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\subsection{Full proofs for Section~\ref{sec:stage1} and Appendix~\ref{app:F}}\label{S:E2}

We use the conventions of Appendix~\ref{app:F}.

\paragraph{Proof of Proposition~\ref{prop:6.11}.}\label{S:E2.611}
In outline, diffusive displacement is needed only up to the time $T_{1}=\min(M^{-2},\eps_{1}^{\mathrm{HK}}\RHK^{2})$. Beyond this time the walk is controlled by blocks of length $T_{1}$, each of which moves it ballistically at speed of order $T_{1}^{-1/2}$, and the mass term outweighs the resulting growth.

We write $\sum_{y}v^{(M)}_{\Gp}(x,y)e^{\theta d(x,y)}=\int_{0}^{\infty}e^{-M^{2}t}\,\mathbb{E}_{x}e^{\theta d(x,X_{t})}dt$. If $T_{1}=M^{-2}$ then $\theta_{M}=\kappa_{W}M=\kappa_{W}T_{1}^{-1/2}$. If $T_{1}=\eps_{1}^{\mathrm{HK}}\RHK^{2}<M^{-2}$ then $\theta_{M}=\kappa_{W}M^{2}T_{1}^{1/2}$. In both cases $\theta_{M}\le\min(T_{1}^{-1/2},c_{\mathrm{ex}}/2)$, as $M\le m_{W}$, and $C_{M}\theta_{M}T_{1}^{1/2}\le C_{M}\kappa_{W}M^{2}T_{1}\le\frac12M^{2}T_{1}\le\frac12$. Hence (HK4) gives $\sup_{w}\mathbb{E}_{w}e^{\theta_{M}d(w,X_{s})}\le q:=1+C_{M}\theta_{M}T_{1}^{1/2}\le\frac32$ for $s\le T_{1}$. For $t=kT_{1}+s$ with $0\le s<T_{1}$, $d(x,X_{t})\le\sum_{i\le k}d(X_{(i-1)T_{1}},X_{iT_{1}})+d(X_{kT_{1}},X_{t})$, and the Markov property gives
\[
\mathbb{E}_{x}e^{\theta_{M}d(x,X_{t})}\le q^{k+1}\le q\exp\bigl(C_{M}\theta_{M}t/T_{1}^{1/2}\bigr)\le q\,e^{M^{2}t/2}.
\]
Hence the integral is at most $2q/M^{2}\le3M^{-2}$, which is \eqref{eq:6.LW}.

At $M=\MY$, $(\eps_{1}^{\mathrm{HK}})^{1/2}\MY\RHK\ge(\eps_{1}^{\mathrm{HK}})^{1/2}K_{0}/\lPV$ by \eqref{eq:6.scales}, which gives \eqref{eq:6.CY}. The bound for $v^{2}_{\Gp}$ follows from the inequalities $0\le v^{2}_{\Gp}\le v_{\Gp}$ of Lemma~\ref{lem:6.2}(iii), and that for $v^{1}_{\Gp}$ follows from \eqref{eq:6.LW} at $M=M_{1}$, since $M\mapsto\theta_{M}$ is nondecreasing. Finally, $\kappa_{0}\le\kappa_{W}$ because $(\eps_{1}^{\mathrm{HK}})^{1/2}K_{0}<1$. If $(\eps_{1}^{\mathrm{HK}})^{1/2}K_{0}\ge\lPV$, then $\kappa_{Y}\MY=\kappa_{W}\mt/\lPV\ge\kappa_{0}\mt/\lPV$. Otherwise $\kappa_{Y}\MY=\kappa_{0}\mt/\lPV^{2}$. In both cases $\kappa_{Y}\MY\ge4\kappa_{0}\mt$, because $\lPV\le\frac14$.
\qed

\paragraph{Proof of Theorem~\ref{thm:6.12}.}\label{S:E2.612}
Since $C_{0}$ is a difference of two resolvents, and hence a positive mixture of heat kernels, every bound on it reduces to the heat kernel. With $r_{\pm}:=\frac12\MY^{2}(1\pm(1-4\lPV^{2})^{1/2})$, the roots of $s^{2}+\MY^{2}s+\MY^{2}\mt^{2}$ are $-r_{\pm}$, and partial fractions give
\begin{gather*}
C_{0}=F(-\Delta_{\Gp}),\qquad F(s)=\frac{\MY^{2}}{(s+r_{-})(s+r_{+})}=\varpi\Bigl(\frac1{s+r_{-}}-\frac1{s+r_{+}}\Bigr),\\
C_{0}=\varpi\bigl(v^{(r_{-}^{1/2})}_{\Gp}-v^{(r_{+}^{1/2})}_{\Gp}\bigr)=\int_{0}^{\infty}k(t)P_{t}\,dt,
\end{gather*}
with $k(t)=\varpi(e^{-r_{-}t}-e^{-r_{+}t})$ and $0\le k(t)\le\min(\MY^{2}t,\varpi)e^{-r_{-}t}$. Since $F(0)=\mt^{-2}$, $C_{0}1_{\KK}=\mt^{-2}1_{\KK}$, that is, the zero modes are part of $C_{0}$. We put $R:=\RHK$, so that $R^{-3}r_{-}^{-1}\le\mt^{-2}R^{-3}\le\mt K_{0}^{-3}=\lPV K_{0}^{-3}\MY$ by \eqref{eq:6.scales}.

(H1). Since $p_{t}\le1$ on $[0,1]$ and $p_{t}\le C_{H}t^{-3/2}+C_{H}R^{-3}$ on $[1,\infty)$ by (HK1), we have
\begin{align*}
C_{0}(x,x)&\le\int_{0}^{1}\MY^{2}t\,dt+C_{H}\int_{1}^{\infty}\min(\MY^{2}t,\varpi)t^{-3/2}dt+C_{H}R^{-3}\int_{0}^{\infty}\varpi e^{-r_{-}t}dt\\
&\le\tfrac12\MY+2C_{H}(1+\varpi)\MY+\varpi C_{H}\lPV K_{0}^{-3}\MY,
\end{align*}
where the middle integral is split at $t=\MY^{-2}$, whatever the order of $\MY^{-2}$ and $R^{2}$.

(H2). Entrywise, $C_{0}\le\varpi v^{(r_{-}^{1/2})}_{\Gp}$, because $v^{(r_{+}^{1/2})}_{\Gp}\ge0$ by Lemma~\ref{lem:6.2}. Since $\mt\le m_{W}/2$, the mass $r_{-}^{1/2}\le1.04\,\mt$ is at most $m_{W}$, and \eqref{eq:6.LW} at this mass has $\theta\ge\kappa_{W}\mt\min(1,(\eps_{1}^{\mathrm{HK}})^{1/2}\mt R)\ge\kappa_{0}\mt$, since $\mt R\ge K_{0}$ and $(\eps_{1}^{\mathrm{HK}})^{1/2}K_{0}<1$. Hence the row sums of $C_{0}e^{\kappa_{0}\mt d}$ are at most $3\varpi/r_{-}=K_{3}\mt^{-2}$, and $C_{0}$ is symmetric.

(H1$'$). First, $C_{0}(x,y)\ge0$ because $k\ge0$ and $p_{t}\ge0$. If $d:=d(x,y)<\MY^{-1}$, then $C_{0}(x,y)\le(C_{0}(x,x)C_{0}(y,y))^{1/2}\le K_{1}\MY\le eK_{1}\MY e^{-\kappa_{0}'\mt d}$, since $\kappa_{0}'\mt d\le\lPV\le1$. If $d\ge\MY^{-1}$, we use $C_{0}(x,y)\le\varpi\int_{0}^{\infty}e^{-r_{-}t}p_{t}(x,y)\,dt$ and split the time axis. (i) For $t\le d$, by (HK0) the contribution is at most $\varpi de^{-d/17}\le313\varpi\MY e^{-d/68}\le313\varpi\MY e^{-\kappa_{0}'\mt d}$, because $de^{-3d/68}\le\MY d^{2}e^{-3d/68}<279\,\MY$ for $d\ge\MY^{-1}$, and $\kappa_{0}'\mt\le1/68$. (ii) For $d\le t\le T_{2}:=2\eps_{1}^{\mathrm{HK}}R^{2}$ we have $t\ge d\ge1$, and (HK3) together with $r_{-}t+c_{\mathrm{gs}}d^{2}/(2t)\ge(2c_{\mathrm{gs}}r_{-})^{1/2}d$ bounds the contribution by
\begin{align*}
\varpi C_{\mathrm{gs}}e^{-(2c_{\mathrm{gs}}r_{-})^{1/2}d}\int_{0}^{\infty}t^{-3/2}e^{-c_{\mathrm{gs}}d^{2}/(2t)}dt&=\varpi C_{\mathrm{gs}}(2\pi/c_{\mathrm{gs}})^{1/2}d^{-1}e^{-(2c_{\mathrm{gs}}r_{-})^{1/2}d}\\
&\le\varpi C_{\mathrm{gs}}(2\pi/c_{\mathrm{gs}})^{1/2}\MY e^{-\kappa_{0}'\mt d}.
\end{align*}
(iii) For $t\ge T_{2}$, as in the proof of (HK3), $p_{t}(x,y)\le2\sup p_{t/2}\cdot\sup_{w}\mathbb{P}_{w}(d(w,X_{t/2})\ge d/2)$. Here $\sup p_{t/2}\le C_{H}(\eps_{1}^{\mathrm{HK}})^{-3/2}R^{-3}$ by (HK1), and Markov's inequality, combined with the blocking bound from the proof of Proposition~\ref{prop:6.11} in \S\ref{S:E2.611} at the mass $r_{-}^{1/2}$ and with its rate $\theta\ge\kappa_{0}\mt$, gives $\mathbb{P}_{w}(d(w,X_{t/2})\ge d/2)\le q\,e^{-\theta d/2}e^{r_{-}t/4}$. Thus the contribution is at most $4\varpi C_{H}(\eps_{1}^{\mathrm{HK}})^{-3/2}R^{-3}r_{-}^{-1}e^{-\kappa_{0}\mt d/2}\le4\varpi C_{H}(\eps_{1}^{\mathrm{HK}})^{-3/2}\lPV K_{0}^{-3}\MY e^{-\kappa_{0}'\mt d}$. The three parts give $K_{1}'$.

(H$_{C}$). By the Schur test, $\|\nu\|_{2\to2}\le\|\nu\|_{0}\le\eps_{\nu}\mt^{2}$, and $C_{0}^{-1}\ge\mt^{2}$. Hence $C_{0}^{-1}+\nu\ge(1-\eps_{\nu})C_{0}^{-1}>0$, $C$ exists, is positive definite, and $C\le(1-\eps_{\nu})^{-1}C_{0}$, which gives $C(x,x)\le2K_{1}\MY$ because $\eps_{\nu}\le1/(2K_{3})<\frac12$. Since $\rcov\le\kappa_{0}\mt$, $\|C_{0}\|_{\rcov}\le K_{3}\mt^{-2}$, and the Neumann series $C=\sum_{n\ge0}(-1)^{n}C_{0}(\nu C_{0})^{n}$ converges in $\|\cdot\|_{\rcov}$ with $\|C\|_{\rcov}\le K_{3}\mt^{-2}\sum_{n}(K_{3}\eps_{\nu})^{n}\le2K_{3}\mt^{-2}$. For the pointwise bound we write $C=C_{0}-C_{0}\nu C$ and use $e^{\rcov d(x,y)}\le e^{\rcov d(x,z)}e^{\rcov d(z,w)}e^{\rcov d(w,y)}$ and the bound $\sup_{z}C_{0}(x,z)e^{\rcov d(x,z)}\le K_{1}''\MY$, which follows from (H1), (H1$'$) and $\rcov\le\kappa_{0}'\mt$. This gives $|C(x,y)|e^{\rcov d(x,y)}\le K_{1}''\MY(1+\|\nu\|_{\rcov}\|C\|_{\rcov})\le K_{1}''\MY(1+2K_{3}\eps_{\nu})\le2K_{1}''\MY$. A matrix $C_{s}$ dominated entrywise by $|C|$, with $C_{s}(x,x)\le C(x,x)$, inherits the three bounds.
\qed

\paragraph{Proof of Lemma~\ref{lem:F.7}.}\label{S:E2.F7}
The heat kernel of $\Zthree$ factorizes into three one-dimensional kernels, and an exponential tilt of each turns the Gaussian bound into a bound with exponential decay. The time integral then produces the factor $(1+|x|)^{-1}$ from times of order $|x|^{2}$ and the decay from the mass.

Let $q_{t}(k):=e^{-2t}I_{k}(2t)$ be the transition probability of the continuous-time walk on $\ZZ$ that jumps to each neighbour at rate $1$, so that the kernel of $e^{t\Delta_{\Zthree}}$ is $p_{t}(x)=\prod_{\mu}q_{t}(x_{\mu})$ and $v^{(M)}(x)=\int_{0}^{\infty}e^{-M^{2}t}p_{t}(x)\,dt$. For $\lambda\ge0$, $q^{\lambda}_{t}(k):=q_{t}(k)e^{\lambda k-2t(\cosh\lambda-1)}$ is the law at time $t$ of $N_{+}-N_{-}$ with independent Poisson variables of means $te^{\lambda}$ and $te^{-\lambda}$. Conditioning on $N_{-}$, we get $\sup_{k}q^{\lambda}_{t}(k)\le\sup_{j}\mathbb{P}(N_{+}=j)\le\sqrt2(1+t)^{-1/2}$, by Stirling's formula. Conditioning on $N_{-}$, $q^{\lambda}_{t}$ is a mixture of translates of the Poisson law $\pi$ of mean $te^{\lambda}$, which is unimodal ($\pi(j)/\pi(j-1)=te^{\lambda}/j$); hence $\sum_{k}|q^{\lambda}_{t}(k+1)-q^{\lambda}_{t}(k)|\le\sum_{k}|\pi(k+1)-\pi(k)|=2\max_{j}\pi(j)\le2\sqrt2(1+t)^{-1/2}$. Writing $q^{\lambda}_{t}=q^{\lambda}_{t/2}*q^{\lambda}_{t/2}$, we obtain $\sup_{k}|q^{\lambda}_{t}(k+1)-q^{\lambda}_{t}(k)|\le8(1+t)^{-1}$. Hence, for $k\ge0$,
\begin{align*}
q_{t}(k)&\le\sqrt2(1+t)^{-1/2}e^{-\lambda k+2t(\cosh\lambda-1)},\\
|q_{t}(k+1)-q_{t}(k)|&\le\bigl(8(1+t)^{-1}+\sqrt2\lambda(1+t)^{-1/2}\bigr)e^{-\lambda k+2t(\cosh\lambda-1)},
\end{align*}
the second because $q_{t}(k+1)-q_{t}(k)=e^{-\lambda k+2t(\cosh\lambda-1)}[e^{-\lambda}q^{\lambda}_{t}(k+1)-q^{\lambda}_{t}(k)]$ and $1-e^{-\lambda}\le\lambda$. For negative $k$ the bounds follow by symmetry, at the cost of a factor $e^{\lambda}$ in the difference bound.

We let $r:=|x|\ge1$, choose $\lambda_{\mu}:=s|x_{\mu}|/r$ with $s\in[0,\frac12]$, and use $\sum_{\mu}2(\cosh\lambda_{\mu}-1)\le\sum_{\mu}\lambda_{\mu}^{2}\cosh\lambda_{\mu}\le s^{2}\cosh s$ and $\sum_{\mu}\lambda_{\mu}|x_{\mu}|=sr$. Then $p_{t}(x)\le2^{3/2}(1+t)^{-3/2}\exp(-sr+ts^{2}\cosh s)$. For $t\le r$ the choice $s=\frac12$ makes the exponent at most $-r/2+r\cosh(\frac12)/4\le-0.218\,r$. For $t\ge r$ the choice $s=r/(2t)\le\frac12$ makes the exponent at most $-r^{2}/(4t)\,(2-\cosh\frac12)\le-0.218\,r^{2}/t$, and
\[
M^{2}t+0.218\,r^{2}/t\ge\bigl(M^{2}t+\tfrac{9}{64}r^{2}/t\bigr)+0.077\,r^{2}/t\ge\tfrac34Mr+0.077\,r^{2}/t.
\]
Hence
\begin{align*}
v^{(M)}(x)&\le2^{3/2}e^{-0.218r}\int_{0}^{r}(1+t)^{-3/2}dt+2^{3/2}e^{-3Mr/4}\int_{0}^{\infty}t^{-3/2}e^{-0.077r^{2}/t}dt\\
&\le2^{5/2}e^{-0.218r}+\frac{2^{3/2}\sqrt\pi}{\sqrt{0.077}\,r}e^{-3Mr/4},
\end{align*}
and $e^{-0.218r}\le e^{-3Mr/4}e^{-0.068r}$ because $\frac34M\le0.15$. This gives the first bound in \eqref{eq:F.L1L2}. For the difference in direction $e=\pm e_{\mu}$ only the factor $q_{t}(x_{\mu})$ changes, and the same choices of $\lambda$ give $|p_{t}(x+e)-p_{t}(x)|\le C(1+t)^{-3/2}\bigl((1+t)^{-1/2}+s\bigr)\exp(-sr+ts^{2}\cosh s)$. For $t\le r$ this is bounded as before. For $t\ge r$ we take $s=r/(2t)$ and use $\int_{0}^{\infty}(t^{-2}+rt^{-5/2})e^{-0.077r^{2}/t}dt\le Cr^{-2}$. For $r<1$ both bounds are trivial.

For the box, the reflections in the six face planes (at half-integer positions) generate a group $\mathcal{G}$ that acts coordinatewise, and $v^{(M)}_{\Lambda'}(y,x)=\sum_{g\in\mathcal{G}}v^{(M)}(y-gx)$. Indeed, the right side is $\mathcal{G}$-invariant, so across each face its two values agree and the Neumann condition holds, and the orbit of $x$ meets $\Lambda'$ only at $x$. Coordinatewise $|y_{\mu}-(gx)_{\mu}|\ge|y_{\mu}-x_{\mu}|$ for $x_{\mu},y_{\mu}\in[1,n_{\mu}]$, so $|y-gx|\ge r_{yx}:=|y-x|$, and $\#\{g:|y-gx|\le\rho\}\le8(\rho M+1)^{3}$ because $n_{\mu}\ge M^{-1}$. For a decreasing $f$, summation by parts gives $\sum_{g}f(|y-gx|)\le8\bigl[(1+Mr_{yx})^{3}f(r_{yx})+\int_{r_{yx}}^{\infty}3M(1+M\rho)^{2}f(\rho)\,d\rho\bigr]$. With $f(\rho)=C_{L}(1+\rho)^{-j}e^{-3M\rho/4}$, $j\in\{1,2\}$, this is at most $C(1+Mr_{yx})^{3}(1+r_{yx})^{-j}e^{-3Mr_{yx}/4}\le C'(1+r_{yx})^{-j}e^{-Mr_{yx}/2}$. Since the differences $u_{x}(y+e)-u_{x}(y)$ are the sums of the differences of the images, (L1) and (L2) follow.
\qed