$\mathbb{P}(U\leqslant\delta\log n,\;T>U,\;S\leqslant
U)\leqslant\mathbb{P}(S\leqslant\delta\log n,\;S\leqslant T)\leqslant
C\delta\log n\cdot n^{-2/3}.$

Putting now all our bounds together, we have proved that the probability in
(68) is smaller than

$n^{-\mathsf{v}\left(1-\frac{1}{2nd}\right)\delta}+n^{-1/2}+2n^{\varepsilon+2\lambda
d\delta-\frac{1}{6}}+C\delta\log n\cdot n^{-2/3}.$

By first choosing $\delta$ small and then choosing $\varepsilon$ much smaller,
this expression is smaller than $n^{-4\varepsilon}$ when $n$ is large enough.
∎

###### Proof of Theorem 1.3.

Let $\varepsilon$ and $\delta$ be as in Lemma 4.3. Define

$\beta:=\frac{4}{|\log\varphi|}+\delta.$

We will prove that

$\mathbb{P}(\xi^{V_{n}}_{\beta\log
n}\neq\varnothing)\xrightarrow{n\to\infty}0.$

By (58), this will follow from proving that

$\lim_{n\to\infty}n\cdot\mathbb{P}(\xi^{\bar{u}}_{\beta\log
n}\neq\varnothing)=0,$

where $\bar{u}$ is a deterministic vertex. In order to prove this, we take the
coupling $(\mathcal{A}_{t},\mathcal{B}_{t})_{t\geqslant 0}$ from Lemma 4.1,
with $\mathcal{A}_{0}=(\xi_{0},\mathscr{E}_{0})=(\\{\bar{u}\\},\varnothing)$
and $\mathcal{B}_{0}=\Psi(\mathcal{A}_{0})$. Recall that
$\sigma:=\inf\\{t:\mathcal{B}_{t}\neq\Psi(\mathcal{A}_{t})\\}$.

Let $\tau$ be the extinction time of $(\xi_{t})$ (which starts from
$\\{\bar{u}\\}$), and also define

$\beta_{0}:=\frac{2}{|\log\varphi|}$

and

$T:=\inf\\{t:\;|\xi_{t}|+|\mathscr{E}_{t}|\geqslant n^{\varepsilon}\\}.$

We bound:

$\displaystyle\mathbb{P}(\tau>\beta\log n)$
$\displaystyle\leqslant\mathbb{P}(\tau>\beta\log n,\;\sigma>\beta_{0}\log n)$
(73) $\displaystyle\quad+\mathbb{P}(\sigma\leqslant\beta_{0}\log
n,\;T\leqslant\sigma)$ (74) $\displaystyle\quad+\mathbb{P}(\tau>\beta\log
n,\;\sigma\leqslant\beta_{0}\log n,\;T>\sigma).$ (75)

To bound (73), we give the same argument as we have used to bound (67); here
it gives:

$\mathbb{P}(\tau>\beta\log n,\;\sigma>\beta_{0}\log n)\leqslant
C\varphi^{\beta_{0}\log n}<n^{-2}.$

To bound (74), we observe again that $|\xi_{t}|+|\mathscr{E}_{t}|$ can only
increase by 2 at any given jump time, and it only increases when there are
contact transmissions generating new births. Moreover, before time $\sigma$,
any time when there is a birth for $(\xi_{t},\mathscr{E}_{t})$, there is also
a particle birth for $\mathscr{B}_{t}$. These considerations allow us to
bound:

$\displaystyle\mathbb{P}(\sigma\leqslant\beta_{0}\log n,\;T\leqslant\sigma)$
$\displaystyle\leqslant\mathbb{P}(\sigma\leqslant\beta_{0}\log
n,\;|\\{t\leqslant\sigma:\;|\xi_{t}|=|\xi_{t-}|+1\\}|\geqslant
n^{\varepsilon}/2)$
$\displaystyle\leqslant\mathbb{P}\left(\sigma\leqslant\beta_{0}\log
n,\;|\\{t<\sigma:\;|\xi_{t}|=|\xi_{t-}|+1\\}|\geqslant\frac{n^{\varepsilon}}{2}-2\right)$
$\displaystyle\leqslant\mathbb{P}\left((\mathscr{B}_{t})\text{ has more than
$\frac{n^{\varepsilon}}{2}-2$ births}\right).$

By Corollary 3.13, this is smaller than

${C}_{p}\left(\frac{n^{\varepsilon}}{2}-2\right)^{-p}$

for any $p\geqslant 1$. Hence, by taking $p>1/(2\varepsilon)$ and $n$ large
enough, it is smaller than $1/n^{2}$.

We now turn to (75). Let us first bound, using Lemma 4.1:

$\mathbb{P}(\sigma\leqslant\beta_{0}\log n,\;T>\sigma)\leqslant
C_{f}\cdot\frac{(n^{\varepsilon})^{2}}{n}\cdot\beta_{0}\log
n=C_{f}\beta_{0}n^{-1+2\varepsilon}\log n.$ (76)

Letting $(\mathcal{F}_{t})_{t\geqslant 0}$ be the natural filtration for
$(\mathcal{A}_{t},\mathcal{B}_{t})_{t\geqslant 0}$, we write

$\displaystyle\mathbb{P}(\tau>\beta\log n,\;\sigma\leqslant\beta_{0}\log
n,\;T>\sigma)$
$\displaystyle\leqslant\mathbb{E}\left[\mathds{1}\\{\sigma\leqslant\beta_{0}\log
n,\;T>\sigma\\}\cdot\mathbb{P}(\tau>\beta\log
n\mid\mathcal{F}_{\sigma})\right]$
$\displaystyle=\mathbb{E}\left[\mathds{1}\\{\sigma\leqslant\beta_{0}\log
n,\;T>\sigma,\;\mathcal{A}_{\sigma}\notin\mathcal{X}_{1}^{\prime}\\}\cdot\mathbb{P}(\tau>\beta\log
n\mid\mathcal{F}_{\sigma})\right]$ (77)
$\displaystyle\quad+\mathbb{E}\left[\mathds{1}\\{\sigma\leqslant\beta_{0}\log
n,\;T>\sigma,\;\mathcal{A}_{\sigma}\in\mathcal{X}_{1}^{\prime}\\}\cdot\mathbb{P}(\tau>\beta\log
n\mid\mathcal{F}_{\sigma})\right].$ (78)

To bound (77), we note that on the event $\\{\sigma\leqslant\beta_{0}\log
n,\;T>\sigma,\;\mathcal{A}_{\sigma}\notin\mathcal{X}_{1}^{\prime}\\}$, we have
that $\mathcal{A}_{\sigma}=(\xi_{\sigma},\mathscr{E}_{\sigma})$ satisfies the
assumptions of Lemma 4.3, and then that lemma implies that, on this event,
$\mathbb{P}(\tau>\beta\log n\mid\mathcal{F}_{\sigma})\leqslant
n^{-4\varepsilon}$. Together with (76), this implies that (77) is smaller than

$C_{f}\beta_{0}n^{-1+2\varepsilon}\log n\cdot
n^{-4\varepsilon}<n^{-1-\varepsilon}$

if $n$ is large enough.

Next, Lemma 4.2 implies that on $\\{\sigma\leqslant\beta_{0}\log
n,\;T>\sigma,\;\mathcal{A}_{\sigma}\in\mathcal{X}_{1}^{\prime}\\}$, we have
$\mathbb{P}(\tau>\beta\log n\mid\mathcal{F}_{\sigma})<n^{-1/2}$; combining
this with (76) shows that (78) is smaller than

$C_{f}\beta_{0}n^{-1+2\varepsilon}\log n\cdot n^{-1/2}<n^{-5/4}$

for $n$ large. This completes the proof. ∎

##  5 Appendix

###  5.1 Proofs of Lemma 3.7 and Lemma 3.9

###### Proof of Lemma 3.7.

The proof is the same for the two functions, so we only treat the first. We
use the simple bounds, that come from comparison with a pure birth process,

$\displaystyle\mathbb{E}[X_{t}\mid\Xi_{0}=\delta_{A}]\leqslant|A|\cdot
e^{d\lambda
t},\quad\mathbb{E}[X_{t}\mid\Xi_{0}=\delta_{A_{e,1}}+\delta_{A_{e,2}}]\leqslant|A|\cdot
e^{d\lambda t},$

together with the expression (18), to obtain

$\displaystyle g_{\mathsf{v}}(\xi,t)$ $\displaystyle\leqslant e^{d\lambda
t}\cdot\sum_{A}\xi(A)\cdot|A|\cdot|\\{\text{active edges of $A$}\\}|$
$\displaystyle\leqslant e^{d\lambda
t}\cdot\left(\sum_{A}\xi(A)\cdot|A|\right)\left(\sum_{A}\xi(A)\cdot|\\{\text{active
edges of $A$}\\}|\right)$ $\displaystyle\leqslant e^{d\lambda
t}\cdot(X(\xi)+\mathscr{E}(\xi))^{2}.$

The statement now readily follows from Corollary 2.3. ∎

The following preliminary result will allow us to perform the exchange of
limit and expectation in (32).

######  Lemma 5.1.

There exist $c_{1},\;c_{2}>0$ (depending on $\lambda,\mathsf{v},\varepsilon)$
such that for any $0\leqslant t<t+s<T$, on $\\{\tau_{\mathrm{sep}}>t\\}$ we
have

$|\widehat{\mathbb{E}}[\mathcal{A}_{t+s}\mid\mathcal{F}_{t}]|\leqslant
c_{1}(X(\mathcal{V}_{t})+\mathscr{E}(\mathcal{V}_{t}))^{c_{2}}\cdot s.$

###### Proof.

We fix $s,t$ as in the statement, and let $N$ denote the number of jumps of
the process $(\mathcal{V},\mathcal{W})$ in $[t,t+s]$. We have

$\displaystyle\mathds{1}_{\\{\tau_{\mathrm{sep}}>t\\}}\cdot|\widehat{\mathbb{E}}\left[\mathcal{A}_{t+s}\mid\mathcal{F}_{t}\right]|$
$\displaystyle\leqslant\mathds{1}_{\\{\tau_{\mathrm{sep}}>t\\}}\cdot\left|\widehat{\mathbb{E}}\left[\left.\mathds{1}_{\\{\tau_{\mathrm{sep}}\leqslant
t+s,\;N=1\\}}\cdot\widehat{\mathbb{E}}[\mathcal{Y}_{T}-\mathcal{X}_{T}\mid\mathcal{F}_{\tau_{\mathrm{sep}}}]\;\right|\;\mathcal{F}_{t}\right]\right|$
(79)
$\displaystyle\quad+\mathds{1}_{\\{\tau_{\mathrm{sep}}>t\\}}\cdot\left|\widehat{\mathbb{E}}\left[\left.\mathds{1}_{\\{\tau_{\mathrm{sep}}\leqslant
t+s,\;N\geqslant
2\\}}\cdot\widehat{\mathbb{E}}[\mathcal{Y}_{T}-\mathcal{X}_{T}\mid\mathcal{F}_{\tau_{\mathrm{sep}}}]\;\right|\;\mathcal{F}_{t}\right]\right|.$
(80)

We will treat the terms (79) and (80) separately. For both, it will be useful
to bound, using domination by a pure birth process,

$\widehat{\mathbb{E}}[\mathcal{X}_{T}\mid\mathcal{F}_{\tau_{\mathrm{sep}}}]\leqslant
e^{d\lambda
T}X(\mathcal{V}_{\tau_{\mathrm{sep}}}),\quad\widehat{\mathbb{E}}[\mathcal{Y}_{T}\mid\mathcal{F}_{\tau_{\mathrm{sep}}}]\leqslant
e^{d\lambda T}X(\mathcal{W}_{\tau_{\mathrm{sep}}})=e^{d\lambda
T}X(\mathcal{V}_{\tau_{\mathrm{sep}}}),$

which gives

$|\widehat{\mathbb{E}}[\mathcal{Y}_{T}-\mathcal{X}_{T}\mid\mathcal{F}_{\tau_{\mathrm{sep}}}]|\leqslant
e^{d\lambda T}X(\mathcal{V}_{\tau_{\mathrm{sep}}}).$ (81)

We start bounding (79). On the event $\\{t<\tau_{\mathrm{sep}}\leqslant
t+s,\;N=1\\}$, we have
$X(\mathcal{V}_{t})=X(\mathcal{W}_{t})=X(\mathcal{V}_{\tau_{\mathrm{sep}}})=X(\mathcal{W}_{\tau_{\mathrm{sep}}})$
(since in this event the only jump of $(\mathcal{V},\mathcal{W})$ in $[t,t+s]$
is a split which causes the separation of the processes, so there is no change
in the number of particles). Then, using (81), we obtain that (79) is smaller
than

$\displaystyle\mathds{1}_{\\{\tau_{\mathrm{sep}}>t\\}}\cdot e^{d\lambda
T}X(\mathcal{V}_{t})\cdot\widehat{\mathbb{P}}(\tau_{\mathrm{sep}}\leqslant
t+s,\;N\geqslant 1\mid\mathcal{F}_{t}).$

The rate with which the process jumps away from the state
$(\mathcal{V}_{t},\mathcal{W}_{t})$ is at most

$\mu_{1}(\mathcal{V}_{t}):=(d\lambda+1)X(\mathcal{V}_{t})+(\mathsf{v}+\varepsilon)\mathscr{E}(\mathcal{V}_{t}),$

so the probability above is at most

$1-\exp\\{-\mu_{1}(\mathcal{V}_{t})\cdot
s\\}\leqslant\mu_{1}(\mathcal{V}_{t})\cdot s.$

We have thus proved that (79) is bounded by the desired expression.

We now turn to (80). We start using (81) to bound (80) by

$\displaystyle\mathds{1}_{\\{\tau_{\mathrm{sep}}>t\\}}\cdot e^{d\lambda
T}\cdot\widehat{\mathbb{E}}\left[\left.\mathds{1}_{\\{\tau_{\mathrm{sep}}\leqslant
t+s,\;N\geqslant 2\\}}\cdot
X(\mathcal{V}_{\tau_{\mathrm{sep}}})\;\right|\;\mathcal{F}_{t}\right]$
$\displaystyle\leqslant\mathds{1}_{\\{\tau_{\mathrm{sep}}>t\\}}\cdot
e^{d\lambda T}\cdot\widehat{\mathbb{E}}\left[\left.\mathds{1}_{\\{N\geqslant
2\\}}\cdot\max_{u\in[t,T]}X(\mathcal{V}_{u})\;\right|\;\mathcal{F}_{t}\right].$

By Hölder’s inequality, the expectation on the right-hand side is smaller than

$\displaystyle\mathds{1}_{\\{\tau_{\mathrm{sep}}>t\\}}\cdot e^{d\lambda
T}\cdot\widehat{\mathbb{P}}\left(N\geqslant
2\mid\mathcal{F}_{t}\right)^{2/3}\cdot\widehat{\mathbb{E}}\left[\left.\max_{u\in[t,T]}X(\mathcal{V}_{u})^{3}\;\right|\;\mathcal{F}_{t}\right]^{1/3}.$
(82)

Corollary 2.3 and the Markov property imply that

$\widehat{\mathbb{E}}\left[\left.\max_{u\in[t,T]}X(\mathcal{V}_{u})^{3}\;\right|\;\mathcal{F}_{t}\right]^{1/3}\leqslant
cX(\mathcal{V}_{t}).$ (83)

Let

$\mu_{2}(\mathcal{V}_{t}):=2(d\lambda+1)(X(\mathcal{V}_{t})+1)+2(\mathsf{v}+\varepsilon)(\mathscr{E}(\mathcal{V}_{t})+1).$

In the event $\\{\tau_{\mathrm{sep}}>t\\}$, the process
$(\mathcal{V},\mathcal{W})$ jumps away from
$(\mathcal{V}_{t},\mathcal{W}_{t})$ with rate smaller than
$\mu_{1}(\mathcal{V}_{t})$ (as previously observed), and after performing a
first jump, it performs a second jump with a rate that is smaller than
$\mu_{2}(\mathcal{V}_{t})$. Indeed, after the first jump the number of
particles or active edges of $\mathcal{V}$ and $\mathcal{W}$ increase by at
most 1, and the factors $2$ in the definition of $\mu_{2}(\mathcal{V}_{t})$
account for the possibility that the first jump is the separation of the two
processes. Letting $({Z}_{t})_{t\geqslant 0}$ be a Poisson process with
constant rate $\mu_{2}(\mathcal{V}_{t})$, we bound (on the event
$\\{\tau_{\mathrm{sep}}>t\\}$):

$\displaystyle\widehat{\mathbb{P}}(N\geqslant
2\mid\mathcal{F}_{t})\leqslant\mathbb{P}({Z}_{s}\geqslant 2)$
$\displaystyle=1-\mathrm{e}^{-\mu_{2}(\mathcal{V}_{t})s}-\mu_{2}(\mathcal{V}_{t})s\cdot\mathrm{e}^{-\mu_{2}(\mathcal{V}_{t})s}$
$\displaystyle\leqslant\mu_{2}(\mathcal{V}_{t})s-\mu_{2}(\mathcal{V}_{t})s\cdot\mathrm{e}^{-\mu_{2}(\mathcal{V}_{t})s}$
$\displaystyle\leqslant(\mu_{2}(\mathcal{V}_{t})s)^{2}.$

Plugging this bound and (83) back in (82) gives the desired bound. ∎

###### Proof of Lemma 3.9.

Fix $t\in[0,T)$. For any $s\in(0,T-t]$, we have

$\mathds{1}\\{\tau_{\mathrm{sep}}>t\\}\cdot\frac{\widehat{\mathbb{E}}[\mathcal{A}_{t+s}\mid\mathcal{F}_{t}]}{s}\leqslant\mathds{1}\\{\tau_{\mathrm{sep}}>t\\}\cdot
c_{1}(X(\mathcal{V}_{t})+\mathscr{E}(\mathcal{V}_{t}))^{c_{2}}.$

The random variable on the right-hand side is integrable, by Corollary 2.3.
This and the Dominated Convergence Theorem justify the exchange of limit in
(32). As explained before (32), this implies that

$\displaystyle\lim_{s\to
0+}\frac{\widehat{\mathbb{E}}[\mathcal{A}_{t+s}]-\widehat{\mathbb{E}}[\mathcal{A}_{t}]}{s}$
$\displaystyle=\widehat{\mathbb{E}}\left[\mathds{1}\\{\tau_{\mathrm{sep}}>t\\}\cdot\lim_{s\to
0+}\frac{\widehat{\mathbb{E}}[\mathcal{A}_{t+s}\mid\mathcal{F}_{t}]}{s}\right]$
$\displaystyle=\varepsilon\cdot\widehat{\mathbb{E}}[\mathds{1}\\{\tau_{\mathrm{sep}}>t\\}\cdot
g_{\mathsf{v},\varepsilon}(\mathcal{V}_{t},T-t)].$

Finally, any function $g:[0,\infty)\to\mathbb{R}$ which is continuous and has
a continuous derivative from the right is necessarily differentiable (with
derivative equal to the derivative from the right), so the proof is complete.
∎

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