# Properties of Certain Subclass of Multivalent Functions with Negative Coefficients

Authors: Jingyu Yang; Shuhai Li
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## Abstract

Making use of a linear operator, which is defined by the Hadamard product, we introduce and study a subclassYa,c+(A,B;p,λ,α) of the class A+(p). In this paper, we obtain the coefficient inequality, distortion theorem, radius of convexity and starlikeness, neighborhood property, modified convolution properties of this class. Furthermore, an application of fractional calculus operator is given. The results are presented here would provide extensions of some earlier works. Several new results are also obtained.


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## Body

## 1. Introduction

LetA+(p) denote the class of functions
(1.1)f(z)=zp-∑k=1∞|ak+p|zk+p(p∈N={1,2,…})
which are analytic and p-valent in the open unit disc U={z:|z|<1} on the complex plane C.We denote bySp*(A,B) and Kp(A,B)(-1≤B<A≤1) the subclass of p-valent starlike functions and the subclass of p-valent convex functions, respectively, that is, (see for details [1, 2])
(1.2)Sp*(A,B)={f(z)∈A(p):zf′(z)pf(z)≺1+Az1+Bz(z∈U;-1≤B<A≤1)},Kp(A,B)={f(z)∈A(p):1p+zf′′(z)pf′(z)≺1+Az1+Bz(z∈U;-1≤B<A≤1)}.A functionf∈A+(p) is said to be analytic starlike of order ξ if it satisfies
(1.3)Re{zf′(z)pf(z)}>ξ
for some ξ(0≤ξ<1) and for all z∈U.Further, a functionf∈A+(p) is said to be analytic convex of order ξ if it satisfies
(1.4)Re{1p+zf′′(z)pf′(z)}>ξ
for some ξ(0≤ξ<1) and for all z∈U.Forf(z)=zp+∑k=1∞ak+pzk+p and g(z)=zp+∑k=1∞bk+pzk+p, the Hadamard product (or convolution) is defined by
(1.5)(f*g)(z)=zp+∑k=1∞ak+pbk+pzk+p=(g*f)(z).The linear operatorLp(a,c) is defined as follows (see Saitoh [3]):
(1.6)Lp(a,c)f(z)=φp(a,c;z)*f(z)(f(z)∈A(p)),
and φp(a,c;z) is defined by
(1.7)φp(a,c;z)=zp+∑k=1∞(a)k(c)kzk+p(a∈R;c∖z0-;z0-:={0,-1,-2,…}),
where (v)k is the pochharmmer symbol defined (in terms of the Gamma function) by
(1.8)(ν)n=Γ(ν+n)Γ(ν)={1,n=0,ν(ν+1)⋯(ν+n-1),n∈N.The operatorLp(a,c) was studied recently by Srivastava and Patel [4]. It is easily verified from (1.6) and (1.7) that
(1.9)z(Lp(a,c)f(z))′=aLp(a+1,c)f(z)-(a-p)Lp(a,c)f(z).Moreover, forf(z)∈A(p),(1.10)Lp(a,a)f(z)=f(z),Lp(p+1,p)f(z)=zf′(z)p,Lp(n+p,1)f(z)=Dn+p-1f(z)(n>-p),
where Dn+p-1f(z) denotes the Ruscheweyh derivative of a function f(z)∈A(z) of order n+p-1 (see [5]).Aouf, Silverman and Srivastava [6] introduced the class of Pa,c+(A,B;p,λ) by use of linear operator Lp(a,c), further investigated the properties of this class. In [7], SokóŁ investigated several properties of the linear Aouf-Silverman-Srivastava operator and furthermore obtained the corresponding characterizations of multivalent analytic functions which were studied by Aouf et al. in paper [6]. The properties of multivalent functions with negative coefficients were studied in [8–10]. References [11, 12] gave the results of the univalent function with negative coefficients.In this paper, we will use operatorLp(a,c) to define a new subclass of A+(p) as follows.Forp∈N,a>0,c>0,α≥0 and for the parameters λ,A,andB such that
(1.11)-1≤B<A≤1,-1≤B<0,0≤λ<p,
we say that a function f(z)∈A+(p) is in the class Ya,c+(A,B;p,λ,α) if it satisfies the following subordination:
(1.12)1p-λ((Lp(a,c)f(z))′zp-1-α|(Lp(a,c)f(z))′pzp-1-1|-λ)≺1+Az1+Bz(z∈U),
or equivalently, if the following inequality holds true:
(1.13)|(Lp(a,c)f(z))′/zp-1-α|(Lp(a,c)f(z))′/pzp-1-1|-pB((Lp(a,c)f(z))′/zp-1-α|(Lp(a,c)f(z))′/pzp-1-1|)-(pB+(A-B)(p-λ))|<1.From the above definition, we can imply that the function classPa,c+(A,B;p,λ) in [6] is the special case of Ya,c+(A,B;p,λ,α) in our present paper because Ya,c+(A,B;p,λ,0)=Pa,c+(A,B;p,λ).SinceYa,c+(A,B;p,λ,0)=Pa,c+(A,B;p,λ), then, like [6], we have the following subclasses which were studied in many earlier works:(1)
Yn+p,1+(-1,1;p,λ,0)=Qn+p-1(λ)(0≤λ<1;n>-p;p∈N)(Aouf and Darwish [8]),(2)
Ya,a+(A,B;p,0,0)=P*(p,A,B)(Shukla and Dashrath [9]),(3)
Ya,a+(1,-1;p,λ,0)=Fp(1,λ)(0≤λ<p;p∈N)(Lee et al. [10]),(4)
Ya,a+(-β,β;1,λ,0)=P*(λ,β)(0≤λ<1;0<β≤1)(Gupta and Jain [11]),(5)
Yn+p,1+(-1,1;1,λ,0)=Qn(λ)(n∈N0:=N∪{0};0≤λ<1) (Uralegaddi and Sarangi [12]).The purpose of this paper is to give various properties of classYa,c+(A,B;p,λ,α). We extend the results of basic paper [6].


## 2. Necessary and Sufficient Condition forf(z)∈Ya,c+(A,B;p,λ,α)

Theorem 2.1.
Let the functionf(z)∈A+(p) be given by (1.1), then  f(z)∈Ya,c+(A,B;p,λ,α) if and only if
(2.1)∑k=1∞ϕk(p,α;B,a,c)|ak+p|≤(A-B),
where
(2.2)ϕk(p,α;B,a,c)=(p+α)(1-B)(k+p)(a)kp(p-λ)(c)k.Proof.
For the sufficient condition, letf(z)=zp-∑k=1∞|ak+p|zk+p, we have
(2.3)|(Lp(a,c)f(z))′/zp-1-α|(Lp(a,c)f(z))′/pzp-1-1|-pB((Lp(a,c)f(z))′/zp-1-α|(Lp(a,c)f(z))′/pzp-1|)-(pB+(A-B)(p-λ))|=|p(Lp(a,c)f(z))′-αeiθ|(Lp(a,c)f(z))′-pzp-1|-p2zp-1B(p(Lp(a,c)f(z))′-αeiθ|(Lp(a,c)f(z))′-pzp-1|)-p(pB+(A-B)(p-λ))zp-1|,
then
(2.4)|p(Lp(a,c)f(z))′-αeiθ|(Lp(a,c)f(z))′-pzp-1|-p2zp-1|-|B(p(Lp(a,c)f(z))′-αeiθ|(Lp(a,c)f(z))′-pzp-1|)-p(pB+(A-B)(p-λ))zp-1|=|-∑k=1∞p(k+p)|ak+p|(a)k(c)kzk+p-1-αeiθ|-∑k=1∞(k+p)|ak+p|(a)k(c)kzk+p-1||-|-∑k=1∞Bp(k+p)|ak+p|(a)k(c)kzk+p-1-αBeiθ|-∑k=1∞(k+p)|ak+p|(a)k(c)kzk+p-1|-p(A-B)(p-λ)zp-1∑k=1∞|≤∑k=1∞p(k+p)|ak+p|(a)k(c)k|z|k+p-1+α|eiθ|∑k=1∞(k+p)|ak+p|(a)k(c)k|z|k+p-1+|B|∑k=1∞p(k+p)|ak+p|(a)k(c)k|z|k+p-1+α|B|∑k=1∞(k+p)|ak+p|(a)k(c)k|z|k+p-1-p(A-B)(p-λ)|z|p-1=(p+α)∑k=1∞(k+p)(1-B)|ak+p|(a)k(c)k-p(A-B)(p-λ)(z∈∂U).
By the maximum modulus theorem, for any z∈U, we have
(2.5)|p(Lp(a,c)f(z))′-αeiθ|(Lp(a,c)f(z))′-pzp-1|-p2zp-1|-|B(p(Lp(a,c)f(z))′-αeiθ|(Lp(a,c)f(z))′-pzp-1|)-p(pB+(A-B)(p-λ))zp-1|≤(p+α)∑k=1∞(k+p)(1-B)|ak+p|(a)k(c)k-p(A-B)(p-λ)=∑k=1∞ϕk(p,α;B,a,c)|ak+p|-(A-B)≤0,
this implies f(z)∈Ya,c+(A,B;p,λ,α).
For necessary condition, letf(z)∈Ya,c+(A,B;p,λ,α) be given by (1.1), then
(2.6)(Lp(a,c)f(z))′zp-1=p-∑k=1∞(k+p)|ak+p|(a)k(c)kzk,
from (1.13) and (2.6), we find that
(2.7)|(Lp(a,c)f(z))′/zp-1-α|(Lp(a,c)f(z))′/pzp-1-1|-pB((Lp(a,c)f(z))′/zp-1-α|(Lp(a,c)f(z))′/pzp-1-1|)-(pB+(A-B)(p-λ))|=|-∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-α|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk|-(A-B)(p-λ)-B∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-αB|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk||<1.
Now, since|Rez|≤|z| for all z, we have
(2.8)|Re-∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-α|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk|-(A-B)(p-λ)-B∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-αB|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk||<1.
We choose value of z on the real axis so that the expression (Lp(a,c)f(z))′/zp-1 is real.
Then, we have(2.9)Re-∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-α|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk|-(A-B)(p-λ)-B∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-αB|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk|=-∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-α|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk|-(A-B)(p-λ)-B∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-αB|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk|,(2.10)|-∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-α|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk|-(A-B)(p-λ)-B∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-αB|-∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)zk||=|-∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-α∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)|z|k-(A-B)(p-λ)-B∑k=1∞(k+p)|ak+p|((a)k/(c)k)zk-αB∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)|z|k|<1.
Lettingz→1- throughout real values in (2.10), we get
(2.11)|-∑k=1∞(k+p)|ak+p|((a)k/(c)k)-α∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)-(A-B)(p-λ)-B∑k=1∞(k+p)|ak+p|((a)k/(c)k)-αB∑k=1∞((k+p)/p)|ak+p|((a)k/(c)k)|<1.
So we have
(2.12)∑k=1∞p(k+p)|ak+p|((a)k/(c)k)+α∑k=1∞(k+p)|ak+p|((a)k/(c)k)(A-B)(p-λ)+B∑k=1∞p(k+p)|ak+p|((a)k/(c)k)+αB∑k=1∞(k+p)|ak+p|((a)k/(c)k)<1,
it is
(2.13)∑k=1∞(p+α)(1-B)(k+p)|ak+p|(a)k(c)k≤p(A-B)(p-λ).
This completes the proof of the theorem.Corollary 2.2.
Let the functionf(z)∈A+(p) be given by (1.1). If  f(z)∈Ya,c+(A,B;p,λ,α),then
(2.14)|ak+p|≤p(A-B)(p-λ)(c)k(p+α)(1-B)(k+p)(a)k(k,p∈N).
The result is sharp for the function given by
(2.15)f(z)=zp-∑k=1∞p(A-B)(p-λ)(c)k(p+α)(1-B)(k+p)(a)kzk+p.


## 3. Distortion Inequality of ClassYa,c+(A,B;p,λ,α)

Theorem 3.1.
If a functionf(z) defined by (1.1) is in the class  Ya,c+(A,B;p,λ,α), then
(3.1)(p!(p-m)!-cp(A-B)(p-λ)p!a(p+α)(1-B)(p+1-m)!|z|)|z|p-m≤|f(m)(z)|≤(p!(p-m)!+cp(A-B)(p-λ)p!a(p+α)(1-B)(p+1-m)!|z|)|z|p-m(z∈U;a≥c>0;m∈N0;p>m).
The result is sharp for the function f(z) given by
(3.2)f(z)=zp-c(A-B)(p-λ)a(1-B)(p+α)z1+p(p∈N).Proof.
Sincef(z)=zp-∑k=1∞|ak+p|zk+p, then
(3.3)f(m)(z)=p!(p-m)!zp-m-∑k=1∞(k+p)!(k+p-m)!|ak+p|zk+p-m.
Fromf(z)∈Ya,c+(A,B;p,λ,α), using Theorem 2.1, we have
(3.4)a(p+α)(1-B)(1+p)cp(A-B)(p-λ)(p+1)!∑k=1∞(k+p)!|ak+p|≤∑k=1∞(p+α)(k+p)(1-B)(a)kp(A-B)(p-λ)(c)k|ak+p|≤1
which readily yields
(3.5)∑k=1∞(k+p)!|ak+p|≤cp(A-B)(p-λ)p!a(p+α)(1-B).
By (3.3) and (3.5), we can imply (3.1).


## 4. Radius of Starlikeness and Convexity

Theorem 4.1.
Let the functionf(z) defined by (1.1) be in the class  Ya,c+(A,B;p,λ,α). Then,  f(z) is starlike of  η(0≤η<1) in  |z|<rη(p,k,α,A,B,a,c) where
(4.1)rη(p,k,α,A,B,a,c)=inf⁡⁡{p(1-η)ϕk(p,α;B,a,c)(A-B)[k+p(1-η)]}1/k
and ϕk(p,α;B,a,c) is defined by (2.1).Proof.
We must show that(4.2)|zf′(z)pf(z)-1|<1-ηfor|z|<rη(p,k,α,A,B,a,c).
Since
(4.3)|zf′(z)pf(z)-1|=|-∑k=1∞k|ak+p|zkp-∑k=1∞p|ak+p|zk|≤∑k=1∞k|ak+p||z|kp-∑k=1∞p|ak+p||z|k,
to prove (4.1), it is sufficient to prove
(4.4)∑k=1∞k|ak+p||z|kp-∑k=1∞p|ak+p||z|k<1-η.
It is equivalent to
(4.5)∑k=1∞[k+p(1-η)]|ak+p|p(1-η)|z|k≤1.
By Theorem2.1, we have
(4.6)∑k=1∞ϕk(p,α;B,a,c)A-B|ak+p|≤1,
hence (4.5) will be true if
(4.7)[k+p(1-η)]|ak+p|p(1-η)|z|k≤ϕk(p,α;B,a,c)A-B|ak+p|.
It is equivalent to
(4.8)|z|≤{p(1-η)ϕk(p,α;B,a,c)(A-B)[k+p(1-η)]}1/k.
This completes the proof.Theorem 4.2.
Let the functionf(z) defined by (1.1) be in the class  Ya,c+(A,B;p,λ,α). Then,  f(z) is convex of  ξ(0≤ξ<1) in  |z|<rξ(p,k,α,A,B,a,c) where
(4.9)rξ(p,k,α,A,B,a,c)=inf⁡⁡{p2(1-ξ)ϕk(p,α;B,a,c)(k+p)(A-B)[k+p(1-ξ)]}1/k
and ϕk(p,α;B,a,c) is defined by (2.1).Proof.
It sufficient to show that(4.10)|zf′′(z)pf′(z)-p-1p|<1-ξfor|z|<rξ(p,k,α,A,B,a,c).
Since
(4.11)|zf′′(z)pf′(z)-p-1p|=|-∑k=1∞k(k+p)|ak+p|zkp2-∑k=1∞p(k+p)|ak+p|zk|≤∑k=1∞k(k+p)|ak+p||z|kp2-∑k=1∞p(k+p)|ak+p||z|k,
to prove (4.9), it is sufficient to prove
(4.12)∑k=1∞k(k+p)|ak+p||z|kp2-∑k=1∞p(k+p)|ak+p||z|k<1-ξ.
It is equivalent to
(4.13)∑k=1∞(k+p)[k+p(1-ξ)]|ak+p|p2(1-ξ)|z|k≤1.
By Theorem2.1, we have
(4.14)∑k=1∞ϕk(p,α;B,a,c)A-B|ak+p|≤1,
hence (4.13) will be true if
(4.15)(k+p)[k+p(1-ξ)]|ak+p|p2(1-ξ)|z|k≤ϕk(p,α;B,a,c)A-B|ak+p|.
It is equivalent to
(4.16)|z|≤{p2(1-ξ)ϕk(p,α;B,a,c)(k+p)(A-B)[k+p(1-ξ)]}1/k.
This completes the proof.


## 5.δ-Neighborhood of Ya,c+(A,B;p,λ,α)

Based on the earlier works by Aouf et al. [6], Altintas et al. [13–15], and Aouf [16], we introduce the δ-neighborhood of a function f(z)∈A+(p) of the form (1.1) and present the relationship between δ-neighborhood and corresponding function class.Definition 5.1.
Forδ>0,a>0,c>0, the δ-neighborhood of a function f(z)∈A+(p) is defined as follows:
(5.1)Nδ+(f)={g:g(z)=zp-∑k=1∞|bk+p|zk+p∈A(p),∑k=1∞tk||bk+p|-|ak+p||<δ},
where
(5.2)tk=(p+α)(1-B)(k+p)(a)kp(A-B)(p-λ)(c)k.Theorem 5.2.
Let the functionf(z) defined by (1.1) be in the class  Ya+1,c+(A,B;p,λ,α). Then,
(5.3)Nδ+(f)⊂Ya,c+(A,B;p,λ,α)(1a+1).
This result is the best possible in the sense that δ cannot be increased.Proof.
Letf(z)∈Ya+1,c+(A,B;p,λ,α), then by Theorem 2.1, we have
(5.4)∑k=1∞(p+α)(1-B)(k+p)(a+1)kp(A-B)(p-λ)(c)k|ak+p|≤1
which is equivalent to
(5.5)∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-λ)(c)k|ak+p|≤aa+1.
For any(5.6)g(z)=zp-∑k=1∞|bk+p|zk+p∈Nδ+(f),
we find from (5.1) that
(5.7)∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-λ)(c)k||bk+p|-|ak+p||≤δ.
By (5.5) and (5.7), we get
(5.8)∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-λ)(c)k|bk+p|≤∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-λ)(c)k|ak+p|+∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-λ)(c)k||bk+p|-|ak+p||≤aa+1+δ=1.
By Theorem 2.1, it implies that g(z)∈Ya,c+(A,B;p,λ,α).
To show the sharpness of the assertion of Theorem5.2, we consider the functions f(z) and g(z) given by
(5.9)f(z)=zp-cp(A-B)(p-λ)(a+1)(p+α)(1-B)(p+1)zp+1∈Ya+1,c+(A,B;p,λ,α),g(z)=zp-(cp(A-B)(p-λ)(a+1)(p+α)(1-B)(p+1)+cp(A-B)(p-λ)δ′a(p+α)(1-B)(1+p))zp+1,
where δ′>δ=1/(a+1).
Since(5.10)(p+α)(p+1)(1-B)(a)1p(A-B)(p-λ)(c)1||b1+p|-|a1+p||=(p+α)(p+1)(1-B)(a)1p(A-B)(p-λ)(c)1⋅cp(A-B)(p-λ)δ′a(p+α)(1-B)(1+p)=δ′,
so g(z)∈Nδ′+(f).
But(5.11)(p+α)(p+1)(1-B)(a)1p(A-B)(p-λ)(c)1[cp(A-B)(p-λ)(a+1)(p+α)(p+1)(1-B)+cp(A-B)(p-λ)δ′a(p+α)(1-B)(1+p)]=aa+1+δ′>1,
by Theorem 2.1g(z)∈¯Ya,c+(A,B;p,λ,α).


## 6. Properties Associated with Modified Hadamard Product

Following early works by Aouf et al. in [6], we provide the properties of modified Hadamard product of Ya,c+(A,B;p,λ,α).For the function(6.1)fj(z)=zp-∑k=1∞|ak+p,j|zk+p(j=1,2;p∈N),
the modified Hadamard product of the functions f1(z) and f2(z) was denoted by (f1·f2)(z) and defined as follows:
(6.2)(f1⋅f2)(z)=zp-∑k=1∞|ak+p,1||ak+p,2|zk+p=(f2⋅f1)(z).Theorem 6.1.
Letfj(z)(j=1,2) given by  (6.1) be in the class  Ya,c+(A,B;p,λ,α), then
(6.3)(f1⋅f2)(z)∈Ya,c+(A,B;p,γ,α),
where
(6.4)γ:=p-cp(A-B)(p-λ)2a(p+α)(1-B)(1+p).
The result is sharp for the functions fj(z)(j=1,2) given by
(6.5)fj(z)=zp-cp(A-B)(p-λ)a(p+α)(1-B)(1+p)zp+1.Proof.
By Theorem2.1, we need to find the largest γ such that
(6.6)∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-γ)(c)k|ak+p,1|⋅|ak+p,2|≤1.
Sincefj(z)∈Ya,c+(A,B;p,λ,α),(j=1,2), then we see that
(6.7)∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-λ)(c)k|ak+p,j|≤1(j=1,2).
By Cauchy-Schwartz inequality, we obtain(6.8)∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-λ)(c)k|ak+p,1|⋅|ak+p,2|≤1.
This implies that we only need to show that(6.9)|ak+p,1|⋅|ak+p,2|p-γ≤|ak+p,1|⋅|ak+p,2|p-λ(k∈N)
or equivalently that
(6.10)|ak+p,1|⋅|ak+p,2|≤p-γp-λ.
By making use of the inequality (6.8), it is sufficient to prove that
(6.11)p(A-B)(p-λ)(c)k(p+α)(1-B)(k+p)(a)k≤p-γp-λ(k∈N).
From (6.11), we have
(6.12)γ≤p-p(A-B)(p-λ)2(c)k(p+α)(1-B)(k+p)(a)k(k∈N).
Define the functionΦ(k) by
(6.13)Φ(k):=p-p(A-B)(p-λ)2(c)k(p+α)(1-B)(k+p)(a)k(k∈N).
We see that Φ(k) is an increasing function of k. Therefore, we conclude that
(6.14)γ≤Φ(1)=p-p(A-B)(p-λ)2(c)1(p+α)(1-B)(k+p)(a)1.
By using arguments similar to those in the proof of Theorem6.1, we can derive the following result.Theorem 6.2.
Letf1(z) defined by  (6.1) be in the class  Ya,c+(A,B;p,λ,α), f2(z) defined by  (6.1) be in the class  Ya,c+(A,B;p,γ,α), then
(6.15)(f1⋅f2)(z)∈Ya,c+(A,B;p,ξ,α),
where
(6.16)ξ:=p-cp(A-B)(p-λ)(p-γ)a(p+α)(1-B)(1+p).
The result is sharp for the functions fj(z)(j=1,2) given by
(6.17)f1(z)=zp-cp(A-B)(p-λ)a(p+α)(1-B)(1+p)zp+1(p∈N),f2(z)=zp-cp(A-B)(p-γ)a(p+α)(1-B)(1+p)zp+1(p∈N).Theorem 6.3.
Letfj(z)(j=1,2) defined by  (6.1) be in the class  Ya,c+(A,B;p,λ,α), then the function  h(z) defined by
(6.18)h(z)=zp-∑k=1∞(|ak+p,1|2+|ak+p,2|2)
belongs to the class Ya,c+(A,B;p,χ,α) where
(6.19)χ:=p-2pc(A-B)(p-λ)a(p+α)(1-B)(1+p).
This result is sharp for the functions given by (6.5).Proof.
By Theorem2.1, we want to find the largest χ such that
(6.20)∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-χ)(c)k(|ak+p,1|2+|ak+p,2|2)≤1.
Sincefj(z)∈Ya,c+(A,B;p,λ,α)(j=1,2), we readily see that
(6.21)∑k=1∞(p+α)(1-B)(k+p)(a)kp(A-B)(p-χ)(c)k|ak+p,j|≤1(j=1,2).
From (6.21), we have
(6.22)∑k=1∞(p+α)2(1-B)2(k+p)2p2(A-B)2(p-χ)2((a)k(c)k)2|ak+p,j|2≤1(j=1,2),
then we have
(6.23)∑k=1∞(p+α)2(1-B)2(k+p)2p2(A-B)2(p-χ)2((a)k(c)k)2(|ak+p,1|2+|ak+p,2|2)≤2.
From (6.23), if we want to prove (6.20), it is sufficient to prove there exists the largest χ such that
(6.24)1p-χ≤(p+α)(k+p)(1-B)(a)k2p(A-B)(p-λ)2(c)k(k∈N),
that is
(6.25)χ≤p-2p(A-B)(p-λ)2(c)k(p+α)(k+p)(1-B)(a)k(k∈N).
Now we defineΨ(k) by
(6.26)Ψ(k)=p-2p(A-B)(p-λ)2(c)k(p+α)(k+p)(1-B)(a)k(k∈N).
We see that Ψ(k) is an increasing function of k. Therefore, we conclude that
(6.27)χ≤Ψ(1)=p-2pc(A-B)(p-λ)2a(p+α)(k+p)(1-B)
which completes the proof of Theorem 6.1.


## 7. Application of Fractional Calculus Operator

References [17–19] have studied the fractional calculus operators extensively. In this part, we only investigate the application of fractional calculus operator which was defined by [6] in the class of Ya,c+(A,B;p,λ,α).Definition 7.1 (see [6]).
The fractional integral of orderμ is defined, for a function f(z),by
(7.1)Dz-μf(z)=1Γ(μ)∫0zf(z)(z-ξ)1-μdξ(μ>0),
where the function f(z) is analytic in a simply connected domain of the complex z-plane containing the origin and the multiplicity of (z-ξ)μ-1 is removed by requiring log(z-ξ) to be real when z-ξ>0.Definition 7.2 (see [6]).
The fractional integral of orderμ is defined, for a function f(z),by
(7.2)Dzμf(z)=1Γ(1-μ)∫0zf(z)(z-ξ)μdξ(0≤μ<1),
where the function f(z) is constrained, the multiplicity of (z-ξ)-μ is removed as in Definition 7.1.In our investigation, we will use the operatorsJδ,p defined by (cf, [20–22])
(7.3)(Jδ,pf)(z):=δ+pzp∫0ztδ-1f(t)dt(f∈A(p);δ>-p;p∈N)
as well as Dzμ for which it is well known that (see [23])
(7.4)Dzμzρ=Γ(ρ+1)Γ(ρ-μ+1)zρ-μ(ρ>-1;μ∈R)
in terms of Gamma.Lemma 7.3 (see Chen et al. [18]).
For a functionf(z)∈A(p),
(7.5)Dzμ{(Jδ,pf)(z)}=Γ(p+1)Γ(p-μ+1)zp-μ+∑k=1∞(δ+p)Γ(k+p+1)(δ+k+p)Γ(k+p-μ+1)ak+pzk+p-μ,Jδ,p(Dzμ{f(z)})=(δ+p)Γ(p+1)(δ+p-μ)Γ(p-μ+1)zp-μ+∑k=1∞(δ+p)Γ(k+p+1)(δ+k+p-μ)Γ(k+p-μ+1)ak+pzk+p-μ(μ∈R;δ>-p;p∈N) provided that no zeros appear in the denominators in (7.5).Remark 7.4.
Throughout this section, we assume further thata≥c>0.Theorem 7.5.
Let function defined by  (1.1) be in the class  Ya,c+(A,B;p,λ,α), then
(7.6)|Dz-μ{(Jδ,pf)(z)}|≥Γ(p+1)Γ(p+μ+1)|z|p+μ(1-cp(δ+p)(A-B)(p-λ)a(p+α)(p+δ+1)(p+μ+1)(1-B)|z|),(7.7)|Dz-μ{(Jδ,pf)(z)}|≤Γ(p+1)Γ(p+μ+1)|z|p+μ(1+cp(δ+p)(A-B)(p-λ)a(p+α)(p+δ+1)(p+μ+1)(1-B)|z|)μ>0;δ>-p;p∈N;z∈U, each of the assertions is sharp.Proof.
Since(7.8)Dz-μ{(Jδ,pf)(z)}=Γ(p+1)Γ(p-μ+1)zp+μ-∑k=1∞(δ+p)Γ(k+p+1)(δ+k+p)Γ(k+p+μ+1)|ak+p|zk+p+μ,
then
(7.9)|Dz-μ{(Jδ,pf)(z)}|=Γ(p+1)Γ(p-μ+1)|zμ||zp-∑k=1∞(δ+p)Γ(k+p+1)Γ(p+μ+1)(δ+k+p)Γ(k+p+μ+1)Γ(p+1)|ak+p|zk+p|.
LetG(z) is defined by
(7.10)G(z)=zp-∑k=1∞(δ+p)Γ(k+p+1)Γ(p+μ+1)(δ+k+p)Γ(k+p+μ+1)Γ(p+1)|ak+p|zk+p=zp-∑k=1∞Q(k)|ak+p|zk+p,
where
(7.11)Q(k)=(δ+p)Γ(k+p+1)Γ(p+μ+1)(δ+k+p)Γ(k+p+μ+1)Γ(p+1)(k,p∈N;μ>0).
SinceQ(k) is a decreasing function of k when μ>0, we get
(7.12)0<Q(k)≤Q(1)=(δ+p)(p+1)(δ+k+p)(1+p+μ),
since f(z)∈Ya,c+(A,B;p,λ,α), by Theorem 2.1, we get
(7.13)a(p+α)(1-B)(p+1)cp(A-B)(p-λ)∑k=1∞|ak+p|≤∑k=1∞(p+α)(k+p)(1-B)(a)kp(A-B)(p-λ)(c)k|ak+p|≤1.
It is that(7.14)∑k=1∞|ak+p|≤cp(A-B)(p-λ)a(α+p)(1-B)(p+1),
then
(7.15)|G(z)|=|zp-∑k=1∞Q(k)|ak+p|zk+p|≥|z|p-Q(1)|z|p+1∑k=1∞|ak+p|≥|z|p-cp(p+δ)(A-B)(p-λ)a(p+α)(δ+k+p)(p+1+μ)(1-B)|z|p+1,|G(z)|=|zp-∑k=1∞Q(k)|ak+p|zk+p|≤|z|p+Q(1)|z|p+1∑k=1∞|ak+p|≤|z|p+cp(p+δ)(A-B)(p-λ)a(p+α)(δ+k+p)(p+1+μ)(1-B)|z|p+1.
From (7.15), we obtain (7.6) and (7.7), respectively.
Equalities in (7.6) and (7.7) are attained for the function f(z) given by
(7.16)Dz-μ{(Jδ,pf)(z)}=Γ(p+1)Γ(p-μ+1)zp+μ(1-cp(δ+p)(A-B)(p-λ)a(p+α)(δ+k+p)(p+μ+1)(1-B)z).Theorem 7.6.
Let function defined by  (1.1) be in the class  Ya,c+(A,B;p,λ,α), then
(7.17)|Dzμ{(Jδ,pf)(z)}|≥Γ(p+1)Γ(p-μ+1)|z|p-μ(1-cp(δ+p)(A-B)(p-λ)a(p+α)(p+δ+1)(p-μ+1)(1-B)|z|),(7.18)|Dzμ{(Jδ,pf)(z)}|≤Γ(p+1)Γ(p-μ+1)|z|p-μ(1+cp(δ+p)(A-B)(p-λ)a(p+α)(p+δ+1)(p-μ+1)(1-B)|z|),μ>0;δ>-p;p∈N;z∈U, each of the assertions is sharp.Proof.
Since(7.19)Dzμ{(Jδ,pf)(z)}=Γ(p+1)Γ(p-μ+1)zp-μ-∑k=1∞(δ+p)Γ(k+p+1)(δ+k+p)Γ(k+p-μ+1)|ak+p|zk+p-μ,
then
(7.20)|Dzμ{(Jδ,pf)(z)}|=Γ(p+1)Γ(p-μ+1)|z-μ||zp-∑k=1∞(δ+p)Γ(k+p+1)Γ(p-μ+1)(δ+k+p)Γ(k+p-μ+1)Γ(p+1)|ak+p|zk+p|.
LetH(z) be defined by
(7.21)H(z)=zp-∑k=1∞(δ+p)Γ(k+p+1)Γ(p-μ+1)(δ+k+p)Γ(k+p-μ+1)Γ(p+1)|ak+p|zk+p=zp-∑k=1∞E(k)(k+p)|ak+p|zk+p,
where
(7.22)E(k)=(δ+p)Γ(k+p)Γ(p-μ+1)(δ+k+p)Γ(k+p-μ+1)Γ(p+1)(k,p∈N;μ>0).
SinceE(k) is a decreasing function of k when 0≤μ<1, then
(7.23)0<E(k)≤E(1)=(δ+p)(δ+1+p)(1+p-μ),
since f(z)∈Ya,c+(A,B;p,λ,α), by Theorem 2.1, we get
(7.24)a(p+α)(1-B)cp(A-B)(p-λ)∑k=1∞(k+p)|ak+p|≤∑k=1∞(p+α)(k+p)(1-B)(a)kp(A-B)(p-λ)(c)k|ak+p|≤1.
It is that(7.25)∑k=1∞(k+p)|ak+p|≤cp(A-B)(p-λ)a(α+p)(1-B),
then
(7.26)|H(z)|=|zp-∑k=1∞E(k)(k+p)|ak+p|zk+p|≥|z|p-E(1)|z|p+1∑k=1∞(k+p)|ak+p|≥|z|p-cp(p+δ)(A-B)(p-λ)a(p+α)(δ+k+p)(p+1-μ)(1-B)|z|p+1,|H(z)|=|zp-∑k=1∞E(k)(k+p)|ak+p|zk+p|≤|z|p+E(1)|z|p+1∑k=1∞(k+p)|ak+p|≤|z|p+cp(p+δ)(A-B)(p-λ)a(p+α)(δ+k+p)(p+1-μ)(1-B)|z|p+1.
From (7.26), we obtain (7.17) and (7.18), respectively.
Equalities in (7.17) and (7.18) are attained for the function f(z) given by
(7.27)Dzμ{(Jδ,pf)(z)}=Γ(p+1)Γ(p-μ+1)zp-μ(1-cp(δ+p)(A-B)(p-λ)a(p+α)(δ+k+p)(p-μ+1)(1-B)z)
or equivalently by
(7.28)(Jδ,pf)(z)=zp-cp(δ+p)(A-B)(p-λ)a(p+α)(δ+k+p)(p-μ+1)(1-B)zp+1.



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*Source: 362312-2012-05-30.xml*