$\displaystyle\qquad\qquad{}+\bigg{|}\theta\int\sum_{ij}(y-x)_{i}(y-x)_{j}\left(\partial_{x_{i}}f(x)\partial_{x_{j}}r(z_{3},\eta)+r(y,\eta)\partial_{x_{i}x_{j}}f(z_{1})\right)q_{\theta}(x,dy)\bigg{|}$

Since $q_{\theta}(x,dy)$ is the density of a Gaussian with mean
$\vec{b}(x)/\theta$ and covariance $\mathbf{C}(x)/\theta$, and both
$\vec{b}(x)$ and $\mathbf{C}(x)$ are uniformly bounded, so that
$\theta\int(y-x)_{i}q_{\theta}(x,dy)$ is bounded as well. Furthermore, a
change of variables that diagonalizes $\mathbf{C}(x)$ shows for any
$g:\mathbb{R}^{d}\to\mathbb{R}^{d+d}$, that if
$C_{g}=\sup_{y}\sup_{\|z\|=1}\sum_{ij}g(y)_{ij}z_{i}z_{j}$ and
$\lambda_{*}=\sup_{y}\sup_{\|z\|=1}\sum_{ij}\mathbf{C}(y)_{ij}z_{i}z_{j}$ then

$\theta\int\sum_{ij}g(y)_{ij}(y-x)_{i}(y-x)_{j}q_{\theta}(x,dy)\leq
C_{g}\lambda_{*}.$

Condition 1 gives uniform bounds on the derivatives of
$r(x,\eta)=r(x,\rho_{r}*\eta(x))$ in this expression and so, provided $f$ also
has uniformly bounded first and second derivatives, we have a bound of the
form

$\displaystyle|B^{\theta}_{f}|\leq K_{1}+K_{2}|f(x)|,$

for suitable constants $K_{1}$, $K_{2}$ that depend only on the derivatives of
$f$.

Now suppose instead that Condition 2 of Lemma 2.9 is satisfied. First note
that

$\displaystyle|B_{f}^{\theta}|=\left|\theta\int_{\mathbb{R}^{n}}\big{\\{}f(y)r\big{(}y,\rho_{r}\\!*\\!\eta(y)\big{)}-f(x)r\big{(}x,\rho_{r}\\!*\\!\eta(x)\big{)}\big{\\}}q_{\theta}(x,dy)\right|$
$\displaystyle\qquad{}\leq\left|\theta\int_{\mathbb{R}^{n}}\big{\\{}f(y)r\big{(}y,\rho_{r}\\!*\\!\eta(y)\big{)}-f(x)r\big{(}x,\rho_{r}\\!*\\!\eta(y)\big{)}\big{\\}}q_{\theta}(x,dy)\right|$
(6.2)
$\displaystyle\qquad\qquad{}+\left|\theta\int_{\mathbb{R}^{n}}\big{\\{}f(x)r\big{(}x,\rho_{r}\\!*\\!\eta(y)\big{)}-f(x)r\big{(}x,\rho_{r}\\!*\\!\eta(x)\big{)}\big{\\}}q_{\theta}(x,dy)\right|.$

(Note the extra term introduced here, $r(x,\rho_{r}\\!*\\!\eta(y))$, has the
two arguments to $r$ “at different locations”, contrary to the usual pattern.)

Writing $K_{3}=\sup_{x,m}\max_{i}|\partial_{x_{i}}f(x)r(x,m)|$ and
$K_{4}=\sup_{x,m}\max_{i,j}|\partial_{x_{i}x_{j}}f(x)r(x,m)|$, the first term
is bounded exactly as above. For the second,

$r(x,\rho_{r}\\!*\\!\eta(y))-r(x,\rho_{r}\\!*\\!\eta(x))\\\
=(\rho_{r}\\!*\\!\eta(y)-\rho_{r}\\!*\\!\eta(x))r^{\prime}(x,\rho_{r}\\!*\\!\eta(x))+\frac{1}{2}(\rho_{r}\\!*\\!\eta(y)-\rho_{r}\\!*\\!\eta(x))^{2}r^{\prime\prime}(x,\overline{m}),$

where
$\overline{m}=\kappa^{\prime}\rho_{r}\\!*\\!\eta(x)+(1-\kappa^{\prime})\rho_{r}\\!*\\!\eta(y)$
for some $0\leq\kappa^{\prime}\leq 1$, and we have used $r^{\prime}$ and
$r^{\prime\prime}$ to denote the first and second derivatives of $r(x,m)$ with
respect to the second argument. So, writing $K_{5}=\|r^{\prime}\|_{\infty}$
and $K_{6}=\|r^{\prime\prime}\|_{\infty}$, the second term in (6.2) is bounded
by $f(x)$ multiplied by

$\displaystyle
K_{5}\left|\theta\int_{\mathbb{R}^{d}}(\rho_{r}\\!*\\!\eta(y)-\rho_{r}\\!*\\!\eta(x))q_{\theta}(x,dy)\right|+K_{6}\left|\theta\int_{\mathbb{R}^{d}}(\rho_{r}\\!*\\!\eta(y)-\rho_{r}\\!*\\!\eta(x))^{2}q_{\theta}(x,dy)\right|.$

Under Condition 2 of Assumptions 2.8, this is bounded by a constant times
$\rho_{\gamma}\\!*\\!\eta(x)+(\rho_{\gamma}\\!*\\!\eta(x))^{2}$ and
$\sup_{x}m^{2}\gamma(x,m)$ is bounded, so therefore
$|\gamma(x,\eta)B^{\theta}_{f}(x,\eta)|\leq K_{7}+K_{8}f(x)$, where $K_{7}$
comes from $K_{3}$, $K_{4}$, and the supremum of $\gamma$, while $K_{8}$ comes
from $K_{5}$, $K_{6}$, and the supremum of $m^{2}\gamma(x,m)$.

###  6.2 Proof of Theorem 2.10: convergence for the nonlocal process

In this section we prove Theorem 2.10. This would be implied by convergence of
the lookdown process (see Kurtz and Rodrigues (2011) and Etheridge and Kurtz
(2019)); however in our setting, because the parameters in the lookdown
process depend on the empirical distribution, we actually use tightness of the
sequence of population processes in the proofs of tightness for the
corresponding lookdown processes.

###### Proof of Theorem 2.10..

The proof follows a familiar pattern. First we extend $\mathbb{R}^{d}$ to its
one-point compactification $\overline{\mathbb{R}}^{d}$ and establish, in Lemma
6.2, compact containment of the sequence of scaled population processes in
$\mathcal{M}_{F}(\overline{\mathbb{R}}^{d})$ (for which, since we have
compactified $\mathbb{R}^{d}$, it suffices to consider the sequence of total
masses); armed with this, tightness of the population processes in
$\mathcal{D}_{[0,\infty)}(\mathcal{M}_{F}(\overline{\mathbb{R}}^{d}))$ follows
from tightness of the real-valued processes $(H(\eta_{t}))_{t\geq 0}$ for a
sufficiently large class of test functions $H$, which we establish through an
application of the Aldous-Rebolledo criterion in Lemma 6.3. These ingredients
are gathered together in Proposition 6.4 to deduce tightness of the scaled
population processes in the larger space
$\mathcal{D}_{[0,\infty)}(\mathcal{M}_{F}(\overline{\mathbb{R}}^{d}))$.

We then characterise limit points as solutions to a martingale problem in
Lemma 6.6; finally in Lemma 6.7 we check that in the process of passing to the
limit, no mass ‘escaped to infinity’, so that in fact the limit points take
values in $\mathcal{D}_{[0,\infty)}(\mathcal{M}_{F}(\mathbb{R}^{d}))$.

As advertised, we work with the one-point compactification of $\mathbb{R}^{d}$
and consider $(\eta^{N}_{t})_{t\geq 0}$ as a sequence of
$\mathcal{M}_{F}(\overline{\mathbb{R}}^{d})$-valued processes. Since, for each
$K>0$, $\\{\eta:\langle 1,\eta\rangle\leq K\\}$ is a compact set in
$\mathcal{M}_{F}(\overline{\mathbb{R}}^{d})$, we shall focus on controlling
$(\langle 1,\eta^{N}_{t}\rangle)_{t\geq 0}$. The key is that Assumptions 2.8
are precisely chosen to guarantee boundedness of the net per-capita
reproduction rate.

######  Lemma 6.1.

Under Assumptions 2.8, for all $f\in C^{2}_{b}(\mathbb{R}^{d})$ with uniformly
bounded first and second derivatives, and all $T>0$, there exists a
$C=C(f,T)<\infty$, independent of $N$, such that

$\mathbb{E}[\langle f,\eta^{N}_{t}\rangle]\leq C\mathbb{E}[\langle
1,\eta^{N}_{0}\rangle]$ (6.3)

for all $N\geq 1$.

###### Proof.

Consider the semimartingale decomposition from equation (2.3):

$\displaystyle\langle f,\eta^{N}_{t}\rangle$ $\displaystyle=\langle
f,\eta^{N}_{0}\rangle+\int_{0}^{t}\int_{\mathbb{R}^{d}}\big{\\{}\gamma(x,\eta^{N}_{s})B^{\theta}_{f}(x,\eta^{N}_{s})+f(x)F(x,\eta^{N}_{s})\big{\\}}\eta^{N}_{s}(dx)ds+M^{N}_{t}(f),$
(6.4)

where $M^{N}_{t}(f)$ is a martingale and $B^{\theta}_{f}$ is defined in (6.1).
First note that Condition 6 of Assumptions 2.8 stipulates that $|\gamma
B^{\theta}_{f}|$ is uniformly bounded by a contant times $1+f$, and so
recalling that $F$ is bounded above, we conclude that under Assumptions 2.8
$\gamma(x,\eta)B^{\theta}_{f}(x,\eta)+f(x)F(x,\eta)\leq C_{f}(1+f(x))$ for
some $C_{f}$.

Now, taking expectations in (6.4),

$\displaystyle\mathbb{E}\left[\langle f,\eta^{N}_{t}\rangle\right]$
$\displaystyle\leq\mathbb{E}\left[\langle
f,\eta^{N}_{0}\rangle\right]+C_{f}\int_{0}^{t}\mathbb{E}\left[\langle
1+f,\eta^{N}_{s}\rangle\right]dt.$ (6.5)

The bound (6.3) then follows by first applying Gronwall’s inequality in the
case $f=1$, which yields

$\mathbb{E}\big{[}\langle 1,\eta_{t}^{N}\rangle\big{]}\leq
e^{Ct}\mathbb{E}\big{[}\langle 1,\eta_{0}^{N}\rangle\big{]},$

with $C$ independent of $N$, and then substituting the resulting bound on
$\mathbb{E}\big{[}\langle 1,\eta_{s}^{N}\rangle\big{]}$ into the expression
above.

With a bound on per-capita net growth rate in hand, bounds on the expectation
of the supremum of the total population size over a finite time interval also
follow easily.

######  Lemma 6.2 (Compact containment for the population process).

Under the assumptions of Theorem 2.10, for each $T>0$, there exists some
constant $C_{T}$, independent of $N$, such that

$\displaystyle\mathbb{E}\left[\sup_{0\leq t\leq T}\langle
1,\eta^{N}_{t}\rangle\right]\leq C_{T}\mathbb{E}[\langle 1,\eta_{0}\rangle].$
(6.6)

In particular, for any $\delta>0$, there exists $K_{\delta}>0$ such that

$\limsup_{N\to\infty}\mathbb{P}\left\\{\sup_{s\in[0,T]}\langle
1,\eta^{N}_{s}\rangle>K_{\delta}\right\\}\leq\frac{C_{T}}{K_{\delta}}<\delta.$
(6.7)

###### Proof.

First note that by Lemma 6.1, $\mathbb{E}[\langle
1,\eta^{N}_{t}\rangle]\leq\mathbb{E}[\langle 1,\eta^{N}_{0}\rangle]e^{Ct}$ for
some $C$ (independent of $N$). Now, let $M^{N*}_{t}(f)=\sup_{0\leq s\leq
t}M^{N}_{t}(f)$, and as before let $\langle M^{N}(f)\rangle_{t}$ be the angle
bracket process of $M^{N}_{t}(f)$. The Burkholder-Davis-Gundy inequality says
that there is a $K$ for which $\mathbb{E}\left[M^{N*}_{t}(1)\right]\leq
K\mathbb{E}[\sqrt{[M^{N}(1)]_{t}}]$, where $[M^{N}(1)]_{t}$ is the quadratic
variation of $M^{N}(1)$. Furthermore, as discussed by Hernández-Hernández and
Jacka (2022), the expectation of the quadratic variation of a local martingale
is bounded by a (universal) constant multiple of the expectation of its angle
bracket process (Barlow et al., 1986, Item (4.b’), Table 4.1, p. 162). Now,
since $\sqrt{x}\leq 1+x$, in the notation of Lemma 6.1, there is a
$C^{\prime}$ such that

$\displaystyle\mathbb{E}\left[M^{N*}_{t}(1)\right]$ $\displaystyle\leq
C^{\prime}\left(1+\mathbb{E}\left[\left\langle
M^{N}(1)\right\rangle_{t}\right]\right)$
$\displaystyle=C^{\prime}\left(1+\frac{\theta}{N}\mathbb{E}\left[\int_{0}^{t}\Big{\langle}\left\\{\gamma(x,\eta^{N}_{s})\int_{\mathbb{R}^{d}}r(y,\eta^{N}_{s})q_{\theta}(x,dy)+\mu_{\theta}(x,\eta^{N}_{s})\right\\},\eta_{s}^{N}(dx)\Big{\rangle}ds\right]\right)$
$\displaystyle=C^{\prime}\Big{(}1+\mathbb{E}\Big{[}\int_{0}^{t}\Big{\langle}\Big{\\{}\frac{2\theta}{N}\gamma(x,\eta^{N}_{s})r(x,\eta^{N}_{s})$
$\displaystyle\qquad\qquad\qquad\qquad\qquad{}+\frac{\gamma(x,\eta_{s}^{N})}{N}B^{\theta}_{1}(x,\eta^{N}_{s})-\frac{1}{N}F(x,\eta^{N}_{s})\Big{\\}},\eta_{s}^{N}(dx)\Big{\rangle}ds\Big{]}\Big{)}.$

We have not assumed that $F$ is bounded below, but to see that the term
involving $-F$ does not cause us problems, we rearrange equation (6.4) with
$f=1$ to see that

$\displaystyle\begin{split}\mathbb{E}\left[\int_{0}^{t}\Big{\langle}-F(x,\eta^{N}_{s}),\eta^{N}_{s}(dx)\Big{\rangle}ds\right]&=\mathbb{E}[\langle
1,\eta^{N}_{0}\rangle]-\mathbb{E}[\langle 1,\eta^{N}_{t}\rangle]\\\
&\qquad{}+\mathbb{E}\left[\int_{0}^{t}\Big{\langle}\gamma(x,\eta_{s}^{N})B^{\theta}_{1}(x,\eta_{s}^{N}),\eta_{s}^{N}(dx)\Big{\rangle}ds\right],\end{split}$
(6.8)

which is bounded since $\gamma(x,\eta)$ and $B^{\theta}_{1}(x,\eta)$ are both
bounded and $\langle 1,\eta_{t}\rangle\geq 0$. Since
$\theta/N\to\alpha<\infty$, combining constants, we obtain that for some
$C^{\prime\prime}$,

$\displaystyle\mathbb{E}\left[M^{N*}_{t}(1)\right]$ $\displaystyle\leq
C^{\prime}+C^{\prime\prime}\mathbb{E}[\langle 1,\eta_{0}^{N}\rangle]e^{tC}.$

Taking suprema and expectations on both sides of equation (6.4), then again
using the fact that $\gamma(x,\eta)B^{\theta}_{1}(x,\eta)+F(x,\eta)\leq C$,

$\displaystyle\mathbb{E}\left[\sup_{0\leq s\leq T}\langle
1,\eta^{N}_{s}\rangle\right]$ $\displaystyle\leq\mathbb{E}[\langle
1,\eta^{N}_{0}\rangle]+\mathbb{E}\left[\sup_{0\leq t\leq
T}\int_{0}^{t}\Big{\langle}\left\\{\gamma(x,\eta^{N}_{s})B^{\theta}_{1}(x,\eta^{N}_{s})+F(x,\eta^{N}_{s})\right\\},\eta^{N}_{s}(dx)\Big{\rangle}ds\right]$
$\displaystyle\qquad\qquad{}+\mathbb{E}[M^{N*}_{t}(1)]$
$\displaystyle\leq\mathbb{E}[\langle
1,\eta^{N}_{0}\rangle]+C\mathbb{E}\left[\int_{0}^{T}\sup_{0\leq s\leq
t}\langle 1,\eta^{N}_{s}\rangle
dt\right]+C^{\prime}+C^{\prime\prime}\mathbb{E}[\langle
1,\eta_{0}^{N}\rangle]e^{tC}.$

Once again applying Gronwall’s inequality,

$\displaystyle\mathbb{E}\left[\sup_{0\leq s\leq T}\langle
1,\eta^{N}_{s}\rangle\right]$ $\displaystyle\leq
C^{\prime\prime\prime}\left(1+\mathbb{E}[\langle
1,\eta^{N}_{0}\rangle]\right)e^{2TC}.$

For any $T$, the quantity on the right is bounded above by a constant $C(T)$
independent of $N$. As a result, for any $K>0$,

$\limsup_{N\to\infty}\mathbb{P}\left[\sup_{0\leq s\leq T}\langle
1,\eta^{N}_{s}\rangle\geq K\right]\leq\frac{C(T)}{K}.$

Our next task is to show tightness of $(\langle f,\eta_{t}^{N}\rangle)_{t\geq
0}$ for $f\in C_{b}^{\infty}(\overline{\mathbb{R}}^{d})$.

######  Lemma 6.3 (Tightness of $(\langle f,\eta^{N}_{t}\rangle)_{t>0}$).

For each $f\in C^{\infty}_{b}(\overline{\mathbb{R}}^{d})$, the collection of
processes $(\langle f,\eta^{N}_{t}\rangle)_{t\geq 0}$ for $N=1,2,\ldots$ is
tight as a sequence of càdlàg, real-valued processes.

###### Proof.

The Aldous-Rebolledo criterion (Theorem B.2) applied to the semimartingale
representation of $\langle f,\eta^{N}_{t}\rangle$ of equation (6.4), tells us
that it suffices to show that for each $T>0$, (a) for each fixed $0\leq t\leq
T$, the sequence $\\{\langle f,\eta^{N}_{t}\rangle\\}_{N\geq 1}$ is tight, and
(b) for any sequence of stopping times $\tau_{N}$ bounded by $T$, and for each
$\nu>0$, there exist $\delta>0$ and $N_{0}>0$ such that

$\displaystyle\sup_{N>N_{0}}\sup_{t\in[0,\delta]}\mathbb{P}\left\\{\left|\int_{\tau}^{\tau+t}\int_{\mathbb{R}^{d}}\left\\{\gamma(x,\eta^{N}_{s})B^{\theta}_{f}(x,\eta^{N}_{s})+f(x)F(x,\eta^{N}_{s})\right\\}\eta^{N}_{s}(dx)ds\right|>\nu\right\\}<\nu,$
(6.9)
$\displaystyle\text{and}\qquad\sup_{N>N_{0}}\sup_{t\in[0,\delta]}\mathbb{P}\left\\{\big{|}[M^{N}(f)]_{\tau+t}-[M^{N}(f)]_{\tau}\big{|}>\nu\right\\}<\nu.$
(6.10)

Tightness of $\langle f,\eta^{N}_{t}\rangle$ for fixed $t$ follows from Lemma
6.1 and Markov’s inequality, so we focus on the remaining conditions.

The proof of Lemma 6.1 provides a uniform bound on $\gamma B^{\theta}_{f}$,
but we only know that $F$ is bounded above. However, by assumption, for each
fixed value of $m$, $\sup_{k\leq m}|F(x,k)|$ is uniformly bounded as a
function of $x$. Noting that $\rho_{F}\\!*\\!\eta\leq\langle
1,\eta\rangle\|\rho_{F}\|_{\infty}$, we can use Lemma 6.2 to choose $N_{0}$
and $K$ such that if $N>N_{0}$, then

$\mathbb{P}\left\\{\sup_{0\leq s\leq T}\langle 1,\eta_{s}^{N}\rangle\geq
K\right\\}<\nu/2,$

we now choose $\delta_{1}$ so that

$\delta_{1}\|f\|_{\infty}\sup\big{\\{}\sup_{x}|F(x,k)|:k\leq
K\|\rho_{F}\|_{\infty}\big{\\}}<\nu/4,\qquad\sup_{x,\eta}\gamma(x,\eta)B^{\theta}_{f}(\,\eta)\delta_{1}<\nu/4,$

so that (6.9) is satisfied with $\delta=\delta_{1}$.

Similarly,

$\displaystyle\big{|}[M^{N}(f)]_{\tau+t}-[M^{N}(f)]_{\tau}\big{|}$
$\displaystyle\qquad=\Big{|}\int_{\tau}^{\tau+t}\frac{\theta}{N}\int_{\mathbb{R}^{d}}\left\\{\gamma(x,\eta^{N}_{s})\int_{\mathbb{R}^{d}}f^{2}(y)r(y,\eta^{N}_{s})q_{\theta}(x,dy)+\mu_{\theta}(x,\eta^{N}_{s})f^{2}(x)\right\\}\eta^{N}_{s}(dx)ds\Big{|}$
$\displaystyle\qquad=\Big{|}\int_{\tau}^{\tau+t}\frac{\theta}{N}\int_{\mathbb{R}^{d}}\left\\{\gamma(x,\eta^{N}_{s})\left(2f^{2}(x)r(x,\eta^{N}_{s})+B^{\theta}_{f^{2}}(x,\eta^{N}_{s})\right)-f^{2}(x)\frac{F(x,\eta^{N}_{s})}{\theta}\right\\}\eta^{N}_{s}(dx)ds\Big{|},$

and so using the fact that $\theta/N\to\alpha<\infty$, an argument entirely
analogous to that for (6.9) yields a $\delta_{2}$ for which (6.10) is
satsified. Taking $\delta=\min\\{\delta_{1},\delta_{2}\\}$, the result
follows.

We collect the implications of the last two lemmas into a proposition.

######  Proposition 6.4 (Tightness of $(\eta^{N}_{t})_{t\geq 0}$).

The collection of measure-valued processes $\\{(\eta^{N}_{t})_{t\geq 0}:N\geq
1\\}$ is tight in
$\mathcal{D}_{[0,\infty)}(\mathcal{M}_{F}(\overline{\mathbb{R}}^{d}))$.

###### Proof.

Theorem 3.9.1 in Ethier and Kurtz (1986) says that if the collection of
$E$-valued processes satisfies a compact containment condition (for any
$\epsilon>0$ and $T>0$, there is a compact set such that the processes stay
within that set up to time $T$ with probability at least $1-\epsilon$), then
the collection is relatively compact (which is equivalent to tightness since
we are working on a Polish space) if and only if $\\{(f(\eta^{N}_{t}))_{t\geq
0}:N\geq 1\\}$ is relatively compact for all $f$ in a dense subset of
$C_{b}(E)$ under the topology of uniform convergence in compact sets.

Since $\\{\nu:\langle 1,\nu\rangle\leq K\\}$ is compact in
$\mathcal{M}_{F}(\overline{\mathbb{R}}^{d})$, Lemma 6.2 gives compact
containment. Lemma 6.3 shows that the real-valued processes $\langle
f,\eta^{N}_{t}\rangle$ are relatively compact for all
$f\in\mathcal{C}^{\infty}_{b}(\overline{\mathbb{R}}^{d})$. Since by the Stone-
Weierstrass theorem, the algebra of finite sums and products of terms of this
form is dense in the space of bounded continuous functions on
$\mathcal{M}_{F}(\overline{\mathbb{R}}^{d})$, and tightness of $\langle
f,\eta_{t}^{N}\rangle$ extends to sums and products of this form by Lemma B.3,
we have relative compactness in
$\mathcal{D}_{[0,\infty)}(\mathcal{M}_{F}(\overline{\mathbb{R}}^{d}))$.

We wish to characterise the limit points of $\\{(\eta^{N}_{t})_{t>0}\\}_{N\geq
1}$ as solutions to a martingale problem with generator $\mathcal{P}^{\infty}$
which we now identify. Most of the work was done in Section 4. First, we
record an equivalent formulation of the martingale problems, which were
essentially laid out in Subsection 4.1.

######  Lemma 6.5.

For $G\in\mathcal{C}^{\infty}(\mathbb{R})$ with
$\|G^{\prime\prime\prime}\|_{\infty}<\infty$, and
$f\in\mathcal{C}_{b}^{\infty}(\overline{\mathbb{R}}^{d})$, define the function
$G_{f}$ by $G_{f}(\eta):=G(\langle f,\eta\rangle)$. Let $\mathcal{P}^{N}$ be
the generator given by

$\displaystyle\mathcal{P}^{N}G_{f}(\eta)$ $\displaystyle:=\theta
N\bigg{\langle}\gamma(x,\eta)\int\left(G(\langle
f,\eta\rangle+f(z)/N)-G(\langle
f,\eta\rangle)\right)r(z,\eta)q_{\theta}(x,dz)$ (6.11)
$\displaystyle\qquad\qquad{}+\left(G(\langle f,\eta\rangle-f(x)/N)-G(\langle
f,\eta\rangle)\right)\mu_{\theta}(x,\eta),\eta(dx)\bigg{\rangle}.$

The process $(\eta^{N}_{t})_{t\geq 0}$ of Definition 2.4 is the unique
solution to the $(\mathcal{P}^{N},\eta_{0})$-martingale problem, i.e., if

$M_{t}:=G_{f}(\eta^{N}_{t})-G_{f}(\eta^{N}_{0})-\int_{0}^{t}\mathcal{P}^{N}G_{f}(\eta^{N}_{s})ds$

is a martingale (with respect to the natural $\sigma$-field).

Furthermore, let $\mathcal{P}^{\infty}$ be the generator given by

$\displaystyle\mathcal{P}^{\infty}G_{f}(\eta)$
$\displaystyle:=G^{\prime}(\langle
f,\eta\rangle)\big{\langle}\gamma(x,\eta)\mathcal{B}\left(f(\cdot)r(\cdot,\eta)\right)(x)+f(x)F(x,\eta),\eta(dx)\big{\rangle}$
(6.12) $\displaystyle\qquad{}+\alpha G^{\prime\prime}(\langle
f,\eta\rangle)\big{\langle}\gamma\left(x,\eta\right)r\left(x,\eta\right)f^{2}(x),\eta(dx)\big{\rangle}.$

A process $(\eta^{\infty}_{t})_{t\geq 0}$ satisifes the martingale
characterization of equations (2.7) and (2.8) if and only if it is a solution
to the $(\mathcal{P}^{\infty},\eta^{\infty}_{0})$-martingale problem, i.e., if
for all such test functions

$M_{t}:=G_{f}(\eta^{\infty}_{t})-G_{f}(\eta^{\infty}_{0})-\int_{0}^{t}\mathcal{P}^{\infty}G_{f}(\eta^{\infty}_{s})ds$

is a martingale (with respect to the natural $\sigma$-field).

######  Lemma 6.6 (Characterisation of limit points).

Suppose that $(\eta^{N}_{0})_{N\geq 1}$ converges weakly to $\eta_{0}$ as
$N\to\infty$. Then any limit point of $\\{(\eta^{N}_{t})_{t\geq 0}\\}_{N\geq
1}$ in $\mathcal{D}_{[0,\infty)}(\mathcal{M}_{F}(\overline{\mathbb{R}}^{d}))$
is a solution to the martingale problem for $(\mathcal{P}^{\infty},\eta_{0})$.

###### Proof.

We use Theorem 4.8.2 in Ethier and Kurtz (1986). First observe that the set of
functions $\\{G_{f}(\eta):=G(\langle
f,\eta\rangle),~{}G\in\mathcal{C}^{\infty}(\mathbb{R}),\|G^{\prime\prime\prime}\|_{\infty}<\infty,~{}f\in\mathcal{C}_{b}^{\infty}(\overline{\mathbb{R}}^{d})\\}$
is separating on $\mathcal{M}_{F}(\overline{\mathbb{R}}^{d})$. Therefore, it
suffices to show that for any $t>0$ and $\tau>0$ that

$\lim_{N\to\infty}\mathbb{E}\left[\left(G_{f}(\eta^{N}_{t+\tau})-G_{f}(\eta^{N}_{t})-\int_{t}^{t+\tau}\mathcal{P}^{\infty}G_{f}(\eta^{N}_{s})ds\right)\prod_{i=1}^{k}h_{i}(\eta^{N}_{t_{i}})\right]=0$
(6.13)

for all $k\geq 0$, $0\leq t_{1}<t_{2}<\ldots,t_{k}\leq t<t+\tau$, and bounded
continuous functions $h_{1},\ldots,h_{k}$ on
$\mathcal{M}_{F}(\overline{\mathbb{R}}^{d})$. Since $(\eta^{N}_{t})_{t\geq 0}$
is Markov, the tower property gives that, for each $N$,

$\mathbb{E}\left[\left(G_{f}(\eta^{N}_{t+\tau})-G_{f}(\eta^{N}_{t})-\int_{t}^{t+\tau}\mathcal{P}^{N}G_{f}(\eta^{N}_{s})ds\right)\prod_{i=1}^{k}h_{i}(\eta^{N}_{t_{i}})\right]=0.$
(6.14)

Therefore, it suffices to show that

$\lim_{N\to\infty}\mathbb{E}\left[\int_{t}^{t+\tau}\left|\mathcal{P}^{N}G_{f}(\eta^{N}_{s})-\mathcal{P}^{\infty}G_{f}(\eta^{N}_{s})\right|ds\prod_{i=1}^{k}h_{i}(\eta^{N}_{t_{i}})\right]=0,$
(6.15)

and, again using the tower property, since the functions $h_{i}$ are bounded,
this will follow if

$\lim_{N\to\infty}\mathbb{E}\left[\left.\int_{t}^{t+\tau}\left|\mathcal{P}^{N}G_{f}(\eta^{N}_{s})-\mathcal{P}^{\infty}G_{f}(\eta^{N}_{s})\right|ds\right|{\cal
F}_{t}\right]=0$ (6.16)

(where $\\{{\cal F}_{t}\\}_{t\geq 0}$ is the natural $\sigma$-field).

We rewrite $\mathcal{P}^{N}G_{f}(\eta^{N}_{s})$ using a Taylor series
expansion up to third order for $G\big{(}\langle f,\eta\rangle\pm
f(y)/N\big{)}$ around $G(\langle f,\eta\rangle)$. As in Section 4 (except that
now we are more explicit about the error term), we find

$\displaystyle\mathcal{P}^{N}G_{f}(\eta):=$ $\displaystyle G^{\prime}(\langle
f,\eta\rangle)\int_{\mathbb{R}^{d}}\theta\Big{\langle}\gamma(x,\eta)\int_{\mathbb{R}^{d}}f(y)r(y,\eta)q_{\theta}(x,dy),-f(x)\mu_{\theta}(x,\eta),\eta(dx)\Big{\rangle}$
(6.17) $\displaystyle+\frac{1}{2}\frac{\theta}{N}G^{\prime\prime}(\langle
f,\eta\rangle)\int_{\mathbb{R}^{d}}\Big{\langle}\gamma(x,\eta)\int_{\mathbb{R}^{d}}f^{2}(y)r(y,\eta)q_{\theta}(x,dy)+f^{2}(x)\mu_{\theta}(x,\eta),\eta(dx)\Big{\rangle}$
$\displaystyle+\frac{1}{6}\frac{\theta}{N^{2}}\Big{\langle}G^{\prime\prime\prime}(w)\gamma(x,\eta)\int_{\mathbb{R}^{d}}f^{3}(y)r(y,\eta)q_{\theta}(x,dy)-G^{\prime\prime\prime}(v)f^{3}(x)\mu_{\theta}(x,\eta),\eta(dx)\Big{\rangle}$

for some $w,v\in[\langle f,\eta\rangle-\|f\|_{\infty}/N,\langle
f,\eta\rangle+\|f\|_{\infty}/N]$.

Combining with equation (4.5), and the fact that $\mu_{\theta}(x,\eta)\to
r(x,\eta)\gamma(x,\eta)$ as $\theta\to\infty$, we have pointwise convergence:

$\lim_{N\to\infty}|\mathcal{P}^{N}G(\langle
f,\eta\rangle)-\mathcal{P}^{\infty}G(\langle f,\eta\rangle)|=0.$ (6.18)

To conclude convergence of the expectation, we would like to apply the
Dominated Convergence Theorem in (6.15). Recall that $f$ and $G$ and their
derivatives are bounded, and so $\gamma(x,\eta)$ is bounded independent of
$\theta$. Since $\theta/N^{2}\to 0$, rearranging as in (4.3) and using the
convergence of (4.4), we deduce that we can dominate
$\left|\mathcal{P}^{N}G_{f}(\eta^{N}_{s})-\mathcal{P}^{\infty}G_{f}(\eta^{N}_{s})\right|$
by a constant multiple of $\langle 1+|F(x)|,\eta_{s}(dx)\rangle$. Since $F$ is
bounded above, there is a constant $K$ such that $|F|\leq K-F$ so that,
exactly as in equation (6.8), we can check that

$\mathbb{E}\Big{[}\left.\int_{t}^{t+\tau}\big{\langle}|F(x,\eta^{N}_{s})|,\eta^{N}_{s}(dx)\big{\rangle}ds\right|{\cal
F}_{t}\Big{]}<\infty,$

which concludes our proof.

The last step in the proof of Theorem 2.10 is to check that any limit point
$(\eta_{t})_{t\geq 0}$ of $\\{(\eta^{N}_{t})_{t\geq 0}\\}_{N\geq 1}$ actually
takes its values in $\mathcal{M}_{F}(\mathbb{R}^{d})$, that is, “no mass has
escaped to infinity”.

######  Lemma 6.7.

Under the assumptions of Theorem 2.10, let $(\eta_{t})_{t\geq 0}$ be a limit
point of $\\{(\eta^{N}_{t})_{t\geq 0}\\}_{N\geq 1}$. For any $\delta>0$,

$\mathbb{P}\big{[}\eta_{t}\big{(}\\{\|x\|>R\\}\big{)}>\delta\big{]}\to
0\qquad\mbox{ as }R\to\infty.$

###### Sketch.

Take $f_{0}(x)$ as in the statement of Theorem 2.10, i.e., $f_{0}$ is
nonnegative, grows to infinity as $x\to\infty$, has uniformly bounded first
and second derivatives, and has $\langle f_{0},\eta^{N}_{0}\rangle$ uniformly
bounded in $N$. We take a sequence of test functions $f_{n}$ that increase to
the function $f_{0}$ and having uniformly bounded first and second
derivatives, so that there is a (single) $C$ from Condition 6 of Assumptions
2.8 such that $\gamma(x,\eta)B_{f_{n}}^{\theta}(x,\eta)\leq C(1+f_{n}(x))$ for
all $x$, $\eta$, and $f_{n}$. Then, just as we arrived at equation (6.5),

$\displaystyle\mathbb{E}[\langle f_{n}(x),\eta^{N}_{t}(dx)\rangle]$
$\displaystyle\leq\mathbb{E}[\langle
f_{n}(x),\eta^{N}_{0}(dx)\rangle]+C\int_{0}^{t}\mathbb{E}\Big{[}\Big{\langle}f_{n}(x),\eta^{N}_{s}(dx)\Big{\rangle}\Big{]}ds,$

with the same constant for all $n$ and all $N$. Gronwall’s inequality then
implies that $\mathbb{E}[\langle f_{n},\eta^{N}_{t}\rangle]\leq C^{\prime}$
for some $C^{\prime}$ independent of $n$, $N$, and $t\in[0,T]$. By first
taking $N\to\infty$ and then $n\to\infty$, we find that $\mathbb{E}[\langle
f_{0},\eta_{t}(dx)\rangle]\leq C^{\prime}$ for $t\in[0,T]$, and since
$f_{0}\to\infty$ as $|x|\to\infty$, an application of Markov’s inequality
tells us that for any $\delta>0$,

$\mathbb{P}\big{\\{}\eta_{t}(\\{x:\|x\|>R\\})>\delta\big{\\}}\leq\frac{\mathbb{E}[\langle
f_{0},\eta_{t}\rangle]}{\inf_{\\{x:\|x\|\geq R\\}}f_{0}(x)}\to 0\qquad\text{as
}R\to\infty.$

###  6.3 Convergence of some nonlocal equations to classical PDEs

It is natural to conjecture that when the limit of the rescaled population
process that we obtained in the previous section solves a nonlocal PDE, if we
further scale the kernels $\rho_{r}$, $\rho_{\gamma}$, and $\rho_{F}$ by
setting $\rho^{\epsilon}(\cdot)=\rho(\cdot/\epsilon)/\epsilon$, as
$\epsilon\to 0$, the corresponding solutions should converge to a limiting
population density that solves the corresponding “classical” PDE. We verify
this in two examples; in the first the nonlocal equation is a reaction-
diffusion equation with the “nonlocality” only appearing in the reaction term;
in the second the nonlocal PDE is a special case of a nonlinear porous medium
equation. These, in particular, capture the examples that we explored in
Section 3.2.

####  6.3.1 Reaction–diffusion equation limits

In this subsection we prove Proposition 2.15. The conditions of the
proposition are in force throughout this subsection. The proof rests on a
Feynman-Kac representation. We write $(Z_{t})_{t\geq 0}$ for a diffusion with
generator $\mathcal{B}^{*}$ and denote its transition density by $f_{t}(x,y)$.
The first step is a regularity result for this density.

######  Lemma 6.8.

Fix $T>0$. There exists a constant $K=K(T)>0$ such that, for any
$x,y\in\mathbb{R}^{d}$ and $t\in[0,T]$,

$\int|f_{t}(x,z)-f_{t}(y,z)|dz\leq\frac{\|x-y\|}{\sqrt{t}}K.$ (6.19)

###### Proof.

We first use the Intermediate Value Theorem to obtain the bound

$\displaystyle\int|f_{t}(x,z)-f_{t}(y,z)|dz$
$\displaystyle\leq\int\|x-y\|\|\nabla f_{t}(w,z)\|dz$

where $\nabla$ acts on the first coordinate only and $w$ is in the line
segment $[x,y]$ joining $x$ to $y$. Under our assumptions on $b$ and $C$,
equation (1.3) of Sheu (1991), gives existence of constants
$\lambda=\lambda(T)>0$ and $K$ such that,

$\|\nabla f_{t}(w,z)\|\leq\frac{K}{\sqrt{t}}p_{\lambda t}(w,z),$

where $p_{s}(x,y)$ is the Brownian transition density. Hence,

$\displaystyle\int|f_{t}(x,z)-f_{t}(y,z)|dz$ $\displaystyle\leq
K\frac{\|x-y\|}{\sqrt{t}}\int p_{\lambda t}(w,z)dz=K\frac{\|x-y\|}{\sqrt{t}}.$

######  Lemma 6.9.

Fix $T>0$. Let $x,y\in\mathbb{R}^{d}$, $t\in[0,T]$, and denote by
$(Z_{t}^{y})_{t\geq 0}$ and $(Z_{t}^{x})_{t\geq 0}$ independent copies of the
diffusion $(Z_{t})_{t\geq 0}$ starting from $y$ and $x$ respectively. There
exists a constant $K=K(T)>0$ such that,

$\mathbb{E}[\|Z_{t}^{y}-Z_{t}^{x}\|]\leq K(\sqrt{t}+\|y-x\|).$

###### Proof.

First we write,

$\mathbb{E}[\|Z_{t}^{y}-Z_{t}^{x}\|]=\int\int\|u-v\|f_{t}(y,u)f_{t}(x,v)dudv.$

Under our regularity assumptions on $C$, $b$, using equation (1.2) of Sheu
(1991), there exist constants $K$, $\lambda=\lambda(T)>0$ for which,

$f_{t}(y,u)\leq Kp_{\lambda t}(y,u).$

It then follows that,

$\displaystyle\mathbb{E}[\|Z_{t}^{y}-Z_{t}^{x}\|]\leq\int\int\|u-v\|K^{2}p_{\lambda
t}(y,u)p_{\lambda t}(x,v)dvdu=K^{2}\mathbb{E}[\|B_{\lambda t}^{y}-B_{\lambda
t}^{x}\|],$ (6.20)

where $(B_{t}^{y})_{t\geq 0}$ and $(B_{t}^{x})_{t\geq 0}$ are independent
Brownian motions starting at $y$ and $x$ respectively. Using the triangle
inequality, and writing $(B^{0}_{t})_{t\geq 0}$ for a Brownian motion started
from the origin,

$\mathbb{E}[\|B_{\lambda t}^{y}-B_{\lambda
t}^{x}\|]\leq\|y-x\|+\mathbb{E}[\|B^{0}_{2\lambda t}\|]\leq\|y-x\|+C\sqrt{t}.$
(6.21)

Substituting (6.21) in (6.20) gives the result.

We use the representations of the solutions to equations (2.12) and (2.11)
respectively:

$\displaystyle\varphi_{t}(x)$
$\displaystyle=\mathbb{E}_{x}\Big{[}\varphi_{0}(Z_{t})+\int_{0}^{t}\varphi_{s}(Z_{s})F(\varphi_{s}(Z_{s}))ds\Big{]},$
(6.22) $\displaystyle\varphi^{\epsilon}_{t}(x)$
$\displaystyle=\mathbb{E}_{x}\Big{[}\varphi_{0}(Z_{t})+\int_{0}^{t}\varphi^{\epsilon}_{s}(Z_{s})F(\rho^{\epsilon}_{F}*\varphi^{\epsilon}_{s}(Z_{s}))ds\Big{]},$
(6.23)

from which

$\varphi_{t}(x)-\varphi^{\epsilon}_{t}(x)=\mathbb{E}_{x}\left[\int_{0}^{t}\Big{(}\varphi_{s}(Z_{s})F\big{(}\varphi_{s}(Z_{s})\big{)}-\varphi^{\epsilon}_{s}(Z_{s})F\big{(}\rho_{\epsilon}*\varphi^{\epsilon}_{s}(Z_{s})\big{)}\Big{)}ds\right],$
(6.24)

where $\mathbb{E}_{x}$ denotes expectation for $Z$ with $Z_{0}=x$. The key to
our proof of Proposition 2.15 will be to replace $F(\varphi_{s}(Z_{s}))$ by
$F(\rho^{\epsilon}_{F}*\varphi_{s}(Z_{s}))$ in this expression. We achieve
this through three lemmas.

First we need a uniform bound on $\varphi$ and $\varphi^{\epsilon}$.

######  Lemma 6.10.

For any $T>0$ there exists $M=M(T,\|\varphi_{0}\|)>0$ such that, for all
$0\leq t\leq T$:

$\max\\{\|\varphi_{t}(\cdot)\|_{\infty},\|\varphi^{\epsilon}_{t}(\cdot)\|_{\infty}\\}<M.$

###### Proof.

Using that $\varphi_{0}$ and $F$ are bounded above, from the representation
(6.22), we have

$\displaystyle\varphi_{t}(x)\leq\|\varphi_{0}\|_{\infty}+K\mathbb{E}\Big{[}\int_{0}^{t}\varphi_{s}(Z_{s})ds\Big{]}.$

In particular,

$\|\varphi_{t}(\cdot)\|_{\infty}\leq\|\varphi_{0}\|_{\infty}+K\int_{0}^{t}\|\varphi_{s}(\cdot)\|_{\infty}ds,$

so, by Gronwall’s inequality,

$\|\varphi_{t}(\cdot)\|_{\infty}\leq\|\varphi_{0}\|_{\infty}\exp\left(KT\right).$

Similarly,
$\|\varphi^{\epsilon}_{t}(\cdot)\|_{\infty}\leq\|\varphi_{0}\|_{\infty}\exp\left(KT\right)$.

We also need a continuity estimate for $\varphi$.

######  Lemma 6.11.

Let $T>0$. There exists a constant $K=K(T,\|\varphi_{0}\|_{\infty})>0$ and
$\delta_{0}=\delta_{0}(T,\|\varphi_{0}\|_{\infty})>0$ such that for all
$0<\delta<\delta_{0}$ and $0\leq t\leq T$,

$\|x-y\|<\delta^{3}\Rightarrow|\varphi_{t}(x)-\varphi_{t}(y)|<K\delta.$

###### Proof.

First we need some notation. Fix $T>0$ and write $M$ for the corresponding
constant from Lemma 6.10. Let $\|F\|_{M}=\sup_{m\in[0,M]}|F(m)|$. We reserve
$\widehat{K}$ for the constant on the right hand side of equation (6.19) and
$\widetilde{K}$ for the constant in Lemma 6.9, and write $K_{\varphi_{0}}$ for
the Lipschitz constant of $\varphi_{0}$. Set

$\delta_{0}=\min\Big{(}\frac{1}{\|F\|_{M}^{2}},\frac{1}{Me\big{(}2\|F\|_{M}+\widehat{K}\big{)}},\frac{1}{\widetilde{K}K_{\varphi_{0}}+2\|F\|_{M}M},1\Big{)}.$

In what follows we take $0<\delta<\delta_{0}$.

We first prove that the result holds if $t<\delta^{2}$. As before let
$Z_{t}^{x}$ and $Z_{t}^{y}$ be independent copies of the diffusion $Z_{t}$
starting at $x$ and $y$ respectively. From our representation (6.22) and Lemma
6.10, we can write:

$\displaystyle|\varphi_{t}(x)-\varphi_{t}(y)|$
$\displaystyle\leq\big{|}\mathbb{E}_{x}[\varphi_{0}(Z_{t})]-\mathbb{E}_{y}[\varphi_{0}(Z_{t})]\big{|}+2\|F\|_{M}Mt$
$\displaystyle\leq\mathbb{E}[|\varphi_{0}(Z_{t}^{x})-\varphi_{0}(Z_{t}^{y})|]+2\|F\|_{M}Mt$
$\displaystyle\leq
K_{\varphi_{0}}\mathbb{E}[\|Z_{t}^{x}-Z_{t}^{y}\|]+2\|F\|_{M}Mt$
$\displaystyle\leq\widetilde{K}K_{\varphi_{0}}(\sqrt{t}+\|y-x\|)+2\|F\|_{M}Mt$
$\displaystyle\leq\widetilde{K}K_{\varphi_{0}}(\delta+\delta^{3})+2\|F\|_{M}M\delta^{2}\leq(\widetilde{K}K_{\varphi_{0}}+1)\delta,$

where we have used Lemma 6.9 in the fourth inequality and the definition of
$\delta_{0}$ in the last inequality.

Suppose now that $\delta^{2}<t$. We will follow the pattern in Lemma 2.2 of
Penington (2017). First, note that by the Feynman-Kac formula we have an
alternative representation for $\varphi_{t}(x)$: for any $t^{\prime}<t$,

$\varphi_{t}(x)=\mathbb{E}_{x}\Big{[}\varphi_{t-t^{\prime}}(Z_{t^{\prime}})\exp\big{(}\int_{0}^{t^{\prime}}F(\varphi_{t-s}(Z_{s}))ds\big{)}\Big{]}.$

Therefore, setting $t^{\prime}=\delta^{2}$ and using Lemma 6.10, for all $z$,

$e^{-\delta^{2}\|F\|_{M}}\mathbb{E}_{z}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}\leq\varphi_{t}(z)\leq
e^{\delta^{2}\|F\|_{M}}\mathbb{E}_{z}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}.$

We can then deduce that

$\displaystyle\varphi_{t}(x)-\varphi_{t}(y)$ $\displaystyle\leq
e^{\delta^{2}\|F\|_{M}}\mathbb{E}_{x}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}-e^{-\delta^{2}\|F\|_{M}}\mathbb{E}_{y}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}$
$\displaystyle=e^{\delta^{2}\|F\|_{M}}\Big{(}\mathbb{E}_{x}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}-\mathbb{E}_{y}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}\Big{)}$
$\displaystyle\qquad\qquad+\big{(}e^{\delta^{2}\|F\|_{M}}-e^{-\delta^{2}\|F\|_{M}}\big{)}\mathbb{E}_{y}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}$
$\displaystyle\leq
e^{\delta^{2}\|F\|_{M}}\Big{(}\mathbb{E}_{x}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}-\mathbb{E}_{y}\big{[}\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]})+M\big{(}e^{\delta^{2}\|F\|_{M}}-e^{-\delta^{2}\|F\|_{M}}\big{)}.$
(6.25)

To bound the differences of the expected values in the last equation note
that, by using again Lemma 6.10,

$\displaystyle\mathbb{E}_{x}\big{[}$
$\displaystyle\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}-\mathbb{E}_{y}[\varphi_{t-\delta^{2}}(Z_{\delta^{2}})\big{]}$
$\displaystyle=\int\varphi_{t-\delta^{2}}(z)(f_{\delta^{2}}(x,z)-f_{\delta^{2}}(y,z))dz$
$\displaystyle\leq
M\int\big{|}f_{\delta^{2}}(x,z)-f_{\delta^{2}}(y,z)\big{|}dz$
$\displaystyle\leq M\widehat{K}\frac{\|x-y\|}{\delta}\leq
M\widehat{K}\delta^{2},$

where we have used Lemma 6.8 and that $\|x-y\|<\delta^{3}$. Substituting in
(6.25),

$\displaystyle\varphi_{t}(x)-\varphi_{t}(y)$ $\displaystyle\leq
e^{\delta^{2}\|F\|_{M}}\left(M\widehat{K}\delta^{2}+M-Me^{-2\delta^{2}\|F\|_{M}}\right)$
$\displaystyle\leq
e^{\delta^{2}\|F\|_{M}}\left(M\widehat{K}\delta^{2}+2M\delta^{2}\|F\|_{M}\right)$
$\displaystyle\leq
e\left(M\widehat{K}+2M\|F\|_{M}\right)\delta^{2}\leq\delta,$

where the last two inequalities follow from the definition of $\delta$.
Interchanging $x$ and $y$ yields the same bound for
$\varphi_{t}(y)-\varphi_{t}(x)$, and the result follows.

We proceed to control the difference between $F(\varphi)$ and
$F(\rho^{\epsilon}_{F}*\varphi)$. Note first that since $\rho_{F}\in L^{1}$,

$I(\epsilon):=\int_{\\{\|y\|>\epsilon^{3/4}\\}}\rho^{\epsilon}_{F}(y)dy=\int_{\\{\|y\|>\epsilon^{-1/4}\\}}\rho_{F}(y)dy\to
0\qquad\mbox{ as }\epsilon\to 0.$

######  Lemma 6.12.

Let $T>0$. There exists a constant $C=C(T,\|\varphi_{0}\|_{\infty})>0$ such
that, for all $0\leq t\leq T$, for all $\epsilon$ small enough,

$\|\varphi_{t}(\cdot)-\rho^{\epsilon}_{F}*\varphi_{t}(\cdot)\|_{\infty}\leq
C(I(\epsilon)+\epsilon^{1/4}).$ (6.26)

Furthermore, there is a constant
$\widetilde{C}(T,\|\varphi_{0}\|_{\infty})=\widetilde{C}$ such that, for all
$0\leq t\leq T$,

$\|F(\varphi_{t}(\cdot))-F(\rho^{\epsilon}_{F}*\varphi_{t}(\cdot))\|_{\infty}\leq\widetilde{C}\big{(}I(\epsilon)+\epsilon^{1/4}\big{)}.$
(6.27)

###### Proof.

Let $\epsilon<\delta_{0}^{4}$, with $\delta_{0}$ from Lemma 6.11. Then,

$\displaystyle|\varphi_{t}(x)-\rho^{\epsilon}_{F}*\varphi_{t}(x)|$
$\displaystyle\leq\int_{\|x-y\|>\epsilon^{3/4}}\rho^{\epsilon}_{F}(x-y)|\varphi_{t}(y)-\varphi_{t}(x)|dy$
$\displaystyle+\int_{\|x-y\|\leq\epsilon^{3/4}}\rho^{\epsilon}_{F}(x-y)|\varphi_{t}(y)-\varphi_{t}(x)|dy$
$\displaystyle\leq
2M\int_{\|x-y\|>\epsilon^{3/4}}\rho^{\epsilon}_{F}(x-y)dy+\int_{\|x-y\|\leq\epsilon^{3/4}}\rho^{\epsilon}_{F}(x-y)K\epsilon^{1/4}dy$
$\displaystyle\leq 2MI(\epsilon)+K\epsilon^{1/4},$

where we used the estimates of Lemma 6.10 and Lemma 6.11. This proves (6.26).
For (6.27), let $L_{M}$ be the (uniform) Lipschitz constant of $F$ on $[0,M]$,
with $M$ still taken from Lemma 6.10. Then,

$\displaystyle\|F(\varphi_{t}(\cdot))-F(\rho^{\epsilon}_{F}*\varphi_{t}(\cdot))\|_{\infty}$
$\displaystyle\leq
L_{M}\|\varphi_{t}(\cdot))-(\rho^{\epsilon}_{F}*\varphi_{t}(\cdot))\|_{\infty}$
$\displaystyle\leq L_{M}(2MI(\epsilon)+K\epsilon^{1/4}),$

which proves (6.27).

###### of Proposition 2.15.

Let $\epsilon$ be small enough that Lemma 6.12 holds. We use the notation
$\widehat{\delta}(\epsilon)$ for the quantity on the right hand side of
(6.27). Then from the representation (6.24) and Lemma 6.12 we can write,

$\displaystyle|$ $\displaystyle\varphi_{t}(x)-\varphi^{\epsilon}_{t}(x)|$
$\displaystyle\leq\mathbb{E}_{x}\left[\int_{0}^{t}\Big{|}\varphi_{s}(Z_{s})F(\rho^{\epsilon}_{F}*\varphi_{s}(Z_{s}))-\varphi^{\epsilon}_{s}(Z_{s})F\big{(}\rho^{\epsilon}_{F}*\varphi^{\epsilon}_{s}(Z_{s})\big{)}\Big{|}ds\right]+Mt\widehat{\delta}(\epsilon)$
$\displaystyle\leq\mathbb{E}_{x}\left[\int_{0}^{t}\big{|}F(\rho^{\epsilon}_{F}*\varphi^{\epsilon}_{s}(Z_{s}))\big{|}\cdot\big{|}\varphi^{\epsilon}_{s}(Z_{s})-\varphi_{s}(Z_{s})\big{|}ds\right]$
$\displaystyle\qquad+\mathbb{E}_{x}\left[\int_{0}^{t}|\varphi_{s}(Z_{s})|\big{|}F(\rho^{\epsilon}_{F}*\varphi^{\epsilon}_{s}(Z_{s}))-F(\rho^{\epsilon}_{F}*\varphi_{s}(Z_{s}))\big{|}ds\right]+Mt\widehat{\delta}(\epsilon)$
$\displaystyle\leq\|F\|_{M}\int_{0}^{t}\|\varphi^{\epsilon}_{s}(\cdot)-\varphi_{s}(\cdot)\|_{\infty}ds+ML_{M}\int_{0}^{t}\|\rho^{\epsilon}_{F}*\varphi^{\epsilon}_{s}(\cdot)-\rho^{\epsilon}_{F}*\varphi_{s}(\cdot)\|_{\infty}ds+Mt\widehat{\delta}(\epsilon)$
$\displaystyle\leq(\|F\|_{M}+ML_{M})\int_{0}^{t}\|\varphi^{\epsilon}_{s}(\cdot)-\varphi_{s}(\cdot)\|_{\infty}ds+Mt\widehat{\delta}(\epsilon),$

where the second inequality is the triangle inequality, and the third is Lemma
6.10. An application of Gronwall’s inequality then yields,

$\displaystyle\|\varphi^{\epsilon}_{t}(\cdot)-\varphi_{t}(\cdot)\|_{\infty}$
$\displaystyle\leq Mt\widehat{\delta}(\epsilon)\exp(t(\|F\|_{M}+ML_{M}))$
$\displaystyle\leq MT\widehat{\delta}(\epsilon)\exp(T(\|F\|_{M}+ML_{M})),$

giving the result, since $\widehat{\delta}\to 0$ as $\epsilon\to 0$.

####  6.3.2 Porous Medium Equation

In this subsection we prove Proposition 2.18. To ease notation, we present the
proof in $d=1$ (although we retain the notation $\nabla$). However, to recall
the dependence on $\epsilon$ we write $\rho^{\epsilon}$ for $\rho_{\gamma}$.
It should be clear that it extends almost without change to higher dimensions.

Recall that we are concerned with non-negative solutions to the equation
(2.13):

$\partial_{t}\psi_{t}^{\epsilon}(x)=\Delta\left(\psi_{t}^{\epsilon}\,\rho^{\epsilon}*\psi_{t}^{\epsilon}\right)(x)+\psi_{t}^{\epsilon}(x)\left(1-\rho^{\epsilon}*\psi_{t}^{\epsilon}(x)\right).$

and we assume that $\rho=\zeta*\check{\zeta}$ with $\zeta$ a rapidly
decreasing function and $\check{\zeta}(x)=\zeta(-x)$. The example we have in
mind is $\zeta$ (and therefore $\rho$) being the density of a mean zero
Gaussian random variable. We shall prove that under the assumptions of
Proposition 2.18, as $\epsilon\to 0$, we have convergence to the solution to
the porous medium equation with logistic growth, equation (1.2):

$\partial_{t}\psi_{t}(x)=\Delta\left(\psi_{t}^{2}\right)(x)+\psi_{t}(x)\left(1-\psi_{t}(x)\right).$

We work on the time interval $[0,T]$. We will require a lower bound on
$\int\psi_{t}^{\epsilon}(x)\log\psi_{t}^{\epsilon}(x)dx$ which we record as a
lemma.

######  Lemma 6.13.

Suppose that there exists $\lambda\in(0,1)$ and $C<\infty$, both independent
of $\epsilon$, such that $\int\exp(\lambda|x|)\psi_{0}^{\epsilon}(x)dx<C$.
Then there exists a constant $K<\infty$, independent of $\epsilon$, such that
$\int\psi_{t}^{\epsilon}(x)\log\psi_{t}^{\epsilon}(x)dx>-K$ for all
$t\in[0,T]$.

###### Proof.

First observe that, since $x\log x$ is bounded below,
$\int_{-1}^{1}\psi_{t}^{\epsilon}(x)\log\psi_{t}^{\epsilon}(x)dx$ is bounded
below, and recall that $\psi_{t}^{\epsilon}(x)\geq 0$.

Now consider

$\frac{d}{dt}\int\exp(\lambda x)\psi_{t}^{\epsilon}(x)dx=\int\exp(\lambda
x)\Delta\big{(}\psi_{t}^{\epsilon}\,\rho^{\epsilon}*\psi_{t}^{\epsilon}\big{)}(x)dx\\\
+\int\exp(\lambda
x)\psi_{t}^{\epsilon}(x)\big{(}1-\rho^{\epsilon}*\psi_{t}^{\epsilon}(x)\big{)}dx\\\
=\int(\lambda^{2}-1)\exp(\lambda
x)\psi_{t}^{\epsilon}(x)\rho_{\epsilon}\\!*\\!\psi_{t}^{\epsilon}(x)dx+\int\exp(\lambda
x)\psi_{t}^{\epsilon}(x)dx\\\ \leq\int\exp(\lambda
x)\psi_{t}^{\epsilon}(x)dx,$ (6.28)

and so, by Gronwall’s inequality, $\int\exp(\lambda
x)\psi_{t}^{\epsilon}(x)dx$ is uniformly bounded on $[0,T]$. In particular,
combining with the Mean Value Theorem, we find

$\int_{x}^{x+1}\psi_{t}^{\epsilon}(y)dy\leq C\exp(-\lambda x),$

where the constant $C$ is independent of $x\geq 1$. A fortiori,

$\int_{x}^{x+1}\psi_{t}^{\epsilon}(y)\mathbf{1}_{\psi_{t}^{\epsilon}(y)\leq
1}dy\leq C\exp(-\lambda x).$ (6.29)

Now the function $\psi\mapsto\mathbf{1}_{0\leq\psi\leq 1}\psi|\log\psi|$ is
concave, and so using Jensen’s inequality and (6.29),

$\int_{x}^{x+1}\psi_{t}^{\epsilon}(y)|\log\psi_{t}^{\epsilon}(y)|\mathbf{1}_{\psi_{t}^{\epsilon}(y)\leq
1}dy\leq C^{\prime}x\exp(-\lambda x).$

Evidently a symmetric argument applies for $x\leq-1$. Summing over $x$, and
using that $\psi\log\psi\geq-\psi|\log\psi|\mathbf{1}_{\psi\leq 1}$, we find

$\int\psi_{t}^{\epsilon}(x)\log\psi_{t}^{\epsilon}(x)dx\geq-C^{\prime\prime}\sum_{x=1}^{\infty}x\exp(-\lambda
x)>-K>-\infty,$

as required.

###### Proof of Proposition 2.18.

First observe that

$\int\psi_{t}^{\epsilon}(x)\,\rho^{\epsilon}*\psi_{t}^{\epsilon}(x)dx=\int\int\int\psi_{t}^{\epsilon}(x)\psi_{t}^{\epsilon}(x-y)\zeta^{\epsilon}(y-z)\check{\zeta}^{\epsilon}(z)dzdydx\\\
=\int\int\int\psi_{t}^{\epsilon}(\widetilde{x}-\widetilde{z})\psi_{t}^{\epsilon}(\widetilde{x}-\widetilde{y})\zeta^{\epsilon}(\widetilde{y})\zeta^{\epsilon}(\widetilde{z})d\widetilde{z}d\widetilde{y}d\widetilde{x}=\int\left(\zeta^{\epsilon}*\psi_{t}^{\epsilon}(x)\right)^{2}dx,$

where we have set $\widetilde{x}=x-z$, $\widetilde{y}=y-z$,
$\widetilde{z}=-z$.

Now note that

$\displaystyle\frac{d}{dt}\int\psi_{t}^{\epsilon}(x)dx$ $\displaystyle=$
$\displaystyle\int\Delta\big{(}\psi_{t}^{\epsilon}\,\rho^{\epsilon}*\psi_{t}^{\epsilon}\big{)}(x)dx+\int\psi_{t}^{\epsilon}(x)\big{(}1-\rho^{\epsilon}*\psi_{t}^{\epsilon}(x)\big{)}dx$
$\displaystyle=$
$\displaystyle\int\psi_{t}^{\epsilon}(x)dx-\int\big{(}\zeta^{\epsilon}*\psi_{t}^{\epsilon}(x)\big{)}^{2}dx.$

Thus, Gronwall’s inequality implies that $\int\psi_{t}^{\epsilon}(x)dx$ is
uniformly bounded above in $\epsilon$ and $t\in[0,T]$. Note that this also
then gives a uniform bound on the rate of change of
$\int\psi_{t}^{\epsilon}(x)dx$, and since we are working on $[0,T]$ this will
be enough to give continuity in time of the $L^{1}$ norm of the limit when we
pass to a convergent subsequence.

Now consider

$\displaystyle\frac{d}{dt}\int\psi_{t}^{\epsilon}\log\psi_{t}^{\epsilon}dx$
$\displaystyle=$
$\displaystyle\int(1+\log\psi_{t}^{\epsilon})\left[\Delta\big{(}\psi_{t}^{\epsilon}\,\rho^{\epsilon}*\psi_{t}^{\epsilon}\big{)}+\psi_{t}^{\epsilon}\big{(}1-\rho^{\epsilon}*\psi_{t}^{\epsilon}\big{)}\right]dx$
(6.30) $\displaystyle=$
$\displaystyle\int(1+\log\psi_{t}^{\epsilon})\left[\nabla\Big{(}\psi_{t}^{\epsilon}\,\nabla(\rho^{\epsilon}*\psi_{t}^{\epsilon})+\nabla\psi_{t}^{\epsilon}\,\rho^{\epsilon}*\psi_{t}^{\epsilon}\Big{)}+\psi_{t}^{\epsilon}\big{(}1-\rho^{\epsilon}*\psi_{t}^{\epsilon}\big{)}\right]dx$
$\displaystyle=$
$\displaystyle\int\left[-\frac{\nabla\psi_{t}^{\epsilon}}{\psi_{t}^{\epsilon}}\Big{(}\psi_{t}^{\epsilon}\,\nabla(\rho^{\epsilon}*\psi_{t}^{\epsilon})+\nabla\psi_{t}^{\epsilon}\,\rho^{\epsilon}*\psi_{t}^{\epsilon}\Big{)}+(1+\log\psi_{t}^{\epsilon})\psi_{t}^{\epsilon}(1-\rho^{\epsilon}*\psi_{t}^{\epsilon})\right]dx$
$\displaystyle=$
$\displaystyle-\int\left(\nabla(\zeta^{\epsilon}*\psi_{t}^{\epsilon})\right)^{2}dx-\int(\nabla\psi_{t}^{\epsilon})^{2}\frac{\rho^{\epsilon}*\psi_{t}^{\epsilon}}{\psi_{t}^{\epsilon}}dx$
$\displaystyle+\int\left[\psi_{t}^{\epsilon}+\psi_{t}^{\epsilon}\log\psi_{t}^{\epsilon}\big{(}1-\rho^{\epsilon}*\psi_{t}^{\epsilon}\big{)}-\psi_{t}^{\epsilon}\,\rho^{\epsilon}*\psi_{t}^{\epsilon}\right]dx$
$\displaystyle=$
$\displaystyle-\int\left(\nabla(\zeta^{\epsilon}*\psi_{t}^{\epsilon})\right)^{2}dx-\int(\nabla\psi_{t}^{\epsilon})^{2}\frac{\rho^{\epsilon}*\psi_{t}^{\epsilon}}{\psi_{t}^{\epsilon}}dx-\int(\zeta^{\epsilon}*\psi_{t}^{\epsilon})^{2}dx$
$\displaystyle+\int\left[\psi_{t}^{\epsilon}+\psi_{t}^{\epsilon}\log\psi_{t}^{\epsilon}\big{(}1-\rho^{\epsilon}*\psi_{t}^{\epsilon}\big{)}\right]dx.$

The first three terms are negative; and we already saw that the $L^{1}$ norm
of $\psi_{t}^{\epsilon}$ is uniformly bounded. Moreover, since
$\psi_{t}^{\epsilon}\log\psi_{t}^{\epsilon}$ is uniformly bounded below and
$\int\rho^{\epsilon}(x)dx=1$,

$-\int\psi_{t}^{\epsilon}\log\psi_{t}^{\epsilon}\,\rho^{\epsilon}*\psi_{t}^{\epsilon}dx\leq
C\int\rho^{\epsilon}*\psi_{t}^{\epsilon}dx=C\int\psi_{t}^{\epsilon}dx.$

From this and (6.30), we see immediately that
$\int\psi_{t}^{\epsilon}\log\psi_{t}^{\epsilon}dx$ is uniformly bounded above
in $\epsilon$ and $t\in[0,T]$. Combining with Lemma 6.13, we deduce that we
have a uniform bound on
$\int\psi_{s}^{\epsilon}(x)|\log\psi_{s}^{\epsilon}(x)|dx$. From (6.30), this
in turn means that both
$\int_{0}^{t}\int(\zeta^{\epsilon}*\psi_{s}^{\epsilon}(x))^{2}dxds$ and
$\int_{0}^{t}\int\big{(}\nabla(\zeta^{\epsilon}*\psi_{s}^{\epsilon}(x))\big{)}^{2}dxds$
are uniformly bounded in $\epsilon$ and $t\in[0,T]$.

We shall next show that $\zeta^{\epsilon}*\psi_{t}^{\epsilon}$ solves (1.2) up
to a remainder of order $\epsilon$. First observe that

$\int\Delta\left((\rho^{\epsilon}*\psi_{t}^{\epsilon})\psi_{t}^{\epsilon}\right)\phi
dx=-\int\nabla(\rho^{\epsilon}*\psi_{t}^{\epsilon})\,\psi_{t}^{\epsilon}\,\nabla\phi
dx-\int\rho^{\epsilon}*\psi_{t}^{\epsilon}\,\nabla\psi_{t}^{\epsilon}\,\nabla\phi
dx.$ (6.31)

We would like to show that this is close to
$\int(\zeta^{\epsilon}*\psi^{\epsilon}_{t})^{2}\Delta\phi dx$. For the first
term

$\displaystyle\int\nabla(\rho^{\epsilon}*\psi_{t}^{\epsilon})\,\psi_{t}^{\epsilon}\,\nabla\phi
dx$ $\displaystyle=$
$\displaystyle\int\int\int\nabla\psi_{t}^{\epsilon}(x-y)\zeta^{\epsilon}(y-z)\check{\zeta}^{\epsilon}(z)\psi_{t}^{\epsilon}(x)\nabla\phi(x)dzdydx$
$\displaystyle=$
$\displaystyle\int\int\int\nabla\psi_{t}^{\epsilon}(\widetilde{x}-\widetilde{y})\zeta^{\epsilon}(\widetilde{y})\zeta^{\epsilon}(\widetilde{z})\psi_{t}^{\epsilon}(\widetilde{x}-\widetilde{z})\nabla\phi(\widetilde{x}-\widetilde{z})d\widetilde{z}d\widetilde{y}d\widetilde{x}$
$\displaystyle=$
$\displaystyle\int(\nabla\zeta^{\epsilon}*\psi_{t}^{\epsilon})\,\left(\zeta^{\epsilon}*(\psi_{t}^{\epsilon}\nabla\phi)\right)dx$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int\nabla((\zeta^{\epsilon}*\psi_{t}^{\epsilon})^{2})\,\nabla\phi
dx$
$\displaystyle+\int\nabla(\zeta^{\epsilon}*\psi_{t}^{\epsilon})\left[\zeta^{\epsilon}*(\psi_{t}^{\epsilon}\,\nabla\phi)-\nabla\phi\,(\zeta^{\epsilon}*\psi_{t}^{\epsilon})\right]dx,$
(6.32)

where, as before, we have substituted $\widetilde{x}=x-z$,
$\widetilde{y}=y-z$, $\widetilde{z}=-z$. To control the term (6.32) we use the
Intermediate Value Theorem to see that

$\left|\int\left[\psi_{t}^{\epsilon}(x-y)\nabla\phi(x-y)\zeta^{\epsilon}(y)-\nabla\phi(x)\psi_{t}^{\epsilon}(x-y)\zeta^{\epsilon}(y)\right]dy\right|\\\
\leq
C\|\Delta\phi\|_{\infty}\int\psi_{t}^{\epsilon}(x-y)y\zeta^{\epsilon}(y)dy.$

Since $\zeta\in\mathcal{S}(\mathbb{R})$, the integral in this expression is
$\mathcal{O}(\epsilon)$.

Similarly, for the second term in (6.31),

$\displaystyle\int(\rho^{\epsilon}*\psi_{t}^{\epsilon})\,\nabla\psi_{t}^{\epsilon}\,\nabla\phi
dx$ $\displaystyle=$
$\displaystyle\int\int\int\psi_{t}^{\epsilon}(x-y)\zeta^{\epsilon}(y-z)\check{\zeta}^{\epsilon}(z)\nabla\psi_{t}^{\epsilon}(x)\nabla\phi(x)dzdydx$
(6.33) $\displaystyle=$
$\displaystyle\int\int\int\psi_{t}^{\epsilon}(\widetilde{x}-\widetilde{y})\zeta^{\epsilon}(\widetilde{y})\zeta^{\epsilon}(\widetilde{z})\nabla\psi_{t}^{\epsilon}(\widetilde{x}-\widetilde{z})\nabla\phi(\widetilde{x}-\widetilde{z})d\widetilde{z}d\widetilde{y}d\widetilde{x}$
$\displaystyle=$
$\displaystyle\int(\zeta^{\epsilon}*\psi_{t}^{\epsilon})\,\left(\zeta^{\epsilon}*(\nabla\psi_{t}^{\epsilon}\,\nabla\phi)\right)dx$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int\nabla((\zeta^{\epsilon}*\psi_{t}^{\epsilon})^{2})\,\nabla\phi
dx$
$\displaystyle+\int(\zeta^{\epsilon}*\psi_{t}^{\epsilon})\left[\zeta^{\epsilon}*(\nabla\psi_{t}^{\epsilon}\,\nabla\phi)-\nabla\phi\,(\nabla\zeta^{\epsilon}*\psi_{t}^{\epsilon})\right]dx,$

and (6.33) is controlled in the same way as (6.32): using the Intermediate
Value Theorem,

$\left|\int\left[\nabla\psi_{t}^{\epsilon}(x-y)\nabla\phi(x-y)\zeta^{\epsilon}(y)-\nabla\phi(x)\nabla\psi_{t}^{\epsilon}(x-y)\zeta^{\epsilon}(y)\right]dy\right|\\\
\leq
C\|\Delta\phi\|_{\infty}\left|\int\nabla\psi_{t}^{\epsilon}(x-y)y\zeta^{\epsilon}(y)dy\right|,$

which again is $\mathcal{O}(\epsilon)$.

We now have the ingredients that we need. The calculations above yield both a
uniform (in $\epsilon$) bound on $\zeta^{\epsilon}*\psi_{t}^{\epsilon}$ in
$L^{1}\cap L^{2}\big{(}[0,T]\times\mathbb{R}\big{)}$, and that

$\int\psi_{t}^{\epsilon}(x)\phi(x)dx-\int\psi_{0}^{\epsilon}(x)\phi(x)dx=\int_{0}^{t}\int(\zeta^{\epsilon}*\psi_{s}^{\epsilon}(x))^{2}\Delta\phi(x)dx\\\
+\int_{0}^{t}\int\zeta^{\epsilon}*\psi_{s}^{\epsilon}(x)\left(1-\zeta^{\epsilon}*\psi_{s}^{\epsilon}(x)\right)\phi(x)dx+\mathcal{O}(\epsilon)$
(6.34)

(for sufficiently regular $\phi$). Since
$\int\psi_{t}^{\epsilon}(x)\phi(x)dx-\int\zeta^{\epsilon}*\psi_{t}^{\epsilon}(x)\phi(x)dx$
is order $\epsilon$, if we replace $\psi^{\epsilon}$ by
$\zeta^{\epsilon}*\psi^{\epsilon}$ on the left hand side, then (6.34) says
that $\zeta^{\epsilon}*\psi^{\epsilon}$ solves (2.14) weakly up to order
$\epsilon$. Therefore, $\zeta^{\epsilon}*\psi^{\epsilon}$ converges weakly to
$\psi$ in $L^{1}$, where $\psi$ is the (unique) solution to equation (2.14)
and, so, therefore, does $\psi^{\epsilon}$. In fact, strong convergence, that
is $\int|\psi^{\epsilon}-\psi|\phi dx\to 0$, follows from the uniform
integrability of $\psi^{\epsilon}$ that we can deduce from the uniform control
of $\int\psi^{\epsilon}|\log\psi^{\epsilon}|dx$ that we proved above.

##  7 Simultaneous scaling with interaction distance

In this section we prove Theorem 2.20, which proves convergence in the case
that the width of the interaction kernel $\rho_{F}$ simultaneously scales
along with the parameters $\theta$ and $N$, in the special case in which
$r\equiv 1\equiv\gamma$, $q_{\theta}(x,dy)$ is isotropic with zero mean, the
kernel $\rho_{F}$ is Gaussian, and the scaling limit is a reaction-diffusion
equation.

To simplify notation, in this section we shall write

$\rho_{\epsilon}*\eta(x)=\rho_{F}^{\epsilon}*\eta(x)=\langle
p_{\epsilon^{2}}(x,y),\eta(dy)\rangle,$

where $p_{t}(x,y)$ denotes the heat semigroup. The assumptions of Theorem 2.20
will be in force throughout, in particular,

$\epsilon^{2}\theta\to\infty,\qquad\mbox{and }\frac{\theta}{N\epsilon^{d}}\to
0.$ (7.1)

That $N,\theta\to\infty$ and $\epsilon\to 0$ simultaneously will be implicit,
so for example if we write $\lim_{\epsilon\to 0}$, it should be understood
that $\theta,N\to\infty$ in such a way that (7.1) is satisfied. Moreover,
where there is no risk of confusion, except where it is helpful for emphasis,
we suppress dependence of $\eta$ on $N$.

The first part of the proof mirrors that of Theorem 2.10: in Subsection 7.1 we
establish bounds on the moments of $\rho_{\epsilon}*\eta_{t}(x)$ that are
sufficient to imply tightness and then apply standard results on convergence
of Markov processes from Ethier and Kurtz (1986). The challenge comes in
identifying the limit points. This is much more intricate than the case in
which we do not scale the interaction kernel, as weak convergence will no
longer be sufficient to guarantee the form of the nonlinear terms in the
limiting equation. Identification of the limit will rest on regularity
inherited from continuity estimates for a random walk with Gaussian jumps
which we prove in Subsection 7.2, before identifying the limit points in
Subsection 7.3.

###  7.1 Moment bounds for $\rho_{\epsilon}*\eta$

Let us write
$\mathcal{L}^{\theta}f(x):=\theta\int(f(y)-f(x))q_{\theta}(x,y)dy$ where
$q_{\theta}$ is a Gaussian kernel of mean $0$ and variance $1/\theta$. We note
that $\mathcal{L}^{\theta}$ is the generator of a continuous (time and space)
random walk, which makes jumps of mean $0$ and variance $1/\theta$ at rate
$\theta$. In what follows we write $\psi_{t}^{\epsilon,x}(y)$ for the solution
of

$\partial_{t}\psi_{t}^{\epsilon,x}=\mathcal{L}^{\theta}\psi_{t}^{\epsilon,x},$
(7.2)

with initial condition
$\psi_{0}^{\epsilon,x}(y)=\rho_{\epsilon}(y-x)=p_{\epsilon^{2}}(x,y)$.

To see why $\psi_{t}^{\epsilon,x}$ is useful, first note that for any time-
dependent function $\phi_{t}(x)$ with time derivative
$\dot{\phi}_{t}(x)=\partial_{t}\phi_{t}(x)$,

$\langle\phi_{t}(x),\eta_{t}(dx)\rangle=\langle\phi_{0}(x),\eta_{0}(dx)\rangle+M_{t}(\phi)+\int_{0}^{t}\big{\langle}\mathcal{L}^{\theta}\phi_{s}(x)+\dot{\phi}_{s}(x),\eta_{s}(dx)\big{\rangle}ds\\\
{}+\int_{0}^{t}\big{\langle}\phi_{s}(x)F(x,\eta_{s}),\eta_{s}(dx)\big{\rangle}ds,$
(7.3)

where $M_{t}(\phi)$ is a martingale (with respect to the natural filtration)
with angle bracket process given by (2.4) with $f$ replaced by
$\phi_{s}(\cdot)$. So, taking $\phi_{s}(\cdot)=\psi_{t-s}^{\epsilon,x}(\cdot)$
for $0\leq s\leq t$,

$\displaystyle\rho_{\epsilon}*\eta_{t}(x)$ $\displaystyle=$
$\displaystyle\langle\psi^{\epsilon,x}_{0}(y),\eta_{t}(dy)\rangle$ (7.4)
$\displaystyle=$
$\displaystyle\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle+\int_{0}^{t}\big{\langle}\psi^{\epsilon,x}_{t-s}(y)F\big{(}\rho_{\epsilon}*\eta_{s}(y)\big{)},\eta_{s}(dy)\big{\rangle}ds+M_{t}(x),$

where $M_{t}(x)$ has mean zero and a second moment we can easily write down.

######  Lemma 7.1.

Fix $t>0$, let $(\Pi(s))_{s\geq 0}$ be a rate one Poisson process, and let
$T(t)=\Pi(\theta t)/\theta$. Then

$\psi^{\epsilon,x}_{t}(y)=\mathbb{E}\left[p_{\epsilon^{2}+T(t)}(x,y)\right],$

and, moreover, since under our assumptions $\theta\epsilon^{2}$ is bounded
below, there is a $C$ independent of $\epsilon$ or $t$ such that

$\|\psi^{\epsilon,x}_{t}\|_{\infty}\leq\frac{C}{(\epsilon^{2}+t)^{d/2}}.$

###### Proof.

The first claim is immediate from the definition of the random walk with
generator $\mathcal{L}^{\theta}$.

For the second claim, first define $\tau(t)=T(t)-t$. Since if
$\tau(t)\geq-(\epsilon^{2}+t)/2$, then $1/(\epsilon^{2}+T(t))\leq
2/(\epsilon^{2}+t)$, while $\epsilon^{2}+T(t)\geq\epsilon^{2}$ always,
partitioning over $\\{\tau(t)\geq-(\epsilon^{2}+t)/2\\}$ and its complement,

$\displaystyle\|\psi^{\epsilon,x}\|_{\infty}$
$\displaystyle=\mathbb{E}\left[\frac{1}{\big{(}2\pi(\epsilon^{2}+T(t))\big{)}^{d/2}}\right]$
$\displaystyle\leq\frac{C}{(\epsilon^{2}+t)^{d/2}}+\frac{C}{\epsilon^{d}}\mathbb{P}\left\\{\tau(t)<-(\epsilon^{2}+t)/2\right\\}.$
(7.5)

Now, observe that since $\mathbb{E}[e^{-\Pi(\theta t)}]=\exp(-\theta
t(1-e^{-1}))$, by Markov’s inequality,

$\displaystyle\mathbb{P}\left\\{\tau(t)<-\frac{\epsilon^{2}+t}{2}\right\\}$
$\displaystyle=\mathbb{P}\left\\{e^{-\Pi(\theta
t)}>e^{-\theta(t-\epsilon^{2})/2}\right\\}$
$\displaystyle\leq\frac{\mathbb{E}[\exp\big{(}-\Pi(\theta
t)\big{)}]}{\exp\big{(}-\theta(t-\epsilon^{2})/2\big{)}}$
$\displaystyle=\frac{\exp(-\theta
t(1-e^{-1}))}{\exp(-\theta(t-\epsilon^{2})/2)}$
$\displaystyle=\exp\left\\{-\chi\theta
t-\frac{\theta\epsilon^{2}}{2}\right\\},$ (7.6)

where $\chi=1/2-e^{-1}>0$. The second term in (7.5) is therefore bounded by

$C\left(1+\frac{t}{\epsilon^{2}}\right)^{d/2}e^{-\chi\theta
t}\frac{1}{(\epsilon^{2}+t)^{d/2}}e^{-\epsilon^{2}\theta/2}.$

Now observe that the derivative (with respect to $t$) of $e^{-\chi\theta
t}(1+t/\epsilon^{2})^{d/2}$ is

$\left(\frac{d}{2\epsilon^{2}}-\left(1+\frac{t}{\epsilon^{2}}\right)\chi\theta\right)\left(1+\frac{t}{\epsilon^{2}}\right)^{d/2-1}e^{-\chi\theta
t},$

which is negative if $\theta(\epsilon^{2}+t)>d/2\chi$. At the maximum,
$(1+t/\epsilon^{2})=d/(2\chi\theta\epsilon^{2})$, and so this quantity is
bounded uniformly over not only $t$ but also $\epsilon$ (since we’ve assumed
that $\theta\epsilon^{2}$ is bounded below). Therefore, we have the bound

$\frac{1}{\epsilon^{d}}\mathbb{P}\left\\{\tau(t)<-(\epsilon^{2}+t)/2\right\\}\leq\frac{C}{(\epsilon^{2}+t)^{d/2}}e^{-\epsilon^{2}\theta/2}.$
(7.7)

Substituting this into (7.5) yields the result.

######  Lemma 7.2.

Let $\\{{\cal F}_{t}\\}_{t\geq 0}$ denote the natural filtration. Under the
assumptions of Theorem 2.20, for each $T\in[0,\infty)$, and $k\in\mathbb{N}$,
there exist constants $C=C(k,T)$ and $\widetilde{C}=\widetilde{C}(k,T)$,
independent of $\epsilon$, such that for all $x\in\mathbb{R}^{d}$ and all
$u,t\in[0,T]$ with $u<t$,

$\mathbb{E}\Big{[}\left.\big{(}\rho_{\epsilon}*\eta_{t}(x)\big{)}^{k}\right|{\cal
F}_{u}\Big{]}\leq
C\langle\psi_{t-u}^{\epsilon,x}(z),\eta_{u}(dz)\rangle^{k}+C\frac{\theta}{N\epsilon^{d}}\langle\psi_{t-u}^{\epsilon,x}(z),\eta_{u}(dz)\rangle;$
(7.8)

and

$\mathbb{E}\Big{[}\left.\int_{u}^{t}\langle\psi_{t-s}^{\epsilon,x}(z),\eta_{s}(dz)\rangle^{k-1}\big{\langle}\psi_{t-s}^{\epsilon,x}(z)|F(\rho_{\epsilon}*\eta_{s}(z))|,\eta_{s}(dz)\big{\rangle}ds\right|{\cal
F}_{u}\Big{]}\\\
\leq\widetilde{C}\langle\psi_{t-u}^{\epsilon,x}(z),\eta_{u}(dz)\rangle^{k}+\widetilde{C}\frac{\theta}{N\epsilon^{d}}\langle\psi_{t-u}^{\epsilon,x}(z),\eta_{u}(dz)\rangle;$
(7.9)

where the function $\psi^{\epsilon,x}_{t}(\cdot)$ was defined in (7.2). In
particular, under the assumptions of Theorem 2.20, the expected values of the
quantities on the right hand side of (7.8) and (7.9) are both integrable with
respect to Lebesgue measure.

###### Proof.

To simplify our expressions, we shall consider the case $u=0$, but the proof
goes through unchanged for other values of $u$.

We proceed by induction. Taking expectations in (7.4), using that $F$ is
bounded above, and applying Gronwall’s inequality to
$\langle\psi^{\epsilon,x}_{t-s},\eta_{s}\rangle$ we obtain
$\mathbb{E}[\langle\psi_{0}^{\epsilon,x},\eta_{t}\rangle]\leq
C\mathbb{E}[\langle\psi_{t}^{\epsilon,x},\eta_{0}\rangle]$, which implies
(7.8) in the case $k=1$. Moreover, rearranging (7.4) we find

$-\int_{0}^{t}\big{\langle}\psi^{\epsilon,x}_{t-s}(y)F(\rho_{\epsilon}*\eta_{s}(y),\eta_{s}(dy)\big{\rangle}ds=\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle-\langle\psi^{\epsilon,x}_{0}(y),\eta_{t}(dy)\rangle+M_{t}(x),$
(7.10)

and taking expectations again, since
$\langle\psi^{\epsilon,x}_{0},\eta_{t}\rangle>0$, and $M_{0}(x)=0$, this
yields

$\mathbb{E}\Big{[}-\int_{0}^{t}\langle\psi^{\epsilon,x}_{t-s}(y)F(\rho_{\epsilon}*\eta_{s}(y)),\eta_{s}(dy)\rangle
ds\Big{|}{\cal
F}_{0}\Big{]}\leq\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle.$

Since $F$ is bounded above, there exists a constant $K$ such that $|F|\leq
K-F$ and so combined with the bound on
$\mathbb{E}[\langle\psi^{\epsilon,x}_{0}(y),\eta_{t}(dy)\rangle]$ just
obtained, this in turn yields

$\mathbb{E}\Big{[}\int_{0}^{t}\big{\langle}\psi^{\epsilon,x}_{t-s}(y)|F(\rho_{\epsilon}*\eta_{s}(y))|,\eta_{s}(dy)\big{\rangle}ds\Big{|}{\cal
F}_{0}\Big{]}\leq\widetilde{C}\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle,$

which is (7.9) in the case $k=1$.

Now suppose that we have established (7.8) and (7.9) for all exponents $j<k$.
First we apply the generator $\mathcal{P}^{N}$ of our scaled population
process to functions of the form $\langle f,\eta\rangle^{k}$. Recalling that
each jump of the process involves the birth or death of a single individual,
and so increments $\langle f,\eta\rangle$ by $\pm f/N$ at the location of that
individual and that $r\equiv\gamma\equiv 1$, we find

$\mathcal{P}^{N}\Big{(}\langle
f,\eta\rangle^{k}\Big{)}=\Big{\langle}\int\theta
N\sum_{j=1}^{k}\binom{k}{j}\frac{f(y)^{j}}{N^{j}}\langle
f,\eta\rangle^{k-j}q_{\theta}(x,dy),\eta(dx)\Big{\rangle}\\\
+\Big{\langle}\theta
N\Big{(}1-\frac{F(\rho_{\epsilon}*\eta(x))}{\theta}\Big{)}\sum_{j=1}^{k}\binom{k}{j}(-1)^{j}\frac{f(x)^{j}}{N^{j}}\langle
f,\eta\rangle^{k-j},\eta(dx)\Big{\rangle}.$ (7.11)

Mimicking what we did above, we set $f(\cdot)=\psi^{\epsilon,x}_{t}(\cdot)$
and write

$\mathbb{E}\Big{[}\langle\psi^{\epsilon,x}_{0},\eta_{t}\rangle^{k}\Big{|}{\cal
F}_{0}\Big{]}=\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle^{k}+\mathbb{E}\Big{[}\int_{0}^{t}\mathcal{P}^{N}\big{(}\langle\psi^{\epsilon,x}_{t-s}(y),\eta_{s}(dy)\rangle^{k}\big{)}ds\\\
-\int_{0}^{t}\langle
k\dot{\psi}^{\epsilon,x}_{t-s}(y),\eta_{s}(dy)\rangle\big{\langle}\psi^{\epsilon,x}_{t-s}(y),\eta_{s}(dy)\big{\rangle}^{k-1}ds\Big{|}{\cal
F}_{0}\Big{]}.$ (7.12)

Since $\dot{\psi}^{\epsilon,x}_{s}=\mathcal{L}_{\theta}\psi^{\epsilon,x}_{s}$,
the $j=1$ terms from
$\mathcal{P}^{N}(\langle\psi^{\epsilon,x}_{t-s}(y),\eta_{s}(dy)\rangle^{k})$
combines with the last term in (7.12) to yield

$\int_{0}^{t}k\langle\psi^{\epsilon,x}_{t-s},\eta\rangle^{k-1}\langle
F(\rho_{\epsilon}*\eta_{s}(y))\psi^{\epsilon,x}_{t-s}(y),\eta_{s}(dy)\rangle
ds.$

As for the remaining terms, using (from Lemma 7.1) that
$\sup_{s}\|\psi^{\epsilon,x}_{s}(\cdot)\|_{\infty}=C/\epsilon^{d}$,
$N\epsilon^{d}>1$, and our inductive hypothesis, we find

$\mathbb{E}\Big{[}\Big{\langle}\int_{0}^{t}\theta
N\sum_{j=2}^{k}\binom{k}{j}\int\frac{\psi_{t-s}^{\epsilon,x}(z)^{j}}{N^{j}}\langle\psi^{\epsilon,x}_{t-s},\eta_{s}\rangle^{k-j}q_{\theta}(y,dz),\eta_{s}(dy)\Big{\rangle}ds\\\
+\Big{\langle}\int_{0}^{t}\theta
N\sum_{j=2}^{k}\binom{k}{j}\frac{\psi_{t-s}^{\epsilon,x}(y)^{j}}{N^{j}}\langle\psi^{\epsilon,x}_{t-s},\eta_{s}\rangle^{k-j}(-1)^{j}\Big{(}1-\frac{F(\rho_{\epsilon}*\eta_{s}(y))}{\theta}\Big{)},\eta_{s}(dy)\Big{\rangle}ds\Big{|}{\cal
F}_{0}\Big{]}\\\ \leq
C\mathbb{E}\Big{[}\Big{\langle}\int_{0}^{t}\sum_{j=2}^{k}\frac{\theta}{N\epsilon^{d}}\Big{(}\frac{1}{(N\epsilon^{d})^{j-2}}\Big{)}\Big{\langle}\psi^{\epsilon,x}_{t-s}(y)\Big{(}2+\frac{|F(\rho_{\epsilon}*\eta_{s}(y))|}{\theta}\Big{)},\eta_{s}(dy)\Big{\rangle}\langle\psi^{\epsilon,x}_{t-s},\eta_{s}\rangle^{k-j}ds\big{|}{\cal
F}_{0}\Big{]}\\\ \leq
C^{\prime}\frac{\theta}{N\epsilon^{d}}\sum_{j=1}^{k-1}\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle^{j}\leq
C^{\prime\prime}\frac{\theta}{N\epsilon^{d}}\Big{(}\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle^{k}+\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle\Big{)}.$

Combining this with (7.11) and (7.12), using once again the fact that $F$ is
bounded above, we find

$\mathbb{E}\Big{[}\langle\psi^{\epsilon,x}_{0},\eta_{t}\rangle^{k}\Big{|}{\cal
F}_{0}\Big{]}\leq\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle^{k}+\widetilde{C}\mathbb{E}\Big{[}\int_{0}^{t}\langle\psi^{\epsilon,x}_{t-s}(y),\eta_{s}(dy)\rangle^{k}ds\Big{|}{\cal
F}_{0}\Big{]}\\\
+C^{\prime\prime}\frac{\theta}{N\epsilon^{d}}\Big{(}\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle^{k}+\langle\psi^{\epsilon,x}_{t}(y),\eta_{0}(dy)\rangle\Big{)},$

and (7.8) follows from Gronwall’s inequality. Rearranging exactly as in the
case $k=1$, we recover (7.9) and the inductive step is complete.

We shall also need the following consequence of the bounds that we obtained in
Lemma 7.2:

######  Corollary 7.3.

Under the assumptions of Theorem 2.20, for each $k\geq 1$, $T>0$, there is a
$C(k,T)$ such that

$\mathbb{E}\Big{[}\big{\langle}(\rho_{\epsilon}*\eta_{t})^{k},\eta_{t}\big{\rangle}\Big{]}<C(k,T)<\infty,\qquad\mbox{
for all }t\in[0,T].$ (7.13)

###### Sketch.

First observe that if $A\in(0,1)$, then

$p_{A\epsilon^{2}}(x,y)=\frac{1}{A^{d/2}}p_{\epsilon^{2}}(x,y)\exp\Big{(}-\frac{\|x-y\|^{2}}{2\epsilon^{2}}\big{(}\frac{1}{A}-1\big{)}\Big{)}\leq\frac{1}{A^{d/2}}p_{\epsilon^{2}}(x,y).$
(7.14)

Now consider

$\displaystyle\mathbb{E}\big{[}\langle\rho_{\epsilon}*\eta_{t}(x),\eta_{t}(dx)\rangle\big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int\int
p_{\epsilon^{2}}(x,z)\eta_{t}(dz)\eta_{t}(dx)\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int\int\int
p_{\epsilon^{2}/2}(x,y)p_{\epsilon^{2}/2}(y,z)dy\eta_{t}(dz)\eta_{t}(dx)\Big{]}$
$\displaystyle=\mathbb{E}\Big{[}\int\left(p_{\epsilon^{2}/2}*\eta_{t}(y)\right)^{2}dy\Big{]}$
$\displaystyle\leq
C\int\mathbb{E}\big{[}\big{(}\rho_{\epsilon}*\eta_{t}(x)\big{)}^{2}\big{]}dx,$

where we used (7.14) in the last line. Using Lemma 7.2 and our assumptions on
$\eta_{0}$, this quantity is finite.

To illustrate the inductive step, now consider

$\mathbb{E}\big{[}\langle\rho_{\epsilon}*\eta_{t}(x)^{2},\eta_{t}(dx)\rangle\big{]}=\mathbb{E}\Big{[}\int\int\int
p_{\epsilon^{2}}(x,z_{1})p_{\epsilon^{2}}(x,z_{2})\eta_{t}(dz_{1})\eta_{t}(dz_{2})\eta_{t}(dx)\Big{]}\\\
=\mathbb{E}\Big{[}\int\cdots\int
p_{\epsilon^{2}/2}(x,y_{1})p_{\epsilon^{2}/2}(x,y_{2})p_{\epsilon^{2}/2}(y_{1},z_{1})p_{\epsilon^{2}/2}(y_{2},z_{2})\eta_{t}(dz_{1})\eta_{t}(dz_{2})dy_{1}dy_{2}\eta_{t}(dx)\Big{]}.$
(7.15)

We use the identity

$p_{\epsilon^{2}/2}(x,y_{1})p_{\epsilon^{2}/2}(x,y_{2})=p_{\epsilon^{2}}(y_{1},y_{2})p_{\epsilon^{2}/4}\Big{(}x,\frac{y_{1}+y_{2}}{2}\Big{)}$

to rewrite (7.15) as

$\displaystyle\mathbb{E}\Big{[}\int\int
p_{\epsilon^{2}/2}*\eta_{t}(y_{1})\,p_{\epsilon^{2}/2}*\eta_{t}(y_{2})\,p_{\epsilon^{2}/4}*\eta_{t}\big{(}\frac{y_{1}+y_{2}}{2}\big{)}\,p_{\epsilon^{2}}(y_{1},y_{2})dy_{1}dy_{2}\Big{]}$
$\displaystyle\leq\mathbb{E}\Big{[}\int\int\Big{\\{}\big{(}p_{\epsilon^{2}/2}*\eta_{t}(y_{1})\big{)}^{3}+\big{(}p_{\epsilon^{2}/2}*\eta_{t}(y_{2})\big{)}^{3}+\big{(}p_{\epsilon^{2}/4}*\eta_{t}\big{(}\frac{y_{1}+y_{2}}{2}\big{)}\big{)}^{3}\Big{\\}}p_{\epsilon^{2}}(y_{1},y_{2})dy_{1}dy_{2}\Big{]},$

where we have used that for any non-negative real numbers $\beta_{1}$,
$\beta_{2}$, $\beta_{3}$,
$\beta_{1}\beta_{2}\beta_{3}\leq\beta_{1}^{3}+\beta_{2}^{3}+\beta_{3}^{3}$.
For the first two terms in the sum we integrate with respect to $y_{2}$ and
$y_{1}$ respectively to reduce to an expression of the form considered in
Lemma 7.2. For the final term, the change of variables $z_{1}=y_{1}+y_{2}$,
$z_{2}=y_{1}-y_{2}$ in the integral similarly allows us to integrate out the
heat kernel, and we conclude that the result holds for $k=2$.

We can proceed in the same way for larger values of $k$, using repeatedly that

$p_{t_{1}}(x,y_{1})p_{t_{2}}(x,y_{2})=p_{\frac{t_{1}t_{2}}{t_{1}+t_{2}}}\Big{(}x,\frac{t_{2}y_{1}+t_{1}y_{2}}{t_{1}+t_{2}}\Big{)}p_{t_{1}+t_{2}}(y_{1},y_{2})$

to write

$\prod_{j=1}^{k}p_{\tau}(y,y_{j})=\prod_{j=2}^{k}p_{\frac{j\tau}{j-1}}\big{(}y_{j},Y_{j-1}\big{)}p_{\frac{\tau}{k}}(y,Y_{k})$

where

$Y_{1}=y_{1},\qquad Y_{j}=\frac{j-1}{j}Y_{j-1}+\frac{1}{j}y_{j},\mbox{for
}j\geq 2.$

Writing $p_{\epsilon^{2}}(x,z_{j})=\int
p_{\epsilon^{2}/2}(x,y_{j})p_{\epsilon^{2}/2}(y_{j},z_{j})dy_{j}$ and using
the above with $\tau=\epsilon^{2}/2$, this yields

$\big{\langle}\big{(}\rho_{\epsilon}*\eta_{t}(x)\big{)}^{k},\eta_{t}(dx)\big{\rangle}=\int\cdots\int\prod_{j=2}^{k}p_{\epsilon^{2}j/2(j-1)}(y_{j},Y_{j-1})\prod_{i=1}^{k}p_{\epsilon^{2}/2}*\eta_{t}(y_{i})p_{\epsilon^{2}/2k}*\eta_{t}(Y_{k})dy_{1}\ldots
dy_{k}\\\
\leq\int\cdots\int\prod_{j=2}^{k}p_{\epsilon^{2}j/2(j-1)}(y_{j},Y_{j-1})\Big{\\{}\sum_{i=1}^{k}\big{(}p_{\epsilon^{2}/2}*\eta_{t}(y_{i})\big{)}^{k+1}+\big{(}p_{\epsilon^{2}/2k}*\eta_{t}(Y_{k})\big{)}^{k+1}\Big{\\}}dy_{1}\ldots
dy_{k},$

and once again we can change variables in the integrals and use (7.14) to
bound this by a constant multiple of
$\int\mathbb{E}\big{[}\big{(}\rho_{\epsilon}*\eta_{t}(x)\big{)}^{k+1}\big{]}dx$,
and the inductive step is complete.

######  Corollary 7.4 (Tightness of
$\\{(\rho_{\epsilon}*\eta_{t}^{N}(x)dx)_{t\geq 0}\\}$).

Under the assumptions of Theorem 2.20, the sequence of measure valued
processes $\\{\rho_{\epsilon}*\eta_{t}^{N}(x)dx\\}_{t\geq 0}$ (taking values
in ${\mathcal{D}}_{[0,T]}(\mathcal{M}_{F}(\mathbb{R}^{d}))$) is tight.

###### Proof.

First observe that the proof, from Lemma 6.2, that $\mathbb{E}[\sup_{0\leq
t\leq T}\langle 1,\eta_{t}^{N}\rangle]$ is bounded goes through unchanged, and
since $\langle 1,\rho_{\epsilon}*\eta_{t}^{N}(x)dx\rangle=\langle
1,\eta_{t}^{N}\rangle$, compact containment follows.

As in the nonlocal case, it suffices to prove that for $T>0$, and any $f\in
C_{b}^{\infty}(\mathbb{R}^{d})$ with bounded second derivatives and
$\int|f(x)|dx<\infty$, the sequence of real-valued processes
$\big{\\{}\big{(}\int f(x)\rho_{\epsilon}*\eta_{t}^{N}(x)dx\big{)}_{t\geq
0}\big{\\}}_{N\geq 1}$ is tight. Let us temporarily write $X_{f}^{N}(t)$ for
$\int f(x)\rho_{\epsilon}*\eta_{t}^{N}(x)dx$ and set

$w^{\prime}\big{(}X_{f}^{N},\delta,T\big{)}=\inf_{\\{t_{i}\\}}\max_{i}\sup_{s,t\in[t_{i-1},t_{i})}\big{|}X_{f}^{N}(t)-X_{f}^{N}(s)\big{|},$

where $\\{t_{i}\\}$ ranges over all partitions of the form
$0=t_{0}<t_{1}<\cdots<t_{n-1}<T\leq t_{n}$ with $\min_{1\leq i\leq
n}(t_{i}-t_{i-1})>\delta$ and $n\geq 1$. Using Corollary 3.7.4 of Ethier and
Kurtz (1986), to prove tightness of the sequence of real-valued processes
$X_{f}^{N}$ it suffices to check compact containment of the sequence $\\{\int
f(x)\rho_{\epsilon}*\eta_{t}^{N}(x)dx\\}_{N\geq 1}$ at any rational time $t$
and that for every $\nu>0$ and $T>0$, there exists $\delta>0$ such that

$\limsup_{N\to\infty}\mathbb{P}\big{[}w^{\prime}\big{(}X_{f}^{N},\delta,T\big{)}>\nu\big{]}<\nu.$

Evidently this will follow if we can show that this condition is satisfied
when we replace the minimum over all partitions with mesh at least $\delta$ in
the definition of $w^{\prime}$, by the partition into intervals of length
exactly $\delta$.

We have

$\left|\langle\rho_{\epsilon}*f,\eta_{t}^{N}\rangle-\langle\rho_{\epsilon}*f,\eta_{s}^{N}\rangle\right|\leq\left|\int_{s}^{t}\Big{\langle}\theta\int\big{(}\rho_{\epsilon}*f(y)-\rho_{\epsilon}*f(x)\big{)}q_{\theta}(x,dy),\eta_{u}^{N}(dx)\Big{\rangle}du\right|\\\
+\int_{s}^{t}\Big{\langle}|F\big{(}\rho_{\epsilon}*\eta_{u}^{N}(x)\big{)}|\rho_{\epsilon}*|f|(x),\eta_{u}^{N}(dx)\Big{\rangle}du+2\sup_{0\leq
u\leq T}|\widehat{M}^{N}(f)_{u}|,$ (7.16)

where $\widehat{M}^{N}(f)$ is the martingale of (6.4) with the test function
$f$ replaced by $\rho_{\epsilon}*f$. We control each of the three terms on the
right hand side separately.

By the Intermediate Value Theorem, using $T_{t}$ to denote the heat semigroup,
there exists $s\in(0,1/\theta)$ such that

$\left|\theta\int\big{(}\rho_{\epsilon}*f(y)-\rho_{\epsilon}*f(x)\big{)}q_{\theta}(x,dy)\right|=\left|\theta\Big{(}T_{\epsilon^{2}+1/\theta}f(x)-T_{\epsilon^{2}}f(x)\Big{)}\right|\\\
=\left|\partial_{s}T_{\epsilon^{2}+s}f(x)\right|=\left|T_{\epsilon^{2}+s}\Delta
f(x)\right|\leq\|\Delta f\|_{\infty}.$

The first term in (7.16) is therefore bounded by

$\|\Delta f\|_{\infty}|t-s|\sup_{0\leq u\leq T}\langle 1,\eta_{u}^{N}\rangle.$

We follow the approach of Lemma 6.2. Consulting (2.4), the angle bracket
process of $\widehat{M}^{N}(f)$ satisfies
$\mathbb{E}[\langle\widehat{M}^{N}_{f}\rangle_{T}]\leq
C(\theta/N)\int_{0}^{T}\mathbb{E}[\langle 1,\eta_{s}\rangle]ds\leq
C^{\prime}\theta/N$ for some constants $C$ and $C^{\prime}$. Now, using the
Burkholder-Davis-Gundy inequality and Barlow et al. (1986),
$\mathbb{E}[\sup_{0\leq u\leq T}|\widehat{M}^{N}(f)_{u}|^{2}]\leq
C^{\prime\prime}\mathbb{E}[\langle\widehat{M}^{N}(f)\rangle_{u}]$, and so
using Markov’s inequality,

$\limsup_{N\to\infty}\mathbb{P}\Big{[}2\sup_{0\leq u\leq
T}|\widehat{M}^{N}(f)_{u}|>\frac{\nu}{3}\Big{]}\leq\limsup_{N\to\infty}\frac{36}{\nu^{2}}C^{\prime\prime}\mathbb{E}\big{[}\langle\widehat{M}^{N}(f)\rangle_{T}\big{]}\leq\limsup_{N\to\infty}\frac{36}{\nu^{2}}\frac{C^{\prime}C^{\prime\prime}\theta}{N\epsilon^{d}}=0.$
(7.17)

Now consider

$\mathbb{E}\Big{[}\Big{(}\int_{s}^{t}\big{\langle}\rho_{\epsilon}*|f|(x)\big{|}F\big{(}\rho_{\epsilon}*\eta_{u}^{N}(x)\big{)}\big{|},\eta_{u}^{N}(dx)\big{\rangle}du\Big{)}^{2}\Big{]}\\\
=2\mathbb{E}\Big{[}\int_{s}^{t}\big{\langle}\rho_{\epsilon}*|f|(x)\big{|}F\big{(}\rho_{\epsilon}*\eta_{u}^{N}(x)\big{)}\big{|},\eta_{u}^{N}(dx)\big{\rangle}\int_{u}^{t}\big{\langle}\rho_{\epsilon}*|f|(x)\big{|}F\big{(}\rho_{\epsilon}*\eta_{r}^{N}(x)\big{)}\big{|},\eta_{r}^{N}(dx)\big{\rangle}drdu\Big{]}.$
(7.18)

Since $F$ is polynomial, we use the approach of Corollary 7.3, the tower
property, and Lemma 7.2, to bound this in terms of sums of terms of the form

$\mathbb{E}\Big{[}\int_{s}^{t}(t-u)\int\rho_{\epsilon}*|f|(x)\rho_{\epsilon}*\eta_{u}^{N}(x)^{j}dx\int\rho_{\epsilon}*|f|(y)\rho_{\epsilon}*\eta_{u}^{N}(y)^{k}dydu\Big{]}.$

Now observe that, again using Lemma 7.2, since for nonnegative $a$ and $b$,
$a^{j}b^{k}\leq a^{j+k}+b^{j+k}$,

$\mathbb{E}\left[\int\int\rho_{\epsilon}*|f|(x)\rho_{\epsilon}*\eta_{u}^{N}(x)^{j}\rho_{\epsilon}*|f|(y)\rho_{\epsilon}*\eta_{u}^{N}(y)^{k}dxdy\right]\\\
\leq\mathbb{E}\left[\int\int\|f\|_{\infty}\rho_{\epsilon}*\eta_{u}^{N}(x)^{j+k}\rho_{\epsilon}*|f|(y)dxdy+\int\int\rho_{\epsilon}*|f|(x)\|f\|_{\infty}\rho_{\epsilon}*\eta_{u}^{N}(y)^{j+k}dxdy\right]\\\
\leq C\int|f|(x)dx.$

Thus the quantity (7.18) is bounded by $C(t-s)^{2}$ for a new constant $C$
which we can take to be independent of $s$, $t$ and $\epsilon$. Markov’s
inequality then gives

$\mathbb{P}\Big{[}\|f\|_{\infty}\int_{s}^{t}\big{\langle}\big{|}F\big{(}\rho_{\epsilon}*\eta_{u}^{N}(x)\big{)}\big{|},\eta_{u}^{N}(dx)\big{\rangle}du\geq\frac{\nu}{3}\Big{]}\leq
C\frac{(t-s)^{2}}{\nu^{2}}.$

A union bound gives that

$\mathbb{P}\Big{[}\max_{i}\|f\|_{\infty}\int_{t_{i-1}}^{t_{i}}\big{\langle}\big{|}F\big{(}\rho_{\epsilon}*\eta_{u}^{N}(x)\big{)}\big{|},\eta_{u}^{N}(dx)\big{\rangle}du\geq\frac{\nu}{3}\Big{]}\leq
C\frac{T\delta}{\nu^{2}}.$ (7.19)

Now using Markov’s inequality, we can choose $K$ so that

$\mathbb{P}\Big{[}\|\Delta f\|_{\infty}\,\sup_{0\leq t\leq T}\langle
1,\eta_{t}^{N}\rangle>K\Big{]}<\frac{\nu}{3},$

and so choosing $\delta$ so that $K\delta<\nu/3$ in this expression and
$C\delta<\nu^{3}/3T$ in (7.19), combining with (7.17), the result follows.

###  7.2 Continuity estimates for $\rho_{\epsilon}*\eta$

To identify the limit point of any convergent subsequence of
$\\{\rho_{\epsilon}*\eta^{N}(x)\\}$, we will require some control on the
spatial continuity of the functions $\rho_{\epsilon}*\eta^{N}(x)$. This will
be inherited from the regularity of the transition density of the Gaussian
random walk with generator ${\cal L}^{\theta}$, which in turn follows from its
representation as that of a Brownian motion evaluated at the random time
$T(t)$ defined in Lemma 7.1. Our approach will be to approximate
$\psi_{t}^{\epsilon,x}(\cdot)$ by $p_{\epsilon^{2}+t}(x,\cdot)$, and to
control the error that this introduces we need to control $T(t)-t$.

######  Lemma 7.5.

In the notation of Lemma 7.1, for any $A>1$,

$\displaystyle\mathbb{P}\left\\{T(t)-t>A(\epsilon^{2}+t)\right\\}\leq\exp\left(-\frac{\theta
A}{4}\big{(}\epsilon^{2}+t\big{)}\right).$

###### Proof.

This is just a Chernoff bound. With $\Pi$ a rate one Poisson process as in
Lemma 7.1, for any $A>1$,

$\displaystyle\mathbb{P}\left\\{T(t)-t>A(\epsilon^{2}+t)\right\\}$
$\displaystyle=\mathbb{P}\Big{\\{}\Pi(\theta
t)>\theta\Big{(}t+A(\epsilon^{2}+t)\Big{)}\Big{\\}}$
$\displaystyle\leq\frac{\mathbb{E}\left[\exp\left(\alpha\Pi(\theta
t)\right)\right]}{\exp\left(\alpha\theta\big{(}t+A(\epsilon^{2}+t)\big{)}\right)}$
$\displaystyle=\exp\left(\theta
t\big{(}e^{\alpha}-1\big{)}-\alpha\theta\big{(}t+A(\epsilon^{2}+t)\big{)}\right)$
$\displaystyle\leq\exp\left(\theta
t\big{(}e^{\alpha}-\alpha-1-\frac{A\alpha}{2}\big{)}-\frac{A\alpha}{2}\theta(\epsilon^{2}+t)\right).$

Now set $\alpha=1/2$. Since $A>1$, $e^{\alpha}-\alpha-1-A\alpha/2<0$ and the
result follows.

As advertised, we wish to control the difference between
$\psi_{t}^{\epsilon,x}(y)$ and $p_{\epsilon^{2}+t}(x,y)$.

######  Lemma 7.6.

In the notation of Lemma 7.1, there exists a $C<\infty$ such that

$\left|\psi_{t}^{\epsilon,x}(y)-p_{\epsilon^{2}+t}(x,y)\right|\leq\frac{C}{(\epsilon^{2}\theta)^{1/2}}p_{6(\epsilon^{2}+t)}(x,y)+\frac{C}{(\epsilon^{2}+t)^{d/2}}\exp(-\epsilon^{2}\theta/2).$
(7.20)

###### Proof.

Still using the notation of Lemma 7.1, we partition into three events
according to the value of $\tau(t)$. Let
$A_{1}=\\{\tau(t)<-(\epsilon^{2}+t)/2\\}$,
$A_{2}=\\{\tau(t)>2(\epsilon^{2}+t)\\}$, and $A_{3}$ the remaining event,
$\\{-(\epsilon^{2}+t)/2\leq\tau(t)\leq 2(\epsilon^{2}+t)\\}$. Then,

$\displaystyle\left|\psi_{t}^{\epsilon,x}(y)-p_{\epsilon^{2}+t}(x,y)\right|$
$\displaystyle=\left|\mathbb{E}\left[p_{\epsilon^{2}+t+\tau(t)}(x,y)-p_{\epsilon^{2}+t}(x,y)\right]\right|$
$\displaystyle\leq\mathbb{E}\left[(1_{A_{1}}+1_{A_{2}}+1_{A_{3}})\left|p_{\epsilon^{2}+t+\tau(t)}(x,y)-p_{\epsilon^{2}+t}(x,y)\right|\right].$

For the first term, note that if $a<b$ then

$\displaystyle|p_{a}(x,y)-p_{b}(x,y)|$
$\displaystyle=\frac{1}{(2\pi)^{d/2}}\left|\frac{1}{a^{d/2}}e^{-\|x-y\|^{2}/2a}-\frac{1}{b^{d/2}}e^{-\|x-y\|^{2}/2b}\right|$
$\displaystyle=\frac{1}{(2\pi
a^{2})^{d/2}}e^{-\|x-y\|^{2}/2b}\left|e^{-\|x-y\|^{2}\left(\frac{1}{2a}-\frac{1}{2b}\right)}-\left(\frac{a}{b}\right)^{d/2}\right|$
$\displaystyle\leq C\left(\frac{b}{a}\right)^{d/2}p_{b}(x,y),$

where the inequality follows because both terms under the absolute value are
less than 1. Since, on the event $A_{1}$, $\tau(t)<0$, we can apply this with
$a=\epsilon^{2}+t+\tau(t)$ and $b=\epsilon^{2}+t$, and, using the bound (7.6),

$\displaystyle\mathbb{E}\left[1_{A_{1}}|p_{a}(x,y)-p_{b}(x,y)|\right]$
$\displaystyle\leq
C\left(\frac{\epsilon^{2}+t}{\epsilon^{2}}\right)^{d/2}p_{\epsilon^{2}+t}(x,y)\mathbb{P}\left\\{\tau(t)<-\frac{\epsilon^{2}+t}{2}\right\\}$
$\displaystyle\leq
C\frac{1}{\epsilon^{d}}\mathbb{P}\left\\{\tau(t)<-\frac{\epsilon^{2}+t}{2}\right\\}$
$\displaystyle\leq\frac{C}{(\epsilon^{2}+t)^{d/2}}\exp\left(-\frac{\theta\epsilon^{2}}{2}\right).$

For the third term, we will first collect some facts. Observe that on the
event $A_{3}$, $\epsilon^{2}+t+\tau(t)$ is between $(\epsilon^{2}+t)/2$ and
$3(\epsilon^{2}+t)$, and for any $s$ in this interval,

$\displaystyle p_{2s}(y)$
$\displaystyle\leq\left(\frac{6(\epsilon^{2}+t)}{\epsilon^{2}+t}\right)^{d/2}p_{6(\epsilon^{2}+t)}(x,y)$
$\displaystyle=6^{d/2}p_{6(\epsilon^{2}+t)}(x,y).$ (7.21)

Moreover, since $ue^{-u}\leq e^{-1}$ for all $u\geq 0$,

$\displaystyle\frac{\|x-y\|^{2}}{s}p_{s}(x,y)$ $\displaystyle=\frac{4}{(2\pi
s)^{d/2}}e^{-\frac{\|x-y\|^{2}}{4s}}\frac{\|x-y\|^{2}}{4s}e^{-\frac{\|x-y\|^{2}}{4s}}$
$\displaystyle\leq Cp_{2s}(x,y).$ (7.22)

Now, by the Intermediate Value Theorem,

$\displaystyle\left|p_{\epsilon^{2}+t+\tau(t)}(x,y)-p_{\epsilon^{2}+t}(x,y)\right|$
$\displaystyle=\left|\tau(t)\right|\left|\frac{\partial p_{s}(x,y)}{\partial
s}\right|$ (7.23)

for some $s$ between $\epsilon^{2}+t+\tau(t)$ and $\epsilon^{2}+t$. Since

$\displaystyle\partial_{s}p_{s}(x,y)$
$\displaystyle=\partial_{s}\left(\frac{1}{(2\pi
s)^{d/2}}\exp\left(-\frac{\|x-y\|^{2}}{2s}\right)\right)$
$\displaystyle=-\frac{d}{2s}p_{s}(x,y)+\frac{\|x-y\|^{2}}{2s^{2}}p_{s}(x,y),$

applying the inequality (7.22), using the fact that $p_{s}(x,y)\leq
2^{d/2}p_{2s}(x,y)$, and then (7.21), we have that for any
$s\in((\epsilon^{2}+t)/2,3(\epsilon^{2}+t))$,

$\displaystyle\left|\frac{\partial}{\partial
s}p_{s}(x,y)\right|\leq\frac{C}{s}p_{2s}(x,y)\leq\frac{C}{\epsilon^{2}+t}p_{6(\epsilon^{2}+t)}(x,y).$

Therefore, recalling that $\mathbb{E}[\tau(t)^{2}]=t/\theta$, substituting
into (7.23),

$\displaystyle\mathbb{E}\left[1_{A_{3}}\left|p_{\epsilon^{2}+t+\tau(t)}(x,y)-p_{\epsilon^{2}+t}(x,y)\right|\right]$
$\displaystyle\leq\frac{C}{\epsilon^{2}+t}p_{6(\epsilon^{2}+t)}(x,y)\mathbb{E}\left[|\tau(t)|\right]$
$\displaystyle\leq\frac{C}{\epsilon^{2}+t}p_{6(\epsilon^{2}+t)}(x,y)\mathbb{E}\left[\tau(t)^{2}\right]^{1/2}$
$\displaystyle=\left(\frac{Ct}{\theta(\epsilon^{2}+t)^{2}}\right)^{1/2}p_{6(\epsilon^{2}+t)}(x,y)$
$\displaystyle\leq\frac{C}{\sqrt{\theta\epsilon^{2}}}p_{6(\epsilon^{2}+t)}(x,y),$

where the last inequality follows from
$2\epsilon^{2}t\leq(\epsilon^{2}+t)^{2}$.

Finally, on the event $A_{2}=\\{\tau(t)>2(\epsilon^{2}+t)\\}$, we simply use

$\displaystyle\left|p_{\epsilon^{2}+t+\tau(t)}(x,y)-p_{\epsilon^{2}+t}(x,y)\right|\leq\frac{C}{(\epsilon^{2}+t)^{d/2}},$

so that

$\displaystyle\mathbb{E}\left[1_{A_{2}}\left|p_{\epsilon^{2}+t+\tau(t)}(x,y)-p_{\epsilon^{2}+t}(x,y)\right|\right]\leq\frac{C}{(\epsilon^{2}+t)^{d/2}}\mathbb{P}\left\\{\tau(t)>2(\epsilon^{2}+t)\right\\},$

and apply Lemma 7.5 with $A=2$.

The last result will be useful when combined with the next bound for the heat
kernel.

######  Lemma 7.7.

Let $s>0$, and $x,y,z\in\mathbb{R}^{d}$. The following estimate holds:

$|p_{s}(x,z)-p_{s}(y,z)|\leq\frac{C\|x-y\|}{\sqrt{s}}\left(p_{2s}(x,z)+p_{2s}(y,z)\right),$

where the constant $C$ does not depend on $x,y,z$ or $s$.

###### Proof.

Expanding the difference of two squares,

$\displaystyle e^{-\frac{\|y-z\|^{2}}{2s}}-e^{-\frac{\|x-z\|^{2}}{2s}}$
$\displaystyle=\left(e^{-\frac{\|y-z\|^{2}}{4s}}-e^{-\frac{\|x-z\|^{2}}{4s}}\right)\left(e^{-\frac{\|y-z\|^{2}}{4s}}+e^{-\frac{\|x-z\|^{2}}{4s}}\right).$

Now, thinking of the first term in brackets as a function of a single variable
$x$ on the line segment $[y,z]$ connecting $y$ to $z$, we can apply the
Intermediate Value Theorem and take the modulus to bound this expression by

$\|y-x\|\left(\frac{2\|w-z\|}{4s}\exp\left(-\frac{\|w-z\|^{2}}{4s}\right)\right)(4\pi
s)^{d/2}\left(p_{2s}(y,z)+p_{2s}(x,z)\right)$

for some $w\in[y,z]$. Using the fact that $xe^{-x^{2}}$ is uniformly bounded,
we can bound the first bracket in the last equation by $C/\sqrt{s}$, and the
result follows.

We now have the ingredients that we need to write down a continuity estimate
for $\rho_{\epsilon}*\eta$. We fix $\delta>0$ and suppose that $s>\delta$. Let
us write

$\widehat{\epsilon}(\delta,\epsilon,\theta):=\frac{1}{(\epsilon^{2}+\delta)^{d/2}}e^{-\epsilon^{2}\theta/2},$

and note that under the assumption that $\epsilon^{2}\theta\to\infty$, for
each fixed $\delta>0$, $\lim_{\epsilon\to
0,\theta\to\infty}\widehat{\epsilon}(\delta,\epsilon,\theta)=0$. Using the
semimartingale decomposition (7.4), and Lemma 7.6, we have

$\displaystyle|\rho_{\epsilon}*\eta_{s}(y)-\rho_{\epsilon}*\eta_{s}(w)|=|\langle
p_{\epsilon^{2}}(y,z)-p_{\epsilon^{2}}(w,z),\eta_{s}(dz)\rangle|$
$\displaystyle\qquad\leq\big{\langle}|p_{\epsilon^{2}+s}(y,z)-p_{\epsilon^{2}+s}(w,z)|,\eta_{0}(dz)\big{\rangle}$
$\displaystyle\qquad\qquad{}+\int_{0}^{s-\delta}\big{\langle}|p_{s-r+\epsilon^{2}}(y,z)-p_{s-r+\epsilon^{2}}(w,z)|||F(\rho_{\epsilon}*\eta_{r}(z)|,\eta_{r}(dz)\big{\rangle}dr$
$\displaystyle\qquad\qquad{}+\Big{\langle}\frac{C}{(\theta\epsilon^{2})^{1/2}}\Big{(}p_{6(\epsilon^{2}+s)}(y,z)+p_{6(\epsilon^{2}+s)}(w,z)\Big{)}+C\widehat{\epsilon}(\delta,\epsilon,\theta),\eta_{0}(dz)\Big{\rangle}$
$\displaystyle\qquad\qquad{}+\int_{0}^{s-\delta}\big{\langle}\Big{\\{}\frac{C}{(\epsilon^{2}\theta)^{1/2}}|p_{6(s-r+\epsilon^{2})}(y,z)+p_{6(s-r+\epsilon^{2})}(w,z)|+\widehat{\epsilon}(\delta,\epsilon,\theta)\Big{\\}}|F(\rho_{\epsilon}*\eta_{r}(z)|,\eta_{r}(dz)\big{\rangle}dr$
$\displaystyle\qquad\qquad{}+\int_{s-\delta}^{s}\big{\langle}|\psi_{s-r}^{\epsilon,y}(z)+\psi_{s-r}^{\epsilon,w}(z)||F(\rho_{\epsilon}*\eta_{r}(z)|,\eta_{r}(dz)\big{\rangle}dr$
$\displaystyle\qquad\qquad{}+|M_{s}(y)|+|M_{s}(w)|$
$\displaystyle\qquad\leq\Big{\langle}\frac{\|y-w\|}{\sqrt{s+\epsilon^{2}}}\big{(}p_{2(s+\epsilon^{2})}(y,z)+p_{2(s+\epsilon^{2})}(w,z)\big{)},\eta_{0}(dz)\Big{\rangle}$
$\displaystyle\qquad\qquad{}+\int_{0}^{s-\delta}\Big{\langle}\frac{\|y-w\|}{\sqrt{s-r+\epsilon^{2}}}\big{(}p_{2(s-t+\epsilon^{2})}(y,z)+p_{2(s-r+\epsilon^{2})}(w,z)\big{)}|F(\rho_{\epsilon}*\eta_{r}(z)|,\eta_{r}(dz)\Big{\rangle}dr$
$\displaystyle\qquad\qquad{}+\Big{\langle}\frac{C}{(\theta\epsilon^{2})^{1/2}}\Big{(}p_{6(\epsilon^{2}+s)}(y,z)+p_{6(\epsilon^{2}+s)}(w,z)\Big{)}+C\widehat{\epsilon}(\delta,\epsilon,\theta),\eta_{0}(dz)\Big{\rangle}$
$\displaystyle\qquad\qquad{}+\int_{0}^{s-\delta}\big{\langle}\Big{\\{}\frac{C}{(\epsilon^{2}\theta)^{1/2}}(p_{6(s-r+\epsilon^{2})}(y,z)+p_{6(s-r+\epsilon^{2})}(w,z))+\widehat{\epsilon}(\delta,\epsilon,\theta)\Big{\\}}|F(\rho_{\epsilon}*\eta_{r}(z)|,\eta_{r}(dz)\big{\rangle}dr$
$\displaystyle\qquad\qquad{}+\int_{s-\delta}^{s}\big{\langle}|\psi_{s-r}^{\epsilon,y}(z)+\psi_{s-r}^{\epsilon,w}(z)||F(\rho_{\epsilon}*\eta_{r}(z)|,\eta_{r}(dz)\big{\rangle}dr$
$\displaystyle\qquad\qquad{}+|M_{s}(y)|+|M_{s}(w)|.$ (7.24)

Although this expression is lengthy, we have successfully isolated the terms
involving $\|y-w\|$, which will control the regularity as we pass to the
limit. Asymptotically, we don’t expect the martingale terms to contribute,
since their quadratic variation scales with $\theta/(N\epsilon^{d})$; under
the assumption that $\epsilon^{2}\theta\to\infty$, for any fixed $\delta>0$,
the terms arising from approximating the transition density
$\psi_{s-r}^{\epsilon,\cdot}(\cdot)$ of the Gaussian walk by
$p_{s-r+\epsilon^{2}}(\cdot,\cdot)$ at times with $s-r>\delta$ will tend to
zero; and the moment bounds of Lemma 7.2 will allow us to control the integral
over $[s-\delta,s]$. There is some technical work to be done to rigorously
identify the limit points of $\rho_{\epsilon}*\eta^{N}$, but it really amounts
to applying the tower property and our moment bounds from Lemma 7.2 and
Corollary 7.3.

###  7.3 Identification of the limit

We now turn to the identification of the limit points of the sequence
$\big{\\{}\big{(}\rho_{\epsilon}*\eta_{t}^{N}(x)dx\big{)}_{t\geq
0}\big{\\}}_{N\geq 1}$. We would like to show that any limit point solves
(2.16) in the limit, i.e.,

$\langle f(x),\varphi(t,x)dx\rangle=\int_{0}^{t}\Big{\langle}\frac{1}{2}\Delta
f(x)+f(x)F(\varphi(s,x)),\varphi(s,x)dx\Big{\rangle}ds.$ (7.25)

Since $\langle
f,\rho_{\epsilon}*\eta_{t}^{N}(x)dx\rangle=\langle\rho_{\epsilon}*f(x),\eta_{t}^{N}(dx)\rangle$,
and the limit is deterministic, this will follow if we can show that each of
the terms in the semimartingale decomposition (2.3), with the test function
$f$ replaced by $\rho_{\epsilon}*f(\cdot)$, converges to the corresponding
term in (7.25).

The linear term is straightforward. Write $T_{\cdot}$ for the heat semigroup,
so that $\rho_{\epsilon}*f(x)=T_{\epsilon^{2}}f(x)$. By a Taylor expansion,

$\displaystyle\int_{0}^{t}\big{\langle}\mathcal{L}^{\theta}T_{\epsilon^{2}}f,\eta_{s}^{N}(dx)\big{\rangle}ds$
$\displaystyle=\int_{0}^{t}\big{\langle}\frac{1}{2}\Delta
T_{\epsilon^{2}}f(x),\eta_{s}^{N}(dx)\big{\rangle}+\mathcal{O}\Big{(}\frac{1}{\theta}\Big{)}$
$\displaystyle=\int_{0}^{t}\big{\langle}\frac{1}{2}\Delta
f(x),\rho_{\epsilon}*\eta_{s}^{N}(x)dx\big{\rangle}ds+\mathcal{O}\Big{(}\frac{1}{\theta}\Big{)}.$

Thus, from weak convergence we can deduce that under our scaling, for any
(weakly) convergent subsequence $\\{\rho_{\epsilon}*\eta^{N}(x)dx\\}_{N\geq
1}$,

$\int_{0}^{t}\big{\langle}\mathcal{L}^{\theta}T_{\epsilon^{2}}f,\eta_{s}^{N}(dx)\big{\rangle}ds\rightarrow\int_{0}^{t}\big{\langle}\frac{1}{2}\Delta
f(x),\varphi(s,x)dx\big{\rangle}ds.$

The nonlinear term in the semimartingale decomposition is more intricate. It
takes the form

$\mathbb{E}\Big{[}\int_{0}^{t}\big{\langle}T_{\epsilon^{2}}f(y)F\big{(}\rho_{\epsilon}*\eta_{s}(y)\big{)},\eta_{s}(dy)\big{\rangle}ds\Big{]}$

and we should like to show that this converges to

$\int_{0}^{t}\int f(y)F(\varphi(s,y))\varphi(s,y)dyds.$

We proceed in stages. First we should like to transfer the heat semigroup from
$T_{\epsilon^{2}}f$ onto $\eta_{s}$. Since $f$ is smooth, this will follow
easily if we can show that

$\displaystyle\mathbb{E}\Big{[}\int_{0}^{t}\big{\langle}T_{\epsilon^{2}}f(y)F\big{(}\rho_{\epsilon}*\eta_{s}(y)\big{)},\eta_{s}(dy)\big{\rangle}ds\Big{]}\sim\mathbb{E}\Big{[}\int_{0}^{t}\big{\langle}T_{\epsilon^{2}}f(y)F\big{(}\rho_{\epsilon}*\eta_{s}(y)\big{)},\rho_{\epsilon}*\eta_{s}(y)dy\big{\rangle}ds\Big{]}.$

This is the content of Proposition 7.8.

######  Proposition 7.8.

Under the conditions of Theorem 2.20,

$\lim_{\epsilon\to
0}\mathbb{E}\Big{[}\Big{|}\int_{0}^{t}\big{\langle}T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y)),\eta_{s}(y)\big{\rangle}-\big{\langle}T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y)),\rho_{\epsilon}*\eta_{s}(y)dy\big{\rangle}ds\Big{|}\Big{]}=0.$
(7.26)

###### Proof.

In fact we are going to fix $\delta>0$, with $t>\delta$, and show that the
expression on the left hand side of (7.26) is less than a constant times
$\delta$, with a constant independent of $\delta$, $N$, and $\epsilon$. Since
$\delta$ is arbitrary, the result will follow.

We first note that,

$\displaystyle\langle
T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y)),\rho_{\epsilon}*\eta_{s}(y)dy\rangle-\langle
T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(dy)),\eta_{s}(dy)\rangle$
$\displaystyle=\langle\int
T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y))\rho_{\epsilon}(y-w)dy,\eta_{s}(dw)\rangle-\langle
T_{\epsilon^{2}}f(w)F(\rho_{\epsilon}*\eta_{s}(w)),\eta_{s}(dw)\rangle$
$\displaystyle=\langle\int\big{\\{}T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y))-T_{\epsilon^{2}}f(w)F(\rho_{\epsilon}*\eta_{s}(w))\big{\\}}\rho_{\epsilon}(w-y)dy,\eta_{s}(dw)\rangle.$

Let us denote the integral against $dy$ in the last expression by $I$, that is

$I:=\int\\{T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y))-T_{\epsilon^{2}}f(w)F(\rho_{\epsilon}*\eta_{s}(w))\\}\rho_{\epsilon}(w-y)dy,$

and note that $|I|$ is bounded by

$\displaystyle\int\Big{\\{}|F(\rho_{\epsilon}*\eta_{s}(y))-F(\rho_{\epsilon}*\eta_{s}(w))|T_{\epsilon^{2}}f(y)+F(\rho_{\epsilon}*\eta_{s}(w))|T_{\epsilon^{2}}f(y)-T_{\epsilon^{2}}f(w))|\Big{\\}}\rho_{\epsilon}(w-y)dy$
$\displaystyle\leq\int\|f\|_{\infty}\big{|}F(\rho_{\epsilon}*\eta_{s}(y))-F(\rho_{\epsilon}*\eta_{s}(w))\big{|}\rho_{\epsilon}(w-y)dy+C\epsilon\|f^{\prime}\|_{\infty}|F(\rho_{\epsilon}*\eta_{s}(w))|,$
(7.27)

where we have used that

$\int|T_{\epsilon^{2}}f(y)-T_{\epsilon^{2}}f(w)|p_{\epsilon^{2}}(w,y)dy\leq\|f^{\prime}\|_{\infty}\int|y-w|p_{\epsilon^{2}}(w,y)dy.$

Now recall that $F$ is a polynomial of degree $n$, and so there exist real
numbers $b_{k}$ such that $F(a)-F(b)=(a-b)\sum_{k=1}^{n-1}b_{k}a^{k}b^{n-1-k}$
and so

$|F(\rho_{\epsilon}*\eta_{s}(y))-F(\rho_{\epsilon}*\eta_{s}(w))|\leq|\rho_{\epsilon}*\eta_{s}(y)-\rho_{\epsilon}*\eta_{s}(w)|\sum_{k=1}^{n-1}|b_{k}|\left(\rho_{\epsilon}*\eta_{s}(y)^{n-1}+\rho_{\epsilon}*\eta_{s}(w)^{n-1}\right).$

Combining the above, we have reduced the problem to showing that for any
$k\geq 0$,

$\lim_{\epsilon\to
0}\mathbb{E}\Big{[}\int_{0}^{t}\big{\langle}\int|\rho_{\epsilon}*\eta_{s}(y)-\rho_{\epsilon}*\eta_{s}(w)|\left(\rho_{\epsilon}*\eta_{s}(y))^{k}+\rho_{\epsilon}*\eta_{s}(w)^{k}\right)p_{\epsilon^{2}}(w,y)dy,\eta_{s}(dw)\big{\rangle}ds\Big{]}=0.$
(7.28)

We are going to use the estimate (7.24). First note that by Lemma 7.2 the
contribution to (7.28) from the integral over the time interval $[0,\delta]$
is ${\mathcal{O}}(\delta)$. We focus instead on the interval $(\delta,t]$.

The first term in (7.24) gives

$\int_{\delta}^{t}\mathbb{E}\Big{[}\Big{\langle}\int\big{\langle}\frac{\|y-w\|}{\sqrt{s+\epsilon^{2}}}\big{(}p_{2(s+\epsilon^{2})}(y,z)+p_{2(s+\epsilon^{2})}(w,z)\big{)},\eta_{0}(dz)\big{\rangle}\\\
\big{(}\rho_{\epsilon}*\eta_{s}(y)^{k}+\rho_{\epsilon}*\eta_{s}(w)^{k}\big{)}p_{\epsilon^{2}}(w,y)dy,\eta_{s}(dw)\Big{\rangle}\Big{]}ds.$

We “borrow” from the exponential term to see that
$\|y-w\|p_{\epsilon^{2}}(w,y)\leq C\epsilon p_{2\epsilon^{2}}(w,y)$ and so
bound this by

$C\int_{\delta}^{t}\frac{\epsilon}{\sqrt{s+\epsilon^{2}}}\mathbb{E}\Big{[}\Big{\langle}\int\big{\langle}\big{(}p_{2(s+\epsilon^{2})}(y,z)+p_{2(s+\epsilon^{2})}(w,z)\big{)},\eta_{0}(dz)\big{\rangle}\\\
\big{(}\rho_{\epsilon}*\eta_{s}(y)^{k}+\rho_{\epsilon}*\eta_{s}(w)^{k}\big{)}p_{2\epsilon^{2}}(w,y)dy,\eta_{s}(dw)\Big{\rangle}ds\Big{]}.$
(7.29)

The four terms in the product are taken separately, according to the
combinations of $w$ and $y$ appearing. First,

$\int_{\delta}^{t}\frac{\epsilon}{\sqrt{s+\epsilon^{2}}}\mathbb{E}\Big{[}\Big{\langle}\int\big{\langle}p_{2(s+\epsilon^{2})}(y,z),\eta_{0}(dz)\big{\rangle}\rho_{\epsilon}*\eta_{s}(y)^{k}p_{2\epsilon^{2}}(w,y)dy,\eta_{s}(dw)\Big{\rangle}\Big{]}ds$

can be rewritten as

$\int_{\delta}^{t}\frac{\epsilon}{\sqrt{s+\epsilon^{2}}}\mathbb{E}\Big{[}\int\int\big{\langle}p_{2(s+\epsilon^{2})}(y,z),\eta_{0}(dz)\big{\rangle}\rho_{\epsilon}*\eta_{s}(y)^{k}p_{\epsilon^{2}}(x,y)\rho_{\epsilon}*\eta_{s}(x)dydxds\Big{]}\\\
\leq\int_{\delta}^{t}\frac{\epsilon}{\sqrt{s+\epsilon^{2}}}\mathbb{E}\Big{[}\int\int\big{\langle}p_{2(s+\epsilon^{2})}(y,z),\eta_{0}(dz)\big{\rangle}\big{(}\rho_{\epsilon}*\eta_{s}(y)^{k+1}+\rho_{\epsilon}*\eta_{s}(x)^{k+1}\big{)}p_{\epsilon^{2}}(x,y)dydx\Big{]}ds,$

and using Lemma 7.2 and the tower property, and integrating with respect to
$s$, under our assumptions on $\eta_{0}$, this is bounded by

$C\epsilon\int_{\delta}^{t}\frac{1}{\sqrt{s+\epsilon^{2}}}\mathbb{E}\Big{[}\int\int\big{\langle}p_{2(s+\epsilon^{2})}(y,z),\eta_{0}(dz)\big{\rangle}\\\
\big{(}\rho_{\epsilon}*\eta_{0}(y)+\rho_{\epsilon}*\eta_{0}(y)^{k+1}+\rho_{\epsilon}*\eta_{0}(x)+\rho_{\epsilon}*\eta_{0}(x)^{k+1}\big{)}p_{\epsilon^{2}}(x,y)dydx\Big{]}ds\\\
\leq C^{\prime}\epsilon\int_{\delta}^{t}\frac{1}{(s+\epsilon^{2})^{d/2}}ds.$

For fixed $\delta$, this bound tends to zero as $\epsilon\to 0$. The term
involving
$\big{\langle}p_{2(s+\epsilon^{2})}(w,z),\eta_{0}(dz)\big{\rangle}\rho_{\epsilon}*\eta_{s}(w)^{k}$
is handled similarly.

On the other hand

$\big{\langle}\int\langle
p_{2(s+\epsilon^{2})}(y,z),\eta_{0}(dz)\rangle\rho_{\epsilon}*\eta_{s}(w)^{k}p_{2\epsilon^{2}}(w,y)dy,\eta_{s}(dw)\big{\rangle}\\\
\leq\frac{C}{(s+\epsilon^{2})^{d/2}}\langle
1,\eta_{0}\rangle\big{\langle}\rho_{\epsilon}*\eta_{s}(w)^{k},\eta_{s}(dw)\big{\rangle},$

and since $\langle 1,\eta_{0}\rangle$ is uniformly bounded we apply Corollary
7.3 to obtain a bound on the contribution to (7.29) from this term of the same
form as the others.

Now consider the contribution to the left hand side of (7.28) from the second
term in (7.24). Since $F$ is a polynomial, it is bounded by a sum of terms of
the form

$\int_{\delta}^{t}\int_{0}^{s-\delta}\Big{\langle}\int\frac{\|y-w\|}{\sqrt{s-r+\epsilon^{2}}}\big{\langle}\big{(}p_{2(s-r+\epsilon^{2})}(y,z)+p_{2(s-r+\epsilon^{2})}(w,z)\big{)}\rho_{\epsilon}*\eta_{r}(z)^{j},\eta_{r}(dz)\big{\rangle}\\\
\rho_{\epsilon}*\eta_{s}(y)^{k}p_{\epsilon^{2}}(y,w)dy,\eta_{s}(dw)\Big{\rangle}drds\\\
\leq
C\epsilon\int_{\delta}^{t}\int_{0}^{s-\delta}\frac{1}{\sqrt{s-r+\epsilon^{2}}}\Big{\langle}\int\big{\langle}\big{(}p_{2(s-r+\epsilon^{2})}(y,z)+p_{2(s-r+\epsilon^{2})}(w,z)\big{)}\rho_{\epsilon}*\eta_{r}(z)^{j},\eta_{r}(dz)\big{\rangle}\\\
\rho_{\epsilon}*\eta_{s}(y)^{k}p_{2\epsilon^{2}}(y,w)dy,\eta_{s}(dw)\Big{\rangle}drds,$

where as usual we have “borrowed” from the exponential term in
$p_{\epsilon}^{2}(y,w)$ to replace $\|y-w\|$ by a constant times $\epsilon$.

Once again, our approach is to rearrange terms so that we can apply Lemma 7.2
or Corollary 7.3 to obtain a bound on the contribution to (7.28) from these
terms of the form $C\epsilon$ (where $C$ may depend on $\delta$ but not
$\epsilon$).

For example, using the Chapman-Kolmogorov equation to rewrite

$\int\Big{\langle}\big{\langle}p_{2(s-r+\epsilon^{2})}(y,z)\rho_{\epsilon}*\eta_{r}(z)^{j},\eta_{r}(dz)\big{\rangle}\rho_{\epsilon}*\eta_{s}(y)^{k}p_{2\epsilon^{2}}(y,w),\eta_{s}(dw)\Big{\rangle}dy$

as

$\int\int\big{\langle}p_{2(s-r+\epsilon^{2})}(y,z)\rho_{\epsilon}*\eta_{r}(z)^{j},\eta_{r}(dz)\big{\rangle}\rho_{\epsilon}*\eta_{s}(y)^{k}p_{\epsilon^{2}}(y,x)\rho_{\epsilon}*\eta_{s}(x)dxdy,$

and using Lemma 7.2 and the tower property, we are led to control terms of the
form

$\mathbb{E}\Big{[}\int\big{\langle}p_{2(s-r+\epsilon^{2})}(y,z)\rho_{\epsilon}*\eta_{r}(z)^{j},\eta_{r}(dz)\big{\rangle}\rho_{\epsilon}*\eta_{r}(y)^{k+1}dy\Big{]}.$

This, in turn, is at most

$\mathbb{E}\Big{[}\big{\langle}\rho_{\epsilon}*\eta_{r}(z)^{j+k+1},\eta_{r}(dz)\rangle\Big{]}+\mathbb{E}\Big{[}\int\int
p_{2(s-r)}(y,x)\rho_{\epsilon}*\eta_{r}(x)\rho_{\epsilon}*\eta_{r}(y)^{j+k+1}dydx\Big{]}\\\
\leq\mathbb{E}\Big{[}\big{\langle}\rho_{\epsilon}*\eta_{r}(z)^{j+k+1},\eta_{r}(dz)\rangle\Big{]}+2\int\mathbb{E}\Big{[}\rho_{\epsilon}*\eta_{r}(x)^{j+k+2}\Big{]}dx,$

which is bounded by Lemma 7.2.

We now turn to the contribution arising from the martingale terms in (7.24):

$\mathbb{E}\Big{[}\int_{\delta}^{t}\Big{\langle}\int\big{(}|M_{s}(y)|+|M_{s}(w)|\big{)}\big{(}\rho_{\epsilon}*\eta_{s}(y)^{k}+\rho_{\epsilon}*\eta_{s}(w)^{k}\big{)}p_{\epsilon^{2}}(w,y)dy,\eta_{s}(dw)\Big{\rangle}ds\Big{]}.$

Since $\psi_{t-s}^{\epsilon,x}(y)=\mathbb{E}[p_{T(t-s)+\epsilon^{2}}(x,y)]$,
rearranging (7.4) we see that we can pull a convolution with
$p_{\epsilon^{2}/2}$ out of our expressions for $M_{s}(y)$ and $M_{s}(w)$ and
so all the manipulations that we used to control terms above with still be
valid. To deal with the two terms in the product involving $|M_{s}(y)|$, we
write the first as $\int|M_{s}(y)|\rho_{\epsilon}*\eta_{s}(y)^{k+1}dy$ and
then use Hölder’s inequality, Lemma 7.2, and the fact that
$\mathbb{E}\big{[}|M_{s}(y)|^{2}\big{]}$ is
${\mathcal{O}}\big{(}\theta/(N\epsilon^{d})\big{)}$ to see that the
contribution from this term tends to zero in the limit. For the second, we use
the idea of the proof of Corollary 7.3 to reduce to a form to which we can
apply Hölder’s inequality.

Control of the terms arising from approximating $\psi^{\epsilon,x}$ by the
heat kernel follows in an entirely analogous way.

Combining the above, we see that given $\delta>0$,

$\lim_{\epsilon\to
0}\mathbb{E}\Big{[}\Big{|}\int_{0}^{t}\big{\langle}T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y)),\eta_{s}(y)\big{\rangle}ds-\int_{0}^{t}\big{\langle}T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y)),\rho_{\epsilon}*\eta_{s}(y)dy\big{\rangle}ds\Big{|}\Big{]}<C\delta,$

where the constant $C$ is independent of $\delta$. Since $\delta$ was
arbitrary, the proof is complete.

Since $f$ is smooth, $T_{\epsilon}^{2}f-f$ is ${\mathcal{O}}(\epsilon)$, with
an application of the triangle inquality,

$\lim_{\epsilon\to
0}\mathbb{E}\Big{[}\Big{|}\int_{0}^{t}\big{\langle}T_{\epsilon^{2}}f(y)F(\rho_{\epsilon}*\eta_{s}(y)),\eta_{s}(y)\big{\rangle}ds-\int_{0}^{t}\big{\langle}f(y)F(\rho_{\epsilon}*\eta_{s}(y)),\rho_{\epsilon}*\eta_{s}(y)dy\big{\rangle}ds\Big{|}\Big{]}<C\delta,$

now follows immediately. Thus to complete the characterisation of the limit,
it remains to show that if we take a convergent subsequence
$\big{\\{}\big{(}\rho_{\epsilon}*\eta_{t}^{N}(dx)\big{)}_{t\geq 0}\big{\\}}$
converging to a limit point $\big{(}\varphi(t,x)dx\big{)}_{t\geq 0}$, then

$\int_{0}^{t}\int
f(x)\rho_{\epsilon}*\eta_{s}(x)F(\rho_{\epsilon}*\eta_{s}^{N}(x))dxds\rightarrow\int_{0}^{t}\int
f(x)\varphi(s,x)F(\varphi(s,x))dxds.$

Since $F$ is a polynomial, we consider powers of $\rho_{\epsilon}*\eta$. To
illustrate the approach, we first prove that

$\int_{0}^{t}\int
f(x)\rho_{\epsilon}*\eta_{s}^{N}(x)^{2}dxds\rightarrow\int_{0}^{t}\int
f(x)\varphi(s,x)^{2}dxds.$ (7.30)

The convergence of higher powers will follow in an entirely analogous manner,
but with more complex expressions.

The approach is standard. We fix $\tau>0$ and, in keeping with our notation
$\rho_{\epsilon}$, in this subsection, use $\rho_{\tau}$ to denote the
symmetric Gaussian kernel with variance parameter $\tau^{2}$. Our strategy is
to show that, up to an error that tends to zero as $\tau\to 0$,

$\int_{0}^{t}\int
f(z)\rho_{\epsilon}*\eta_{s}(z)^{2}dzds\sim\int_{0}^{t}\int\int
f(z)(\rho_{\epsilon}*\eta_{s})(z)\rho_{\tau}(z-y)(\rho_{\epsilon}*\eta_{s})(y)dzdyds.$
(7.31)

Analogously, also up to an error that vanishes as $\tau\to 0$,

$\int_{0}^{t}\int f(z)\varphi(s,z)^{2}dzds\sim\int_{0}^{t}\int\int
f(z)\varphi(s,z)\rho_{\tau}(z-y)\varphi(s,y)dzdyds.$ (7.32)

On the other hand, weak convergence of $\rho_{\epsilon}*\eta$ (plus continuity
of the mapping $(z,y)\rightarrow f(z)\rho_{\tau}(z-y)$) gives that

$\int_{0}^{t}\int\int
f(z)(\rho_{\epsilon}*\eta_{s})(z)\rho_{\tau}(z-y)(\rho_{\epsilon}*\eta_{s})(y)dzdyds\to\int_{0}^{t}\int\int
f(z)\varphi(s,z)\rho_{\tau}(z-y)\varphi(s,y)dzdyds.$ (7.33)

Since $\tau$ is arbitrary, the convergence (7.30) will follow.

######  Proposition 7.9.

Under the conditions of Theorem 2.20, we have that along any convergent
subsequence,

$\limsup_{\epsilon\to 0}\mathbb{E}\Big{[}\Big{|}\int_{0}^{t}\int
f(y)\rho_{\epsilon}*\eta_{s}(y)^{2}dyds\\\ -\int_{0}^{t}\int\int
f(z)(\rho_{\epsilon}*\eta_{s})(z)\rho_{\tau}(z-y)(\rho_{\epsilon}*\eta_{s})(y)dzdyds\Big{|}\Big{]}\leq
C\tau,$ (7.34)

where $C$ is independent of $\tau$.

###### Proof.

First note,

$\displaystyle\int_{0}^{t}\mathbb{E}\left[\big{|}\langle
f(y),(\rho_{\epsilon}*\eta_{s}(y))^{2})dy\rangle-\int\int
f(y)(\rho_{\epsilon}*\eta_{s})(z)\rho_{\tau}(z-y)(\rho_{\epsilon}*\eta_{s})(y)dzdy\big{|}\right]ds$
$\displaystyle\leq\|f\|_{\infty}\int\int_{0}^{t}\mathbb{E}\left[\int\left\\{\left|(\rho_{\epsilon}*\eta_{s})(y)-(\rho_{\epsilon}*\eta_{s})(z)\right|\rho_{\tau}(z-y)dz\right\\}(\rho_{\epsilon}*\eta_{s})(y)\right]dsdy.$
(7.35)

Now proceed exactly as in the proof of Proposition 7.8. The only distinction
is that $|p_{\epsilon^{2}}(y,z)-p_{\epsilon^{2}}(w,z)|$ is replaced by
$|p_{\tau}(y,z)-p_{\tau}(w,z)|$ and the estimate $\|y-w\|p_{\tau^{2}}(y,w)\leq
C\tau p_{2\tau^{2}}(y,w)$ replaces the corresponding statement with
$\epsilon^{2}$ replacing $\tau^{2}$ in our previous argument.

The extension of Proposition 7.9 to higher moments is straightforward, if
notationally messy. For fixed (but arbitrary) $\tau$, one shows that

$\limsup_{\epsilon\to 0}\mathbb{E}\Big{[}\Big{|}\int_{0}^{t}\int
f(y)\rho_{\epsilon}*\eta_{s}^{N}(y)^{k}dyds\\\ -\int_{0}^{t}\int\cdots\int
f(y_{1})\rho_{\epsilon}*\eta_{s}^{N}(y_{1})\prod_{i=2}^{k}\rho_{\tau}(y_{i}-y_{i-1})\rho_{\epsilon}*\eta_{s}^{N}(y_{i})dy_{k}\ldots
dy_{1}ds\Big{|}\Big{]}\leq C\tau,$

as well as a corresponding statement with $\rho_{\epsilon}*\eta_{s}^{N}(x)$
replaced by $\varphi(s,x)$ and then use weak convergence to see that, up to an
error of order $\tau$, any limit point of the sequence
$\\{\rho_{\epsilon}*\eta^{N}(x)dx\\}$ solves (the weak form of) equation
(2.16). Since $\tau$ was arbitrary, the proof of Theorem 2.20 is complete.

##  8 Proofs of results for the lookdown process and ancestral lineages

Now we turn to results about the lookdown process, first establishing the
basic connection between the population process $\eta^{N}$ and the lookdown
process $\xi^{N}$, Proposition 5.3, and then in the next section, convergence
of the lookdown process itself.

###### of Proposition 5.3.

This proposition is the content of the Markov Mapping Theorem, reproduced from
Etheridge and Kurtz (2019) as Theorem A.1, applied to our situation. The
function $\gamma$ of that theorem is what we have called $\kappa$ above, and
the kernel $\alpha$ of that theorem is the transition function that assigns
levels uniformly on $[0,N]$ (in the first case) or as a Poisson process with
Lesbegue intensity (in the limiting case). We need a continuous
$\psi^{N}(\xi)\geq 1$ such that $|A^{N}f(\xi)|\leq c_{f}\psi^{N}(\xi)$ for all
$f$ in the domain of $A^{N}$ (and similarly a function $\psi$ for $A$). We
also need that applying the lookdown generator to a function and averaging
over levels is equivalent to applying the population process generator to the
function whose dependence on levels has been averaged out, a condition which
we precisely state, and verify, in Lemmas A.2 and A.3 of the Appendix.

For finite $N$, taking $f(\xi)$ of the form (5.6), we can use
$\psi^{N}(\xi)=\langle C(1+u|F(x,\eta)|),\xi(dx,du)\rangle$ for an appropriate
constant $C$. For the scaling limit, recall that the test functions $f$ are of
the form $f(\xi)=\prod_{(x,u)\in\xi}g(x,u)$ with $g(x,u)=1$ for $u\geq u_{0}$,
and consulting (5.11), we see that most terms in $Af(\xi)$ can be bounded as
above by constant multiples of $\langle 1,\eta\rangle$. However, the term
involving $F$ is, as usual, more troublesome. Since $0\leq f(\xi)/g(x,u)\leq
1$ for any $(x,u)\in\xi$,

$\displaystyle\big{|}f(\xi)\sum_{(x,u)\in\xi}F(x,\eta)u\frac{\partial_{u}g(x,u)}{g(x,u)}\big{|}$
$\displaystyle\leq\|\partial_{u}g\|_{\infty}\sum_{(x,u)\in\xi}|F(x,\eta)u\mathbf{1}_{u\leq
u_{0}}|$
$\displaystyle\leq\|\partial_{u}g\|_{\infty}e^{u_{0}}\sum_{(x,u)\in\xi}|F(x,\eta)ue^{-u}|.$

The first line would be just what we want, except that $\psi(\xi)$ cannot
depend on $f$, and hence neither on $u_{0}$. So, the second line provides us
with the required bound: we absorb $\|\partial_{u}g\|_{\infty}e^{u_{0}}$ into
$c_{f}$ and take $\psi(\xi)=1+\langle 1+F(x,\eta)ue^{-u},\xi(dx,du)\rangle$.

###  8.1 Tightness of the Lookdown Process

Now we turn to the main theorem on convergence of the lookdown process,
Theorem 5.4, whose proof follows a similar pattern to that of convergence for
the population processes in Section 6.2.

We first give a description of the lookdown process $\xi^{N}$ in terms of the
lines of descent introduced in Section 5.2. Each line of descent gives birth
to lines at higher levels at rate $2(N-u)c_{\theta}(x,\eta)$, and each such
new line chooses a level uniformly from $[u,N]$, a spatial location $y$ from
the kernel

$q^{m}(x,dy,\eta)=r(y,\eta)q(x,dy)/\int_{\mathbb{R}^{d}}r(y,\eta)q(x,dy),$
(8.1)

and the two lines swap spatial locations with probability 1/2; the level of
each line of descent evolves according to equation (5.21).

It is evident from the description of the process (or, by differentiating in
Definition 5.1) that

$\displaystyle\begin{split}\langle f,\xi_{t}^{N}\rangle&=\langle
f,\xi_{0}^{N}\rangle+M^{f}_{t}\\\
&\qquad{}+\int_{0}^{t}\bigg{\langle}c_{\theta}(x,\eta^{N}_{s})\int_{u}^{N}\int_{\mathbb{R}^{d}}\left(f(y,u_{1})+f(x,u_{1})+f(y,u)-f(x,u)\right)q^{m}(x,dy,\eta^{N}_{s})du_{1}\\\
&\qquad\qquad\qquad\qquad{}+\left(c_{\theta}(x,\eta^{N}_{s})u^{2}-b_{\theta}(x,\eta^{N}_{s})u\right)\frac{d}{du}f(x,u),\xi_{s}^{N}(dx,du)\bigg{\rangle}ds,\end{split}$
(8.2)

where $M^{f}$ is a martingale with angle bracket process

$\displaystyle\begin{split}\left\langle
M^{f}\right\rangle_{t}&=\int_{0}^{t}\bigg{\langle}c_{\theta}(x,\eta_{s}^{N})\int_{u}^{N}\int_{\mathbb{R}^{d}}\big{[}f(y,u_{1})^{2}\\\
&\qquad\qquad{}+\left(f(x,u_{1})+f(y,u)-f(x,u)\right)^{2}\big{]}du_{1}q^{m}(x,dy,\eta^{N}_{s}),\xi_{s}^{N}(dx,du)\bigg{\rangle}ds.\end{split}$
(8.3)

######  Remark 8.1.

In addition to tightness of the measure-values processes $\xi^{N}$, the bounds
used in the proofs below also imply tightness of the number of lines of
descent and the number of births below a fixed level, and of the motion of
individual lines of descent. In other words, the limiting “line of descent”
construction of Section 5.2 holds.

###### Proof of Theorem 5.4.

As in Section 6.2, the theorem will follow from tightness and characterization
of the limit points. This time, the processes $\xi^{N}$ take values in
$\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty))$, the space of locally
finite measures on space $\times$ levels. (They will in fact be point
measures, including the limit, but that is a consequence of this theorem.)
Again, tightness follows from a compact containment condition, tightness of
one-dimensional distributions, and an application of Ethier and Kurtz (1986)
Theorem 3.9.1.

Lines of descent can escape to infinite level in finite time, and so we endow
$\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty))$ with the vague
topology “in the level coordinate”, induced by test functions on
$\overline{\mathbb{R}^{d}}\times[0,\infty)$ of the form $g(x)h(u)$, where
$g\in C_{b}(\overline{\mathbb{R}^{d}})$ is bounded and continuous and $h\in
C_{c}([0,\infty))$ is compactly supported (following, e.g., Etheridge and
Kurtz (2019), Condition 2.1). In several places below we require a dense
subset of $C_{b}(\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty)))$, the
bounded, continuous functions on
$\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty))$. The functions
$\xi\mapsto\langle f,\xi\rangle$ do not form not a dense subset of
$C_{b}(\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty)))$, but they do
separate points and vanish nowhere, i.e., for any $\xi_{1}$ and $\xi_{2}$
there is an $f$ with $\langle f,\xi_{1}\rangle\neq\langle f,\xi_{2}\rangle$,
and a $g$ such that $\langle g,\xi_{1}\rangle\neq 0$. Therefore, by the Stone-
Weierstrass theorem, the algebra they generate is dense in
$C_{b}(\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty)))$ (with respect
to uniform convergence on compact subsets). Topologized in this way, the space
$\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty))$ is completely
metrizable, and we may choose a countable set of nonnegative $f_{k}$, each
supported on $\mathbb{R}^{d}\times[0,u_{k}]$ for some $u_{k}<\infty$, such
that a subset $K\subset\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty))$
is relatively compact if and only if $\sup_{\xi\in K}\langle
f_{k},\xi\rangle<\infty$ for each $k$. (To see this, use Theorem A.2.3 of
Kallenberg (1997) and the first argument made in the proof of Lemma B.3.)
Below, Lemma 8.4 proves exactly this, and therefore compact containment. Here
we have compactified $\mathbb{R}^{d}$ for convenience (since it turned out to
be straightforward to show that mass does not escape to infinity in space);
however, we need to use the vague topology “in the level direction” because
_levels_ may escape to infinity in finite time in the limit.

In order to apply Ethier and Kurtz (1986) Theorem 3.9.1 we require that
$\\{(F(\xi^{N}_{t}))_{t\geq 0}\\}_{N}$ is tight as a sequence of real-valued
càdlàg processes, for all $F$ in a subset of
$C_{b}(\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty)))$ that is dense
with respect to uniform convergence on compact subsets. Lemma 8.5 shows that
$\\{\langle f,\xi^{N}_{t}\rangle\\}_{N}$ is a tight sequence for any
$f:\overline{\mathbb{R}^{d}}\times[0,\infty)\to\mathbb{R}$ with compact
support in the level direction. Since as above the algebra generated by the
functions $\xi\mapsto\langle f,\xi\rangle$ is dense in
$C_{b}(\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty)))$, it suffices
to show that tightness for the processes $\langle f,\xi^{N}_{t}\rangle$
extends to finite sums and products of these processes, which is shown in
Lemma B.3. The fact that martingale properties are preserved under passage to
the limit is straightforward, and can be proved in a way analogous to Lemma
6.6; we omit the proof. Finally, we must show that the limiting lookdown
process $\xi$ projects to the limiting process $\eta$, i.e., a solution of the
martingale problem in Theorem 2.10. Let $N_{k}\to\infty$ be a sequence along
which $\xi^{N_{k}}$ converges. By Theorem 2.10, there is a subsequence
$N_{k(j)}$ along which the projected population processes $\eta^{N_{k(j)}}$
converge, and the limit solves the martingale problem. Thus any limit point of
$\xi^{N}$ projects to a population process $\eta$ solving the martingale
problem of Theorem 2.10.

What we need for compact containment will come from the following Lemma. The
generality is unimportant – for concreteness one may take $h(u)=e^{-u}$.

######  Lemma 8.2.

Let $h$ be a positive, continuous, nonincreasing, differentiable function on
$[0,\infty)$ such that $\int_{0}^{\infty}\int_{u}^{\infty}h(v)dvdu$,
$\int_{0}^{\infty}u^{2}|h^{\prime}(u)|du$, and $\int_{0}^{\infty}h(u)^{2}du$
are all finite. Suppose that Assumptions 2.8 hold, and that
$\theta/N\to\alpha$ and $\xi_{0}^{N}\to\xi_{0}$ weakly as $N\to\infty$, where
each $\xi_{0}^{N}$ is conditionally uniform given $\eta_{0}^{N}$ in the sense
of (5.12) and $\xi_{0}$ is conditionally Poisson given $\eta_{0}$ in the sense
of (5.13). Then for any $T$ there exists a constant $K(T)$ such that for all
$M>0$,

$\limsup_{N\to\infty}\mathbb{P}\left\\{\sup_{0\leq t\leq T}\langle
h,\xi^{N}_{t}\rangle>M\right\\}<\frac{K(T)}{M}.$

We postpone the proof of this Lemma until we have shown how it yields compact
containment. First, we show that this implies compact containment of the
processes $(\langle f,\xi^{N}_{t}\rangle)_{0\leq t\leq T}$ for arbitrary
compactly supported $f$.

######  Lemma 8.3.

Suppose $f\in C(\overline{\mathbb{R}}^{d}\times[0,\infty))$ and there is a
$u_{f}$ such that if $u\geq u_{f}$ then $\sup_{x}f(x,u)=0$. Under the
assumptions of Lemma 8.2, for any $T$ there exists a constant $K(f,T)$ such
that for all $M>0$,

$\limsup_{N\to\infty}\mathbb{P}\left\\{\sup_{0\leq t\leq T}\langle
f,\xi^{N}_{t}\rangle>M\right\\}<\frac{K(f,T)}{M}.$

###### of Lemma 8.3.

Let $h$ be as in Lemma 8.2, so there is a $c_{f}<\infty$ such that $f(x,u)\leq
c_{f}h(u)$ for all $x$ and $u$. Therefore, $\langle f,\xi\rangle\leq\langle
h,\xi\rangle$, and so by Lemma 8.2,

$\displaystyle\limsup_{N\to\infty}\mathbb{P}\left\\{\sup_{0\leq t\leq
T}\langle
f,\xi^{N}_{t}\rangle>M\right\\}\leq\limsup_{N\to\infty}\mathbb{P}\left\\{\sup_{0\leq
t\leq T}\langle h,\xi^{N}_{t}\rangle>M/c_{f}\right\\}<\frac{K(T)c_{f}}{M}.$

######  Lemma 8.4 (Compact containment for $\xi$).

Let $f_{1},f_{2},\ldots$ be a sequence of functions each satisfying the
conditions of Lemma 8.3. Under the assumptions of Lemma 8.2, for any $T$ and
$\delta>0$ there exists a sequence $(C_{1},C_{2},\ldots)$ of finite constants
such that

$\displaystyle\limsup_{N\to\infty}\mathbb{P}\left\\{\sup_{0\leq t\leq
T}\langle f_{k},\xi^{N}_{t}\rangle>C_{k}\text{ for some }k\geq
1\right\\}<\delta.$ (8.4)

In other words, the processes $\xi^{N}$ stay in the set

$\left\\{\xi\in\mathcal{M}(\overline{\mathbb{R}^{d}}\times[0,\infty))\;:\;\langle
f_{k},\xi\rangle\leq C_{k}\text{ for all }k\geq 1\right\\},$

for all $0\leq t\leq T$ with uniformly high probability, a set which (as
discussed in the proof of Theorem 5.4) is relatively compact for an
appropriate choice of $\\{f_{k}\\}_{k\geq 1}$.

###### of Lemma 8.4.

By a union bound,

$\mathbb{P}\left\\{\sup_{0\leq t\leq T}\langle
f_{k},\xi^{N}_{t}\rangle>C_{k}\text{ for some }k\geq 1\right\\}\leq\sum_{k\geq
1}\mathbb{P}\left\\{\sup_{0\leq t\leq T}\langle
f_{k},\xi^{N}_{t}\rangle>C_{k}\right\\},$

so (8.4) follows by taking $C_{k}=2^{k-1}K(f_{k},T)/\delta$ and using Lemma
8.3.

Finally, we prove the key lemma.

###### of Lemma 8.2.

Applied to $f(x,u)=h(u)$, the martingale representation (8.2) is

$\displaystyle\begin{split}\langle h,\xi_{t}^{N}\rangle&=\langle
h,\xi_{0}^{N}\rangle+M^{h}_{t}\\\
&\qquad{}+\int_{0}^{t}\big{\langle}2c_{\theta}(x,\eta^{N}_{s})\int_{u}^{N}h(v)dv,\xi_{s}^{N}(dx,du)\big{\rangle}ds\\\
&\qquad\qquad{}+\int_{0}^{t}\big{\langle}\left(c_{\theta}(x,\eta^{N}_{s})u^{2}-b_{\theta}(x,\eta^{N}_{s})u\right)h^{\prime}(u),\xi_{s}^{N}(dx,du)\big{\rangle}ds,\end{split}$

where $M^{h}_{t}$ is a martingale with angle bracket process

$\displaystyle\left\langle M^{h}\right\rangle_{t}$
$\displaystyle=\int_{0}^{t}\langle
2c_{\theta}(x,\eta_{s}^{N})\int_{u}^{N}h(v)^{2}dv,\xi_{s}^{N}(dx,du)\big{\rangle}ds.$

Now, note that $0\leq c_{\theta}(x,\eta_{x}^{N})\leq C_{a}<\infty$ and
$b_{\theta}(x,\eta_{s}^{N})\leq C_{b}<\infty$, and we have assumed that
$h^{\prime}(u)\leq 0$ (since $h$ is nonincreasing), so we may bound

$\displaystyle\begin{split}\langle h,\xi_{t}^{N}\rangle&\leq\langle
h,\xi_{0}^{N}\rangle+M^{h}_{t}\\\
&\qquad{}+\int_{0}^{t}\big{\langle}2C_{a}\int_{u}^{\infty}h(v)dv+\left(C_{a}u^{2}+C_{b}u\right)|h^{\prime}(u)|,\xi_{s}^{N}(dx,du)\big{\rangle}ds.\end{split}$
(8.5)

Now, since $\xi_{t}^{N}$ is conditionally uniform given $\eta_{t}^{N}$ in the
sense of (5.12), we know that for compactly supported $f$, $\mathbb{E}[\langle
f,\xi^{N}_{t}\rangle]=\mathbb{E}[\langle\widetilde{f}_{N},\eta^{N}_{t}\rangle]$,
where $\widetilde{f}_{N}(x)=\int_{0}^{N}f(x,u)du$. By our assumptions on $h$,
we know that

$\int_{0}^{\infty}\left(2C_{a}\int_{u}^{\infty}h(v)dv+\left(C_{a}u^{2}+C_{b}u\right)|h^{\prime}(u)|\right)du<C$

for some $C<\infty$, and so (by dominated convergence)

$\displaystyle\begin{split}\mathbb{E}\left[\langle
h,\xi_{t}^{N}\rangle\right]&\leq\mathbb{E}\left[\langle
h,\xi_{0}^{N}\rangle\right]+C\int_{0}^{t}\mathbb{E}\bigg{[}\big{\langle}1,\eta_{s}^{N}\big{\rangle}\bigg{]}ds,\end{split}$

which we know by Lemma 6.1 is bounded by $C_{0}e^{C_{1}t}$ for some other
constants $C_{0}$ and $C_{1}$.

Now consider the maximum. By (8.5), using that the integrand is nonnegative,

$\displaystyle\sup_{0\leq t\leq T}\langle h,\xi_{t}^{N}\rangle$
$\displaystyle\leq\langle h,\xi_{0}^{N}\rangle+\sup_{0\leq t\leq T}M^{h}_{t}$
$\displaystyle\qquad{}+\int_{0}^{T}\big{\langle}2C_{a}\int_{u}^{\infty}h(v)dv+\left(C_{a}u^{2}+C_{b}u\right)|h^{\prime}(u)|,\xi_{s}^{N}(dx,du)\big{\rangle}ds.$

Since $\sqrt{x}\leq 1+x$ for $x\geq 0$, the Burkholder-Davis-Gundy inequality
tells us that there is a $C^{\prime}$ such that

$\displaystyle\mathbb{E}\left[\sup_{0\leq t\leq T}M^{h}_{t}\right]$
$\displaystyle\leq
C^{\prime}\left(1+\mathbb{E}\left[[M^{h}]_{T}\right]\right)$
$\displaystyle\leq C^{\prime}\left(1+\int_{0}^{T}\mathbb{E}\left[\langle
2c_{\theta}(x,\eta_{s}^{N})\int_{u}^{\infty}h(v)^{2}dv,\xi_{s}^{N}(dx,du)\big{\rangle}\right]ds\right)$
$\displaystyle\leq
C^{\prime}\left(1+2C_{a}\int_{0}^{\infty}h(v)^{2}dv\int_{0}^{T}\mathbb{E}\left[\langle
1,\xi_{s}^{N}(dx,du)\big{\rangle}\right]ds\right)$ $\displaystyle\leq
C_{2}e^{C_{1}T},$

for a constant $C_{2}$ which is finite by our assumption that
$\int_{0}^{\infty}h(v)^{2}dv<\infty$.

Therefore,

$\displaystyle\mathbb{E}\left[\sup_{0\leq t\leq T}\langle
h,\xi_{t}^{N}\rangle\right]$ $\displaystyle\leq\mathbb{E}\left[\langle
h,\xi_{0}^{N}\rangle\right]+(C_{2}+C_{0}/C_{1})e^{C_{1}T},$

and so

$\displaystyle\mathbb{P}\left\\{\sup_{0\leq t\leq T}\langle
h,\xi_{t}^{N}\rangle>K\right\\}$
$\displaystyle\leq\frac{\mathbb{E}\left[\langle
h,\xi_{0}^{N}\rangle\right]+(C_{2}+C_{0}/C_{1})e^{C_{1}T}}{K}.$

######  Lemma 8.5.

Let $f$ be a bounded, continuous real-valued function on
$\mathbb{R}^{d}\times[0,\infty)$ with uniformly bounded first and second
derivatives for which there exists a $u_{0}$ such that if $u>u_{0}$ then
$f(x,u)=0$. Then, the sequence of real-valued processes $(\langle
f,\xi_{t}^{N}\rangle)_{t\geq 0}$ for $N\geq 1$ is tight in ${\cal
D}_{[0,\infty)}(\mathbb{R})$.

###### of Lemma 8.5.

Again, we use the Aldous-Rebolledo criterion. Tightness of $\langle
f,\xi_{t}\rangle$ for a fixed $t$ follows from Lemma 8.3, so we need only
prove conditions analogous to (6.9) and (6.10) applied to the martingale
representation of equations (8.2) and (8.3). Rewriting (8.2) with
$c_{\theta}=c_{\theta}(x,\eta_{s})$,

$\displaystyle\langle f,\xi_{t}\rangle$ $\displaystyle=\langle
f,\xi_{0}\rangle+M^{f}_{t}+\int_{0}^{t}\Big{\langle}c_{\theta}\int_{u}^{N}\int(f(y,u_{1})+f(x,u_{1}))q^{m}(x,dy,\eta)du_{1}$
$\displaystyle\qquad+c_{\theta}(N-u)\int_{0}^{t}(f(y,u)-f(x,u))q^{m}(x,dy,\eta)+(c_{\theta}u^{2}-b_{\theta}u)\frac{d}{du}f,\xi_{s}\Big{\rangle}ds.$

The bounds analogous to (6.9) and (6.10) follow as in the proof of Lemma 6.3:
for instance, observe that using that $c_{\theta}\leq C_{a}$ for some $C_{a}$,
the predictable part of this semimartingale decomposition is bounded by

$\displaystyle\Big{\langle}2C_{a}\|f\|_{\infty}u_{f}+(1-u/N)\gamma
B^{\theta}_{f}+(C_{a}u^{2}-b_{\theta}u)\frac{d}{du}f,\xi_{s}\Big{\rangle},$

the last term of which is bounded by

$\displaystyle\langle
C_{a}u_{f}^{2}+\sup_{x}|b_{\theta}(x,\eta_{s})|u_{f}\|\frac{d}{du}f\|_{\infty}\rangle,$

which can be bounded as we did for (6.9).

###  8.2 Motion of ancestral lineages

In this section we prove Theorem 2.23. The argument follows directly from the
discussion in Section 5.3.

###### of Theorem 2.23.

For brevity, in the proof we write $\gamma(x)$ or $\gamma$ for
$\gamma(x,\eta)$.

Here we have taken the high-density, deterministic limit (so,
$\theta,N\to\infty$ and $\theta/N\to 0$). We first proceed informally, as if
the limiting process has a density $\varphi_{t}(x)$ at location $x$ and time
$t$ (which it may not), and follow this with an integration against test
functions to make the argument rigorous. Let $Y$ denote the spatial motion
followed by a single line of descent. Above equation (5.19), we showed that
$Y$ is a diffusion with generator at time $s$

$\mathcal{L}^{Y}_{s}g(x)=\gamma(x,\eta_{s})(\mathcal{B}(r(\cdot,\eta_{s})g(\cdot))(x)-g(x)\mathcal{B}r(x,\eta_{s})).$

The diffusion is time-inhomogeneous if the density is not constant in time.
Let $\varphi_{t}(x)$ be the limiting density, which is a weak solution to
(1.1),
$\partial_{t}\varphi_{t}=r\mathcal{B}^{*}[\varphi_{t}\gamma]+\varphi_{t}F$.
Formally, the intensity of individuals at $y$ at time $t$ that are descended
from individuals that were at $x$ at time $s$ (with $s<t$) is

$\varphi_{s}(x)\mathbb{E}_{s,x}\left[\exp\left(\int_{s}^{t}(F+\gamma\mathcal{B}r)(Y_{u})du\right)\mathbf{1}_{Y_{t}=y}\right]dy,$
(8.6)

where the subscript $s,x$ in the expectation indicates that $Y_{s}=x$. To see
why this should be true, suppose that an ancestor at time $s$ has level $v$.
Conditional on its spatial motion $\\{Y_{u}\\}_{s\leq u\leq t}$, its level at
time $t$ will be $v\exp(-\int_{s}^{t}(F+\gamma\mathcal{B}r)(Y_{u})du)$. This
will be less than a given level $\lambda$ if
$v<\lambda\exp(\int_{s}^{t}(F+\gamma\mathcal{B}r)(Y_{u})du)$. The intensity of
levels at $y$ that are descended from individuals at $x$ can therefore be
obtained as the limit as $\lambda\to\infty$ of $1/\lambda$ times the number of
levels at $x$ at time $s$ with
$u<\lambda\exp(\int_{s}^{t}(F+\gamma\mathcal{B}r)(Y_{u})du)$ and for which the
corresponding individual is at $y$ at time $t$, which is precisely the
quantity in (8.6).

By our construction in Section 5.3, when we integrate (8.6) with respect to
$x$ we recover $\varphi_{t}(y)dy$. Consider an individual sampled at location
$y$ at time $t$, and write $p(t,s,y,x)$ for the probability density that their
ancestor at time $s$ was at $x$. As a consequence of (8.6), still formally,

$p(t,s,y,x)=\frac{\varphi_{s}(x)}{\varphi_{t}(y)}\mathbb{E}_{s,x}\left[\exp\left(\int_{s}^{t}(F+\gamma\mathcal{B}r)(Y_{u})du\right)\mathbf{1}_{Y_{t}=y}\right]\qquad\text{
for }s<t.$ (8.7)

To make (8.7) meaningful, we multiply by suitable test functions $f$ and $g$
and integrate.

$\displaystyle\int\int f(y)\varphi_{t}(y)p(t,s,y,x)g(x)dydx$
$\displaystyle\qquad=\int
g(x)\varphi_{s}(x)\mathbb{E}_{x,s}\left[\exp\left(\int_{s}^{t}(F+\gamma\mathcal{B}r)(Y_{u})du\right)f(Y_{t})\right]dx.$

Writing $\widehat{T}_{t,s}$ for the time-inhomogeneous semigroup corresponding
to the motion of ancestral lineages backwards in time (that is,
$\widehat{T}_{t,s}f(y)=\int p(t,s,x,y)f(x)dy$), we can write this as

$\displaystyle\int f(y)\varphi_{t}(y)\widehat{T}_{t,s}g(y)dy=\int
g(x)\varphi_{s}(x)\mathbb{E}_{s,x}\left[\exp\left(\int_{s}^{t}(F+\gamma\mathcal{B}r)(Y_{u})du\right)f(Y_{t})\right]dx.$
(8.8)

Next, we will differentiate this equation with respect to $t$. There are two
terms in the product on the left-hand side that depend on $t$, so if we use
that $\partial_{t}\varphi_{t}=r\mathcal{B}^{*}[\varphi_{t}\gamma]+\varphi F$
(in a weak sense), and write $\mathcal{L}_{u}$ for the generator of
$\widehat{T}_{t,s}$ at time $t=u$ so that
$\partial_{t}\widehat{T}_{t,s}g(y)\Big{|}_{t=s}=\mathcal{L}_{s}g(y)$, then

$\qquad\qquad\frac{d}{dt}\int
f(y)\varphi_{t}(y)\widehat{T}_{t,s}g(y)dy\Big{|}_{t=s}\\\ {}=\int
f(y)\left\\{\varphi_{s}(y)\mathcal{L}_{s}g(y)+\left[r(y)\mathcal{B}^{*}(\gamma\varphi_{s})(y)+\varphi_{s}(y)F(y)\right]g(y)\right\\}dy.$

As for the right-hand side, since $Y_{s}=x$ under $\mathbb{E}_{x,s}$,

$\frac{d}{dt}\mathbb{E}_{x,s}\left[\exp\left(\int_{s}^{t}(F+\gamma\mathcal{B}r)(Y_{u})du\right)f(Y_{t})\right]\Bigg{|}_{t=s}=\left[F(x)+\gamma(x)\mathcal{B}r(x)\right]f(x)+\mathcal{L}^{Y}_{s}f(x).$

Therefore, the derivative of (8.8) (with respect to $t$, evaluated at $t=s$)
is

$\displaystyle\int
f(y)\left\\{\varphi_{s}(y)\mathcal{L}_{s}g(y)+\left(r(y)\mathcal{B}^{*}(\gamma\varphi_{s})(y)+\varphi_{s}(y)F(y)\right)g(y)\right\\}dy$
$\displaystyle\qquad=\int
g(x)\varphi_{s}(x)\left(\mathcal{L}^{Y}_{s}f(x)+\left[F(x)+\gamma(x)\mathcal{B}r(x)\right]f(x)\right)dx$
$\displaystyle\qquad=\int
f(x)\left((\mathcal{L}^{Y}_{s})^{*}(\varphi_{s}g)(x)+\left[F(x)+\gamma(x)\mathcal{B}r(x)\right]\varphi_{s}(x)g(x)\right)dx,$

where $(\mathcal{L}^{Y}_{s})^{*}$ is the adjoint of $\mathcal{L}^{Y}_{s}$ .
Since $f$ was arbitrary,

$\displaystyle\mathcal{L}_{s}g$ $\displaystyle=$
$\displaystyle\frac{1}{\varphi_{s}}\left[(\mathcal{L}^{Y}_{s})^{*}(\varphi_{s}g)+\gamma\varphi_{s}g\mathcal{B}(r)-rg\mathcal{B}^{*}(\gamma\varphi_{s})\right].$

(Note that the $\varphi_{s}Fg$ terms have cancelled.) Since the adjoint of
${\mathcal{L}}^{Y}_{s}$ is

$\displaystyle({\mathcal{L}}^{Y}_{s})^{*}f$
$\displaystyle=r\mathcal{B}^{*}(\gamma f)-\gamma f\mathcal{B}r,$

we can rewrite the generator of a lineage as

$\displaystyle\mathcal{L}_{s}g$ $\displaystyle=$
$\displaystyle\frac{r}{\varphi_{s}}\left[\mathcal{B}^{*}(\gamma\varphi_{s}g)-g\mathcal{B}^{*}(\gamma\varphi_{s})\right].$

This is equation (2.17).

To simplify to equation (2.18), first define
$\mathcal{D}f(x)=\sum_{ij}\mathbf{C}_{ij}\partial_{ij}f(x),$ and so the
adjoint of $\mathcal{D}$ is

$\mathcal{D}^{*}f(x)=\sum_{ij}\partial_{ij}(\mathbf{C}_{ij}f(x)).$

Note that $\mathcal{D}^{*}$ satisfies the following identity:

$\displaystyle\mathcal{D}^{*}(fg)$
$\displaystyle=\sum_{ij}\left\\{g\partial_{ij}(\mathbf{C}_{ij}f)+2f\partial_{i}(\mathbf{C}_{ij})\partial_{j}(g)+2\mathbf{C}_{ij}\partial_{i}(f)\partial_{j}(g)+\mathbf{C}_{ij}f\partial_{ij}g\right\\}$
$\displaystyle=g\mathcal{D}^{*}f+2f\vec{c}\cdot\nabla g+2(\mathbf{C}\nabla
f)\cdot\nabla g+f\mathcal{D}g,$

where $\vec{c}_{j}=\sum_{i}\partial_{j}\mathbf{C}_{ij}$. So, with
$f=\gamma\varphi_{s}$,

$\displaystyle\mathcal{L}_{s}g$ $\displaystyle=$
$\displaystyle\frac{r}{\varphi_{s}}\left[\mathcal{D}^{*}(\gamma\varphi_{s}g)-\nabla\cdot(\gamma\varphi_{s}g\vec{b})-g\mathcal{D}^{*}(\gamma\varphi_{s})+g\nabla\cdot(\gamma\varphi_{s}\vec{b})\right]$
$\displaystyle=$
$\displaystyle\frac{r}{\varphi_{s}}\left[\gamma\varphi_{s}\mathcal{D}g+2\gamma\varphi_{s}\vec{c}\cdot\nabla
g+2(\mathbf{C}\nabla(\gamma\varphi_{s}))\cdot\nabla
g-\gamma\varphi_{s}\vec{b}\cdot\nabla g\right]$ $\displaystyle=$
$\displaystyle r\gamma\left[\mathcal{D}g+2\vec{c}\cdot\nabla
g+2(\mathbf{C}\nabla\log(\gamma\varphi_{s}))\cdot\nabla g-\vec{b}\cdot\nabla
g\right],$

which is equation (2.18).

###### of Corollary 2.26.

For the moment, we will write $r(x)$ for $r(x,\eta)$ and $\gamma(x)$ for
$\gamma(x,\eta)$. First note that since in this case the semigroup does not
depend on time, we can write $\mathcal{L}=\mathcal{L}_{s}$, and

$\mathcal{L}f=\sigma^{2}r\gamma\left(\Delta
f+\nabla(2\log(\gamma\varphi)-h/\sigma^{2})\cdot\nabla f\right).$

Now, observe that

$\int_{\mathbb{R}^{d}}e^{H(x)}f(x)(\Delta+\nabla
H(x)\cdot\nabla)g(x)dx=-\int_{\mathbb{R}^{d}}e^{H(x)}\left\\{\nabla
f(x)\cdot\nabla g(x)\right\\}dx,$

so that by choosing $H(x)=2\log(\gamma(x)\varphi(x))-h(x)/\sigma^{2}$ and

$\pi(x)=e^{H(x)}/(\sigma^{2}r(x)\gamma(x))=\frac{\gamma(x)\varphi(x)^{2}e^{-h(x)/\sigma^{2}}}{\sigma^{2}r(x)},$

we have that

$\int_{\mathbb{R}^{d}}\pi(x)f(x)\mathcal{L}g(x)dx=-\int_{\mathbb{R}^{d}}e^{H(x)}\nabla
f(x)\cdot\nabla g(x)dx.$

Since this is symmetric in $f$ and $g$, the process $Y$ is reversible with
respect to $\pi$; the constant factor of $\sigma^{2}$ does not affect the
result.

##  Appendix A Markov Mapping Theorem

The following appears as Theorem A.2 in Etheridge and Kurtz (2019),