1 Introduction

Denote by \({\mathcal {P}}\) the class of functions analytic in \({\mathbb {D}}=\{z\in {\mathbb {C}}: |z|<1\}\), given by

$$\begin{aligned} h(z) = 1+\sum _{n=1}^\infty p_n z^n \end{aligned}$$
(1)

and having a positive real part.

It is clear that

$$\begin{aligned} |p_{n-1}p_{n+1}-p_n{}^2|\le 4 \end{aligned}$$
(2)

is true for all \(h\in {\mathcal {P}}\) and positive integers \(n\ge 2\). It immediately follows from an inequality

$$\begin{aligned} |p_{n-1}p_{n+1}-p_n{}^2|\le |p_{n-1}p_{n+1}-p_{2n}| + |p_{2n}-p_n{}^2| \end{aligned}$$

and the inequality \(|p_{k}p_{n-k}-p_{n}|\le 2\) proved by Livingston ( [10]). Equality in (2) holds for example for \(h(z)=\frac{1+z^n}{1-z^n}\) which means that Inequlity (2) is sharp.

On the other hand, Inequality (2) is very often not strong enough to derive good results in many extremal problems concerning coefficients of analytic functions. For this reason we shall improve it by substituting the constant 4 by an expression depending on the first coefficient of h, i.e. \(|p_1|\).

This idea appears in many papers concerning functions with positive real part. As an example it is worth recalling the result of Robertson ( [12])

$$\begin{aligned} |p_{n+1}-p_{n}| \le (2n+1)|2-p_1| \end{aligned}$$

or results obtained by Brown ( [1]) and Lecko ( [7]).

The functional \(F(h)=|p_{n-1}p_{n+1}-p_n{}^2|\) defined for \(h\in {\mathcal {P}}\) is rotationally invariant. It means that \(F(h)=F(h_\varphi )\), where \(h_\varphi (z)=h(e^{i\varphi } z)\). For this reason, if necessary, we can assume that \(p_1\in [0,2]\).

To prove our results we need the following lemma (see, [6]).

Lemma 1

If \(p \in {\mathcal {P}}\) is of the form (1) with \(p_1\ge 0\), then

$$\begin{aligned} 2p_2= & {} p_1^2 + (4-p_1^2)x, \end{aligned}$$
(3)
$$\begin{aligned} 4p_3= & {} p_1^3 +(4-p_1^2)p_1 x (2-x) + 2(4-p_1^2)( 1 - |x|^2) y \end{aligned}$$
(4)

and

$$\begin{aligned} \begin{aligned} 8p_4&= p_1^4+(4-p_1^2) x \left[ p_1^2(x^2-3x+3)+4x \right] \\&\quad -4(4-p_1^2)(1-|x|^2)\left[ p_1(x-1)y+\overline{x}y^2-\left( 1-|y|^2\right) w\right] \end{aligned} \end{aligned}$$
(5)

for some x, y, \(w\in \overline{{\mathbb {D}}}:=\{z\in {\mathbb {C}}:|z|\le 1 \}.\)

At the begining we shall estimate \(|p_{1}p_{3}-p_2{}^2|\) in terms of \(p_1\).

Theorem 2

If \(h\in {\mathcal {P}}\) is given by (1), then

$$\begin{aligned} |p_{1}p_{3}-p_2{}^2| \le 4-|p_1|^2\ . \end{aligned}$$
(6)

Equality holds for rotations \(h_t(\varepsilon z)\) and \(g_t(\varepsilon z)\), \(|\varepsilon |=1\) of

$$\begin{aligned} h_t(z)=\frac{1-z^2}{1-2tz+z^2} \ ,\ t\in [-1,1] \end{aligned}$$
(7)

and

$$\begin{aligned} g_t(z)=\tfrac{1-t}{2} h_{-1}(z)+\tfrac{1+t}{2} h_1(z)=\frac{1+2tz+z^2}{1-z^2}\ ,\ t\in [-1,1]\ . \end{aligned}$$
(8)

Proof

From Lemma 1,

$$\begin{aligned} 4\left( p_{1}p_{3}-p_2{}^2\right) = (4-p_1{}^2)\left[ -4x^2+2p_1(1-|x|^2)y\right] . \end{aligned}$$

Assuming \(p=p_1\in [0,2]\) and writing \(r=|x|\in [0,1]\), we obtain

$$\begin{aligned} 4\left| p_{1}p_{3}-p_2{}^2\right| \le (4-p^2)\left[ 4r^2+2p(1-r^2)\right] \le (4-p^2)\left[ 4r^2+4(1-r^2)\right] \end{aligned}$$

which results in (6). \(\square \)

It is an easy exercise to show that equality in (6) holds for \(h_t\) and \(g_t\) defined by (7) and (8), respectively. In general case, if

$$\begin{aligned} |p_{n-1}p_{n+1}-p_n{}^2| \le 4-|p_1|^2 \end{aligned}$$
(9)

holds true, then equality in (9) will be obtained also by \(h_t\) and \(g_t\). Indeed, for \(h_t\), \(t\in [-1,1]\) given by (7) we have

$$\begin{aligned} h_t(z)=1+\sum _{n=1}^\infty 2T_n(t) z^n\ , \end{aligned}$$
(10)

where \(T_n(t)\) stands for Chebyshev polynomials of the first kind. It is known for these polynomials that the inequality

$$\begin{aligned} T_{n-1}(t)T_{n+1}(t)-[T_n(t)]^2 = t^2-1\, \end{aligned}$$

called Turan’s inequality, holds for all integers \(n\ge 2\). This results in the equality in (9).

Analogously, for \(g_t\), \(t\in [-1,1]\) given by (8) there is

$$\begin{aligned} g_t(z)=1+2tz+2z^2+2tz^3+\ldots \ ,\ t\in [-1,1]\ \end{aligned}$$
(11)

and so

$$\begin{aligned} p_{n-1}p_{n+1}-p_n{}^2 = 4(-1)^{n}(t^2-1) = (-1)^n(p_1{}^2-4) . \end{aligned}$$

Unfortunately, we are not able to prove the conjecture (9), but we can do it in one more case, i.e. for \(n=3\).

2 Main Result

Theorem 3

If \(h\in {\mathcal {P}}\) is given by (1), then

$$\begin{aligned} |p_{2}p_{4}-p_3{}^2| \le 4-|p_1|^2\ . \end{aligned}$$
(12)

Equality holds for the same functions as in Theorem 2.

Proof

From Lemma 1, after tidious but elementary calculation,

$$\begin{aligned} 16\left( p_{2}p_{4}-p_3{}^2\right)= & {} (4-p_1{}^2)\left[ 16x^3-(1-|x|^2)\left[ 16p_1 xy+4\left( 4-p_1{}^2+p_1{}^2\overline{x}\right) y^2\right] \right. \\{} & {} +\left. 4(1-|x|^2)(1-|y|^2)\left[ (4-p_1{}^2)x+p_1{}^2\right] w\right] . \end{aligned}$$

From rotational invariance of \(|p_{2}p_{4}-p_3{}^2|\) we can assume that \(p=p_1\in [0,2]\) and write \(p=2t\), \(t\in [0,1]\). Therefore,

$$\begin{aligned} \left| p_{2}p_{4}-p_3{}^2\right| = (4-p^2)\cdot |W|\ , \end{aligned}$$
(13)

where

$$\begin{aligned} W= & {} x^3-(1-|x|^2)\left[ 2txy+\left( 1-t^2+t^2\overline{x}\right) y^2\right] \\{} & {} + (1-|x|^2)(1-|y|^2)\left[ (1-t^2)x+t^2\right] w. \end{aligned}$$

After rearranging terms and applying the triangle inequality we have

$$\begin{aligned} |W|\le & {} \left| x^3-(1-|x|^2)\left( 2txy+t^2\overline{x}y^2\right) \right| \\{} & {} + (1-|x|^2)(1-t^2)|y|^2 + (1-|x|^2)(1-|y|^2)\left| (1-t^2)x+t^2\right| \end{aligned}$$

Let \(x=re^{i\varphi }\), \(y=\varrho e^{i\theta }\); hence

$$\begin{aligned} |W|\le & {} \left| r^3e^{3i\varphi }-(1-r^2)\left[ 2tr\varrho e^{i(\varphi +\theta )} +t^2r\varrho ^2 e^{i(2\theta -\varphi )}\right] \right| \\{} & {} + (1-r^2)\varrho ^2(1-t^2) + (1-r^2)(1-\varrho ^2)\left| (1-t^2)re^{i\varphi }+t^2\right| \\\le & {} \left| e^{i(\varphi +\theta )}\left[ r^3e^{i(2\varphi -\theta )} -2r(1-r^2)\varrho t-r(1-r^2)\varrho ^2 t^2 e^{i(\theta -2\varphi )}\right] \right| \\{} & {} + (1-r^2)\varrho ^2(1-t^2) + (1-r^2)(1-\varrho ^2)\left[ (1-t^2)r+t^2\right] . \end{aligned}$$

Denote \(\alpha =2\varphi -\theta \). The last inequality can be rewritten as follows

$$\begin{aligned} |W|\le & {} r \left| r^2e^{i\alpha }-2(1-r^2)\varrho t-(1-r^2)\varrho ^2 t^2 e^{-i\alpha }\right| \\{} & {} + (1-r^2)\varrho ^2(1-t^2) + (1-r^2)(1-\varrho ^2)\left[ (1-t^2)r+t^2\right] . \end{aligned}$$

Now, we shall maximize the expression

$$\begin{aligned} V = \left| r^2e^{i\alpha }-2(1-r^2)\varrho t-(1-r^2)\varrho ^2 t^2 e^{-i\alpha }\right| \end{aligned}$$

with respect to \(\alpha \). Consequently,

$$\begin{aligned} V^2= & {} \left[ [r^2-(1-r^2)\varrho ^2 t^2]\cos \alpha -2(1-r^2)\varrho t\right] ^2 + [r^2+(1-r^2)\varrho ^2 t^2]^2\sin ^2\alpha \\= & {} -4(1-r^2)\varrho t g(\cos \alpha ) +r^4+(1-r^2)^2\varrho ^4 t^4+2r^2(1-r^2)\varrho ^2 t^2+4(1-r^2)^2\varrho ^2 t^2 \, \end{aligned}$$

where

$$\begin{aligned} g(\tau )=r^2\varrho t \tau ^2+\left( r^2-(1-r^2)\varrho ^2 t^2\right) \tau \,\ \tau \in [-1,1]. \end{aligned}$$

Since \(g'(\tau )=2r^2\varrho t \tau +\left( r^2-(1-r^2)\varrho ^2 t^2\right) \), we discuss three cases.

I. If \(r\le \frac{\varrho t}{1+\varrho t}\), then \(g'(1)\le 0\), so \(g'(\tau )\le 0\) for all \(\tau \in [-1,1]\). Hence, g is a decreasing function of \(\tau \), so the lowest value of g is obtained for \(\tau =1\). Consequently, the lowest values of V and W are obtained for \(\alpha =0\). Moreover, for \(\alpha =0\) there is

$$\begin{aligned} V= & {} \left| r^2-2(1-r^2)\varrho t-(1-r^2)\varrho ^2 t^2\right| \\= & {} (1+\varrho t)^2 \left| r^2-\frac{\varrho t(2 +\varrho t)}{(1+\varrho t)^2}\right| \\= & {} -r^2(1+\varrho t)^2+2\varrho t+\varrho ^2 t^2. \end{aligned}$$

For this reason,

$$\begin{aligned} |W| \le -r^3+(1-r^2)h(\varrho ,t) \equiv H(r,\varrho ,t)\, \end{aligned}$$

where

$$\begin{aligned} h(\varrho ,t) = -2(1-r)\varrho ^2 t^2+(1-r)(\varrho ^2 +t^2)+2r\varrho t+r\,\ \varrho ,t\in [0,1]. \end{aligned}$$

The function h has at most one critical point \((\varrho _0,t_0)\) inside \([0,1]\times [0,1]\) depending on r, where \(\varrho _0=t_0=\sqrt{\frac{1}{2(1-r)}}\) and \(h(\varrho _0,t_0)=\frac{1+2r-2r^2}{2(1-r)}\). For this critical point we have

$$\begin{aligned} H(\varrho _0,t_0)=\tfrac{1}{2}(1+3r-4r^3) \le 1 \end{aligned}$$

with equality for \(r=1/2\).

Considering values of h on the boundary of \([0,1]\times [0,1]\) results in

$$\begin{aligned} h(0,t)= & {} r+t^2(1-r) \le 1 \\ h(1,t)= & {} -(1-r)t^2+2tr+1 \le \frac{1-r+r^2}{1-r} \end{aligned}$$

and so

$$\begin{aligned} H(0,t)\le & {} -r^3+(1-r^2)\cdot 1 \le 1 \\ H(1,t)\le & {} -r^3+(1-r^2)\cdot \frac{1-r+r^2}{1-r} =1 \end{aligned}$$

Since \(h(\varrho ,0) = h(0,\varrho )\) and \(h(\varrho ,1) = h(1,\varrho )\), thus \(H(\varrho ,0)\) and \(h(\varrho ,1)\) are also bounded by 1.

Summing up, if \(r\le \frac{\varrho t}{1+\varrho t}\), then \(|W|\le 1\). Observe that here and on it is not necessary to check whether the greatest value of particular functions discussed above are obtained for \(\varrho \) or t in the interval [0, 1].

II. If \(r\ge \frac{\varrho t}{1-\varrho t}\), then \(g'(-1)\ge 0\), so \(g'(\tau )\ge 0\) for all \(\tau \in [-1,1]\). Hence, g is a increasing function of \(\tau \), so the lowest value of g is obtained for \(\tau =-1\). Consequently, the lowest values of V and W are obtained for \(\alpha =\pi \). Moreover, for \(\alpha =\pi \) there is

$$\begin{aligned} V= & {} \left| -r^2-2(1-r^2)\varrho t+(1-r^2)\varrho ^2 t^2\right| \\= & {} (1-\varrho t)^2 \left| r^2+\frac{\varrho t(2 -\varrho t)}{(1-\varrho t)^2}\right| \\= & {} r^2(1-\varrho t)^2+2\varrho t-\varrho ^2 t^2. \end{aligned}$$

For this reason,

$$\begin{aligned} |W| \le r^3+(1-r^2)h(\varrho ,t) \equiv H(r,\varrho ,t)\, \end{aligned}$$

where

$$\begin{aligned} h(\varrho ,t) = -2\varrho ^2 t^2+(1-r)(\varrho ^2 +t^2)+2r\varrho t+r\,\ \varrho ,t\in [0,1]. \end{aligned}$$

The only critical point of the function h inside \([0,1]\times [0,1]\) is \((\varrho _0,t_0)\), where \(\varrho _0=t_0=\frac{\sqrt{2}}{2}\) and \(h(\varrho _0,t_0)=\tfrac{1}{2}+r\). Additionally,

$$\begin{aligned} H(\varrho _0,t_0)=\tfrac{1}{2}(1+2r-r^2) \le 1 \end{aligned}$$

with equality for \(r=1\).

Considering values of h on the boundary of \([0,1]\times [0,1]\) results in

$$\begin{aligned} h(0,t)= & {} r+t^2(1-r) \le 1 \\ h(1,t)= & {} -(1+r)t^2+2tr+1 \le \frac{1+r+r^2}{1+r} \end{aligned}$$

and so

$$\begin{aligned} H(0,t)\le & {} r^3+(1-r^2)\cdot 1 = 1-r^2(1-r) \le 1 \\ H(1,t)\le & {} r^3+(1-r^2)\cdot \frac{1+r+r^2}{1+r} =1 \end{aligned}$$

Since \(h(\varrho ,0) = h(0,\varrho )\) and \(h(\varrho ,1) = h(1,\varrho )\), thus \(H(\varrho ,0)\) and \(h(\varrho ,1)\) are also bounded by 1.

Summing up, if \(r\ge \frac{\varrho t}{1-\varrho t}\), then \(|W|\le 1\).

III. Let \(\frac{\varrho t}{1+\varrho t}<r<\frac{\varrho t}{1-\varrho t}\). In this case the lowest value of g is equal to

$$\begin{aligned} g(\tau _0)=-\frac{\left[ r^2-(1-r^2)\varrho ^2 t^2\right] ^2}{4r^2\varrho t}\, \end{aligned}$$

where \(\tau _0=\frac{(1-r^2)\varrho ^2 t^2-r^2}{2r^2\varrho t}\). Consequently,

$$\begin{aligned} V \le \frac{(1-r^2)\varrho ^2 t^2+r^2}{r} \end{aligned}$$

and

$$\begin{aligned} |W|\le & {} (1-r^2)\varrho ^2 t^2+r^2 + (1-r^2)\varrho ^2(1-t^2) + (1-r^2)(1-\varrho ^2)\left[ (1-t^2)r+t^2\right] \\= & {} 1-(1-r)(1-r^2)(1-\varrho ^2)(1-t^2) \le 1. \end{aligned}$$

Combining cases I - III, we have \(|W|\le 1\) for all \(r, \varrho , t\in [0,1]\). This and (13) lead to (12). The sharpness of this result follows from the argument presented before Theorem 3. \(\square \)

3 Relative Inequalities for Schwarz Functions

Let \({\mathcal {B}}_0\) be the class of Schwarz functions, i.e., analytic functions \(\omega :{\mathbb {D}}\rightarrow {\mathbb {D}}\), \(\omega (0)=0\), where \({\mathbb {D}}\) stands for the open unit disk \(\{z\in {\mathbb {C}}: |z|<1\}\). The function \(\omega \in {\mathcal {B}}_0\) can be written as a power series

$$\begin{aligned} \omega (z)=\sum _{n=1}^\infty c_nz^n\ ,\ z\in {\mathbb {D}}\ . \end{aligned}$$
(14)

It is clear that if

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

then

$$\begin{aligned} p\in {\mathcal {P}} \quad \text {if and only if}\quad \omega \in {\mathcal {B}}_0. \end{aligned}$$

This property makes it possible to discuss problems in \({\mathcal {B}}_0\) considering the class \({\mathcal {P}}\) and vice versa. Further, we apply this property to establish a relation between the initial coefficients of \(\omega \in {\mathcal {B}}_0\) and \(p\in {\mathcal {P}}\).

Let p(z) and \(\omega (z)\) be of the form (1) and (14), respectively. Comparing coefficients at powers of z in

$$\begin{aligned} \left[ 1-\omega (z)\right] p(z)=1+\omega (z)\, \end{aligned}$$

we obtain

$$\begin{aligned} p_1= & {} 2c_1 \nonumber \\ p_2= & {} 2c_2+2c_1^2 \nonumber \\ p_3= & {} 2c_3+4c_1c_2+2c_1^3 \nonumber \\ p_4= & {} 2c_4+4c_1c_3+2c_2^2+6c_1^2c_2+2c_1^4 . \end{aligned}$$
(15)

Coefficient problems for functions \(\omega \in {\mathcal {B}}_0\) were studied in numerous papers; for details, see ( [2, 3, 11, 13, 15]). Here, we need the inequality obtained in [15].

Theorem 4

If \(\omega \in {\mathcal {B}}_0\) is given by (14), then

$$\begin{aligned} \left| c_2c_4 - c_3^2\right| \le (1-|c_1|^2)^2\ . \end{aligned}$$
(16)

Equality holds for rotations \(\varepsilon ^{-1} \omega (\varepsilon z)\), \(|\varepsilon |=1\) of

$$\begin{aligned} \omega (z)=\frac{z(t+z^2)}{1+tz^2}=tz+(1-t^2)z^3-t(1-t^2)z^5+\ldots \,\ t\in [-1,1]. \end{aligned}$$

Directly from Theorem 3 and (15) we get the following theorem.

Theorem 5

If \(\omega \in {\mathcal {B}}_0\) is given by (14), then

$$\begin{aligned} |c_{2}c_{4}-c_3{}^2+c_1{}^2c_4-2c_1c_2c_3+c_2{}^3| \le 1-|c_1|^2\ . \end{aligned}$$
(17)

Equality holds for rotations \(\varepsilon ^{-1} \omega (\varepsilon z)\), \(|\varepsilon |=1\) of

$$\begin{aligned} \omega (z)=\frac{(t+z)z}{1+tz} = tz-(1-t^2)z^2-t(1-t^2)z^3-t^2(1-t^2)z^4+\ldots \ ,\ t\in [-1,1]\ . \end{aligned}$$
(18)

Finally, we can combine the inequalities from Theorem 5 and Theorem 4 to obtain more general inequality.

Theorem 6

If \(\omega \in {\mathcal {B}}_0\) is given by (14), then for all \(\alpha \in [0,1]\)

$$\begin{aligned} |c_{2}c_{4}-c_3{}^2+\alpha \left( c_1{}^2c_4-2c_1c_2c_3+c_2{}^3\right) | \le (1-|c_1|^2)(1-(1-\alpha )|c_1|^2)\ . \end{aligned}$$
(19)

4 Applications

As an application of the theorems presented in the two previous sections we derive the sharp bounds of Hankel determinants for two classes of analytic functions, namely for the class \({\mathcal {S}}^*(1/2)\) of starlike functions of order 1/2 and the class \({\mathcal {M}}\) which consists of functions which are not necessarily univalent. Let us recall the definitions of these classes.

Denote by \({\mathcal {S}}^*(\alpha )\), \(\alpha \in [0,1)\) the class of functions analytic in \({\mathbb {D}}\) such that

$$\begin{aligned} f(z) = z+\sum _{n=2}^\infty a_n z^n \end{aligned}$$
(20)

and satisfying the condition

$$\begin{aligned} \textrm{Re}\left( \frac{zf'(z)}{f(z)}\right) > \alpha . \end{aligned}$$

In a similar way the class \({\mathcal {M}}(\alpha )\), \(\alpha >0\) is defined. Namely, this family consists of functions analytic in \({\mathbb {D}}\) with a power series expanion (20) such that

$$\begin{aligned} \textrm{Re}\left( \frac{zf'(z)}{f(z)}\right) < 1+\frac{\alpha }{2}. \end{aligned}$$

Let \({\mathcal {M}}={\mathcal {M}}(1)\). Clearly, if \(f\in {\mathcal {M}}\), then

$$\begin{aligned} \textrm{Re}\left( \frac{zf'(z)}{f(z)}\right) < \frac{3}{2}. \end{aligned}$$

The Hankel determinants which we shall discuss are defined as follows:

$$\begin{aligned} H_{3,1} = a_3(a_2a_4-a_3{}^2) + a_4(a_2a_3-a_4) + a_5(a_3-a_2{}^2)\ \end{aligned}$$
(21)

and

$$\begin{aligned} H_{2,3} = a_3a_5-a_4{}^2 \ . \end{aligned}$$
(22)

The problem of deriving bounds of Hankel determinants has focused the attention of many mathematicians in the recent years. It was solved for the majority of classes of analytic functions in case of the second Hankel determinant given by \(H_{2,2} = a_2a_4-a_3{}^2\). Generally speaking, this problem for \(H_{3,1}\) is difficult, but today it is solved, among others, for main classes of function, for example for \({\mathcal {S}}^*\) or \({\mathcal {K}}\) (see, [4, 5]). Usually, the calculation is very hard. The studies on \(H_{2,3}\) are not so intensive as in case of \(H_{3,1}\), only a few sharp results are known (see, [14]).

Theorem 7

If \(f\in {\mathcal {S}}^*(1/2)\) is given by (20), then

$$\begin{aligned} |H_{2,3}| \le \frac{3}{16}\ . \end{aligned}$$
(23)

Equality holds for

$$\begin{aligned} f(z)=\frac{z}{\sqrt{1-z^2}} = z+\tfrac{1}{2} z^3+\tfrac{3}{8} z^5+\ldots . \end{aligned}$$

Proof

Each function \(f\in {\mathcal {S}}^*(1/2)\) can be written in terms of \(h\in {\mathcal {P}}\)

$$\begin{aligned} \frac{zf'(z)}{f(z)}=\tfrac{1}{2}\left( 1+h(z)\right) \end{aligned}$$

or equivalently, in terms of \(\omega \in {\mathcal {B}}_0\)

$$\begin{aligned} \frac{zf'(z)}{f(z)}=\frac{1}{1-\omega (z)}\ . \end{aligned}$$
(24)

Hence, comparing coefficients in the equality \(zf'(z)(1-\omega (z))=f(z)\), where f and \(\omega \) are of the form (20) and (14), respectively, we obtain

$$\begin{aligned} a_3=\tfrac{1}{2} c_2+c_1{}^2\ ,\ a_4=\tfrac{1}{3} c_3+\tfrac{7}{6} c_1c_2+c_1{}^3\ ,\ a_5=\tfrac{1}{4} c_4+\tfrac{5}{6} c_1c_3+\tfrac{3}{8} c_2{}^2+\tfrac{23}{12}c_1{}^2c_2+c_1{}^3\ . \end{aligned}$$
(25)

Consequently,

$$\begin{aligned} H_{2,3}= & {} \frac{1}{144}\left( 18c_2c_4+36c_1{}^2c_4-16c_3^2+24c_1{}^3c_3-52c_1c_2c_3\right. \\{} & {} \left. +27c_2{}^3 +12c_1{}^4c_2-4c_1{}^2c_2{}^2\right) \\= & {} \frac{1}{144}\left[ 16\left( c_2c_4-c_3^2+c_1{}^2c_4-2c_1c_2c_3+c_2{}^3\right) + (2c_2+20c_1{}^2)c_4 \right. \\{} & {} +\left. (24c_1{}^3-20c_1c_2)c_3+11c_2{}^3+12c_1{}^4c_2-4c_1{}^2c_2{}^2\right] \end{aligned}$$

and

$$\begin{aligned} |H_{2,3}| \le \frac{1}{144} G(|c_1|,|c_2|)\, \end{aligned}$$

where

$$\begin{aligned} G(c,d)= & {} 16(1-c^2)+(2d+20c^2)(1-c^2-d^2)\\{} & {} +(24c^3+20cd) \left( 1-c^2-\tfrac{d^2}{1+c}\right) +11d^3+12c^4d+4c^2d^2 \\= & {} \tfrac{9-11c}{1+c}d^3-\tfrac{8(2+5c)c^2}{1+c}d^2 +\left[ 2(1-c^2)(1+10c)+12c^4\right] d\\{} & {} +4(1-c^2)(4+5c^2+6c^3). \end{aligned}$$

If \(c\in [0,9/11)\), then

$$\begin{aligned} G(c,d)\le & {} \tfrac{9-11c}{1+c}d^3+\left[ 2(1-c^2)(1+10c) +12c^4\right] d\\{} & {} +4(1-c^2)(4+5c^2+6c^3) \\\le & {} \tfrac{9-11c}{1+c}(1-c^2)^3+\left[ 2(1-c^2)(1+10c) +12c^4\right] (1-c^2)\\{} & {} +4(1-c^2)(4+5c^2+6c^3) \\= & {} 27-7c^2+24c^3-19c^4-24c^5-c^6. \end{aligned}$$

Denoting the latter by \(g_1(c)\) we can see that

$$\begin{aligned} g_1'(c)=-2c(7-36c+38c^2+60c^3+3c^4) \le 0\quad \text {for all}\ c\in [0,9/11) \end{aligned}$$

and so in this case

$$\begin{aligned} G(c,d) \le g_1(0) = 27. \end{aligned}$$

If \(c\in [9/11,1]\), then

$$\begin{aligned} G(c,d)\le & {} \left[ 2(1-c^2)(1+10c)+12c^4\right] d+4(1-c^2)(4+5c^2+6c^3) \\\le & {} \left[ 2(1-c^2)(1+10c)+12c^4\right] (1-c^2)+4(1-c^2)(4+5c^2+6c^3) \\= & {} 18+20c-16c^3-6c^4-4c^5-12c^6. \end{aligned}$$

Denoting the latter by \(g_2(c)\) we obtain

$$\begin{aligned} g_2'(c)=20-48c^2-24c^3-20c^4-72c^5 \le 0\quad \text {for all}\ c\in [9/11,1] \end{aligned}$$

and in this case

$$\begin{aligned} G(c,d) \le g_2\left( \tfrac{9}{11}\right) < 18. \end{aligned}$$

Combining both cases we have proven that \(G(c,d)\le 27\) and so \(|H_{2,3}| \le \frac{3}{16}\) with equality for \(c=0\) and \(d=1\). It means that the extremal function \(f\in {\mathcal {S}}^*(1/2)\) is obtained when in (24) we put \(\omega (z)=z^2\). \(\square \)

Theorem 8

If \(f\in {\mathcal {M}}\) is given by (20), then

$$\begin{aligned} |H_{3,1}| \le \frac{3}{16}\ . \end{aligned}$$
(26)

Equality holds for

$$\begin{aligned} f(z)=\frac{z}{\sqrt{1-z^2}} = z+\tfrac{1}{2} z^3+\tfrac{3}{8} z^5+\ldots . \end{aligned}$$

Proof

Each function \(f\in {\mathcal {M}}\) can be written in terms of \(h\in {\mathcal {P}}\)

$$\begin{aligned} \frac{zf'(z)}{f(z)}=\tfrac{1}{2}\left( 3-h(z)\right) \end{aligned}$$

or equivalently, in terms of \(\omega \in {\mathcal {B}}_0\)

$$\begin{aligned} \frac{zf'(z)}{f(z)}=\frac{1-2\omega (z)}{1-\omega (z)}\ . \end{aligned}$$
(27)

Hence, comparing coefficients in the equality \(zf'(z)(1-\omega (z))=f(z)(1-2\omega (z))\), where f and \(\omega \) are of the form (20) and (14), respectively, we obtain

$$\begin{aligned}{} & {} a_2=-c_1\,\ a_3=-\tfrac{1}{2} c_2\,\ a_4=-\tfrac{1}{3} c_3-\tfrac{1}{6} c_1c_2\,\nonumber \\{} & {} a_5=-\tfrac{1}{4} c_4-\tfrac{1}{6} c_1c_3-\tfrac{1}{8} c_2{}^2-\tfrac{1}{12}c_1{}^2c_2. \end{aligned}$$
(28)

Consequently,

$$\begin{aligned} H_{3,1}= & {} \frac{1}{144}\left( 18c_2c_4+36c_1{}^2c_4-16c_3^2+24c_1{}^3c_3\right. \\{} & {} \left. -52c_1c_2c_3+27c_2{}^3+12c_1{}^4c_2-4c_1{}^2c_2{}^2\right) . \end{aligned}$$

Note that \(H_{3,1}\) for \({\mathcal {M}}\) is exactly the same as \(H_{2,3}\) for \({\mathcal {S}}^*(1/2)\). From the proof of Theorem 7 it follows our result. \(\square \)