# Certain Subclasses of Multivalent Analytic Functions

Authors: Yi-Ling Cang; Jin-Lin Liu
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Two new subclassesHp,k(λ,A,B) and Qp,k(λ,A,B) of multivalent analytic functions are introduced. Distortion inequalities and inclusion relation for Hp,k(λ,A,B) and Qp,k(λ,A,B) are obtained. Some results of the partial sums of functions in these classes are also given.


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## Body

## 1. Introduction

Throughout this paper, we assume that(1)N={1,2,3,…},k∈N∖{1},-1≤B<0,B<A≤1,0≤λ≤1.
Let A(p) denote the class of functions of the form
(2)f(z)=zp+∑n=2p∞‍anzn(p∈N),
which are analytic in the unit disk U={z:|z|<1}.For functionsf and g analytic in U, we say that f is subordinate to g in U and write f(z)≺g(z)(z∈U), if there exists a Schwarz function w(z) in U such that
(3)|w(z)|≤|z|,f(z)=g(w(z))(z∈U).Let(4)fj(z)=zp+∑n=2p∞‍an,jzn∈A(p)(j=1,2).
Then the Hadamard product (or convolution) of f1(z) and f2(z) is defined by
(5)(f1*f2)(z)=zp+∑n=2p∞‍an,1an,2zn=(f2*f1)(z).
The following lemma will be required in our investigation.Lemma 1.
Letf∈A(p) defined by (2) satisfy
(6)∑n=2p∞‍[n(1-B)-pλ(1-A)δn,p,k]|an|≤p(A-B).
Then
(7)zf′(z)(1-λ)zp+λfp,k(z)≺p1+Az1+Bz(z∈U),
where
(8)fp,k(z)=1k∑j=0k-1‍εk-jpf(εkjz),εk=exp(2πik),(9)δn,p,k={0(n-pk∉N),1(n-pk∈N).Proof.
Forf∈A(p) defined by (2), the function fp,k(z) in (8) can be expressed as
(10)fp,k(z)=zp+∑n=2p∞‍δn,p,kanzn
with
(11)δn,p,k=1k∑j=0k-1‍εkj(n-p)={0(n-pk∉N),1(n-pk∈N).
In view of (1) and (9), we see that
(12)pAλδn,p,k-nB≥-B(n-pλδn,p,k)≥0(n≥2p).
Let inequality (6) hold. Then from (10) and (12) we deduce that
(13)|zf′(z)/((1-λ)zp+λfp,k(z))-ppA-Bzf′(z)/((1-λ)zp+λfp,k(z))|=|∑n=2p∞‍(n-pλδn,p,k)anzn-pp(A-B)+∑n=2p∞‍(pAλδn,p,k-nB)anzn-p|≤∑n=2p∞‍(n-pλδn,p,k)|an|p(A-B)-∑n=2p∞‍(pAλδn,p,k-nB)|an|≤1(|z|=1).
Hence, by the maximum modulus theorem, we arrive at (7).We now consider the following two subclasses ofA(p).Definition 2.
A functionf∈A(p) defined by (2) is said to be in the class Hp,k(λ,A,B) if and only if it satisfies the coefficient inequality (6).Definition 3.
A functionf∈A(p) defined by (2) is said to be in the class Qp,k(λ,A,B) if and only if it satisfies
(14)∑n=2p∞‍n[n(1-B)-pλ(1-A)δn,p,k]|an|≤p2(A-B).
It is obvious from Definitions2 and 3 that
(15)f(z)∈Qp,k(λ,A,B)iffzf′(z)p∈Hp,k(λ,A,B).
If we write(16)αn=n(1-B)-pλ(1-A)δn,p,kp(A-B),βn=npαn(n≥2p),
then it is easy to verify that
(17)∂βn∂λ=np∂αn∂λ≤0,∂βn∂A=np∂αn∂A<0,∂βn∂B=np∂αn∂B≥0.
Thus we obtain the following inclusion relations:
(18)Qp,k(λ,A,B)⊂Hp,k(λ,A,B)⊆Hp,k(1,1,-1),Qp,k(λ,A,B)⊆Qp,k(1,1,-1).
Therefore, by Lemma 1, we see that each function in the classes Hp,k(λ,A,B) and Qp,k(λ,A,B) is starlike with respect to k-symmetric points. Analytic functions which are starlike with respect to symmetric points and related functions have been extensively investigated in [1–6]. Recently, several authors have  obtained many important properties and characteristics of multivalent analytic functions (see, e.g., [7–11]).
The main object of this paper is to present some distortion inequalities of functions in the classesHp,k(λ,A,B) and Qp,k(λ,A,B) which we have introduced here. In particular some results of inclusion relation and convolution of functions in these classes are also given. Further we derive several interesting results of the partial sums of functions in these classes.


## 2. Main Results

Theorem 4.
Letp/k∉N and suppose that either(a)
1-B≥p(1-A) and 0≤λ≤1, or(b)
1-B<p(1-A) and 0≤λ≤(1-B)/p(1-A).
(i)
Iff∈Hp,k(λ,A,B), then, for z∈U,
(19)
|
z
|
p
-
A
-
B
2
(
1
-
B
)
|
z
|
2
p
≤
|
f
(
z
)
|
≤
|
z
|
p
+
A
-
B
2
(
1
-
B
)
|
z
|
2
p
.
The bounds in (19) are best possible for the function f defined by
(20)f(z)=zp+A-B2(1-B)z2p.(ii)
Iff∈Qp,k(λ,A,B), then, for z∈U,(21)p|z|p-1-p(A-B)2(1-B)|z|2p-1≤|f′(z)|≤p|z|p-1+p(A-B)2(1-B)|z|2p-1.
The bounds in (21) are best possible for the function f defined by
(22)f(z)=zp+A-B4(1-B)z2p.Proof.
Letp/k∉N. For n≥2p(n∈N) and  (n-p)/k∉N, we have δn,p,k=δ2p,p,k=0, and so
(23)n(1-B)-pλ(1-A)δn,p,kp(A-B)≥2(1-B)A-B.
For n≥2p(n∈N) and (n-p)/k∈N, we have δn,p,k=1 and
(24)n(1-B)-pλ(1-A)δn,p,kp(A-B)≥(2p+1)(1-B)-pλ(1-A)p(A-B).
If either (a) or (b) is satisfied, then
(25)(2p+1)(1-B)-pλ(1-A)p(A-B)≥2(1-B)A-B>1.(i)
If(26)f(z)=zp+∑n=2p∞‍anzn∈Hp,k(λ,A,B),
then it follows from (23) to (25) that
(27)2(1-B)A-B∑n=2p∞‍|an|≤1.
Hence we have
(28)|f(z)|≤|z|p+|z|2p∑n=2p∞‍|an|≤|z|p+A-B2(1-B)|z|2p,(29)|f(z)|≥|z|p-|z|2p∑n=2p∞|an|‍≥|z|p-A-B2(1-B)|z|2p>0
for z∈U.(ii)
If(30)f(z)=zp+∑n=2p∞‍anzn∈Qp,k(λ,A,B),
then (23) to (25) yield
(31)2(1-B)p(A-B)∑n=2p∞‍n|an|≤1.
This leads to (21). The proof of the theorem is complete.Theorem 5.
Let(32)pk∉N,1-B<p(1-A),1-Bp(1-A)<λ≤1.(i)
Iff(z)=zp+a2pz2p+⋯∈Hp,k(λ,A,B), then, for z∈U,(33)|f(z)|≤|z|p+|a2p||z|2p+p(A-B)-2p(1-B)|a2p|(2p+1)(1-B)-pλ(1-A)|z|2p+1,(34)|f(z)|≥|z|p-|a2p||z|2p-p(A-B)-2p(1-B)|a2p|(2p+1)(1-B)-pλ(1-A)|z|2p+1.
Equalities in (33) are attained, for example, by
(35)f(z)=zp+p(A-B)(2p+1)(1-B)-pλ(1-A)z2p+1.(ii)
Iff(z)=zp+a2pz2p+⋯∈Qp,k(λ,A,B), then, for z∈U,(36)|f′(z)|≤p|z|p-1+2p|a2p||z|2p-1+p2[(A-B)-4(1-B)|a2p|](2p+1)(1-B)-pλ(1-A)|z|2p,|f′(z)|≥p|z|p-1-2p|a2p||z|2p-1-p2[(A-B)-4(1-B)|a2p|](2p+1)(1-B)-pλ(1-A)|z|2p.
Equalities in (36) are attained, for example, by
(37)f(z)=zp+p2(A-B)(2p+1)[(2p+1)(1-B)-pλ(1-A)]×z2p+1.Proof.
Note that(1-B)/p(1-A)<λ≤1 implies that
(38)2(1-B)A-B>(2p+1)(1-B)-pλ(1-A)p(A-B).(i)
Forf(z)=zp+a2pz2p+⋯∈Hp,k(λ,A,B), it follows from (23), (24), and (38) that(39)2(1-B)A-B|a2p|+(2p+1)(1-B)-pλ(1-A)p(A-B)∑n=2p+1∞‍|an|≤1.
From this we can get (33).(ii)
Forf(z)=zp+a2pz2p+⋯∈Qp,k(λ,A,B), from (23), (24), and (38) we deduce that(40)4(1-B)A-B|a2p|+(2p+1)(1-B)-pλ(1-A)p2(A-B)∑n=2p+1∞‍n|an|≤1.
Hence we have (36). The proof of the theorem is complete.Theorem 6.
Letp/k∈N.(i)
Iff∈Hp,k(λ,A,B), then, for z∈U,
(41)|z|p-A-B2(1-B)-λ(1-A)|z|2p≤|f(z)|≤|z|p+A-B2(1-B)-λ(1-A)|z|2p.The bounds in (41) are sharp for the function f defined by
(42)f(z)=zp+A-B2(1-B)-λ(1-A)z2p.(ii)
Iff∈Qp,k(λ,A,B), then, for z∈U,(43)p|z|p-1-p(A-B)2(1-B)-λ(1-A)|z|2p-1≤|f′(z)|≤p|z|p-1+p(A-B)2(1-B)-λ(1-A)|z|2p-1.
The bounds in (43) are sharp for the function f defined by
(44)f(z)=zp+A-B4(1-B)-2λ(1-A)z2p.Proof.
Letp/k∈N. For n≥2p(n∈N) and (n-p)/k∈N, we have n=2p+k(l-1)(l∈N), δn,p,k=δ2p,p,k=1,  and
(45)n(1-B)-pλ(1-A)δn,p,kp(A-B)≥2(1-B)-λ(1-A)A-B.
For n≥2p(n∈N) and (n-p)/k∉N, we have δn,p,k=δ2p+1,p,k=0, and so
(46)n(1-B)-pλ(1-A)δn,p,kp(A-B)≥(2p+1)(1-B)p(A-B)≥2(1-B)-λ(1-A)A-B.(i)
If(47)f(z)=zp+∑n=2p∞‍anzn∈Hp,k(λ,A,B),
then it follows from (45) and (46) that
(48)2(1-B)-λ(1-A)A-B∑n=2p∞‍|an|≤1.
Hence we have
(49)|f(z)|≤|z|p+|z|2p∑n=2p∞‍|an|≤|z|p+A-B2(1-B)-λ(1-A)|z|2p,|f(z)|≥|z|p-|z|2p∑n=2p∞‍|an|≥|z|p-A-B2(1-B)-λ(1-A)|z|2p
for z∈U.(ii)
If(50)f(z)=zp+∑n=2p∞‍anzn∈Qp,k(λ,A,B),then (45) and (46) yield
(51)2(1-B)-λ(1-A)p(A-B)∑n=2p∞‍n|an|≤1.
This leads to (43). Thus we complete the proof.Next, we generalize the inclusion relationQp,k(λ,A,B)⊂Hp,k(λ,A,B) which is mentioned in (18).Theorem 7.
If-1≤D<0, then
(52)Qp,k(λ,A,B)⊂Hp,k(λ,C(D),D),
where
(53)C(D)=D+(1-D)(A-B)2(1-B).Proof.
SinceB<A≤1 and -1≤D<0, we see that
(54)D<C(D)<1.
Letf∈Qp,k(λ,A,B). In order to prove that  f∈Hp,k(λ,C(D),D), we need only to find the smallest C(D<C≤1) such that
(55)n(1-D)-pλ(1-C)δn,p,kp(C-D)≤n[n(1-B)-pλ(1-A)δn,p,k]p2(A-B)
for all n≥2p, that is, that
(56)(1-D)(n-pλδn,p,k)p(C-D)+λδn,p,k≤np{(1-B)(n-pλδn,p,k)p(A-B)+λδn,p,k}.
Forn≥2p and (n-p)/k∈N, (56) is equivalent to
(57)C≥D+1-D(n-p)λ/(n-pλ)+n(1-B)/p(A-B)=φ(λ,n).
Noting (1), a simple calculation shows that (∂φ(λ,x))/∂x<0 for all real x≥2p and 0≤λ≤1, and so the function φ(λ,n) is decreasing in n(n≥2p). Therefore
(58)φ(λ,n)≤{φ(λ,2p)(pk∈N),φ(λ,k([pk]+1)+p)(pk∉N).
For n≥2p and (n-p)/k∉N, (56) becomes
(59)C≥D+p(1-D)(A-B)n(1-B)=φ(0,n),φ(0,n)≤{φ(0,2p+1)(pk∈N),φ(0,2p)(pk∉N).
Consequently, by taking(60)C(D)=φ(0,2p)=D+(1-D)(A-B)2(1-B),
it follows from (55) to (60) that f∈Hp,k(λ,C(D),D). The proof is complete.Remark 8.
If we takeD=B in Theorem 7, then from (1) we have C(D)=(A+B)/2<A. This shows that
(61)Qp,k(λ,A,B)⊂Hp,k(λ,A+B2,B)⊂Hp,k(λ,A,B).Theorem 9.
Letf∈Hp,k(λ,A,B). Then
(62)(f*hσ)(z)≠0(z∈U;σ∈C,|σ|=1),
where
(63)hσ(z)=zp-2(1+Bσ)σ(A-B)·z2p1-z-1+Bσpσ(A-B)·z2p+1(1-z)2+λ(1+Aσ)σ(A-B)gp,k(z),gp,k(z)={z2p1-zk(pk∈N),zk([p/k]+1)+p1-zk(pk∉N).Proof.
Forf∈Hp,k(λ,A,B), from Lemma 1 we have (7), which is equivalent to
(64)zf′(z)(1-λ)zp+λfp,k(z)≠p1+Aσ1+Bσ(z∈U;σ∈C,|σ|=1,1+Bσ≠0)
or to
(65)p(1+Aσ)[(1-λ)zp+λfp,k(z)]-(1+Bσ)zf′(z)≠0(z∈U;σ∈C,|σ|=1).
Obviously
(66)zp=f(z)*zp,zf′(z)p=f(z)*(zp+1p∑n=2p∞‍nzn)=f(z)*(zp+2z2p1-z+z2p+1p(1-z)2).
If we put
(67)fp,k(z)=f(z)*(zp+gp,k(z)),
then, for p/k∈N,
(68)gp,k(z)=∑n=2p∞‍δn,p,kzn=∑l=0∞‍z2p+lk=z2p1-zk,
and, for p/k∉N,
(69)gp,k(z)=∑l=1∞‍zk([p/k]+l)+p=zk([p/k]+1)+p1-zk.
Now, making use of (65) to (69), we arrive at
(70)f(z)*{z2p+1(1-z)2p(1+Aσ)[(1-λ)zp+λ(zp+gp,k(z))]1+Bσ--(1+Bσ)(pzp+2pz2p1-z+z2p+1(1-z)2)}≠0
for z∈U,σ∈C, and |σ|=1. This gives the desired result (62). The proof of the theorem is complete.Corollary 10.
Letf∈Qp,k(λ,A,B). Then
(71)f(z)*zhσ′(z)≠0(z∈U;σ∈C,|σ|=1),
where hσ(z) is the same as in Theorem 9.Proof.
Sincef∈Qp,k(λ,A,B) if and only if
(72)zf′(z)p∈Hp,k(λ,A,B),
it follows from Theorem 9 that
(73)f(z)*zhσ′(z)p=zf′(z)p*hσ(z)≠0(z∈U;σ∈C,|σ|=1).
Thus we complete the proof.Finally, we derive certain results of the partial sums of functions in the classesHp,k(λ,A,B) and Qp,k(λ,A,B).Letf∈A(p) be given by (2) and define the partial sums s1(z) and sm(z) by
(74)s1(z)=zp,sm(z)=zp+∑n=2p2p+m-2‍anzn(m∈N∖{1}).
For simplicity we use the notation αn(n≥2p) defined by (16).Theorem 11.
Letf∈Hp,k(λ,A,B) and let either(a)
1-B≥p(1-A) and 0≤λ≤1, or(b)
1-B<p(1-A) and 0≤λ≤(1-B)/p(1-A).Then, for m∈N, we have
(75)Re(f(z)sm(z))>1-1α2p+m-1(z∈U),(76)Re(sm(z)f(z))>α2p+m-11+α2p+m-1(z∈U).
The bounds in (75) and (76) are best possible for each m.Proof.
If either (a) or (b) is satisfied, then, forn≥2p,
(77)αn=n(1-B)-pλ(1-A)δn,p,kp(A-B)≥1-BA-B≥1,αn+1=(n+1)(1-B)-pλ(1-A)δn+1,p,kp(A-B)=αn+1-B-pλ(1-A)(δn+1,p,k-δn,p,k)p(A-B)≥αn+1-B-pλ(1-A)p(A-B)≥αn.
Letf∈Hp,k(λ,A,B). Then it follows from (77) that
(78)∑n=2p2p+m-2‍|an|+α2p+m-1·∑n=2p+m-1∞‍|an|≤∑n=2p∞‍αn|an|≤1(m∈N∖{1}).
If we put(79)p1(z)=1+α2p+m-1(f(z)sm(z)-1)
for z∈U and m∈N∖{1}, then p1(0)=1 and we deduce from (78) that
(80)|p1(z)-1p1(z)+1|=|α2p+m-1∑n=2p+m-1∞‍anzn-p2(1+∑n=2p2p+m-2‍anzn-p)+α2p+m-1∑n=2p+m-1∞‍anzn-p|≤α2p+m-1∑n=2p+m-1∞‍|an|2-2∑n=2p2p+m-2‍|an|-α2p+m-1∑n=2p+m-1∞‍|an|≤1(z∈U;m∈N∖{1}).
This implies that Re{p1(z)}>0 for z∈U, and so (75) holds true for m∈N∖{1}.
Similarly, by setting(81)p2(z)=(1+α2p+m-1)sm(z)f(z)-α2p+m-1,
it follows from (78) that
(82)|p2(z)-1p2(z)+1|=|×××++++(1-α2p+m-1)∑n=2p+m-1∞‍anzn-p)-1(-(1+α2p+m-1)∑n=2p+m-1∞‍anzn-p)××××(2(1+∑n=2p2p+m-2‍anzn-p)×××++++(1-α2p+m-1)∑n=2p+m-1∞‍anzn-p)-1|≤(1+α2p+m-1)∑n=2p+m-1∞‍|an|2-2∑n=2p2p+m-2‍|an|-(α2p+m-1-1)∑n=2p+m-1∞‍|an|≤1(z∈U;m∈N∖{1}).
Hence we have (76) for m∈N∖{1}.
Form=1, replacing (78) by
(83)α2p∑n=2p∞‍|an|≤∑n=2p∞‍αn|an|≤1
and proceeding as the above, we see that (75) and (76) are also true.
Furthermore, taking the functionf defined by
(84)f(z)=zp+z2p+m-1α2p+m-1∈Hp,k(λ,A,B),
we have sm(z)=zp,
(85)Re(f(z)sm(z))⟶1-1α2p+m-1asz⟶exp(πip+m-1),Re(sm(z)f(z))⟶α2p+m-11+α2p+m-1asz⟶1.
Thus the proof of Theorem 11 is completed.Remark 12.
ReplacingHp,k(λ,A,B) by Qp,k(λ,A,B), it follows from Theorem 11 that inequalities (75) and (76) are true. In Theorem 13 we improve the bounds in (75) and (76) for f∈Qp,k(λ,A,B).Theorem 13.
Letf∈Qp,k(λ,A,B) and let either (a) or (b) in Theorem 11 be satisfied. Then, for m∈N, one has
(86)Re(f(z)sm(z))>1-p(2p+m-1)α2p+m-1(z∈U),Re(sm(z)f(z))>(2p+m-1)α2p+m-1p+(2p+m-1)α2p+m-1(z∈U).
The bounds in (86) are sharp for the function f defined by
(87)f(z)=zp+pz2p+m-1(2p+m-1)α2p+m-1∈Qp,k(λ,A,B).Proof.
In view of the assumptions of Theorem13, it follows from (77) that
(88)∑n=2p2p+m-2‍|an|+(2p+m-1)α2p+m-1p∑n=2p+m-1∞‍|an|≤∑n=2p∞‍npαn|an|≤1(m∈N∖{1}),α2p∑n=2p∞‍|an|≤∑n=2p∞‍npαn|an|≤1.
If we put
(89)p1(z)=1+(2p+m-1)α2p+m-1p(f(z)sm(z)-1),p2(z)=(1+(2p+m-1)α2p+m-1p)sm(z)f(z)-(2p+m-1)α2p+m-1p,
then (88) leads to Re(pj(z))>0(z∈U;m∈N;j=1,2). Hence we have (86). Sharpness can be verified easily.



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