KYUNGPOOK Math. J. 2019; 59(3): 481-491  
Some Coefficient Inequalities Related to the Hankel Determinant for a Certain Class of Close-to-convex Functions
Yong Sun∗, Zhi-Gang Wang
School of Science, Hunan Institute of Engineering, Xiangtan, 411104, Hunan, People’s Republic of China
e-mail :
Mathematics and Computing Science, Hunan First Normal University, Changsha, 410205, Hunan, People’s Republic of China
e-mail :
* Corresponding Author.
Received: November 14, 2017; Revised: February 1, 2019; Accepted: March 4, 2019; Published online: September 23, 2019.
© Kyungpook Mathematical Journal. All rights reserved.

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In the present paper, we investigate the upper bounds on third order Hankel determinants for certain class of close-to-convex functions in the unit disk. Furthermore, we obtain estimates of the Zalcman coefficient functional for this class.

Keywords: coefficient inequality, Hankel determinant, Zalcman’s conjecture, close-to-convex functions.

Let be the class of functions analytic in the unit disk of the form f(z)=z+n=2anzn.We denote by the subclass of consisting of univalent functions.

A function is said to be starlike of order α (0 ≤ α < 1), if it satisfies (zf(z)f(z))>α(zD).We denote by the class of starlike functions of order α. In particular, .

Recall that a function is close-to-convex in if it is univalent and the range is a close-to-convex domain, i.e., the complement of can be written as the union of nonintersecting half-lines. A normalized analytic function f in is close-to-convex in if there exists a function , such that the following inequality (zf(z)g(z))>0(zD)holds. Denote by the class of close-to-convex functions. We refer to [8, 16, 17, 28] for discussion and basic results on close-to-convex functions.

In [11], Gao and Zhou investigated the following class of close-to-convex functions.

Definition 1.1

Suppose that is analytic in of the form (1.1). We say that , if there exists a function , such that (z2f(z)g(z)g(z))<0(zD).Let g(z)=z+n=2bnznS*(1/2)(zD)and G(z)=g(z)g(z)z=z+n=2B2n1z2n1(zD).Then G(−z) = G(z), so G(z) is an odd starlike function. It is well-known that |B2n1|1(n=2,3,).Substituting the series expressions of g(z), G(z) in (1.4) and (1.5), and using (1.6), then the following result holds.

Theorem A. ([11])

Let. Then for n ≥ 2, |B2n1|=|2b2n12b2b2n2++(1)n2bn1bn+1+(1)n+1bn2|1.The estimates are sharp, with the extremal function given by g(z) = z/(1− z).

Theorem B. ([11])

Letbe of the form(1.1). Then|an|1(n=2,3,).The estimates are sharp, with the extremal function given by f(z) = z/(1 − z).

Noonan and Thomas [24] studied the Hankel determinant Hq,n(f) defined as Hq,n(f)=|anan+1an+q1an+1an+2an+qan+q1an+qan+2(q1)|(q,n).Problems involving Hankel determinants Hq,n(f) in geometric function theory originate from the work of such authors as Hadamard, Polya and Edrei (see [7, 9]), who used them in study of singularities of meromorphic functions. For example, Hankel determinants can be used in showing that a function of bounded characteristic in , i.e., a function which is a ratio of two bounded analytic functions with its Laurent series around the origin having integral coefficients, is rational [5]. Pommerenke [25] proved that the Hankel determinants of univalent functions satisfy the inequality |Hq,n(f)|<Kn(12+β)q+32, where β > 1/4000 and K depends only on q. Furthermore, Hayman [12] proved a stronger result for areally mean univalent functions, i.e., he showed that H2,n(f) < An1/2, where A is an absolute constant.

We note that H2,1(f) is the well-known Fekete-Szegő functional, see [10, 16, 17]. The sharp upper bounds on H2,2(f) were obtained in the articles [2, 14, 15, 18] for various classes of functions.

By the definition, H3,1(f) is given by H3,1(f)=|a1a2a3a2a3a4a3a4a5|.Note that for , a1 = 1 so that H3,1(f)=a3(a2a4a32)+a4(a2a3a4)+a5(a3a22),by the triangle inequality, we have |H3,1(f)||a3a2a4a32|+|a4a2a3a4|+|a5||a3a22|.Obviously, the case of the upper bound of H3,1(f) is much more difficult than the cases of H2,1(f) and H2,2(f). Recently, Prajapat et al.[26] studied the upper bounds on the Hankel determinants for the class of close-to-convex functions.

Theorem C

Letbe of the form(1.1). Then|a2a3a4|3,|a2a4a32|8536and|H3,1(f)|28912.For further information about the upper bounds of the third Hankel determinants for some classes of univalent functions, see e.g. [1, 3, 6, 27, 29].

In 1960, Lawrence Zalcman posed a conjecture that the coefficients of satisfy the sharp inequality |an2a2n1|(n1)2(n),with equality only for the Koebe function k(z) = z/(1− z)2 and its rotations. We call Jn(f)=an2a2n1 the Zalcman functional for . Clearly, for , we have |J2(f)| = |H2,1(f)|. The Zalcman conjecture was proved for certain special subclasses of in [4, 19, 22, 23].

In the present investigation, our purpose is to develop similar results on the Hankel determinants in the context the close-to-convex functions . Further-more, the upper bounds to the Zalcman functional for this class are obtained.

Preliminary Results
y Results

Denote by the class of Carathéodory functions p normalized by p(z)=1+n=1cnznand(p(z))>0(zD).The following results are well known for functions belonging to the class .

Lemma 2.1. ([8])

Ifis of the form(2.1), then|cn|2(n).The inequality(2.2)is sharp and the equality holds for the functionφ(z)=1+z1z=1+2n=1zn.

Lemma 2.2.([13])

Ifis of the form(2.1), then the sharp estimate(2.3)is vaild|cnμckcnk|2(n,k,n>k;0μ1).

Lemma 2.3.([20, 21])

Ifis of the form(2.1), then there exist x, z such that |x| ≤ 1 and |z| ≤ 1, 2c2=c12+(4c12)x,and4c3=c13+2c1(4c12)xc1(4c12)x2+2(4c12)(1|x|2)z.

The Upper Bounds of the Hankel Determinant

In this section, we first give an upper bound of the functional |a2a3 a4| for functions .

Theorem 3.1

Letbe of the form(1.1). Then



Let g be given by (1.4), and p(z)=z2f(z)g(z)g(z)=1+c1z+c2z2+(zD).Then, we have ℜ(p(z)) > 0, and z2f(z)=g(z)g(z)p(z).Substituting the expansions of f(z), g(z) and p(z) in (3.2), and equating the coefficients, we obtain {a2=12c1,a3=13(c2+2b3b22),a4=14[c3+(2b3b22)c1].Hence, by using the above values of a2, a3 and a4 from (3.3), and the relations of (2.4) and (2.5) we obtain, for some x and z such that |x| ≤ 1 and |z| ≤ 1, |a2a3a4|=112|(2b3b22)c1+2c1c23c3|=148|c134(2b3b22)c1+(4c12)[2c1x+3c1x26(1|x|2)z]|.By Lemma 2.1, we have |c1| ≤ 2. By setting c := c1, we may assume without loss of generality that c ∈ [0, 2]. Thus, by applying the triangle inequality in (3.4) with μ = |x|, we obtain |a2a3a4|148{c3+4c+(4c2)[3(c2)μ2+2cμ+6]}=:F(c,μ).Let ϕ(μ)=3(c2)μ2+2cμ+6(c[0,2];μ[0,1]).In particular, for the case of c = 2, we have ϕ(μ)=4μ+6ϕ(1)=10.For the case of 0 ≤ c < 2, then φ(μ) is a quadratic function of μ ∈ [0, 1], and we can get ϕ(μ)=3(c2)(μc3(2c))2+c218c+363(2c).If μ0=c3(2c)1, that is, 0c32, we obtain ϕ(μ)ϕ(μ0)+c218c+363(2c).If μ0=c3(2c)1, that is, 32c<2, we get ϕ(μ)ϕ(1)=5c.Thus, we have F(c,μ)G(c)={G1(c)=136(c34c2+3c+18)(0c3/2),G2(c)=112(c3+6c)(3/2c2).For G1(c), we have G1(c)=136(3c28c+3)andG1(c)=118(3c4).Let C0=473[0,32],then, we obtain G1(C0)=0andG1(C0)<0.For G2(c), we have G2(c)=14(2c2)<0,(32c2).Obviously, G2(c) is an decreasing function of c on [3/2, 2] and, hence, G2(c)G2(32)=1532.Since G(c) is a continuous function of c on the closed interval [0, 2], it follows that |a2a3a4|G(c)max{G1(0),G1(C0),G2(32)}=12.Now, we are ready to give an upper bound of |a2a4a32| for functions .

Theorem 3.2

Letbe of the form(1.1). Then|a2a4a32|1.


Using the values of a2, a3 and a4 from (3.3), and using (2.4) and (2.5) for some x and z such that |x| ≤ 1 and |z| ≤ 1, we get a2a4a32=1288{c14+(4c12)[2c12x(32+c12)x2+18(1|x|2)c1z]}29(2b3b22)(c2916c12)19(2b3b22)2.By Lemma 2.1, we may assume that |c1| = c ∈ [0, 2]. By applying Theorem A, Lemma 2.1, Lemma 2.2 and the triangle inequality in above relation with µ = |x|, we obtain |a2a4a32|1288{c4+(4c2)[(c218c+32)μ2+2c2μ+18c]}+59.Let ψ(μ)=(c218c+32)μ2+2c2μ+18c,S(c,μ)=1288[c4+(4c2)ψ(μ)].Therefore, ψ(μ)=2(c2)(c16)μ+2c20,which implies that ψ(µ) is an increasing function of µ on [0, 1]. Hence, we have ψ(μ)ψ(1)=3c2+32,which yields that S(c,μ)S(c,1)=1144(c410c2+64)49,(0c2).Thus, we obtain the bound of |a2a4a32|.

Let . Then using the above results in theorem B, Theorem 3.1 and Theorem 3.2, together with the known inequality |a22a3|1 (see [16]), we obtain the upper bound of the third Hankel determinant for close-to-convex functions .

Theorem 3.3

Let be of the form(1.1). Then|H3,1(f)|52.

Remark 3.1

In Theorem 3.1, Theorem3.2 and Theorem 3.3, we have obtained the upper bounds for the Hankel determinant. However, these results are far from sharp.

The Upper Bounds of the Zalcman Functional

In this section, we consider the Zalcman functional for functions .

Theorem 4.1

Letbe of the form(1.1). Then|a22a3|1,|a32a5|3445,and|an2a2n1|{24(n1)n2(n=2k4),24n(n=2k+15).


Let g(z), G(z) and p(z) be given by (1.4), (1.5) and (3.1), respectively. Then, we have zf(z)=p(z)G(z)(zD).Comparing the coefficients of two sides of this equation, we obtain an={12k(c2k1B1+c2k3B3++c1B2k1)(n=2k),12k+1(c2kB1+c2k2B3++c0B2k+1)(n=2k+1),where k ∈ ℕ and B1 = c0 = 1.

For the case of n = 2k, we have |an2a2n1|=|a4k1a2k2|=|14k1(c4k2B1+c4k4B3++c2kB2k1+c2k2B2k+1++c0B4k1)14k2(c2k1B1+c2k3B3++c1B2k1)2|=|14k1(c4k24k14k2c2k12)+B34k1(c4k44k12k2c2k1c2k3)++B2k14k1(c2k4k12k2c2k1c1)+14k1(c2k2B2k+1++c2B4k3+c0B4k1)14k2(c2k3B3++c1B2k1)2|.If k = 1, using Theorem B and Lemma 2.2, we have |a22a3|=|13(c234c12)+13B3|13|c234c12|+13|B3|1.If k ≥ 2 we note that 4k14k21and4k12k21(k2),by Theorem B, Lemma 2.1, Lemma 2.2 and the triangle inequality, we obtain |an2a2n1|2k4k1+2(k1)+14k1+[2(k1)]24k2=24(n1)n2.For the case of n = 2k + 1, we have |an2a2n1|=|a4k+1a2k+12|=|14k+1(c4kB1+c4k2B3++c2k+2B2k1+c2kB2k+1+c2k2B2k+3++c0B4k+1)1(2k+1)2(c2kB1+c2k2B3++c2B2k1+c0B2k+1)2|=|14k+1(c4k4k+1(2k+1)2c2k2)+B34k+1(c4k22(4k+1)(2k+1)2c2kc2k2)++B2k14k+1(c2k+22(4k+1)(2k+1)2c2kc2)+(14k+12(2k+1)2)c2kB2k+1+14k+1(c2k2B2k+3++c2B4k1+c0B4k+1)1(2k+1)2(c2k2B3++c2B2k1+c0B2k+1)2|.If k = 1, using Theorem B, Lemma 2.1 and Lemma 2.2, we have |a32a5|15|c459c22|+|1529||c2B3|+15|B5|+19|B32|3445.If k ≥ 2 we note that 14k+12(2k+1)20,4k+1(2k+1)21and2(4k+1)(2k+1)21(k2),by Theorem B, Lemma 2.1, Lemma 2.2 and the triangle inequality, we obtain |an2a2n1|2k4k+1+(24k+14(2k+1)2)+2k14k+1+(2k1)2(2k+1)2=24n.This completes the proof.

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