\section{Near-extremal structure}\label{sec:structure}

In this section we work in the base-$0$ frame: $(W,B)$ satisfies (M1) and (M2) with $k=\deg B\ge1$ and zero set $A$, normalised by
$B(0)>0$, the gain
is $K_0:=1/|B(0)|$, $\Omega=W(\mathbb D)$ and $r_{\rm in}:=\operatorname{dist}(0,\partial\Omega)\ge1/4$ (Koebe). A statement for
Blaschke models\label{term:model} assumes (M3) as well; several statements need $(W B)'\neq0$ only on the open disk, and we say so. Put\label{term:F4}
\[
  \Phi(z):=z\,B(W^{-1}(z)),
\]
holomorphic near $\overline\Omega$, with $\Phi(0)=0$, $\Phi'(0)=B(0)$, $\Phi'(W(w))W'(w)=(W B)'(w)$ and $|\Phi(z)|=|z|$ on $\partial\Omega$.
Lemma~\ref{lem:branch}, Theorem~\ref{thm:C}, Proposition~\ref{prop:green}, Corollary~\ref{cor:offtip} and the first statement of
Proposition~\ref{prop:Ninf} hold for all admissible pairs: there $W\in\mathcal S$ need not extend past the circle, and $\Phi$ is
defined on $\Omega$ only.
On the circle, $v=W/(wW')$, $\sigma=wB'/B=|B'|>0$ and $\gamma=1+\sigma v$ (Section~\ref{sec:criterion} with $w_0=0$); ``mean''
is the integral against $d\theta/2\pi$. Since $v$ is holomorphic and zero-free near the closed disk with $v(0)=1$, the mean of
$\log|v|$ is $0$. The Koebe function is $\kappa(w)=w/(1-w)^2$; it maps $\mathbb D$ onto $\mathbb C\setminus(-\infty,-1/4]$, and the tip
$-1/4=\kappa(-1)$ corresponds to $\theta=\pi$.

\subsection{The inverse branch and the limit}

\begin{lemma}[inverse branches]\label{lem:branch}
Let $(W,B)$ be an admissible pair. For every zero $b$ of $\Phi$ in $\Omega$, i.e.\ $b\in\{0\}\cup W(A)$, let $E_b$ be the
component of $\{z\in\Omega:|\Phi(z)|<r_{\rm in}\}$ that contains $b$. Then $\Phi$ maps $E_b$ bijectively onto $D(0,r_{\rm in})$. In
particular $E_b$ contains no other zero of $\Phi$, and the components $E_b$ are pairwise disjoint.
\end{lemma}

\begin{proof}
$\Phi'\neq0$ in $\Omega$. Let $z_n\in E_b$ leave every compact subset of $E_b$. We show that $|\Phi(z_n)|\to r_{\rm in}$
along a subsequence; applied to every subsequence, this gives $|\Phi(z_n)|\to r_{\rm in}$. Pass to a subsequence that
converges on the Riemann sphere, to $z$. If $z\in\Omega$, then $z\notin E_b$ and $|\Phi(z)|=r_{\rm in}$, since
$|\Phi(z)|<r_{\rm in}$ would put a disk about $z$ into the component $E_b$. Otherwise the points $w_n=W^{-1}(z_n)$ tend to
the circle (a limit point $w\in\mathbb D$ would give $z=W(w)\in\Omega$), so $|B(w_n)|\to1$, and
$|\Phi(z_n)|=|z_n|\,|B(w_n)|$ tends to $|z|\ge r_{\rm in}$ if $z\in\partial\Omega$ and to $\infty$ if $z=\infty$. As
$|\Phi|<r_{\rm in}$ on $E_b$, the case $z=\infty$ does not occur, and in the other cases $|\Phi(z_n)|\to r_{\rm in}$. So
$\Phi:E_b\to D(0,r_{\rm in})$ is proper, and it is a
local homeomorphism, hence a covering of the simply connected disk by the connected set $E_b$, i.e.\ a bijection. Since $\Phi$ is
injective on $E_b$ and vanishes at $b$, $E_b$ contains no other zero of $\Phi$; two components containing different zeros are
therefore different, hence disjoint.
\end{proof}

For $b=0$ this is the argument in the proof of~\cite[Thm.~7.1]{JatarNg2026b}.

\begin{theorem}\label{thm:C}
Let $(W_n,B_n)$ be admissible pairs with gains $K_{0,n}=1/B_n(0)\to4$. Then, after rotations
$W_n\mapsto e^{i\theta_n}W_n(e^{-i\theta_n}w)$, $B_n\mapsto B_n(e^{-i\theta_n}w)$,
\[
  W_n\to\kappa\quad\text{and}\quad B_n\to(1-w)^2/4\qquad\text{locally uniformly in }\mathbb D,
\]
$\Phi_n/B_n(0)\to\kappa^{-1}$ locally uniformly on $\kappa(\mathbb D)$, the zeros of $B_n$ leave every compact subset of $\mathbb D$, and
$\deg B_n\to\infty$.
\end{theorem}

\begin{proof}
We first pass to a subsequence; the last step removes it. Write $r_n:=r_{\rm in}(\Omega_n)$ and let $g_n:D(0,r_n)\to E_0$ be the inverse of Lemma~\ref{lem:branch} at $b=0$.

\emph{Schwarz step.} $h_n(u):=W_n^{-1}(g_n(r_nu))$ is univalent from $\mathbb D$ into $\mathbb D$, with $h_n(0)=0$ and
$h_n'(0)=r_n/B_n(0)=r_nK_{0,n}>0$. By Schwarz's lemma $r_nK_{0,n}\le1$, so $r_n\le1/K_{0,n}\to1/4$, while $r_n\ge1/4$; hence
$r_n\to1/4$ and $h_n'(0)\to1$. By compactness of $\mathcal S$, $W_n\to W\in\mathcal S$ along a subsequence. Points
$\zeta_n\in\partial\Omega_n$ with $|\zeta_n|=r_n$ have a limit point $\zeta$ with $|\zeta|=1/4$, and $\zeta\notin W(\mathbb D)$: otherwise,
by Hurwitz's theorem, $W_n$ would take the value $\zeta_n$ in $\mathbb D$ for large $n$, although $\zeta_n\notin\Omega_n$. So
$\operatorname{dist}(0,\partial W(\mathbb D))\le1/4$, which is the case of equality in Koebe's one-quarter theorem: $W$ is a
rotation of $\kappa$, and suitable rotations $\theta_n$ make it $\kappa$.

\emph{Limit.} The self-maps $h_n$ fix $0$ and $h_n'(0)\to1$, so by normality and Schwarz's lemma $h_n\to\mathrm{id}$. From
$\Phi_n(g_n(r_nu))=r_nu$ and $h_n(u)\neq0$ for $u\neq0$,
\[
  B_n(h_n(u))=\frac{r_nu}{W_n(h_n(u))}\ \longrightarrow\ \frac{u}{4\kappa(u)}=\frac{(1-u)^2}4\qquad(u\neq0),
\]
and also at $u=0$ by Vitali's theorem, both sides being bounded and holomorphic. The family $(B_n)$ is equicontinuous on compact
sets, so $B_n\to(1-w)^2/4$. The limit has no zero in $\mathbb D$, so by Hurwitz the zeros of $B_n$ leave every compact set. If
$\deg B_n=d$ along a subsequence, then for every $R<1$ all zeros eventually satisfy $|a|>R$, so $B_n(0)=\prod|a_j|\ge R^d$, which
contradicts $B_n(0)\to1/4$ when $R^d>1/4$; hence $\deg B_n\to\infty$. Finally, by Carath\'eodory's kernel theorem
$W_n^{-1}\to\kappa^{-1}$ locally uniformly on $\kappa(\mathbb D)$, so $\Phi_n(z)=zB_n(W_n^{-1}(z))\to z(1-u)^2/4=u/4$ with
$u=\kappa^{-1}(z)$, and $B_n(0)\to1/4$.

\emph{The whole sequence.} The limits do not depend on the subsequence, so it can be dropped: choose $\theta_n$ to minimise
$\max_{|w|\le1/2}|e^{i\theta}W_n(e^{-i\theta}w)-\kappa(w)|$. Every subsequence has a further subsequence along which
the two steps above apply with some rotations $\theta_n'$; along it the minimum tends to $0$, and since no rotation other than the
identity fixes $\kappa$, $\theta_n-\theta_n'\to0$ modulo $2\pi$, so the conclusions hold with $\theta_n$. Hence they hold for the whole
sequence.
\end{proof}

The Schwarz step alone gives $K_0\le1/r_{\rm in}\le4$ for each $(W,B)$ with $(W B)'\neq0$ on $\mathbb D$ (compare~\cite[Thm.~7.1]{JatarNg2026b}:
$B(0)\ge r_{\rm in}\ge1/4$), with equality only if $r_{\rm in}=1/4$, i.e.\ if $W$ is
a rotation of $\kappa$, which is excluded by (M1). So every single Blaschke model has gain strictly below $4$; Section~\ref{sec:permodel}
gives quantitative forms.

\begin{remark}\label{rem:Cadmissible}
The statements of Theorem~\ref{thm:C} on $W_n$, on the zeros of $B_n$ and on $\deg B_n$
are in~\cite[Rem.~7.3]{JatarNg2026b}.
\end{remark}

\begin{proposition}[Green density]\label{prop:green}
Under Theorem~\ref{thm:C}, the measures $\Theta_n:=\sum_j\log(1/|a_j|)\,\delta_{a_j/|a_j|}$ on the circle (zeros of $B_n$, with
multiplicity) have mass $\log K_{0,n}\to\log4$ and converge weak-$*$ to $\Theta(s)\,ds$ with
\[
  \Theta(s)=-\frac1\pi\log\Bigl|\sin\frac s2\Bigr| .
\]
\end{proposition}

\begin{proof}
As $|a|\to1$, $g_{\mathbb D}(w,a)=(1+o(1))\log(1/|a|)\,(1-|w|^2)/\bigl|a/|a|-w\bigr|^2$ uniformly for $w$ in a compact subset of $\mathbb D$. So the Poisson integral of $\Theta_n$ tends to $\lim_n\log(1/|B_n(w)|)=\log(4/|1-w|^2)$ for every
$w\in\mathbb D$. Every weak-$*$ limit of $(\Theta_n)$ has mass $\log4$ and this Poisson integral, and a measure is determined by its
Poisson integral. The positive harmonic function $\log(4/|1-w|^2)$ is the Poisson integral of $-2\log|\sin(s/2)|\,ds/2\pi$.
\end{proof}

\subsection{\texorpdfstring{$\sigma$}{sigma} away from the tip}

\begin{lemma}\label{lem:H}
Let $0\le\rho<1$ and $t\in\mathbb R$. If every zero $a$ of $B$ satisfies
$X(a):=(1-|a|^2)(1-\rho^2)/|1-\bar a\rho e^{it}|^2\le1/2$, then
\[
  \log\frac1{|B(\rho e^{it})|}\le4(1-\rho^2)\,\sigma(e^{it}).
\]
\end{lemma}

\begin{proof}
With $w=\rho e^{it}$ we have $1-|B_a(w)|^2=X(a)$, so $g_{\mathbb D}(w,a)=-\frac12\log(1-X(a))\le X(a)$ for $X(a)\le1/2$. Put
$b=\bar ae^{it}$. Then $|1-b|\le|1-\rho b|+(1-\rho)\le2|1-\rho b|$, since $|1-\rho b|\ge1-\rho$; so
$X(a)\le4(1-\rho^2)(1-|a|^2)/|1-b|^2=4(1-\rho^2)P_a(e^{it})$. Sum over the zeros.
\end{proof}

\begin{corollary}\label{cor:offtip}
Under Theorem~\ref{thm:C}, $\sigma_n\to\infty$ uniformly on every closed arc $I$ of the circle that avoids $\theta=\pi$.
\end{corollary}

\begin{proof}
Let $\delta=\min(\operatorname{dist}(I,\pi),\pi/2)$ and $c=\cos(\delta/2)<1$. On $I$, $|1-\rho e^{it}|\le2|\sin(t/2)|+(1-\rho)\le2c+(1-\rho)$,
and for $1-\rho\le2(\sqrt c-c)$ this gives $\log4-2\log|1-\rho e^{it}|\ge-\log c>0$. Since $X(a)\le(1-|a|^2)(1+\rho)/(1-\rho)$ and the
zeros leave compact sets, Lemma~\ref{lem:H} applies for each such $\rho$ once $n$ is large, and
$\log(1/|B_n(\rho e^{it})|)\to\log4-2\log|1-\rho e^{it}|$ uniformly in $t$. Hence
$\liminf_n\min_I\sigma_n\ge-\log c/(4(1-\rho^2))$ for every $\rho\in[1-2(\sqrt c-c),1)$; let $\rho\to1$.
\end{proof}

\input{sections/sec6_NI_ninf}

\begin{lemma}\label{lem:tanh}
For every finite Blaschke product $B$ with $B(0)\neq0$, $\sigma\ge(K_0-1)/(K_0+1)$ on the whole circle. Consequently every pocket of
depth greater than $(K_0+1)/(K_0-1)$ is on; for a Blaschke model this threshold exceeds $5/3$.
\end{lemma}

\begin{proof}
$P_a\ge(1-|a|)/(1+|a|)=\tanh(\ell_a/2)$ with $\ell_a=\log(1/|a|)$, and $\ell\mapsto\tanh(\ell/2)$ is concave with value $0$ at $0$, hence
subadditive; so $\sigma\ge\tanh(\sum_j\ell_{a_j}/2)=\tanh(\frac12\log K_0)=(K_0-1)/(K_0+1)$. For a Blaschke model $K_0<4$.
\end{proof}

\subsection{The cost of the gain}

\begin{lemma}[cost]\label{lem:cost}
For every $\sigma>0$, $|1+\sigma v|\ge\max(1,\sigma|v|/2)$ if $\operatorname{Re}v\ge0$, and $|1+\sigma v|\ge\sin\alpha\,\max(1,\sigma|v|/2)$ if
$\operatorname{Re}v<0$. Hence every Blaschke model satisfies
\[
  \log K_0\ \le\ \mathcal I(\Omega,0)-\operatorname{mean}\log^+\frac{\sigma|v|}2\ \le\ \mathcal I(\Omega,0)+\log2-\operatorname{mean}\log\sigma .
\]
\end{lemma}

\begin{proof}
If $\operatorname{Re}v\ge0$, then $|1+\sigma v|\ge\max(1,\sigma|v|)$. If $\operatorname{Re}v<0$, then $|1+\sigma v|\ge\sin\alpha$ (proof of
Theorem~\ref{thm:sin}), and $|1+\sigma v|\ge\sigma|v|-1\ge\sigma|v|/2$ when $\sigma|v|\ge2$. Take logarithms and use
Proposition~\ref{prop:gain}(a); then $\log^+\ge\log$ and $\operatorname{mean}\log|v|=0$.
\end{proof}

\begin{corollary}\label{cor:cost}
Let $(W_n,B_n)$ be Blaschke models with $K_{0,n}\to4$. Then $\operatorname{mean}\log\sigma_n\to\infty$, hence $\mathcal I(\Omega_n,0)\to\infty$.
Moreover, with $\ell_n=-\log|\gamma_n|$, the mean of $\ell_n^-=\log^+|\gamma_n|$ is at least
$\operatorname{mean}\log\sigma_n-\log2\to\infty$, and the mean of the positive part $\ell_n^+$ is at least
$\log K_{0,n}+\operatorname{mean}\log\sigma_n-\log2$.
\end{corollary}

\begin{proof}
These quantities do not change under rotations or unimodular factors, so
Theorem~\ref{thm:C} and Corollary~\ref{cor:offtip} apply. Thus $\sigma_n\to\infty$ uniformly on a fixed closed
arc $I$ avoiding $\pi$, and $\sigma_n\ge(K_{0,n}-1)/(K_{0,n}+1)\to3/5$ off $I$ (Lemma~\ref{lem:tanh}); so
$\operatorname{mean}\log\sigma_n\to\infty$, and Lemma~\ref{lem:cost} gives $\mathcal I(\Omega_n,0)\to\infty$. Pointwise
$\log^+|\gamma|\ge\log(\sigma|v|/2)$ (if $\sigma|v|\ge2$, then $|\gamma|\ge\sigma|v|-1\ge\sigma|v|/2\ge1$; otherwise the right side is
negative). Take means, use $\operatorname{mean}\log|v|=0$ and $\operatorname{mean}\ell_n=\log K_{0,n}$ (Proposition~\ref{prop:gain}).
\end{proof}

\begin{proposition}[where $|\gamma|<1$]\label{prop:gainwhere}
For every Blaschke model, $|\gamma(w)|<1$ implies $\operatorname{Re}v(w)<0$ and $|W'(w)|>\sigma(w)/8$. In particular, for Blaschke models as in
Theorem~\ref{thm:C}, the positive part of $-\log|\gamma_n|$ on a closed arc $I$ avoiding $\pi$ is carried by a part of $\partial\Omega_n$
where the harmonic density $d\omega/ds=1/(2\pi|W_n'|)$ tends to $0$ uniformly.
\end{proposition}

\begin{proof}
$|1+\sigma v|<1$ means $\sigma|v|^2<-2\operatorname{Re}v$, i.e.\ $\operatorname{Re}v<0$ and $\sigma|v|<-2\cos\alpha\le2$. As
$|W|\ge r_{\rm in}\ge1/4$ on the circle, $|W'|=|W|/|v|>\sigma/8$. Then apply Corollary~\ref{cor:offtip}.
\end{proof}

\begin{remark}\label{rem:relaxed}
For $v=v_W$, by Propositions~\ref{prop:criterion} and~\ref{prop:gain}, the supremum of $\log K_0$ over the Blaschke models on $W$ with
$\deg B\ge1$ is the supremum of $-\operatorname{mean}\log|1+\sigma v|$ over $\sigma=\sum_{j\le k}P_{a_j}$ with $k\ge1$, $1+\sigma v\neq0$ on the
circle and $\operatorname{wind}(1+\sigma v)=-k$.
\end{remark}

\subsection{Exit points and an identity at the first exit}\label{sec:exit}

For a Blaschke model, every zero $b\in\{0\}\cup W(A)$ of $\Phi$ in $\Omega$ is simple ($\Phi'\neq0$ on $\overline\Omega$), so $\Phi$ has a holomorphic
inverse $g_b$ near $0$ with $g_b(0)=b$. Let $R_b$ be the supremum of the $r>0$ such that $g_b$ extends holomorphically to
$D(0,r)$ with values in $\Omega$. Since $\Phi(g_b(w))=w$ there, $g_b$ is injective on $D(0,R_b)$.

\begin{lemma}[exit points]\label{lem:exit}
Let $(W,B)$ be a Blaschke model and $b\in\{0\}\cup W(A)$. Then $r_{\rm in}\le R_b<\infty$. Call $z^*\in\partial\Omega$ an
\emph{exit point} of $b$ if $z^*=\lim g_b(w_n)$ for some $w_n\in D(0,R_b)$ with $|w_n|\to R_b$. Exit points exist, and every exit
point $z^*$ satisfies $|z^*|=|\Phi(z^*)|=R_b$ and the following properties, where $q^*:=\Phi(z^*)$, $u^*:=q^*/R_b$,
$w^*:=W^{-1}(z^*)$ and $h_b(u):=W^{-1}(g_b(R_bu))$.
\begin{enumerate}\renewcommand{\labelenumi}{\textup{(\roman{enumi})}}
\item $g_b$ extends holomorphically to a neighbourhood of $q^*$ with $g_b(q^*)=z^*$; so $h_b$ is a holomorphic map of $\mathbb D$ into
$\mathbb D$, holomorphic at $u^*$, with $h_b(u^*)=w^*$ and $h_b(0)=W^{-1}(b)$.
\item $|z|\ge|z^*|$ for all $z\in\partial\Omega$ near $z^*$: the point $z^*$ is a local minimum of $|z|$ along $\partial\Omega$.
\item $u^*h_b'(u^*)/h_b(u^*)=|h_b'(u^*)|>0$, $v(w^*)$ is real, and
\[
  \sigma(w^*)+\frac1{v(w^*)}=\frac1{|h_b'(u^*)|}.
\]
\end{enumerate}
\end{lemma}

\begin{proof}
The translated pair $(\Omega-b,\,\Phi(b+\cdot)/\Phi'(b))$ is a domain pair, and its inverse germ at $0$ is $w\mapsto g_b(\Phi'(b)w)-b$.
Proposition~\ref{prop:smale} and its Steps~1--2 give $R_b<\infty$ and an exit point; for any exit point $z^*$, a subsequence of the
$w_n$ tends to some $q^*$, and $z^*$ is a limit of $g_b(w_n)$ with $w_n\to q^*$, $|q^*|=R_b$ and $\Phi(z^*)=q^*$; then $|z^*|=|\Phi(z^*)|=R_b$, and $R_b\ge r_{\rm in}$ as $z^*\in\partial\Omega$.

(i) As $\Phi'(z^*)\neq0$, $\Phi$ is injective on a neighbourhood $V_0$ of $z^*$ and $(\Phi|_{V_0})^{-1}$ is defined on a disk $D(q^*,s)$. For
large $n$, $g_b(w_n)\in V_0$, so the open set $\{w\in D(0,R_b)\cap D(q^*,s):g_b(w)\in V_0\}$ is non-empty, and there
$g_b=(\Phi|_{V_0})^{-1}$; by the identity theorem this holds on the convex set $D(0,R_b)\cap D(q^*,s)$. This is the identity-theorem
part of Step~1 of Proposition~\ref{prop:smale}, used at a boundary point ($V_0$ need not lie in $\Omega$). The rest of (i) follows,
as $W^{-1}$ is holomorphic near $\overline\Omega$.

(ii) Let $z\in\partial\Omega$ be so close to $z^*$ that $z\in V_0$ and $\Phi(z)\in D(q^*,s)$. If $|\Phi(z)|<R_b$, then
$z=(\Phi|_{V_0})^{-1}(\Phi(z))=g_b(\Phi(z))\in\Omega$, which is impossible. So $|z|=|\Phi(z)|\ge R_b$.

(iii) $\Phi(g_b(R_bu))=R_bu$ reads $W(h_b(u))\,B(h_b(u))=R_bu$. Taking $u\,\frac d{du}\log$ at $u^*$ and using $wW'/W=1/v$ and
$wB'/B=\sigma$ on the circle,
\[
  \frac{u^*h_b'(u^*)}{h_b(u^*)}\Bigl(\frac1{v(w^*)}+\sigma(w^*)\Bigr)=1;
\]
in particular $h_b'(u^*)\neq0$. Write $u=e^{s+it}$ and $X=\log|h_b(e^{s+it})|$ near $(0,\theta^*)$, $u^*=e^{i\theta^*}$. Then $X\le0$
for $s\le0$ and $X(0,\theta^*)=0$, so $X_t=0$ and $X_s\ge0$ there, and by the Cauchy--Riemann equations for $\log h_b$,
$uh_b'/h_b=X_s-iX_t$ is real and $\ge0$ at $u^*$; it is non-zero by the display. Since $|u^*|=|h_b(u^*)|=1$, it equals $|h_b'(u^*)|$.
\end{proof}

\begin{theorem}[the first exit]\label{thm:T1}
Let $(W,B)$ be a Blaschke model, $b=0$, and let $z^*$, $w^*$, $u^*=e^{i\theta^*}$, $h=h_0$ be as in Lemma~\ref{lem:exit}. Then:
\begin{enumerate}\renewcommand{\labelenumi}{\textup{(\roman{enumi})}}
\item $R_0K_0\le\min_{a\in A}{\hat c}(|a|)<1$, with $\hat c(r)=2r\log(1/r)/(1-r^2)$.
\item Let $\nu$ be the Herglotz measure of the positive harmonic function $\log|u/h(u)|$; its mass is $\log(1/(R_0K_0))$. Then
\[
  \sigma(w^*)+\frac1{v(w^*)}=\frac1{1+\mathcal U},\qquad {\mathcal U}:=\int\frac{d\nu(\varphi)}{2\sin^2((\varphi-\theta^*)/2)}\ \ge\ \frac12\log\frac1{R_0K_0}>0,
\]
and hence $\sigma(w^*)+1/v(w^*)\le2/(2+\log(1/(R_0K_0)))\le(1+R_0K_0)/2<1$.
\item Either $v(w^*)>0$ \textup{(}front case\textup{)}, and then $\sigma(w^*)<1-1/v(w^*)<1$; or $v(w^*)<0$ \textup{(}back case\textup{)},
and then $w^*$ is an on pocket with $\varepsilon(w^*)=-1$.
\end{enumerate}
\end{theorem}

\begin{proof}
(i) $h$ maps $\mathbb D$ into $\mathbb D$ with $h(0)=0$ and $h'(0)=R_0/B(0)=R_0K_0>0$, and $h$ omits every $a\in A$: if
$g_0(w)=W(a)$, then $w=\Phi(W(a))=W(a)B(a)=0$, but $g_0(0)=0\neq W(a)$. With hyperbolic densities of curvature $-4$, $\lambda_{\mathbb D}(z)=1/(1-|z|^2)$
and, on the punctured disk $\mathbb D^*=\mathbb D\setminus\{0\}$, $\lambda_{\mathbb D^*}(z)=1/(2|z|\log(1/|z|))$ (cover $\mathbb D^*$ by the upper half-plane
through $\tau\mapsto e^{i\tau}$); the automorphism $B_a$ maps $\mathbb D\setminus\{a\}$ onto $\mathbb D^*$
with $|B_a(0)|=|a|$ and $|B_a'(0)|=1-|a|^2$, so the density of $\mathbb D\setminus\{a\}$ at $0$ is
$(1-|a|^2)/(2|a|\log(1/|a|))=1/\hat c(|a|)$. Schwarz--Pick gives $h'(0)\le {\hat c}(|a|)$, and $\hat c(r)=\ell/\sinh \ell<1$ with $\ell=\log(1/r)$.

(ii) $h$ is univalent with no zero besides $0$, so $\log(u/h(u))$ is holomorphic in $\mathbb D$, with real part $\ge0$ (Schwarz)
and value $\log(1/(R_0K_0))$ at $0$; hence $\log(u/h(u))=\int\frac{e^{i\varphi}+u}{e^{i\varphi}-u}\,d\nu(\varphi)$. Apply
$u\,\frac d{du}$ at $u=ru^*$, $0<r<1$:
\[
  1-\frac{ru^*h'(ru^*)}{h(ru^*)}=\int\frac{2ru^*e^{i\varphi}}{(e^{i\varphi}-ru^*)^2}\,d\nu(\varphi).
\]
As $r\to1$, the left side tends to $1-u^*h'(u^*)/h(u^*)$, the kernel tends to $-1/(2\sin^2((\varphi-\theta^*)/2))$, and
its modulus is $2r/((1-r)^2+4r\sin^2((\varphi-\theta^*)/2))\le1/(2\sin^2((\varphi-\theta^*)/2))$. This bound is $\nu$-integrable: on an
arc $I$ about $\theta^*$ on which $h$ is holomorphic and zero-free, $\nu=H\,d\varphi/2\pi$ with $H(\varphi)=\log(1/|h(e^{i\varphi})|)$
($\nu$ is the weak-$*$ limit of $\log|u/h|(re^{i\varphi})\,d\varphi/2\pi$, which converges uniformly on compact subarcs of $I$);
$H\ge0$ is real-analytic with $H(\theta^*)=0$, so it has a zero of even order $\ge2$ there, and $H/\sin^2((\varphi-\theta^*)/2)$ is
bounded near $\theta^*$; off $I$ the kernel is bounded. ($H$ does not vanish identically near $\theta^*$: otherwise $|z|=R_0$ on an
arc of $\partial\Omega$, so $|W(e^{it})|=R_0$ on an arc and hence, by real-analyticity, on the whole circle; then $\Omega=D(0,R_0)$,
$W=\mathrm{id}$, and $(wB)'$, the derivative of a Blaschke product of degree $k+1\ge2$, has $k$ zeros in $\mathbb D$, against
(M3).) By dominated convergence $u^*h'(u^*)/h(u^*)=1+\mathcal U$, so $|h_0'(u^*)|=1+\mathcal U$ in Lemma~\ref{lem:exit}(iii), and $\mathcal U\ge\nu(\partial\mathbb
D)/2$ as $\sin^2\le1$. Finally $2/(2+m)\le(1+e^{-m})/2$ for $m\ge0$, since $f(m)=(2+m)(1+e^{-m})-4$ has $f(0)=0$ and
$f'(m)=1-(1+m)e^{-m}\ge0$; and $m=\log(1/(R_0K_0))>0$ by (i).

(iii) By (ii), $\sigma(w^*)+1/v(w^*)<1$, which gives the front case. In the back case $\gamma(w^*)=v(w^*)/|h_0'(u^*)|<0$, so the pocket
$w^*$ is on. $\operatorname{Im}v$ is not identically zero (else $W=\mathrm{id}$ and $v=1$), so its zeros are isolated
(Lemma~\ref{lem:Zfinite}), and by Lemma~\ref{lem:exit}(ii) and $\frac d{d\theta}|W(e^{i\theta})|^2=2|W'|^2\operatorname{Im}v$,
$\operatorname{Im}v$ passes from negative to positive at $w^*$: $\varepsilon(w^*)=-1$.
\end{proof}

\subsection{The tip}

\begin{theorem}\label{thm:T2}
\textup{(a)} For the limit density of Proposition~\ref{prop:green},
\[
  \int_0^{2\pi}\frac{\Theta(s)\,ds}{2\cos^2(s/2)}=1.
\]

\textup{(b)} Along every sequence of Blaschke models with $K_{0,n}\to4$, $\liminf_n\min_{\partial\mathbb D}\sigma_n\ge1$.
\end{theorem}

\begin{proof}
(a) With $x=s/2$ the integral is $\frac1\pi\int_0^\pi(-\log\sin x)\,\frac{dx}{\cos^2x}=\frac1\pi\bigl([-\log(\sin x)\tan x]+
\int_0^\pi1\,dx\bigr)=1$, integrating by parts on $(0,\pi/2)$ and $(\pi/2,\pi)$; the bracket vanishes at $0$, $\pi/2$ and $\pi$.

(b) $\min\sigma_n$ does not change under rotations or unimodular factors, so it suffices to treat a subsequence normalised as in
Theorem~\ref{thm:C}. Let $t_n\to\pi$ and $\eta>0$, and let $\psi$ be continuous on the circle, $0\le\psi\le1$, $\psi=0$ on
$|s-\pi|\le\eta$ and $\psi=1$ on $|s-\pi|\ge2\eta$. For a zero $a=\rho e^{is}$ with $|s-\pi|\ge\eta$ and for $|t_n-\pi|<\eta/2$,
\[
  P_a(e^{it_n})=\frac{1-\rho^2}{1-2\rho\cos(t_n-s)+\rho^2}=(1+o(1))\log\frac1\rho\cdot\frac1{2\sin^2((t_n-s)/2)}
\]
uniformly in such $a$, since all zeros tend to the circle. So $\sigma_n(e^{it_n})\ge(1+o(1))\int\psi(s)\,(2\sin^2((t_n-s)/2))^{-1}\,
d\Theta_n(s)$; the kernel converges uniformly on the support of $\psi$ to $1/(2\cos^2(s/2))$, and $\Theta_n\to\Theta(s)\,ds$ weak-$*$ with
bounded masses (Proposition~\ref{prop:green}). Hence
\[
  \liminf_n\sigma_n(e^{it_n})\ \ge\ \int\psi(s)\,\frac{\Theta(s)\,ds}{2\cos^2(s/2)},
\]
and letting $\eta\to0$ gives $\ge1$ by (a). A sequence of minimum points of $\sigma_n$ has a convergent subsequence; if its limit is
not $\pi$, Corollary~\ref{cor:offtip} gives $\sigma_n\to\infty$ there, and otherwise the argument above applies.
\end{proof}

Part (a) is a statement about the limit object; its value $1$ is $|B_\infty'(-1)|$ for $B_\infty(w)=(1-w)^2/4$, which is
unimodular at the tip $w=-1$, as $B_\infty'(w)=-(1-w)/2$. Part (a) does not say that $\sigma_n(-1)\to1$: a zero $a_n$ close to $-1$ adds
$P_{a_n}(-1)=(1-|a_n|^2)/|1+a_n|^2$, which part (b) does not bound from above. For comparison, Lemma~\ref{lem:tanh} gives only
$\sigma\ge(K_0-1)/(K_0+1)$, which tends to $3/5$.

\begin{corollary}\label{cor:tip}
Let $(W_n,B_n)$ be Blaschke models with $K_{0,n}\to4$ and $k_n=\deg B_n$.
\begin{enumerate}\renewcommand{\labelenumi}{\textup{(\roman{enumi})}}
\item For every $\eta>0$ and all large $n$, $(W_n,B_n)$ has at least $k_n$ on pockets with $\varepsilon=-1$ and depth less than
$1+\eta$; and $k_n\to\infty$.
\item Along every subsequence of front cases \textup{(}Theorem~\ref{thm:T1}(iii)\textup{)}, $\min\sigma_n\to1$ and
$\sigma_n(w_n^*)\to1$; hence $\mathcal U_n\to0$ and $v_n(w_n^*)\to+\infty$, i.e.\ $|W_n'(w_n^*)|\to0$. After the rotations of
Theorem~\ref{thm:C}, $w_n^*\to-1$, the tip.
\end{enumerate}
\end{corollary}

\begin{proof}
(i) Each Blaschke model has finitely many pockets, so the depths of all the Blaschke models form a countable set; choose $\eta'\in(0,\eta)$ such that
$T=1+\eta'$ is a depth of no Blaschke model of the sequence. By Theorem~\ref{thm:T2}(b), every pocket of depth greater than $T$ is on for
large $n$. By Corollary~\ref{cor:constant} the pockets of depth greater than $T$ have signs adding up to $\ge0$, and by
Theorem~\ref{thm:pocket} the on pockets have signs adding up to $-k_n$. So the on pockets of depth less than $T$ have signs adding
up to at most $-k_n$, and at least $k_n$ of them have $\varepsilon=-1$. $k_n\to\infty$ by Theorem~\ref{thm:C}.

(ii) In the front case $\min\sigma_n\le\sigma_n(w_n^*)\le1/(1+\mathcal U_n)-1/v_n(w_n^*)<1$, and Theorem~\ref{thm:T2}(b) gives the limits; then
$1/(1+\mathcal U_n)\to1$ and $1/v_n(w_n^*)\to0$, and $|W_n'(w_n^*)|=|W_n(w_n^*)|/v_n(w_n^*)$ with $|W_n(w_n^*)|=R_0\le1$. Since
$\sigma_n(w_n^*)<1$, Corollary~\ref{cor:offtip} forces $w_n^*\to-1$ after the rotations.
\end{proof}

In the back case Theorem~\ref{thm:T1} gives only $|\gamma(w^*)|\le|v(w^*)|$, and Theorem~\ref{thm:T2}(b) gives only $\liminf\min\sigma_n\ge1$.