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Theorem 1 · coeff
Theorem 1. Let is given by (1) and let. Then - (a) if; (b) if; (c) if; (d) if. All these results are sharp. Proof. (a) Since is equivalent…
Theorem 1. Let $f \in A$ is given by (1) and let $a_2 = 0$ . Then
- (a) $|H_3(2)(f^{-1})| \leq \frac{28}{45}$ if $f \in \mathcal{R}$ ; (b) $|H_3(2)(f^{-1})| \leq \frac{2}{45}$ if $f \in \mathcal{C}$ ; (c) $|H_3(2)(f^{-1})| \leq 2$ if $f \in \mathcal{S}$ ; (d) $|H_3(2)(f^{-1})| \leq 2$ if $f \in \mathcal{S}^_s$ .
All these results are sharp.
Proof.
(a) Since $f \in \mathcal{R}$ is equivalent to
$$f'(z) = \frac{1 + \omega(z)}{1 - \omega(z)},$$
for certain Schwartz function $\omega$ , we receive that
(6)
$$f'(z) = 1 + 2\omega(z) + 2\omega^{2}(z) + \cdots$$
Using the notations for f and $\omega$ given by (1) and (5), and equating the coefficients in (6), we receive
(7)
$$\begin{cases} a_2 = c_1, \\ a_3 = \frac{2}{3}(c_2 + c_1^2), \\ a_4 = \frac{1}{2}(c_3 + 2c_1c_2 + c_1^3), \\ a_5 = \frac{2}{5}(c_4 + 2c_1c_3 + 3c_1^2c_2 + c_2^2 + c_1^4). \end{cases}$$
Since $a_2 = 0$ , by (7) we have $c_1 = 0$ , and the appropri
Since $a_2 = 0$ , by (7) we have $c_1 = 0$ , and the appropriate coefficients have the next form:
(8)
$$a_3 = \frac{2}{3}c_2, \quad a_4 = \frac{1}{2}c_3, \quad a_5 = \frac{2}{5}(c_4 + c_2^2).$$
Now, from (3) and (8), after simple computation, we obtain
$$H_3(2)(f^{-1}) = \frac{4}{15}c_2c_4 - \frac{1}{4}c_3^2 - \frac{28}{45}c_2^3,$$
and further,
$$|H_3(2)(f^{-1})| \le \frac{4}{15}|c_2||c_4| + \frac{1}{4}|c_3|^2 + \frac{28}{45}|c_2|^3.$$
Applying Lemma 1 (with $c_1 = 0$ ) we receive
$$|H_3(2)(f^{-1})| \le \frac{4}{15}|c_2|(1-|c_2|^2) + \frac{1}{4}(1-|c_2|^2)^2 + \frac{28}{45}|c_2|^3$$
and, finally,
$$(9) \qquad |H_3(2)(f^{-1})| \leq \frac{1}{4} + \frac{4}{15}|c_2| - \frac{1}{2}|c_2|^2 + \frac{16}{45}|c_2|^3 + \frac{1}{4}|c_2|^4 =: \varphi_1(|c_2|),$$
where $0 \le |c_2| \le 1$ . Since
$$\varphi_1'(|c_2|) = \frac{4}{15} - |c_2| + \frac{16}{15}|c_2|^2 + |c_2|^3$$
$$= \frac{4}{15}(1 - 2|c_2|)^2 + \frac{1}{15}|c_2| + |c_2|^3 > 0,$$
we have $\varphi_1(|c_2|) \leq \varphi_1(1) = \frac{28}{45}$ , and from (9),
$$|H_3(2)(f^{-1})| \le \frac{28}{45} = 0.622....$$
This result is best possible as the function $f_1(z) = \ln \frac{1+z}{1-z} - z$ defined by $f'_1(z) = \frac{1+z^2}{1-z^2}$ , shows.
(b) We apply the same method as in the previous case. Namely, from the definition of the class $\mathcal C$ we have
$$1 + \frac{zf''(z)}{f'(z)} = \frac{1 + \omega(z)}{1 - \omega(z)},$$
where $\omega$ is a Schwartz function, and from here
(10)
$$(zf'(z))' = \left[1 + 2\left(\omega(z) + \omega^2(z) + \cdots\right)\right] \cdot f'(z).$$
Using the notations (1) and (5), and comparing the coefficients in the relation (10), after some simple calculations, we obtain
(11)
$$\begin{cases}
a_2 = c_1, \\
a_3 = \frac{1}{3} (c_2 + 3c_1^2), \\
a_4 = \frac{1}{6} (c_3 + 5c_1c_2 + 6c_1^3) \\
a_5 = \frac{1}{30} (3c_4 + 14c_1c_3 + 43c_1^2c_2 + 30c_1^4 + 6c_2^2).
\end{cases}$$
If $a_2 = 0$ , then by (11) we have $c_1 = 0$ , which implies
(12)
$$a_3 = \frac{1}{3}c_2, \quad a_4 = \frac{1}{6}c_3, \quad a_5 = \frac{1}{10}(c_4 + 2c_2^2).$$
Using (3) and (12) we obtain
$$H_3(2)(f^{-1}) = \frac{1}{180} \left( 6c_2c_4 - 5c_3^2 - 8c_2^3 \right).$$
From the last relation we get
$$|H_3(2)(f^{-1})| \le \frac{1}{180} \left( 6|c_2||c_4| + 5|c_3|^2 + 8|c_2|^3 \right)$$
and further, after applying Lemma 1 (with $c_1 = 0$ ),
$$|H_3(2)(f^{-1})| \le \frac{1}{180} \left( 6|c_2|(1-|c_2|^2) + 5(1-|c_2|^2)^2 + 8|c_2|^3 \right),$$
i.e.,
$$(13) |H_3(2)(f^{-1})| \le \frac{1}{180} \left( 5 + 6|c_2| - 10|c_2|^2 + 2|c_2|^3 + 5|c_2|^4 \right) =: \varphi_2(|c_2|),$$
where $0 < |c_2| < 1$ . Since
$$\varphi_2'(|c_2|) = \frac{1}{90} \left( 3 - 10|c_2| + 3|c_2|^2 + 10|c_2|^3 \right),$$
which, after considering this polynomial in the interval [0,1], gives $\varphi_1(|c_2|) \le \varphi_2(1) = \frac{2}{45}$ , and further, from (13),
$$|H_3(2)(f^{-1})| \le \frac{2}{45} = 0.044...$$
The function $f_2(z) = \operatorname{artanh} z$ satisfying $1 + \frac{zf_2''(z)}{f_2'(z)} = \frac{1+z^2}{1-z^2}$ shows that the result is the best possible.
(c) From the definition of the class $\mathcal{S}^{\star}$ we have that there exists a Schwartz function $\omega$ such that
$$\frac{zf'(z)}{f(z)} = \frac{1 + \omega(z)}{1 - \omega(z)}$$
and from here
(14)
$$zf'(z) = \left[1 + 2\left(\omega(z) + \omega^2(z) + \cdots\right)\right] \cdot f(z).$$
As in the two previous cases ((a) and (b)), by comparing the coefficients in the relation (14), and some simple calculations, we have
$$\begin{cases} a_2 = 2c_1 \\ a_3 = c_2 + 3c_1^2 \\ a_4 = \frac{2}{3} \left( c_3 + 5c_1c_2 + 6c_1^3 \right) \\ a_5 = \frac{1}{2} \left( c_4 + \frac{14}{3}c_1c_3 + \frac{43}{3}c_1^2c_2 + 10c_1^4 + 2c_2^2 \right). \end{cases}$$
For the case $a_2 = 0$ we have the next
(15)
$$a_3 = c_2, \quad a_4 = \frac{2}{3}c_3, \quad a_5 = \frac{1}{2}(c_4 + 2c_2^2).$$
So, from (3) and (15) we obtain
$$H_3(2)(f^{-1}) = \frac{1}{18} \left( 9c_2c_4 - 8c_3^2 - 36c_2^3 \right),$$
and from here
$$|H_3(2)(f^{-1})| \le \frac{1}{18} \left( 9|c_2||c_4| + 8|c_3|^2 + 36|c_2|^3 \right).$$
Using estimates for $|c_4|$ and $|c_3|$ from Lemma1 (with $c_1=0$ ) from the last relation we receive
(16)
$$|H_3(2)(f^{-1})| \le \frac{1}{18} \left( 8 + 9|c_2| - 16|c_2|^2 + 27|c_2|^3 + 8|c_2|^4 \right) =: \varphi_3(|c_2|),$$
where $0 < |c_2| < 1$ . Since
$$\varphi_3'(|c_2|) = \frac{1}{18} \left( 9 - 32|c_2| + 81|c_2|^2 + 32|c_2|^3 \right)$$
$$= \frac{1}{18} \left[ 9(1 - 3|c_2|)^2 + 22|c_2| + 32|c_2|^3 \right] > 0,$$
then $\varphi_3(|c_2|) \leq \varphi_3(1) = 2$ , and from (16),
$$|H_3(2)(f^{-1})| \le 2.$$
The result is the best possible as the function $f_3(z) = \frac{z}{1-z^2}$ shows.
(d) From the definition of the class $S_s^{\star}$ we have that there exists a Schwartz function $\omega$ such that
$$\frac{2zf'(z)}{f(z) - f(-z)} = \frac{1 + \omega(z)}{1 - \omega(z)},$$
and from here
(17)
$$2zf'(z) = [1 + 2(\omega(z) + \omega^2(z) + \cdots)] \cdot [f(z) - f(-z)].$$
Similarly as in previous cases, by comparing the coefficients in the relation (17), after some simple calculations, we receive
(18)
$$\begin{cases} a_2 = c_1 \\ a_3 = c_2 + c_1^2 \\ a_4 = \frac{1}{2} \left( c_3 + 3c_1c_2 + 2c_1^3 \right) \\ a_5 = \frac{1}{2} \left( c_4 + 2c_1c_3 + 5c_1^2c_2 + 2c_1^4 + 2c_2^2 \right). \end{cases}$$
For $a_2 = 0$ $(c_1 = 0)$ , from (18) we get
$$a_3 = c_2$$
, $a_4 = \frac{1}{2}c_3$ , $a_5 = \frac{1}{2}(c_4 + 2c_2^2)$ ,
and using (3),
$$H_3(2)(f^{-1}) = \frac{1}{4} \left( 2c_2c_4 - c_3^2 - 8c_2^3 \right),$$
and from here
$$|H_3(2)(f^{-1})| \le \frac{1}{4} (2|c_2||c_4| + |c_3|^2 + 8|c_2|^3).$$
Using the estimates for $|c_4|$ and $|c_3|$ from Lemma 1 (with $c_1 = 0$ ) from the last relation we have
$$(19) |H_3(2)(f^{-1})| \le \frac{1}{4} \left( 1 + 2|c_2| - 2|c_2|^2 + 6|c_2|^3 + |c_2|^4 \right) =: \varphi_4(|c_2|),$$
where $0 \le |c_2| \le 1$ . Since
$$\varphi_4'(|c_2|) = \frac{1}{2} \left( 1 - 2|c_2| + 9|c_2|^2 + 2|c_2|^3 \right)$$
= $\frac{1}{2} \left[ (1 - |c_2|)^2 + 8|c_2| + 2|c_2|^3 \right] > 0,$
then $\varphi_4$ is an increasing function and $\varphi_4(|c_2|) \leq \varphi_4(1) = 2$ . So, from (19),
$$|H_3(2)(f^{-1})| \le 2.$$
This result is the best possible as the function $f_4$ defined by
$$\frac{2zf_4'(z)}{f_4(z) - f_4(-z)} = \frac{1+z^2}{1-z^2}$$
shows.
Remark 1. From the relation (4) we get the following.
(a) For $f \in \mathcal{R}$ ,
$$|H_3(2)(f^{-1}) - H_3(2)(f)| = 3|a_3|^3 = 3\left(\frac{2}{3}|c_2|\right)^3 \le \frac{8}{9},$$
and the result is the best possible as the function $f_1$ shows (in this case $H_3(2)(f_1) = \frac{4}{15}$ and $H_3(2)(f_1^{-1}) = -\frac{28}{45}$ ).
(b) For $f \in \mathcal{C}$ ,
$$|H_3(2)(f^{-1}) - H_3(2)(f)| = 3|a_3|^3 = 3\left(\frac{|c_2|}{3}\right)^3 \le \frac{1}{9},$$
and the result is the best possible as the function $f_2$ shows.
(c) For $f \in \mathcal{S}^*$ ,
$$|H_3(2)(f^{-1}) - H_3(2)(f)| = 3|a_3|^3 = 3|c_2|^3 \le 3,$$
and the result is the best possible as the function $f_3$ shows.
(d) For $f \in \mathcal{S}_s^{\star}$ ,
$$|H_3(2)(f^{-1}) - |H_3(2)(f)| = 3|a_3|^3 = 3|c_2|^3 \le 3,$$
and the result is the best possible for the function $f_4$ .
For obtaining the corresponding result for the whole class S we will use method based on Grunsky coefficients. In the proof we will use mainly the notations and results given in the book of N.A. Lebedev ([3]).
Here are basic definitions and results.
Let $f \in \mathcal{S}$ and let
$$\log \frac{f(t) - f(z)}{t - z} = \sum_{p,q=0}^{\infty} \omega_{p,q} t^p z^q,$$
where $\omega_{p,q}$ are the Grunsky's coefficients with property $\omega_{p,q} = \omega_{q,p}$ . For those coefficients we have the next Grunsky's inequality ([2, 3]):
(20)
$$\sum_{q=1}^{\infty} q \left| \sum_{p=1}^{\infty} \omega_{p,q} x_p \right|^2 \le \sum_{p=1}^{\infty} \frac{|x_p|^2}{p},$$
where $x_p$ are arbitrary complex numbers such that last series converges.
Further, it is well-known that if the function f given by (1) belongs to S, then also
(21)
$$\tilde{f}_2(z) = \sqrt{f(z^2)} = z + c_3 z^3 + c_5 z^5 + \cdots$$
belongs to the class S. Then, for the function $\tilde{f}_2$ we have the appropriate Grunsky's coefficients of the form $\omega_{2p-1,2q-1}^{(2)}$ and the inequality (20) has the form:
(22)
$$\sum_{q=1}^{\infty} (2q-1) \left| \sum_{p=1}^{\infty} \omega_{2p-1,2q-1} x_{2p-1} \right|^2 \le \sum_{p=1}^{\infty} \frac{|x_{2p-1}|^2}{2p-1}.$$
Here, and further in the paper we omit the upper index (2) in $\omega_{2p-1,2q-1}^{(2)}$ if compared with Lebedev's notation.
If in the inequality (22) we put $x_1 = 1$ and $x_{2p-1} = 0$ for p = 2, 3, ..., then we receive
(23)
$$|\omega_{11}|^2 + 3|\omega_{13}|^2 + 5|\omega_{15}|^2 + 7|\omega_{17}|^2 \le 1.$$
As it has been shown in [3, p.57], if f is given by (1) then the coefficients $a_2$ , $a_3$ , $a_4$ and $a_5$ are expressed by Grunsky's coefficients $\omega_{2p-1,2q-1}$ of the function $\tilde{f}_2$ given by (21) in the following way:
$$a_{2} = 2\omega_{11},$$
$$a_{3} = 2\omega_{13} + 3\omega_{11}^{2},$$
$$a_{4} = 2\omega_{33} + 8\omega_{11}\omega_{13} + \frac{10}{3}\omega_{11}^{3},$$
$$a_{5} = 2\omega_{35} + 8\omega_{11}\omega_{33} + 5\omega_{13}^{2} + 18\omega_{11}^{2}\omega_{13} + \frac{7}{3}\omega_{11}^{4},$$
$$0 = 3\omega_{15} - 3\omega_{11}\omega_{13} + \omega_{11}^{3} - 3\omega_{33},$$
$$0 = \omega_{17} - \omega_{35} - \omega_{11}\omega_{33} - \omega_{13}^{2} + \frac{1}{3}\omega_{11}^{4}.$$
We note that in the cited book of Lebedev there exists a typing mistake for the coefficient $a_5$ . Namely, instead of the term $5\omega_{13}^2$ , there is $5\omega_{15}^2$ .