$\displaystyle=2\mathop{}\\!\mathbb{P}\bigl{(}|X_{(r+W_{G_{k}})^{2}\log^{6}(1/r)}|<W_{G_{k}}+r\bigr{)}$
$\displaystyle=2\mathop{}\\!\mathbb{P}(|X_{1}|<\log^{-3}(1/r))$
$\displaystyle=O(\log^{-3}(1/r)).$

Since $K_{r}=o(\log(1/r))$, then a union bound combined with the
Borel–Cantelli lemma imply that almost surely, all the events for $k\leq
K_{r}$ fail for $r$ small enough (of the form $r=2^{-n}$). Thus, in order to
avoid that both $X^{L,k}$ and $X^{R,k}$ reach $-r$, we must have
$\Delta_{k}\leq 2(r+W_{G_{k}})^{2}\log^{6}(1/r)$ for every $k\leq K_{r}$.
Since $G_{k}$ is smaller than the first hitting time of $-r\log^{10k}(1/r)$ by
definition, then combined with Lemma 3.12, we have:

$-r<W_{G_{k}}\leq r\log^{10k+3}(1/r)\qquad\text{and so}\qquad\Delta_{k}\leq
3(r\log^{10k+6}(1/r))^{2}\qquad\text{for each}\enskip k\leq K_{r}.$

According to the strong Markov property applied at the first hitting time of
$-r\log^{10k}(1/r)$, the path $W$ behaves after this time as an unconditioned
Brownian motion started from this value during the remaining time of the
interval $[G_{k},D_{k}]$, which is at most $3(r\log^{10k+6}(1/r))^{2}$ on the
previous event. This is not enough for $W$ to further reach
$-r\log^{10(k+1)}(1/r)$. Indeed, if $X$ is a generic Brownian motion, then:

$\mathop{}\\!\mathbb{P}\Bigl{(}\inf_{[0,4(r\log^{10k+6}(1/r))^{2}]}X<-r\log^{10k+8}(1/r)\Bigr{)}=\mathop{}\\!\mathbb{P}\Bigl{(}|X_{1}|>\frac{1}{2}\log^{2}(1/r)\Bigr{)}\leq
2\exp\Bigl{(}-\frac{1}{4}\log^{2}(1/r)\Bigr{)},$

and the right-hand side, multiplied by $K_{r}$, is again a convergent series
when summing over all values $r\in\\{2^{-n},n\in\mathbb{N}\\}$. Hence, we may
also assume that the intervals for different values of $k\leq K_{r}$ are
disjoint, and so the Brownian motions $X^{L,k}$ and $X^{R,k}$ that we sample
are independent.

Finally, by the last claim of Lemma 3.12, for a positive proportion of the
indices $k\leq K_{r}$ we have both $W_{G_{k}}\leq r\log^{10k}(1/r)$ and
$\Delta_{k}>(r\log^{10k}(1/r))^{2}$, so there is a probability uniformly
bounded away from $0$ that both $X^{L,k}$ and $X^{R,k}$ reach $-r$, in the
first half of the interval. We conclude by independence that this occurs for a
positive proportion of the indices $k\leq K_{r}$.

Step 2: Proof for Brownian bridges. Let us next prove the actual statement by
comparing the Brownian motions $X^{L,k}$ and $X^{R,k}$ with Brownian bridges
$W^{L,k}$ and $W^{R,k}$. We aim at proving that the probability that one of
these bridges stays above $-r$ in the first half of the interval when
$\Delta_{k}>(r+W_{G_{k}})^{2}\log^{6}(1/r)$ is summable when $r$ ranges over
$\\{2^{-n},n\in\mathbb{N}\\}$, and then that if both $W_{G_{k}}\leq
r\log^{10k}(1/r)$ and $\Delta_{k}>(r\log^{10k}(1/r))^{2}$, then there is a
probability bounded away from $0$ uniformly in $r$ and $k$ that the two
bridges both reach $-r$. Let us only consider $X^{L,k}$ and $W^{L,k}$ since
the same argument applies to $X^{R,k}$ and $W^{R,k}$. Since, conditionally on
$W_{G_{k}}$, $W_{D_{k}}$, $\Delta_{k}$, and $\Delta^{L}_{k}$, the path
$W^{L,k}$ is a Brownian bridge from $W_{G_{k}}$ to $W_{D_{k}}$ with duration
$\Delta^{L}_{k}\geq\Delta_{k}$, then the following absolute continuity
relation holds: for any measurable and nonnegative function $F$,

$\mathop{}\\!\mathbb{E}\bigl{[}F(W^{L,k}_{t},t\leq\Delta^{L}_{k}/2)\mid
W_{G_{k}},W_{D_{k}},\Delta_{k},\Delta^{L}_{k}\bigr{]}=\mathop{}\\!\mathbb{E}\Bigl{[}F(X^{L,k}_{t},t\leq\Delta^{L}_{k}/2)\,\frac{p_{\Delta^{L}_{k}/2}(W_{D_{k}}-X^{L}_{\Delta^{L}_{k}/2})}{p_{\Delta^{L}_{k}}(W_{D_{k}}-W_{G_{k}})}\bigm{|}W_{G_{k}},W_{D_{k}},\Delta_{k},\Delta^{L}_{k}\Bigr{]},$

where $p_{t}(x)=(2\pi t)^{-1/2}\exp(-x^{2}/(2t))$. We then integrate with
respect to $W_{D_{k}}$. Precisely, we claim that there exists a constant $C>0$
such that on the event
$\\{W_{G_{k}}>-r,W_{D_{k}}>-r,\Delta_{k}>2(r+W_{G_{k}})^{2}\log^{6}(1/r)\\}$,
we have

$\mathop{}\\!\mathbb{E}\Bigl{[}\frac{p_{\Delta^{L}_{k}/2}(W_{D_{k}}-X^{L}_{\Delta^{L}_{k}/2})}{p_{\Delta^{L}_{k}}(W_{D_{k}}-W_{G_{k}})}\bigm{|}X^{L}_{\Delta^{L}_{k}/2},W_{G_{k}},\Delta_{k},\Delta^{L}_{k}\Bigr{]}\leq
C.$ (3.8)

This allows to infer from the first part of the proof that

$\mathop{}\\!\mathbb{P}\Bigl{(}\Delta_{k}>2(r+W_{G_{k}})^{2}\log^{6}(1/r)\enskip\text{and}\enskip\max\Bigl{\\{}\inf_{t\in[0,\Delta_{k}/2]}W^{L,k}_{t},\inf_{t\in[0,\Delta_{k}/2]}W^{R,k}_{t}\Bigr{\\}}>-r\Bigr{)}=O(\log^{-3}(1/r)),$

and then that, as in the first part, we must impose $\Delta_{k}\leq
2(r+W_{G_{k}})^{2}\log^{6}(1/r)\leq 3(r\log^{10k+6}(1/r))^{2}$ for every
$k\leq K_{r}$ in order to prevent the bridges $W^{L,k}$ and $W^{R,k}$ to reach
$-r$.

Let us prove the upper bound (3.8). Conditionally on $W_{G_{k}}$ and
$\Delta_{k}$, the path $(W_{t};t\in[G_{k},D_{k}])$ has the law of a Brownian
motion $Y$ with duration $\Delta_{k}$, started from $W_{G_{k}}$, and
conditioned on the event
$E_{k}=\\{\inf\\{Y_{t};t\in[G_{k},D_{k}]\\}\leq-r\log^{10k}(1/r)\\}$. Then the
law of $W_{D_{k}}$ can be obtained by splitting at the first hitting time of
this value; by the Markov, property, the remaining path after is an
unconditioned Brownian motion. Recall also that we work implicitly on the
event where $W_{G_{k}},W_{D_{k}}>-r$. For $z>0$, let us write $q_{z}$ for the
density of the first hitting time of $-z$ by a standard Brownian motion, then:

$\displaystyle\mathop{}\\!\mathbb{E}\Bigl{[}\frac{p_{\Delta^{L}_{k}/2}(W_{D_{k}}-X^{L}_{\Delta^{L}_{k}/2})}{p_{\Delta^{L}_{k}}(W_{D_{k}}-W_{G_{k}})}\mathbf{1}_{\\{W_{G_{k}},W_{D_{k}}>-r\\}}\bigm{|}X^{L}_{\Delta^{L}_{k}/2},W_{G_{k}},\Delta_{k},\Delta^{L}_{k}\Bigr{]}$
$\displaystyle=\frac{1}{\mathop{}\\!\mathbb{P}(E_{k})}\int_{0}^{\Delta_{k}}\mathop{}\\!\mathrm{d}t\int_{-r}^{\infty}\mathop{}\\!\mathrm{d}y\
q_{r\log^{10k}(1/r)+W_{G_{k}}}(t)p_{\Delta_{k}-t}(y+r\log^{10k}(1/r))\frac{p_{\Delta^{L}_{k}/2}(y-X^{L}_{\Delta^{L}_{k}/2})}{p_{\Delta^{L}_{k}}(y-W_{G_{k}})}\mathbf{1}_{\\{W_{G_{k}}>-r\\}}$
$\displaystyle=\frac{1}{\mathop{}\\!\mathbb{P}(E_{k})}\int_{0}^{\Delta_{k}}\mathop{}\\!\mathrm{d}t\frac{q_{r\log^{10k}(1/r)+W_{G_{k}}}(t)}{2\sqrt{\pi(\Delta_{k}-t)}}\int_{-r}^{\infty}\mathop{}\\!\mathrm{d}y\exp\Bigl{(}\frac{(y-W_{G_{k}})^{2}}{2\Delta^{L}_{k}}-\frac{(y-X^{L}_{\Delta^{L}_{k}/2})^{2}}{\Delta^{L}_{k}}-\frac{(y+r\log^{10k}(1/r))^{2}}{2(\Delta_{k}-t)}\Bigr{)}\mathbf{1}_{\\{W_{G_{k}}>-r\\}}.$

We have
$\mathop{}\\!\mathbb{P}(E_{k})=\int_{0}^{\Delta_{k}}\mathop{}\\!\mathrm{d}tq_{r\log^{10k}(1/r)+W_{G_{k}}}(t)$,
it thus remains to prove that the integral in $y$ of the rest of the
expression is bounded, when in addition
$\Delta_{k}>2(r+W_{G_{k}})^{2}\log^{6}(1/r)$. Let us first consider the
integral over $[W_{G_{k}},\infty)$. Indeed, if $y>W_{G_{k}}>-r$, then
$(y-W_{G_{k}})^{2}\leq y^{2}\leq(y+r\log^{10k}(1/r))^{2}$ and consequently

$\displaystyle\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}\frac{(y-W_{G_{k}})^{2}}{2\Delta^{L}_{k}}-\frac{(y-X^{L}_{\Delta^{L}_{k}/2})^{2}}{\Delta^{L}_{k}}-\frac{(y+r\log^{10k}(1/r))^{2}}{2(\Delta_{k}-t)}\Bigr{)}$
$\displaystyle\leq\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}-\frac{(y+r\log^{10k}(1/r))^{2}}{2}\Bigl{(}\frac{1}{\Delta_{k}-t}-\frac{1}{\Delta^{L}_{k}}\Bigr{)}-\frac{(y-X^{L}_{\Delta^{L}_{k}/2})^{2}}{\Delta^{L}_{k}}\Bigr{)}.$

Suppose first that $\Delta^{L}_{k}\geq 2\Delta_{k}$ so $\Delta^{L}_{k}\geq
2(\Delta_{k}-t)$ for any $t\in(0,\Delta_{k})$ and thus
$\frac{1}{\Delta_{k}-t}-\frac{1}{\Delta^{L}_{k}}\geq\frac{1}{2(\Delta_{k}-t)}$.
In this case, by simply removing the last term in the exponential, we have:

$\displaystyle\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}-\frac{(y+r\log^{10k}(1/r))^{2}}{2}\Bigl{(}\frac{1}{\Delta_{k}-t}-\frac{1}{\Delta^{L}_{k}}\Bigr{)}-\frac{(y-X^{L}_{\Delta^{L}_{k}/2})^{2}}{\Delta^{L}_{k}}\Bigr{)}$
$\displaystyle\leq\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}-\frac{(y+r\log^{10k}(1/r))^{2}}{4(\Delta_{k}-t)}\Bigr{)},$

which is the density of a Gaussian law, in particular its integral over the
entire real line equals $1$. If $\Delta^{L}_{k}$ is close to $\Delta_{k}$ and
$t$ is small, one has to be more careful. Suppose now that
$\Delta_{k}\leq\Delta^{L}_{k}\leq 2\Delta_{k}$, then,

$\displaystyle\leq\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}-\frac{(y+r\log^{10k}(1/r))^{2}}{2}\Bigl{(}\frac{1}{\Delta_{k}-t}-\frac{1}{\Delta^{L}_{k}}\Bigr{)}-\frac{(y-X^{L}_{\Delta^{L}_{k}/2})^{2}}{\Delta^{L}_{k}}\Bigr{)}$
$\displaystyle\leq\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}-\frac{(y+r\log^{10k}(1/r))^{2}}{2}\frac{t}{\Delta_{k}(\Delta_{k}-t)}-\frac{(y-X^{L}_{\Delta^{L}_{k}/2})^{2}}{2\Delta_{k}}\Bigr{)}$
$\displaystyle\leq\frac{1}{2\sqrt{\pi\Delta_{k}/2}}\exp\Bigl{(}-\frac{(y-X^{L}_{\Delta^{L}_{k}/2})^{2}}{2\Delta_{k}}\Bigr{)}\mathbf{1}_{\\{0<t\leq\Delta_{k}/2\\}}+\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}-\frac{(y+r\log^{10k}(1/r))^{2}}{4(\Delta_{k}-t)}\Bigr{)}\mathbf{1}_{\\{\Delta_{k}/2<t<\Delta_{k}\\}}.$

Again the integral over the entire real line is constant. Finally, if
$-r<y<W_{G_{k}}$ then $(y-W_{G_{k}})^{2}\leq(r+W_{G_{k}})^{2}$. Here we use
that we work on the event $\Delta_{k}>2(r+W_{G_{k}})^{2}\log^{6}(1/r)\geq
2(y-W_{G_{k}})^{2}\log^{6}(1/r)$. Indeed, this implies:

$\displaystyle\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}\frac{(y-W_{G_{k}})^{2}}{2\Delta^{L}_{k}}-\frac{(y-X^{L}_{\Delta^{L}_{k}/2})^{2}}{\Delta^{L}_{k}}-\frac{(y+r\log^{10k}(1/r))^{2}}{2(\Delta_{k}-t)}\Bigr{)}$
$\displaystyle\leq\frac{1}{2\sqrt{\pi(\Delta_{k}-t)}}\exp\Bigl{(}\frac{1}{4\log^{6}(1/r)}-\frac{(y+r\log^{10k}(1/r))^{2}}{2(\Delta_{k}-t)}\Bigr{)},$

and again the integral over $y$ of the last line is uniformly bounded above.
This completes the proof the bound (3.8).

Let us finally lower bound the probability that the Brownian bridge $W^{L,k}$
reaches $-r$ in the first half of the interval when both $-r<W_{G_{k}}\leq
r\log^{10k}(1/r)$ and $\Delta_{k}>(r\log^{10k}(1/r))^{2}$. We argue similarly:
conditionally on $\Delta^{L}_{k}$, $W_{G_{k}}$, and $W_{D_{k}}$, this
probability equals

$\displaystyle\int_{0}^{\Delta^{L}_{k}/2}\mathop{}\\!\mathrm{d}t\
q_{-(r+W_{G_{k}})}(t)\frac{p_{\Delta^{L}_{k}-t}(W_{D_{k}}+r)}{p_{\Delta^{L}_{k}}(W_{D_{k}}-W_{G_{k}})}$
$\displaystyle=\int_{0}^{\Delta^{L}_{k}/2}\mathop{}\\!\mathrm{d}t\
q_{-(r+W_{G_{k}})}(t)\sqrt{\frac{\Delta^{L}_{k}}{\Delta^{L}_{k}-t}}\exp\Bigl{(}\frac{(W_{D_{k}}-W_{G_{k}})^{2}}{2\Delta^{L}_{k}}-\frac{(W_{D_{k}}+r)^{2}}{2(\Delta^{L}_{k}-t)}\Bigr{)}$
$\displaystyle\geq\int_{0}^{\Delta_{k}/2}\mathop{}\\!\mathrm{d}t\
q_{-(r+W_{G_{k}})}(t)\sqrt{2}\exp\Bigl{(}-\frac{(W_{D_{k}}+r)^{2}}{2(\Delta_{k}-t)}\Bigr{)}.$

Without the exponential, the last integral is bounded away from $0$. We then
lower bound the expectation of the exponential with respect to $W_{D_{k}}$.
Recall that conditionally on $W_{G_{k}}$ and $\Delta_{k}$, the path
$(W_{t};t\in[G_{k},D_{k}])$ has the law of a Brownian motion started from
$W_{G_{k}}$ and conditioned on the event $E_{k}$ that it reaches
$-r\log^{10k}(1/r)$ at some point in the time interval, so by splitting at the
first hitting time of this value, we have since we restrict our attention to
$t\leq\Delta_{k}/2$,

$\displaystyle\mathop{}\\!\mathbb{E}\Bigl{[}\exp\Bigl{(}-\frac{(W_{D_{k}}+r)^{2}}{2(\Delta_{k}-t)}\Bigr{)}\mathbf{1}_{\\{W_{G_{k}},W_{D_{k}}>-r\\}}\bigm{|}W_{G_{k}},\Delta_{k}\Bigr{]}$
$\displaystyle=\frac{1}{\mathop{}\\!\mathbb{P}(E_{k})}\int_{0}^{\Delta_{k}}\mathop{}\\!\mathrm{d}s\int_{-r}^{\infty}\mathop{}\\!\mathrm{d}y\
\frac{q_{-(r\log^{10k}(1/r)+W_{G_{k}})}(s)}{\sqrt{2\pi(\Delta_{k}-s)}}\exp\Bigl{(}-\frac{(y+r\log^{10k}(1/r))^{2}}{2(\Delta_{k}-s)}-\frac{(y+r)^{2}}{2(\Delta_{k}-t)}\Bigr{)}\mathbf{1}_{\\{W_{G_{k}}>-r\\}}$
$\displaystyle\geq\int_{0}^{\Delta_{k}}\mathop{}\\!\mathrm{d}s\int_{-r}^{\infty}\mathop{}\\!\mathrm{d}y\
\frac{q_{-(r\log^{10k}(1/r)+W_{G_{k}})}(s)}{\sqrt{2\pi(\Delta_{k}-s)}}\exp\Bigl{(}-\frac{(y+r\log^{10k}(1/r))^{2}}{2}\Bigl{(}\frac{1}{\Delta_{k}-s}+\frac{1}{\Delta_{k}-t}\Bigr{)}\Bigr{)}\mathbf{1}_{\\{W_{G_{k}}>-r\\}}$
$\displaystyle\geq\int_{0}^{\Delta_{k}/2}\mathop{}\\!\mathrm{d}s\int_{-r}^{\infty}\mathop{}\\!\mathrm{d}y\
\frac{q_{-(r\log^{10k}(1/r)+W_{G_{k}})}(s)}{\sqrt{2\pi\Delta_{k}}}\exp\Bigl{(}-2\frac{(y+r\log^{10k}(1/r))^{2}}{\Delta_{k}}\Bigr{)}\mathbf{1}_{\\{W_{G_{k}}>-r\\}}.$

Consider the integral over $y$ only in the last line, and without de density
$q$. It equals half the probability that a Gaussian random variable with mean
$-r\log^{10k}(1/r)$ and variance $\Delta_{k}/4$ is larger than $-r$, or
equivalently that a Gaussian random variable with mean $0$ and variance $1$ is
larger than $2(r\log^{10k}(1/r)-r)/\sqrt{\Delta_{k}}$. Since
$\Delta_{k}>(r\log^{10k}(1/r))^{2}$, this probability is indeed bounded away
from $0$ uniformly in $k$ and $r$. Then the integral of $q$ over $s$ equals
the probability that a standard Brownian motion reaches
$-(W_{G_{k}}+r\log^{10k}(1/r))$ in less than $\Delta_{k}/2$ when
$W_{G_{k}}\leq r\log^{10k}(1/r)\leq\sqrt{\Delta_{k}}$, which is again bounded
away from $0$.

Combined with the results on $X^{L,k}$, we infer that there exists some
constant $c>0$ such that for all $r<1$, on each interval $[G_{k},D_{k}]$ with
$W_{G_{k}}\leq r\log^{10k}(1/r)$ and $\Delta_{k}\geq(r\log^{10k}(1/r))^{2}$,
there is a probability at least $c$ that $W^{L,k}$ reaches the value $-r$ in
the first half of the interval. The same holds for $X^{L,k}$ and since they
are independent, they both reach the value $-r$ in the first half with
probability at least $c^{2}$. We can then conclude from Lemma 3.12 that this
occurs for a positive proportion of indices $k\leq K_{r}$. ∎

##  4. Computation of the dimensions

Throughout this section, we let $\mathscr{L}(X^{\mathrm{ex}})$ denote the
looptree constructed from the excursion $X^{\mathrm{ex}}$ and as in the
previous section, we let $(M,d_{M},p_{M})$ denote a subsequential limit of
discrete random maps, related to a pair $(X^{\mathrm{ex}},Z^{\mathrm{ex}})$,
from Theorem 2.1 and Theorem 2.4. In addition, thanks to Skorokhod’s
representation theorem, we assume that all convergences in these theorems hold
in the almost sure sense. Our main goal is to compute the dimensions of these
looptrees and maps stated in Theorem 1.1 and Theorem 1.2 respectively. The
upper bound on the dimensions will follow from Hölder continuity estimates
established in Section 4.1 while the lower bounds will follow in Section 4.2
from the estimates on the volume of balls with small radius established in the
previous section.

###  4.1. Hölder continuity of the distances

Recall from (1.1) the upper and lower exponents $\boldsymbol{\eta}$ and
$\boldsymbol{\gamma}$ of $\psi$ at infinity. The main goal of this subsection
is to prove that the canonical projections
$[0,1]\to\mathscr{L}(X^{\mathrm{ex}})$ and $[0,1]\to M$ are almost surely
Hölder continuous with exponent $a$ and $a/2$ respectively for any
$a\in(0,1/\boldsymbol{\eta})$. We shall need the following tail bound.

######  Lemma 4.1.

There exists a function $h\colon(0,\infty)\to(0,\infty)$ and a constant $C>0$
such that for every $t,x>0$,

$\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{s\in[0,t]}|X_{s}|>x\Bigr{)}\leq C\cdot
t\cdot h(x).$

Furthermore, it has the following asymptotic behaviour near $0$:

$\inf\Bigl{\\{}c\geq 0\colon\limsup_{x\downarrow
0}x^{c}h(x)=0\Bigr{\\}}=\boldsymbol{\eta}\qquad\text{and}\qquad\inf\Bigl{\\{}c\geq
0\colon\liminf_{x\downarrow 0}x^{c}h(x)=0\Bigr{\\}}=\boldsymbol{\gamma}.$
(4.1)

###### Proof.

Let us first assume that $X$ has no drift and no Gaussian component. We then
deduce the general case in the end. The bound on the tail probability is given
by Pruitt [Pru81], even in the case where $X$ may also have negative jumps,
see Equation 3.2 there. By combining [Pru81, Equation 3.4] with the work
[BG61], one can derive the upper exponent $\boldsymbol{\eta}$ for $h$, but
only lower bound the lower exponent of $h$ by $\boldsymbol{\gamma}$. Let us
give an alternative proof and prove that $\boldsymbol{\gamma}$ is precisely
the lower exponent of $h$. The function $h$ is given in [Pru81] (up to the
choice of the convention for the Lévy–Khintchine formula) by:

$h(x)=\overline{\pi}(x)+x^{-2}\int_{(0,x]}r^{2}\pi(\mathop{}\\!\mathrm{d}r)+x^{-1}\int_{(x,\infty)}r\pi(\mathop{}\\!\mathrm{d}r)$

where we have set $\overline{\pi}(x)=\pi((x,\infty))$. We claim that

$h(x)\asymp\psi(1/x)\qquad\text{in }(0,1),$ (4.2)

by which we mean that there exist two constants $c_{1},c_{2}>0$ such that
$c_{1}\psi(1/x)\leq h(x)\leq c_{2}\psi(1/x)$ for every $x\in(0,1)$. The
exponents of $h$ will then follow from those of $\psi$ in (1.1).

To establish (4.2) we rely on [Duq12, Section 2.3]. Indeed, recall that
$\phi=\psi^{\prime}$ is the Laplace exponent of a subordinator which has Lévy
measure $\nu(\mathop{}\\!\mathrm{d}r)=r\pi(\mathop{}\\!\mathrm{d}r)$. Then
$h(x)-\overline{\pi}(x)$ equals precisely $x^{-2}J_{\psi^{\prime}}(x)$ in the
notation of [Duq12, Equation 47]. Recall the notation
$\widetilde{\psi}(u)=\psi(u)/u$. By combining Equations 47 and 50 in [Duq12],
we infer that

$h(x)-\overline{\pi}(x)=x^{-2}J_{\psi^{\prime}}(x)\asymp
x^{-1}\psi^{\prime}(1/x)\asymp x^{-1}\widetilde{\psi}(1/x)=\psi(1/x),$

in $(0,1)$. This already gives the lower bound in (4.2) by simply lower
bounding $\overline{\pi}$ by $0$. As for the upper bound, simply observe that
for every $x>0$, we have $\overline{\pi}(x)\leq
x^{-1}\int_{(x,\infty)}r\pi(\mathop{}\\!\mathrm{d}r)$ so $\overline{\pi}$ is
also upper bounded by some constant times $\psi(1/\cdot)$ by the previous
argument. This concludes the proof of (4.2) when $\beta=0$.

In the case where $X$ has a drift $d\neq 0$ and a Gaussian parameter
$\beta>0$, the Lévy–Itō decomposition reads $X=D+W+Y$ in distribution, where
$D_{t}=dt$, where $W$ is $\sqrt{\beta}$ times a standard Brownian motion, and
where $Y$ is independent of $W$ and has no drift nor Gaussian component.
Recall that in this case, we have $\psi(\lambda)\sim\beta\lambda^{2}$ as
$\lambda\to\infty$. By the first part of the proof, $Y$ satisfies the claim
with some $1\leq\boldsymbol{\gamma}\leq\boldsymbol{\eta}\leq 2$. So does $W$,
with $h(x)=x^{-2}$, by e.g. an application of the reflection principle and a
well-known tail bound for Gaussian random variables:

$\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{s\in[0,t]}|W_{s}|>x\Bigr{)}\leq
4\mathop{}\\!\mathbb{P}(W_{t}>x)\leq 4\exp\Bigl{(}-\frac{x^{2}}{2\beta
t}\Bigr{)}\leq\frac{4}{\mathrm{e}}\frac{2\beta t}{x^{2}}.$

The claim, with $h(x)=x^{-2}$, then easily follows from a union bound. ∎

######  Remark 4.2.

Let us mention that the lower exponent of $\overline{\pi}$ near $0$ may be
smaller than $\boldsymbol{\gamma}$, even in the case of a subordinator, see
e.g. Theorems 6.1 and 7.1 in [BG61] for some more information.

Recall from [Mar22b, Proposition 3.2] that the distance function
$d_{\mathscr{L}(X^{\mathrm{ex}})}$ from (2.6) is almost surely a continuous
pseudo-distance, so the canonical projection
$\mathbf{p}_{\mathscr{L}(X^{\mathrm{ex}})}\colon[0,1]\to\mathscr{L}(X^{\mathrm{ex}})$
is almost surely continuous. The next proposition shows that it is Hölder
continuous; we will easily derive from this the upper bound for the Hausdorff
and upper Minkowski dimensions of the looptree.

######  Proposition 4.3.

For any $a\in(0,1/\boldsymbol{\eta})$, the projection
$\mathbf{p}_{\mathscr{L}(X^{\mathrm{ex}})}$ is almost surely Hölder continuous
with exponent $a$ in the sense that almost surely, there exists
$C_{a}\in(0,\infty)$ such that:

$d_{\mathscr{L}(X^{\mathrm{ex}})}(s,t)\leq C_{a}\cdot|t-s|^{a}\qquad\text{for
all }s,t\in[0,1].$ (4.3)

We shall see that the exponent $1/\boldsymbol{\eta}$ is optimal in the sense
that $\mathbf{p}_{\mathscr{L}(X^{\mathrm{ex}})}$ is almost surely not Hölder
continuous with exponent larger than $1/\boldsymbol{\eta}$, see Remark 4.7. As
the proof will show, the claim also holds for the unconditioned Lévy process
$X$ and its bridge version.

###### Proof.

We claim that given $a<1/\boldsymbol{\eta}$, there exists a deterministic
constant $K_{a}\in(0,\infty)$ which only depends on $a$, not even on the law
of the Lévy process, such that almost surely, there exists $n_{0}\geq 1$
(random, and which may depend on the law of the Lévy process) such that for
every $n\geq n_{0}$, we have:

$\sup_{0\leq i\leq
2^{n}-1}d_{\mathscr{L}(X^{\mathrm{ex}})}(i2^{-n},(i+1)2^{-n})\leq
K_{a}2^{-an}.$ (4.4)

For $n\leq n_{0}$, there are only finitely many quantities involved, so we
infer that almost surely, there exists a now random $C_{a}\in(0,\infty)$ such
that (4.3) holds for every pair $(s,t)$ of the form $(j2^{-n},(j+1)2^{-n})$
with $n\geq 1$ and $j\in\\{0,\dots,2^{n}-1\\}$. It then extends to all dyadic
numbers by a routing argument, after multiplying $C_{a}$ by $2/(1-2^{-a})$,
and then to the whole interval $[0,1]$ by density, using that
$d_{\mathscr{L}(X^{\mathrm{ex}})}$ is continuous almost surely. It remains to
prove (4.4). Let us first replace $X^{\mathrm{ex}}$ by the unconditioned
process $X$ and then argue by absolute continuity. According to (2.7), almost
surely, for $s<t$, we have:

$d_{\mathscr{L}(X^{\mathrm{ex}})}(s,t)\leq
X_{s}+X_{t-}-2\inf_{r\in[s,t]}X_{r}\leq 4\sup_{r\in[s,t]}|X_{r}-X_{s}|.$

Indeed, informally the term in the middle is the length of the path if one
goes from $s$ to $t$ in the looptree by always following a cycle on its right,
instead of optimising between left and right. Note that the Markov property
implies that the right-hand side has the same law as
$4\sup_{r\in[0,t-s]}|X_{r}|$. This bound is too crude to be used directly
because of the jumps of $X$, but we shall use it between two consecutive large
jumps.

Precisely, fix $\delta>0$ and let
$\theta=(1+\delta)(\boldsymbol{\eta}+\delta)$; we shall take $a=1/\theta$. Fix
$n\geq 1$ and $i\in\\{0,\dots,2^{n}-1\\}$. Let
$(t_{n,i}(1),\dots,t_{n,i}(N_{n,i}))$ denote the increasing enumeration of the
times $\\{t\in[i2^{-n},(i+1)2^{-n}]\colon\Delta X_{t}>2^{-n/\theta}\\}$. Set
also $t_{n,i}(0)=i2^{-n}$ and $t_{n,i}(N_{n,i}+1)=(i+1)2^{-n}$. Then combining
the last display with the triangle inequality, we have:

$d_{\mathscr{L}(X^{\mathrm{ex}})}(i2^{-n},(i+1)2^{-n})\leq
4\sum_{j=1}^{N_{n,i}}\sup_{r\in[t_{n,i}(j),t_{n,i}(j+1)]}|X_{r}-X_{t_{n,i}(j)}|.$
(4.5)

Let us first bound $N_{n,i}$. The latter has the Poisson distribution with
rate $2^{-n}\overline{\pi}(2^{-n/\theta})$ and the Chernoff bound shows that
the tail probability of a Poisson random variable $Y$ with rate $\lambda$
satisfies $\mathop{}\\!\mathbb{P}(Y\geq
k)\leq\operatorname{e}^{-\lambda}(\operatorname{e}\lambda/k)^{k}$ for every
$k>\lambda$. Hence for any integer $N\geq 1$, there exists a constant $C_{N}$,
which depends on $N$, such that:

$\mathop{}\\!\mathbb{P}(N_{n,i}\geq N)\leq
C_{N}(2^{-n}\overline{\pi}(2^{-n/\theta}))^{N}.$

Now recall from the proof Lemma 4.1 that $\overline{\pi}$ is upper bounded by
some constant times $\psi(1/\cdot)$, so there exists a constant $C_{\delta}>0$
such that $\overline{\pi}(x)\leq
C_{\delta}x^{-(\boldsymbol{\eta}+\delta)}=C_{\delta}x^{-\theta/(1+\delta)}$
for every $x\in(0,1)$. We infer that

$\mathop{}\\!\mathbb{P}(N_{n,i}\geq N)\leq
C_{N}C_{\delta}^{N}2^{-nN\delta/(1+\delta)}.$

Let us next bound the variation of $X$ between $t_{n,i}(j)$ and
$t_{n,i}(j+1)$. By definition, between these times, the process $X$ makes no
jump larger than $2^{-n/\theta}$, therefore, at the first passage time above
this level, the value is at most $2\times 2^{-n/\theta}$. Using the strong
Markov property, one then easily gets by induction that for every integer
$M\geq 1$, it holds:

$\displaystyle\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{r\in[0,t_{n,i}(j+1)-t_{n,i}(j)]}|X_{r}|>2M2^{-n/\theta}\bigm{|}t_{n,i}(j),t_{n,i}(j+1)\Bigr{)}\leq\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{r\in[0,t_{n,i}(j+1)-t_{n,i}(j)]}|X_{r}|>2^{-n/\theta}\bigm{|}t_{n,i}(j),t_{n,i}(j+1)\Bigr{)}^{M}.$

We next upper bound the probability on the right thanks to Lemma 4.1. There
exists a constant $C_{\delta}>0$ such that, with the constant $C$ and the
function $h$ from this lemma, we have $Ch(x)\leq
C_{\delta}x^{-(\boldsymbol{\eta}+\delta)}=C_{\delta}x^{-\theta/(1+\delta)}$
for every $x\in(0,1)$. Hence

$\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{r\in[0,t_{n,i}(j+1)-t_{n,i}(j)]}|X_{r}|>2^{-n/\theta}\bigm{|}t_{n,i}(j),t_{n,i}(j+1)\Bigr{)}\leq
C2^{-n}h(2^{-n/\theta})\leq C_{\delta}2^{-n\delta/(1+\delta)}.$

Combined with union bounds, we obtain:

$\displaystyle\mathop{}\\!\mathbb{P}\Bigl{(}N_{n,i}<N\quad\text{and}\quad
4\sum_{j=0}^{N_{n,i}}\sup_{r\in[t_{n,i}(j),t_{n,i}(j+1)]}|X_{r}-X_{t_{n,i}(j)}|>8N^{2}2^{-n/\theta}\Bigr{)}$
$\displaystyle\qquad\leq\mathop{}\\!\mathbb{P}\Bigl{(}N_{n,i}<N\quad\text{and}\quad\exists
j\leq
N_{n,i}\colon\sup_{r\in[t_{n,i}(j),t_{n,i}(j+1)]}|X_{r}-X_{t_{n,i}(j)}|>2N2^{-n/\theta}\Bigr{)}$
$\displaystyle\qquad\leq NC_{\delta}^{M}2^{-nN\delta/(1+\delta)}.$

We have thus shown that for every $N\geq 1$, there exists a constant $K$ which
depends on $N$ and $\delta$, but neither on $i$ nor $n$, such that

$\mathop{}\\!\mathbb{P}(d_{\mathscr{L}(X^{\mathrm{ex}})}(i2^{-n},(i+1)2^{-n})>8N^{2}2^{-n/\theta})\leq
K2^{-nN\delta/(1+\delta)}.$

A further union bound implies then:

$\mathop{}\\!\mathbb{P}(\exists i\leq 2^{n}-1\colon
d_{\mathscr{L}(X^{\mathrm{ex}})}(i2^{-n},(i+1)2^{-n})>8N^{2}2^{-n/\theta})\leq
K2^{n-nN\delta/(1+\delta)}.$

If we take $N$ such that $N\geq 2+\frac{1}{\delta}$, then the sum over all
$n\geq 0$ of the right-hand side is finite; the Borel–Cantelli lemma then
yields (4.4) for the unconditioned Lévy process.

Actually, more than (4.4) we have shown that its upper bound (4.5), which is
expressed in a simple way in terms of the Lévy process, is itself bounded as
in (4.4). This immediately extends to the first half the bridge
$(X^{\mathrm{br}}_{t};0\leq t\leq 1/2)$, taking only the supremum over
$i\in\\{0,\dots,2^{n-1}\\}$, by absolutely continuity with respect to the
unconditioned process $(X_{t};0\leq t\leq 1/2)$. In addition, the time- and
space-reversal of the second half of the bridge has the same law as the first
half, so the upper bound also holds for the second half and thus for the
entire bridge. Finally we transfer it to the excursion by applying the Vervaat
transform. Notice that the latter changes the starting point and thus shifts
the intervals, but this only mildly affects the constant $C_{a}$. ∎

######  Remark 4.4.

Note that if one wanted to replace the exponent $1/\boldsymbol{\eta}$ by the
smaller one $1/\boldsymbol{\gamma}$, then instead of the uniform upper bound
$h(x)\leq C_{\delta}x^{-(\boldsymbol{\eta}+\delta)}$ for every $x\in(0,1)$,
one would only have the existence of a sequence $x_{n}\downarrow 0$ along
which $\overline{\pi}(x_{n})\leq h(x_{n})\leq
C_{\delta}x_{n}^{-(\boldsymbol{\gamma}+\delta)}$. Taking a subsequence if
necessary, we may assume that $\sum_{n}x_{n}<\infty$ and then the preceding
argument shows that instead of (4.4) for any $a\in(0,1/\boldsymbol{\gamma})$,
almost surely there exist $K_{a}>0$ and a sequence $\varepsilon_{n}\downarrow
0$ such that, for every $n$ large enough,

$d_{\mathscr{L}(X^{\mathrm{ex}})}(i\varepsilon_{n},(i+1)\varepsilon_{n})\leq
K_{a}\varepsilon_{n}^{a}\qquad\text{for all}\enskip
i\in\\{0,\dots,\lfloor\varepsilon_{n}^{-1}\rfloor-1\\}.$

This will be used to upper bound the packing and lower Minkowski dimensions.

We then turn to the map $M$ and the label process $Z^{\mathrm{ex}}$.
Informally, we simply loose a factor $1/2$ due to the regularity of the
Brownian labels.

######  Corollary 4.5.

For any $a\in(0,1/(2\boldsymbol{\eta}))$, the label process
$(Z^{\mathrm{ex}}_{t})_{t\in[0,1]}$ has a modification that is Hölder
continuous with exponent $a$.

As for looptree distances, the claim also holds for the label process
associated with the unconditioned Lévy process or its bridge version.

###### Proof.

Recall the moment bound on the Brownian labels (2.9), then if the Hölder
continuity (4.3) on the looptree distances held true with a deterministic
constant $C_{a}$, then we could apply the classical Kolmogorov continuity
criterion, showing that for any $a<1/\boldsymbol{\eta}$ and any $q>0$, the
process $(Z^{\mathrm{ex}}_{t})_{t}$ has a modification that is Hölder
continuous with any exponent $b<q^{-1}(aq/2-1)$, which is arbitrarily close to
$1/(2\boldsymbol{\eta})$ when $q$ is large. The idea then is to adapt the
proof this criterion, relying on (4.4) where the constant $K_{a}$ is indeed
deterministic.

As in the previous proof, it suffices to prove the claim in the infinite
volume setting by replacing $X^{\mathrm{ex}}$ by the unconditioned Lévy
process $X$ and constructing the associated looptree and label process $Z$. By
absolute continuity, it then extends to the bridge, and finally to the
excursion by the Vervaat transform. Fix $a<1/\boldsymbol{\eta}$ and $q>0$ and
let $b<q^{-1}(aq/2-1)$ and let $A_{n}$ be the event

$A_{n}=\Bigl{\\{}\sup_{0\leq i\leq
2^{n}-1}d_{\mathscr{L}(X)}(i2^{-n},(i+1)2^{-n})\leq K_{a}2^{-an}\Bigr{\\}}$

from (4.4), where $K_{a}$ is a constant chosen such that
$\sum_{n}\mathop{}\\!\mathbb{P}(A_{n}^{c})<\infty$. Then using a union bound,
the Markov inequality for the conditional expectation given the process $X$,
and the moment bound (2.9), we have for every $n\geq 1$:

$\displaystyle\mathop{}\\!\mathbb{P}\Bigl{(}A_{n}\text{ and }\sup_{0\leq i\leq
2^{n}-1}|Z_{i2^{-n}}-Z_{(i+1)2^{-n}}|>2^{-nb}\Bigr{)}$ $\displaystyle\leq
2^{n}\sup_{0\leq i\leq 2^{n}-1}\mathop{}\\!\mathbb{P}(A_{n}\text{ and
}|Z_{i2^{-n}}-Z_{(i+1)2^{-n}}|^{q}>2^{-nbq})$ $\displaystyle\leq
2^{n}\sup_{0\leq i\leq
2^{n}-1}2^{nbq}\mathop{}\\!\mathbb{E}[\mathbf{1}_{\\{A_{n}\\}}\mathop{}\\!\mathbb{E}[|Z_{i2^{-n}}-Z_{(i+1)2^{-n}}|^{q}\mid
X]]$ $\displaystyle\leq 2^{n}\sup_{0\leq i\leq
2^{n}-1}2^{nbq}K_{q}\mathop{}\\!\mathbb{E}[\mathbf{1}_{\\{A_{n}\\}}d_{\mathscr{L}(X)}(i2^{-n},(i+1)2^{-n})^{q/2}]$
$\displaystyle\leq 2^{n}2^{nbq}K_{q}K_{a}^{q/2}2^{-anq/2}$ $\displaystyle\leq
K_{q}K_{a}^{q/2}2^{n(1+bq-aq/2)},$

which is a convergent series since $1+bq-aq/2<0$. Since
$\sum_{n}\mathop{}\\!\mathbb{P}(A_{n}^{c})<\infty$, then we infer that

$\sum_{n\geq 1}\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{0\leq i\leq
2^{n}-1}|Z_{i2^{-n}}-Z_{(i+1)2^{-n}}|>2^{-nb}\Bigr{)}<\infty.$

From here we can catch up with the proof the classical Kolmogorov continuity
criterion: the Borel–Cantelli lemma shows that almost surely, these events
fail eventually; we then extend the bound to any pair of dyadic numbers and
then to any times by density. ∎

From now on, we always deal with this Hölder continuous modification of $Z$
and $Z^{\mathrm{ex}}$. This further implies that the distances in the map are
Hölder continuous.

######  Proposition 4.6.

For any $a\in(0,1/(2\boldsymbol{\eta}))$, the projection $\mathbf{p}_{M}$ is
almost surely Hölder continuous with exponent $a$ in the sense that almost
surely, there exists $C_{a}\in(0,\infty)$ such that:

$d_{M}(s,t)\leq C_{a}\cdot|t-s|^{a}\qquad\text{for all }s,t\in[0,1].$ (4.6)

As for Proposition 4.3, we shall see that $\mathbf{p}_{M}$ is almost surely
not Hölder continuous with exponent larger than $1/(2\boldsymbol{\eta})$, see
again Remark 4.7.

###### Proof.

Define for every $0\leq s\leq t\leq 1$ a pseudo-distance by:

$D^{\circ}(s,t)=D^{\circ}(t,s)=Z^{\mathrm{ex}}_{s}+Z^{\mathrm{ex}}_{t}-2\inf_{[s,t]}Z^{\mathrm{ex}}.$
(4.7)

Then the Hölder continuity of $Z^{\mathrm{ex}}$ from Corollary 4.5 transfers
to $D^{\circ}$ which thus satisfies the upper bound (4.6). Our claim follows
then from the bound

$d_{M}\leq D^{\circ}.$

Indeed, a similar bound holds in the discrete setting, see e.g. Le Gall [LG07,
Lemma 3.1] in a slightly different context, and it passes to the limit. To
explain it, in the construction of a discrete map from a labelled looptree,
the labels on the looptree codes the distances in the map to a distinguished
vertex. Precisely, the chain of successors define for every vertex a special
geodesic path to the distinguished vertex. Then if one takes two arbitrary
vertices and follows these two special geodesic paths starting from each of
them until they meet, then the sum of the two lengths is precisely the
discrete analogue of $D^{\circ}$, and is clearly not smaller than the graph
distance in the map between the two vertices. ∎

###  4.2. Fractal dimensions

We are now ready to compute the dimensions of our spaces. Recall that Theorem
1.1 about the looptrees precisely states that almost surely, we have:

$\dim_{H}\mathscr{L}(X^{\mathrm{ex}})=\operatorname{\underline{\dim}}\mathscr{L}(X^{\mathrm{ex}})=\boldsymbol{\gamma}\qquad\text{and}\qquad\dim_{p}\mathscr{L}(X^{\mathrm{ex}})=\operatorname{\overline{\dim}}\mathscr{L}(X^{\mathrm{ex}})=\boldsymbol{\eta}.$

For the maps, Theorem 1.2 states that the dimensions are twice these
quantities.

###### Proof Theorem 1.1 and Theorem 1.2.

Let us refer to [Mat95] for the definition and basic properties of these
dimensions. Let us only recall here the following relations between these
notions of dimension:

$\dim_{H}\leq\operatorname{\underline{\dim}}\qquad\text{and}\qquad\dim_{H}\leq\dim_{p}\leq\operatorname{\overline{\dim}}.$

It thus suffices to upper bound $\operatorname{\underline{\dim}}$ and
$\operatorname{\overline{\dim}}$ and to lower bound $\dim_{H}$ and $\dim_{p}$.

The upper bound for $\operatorname{\overline{\dim}}$ is an immediate
consequence of the Hölder continuity provided by Proposition 4.3 and
Proposition 4.6. Indeed, fix $\delta>0$ and let $N_{n}=\lfloor
2^{n(\boldsymbol{\eta}+\delta)}\rfloor$; then by the former result, almost
surely, for every $1\leq i\leq N_{n}$, the diameter of the image of
$[iN_{n}^{-1},(i-1)N_{n}^{-1}]$ in the looptree is small compared to $2^{-n}$.
Hence, almost surely, for every $n$ large enough, the space
$\mathscr{L}(X^{\mathrm{ex}})$ can be covered by $N_{n}$ balls with radius
$2^{-n}$ and this implies that
$\operatorname{\overline{\dim}}\mathscr{L}(X^{\mathrm{ex}})\leq\boldsymbol{\eta}+\delta$
almost surely with $\delta$ arbitrarily small. The upper bound
$\operatorname{\underline{\dim}}\mathscr{L}(X^{\mathrm{ex}})\leq\boldsymbol{\gamma}$
is similar, using Remark 4.4: with the sequence $(\varepsilon_{n})_{n\geq 1}$
from this remark, we may apply the previous reasoning with
$N_{n}=\lfloor\varepsilon_{n}^{-\boldsymbol{\gamma}-\delta}\rfloor$. The upper
bounds for the maps are proved similarly using Proposition 4.6 and the
analogue of Remark 4.4, which is derived from the latter.

The lower bounds on the Hausdorff and packing dimensions follow from our
estimate on the volume of small balls. Recall that we associate times on
$[0,1]$ by their projection in the looptree; let then $B(u,r)$ denote the ball
centred at $u\in[0,1]$ and with radius $r\geq 0$ for the looptree distance.
Let also $|\cdot|$ denote the Lebesgue measure on $[0,1]$. Let $U$ be an
independent uniform random time on $[0,1]$ and $\delta>0$. We read from
Proposition 3.1 that almost surely, for every $n$ large enough, we have:

$|B(U,2^{-n})|\leq\psi(2^{n(1-\delta)})^{-1}.$

By definition of $\boldsymbol{\gamma}$, we know that
$2^{-n(1-\delta)(\boldsymbol{\gamma}-\delta)}\psi(2^{n(1-\delta)})\to\infty$.
Therefore, almost surely, for every $r>0$ small enough, if $2^{-(n+1)}\leq
r\leq 2^{-n}$, then

$|B(U,r)|\leq|B(U,2^{-n})|=o(2^{-n(1-\delta)(\boldsymbol{\gamma}-\delta)})=o(r^{(1-\delta)(\boldsymbol{\gamma}-\delta)}).$

On the other hand, by definition of $\boldsymbol{\eta}$, we know that
$2^{-n(1-\delta)(\boldsymbol{\eta}-\delta/2)}\psi(2^{n(1-\delta)})$ does not
tend to $0$, so there is exists a subsequence which is bounded away from $0$
and we infer similarly that for some $r_{n}\to 0$, it holds:

$|B(U,\varepsilon_{n})|=o(r_{n}^{(1-\delta)(\boldsymbol{\eta}-\delta)}).$

This shows that almost surely, for Lebesgue almost every $u\in[0,1]$, we have
both:

$\limsup_{r\downarrow
0}\frac{|B(u,r)|}{r^{(1-\delta)(\boldsymbol{\gamma}-\delta)}}=0\qquad\text{and}\qquad\liminf_{r\downarrow
0}\frac{|B(u,r)|}{r^{(1-\delta)(\boldsymbol{\eta}-\delta)}}=0.$

We conclude from density theorems that
$\dim_{H}\mathscr{L}(X^{\mathrm{ex}})\geq\boldsymbol{\gamma}-\delta$ and
$\dim_{p}\mathscr{L}(X^{\mathrm{ex}})\geq\boldsymbol{\eta}-\delta$, see e.g.
[Mat95, Theorem 6.9 and 6.11]. Again, this is adapted to the maps by simply
replacing the estimate from Proposition 3.1 by that from Proposition 3.2. ∎

######  Remark 4.7.

Notice that the upper bounds on the Hausdorff dimensions only rely on the
Hölder continuity results and extend as far as the latter do. Therefore if the
canonical projection on the looptree (resp. on the map) was Hölder continuous
with exponent $a>1/\boldsymbol{\eta}$ (resp. $a>1/(2\boldsymbol{\eta}))$, then
we would upper bound the dimension by $1/a<\boldsymbol{\eta}$ (resp.
$1/a<2\boldsymbol{\eta})$, and this would contradict the lower bounds. In
other words, if the bound (4.3) held with $a>1/\boldsymbol{\eta}$, then by
taking $\delta>0$ small enough, this would contradict the liminf in the last
display of the previous proof. Therefore the exponents in Proposition 4.3 and
Proposition 4.6 are optimal.

##  5. Approximations in the continuum world

In this section we ask ourselves the following question. Suppose that for all
$\lambda>0$ we are given a Lévy process $X^{(\lambda)}$ with no negative jump
and infinite variation paths, with exponent $\psi^{(\lambda)}$, and another
one $X$ with exponent $\psi$. Suppose that $X^{(\lambda)}$ and $X$ satisfy the
integrability condition (1.2), so they all have continuous transition
densities $p^{(\lambda)}_{t}(x)$ and $p_{t}(x)$ respectively, and we can
define bridges and excursions as before. If these transition densities
converge, then it is well known that the Lévy processes converge in law; but
do the corresponding looptrees, Gaussian labels, and continuum maps converge
as well? If the limit $X$ has no Gaussian part, then it has the pure jump
property from [Mar22b] and Section 4 there answers positively the question.
Here we consider the general setting: we prove the convergence of labelled
looptrees in Section 5.1 and that of maps (along a subsequence) in Section
5.2.

A typical example of application we have in mind is the case where
$X^{(\lambda)}$ is an $\alpha$-stable spectrally positive Lévy process with
drift $\lambda$, with varying $\alpha$ and $\lambda$, studied in Section 6.

Let us start as a warmup with an easy lemma.

######  Lemma 5.1.

Suppose that for every $t\geq 0$, the transition density $p^{(\lambda)}_{t}$
converges uniformly as $\lambda\to\infty$ to $p_{t}$. Then $X^{(\lambda)}\to
X$ in distribution for the Skorokhod $J_{1}$ topology under the unconditional
law as well as under $\mathop{}\\!\mathbb{P}(\,\cdot\mid
X^{(\lambda)}_{1}=\vartheta)$ for any $\vartheta\in\mathbb{R}$, and finally
under the excursion law.

###### Proof.

The convergence of the densities implies the convergence in distribution
$X^{(\lambda)}_{t}\to X_{t}$ for any $t\geq 0$ fixed under the unconditional
law. Since they are Lévy processes, this implies the convergence in
distribution for the Skorokhod $J_{1}$ topology, see e.g. [JS03, Chapter VII]
for details. Next, fix $\vartheta\in\mathbb{R}$, then for every
$\varepsilon\in(0,1)$, recall that the bridges satisfy the absolute continuity
relation:

$\mathop{}\\!\mathbb{E}[F(X^{(\lambda)}_{t};t\in[0,1-\varepsilon])\mid
X^{(\lambda)}_{1}=\vartheta]=\mathop{}\\!\mathbb{E}\Bigl{[}F(X^{(\lambda)}_{t};t\in[0,1-\varepsilon])\cdot\frac{p^{(\lambda)}_{\varepsilon}(\vartheta-X^{(\lambda)}_{1-\varepsilon})}{p^{(\lambda)}_{1}(\vartheta)}\Bigr{]},$

for any continuous and bounded function $F$. We infer from the convergence in
distribution of the unconditioned processes and the uniform convergence
$p^{(\lambda)}_{\varepsilon}\to p_{\varepsilon}$ that the right-hand side
converges to the same quantity for $X$, hence the bridges converge on any
interval $[0,1-\varepsilon]$. Notice then that the bridges are invariant under
space- and time-reversal (since Lévy processes are), so the part on the
interval $[1-\varepsilon,1]$ converges as well, hence both are tight and the
entire bridge on $[0,1]$ is tight. The convergence on any interval
$[0,1-\varepsilon]$ then characterises uniquely the subsequential limits.
Finally, the convergence of bridges for $\vartheta=0$ is transferred to the
excursions by the Vervaat transform. ∎

In practice, the uniform convergence of transition densities can be often be
checked by using the inverse Fourier transform. Precisely, we may write:

$\sup_{x\in\mathbb{R}}|p^{(\lambda)}_{t}(x)-p_{t}(x)|\leq\frac{1}{2\pi}\int_{-\infty}^{\infty}|\mathop{}\\!\mathbb{E}[\operatorname{e}^{iuX^{(\lambda)}_{t}}]-\mathop{}\\!\mathbb{E}[\operatorname{e}^{iuX_{t}}]|\mathop{}\\!\mathrm{d}u.$

The right-hand side tends to $0$ provided the integrand does and one can e.g.
apply the dominated convergence theorem. This will be the case in our
application in Section 6.

We assume for the rest of this section that the assumption of Lemma 5.1 is in
force.

###  5.1. Convergence of looptrees and labels

Let us first study the convergence of the looptrees and Gaussian labels
associated with $X^{(\lambda)}$ and $X$. Recall from (2.6) and (2.8)
respectively that they are defined using a parameter which acts as a
multiplicative constant on the continuous part. In the next theorem, we fix
the value of this constant: $1/2$ for looptrees and $1/3$ for the labels. We
discuss this restriction after the statement.

######  Theorem 5.2.

Suppose that for every $t\geq 0$, the transition density $p^{(\lambda)}_{t}$
converges uniformly as $\lambda\to\infty$ to $p_{t}$. Under the unconditional
law or the bridge or excursion law, we have jointly with the convergence in
distribution $X^{(\lambda)}\to X$:

$\mathscr{L}^{1/2}(X^{(\lambda)})\enskip\mathop{\longrightarrow}^{(d)}_{\lambda\to\infty}\enskip\mathscr{L}^{1/2}(X)\qquad\text{and}\qquad
Z^{(\lambda,1/3)}\enskip\mathop{\longrightarrow}^{(d)}_{\lambda\to\infty}\enskip
Z^{1/3}$

for the Gromov–Hausdorff–Prokhorov topology and for the uniform topology
respectively.

Recall that the factor $1/2$ in the looptree above refers to the
multiplicative constant applied to the continuous part in (2.6) and similarly
in (2.8) for $1/3$ in the label process. First notice that in absence of jumps
in the limit, namely if $X$ is a scaled Brownian excursion, then for any
$c>0$, the looptree $\mathscr{L}^{c}(X)=c\,\mathscr{L}^{1}(X)$ is a scaled
Brownian tree and $Z^{c}=\sqrt{c}\,Z$ is a scaled Brownian snake so the choice
of $c$ amounts to fix a normalisation. In the opposite case, if this
continuous part in the limit vanishes, then the choice of this constant is
irrelevant. Similarly, and this will be the case in our application, if the
continuous part in $X^{(\lambda)}$ vanishes, then this constant is irrelevant
in the sequence. The reason for this restriction is the following. The
continuous part for $X$ appears as the limit of the continuous part of
$X^{(\lambda)}$ together with sum of quantities of the form $R^{t}_{s}$ as
defined in (2.3), corresponding to small jumps of $X^{(\lambda)}$. Now
typically, the contribution to the looptree distances of these small jumps
will be half of the sum of these $R^{t}_{s}$’s by symmetry, since the latter
code the right-length of the cycles. Hence in order to control the looptrees
with a different multiplicative constant than $1/2$, we must understand the
precise contribution of the continuous part of $X^{(\lambda)}$ and its small
jumps to the continuous part of $X$, whereas with the constant $1/2$, in the
looptree, they both contribute equally.

Let us first provide a tightness argument. We note that the latter does not
require to restrict to the parameters $1/2$ and $1/3$.

###### Proof tightness.

Thanks to the upper bound (2.7) on looptree distances, the convergence of the
excursions implies tightness of the associated looptrees. Indeed, for any
$s,s^{\prime},t,t^{\prime}\in[0,1]$ we have by the triangle inequality:

$\displaystyle|d^{1/2}_{\mathscr{L}(X^{(\lambda)})}(s,t)-d^{1/2}_{\mathscr{L}(X^{(\lambda)})}(s^{\prime},t^{\prime})|$
$\displaystyle\leq
d^{1/2}_{\mathscr{L}(X^{(\lambda)})}(s,s^{\prime})+d^{1/2}_{\mathscr{L}(X^{(\lambda)})}(t,t^{\prime})$
$\displaystyle\leq
d_{X^{(\lambda)}}(s,s^{\prime})+d_{X^{(\lambda)}}(t,t^{\prime}).$

The last line converges in distribution to
$d_{X}(s,s^{\prime})+d_{X}(t,t^{\prime})$ under the unconditional law or the
bridge or excursion law by Lemma 5.1. This limit in turn is arbitrarily small
when the pairs $(s,t)$ and $(s^{\prime},t^{\prime})$ are arbitrarily close to
each other (for the Euclidean distance). Tightness of the looptree distances
follows then easily from this control on the modulus of continuity.

Tightness of the label process is less immediate. Let us prove it providing
that $d^{1/2}_{\mathscr{L}(X^{(\lambda)})}\to d^{1/2}_{\mathscr{L}(X)}$ in
distribution, which we shall prove below. Our aim is again to control the
modulus of continuity. Precisely, let us fix $\varepsilon,\theta>0$ and let us
prove that there exists $\delta>0$ such that for every $\lambda$ large enough,
we have:

$\mathop{}\\!\mathbb{P}(\exists(s,t)\in[0,1]^{2}\colon|t-s|\leq\delta\quad\text{and}\quad|Z^{(\lambda,1/3)}_{s}-Z^{(\lambda,1/3)}_{t}|>\theta)\leq
10\varepsilon.$

Let $C_{a}$ be the random variable as in (4.3) for the limit process $X$, for
some fixed admissible $a>0$. Since $C_{a}<\infty$ almost surely then there
exists a deterministic constant $M\in(0,\infty)$ such that $C_{a}\leq M$ with
probability at least $1-\varepsilon$. Let $C>0$ which will be chosen large
enough, depending only on $\theta$ and $M$ (and thus $\varepsilon$), and let
us set $\kappa=(\theta/C)^{2}$ and then
$N=\lceil(M/\kappa)^{1/a}\rceil=\lceil(MC^{2}/\theta^{2})^{1/a}\rceil$; we
shall take $\delta=1/N$. Consider the event:

$G(N,\kappa,\lambda)=\\{\forall(s,t)\in[0,1]^{2}\enskip\text{such
that}\enskip|t-s|\leq 1/N\enskip\text{it holds}\enskip
d^{1/2}_{\mathscr{L}(X^{(\lambda)})}(s,t)\leq\kappa\\},$

and let us write $G(N,\kappa)$ for the same event for the process $X$. It
follows from the convergence of looptree distances, the bound (4.3), and our
choice of constants that

$\mathop{}\\!\mathbb{P}(G(N,\kappa,\lambda))\enskip\mathop{\longrightarrow}_{\lambda\to\infty}\enskip\mathop{}\\!\mathbb{P}(G(N,\kappa))\geq
1-\varepsilon.$

We shall work on the event $G(N,\kappa,\lambda)$ and upper bound

$\mathop{}\\!\mathbb{P}(\\{\exists(s,t)\in[0,1]^{2}\colon|t-s|\leq
1/N\quad\text{and}\quad|Z^{(\lambda,1/3)}_{s}-Z^{(\lambda,1/3)}_{t}|>C\sqrt{\kappa}\\}\cap
G(N,\kappa,\lambda)).$

By the triangle inequality and a union bound, this probability is less than or
equal to

$N\max_{0\leq i<N}\mathop{}\\!\mathbb{P}(\\{\exists
s\in[0,1/N]\colon|Z^{(\lambda,1/3)}_{s+i/N}-Z^{(\lambda,1/3)}_{i/N}|>C\sqrt{\kappa}/4\\}\cap
G(N,\kappa,\lambda)).$

Recall that conditional on $X^{(\lambda)}$, the label process
$Z^{(\lambda,1/3)}$ is a centred Gaussian process with a so-called canonical
or intrinsic pseudo-distance given by:

$\rho^{(\lambda)}(s,t)=\sqrt{\mathop{}\\!\mathbb{E}[(Z^{(\lambda,1/3)}_{t}-Z^{(\lambda,1/3)}_{s})^{2}\mid
X^{(\lambda)}]}\leq\sqrt{d^{1/2}_{\mathscr{L}(X^{(\lambda)})}(s,t)},$

where the upper bound follows by replacing the product of left and right
length along cycles by the minimum. Let then $N^{(\lambda)}(r,N,i)$ denote the
number of $\rho^{(\lambda)}$-balls of radius $r$ needed to cover the interval
$[i/N,(i+1)/N]$. Then on the event $G(N,\kappa,\lambda)$ we have
$N^{(\lambda)}(r,N,i)=1$ for $r\geq\sqrt{\kappa}$ and
$N^{(\lambda)}(r,N,i)\leq\lceil\sqrt{\kappa}/r\rceil$ for $r<\sqrt{\kappa}$.
Then Dudley’s entropy bound [BLM13, Corollary 13.2] shows that there exists a
universal constant $K$ such that

$\displaystyle\mathop{}\\!\mathbb{E}\Bigl{[}\sup_{s\in[0,1/N]}|Z^{(\lambda,1/3)}_{s+i/N}-Z^{(\lambda,1/3)}_{i/N}|\Bigm{|}X^{(\lambda)}\Bigr{]}\,\mathbf{1}_{\\{G(N,\kappa,\lambda)\\}}$
$\displaystyle\leq K\int_{0}^{\infty}\sqrt{\log
N^{(\lambda)}(r,N,i)}\mathop{}\\!\mathrm{d}r\,\mathbf{1}_{\\{G(N,\kappa,\lambda)\\}}$
$\displaystyle\leq
K\int_{0}^{\sqrt{\kappa}}\sqrt{\log\lceil\sqrt{\kappa}/r\rceil}\mathop{}\\!\mathrm{d}r$
$\displaystyle\leq K\sqrt{\kappa}\int_{0}^{1}\sqrt{\log\lceil
1/x\rceil}\mathop{}\\!\mathrm{d}x,$

and the integral is some universal constant as well. Let us write
$L\sqrt{\kappa}$ the last line. Recall the exponent $a<1/\boldsymbol{\eta}$
which we have fixed. If we were able to take $a>1/2$ then we could conclude by
the Markov inequality. However in the case $\boldsymbol{\eta}=2$, in
particular in the Brownian case, we need a slightly better bound. A very
strong bound [BLM13, Theorem 5.8] which follows from standard Gaussian
concentration results is the following: for every $u>0$, we have

$\displaystyle\mathop{}\\!\mathbb{P}\Bigl{(}G(N,\kappa,\lambda)\cap\Bigl{\\{}\sup_{s\in[0,1/N]}|Z^{(\lambda,1/3)}_{s+i/N}-Z^{(\lambda,1/3)}_{i/N}|>u+\mathop{}\\!\mathbb{E}\Bigl{[}\sup_{s\in[0,1/N]}|Z^{(\lambda,1/3)}_{s+i/N}-Z^{(\lambda,1/3)}_{i/N}|\Bigm{|}X^{(\lambda)}\Bigr{]}\Bigr{\\}}\Bigr{)}$
$\displaystyle\leq\mathop{}\\!\mathbb{E}\Bigl{[}\mathbf{1}_{\\{G(N,\kappa,\lambda)\\}}\exp\Bigl{(}-\frac{u^{2}}{2\sup_{s\in[0,1/N]}\mathop{}\\!\mathbb{E}[(Z^{(\lambda,1/3)}_{s+i/N}-Z^{(\lambda,1/3)}_{i/N})^{2}]}\Bigr{)}\Bigr{]}$
$\displaystyle\leq\exp\Bigl{(}-\frac{u^{2}}{2\kappa}\Bigr{)}.$

Now recall that the expectation in the first line is bounded above by
$L\sqrt{\kappa}$ where $L$ is a universal constant, hence

$\mathop{}\\!\mathbb{P}\Bigl{(}G(N,\kappa,\lambda)\cap\Bigl{\\{}\sup_{s\in[0,1/N]}|Z^{(\lambda,1/3)}_{s+i/N}-Z^{(\lambda,1/3)}_{i/N}|>u+L\sqrt{\kappa}\Bigr{\\}}\Bigr{)}\leq\exp\Bigl{(}-\frac{u^{2}}{2\kappa}\Bigr{)},$

for every $u>0$. Combining all the previous displays, we infer with
$u=(C/4-L)\sqrt{\kappa}$ that for every $\lambda$ large enough, it holds:

$\mathop{}\\!\mathbb{P}(\exists(s,t)\in[0,1]^{2}\colon|t-s|\leq
1/N\quad\text{and}\quad|Z^{(\lambda,1/3)}_{s}-Z^{(\lambda,1/3)}_{t}|>\theta)\leq
2\varepsilon+N\exp\Bigl{(}-\frac{(C/4-L)^{2}}{2}\Bigr{)}.$

Recall that $\theta=C\sqrt{\kappa}$ and
$N=\lceil(M/\kappa)^{1/a}\rceil=\lceil(MC^{2}/\theta^{2})^{1/a}\rceil$ where
$M$ was chosen only in terms of $\varepsilon$ in order to get the upper bound
$2\varepsilon$. Since $L$ is a universal constant, then we can choose $C$
depending only on $\theta$ and $M$ (and thus $\varepsilon$) such that the
right-hand side above is less than $3\varepsilon$. This fixes the value of $N$
in terms of $\theta$ and $\varepsilon$ only and the proof tightness is
complete. ∎

To complete the proof Theorem 5.2, it remains to characterise the
subsequential limits. We first consider the convergence under the
unconditional law of Lévy processes.

###### Proof uniqueness of the limits for the unconditioned processes.

Let us here consider the unconditioned Lévy processes and let us first prove
that for any $t>0$ fixed, the looptree distance from $0$ to $t$ for
$X^{(\lambda)}$ converges in distribution to that for $X$. Then we discuss the
convergence of the distance between any $s$ and $t$ fixed, the argument for
more general finite dimensional distributions is the same. We then briefly
mention the modifications to control the labels.

Step 1: distance from $0$ to $t$. Fix $t>0$. Let us denote by
$C^{(\lambda)}_{t}$ and $R^{(\lambda),t}_{s}$ respectively the continuous part
and jump part that appear in the looptree distance (2.6) associated with
$X^{(\lambda)}$. By standard properties of the Skorokhod $J_{1}$ topology, the
convergence $X^{(\lambda)}\to X$ implies that for any $\delta>0$, the jumps
larger than $\delta$ of $X$ are the limits of large jumps of $X^{(\lambda)}$,
and quantities such that $R^{(\lambda),t}_{s}$ for such jump times $s$
converge as well to their analogue for $X$. Hence, the contribution of the
times of jumps larger than $\delta$ in the looptree distance associated with
$X^{(\lambda)}$ converges to that for $X$. It remains to consider the joint
contribution of the continuous part and the small jumps, namely to prove that:

$\frac{1}{2}C^{(\lambda)}_{t}+\sum_{s\prec t}\min(R^{(\lambda),t}_{s},\Delta
X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\enskip\mathop{\longrightarrow}^{(d)}\enskip\frac{1}{2}C_{t}$
(5.1)

as first $\lambda\to\infty$ and then $\delta\to 0$.

First note that the convergence in the Skorokhod $J_{1}$ topology also implies
that $X^{(\lambda)}_{t-}$ converges to $X_{t-}$, hence as a consequence of the
previous lines, we have for any $\delta>0$ fixed:

$C^{(\lambda)}_{t}+\sum_{s\prec t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}=X^{(\lambda)}_{t-}-\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}>\delta\\}}\enskip\mathop{\longrightarrow}^{(d)}_{\lambda\to\infty}\enskip
X_{t-}-\sum_{s\prec t}R^{t}_{s}\mathbf{1}_{\\{\Delta
X_{s}>\delta\\}}=C_{t}+\sum_{s\prec t}R^{t}_{s}\mathbf{1}_{\\{\Delta
X_{s}\leq\delta\\}},$

and the right-hand side further converges to $C_{t}$ as $\delta\to 0$. We
shall compare the left-hand side of this display with (twice) that of (5.1).
First note that as a consequence of this last convergence, if for
$\varepsilon>0$ we let

$A_{\lambda,\delta,\varepsilon}=\Bigl{\\{}\Bigl{|}C^{(\lambda)}_{t}+\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{|}<\varepsilon^{-1/2}\Bigr{\\}},$

then the probability of this event goes to $1$ as first $\lambda\to\infty$,
then $\delta\to 0$, and finally $\varepsilon\to 0$. Unless specified, we shall
below consider limits in this sense.

Let us introduce a second event:

$B_{\lambda,\delta,\varepsilon}=\Bigl{\\{}\sum_{s\prec t}\Delta
X^{(\lambda)}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\leq\varepsilon\Bigr{\\}}.$

Recall that $\min(R^{(\lambda),t}_{s},\Delta
X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})\leq R^{(\lambda),t}_{s}\leq\Delta
X^{(\lambda)}_{s}$. Fix a bounded and $K$-Lipschitz function $F$ for some
$K>0$, then we have:

$\displaystyle\mathop{}\\!\mathbb{E}\Bigl{[}\Bigl{|}F\Bigl{(}\frac{1}{2}C^{(\lambda)}_{t}+\sum_{s\prec
t}\min(R^{(\lambda),t}_{s},\Delta
X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{)}-F\Bigl{(}\frac{1}{2}\Bigl{(}C^{(\lambda)}_{t}+\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{)}\Bigr{)}\Bigr{|}\mathbf{1}_{\\{B_{\lambda,\delta,\varepsilon}\\}}\Bigr{]}$
$\displaystyle\leq\mathop{}\\!\mathbb{E}\Bigl{[}K\Bigl{(}\sum_{s\prec
t}\min(R^{(\lambda),t}_{s},\Delta
X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}+\frac{1}{2}\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{)}\mathbf{1}_{\\{B_{\lambda,\delta,\varepsilon}\\}}\Bigr{]}$
$\displaystyle\leq K\cdot\frac{3\varepsilon}{2}.$

On the complementary event $B_{\lambda,\delta,\varepsilon}^{c}$, we may adapt
the argument in the proof Theorem 1.2 in [CK14]. Precisely, we can write
$R^{(\lambda),t}_{s}=\Delta X^{(\lambda)}_{s}\times u^{t}_{s}$, where $\Delta
X^{(\lambda)}_{s}$ is the size of the jump of $X^{(\lambda)}$ at time $s$, and
in the unconditioned process, the random variables $(u^{t}_{s};s\prec t)$ are
i.i.d. with the uniform law on $[0,1]$ and are independent from $(\Delta
X^{(\lambda)}_{s};s\prec t)$. Consequently,

$\mathop{}\\!\mathbb{E}\biggl{[}\frac{\sum_{s\prec t}\Delta
X^{(\lambda)}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}u^{t}_{s}}{\sum_{s\prec t}\Delta
X^{(\lambda)}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}}\bigm{|}\Delta X^{(\lambda)}_{s};s\prec
t\biggr{]}=\frac{1}{2}$

and

$\mathrm{Var}\biggl{(}\frac{\sum_{s\prec t}\Delta
X^{(\lambda)}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}u^{t}_{s}}{\sum_{s\prec t}\Delta
X^{(\lambda)}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}}\bigm{|}\Delta X^{(\lambda)}_{s};s\prec
t\biggr{)}=\frac{\sum_{s\prec t}(\Delta
X^{(\lambda)}_{s})^{2}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}}{12(\sum_{s\prec t}\Delta
X^{(\lambda)}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}})^{2}}\leq\frac{\sup_{s\prec t}\Delta
X^{(\lambda)}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}}{12\sum_{s\prec t}\Delta
X^{(\lambda)}_{s}\mathbf{1}_{\\{\Delta X^{(\lambda)}_{s}\leq\delta\\}}},$

which is bounded by $\delta/(12\varepsilon)$ on the event
$B_{\lambda,\delta,\varepsilon}^{c}$. The same holds if we replace $u^{t}_{s}$
by $\min(u^{t}_{s},1-u^{t}_{s})$, in which case the numerator on left
represents the contribution to the looptree distance of the small jumps, the
only difference is that the expectation is now $1/4$. We infer that if we let

$C_{\lambda,\delta,\varepsilon}=\biggl{\\{}\frac{\frac{1}{2}C^{(\lambda)}_{t}+\sum_{s\prec
t}\min(R^{(\lambda),t}_{s},\Delta
X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}}{C^{(\lambda)}_{t}+\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}}\in\Bigl{[}\frac{1/4-\varepsilon}{1/2+\varepsilon},\frac{1/4+\varepsilon}{1/2-\varepsilon}\Bigr{]}\biggr{\\}},$

then

$\mathop{}\\!\mathbb{P}(B_{\lambda,\delta,\varepsilon}^{c}\cap
C_{\lambda,\delta,\varepsilon}^{c})\leq\frac{\delta}{L\,\varepsilon^{3}},$

for some universal constant $L$ coming from the variances. Let
$g(\varepsilon)=\max(\frac{1}{2}-\frac{1/4-\varepsilon}{1/2+\varepsilon},\frac{1/4+\varepsilon}{1/2-\varepsilon}-\frac{1}{2})$.
Recall the event
$A_{\lambda,\delta,\varepsilon}=\\{|C^{(\lambda)}_{t}+\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}|<\varepsilon^{-1/2}\\}$ and recall that $F$ is
$K$-Lipschitz. Then we have:

$\displaystyle\mathop{}\\!\mathbb{E}\Bigl{[}\Bigl{|}F\Bigl{(}\frac{1}{2}C^{(\lambda)}_{t}+\sum_{s\prec
t}\min(R^{(\lambda),t}_{s},\Delta
X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{)}-F\Bigl{(}\frac{1}{2}\Bigl{(}C^{(\lambda)}_{t}+\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{)}\Bigr{)}\Bigr{|}\mathbf{1}_{\\{A_{\lambda,\delta,\varepsilon}\cap
C_{\lambda,\delta,\varepsilon}\\}}\Bigr{]}$
$\displaystyle\leq\mathop{}\\!\mathbb{E}\Bigl{[}K\cdot\Bigl{|}\frac{1}{2}C^{(\lambda)}_{t}+\sum_{s\prec
t}\min(R^{(\lambda),t}_{s},\Delta
X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}-\frac{1}{2}\Bigl{(}C^{(\lambda)}_{t}+\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{)}\Bigr{|}\mathbf{1}_{\\{A_{\lambda,\delta,\varepsilon}\cap
C_{\lambda,\delta,\varepsilon}\\}}\Bigr{]}$
$\displaystyle\leq\mathop{}\\!\mathbb{E}\Bigl{[}K\cdot
g(\varepsilon)\cdot\Bigl{|}C^{(\lambda)}_{t}+\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{|}\mathbf{1}_{\\{A_{\lambda,\delta,\varepsilon}\cap
C_{\lambda,\delta,\varepsilon}\\}}\Bigr{]}$ $\displaystyle\leq K\cdot
g(\varepsilon)/\sqrt{\varepsilon},$

which converges to $0$ as $\varepsilon\to 0$ since $g$ is approximately linear
near $0$.

Recall that $F$ is $K$-Lipschitz and bounded, say $|F|\leq K^{\prime}$. Then
by partitioning the space into the four events:
$B_{\lambda,\delta,\varepsilon}\cup\\{A_{\lambda,\delta,\varepsilon}^{c}\cap
B_{\lambda,\delta,\varepsilon}^{c}\\}\cup\\{A_{\lambda,\delta,\varepsilon}\cap
B_{\lambda,\delta,\varepsilon}^{c}\cap
C_{\lambda,\delta,\varepsilon}\\}\cup\\{A_{\lambda,\delta,\varepsilon}\cap
B_{\lambda,\delta,\varepsilon}^{c}\cap C_{\lambda,\delta,\varepsilon}^{c}\\}$,
we conclude that

$\Bigl{|}\mathop{}\\!\mathbb{E}\Bigl{[}F\Bigl{(}\frac{1}{2}C^{(\lambda)}_{t}+\sum_{s\prec
t}\min(R^{(\lambda),t}_{s},\Delta
X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{)}\Bigr{]}-\mathop{}\\!\mathbb{E}\Bigl{[}F\Bigl{(}\frac{1}{2}\Bigl{(}C^{(\lambda)}_{t}+\sum_{s\prec
t}R^{(\lambda),t}_{s}\mathbf{1}_{\\{\Delta
X^{(\lambda)}_{s}\leq\delta\\}}\Bigr{)}\Bigr{)}\Bigr{]}\Bigr{|}$

is upper bounded by:

$K\frac{3\varepsilon}{2}+2K^{\prime}\mathop{}\\!\mathbb{P}(A_{\lambda,\delta,\varepsilon}^{c})+K\frac{g(\varepsilon)}{\sqrt{\varepsilon}}+2K^{\prime}\frac{\delta}{L\,\varepsilon^{3}},$

which goes to $0$ as first $\lambda\to\infty$, then $\delta\to 0$, and finally
$\varepsilon\to 0$. Recall from the display below (5.1) that the second
expectation above tends to $\mathop{}\\!\mathbb{E}[F(C_{t})]$ along these
limits, hence so does the first one. This proves (5.1) by the Portmanteau
theorem.

Step 2: distance from an ancestor to $t$. So far, we have proved the
convergence of the looptree distance from $0$ to any given $t>0$. Let
$s\in(0,t)$ and suppose first that $s\prec t$ in the looptree coded by $X$,
namely that $X_{s-}\leq X_{r}$ for all $r\in[s,t]$. There are two cases
according as wether $X$ jumps at time $s$ or not.

If $\Delta X_{s}>0$, then recall from the proof Lemma 3.3 that $X$ cannot make
a local minimum at the height $X_{s-}$. In particular, either $X_{t}>X_{s-}$
or $t=\inf\\{r>s\colon X_{r}=X_{s-}\\}$. In the second case, this means that
we can find $t^{\prime}<t$ arbitrarily close to $t$, with
$X_{t^{\prime}}>X_{t}=X_{s-}$ and $X_{t^{\prime}}$ is arbitrarily close to
$X_{t}$; this implies that the looptree distance between $t$ and $t^{\prime}$
is arbitrarily small and $t^{\prime}$ falls into the first case. Therefore we
may assume when $\Delta X_{s}>0$ that $X_{t}>X_{s-}$ and actually
$X_{r}>X_{s-}$ for all $r\in[s,t]$. In this case, as before, there are times
$s^{(\lambda)}\prec t^{(\lambda)}$ for $X^{(\lambda)}$, such that
$X^{(\lambda)}$ jumps at time $s^{(\lambda)}$, and such that $s$, $t$, $\Delta
X_{s}$ and $R^{t}_{s}$ are the limits of the analogous quantities in
$X^{(\lambda)}$ and the previous argument extends readily to prove the
convergence of the looptree distance from $s^{(\lambda)}$ to $t^{(\lambda)}$
for $X^{(\lambda)}$ to that from $s$ to $t$ for $X$.

Suppose next that $\Delta X_{s}=0$. We aim at finding similarly times
$s^{(\lambda)}\prec t^{(\lambda)}$ that converge to $s$ and $t$ to which we
can apply the previous argument. Again there are two cases. In the case where
$X_{r}>X_{s}$ for all $r\in(s,t]$, there exists $\varepsilon>0$ small such
that $X_{r}>X_{s}+2\varepsilon$ for $r\in(s+2\varepsilon,t]$. The Skorokhod
convergence implies that for all $\lambda$ sufficiently large, there exists
$r^{(\lambda)}\in(s-\varepsilon,s+\varepsilon)$ such that
$X^{(\lambda)}_{r}>X^{(\lambda)}_{r^{(\lambda)}}+\varepsilon$ for all
$r\in(r^{(\lambda)}+4\varepsilon,t]$. This further implies this existence of
some time
$s^{(\lambda)}\in(r^{(\lambda)},r^{(\lambda)}+4\varepsilon)\subset(s-\varepsilon,s+5\varepsilon)$
with $s^{(\lambda)}\prec t$ and we conclude in this case. A last case is when
both $\Delta X_{s}=0$ and $X_{r}=X_{s}$ for some $r\in(s,t)$. Then necessarily
this time $r$ is a time of local minimum and since the latter are unique, then
$s$ is not a time of local minimum so we can find $s^{\prime}<s$ arbitrarily
close to $s$ which falls into the previous case.

Step 3: pairwise distances. Assume now that $s\not\prec t$ in the looptree
coded by $X$. Then we can cut at their last common ancestor $s\wedge t$ and
apply the same reasoning as above: if this last common ancestor is a jump
time, then the quantities of interest associated with this macroscopic loop
are limits of those for $X^{(\lambda)}$ and the two branches from $s\wedge t$
to $s$ on the one hand and from $s\wedge t$ to $t$ on the other hand converge
by the previous paragraph. The case when $s\wedge t$ is not a jump time is
controlled similarly. The same argument extends to control the pairwise
distances between any finite collection of times, cutting at the corresponding
branch-points. Details are left to the reader.

Step 4: labels. Let us end with a few words on the Gaussian labels on the
looptree. Recall that the conditional covariance function given the Lévy
process is very close to be the looptree distance: one just replaces the
infimum between the left and right length of the loops by their product. Thus
the only modification in Step 2 is that one should replace
$\min(R^{(\lambda),t}_{s},\Delta X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})$ by
$R^{(\lambda),t}_{s}(\Delta X^{(\lambda)}_{s}-R^{(\lambda),t}_{s})$. If $U$
has the uniform law on $[0,1]$, we used previously that
$\mathop{}\\!\mathbb{E}[\min(U,1-U)]=1/4$, then this value is now replaced by
$\mathop{}\\!\mathbb{E}[U(1-U)]=1/6$. Once divided by
$\mathop{}\\!\mathbb{E}[U]=1/2$, we obtain the constant $1/3$, which explains
why we consider $Z^{1/3}$ where we considered $\mathscr{L}^{1/2}(X)$. Step 2
and 3 are easily adapted. ∎

We finally turn to conditioned processes.

###### Proof uniqueness of the limits for bridges and excursions.

From the two previous proofs, we know that the looptree distances and labels
associated with the bridges are tight and those associated with the
unconditioned processes converge. Fix $\varepsilon\in(0,1)$ and notice that
the looptree distance between all pairs $s,t\in[0,1-\varepsilon]$ is a
measurable function of the trajectory of the process restricted to the time
interval $[0,1-\varepsilon]$. The same holds for the covariance function of
the labels on the time interval $[0,1-\varepsilon]$. Then the convergence of
these looptree distances and labels for the bridge follows from the absolute
continuity relation as in the proof Lemma 5.1. This characterises the
subsequential limits and we conclude from tightness.

We next turn to the excursion versions. Recall that we recover the bridge
$X^{\mathrm{br}}$ by cyclicly shifting the excursion $X^{\mathrm{ex}}$ at an
independent uniform random time $U$. Notice that the looptree distance in the
excursion between any two times $s,t\in[0,U]$ is the same as the looptree
distance in the bridge between their images $1-U+s$ and $1-U+t$ after the
cyclic shift. However they differ if one or both times lie after $U$. Using
the same time $U$ to relate the excursion and bridge versions of each
$X^{(\lambda)}$, we infer from the case of bridges that the looptree distances
restricted to $[0,U]$ in the excursions converge in distribution to those in
$X^{\mathrm{ex}}$.

Assume next that $s<U<t$ and consider $q=\inf\\{r>U\colon
X^{\mathrm{ex}}_{r-}\leq\inf_{[r,t]}X^{\mathrm{ex}}\\}$ the first ancestor of
$t$ that lies after $U$ and $p=\sup\\{r<U\colon X^{\mathrm{ex}}_{r-}\leq
X^{\mathrm{ex}}_{q-}\\}$ the last ancestor of $t$ before $U$. Then the
looptree distance between $t$ and $q$ in the excursion is the same as that of
their images $t-U$ and $q-U$ in the bridge, and all the other contributions to
the distance between $s$ and $t$ depend only on the trajectory up to time $U$
as well as the value $X^{\mathrm{ex}}_{U}-X^{\mathrm{ex}}_{q-}$. Finally if
both $s,t>U$, then it may be that their last common ancestor also satisfies
$s\wedge t\geq U$, in which case the looptree distance in the excursion is
that of their images in the bridge. If $s\wedge t\geq U$, then we easily adapt
the previous reasoning. In any case we infer the convergence in distribution
of the looptree distances in the excursion from the joint convergence of the
bridges and the associated looptree distances.

The same reasoning shows the convergence of the covariance function of the
label processes, which combined with the tightness argument implies the
convergence in distribution. ∎

###  5.2. Convergence of maps

Let us suppose that for every $\lambda>0$, the pair
$(X^{(\lambda)},Z^{(\lambda,1/3)})$ arises as the scaling limit of discrete
paths $(W^{(\lambda,n)},Z^{(\lambda,n)})$ in the sense of Theorem 2.1. Suppose
that these paths encode a bipartite planar map
$M^{(\lambda,n)}=(M^{(\lambda,n)},d^{(\lambda,n)},p^{(\lambda,n)})$. Then
Theorem 2.4 implies the convergence of the rescaled map along a subsequence to
a limit space $M^{(\lambda)}=(M^{(\lambda)},d^{(\lambda)},p^{(\lambda)})$. Now
assume that for every $t\geq 0$, the transition density $p^{(\lambda)}_{t}$
converges uniformly as $\lambda\to\infty$ to $p_{t}$, so both Lemma 5.1 and
Theorem 5.2 are satisfied, and let us study the convergence of $M^{(\lambda)}$
as $\lambda\to\infty$, again along a subsequence.

######  Theorem 5.3.

Suppose that the previous assumptions are satisfied. Then from every sequence
tending to infinity, one can extract a subsequence along which the convergence
in distribution

$d^{(\lambda)}\enskip\mathop{\longrightarrow}^{(d)}_{\lambda\to\infty}\enskip
D,$

holds for the uniform topology on $[0,1]^{2}$, jointly with Theorem 5.2. The
limit $D$ is a random continuous pseudo-distance which satisfies:

$D(U,\cdot)\enskip\mathop{=}^{(d)}\enskip Z^{1/3}-\min Z^{1/3},$

where $U$ is sampled uniformly at random on $[0,1]$ and independently of the
rest. Furthermore, along this subsequence, we have

$M^{(\lambda)}\enskip\mathop{\longrightarrow}^{(d)}_{n\to\infty}\enskip{}M=[0,1]/\\{D=0\\}$

in the Gromov–Hausdorff–Prokhorov topology. Finally, if the excursion of $X$
reduces to a constant $\beta$ times the Brownian excursion, then these
subsequential limits agree with $\sqrt{\beta/3}$ times the standard Brownian
sphere.

###### Proof.

The argument is close to that of Theorem 2.4. Indeed, from this theorem we
have the identity in law $d^{(\lambda)}(U,\cdot)=Z^{(\lambda,1/3)}-\min
Z^{(\lambda,1/3)}$ where $U$ is sampled uniformly at random on $[0,1]$ and
independently of the rest. Since $Z^{(\lambda,1/3)}$ converges in distribution
to $Z^{1/3}$ by Theorem 5.2, then tightness of $d^{(\lambda)}$ follows from
the triangle inequality as in the discrete setting and so does the identity in
law $D(U,\cdot)=Z^{1/3}-\min Z^{1/3}$ for any subsequential limit $D$, and the
convergence in distribution $d^{(\lambda)}\to D$ holds jointly with that of
$(X^{(\lambda)},Z^{(\lambda,1/3)})\to(X,Z^{1/3})$ under the excursion law.

Suppose now that the limit excursion $X$ equals $\beta$ times the standard
Brownian excursion, so $Z^{1/3}$ has the law of $\sqrt{\beta/3}$ times the
standard Brownian snake driven by the standard Brownian excursion. We know
from [LG13, Mie13] that the distance $D^{\ast}$ in the Brownian sphere
satisfies the identity in law: $Z^{1/3}-\inf Z^{1/3}=\sqrt{\beta/3}\cdot
D^{\ast}(U,\cdot)$. We thus have the identity in law:

$D(U,V)\enskip\mathop{=}^{(d)}\enskip\sqrt{\beta/3}\cdot D^{\ast}(U,V)$

for every subsequential limit $D$ where $U$ and $V$ are sampled uniformly at
random on $[0,1]$ and independently of the rest and of each other. Following
Le Gall’s trick from [LG13], we claim that this identity holds almost surely.
If this holds true, then we can sample countably many uniform random variables
on $[0,1]$, say $(U_{i})_{i\geq 1}$ and infer that almost surely
$D(U_{i},U_{j})=\sqrt{\beta/3}\cdot D^{\ast}(U_{i},U_{j})$ for every $i,j\geq
1$. By continuity, this implies that $D=D^{\ast}$ almost surely.

Let us recall from [LG13, Mie13] that $D^{\ast}$ is a pseudo-distance on
$[0,1]$ constructed from another one $D^{\circ}$ defined in (4.7). As opposed
to $D^{\circ}$, the pseudo-distance $D^{\ast}$ can be seen as a pseudo-
distance on the Brownian tree, in the sense of Lemma 3.8. In fact, it is the
largest such pseudo-distance bounded above by $D^{\circ}$. Now by this lemma,
the pseudo-distance $d^{(\lambda)}$ can also be seen as a pseudo-distance on
the looptree $\mathscr{L}(X^{(\lambda)})$. Then the same argument used to
prove this lemma, using the approximation $X^{(\lambda)}\to X$ instead of
discrete paths, shows further that the subsequential limit $D$ can be seen as
a pseudo-distance on the looptree $\mathscr{L}(X)$, which reduces to a scaled
Brownian tree. Moreover, using the discrete approximation, for every
$\lambda>0$, the pseudo-distance $d^{(\lambda)}$ is upper bounded by the same
function as $D^{\circ}$ but constructed from $Z^{(\lambda,1/3)}$. Letting
$\lambda\to\infty$ along our subsequence, we infer that $D$ is upper bounded
by the same function as $D^{\circ}$ but constructed from $Z^{1/3}$, which we
recall is $\sqrt{\beta/3}$ times the standard Brownian snake. We infer that
$D\leq\sqrt{\beta/3}D^{\ast}$ almost surely by the maximality property of the
latter. This allows to extend the equality in law $D(U,V)=\sqrt{\beta/3}\cdot
D^{\ast}(U,V)$ to an almost sure one. ∎

##  6. Looptrees and maps driven by stable excursions with a drift

Let us present in this section applications of our general results in the case
where the coding process is a spectrally positive stable Lévy process with a
drift. In Section 6.1 we first state the general version of Theorem 1.3 in the
introduction, and then explain how these objects naturally appear as scaling
limits of multi-conditioned random planar maps introduced in [KM23]. The
convergence in the Brownian regime is then proved in Section 6.3, relying on
an exponential change of law briefly discussed in Section 6.2. The
condensation regime is treated in Section 6.5, relying on some technical
estimates on stable densities stated in Section 6.4 and proved in Section 6.7.
In Section 6.6 we also briefly discussed a missing case in Therorem 6.1.

###  6.1. Main convergence results

For every $\alpha\in(1,2)$, let us denote by $X^{\alpha}$ the stable Lévy
process characterised by the Laplace exponent
$\psi(\lambda)=\lambda^{\alpha}$, or equivalently by the characteristic
function:

$\mathop{}\\!\mathbb{E}[\exp(iuX^{\alpha}_{t})]=\exp\Bigl{(}t|u|^{\alpha}\Bigl{(}\cos\Bigl{(}\frac{\pi\alpha}{2}\Bigr{)}-i\sigma(u)\sin\Bigl{(}\frac{\pi\alpha}{2}\Bigr{)}\Bigr{)}\Bigr{)},$
(6.1)

for every $u\in\mathbb{R}$ and $t>0$, where we used the notation
$\sigma(u)=\mathbf{1}_{\\{u>0\\}}-\mathbf{1}_{\\{u<0\\}}$ for the sign
function to distinguish it from the sine function. The right-hand side
satisfies the integrability condition (1.2), and thus it admits positive, bi-
continuous, transition densities $p^{\alpha}_{t}$ given by inverse Fourier
transform, from which one can define the conditional law
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha}_{1}=\vartheta)$ for any
$\vartheta\in\mathbb{R}$, characterised by (2.1). For any
$\vartheta\in\mathbb{R}$, let us also denote by
$X^{\alpha,\vartheta}_{t}=X^{\alpha}_{t}-\vartheta t$ the stable process with
drift $-\vartheta$, which has transition densities
$p^{\alpha}_{t}(\cdot+\vartheta t)$. One can also define bridges of this
process and we let $X^{\alpha,\vartheta,\mathrm{br}}$ denote the version of
$X^{\alpha,\vartheta}$ under the conditional law
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha,\vartheta}_{1}=0)$. A
straightforward consequence of the absolute continuity relation (2.1) is that
$X^{\alpha,\vartheta,\mathrm{br}}$ has the same law as $(X^{\alpha}-\vartheta
t)_{t\in[0,1]}$ under $\mathop{}\\!\mathbb{P}(\,\cdot\mid
X^{\alpha}_{1}=\vartheta)$. We shall rather use the latter representation of
$X^{\alpha,\vartheta,\mathrm{br}}$. Finally let
$X^{\alpha,\vartheta,\mathrm{ex}}$ denote the excursion obtained by applying
the Vervaat transform (2.2) to $X^{\alpha,\vartheta,\mathrm{br}}$.

#### Varying the index of stability and the drift

Our main result in this section deals with the asymptotic behaviour of
$X^{\alpha,\vartheta,\mathrm{ex}}$ in the Skorokhod $J_{1}$ topology as
$\alpha$ and $\vartheta$ vary. It extends Theorem 1.3 in the introduction
which focused on a fixed $\alpha$.

######  Theorem 6.1.

Assume that the pair $(\alpha,\vartheta)$ converges to a limit
$(\alpha_{\ast},\vartheta_{\ast})\in[1,2]\times[-\infty,\infty]$.

  1. (i)

In the three cases when either both $\alpha_{\ast}=2$ and
$\vartheta_{\ast}<\infty$, or $\vartheta_{\ast}=-\infty$, or both
$\alpha_{\ast}=1$ and $\vartheta_{\ast}<-1$, it holds:

$\biggl{(}\frac{|\vartheta|^{(2-\alpha)/(\alpha-1)}}{(\alpha-1)\alpha^{1/(\alpha-1)}}\biggr{)}^{1/2}X^{\alpha,\vartheta,\mathrm{ex}}\enskip\mathop{\longrightarrow}^{(d)}\enskip\boldsymbol{e},$

where $\boldsymbol{e}$ is the standard Brownian excursion.

  2. (ii)

In the two cases when either both $\alpha_{\ast}<2$ and
$\vartheta_{\ast}=\infty$ or both $\alpha_{\ast}=1$ and $\vartheta_{\ast}>-1$
it holds:

$\Bigl{(}\frac{1}{\vartheta+1}\cdot
X^{\alpha,\vartheta,\mathrm{ex}}_{(1-t)-}\Bigr{)}_{t\in[0,1]}\enskip\mathop{\longrightarrow}^{(d)}\enskip(t\,\mathbf{1}_{\\{t\neq
1\\}})_{t\in[0,1]}.$

The time-reversal in the second statement is only a way to formally state that
$(\vartheta+1)^{-1}X^{\alpha,\vartheta,\mathrm{ex}}_{t}\to(1-t)\mathbf{1}_{\\{0<t\leq
1\\}}$ since this limit is not right-continuous at $0$. Alternatively, one
could extend all paths on the negative half-line by setting them equal to $0$
on $(-\infty,0)$, allowing for a jump at $0$. Let us recall Figure 4 in the
introduction for simulations of $X^{\alpha,\vartheta,\mathrm{ex}}$ displaying
four different regimes from this theorem and let us refer to Figure 11 for a
diagram for these regimes. We leave open the two regimes in which either
$(\alpha,\vartheta)\to(2,\infty)$ or $(\alpha,\vartheta)\to(1,-1)$, for which
the behaviour depends more precisely on the speed of each convergence.

Figure 11: Diagram representing the different regimes in Theorem 6.1. At each
point $(\alpha,\vartheta)$ in the open strip lies an excursion of an
$\alpha$-stable process with drift $-\vartheta$. On the plain line part of the
boundary, these excursions converge to the Brownian excursion; on the dashed
line part, they converge to the linear drift. The two points at the
intersection of these regimes are excluded. Finally the dotted vertical line,
corresponding to $\vartheta=0$ was first considered in [CK14].

Theorem 6.1 extends [CK14, Proposition 3.5 and Proposition 3.6], where it is
shown that when $\vartheta=0$, the excursion $X^{\alpha,0,\mathrm{ex}}$
converges to the Brownian excursion as $\alpha\uparrow 2$ and to the linear
drift $(1-t)\mathbf{1}_{\\{0<t\leq 1\\}}$ as $\alpha\downarrow 1$
respectively. When $\alpha\uparrow 2$, the limit is always the Brownian
excursion, whatever the (finite) value of $\vartheta$. However the behaviour
when $\alpha\downarrow 1$ depends on the value of $\vartheta$: we recover the
linear drift when $\vartheta>-1$, but the Brownian excursion instead when
$\vartheta<-1$. To understand intuitively this change of behaviour, one can
notice that the unconditioned process $X^{\alpha}$ converges in probability to
the linear drift $t\mapsto-t$ as $\alpha\downarrow 1$. This can be easily
shown using the explicit expression of the characteristic function given in
(6.1). Then informally, $X^{\alpha}$ naturally wants to end near $-1$ as
$\alpha\downarrow 1$ and without jumping. Thus if we impose to end at a
$X^{\alpha}_{1}=\vartheta<-1$, then it will follow the corresponding drift and
avoid any positive jump, hence Brownian fluctuations, whereas if we impose to
end at $X^{\alpha}_{1}=\vartheta>-1$, then the easiest way for it is to make a
single jump of size about $\vartheta+1$ and drift approximately like $-t$
otherwise.

This second case is a new instance of the phenomenon known as the “one big
jump principle”, well-known for discrete conditioned random walks such as in
[DDS08, AL11] and more recently [BBY23]. Closer to this work, Konstantopoulos
& Richardson [KR02] considered also spectrally positive Lévy processes, with a
stable flavour, but rather somehow conditioned in the regular sense on the
event $\\{X^{\alpha}_{1}>\vartheta\\}$ as $\vartheta\to\infty$, see also
[DJ12, Theorem 6].

#### Limits of biconditioned planar maps

Let us now explain how Lévy looptrees and Lévy maps driven by stable
excursions with a drift appear as scaling limits of multi-conditioned random
planar maps.

Let $W=(W_{k};k\geq 0)$ denote a random walk, started at $0$, with increments
in $Z_{\geq-1}=\\{-1,0,1,\dots\\}$, and whose step distribution is aperiodic
and centred. Suppose furthermore that these increments belong to the domain of
attraction of a stable law with index $\alpha\in(1,2)$, which can be
characterized by the fact that the tail distribution takes the form
$\mathop{}\\!\mathbb{P}(W_{1}>x)=x^{-\alpha}L_{1}(x)$ with $L_{1}$ a slowly
varying function, that is, such that $L_{1}(cx)/L_{1}(x)\to 1$ as $x\to\infty$
for any $c>0$ fixed. It is classical (see e.g. [Fel71, Theorem XVII.5.2] or
[Kal02, Theorem 16.14]) that if $(r_{n})_{n\geq 1}$ is a sequence such that
$r_{n}^{\alpha}\sim\alpha\Gamma(-\alpha)nL_{1}(r_{n})$ as $n\to\infty$, then
the convergence in distribution:

$(r_{n}^{-1}W_{\lfloor nt\rfloor})_{t\geq
0}\enskip\mathop{\longrightarrow}^{(d)}_{n\to\infty}\enskip X^{\alpha}$ (6.2)

holds for the Skorokhod $J_{1}$ topology.

Recently in [KM23] we extended this convergence to excursions $W^{n,K_{n}}$ of
$W$ conditioned both to have duration $n$ and a total $K_{n}$ of negative
increments. Precisely, Theorem 4.4(i) in [KM23] shows that if there exists
$\vartheta\in\mathbb{R}$ such that

$\frac{K_{n}-pn}{r_{n}(1-p)}\enskip\mathop{\longrightarrow}_{n\to\infty}\enskip\vartheta,$

with $r_{n}$ as in (6.2) and $p=\mathop{}\\!\mathbb{P}(W_{1}=-1)$, then

$(r_{n}^{-1}W^{n,K_{n}}_{\lfloor
nt\rfloor})_{t\in[0,1]}\enskip\mathop{\longrightarrow}^{(d)}_{n\to\infty}\enskip
X^{\alpha,\vartheta,\mathrm{ex}}.$ (6.3)

Let us mention that the statement in [KM23] actually applies more generally to
non-centred distributions. The case $\alpha=2$ is also treated there, although
in this regime the limit $X^{2,\vartheta,\mathrm{ex}}$ is simply the Brownian
excursion independently of $\vartheta$, as it can be checked using Girsanov’s
formula (see Section 6.2 below).

Items (ii) and (iii) in [KM23, Theorem 4.4] focus on the limit when
$(K_{n}-pn)/r_{n}$ converges instead to $-\infty$ or $+\infty$. In this case,
under a different scaling which depends on the speed of this convergence, the
discrete path $W^{n,K_{n}}$ converges in distribution respectively towards the
Brownian excursion and to the linear drift $(1-t)\mathbf{1}_{\\{0<t\leq
1\\}}$. Theorem 6.1 thus complements this result by showing that, once
suitably rescaled, the excursion $X^{\alpha,\vartheta,\mathrm{ex}}$ further
converges in distribution to these two limits as $\vartheta\to\pm\infty$. This
provides a way to interpolate continuously between these extreme models,
passing through a stable regime for a given $\alpha\in(1,2)$. For example we
believe that these objects hide in the recent work [FS23].

Let $\mathsf{L}(W^{n,K_{n}})$ be the discrete looptree coded by the
Łukasiewicz path $W^{n,K_{n}}$ in the sense of Section 2.1. Conditionally on
$W^{n,K_{n}}$, sample a good labelling of $\mathsf{L}(W^{n,K_{n}})$ uniformly
at random and let $M^{n,K_{n}}$ be the corresponding (pointed) bipartite
planar map. Recall from Section 2.1 that, alternatively, $M^{n,K_{n}}$ is a
Boltzmann random planar map with weight sequence given by
$q_{k}=\mathop{}\\!\mathbb{P}(W_{1}=k-1)/\binom{2k-1}{k}$ for every $k\geq 1$,
conditioned on having $n$ edges and $K_{n}$ vertices (hence $n-K_{n}+2$
faces).

By a direct application of our previous results, namely theorems 2.1, 1.1, and
5.2 for looptrees and theorems 2.4, 1.2, and 5.3, for maps we infer the
following results.

######  Corollary 6.2.

The following convergences hold for the Gromov–Hausdorff–Prokhorov topology.

  1. (i)

Let $(\alpha,\vartheta)\in(1,2)\times(-\infty,\infty)$, then under the
assumptions leading to (6.3), jointly with this convergence, we have

$r_{n}^{-1}\,\mathsf{L}(W^{n,K_{n}})\enskip\mathop{\longrightarrow}^{(d)}_{n\to\infty}\enskip\mathscr{L}(X^{\alpha,\vartheta,\mathrm{ex}}),$

and the limit satisfies almost surely:

$\dim_{H}\mathscr{L}(X^{\alpha,\vartheta,\mathrm{ex}})=\dim_{p}\mathscr{L}(X^{\alpha,\vartheta,\mathrm{ex}})=\operatorname{\underline{\dim}}\mathscr{L}(X^{\alpha,\vartheta,\mathrm{ex}})=\operatorname{\overline{\dim}}\mathscr{L}(X^{\alpha,\vartheta,\mathrm{ex}})=\alpha.$

In addition, the sequence $((2r_{n})^{-1/2}M^{n,K_{n}})_{n}$ is tight and
every subsequential limit, say $M^{\alpha,\vartheta}$, has almost surely:

$\dim_{H}M^{\alpha,\vartheta}=\dim_{p}M^{\alpha,\vartheta}=\operatorname{\underline{\dim}}M^{\alpha,\vartheta}=\operatorname{\overline{\dim}}M^{\alpha,\vartheta}=2\alpha.$

  2. (ii)

Assume further that the pair $(\alpha,\vartheta)$ converges to
$(\alpha_{\ast},\vartheta_{\ast})\in[1,2]\times[-\infty,\infty]$. In the three
cases when either both $\alpha_{\ast}=2$ and $\vartheta_{\ast}<\infty$, or
$\vartheta_{\ast}=-\infty$, or both $\alpha_{\ast}=1$ and
$\vartheta_{\ast}<-1$, it holds:

$\biggl{(}\frac{4|\vartheta|^{(2-\alpha)/(\alpha-1)}}{(\alpha-1)\alpha^{1/(\alpha-1)}}\biggr{)}^{1/2}\mathscr{L}(X^{\alpha,\vartheta,\mathrm{ex}})\enskip\mathop{\longrightarrow}^{(d)}\enskip\mathscr{T}\qquad\text{and}\qquad\biggl{(}\frac{9|\vartheta|^{(2-\alpha)/(\alpha-1)}}{(\alpha-1)\alpha^{1/(\alpha-1)}}\biggr{)}^{1/4}M^{\alpha,\vartheta}\enskip\mathop{\longrightarrow}^{(d)}\enskip\mathscr{S},$

where $\mathscr{T}$ is the standard Brownian tree and $\mathscr{S}$ the
standard Brownian sphere.

  3. (iii)

If the pair $(\alpha,\vartheta)$ converges to
$(\alpha_{\ast},\vartheta_{\ast})$ and either both $\alpha_{\ast}<2$ and
$\vartheta_{\ast}=\infty$ or both $\alpha_{\ast}=1$ and $\vartheta_{\ast}>-1$,
then it holds instead:

$|\vartheta+1|^{-1}\mathscr{L}(X^{\alpha,\vartheta,\mathrm{ex}})\enskip\mathop{\longrightarrow}^{(d)}\enskip\mathscr{C}\qquad\text{and}\qquad|\vartheta+1|^{-1/2}M^{\alpha,\vartheta}\enskip\mathop{\longrightarrow}^{(d)}\enskip\mathscr{T},$

where $\mathscr{C}$ is the circle with unit length and $\mathscr{T}$ is again
the standard Brownian tree.

In the first claim, one can also derive the convergence without extraction of
the profile and some other statistics on the distances in the map by
application of Theorem 2.4. In the second claim, let us notice that the
continuous part associated with $X^{\alpha,\vartheta,\mathrm{ex}}$, as defined
in (2.4), vanishes since the Lévy process has no Gaussian component (recall
the discussion a few lines below (2.4)). The limit of the excursion in this
regime is the Brownian excursion, whose associated looptree is the Brownian
tree, and the parameters $1/2$ and $1/3$ in Theorem 5.2 simply translate into
multiplicative constants $1/2$ and $1/\sqrt{3}$ respectively (recall the
discussion after this theorem). This explains the difference of constants in
the second claim above compared to Theorem 6.1. Finally, let us note that
actually, the last claim is not included in Theorem 5.2 and 5.3 since the
limit path $(1-t)\mathbf{1}_{\\{0<t<1\\}}$ is not an excursion of a Lévy
process, but this degenerate regime can be studied by hand, see also [Mar22b,
Section 4] for more general results under a “pure jump” property.

###  6.2. Exponential tilting

Lut us here recall a change of measure in Lévy processes using the exponential
martingale, called the Esscher transform, which we shall use in the next
subsection. We refer the reader to [Kyp14, Chapter 3.3] for details. Let $X$
be any Lévy process and let $\kappa\in\mathbb{R}$ be such that
$\psi(\kappa)=\log\mathop{}\\!\mathbb{E}[\operatorname{e}^{-\kappa X_{1}}]$ is
well-defined. Then one can define a process $X^{(\kappa)}$ characterised by
the absolute continuity relation: for every $t\geq 0$ and for every continuous
and bounded functional $F$,

$\mathop{}\\!\mathbb{E}[F(X^{(\kappa)}_{s},s\leq
t)]=\mathop{}\\!\mathbb{E}[F(X_{s},s\leq t)\operatorname{e}^{-\kappa
X_{t}-\psi(\kappa)t}].$ (6.4)

By [Kyp14, Theorem 3.9], the process $X^{(\kappa)}$ remains a Lévy process.
Further, as noted in [Kyp14, Corollary 3.10], if $X$ has no negative jump,
then $X^{(\kappa)}$ is well-defined for any $\kappa\geq 0$ and $X^{(\kappa)}$
also has no negative jump, and for every $\kappa,\lambda\geq 0$, we have:

$\psi^{(\kappa)}(\lambda)=\log\mathop{}\\!\mathbb{E}[\operatorname{e}^{-\lambda
X^{(\kappa)}_{1}}]=\log\mathop{}\\!\mathbb{E}[\operatorname{e}^{-\lambda
X_{1}-\kappa X_{1}-\psi(\kappa)}]=\psi(\lambda+\kappa)-\psi(\kappa),$

so $X^{(\kappa)}$ has drift $-\psi^{\prime}(\kappa)$. In the particular case
of the standard Brownian motion, so $\psi(\lambda)=\lambda^{2}/2$, the process
$X^{(\kappa)}$ is well-defined for any $\kappa\in\mathbb{R}$ and reduces to
the Brownian motion minus the drift with slope $\kappa$.

The following result shows that this exponential tilting has no effect on a
bridge-type conditioning. We let $X$ be a general spectrally positive Lévy
process with infinite variation paths and whose characteristic function is
integrable (1.2).

######  Lemma 6.3.

Let $\kappa>0$ and define $X^{(\kappa)}$ as in (6.4). Then for every
$\vartheta\in\mathbb{R}$, the laws of $(X_{t})_{t\in[0,1]}$ under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X_{1}=\vartheta)$ and of
$(X^{(\kappa)}_{t})_{t\in[0,1]}$ under $\mathop{}\\!\mathbb{P}(\,\cdot\mid
X^{(\kappa)}_{1}=\vartheta)$ are equal.

The point of this change of measure is to make the endpoint $\vartheta$
“typical” by matching it with the drift $-\psi^{\prime}(\kappa)$, when
possible. We shall use it on our stable Lévy process $X^{\alpha}$ with
$\psi(\lambda)=\lambda^{\alpha}$ for $\lambda\geq 0$, so one must impose
$\kappa\geq 0$ and so $\vartheta=-\alpha\kappa^{\alpha-1}\leq 0$.

###### Proof.

Recall from the discussion in the beginning of Section 2.2 that the
integrability condition (1.2) on $(X_{t})_{t}$ ensures that it indeed admits
transition densities $(p_{t})_{t}$ which are bi-continuous and positive
everywhere, and which are used to define bridge versions, via the absolute
continuity relation (2.1). We easily derive from (6.4) transition densities
$(p^{(\kappa)}_{t})_{t}$ for $(X^{(\kappa)})_{t}$, namely:

$p^{(\kappa)}_{t}(x)=p_{t}(x)\operatorname{e}^{-\kappa x-\psi(\kappa)t}.$

They inherit the continuity and positivity properties of $(p_{t})_{t}$, so we
may define bridges associated with $X^{(\kappa)}$ in the same way as $X$.
Also, note that for $0<t<1$ and $\vartheta\in\mathbb{R}$, we have:

$\frac{p^{(\kappa)}_{1-t}(\vartheta-X^{(\kappa)}_{t})}{p^{(\kappa)}_{1}(\vartheta)}=\frac{p_{1-t}(\vartheta-X^{(\kappa)}_{t})\operatorname{e}^{-\kappa(\vartheta-X^{(\kappa)}_{t})-\psi(\kappa)(1-t)}}{p_{1}(\vartheta)\operatorname{e}^{-\kappa\vartheta-\psi(\kappa)}}=\frac{p_{1-t}(\vartheta-X^{(\kappa)}_{t})}{p_{1}(\vartheta)}\operatorname{e}^{\kappa
X^{(\kappa)}_{t}+\psi(\kappa)t}.$

The exponential term on the right is exactly the inverse of that in (6.4).
Combined with the characterisation (2.1) of the bridges, we infer for any
$t\in(0,1)$ and any continuous and bounded function $F$:

$\displaystyle\mathop{}\\!\mathbb{E}[F(X^{(\kappa)}_{s},s\leq t)\mid
X^{(\kappa)}_{1}=\vartheta]$
$\displaystyle=\mathop{}\\!\mathbb{E}\Bigl{[}F(X^{(\kappa)}_{s},s\leq
t)\frac{p^{(\kappa)}_{1-t}(\vartheta-X^{(\kappa)}_{t})}{p^{(\kappa)}_{1}(\vartheta)}\Bigr{]}$
$\displaystyle=\mathop{}\\!\mathbb{E}\Bigl{[}F(X^{(\kappa)}_{s},s\leq
t)\frac{p_{1-t}(\vartheta-X^{(\kappa)}_{t})}{p_{1}(\vartheta)}\operatorname{e}^{\kappa
X^{(\kappa)}_{t}+\psi(\kappa)t}\Bigr{]}$
$\displaystyle=\mathop{}\\!\mathbb{E}\Bigl{[}F(X_{s},s\leq
t)\frac{p_{1-t}(\vartheta-X_{t})}{p_{t}(\vartheta)}\Bigr{]}$
$\displaystyle=\mathop{}\\!\mathbb{E}[F(X_{s},s\leq t)\mid X_{1}=\vartheta].$

This indeed identifies the two bridge laws. ∎

######  Corollary 6.4.

For any $\vartheta\in\mathbb{R}$, a unit duration Brownian bridge from $0$ to
$\vartheta$ has the same law as the sum of a standard Brownian bridge and the
linear drift with slope $\vartheta$.

###### Proof.

We have already observed that if $X=B$ is a standard Brownian motion, then
$B^{(\kappa)}$ is well-defined for any $\kappa\in\mathbb{R}$ and reduces to a
Brownian motion minus the drift with slope $\kappa$. For $\kappa=-\vartheta$,
Lemma 6.3 thus shows that the law of $(B_{t}+\vartheta t)_{t\in[0,1]}$ under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid B_{1}+\vartheta=\vartheta)$ does not
depend on $\vartheta\in\mathbb{R}$. On the other hand, we have already argued
that is the same as the law of $(B_{t}+\vartheta t)_{t\in[0,1]}$ under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid B_{1}=0)$. ∎

###  6.3. The Brownian regime

Let us first prove Theorem 6.1(i) on the convergence of the rescaled excursion
path $X^{\alpha,\vartheta,\mathrm{ex}}$ to a Brownian excursion. According to
Lemma 5.1, it suffices to prove the convergence of the transition densities.
In the case $\alpha\uparrow 2$ and $\vartheta$ has a finite limit, we can
easily do it directly, but otherwise we adapt this reasoning using the above
Esscher transform first. Let us start with the first case.

###### Proof Theorem 6.1(i) in the case $\alpha_{\ast}=2$ and
$\vartheta_{\ast}\in(-\infty,\infty)$.

We aim at proving that under the excursion law, the process defined for
$t\in[0,1]$ by:

$\Bigl{(}\frac{|\vartheta|^{(2-\alpha)/(\alpha-1)}}{(\alpha-1)\alpha^{1/(\alpha-1)}}\Bigr{)}^{1/2}(X^{\alpha}_{t}-\vartheta
t)$ (6.5)

converges in distribution to the standard Brownian excursion. Notice that the
scaling factor converges to $1/\sqrt{2}$, so we may as well replace it by this
value.

For every $t>0$, let $p^{\alpha,\vartheta}_{t}$ denote the density of
$X^{\alpha}_{t}-\vartheta t$ and let $p^{2,\vartheta_{\ast}}_{t}$ that of
$X^{2}_{t}-\vartheta_{\ast}t$ where $X^{2}$ is $\sqrt{2}$ times a Brownian
motion. Then by inverse Fourier transform, we have the following uniform
bound:

$\sup_{x\in\mathbb{R}}|p^{\alpha,\vartheta}_{t}(x)-p^{2,\vartheta_{\ast}}_{t}(x)|\leq\frac{1}{2\pi}\int_{-\infty}^{\infty}|\mathop{}\\!\mathbb{E}[\operatorname{e}^{iu(X^{\alpha}_{t}-\vartheta
t)}]-\mathop{}\\!\mathbb{E}[\operatorname{e}^{iu(X^{2}_{t}-\vartheta_{\ast}t)}]|\mathop{}\\!\mathrm{d}u.$

One can easily check from the explicit expression (6.1) of the characteristic
function of $X^{\alpha}_{t}$ that the integrand on the right-hand side
converges pointwise to $0$ as $(\alpha,\vartheta)\to(2,\vartheta_{\ast})$.
Further it is easily seen to the be dominated by an integrable function
uniformly for $\alpha$ bounded away from $1$. We conclude from the dominated
convergence theorem that $p^{\alpha,\vartheta}_{t}$ converges uniformly to
$p^{2,\vartheta_{\ast}}_{t}$. By applying Lemma 5.1 with the endpoint $0$,
this implies after a shift in both conditionings that
$(X^{\alpha}_{t}-\vartheta t)_{t\in[0,1]}$ under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha}_{1}=\vartheta)$ converges in
law to $(X^{2}-\vartheta_{\ast}t)_{t\in[0,1]}$ under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{2}_{1}=\vartheta_{\ast})$. By Corollary
6.4, the latter has the same law as $\sqrt{2}$ times the standard Brownian
bridge. The excursions further converge by the Vervaat transform. ∎

We know turn to the second case, using the Esscher transform presented in
Section 6.2.

###### Proof Theorem 6.1(i) when either $\vartheta_{\ast}=-\infty$ or both
$\alpha_{\ast}=1$ and $\vartheta_{\ast}<-1$.

We may assume that $\vartheta<0$. Let
$\kappa=(|\vartheta|/\alpha)^{1/(\alpha-1)}>0$ be the solution to
$\vartheta=-\alpha\kappa^{\alpha-1}$ and observe that under either assumption
the ratio $|\vartheta|/\alpha$ is asymptotically larger than $1$. In
particular $\kappa$ tends to infinity and even faster than any negative power
of $\alpha-1$. Let $X^{(\alpha,\kappa)}$ be defined by (6.4) from the stable
process $X^{\alpha}$ and let then $Y^{(\alpha,\kappa)}$ be the process given
by:

$Y^{(\alpha,\kappa)}_{t}=\biggl{(}\frac{|\vartheta|^{(2-\alpha)/(\alpha-1)}}{(\alpha-1)\alpha^{1/(\alpha-1)}}\biggr{)}^{1/2}(X^{(\alpha,\kappa)}_{t}-\vartheta
t)=\frac{\kappa}{\sqrt{\alpha(\alpha-1)\kappa^{\alpha}}}(X^{(\alpha,\kappa)}_{t}+\alpha\kappa^{\alpha-1}t),$

for every $t\geq 0$. It is enough to check that $Y^{(\alpha,\kappa)}$ under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid Y^{(\alpha,\kappa)}_{1}=0)$ converges in
distribution to the Brownian bridge. Indeed, by Lemma 6.3, the law of
$(X^{(\alpha,\kappa)}_{t}-\vartheta t)_{t\in[0,1]}$ under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{(\alpha,\kappa)}_{1}=\vartheta)$ is the
same as that of $X^{\alpha,\vartheta,\mathrm{br}}$, and the desired result
then follows by applying the Vervaat transform. As for the case $\alpha\to 2$,
by Lemma 5.1, the convergence of the bridge of $Y^{(\alpha,\kappa)}$ will
follow from the convergence of its densities, and relying on inverse Fourier
transform, it suffices to show that the characteristic function of
$Y^{(\alpha,\kappa)}_{t}$ converges to that of the centred Gaussian with
variance $t$ and is dominated by an integrable function independent of
$\alpha$ and $\kappa$.

First, recall that the Laplace transform of $X^{\alpha}$ extends to the
complex half plane, namely for any complex number $z=(a+ib)$ with $a>0$, we
have:

$\displaystyle\mathop{}\\!\mathbb{E}[\exp(-zX^{\alpha}_{t})]$
$\displaystyle=\exp\bigl{(}t|z|^{\alpha}\operatorname{e}^{i\alpha\arg(z)}\bigr{)}$
$\displaystyle=\exp\bigl{(}t(a^{2}+b^{2})^{\alpha/2}\bigl{(}\cos(\alpha\arctan(b/a))+i\sin(\alpha\arctan(b/a))\bigr{)}\bigr{)}.$

Consequently, fix $u\in\mathbb{R}$ and let
$x=(\alpha(\alpha-1)\kappa^{\alpha})^{-1/2}u$ to ease notation, then we have:

$\displaystyle\mathop{}\\!\mathbb{E}[\exp(iuY^{(\alpha,\kappa)}_{t})]$
$\displaystyle=\mathop{}\\!\mathbb{E}\bigl{[}\exp\bigl{(}i\kappa
x(X^{\alpha}_{t}+\alpha\kappa^{\alpha-1}t)-\kappa
X^{\alpha}_{t}-\kappa^{\alpha}t\bigr{)}\bigr{]}$
$\displaystyle=\mathop{}\\!\mathbb{E}\bigl{[}\exp\bigl{(}-(\kappa-i\kappa
x)X^{\alpha}_{t}\bigr{)}\bigr{]}\exp(i\alpha
x\kappa^{\alpha}t-\kappa^{\alpha}t)$
$\displaystyle=\exp\bigl{(}t(\kappa^{2}+(\kappa
x)^{2})^{\alpha/2}\bigl{(}\cos(\alpha\arctan(-x))+i\sin(\alpha\arctan(-x))\bigr{)}\bigr{)}\exp(i\alpha
x\kappa^{\alpha}t-\kappa^{\alpha}t)$
$\displaystyle=\exp\bigl{(}t\kappa^{\alpha}\bigl{(}(1+x^{2})^{\alpha/2}\cos(\alpha\arctan(x))-1\bigr{)}-it\kappa^{\alpha}\bigl{(}(1+x^{2})^{\alpha/2}\sin(\alpha\arctan(x))-\alpha
x\bigr{)}\bigr{)}.$

Notice that $x\to 0$, then elementary calculus shows that

$(1+x^{2})^{\alpha/2}\cos(\alpha\arctan(x))-1\sim-\frac{\alpha(\alpha-1)x^{2}}{2}=-\frac{u^{2}}{2\kappa^{\alpha}},$

as well as

$(1+x^{2})^{\alpha/2}\sin(\alpha\arctan(x))-\alpha x\sim\frac{\alpha
x^{3}}{2}=\frac{u^{3}}{2\sqrt{\alpha}((\alpha-1)\kappa^{\alpha})^{3/2}}=o(\kappa^{-\alpha}),$

since $\kappa$ grows faster than any negative power of $\alpha-1$. We infer
that

$\mathop{}\\!\mathbb{E}[\exp(iuY^{(\alpha,\kappa)}_{t})]\enskip\mathop{\longrightarrow}_{\kappa\to\infty}\enskip\exp\Bigl{(}-\frac{u^{2}t}{2}\Bigr{)}.$

Next we claim that there exist universal constants $C,q>0$ such that for all
$u\in\mathbb{R}$,

$|\mathop{}\\!\mathbb{E}[\exp(iuY^{(\alpha,\kappa)}_{t})]|\leq\exp(-C|u|^{q}t).$
(6.6)

Let us again put $x=u/\sqrt{\alpha(\alpha-1)\kappa^{\alpha}}$ to ease notation
and recall the explicit expression of the characteristic function above,
namely:

$\frac{1}{t}\log|\mathop{}\\!\mathbb{E}[\exp(iuY^{(\alpha,\kappa)}_{t})]|=\kappa^{\alpha}(1+x^{2})^{\alpha/2}\cos(\alpha\arctan
x)-\kappa^{\alpha}.$

Since the function is even, we focus on the case $x>0$. First, one can refine
the preceding asymptotic equivalence to show that

$-\frac{\alpha(\alpha-1)x^{2}}{2}\leq(1+x^{2})^{\alpha/2}\cos(\alpha\arctan
x)-1\leq-\frac{\alpha(\alpha-1)x^{2}}{2}(1-12x^{2}).$

Hence for $|x|\leq 1/\sqrt{24}$, the second inequality yields:

$\kappa^{\alpha}(1+x^{2})^{\alpha/2}\cos(\alpha\arctan
x)-\kappa^{\alpha}\leq-\frac{\alpha(\alpha-1)\kappa^{\alpha}x^{2}}{4},$

and the claimed bound (6.6) thus holds in this case, with $q=2$. On the other
hand, letting $y=\arctan x\in(0,\pi/2)$, the derivative of
$x^{-\alpha}((1+x^{2})^{\alpha/2}\cos(\alpha\arctan x)-1)$ has the same sign
as

$\cos(\alpha y)-\tan(y)\sin(\alpha y)-1=\cos(y)\cos(\alpha
y)-\sin(y)\sin(\alpha y)-1=\cos((\alpha+1)y)-1<0.$

Therefore the function is decreasing, and since it vanishes at $0$ by the
preceding bounds, then letting $-c<0$ denote its value at $x=1/\sqrt{24}$, we
infer that for every $x\geq 1/\sqrt{24}$, we have:

$\kappa^{\alpha}(1+x^{2})^{\alpha/2}\cos(\alpha\arctan
x)-\kappa^{\alpha}\leq-c(\kappa
x)^{\alpha}=-c\alpha^{-\alpha/2}(\alpha-1)^{-\alpha/2}\kappa^{\alpha(1-\alpha/2)}u^{\alpha}.$

Since $\kappa$ grows faster than any negative power of $\alpha-1$, then the
claimed bound (6.6) also holds in this case, with $q=\alpha$.

We conclude from Lebesgue’s theorem that

$\int_{\mathbb{R}}|\mathop{}\\!\mathbb{E}[\exp(iuY^{(\alpha,\kappa)}_{t})]-\operatorname{e}^{-u^{2}t/2}|\mathop{}\\!\mathrm{d}u$

converges to $0$, so by inverse Fourier transform, the density of
$Y^{(\alpha,\kappa)}_{t}$ converges uniformly to that of a Brownian motion at
time $t$, and so do the bridges and excursions from Lemma 5.1. ∎

###  6.4. Technical estimates on stable laws

We next want to consider the behaviour when either $\alpha\downarrow 1$ or
$\vartheta\to\infty$. In order to transfer estimates from the unconditioned
process $X^{\alpha}$ to a bridge conditioned on $X^{\alpha}_{1}=\vartheta$, we
need some estimates on the transition densities of this process. Recall that
the characteristic function of $X^{\alpha}_{t}$ is given by:

$\mathop{}\\!\mathbb{E}[\exp(iuX^{\alpha}_{t})]=\exp\Bigl{(}t|u|^{\alpha}\Bigl{(}\cos\Bigl{(}\frac{\pi\alpha}{2}\Bigr{)}-i\sigma(u)\sin\Bigl{(}\frac{\pi\alpha}{2}\Bigr{)}\Bigr{)}\Bigr{)},$

for every $u\in\mathbb{R}$, so by inverse Fourier, the density
$p^{\alpha}_{t}$ of $X^{\alpha}_{t}$ is precisely given by:

$\displaystyle p^{\alpha}_{t}(x)$
$\displaystyle=\frac{1}{2\pi}\int_{-\infty}^{\infty}\operatorname{e}^{-iux}\mathop{}\\!\mathbb{E}[\operatorname{e}^{iuX^{\alpha}_{t}}]\mathop{}\\!\mathrm{d}u$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}\operatorname{e}^{tu^{\alpha}\cos(\pi\alpha/2)}\cos\bigl{(}ux+tu^{\alpha}\sin(\pi\alpha/2)\bigr{)}\mathop{}\\!\mathrm{d}u.$

It is bounded, and precisely the change of variable
$v=-\cos(\pi\alpha/2)tu^{\alpha}$ yields:

$p^{\alpha}_{t}(x)\leq\frac{1}{\pi}\int_{0}^{\infty}\operatorname{e}^{tu^{\alpha}\cos(\pi\alpha/2)}\mathop{}\\!\mathrm{d}u\leq\frac{1}{\pi}\int_{0}^{\infty}\frac{1}{\alpha|t\cos(\pi\alpha/2)|^{1/\alpha}}v^{\frac{1}{\alpha}-1}\operatorname{e}^{-v}\mathop{}\\!\mathrm{d}v=\frac{\Gamma(1/\alpha)}{\alpha\pi|t\cos(\pi\alpha/2)|^{1/\alpha}}.$

This upper bound is finite for any $\alpha\in(1,2)$ and as $\alpha\downarrow
1$, we have $\cos(\pi\alpha/2)=-\sin(\pi(\alpha-1)/2)\sim-\pi(\alpha-1)/2$,
so:

$\sup_{x\in\mathbb{R}}p^{\alpha}_{t}(x)\leq\frac{\Gamma(1/\alpha)}{\alpha\pi|t\cos(\pi\alpha/2)|^{1/\alpha}}\enskip\mathop{\sim}_{\alpha\downarrow
1}\enskip\frac{2}{\pi^{2}t(\alpha-1)},$ (6.7)

where we used that $(\alpha-1)^{1-1/\alpha}\to 1$ as $\alpha\to 1$. Note that
the scaling property (or a simple change of variable in the inverse Fourier
transform) implies that
$p^{\alpha}_{t}(x)=t^{-1/\alpha}p^{\alpha}_{1}(t^{-1/\alpha}x)$. For this
reason, below we restrict to $t=1$.

When $\alpha\in(1,2)$ is fixed, Zolotarev [Zol86, Chapter 2.5] provides
precise series expansions as $x\to\pm\infty$ of the density
$p^{\alpha}_{1}(x)$, which corresponds to $g(x,\alpha,1)$ in the notation
there. In particular, by [Zol86, Equation 2.5.4] we have:

$p^{\alpha}_{1}(x)\enskip\mathop{\sim}_{x\to\infty}\enskip\frac{\alpha\Gamma(\alpha)}{\pi}\sin(\pi(2-\alpha))x^{-\alpha-1}.$

Note that as $\alpha\downarrow 1$, we have
$\sin(\pi(2-\alpha))=\sin(\pi(\alpha-1))\sim\pi(\alpha-1)$. In the next lemma,
we let both $x$ converge to a finite or infinite limit and $\alpha\downarrow
1$.

######  Lemma 6.5.

Let $x_{\ast}\in(-1,\infty]$, then as the pair $(\alpha,x)$ converge to
$(1,x_{\ast})$, we have:

$\frac{(1+x)^{\alpha+1}p^{\alpha}_{1}(x)}{\alpha-1}\enskip\mathop{\longrightarrow}\enskip
1.$

In addition, for every $\varepsilon>0$, there exists $C>0$ such that the upper
bound

$\frac{(1+x)^{\alpha+1}p^{\alpha}_{1}(x)}{\alpha-1}\leq C$

holds uniformly for all $\alpha\in(1,2-\varepsilon]$ and
$x\geq-1+\varepsilon$.

The proof involves heavy computations relying on some results from [Zol86],
and is deferred to Section 6.7. Combined with (6.7), we infer that when either
$\alpha\downarrow 1$ or $\vartheta\to\infty$, we have for any $t\in(0,1)$
fixed:

$\frac{\sup_{x\in\mathbb{R}}p^{\alpha}_{t}(x)}{p^{\alpha}_{1}(\vartheta)}\leq\frac{2}{\pi^{2}t}\frac{(\vartheta+1)^{\alpha+1}}{(\alpha-1)^{2}}(1+o(1)).$
(6.8)

In addition, using also the scaling property
$p^{\alpha}_{t}(x)=t^{-1/\alpha}p^{\alpha}_{1}(t^{-1/\alpha}x)$, for every
$\varepsilon>0$, there exists $C_{\varepsilon}>0$ such that the bound

$\frac{p^{\alpha}_{t}(x)}{p^{\alpha}_{1}(\vartheta)}\leq
t^{-1/\alpha}C_{\varepsilon}\Bigl{(}\frac{\vartheta+1}{t^{-1/\alpha}x+1}\Bigr{)}^{\alpha+1}$
(6.9)

holds for every $x\geq t^{1/\alpha}(-1+\varepsilon)$ and $\alpha\leq
2-\varepsilon$. This will enable us to transfer bounds on probabilities from
the unconditioned process to the bridge version.

###  6.5. The condensation regime

Let us next turn to Theorem 6.1(ii) and prove that when either
$\vartheta\to\infty$ while $\alpha$ stays bounded away from $2$, or $\alpha\to
1$ and $\vartheta>-1$ stays bounded away from $-1$, we have:

$(|\vartheta+1|^{-1}(X^{\alpha}_{t}-\vartheta
t))_{t\in[0,1]}\enskip\mathop{\longrightarrow}^{(d)}\enskip(\mathbf{1}_{\\{U\leq
t\\}}-t)_{t\in[0,1]},$ (6.10)

under the conditional law $\mathop{}\\!\mathbb{P}(\,\cdot\mid
X^{\alpha}_{1}=\vartheta)$, where $U$ has the uniform distribution on $[0,1]$.
The actual claim of the theorem, under the excursion law, then follows from
the Vervaat transform as before.

The idea is to cut the trajectory into three components as follows. Let
$\pi^{\alpha}(\mathop{}\\!\mathrm{d}x)=\frac{\alpha(\alpha-1)}{\Gamma(2-\alpha)}x^{-\alpha-1}\mathop{}\\!\mathrm{d}x$
denote the Lévy measure of $X^{\alpha}$ and set

$c=(\alpha-1)^{p}(\vartheta+1)^{q}$

with some $p\in(0,1)$ small and $q\in(0,1)$ close to $1$. Then let us write:

$X^{\alpha}_{t}=\sum_{s\leq t}\Delta X^{\alpha}_{s}\mathbf{1}_{\\{\Delta
X^{\alpha}_{s}>c\\}}+X^{\alpha,\leq
c}_{t}-t\int_{c}^{\infty}x\pi^{\alpha}(\mathop{}\\!\mathrm{d}x).$ (6.11)

On the right, the integral equals the drift of the sum, therefore the middle
term $X^{\alpha,\leq c}$ is the process $X^{\alpha}$ to which we remove every
jump larger than $c$ and that we recenter to give it mean $0$.

Let us first notice that

$\int_{c}^{\infty}x\pi^{\alpha}(\mathop{}\\!\mathrm{d}x)=\frac{\alpha}{\Gamma(2-\alpha)}c^{-(\alpha-1)}=\frac{\alpha}{\Gamma(2-\alpha)}(\alpha-1)^{-p(\alpha-1)}(\vartheta+1)^{-q(\alpha-1)},$

which converges to $1$ as both $\alpha\downarrow 1$ and $\vartheta$ is bounded
away from $\infty$ and it is small compared to $\vartheta$ when the latter
tends to infinity. In any case, in the regimes of Theorem 6.1(ii), we have:

$\frac{\vartheta+\int_{c}^{\infty}x\pi^{\alpha}(\mathop{}\\!\mathrm{d}x)}{\vartheta+1}\enskip\longrightarrow\enskip
1.$ (6.12)

Thus if we decompose $(X^{\alpha}_{t}-\vartheta t)/(\vartheta+1)$ as in
(6.11), then the drift component converges to $-t$. We shall prove that, for
some well-chosen values of $p$ and $q$, under the conditional law
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha}_{1}=\vartheta)$, with high
probability the path $X^{\alpha,\leq c}$ is small compared to $\vartheta+1$
and the path $\sum_{s\leq t}\Delta X^{\alpha}_{s}\mathbf{1}_{\\{\Delta
X^{\alpha}_{s}>c\\}}$ is composed of a single jump, with size about
$\vartheta+1$. This implies (6.10).

Let us first prove that the small jumps component $X^{\alpha,\leq c}$ from
(6.11) is small compared to $\vartheta+1$. This relies on the following lemma
inspired from [KR02].

######  Lemma 6.6.

Fix $0<r<p<1$ and $0<q<s<1$ and let $c=(\alpha-1)^{p}(\vartheta+1)^{q}$ and
$K=(\alpha-1)^{r}(\vartheta+1)^{s}$. Then

$\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{s\leq 1}|X^{\alpha,\leq
c}_{s}|>K\bigm{|}X^{\alpha}_{1}=\vartheta\Bigr{)}\enskip\mathop{\longrightarrow}\enskip
0$

as either $\vartheta\to\infty$ and $\alpha\to\alpha_{\ast}<2$ or
$\vartheta\to\vartheta_{\ast}>-1$ and $\alpha\to 1$.

Notice that

$\frac{K}{\vartheta+1}=(\alpha-1)^{r}(\vartheta+1)^{s-1}\enskip\longrightarrow\enskip
0,$

as either $\alpha\downarrow 1$ or $\vartheta\to\infty$. Combined with (6.12),
we deduce that

$(\vartheta+1)^{-1}\Bigl{(}X^{\alpha,\leq c}_{t}-\vartheta
t-t\int_{c}^{\infty}x\pi^{\alpha}(\mathop{}\\!\mathrm{d}x)\Bigr{)}\enskip\mathop{\longrightarrow}^{\mathop{}\\!\mathbb{P}}\enskip-t,$
(6.13)

under $\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha}_{1}=\vartheta)$ and for
the uniform topology on $[0,1]$.

###### Proof.

Let us start with the unconditioned process. Note that $X^{\alpha,\leq c}$
remains a Lévy process, and since the increments are bounded, then the Laplace
transform is well-defined on the whole real axis and for every $t>0$ and
$\lambda>0$, we have:

$\log\mathop{}\\!\mathbb{E}[\exp(\lambda X^{\alpha,\leq
c}_{1})]=\int_{0}^{c}(\operatorname{e}^{\lambda u}-1-\lambda
u)\pi^{\alpha}(\mathop{}\\!\mathrm{d}u).$

Recall the bound $\operatorname{e}^{x}-1-x\leq x^{2}$ for $0\leq x\leq 1$,
then for $\lambda=1/c$ we get:

$\log\mathop{}\\!\mathbb{E}[\exp(c^{-1}X^{\alpha,\leq
c}_{1})]\leq\int_{0}^{c}(u/c)^{2}\frac{\alpha(\alpha-1)}{\Gamma(2-\alpha)}u^{-\alpha-1}\mathop{}\\!\mathrm{d}u=\frac{\alpha(\alpha-1)}{(2-\alpha)\Gamma(2-\alpha)}c^{-\alpha},$

Applying the maximal inequality to the exponential martingale with parameter
$c^{-1}$, we obtain that for any $K>0$, we have

$\displaystyle\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{s\leq 1/2}|X^{\alpha,\leq
c}_{s}|>K\Bigr{)}$ $\displaystyle\leq
2\exp\Bigl{(}-\frac{K}{c}+\frac{1}{2}\log\mathop{}\\!\mathbb{E}[\exp(c^{-1}X^{\alpha,\leq
c}_{1})]\Bigr{)}$ $\displaystyle\leq
2\exp\Bigl{(}-\frac{K}{c}+\frac{\alpha(\alpha-1)c^{-\alpha}}{2(2-\alpha)\Gamma(2-\alpha)}\Bigr{)},$

see [KR02, Proposition 1] for details. Let us next transfer this bound to the
bridge version by absolute continuity. Since the bridge is invariant under
time- and space-reversal, then we have, using (6.8) for the last inequality,

$\displaystyle\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{s\leq 1}|X^{\alpha,\leq
c}_{s}|>K\bigm{|}X^{\alpha}_{1}=\vartheta\Bigr{)}$ $\displaystyle\leq
2\mathop{}\\!\mathbb{E}\Bigl{[}\mathbf{1}_{\\{\sup_{s\leq 1/2}|X^{\alpha,\leq
c}_{s}|>K\\}}\frac{p^{\alpha}_{1/2}(\vartheta-X^{\alpha}_{1/2})}{p^{\alpha}_{1}(\vartheta)}\Bigr{]}$
$\displaystyle\leq 2\mathop{}\\!\mathbb{P}\Bigl{(}\sup_{s\leq
1/2}|X^{\alpha,\leq c}_{s}|>K\Bigr{)}\frac{\sup
p^{\alpha}_{1/2}}{p^{\alpha}_{1}(\vartheta)}$ $\displaystyle\leq
4\exp\Bigl{(}-\frac{K}{c}+\frac{\alpha(\alpha-1)c^{-\alpha}}{2(2-\alpha)\Gamma(2-\alpha)}\Bigr{)}\frac{4}{\pi^{2}}\frac{(\vartheta+1)^{\alpha+1}}{(\alpha-1)^{2}}(1+o(1)).$

Our choice of $c$ and $K$ yields:

$\frac{K}{c}=(\alpha-1)^{r-p}(\vartheta+1)^{s-q}\qquad\text{and}\qquad(\alpha-1)c^{-\alpha}=(\alpha-1)^{1-p}(\alpha-1)^{-p(\alpha-1)}(\vartheta+1)^{-q\alpha}.$

From $r<p<1$ and $0<q<s$ we infer that $(\alpha-1)c^{-\alpha}\to 0$ when
either $\alpha\downarrow 1$ or $\vartheta\to\infty$, and that $\exp(-K/c)$
goes to $0$ faster than $(\alpha-1)^{2}(\vartheta+1)^{-(\alpha+1)}$. ∎

We finally consider the jumps of $X^{\alpha}$ larger than $c$ in the
decomposition (6.11). By (6.13), if we remove these jumps from
$(X^{\alpha}_{t}-\vartheta t)/(\vartheta+1)$, then the remaining path under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha}_{1}=\vartheta)$ converges to
the linear drift $-t$. Since the path has to end at
$X^{\alpha}_{1}=\vartheta\to\vartheta_{\ast}>-1$, then it necessarily has at
least one jump larger than $c$, and the sum of these jumps, rescaled by
$\vartheta+1$, converges to $1$. Let us prove that with high probability,
there exists only one such jump, therefore necessarily of size
$(\vartheta+1)(1+o(1))$. Let us first rule out the possibility of more than
two jumps larger than $c$.

######  Lemma 6.7.

Suppose that either $\alpha\to\alpha_{\ast}\in(1,2)$ and $\vartheta\to\infty$
or $\alpha\downarrow 1$ and $\vartheta\to\vartheta_{\ast}\in(-1,\infty]$. Let
$p\in(0,1/6)$ and $1>q>(\alpha+1)/(2\alpha)$, which is indeed asymptotically
smaller than $1$. Let $c=(\alpha-1)^{p}(\vartheta+1)^{q}$, then

$\mathop{}\\!\mathbb{P}(\\#\\{t\in(0,1)\colon\Delta X^{\alpha}_{t}>c\\}\geq
3\mid X^{\alpha}_{1}=\vartheta)\enskip\longrightarrow\enskip 0.$

###### Proof.

Let us first consider the unconditioned process. The jump sizes of
$X^{\alpha}$ are then given by a Poisson random measure, hence:

$\mathop{}\\!\mathbb{P}(\\#\\{t\in(0,1)\colon\Delta X^{\alpha}_{t}>c\\}\geq
3)=\mathop{}\\!\mathbb{P}(\mathrm{Poisson}(\overline{\pi}^{\alpha}(c))\geq
3)\sim\frac{\overline{\pi}^{\alpha}(c)^{3}}{6}.$

The tail of the Lévy measure $\pi^{\alpha}$ is given by:

$\overline{\pi}^{\alpha}(c)=\int_{c}^{\infty}\frac{\alpha(\alpha-1)}{\Gamma(2-\alpha)}x^{-\alpha-1}\mathop{}\\!\mathrm{d}x=\frac{\alpha-1}{\Gamma(2-\alpha)}c^{-\alpha}=\frac{\alpha-1}{\Gamma(2-\alpha)}(\alpha-1)^{-p\alpha}(\vartheta+1)^{-q\alpha},$

which converges to $0$. Let us transfer this result to the bridge by absolute
continuity and the bound (6.8) on the ratio of densities. Precisely, for every
$\varepsilon>0$, it holds:

$\displaystyle\mathop{}\\!\mathbb{P}(\\#\\{t\in(0,1-\varepsilon)\colon\Delta
X^{\alpha}_{t}>c\\}\geq 3\mid X^{\alpha}_{1}=\vartheta)$
$\displaystyle=\mathop{}\\!\mathbb{E}\Bigl{[}\mathbf{1}_{\\{\\#\\{t\in(0,1-\varepsilon)\colon\Delta
X^{\alpha}_{t}>c\\}\geq
3\\}}\frac{p^{\alpha}_{\varepsilon}(\vartheta-X^{\alpha}_{1-\varepsilon})}{p^{\alpha}_{1}(\vartheta)}\Bigr{]}$
$\displaystyle\leq\mathop{}\\!\mathbb{P}(\\#\\{t\in(0,1)\colon\Delta
X^{\alpha}_{t}>c\\}\geq 3)\frac{\sup
p^{\alpha}_{\varepsilon}}{p^{\alpha}_{1}(\vartheta)}$
$\displaystyle\leq\frac{\overline{\pi}^{\alpha}(c)^{3}}{6}\frac{\sup
p^{\alpha}_{\varepsilon}}{p^{\alpha}_{1}(\vartheta)}(1+o(1)).$

Recall that
$\overline{\pi}^{\alpha}(c)=\frac{\alpha-1}{\Gamma(2-\alpha)}(\alpha-1)^{-p\alpha}(\vartheta+1)^{-q\alpha}$
and recall the bound (6.8) on the ratio of densities, so the last line is
bounded by some constant times

$(\alpha-1)^{1-3p\alpha}(\vartheta+1)^{\alpha+1-3q\alpha},$

and we have $1-3p\alpha>1-\alpha/2$ is bounded away from $0$ and
$\alpha+1-3q\alpha<-(\alpha+1)/2<-1$. This shows that with high probability
the bridge does not make more than two jumps larger than $c$ before time
$1-\varepsilon$. By exchangeability, the probability that it makes one of its
two largest jumps after time $1-\varepsilon$ is small when $\varepsilon$ is.
Hence it does not make more than two jumps at all, with high probability. ∎

######  Remark 6.8.

Let us point out that when $\alpha$ stays bounded away from $1$ and
$\vartheta\to\infty$, the preceding argument can be adapted to show directly
that

$\mathop{}\\!\mathbb{P}(\\#\\{t\in(0,1)\colon\Delta X^{\alpha}_{t}>c\\}\geq
2\mid X^{\alpha}_{1}=\vartheta)\enskip\longrightarrow\enskip 0.$

Indeed, the probability to make at least two jumps larger than $c$ before time
$1-\varepsilon$ is now bounded above by a constant times
$(\alpha-1)^{-2p\alpha}(\vartheta+1)^{\alpha+1-2q\alpha}$, which converges to
$0$ in this case. However this fails when $\alpha\downarrow 1$.

Let us finally conclude that there exists a single jump, which completes the
proof the theorem.

###### Proof Theorem 6.1(ii).

Let us introduce some notation: let us $c$ be as in Lemma 6.7 and set

$J_{t}=(\vartheta+1)^{-1}\sum_{s\leq t}\Delta
X^{\alpha}_{s}\mathbf{1}_{\\{\Delta X^{\alpha}_{s}>c\\}}\qquad{and}\qquad
D_{t}=(\vartheta+1)^{-1}(X^{\alpha}_{t}-\vartheta t)-J_{t},$

and then

$Y_{t}=J_{t}+D_{t}=(\vartheta+1)^{-1}(X^{\alpha}_{t}-\vartheta t),$

which corresponds to the decomposition (6.11). By (6.13), under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha}_{1}=\vartheta)$, the path
$(D_{t})_{t}$ converges to the linear drift $t~\mapsto-t$. Hence the path
$(J_{t})_{t}$ has to contain at least one jump. We just proved in Lemma 6.7
that with high probability, it has at most two, and actually only one when
$\alpha_{\ast}>1$. Let us prove that this remains true when $\alpha\downarrow
1$.

Let us work under $\mathop{}\\!\mathbb{P}(\,\cdot\mid
X^{\alpha}_{1}=\vartheta)$. Fix an integer $K\geq 1$, let $\varepsilon=1/K$
and let us split the interval $(0,1)$ into subintervals of length
$\varepsilon$. Our aim is to prove that with high probability, the path
$(J_{t})_{t}$ does not have jumps larger than $c$ in two different
subintervals. Indeed, on the other hand by exchangeability, the probability
that the two largest jumps occur in a time-frame $\varepsilon$ is small, so we
can conclude that with high probability, it has at most (and hence exactly)
one jump larger than $c$, necessarily of size approximately $\vartheta+1$. Fix
$\delta>0$ and let us assume that $\sup_{t\in[0,1]}|D_{t}+t|\leq\delta/2$.
Then $Y_{t}\geq D_{t}\geq-t-\delta/2$ for all $t\in[0,1]$. Also, since
$Y_{1}=0$ almost surely then $Y_{1-t}\leq t+\delta/2$.

Let $1\leq k\leq K-1$ and denote by $1-t=k\varepsilon$ the right-edge of the
$k$’th subinterval. Then

$\mathop{}\\!\mathbb{P}(Y_{1-t}\in(-1+t+\delta,t-\delta)\mid
X^{\alpha}_{1}=\vartheta)=\mathop{}\\!\mathbb{E}\Bigl{[}\mathbf{1}_{\\{Y_{1-t}\in(-1+t+\delta,t-\delta)\\}}\frac{p^{\alpha}_{t}(\vartheta-X^{\alpha}_{1-t})}{p^{\alpha}_{1}(\vartheta)}\Bigr{]}.$

Notice that $\vartheta-X^{\alpha}_{1-t}=\vartheta
t-(\vartheta+1)Y_{1-t}\geq-t+\delta(\vartheta+1)$ on the event in the
indicator function. We may then appeal to (6.9) to upper bound the ratio of
the densities and infer that to for some constant $C$, which depends only on
$\delta$, we have:

$\displaystyle\mathop{}\\!\mathbb{P}(Y_{1-t}\in(-1+t+\delta,t-\delta)\mid
X^{\alpha}_{1}=\vartheta)$ $\displaystyle\qquad\leq
C\,t^{-1/\alpha}\mathop{}\\!\mathbb{E}\Bigl{[}\mathbf{1}_{\\{Y_{1-t}\in(-1+t+\delta,t-\delta)\\}}\Bigl{(}\frac{\vartheta+1}{t^{-1/\alpha}(\vartheta-X^{\alpha}_{1-t})+1}\Bigr{)}^{\alpha+1}\Bigr{]}$
$\displaystyle\qquad\leq
C\,t^{-1/\alpha}\Bigl{(}\frac{\vartheta+1}{t^{-1/\alpha}\delta(\vartheta+1)-t^{-(\alpha-1)/\alpha}+1}\Bigr{)}^{\alpha+1}\mathop{}\\!\mathbb{P}(Y_{1-t}\in(-1+t+\delta,t-\delta)).$

The ratio in the last line is bounded above uniformly in $\alpha$ and
$\vartheta$. As for the last probability, recall that
$Y_{1-t}=D_{1-t}+J_{1-t}$ with $\sup_{t\in[0,1]}|D_{t}+t|\leq\delta/2$ so
$Y_{1-t}>-1+t+\delta$ implies that $J_{1-t}$ is nonzero, namely that the
process $X^{\alpha}$ has at least one jump larger than $c$ before time $1-t$.
However in this unconditioned model the jumps are given by a Poisson measure,
namely:

$\mathop{}\\!\mathbb{P}(J_{1-t}>0)=\mathop{}\\!\mathbb{P}(\\#\\{s\in(0,1-t)\colon\Delta
X^{\alpha}_{s}>c\\}\geq
1)=\mathop{}\\!\mathbb{P}(\mathrm{Poisson}((1-t)\overline{\pi}^{\alpha}(c))\geq
1)\sim(1-t)\overline{\pi}^{\alpha}(c).$

Recall that
$\overline{\pi}^{\alpha}(c)=\frac{1}{\Gamma(2-\alpha)}(\alpha-1)^{1-p\alpha}(\vartheta+1)^{-q\alpha}\to
0$ as either $\alpha\to 1$ or $\vartheta\to\infty$ (or both) since $p<1$.
Hence

$\mathop{}\\!\mathbb{P}(Y_{1-t}\in(-1+t+\delta,t-\delta)\mid
X^{\alpha}_{1}=\vartheta)\enskip\longrightarrow\enskip 0,$

so with high probability, either $|Y_{1-t}+(1-t)|\leq\delta$, or
$|Y_{1-t}-t|\leq\delta$. Recall that $1-t=k\varepsilon$ is the right-edge of
the $k$’th subinterval of length $\varepsilon=1/K$. Since $K$ is fixed, then
this holds jointly for all these intervals: with high probability under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha}_{1}=\vartheta)$, at the end of
each such interval, the process $Y$ is either $\delta$-close to the linear
drift $-t$, or $\delta$-close to this drift plus $1$.

Recall one last time that high probability under
$\mathop{}\\!\mathbb{P}(\,\cdot\mid X^{\alpha}_{1}=\vartheta)$, we may
decompose $Y=J+D$ with $D$ being $\delta/2$-close to the drift $t\mapsto-t$,
with $J$ having only (at most two) positive jumps and remaining constant
otherwise, and with $Y_{1}=Y_{0}=0$. Then if $J$ does not make any jump larger
than $\delta/2$ before time $t$, then the path $(Y_{s})_{s\leq t}$ remains
$\delta$-close to $(-s)_{s\leq t}$. Also if $Y_{t}$ is $\delta$-close to
$1-t$, then $J$ does not make jumps larger than $2\delta$ after time $t$ since
otherwise $Y_{1}>0$. We infer from the previous paragraph that with high
probability, the process $J$ only has jumps larger than $2\delta$ in a single
interval of length $\varepsilon$, in which $Y$ transitions from being close to
$t\mapsto-t$ to being close to $t\mapsto 1-t$. By exchangeability, the
probability that the two largest jumps of $Y$ occur in a time-frame at most
$\varepsilon$ is small when $\varepsilon$ is; hence the transition of $Y$
occurs by a single jump with high probability and this concludes the proof. ∎

###  6.6. Remarks on the regime $\vartheta=-1$ and $\alpha\downarrow 1$

Recall that we have excluded the two cases $(\alpha,\vartheta)\to(1,-1)$ and
$(\alpha,\vartheta)\to(2,\infty)$, in which case the asymptotic behaviour
depends more precisely on the rate of convergence. For example, the proof in
the Brownian regime extends as soon as the parameter in the Esscher transform
$\kappa=(|\vartheta|/\alpha)^{1/(\alpha-1)}$ tends to infinity faster than any
negative power of $\alpha-1$, which holds e.g. for
$\vartheta=\pm(\alpha-(\alpha-1)^{2})$. Let us here briefly discuss the case
when $\vartheta$ is fixed equal to $-1$ and $\alpha\downarrow 1$, which seems
to display an interesting behaviour.

Recall the small discussion just after the statement of Theorem 6.1 claiming
that the unconditioned process $X^{\alpha}$ converges in probability to the
drift $t\mapsto-t$ as $\alpha\downarrow 1$. More precisely, from the explicit
expression (6.1) of the characteristic function, it is straightforward to
check that

$(\lambda_{\alpha}(X^{\alpha}_{t}+t),t\geq
0)\enskip\mathop{\longrightarrow}^{\mathop{}\\!\mathbb{P}}_{\alpha\downarrow
1}\enskip(-t,t\geq
0)\qquad\text{where}\qquad\lambda_{\alpha}\log\lambda_{\alpha}=\frac{1}{\alpha-1}.$

In particular, the endpoint $\vartheta=-1$ looks typical for $X^{\alpha}_{1}$
at first glance, but actually the process wants to end near
$-1-\lambda_{\alpha}^{-1}$. We then conjecture as in the case $\vartheta>-1$
that the easiest way to end exactly at $-1$ is to make a single jump of size
approximately $\lambda_{\alpha}^{-1}\to 0$ and drift linearly otherwise, in a
“one-small jump principle”.

######  Conjecture 6.9.

Under the coniditional law $\mathop{}\\!\mathbb{P}(\,\cdot\mid
X^{\alpha}_{1}=-1)$, we have

$(\lambda_{\alpha}(X^{\alpha}_{t}+t))_{t\in[0,1]}\enskip\mathop{\longrightarrow}^{(d)}_{\alpha\downarrow
1}\enskip(\mathbf{1}_{\\{U\leq t\\}}-t)_{t\in[0,1]}$

where $\lambda_{\alpha}>0$ is the solution of
$\lambda_{\alpha}\log\lambda_{\alpha}=1/(\alpha-1)$.

We believe that the argument of the previous subsection can be adapted, after
simply replacing $\vartheta+1$ by $\lambda_{\alpha}^{-1}$, and taking
$p=1-q>0$ small. Indeed, define now

$c=(\alpha-1)^{p}\lambda_{\alpha}^{-(1-p)}=(\alpha-1)(\log\lambda_{\alpha})^{1-p}=\lambda_{\alpha}^{-1}(\log\lambda_{\alpha})^{-p},$

and let us split the process $X^{\alpha}$ as in (6.11), namely write:

$\lambda_{\alpha}(X^{\alpha}_{t}+t)=\lambda_{\alpha}\Bigl{(}\sum_{s\leq
t}\Delta X^{\alpha}_{s}\mathbf{1}_{\\{\Delta
X^{\alpha}_{s}>c\\}}+X^{\alpha,\leq
c}_{t}-t\Bigl{(}\int_{c}^{\infty}x\pi^{\alpha}(\mathop{}\\!\mathrm{d}x)-1\Bigr{)}\Bigr{)}.$

One can easily check that the drift coefficient does indeed converge to $-t$.
Then the argument to control the first two terms on the right for the
unconditioned process are the same as before. The key missing point is to
extend Lemma 6.5 and find the asymptotic behaviour of the density
$p^{\alpha}_{1}(-1)$ as $\alpha\downarrow 1$ to transfer the results to the
bridge.

###  6.7. Proof the technical estimates on stable laws

Let us finally prove Lemma 6.5. Recall that it claims that if $p^{\alpha}_{1}$
is the density at time $1$ of the stable process $X^{\alpha}$ then for every
$x_{\ast}\in(-1,\infty]$, we have:

$\frac{(1+x)^{\alpha+1}p^{\alpha}_{1}(x)}{\alpha-1}\enskip\mathop{\longrightarrow}\enskip
1$ (6.14)

as the pair $(\alpha,x)$ converge to $(1,x_{\ast})$ and that moreover for
every $\delta>0$, there exists $C>0$ such that the upper bound

$\frac{(1+x)^{\alpha+1}p^{\alpha}_{1}(x)}{\alpha-1}\leq C$ (6.15)

holds uniformly for all $\alpha\leq 2-\delta$ and $x\geq-1+\delta$.

###### Proof Lemma 6.5.

The density $p^{\alpha}_{1}(x)$ corresponds to $g(x,\alpha,1)$ in the notation
of Zolotarev [Zol86]. Equation 2.4.6 there provides the following series
expansion:

$g(x,\alpha,1)=\frac{1}{\pi}\sum_{n\geq
1}(-x)^{n-1}\frac{\Gamma(n/\alpha+1)}{\Gamma(n+1)}\sin(n\pi(\alpha-1)/\alpha).$

Once divided by $\alpha-1$, the $n$’th term converges to $\pi n(-x)^{n-1}$ as
$\alpha\downarrow 1$ and its absolute value is less than
$\alpha^{-1}n\pi|x|^{n-1}$ for all $\alpha$. Thus, when $|x_{\ast}|<1$, we
infer from dominated convergence that

$\frac{(1+x)^{\alpha+1}p^{\alpha}_{1}(x)}{\alpha-1}\enskip\mathop{\longrightarrow}\enskip(1+x_{\ast})^{2}\sum_{n\geq
1}n(-x_{\ast})^{n-1}=1.$

In addition, for every $\delta>0$ the left-hand side is bounded above by a
constant, which depends on $\delta$ but is uniform in $\alpha\in(1,2)$ and
$x\in(-1+\delta,1-\delta)$. Our claims (6.14) and (6.15) thus hold for
$|x|<1-\delta$.

Let us henceforth assume $x>0$ is bounded away from $0$. An asymptotic result
as $x\to\infty$ when $\alpha>1$ is fixed is proved in [Zol86, Equation 2.5.4]
by relating the $\alpha$-stable law to an $\alpha^{\prime}=1/\alpha<1$ stable
law. Precisely [Zol86, Equation 2.5.5] reads: for $x>0$,

$g(x,\alpha,1)=x^{-\alpha-1}g(x^{-\alpha},\alpha^{\prime},\beta)\qquad\text{where}\qquad\alpha^{\prime}=1/\alpha\enskip\text{and}\enskip\beta=2(\alpha-1)-1,$
(6.16)

where we recall that the left-hand side equals $p^{\alpha}_{1}(x)$. It thus
remains to study the behaviour of $g(y,\alpha^{\prime},\beta)$, with $y>0$
bounded away from $\infty$; we recall the definition of the latter just a few
lines below. It seems that the proof of [Zol86, Theorem 2.5.1] (with $n=1$)
when $\alpha$ is fixed and $y\to 0$ does not extend here, because the term
$\cos(\pi\alpha\beta/2)^{-2/\alpha}$ in [Zol86, Equation 2.5.2] blows up as
$\alpha\to 1$. We shall somehow use a tighter bound on the rest, without
bounding the real part by the modulus.

Let $\gamma=\pi\alpha^{\prime}\beta/2$ and $\varepsilon=\pi(\alpha-1)\to 0$.
The change of variable $v=u^{\alpha^{\prime}}\operatorname{e}^{i\gamma}$
yields:

$\displaystyle g(y,\alpha^{\prime},\beta)$
$\displaystyle=\frac{1}{\pi}\Re\int_{0}^{\infty}\operatorname{e}^{iuy}\operatorname{e}^{-u^{\alpha^{\prime}}\exp(i\gamma)}\mathop{}\\!\mathrm{d}u$
$\displaystyle=\frac{1}{\pi\alpha^{\prime}}\int_{0}^{\infty}\Re\bigl{(}\operatorname{e}^{-i\gamma/\alpha^{\prime}}\operatorname{e}^{iyv^{1/\alpha^{\prime}}\exp(-i\gamma/\alpha^{\prime})}\bigr{)}\operatorname{e}^{-v}v^{\frac{1}{\alpha^{\prime}}-1}\mathop{}\\!\mathrm{d}v$
$\displaystyle=\frac{\alpha}{\pi}\int_{0}^{\infty}\Re\bigl{(}\operatorname{e}^{-i\pi\beta/2}\operatorname{e}^{iyv^{\alpha}\exp(-i\pi\beta/2)}\bigr{)}\operatorname{e}^{-v}v^{\alpha-1}\mathop{}\\!\mathrm{d}v.$

Observe that $\cos(\pi\beta/2)=\sin(\varepsilon)$ and
$\sin(\pi\beta/2)=-\cos(\varepsilon)$, which implies that the real part in the
integral equals:

$\displaystyle\Re\big{(}\operatorname{e}^{-i\pi\beta/2}\operatorname{e}^{iyv^{\alpha}\exp(-i\pi\beta/2)}\bigr{)}$
$\displaystyle=\Re\bigl{(}(\sin(\varepsilon)+i\cos(\varepsilon))\operatorname{e}^{-yv^{\alpha}\cos(\varepsilon)+iyv^{\alpha}\sin(\varepsilon)}\bigr{)}$
$\displaystyle=\operatorname{e}^{-yv^{\alpha}\cos(\varepsilon)}\Re\bigl{(}(\sin(\varepsilon)+i\cos(\varepsilon))(\cos(yv^{\alpha}\sin(\varepsilon))+i\sin(yv^{\alpha}\sin(\varepsilon))\bigr{)}$
$\displaystyle=\operatorname{e}^{-yv^{\alpha}\cos(\varepsilon)}\bigl{(}\sin(\varepsilon)\cos(yv^{\alpha}\sin(\varepsilon))-\cos(\varepsilon)\sin(yv^{\alpha}\sin(\varepsilon))\bigr{)}.$

Notice that for $z>0$ it holds both $1-z^{2}/2\leq\cos(z)\leq 1$ and
$z\geq\sin(z)\geq z-z^{3}/6$. We infer that:

$-\sin(\varepsilon)z^{2}/2\leq(\sin(\varepsilon)\cos(z)-\cos(\varepsilon)\sin
z)-(\sin(\varepsilon)-\cos(\varepsilon)z)\leq\cos(\varepsilon)z^{3}/6.$

In addition, by dominated convergence, for any integer $k\geq 1$, as $y\to
y_{\ast}\in[0,\infty)$ and $\alpha\to 1$ it holds: