# A New Class of Analytic Functions Defined by Using Salagean Operator

Authors: R. M. El-Ashwah; M. K. Aouf; A. A. M. Hassan; A. H. Hassan
Journal: International Journal of Analysis
 (2013)

---

## Abstract

We derive some results for a new class of analytic functions defined by using Salagean operator. We give some properties of functions in this class and obtain numerous sharp results including for example, coefficient estimates, distortion theorem, radii of star-likeness, convexity, close-to-convexity, extreme points, integral means inequalities, and partial sums of functions belonging to this class. Finally, we give an application involving certain fractional calculus operators that are also considered.


---

## Body

## 1. Introduction

Let𝒜 denote the class of functions of the form
(1)f(z)=z+∑k=2∞akzk
that are analytic and univalent in the open unit disc U={z∈ℂ:|z|<1}.Forf(z)∈𝒜, Salagean [1] introduced the following differential operator: D0f(z)=f(z), D1f(z)=zf′(z),…, Dnf(z)=D(Dn-1f(z))(n∈N={1,2,…}).We note that(2)Dnf(z)=z+∑k=2∞knakzk(n∈ℕ0=ℕ∪{0}).Definition 1 (subordination principle).
For two functionsf and g, analytic in U, we say that the function f(z) is subordinate to g(z) in U and write f(z)≺g(z), if there exists a Schwarz function w(z), which (by definition) is analytic in U with w(0)=0 and |w(z)|<1, such that f(z)=g(w(z))(z∈U). Indeed it is known that
(3)f(z)≺g(z)⇒f(0)=g(0),f(U)⊂g(U).
Furthermore, if the function g is univalent in U, then we have the following equivalence [2, page 4]:
(4)f(z)≺g(z)⟺f(0)=g(0),f(U)⊂g(U).Definition 2 (see [3]).
LetUm,n(β,A,B) denote the subclass of 𝒜 consisting of functions f(z) of the form (1) and satisfy the following subordination:
(5)Dmf(z)Dnf(z)-β|Dmf(z)Dnf(z)-1|≺1+Az1+Bz,(-1≤B<A≤1;-1≤B<0;β≥0;m∈ℕ;n∈ℕ0,m>n;z∈U).
Specializing the parameters A,B,β,m, and n, we obtain the following subclasses studied by various authors:
(i)(6)Um,n(β,1-2α,-1)=Nm,n(α,β)={f∈𝒜:Re{Dmf(z)Dnf(z)-α}>β|Dmf(z)Dnf(z)-1|{{Dmf(z)Dnf(z)-α}>β|Dmf(z)Dnf(z)-1|}(0≤α<1;β≥0;m∈ℕ;n∈ℕ0;m>n;z∈U)},
(see Eker and Owa [4]);
(ii)(7)U1,0(β,1-2α,-1)=US(α,β)={f∈𝒜:Re{zf′(z)f(z)-α}>β|zf′(z)f(z)-1|(0≤α<1;β≥0;z∈U)|zf′(z)f(z)-1|},U2,1(β,1-2α,-1)=UK(α,β)={f∈𝒜:Re{1+zf′′(z)f′(z)-α}>β|zf′′(z)f′(z)|(0≤α<1;β≥0;z∈U)|zf′(z)f(z)-1|},
(see Shams et al. [5, 6]);
(iii)(8)U1,0(0,A,B)=S*(A,B)={f∈𝒜:zf′(z)f(z)≺1+Az1+Bz(-1≤B<A≤1;z∈U)zf′(z)f(z)},U2,1(0,A,B)=K(A,B)={f∈𝒜:1+zf′′(z)f′(z)≺1+Az1+Bz(-1≤B<A≤1;z∈U)zf′′(z)f′(z)},
(see Janowski [7] and Padmanabhan and Ganesan [8]).Also we note that(9)Um,n(0,A,B)=U(m,n;A,B)={f(z)∈𝒜:Dmf(z)Dnf(z)≺1+Az1+Bz(-1≤B<A≤1;m∈ℕ;n∈ℕ0;m>n;z∈U)Dmf(z)Dnf(z)}.LetT denote the subclass of functions of 𝒜 of the form
(10)f(z)=z-∑k=2∞akzk,ak≥0.Further, we define the classTSγ(f,g;α,β) by
(11)TUm,n(β,A,B)=Um,n(β,A,B)∩T.For suitable choices of the parametersA,B,β,m, and n, we can get various known or new subclasses of T. For example, we have the following:(i)
TUn+1,n(β,1-2α,-1)=TS(n,α,β)(0≤α<1,β≥0,n∈ℕ0) (see Rosy and Murugusundaramoorthy [9] and Aouf [10]);(ii)
TU1,0(1,1-2α,-1)=SpT(α) and TU2,1(1,1-2α,-1)=UCT(α)(0≤α<1) (see Bharati et al. [11]);(iii)
TU1,0(0,1-2α,-1)=T*(α) and TU2,1(0,1-2α,-1)=C(α)(0≤α<1) (see Silverman [12]).


## 2. Coefficient Estimates

Unless otherwise mentioned, we assume in the reminder of this paper that-1≤B<A≤1,-1≤B<0,β≥0,m∈ℕ,n∈ℕ0,m>n and z∈U.Now, we will need the following lemma which gives a sufficient condition for functions belonging to the classUm,n(β,A,B).Lemma 3 (see [13]).
A functionf(z) of the form (1) is in the class Um,n(β,A,B) if
(12)∑k=2∞[(1+β(1+|B|))(km-kn)+|Bkm-Akn|]|ak|≤A-B.In Theorem4, it is shown that the condition (12) is also necessary for functions f(z) of the form (10) to be in the class TUm,n(β,A,B).Theorem 4.
Letf(z)∈T. Then f(z)∈TUm,n(β,A,B) if and only if
(13)∑k=2∞[(1+β(1+|B|))(km-kn)+|Bkm-Akn|]ak≤A-B.Proof.
In view of Lemma3, we only need to prove the only if part of Theorem 4. Since TUm,n(β,A,B)⊂Um,n(β,A,B), for functions f(z)∈TUm,n(β,A,B), we can write
(14)|p(z)-1A-Bp(z)|<1,wherep(z)=Dmf(z)Dnf(z)-β|Dmf(z)Dnf(z)-1|.
then
(15)|(∑k=2∞(km-kn)akzk+βeiθ|∑k=2∞(km-kn)akzk|){+Bβeiθ|∑k=2∞(km-kn)akzk|)-1}×((A-B)z+∑k=2∞(Bkm-Akn)akzk+Bβeiθ|∑k=2∞(km-kn)akzk|)-1|<1.
Since Re(z)≤|z|(z∈U), then we obtain
(16)Re{+Bβeiθ|∑k=2∞(km-kn)akzk|)-1(∑k=2∞(km-kn)akzk+βeiθ|∑k=2∞(km-kn)akzk|)×((A-B)z+∑k=2∞(Bkm-Akn)akzk+Bβeiθ|∑k=2∞(km-kn)akzk|)-1}<1.
Now choosing z to be real and letting z→1-, we obtain
(17)∑k=2∞[(1+β(1-B))(km-kn)-(Bkm-Akn)]ak≤A-B.
Or, equivalently
(18)∑k=2∞[(1+β(1+|B|))(km-kn)+|Bkm-Akn|]ak≤A-B.
This completes the proof of Theorem 4.Remark 5.
(i) The result obtained by Theorem4 corrects the result obtained by Li and Tang [3, Theorem 1].
(ii) PuttingA=1-2α(0≤α<1) and B=-1 in Theorem 4, we correct the result obtained by Eker and Owa [4, Theorem  2.1].
(iii) PuttingA=1-2α(0≤α<1),B=-1, and m=n+1(n∈ℕ0) in Theorem 4, we obtain the result obtained by Rosy and Murugusudaramoorthy [9, Theorem 2].Corollary 6.
Let the functionf(z) be defined by (10) and let it be in the class TUm,n(β,A,B). Then
(19)ak≤A-B(1+β(1+|B|))(km-kn)+|Bkm-Akn|,(k≥2).
The result is sharp for the function
(20)f(z)=z-A-B(1+β(1+|B|))(km-kn)+|Bkm-Akn|zk,(k≥2).


## 3. Distortion Theorems

Theorem 7.
Let the functionf(z) defined by (10) be in the class TUm,n(β,A,B). Then
(21)|f(z)|≥|z|-A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n||z|2≤|z|+A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n||z|2.
The result is sharp.Proof.
In view of Theorem4, since
(22)Φ(k)=(1+β(1+|B|))(km-kn)+|Bkm-Akn|
is an increasing function of k(k≥2), we have
(23)Φ(2)∑k=2∞|ak|≤∑k=2∞Φ(k)|ak|≤A-B,
that is
(24)∑k=2∞|ak|≤A-BΦ(2).
Thus we have
(25)|f(z)|≤|z|+|z|2∑k=2∞|ak|,|f(z)|≤|z|+A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n||z|2.
Similarly, we get
(26)|f(z)|≥|z|-∑k=2∞|ak||z|k≥|z|-|z|2∑k=2∞|ak|≥|z|-A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n||z|2.
Finally the result is sharp for the function
(27)f(z)=z-A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n|z2,
at z=r and z=rei(2k+1)π(k∈ℤ={…,-2,-1,0,1,2,…}). This completes the proof of Theorem 7.Theorem 8.
Let the functionf(z) defined by (10) be in the class TUm,n(β,A,B). Then
(28)|f′(z)|≥1-2(A-B)(1+β(1+|B|))(2m-2n)+|B2m-A2n||z|≤1+2(A-B)(1+β(1+|B|))(2m-2n)+|B2m-A2n||z|.
The result is sharp.Proof.
SimilarlyΦ(k)/k is an increasing function of k(k≥2), in view of Theorem 4, we have
(29)Φ(2)2∑k=2∞k|ak|≤∑k=2∞Φ(k)kk|ak|=∑k=2∞Φ(k)|ak|≤A-B,
that is
(30)∑k=2∞k|ak|≤2(A-B)Φ(2).
Thus we have
(31)|f′(z)|≤1+|z|∑k=2∞k|ak|≤1+2(A-B)(1+β(1+|B|))(2m-2n)+|B2m-A2n||z|.
Similarly
(32)|f′(z)|≥1-|z|∑k=2∞k|ak|≥1-2(A-B)(1+β(1+|B|))(2m-2n)+|B2m-A2n||z|.
Finally, we can see that the assertions of Theorem 8 are sharp for the function f(z) defined by (27). This completes the proof of Theorem 8.


## 4. Radii of Starlikeness, Convexity, and Close-to-Convexity

In this section radii of close-to-convexity, starlikeness, and convexity for functions belonging to the classTUm,n(β,A,B) are obtained.Theorem 9.
Let the functionf(z) defined by (10) be in the class TUm,n(β,A,B); then(i)
f(z) is starlike of order φ(0≤φ<1) in |z|<r1, where
(33)r1=infk≥2{(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B×(1-φk-φ)(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B}1/(k-1),(ii)
f(z) is convex of order φ(0≤φ<1) in |z|<r2, where
(34)r2=infk≥2{(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B×(1-φ)k(k-φ)(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B}1/(k-1),(iii)
f(z)is close-to-convex of order φ(0≤φ<1) in |z|<r3, where
(35)r3=infk≥2{(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B×(1-φk)(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B}1/(k-1).Each of these results is sharp for the function f(z) given by (20).Proof.
It is sufficient to show that(36)|zf′(z)f(z)-1|≤1-φ,for|z|<r1,
where r1 is given by (33). Indeed we find from (10) that
(37)|zf′(z)f(z)-1|≤∑k=2∞(k-1)ak|z|k-11-∑k=2∞ak|z|k-1.
Thus we have
(38)|zf′(z)f(z)-1|≤1-φ,
if and only if
(39)∑k=2∞(k-φ)ak|z|k-1(1-φ)≤1.
But, by Theorem 4, (39) will be true if
(40)(k-φ)|z|k-1(1-φ)≤(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B,
that is, if
(41)|z|≤{(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B×(1-φk-φ)}1k-1(k≥2).
Or
(42)r1=infk≥2{(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B×(1-φk-φ)}1/(k-1).
This completes the proof of (33).To prove (34) and (35) it is sufficient to show that
(43)|1+zf′′(z)f′(z)-1|≤1-φ(|z|<r2;0≤φ<1),(44)|f′(z)-1|≤1-φ(|z|<r3;0≤φ<1),
respectively.


## 5. Extreme Points

Theorem 10.
Let(45)f1(z)=z,fk(z)=z-A-B(1+β(1+|B|))(km-kn)+|Bkm-Akn|zk,(k=2,3,…).
Then f(z)∈TUm,n(β,A,B) if and only if it can be expressed in the following form:
(46)f(z)=∑k=1∞ηkfk(z),
where
(47)ηk≥0,∑k=1∞ηk=1.Proof.
Suppose that(48)f(z)=∑k=1∞ηkfk(z)=z-∑k=2∞ηkA-B(1+β(1+|B|))(km-kn)+|Bkm-Akn|zk.
Then, from Theorem 4, we have
(49)∑k=2∞[A-B(1+β(1+|B|))(km-kn)+|Bkm-Akn|{(1+β(1+|B|))(km-kn)+|Bkm-Akn|}·A-B(1+β(1+|B|))(km-kn)+|Bkm-Akn|ηk]=(A-B)∑k=2∞ηk=(A-B)(1-η1)≤A-B.
Thus, in view of Theorem 4, we find that f(z)∈TUm,n(β,A,B).
Conversely, let us suppose thatf(z)∈TUm,n(β,A,B), then, since
(50)ak≤A-B(1+β(1+|B|))(km-kn)+|Bkm-Akn|,(k=2,3,…).
Set
(51)ηk=(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-Bak,(k=2,3,…),η1=1-∑k=2∞ηk.
Thus clearly, we have
(52)f(z)=∑k=1∞ηkfk(z).
This completes the proof of Theorem 10.Corollary 11.
The extreme points of the classTUm,n(β,A,B) are given by
(53)f1(z)=z,fk(z)=z-A-B(1+β(1+|B|))(km-kn)+|Bkm-Akn|zk,(k=2,3,…).


## 6. Integral Means Inequalities

In 1925, Littlewood [14] proved the following subordination lemma.Lemma 12.
If the functionsf and g are analytic in U with
(54)f(z)≺g(z)(z∈U),
then for p>0and z=reiθ(0<r<1),
(55)∫02π|f(z)|pdθ≤∫02π|g(z)|pdθ.
We now make use of Lemma 12 to prove Theorem 13.Theorem 13.
Suppose thatf(z)∈TUm,n(β,A,B),p>0,-1≤B<A≤1,β>0,m∈ℕ,n∈ℕ0,m>n, and f2(z) is defined by
(56)f2(z)=z-A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n|z2.
Then for z=reiθ(0<r<1), we have
(57)∫02π|f(z)|pdθ≤∫02π|f2(z)|pdθ.Proof.
Forf(z)=z-∑k=2∞akzk(ak≥0), (55) is equivalent to prove that
(58)∫02π|1-∑k=2∞akzk-1|pdθ≤∫02π|1-A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n|z|pdθ.
By applying Littlewood’s subordination lemma (Lemma 12), it would suffice to show that
(59)1-∑k=2∞akzk-1≺1-A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n|z.
By setting
(60)1-∑k=2∞akzk-1=1-A-B(1+β(1+|B|))(2m-2n)+|B2m-A2n|w(z)
and using (13), we obtain
(61)|w(z)|=|∑k=2∞(1+β(1+|B|))(2m-2n)+|B2m-A2n|A-Bakzk-1|≤|z|∑k=2∞(1+β(1+|B|))(2m-2n)+|B2m-A2n|A-Bak≤|z|∑k=2∞(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-Bak≤|z|<1.
This completes the proof of Theorem 13.


## 7. Partial Sums

In this section partial sums of functions in the classUm,n(β,A,B) are obtained, also we will obtain sharp lower bounds for the ratios of real part of f(z) to fn(z).Theorem 14.
Define the partial sumsf1(z) and fn(z) by
(62)f1(z)=z,fn(z)=z+∑k=2nakzk(n∈ℕ∖{1}).
Let the function f(z)∈Um,n(β,A,B) be given by (1) and let it satisfy the condition (12) and
(63)ck≥{1,k=2,3,…,n,cn+1,k=n+1,n+2,…,
where, for convenience,
(64)ck=(1+β(1+|B|))(km-kn)+|Bkm-Akn|A-B,
Then
(65)Re{f(z)fn(z)}>1-1cn+1(z∈U;n∈ℕ),(66)Re{fn(z)f(z)}>cn+11+cn+1.Proof.
For the coefficientsck given by (64) it is not difficult to verify that
(67)ck+1>ck>1.
Therefore we have
(68)∑k=2n|ak|+cn+1∑k=n+1∞|ak|≤∑k=2∞ck|ak|≤1.
By setting
(69)g1(z)=cn+1{f(z)fn(z)-(1-1cn+1)}=1+cn+1∑k=n+1∞akzk-11+∑k=2nakzk-1
and applying (68), we find that
(70)|g1(z)-1g1(z)+1|<cn+1∑k=n+1∞|ak|2-2∑k=2n|ak|-cn+1∑k=n+1∞|ak|.
Now
(71)|g1(z)-1g1(z)+1|<1,
if
(72)∑k=2n|ak|+cn+1∑k=n+1∞|ak|≤1.
From the condition (12), it is sufficient to show that
(73)∑k=2n|ak|+cn+1∑k=n+1∞|ak|≤∑k=2∞ck|ak|,
which is equivalent to
(74)∑k=2n(ck-1)|ak|+∑k=n+1∞(ck-cn+1)|ak|≥0,
which readily yields the assertion (65) of Theorem 14. In order to see that
(75)f(z)=z+zn+1cn+1
gives sharp result, we observe that for z=reiπ/n that f(z)/fn(z)=1+zn/cn+1→1-1/cn+1 as z→1-. Similarly, if we take
(76)g2(z)=(1+cn+1){fn(z)f(z)-cn+11+cn+1}=1-(1+cn+1)∑k=n+1∞akzk-11+∑k=2∞akzk-1
and making use of (68), we can deduce that
(77)|g2(z)-1g2(z)+1|<(1+cn+1)∑k=n+1∞|ak|2-2∑k=2n|ak|-(1-cn+1)∑k=n+1∞|ak|,
which leads us immediately to the assertion (66) of Theorem 14.
The bound in (66) is sharp for each n∈ℕ with the extremal function f(z) given by (75). Then the proof of Theorem 14 is completed.


## 8. Distortion Theorems Involving Fractional Calculus

In this section, we will prove several distortion theorems for functions belonging to the classTUm,n(β,A,B). Each of these theorems would involve certain operators of fractional calculus (i.e., fractional integrals and fractional derivatives), which are defined as follows (see, for details, [15–18]). For our present investigation, we recall the following definitions.Definition 15.
The fractional integral of orderδ is defined, for a function f(z), by
(78)Dz-δf(z)=1Γ(δ)∫0zf(ζ)(z-ζ)1-δdζ(δ>0),
where the functionf(z) is analytic in a simply connected domain of the complex z-plane containing the origin, and the multiplicity of (z-ζ)δ-1 is removed by requiring log(z-ζ) to be real when z-ζ>0.Definition 16.
The fractional derivative of orderδ is defined, for a function f(z), by
(79)Dzδf(z)=1Γ(1-δ)ddz∫0zf(ζ)(z-ζ)δdζ(0≤δ<1),
where the function f(z) is constrained, and the multiplicity of(z-ζ)-δ is removed as in Definition 15.Definition 17.
Under the hypotheses of Definition16, the fractional derivative of order δ is defined, for a function f(z), by
(80)Dzn+δf(z)=dndzn{Dzδf(z)}(0≤δ<1;n∈ℕ0).
Using Definitions 15, 16, and 17, we obtain
(81)Dzδ{zk}=Γ(k+1)Γ(k+1-δ)zk-δ(k∈ℕ;0≤δ<1),Dz-δ{zk}=Γ(k+1)Γ(k+1+δ)zk+δ(k∈ℕ;δ>0),
in terms of Gamma functions.Theorem 18.
Let the functionf(z) defined by (10) be in the class TUm,n(β,A,B). Then
(82)|Dz-δf(z)|≥|z|1+δΓ(2+δ)×(1-(A-B)(2+δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]|z|),(δ>0;z∈U),|Dz-δf(z)|≤|z|1+δΓ(2+δ)×(1+(A-B)(2+δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]|z|),(δ>0;z∈U).
The results are sharp.Proof.
Let(83)F(z)=Γ(2+δ)z-δDz-δf(z)=z-∑k=2∞Γ(k+1)Γ(2+δ)Γ(k+1+δ)akzk=z-∑k=2∞Λ(k)akzk,
where
(84)Λ(k)=Γ(k+1)Γ(2+δ)Γ(k+1+δ),(k=2,3,…).
Since Λ(k) is a decreasing function of k, we can write
(85)0<Λ(k)≤Λ(2)=22+δ.
Furthermore, in view of Theorem 4, we have
(86)[(1+β(1+|B|))(2m-2n)+|B2m-A2n|]∑k=2∞ak≤∑k=2∞[(1+β(1+|B|))(km-kn)+|Bkm-Akn|]ak≤A-B.
Then
(87)∑k=2∞ak≤A-B[(1+β(1+|B|))(2m-2n)+|B2m-A2n|].
Therefore, by using (85) and (87), we can see that
(88)|F(z)|≥|z|-Λ(2)|z|2∑k=2∞ak≥|z|-(A-B)(2+δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]|z|2
and similarly
(89)|F(z)|≤|z|+Λ(2)|z|2∑k=2∞ak≤|z|+(A-B)(2+δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]|z|2,
which prove Theorem 18.
Finally, the equalities are attained for the functionf(z) defined by
(90)Dz-δf(z)=z1+δΓ(2+δ)×(1-(A-B)(2+δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]z)
or, equivalently, by f(z) given by (27).
Then the results are sharp, and the proof of Theorem18 is completed.Corollary 19.
Under the hypothesis of Theorem20, Dz-δf(z) is included in a disk with its center at the origin and radius R1 given by
(91)R1=1Γ(2+δ)×(1-(A-B)(2+δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]).Theorem 20.
Let the functionf(z) defined by (10) be in the class TUm,n(β,A,B). Then
(92)|Dzδf(z)|≥|z|1-δΓ(2-δ)×(1-(A-B)(2-δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]|z|),(0≤δ<1;z∈U),|Dzδf(z)|≤|z|1-δΓ(2-δ)×(1+(A-B)(2-δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]|z|),(0≤δ<1;z∈U).
Each of these results is sharp.Proof.
Let(93)G(z)=Γ(2-δ)zδDzδf(z)=z-∑k=2∞Γ(k)Γ(2-δ)Γ(k+1-δ)kakzk=z-∑k=2∞Ω(k)kakzk,
where
(94)Ω(k)=Γ(k)Γ(2-δ)Γ(k+1-δ)(k=2,3,…).
Since Ω(k) is a decreasing function of k, we can write
(95)0<Ω(k)≤Ω(2)=12-δ.
Furthermore, in view of Theorem 4, we have
(96)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]∑k=2∞kak≤∑k=2∞[(1+β(1+|B|))(km-kn)+|Bkm-Akn|]ak≤A-B.
Then
(97)∑k=2∞kak≤A-B[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|].
Therefore, by using (95) and (97), we can see that
(98)|G(z)|≥|z|-Ω(2)|z|2∑k=2∞kak≥|z|-(A-B)(2-δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]|z|2
and similarly
(99)|G(z)|≤|z|+Ω(2)|z|2∑k=2∞kak≤|z|+(A-B)(2-δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]|z|2,
which together prove the two assertions of Theorem 20.
Finally, the equalities are attained for the functionf(z) defined by
(100)Dzδf(z)=z1-δΓ(2-δ)×(1-(A-B)(2-δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]z)
or, equivalently, by f(z) given by (27).
Then the result is sharp, and the proof of Theorem20 is completed.Corollary 21.
Under the hypothesis of Theorem20, Dzδf(z) is included in a disk with its center at the origin and radius R2 given by
(101)R2=1Γ(2-δ)×(1-(A-B)(2-δ)[(1+β(1+|B|))(2m-1-2n-1)+|B2m-1-A2n-1|]).



---

*Source: 153128-2013-02-05.xml*