# General Univalence Criterion Associated with thenth Derivative

Authors: Oqlah Al-Refai; Maslina Darus
Journal: Abstract and Applied Analysis
 (2012)

---

## Abstract

For normalized analytic functionsf(z)
with f(z)≠0 for 0<|z|<1, we introduce a univalence
criterion defined by sharp inequality associated with the nth derivative of z/f(z), where n∈{3,4,5,…}.


---

## Body

## 1. Introduction

Let𝒜 denote the class of functions of the following form:(1.1)f(z)=z+∑n=2∞‍anzn,
which are normalized analytic in the open unit disk 𝕌:={z:|z|<1}.In [1], Aksentev proved that the condition
(1.2)|z2f′(z)f2(z)-1|≤1
or equivalently Re(f2(z)/z2f′(z))≥1/2, for z∈𝕌, is sufficient for f(z)∈𝒜 to be univalent in 𝕌. By virtue of the aforementioned result of Aksentev, the class of functions defined by (1.2) was extensively studied by Obradović and Ponnusamy [2, 3], Ozaki and Nunokawa [4], Obradović et al. [5], and others. Afterwards, Nunokawa et al. [6] proved for f(z)∈𝒜 with f(z)≠0 when 0<|z|<1 that
(1.3)|(zf(z))′′|≤1
implies |z2f′(z)/f2(z)-1|≤1 for z∈𝕌, and hence f(z) is univalent in 𝕌. Later, Yang and Liu [7] extended this result for f(z)∈𝒜:
(1.4)|(zf(z))′′|≤2
with f(z)≠0 when 0<|z|<1 implies that f(z) is univalent in 𝕌 and the bound 2 is best possible for univalence. This result was also given first in the preprint of reports of the Department of Mathematics, University of Helsinki: M. Obradović, S. Ponnusamy, New criteria, and distortion theorems for univalent functions, Preprint 190, June 1998. Later, under the same name, the paper was published in Complex Variables Theory Application (see [3]). Corresponding to the functions defined by (1.4), Yang and Liu in [7] studied a class of analytic univalent functions f(z) satisfying |(z/f(z))′′|≤β(0<β≤2) and denoted by S(β). The class S(β) is extensively studied in the recent years (see [2, 3, 8–10]).In this work, we introduce a univalence criteria defined by the conditionsf(z)≠0 for 0<|z|<1 and
(1.5)∑k=2n-1‍k-1k!|βk|+n-1n!|dndzn(zf(z))|≤1for|z|<1,
where f(z) is normalized analytic in 𝕌 and βk=(dk/dzk)(z/f(z))|z=0, n∈{3,4,…}. The sharpness occurs for the Koebe function. Indeed, all functions satisfying the condition (1.5) are univalent in 𝕌 and the bound 1 in the inequality is best possible for univalence. Letting n=2 in (1.5) gives the univalence criterion defined by (1.4). Some special cases and examples for functions satisfying (1.5) are given.


## 2. Sufficient Conditions for Univalence

Let us prove the following theorem.Theorem 2.1.
Letf(z)∈𝒜 with f(z)≠0 for 0<|z|<1 and let g(z)∈𝒜 be bounded in 𝕌 and satisfy
(2.1)m=inf⁡{|g(z1)-g(z2)z1-z2|:z1,z2∈U}>0.
For any n∈{3,4,…}, if
(2.2)|dndzn(zf(z)-zg(z))|≤K(z∈U),
where
(2.3)K=n!n-1(mM2-∑k=2n-1‍k-1k!|αk|),αk=dkdzk(zg(z)-zf(z))|z=0,
and M=sup{|g(z)|:z∈𝕌}, then f(z) is univalent in 𝕌.Proof.
If we put(2.4)h(z)=dndzn(zf(z)-zg(z)),
then the function h is analytic in 𝕌 and, by integration from 0 to z, we get
(2.5)dn-1dzn-1(zf(z)-zg(z))=αn-1+∫0z‍h(u1)du1.
Integrating both sides of the previous equation (n-1)-times from 0 to z gives
(2.6)zf(z)-zg(z)=∑k=1n-1‍αkk!zk+∫0z‍dun∫0un‍dun-1⋯∫0u3‍du2∫0u2‍h(u1)du1.
Thus, we have
(2.7)f(z)=g(z)1+g(z)∑k=1n-1(αk/k!)zk-1+g(z)(ψ(z)/z),
where
(2.8)ψ(z)=∫0z‍dun∫0un‍dun-1⋯∫0u3‍du2∫0u2‍h(u1)du1.
Next, for n=3, we have
(2.9)z2(ψ(z)z)′=∫0z‍uψ′′(u)du=∫0z‍udu∫0u‍h(u1)du1,
and for n=4,
(2.10)z2(ψ(z)z)′=∫0z‍uψ′′(u)du=∫0z‍udu∫0u‍du2∫0u2‍h(u1)du1.
In general, for n∈{3,4,…},
(2.11)z2(ψ(z)z)′=∫0z‍uψ′′(u)du=∫0z‍udu∫0u‍dun-2∫0un-2‍dun-3⋯∫0u2‍h(u1)du1=∫01‍z2tdt∫0zt‍dun-2∫0un-2‍dun-3⋯∫0u2‍h(u1)du1(bysettingu=zt)=∫01‍z3t2dt∫01‍ds1∫0un-2‍dun-3⋯∫0u2‍h(u1)du1(bysettingun-2=zts1)=∫01‍z4t3dt∫01‍s1ds1∫01‍ds2⋯∫0u2‍h(u1)du1(bysettingun-3=zts1s2)=∫01‍zntn-1dt∫01‍s1n-3ds1∫01‍s2n-4ds2⋯∫01‍sn-3dsn-3∫01‍h(zts1⋅sn-2)dsn-2(bysettingu1=zts1s2⋯sn-2),
therefore
(2.12)|(ψ(z)z)′|≤|z|n-2n⋅1n-2⋅1n-3⋯12∫01‍|h(zts1s2⋯sn-2)|dsn-2≤n-1n!K,
and so
(2.13)|ψ(z2)z2-ψ(z1)z1|=|∫z1z2‍(ψ(z)z)′dz|≤n-1n!K|z2-z1|
for z1,z2∈𝕌 and z1≠z2. If z1≠z2, then g(z1)≠g(z2), and it follows, from (2.7) and (2.13), that
(2.14)|f(z1)-f(z2)|=|g(z1)-g(z2)+g(z1)g(z2)∑k=2n-1(αk/k!)(z2k-1-z1k-1)+g(z1)g(z2)(ψ(z2)/z2-ψ(z1)/z1)||1+g(z1)∑k=1n-1(αk/k!)z1k-1+g(z1)(ψ(z1)/z1)||1+g(z2)∑k=1n-1(αk/k!)z2k-1+g(z2)(ψ(z2)/z2)|>|g(z1)-g(z2)|-M2|z1-z2|∑k=2n-1(|αk|/k!)|∑t=0k-2z1tz2k-1-t|-((n-1)/n!)KM2|z1-z2||1+g(z1)∑k=1n-1(αk/k!)z1k-1+g(z1)(ψ(z1)/z1)||1+g(z2)∑k=1n-1(αk/k!)z2k-1+g(z2)(ψ(z2)/z2)|>|g(z1)-g(z2)|-M2|z1-z2|∑k=2n-1(|αk|(k-1)/k!)-((n-1)/n!)KM2|z1-z2||1+g(z1)∑k=1n-1(αk/k!)z1k-1+g(z1)(ψ(z1)/z1)||1+g(z2)∑k=1n-1(αk/k!)z2k-1+g(z2)(ψ(z2)/z2)|≥0.
Hence, f(z) is univalent in 𝕌.Corollary 2.2.
Letf(z)∈𝒜 with f(z)≠0 when 0<|z|<1. For any n∈{3,4,…}, if
(2.15)∑k=2n-1‍k-1k!|βk|+n-1n!|dndzn(zf(z))|≤1(z∈U),
where βk=(dk/dzk)(z/f(z))|z=0, then f(z) is univalent in 𝕌. The result is sharp, where equality occurs for the Koebe function k(z)=z/(1-z)2 and also for functions of the following form:
(2.16)f(z)=z1+az+z2,(|a|≤2),fn(z)=z(1±(1/(n-2))z)n-1.Proof.
Settingg(z)=z in Theorem 2.1 immediately yields (2.15). To show that the result is sharp for n≥3, we consider
(2.17)f(z)=z(1+(1/(n-2))z)n+ϵ-1(ϵ>0).
A computation shows, for 1≤k≤n-1, that
(2.18)dkdzk(zf(z))=(n-2)-k(ϵ+n-1)(ϵ+n-2)⋯(ϵ+n-k)(1+1n-2z)ϵ+n-k-1.
Letting ϵ=0 in (2.17) and (2.18) implies, respectively, that (dn/dzn)(z/f(z))=0 and
(2.19)|βk|=(n-1)!(n-k-1)!(n-2)k.
This satisfies the equality in (2.15), because for x∈ℝ and n≥3, an application of the binomial theorem gives
(2.20)(1+x)n-1=∑k=0n-1‍(n-1k)xk,
and so
(2.21)∑k=2n-1‍(k-1)(n-1k)xk=1+(n-1)(1+x)n-2x-(1+x)n-1=1+(1+x)n-2[x(n-2)-1].
Choosing x=1/(n-2) in assertion (2.21) gives the equality. However, for every ϵ>0, we have
(2.22)f′(n-2n-2+ϵ)=0.
Hence f is not univalent in 𝕌 and the result is sharp. Moreover it can be easily checked that the equality in (2.15) holds for the given functions and the proof is complete.


## 3. Special Cases and Examples

Lettingn=2 in inequality (2.15) gives the univalence criterion defined by (1.4), which is due to Yang and Liu [7]. Next, we reduce the result for some values of n by computing the corresponding values of βk in terms of the coefficients. More precisely, for n=3 and n=4, Corollary 2.2 reduces at once to the following two remarks.Remark 3.1.
Letf(z)=z+∑k=2∞‍akzk with f(z)≠0 when 0<|z|<1 satisfy
(3.1)|(zf(z))′′′|≤3-3|a22-a3|(z∈U).
Then f(z) is univalent in 𝕌. The bound in (3.1) is best possible, where equality occurs for the Koebe function and for functions of the following form:
(3.2)f(z)=z1+az+z2(|a|≤2).Proof.
The result follows from takingn=3 in Corollary 2.2 and that |β2|=2|a22-a3|.Remark 3.2.
Letf(z)=z+∑k=2∞‍akzk with f(z)≠0 for 0<|z|<1 satisfy
(3.3)|d4dz4(zf(z))|≤8-8|a22-a3|-16|a4-2a2a3+a23|(z∈U).
Then f(z) is univalent in 𝕌. The bound in (3.3) is best possible, where equality occurs for the Koebe function and also for functions of the following form:
(3.4)f(z)=z1+az+z2(|a|≤2),f(z)=z(1±(1/2)z)3.Proof.
The result follows from takingn=4 in Corollary 2.2 and that |β3|=6|a4-2a2a3+a23|, and |β2|=2|a22-a3|.To understand the behavior of the extremal functions for our criterion (2.15), let us consider, for example, f(z)=z/(1-(1/2)z)3, which is an extremal function for the case n=4. Figures 1(a) and 1(b) show the images of the unit circle under the functions f(z) and g(z)=z/(1-(1/2)z)3.05, respectively. If we restrict the images around the cusps as shown in Figures 1(c) and 1(d), we see that the image of g is a curve that intersects itself in some purely real point u. This means that there are two different points z1 and z2 that lie on the unit circle such that g(z1)=g(z2)=u. In fact, each purely real point lies inside the closed curve of Figures 1(c) and 1(d) which is an image for two different points in 𝕌 having the same modulus but different arguments. However, we cannot find such points for the function f, and this interprets why f is an extremal function for univalence, since the closed curve of Figure 1(d) vanishes whenever the power in the function g approaches to 3 as shown in Figure 1(c).Geometric description for the sharpness of the casen=4.
(a)
(b)
(c)
(d)From Corollary2.2, we have the following.Corollary 3.3.
Let(3.5)f(z)=z1+∑k=1∞bkzk∈A
with f(z)≠0 for 0<|z|<1 and
(3.6)∑k=2n‍(k-1)|bk|+(n-1)∑k=n+1∞‍(kn)|bk|≤1,
for some n∈{2,3,…}. Then f(z) is univalent in 𝕌.Proof.
In view of (3.5) and by simple computation we have
(3.7)dndzn(zf(z))=n!bn+∑k=n+1∞‍k!(k-n)!bkzk-n,
and so βm=m!bm, for 1≤m≤n-1. It follows that
(3.8)|dndzn(zf(z))|≤∑k=n∞‍k!|bk|(k-n)!.
Hence, by applying Corollary 2.2, we get the desired result.Remark 3.4.
Takingn=2 in Corollary 3.3 gives a result of Yang and Liu [7].Example 3.5.
From Corollary3.3, it can be easily seen that the functions
(3.9)f(z)=z1+∑k=1nbkzk,
with f(z)≠0 for 0<|z|<1 and ∑k=2n‍(k-1)|bk|≤1, are univalent in 𝕌.



---

*Source: 307526-2012-06-21.xml*