\section{Fixed-degree optima}\label{sec:fixeddeg}

The pairs of Section~\ref{sec:onezero} with $B$ of degree one come from the construction of~\cite[Lemma~6.2]{JatarNg2026b} with a polynomial of some
degree $n$. Theorem~\ref{thm:S} improves such a pair, but its perturbation is a polynomial whose degree grows with the number of its
Hermite conditions. We show that the improvement can be made without raising the degree as long as the pair has few boundary
defects. So a pair that is best among those of degree $n$ must have many defects (Theorem~\ref{thm:fixeddeg}), in the way a best
approximation from an $(n+1)$-dimensional space attains its error at $n+2$ points (Chebyshev's alternation theorem).

We use the notation of Section~\ref{sec:onezero}, with $z$ the variable of $Q$ and $H$ and $w$ that of $W$. Let $\mathcal L_n$\label{term:Ln} be the
set of pairs $(W,B)$ given there by a polynomial $Q$ of degree at most $n$: $a\neq b$ in $\mathbb D$, $Q(a)=Q(b)=0$, $Q'\neq0$ in
$\mathbb D$, $H=(1-\bar az)Q/(z-a)$ univalent in $\mathbb D$, $W=(H\circ\varphi)/N_b$, and $B$ the multiple of $B_a\circ\varphi$ by the
constant of modulus $1$ with $B(0)>0$. Let $\mathcal G_n$ be the set of pairs given in the same way, $W=(H\circ\varphi)/N_b$ and $B$ the multiple of
$B_a\circ\varphi$ by the constant of modulus $1$ with $B(0)>0$, where $H$
is a polynomial of degree at most $n$, $a,b\in\mathbb D$, $H(b)=0$, $H$ is univalent in $\mathbb D$ and $(HB_a)'\neq0$ in $\mathbb
D$; here $HB_a$ need not be a polynomial. This is the first of the two approaches described in~\cite[Sect.~6]{JatarNg2026b}. In
$\mathcal G_n$, $a\neq b$, since otherwise $HB_a$ has a double zero at $b$; and $\mathcal L_n\subset\mathcal G_n$. Both kinds of
pairs are admissible, with gain $\sqrt{J(a,b)}$, and $F=WB$ is a constant multiple of $(HB_a)\circ\varphi$, so the zeros of $F'$ on
$\mathbb T$ are the points $\varphi^{-1}(\xi)$ for the zeros $\xi$ of $(HB_a)'$ on $\mathbb T$, with the same orders. $W$ is rational,
with the only pole $-1/\bar b$ if $b\neq0$, and a polynomial if $b=0$; so $W$ is holomorphic past the circle, and
Lemma~\ref{lem:D} applies to it.

For such a pair let $z_1,\dots,z_p$ be the zeros of $F'$ on $\mathbb T$, of orders $m_1,\dots,m_p$; let
$\mathrm{Crit}=\{c_1,\dots,c_q\}$, and let $\{\zeta_l,\zeta_l'\}$ $(1\le l\le r)$ be the contact pairs of $W$; and let $e$ be the
number of contact points that are also some $w_i$. Put
\[
  n_H=2+\sum_{i=1}^p(m_i+1)+2q+2r-e,
\]
the number of Hermite conditions in the proof of Theorem~\ref{thm:S}. We call $n_H-2$ the weighted number of boundary defects\label{term:defects} of the pair. Here ``local'', as in ``local maximiser'',
refers to the topology of the coefficients of the triples $(H,a,b)$ for $\mathcal G_n$, resp.\ $(Q,a,b)$ for $\mathcal L_n$.

\begin{theorem}\label{thm:fixeddeg}
Let $(W,B)$ be a pair in $\mathcal G_n$, resp.\ in $\mathcal L_n$, with data $(H,a,b)$, resp.\ $(Q,a,b)$.
\begin{enumerate}
\item[(G)] If $n_H\le n+2$, then for every small $\delta>0$ there is a pair in $\mathcal G_n$ with data $(H_\delta,a_\delta,b)$
  and gain $(1-\delta)^{-1}/B(0)$, where $H_\delta\to H$ coefficientwise and $a_\delta\to a$ as $\delta\to0$.
\item[(L)] If $n_H\le n+1$, then for every small $\delta>0$ there is a pair in $\mathcal L_n$ with data $(Q_\delta,a_\delta,b)$
  and gain $(1-\delta)^{-1}/B(0)$, where $Q_\delta\to Q$ coefficientwise and $a_\delta\to a$ as $\delta\to0$.
\end{enumerate}
In particular a local maximiser of the gain over $\mathcal G_n$ has $n_H\ge n+3$, and a local maximiser over $\mathcal L_n$ has
$n_H\ge n+2$; a local maximiser over $\mathcal L_n$ such that $F'$ has no zero on $\mathbb T$ has $q+r\ge n/2$.
\end{theorem}

\begin{corollary}\label{cor:defectgrowth}
Suppose that for each $n$ the gain has a local maximiser $(W_n,B_n)$ over $\mathcal L_n$, and that $\sum_i(m_i+1)\le {\mathcal X}$ for all $n$.
Then $q_n+r_n\ge(n-\mathcal X)/2$: the number of boundary critical points plus the number of contact pairs grows at least linearly in $n$.
\end{corollary}

By Lemma~\ref{lem:D} each single pair has finitely many defects; the growth is what Theorem~\ref{thm:fixeddeg} forces. The
corollary follows from $n_H\ge n+2$ with $e\ge0$, and the last statement of the theorem is the case $p=e=0$. When $\deg Q=n$,
Theorem~\ref{thm:degup} below gives $n_H\ge n+3$.

\begin{proof}[Proof of Theorem~\ref{thm:fixeddeg}]
Let $\alpha=\varphi^{-1}(a)$, the zero of $B$; $\alpha\neq0$ as $a\neq b$.

\emph{Step 1: the conditions.} Let $U_1$ and the conditions be as in the proof of Theorem~\ref{thm:S}, with $U_1$ also free of
$-1/\bar b$. Here $B_\delta$ is $B$ with its zero $\alpha$ replaced by $(1-\delta)\alpha$, so $B_\delta(0)=(1-\delta)B(0)$, and
$B_\delta$ is a unimodular multiple of $B_{a_\delta}\circ\varphi$ with $a_\delta=\varphi((1-\delta)\alpha)$, a smooth function of
$\delta$ with $a_0=a$. We impose the Hermite conditions of that proof at the distinct nodes $0$, $w_i$, $c_j$, $\zeta_l$, $\zeta_l'$
except $\upsilon'(0)=0$: there are $n_H-1$ of them, with data bounded uniformly in $\delta$.

\emph{Step 2: $\upsilon=\mathfrak p\circ\varphi$.} For a polynomial $\mathfrak p$, the derivatives of orders $0,\dots,s$ of $\mathfrak p\circ\varphi$ at a node $w_0$
are an invertible triangular transform (with diagonal $\varphi'(w_0)^k\neq0$) of those of $\mathfrak p$ at $\varphi(w_0)$. So the conditions on
$\upsilon$ are $n_H-1$ Hermite conditions on $\mathfrak p$ at the distinct fixed nodes $\varphi(0)=b$ (of order $0$ only),
$\varphi(w_i)$, $\varphi(c_j)$, $\varphi(\zeta_l)$, $\varphi(\zeta_l')$, with data bounded uniformly in $\delta$.

In case (G), $n_H-1\le n+1$; let $\mathfrak p_\delta$ be the interpolant of degree at most $n_H-2\le n$. As the nodes are fixed
and the data bounded, its coefficients are bounded uniformly in $\delta$.

In case (L), $\check Q=Q/(z-a)$ is a polynomial of degree at most $n-1$ with $\check Q(b)=0$, and $H=(1-\bar az)\check Q$. Seek
\[
  \mathfrak p_\delta=g_\delta+(1-\bar a_\delta z)\,r_\delta,\qquad g_\delta=\frac{\bar a-\bar a_\delta}{\delta}\,\frac{z\check Q}{N_b},\qquad
  \deg r_\delta\le n-1 .
\]
The coefficients of $g_\delta$ are bounded uniformly in $\delta$. On the closed disk $1-\bar a_\delta z\neq0$, so at each node the
multiplication by $1-\bar a_\delta z$ is an invertible triangular map on jets, with inverse bounded uniformly in $\delta$. Subtracting
the jets of $g_\delta$ and dividing, the conditions on $\mathfrak p$ become $n_H-1\le n$ Hermite conditions on $r_\delta$ at the same
nodes, with bounded data (at $b$: $r_\delta(b)=0$, as $g_\delta(b)=0$). Let $r_\delta$ be the interpolant of degree at most
$n_H-2\le n-1$; its coefficients, and those of $\mathfrak p_\delta$, are bounded uniformly in $\delta$, and $\deg {\mathfrak p}_\delta\le n$.

In both cases $\upsilon_\delta=\mathfrak p_\delta\circ\varphi$ satisfies the conditions and is bounded in every $C^m(U_1)$ uniformly in $\delta$.

\emph{Step 3: the scale.} The rest of the proof of Theorem~\ref{thm:S} uses only these conditions and these bounds; the condition
$\upsilon'(0)=0$ served there only to get $W_\delta'(0)=1$. Here $W_\delta=W+\delta\upsilon_\delta$ has $W_\delta(0)=0$ and
$W_\delta'(0)=1+\delta\upsilon_\delta'(0)\neq0$ for small $\delta$. So $\widetilde W_\delta=W_\delta/(1+\delta\upsilon_\delta'(0))$ is
univalent on a neighbourhood of the closed disk, $\widetilde W_\delta\in\mathcal S$, and $(\widetilde W_\delta B_\delta)'$ has the zeros
of $(W_\delta B_\delta)'$, so none in $\mathbb D$, for all small $\delta>0$.

\emph{Step 4: the pair stays in the family.} Put $H_\delta=H+\delta N_b\mathfrak p_\delta$, a polynomial of degree at most $n$, and
$N_\delta=(1-|b|^2)H_\delta'(b)=N_b(1+\delta\upsilon_\delta'(0))$. As $\mathfrak p_\delta(b)=\upsilon_\delta(0)=0$, $H_\delta(b)=0$ and
$(H_\delta\circ\varphi)/N_\delta=\widetilde W_\delta$. So $H_\delta$ is univalent in $\mathbb D$, and $(H_\delta B_{a_\delta})\circ\varphi$
is a non-zero constant multiple of $\widetilde W_\delta B_\delta$, so $(H_\delta B_{a_\delta})'\neq0$ in $\mathbb D$. Hence
$(H_\delta,a_\delta,b)$ defines a pair in $\mathcal G_n$, namely $(\widetilde W_\delta,B_\delta)$, with gain
$1/B_\delta(0)=(1-\delta)^{-1}/B(0)$. In case (L), $\delta N_bg_\delta=(\bar a-\bar a_\delta)z\check Q$, so
\[
  H_\delta=(1-\bar az)\check Q+(\bar a-\bar a_\delta)z\check Q+\delta N_b(1-\bar a_\delta z)r_\delta=(1-\bar a_\delta z)(\check Q+\delta N_br_\delta),
\]
and $Q_\delta:=H_\delta B_{a_\delta}=(z-a_\delta)(\check Q+\delta N_br_\delta)$ is a polynomial of degree at most $n$ with
$Q_\delta(a_\delta)=Q_\delta(b)=0$, $Q_\delta'\neq0$ in $\mathbb D$ and $H_\delta=(1-\bar a_\delta z)Q_\delta/(z-a_\delta)$; with
$a_\delta\neq b$ for small $\delta$, this is a pair in $\mathcal L_n$. No node outside the disk and no condition on $a$ is needed.

\emph{Step 5: nearness.} $H_\delta-H=\delta N_b\mathfrak p_\delta=O(\delta)$ coefficientwise, $a_\delta\to a$, and $b$ is fixed; in case (L),
$Q_\delta-Q=(a-a_\delta)\check Q+\delta N_b(z-a_\delta)r_\delta=O(\delta)$.
\end{proof}

\begin{remark}\label{rem:OTimprove}
For pair~\ref{pair:OT}, read as a pair in $\mathcal L_{13}$, Theorem~\ref{thm:tablepairs} and Lemma~\ref{lem:certify} show that $H$ is
univalent past the circle, so $q=r=e=0$, and, by (C), that $F'$ has on $\mathbb T$ exactly the two zeros of orders $5$ and $1$. So $n_H=2+6+2=10\le n+1=14$,
and Theorem~\ref{thm:fixeddeg}(L) gives pairs in $\mathcal L_{13}$ near pair~\ref{pair:OT} with larger gains, with no explicit
constant. Raising the degree of $Q$ by one gains nothing to first order:
by Lemma~\ref{lem:gauge}, an automorphism of the disk removes the new top coefficient to first order and keeps the gain.
\end{remark}

\begin{remark}\label{rem:Gn}
A pair in $\mathcal G_n$ such that $F'$ has no zero on $\mathbb T$ can be improved within $\mathcal G_n$ by moving $a$ alone: keep $H$
and replace $a$ by $a_\delta$. Then $F'\neq0$ on the closed disk, and $B_\delta\to B$ in $C^1$ on the closed disk, so
$(WB_\delta)'\neq0$ there for small $\delta$. So every local maximiser over $\mathcal G_n$ has $p\ge1$, in addition to $n_H\ge n+3$; in
particular the Blaschke models~\ref{pair:F12c}, \ref{pair:F8s}, \ref{pair:F8p} and~\ref{pair:F8} of Table~\ref{tab:onezero} can be improved in this way. In $\mathcal L_n$
this argument fails, since $H$ must keep the factor $1-\bar az$.
\end{remark}

\medskip\noindent\emph{Raising the degree of $Q$ by one.} Theorem~\ref{thm:fixeddeg}(L) keeps $b$ fixed. Letting $b$ move as well, we show that a
local maximiser with $\deg Q=n$ has a defect count $n_H$ one larger: to first order, raising the degree by one is a change of variables by an
automorphism of the disk. For $y\in\mathbb D$ put $\varphi_y(w)=(w+y)/(1+\bar yw)$, so that $\varphi=\varphi_b$, and write $[z^j]P$ for the
coefficient of $z^j$ in a polynomial $P$.

\begin{lemma}\label{lem:gauge}
Let $Q$ be holomorphic near the closed disk, let $a\neq b$ in $\mathbb D$ with $Q(a)=Q(b)=0$, suppose that $H=(1-\bar az)Q/(z-a)$
has $H'(b)\neq0$, and let $(W,B)$ be given by the formulas of Section~\ref{sec:onezero}, with $B(0)>0$. Let $\psi$ be an
automorphism of $\mathbb D$. Then the data $(Q\circ\psi,\psi^{-1}(a),\psi^{-1}(b))$ give the pair
$w\mapsto\bigl(e^{-it}W(e^{it}w),\,B(e^{it}w)\bigr)$ for some $t\in\mathbb R$, and their function $H$ is a unimodular multiple of
$H\circ\psi$. In particular the two pairs have the same gain, they are admissible together, and their boundary defects correspond
under $w\mapsto e^{-it}w$, with the same orders.
\end{lemma}

\begin{proof}
Since $HB_a=Q$, $F=WB$ is a constant multiple of $Q\circ\varphi_b$, and $B$ is the
Blaschke factor with $B(0)>0$ whose zero is $\varphi_b^{-1}(a)$; with $W'(0)=1$, the pair is determined by $Q\circ\varphi_b$ and
$\varphi_b^{-1}(a)$. The automorphisms $\psi\circ\varphi_{\psi^{-1}(b)}$ and $\varphi_b$ both take $0$ to $b$, so
$\psi\circ\varphi_{\psi^{-1}(b)}=\varphi_b\circ\rho$ with $\rho(w)=e^{it}w$. So the data $(Q\circ\psi,\psi^{-1}(a),\psi^{-1}(b))$
give $(Q\circ\varphi_b)\circ\rho$ and $\rho^{-1}(\varphi_b^{-1}(a))$, that is, a constant multiple of $F\circ\rho$ and the factor $B\circ\rho$, and $W'(0)=1$ fixes the
constant. Finally $B_a\circ\psi$ is a unimodular multiple of $B_{\psi^{-1}(a)}$, which gives the statement on $H$.
\end{proof}

\begin{lemma}\label{lem:firstorder}
Let $(W,B)$ be as in Theorem~\ref{thm:S}, with the points $c_j$ and the contact pairs $\{\zeta_l,\zeta_l'\}$ of Lemma~\ref{lem:D},
and let $U_1$ be a compact neighbourhood of the closed disk on which $W$ is holomorphic. Let $\upsilon_\delta$ $(0<\delta<\delta_0)$ be
holomorphic on $U_1$, bounded in every $C^m(U_1)$ uniformly in $\delta$, with $\upsilon_\delta(0)=0$, and suppose that
\[
  \varpi_j(\upsilon_\delta):=2\operatorname{Re}\Bigl(\bar c_j\,\frac{-\upsilon_\delta'(c_j)}{W''(c_j)}\Bigr)\ge\varpi_0,\qquad
  \varpi_l(\upsilon_\delta):=\operatorname{Re}\Bigl(\bar\zeta_l'\,\frac{\upsilon_\delta(\zeta_l)-\upsilon_\delta(\zeta_l')}{W'(\zeta_l')}\Bigr)\ge\varpi_0
\]
for all $j$, $l$ and $\delta$, with $\varpi_0>0$ fixed. Then $W+\delta\upsilon_\delta$ is univalent on a neighbourhood of the closed disk for all small
$\delta$.
\end{lemma}

\begin{proof}
We use the notation of the proof of Theorem~\ref{thm:S}. There the paragraphs on the defects, on local injectivity and on global
injectivity use $\upsilon_\delta$ only through these bounds and through $c_\delta=c-\delta\upsilon_\delta'(c)/W''(c)+O(\delta^2)$ and
$\tau_\delta(\zeta_1)-\zeta_2=\delta(\upsilon_\delta(\zeta_1)-\upsilon_\delta(\zeta_2))/W'(\zeta_2)+O(\delta^2)$. In general these give
$\log|c_\delta|=\varpi_j\delta/2+O(\delta^2)$; $\log|\beta_\delta(c)|=\varpi_j\delta+O(\delta^2)$, since
$\beta_\delta(c)=2c_\delta-c+O(\delta^2)$; and $\log|\tau_\delta(\zeta_1)|=\varpi_l\delta+O(\delta^2)$. So $|c_\delta|>1$ and
$G_\delta(1,0)\ge\varpi_0\delta-C\delta^2$, and the inequality of the local injectivity holds with $\varpi_0$ in place of $\varpi$;
the rest of those paragraphs is unchanged. They do not use the conditions at the $w_i$, which serve only to keep $F_\delta'\neq0$ in
$\mathbb D$: near a $w_i$ that is not a contact point, $W'\neq0$, since $F'(c)\neq0$ at every $c\in\mathrm{Crit}$, and the segment
argument applies there.
\end{proof}

\begin{theorem}\label{thm:degup}
Let $(W,B)$ be a pair in $\mathcal L_n$ with data $(Q,a,b)$, where $\deg Q=n$, and suppose that $n_H\le n+2$. Then for every
small $\delta>0$ there is a pair in $\mathcal L_n$ with data $(Q^\sharp,a^\sharp,b^\sharp)$ and gain at least
$(1-\delta)^{-1}/B(0)-C\delta^2$, with $C$ independent of $\delta$, where $Q^\sharp\to Q$ coefficientwise, $a^\sharp\to a$ and
$b^\sharp\to b$ as $\delta\to0$. In particular, a local maximiser of the gain over $\mathcal L_n$ with $\deg Q=n$ has
$n_H\ge n+3$; if moreover $F'$ has no zero on $\mathbb T$, then $q+r\ge(n+1)/2$.
\end{theorem}

\begin{proof}
The plan. The proof of Theorem~\ref{thm:fixeddeg}(L) gives the gain factor $(1-\delta)^{-1}$, but in degree $n+1$ (Step~1). A disk
automorphism $\psi_\delta$ removes the new top coefficient up to $O(\delta^2)$ (Step~2, with Lemma~\ref{lem:gauge}); the result is not a
polynomial, so Step~3 replaces it by a polynomial $Q^\sharp$ of degree at most $n$ whose derivative has the same zeros on $\mathbb T$.
Step~4 shows that the new map is still univalent, and Step~5 compares the gains.

Below, $C$ is independent of $\delta$. Put $\xi_i=\varphi(w_i)$; then $Q'$ has on the closed disk exactly the zeros $\xi_i\in\mathbb
T$, of orders $m_i$, and $\sum_im_i\le n-1$ as $\deg Q'=n-1$.

\emph{Step 1: one degree up.} The pair lies in $\mathcal L_{n+1}$, and there $n_H\le(n+1)+1$. So the proof of
Theorem~\ref{thm:fixeddeg}(L) in $\mathcal L_{n+1}$ gives, for small $\delta$, the pair $(\widetilde W_\delta,B_\delta)$ in
$\mathcal L_{n+1}$ with data $(Q_\delta,a_\delta,b)$, where $a_\delta=\varphi((1-\delta)\alpha)$,
$Q_\delta=(z-a_\delta)(\check Q+\delta N_br_\delta)$ with $\deg {\check Q}=n-1$, $\deg r_\delta\le {n_H}-2\le n$ and $r_\delta$ bounded, and
$\widetilde W_\delta=(W+\delta\upsilon_\delta)/(1+\delta\upsilon_\delta'(0))$ with $\upsilon_\delta$ bounded in every $C^m(U_1)$,
$\upsilon_\delta(0)=0$, $\varpi_j(\upsilon_\delta)=2$ and $\varpi_l(\upsilon_\delta)=1$. So $Q_\delta-Q=O(\delta)$, and the
coefficient of degree $n+1$ is $[z^{n+1}]Q_\delta=\delta N_b[z^n]r_\delta$. The conditions at the $w_i$ make
$(\widetilde W_\delta B_\delta)'$ vanish to order at least $m_i$ at $w_i$, and its derivative of order $m_i$ there tends to that of
$F'$, which is not $0$. As $F'\neq0$ elsewhere on the closed disk, for small $\delta$ the zeros of $Q_\delta'$ on the closed disk are
the $\xi_i$, with the orders $m_i$.

\emph{Step 2: the gauge.} Define $s_\delta$ by $\bar s_\delta\,n\,[z^n]Q=[z^{n+1}]Q_\delta$, so that $s_\delta=O(\delta)$, and put
$\psi_\delta=\varphi_{s_\delta}$. On $D_1=\{|z|\le1+\eta\}$, with $\eta\in(0,1)$ fixed below, $\psi_\delta(z)=z+s_\delta-\bar
s_\delta z^2+O(\delta^2)$, so $Q_\delta\circ\psi_\delta=\widetilde Q_\delta+O(\delta^2)$ in $C^k(D_1)$ for every $k$, where
$\widetilde Q_\delta=Q_\delta+(s_\delta-\bar s_\delta z^2)Q'$ (by Taylor's formula $Q_\delta\circ\psi_\delta=Q_\delta+(\psi_\delta-z)Q_\delta'+O(\delta^2)$,
as $Q_\delta-Q=O(\delta)$; Cauchy's estimates on a slightly larger disk give the derivatives). As $[z^{n+1}](z^2Q')=n\,[z^n]Q$, $\deg\widetilde Q_\delta\le n$. By
Lemma~\ref{lem:gauge}, the data $(Q_\delta\circ\psi_\delta,\hat a,\hat b)$ with $\hat a=\psi_\delta^{-1}(a_\delta)$ and
$\hat b=\psi_\delta^{-1}(b)$ give the pair $(\widetilde W_\delta,B_\delta)$ turned by $w\mapsto e^{it_\delta}w$, but
$Q_\delta\circ\psi_\delta$ is not a polynomial.

\emph{Step 3: the same boundary zeros in degree $n$.} As $Q_\delta\circ\psi_\delta$ is not a polynomial, we replace it by a
polynomial $Q^\sharp$ of degree at most $n$ whose derivative has the same zeros on $\mathbb T$, at a cost $O(\delta^2)$. Put $\hat z_i=\psi_\delta^{-1}(\xi_i)\in\mathbb T$ and
$\Pi=\prod_i(z-\hat z_i)^{m_i}$. By Step~1, $(Q_\delta\circ\psi_\delta)'=\Pi Y_\delta$ with $Y_\delta$ holomorphic on $D_1$, and
$Y_\delta\to Y_0=Q'/\prod_i(z-\xi_i)^{m_i}$ uniformly on $D_1$, by the maximum principle on circles of a fixed radius about the
$\xi_i$. $Y_0$ has no zero on the closed disk; we fix $\eta$ so that it has none on $D_1$, and then $Y_\delta$ has none for small
$\delta$. Divide: $\widetilde Q_\delta'=\Pi A_\delta+V_\delta$ with $\deg V_\delta<\sum_im_i$. By Step~2, the derivatives of
$\widetilde Q_\delta'-(Q_\delta\circ\psi_\delta)'$ are $O(\delta^2)$ on $D_1$; at $\hat z_i$, those of orders below $m_i$ are the
derivatives of $V_\delta$, and the nodes $\hat z_i$ stay apart, so $V_\delta=O(\delta^2)$. Then $\Pi(A_\delta-Y_\delta)=O(\delta^2)$
on $D_1$, so $A_\delta-Y_\delta=O(\delta^2)$ on $D_2=\{|z|\le1+\eta/2\}$ by the maximum principle again, and $A_\delta$ has no zero on
$D_2$. Put
\[
  Q^\sharp(z)=\int_{\hat b}^z\Pi A_\delta,\qquad b^\sharp=\hat b .
\]
Then $\deg Q^\sharp\le n$, $Q^\sharp(b^\sharp)=0$, and $Q^\sharp-Q_\delta\circ\psi_\delta=O(\delta^2)$ in $C^k(D_2)$, as both vanish
at $\hat b$ and $D_2$ is convex. $Q_\delta\circ\psi_\delta$ has a simple zero at $\hat a$, with derivative bounded away from $0$ as
$Q'(a)\neq0$, so by Rouch\'e's theorem $Q^\sharp$ has a simple zero $a^\sharp=\hat a+O(\delta^2)$, and $a^\sharp\neq b^\sharp$. The
derivative $\Pi A_\delta$ of $Q^\sharp$ has no zero in $\mathbb D$, and on $\mathbb T$ exactly the zeros $\hat z_i$, of orders $m_i$.

\emph{Step 4: univalence.} If $f(z_0)=0$, then $f(z)/(z-z_0)=\int_0^1f'(z_0+x(z-z_0))\,dx$ on the convex set $D_2$. So
$H^\sharp=(1-\bar a^\sharp z)Q^\sharp/(z-a^\sharp)$ differs by $O(\delta^2)$ in $C^{k-1}(D_2)$ from the function
$(1-\bar{\hat a}z)(Q_\delta\circ\psi_\delta)/(z-\hat a)$, and $N^\sharp=(1-|b^\sharp|^2)H^{\sharp\prime}(b^\sharp)$ differs by $O(\delta^2)$ from the
corresponding constant, which is not $0$. Let $U_2\subset U_1$ be a closed disk about $0$ whose interior contains the closed disk,
with $\varphi_{\hat b}(U_2)\subset D_2$ for small $\delta$. Then $W^\sharp=(H^\sharp\circ\varphi_{b^\sharp})/N^\sharp$ differs by
$O(\delta^2)$ in $C^m(U_2)$ (take $k$ large) from the map given by the data $(Q_\delta\circ\psi_\delta,\hat a,\hat b)$, which is
$e^{-it_\delta}\widetilde W_\delta(e^{it_\delta}w)$ by Lemma~\ref{lem:gauge}; so $e^{it_\delta}W^\sharp(e^{-it_\delta}w)=\widetilde
W_\delta+O(\delta^2)$ in $C^m(U_2)$.
So $(1+\delta\upsilon_\delta'(0))\,e^{it_\delta}W^\sharp(e^{-it_\delta}w)=W+\delta\upsilon^\sharp_\delta$ with
$\upsilon^\sharp_\delta=\upsilon_\delta+O(\delta)$ in $C^m(U_2)$: $\upsilon^\sharp_\delta$ is bounded, $\upsilon^\sharp_\delta(0)=0$,
$\varpi_j(\upsilon^\sharp_\delta)=2+O(\delta)$ and $\varpi_l(\upsilon^\sharp_\delta)=1+O(\delta)$. By Lemma~\ref{lem:firstorder} on
$U_2$, $W+\delta\upsilon^\sharp_\delta$ is univalent near the closed disk for small $\delta$, and so are $W^\sharp$ and $H^\sharp$. With
Step~3, $(Q^\sharp,a^\sharp,b^\sharp)$ gives a pair in $\mathcal L_n$.

\emph{Step 5: gain and nearness.} By Lemma~\ref{lem:gauge}, the zero $\varphi_{\hat b}^{-1}(\hat a)$ of $B_\delta(e^{it_\delta}w)$ is
$e^{-it_\delta}(1-\delta)\alpha$. As $a^\sharp=\hat a+O(\delta^2)$ and $b^\sharp=\hat b$, the zero of the factor $B^\sharp$ of the pair
of $(Q^\sharp,a^\sharp,b^\sharp)$ has modulus $(1-\delta)|\alpha|+O(\delta^2)$, and $B(0)=|\alpha|$; so $1/B^\sharp(0)$ is at
least $(1-\delta)^{-1}/B(0)-C\delta^2$, which exceeds $1/B(0)$ for small $\delta$. As $s_\delta\to0$, $Q^\sharp\to Q$, $a^\sharp\to a$ and $b^\sharp\to b$. The last statement of the
theorem follows as in Theorem~\ref{thm:fixeddeg}: if $F'$ has no zero on $\mathbb T$, then $p=e=0$ and $n_H=2+2(q+r)$.
\end{proof}

So a pair in $\mathcal L_n$ with $\deg Q=n$, no zero of $F'$ on $\mathbb T$ and $q+r=n/2$ has $n_H=n+2$, and
Theorem~\ref{thm:degup} improves it directly. For odd $n$ the bound $q+r\ge(n+1)/2$ is that of Theorem~\ref{thm:fixeddeg}; for even
$n$ it is larger by one. If the maximisers of Corollary~\ref{cor:defectgrowth} have $\deg Q_n=n$, then $q_n+r_n\ge(n+1-\mathcal X)/2$. In
$\mathcal G_n$, Lemma~\ref{lem:gauge} and Step~2 apply to $H$, but Step~3 would have to keep the boundary zeros of $(HB_a)'$, which is
not a polynomial; whether Theorem~\ref{thm:degup} has a form for $\mathcal G_n$ is open. By Lemma~\ref{lem:gauge} the construction
of~\cite[Lemma~6.2]{JatarNg2026b} is invariant under the automorphisms of the disk, while the condition $\deg Q\le n$ is not. The
mechanism of Lemma~\ref{lem:gauge} is that of the proof of~\cite[Lemma~6.2]{JatarNg2026b}, where $WB$ is a constant multiple of
$Q\circ\varphi$.

\begin{remark}\label{rem:offTzeros}
Let $(W,B)$ be a local maximiser of the gain over $\mathcal L_n$ with data $(Q,a,b)$ and $\deg Q=n$. By Theorem~\ref{thm:degup},
$n_H\ge n+3$, that is, $\sum_{i=1}^pm_i\ge n+1-p-(2q+2r-e)$. The zeros of $F'$ on $\mathbb T$ correspond to those of $Q'$ on
$\mathbb T$, with the same orders, and $\deg Q'=n-1$. So $Q'$ has at most
\[
  n-1-\sum_{i=1}^pm_i\ \le\ p-2+2q+2r-e
\]
zeros off the circle, counted with multiplicity. In particular, if $q=r=0$ (so $e=0$), then $p\ge2$, and all zeros of $Q'$ but at
most $p-2$ lie on the circle. Pairs~\ref{pair:O14} and~\ref{pair:O12F} have $p=1$ and $q=r=0$, so they are not local maximisers over
$\mathcal L_n$ with $\deg Q=n$, as Theorem~\ref{thm:fixeddeg} also shows.
\end{remark}

\medskip\noindent\emph{Releasing boundary zeros.} In $\mathcal L_n$ the gain $\sqrt{J(a,b)}$ depends on $a$ and $b$ only. Keeping them
fixed, the zeros of $Q'$ on $\mathbb T$ can be moved off the circle without changing the gain; we use the following openness lemma.

\begin{lemma}\label{lem:openuniv}
Let $H$ be injective on $\{|z|<r_0\}$ with $r_0>1$, let $1<r<r_0$ and $\overline D_r=\{|z|\le r\}$. If $G_k\to H$ in $C^1(\overline D_r)$, then $G_k$ is
injective on $\overline D_r$ for all large $k$.
\end{lemma}

\begin{proof}
$H'\neq0$ on $\overline D_r$, as $H$ is injective. If the claim fails, then along a subsequence, $G_k(z_k)=G_k(z_k')$ with $z_k\neq z_k'$ in $\overline D_r$, and we may
assume $z_k\to z$ and $z_k'\to z'$ in $\overline D_r$. If $z\neq z'$, then $H(z)=H(z')$, against injectivity. If $z=z'$, then
\[
  0=\frac{G_k(z_k)-G_k(z_k')}{z_k-z_k'}=\int_0^1G_k'\bigl(z_k'+s(z_k-z_k')\bigr)\,ds\ \longrightarrow\ H'(z)\neq0,
\]
since the segment lies in the convex set $\overline D_r$, $G_k'\to H'$ uniformly there and $H'$ is uniformly continuous on $\overline D_r$.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:flatnotsharp}]
The last sentence is Proposition~\ref{prop:flatmax}(i) with $k_\zeta=5$. We work in the variable $z=\varphi(w)$ of $H$, as in Section~\ref{sec:onezero}; by
Remark~\ref{rem:flatframe} it suffices to find $H$ with the stated properties at $z=1$.

\emph{The start.} Let $a$, $b$ and $Q$ be the exact data of pair~\ref{pair:O14} (Theorem~\ref{thm:tablepairs}), and
$H_0=(1-\bar az)Q/(z-a)$. By Theorem~\ref{thm:tablepairs} and Lemma~\ref{lem:certify}, $H_0$ is univalent on a disk $\{|z|<r\}$
with $r>1$, and by (C) of Lemma~\ref{lem:certify} (checked in Theorem~\ref{thm:tablepairs}), $Q'=(z-1)^5R$ with $R(1)\neq0$. Put $G_0(t)=g_0(e^{it})$
with $g_0=zH_0'/H_0$. On the circle $zB_a'/B_a=P_a$, so $g_0+P_a=zQ'/Q$, and as $(e^{it}-1)^5=(it)^5(1+O(t))$,
\[
  G_0(t)+P_a(e^{it})=c\,t^5+O(t^6),\qquad c=\frac{iR(1)}{Q(1)} .
\]
In exact rational arithmetic over $\mathbb Q(i)$, $\operatorname{Re}(R(1)/Q(1))>0$, so $\operatorname{Im}c>0$ (numerically
$c\approx54.084+12.880\,i$). As $P_a$ is real on the circle, $u_0'=-\operatorname{Im}G_0=-\operatorname{Im}(c)\,t^5+O(t^6)$: the
function $u_0=\log|H_0|$ has a strict local maximum at $z=1$ with $k_\zeta=5$. Also $\operatorname{Re}G_0(0)=-P_a(1)<0$, so
$f_0=-\operatorname{Re}G_0$ is positive near $0$; and $f_0-P_a(e^{it})=O(t^5)$, so $\Upsilon(f_0)(0)=\Upsilon(P_a)(1)=0$ by
Lemma~\ref{lem:circschw}(a).

\emph{The family.} For $y=(y_0,\dots,y_5)\in\mathbb C^6$ put $H_y=H_0+(z-b)\sum_{l=0}^5y_l(z-1)^l$. Then $H_y(b)=0$, $\deg H_y=14$, and
$H_y'(b)\neq0$ for small $y$. Let $G_y(t)=e^{it}H_y'(e^{it})/H_y(e^{it})$,
\[
  \operatorname{jet}(y)=\bigl(\operatorname{Im}G_y^{(j)}(0)\bigr)_{0\le j\le4}\in\mathbb R^5,\qquad \Upsilon_y=\Upsilon(-\operatorname{Re}G_y)(0).
\]
Both are real-analytic near $y=0$, $\operatorname{jet}(0)=0$ since $u_0'=O(t^5)$, and $\Upsilon_0=0$.

\emph{The derivative is onto.} The first-order change of $g=zH'/H$ is $\delta g=z\,(\delta H/H_0)'$, with $\delta H=(z-b)\sum_l\delta
y_l(z-1)^l$. As $b\neq1$, the derivatives of orders $0,\dots,5$ of $\delta H$ at $z=1$ are arbitrary; as $H_0(1)\neq0$, so are those
of $\delta H/H_0$ (a triangular change); hence the derivatives of orders $0,\dots,4$ of $\delta g$ at $z=1$ are arbitrary (again a
triangular change), and so are the $t$-derivatives of orders $0,\dots,4$ of $\delta G$ at $0$, since $d/dt=iz\,d/dz$ at
$z=e^{it}$. So $d\operatorname{jet}(0)$ maps onto $\mathbb R^5$, and there is a first-order change $\delta y$ with $\delta G^{(j)}(0)=0$ for
$j\neq2$ and $\delta G''(0)=\rho$ real and nonzero. It lies in the kernel of $d\operatorname{jet}(0)$, and along it only $f''(0)$ changes, so
$\delta\Upsilon=-\rho/f_0(0)$, of either sign.

\emph{The point.} By the implicit function theorem, $\operatorname{jet}^{-1}(0)$ is, near $y=0$, a real-analytic submanifold of $\mathbb C^6=\mathbb
R^{12}$ through $0$ with tangent space $\ker d\operatorname{jet}(0)$. Since $\Upsilon_0=0$ and the derivative of $y\mapsto\Upsilon_y$ along this
submanifold is nonzero at $0$, there are $y\in\operatorname{jet}^{-1}(0)$ arbitrarily close to $0$ with $\Upsilon_y<0$. Fix such a $y$, close
enough to $0$. By Lemma~\ref{lem:openuniv}, $H_y$ is injective on $\{|z|\le r'\}$ for a fixed $r'\in(1,r)$. As $\operatorname{jet}(y)=0$,
$u_y'=-\operatorname{Im}G_y=O(t^5)$, and $\operatorname{Im}G_y^{(5)}(0)$ has the sign of $\operatorname{Im}G_0^{(5)}(0)=5!\operatorname{Im}c>0$;
so $u_y=\log|H_y|$ has a strict local maximum at $z=1$ with $k_\zeta=5$. Also $\operatorname{Re}G_y(0)<0$ and
$\Upsilon(-\operatorname{Re}G_y)(0)=\Upsilon_y<0$. Finally $W=(H_y\circ\varphi)/N_b$ with $N_b=(1-|b|^2)H_y'(b)$ is in $\mathcal S$ and
univalent on a neighbourhood of the closed disk, as in the proof of Lemma~\ref{lem:recipek}, and by Remark~\ref{rem:flatframe} the
point $\zeta=\varphi^{-1}(1)$ has the stated properties.
\end{proof}

\begin{proposition}\label{prop:split}
Let $(W,B)$ be a pair in $\mathcal L_n$ with data $(Q,a,b)$ such that $H$ is univalent on a disk $\{|z|<r\}$ with $r>1$. Let
$\xi_1,\dots,\xi_p$ be the zeros of $Q'$ on $\mathbb T$, of orders $m_1,\dots,m_p$, put $M=\sum_im_i$ and $\Pi=(z-\xi_1)^{m_1}\cdots(z-\xi_p)^{m_p}$, and
write $Q'=\Pi R$, so that $R$ has no zero on the closed disk. Suppose that
\[
  \textstyle\int_a^b\Pi\tilde R\,dt\neq0\quad\text{for some polynomial $\tilde R$ of degree at most $n-1-M$.}\tag{R$_{\mathcal L}$}
\]
Let $\hat\xi=(\hat\xi_1,\dots,\hat\xi_M)$ be close to the list $\xi$ in which each $\xi_i$ appears $m_i$ times, with $|\hat\xi_j|\ge1$ for all $j$. Then there
is a polynomial $Q_{\hat\xi}$ of degree at most $n$ with $Q_{\hat\xi}(a)=Q_{\hat\xi}(b)=0$ and $Q_{\hat\xi}'=\prod_j(z-\hat\xi_j)\,R_{\hat\xi}$, where $R_{\hat\xi}$ has no zero on the
closed disk, such that $Q_{\hat\xi}\to Q$ coefficientwise as $\hat\xi\to\xi$, and, for $\hat\xi$ close enough to $\xi$, $(Q_{\hat\xi},a,b)$ gives a pair
$(W_{\hat\xi},B)$ in $\mathcal L_n$, with the same $B$ and so the same gain $\sqrt{J(a,b)}$, whose $H_{\hat\xi}$ is univalent on $\{|z|<r'\}$ for a
given $r'\in(1,r)$. If all $|\hat\xi_j|>1$, then $(W_{\hat\xi},B)$ is a Blaschke model. If $p=1$ and $m_1\le n-2$, then (R$_{\mathcal L}$) holds.
\end{proposition}

\begin{proof}
A polynomial $P$ of degree at most $n-1$ is the derivative of a polynomial of degree at most $n$ that vanishes at $a$ and $b$ if and
only if $\int_a^bP=0$, and that polynomial is then $\int_a^zP$. Put $\Pi_{\hat\xi}=\prod_j(z-\hat\xi_j)$ and, with $\tilde R$ from (R$_{\mathcal L}$),
\[
  \hat y(\hat\xi)=-\frac{\int_a^b\Pi_{\hat\xi}R\,dt}{\int_a^b\Pi_{\hat\xi}\tilde R\,dt},\qquad Q_{\hat\xi}(z)=\int_a^z\Pi_{\hat\xi}\bigl(R+\hat y(\hat\xi)\tilde R\bigr)\,dt .
\]
The denominator is continuous in $\hat\xi$ and not $0$ at $\hat\xi=\xi$, and $\int_a^b\Pi R=\int_a^bQ'=0$; so $\hat y$ is defined and continuous near
$\xi$, with $\hat y(\xi)=0$. Then $\deg Q_{\hat\xi}\le n$, $Q_{\hat\xi}(a)=Q_{\hat\xi}(b)=0$, $Q_{\hat\xi}'=\Pi_{\hat\xi}R_{\hat\xi}$ with $R_{\hat\xi}=R+\hat y(\hat\xi)\tilde R$, and $Q_{\hat\xi}\to Q$
coefficientwise. So $R_{\hat\xi}\to R$ uniformly on the closed disk, where $\min|R|>0$, and $R_{\hat\xi}$ has no zero there for $\hat\xi$ close. The
polynomial $H_{\hat\xi}=(1-\bar az)Q_{\hat\xi}/(z-a)$ has $H_{\hat\xi}(b)=0$ and tends to $H$ coefficientwise, so in $C^1$ on $\{|z|\le r'\}$, and by
Lemma~\ref{lem:openuniv} it is univalent on $\{|z|<r'\}$ for $\hat\xi$ close. As $|\hat\xi_j|\ge1$, $Q_{\hat\xi}'\neq0$ in $\mathbb D$, so $(Q_{\hat\xi},a,b)$
gives a pair in $\mathcal L_n$; its factor $B$, the multiple of $B_a\circ\varphi$ by the constant of modulus $1$ with $B(0)>0$, depends on $a$
and $b$ only. If all $|\hat\xi_j|>1$, $Q_{\hat\xi}'$ has
no zero on the closed disk, and Lemma~\ref{lem:recipek} (with $k=1$) applies. For $p=1$ and $\Pi=(z-\xi)^m$, take $\tilde R=1$ and $\tilde R=z-\xi$,
of degree at most $n-1-m$ when $m\le n-2$. If both integrals vanish, then $(b-\xi)^{m+1}=(a-\xi)^{m+1}$ and
$(b-\xi)^{m+2}=(a-\xi)^{m+2}$; as $a-\xi$ and $b-\xi$ are not $0$, dividing gives $b=a$, which is excluded.
\end{proof}

\begin{corollary}\label{cor:recordBM}
Let $(a,b,Q)$ be the data of one of pairs~\ref{pair:N18}, \ref{pair:N17}, \ref{pair:N16}, \ref{pair:O14}, \ref{pair:O12F},
\ref{pair:O12}, \ref{pair:T6p} and~\ref{pair:T6}, of degree $n=18$, $17$, $16$, $14$, $12$, $12$, $6$ and $6$. Then there are Blaschke models with one zero and gain exactly $\sqrt{J(a,b)}$, given by
polynomials $Q_{\hat\xi}$ of degree $n$ with $Q_{\hat\xi}(a)=Q_{\hat\xi}(b)=0$.
\end{corollary}

\begin{proof}
By Theorems~\ref{thm:onezeroN18} and~\ref{thm:tablepairs}, facts (R), (S) and (C) of Lemma~\ref{lem:certify} hold for these
pairs, with $\Pi=(z-1)^m$, where $m=3$ for pairs~\ref{pair:T6p} and~\ref{pair:T6} and $m=5$ for the others. So $H$ is univalent on a
disk $\{|z|<r\}$ with $r>1$ (Lemma~\ref{lem:certify}), and $Q'=(z-1)^mR$ with $R$ zero-free on the closed disk: $p=1$ and
$m_1=m\le n-2$.
Proposition~\ref{prop:split} with all $|\hat\xi_j|>1$ gives the Blaschke models; their degree is that of $Q$, since $Q_{\hat\xi}\to Q$.
\end{proof}

\begin{remark}\label{rem:split}
(a) \emph{Order of the zero.} Among the pairs in $\mathcal L_n$ with $H$ univalent past the circle and one zero of $Q'$ on $\mathbb T$, of
order $m\le n-2$, every gain reached with order $m$ is reached, at the same $(a,b)$ and degree, with every lower order and by a
Blaschke model. So in this class a zero of order $7$ (which needs $n\ge9$) gives no larger supremum than order $5$. The statement
does not cover pairs with boundary critical points or contact pairs.
(b) \emph{Same gain.} Each pair of Proposition~\ref{prop:split} is a limit of Blaschke models in $\mathcal L_n$ with exactly its gain, whereas
\cite[Lemma~2.1]{JatarNg2026b} gives regular pairs with the smaller gain $t/B(0)$, $t<1$.
(c) \emph{Local maximisers.} If a local maximiser over $\mathcal L_n$ had $H$ univalent past the circle and (R$_{\mathcal L}$), the Blaschke
models of (b), arbitrarily close to it, have $n_H=2\le n+1$, so Theorem~\ref{thm:fixeddeg}(L) would give pairs of larger gain
arbitrarily close to it. So a local maximiser over $\mathcal L_n$ has $q+r\ge1$, or (R$_{\mathcal L}$) fails; for $p=1$ the latter means
$m_1=n-1$. This complements Theorem~\ref{thm:fixeddeg}, which needs neither $q=r=0$ nor (R$_{\mathcal L}$) and, when $q=r=0$, excludes
$M\le n-1-p$.
(d) \emph{Sharpness.} If $M=n-1$, then $R$ is a constant, so every $\tilde R$ of degree $0$ is a multiple of $R$, and
$\int_a^b\Pi R=\int_a^bQ'=0$; so (R$_{\mathcal L}$) needs $M\le n-2$. It fails, for instance, for the pair of~\cite[Thm.~6.1]{JatarNg2026b}, where
$Q'=4(z-1)^3$ and $n=4$.
\end{remark}

The same argument works with several zeros, for the pairs of Lemma~\ref{lem:kzero}; now $k$ corrections are needed.

\begin{proposition}\label{prop:splitk}
Let $k\ge1$, and let $Q$, $b,a_1,\dots,a_k$ and $H$ be as in Lemma~\ref{lem:kzero}, with $\deg Q\le n$ and with $H$
univalent on a disk $\{|z|<r\}$ with $r>1$. Let $\xi_1,\dots,\xi_p$ be the zeros of $Q'$ on $\mathbb T$, of orders $m_1,\dots,m_p$, put
$M=\sum_im_i$ and $\Pi=(z-\xi_1)^{m_1}\cdots(z-\xi_p)^{m_p}$, and write $Q'=\Pi R$, so that $R$ has no zero on the closed disk. Suppose that
\[
  \det\Bigl[\int_b^{a_i}\Pi\tilde R_j\,dt\Bigr]_{i,j=1}^{k}\neq0\quad\text{for some polynomials $\tilde R_1,\dots,\tilde R_k$ of degree at
  most $n-1-M$.}\tag{R$_{\mathcal L}^k$}
\]
Let $\hat\xi=(\hat\xi_1,\dots,\hat\xi_M)$ be close to the list $\xi$ in which each $\xi_i$ appears $m_i$ times, with $|\hat\xi_j|\ge1$ for all $j$. Then there
is a polynomial $Q_{\hat\xi}$ of degree at most $n$ with $Q_{\hat\xi}(b)=Q_{\hat\xi}(a_1)=\dots=Q_{\hat\xi}(a_k)=0$ and $Q_{\hat\xi}'=\prod_j(z-\hat\xi_j)\,R_{\hat\xi}$, where $R_{\hat\xi}$ has no
zero on the closed disk, such that $Q_{\hat\xi}\to Q$ coefficientwise as $\hat\xi\to\xi$, and, for $\hat\xi$ close enough to $\xi$,
Lemma~\ref{lem:kzero} applied to $(Q_{\hat\xi},b,a_1,\dots,a_k)$ gives an admissible pair $(W_{\hat\xi},B)$, with the same $B$ and so the same gain
$\prod_jJ(a_j,b)^{1/2}$, whose $H_{\hat\xi}$ is univalent on $\{|z|<r'\}$ for a given $r'\in(1,r)$. If all $|\hat\xi_j|>1$, then $(W_{\hat\xi},B)$ is a
Blaschke model.
\end{proposition}

\begin{proof}
A polynomial $P$ of degree at most $n-1$ is the derivative of a polynomial of degree at most $n$ that vanishes at $b,a_1,\dots,a_k$ if
and only if $\int_b^{a_i}P=0$ for $i=1,\dots,k$, and that polynomial is then $\int_b^zP$. Put $\Pi_{\hat\xi}=\prod_j(z-\hat\xi_j)$; let $\hat G(\hat\xi)$ be the
$k\times k$ matrix with entries $\hat G_{ij}(\hat\xi)=\int_b^{a_i}\Pi_{\hat\xi}\tilde R_j\,dt$, with $\tilde R_j$ from (R$_{\mathcal L}^k$), and $\hat g(\hat\xi)$ the vector with
entries $\hat g_i(\hat\xi)=\int_b^{a_i}\Pi_{\hat\xi}R\,dt$. Both are continuous in $\hat\xi$, $\det {\hat G}(\xi)\neq0$ by (R$_{\mathcal L}^k$), and
$\hat g_i(\xi)=\int_b^{a_i}Q'=Q(a_i)-Q(b)=0$. So for $\hat\xi$ near $\xi$ the vector $\hat y(\hat\xi)=-\hat G(\hat\xi)^{-1}{\hat g}(\hat\xi)$ is defined and continuous, with
$\hat y(\xi)=0$. Put
\[
  R_{\hat\xi}=R+\sum_{j=1}^k\hat y_j(\hat\xi)\,\tilde R_j,\qquad Q_{\hat\xi}(z)=\int_b^z\Pi_{\hat\xi}R_{\hat\xi}\,dt .
\]
Then $\deg Q_{\hat\xi}\le n$, since $\deg \Pi_{\hat\xi}R=\deg Q'\le n-1$ and $\deg \Pi_{\hat\xi}\tilde R_j\le n-1$; $Q_{\hat\xi}(b)=0$; and
$Q_{\hat\xi}(a_i)=\hat g_i(\hat\xi)+(\hat G(\hat\xi)\hat y(\hat\xi))_i=0$ for each $i$. Also $Q_{\hat\xi}'=\Pi_{\hat\xi}R_{\hat\xi}$ and $Q_{\hat\xi}\to Q$ coefficientwise. So $R_{\hat\xi}\to R$ uniformly on the
closed disk, where $\min|R|>0$, and $R_{\hat\xi}$ has no zero there for $\hat\xi$ close. The polynomial $H_{\hat\xi}=Q_{\hat\xi}\prod_j(1-\bar a_jz)/(z-a_j)$ has
$H_{\hat\xi}(b)=0$; the division of $Q_{\hat\xi}$ by $\prod_j(z-a_j)$ is exact, and its quotient depends continuously on $Q_{\hat\xi}$, so $H_{\hat\xi}\to H$
coefficientwise, hence in $C^1$ on $\{|z|\le r'\}$, and by Lemma~\ref{lem:openuniv} $H_{\hat\xi}$ is univalent on $\{|z|<r'\}$ for $\hat\xi$ close.
As $|\hat\xi_j|\ge1$, $Q_{\hat\xi}'\neq0$ in $\mathbb D$, so Lemma~\ref{lem:kzero} applies to $(Q_{\hat\xi},b,a_1,\dots,a_k)$; its factor $B$ depends on
$b,a_1,\dots,a_k$ only. If all $|\hat\xi_j|>1$, $Q_{\hat\xi}'$ has no zero on the closed disk, and Lemma~\ref{lem:recipek} applies.
\end{proof}

For $k=1$ the condition (R$_{\mathcal L}^k$) is (R$_{\mathcal L}$), up to sign. For $k\ge2$ the determinant is computed directly, as for pair~\ref{pair:K2N18} below.

\begin{corollary}\label{cor:K2N18BM}
Let $b$, $a_1$, $a_2$ and $Q$ be the data of pair~\ref{pair:K2N18}, and let $J_{\rm I}$ be as in Theorem~\ref{thm:twozeroK2N18}. Then
there are Blaschke models with two zeros and gain exactly $\sqrt{J_{\rm I}}$, given by polynomials $Q_{\hat\xi}$ of degree $18$ with
$Q_{\hat\xi}(b)=Q_{\hat\xi}(a_1)=Q_{\hat\xi}(a_2)=0$.
\end{corollary}

\begin{proof}
By the proof of Theorem~\ref{thm:twozeroK2N18}, facts (R), (S) and (C) of Lemma~\ref{lem:certify} hold, with $\Pi=(z-1)^5$. So $H$
is univalent on a disk $\{|z|<r\}$ with $r>1$ (Lemma~\ref{lem:certify}), and $Q'=(z-1)^5R$ with $R$ zero-free on the closed disk:
$p=1$, $\xi_1=1$, $M=m_1=5$ and $n=18$. Take $\tilde R_1=1$ and $\tilde R_2=z-1$, of degree at most $n-1-M=12$. Then
\[
  \int_b^{a_i}(t-1)^5\tilde R_j(t)\,dt=\frac{(a_i-1)^{5+j}-(b-1)^{5+j}}{5+j}\qquad(i,j=1,2),
\]
and the determinant of this $2\times2$ matrix, computed in exact arithmetic over $\mathbb Q(i)$, is
$-0.7186277\ldots+0.6643654\ldots\,i$. It is not $0$, so (R$_{\mathcal L}^k$) holds with $k=2$. Proposition~\ref{prop:splitk} with all $|\hat\xi_j|>1$
gives the Blaschke models; their degree is that of $Q$, since $Q_{\hat\xi}\to Q$.
\end{proof}

It is open whether the supremum of the gain over $\mathcal L_n$ is attained, and how its maximisers behave as $n\to\infty$. By Theorem~\ref{thm:fixeddeg} and
Corollary~\ref{cor:defectgrowth}, if local maximisers exist for all $n$ and $\sum_i(m_i+1)$ stays bounded, their numbers $q_n+r_n$ of
defects grow at least linearly in $n$, while by Lemmas~\ref{lem:D} and~\ref{lem:Dplus} a single $W$ meromorphic
past the circle has finitely many defects unless $W(\mathbb D)$ is a slit plane.