Logarithmic Coefficients for Univalent Functions Defined by Subordination

Abstract

In this work, the bounds for the logarithmic coefficients ${\gamma }_n$ of the general classes ${\mathcal{S}}^{*}\left(\phi \right)$ and $\mathcal{K}\left(\phi \right)$ were estimated. It is worthwhile mentioning that the given bounds would generalize some of the previous papers. Some consequences of the main results are also presented, noting that our method is more general than those used by others.

1. Introduction

Let $\mathcal{H}$ denote the class of analytic functions in the open unit disk $\mathbb{U}:=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ and $\mathcal{A}$ denote the subclass of $\mathcal{H}$ consisting of functions of the form

$$\begin{array}{c}\hfill f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n.\end{array}$$

(1)

Also, let $\mathcal{S}$ be the subclass of $\mathcal{A}$ consisting of all univalent functions in $\mathbb{U}.$ Then the logarithmic coefficients ${\gamma }_n$ of $f\in \mathcal{S}$ are defined with the following series expansion:

$$\begin{array}{c}\hfill log\left({\displaystyle \frac{f\left(z\right)}{z}}\right)=2\sum _{n=1}^{\infty }{\gamma }_n\left(f\right)z^n,z\in \mathbb{U}.\end{array}$$

(2)

These coefficients play an important role for various estimates in the theory of univalent functions. Note that we use ${\gamma }_n$ instead of ${\gamma }_n\left(f\right)$. The idea of studying the logarithmic coefficients helped Kayumov [1] to solve Brennan’s conjecture for conformal mappings.

Recall that we can rewrite (2) in the series form as follows:

$$\begin{array}{cc}\hfill 2\sum _{n=1}^{\infty }{\gamma }_n{z}^n=& a_{2}z+a_3{z}^2+a_4{z}^3+\dots -{\displaystyle \frac{1}{2}}{[a_{2}z+a_3{z}^2+a_4{z}^3+\dots ]}^2\hfill \\ & +{\displaystyle \frac{1}{3}}{[a_{2}z+a_3{z}^2+a_4{z}^3+\dots ]}^3+\dots .\hfill \end{array}$$

Now, considering the coefficients of $z^n$ for $n=1,2,3$, it follows that

$$\left\{\begin{array}{c}2{\gamma }_1=a_2,\hfill \\ 2{\gamma }_2=a_3-{\displaystyle \frac{1}{2}}{a}_2^2,\hfill \\ 2{\gamma }_3=a_4-a_2{a}_3+{\displaystyle \frac{1}{3}}{a}_2^3.\hfill \end{array}\right.$$

(3)

For two functions f and g that are analytic in $\mathbb{U},$ we say that the function f is subordinate to g in $\mathbb{U}$ and write $f\left(z\right)\prec g\left(z\right)$ if there exists a Schwarz function $\omega$ that is analytic in $\mathbb{U}$ with $\omega \left(0\right)=0$ and $\left|\omega \left(z\right)\right|<1$ such that

$$f\left(z\right)=g\left(\omega \left(z\right)\right)\left(z\in \mathbb{U}\right).$$

In particular, if the function g is univalent in $\mathbb{U},$ then $f\prec g$ if and only if $f\left(0\right)=g\left(0\right)$ and $f\left(\mathbb{U}\right)\subseteq g\left(\mathbb{U}\right)$.

Using subordination, different subclasses of starlike and convex functions were introduced by Ma and Minda [2], in which either of the quantity $\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}$ or $1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}$ is subordinate to a more general superordinate function. To this aim, they considered an analytic univalent function $\phi$ with positive real part in $\mathbb{U}$. $\phi \left(\mathbb{U}\right)$ is symmetric respecting the real axis and starlike considering $\phi \left(0\right)=1$ and ${\phi }^{\prime }\left(0\right)>0$. They defined the classes consisting of several well-known classes as follows:

$${\mathcal{S}}^{*}\left(\phi \right):=\left\{f\in \mathcal{S}:{\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\prec \phi \left(z\right),z\in \mathbb{U}\right\},$$

and

$$\mathcal{K}\left(\phi \right):=\left\{f\in \mathcal{S}:1+{\displaystyle \frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}}\prec \phi \left(z\right),z\in \mathbb{U}\right\}.$$

For example, the classes ${\mathcal{S}}^{*}\left(\phi \right)$ and $\mathcal{K}\left(\phi \right)$ reduce to the classes ${\mathcal{S}}^{*}[A,B]:={\displaystyle \frac{1+Az}{1+Bz}}$ and $\mathcal{K}[A,B]:={\displaystyle \frac{1+Az}{1+Bz}}$ of the well-known Janowski starlike and Janowski convex functions for $-1\le B<A\le 1$, respectively. By replacing $A=1-2\alpha$ and $B=-1$ where $0\le \alpha <1$, we conclude the classes ${\mathcal{S}}^{*}\left(\alpha \right)$ and $\mathcal{K}\left(\alpha \right)$ of the starlike functions of order $\alpha$ and convex functions of order $\alpha$, respectively. In particular, ${\mathcal{S}}^{*}:={\mathcal{S}}^{*}\left(0\right)$ and $\mathcal{K}:=\mathcal{K}\left(0\right)$ are the class of starlike functions and of convex functions in the unit disk $\mathbb{U}$, respectively. The Koebe function $k\left(z\right)=z/{(1-z)}^2$ is starlike but not convex in $\mathbb{U}$. Thus, every convex function is starlike but not conversely; however, each starlike function is convex in the disk of radius $2-\sqrt{3}$.

Lately, several researchers have subsequently investigated similar problems in the direction of the logarithmic coefficients, the coefficient problems, and differential subordination [3,4,5,6,7,8,9,10,11], to mention a few. For example, the rotation of Koebe function $k\left(z\right)=z{(1-e^{i\theta })}^{-2}$ for each $\theta$ has logarithmic coefficients ${\gamma }_n=e^{i\theta n}/n,n\ge 1.$ If $f\in \mathcal{S},$ then by using the Bieberbach inequality for the first equation of (3) it concludes $|{\gamma }_1|\le 1$ and by utilizing the Fekete–Szegö inequality for the second equation of (3), (see [12] (Theorem 3.8)),

$$|{\gamma }_2|={\displaystyle \frac{1}{2}}|a_3-{\displaystyle \frac{1}{2}}{a}_2^2|\le {\displaystyle \frac{1}{2}}\left(1+2e^{-2}\right)=0.635\dots .$$

It was shown in [12] (Theorem 4) that the logarithmic coefficients ${\gamma }_n$ of every function $f\in \mathcal{S}$ satisfy

$$\sum _{n=1}^{\infty }{\left|{\gamma }_n\right|}^2\le {\displaystyle \frac{{\pi }^2}{6}},$$

and the equality is attained for the Koebe function. For $f\in {\mathcal{S}}^{*},$ the inequality $|{\gamma }_n|\le 1/n$ holds but is not true for the full class $\mathcal{S}$, even in order of magnitude (see [12] (Theorem 8.4)). In 2018, Ali and Vasudevarao [3] and Pranav Kumar and Vasudevarao [6] obtained the logarithmic coefficients ${\gamma }_n$ for certain subclasses of close-to-convex functions. Nevertheless, the problem of the best upper bounds for the logarithmic coefficients of univalent functions for $n\ge 3$ is presumably still a concern.

Based on the results presented in previous research, in the current study, the bounds for the logarithmic coefficients ${\gamma }_n$ of the general classes ${\mathcal{S}}^{*}\left(\phi \right)$ and $\mathcal{K}\left(\phi \right)$ were estimated. It is worthwhile mentioning that the given bounds in this paper would generalize some of the previous papers and that many new results are obtained, noting that our method is more general than those used by others. The following lemmas will be used in the proofs of our main results.

For this work, let $\mathsf{\Omega }$ represent the class of all analytic functions $\omega$ in $\mathbb{U}$ that equips with conditions $\omega \left(0\right)=0$ and $\left|\omega \right(z\left)\right|<1$ for $z\in \mathbb{U}$. Such functions are called Schwarz functions.

Lemma 1.

[13] (p. 172) Assume that ω is a Schwarz function so that $\omega \left(z\right)={\sum }_{n=1}^{\infty }{p}_n{z}^n.$ Then

$$|p_1|\le 1,|p_n|\le 1-|p_1{|}^{2}n=2,3,\dots .$$

Lemma 2.

[14] Let $\psi ,\varpi \in \mathcal{H}$ be any convex univalent functions in $\mathbb{U}$. If $f\left(z\right)\prec \psi \left(z\right)$ and $g\left(z\right)\prec \varpi \left(z\right)$, then $f\left(z\right)\ast g\left(z\right)\prec \psi \left(z\right)\ast \varpi \left(z\right)$ where $f,g\in \mathcal{H}$.

We observe that in the above lemma, nothing is assumed about the normalization of $\psi$ and $\varpi$, and “∗” represents the Hadamard (or convolution) product.

Lemma 3.

[12,15] (Theorem 6.3, p. 192; Rogosinski’s Theorem II (i)) Let $f\left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{a}_n{z}^n$ and $g\left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{b}_n{z}^n$ be analytic in $\mathbb{U}$, and suppose that $f\prec g$ where g is univalent in $\mathbb{U}$. Then

$$\sum _{k=1}^n|a_k{|}^2\le \sum _{k=1}^n|b_k{|}^2,n=1,2,\dots .$$

Lemma 4.

[12,15] (Theorem 6.4 (i), p. 195; Rogosinski’s Theorem X) Let $f\left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{a}_n{z}^n$ and $g\left(z\right)={\displaystyle \sum _{n=1}^{\infty }}{b}_n{z}^n$ be analytic in $\mathbb{U}$, and suppose that $f\prec g$ where g is univalent in $\mathbb{U}$. Then

(i) 

If g is convex, then $|a_n|\le |g^{\prime }\left(0\right)|=|b_1|,n=1,2,\dots$.

(ii) 

If g is starlike (starlike with respect to 0), then $|a_n|\le n|g^{\prime }\left(0\right)|=n|b_1|,n=2,3,\dots$.

Lemma 5.

[16] If $\omega \left(z\right)={\sum }_{n=1}^{\infty }{p}_n{z}^n\in \mathsf{\Omega }$, then for any real numbers $q_1$ and $q_2$, the following sharp estimate holds:

$$|p_3+q_1{p}_1{p}_2+q_2{p}_1^3|\le H(q_1;q_2),$$

where

$$\begin{array}{c}\hfill H(q_1;q_2)=\left\{\begin{array}{ccc}1\hfill & \mathrm{if}\hfill & (q_1,q_2)\in D_1\cup D_2\cup \left\{(2,1)\right\},\hfill \\ |q_2|\hfill & \mathrm{if}\hfill & (q_1,q_2)\in {\cup }_{k=3}^7{D}_k,\hfill \\ \frac{2}{3}\left(\right|q_1|+1){\left(\frac{|q_1|+1}{3\left(\right|q_1|+1+q_2)}\right)}^{\frac{1}{2}}\hfill & \mathrm{if}\hfill & (q_1,q_2)\in D_8\cup D_9,\hfill \\ \frac{q_2}{3}\left(\frac{q_1^2-4}{q_1^2-4q_2}\right){\left(\frac{q_1^2-4}{3(q_2-1)}\right)}^{\frac{1}{2}}\hfill & \mathrm{if}\hfill & (q_1,q_2)\in D_{10}\cup D_{11}\backslash \left\{(2,1)\right\},\hfill \\ \frac{2}{3}\left(\right|q_1|-1){\left(\frac{|q_1|-1}{3\left(\right|q_1|-1-q_2)}\right)}^{\frac{1}{2}}\hfill & \mathrm{if}\hfill & (q_1,q_2)\in D_{12}.\hfill \end{array}\right.\end{array}$$

While the sets $D_k,k=1,2,\dots ,12$ are defined as follows:

$$\begin{array}{c}{D}_1=\left\{(q_1,q_2):|q_1|\le \frac{1}{2},\left|q_2\right|\le 1\right\},\hfill \\ D_2=\left\{(q_1,q_2):\frac{1}{2}\le |q_1|\le 2,\frac{4}{27}\left(\left(\right|q_1{|+1)}^3\right)-\left(\right|q_1|+1)\le |q_2|\le 1\right\},\hfill \\ D_3=\left\{(q_1,q_2):|q_1|\le \frac{1}{2},\left|q_2\right|\le -1\right\},\hfill \\ D_4=\left\{(q_1,q_2):|q_1|\ge \frac{1}{2},|q_2|\le -\frac{2}{3}\left(\right|q_1|+1)\right\},\hfill \\ D_5=\left\{(q_1,q_2):|q_1|\le 2,|q_2|\ge 1\right\},\hfill \\ D_6=\left\{(q_1,q_2):2\le |q_1|\le 4,|q_2|\ge \frac{1}{12}(q_1^2+8)\right\},\hfill \\ D_7=\left\{(q_1,q_2):|q_1|\ge 4,|q_2|\ge \frac{2}{3}\left(\right|q_1|-1)\right\},\hfill \\ D_8=\left\{(q_1,q_2):\frac{1}{2}\le |q_1|\le 2,-\frac{2}{3}\left(\right|q_1|+1)\le q_2\le \frac{4}{27}\left(\left(\right|q_1{|+1)}^3\right)-\left(\right|q_1|+1)\right\},\hfill \\ D_9=\left\{(q_1,q_2):|q_1|\ge 2,-\frac{2}{3}\left(\right|q_1|+1)\le q_2\le \frac{2|q_1\left|\right(|q_1+1\left|\right)}{q_1^2+2\left|q_1\right|+4}\right\},\hfill \\ D_{10}=\left\{(q_1,q_2):2\le \left|q_1\right|\le 4,\frac{2|q_1\left|\right(|q_1+1\left|\right)}{q_1^2+2\left|q_1\right|+4}\le q_2\le \frac{1}{12}(q_1^2+8)\right\},\hfill \\ D_{11}=\left\{(q_1,q_2):\left|q_1\right|\ge 4,\frac{2|q_1\left|\right(|q_1+1\left|\right)}{q_1^2+2\left|q_1\right|+4}\le q_2\le \frac{2|q_1\left|\right(|q_1-1\left|\right)}{q_1^2-2\left|q_1\right|+4}\right\},\hfill \\ D_{12}=\left\{(q_1,q_2):|q_1|\ge 4,\frac{2|q_1\left|\right(|q_1-1\left|\right)}{q_1^2-2\left|q_1\right|+4}\le q_2\le \frac{2}{3}\left(\right|q_1|-1)\right\}.\hfill \end{array}$$

2. Main Results

Throughout this paper, we assume that $\phi$ is an analytic univalent function in the unit disk $\mathbb{U}$ satisfying $\phi \left(0\right)=1$ such that it has series expansion of the form

$$\begin{array}{c}\hfill \phi \left(z\right)=1+B_{1}z+B_2{z}^2+B_3{z}^3+\dots ,B_1\ne 0.\end{array}$$

(4)

Theorem 1.

Let the function $f\in {\mathcal{S}}^{*}\left(\phi \right)$. Then the logarithmic coefficients of f satisfy the inequalities:

(i) 

If φ is convex, then

$$|{\gamma }_n|\le {\displaystyle \frac{|B_1|}{2n}},n\in \mathbb{N},$$

(5)

$$\sum _{n=1}^k{\left|{\gamma }_n\right|}^2\le {\displaystyle \frac{1}{4}}\sum _{n=1}^k{\displaystyle \frac{|B_n{|}^2}{n^2}},k\in \mathbb{N},$$

(6)

and

$$\sum _{n=1}^{\infty }{\left|{\gamma }_n\right|}^2\le {\displaystyle \frac{1}{4}}\sum _{n=1}^{\infty }{\displaystyle \frac{|B_n{|}^2}{n^2}}.$$

(7)

(ii) 

If φ is starlike with respect to 1, then

$$|{\gamma }_n|\le {\displaystyle \frac{|B_1|}{2}},n\in \mathbb{N}.$$

(8)

All inequalities in (5), (7), (8) and are sharp such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}=\phi \left(z^n\right)$ and the function f given by ${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}=\phi \left(z\right)$, respectively.

Proof. 

Suppose that $f\in {\mathcal{S}}^{*}\left(\phi \right)$. Then considering the definition of ${\mathcal{S}}^{*}\left(\phi \right)$, it follows that

$$z{\displaystyle \frac{d}{dz}}\left(log\left({\displaystyle \frac{f\left(z\right)}{z}}\right)\right)={\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\prec \phi \left(z\right)-1=:\varphi \left(z\right),z\in \mathbb{U},$$

which according to the logarithmic coefficients ${\gamma }_n$ of f given by (1), concludes

$$\sum _{n=1}^{\infty }2n{\gamma }_n{z}^n\prec \varphi \left(z\right),z\in \mathbb{U}.$$

Now, for the proof of inequality (5), we assume that $\phi$ is convex in $\mathbb{U}$. This implies that $\varphi \left(z\right)$ is convex with ${\varphi }^{\prime }\left(0\right)=B_1$, and so by Lemma 4(i) we get

$$2n|{\gamma }_n|\le |{\varphi }^{\prime }\left(0\right)|=|B_1|,n\in \mathbb{N},$$

and concluding the result.

Next, for the proof of inequality (6), we define $h\left(z\right):={\displaystyle \frac{f\left(z\right)}{z}}$, which is an analytic function, and it satisfies the relation

$${\displaystyle \frac{z{h}^{\prime }\left(z\right)}{h\left(z\right)}}={\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1\prec \varphi \left(z\right),z\in \mathbb{U},$$

(9)

as $\varphi$ is convex in $\mathbb{U}$ with $\varphi \left(0\right)=0$.

On the other hand, it is well known that the function (see [17])

$$b_0\left(z\right)=log\left(\frac{1}{1-z}\right)=\sum _{n=1}^{\infty }{\displaystyle \frac{z^n}{n}}$$

belongs to the class $\mathcal{K}$, and for $f\in \mathcal{H}$,

$$f\left(z\right)\ast b_0\left(z\right)={\int }_0^z{\displaystyle \frac{f\left(t\right)}{t}}dt.$$

(10)

Now, by Lemma 2 and from (9), we obtain

$${\displaystyle \frac{z{h}^{\prime }\left(z\right)}{h\left(z\right)}}\ast b_0\left(z\right)\prec \varphi \left(z\right)\ast b_0\left(z\right).$$

Considering (10), the above relation becomes

$$log\left({\displaystyle \frac{f\left(z\right)}{z}}\right)\prec {\int }_0^z{\displaystyle \frac{\varphi \left(t\right)}{t}}dt.$$

In addition, it has been proved in [18] that the class of convex univalent functions is closed under convolution. Therefore, the function ${\displaystyle {\int }_0^z{\displaystyle \frac{\varphi \left(t\right)}{t}}dt}$ is convex univalent. In addition, the above relation considering the logarithmic coefficients ${\gamma }_n$ of f given by (1) is equivalent to

$$\sum _{n=1}^{\infty }2{\gamma }_n{z}^n\prec \sum _{n=1}^{\infty }{\displaystyle \frac{B_n{z}^n}{n}}.$$

Applying Lemma 3, from the above subordination this gives

$$4\sum _{n=1}^k{\left|{\gamma }_n\right|}^2\le \sum _{n=1}^k{\displaystyle \frac{|B_n{|}^2}{n^2}},$$

which yields the inequality in (6). Supposing that $k\to \infty$, we deduce that

$$4\sum _{n=1}^{\infty }{\left|{\gamma }_n\right|}^2\le \sum _{n=1}^{\infty }{\displaystyle \frac{|B_n{|}^2}{n^2}},$$

and it concludes the inequality (7).

Finally, we suppose that $\phi$ is starlike with respect to 1 in $\mathbb{U}$, which implies $\varphi \left(z\right)$ is starlike, and thus by Lemma 4(ii), we obtain

$$2n|{\gamma }_n|\le n|{\varphi }^{\prime }\left(0\right)|=n|B_1|,n\in \mathbb{N},$$

This implies the inequality in (8).

For the sharp bounds, it suffices to use the equality

$$z{\displaystyle \frac{d}{dz}}\left(log\left({\displaystyle \frac{f\left(z\right)}{z}}\right)\right)={\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}-1,$$

and so these results are sharp in inequalities (5), (6), and (8) such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}=\phi \left(z^n\right)$ and the function f given by ${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}=\phi \left(z\right)$, respectively. This completes the proof.  □

In the following corollaries, we obtain the logarithmic coefficients ${\gamma }_n$ for two subclasses ${\mathcal{S}}^{*}(\alpha +(1-\alpha )e^z)$ and ${\mathcal{S}}^{*}(\alpha +(1-\alpha )\sqrt{1+z})$, which were defined by Khatter et al. in [19], and $\alpha +(1-\alpha )e^z$ and $\alpha +(1-\alpha )\sqrt{1+z}$ are the convex univalent functions in $\mathbb{U}$. For $\alpha =0$, these results reduce to the logarithmic coefficients ${\gamma }_n$ for the subclasses ${\mathcal{S}}^{*}\left(e^z\right)$ and ${\mathcal{S}}^{*}\left(\sqrt{1+z}\right)$ (see [20,21]).

Corollary 1.

For $0\le \alpha <1$, let the function $f\in {\mathcal{S}}^{*}(\alpha +(1-\alpha )e^z)$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_n|\le {\displaystyle \frac{1-\alpha }{2n}},n\in \mathbb{N}$$

and

$$\sum _{n=1}^{\infty }{\left|{\gamma }_n\right|}^2\le {\displaystyle \frac{1}{4}}\sum _{n=1}^{\infty }{\displaystyle \frac{{(1-\alpha )}^2/{(n!)}^2}{n^2}}.$$

These results are sharp such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}=\alpha +(1-\alpha )e^{z^n}$ and the function f given by ${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}=\alpha +(1-\alpha )e^z$.

Corollary 2.

For $0\le \alpha <1$, let the function $f\in {\mathcal{S}}^{*}(\alpha +(1-\alpha )\sqrt{1+z})$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_n|\le {\displaystyle \frac{1-\alpha }{4n}},n\in \mathbb{N}$$

and

$$\sum _{n=1}^{\infty }{\left|{\gamma }_n\right|}^2\le {\displaystyle \frac{1}{4}}\sum _{n=1}^{\infty }{\displaystyle \frac{{\left((1-\alpha )\left(\genfrac{}{}{0pt}{}{\frac{1}{2}}{n}\right)\right)}^2}{n^2}}.$$

These results are sharp such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}=\alpha +(1-\alpha )\sqrt{1+z^n}$ and the function f given by ${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}=\alpha +(1-\alpha )\sqrt{1+z}$.

The following corollary concludes the logarithmic coefficients ${\gamma }_n$ for a subclass ${\mathcal{S}}^{*}(1+sinz)$ defined by Cho et al. in [22], in which considering the proof of Theorem 1 and Corollary 1, the convexity radius for $q_0\left(z\right)=1+sinz$ is given by $r_0\approx 0.345$.

Corollary 3.

Let the function $f\in {\mathcal{S}}^{*}(1+sinz)$ where $q_0\left(z\right)$ is a convex univalent function for $r_0\approx 0.345$ in $\mathbb{U}$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_n|\le {\displaystyle \frac{1}{2n}},n\in \mathbb{N}$$

and

$$\sum _{n=1}^{\infty }{\left|{\gamma }_n\right|}^2\le {\displaystyle \frac{1}{4}}\sum _{n=1}^{\infty }{\displaystyle \frac{1}{{((2n+1)!n)}^2}}.$$

These results are sharp such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}=q_0\left(z^n\right)$ and the function f given by ${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}=q_0\left(z\right)$.

In the following result, we get the logarithmic coefficients ${\gamma }_n$ for a subclass ${\mathcal{S}}^{*}\left(p_k\left(z\right)\right)$ defined by Kanas and Wisniowska in [23] (see also [24,25]), in which

$$p_k\left(z\right)=1+P_1\left(k\right)z+P_2\left(k\right)z^2+\dots ,$$

where $p_k\left(z\right)$ is a convex univalent function in $\mathbb{U}$ and

$$\begin{array}{c}\hfill P_1\left(k\right)=\left\{\begin{array}{ccc}\frac{2A^2}{1-k^2}\hfill & \mathrm{if}\hfill & 0\le k<1,\hfill \\ \frac{8}{{\pi }^2}\hfill & \mathrm{if}\hfill & k=1,\hfill \\ \frac{{\pi }^2}{4{\kappa }^2\left(t\right)(k^2-1)(1+t)\sqrt{t}}\hfill & \mathrm{if}\hfill & k>1.\hfill \end{array}\right.\end{array}$$

$A=\frac{2}{\pi }arccosk$ and $\kappa \left(t\right)$ is the complete elliptic integral of the first kind.

Corollary 4.

For $0\le k<\infty$, let the function $f\in {\mathcal{S}}^{*}\left(p_k\left(z\right)\right)$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_n|\le {\displaystyle \frac{P_1\left(k\right)}{2n}},n\in \mathbb{N}.$$

This result is sharp such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}=p_k\left(z^n\right)$.

The following result concludes the logarithmic coefficients ${\gamma }_n$ for a subclass ${\mathcal{S}}^{*}\left(\sqrt{2}-(\sqrt{2}-1)\sqrt{\frac{1-z}{1+2(\sqrt{2}-1)z}}\right)$ defined by Mendiratta et al. in [26], in which

$${\phi }_0\left(z\right)=\sqrt{2}-(\sqrt{2}-1)\sqrt{{\displaystyle \frac{1-z}{1+2(\sqrt{2}-1)z}}}=1+{\displaystyle \frac{5-3\sqrt{2}}{2}}z+{\displaystyle \frac{71-51\sqrt{2}}{8}}{z}^2+\dots ,$$

where ${\phi }_0$ is a convex univalent function in $\mathbb{U}$.

Corollary 5.

Let the function $f\in {\mathcal{S}}^{*}\left(\sqrt{2}-(\sqrt{2}-1)\sqrt{\frac{1-z}{1+2(\sqrt{2}-1)z}}\right)$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_n|\le {\displaystyle \frac{5-3\sqrt{2}}{4n}},n\in \mathbb{N}.$$

This result is sharp such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}={\phi }_0\left(z^n\right)$.

The following results conclude the logarithmic coefficients ${\gamma }_n$ for two subclasses ${\mathcal{S}}^{*}(z+\sqrt{1+z^2})$ and ${\mathcal{S}}^{*}(1+\frac{z}{(1-\alpha z^2)})$ defined by Krishna Raina and Sokół in [27] and Kargar et al. in [28], where

$$z+\sqrt{1+z^2}=1+z+\sum _{n=1}^{\infty }{\left(\genfrac{}{}{0pt}{}{\frac{1}{2}}{n}\right)}^2{z}^{2n},$$

and

$$1+{\displaystyle \frac{z}{(1-\alpha z^2)}}=1+z+\sum _{n=1}^{\infty }{\alpha }^n{z}^{2n+1},(0\le \alpha <1),$$

respectively. These functions are univalent and starlike with respect to 1 in $\mathbb{U}$.

Corollary 6.

Let the function $f\in {\mathcal{S}}^{*}(z+\sqrt{1+z^2})$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_n|\le {\displaystyle \frac{1}{2}},n\in \mathbb{N}.$$

This result is sharp such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}=z^n+\sqrt{1+z^{2n}}$.

Corollary 7.

Let the function $f\in {\mathcal{S}}^{*}(1+\frac{z}{(1-\alpha z^2)})$, where $0\le \alpha <1$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_n|\le {\displaystyle \frac{1}{2}},n\in \mathbb{N}.$$

This result is sharp such that for any $n\in \mathbb{N}$, there is the function $f_n$ given by ${\displaystyle \frac{z{f}_n^{\prime }\left(z\right)}{f_n\left(z\right)}}=1+{\displaystyle \frac{z}{(1-\alpha z^{2n})}}$.

Remark 1.

1. Letting

$$\begin{array}{cc}\hfill \phi \left(z\right)=& {\displaystyle \frac{1+Az}{1+Bz}}\hfill \\ \hfill =& 1+(A-B)z-B(A-B)z^2+B^2(A-B)z^3+\dots \hfill \\ \hfill =& 1+\left\{\begin{array}{ccc}{\displaystyle \frac{A-B}{B}}\sum _{n=1}^{\infty }{(-1)}^{n-1}{B}^n{z}^n,\hfill & \mathrm{if}\hfill & B\ne 0\hfill \\ Az,\hfill & \mathrm{if}\hfill & B=0,\hfill \end{array}\right.(-1\le B<A\le 1),\hfill \end{array}$$

which is convex univalent in $\mathbb{U}$ in Theorem 1, then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.1 and Corollary 2.3).

2. For $A=e^{i\alpha }(e^{i\alpha }-2\beta cos\alpha )$, where $\beta \in [0,1)$ and $\alpha \in (-\pi /2,\pi /2)$ in the above expression, then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.5).

3. Taking

$$\begin{array}{cc}\hfill \phi \left(z\right)={\left(\frac{1+z}{1-z}\right)}^{\alpha }=& 1+2\alpha z+2{\alpha }^2{z}^2+\frac{8{\alpha }^3+4\alpha }{6}{z}^3+\dots \hfill \\ \hfill =& 1+{\displaystyle \sum _{n=1}^{\infty }}{A}_n\left(\alpha \right)z^n,(0<\alpha \le 1),\hfill \end{array}$$

which is convex univalent in $\mathbb{U}$, and $A_n\left(\alpha \right)={\displaystyle \sum _{k=1}^n}\left(\genfrac{}{}{0pt}{}{n-1}{k-1}\right)\left(\genfrac{}{}{0pt}{}{\alpha }{k}\right)2^k$ in Theorem 1, then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.6).

4. Setting

$$\begin{array}{cc}\hfill \phi \left(z\right)=1+{\displaystyle \frac{\beta -\alpha }{\pi }}ilog\left(\frac{1-e^{2\pi i\frac{1-\alpha }{\beta -\alpha }z}}{1-z}\right)=& 1+\sum _{n=1}^{\infty }{C}_n{z}^n,(\alpha >1,\beta <1),\hfill \end{array}$$

which is convex univalent in $\mathbb{U}$, and $C_n={\displaystyle \frac{\beta -\alpha }{n\pi }}i(1-e^{2n\pi i\frac{1-\alpha }{\beta -\alpha }})$ in Theorem 1, then we get the results obtained by Kargar [5] (Theorems 2.2 and 2.3).

5. Letting

$$\begin{array}{cc}\hfill \phi \left(z\right)=1+{\displaystyle \frac{1}{2isin\delta }}log\left(\frac{1+z{e}^{i\delta }}{1+z{e}^{-i\delta }}\right)=& 1+\sum _{n=1}^{\infty }{D}_n{z}^n,(\pi /2\le \delta <\pi ),\hfill \end{array}$$

which is convex univalent in $\mathbb{U}$, and $D_n={\displaystyle \frac{{(-1)}^{n-1}sinn\delta }{nsin\delta }}$ in Theorem 1, then we get the results obtained by Kargar [5] (Theorems 2.5 and 2.6).

6. Letting

$$\begin{array}{cc}\hfill \phi \left(z\right)={\left({\displaystyle \frac{1+cz}{1-z}}\right)}^{({\alpha }_1+{\alpha }_2)/2}=& 1+\sum _{n=1}^{\infty }{\lambda }_n{z}^n,\left(0<{\alpha }_1,{\alpha }_2\le 1,c=e^{\pi i\theta },\theta =\frac{{\alpha }_2-{\alpha }_1}{{\alpha }_2+{\alpha }_1}\right),\hfill \end{array}$$

which is convex univalent in $\mathbb{U}$, and

$${\lambda }_n=\sum _{k=1}^n\left(\genfrac{}{}{0pt}{}{n-1}{k-1}\right)\left(\genfrac{}{}{0pt}{}{({\alpha }_1+{\alpha }_2)/2}{k}\right){(1+c)}^k$$

in Theorem 1, then we get the results obtained for $|{\gamma }_n|$ by Kargar et al. [29] (Theorem 3.1). Moreover, for ${\alpha }_1={\alpha }_2=\beta$, we get the result presented by Thomas in [30] (Theorem 1).

7. Let the function $f\in \mathcal{K}\left(1-{\displaystyle \frac{cz}{1-z}}\right)=\mathcal{K}(1-cz-c{z}^2-c{z}^3+\dots )$, where $c\in (0,1]$. It is equivalent to

$$Re\left(1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}\right)<1+{\displaystyle \frac{c}{2}}.$$

Then we have (see e.g., [31] (Theorem 1))

$${\displaystyle \frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}}\prec {\displaystyle \frac{(1+c)(1-z)}{1+c-z}},$$

where ${\displaystyle \frac{(1+c)(1-z)}{1+c-z}}$ is a convex univalent function in $\mathbb{U}$, and

$${\displaystyle \frac{(1+c)(1-z)}{1+c-z}}=1-{\displaystyle \frac{c}{c+1}}z-{\displaystyle \frac{c}{{(c+1)}^2}}{z}^2+\dots =1-c\sum _{n=1}^{\infty }{\displaystyle \frac{z^n}{{(1+c)}^n}}.$$

Thus, applying Theorem 1, we get the results obtained by Obradović et al. [4] (Theorem 2 and Corollary 2).

Theorem 2.

Let the function $f\in \mathcal{K}\left(\phi \right)$. Then the logarithmic coefficients of f satisfy the inequalities

$$|{\gamma }_1|\le {\displaystyle \frac{|B_1|}{4}},$$

(11)

$$\begin{array}{cc}\hfill |{\gamma }_2|\le & \left\{\begin{array}{ccc}{\displaystyle \frac{|B_1|}{12}}\hfill & \mathrm{if}\hfill & |4B_2+B_1^2|\le 4|B_1|\hfill \\ {\displaystyle \frac{|4B_2+B_1^2|}{48}}\hfill & \mathrm{if}\hfill & |4B_2+B_1^2|>4|B_1|\hfill \end{array},\right.\hfill \end{array}$$

(12)

and if $B_1,B_2$, and $B_3$ are real values,

$$\begin{array}{c}\hfill |{\gamma }_3|\le {\displaystyle \frac{|B_1|}{24}}H(q_1;q_2),\end{array}$$

(13)

where $H(q_1;q_2)$ is given by Lemma 5, $q_1=\frac{B_1+\frac{4B_2}{B_1}}{2}$, and $q_2=\frac{B_2+\frac{2B_3}{B_1}}{2}$. The bounds (11) and (12) are sharp.

Proof. 

Let $f\in \mathcal{K}\left(\phi \right)$. Then by the definition of the subordination, there is a $\omega \in \mathsf{\Omega }$ with $\omega \left(z\right)={\sum }_{n=1}^{\infty }{c}_n{z}^n$ so that

$$\begin{array}{cc}\hfill 1+\frac{z{f}^{″}\left(z\right)}{f^{\prime }\left(z\right)}=& \phi \left(\omega \right(z\left)\right)\hfill \\ \hfill =& 1+B_1{c}_{1}z+(B_1{c}_2+B_2{c}_1^2)z^2+(B_1{c}_3+2c_1{c}_2{B}_2+B_3{c}_1^3)z^3+\dots .\hfill \end{array}$$

From the above equation, we get that

$$\begin{array}{c}\hfill \left\{\begin{array}{c}2a_2=B_1{c}_1\hfill \\ 6a_3-4a_2^2=B_1{c}_2+B_2{c}_1^2\hfill \\ 12a_4-18a_2{a}_3+8a_2^3=B_1{c}_3+2c_1{c}_2{B}_2+B_3{c}_1^3.\hfill \end{array}\right.\end{array}$$

(14)

By substituting values $a_n$ (n = 1, 2, 3) from (14) in (3), we have

$$\begin{array}{c}\hfill \left\{\begin{array}{c}2{\gamma }_1={\displaystyle \frac{B_1{c}_1}{2}}\hfill \\ 2{\gamma }_2={\displaystyle \frac{8B_1{c}_2+(8B_2+2B_1^2)c_1^2}{48}}\hfill \\ 2{\gamma }_3={\displaystyle \frac{B_1}{12}}[c_3+{\displaystyle \frac{B_1+{\displaystyle \frac{4B_2}{B_1}}}{2}}{c}_1{c}_2+{\displaystyle \frac{B_2+{\displaystyle \frac{2B_3}{B_1}}}{2}}{c}_1^3].\hfill \end{array}\right.\end{array}$$

Firstly, for ${\gamma }_1$, by applying Lemma 1 we get $|{\gamma }_1|\le {\displaystyle \frac{|B_1|}{4}}$, and this bound is sharp for $|c_1|=1$. Next, applying Lemma 1 for ${\gamma }_2$, we have

$$\begin{array}{cc}\hfill |{\gamma }_2|\le & {\displaystyle \frac{4|B_1\left|\right(1-|c_1{|}^2)+|4B_2+B_1^2\left|\right|c_1{|}^2}{48}}\hfill \\ \hfill =& {\displaystyle \frac{4|B_1|+\left[|4B_2+B_1^2|-4|B_1|\right]{\left|c_1\right|}^2}{48}}\hfill \\ \hfill \le & \left\{\begin{array}{ccc}{\displaystyle \frac{4|B_1|}{48}}\hfill & \mathrm{if}\hfill & |4B_2+B_1^2|\le 4|B_1|\hfill \\ {\displaystyle \frac{|4B_2+B_1^2|}{48}}\hfill & \mathrm{if}\hfill & |4B_2+B_1^2|>4|B_1|.\hfill \end{array}\right.\hfill \end{array}$$

These bounds are sharp for $c_1=0$ and $|c_1|=1$, respectively.

Finally, using Lemma 5 for ${\gamma }_3$, we obtain

$$\begin{array}{cc}\hfill 2|{\gamma }_3|\le & {\displaystyle \frac{|B_1|}{12}}|c_3+{\displaystyle \frac{B_1+{\displaystyle \frac{4B_2}{B_1}}}{2}}{c}_1{c}_2+{\displaystyle \frac{B_2+{\displaystyle \frac{2B_3}{B_1}}}{2}}{c}_1^3|\le H(q_1;q_2),\hfill \end{array}$$

where $q_1=\frac{B_1+\frac{4B_2}{B_1}}{2}$ and $q_2=\frac{B_2+\frac{2B_3}{B_1}}{2}$. Therefore, this completes the proof.  □

Remark 2.

1. Letting

$$\begin{array}{cc}\hfill \phi \left(z\right)=& 1+\frac{cz}{1-z}\hfill \\ \hfill =& 1+cz+c{z}^2+c{z}^3+\dots (c\in (0,3])\hfill \end{array}$$

in Theorem 2, (for $|{\gamma }_3|$ with respect to $D_6$) then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.7 and Corollary 2.8).

2. Taking

$$\begin{array}{cc}\hfill \phi \left(z\right)=& 1-\frac{cz}{1-z}\hfill \\ \hfill =& 1-cz-c{z}^2-c{z}^3+\dots (c\in (0,1])\hfill \end{array}$$

in Theorem 2, (for $|{\gamma }_3|$ respect to $D_2$) then we get the results obtained by Ponnusamy et al. [7] (Theorem 2.10).