\section{Preliminaries}\label{sec:prelim}

\subsection{The constant}\label{sec:extremal}

\begin{definition}\label{def:K}
For $n\ge 2$ let $\mathcal P_n$ be the set of complex polynomials $p$ of degree $n$ with $p(0)=0$ and $p'(0)=1$. Every critical
point $b$ of $p\in\mathcal P_n$ is non-zero, and we put
\[
  m(p):=\min_{p'(b)=0}\frac{|p(b)|}{|b|},\qquad K(n):=\sup_{p\in\mathcal P_n} m(p),\qquad K^*:=\sup_{n\ge 2}K(n).
\]
\end{definition}

For a polynomial $P$ and a point $z$ with $P'(z)\neq 0$, the polynomial $p(u)=(P(z+u)-P(z))/P'(z)$ is normalised, its critical
points are the points $b$ with $z+b$ critical for $P$, and $|P(z)-P(z+b)|/(|b|\,|P'(z)|)=|p(b)|/|b|$. Hence $K(n)$ is the best
constant in Smale's inequality in degree $n$, and $K^*$ is Smale's mean value constant (denoted $\mathcal M$ in~\cite{JatarNg2026b}).
Smale proved $K^*\le4$~\cite{Smale1981}; the polynomial $z-z^n/n$ has $m=(n-1)/n$, so $K^*\ge1$.

\subsection{Domain pairs}

Besides the admissible pairs $(W,B)$ of Section~\ref{sec:intro}, we use a second kind of pair, which belongs to the domain
formulation.

\begin{definition}\label{def:admissible}
A \emph{domain pair} $(\Omega,\Phi)$ consists of a bounded domain $\Omega$ with $0\in\Omega$ and $\mathbb C\setminus\overline\Omega$
connected, and a function $\Phi$ holomorphic on an open set containing $\overline\Omega$, with $\Phi(0)=0$, $\Phi'(0)=1$ and $\Phi'(z)\neq0$
for every $z\in\overline\Omega$. Its \emph{ratio} is
\[
  \mu(\Omega,\Phi):=\min_{z\in\partial\Omega}\frac{|\Phi(z)|}{|z|},
\]
and $\Lambda$ is the supremum of $\mu(\Omega,\Phi)$ over all domain pairs.
\end{definition}

Jatar and Ng~\cite[Thm.~4.4]{JatarNg2026b} proved that $K^*$ equals the supremum $\Lambda_{\rm JN}$ of the ratio over the pairs in which $\Omega$
is a bounded \emph{analytic Jordan} domain containing $0$ and $\Phi$ is locally univalent near $\overline\Omega$. Such pairs are
domain pairs, so $\Lambda_{\rm JN}\le\Lambda$. The next lemma gives the reverse inequality, so the wider class has the same supremum; Theorem~\ref{thm:A} is not used below.

\begin{theorem}\label{thm:A}
$K^*=\Lambda$.
\end{theorem}

\begin{lemma}\label{lem:analytic}
For every domain pair $(\Omega,\Phi)$ and every $\varepsilon>0$ there is a bounded analytic Jordan domain $\Omega'$ with
$\overline\Omega\subset\Omega'$, such that $\Phi$ is holomorphic near $\overline{\Omega'}$, $\Phi'\neq0$ on $\overline{\Omega'}$, and
$\min_{\partial\Omega'}|\Phi(z)/z|\ge\mu(\Omega,\Phi)-\varepsilon$.
\end{lemma}

\begin{proof}
Write $\mu=\mu(\Omega,\Phi)$ and $E:=\overline\Omega$, a non-degenerate continuum with $\mathbb C\setminus E$ connected; let
$r_0:=\operatorname{dist}(0,\partial\Omega)>0$. Fix $\delta_0\in(0,r_0/2)$ such that $\{z:\operatorname{dist}(z,E)\le\delta_0\}$
lies in the domain of $\Phi$. On the compact set $\{\operatorname{dist}(z,\partial\Omega)\le\delta_0\}$, where $|z|\ge r_0/2$, the
function $|\Phi(z)/z|$ is uniformly continuous, and it is $\ge\mu$ on $\partial\Omega$. Choose $\delta\in(0,\delta_0]$ such that, with
$V:=\{z:\operatorname{dist}(z,E)<\delta\}$, $\Phi'\neq0$ on $\overline V$ and $|\Phi(z)/z|\ge\mu-\varepsilon$ for every
$z\in\overline V\setminus\Omega$. (Such a $z$ lies within $\delta$ of $\partial\Omega$: if its nearest point of $E$ lies in
$\Omega$, the segment to it crosses $\partial\Omega$.)

Let $g_E$ be the Green function of $\mathbb C\setminus E$ with pole at infinity; it is continuous on $\mathbb C$ (every boundary point of $\mathbb C\setminus E$ is regular, as $E$ is a non-degenerate
continuum~\cite[Sect.~4.2]{Ransford1995}), zero on $E$ and positive off $E$, and $\operatorname{cap}(E)>0$. Let $\mathcal F_n$ be Fekete
polynomials of $E$; their zeros, the Fekete points, lie in $E$. Then
$\|\mathcal F_n\|_E^{1/n}\to\operatorname{cap}(E)$, and $\tfrac1n\log|\mathcal F_n(z)|\to\log\operatorname{cap}(E)+g_E(z)$ locally uniformly on
$\mathbb C\setminus E$ \cite{Ransford1995,SaffTotik1997}. Let $g_1>0$ be the minimum of $g_E$ on $\{\operatorname{dist}(z,E)=\delta\}$,
fix $\tau>0$ with $3\tau<g_1$, then $n$ so large that $\|\mathcal F_n\|_E^{1/n}<\operatorname{cap}(E)e^{\tau}$ and
$|\mathcal F_n|^{1/n}>\operatorname{cap}(E)e^{g_1-\tau}$ on $\{\operatorname{dist}(z,E)=\delta\}$, and put $q:=\mathcal F_n/(\operatorname{cap}(E)e^{2\tau})^n$.
Then $|q|<1$ on $E$, and $|q|>1$ on $\{\operatorname{dist}(z,E)\ge\delta\}$ ($\log|q|$ is harmonic there, positive on its boundary
and tends to $+\infty$). Moving $\tau$ slightly, we may assume that no critical value of $q$ has modulus $1$.

Put $\Omega':=\{|q|<1\}$. Every component of $\Omega'$ contains a zero of $q$ (minimum modulus principle), and all zeros of $q$ lie
in the connected set $E\subset\Omega'$; so $\Omega'$ is a domain, and $E\subset\Omega'\subset V$. As $q'\neq0$ on $\{|q|=1\}$, this set
is a finite union of disjoint analytic Jordan curves, each point of which is a limit of points with $|q|>1$. The set $\{|q|>1\}$ is
connected (a bounded component $H$ would have $|q|=1$ on $\partial H$, hence $|q|\le1$ on $H$ by the maximum principle), so its
closure $\mathbb C\setminus\Omega'$ is connected and $\Omega'$ is simply connected. A bounded simply connected plane domain has
connected boundary~\cite{Newman1951}; so $\partial\Omega'$ is one analytic Jordan curve and $\Omega'$ is its interior. Finally
$\overline{\Omega'}\subset\overline V$ lies in the domain of $\Phi$ with $\Phi'\neq0$ there, and $\partial\Omega'\subset\overline V\setminus\Omega$
(as $|q|<1$ on $E$), where $|\Phi(z)/z|\ge\mu-\varepsilon$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:A}]
$K^*=\Lambda_{\rm JN}\le\Lambda$ by~\cite[Thm.~4.4]{JatarNg2026b}, and $\Lambda\le\Lambda_{\rm JN}$ by Lemma~\ref{lem:analytic}.
\end{proof}

\begin{remark}[Domain pairs]\label{rem:domainonly}
Jatar and Ng~\cite[Sect.~9]{JatarNg2026b} mention the estimate $\Lambda_{\rm JN}>1.00114$ and ask for larger bounds through
$\Lambda_{\rm JN}$. By Corollary~\ref{cor:K2N18BM} and the paragraph after Theorem~\ref{thm:twozeroK2N18}, the Blaschke models behind
Theorem~\ref{thm:intro} give pairs of the kind defining $\Lambda_{\rm JN}$ with ratio $\sqrt{J_{\rm I}}>1.1559837$.
\end{remark}

We also use the inverse-branch argument behind Smale's bound, written for pairs.

\begin{proposition}\label{prop:smale}
$\mu(\Omega,\Phi)\le4$ for every domain pair.
\end{proposition}

\begin{proof}
Let $G$ be the inverse germ of $\Phi$ at $0$, so $G(0)=0$ and $G'(0)=1$. Call $r>0$ \emph{good} if $G$ extends holomorphically to
$D_r=\{|w|<r\}$ with $G(D_r)\subset\Omega$; then $\Phi(G(w))=w$ on $D_r$ by the identity theorem, so $G$ is injective there.
Extensions to two disks agree on the smaller one, so $G$ is defined on $D_\rho$, $\rho:=\sup\{r\text{ good}\}>0$. Koebe's
one-quarter theorem for $G(rw)/r$ gives $G(D_r)\supset D_{r/4}$, and $G(D_r)$ lies in the bounded set $\Omega$; so $\rho<\infty$.

\emph{Step 1: $\overline{G(D_\rho)}$ meets $\partial\Omega$.} If not, $Q_0:=\overline{G(D_\rho)}$ is a compact subset of
$\Omega$. Fix $\zeta$ with $|\zeta|=\rho$ and take $w_n\to\zeta$ in $D_\rho$ with $G(w_n)\to z_0\in Q_0$ (a subsequence). Then
$\Phi(z_0)=\zeta$ and $\Phi'(z_0)\neq0$, so $\Phi$ maps an open set $V_0$ with $z_0\in V_0\subset\Omega$ biholomorphically onto a disk
$D(\zeta,s)$; let $G_\zeta=(\Phi|_{V_0})^{-1}$. On the convex set $D(\zeta,s)\cap D_\rho$ the set $\{w:G(w)\in V_0\}$ is open and
non-empty, and there $G=G_\zeta$, because both take values in $V_0$, where $\Phi$ is injective, and $\Phi(G(w))=w=\Phi(G_\zeta(w))$. By the
identity theorem $G=G_\zeta$ on $D(\zeta,s)\cap D_\rho$. Cover the circle $|w|=\rho$ by finitely many such disks; two overlapping
disks centred on the circle meet inside $D_\rho$, so the extensions agree on overlaps. This extends $G$ to $D_{\rho+\varepsilon}$
with image in $\Omega$, contrary to the choice of $\rho$.

\emph{Step 2.} Take $z^*\in\overline{G(D_\rho)}\cap\partial\Omega$, $z^*=\lim G(w_n)$ with $w_n\in D_\rho$, $w_n\to w^*$ (a
subsequence). As $\Phi$ is continuous on $\overline\Omega$, $\Phi(z^*)=w^*$, and $|w^*|=\rho$ (if $|w^*|<\rho$, then $z^*=G(w^*)$ would
lie in $\Omega$). Since $z^*\notin\Omega$ and $G(D_\rho)\supset D_{\rho/4}$, we have $|z^*|\ge\rho/4$. Therefore
$\mu(\Omega,\Phi)\le|\Phi(z^*)|/|z^*|\le\rho/(\rho/4)=4$.
\end{proof}

\begin{remark}\label{rem:zeros}
If $\mu(\Omega,\Phi)>1$, then $\Phi(z)/z$ has zeros in $\Omega$: otherwise $\Phi(z)/z$ is zero-free on $\Omega$ with value $1$ at $0$,
and the minimum principle gives $\min_{\partial\Omega}|\Phi(z)/z|\le1$.
\end{remark}

\subsection{Blaschke models}\label{sec:blaschke}

Let $\mathbb D$ be the open unit disk, $\mathcal S$ the class of univalent $W$ on $\mathbb D$ with $W(0)=0$, $W'(0)=1$, and
$B_a(w)=(w-a)/(1-\bar a w)$ for $a\in\mathbb D$.

\begin{definition}\label{def:model}
A \emph{Blaschke model} is a pair $(W,B)$ such that
\begin{enumerate}
\item[(M1)] $W\in\mathcal S$, and $W$ extends to a holomorphic univalent map on a disk $\{|w|<R\}$ for some $R>1$;
\item[(M2)] $B=cB_{a_1}\cdots B_{a_k}$ with $|c|=1$, $k\ge0$ and $a_j\in\mathbb D$ (a finite Blaschke product);
\item[(M3)] $(W B)'$ has no zero on the closed disk $\{|w|\le1\}$.
\end{enumerate}
Its \emph{gain} is $1/|B(0)|$.
\end{definition}

In the language of~\cite{JatarNg2026b}, the Blaschke models with $B(0)>0$ are exactly the \emph{regular}\label{term:regular} pairs
(admissible pairs whose $W$ extends to a univalent function on a neighbourhood of the closed disk, with $(WB)'\neq0$ there);
a Blaschke model differs from a regular pair only in that $B(0)$ need not be positive. Jatar and Ng~\cite[Thm.~1.2]{JatarNg2026b} proved that $K^*$ is the supremum of $1/B(0)$ over the pairs with $(WB)'\neq0$
in the open disk, and~\cite[Lemma~2.1]{JatarNg2026b} that regular pairs give the same supremum. Hence:

\begin{theorem}[{\cite[Thm.~1.2, Lemma~2.1]{JatarNg2026b}}]\label{thm:BF}
$K^*$ is the supremum of the gains $1/|B(0)|$ over all Blaschke models.
\end{theorem}

Sections~\ref{sec:pockets}--\ref{sec:own} use this closed-disk form, with an arbitrary base point $w_0$:

\begin{corollary}\label{cor:BF}
$K^*$ is the supremum of $1/|B(w_0)|$ over triples $(W,B,w_0)$ with $W$ holomorphic and univalent on a disk $\{|w|<R\}$,
$R>1$, $w_0\in\mathbb D$, $B$ as in \textup{(M2)}, and $\bigl((W-W(w_0))B\bigr)'$ zero-free on $\{|w|\le1\}$.
\end{corollary}

\begin{proof}[Proof of Corollary~\ref{cor:BF}]
Given $(W,B,w_0)$, let $\varphi_{w_0}(u)=(u+w_0)/(1+\bar w_0u)$ and $G=(W-W(w_0))B$. Since $G'(w_0)=W'(w_0)B(w_0)\neq0$,
$B(w_0)\neq0$. Put $\widetilde W=(W\circ\varphi_{w_0}-W(w_0))/c_1$ with $c_1=W'(w_0)\varphi_{w_0}'(0)$, and $\widetilde B=B\circ\varphi_{w_0}$, a
finite Blaschke product of the same degree. Then $\widetilde W$ satisfies (M1), because $\varphi_{w_0}$ maps a disk $\{|u|<R''\}$, $R''>1$,
into $\{|w|<R\}$, and $(\widetilde W\widetilde B)'=G'(\varphi_{w_0})\varphi_{w_0}'/c_1$ has no zero on $\{|u|\le1\}$. So
$1/|B(w_0)|=1/|\widetilde B(0)|\le K^*$ by Theorem~\ref{thm:BF}; the case $w_0=0$ gives the converse.
\end{proof}

\subsection{The winding criterion}\label{sec:criterion}
Let $W$ be holomorphic and univalent on a disk $\{|w|<R\}$, $R>1$, let $w_0\in\mathbb D$, and let $B$ be as in \textup{(M2)}
with $\deg B=k$; we call $(W,B,w_0)$ a \emph{triple}\label{term:triple} (the condition of Corollary~\ref{cor:BF} on $((W-W(w_0))B)'$ is not
assumed). Put $G=(W-W(w_0))B$, so that $G'=W'N$ with
\[
  N=B+\frac{W-W(w_0)}{W'}\,B'.
\]
On the circle $N=B\gamma$, where
\[
  \gamma=1+\sigma v,\qquad \sigma=\frac{wB'}{B}=\sum_{j=1}^k\frac{1-|a_j|^2}{|w-a_j|^2}\ \ (\ge0,\ >0\text{ if }k\ge1),\qquad
  v=\frac{W-W(w_0)}{wW'}.
\]
The following is the argument principle, as used in~\cite[Sect.~8]{JatarNg2026b} (see the display before their Lemma~8.1; for $w_0=0$ their $g$ is our
$1/v$); we record it in the form used below.

\begin{proposition}\label{prop:criterion}
$G'$ has no zero on $\{|w|\le1\}$ if and only if $\gamma\neq0$ on the circle and $\operatorname{wind}(\gamma)=-k$.
\end{proposition}

\begin{proof}
$W'\neq0$ near the closed disk, so the zeros of $G'$ there are those of $N$, which is holomorphic near the closed disk. On the
circle $|B|=1$, so $N\neq0$ there iff $\gamma\neq0$, and then the number of zeros of $N$ in $\mathbb D$ is
$\operatorname{wind}(N)=\operatorname{wind}(B)+\operatorname{wind}(\gamma)=k+\operatorname{wind}(\gamma)$.
\end{proof}