ISSN 2307-7743 http://scienceasia.asia
FEKETE-SZEGO¨ PROBLEM FOR SOME SUBCLASSES
OF COMPLEX ORDER RELATED TO SA˘LA˘GEAN
OPERATOR
C.SELVARAJ, T.R.K.KUMAR
Abstract. In the present investigation,sharp upper bounds of
a3−µa22
for function f(z) belonging to certain subclasses of
Reh1 +1
b n
(1−α)f(zz)+αf0(z)−1oi0 are obtained.Also cer-tain applications of the main results for subclasses of functions defined by convolution with a normalized analytic function are given.In particular,F ekete−Szeg¨oinequalities for certain classes of functions defined through fractional derivatives are obtained.
1. Introduction
We let A to denote the class of all analytic functions f(z) of the form
(1.1) f(z) = z+
∞
X
n=2
anzn,(z ∈D={z ∈C:|z|<1})
and S be the subclass of A consisting of univalent functions. For t-wo analytic functions f(z) given by (1.1) and g(z) = z +P∞
n=2bnz
n ,
their convolution (or Hadamard product ) is defined to be the function (f∗g) (z) given by (f∗g) (z) =z+P∞
n=2anbnz
n.
For two functions f, g ∈ A,we say that the function f(z) is subor-dinate to g(z) in D and write f ≺ g or f(z) ≺ g(z) (z ∈D), if there exists an analytic function w(z) with w(0) = 0 and |w(z)| <1 (z ∈D) ,such that
f(z) = g(w(z)) (z ∈D).
2010Mathematics Subject Classification. 30C45.
Key words and phrases. analytic functions, subordination, coefficient problem, Fekete-Szeg¨oinequality.
c
2014 Science Asia
In particular , if the functiong is univalent in D ,the above subordi-nation is equivalent to f(0) =g(0) andf(D)⊂g(D).
Throughout this paper, we assume that φ is an analytic univalent function with positive real part in D ,φ(D) is symmetric with respect to the real axis and starlike with respect to φ(0) = 1, andφ0(0)>0.The Taylor’s series expansion of such function is of the form
(1.2) φ(z) = 1 +B1z+B2z2 +B3z3+· · ·with B1 >0.
In the present investigation,we obtain Fekete-Szego¨inequality for func-tion in a more general class <(α, φ) which we define below.We also give applications of our results to certain functions defined through Hadamard product and functions defined by fractional derivatives.
Definition 1.1. Let α ≥ 0. A function f ∈ A given by(1.1) is in the class<(α, φ),if it satisfies
(1.3) Re
1 + 1
b
(1−α)f(z)
z +αf
0
(z)−1
0
Lemma 1.1. If p1(z) = 1 +c1z+c2z2+c3z3+· · · is analytic function
with positive real part in D,then
c2−vc21
≤
−4v + 2 if v ≤0,
2 if 0≤v ≤1,
4v −2 if v ≥1.
when v < 0 or v > 1, the equality holds if and only if p1(z) is
(1 +z)/(1−z) or one of its rotations.If 0 < v <1 ,then the equality holds if and only if p1(z) is (1 +z2)/(1−z2) or one of its rotations.If
v = 0 ,the equality holds if and only if
p1(z) =
1 2 +
1 2λ
1 +z
1−z +
1 2−
1 2λ
1−z
1 +z ,(0≤λ ≤1)
or one of its rotations.If v = 1 , the equality holds if and only ifp1(z)
is the reciprocal of one of the functions such that the equality holds in the case v = 0.Also the above upper bound is sharp and it can be improved as follows:
when 0< v <1,
c2−vc21
+v|c1|
2
and
c2−vc21
+ (1−v)|c1|2 ≤2 (1/2< v ≤1).
Let a differential operator be defined S˘al˘agean [10] on a class of analytic functions of the form (1.1) as follows
D0f(z) = f(z), D1f(z) =Df(z) = zf0(z),
and in general
Dnf(z) =D Dn−1f(z),(n∈N0 =N∪ {0}).
We easily find that
(1.4) Dnf(z) =z+
∞
X
k=2
knakzn (n∈N0).
2. Fekete-SzegO¨ Poroblem for the Function class <(α, φ) By using Lemma 1.1 ,we prove the following Fekete-Szego¨ inequali-ties.
Theorem 2.1. Let b be a non zero complex number.If f(z) given by (1.1) belongs to Nb
n(α, φ), then
a3−µa22 ≤
b
3n
h
B2
1+2α − µB2
1b
(1+α)2 3 4
ni
if µ≤σ1,
bB1
3n(1+2α) if σ1 ≤µ≤σ2,
− b
3n
h
B2
1+2α − µB2
1b
(1+α)2 3 4
ni
if µ≥σ2.
where
σ1 :=
(1 +α)2(B2−B1)
b(1 + 2α)B2 1
4 3
n
, σ2 :=
(1 +α)2(B2+B1)
b(1 + 2α)B2 1
4 3
n
The result is sharp.
If f(z) ∈ <(α, φ), then there exists a Schwarz functions w(z) ana-lytic in D with
w(0) = 0 and |w(z)|<1 (z ∈D) ,such that
(2.1) 1 + 1
b
(1−α)f(z)
z +αf
0
(z)−1
Define the functionp1 by
(2.2) p1 =
1 +w(z)
1−w(z) = 1 +c1z+c2z
2+c
3z3+· · ·
since w(z) is a Schwarz function,we see that <(p1(z)) > 0 (z ∈D)
and p1(0) = 1.Now, defining the function p(z) by
(2.3)
p(z) = 1 +1
b
(1−α)f(z)
z +αf
0
(z)−1
= 1 +b1z+b2z2+b3z3+· · ·
we find from (2.1) and (2.2) that
(2.4) p(z) = φ
p1(z)−1
p1(z) + 1
.
Thus by using (2.2) in (2.4), we obtain
b1 =
1
2B1c1 b2 = 1 2B1
c2−
1 2c
2 1
+ 1
4B2c
2 1.
Also, from (2.3) we obtain
b1 =
1
b (1 +α) 2
n
a2 b2 =
1
b (1 + 2α) 3
n
a3.
Therefore we have
(2.5) a3−µa22 =
bB1
3n2 (1 + 2α)
c2−vc21
.
where
v := 1 2
1− B2
B1
+ bµB1(1 + 2α) (1 +α)2
3 4
n
.
Our result now follows by an application of lemma 1.1 .To show that the bounds are sharp,we define the functionsKα
φn (n = 2,3,4, . . .) by
1 + 1
b
(
(1−α)
Kα φn(z)
z +α
Kφα n
0
(z)−1 )
=φ zn−1
,
Kφαn(0) = 0 = Kφαn 0
(0)−1 and the function Fα
λ and Gαλ (0≤λ≤1) by
1 + 1
b
(1−α)[F
α λ (z)]
z +α[F
α λ]
0
(z)−1
1 + 1
b
(1−α)[G
α λ(z)]
z +α[G
α λ]
0
(z)−1
=φ zn−1,
Gαλ(0) = 0 = [Gαλ]0(0)−1 Clearly the functions Kα
φn , F
α
λ and Gαλ ∈ <(α, φ).Also we writeKφα:=
Kα φ2.
Ifµ≤σ1 orµ≥σ2, then the equality holds if and only if f isKφα or
one of its rotations.When σ1 ≤µ≤σ2 , the equality holds if and only
if f is Kα
φ3 or one of its rotations.If µ=σ1 , then the equality holds if and only if f is Fα
λ or one of its rotations.If µ=σ2, then the equality
holds if and only if f is Gα
λ or one of its rotations.
Remark 2.1. Ifσ1 ≤µ≤σ2, then,in view of Lemma 1.1 , Theorem
2.1 can be improved. Let σ3 be given by
σ3 :=
(1 +α)2B2
b(1 + 2α)B2 1
4 3
n
.
Letf ∈ <(α, φ).If σ1 ≤µ≤σ3 , then
a3−µa22 +
1
b(1 + 2α)B2 1
4 3
n
(1 +α)2(B2−B1)
3 4
n
+µb(1 + 2α)B12
4 3
n |a2|
2
≤ bB1 3n(1 + 2α).
If σ3 ≤µ≤σ2 , then
a3 −µa22
+
1
b(1 + 2α)B2 1
4 3
n
(1 +α)2(B2+B1)
3 4
n
−µb(1 + 2α)B12
4 3
n
|a2|2 ≤
bB1
3n(1 + 2α).
For φ(z) = (1 +Cz)/(1 +Dz) , −1 ≤ D < C ≤ 1.Theorem 2.1 leads to the following results:
a3−µa22 ≤
b(D−C) 3n(1+2α)
h
D− bµ(D−C)(1+2α)
2(1+α)2 3 4
ni
if µ≤ −2
b
h
(1+D)(1+α)2 (1+2α)(C−D)
4 3
ni
,
b(C−D)
3n(1+2α) if − 2b
h
(1+D)(1+α)2 (1+2α)(C−D)
4 3
ni
≤µ≤2
b
h
(1−D)(1+α)2 (1+2α)(C−D)
4 3
ni
,
−3bn((1+2D−Cα))
h
D− bµ(D−C)(1+2α)
2(1+α)2 3 4
ni
if µ≤ 2
b
h
(1−D)(1+α)2 (1+2α)(C−D)
4 3
ni
.
For φ(z) = (1 +z)/(1−z) ,Theorem 2.1 leads to the following re-sults:
Corollary 2.3. Let−1≤D < C ≤1.Iff ∈ <(α,(1 +z)/(1−z)),then
a3−µa22
≤ 2b
3n(1+2α)
h
1− µb(1+2α)
(1+α)2 3 4
ni
if µ≤0,
2b
3n(1+2α) if 0≤µ≤
2(1+α)2
b(1+2α) 4 3
n
,
− 2b
3n(1+2α)
h
1− µb(1+2α)
(1+α)2 3 4
ni
if µ≥ 2(1+b(1+2αα)2) 4 3
n
.
For C = 1−2β with 0 ≤β <1 and D=−1, Corollary 2.2 reduces to the following result:
Corollary 2.4.
a3−µa22 ≤
2b(1−β) 3n(1+2α)
h
1− µb(1−β)(1+2α)
(1+α)2 3 4
ni
if µ≤0,
2b(1−β)
3n(1+2α) if 0≤µ≤
2(1+α)2
b(1−β)(1+2α) 4 3
n
,
−32nb(1+2(1−βα))
h
1− µb(1−β)(1+2α)
(1+α)2 3 4
ni
if µ≥ b(12(1+−β)(1+2α)2α) 4 3
n
.
3. Application to Functions Defined by Fractional Derivatives
Dλzf(z) := 1 Γ (1−λ)
d dz
Z z
0
f(ζ)
(z−ζ)λdζ (0≤λ <1),
where the multiplicity of (z−ζ)λis removed by requiring thatlog(z−ζ) is real for z−ζ >0. Using the above Definition 3.1 and its known ex-tensions involving fractional derivatives and fractional integrals, Owa and Srivastava [3] introduced the operator Ωλ :A → A defined by
Ωλf
(z) = Γ (2−λ)zλDλzf(z) (λ6= 2,3,4, . . .).
The class <λ(α, φ) consists of functions f ∈ A for which Ωλf ∈
<(α, φ). Note that <λ(α, φ) is the special case of the class <g(α, φ)
when
(3.1) g(z) = z+
∞
X
n=2
Γ (n+ 1) Γ (2−λ) Γ (n+ 1−λ) z
n.
Let g(z) = z +P∞
n=2gnz
n (g
n>0).Since f(z) = z +
P∞
n=2anz
n ∈
<g(α, φ) if and only if (f∗g) (z) = z +P∞
n=2gnanz
n ∈ <(α, φ),we
obtain the coefficient estimate for functions in the class <g(λ, φ), from
the corresponding estimate for functions in the class <(λ, φ).Applying Theorem 2.1 for the function (f∗g) (z) =z+g2a2z2+g3a3z3+· · ·,we
get the following theorem after an obvious change of the parameter µ:
Theorem 3.1. Let the function φ(z) be given by φ(z) = 1 +B1z+
B2z2+B3z3+· · ·.If f(z) given by (1.1) belongs to <g(α, φ) , then
a3−µa22 ≤
b g3
h
B2
1+2α −
µbg3B21
(1+α)2g2 2 i
if µ≤σ1,
bB1
g3(1+2α) if σ1 ≤µ≤σ2, −b
g3 h
B2
1+2α −
µbg3B21
(1+α)2g2 2 i
if µ≥σ2.
where
σ1 :=
g2
2(1 +α) 2
(B2−B1)
bg3(1 + 2α)B12
, σ1 :=
g2
2(1 +α) 2
(B2+B1)
bg3(1 + 2α)B12
Since
Ωλf(z) =z+
∞
X
n=2
Γ (n+ 1) Γ (2−λ) Γ (n+ 1−λ) anz
n ,
we have
(3.2) g2 :=
Γ (3) Γ (2−λ) Γ (3−λ) =
2 2−λ
and
(3.3) g3 :=
Γ (4) Γ (2−λ) Γ (4−λ) =
6
(2−λ) (3−λ).
For g2 and g3 given by (3.2) and (3.3) , Theorem 3.1 reduces to the
following:
Theorem 3.2. Let the function φ(z) be given by φ(z) = 1 +B1z+
B2z2+B3z3+· · ·.If f(z) given by (1.1) belongs to <λ(α, φ) , then
a3−µa22 ≤
b(2−λ)(3−λ) 6
h
B2
1+2α −
µb3(2−λ)B2 1
2(1+α)2(3−λ)
i
if µ≤σ1,
b(2−λ)(3−λ)B1
6(1+2α) if σ1 ≤µ≤σ2,
−b(2−λ6)(3−λ)h B2
1+2α −
µb3(2−λ)B2 1
2(1+α)2(3−λ)
i
if µ≥σ2.
where
σ1 :=
2 (3−λ) (1 +α)2(B2−B1)
3b(2−λ) (1 + 2α)B2 1
, σ2 :=
2 (3−λ) (1 +α)2(B2+B1)
3b(2−λ) (1 + 2α)B2 1
The result is sharp.
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C.Selvaraj, Presidency College, Chennai-600 005, Tamilnadu, India
T.R.K.Kumar∗, R.M.K.Engineering College, R.S.M.Nagar, Kavaraipettai-601 206, Tamilnadu, India
∗
