we know that $\Prb{\abs{\bullet_{\Ts^{n}}}=n-K_n, \abs{\circ_{\Ts^{n}}} =K_n+1} \sim\ C n^{-9/4}$ as $n \rightarrow \infty$, for a certain constant $C>0$.
We have therefore proved that

\[  \Pr{\exists v \in \bullet_{\Ts_{n}}: \, Q_\eps(\Ts_{n},v)}= \mathcal O(n^{-1/4}),\]

which tends to $0$ as $n \rightarrow \infty$.  This completes the proof.

§.§ Local limit theorems

We now establish two local limit estimates concerning the random walks $\Sbn$ and $\Scn$
(Lemmas <ref> and <ref>).

*Local limit theorem for $\Sbn$.

Let $\phibn(t)=\Esb{e^{it\Sbn_{1}}}= \tfrac{F( \bbn e^{it})}{F(\bbn)}$  be the characteristic function of $\Sbn_{1}$,
i.e. of the distribution $\mubn$.
(Recall that $F(z)=\sum_{k \ge 0} \tfrac{(k+1)^{k-1}}{k!} z^k$ was introduced in the proof of Lemma <ref>.)

Assume that $\tfrac{K_{n}}{n} \rightarrow 0$ and $K_{n} \rightarrow \infty$ as $n \rightarrow \infty$. For every $n$ sufficiently large, for every $ 0 \leq |t| \leq \pi$  we have

\begin{equation}
\label{eq:phib}  \ln | \phibn (t)| \leq -  \frac{K_{n}}{ 8n} t^{2}.
\end{equation}

We take two different approaches depending on the distance of $|t|$ to $0$.

First assume that $|t| \leq 1$. We have (see e.g. <cit.>)

\[\phibn(t)=1+\textrm{i}\E[\Sbn_{1}]t- \E[(\Sbn_{1})^{2}] \frac{t^{2}}{2}+r^{n}_{1}(t),\]

with $|r^{n}_{1}(t)| \leq |t|^{3} \E[(\Sbn_{1})^{3}]/6$ for every $n \geq 1$ and $t \in \R$. By using the expansion of $F$ around $0$, we see that $ \E[\Sbn_{1}] \sim \E[(\Sbn_{1})^{2}] \sim \E[(\Sbn_{1})^{3}] \sim \frac{K_{n}}{n} \rightarrow 0$ as $n \rightarrow \infty$, so that

\begin{align*}
\hspace{ -2mm} |\phibn(t)|^{2} &= \left( 1+\textrm{i}\E[\Sbn_{1}]t- \E[(\Sbn_{1})^{2}] \frac{t^{2}}{2}+r^{n}_{1}(t) \right) \left( 1-\textrm{i}\E[\Sbn_{1}]t- \E[(\Sbn_{1})^{2}] \frac{t^{2}}{2}+\overline{r^{n}_{1}(t)} \right)\\
&= 1 - (\sbn)^{2} t^{2}+r^{n}_{2}(t),
\end{align*}

where $|r^{n}_{2}(t)| \leq  |t|^{3} K_{n}/(2n)$ for every $n$ sufficiently large, uniformly for $t \in [-1,1]$.

\[ 2 \ln |\phibn(t)| \leq - (\sbn)^{2} t^{2}+ |t|^{3} K_{n}/(2n)= - \frac{K_{n}}{n} t^{2} \left(  \frac{(\sbn)^{2} n}{K_{n}}-|t|/2 \right)
\]

for every $n$ sufficiently large, uniformly for $t \in [-1,1]$.
Since $(\sbn)^{2} \sim K_{n}/n$ as $n \rightarrow \infty$ by Lemma <ref> (ii), it follows that for every $n$ sufficiently large and $0 \leq |t| \leq 1$ we have $ \ln | \phibn (t)| \leq -  \frac{K_{n}}{ 8n} t^{2}$.

Now assume that $1 \leq |t| \leq \pi$. Since $F(z)=1+z+o(z)$ as $z \rightarrow 0$ and, by definition, $\phibn(t)=\tfrac{F( \bbn e^{it})}{F(\bbn)}$,
we have

\[|\phibn(t)|^{2}=(1+\bbn(e^{it}-1)+o(\bbn))(1+\bbn(e^{-it}-1)+o(\bbn))=1+2\bbn(\cos(t)-1)+o(\bbn),\]

where the $o(\bbn)$ is uniform in $1\leq |t| \leq \pi$.
Therefore, for every $1 \leq |t| \leq \pi$, for every $n$ sufficiently large,

\[ 2 \ln |\phibn(t)| \leq - \bbn \frac{t^{2}}{3}+o(\bbn) \leq - \frac{K_{n}}{4n} t^{2},\]

where for the last inequality we have used the fact that $\bbn \sim \frac{K_{n}}{n}$ as $n \rightarrow \infty$ by Lemma <ref> (ii) and that $t$ is not too close to $0$.

We first check that the convergence

\begin{equation}
\label{eq:cv1b}\E\bigg[e^{i t \frac{\Sbn_{N}-N \frac{K_{n}+1}{n-K_{n}} }{ \DbnN }} \bigg]  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad e^{- \frac{t^{2}}{2}}
\end{equation}

holds uniformly for $un \leq N \leq n$ and $t$ in compact subsets of $\R$.  As in the proof of Lemma <ref>, since $\tfrac{\E[(\Sbn_{1})^{3}]}{(\DbnN)^{3}}=\mathcal O(\tfrac{1}{n \sqrt{K_{n}}})= o(\frac{1}{n})$, we have

\[\phibn\left( \frac{t}{ \DbnN } \right)=1+\textrm{i}\E[\Sbn_{1}] \frac{t}{ \DbnN } - \E[(\Sbn_{1})^{2}] \frac{t^{2}}{2(\DbnN)^{2}}+ o \left( \frac{1}{n} \right),\]

where the $o$ is uniform when $un \leq N \leq n$ and $t$ belongs to a compact subset of $\R$, so that

\[\ln \phibn \left( \frac{t}{ \DbnN } \right)=it  \frac{K_{n}+1}{(n-K_{n})\DbnN} - \frac{t^{2}}{2} \cdot \frac{(\sbn)^{2}}{(\DbnN )^{2}}+ o \left( \frac{1}{n} \right).\]

\[\Es{e^{i t \frac{\Sbn_{N}-N \frac{K_{n}+1}{n-K_{n}} }{ \DbnN }}}= \exp \left(- \frac{t^{2}}{2} \cdot  \frac{ N (\sbn)^{2} }{(\DbnN)^{2}} + o \left( \frac{N}{n} \right)   \right).\]

Since $(\DbnN)^{2}= N (\sbn)^{2}$, this  implies (<ref>).

We then follow the steps of the analytic proof of the standard local limit theorem for a sequence of independent identically distributed random variables. The main difficulty is that the distribution of $\Sbn_{1}$ depends on $n$. To simplify notation, set $f(t)= e^{- \frac{t^{2}}{2}}$ for $t \in \R$. By Fourier inversion we have, for every $x \in \mathbb{R}$,

\begin{equation}
\label{eq:fourierb}  p(x)= \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-itx} f(t) {\d}t.
\end{equation}

Also, for $k \in \mathbb{Z}$,

\[ \Pr{\Sbn_{N}=k} = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-itk} \phibn(t)^{ N} {\mathrm{d}}t.\]

In the following, we only consider values of $x \in \R$ such that $N \frac{K_{n}+1}{n-K_{n}}+x  \DbnN$ is an integer.
For such $x$,

\begin{multline*}
\DbnN \cdot \Pr{\Sbn_{ N}= N \frac{K_{n}+1}{n-K_{n}}+x  \DbnN}\\
=  \frac{1}{2\pi} \int^{\pi \DbnN}_{-\pi \DbnN} e^{- i t x} \left(  \phibn \left(  \frac{t}{\DbnN} \right)   e^{-it \frac{K_{n}+1}{(n-K_{n}) \DbnN}} \right)^{N} {\mathrm{d}}t.
\end{multline*}

Therefore for every fixed $A>0$, for $n$ sufficiently large,

\begin{multline*}
\left| \DbnN \cdot \Pr{\Sbn_{N}= N \frac{K_{n}+1}{n-K_{n}}+x  \DbnN} -p( x)\right| \\
\leq \frac{1}{2 \pi} \big(|I^{(1)}_{A}(x,N,n)| + |I^{(2)}_{A}(x,N,n)|+|I^{(3)}(x)|\big),
\end{multline*}

\begin{align*}
I^{(1)}_{A}(x,N,n)&=\int_{-A}^{A} e^{-i t x} \left(   \left(  \phibn \left(  \frac{t}{\DbnN} \right)   e^{-it \frac{K_{n}+1}{(n-K_{n}) \DbnN}} \right)^{N}- f(t) \right) {\mathrm{d}}t,\\
I^{(2)}_{A}(x,N,n)&= \int_{A<|t|< \pi \DbnN} e^{-itx - it \frac{(K_{n}+1) N}{(n-K_{n}) \DbnN}} \phibn \left(  \frac{t}{\DbnN} \right) ^{ N}{\mathrm{d}}t, \\
I^{(3)}_{A}(x)&= \int_{|t|>A} e^{-itx} f(t) {\mathrm{d}}t.
\end{align*}

We shall check that for every fixed $\epsilon>0$, there exists $A>0$ such that for every $n$ sufficiently large, for every $ un \leq N \leq  n$, for every $x \in \R$ (with the above integrality condition),  $|I^{(1)}_{A}(x,N,n)| \leq \epsilon$, $|I^{(2)}_{A}(x,N,n)| \leq \epsilon$ and $|I^{(3)}_{A}(x)| \leq\epsilon$.

We choose $A >0$ such that

\begin{equation}
\label{eq:Ab} 2 \int_{A}^{\infty } e^{-  \frac{1}{9} t^{2}} {\d}t < \epsilon.
\end{equation}

Bounding $|I^{(1)}_{A}(x,N,n)|$. Since the convergence (<ref>) holds  uniformly on compact subsets of $\R$, for   $n$ sufficiently large, for every $ un \leq N \leq n$ and $x \in \R$, we have $|I^{(1)}_{A}(x,N,n)| \leq \epsilon$.

Bounding $|I^{(3)}_{A}(x)|$. By the choice of $A$ in (<ref>)
since $|f(t)|=e^{-  \frac{1}{2} t^{2}} \le e^{-  \frac{1}{9} t^{2}}$,
for every $x \in \R$ we have $|I^{(3)}_{A}(x)| \leq \epsilon$.

Bounding $|I^{(2)}_{{A} }(x,N,n)|$.  By Lemma <ref>,  for every $n$ sufficiently large, for every $ un \leq N \leq n$, $A<|t| \leq \pi  \DbnN$ we have

\[|I^{(2)}(x,N,n)| \leq 2 \int_{A}^{\pi \DbnN} e^{-  \frac{ K_{n} N }{ 8 n (\DbnN)^{2}} \cdot t^{2} } {\d}t \leq 2 \int_{A}^{\infty } e^{-  \frac{1}{9} t^{2}} {\d}t,\]

which is less than $\epsilon$ by (<ref>). This completes the proof.

*Local limit theorem for $\Scn$.

In the case of $\Scn$, the analysis is more subtle.
As above, we start with an estimate on the characteristic function
of the step distribution of the random walk

\[\phicn(t)\coloneqq\Es{e^{it\Scn_{1}}}=\frac{F( \bcn e^{it})}{F(\bcn)}.\]

To simplify notation, set  $y_{n}=1/e-\bcn$.

Assume that $\tfrac{K_{n}}{\sqrt{n}} \rightarrow c>0$ as $n \rightarrow \infty$.  There exist constants $A_{0},\epsilon,\kappa>0$, which may only depend on $c$, such that the following holds.
For every $n$ sufficiently large and for every $|t| \in (\frac{A_{0}}{\Dcn},\epsilon)$, we have $|\phicn(t)| \leq e^{- \kappa |t|^{1/2}}$.

We start with some preliminary observations which fix the values of the different constants.

–  Since $\Dcn \sim n /c$ and $y_{n} \sim \frac{c^{2}}{2en}$ (as was seen in the proof of Lemma <ref>),
we can find a constant $\gamma>0$ depending only on $c$ such that
the conditions $t \ge  {A_0}/{\Dcn}$ and $A_{0} \geq 1$ imply
$t \ge {\gamma A_0}/{n}$ and $t \ge \gamma y_{n}$,
for $n$ sufficiently large.

– We then fix $A_{0} \geq 1$ such that $\frac{2c}{\sqrt{\gamma A_{0}}} \leq  \frac{1}{4\sqrt{e}}$.

– From the expansion $ \ln F \big( \tfrac{1}{e}-z \big)= 1-\sqrt{2e} \sqrt{z}+o(\sqrt{z})$, there exists $\eta>0$ such that

\begin{equation}
\label{eq:z} \Re(z) \ge 0, \, |z| \leq \eta  \qquad \implies  \qquad  \Re \ln F \left( \frac{1}{e}-z \right) \leq  1-  \Re \sqrt{z}.
\end{equation}

Then set $\epsilon= \sqrt{\eta^{2}/(e^{-2}+\gamma^{-2})}$.

We now turn to the main part of the proof. For $t \in \R$, since
$ \ln |\phicn(t)| =\Re \ln \phicn(t)$, we have

\begin{equation}
\label{eq:lnphi} \ln |\phicn(t)| =\Re \ln F( \bcn e^{it})- \ln F( \bcn).
\end{equation}

We first study the term $\ln F( \bcn)$.
By (<ref>) and (<ref>), we have $\ln F( \bcn)=1- \tfrac{c}{\sqrt{n}}+ o \big(  \frac{1}{\sqrt{n}} \big)$.  As a consequence, for every $n$ sufficiently large, for every $  {A_{0}}/{\Dcn} \leq |t| \leq 1$, we have

\begin{equation}
\label{eq:ln}\ln F( \bcn) \geq 1- \frac{2c}{\sqrt{n}} \geq 1- \frac{2c}{\sqrt{\gamma A_{0}}} \sqrt{|t|}.
\end{equation}

Consider now the term $\Re \ln F( \bcn e^{it})$.
To this end, we set $z=z_{n}(t)=\tfrac{1}{e}-\bcn e^{it}$.
Note that we always have $\Re(z_n(t)) \ge 0$.
Observing that $e^{-it} z_{n}(t)= \tfrac{1}{e}(e^{-it}-1)+y_n$, we have

\begin{equation}
\label{eq:module} |z_{n}(t)|^{2}=  \tfrac{1}{e^2} \big|e^{-it}-1\big|^2
+\tfrac{2}{e} \big\langle y_n, e^{-it}-1 \big\rangle +|y_n|^2
= \frac{2(1-\cos(t))}{e^{2}} -\frac{2 y_{n}(1-\cos(t))}{e}+y_{n}^{2}.
\end{equation}

For $n$ sufficiently large and $|t| \leq \epsilon$, it follows that

\begin{equation}
\label{ineq:module} |z_{n}(t)|^{2} \leq  \left( \frac{1}{e^{2}} + \frac{1}{\gamma^{2}} \right)  t^{2} \leq \eta^{2}.
\end{equation}

Using (<ref>), we have

\[\Re \ln F( \bcn e^{it}) \leq 1-   \Re \sqrt{z_{n}(t)}.\]

Since $\Re z_{n}(t) \ge 0$, we have

\[ \Re \sqrt{z_{n}(t)}=\sqrt{\frac{\Re(z_n(t))+|z_n(t)|}{2}} \ge \sqrt{\frac{|z_n(t)|}{2}}. \]

But, by (<ref>), for $n$ sufficiently large and  for every $|t|$ in $(A_{0}/\Dcn,1)$,

\[|z_{n}(t)|^{2} = \frac{2(1-\cos(t))}{e^{2}} (1-y_{n} e)+y_{n}^{2} \geq \frac{1-\cos(t)}{e^{2}} \geq  \frac{t^{2}}{4e^{2}}.\]

As a consequence $\Re \sqrt{z_{n}(t)} \geq \frac{\sqrt{|t|}}{2\sqrt{e}}$, so that

\begin{equation}
\label{eq:reln}\Re \ln F( \bcn e^{it}) \leq 1- \frac{\sqrt{|t|}}{2\sqrt{e}}.
\end{equation}

To conclude, by combining (<ref>) with (<ref>) and (<ref>),
we get that for every $n$ sufficiently large, for every $|t|$ in $(A_{0}/\Dcn, \epsilon)$,

\[\ln|\phicn(t)| \leq \left(  \frac{2c}{\sqrt{\gamma A_{0}}}- \frac{1}{2\sqrt{e}} \right) \cdot \sqrt{|t|} \leq -\frac{1}{4\sqrt{e}} \sqrt{|t|}.\]

This completes the proof, with $\kappa= \frac{1}{4\sqrt{e}}$.

We are now ready to tackle the proof of Lemma <ref>.

The first step is to show the convergence

\begin{equation}
\label{eq:cv1}\Es{e^{i t \frac{\Scn_{ u K_{n} +j}}{ \Dcn }}}  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad \exp \left( u c^{2} \left( 1 -\sqrt{1-  \frac{2it}{c}}  \right)  \right),
\end{equation}

uniformly for $t$ in compact subsets of $\R$ and $|j| \leq n^{3/8}$.

The characteristic function $\phicn(t)=\Es{e^{it\Scn_{1}}}$  of $\Scn_{1}$
is given as $\phicn(t)= \frac{F( \bcn e^{it})}{F(\bcn)}$,
where $\bcn$ has been chosen such that $ \frac{\bcn F'(\bcn)}{F(\bcn)}= \frac{n-K_{n}}{K_{n}+1}$.
Recall that $\bcn=1/e-y_{n}$ with
$y_{n} \sim \frac{1}{2e}  \big( \frac{K_{n}}{n} \big)^{2} \sim \frac{c^{2}}{2e} \cdot \frac{1}{n}$
(from (<ref>)).
Since $ \Dcn \sim  n /c$, we have, uniformly on compact subsets of $\R$ as $n \rightarrow \infty$,

\[\bcn e^{ \frac{it}{\Dcn}}=(1/e-y_{n})e^{ \frac{it}{\Dcn}}=1/e-  \left( \frac{c^{2}}{2e} - \frac{it c}{e } \right) \cdot \frac{1}{{n}}+o \left(  \frac{1}{n} \right).\]

By (<ref>), we have $ \ln F \left( \frac{1}{e}-z \right)= 1-\sqrt{2e} \sqrt{z}+o(\sqrt{z})$ as $z \rightarrow 0$.
As a consequence as $n \rightarrow \infty$,
\[\ln \phicn \left( \frac{t}{ \Dcn } \right)=-\sqrt{2e} \sqrt{ \frac{c^{2}}{2e} - \frac{it c}{e}} \cdot \frac{1}{\sqrt{n}} + \sqrt{2e} \sqrt{ \frac{c^{2}}{2e}} \cdot \frac{1}{\sqrt{n}}+o \left(  \frac{1}{\sqrt{n}} \right),\]

uniformly on compact subsets of $\R$. Therefore

\[( uK_{n}  +j) \ln \phicn \left( \frac{t}{\Dcn} \right)  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad u c^{2}-uc \sqrt{c^{2}- 2itc},\]

uniformly when $t$ belongs to compact subsets of $\R$ and $|j| \leq n^{3/8}$, which implies (<ref>).

As in the proof of Lemma <ref>, the second step consists in estimating $\Pr{\Scn_{ uK_{n}  +j}=k}$ by Fourier inversion.
We first let, for $t \in \mathbb{R}$,

\[f(t)=\exp \left( u c^{2} \left( 1 -\sqrt{1-  \frac{2it}{c}}  \right)  \right)\]

be the expression appearing in (<ref>).
By <cit.>),  $f$ is the characteristic function of a random variable with explicit density

\[q_u(x)=\left( \frac{u^{2} c^{3}}{2 \pi x^{3}} \right)^{1/2} \exp \left(  - \frac{c (x-uc)^{2} }{2x} \right) \mathbbm{1}_{x>0},\]

so that by Fourier inversion we have, for every $x \in \mathbb{R}$,

\begin{equation}
\label{eq:fourier}
q_u(x)=  \frac{1}{2\pi}\int_{-\infty}^{\infty}  {\mathrm{d}t} e^{-itx} \exp \left( u c^{2} \left( 1 -\sqrt{1-  \frac{2it}{c}}  \right)  \right).
\end{equation}

On the other hand,
for $k \in \mathbb{Z}$,

\[ \Pr{\Scn_{ uK_{n}  +j}=k} = \frac{1}{2\pi} \int_{-\pi}^{\pi} e^{-itk} \phicn(t)^{ uK_{n}  +j} {\mathrm{d}}t.\]

Therefore, assuming that $x \in \R$ is chosen so that $x \Dcn$ is an integer, we get that

\[ \Dcn \cdot \Pr{\Scn_{ uK_{n}  +j}=x  \Dcn}=  \frac{1}{2\pi} \int^{\pi \Dcn}_{-\pi \Dcn} e^{- i t x} \phicn \left(  \frac{t}{\Dcn} \right) ^{ uK_{n}  +j} {\mathrm{d}}t.\]

We deduce that for every fixed $A>0$ and $0< \epsilon \leq 1$, for $n$ sufficiently large, it holds

\begin{multline*}
\left| \Dcn \cdot \Pr{\Scn_{ uK_{n}  +j}=x  \Dcn} - q_{u}(x)\right| \\
\leq \frac{1}{2 \pi} (|I^{(1)}_{A}(x,j,n)| + |I^{(2)}_{\epsilon,A}(x,j,n)|+|I^{(3)}_{\epsilon}(x,j,n)|+|I^{(4)}_{A}(x)|),
\end{multline*}

\[I^{(1)}_{A}(x,j,n)=\int_{-A}^{A} e^{-i t x} \left(  \phicn \left(  \frac{t}{\Dcn} \right) ^{ uK_{n}  +j}- f(t) \right) {\mathrm{d}}t\]

\[ I^{(2)}_{\epsilon,A}(x,j,n)= \int_{A<|t|< \epsilon \Dcn} e^{-itx} \phicn \left(  \frac{t}{\Dcn} \right) ^{ uK_{n}  +j}{\mathrm{d}}t,\]

\[ I^{(3)}_{\epsilon}(x,j,n)=\int_{\epsilon \Dcn<|t|<\pi \Dcn} e^{-itx} \phicn  \left(  \frac{t}{\Dcn} \right) ^{ uK_{n}  +j} {\mathrm{d}}t, \qquad I^{(4)}_{A}(x)= \int_{|t|>A} e^{-itx} f(t) {\mathrm{d}}t .\]

We shall check that for every fixed $\epsilon'>0$, there exists $A>0$ and $0 < \epsilon \leq 1$ such that for every $n$ sufficiently large, for every $|j| \leq n^{3/8}$, for every $x \in \R$,
$|I^{(1)}_{A}(x,j,n)| \leq \epsilon', |I^{(2)}_{\epsilon,A}(x,j,n)| \leq \epsilon', |I^{(3)}_{\epsilon}(x,j,n)| \leq \epsilon', |I^{(4)}_{A}(x)| \leq\epsilon'$.

Fix $\epsilon'>0$. We first explain how to choose $A$ and $\epsilon$.
As a preliminary observation, note that there exists a constant $C_{1}>0$
(which only depend on $u$ and $c$) such that $|f(t)| \leq C_{1} e^{-C_{1}^{-1} \sqrt{t}}$ for $t \geq 0$
(this is straightforward, using the explicit expression of $f$).
Let $A_{0}$ and $\kappa$ be the constants given by Lemma <ref>.
We may choose $A \geq A_{0}$ such that

\begin{equation}
\label{eq:A}2\int_{A}^{\infty} |f(t)|  \mathrm{d}t <\epsilon' \qquad \textrm{and} \qquad  2 \int_{A}^{\infty} e^{- \frac{\kappa u c^{3/2}}{2} t^{1/2}} \mathrm{d}t < \epsilon'.
\end{equation}

We then let $\epsilon$ be also given by Lemma <ref>.

Bounding $|I^{(1)}_{A}(x,j,n)|$. Since the convergence (<ref>) holds  uniformly on compact subsets of $\R$ and $|j| \leq n^{3/8}$, for   $n$ sufficiently large, for every $|j| \leq n^{3/8}$ and $x \in \R$, we have $|I^{(1)}_{A}(x,j,n)| \leq \epsilon'$.

Bounding $|I^{(4)}_{A}(x)|$. By the choice of $A$ in (<ref>), for every $n$ sufficiently large, for every $x \in \R$ we have $|I^{(4)}_{A}(x,n)| \leq \epsilon'$.

Bounding $|I^{(3)}_{\epsilon}(x,j,n)|$.  Set $\phi(t)=\frac{F(e^{it}/e)}{F(1/e)}$, which is the characteristic function of  the probability distribution $ \frac{(i+1)^{(i-1)}}{e^{i+1} i!}$ on $\Z_{+}$. Since the latter is non lattice, we have $|\phi(t)|<1$ for $t \in (0,2\pi)$ (see e.g. <cit.>).
Therefore, there exists $C_2>0$ such that $|\phi(t)|<e^{-2C_2}$ for $ \epsilon \leq |t| \leq \pi$.
Since $F$ is a power series with nonnegative coefficients and converges in $1/e$,
it converges uniformly on $\{z: |z| \le 1/e\}$ and is continuous on this compact set.
It is therefore uniformly continuous and
we have $\phicn(t) \rightarrow \phi(t)$, uniformly for $t \in \R$.
In particular, for $n$ sufficiently large and $t$ with $ \epsilon \leq |t| \leq \pi$,
we have $|\phicn(t)|<e^{-C_2}$.
Therefore, for every $n$ sufficiently large, for every $|j| \leq n^{3/8}$ and  $x \in \R$,

\[ \left|I^{(3)}_{\epsilon}(x,j,n) \right|  \leq 2 \int_{\epsilon \Dcn}^{\pi \Dcn} e^{-C_2 (uK_{n}  +n^{3/8})} \mathrm{d}t
\leq 2 \pi \Dcn \cdot (e^{- C_2 u c \sqrt{n}/2} ) \le \epsilon'.\]

Bounding $|I^{(2)}_{\epsilon, {A} }(x,j,n)|$.  By Lemma <ref>, for every $A<|t| \leq \epsilon \Dcn$ we have
\[  \left| \phicn \left(  \frac{t}{\Dcn} \right) \right|  \leq e^{-\kappa \left(  \frac{|t|}{\Dcn} \right) ^{1/2}}.\]

Hence, for $n$ sufficiently large, for every $|j| \leq n^{3/8}$ and $x \in \R$,

\[ \left| I^{(2)}_{\epsilon,A}(x,j,n)  \right| \leq 2 \int_{A}^{\epsilon \Dcn}\left( e^{-\kappa  \left(  \frac{t}{\Dcn} \right) ^{1/2}} \right) ^{ uK_{n}  -n^{3/8}} \mathrm{d}t .\]

Since $K_{n} \sim c \sqrt{n}$ and $ \Dcn\sim  n /c$, it follows that for $n$ sufficiently large, for every $|j| \leq n^{3/8}$ and $x \in \R$,

\[ \left| I^{(2)}_{\epsilon,A}(x,j,n) \right| \leq 2 \int_{A}^{\epsilon \Dcn}  e^{- \frac{\kappa u  c^{3/2}}{2} t^{1/2}} \mathrm{d}t  \leq 2 \int_{A}^{\infty} e^{- \frac{\kappa u c^{3/2}}{2} t^{1/2}} \mathrm{d}t,\]

which is less than $\epsilon'$ by (<ref>).   This completes the proof.

§ MINIMAL FACTORIZATIONS AND  LAMINATIONS

Table of the main notation and symbols appearing in Section <ref>.

$\mathbf{L}_{\infty} $         The Brownian triangulation.

$\mathbf{L}_{c} $        The lamination coded by an excursion of a Lévy process with characteristic exponent given by (<ref>).

$(t_{1}^{(n)}, \ldots, t_{n-1}^{(n)})$         A minimal factorization of $\mathfrak{M}_{n}$.

$ \PPP_{n}$, $ \dot{\PPP}_{n}$         The non-crossing partition $ \PPP(  t_{1}^{(n)}, \ldots, t_{K_{n}}^{(n)})$ and its associated compact set.

$ \FFF_{n}$, $\dot{ \FFF}_{n}$        The non-crossing forest $ \FFF(  t_{1}^{(n)}, \ldots, t_{K_{n}}^{(n)})$ and its associated compact set.

$T_{n}$           The tree $ \mathcal{T}(  \PPP_{n})$   coding the non-crossing partition $ \PPP_{n}$.

A geodesic lamination $L$ of $\overline{\D}$ is a closed subset of $\overline{\D}$ such that $L$  can be written as the union non-crossing chords, i.e. which do not intersect in $\D$.
Recall that by convention, $ \{x\}$ is a chord for every $x \in \S$. In particular, $\S$ is a geodesic lamination. In the sequel, by lamination we will always mean geodesic lamination of $\overline{\D}$.  It is well known (and simple to see) that the set of all geodesic laminations of $\overline{\D}$ equipped with Hausdorff topology is compact.

§.§ Definition of the random lamination $\mathbf{L}_{c}$

We start with defining $\mathbf{L}_{c}$ in the three cases $c=0$, $c=\infty$ and $c \in \R_{+}$. First, for $c=0$, we simply let $\mathbf{L}_{0}$ be the unit circle $\mathbb{S}$.

*Case $c=\infty$.  Let $Z=(Z_{u})_{0 \leq u \leq 1} \in \mathbb{C}([0,1],\R)$ be a continuous function.
We assume $Z_{0}=Z_{1}=0$, as well as the following property:

*  Local minima of $Z$ have distinct values,
meaning that for every $0 \le s < t \le 1$, there exists at most one value $r \in (s, t)$ such that $Z_{r} = \inf_{[s, t]} Z$.

Following [4, 55], we construct a lamination $L(Z)$ from $Z$ as follows.
every $s,t \in [0,1]$, we set $s \, {\sim}^{Z}\,  t$ if we have $Z_{s}
=Z_{t} = \min_{[s\wedge t, s \vee t]} Z$. We  set

\begin{equation}
\label{eq:defbl}
L(Z) \quad \coloneqq \quad  \bigcup_{s \sim^{Z} t} \big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big].
\end{equation}

Then $L(Z) $ is a geodesic lamination, which is a triangulation (in the sense that the
complement of $ L(Z) $ in $\overline{ \mathbb{D}}$ is a
disjoint union of open Euclidean triangles whose vertices belong to
the unit circle).  In particular,  $L(Z) $ is maximal with respect to inclusion for geodesic laminations (see <cit.>). Observe that $\S \subset  L(Z)$.
Using a continuity argument and the fact that the set of local minima is countable, it is a simple matter to see that

\begin{equation}
\label{eq:adherence}
L(Z)= \overline{ \bigcup_{\substack{s \sim^{Z} t, s \neq t\\ s,t \textrm{ are not local minima}}} \big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big]}.
\end{equation}

(Roughly speaking, this means that $ L(Z)$ is the closure of all non-trivial chords which do not belong to triangles.)
In a similar way, since $Z$ is continuous, we have

\begin{equation}
\label{eq:isolated} s \sim^{Z} t \ \implies \  \forall \epsilon>0,\, \exists s',t' \textrm{ such that }  s' \sim^{Z} t'  \textrm{ with } 0<|s-s'| \leq \epsilon, 0<|t-t'| \leq \epsilon.
\end{equation}

(Roughly speaking, this means that  chords are not isolated).

We now consider the Brownian excursion $\mathbbm{e}$,
which is informally defined as a Brownian motion
conditioned to return to the origin at time $1$ and to stay positive on the time interval $(0,1)$.
It is well known that it satisfies <ref>
a.s.  We then define

\[\mathbf{L}_{\infty} \coloneqq L(\mathbbm{e})\]

to be the Brownian triangulation, see Figure <ref> for a simulation.

\[\begin{array}{c}
\includegraphics[scale=.45]{Excursion_Brownienne.pdf}
\end{array} \qquad
\begin{array}{c}
\includegraphics[scale=.35]{Triangulation_Brownienne.pdf}
\end{array}\]

A realization of the Brownian excursion $\mathbbm{e}$ and its associated lamination.
The latter is therefore a realization of the Brownian triangulation $\mathbf{L}_{\infty}$.

The Brownian triangulation has been introduced by Aldous [4, 5], and it has been shown to be the universal limit of many different families of random non-crossing configurations [50, 19, 49].

*Case $c >0$.

Let $Z \in {\D}([0, 1], \R)$ be a càdlàg function.
For $t  \in [0,1]$, we denote $Z_{t-}$ the left-hand limit of $Z$ in $t$,
and set $\Delta Z_{t}=Z_{t}-Z_{t-}$, with the convention $Z_{0-}=0$.
We fix a function $Z \in {\D}([0, 1], \R)$ such that $Z_{1} = 0$, $Z_{t} > 0$ and $\Delta Z_{t} \ge 0$ for every $t \in (0,1)$,
and we define the following four properties:

*  $ \{t \in (0,1) : \Delta Z_{t}>0\}$ is dense in $[0,1]$.

For every $0 \le s < t \le 1$, there exists at most one value $r \in (s, t)$ such that $Z_{r} = \inf_{[s, t]} Z$.

For every $t \in [0,1)$ such that $\Delta Z_{t} > 0$, we have $\inf_{[t, t+\varepsilon]} Z < Z_{t}$ for every $0 < \varepsilon \le 1-t$;

For every $t \in (0,1)$ such that $Z$ attains a local minimum at $t-$ (meaning that there exists $\epsilon>0$ such that $Z_{t-}=\inf_{[t-\epsilon,t+\epsilon]} Z$), if $s = \sup\{ u \in [0,t] : Z_{u} < Z_{t-}\}$, then $\Delta Z_{s} > 0$ and $Z_{s-} < Z_{t-} < Z_{s}$.

Following [50],
we construct a lamination $L(Z)$ from a function $Z$ with the above 4 properties
(note that in contrast with [50], we have one less property,
and the analogue of <ref> differs since we accept here the case $Z_{0} > 0$).
To this end, we define a relation (which is not an equivalence relation in general) on $[0, 1]$ as follows:
for every $0 \le s < t \le 1$, we set

\[
s \simeq^Z t \qquad\text{if}\qquad t = \inf \left\{u > s : Z_{u} \le Z_{s-} \right\},
\]

then for $0 \le t < s \le 1$, we set $s \simeq^Z t$ if $t \simeq^Z s$, and we agree that $s \simeq^Z s$ for every $s \in [0,1]$. We finally define a subset of $\overline{\D}$ by

\begin{equation}
L(Z) \quad \coloneqq \quad \bigcup_{s \simeq^Z t} \left[\e^{-2\pi \i s}, \e^{-2\pi\i t}\right].
\label{eq:defl_cadlag}
\end{equation}

Observe that $\S \subset L(Z)$ by definition. By using <ref> and <ref>, it is a simple matter to check that

\begin{equation}\label{eq:lamination_stable2}
L(Z) \quad = \quad \overline{\bigcup_{\substack{s \simeq^Z t \\ s \neq t}} \left[\e^{-2\pi \i s}, \e^{-2\pi\i t}\right]}.
\end{equation}

Note that, if $s \simeq^Z t$ with $s<t$, we automatically have $Z_{s-}=Z_{t}$.

*Comment on notation. Observe that the definition of the lamination $L(Z)$ for càdlàg functions  (Eq. (<ref>)) with the above properties differs from the one
for continuous functions (Eq. (<ref>)). Since these two sets of functions are disjoint, this should not create any confusion.

Assume that  $Z \in {\D}([0, 1], \R)$ satisfies $Z_{1} = 0$, $Z_{t} > 0$ and $\Delta Z_{t} \ge 0$ for every $t \in (0,1)$, <ref> and <ref>. Then the lamination $L(Z)$ is a geodesic lamination of $\overline{\D}$, called the lamination coded by $Z$.

This can be proved by using the same arguments as <cit.>, we leave the details to the reader.
Note that the assumptions <ref> and <ref> are not required in Lemma <ref>,
but will be later needed in Proposition <ref>.

Recall from Section <ref> that $\Xexc$ denotes the excursion process,
associated with the Lévy process with characteristic exponent given by (<ref>).
Almost surely $\Xexc$  satisfies <ref>, <ref>, <ref> and <ref>.

We mimic the proof of <cit.>.
By the construction of $\Xexc$ as the Vervaat transform of $\Xbr$, and by the absolute continuity relation (<ref>), it is sufficient to prove that analogous properties hold
for the Lévy process $X$.  The property <ref> follows from the fact that the Lévy measure of $X$ is infinite. The properties <ref> and <ref> are consequences
of the (strong) Markov property of $X$ and the fact that $0$ is regular for $(-\infty,0)$.

For <ref>, we will use the time-reversal property of $X$, which
states that if $t > 0$ and $\widehat{X}^{(t)}$ is the process defined by $\widehat{X}^{(t)}_{s}=X_{t}-X_{(t-s)-}$ for $0 \leq s<t$ and $\widehat{X}^{(t)}_{t}=X_{t}$, then the two processes $(X_{s},0 \leq s \leq t)$ and $(\widehat{X}^{(t)}_{s},0 \leq s \leq t)$ have the same law.
Set $S_{t}= \sup_{0 \leq s \leq t} X_{s}$. By the time-reversal property of $X$,
it is sufficient to prove that if $q > 0$ is rational and $T = \inf\{t  \geq  q : X_{t} > S_{q}\}$, then
$X_{T} > S_{q}  > X_{T-}$ almost surely. This follows from the Markov property at time $q$
and the fact that for any $a > 0$, $X$ jumps a.s. across $a$ at its first passage time
above $a$ (Lévy processes which can creep upwards
are characterized e.g. in <cit.>;
since the Lévy process we are considering has paths of bounded variations,
a negative drift and no Gaussian component, it cannot creep upwards
and it jumps a.s. across any level $a$ at its first passage).

Almost surely $\Xexc$ also satisfies the fact that for every $t \in (0,1)$ such that $\Delta \Xexc(t) > 0$, we have $\inf_{[t-\varepsilon, t]} \Xexc > \Xexc(t-)$ for every $0 < \varepsilon \le t$ (this follows from the time-reversal property of $X$ and the fact that $0$ is irregular for $(0,\infty)$), but we will not use this property.

For $c>0$,
we now define $\mathbf{L}_{c}$ as

\[\mathbf{L}_{c} \coloneqq L(\Xexc),\]

where $\Xexc$ is the Lévy excursion process introduced in Section <ref>;
see Figure <ref> for a simulation of $\mathbf{L}_{5}$.

\[\begin{array}{c}
\includegraphics[scale=.45]{Excursion_C5.pdf}
\end{array} \qquad
\begin{array}{c}
\includegraphics[scale=.35]{Lamination_C5.pdf}
\end{array}\]

A realization of the Lévy excursion process $\Xexc$ from Section <ref> for $c=5$
and its associated lamination.
The latter is therefore a realization of the random lamination $\mathbf{L}_{5}$.

In what follows, we first concentrate on the case $c\in \R_{+}$, and discuss the case $c=\infty$ in Section <ref>.

§.§ A symmetry result.
As a warm-up, we establish a result that will be useful twice: first, it will allow us to restrict ourselves to the case where $K_{n} \leq n/2$ in the proof of Theorem <ref> (ii), and it will also allow us to immediately lift, for $ \dot{\mathcal{P}}^{(n)}_{K_{n}}$, (i) in  Theorem <ref> to (iii).

Recall that if $P$ is a non-crossing partition, singleton blocks of $P$ do not appear in $\dot{P}$.  We denote by $\hat{P}$  the compact subset of $\overline{\D}$ obtained from $\dot{P}$ by adding the vertices $\{e^{-2 \i \pi k/n}, 1 \leq k \leq n\}$. This definition has its importance in view of the next result.

Set $\hat{\PPP}^{(n)}_{i}=\hat{\PPP}(\tb_{1}^{(n)} \tb_{2}^{(n)} \cdots \tb_{i}^{(n)})$ for $1 \leq i \leq n-1$.
Assume that there exists a random lamination $L$ such that the convergence $\hat{\PPP}^{(n)}_{n-K_{n}-1} \rightarrow  L$ holds in distribution.

\[  \hat{\PPP}^{(n)}_{K_{n}} \quad \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \quad L.\]

Recall that $ \mathfrak{P}_{n}$ denotes the set of all non-crossing partition of $[n]$ and that $\mathcal{K}(P)$ is the Kreweras complement of a non-crossing partition $P$.
Using the visual description of Kreweras complement (Figure <ref>),
it is a simple matter to see that $ \max_{P \in \mathfrak{P}_{n}} d_{H} \big(\hat{P}, \widehat{\mathcal{K}(P)} \big) \rightarrow 0$ as $n \rightarrow \infty$.  As a consequence,

\[ \widehat{\mathcal{K} \big(\PPP^{(n)}_{n-K_{n}-1}\big)}    \quad \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \quad  L.\]

But by Lemma <ref>, $\PPP^{(n)}_{K_{n}}$ and $\mathcal{K}( \PPP^{(n)}_{n-K_{n}-1})$ have the same distribution. This entails the desired result.

§.§ Convergence to $\mathbf{L}_{c}$ for $c >0$ and proof of Theorem <ref> (i)$_{c>0}$.

The goal of this subsection is to prove Theorem <ref> (i)$_{c>0}$,
which deals with the case $\tfrac{K_{n}}{\sqrt{n}} \rightarrow c>0$.
Recall that if $\tau$ is a tree, $(\oB^{n}_{i}(\tau))_{1 \leq i \leq |\tau|}$ is the Łukasiewicz path of the reduced black subtree of $\tau$, as introduced in Section <ref>. We also set $\oB^{n}_{i}(\tau)=0$ for $i \leq 0$ or $i>|\tau|$.

For every $n \geq 1$, we let $(t_{1}^{(n)}, \ldots, t_{n-1}^{(n)}) \in \mathfrak{M}_{n}$  be a minimal factorization. If $1 \leq K_{n} \leq n-1$, to simplify notation we set

\[  \FFF_{n}= \FFF \big( t_{1}^{(n)}, \ldots, t_{K_{n}}^{(n)}\big) \qquad \text{and} \qquad \PPP_{n}= \PPP\big(  t_{1}^{(n)}  t_{2}^{(n)}  \cdots t_{K_{n}}^{(n)}\big),\]

and let $\dFn$ and respectively  $\dPn$ be their associated compact subsets of $\overline{\D}$.
Finally, we denote by $T_{n}= \mathcal{T}( \PPP_{n})$   the tree coding the non-crossing partition  $\PPP_{n}$ as in Section <ref>.

Roughly speaking,  Theorem <ref> (i)$_{c>0}$ follows from the following deterministic convergence result
(the proof of this implication is given at the end of Section <ref>).

Let $Z \in {\D}([0, 1], \R)$ be a càdlàg function satisfying <ref>, <ref>, <ref> and <ref>
and let $\FFF_{n}$, $\PPP_{n}$, $\dFn$, $\dPn$ and $T_{n}$ be as above.
Assume that the following four properties hold.

*   We have $\tfrac{K_n}{\sqrt{n}} \rightarrow c>0$ as $n \rightarrow \infty$.

*  The convergence

\begin{equation}
\left(  c \frac{\oB^{n}_{\lfloor u (n-K_{n}) \rfloor}(T_n)}{ n} \right)_{-1 \leq u \leq 1}
\quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad (Z_{u})_{- 1\leq u \leq 1}
\label{eq:cv_HB_Skorokhod}
\end{equation}

holds for the Skorokhod $J_{1}$ topology, where we set $Z_{u}=0$ for $u<0$.

*  For $n$ sufficiently large, $T_n$ has no black vertex  with at least three children.

*  For every $\eps>0$, there exists $n_0(\eps)>0$ such that  for every $n \ge n_0(\eps)$, the tree $T_n$ does not contain a black
vertex $v$ that has two children, both with at least $\eps \, n$ descendants (counting both white and black vertices).

Then we have the  convergence

\begin{equation}
\left(  \dFn, \dPn   \right)   \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad ( L(Z), L(Z))
\label{eq:cv_FP_Skorokhod}
\end{equation}

in the Hausdorff topology.

Before tackling the proof of Proposition <ref>, we state and prove two lemmas. The first one is a simple geometric lemma and the second one concerns approximation properties of the Skorokhod $J_{1}$ metric.

Fix $\epsilon>0$ and assume that <ref>, <ref> hold. Then, for every $n$ sufficiently large,

\[d_{H}\left( \dFn  , \dPn  \right)  \leq 7\epsilon,\]

where $d_{H}$ denotes the Hausdorff distance.

Assume that $n$ is chosen sufficiently large so that <ref> and <ref> for our fixed value of $\epsilon$ are in force.
By <ref>, $ \dPn$ is a disjoint union of line segments and of triangles. Also, by Lemma <ref>, $ \dFn \subset \dPn$ and  $\dFn$ is obtained from $\dPn$ by removing an edge from every triangle. Now, if $ \Delta$ is a triangle of $\dPn$, it splits the unit circle into three arcs, and by <ref>, one of these arcs has length at most $7 \epsilon$. If $ {\Delta}'$ is obtained from $\Delta$ by removing one of the three edges of $\Delta$, this implies that $d_{H}(\Delta,{\Delta}') \leq 7 \epsilon$. The desired result readily follows.

Suppose that the assumptions of Proposition <ref> are in force. Fix $0<s<t < 1$.
Then the following assertions are equivalent:

(i) We have $s \simeq^Z t$;

(ii) There exists integers $i_{n}< j_{n}$ such that $ \tfrac{i_{n}}{n-K_{n}} \rightarrow s$, $ \tfrac{j_{n}}{n-K_{n}} \rightarrow t$ and

\[ j_{n}= \inf \left\{k>i_{n}: \,  \oB^{n}_k(T_n)=\oB^{n}_{i_{n}}(T_n)-1 \right\}.\]

The implication $(i) \implies (ii)$ follows from standard approximation properties of the Skorokhod $J_{1}$ topology (see e.g. the proof of Lemma 4.6. in [49]).

Now assume (ii). To simplify notation, set $ \hB^{(n)}(u)=c \tfrac{\oB^{n}_{\lfloor u (n-K_{n}) \rfloor}(T_n)}{ n}$, so that

\begin{equation}
\label{eq:ineg} \hB^{(n)}(r) \geq \hB^{(n)} \left(  \frac{i_{n}}{n-K_{n}} \right) =\hB^{(n)} \left(  \frac{j_{n}}{n-K_{n}} \right)+ \frac{c}{n}  \quad \textrm{for every } r \in \left[  \frac{i_{n}}{n-K_{n}}, \frac{j_{n}}{n-K_{n}}  \right).
\end{equation}

First assume that  $Z_{s}=Z_{s-}$.
Then $\hB^{(n)} \big(  \tfrac{i_{n}}{n-K_{n}} \big) \rightarrow Z_{s}$
and it follows from (<ref>) and from <ref> that  $Z_r \ge Z_{t-}=Z_{s}$ for every $r \in (s,t)$.
In addition, we claim that $Z_{t-}=Z_{t}$.
Indeed, if $Z_{t-}<Z_{t}$ and if $Z_{u}\geq Z_{s}$ for $u$ belonging to a small neighborhood of the left of $s$,
then <ref> is not fulfilled.
On the other hand, if $Z_{t-}<Z_{t}$ and if $s=\sup\{ u \in [0,t] : Z_{u} < Z_{t-}\}$, then <ref>
is not fulfilled.
Therefore, we must have $Z_{t-}=Z_{t}$.
Similarly, we necessarily have $Z_{r}>Z_{t}$ for every $r \in (s,t)$.
Indeed, assuming $Z_r=Z_t$ yields a contradiction with either <ref> or <ref>,
using the same case distinction as above.
We conclude that $s \simeq^Z t$.

Next assume that $Z_{s}>Z_{s-}$. Then, by <ref>, we can find $r \in (s,t)$ such that $Z_{r}<Z_{s}$.
As a consequence,  $\hB^{(n)} \big(  \tfrac{i_{n}}{n-K_{n}} \big) \rightarrow Z_{s-}$
(it cannot converge to $Z_s$ because of (<ref>)), and $Z_{t-}=Z_{s-}$.
Similarly as above, we can prove that $Z_{t}=Z_{t-}$.
We thus have $Z_{s-}=Z_{t} \leq Z_{r}$ for every $r \in (s,t)$,
and this inequality is strict (otherwise, as above, this would contradict either <ref> or <ref>).
Therefore $s \simeq^Z t$, and this completes the proof.

We are now ready to establish Proposition <ref>.

Since the set of all geodesic laminations of $\overline{\D}$ equipped with Hausdorff topology is compact, up to extraction,
we may assume that $(\dFn, \dPn)$  converges to a pair of geodesic laminations $(K,K')$ of $\overline{\D}$ and we aim at proving that $(K,K')=( L(Z), L(Z))$. By Lemma <ref>, $K=K'$ and it is therefore enough to show that $K=L(Z)$. Denote by $(v^{\bullet,n}_{i})_{0 \leq i < n-K_{n}}$ the black vertices of $T_{n}$ listed in lexicographical order.

First note that there can not exist $s \in (0,1)$ such that $0 \simeq^{Z} s$,
and since $Z_{t}>0$ for $t \in (0,1)$, there does not exist $s \in (0,1)$ such that $ s \simeq^{Z} 1$ either.

*Step $1$: $L(Z) \subset K$.  Fix $0 < s \le t < 1$ such that  $s \simeq^Z t$.
We need to prove that $[e^{-2\pi\, \textrm{i}\, s}, e^{-2\pi\, \textrm{i}\, t}] \subset K$.
By (<ref>), it is enough to prove it for $s<t$.
Since $K$ is a limit point of the sequence $(\dPn)_{n \ge 0}$
it is enough to check that for fixed $\epsilon>0$, for every $n$ sufficiently large,
there exists a chord of $\dPn$  within distance $12\epsilon$ of the chord $[e^{-2\pi\, \textrm{i}\, s}, e^{-2\pi\, \textrm{i}\, t}]$.

By Lemma <ref>, for $n$ sufficiently large, there exist integers $i_{n},j_{n}$ such that

\begin{equation}
\label{eq:injn}
\left|\frac{i_{n}}{n-K_n} -s \right|  \le \eps, \quad \left|\frac{j_{n}}{n-K_n} -t \right|  \le \eps, \quad
j_{n}= \inf \left\{k>i_{n}: \,  \oB^{n}_k(T_n)=\oB^{n}_{i_{n}}(T_n)-1 \right\}.
\end{equation}

Note that the last equality means that $v^{\bullet,n}_{j_{n}}$ is the first vertex greater (in the lexicographical order) than $v^{\bullet,n}_{i_{n}}$ and than all the descendants of $v^{\bullet,n}_{i_{n}}$.

We now find a chord of $ \dPn $ which is within distance $12\epsilon$ of the chord $[e^{-2\pi\, \textrm{i}\, s}, e^{-2\pi\, \textrm{i}\, t}]$ as follows. Denote by  $x^{\bullet,n}_{i_{n}}$ (resp. $y ^{\bullet,n}_{i_{n}}$)  the label of the first (resp. the last) corner of $v^{\bullet,n}_{i_{n}}$ visited by the contour sequence.
By construction (see in Section <ref> how the partition $\PPP_{n} $ is reconstructed from $T_{n}$),
we have $[e^{-2\pi\, \textrm{i} {x^{\bullet,n}_{i_{n}}}/{n}}, e^{-2\pi\, \textrm{i} {y ^{\bullet,n}_{i_{n}}}/{n}}] \in \dPn$.
Lemma <ref>  (iii) then implies that

\begin{equation}
i_{n} \leq x^{\bullet,n}_{i_{n}} \leq i_{n}+ K_{n}+1, \qquad  i_{n}+n^{\bullet}_{i_{n}} \leq y^{\bullet,n}_{i_{n}} \leq   i_{n}+n^{\bullet}_{i_{n}} + 2 K_{n}+2,
\label{eq:Control_xy}
\end{equation}

where $n^{\bullet}_{i_{n}} $ is the number of black descendants of $v^{\bullet,n}_{i_{n}}$.
But, because of (<ref>),  the black vertices visited between $v^{\bullet,n}_{i_{n}}$ and $v^{\bullet,n}_{j_{n}}$
in the contour walk of $T_{n}$ are exactly the black descendants of $v^{\bullet,n}_{i_{n}}$,
so that $n^{\bullet}_{i_{n}}= j_{n}-i_{n}-1$. Thus

\[i_{n} \leq x^{\bullet,n}_{i_{n}} \leq i_{n}+ K_{n}+1, \qquad  j_{n}-1 \leq y^{\bullet,n}_{i_{n}} \leq    j_{n}-1 + 2 K_{n}+2.\]

Since $K_{n} \sim c \sqrt{n}$, this implies that for every $n$ sufficiently large,

\[\left| \frac{x^{\bullet,n}_{i_{n}}}{n}-s \right| \leq 2 \epsilon, \qquad   \left| \frac{y^{\bullet,n}_{i_{n}}}{n}-t \right|  \leq 2 \epsilon.\]

As a consequence, for $n$ sufficiently large, $[e^{-2\pi\, \textrm{i} {x^{\bullet,n}_{i_{n}}}/{n}}, e^{-2\pi\, \textrm{i} {y ^{\bullet,n}_{i_{n}}}/{n}}] \in\dPn$  is within distance $12\epsilon$ of the chord $[e^{-2\pi\, \textrm{i}\, s}, e^{-2\pi\, \textrm{i}\, t}]$. This shows that $ L(Z)   \subset K$.

*Step $2$: $ K \subset L(Z)$.
Since $K$ is a geodesic lamination (i.e. a union of non crossing chords) and since $L(Z)$ contains the circle $\S$,
it is enough to show that nontrivial chords of $K$ are included in $L(Z)$.
Namely, we need to prove that if $s<t$ are such that $[e^{-2\pi\, \textrm{i}\, s}, e^{-2\pi\, \textrm{i}\, t}] \subset K$,
then $s \simeq^Z t$.
By definition, $K$ is a limit point of $(\dPn)_{n \ge 0}$
so, up to extraction, we may assume that there exist sequences $(s_{n})$ and $(t_{n})$
such that $s_{n} \rightarrow s$, $t_{n} \rightarrow t$, $n s_{n}$ and $n t_{n}$ are integers, and $[e^{-2\pi\, \textrm{i}\, s_{n}}, e^{-2\pi\, \textrm{i}\, t_{n}}] \subset \dPn$.
By construction,
the condition $[e^{-2\pi\, \textrm{i}\, s_{n}}, e^{-2\pi\, \textrm{i}\, t_{n}}] \subset \dPn$
means that
$ns_n$ and $nt_n$ are the labels of two consecutive corners of the same black vertex of $T_n$;
call $v^{\bullet,n}_{i_{n}}$ this black vertex.
Combining this with the hypothesis <ref>, there are two cases:
$v^{\bullet,n}_{i_{n}}$ may have 1 or 2 children, but not more.

Let $n s'_{n}$ (resp. $n t'_{n}$) is the index of the first (resp. last) black corner of $v^{\bullet,n}_{i_{n}}$ visited by the contour sequence (if $v^{\bullet,n}_{i_{n}}$ has one child, we simply have $(s_{n},t_{n})=(s'_{n},t'_{n})$).
By <ref>, for $n \ge n_0(\eps)$, only one of the quantities $t'_n-t_n$, $t_n-s_n$ and $s_n-s'_n$
can be bigger than $\eps$.
Since $t>s$, for $\eps$ small enough and $n$ sufficiently large, we have $t_n-s_n>\eps$.
Therefore $t'_n-t_n$ and $s_n-s'_n$ tend to $0$ and
we have $(s'_{n},t'_{n}) \rightarrow (s,t)$.
Note that we crucially need <ref> to rule out the possibility that  $v^{\bullet,n}_{i_{n}}$ has two white children both having a macroscopic descendance, in which case the last convergence may fail.
Then, by Lemma <ref>   (iii),

\[ i_{n} \leq n s'_{n} \leq i_{n}+  K_{n}+1, \qquad i_{n}+n^{\bullet}_{i_{n}} \leq n t'_{n} \leq   i_{n}+n^{\bullet}_{i_{n}} +2 K_{n}+2,\]

where $n^{\bullet}_{i_{n}}$ is the number of black descendants of $ v^{\bullet,n}_{i_{n}}$.
Hence, since $K_n/n$ tends to $0$, we have $i_{n}/(n-K_{n}) \rightarrow s$ and $  (i_{n}+n^{\bullet}_{i_{n}})/(n-K_{n}) \rightarrow t$.
Besides, by construction of $\overline{B}$, it holds that

\[i_{n}+n^{\bullet}_{i_{n}}+1= \min \{k>i_{n}: \oB^{n}_{k}(T_{n}) =  \oB^{n}_{i_{n}}(T_{n})-1\}.\]

Lemma <ref> implies that $s \simeq^Z t$, and this completes the proof.

We now conclude the proof of Theorem <ref> (i) in the case $c>0$.

Since the space of laminations of the disk is compact,
it is enough to show that the sequence $\big( \dFKn  ,  \dPKn \big) $ has a unique  limit point in distribution. Up to extraction, we may therefore assume that $\big( \dFKn  ,  \dPKn \big)$  converges in distribution. We shall show that the limiting distribution is the law of $(\mathbf{L}_c,\mathbf{L}_c)$.

For each $n \ge 1$, we denote by $\tTs_n= \mathcal{T}( \PPP( \tb_{1}^{(n)} \tb_{2}^{(n)} \cdots \tb_{K_{n}}^{(n)}))$
the tree associated with $\PKn$.
From Proposition <ref>, it has the distribution of an alternating BGW tree with our favorite offspring distributions
and a modified offspring distribution at the root (as in Theorem <ref> (ii)).
We consider the following events:

* for $\epsilon >0$, we let $A_{n}(\epsilon)$ be the event “the tree $\tTs_{n}$ does not contain a black
vertex $v$ that has two children, both with at least $\eps \, n$ descendants”;

* $B_{n}$ is the event “$\tTs_{n}$ does not contain any black vertex with three or more children”.

From Lemmas <ref> and <ref> combined with Lemma <ref>, we know that
as $n \rightarrow \infty$, $\Pr{B_{n}} \rightarrow 0$ and, for each $\eps >0$, $\Pr{A_{n}(\epsilon)} \rightarrow 0$ as well.
Therefore, we can find a subsequence $(\phi(n))_{n \geq 1}$ such that, for every $n \geq 1$,

\[ \Pr{{B_{\phi (n)}}} \leq \tfrac{1}{2^{n}}, \qquad \Pr{{A_{ \phi(n)} (\tfrac1n)}} \leq  \tfrac{1}{2^{n}}.\]

The Borel-Cantelli lemma ensures the almost sure convergence of
$\One_{B_{\phi (n)}}$ and $\One_{A_{ \phi(n)} (1/n)}$ to $0$.
On the other hand, from Theorem <ref> (ii), we know that the following convergence in distribution holds

\[  \left(  c \frac{\oB_{\lfloor u (n-K_{n}) \rfloor}(\tilde{\Ts}_{n})}{ n} : -1 \leq u \leq 1\right)  \quad \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \quad (\Xexc_{u}: -1 \leq u \leq 1).\]

The three previous convergences hold jointly in distribution, so that by Skorokhod representation theorem,
we can find a coupling of  $(\tTs_{\phi(n)})_{n \ge 1}$ such that these convergences hold almost surely.
The sequence of associated random factorizations almost surely satisfies the assumptions
<ref>, <ref> and <ref> of Proposition <ref>.
Besides, from Lemma <ref>, the random process $\Xexc$ almost surely
satisfies conditions <ref>, <ref>, <ref> and <ref>.
From Proposition <ref>,
we conclude that $\big( \dFKphin  ,  \dPKphin \big)$ converges in distribution
to $(\mathbf{L}_c,\mathbf{L}_c)$, since $\mathbf{L}_c=L(\Xexc)$ by definition.
Therefore the limiting distribution of the whole sequence $\big( \dFKn, \dPKn\big)$
must be $(\mathbf{L}_c,\mathbf{L}_c)$ and this completes the proof of Theorem <ref> (i) in the case $c>0$.

§.§ Convergence to $\mathbf{L}_{\infty}$ and proof of Theorem <ref> (ii)

Our goal is now to establish Theorem <ref> (ii), when  both $\frac{K_{n}}{\sqrt{n}} \rightarrow \infty$ and  $ \frac{n-K_{n}}{\sqrt{n}} \rightarrow \infty$ as $n \rightarrow \infty$.
The main difference  with the case $ \frac{K_{n}}{\sqrt{n}}  \rightarrow c>0$ occurs when we assume $ \tfrac{K_{n}}{n} \rightarrow \gamma \in (0,1)$.
In the latter case,  both the  number of white and black vertices then have a positive proportion in the dual tree coding $ \PPP(\tb_{1}^{(n)}\tb_{2}^{(n)} \cdots \tb_{K_{n}}^{(n)}) $.

As before, $\mubn$ and $\mucn$  will denote the probability distributions satisfying the conditions of Lemma <ref>.
We also set $\widetilde{\mu}^{n}_{\bullet}(i)=\mubn(i-1)$ for $i \geq 1$.
We let $\tilde{\Ts}^{n}$ be a random tree with distribution ${\BGW}^{\widetilde{\mu}_{\bullet}^{n},\mubn,\mucn}$, as defined in (<ref>).
In words, $\tilde{\Ts}^{n}$ is
an alternating two-type BGW tree (with a black root), with offspring distributions $\mubn$ and $\mucn$ except the root which has offspring distribution $\widetilde{\mu}^{n}_{\bullet}$.
We then define $\tTs_n$, as $\tTs^{n}$ conditioned on having $n-K_{n}$ black vertices and $K_{n}+1$ white vertices.
Recall from the beginning of Section <ref> that we associate
two paths $(\oB^{n}_{i}(T))_{0 \leq i \leq n-K_{n}}$ and  $(\oH^{n}_{i}(T))_{0 \leq i \leq n-K_{n}}$ with each alternating tree $T$
with $n-K_n$ black vertices.

We first give the analogue of  Theorem <ref> in the setting $c=\infty$.
In the following statement, as in Section <ref>,
$\mathbbm{e}$ is the Brownian excursion.
We also use the notation of Lemma <ref> and Section <ref>.
In particular,
we denote by  $(v^{\bullet,n}_{i})_{0 \leq i < n-K_{n}}$ are the black vertices $\tTs_n$, listed in lexicographical order,
and, for $0 \leq i < n-K_{n}$,
we let $\ell^{\bullet,n}_{i}$ be  the number of black corners
branching on the right of  $\llbracket \emptyset, v^{\bullet,n}_{i} \llbracket$ for $0 \leq i < n-K_{n}$.

Assume that $\tfrac{K_{n}}{\sqrt{n}} \rightarrow \infty$ and $ \tfrac{K_{n}}{n}  \rightarrow \gamma \in [0,1)$ as $n \rightarrow \infty$. Set $\Dcn=\sqrt{ (\scn)^{2} K_{n} }$. Then

\[  \left(  \frac{\oB^{n}_{\lfloor u (n-K_{n}) \rfloor}(\tTs_{n})}{ \Dcn} : 0\leq u \leq 1\right)  \quad \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \quad\mathbbm{e}.\]

In addition, if $\gamma>0$, then

\[  \sup_{0 \leq u \leq 1} \left|   \frac{\oH^{n}_{ \lfloor u (n-K_{n}) \rfloor }(\tTs_{n})}{K_{n}}-u\right|  \quad \mathop{\longrightarrow}^{(\P)}_{n \rightarrow \infty} \quad 0 \qquad \textrm{and} \qquad \frac{1}{n} \max_{1 \leq i \leq n} \ell^{\bullet,n}_{i}  \quad \mathop{\longrightarrow}^{(\P)}_{n \rightarrow \infty} \quad 0.\]

The condition $ \tfrac{K_{n}}{n}  \rightarrow \gamma \in (0,1)$ means the both the number of white and black vertices have a positive proportion in $\tilde{\Ts}_{n}$. The proof of Proposition <ref>, very similar in spirit to that of Theorem <ref>, is postponed  to the end of Section <ref>.

In the next Section <ref>, we develop deterministic tools that allow us to show that Proposition <ref> implies Theorem <ref> (ii).

§.§.§ Convergence to the Brownian triangulation

As in Section <ref>, for every $n \geq 1$, we let $(t_{1}^{(n)}, \ldots, t_{n-1}^{(n)}) \in \mathfrak{M}_{n}$  be a minimal factorization. If $1 \leq K_{n} \leq n-1$, we set

\[  \FFF_{n}= \FFF \big( t_{1}^{(n)}, \ldots, t_{K_{n}}^{(n)}\big) \qquad \text{and} \qquad \PPP_{n}= \PPP\big(  t_{1}^{(n)}  t_{2}^{(n)}  \cdots t_{K_{n}}^{(n)}\big)\]

and we denote by $T_{n}= \mathcal{T}( \PPP_{n})$   the tree coding the non-crossing partition  $\PPP_{n}$ as in Section <ref>.

Our goal is to prove some convergence result for the pair $\big(\dFn , \dPn  \big)$,
but it turns out to be enough to consider
$ \dPn$, as ensured by the following lemma.

Let $Z \in {\mathbb{C}}([0, 1], \R)$ be a continuous function with $Z_{0}=Z_{1}=0$ satisfying <ref>.
We assume that $\dPn \rightarrow   L(Z)$ as $n$ tends to infinity.
Then we also have  $\dFn \rightarrow   L(Z)$.

Since the set of all geodesic laminations of $\overline{\D}$ equipped with Hausdorff topology is compact, we may assume that the convergence $\big( \dFn, \dPn  \big)   \rightarrow   (K,L(Z))$ holds, where $K$ is a lamination, and we aim at showing that $K=L(Z)$. By maximality of $L(Z)$, it is enough to show that $L(Z) \subset K$.
Consider a chord  $\big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big]$ of $L(Z)$ (with $s \neq t$),
we want to show that $\big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big] \subset K$.

By (<ref>), for every $\epsilon>0$, we can find another chord  $\big[e^{-2\textrm{i} \pi s'},e^{-2\textrm{i} \pi t'}\big] \subset L(Z)$ such that  $0<|s-s'| \leq \epsilon$ and $0<|t-t'| \leq \epsilon$.
By symmetry, we can assume that  $s<s'<t'<t$.
The following argument having some geometric flavor,
looking at Figure <ref> while reading the proof might help.

These two chords form a region of the disk denoted by $R$.
Again by (<ref>),  there exists a chord $\big[e^{-2\textrm{i} \pi u},e^{-2\textrm{i} \pi v}\big] \subset L(Z)$ such that $s<u<s'$ and $t'<v<t$.
This implies that for each $\eta>0$ small enough and $n$ larger to some threshold $n_0(\eta)$,
we can find a block $\dot{B}_{n}$ of $\dPn$ such that:

* $\dot{B}_{n}$ has a chord with endpoints at distance at most $\eta$ of
$e^{-2\textrm{i} \pi u}$ and $e^{-2\textrm{i} \pi v}$;

* moreover, all vertices of $\dot{B}_{n}$ are in the $\eta$ enlargement
of one of the arcs $(e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi s'})$ or
$(e^{-2\textrm{i} \pi t},e^{-2\textrm{i} \pi t'})$ (we denote arcs with parentheses to
distinguish them from chords).

We recall that $\dFn$ is obtained from $\dPn$ by replacing each polygon
by a spanning tree of its vertices. It is clear, that for the block $B_n$
with the above property, such a spanning tree must contain a chord that
has one extremity on each side, i.e. one in the $\eta$ enlargement
of $(e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi s'})$
and one in the $\eta$ enlargement
of $(e^{-2\textrm{i} \pi t},e^{-2\textrm{i} \pi t'})$.
Therefore $\dFn$ contains a chord at distance $\eps+\eta$
of our original chord $\big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big]$.
Since this is true for all $\eps$ and $\eta$ sufficiently small,
this implies that  $\big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big] \subset K$ and shows that $K=L(Z)$.

An illustration of the second part of the proof. The region $R$ is in gray,
the block $\dot{B}_{n}$ is in dashed lines.
This block is entirely contained in a small enlargement of $R$
and has at least one chord joining a small enlargement of the $(e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi s'})$ arc
with that of the $(e^{-2\textrm{i} \pi t},e^{-2\textrm{i} \pi t'})$ arc.
No matter which spanning tree of the vertices of $\dot{B}_{n}$ is present in $\dFn$
it will contain a chord close to $\big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big]$.

Let us point out that, although the above lemma is similar in spirit to Lemma <ref>
(both show that $\dPn$ and $\dFn$ are in some sense close),
the arguments are quite different.
One the one hand, Lemma <ref>, which is the base of Lemma <ref>,
has certainly no analogue when there is a constant proportion of vertices of both colors in the tree.
On the other hand, here, we need to assume the convergence of $\dPn$ to some lamination
encoded by a continuous function satisfying <ref> to prove that it is close to $\dFn$.

The gist of the proof of Theorem <ref> (ii) is contained in the next result.
The notation is the same as in Proposition <ref>.

Let $Z \in {\mathbb{C}}([0, 1], \R)$ be a continuous function such that $Z_{0}=Z_{1}=0$ satisfying <ref>. For every $n \geq 1$, let $(t_{1}^{(n)}, \ldots, t_{n-1}^{(n)}) \in \mathfrak{M}_{n}$  be a minimal factorization, and denote by $T_{n}= \mathcal{T}(  \PPP_{n})$ the tree coding the non-crossing partition $  \PPP_{n}=\PPP(  t_{1}^{(n)}, \ldots, t_{K_{n}}^{(n)})$. Assume that the following two properties hold.

*  We have $\frac{K_{n}}{\sqrt{n}} \rightarrow \infty$ and $ \tfrac{K_{n}}{n} \rightarrow \gamma \in [0,1)$ as $n \rightarrow \infty$.

*  There exists a sequence $\Dcn \rightarrow \infty$ such that

\[  \left(  \frac{\oB^{n}_{\lfloor u (n-K_{n}) \rfloor} (T_{n})}{ \Dcn} : 0\leq u \leq 1\right)  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad Z.\]

*  If $\gamma>0$, we require that

\[ \sup_{0 \leq u \leq 1} \left|   \frac{\oH^{n}_{ \lfloor u (n-K_{n}) \rfloor  }(T_{n})}{K_{n}}-u\right|  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad 0 \qquad \textrm{and} \qquad \frac{1}{n} \max_{0 \leq i < n-K_{n}} \ell^{\bullet,n}_{i}    \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad 0.\]

Then we have the  convergence

\begin{equation}
\dot{\PPP}_{n}    \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad L(Z)
\label{eq:cv_FP_SkorokhodBrownien}
\end{equation}

in the Hausdorff topology.

Before proving Proposition <ref>, let us explain how Theorem <ref> (ii)  follows from a combination of Propositions <ref> and <ref> and Lemmas <ref> and <ref>.

First assume that $\frac{K_{n}}{\sqrt{n}} \rightarrow \infty$ as $n \rightarrow \infty$, and that $K_{n} \leq \frac{n}{2}$ for every $n \geq 1$. Up to extraction, we can suppose that $\tfrac{K_{n}}{n} \rightarrow \gamma \in [0,\tfrac{1}{2}]$. By Skorokhod's representation theorem, we may also suppose that the convergence in  Proposition <ref> holds almost surely. Then, by Proposition <ref>, the convergence

\[ \dPKn    \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad \mathbf{L}_{\infty},\]

holds almost surely. In turn, by Lemma <ref>, this implies that the joint convergence

\[ \left( \dFKn , \dPKn \right)   \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad ( \mathbf{L}_{\infty},\mathbf{L}_{\infty})\]

holds almost surely, and the desired conclusion follows.

Now assume that $\frac{n-K_{n}}{\sqrt{n}} \rightarrow \infty$ and $K_{n} > \frac{n}{2}$ as $n \rightarrow \infty$.
The previous paragraph tells us that

\[   \dot{\mathcal{P}}^{(n)}_{n-K_{n}-1}  \quad \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \quad \mathbf{L}_{\infty}.\]

Lemma <ref> entails that the convergence $\hat{\mathcal{P}}^{(n)}_{K_{n}} \rightarrow  \mathbf{L}_{\infty}$ holds in distribution. As a consequence,  since $\hat{\mathcal{P}}^{(n)}_{K_{n}}$ is obtained from  $\dot{\mathcal{P}}^{(n)}_{K_{n}}$ by adding points of $\S$, any sub-sequential distributional limit of $\dot{\mathcal{P}}^{(n)}_{K_{n}}$ must contain all the non-trivial chords of $\mathbf{L}_{\infty}$, and by (<ref>) must be equal to $\mathbf{L}_{\infty}$. In other words, the convergence $\dot{\mathcal{P}}^{(n)}_{K_{n}} \rightarrow  \mathbf{L}_{\infty}$ holds in distribution, and  another application of Lemma <ref> implies that  the joint convergence

\[ \left( \dFKn,   \dPKn \right)   \quad \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \quad ( \mathbf{L}_{\infty},\mathbf{L}_{\infty})\]

holds in distribution. This completes the proof.

Since the set of all geodesic laminations of $\overline{\D}$ equipped with Hausdorff topology is compact, up to extraction, we may therefore assume that  $ \dot{\PPP}_{n}$  converges to a geodesic lamination $K$ of $\overline{\D}$ and we aim at proving that $K=L(Z)$.
By maximality, it is enough to check that $ L(Z) \subseteq K$.
By (<ref>),  it suffices to show that if $s \sim^{Z} t$  and if $s$ nor $t$ are local minima of $Z$, with $0<s < t<$, then $\big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big] \subset K$.
Using the convergence <ref>, similarly as in Eq. (<ref>), we can find black vertices
$v^{\bullet,n}_{i_{n}}$ and $v^{\bullet,n}_{j_{n}}$ in $T_{n}$
such that:

* $i_{n}/(n-K_{n}) \rightarrow s$  and $j_{n}/(n-K_{n}) \rightarrow t$;

* all the black vertices between $v^{\bullet,n}_{i_{n}}$ and $ v^{\bullet,n}_{j_{n}}$
in the lexicographical order are descendants of  $v^{\bullet,n}_{i_{n}}$.

Denote by  $x^{\bullet,n}_{i_{n}}$ (resp. $y ^{\bullet,n}_{i_{n}}$)
the label of the first (resp. the last) corner of $v^{\bullet,n}_{i_{n}}$ visited by the contour sequence. We claim that

\begin{equation}
\label{eq:cvij} \frac{x^{\bullet,n}_{i_{n}}}{n}  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad s, \qquad \frac{y ^{\bullet,n}_{i_{n}}}{n}  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad t.
\end{equation}

Since $[e^{-2\pi\, \textrm{i} {x^{\bullet,n}_{i_{n}}}/{n}}, e^{-2\pi\, \textrm{i} {y ^{\bullet,n}_{i_{n}}}/{n}}] \subset \dot{\PPP}_{n}$, this will then imply $\big[e^{-2\textrm{i} \pi s},e^{-2\textrm{i} \pi t}\big] \subset K$ and the proof will be complete.

*First case: $\gamma=0$. By  Lemma <ref>  (iii),
we have $i_{n} \leq x^{\bullet,n}_{i_{n}} \leq i_{n}+ K_{n}+1$ and

\[x^{\bullet,n}_{i_{n}}+ (j_n-i_n-1) \leq y^{\bullet,n}_{i_{n}} \leq x^{\bullet,n}_{i_{n}}+(i_n-j_n-1) +2 K_{n}+2,\]

since the number of black descendants of $v^{\bullet,n}_{i_{n}}$ is $j_{n}-i_{n}-1$.
But we have ${K_{n}}/{n} \rightarrow 0$, $i_{n}/(n-K_{n}) \rightarrow s$  and $j_{n}/(n-K_{n}) \rightarrow t$,
which implies (<ref>).

*Second case: $\gamma>0$. In this case, the number of white vertices is no longer negligible, and we use the precise identity of   Lemma <ref> (ii). Recalling that $k_j$ is the number of children of $v^{\bullet,n}_j$
and that $\ell_{i_{n}}^{\bullet,n}$ denotes the number of black corners
branching on the right of  $\llbracket \emptyset, v^{\bullet,n}_{i_{n}} \llbracket$,
we have

\begin{equation}
\label{eq:est1}x^{\bullet,n}_{i_{n}}= i_{n}+\sum_{j=0}^{i_{n}-1} k_{j} - \ell_{i_{n}}^{\bullet,n}=i_{n}+\oH^{n}_{i_{n}}(T_{n})- \ell_{i_{n}}^{\bullet,n}.
\end{equation}

Since $i_{n}/(n-K_{n}) \rightarrow s$ and $K_{n} \sim \gamma n$,  using <ref>, we get

\begin{equation}
\label{eq:est2}i_{n}+ \oH^{n}_{i_{n}}(T_{n})+ \ell_{i_{n}}^{\bullet,n}=  s(1-\gamma) n + s \gamma n + o(n)=sn+o(n).
\end{equation}

This proves that $\tfrac{x^{\bullet,n}_{i_{n}}}{n} \rightarrow s$.

In order to show that $\tfrac{y ^{\bullet,n}_{i_{n}}}{n} \rightarrow t$, note that the number of black descendants of $v^{\bullet,n}_{i_{n}}$ is again $j_{n}-i_{n}-1$ while its number of white descendants is $\oH^{n}_{j_{n}}(T_{n})-\oH^{n}_{i_{n}}(T_{n})$. Hence, by Lemma <ref> (i) and using <ref>, we get

\[y ^{\bullet,n}_{i_{n}}=x ^{n}_{i_{n}}+\big(j_{n}-i_{n}-1\big)+\big(\oH^{n}_{j_{n}}(T_{n})-\oH^{n}_{i_{n}}(T_{n})\big)=j_{n}+\oH^{n}_{j_{n}}(T_{n})- \ell_{i_{n}}^{\bullet,n}-1= tn+o(n),\]

where the last equality is obtained exactly as before. This completes the proof.

§.§.§ Proof of the technical results

We now turn to the proof of Proposition <ref>.
The strategy is similar to the case $c>0$: we first establish convergence of unconditioned processes using local limit theorems, then lift them to the bridge version by absolute continuity and finally conclude by using a Vervaat transform.

As before, we let
$\mucn$ and $\mubn$ be the offspring distributions defined in Lemma <ref> (i), and we let $\Scn_{k}$, $\Sbn_{k}$ denote respectively the sum of $k$ i.i.d. random variables distributed according to $\mucn,\mubn$.

*Local limit theorems. We first state the analogue of Lemmas <ref> and <ref> in our new regime.
First set

\[\DbnN \coloneqq \sqrt{ (\sbn)^{2}  N}, \qquad \Dcn \coloneqq\sqrt{ (\scn)^{2} K_{n} },\]

and $\Dbn \coloneqq D^{\bullet}_{n,n}= \sqrt{ (\sbn)^{2}  n}$.

Assume that $\tfrac{K_{n}}{\sqrt{n}} \rightarrow  \infty$ and $\tfrac{K_{n}}{n}   \rightarrow  \gamma \in [0,1)$.
For every $u \in (0,1]$,

\begin{equation}
\label{eq:ext1}\sup_{un \leq N \leq n} \sup_{k \in \Z  } \left|  \DbnN  \cdot \Pr{\Sbn_N=k}
-    \frac{1}{\sqrt{2 \pi}}\exp \left(-   \frac{1}{2}\left( \frac{k-  N \cdot  \frac{K_{n}+1}{n-K_{n}}}
{  \DbnN  } \right)^{2}\right)\right|  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad 0
\end{equation}

\begin{equation}
\label{eq:ext2} \hspace{-3.5mm} \sup_{|j| \leq K_{n}^{3/4}} \sup_{k \in \Z} \left|  \Dcn \cdot \Pr{\Scn_{ uK_{n}  + j }=k} -  \frac{1}{\sqrt{2 \pi u}} \exp \left(  - \frac{1}{2u} \left( \frac{k- (uK_{n}+j) \frac{n-K_{n}}{K_{n}+1}}{\Dcn} \right)^{2} \right)\right|  \ \mathop{\longrightarrow}_{n \rightarrow \infty} \ 0.
\end{equation}

We only give the main steps, since the structure of the proof is very similar to those of Lemmas <ref> and <ref>. One separately treats two cases. The first one is $\gamma>0$. In this case, $\bbn$ and $\bcn$ converge to values belonging to $(0,1/e)$. In particular, there are no singularities, and one may follow the standard proof of the local limit theorem (see e.g. <cit.>). The second case is $\gamma=0$, in which case  $ \Dbn \sim \sqrt{K_{n}}$ and $\Dcn \sim n^{3/2} /K_{n}$. The first estimate (<ref>) in contained in Lemma <ref>. For the second estimate, the main idea is to write, as in the proof of Lemma <ref>,

\[\phicn(t)=1+\textrm{i}\E[\Scn_{1}]t- \E[(\Scn_{1})^{2}] \frac{t^{2}}{2}+r^{n}(t),\]

where $|r^{n}(t)| \leq |t|^{3}  \E[(\Scn_{1})^{3}]/6$ for every $n \geq 1$ and $t \in \R$. In particular, by using the expansion of $F(1/e-z)$ around $z=0$, one sees that $ \E[(\Scn_{1})^{3}] \sim 3 (n/K_{n})^{5}$, so that for every $n$ sufficiently large and $t \in \R$, $|r^{n}(t/\Dcn)| \leq |t|^{3} \tfrac{\sqrt{n}}{K_{n}^{2}}$, and the key point is that $ \tfrac{\sqrt{n}}{K_{n}^{2}}= o(\tfrac{1}{K_{n}})$ since $K_{n}/\sqrt{n} \rightarrow \infty$. From this follows an analogue of Lemma <ref>, namely that there exist constants $A_{0},\epsilon,\kappa>0$ such that  for every $n$ sufficiently large and for every $|t|$ in $(A_{0}, \epsilon \Dcn)$, we have $|\phicn( \tfrac{t}{\Dcn})| \leq e^{- \kappa  \frac{t^{2}}{K_{n}}}$. This then allows us to follow the same steps as the proof of Lemma <ref>. We leave the details to the reader.

*Convergence of the unconditioned processes.  Recall that $(H^{n}_{k},B^{n}_{k})_{k \geq 1}$ denotes a sequence of i.i.d. random variables with distribution given by: for $i, j \geq 0$,

\[ \Prb{H^{n}=i,B^{n}=j}=\mubn(i) \,\Prb{\Scn_{i}=j}.\]

Also, for $i \geq 0$, we set  $\oH^{n}_{i}=H^{n}_{1}+H^{n}_{2}+ \cdots+H^{n}_{i}$ and  $\oB^{n}_{i}=B^{n}_{1}+B^{n}_{2}+ \cdots+B^{n}_{i}-i$.
To simplify notation, set
\[\hH^{(n)}_{u}=\frac{\oH^{n}_{ \lfloor u (n-K_{n}) \rfloor } - u (K_{n}+1) }{\Dbn}, \qquad \hB^{(n)}_{u}= \frac{\oB^{n}_{ \lfloor u (n-K_{n}) \rfloor }  - \frac{n-K_{n}}{K_{n}+1} \hH^{(n)}_{u}  }{\Dcn}.\]

Assume that $\frac{K_{n}}{\sqrt{n}} \rightarrow   \infty$ and   $\tfrac{K_{n}}{n}  \rightarrow \gamma \in [0,1)$ as $n \rightarrow \infty$.
The following convergence holds in distribution in $\mathbb{D}([0,1],\R^{2})$:

\[ \left( \hH^{(n)}_{u} ,  \hB^{(n)}_{u} \right)_{0 \leq u \leq 1}  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad  (W_{u},X_{u})_{0 \leq u \leq 1},\]

where  $W,X$ are two independent standard Brownian motions.

As the proof will show, the appearance of the normalized quantity $\hH^{(n)}_{u}$ is important in the definition of $\hB^{(n)}_{u}$.

The proof is similar to that of Lemma <ref>. In virtue of <cit.>, it is enough to check that the one-dimensional convergence holds for $u=1$.  For fixed $x,y \in \R$, by Lemma <ref>,

\begin{multline*}
\Pr{\oH^{n}_{  n-K_{n} }= \lfloor x \Dbn+K_{n} \rfloor ,\oB^{n}_{  n-K_{n}}= \lfloor y  \Dcn + \frac{n-K_{n}}{K_{n}+1} \lfloor x \Dbn \rfloor \rfloor} \\
=\Pr{\Sbn_{ n-K_{n} }=  \lfloor x \Dbn +K_{n} \rfloor} \Pr{\Scn_{ \lfloor x \Dbn +K_{n} \rfloor }= n-K_{n}   + \left\lfloor  y  \Dcn + \frac{n-K_{n}}{K_{n}+1} \lfloor x \Dbn \rfloor \right\rfloor}.
\end{multline*}

By (<ref>) and (<ref>), as $n \rightarrow \infty$, this quantity is asymptotic to $ \tfrac{1}{ \Dbn  }    p (x)  \cdot  \tfrac{1}{\Dcn} p (y)$,
where $p(x)= \tfrac{1}{\sqrt{2\pi}} e^{- {x^{2}}/{2}}$ denotes the Gaussian density.
It is then standard (see e.g. <cit.>) that this implies that the convergence

\[ \left( \hH^{(n)}_{1} ,  \frac{\oB^{n}_{ n-K_{n} }  - \frac{n-K_{n}}{K_{n}+1} \hH^{(n)}_{1}  }{\Dcn} \right)  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad  (W_{1},X_{1}),\]

holds in distribution. This completes the proof.

*Convergence of the bridge version. The proof of the following lemma is similar to that of Proposition <ref> and is left to the reader.

Assume that $\frac{K_{n}}{\sqrt{n}} \rightarrow   \infty$ and   $\tfrac{K_{n}}{n}  \rightarrow \gamma \in [0,1)$ as $n \rightarrow \infty$. Conditionally given the event
$ \{\oH^{n}_{n-K_{n}}=K_{n}+1, \oB^{n}_{n-K_{n}}=-1 \}$, the following  convergence in distribution holds jointly

\[  \left(  \hH^{(n)}_{u},  \hB^{(n)}_{u} \right)_{0 \leq u \leq 1}  \quad \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \quad (W^{\mathrm{br}},\Xbr),\]

where $W^{\mathrm{br}}$ and $\Xbr$ are two independent Brownian bridges.

Assume that, as $n\rightarrow \infty$, we have $\frac{K_{n}}{\sqrt{n}} \rightarrow   \infty$ and   $\tfrac{K_{n}}{n}  \rightarrow \gamma$
with $\gamma$ in $[0,1)$. Conditionally given the event
$ \{\oH^{n}_{n-K_{n}}=K_{n}+1, \oB^{n}_{n-K_{n}}=-1 \}$, the following  convergence in distribution holds jointly

\[\left(  \frac{\oB^{n}_{ \lfloor u (n-K_{n}) \rfloor }}{\Dcn} \right)_{0 \leq u \leq 1}  \ \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \  \Xbr,  \qquad  \sup_{0 \leq u \leq 1} \left|   \frac{\oH^{n}_{ \lfloor u (n-K_{n}) \rfloor }}{K_{n}}-u\right|  \ \mathop{\longrightarrow}_{n \rightarrow \infty} \ 0.\]

Since $  \hH^{(n)}_{u}$
converges in distribution and since $ \frac{1}{\Dcn} \cdot \frac{n-K_{n}}{K_{n}+1} \rightarrow 0$, we have that

\[  \frac{n-K_{n}}{\Dcn(K_{n}+1)} \cdot \sup_{0 \leq u \leq 1}| \hH^{(n)}_{u}| \rightarrow 0.\]

Therefore, $\frac{\oB^{n}_{ \lfloor u (n-K_{n}) \rfloor }}{\Dcn}$
has the same limit as $\hB^{(n)}_{u}$, namely $\Xbr$.

For the second convergence, we simply write

\[\frac{\oH^{n}_{ \lfloor u (n-K_{n}) \rfloor }}{K_{n}}-u= \frac{\Dbn}{K_{n}} \cdot \hH^{(n)}_{u}\]

and the result follows from the fact that $ \frac{\Dbn}{K_{n}} \rightarrow 0$.

*Convergence of the excursion version.

Using the (bivariate) Vervaat transform and Lemma <ref>,
we can now state an invariance principle for $\oB^{n}$ and $\oH^{n}$
conditionally given the event

\[\mathcal C^+=\{\oH^{n}_{n-K_{n}}=K_{n}+1,\, \oB^{n}_{n-K_{n}}=-1,\, \oB^{n}_i \ge 0\text{ for all }i <n-K_n \}.\]

From Lemma <ref>,
under this conditioning $(\oB^{n}, \oH^{n})$ has the same distribution
as $(\oB(\Ts_n),\oH(\Ts_n))$,
where $\Ts_n$ is an alternating BGW tree with offspring distributions $\mubn$ and $\mucn$
conditioned on having $n-K_n$ black vertices and $K_n+1$ white vertices.
We will thus state our invariance principle in terms of $(\oB(\Ts_n),\oH(\Ts_n))$.

Assume that $\frac{K_{n}}{\sqrt{n}} \rightarrow   \infty$ and   $\tfrac{K_{n}}{n}  \rightarrow \gamma \in [0,1)$ as $n \rightarrow \infty$. The following convergences hold jointly in distribution
\[  \left(  \frac{\oB_{\lfloor u (n-K_{n}) \rfloor}(\Ts_{n})}{ \Dcn} : 0\leq u \leq 1\right)  \ \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \ \mathbbm{e}, \qquad  \sup_{0 \leq u \leq 1} \left|   \frac{\oH_{ \lfloor u (n-K_{n}) \rfloor  }(\Ts_{n})}{K_{n}}-u\right|  \ \mathop{\longrightarrow}_{n \rightarrow \infty} \ 0.\]

This proposition follows from Corollary <ref>,
in the same way that Theorem <ref> (i) was established using Proposition <ref>.
Again, we leave details to the reader.

*End of the proof of Proposition <ref>

We consider separately the cases $\gamma=0$ and $\gamma>0$.
The first one is effortless.

In this case, the conclusions of Lemma <ref> are true with the same proof, where the occurrences of the quantity $p(0) q_{1}(c)$ should be replaced by $p(0)^{2}$, and where instead of using the local limit Lemmas <ref> and <ref>, one uses (<ref>) and (<ref>).  This shows that as before,   $\Ts_{n}$ and $\tTs _{n}$ can be coupled so that $\Ts_{n}=\tTs_{n}$ with probability tending to $1$.  Proposition <ref> then readily follows from Proposition <ref>.

We now concentrate on the case  $\tfrac{K_{n}}{\sqrt{n}} \rightarrow \infty$ and $ \tfrac{K_{n}}{n}  \rightarrow \gamma \in (0,1)$ as $n \rightarrow \infty$, which requires new ideas.
In this setting, we set $m_\bullet:=\frac{\gamma}{1-\gamma}$ and $m_\circ:=\frac{1-\gamma}{\gamma}$,
which are the limits of the means $m_\bullet^n$ and $m_\circ^n$, of $\mubn$ and $\mucn$ respectively.
By an argument similar to that used in Lemma <ref>, there exist unique parameters $(\ab,\bb,\ac,\bc)$ with $\bb,\bc \in (0,1/e)$
such that the following equations define probability distributions with means $m_\bullet$ and $m_\circ$:

\begin{equation}
\label{eq:mubc}\mub(i)= \ab \cdot  (\bb)^{i} \cdot \frac{(i+1)^{i-1}}{i!}  \quad (i \geq 0), \qquad \muc(i)= \ac \cdot  (\bc)^{i} \cdot \frac{(i+1)^{i-1}}{i!}  \quad (i \geq 0).
\end{equation}

It is then immediate that $(\abn,\bbn,\acn,\bcn)$ converges as $n \rightarrow \infty$ to the vector $(\ab,\bb,\ac,\bc)$.
As a consequence, one easily check that the distributions $\mubn$ and $\mucn$
have uniform exponential moments, in the sense that

\begin{equation}
\label{eq:expmoment} \exists \delta>0,\,  \qquad  \sup_{n \ge 1} \, \sum_{i=0}^{\infty} e^{\delta i} \mubn(i) < \infty
\quad \textrm{and} \quad \sup_{n \ge 1}\, \sum_{i=0}^{\infty} e^{\delta i} \mucn(i) < \infty.
\end{equation}

This result will be useful later.

We first establish a result which will allow us to replace $\tTs_{n}$ with $\Ts_{n}$,
that is to forget about the specific offspring distribution of the root.
As before, we denote $\tilde{\Ts} ^{n}$ the unconditioned version of $\tTs_n$
with distribution ${\BGW}^{\widetilde{\mu}_{\bullet}^{n},\mubn,\mucn}$, as defined in (<ref>).
Recall that the superscript $n$ indicates that the offspring distributions depend on $n$, and not a conditioning.

The following assertions are satisfied.

(i) We have

\[\Pr{\abs{\circ_{\tTs^{n}}}=K_{n}+1, \abs{\bullet_{\tTs^{n}}}=n-K_{n}}   \ \mathop{\sim}_{n \rightarrow \infty} \ \left( \sum_{a=1}^{\infty} a \mub(a-1) \right)\cdot \frac{1}{ 2\pi \sb \sc  \gamma \sqrt{\gamma(1-\gamma)} } \cdot \frac{1}{n^{2}},\]

where $(\sb)^{2}$, $(\sc)^{2}$ denote respectively the variance of $\mub$, $\muc$.

(ii) The couple $\big(\oH^{n}_{1}(\tTs_{n}),\oB^{n}_{1}(\tTs_{n})\big)$ converges in distribution as $n \rightarrow \infty$.

(iii) Let  $ \mathscr{F}^{\bullet,n}_{k} =(\Ts^{\bullet,n}_{1}, \ldots, \Ts^{\bullet,n}_{k})$ denote a collection of $k$ i.i.d. BGW trees distributed as  $ \Tsbn$ (with a black root). Then, for fixed $i_{0}$, conditionally given $ \{ \abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}=K_{n}+1-i_{0},\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}=n-K_{n}-1 \} $, the random variable $ \min_{1 \leq i \leq k} (\abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}- \abs{\circ_{\Ts^{\bullet,n}_{i}}},\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}- \abs{\bullet_{\Ts^{\bullet,n}_{i}}})$ converges in distribution, where the minimum refers to the lexicographical order on $\Z_{+}^{2}$.

We follow the strategy of <cit.>.
For $j \geq 0$, let $ \mathscr{F}^{\circ,n}_{j}$ denote a collection of $j$ i.i.d. BGW trees distributed as  $ \Tscn$ (with a white root). By considering the trees (with white roots) grafted on the root of $\tTs^{n}$  and using (<ref>), write

\begin{align*}
&\Pr{\abs{\circ_{\tTs^{n}}}=K_{n}+1, \abs{\bullet_{\tTs^{n}}}=n-K_{n}} \\
& \qquad\qquad = \sum_{a=1}^{\infty} \mubn(a-1) \Pr{\abs{ \circ_{ \mathscr{F}^{\circ,n}_{a}}}=K_{n}+1, \abs{\bullet_{ \mathscr{F}^{\circ,n}_{a}}}=n-K_n-1} \\
& \qquad \qquad =\sum_{a=1}^{\infty} a \mubn(a-1) \frac{1}{K_{n}+1} \Pr{\Sbn_{n-K_{n}-1}=K_{n}+1-a} \Pr{\Scn_{K_{n}+1}=n-K_{n}-1}.
\end{align*}

The desired result then follows by combining (<ref>) and (<ref>) with the dominated convergence theorem.

For (ii), fix $a,b \geq 1$. Since $(\oH^{n}_{1}(\tTs_{n}),\oB^{n}_{1}(\tTs_{n}))=(a,b-1)$ if and only if the root of $\tTs_{n}$ has $a$ (white) children, which altogether have $b$ (black) children, using (<ref>),  the quantity $\Prb{\oH^{n}_{1}(\tTs_{n}),\oB^{n}_{1}(\tTs_{n})=(a,b-1)}$ is equal to

\[
\frac{\mubn(a-1) \Pr{\Scn_{a}=b}  \Pr{\abs{ \circ_{ \mathscr{F}^{\bullet,n}_{b}}}=K_{n}+1-a, \abs{\bullet_{ \mathscr{F}^{\bullet,n}_{b}}}=n-K_n-1}}{\Pr{\abs{\circ_{\tTs^{n}}}=K_{n}, \abs{\bullet_{\tTs^{n}}}=n-K_{n}}},
\]

where, for $j \geq 1$,  $ \mathscr{F}^{\bullet,n}_{j}$ denotes a collection of $j$ i.i.d. BGW trees distributed as  $ \Tsbn$ (with a black root).
By (<ref>), the quantity $\Prb{\abs{ \circ_{ \mathscr{F}^{\bullet,n}_{b}}}=K_{n}+1-a, \abs{\bullet_{ \mathscr{F}^{\bullet,n}_{b}}}=n-K_n-1}$ is equal to

\[\ \frac{b}{n-K_{n}-1} \Pr{\Sbn_{n-K_{n}-1}=K_{n}+1-a} \Pr{\Scn_{K_{n}+1-a} =n-K_{n}-1-b}.\]

By  combining (<ref>) and (<ref>) with (i), we finally get that

\[\Prb{\oH^{n}_{1}(\tTs_{n}),\oB^{n}_{1}(\tTs_{n})=(a,b-1)}  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad  \frac{\gamma}{1-\gamma} \cdot \frac{\mub(a-1)  b\Pr{S^{\circ}_{a}=b}}{\sum_{i=1}^{\infty} i \mub(i-1)},\]

where $(S^{\circ}_{i})_{i \geq 1}$ is a random walk with jump distribution $\muc$.
Since $\Es{S^{\circ}_{1}}= \frac{1-\gamma}{\gamma}$, the limiting quantity indeed defines a probability distribution.

For (iii), fix integers $u,v \geq 0$. Observe that for $n$ sufficiently large, the following union is disjoint

\begin{align*}
& \scalebox{.8}{$\left\{  \min_{1 \leq i \leq k} (\abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}- \abs{\circ_{\Ts^{\bullet,n}_{i}}},\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}- \abs{\bullet_{\Ts^{\bullet,n}_{i}}})=(u,v), \ \abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}=K_{n}+1-i_{0},\ \abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}=n-K_{n}-1  \right\}$} \\
& \scalebox{.8}{$ \quad =\bigcup_{i=1}^{k} \left\{ \abs{\circ_{\Ts^{\bullet,n}_{i}}}=\abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}-u, \abs{\bullet_{\Ts^{\bullet,n}_{i}}}=\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}-v,  \abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}=K_{n}+1-i_{0},\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}=n-K_{n}-1\right\}. $}
\end{align*}

Hence, by exchangeability, the probability of the latter event is

\[\!\!
k \Pr{\abs{\circ_{\Ts^{\bullet,n}_{1}}}=K_{n}+1-i_{0}-u, \abs{\bullet_{\Ts^{\bullet,n}_{1}}}=n-K_{n}-1-v}\Pr{\abs{\circ_{\mathscr{F}^{\bullet,n}_{k-1}}}=u,\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k-1}}}=v}\]

Now, using  (<ref>),

\begin{multline*}
\frac{k \Pr{\abs{\circ_{\Ts^{\bullet,n}_{1}}}=K_{n}+1-i_{0}-u, \abs{\bullet_{\Ts^{\bullet,n}_{1}}}=n-K_{n}-1-v}}{\Pr{\abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}=K_{n}+1-i_{0},\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}=n-K_{n}-1}}\\
= \frac{ \frac{k}{n-K_{n}-1-v} \Pr{\Sbn_{n-K_{n}-1-v}=K_{n}+1-i_{0}-u} \Pr{\Scn_{K_{n}+1-i_{0}-u} =n-K_{n}-2-v} }{ \frac{k}{n-K_{n}-1} \Pr{\Sbn_{n-K_{n}-1}=K_{n}+1-i_{0} } \Pr{\Scn_{K_{n}+1-i_{0}}=n-K_{n}-2} }.
\end{multline*}

By  (<ref>) and (<ref>), this quantity tends to $1$ as $n \rightarrow \infty$. By combining Eqs. (<ref>), (<ref>) and (<ref>) we also get that

\[\Pr{\abs{\circ_{\mathscr{F}^{\bullet,n}_{k-1}}}=u,\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k-1}}}=v}  \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad \Pr{\abs{\circ_{\mathscr{F}^{\bullet}_{k-1}}}=u,\abs{\bullet_{\mathscr{F}^{\bullet}_{k-1}}}=v},\]

where $ \mathscr{F}^{\bullet}_{k-1}$ denotes a collection of $k-1$ i.i.d. alternating BGW trees with offspring distribution $\mub$ and $\muc$ and black roots.

We conclude that

\[\scalebox{.8}{$\Pr{\min_{1 \leq i \leq k} (\abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}- \abs{\circ_{\Ts^{\bullet,n}_{i}}},\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}- \abs{\bullet_{\Ts^{\bullet,n}_{i}}}) =(u,v) \, \Big| \, \abs{\circ_{\mathscr{F}^{\bullet,n}_{k}}}=K_{n}+1-i_{0},\abs{\bullet_{\mathscr{F}^{\bullet,n}_{k}}}=n-K_{n}-1 }$} \]

converges to $\Pr{\abs{\circ_{\mathscr{F}^{\bullet}_{k-1}}}=u,\abs{\bullet_{\mathscr{F}^{\bullet}_{k-1}}}=v}$ as $n \rightarrow \infty$.
This completes the proof.

We are now ready to establish Proposition <ref> in the final case.

Lemma <ref> (ii) and (iii) shows that with probability tending to $1$ as $n \rightarrow \infty$, among all the trees grafted on a (black) grandchild of the root of $\tTs_{n}$, only one has $n-K_{n}+o(n)$ black vertices and $K_{n}+o(n)$ white vertices, while all the others have $o(n)$ vertices. It is therefore sufficient to establish the desired result with $\tTs_{n}$ replaced with $\Ts_{n}$.  The first two convergences in  Proposition <ref>  are then obtained from Proposition <ref>.

It remains to check that

\[ \frac{1}{n} \min_{0 \leq i < n-K_{n}} \ell^{\bullet,n}_{i}  \quad \mathop{\longrightarrow}^{(\P)}_{n \rightarrow \infty} \quad 0,\]

where $\ell^{\bullet,n}_{i}$ denotes  the number of black corners  branching on the right of  $\llbracket \emptyset, v^{\bullet,n}_{i} \llbracket$ for $0 \leq i < n-K_{n}$ (recall that $v^{\bullet,n}_{i} $ is the $i$-th black vertex of $\Ts_{n}$ in lexicographical order).
The proof is divided in several steps: (i) Reduction to a one-type tree (ii) Reduction to a non-conditioned statement (iii) Large deviation estimates.

*Step 1: reduction to a one-type tree.

(0) at (0,0);
(1) at (-1.5,1);
(11) at (-1.5,2);
(111) at (-2,3);
(1111) at (-2,4);
(112) at (-1,3);
(1121) at (-1,4);
(2) at (-.5,1);
(3) at (.5,1);
(31) at (-0.5,2);
(32) at (0.5,2);
(321) at (0.5,3);
(3211) at (0.5,4);
(33) at (1.5,2);
(331) at (1.5,3);
(4) at (1.5,1);

(0) – (1)	(0) – (2)	(0) – (3)	(0) – (4)
(1) – (11)
(11) – (111) – (1111)	(11) – (112) – (1121)
(3) – (31)	(3) – (32) – (321) –(3211)	(3) – (33) – (331)

(0) circle (1.5pt)
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(321)++(.1,0) node[right] $w_{2}$
(111)  node[right] $w_{1}$
(0) node[below] $\emptyset$
[domain=-33.69:-270] plot (-1.5+0.2*cos(), 1+0.2*sin());
[domain=-33.69:-270] plot (-1.5+0.3*cos(), 1+0.3*sin());

(0) at (0,0);
(1) at (-1,1);
(11) at (-1.5,2);
(111) at (-1.5,3);
(12) at (-0.5,2);
(121) at (-0.5,3);
(2) at (1.5,1);
(21) at (0.5,2);
(22) at (1.5,2);
(221) at (0.5,3);
(222) at (1.5,3);
(2221) at (1.5,4);
(22211) at (1.5,5);
(223) at (2.5,3);
(2231) at (2.5,4);
(23) at (2.5,2);

(0) – (1)	(0) – (2)
(1) – (11)	(1) – (12)
(11) – (111)
(12) – (121)
(2) – (21)	(2) – (22)	(2) – (23)
(22) – (221)	(22) – (222) – (2221) –(22211)	(22) – (223) – (2231)

[domain=33.69:-225] plot (0.2*cos(), 0.2*sin());
[domain=33.69:-225] plot (0.3*cos(), 0.3*sin());

(11) node[right] $w_{1}$
(2221)++(.1,0) node[right] $w_{2}$
(2231)++(.1,0) node[right] $z$

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(0) at (0,0);
(1) at (-2,1);
(2) at (-1,1);
(3) at (0,1);
(4) at (1,1);
(41) at (0.5,2);
(42) at (1.5,2);
(5) at (2,1);

(0) – (1)	(0) – (2)	(0) – (3)	(0) – (4)	(0) – (5)
(4) – (41)	(4)–(42);

(1) node[above] $w_{1}$
(41)++(0,0.1) node[above] $w_{2}$

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From left to right: a two-type tree $T$ with black root, the two-type tree $T^{\circ}$ with white root obtained be re-rooting $T$ at its first white corner (which is highlighted), and finally its associated reduced white subtree $\hat{T}^{\circ}$. The  number of black corners branching on the right of $\llbracket \emptyset, w_{2} \llbracket$ in $T$ is $3$, which is   the same as the number of black corners branching on the right of $\llbracket \emptyset, w_{2} \llbracket$ in $T^{\circ}$, and which is less or equal than the number of corners branching to the right of $\llbracket \emptyset, w_{2} \llbracket$ in $\hat{T^{\circ}}$ (which is $4$).  The  number of black corners branching on the right of $\llbracket \emptyset, w_{1} \llbracket$ in $T$ is $6$, which is indeed  the same as the number of black corners branching on the right of $\llbracket \emptyset, w_{1} \llbracket$ in $T^{\circ}$ plus the number of children of $\emptyset$ in $T$, and which is less or equal than the number of corners branching to the right of $\llbracket \emptyset, w_{1} \llbracket$ in $\hat{T^{\circ}}$ (which is $5$) plus one.

The first idea is to reduce the statement to a one-type tree, a reduced white subtree.  Specifically, we proceed in several elementary steps:

* For $0 \le j\le K_n$, we denote $v^{\circ,n}_{j} $ the $j$-th white vertex of $\Ts_{n}$ in lexicographical order
and $\ell^{\circ,n}_{j}$ the number of black corners  branching on the right of  $\llbracket \emptyset, v^{\circ,n}_{j} \llbracket$.
If $v^{\circ,n}_{j}$ is the parent of $v^{\bullet,n}_{i}$, we have $\ell^{\circ,n}_{j}=\ell^{\bullet,n}_{i}$.
We conclude that

\[ \max_{0 \leq i < n-K_{n}} \ell^{\bullet,n}_{i}= \max_{0 \leq j \leq  K_{n}} \ell^{\circ,n}_{j}.\]

* Denote by $\Ts_{n}^{\circ}$ the tree obtained by re-rooting $\Ts_{n}$ at its first white corner (see Figure <ref>). Then for every white vertex $w$, the number of black corners branching on the right of $\llbracket \emptyset, w \llbracket$ is the same in $ \Ts_{n}$ and in  $\Ts_{n}^{\circ}$ (except for white vertices which are descendants of the first child of $\emptyset$ in $\Ts_{n}$, for which one has to add the number of children of $\emptyset$ in $\Ts_{n}$). See Figure <ref> for an example.

* Denote by $\hat{\Ts_{n}^{\circ}}$ the reduced white-subtree of $\Ts_{n}^{\circ}$. Then for every white vertex $w$, the number of black corners branching on the right of $\llbracket \emptyset, w \llbracket$ in $\Ts_{n}^{\circ}$ is less than or equal to   the number  of  corners branching on the right of $\llbracket \emptyset, w \llbracket$  in  $\hat{\Ts_{n}^{\circ}}$ (except for white vertices which are the descendants of the first child of $\emptyset$ in $\Ts_{n}$, for which one has to add $1$).  See Figure <ref> for an example. The reason why there may not be equality is that the reduction operation forgets the genealogy of black vertices: for instance, if in Figure <ref> the vertex $z$ were a younger sibling of $w_{2}$, the reduced white subtree would not have changed, but  the number of black corners branching on the right of $\llbracket \emptyset, w \llbracket$ in $\Ts_{n}^{\circ}$ would have increased.

Thanks to the previous observations, it therefore remains to check that

\[ \frac{1}{n}  \max_{0 \leq i  \leq K_{n}} \hat{\ell}^{\circ,n}_{i}  \quad \mathop{\longrightarrow}^{(\P)}_{n \rightarrow \infty} \quad 0,\]

where $\hat{\ell}^{\circ,n}_{i}$ is the number of  corners  branching on the right of  $\llbracket \emptyset, v^{\circ,n}_{i} \llbracket$, where $v^{\circ,n}_{i} $ is the $i$-th  vertex in lexicographical order of the white reduced tree $\hat{\Ts^{\circ}_{n}}$.

In turn, if $\hat{R}^{\circ,n}_{i}$ denotes the number of white vertices  branching on the right of  $\llbracket \emptyset, v^{\circ,n}_{i} \llbracket$ in $\hat{\Ts^{\circ}_{n}}$, we have

\[\max_{0 \leq i  \leq K_{n}} \hat{\ell}^{\circ,n}_{i} \leq \mathsf{Height}(\hat{\Ts^{\circ}_{n}})+ \max_{0 \leq i  \leq K_{n}} \hat{R}^{\circ,n}_{i}.\]

The advantage of considering  $\hat{R}^{\circ,n}_{i}$ is that this quantity is simply the value at time $i$ of the Łukasiewicz path of $\hat{\Ts^{\circ}_{n}}$ (this  is a well-known fact of the depth-first search of trees). But Proposition <ref>, applied with colors exchanged, shows that  the Łukasiewicz path of $\hat{\Ts^{\circ}_{n}}$, scaled in time by a factor $n$ and in space by a factor $\sqrt{n}$ possesses a functional scaling limit. As a consequence, $\tfrac{1}{\sqrt{n}} \max_{0 \leq i \leq  K_{n}} \hat{R}^{\circ,n}_{i}$ converges in distribution. Therefore, it remains to check that

\[\frac{\mathsf{Height}(\hat{\Ts^{\circ}_{n}})}{n}  \quad \mathop{\longrightarrow}^{(\P)}_{n \rightarrow \infty} \quad 0.\]

To this end, we will establish that for every fixed $\epsilon>0$,

\begin{equation}
\label{eq:height}\Pr{\mathsf{Height}(\hat{\Ts^{\circ}_{n}}) \geq n^{ \tfrac{1}{2} +\epsilon}}   \quad \mathop{\longrightarrow}_{n \rightarrow \infty} \quad 0,
\end{equation}

which is a stronger result. This is classical (see e.g. [17, 60]) for one-type (critical, finite variance) BGW trees conditioned on having a fixed size. The main issue here is that even though $\hat{\Ts^{\circ}_{n}}$ may be seen as a “one-type tree”, $\hat{\Ts^{\circ}_{n}}$ is not a BGW tree conditioned on having a fixed size (this would have been the case if $\Ts_{n}$ was obtained by  conditioning $\Ts^{n}$ on having a fixed number of white vertices only). Another (smaller) issue is that  offspring distributions depend on $n$.

*Step 2: Reduction to a non-conditioned statement.
In order to get rid of the conditioning, we start by adapting Section <ref> in the setting where the root of the two-type tree is white. We fix a tree $\tau$, and we now assume that white vertices are at even generation and black vertices are at odd generation (this is the case of $\Ts^{\circ}_{n}$).  Denote by $(v^{\circ}_{i}(\tau))_{0 \leq i < \abs{\circ_{\tau}}} $ the white vertices of $\tau$ listed in lexicographical order.  For $1 \leq i \leq \abs{\circ_{\tau}}$, let $H_{i}(\tau)$ be the number of black children of $v^{\circ}_{i-1}(\tau)$ and let $W_{i}(\tau)$ be the number of white grandchildren of $v^{\circ}_{i-1}(\tau)$. Finally, set $\oH_{0}(\tau)=\oH_{0}(\tau)=0$ and  for $1 \leq i \leq \abs{\circ_{\tau}}$ set

\[\oH_{i}(\tau)=H_{1}(\tau)+ \cdots+H_{i}(\tau), \qquad  \overline{W}_{i}(\tau)=W_{1}(\tau)+ \cdots+W_{i}(\tau)-i.\]

Now note that the tree whose Łukasiewicz path is $\overline{W}(\tau)$ is precisely the reduced white subtree of $\tau$.
Let $(H^{n},W^{n})$ be a random variable such that

\begin{equation}
\Pr{H^{n}=i,W^{n}=j}=\mucn(i) \Pr{\Sbn_{i}=j}, \qquad  i, j \geq 0.
\label{eq:HW}
\end{equation}

Let $(H^{n}_{i},W^{n}_{i})_{i \geq 1}$ be a sequence of i.i.d. random variables distributed as $(H^{n},W^{n})$. Set $\overline{H}^{n}_{i}=H^{n}_{1}+H^{n}_{2}+ \cdots+H^{n}_{i}$ and  $\overline{W}^{n}_{i}=W_{1}^{n}+W^{n}_{2}+ \cdots+W^{n}_{i}-i$. Then Lemma <ref> (ii) (applied with colors switched) tells us that $(H_{i}(\Ts_{n}^{\circ}) ,W_{i}(\Ts_{n}^{\circ}))_{1 \leq i \leq K_{n}+1}$ has the same distribution as $(H^{n}_{i},W^{n}_{i})_{1 \leq i \leq K_{n}+1}$ conditioned on the event

\[ \mathcal{C}^{+}_{n} \coloneqq \{\oH^{n}_{K_{n}+1}=n-K_{n},\overline{W}^{n}_{K_{n}+1}=-1 \textrm{ and } \overline{W}^{n}_{i} \geq 0 \textrm{ for every } 1 \leq i < K_{n}+1\}.\]

We now observe that $u^{\circ}_{i}(\Ts^{\circ}_{n})$ is an ancestor of $u^{\circ}_{k}(\Ts^{\circ}_{n})$ in $\Ts^{\circ}_{n}$
(or equivalently in the white reduced tree $\hat{\Ts^{\circ}_{n}}$)
if and only if $\overline{W}_{i}(\hat{\Ts^{\circ}_{n}})=\min_{i \leq j \leq k}  \overline{W}_{j}(\hat{\Ts^{\circ}_{n}})$.
Therefore the height of $u^{\circ}_{k}(\Ts^{\circ}_{n})$ in the white reduced tree $\hat{\Ts^{\circ}_{n}}$
is equal to $ \# \{0 \leq i \leq k-1: \overline{W}_{i}(\hat{\Ts^{\circ}_{n}})=\min_{i \leq j \leq k}  \overline{W}_{j}(\hat{\Ts^{\circ}_{n}}) \}$. As a consequence, we can write

\[\mathsf{Height}(\hat{\Ts^{\circ}_{n}})= \Psi\big(  \overline{W}_{i}(\hat{\Ts^{\circ}_{n}}) :0 \leq i \leq K_{n}+1 \big),\]

where for any discrete trajectory $\omega=(\omega_{0},\omega_{1}, \ldots, \omega_{K_{n}+1})$ we have set

\[\Psi(\omega)= \max_{0 \leq k \leq K_{n}+1} \# \left\{0 \leq i \leq k-1: \omega_{i}=\min_{i \leq j \leq k}  \omega_{j} \right \}.\]

As a consequence, for fixed $\epsilon>0$,

\begin{align*}
\Pr{\mathsf{Height}(\hat{\Ts^{\circ}_{n}}) \geq n^{ \tfrac{1}{2} +\epsilon}} &= \frac{\Pr{\Psi( \overline{W}_{i} :0 \leq i \leq K_{n}+1) \geq n^{{1}/{2}+\epsilon},{ \mathcal{C}^{+}_{n} }}}{\Pr{ \mathcal{C}^{+}_{n} }} \\
& \leq  \frac{\Pr{\Psi( \overline{W}^{n}_{i} :0 \leq i \leq K_{n}+1) \geq n^{{1}/{2}+\epsilon}}}{\Pr{ \mathcal{C}^{+}_{n} }}
\end{align*}

We now claim that for every $\eps>0$, there exists $\epsilon'>0$ such that for every $n$ sufficiently large

\begin{equation}
\label{eq:decroissance} \P\big(\Psi( \overline{W}^{n}_{i} :0 \leq i \leq K_{n}+1) \geq n^{{1}/{2}+\epsilon}
\big) \leq e^{-n^{\epsilon'}}.
\end{equation}

The same reasoning that led to Lemma <ref> shows that there exists a constant $C>0$ such that $\Pr{ \mathcal{C}^{+}_{n} } \sim \tfrac{C}{n^{2}}$ as $n \rightarrow \infty$, so that the estimate (<ref>) indeed implies (<ref>).

*Step 3: large deviation estimates.

Our goal is now to prove the large deviation estimate (<ref>).
For a discrete trajectory
in addition to $\Psi(\omega)$, we also let

\[\Phi(\omega)=\# \left\{0 < i \leq k: \omega_{i}=\max_{0 \leq j \leq i} \, \omega_{j} \right \}\]

be the number of weak records of $\omega$.
By a standard time-reversal argument, we have, for every fixed $k \geq 0$,

\[
\# \left\{0 \le i <k:  \overline{W}^{n}_{i}=\min_{i \leq j \leq k}  \overline{W}^{n}_{j} \right\}
\quad \stackrel{(d)}{=} \quad
\Phi(\overline{W}^{n}_0,\dots,\overline{W}^{n}_k).
\]

By definition, $\Psi( \overline{W}^{n}_{i} :0 \leq i \leq K_{n}+1)$ is the maximum
of the left-hand side of the above display for $k$ in $\{0,\dots,K_n\}$.
Thus we have

\begin{multline*}
\P\Big(\Psi( \overline{W}^{n}_{i} :0 \leq i \leq K_{n}+1) \geq n^{1/2+\epsilon}
\Big)  \leq  \sum_{0 \leq k \leq K_{n+1}} \P\Big(\Phi( \overline{W}^{n}_{i} :0 \leq i \leq k) \geq n^{1/2+\epsilon} \Big)\\
\leq (K_n+2) \cdot \P\Big(\Phi( \overline{W}^{n}_{i} :0 \leq i \leq K_{n}+1) \geq n^{1/2+\epsilon} \Big),
\end{multline*}

where the second inequality follows form the monotonicity of $\Phi$.
As a consequence, (<ref>) will follow if we establish that
for every $\eps>0$, there exists $\epsilon'>0$ such that for every $n$ sufficiently large

\begin{equation}
\label{eq:decroissancePhi} \P\big(\Phi( \overline{W}_{i} :0 \leq i \leq K_{n}+1) \geq n^{1/2+\epsilon}
\big) \leq e^{-n^{\epsilon'}}.
\end{equation}

To this end, we closely follow [60] (see also <cit.>). Set

\[\nu^{n}(k)=\sum_{ i=0}^{\infty}\mucn(i) \Pr{\Sbn_{i}=k},\qquad k \geq 0,\]

so that  $(\overline{W}^{n}_{i})_{i \geq 0}$ is a random walk  with jump distribution
Observe that $\nu^{n}$ has expectation $1$,
so that $\overline{W}^{n}$ is centered.
We denote by $\sigma_{n}^{2}$  the variance of $\overline{W}^{n}_1$,
which converges to a positive constant when $n$ tends to infinity
(since the parameters $a_\bullet^n,b_\bullet^n,a_\circ^n,b_\circ^n$
of the laws $\mubn$ and $\mucn$ converge to those of $\mub$ and $\muc$).
Finally we let

\[
T^{n}= \inf \Big\{k \geq 1 : \overline{W}^{n}_{k}= \max_{0 \leq i \leq k} \overline{W}^{n}_{i} \Big\}, \quad M^n = \max_{0 \le i \le {K_n+1}} \overline{W}^{n}_i.
\]

be respectively the first ladder time and the maximum of $\overline{W}^{n}$ up to time $K_{n}+1$.

Observe that the quantity $ \P\big(\Phi( \overline{W}_{i} :0 \leq i \leq K_{n}+1) \geq n^{1/2+\epsilon}
\big) $ is bounded from above by

\begin{equation}
\label{eq:somme} \P\bigg( \big|M^n-\frac{\sigma_n^2}{2}\Phi(\overline{W}^{n}_0,\dots,\overline{W}^{n}_{K_n+1})\big|
\geq n^{1/4+\eps} \bigg) + \P\bigg( M^n  \geq n^{1/2+\eps}- \frac{2}{\sigma_{n}^{2}} n^{1/4+\epsilon} \bigg)
\end{equation}

In order to bound from above the first term of (<ref>), we use the fact that there exists $\epsilon_{1}>0$ such that for every $n$ sufficiently large

\begin{equation}
\P\bigg( \big|M^n-\frac{\sigma_n^2}{2}\Phi(\overline{W}^{n}_0,\dots,\overline{W}^{n}_{K_n+1})\big|
>n^{1/4+\eps} \bigg) < e^{-n^{\eps_{1}}},
\label{eq:Phi_Close_M}
\end{equation}

as in <cit.> (we warn the reader that in loc. cit.,
$\Phi(\overline{W}^{n}_0,\dots,\overline{W}^{n}_{K_n+1})$ is denoted by $K_n$).
We cannot directly apply <cit.> because our offspring distributions depend on $n$;
we note however that the same proof can be applied,
provided that we have uniform deviation estimates for sums of i.i.d. variables
distributed as $\overline{W}^{n}_{1}$ or $\overline{W}^{n}_{T^n}-\tfrac12 \sigma_n^2$;
here, uniform means that the quantities $\eps$ and $\eps_{1}$ appearing
in the exponents should be independent of $n$.
Such estimates directly follow from the estimate for its Laplace transform,
given in Lemma <ref> (ii) and (iii) below (see, e.g., the proof of <cit.>).
Therefore (<ref>) is proved.

In order to bound from above the second term of (<ref>), we use the fact that there exists $\epsilon_{2} \in (0,\epsilon/4)$ such that for every $n$ sufficiently large

\begin{equation}
\label{eq:Mpasgrand}
\P\bigg( M^n  >n^{1/2+\eps}- \frac{2}{\sigma_{n}^{2}} n^{1/4+\epsilon} \bigg) \leq  n \max_{1 \leq k \leq n} \P\bigg( \overline{W}^{n}_{k}>n^{1/2+\eps}- \frac{2}{\sigma_{n}^{2}} n^{1/4+\epsilon} \bigg) \leq n e^{-n^{\epsilon_{2}}}.
\end{equation}
This similarly follows the proof of <cit.> combined with Lemma <ref> (ii) below.

By combing Eqs. (<ref>) and (<ref>) with (<ref>), we get (<ref>) and this completes the proof.

The following assertions hold.

(i) Let $(Y_{n})_{n \geq 1}$ be a sequence of centered random variables such that there exists $a>0$ and $C_{1}>0$ such that $\Es{e^{a|Y_{n}|}} \leq C_{1}$ for every $n \geq 1$. Then there exists $C_{2}>0$ such that for $\lambda>0$ sufficiently small, for every $n \geq 1$, $\E\big[e^{\lambda Y_{n}}\big] \leq e^{C_{2} \lambda ^{2}}$.

(ii)  There exists a constant $C>0$ such that for $\lambda>0$ sufficiently small, for every $n  \geq 1$, $\E\big[e^{\lambda \overline{W}^{n}_{1}}\big] \leq e^{C \lambda ^{2}}$.

(iii) There exists a constant $C>0$ such that for $\lambda>0$ sufficiently small, for every $n \geq 1$, $\E\big[{e^{\lambda(\overline{W}^{n}_{T^{n}} - {\sigma_{n}^{2}}/{2} ) }}\big] \leq e^{C \lambda ^{2}}$.

The first assertion is a standard result concerning sub-exponential distributions. However, since we need estimates uniform in $n$, we give a proof for completeness. First,  using Markov's inequality, write, for $t \geq 0$ and $n \geq 1$,

\[\P(|Y_{n}| \geq t)= \P(e^{a|Y_{n}|} \geq e^{at}) \leq C_{1} e^{-at}.\]

Hence, for $n,k \geq 1$,

\[\Es{|Y_{n}|^{k}}=\int_{0}^{\infty} k t^{k-1} \Pr{|Y_{n}| \geq t} {\d}t \leq  C_{1} k \int_{0}^{\infty} t^{k-1} e^{-a t} {\d}t.\]

In particular, $\Es{|Y_{n}|^{2}} \leq 2C_{1}/a^{2}$.
Now, for $\lambda>0$ small enough, since $\E[Y_{n}]=0$,

\[ \Es{e^{\lambda Y_{n}}}=1+ \sum_{k=2}^{\infty} \Es{Y_{n}^{k}} \frac{\lambda^{k}}{k!}  \leq 1+ \frac{C_{1}}{a^{2}} \lambda^{2}+  C_{1}  \int_{0}^{\infty}  \left( \sum_{k=3}^{\infty}  kt^{k-1} \frac{\lambda^{k}}{k!} \right) e^{-a t} {\d}t.\]

By calculating explicitly the last sum and then the integral, we finally get that for $\lambda \in (0,a/2)$, for every $n \geq 1$, $\E\big[e^{\lambda Y_{n}} \big] \leq  1+ \tfrac{C_{1}}{a^{2}} \lambda^{2}+  \tfrac{2C_{1}}{a^{3}} \lambda^{3} \leq 1+ \big( \tfrac{C_{1}}{a^{2}}+ \tfrac{2C_{1}}{a^{3}} \big) \lambda^{2}$, which gives the desired result with $C_{2}= \tfrac{C_{1}}{a^{2}}+ \tfrac{2C_{1}}{a^{3}}$.

For the second assertion, by (i), it is enough to show that there exists $a>0$ and $C_{1}>0$ such that $\E\big[e^{a |\overline{W}^{n}_{1}|}\big] \leq C_{1}$ for every $n \geq 1$.
Since $\overline{W}^{n}_{1}$ takes value in $\{-1,0,1,\dots\}$, it is equivalent to prove
$\E\big[e^{a (\overline{W}^{n}_{1}+1)}\big] \leq C'_{1}$ for every $n \geq 1$
(where $C'_1$ is another constant).
To this end, write

\begin{equation}
\E\big[e^{a (\overline{W}^{n}_{1}+1)}\big]  \leq  \sum_{ i=0}^{\infty}\mucn(i)  \E \big[ e^{a \Sbn_{i}} \big] =   \sum_{ i=0}^{\infty}\mucn(i)  \E \big[e^{a \Sbn_{1}} \big]^{i}.
\label{eq:Tech5}
\end{equation}

Let $\delta>0$ be such that  (<ref>)  holds.
Hence $\sup_{n \ge 1} \E[e^{\delta \Sbn_{1}}]<\infty$
and from the first assertion,
for any $A>1$, we have $\sup_{n \ge 1} \E[e^{a \Sbn_{1}}] <A$ for $a$ sufficiently small.
Choosing $A=e^\delta$, and going back to (<ref>), we get

\[\E\big[e^{a (\overline{W}^{n}_{1}+1)}\big]  \leq \sum_{ i=0}^{\infty}\mucn(i) e^{\delta i}.\]

By (<ref>), the right-hand side is finite and bounded, as a function of $n$,
which concludes the proof of (ii).

For the last assertion, we first observe that by <cit.>, the law of $\overline{W}^{n}_{T^{n}}$ is given by

\[\Pr{\overline{W}^{n}_{T^{n}}=k}=\nu^{n}([k,\infty)), \qquad k \geq 0.\]

Since $\sigma_{n}^{2}$ converges as $n \rightarrow \infty$,
it is enough to show that there exists $a>0$ and $C>0$
such that $\E\big[e^{a |\overline{W}^{n}_{T^{n}}|}\big] \leq C$ for every $n \geq 1$.
Note that the absolute values are here superfluous since $\overline{W}^{n}_{T^{n}}$
is nonnegative.

The proof of (ii) shows that there exists two constants $C',b>0$ such that

\[\Pr{\overline{W}^{n}_{T^{n}}=k}=\nu^{n}([k,\infty)) = \P(\overline{W}^{n}_{1} \geq k-1)  \leq e^{-b(k-1)} \E\big[ e^{b\, \overline{W}^{n}_1} \big]\le  C' e^{-bk}\]

for every $n \geq 1$ and $ k \geq 0$.
This implies the existence of $a>0$ and $C>0$ such that $\E\big[e^{a \overline{W}^{n}_{T^{n}}}\big] \leq C$ for every $n \geq 1$. This concludes the proof.

§.§ Completing the proof of Theorem <ref>.
So far, we have established Theorem <ref> in the following cases:

* (i)$_{c=0}$ only for $ \dFKn$ in Section <ref>;

* (i)$_{c>0}$ in Section <ref>;

* (ii) in Section <ref>.

In this Section, we treat the missing cases. More precisely, in Section <ref>, we show that the convergence $\dFKn \rightarrow \mathbf{L}_{\infty}$  in case (iii)  follows from (ii) by a maximality argument. Then, in Section <ref>, we establish, in case (i)$_{c=0}$, the  convergence $\dPKn \rightarrow \mathbf{L}_{0}$ by using, surprisingly, the convergence $ \dFKn \rightarrow \mathbf{L}_{\infty}$ just established in Section <ref>. We conclude in Section <ref> that   the last missing convergence $\dPKn \rightarrow \mathbf{L}_{c}$ in case (iii) holds by using a short symmetry argument,  as well as again the convergence $ \dFKn \rightarrow \mathbf{L}_{\infty}$  established in Section <ref>.

§.§.§ Proof of  Theorem <ref> (iii) for $ \dFKn$.

Suppose that $ \frac{n-K_{n}}{\sqrt{n}} \rightarrow c$ as $n \rightarrow \infty$, with $c \geq 0$. We choose $K'_{n}$ such that $ \frac{K'_{n}}{\sqrt{n}} \rightarrow \infty$ and $ \frac{n-K'_{n}}{\sqrt{n}} \rightarrow \infty$ as $n \rightarrow \infty$, and $K'_{n} \leq K_{n}$ for $n$ sufficiently large. Then by  Theorem <ref> (ii), which has been established in Section <ref>, $ \dot{\FFF}^{(n)}_{K'_{n}}$ converges in distribution to $\mathbf{L}_{\infty}$, which is maximal for inclusion. Since  $ \dot{\FFF}^{(n)}_{K'_{n}}\subset\dFKn$, it readily follows by a compactness argument that $ \dFKn$ also converges in distribution to $\mathbf{L}_{\infty}$ (and in fact we have the joint convergence $ (\dot{\FFF}^{(n)}_{K'_{n}},\dFKn) \to (Z,Z)$, where $Z$ has the distribution of $\mathbf{L}_{\infty}$). This completes the proof.

The same argument can be used to deduce the convergence of $\dFKn$ to $\mathbf L_\infty$ in the case
$c=\infty,\, \gamma>0$ from the case $c=\infty,\, \gamma=0$.
Since in Section <ref>, the most difficult part is the case $\gamma>0$,
we could have shortened significantly the proof if we were only interested in $\dFKn$
and not in $\dPKn$. However, it is not true in general that  if $\dFKn \rightarrow \mathbf{L}_{\infty}$, then    $(\dFKn,\dPKn) \rightarrow ( \mathbf{L}_{\infty}, \mathbf{L}_{\infty})$ (but it is true that if $\dPKn \rightarrow \mathbf{L}_{\infty}$ then $(\dFKn,\dPKn) \rightarrow ( \mathbf{L}_{\infty}, \mathbf{L}_{\infty})$, see Lemma <ref>).

§.§.§ Proof of Theorem <ref> (i)$_{c=0}$ for $ \dPKn$.

Assume that $K_{n} \rightarrow \infty$ and that $\frac{K_n}{\sqrt{n}} \to 0$.
We want to prove that $\dPKn$ tends to $\mathbb{L}_0=\S$.
Recall from Lemma <ref> that
the blocks of $\PKn$ are the connected components of $\FKn$; in particular, $ \S \cap \dPKn= \S \cap \dFKn$).
Since $\S \cap \dFKn$ converges to $\S$ in distribution
(this was established in the second part of the convergence in distribution
$\dFKn \rightarrow \mathbf{L}_0$ in Section <ref>),
any limit point of $\dPKn$ contain $\S$ and
it is enough to check that
$\Lambda(\dPKn) \rightarrow 0$  in probability as $n \rightarrow \infty$,
where $\Lambda(L)$ is the Euclidean length of the longest chord of the lamination $L$.

In the first part of the proof of the convergence in distribution $\dFKn \rightarrow \mathbf{L}_0$ in Section <ref>, it was shown that  $\Lambda(\dFKn) \rightarrow 0$ in probability as $n \rightarrow \infty$.
The problem is that $\Lambda(\dFKn) \rightarrow 0$ does not automatically imply that $\Lambda(\dPKn) \rightarrow 0$,
since $\dFKn$ may have a connected component made of many small chords
that creates a long chord in $\dPKn$ (for instance, if $\dFKn= \cup_{k=1}^{\lfloor n/2 \rfloor} [e^{-2 \i \pi k/n},e^{-2 \i \pi (k+1)/n}]$, then $\dPKn$ has a chord of length approximately $2$ while $\dFKn$ only has chords of length order $1/n$).

To prove $\Lambda(\dPKn) \rightarrow 0$, we shall additionally use the convergence
of $\dFKn \rightarrow \mathbf{L}_0$ (proved in Section <ref> since we are in the case $c=0$)
and of $\dot{\mathcal{F}}^{(n)}_{n-1}$ to the Brownian triangulation $\mathbf{L}_{\infty}$ (proved in Section <ref>).
From Skorokhod's representation theorem, we can assume that the convergence
$\big(\dFKn, \Lambda(\dFKn), \dot{\mathcal{F}}^{(n)}_{n-1}\big) \rightarrow (\S,0,\mathbf{L}_{\infty})$ holds almost surely.
Note also that, almost surely, for every $\epsilon>0$, for every arc ${A} \subset \S$ of length at least $\epsilon$,
$\mathbf{L}_{\infty}$ contains a chord of positive Euclidean length with endpoints belonging to $ {A}$ (indeed, this easily follows from the fact that local minima times are almost surely dense for the Brownian excursion).

We argue by contradiction, and assume the existence of $\epsilon>0$ such that,
up to extraction,  $\Lambda(\dPKn) \geq 2\epsilon$ for every $n \geq 1$.
By compactness, up to extraction, we may assume that
$\dPKn$ contains a chord $ [e^{-2 \i \pi s^{n}},e^{-2 \i \pi t^{n}}]$
such that $s^{n} \rightarrow s$ and $t^{n} \rightarrow t$,
with $\min(t-s,1-t+s) \geq \epsilon$.
Since $\Lambda(\dFKn) \rightarrow 0$, for $n$ sufficiently large
the chord $[e^{-2 \i \pi s^{n}},e^{-2 \i \pi t^{n}}]$ is not in $\FKn$. But since the blocks of $\PKn$ are the connected components of $\FKn$, $n s^{n}$ and $n t^{n}$ must belong to the same connected component $ \mathcal{C}_{n}$  of $\FKn$. By considering the path that joins them in $\dFKn$, we get the existence of $s_{1}^{n}, \ldots, s_{d_{n}}^{n}$ with $s_{1}^{n}=s^{n}$, $s_{d_{n}}=t^{n}$, and $n s_{i}^{n} \in \mathcal{C}_{n}$ for every $1 \leq i \leq d_{n}$, with either $ \{s_{1}^{n}, \ldots, s_{d_{n}}^{n}\}  \subset[s^{n},t^{n}]$, or $ \{s_{1}^{n}, \ldots, s_{d_{n}}^{n}\}  \subset [0,1] \backslash [s^{n},t^{n}]$. To simplify, let us treat the first case (the argument in the second case is exactly the same). Then $s_{1}^{n} \leq s_{2}^{n} \leq \cdots \leq s_{d_{n}}^{n}$, and since $\Lambda(\dFKn) \rightarrow 0$, we have $\max_{1 \leq i \leq d_{n}-1} (s^{n}_{i+1}-s_{i}^{n}) \rightarrow 0$ as $n \rightarrow \infty$.
Since $\dFKn$ is a non-crossing tree,  this implies that for every fixed $\eta>0$, for $n$ sufficiently large, in $\dFKn$ there is no chord of Euclidean length at least $\eta$ with endpoints belonging to $[s^{n},t^{n}]$. Since $  \dot{\mathcal{F}}^{(n)}_{n-1}$ is still a non-crossing tree obtained from $\dFKn$ by adding chords, so the latter property also holds for $ \dot{\mathcal{F}}^{(n)}_{n-1}$ (see Figure <ref> for an illustration).
But $ \dot{\mathcal{F}}^{(n)}_{n-1} \rightarrow \mathbf{L}_{\infty}$. Therefore, in $\mathbf{L}_{\infty}$,  there is no chord of positive Euclidean length with endpoints belonging to the arc $ \{ e^{-2 \i \pi u} : u \in (s,t) \}$. This is a contradiction, and this completes the proof.

Illustration of the proof of Theorem <ref> (i)$_{c=0}$ for $ \dPKn$: the blue “macroscopic” chord belongs to $\dPKn$. Since  the blocks of $\PKn$ are the connected components of $\FKn$ and since all the chords of $\dFKn$ are “small”, there exists a path of such chords (in black) joining the two endpoints of the “macroscopic” chord. It is then not possible to add a new  “macroscopic” chord with endpoints  belonging to the bold dashed red arc.

§.§.§ Proof of  Theorem <ref> (iii) for $ \dPKn$

Assume that $ \frac{n-K_{n}}{\sqrt{n}} \rightarrow c$ as $n \rightarrow \infty$, with $c \geq  0$.
We first establish that

\begin{equation}
\label{eq:cvhat} \hat{\PPP}^{(n)}_{K_{n}}  \quad \mathop{\longrightarrow}^{(d)}_{n \rightarrow \infty} \quad \mathbf{L}_{c}.
\end{equation}

By Lemma <ref>, it suffices to show that $ \hat{\PPP}^{(n)}_{n-K_{n}} \rightarrow \mathbf{L}_{c}$ as $n \rightarrow \infty$. If $n-K_{n} \rightarrow \infty$,  this follows from Theorem <ref> (i) for $\dot{\PPP}^{(n)}_{n-K_{n}}$ (which was established in Section <ref> for $c>0$ and in Section <ref> for $c=0$). If $ \limsup_{n \rightarrow \infty} (n-K_{n})<\infty$, then   $ \hat{\FFF}^{(n)}_{n-K_{n}}$ has a bounded number of chords, whose Euclidean length, divided by $n$, tends to $0$ with probability tending to one as $n \rightarrow \infty$ (this was established in Section <ref>), and this readily implies the convergence $ \hat{\PPP}^{(n)}_{n-K_{n}} \rightarrow \mathbf{L}_{0}$. This establishes (<ref>).

Now assume that $c>0$. Since $\hat{\mathcal{P}}^{(n)}_{K_{n}}$ is obtained from  $\dPKn$ by adding points of $\S$ and since $\hat{\mathcal{P}}^{(n)}_{K_{n}} \rightarrow \mathbf{L}_{c}$ in distribution, any sub-sequential distributional limit of $\dPKn$ must contain all the chords of $\mathbf{L}_{c}$, and by (<ref>) must be equal to $\mathbf{L}_{c}$. In other words, the convergence $\dPKn \rightarrow  \mathbf{L}_{c}$ holds in distribution.

Finally assume that $c=0$.  Since $\hat{\mathcal{P}}^{(n)}_{K_{n}}$ is obtained from  $\dPKn$ by adding points of $\S$, (<ref>) shows that any subsequential distribution limit of $\dPKn$ is included in $\mathbf{L}_{0}=\S$. Therefore, it is enough to show that the convergence $\S \cap \dPKn \rightarrow \S$  holds in distribution as $n \rightarrow \infty$. To this end, the key idea is to use the fact that $\S \cap \dPKn=\S \cap \dFKn$. Indeed, by  Theorem <ref> (iii) for $ \dFKn$ (this was established in Section <ref>), $\dot{\mathcal{F}}^{(n)}_{K_{n}} \rightarrow \mathbf{L}_{\infty}$. By (<ref>),
there exists an endpoint of some non-trivial chords of $\mathbf{L}_{\infty}$ arbitrarily close to any fixed point of $\S$,
which implies that   $\S \cap \dot{\mathcal{F}}^{(n)}_{K_{n}} \rightarrow \S$ in distribution as $n \rightarrow \infty$, and hence  $\S \cap \dot{\mathcal{P}}^{(n)}_{K_{n}} \rightarrow \S$ in distribution as $n \rightarrow \infty$. This completes the proof.

* In view of the previous results, we have established that a condensation phenomenon occurs in the dual tree $ \mathcal{T}(\PKn)$ when either $K_{n} \rightarrow \infty$ and $\tfrac{K_{n}}{\sqrt{n}} \rightarrow 0$, or $\tfrac{n-K_{n}}{\sqrt{n}} \rightarrow 0$. Indeed, when  $\tfrac{n-K_{n}}{\sqrt{n}} \rightarrow 0$, the fact that the convergence $\dPKn {\rightarrow} \S$ holds in distribution indicates that the dual tree  $ \mathcal{T}(\PKn)$ contains a unique (black) vertex of macroscopic degree (of order $n$)  such that the subtrees grafted on it all have size $o(n)$.  When $K_{n} \rightarrow \infty$ and $\tfrac{K_{n}}{\sqrt{n}} \rightarrow 0$, the same phenomenon happens with condensation on a white vertex.  It seems difficult to establish these facts directly (such as in [3, 44, 51]).

* If $K \geq 1$ is fixed, as $n \rightarrow \infty$, we believe that $(\dot{\mathcal{P}}^{(n)}_{K},\dot{\mathcal{F}}^{(n)}_{K})$ converge jointly in distribution to the same limit, a collection of $K$ i.i.d. uniformly distributed points on $\S$, and that this could be established by using Proposition <ref>.

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