# Logarithmic Coefficient Bounds and Coefficient Conjectures for Classes Associated with Convex Functions

Authors: Davood Alimohammadi; Ebrahim Analouei Adegani; Teodor Bulboacă; Nak Eun Cho
Journal: Journal of Function Spaces
 (2021)

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## Abstract

It is well-known that the logarithmic coefficients play an important role in the development of the theory of univalent functions. IfS denotes the class of functions fz=z+∑n=2∞anzn analytic and univalent in the open unit disk U, then the logarithmic coefficients γnf of the function f∈S are defined by logfz/z=2∑n=1∞γnfzn. In the current paper, the bounds for the logarithmic coefficients γn for some well-known classes like C1+αz for α∈0,1 and CVhpl1/2 were estimated. Further, conjectures for the logarithmic coefficients γn for functions f belonging to these classes are stated. For example, it is forecasted that if the function f∈C1+αz, then the logarithmic coefficients of f satisfy the inequalities γn≤α/2nn+1,n∈ℕ. Equality is attained for the function Lα,n, that is, logLα,nz/z=2∑n=1∞γnLα,nzn=α/nn+1zn+⋯,z∈U.


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## Body

## 1. Introduction

LetU≔z∈ℂ:z<1 denote the open unit disk in the complex plane ℂ. Let A be the category of analytic functions f in U for which f has the following representation:
(1)fz=z+∑n=2∞anzn,z∈U.Also, letS be the subclass of A consisting of all univalent functions in U. Then, the logarithmic coefficientsγn of the function f∈S are defined with the aid of the following series expansion:
(2)logfzz=2∑n=1∞γnfzn,z∈U.These coefficients play an important role for different estimates in the theory of univalent functions, and note that we useγn instead of γnf. Kayumov [1] solved Brennan’s conjecture for conformal mappings with the help of studying the logarithmic coefficients. The significance of the logarithmic coefficients follows from Lebedev-Milin inequalities ([2], chapter 2; see also [3, 4]), where estimates of the logarithmic coefficients were applied to obtain bounds on the coefficients of f. Milin [2] conjectured the inequality
(3)∑m=1n∑k=1mkγk2−1k≤0,n=1,2,3,⋯,that implies Robertson’s conjecture [5] and hence Bieberbach’s conjecture [6], which was the well-known coefficient problem in the theory of univalent functions. De Branges [7] proved Bieberbach’s conjecture by establishing Milin’s conjecture.Recall that we can rewrite (2) in the power series form as follows:
(4)2∑n=1∞γnzn=a2z+a3z2+a4z3+⋯−12a2z+a3z2+a4z3+⋯2+13a2z+a3z2+a4z3+⋯3+⋯,z∈U,and equating the coefficients of zn for n=1,2,3, it follows that
(5)2γ1=a2,2γ2=a3−12a22,2γ3=a4−a2a3+13a23.If the functionsf and g are analytic in U, the function f is called to be subordinate to the function g, written fz≺gz, if there exists a function w analytic in U with wz<1, z∈U, and w0=0, such that f=g≺w. In particular, if g is univalent in U, then the following equivalence relationship holds true:
(6)fz≺gz⇔f0=g0,fU⊂gU.Using the principle of subordination, Ma and Minda [8] introduced the classes S∗φ and Cφ, where we make here the weaker assumptions that the function φ is analytic in the open unit disk U and satisfies φ0=1, such that it has a series expansion of the form
(7)φz=1+B1z+B2z2+B3z3+⋯,z∈U,withB1≠0.They considered the abovementioned classes as follows:(8)S∗φ≔f∈A:zf′zfz≺φz,Cφ≔f∈A:1+zf′′zf′z≺φz.Some special subclasses of the classS∗φ and Cφ play a significant role in the Geometric Function Theory because of their geometric properties.For example, takingφz=1+Az/1+Bz where A∈ℂ, −1≤B≤0, and A≠B, we get the classes S∗A,B and CA,B, respectively (see also [9, 10]). The mentioned classes with the restriction −1≤B<A≤1 reduce to the popular Janowski starlike and Janowski convex functions, respectively. By replacing A=1−2α and B=−1, where 0≤α<1, we obtain the classes S∗α and Cα of the starlike functions of orderα and convex functions of orderα, respectively. In particular, S∗≔S∗0 and C≔C0 are the class of starlike functions and of convex functions in the open unit disk U, respectively. Further, by altering A=α and B=0, where 0≤α<1, we get the classes S∗1+αz and C1+αz, which are the extensions of the classes S∗1+z and C1+z, respectively (see [11]), that is,
(9)S∗1+αz≔f∈A:zf′zfz−1<α,C1+αz≔f∈A:zf′′zf′z<α,where 0<α≤1.Supposing thatΨα,n∈S∗1+αz is such that
(10)zΨα,n′zΨα,nz=1+αzn,n∈ℕ,each function Ψα,n is of the form
(11)Ψα,nz=zexp∫0z1+αtn−1tdt=z+αnzn+1+⋯,z∈U,and is the extremal function for various problems in S∗1+αz. Also, suppose that Lα,n∈C1+αz is such that
(12)1+zLα,n′′zLα,n′z=1+αzn,n∈ℕ.Then, each functionLα,n is of the form
(13)Lα,nz=∫0zexp∫0x1+αtn−1tdtdx=z+αnn+1zn+1+⋯,z∈U,and plays as extremal function for some extremal problems in the set C1+αz.Lately, Kanas et al. [12] introduced the categories SThpls and CVhpls by
(14)SThpls≕f∈A:zf′zfz≺qsz≔11−zs,0<s≤1,CVhpls≔f∈A:1+zf′′zf′z≺qsz≔11−zs,0<s≤1,and obtained some geometric properties in these categories. Further, the functions
(15)Φs,nz=zexp∫0zqstn−1tdt=z+snzn+1+⋯,z∈U,n∈ℕ,Ks,nz=∫0zexp∫0xqstn−1tdtdx=z+snn+1zn+1+⋯,z∈U,n∈ℕ,play as extremal functions for some issues of the families SThpls and CVhpls, respectively.Lately, several researchers have subsequently investigated same problems regarding the logarithmic coefficients and the coefficient problems [9, 13–23], to mention a few of them. For instance, the rotation of the Koebe functionkz=z1−eiθz−2 for each θ∈ℝ has the logarithmic coefficients γn=eiθn/n, n≥1. If f∈S, then by applying the Bieberbach inequality for the first relation of (5), it follows that γ1≤1, and using the Fekete-Szegö inequality for the second relation of (5) (see [24], Theorem 3.8) leads to
(16)γ2=12a3−12a22≤121+2e−2≃0.635⋯.It was established in ([25], Theorem 4) that the logarithmic coefficients γn of f∈S satisfy the inequality
(17)∑n=1∞γn2≤π26,and the equality is obtained for the Koebe function. For f∈S∗, the inequality γn≤1/n holds but is not true for the full class S, even in order of magnitude (see [24], Theorem 8.4). In 2018, some first logarithmic coefficients γn were estimated for special subclasses of close-to-convex functions in [15, 20]. However, the problem of the best upper bounds for the logarithmic coefficients of univalent functions for n≥3 is presumably still a concern. In [13], the authors obtained the bounds of logarithmic coefficients γn, n∈ℕ, for the general class S∗φ, and the bounds of the logarithmic coefficients γn when n=1,2,3 for the class Kφ, while the estimated bounds would generalize many of the previous outcomes.In the present study, which is motivated essentially by the recent works [13, 16], the bounds for the logarithmic coefficients γn, n∈ℕ, of the class C1+αz for α∈0,1 and CVhpl1/2 were estimated. Further, conjectures for the logarithmic coefficients γn for f belonging to these classes are stated.


## 2. Main Results

First, we will obtain the bounds forγn of the classes S∗1+αz and C1+αz for α∈0,1. In this regard, the following outcomes will be employed in the key results.Lemma 1 (see [13], Theorem 1).
Letf∈S∗φ. If φ is convex univalent, then the logarithmic coefficients of f satisfy the following inequalities:
(18)γn≤B12n,n∈ℕ,(19)∑n=1∞γn2≤14∑n=1∞Bn2n2.
The inequalities in (18) and (19) are sharp, such that for any n∈ℕ, there exist the function fn given by zfn′z/fnz=φzn and the function f given by zf′z/fz=φz, respectively, for those equalities we obtain.Lemma 2 (see [13], Theorem 2).
Letf∈Cφ. Then, the logarithmic coefficients of f satisfy the inequalities
(20)γ1≤B14,(21)γ2≤∣B1∣12,if4B2+B12≤4B1,4B2+B1248,if4B2+B12>4B1,and if B1,B2, and B3 are real values, then
(22)γ3≤B124Hq1;q2,where Hq1;q2 is given in ([26], Lemma 2) (or [9], Lemma 5), q1=B1+4B2/B1/2, and q2=B2+2B3/B1/2. The bounds (20) and (21) are sharp.Lemma 3 (see [18], Theorem 30).
Iff∈CVhpl1/2, then
(23)γ1≤18,γ2≤124,γ3≤148.
The first two bounds are sharp forf=K1/2,1 and f=K1/2,2, respectively.If we consider Lemma1 with the function φz≕1+αz, then we immediately get the next result:Theorem 4.
Iff∈S∗1+αz, then
(24)γn≤α2n,n∈ℕ,∑n=1∞γn2≤α4.
These inequalities are sharp forf=Ψα,n and f=Ψα,1, respectively.Corollary 5.
Letf∈C1+αz. Then, the logarithmic coefficients of f satisfy the inequalities
(25)γ1≤α4,γ2≤α12,γ3≤α24.Equalities in these inequalities are attained for the functionsLα,n for n=1,2,3, respectively.Proof.
Forφz=1+αz, where B1=α,B2=B3=0, in Theorem 6, we obtain the required result. Also, since
(26)logLα,1zz=2∑n=1∞γnLα,1zn=α2z+⋯,z∈U,logLα,2zz=2∑n=1∞γnLα,2zn=α6z2+⋯,z∈U,logLα,3zz=2∑n=1∞γnLα,3zn=α12z3+⋯,z∈U,it follows that these inequalities are attained for the functions Lα,n for n=1,2,3, respectively.☐Theorem 6.
Letf∈C1+αz. Then, the logarithmic coefficients of f satisfy the inequalities
(27)γn≤α4n,n∈ℕ.
This inequality is sharp for∣γ1∣ for the function Lα,1.Proof.
Iff∈C1+αz, this is equivalent to f∈A and
(28)1+zf′′zf′z≺1+αz≕φαz.
If we definepz≔zf′z/fz, then p0=1, and the above subordination relation can be written as
(29)pz+zp′zpz≺φαz.
Supposing that the functionψα satisfies the differential equation
(30)ψαz+zψα′zψαz=φαz,ψα0=1,we will prove that ψα is a convex univalent function in U.
The functionφα has positive real part in U whenever α∈0,1. Therefore, using ([27], Theorem 1) for β=1, γ=0, and c=1, it follows that the solution ψα of the differential equation (30) is analytic in U, with Reψαz>0 for all z∈U, and
(31)ψαz=Hz∫0zHttdt−1=αzexpαzexpαz−1=1+α2z+⋯,z∈U,where
(32)Hz=zexp∫0zφαt−1tdt=zexpαz,and all powers are considered at the principal branch, that is, log1=0.
Sinceφα is convex and ψα is analytic with Reψαz>0 for all z∈U, using [28] (Theorem 3.2i) for n=1, we deduce that ψα is univalent in U. Moreover, from Figure 1 made with MAPLE software, we get
(33)Ψz≔Re1+zψα′′zψα′z>0,z∈U,and ψα′0=α/2≠0, so ψα is a convex function. Hence, it follows that ψα is a convex univalent function in U.
Therefore, according to [28] (Theorem 3.2i), the differential subordination (29) implies
(34)pz≺ψαz,for all 0<α≤1, and ψα is the best dominant. Thus,
(35)zf′zfz≺ψαz,for all 0<α≤1. Hence,
(36)C1+αz⊂S∗ψα.
From the above relation, we get(37)supγnf:f∈C1+αz≤supγnf:f∈S∗ψα.
Hence, from Lemma1, we obtain
(38)supγnf:f∈C1+αz≤α4n.
Therefore, forf∈C1+αz and for all n∈ℕ, we conclude that
(39)γnf≤α4n.☐Remark 7.
If we compare the results of Corollary5 with those of Theorem 6, then we conclude that the results of Theorem 6 are not the best possible. We conjecture that if the function f∈C1+αz, then the logarithmic coefficients of f satisfy the inequalities
(40)γn≤α2nn+1,n∈ℕ.
Equality is attained for the functionLα,n, that is,
(41)logLα,nzz=2∑n=1∞γnLα,nzn=αnn+1zn+⋯,z∈U.Theorem 8.
Letf∈CVhpl1/2. Then, the logarithmic coefficients of f satisfy the inequalities
(42)γn≤18n,n∈ℕ.
This inequality is sharp forγ1 for the function K1/2,1.Proof.
Lettingf∈CVhpl1/2, it follows that
(43)1+zf′′zf′z≺11−z≕q12z.
Suppose thatp satisfies the differential equation
(44)pz+zp′zpz=11−z.
If we definepz≔zf′z/fz, then the subordination (43) can be rewritten as
(45)pz+zp′zpz≺q12z.
According to the inequality (20) of [12] (Theorem 2.3), the function q1/2 is analytic with positive real part in U. Therefore, using [27] (Theorem 1) for β=1, γ=0, and c=1, it follows that the solution p of the differential equation (44) is analytic in U with Repz>0, z∈U, and
(46)pz=Hz∫0zHttdt−1=4z1+1−z21−8/1+1−z−8ln1+1−z+4+8ln2=1+14z+⋯,z∈U,where
(47)Hz=zexp∫0zq1/2t−1tdt=4z1+1−z2,and all powers are considered at the principal branch, that is, log1=0. Moreover, from Figure 2 made with MAPLE software, we get
(48)Ψz≔Re1+zp′′zp′z>0,z∈U,and p′0=1/4≠0. Hence, ψ is convex in U.
Sincep satisfies in the subordination (45), using [28] (Theorem 3.2i), we conclude that pz≺ψz, that is,
(49)zf′zfz≺pz,and p is the best dominant. Thus, f∈CVhpl1/2 implies f∈S∗p, that is,
(50)CVhpl1/2⊂S∗p.
Therefore, sincep is convex univalent, from Lemma 1, it follows that
(51)supγnf:f∈CVhpl12≤supγnf:f∈S∗p=18n,and we obtain the result. This completes the proof.□Remark 9.
If we compare the results of Lemma1 with those of Theorem 8, then we conclude that the results of Theorem 8 are not the best possible. We conjecture that if the function f∈CVhpl1/2, then the logarithmic coefficients of f satisfy the inequalities
(52)γn≤14nn+1,n∈ℕ.
Equality is attained for the functionK1/2,n, that is,
(53)logK1/2,nzz=2∑n=1∞γnK1/2,nzn=12nn+1zn+⋯,z∈U.Figure 1
The plot ofΨReit for t∈0,2π, R=1, and α∈10−6,1.Figure 2
The plot ofΨReit for t∈0,2π and R=1.



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