(z) =
z
1−z ,
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
zn.
[2n − (α + 1)]bn
Proof. If f ∈ T Spg (α), then
[2n − (α + 1)]an bn ≤
∞
X
[2n − (α + 1)]an bn
n=2
≤ 1−α
implies an ≤
1−α
[2n−(α+1)]bn .
We have equality for
fn (z) = z −
(1 − α)z n
∈ T Spg (α).
[2n − (α + 1)]bn
Theorem 6. If f ∈ T Spg (α), then
r−
1−α
1−α
r2 ≤ |f (z)| ≤ r +
r2 ,
min[(2n − 1 − α)bn ]
min[(2n − 1 − α)bn ]
|z| = r < 1. The result is sharp for f (z) = z −
1−α
2
min[(2n−1−α)bn ] z .
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
http://www.utgjiu.ro/math/sma
288
K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
P∞
Proof. Let |z| = r < 1. For f (z) = z −
n=2 an z
∞
X
|f (z)| ≤ |z| +
n,
we have
|an ||z|n
n=2
≤ r+
∞
X
an rn ≤ r +
n=2
∞
X
2
= r+r
∞
X
an r2
n=2
(2.1)
an
n=2
Since min[(2n − 1 − α)bn ] ≤ [2n − (α + 1)]bn , in view of Theorem 3,
min[(2n − 1 − α)bn ]
∞
X
an ≤
n=2
∞
X
[2n − (α + 1)]an bn
n=2
≤ 1−α
Hence
|f (z)| ≤ r + r2
1−α
.
min[(2n − 1 − α)bn ]
1−α
.
Similarly we can show that |f (z)| ≥ r − r2 min[(2n−1−α)b
n]
We now prove a convolution result for the family T Spg (α).
Definition 7. Let
f1 (z) = z −
∞
X
an z n , an ≥ 0
n=2
and
f2 (z) = z −
∞
X
bn z n , bn > 0.
n=2
Then the Quasi Hadamard product (f1 ∗ f2 )(z) is defined by
(f1 ∗ f2 )(z) = f1 (z) ∗ f2 (z) = z −
∞
X
an bn z n
n=2
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
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289
Analytic functions based on convolution
Theorem 8. Let f ∈ T Spk1 (α), g ∈ T Spk2 (α) where, for z ∈ ∆,
ki (z) = z +
h(z) = z +
f (z) = z −
g(z) = z −
∞
X
n=2
∞
X
n=2
∞
X
n=2
∞
X
bin z n , bin > 0 for i = 1, 2
hn z n , hn > 0
fn z n , fn > 0
gn z n , gn > 0
n=2
then
f ∗ g ∈ T Sph (γ), γ = min G(n)
n
where
G(n) =
[2n − (α + 1)][2n − (β + 1)]b1n b2n − (2n − 1)hn (1 − α)(1 − β)
[2n − (α + 1)][2n − (β + 1)]b1n b2n − hn (1 − α)(1 − β)
provided b1n b2n > (2n − 1)hn .
Proof. Since f ∈ T Spk1 (α), we have
∞
X
[2n − (α + 1)]
n=2
Also g ∈ T Spk2 (β) implies
∞
X
[2n − (β + 1)]
n=2
Now, we will show that
f ∗g =z−
∞
X
n=2
1−α
1−β
fn gn z n ∈ T Sph (γ) i.e.
fn b1n ≤ 1
(2.2)
gn b2n ≤ 1
(2.3)
∞
X
[2n − (γ + 1)]
n=2
1−γ
fn gn hn ≤ 1.
In order to prove that (2.2) and (2.3) imply (2.4), we note that
1/2
∞
X
[2n − (α + 1)][2n − (β + 1)]
fn gn b1n b2n
(1 − α)(1 − β)
n=2
!1/2 ∞
!1/2
∞
X [2n − (β + 1)]
X
[2n − (α + 1)]
fn b1n
gn b2n
≤
1−α
1−β
n=2
≤ 1
(2.4)
n=2
(2.5)
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
http://www.utgjiu.ro/math/sma
290
K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
In order to prove (2.4), it suffices to show that
1/2
[2n − (γ + 1)]
[2n − (α + 1)][2n − (β + 1)]
fn gn hn ≤
fn gn b1n b2n
1−γ
(1 − α)(1 − β)
or equivalently
p
1/2
[2n − (α + 1)][2n − (β + 1)]
1−γ
b1n b2n
fn gn ≤
[2n − (γ + 1)]hn
(1 − α)(1 − β)
(2.6)
From (2.5), we have
p
fn gn ≤
s
(1 − α)(1 − β)
[2n − (α + 1)][2n − (β + 1)]b1n b2n
(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)]b1n b2n
1/2
[2n − (α + 1)][2n − (β + 1)]
1−γ
b1n b2n
≤
[2n − (γ + 1)]hn
(1 − α)(1 − β)
s
[2n − (α + 1)][2n − (β + 1)]
2n − (γ + 1)
1
⇔
≤
b1n b2n
1−γ
hn
(1 − α)(1 − β)
s
[2n − (α + 1)][2n − (β + 1)]
×
b1n b2n
(1 − α)(1 − β)
⇔
1 [2n − (α + 1)][2n − (β + 1)]
2n − (γ + 1)
≤
b1n b2n
1−γ
hn
(1 − α)(1 − β)
⇔ 2n − (γ + 1) ≤ B − Bγ
(2.8)
where
B=
1 [2n − (α + 1)][2n − (β + 1)]
b1n b2n .
hn
(1 − α)(1 − β)
Then B ≥ 2n − 1 > 1 and (2.8) is equivalent to
γ ≤
=
=
B − (2n − 1)
B−1
1 − 2n−1
B
1 − B1
1−
1−
(2n−1)hn (1−α)(1−β)
[2n−(α+1)][2n−(β+1)]b1n b2n
hn (1−α)(1−β)
[2n−(α+1)][2n−(β+1)]b1n b2n
(2.9)
(2.10)
= G(n).
(2.11)
This proves the result.
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
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291
Analytic functions based on convolution
Theorem 9. If
f1 (z) = z −
∞
X
n
an z , f2 (z) = z −
n=2
and
where f1 (z), f2 (z) ∈
β = min
n
cn z n
n=2
f3 (z) = z −
T Spg (α),
∞
X
then f3 ∈
∞
X
(a2n + c2n )z n
n=2
T Spg1 (β)
where
h [2n − (α + 1)]2 − 2(2n − 1)(1 − α)2 i
[2n − (α + 1)]2 − 2(1 − α)2
and g1 (z) = (g ∗ g)(z) = z +
.
2 n
n=2 bn z .
P∞
Proof. Since f1 (z) ∈ T Spg (α), we have
∞
X
2n − (α + 1)
n=2
and hence
Also since f2 (z) ∈
n=2
g
T Sp (α), we
1−α
a2n b2n ≤ 1.
have
∞
X
2n − (α + 1)
1−α
cn bn ≤ 1
∞
X
2n − (α + 1) 2
n=2
Therefore
an bn ≤ 1
∞
X
2n − (α + 1) 2
n=2
and hence
1−α
1−α
c2n b2n ≤ 1.
∞
1 X 2n − (α + 1) 2 2
(an + c2n )b2n ≤ 1.
2
1−α
n=2
In view of Theorem 3, we have to show that
∞
X
2n − (β + 1)
(a2n + c2n )b2n ≤ 1
1−β
n=2
The last inequality will be satisfied if
1 2n − (α + 1) 2
2n − (β + 1)
≤
.
1−β
2
1−α
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
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292
K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
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
zn,
[2n − (α + 1)]bn
n = 2, 3, . . . then f ∈ T Spg (α) if and only if it can be expressed in the form
f (z) =
∞
X
µn fn (z);
n=2
P
g
where µn ≥ 0 and ∞
n=1 µn = 1. In particular the extreme points of T Sp (α) are the
functions f1 (z) = z and
fn (z) = z −
1−α
zn,
[2n − (α + 1)]bn
n = 2, 3, . . . .
Proof. First let f be expressed as in the above theorem. This means that
we can write
f (z) =
∞
X
µn fn (z) = z −
n=2
n=1
= z−
∞
X
∞
X
(1 − α)µn
zn
[2n − (α + 1)]bn
tn z n
n=2
Therefore f ∈ T Spg (α), since
P∞
n=2
2n−(α+1)
tn b n
1−α
=
P∞
n=2 µn
= 1 − µ1 < 1
Conversely, If f ∈ T Spg (α), by Theorem 3, we have
an ≤
so we may set µn =
[2n−(α+1)]an bn
1−α
1−α
[2n − (α + 1)]bn
and µ1 = 1 −
P∞
n=2 µn .
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
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293
Analytic functions based on convolution
Then
f (z) = z −
= z−
=
∞
X
n=2
∞
X
an z n = z −
∞
X
n=2
1−α
µn z n
[2n − (α + 1)]bn
µn [z − fn (z)]
n=2
∞
X
1−
n=2
µn
!
z+
∞
X
µn fn (z) =
n=2
∞
X
µn fn (z).
n=1
Remark 11.
1. For g(z) =
z
1−z ,
we obtain corollary 1 in [8].
z
2. For g(z) = 1−z
and replacing α by
corresponding result in [7].
1+α
2 ,
in T ∗ (α), we obtain the
Theorem 12. If f is in T Spg (α), then
(f ∗ g)(z) = exp
Z
0
z
1 + αρ(t)
dt
t(1 − ρ(t))
for some ρ(z), |ρ(z)| < 1, z ∈ ∆ where
g(z) = z +
∞
X
bn z n ,
n=2
bn > 0 for n = 2, 3, . . .. Also
(f ∗ g)(z) = z exp
Z
−(1+α)
ln(1 − xz)
x
dµ(x)
where µ(x) is probability measure on X = {x : |x| = 1}.
Proof. Let f ∈ T Spg (α) and
ω=
z(f ∗ g)′ (z)
.
(f ∗ g)(z)
We have Re ω − α > |ω − 1|. Therefore
ω −1
−1, T Sp (α) reduces to the class D(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 Theorem 9 and to improve the paper.
References
[1] O.P. Ahuja, Hadamard products of analytic functions defined by Ruscheweyh
derivatives, Current topics in Analytic Funtion Theory (H.M. Srivastava, S. Owa
editors) World Scientific, Singapore 1992,
13-29. MR1232426(94j:30006). Zbl 1002.30007.
[2] K. Al-Shaqsi and M. Darus, Fekete-Szeg¨
o Problem for Univalent Functions with
Respect to k-Symmetric Points, The Austral. J. Math. Anal. Appl., 5(2) (2008)
Art. No.6. MR2461677(2009m:30035). Zbl 1165.30308.
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
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Analytic functions based on convolution
295
[3] M. Haji Mohd, R. M. Ali, S.K. Lee and V. Ravichandran, Subclasses of Meromorphic Functions Associated with Convolution, J. Ineq. Appl. Vol. 2009 (2009),
Article ID 190291, 10 pages MR2496271(2010k:30009). Zbl 1176.30044.
[4] G. Murugusundaramoorthy and R.K. Raina, On a certain class of analytic functions associated with a convolution structure, Demonstratio Mathematica, 41(3)
(2008) 551-559. MR2433307(2009h:30022). Zbl 1176.30044.
[5] St. Ruscheweyh, New criteria for univalent functions, Proc. Amer. Math. Soc.
49 (1975), 109-115. MR0367176(51 #3418). Zbl 0303.30006.
[6] S. Shams, S.R. Kulkarni and J.M. Jahangiri, On a class of univalent functions
defined by Ruscheweyh derivatives, Kyungpook Math. J. 43 (2003), 579-585.
MR2026116(2004j:30026). Zbl 1067.30032.
[7] H. Silverman, Univalent function with negative coefficients, Proc. Amer. Math.
Soc. 51 (1975), pp. 109-116. MR0369678(51#5910). Zbl 0311.30007.
[8] K.G. Subramanian, T.V. Sudharsan, P. Balasubrahmanyam and
H. Silverman, Classes of uniformly starlike functions, Publ. Math, Debrecen
53/3-4 (1998), 309-315. MRMR1657542(99i:30028). Zbl 0921.30007.
[9] S. Sivaprasad Kumar , V. Ravichandran and H. C. Taneja, Meromorphic Functions with positive coefficients defined using convolution, J. Ineq. Pure and Appl.
Math., 6(2) (2005), Article No. 58. MR2150912. Zbl 1080.30012.
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
Department of Mathematics,
Department of Mathematics,
Government Arts College,
College of Charleston,
Chennai 600 035 India.
Charleston SC 29424, USA.
e-mail: [email protected]
e-mail: [email protected]
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Surveys in Mathematics and its Applications 5 (2010), 285 – 295
http://www.utgjiu.ro/math/sma
ISSN 1842-6298 (electronic), 1843-7265 (print)
Volume 5 (2010), 285 – 295
A CLASS OF ANALYTIC FUNCTIONS BASED ON
CONVOLUTION
K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
P
Abstract. We introduce a class T Spg (α) of analytic functions with negative coefficients defined
by convolution with a fixed analytic function g(z) = z +
∞
n=2
bn z n , bn > 0, |z| < 1. We obtain the
coefficient inequality, coefficient estimate, distortion theorem, a convolution result, extreme points
and integral representation for functions in the class T Spg (α).
1
Introduction
Let S denote the class of functions
f (z) = z +
∞
X
an z n
(1.1)
n=2
that are analytic and univalent in the unit disk ∆ = {z : |z| < 1}. Let S ∗ (α) denote
the subfamily of S consisting of functions starlike of order α, 0 ≤ α < 1.
Let T be the class of all analytic functions with negative coefficients of the form
f (z) = z −
∞
X
an z n
(an ≥ 0, z ∈ ∆)
(1.2)
n=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
′
′
zf (z)
zf (z)
Re
−α ≥
− 1 , −1 ≤ α < 1
f (z)
f (z)
has been studied. We need the following theorem proved in [8].
Theorem 1.
and sufficient condition for f of the form
P A necessary
n , a ≥ 0 to be in T S (α), −1 ≤ α < 1 is that
fP(z) = z − ∞
a
z
n
p
n=2 n
∞
[2n
−
(α
+
1)]a
≤
1
−
α.
n
n=2
2010 Mathematics Subject Classification: 30C45; 30C50.
Keywords: Analytic functions; Starlike functions; Convolution.
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286
K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
In [1], Ahuja has studied the class Tλ (α), consisting of functions f (z) ∈ T satλ f (z))′
isfying the condition Re z(D
> α, z ∈ ∆, λ > −1, α < 1, where the operator
Dλ f (z)
Dλ f is the Ruscheweyh derivative [5] of f defined by
z
Dλ f (z) = f (z) ∗ (1−z)
λ+1 . In fact we can write
z(f ∗ g)′ (z)
z
Tλ (α) = f ∈ T : Re
> α, g(z) =
,
(f ∗ g)(z)
(1 − z)λ+1
where f ∗ g is the Hadamard product of analytic functions.
Recall that for any two functions f (z) and g(z) given by
f (z) = z +
∞
X
an z n ,
n=2
g(z) = z +
∞
X
bn z n ,
n=2
the Hadamard product (or convolution) of f and g is defined by
(f ∗ g)(z) := z +
∞
X
an bn z n =: (g ∗ f )(z).
n=2
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
we introduce and study a class T Spg (α),
P∞ by [1],
n
−1 ≤ α < 1 where g(z) = z + n=2 bn z , bn > 0, z ∈ ∆ is a fixed analytic function.
P
n
Definition 2. Let g(z) = z + ∞
n=2 bn z be a fixed analytic function in ∆ = {z :
|z| < 1} and for n ≥P
2, bn > 0,. The class T Spg (α) consists of analytic functions of
n
the form f (z) = z − ∞
n=2 an z , where z ∈ ∆ and for n ≥ 2, an ≥ 0, satisfying the
inequality,
z(f ∗ g)′ (z)
z(f ∗ g)′ (z)
Re
−α ≥
− 1 , −1 ≤ α < 1
(f ∗ g)(z)
(f ∗ g)(z)
We obtain coefficient inequality, coefficient estimate, distortion theorem, convolution 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 Spg (α)
We begin with a necessary and sufficient condition to be in the class T Spg (α).
P∞
g
n
Theorem
3.
A
function
f
(z)
=
z
−
n=2 an z ∈ T Sp (α) if and only if
P∞
n=2 [2n − (α + 1)]an bn ≤ 1 − α.
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Analytic functions based on convolution
Proof. From the definitions of the classes T Spg (α) and T Sp (α) it follows on using
Theorem 1 that
f ∈ T Spg (α) ⇔ f ∗ g ∈ T Sp (α) ⇔
∞
X
[2n − (α + 1)]an bn ≤ 1 − α.
n=2
Remark 4. For −1 ≤ α < 1 and g(z) =
z
1−z ,
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
zn.
[2n − (α + 1)]bn
Proof. If f ∈ T Spg (α), then
[2n − (α + 1)]an bn ≤
∞
X
[2n − (α + 1)]an bn
n=2
≤ 1−α
implies an ≤
1−α
[2n−(α+1)]bn .
We have equality for
fn (z) = z −
(1 − α)z n
∈ T Spg (α).
[2n − (α + 1)]bn
Theorem 6. If f ∈ T Spg (α), then
r−
1−α
1−α
r2 ≤ |f (z)| ≤ r +
r2 ,
min[(2n − 1 − α)bn ]
min[(2n − 1 − α)bn ]
|z| = r < 1. The result is sharp for f (z) = z −
1−α
2
min[(2n−1−α)bn ] z .
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K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
P∞
Proof. Let |z| = r < 1. For f (z) = z −
n=2 an z
∞
X
|f (z)| ≤ |z| +
n,
we have
|an ||z|n
n=2
≤ r+
∞
X
an rn ≤ r +
n=2
∞
X
2
= r+r
∞
X
an r2
n=2
(2.1)
an
n=2
Since min[(2n − 1 − α)bn ] ≤ [2n − (α + 1)]bn , in view of Theorem 3,
min[(2n − 1 − α)bn ]
∞
X
an ≤
n=2
∞
X
[2n − (α + 1)]an bn
n=2
≤ 1−α
Hence
|f (z)| ≤ r + r2
1−α
.
min[(2n − 1 − α)bn ]
1−α
.
Similarly we can show that |f (z)| ≥ r − r2 min[(2n−1−α)b
n]
We now prove a convolution result for the family T Spg (α).
Definition 7. Let
f1 (z) = z −
∞
X
an z n , an ≥ 0
n=2
and
f2 (z) = z −
∞
X
bn z n , bn > 0.
n=2
Then the Quasi Hadamard product (f1 ∗ f2 )(z) is defined by
(f1 ∗ f2 )(z) = f1 (z) ∗ f2 (z) = z −
∞
X
an bn z n
n=2
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Analytic functions based on convolution
Theorem 8. Let f ∈ T Spk1 (α), g ∈ T Spk2 (α) where, for z ∈ ∆,
ki (z) = z +
h(z) = z +
f (z) = z −
g(z) = z −
∞
X
n=2
∞
X
n=2
∞
X
n=2
∞
X
bin z n , bin > 0 for i = 1, 2
hn z n , hn > 0
fn z n , fn > 0
gn z n , gn > 0
n=2
then
f ∗ g ∈ T Sph (γ), γ = min G(n)
n
where
G(n) =
[2n − (α + 1)][2n − (β + 1)]b1n b2n − (2n − 1)hn (1 − α)(1 − β)
[2n − (α + 1)][2n − (β + 1)]b1n b2n − hn (1 − α)(1 − β)
provided b1n b2n > (2n − 1)hn .
Proof. Since f ∈ T Spk1 (α), we have
∞
X
[2n − (α + 1)]
n=2
Also g ∈ T Spk2 (β) implies
∞
X
[2n − (β + 1)]
n=2
Now, we will show that
f ∗g =z−
∞
X
n=2
1−α
1−β
fn gn z n ∈ T Sph (γ) i.e.
fn b1n ≤ 1
(2.2)
gn b2n ≤ 1
(2.3)
∞
X
[2n − (γ + 1)]
n=2
1−γ
fn gn hn ≤ 1.
In order to prove that (2.2) and (2.3) imply (2.4), we note that
1/2
∞
X
[2n − (α + 1)][2n − (β + 1)]
fn gn b1n b2n
(1 − α)(1 − β)
n=2
!1/2 ∞
!1/2
∞
X [2n − (β + 1)]
X
[2n − (α + 1)]
fn b1n
gn b2n
≤
1−α
1−β
n=2
≤ 1
(2.4)
n=2
(2.5)
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K.G. Subramanian, T.V. Sudharsan, R. Thirumalaisamy and H. Silverman
In order to prove (2.4), it suffices to show that
1/2
[2n − (γ + 1)]
[2n − (α + 1)][2n − (β + 1)]
fn gn hn ≤
fn gn b1n b2n
1−γ
(1 − α)(1 − β)
or equivalently
p
1/2
[2n − (α + 1)][2n − (β + 1)]
1−γ
b1n b2n
fn gn ≤
[2n − (γ + 1)]hn
(1 − α)(1 − β)
(2.6)
From (2.5), we have
p
fn gn ≤
s
(1 − α)(1 − β)
[2n − (α + 1)][2n − (β + 1)]b1n b2n
(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)]b1n b2n
1/2
[2n − (α + 1)][2n − (β + 1)]
1−γ
b1n b2n
≤
[2n − (γ + 1)]hn
(1 − α)(1 − β)
s
[2n − (α + 1)][2n − (β + 1)]
2n − (γ + 1)
1
⇔
≤
b1n b2n
1−γ
hn
(1 − α)(1 − β)
s
[2n − (α + 1)][2n − (β + 1)]
×
b1n b2n
(1 − α)(1 − β)
⇔
1 [2n − (α + 1)][2n − (β + 1)]
2n − (γ + 1)
≤
b1n b2n
1−γ
hn
(1 − α)(1 − β)
⇔ 2n − (γ + 1) ≤ B − Bγ
(2.8)
where
B=
1 [2n − (α + 1)][2n − (β + 1)]
b1n b2n .
hn
(1 − α)(1 − β)
Then B ≥ 2n − 1 > 1 and (2.8) is equivalent to
γ ≤
=
=
B − (2n − 1)
B−1
1 − 2n−1
B
1 − B1
1−
1−
(2n−1)hn (1−α)(1−β)
[2n−(α+1)][2n−(β+1)]b1n b2n
hn (1−α)(1−β)
[2n−(α+1)][2n−(β+1)]b1n b2n
(2.9)
(2.10)
= G(n).
(2.11)
This proves the result.
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Analytic functions based on convolution
Theorem 9. If
f1 (z) = z −
∞
X
n
an z , f2 (z) = z −
n=2
and
where f1 (z), f2 (z) ∈
β = min
n
cn z n
n=2
f3 (z) = z −
T Spg (α),
∞
X
then f3 ∈
∞
X
(a2n + c2n )z n
n=2
T Spg1 (β)
where
h [2n − (α + 1)]2 − 2(2n − 1)(1 − α)2 i
[2n − (α + 1)]2 − 2(1 − α)2
and g1 (z) = (g ∗ g)(z) = z +
.
2 n
n=2 bn z .
P∞
Proof. Since f1 (z) ∈ T Spg (α), we have
∞
X
2n − (α + 1)
n=2
and hence
Also since f2 (z) ∈
n=2
g
T Sp (α), we
1−α
a2n b2n ≤ 1.
have
∞
X
2n − (α + 1)
1−α
cn bn ≤ 1
∞
X
2n − (α + 1) 2
n=2
Therefore
an bn ≤ 1
∞
X
2n − (α + 1) 2
n=2
and hence
1−α
1−α
c2n b2n ≤ 1.
∞
1 X 2n − (α + 1) 2 2
(an + c2n )b2n ≤ 1.
2
1−α
n=2
In view of Theorem 3, we have to show that
∞
X
2n − (β + 1)
(a2n + c2n )b2n ≤ 1
1−β
n=2
The last inequality will be satisfied if
1 2n − (α + 1) 2
2n − (β + 1)
≤
.
1−β
2
1−α
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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
zn,
[2n − (α + 1)]bn
n = 2, 3, . . . then f ∈ T Spg (α) if and only if it can be expressed in the form
f (z) =
∞
X
µn fn (z);
n=2
P
g
where µn ≥ 0 and ∞
n=1 µn = 1. In particular the extreme points of T Sp (α) are the
functions f1 (z) = z and
fn (z) = z −
1−α
zn,
[2n − (α + 1)]bn
n = 2, 3, . . . .
Proof. First let f be expressed as in the above theorem. This means that
we can write
f (z) =
∞
X
µn fn (z) = z −
n=2
n=1
= z−
∞
X
∞
X
(1 − α)µn
zn
[2n − (α + 1)]bn
tn z n
n=2
Therefore f ∈ T Spg (α), since
P∞
n=2
2n−(α+1)
tn b n
1−α
=
P∞
n=2 µn
= 1 − µ1 < 1
Conversely, If f ∈ T Spg (α), by Theorem 3, we have
an ≤
so we may set µn =
[2n−(α+1)]an bn
1−α
1−α
[2n − (α + 1)]bn
and µ1 = 1 −
P∞
n=2 µn .
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Analytic functions based on convolution
Then
f (z) = z −
= z−
=
∞
X
n=2
∞
X
an z n = z −
∞
X
n=2
1−α
µn z n
[2n − (α + 1)]bn
µn [z − fn (z)]
n=2
∞
X
1−
n=2
µn
!
z+
∞
X
µn fn (z) =
n=2
∞
X
µn fn (z).
n=1
Remark 11.
1. For g(z) =
z
1−z ,
we obtain corollary 1 in [8].
z
2. For g(z) = 1−z
and replacing α by
corresponding result in [7].
1+α
2 ,
in T ∗ (α), we obtain the
Theorem 12. If f is in T Spg (α), then
(f ∗ g)(z) = exp
Z
0
z
1 + αρ(t)
dt
t(1 − ρ(t))
for some ρ(z), |ρ(z)| < 1, z ∈ ∆ where
g(z) = z +
∞
X
bn z n ,
n=2
bn > 0 for n = 2, 3, . . .. Also
(f ∗ g)(z) = z exp
Z
−(1+α)
ln(1 − xz)
x
dµ(x)
where µ(x) is probability measure on X = {x : |x| = 1}.
Proof. Let f ∈ T Spg (α) and
ω=
z(f ∗ g)′ (z)
.
(f ∗ g)(z)
We have Re ω − α > |ω − 1|. Therefore
ω −1
−1, T Sp (α) reduces to the class D(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 Theorem 9 and 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
Department of Mathematics,
Department of Mathematics,
Government Arts College,
College of Charleston,
Chennai 600 035 India.
Charleston SC 29424, USA.
e-mail: [email protected]
e-mail: [email protected]
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