fact, fails for $n=3$. Note that the two estimates have tangency of infinite order at $k=0$, while the difference from the r.h.s. of  behaves asymptotically as $4k^2$. The three estimates are shown in Figure \[FI\_graph\].

![Estimates for $|a_3|$ mentioned in Remark \[RM\_Krushkal\].[]{data-label="FI_graph"}](graphM.pdf){width="9.5cm"}

2ex

The class ${\mathcal{S}}_k^B$, ${k\in(0,1)}$, admits a Loewner-type parametric representation. Denote by ${\mathcal{H}}_k$ the class of all normalized Herglotz functions $p$ satisfying $p({\mathbb{D}},t)\subset U(k)$ for a.e. $t\ge0$, where $U(k)$ is the closed disk defined in Theorem \[TH\_Becker\]. As it follows from the very definition, ${\mathcal{S}}_k^B$ coincides with the image of the map $${\mathcal{H}}_k\ni p~\mapsto~f:=\lim_{t\to+\infty}e^{t}w(z,t)~\in\,{\mathcal{S}},$$ where for each $z\in{\mathbb{D}}$ the function $[0,+\infty)\ni t\mapsto w(z,t)\in{\mathbb{D}}$ is defined as the unique solution to the initial value problem . Write $p(z,t)=1+p_1(t)z+p_2(t)z^2+\ldots$ for all $z\in{\mathbb{D}}$ and a.e. $t\ge0$ and let $$f(z,t):=e^{t}w(z,t)=z+a_2(t)z^2+a_3(t)z^3+\ldots$$ There is one-to-one correspondence between the class of all normalized Herglotz functions and ${\mathcal{H}}_k$. Indeed, $p\in{\mathcal{H}}_k$ if and only if it can be written as $p(\cdot,t)=L\circ p_0(\cdot,t)$ for a.e. $t\ge0$, where $p_0(z,t)=1+c_1(t)z+c_2(t)z^2+\ldots$ is an arbitrary normalized Herglotz function and $$L(z):=\frac{1+Kz}{K+z},\quad K:=\frac{1+k}{1-k},$$ is a conformal map of ${\mathbb{H}}:=\{z\colon\Re z>0\}$ onto $U(k)$ with $L(1)=1$.

As usual, from  we obtain the initial value problem for the coefficients $a_2$ and $a_3$, $$\begin{aligned}
\frac{{\mathrm{d}}a_2}{{\mathrm{d}}t}&=-e^{-t} p_1(t)=-ke^{-t} c_1(t), & 