# A Class of Analytic Functions with Missing Coefficients

Authors: Ding-Gong Yang; Jin-Lin Liu
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

LetTn(A,B,γ,α)(−1≤B<1,B<A,0<γ≤1 and α>0) denote the class of functions of the form f(z)=z+∑k=n+1∞akzk(n∈N={1,2,3,…}), which are analytic in the open unit disk U and satisfy the following subordination condition f′(z)+αzf′′(z)≺((1+Az)/(1+Bz))γ, for(z∈U;A≤1;0<γ<1),(1+Az)/(1+Bz), for(z∈U;γ=1). We obtain sharp bounds on Ref'(z),Ref(z)/z,|f(z)|, and coefficient estimates for functions f(z) belonging to the class Tn(A,B,γ,α). Conditions for univalency and starlikeness, convolution properties, and the radius of convexity are also considered.


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## Body

## 1. Introduction

LetAn denote the class of functions of the form (1.1)f(z)=z+∑k=n+1∞akzk(n∈N={1,2,3,…}),
which are analytic in the open unit disk U={z:z∈Cand|z|<1}. Let Sn and Sn* denote the subclasses of An whose members are univalent and starlike, respectively.For functionsf(z) and g(z) analytic in U, we say that f(z) is subordinate to g(z) in U and we write f(z)≺g(z)(z∈U), if there exists an analytic function w(z) in U such that (1.2)|w(z)|≤|z|,f(z)=g(w(z))(z∈U).
Furthermore, if the function g(z) is univalent in U, then (1.3)f(z)≺g(z)(z∈U)⟺f(0)=g(0),f(U)⊂g(U).
Throughout our present discussion, we assume that (1.4)n∈N,-1≤B<1,B<A,α>0,β<1,0<γ≤1.
We introduce the following subclass of An.Definition 1.1.
A functionf(z)∈An is said to be in the class Tn(A,B,γ,α) if it satisfies
(1.5)f′(z)+αzf′′(z)≺h(z)(z∈U),
where
(1.6)h(z)={(1+Az1+Bz)γ,(A≤1;0<γ<1),1+Az1+Bz,(γ=1).
The classes
(1.7)T1(1-2β,-1,1,1)=R(β)(β=0orβ<1),T1(A,0,1,α)=R̃(α,A)(A>0)
have been studied by several authors (see [1–5]). Recently, Gao and Zhou [6] showed some mapping properties of the following subclass of A1:
(1.8)R(β,α)={f(z)∈A1:Re{f′(z)+αzf′′(z)}>β(z∈U)}.
Note that
(1.9)R(β,1)=R(β),T1(1-2β,-1,1,α)=R(β,α).
For further information of the above classes (with γ=1) and related analytic function classes, see Srivastava et al. [7], Yang and Liu [8], Kim [9], and Kim and Srivastava [10].In this paper, we obtain sharp bounds onRef′(z),Re(f(z)/z),|f(z)|, and coefficient estimates for functions f(z) belonging to the class Tn(A,B,γ,α). Conditions for univalency and starlikeness, convolution properties, and the radius of convexity are also presented. One can see that the methods used in [6] do not work for the more general class Tn(A,B,γ,α) than R(β,α).


## 2. The bounds onRef′(z), Re(f(z)/z), and |f(z)| in Tn(A,B,γ,α)

In this section, we let(2.1)λm(A,B,γ)={∑j=0m(γj)(-γm-j)AjBm-j,(A≤1;0<γ<1),(A-B)(-B)m-1,(γ=1),
where m∈N and (2.2)(γj)={γ(γ-1)⋯(γ-j+1)j!,(j=1,2,…,m),1,(j=0).
With (2.1), it is easily seen that the function h(z) given by (1.6) can be expressed as (2.3)h(z)=1+∑m=1∞λm(A,B,γ)zm(z∈U).Theorem 2.1.
Letf(z)∈Tn(A,B,γ,α). Then, for |z|=r<1,
(2.4)Ref′(z)≥1+∑m=1∞(-1)mλm(A,B,γ)αnm+1rnm,Ref′(z)≤1+∑m=1∞λm(A,B,γ)αnm+1rnm.
The bounds in (2.4) are sharp for the function fn(z) defined by
(2.5)fn(z)=z+∑m=1∞λm(A,B,γ)(nm+1)(αnm+1)znm+1(z∈U).Proof.
The analytic functionh(z) given by (1.6) is convex (univalent) in U (cf. [11]) and satisfies h(z¯)=h(z)¯(z∈U). Thus, for |ζ|≤σ(ζ∈Candσ<1),
(2.6)h(-σ)≤Reh(ζ)≤h(σ).
Letf(z)∈Tn(A,B,γ,α). Then, we can write
(2.7)f′(z)+αzf′′(z)=h(w(z))(z∈U),
where w(z)=wnzn+wn+1zn+1+⋯ is analytic and |w(z)|<1 for z∈U. By the Schwarz lemma, we know that |w(z)|≤|z|n(z∈U). It follows from (2.7) that
(2.8)(z1/αf′(z))′=1αz(1/α)-1h(w(z)),
which leads to
(2.9)f′(z)=1αz-1/α∫0zζ(1/α)-1h(w(ζ))dζ
or to
(2.10)f′(z)=1α∫01t(1/α)-1h(w(tz))dt(z∈U).
Since
(2.11)|w(tz)|≤(tr)n(|z|=r<1;0≤t≤1),
we deduce from (2.6) and (2.10) that
(2.12)1α∫01t(1/α)-1h(-(tr)n)dt≤Ref′(z)≤1α∫01t(1/α)-1h((tr)n)dt.
Now, by using (2.3) and (2.12), we can obtain (2.4).
Furthermore, for the functionfn(z) defined by (2.5), we find that
(2.13)fn′(z)=1+∑m=1∞λm(A,B,γ)αnm+1znm,(2.14)fn′(z)+αzfn′′(z)=1+∑m=1∞λm(A,B,γ)znm=h(zn)≺h(z)(z∈U).
Hence, fn(z)∈Tn(A,B,γ,α) and from (2.13), we see that the bounds in (2.4) are the best possible.
Hereafter, we write(2.15)Tn(A,B,1,α)=Tn(A,B,α).Corollary 2.2.
Letf(z)∈Tn(A,B,α). Then, for z∈U,
(2.16)Ref′(z)>1-(A-B)∑m=1∞Bm-1αnm+1,(2.17)Ref′(z)<1+(A-B)∑m=1∞(-B)m-1αnm+1(B≠-1).
The results are sharp.Proof.
Forγ=1, it follows from (2.12) (used in the proof of Theorem 2.1) that
(2.18)Ref′(z)>1α∫01t(1/α)-1(1-Atn1-Btn)dt,Ref′(z)<1α∫01t(1/α)-1(1+Atn1+Btn)dt(B≠-1),
for z∈U. From these, we have the desired results.The bounds in (2.16) and (2.17) are sharp for the function (2.19)fn(z)=z+(A-B)∑m=1∞(-B)m-1(nm+1)(αnm+1)znm+1∈Tn(A,B,α).Theorem 2.3.
Letf(z)∈Tn(A,B,γ,α). Then, for |z|=r<1,
(2.20)Ref(z)z≥1+∑m=1∞(-1)mλm(A,B,γ)(nm+1)(αnm+1)rnm,Ref(z)z≤1+∑m=1∞λm(A,B,γ)(nm+1)(αnm+1)rnm.
The results are sharp.Proof.
Noting that(2.21)f(z)=z∫01f′(uz)du,Ref(z)z=∫01Ref′(uz)du(z∈U),
an application of Theorem 2.1 yields (2.20). Furthermore, the results are sharp for the function fn(z) defined by (2.5).Corollary 2.4.
Letf(z)∈Tn(A,B,α). Then, for z∈U,
(2.22)Ref(z)z>1-(A-B)∑m=1∞Bm-1(nm+1)(αnm+1),Ref(z)z<1+(A-B)∑m=1∞(-B)m-1(nm+1)(αnm+1).
The results are sharp for the function fn(z) defined by (2.19).Proof.
Forf(z)∈Tn(A,B,α), it follows from (2.6) and (2.10) (with γ=1) that
(2.23)1α∫01t(1/α)-1(1-A(ut)n1-B(ut)n)dt<Ref′(uz)<1α∫01t(1/α)-1(1+A(ut)n1+B(ut)n)dt,
for z∈U and 0<u≤1. Making use of (2.21) and (2.23), we can obtain (2.22).Theorem 2.5.
Letf(z)∈T1(A,B,α) and g(z)∈T1(A0,B0,α0)(-1≤B0<1,B0<A0 and α0>0). If
(2.24)(A0-B0)∑m=1∞B0m-1(m+1)(α0m+1)≤12,
then (f*g)(z)∈T1(A,B,α), where the symbol * stands for the familiar Hadamard product (or convolution) of two analytic functions in U.Proof.
Sinceg(z)∈T1(A0,B0,α0)(-1≤B0<1,B0<A0 and α0>0), it follows from Corollary 2.4 (with n=1) and (2.24) that
(2.25)Reg(z)z>1-(A0-B0)∑m=1∞B0m-1(m+1)(α0m+1)≥12(z∈U).
Thus, g(z)/z has the Herglotz representation
(2.26)g(z)z=∫|x|=1dμ(x)1-xz(z∈U),
where μ(x) is a probability measure on the unit circle |x|=1 and ∫|x|=1dμ(x)=1.
Forf(z)∈T1(A,B,α), we have
(2.27)(f*g)′(z)+αz(f*g)′′(z)=F(z)*g(z)z(z∈U),
where
(2.28)F(z)=f′(z)+αzf′′(z)≺1+Az1+Bz(z∈U).
In view of the function (1+Az)/(1+Bz) is convex (univalent) in U, we deduce from (2.26) to (2.28) that
(2.29)(f*g)′(z)+αz(f*g)′′(z)=∫|x|=1F(xz)dμ(x)≺1+Az1+Bz(z∈U).
This shows that (f*g)(z)∈T1(A,B,α).Corollary 2.6.
Letf(z)∈T1(A,B,α), g(z)∈R(β,1) and
(2.30)β≥-π2-912-π2.
Then, (f*g)(z)∈T1(A,B,α).Proof.
By takingA0=1-2β, B0=-1 and α0=1, (2.24) in Theorem 2.5 becomes
(2.31)2(1-β)∑m=1∞(-1)m-1(m+1)2=2(1-β)(1-π212)≤12,
that is,
(2.32)β≥-π2-912-π2.
Hence, the desired result follows as a special case from Theorem 2.5.Remark 2.7.
R. Singh and S. Singh [4, Theorem 3] proved that, if f(z) and g(z) belong to R(0,1), then (f*g)(z)∈R(0,1). Obviously, for
(2.33)-π2-912-π2≤β<0,
Corollary 2.6 generalizes and improves Theorem 3 in [4].Theorem 2.8.
Letf(z)∈Tn(A,B,γ,α) and AB≤1. Then, for |z|=r<1,
(2.34)|f(z)|≤r+∑m=1∞λm(A,B,γ)(nm+1)(αnm+1)rnm+1.
The result is sharp, with the extremal function fn(z) defined by (2.5).Proof.
It is well known that forζ∈C and |ζ|≤σ<1,
(2.35)|1+Aζ1+Bζ-1-ABσ21-B2σ2|≤(A-B)σ1-B2σ2.
Since AB≤1, we have 1-ABσ2>0 and so (2.35) leads to
(2.36)|1+Aζ1+Bζ|γ≤(|1-ABσ21-B2σ2|+(A-B)σ1-B2σ2)γ=(1+Aσ1+Bσ)γ(|ζ|≤σ<1).
By virtue of (1.6), (2.10), and (2.36), we have
(2.37)|f′(uz)|≤1α∫01t(1/α)-1|h(w(utz))|dt≤1α∫01t(1/α)-1h((ut|z|)n)dt,
for z∈U and 0≤u≤1. Now, by using (2.3), (2.21) and (2.37), we can obtain (2.34).Theorem 2.9.
Let(2.38)f(z)=z+∑k=n+1∞akzk∈Tn(A,B,γ,α).
Then,
(2.39)|ak|≤γ(A-B)k(α(k-1)+1)(k≥n+1).
The result is sharp for each k≥n+1.Proof.
It is known (cf. [12]) that, if
(2.40)φ(z)=∑k=1∞bkzk≺ψ(z)(z∈U),
where φ(z) is analytic in U and ψ(z)=z+⋯ is analytic and convex univalent in U, then |bk|≤1(k∈N).
By (2.38), we have
(2.41)f′(z)+αzf′′(z)-1γ(A-B)=∑k=n+1∞k(α(k-1)+1)γ(A-B)akzk-1≺ψ(z)(z∈U),
where
(2.42)ψ(z)=h(z)-1γ(A-B)=z+⋯
and h(z) is given by (1.6). Since the function ψ(z) is analytic and convex univalent in U, it follows from (2.41) that
(2.43)k(α(k-1)+1)γ(A-B)|ak|≤1(k≥n+1),
which gives (2.39).
Next, we consider the function(2.44)fk-1(z)=z+∑m=1∞λm(A,B,γ)(m(k-1)+1)(αm(k-1)+1)zm(k-1)+1(z∈U;k≥n+1).
It is easy to verify that
(2.45)fk-1′(z)+αzfk-1′′(z)=h(zk-1)≺h(z)(z∈U),fk-1(z)=z+γ(A-B)k(α(k-1)+1)zk+⋯.
The proof of Theorem 2.9 is completed.


## 3. The Univalency and Starlikeness ofTn(A,B,α)

Theorem 3.1.
Tn(A,B,α)⊂Sn if and only if
(3.1)(A-B)∑m=1∞Bm-1αnm+1≤1.Proof.
Letf(z)∈Tn(A,B,α) and (3.1) be satisfied. Then, by (2.16) in Corollary 2.2, we see that Ref′(z)>0(z∈U). Thus, f(z) is close-to-convex and univalent in U.
On the other hand, if(3.2)(A-B)∑m=1∞Bm-1αnm+1>1,
then the function fn(z) defined by (2.19) satisfies fn′(0)=1>0 and
(3.3)fn′(reπi/n)=1-(A-B)∑m=1∞Bm-1αnm+1rnm⟶1-(A-B)∑m=1∞Bm-1αnm+1<0
as r→1. Hence, there exists a point zn=rneπi/n(0<rn<1) such that fn′(zn)=0. This implies that fn(z) is not univalent in U and so the theorem is proved.Theorem 3.2.
Let (3.1) in Theorem 3.1 be satisfied. If α≥1 and
(3.4)(α-1)(1-(A-B)∑m=1∞Bm-1αnm+1)+nα2(1-(A-B)∑m=1∞Bm-1(nm+1)(αnm+1))≥A-11-B,
then Tn(A,B,α)⊂Sn*.Proof.
We first show that(3.5)∑m=1∞Bm-1αnm+1≥∑m=1∞Bm-1(nm+1)(αnm+1)(α≥1).
Equation (3.5) is obvious when B≥0. For 0>B≥-1, we have
(3.6)∑m=1∞Bm-1αnm+1-∑m=1∞Bm-1(nm+1)(αnm+1)=μ1-μ2+μ3-μ4+⋯+(-1)m-1μm+⋯,
where
(3.7)μm=nm|B|m-1(nm+1)(αnm+1)>0.
Since |B|≤1 and
(3.8)ddx(x(x+1)(αx+1))=1-αx2(x+1)2(αx+1)2≤0(x≥1;α≥1),{μm} is a monotonically decreasing sequence. Therefore, the inequality (3.5) follows from (3.6).
Letf(z)∈Tn(A,B,α). Then,
(3.9)Re{f′(z)+αzf′′(z)}>1-A1-B(z∈U).
Define p(z) in U by
(3.10)p(z)=zf′(z)f(z).
In view of (3.1) in Theorem 3.1 is satisfied, the function f(z) is univalent in U, and so p(z)=1+pnzn+pn+1zn+1+⋯ is analytic in U. Also it follows from (3.10) that
(3.11)f′(z)+αzf′′(z)=(1-α)f′(z)+αf(z)z[zp′(z)+(p(z))2].We want to prove now thatRep(z)>0 for z∈U. Suppose that there exists a point z0∈U such that (3.12)Rep(z)>0(|z|<|z0|),Rep(z0)=0.
Then, applying a result of Miller and Mocanu [13, Theorem 4], we have(3.13)z0p′(z0)+(p(z0))2≤-n2Re(1-p(z0))-(Im⁡p(z0))2≤-n2.
For α≥1, we deduce from Corollaries 2.2 and 2.4, (3.1), (3.5), (3.11), (3.13), and (3.4) that (3.14)Re{f′(z0)+αz0f′′(z0)}≤(1-α)(1-(A-B)∑m=1∞Bm-1αnm+1)-nα2(1-(A-B)∑m=1∞Bm-1(nm+1)(αnm+1))≤1-A1-B.
But this contradicts (3.9) at z=z0. Therefore, we must have Rep(z)>0(z∈U) and the proof of Theorem 3.2 is completed.Remark 3.3.
In [6, Theorem 4(ii)], the authors gave the following: if 0<α<1 and β1 is the solution of the equation
(3.15)1-3α2=β+(1-β)∑m=2∞(-1)m-1α+2(α-1)mm(α(m-1)+1),
then R(β,α)⊂S1* for β≥β1. However, this result is not true because the series in (3.15) diverges.


## 4. The Radius of Convexity

Theorem 4.1.
Letf(z) belong to the class Tn(γ) defined by
(4.1)Tn(γ)=Tn(1,-1,γ,0)={f(z)∈An:f′(z)≺(1+z1-z)γ,(z∈U)},0<δ≤1 and 0≤ρ<1. Then,
(4.2)Re{(1-δ)(f′(z))1/γ+δ(1+zf′′(z)f′(z))}>ρ(|z|<rn(γ,δ,ρ)),
where rn(γ,δ,ρ) is the root in (0,1) of the equation
(4.3)(1-2δ+ρ)r2n-2(1-δ+nδγ)rn+1-ρ=0.
The result is sharp.Proof.
Forf(z)∈Tn(γ), we can write
(4.4)(f′(z))1/γ=1+znφ(z)1-znφ(z),
where φ(z) is analytic and |φ(z)|≤1 in U. Differentiating both sides of (4.4) logarithmically, we arrive at
(4.5)1+zf′′(z)f′(z)=1+2nγznφ(z)1-(znφ(z))2+2γzn+1φ′(z)1-(znφ(z))2(z∈U).
Put |z|=r<1 and (f′(z))1/γ=u+iv(u,v∈R). Then, (4.4) implies that
(4.6)znφ(z)=u-1+ivu+1+iv,(4.7)1-rn1+rn≤u≤1+rn1-rn.
With the help of the Carathéodory inequality
(4.8)|φ′(z)|≤1-|φ(z)|21-r2,
it follows from (4.5) and (4.6) that
(4.9)Re{(1-δ)(f′(z))1/γ+δ(1+zf′′(z)f′(z))}≥(1-δ)u+δ+2nδγRe{znφ(z)1-(znφ(z))2}-2δγ|zn+1φ′(z)1-(znφ(z))2|≥(1-δ)u+δ+nδγ2(u-uu2+v2)+δγ2(u-1)2+v2-r2n((u+1)2+v2)rn-1(1-r2)(u2+v2)1/2=Fn(u,v)(say),(4.10)∂∂vFn(u,v)=δγvGn(u,v),
where 0<r<1,0<δ≤1 and
(4.11)Gn(u,v)=nu(u2+v2)2+1-r2nrn-1(1-r2)(u2+v2)1/2+r2n((u+1)2+v2)-((u-1)2+v2)2rn-1(1-r2)(u2+v2)3/2>0
because of (4.6) and (4.7). In view of (4.10) and (4.11), we see that
(4.12)Fn(u,v)≥Fn(u,0)=(1-δ)u+δ+nδγ2(u-1u)+δγ2rn-1(1-r2)×{(1-r2n)(u+1u)-2(1+r2n)}.Let us now calculate the minimum value ofFn(u,0) on the closed interval [(1-rn)/(1+rn),(1+rn)/(1-rn)]. Noting that (4.13)1-r2nrn-1(1-r2)≥n(see[8])
and (4.7), we deduce from (4.12) that (4.14)dduFn(u,0)=1-δ+δγ2[(1-r2nrn-1(1-r2)+n)-1u2(1-r2nrn-1(1-r2)-n)]≥1-δ+δγ2[(1-r2nrn-1(1-r2)+n)-(1+rn1-rn)2(1-r2nrn-1(1-r2)-n)]=1-δ+2δγIn(r)(1-rn)2,
where (4.15)In(r)=n2(1+r2n)-r(1+r2+⋯+r2n-2).
Also (4.16)In′(r)=n2r2n-1-(1+3r2+⋯+(2n-1)r2n-2)
and I1′(r)=r-1<0. Suppose that In′(r)<0. Then,(4.17)In+1′(r)=(n+1)2r2n+1-(2n+1)r2n-(1+3r2+⋯+(2n-1)r2n-2)<n2r2n-(1+3r2+⋯+(2n-1)r2n-2)<In′(r)<0.
Hence, by virtue of the mathematical induction, we have In′(r)<0 for all n∈N and 0≤r<1. This implies that (4.18)In(r)>In(1)=0(n∈N;0≤r<1).
In view of (4.14) and (4.18), we see that (4.19)dduFn(u,0)>0(1-rn1+rn≤u≤1+rn1-rn).
Further it follows from (4.9), (4.12), and (4.19) that(4.20)Re{(1-δ)(f′(z))1/γ+δ(1+zf′′(z)f′(z))}-ρ≥Fn(1-rn1+rn,0)-ρ=(1-δ)1-rn1+rn+δ1-2nγrn-r2n1-r2n-ρ=Jn(r)1-r2n,
where 0≤ρ<1 and (4.21)Jn(r)=(1-2δ+ρ)r2n-2(1-δ+nδγ)rn+1-ρ.
Note that Jn(0)=1-ρ>0 and Jn(1)=-2nδγ<0. If we let rn(γ,δ,ρ) denote the root in (0,1) of the equation Jn(r)=0, then (4.20) yields the desired result (4.2).To see that the boundrn(γ,δ,ρ) is the best possible, we consider the function (4.22)f(z)=∫0z(1-tn1+tn)γdt∈Tn(γ).
It is clear that for z=r∈(rn(γ,δ,ρ),1), (4.23)(1-δ)(f′(r))1/γ+δ(1+rf′′(r)f′(r))-ρ=Jn(r)1-r2n<0,
which shows that the bound rn(γ,δ,ρ) cannot be increased.Settingδ=1, Theorem 4.1 reduces to the following result.Corollary 4.2.
Letf(z)∈Tn(γ) and 0≤ρ<1. Then, f(z) is convex of order ρ in
(4.24)|z|<[((nγ)2+(1-ρ)2)1/2-nγ1-ρ]1/n.
The result is sharp.



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