Initial logarithmic coefficients for functions starlike with respect to symmetric points

Abstract

In this paper, we obtain the bounds of the initial logarithmic coefficients for functions in the classes ${\mathcal {S}}_S^*$ and ${\mathcal {K}}_S$ of functions which are starlike with respect to symmetric points and convex with respect to symmetric points, respectively. In our research, we use a different approach than the usual one in which the coeffcients of f are expressed by the corresponding coeffcients of functions with positive real part. In what follows, we express the coeffcients of f in ${\mathcal {S}}_S^*$ and ${\mathcal {K}}_S$ by the corresponding coeffcients of Schwarz functions. In the proofs, we apply some inequalities for these functions obtained by Prokhorov and Szynal, by Carlson and by Efraimidis. This approach offers a additional benefit. In many cases, it is easily possible to predict the exact result and to select extremal functions. It is the case for ${\mathcal {S}}_S^*$ and ${\mathcal {K}}_S$.

1 Introduction

Let ${{\mathbb {D}}}$ be the unit disk $\{z\in {\mathbb {C}}:|z|<1\}$ and ${\mathcal {A}}$ be the family of all functions f analytic in ${{\mathbb {D}}}$, normalized by the condition $f(0)=f'(0)-1=0$. It means that f has the expansion

$$\begin{aligned} f(z)=z+\sum _{n=2}^{\infty } a_n z^n\ . \end{aligned}$$

(1.1)

Let ${\mathcal {S}}$ be the class of all functions in ${\mathcal {A}}$ which are univalent and ${\mathcal {S}}^*$ be the subset of ${\mathcal {S}}$ consisting of those functions which are starlike in ${{\mathbb {D}}}$. The logarithmic coefficients of $f\in {\mathcal {S}}$, denoted by $\gamma _n=\gamma _n(f)$, are defined by

$$\begin{aligned} \tfrac{1}{2}\log \frac{f(z)}{z} = \sum _{n=1}^\infty \gamma _n z^n\ . \end{aligned}$$

(1.2)

If f is given by (1.1), then its logarithmic coefficients are given by

$$\begin{aligned} \gamma _1= & {} \tfrac{1}{2} a_2 \nonumber \\ \gamma _2= & {} \tfrac{1}{2} \left( a_3 - \tfrac{1}{2} a_2^2\right) \nonumber \\ \gamma _3= & {} \tfrac{1}{2} \left( a_4 - a_2a_3 + \tfrac{1}{3} a_2^3\right) \nonumber \\ \gamma _4= & {} \tfrac{1}{2} \left( a_5 - a_2a_4 + a_2^2a_3 - \tfrac{1}{2} a_3^2 - \tfrac{1}{4} a_2^4\right) \nonumber \\ \gamma _5= & {} \tfrac{1}{2} \left( a_6 - a_2a_5 - a_3a_4 + a_2a_3^2 + a_2^2a_4 - a_2^3a_3 + \tfrac{1}{5} a_2^5\right) \ . \end{aligned}$$

(1.3)

It is known that for the Koebe function $f(z)=\frac{z}{(1-z)^2}$ there is $\gamma _n=\tfrac{1}{n}$ for each positive integer n. Since the Koebe function appears as an extremal function in many problems of geometric theory of analytic functions, one could expect that $|\gamma _n|\le \tfrac{1}{n}$ for each $f\in {\mathcal {S}}$ and $n\in {\mathbb {N}}$. It is not true even for $\gamma _2$. It is enough to use the Fekete-Szegö inequality

$$\begin{aligned} |a_3-\mu a_2^2| \le 1+2e^{-\frac{2\mu }{1-\mu }} \end{aligned}$$

which holds for all $f\in {\mathcal {S}}$ and $0\le \mu <1$. Consequently, the following sharp bound $|\gamma _2|\le \tfrac{1}{2} (1+2e^{-2}) = 0.635\ldots$ is valid for ${\mathcal {S}}$. A little is known about succeeding logarithmic coefficients of univalent functions. In very recent paper [5], Obradović and Tuneski proved that $|\gamma _3|\le \tfrac{\sqrt{133}}{15}$ for all $f\in {\mathcal {S}}$.

Girela in [3] shown that also in the class ${\mathcal {C}}$ of close-to-convex functions there exist functions such that their logarithmic coefficients are greater than $\tfrac{1}{n}$. The similar fact, but for the class ${\mathcal {U}}$ of univalent functions satisfying the condition

$$\begin{aligned} \left| \left( \frac{z}{f(z)}\right) ^2 f'(z)-1\right| < 1\ , \end{aligned}$$

was proved by Obradović et al. in [4]. On the other hand, the inequality $|\gamma _n|\le \tfrac{1}{n}$ holds for each $f\in {\mathcal {S}}^*$. For a summary of some of the significant results concerning the logarithmic coefficients for univalent functions, see [8].

In this paper, we consider two subclasses of ${\mathcal {S}}$: the class ${\mathcal {S}}_S^*$ of functions starlike with respect to the symmetric points and the relative class ${\mathcal {K}}_S$ of functions convex with respect to the symmetric points. The definitions are as follows

$$\begin{aligned} {\mathcal {S}}_S^*= \left\{ f\in {\mathcal {S}}: Re\left\{ \frac{2zf'(z)}{f(z)-f(-z)} \right\} >0, \quad z \in \varDelta \right\} \end{aligned}$$

and

$$\begin{aligned} {\mathcal {K}}_S = \left\{ f\in {\mathcal {S}}: Re\left\{ \frac{2\left[ zf'(z)\right] '}{\left[ f(z)-f(-z)\right] '} \right\} >0, \quad z \in \varDelta \right\} \ . \end{aligned}$$

It is known (see, [7]) that ${\mathcal {K}}\subset {\mathcal {S}}_S^*$ and ${\mathcal {S}}^{*^{(2)}}\subset {\mathcal {S}}_S^*$, where ${\mathcal {S}}^{*^{(2)}}$ stands for the class of odd starlike functions. On the other hand, ${\mathcal {S}}_S^*\subset {\mathcal {C}}$.

Our aim is to derive the bounds of the initial logarithmic coefficients for functions in the both classes defined above. It is worth observing that the Koebe function does not belong to ${\mathcal {S}}_S^*$, so it cannot play a role of extremal function for ${\mathcal {S}}_S^*$.

In our research, we use a different approach than the usual one in which the coeffcients of f are expressed by the corresponding coeffcients of functions with positive real part. In what follows, we express the coeffcients of f in ${\mathcal {S}}_S^*$ and ${\mathcal {K}}_S$ by the corresponding coeffcients of Schwarz functions. This makes the calculation easier. Additionally, this approach offers a valuable benefit. In many cases, it is easily possible to predict the exact result and to select extremal functions. It is the case for ${\mathcal {S}}_S^*$ and ${\mathcal {K}}_S$.

Let ${\mathcal {B}}_0$ be the class of Schwarz functions, i.e., analytic functions $\omega :{{\mathbb {D}}}\rightarrow {{\mathbb {D}}}$, $\omega (0)=0$. A function $\omega \in {\mathcal {B}}_0$ can be written as a power series

$$\begin{aligned} \omega (z)=\sum _{n=1}^\infty c_nz^n\ . \end{aligned}$$

(1.4)

To prove our results, we need the following lemmas for Schwarz functions obtained by Prokhorov and Szynal [6], by Carlson ([1]) and by Efraimidis [2].

Lemma 1

Let $\omega (z)=c_{1}z+c_{2}z^{2}+\cdots$ be a Schwarz function. Then, for any real numbers $\mu$ and $\nu$ such that

$$\begin{aligned} (\mu ,\nu ) \in \left\{ \frac{1}{2}\le |\mu |\le 2\ ,\ \frac{4}{27} (|\mu |+1)^3-(|\mu |+1)\le \nu \le 1 \right\} \ , \end{aligned}$$

the following sharp estimate holds

$$\begin{aligned} \left| c_{3}+\mu c_{1}c_{2}+\nu c_{1}^{3}\right| \le 1. \end{aligned}$$

Lemma 2

Let $\omega (z)=c_{1}z+c_{2}z^{2}+\cdots$ be a Schwarz function. Then

$$\begin{aligned} |c_2|\le 1-|c_1|^2 \quad ,\quad |c_3|\le 1-|c_1|^2 -\tfrac{|c_2|^2}{1+|c_1|} \quad ,\quad |c_4|\le 1-|c_1|^2 -|c_2|^2\ . \end{aligned}$$

Lemma 3

Let $\omega (z)=c_{1}z+c_{2}z^{2}+\cdots$ be a Schwarz function. Then

$$\begin{aligned} |c_4+2c_1c_3+c_2^2+3c_1^2c_2+c_1^4|\le 1\ . \end{aligned}$$

Lemma 3 is a particular case of more general theorem which is needed in our proofs. For p in ${\mathcal {P}}$, the class of analytic functions p such that $Rep(z)>0$ and $p(0)=1$, and for $w\in {\mathbb {C}}$, Efraimidis in [2] defined the determinant

$$\begin{aligned} A_{k,n}(w) = \left| \begin{array}{cccccc} p_{n+k} &{} p_{n+k-1} &{} p_{n+k-2} &{} \ldots &{} p_{n+1} &{} p_{n} \\ w p_{1} &{} 1 &{} 0 &{} \ldots &{} 0 &{} 0 \\ w p_{2} &{} w p_{1} &{} 1 &{} \ldots &{} 0 &{} 0 \\ \ddots &{} \ddots &{} \ddots &{} \vdots &{} \ddots &{} \ddots \\ w p_{k-1} &{} w p_{k-2} &{} w p_{k-3} &{} \ldots &{} 1 &{} 0 \\ w p_{k} &{} w p_{k-1} &{} w p_{k-2} &{} \ldots &{} w p_{1} &{} 1 \end{array} \right| \end{aligned}$$

(1.5)

and proved the following theorem.

Theorem 1

If $p\in {\mathcal {P}}$ and $w\in {\mathbb {C}}$, then

$$\begin{aligned} |A_{k,n}(w)| \le 2\max \{1, |1-2w|^k\} \end{aligned}$$

for all integers $k\ge 0$ and $n\ge 1$.

Applying the correspondence between $p\in {\mathcal {P}}$ and $\omega \in {\mathcal {B}}_0$,

$$\begin{aligned} p(z) = \frac{1+\omega (z)}{1-\omega (z)} \ , \end{aligned}$$

(1.6)

it is possible to obtain the analogous theorem for Schwarz functions. As a corollary, putting $w=0$ and $k=3$, $n=1$, Lemma 3 follows.

Consider now the case $k+n=5$ in Theorem 1. Formula (1.5) and (1.6) result in

Corollary 1

If $\omega \in {\mathcal {B}}_0$ is of the form (1.4) and $\mu \in {\mathbb {C}}$, $|\mu |\le 1$, then

$$\begin{aligned}&\vert c_5 + 2\mu c_1c_4 + 2\mu c_2c_3 + 3\mu ^2 c_1c_2^2 + 3\mu ^2 c_1^2c_3 + 4\mu ^3 c_1^3c_2 + \mu ^4 c_1^5\vert \le 1 \ , \end{aligned}$$

(1.7)

$$\begin{aligned}&\vert c_5 + (1+\mu ) c_1c_4 + 2\mu c_2c_3 + \mu (1+2\mu ) c_1c_2^2 + \mu (2+\mu ) c_1^2c_3 \nonumber \\&\quad + \mu ^2(3+\mu ) c_1^3c_2 + \mu ^3 c_1^5\vert \le 1 \ , \end{aligned}$$

(1.8)

$$\begin{aligned}&\vert c_5 + (1+\mu ) c_1c_4 + (1+\mu ) c_2c_3 + 3\mu c_1c_2^2 + (1+\mu +\mu ^2) c_1^2c_3 \nonumber \\&\quad + 2\mu (1+\mu ) c_1^3c_2 + \mu ^2 c_1^5\vert \le 1 \ . \end{aligned}$$

(1.9)

Combining Formulae (1.7–1.9) for suitably chosen $\mu$ we can get other relative inequalities; some of them will be useful in proving theorems from the two next sections.

Taking $\mu =0$ and $\mu =-1$ in (1.9), we have

$$\begin{aligned} \left| c_5 + c_1c_4 + c_2c_3 + c_1^2c_3\right| \le 1 \end{aligned}$$

(1.10)

and

$$\begin{aligned} \left| c_5 - 3c_1c_2^2 + c_1^2c_3 + c_1^5\right| \le 1 \ . \end{aligned}$$

(1.11)

From (1.7) with $\mu =1$ and $\mu =-1$, it follows that

$$\begin{aligned} \left| c_5 + 3c_1c_2^2 + 3c_1^2c_3 + c_1^5\right| \le 1 \ . \end{aligned}$$

(1.12)

Finally, (1.11) and (1.12) results in

$$\begin{aligned} \left| c_5 + 2c_1^2c_3 + c_1^5\right| \le 1 \ . \end{aligned}$$

(1.13)

2 Logarithmic coefficients for functions in ${\mathcal {S}}_S^*$

Although the first three results in the theorem below are easy to obtain in other way, for completeness of results, we prove all of them in a uniform way.

Theorem 2

If $f\in {\mathcal {S}}_S^*$, then

$$\begin{aligned} |\gamma _1|\le \tfrac{1}{2} \quad ,\quad |\gamma _2|\le \tfrac{1}{2} \quad ,\quad |\gamma _3|\le \tfrac{1}{4} \quad ,\quad |\gamma _4|\le \tfrac{1}{4} \quad ,\quad |\gamma _5|\le \tfrac{1}{6}\ . \end{aligned}$$

All bounds are sharp.

Proof

If $f\in {\mathcal {S}}_S^*$, then

$$\begin{aligned} \frac{2zf'(z)}{f(z)-f(-z)} = \frac{1+\omega (z)}{1-\omega (z)}\ , \end{aligned}$$

where $\omega \in {\mathcal {B}}_0$. Using (1.1) and (1.4) and comparing coefficients at powers of z in

$$\begin{aligned} 2zf'(z)\left( 1-\omega (z)\right) = \tfrac{1}{2}\left[ f(z)-f(-z)\right] \left( 1+\omega (z)\right) \end{aligned}$$

we get

$$\begin{aligned} a_2= & {} c_1 \nonumber \\ a_3= & {} c_2 + c_1^2 \nonumber \\ a_4= & {} \tfrac{1}{2} \left( c_3 + 3c_1c_2 + 2c_1^3\right) \nonumber \\ a_5= & {} \tfrac{1}{2} \left( c_4 + 2c_1c_3 + 2c_2^2 + 5c_1^2c_2 + 2c_1^4\right) \nonumber \\ a_6= & {} \tfrac{1}{3} \left( c_5 + \tfrac{5}{2} c_1c_4 + 3c_2c_3 + 6c_1c_2^2 + 5c_1^2c_3 + \tfrac{19}{2}c_1^3c_2 + 3c_1^5\right) \ . \end{aligned}$$

(2.1)

Applying (2.1) in (1.3), we obtain

$$\begin{aligned} \gamma _1= & {} \tfrac{1}{2} c_1 \nonumber \\ \gamma _2= & {} \tfrac{1}{2} \left( c_2 + \tfrac{1}{2} c_1^2\right) \nonumber \\ \gamma _3= & {} \tfrac{1}{4} \left( c_3 + c_1c_2 + \tfrac{2}{3} c_1^3\right) \nonumber \\ \gamma _4= & {} \tfrac{1}{4} \left( c_4 + c_1c_3 + c_2^2 + 2c_1^2c_2 + \tfrac{1}{2} c_1^4\right) \nonumber \\ \gamma _5= & {} \tfrac{1}{6} \left( c_5 + c_1c_4 + \tfrac{3}{2} c_2c_3 + \tfrac{3}{2} c_1c_2^2 + 2c_1^2c_3 + 2c_1^3c_2 + \tfrac{3}{5} c_1^5\right) \end{aligned}$$

(2.2)

Since $|c_1|\le 1$ and $|c_2|\le 1-|c_1|^2$, the bounds of $\gamma _1$ and $\gamma _2$ are obvious. Taking $\mu =1$ and $\nu =2/3$ in Lemma 1, the bound of $\gamma _3$ follows.

To derive the bound of $\gamma _4$, we can write

$$\begin{aligned} \gamma _4 = \tfrac{1}{8} \left[ \left( c_4 + 2c_1c_3 + c_2^2 + 3c_1^2c_2 + c_1^4\right) +\left( c_4+c_2^2+c_1^2c_2\right) \right] \ . \end{aligned}$$

(2.3)

By Lemma 3, the first expression in square brackets is bounded by 1. Now, it is enough to apply Lemma 2 together with the triangle inequality to obtain

$$\begin{aligned} \left| c_4+c_2^2+c_1^2c_2\right| \le (1-|c_1|^2-|c_2|^2) + |c_2|^2 + |c_1|^2(1-|c_1|^2) = 1-|c_1|^4\le 1\ . \end{aligned}$$

In this way, $|\gamma _4|\le \tfrac{1}{4}$.

The coefficient $\gamma _5$ can be rewritten as follows

$$\begin{aligned} \gamma _5= & {} \tfrac{1}{12} \left[ \left( c_5 + 2c_1c_4 + 2c_2c_3 + 3c_1c_2^2 + 3c_1^2c_3 + 4c_1^3c_2 + c_1^5\right) \right. \\&+ \left. \left( c_5 + c_2c_3 + c_1^2c_3 + \tfrac{1}{5} c_1^5\right) \right] \ . \end{aligned}$$

The first component in the square brackets is bounded by 1. It is a simple consequence of Formula (1.7) with $\mu =1$.

The second component is equal to $W_1 + W_2$, where

$$\begin{aligned} W_1 = \tfrac{1}{5} \left( c_5+2c_1^2c_3+c_1^5\right) \quad \text {and}\quad W_2 = \tfrac{4}{5} \left( c_5+\tfrac{3}{4} c_1^2c_3+\tfrac{5}{4} c_2c_3\right) \ . \end{aligned}$$

From (1.13),

$$\begin{aligned} |W_1| \le \tfrac{1}{5}\ . \end{aligned}$$

(2.4)

We shall estimate $W_2$ using Lemma 2. Namely,

$$\begin{aligned} |W_2| = \tfrac{4}{5} \left| c_5+\tfrac{3}{4} c_1^2c_3+\tfrac{5}{4} c_2c_3\right| \le \tfrac{4}{5} H(x,y,z)\ , \end{aligned}$$

(2.5)

where

$$\begin{aligned} H(x,y,z) = 1-x^2-y^2-\frac{z^2}{1+x} +\tfrac{3}{4} x^2z+\tfrac{5}{4} yz \end{aligned}$$

and $x=|c_1|$, $y=|c_2|$, $z=|c_3|$. Clearly, all three variables x, y and z are in [0, 1].

The function H as a quadratic function of a variable y achieves its greatest value when $y=\tfrac{5}{8} z$. Consequently,

$$\begin{aligned} H(x,y,z) \le H(x,\tfrac{5}{8} z,z) = 1-x^2+\tfrac{25}{64} z^2 -\frac{z^2}{1+x}+\tfrac{3}{4} x^2z = 1 - g(x,z)\ , \end{aligned}$$

with

$$\begin{aligned} g(x,z) = \left( \tfrac{1}{1+x}-\tfrac{25}{64}\right) z^2 + x^2\left( 1-\tfrac{3}{4} z\right) \ . \end{aligned}$$

Since $g(x,z)\ge 0$ for $x\in [0,1]$ and $z\in [0,1]$, so

$$\begin{aligned} H(x,y,z) \le 1\ . \end{aligned}$$

(2.6)

Taking into account (2.4–2.6),

$$\begin{aligned} |W_1 + W_2| \le 1\ , \end{aligned}$$

and consequently,

$$\begin{aligned} |\gamma _5| \le \tfrac{1}{6}\ . \end{aligned}$$

For the sharpness of the results, it is enough to observe that taking $\omega (z)=z^k$ leads to equalities in each inequality in Theorem 2. $\square$

It is worth to write the extremal functions $f\in {\mathcal {S}}_S^*$ and the corresponding functions

$$\begin{aligned} F(z)=\tfrac{1}{2}\log \left( \frac{f(z)}{z}\right) \end{aligned}$$

(2.7)

explicitly. This is easy for even logarithmic coefficients. One can check that

$$\begin{aligned} f_{2n}(z)=\frac{z}{\root n \of {1-z^{2n}}}\in {\mathcal {S}}_S^*\ . \end{aligned}$$

(2.8)

Indeed, this function is generated from the odd starlike function $f(z)=\frac{z}{1-z^2}$ under the n-th root transformation. Since this transformation preserves starlikeness, we can see that $f_{2n}$ given by (2.8) is in ${\mathcal {S}}^{*^{(2)}}$. Hence, it is in ${\mathcal {S}}_S^*$. For $f_{2n}$, we derive

$$\begin{aligned} F_{2n}(z) = -\tfrac{1}{2n}\log (1-z^{2n}) = \sum _{j=1}^\infty \tfrac{1}{2jn} z^{2jn} = \tfrac{1}{2n} z^{2n} + \tfrac{1}{4n} z^{4n} + \tfrac{1}{6n} z^{6n} + \ldots \ . \end{aligned}$$

For $n=1$, the extremal function is $f_1(z)=\frac{z}{1-z}$; for this function

$$\begin{aligned} F_1(z)=-\tfrac{1}{2} \log (1-z) = \tfrac{1}{2} z + \tfrac{1}{4} z^2 + \tfrac{1}{6} z^3+\ldots \ . \end{aligned}$$

Consider now other positive odd integers and define a function

$$\begin{aligned} f_{2n+1}(z) = \frac{z(1+z^{2n+1})}{\root 2n+1 \of {1-z^{4n+2}}} - \tfrac{n}{n+1}z^{2n+2} {}_2F_1\left( \tfrac{1}{2n+1},\tfrac{n+1}{2n+1};1+\tfrac{1}{2n+1};z^{4n+2}\right) \ . \end{aligned}$$

(2.9)

Observe that $f_{2n+1}$ is a difference of the two components. Denote them by g and h, respectively. Clearly,

$$\begin{aligned} \tfrac{1}{2}\left[ g(z)-g(-z)\right] = \frac{z}{\root 2n+1 \of {1-z^{4n+2}}}\quad \text {and}\quad \tfrac{1}{2}\left[ h(z)-h(-z)\right] = 0\ . \end{aligned}$$

(2.10)

Moreover,

$$\begin{aligned} zg'(z) = \frac{z}{(1-z^{2n+1})\root 2n+1 \of {1-z^{4n+2}}}\cdot \left[ 1+z^{2n+1}+2nz^{2n+1} \left( 1-z^{2n+1}\right) \right] \ ,\end{aligned}$$

so

$$\begin{aligned} \frac{2zg'(z)}{g(z)-g(-z)} = \frac{1+z^{2n+1}}{1-z^{2n+1}} + 2nz^{2n+1} \ . \end{aligned}$$

(2.11)

On the other hand, since

$$\begin{aligned} h(z)= & {} \tfrac{n}{n+1} \sum _{k=0}^\infty \frac{\left( \tfrac{1}{2n+1}\right) _k \left( \tfrac{n+1}{2n+1}\right) _k z^{(4n+2)k+2n+2}}{k! \left( 1+\tfrac{1}{2n+1}\right) _k} \\= & {} \tfrac{n}{n+1} \sum _{k=0}^\infty \frac{\left( \tfrac{1}{2n+1}\right) _k z^{(4n+2)k+2n+2}}{k! \tfrac{2n+1}{n+1} \left( k+\tfrac{n+1}{2n+1}\right) } \\= & {} n \sum _{k=0}^\infty \frac{\left( \tfrac{1}{2n+1}\right) _k z^{(4n+2)k+2n+2}}{k! \left[ (2n+1)k+n+1\right] }\ , \end{aligned}$$

we have

$$\begin{aligned} h'(z) = 2n \sum _{k=0}^\infty \frac{1}{k!} \left( \tfrac{1}{2n+1}\right) _k z^{(4n+2)k+2n+1}\ . \end{aligned}$$

(2.12)

Combining (2.10-2.12) leads to

$$\begin{aligned}&\frac{2zf_{2n+1}'(z)}{f_{2n+1}(z)-f_{2n+1}(-z)} = \frac{2zg'(z)-2zh'(z)}{g(z)-g(-z)} \\&\quad = \frac{1+z^{2n+1}}{1-z^{2n+1}} + 2nz^{2n+1} - 2n \frac{\sum _{k=0}^\infty \frac{1}{k!} \left( \tfrac{1}{2n+1}\right) _k z^{(4n+2)k+2n+1}}{\left( 1-z^{4n+2}\right) ^{-\tfrac{1}{2n+1}}} \\&\quad = \frac{1+z^{2n+1}}{1-z^{2n+1}} + \frac{2nz^{2n+1}}{\left( 1-z^{4n+2}\right) ^{-\tfrac{1}{2n+1}}} \left[ \sum _{k=0}^\infty \frac{1}{k!} \left( 1-\tfrac{1}{2n+1} -k\right) _k \left( -z^{(4n+2)}\right) ^k \right. \\&\qquad - \left. \sum _{k=0}^\infty \frac{1}{k!} \left( \tfrac{1}{2n+1}\right) _k z^{(4n+2)k}\right] \\&\quad = \frac{1+z^{2n+1}}{1-z^{2n+1}} + 2nz^{2n+1} \root 2n+1 \of {1-z^{4n+2}} \\&\qquad \times \sum _{k=0}^\infty \frac{1}{k!} \left[ (-1)^k \left( 1-\tfrac{1}{2n+1}-k\right) _k - \left( \tfrac{1}{2n+1}\right) _k \right] z^{(4n+2)k} \end{aligned}$$

But

$$\begin{aligned} (-1)^k\left( 1-\tfrac{1}{2n+1}-k\right) _k = \left( \tfrac{1}{2n+1}\right) _k\ , \end{aligned}$$

so it results in

$$\begin{aligned} \frac{2zf_{2n+1}'(z)}{f_{2n+1}(z)-f_{2n+1}(-z)} = \frac{1+z^{2n+1}}{1-z^{2n+1}}\ . \end{aligned}$$

This means that $f_{2n+1}$ belongs to the class ${\mathcal {S}}_S^*$.

From (2.9), we conclude that $f_{2n+1}$ is $(2n+1)$-fold symmetric function, i.e. $f_{2n+1}(\varepsilon z)=\varepsilon f_{2n+1}(z)$ with $\varepsilon =\exp (2\pi /(2n+1))$ being a root of order $2n+1$ of unity, and

$$\begin{aligned} f_{2n+1}(z) = z+\tfrac{1}{n+1}z^{2n+2}+\ldots \ . \end{aligned}$$

Hence, the corresponding function $F_{2n+1}$ is also $(2n+1)$-fold symmetric and

$$\begin{aligned} F_{2n+1}(z) = 1+\tfrac{1}{2(n+1)}z^{2n+1}+\ldots \ . \end{aligned}$$

The above proves that the third logarithmic coefficient is equal to 1/4 for $f_3$ and the fifth logarithmic coefficient is equal to 1/6 for $f_5$. The natural conjecture is following

$$\begin{aligned} |\gamma _{2n}|\le \tfrac{1}{2n} \quad \text {and}\quad |\gamma _{2n-1}|\le \tfrac{1}{2n} \end{aligned}$$

(2.13)

for all positive integers n with extremal functions of the form (2.8) and (2.9) depending on parity of n.

3 Logarithmic coefficients for functions in ${\mathcal {K}}_S$

Theorem 3

If $f\in {\mathcal {K}}_S$, then

$$\begin{aligned} |\gamma _1|\le \tfrac{1}{4} \quad ,\quad |\gamma _2|\le \tfrac{1}{6} \quad ,\quad |\gamma _3|\le \tfrac{1}{16} \quad ,\quad |\gamma _4|\le \tfrac{13}{180} \quad ,\quad |\gamma _5|\le \tfrac{19}{576}+0.0017...\ . \end{aligned}$$

The first four bounds are sharp.

Proof

Applying the Alexander relation, (2.1) and (1.3) we obtain

$$\begin{aligned} \gamma _1= & {} \tfrac{1}{4} c_1 \nonumber \\ \gamma _2= & {} \tfrac{1}{6} \left( c_2 + \tfrac{5}{8} c_1^2\right) \nonumber \\ \gamma _3= & {} \tfrac{1}{16} \left( c_3 + \tfrac{5}{3} c_1c_2 + c_1^3\right) \nonumber \\ \gamma _4= & {} \tfrac{1}{20} \left( c_4 + \tfrac{11}{8}c_1c_3 + \tfrac{13}{9}c_2^2 + \tfrac{205}{72}c_1^2c_2 + \tfrac{251}{288} c_1^4\right) \nonumber \\ \gamma _5= & {} \tfrac{1}{36} \left( c_5 + \tfrac{8}{5}c_1c_4 + \tfrac{9}{4} c_2c_3 + \tfrac{59}{20} c_1c_2^2 + \tfrac{241}{80}c_1^2c_3 + \tfrac{67}{16}c_1^3c_2 + \tfrac{19}{16} c_1^5\right) \end{aligned}$$

(3.1)

The first two bounds of $\gamma _1$ and $\gamma _2$ are clear. By Lemma 1 with $\mu =5/3$ and $\nu =1$, there is $|\gamma _3|\le 1/16$.

Now, we write $\gamma _4$ as follows

$$\begin{aligned} \gamma _4= & {} \tfrac{1}{5760} \left[ 198\left( c_4 + 2c_1c_3 + c_2^2 + 3c_1^2c_2 + c_1^4\right) \right. \\&\qquad + \left. \left( 90c_4+218c_2^2+226c_1^2c_2+53c_1^4\right) \right] \ . \end{aligned}$$

By Lemma 3, $|c_4 + 2c_1c_3 + c_2^2 + 3c_1^2c_2 + c_1^4|\le 1$. From Lemma 2 and the triangle inequality, we obtain

$$\begin{aligned}&\left| 90c_4 \right. + \left. 218c_2^2+226c_1^2c_2+53c_1^4\right| \\&\quad \le 90(1-|c_1|^2-|c_2|^2) + 218|c_2|^2 + 226|c_1|^2|c_2| + 53|c_1|^4 \\&\quad = 90 - 90|c_1|^2 + 128|c_2|^2 + 226|c_1|^2|c_2| + 53|c_1|^4 \\&\quad \le 90 - 90|c_1|^2 + 128(1-|c_1|^2)^2 + 226|c_1|^2(1-|c_1|^2) + 53|c_1|^4 \\&\quad = 218 - 120|c_1|^2-45|c_1|^4 \le 218\ . \end{aligned}$$

Consequently,

$$\begin{aligned} |\gamma _4| \le \tfrac{1}{5760} \left( 198 + 218\right) = \tfrac{13}{180}\ . \end{aligned}$$

The coefficient $\gamma _5$ can be rewritten as

$$\begin{aligned} \gamma _5= & {} \tfrac{1}{2880} \left[ 80\left( c_5 + 2c_1c_4 + 2c_2c_3 + 3c_1c_2^2 + 3c_1^2c_3 + 4c_1^3c_2 + c_1^5\right) \right. \nonumber \\&\qquad + \left. \left( -32c_1c_4 + 20c_2c_3 + 15c_1^3c_2 + c_1^2c_3 -4c_1c_2^2+ 15 c_1^5\right) \right] \ . \end{aligned}$$

(3.2)

From Formula (1.7) with $\mu =1$, the first component in the square brackets is bounded by 80. Let us denote the second component by V.

Applying the inequalities from Lemma 2 and writing $x=|c_1|$, $y=|c_2|$, $x,y\in [0,1]$, the expression V can be estimated as follows

$$\begin{aligned} |V|\le & {} 32x(1-x^2-y^2)+(20y+x^2)\left( 1-x^2-\tfrac{y^2}{1+x}\right) +15x^3y+4xy^2+15x^5 \\\le & {} 32x(1-x^2-y^2)+(20y+x^2)(1-x^2)+15x^3y+4xy^2+15x^5\ . \end{aligned}$$

Assume that

$$\begin{aligned} G(x,y) = -28xy^2+5(4-4x^2+3x^3)y+x(32+x)(1-x^2)+15x^5; \end{aligned}$$

hence

$$\begin{aligned} |V| \le G(x,y)\ . \end{aligned}$$

In a view of Lemma 2, the region of variability of a pair $(|c_1|,|c_2|)$ coincides with a set

$$\begin{aligned} \varOmega =\{(x,y): 0\le x\le 1, 0\le y\le 1-x^2\}\ . \end{aligned}$$

The critical points of G satisfy the conditions

$$\begin{aligned} -28y^2-5x(8-9x)y+32+2x-96x^2-4x^3+75x^4=0 \quad \text {and}\quad y=\tfrac{5(4-4x^2+3x^3)}{56x} \ . \end{aligned}$$

Solving this system leads to an equality

$$\begin{aligned} -400+2784x^2+1424x^3-9552x^4-2848x^5+9525x^6 = 0\ , \end{aligned}$$

which has two solutions: $x_1=-0.770\ldots$ and $x_2=0.898\ldots$. Hence, we obtain two critical points: $(-0.770\ldots ,-0.029\ldots )$ and $(0.898\ldots ,0.292\ldots )$. Both points do not belong to $\varOmega$.

On the boundary of $\varOmega$, we have

$$\begin{aligned} G(0,y)= & {} 20y\le 20\ ,\\ G(x,0)= & {} x\left[ 32(1-x^2)+x(1-x^2)+15x^4\right] = g_1(x)\ ,\\ G(x,1-x^2)= & {} 20+4x-39x^2+39x^3+19x^4-28x^5 = g_2(x)\ . \end{aligned}$$

For $g_1$, we have

$$\begin{aligned} g_1'(x)=8(2-3x^2)^2+h(x)\ , \end{aligned}$$

with $h(x)=2x-4x^3+3x^4$. Since h is increasing for $x\in [0,1]$, so $h(x)\ge h(0)=0$. In this way, we have proven that $g_1'(x)\ge 0$ for $x\in [0,1]$. Consequently,

$$\begin{aligned} G(x,0) \le g_1(1) = 15\ . \end{aligned}$$

The function $g_2$ has in [0, 1] the only critical point $x_0=0.056\ldots$, so

$$\begin{aligned} \max \{\ g_2(x): x\in [0,1]\ \}=\max \{\ g_2(0), g_2(x_0), g_2(1)\ \} = 20.108\ldots \ . \end{aligned}$$

Summing the bounds in (3.2), we get

$$\begin{aligned} |\gamma _5| \le 0.0347\ldots \end{aligned}$$

or, in other way,

$$\begin{aligned} |\gamma _5| \le \tfrac{19}{576} + 0.0017\ldots \ . \end{aligned}$$

Finally, note that the equalities $|\gamma _1|=\tfrac{1}{4}$ and $|\gamma _3|=\tfrac{1}{16}$ hold if we take $c_1=1$ in (3.2). Similarly, $|\gamma _2|=\tfrac{1}{6}$ and $|\gamma _4|=\tfrac{13}{180}$ if we put $c_2=1$ into (3.2). This means that the extremal convex functions are

$$\begin{aligned} f_1(z)=-\log (1-z) \quad \text {and}\quad f_2(z)=\tfrac{1}{2}\log \frac{1+z}{1-z}\ . \end{aligned}$$

For these functions, the corresponding functions F are following

$$\begin{aligned} F_1(z) = \tfrac{1}{4} z + \tfrac{5}{48} z^2 + \tfrac{1}{16} z^3 + \tfrac{251}{5760} z^4 + \tfrac{19}{576} z^5 + \ldots \end{aligned}$$

and

$$\begin{aligned} F_2(z) = \tfrac{1}{6} z^2 + \tfrac{13}{180} z^4 + \tfrac{251}{5760} z^6 + \ldots \ , \end{aligned}$$

respectively. $\square$

For $f_1$, we can also see that $\gamma _5=\tfrac{19}{576}$ which is expected sharp bound of $\gamma _5$ for all functions in ${\mathcal {K}}_S$.