# Second Hankel Determinant for a Class of Analytic Functions Defined by Fractional Derivative

Authors: A. K. Mishra; P. Gochhayat
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008-02-03)

---

## Abstract

By making use of the fractional differential
operatorΩzλ due to Owa and Srivastava, a class of
analytic functions ℛλ(α,ρ)(0≤ρ≤1,0≤λ<1,|α|<π/2) is introduced. The sharp bound for the nonlinear functional |a2a4−a32| is found. Several basic properties such as inclusion, subordination,
integral transform, Hadamard product are also studied.


---

## Body

## 1. Introduction

Let𝒜 denote the
class of functions analytic in the open unit disc(1.1)𝒰:={z:z∈ℂ,|z|<1} and let 𝒜0 be the class of
functions f in 𝒜 given by the normalized power series(1.2)f(z)=z+∑n=2∞anzn(z∈𝒰).Also let𝒮,𝒮*(β),𝒞𝒱(β), and 𝒦 denote,
respectively, the subclasses of 𝒜0 consisting of
functions which are univalent, starlike of order β, convex of order β (cf. [1]), and close-to-convex (cf. [2]) in 𝒰. In particular, 𝒮*(0)=𝒮* and 𝒞𝒱(0)=𝒞𝒱  are the familiar
classes of starlike and convex functions in 𝒰 (cf. [2]).Givenf and g in 𝒜, the function f is said to be subordinate to g in 𝒰 if there exits
a function ω∈𝒜  satisfying
the conditions of the Schwarz Lemma such that f(z)=g(ω(z)),(z∈𝒰). We denote the subordination by(1.3)f(z)≺g(z)(z∈𝒰)orf≺gin𝒰. It is well
known [2] that if g is univalent in 𝒰, then f≺g in 𝒰 is equivalent
to f(0)=g(0) and f(𝒰)⊂g(𝒰).For the functionsf and g given by the
power series(1.4)f(z)=∑n=0∞anzn,g(z)=∑n=0∞bnzn(z∈𝒰), their Hadamard
product (or convolution), denoted by f*g, is defined by(1.5)(f*g)(z)=∑n=0∞anbnzn=(g*f)(z)(z∈𝒰). Note that f*g∈𝒜.By making use of the Hadamard product,
Carlson-Shaffer [3] defined the linear operator ℒ(a,c):𝒜→𝒜 by(1.6)(ℒ(a,c)f)(z):=Φ(a,c;z)*f(z)(z∈𝒰,f∈𝒜),  where(1.7)Φ(a,c;z):=∑k=0∞(a)k(c)kzk+1(z∈𝒰,c∉ℤ0−={0}∪{−1,−2,−3,…})  and (λ)k is the
Pochhammer symbol (or shifted factorial) defined in terms of the gamma
function by(1.8)(λ)k=Γ(λ+k)Γ(λ)={1(k=0),λ(λ+1)(λ+2)⋯(λ+k−1)(k∈ℕ:={1,2,…}).It can be readily verified thatℒ(a,a)(a∉ℤ0−) is the identity
operator; the operators ℒ(a,b),ℒ(c,d) commute, where b,d∉ℤ0−, that is,(1.9)ℒ(a,b)ℒ(c,d)f=ℒ(c,d)ℒ(a,b)f(f∈𝒜), and the
transitive property, that is,(1.10)ℒ(a,b)ℒ(b,c)f=ℒ(a,c)f(b,c∉ℤ0−,f∈𝒜), holds. Each of
the following definitions will also be required in our present
investigation.Definition 1.1
(cf. [4, 5], see also [6]). Let the function f
be analytic in
a simply connected region of the z-plane containing the origin.
The fractional
derivative of
fof orderλ is defined by (1.11)(Dzλf)(z)=1Γ(1−λ)ddz∫0zf(ζ)(z−ζ)λdζ(0≤λ<1), where the
multiplicity of (z−ζ)λ is removed by
requiring log(z−ζ) to be real when (z−ζ)>0.Using
Definition1.1 and its known extensions involving fractional derivatives and
fractional integrals, Owa and Srivastava [5] introduced the
fractional
differintegral operator
Ωzλ:𝒜0→𝒜0 defined by(1.12)(Ωzλf)(z)=Γ(2−λ)zλ(Dzλf)(z)(λ≠2,3,…,z∈𝒰). Note that Ωz0f(z)=f(z), Ωz1f(z)=zf′(z), and(1.13)(Ωzλf)(z)=(ℒ(2,2−λ)f)(z)(0≤λ<1,z∈𝒰).Definition 1.2
(cf. [7]). For the function   f
given by (1.2)
and   q∈ℕ:={1,2,3,…}, the   qth   Hankel
determinant of   f   is defined by (1.14)|anan+1…an+q−1an+1an+2…an+q⋮⋮⋮⋮an+q−1an+q…an+2q−2|.We now
introduce the following class of functions.Definition 1.3.
The functionf∈𝒜0   is said to be
in the class   ℛλ(α,ρ)(0≤λ<1,|α|<π/2,0≤ρ≤1)
if it satisfies
the inequality (1.15)ℜ{eiαΩzλf(z)z}>ρcosα(z∈𝒰).Write(1.16)ℛλ(0,ρ):=ℛλ(ρ).  Let 𝒫  be the family of functions p∈𝒜 satisfying p(0)=1 and ℜ(p(z))>0(z∈𝒰).It follows from (1.15) that(1.17)f∈ℛλ(α,ρ)⟺eiαΩzλf(z)z=[(1−ρ)p(z)+ρ]cosα+isinα,  where α is real, |α|<π/2,andp(z)∈𝒫.We note that(1.18)ℛ0(α,ρ):={f∈𝒜0∣ℜ{eiαf(z)z}>ρcosα},ℛ1(α,ρ):={f∈𝒜0∣ℜ{eiαf′(z)}>ρcosα},   and the class ℛλ(ρ) has been
studied in [8].It is well known (cf. [2]) that for f∈𝒮 and given by
(1.2), the sharp inequality |a3−a22|≤1 holds. This
corresponds to the Hankel determinant with q=2 and n=1. For a given family ℱ of functions in 𝒜0, the more general problem of finding sharp estimates
for |μa22−a3|(μ∈ℝorμ∈ℂ)   is popularly
known as the Fekete-Szegö problem for ℱ. The Fekete-Szegö problem for the families 𝒮,𝒮*,𝒞𝒱,𝒦 has been
completely solved by many authors including [9–12].In the present paper, we consider the Hankel
determinant forq=2 and n=2 and we find the
sharp bound for the functional |a2a4−a32|(f∈ℛλ(α,ρ)). We also obtain some basic properties of the class ℛλ(α,ρ). Our investigation includes a recent result of
Janteng et al. [13]. We also generalize some results of Ling and Ding [8].


## 2. Preliminaries

To establish
our results, we recall the following.Lemma 2.1  (See [2]).
Let the
functionp∈𝒫and be given by the series(2.1)p(z)=1+c1z+c2z2+…(z∈𝒰).  Then, the sharp
estimate(2.2)|ck|≤2(k∈ℕ) holds.Lemma 2.2  (cf. [14, Page 254], See Also [15]).
Let the
functionp∈𝒫be given by the power series
(2.1). Then,(2.3)2c2=c12+x(4−c12)  for some x,|x|≤1, and(2.4)4c3=c13+2(4−c12)c1x−c1(4−c12)x2+2(4−c12)(1−|x|2)z  for some z,|z|≤1.Lemma 2.3  (See [16]).
LetFandGbe univalent
convex functions in𝒰.Then, the Hadamard productF*Gis also a
univalent convex function in𝒰.Lemma 2.4  (See [17]).
LetFandGbe univalent
convex functions in𝒰.Also letf≺Fandg≺Gin𝒰.Then,f*g≺F*Gin𝒰.Lemma 2.5  (See [16], Also See [8]).
Letfandgbe starlike of
order1/2.Then, for each functionF(z),satisfyingℜ(F(z))>α(0≤α<1,z∈𝒰),one has(2.5)ℜ(f(z)*F(z)g(z)f(z)*g(z))>α(z∈𝒰).Lemma 2.6  (See [8]).
Let the functionh(z)=1+h1z+h2z2+…be univalent
convex in𝒰.For0≤λ<1ifΩzλf(z)/z≺h(z)(z∈𝒰),then(2.6)f(z)z≺{ℒ(2−λ,2)[zh(z)]}(z∈𝒰).


## 3. Main Results

We prove the
following.Theorem 3.1.
Let the
functionfgiven by (1.2)
be in the classℛλ(α,ρ)(0≤λ<1,−π/2<α<π/2,and0≤ρ≤1).Then,(3.1)|a2a4−a32|≤(1−ρ)2(2−λ)2(3−λ)2cos2α9.  The estimate
(3.1) is sharp.Proof.
Letf∈ℛλ(α,ρ)(0≤λ<1,−π/2<α<π/2,and0≤ρ≤1). Then, by
(1.17),(3.2)eiαΩzλf(z)z=[(1−ρ)p(z)+ρ]cosα+isinα(z∈𝒰), where p∈𝒫 and is given by (2.1). Using
(1.6), (1.7),
and (1.13), we write(3.3)Ωzλf(z)=z+∑n=2∞Γ(n+1)Γ(2−λ)Γ(n+1−λ)anzn,(z∈𝒰). Comparing the coefficients, we get(3.4)eiα2(2−λ)a2=(1−ρ)c1cosα,eiα6(2−λ)(3−λ)a3=(1−ρ)c2cosα,eiα24(2−λ)(3−λ)(4−λ)a4=(1−ρ)c3cosα.  Therefore,
(3.4) yields(3.5)|a2a4−a32|=(1−ρ)2(2−λ)2(3−λ)(cos2α)12|((4−λ)c1c34−(3−λ)c223)|.  Since the
functions p(z) and p(eiθz),(θ∈ℝ) are members of
the class 𝒫 simultaneously, we assume
without loss of generality that c1>0. For convenience of notation, we take c1=c(c∈[0,2]).
Using (2.3) along with (2.4), we get(3.6)|a2a4−a32|=(1−ρ)2(2−λ)2(3−λ)(cos2α)12×|(4−λ)c16{c3+2(4−c2)cx−c(4−c2)x2+2(4−c2)(1−|x|2)z}|=(1−ρ)2(2−λ)2(3−λ)(cos2α)48×|((4−λ)4−3−λ3)c4+((4−λ)(4−c2)c22−2c2(3−λ)(4−c2)3)x−((4−λ)(4−c2)c24+(3−λ)(4−c2)23)x2+(4−λ)(4−c2)c(1−|x|2)z2|=(1−ρ)2(2−λ)2(3−λ)(cos2α)48×|λc412+λ(4−c2)c2x6−(48−λ(16−c2)12)(4−c2)x2+(4−λ)(4−c2)c(1−|x|2)z2|. An application
of triangle inequality and replacement of |x| by μ give(3.7)|a2a4−a32|≤(1−ρ)2(2−λ)2(3−λ)(cos2α)48×[λc412+λ(4−c2)c2μ6+(4−c2)[48−λ(16−c2)]μ212+(4−λ)(4−c2)c2−(4−λ)(4−c2)cμ22]=(1−ρ)2(2−λ)2(3−λ)(cos2α)48×[λc412+(4−λ)(4−c2)c2+λ(4−c2)c2μ6+λ[c2−6(4−λ)c/λ+16(3−λ)/λ](4−c2)μ212]=(1−ρ)2(2−λ)2(3−λ)(cos2α)48×[λc412+(4−λ)(4−c2)c2+λ(4−c2)c2μ6+λ(c−β1)(c−β2)(4−c2)μ212]:=F(c,μ)(say),  where(3.8)β1=2,β2=8(3−λ)λ,0≤c≤2,0≤μ≤1. We next maximize the function F(c,μ) on the closed
square [0,2]×[0,1]. Since (3.9)∂F∂μ=(1−ρ)2(2−λ)2(3−λ)cos2α48[λ(4−c2)c26+λ(4−c2)(c−2)(c−8(3−λ)/λ)μ6],c−2<0, and c−8(3−λ)/λ<0, we have ∂F/∂μ>0 for 0<c<2,0<μ<1. Thus F(c,μ) cannot have a
maximum in the interior of the closed square [0,2]×[0,1]. Moreover, for fixed c∈[0,2],(3.10)max0≤μ≤1F(c,μ)=F(c,1)=G(c)(say). Next,(3.11)G'(c)=−(1−ρ)2(2−λ)2(3−λ)(c2−(7λ−12))ccos2α72, so that G'(c)<0 for 0<c<2 and has real
critical point at c=0. Also G(c)>G(2). Therefore, max0≤c≤2 occurs at c=0. Therefore, the upper bound of (3.7) corresponds to μ=1 and c=0. Hence,(3.12)|a2a4−a32|≤(1−ρ)2(2−λ)2(3−λ)2cos2α9 which is the assertion
(3.1). Equality holds for the
function(3.13)f(z)=Φ(2−λ,2;z)*e−iα[z(1+(1−2ρ)z21−z2cosα+isinα)]. The proof of
Theorem 3.1 is complete.The choice ofα=0 yields what follows.Corollary 3.2.
Let the functionfgiven by (1.2)
be a member of the classℛλ(ρ).Then,(3.14)|a2a4−a32|≤(1−ρ)2(2−λ)2(3−λ)29. Equality holds for the function(3.15)f(z)=ℒ(2−λ,2)*z(1+(1−2ρ)z2)1−z2.Remark 3.3.
Takingλ→1,α=0, and   ρ=0, we get a recent result due to Janteng et al. [13].Theorem 3.4.
Suppose−π/2<α<π/2,0≤ρ<1, and 0≤μ<λ<1.Then,(3.16)ℛλ(α,ρ)⊂ℛμ(α,ρ).Proof.
Let(3.17)f∈ℛλ(α,ρ)(0≤μ<λ<1,−π2<α<π2,0≤ρ≤1). Using the associative and commutative properties of
the operator ℒ, we write(3.18)Ωzμf(z)=ℒ(2,2−μ)f(z)=ℒ(2−λ,2)ℒ(2,2−λ)ℒ(2,2−μ)f(z)=ℒ(2−λ,2−μ)Ωzλf(z)=Φ(2−λ,2−μ;z)*Ωzλf(z), where the
function Φ is defined by
(1.7). Therefore,(3.19)eiαΩzμf(z)z=Φ(2−λ,2−μ;z)*(eiαΩzλf(z)/z)⋅zΦ(2−λ,2−μ;z)*z=f(z)*F(z)g(z)f(z)g(z), where f(z)=Φ(2−λ,2−μ;z),g(z)=z,F(z)=eiαΩzλf(z)/z. We note that g∈𝒮*(1/2), and ℜ(F(z))>ρcosα(0≤ρ≤1,−π/2<α<π/2). Moreover,  it is well known (cf. [18]) that Φ(2−λ,2−μ;z)∈𝒮*(1/2). Therefore, by Lemma 2.5,(3.20)ℜ(eiαΩzμf(z)z)>ρcosα(−π2<α<π2,z∈𝒰,0≤ρ≤1). Hence, f(z)∈ℛμ(α,ρ), and the proof of Theorem 3.4 is complete.Theorem 3.5.
Letf∈𝒮*(1/2)andg∈ℛλ(α,ρ)(0≤ρ≤1,−π/2<α<π/2,0≤λ<1).Then the Hadamard product(3.21)f*g∈ℛλ(α,ρ).Proof.
Since
the Hadamard product is associative and commutative, we have(3.22)Ωzλ(f*g)(z)=f(z)*Ωzλg(z). Therefore,(3.23)eiαΩzλ(f*g)(z)z=f(z)*(eiαΩzλg(z)/z)⋅zf(z)*z. Now applying
Lemma 2.5, we get(3.24)ℜ(eiαΩzλ(f*g)(z)z)>ρcosα. Hence, f*g∈ℛλ(α,ρ), and the proof
of Theorem 3.5 is complete.Theorem 3.6.
Letf∈ℛλ(α,ρ)(0≤λ<1,−π/2<α<π/2,0≤ρ≤1).Then, the functionℐ(f)defined by the
integral transform(3.25)ℐ(f)(z)=γ+1zγ∫0ztγ−1f(t)dt(z∈𝒰,γ>−1) is also in ℛλ(α,ρ).Proof.
The
Integral transformℐ(f) can be  written
in terms of Carlson-Shaffer operator as(3.26)(ℐ(f))(z)=(ℒ(γ+1,γ+2)f)(z). Hence,(3.27)(Ωzλℐ(f))(z)=ℒ(γ+1,γ+2)Ωzλf(z)=Φ(γ+1,γ+2;z)*Ωzλf(z). Therefore,(3.28)eiα(Ωzλℐ(f))(z)z=Φ(γ+1,γ+2;z)*(eiαΩzλf(z)/z)zΦ(γ+1,γ+2;z)*z. Using a result
of Bernardi [19], it can be verified that Φ(γ+1,γ+2;z)∈𝒮*(1/2). Thus by applying Lemma 2.5, the proof of Theorem
3.6
is complete.Theorem 3.7.
Letf∈ℛλ(α,ρ),(0≤λ<1,−π/2<α<π/2,0≤ρ≤1).Then,(3.29)f(z)z≺𝒢(z)(z∈𝒰),  where(3.30)𝒢(z)=e−iαz{Φ(2−λ,2;z)*[zh(z)]},h(z)=(1+(1−2ρ)z1−zcosα+isinα), and Φ is defined by
(1.7). Moreover, 𝒢 is a univalent
convex function in 𝒰.Proof.
SinceΩzλf(z)/z≺e−iαh(z), by an application of Lemma 2.6, we get(3.31)f(z)z≺e−iαz{ℒ(2−λ,2)*[zh(z)]}=𝒢(z). The assertion
(3.29) is proved.
It is well known (cf. [18]) that Φ(2−λ,2;z)/z is a univalent
convex function. Therefore, by Lemma 2.4, 𝒢(z) is univalent
convex function.Remark 3.8.
Forα=0, Theorem 3.7(i) gives a result of Ling and Ding [8, Theorem 2].



---

*Source: 153280-2008-03-11.xml*