# Convergences of looptrees coded by excursions

Robin Khanfir Sorbonne Université, Campus Pierre et Marie Curie, LPSM, Case
courrier 158, 4, place Jussieu, 75252 Paris Cedex 05 France. Email:
<EMAIL_ADDRESS>
###### Abstract

In order to study convergences of looptrees, we construct continuum trees and
looptrees from real-valued càdlàg functions without negative jumps called
excursions. We then provide a toolbox to manipulate the two resulting codings
of metric spaces by excursions and we formalize the principle that jumps
correspond to loops and that continuous growths correspond to branches.
Combining these codings creates new metric spaces from excursions that we call
vernation trees. They consist of a collection of loops and trees glued along a
tree structure so that they unify trees and looptrees. We also propose a
topological definition for vernation trees, which yields what we argue to be
the right space to study convergences of looptrees. However, those first
codings lack some functional continuity, so we adjust them. We thus obtain
several limit theorems. Finally, we present some probabilistic applications,
such as proving an invariance principle for random discrete looptrees.

Keywords Looptree $\cdot$ Tree $\cdot$ Coding by real-valued functions $\cdot$
Limit theorem $\cdot$ Scaling limit $\cdot$ Random metric space $\cdot$
Geodesic space

Mathematics Subject Classification 60F17 $\cdot$ 54C30 $\cdot$ 54E70 $\cdot$
05C05 $\cdot$ 54F50

###### Contents

  1. 1 Introduction
  2. 2 Tree, looptree, and vernation tree coded by an excursion
    1. 2.1 Genealogy associated with an excursion
    2. 2.2 Pseudo-distances coded by an excursion
    3. 2.3 Quotient metric spaces induced by those pseudo-distances
    4. 2.4 First properties and an example
  3. 3 Unification and relations between trees, looptrees, and vernation trees
  4. 4 Limit theorems
    1. 4.1 Continuity for the shuffled vernation tree pseudo-distance
    2. 4.2 Two particular cases for convergences of unshuffled pseudo-distances
    3. 4.3 From convergences of pseudo-distances to convergences of quotient metric spaces
  5. 5 Topological space of vernation trees
  6. 6 Probabilistic applications
    1. 6.1 Metric asymptotics for uniform random mappings
    2. 6.2 Continuity in distribution of random stable looptrees
    3. 6.3 Invariance principle for random discrete looptrees

##  1 Introduction

#### Background

Informally, a looptree consists of a collection of loops tangently glued
together along some genealogical structure. Perhaps more clearly, one can
associate a looptree with a finite ordered tree by replacing each vertex with
a circle of length proportional to its degree and each edge with a unique
point of tangency between the corresponding circles, thereby preserving the
ordered tree structure. An example will be given with Figure 3 in Section 6.3.
Ever since their introduction by Curien and Kortchemski in [18], looptrees
have generated a growing interest thanks to their natural appearances in some
random geometric models and because they can provide useful tools or insights
to study other objects. Let us mention some of them. To begin with, when a
sequence of trees admits a scaling limit and if the maximum degree becomes
negligible against the height, then the scaling limit is often a continuum
tree whose all branch points are of infinite degree. In this situation, it is
then difficult to recover the limiting joint distribution of the scaled
degrees from that object. However, a scaling limit of the associated looptrees
would not only give such information (merely via the lengths of the loops) but
also the whole asymptotic degree structure. This is exactly what happens for
large critical Galton-Watson trees whose offspring distribution belongs to the
domain of attraction of an $\alpha$-stable law, with $\alpha\in(1,2)$. While
Duquesne showed in [21] these random trees converge, after suitable scaling,
towards the $\alpha$-stable tree introduced in [32, 23], Curien and
Kortchemski proved the rescaled associated looptrees converge towards a random
compact metric space they called the $\alpha$-stable looptree. There are even
more striking cases where the associated looptrees admit a non-degenerate
scaling limit but the trees themselves do not so they are an interesting means
to give sense to scaling limits of highly dense trees. An instance of such a
situation is studied in [16] by Curien, Duquesne, Kortchemski, and Manolescu
with the model of random trees built by linear preferential attachment, which
was introduced and popularized in [41, 7, 11]. The scaling limit for this
model is called the Brownian looptree and is distinct from the stable
looptrees.

In addition to their help to study trees, looptrees spontaneously appear as
scaling limits of other random structures. For example, random Boltzmann
dissections of a regular polygon, that were introduced in [29], have a similar
geometry as looptrees in some regimes. Indeed, Curien and Kortchemski proved
in [18] that their scaling limit as the number of sides of the polygon tends
to infinity is the $\alpha$-stable looptree, depending on the regime. Let us
mention links between looptrees and dissections were also detected in [17].
Several instances of looptrees can be found within random planar maps too. The
goal of this active area of research is to understand universal large-scale
properties of graphs or their proper embeddings in the two-dimensional sphere.
We refer to [1, 37] for surveys of this field. In [19], Curien and Kortchemski
considered boundaries of percolation clusters on the uniform infinite planar
triangulation introduced in [6]. These boundaries are almost looptrees and the
authors showed that in the critical case of the percolation, their scaling
limit is the $3/2$-stable looptree. Similarly, in [30, 39], Kortchemski and
Richier discussed asymptotics for the boundaries of Boltzmann planar maps
conditioned on having a large perimeter. Once again, stable looptrees arise as
scaling limits in some regimes.

Nevertheless, the most important application of looptrees may be their
substantial connections with scaling limits of random planar maps. The
bijections of Bouttier, Di Fransesco, Guitter [12] and of Janson, Stefánsson
[26] yield a one-to-one correspondence between planar maps and some labeled
trees (where each vertex is equipped with an integer), which can be
reformulated into a bijection between planar maps and some labeled looptrees.
Looptrees seem to be the right objects to consider here because the scaling
limits of random planar maps with large faces introduced by Le Gall and
Miermont in [33] are implicitly constructed from a Gaussian field on stable
looptrees. Furthermore, Marzouk obtained limit theorems for planar maps in
[34] without a strong control on the corresponding trees, but by observing
their Lukasiewicz walks instead. As we will see later, the Lukasiewicz walk
directly codes the geometry of the associated looptree. For more detailed
discussions on the relations between maps and looptrees, we refer to Marzouk
[35].

#### Motivations and main results

The purpose of the present work is to build a framework and to provide a
toolbox to construct, manipulate, and demonstrate convergences involving
looptrees under the most general setting possible. A first issue is how to
define compact continuum fractal looptrees that may appear as scaling limits.
Although compact continuum fractal trees are well-known since the introduction
of the Brownian Continuum Random Tree (CRT) by Aldous in [3], mimicking the
discrete setting to define continuum looptrees associated with them is not so
easy. On the one hand, it could be not clear how to choose the lengths of the
loops while ensuring the compactness of the resulting space. On the other
hand, when the branch points are dense in the tree, it might be possible that
two different cycles in the associated looptree would never share a common
point. It is indeed what happens for the $\alpha$-stable looptree
$\mathscr{L}_{\alpha}$ of Curien and Kortchemski, with any $\alpha\in(1,2)$.
They avoided the above difficulties [18] by building directly
$\mathscr{L}_{\alpha}$ from the excursion $X^{\mathsf{exc},(\alpha)}$ of an
$\alpha$-stable spectrally positive Lévy process. This process also codes the
$\alpha$-stable tree, and Curien and Kortchemski justified that
$\mathscr{L}_{\alpha}$ can be interpreted as its associated looptree. The idea
to encode metric spaces by real-valued functions is in line with the founding
work of Le Gall [25] which studies and constructs compact real trees from
continuous excursions. Namely, if $f:[0,1]\longrightarrow[0,\infty)$ is
continuous with $f(0)=f(1)=0$ then setting

$d^{\mathrm{clas}}_{f}(t,s)=d^{\mathrm{clas}}_{f}(s,t)=f(s)+f(t)-2\inf_{[s,t]}f$
(1)

for all $s,t\in[0,1]$ with $s\leq t$ defines a pseudo-distance on $[0,1]$,
which in turn induces a compact real tree by quotient. See [25, 22] for an
extensive study of this coding. The definition of real trees will be reminded
in Section 5 and the quotient metric space induced by a pseudo-distance is
defined in Section 2.3. Similarly, the $\alpha$-stable looptree is induced by
another pseudo-distance inherited from $X^{\mathsf{exc},(\alpha)}$.

######  Definition 1.

A càdlàg function $f:[0,1]\longrightarrow\mathbb{R}$ is a right-continuous
function that has left limits everywhere. In that case, we write $f(0-)=0$,
$\Delta_{0}(f)=0$, $f(t-)=\lim_{s\rightarrow t-}f(s)$ and we denote
$\Delta_{t}(f)=f(t)-f(t-)$ for all $t\in(0,1]$. If no confusion over the
function is possible, we will simply write $\Delta_{t}$. We say a non-negative
càdlàg function $f:[0,1]\longrightarrow[0,\infty)$ is an excursion when
$f(1)=0$ and when all jumps of $f$ are non-negative, namely $\Delta_{t}\geq 0$
for all $t\in[0,1]$. We denote by $\mathcal{H}$ the set of excursions.

We straightforwardly extend the work of Curien and Kortchemski in [18] to
construct a pseudo-distance $d^{\mathsf{L}}_{f}$ from any excursion $f$. We
denote by $\mathscr{L}_{f}$ and we call the looptree coded by $f$ the quotient
metric space induced by $d^{\mathsf{L}}_{f}$. In particular, we have
$\mathscr{L}_{\alpha}=\mathscr{L}_{X^{\mathsf{exc},(\alpha)}}$. Without
immediately giving the definition of $d^{\mathsf{L}}_{f}$, let us already
state that each jump of $f$ of height $\Delta$ corresponds to a circle of
length $\Delta$ in $\mathscr{L}_{f}$, that those circles are dense in
$\mathscr{L}_{f}$, and that if $g$ is a "sub-excursion" of $f$ then
$\mathscr{L}_{g}$ is a subset of $\mathscr{L}_{f}$. This method has many
benefits. While it allows an automatic construction of a large diversity of
complex looptrees, it is also fairly simple to find an excursion coding a
given discrete looptree. This kind of behavior is especially useful to tackle
problems involving transitions from discrete to continuous worlds.

However, $d^{\mathsf{L}}$ alone is not enough to understand all convergences
of looptrees. Indeed, a looptree consisting of a chain of $n$ loops of lengths
$1/n$ put back to back is asymptotically close to a segment that has no loops.
It is not an unusual situation, as [17], [18], [19], and [30] provide as many
examples of looptrees that converge towards loopless compact real trees. In
particular, it ensures $d^{\mathsf{L}}$ lacks the key property of being
continuous with respect to the coding function, contrary to
$d^{\mathrm{clas}}$. Furthermore, the coding brought by $d^{\mathsf{L}}$ does
not fully exploit the diversity of excursions because it only cares about the
jumps: if $f$ is a continuous excursion with $f(0)=0$, then
$d^{\mathsf{L}}_{f}=0$ and the looptree coded by $f$ is reduced to a single
point. In fact, $d^{\mathsf{L}}$ disregards continuous growth that would
classically code a real tree with $d^{\mathrm{clas}}$. In contrast, applying
$d^{\mathrm{clas}}$ to a discontinuous excursion would still give a tree but
without really distinguishing between jumps and continuous growth. These
observations motivate us to define a new pseudo-distance $d^{\mathsf{T}}_{f}$
that induces a real tree, and thus codes additional limits of looptrees we
were lacking, but that only harnesses the continuous growth of $f$.
Informally, $d^{\mathsf{T}}_{f}$ is obtained by stripping $f$ of its jumps
then by taking $d^{\mathrm{clas}}$. We denote by $\mathscr{T}_{f}$ and we call
the tree coded by $f$ the quotient metric space induced by
$d^{\mathsf{T}}_{f}$. Plus, we clarify that it naturally holds
$d^{\mathsf{L}}_{0}=d^{\mathsf{T}}_{0}=0$. Our first main result describes the
relations between trees and looptrees coded by excursions. It will be
specified by Theorems 4 and 5.

######  Theorem 1.

Let us set $I:f\in\mathcal{H}\longmapsto f\in\mathcal{H}$. There exists an
operator $J:\mathcal{H}\longrightarrow\mathcal{H}$ such that $J\circ J=J$ and
$J\circ(I-J)=0$. Moreover, it holds $d^{\mathsf{L}}_{f}=d^{\mathsf{L}}_{Jf}$
and $d^{\mathsf{T}}_{f}=d^{\mathsf{T}}_{f-Jf}=d^{\mathrm{clas}}_{f-Jf}$ for
all excursions $f$. Furthermore, $f$ is a continuous excursion with $f(0)=0$
if and only if $Jf=0$.

An excursion $f$ such that $Jf=f$ or $Jf=0$ will be respectively called a pure
jump growth (PJG) excursion or a continuous excursion. These two classes of
excursions and the operator $J$ will be properly defined at the beginning of
Section 3. A linear combination of $d^{\mathsf{T}}_{f}$ and
$d^{\mathsf{L}}_{f}$ gives a pseudo-distance that induces a new hybrid metric
space, denoted by $\mathscr{V}_{f}$, composed of loops and trees tangently
glued together along some genealogical structure. More precisely, we choose
$d^{\mathsf{V}}_{f}=d^{\mathsf{T}}_{f}+2d^{\mathsf{L}}_{f}$ because the
diameter of a metric circle is half of its length. We call $\mathscr{V}_{f}$
the vernation tree coded by $f$. The name vernation tree was chosen to
distinguish this new object from the looptree $\mathscr{L}_{f}$. It refers to
the arrangement of bud scales or young leaves in a leaf bud before it opens.
Figure 1 presents some examples and non-examples of vernation trees. Let us
already mention that independently, Blanc-Renaudie in [10] and Marzouk in [35]
construct the same kind of spaces as part of their studies of scaling limits
of models with prescribed degrees. While Blanc-Renaudie took the point of view
of stick-breaking constructions and managed to mimic the association of a
looptree with a discrete tree, Marzouk found the same coding by excursions as
us and also identified the significance of PJG excursions. However, our
methods to study convergences are completely different and we truly believe
both should be useful according to context.

Figure 1: All eight figures have to be understood as intrinsic one-dimensional
spaces. Left: Four examples of vernation trees. Right: Four non-examples of
vernation trees.

We argue vernation trees yield the right notion to study convergences of
looptrees. Theorem 1 justifies this notion unifies the classical encoding of
real trees and the encoding of looptrees introduced in [18] by Curien and
Kortchemski. Moreover, vernation trees can naturally appear as limits of
looptrees. Indeed, looptrees may converge to trees or other looptrees, and
gluing two looptrees together still gives a looptree. Conversely, the limits
of vernation trees are also vernation trees in some sense. To make this point
rigorous, we will provide an intrinsic and topological definition for
vernation trees in Section 5. Informally, a metric space is a vernation tree
when any two points at distance $r$ are joined by a path of length $r$ and any
two distinct loops do not intersect at more than one point. We let the reader
compare this definition with Figure 1. Then, Theorems 9, 10, 11 imply the
following result.

######  Theorem 2.

The space of compact vernation trees is the closure for the Gromov-Hausdorff
topology of the set of vernation trees coded by excursions.

The definition of the Gromov-Hausdorff topology, which formalizes convergence
of (isometry classes of) compact metric spaces, is recalled in Section 4.3.
Vernation trees reveal another interesting observation. Again in [18], Curien
and Kortchemski have proved the $\alpha$-stable looptrees converge in
distribution to the Brownian CRT (up to a constant multiplicative factor) as
$\alpha$ tends to $2$, while the excursion $X^{\mathsf{exc},(\alpha)}$
converge in distribution to a multiple of the Brownian excursion. However, we
will see in Section 6.2 that the excursions $X^{\mathsf{exc},(\alpha)}$ are
PJG and the Brownian excursion is continuous. Plus, the multiplicative
constants match. Hence, the vernation trees coded by the
$X^{\mathsf{exc},(\alpha)}$ converge to the vernation tree coded by the
limiting excursion. Furthermore, [18] and [30] exhibit several instances where
rescaled looptrees associated with discrete trees converge to the looptree or
the tree coded by the scaling limit of their Lukasiewicz walks. Thereby, it is
natural to hope the coding $f\in\mathcal{H}\longmapsto\mathscr{V}_{f}$ enjoys
some kind of functional continuity. If it were true, understanding
convergences of looptrees and vernation trees would merely be reduced to
studying the convergences of their coding excursions. In general, it is a much
simpler problem thanks to a lot of tools and methods that were developed over
the years. In fact, one could argue the functional continuity of the classical
coding $f\longmapsto d^{\mathrm{clas}}_{f}$ is the main reason for its
success. Sadly, it does not hold for $d^{\mathsf{V}}$ or $\mathscr{V}$. We
will discuss in more depth why this fails at the beginning of Section 4, but
[17, Theorem 13] and [30, Theorem 2] already provide a counterexample where
the limit differs from a strange multiplicative factor from what one could
expect. In order to address this issue, we define another metric space
$\widetilde{\mathscr{V}}_{f}$, induced by a pseudo-distance
$\widetilde{d}^{\mathsf{V}}_{f}$, called the shuffled vernation tree coded by
$f$. Informally, $\widetilde{\mathscr{V}}_{f}$ is obtained by shuffling the
positions on which the components of $\mathscr{V}_{f}$ are glued on each loop
while keeping the loops and the trees arranged in the same genealogical
structure as $\mathscr{V}_{f}$. See Figure 2 for an example. Shuffling
instructions are contained in an object $\Phi$ we will call shuffle and that
one can choose with some freedom. The coding
$f\longmapsto\widetilde{\mathscr{V}}_{f}$ is continuous in some sense, which
justifies its definition. Let us denote by $D([0,1])$ the space of càdlàg
functions on $[0,1]$ and let us endow it with the Skorokhod distance $\rho$
defined by

$\rho(f_{1},f_{2})=\inf_{\lambda\in\Lambda}\max\left(||\lambda-\mathsf{id}||_{\infty},||f_{1}-f_{2}\circ\lambda||_{\infty}\right)$

for all $f_{1},f_{2}\in D([0,1])$, where $\Lambda$ is the set of increasing
bijections from $[0,1]$ into itself. The distance $\rho$ is not complete but
induces a separable and completely metrizable topology on $\mathcal{D}([0,1])$
called the Skorokhod topology, see [9, Chapter 3]. We set

$\displaystyle\Theta:D([0,1])$ $\displaystyle\longrightarrow D([0,1])$
$\displaystyle f$ $\displaystyle\longmapsto\Theta f:t\mapsto{\bf 1}_{2t\geq
1}f(2t-1)$

and we write $f_{n}\longrightarrow f$ for the relaxed Skorokhod topology when
$\Theta f_{n}\longrightarrow\Theta f$ for the Skorokhod topology on
$D([0,1])$. This induces a weaker separable and completely metrizable topology
than the Skorokhod topology. It is clear that the set of excursions is closed
for both of those topologies. We are now ready to state our limit theorem.

######  Theorem 3.

We denote by $\partial$ the metric space with a unique point. Let $f$ be an
excursion and let $(f_{n})$ be a sequence of excursions such that
$f_{n}\longrightarrow f$ for the relaxed Skorokhod topology.

  1. (i)

If $B_{f}\cap\mathbb{B}(\Phi)=\emptyset$, then the convergence
$\widetilde{\mathscr{V}}_{f_{n}}\longrightarrow\widetilde{\mathscr{V}}_{f}$
holds for the pointed Gromov-Hausdorff-Prokhorov topology.

  2. (ii)

If there is $N\geq 1$ such that all the $f_{n}$ have at most $N$ jumps, then
the convergences $\mathscr{L}_{f_{n}}\longrightarrow\mathscr{L}_{f}$,
$\mathscr{T}_{f_{n}}\longrightarrow\mathscr{T}_{f}$, and
$\mathscr{V}_{f_{n}}\longrightarrow\mathscr{V}_{f}$ hold for the pointed
Gromov-Hausdorff-Prokhorov topology.

  3. (iii)

If $f$ is PJG, then the convergences
$\mathscr{L}_{f_{n}}\longrightarrow\mathscr{L}_{f}$,
$\mathscr{T}_{f_{n}}\longrightarrow\partial$, and
$\mathscr{V}_{f_{n}}\longrightarrow\mathscr{V}_{f}$ hold for the pointed
Gromov-Hausdorff-Prokhorov topology.

The pointed Gromov-Hausdorff-Prokhorov topology formalizes convergence of
(equivalence classes of) compact metric spaces endowed with a distinguished
point and a Borel probability measure. We will give a proper definition in
Section 4.3. The assumption $B_{f}\cap\mathbb{B}(\Phi)=\emptyset$ is a minimal
condition that is presented in Section 4.1. The price for the functional
continuity of $f\longmapsto\widetilde{\mathscr{V}}_{f}$ is that there is no
simple or canonical choice for the shuffle $\Phi$. Moreover, it lacks some
other useful properties that $f\longmapsto\mathscr{V}_{f}$ shares with
$f\longmapsto d^{\mathrm{clas}}_{f}$. That makes $\widetilde{\mathscr{V}}_{f}$
inconvenient to use with explicit or discrete examples. Thus, we think it is
best to first define a model or to express a given vernation tree by using
(unshuffled) $\mathscr{V}_{f}$, and then to study the asymptotic behavior
thanks to the point $(ii)$ and $(iii)$ of Theorem 3 or by comparing
$\mathscr{V}_{f}$ with $\widetilde{\mathscr{V}}_{f}$. While this technical
task will not be automatic, a nice choice of shuffle may ease it. For example,
if $X$ is a random excursion, it may be possible to select $\Phi$ such that
$\widetilde{\mathscr{V}}_{X}$ and $\mathscr{V}_{X}$ have the same
distribution. Another advantage of the flexibility of the choice for the
shuffle is that it is easy to ensure the condition
$B_{f}\cap\mathbb{B}(\Phi)=\emptyset$ holds. We will give three concrete
probabilistic applications of this method.

The first one reformulates in metric terms the asymptotics in terms of
processes found by Aldous, Miermont, and Pitman in [5] for uniform random
mappings. Indeed, such a mapping $M$ can be represented by a graph on
$\\{1,2,...,n\\}$ with edges $(i,M(i))$ and each connected component of this
graph is a vernation tree, as a collection of trees rooted on the same cycle.
We give scaling limits for these components. The second application retrieves
and completes [18, Theorem 1.2] of Curien and Kortchemski about the
convergences of $\alpha$-stable looptrees when $\alpha$ tends to $1$ or $2$ by
showing that their whole family is continuous in distribution with respect to
$\alpha$. The strategy of the proof is to prove the coding excursions
$X^{\mathsf{exc},(\alpha)}$ are PJG and to find a shuffle such that
$\widetilde{\mathscr{V}}_{X^{\mathsf{exc},(\alpha)}}$ is distributed as
$\mathscr{V}_{X^{\mathsf{exc},(\alpha)}}$. The last application provides two
invariance principles for discrete looptrees associated with finite ordered
trees. We first retrieve [18, Theorem 4.1] of Curien and Kortchemski under the
PJG case. In the general case, we prove that if a sequence of exchangeable
random ordered trees admits $X$ as the scaling limit of their Lukasiewicz
walks and if their heights become negligible against the scaling, then the
scaling limit of their associated looptrees is $1/2\cdot\mathscr{V}_{X}$,
where $c\cdot E$ stands for the metric space obtained from $E$ by multiplying
all distances by $c>0$. Of course, we will clarify what it means for a random
ordered tree to be exchangeable, but informally, it means its distribution is
invariant by changing the order.

#### Outline

In Section 2, we construct the mentioned pseudo-distances and the metric
spaces they induce by quotient. Section 3 gives some tools to better
understand the relations between trees, looptrees, and excursions, and shows
vernation trees generalize both trees and looptrees. Section 4 studies
convergences of vernation trees by first stating the functional continuity of
$f\longmapsto\widetilde{d}^{\mathsf{V}}_{f}$ and then deducing other limit
theorems under some particular cases. This results in a proof of Theorem 3. In
Section 5, we introduce our topological notion of vernation trees and
demonstrate Theorem 2. Section 6 is devoted to the three probabilistic
applications of our work.

##  2 Tree, looptree, and vernation tree coded by an excursion

In all this Section, we fix an excursion denoted by $f$.

###  2.1 Genealogy associated with an excursion

In order to encode a looptree by an excursion $f$, we begin to construct a
genealogy on $[0,1]$ from $f$, which will give the tree structure of the loops
in the associated looptree. If we take as a principle that $f$ codes a
looptree $\mathscr{L}$ with root $0$, then if $f(s-)\leq\inf_{[s,t]}f$, $s\leq
t\leq r$, and $f(s-)=f(r)$, the excursion $u\mapsto f(s+u(r-s))-f(u-)$ should
encode a sub-looptree of $\mathscr{L}$, with root $s$, and containing $t$.
Hence, the point $s$ is an ancestor of $t$ according to the genealogy of
$\mathscr{L}$. This motivates the following definition, which is a direct
extension of the genealogy defined by Curien and Kortchemski in [18] for the
stable looptrees. We set for all $s,t\in[0,1]$,

$s\preceq_{f}t\text{ when both }s\leq t\text{ and }f(s-)\leq\inf_{[s,t]}f.$

We also write $s\prec_{f}t$ when $s\preceq_{f}t$ and $s<t$. If the excursion
$f$ is obvious according to context, we would just write $\preceq$ and
$\prec$.

######  Proposition 1.

The relation $\preceq$ enjoys the following properties.

  1. (i)

The relation $\preceq$ is a partial order on $[0,1]$.

  2. (ii)

The genealogy admits $0$ as its root, namely $0\preceq t$ for all $t\in[0,1]$.

  3. (iii)

Two points have a most recent common ancestor. For any $s,t\in[0,1]$, there is
a unique $u\in[0,1]$ such that for all $r\in[0,1]$, $r\preceq
u\Longleftrightarrow r\preceq s,t$. We denote it by $s\wedge_{f}t$, or just by
$s\wedge t$ if the excursion $f$ is obvious according to context.

  4. (iv)

The ancestral lineages are totally ordered. For any $t\in[0,1]$, the relation
$\preceq$ induces a total order on $\\{s\in[0,1]\ :\ s\preceq t\\}$.

###### Proof.

Let $t,s,r\in[0,1]$. Obviously, we have $t\leq t$ and $f(t-)\leq
f(t)=\inf_{[t,t]}f$ because $f$ is an excursion. So $t\preceq t$ and the
relation $\preceq$ is reflexive. If $s\preceq t$ and $t\preceq s$, then by
definition $s\leq t\leq s$, so $t=s$, and $\preceq$ is antisymmetric. To prove
the transitivity, we suppose that $s\preceq t$ and $t\preceq r$ and we can
assume $s<t<r$. We find $f(s-)\leq\inf_{[s,t]}f\leq f(t-)\leq\inf_{[t,r]}f$,
which entails $s\preceq r$ and $(i)$. The point $(ii)$ is simply deduced from
the fact that $f$ is an excursion. Let us prove $(iii)$. We set
$u=\sup\\{r\in[0,1]\ :\ r\preceq s,t\\}$. By definition, $u\leq s$ and $u\leq
t$. We give ourselves a sequence $(r_{n})_{n\geq 1}$ that converges towards
$u$ and such that $0\leq r_{n}\leq r_{n+1}\leq u$ and $r_{n}\preceq s,t$ for
all $n\geq 1$. We have $\displaystyle{f(r_{n}-)\longrightarrow f(u-)}$ and
thanks to the inequality

$f(r_{n}-)\leq\inf_{[r_{n},t]}f\leq\inf_{[u,t]}f,$

we obtain $u\preceq t$. Likewise, $u\preceq s$. Conversely, if $r\preceq t$
and $r\preceq s$, then $r\leq u$ by definition and
$f(r-)\leq\inf_{[r,t]}f\leq\inf_{[r,u]}f$, so $r\preceq u$. The uniqueness of
$s\wedge t$ immediately follows from the antisymmetry of $\preceq$. Let us
prove $(iv)$. We suppose $s,r\preceq t$ and without loss of generality, we
assume $r\leq s$. Then, $f(r-)\leq\inf_{[r,t]}f\leq\inf_{[r,s]}f$, so
$r\preceq s$, which concludes the proof. ∎

######  Lemma 1.

Let $s,t\in[0,1]$. If $s\leq t$, then it holds that $\inf_{[s\wedge
t,t]}f=\inf_{[s,t]}f$ and

$s\wedge t=\sup\left\\{r\leq s\ :\ r\preceq t\right\\}=\sup\left\\{r\leq s\ :\
f(r-)\leq\inf_{[s,t]}f\right\\}.$

###### Proof.

If $r\leq s$ and $r\preceq t$, then $f(r-)\leq\inf_{[r,t]}f\leq\inf_{[r,s]}f$,
and so $r\preceq s$. Thus, $r\preceq s\wedge t$. Conversely, if $r\preceq
s\wedge t$, then $r\leq s$ and $r\preceq t$. So, we have shown $s\wedge
t=\sup\\{r\leq s\ :\ r\preceq t\\}$. In particular, if $s\wedge t<r\leq s$,
then we cannot have $r\preceq t$. It follows that for all $r\in(s\wedge t,s]$,
it holds

$f(r)\geq f(r-)>\inf_{[r,t]}f$

Since $f$ is càdlàg, this leads to $\inf_{[r,s]}f>\inf_{[r,t]}f$. Hence,
$\inf_{[r,t]}f=\inf_{[s,t]}f$ for all $r\in(s\wedge t,s]$ and we end the proof
by making $r$ tend towards $s\wedge t$. ∎

As same as in [18], we define a quantity $x_{s}^{t}$ which will represent the
relative position of the ancestor of $t$ on the loop containing $s$ on the
looptree associated with $f$, when $s\preceq t$. For all $s,t\in[0,1]$, we set

$x_{s}^{t}(f)={\bf 1}_{s\leq t}\max\left(\inf_{[s,t]}f-f(s-),0\right).$

When the excursion $f$ is obvious according to context, we would just write
$x_{s}^{t}$. Let us remark that $x_{s}^{t}\in[0,\Delta_{s}]$,
$x_{s}^{s}=\Delta_{s}$, and $x_{s}^{t}>0\Longrightarrow s\preceq t$. We finish
this first step by providing two convenient lemmas about $x_{s}^{t}$.

######  Lemma 2.

Let $u,s,t\in[0,1]$. If $u<t\wedge s$, then
$x_{u}^{t}=x_{u}^{s}=x_{u}^{t\wedge s}$.

###### Proof.

We merely write $\inf_{[u,t\wedge s]}f\leq f(t\wedge
s-)\leq\min\left(\inf_{[t\wedge s,t]}f,\inf_{[t\wedge s,s]}f\right)$. ∎

######  Lemma 3.

Let $s,t\in[0,1]$. If $s<t$, then

$\sum_{\begin{subarray}{c}s<r\\\ r\prec t\end{subarray}}x_{r}^{t}\leq
f(t-)-\inf_{[s,t]}f.$ (2)

###### Proof.

Let $\\{r_{i}\ :\ 1\leq i\leq n\\}$ be a finite subset of $\\{r\in[0,1]\ :\
r\prec t,s<r\\}$ such that $r_{i}<r_{i+1}$ for any $1\leq i\leq n-1$. Let us
also set $r_{n+1}=t$. With these notations, we have $r_{i+1}\in(r_{i},t]$ for
any $1\leq i\leq n$, so we can write

$\sum_{i=1}^{n}x_{r_{i}}^{t}=\sum_{i=1}^{n}\left(\inf_{[r_{i},t]}f-f(r_{i}-)\right)\leq\sum_{i=1}^{n}f(r_{i+1}-)-f(r_{i}-)=f(t-)-f(r_{1}-).$

Eventually, $f(r_{1}-)\geq\inf_{[s,t]}f$ because $r_{1}\in(s,t]$. The lemma
follows. ∎

###  2.2 Pseudo-distances coded by an excursion

We denote by $(\mathcal{C},\delta)$ the metric circle of perimeter $1$, seen
as $[0,1]$ endowed with the pseudo-distance $\delta$ given by
$\delta(a,b)=\min(|a-b|,1-|a-b|)$ so that $0$ and $1$ are identified. We say
that a function is càglàd when it is left-continuous with right limits
(continue à gauche, limite à droite in French). We denote by
$\overleftarrow{D}([0,1],\mathcal{C})$ the set of càglàd functions from
$[0,1]$ to $\mathcal{C}$ and we endow it with the Skorokhod distance
$\overleftarrow{\rho}$ given by

$\overleftarrow{\rho}(\phi_{1},\phi_{2})=\inf_{\lambda\in\Lambda}\max\left(||\lambda-\mathsf{id}||_{\infty},\sup_{x\in[0,1]}\delta(\phi_{1}(x),\phi_{2}\circ\lambda(x))\right)$

for all $\phi_{1},\phi_{2}\in\overleftarrow{D}([0,1],\mathcal{C})$, where
$\Lambda$ is the set of increasing bijections from $[0,1]$ into itself.

######  Definition 2.

A shuffle is a Borel application
$\Phi:\Delta>0\longmapsto\phi_{\Delta}\in\overleftarrow{D}([0,1],\mathcal{C})$
such that for all $\Delta>0$, $\phi_{\Delta}$ is surjective,

$\sup_{u\in(0,\Delta]}\sup_{x\in(0,1]}\left|\frac{2}{x}\delta\left(\phi_{u}(0),\phi_{u}(x)\right)-1\right|<\infty,\text{
and
}\sup_{x\in(0,1]}\left|\frac{2}{x}\delta\left(\phi_{\Delta}(0),\phi_{\Delta}(x)\right)-1\right|\underset{\Delta\rightarrow
0+}{\longrightarrow}0.$ (3)

######  Remark 1.

Our requirements for the $\phi_{\Delta}$ to be surjective with the convergence
in (3) lead to the fact that the $\phi_{\Delta}$ have to oscillate more and
more quickly near $x=0$ when $\Delta$ is small enough. Moreover, when $\Delta$
gets smaller, $\phi_{\Delta}$ needs to jump an infinite number of times near
$0$. As a result, there is no simple or canonical choice for $\Phi$.
Nevertheless, it is not too difficult to construct various examples of
shuffles. For instance, we could choose $\phi_{\Delta}(x)=x$ when $\Delta\geq
1$ and

$\phi_{1-\Delta}(x)=\begin{cases}x-\frac{\Delta^{k}}{2}&\text{ if
}x\in\left(\frac{\Delta^{k}+\Delta^{k+1}}{2},\Delta^{k}\right]\text{ with an
integer }k\\\ 1-x+\frac{\Delta^{k+1}}{2}&\text{ if
}x\in\left(\Delta^{k+1},\frac{\Delta^{k}+\Delta^{k+1}}{2}\right]\text{ with an
integer }k\\\ 0&\text{ if }x=0\end{cases}$

when $\Delta\in(0,1)$, for all $x\in[0,1]$.

For the following, we fix a shuffle and let the dependence on it be implicit.
As we said before, the loops of our space will match the jumps of $f$, and the
lengths of the loops will be equal to the heights of the jumps. For
$t\in[0,1]$, we provide the pseudo-distances

$\displaystyle\delta_{t}(a,b)$
$\displaystyle=\Delta_{t}\delta\left(\frac{a}{\Delta_{t}},\frac{b}{\Delta_{t}}\right)=\min\left(|a-b|,\Delta_{t}-|a-b|\right),$
$\displaystyle\tilde{\delta}_{t}(a,b)$
$\displaystyle=\Delta_{t}\delta\left(\phi_{\Delta_{t}}\left(\frac{a}{\Delta_{t}}\right),\phi_{\Delta_{t}}\left(\frac{b}{\Delta_{t}}\right)\right)$

to the segment $[0,\Delta_{t}]$ when $\Delta_{t}>0$. When $\Delta_{t}=0$, we
set $\delta_{t}(0,0)=\tilde{\delta}_{t}(0,0)=0$. If needed, we would precise
the dependence on $f$ by writing $\delta_{t}^{f}(\cdot,\cdot)$ or
$\tilde{\delta}_{t}^{f}(\cdot,\cdot)$. Let us notice that
$([0,\Delta_{t}],\delta_{t})$ and $([0,\Delta_{t}],\tilde{\delta}_{t})$ are
isometric to a circle of length $\Delta_{t}$.

We are now ready to define our pseudo-distances of interest. For $s,t\in[0,1]$
with $s\preceq t$, we set

$\displaystyle d^{\circ}_{f}(s,t)$ $\displaystyle=\sum_{s\prec r\preceq
t}\delta_{r}(0,x_{r}^{t}),$ $\displaystyle\widetilde{d}^{\circ}_{f}(s,t)$
$\displaystyle=\sum_{s\prec r\preceq t}\tilde{\delta}_{r}(0,x_{r}^{t}).$

It is the distance to run through between $t$ and its ancestor on the loop of
$s$. Indeed, each term of the sum corresponds to the length of the path
getting through one of the loops between $s$ and $t$. For
$\widetilde{d}^{\circ}$, the points where two loops are glued were shuffled.
We reach $t$ from $s$ by going through the loop of $t\wedge s$. Explicitly, we
set

$\displaystyle d^{\mathsf{L}}_{f}(s,t)$ $\displaystyle=\delta_{s\wedge
t}(x_{s\wedge t}^{s},x_{s\wedge t}^{t})+d^{\circ}_{f}(s\wedge
t,s)+d^{\circ}_{f}(s\wedge t,t),$
$\displaystyle\widetilde{d}^{\mathsf{L}}_{f}(s,t)$
$\displaystyle=\tilde{\delta}_{s\wedge t}(x_{s\wedge t}^{s},x_{s\wedge
t}^{t})+\widetilde{d}^{\circ}_{f}(s\wedge
t,s)+\widetilde{d}^{\circ}_{f}(s\wedge t,t),$

for all $s,t\in[0,1]$. The pseudo-distance $d^{\mathsf{L}}$ is exactly the one
constructed by Curien and Kortchemski in [18] for the stable looptrees. Now,
we want to define a pseudo-distance from $f$ that should induce a tree. For
$s,t\in[0,1]$ with $s\preceq t$, we set

$d^{\mathsf{T}}_{f}(s,t)=f(t)-\inf_{[s,t]}f-\sum_{s\prec r\preceq
t}x_{r}^{t}.$

While the term $f(t)-\inf_{[s,t]}f$ should correspond to the distance to
travel if the metric space was a tree, the excursion has jumps we have to
erase because we do not want to count the distance traveled on loops. Thus, we
subtract the terms $x_{r}^{t}$ that are already counted in the term
$f(t)-\inf_{[s,t]}f$. More generally, if $s,t\in[0,1]$, we set

$d^{\mathsf{T}}_{f}(t,s)=d^{\mathsf{T}}_{f}(t\wedge
s,t)+d^{\mathsf{T}}_{f}(t\wedge s,s).$

Observe the definition of $d^{\mathsf{T}}$ is consistent because
$d^{\mathsf{T}}(s,s)=f(s)-f(s)=0$. In order to combine the paths on the loops
and on the tree branches, we set

$\displaystyle d^{\mathsf{V}}_{f}$
$\displaystyle=d^{\mathsf{T}}_{f}+2d^{\mathsf{L}}_{f},$
$\displaystyle\widetilde{d}^{\mathsf{V}}_{f}$
$\displaystyle=d^{\mathsf{T}}_{f}+2\widetilde{d}^{\mathsf{L}}_{f}.$

The constant $2$ before $d^{\mathsf{L}}$ (or $\widetilde{d}^{\mathsf{L}}$)
could be replaced with another one but we will justify this choice by the fact
that $f\longmapsto\widetilde{d}^{\mathsf{V}}_{f}$ enjoys a functional
continuity with the constant $2$. It would be not the case with another
constant. Nevertheless, Marzouk showed in [35] that metrics
$d^{\mathsf{L}}+ad^{\mathsf{T}}$ with $a\neq 1/2$ can naturally appear as
scaling limits of uniformly random discrete looptrees with prescribed degrees.
When the function $f$ is obvious according to context, we will just write
$d^{\circ}$, $\widetilde{d}^{\circ}$, $d^{\mathsf{L}}$,
$\widetilde{d}^{\mathsf{L}}$, $d^{\mathsf{T}}$, $d^{\mathsf{V}}$, and
$\widetilde{d}^{\mathsf{V}}$. It is important to be aware that the shuffle
$\Phi$ was used to construct $\widetilde{d}^{\mathsf{L}}$ and
$\widetilde{d}^{\mathsf{V}}$, but not for $d^{\mathsf{L}}$, $d^{\mathsf{T}}$,
and $d^{\mathsf{V}}$.

We now show our quantities are finite. The functions $d^{\circ}$,
$\widetilde{d}^{\circ}$, $d^{\mathsf{L}}$, and $\widetilde{d}^{\mathsf{L}}$
obviously are non-negative because they are series of non-negative terms.
Keeping only the term for $r=t$ yields that if $s\prec t$ then
$d^{\mathsf{T}}(s,t)\leq f(t-)-\inf_{[s,t]}f$ . Moreover, the bound (2)
implies that if $s\prec t$ then

$\displaystyle 0\leq d^{\circ}(s,t)$ $\displaystyle\leq f(t-)-\inf_{[s,t]}f,$
(4) $\displaystyle 0\leq d^{\mathsf{T}}(s,t)$ $\displaystyle\leq
f(t-)-\inf_{[s,t]}f,$ (5)

because $\delta_{t}(0,x_{t}^{t})=\delta_{t}(0,\Delta_{t})=0$ and
$\delta(0,x)\leq x$ for all $x\in[0,1]$. If we only have $s\preceq t$, the
upper bound given by $f(t)-\inf_{[s,t]}f$ stays true. Now, we only assume
$s<t$. We observe $t\wedge s\leq s<t$ and $\delta_{t\wedge s}(x_{s\wedge
t}^{s},x_{s\wedge t}^{t})\leq x_{t\wedge s}^{s}-x_{t\wedge s}^{t}$. Recall we
know $\inf_{[s,t]}f=\inf_{[t\wedge s,t]}f$ from Lemma 1, so if $s<t$ then

$\displaystyle 0\leq d^{\mathsf{L}}(s,t)$ $\displaystyle\leq
f(s)+f(t-)-2\inf_{[s,t]}f,$ (6) $\displaystyle 0\leq d^{\mathsf{T}}(s,t)$
$\displaystyle\leq f(s)+f(t-)-2\inf_{[s,t]}f.$ (7)

Remark we cannot find these bounds for $\widetilde{d}^{\mathsf{L}}$ because
$\tilde{\delta}_{t}(0,\Delta_{t})\neq 0$ and
$\tilde{\delta}_{t}(a,b)\nleq|a-b|$. Nethertheless, the bound in (3) ensures
there exists a constant $K\in(0,\infty)$ which only depends on
$||f||_{\infty}$ and $\Phi$ such that $\tilde{\delta}_{t}(0,x)\leq Kx$ for any
$t\in[0,1]$ and $x\in[0,\Delta_{t}]$. In particular, if $s\preceq t$, then the
bound (2) implies

$0\leq\widetilde{d}^{\circ}(s,t)\leq K\left(f(t)-\inf_{[s,t]}f\right).$ (8)

While the definitions of these quantities are intuitive and natural, the fact
that $s\wedge t$ appears could make them a little troublesome to manipulate,
from a computational point of view. To handle this issue, we verify that for
all $s,t\in[0,1]$ with $s\leq t$, it holds

$\displaystyle d^{\mathsf{L}}(s,t)$
$\displaystyle=\sum_{r\in[0,1]}\delta_{r}(x_{r}^{s},x_{r}^{t}),$ (9)
$\displaystyle\widetilde{d}^{\mathsf{L}}(s,t)$
$\displaystyle=\sum_{r\in[0,1]}\tilde{\delta}_{r}(x_{r}^{s},x_{r}^{t}),$ (10)
$\displaystyle d^{\mathsf{T}}(s,t)$
$\displaystyle=f(s)+f(t)-2\inf_{[s,t]}f-\sum_{r\in[0,1]}|x_{r}^{s}-x_{r}^{t}|.$
(11)

Indeed, Lemma 2 ensures $x_{r}^{s}=x_{r}^{t}$ when $r\prec t\wedge s$, which
yields

$\displaystyle\delta_{r}(x_{r}^{s},x_{r}^{t})$
$\displaystyle=\delta_{r}(0,x_{r}^{s}){\bf 1}_{s\wedge t\prec r\preceq
s}+\delta_{r}(0,x_{r}^{t}){\bf 1}_{s\wedge t\prec r\preceq
t}+\delta_{r}(x_{r}^{s},x_{r}^{t}){\bf 1}_{s\wedge t=r}$
$\displaystyle\tilde{\delta}_{r}(x_{r}^{s},x_{r}^{t})$
$\displaystyle=\tilde{\delta}_{r}(0,x_{r}^{s}){\bf 1}_{s\wedge t\prec r\preceq
s}+\tilde{\delta}_{r}(0,x_{r}^{t}){\bf 1}_{s\wedge t\prec r\preceq
t}+\tilde{\delta}_{r}(x_{r}^{s},x_{r}^{t}){\bf 1}_{s\wedge t=r},$
$\displaystyle|x_{r}^{s}-x_{r}^{t}|$ $\displaystyle=x_{r}^{s}{\bf 1}_{s\wedge
t\prec r\preceq s}+x_{r}^{t}{\bf 1}_{s\wedge t\prec r\preceq
t}+\left(\inf_{[r,s]}f-\inf_{[r,t]}f\right){\bf 1}_{s\wedge t=r}.$

Thus, (9) and (10) follow and Lemma 1 leads to (11). Let us show our
quantities are pseudo-distances.

######  Proposition 2.

The functions $d^{\mathsf{L}}$, $\widetilde{d}^{\mathsf{L}}$, and
$d^{\mathsf{T}}$ are pseudo-distances on $[0,1]$. As linear combinations of
pseudo-distances, $d^{\mathsf{V}}$ and $\widetilde{d}^{\mathsf{V}}$ are also
pseudo-distances on $[0,1]$.

###### Proof.

We already proved these functions are well-defined and non-negative. The
symmetries are obvious. Let $t\in[0,1]$, it holds clearly
$d^{\mathsf{L}}(t,t)=\widetilde{d}^{\mathsf{L}}(t,t)=0$ and we already saw
$d^{\mathsf{T}}(t,t)=0$. Let $s,r\in[0,1]$. We want to prove the triangular
inequalities on $d^{\mathsf{L}}(t,r)$, $\widetilde{d}^{\mathsf{L}}(t,r)$, and
$d^{\mathsf{T}}(t,r)$. The formula (9) makes it easy for $d^{\mathsf{L}}$ as a
simple application of the triangular inequality of $\delta_{u}$:

$d^{\mathsf{L}}(t,r)=\sum_{u\in[0,1]}\delta_{u}(x_{u}^{t},x_{u}^{r})\leq\sum_{u\in[0,1]}\left(\delta_{u}(x_{u}^{t},x_{u}^{s})+\delta_{u}(x_{u}^{s},x_{u}^{r})\right)=d^{\mathsf{L}}(t,s)+d^{\mathsf{L}}(s,t).$

The proof for $\widetilde{d}^{\mathsf{L}}$ is identical with (10) but the
proof for $d^{\mathsf{T}}$ requires some more care. We know either $s\wedge
r\preceq t\wedge s$ or $t\wedge s\preceq s\wedge r$, because the ancestral
lineage of $s$ is totally ordered. By symmetry, we can assume $s\wedge
r\preceq t\wedge s$ without loss of generality. Then, we also have $s\wedge
r\preceq t\wedge r$. Thanks to the total order of the ancestral lineage of
$t$, it holds either $t\wedge r\preceq t\wedge s$ or $t\wedge s\prec t\wedge
r$. If $t\wedge r\preceq t\wedge s$, then $t\wedge r\preceq s\wedge r$, and by
antisymmetry of $\preceq$, we have $t\wedge r=s\wedge r$. Likewise, if
$t\wedge s\prec t\wedge r$ then $t\wedge s\preceq s\wedge r$ and $t\wedge
s=s\wedge r$. From here, we treat the two cases separately. Recall Lemma 2
which we will use several times.

  * •

If $t\wedge r=s\wedge r\preceq t\wedge s$: We can write
$d^{\mathsf{T}}(t,r)=d^{\mathsf{T}}(t\wedge r,t)+d^{\mathsf{T}}(s\wedge r,r)$.
Let us begin by bounding the distance between $t\wedge r$ and $t$. Because
$t\wedge r\preceq t\wedge s\preceq t$, we can easily make the distance between
$t\wedge s$ and $t$ appear, which gives

$d^{\mathsf{T}}(t\wedge r,t)=d^{\mathsf{T}}(t\wedge s,t)+\left(\inf_{[t\wedge
s,t]}f-\inf_{[t\wedge r,t]}f\right)-\sum_{t\wedge r\prec u\preceq t\wedge
s}x_{u}^{t}.$

Now, we use Lemma 2 to transform the sum indexed by $s\wedge r=t\wedge r\prec
u\prec t\wedge s$. We point out that the lemma does not allow us to transform
the term for $u=t\wedge s$, thus we isolate it. Then, we recognize some terms
of the distance between $s\wedge r$ and $s$, namely

$\sum_{t\wedge r\prec u\preceq t\wedge s}x_{u}^{t}=\sum_{s\wedge r\prec u\prec
t\wedge s}x_{u}^{s}+x_{t\wedge s}^{t}{\bf 1}_{s\wedge r<t\wedge s}.$

We make $d^{\mathsf{T}}(s\wedge r,s)$ appear by adding and subtracting the
$x_{u}^{s}$ for $t\wedge s\preceq u\preceq s$, so that

$-\sum_{t\wedge r\prec u\preceq t\wedge s}x_{u}^{t}=d^{\mathsf{T}}(s\wedge
r,s)+\inf_{[s\wedge r,s]}f-d^{\mathsf{T}}(t\wedge s,s)-\inf_{[t\wedge
s,s]}f-(x_{t\wedge s}^{t}-x_{t\wedge s}^{s}){\bf 1}_{s\wedge r<t\wedge s}.$

Since $d^{\mathsf{T}}(t\wedge s,s)\geq 0$, it follows

$d^{\mathsf{T}}(t\wedge r,t)\leq d^{\mathsf{T}}(t\wedge
s,t)+d^{\mathsf{T}}(s\wedge r,s)-(x_{s\wedge r}^{t}-x_{s\wedge
r}^{s})+(x_{t\wedge s}^{t}-x_{t\wedge s}^{s}){\bf 1}_{s\wedge r=t\wedge s}.$

Then, we remark that $-(x_{s\wedge r}^{t}-x_{s\wedge r}^{s})+(x_{t\wedge
s}^{t}-x_{t\wedge s}^{s}){\bf 1}_{s\wedge r=t\wedge s}=-(x_{s\wedge
r}^{t}-x_{s\wedge r}^{s}){\bf 1}_{s\wedge r<t\wedge s}$, so Lemma 2 ensures
$d^{\mathsf{T}}(t\wedge r,t)\leq d^{\mathsf{T}}(t\wedge
s,t)+d^{\mathsf{T}}(s\wedge r,s)$. Eventually, we find

$d^{\mathsf{T}}(t,r)\leq d^{\mathsf{T}}(s\wedge r,r)+d^{\mathsf{T}}(t\wedge
s,t)+d^{\mathsf{T}}(s\wedge r,s)\leq d^{\mathsf{T}}(t,s)+d^{\mathsf{T}}(s,r).$

  * •

If $t\wedge s=s\wedge r\prec t\wedge r$: In this case, the triangular
inequality is rougher, so its proof is simpler. To bound
$d^{\mathsf{T}}(t\wedge r,t)$, we first make $d^{\mathsf{T}}(t\wedge s,t)$
appear by writing

$d^{\mathsf{T}}(t\wedge r,t)=d^{\mathsf{T}}(t\wedge s,t)+\inf_{[t\wedge
s,t]}f-\inf_{[t\wedge r,t]}f+\sum_{t\wedge s\prec u\preceq t\wedge
r}x_{u}^{t}.$

We transform the sum thanks to Lemma 2 once again and we see $\inf_{[t\wedge
r,t]}f=x_{t\wedge r}^{t}-x_{t\wedge r}^{t\wedge r}+f(t\wedge r)$, so that we
get

$d^{\mathsf{T}}(t\wedge r,t)=d^{\mathsf{T}}(t\wedge s,t)+\inf_{[t\wedge
s,t]}f-\inf_{[t\wedge s,t\wedge r]}f-d^{\mathsf{T}}(t\wedge s,t\wedge r).$

Lemma 2 also implies $\inf_{[t\wedge s,t]}f=\inf_{[t\wedge s,t\wedge r]}f$.
Plus, $d^{\mathsf{T}}(t\wedge s,t\wedge r)$ is non-negative, so it follows
$d^{\mathsf{T}}(t\wedge r,t)\leq d^{\mathsf{T}}(t\wedge s,t)$. Thanks to the
symmetry between $t$ and $r$ in this case, we find

$d^{\mathsf{T}}(t,r)\leq d^{\mathsf{T}}(t\wedge s,t)+d^{\mathsf{T}}(s\wedge
r,r)\leq d^{\mathsf{T}}(t,s)+d^{\mathsf{T}}(s,r).$

∎

######  Remark 2.

The triangular inequality for $d^{\mathsf{T}}$ will become clearer later.
Indeed, we will prove with Theorem 4 that there exists a continuous excursion
$g$ with $g(0)=0$ such that $d^{\mathsf{T}}_{f}=d^{\mathrm{clas}}_{g}$, where
$d^{\mathrm{clas}}$ is the classical tree pseudo-distance defined by (1). In
fact, Marzouk defines the same pseudo-distance $d^{\mathsf{T}}$ as such in
[35].

###  2.3 Quotient metric spaces induced by those pseudo-distances

A pointed metric space is a triple $(X,d,a)$ where $(X,d)$ is a metric space
equipped with a distinguished point, also called a root, $a\in X$. We say two
pointed metric spaces $(X_{1},d_{1},a_{1})$ and $(X_{2},d_{2},a_{2})$ are
pointed-isometric when there exists an bijective isometry $\lambda$ from
$X_{1}$ to $X_{2}$ such that $\lambda(a_{1})=a_{2}$. A pointed weighted metric
space is a quadruple $(X,d,a,\mu)$ where $(X,d,a)$ is a pointed metric space
also equipped with a Borel probability measure $\mu$ on $(X,d)$. We say two
pointed weighted metric spaces $(X_{1},d_{1},a_{1},\mu_{1})$ and
$(X_{2},d_{2},a_{2},\mu_{2})$ are GHP-isometric when there exists an bijective
isometry $\lambda$ from $X_{1}$ to $X_{2}$ such that $\lambda(a_{1})=a_{2}$
and $\lambda_{*}\mu_{1}=\mu_{2}$. When no confusion is possible, we will
denote a metric space (possibly endowed with a root and/or a Borel probability
measure) by its underlying set. Finally, when $c>0$, we write
$c\cdot(X,d,a,\mu)=(X,cd,a,\mu)$ for the pointed weighted metric space
obtained after multiplying all distances by $c$. As same, we write
$c\cdot(X,d,a)=(X,cd,a)$ and $c\cdot(X,d)=(X,cd)$.

Let $d$ be a pseudo-distance on a set $X$. We denote by $X/\\{d=0\\}$ the
quotient space obtained by identifying the points $x,y\in X$ such that
$d(x,y)=0$ and we endow it with the genuine distance (still denoted by $d$
with a slight abuse of notation) induced by $d$. When $X=[0,1]$, we equip the
metric space $[0,1]/\\{d=0\\}$ with the canonical projection of $1$ and with
the image of the Lebesgue measure on $[0,1]$, so that it is a pointed weighted
metric space. We define the following pointed weighted metric spaces:

  * •

the tree coded by $f$, denoted by $\mathscr{T}_{f}$, as
$[0,1]/\\{d^{\mathsf{T}}_{f}=0\\}$,

  * •

the looptree coded by $f$, denoted by $\mathscr{L}_{f}$, as
$[0,1]/\\{d^{\mathsf{L}}_{f}=0\\}$,

  * •

the shuffled (with $\Phi$) looptree coded by $f$, denoted by
$\widetilde{\mathscr{L}}_{f}$, as
$[0,1]/\\{\widetilde{d}^{\mathsf{L}}_{f}=0\\}$,

  * •

the vernation tree coded by $f$, denoted by $\mathscr{V}_{f}$, as
$[0,1]/\\{d^{\mathsf{V}}_{f}=0\\}$,

  * •

the shuffled (with $\Phi$) vernation tree coded by $f$, denoted by
$\widetilde{\mathscr{V}}_{f}$, as
$[0,1]/\\{\widetilde{d}^{\mathsf{V}}_{f}=0\\}$.

######  Proposition 3.

The functions $d^{\mathsf{L}},d^{\mathsf{T}},d^{\mathsf{V}}$ are continuous on
$[0,1]^{2}$. The metric spaces $\mathscr{L}_{f}$, $\mathscr{T}_{f}$, and
$\mathscr{V}_{f}$ are thus compact.

###### Proof.

Since $d^{\mathsf{L}}$ is a pseudo-distance, we only need to show that
$d^{\mathsf{L}}(t,t_{n})\longrightarrow 0$ when $t_{n}\rightarrow t-$ or
$t_{n}\rightarrow t+$. We use the inequality (6) and the fact that $f$ is an
excursion. If $t_{n}<t<t_{m}$ then

$\displaystyle d^{\mathsf{L}}(t,t_{n})\leq f(t-)+f(t_{n})-2\inf_{[t_{n},t]}f$
$\displaystyle\underset{t_{n}\rightarrow
t-}{\longrightarrow}f(t-)+f(t-)-2f(t-)=0,$ $\displaystyle
d^{\mathsf{L}}(t,t_{m})\leq f(t)+f(t_{m}-)-2\inf_{[t,t_{m}]}f$
$\displaystyle\underset{t_{m}\rightarrow
t+}{\longrightarrow}f(t)+f(t)-2f(t)=0.$

We prove the continuity of $d^{\mathsf{T}}$ in the same way with (7). The
continuity of $d^{\mathsf{V}}$ follows. ∎

The pseudo-distance $\widetilde{d}^{\mathsf{L}}$ (as well as
$\widetilde{d}^{\mathsf{V}}$) is not continuous in general because of the
possible discontinuities of the $\phi_{\Delta}$. However, it still enjoys some
regularity.

######  Definition 3.

We denote by $\mathcal{D}([0,1]^{2})$ the set of functions
$\psi:[0,1]^{2}\longrightarrow\mathbb{R}$ such that for every monotonous
sequences $(s_{n})$ and $(t_{n})$ of elements of $[0,1]$, the sequence
$(\psi(s_{n},t_{n}))$ converges, and such that its limit is $\psi(\lim
s_{n},\lim t_{n})$ when $(s_{n})$ and $(t_{n})$ are non-increasing. We provide
a distance $\rho_{2}$ to $\mathcal{D}([0,1]^{2})$ by setting

$\rho_{2}(\psi_{1},\psi_{2})=\inf_{\lambda,\mu\in\Lambda}\max\left(||\psi_{1}-\psi_{2}\circ(\lambda,\mu)||_{\infty},||\lambda-\mathsf{id}||_{\infty},||\mu-\mathsf{id}||_{\infty}\right)$

for all $\psi_{1},\psi_{2}\in\mathcal{D}([0,1]^{2})$, where $\Lambda$ is the
set of increasing bijections from $[0,1]$ into itself. The distance $\rho_{2}$
is not complete but induces a separable and completely metrizable topology on
$\mathcal{D}([0,1]^{2})$, that we call the Skorokhod topology.

This is a generalization of the Skorokhod space $\mathcal{D}([0,1])$ for
bivariate functions, and is more precisely presented in [40]. Let us at least
mention that if $(\psi_{n})$ converges to $\psi$ for this topology and if
$\psi$ is continuous on $[0,1]^{2}$, then the convergence also happens
uniformly on $[0,1]^{2}$. Indeed, such $\psi$ would be continuous on a
compact, so uniformly continuous, which is enough to conclude.

######  Proposition 4.

The metric spaces $\widetilde{\mathscr{L}}_{f}$ and
$\widetilde{\mathscr{V}}_{f}$ are compact and the functions
$\widetilde{d}^{\mathsf{L}}$ and $\widetilde{d}^{\mathsf{V}}$ are in
$\mathcal{D}([0,1]^{2})$. Equivalently, the canonical projection maps from
$[0,1]$ to $\widetilde{\mathscr{L}}_{f}$ or $\widetilde{\mathscr{V}}_{f}$ are
càdlàg.

###### Proof.

By triangular inequality, it is enough to show that if $s_{n}\uparrow s$ and
$t_{n}\downarrow t$, then $\widetilde{d}^{\mathsf{L}}(t_{n},t)\longrightarrow
0$ and there is $\bar{s}\in[0,1]$ such that
$\widetilde{d}^{\mathsf{L}}(s_{n},\bar{s})\longrightarrow 0$ and
$d^{\mathsf{T}}(s,\bar{s})=0$. We begin by proving
$\widetilde{d}^{\mathsf{L}}(t,t_{n})\longrightarrow 0$. Thanks to the
inequality (8) and Lemma 1, there exists a constant $K\in(0,\infty)$ such that

$\widetilde{d}^{\circ}(t\wedge t_{n},t)+\widetilde{d}^{\circ}(t\wedge
t_{n},t_{n})\leq K\left(f(t)+f(t_{n})-2\inf_{[t,t_{n}]}f\right),$

but $f(t)+f(t_{n})-2\inf_{[t,t_{n}]}f\longrightarrow 0$ because $f$ is càdlàg.
Hence, we only need to show $\tilde{\delta}_{t\wedge t_{n}}(x_{t\wedge
t_{n}}^{t},x_{t\wedge t_{n}}^{t_{n}})$ tends to $0$. Thanks to Lemma 1, we
observe the sequence $(t\wedge t_{n})$ is non-decreasing. Moreover, the set
$\\{r\in[0,1]\ :\ \Delta_{r}\geq\varepsilon\\}$ is finite for all
$\varepsilon>0$, so only two cases are possible: either $\Delta_{t\wedge
t_{n}}\longrightarrow 0$, or there exists $\tau\in[0,1]$ with
$\Delta_{\tau}>0$ such that $t\wedge t_{n}=\tau$ for $n$ large enough.

  * •

In the first case: We have $\tilde{\delta}_{t\wedge t_{n}}(x_{t\wedge
t_{n}}^{t},x_{t\wedge t_{n}}^{t_{n}})\leq\Delta_{t\wedge t_{n}}\longrightarrow
0$.

  * •

In the second case: When $n$ is large enough, we have $\tilde{\delta}_{t\wedge
t_{n}}(x_{t\wedge t_{n}}^{t},x_{t\wedge
t_{n}}^{t_{n}})=\tilde{\delta}_{\tau}(x_{\tau}^{t},x_{\tau}^{t_{n}})$. But,
$x_{\tau}^{t_{n}}\uparrow x_{\tau}^{t}$ because $f$ is càdlàg. The function
$\phi_{\Delta_{\tau}}$ is càglàd, so
$\tilde{\delta}_{\tau}(x_{\tau}^{t},x_{\tau}^{t_{n}})\longrightarrow\hat{\delta}_{\tau}(x_{\tau}^{t},x_{\tau}^{t})=0$.

This concludes the proof of
$\widetilde{d}^{\mathsf{L}}(t,t_{n})\longrightarrow 0$.

We set $\displaystyle{s^{\prime}=\inf\\{r\geq s\ :\ f(r)\leq f(s-)\\}}$, so
that $\inf_{[s,s^{\prime}]}f=f(s-)=f(s^{\prime})=f(s^{\prime}-)$ because $f$
is an excursion. Thus, $s\preceq s^{\prime}$ and
$d^{\mathsf{T}}(s,s^{\prime})=0$ by the inequality (5). The sequence
$(s_{n}\wedge s^{\prime})$ is non-decreasing according to Lemma 1, so as
before, either $\Delta_{s_{n}\wedge s^{\prime}}\longrightarrow 0$, or there
exists $\sigma\in[0,1]$ with $\Delta_{\sigma}>0$ such that $s_{n}\wedge
s^{\prime}=\sigma$ for $n$ large enough.

  * •

In the first case: We set $\bar{s}=s^{\prime}$. As $f$ is càdlàg, it holds
$f(\bar{s})+f(s_{n})-2\inf_{[s_{n},\bar{s}]}f\longrightarrow 0$ by definition
of $s^{\prime}=\bar{s}$. Plus, we have $\tilde{\delta}_{\bar{s}\wedge
s_{n}}(x_{\bar{s}\wedge s_{n}}^{s_{n}},x_{\bar{s}\wedge
s_{n}}^{\bar{s}})\leq\Delta_{s_{n}\wedge s^{\prime}}\longrightarrow 0$ by
asssumption. Then, recall the use of (8) and of Lemma 1 to conclude
$\widetilde{d}^{\mathsf{L}}(s_{n},\bar{s})\longrightarrow 0$. We already know
$d^{\mathsf{T}}(s,\bar{s})=0$.

  * •

In the second case: When $n$ is large enough, it holds $\sigma=s_{n}\wedge
s^{\prime}\preceq s^{\prime}$ so
$\inf_{[\sigma,s^{\prime}]}f=\inf_{[s_{n},s^{\prime}]}f$ according to Lemma 1.
But, we see $\inf_{[s_{n},s^{\prime}]}f\longrightarrow f(s^{\prime})$ by
definition of $s^{\prime}$, so the inequality (5) yields
$d^{\mathsf{T}}(\sigma,s^{\prime})=0$. We already know
$d^{\mathsf{T}}(s,s^{\prime})=0$ so $d^{\mathsf{T}}(\sigma,s)=0$. Let us set

$\bar{s}=\inf\left\\{r\geq\sigma\ :\
\phi_{\Delta_{\sigma}}\left(\frac{f(r)-f(\sigma-)}{\Delta_{\sigma}}\right)=\phi_{\Delta_{\sigma}}\left(\frac{x_{\sigma}^{s}}{\Delta_{\sigma}}+\right)\right\\},$

which is well-defined because $\phi_{\Delta_{\sigma}}$ is surjective and
càglàd, and because $f$ is an excursion. Using properties of excursions, we
find $\sigma\preceq\bar{s}$ and $f(\bar{s})=\inf_{[\sigma,\bar{s}]}f$. As a
result, $\sigma\preceq s_{n}\wedge\bar{s}$ when $n$ is large enough and if
$r\in(\sigma,\bar{s}]$ then $x_{r}^{\bar{s}}=0$. It follows
$d^{\mathsf{T}}(\sigma,\bar{s})=0=d^{\mathsf{T}}(s,\bar{s})$ and

$\widetilde{d}^{\mathsf{L}}(s_{n},\bar{s})\leq\widetilde{d}^{\circ}(s_{n}\wedge
s^{\prime},s_{n})+\tilde{\delta}_{\sigma}(x_{\sigma}^{s_{n}},x_{\sigma}^{\bar{s}})$

when $n$ is large enough. As seen in the previous case,
$f(s_{n})-\inf_{[s_{n},s^{\prime}]}f\longrightarrow 0$ so (8) and Lemma 1
ensure $\widetilde{d}^{\circ}(s_{n}\wedge s^{\prime},s_{n})\longrightarrow 0$.
Then, we check
$\phi_{\Delta_{\sigma}}(x_{\sigma}^{\bar{s}}/\Delta_{\sigma})=\phi_{\Delta_{\sigma}}(x_{\sigma}^{s}/\Delta_{\sigma}+)$
using the left-continuity of $\phi_{\Delta_{\sigma}}$. Since $f(s)\geq f(s-)$,
$x_{\sigma}^{s_{n}}$ non-increasingly tends towards $x_{\sigma}^{s}$. Thus,
$\tilde{\delta}_{\sigma}(x_{\sigma}^{s_{n}},x_{\sigma}^{\bar{s}})\longrightarrow
0$ and the proof is completed.

∎

###  2.4 First properties and an example

Here, we provide some basic properties of the codings
$f\in\mathcal{H}\longmapsto
d^{\mathsf{L}}_{f},\widetilde{d}^{\mathsf{L}}_{f},d^{\mathsf{T}}_{f},d^{\mathsf{V}}_{f},\widetilde{d}^{\mathsf{V}}_{f}$,
that they share with the classical coding of a tree (1). These properties make
the codings somewhat easy to manipulate.

######  Proposition 5.

The application $f\in\mathcal{H}\longmapsto d^{\mathsf{L}}_{f}$ (respectively
$d^{\mathsf{T}}_{f}$, $d^{\mathsf{V}}_{f}$) is homogeneous. Namely, if
$\alpha>0$, then it holds $d^{\mathsf{L}}_{\alpha f}=\alpha
d^{\mathsf{L}}_{f}$. Thus, the pointed weighted metric spaces
$\mathscr{L}_{\alpha f}$ and $\alpha\cdot\mathscr{L}_{f}$ are GHP-isometric.

The application $f\in\mathcal{H}\longmapsto d^{\mathsf{L}}_{f}$ (respectively
$d^{\mathsf{T}}_{f}$, $\widetilde{d}^{\mathsf{L}}_{f}$, $d^{\mathsf{V}}_{f}$,
$\widetilde{d}^{\mathsf{V}}_{f}$) enjoys the time-changing property. Namely,
for any increasing bijection $\lambda:[0,1]\longrightarrow[0,1]$, it holds
$d^{\mathsf{L}}_{f\circ\lambda}(s,t)=d^{\mathsf{L}}_{f}(\lambda(s),\lambda(t))$
for all $s,t\in[0,1]$. Thus, the pointed metric spaces
$\mathscr{L}_{f\circ\lambda}$ and $\mathscr{L}_{f}$ are pointed-isometric.

###### Proof.

The first thing to see is that $\alpha f$ and $f\circ\lambda$ are indeed
excursions. Then, it is obvious that $x_{s}^{t}(\alpha f)=\alpha x_{s}^{t}(f)$
and $x_{s}^{t}(f\circ\lambda)=x_{\lambda(s)}^{\lambda(t)}(f)$, for all
$s,t\in[0,1]$. In particular, $\Delta_{s}(\alpha f)=\alpha\Delta_{s}(f)$ and
$\Delta_{s}(f\circ\lambda)=\Delta_{\lambda(s)}(f)$. By definition of
$\delta_{s}$ and $\tilde{\delta}_{s}$ and thanks to the identities (9), (10),
and (11), the proposition follows. Let us point out that we have used
$\delta_{s}(\alpha a,\alpha b)=\alpha\delta_{s}(a,b)$ which does not hold for
$\tilde{\delta}_{s}$, which explains why
$f\longmapsto\widetilde{d}^{\mathsf{L}}_{f}$ and
$f\longmapsto\widetilde{d}^{\mathsf{V}}_{f}$ are not homogeneous. ∎

Another important property of the codings is the branching property. It tells
that a "sub-excursion" codes a subspace of the looptree or the vernation tree.
It allows observing directly the structure of the coded space on the graph of
the excursion. We have already presented this idea when we defined the
genealogy $\preceq$.

######  Definition 4.

Let $(X_{0},d_{0},a_{0})$ and $(X_{1},d_{1},a_{1})$ be two pointed metric
spaces and let $a\in X_{0}$. We write $a_{0}^{*}=a$ and $a_{1}^{*}=a_{1}$. The
gluing of $X_{1}$ on $X_{0}$ at $a$ is the pointed metric space denoted by
$(X_{0}\vee_{a}X_{1},d,a_{0})$ that is obtained from the quotient of the
disjoint union $X_{0}\sqcup X_{1}$ by the identification $a\sim a_{1}$,
endowed with the distance $d$ defined by setting $d(x,y)=d_{i}(x,y)$ if
$x,y\in X_{i}$ with $i\in\\{0,1\\}$, and by setting
$d(x,y)=d_{i}(x,a_{i}^{*})+d_{1-i}(a_{1-i}^{*},y)$ if $x\in X_{i}$ and $y\in
X_{1-i}$ with $i\in\\{0,1\\}$. We see $X_{0}$ and $X_{1}$ as closed subsets of
$X_{0}\vee_{a}X_{1}$, so that $X_{0}\cup X_{1}=X_{0}\vee_{a}X_{1}$, $X_{0}\cap
X_{1}=\\{a\\}=\\{a_{1}\\}$, and the distinguished points of $X_{0}$ and
$X_{0}\vee_{a}X_{1}$ are the same.

######  Proposition 6.

The application $f\in\mathcal{H}\longmapsto d^{\mathsf{L}}_{f}$ (respectively
$d^{\mathsf{T}}_{f}$, $\widetilde{d}^{\mathsf{L}}_{f}$, $d^{\mathsf{V}}_{f}$,
$\widetilde{d}^{\mathsf{V}}_{f}$) enjoys the branching property. Namely, let
$u,v\in[0,1]$ be such that $u\prec v$ and $f(u-)=f(v-)$, and let us set

$\displaystyle g$ $\displaystyle:t\in[0,1]\longmapsto\begin{cases}f(t)&\text{
if }t\notin[u,v)\\\ f(u-)&\text{ if }t\in[u,v)\end{cases},$ $\displaystyle h$
$\displaystyle:t\in[0,1]\longmapsto\begin{cases}f(u+t(v-u))-f(u-)&\text{ if
}t<1\\\ 0&\text{ if }t=1\end{cases}.$

If $s,t\in[0,1)$ and $a,b\in[0,1]\backslash[u,v)$, then

$\displaystyle d^{\mathsf{L}}_{f}(a,b)$
$\displaystyle=d^{\mathsf{L}}_{g}(a,b),$ $\displaystyle
d^{\mathsf{L}}_{f}(u+s(v-u),u+t(v-u))$
$\displaystyle=d^{\mathsf{L}}_{h}(s,t),$ $\displaystyle
d^{\mathsf{L}}_{f}(u+s(v-u),a)$
$\displaystyle=d^{\mathsf{L}}_{g}(a,u)+d^{\mathsf{L}}_{h}(s,1).$

Thus, the pointed metric space $\mathscr{L}_{f}$ is pointed-isometric to the
gluing of $\mathscr{L}_{h}$ on $\mathscr{L}_{g}$ at the canonical projection
of $u$.

###### Proof.

First, we observe that $g$ and $h$ are indeed excursions. We only show the
result for $d^{\mathsf{L}}$ and $d^{\mathsf{T}}$, as the same proof holds for
$\widetilde{d}^{\mathsf{L}}$ and because the branching property is preserved
by linear combination. Let $a,b\in[0,1]\backslash[u,v)$ and $s,t\in[0,1)$. We
assume $a\leq b$ and $s\leq t$, and we set $w=u+s(v-u)$ to lighten the
notations. We immediately observe that
$x_{u+s(v-u)}^{u+t(v-u)}(f)=x_{s}^{t}(h)$, and if $[u,v)\cap[a,b]=\emptyset$
then $x_{a}^{b}(f)=x_{a}^{b}(g)$. If $[a,b]\subset[u,v)$ then we still have
$x_{a}^{b}(f)=x_{a}^{b}(g)$ because $\inf_{[u,v)}f=f(u-)=\inf_{[u,v)}g$.
Concerning $x_{a}^{w}$, if $w<a$ then $x_{a}^{w}(f)=x_{a}^{u}(g)=0$ by
definition, since $u\leq w$. If $a\leq w$ then $a<u\leq w<v$, so
$x_{a}^{w}(f)=x_{a}^{u}(g)$ still holds, because

$\inf_{[u,w]}f\geq\inf_{[u,v]}f\geq
f(u-)=\inf_{[u,w]}g\geq\inf_{[a,u]}f=\inf_{[a,u]}g.$

Concerning $x_{w}^{a}$, if $a<w$ then $x_{w}^{a}(f)=x_{w}^{a}(g)=0$ by
definition. If $w\leq a$ then $u\leq w<v\leq a$, so
$x_{w}^{a}(f)=x_{w}^{a}(g)=0$ still holds because

$f(w-)\geq\min\left(f(u-),\inf_{[u,v]}f\right)\geq
f(u-)=g(w-)=f(v-)=g(v-)\geq\max\left(\inf_{[w,a]}f,\inf_{[w,a]}g\right).$

Moreover, $x_{w}^{u}(g)\leq\Delta_{w}(g)=0$ and $x_{s}^{1}(h)\leq h(1)=0$, so
that $x_{w}^{u}(g)=x_{s}^{1}(h)=0$. To sum up, we have

$\displaystyle\Delta_{a}(f)=\Delta_{a}(g)$ $\displaystyle\text{ and
}\Delta_{u+s(v-u)}(f)=\Delta_{s}(h),$ $\displaystyle
x_{a}^{b}(f)=x_{a}^{b}(g)$ $\displaystyle\text{ and
}x_{u+s(v-u)}^{u+t(v-u)}(f)=x_{s}^{t}(h),$ $\displaystyle
x_{a}^{w}(f)=x_{a}^{u}(g)$ $\displaystyle\text{ and
}x_{w}^{a}(f)=x_{w}^{a}(g)=x_{w}^{u}(g)=x_{s}^{1}(h)=0.$

Combining these identities with the formula (9), we obtain the branching
property for $d^{\mathsf{L}}$. Some easy other observations together with the
formula (11) and the above identities lead to the branching property for
$d^{\mathsf{T}}$. ∎

$1$$0.5$$1$$1.5$$2$$2.5$$3$$3.5$$4$$4.5$$5$$5.5$ (a) An excursion $f$.

(b) Its coded tree $\mathscr{T}_{f}$.

(c) Its coded looptree $\mathscr{L}_{f}$.

(d) Its coded shuffled looptree $\widetilde{\mathscr{L}}_{f}$.

(e) Its coded vernation $\mathscr{V}_{f}$.

(f) Its coded shuffled vernation tree $\widetilde{\mathscr{V}}_{f}$.

Figure 2: An example of an excursion with simple variations and of the
quotient metric spaces that it codes. We point out that the representations of
the shuffled looptree and the shuffled vernation tree are only likely examples
because they could differ depending on the chosen shuffle $\Phi$. The
representations given are isometric with respect to the scale of the $y$-axis
of the graph of the excursion.

######  Corollary 1.

Let $s,t\in[0,1]$. If $s\preceq t$, then
$d^{\mathsf{T}}(t,1)=d^{\mathsf{T}}(s,1)+d^{\mathsf{T}}(s,t)$.

###### Proof.

Let $\bar{s}=\inf\\{r\geq t\ :\ f(r)=f(s-)\\}$, we have $t\leq\bar{s}$,
$s\preceq\bar{s}$, and $f(s-)=f(\bar{s})=f(\bar{s}-)$. Observe
$f(\bar{s})=\inf_{[t,\bar{s}]}f\leq\inf_{[s,t]}f$ so
$d^{\mathsf{T}}_{f}(s,\bar{s})=0$ thanks to (5). If $t=\bar{s}$ then the
result becomes obvious. Let us assume that $t<\bar{s}$. We use the notations
of Proposition 6 with $u=s$ and $v=\bar{s}$. The inequality (5) gives
$d^{\mathsf{T}}_{f}(0,1)=d^{\mathsf{T}}_{h}(0,1)=d^{\mathsf{T}}_{f}(s,\bar{s})=d^{\mathsf{T}}_{g}(s,\bar{s})=0$.
Then, the branching property implies the desired result. ∎

Let us discuss how one can intuitively understand the looptree or the
vernation tree coded by an excursion $f$ from a glance at its graph. On the
one hand, if $f$ does not have any jump, then (9) and (11) ensure that
$d^{\mathsf{L}}=0$ and that $d^{\mathsf{T}}$ is equal to the classical tree
pseudo-distance $d^{\mathrm{clas}}$ defined by (1). On the other hand, each
jump of $f$ indeed corresponds to a loop in the associated looptree or
vernation tree. Let $s\in[0,1]$ such that $\Delta_{s}>0$ and let
$x\in\mathcal{C}$, the times

$\displaystyle\tau(x)$ $\displaystyle=\inf\left\\{t\geq s\ :\
\frac{f(t)-f(s-)}{\Delta_{s}}=x\right\\},$ $\displaystyle\tilde{\tau}(x)$
$\displaystyle=\inf\left\\{t\geq s\ :\
\phi_{\Delta_{s}}\left(\frac{f(t)-f(s-)}{\Delta_{s}}\right)=x\right\\}$

are well-defined, with $x_{s}^{\tau(x)}/\Delta_{s}=x$ and
$\phi_{\Delta_{s}}(x_{s}^{\tilde{\tau}(x)}/\Delta_{s})=x$. Then, we let the
reader check that
$d^{\mathsf{T}}(\tau(x),\tau(y))=d^{\mathsf{T}}(\tilde{\tau}(x),\tilde{\tau}(y))=0$
and

$d^{\mathsf{L}}(\tau(x),\tau(y))=\delta_{s}(x_{s}^{\tau(x)},x_{s}^{\tau(y)})=\Delta_{s}\delta(x,y)=\tilde{\delta}_{s}(x_{s}^{\tilde{\tau}(x)},x_{s}^{\tilde{\tau}(y)})=\widetilde{d}^{\mathsf{L}}(\tilde{\tau}(x),\tilde{\tau}(y))$

for any $x,y\in\mathcal{C}$. Hence, the metric subspaces of $\mathscr{L}_{f}$
and $\widetilde{\mathscr{L}}_{f}$ respectively induced by $\\{\tau(x)\ :\
x\in\mathcal{C}\\}$ and $\\{\tilde{\tau}(x)\ :\ x\in\mathcal{C}\\}$ are both
isometric to $\Delta_{s}\cdot\mathcal{C}$, namely a metric circle of length
$\Delta_{s}$. The metric subspaces of $\mathscr{V}_{f}$ and
$\widetilde{\mathscr{V}}_{f}$ induced by the same previous subsets of $[0,1]$
are isometric to $2\Delta_{s}\cdot\mathcal{C}$, namely a metric circle of
length $2\Delta_{s}$. These observations together with the homogeneity, the
time-changing property, and the branching property are generally enough to
accurately draw $\mathscr{T}_{f}$, $\mathscr{L}_{f}$, or $\mathscr{V}_{f}$
from the graph of $f$ when it has simple variations. We give an example with
Figure 2. For $\widetilde{\mathscr{L}}_{f}$ and $\widetilde{\mathscr{V}}_{f}$,
the lack of homogeneity and the complexity of the shuffle can make them hard
to be precisely drawn. Nevertheless, we can understand them as transformed
versions of $\mathscr{L}_{f}$ and $\mathscr{V}_{f}$ where the positions of the
joint points between loops or trees were shuffled, see Figure 2 once again.

##  3 Unification and relations between trees, looptrees, and vernation trees

In this section, we inspect more in-depth the intuitive principle that jumps
code for loops and continuous growths code for branches in the associated
vernation tree. The vernation distance $d^{\mathsf{V}}$, or
$\widetilde{d}^{\mathsf{V}}$, is the combinaison of two distances
$d^{\mathsf{L}}$, or $\widetilde{d}^{\mathsf{L}}$, and $d^{\mathsf{T}}$.
Similarly, we will see that we can always decompose an excursion into two
others whose associated vernation pseudo-distances are respectively reduced to
$2d^{\mathsf{L}}$ and $d^{\mathsf{T}}$. Let us begin with the definition of
these two kinds of excursions.

######  Definition 5.

Let $f$ be an excursion. We say $f$ is a continuous excursion when
$\Delta_{t}=0$ for all $t\in[0,1]$. Notice that it do not only require $f$ to
be continuous on $[0,1]$ but also that $f(0)=0$. We say $f$ is a pure jump
growth (PJG) excursion when $\displaystyle{f(t)=\sum_{r\preceq t}x_{r}^{t}}$
for all $t\in[0,1]$.

The following lemma is practical to both wrap one’s head around the notion of
PJG excursions and to identify them in simple cases.

######  Lemma 4.

Let $f$ be an excursion. If there exists a partition
$0=r_{0}<r_{1}<...<r_{n}<r_{n+1}=1$ such that $f$ is non-increasing on
$[r_{i},r_{i+1})$ for all $0\leq i\leq n$, then $f$ is PJG.

###### Proof.

First, observe that $f$ is PJG if and only if $d^{\mathsf{T}}(t,1)=0$ for all
$t\in[0,1]$. Indeed, $x_{s}^{1}\leq f(1)=0$ for all $s\in[0,1]$ so the
identity (11) becomes $\displaystyle{d^{\mathsf{T}}(t,1)=f(t)-\sum_{r\preceq
t}x_{r}^{t}}$. We prove the lemma by induction on $n$. If $n=0$, then $f$ is
non-increasing on $[0,1]$ so

$d^{\mathsf{T}}(t,1)\leq f(t)-x_{0}^{t}=f(t)-\inf_{[0,t]}f=0,$

and it follows that $f$ is PJG. If $n\geq 1$, we set $\bar{r}=\inf\\{r\geq
r_{n}\ :\ f(r)=f(r_{n}-)\\}$ so that $r_{n}\preceq\bar{r}$ and
$f(r_{n}-)=f(\bar{r})=f(\bar{r}-)$. We can assume $r_{n}<\bar{r}$ because
otherwise, $f(r_{n})=f(r_{n}-)$ which means that $f$ is in fact non-increasing
on $[r_{n-1},1]$ and so $f$ is PJG by induction. We apply the branching
property as given by Proposition 6 with $u=r_{n}$ and $v=\bar{r}$ and we keep
the same notations. It is easy to see that $h$ is non-increasing on $[0,1]$
and that $g$ is non-increasing on $[r_{n-1},1]$ and on $[r_{i},r_{i+1})$ for
all $0\leq i\leq n-2$. By induction, $g$ and $h$ are PJG, so
$d^{\mathsf{T}}_{g}(t,1)=d^{\mathsf{T}}_{h}(t,1)=0$ for all $t\in[0,1]$. The
branching property then ensures $d^{\mathsf{T}}_{f}(t,1)=0$ for all
$t\in[0,1]$, so $f$ is PJG. ∎

For any $\varepsilon>0$, we define the two following operators.

$\displaystyle J:\mathcal{H}$ $\displaystyle\longrightarrow\mathcal{H}$
$\displaystyle J^{\varepsilon}:\mathcal{H}$
$\displaystyle\longrightarrow\mathcal{H}$ $\displaystyle f$
$\displaystyle\longmapsto Jf:t\mapsto\sum_{r\preceq t}x_{r}^{t}$
$\displaystyle f$ $\displaystyle\longmapsto
J^{\varepsilon}f:t\mapsto\sum_{\begin{subarray}{c}r\preceq t\\\
\Delta_{r}\geq\varepsilon\end{subarray}}x_{r}^{t}$

The operator $J$ is the same as in Theorem 1 and the operator
$J^{\varepsilon}$ will be necessary for some approximations later. We are now
ready to formulate the decomposition of an excursion into a sum of a
continuous excursion with a PJG excursion, and how this decomposition
translates into the associated pseudo-distances.

######  Theorem 4.

Let $f$ be an excursion and let $\varepsilon>0$. The following points hold.

  1. (i)

The function $f-Jf$ is a continuous excursion and the functions $Jf$ and
$J^{\varepsilon}f$ are PJG excursions. Moreover, for any $t\in[0,1]$,
$\Delta_{t}(J^{\varepsilon}f)>0\Longrightarrow\Delta_{t}(J^{\varepsilon}f)\geq\varepsilon$.

  2. (ii)

The excursion $f$ is PJG if and only if $Jf=f$.

  3. (iii)

The excursion $f$ is a continuous excursion if and only if $Jf=0$.

  4. (iv)

We have $Jf=J^{\varepsilon}f$ if and only if for any $t\in[0,1]$,
$\Delta_{t}(f)>0\Longrightarrow\Delta_{t}(f)\geq\varepsilon$.

  5. (v)

The function $f-Jf+J^{\varepsilon}f$ is an excursion and
$J(f-Jf+J^{\varepsilon}f)=J^{\varepsilon}f$.

###### Proof.

For any $t\in[0,1]$, the quantities $Jf(t)$ and $J^{\varepsilon}f(t)$ are sums
of non-negative terms, so $Jf(t)\geq 0$ and $J^{\varepsilon}f(t)\geq 0$. Plus,
$x_{t}^{1}(f)\leq f(1)=0$ so $Jf(1)=J^{\varepsilon}f(1)=0=f(1)$. It is clear
$f(0)-J(0)=0$. As seen in the proof of the above lemma, we have

$d^{\mathsf{T}}_{f}(t,1)=f(t)-Jf(t).$ (12)

Since $d^{\mathsf{T}}_{f}$ is a continuous pseudo-distance, this immediately
gives that $f-Jf$ is continuous and non-negative on $[0,1]$, that $Jf$ is
càdlàg, and that $\Delta_{t}(Jf)=\Delta_{t}(f)$ for all $t\in[0,1]$. We remark
the fact that $f$ is an excursion implies that for any $r\in[0,1]$, the
function $t\in[0,1]\longmapsto x_{r}^{t}(f)$ is càdlàg with a unique jump at
$r$, which has height $\Delta_{r}(f)$. Hence, the expression

$J^{\varepsilon}f(t)=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)\geq\varepsilon\end{subarray}}x_{r}^{t}(f)$

shows $J^{\varepsilon}f$ is càdlàg as a finite sum of càdlàg functions and
$\Delta_{t}(J^{\varepsilon}f)=\Delta_{t}(f){\bf
1}_{\Delta_{t}(f)\geq\varepsilon}$. Let us show that $Jf$ and
$J^{\varepsilon}f$ are PJG to complete the proof of the point $(i)$. The set
$\\{s\in[0,1]\ :\ \Delta_{s}(f)\geq\varepsilon\\}$ is finite, so it provides a
partition $0=s_{0}<s_{1}<...<s_{n-1}<s_{n}=1$ such that $J^{\varepsilon}f$ is
non-increasing on $[s_{i},s_{i+1})$ for all $0\leq i\leq n-1$, as a finite sum
of non-increasing functions. Hence, $J^{\varepsilon}f$ is PJG according to
Lemma 4. Let $s,t,u\in[0,1]$ such that $s\leq u\leq t$. If $s\preceq_{f}t$
then $s\preceq_{f}u$ and

$Jf(u)=Jf(s-)+f(u)-f(s-)-d^{\mathsf{T}}_{f}(s,u)$

thanks to (12) and Corollary 1. The inequality (5) leads to

$x_{s}^{u}\leq Jf(u)-Jf(s-)\leq f(u)-f(s-).$

By taking the infimum over $u$, we can deduce that if $s\preceq_{f}t$ then
$x_{s}^{t}(Jf)=x_{s}^{t}(f)$. Now, we set $\bar{s}=\inf\\{r\geq s\ :\
f(r)=f(s-)\\}$, so that $s\preceq_{f}\bar{s}$ but $x_{s}^{\bar{s}}(f)=0$.
Then, if we do not have $s\preceq_{f}t$ then $\bar{s}<t$, and
$x_{s}^{t}(Jf)\leq x_{s}^{\bar{s}}(Jf)=x_{s}^{\bar{s}}(f)=0=x_{s}^{t}(f)$.
Thus, we have proven that for all $s,t\in[0,1]$,

$x_{s}^{t}(Jf)=x_{s}^{t}(f).$ (13)

It follows that $Jf$ is PJG because

$Jf(t)=\sum_{r\in[0,1]}x_{r}^{t}(f)=\sum_{r\in[0,1]}x_{r}^{t}(Jf)=\sum_{r\preceq_{Jf}t}x_{r}^{t}(Jf).$

The point $(ii)$ is a rewording of the definition of PJG excursions. The point
$(iii)$ follows from the definition of $J$ and from (12), by continuity of
$d^{\mathsf{T}}$. Proving the point $(iv)$ is easy by using the identities
$\Delta_{t}(Jf)=\Delta_{t}(f)$ and
$\Delta_{t}(J^{\varepsilon}f)=\Delta_{t}(f){\bf
1}_{\Delta_{t}(f)\geq\varepsilon}$.

It remains to prove the point $(v)$. We set $g=f-Jf+J^{\varepsilon}f$. First,
it is easy to see that $g$ is indeed an excursion with the expression
$g(t)=d^{\mathsf{T}}_{f}(t,1)+J^{\varepsilon}f(t)$ inherited from (12). Plus,
it also gives $\Delta_{t}(g)=\Delta_{t}(J^{\varepsilon}f)=\Delta_{t}(f){\bf
1}_{\Delta_{t}(f)\geq\varepsilon}$. Also recall
$\Delta_{t}(Jf)=\Delta_{t}(f)$. Let $s,t\in[0,1]$ with $s\preceq_{f}t$ and
$u\in[s,t]$, so that $s\preceq_{f}u$. We obtain

$Jf(u)-J^{\varepsilon}f(u)\geq\left(Jf(s)-\Delta_{s}(f)\right)-\left(J^{\varepsilon}f(s)-\Delta_{s}(f){\bf
1}_{\Delta_{s}(f)\geq\varepsilon}\right)$ (14)

by expressing the left-hand-side as a series over $r\in[0,1]$, then by cutting
the series at $s$, and by applying Lemma 2. A similar trick together with
Corollary 1 leads to

$d^{\mathsf{T}}_{f}(u,1)+J^{\varepsilon}f(u)\geq
d^{\mathsf{T}}_{f}(s,1)+\left(J^{\varepsilon}f(s)-\Delta_{s}(f){\bf
1}_{\Delta_{s}(f)\geq\varepsilon}\right)+x_{s}^{u}(f){\bf
1}_{\Delta_{s}(f)\geq\varepsilon}.$ (15)

The expressions for $\Delta_{s}(Jf)$, $\Delta_{s}(J^{\varepsilon}f)$, and $g$
and the bounds (14) and (15) eventually get us

$x_{s}^{u}(f){\bf 1}_{\Delta_{s}(f)\geq\varepsilon}\leq g(u)-g(s-)\leq
f(u)-f(s-),$

which yields $x_{s}^{t}(f){\bf 1}_{\Delta_{s}(f)\geq\varepsilon}\leq
x_{s}^{t}(g)\leq x_{s}^{t}(f)$ by taking the infimum over $u$. But as
$x_{s}^{t}(g)\leq\Delta_{s}(g)=\Delta_{t}(f){\bf
1}_{\Delta_{s}(f)\geq\varepsilon}$, we deduce that if $s\preceq_{f}t$ then
$x_{s}^{t}(g)=x_{s}^{t}(f){\bf 1}_{\Delta_{s}(f)\geq\varepsilon}$. Conversely,
the same argument as for $Jf$ gives that if we do not have $s\preceq_{f}t$
then $x_{s}^{t}(g)=x_{s}^{t}(f)=0$. Thus, we have shown that for all
$s,t\in[0,1]$,

$x_{s}^{t}(f-Jf+J^{\varepsilon}f)=x_{s}^{t}(f){\bf
1}_{\Delta_{s}(f)\geq\varepsilon}.$ (16)

Finally, we end the proof by writing

$J(f-Jf+J^{\varepsilon}f)(t)=\sum_{r\in[0,1]}x_{r}^{t}(f-Jf+J^{\varepsilon}f)=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)\geq\varepsilon\end{subarray}}x_{r}^{t}(f)=J^{\varepsilon}f(t).$

∎

######  Theorem 5.

Let $f$ be an excursion. It holds
$d^{\mathsf{T}}_{f}=d^{\mathsf{T}}_{f-Jf}=d^{\mathrm{clas}}_{f-Jf}$,
$d^{\mathsf{L}}_{f}=d^{\mathsf{L}}_{Jf}$, and
$\widetilde{d}^{\mathsf{L}}_{f}=\widetilde{d}^{\mathsf{L}}_{Jf}$. Moreover,

$\displaystyle d^{\mathsf{L}}_{J^{\varepsilon}f}(s,t)$
$\displaystyle=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)\geq\varepsilon\end{subarray}}\delta_{r}^{f}\left(x_{r}^{s}(f),x_{r}^{t}(f)\right),$
(17) $\displaystyle\widetilde{d}^{\mathsf{L}}_{J^{\varepsilon}f}(s,t)$
$\displaystyle=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)\geq\varepsilon\end{subarray}}\tilde{\delta}_{r}^{f}\left(x_{r}^{s}(f),x_{r}^{t}(f)\right),$
(18)

for all $s,t\in[0,1]$.

Before proving the latter theorem, we sum up why vernation trees can be
considered as a natural unification of trees and looptrees. On the one hand,
if $f$ is a continuous excursion then $d^{\mathsf{L}}=0$ and
$d^{\mathsf{V}}=d^{\mathsf{T}}=d^{\mathrm{clas}}$, so $f$ naturally codes a
real tree. On the other hand, if $f$ is a PJG excursion then
$d^{\mathsf{T}}=0$ and $d^{\mathsf{V}}=2d^{\mathsf{L}}$, so $f$ naturally
codes a looptree. Conversely, each excursion has a natural continuous part,
which codes its tree distance $d^{\mathsf{T}}$, and a natural PJG part, which
codes its looptree distance $d^{\mathsf{L}}$. Observe Theorem 1 is a direct
consequence of Theorems 4 and 5.

###### Proof.

Recall (1) for the definition of $d^{\mathrm{clas}}$. Let $s,t\in[0,1]$ with
$s<t$. The identities (12) and (11) yield

$d^{\mathsf{T}}_{f-Jf}(s,t)=d^{\mathrm{clas}}_{f-Jf}(s,t)=d^{\mathsf{T}}_{f}(s,1)+d^{\mathsf{T}}_{f}(t,1)-2\inf_{u\in[s,t]}d^{\mathsf{T}}_{f}(u,1)$

because $f-Jf$ is a continuous excursion. If $u\in[s,t]$ then
$s\wedge_{f}t\preceq_{f}u$ so $d^{\mathsf{T}}_{f}(u,1)\geq
d^{\mathsf{T}}_{f}(s\wedge_{f}t,1)$ by Corollary 1. Moreover, Lemma 1 and the
inequality (5) allow us to bound

$\inf_{u\in[s,t]}d^{\mathsf{T}}_{f}\left(s\wedge_{f}t,u\right)\leq\inf_{u\in[s,t]}f(u)-\inf_{[s,t]}f=0.$

Hence, we conclude

$d^{\mathsf{T}}_{f-Jf}(s,t)=d^{\mathsf{T}}_{f}(s,1)+d^{\mathsf{T}}_{f}(t,1)-2d^{\mathsf{T}}_{f}(s\wedge_{f}t,1)=d^{\mathsf{T}}_{f}(s\wedge_{f}t,s)+d^{\mathsf{T}}_{f}(s\wedge_{f}t,t)=d^{\mathsf{T}}_{f}(s,t)$

with another application of Corollary 1. Thanks to (13), it holds
$\Delta_{r}(Jf)=\Delta_{r}(f)$ and $x_{r}^{t}(Jf)=x_{r}^{t}(f)$ for all
$r\in[0,1]$. The formula (9) then implies
$d^{\mathsf{L}}_{Jf}=d^{\mathsf{L}}_{f}$, and the formula (10) implies
$\widetilde{d}^{\mathsf{L}}_{Jf}=\widetilde{d}^{\mathsf{L}}_{f}$. The same
argument can be applied to $f-Jf+J^{\varepsilon}f$, so that
$\Delta_{r}(J^{\varepsilon}f)=\Delta_{r}(f-Jf+J^{\varepsilon}f)$ and
$x_{r}^{t}(J^{\varepsilon}f)=x_{r}^{t}(f-Jf+J^{\varepsilon}f)$ thanks to the
point $(v)$ of Theorem 4. Furthermore, the identity (16) ensures
$\Delta_{r}(f-Jf+J^{\varepsilon}f)=\Delta_{r}(f){\bf
1}_{\Delta_{r}(f)\geq\varepsilon}$ and
$x_{r}^{t}(f-Jf+J^{\varepsilon}f)=x_{r}^{t}(f){\bf
1}_{\Delta_{r}(f)\geq\varepsilon}$. Eventually, the formulas (9) and (10) lead
to

$\displaystyle d^{\mathsf{L}}_{J^{\varepsilon}f}(s,t)$
$\displaystyle=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(J^{\varepsilon}f)>0\end{subarray}}\delta_{r}^{J^{\varepsilon}f}\left(x_{r}^{s}(J^{\varepsilon}f),x_{r}^{t}(J^{\varepsilon}f)\right)=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)\geq\varepsilon\end{subarray}}\delta_{r}^{f}\left(x_{r}^{s}(f),x_{r}^{t}(t)\right),$
$\displaystyle\widetilde{d}^{\mathsf{L}}_{J^{\varepsilon}f}(s,t)$
$\displaystyle=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(J^{\varepsilon}f)>0\end{subarray}}\tilde{\delta}_{r}^{J^{\varepsilon}f}\left(x_{r}^{s}(J^{\varepsilon}f),x_{r}^{t}(J^{\varepsilon}f)\right)=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)\geq\varepsilon\end{subarray}}\tilde{\delta}_{r}^{f}\left(x_{r}^{s}(f),x_{r}^{t}(t)\right).$

∎

######  Corollary 2.

For all excursion $f$ and $\varepsilon\in(0,1)$, it holds

$\displaystyle||d^{\mathsf{L}}_{f}-d^{\mathsf{L}}_{J^{\varepsilon}f}||_{\infty}$
$\displaystyle\leq 2||Jf-J^{\varepsilon}f||_{\infty},$
$\displaystyle||\widetilde{d}^{\mathsf{L}}_{f}-\widetilde{d}^{\mathsf{L}}_{J^{\varepsilon}f}||_{\infty}$
$\displaystyle\leq 2K||Jf-J^{\varepsilon}f||_{\infty},$

where $K$ is a finite constant which only depends on the shuffle $\Phi$, as

$K=\sup_{\Delta\in(0,1]}\sup_{x\in(0,1]}\frac{1}{x}\delta\left(\phi_{\Delta}(0),\phi_{\Delta}(x)\right).$

###### Proof.

From the definition of a shuffle, we know $K$ is indeed finite. By definition,
we have $\tilde{\delta}_{r}^{f}(0,x)\leq Kx$ for any $r\in[0,1]$ such that
$\Delta_{r}(f)\leq 1$ and for any $x\in[0,\Delta_{r}(f)]$. Plus, it is clear
that $\delta_{r}^{f}(0,x)\leq x$. It follows

$\displaystyle\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\varepsilon\end{subarray}}\delta_{r}^{f}\left(0,x_{r}^{t}(f)\right)$
$\displaystyle\leq\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\varepsilon\end{subarray}}x_{r}^{t}(f)=Jf(t)-J^{\varepsilon}f(t),$
$\displaystyle\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\varepsilon\end{subarray}}\tilde{\delta}_{r}^{f}\left(0,x_{r}^{t}(f)\right)$
$\displaystyle\leq\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\varepsilon\end{subarray}}Kx_{r}^{t}(f)=K\left(Jf(t)-J^{\varepsilon}f(t)\right).$

We conclude by applying the formulas (9), (10), (17), and (18), as well as the
triangular inequalities of the $\delta_{r}^{f}$ and the
$\tilde{\delta}_{r}^{f}$. ∎

We finish this section by providing some tools to study convergences involving
$J$ or $J^{\varepsilon}$.

######  Proposition 7.

Let $f$ be an excursion, $\varepsilon>0$, and let
$\lambda:[0,1]\longrightarrow[0,1]$ be an increasing bijection. The following
points hold.

  1. (i)

$J(\Theta f)=\Theta(Jf)$ and $J^{\varepsilon}(\Theta
f)=\Theta(J^{\varepsilon}f)$,

  2. (ii)

$J(f\circ\lambda)=(Jf)\circ\lambda$ and
$J^{\varepsilon}(f\circ\lambda)=(J^{\varepsilon}f)\circ\lambda$,

  3. (iii)

If $\varepsilon\notin\\{\Delta_{t}(f)\ :\ t\in[0,1]\\}$, then
$J^{\varepsilon}$ is continuous at $f$ with respect to the topology of uniform
convergence,

  4. (iv)

$\left|\left|Jf-J^{\varepsilon}f\right|\right|_{\infty}\underset{\varepsilon\rightarrow
0}{\longrightarrow}0$.

###### Proof.

When $s,t\in[1/2,1]$, recall $\Theta f(s)=f(2s-1)$ so $x_{s}^{t}(\Theta
f)=x_{2s-1}^{2t-1}(f)$. Then, the point $(i)$ comes from the fact that
$\Delta_{r}(\Theta f)=0$ when $r\in[0,1/2)$. As for the proof of the time-
changing property, we deduce $(ii)$ from the strict monotony of $\lambda$ and
from the equality $x_{s}^{t}(f\circ\lambda)=x_{\lambda(s)}^{\lambda(t)}(f)$
for all $s,t\in[0,1]$. Let us prove $(iii)$. The excursion $f$ is càdlàg so
the set $\\{t\in[0,1]\ :\ \Delta_{t}(f)>\varepsilon/2\\}$ is finite. Thus,

$\displaystyle\eta_{1}$ $\displaystyle=\max\\{\Delta_{t}(f)\ :\
t\in[0,1]\text{ such that
}\varepsilon/2<\Delta_{t}(f)<\varepsilon\\}\cup\\{\varepsilon/2\\},$
$\displaystyle\eta_{2}$ $\displaystyle=\min\\{\Delta_{t}(f)\ :\
t\in[0,1]\text{ such that
}\Delta_{t}(f)\geq\varepsilon\\}\cup\\{2\varepsilon\\}$

are well-defined and $\eta_{1}<\varepsilon<\eta_{2}$. Let $g$ be an excursion
such that
$||f-g||_{\infty}<\min(\varepsilon-\eta_{1},\eta_{2}-\varepsilon)/2$. If
$\Delta_{r}(g)\geq\varepsilon$ then $\Delta_{r}(f)>\eta_{1}$ and we obtain
$\Delta_{r}(f)\geq\varepsilon$ by definition of $\eta_{1}$. Likewise, if
$\Delta_{r}(g)<\varepsilon$ then $\Delta_{r}(f)<\eta_{2}$ and we obtain
$\Delta_{r}(f)<\varepsilon$ by definition of $\eta_{2}$. Hence,

$J^{\varepsilon}g(t)=\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)\geq\varepsilon\end{subarray}}x_{r}^{t}(g)$

for all $t\in[0,1]$, so
$\left|\left|J^{\varepsilon}f-J^{\varepsilon}g\right|\right|_{\infty}\leq
2N(f)\times||f-g||_{\infty}$ where we denote by $N(f)$ the number of jumps of
$f$ of height at least $\varepsilon$. The point $(iii)$ follows. To show the
point $(iv)$, we mimic the proof of Dini’s theorem. We fix $\nu>0$ and we set

$U_{n}=\\{t\in[0,1]\ :\ Jf(t)-J^{1/n}f(t)<\nu\text{ and
}Jf(t-)-J^{1/n}f(t-)<\nu\\}$

for all $n\geq 1$. We remark that $U_{n}\subset U_{n+1}$ because $J^{1/n}f\leq
J^{1/(n+1)}f\leq Jf$. The functions $Jf$ and $J^{1/n}f$ are càdlàg, so $U_{n}$
is open for all $n\geq 1$. Furthermore, the monotone convergence theorem
implies that $J^{1/n}f(t)\underset{n\rightarrow\infty}{\longrightarrow}Jf(t)$
for all $t\in[0,1]$. When $n$ is large enough (depending on $t$), we have

$\Delta_{t}(J^{1/n}f)=\Delta_{t}(f){\bf 1}_{\Delta_{t}(f)\geq
1/n}=\Delta_{t}(f)=\Delta_{t}(Jf)$

which yields $Jf(t-)-J^{1/n}f(t-)=Jf(t)-J^{1/n}f(t)$. Therefore, the family
$(U_{n})_{n\geq 1}$ covers $[0,1]$, so there exists $N\geq 1$ such that
$U_{N}=[0,1]$ by compactness. As a result, $||Jf-J^{1/n}f||_{\infty}\leq\nu$
when $n\geq N$. ∎

##  4 Limit theorems

The main flaw of the pseudo-distances $d^{\mathsf{L}}$, $d^{\mathsf{T}}$, and
$d^{\mathsf{V}}$ is that they do not enjoy any good property of functional
continuity, unlike the classical pseudo-distance $d^{\mathrm{clas}}$ which
induces a real tree from a continuous excursion $f$. We have already explained
why a PJG excursion should code a looptree. However, it is possible to show
every encoding of a pseudo-distance from an excursion cannot be continuous if
it is equal to $d^{\mathsf{L}}_{f}$ whenever $f$ is PJG. For example, we
observe the two sequences of functions defined by

$\left(f_{n}(t),g_{n}(t)\right)=\begin{cases}\left(k/2n+1/n-(t-k/2n),(k+1)/2n\right)&\text{
if }\frac{k}{2n}\leq t<\frac{k+1}{2n}\text{ with }0\leq k\leq n-1\\\
\left(1/2-(t-1/2),1/2-(t-1/2)\right)&\text{ if }1/2\leq t\leq 1\end{cases}$

are PJG thanks to Lemma 4. Both have the same asymptotic behavior because both
uniformly converge to the same continuous excursion $f:t\in[0,1]\longmapsto
1/2-|t-1/2|$, while their associated looptrees have very different limits.
Indeed, on the one hand, $\mathscr{L}_{f_{n}}$ consists of a chain of $n$
loops of length $1/n$ put back to back. Intuitively, this sequence converges
to a segment of length $1/2$, which is also $\mathscr{T}_{f}$. On the other
hand, $\mathscr{L}_{g_{n}}$ consists of a bouquet of $n$ loops of length
$1/2n$ all glued at the same point. This sequence converges to a single point,
which is also $\mathscr{L}_{f}$, as the sequence of the diameters tends to
$0$.

###  4.1 Continuity for the shuffled vernation tree pseudo-distance

We fix a shuffle $\Phi$ and we set
$\mathbb{B}(\Phi)=\mathbb{B}^{1}(\Phi)\cup\mathbb{B}^{2}(\Phi)$, where

$\displaystyle\mathbb{B}^{1}(\Phi)$ $\displaystyle=\left\\{\Delta>0\ :\
\Delta\text{ discontinuity point of }\Phi\text{ for the Skorokhod
topology}\right\\},$ $\displaystyle\mathbb{B}^{2}(\Phi)$
$\displaystyle=\left\\{(\Delta,x)\ :\ \Delta>0,\ x\in[0,1],\ x\text{
discontinuity point of }\phi_{\Delta}\right\\}.$

We are going to show that $\widetilde{d}^{\mathsf{V}}$ enjoys a property of
functional continuity, which justifies its construction. The set
$\mathbb{B}(\Phi)$ will represent the imperfections of that continuity. Let us
remark the definition of a shuffling implies
$\mathbb{B}(\Phi)\cap\left((0,\infty)\times\\{0,1\\}\right)=\emptyset$. For
any excursion $f$, we also define $B_{f}=B_{f}^{1}\cup B_{f}^{2}$, where

$\displaystyle B_{f}^{1}$ $\displaystyle=\left\\{\Delta_{t}\ :\
t\in[0,1]\text{ such that }\Delta_{t}>0\right\\},$ $\displaystyle B_{f}^{2}$
$\displaystyle=\left\\{\left(\Delta_{r},\frac{x_{r}^{t}}{\Delta_{r}}\right)\
:\ r,s,t\in[0,1]\text{ such that }s\neq t,\Delta_{r}>0,\text{ and
}x_{r}^{t}=x_{r}^{s}>0\right\\},$

which will indicate if the functional
$g\longmapsto\widetilde{d}^{\mathsf{V}}_{g}$ is continuous at $f$. We remind
that the set of excursions $\mathcal{H}$ is a closed subset of the Skorokhod
space $\mathcal{D}([0,1])$, endowed with the Skorokhod distance $\rho$. Also
recall from Definition 3 the Skorokhod topology on $\mathcal{D}([0,1]^{2})$
induced by the distance $\rho_{2}$. Furthermore, we already know from
Proposition 4 that $\widetilde{d}^{\mathsf{V}}_{f}\in\mathcal{D}([0,1]^{2})$
for any $f\in\mathcal{H}$.

######  Theorem 6.

Let $f$ be an excursion. If $B_{f}\cap\mathbb{B}(\Phi)=\emptyset$, then the
functional
$g\in\mathcal{H}\longmapsto\widetilde{d}^{\mathsf{V}}_{g}\in\mathcal{D}([0,1]^{2})$
is continuous at $f$ for the Skorokhod topologies.

###### Proof.

The proof requires a lot of care and can be tedious sometimes. The main idea
is to use the convergence in (3) to forget the small jumps of $f$. Then, the
hypothesis $B_{f}\cap\mathbb{B}(\Phi)=\emptyset$ and the continuity of $\Phi$
allow us to control the variations generated by the big jumps.

#### Choice of some parameters.

Let $\varepsilon\in(0,1)$. We are able to choose some
$\Delta\in(0,\varepsilon)$ such that

$\text{if }u\in(0,2\Delta)\text{ then
}\sup_{x\in(0,1]}\left|\frac{2}{x}\delta\left(\phi_{u}(0),\phi_{u}(x)\right)-1\right|\leq\varepsilon$
(19)

thanks to (3). The set $\\{s\in[0,1]\ :\ \Delta_{s}\geq\Delta\\}$ is finite
because $f$ is càdlàg. We denote by $N$ its size and by
$r_{1}<r_{2}<...<r_{N}$ its elements. The functions $\phi_{\Delta_{r_{n}}}$
are càglàd and continuous at $0$. Hence, we have some $\nu>0$ and some
partitions of $[0,1]$ denoted by $0<y_{n,1}<y_{n,2}<...<y_{n,k_{n}}<1$, where
the $y_{n,i}$ are discontinuity points of $\phi_{\Delta_{r_{n}}}$ for any
$1\leq i\leq k_{n}$, such that for all $1\leq n\leq N$,

$\displaystyle\max_{1\leq i\leq
k_{n}}\sup_{\begin{subarray}{c}y,y^{\prime}\in(y_{n,i},y_{n,i+1}]\\\
|y-y^{\prime}|\leq\nu\end{subarray}}\delta(\phi_{\Delta_{r_{n}}}(y),\phi_{\Delta_{r_{n}}}(y^{\prime}))$
$\displaystyle\leq\frac{\varepsilon}{N+1},$ (20)
$\displaystyle\sup_{\begin{subarray}{c}y,y^{\prime}\in[0,y_{n,1}]\\\
|y-y^{\prime}|\leq\nu\end{subarray}}\delta(\phi_{\Delta_{r_{n}}}(y),\phi_{\Delta_{r_{n}}}(y^{\prime}))$
$\displaystyle\leq\frac{\varepsilon}{N+1},$ (21)

with $y_{n,k_{n}+1}=1$ by convention. We now take a parameter $\eta>0$ that
respects the following inequalities:

$\eta\leq\min\left(\frac{\Delta}{4},\frac{\varepsilon}{N+1}\right)\text{ and
}\left(1+\frac{4}{\Delta}+\frac{4}{\Delta^{2}}||f||_{\infty}\right)\eta\leq\nu.$
(22)

The fact that $f$ is an excursion ensures that for $1\leq n\leq N$ and $1\leq
i\leq k_{n}$,

$t_{n,i}=\inf\\{t\geq r_{n}\ :\ f(t)-f(r_{n}-)=y_{n,i}\Delta_{r_{n}}\\}$

is well-defined, $f(t_{n,i})=f(t_{n,i}-)$, and
$x_{r_{n}}^{t_{n,i}}=y_{n,i}\Delta_{r_{n}}>0$. In particular, the $t_{n,i}$,
for $1\leq n\leq N$ and $1\leq i\leq k_{n}$, are continuity points of $f$, so
they are distinct from the $r_{n}$, for $1\leq n\leq N$. Moreover, if
$t_{n,i}=t_{m,j}$ with $n<m$, $1\leq i\leq k_{n}$ and $1\leq j\leq k_{m}$,
then $x_{r_{n}}^{t_{n,i}}=x_{r_{n}}^{r_{m}}$ because
$f(r_{m}-)\leq\inf_{[r_{m},t_{m,j}]}f$. Since $t_{n,i}\neq r_{m}$, we find
$(\Delta_{r_{n}},y_{n,i})\in B_{f}$. By choice of $y_{n,i}$, we also have
$(\Delta_{r_{n}},y_{n,i})\in\mathbb{B}(\Phi)$, which contradicts the
assumption $B_{f}\cap\mathbb{B}(\Phi)=\emptyset$. If $t_{n,i}=t_{n,j}$, with
$1\leq i,j\leq k_{n}$, then
$y_{n,i}\Delta_{r_{n}}=x_{r_{n}}^{t_{n,i}}=x_{r_{n}}^{t_{n,j}}=y_{n,j}\Delta_{r_{n}}$
so $i=j$. Thus, the $t_{n,i}$ are distinct.

Eventually, we obtain the existence of some $\gamma\in(0,\varepsilon)$ such
that the intervals $[t_{n,i}-\gamma,t_{n,i}+\gamma]$, for $1\leq n\leq N$ and
$1\leq i\leq k_{n}$, are included in $[0,1]$, disjoint, do not contain any of
the $r_{m}$, and such that

$\text{if }t\in[0,1]\text{ with }|t-t_{n,i}|\leq\gamma\text{ then
}|f(t)-f(t_{n,i})|\leq\eta/3.$ (23)

We set

$\omega=\min_{\begin{subarray}{c}1\leq n\leq N\\\ 1\leq i\leq
k_{n}\end{subarray}}\min(x_{r_{n}}^{t_{n,i}-\gamma}-x_{r_{n}}^{t_{n,i}},x_{r_{n}}^{t_{n,i}}-x_{r_{n}}^{t_{n,i}+\gamma}),$

and we know $\omega>0$ thanks to the assumption
$B_{f}\cap\mathbb{B}(\Phi)=\emptyset$. Then, we choose $\xi\in(0,\eta/3)$ such
that

$\left(\frac{8}{\Delta}+1\right)\xi<\frac{\omega}{2+||f||_{\infty}}.$ (24)

Recall the definition of the Skorokhod distance $\overleftarrow{\rho}$ for
$\mathcal{C}$-valued càglàd functions given just before Definition 2. Because
$\Phi:u>0\longmapsto\phi_{u}$ is continuous at $\Delta_{r_{n}}$ for $1\leq
n\leq N$ thanks to the assumption $B_{f}\cap\mathbb{B}(\Phi)=\emptyset$, we
can choose some $\nu^{\prime}\in(0,\Delta)$ such that for all $1\leq n\leq N$,

$\sup_{|u-\Delta_{r_{n}}|\leq\nu^{\prime}}\overleftarrow{\rho}(\phi_{u},\phi_{\Delta_{r_{n}}})\leq\xi/2.$
(25)

#### Correspondence between big discontinuities.

Let $g\in\mathcal{H}$ such that
$||f-g||_{\infty}\leq\min(\xi,\nu^{\prime}/2)$. We are going to construct an
increasing bijection from $[0,1]$ to itself which makes correspond the
discontinuity points of $\widetilde{d}^{\mathsf{V}}_{f}$ and
$\widetilde{d}^{\mathsf{V}}_{g}$. If $s,t\in[0,1]$ then
$|x_{s}^{t}(f)-x_{s}^{t}(g)|\leq 2||f-g||_{\infty}$. In particular for $s=t$,
we have $|\Delta_{s}(f)-\Delta_{s}(g)|\leq\nu^{\prime}$, so (25) gives us some
increasing bijections from $[0,1]$ to itself, denoted by $\mu_{n}$ for $1\leq
n\leq N$, such that

$||\mu_{n}-\mathsf{id}||_{\infty}\leq\xi\text{ and
}\sup_{x\in[0,1]}\delta\left(\phi_{\Delta_{r_{n}}(g)}(x),\phi_{\Delta_{r_{n}}(f)}\circ\mu_{n}(x)\right)\leq\xi.$
(26)

Let $\lambda$ be the continuous function which is affine on the intervals
generated by the points

$\\{t_{n,i},\ t_{n,i}+\gamma,\ t_{n,i}-\gamma\ :\ 1\leq n\leq N\text{ and
}1\leq i\leq k_{n}\\}$

such that $\lambda(0)=0,\ \lambda(1)=1,\
\lambda(t_{n,i}-\gamma)=t_{n,i}-\gamma,\
\lambda(t_{n,i}+\gamma)=t_{n,i}+\gamma$, and

$\lambda(t_{n,i})=\inf\left\\{t\geq r_{n}\ :\
\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(g)\leq\mu_{n}^{-1}(y_{n,i})\right\\}.$

It is well-defined because $g$ is an excursion. Observe that
$x_{r_{n}}^{\lambda(t_{n,i})}(g)=\Delta_{r_{n}}(g)\times\mu_{n}^{-1}(y_{n,i})$.
Let us prove that $\lambda$ is an increasing function, close to the identity.
Let $1\leq n\leq N$ and $1\leq i\leq k_{n}$. We use $||f-g||_{\infty}\leq\xi$
then we recall the definitions of $t_{n,i}$ and $\omega$ to find

$\frac{x_{r_{n}}^{t_{n,i}-\gamma}}{\Delta_{r_{n}}}(g)\geq\frac{x_{r_{n}}^{t_{n,i}-\gamma}(f)-2\xi}{\Delta_{r_{n}}(g)}\geq\frac{y_{n,i}\Delta_{r_{n}}(f)+\omega-2\xi}{\Delta_{r_{n}}(g)}.$

Plus, we found $\Delta_{r_{n}}(g)\geq\Delta_{r_{n}}(f)-2\xi\geq\Delta-2\eta$,
so $\Delta_{r_{n}}(g)\geq\Delta/2$ according to (22). We also have
$\Delta_{r_{n}}(g)\leq\Delta_{r_{n}}(f)+2\xi\leq 2+||f||_{\infty}$. It follows

$\frac{x_{r_{n}}^{t_{n,i}-\gamma}}{\Delta_{r_{n}}}(g)\geq
y_{n,i}+\frac{\omega}{2+||f||_{\infty}}-\frac{8\xi}{\Delta},$

then successive applications of (26) and (24) lead to

$\frac{x_{r_{n}}^{t_{n,i}-\gamma}}{\Delta_{r_{n}}}(g)\geq\mu_{n}^{-1}(y_{n,i})+\frac{\omega}{2+||f||_{\infty}}-\left(\frac{8}{\Delta}+1\right)\xi>\mu_{n}^{-1}(y_{n,i}).$

We prove
$x_{r_{n}}^{t_{n,i}+\gamma}(g)/\Delta_{r_{n}}(g)<\mu_{n}^{-1}(y_{n,i})$ in the
same way. Thus, $|\lambda(t_{n,i})-t_{n,i}|<\gamma$. We deduce that $\lambda$
is strictly increasing and

$||\lambda-\mathsf{id}||_{\infty}\leq\gamma\leq\varepsilon.$ (27)

Moreover, $\lambda$ is a bijection from $[0,1]$ to itself because
$\lambda(0)=0$, $\lambda(1)=1$, and $\lambda$ is continuous.

#### Control on the big loops.

We easily see that if $t$ is outside all the intervals
$[t_{n,i}-\gamma,t_{n,i}+\gamma]$ then $\lambda(t)=t$. But if
$t\in[t_{n,i}-\gamma,t_{n,i}+\gamma]$ then
$\lambda(t)\in[t_{n,i}-\gamma,t_{n,i}+\gamma]$ too. Therefore, (23) allows us
to deduce

$||f-g\circ\lambda||_{\infty}\leq||f-f\circ\lambda||_{\infty}+||f-g||_{\infty}\leq
2\eta/3+\eta/3=\eta.$

Now, we set $h=g\circ\lambda$ so that $||f-h||_{\infty}\leq\eta$. Our goal is
to show that
$||\widetilde{d}^{\mathsf{V}}_{f}-\widetilde{d}^{\mathsf{V}}_{h}||_{\infty}$
is small, but before that, we are going to bound the difference of the
distances on the big loops. We begin by recalling that the point $r_{n}$ is
outside all the intervals $[t_{n,i}-\gamma,t_{n,i}+\gamma]$ so
$\lambda(r_{n})=r_{n}$, which implies $\Delta_{r_{n}}(h)=\Delta_{r_{n}}(g)$
and $x_{r_{n}}^{t}(h)=x_{r_{n}}^{\lambda(t)}(g)$. Let $t\geq r_{n}$ and $1\leq
i\leq k_{n}$. Recall $(\Delta_{r_{n}}(f),y_{n,i})\notin B_{f}$ because
$(\Delta_{r_{n}}(f),y_{n,i})\in\mathbb{B}(\Phi)$, so

$\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(f)\leq y_{n,i}\Longleftrightarrow
t_{n,i}\leq t.$

Then, $\lambda(r_{n})=r_{n}\leq\lambda(t)$ because $\lambda$ is increasing. By
definition of $\lambda(t_{n,i})$ together with the observation that
$x_{r_{n}}^{\lambda(t_{n,i})}(g)=\Delta_{r_{n}}(g)\times\mu_{n}^{-1}(y_{n,i})$,
we have
$x_{r_{n}}^{\lambda(t)}(g)/\Delta_{r_{n}}(g)\leq\mu_{n}^{-1}(y_{n,i})\Longleftrightarrow\lambda(t_{n,i})\leq\lambda(t)$.
The functions $\lambda$ and $\mu_{n}$ are strictly increasing so it follows

$\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(f)\leq
y_{n,i}\Longleftrightarrow\mu_{n}\left(\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(h)\right)\leq
y_{n,i}.$

Hence, the points $x_{r_{n}}^{t}(f)/\Delta_{r_{n}}(f)$ and
$\mu_{n}\left(x_{r_{n}}^{t}(h)/\Delta_{r_{n}}(h)\right)$ are either both in
the interval $[0,y_{n,1}]$ or both in the same interval $(y_{n,i},y_{n,i+1}]$
with $1\leq i\leq k_{n}$. Furthermore, we can bound

$\left|\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(f)-\mu_{n}\left(\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(h)\right)\right|\leq||\mu_{n}-\mathsf{id}||_{\infty}+\frac{2||f-h||_{\infty}}{\Delta_{r_{n}}(g)}+|x_{r_{n}}^{t}(f)|\frac{2||f-h||_{\infty}}{\Delta_{r_{n}}(f)\Delta_{r_{n}}(g)},$

so that the inequalities $\Delta_{r_{n}}(g)\geq\Delta/2$, (26), and (22) give

$\left|\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(f)-\mu_{n}\left(\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(h)\right)\right|\leq\left(1+\frac{4}{\Delta}+\frac{4}{\Delta^{2}}||f||_{\infty}\right)\eta\leq\nu.$

Therefore, it follows from (20) and (21) that

$\delta\left(\phi_{\Delta_{r_{n}}(f)}\left(\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(f)\right),\phi_{\Delta_{r_{n}}(f)}\left(\mu_{n}\left(\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(h)\right)\right)\right)\leq\frac{\varepsilon}{N+1}$
(28)

for all $t\in[0,1]$, the case $t<r_{n}$ being obvious. Now, let $s,t\in[0,1]$,
the triangular inequality on $\delta$ implies

$\left|\tilde{\delta}_{r_{n}}^{f}\left(x_{r_{n}}^{s}(f),x_{r_{n}}^{t}(f)\right)-\tilde{\delta}_{r_{n}}^{h}\left(x_{r_{n}}^{s}(h),x_{r_{n}}^{t}(h)\right)\right|\leq|\Delta_{r_{n}}(f)-\Delta_{r_{n}}(h)|\\\
+\Delta_{r_{n}}(f)\times\delta\left(\phi_{\Delta_{r_{n}}(f)}\left(\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(f)\right),\phi_{\Delta_{r_{n}}(h)}\left(\frac{x_{r_{n}}^{t}}{\Delta_{r_{n}}}(h)\right)\right)\\\
+\Delta_{r_{n}}(f)\times\delta\left(\phi_{\Delta_{r_{n}}(f)}\left(\frac{x_{r_{n}}^{s}}{\Delta_{r_{n}}}(f)\right),\phi_{\Delta_{r_{n}}(h)}\left(\frac{x_{r_{n}}^{s}}{\Delta_{r_{n}}}(h)\right)\right).$

Recall that $\Delta_{r_{n}}(h)=\Delta_{r_{n}}(g)$, $||f-h||_{\infty}\leq\eta$,
and $\xi\leq\eta$. We apply the inequalities (22), (26), and (28) into the
previous bound to conclude that for all $s,t\in[0,1]$,

$\left|\tilde{\delta}_{r_{n}}^{f}\left(x_{r_{n}}^{s}(f),x_{r_{n}}^{t}(f)\right)-\tilde{\delta}_{r_{n}}^{h}\left(x_{r_{n}}^{s}(h),x_{r_{n}}^{t}(h)\right)\right|\leq(2+4||f||_{\infty})\frac{\varepsilon}{N+1}.$
(29)

#### Uniform control.

We are now ready to bound
$||\widetilde{d}^{\mathsf{V}}_{f}-\widetilde{d}^{\mathsf{V}}_{h}||_{\infty}$.
Let $s,t\in[0,1]$ with $s\leq t$. With the help of Lemma 1 and of the
identities (10) and (11), we verify we can write

$\displaystyle\left|\widetilde{d}^{\mathsf{V}}_{f}(s,t)-\widetilde{d}^{\mathsf{V}}_{h}(s,t)\right|\leq$
$\displaystyle|f(s)-h(s)|+|f(t)-h(t)|+2\left|\inf_{[s,t]}f-\inf_{[s,t]}h\right|$
$\displaystyle+\sum_{n=1}^{N}\left|x_{r_{n}}^{s}(f)-x_{r_{n}}^{s}(h)\right|+\sum_{n=1}^{N}\left|x_{r_{n}}^{t}(f)-x_{r_{n}}^{t}(h)\right|$
$\displaystyle+2\sum_{n=1}^{N}\left|\tilde{\delta}_{r_{n}}^{f}\left(x_{r_{n}}^{s}(f),x_{r_{n}}^{t}(f)\right)-\tilde{\delta}_{r_{n}}^{h}\left(x_{r_{n}}^{s}(h),x_{r_{n}}^{t}(h)\right)\right|$
$\displaystyle+3\Delta_{s\wedge_{f}t}(f){\bf
1}_{\Delta_{s\wedge_{f}t}(f)<\Delta}+3\Delta_{s\wedge_{h}t}(h){\bf
1}_{\Delta_{s\wedge_{h}t}(f)<\Delta}$
$\displaystyle+\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\Delta\end{subarray}}\left|2\tilde{\delta}_{r}^{f}\left(0,x_{r}^{s}(f)\right)-x_{r}^{s}(f)\right|+\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\Delta\end{subarray}}\left|2\tilde{\delta}_{r}^{f}\left(0,x_{r}^{t}(f)\right)-x_{r}^{t}(f)\right|$
$\displaystyle+\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\Delta\end{subarray}}\left|2\tilde{\delta}_{r}^{h}\left(0,x_{r}^{s}(h)\right)-x_{r}^{s}(h)\right|+\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\Delta\end{subarray}}\left|2\tilde{\delta}_{r}^{h}\left(0,x_{r}^{t}(h)\right)-x_{r}^{t}(h)\right|.$

The sum of all the terms in the first two rows of the right-hand side can be
easily bounded by $8\varepsilon$ because
$||f-h||_{\infty}\leq\eta\leq\varepsilon/(N+1)$, thanks to (22). The term of
the third row is smaller than $(4+8||f||_{\infty})\varepsilon$ by a direct
application of (29). Next, if $\Delta_{r}(f)<\Delta$ then
$\Delta_{r}(h)<\Delta+2\eta<2\Delta$, since the inequality (22) justifies
$\eta\leq\Delta/4$. As we chose $\Delta$ so that $\Delta<\varepsilon$, the sum
of the terms in the fourth row is bounded by $9\varepsilon$. Moreover, the
fact (19) ensures that if $\Delta_{r}(f)<\Delta$, then it holds

$\left|2\tilde{\delta}_{r}^{f}(0,x)-x\right|\leq\varepsilon x\text{ and
}\left|2\tilde{\delta}_{r}^{h}(0,y)-y\right|\leq\varepsilon y$

for all $x\in[0,\Delta_{r}(f)]$ and for all $y\in[0,\Delta_{r}(h)]$. It allows
us to control the last four terms of the right-hand side. For instance,

$\displaystyle\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\Delta\end{subarray}}\left|2\tilde{\delta}_{r}^{f}\left(0,x_{r}^{s}(f)\right)-x_{r}^{s}(f)\right|$
$\displaystyle\leq\varepsilon\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\Delta\end{subarray}}x_{r}^{s}(f)\leq\varepsilon
Jf(s)\leq||f||_{\infty}\varepsilon,$
$\displaystyle\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\Delta\end{subarray}}\left|2\tilde{\delta}_{r}^{h}\left(0,x_{r}^{s}(h)\right)-x_{r}^{s}(h)\right|$
$\displaystyle\leq\varepsilon\sum_{\begin{subarray}{c}r\in[0,1]\\\
\Delta_{r}(f)<\Delta\end{subarray}}x_{r}^{s}(h)\leq\varepsilon
Jh(s)\leq\left(1+||f||_{\infty}\right)\varepsilon,$

because we know from Theorem 4 that $f-Jf\geq 0$. Eventually, we find

$\left|\widetilde{d}^{\mathsf{V}}_{f}(s,t)-\widetilde{d}^{\mathsf{V}}_{h}(s,t)\right|\leq\left(23+12||f||_{\infty}\right)\varepsilon.$

#### Conclusion.

To sum up, we have shown that for all $\varepsilon>0$, there exists $\eta>0$
such that if $||f-g||_{\infty}\leq\eta$ then there exists $\lambda$ an
increasing bijection from $[0,1]$ to itself with
$||\widetilde{d}^{\mathsf{V}}_{f}-\widetilde{d}^{\mathsf{V}}_{g\circ\lambda}||_{\infty}\leq\varepsilon$
and $||\lambda-\mathsf{id}||_{\infty}\leq\varepsilon$. Indeed, recall (27).
Now, if the Skorokhod distance between $f$ and $g$ is smaller than
$\min(\eta,\varepsilon)/2$, then we have $\mu$ an increasing bijection from
$[0,1]$ to itself such that $||\mu-\mathsf{id}||_{\infty}\leq\varepsilon$ and
$||f-g\circ\mu||_{\infty}\leq\eta$. Hence, there exists $\lambda$ an
increasing bijection from $[0,1]$ to itself such that
$||\mu\circ\lambda-\mathsf{id}||_{\infty}\leq 2\varepsilon$ and
$||\widetilde{d}^{\mathsf{V}}_{f}-\widetilde{d}^{\mathsf{V}}_{g\circ\mu\circ\lambda}||_{\infty}\leq\varepsilon$.
Eventually, thanks to the time-changing property,
$\widetilde{d}^{\mathsf{V}}_{g\circ\mu\circ\lambda}=\widetilde{d}^{\mathsf{V}}_{g}\circ(\mu\circ\lambda,\mu\circ\lambda)$.
Thereby, if $\rho(f,g)\leq\min(\eta,\varepsilon)/2$ then
$\rho_{2}\left(\widetilde{d}^{\mathsf{V}}_{f},\widetilde{d}^{\mathsf{V}}_{g}\right)\leq
2\varepsilon$, where we remind that $\rho$ is the Skorokhod distance on
$\mathcal{H}$ and $\rho_{2}$ is the Skorokhod distance for bivariate
functions. ∎

###  4.2 Two particular cases for convergences of unshuffled pseudo-distances

Even if $d^{\mathsf{L}}$, $d^{\mathsf{T}}$, and $d^{\mathsf{V}}$ do not enjoy
a property of functional continuity, we are still able to provide two
particular cases where it is possible to automatically deduce a convergence
for $(d^{\mathsf{L}}_{f_{n}})$, $(d^{\mathsf{T}}_{f_{n}})$, or
$(d^{\mathsf{V}}_{f_{n}})$ from the convergence of $(f_{n})$.

######  Theorem 7.

If $f_{n}\longrightarrow f$ for the Skorokhod topology on $\mathcal{H}$ and if
there exists $N\geq 1$ such that all the $f_{n}$ have at most $N$ jumps, then
$d^{\mathsf{L}}_{f_{n}}\longrightarrow d^{\mathsf{L}}_{f}$ and
$d^{\mathsf{T}}_{f_{n}}\longrightarrow d^{\mathsf{T}}_{f}$ uniformly on
$[0,1]^{2}$. It follows $d^{\mathsf{V}}_{f_{n}}\longrightarrow
d^{\mathsf{V}}_{f}$ uniformly on $[0,1]^{2}$.

###### Proof.

The limiting excursion $f$ also has at most $N$ jumps. Let
$\varepsilon\in(0,1)$ such that all the jumps of $f$ are higher than
$2\varepsilon$. Hence, we get $J^{\varepsilon}f_{n}\longrightarrow
J^{\varepsilon}f=Jf$ for the Skorokhod topology thanks to Theorem 4 and to
Proposition 7. Let $s\in[0,1]$, we know there is $t\in[0,1]$ such that
$|\Delta_{t}(f)-\Delta_{s}(f_{n})|\leq 2\rho(f_{n},f)$. If
$\rho(f_{n},f)<\varepsilon/2$ and $\Delta_{s}(f_{n})<\varepsilon$, then
$\Delta_{t}(f)<2\varepsilon$, so that $\Delta_{t}(f)=0$ and
$\Delta_{s}(f_{n})\leq 2\rho(f_{n},f)$. Hence, when $n$ is large enough, we
find

$||Jf_{n}-J^{\varepsilon}f_{n}||_{\infty}\leq\sum_{s\in[0,1]}\Delta_{s}(f_{n}){\bf
1}_{0<\Delta_{s}(f_{n})<\varepsilon}\leq
2N\rho(f_{n},f)\underset{n\rightarrow\infty}{\longrightarrow}0.$

In particular, the sequence of excursions
$g_{n}:=f_{n}-Jf_{n}+J^{\varepsilon}f_{n}$ converges to $f$ for the Skorokhod
topology. Let us fix $\Phi:\Delta>0\longmapsto\phi_{\Delta}$ such that
$\phi_{\Delta}(x)=x$ for all $\Delta\geq\varepsilon$ and $\phi_{1-\Delta}(x)$
is expressed as in Remark 1 for all $1-\Delta\in(0,\varepsilon)$, for all
$x\in[0,1]$. It is straightforward to check that $\Phi$ is a shuffle. We
observe $B_{f}\cap\mathbb{B}(\Phi)=\emptyset$ and
$B_{Jf}\cap\mathbb{B}(\Phi)=\emptyset$, so Theorem 6 justifies

$\widetilde{d}^{\mathsf{V}}_{J^{\varepsilon}f_{n}}\underset{n\rightarrow\infty}{\longrightarrow}\widetilde{d}^{\mathsf{V}}_{Jf}\text{
and
}\widetilde{d}^{\mathsf{V}}_{g_{n}}\underset{n\rightarrow\infty}{\longrightarrow}\widetilde{d}^{\mathsf{V}}_{f}$

for the Skorokhod topology on $\mathcal{D}([0,1]^{2})$. Then, for all $n\geq
1$,
$\widetilde{d}^{\mathsf{L}}_{J^{\varepsilon}f_{n}}=d^{\mathsf{L}}_{J^{\varepsilon}f_{n}}$
and $\widetilde{d}^{\mathsf{L}}_{f}=d^{\mathsf{L}}_{f}$ because
$\Delta_{t}(J^{\varepsilon}f_{n})$ and $\Delta_{t}(f)$ are not in
$(0,\varepsilon)$ for all $t\in[0,1]$. Recall that $J^{\varepsilon}f_{n}$ and
$Jf$ are PJG, which yields that $d^{\mathsf{L}}_{J^{\varepsilon}f_{n}}$
converges to $d^{\mathsf{L}}_{f}$ for the Skorokhod topology, by applying
Theorem 5. Since $d^{\mathsf{L}}_{f}$ is continuous on $[0,1]^{2}$, this
convergence also happens uniformly on $[0,1]^{2}$. Similarly, Theorems 4 and 5
ensure that
$\widetilde{d}^{\mathsf{L}}_{g_{n}}=\widetilde{d}^{\mathsf{L}}_{Jg_{n}}=\widetilde{d}^{\mathsf{L}}_{J^{\varepsilon}f_{n}}=d^{\mathsf{L}}_{J^{\varepsilon}f_{n}}$
and
$d^{\mathsf{T}}_{g_{n}}=d^{\mathsf{T}}_{g_{n}-Jg_{n}}=d^{\mathsf{T}}_{f_{n}-Jf_{n}}=d^{\mathsf{T}}_{f_{n}}$.
Hence,

$d^{\mathsf{T}}_{f_{n}}+2d^{\mathsf{L}}_{J^{\varepsilon}f_{n}}\underset{n\rightarrow\infty}{\longrightarrow}d^{\mathsf{T}}_{f}+2d^{\mathsf{L}}_{f}$

for the Skorokhod topology on $\mathcal{D}([0,1]^{2})$, which implies that
$d^{\mathsf{T}}_{f_{n}}$ uniformly converges to $d^{\mathsf{T}}_{f}$ on
$[0,1]^{2}$, because $d^{\mathsf{T}}_{f}$ is continuous too. To conclude,
recall that
$||Jf_{n}-J^{\varepsilon}f_{n}||_{\infty}\underset{n\rightarrow\infty}{\longrightarrow}0$.
We deduce the uniform convergence of $(d^{\mathsf{L}}_{f_{n}})$ to
$d^{\mathsf{L}}_{f}$ on $[0,1]^{2}$ with the help of Corollary 2. ∎

######  Theorem 8.

If $f_{n}\longrightarrow f$ for the Skorokhod topology on $\mathcal{H}$ and if
$f$ is PJG, then $d^{\mathsf{L}}_{f_{n}}\longrightarrow d^{\mathsf{L}}_{f}$
and $d^{\mathsf{T}}_{f_{n}}\longrightarrow 0$ uniformly on $[0,1]^{2}$. It
follows $d^{\mathsf{V}}_{f_{n}}\longrightarrow
d^{\mathsf{V}}_{f}=2d^{\mathsf{L}}_{f}$ uniformly on $[0,1]^{2}$.

###### Proof.

Let $\varepsilon>0$ such that $\varepsilon\notin\\{\Delta_{t}(f)\ :\
t\in[0,1]\\}$. This condition ensures $J^{\varepsilon}f_{n}\longrightarrow
J^{\varepsilon}f$ for the Skorokhod topology on $\mathcal{H}$ by Proposition
7, and even more, it holds
$||f_{n}-J^{\varepsilon}f_{n}||_{\infty}\longrightarrow||f-J^{\varepsilon}f||_{\infty}$.
The triangular inequality of $d^{\mathsf{T}}_{f_{n}}$ and the formula (12)
lead to $||d^{\mathsf{T}}_{f_{n}}||_{\infty}\leq
2||f_{n}-J^{\varepsilon}f_{n}||_{\infty}$, because $J^{\varepsilon}f_{n}\leq
Jf_{n}\leq f_{n}$. Moreover, Corollary 2 implies

$||d^{\mathsf{L}}_{f_{n}}-d^{\mathsf{L}}_{f}||_{\infty}\leq
2||f_{n}-J^{\varepsilon}f_{n}||_{\infty}+2||f-J^{\varepsilon}f||_{\infty}+||d^{\mathsf{L}}_{J^{\varepsilon}f_{n}}-d^{\mathsf{L}}_{J^{\varepsilon}f}||_{\infty}.$

All the jumps of the $J^{\varepsilon}f_{n}$ are higher or equal to
$\varepsilon$ and $J^{\varepsilon}f_{n}$ converges to $J^{\varepsilon}f$, thus
there is $N\geq 1$ such that all the $J^{\varepsilon}f_{n}$ have at most $N$
jumps. By the previous theorem,
$||d^{\mathsf{L}}_{J^{\varepsilon}f_{n}}-d^{\mathsf{L}}_{J^{\varepsilon}f}||_{\infty}\longrightarrow
0$, so we get

$\limsup||d^{\mathsf{T}}_{f_{n}}||_{\infty}+\limsup||d^{\mathsf{L}}_{f_{n}}-d^{\mathsf{L}}_{f}||_{\infty}\leq
6||f-J^{\varepsilon}f||_{\infty}.$

Since the excursion $f$ is PJG, it holds $f=Jf$ and
$||f-J^{\varepsilon}f||_{\infty}\underset{\varepsilon\rightarrow
0}{\longrightarrow}0$ thanks to Proposition 7. The set $\\{\Delta_{t}(f)\ :\
t\in[0,1]\\}$ is countable, so we can make $\varepsilon\rightarrow 0$ within
the above inequality. ∎

###  4.3 From convergences of pseudo-distances to convergences of quotient
metric spaces

Recall the notions presented at the start of Section 2.3. To see how Theorems
6, 7, and 8 can bring us some convergences of pointed weighted metric spaces
under the form of Theorem 3, we have to present the definition of the pointed
Gromov-Hausdorff-Prokhorov distance, which truly induces a topology on the set
of the GHP-isometry classes of pointed weighted compact metric spaces. Recall
that when no confusion is possible, we simply denote a metric space (possibly
endowed with some additional structure) by its underlying set.

A correspondence between two compact metric spaces $(X_{1},d_{1})$ and
$(X_{2},d_{2})$ is a subset $\mathcal{R}$ of $X_{1}\times X_{2}$ such that for
all $x_{1}\in X_{1}$ and $y_{2}\in X_{2}$, there exists $x_{2}\in X_{2}$ and
$y_{1}\in X_{1}$ such that $(x_{1},x_{2}),(y_{1},y_{2})\in\mathcal{R}$. We say
a correspondence between $X_{1}$ and $X_{2}$ is compact if it is compact in
the product space $X_{1}\times X_{2}$. The distortion of a correspondence
$\mathcal{R}$ is defined by

$\mathrm{dis}(\mathcal{R})=\sup\left\\{\left|d_{1}(x_{1},y_{1})-d_{2}(x_{2},y_{2})\right|\
:\ (x_{1},x_{2}),(y_{1},y_{2})\in\mathcal{R}\right\\}.$

The Gromov-Hausdorff distance between $X_{1}$ and $X_{2}$ is then expressed by
the formula

$\mathtt{d}_{\mathrm{GH}}(X_{1},X_{2})=\frac{1}{2}\inf_{\mathcal{R}}\mathrm{dis}(\mathcal{R}),$

where the infimum is taken over all correspondences $\mathcal{R}$ between
$X_{1}$ and $X_{2}$. Similarly, the pointed Gromov-Hausdorff distance
$\mathtt{d}_{\mathrm{GH}}^{\bullet}(X_{1},X_{2})$ between two pointed compact
metric spaces $(X_{1},d_{1},a_{1})$ and $(X_{2},d_{2},a_{2})$ is expressed by
the same formula up to the difference that the infimum is instead taken over
all correspondences $\mathcal{R}$ between $X_{1}$ and $X_{2}$ such that
$(a_{1},a_{2})\in\mathcal{R}$. For both distances, we may restrict the infimum
to compact correspondences without modifying the value. Indeed, a
correspondence and its closure have the same distortion, and the latter is a
compact correspondence because $X_{1}\times X_{2}$ is compact. Let
$(X_{1},d_{1},a_{1},\mu_{1})$ and $(X_{2},d_{2},a_{2},\mu_{2})$ be two pointed
weighted compact metric spaces. A coupling between $\mu_{1}$ and $\mu_{2}$ is
a Borel probability measure on $X_{1}\times X_{2}$ whose marginals on $X_{1}$
and $X_{2}$ are $\mu_{1}$ and $\mu_{2}$. The pointed Gromov-Hausdorff-
Prokhorov distance between $X_{1}$ and $X_{2}$ is then defined by the formula

$\mathtt{d}_{\mathrm{GHP}}^{\bullet}(X_{1},X_{2})=\inf_{\mathcal{R},\nu}\max\left(\frac{1}{2}\mathrm{dis}(\mathcal{R}),1-\nu(\mathcal{R})\right)$

where the infimum is taken over all couplings $\nu$ between $\mu_{1}$ and
$\mu_{2}$ and all compact correspondences $\mathcal{R}$ between $X_{1}$ and
$X_{2}$ such that $(a_{1},a_{2})\in\mathcal{R}$.

The functions $\mathtt{d}_{\mathrm{GH}}$,
$\mathtt{d}_{\mathrm{GH}}^{\bullet}$, and
$\mathtt{d}_{\mathrm{GHP}}^{\bullet}$ are only pseudo-distances but
$\mathtt{d}_{\mathrm{GH}}(X_{1},X_{2})=0$ if and only if $X_{1}$ and $X_{2}$
are isometric, $\mathtt{d}_{\mathrm{GH}}^{\bullet}(X_{1},X_{2})=0$ if and only
if $X_{1}$ and $X_{2}$ are pointed-isometric, and
$\mathtt{d}_{\mathrm{GHP}}^{\bullet}(X_{1},X_{2})=0$ if and only if $X_{1}$
and $X_{2}$ are GHP-isometric. Hence, $\mathtt{d}_{\mathrm{GH}}$ defines a
genuine distance on the space $\mathbb{K}$ of isometry classes of compact
metric spaces, $\mathtt{d}_{\mathrm{GH}}^{\bullet}$ defines a genuine distance
on the space $\mathbb{K}^{\bullet}$ of pointed-isometry classes of pointed
compact metric spaces, and $\mathtt{d}_{\mathrm{GHP}}^{\bullet}$ defines a
genuine distance on the space $\mathbb{K}_{\mathrm{w}}^{\bullet}$ of GHP-
isometry classes of pointed weighted compact metric spaces. We call their
respective topologies the Gromov-Hausdorff, the pointed Gromov-Hausdorff, and
the pointed Gromov-Hausdorff-Prokhorov topologies. Furthermore, the metric
spaces $\mathbb{K}$, $\mathbb{K}^{\bullet}$, and
$\mathbb{K}_{\mathrm{w}}^{\bullet}$ are separable and complete. See [28] and
[2] for proof of the stated facts and for more information. Now, recall from
Section 2.3 the definition of quotient metric spaces and Definition 3.

######  Lemma 5.

For all $n\geq 1$, let $d_{n},d\in\mathcal{D}([0,1]^{2})$ be two pseudo-
distances on $[0,1]$ such that the quotient pointed weighted metric spaces
$X_{n}=[0,1]/\\{d_{n}=0\\}$ and $X=[0,1]/\\{d=0\\}$ are compact. It holds

$\displaystyle\mathtt{d}_{\mathrm{GH}}^{\bullet}(X_{n},X)$
$\displaystyle\leq\rho_{2}(d_{n},d),$ (30)
$\displaystyle\mathtt{d}_{\mathrm{GHP}}^{\bullet}(X_{n},X)$
$\displaystyle\leq||d_{n}-d||_{\infty}.$ (31)

Furthermore, if $d_{n}\longrightarrow d$ for the Skorokhod topology on
$\mathcal{D}([0,1]^{2})$, then $X_{n}\longrightarrow X$ for the pointed
Gromov-Hausdorff-Prokhorov topology.

###### Proof.

Let us denote by $[s]_{n}$ and $[s]$ the canonical projections of $s\in[0,1]$
respectively on $X_{n}$ and $X$, and by $\mu_{n}$ and $\mu$ the push-forward
measures of the Lebesgue measure on $[0,1]$ by the respective canonical
projections. Let $\lambda_{n},\mu_{n}:[0,1]\longrightarrow[0,1]$ be two
increasing bijections. The triangular inequality yields
$||d_{n}-d\circ(\lambda_{n},\lambda_{n})||_{\infty}\leq
2||d_{n}-d\circ(\lambda_{n},\mu_{n})||_{\infty}$. But the quantity
$||d_{n}-d_{2}\circ(\lambda_{n},\lambda_{n})||_{\infty}$ is the distortion of
the correspondence $\mathcal{R}_{n}=\\{([s]_{n},[\lambda_{n}(s)])\ :\
s\in[0,1]\\}$. This proves (30) by taking the infimum over
$\lambda_{n},\mu_{n}$. When $\lambda_{n}=\mathsf{id}$, the correspondence
$\mathcal{R}_{n}$ has total mass for the coupling $\nu_{n}$ between $\mu_{n}$
and $\mu$ defined by

$\int_{X_{n}\times X}g{\rm d}\nu=\int_{0}^{1}g([s]_{n},[s]){\rm d}s$

for any bounded measurable $g$. This shows (31). If $d_{n}\longrightarrow d$
for the Skorokhod topology on $\mathcal{D}([0,1]^{2})$ then we can assume
$||\lambda_{n}-\mathsf{id}||_{\infty}+||d_{n}-d\circ(\lambda_{n},\lambda_{n})||_{\infty}\longrightarrow
0$. For all $\varepsilon>0$, we define another compact correspondence
$\mathcal{R}_{n}^{\varepsilon}=\\{([s]_{n},x)\ :\ s\in[0,1],x\in X\text{ such
that }d([\lambda_{n}(s)],x)\leq\varepsilon\\}$ between $X_{n}$ and $X$. It
holds $\limsup\mathrm{dis}(\mathcal{R}_{n}^{\varepsilon})\leq 2\varepsilon$.
Remark the compactness of $[0,1]/\\{d=0\\}$ when $d\in\mathcal{D}([0,1]^{2})$
implies the canonical projection $[0,1]\longrightarrow X$ is càdlàg, so it is
continuous almost everywhere. Thus, $[\lambda_{n}(s)]\longrightarrow[s]$
almost everywhere and $\nu_{n}(\mathcal{R}_{n}^{\varepsilon})\longrightarrow
1$ according to the dominated convergence theorem. ∎

###### Proof of Theorem 3.

By definition of the relaxed Skorokhod topology, $\Theta
f_{n}\longrightarrow\Theta f$ for the (rigid) Skorokhod topology. Observe that
if $t\in[0,1/2)$ then $\widetilde{d}^{\mathsf{V}}_{\Theta f}(t,1)=0$, and that
if $s,t\in[1/2,1]$ then $\widetilde{d}^{\mathsf{V}}_{\Theta
f}(s,t)=\widetilde{d}^{\mathsf{V}}_{f}(2s-1,2t-1)$ thanks to the branching
property. It is then a small exercise to check the convergences
$\widetilde{\mathscr{V}}_{f_{n}}\longrightarrow\widetilde{\mathscr{V}}_{f}$
and $\widetilde{\mathscr{V}}_{\Theta
f_{n}}\longrightarrow\widetilde{\mathscr{V}}_{\Theta f}$ for the pointed
Gromov-Hausdorff-Prokhorov topology are equivalent. Finally, the previous
lemma and Theorem 6 implies the point $(i)$ because $B_{\Theta f}=B_{f}$.
Likewise, we show the two other points using Theorems 7 and 8. Indeed, $f_{n}$
has at most $N$ jumps if and only if $\Theta f_{n}$ has at most $N$ jumps.
When $f$ is PJG, $J(\Theta f)=\Theta f$ by Proposition 7, thus $\Theta f$ is
PJG. ∎

##  5 Topological space of vernation trees

Here, we propose a topological notion of vernation trees that does not require
coding by excursions. First, let us fix our framework by giving some
definitions. Let $(X,d)$ be a metric space and let $x,y\in X$. We say that
$\gamma$ is a path on $X$ from $x$ to $y$ when $\gamma:[0,1]\longrightarrow X$
is continuous such that $\gamma(0)=x$ and $\gamma(1)=y$. In that case, we will
denote by $\operatorname{Im}\gamma=\gamma([0,1])$ its image. If $\gamma$ is
also injective, we say that $\gamma$ is an arc on $X$ from $x$ to $y$. By
compactness of $[0,1]$, an arc provides a topological embedding of the segment
$[0,1]$ into $X$. Moreover, we say that $\gamma$ is a geodesic on $X$ from $x$
to $y$ when $d(\gamma(t),\gamma(s))=|t-s|d(x,y)$ for all $t,s\in[0,1]$. A
geodesic from $x$ to $y$ is an arc if and only if $x\neq y$. Finally, we say
that $\gamma$ is a loop on $X$ based at $x$ when $\gamma:[0,1]\longrightarrow
X$ is continuous, injective on $[0,1)$, and such that $\gamma(0)=\gamma(1)=x$.
A loop provides a topological embedding of the circle $\mathcal{C}$ into $X$.

######  Definition 6.

A metric space $(X,d)$ is geodesic when there exists a geodesic on $X$ from
$x$ to $y$, for all $x,y\in X$. A metric space $(X,d)$ is a real tree when it
is geodesic and when all arcs on $X$ from $x$ to $y$ have the same image, for
all $x,y\in X$. We say a metric space $(X,d)$ is a vernation tree when it is
geodesic and when it satisfies that for any $\gamma_{1},\gamma_{2}$, two loops
on $X$ based at $x$, then either
$\operatorname{Im}\gamma_{1}=\operatorname{Im}\gamma_{2}$ or
$\operatorname{Im}\gamma_{1}\cap\operatorname{Im}\gamma_{2}=\\{x\\}$, for all
$x\in X$.

We point out that any geodesic space is arc-connected by definition. For our
work, we exclusively are interested in compact metric spaces. When a metric
space is indeed compact, it is geodesic if and only if it is a length space.
See [20, Chapter 2] for a definition of length spaces, a proof of the just
mentioned fact, and an extensive study of the notion. For a detailed study of
real trees, we refer to Evans [24] or to Paulin [38]. Observe a geodesic space
is locally path-connected. Indeed, for any $x\in X$ and $r>0$, if $d(x,y)<r$
then a geodesic between $x$ and $y$ stays inside the ball of radius $r$ and of
center $x$. Moreover, it is well-known that a Hausdorff space, so a metric
space, is pathwise connected if and only if it is arcwise connected, see [42,
Chapter 31] or [13] for example. We deduce the following convenient lemma.

######  Lemma 6.

Let $(X,d)$ be a compact geodesic space. If $F$ is a finite subset of $X$,
then the connected components of $X\backslash F$ are open and arcwise
connected.

######  Proposition 8.

When $(X,d)$ is a compact geodesic space, the following statements are
equivalent.

  1. (i)

The space $(X,d)$ is a real tree.

  2. (ii)

For all $x,y\in X$, if $x\neq y$ then there exists $u\in X\backslash\\{x,y\\}$
such that $x$ and $y$ are in different connected components of
$X\backslash\\{u\\}$.

  3. (iii)

There are no loops on $X$.

The implications $(i)\Longrightarrow(ii)\Longrightarrow(iii)$ are obvious. The
equivalence between $(i)$ and $(iii)$ is quite easy to show and can be found
in [15, Proposition 2.3]. We observe that the point $(iii)$ obviously ensures
that a real tree is also a vernation tree, which might not be so clear by only
looking at the definitions. The interest of the point $(ii)$ is that it can be
naturally adapted into an equivalent definition for vernation trees.

######  Proposition 9.

When $(X,d)$ is a compact geodesic space, the following statements are
equivalent.

  1. (i)

The space $(X,d)$ is a vernation tree.

  2. (ii)

For all $x,y\in X$, if $x\neq y$ then there exists $u,v\in
X\backslash\\{x,y\\}$ such that $x$ and $y$ are in different connected
components of $X\backslash\\{u,v\\}$.

  3. (iii)

For all $x,y\in X$, if $x\neq y$ then there exists $u,v\in
X\backslash\\{x,y\\}$ such that $x$ and $y$ are in different connected
components of $X\backslash\\{u,v\\}$ and such that
$2\min\left(d(x,u),d(x,v),d(y,u),d(y,v)\right)\geq d(x,y)$.

###### Proof.

The implication $(iii)\Longrightarrow(ii)$ is obvious. Let us assume that
$(ii)$ holds. Let $\gamma_{1},\gamma_{2}$ be two loops on $X$ based at $x$ and
let $s\in(0,1)$ such that $\gamma_{1}(s)\notin\operatorname{Im}\gamma_{2}$. We
set

$a_{1}=\sup\\{t\in[0,s]\ :\
\gamma_{1}(t)\in\operatorname{Im}\gamma_{2}\\}\text{ and
}b_{1}=\inf\\{t\in[s,1]\ :\ \gamma_{1}(t)\in\operatorname{Im}\gamma_{2}\\},$

so that $\gamma_{1}(a_{1}),\gamma_{2}(b_{1})\in\operatorname{Im}\gamma_{2}$
because $\gamma_{1}$ is continuous and $\operatorname{Im}\gamma_{2}$ is
closed. We choose $a_{2},b_{2}\in[0,1]$ such that
$\gamma_{2}(a_{2})=\gamma_{1}(a_{1})$ and
$\gamma_{2}(b_{2})=\gamma_{1}(b_{1})$. Even if it means reversing
$\gamma_{2}$, we can assume that $a_{2}\leq b_{2}$ without loss of generality.
We can define three paths on $X$ from $\gamma_{1}(a_{1})$ to
$\gamma_{1}(b_{1})$ by setting

$\displaystyle\gamma_{1}^{\prime}:t\in[0,1]$
$\displaystyle\longmapsto\gamma_{1}(a_{1}+t(b_{1}-a_{1})),$
$\displaystyle\gamma_{2}^{\prime}:t\in[0,1]$
$\displaystyle\longmapsto\gamma_{2}(a_{2}+t(b_{2}-a_{2})),$
$\displaystyle\gamma_{3}^{\prime}:t\in[0,1]$
$\displaystyle\longmapsto\begin{cases}\gamma_{2}(a_{2}-t)&\text{ if }t\leq
a_{2}\\\ x&\text{ if }a_{2}\leq t\leq b_{2}\\\ \gamma_{2}(b_{2}+1-t)&\text{ if
}b_{2}\leq t\leq 1\end{cases}.$

By definition of $a_{1}$ and $b_{1}$, it is clear that
$\operatorname{Im}\gamma_{1}^{\prime}\cap\operatorname{Im}\gamma_{2}=\\{\gamma_{1}(a_{1}),\gamma_{1}(b_{1})\\}$,
so it easily follows that
$\operatorname{Im}\gamma_{1}^{\prime}\cap\operatorname{Im}\gamma_{2}^{\prime}=\operatorname{Im}\gamma_{1}^{\prime}\cap\operatorname{Im}\gamma_{3}^{\prime}=\\{\gamma_{1}(a_{1}),\gamma_{1}(b_{1})\\}$.
Moreover,
$\operatorname{Im}\gamma_{2}^{\prime}\cap\operatorname{Im}\gamma_{3}^{\prime}=\\{\gamma_{1}(a_{1}),\gamma_{1}(b_{1})\\}$
because $\gamma_{2}$ is injective on $[0,1)$. In the case where
$\gamma_{1}(a_{1})\neq\gamma_{1}(b_{1})$, there exists $u,v\in
X\backslash\\{\gamma_{1}(a_{1}),\gamma_{1}(b_{1})\\}$ such that
$\gamma_{1}(a_{1})$ and $\gamma_{1}(b_{1})$ are in different pathwise
connected components of $X\backslash\\{u,v\\}$ thanks to $(ii)$. In
particular, we would have
$\operatorname{Im}\gamma_{i}^{\prime}\cap\\{u,v\\}\neq\emptyset$ for
$i=1,2,3$, which is impossible regarding the intersections between the
$\operatorname{Im}\gamma_{i}^{\prime}$. Hence, we find that
$\gamma_{1}(a_{1})=\gamma_{1}(b_{1})$ and since $a_{1}<s<b_{1}$, we conclude
that $a_{1}=0$ and $b_{1}=1$. This means
$\operatorname{Im}\gamma_{1}\cap\operatorname{Im}\gamma_{2}=\\{x\\}$. By
contraposition, if
$\operatorname{Im}\gamma_{1}\cap\operatorname{Im}\gamma_{2}\neq\\{x\\}$ then
$\operatorname{Im}\gamma_{1}\subset\operatorname{Im}\gamma_{2}$ and
$\operatorname{Im}\gamma_{1}=\operatorname{Im}\gamma_{2}$ by symmetry, which
ends the proof of $(ii)\Longrightarrow(i)$.

Let us assume that $(i)$ holds. Let $g$ be a geodesic on $X$ from $x$ to $y$
with $x\neq y$, and let us set $u=g(1/2)$ so that $d(x,u)=d(y,u)=d(x,y)/2$.
Let $\gamma_{1}$ be an arc on $X\backslash\\{u\\}$ from $x$ to $y$. It is also
an arc on $X$ such that $u\notin\operatorname{Im}\gamma_{1}$. We set

$a_{1}=\sup\\{t\in[0,1]\ :\ \gamma_{1}(t)\in g([0,1/2])\\}\text{ and
}b_{1}=\inf\\{t\in[a_{1},1]\ :\ \gamma_{1}(t)\in g([1/2,1])\\},$

which are well-defined because $\gamma_{1}(0)=g(0)$ and $\gamma_{1}(1)=g(1)$.
Since $\gamma_{1}$ and $g$ are continuous, there exists
$\alpha_{1},\beta_{1}\in[0,1]$ such that $\alpha_{1}\leq 1/2\leq\beta_{1}$,
$g(\alpha_{1})=\gamma_{1}(a_{1})$, and $g(\beta_{1})=\gamma_{1}(b_{1})$. In
fact, it even holds $\alpha_{1}<1/2<\beta_{1}$ because
$u\notin\operatorname{Im}\gamma_{1}$, so $a_{1}<b_{1}$ because $g$ is
injective. The real-valued function $t\in[a_{1},b_{1}]\longmapsto
d(x,\gamma_{1}(t))$ is continuous and

$d(x,\gamma_{1}(a_{1}))=\alpha_{1}d(x,y)<d(x,y)/2<\beta_{1}d(x,y)=d(x,\gamma_{1}(b_{1})),$

so there exists $s\in(a_{1},b_{1})$ such that $d(x,\gamma_{1}(s))=d(x,y)/2$.
We set $v=\gamma_{1}(s)$, and we observe that $d(x,y)\leq d(x,y)/2+d(y,v)$. We
only need to verify that $x$ and $y$ are in different connected components of
$X\backslash\\{u,v\\}$ in order to show $(i)\Longrightarrow(iii)$. Let
$\gamma_{2}$ be an arc on $X\backslash\\{u,v\\}$ from $x$ to $y$ by
contradiction with Lemma 6. It is also an arc on $X$ such that
$\operatorname{Im}\gamma_{2}\cap\\{u,v\\}=\emptyset$. Similarly as with
$\gamma_{1}$, we set

$a_{2}=\sup\\{t\in[0,1]\ :\ \gamma_{2}(t)\in g([0,1/2])\\}\text{ and
}b_{2}=\inf\\{t\in[a_{2},1]\ :\ \gamma_{2}(t)\in g([1/2,1])\\}$

and we fix $\alpha_{2},\beta_{2}\in[0,1]$ such that
$\alpha_{2}<1/2<\beta_{2}$, $g(\alpha_{2})=\gamma_{2}(a_{2})$, and
$g(\beta_{2})=\gamma_{2}(b_{2})$. Moreover, $a_{2}<b_{2}$. Then, we define two
paths $\gamma_{1}^{\prime},\gamma_{2}^{\prime}$ on $X$ from $u$ to $u$ by
setting for all $t\in[0,1]$,

$\gamma_{1}^{\prime}(t),\gamma_{2}^{\prime}(t)=\begin{cases}g(1/2+3t(\beta_{1}-1/2)),g(1/2+3t(\beta_{2}-1/2))&\text{
if }t\leq 1/3\\\
\gamma_{1}(b_{1}+3(t-1/3)(a_{1}-b_{1})),\gamma_{2}(b_{2}+3(t-1/3)(a_{2}-b_{2}))&\text{
if }1/3\leq t\leq 2/3\\\
g(\alpha_{1}+3(t-2/3)(1/2-\alpha_{1})),g(\alpha_{2}+3(t-2/3)(1/2-\alpha_{2}))&\text{
if }2/3\leq t\end{cases}.$

It is clear that $\gamma_{1}^{\prime}$ and $\gamma_{2}^{\prime}$ are paths on
$X$ from $u$ to $u$. By definition of $a_{1},b_{1},a_{2},b_{2}$, if
$t_{1}\in(a_{1},b_{1})$ and $t_{2}\in(a_{2},b_{2})$, then
$\gamma_{1}(t_{1})\notin\operatorname{Im}g$ and
$\gamma_{2}(t_{2})\notin\operatorname{Im}g$. Since $g,\gamma_{1},\gamma_{2}$
are injective, it follows that $\gamma_{1}^{\prime}$ and $\gamma_{2}^{\prime}$
are loops on $X$ based at $u$. We observe that if
$1/2<t<\min(\beta_{1},\beta_{2})$ then $g(t)\neq u$ and
$g(t)\in\operatorname{Im}\gamma_{1}^{\prime}\cap\operatorname{Im}\gamma_{2}^{\prime}$,
so it ensures that
$\operatorname{Im}\gamma_{1}^{\prime}=\operatorname{Im}\gamma_{2}^{\prime}$.
However, $s\in(a_{1},b_{1})$ so
$v=\gamma_{1}(s)\in\operatorname{Im}\gamma_{1}^{\prime}$, while
$v\notin\operatorname{Im}g$ and $v\notin\operatorname{Im}\gamma_{2}$, so
$v\notin\operatorname{Im}\gamma_{2}^{\prime}$. That is a contradiction, which
concludes the proof. ∎

When $(X,d)$ is a compact metric space, we write
$\mathrm{diam}(X)=\sup_{x,y\in X}d(x,y)$. According to the next lemma, a
compact vernation tree has only a countable number of images of loops.

######  Lemma 7.

If $(X,d)$ is a compact vernation tree, then the set
$\\{\operatorname{Im}\gamma\ :\ \gamma\text{ loop on
}X,\mathrm{diam}(\operatorname{Im}\gamma)\geq\varepsilon\\}$ is finite for all
$\varepsilon>0$.

###### Proof.

We assume by contradiction the set of interest is infinite. Then, the
compactness of $X$ yields there exists two loops $\gamma_{1},\gamma_{2}$ on
$X$ such that $\operatorname{Im}\gamma_{1}\neq\operatorname{Im}\gamma_{2}$,
and there exists $x_{1},y_{1}\in\operatorname{Im}\gamma_{1}$ and
$x_{2},y_{2}\in\operatorname{Im}\gamma_{2}$ such that
$d(x_{1},y_{1})\geq\varepsilon$, $d(x_{1},x_{2})\leq\varepsilon/4$, and
$d(y_{1},y_{2})\leq\varepsilon/4$. We give ourselves two geodesics
$g_{x},g_{y}$ on $X$ respectively from $x_{1}$ to $x_{2}$ and from $y_{1}$ to
$y_{2}$. Observe that
$\operatorname{Im}g_{x}\cap\operatorname{Im}g_{y}=\emptyset$ because
$d(\operatorname{Im}g_{x},\operatorname{Im}g_{y})\geq
d(x_{1},y_{1})-d(x_{1},x_{2})-d(y_{1},y_{2})>0$. Considering $a_{z}=\sup
g_{z}^{-1}(\operatorname{Im}\gamma_{1})$ and $b_{z}=\inf
g_{z}^{-1}(\operatorname{Im}\gamma_{2})\cap[a_{z},1]$ for $z\in\\{x,y\\}$
allows us to assume
$\operatorname{Im}g_{x}\cap\operatorname{Im}\gamma_{1}=\\{x_{1}\\}$,