\begin{document}
\maketitle

\newpage
\begin{abstract}
We consider the spatial $\Lambda$-Fleming-Viot process model (\cite{BEV2010}) for
frequencies of genetic types in
a population living in $\R^d$, with two types of individuals ($0$ and $1$) and natural
selection favouring individuals of type $1$. We first prove that the model with selection is well-defined.
Next, we consider two cases, one in which the dynamics of the process are driven by
purely `local' events and one incorporating large-scale extinction-recolonisation events.
In both cases, we consider a sequence of spatial $\Lambda$-Fleming-Viot processes indexed by $n$,
and we assume that the fraction of individuals replaced during a reproduction event and the relative
frequency of events during which natural selection acts tend to $0$ as $n$ tends to infinity.
We choose the decay of these parameters in such a way that the frequency of the less favoured type
converges in distribution to the solution to the Fisher-KPP equation (with noise in one dimension)
when reproduction is only local, or to the solution to an analogous equation in which the Laplacian
is replaced by a fractional Laplacian when large-scale extinction-recolonisation events occur.
We also define the process of `potential ancestors' of a sample of individuals taken from these
populations, and show their convergence in distribution towards a system of Brownian or stable motions
which branch at some finite rate. In one dimension, in the limit, pairs of particles also coalesce at a
rate proportional to their local time together. In contrast to previous proofs of scaling limits for the
spatial $\Lambda$-Fleming-Viot process
here the convergence of the more complex \emph{forwards in time} processes is used to prove the
convergence of the potential ancestries.

\bigskip
\noindent\textbf{AMS 2010 subject classifications.}  {\em Primary:} 60G57, 60J25, 92D10 ; {\em Secondary:} 60J75, 60G52.

\noindent{\bf Key words and phrases:} Generalised Fleming-Viot process, natural selection, limit theorems, duality, symmetric stable processes, population genetics.
\end{abstract}

\section{Introduction}
\label{introduction}

The {\em spatial $\Lambda$-Fleming-Viot process} (SLFV) was introduced in
\cite{etheridge:2008, BEV2010}.  In fact, it is not so
much a process as a general framework for modelling frequencies of
different genetic types in populations which evolve in a spatial continuum.
For example, it is readily adapted to include things like the large-scale
extinction/recolonisation events which have dominated the demographic
history of many species.  In this paper we shall be concerned with
extending the framework to incorporate an important
form of natural selection and
investigating various rescaling limits which capture the resultant
patterns of genetic diversity over large spatial and temporal scales.
In particular, we recover the Fisher-KPP equation and, in one dimension,
its stochastic counterpart. In the presence of large-scale demographic
events we obtain analogous equations with the Laplacian replaced by the
fractional Laplacian, but, intriguingly, no other trace of the large-scale
events survives.  The limits obtained here assume that the `neighbourhood
size', which corresponds to the local population density, is high, thus
complementing results of
\cite{etheridge/freeman/straulino:2014} which
address the interaction of natural selection and genetic drift when
neighbourhood size is small.

\subsection{The spatial $\Lambda$-Fleming-Viot process with selection}

The main innovation in the SLFV is that reproduction in the population is
based on `events', rather than on individuals.  It is this which overcomes
Felsenstein's {\em pain in the torus} (\cite{felsenstein:1975}). This is
discussed in detail in \cite{BEV2010} and we don't repeat the motivation here.
Each event determines the region of space in which reproduction (or
extinction/recolonisation) will take place and an `impact' $u$.  As a
result of the event, a proportion $u$ of the individuals living in the
region are replaced by offspring of a parent chosen from the population
immediately before the event.  Events occur according to a Poisson point
process.  The Poisson structure renders the process particularly
amenable to analytic study.
In the neutral setting which has been studied thus far in
the literature, the parent is chosen uniformly at random from the affected
region, irrespective of type.  There are many possible ways to incorporate
natural selection. Here we shall focus on one of the simplest, but also
most important, in which in the selection of the parent, individuals
are weighted according to their genetic type.

To motivate our definition of the process with (fecundity) selection,
suppose that there are two possible types in the population, which we shall
denote by $0$ and $1$.  In order to give a slight selective advantage to
type $1$, we fix a {\em selection coefficient} $s>0$
and suppose that, when an event falls, if the proportion of type $0$
individuals in the affected region immediately before the event is $\bar{w}$,
then the probability of picking a type $0$ parent is
$p(\bar{w},s)=\bar{w}/(1+s(1-\bar{w}))$.  Typically one is
interested in weak selection, so
that $s\ll 1$ and, in this case, we can estimate this
probability by $(1-s)\bar{w}+s\bar{w}^2$.  Here again we reap the
benefit of the Poisson structure of events: we can think of events as
being of one of two types. A proportion $(1-s)$ of events are `neutral':
the parent is selected exactly as in the neutral setting and has
probability $\bar{w}$ of being of type $0$.  On the other hand a
proportion $s$ of events are `selective' and then the probability of
a type $0$ parent is $\bar{w}^2$.  One way to achieve this is to dictate
that at selective events we choose {\em two} potential parents and only if
both are type $0$ will the offspring be type $0$. The Poisson structure
allows us to view neutral and selective events as being driven by
independent Poisson processes. This approach exactly
parallels that usually adopted to incorporate genic selection into the
classical Moran model.

Of course there are many ways to modify the selection mechanism. For
example, in Definition~\ref{defn of process} below, we allow both
the distribution of
the size of the region affected and of the impact to differ between selective
and neutral events.

Let us turn to a precise definition. First we describe the state space of
the process. We suppose that the population is
evolving in a geographical space $E\subseteq \IR^d$. We borrow some results
from \cite{VW2012} in the special case when the space of
genetic types is $K=\{0,1\}$.
At each time $t$, the population
will be represented by a random function $\{w_t(x),\, x\in E\}$ defined, up to
a Lebesgue null set of $E$ by
$$
w_t(x):= \hbox{ proportion of type }0\hbox{ at spatial position }x\hbox{ at time }t.
$$
The space in which it takes its value is in one-to-one correspondence
with the space $\cM_\lambda(E\times \{0,1\})$ of all measures on
$E\times \{0,1\}$ whose `spatial' marginal is Lebesgue measure,
through the identification
$$
\int_{E\times \{0,1\}} f(x,\kappa) M(dx,d\kappa) = \int_E \big\{w(x)f(x,0)+ (1-w(x))f(x,1)\big\}\, dx.
$$
By Lemma~1.1 in \cite{VW2012}, the space $\cM_\lambda(E\times \{0,1\})$
endowed with the topology of vague convergence is compact. Therefore, this
property is inherited by the state space of the process $(w_t)_{t\geq 0}$
which we shall also, by a slight abuse of notation,
call $\cM_\lambda(E\times \{0,1\})$.
In what follows we shall write, for every $f\in C_c(E)$
(the space of all continuous functions on $E$ with compact support),
\begin{equation}\label{notation}
\la w,f\ra := \int_E f(x)w(x)\, dx.
\end{equation}

\begin{defn}[SLFV with (fecundity) selection (SLFVS)]
\label{defn of process}
Let $\mu,\mu'$ be two $\sigma$-finite measures on $(0,\infty)$, and
let $\nu=\{\nu_r,\, r>0\},\, \nu'=\{\nu_r',\, r>0\}$ be two collections of
probability measures on $[0,1]$ such that the map $r\mapsto \nu_r$
(resp., $r\mapsto \nu'_r$) is measurable with respect to $\mu$ (resp., $\mu'$)
and
\begin{equation} \label{cond existence}
\int_{(0,\infty)}r^d \int_0^1 u\nu_r(du)\mu(dr) <\infty,\qquad \hbox{and}\qquad \int_{(0,\infty)}r^d \int_0^1 u\nu'_r(du)\mu'(dr) <\infty.
\end{equation}
Further, let $\Pi^N$ and $\Pi^S$ be two independent Poisson point processes on
$\R\times E\times (0,\infty)\times [0,1]$ with respective intensity
measures $dt\otimes dx\otimes \mu(dr)\nu_r(du)$ and
$dt\otimes dx\otimes \mu'(dr)\nu'_r(du)$.

The {\em spatial $\Lambda$-Fleming-Viot process with selection} (SLFVS) with
driving noises $\Pi^N$ and $\Pi^S$ is the
$\cM_\lambda(E\times \{0,1\})$-valued process
$(w_t)_{t\geq 0}$ with dynamics given as follows.
If $(t,x,r,u)\in \Pi^N$,
a neutral event occurs at time $t$, within the closed ball $B(x,r)$:
\begin{enumerate}
\item Choose a parental location $z$ uniformly at random within $B(x,r)$,
and a parental type, $\kappa$, according to $w_{t-}(z)$, that is
$\kappa=0$ with probability $w_{t-}(z)$
and $\kappa=1$ with probability $1-w_{t-}(z)$.
\item For every $y\in B(x,r)$, set
$w_t(y) = (1-u)w_{t-}(y) + u\ind_{\{\kappa=0\}}$.
\end{enumerate}
Similarly, if $(t,x,r,u)\in \Pi^S$, a selective event occurs at time $t$,
within the closed ball $B(x,r)$:
\begin{enumerate}
\item Choose two `potential' parental locations $z,z',$ independently and
uniformly at
random within $B(x,r)$, and at each of these sites `potential' parental types
$\kappa$, $\kappa'$, according to $w_{t-}(z), w_{t-}(z')$ respectively.
\item
For every $y\in B(x,r)$, set
$w_t(y) = (1-u)w_{t-}(y) + u\ind_{\{\kappa =\kappa'=0\}}$.
Declare the parental location to be $z$ if $\kappa=\kappa'=0$ or
$\kappa=\kappa'=1$ and to be $z$ (resp. $z'$) if $\kappa=1-\kappa'=1$
(resp. $1-\kappa=\kappa'=1$).
\end{enumerate}
\end{defn}
Since we only consider this particular form of selection in this paper, there
should be no ambiguity in simply calling this process the SLFV
{\em with selection},
but we emphasize that, although this is certainly one of the most natural,
there are many alternative models.
For example, one could
modify the construction so that one first selects a parental type and then an
impact depending on that type, or one could `kill' with differential weights
(c.f.~\cite{bah/pardoux:2013, foucart:2013} in the non-spatial setting).

Two different constructions of SLFV are given in \cite{etheridge/kurtz:2014}.
Both incorporate forms of selection.  The first establishes, under somewhat
weaker conditions than~(\ref{cond existence}), the existence of an SLFV with
differential killing.  The second, under precisely
conditions~(\ref{cond existence}), allows for the sort of selection which we
consider here. Notwithstanding the proof of existence provided by that paper,
in \S\ref{s: proof existence} we shall provide a direct proof of
existence. Although our approach does not yield the `lookdown' construction
of \cite{etheridge/kurtz:2014}, and nor does it allow us to relax
conditions~(\ref{cond existence}), it is extremely flexible and can,
under conditions analogous to~(\ref{cond existence}),
be readily modified to incorporate models with many different
forms of selection.  Thus our first result is the following.
\begin{theorem}\label{th: existence}
Let $E$ be a connected subset of $\R^d$, and $\mu$, $\mu'$, $\nu$ and $\nu'$
satisfy (\ref{cond existence}).
Write $\cL$ for the generator of the SLFVS acting on test functions from
$\cD(\cL)=\{F(\la w,f\ra); f\in C_c(E), F\in C^1(\R)\}$.
Then the martingale problem for $(\cL,\cD(\cL))$ is well-posed.
\end{theorem}

\subsection{Statement of the main results}

Our main results concern the patterns of variation that we see under this model
if we look over large spatial and temporal scales. We are concerned with
the regime of large neighbourhood size, corresponding to small impact. We
shall concentrate on the particular case in which $E=\IR^d$ and
$\mu'(dr)\nu_r'(du)=s\mu(dr)\nu_r(du)$ for a small parameter $s>0$.
This corresponds to the weighting of the
selection of the parent which motivated our definition of the SLFVS.
In fact we shall choose very
special forms for the measures $\mu(dr)$ and $\nu_r(du)$.  Our
results will certainly hold under much more general conditions, but the
proofs become obscured by notation.

More precisely, for each
$n\in\IN$, we fix a number $u_n\in (0,1)$ and assume that all events
(neutral and selective) have impact $u_n$, that is,
$\nu_r(du)=\delta_{u_n}(du)$ for every $r>0$ and that
$\mu' = s_n\mu$ for some $s_n>0$.
We consider the regime in which $u_n$ and $s_n$ go to $0$ as
$n\rightarrow \infty$. That is, the neighbourhood
size (or local population density) tends to infinity while
selection is weak. This mirrors the usual assumptions in the Moran
and Wright-Fisher models, in the absence of spatial
structure, in which one is interested in the scaling limits that
are obtained as population size $N$ tends to infinity while $Ns_N$ remains
${\mathcal O}(1)$.
We shall find scalings of time and space for which the rescaled
SLFVS converges to a non-trivial limit as $n\rightarrow \infty$.
Specifically, we identify a parameter regime in which
the SLFVS behaves like the Fisher-KPP equation (with noise in $d=1$),
or its analogue with long-range dispersal.
In particular of course, when neighbourhood size is large, this will tell us
how strong selection must be (relative to
neighbourhood size) if we are
to see its effect over large spatial and temporal scales.
With this in mind, let us assume that for some $\gamma,\delta >0$,
$$
u_n = \frac{u}{n^\gamma}, \qquad \hbox{and} \qquad s_n=\frac{\sigma}{n^\delta}.
$$
Further, let $V_r$ denote the volume of a ball of radius $r$
(which, of course, depends on the dimension $d$, but we suppress that in
our notation).

We shall consider the following two cases:
\begin{itemize}
\item {\bf Fixed radius:} $\mu(dr)= \delta_R(dr)$, for some fixed $R>0$.
In this case, we choose $\gamma = 1/3$, $\delta=2/3$, and set (in any
dimension)
$$
w^n_t(x) := w_{nt}(n^{1/3}x), \qquad \hbox{and}\qquad \bw^n_t(x):=\frac{n^{d/3}}{V_R}\, \int_{B(x,n^{-1/3}R)}w^n_t(y)\, dy.
$$
\item {\bf Stable radii:} For some $\alpha\in (1,2)$, we set
$$
\mu(dr)=\frac{\ind_{\{r\geq 1\}}}{r^{d+\alpha+1}}\, dr,
$$
$$
w^n_t(x) := w_{nt}(n^{\beta}x), \qquad \hbox{and}\qquad  \bw^n_t(x):=\frac{n^{d\beta}}{V_1}\, \int_{B(x,n^{-\beta})}w^n_t(y)\, dy,
$$
where
\begin{equation}\label{def constants}
\beta = \frac{1}{2\alpha -1},\quad \gamma= \frac{\alpha-1}{2\alpha-1}\quad \hbox{and}\quad \delta = \frac{\alpha}{2\alpha-1}.
\end{equation}
\end{itemize}
Not surprisingly, we recover the parameters for the fixed radius case from those
for stable radii on setting $\alpha =2$.
We provide an informal argument which explains why this is the
appropriate choice of the parameters $\beta$, $\gamma$, $\delta$ in
\S\ref{heuristics}.

To simplify notation, we write $\cM$ for $\cM_\lambda(\R^d \times\{0,1\})$,
and $D_\cM[0,\infty)$ for the set of all c\`adl\`ag paths with values in
$\cM$. We also write $C_c^\infty(\R^d)$ for the set of smooth
compactly supported functions on $\R^d$.
Our main results are as follows.
\begin{theorem}\label{th:fixed}{\bf (Fixed radius)}
Suppose that $\bw^n_0$ converges weakly to some $w^0\in \cM$.  Then, as $n\rightarrow \infty$, the process
$(\bw_t^n)_{t\geq 0}$
converges weakly in $D_{\cM}[0,\infty)$ towards a process
$(w_t^\infty)_{t\geq 0}$
with initial value
$w^{\infty}_0=w^0$. Furthermore,

$(i)$ When $d=1$, $(w_t^\infty)_{t\geq 0}$ is the unique process for which,
for every $f\in C^{\infty}_c(\R)$,
$$
\la w^\infty_t,f\ra - \la w_0^\infty,f\ra -\int_0^t
\bigg\{\frac{u\Gamma_R}{2}\, \la w_s^\infty ,\Delta f\ra - 2Ru\sigma\,
\la w_s^\infty (1-w_s^\infty),f\ra\bigg\}\, ds
$$
is a zero-mean martingale with quadratic variation
$$
4R^2 u^2 \int_0^t \la w_s^\infty (1-w_s^\infty),f^2\ra\, ds,
$$
where $\Gamma_R>0$ depends only on $R$ and is defined in (\ref{def Gamma}).

$(ii)$ When $d\geq 2$, $(w_t^\infty)_{t\geq 0}$
is the unique deterministic process
for which, for every $f\in C^{\infty}_c(\R^d)$,
$$
\la w^\infty_t,f\ra = \la w_0^\infty,f\ra +
\int_0^t \bigg\{\frac{u\Gamma_R}{2}\, \la w_s^\infty ,\Delta f\ra - u\sigma
V_R\, \la w_s^\infty (1-w_s^\infty),f\ra\bigg\}\, ds,
$$
where, again, $\Gamma_R>0$ depends only on $R$ and is defined
in (\ref{def Gamma}).
\end{theorem}

In other words, in one space dimension, the limiting process
$(w_t^\infty)_{t\geq 0}$ is a weak solution to the
stochastic partial differential equation
$$
\frac{\partial w}{\partial t} =
\frac{u\Gamma_R}{2}\, \Delta w - 2Ru\sigma w(1-w)
+ 2Ru\sqrt{w(1-w)}\, \dot{\mathcal{W}},
$$
with $w_0=w^0$, and ${\mathcal{W}}$ a space-time white noise.
In dimension $d\geq 2$, on the other hand, the noise term disappears
in the limit and $(w_t^\infty)_{t\geq 0}$ is a weak solution of the
deterministic Fisher-KPP equation
$$
\frac{\partial w}{\partial t} = \frac{u\Gamma_R}{2}\, \Delta w - u\sigma
V_R\, w(1-w), \qquad w_0=w^0.
$$
\begin{rmk}
We  emphasize that we are taking the limit of $\bw^n$, not that of $w^n$.
\end{rmk}
To state the corresponding result for stable radii, we need some more
notation.
We write $V_r(x,y)$ for the volume of $B(x,r)\cap B(y,r)$ and define
$$
\Phi(|z-y|):= \int_{\frac{|z-y|}{2}}^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(y,z)}{V_r}\, dr.
$$
Now set
\begin{equation}\label{generator of stable motion}
\Da f(y) = u \int_{\R^d} \Phi(|z-y|) (f(z)-f(y)) dz.
\end{equation}
We shall check in Lemma~\ref{lemma generator alpha-stable}
that this defines the
generator of a symmetric stable process (that is, it is a constant
multiple of the fractional Laplacian).  Our result for stable radii is
then as follows.
\begin{theorem}\label{th:stable}{\bf (Stable radii)}
Suppose that $\bw^n_0$ converges weakly to some $w^0\in \cM$. Then,
as $n\rightarrow \infty$, the process $(\bw_t^n)_{t\geq 0}$
converges weakly in
$D_{\cM}[0,\infty)$ towards a process $(w_t^\infty)_{t\geq 0}$
with initial value $w^0$.
Furthermore, if $\Da$ denotes the generator of the symmetric $\alpha$-stable
process defined in (\ref{generator of stable motion}), then

$(i)$ When $d=1$, $(w_t^\infty)_{t\geq 0}$ is the unique process for which,
for every $f\in C_c^{\infty}(\R)$,
$$
\la w_t^\infty , f\ra - \la w_0^\infty , f\ra -\int_0^t
\bigg\{\la w_s^\infty, \Da f\ra - \frac{2u\sigma}{\alpha}\,
\la w_s^\infty(1-w_s^\infty),f\ra \bigg\}\, ds
$$
is a zero-mean martingale with quadratic variation
$$
\frac{4u^2}{\alpha-1}\int_0^t  \la w_s^\infty(1-w_s^\infty), f^2\ra\, ds.
$$

$(ii)$ When $d\geq 2$, $(w_t^\infty)_{t\geq 0}$ is the deterministic process for which,
for every $f\in C_c^\infty(\R^d)$,
$$
\la w_t^\infty , f\ra = \la w_0^\infty , f\ra +\int_0^t
\bigg\{\la w_s^\infty, \Da f\ra - \frac{u\sigma V_1}{\alpha}\,
\la w_s^\infty(1-w_s^\infty),f\ra \bigg\}\, ds.
$$
\end{theorem}
\begin{rmk}
Observe from (\ref{generator of stable motion}) that here again, the
deterministic component of the limiting motion is proportional to $u$
and the quadratic variation is proportional to $u^2$ so that $u$ can be thought
of as scaling time
({\em c.f.}~Remark~\ref{timescales}).  Moreover, the limiting process
that we have obtained under this rescaling is a weak solution
to a (stochastic) pde which only differs from that obtained in the fixed radius
case in that the Laplacian has been replaced by the generator of a
symmetric stable process. This is, perhaps, at first sight rather surprising.
The only effect of the large scale events is on the spatial motion of
individuals in the population, and we see no trace of the correlations in
their movement, or of the selection or `genetic drift' acting over large scales,
that we have in the prelimiting model. Notice also that the scaling of $s_n$
(relative to $u_n$) that leads to a nontrivial limit is independent of spatial
dimension. This is in contrast to the case of bounded neighbourhood size considered in
\cite{etheridge/freeman/straulino:2014}.
\end{rmk}
As remarked above, we would obtain the same results under much more general
conditions. For example, in selecting the regions to be affected by events,
not only
could one take more general measures $\mu$ (it is the tail behaviour of
$\mu(dr)$ that we see in our limits), but also reproduction events don't need
to be based on discs.  We anticipate that this robustness will also
be maintained if one replaces our selection mechanism with any other in
which one type is favoured over the other (with appropriate modification
if the strength of selection is `density dependent', that is
the parameter $s_n$ depends on
the local frequencies of the different types in the population),
and it should be clear how to modify our proofs in such cases.

\subsection{Duality}

In the same way as the Wright-Fisher diffusion is dual to a Kingman
coalescent, the neutral SLFV is dual to a system of coalescing random walks
that describe the ancestry of a sample from the population.
In the presence of selection, the Kingman coalescent is replaced by the
branching and coalescing structure known as the {\em ancestral selection
graph} (ASG), \cite{krone/neuhauser:1997, neuhauser/krone:1997}. In
\S\ref{s: existence} we shall see that the SLFVS is dual to a spatial
analogue of the ASG, described in terms of a system
of `branching and coalescing' random walks (although the mechanism of
branching and coalescence differs substantially from that in the classical ASG).
Indeed, this duality will guarantee
uniqueness of the SLFVS.
We shall use knowledge of the limiting forwards in time model to
recover corresponding limiting results for our rescaled branching and
coalescing duals.  These will be stated in
Theorems~\ref{th: conv duals} and \ref{th: conv duals alpha}.
The difficulty with proving these results directly
stems from problems with identifying the limiting coalescence mechanism in one
dimension.  This contrasts with the situation of uniformly bounded neighbourhood
size, considered in \cite{BEV2012} in the neutral case and in
\cite{etheridge/freeman/straulino:2014} in the selective case,
where it is the ability to identify the
limiting behaviour of the (analytically tractable)
coalescent dual that allows us to prove results about
the forwards in time model.

\subsection{Structure of the paper}

The rest of the paper is laid out as follows. In \S\ref{s: existence}
we provide a very direct proof of existence of the SLFVS and prove the
duality with an analogue of the ASG. We also state the counterparts of
Theorems~\ref{th:fixed} and~\ref{th:stable} for the rescaled dual
processes. In \S\ref{heuristics}, we provide heuristic arguments to explain our rescalings.
In \S\ref{s: convergence 1},
we turn to proving Theorem~\ref{th:fixed}, the
scaling limit in the case of fixed radii, and Theorem~\ref{th: conv duals}
which provides the
corresponding result for the rescaled duals. In \S\ref{s: convergence 2}, we
prove Theorems~\ref{th:stable} and~\ref{th: conv duals alpha},
the analogous results for stable radii.
In Appendix~\ref{continuity stable}, we obtain
continuity estimates for the rescaled SLFVS of Section~\ref{s: convergence 2}.
In particular, these rather technical estimates are a key ingredient in
(and a nice complement to) the
proof of Theorem~\ref{th:stable}.

\section{The spatial $\Lambda$-Fleming-Viot process with selection}
\label{s: existence}

\subsection{A dual process of branching and coalescing jump processes}
\label{ss: general dual}

In this subsection, assuming existence of the SLFVS, we identify a
dual system of branching and coalescing lineages. The idea is that we sample a finite collection of
points at time zero. At any given time in
the past, our dual describes the smallest subset of spatial points at which
we must know the distribution of types in the ancestral population, in order
to determine the current distribution of types at our sampled points.
Put another way, it can be understood as the collection of
`potential ancestors' of a sample from the current population.
Only by knowing
the types of all these potential ancestors are we able to extract the
true ancestry
of the sample.

The dynamics of the dual are driven by the same Poisson processes of events
$\Pi^N$ and $\Pi^S$ that drove the forwards in time process. These processes
are reversible and we shall abuse notation by indexing events by
`backwards time' when discussing our dual. We suppose that
at time $0$, `the present', we sample $k$ individuals from locations
$x_1,\ldots ,x_k$ and we write $\xi_s^1,\ldots ,\xi_s^{N_s}$ for
the locations of the $N_s$ `potential ancestors' that make up our
dual at time $s$ before the present.
\begin{defn}[Branching and coalescing dual]
\label{defn of dual}
The branching and coalescing dual process $(\Xi_t)_{t\geq 0}$ is the
$\bigcup_{n\geq 1}E^n$-valued Markov process with dynamics defined as follows.
At each event $(t,x,r,u)\in \Pi^N$:
\begin{enumerate}
\item for each $\xi_{t-}^i\in B(x,r)$, independently mark the corresponding
potential ancestral lineage with probability $u$;
\item if at least one lineage is marked,
all marked lineages disappear and are replaced by a single
potential ancestor, whose location is drawn uniformly at random
from within $B(x,r)$.
\end{enumerate}
At each event $(t,x,r,u)\in \Pi^S$:
\begin{enumerate}
\item for each $\xi_{t-}^i\in B(x,r)$, independently mark the corresponding
potential ancestral lineage with probability $u$;
\item if at least one lineage is marked,
	all marked lineages disappear and are replaced by {\em two} potential
ancestors, whose locations are drawn independently and uniformly from
within $B(x,r)$.
\end{enumerate}
In both cases, if no particles are marked, then nothing happens.
\end{defn}
Since we only consider finitely many initial individuals in the sample,
the integrability condition (\ref{cond existence}) guarantees that the
jump rate in this process is finite and so this description gives rise to
a well-defined process.


Note that the process never becomes extinct,
since any death is accompanied by the birth of at least one new potential
ancestor.

The difficulty that we face in establishing a duality between this system
of branching and coalescing lineages and the SLFVS is that, just as in the
neutral setting, the SLFVS will only be defined, as a function,
Lebesgue a.e.~and so the usual test functions used to establish such dualities
in population genetics, which take the form
$\prod_{i=1}^kw_t(x_i)$ for fixed points $x_1,\ldots ,x_k\in E$,
will not make sense. However, if, instead
of taking deterministic points $x_1,\ldots, x_k$, we take random points,
with a distribution which has a density $\psi$ with respect to
Lebesgue measure on $E^k$,
then the test function is replaced by
$$\int_{E^k}\psi(x_1,\ldots ,x_k)\bigg\{\prod_{j=1}^kw_t(x_j)\bigg\}
dx_1\ldots dx_k,$$
which {\em is} well-defined.

Assuming that the spatial $\Lambda$-Fleming-Viot process with selection exists,
we have the following property.
\begin{prop}\label{prop: dual}
The spatial $\Lambda$-Fleming-Viot process with selection is
dual to the process $(\Xi_t)_{t\geq 0}$ in the sense that
for every $k\in \N$ and $\psi\in C(E^k)\cap L^1(E^k)$, we have
\begin{align}
\E_{w_0}\bigg[\int_{E^k} & \psi(x_1,\ldots,x_k)\bigg\{\prod_{j=1}^k w_t(x_j)\bigg\}\, dx_1\ldots dx_k\bigg] \nonumber\\
& = \int_{E^k} \psi(x_1,\ldots,x_k)\E_{\{x_1,\ldots,x_k\}}\bigg[\prod_{j=1}^{N_t} w_0\big( \xi_t^j\big)\bigg]\, dx_1 \ldots dx_k. \label{dual formula}
\end{align}
\end{prop}
\begin{rmk}
Of course, we are abusing notation here:
the expectations on the left and right of this equation are
taken with respect to different measures. The subscripts on the expectations
are the initial values for the processes on each side.
\end{rmk}
\noindent {\bf Proof of Proposition~\ref{prop: dual}.}

By linearity, there is no loss of generality in supposing that $\psi$ is a
probability density on $E^k$, which we shall think of as the distribution
of $\Xi_0$. Again by linearity, setting $\prod_{j=1}^{N_t}w_0(\xi_t^j)$ to
be zero on the Lebesgue-null set where it is not defined, we then have
$$\E\bigg[\prod_{j=1}^{N_t}w_0\big(\xi_t^j\big)\, \Big|\, \Xi_0\sim\psi\bigg]
 = \int_{E^k} \psi(x_1,\ldots,x_k)\E_{\{x_1,\ldots,x_k\}}\bigg[\prod_{j=1}^{N_t} w_0\big( \xi_t^j\big)\bigg]\, dx_1 \ldots dx_k.$$

Now note that if the distribution of
$\Xi_0$ has a density with respect to Lebesgue measure
on $E^k$, then conditional on $N_s=l$, the distribution of
$\Xi_s$ has a density with respect to Lebesgue measure on $E^{l}$.
Thus, for such random initial conditions for $\Xi_0$, and
any times $0<s<t$, $\E[\prod_{j=1}^{N_s} w_{t-s}(\xi^j_{s})]$ will be
well defined.

To complete the proof, it
suffices to evaluate the generator, $\mathcal{G}$, of $(\Xi_t)_{t\geq 0}$
on an appropriate class of
test functions.
To simplify notation, in the calculations
below, we write $I$ for the set of indices of the lineages in $\Xi_s$

that lie in the region affected by an event. In addition, $|D|$ denotes
the cardinality of the set $D$ and $V_{x,r}$ denotes the volume of the ball
$B(x,r)\subseteq E$.
Fix $w\in \cM_{\lambda}(E\times \{0,1\})$.
Then for any $l\in \N$ and
$x_1,\ldots,x_l\in E$,
\begin{align*}
& \mathcal{G}\bigg(\prod_{j=1}^l w(x_j)\bigg)\\
& = \int_E \int_{(0,\infty)}\int_0^1 \int_{B(x,r)}\frac{1}{V_{x,r}}\prod_{j\in I^c} w(x_j)  \\
& \qquad \qquad  \times \Bigg[\sum_{D\subset I;|D|\geq 1}u^{|D|}(1-u)^{|I\setminus D|}\bigg(w(z)\prod_{i\in D^c}w(x_i) - \prod_{i\in I}w(x_i)\bigg)\Bigg]dz \nu_r(du)\mu(dr)dx\\
&\qquad + \int_E \int_{(0,\infty)}\int_0^1 \int_{B(x,r)^2}\frac{1}{V_{x,r}^2}\prod_{j\in I^c} w(x_j)  \\
& \qquad \quad  \times \Bigg[\sum_{D\subset I;|D|\geq 1}u^{|D|}(1-u)^{|I\setminus D|}\bigg(w(z)w(z')\prod_{i\in D^c}w(x_i) - \prod_{i\in I}w(x_i)\bigg)\Bigg]dzdz' \nu'_r(du)\mu'(dr)dx \\
&= \int_E \int_{(0,\infty)}\int_0^1 \int_{B(x,r)}\frac{1}{V_{x,r}}\prod_{j\in
I^c} w(x_j) \Bigg[w(z)\Bigg(\sum_{D\subset I;|D|\geq 1}u^{|D|}(1-u)^{|I\setminus D|}\prod_{i\in D^c}w(x_i)\Bigg)\\
& \qquad \qquad - \big(1-(1-u)^{|I|}\big)\prod_{i\in I}w(x_i)\Bigg] dz\nu_r(du)\mu(dr)dx\\
& \quad  + \int_E \int_{(0,\infty)}\int_0^1 \int_{B(x,r)^2}\frac{1}{V_{x,r}^2}\prod_{j\in I^c} w(x_j)\Bigg[w(z)w(z')\Bigg(\sum_{D\subset I;|D|\geq 1}u^{|D|}(1-u)^{|I\setminus D|}\prod_{i\in D^c}w(x_i)\Bigg)\\
& \qquad \qquad  - \big(1-(1-u)^{|I|}\big)\prod_{i\in I}w(x_i)\Bigg] dzdz'\nu'_r(du)\mu'(dr)dx.
\end{align*}
On the other hand, the generator $\mathcal{L}$ of the SLFVS
applied to the function
$$
F_\phi (w):= \int_{E^{l}}\phi(x_1,\ldots,x_l) \, \prod_{j=1}^l w(x_j)\, dx_1\ldots dx_l,
$$
is equal to
\begin{align}
&\int_E \int_0^\infty \int_0^1 \int_{B(x,r)}\frac{1}{V_{x,r}}\int_{E^l} \phi(x_1,\ldots,x_l) \prod_{j\in I^c}w(x_j) \bigg[w(z)\prod_{j\in I}\big((1-u)w(x_j)+u\big)\label{def L}\\
& \quad  + (1-w(z))\prod_{j\in I}\big((1-u)w(x_j)\big) - \prod_{j\in I}w(x_j)\bigg]dx_1\ldots dx_l dz\nu_r(du)\mu(dr)dx\nonumber\\
& + \int_E \int_0^\infty \int_0^1 \int_{B(x,r)^2}\frac{1}{V_{x,r}^2}\int_{E^l} \phi(x_1,\ldots,x_l) \prod_{j\in I^c}w(x_j) \bigg[w(z)w(z')\prod_{j\in I}\big((1-u)w(x_j)+u\big)\nonumber\\
& \quad  + (1-w(z)w(z'))\prod_{j\in I}\big((1-u)w(x_j)\big) - \prod_{j\in I}w(x_j)\bigg] dx_1\ldots dx_l dzdz'\nu'(dr)\mu'(dr) dx \nonumber\\
& =: A+B.  \nonumber
\end{align}
Now,
\begin{align*}
A & = \int_E \int_{(0,\infty)} \int_0^1 \int_{B(x,r)}\frac{1}{V_{x,r}}\int_{E^l}\, \phi(x_1,\ldots,x_l) \prod_{j\in I^c}w(x_j) \Bigg[w(z) \sum_{D\subset I;|D|\geq 1}u^{|D|}(1-u)^{|I\setminus D|}\prod_{j\in D^c}w(x_j)\\
& \qquad  - \big(1-(1-u)^{|I|}\big)\prod_{j\in I}w(x_j)\Bigg] dx_1\ldots dx_ldz \nu_r(du)\mu(dr) dx,
\end{align*}
which, by Fubini's Theorem, is equal to the first term of
$ \mathcal{G}\big(\prod_{j=1}^l w(x_j)\big)$ integrated against the
measure $\phi(x_1,\ldots,x_l)dx_1\ldots dx_l$.  In the same way, expanding the
product, we see that the `selection' term in $\mathcal{L}F_\phi$ is equal
to the integral of the second term of
$\mathcal{G}\big(\prod_{j=1}^l w(x_j)\big)$ against the same measure.

For each $0\leq s\leq t$, we now partition on the events $\{N_s=l\}$ and
apply the calculation above with $\phi\in C(E^l)$ the density function of $\xi_s^1,\ldots ,\xi_s^l$
conditional on $N_s=l$, to deduce that for every $t\geq 0$, provided
that the distribution of
$\Xi_0$ has a density with respect to Lebesgue measure,
$$
\frac{d}{ds}\E\Bigg[\prod_{j=1}^{N_{t-s}} w_s(\xi_{t-s}^j)\Bigg] = 0,
$$
(where the expectation is with respect to the joint distribution of $w_{t-s}$
and
$\Xi_s$). This proves the duality between $(w_t)_{t\geq 0}$ and $(\Xi_t)_{t\geq 0}$. \hfill $\Box$

\medskip

Later it will be convenient to use a different set of test functions to
characterise the SLFVS,
namely the set of all functions of the form $F(\la w,f \ra)$,
with $f\in C_c(E)$ and $F\in C^3(\R)$.
To ease notation, we define
\begin{align}
\Theta^+_{x,r,u}(w)& = \ind_{B(x,r)^c}w + \ind_{B(x,r)}((1-u)w+u),\qquad \hbox{and}\nonumber\\
\Theta^-_{x,r,u}(w)& = \ind_{B(x,r)^c}w + \ind_{B(x,r)}(1-u)w. \label{notation Delta}
\end{align}
These quantities describe the value of the process immediately after an event
$(t,x,r,u)$ if the parent is of type $0$ or type $1$ respectively.
The generator of the SLFVS on our new set of test functions then takes the form
\begin{align}
{\mathcal L}F(\la \omega,f\ra) & =
\int_E \int_0^\infty\int_0^1 \int_{B(x,r)}
\frac{1}{V_{x,r}}\, \Big[\omega(y)
F(\la \Theta^+_{x,r,u}(\omega), f \ra)\nonumber \\
& \qquad \quad  + (1-\omega(y))F(\la \Theta^-_{x,r,u}(\omega),f\ra) -
F(\la \omega,f \ra)\Big] dy\,\nu_r(du)\,\mu(dr)\,dx\nonumber \\
& \quad  + \int_E \int_0^\infty\int_0^1  \int_{B(x,r)^2}
\frac{1}{V_{x,r}^2}\, \Big[\omega(y)\omega(z)F\big(\la \Theta^+_{x,r,u}(\omega), f \ra\big)\nonumber \\
& \qquad \quad   + (1-\omega(y)\omega(z))F(\la \Theta^-_{x,r,u}(\omega), f\ra) - F(\la \omega,f \ra)\Big] dy\,dz\,\nu'(dr)\,\mu'(dr)\,dx. \label{other def L}
\end{align}
To see that this corresponds to our previous expression for the generator,
observe first that, by a density argument,
in Proposition~\ref{prop: dual} we could restrict our attention to
functions $\psi$ of the product form $\psi_1(x_1)\cdots \psi_k(x_k)$,
with $\psi_i\in C_c(E)$ for every $i$. The corresponding test function
then reads $\prod_{i=1}^k \la \omega,\psi_i\ra$. But by polarization,
such a function can be written as a linear combination of functions of the
form $\la \omega, f\ra^m$, with $m\in \N$ and $f\in C_c(E)$. An easy
calculation then shows that our two expressions for $\mathcal{L}$
coincide on these test functions.
Either of these two sets of test functions characterizes the law of the
SLFVS (see e.g. Lemma~2.1 in \cite{VW2012} for the first set, while the second
assertion is a standard result on random measures), and so we can use them
interchangeably.
In particular, this
guarantees the following corollary to Proposition~\ref{prop: dual}.
\begin{coro}\label{prop: same dual}
Suppose that $(\omega_t)_{t\geq 0}$ is a Markov process with generator
given by equation~(\ref{other def L}) for every $f\in C_c(E)$ and $F\in C^1(\R)$.
Then $(\omega_t)_{t\geq 0}$ is dual to the system of potential
ancestral lineages $(\Xi_t)_{t\geq 0}$ with parameters $\mu,\mu',\nu,\nu'$.
\end{coro}

Theorems~\ref{th:fixed},~\ref{th:stable}
have counterparts for the corresponding rescaled dual processes.
\begin{theorem}\label{th: conv duals}
For every $n\in \N$, let $(\xi_t^1,\ldots,\xi_t^{N_t})_{t\geq 0}$ be the
process of branching and coalescing jump processes which is dual to
the unscaled process $(w_t)_{t\geq 0}$ with parameters $\mu=\delta_R$,
$\mu'= s_n\delta_R$, $\nu_R=\nu_R'=\delta_{u_n}$, where $s_n=\sigma/n^{2/3}$
and $u_n=u/n^{1/3}$.
Define the rescaled process $(\Xi_t^n)_{t\geq 0}$ so that for every $t\geq 0$,
$$
\Xi^n_t = \big(\xi_t^{n,1},\ldots, \xi_t^{n,N^n_t}\big):= (n^{-1/3}\xi_{nt}^1,\ldots, n^{-1/3}\xi^{N_{nt}}_{nt}).
$$
Then, if $d\geq 2$, as $n\rightarrow \infty$, $(\Xi_t^n)_{t\geq 0}$
converges in distribution
(as a c\`adl\`ag process) to a branching Brownian motion
$(\Xi_t^\infty)_{t\geq 0}$,
in which individuals follow independent Brownian motions with variance
parameter $u\Gamma_R$, which branch at rate $u\sigma V_R$ into two new
particles, started at the location of the parent.
When $d=1$, the corresponding object is a branching and coalescing system
with the same diffusion constant and branching rate, but, in addition,
each pair of particles, independently, also coalesces at
rate $4R^2u^2$ times their local time together.


\end{theorem}
The result for stable radii has the same flavour:
\begin{theorem} \label{th: conv duals alpha}
For every $n\in \N$, let $(\xi_t^1,\ldots, \xi_t^{N_t})_{t\geq 0}$ be the
system of branching and coalescing jump processes which is dual to the
unscaled process $(w_t)_{t\geq 0}$ corresponding to the case of stable radii.
Define the rescaled process $(\Xi_t^n)_{t\geq 0}$ with parameters $u_n=u/n^\gamma$ and
$s_n=\sigma/n^\delta$ in such a way that for every $t\geq 0$,
$$
\Xi^n_t = \big(\xi_t^{n,1}, \ldots, \xi_t^{n,N^n_t}\big) := \big(n^{-\beta}\xi_{nt}^1,\ldots,n^{-\beta}\xi_{nt}^{N_{nt}}\big).
$$
Then, if the initial condition $\Xi^n_0$ converges weakly towards some
$\Xi_0$ as $n\rightarrow \infty$, $(\Xi_t^n)_{t\geq 0}$ converges in distribution
(as a c\`adl\`ag process) to a system $(\Xi_t^\infty)_{t\geq 0}$ of independent
symmetric $\alpha$-stable processes, which branch at rate $u\sigma V_1/\alpha$
into two particles starting at the location of their parent.
The motion of a single particle is described by the generator $\Da$ defined
in (\ref{generator of stable motion}). In addition, when $d=1$, each pair of
particles, independently, coalesces at rate $4u^2/(\alpha-1)$ times
their local time together.
\end{theorem}
We shall prove these results in \S\ref{s: convergence 1} and
\S\ref{s: convergence 2} respectively.


\subsection{Proof of Theorem~\ref{th: existence}}
\label{s: proof existence}

In this section we provide a rather elementary proof of existence of the SLFVS as
the unique solution to a martingale problem.
We include it here as what it lacks in sophistication, it makes up for in
flexibility.

If $E$ has finite volume and $\mu$ and $\mu'$ have finite masses, then
the events fall globally at a finite rate and the SLFVS
is well-defined. To extend to arbitrary measures that satisfy
condition~(\ref{cond existence}) it is convenient to
proceed in two steps:
\begin{itemize}
\item[$(i)$] We show existence when $E$ is relatively compact but
$\mu$ and $\mu'$ are only $\sigma$-finite.
\item[$(ii)$] Given $(i)$, we extend to (an infinite subset of) $\IR^d$ by
proving tightness of a sequence of processes obtained by restricting to
an increasing family of sets $(E_n)_{n\geq 1}$ which exhaust the space.
\end{itemize}


\subsubsection*{Proof of $(i)$.}
Let $E$ be a compact subset of $\R^d$ and let $\mu$, $\mu'$ be
$\sigma$-finite measures on $(0,\infty)$. Let also $(\mu^n)_{n\geq 1}$,
$(\mu^{'n})_{n\geq 1}$ be two sequences of finite measures on $(0,\infty)$,
and $(\{\nu_r^n,\, r>0\})_{n\geq 1}$, $(\{\nu^{'n}_r,\, r>0\})_{n\geq 1}$
two sequencs of probability measures on $[0,1]$ for which, as $n\rightarrow \infty$,
\begin{equation}\label{approx}
\mu^n(dr)\nu_r^n(du) \nearrow \mu(dr)\nu_r(du) \qquad \hbox{and}\qquad \mu^{'n}(dr)\nu^{'n}_r(du) \nearrow \mu'(dr)\nu'_r(du).
\end{equation}
Finally, let us write $(w_t^{(n)})_{t\geq 0}$
for the SLFVS with parameters $\mu^n$,
$\mu^{'n}$, $\nu^n$ and $\nu^{'n}$.

The strategy of the proof is as follows. First we show that the
sequence $((w_t^{(n)})_{t\geq 0})_{n\in \N}$ is tight in the space
$D_{\cM_{\lambda}(E\times \{0,1\})}[0,\infty)$ of all c\`adl\`ag paths
with values in $\cM_{\lambda}(E\times \{0,1\})$. Second,
writing $\mathcal{L}^{(n)}$ for the generator of $(w_t^{(n)})_{t\geq 0}$,
we check that $\mathcal{L}^{(n)}(\la w,f\ra)\rightarrow \mathcal{L}(\la w,f\ra)$
as $n\rightarrow\infty$, uniformly in $w$ for each $f\in C_c(E)$, $F\in C^1(\R)$.
Now if there is a solution to the martingale problem associated with $(\mathcal{L}, \cD(\cL))$,
by Corollary~\ref{prop: same dual} it is dual to the system of branching and coalescing lineages of
Proposition~\ref{prop: dual} and (since by Lemma~1.1 of \cite{VW2012} the test functions of
Proposition~\ref{prop: dual} form a separating class) it is unique.
This uniqueness, plus the uniform (in $w$) convergence of $\cL^{(n)}(\la w,f\ra)$ to
$\cL(\la w,f\ra)$ is enough to apply Theorem~4.8.10 of \cite{EK1986} to deduce that in fact
$(w_t^{(n)})_{t\geq 0}$ converges as $n\rightarrow\infty$ to a
solution to the $(\cL,\cD(\cL))$-martingale problem.

Let us check then tightness and uniform convergence.


\medskip

\noindent
{\em Tightness of $((w_t^{(n)})_{t\geq 0})_{n\geq 1}$.}

First recall that $\cM_{\lambda}(E\times \{0,1\})$,
equipped with the topology of vague convergence, is a compact space
(c.f.~Lemma~1.1 in \cite{VW2012}). Therefore, by the Aldous-Rebolledo
criterion, we only have to show that for every $f\in C_c(E)$ and every
$F\in C^1(\R)$, both the finite variation part and the quadratic variation
part of the real-valued processes $((F(\la w^{(n)}_t, f\ra))_{t\geq 0})_{n\geq 1}$
are tight. Since each $(w_t^{(n)})_{t\geq 0}$
is a pure jump process, the finite variation part of $F(\la w_t^{(n)},f\ra)$ is given by
$$
\Phi^n(t)=\int_0^t  \mathcal{L}^{(n)}F(\la w^{(n)}_s, f\ra)\, ds
$$
and, recalling the definitions
of $\Theta^+_{x,r,u}(w)$ and $\Theta^-_{x,r,u}(w)$
from (\ref{notation Delta}), the quadratic variation is
\begin{align*}
& \Psi^n(t) \\
& = \int_0^t \int_E \int_0^{\infty} \int_0^1 \int_{B(x,r)}\frac{1}{V_{x,r}}\,
\Big\{w^{(n)}_{s-}(y)\big[F(\la \Theta^+_{x,r,u}(w^{(n)}_{s-}),f\ra)-
F(\la w^{(n)}_{s-},f \ra)\big]^2\\
& \qquad \qquad   + (1-w^{(n)}_{s-}(y))\big[F(\la
\Theta^-_{x,r,u}(w^{(n)}_{-s}),f\ra) - F(\la w^{(n)}_{s-},f \ra)\big]^2\Big\}\,
 dy\,\nu^n_r(du)\,\mu^n(dr)\,dx\, ds \\
& \quad + \int_0^t \int_E \int_0^{\infty} \int_0^1 \int_{B(x,r)^2}
\frac{1}{V_{x,r}^2}\, \Big\{w^{(n)}_{s-}(y)w^{(n)}_{s-}(z)
\big[F(\la \Theta^+_{x,r,u}(w^{(n)}_{s-}),f\ra) - F(\la w^{(n)}_{s-},f \ra)
\big]^2  \\ & \quad  + (1-w^{(n)}_{s-}(y)w^{(n)}_{s-}(z))
\big[F(\la \Theta^-_{x,r,u}(w^{(n)}_{s-}),f\ra) - F(\la w^{(n)}_{s-},f \ra)
\big]^2\Big\}\, dy\,dz\,\nu^{'n}(du)\,\mu^{'n}(dr)\,dx\,ds,
\end{align*}
where, as before, $V_{x,r}$ is the volume of $B(x,r)$. Using the expression
for $\mathcal{L}^{(n)}$ given in (\ref{other def L}), a Taylor expansion of
$F$ (of class $C^1$), and the fact that the increments of $\la w^{(n)},f\ra$
during an event are proportional to $u$, we obtain that for every
$w\in \cM_{\lambda}(E\times \{0,1\})$,
\begin{align}
\label{bound on LN}
\big|\mathcal{L}^{(n)}F(\la w, f\ra)\big| & \leq  \|F'\|\|f\|
\bigg(\int_E \int_0^\infty \int_0^1 \, ur^d \nu^n_r(du) \mu^n(dr)dx+
\int_E \int_0^\infty \int_0^1 \, ur^d \nu^{'n}_r(du)\mu^{'n}(dr)dx\bigg) \\
& \leq C \bigg( \int_0^\infty \int_0^1 \, ur^d \nu_r(du)\mu(dr) + \int_0^\infty \int_0^1 \, ur^d\nu'_r(du)\mu'(dr)\bigg),
\end{align}
where we have used that $E$ has finite volume and, by assumption,
$\mu^n(dr)\nu_r^n(du)\nearrow \mu(dr)\nu_r(du)$ (and the corresponding statement with primes).
By condition (\ref{cond existence}), the expression on
the right-hand side is finite (and independent of $w$),
and so we conclude that for every $\e$, there exists $\delta>0$ such that
for every $T>0$
$$
\limsup_{n\rightarrow \infty}\ \P\big[\omega'(\Phi^n,\delta,T)>\e\big]\leq \e,
$$
where $\omega'$ is the Skorokhod modulus of continuity defined by
$$
\omega'(f,\delta,T) = \inf_{\{t_i\}}\max_i\sup_{s,t\in [t_{i-1},t_i)}|f(s)-f(t)|
$$
and the infimum is taken over all finite partitions of $[0,T]$ such
that $t_i-t_{i-1}>\delta$ for every $i$. The sequence
$((\Phi^n(t))_{t\geq 0})_{n\geq 1}$ is thus tight.


Similarly, the increments of $\Psi^n(t)$ are bounded
by $\|F'\|\|f\|\, u^2 r^d$. But $u^2\leq u$ and so
the same reasoning shows that the sequence $((\Psi^n(t))_{t\geq 0})_{n\geq 1}$
is also tight.


Tightness of
$((w_t^{(n)})_{t\geq 0})_{n\geq 1}$
in $D_{\cM_{\lambda}(E\times \{0,1\})}[0,\infty)$ now follows from the Aldous-Rebolledo
criterion.

\medskip

\noindent
{\em Convergence of $\cL^{(n)}$ to $\cL$.}

Using (\ref{approx}) and the Dominated Convergence Theorem,
we see that $\mathcal{L}^{(n)}$ converges to the operator
$\mathcal{L}$ given by
\begin{align}
\mathcal{L}F(\la w,f\ra)& = \int_E  \int_0^\infty\int_0^1  \int_{B(x,r)}\frac{1}{V_{x,r}}\, \Big[w(y)F(\la \Theta^+_{x,r,u}(w), f \ra) + (1-w(y))F(\la \Theta^-_{x,r,u}(w),f\ra)\nonumber \\
& \qquad \qquad \qquad \qquad \qquad \qquad  - F(\la w,f \ra)\Big] \, dy_, \nu_r(du)\,\mu(dr)\,dx\nonumber \\
& \quad  + \int_E  \int_0^\infty\int_0^1  \int_{B(x,r)^2}\frac{1}{V_{x,r}^2}\, \Big[w(y)w(z)F\big(\la \Theta^+_{x,r,u}(w), f \ra\big)\nonumber \\
& \qquad \quad + (1-w(y)w(z))F(\la \Theta^-_{x,r,u}(w), f\ra) - F(\la w,f \ra)\Big]\, dy\,dz\,\nu'_r(du)\,\mu'(dr)\,dx. \label{limit gen}
\end{align}
Applying the estimate~(\ref{bound on LN}) with the measures $\mu(dr)\nu_r(du)-\mu^n(r)\nu^n_r(du)$
and $\mu'(dr)\nu'_r(du)-\mu'^n(r)\nu'^n_r(du)$, we see that the convergence is uniform, as required,
and the proof of $(i)$ is complete.


\subsubsection*{Proof of $(ii)$.}

The proof of $(ii)$ follows exactly the same pattern, but now the task of bounding the integrals
defining $\Phi^n$ and $\Psi^n$ becomes more delicate. The resolution is to exploit the fact
that $f$ has compact support $S_f$.


Fix $\mu,\mu', \nu$ and $\nu'$ satisfying (\ref{cond existence}) and
let $\{E_n\}_{n\geq 1}$ be a sequence of compact sets increasing to $E$. 
We embed each $\cM_{\lambda}(E_n \times\{0,1\})$ into
$\cM_{\lambda}(\R^d\times \{0,1\})$ by setting $w(x)\equiv 0$ outside $E_n$.
In this way, we have a sequence of SLFVS processes,

which by an abuse of notation we also denote $(w_t^{(n)})_{t\geq 0}$,
with generators $\cL_{(n)}$ given by


\begin{align}
\mathcal{L}_{(n)}F(\la w,f\ra)& = \int_0^\infty\int_{(S_f+B(0,r))\cap E_n}
\int_0^1  \int_{B(x,r)}\frac{1}{V_{x,r}}
\Big[w(y)F(\la \Theta^+_{x,r,u}(w),f\ra) \nonumber\\
& \qquad \qquad     + (1-w(y))F(\la \Theta^-_{x,r,u}(w),f\ra) - F(\la w,f\ra)\Big]\, dy\,\nu_r(du)\,dx\,\mu(dr) \nonumber\\
&\quad +
\int_0^\infty\int_{(S_f+B(0,r))\cap E_n} \int_0^1  \int_{B(x,r)^2}
\frac{1}{V_{x,r}}\Big[w(y)w(z)F\big(\la \Theta^+_{x,r,u}(w),f\ra\big)
\nonumber \\
& \qquad + (1-w(y)w(z))F\big(\la \Theta^-_{x,r,u}(w),f\ra\big) -
F(\la w,f\ra)\Big]\, dy\,dz\,\nu'_r(du)\,dx\,\mu'(dr). \label{generator}
\end{align}
The key observation is that
\begin{equation}
\label{bound for increments}
|\la \ind_{B(x,r)}w,f\ra|\leq \|f\| \mathrm{Vol}(S_f\cap B(x,r))\leq C_1\|f\| (r^d\wedge 1),
\end{equation}
and
\begin{equation}
\label{bound for integrating increments}
\mathrm{Vol}\{x:S_f\cap B(x,r)\neq\emptyset\}\leq C_2 (r^d \vee 1),
\end{equation}
where $C_1$ and $C_2$ are independent of $r$ and depend only on the support of $f$.
Moreover, the estimate~(\ref{bound for increments})
is uniform in $w$ and, in particular, the same bound holds if we replace
$w$ by $1-w$.

To see how to apply this, consider the neutral part of~(\ref{generator}). We split
the integral over $(0,\infty)$ at some radius $R_0>1$. We have that


\begin{align}
\bigg|&\int_{R_0}^\infty \int_{(S_f+B(0,r))\cap E_n}
\int_0^1\int_{B(x,r)}\frac{1}{V_{x,r}}\Big[w(y)F(\la \Theta^+_{x,r,u}(w),f\ra) \nonumber \\
& \qquad \qquad\qquad\qquad+ (1-w(y))F(\la \Theta^-_{x,r,u}(w),f\ra) -
F(\la w,f\ra)\Big]\, dy\,\nu_r(du)\,dx\,\mu(dr)\bigg|\nonumber \\
& \leq \|F'\|\|f\| \int_{R_0}^\infty \int_{(S_f+B(0,r))\cap E_n}
\int_0^1 \, u \mathrm{Vol}(B(x,r)\cap S_f) \, \nu_r(du)\,dx\,\mu(dr)\nonumber\\
& \leq C\|F'\|\|f\| \mathrm{Vol}(S_f)\int_{R_0}^\infty \int_0^1\, ur^d\, \nu_r(du)\mu(dr). \label{vanishing bound}
\end{align}


To control the second part of the integral corresponding to the neutral part,
notice that a simple estimate using the fact that the corresponding events
have radius bounded above by $R_0$, yields
\begin{align*}
\bigg|&\int_0^{R_0} \int_{(S_f+B(0,r))\cap E_n}
\int_0^1\int_{B(x,r)}\frac{1}{V_{x,r}}\Big[w(y)F(\la \Theta^+_{x,r,u}(w),f\ra) \nonumber \\
& \qquad \qquad\qquad\qquad+ (1-w(y))F(\la \Theta^-_{x,r,u}(w),f\ra) -
F(\la w,f\ra)\Big]\, dy\,\nu_r(du)\,dx\,\mu(dr)\bigg|\nonumber \\
& \leq \|F'\|\|f\| \int_0^{R_0} \int_{(S_f+B(0,r))\cap E_n}
\int_0^1 \, u \mathrm{Vol}(B(x,r)\cap S_f) \, \nu_r(du)\,dx\,\mu(dr)\nonumber\\
& \leq C R_0^{d+1}\|F'\|\|f\|.
\end{align*}
Exactly the same arguments control the selection part of the generator $\cL_{(n)}$ and,
combining with the above, this gives the tightness of the finite variation parts of the
processes. As in $(i)$, since $u^2<u$, tightness of the quadratic variation parts follows
easily and the Aldous-Rebolledo criterion yields tightness of the sequence of
processes $((w_t^{(n)})_{t\geq 0})_{n\geq 1}$.

To check uniform convergence of $\cL_{(n)}F(\la w,f\ra)$ to $\cL F(\la w,f\ra)$
(which takes the form of $\cL_{(n)}$ but with $E_n$ replaced by $E$ in the domain of
integration), notice that
by Condition~(\ref{cond existence}), by taking $R_0$ sufficiently large, the right
hand side of~(\ref{vanishing bound}) can be made arbitrarily small, independent of $w$,
and this is enough to ensure that
\begin{align}
\bigg|&\int_{0}^\infty \int_{(S_f+B(0,r))\cap E_n^c}
\int_0^1\int_{B(x,r)}\frac{1}{V_{x,r}}\Big[w(y)F(\la \Theta^+_{x,r,u}(w),f\ra) \nonumber \\
& \qquad \qquad\qquad\qquad+ (1-w(y))F(\la \Theta^-_{x,r,u}(w),f\ra) -
F(\la w,f\ra)\Big]\, dy\,\nu_r(du)\,dx\,\mu(dr)\bigg|\nonumber \\
\end{align}
converges to zero, uniformly in $w$ as $n\rightarrow\infty$.


By the same duality argument (based on Corollary~\ref{prop: same dual}) that we used in $(i)$,
the martingale problem associated with $\mathcal{L}$ has at most one solution,
and so Theorem~4.8.10 in \cite{EK1986} yields the desired convergence.
Theorem~\ref{th: existence} is proved. \hfill $\Box$

\section{Heuristics}
\label{heuristics}

In this section we provide an informal
justification of our choices for the parameters $\beta$, $\gamma$ and
$\delta$.
To understand why they lead to nontrivial limits, it is convenient
to think about our branching and coalescing dual process.

First consider the case of fixed (or more generally bounded variance) radius
events.
Ignoring for a moment the selective (branching) events, a single ancestral
lineage in the rescaled dual makes mean zero, finite variance, jumps of size of
order $1/n^\beta$ at rate proportional to $nu_n=n^{1-\gamma}$. Thus, provided
that $1-\gamma=2\beta$, its spatial motion will converge to a Brownian motion
as $n\rightarrow\infty$.

Now consider what happens at a selective event. The two new lineages
are born at a separation of order $1/n^\beta$. If we are to `see' the event,
they must move apart to a separation of order one before (perhaps)
coalescing. The number of excursions they must make away from the region in which they
can both be affected by an event (and thus coalesce) before we can expect to
see such a `long' excursion is order $1$ in $d\geq 3$, order $\log n$ in $d=2$
and order $n^{\beta}$ in $d=1$. On the other hand, when they are sufficiently close
together that they can be hit by the same event, given that one of them jumps,
there is a probability of order $u_n=u/n^\gamma$ that the other one is
affected by the same event and so they coalesce. So the number of times they come close
to one another
before they coalesce is order $n^\gamma$. Thus, in the limit as
$n\rightarrow\infty$, for each branching event
in the dual, in dimensions at least $2$, the probability that there is a long excursion
before coalescence (and so we `see' the event) tends to
one. Moreover, the same argument tells us that we will never see coalescence
of any other lineages in our system. In one dimension, we can expect to see both branching and
coalescence provided that the number of excursions we expect to wait before
seeing a coalescence and the number we expect to wait before the lineages
escape to a distance of order one, that is
$1/u_n$ and $n^{\beta}$, are comparable. This gives $\beta=\gamma$ and, combining
with the condition $1-\gamma=2\beta$ above, we find $\beta=\gamma=1/3$.
Finally, selection events occur
at a rate proportional to $nu_ns_n$ in the rescaled process, and so
we choose $\delta=2/3$ to make this order one.

We now turn to the stable case. As before, we first consider the motion of
a single rescaled lineage and we see that if we choose $nu_n=n^{\alpha\beta}$,
then in the limit as $n\rightarrow\infty$ its motion will converge to a
symmetric stable process.
Now consider selection. Since $u_n\rightarrow 0$ as $n\rightarrow\infty$,
although it is now the case that two lineages can always be affected
by the same event, `most of the time' they will not and the motions are
almost independent. Moreover, since `small' events are so much more
frequent than `big' events, selection events are almost always `small' and,
moreover, lineages only have a realistic chance of
coalescing when they are close together. We now use the same argument as
before. The number of excursions away from each other before they are
`visible' under our rescaling is order $1$ in $d\geq 2$ and order $n^{(\alpha -1)\beta}$
in $d=1$. Equating this to the number of visits together before we expect
to see a coalescence event yields $\gamma=(\alpha -1)\beta$. In order
to see any selection events at all, we need $nu_ns_n$ to be order one,
so $1-\gamma-\delta=0$. We now have three equations in three unknowns (in
one dimension) and solving gives the values in equation~(\ref{def constants}).

\begin{rmk}
\label{timescales}
At first sight, these scalings do not perhaps look altogether natural.
The reason for this is that the `timescale' of the SLFV process is not
one of generations. Suppose that one thinks of a generation as being
the time that it takes for an `individual' in
the SLFV to be affected by a reproduction event. Then a generation is
proportional to $1/u$ units of SLFV time. In the `generation timescale', we
are speeding up time by a factor of $nu_n$ and then we recognise the scaling
in the fixed radius case as exactly the diffusive rescaling
and the scaling in the stable case as its
natural analogue when we have long-range dispersal.

Note also that
by choosing $1-\gamma=\alpha\beta$ and $1-\gamma-\delta =0$ but with $\gamma>(\alpha -1)\beta$,
we could eliminate the coalescence term in one dimension, corresponding to removing the noise
term in the forwards in time description of allele frequencies.
\end{rmk}

It turns out to be highly non-trivial to turn these heuristics into a
rigorous proof and so, instead, we work with the forwards in time
model and deduce convergence of the dual processes as a corollary.

\section{Convergence of the rescaled SLFVS and its dual - the fixed radius case} \label{s: convergence 1}

In this section, we prove Theorem~\ref{th:fixed} and, from it, deduce
Theorem~\ref{th: conv duals}. From now on, we shall be concerned with
$E=\IR^d$ and so there will be no ambiguity in writing $V_r$ for the
volume of a ball of radius $r$.
Recall our notation $V_r(x,y)$ for the volume of the intersection
$B(x,r)\cap B(y,r)$ and that in the fixed radius case, all reproduction
events have the same radius $R>0$.

\subsection{Proof of Theorem~\ref{th:fixed}.}

The proof proceeds in the usual way.  First we show that the sequence of
rescaled processes is tight, then we identify the possible limit points
and finally uniqueness of the limit point guarantees that the whole sequence
in fact converges.

\medskip
{\bf 1) Tightness.}

\noindent
Recall the notation $\Theta^+$ and $\Theta^-$ from~(\ref{notation Delta}).
Before scaling, the generator of the SLFVS with reproduction events of
fixed radius $R$, and parameters $u_n$, $s_n$,
acting on functions of the form
$F(\la w, \varphi\ra)$ with $\varphi\in C_c(\R^d)$ and $F\in C^1(\R)$,
takes the form
\begin{align*}
{\cL} F(\la w,\varphi\ra)& = \int_{\R^d}\int_{B(x,R)^2}\frac{1}{V_R^2}\, \Big\{w(y)(1+s_nw(z))\big[ F(\la \Theta^+_{x,R,u_n}(w),\varphi \ra) - F(\la w,\varphi\ra)\big] \\
 &\qquad + (1-w(y)+s_n(1-w(y)w(z)))\big[ F(\la \Theta^-_{x,R,u_n}(w),\varphi \ra) - F(\la w,\varphi\ra)\big]\Big\}dydzdx.
\end{align*}
To identify the generator of the rescaled process
$(w_{nt}(n^{1/3}\cdot))_{t\geq 0}$,
observe that for a reproduction event with centre $x$, the value of $w_t^n(y)$
will change iff $n^{1/3}y\in B(x,R)$. Since we also accelerate time by a
factor $n$, the generator of $(w_t^n)_{t\geq 0}$ is given by
\begin{align}
&{\cL}^n F(\la w,\varphi\ra) \nonumber\\
& = n\int_{\R^d}\int_{B(x,R)^2}\frac{1}{V_R^2}\, \Big\{w(n^{-1/3}y)\big(1+s_n w(n^{-1/3}z))\big[ F(\la \Theta^+_{n^{-1/3}x,n^{-1/3}R,u_n}(w),\varphi \ra) \nonumber\\
&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad   - F(\la w,\varphi\ra)\big] \nonumber\\
& \quad + (1-w(n^{-1/3}y)+s_n(1-w(n^{-1/3}y)w(n^{-1/3}z)))\big[ F(\la \Theta^-_{n^{-1/3}x,n^{-1/3}R,u_n}(w),\varphi \ra) \nonumber\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad   - F(\la w,\varphi\ra)\big]\Big\}dydz dx \nonumber \\
& = n^{1+\frac{d}{3}}\int_{\R^d} \Big\{\bw(x)(1+s_n\bw(x))\big[F(\la \Theta^+_{x,n^{-1/3}R,u_n}(w),\varphi\ra)-F(\la w,\varphi\ra)\big] \nonumber \\
& \qquad \qquad  +\big(1-\bw(x)+s_n(1-\bw(x)^2)\big)\big[F(\la \Theta^-_{x,n^{-1/3}R,u_n}(w),\varphi\ra)-F(\la w,\varphi\ra)\big]\Big\}dx, \label{generator wn}
\end{align}
where we have performed several changes of variables and we have set
$B_n(x)=B(x,n^{-1/3}R)$. The generator of the process
$(\bw^n_t)_{t\geq 0}$ in which we are interested, acting on $F(\la \bw,f\ra)$ for $f\in C_c(\R^d)$
and $F\in C^1(\R)$, can be recovered by evaluating the expression above for
the generator of $(w_t^n)_{t\geq 0}$ with $\varphi$ of the form
\begin{equation}\label{def varphif}
\varphi_f(x)=\frac{n^{d/3}}{V_R}\int_{B_n(x)} f(y)dy,
\end{equation}
and observing that, by Fubini's Theorem,
$$
\la w^n, \varphi_f\ra = \int_{\R^d}\int_{\R^d}\, w^n(y)\frac{n^{d/3}}{V_R}f(z) \ind_{\{|z-y|\leq n^{-1/3}R\}} dydz = \la \bw^n,f\ra.
$$

\begin{rmk}\label{relation f-varphif}
Assuming that $f$ is twice continuously differentiable, and
writing $S_f$ for its compact support, we have
\begin{align*}
\varphi_f(x) & = f(x) + \frac{n^{d/3}}{V_R}\int_{B_n(x)}(y-x).\nabla f(x)dy + \mathcal{O}\bigg(\frac{n^{d/3}}{V_R}\int_{B_n(x)}\|x-y\|^2dy\bigg)\ind_{\{B_n(x)\cap S_f \neq \emptyset\}} \\
& = f(x) + \mathcal{O}(n^{-2/3})\ind_{\{B_n(x)\cap S_f \neq \emptyset\}},
\end{align*}
where the error term is bounded uniformly in $x$. Consequently, we can write
\begin{equation}\label{approximation}
\la \bw^n ,f\ra = \la w^n,\varphi_f \ra = \la w^n, f\ra + \mathcal{O}(n^{-2/3}).
\end{equation}
We shall sometimes use this approximation to simplify our computations.
\end{rmk}

First we must show that the sequence of processes $((\bw_t^n)_{t\geq 0})_{n\geq 1}$ is
tight. Since the state space of the processes is compact (in the topology
of vague convergence), as in Section~\ref{s: proof existence}, we use the
Aldous-Rebolledo criterion to reduce the problem to tightness of
the finite and quadratic variation parts of the sequences
$(F(\la w_t^n,f\ra))_{t\geq 0}$ for every $F\in C^3(\R)$ and $f\in C^2_c(\R^d)$.
Fix such $F$ and $f$. Since $(w_t^n)_{t\geq 0}$ is a jump process,
it is a simple matter
to identify the finite variation part, $(\Phi^n(t))_{t\geq 0}$,
of $(F(\la \bw_t^n,f\ra))_{t\geq 0}$.
\begin{equation}\label{n-finite variation}
\Phi^n(t) = \int_0^t {\cL}^n F(\la w^n_s,\varphi_f\ra)\, ds
\end{equation}
and its quadratic variation $(\Psi^n(t))_{t\geq 0}$ is given by
\begin{align*}
& n^{1+\frac{d}{3}}\int_0^t \int_{\R^d} \Big\{\bw^n_s(x)(1+s_n\bw^n_s(x))\big[F(\la \Theta^+_{x,n^{-1/3}R,u_n}(w^n_s),\varphi_f\ra)-F(\la w^n_s,\varphi_f\ra)\big]^2 \\
& \quad +\big(1-\bw^n_s(x)+s_n(1-\bw^n_s(x)^2)\big)\big[F(\la \Theta^-_{x,n^{-1/3}R,u_n}(w_s^n),\varphi_f\ra)-F(\la w^n_s,\varphi_f\ra)\big]^2\Big\}dxds.
\end{align*}

Note that
\begin{align*}
\la \Theta^+_{x,n^{-1/3}R,u_n}(w),\varphi_f\ra - \la w,\varphi_f\ra & = u_n\, \la \ind_{B_n(x)}(1-w),\varphi_f \ra \\
\la \Theta^-_{x,n^{-1/3}R,u_n}(w),\varphi_f\ra - \la w,\varphi_f\ra & = -u_n\, \la \ind_{B_n(x)}w,\varphi_f \ra,
\end{align*}
so that both increments are of the order of $u_n n^{-d/3}$. Moreover we can use the
bounds~(\ref{bound for increments}) and~(\ref{bound for integrating increments}) to control their
integrals.

Using this observation, we first show that
$|\Phi^n(t)|$ is bounded by a constant independent of $n$ and $w$. To this end, we
write it as the sum of a neutral term and a selective term and perform a Taylor expansion of
$F$ (truncating at second order in the neutral term and at first order in the selective term).
This yields
$$
\Phi^n(t) = \int_0^t \big(A_n(s)+B_n(s)+C_n(s)+D_n(s)+E_n(s)\big)\, ds,
$$
where
\begin{align*}
A_n(s) &= u_nn^{1+\frac{d}{3}}F'(\la \bw^n_s,f\ra)\int_{\R^d}  \Big[\bw^n_s(x)\la \ind_{B_n(x)}(1-w^n_s),\varphi_f\ra \\
& \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad\qquad \qquad - (1-\bw^n_s(x))\la \ind_{B_n(x)}w^n_s,\varphi_f\ra\Big] dx, \\
B_n(s) &= u_n^2 n^{1+\frac{d}{3}}\frac{F''(\la \bw^n_s,f\ra)}{2}\int_{\R^d} \Big[\bw^n_s(x)\la \ind_{B_n(x)}(1-w^n_s),\varphi_f\ra^2 \\
& \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad\qquad \qquad + (1-\bw^n_s(x)) \la \ind_{B_n(x)}w^n_s,\varphi_f\ra^2 \Big] dx , \\
C_n(s) & \leq \mathcal{C}n^{1+\frac{d}{3}} \int_{\R^d}  \big(u_n \mathrm{Vol}(B_n(x))\big)^3 \ind_{\{B_n(x)\cap S_f\neq \emptyset\}} dx , \\
D_n(s) & = u_ns_nn^{1+\frac{d}{3}} F'(\la \bw^n_s,f\ra)\int_{\R^d} \Big[\bw^n_s(x)^2\la \ind_{B_n(x)}(1-w^n_s),\varphi_f\ra \\
& \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad- (1-\bw^n_s(x)^2)\la \ind_{B_n(x)}w^n_s,\varphi_f\ra\Big]dx ,\\
E_n(s) & \leq \mathcal{C'} n^{1+\frac{d}{3}}s_n u_n^2 \int_{\R^d}  \mathrm{Vol}(B_n(x))^2 \ind_{\{B_n(x)\cap S_f\neq \emptyset\}}dx.
\end{align*}
To control these expressions, one takes a Taylor expansion of $\varphi_f$. We illustrate with
the term $A_n(s)$.
In fact, in identifying the limiting process we shall need a precise expression
for the limit of $A_n(s)$ and so we perform the expansion slightly more carefully than would
be required to simply conclude boundedness.

Let us write $D\varphi_f$ for the vector of first derivatives of $\varphi_f$ and $H\varphi_f$ for
the corresponding Hessian ($H\varphi_f=DD\varphi_f$). Then
\begin{align*}
A_n(s)& = u_nn^{1+\frac{d}{3}}F'(\la \bw^n_s,f\ra)\int_{\R^d}  \Big[\bw^n_s(x)\la \ind_{B_n(x)},\varphi_f\ra - \la \ind_{B_n(x)}w^n_s,\varphi_f\ra\Big]dx\\
& = u_nn^{1+\frac{d}{3}}F'(\la \bw^n_s,f\ra) \int_{\R^d} \frac{n^{d/3}}{V_R}\int \int \ind_{\{|y-x|\leq n^{-1/3}R\}}\ind_{\{|z-x|\leq n^{-1/3}R\}} w^n_s(y)(\varphi_f(z)-\varphi_f(y)) dz dydx  \\
& = u_nn^{1+\frac{d}{3}}F'(\la \bw^n_s,f\ra) \int_{\R^d} \frac{n^{d/3}}{V_R}\int_{\R^d} \int_{\R^d} \ind_{\{|y-x|\leq n^{-1/3}R\}}\ind_{\{|z-x|\leq n^{-1/3}R\}} w^n_s(y)\\
& \qquad \qquad \times \big[D\varphi_f(y)(z-y) +
\frac{1}{2}(z-y)H\varphi_f(y)(z-y)+\mathcal{O}(|z-y|^3)\ind_{\{y\in S_f\}}\big]dz  dy dx.
\end{align*}
Consider the first term on the right. Integrating first with respect to $x$ (using Fubini's Theorem)
this term is
\begin{equation*}
\frac{u_nn^{1+\frac{2d}{3}}}{V_R} F'(\la \bw^n_s,f\ra) \int_{\R^d}  w^n_s(y) \int_{\R^d}
\mathrm{Vol}(B_n(y)\cap B_n(z))D\varphi_f(y)(z-y) dz dy,
\end{equation*}
and since $\mathrm{Vol}(B_n(y)\cap B_n(z))$ is a function of $\|z-y\|$ alone, the integrand is
antisymmetric as a function of $z-y$ and so the integral with respect to $z$ vanishes.

Similary, the integrals corresponding to the off-diagonal terms in the Hessian will vanish, leaving
\begin{equation*}
u_nn^{1+\frac{d}{3}}F'(\la \bw^n_s,f\ra) \int_{\R^d}
\frac{n^{d/3}}{V_R}\int_{\R^d} \int_{\R^d}
\ind_{\{|y-x|\leq n^{-1/3}R\}}\ind_{\{|z-x|\leq n^{-1/3}R\}} w^n_s(y)
\frac{1}{2}\sum_{i=1}^d(z_i-y_i)^2\frac{\partial^2}{\partial y_i^2}\varphi_f(y)dydzdx
\end{equation*}
plus a lower order term. Now observe that (from Remark~\ref{relation f-varphif})
$$\frac{\partial^2}{\partial y_i^2}\varphi_f(y)
=\varphi_{\frac{\partial^2 f}{\partial y_i^2}}(y)
=\frac{\partial^2f}{\partial y_i^2}(y)+\cO(n^{-2/3})\ind_{\{B_n(x)\cap S_f\neq \emptyset\}}$$
which yields
\begin{align}
A_n(s) &=
u_nn^{1+\frac{d}{3}}F'(\la \bw^n_s,f\ra) \int_{\R^d}
\frac{n^{d/3}}{V_R}\int_{\R^d} \int_{\R^d}
\ind_{\{|y-x|\leq n^{-1/3}R\}}\ind_{\{|z-x|\leq n^{-1/3}R\}} w^n_s(y)\nonumber
\\ &\qquad\qquad\qquad\qquad\qquad\qquad
\times\frac{1}{2}\sum_{i=1}^d(z_i-y_i)^2\frac{\partial^2f}{\partial y_i^2}(y)dydzdx
\nonumber
\\ &\qquad
+\cO(n^{-2/3})n^{2/3(1+d)}
\int_{\R^d}\int_{\R^d}\int_{\R^d}
\ind_{\{|y-x|\leq n^{-1/3}R\}}\ind_{\{|z-x|\leq n^{-1/3}R\}}
\|z-y\|^2\ind_{\{B_n(x)\cap S_f\neq \emptyset\}}\nonumber\\
&=
\frac{un^{\frac{2}{3}(1+d)}}{2V_R}F'(\la \bw^n_s,f\ra) \int_{\R^d}
\int_{B_n(x)^2}\, w^n(y) \sum_{i=1}^d (z_i-y_i)^2 \frac{\partial^2 f}{\partial y_i^2}(y)dydz dx + \mathcal{O}(n^{-2/3})\nonumber\\
& = \frac{u\Gamma_R}{2} F'(\la \bw^n_s,f\ra) \int_{\R^d}
\, w^n(y) \Delta f(y) dy + \mathcal{O}(n^{-2/3})\nonumber\\
& = \frac{u\Gamma_R}{2} F'(\la \bw^n_s,f\ra)
\, \la \bw^n,\Delta f\ra + \mathcal{O}(n^{-2/3}). 
\label{careful form of AN}
\end{align}
In particular, we conclude that $|A_n(s)|\leq \mathcal{C}_A$ for
every time $s$ and every large $n$.

Very similar arguments allow us to control the other terms:
$$
|B_n(s)| \leq  \frac{u_n^2n^{1+\frac{d}{3}}}{2}\, |F''(\la \bw^n_s,f \ra)|\int  \, 2\mathrm{Vol}(B_n(x))^2 \ind_{\{x\in S_f\}} \|f\|^2 dx \leq  \mathcal{C}_B\, n^{\frac{1-d}{3}},
$$
and, again by the same arguments,
$$
|C_n(s)| \leq \mathcal{C}_C\, n^{-\frac{2d}{3}}, \quad |D_n(s)| \leq \mathcal{C}_D \quad \hbox{and}\quad |E_n(s)| \leq \mathcal{C}_E\, n^{-\frac{1+d}{3}}.
$$
Consequently, for every $s<t$ we have
$$
|\Phi^n(t)-\Phi^n(s)| \leq \big(\mathcal{C}_A + \mathcal{C}_B\, n^{\frac{1-d}{3}} + \mathcal{C}_C\, n^{-\frac{2d}{3}} + \mathcal{C}_D + \mathcal{C}_E\, n^{-\frac{1+d}{3}} \big) (t-s),
$$
which shows that the finite variation part of $((F(\la \bw_t^n,f \ra))_{t\geq 0})_{n\geq 1}$ is tight.
(In fact, its modulus of continuity is uniformly bounded
and so we actually have tightness in the topology of
uniform convergence over compact sets).

Similarly, an elementary application of~(\ref{bound for increments}) yields
$$
\big[F(\la \Theta^{\pm}_{x,n^{-1/3}R,u_n}(w^n_s),\varphi_f\ra)-F(\la w^n_s,\varphi_f\ra)\big]^2 \leq \mathcal{C}''\|F'\|^2 \|f\|^2 u_n^2 \hbox{Vol}(B_n(x))^2 \ind_{\{B_n(x)\cap S_f\neq \emptyset\}}.
$$
Notice that this bound is independent of the value of $w_s^n$.
Substituting into the definition of $\Psi^n(t)$, we obtain that for every $s<t$,
$$
|\Psi^n(t)-\Psi^n(s)|\leq \mathcal{C}_F\, n^{\frac{1-d}{3}}(t-s),
$$
and the sequence of quadratic variations of $((F(\la \bw_t^n,f\ra))_{t\geq 0})_{n\geq 1}$ is not only tight,
but also when $d\geq 2$ it tends to $0$ uniformly over compact time intervals.

Combining these results with the Aldous-Rebolledo criterion, we
conclude that $((\bw_t^n)_{t\geq 0})_{n\geq 1}$ is
tight in $D_{\cM}[0,\infty)$, as required.

\medskip
{\bf 2) Limiting process.}

\noindent
We now identify the limiting process.
We begin with the case $d\geq 2$. Since the quadratic variation vanishes
(uniformly in $w$)
as $n\rightarrow \infty$, any limit of $((\bw_t^n)_{t\geq 0})_{n\geq 1}$
should be
deterministic. It remains to identify the limit of the finite variation part.
Specialising the computation of $\Phi^n$ above to the case
$F=\mathrm{Id}$, we have
\begin{align}
A_n(s)


& = \frac{u\Gamma_R}{2} \, \la \bw^n,\Delta f\ra + \mathcal{O}(n^{-2/3}), \label{first term}
\end{align}
where
\begin{equation}\label{def Gamma}
\Gamma_R= \frac{n^{\frac{2}{3}(1+d)}}{V_R} \int_{B_n(y)} \int_{B_n(x)} (z_1-y_1)^2 dzdx = \frac{1}{V_R}\int_{B(0,R)}\int_{B(x,R)}(z_1)^2dz\, dx
\end{equation}
depends only on $R$ and $d$.

Next,
\begin{align}
D_n(s) & = \sigma u n^{d/3} \int_{\R^d}  \Big\{\bw_s^n(x)^2 \la \ind_{B_n(x)}(1-w^n_s),(f(x) + \mathcal{O}(\|y-x\|))\ra \nonumber \\
&\qquad \qquad \qquad \qquad \qquad \qquad - (1-\bw_s^n(x)^2)\la \ind_{B_n(x)}w^n_s,(f(x) + \mathcal{O}(\|y-x\|))\ra \Big\}dx\nonumber \\
& = \sigma uV_R \int_{\R^d}  \big\{\bw_s^n(x)^2(1-\bw_s^n(x)) - (1-\bw_s^n(x)^2)\bw^n_s(x)\big\}\, f(x) dx + \mathcal{O}(n^{-1/3}) \nonumber\\
& = - \sigma uV_R\, \la \bw_s^n(1- \bw_s^n), f\ra + \mathcal{O}(n^{-1/3}). \label{second term}
\end{align}

Combining (\ref{first term}) and (\ref{second term}), and using the fact
that the terms $B_n(s)$, $C_n(s)$ and $E_n(s)$ tend to zero uniformly in $w$,
we conclude that any limit point of $((\bw_t^n)_{t\geq 0})$ satisfies, for every
$f\in C_c^\infty(\R^d)$ and every $t\geq 0$,
$$
\la w_t,f\ra = \la w^0,f\ra + \int_0^t  \bigg\{\frac{u\Gamma_R}{2}\,
\la w_s,\Delta f\ra - \sigma uV_R\, \la w_s(1- w_s), f\ra \bigg\}\, ds.
$$
As a consequence, such a limit is continuous and since the solution to this
problem is unique, we conclude that $((\bw_t^n)_{t\geq 0})_{n\geq 1}$
converges in
distribution to the process described in $(ii)$ of Theorem~\ref{th:fixed}.

We now turn to the case $d=1$. Observe first that (\ref{first term}) and
(\ref{second term}) still hold. Consequently, we know that for every $f\in C_c^\infty(\R^d)$ (taking again $F=\mathrm{Id}$ so that $F''=0$),
\begin{align*}
M^n_t(f)& := \la w^n_t,f\ra - \la w^n_0,f\ra -\int_0^t \,
\cL^n\la w^n_s,f\ra \, ds \\
& = \la \bw^n_t,f\ra - \la \bw^n_0,f\ra -
\int_0^t \bigg\{\frac{u\Gamma_R}{2} \, \la \bw^n_s,\Delta f\ra - \sigma uV_R\,
\la \bw_s^n(1- \bw_s^n), f\ra \bigg\} \, ds+ \mathcal{O}(n^{-2/3})
\end{align*}
is a zero-mean martingale with quadratic variation
\begin{align*}
u_n^2n^{4/3}&\int_0^t \int_{\R^d} \Big\{\bw^n_s(x)(1+s_n\bw^n_s(x))\la \ind_{B_n(x)}(1-w^n_s),f\ra^2 \\
& \qquad \qquad \qquad  + (1-\bw^n_s(x)+s_n(1-\bw^n_s(x)^2))\la \ind_{B_n(x)}w^n_s,f\ra^2\Big\}\,  dxds \\
& = u^2V_R^2\int_0^t \, \la \bw^n_s(1-\bw^n_s),f^2\ra ds  + \mathcal{O}(n^{-1/3}).
\end{align*}
Recall that the sequence $((\bw_t^n)_{t\geq 0})_{n\geq 1}$ is tight. Theorem~6.3.3 in \cite{EK1986}
implies that if some subsequence $((\bw_t^{\phi(n)})_{t\geq 0})_{n\geq 1}$
converges
in distribution to a process $(w_t)_{t\geq 0}$,
then the sequence $((M_t^{\phi(n)}(f))_{t\geq 0})_{n\geq 1}$
converges to the (time-changed Brownian motion) solution to
$$
dM_t = uV_R\, \sqrt{\la w_t(1-w_t), f^2\ra}\, dB^f_t,
$$
where $(B^f_t)_{t\geq 0}$ denotes standard Brownian motion.
As a consequence, any limit point $(w_t)_{t\geq 0}$ of
$((\bw_t^n)_{t\geq 0})_{n\geq 1}$ satisfies the following system of
stochastic differential equations: for every $f\in C_c^\infty(\R^d)$,
$$
d\la w_t,f\ra = \bigg\{\frac{u\Gamma_R}{2} \, \la w_t,\Delta f\ra - \sigma
uV_R\, \la w_t(1- w_t), f\ra \bigg\}\, dt + uV_R\, \sqrt{\la w_t(1-w_t),
f^2\ra}\, dB^f_t,
$$
with $w_0=w^0$. Since this system has a unique solution, Theorem~\ref{th:fixed} is proved. \hfill $\Box$

\medskip

\subsection{Proof of Theorem~\ref{th: conv duals}.}

We divide the proof into two parts.  The first, and simpler, shows that
the only possible limit for $((\Xi_t^n)_{t\geq 0})_{n\geq 1}$
is the system of branching and coalescing
Brownian motions $(\Xi_t^\infty)_{t\geq 0}$.
The second part, tightness of the sequence
$((\Xi_t^n)_{t\geq 0})_{n\geq 1}$, is rather more
involved and will be broken into a number of smaller steps.

\begin{lemma}
The finite dimensional distributions of the system of scaled processes
$((\Xi_t^n)_{t\geq 0})_{n\geq 1}$ converge as $n\rightarrow\infty$ to those of
the system of
branching and coalescing Brownian motions $(\Xi_t^\infty)_{t\geq 0}$,
described in the
statement of Theorem~\ref{th: conv duals}. In particular, the only possible
limit point for the sequence $((\Xi_t^n)_{t\geq 0})_{n\geq 1}$ is
$(\Xi_t^\infty)_{t\geq 0}$.
\end{lemma}
\noindent
{\bf Proof}

By Theorem~\ref{th:fixed}, the rescaled forwards-in-time processes
$(\bw_t^n)_{t\geq 0}$ converge to the process $(w_t^\infty)_{t\geq 0}$ for which,
for every $f\in C_c^\infty(\R^d)$,
$$
\la w_t^\infty,f\ra - \la w_0^\infty, f\ra - \int_0^t
\bigg\{ \frac{u\Gamma_R}{2}\, \la w_s^\infty ,\Delta f\ra -
u\sigma V_R\, \la w_s^\infty (1-w_s^\infty),f\ra \bigg\}\, ds
$$
is a martingale, with quadratic variation $0$ when $d\geq 2$, and
$$
u^2V_R^2 \int_0^t \la w_s^\infty (1-w_s^\infty),f^2\ra\, ds
$$
when $d=1$. By It\^o's formula, this description suffices to characterize
the evolution of any product of the form
$\la w_t^\infty ,f_1\ra\la w_t^\infty, f_2\ra \cdots \la w_t^\infty, f_k\ra$,
for any $k\geq 1$ and $f_1,\ldots,f_k \in C_c^\infty (\R^d)$.

Now, in Chapter 7 of \cite{LIA2009}, Liang shows that when $\sigma=0$,
any solution to these
equations is dual, through the set of functions (\ref{dual formula}), to a
system of independent Brownian motions with variance parameter $u\Gamma_R$,
in which, when $d=1$, individuals coalesce (pairwise), at a rate proportional to
$u^2V_R^2$ times the local time at $0$ of their separation.
This is easily modified to $\sigma>0$, in which case individuals
branch into two at rate $u\sigma V_R$.

To be slightly more rigorous, let us describe a particle configuration
$\Xi$ by an element of ${\mathcal N}(\R^d)$, the set of all point
measures on $\R^d$, through the identification
$$
\Xi = \sum_{i=1}^{|\Xi|} \delta_{\xi^i}
$$
(where $|\Xi|$ is the number of particles in $\Xi$). The set of test functions
$$
\Xi \mapsto \prod_{i=1}^{|\Xi|} f(\xi^i) = e^{\la \Xi ,\log f\ra},
$$
where $f$ is measurable and takes values in $(0,1]$,
is sufficient to characterize
the law of $\Xi$. Now we use the duality formula (\ref{dual formula}) and
the approximation~(\ref{approximation})
to write that, for every $k\in \N$ and $\psi_1,\ldots,\psi_k \in C^2_c(\R^d)$,
\begin{align}
\E\bigg[& \int_{(\R^d)^k} \psi_1(x_1)\cdots \psi_k(x_k)\bigg\{\prod_{i=1}^k
\bw_t^n(x_i)\bigg\}dx_1\ldots dx_k\bigg] + \mathcal{O}(n^{-2/3}) \nonumber\\
& = \E\bigg[\int_{(\R^d)^k} \psi_1(x_1)\cdots \psi_k(x_k)\bigg\{\prod_{i=1}^k w_{nt}(n^{1/3}x_i)\bigg\}dx_1\ldots dx_k\bigg] + \mathcal{O}(n^{-2/3}) \nonumber\\
& = n^{-dk/3}\E\bigg[\int_{(\R^d)^k} \psi_1(n^{-1/3}x_1)\cdots \psi_k(n^{-1/3}x_k)\bigg\{\prod_{i=1}^k w_{nt}(x_i)\bigg\}dx_1\ldots dx_k\bigg]+ \mathcal{O}(n^{-2/3}) \nonumber\\
& = n^{-dk/3} \int_{(\R^d)^k} \psi_1(n^{-1/3}x_1)\cdots \psi_k(n^{-1/3}x_k) \E_{\{x_1,\ldots, x_k\}}\bigg[\prod_{j=1}^{N_{nt}}w_0(\xi_{nt}^j)\bigg]dx_1\ldots dx_k + \mathcal{O}(n^{-2/3}) \nonumber\\
& = \int_{(\R^d)^k} \psi_1(x_1) \cdots \psi_k(x_k) \E_{\{n^{1/3}x_1,\ldots, n^{1/3}x_k\}}\bigg[\prod_{j=1}^{N_{nt}}w_0(n^{1/3}(n^{-1/3}\xi_{nt}^j))\bigg]dx_1\ldots dx_k + \mathcal{O}(n^{-2/3}) \nonumber\\
& = \int_{(\R^d)^k} \psi_1(x_1) \cdots \psi_k(x_k) \E_{\{x_1,\ldots,x_k\}} \bigg[\prod_{j=1}^{N^n_{t}}w_0^n(\xi_{t}^{n,j})\bigg]dx_1\ldots dx_k
+ \mathcal{O}(n^{-2/3}).
\label{scaled dual}
\end{align}
Letting $n\rightarrow \infty$ and applying the Dominated Convergence Theorem,
we deduce that the only possible limit for $((\Xi_t^n)_{t\geq 0})_{n\geq 1}$
is the system of particles
$(\Xi_t^\infty)_{t\geq 0}$. In addition, since (\ref{scaled dual}) holds for every
$\psi\in C^2_c((\R^d)^k)$ and every $w_0$, we also obtain convergence of
the one-dimensional distributions of $(\Xi_t^n)_{t\geq 0}$
to those of $(\Xi_t^\infty)_{t\geq 0}$.
The generalization to the finite-dimensional distributions is straightforward
since the duality formula (\ref{dual formula}) holds on any time interval
$[s,t]$ (if we replace $w_0$ by $w_s$ and $\xi_t^j$ by $\xi_{t-s}^j$).
\hfill $\Box$

\medskip
\noindent
{\bf Tightness}

We now show tightness of the sequence $((\Xi_t^n)_{t\geq 0})_{n\geq 1}$,
for every
initial value $\mathbf{x}=\{x_1,\ldots,x_k\}\in (\R^d)^k$. We apply
Aldous's criterion based on stopping times (see Theorem~1 in \cite{ALD1978}).
Fix $T>0$ and $f\in C_c^2(\R^d)$ with values in $[0,1]$, and suppose that
$(\tau_n)_{n\geq 1}$ is any sequence of stopping times bounded by $T$. We must
show that for every $\e>0$ there exists $\delta=\delta(f,T,\mathbf{x})$
such that
\begin{equation}\label{aim tightness}
\limsup_{n\rightarrow \infty}\, \P_{\mathbf{x}}\Bigg[\sup_{0\leq t\leq \delta}\bigg|\prod_{i=1}^{N^n_{\tau_n+t}}f(\xi^{n,i}_{\tau_n+t}) - \prod_{i=1}^{N^n_{\tau_n}}f(\xi^{n,i}_{\tau_n})\bigg|>\e \Bigg] \leq \e.
\end{equation}
We shall proceed in a number of steps.  First we control the maximum
number of particles in $\Xi_t^n$ up to time $T$. Conditional on this, it is
easy to control the probability that there is a branch in an interval of
length $\delta$.  If we can also show that with high probability
there is no coalescence (so that the number of particles in the system does
not change), then the problem is reduced to controlling
the jumps in a random walk. The most involved step, which is the substance
of Proposition~\ref{prop: controlling coalescence},
is showing that there is no accumulation of coalescence events.

\begin{lemma}
Given $\e>0$, there is $K >0$ such that
\begin{equation}\label{total number}
\P_{\mathbf{x}}\bigg[\sup_{0\leq t\leq T} |\Xi^n_t| >K \bigg] \leq \frac{\e}{4}.
\end{equation}
\end{lemma}
\noindent
{\bf Proof}

Recall that two particles are created when at least one of the extant
particles is affected by a selective event. For a given particle of
$\Xi_t^n$, this happens at rate $ns_n V_R u_n = u\sigma V_R$.
Furthermore, the presence of more than one particle in the area affected
by the event does not speed up the branching.  Consequently, the number of
particles in $(\Xi_t^n)_{t\geq 0}$ is stochastically bounded by the
number of particles
in a continuous-time branching process in which particles split
(independently of one another) into two offspring at rate $u\sigma V_R$.
Since the initial value, $\Xi_0^n$, has finitely many particles,
we conclude that there
exists $K\in \N$ such that for every $n\in \N$,
\begin{equation}\nonumber
\P_{\mathbf{x}}\bigg[\sup_{0\leq t\leq T} |\Xi^n_t| >K \bigg] \leq \frac{\e}{4},
\end{equation}
as required. \hfill $\Box$

\medskip

>From now on, all our calculations will proceed conditional on
the event $A_n= \{\sup_{0\leq t\leq T} |\Xi^n_t| \leq K\}$.
>From our calculations above, we already see that for any $t\in [0,T]$,
conditional on $A_n$, the probability that at least one particle is
created during the time interval $[t,t+\delta]$ is bounded by
$$
K\, \P\big[\hbox{a given particle branches in }[t,t+\delta]\big] \leq K
\big(1- e^{-u\sigma V_R \delta}\big) \leq u\sigma KV_R\, \delta.
$$
This bound is uniform and so we see that there exists $\delta_1>0$ such that
\begin{equation}\label{proba branching}
\P_{\mathbf{x}}\big[ \hbox{at least 1 particle created in }[\tau_n,\tau_n+\delta_1] \, ;\, A_n\big] \leq \frac{\e}{4}.
\end{equation}

We also want to control the probability of coalescence events.
Because of the calculation above, it is enough to do so in the absence of
branching.
\begin{prop}
\label{prop: controlling coalescence}
Let $B_\delta^c$ denote the event that there is no branching event in
$[\tau_n,\tau_n+\delta]$.
There exists $\delta_3>0$ such that
\begin{equation}\label{proba coalescing}
\P[\hbox{at least 1 coalescence in }[\tau_n,\tau_n+\delta_3]\, ;\, A_n, B_{\delta_3}^c] \leq \frac{K(K-1)}{2} \frac{\e}{2K(K-1)} = \frac{\e}{4}.
\end{equation}
\end{prop}

Before proving Proposition~\ref{prop: controlling coalescence}, let us turn
to the final ingredient in the proof and
control the jumps of a single lineage.

Conditional on the number of individuals not changing during a time interval
of length $\delta$, we can index the particles of $\Xi^n_{\tau_n}$ and
$\Xi^n_{\tau_n+\delta}$ by a common indexing set which we denote $I_n$.
Under this assumption, recalling that $\|f\|\leq 1$, a Taylor expansion yields
\begin{equation}\label{diff products}
\bigg|\prod_{i\in I_n}f(\xi^{n,i}_{\tau_n+t}) - \prod_{i\in I_n}f(\xi^{n,i}_{\tau_n})\bigg| \leq C \|\nabla f\| \sum_{i\in I_n}\big|\xi^{n,i}_{\tau_n+t} - \xi^{n,i}_{\tau_n}\big|,
\end{equation}
for some $C>1$. Since, on the event $A_n$, $|I_n|\leq K$, it suffices to
consider the motion of each lineage separately.

>From the description in Section~\ref{ss: general dual}, after rescaling
of time and space, $\xi^{n,1}$ jumps at rate
$nu_n V_R(1+s_n) = n^{2/3}uV_R (1+o(1))$, to a new location whose
distribution is symmetric about its current location. Since the locations of
the lineage both before and after the jump belong to the same ball of
radius $Rn^{-1/3}$, the length of the jump is bounded by $2Rn^{-1/3}$.
Moreover, since the distribution of the new location depends on $n$
only through the spatial scaling by $n^{-1/3}$, it follows that the motion
$(\xi^{n,1}_t)_{t\geq 0}$ converges to Brownian motion. Doob's Maximal
Inequality then implies that there exists $C_1>0$ such that for $n$
sufficiently large and any $\delta,\eta >0$,
$$
\P\bigg[\sup_{t\in [0,\delta]}\big|\xi^{n,1}_{\tau_n+t} - \xi^{n,1}_{\tau_n}\big| > \eta \bigg] \leq \frac{C_1}{\eta^2}\, \delta,
$$
where we have used the strong Markov property of $\xi^{n,1}$ at time
$\tau_n$. Together with (\ref{diff products}) and the choice
$\eta = \e/(K C\|\nabla f\|)$, this shows that there exists $\delta_2>0$ such that
for $n$ large enough, writing $C_{\delta}^c$ for the event that there is no coalescence
in $[\tau_n, \tau_n+\delta]$,
\begin{equation}\label{proba moving}
\P_{\mathbf{x}}\bigg[\sup_{t\in [0,\delta_3]}\bigg|\prod_{i\in I_n}
f(\xi^{n,i}_{\tau_n+t}) - \prod_{i\in I_n}f(\xi^{n,i}_{\tau_n})\bigg|>\e\, ;\,
A_n, B_{\delta_2}^c, C_{\delta_2}^c\bigg]\leq \frac{KC_1}{\eta^2}\, \delta_2 \leq \frac{\e}{4}.
\end{equation}
\begin{rmk} Before proving Proposition~\ref{prop: controlling coalescence},
let us remark that it is not enough to consider lineages at an
initial separation of order $\mathcal{O}(1)$ (or $\mathcal{O}(n^{1/3})$
before rescaling). In particular, when two particles are created through a
selective event, their (rescaled) initial distance is of order
$\mathcal{O}(n^{-1/3})$ and so we also need to control the
coalescence of particles starting from very small initial separations.
\end{rmk}

\noindent
{\bf Proof of Proposition~\ref{prop: controlling coalescence}.}

It suffices to consider just two particles and find $\delta_3>0$
such that the probability that they coalesce in a time interval
of length $\delta_3$ is bounded by $\e/(2K(K-1))$, irrespective of their
initial separation. Once this bound has been established, we can write
\begin{equation}
\P[\hbox{at least 1 coalescence in }[\tau_n,\tau_n+\delta_3]\, ;\, A_n, B_{\delta_3}^c] \leq \frac{K(K-1)}{2} \frac{\e}{2K(K-1)} = \frac{\e}{4},
\end{equation}
since, on the event $A_n$, there are at most $K(K-1)/2$ pairs of particles at
any time.

Recall that before scaling, each lineage jumps at rate proportional to $u_n$.
This makes it convenient to work in the timescale
$(n^{1/3}t,\, t\geq 0)$ and without rescaling space.
We shall write $\txi^{n,i}_{t} = \xi_{n^{1/3}t}^i$, $i\in \{1,2\}$.

When $\txi^{n,1}$ and $\txi^{n,2}$ are separated by more than $2R$,
they cannot be contained in the same reproduction event, and so they
evolve independently of one another.
The $i$th lineage jumps at rate $n^{1/3}u_nV_R(1+s_n) = uV_R(1+o(1))$
to a new location, which is uniformly
distributed over the ball $B(Z,R)$, where $Z$ itself is chosen uniformly
at random from $B(\txi^{n,i,}R)$. In what follows, we only need
that the jump made by each lineage is an independent realization
of a random variable $X$ taking values in $B(0,2R)$, whose distribution is
symmetric about the origin.

On the other hand, when $|\txi^{n,1}-\txi^{n,2}|<2R$, the two particles can
both lie in a region affected by a given reproduction event and their jumps
become correlated.
In particular, if they are both affected by this event, they merge together.
The generator of $((\txi_t^{n,1},\txi_t^{n,2}))_{t\geq 0}$ takes the form
\begin{align*}
&u (1+s_n) \int_{B(\txi^1,R)\setminus B(\txi^2,R)} \int_{B(x,R)} \frac{1}{V_R}\, \big(f(z,\txi^{n,2})-f(\txi^{n,1},\txi^{n,2})\big)dzdx \\
& +  u (1+s_n)\int_{B(\txi^2,R)\setminus B(\txi^1,R)} \int_{B(x,R)} \frac{1}{V_R}\, \big(f(\txi^{n,1},z)-f(\txi^{n,1},\txi^{n,2})\big)dzdx \\
& + u(1-un^{-1/3})(1+s_n)\int_{B(\txi^1,R)\cap B(\txi^2,R)} \int_{B(x,R)} \frac{1}{V_R}\, \big(f(z,\txi^{n,2})+f(\txi^{n,1},z)-2f(\txi^{n,1},\txi^{n,2})\big)dzdx \\
& + u^2n^{-1/3}(1+s_n) \int_{B(\txi^1,R)\cap B(\txi^2,R)} \int_{B(x,R)} \frac{1}{V_R}\, \big(f(z,z)-f(\txi^{n,1},\txi^{n,2})\big)dzdx.
\end{align*}
We can think of this as composed of two parts: the process
$((\hat{\xi}_t^{n,1},\hat{\xi}_t^{n,2}))_{t\geq 0}$ whose generator is
determined by the first three lines above,
on top of which a coalescence event occurs at instantaneous rate
$u^2n^{-1/3}(1+s_n)V_R(0,\hat{\xi}^{n,1}_t - \hat{\xi}^{n,2}_t)$
(recall that $V_R(0,a)$ is the volume of the intersection $B(0,R)\cap B(a,R)$).

With this description, the probability that the two particles have not
coalesced by time $\delta n^{2/3}$ (which corresponds to a time span of
$\delta$ on the timescale of $\xi^{n,i}$) is given by
\begin{equation}\label{coal proba}
\P_{\mathbf{x}}\big[\tilde{T}> \delta n^{2/3}\big] =
\E_{\mathbf{x}}\bigg[\exp\bigg\{-\frac{u^2(1+s_n)}{n^{1/3}}
\int_0^{\delta n^{2/3}}V_R(0,\hat{\xi}_s^{n,1}-\hat{\xi}_s^{n,2})\,
ds\bigg\}\bigg],
\end{equation}
where we have written $\tilde{T}$ for the coalescence time of the two
particles.

Since $V_R(0,x)=0$ when $x\geq 2R$, it just remains to
establish how much time $\hat{\xi}^{n,1}-\hat{\xi}^{n,2}$ spends in the
ball $B(0,2R)$ by time $\delta n^{2/3}$. To do this, we define two
sequences of stopping times, $(\sigma^n_k)_{k\geq 1}$ and
$(\tau^n_k)_{k\geq 1}$ by
$$
\sigma^n_1 = \inf\{t\geq 0\, :\, |\hat{\xi}_t^{n,1}-\hat{\xi}_t^{n,2}|\leq 2R\},
 \qquad \tau^n_1 = \inf\{t\geq \sigma^n_1\, :\, |\hat{\xi}_t^{n,1}-
\hat{\xi}_t^{n,2}|>2R\},
$$
and for every $k\geq 1$,
$$
\sigma^n_k = \inf\{t\geq \tau^n_{k-1}\, :\, |\hat{\xi}_t^{n,1}-
\hat{\xi}_t^{n,2}|\leq 2R\}, \qquad
\tau^n_k = \inf\{t\geq \sigma^n_k\, :\, |\hat{\xi}_t^{n,1}-
\hat{\xi}_t^{n,2}|>2R\}.
$$
Now, we have the following result.
\begin{lemma}\label{lem: incursions}
There exists $\mathcal{C}>0$ such that for every $n,k\geq 1$,
$$
\E[\tau^n_k - \sigma^n_k] \leq \mathcal{C}.
$$
\end{lemma}
In words, although the two particles are correlated when they are close
together, each `incursion' of $\hat{\xi}^{n,1}-\hat{\xi}^{n,2}$ inside
$B(0,2R)$ lasts only $\mathcal{O}(1)$ units of time, uniformly in $n$.
The proof of Lemma~\ref{lem: incursions} is similar to that of Lemma~6.6
in \cite{BEV2010} (based on the facts that the difference walk jumps at a
rate bounded from below by a positive constant, independent of its current
value, and that the probability that this jump leads to a sufficient
increase of their separation for $\hat{\xi}_t^{n,1}-\hat{\xi}_t^{n,2}$
to leave $B(0,2R)$ is
also bounded from below by a positive constant). Therefore, we omit it here.

Outside $B(0,2R)$, the difference $\hat{\xi}_t^{n,1}-\hat{\xi}_t^{n,2}$ has
the same law as a symmetric random walk, with jumps of size at most $2R$,
jumping at rate $2uV_R(1+s_n)$. Its behaviour will be determined by the
spatial dimension.

\medskip

\noindent
{${\mathbf d\geq 3}$:}
When $d\geq 3$, transience of the random walk guarantees that the number
of times $\hat{\xi}^{n,1}-\hat{\xi}^{n,2}$ returns to $B(0,2R)$ is a.s.~finite.
Since the parameter $n$ appears only in the jump rates and not in the embedded
chain of locations (during an excursion outside $B(0,2R)$), the probability
that the difference walk enters $B(0,2R)$ at least $k$ times
decays to $0$, uniformly in $n$, as $k\rightarrow \infty$. Together
with Lemma~\ref{lem: incursions} and the fact that $V_R(0,\cdot)$ is bounded,
this shows that for every $\eta>0$,
$$
\lim_{n\rightarrow \infty}\P_{\mathbf{x}}\bigg[\int_0^{\delta n^{2/3}}
V_R(0,\hat{\xi}_s^{n,1}-\hat{\xi}_s^{n,2})\, ds > \eta\, \frac{n^{1/3}}{u^2(1+s_n)} \bigg] =0
$$
As a consequence, coming back to~(\ref{coal proba}) and choosing $\eta$
small enough that $\P[\mathrm{Exp}(1)\leq \eta]\leq \e/12$, we can conclude
that for any $\delta>0$,
\begin{equation}\label{proba of coal}
\limsup_{n\rightarrow \infty}\, \P_{\mathbf{x}}\big[\tilde{T}\leq  \delta n^{2/3}\big] \leq \frac{\e}{12},
\end{equation}
uniformly in $\mathbf{x}\in (\R^d)^2$.

\medskip

\noindent
{$\mathbf d=2$:}
When $d=2$, we claim that there exists $\mathcal{C}'>0$, independent of $n$,
such that for every $x_1,x_2$ with $|x_1-x_2|> 2R$,
$$
\P_{\{x_1,x_2\}}\big[\sigma^n_1> \delta n^{2/3}\big] \geq \frac{\mathcal{C}'}{\log(\delta n^{2/3})}.
$$
The proof of this claim is very similar to the beginning of the proof of
Lemma~4.2 in \cite{BEV2012}, and so we only sketch the main ideas. We can
a.s.~embed the trajectories of the difference process
$\hat{\xi}_t^{n,1}-\hat{\xi}_t^{n,2}$ into the trajectories of a
two-dimensional Brownian motion, in the same spirit as Skorokhod's embedding in
one dimension (see e.g.~\cite{BIL1995}). Now, since the jumps of the difference
process (when outside $B(0,2R)$) are rotationally invariant, we have
$$
\inf_{|x_1-x_2|>2R} \, \P_{\{x_1,x_2\}}\big[\hat{\xi}^{n,1}-\hat{\xi}^{n,2}
\hbox{ leaves }B(0,4R)\hbox{ before entering }B(0,2R)\big] >0,
$$
and the result then follows from that for Brownian motion, namely Theorem~2 in \cite{RR1966} applied with $a=2R$ and $r\geq 4R$.

As a consequence, the number $N_E^n$ of excursions outside $B(0,2R)$ that
the difference walk makes before starting an excursion of (time) length
at least $\delta n^{2/3}$ is stochastically bounded by a geometric random
variable with
success probability $C/\log(\delta n^{2/3})$. Now, once the difference walk
has started such a long excursion (say, the $k$th one), it is sure not to
come back within $B(0,2R)$ before time $\delta n^{2/3}$ and the number of
incursions in $B(0,2R)$ in the time interval $[0,\delta n^{2/3}]$ is
bounded by $k$. Thus, fixing $\eta>0$ as before, we obtain that
\begin{align*}
\P_{\mathbf{x}} \bigg[\int_0^{\delta n^{2/3}}& V_R(0,\hat{\xi}_s^{n,1}-
\hat{\xi}_s^{n,2})\, ds > \eta\, \frac{n^{1/3}}{u^2(1+s_n)} \bigg]\\
 & \leq \P_{\mathbf{x}}\big[N_E^n > C_E^n \log (\delta n^{2/3})\big] + \P_{\mathbf{x}}\bigg[\sum_{k=1}^{\lceil C_E^n \log (\delta n^{2/3})\rceil} (\tau^n_k -\sigma^n_k) >\eta\, \frac{n^{1/3}}{u^2(1+s_n)\|V_R\|}\bigg] \\
& \leq e^{-C_E^n \mathcal{C}'} + \frac{u^2(1+s_n)\|V_R\|}{\eta n^{1/3}}\, C_E^n\log(\delta n^{2/3}) \mathcal{C},
\end{align*}
where the last inequality uses the stochastic bound of $N_E^n$ first,
and then Markov's inequality. Choosing $C_E^n = \log n$, for instance, we
deduce that for any $\delta >0$,
$$
\lim_{n\rightarrow \infty} \P_{\mathbf{x}} \bigg[\int_0^{\delta n^{2/3}}
V_R(0,\hat{\xi}_s^{n,1}-\hat{\xi}_s^{n,2})\, ds > \eta\, \frac{n^{1/3}}{u^2(1+s_n)} \bigg] =0,
$$
and we conclude as in (\ref{proba of coal}).

\medskip

\noindent
{$\mathbf d=1$:}
Finally, when $d=1$ it is shown in \cite{PS1971} 
that there exists $\mathcal{C}'>0$ such that for every $x_1,x_2$ such
that $|x_1-x_2|> 2R$,
$$
\P_{\{x_1,x_2\}}[\sigma^n_1> \delta n^{2/3}] \geq \frac{\mathcal{C}'}{\sqrt{\delta}\, n^{1/3}}.
$$
Proceeding as before, and with the same notation, we therefore have
\begin{align*}
\P_{\mathbf{x}} \bigg[\int_0^{\delta n^{2/3}}&
V_R(0,\hat{\xi}_s^{n,1}-\hat{\xi}_s^{n,2})\, ds >
\eta\, \frac{n^{1/3}}{u^2(1+s_n)} \bigg]\\
 & \leq \P_{\mathbf{x}}\big[N_E^n > C_E^n \sqrt{\delta} n^{1/3}\big] + \P_{\mathbf{x}}\bigg[\sum_{k=1}^{\lceil C_E^n \sqrt{\delta} n^{1/3}\rceil} (\tau^n_k -\sigma^n_k) >\eta\, \frac{n^{1/3}}{u^2(1+s_n)\|V_R\|}\bigg] \\
& \leq e^{-C_E^n \mathcal{C}'} + \frac{u^2(1+s_n)\|V_R\|}{\eta n^{1/3}}\, C_E^n \sqrt{\delta} n^{1/3} \mathcal{C}.
\end{align*}
Choosing $C_E^n$ to be a constant large enough for the first term to be less
than $\e/12$, and then $\delta_3>0$ small enough for the second term to be
less than $\e/12$, we obtain that for any $\delta\leq \delta_3$,
\begin{align}
\P_{\mathbf{x}}\big[\tilde{T}\leq  \delta n^{2/3}\big] & \leq
\P_{\mathbf{x}} \bigg[\int_0^{\delta n^{2/3}}
V_R(0,\hat{\xi}_s^{n,1}-\hat{\xi}_s^{n,2})\, ds > \eta\, \frac{n^{1/3}}{u^2(1+s_n)} \bigg] + \P[\mathrm{Exp}(1)<\eta]\nonumber \\
& \leq \frac{\e}{12}+\frac{\e}{12} + \frac{\e}{12} = \frac{\e}{4}.\label{proba of coal 2}
\end{align}
We have now proved the desired bound for (\ref{proba coalescing}) in
any dimension and the proof of Proposition~\ref{prop: controlling coalescence}
is complete.
\hfill $\Box$

\medskip

Finally we return to~(\ref{aim tightness}). Choosing
$\delta_4 = \min\{\delta_1,\delta_2,\delta_3\}$ and recalling that
$A_n =\{\sup_{0\leq s\leq T} |\Xi^n_s| \leq K\}$, we have
\begin{align*}
\P_{\mathbf{x}}& \Bigg[\sup_{0\leq t\leq \delta_4}\bigg|\prod_{i=1}^{N^n_{\tau_n+t}}f(\xi^{n,i}_{\tau_n+t}) - \prod_{i=1}^{N^n_{\tau_n}}f(\xi^{n,i}_{\tau_n})\bigg|>\e \Bigg] \\
& \leq \P_{\mathbf{x}}\bigg[\sup_{0\leq s\leq T} |\Xi^n_s| >K \bigg] + \P_{\mathbf{x}}\big[ \hbox{at least 1 particle created in }[\tau_n,\tau_n+\delta_4] \, ;\, A_n\big] \\
& \qquad + \P[\hbox{at least 1 coalescence in }[\tau_n,\tau_n+\delta_4]\, ;\, A_n, B_{\delta_4}^c] \\
& \qquad + \P_{\mathbf{x}}\bigg[\sup_{t\in [0,\delta_4]}\bigg|\prod_{i\in I_n}f(\xi^{n,i}_{\tau_n+t}) - \prod_{i\in I_n}f(\xi^{n,i}_{\tau_n})\bigg|>\e\, ;\, A_n, B_{\delta_4}^c, C_{\delta_4}^c \bigg] \\
& \leq \e,
\end{align*}
where the last line uses (\ref{total number}), (\ref{proba branching}), (\ref{proba moving}) and (\ref{proba coalescing}). The tightness of
$((\Xi_t^n)_{t\geq 0})_{n\in \N}$ is shown, and the proof of Theorem~\ref{th: conv duals} is complete. \hfill $\Box$


\section{Convergence of the SLFVS and its dual - the stable radii case} \label{s: convergence 2}
Recall the notation $V_r$ for the volume of a ball of radius $r$ in $\R^d$, and $V_r(x,y)$ for the volume of the intersection $B(x,r)\cap B(y,r)$.

\medskip
\noindent {\bf Proof of Theorem~\ref{th:stable}.}

\medskip
{\bf 1) Tightness.}

We shall use the same method as in the proof of Theorem~\ref{th:fixed}, but the computations required will be different. Recall the notation $\Theta^+_{x,r,u_n}(w)$ and $\Theta^-_{x,r,u_n}(w)$ from (\ref{notation Delta}). The generator of the unscaled process is
\begin{align*}
\tilde{\cG}^n F(\la w,f\ra)& = \int_{\R^d}\int_1^{\infty}\int_{B(x,r)^2}\frac{1}{V_r^2}\, \Big\{w(y)(1+s_nw(z))\big[ F(\la \Theta^+_{x,r,u_n}(w),f \ra) - F(\la w,f\ra)\big] \\
 &\ + (1-w(y)+s_n(1-w(y)w(z)))\big[ F(\la \Theta^-_{x,r,u_n}(w),f \ra) - F(\la w,f\ra)\big]\Big\}dydz\mu(dr)dx.
\end{align*}
To make the expressions easier to read, we retain the notation $\beta$, $\gamma$ and $\delta$
from (\ref{def constants}). The generator of the rescaled process is then given by
\begin{align}
&\cG^n F(\la w,f\ra) \nonumber\\
& = n\int_{\R^d}\int_1^{\infty}\int_{B(x,r)^2}\frac{1}{V_r^2}\, \Big\{w(n^{-\beta}y)\big(1+s_n w(n^{-\beta}z))\big[ F(\la \Theta^+_{n^{-\beta}x,n^{-\beta}r,u_n}(w),f \ra)  - F(\la w,f\ra)\big]\big) \nonumber\\
& \quad + (1-w(n^{-\beta}y)+s_n(1-w(n^{-\beta}y)w(n^{-\beta}z)))\big[ F(\la \Theta^-_{n^{-\beta}x,n^{-\beta}r,u_n}(w),f \ra) - F(\la w,f\ra)\big]\Big\}\nonumber \\ & \qquad \qquad\qquad \qquad\qquad \qquad\qquad \qquad \qquad \qquad\qquad \qquad\qquad \qquad \qquad \qquad dydz\mu(dr) dx  \nonumber \\
& = n^{1-\beta\alpha}\int_{\R^d}\int_{n^{-\beta}}^{\infty}\frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^2}\frac{1}{V_r^2}\, \Big\{w(y)(1+s_nw(z))\big[F(\la \Theta^+_{x,r,u_n}(w),f\ra)-F(\la w,f\ra)\big] \nonumber \\
& \qquad \qquad  +\big(1-w(y)+s_n(1-w(y)w(z))\big)\big[F(\la \Theta^-_{x,r,u_n}(w),f\ra)-F(\la w,f\ra)\big]\Big\}\, dydzdrdx. \label{generator wn 2}
\end{align}

To simplify notation, we shall show the tightness of $((F(\la w_t^n,f\ra))_{t\geq 0})_{n\geq 1}$.
This implies that the sequence $((F(\la \bw_t^n,f\ra))_{t\geq 0})_{n\geq 1}$ in which we are
interested is tight on replacing $f$ by $\phi_f$ defined by
\begin{equation}
\label{stable varphi}
\varphi_f(x) = \frac{n^{d\beta}}{V_1}\int_{B(x,n^{-\beta})} f(y)\, dy,
\end{equation}
and exploiting the bound
\begin{equation}\label{approximation2}
\la \bw^n,f\ra = \la w^n, \varphi_f\ra=\la w^n,f\ra + \delta_n(w,f)\, n^{-2\beta} ,
\end{equation}
for every $f\in C_c^\infty(\R^d)$,
where $|\delta_n(w,f)|$ is bounded by a constant $\eta(f)>0$ uniformly in
$n,w$.

Just as in the previous section, we write
$(\Phi^n(t))_{t\geq 0}$ for the finite variation part of $(F(\la w_t^n,f\ra))_{t\geq 0}$ and
$(\Psi^n(t))_{t\geq 0}$ for its quadratic variation. As before, it is convenient to
split $\cG^n F(\la w,f\ra)$ into its neutral and selective components.
Using a Taylor expansion of the function $F$, we obtain that the neutral part is equal to
\begin{align}
& n^{1-\beta\alpha}F'(\la w,f\ra)\int_{\R^d} \int_{n^{-\beta}}^{\infty}\frac{1}{r^{d+1+\alpha}}\int_{B(x,r)}\frac{1}{V_r}\Big[w(y)\big\la \Theta^+_{x,r,u_n}(w)-w,f \big\ra \nonumber\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad + (1-w(y))\big\la \Theta^-_{x,r,u_n}(w)-w,f \big\ra \Big]\, dy dr dx  \nonumber\\
& \quad + n^{1-\beta\alpha}\frac{F''(\la w,f\ra)}{2} \int_{\R^d} \int_{n^{-\beta}}^{\infty}\frac{1}{r^{d+1+\alpha}}\int_{B(x,r)}\frac{1}{V_r}\Big[w(y)\big\la \Theta^+_{x,r,u_n}(w)-w,f \big\ra^2  \nonumber\\
& \qquad \qquad \qquad \qquad \qquad \qquad  \qquad \qquad \qquad \qquad + (1-w(y))\big\la \Theta^-_{x,r,u_n}(w)-w,f \big\ra^2 \Big]\, dy dr dx  + \e_n \nonumber\\
& = n^{1-\beta\alpha-\gamma}u F'(\la w,f\ra)\int_{\R^d} \int_{n^{-\beta}}^{\infty}\frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^2}\frac{1}{V_r}\, w(y)(f(z)-f(y))\, dydz dr dx \nonumber\\
& \quad + n^{1-\beta\alpha-2\gamma}\frac{u^2}{2}\, F''(\la w,f\ra)\int_{\R^d} \int_{n^{-\beta}}^{\infty}\frac{1}{r^{d+1+\alpha}}\int_{B(x,r)}\frac{1}{V_r}\, \big\{w(y) \la \ind_{B(x,r)}(1-w),f \ra^2 \nonumber\\
& \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad + (1-w(y))\la \ind_{B(x,r)}w,f\ra^2 \big\}\, dy dr dx +\e_n, \label{finite variation}
\end{align}
where
$$|\e_n|\leq n^{1-\alpha\beta -3 \gamma}\frac{u^3\|F^{(3)}\|_{\infty}}{3!}
\int_{\R^d}\int_{n^{-\beta}}^\infty\frac{1}{r^{d+1+\alpha}}
\int_{B(x,r)}\frac{2}{V_r}\la\ind_{B(x,r)},f\ra^3 dy dr dx.
$$
Consider the first term on the right hand side of~(\ref{finite variation}).
Since $1-\alpha\beta -\gamma=0$, $n^{1-\beta\alpha-\gamma}=1$. We split the integral over
the radii into the sum of the integrals over $[n^{-\beta},1]$ and $[1,\infty]$.
By using a Taylor expansion of $f$ and a symmetry argument to cancel the integral of $(z-y)dz$,
we obtain that
\begin{align*}
\bigg|u F'(\la w,f\ra)\int_{\R^d} & \int_{n^{-\beta}}^1\frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^2}\frac{1}{V_r}\, w(y)(f(z)-f(y))\, dydz drdx\bigg| \\
& \leq C \, \bigg|\int_{\R^d} \int_{n^{-\beta}}^1\frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^2}\frac{1}{V_r}\, \|z-y\|^2 \ind_{\{B(x,r)\cap S_f\neq \emptyset\}}\, dydzdrdx\bigg| \\
& \leq C' \hbox{Vol}(S_f+B(0,1))\int_{n^{-\beta}}^1 \frac{1}{r^{d+1+\alpha}}\, r^{d+2} dr = C''(1-n^{-\beta(2-\alpha)})
\end{align*}
for some constants $C,C',C''>0$.
To control the integral over radii in $[1,\infty)$, the
cruder bound $|f(y)-f(z)|\leq2\|f\|$ suffices and, using~(\ref{bound for integrating increments}),
we have
\begin{align*}
\bigg|u F'(\la w,f\ra)\int_{\R^d} &\int_1^\infty\frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^2}\frac{1}{V_r}\, w(y)(f(z)-f(y)) dydz drdx\bigg| \\
& \leq C \int  \int_1^\infty \frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^2}\frac{1}{V_r}\big(\ind_{\{y\in S_f\}}+ \ind_{\{z\in S_f\}}\big)  dydzdr dx\\
& \leq C' \int_1^\infty \frac{1}{r^{d+1+\alpha}}\int  \ \ind_{\{B(x,r)\cap S_f\neq \emptyset\}}\hbox{Vol}(S_f) dx dr\\
& \leq C'' \int_1^\infty \frac{1}{r^{d+1+\alpha}} \ r^d dr\leq C''',
\end{align*}
again for some constants $C,C',C''$ and $C'''$ which depend only on $d$, $F$ and $f$.

To control the second term on the right hand side of~(\ref{finite variation}) we use the
bounds~(\ref{bound for increments}), (\ref{bound for integrating increments}) to see that it
is bounded by
\begin{align*}
n^{-\gamma}& \frac{u^2}{2}\|F''\| \times 2C_1^2\|f\|^2 \int_{\R^d}\int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}(r^d \wedge 1)^2 \ind_{\{S_f\cap B(x,r)\neq \emptyset\}} dr dx \\
& = C_3 n^{-\gamma} \int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}(r^d \wedge 1)^2 (r^d \vee 1) dr \\
& = C_3 n^{-\gamma} \int_{n^{-\beta}}^1\frac{r^{2d}}{r^{d+1+\alpha}} dr + C_3 n^{-\gamma} \int_1^\infty \frac{r^d}{r^{d+1+\alpha}} dr = C_4 n^{-\gamma}\big(1- n^{-\beta(d-\alpha)}\big).
\end{align*}
When $d\geq 2$, $d-\alpha>0$ and so this bound tends to $0$ as $n\rightarrow \infty$.
When $d=1$, $(\alpha -1)\beta -\gamma = 0$, and so this term is bounded by a constant
as $n\rightarrow \infty$. The same calculation shows that $\e_n\rightarrow 0$, uniformly in $w$,
as $n\rightarrow\infty$. As a consequence, in any dimension the absolute value of the neutral term of $\cG^nF(\la w,f \ra)$ is bounded by a constant independent of $n$ and $w$.

Proceeding in the same way as for the second term above, we obtain that the selection term of the generator is bounded by (recall that $1-\alpha\beta - \gamma=0$)
\begin{align*}
2u\sigma n^{1-\beta\alpha -\gamma -\delta}& \|F'\|\int_{\R^d}  \int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^3}\frac{1}{V_r^2}\, |f(z')| dy\,dz\,dz' dr dx \\
& \leq Cn^{-\delta}\int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}\int_{\R^d}  \big(1\wedge r^d\big)\, \ind_{\{B(x,r)\cap S_f\neq \emptyset\}} dx dr \\
& \leq  C'n^{-\delta}\int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}\, \big(1\wedge r^d\big)\big(1\vee r^d\big) dr \leq C'' n^{-\delta+\alpha\beta} = C'',
\end{align*}
since $\alpha\beta -\delta =0$. Together with our bounds on the neutral term, just as in
the corresponding part of the proof of Theorem~\ref{th:fixed},
this shows the tightness of the sequence of finite variation parts of
$(F(\la w_t^n,f \ra))_{t\geq 0}$.

For the quadratic variation part, a similar analysis yields that the integrand in
$\Psi^n(t)$ is bounded by
\begin{align*}
Cn^{1-\beta\alpha-2\gamma} & \int_{\R^d}  \int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}} \int_{B(x,r)^3}\frac{1}{V_r}\, f(z)f(z')  dy\, dz\, dz'dr dx \\
& \leq C'n^{-\gamma}\int_{\R^d}  \int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}\, \big(1\wedge r^d\big)^2 \ind_{\{B(x,r)\cap S_f\neq \emptyset\}} dr dx \\
& \leq C''n^{-\gamma} \int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}\, \big(1\wedge r^d\big)^2 \big(1\vee r^d\big) dr
\leq  C''' n^{-\gamma}\big(1+n^{-\beta(d-\alpha)}\big),
\end{align*}
which is bounded by a constant independent of $n$ and $w$.
As before, we conclude that the quadratic variation part of
$(F(\la w_t^n ,f\ra))_{t\geq 0}$ is tight.

By the same arguments as in the proof of Theorem~\ref{th:fixed},
we conclude that the sequence
$((\bw_t^n)_{t\geq 0})_{n\geq 1}$ is tight in $D_{\cM}[0,\infty)$.


\medskip
{\bf 2) Identifying the limit.}

Recall that the generator of $(\bw_t^n)_{t\geq 0}$ applied to the test
function $\la \cdot,f\ra$ can be obtained by applying the generator of
$(w_t^n)_{t\geq 0}$ (see (\ref{generator wn 2})) to the function $\la \cdot,\varphi_f\ra$, where
$\varphi$ is given by~(\ref{stable varphi}).

First, let us find the limit of its neutral component. 
Since $1-\alpha\beta-\gamma =0$, the prelimit takes the form
\begin{align*}
& un^{1-\alpha\beta -\gamma}\int_{\R^d} \int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^2}\frac{1}{V_r}\, w^n(y)(\varphi_f(z)-\varphi_f(y))\, dydz dr dx \\

&= u\int_{\R^d}  w^n(y)\int_{\R^d} \bigg(\int_{n^{-\beta}\vee \frac{|z-y|}{2}}^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(y,z)}{V_r}\ dr\bigg)(\varphi_f(z)-\varphi_f(y)) \, dz dy.
\end{align*}
But just as in Remark~\ref{relation f-varphif},
a simple Taylor expansion gives us that
$$
\varphi_f(z)-\varphi_f(y) = f(z)-f(y)+ \mathcal{O}(n^{-2\beta})
\big(\ind_{\{B_n(z)\cap S_f\neq\emptyset\}}+ \ind_{\{B_n(y)\cap S_f\neq\emptyset\}}\big),
$$
where $B_n(\cdot)=B(\cdot, n^{-\beta})$ and
the error term is uniform in $y$ and $z$. Since
\begin{align*}
& n^{-2\beta}\int_{\R^d}  w^n(y)\int_{\R^d}
\bigg(\int_{n^{-\beta}\vee \frac{|z-y|}{2}}^\infty
\frac{1}{r^{d+1+\alpha}}\frac{V_r(y,z)}{V_r}\, dr\bigg)\,
\big(\ind_{\{B_n(z)\cap S_f\neq\emptyset\}}+
\ind_{\{B_n(y)\cap S_f\neq\emptyset\}}\big) dz dy\\
& \qquad \qquad \leq C n^{-2\beta} \int_{S_f+B_n(0)}\int_{\R^d} \bigg(n^{-\beta} \vee \frac{|z-y|}{2}\bigg)^{-d-\alpha} dz dy \leq C' n^{-\beta(2-\alpha)} \rightarrow 0
\end{align*}
as $n\rightarrow \infty$, we can conclude that up to a vanishing error term,
the neutral part of the generator is given by
\begin{equation}\label{first approx}
u\int  w^n(y)\int \bigg(\int_{n^{-\beta}\vee \frac{|z-y|}{2}}^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(y,z)}{V_r}\ dr\bigg)(f(z)-f(y)) \, dz dy.
\end{equation}
Now, our computations in the paragraph on tightness imply that the function
$$
a_n(y) : y\mapsto \int \bigg(\int_{n^{-\beta}\vee \frac{|z-y|}{2}}^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(y,r)}{V_r}\ dr\bigg)(f(z)-f(y)) \, dz
$$
is a continuous function, uniformly bounded in $y$ and $n$. Hence,
up to a vanishing error term we can first replace $w^n$ by $\bw^n$ in
(\ref{first approx}) and, second,  use dominated convergence to pass to
the limit as $n\rightarrow \infty$ in the expression for $a_n$. Doing so,
we obtain that the limit of the neutral term in the generator of
$(\bw_t^n)_{t\geq 0}$, applied to some $\bw$, is equal to
$$
u\int dy\, \bw(y)\int  \Phi(|z-y|)(f(z)-f(y))\, dz,
$$
where, as in~(\ref{generator of stable motion}),
$$
\Phi(|z-y|):= \int_{\frac{|z-y|}{2}}^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(y,z)}{V_r}\, dr.
$$
\begin{lemma}
\label{lemma generator alpha-stable}
Writing
\begin{equation}\label{generator stable motion}
\Da f(y) = u \int_{\R^d} \Phi(|z-y|) (f(z)-f(y)) dz,
\end{equation}
$\Da$ is the generator of a symmetric $\alpha$-stable process
$(\zeta_t)_{t\geq 0}$.
\end{lemma}
\noindent
{\bf Proof}

It is reassuring to first check that this is the generator of a well-defined L\'evy process:
\begin{multline*}
\int_{\R^d}(1\wedge |y|^2) \int_0^\infty \frac{1}{r^{d+1+\alpha}}\, \frac{V_r(0,y)}{V_r}\, dr\, dy
\\
\leq C \int_0^1\frac{1}{r^{d+1+\alpha}}\int_{B(0,2r)}|y|^2\, dy dr
+
C' \int_1^\infty\frac{1}{r^{d+1+\alpha}}dr
<\infty.
\end{multline*}

To verify that the associated L\'evy process is a symmetric stable process,
we check the scaling property.
The generator of $(b^{-1/\alpha}\zeta_{bt})$ is given by
\begin{align*}
\Da_bf(y) & = b u \int_{\R^d} \Phi(|z-b^{1/\alpha}y|) (f(b^{-1/\alpha}z)-f(y))\, dz \\
& = u b^{1+d/\alpha} \int_{\R^d} \Phi(|b^{1/\alpha}z-b^{1/\alpha}y|) (f(z)-f(y))\, dz.
\end{align*}
But a simple change of variables gives us that
\begin{align*}
\Phi(|b^{1/\alpha}z-b^{1/\alpha}y|) &= \int_{\frac{b^{1/\alpha}|z-y|}{2}}^\infty \frac{1}{r^{d+1+\alpha}} \frac{V_r(b^{1/\alpha}y, b^{1/\alpha}z)}{V_r} \, dr\\
& = b^{-1-d/\alpha} \int_{\frac{|z-y|}{2}}^\infty \frac{1}{r^{d+1+\alpha}} \frac{V_r(y,z)}{V_r}\, dr,
\end{align*}
and so $\Da_b = \Da$ for all $b>0$. This shows the desired property of $\Da$.
\hfill$\Box$

Having identified the neutral part of the limit, we now turn to the selection
part. It is given by
\begin{equation} \label{stable selection}
u\sigma n^{1-\beta\alpha-\gamma-\delta}\int_{\R^d} \int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}\int_{B(x,r)^3}\frac{1}{V_r^2}\, (w^n(y)w^n(z)-w^n(z'))\varphi_f(z')\, dydzdz' drdx.
\end{equation}
Now, the term which is linear in $w^n$ is easy to deal with: by Fubini's theorem, it is equal to
\begin{align}
& u\sigma n^{-\delta} \int_{\R^d} \int_{n^{-\beta}}^\infty \frac{1}{r^{d+1+\alpha}}\int_{B(x,r)}w^n(z') \varphi_f(z')\, dz'drdx \nonumber \\
& \qquad= u\sigma n^{-\delta} \int_{\R^d} w^n(z')\varphi_f(z')
\bigg(\int_{n^{-\beta}}^\infty \frac{V_1r^d}{r^{d+1+\alpha}}\, dr\bigg)\, dz'
= \frac{u\sigma V_1}{\alpha}\, \la \bw^n,f\ra, \label{linear part selection}
\end{align}
where the last equality uses the fact that $\alpha\beta-\delta = 0$.

Similar calculations show that the `quadratic' term in (\ref{stable selection}) is equal to
$$
u\sigma n^{-\delta}\int_{\R^d} \int_{n^{-\beta}}^{n^{-\beta}\log n} \frac{1}{r^{d+1+\alpha}}\bigg(\int_{B(x,r)}\frac{1}{V_r}\, w^n(y)dy\bigg)^2 \int_{B(x,r)} \varphi_f(z)dz drdx+ \mathcal{O}\big( (\log n)^{-\alpha}\big).
$$
Now
$$
\bigg(\int_{B(x,r)}\frac{1}{V_r}\, w^n(y)dy\bigg)^2 = \bigg(\int_{B(x,r)}\frac{1}{V_r}\, w^n(y)dy + \bw^n(x)\bigg)\bigg(\int_{B(x,r)}\frac{1}{V_r}\, w^n(y)dy - \bw^n(x)\bigg) + \bw^n(x)^2.
$$
Suppose we have the following lemma (whose proof is quite technical and is given in Appendix~\ref{continuity stable}).
\begin{lemma}\label{lem: strong wish}
Under the conditions of Theorem~\ref{th:stable},
for every $x\in \R^d$, $t\geq 0$ and $r\in [n^{-\beta},n^{-\beta}\log n]$,
$$
\lim_{n\rightarrow \infty} \E\bigg[\bigg|\int_{B(x,r)}\frac{w^n_t(y)}{V_r}\, dy - \bw^n_t(x) \bigg|\bigg] =0,
$$
\end{lemma}

>From this result, we can conclude from a simple dominated convergence argument,
and a Taylor expansion of $\varphi_f$, that along any trajectory of $w^n$, the
`quadratic' part of (\ref{stable selection}) is equal to
\begin{align*}
& u\sigma n^{-\delta} \int \int_{n^{-\beta}}^{n^{-\beta}\log n}
\frac{V_r}{r^{d+1+\alpha}}\,
\bw^n(x)^2 f(x)\, drdx + o(1)= \frac{u\sigma V_1}{\alpha}\, \la (\bw^n)^2,f\ra + o(1).
\end{align*}

So far we have shown that
$$
\la \bw^n_t,f\ra - \la \bw^n_0,f\ra - \int_0^t \bigg\{\la \bw_s^n , \Da f\ra
-\frac{u\sigma V_1}{\alpha}\, \la \bw_s^n(1-\bw_s^n),f\ra  \bigg\}\, ds + o(1)
$$
is a martingale.


As a consequence, any limit point $(w_t^\infty)_{t\geq 0}$ of
$(\bw_t^n)_{t\geq 0}$ should satisfy: for every $f\in C_c^\infty$ and every $t\geq 0$,
\begin{equation}\label{semi-conclusion}
\la w^\infty_t,f\ra - \la w^\infty_0,f\ra = \int_0^t \bigg\{ \la w_s^\infty , \Da f\ra -\frac{u\sigma V_1}{\alpha}\, \la w_s^\infty(1-w_s^\infty),f\ra  \bigg\}\, ds + M_t(f),
\end{equation}
where $(M_t(f))_{t\geq 0}$ is a zero-mean martingale. As in the fixed
radius case, when $d\geq 2$ our computations in the paragraph on
tightness show that $M(f)\equiv 0$ and (\ref{semi-conclusion}) characterizes
any limit point of $((\bw_t^n)_{t\geq 0})_{n\geq 1}$.

When $d=1$, the instantaneous quadratic variation of $\la \bw_t^n,f\ra$ is given by
\begin{align*}
& u^2n^{-\gamma} \int_{\R}\int_{n^{-\beta}}^{n^{-\beta}\log n} \frac{1}{r^{2+\alpha}V_r}\int_{B(x,r)^3} \big\{w^n(y)(1-w^n(z))(1-w^n(z'))+ (1-w^n(y))w^n(z)w^n(z')\big\}\\
& \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \times  \varphi_f(z)\varphi_f(z')dydzdz'drdx + \mathcal{O}\big((\log n)^{-\alpha}\big),
\end{align*}
where we have used the estimates obtained in the paragraph on tightness, which show
that selective events and neutral events with radii greater than $n^{-\beta}\log n$ do not
contribute.

Writing $\bw^{n,r}(x)$ for the average value of $w^n$ over $B(x,r)$
(so that $\bw^{n,n^{-\beta}}(x)=\bw^n(x)$), essentially the same techniques show that the integral
above is equal to
\begin{align}
&u^2n^{-\gamma}\int_\R f(x)^2\int_{n^{-\beta}}^{n^{-\beta}\log n} \frac{V_r^2}{r^{2+\alpha}}\, \big\{\bw^{n,r}(x)(1-\bw^{n,r}(x))^2 + (1-\bw^{n,r}(x))\bw^{n,r}(x)^2 \big\}\, drdx + o(1)\nonumber \\
& = u^2n^{-\gamma}\int_\R f(x)^2\int_{n^{-\beta}}^{n^{-\beta}\log n} \frac{V_r^2}{r^{2+\alpha}}\, \bw^n(x)(1-\bw^n(x))\, drdx \nonumber\\
& \quad + u^2n^{-\gamma}\int_\R f(x)^2\int_{n^{-\beta}}^{n^{-\beta}\log n} \frac{V_r^2}{r^{2+\alpha}}\, \big(\bw^{n,r}(x)- \bw^n(x)\big)\big(1-\bw^{n,r}(x)\big) \, drdx\nonumber \\
& \quad + u^2n^{-\gamma}\int_\R f(x)^2\int_{n^{-\beta}}^{n^{-\beta}\log n} \frac{V_r^2}{r^{2+\alpha}}\, \bw^n(x)\big(\bw^n(x)- \bw^{n,r}(x)\big) \, drdx + o(1).  \label{stable quadratic form}
\end{align}
But the first term on the right hand side of (\ref{stable quadratic form}) is equal to
$$
\frac{4u^2}{\alpha-1}\, \la \bw^n(1-\bw^n),f^2 \ra,
$$
while the other two terms tend to $0$ by Lemma~\ref{lem: strong wish} and dominated convergence. As a consequence,
any limit of $((\bw_t^n)_{t\geq 0})_{n\geq 1}$ satisfies the martingale problem stated in Theorem~\ref{th:stable}.

Uniqueness follows from duality with the limiting process of Theorem~\ref{th: conv duals alpha}, as before.

This completes the proof of Theorem~\ref{th:stable}. \hfill $\Box$

\medskip
Finally, let us prove the convergence of the rescaled dual process.

\medskip
\noindent {\bf Proof of Theorem~\ref{th: conv duals alpha}.}

Most of the proof is identical to that of Theorem~\ref{th: conv duals}.
That the only possible limit for $(\Xi_t^n)_{t\geq 0}$
is the system of branching (and in one dimension coalescing) symmetric
$\alpha$-stable processes described in the theorem, again
follows from an adaptation of Chapter~7 of \cite{LIA2009}, in which the only change is
that Brownian motion is replaced by the stable process generated by $\Da$
(see (\ref{generator stable motion})).

Next, we have to show that the sequence $((\Xi_t^n)_{t\geq 0})_{n\geq 1}$
is tight. As in the proof of Theorem~\ref{th: conv duals}, we shall use Aldous' criterion based on stopping times. That is, we fix an initial value $\mathbf{x}=\{x_1,\ldots,x_k\}\in (\R^d)^k$, $T>0$, $f\in C_c^2(\R^d)$ with values in $[0,1]$ and show that if $(\tau_n)_{n\geq 1}$ is a sequence of stopping times bounded by $T$, then for every $\e>0$ there exists $\delta= \delta(f,T,\mathbf{x})>0$ such that
\begin{equation}\label{tightness again}
\limsup_{n\rightarrow \infty} \, \P_{\mathbf{x}}\bigg[\sup_{0\leq t \leq \delta} \bigg|\prod_{i=1}^{N^n_{\tau_n+t}} f\big(\xi_{\tau_n+t}^{n,i}\big) - \prod_{i=1}^{N^n_{\tau_n}} f\big(\xi_{\tau_n}^{n,i}\big)\bigg|>\e\bigg]\leq \e.
\end{equation}

Again, we proceed in three steps. First, by exactly the same arguments
as in the proof of Theorem~\ref{th: conv duals}, and retaining the notation
from that setting, there exists $K>0$ such that for every $n\in \N$,
\begin{equation}\label{tight 1}
\P_{\mathbf{x}}(A_n):=\P_{\mathbf{x}}\bigg[\sup_{0\leq s\leq T} |\Xi^n|\leq K \bigg] \geq 1-\frac{\e}{4}.
\end{equation}
Furthermore, there exists $\delta_1>0$, independent of the subinterval of $[0,T]$ considered, such that
\begin{equation}\label{tight 2}
\P_{\mathbf{x}}\big[\hbox{at least }1\hbox{ particle created in }[\tau_n,\tau_n+\delta_1]\, ;\, A_n\big] \leq \frac{\e}{4}.
\end{equation}

As before, the difficulty will be to control the coalescence, but suppose for a moment that there is no
change in the number of particles in the interval $[\tau_n, \tau_n+\delta]$ and write
$I_n$ for the indexing set of the particles in $\Xi_{\tau_n}^n$. Then, exactly as before,
\begin{equation}\label{diff products 2}
\bigg|\prod_{i \in I_n} f\big(\xi_{\tau_n+t}^{n,i}\big) - \prod_{i\in I_n} f\big(\xi_{\tau_n}^{n,i}\big)\bigg|\leq C \|\nabla f\| \sum_{i\in I_n}\big|\xi_{\tau_n+t}^{n,i} - \xi_{\tau_n}^{n,i}\big|,
\end{equation}
and so it suffices to consider the motion of a single lineage. This is slightly more involved
than in the fixed radius case.

Let $(Z^n_t)_{t\geq 0}$ be a L\'evy process, independent of $(\xi_t^n)_{t\geq 0}$ and with generator
$$
D^nf(x) := u (1+s_n)\int_0^{n^{-\beta}} \frac{1}{r^{d+1+\alpha}} \int_{\R^d}\frac{V_r(x,y)}{V_r}\, \big(f(y)-f(x)\big)\, dy\, dr,
$$
then the process $(X_t)_{t\geq 0}$ defined by $X_t=\xi_t^n+Z_t^n$ has generator $(1+s_n)\Da$, where $\Da$ was shown in
Lemma~\ref{lemma generator alpha-stable} to be the generator of a symmetric stable process.
Using the strong Markov property and standard results on the growth of L\'evy
processes, see e.g.~\cite{pruitt:1981}, we have for $\eta$ and $\delta>0$,
\begin{equation*}
\P\bigg[\sup_{t\in [0,\delta]}\, \big|X_{\tau_n+t} - X_{\tau_n}\big| >\eta\bigg]<C\frac{\delta}{\eta^\alpha}
\end{equation*}
for a constant $C$ which is independent of $\eta$ and $\delta$.

Since
\begin{equation}\label{decomposition}
\P\bigg[\sup_{t\in [0,\delta]}\, \big|\xi_{\tau_n+t}^n - \xi_{\tau_n}^n\big| >\eta \bigg]
\leq \P\bigg[\sup_{t\in [0,\delta]}\, \big|X_{\tau_n+t} - X_{\tau_n}\big| >\eta \bigg]
+ \P\bigg[\sup_{t\in [0,\delta]}\, \big|Z_{\tau_n+t}^n - Z_{\tau_n}^n\big| >\eta \bigg],
\end{equation}
it remains to show that
\begin{equation*}
\P\bigg[\sup_{t\in [0,\delta]}\, \big|Z_{\tau_n+t}^n - Z_{\tau_n}^n\big| >\eta \bigg]\rightarrow 0,
\qquad\mbox{as }\delta\rightarrow 0.
\end{equation*}
Now, by construction, $(Z^n_t)_{t\geq 0}$
is a L\'evy process whose generator $D^n$ satisfies, for $f\in C^2(\R^d)$,
\begin{align}
|D^nf(x)| & = \bigg|u\int_0^{n^{-\beta}} \frac{1}{r^{d+1+\alpha}}\int_{\R^d}\frac{V_r(x,y)}{V_r}
\big[(y-x).\nabla f(x) + \mathcal{O}(|y-x|^2)\big]\, dydr\bigg| \label{bound dn}\\
& \leq C \int_0^{n^{-\beta}}\frac{1}{r^{d+1+\alpha}}\int_{B(x,2r)}|y-x|^2\, dydr = C'n^{-\beta (2-\alpha)},
\nonumber
\end{align}
where the Taylor expansion is justified since $V_r(x,y)=0$ if $|x-y|>2r$ and we are concentrating
on radii $r\leq n^{-\beta}$, and the first integral on the right hand side of~(\ref{bound dn}) vanishes by
rotational symmetry.

The process $(Z_t^n)_{t\geq 0}$ has finite quadratic variation, whose time derivative when $Z_t^n=x$ is
\begin{align*}
(1+s_n) u & \int_0^{n^{-\beta}}\frac{1}{r^{d+1+\alpha}} \int_{\R^d} \frac{V_r(x,y)}{V_r}\, \big(f(y)-f(x)\big)^2 \, dydr \\
& = (1+s_n) u \int_0^{n^{-\beta}}\frac{1}{r^{d+1+\alpha}} \int_{\R^d} \frac{V_r(x,y)}{V_r}\, \big((y-x).\nabla f(x) + \mathcal{O}(|y-x|^2)\big)^2\, dydr \\
& \leq C \int_0^{n^{-\beta}} \frac{1}{r^{d+1+\alpha}} \int_{B(x,2r)}|y-x|^2\, dydr = C'n^{-\beta(2-\alpha)}.
\end{align*}
Hence, we can also conclude that for a fixed $\eta$ and any $\delta$,
$$
\lim_{n\rightarrow \infty}\P\bigg[\sup_{t\in [0,\delta]}\, \big|Z_{\tau_n+t}^n - Z_{\tau_n}^n\big|
>\eta \bigg] =0.
$$
Coming back to (\ref{decomposition}), this means that we can
find $\delta_2>0$ such that for $n$ large enough,
$$
\P\bigg[\sup_{t\in [0,\delta]}\, \big|\xi_{\tau_n+t}^n - \xi_{\tau_n}^n\big| >\eta \bigg] \leq \frac{\e}{4K}.
$$
Choosing $\eta = \e/(KC \|\nabla f\|)$ and recalling (\ref{diff products 2}), we obtain that for
all sufficiently large $n$,
\begin{equation}\label{tight 3}
\P_{\mathbf{x}}\bigg[\sup_{t\in [0,\delta_2]}\bigg|\prod_{i \in I_n} f\big(\xi_{\tau_n+t}^{n,i}\big)
- \prod_{i\in I_n} f\big(\xi_{\tau_n}^{n,i}\big)\bigg|> \e\, ;\, A_n, B_{\delta_2}^c, C_{\delta_2}^c\bigg]\leq \frac{\e}{4}.
\end{equation}

Exactly as in the fixed radius case, tightness will be proven if we can show that coalescence
events cannot accumulate. In particular, since we have controlled the total number of particles and
the probability of branching, we just need to control the probability that two lineages
coalesce. The result will be based on the following lemma.
\begin{lemma}\label{lem: limit coal}
Let $(\tilde{\xi}^1_{n^\gamma t})_{t\geq 0}$ and $(\tilde{\xi}^2_{n^\gamma t})_{t\geq 0}$ be two independent copies of the
motion of a single (unscaled) lineage on the timescale $(n^\gamma t,\, t\geq 0)$,
and let $\zeta^n_t=\tilde{\xi}^2_{n^\gamma t}-\tilde{\xi}^1_{n^\gamma t}$
denote their difference. 
Then, for every $t\geq 0$ we have:

$(i)$ When $d=1$, there exists $C(t)>0$ such that
$$
\limsup_{n\rightarrow \infty} \, \E\bigg[\frac{1}{n^\gamma}\, \int_0^{n^{1-\gamma}t} \frac{1}{2^\alpha \vee |\zeta^n_s|^\alpha}\, ds\bigg] \leq C(t).
$$
Furthermore, the function $t\mapsto C(t)$ can be chosen such that $C(t)\downarrow 0$ as $t\rightarrow 0$.

$(ii)$ When $d\geq 2$,
$$
\lim_{n\rightarrow \infty} \E\bigg[\frac{1}{n^\gamma}\, \int_0^{n^{1-\gamma}t} \frac{1}{2^\alpha \vee |\zeta^n_s|^\alpha}\, ds\bigg] =0.
$$
\end{lemma}
We defer the proof of Lemma~\ref{lem: limit coal} until after the end of the proof
of Theorem~\ref{th: conv duals alpha}.

Suppose that we start with two lineages at some (unscaled) separation $z_0\in \R^d$.

As before, we work on the timescale $n^\gamma$ so that a single lineage jumps at
rate $\mathcal{O}(1)$ and suppose the two lineages $\xi^1$ and $\xi^2$ are currently at
locations $0$ and $z$ (in fact, only their separation matters). Then, the generator
$\mathcal{A}$ of the difference walk $(\xi_{n^\gamma t}^2-\xi_{n^\gamma t}^1)_{t\geq 0}$
is equal to
\begin{align}
\mathcal{A}f(z) & = 2 u(1+s_n) \int_{\R^d} \bigg\{ \int_1^{\infty} \frac{1}{r^{d+1+\alpha}}\int_{\R^d} \ind_{\{0\notin B(x,r)\}}\ind_{\{z\in B(x,r)\}}\frac{ \ind_{\{y\in B(x,r)\}}}{V_r}\, dx dr \nonumber \\
& \quad + (1-u_n) \int_1^{\infty} \frac{1}{r^{d+1+\alpha}}\int_{\R^d} \ind_{\{0\in B(x,r)\}}\ind_{\{z\in B(x,r)\}}\frac{ \ind_{\{y\in B(x,r)\}}}{V_r}\, dx dr  \bigg\}(f(y)-f(z))dy \nonumber \\
& = 2u(1+s_n) \int_{\R^d} \bigg\{\int_1^\infty \frac{1}{r^{d+1+\alpha}}\, \frac{V_r(y,z)}{V_r}\, dr\bigg\}(f(y)-f(z))\, dy \nonumber \\
 & \quad - 2u^2n^{-\gamma}(1+s_n) \int_{\R^d}\bigg\{\int_1^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(0,y,z)}{V_r}\, dr \bigg\}(f(y)-f(z))\, dy
\label{perturbation}
\\ & \quad +u^2n^{-\gamma}(1+s_n) \int_{\R^d}\bigg\{\int_1^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(0,y,z)}{V_r}\, dr \bigg\}(f(\Delta)-f(z))\, dy,
\label{cemetry}
\end{align}
where $\Delta$ is a cemetry state, corresponding to the two walks having coalesced.

As a consequence, until coalescence
we can couple the difference walk (on the timescale $n^\gamma$) with the
difference $(\zeta_t^n)_{t\geq 0}$ between two independent random walks, each jumping according to the law
of a single lineage but with each jump $z\mapsto y$ `cancelled' with probability
$$
\Delta_n(z,y) = \frac{2u^2n^{-\gamma}(1+s_n) \int_1^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(0,y,z)}{V_r}\, dr}{2u(1+s_n)\int_1^\infty \frac{1}{r^{d+1+\alpha}}\, \frac{V_r(y,z)}{V_r}\, dr}.
$$
(One can check that these two descriptions give rise to the same jump times and embedded chain.)
Each time we cancel a jump, with probability one half it was a coalescence in the original system,
but the key point is that if there are no cancelled jumps, then there was no coalescence.

It therefore suffices to show that we can find $\delta_3>0$ such that, for sufficiently large $n$,
the probability that an event is cancelled in the interval $[0,\delta_3 n^{1-\gamma}]$ is
smaller than $\e/(4K(K-1))$.

Now, according to the expression on the right-hand side of (\ref{perturbation}),
when the two lineages lie at separation $z\in \R^d$, a cancelled event occurs at instantaneous rate
\begin{align*}
2u^2n^{-\gamma}&(1+s_n) \int_{\R^d}\bigg\{\int_1^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(0,y,z)}{V_r}\, dr \bigg\} dy \\
& \leq 2u^2n^{-\gamma}(1+s_n) \int_{\R^d}\int_{1\vee \frac{|z|}{2}\vee \frac{|y|}{2}} \frac{1}{r^{d+1+\alpha}}\, drdy = C_1n^{-\gamma} \big(2\vee |z|\big)^{-\alpha}.
\end{align*}
Hence, (using the coupling with $(\zeta_t^n)_{t\geq 0}$), the probability of having no event
cancelled up to time $n^{1-\gamma}t$ (corresponding to time $nt$ in original units) is equal to
\begin{align*}
\E& \bigg[\exp\bigg\{-\int_0^{n^{1-\gamma}t} 2u^2n^{-\gamma}(1+s_n) \int_{\R^d}\bigg\{\int_1^\infty \frac{1}{r^{d+1+\alpha}}\frac{V_r(0,y,\zeta_s^n)}{V_r\, }\, dr \bigg\} dyds\bigg\}\bigg] \\
& \geq \E\bigg[\exp\bigg\{-C_1n^{-\gamma}\int_0^{n^{1-\gamma}t} \big(2\vee |\zeta^n_s|\big)^{-\alpha}\, ds \bigg\}\bigg]\geq 1- C_1\, \E\bigg[n^{-\gamma} \int_0^{n^{1-\gamma}t} \frac{ds}{\big(2\vee |\zeta^n_s|\big)^{\alpha}}\bigg].
\end{align*}
But Lemma~\ref{lem: limit coal} shows that we can find $\delta_3>0$ such that
\begin{equation}\label{events disappear}
\limsup_{n\rightarrow \infty}\, \E\bigg[n^{-\gamma} \int_0^{n^{1-\gamma}\delta_3} \frac{ds}{\big(2\vee |\zeta^n_s|\big)^{\alpha}}\bigg] \leq \frac{\e}{4C_1K(K-1)}.
\end{equation}


This completes the proof of tightness and therefore of Theorem~\ref{th: conv duals alpha}. \hfill $\Box$

\medskip

\noindent {\bf Proof of Lemma~\ref{lem: limit coal}.}


As before, we shall exploit the fact that $(\zeta_t^n)_{t\geq 0}$ is `nearly' a symmetric
$\alpha$-stable process.
Indeed, the intensity at which $(\zeta_t^n)_{t\geq 0}$ jumps by some vector $y$ is independent of its current location and equal to
$$
2(1+s_n) \bigg(\int_1^\infty \frac{1}{r^{d+1+\alpha}}\, \frac{V_r(0,y)}{V_r}\, dr\bigg)\, dy.
$$
Writing $(Z_t^n)_{t\geq 0}$ for a jump process, independent of $(\zeta_t^n)_{t\geq 0}$, and with jump intensity
$$
2(1+s_n) \bigg(\int_0^1 \frac{1}{r^{d+1+\alpha}}\, \frac{V_r(0,y)}{V_r}\, dr\bigg)\, dy,
$$
then the generator of the process $(X_t)_{t\geq 0}$, where $X_t = \zeta_t^n + Z_t$,
is precisely $2(1+s_n)$ times the
operator $\Da$ defined in (\ref{generator stable motion}), which we already checked
corresponds to a symmetric $\alpha$-stable process.
Once again, the idea is that the jumps of $(Z_t)_{t\geq 0}$ (which are bounded by $2$) do not
contribute much to the evolution of $(X_t)_{t\geq 0}$. More precisely, let us show that there exists
$C>0$ such that for every $n$ large enough and every $s\geq 1$,
\begin{equation}\label{control Z}
\P\bigg[\frac{|Z_s|}{\sqrt{s}} > (\log n)^2\bigg] \leq C e^{-(\log n)^2/d}.
\end{equation}
To this end, observe first that since the law of $Z_s$ is invariant under rotation, we can write that
\begin{equation}\label{one dim}
\P\bigg[\frac{|Z_s|}{\sqrt{s}} > (\log n)^2\bigg] \leq d\, \P\bigg[\frac{|Z_s^{(1)}|}{\sqrt{s}} > \frac{(\log n)^2}{d}\bigg] = 2d\, \P\bigg[\frac{Z_s^{(1)}}{\sqrt{s}} > \frac{(\log n)^2}{d}\bigg],
\end{equation}
where $Z_s^{(1)}$ denotes the first coordinate of $Z_s$. Now, $(Z_s^{(1)})_{s\geq 0}$
is again a symmetric L\'evy process with jumps bounded by $2$, and so Theorem 25.3 in \cite{SAT1999}
shows that for every $s,q\geq 0$, $\E[\exp(q Z_s^{(1)})]<\infty$. In this case, it is known that
the characteristic exponent $\Psi$ of $(Z_s^{(1)})_{s\geq 0}$, given here by a formula of the form
$$
\Psi(q) = \int_{[-2,2]}\big(1-e^{iqx} + iqx\ind_{\{|x|< 1\}}\big) \, \nu(dx),
$$
has an analytic extension to the half-plane with negative imaginary part, and we have
$$
\E\big[e^{qZ_s^{(1)}}\big] = e^{s\psi(q)}, \qquad \hbox{with}\quad \psi(q) = -\Psi(-iq).
$$
As a consequence, the Markov inequality gives us that
\begin{equation}\label{step 2}
\P\bigg[\frac{Z_s^{(1)}}{\sqrt{s}} > \frac{(\log n)^2}{d}\bigg] \leq e^{-(\log n)^2/d + s\psi(1/\sqrt{s})}.
\end{equation}
Since the measure $\nu$ has support in $[-2,2]$, we can write that when $q$ is small
\begin{align*}
\psi(q) & = -\int_{[-2,2]} \big(1- [1+qx +q^2x^2/2 + \mathcal{O}(q^3x^3)] + qx\ind_{\{|x|<1\}}\big) \, \nu(dx) \\
& = q \int_{[-2,2]} x\ind_{\{|x|\geq 1\}}\nu(dx) + \frac{q^2}{2}\int_{[-2,2]}x^2\nu(dx) + \mathcal{O}(q^3),
\end{align*}
where the first term on the right is zero, by symmetry. Consequently, there exists a constant $C>0$ such that for every $s\geq 1$, $\psi(1/\sqrt{s})\leq C/s$. Together with (\ref{one dim}) and (\ref{step 2}), this gives us (\ref{control Z}).

It will be convenient to suppose that $\zeta_0=0$, but notice that there will be no loss of
generality in so-doing, since for $n$ sufficiently
large, $\zeta_0$ will be bounded by $(\log n)^2$ and so, for $s>1$, can be absorbed into our bound for $Z_s$.
Similarly, we can, and do,
replace $2^\alpha\wedge |\zeta_s^n|^\alpha$ by $1\wedge |\zeta_s^n|^\alpha$ in the
denominator of our integrand.

Based on these considerations, let us return to the integral of interest when $d\geq 2$.
Fixing $a\in (0,\gamma)$ and splitting the integral with respect to time into $\int_{[0,n^a]}+ \int_{[n^a, n^{1-\gamma}t]}$, we obtain
\begin{align}
\E& \bigg[\frac{1}{n^\gamma}\, \int_0^{n^{1-\gamma}t} \frac{1}{1 \vee |\zeta^n_s|^\alpha}\, ds\bigg]  = \mathcal{O}(n^{a-\gamma}) + \frac{1}{n^\gamma}\int_{n^a}^{n^{1-\gamma}t}\E\bigg[\frac{1}{1\vee |X_s - Z_s|^\alpha}\bigg]\, ds \nonumber\\
& \leq  C n^{a-\gamma} + n^{-\gamma}\int_{n^a}^{n^{1-\gamma}t}\P\bigg[\frac{|Z_s|}{\sqrt{s}}>(\log n)^2\bigg] \, ds + n^{-\gamma}\int_{n^a}^{n^{1-\gamma}t}\E\bigg[\frac{\ind_{\{|Z_s|\leq (\log n)^2\sqrt{s}\}}}{1\vee |X_s -Z_s|^\alpha}\bigg]\, ds \nonumber\\
& \leq C n^{a-\gamma} + Cn^{1-2\gamma}te^{-(\log n)^2/d} +  n^{-\gamma}\int_{n^a}^{n^{1-\gamma}t}\E\bigg[\frac{\ind_{\{|Z_s|\leq (\log n)^2\sqrt{s}\}}}{1\vee |X_s -Z_s|^\alpha}\bigg]\, ds.  \label{estimate coal rate}
\end{align}
Since the first two terms on the right tend to $0$ as $n\rightarrow \infty$,
it now suffices to show that the last term remains bounded when $n$ is large.

By Lemma 5.3 in \cite{BW1998}, if $(p_s^\alpha)_{s\geq 0}$ denotes the transition density of
$(X_t)_{s\geq 0}$, we have, for every $s>0$ and $x\in \R^d$,
\begin{equation}\label{alpha semigroup}
p_s^\alpha(0,x) =: p_s^\alpha(x) = s^{-d/\alpha} p_1^\alpha(xs^{-1/\alpha})
\end{equation}
and there exists $C_{d,\alpha}>0$ (independent of $x$) such that
\begin{equation}\label{bound semigroup}
0\leq p_1^\alpha(x) \leq C_{d,\alpha}\big(1+|x|^{d+\alpha}\big)^{-1}.
\end{equation}
Hence, for any $s\geq n^a$ and any $z\in \R^d$ such that $|z|\leq (\log n)^2 \sqrt{s}$, we can write
\begin{align*}
\E &\bigg[\frac{1}{1\vee |X_s - z|^\alpha}\bigg]  \leq s^{-d/\alpha}\int_{\R^d} \frac{1}{(1\vee |x-z|^\alpha)(1+|xs^{-1/\alpha}|^{d+\alpha})}\, dx \\
& \leq s^{-d/\alpha} \int_{B(z,1)}\frac{1}{1+|xs^{-1/\alpha}|^{d+\alpha}}\, dx + s^{-d/\alpha}\int_{B(z,1)^c}\frac{1}{|x-z|^\alpha (1+|xs^{-1/\alpha}|^{d+\alpha})}\, dx \\
& \leq Cs^{-d/\alpha} + C's^{-d/\alpha}\int_{B(0,s^{1/\alpha})\setminus B(z,1)} \frac{dx}{|x-z|^\alpha} + C''s^{-d/\alpha}\int_{B(0,s^{1/\alpha})^c}\frac{dx}{|x-z|^\alpha |xs^{-1/\alpha}|^{d+\alpha}}.
\end{align*}
But since $s \geq n^a$ and $|z|\leq (\log n)^2\sqrt{s}$, we have
$$
|z|s^{-1/\alpha}\leq (\log n)^2 s^{\frac{1}{2}-\frac{1}{\alpha}}\leq (\log n)^2 n^{-a(2-\alpha)/(2\alpha)}\rightarrow 0,
$$
and so the second term on the right is bounded (after a change to polar coordinates) by
$$
C' s^{-d/\alpha}\int_1^{s^{1/\alpha}} \rho^{d-1 -\alpha} d\rho = C' s^{-1},
$$
while the third term is bounded by
$$
C'' s^{-d/\alpha} s^{1 + d/\alpha}\int_{s^{1/\alpha}}^\infty \rho^{d-1 -2\alpha -d}d\rho = C'' s^{-1}.
$$
Since all the constants depend on neither $z$ (in the range considered) nor $s$, we deduce
that the right hand side of (\ref{estimate coal rate}) is bounded by
$$
C'n^{a-\gamma} + Cn^{1-2\gamma}t e^{-(\log n)^2/d} + C''n^{-\gamma}\big(n^{-a(d-\alpha)/d} +\log n + \log t\big) \rightarrow 0 \qquad \hbox{as }n\rightarrow \infty,
$$
which proves $(ii)$. \hfill

\medskip
The only point that differs when $d=1$ is that $1-d/\alpha >0$ and so
$$
n^{-\gamma} \int_{n^a}^{n^{1-\gamma}t} s^{-1/\alpha}ds \leq Cn^{-\gamma}n^{(1-\frac{1}{\alpha})(1-\gamma)} t^{1-\frac{1}{\alpha}}.
$$
An easy check confirms that $(1-\frac{1}{\alpha})(1-\gamma)-\gamma = 0$, and so $C(t)$ exists and is
proportional to $t^{1-\frac{1}{\alpha}}$. Since $\alpha>1$, we also have that
$C(t)\downarrow 0$ as $t\rightarrow 0$. \hfill $\Box$
{"config": "arxiv", "file": "1406.5884.tex"}