# New Subclasses of Multivalent Analytic Functions Associated with a Linear Operator

Authors: Ding-Gong Yang; Jin-Lin Liu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Making use of a linear operator, which is defined here by means of the Hadamard product (or convolution), we consider two subclassesFp,n(a,c,λ,A,B) and Gp,n(a,c,λ,A,B) of multivalent analytic functions with negative coefficients in the open unit disk. Some modified Hadamard products, integral transforms, and the partial sums of functions belonging to these classes are studied.


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## Body

## 1. Introduction

LetAp,n denote the class of functions of the form
(1)f(z)=zp+∑k=p+n∞‍akzk,(p,n∈N={1,2,3,…}),
which are analytic in the open unit disk U={z:|z|<1}.For functionsf(z) and g(z) analytic in U, we say that f(z) is subordinate to g(z) in U, written f(z)≺g(z)(z∈U), if there exists an analytic function w(z) in U such that
(2)|w(z)|≤|z|,f(z)=g(w(z)),(z∈U).
Furthermore, if the function g(z) is univalent in U, then
(3)f(z)≺g(z)(z∈U)⟺f(0)=g(0),f(U)⊆g(U).In terms of the Pochhammer symbol(b)n given by (b)n=b(b+1)⋯(b+n-1)(n∈N), we define the function φp,n(a,c;z) by
(4)φp,n(a,c;z)=zp+∑k=p+n∞‍(a)k-p(c)k-pzkφp,nzk,(z∈U;c∉{0,-1,-2,…}).
Corresponding to φp,n(a,c;z), we consider here a linear operator Lp,n(a,c) on Ap,n by the following usual Hadamard product (or convolution):
(5)Lp,n(a,c)f(z)=φp,n(a,c;z)*f(z)=zp+∑k=p+n∞‍(a)k-p(c)k-pakzk,
for f(z) given by (1). For p=n=1, L1,1(a,c) on A1,1 was first defined by Carlson and Shaffer [1]. Its differential-integral representation can be found in [2]. We remark in passing that a much more general convolution operator than the operator Lp,1(a,c) was introduced by Dziok and Srivastava [3].LetTp,n denote the subclass of Ap,n consisting of functions of the form
(6)f(z)=zp-∑k=p+n∞‍akzk,(ak≥0;p,n∈N).
We now consider the following two subclasses of the class Tp,n.Definition 1.
A functionf(z)∈Tp,n is said to be in the class Fp,n(a,c,λ,A,B) if and only if
(7)(1-λ)Lp,n(a,c)f(z)zp+λ(Lp,n(a,c)f(z))′pzp-1≺1+Az1+Bz(z∈U),
where
(8)a>0,c>0,λ≥0,-1≤B≤0,B<A≤1.Definition 2.
A functionf(z)∈Tp,n is said to be in the class Gp,n(a,c,λ,A,B) if and only if
(9)zf′(z)p∈Fp,n(a,c,λ,A,B).For functionsfj(z)∈Tp,n given by
(10)fj(z)=zp-∑k=p+n∞‍ak,jzk,(ak,j≥0;j=1,2),
we denote by (f1·f2)(z) the modified Hadamard (or quasi-Hadamard) product of f1(z) and f2(z); that is,
(11)(f1·f2)(z)=zp-∑k=p+n∞‍ak,1ak,2zk=(f2·f1)(z).The class(12)Fp,n(1,1,λ,1-2αp,-1)=Fp(n,λ,α),Fp(n,λhh,α),(λ≥0;0≤α<p)
with n=1 was introduced and studied earlier by Lee et al. [4] (and was further investigated by Aouf and Darwish [5], Aouf et al. [6], and Yaguchi et al. [7]). The class
(13)Gp,n(1,1,λ,1-2αp,-1)=Gp(n,λ,α),Gp(n,λ,hα),(λ≥0;0≤α<p)
with n=1 was studied by Aouf [8] and Aouf et al. [6]. Recently, Aouf [9] investigated the modified Hadamard products of several functions in the classes Fp(n,λ,α) and Gp(n,λ,α) for n∈N.In the present paper, we prove a number of theorems involving the modified Hadamard products, integral transforms, and the partial sums of functions in the classesFp,n(a,c,λ,A,B) and Gp,n(a,c,λ,A,B). Some of our results are generalizations of the corresponding results in [4–9].In proving our main results, we need the following lemmas.Lemma 3 (see [10, 11]).
A functionf(z)∈Tp,n defined by (6) is in the class Fp,n(a,c,λ,A,B) if and only if
(14)∑k=p+n∞‍(p+(k-p)λ)(a)k-p(c)k-pak≤pA-B1-B.Lemma 4 (see [10, 11]).
A functionf(z)∈Tp,n defined by (6) is in the class Gp,n(a,c,λ,A,B) if and only if
(15)∑k=p+n∞‍k(p+(k-p)λ)(a)k-p(c)k-pak≤p2A-B1-B.Making use of Lemmas3 and 4, we can show the following two results.Corollary 5.
Let(16)0<a1≤a0,0<c0≤c1,0≤λ1≤λ0,-1≤B1≤B0<A0≤A1≤1,B0≤0.
Then
(17)Fp,n(a0,c0,λ0,A0,B0)⊆Fp,n(a1,c1,λ1,A1,B1),Gp,n(a0,c0,λ0,A0,B0)⊆Gp,n(a1,c1,λ1,A1,B1).Corollary 6.
Letf(z)∈Tp,n. Then f(z)∈Fp,n(a0,c0,λ0,A,B)(Gp,n(a0,c0,λ0,A,B)) if and only if (f*g)(z)∈Fp,n(a1,c1,λ1,A,B)(Gp,n(a1,c1,λ1,A,B)), where λ0≥0,  λ1≥0,aj>0  and cj>0(j=0,1), and
(18)g(z)=zp+∑k=p+n∞‍(p+(k-p)λ0)(a0)k-p(c1)k-p(p+(k-p)λ1)(a1)k-p(c0)k-pzk.If we let(19)a=c=1,A=1-2αp,0≤α<p,B=-1,
then Lemmas 3 and 4 reduce to the following result.Corollary 7.
Letf(z)∈Tp,n be defined by (6). Then(i)
f(z) is in the class Fp(n,λ,α) if and only if
(20)∑k=p+n∞‍(p+(k-p)λ)ak≤p-α.(ii)
f(z) is in the class Gp(n,λ,α) if and only if
(21)∑k=p+n∞‍k(p+(k-p)λ)ak≤p(p-α).


## 2. Modified Hadamard Products

Hereafter in this paper we assume that (8) is satisfied:
(22)0≤α<p,0≤αj<p,-1≤Bj≤0,Bj<Aj≤1,(j=1,2,…,m).Theorem 8.
Letfj(z)∈Fp,n(a,c,λ,Aj,Bj)(j=1,2,…,m) and a≥c>0. Then
(23)(f1·f2·…·fm)(z)∈Fp,n(a,c,λ,A(B),B),
where m≥2 and
(24)A(B)=B+(1-B)(p(c)n(p+nλ)(a)n)m-1∏j=1m‍(Aj-Bj1-Bj).
The result is sharp; that is, A(B) cannot be decreased for each B∈[-1,0].Proof.
By (24) we have B<A(B)≤1 for a≥c>0. Let
(25)fj(z)=zp-∑k=p+n∞‍ak,jzk∈Fp,n(a,c,λ,Aj,Bj)Aj,fBj(ak,j≥0;j=1,2,…,m).
Then Lemma 3 gives
(26)(1-BjAj-Bj)∑k=p+n∞‍(p+(k-p)λ)(a)k-pp(c)k-pak,j≤1(j=1,2,…,m),
and hence
(27)∏j=1m‍(1-BjAj-Bj)×∑k=p+n∞‍((p+(k-p)λ)(a)k-pp(c)k-p)mak,1ak,2⋯ak,m≤∏j=1m‍{(1-BjAj-Bj)∑k=p+n∞‍(p+(k-p)λ)(a)k-pp(c)k-pak,j}≤1.
Also, using Lemma 3,
(28)h(z)=(f1·f2·…·fm)(z)=zp-∑k=p+n∞‍ak,1ak,2⋯ak,mzk∈Fp,n(a,c,λ,A,B)
if and only if
(29)(1-BA-B)∑k=p+n∞‍(p+(k-p)λ)(a)k-pp(c)k-pak,1ak,2⋯ak,m≤1.
To prove the result of Theorem8, it follows from (27) and (29) that we need to find the smallest A such that
(30)(p+(k-p)λ)(a)k-p(1-B)p(c)k-p(A-B)≤((p+(k-p)λ)(a)k-pp(c)k-p)m∏j=1m‍(1-BjAj-Bj)((p+(k-p)λ)(a)k-pp(c)k-p)(k≥p+n;m≥2);
that is, that
(31)A≥B+(1-B)(p(c)k-p(p+(k-p)λ)(a)k-p)m-1×∏j=1m‍(Aj-Bj1-Bj)(k≥p+n;m≥2).
Since
(32)(c)k+1-p(a)k+1-p≤(c)k-p(a)k-p(k≥p+n;a≥c>0),
we see that the right-hand side of (31) is a decreasing function of k. Consequently, taking k=p+n in (31), we have h(z)∈Fp,n(a,c,λ,A(B),B), where A(B) is given by (24).
Furthermore, by considering the functionsfj(z) defined by:
(33)fj(z)=zp-p(c)n(Aj-Bj)(p+nλ)(a)n(1-Bj)zp+n0000000000000(j=1,2,…,m),
we have fj(z)∈Fp,n(a,c,λ,Aj,Bj) and
(34)(f1·f2·…·fm)(z)=zp-(p(c)n(p+nλ)(a)n)m∏j=1m‍(Aj-Bj1-Bj)zp+n.
Noting that
(35)(p+nλ)(a)n(c)n(p(c)n(p+nλ)(a)n)m∏j=1m‍(Aj-Bj1-Bj)=pA(B)-B1-B,
we conclude that A(B) cannot be decreased for each B.By using Lemma4 instead of Lemma 3, the following theorem can be proved on the lines of the proof of Theorem 8. We omit the details involved.Theorem 9.
Letfj(z)∈Gp,n(a,c,λ,Aj,Bj)(j=1,2,…,m) and a≥c>0. Then
(36)(f1·f2·…·fm)(z)∈Gp,n(a,c,λ,A(B),B),
where m≥2 and
(37)A(B)=B+(1-B)(p2(c)n(p+n)(p+nλ)(a)n)m-1×∏j=1m‍(Aj-Bj1-Bj)∈(B,1).
The result is sharp for the functions fj(z) defined by
(38)fj(z)=zp-p2(c)n(Aj-Bj)(p+n)(p+nλ)(a)n(1-Bj)zp+n∈Gp,n(a,c,λ,Aj,Bj)(j=1,2,…,m).Theorem 10.
Letfj(z)∈Fp,n(a,c,λ,Aj,Bj)(j=1,2,…,m) with a≥c>0 and λ≥1. Then
(39)(f1·f2·…·fm)(z)∈Gp,n(a,c,λ,A(B),B),
where m≥2 and
(40)A(B)=B+(1-B)(p+np)(p(c)n(p+nλ)(a)n)m-1×∏j=1m‍(Aj-Bj1-Bj).
The result is sharp for the functions fj(z)(j=1,2,…,m) defined by (33).Proof.
Obviously,B<A(B)≤1 for m≥2,  a≥c>0, and   λ≥1. By applying Lemma 4, we know that
(41)(f1·f2·…·fm)(z)=zp-∑k=p+n∞‍ak,1ak,2⋯ak,mzk∈Gp,n(a,c,λ,A,B),
if and only if
(42)(1-BA-B)∑k=p+n∞‍k(p+(k-p)λ)(a)k-pp2(c)k-pak,1ak,2⋯ak,m≤1.
Proceeding as in the proof of Theorem 8, we need to find the smallest  A such that
(43)A≥B+(1-B)kp(p(c)k-p(p+(k-p)λ)(a)k-p)m-1×∏j=1m‍(Aj-Bj1-Bj)(k≥p+n).
Defining the function φ(x) by
(44)φ(x)=x(p+(x-p)λ)m-1(x≥p+n),
we see that
(45)φ′(x)=p(1-λ)-(m-2)λx(p+(x-p)λ)m≤0(x≥p+n),
for m≥2 and λ≥1. Hence, the right-hand side of (43) is a decreasing function of k. Thus, we arrive at (f1·f2·…·fm)(z)∈Gp,n(a,c,λ,A(B),B), where A(B) is given by (40).Sharpness can be verified easily.By putting(46)a=c=1,A(B)=1-2αp,B=-1,Aj=1-2αjp,Bj=-1(j=1,2,…,m).
Theorem 10 reduces to the following.Corollary 11.
Letfj(z)∈Fp(n,λ,αj)(j=1,2,…,m). Then (f1·f2·…·fm)(z)∈Gp(n,λ,α), where m≥2,  λ≥1, and
(47)α=p-p+np(p+nλ)m-1∏j=1m‍(p-αj).
The result is sharp for the functions
(48)fj(z)=zp-p-αjp+nλzp+n∈Fp(n,λ,αj)∈Fp(λαj)(j=1,2,…,m).Theorem 12.
Letfj(z)∈Fp,n(a,c,λ,Aj,Bj)(j=1,2,…,m),
(49)gj(z)∈Fp,n(a,c,λ,Cj,Dj),(-1≤Dj≤0;Dj<Cj≤1;j=1,2,…,m)
and a≥c>0. Then one has(i)
∑j=1m(fj·gj)(z)-(m-1)zp∈Fp,n(a,c,λ,A(B),B), where
(50)A(B)=B+(1-B)mp(c)n(A0-B0)(C0-D0)(p+nλ)(a)n(1-B0)(1-D0),A0=max1≤j≤m{Aj},B0=min1≤j≤m{Bj},C0=max1≤j≤m{Cj},D0=min1≤j≤m{Dj},provided thatA(B)≤1.(ii)
(1/m)∑j=1m(fj·gj)(z)∈Fp,n(a,c,λ,A~(B),B), where
(51)A~(B)=B+(1-B)×p(c)n(A0-B0)(C0-D0)(p+nλ)(a)n(1-B0)(1-D0)∈(B,1].Proof.
It is clear that-1≤B0≤Bj<Aj≤A0≤1,  B0≤0,  -1≤D0≤Dj<Cj≤C0≤1,  D0≤0,
(52)1-BjAj-Bj≥1-B0A0-B0>0,1-DjCj-Dj≥1-D0C0-D0>0,
for j=1,2,…,m. Let
(53)fj(z)=zp-∑k=p+n∞‍ak,jzk∈Fp,n(a,c,λ,Aj,Bj)gj(z)=zp-∑k=p+n∞‍bk,jzk∈Fp,n(a,c,λ,Cj,Dj),000Cj,Dj0000000000(j=1,2,…,m).
Then Lemma 3 gives
(54)∑j=1m‍(fj·gj)(z)-(m-1)zp=zp-∑k=p+n∞‍(∑j=1m‍ak,jbk,j)zk∈Fp,n(a,c,λ,A,B),
if and only if
(55)(1-BA-B)∑k=p+n∞‍{((p+(k-p)λ)(a)k-pp(c)k-p)(∑j=1m‍ak,jbk,j)}≤1.
Also, by Lemma 3 and (52), we deduce that
(56)(1-B0A0-B0)(1-D0C0-D0)×∑k=p+n∞‍((p+(k-p)λ)(a)k-pp(c)k-p)2ak,jbk,j≤{(1-BjAj-Bj)∑k=p+n∞‍(p+(k-p)λ)(a)k-pp(c)k-pak,j}×{(1-DjCj-Dj)∑k=p+n∞‍(p+(k-p)λ)(a)k-pp(c)k-pbk,j}≤1(j=1,2,…,m),
and hence
(57)1m(1-B0A0-B0)(1-D0C0-D0)×∑k=p+n∞‍{((p+(k-p)λ)(a)k-pp(c)k-p)2(∑j=1m‍ak,jbk,j)}≤1.
To prove Theorem12(i), it follows from (55) and (57) that we need to find the smallest A such that
(58)A≥B+(1-B)(mp(c)k-p(p+(k-p)λ)(a)k-p)×(A0-B01-B0)(C0-D01-D0)(k≥p+n),
for a≥c>0. This leads to the assertion of Theorem 12(i).
Analogously, we can prove Theorem12(ii).In the special case when(59)a=c=1,A(B)=1-2αp,A~(B)=1-2α~p,B=-1,Aj=Cj=1-2αjp,Bj=Dj=-1(j=1,2,…,m),
Theorem 12 reduces to the following.Corollary 13.
Letfj(z)∈Fp(n,λ,αj)(j=1,2,…,m). Then(i)
∑j=1m(fj·fj)(z)-(m-1)zp∈Fp(n,λ,α), where
(60)α=p-mp+nλ(p-min1≤j≤m{αj})2,provided thatα≥0.(ii)
(1/m)∑j=1m(fj·fj)(z)∈Fp(n,λ,α~), where
(61)α~=p-1p+nλ(p-min1≤j≤m{αj})2∈[0,p).
Replacing Lemma3 by Lemma 4 in the proof of Theorem 12, one can prove the following.Theorem 14.
Letfj(z)∈Gp,n(a,c,λ,Aj,Bj)(j=1,2,…,m),
(62)gj(z)∈Gp,n(a,c,λ,Cj,Dj),(-1≤Dj≤0;Dj<Cj≤1;j=1,2,…,m)
and a≥c>0. Then(i)
∑j=1m‍(fj·gj)(z)-(m-1)zp∈Gp,n(a,c,λ,A(B),B), where
(63)A(B)=B+(1-B)mp2(c)n(A0-B0)(C0-D0)(p+n)(p+nλ)(a)n(1-B0)(1-D0),andA0,B0,C0, and D0 are given as in Theorem 12, provided that A(B)≤1.(ii)
(1/m)∑j=1m(fj·gj)(z)∈Gp,n(a,c,λ,A~(B),B), where
(64)A~(B)=B+(1-B)×p2(c)n(A0-B0)(C0-D0)(n+p)(p+nλ)(a)n(1-B0)(1-D0)∈(B,1).
As a special case of Theorem14, one has the following.Corollary 15.
Letfj(z)∈Gp(n,λ,αj)(j=1,2,…,m). Then(i)
∑j=1m‍(fj·fj)(z)-(m-1)zp∈Gp(n,λ,α), where
(65)α=p-mp(p+n)(p+nλ)(p-min1≤j≤m{αj})2,provided thatα≥0.(ii)
(1/m)∑j=1m(fj·fj)(z)∈Gp(n,λ,α~), where
(66)α~=p-p(p+n)(p+nλ)(p-min1≤j≤m{αj})2∈(0,p).


## 3. Integral Operator

Theorem 16.
Let(67)0≤λ1≤λ0,0<c0≤a0,0<a1≤c1(or0<a1≤a0and0<c0≤c1).
If f(z)∈Fp,n(a0,c0,λ0,A,B), then the function I(z) defined by
(68)I(z)=μ+pzμ∫0ztμ-1f(t)dt(μ>-p)
belongs to Fp,n(a1,c1,λ1,C(D),D), where -1≤D≤0 and
(69)C(D)=D+(1-D)×(p+nλ1)(a1)n(c0)n(μ+p)(A-B)(p+nλ0)(c1)n(a0)n(μ+p+n)(1-B).
The result is sharp; that is, the number C(D) cannot be decreased for each D.Proof.
Note thatD<C(D)<1. For
(70)f(z)=zp-∑k=p+n∞‍akzk∈Fp,n(a0,c0,λ0,A,B),
it follows from (68) that
(71)I(z)=zp-∑k=p+n∞‍μ+pμ+kakzk(z∈U;μ>-p).
To prove the result of Theorem16, we need to find the smallest C such that
(72)C≥D+(1-D)×(p+(k-p)λ1)(a1)k-p(c0)k-p(μ+p)(A-B)(p+(k-p)λ0)(c1)k-p(a0)k-p(μ+k)(1-B)0000000(μ+k)(1-B)00000000000(k≥p+n),
where we have used Lemma 3. In view of (67), it is easy to know that the right-hand side of (72) is a decreasing function of k. Therefore, we conclude that
(73)I(z)∈Fp,n(a1,c1,λ1,C(D),D),
where C(D) is given by (69).
Furthermore, it can easily be verified that the result is sharp, with the extremal function(74)f(z)=zp-p(c0)n(A-B)(p+nλ0)(a0)n(1-B)×zp+n∈Fp,n(a0,c0,λ0,A,B).With the aid of Lemma4 (instead of Lemma 3) and using the same steps as in the proof of Theorem 16, we can prove the following.Theorem 17.
Let (67) in Theorem 16 be satisfied. If f(z)∈Gp,n(a0,c0,λ0,A,B), then the function I(z) defined by (68) belongs to Gp,n(a1,c1,λ1,C(D),D), where C(D)(-1≤D≤0) is the same as in Theorem 16. The result is sharp for the function
(75)f(z)=zp-p2(c0)n(A-B)(p+n)(p+nλ0)(a0)n(1-B)×zp+n∈Gp,n(a0,c0,λ0,A,B).If we let(76)a0=c0=a1=c1=1,λ0=λ1=λ≥0,A=1-2αp,B=D=-1,C(D)=1-2β(α)p,
then Theorem 17 yields the following.Corollary 18.
Letf(z)∈Gp(n,λ,α). Then the function I(z) defined by (68) belongs to Gp(n,λ,β(α)), where
(77)β(α)=α(μ+p)+npμ+p+n.
The number β(α) cannot be increased for each α∈[0,p).


## 4. Partial Sums

In this section, we letf(z)∈Tp,n be given by (6) and define the partial sums s1(z) and sm(z) by
(78)s1(z)=zp,sm(z)=zp-∑k=p+np+n+m-2‍akzk,sm(z)=zp-∑k=p+np+n+m-2‍akzk(m∈N∖{1}).
Also we make the notation simple by writing
(79)βk=(p+(k-p)λ)(a)k-p(1-B)p(c)k-p(A-B)0(k=p+n,p+n+1,…).Theorem 19.
Letf(z)∈Fp,n(a,c,λ,A,B) and a≥c>0. Then for z∈U, one has the following. 							(80)Ref(z)sm(z)>1-1βp+n+m-1(m∈N),(81)Resm(z)f(z)>βp+n+m-11+βp+n+m-1(m∈N).
The results are sharp for each m∈N.Proof.
Leta≥c>0 and βk be given by (79). Then βk+1≥βk≥1 for k≥p+n, and so it follows from Lemma 3 that
(82)∑k=p+np+n+m-2‍ak+βp+n+m-1·∑k=p+n+m-1∞‍ak≤∑k=p+n∞‍βkak≤1(m≥2),
for f(z)∈Fp,n(a,c,λ,A,B).
If we put(83)p1(z)=1+βp+n+m-1(f(z)sm(z)-1)=1-βp+n+m-1·∑k=p+n+m-1∞akzk-p1-∑k=p+np+n+m-2akzk-p0000000000(z∈U;m≥2),
then p(0)=1 and
(84)|p1(z)-1p1(z)+1|≤βp+n+m-1·∑k=p+n+m-1∞ak2-2∑k=p+np+n+m-2ak-βp+n+m-1·∑k=p+n+m-1∞ak≤1(z∈U;m≥2)
because of (82). Hence, we have Rep1(z)>0 for z∈U, which implies that (80) holds true for m≥2.
Similarly, by setting(85)p2(z)=(1+βp+n+m-1)sm(z)f(z)-βp+n+m-1(z∈U),
it follows from (82) that
(86)|p2(z)-1p2(z)+1|≤(1+βp+n+m-1)·∑k=p+n+m-1∞ak2-2∑k=p+np+n+m-2ak-(βp+n+m-1-1)·∑k=p+n+m-1∞ak≤1(z∈U;m≥2).
Therefore, we see that Rep2(z)>0 for z∈U, that is, that (81) holds for m≥2.
Form=1, replacing (82) by
(87)βp+n∑k=p+n∞‍ak≤∑k=p+n∞‍βkak≤1
and proceeding as the above, we know that (80) and (81) are also true.
Furthermore, by taking the function(88)f(z)=zp-zp+n+m-1βp+n+m-1,
we find that sm(z)=zp,
(89)Ref(z)sm(z)⟶1-1βp+n+m-1asz⟶1,Resm(z)f(z)⟶βp+n+m-11+βp+n+m-1asz⟶exp(πin+m-1).
The proof of Theorem 19 is thus completed.By virtue of Theorem19 and Definition 2, we easily have the following.Corollary 20.
Letf(z)∈Gp,n(a,c,λ,A,B) and a≥c>0. Then we have
(90)Ref′(z)sm′(z)>1-1βp+n+m-1(z∈U;m∈N),Resm′(z)f′(z)>βp+n+m-11+βp+n+m-1(z∈U;m∈N).
The results are sharp for each m∈N.Theorem 21.
Letf(z)∈Fp,n(a,c,λ,A,B) with a≥c>0 and λ≥1. Then one has
(91)Ref′(z)sm′(z)>1-p+n+m-1pβp+n+m-1(z∈U;m∈N),(92)Resm′(z)f′(z)>pβp+n+m-1p+n+m-1+pβp+n+m-1(z∈U;m∈N).
The results are sharp for each m∈N.Proof.
Leta≥c>0, λ≥1 and βk be given by (79). Then it is easy to verify that
(93)βk+1k+1≥βkk≥1p(k≥p+n),
and hence we deduce from Lemma 3 that
(94)∑k=p+np+n+m-2‍kpak+βp+n+m-1p+n+m-1·∑k=p+n+m-1∞‍kak≤∑k=p+n∞‍βkak≤1(m≥2),
for f(z)∈Fp,n(a,c,λ,A,B).
Defining the functionq1(z) by
(95)q1(z)=1+pβp+n+m-1p+n+m-1(f′(z)sm′(z)-1)(z∈U),
it follows from (94) that
(96)|q1(z)-1q1(z)+1|≤(pβp+n+m-1p+n+m-1·∑k=p+n+m-1∞kak)×(2p-2∑k=p+np+n+m-2kak-pβp+n+m-1p+n+m-1-·∑k=p+n+m-1∞kak)-1≤1(z∈U;m≥2).
This leads to the inequality (91) for m≥2.
Similarly, for the functionq2(z) defined by
(97)q2(z)=(1+pβp+n+m-1p+n+m-1)sm′(z)f′(z)-pβp+n+m-1p+n+m-1,
we deduce from (94) that
(98)|q2(z)-1q2(z)+1|≤((1+pβp+n+m-1p+n+m-1)·∑k=p+n+m-1∞kak)×(2p-2∑k=p+np+n+m-2kak-(pβp+n+m-1p+n+m-1-1)·∑k=p+n+m-1∞kak)-1≤1(z∈U;m≥2).
This yields the inequality (92) for m≥2.
Form=1, replacing (94) by
(99)βp+np+n∑k=p+n∞‍kak≤∑k=p+n∞‍βkak≤1,
we know that (91) and (92) are also true.
Furthermore, the bounds in (91) and (92) are the best possible for the function f(z) defined by (88).Finally, Theorem21 yields the following.Corollary 22.
Letf(z)∈Gp,n(a,c,λ,A,B) with a≥c>0 and λ≥1. Then
(100)Ref′(z)+zf′′(z)sm′(z)+zsm′′(z)>1-p+n+m-1pβp+n+m-1-p+n+m-1βp+n+m-1,(z∈U;m∈N),Resm′(z)+zsm′′(z)f′(z)+zf′′(z)>pβp+n+m-1p+n+m-1+pβp+n+m-1>pβp+n+m-1p+m-1+pβp+n+m-1,(z∈U;m∈N).
The results are sharp for each m∈N.



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*Source: 849747-2013-11-21.xml*