# Differential Subordination for Certian Subclasses ofp-Valent Functions Assoicated with Generalized Linear Operator

Authors: R. M. El-Ashwah; M. K. Aouf; S. M. El-Deeb
Journal: Journal of Mathematics
 (2013)

---

## Abstract

By making use of the differential subordination analytic functions, we investigate inclusion relationships among certain classes of analytic andp-valent functions defined by generalized linear operator.


---

## Body

## 1. Introduction

Let𝒜(p) denote the class of functions of the form
(1)f(z)=zp+∑k=1∞‍ak+pzk+p(p∈ℕ={1,2,…}),
which are analytic and p-valent in the open unit disc U={z∈ℂ:|z|<1}.For functionsf∈𝒜(p), given by (1) and g given by
(2)g(z)=zp+∑k=1∞‍bk+pzk+p(p∈ℕ),
the Hadamard product (or convolution) of f and g is defined by
(3)(f*g)(z)=zp+∑k=1∞‍ak+pbk+pzk+p=(g*f)(z)(z∈U;p∈ℕ).Forf,g∈𝒜(p), we say that the function f is subordinate to g denoted by f(z)≺g(z), if there exists a Schwarz function w, that is, w∈𝒜 with w(0)=0 and |w(z)|<1,z∈U, such that f(z)=g(w(z)) for all z∈U. It is well known that, if the function g is univalent in U, then f(z)≺g(z) is equivalent to f(0)=g(0) and f(U)⊂g(U) (see [1, 2]).Now, we introduce the linear multiplier operatorℒc,p,λm,δ:𝒜(p)→𝒜(p) by
(4)ℒc,p,λm,δf(z)=zp+∑k=1∞‍(1+λkp)m(c+pc+k+p)δak+pzk+p(δ≥0;c>-p;λ≥0;m∈ℕ0).
It is easily verified from (4) that
(5)z(ℒc,p,λm,δf(z))′=(c+p)ℒc,p,λm,δ-1f(z)-cℒc,p,λm,δf(z),(6)λz(ℒc,p,λm,δf(z))′=pℒc,p,λm+1,δf(z)-p(1-λ)ℒc,p,λm,δf(z).
By specializing the parameters c,  λ,  δ,  p, and m, we obtain the following operators:(i)
ℒc,p,01,δf(z)=Kc,pδf(z) (see [3]);(ii)
ℒa-1,1,λ0,δf(z)=Laδf(z) (see [4, 5]);(iii)
ℒ1,p,01,δf(z)=ℐpδf(z) (see [6, 7]);(iv)
ℒc,p,λm,0f(z)=Dλ,pmf(z) (see [8]);(v)
ℒc,1,01,δf(z)=𝒫cδf(z) (see [9, 10]);(vi)
ℒc,1,01,1f(z)=ℒcf(z) (see [11]);(vii)
ℒ1,1,01,δf(z)=ℐδf(z) (see [12]);(viii)
ℒc,1,λm,0f(z)=𝒟λmf(z) (see [13]);(ix)
ℒc,1,1m,0f(z)=𝒟mf(z) (see [14]).By using the multiplier operatorℒc,p,λm,δf(z), we define the following classes of functions.Definition 1.
For fixed parametersA and B, with -1≤B<A≤1 and p>α, we say that the function f∈𝒜(p) is in the class 𝒮c,p,λm,δ(α;A,B) if it satisfies the following subordination condition:
(7)1p-α(z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)-α)≺1+Az1+Bz.
We note that(i)
𝒮1,p,01,0(α;A,B)=𝒮p*(α;A,B) was studied by Cho et al. [15];(ii)
𝒮1,p,01,δ(α;A,B)=𝒮pδ(α;A,B) was studied by Aouf et al. [16];(iii)
𝒮c,1,λm,0(0;A,B)=𝒮λm(A,B) was studied by Patel [17].
For complex numbersa, b, and c, the Gaussian hypergeometric function is defined by
(8)2F1(a,b,d;z)=1+a·bdz1!+a(a+1)·b(b+1)d(d+1)z22!+⋯=∑k=0∞(a)k(b)k(d)kzkk!,a,b∈ℂ,d∈ℂ∖{0,-1,-2,…},
where (λ)k=λ(λ+1)⋯(λ+k-1) and (λ)0=1. The series (8) converges absolutely for z∈U, hence it represents an analytic function in U (see [18, Chapter 14]).
IfB≠-1, from the fact that h(z¯)=h(z)¯,z∈U, we deduce that the image h(U) is symmetric with respect to the real axis, and that h maps the unit disc U onto the disc |w-(1-AB)/(1-B2)|<(A-B)/(1-B2). If B=-1, the function h maps the unit disc U onto the half plan Rew>(1-A)/2, hence we obtain the following.Remark 2.
The functionf∈𝒜(p) is in the class 𝒮c,p,λm,δ(α;A,B) if and only if
(9)|1p-α(z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)-α)-1-AB1-B2|<A-B1-B2,z∈U,forB≠-1,(10)Re1p-α(z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)-α)>1-A2,z∈U,forB=-1.Definition 3.
The functionf∈𝒜(p)is in the class 𝒮c,p,λm,δ(α;ρ) if it satisfies the following the inequality:
(11)Re1p-α(z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)-α)>ρ,z∈U,
where ρ<1; from (9) and (10) it follows, respectively, that
(12)𝒮c,p,λm,δ(α;A,B)⊂𝒮c,p,λm,δ(α;1-A1-B),𝒮c,p,λm,δ(α;A,B)=𝒮c,p,λm,δ(α;1-A1-B)⟺B=-1.
We note that when c=1 and m=0, the class 𝒮1,p,01,δ(α;ρ)=𝒮pδ(α;ρ) was studied by Aouf et al. [16].
Let us consider the first-order differential subordination(13)H(φ(z),zφ′(z))≺h(z).
A univalent functionq is called its dominant, if φ(z)≺q(z) for all analytic functions φ that satisfy this differential subordination. A dominant q~ is called the best dominant, if q~(z)≺q(z) for all dominants q. For the general theory of the first-order differential subordination and its applications, we refer the reader to [1, 2].
The object of the present paper is to obtain several inclusion relationships and other interesting properties of functions belonging to the subclasses𝒮c,p,λm,δ(α;A,B) and 𝒮c,p,λm,δ(α;ρ) by using the theory of differential subordination.
To establish our main results, we will require the following lemmas.Lemma 4 (see [19]).
Let
β
,
γ
∈
ℂ, and let h be a convex function with(14)Re[βh(z)+γ]>0,z∈U.Ifp is analytic in U, with p(0)=h(0), then
(15)p(z)+zp′(z)βp(z)+γ≺h(z)⇒p(z)≺h(z).Lemma 5 (see [20]).
Letβ>0,β+γ>0,  and consider the integral operator Jβ,γ defined by
(16)Jβ,γ(f)(z)=[β+γzγ∫0zfβ(t)tγ-1dt]1/β,
where the powers are the principal ones.
Ifσ∈[-γ/β,1) then the order of starlikeness of the class Jβ,γ(S*(σ)), that is, the largest number δ(σ;β,γ) such that Jβ,γ(S*(σ))⊂S*(δ), is given by the number δ(σ;β,γ)=inf{Req(z):z∈U}, where
(17)q(z)=1βQ(z)-γβ,Q(z)=∫01(1-z1-tz)2β(1-σ)tβ+γ-1dt.
Moreover, ifσ∈[σ0,1), where σ0=max{(β-γ-1)/2β;-γ/β} and g=Jβ,γ(f)(z)  with f∈S*(σ), then
(18)Rezg′(z)g(z)>δ(σ;β,γ)(z∈U),
where
(19)δ(σ;β,γ)=1β[β+γF12(1,2β(1-σ),β+γ+1;1/2)-γ].Lemma 6 (see [21]).
Let
ϕ  beanalyticinUwithϕ(0)=1 and ϕ(z)≠0 for 0<|z|<1, and let A,B∈ℂ with A≠B, |B|≤1.(i) LetB≠0 and γ∈ℂ*=ℂ∖{0} satisfy either |(γ(A-B)/B)-1|≤1 or |(γ(A-B)/B)+1|≤1. If ϕ satisfies
(20)1+zϕ′(z)γϕ(z)≺1+Az1+Bz
then
(21)ϕ(z)≺(1+Bz)γ(A-B)/B,
and this is the best dominant.(ii) Letγ∈ℂ* be such that |γA|<π, and if ϕ satisfies
(22)1+zϕ′(z)γϕ(z)≺1+Az
then
(23)ϕ(z)≺eγAz,
and this is the best dominant.


## 2. Inclusion Relationships

Unless otherwise mentioned, we assume throughout this paper that(24)-1≤B<A≤1,p>α,δ≥0,c>-p,λ≥0,p∈ℕ,m∈ℤ
and the power is the principal one.Theorem 7.
Let(25)(p-α)(1-A)+(α+c)(1-B)≥0.(1)
Supposing thatℒc,p,λm,δf(z)≠0 for all z∈U.=U∖{0}, then 𝒮c,p,λm,δ-1(α;A,B)⊂𝒮c,p,λm,δ(α;A,B).(2)
Moreover, if we suppose in addition that(26)A≤1+(1-Bp-α)(α+c-p2+12),then
(27)𝒮c,p,λm,δ-1(α;A,B)⊂𝒮c,p,λm,δ(α;ρ(p,α,c,A,B)),
where the bound
(28)ρ(p,α,c,A,B)=1p-α×[p+cF12(1,2(p-α)(A-B)/(1-B),p+c+1;1/2)-(α+c)p+cF12],
is the best possible.Proof.
Letf∈𝒮c,p,λm,δ-1(α;A,B), and put
(29)g(z)=z(ℒc,p,λm,δf(z)zp)1/(p-α)(z∈U),
the function g is analytic in U, with g(0)=0 and g′(0)=1. Differentiating (29) logarithmically with respect to z, we have
(30)ϕ(z)=zg′(z)g(z)=1p-α(z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)-α)(z∈U),
then, using (5) in (30), we obtain
(31)(c+p)ℒc,p,λm,δ-1f(z)ℒc,p,λm,δf(z)=(p-α)ϕ(z)+(α+c).
By differentiating both sides of (31) logarithmically with respect to z and multiplying by z, we have
(32)1p-α(z(ℒc,p,λm,δ-1f(z))′ℒc,p,λm,δ-1f(z)-α)=ϕ(z)+zϕ′(z)(p-α)ϕ(z)+(α+c).
Combining (32) together with f∈𝒮c,p,λm,δ-1(α;A,B), we obtain that the function ϕ satisfies the Briot-Bouquet differential subordination as follows:
(33)ϕ(z)+zϕ′(z)(p-α)ϕ(z)+(α+c)≺1+Az1+Bz≡h(z).
Now we will use Lemma4 for the special case β=p-α and γ=α+c. Since h is a convex function in U, a simple computation shows that
(34)Re{(p-α)h(z)+(α+c)}>0,z∈U,
whenever (25) holds, we have ϕ(z)≺h(z); that is, f∈𝒮c,p,λm,δ(α;A,B). If in addition, we suppose that the inequality (26) holds, then all the assumptions of Lemma 5 are verified for the above values of β, γ, and σ=(1-A)/(1-B). Then it follows the inclusion 𝒮c,p,λm,δ-1(α;A,B)⊂𝒮c,p,λm,δ(α;ρ(p,α,c,A,B)), where the bound ρ(p,α,c,A,B) given by (28) is the best possible.
From Theorem7, according to the definitions (7) and (11), we deduce the next inclusions.Corollary 8.
Letp>α, such that (25) holds.(1)
Supposing thatℒc,p,λm,δf(z)≠0 for all z∈U., then
(35)𝒮c,p,λm,δ-1(α;A,B)⊂𝒮c,p,λm,δ(α;A,B)⊂𝒮c,p,λm,δ(α;1-A1-B).(2)
If we suppose in addition that (26) holds, then
(36)𝒮c,p,λm,δ-1(α;A,B)⊂𝒮c,p,λm,δ(α;A,B)⊂𝒮c,p,λm,δ(α;ρ(p,α,c,A,B)),where ρ(p,α,c,A,B) is given by (28). As a consequence of the last inclusion, one has ρ(p,α,c,A,B)≥(1-A)/(1-B).
For the special caseB=-1, Theorem 7 reduces to the following.Corollary 9.
Leta≥-(c+α)/(p-α).(1)
Supposing thatℒc,p,λm,δf(z)≠0 for all z∈U., then(37)𝒮c,p,λm,δ-1(α;a)⊂𝒮c,p,λm,δ(α;a).(2)
If we suppose in addition that(38)a≥max{-2α+c-p+12(p-α);-α+c(p-α)},then
(39)𝒮c,p,λm,δ-1(α;a)⊂𝒮c,p,λm,δ(α;ρ(p,α,c,a)),
where the bound
(40)ρ(p,α,c,a)=1p-α×[p+cF12(1,2(p-α)(1-a),p+c+1;1/2)-(α+c)p+cF12],
is the best possible.Theorem 10.
Let
f
∈
𝒮
c
,
p
,
λ
m
,
δ
(
α
;
ρ
), where ρ<1 and ρ>α, then f∈𝒮c,p,λm,δ-1(α;ρ) for |z|<R(p,α,c,ρ), where
(41)R(p,α,c,ρ)=min{r>0:θ(p,α,c,ρ,r)=0},θ(p,α,c,ρ,r)=2r((1-r)(p-α)×|(1-ρ)(1-r)-|ρ+(α+c)/(p-α)|(1+r)|)-1.Proof.
Sincef∈𝒮c,p,λm,δ(α;ρ), the function u given by
(42)u(z)=1p-α(z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)-α)(z∈U),
is analytic in U with u(0)=1 and Reu(z)>ρ. Using (5) in (42) and taking the logarithmic differentiation in the resulting equation, we obtain
(43)1p-α(z(ℒc,p,λm,δ-1f(z))′ℒc,p,λm,δ-1f(z)-α)=u(z)+zu′(z)(p-α)u(z)+(α+c).
If we denote k(z)=(u(z)-ρ)/(1-ρ), then k(0)=1 and Rek(z)>0,z∈U and substituting in (43) we obtain
(44)1p-α(z(ℒc,p,λm,δ-1f(z))′ℒc,p,λm,δ-1f(z)-α)-ρ=(1-ρ)×[k(z)zk′(z)(1-ρ)k(z)+zk′(z)(p-α)(1-ρ)k(z)+ρ(p-α)+(α+c)],
hence
(45)Re1p-α(z(ℒc,p,λm,δ-1f(z))′ℒc,p,λm,δ-1f(z)-α)-ρ≥(1-ρ)×[Rek(z)(|(p-α)||)-1-(|zk′(z)|)((p-α)×|(1-ρ)k(z)-|ρ+((α+c)/(p-α))||)-1].
By using the well-known results [22]
(46)|zk′(z)|≤2r1-r2Rek(z),Rek(z)≥1-r1+r,|z|=r<1,
together with the inequality (45), we get
(47)Re1p-α(z(ℒc,p,λm,δ-1f(z))′ℒc,p,λm,δ-1f(z)-α)-ρ≥(1-ρ)[1-θ(p,α,c,ρ,r)]Rek(z),|z|=r.
Since the right hand side term of the inequality (47) is nonnegative whenever |z|≤R(p,α,c,ρ) is given by (41), using the fact that the real part of an analytic function is harmonic, we deduce that f∈𝒮c,p,λm,δ-1(α;ρ) for |z|<R(p,α,c,ρ).
For a functionf∈𝒜(p), let the integral operator Fμ,p:𝒜(p)→𝒜(p) defined by Saitoh [23] and Saitoh et al. [24]
(48)Fμ,p(f(z))=μ+pzμ∫0ztμ-1f(t)dt=zp+∑k=p+1∞‍μ+pk+μakzk=(zp+∑k=p+1∞‍μ+pk+μzk)*f(z)=z2pF1(1,μ+p;μ+p+1;z)*f(z)(z∈U;μ>-p).
From (4) and (48), we have
(49)z(ℒc,p,λm,δFμ,p(f(z)))′=(μ+p)ℒc,p,λm,δf(z)-μℒc,p,λm,δFμ,p(f(z)),ℒc,p,λm,δFμ,p(f(z))=Fμ,p(ℒc,p,λm,δf(z)),f∈A(p).We now prove the next result.Theorem 11.
Let
α
+
μ
>
0 and
(50)(α+μ)(1-B)+(p-α)(1-A)≥0.(i)
Supposing thatFμ,p(f(z))≠0 for all z∈U.=U∖{0}, then
(51)Fμ,p(𝒮c,p,λm,δ(α;A,B))⊂𝒮c,p,λm,δ(α;A,B).(ii)
Moreover, if we suppose in addition that(52)A≤1+(1-Bp-α)min{μ+2α-p+12;μ+α},then
(53)Fμ,p(𝒮c,p,λm,δ(α;A,B))⊂𝒮c,p,λm,δ(α;r(p,α,μ,A,B)),
where the bound
(54)r(p,α,μ,A,B)=1p-α×[p+μF12(1,2(p-α)(A-B)/(1-B),μ+p+1;1/2)-(μ+α)pF12],
is the best possible.Proof.
Letf∈𝒮c,p,λm,δ(α;A,B), and suppose that Fμ,p(f(z))≠0 for all z∈U.. Let
(55)g(z)=z(ℒc,p,λm,δFμ,p(f(z))zp)1/(p-α)(z∈U),
then g is analytic in U, with g(0)=0 and g′(0)=1. Taking the logarithmic differentiation in (55), we have
(56)ϕ(z)=zg′(z)g(z)=1p-α(z(ℒc,p,λm,δFμ,p(f(z)))′ℒc,p,λm,δFμ,p(f(z))-α)(z∈U).
Now, by using (49) in (56), we obtain
(57)(p+μ)ℒc,p,λm,δf(z)ℒc,p,λm,δFμ,p(f(z))=(p-α)ϕ(z)+(μ+α).
By differentiating in both sides of (57) logarithmical with respect to z and multiplying by z, we have
(58)1p-α(z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)-α)=ϕ(z)+zϕ′(z)(p-α)ϕ(z)+(μ+α).
Since f∈𝒮c,p,λm,δ(α;A,B), from (58), we obtain that the function ϕ satisfies the Briot-Bouquet differential subordination
(59)ϕ(z)+zϕ′(z)(p-α)ϕ(z)+(μ+α)≺1+Az1+Bz≡h(z).
Now we will use Lemma4 for the special case β=p-α and γ=μ+α; we have ϕ(z)≺h(z), that is, Fμ,p∈𝒮c,p,λm,δ(α;A,B). If we suppose in addition that the inequality (52) holds, then all the assumptions of the Lemma 5 are satisfied for β,  γ, and σ=(1-A)/(1-B), hence it follows the inclusion Fμ,p(𝒮c,p,λm,δ(α;A,B))⊂𝒮c,p,λm,δ(α;r(p,α,μ,A,B)), and the bound r(p,α,μ,A,B) given by (54) is the best possible.TakingB=-1 in Theorem 11, we obtain the next corollary.Corollary 12.
Letα+μ>0 and a≥-(μ+α)/(p-α).(1)
Supposing thatℒc,p,λm,δFμ,p(f(z))≠0 for all z∈U., then(60)Fμ,p(𝒮c,p,λm,δ(α;a))⊂𝒮c,p,λm,δ(α;a).(2)
If we suppose in addition that(61)a≥max{-μ+2α-p+12(p-α);-μ+αp-α},then
(62)Fμ,p(𝒮c,p,λm,δ(α;a))⊂𝒮c,p,λm,δ(α;r(p,α,μ,a)),
where the bound
(63)r(p,α,μ,a)=1p-α×[μ+pF12(1,2(p-α)(1-a),μ+p+1;1/2)-(μ+α)pF12],
is the best possible.Theorem 13.
Letν∈ℂ*, and let A,B∈ℂ with A≠B and |B|≤1. Suppose that
(64)|ν(c+p)(A-B)B-1|≤1or|ν(c+p)(A-B)B+1|≤1,ifB≠0,|νA|≤πc+p,ifB=0.
If f∈𝒜(p) with ℒc,p,λm,δf(z)≠0 for all z∈U., then
(65)ℒc,p,λm,δ-1f(z)ℒc,p,λm,δf(z)≺1+Az1+Bz
implies
(66)(ℒc,p,λm,δf(z)zp)ν≺q1(z),
where
(67)q1(z)={(1+Bz)ν(c+p)(A-B)/BifB≠0,eν(c+p)AzifB=0,
is the best dominant.Proof.
Let us put(68)ϕ(z)=(ℒc,p,λm,δf(z)zp)ν(z∈U),
then ϕ is analytic in U, with ϕ(0)=1 and ϕ(z)≠0for all z∈U.. By differentiating both sides of (68) logarithmical with respect to z and using (5), we have
(69)1+zϕ′(z)ν(c+p)ϕ(z)=ℒc,p,λm,δ-1f(z)ℒc,p,λm,δf(z)≺1+Az1+Bz.
Now the assertions of Theorem 13 follows by using Lemma 6 for the special case γ=ν(c+p).PuttingB=-1 and A=1-2ρ,0≤ρ<1, in Theorem 13, we obtain the following corollary.Corollary 14.
Assume thatν∈ℂ* satisfies either
(70)|2ν(c+p)(1-ρ)-1|≤1or|2ν(c+p)(1-ρ)+1|≤1,ifB≠0,
If f∈𝒜(p) with ℒc,p,λm,δf(z)≠0 for all z∈U., then
(71)Reℒc,p,λm,δ-1f(z)ℒc,p,λm,δf(z)>ρ
implies
(72)(ℒc,p,λm,δf(z)zp)ν≺q2(z)=(1-z)-2ν(c+p)(1-ρ),
and q2 is the best dominant.


## 3. Properties Involving the Multiplier Operatorℒc,p,λm,δ

Theorem 15.
Iff∈𝒮c,p,λm,δ(α;A,B), then for all s, t∈ℂ with |s|≤1,  |t|≤1 and s≠t, the next subordination holds
(73)tpℒc,p,λm,δf(zs)spℒc,p,λm,δf(zt)≺{(1+Bzs1+Bzt)(p-α)(A-B)/B,forB≠0exp[(p-α)Az(s-t)],forB=0.Proof.
Iff∈𝒮c,p,λm,δ(α;A,B), from (7) it follows that
(74)z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)≺p+[pB+(p-α)(A-B)]z1+Bz≡k(z).
Moreover, the function k defined by (74) and the function h given by
(75)h(z)≡h(z;s,t)=∫0z(s1-su-t1-tu)du
are convex in U. By combining a general subordination theorem [25, Theorem 4] with (74), we get
(76)(z(ℒc,p,λm,δf(z))′ℒc,p,λm,δf(z)-p)*h(z)≺(p-α)(A-B)z1+Bz*h(z).
For every analytic function ϕ in U with ϕ(0)=0, we have
(77)ϕ(z)*h(z)=∫tzszϕ(u)udu,
and thus, from (76) and (77), we deduce
(78)∫tzsz((ℒc,p,λm,δf(u))′ℒc,p,λm,δf(u)-pu)du≺(p-α)(A-B)∫tzszdu1+Bu.
This last subordination implies
(79)exp(∫tzsz((ℒc,p,λm,δf(u))′ℒc,p,λm,δf(u)-pu)du)≺exp((p-α)(A-B)∫tzszdu1+Bu),
and by simplification, we get the assertion of Theorem 15.Corollary 16.
Iff∈𝒮c,p,λm,δ(α;A,B), then for |z|=r<1, the next inequalities hold
(80)|ℒc,p,λm,δf(z)|≤{rp(1+Br)(p-α)(A-B)/B,forB≠0,rpexp[(p-α)Ar],forB=0,(81)|ℒc,p,λm,δf(z)|≥{rp(1-Br)(p-α)(A-B)/B,forB≠0,rpexp[-(p-α)Ar],forB=0,(82)|argℒc,p,λm,δf(z)zp|≤{(p-α)(A-B)|B|sin-1(|B|r),forB≠0,(p-α)Ar,forB=0.
All of the estimates asserted here are sharp.Proof.
Takings=1 and t=0 in (73) and using the definition of subordination, we obtain
(83)ℒc,p,λm,δf(z)zp={(1+Bw(z))(p-α)(A-B)/B,forB≠0,exp[(p-α)Aw(z)],forB=0,
where w is analytic function in U with w(0)=0 and |w(z)|≤1 for z∈U. According to the well-known Schwarz's theorem, we have |w(z)|≤|z| for all z∈U.(i)
IfB>0, then we find from (83) that
(84)|ℒc,p,λm,δf(z)zp|=exp[(p-α)(A-B)Blog|1+Bw(z)|]=|1+Bw(z)|(p-α)(A-B)/B≤(1+Br)(p-α)(A-B)/B.(ii)
IfB<0, we can easily obtain
(85)|ℒc,p,λm,δf(z)zp|=|1+Bw(z)|(p-α)(A-B)/-B≤[(1+Br)-1](p-α)(A-B)/-B=(1+Br)(p-α)(A-B)/B.This proves the inequality (80) for B≠0. Similarly, we can prove the other inequalities in (80) and (81). Now, for |z|=r and B≠0, we observe from (83) that
(86)|argℒc,p,λm,δf(z)zp|=(p-α)(A-B)|B||arg(1+Bw(z))|≤(p-α)(A-B)|B|sin-1(|B|r),
and for B=0, (82) is a direct consequence of (83).
It is easy to see that all of the estimates in Corollary16 are sharp, being attained by the function f0 defined by
(87)ℒc,p,λm,δf0(z)={zp(1+Bz)(p-α)(A-B)/B,forB≠0,zpexp[(p-α)Az],forB=0.Remark 17.
(i) Puttingc=1 and m=0 in our results, we obtain the results obtained by Aouf et al. [16].
(ii) By specializing the parametersc,λ,δ,p, and m, we obtain various results for different operators defined in Section 1.



---

*Source: 692045-2013-03-21.xml*