Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 3, Article 86, 2005
A SUBCLASS OF UNIFORMLY CONVEX FUNCTIONS ASSOCIATED WITH
CERTAIN FRACTIONAL CALCULUS OPERATORS
G. MURUGUSUNDARAMOORTHY, THOMAS ROSY, AND MASLINA DARUS
D EPARTMENT OF M ATHEMATICS
V ELLORE I NSTITUTE OF T ECHNOLOGY, D EEMED U NIVERSITY
V ELLORE - 632014, I NDIA .
gmsmoorthy@yahoo.com
D EPARTMENT OF M ATHEMATICS
M ADRAS C HRISTIAN C OLLEGE ,
C HENNAI - 600059, I NDIA .
S CHOOL OF M ATHEMATICAL S CIENCES
FACULTY OF S CIENCE AND T ECHNOLOGY,
U NIVERSITI K EBANGSAAN M ALAYSIA ,
BANGI 43600 S ELANGOR , M ALAYSIA .
maslina@pkrisc.cc.ukm.my
Received 03 March, 2005; accepted 26 July, 2005
Communicated by G. Kohr
A BSTRACT. In this paper, we introduce a new class of functions which are analytic and univalent with negative coefficients defined by using a certain fractional calculus and fractional
calculus integral operators. Characterization property,the results on modified Hadamard product
and integrals transforms are discussed. Further, distortion theorem and radii of starlikeness and
convexity are also determined here.
Key words and phrases: Univalent, Convex, Starlike, Uniformly convex , Fractional calculus integral operator.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION AND P RELIMINARIES
Fractional calculus operators have recently found interesting applications in the theory of
analytic functions. The classical definition of fractional calculus and its other generalizations
have fruitfully been applied in obtaining, the characterization properties, coefficient estimates
and distortion inequalities for various subclasses of analytic functions.
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
062-05
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G. M URUGUSUNDARAMOORTHY , T HOMAS ROSY , AND M ASLINA DARUS
Denote by A the class of functions of the form
f (z) = z +
(1.1)
∞
X
an z n
n=2
which are analytic and univalent in the open disc E = {z : z ∈ C and |z| < 1}. Also denote by
T [11] the subclass of A consisting of functions of the form
f (z) = z −
(1.2)
∞
X
an z n ,
(an ≥ 0).
n=2
A function f ∈ A is said to be in the class of uniformly convex functions of order α, denoted
by U CV (α) [9] if
00
zf (z)
zf 00 (z)
− α ≥ β 0
− 1 ,
(1.3)
Re 1 + 0
f (z)
f (z)
and is said to be in a corresponding subclass of U CV (α) denote by Sp (α) if
0
0
zf (z)
zf (z)
(1.4)
Re
− α ≥ β − 1 ,
f (z)
f (z)
where −1 ≤ α ≤ 1 and z ∈ E.
The class of uniformly convex and uniformly starlike functions has been extensively studied
by Goodman [3, 4] and Ma and Minda [6].
P
n
If f of the form (1.1) and g(z) = z + ∞
n=2 bn z are two functions in A, then the Hadamard
product (or convolution) of f and g is denoted by f ∗ g and is given by
(f ∗ g)(z) = z +
(1.5)
∞
X
an b n z n .
n=2
Let φ(a, c; z) be the incomplete beta function defined by
(1.6)
φ(a, c; z) = z +
∞
X
(a)n
n=2
(c)n
zn,
c 6= 0, −1, −2, . . . ,
where (λ)n is the Pochhammer symbol defined in terms of the Gamma functions, by
(
)
1
n=0
Γ(λ + n)
(λ)n =
=
Γ(λ)
λ(λ + 1)(λ + 2) · · · (λ + n − 1), n ∈ N }
Further suppose
L(a, c)f (z) = φ(a, c; z) ∗ f (z),
for
f ∈A
where L(a, c) is called Carlson - Shaffer operator [2].
For real number µ (−∞ < µ < 1) and γ (−∞ < γ < 1) and a positive real number η, we
define the operator
µ,γ,η
U0,z
: A −→ A
by
(1.7)
µ,γ,η
U0,z
=z+
∞
X
n=2
J. Inequal. Pure and Appl. Math., 6(3) Art. 86, 2005
(2 − γ + η)n−1 (2)n−1
an z n ,
(2 − γ)n−1 (2 − µ + η)n−1
http://jipam.vu.edu.au/
U NIFORMLY C ONVEX F UNCTIONS
which for f (z) 6= 0 may be written as

Γ(2 − γ)Γ(2 − µ + γ) γ µ,γ,η


z J0,z f (z);


Γ(2 − γ + η)
µ,γ,η
(1.8)
U0,z f (z) =

Γ(2 − γ)Γ(2 − µ + γ) γ −µ,γ,η


z I0,z f (z);

Γ(2 − γ + η)
3
0≤µ<1
−∞ ≤ µ < 0
µ,γ,η
−µ,γ,η
where J0,z
and I0,z
are fractional differential and fractional integral operators [12] respectively.
It is interesting to observe that
Uzµ f (z) = Γ(2 − µ)z µ Dzµ f (z),
=
(1.9)
−∞<µ<1
Ωµz f (z)
and Dzµ is due to Owa [7]. Uzµ is called a fractional integral operator of order µ, if −∞ < µ < 0
and is called fractional differential operator of order µ if 0 ≤ µ < 1.
Further note that
µ,γ,η
U0,z
f (z) = f (z) if
µ=γ=0
µ,γ,η
U0,z
f (z) = zf 0 (z) if µ = γ = 1.
∗
For −1 ≤ α < 1, a function f ∈ A is said to be in Sµ,γ,η
(α), if and only if
µ,γ,η
µ,γ,η
z(U0,z
f (z))0
z(U0,z
f (z))0
(1.10)
Re
− α ≥ − 1 , z ∈ ∆.
µ,γ,η
µ,γ,η
U0,z f (z)
U0,z f (z)
where −∞ < µ < 1, −∞ < γ < 1, and η ∈ R+ .
∗
Now let us write T R(µ, γ, η, α) = Sµ,γ,η
(α) ∩ T.
It follows from the statement, that for µ = γ = 0, we have
∗
Sµ,γ,η
(α) = Sp (α)
and for µ = γ −→ 1, we have
∗
Sµ,γ,η
(α) = U CV (α).
The classes Sp (α) and U CV (α) are introduced and studied by various authors including [8],
[9] and [1].
2. C HARACTERIZATION P ROPERTY
We now investigate the characterization property for the function f to belong to the class
∗
Sµ,γ,η
(α), by obtaining the coefficient bounds.
Definition 2.1. A function f is in T R(µ, γ, η, α) if f satisfies the analytic characterization
µ,γ,η
µ,γ,η
z(U0,z
f (z))0
f (z))0
z(U0,z
(2.1)
Re
− α > − 1 ,
µ,γ,η
µ,γ,η
U0,z f (z)
U0,z f (z)
where 0 ≤ α < 1, −∞ < µ < 1, −∞ < γ < 1, and η ∈ R.
Theorem 2.1 (Coefficient Bounds). A function f defined by (1.2) is in the class T R(µ, γ, η, α)
if and only if
∞
X
(2 − γ + η)n−1 (2)n−1
2n − 1 − α
(2.2)
·
|an | ≤ 1
(2 − γ)n−1 (2 − µ + η)n−1
1−α
n=2
where 0 ≤ α < 1, −∞ < µ < 1, −∞ < γ < 1, and η ∈ R.
J. Inequal. Pure and Appl. Math., 6(3) Art. 86, 2005
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4
G. M URUGUSUNDARAMOORTHY , T HOMAS ROSY , AND M ASLINA DARUS
Proof. It suffices to show that
µ,γ,η
µ,γ,η
z(U0,z
f (z))0
z(U0,z
f (z))0
−α ,
µ,γ,η
U µ,γ,η f (z) − 1 ≤ Re
U0,z
f (z)
0,z
and we have
µ,γ,η
µ,γ,η
z(U0,z
f (z))0
z(U0,z
f (z))0
− 1 + (1 − α).
µ,γ,η
U µ,γ,η f (z) − 1 ≤ Re
U0,z
f (z)
0,z
That is
µ,γ,η
µ,γ,η
µ,γ,η
z(U0,z
z(U0,z
z(U0,z
f (z))0
f (z))0
f (z))0
− 1 ≤ 2
− 1
µ,γ,η
µ,γ,η
U µ,γ,η f (z) − 1 − Re
U0,z f (z)
U0,z f (z)
0,z
P∞
(n − 1)ψ(n)|an |
≤ n=2P∞
1 − n=2 ψ(n)|an |
where
(2 − γ + η)n−1 (2)n−1
.
(2 − γ)n−1 (2 − µ + η)n−1
The above expression is bounded by (1 − α) and hence the assertion of the result.
Now we need to show that f ∈ T R(µ, γ, η, α) satisfies the coefficient inequality. If f ∈
T R(µ, γ, η, α) and z is real then (2.1) yields
P
P∞
n−1
1− ∞
nψ(n)a
z
1
−
(n − 1)ψ(n)an z n−1
n
n=2
Pn=2
P
−
α
≥
.
∞
n−1
1 − n=2 ψ(n)an z n−1
1− ∞
n=2 ψ(n)an z
ψ(n) =
Letting z → 1 along the real axis leads to the desired inequality
∞
X
(2n − 1 − α)ψ(n)an ≤ 1 − α.
n=2
Corollary 2.2. Let a function f defined by (1.2) belong to the class T R(µ, γ, η, α). Then
an ≤
1−α
(2 − γ)n−1 (2 − µ + η)n−1
·
,
(2 − γ + η)n−1 (2)n−1
2n − 1 − α
n ≥ 2.
Next we consider the growth and distortion theorem for the class T R(µ, γ, η, α). We shall
omit the proof as the techniques are similar to various other papers.
Theorem 2.3. Let the function f defined by (1.2) be in the class T R(µ, γ, η, α). Then
(2.3)
|z| − |z|2
(2 − γ)(2 − µ + η)(1 − α)
µ,γ,η
≤ |U0,z
f (z)|
2(2 − γ + η)(3 − α)
(2 − γ)(2 − µ + η)(1 − α)
≤ |z| + |z|2
2(2 − γ + η)(3 − α)
and
(2.4)
1 − |z|
(2 − γ)(2 − µ + η)(1 − α)
µ,γ,η
≤ |(U0,z
f (z))0 |
(2 − γ + η)(3 − α)
(2 − γ)(2 − µ + η)(1 − α)
≤ 1 + |z|
.
(2 − γ + η)(3 − α)
J. Inequal. Pure and Appl. Math., 6(3) Art. 86, 2005
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U NIFORMLY C ONVEX F UNCTIONS
5
The bounds (2.3) and (2.4) are attained for functions given by
(2.5)
f (z) = z −
(2 − γ)(2 − µ + η)(1 − α)z 2
.
2(2 − γ + η)(3 − α)
Theorem 2.4. Let a function f be defined by (1.2) and
g(z) = z −
(2.6)
∞
X
bn z n
n=2
be in the class T R(µ, γ, η, α). Then the function h defined by
(2.7)
h(z) = (1 − λ)f (z) + λg(z) = z −
∞
X
qn z n
n=2
where qn = (1 − λ)an + λbn , 0 ≤ λ ≤ 1 is also in the class T R(µ, γ, η, α).
Proof. The result follows easily by using (2.2) and (2.7).
We prove the following theorem by defining functions fj (z) (j = 1, 2, . . . , m) of the form
(2.8)
fj (z) = z −
∞
X
an,j z n
an,j ≥ 0, z ∈ U.
for
n=2
Theorem 2.5 (Closure theorem). Let the functions fj (z) (j = 1, 2 . . . , m) defined by (2.8) be
in the classes T R(µ, γ, η, αj ) (j = 1, 2, . . . , m) respectively. Then the function h(z) defined by
!
∞
m
1 X X
an,j z n
h(z) = z −
m n=2 j=1
is in the class T R(µ, γ, η, α) where
(2.9)
α = min {αj }
1≤j≤m
with 0 ≤ αj < 1.
Proof. Since fj ∈ T R(µ, γ, η, αj ) (j = 2, . . . , m) by applying Theorem 2.1, we observe that
!
!
∞
m
m
∞
X
1 X
1 X X
ψ(n)(2n − 1 − α)
an,j =
ψ(n)(2n − 1 − α)an,j
m
m
n=2
n=2
j=1
j=1
m
1 X
≤
(1 − αj ) ≤ 1 − α,
m j=1
which in view of Theorem 2.1, again implies that h ∈ T R(µ, γ, η, α) and the proof is complete.
3. R ESULTS I NVOLVING M ODIFIED H ADAMARD P RODUCTS
We let
(f ∗ g)(z) = z −
∞
X
an bn z n
n=2
be the modified Hadamard product of functions f and g defined by (1.2) and (2.6) respectively.
The following results are proved using the techniques of Schild and Silverman [10].
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G. M URUGUSUNDARAMOORTHY , T HOMAS ROSY , AND M ASLINA DARUS
Theorem 3.1. For functions fj (z) (j = 1, 2) defined by (2.8), let f1 (z) ∈ T R(µ, γ, η, α) and
f2 (z) ∈ T R(µ, γ, η, β). Then f1 ∗ f2 ∈ T R(µ, γ, η, ξ) where
2(1 − α)(1 − β)
(3.1)
ξ = ξ(µ, γ, η, β) = 1 −
,
(3 − α)(3 − β)ψ(2) − (1 − α)(1 − β)
where ψ(2) =
(2−γ+η)(2)
.
(2−γ)(2−µ+η)
The result is the best possible for
1−α
z2,
(3 − α)ψ(2)
1−β
f2 (z) = z −
z2,
(3 − β)ψ(2)
f1 (z) = z −
where ψ(2) =
(2−γ+η)(2)
.
(2−γ)(2−µ+η)
Proof. In the view of Theorem 2.1, it suffices to prove that
∞
X
2n − 1 − ξ
ψ(n)an,1 an,2 ≤ 1,
1
−
ξ
n=2
where ξ is defined by (3.1) under the hypothesis, it follows from (2.1) and the Cauchy-Schwarz
inequality that
∞
X
[2n − 1 − α]1/2 [2n − 1 − β]1/2
√
p
(3.2)
ψ(n) an,1 an,2 ≤ 1.
(1 − α)(1 − β)
n=2
Thus we need to find largest ξ such that
∞
∞
X
X
[2n − 1 − α]1/2 [2n − 1 − β]1/2
2n − 1 − ξ
√
p
ψ(n)an,1 an,2 ≤
ψ(n) an,1 an,2 ≤ 1
1−ξ
(1 − α)(1 − β)
n=2
n=2
or, equivalently that
[2n − 1 − α]1/2 [2n − 1 − β]1/2
1−ξ
p
for n ≥ 2.
·
2n − 1 − ξ
(1 − α)(1 − β)
By virtue of (3.2) it is sufficient to find the largest ψ such that
p
(1 − α)(1 − β)
[2n − 1 − α]1/2 [2n − 1 − β]1/2 ψ(n)
1−ξ
[2n − 1 − α]1/2 [2n − 1 − β]1/2
p
·
≤
for n ≥ 2
2n − 1 − ξ
(1 − α)(1 − β)
√
an,1 an,2 ≤
which yields
ξ ≤1−
2(n − 1)(1 − α)(1 − β)
,
(2n − 1 − α)(2n − 1 − β)ψ(n) − (1 − α)(1 − β)
where
(2 − γ + η)n−1 (2)n−1
for n ≥ 2.
(2 − γ)n−1 (2 − µ + η)n−1
Since ψ(n) is a decreasing function of n (n ≥ 2), we have
2(1 − α)(1 − β)
ξ = ξ(µ, γ, η, α, β) = 1 −
,
(3 − α)(3 − β)ψ(2) − (1 − α)(1 − β)
(3.3)
where ψ(2) =
ψ(n) =
(2−γ+η)(2)
.
(2−γ)(2−µ+η)
Thus completes the proof.
J. Inequal. Pure and Appl. Math., 6(3) Art. 86, 2005
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U NIFORMLY C ONVEX F UNCTIONS
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Theorem 3.2. Let the functions fj (z) (j = 1, 2) defined by (2.8) be in the class T R(µ, γ, η, α).
Then (f1 ∗ f2 )(z) ∈ T R(µ, γ, η, δ), where
δ =1−
with ψ(2) =
2(1 − α)2
,
(3 − α)2 ψ(2) − (1 − α)2
(2−γ+η)(2)
.
(2−γ)(2−µ+η)
Proof. By taking β = α in the above theorem, the results follows.
Theorem 3.3. Let the function f defined by (1.2) be in the class T R(µ, γ, η, α). Also let
g(z) = z −
∞
X
bn z n
for |bn | ≤ 1.
n=2
Then (f ∗ g)(z) ∈ T R(µ, γ, η, α).
Proof. Since
∞
X
ψ(n)(2n − 1 − α)|an bn | =
n=2
∞
X
ψ(n)(2n − 1 − α)an |bn |
n=2
≤
∞
X
ψ(n)(2n − 1 − α)an
n=2
≤1−α
(by Theorem 2.1),
where ψ(n) is defined by (3.3). Hence it follows that (f ∗ g)(z) ∈ T R(µ, γ, η, α).
Corollary
3.4. Let the function f defined by (1.2) be in the class T R(µ, γ, η, α). Also let g(z) =
P
n
b
z− ∞
n=2 n z for 0 ≤ bn ≤ 1. Then (f ∗ g)(z) ∈ T R(µ, γ, η, α).
For functions in the class T R(µ, γ, η, α) we can prove the following inclusion property also.
Theorem 3.5. Let the functions fj (z) (j = 1, 2) defined by (2.5) be in the class T R(µ, γ, η, α).
Then the function h defined by
h(z) = z −
∞
X
(a2n,1 + a2n,2 )z n
n=2
is in the class T R(µ, γ, η, ∆) where
4(1 − α)2
∆=1−
(3 − α)2 ψ(2) − 2(1 − α)2
2(2 − γ + η)
ψ(2) =
.
(2 − γ)(2 − µ + η)
with
Proof. In view of Theorem 2.1, it is sufficient to prove that
(3.4)
∞
X
ψ(n)
n=2
2n − 1 − ∆ 2
(an,1 + a2n,2 ) ≤ 1
1−∆
where fj (z) ∈ T R(µ, γ, η, α) (j = 1, 2). We find from (2.8) and Theorem 2.1, that
2
2
∞ ∞ X
X
2n − 1 − α
2n − 1 − α
2
(3.5)
an,j ≤
ψ(n)
an,j ≤ 1,
ψ(n)
1−α
1−α
n=2
n=2
J. Inequal. Pure and Appl. Math., 6(3) Art. 86, 2005
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8
G. M URUGUSUNDARAMOORTHY , T HOMAS ROSY , AND M ASLINA DARUS
which would yield
(3.6)
2
∞
X
1
2n − 1 − α
ψ(n)
(a2n,1 + a2n,2 ) ≤ 1.
2
1
−
α
n=2
On comparing (3.5) and (3.6) it can be seen that inequality (3.4) will be satisfied if
2
1
2n − 1 − α
2n − 1 − ∆ 2
2
(a2n,1 + a2n,2 ).
ψ(n)
(an,1 + an,2 ) ≤
ψ(n)
1−∆
2
1−α
That is, if
(3.7)
∆≤1−
4(1 − α)2
,
(2n − 1 − α)2 ψ(n) − 2(1 − α)2
where ψ(n) is given by (3.3). Hence we conclude from (3.7)
∆ = ∆(µ, γ, η, α) = 1 −
where ψ(2) =
2(2−γ+η)
(2−γ)(2−µ+η)
4(1 − α)2
ψ(2) − 2(1 − α)2 ,
2
(3 − α)
which completes the proof.
4. I NTEGRAL T RANSFORM
OF THE
C LASS T R(µ, γ, η, α)
For f ∈ T R(µ, γ, η, α) we define the integral transform
Z 1
f (tz)
Vλ (f )(z) =
λ(t)
dt,
t
0
R1
where λ is a real valued, non-negative weight function normalized so that 0 λ(t)dt = 1. Since
special cases of λ(t) are particularly interesting such as λ(t) = (1 + c)tc , c > −1, for which Vλ
is known as the Bernardi operator, and
δ−1
(c + 1)δ c
1
λ(t) =
t log
, c > −1, δ ≥ 0
λ(δ)
t
which gives the Komatu operator. For more details see [5].
First we show that the class T R(µ, γ, η, α) is closed under Vλ (f ).
Theorem 4.1. Let f ∈ T R(µ, γ, η, α). Then Vλ (f ) ∈ T R(µ, γ, η, α).
Proof. By definition, we have
!
Z
∞
X
(c + 1)δ 1
Vλ (f ) =
(−1)δ−1 tc (log t)δ−1 z −
an z n tn−1 dt
λ(δ)
0
n=2
"Z
! #
∞
1
δ−1
δ
X
(−1) (c + 1)
=
lim+
tc (log t)δ−1 z −
an z n tn−1 dt ,
r→0
λ(δ)
r
n=2
and a simple calculation gives
Vλ (f )(z) = z −
δ
∞ X
c+1
n=2
c+n
an z n .
We need to prove that
δ
∞
X
2n − 1 − α
(2 − γ + η)n−1 (2)n−1
c+1
(4.1)
·
an < 1.
1−α
(2 − γ)n−1 (2 − µ + η)n−1 c + n
n=2
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U NIFORMLY C ONVEX F UNCTIONS
9
On the other hand by Theorem 2.1, f ∈ T R(µ, γ, η, α) if and only if
∞
X
2n − 1 − α
(2 − γ + η)n−1 (2)n−1
·
< 1.
1
−
α
(2
−
γ)
(2
−
µ
+
η)
n−1
n−1
n=2
Hence
c+1
c+n
< 1. Therefore (4.1) holds and the proof is complete.
Next we provide a starlike condition for functions in T R(µ, γ, η, α) and Vλ (f ).
Theorem 4.2. Let f ∈ T R(µ, γ, η, α). Then Vλ (f ) is starlike of order 0 ≤ γ < 1 in |z| < R1
where
1
"
# n−1
δ
c+n
1 − γ(2n − 1 − α)
.
R1 = inf
·
φ(n)
n
c+1
(n − γ)(1 − α)
Proof. It is sufficient to prove
z(Vλ (f )(z))0
(4.2)
Vλ (f )(z) − 1 < 1 − γ.
For the left hand side of (4.2) we have
P∞
c+1 δ
n=2 (1 − n)( c+n
z(Vλ (f )(z))0
) an z n−1 Vλ (f )(z) − 1 = 1 − P∞ ( c+1 )δ a z n−1 n
n=2 c+n
P∞
c+1 δ
n−1
n=2 (n − 1)( c+n ) an |z|
P
≤
.
c+1 δ
n−1
1− ∞
n=2 ( c+n ) an |z|
This last expression is less than (1 − γ) since
δ
(1 − γ)[2n − 1 − α]
c+1
n−1
φ(n).
|z|
<
c+n
(n − γ)(1 − α)
Therefore the proof is complete.
Using the fact that f is convex if and only if zf 0 is starlike, we obtain the following:
Theorem 4.3. Let f ∈ T R(µ, γ, η, α). Then Vλ (f ) is convex of order 0 ≤ γ < 1 in |z| < R2
where
1
"
# n−1
δ
c + n (1 − γ)[2n − 1 − α]
R2 = inf
φ(n)
.
n
c+1
n(n − γ)(1 − α)
We omit the proof as it is easily derived.
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