2.1. Some notation

We first introduce some notation used in this paper. In Section 1 we defined the discrete torus $\mathbb{T}_N = [{-}N/2, N/2)^d \cap \mathbb{Z}^d$ , $d\geq 3$ , and the canonical projection map $\varphi_N: \mathbb{Z}^d \to \mathbb{T}_N$ . From here on, we will omit the N-dependence and write $\varphi$ and $\tau_{\textbf{x}}$ instead.

We write vertices and subsets of the torus in bold, i.e. $\textbf{x} \in \mathbb{T}_N$ and $\textbf{K} \subset \mathbb{T}_N$ . In order to simplify notation, in the rest of the paper we abbreviate $\textbf{K} = \tau_{\textbf{x}} \varphi(K)$ .

Let $(Y_t)_{t \ge 0}$ be a discrete-time lazy simple random walk on $\mathbb{Z}^d$ ; that is,

\begin{align*} \textbf{P} \big[ Y_{t+1} = y^{\prime} \,|\, Y_t = x^{\prime} \big] = \begin{cases} \frac{1}{2} & \text{when $y^{\prime} = x^{\prime}$;} \\[3pt] \frac{1}{4d} & \text{when $|y^{\prime} - x^{\prime}| = 1$.} \end{cases} \end{align*}

We denote the corresponding lazy random walk on $\mathbb{T}_N$ by $(\textbf{Y}_t)_{t\geq 0} = (\varphi(Y_t))_{t\geq 0}$ . Let $\textbf{P}_{x^{\prime}}$ denote the distribution of the lazy random walk on $\mathbb{Z}^d$ started from $x^{\prime} \in \mathbb{Z}^d$ , and write $\textbf{P}_{\textbf{x}}$ for the distribution of the lazy random walk on $\mathbb{T}_N$ started from $\textbf{x} = \varphi(x^{\prime}) \in \mathbb{T}_N$ . We write $p_t(x^{\prime},y^{\prime}) = \textbf{P}_{x^{\prime}} [ Y_t = y^{\prime} ]$ for the t-step transition probability. Further notation we will use includes the following:

• $L = m N$ , where $L^2 \sim A N^d$ as $N \to \infty$ for some constant $A \in (0,\infty)$ ;

• $D = ({-}m,m)^d$ , the rescaled box, indicates which copy of the torus the walk is in;

• $n = \lfloor N^\delta \rfloor$ for some $2 \lt \delta \lt d$ , long enough for the mixing property on the torus, but short compared to $L^2$ ;

• $\textbf{x}_\textbf{0} \in \textbf{K}$ is a fixed point of $\textbf{K}$ ;

• we write points in the original lattice $\mathbb{Z}^d$ with a prime, such as $y^{\prime}$ , and decompose a point $y^{\prime}$ as $y N+\textbf{y}$ with y in another lattice isomorphic to $\mathbb{Z}^d$ and $\textbf{y} = \varphi(y^{\prime}) \in \mathbb{T}_N$ ;

• $T = \inf \big\{ t \ge 0\, :\,Y_t \not\in ({-}L,L)^d \big\}$ , the first exit time from $({-}L,L)^d$ ;

• $S = \inf \big\{ \ell \ge 0\, :\,Y_{n \ell} \not\in ({-}L,L)^d \big\}$ , so that the first multiple of n when the rescaled point $Y_{n \ell} / N$ is not in $({-}m,m)^d$ equals $S\cdot n$ .

For simplicity, we omit the dependence on d and N from some of the notation above.

2.2. Some auxiliary results on random walks

In this section, we collect some known results required for the proof of Theorem 1.1. We will rely heavily on the local central limit theorem (LCLT) [Reference Lawler and Limic9, Chapter 2], with error term, and the martingale maximal inequality [Reference Lawler and Limic9, Equation (12.12) of Corollary 12.2.7]. We will also use [Reference Lawler and Limic9, Equation (6.31)], which relates $\textrm{Cap}(K)$ to the probability that a random walk started from the boundary of a large ball with radius n hits the set K before exiting the ball. In estimating some error terms in our arguments, sometimes we will use the Gaussian upper and lower bounds [Reference Hebisch and Saloff-Coste5]. We also need to derive a lemma related to the mixing property on the torus [Reference Levin, Peres and Wilmer10, Theorem 5.6] to show that the starting positions of different stretches are not far from uniform on the torus; see Lemma 2.1.

We recall the LCLT from [Reference Lawler and Limic9, Chapter 2]. The following is a specialization of [Reference Lawler and Limic9, Theorem 2.3.11] to lazy simple random walks. The covariance matrix $\Gamma$ and the square root $J^{*}(x)$ of the associated quadratic form are given by

\begin{align*}\Gamma = (2d)^{-1}I,\quad J^{*}(x) = (2d)^{\frac{1}{2}}|x|,\end{align*}

where I is the $(d \times d)$ -unit matrix.

Let $\bar{p}_t(x^{\prime})$ denote the estimate of $p_t(x^{\prime})$ that one obtains by the LCLT, for a lazy simple random walk. We have

\begin{align*}\bar{p}_t(x^{\prime})&= \frac{1}{(2\pi t)^{d/2}\sqrt{\det\Gamma}} \exp\left({-}\frac{J^{*}(x^{\prime})^2}{2t}\right) \\&= \frac{1}{(2\pi t)^{d/2}(2d)^{-d/2}} \exp\left({-}\frac{2d\,|x^{\prime}|^2}{2t}\right) \\&= \frac{\bar{C}}{t^{d/2}}\exp\left({-}\frac{d\,|x^{\prime}|^2}{t}\right).\end{align*}

The lazy simple random walk $(Y_t)_{t\geq 0}$ in $\mathbb{Z}^d$ is aperiodic and irreducible with mean zero, finite second moment, and finite exponential moments. All joint third moments of the components of $Y_1$ vanish.

Theorem 2.1. ([Reference Lawler and Limic9, Theorem 2.3.11].) For a lazy simple random walk $(Y_t)_{t\geq 0}$ in $\mathbb{Z}^d$ , there exists $\rho\gt 0$ such that for all $t\geq 1$ and all $x^{\prime}\in\mathbb{Z}^d$ with $|x^{\prime}|\lt\rho t$ ,

\begin{align*}p_t(x^{\prime})= \bar{p}_t(x^{\prime})\exp\left\{O\left(\frac{1}{t}+\frac{|x^{\prime}|^4}{t^3}\right)\right\}.\end{align*}

The martingale maximal inequality in [Reference Lawler and Limic9, Equation (12.12) of Corollary 12.2.7] is stated as follows. Let $\Big(Y^{(i)}_t\Big)_{t \ge 0}$ denote the ith coordinate of $(Y_t)_{t \ge 0}$ ( $1 \le i \le d$ ). The standard deviation $\sigma$ of $Y^{(i)}_1$ is given by $\sigma^2 = (2d)^{-1}$ . For all $t \ge 1$ and all $r \gt 0$ we have

(3) \begin{equation}\textbf{P}_o \left[\max_{0\leq j\leq t} Y^{(i)}_j \geq r\sigma\sqrt{t} \right]\le e^{-r^2/2}\exp\left\{O\left(\frac{r^3}{\sqrt{t}}\right)\right\}.\end{equation}

Now we state the result of [Reference Lawler and Limic9, Equation (6.31)]. Recall that $B_r(o)$ is the discrete ball centred at o with radius r. For any subset $B \subset \mathbb{Z}^d$ , let

\begin{align*} \xi_B = \inf \big\{ t \ge 1\, :\,Y_t \not\in B) \big\}. \end{align*}

Let $\partial B_r(o) = \big\{ y^{\prime} \in \mathbb{Z}^d \setminus B_r(o)\, :\,\exists x^{\prime} \in B_r(o)$ such that $|x^{\prime} - y^{\prime}| = 1 \big\}$ . For a given finite set $K \subseteq\mathbb{Z}^d$ , let $H_K$ denote the hitting time

\begin{align*} H_K = \inf \big\{ t \ge 1\, :\,Y_t \in K \big\}. \end{align*}

Then we have

(4) \begin{equation}\frac{1}{2}\textrm{Cap}(K)= \lim_{r\to\infty}\sum_{y^{\prime} \in \partial B_r(o)} \textbf{P}_{y^{\prime}}\big[H_{K} \lt \xi_{B_r(o)}\big].\end{equation}

Here $\textrm{Cap}(K)$ is the capacity of K; see [Reference Lawler and Limic9, Section 6.5], which states the analogous statement for the simple random walk. Since we consider the lazy random walk, this introduces a factor of $1/2$ .

In estimating some error terms in our arguments, sometimes we will use the Gaussian upper and lower bounds [Reference Hebisch and Saloff-Coste5]: there exist constants $C = C(d)$ and $c = c(d)$ such that

(5) \begin{equation}\begin{split} p_t\big(x^{\prime},y^{\prime}\big) &\le \frac{C}{t^{d/2}} \exp \left({-}c \frac{|y^{\prime}-x^{\prime}|^2}{t} \right), \quad \text{for $x^{\prime}, y^{\prime} \in \mathbb{Z}^d$ and $t \ge 1$;} \\ p_t\big(x^{\prime},y^{\prime}\big) &\ge \frac{c}{t^{d/2}} \exp \left({-}C \frac{|y^{\prime}-x^{\prime}|^2}{t} \right), \quad\text{for $|y^{\prime}-x^{\prime}| \le ct$.} \end{split}\end{equation}

Recall that the norm $|\cdot|$ refers to the Euclidean norm.

Regarding mixing times, recall that the lazy simple random walk on the N-torus mixes in time $N^2$ [Reference Levin, Peres and Wilmer10, Theorem 5.6]. With this in mind we derive the following simple lemma.

Recall that $2\lt\delta\lt d$ and $n=\lfloor N^{\delta} \rfloor$ .

Lemma 2.1. There exists $C=C(d)$ such that for any $N \geq 1$ and any $t \ge n$ we have

\begin{equation*}\textbf{P}_{\textbf{o}}[\textbf{Y}_t = \textbf{x}]\le \frac{C}{N^d},\quad \textbf{x} \in \mathbb{T}_N.\end{equation*}

Proof. Using the Gaussian upper bound, the left-hand side can be bounded by

\begin{equation*}\begin{split} \sum_{x \in \mathbb{Z}^d} p_t(o,xN+\textbf{x}) &\le \frac{C}{t^{d/2}} \sum_{x \in \mathbb{Z}^d} \exp \left({-}c \frac{|xN + \textbf{x}|^2}{t} \right) \le \frac{C}{t^{d/2}} \sum_{x \in \mathbb{Z}^d} \exp \left({-}c \frac{|xN|^2}{t} \right) \\ &\le \frac{C}{t^{d/2}} \sum_{k=0}^{\infty} (k+1)^{d-1} \frac{t^{d/2}}{N^{d}}\exp\big({-}ck^2\big) \\ &\le \frac{C}{N^{d}} \sum_{k=0}^{\infty} (k+1)^{d-1} \exp\big({-}ck^2\big) \le \frac{C}{N^d}.\end{split}\end{equation*}

Here we bounded the number of x in $\mathbb{Z}^d$ satisfying $k \sqrt{t}/N \le |x| \lt (k+1) \sqrt{t}/N$ by $C(k+1)^{d-1} t^{d/2}/ N^{d}$ , where $k=0, 1, 2, \dots$ .

2.3. Key propositions

In this section we state some propositions to be used in Section 3 to prove Theorem 1.1. The propositions will be proved in Section 4.

The strategy of the proof is to consider stretches of length n of the walk, and estimate the small probability in each stretch that $\textbf{K}$ is hit by the projection. What makes this strategy work is that we can estimate, conditionally on the starting point and endpoint of a stretch, the probability that $\textbf{K}$ is hit, and this event asymptotically decouples from the number of stretches. The number of stretches will be the random variable S. Since $n S \approx T$ , and T is $\Theta\big(N^d\big)$ in probability, we have that S is $\Theta\big(N^d/n\big)$ . In Lemma 2.2 below we show a somewhat weaker estimate for S (which suffices for our needs).

The main part of the proof will be to show that during a fixed stretch, $\textbf{K}$ is not hit with probability

(6) \begin{equation} 1 - \frac{1}{2} \textrm{Cap}(\textbf{K}) \frac{n}{N^d} (1+o(1)).\end{equation}

Heuristically, conditionally on S this results in the probability

\begin{align*} \left( 1 - \frac{1}{2} \textrm{Cap}(\textbf{K}) \frac{n}{N^d} (1+o(1)) \right)^{S} \approx \exp \left({-}\frac{1}{2} \textrm{Cap}(\textbf{K}) \frac{n}{N^d} S \right), \end{align*}

and we will conclude by showing that $\left(n/N^d\right ) S$ converges in distribution to a constant multiple of the Brownian exit time $\sigma_1$ .

The factor $n/N^d$ in (6) arises as the expected time spent by the projected walk at a fixed point of the torus during a given stretch. The capacity term arises as we pass from expected time to hitting probability.

For the above approach to work, we need a small-probability event on which the number of stretches or endpoints of stretches are not sufficiently well-behaved. First, we will need to restrict to realizations where $\big(\sqrt{\log \log n}\big)^{-1} \left(N^d/n\right) \le S \le \log N \left(N^d/n\right )$ , which occurs with high probability as $N \to \infty$ (see Lemma 2.2 below). Second, suppose that the $\ell$ th stretch starts at the point $y^{\prime}_{\ell-1}$ and ends at the point $y^{\prime}_{\ell}$ ; that is, $y^{\prime}_{\ell-1}$ and $y^{\prime}_{\ell}$ are realizations of $Y_{(\ell-1) n}$ and $Y_{\ell n}$ . In order to have a good estimate of the probability that $\textbf{K}$ is hit during this stretch, we will need to impose a few conditions on $y^{\prime}_{\ell-1}$ and $y^{\prime}_{\ell}$ . One of these is that the displacement $|y^{\prime}_{\ell} - y^{\prime}_{\ell-1}|$ is not too large: we will require that for all stretches, it is at most $f(n) \sqrt{n}$ , for a function to be chosen later that increases to infinity with N. We will be able to choose f(n) of the form $f(n) = C_1 \sqrt{\log N}$ in such a way that this restriction holds for all stretches with high probability. A third condition we need to impose, that will also hold with high probability, is that $y^{\prime}_{\ell-1}$ is at least a certain distance $N^\zeta$ from $\varphi^{-1}(\textbf{K})$ for a parameter $0 \lt \zeta \lt 1$ (this will only be required for $\ell \ge 1$ , and is not needed for the first stretch starting with $y^{\prime}_0 = o$ ). The reason we need this is to be able to appeal to (4) to extract the $\textrm{Cap}(\textbf{K})$ contribution, when we know that $\textbf{K}$ is hit from a long distance (we will take $r = N^\zeta$ in (4)). The larger the value of $\zeta$ , the better error bound we get on the approach to $\textrm{Cap}(\textbf{K})$ . On the other hand, $\zeta$ should not be too close to 1, because we want the separation of $y^{\prime}_{\ell-1}$ from $\textbf{K}$ to occur with high enough probability.

The set $\mathcal{G}_{\zeta,C_1}$ defined below represents realizations of S and the sequence $Y_{n}$ , $Y_{2n}$ , $\dots$ , $Y_{Sn}$ satisfying the above restrictions. Proposition 2.1 below implies that these restrictions hold with high probability. First, we will need $2 \lt \delta \lt d$ to satisfy the inequality

(7) \begin{equation} \delta - \frac{2\delta}{d} \gt d - \delta \qquad \Leftrightarrow \qquad 2\delta \gt \frac{d^2}{d-1}.\end{equation}

This can be satisfied if $d \ge 3$ and $\delta$ is sufficiently close to d, say $\delta = \frac{7}{8} d$ . Since the left-hand side of the left-hand inequality in (7) equals $(\delta/d)(d-2)$ , we can subsequently choose $\zeta$ such that we also have

(8) \begin{equation} 0 \lt \zeta \lt \frac{\delta}{d}, \qquad\qquad \zeta(d - 2) \gt d - \delta.\end{equation}

With the parameter $\zeta$ fixed satisfying the above, we now define

(9) \begin{align}\mathcal{G}_{\zeta,C_1}= \left\{ \begin{array}{l@{\quad}l} & \left(\sqrt{\log \log n}\right)^{-1} \frac{N^d}{n} \le \tau \le \log N \frac{N^d}{n}; \\ & y_\ell \in D, \textbf{y}_\ell \in \mathbb{T}_N \setminus B\left(\textbf{x}_\textbf{0}, N^{\zeta}\right) \text{ for}\ 1 \le \ell \lt \tau;\\[-10pt]\left( \tau, (y_\ell, \textbf{y}_\ell)_{\ell=1}^\tau \right)\, :\, & \\ & y_\tau \in D^c \text{ and}\ \textbf{y}_\tau \in \mathbb{T}_N \setminus B\left(\textbf{x}_\textbf{0}, N^{\zeta}\right); \\ & |y^{\prime}_\ell - y^{\prime}_{\ell-1}| \leq f(n)n^{\frac{1}{2}} \text{ for}\ 1 \le \ell \le \tau \end{array}\right\}\kern-2pt,\end{align}

where $f(n) = C_1 \,\sqrt{\log N}$ and recall that we write $y^{\prime}_\ell = y_\ell N + \textbf{y}_\ell$ and $y^{\prime}_{\ell-1} = y_{\ell-1} N + \textbf{y}_{\ell-1}$ , and we define $y^{\prime}_0 = o$ . The time $\tau$ is corresponding to a particular value of the exit time S, so $y_\ell \in D$ for $1 \le \ell \lt \tau$ and $y_{\tau} \notin D$ . The parameter $C_1$ will be chosen in the course of the proof. See Figure 1 for a visual illustration of the sequence $y^{\prime}_0, y^{\prime}_1, \dots$ in the definition of $\mathcal{G}_{\zeta,C_1}$ .

Figure 1. This figure explains the properties of the set $\mathcal{G}_{\zeta,C_1}$ (not to scale). The shaded regions represent the balls of radius $N^\zeta$ in each copy of the torus. None of the $y^{\prime}_\ell$ , for $\ell \ge 1$ , is in a shaded region.

The next lemma shows that the restriction made on the time-parameter $\tau$ in the definition of $\mathcal{G}_{\zeta,C_1}$ holds with high probability for S.

Lemma 2.2. We have

\begin{equation*}\textbf{P}_{o} \left[ \left\{\frac{S n}{N^d} \lt \left(\sqrt{\log\log n}\right)^{-1} \right\} \cup \left\{\frac{S n}{N^d} \gt \log N \right\}\right] \to 0\end{equation*}

as $N \to \infty$ .

Proof. By the definitions of S and T, we first notice that

\begin{equation*}\begin{split}\textbf{P}_{o}\left[S \lt \left(\sqrt{\log\log n}\right)^{-1}\frac{N^d}{n} \right]&\leq \textbf{P}_{o}\left[ T \lt \left(\sqrt{\log\log n}\right)^{-1} N^d \right] \\&\leq \sum_{1\leq i\leq d} \Bigg(\textbf{P}_{o}\left[\max_{0\leq j \leq \left(\sqrt{\log\log n}\right)^{-1}N^d} Y^{(i)}_{j}\geq L\right]\\ &\qquad\qquad+\textbf{P}_{o}\left[\max_{0\leq j \leq \left(\sqrt{\log\log n}\right)^{-1}N^d} -Y^{(i)}_{j}\geq L\right]\Bigg),\end{split}\end{equation*}

where $Y^{(i)}$ denotes the ith coordinate of the d-dimensional lazy random walk.

We are going to use (3). Setting $t= \left(\sqrt{\log\log n}\right)^{-1} N^d$ and $r\sigma\sqrt{t} = L$ , we can evaluate each term $\Big($ similarly for the event with $-Y^{(i)}_j\Big)$ in the sum

(10) \begin{equation}\begin{split} &\textbf{P}_{o}\left[\max_{0\leq j \leq \left(\sqrt{\log\log n}\right)^{-1}N^d} Y^{(i)}_{j}\geq L\right] \\ &\quad\leq \exp\left\{-\frac{1}{2}\frac{L^2}{\sigma^2\left(\sqrt{\log\log n}\right)^{-1} N^d} + O\left(\frac{L^3}{\sigma^3\left(\sqrt{\log\log n}\right)^{-2}N^{2d}}\right)\right\}\kern-2pt.\end{split}\end{equation}

Recall that $L^2 \sim A N^d$ and $\sigma^2 = 1/2d$ . For the main term in the exponential in (10), we have the upper bound

\begin{equation*}\begin{split}\exp\left( -(1+o(1)) \frac{1}{2}\frac{2d\cdot A N^d}{ \left(\sqrt{\log\log n}\right)^{-1} N^d}\right)&= \exp\left({-}(1+o(1)) Ad\sqrt{\log\log n}\right)\\&\to 0, \quad\text{as $N\to\infty$}.\end{split}\end{equation*}

The big-O term in the exponential in (10) produces an error term because

\begin{equation*}\begin{split} \exp\left\{O\left(\frac{(A N^d)^{3/2}}{\sigma^3\left(\sqrt{\log\log n}\right)^{-2}N^{2d}}\right)\right\} &= \exp \left\{O\left(N^{-d/2}(\log\log n)\right)\right\}\\ &= 1+o(1), \quad\text{as $N\to\infty$.}\end{split}\end{equation*}

Coming to the second event $\left\{\frac{S n}{N^d} \gt \log N \right\}$ , observe that the central limit theorem applied to $\Big(Y_{k n \lfloor N^d/n \rfloor) + t} - Y_{k n \lfloor N^d/n \rfloor}\Big)_{t \ge 0}$ implies that

\begin{align*} \textbf{P}_o \bigg[ S \gt (k+1) \Big\lfloor \frac{N^d}{n} \Big\rfloor \,\Big|\, S \gt k \Big\lfloor \frac{N^d}{n} \Big\rfloor \bigg] \le \max_{z \in ({-}L,L)^d} \big(1 - \textbf{P}_z \big[ Y_{n \lfloor N^d/n \rfloor} \not\in ({-}2L,2L)^d \big]\big) \le 1-c \end{align*}

for some $c \gt 0$ . Hence we have $\textbf{P}_{o}\left[S \gt k\frac{N^d}{n}\right]\leq e^{-c k}$ for all $k \ge 0$ .

Applying this with $k=\log N$ , we obtain

\begin{equation*} \textbf{P}_{o}\left[S \gt \log N\frac{N^d}{n}\right] \leq e^{-c \log N} \to 0\end{equation*}

as required.

The starting point for the proof of Theorem 1.1 is the following proposition, which decomposes the probability we are interested in into terms involving single stretches of duration n.

Proposition 2.1. For a sufficiently large value of $C_1$ , we have that

(11) \begin{equation}\textbf{P}_o \left[ \left(S, (Y_{\ell n}, \varphi(Y_{\ell n})_{\ell=1}^S \right) \not\in \mathcal{G}_{\zeta,C_1} \right] = o(1) \quad \text{as $N \to \infty$}.\end{equation}

Furthermore,

(12) \begin{equation}\begin{split} &\textbf{P}_{o} \left[ Y_t \not\in \varphi^{-1}(\textbf{K}),\, 0 \le t \lt T \right] \\ &\quad = \sum_{( \tau, (y_\ell, \textbf{y}_\ell)_{\ell=1}^\tau ) \in \mathcal{G}_{\zeta,C_1}} \prod_{\ell=1}^\tau \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell,\, Y_t \not\in \varphi^{-1}(\textbf{K})\ \textit{for}\,\, 0 \le t \lt n \right] + o(1),\end{split}\end{equation}

where $o(1) \to 0$ as $N \to \infty$ .

Central to the proof of Theorem 1.1 is the following proposition, which estimates the probability of hitting a copy of $\textbf{K}$ during a ‘good stretch’ where the displacement $|y^{\prime}_\ell - y^{\prime}_{\ell-1}|$ is almost of order $\sqrt{n}$ . This will not hold for all stretches with high probability, but the fraction of stretches for which it fails will vanish asymptotically.

Proposition 2.2. There exists a sufficiently large value of $C_1$ so that the following holds. Let $\big( \tau, (y_\ell, \textbf{y}_\ell)_{\ell=1}^\tau \big) \in \mathcal{G}_{\zeta,C_1}$ . Then for all $2 \le \ell \le \tau$ such that $|y^{\prime}_\ell - y^{\prime}_{\ell-1}| \le 10\sqrt{n}\log\log n$ we have

(13) \begin{equation}\begin{split} &\textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell,\, Y_t \not\in \varphi^{-1}(\textbf{K})\ \textit{for}\,\, 0 \le t \lt n \right] \\ &\qquad = \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell \right] \, \left( 1 - \frac{1}{2} \, \frac{\textrm{Cap} (\textbf{K}) \, n}{N^d} \, (1 + o(1)) \right)\kern-2pt.\end{split}\end{equation}

In addition to the above proposition (that we prove in Section 4.2), we will need a weaker version for the remaining ‘bad stretches’ that have less restriction on the distance $|y^{\prime}_\ell - y^{\prime}_{\ell-1}|$ . This will be needed to estimate error terms arising from the ‘bad stretches’, and it will also be useful in demonstrating some of our proof ideas for other error terms arising later in the paper. It will be proved in Section 4.1.

Proposition 2.3. Let $\big( \tau, (y_\ell, \textbf{y}_\ell)_{\ell=1}^\tau \big) \in \mathcal{G}_{\zeta,C_1}$ . For all $2 \le \ell \le \tau$ we have

(14) \begin{equation} \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell,\, Y_t \not\in \varphi^{-1}(\textbf{K})\ \textit{for all}\ 0 \le t \lt n \right] = \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell \right] \, \left( 1 - O\left(\frac{n}{N^d}\right) \right)\kern-2pt,\end{equation}

and for the first stretch we have

(15) \begin{equation} \textbf{P}_{o} \left[ Y_n = y^{\prime}_1,\, Y_t \not\in \varphi^{-1}(\textbf{K})\ \textit{for all}\ 0 \le t \lt n \right] = \textbf{P}_{o} \left[ Y_n = y^{\prime}_1 \right] \, \left( 1 - o(1) \right)\kern-2pt.\end{equation}

Here the $-O\big(n/N^d\big)$ and $-o(1)$ error terms are negative.

Our final proposition is needed to estimate the number of stretches that are ‘bad’.

Proposition 2.4. We have

(16) \begin{equation}\begin{split} &\textbf{P}_o \left[ \# \left\{ 1 \le \ell \le \frac{N^d}{n} C_1 \log N\, :\, |Y_{n \ell} - Y_{n (\ell-1)}| \gt 10\sqrt{n}\log\log n \right\} \ge \frac{N^d}{n} \frac{1}{\log \log n} \right] \\ &\qquad \to 0, \end{split}\end{equation}

as $N \to \infty$ .

4.1. Proof of Proposition 2.3

In the proof of the proposition we will need the following lemma, which bounds the probability of hitting some copy of $\textbf{K}$ in terms of the Green’s function of the random walk. Recall that the Green’s function is defined by

\begin{align*} G(x^{\prime},y^{\prime}) = \sum_{t\,=\,0}^\infty p_t(x^{\prime},y^{\prime}), \end{align*}

and in all dimensions $d \ge 3$ it satisfies the bound [Reference Lawler and Limic9]

\begin{align*} G(x^{\prime},y^{\prime}) \le \frac{C_G}{|y^{\prime}-x^{\prime}|^{d\,-\,2}} \end{align*}

for a constant $C_G = C_G(d)$ . For Part (ii) of the lemma recall that $\textbf{K} \cap \varphi(B_{g(N)}(o)) = \emptyset$ . We also define $\textrm{diam}(\textbf{K})$ as the maximum Euclidean distance between two points in $\textbf{K}$ .

Proof. (i) We split the sum according to whether $|y^{\prime}-x^{\prime}| \gt N^{1+4\varepsilon}$ or $\le N^{1+4\varepsilon}$ . In the first case we use (5) and write $r = \lfloor |x^{\prime}-y^{\prime}| \rfloor$ to get

\begin{equation*}\begin{split} \sum_{t=0}^{N^{2+6\varepsilon}}\sum_{\substack{x^{\prime} \in \varphi^{-1}(\textbf{K}) \\|x^{\prime}-y^{\prime}|\gt N^{1+4\varepsilon}}} p_t(y^{\prime}, x^{\prime}) &\le \sum_{t=0}^{N^{2+6\varepsilon}}\sum_{r=\lfloor N^{1+4\varepsilon} \rfloor}^{\infty} C\,r^{d-1} \exp\left({-}\frac{c\, r^2}{N^{2+6\varepsilon}}\right)\\ &\le N^{2+6\varepsilon}\sum_{r=\lfloor N^{1+4\varepsilon}\rfloor}^{\infty} C\, r^{d-1} \exp\left({-}\frac{c \, r^2}{N^{2+6\varepsilon}}\right)\\ &\le N^{O(1)}\exp({-}cN^{2\varepsilon{}}).\end{split}\end{equation*}

For the remaining terms, we have the upper bound

\begin{equation*} \sum_{t=0}^{N^{2+6\varepsilon}}\sum_{\substack{x^{\prime} \in \varphi^{-1}(\textbf{K}) \\|x^{\prime}-y^{\prime}|\le N^{1+4\varepsilon}}} p_t(y^{\prime}, x^{\prime}) \le \sum_{\substack{x^{\prime} \in \varphi^{-1}(\textbf{K}) \\|x^{\prime}-y^{\prime}|\leq N^{1+4\varepsilon}}} G(y^{\prime},x^{\prime}).\end{equation*}

Let Q(k N) be the cube with radius kN centred at o. Then $y^{\prime} + ( Q(k N)\backslash Q((k-1)N))$ are disjoint annuli for $k = 1, 2, \dots$ . We decompose the sum over $x^{\prime}$ according to which annulus $x^{\prime}$ falls into. For $k \ge 2$ we have

\begin{align*}\sum_{\substack{x^{\prime}\in\varphi^{-1}(\textbf{K}) \\x^{\prime}-y^{\prime}\in Q(k N)\backslash Q\left((k -1)N\right)}} \frac{C_G}{|y^{\prime}-x^{\prime}|^{d-2}}\leq |\textbf{K}| C k^{d-1} C_G (N k)^{2-d}\leq |\textbf{K}| C k N^{2-d},\end{align*}

where $C_G$ is the Green’s function constant. The contribution from any copy of $\textbf{K}$ in $y^{\prime} + Q(N)$ will be of order $N^{2-d}$ if its distance from $y^{\prime}$ is at least $N/3$ , say. Note that there is at most one copy of $\textbf{K}$ within distance $N/3$ of $y^{\prime}$ , which may have a distance as small as $N^\zeta$ .

We have to sum over the following values of k:

\begin{align*}k=1,\dots, \frac{N^{1+4\varepsilon}}{N}= N^{4\varepsilon}.\end{align*}

Since $x^{\prime}\in\varphi^{-1}(\textbf{K})$ and $y^{\prime}\notin \varphi^{-1}\left(B(\textbf{x}_0, N^{\zeta})\right)$ for $\textbf{x}_0\in \textbf{K}$ , the distance between $x^{\prime}$ and $y^{\prime}$ is at least $N^{\zeta}$ . Therefore, we get the upper bound as follows:

\begin{align*}\sum_{\substack{x^{\prime}\in\varphi^{-1}(\textbf{K})\\|x^{\prime}-y^{\prime}|\leq N^{1+4\varepsilon}}}G(y^{\prime},x^{\prime})&\leq |\textbf{K}|N^{\zeta(2-d)} + \sum_{k=1}^{N^{4 \varepsilon}} |\textbf{K}|CkN^{2-d}\\&\leq |\textbf{K}|N^{\zeta(2-d)} + C |\textbf{K}|N^{2-d}\times N^{8\varepsilon}\leq C|\textbf{K}|N^{\zeta(2-d)}.\end{align*}

Here the last inequality follows from the choice of $\zeta$ , (8), for sufficiently small $\varepsilon \gt 0$ .

1. (ii) The proof is essentially the same, except for the contribution of the ‘nearest’ copy of $\textbf{K}$ , which is now $C |\textbf{K}| g(N)^{2-d}$ .

2. (iii) The proof is very similar to that in Part (i). Recall that $n=\lfloor N^{\delta} \rfloor$ . This time we need to sum over $k = 1, \dots, n^{\frac{1}{2}+\varepsilon} / N$ , which results in the bound

   \begin{align*} C |\textbf{K}| N^{-\zeta(d-2)} + C|\textbf{K}| N^{2-d} \times N^{\delta + 2 \delta \varepsilon - 2} = C|\textbf{K}| \left[ N^{-\zeta(d-2)} + N^{\delta - d + 2 \delta \varepsilon} \right]\kern-2pt. \end{align*}
   Here, for $\varepsilon \gt 0$ small enough, the second term dominates thanks to the choice of $\zeta$ ; see (8).

Proof of Proposition 2.3. Since

\begin{equation*}\begin{split} &\textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell,\, \text{$Y_t \not\in \varphi^{-1}(\textbf{K})$ for $0 \le t \lt n$} \right] \\ &\qquad = \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell \right] - \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell,\, \text{$Y_t \in \varphi^{-1}(\textbf{K})$ for some $0 \le t \lt n$} \right]\kern-2pt,\end{split}\end{equation*}

we need to show that

\begin{equation*} \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell,\, \text{$Y_t \in \varphi^{-1}(\textbf{K})$ for some $0 \le t \lt n$} \right] = O\left(\frac{n}{N^d}\right)\textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell \right]\kern-2pt.\end{equation*}

Define $A(x)= \left\{Y_n=y^{\prime}_{\ell},\, Y_t \in x N + \textbf{K} \text { for some } 0 \le t \lt n \right\}$ , so that

(28) \begin{equation} \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_\ell,\, \text{$Y_t \in \varphi^{-1}(\textbf{K})$ for some $0 \le t \lt n$} \right] \le \sum_{x \in \mathbb{Z}^d} \textbf{P}_{y^{\prime}_{\ell-1}} [ A(x) ].\end{equation}

We have

(29) \begin{equation}\begin{split}\textbf{P}_{y^{\prime}_{\ell-1}}[A(x)] \leq \sum_{n_1+n_2=n}\sum_{x^{\prime}\in \textbf{K}+xN} p_{n_1}\big(y^{\prime}_{\ell-1},x^{\prime}\big) p_{n_2}\big(x^{\prime},y^{\prime}_{\ell}\big).\end{split}\end{equation}

We bound this by splitting up the sum into different contributions. Let $\varepsilon \gt 0$ ; this will be chosen sufficiently small in the course of the proof.

Case 1: $n_1, n_2\geq N^{2+6\varepsilon}$ and $|y^{\prime}_{\ell-1}-x^{\prime}|\leq n^{\frac{1}{2}+\varepsilon}_1$ , $|x^{\prime}-y^{\prime}_{\ell}|\leq n^{\frac{1}{2}+\varepsilon}_2$ . By the LCLT we have that

(30) \begin{equation}\begin{split} p_{n_1}\big(y^{\prime}_{\ell-1},x^{\prime}\big) &\leq Cp_{n_1}\big(y^{\prime}_{\ell-1},u^{\prime}\big) \quad \text{for any $u^{\prime} \in \mathbb{T}_N+xN$}, \\ p_{n_2}\big(x^{\prime},y^{\prime}_{\ell}\big) &\leq Cp_{n_2}\big(u^{\prime},y^{\prime}_{\ell}\big) \quad \text{for any $u^{\prime} \in \mathbb{T}_N+xN$}.\end{split}\end{equation}

For this note that we have

\begin{equation*}\begin{split}\left|\frac{d|y^{\prime}_{\ell-1}-x^{\prime}|^2}{n_1} - \frac{d|y^{\prime}_{\ell-1}-u^{\prime}|^2}{n_1}\right|&\leq \frac{d|x^{\prime}-u^{\prime}|^2}{n_1}+ \frac{2d|\langle x^{\prime}-u^{\prime},y^{\prime}_{\ell-1}-x^{\prime}\rangle|}{n_1} \\&\leq C\frac{N^2}{n_1} + \frac{CN\cdot n_1^{\frac{1}{2}+\varepsilon}}{n_1},\end{split}\end{equation*}

where the first term tends to 0 and the rest equals

\begin{equation*}\begin{split}CN n_1^{-\frac{1}{2}+\varepsilon}\leq CN\cdot N^{(2+6\varepsilon)\big({-}\frac{1}{2}+\varepsilon\big)}= CN^{-\varepsilon+6\varepsilon^2}\to 0, \quad\text{as $N\to \infty$.}\end{split}\end{equation*}

Here we choose $\varepsilon$ so that $-\varepsilon+6\varepsilon^2 \lt 0$ . A similar observation shows the estimate for $p_{n_2}\big(x^{\prime},y^{\prime}_\ell\big)$ .

The way we are going to use (30) is to replace the summation over $x^{\prime}$ by a summation over all $u^{\prime} \in \mathbb{T}_N + xN$ , at the same time inserting a factor $|\textbf{K}|/N^d$ . Hence the contribution of the values of $n_1, n_2$ and x in Case 1 to the right-hand side of (28) is at most

\begin{equation*}\begin{split} \frac{C|\textbf{K}|}{N^d} \sum_{n_1 + n_2 = n} \sum_{u^{\prime} \in \mathbb{Z}^d} p_{n_1}\big(y^{\prime}_{\ell-1},u^{\prime}\big) p_{n_2}(u^{\prime}, y^{\prime}_\ell) &= \frac{C|\textbf{K}|}{N^d} \sum_{n_1 + n_2 = n} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_{\ell}\big) \\ &\le \frac{C|\textbf{K}| n}{N^d} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{split}\end{equation*}

This completes the bound in Case 1. For future use, note that if $\varepsilon_n \to 0$ is any sequence, and we add the restriction $n_1 \le \varepsilon_n n$ to the conditions in Case 1, we obtain the upper bound

(31) \begin{equation}\begin{split} C |\textbf{K}| \frac{\varepsilon_n n}{N^d} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) = o(1) \frac{n}{N^d} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{split}\end{equation}

Case 2a: $n_1, n_2 \ge N^{2+6\varepsilon}$ but $\big|x^{\prime} - y^{\prime}_{\ell-1}\big| \gt n_1^{\frac{1}{2}+\varepsilon}$ . In this case we bound $p_{n_2}\big(x^{\prime}, y^{\prime}_\ell\big) \le 1$ and have that the contribution of this case to the right-hand side of (28) is at most

\begin{equation*}\begin{split} \sum_{\substack{n_1+n_2 = n \\ n_1,n_2 \ge N^{2+6\varepsilon}}} \textbf{P}_{y^{\prime}_{\ell-1}} \Big[ \big|Y_{n_1} - y^{\prime}_{\ell-1}\big| \gt n_1^{1/2+\varepsilon} \Big] &\le \sum_{\substack{n_1+n_2 = n \\ n_1,n_2 \ge N^{2+6\varepsilon}}} C \exp \big({-}c n_1^{2\varepsilon} \big)\\ &\le C n \exp \big({-}c N^{4 \varepsilon} \big) = o \left(\frac{n}{N^d}\right) p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big),\end{split}\end{equation*}

where in the first step we used (3) and in the last step we used the Gaussian lower bound (5) for $p_n$ . Indeed, the requirement for the Gaussian lower bound is satisfied for sufficiently large N because $\big|y^{\prime}_{\ell}-y^{\prime}_{\ell+1}\big| \le C_1 \sqrt{\log n}\sqrt{n} \le c\,n$ . Therefore, we have

(32) \begin{equation} p_{n}\big(y^{\prime}_{\ell-1},y^{\prime}_{\ell}\big) \ge \frac{c}{n^{d/2}}\exp\left({-}\frac{C\big|y^{\prime}_{\ell}-y^{\prime}_{\ell-1}\big|^2}{n}\right) \ge \frac{c}{n^{d/2}}\exp\left({-}C\,\log n\right)\kern-2pt.\end{equation}

Then we have

\begin{equation*}\begin{split}\frac{Cn\exp ({-}cN^{4\varepsilon})}{c n^{-d/2}\exp\left({-}C\,\log n\right)}\le Cn^{1+d/2}\exp\left({-} c N^{4\varepsilon} + C\,\log n\right)= o \left( \frac{n}{N^d} \right)\kern-2pt, \,\text{as $N\to\infty$.}\end{split}\end{equation*}

Case 3a: $n_1 \lt N^{2+6\varepsilon}$ and $\big|x^{\prime}-y^{\prime}_{\ell-1}\big|\le N^{\frac{\delta}{2}-\varepsilon}$ . By the LCLT we have

\begin{equation*}\begin{split} p_{n_2}\big(x^{\prime},y^{\prime}_{\ell}\big) &= \frac{C}{n_2^{d/2}}\exp\left({-}\frac{d|y^{\prime}_{\ell}-x^{\prime}|^2}{n_2}\right) (1+o(1)),\\ p_{n}\big(y^{\prime}_{\ell-1},y^{\prime}_{\ell}\big) &= \frac{C}{n^{d/2}}\exp\left({-}\frac{d|y^{\prime}_{\ell}-y^{\prime}_{\ell-1}|^2}{n}\right) (1+o(1)).\end{split}\end{equation*}

We claim that

(33) \begin{equation}p_{n_2}\big(x^{\prime},y^{\prime}_{\ell}\big)\le C\,p_n\big(y^{\prime}_{\ell-1},y^{\prime}_\ell\big).\end{equation}

We first note that $n_2=n-n_1 = n(1+o(1))$ , then deduce that $n_2^{-d/2}= O\big(n^{-d/2}\big)$ and

\begin{equation*}\exp\left({-}\frac{d|y^{\prime}_{\ell}-y^{\prime}_{\ell-1}|^2}{n}\right)\ge \exp\left({-}\frac{d|y^{\prime}_{\ell}-y^{\prime}_{\ell-1}|^2}{n_2}\right)\kern-2pt.\end{equation*}

Since we have $\big|x^{\prime}-y^{\prime}_{\ell-1}\big|\le N^{\frac{\delta}{2}-\varepsilon}$ in the exponent, as $N\to\infty$ we have

\begin{equation*}\frac{|x^{\prime}-y^{\prime}_{\ell-1}|^2}{n_2}\le \frac{N^{\delta-2\varepsilon}}{n_2}\to 0\end{equation*}

and

\begin{equation*}\frac{|y^{\prime}_{\ell}-y^{\prime}_{\ell-1}||x^{\prime}-y^{\prime}_{\ell-1}|}{n_2}\le \frac{n^{\frac{1}{2}}C_1\sqrt{\log n}N^{\frac{\delta}{2}-\varepsilon}}{n_2}\to 0.\end{equation*}

These imply that

\begin{equation*}\begin{split} \left|\frac{|y^{\prime}_{\ell}-y^{\prime}_{\ell-1}|^2-|y^{\prime}_{\ell}-x^{\prime}|^2}{n_2}\right| &\le \left|\frac{|y^{\prime}_{\ell}-y^{\prime}_{\ell-1}|^2 -|(y^{\prime}_{\ell}-y^{\prime}_{\ell-1})+(y^{\prime}_{\ell-1}-x^{\prime})|^2}{n_2}\right|\\ &\le \frac{|x^{\prime}-y^{\prime}_{\ell-1}|^2}{n_2} +\frac{2|y^{\prime}_{\ell}-y^{\prime}_{\ell-1}||x^{\prime}-y^{\prime}_{\ell-1}|}{n_2} \to 0.\end{split}\end{equation*}

Thus (33) follows from comparing the LCLT approximations of the two sides.

We now have that the contribution of this case to the right-hand side of (28) is at most

\begin{equation*}\begin{split} C p_{n}(y^{\prime}_{\ell-1},y^{\prime}_{\ell}) \sum_{\substack{n_1 \lt N^{2+6\varepsilon}}} \sum_{x \in \mathbb{Z}^d} \sum_{x^{\prime}\in \textbf{K}+xN} p_{n_1}\big(y^{\prime}_{\ell-1},x^{\prime}\big) &\le \frac{C}{N^{\zeta(d-2)}} p_{n}\big(y^{\prime}_{\ell-1},y^{\prime}_{\ell}\big) \\ &\le o(1) \frac{n}{N^d} p_{n}\big(y^{\prime}_{\ell-1},y^{\prime}_{\ell}\big),\end{split}\end{equation*}

where in the first step we used Lemma 4.1(i) and the last step holds for the value of $\zeta$ we chose; cf. (8).

Case 3b: $n_1 \lt N^{2+6\varepsilon}$ but $\big|x^{\prime}-y^{\prime}_{\ell-1}\big|\gt N^{\frac{\delta}{2}-\varepsilon}$ . Use the Gaussian upper bound (5) to bound $p_{n_1}$ , and bound the sum over all $x^{\prime}\in\mathbb{Z}^d$ of $p_{n_2}$ by 1 using symmetry of $p_{n_2}$ , to get

\begin{equation*}\begin{split} &\sum_{\substack{n_1+n_2=n\\n_1 \lt N^{2+6\varepsilon}}} \sum_{x \in \mathbb{Z}^d} \sum_{x^{\prime}\in\textbf{K}+xN} p_{n_1}\big(y^{\prime}_{\ell-1},x^{\prime}\big)p_{n_2}\big(x^{\prime},y^{\prime}_\ell\big)\\ &\qquad\le \sum_{n_1 \lt N^{2+6\varepsilon}} \frac{C}{n_1^{d/2}}\exp\left({-}\frac{N^{\delta-2\varepsilon}}{N^{2+6\varepsilon}}\right) \sum_{x^{\prime}\in \mathbb{Z}^d} p_{n-n_1}\big(x^{\prime},y^{\prime}_\ell\big)\\ &\qquad\le \sum_{n_1 \lt N^{2+6\varepsilon}} \frac{C}{n_1^{d/2}}\exp\left({-}\frac{N^{\delta-2\varepsilon}}{N^{2+6\varepsilon}}\right)\\ &\qquad \le C N^{O(1)} \exp\big({-}N^{\delta-2-8\varepsilon}\big) = o\left(\frac{n}{N^d}\right) p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big), \quad\text{as $N\to\infty$.}\end{split}\end{equation*}

In the last step we used a Gaussian lower bound for $p_n$ ; cf. (32).

Therefore, we have discussed all possible cases and proved statement (14) of the proposition as required.

The proof of (15) is similar to that of the first part, with only a few modifications. In this part we have to show that

\begin{equation*} \textbf{P}_{o} \left[ Y_n = y^{\prime}_1,\, \text{$Y_t \in \varphi^{-1}(\textbf{K})$ for some $0 \le t \lt n$} \right] = o(1)\textbf{P}_{o} \left[ Y_n = y^{\prime}_1 \right].\end{equation*}

Define $A_0(x)= \left\{Y_n=y^{\prime}_1, Y_t\in x N + \textbf{K}\right.$ for some $\left.0\leq t \lt n \right\}$ , so that

(34) \begin{equation} \textbf{P}_{o} \left[ Y_n = y^{\prime}_1,\, \text{$Y_t \in \varphi^{-1}(\textbf{K})$ for some $0 \le t \lt n$} \right] \le \sum_{x \in \mathbb{Z}^d} \textbf{P}_{o} [ A_0(x) ].\end{equation}

We have

(35) \begin{equation}\begin{split}\textbf{P}_{o}[A_0(x)] \leq \sum_{n_1+n_2=n}\sum_{x^{\prime}\in \textbf{K}+xN} p_{n_1}(o,x^{\prime}) p_{n_2}\big(x^{\prime},y^{\prime}_{1}\big).\end{split}\end{equation}

We bound the term above by splitting up the sum into the same cases as in the proof of (14). The different cases can be handled very similarly to the first part. The difference is only in Case 3a while applying the Green’s function bound Lemma 4.1.

In Case 3a, by the LCLT, we can deduce that

\begin{equation*} p_{n_2}\big(x^{\prime},y^{\prime}_1\big) \le C\,p_n\big(o,y^{\prime}_1\big).\end{equation*}

If $g(N) \gt N^\zeta$ , the bound of Lemma 4.1(i) can be used as before. If $g(N) \le N^\zeta$ , by Lemma 4.1(ii), we have that the contribution of this case to the right-hand side of (34) is at most

\begin{equation*}\begin{split} C p_{n}\big(o,y^{\prime}_1\big) \sum_{\substack{n_1 \lt N^{2+6\varepsilon}}} \sum_{x \in \mathbb{Z}^d} \sum_{x^{\prime}\in \textbf{K}+xN} p_{n_1}(o,x^{\prime}) \le \frac{C}{g(N)^{d-2}} p_{n}\big(o,y^{\prime}_1\big) = o(1) p_{n}\big(o,y^{\prime}_1\big).\end{split}\end{equation*}

Here we used that $g(N) \to \infty$ .

Note that Case 4a can be handled in the same way as in the proof of (14), since the distance between $y^{\prime}_1$ and any copy of $\textbf{K}$ is at least $N^\zeta$ .

Therefore, we have discussed all possible cases and proved (15) as required.

For future use, we extract a few corollaries of the proof of Proposition 2.3.

Corollary 4.1. Assume that $y^{\prime}, y^{\prime\prime} \in \mathbb{Z}^d$ are points such that $|y^{\prime\prime} - y^{\prime}| \le 2 C_1 \sqrt{\log n} \sqrt{n}$ . Then for all $n/2 \le m \le n$ we have

(36) \begin{equation} \sum_{n_1+n_2=m} \sum_{x \in \mathbb{Z}^d} \sum_{\substack{x^{\prime}\in \textbf{K}+xN \\ |y^{\prime} - x^{\prime}|, |y^{\prime\prime} - x^{\prime}| \gt N^\zeta}} p_{n_1}(y^{\prime},x^{\prime}) p_{n_2}(x^{\prime},y^{\prime\prime}) = O\left(\frac{n}{N^d}\right) p_{m}(y^{\prime},y^{\prime\prime}).\end{equation}

The following is merely a restatement of what was observed in (31) (with Part (ii) below holding by symmetry).

The following is a restatement of the bounds of Cases 2a and 2b.

The following is a restatement of the bounds of Cases 3a and 3b combined.

Corollary 4.4. For $\ell \ge 2$ we have

(40) \begin{equation}\begin{split} \sum_{\substack{n_1+n_2=n \\ n_1 \lt N^{2+6\varepsilon}}} \ \sum_{x \in \mathbb{Z}^d} \ \sum_{\substack{x^{\prime} \in \textbf{K} +xN}} p_{n_1}\big(y^{\prime}_{\ell-1},x^{\prime}\big) p_{n_2}\big(x^{\prime},y^{\prime}_{\ell}\big) = o(1) \frac{n}{N^d} p_n\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{split}\end{equation}

4.2. Proof of Proposition 2.2

In this section we need $C_1$ large enough so that we have

(41) \begin{equation} e^{-d f(n)^2} N^d n^{1+3d/2} \to 0.\end{equation}

We have

\begin{align*} \textbf{P}_{y^{\prime}_{\ell-1}} \big[ Y_n = y^{\prime}_\ell,\, \text{$Y_t \in \varphi^{-1}(\textbf{K})$ for some $0 \le t \lt n$} \big] = \textbf{P}_{y^{\prime}_{\ell-1}} \left[ \cup_{x \in \mathbb{Z}^d} A(x) \right]\kern-2pt, \end{align*}

where

\begin{align*} A(x) = \left\{ Y_n = y^{\prime}_\ell, \text{$Y_t \in x N + \textbf{K}$ for some $0 \le t \lt n$} \right\}\kern-2pt. \end{align*}

The strategy is to estimate the probability via the Bonferroni inequalities:

(42) \begin{equation}\begin{split} \sum_{x} \textbf{P}_{y^{\prime}_{\ell-1}} [ A(x) ] - \sum_{x_1 \not= x_2} \textbf{P}_{y^{\prime}_{\ell-1}} [ A(x_1) \cap A(x_2) ] &\le \textbf{P}_{y^{\prime}_{\ell-1}} \left[ \cup_{x \in \mathbb{Z}^d} A(x) \right] \\ &\le \sum_{x} \textbf{P}_{y^{\prime}_{\ell-1}} [ A(x) ].\end{split}\end{equation}

We are going to use a parameter $A_n$ that we choose as $A_n = 10 \log \log n$ , so that in particular $A_n \to \infty$ .

4.2.1. The main contribution

In this section, we consider only stretches with $|y^{\prime}_\ell - y^{\prime}_{\ell-1}| \le A_n \sqrt{n}$ . We will show that the main contribution in (42) comes from x in the set

\begin{align*} G = \left\{ x \in \mathbb{Z}^d\, :\,|y^{\prime}_{\ell-1} - x N| \le A_n^2 \sqrt{n},\, |x N - y^{\prime}_{\ell}| \le A_n^2 \sqrt{n} \right\}\kern-2pt. \end{align*}

We first examine $\textbf{P}_{y^{\prime}_{\ell-1}} [ A(x) ]$ for $x \in G$ . Putting $B_{0,x} = B(\textbf{x}_0 + x N, N^{\zeta})$ , let $n_1$ be the time of the last visit to $\partial B_{0,x}$ before hitting $\textbf{K}+xN$ , let $n_1 + n_2$ be the time of the first hit of $\textbf{K}+xN$ , and let $n_3 = n - n_1 - n_2$ . See Figure 2 for an illustration of this decomposition.

Figure 2. The decomposition of a path hitting a copy of $\textbf{K}$ into three subpaths (not to scale).

Then we can write

(43) \begin{equation}\begin{split} \textbf{P}_{y^{\prime}_{\ell-1}} [ A(x) ] &= \sum_{n_1 + n_2 + n_3 = n} \sum_{z^{\prime} \in \partial B_{0,x}} \sum_{x^{\prime} \in \textbf{K} + x N} \widetilde{p}^{(x)}_{n_1} \big(y^{\prime}_{\ell-1},z^{\prime}\big) \\ &\qquad\quad \times \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}},\, Y_{H_{\textbf{K}+xN}} = x^{\prime} \big] \, p_{n_3}\big(x^{\prime},y^{\prime}_\ell\big),\end{split}\end{equation}

where

\begin{align*} \widetilde{p}^{(x)}_{n_1} \big(y^{\prime}_{\ell-1},z^{\prime}\big) = \textbf{P}_{y^{\prime}_{\ell-1}} \big[ Y_{n_1} = z^{\prime},\, \text{$Y_t \not\in \textbf{K}+xN$ for $0 \le t \le n_1$} \big]. \end{align*}

We are going to use another parameter $\varepsilon_n$ that will need to go to 0 slowly. We choose it as $\varepsilon_n = (10 \log \log n)^{-1} \to 0$ . The main contribution to (43) will be when $n_1 \ge \varepsilon_n n$ , $n_3 \ge \varepsilon_n n$ , and $n_2 \le N^{2\delta/d} \sim n^{2/d}$ . Therefore, we split the sum over $n_1, n_2, n_3$ in (43) into a main contribution I(x) and an error term II(x). In order to define these, let

(44) \begin{equation}\begin{split} &F\big(n_1,n_2,n_3,x,y^{\prime}_{\ell-1},y^{\prime}_\ell\big) \\ &\quad = \sum_{z^{\prime} \in \partial B_{0,x}} \sum_{x^{\prime} \in \textbf{K} + x N}\widetilde{p}^{(x)}_{n_1} \big(y^{\prime}_{\ell-1},z^{\prime}\big) \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}},\, Y_{H_{\textbf{K}+xN}} = x^{\prime} \big] p_{n_3}\big(x^{\prime},y^{\prime}_\ell\big).\end{split}\end{equation}

Then with

(45) \begin{equation}\begin{split} I(x) &\,:\!= \sum_{\substack{n_1 + n_2 + n_3 = n \\ n_1, n_3 \ge \varepsilon_n n,\, n_2 \le N^{2 \delta/d}}} F\big(n_1,n_2,n_3,x,y^{\prime}_{\ell-1},y^{\prime}_\ell\big), \\ II(x) &\,:\!= \sum_{\substack{n_1 + n_2 + n_3 = n \\ \text{$n_1 \lt \varepsilon_ n n$ or $n_3 \lt \varepsilon_n n$} \\ \text{or $n_2 \gt N^{2 \delta/d}$}}} F\big(n_1,n_2,n_3,x,y^{\prime}_{\ell-1},y^{\prime}_\ell\big),\end{split}\end{equation}

we have

\begin{equation*}\begin{split} \textbf{P}_{y^{\prime}_{\ell-1}} [ A(x) ] = I(x) + II(x).\end{split}\end{equation*}

Lemma 4.2. When $x \in G$ and $n_3 \ge \varepsilon_n n$ , we have

\begin{align*} p_{n_3}\big(x^{\prime},y^{\prime}_\ell\big) = (1 + o(1)) p_{n_3}\big(u^{\prime},y^{\prime}_\ell\big) \quad {for \,all \,x^{\prime} \in \boldsymbol{K}+xN \, and \,\, all\, u^{\prime} \in \mathbb{T}_N + xN}, \end{align*}

with the o(1) term uniform in $x^{\prime}$ and $u^{\prime}$ .

Proof. By the LCLT, we have

\begin{equation*}\begin{split} p_{n_3}\big(x^{\prime},y^{\prime}_\ell\big) &= \frac{C}{n_3^{d/2}}\exp\left({-}\frac{d|y^{\prime}_{\ell}-x^{\prime}|^2}{n_3}\right) (1+o(1)),\\ p_{n_3}\big(u^{\prime},y^{\prime}_\ell\big) &= \frac{C}{n_3^{d/2}}\exp\left({-}\frac{d|y^{\prime}_{\ell}-u^{\prime}|^2}{n_3}\right) (1+o(1)).\end{split}\end{equation*}

We compare the exponents

\begin{equation*}\begin{split}\left|\frac{d\big|y^{\prime}_{\ell}-x^{\prime}\big|^2}{n_3} - \frac{d\big|y^{\prime}_{\ell}-u^{\prime}\big|^2}{n_3}\right|&\leq \frac{d|x^{\prime}-u^{\prime}|^2}{n_3}+ \frac{2d\big|\big\langle x^{\prime}-u^{\prime},y^{\prime}_{\ell}-x^{\prime}\big\rangle\big|}{n_3} \\&\leq C\frac{N^2}{n_3} + \frac{CN\cdot A_n^2\sqrt{n}}{n_3}\to 0,\end{split}\end{equation*}

as $N\to\infty$ .

Lemma 4.3. When $x \in G$ and $n_1 \ge \varepsilon_n n$ , we have

\begin{align*} \widetilde{p}^{(x)}_{n_1}\big(y^{\prime}_{\ell-1},z^{\prime}\big) = (1 + o(1)) p_{n_1}\big(y^{\prime}_{\ell-1},u^{\prime}\big) \quad \textit{ for all } z^{\prime} \in \partial B_{0,x}\textit{ and all } u^{\prime} \in \mathbb{T}_N + xN, \end{align*}

with the o(1) term uniform in $z^{\prime}$ and $u^{\prime}$ .

Proof. The statement will follow if we show the following claim:

\begin{align*} \textbf{P}_{y^{\prime}_{\ell-1}} \big[ Y_{n_1} = z^{\prime},\, \text{$Y_t \in \textbf{K}+xN$ for some $0 \le t \le n_1$} \big] = o(1) p_{n_1} \big(y^{\prime}_{\ell-1}, z^{\prime}\big). \end{align*}

For this, observe that by (5) we have

(46) \begin{equation}\begin{split} p_{n_1}\big(y^{\prime}_{\ell-1}, z^{\prime}\big) &\ge \frac{c}{n_1^{d/2}} \exp \left( -C \frac{|z^{\prime} - y^{\prime}_{\ell-1}|^2}{n_1} \right) \\ &\ge \frac{c}{n^{d/2}} \exp \left( -c \frac{A_n^2 n + N^{\zeta}}{\varepsilon_n n} \right) \\ &\ge \frac{c}{n^{d/2}} \exp \left( -C (\log \log n)^{O(1)} \right) \\ &= n^{-d/2+o(1)}.\end{split}\end{equation}

On the other hand, using the Markov property, (5), and the fact that for $x^{\prime} \in \textbf{K}+xN$ we have $\big|y^{\prime}_{\ell-1} - x^{\prime}\big| \ge c N^{\zeta}$ and $|x^{\prime} - z^{\prime}| \ge c N^{\zeta}$ , we get

(47) \begin{equation}\begin{split} &\textbf{P}_{y^{\prime}_{\ell-1}} \big[ Y_{n_1} = z^{\prime},\, Y_t \in \textbf{K}+xN \text{ for some } 0 \le t \le n_1 \big] \\ &\qquad \le \sum_{1 \le m \le n_1-1} \sum_{x^{\prime} \in \textbf{K}+xN} p_m\big(y^{\prime}_{\ell-1},x^{\prime}\big) \, p_{n_1-m}(x^{\prime},z^{\prime}) \\ &\qquad \le C \sum_{1 \le m \le n_1-1} \frac{1}{m^{d/2}} \frac{1}{(n_1-m)^{d/2}} \exp \left({-}c \frac{N^{2\zeta}}{m} \right) \exp \left( -c \frac{N^{2\zeta}}{n_1-m} \right)\\ &\qquad \le C \sum_{1 \le m \le n_1/2} \frac{1}{m^{d/2}} \frac{1}{(n_1-m)^{d/2}} \exp \left({-}c \frac{N^{2\zeta}}{m} \right) \exp \left( -c \frac{N^{2\zeta}}{n_1-m} \right)\kern-2pt.\end{split}\end{equation}

We note here that the sum over $1 \le m \le n_1/2$ and the sum over $n_1/2 \le m \le n_1-1$ are symmetric. Bounding the sum over $1 \le m \le n_1/2$ gives

(48) \begin{equation}\begin{split} &\frac{C}{n_1^{d/2}} \sum_{1 \le m \le n_1/2} \frac{1}{m^{d/2}} \exp \left({-}c \frac{N^{2\zeta}}{m} \right) \\ &\quad = \frac{C}{n_1^{d/2}} \left[ \sum_{1 \le m \le N^{2 \zeta}} \frac{1}{m^{d/2}}\exp \left({-}c \frac{N^{2\zeta}}{m} \right) + \sum_{N^{2 \zeta} \lt m \le n_1/2} \frac{1}{m^{d/2}}\exp \left({-}c \frac{N^{2\zeta}}{m} \right)\right]\kern-2pt.\end{split}\end{equation}

In the second sum we can bound the exponential by 1 and get the upper bound

\begin{align*} \frac{C}{n_1^{d/2}} N^{\zeta(2-d)} = o\big(n^{-d/2+o(1)}\big). \end{align*}

In the first sum, we group terms on dyadic scales k so that $2^k \le N^{2 \zeta}/m \le 2^{k+1}$ , $k = 0, \dots, \lfloor \log_2 N^{2 \zeta}\rfloor +1 $ . This gives the bound

\begin{align*} \frac{C}{n_1^{d/2}} \sum_{k=0}^{\lfloor\log_2 N^{2 \zeta}\rfloor +1 } \frac{(2^{k+1})^{d/2}}{(N^{2 \zeta})^{d/2}} \exp ({-}c 2^k ) \le \frac{C}{n_1^{d/2}} \frac{1}{N^{\zeta d}}, \end{align*}

which is also $o\big(n^{-d/2+o(1)}\big)$ .

In order to apply the previous two lemmas to analyse I(x) in (45), we first define a modification of F in (44), where $z^{\prime}$ and $x^{\prime}$ are both replaced by a vertex $u^{\prime} \in \mathbb{T}_N + xN$ . That is, we define

\begin{equation*}\begin{split} &\widetilde{F}\big(n_1,n_2,n_3,u^{\prime},x,y^{\prime}_{\ell-1},y^{\prime}_\ell\big) \\ &\quad = \sum_{z^{\prime} \in \partial B_{0,x}} \sum_{x^{\prime} \in \textbf{K} + x N} p_{n_1} \big(y^{\prime}_{\ell-1},u^{\prime}\big) \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}},\, Y_{H_{\textbf{K}+xN}} = x^{\prime} \big] p_{n_3}\big(u^{\prime},y^{\prime}_\ell\big).\end{split}\end{equation*}

Then Lemmas 4.2 and 4.3 allow us to write, for $x \in G$ , the main term I(x) in (45) as

(49) \begin{equation}\begin{split} I(x) & = \sum_{\substack{n_1 + n_2 + n_3 = n \\ n_1, n_3 \ge \varepsilon_n n,\, n_2 \le N^{2\delta/d}}} F\big(n_1,n_2,n_3,x,y^{\prime}_{\ell-1},y^{\prime}_\ell\big) \\ & = \frac{1 + o(1)}{N^d} \sum_{u^{\prime} \in \mathbb{T}_N+xN} \sum_{\substack{n_1 + n_2 + n_3 = n \\ n_1, n_3 \ge \varepsilon_n n,\, n_2 \le N^{2 \delta/d}}} \widetilde{F}\big(n_1,n_2,n_3,u^{\prime},x,y^{\prime}_{\ell-1},y^{\prime}_\ell\big)\\ &= \frac{1 + o(1)}{N^d} \sum_{u^{\prime} \in \mathbb{T}_N+xN} \sum_{\substack{n_1 + n_2 + n_3 = n \\ n_1, n_3 \ge \varepsilon_n n \\ n_2 \le N^{2 \delta/d}}} p_{n_1} \big(y^{\prime}_{\ell-1}, u^{\prime}\big) \, p_{n_3}\big(u^{\prime}, y^{\prime}_\ell\big) \\ &\times\sum_{z^{\prime} \in \partial B_{0,x}} \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}} \big],\end{split}\end{equation}

where the sum over $x^{\prime}$ is removed since

\begin{equation*} \sum_{x^{\prime} \in \textbf{K} + x N} \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}},\, Y_{H_{\textbf{K}+xN}} = x^{\prime} \big] = \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}} \big].\end{equation*}

Lemma 4.4. Assume that $n_1, n_3 \ge \varepsilon_n n$ and $n_2 \le N^{2 \delta/d}$ .

1. (i) We have

   \begin{align*} p_{n_1+n_3} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) = (1 + o(1)) p_n\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big). \end{align*}

2. (ii) We have

   (50) \begin{equation} \sum_{x \in G} \sum_{u^{\prime} \in \mathbb{T}_N+xN} p_{n_1} \big(y^{\prime}_{\ell-1}, u^{\prime}\big) \, p_{n_3}\big(u^{\prime}, y^{\prime}_\ell\big) = (1 + o(1)) p_{n_1+n_3} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{equation}

Proof.

1. (i) When $n_2 \le N^{2 \delta/d} \sim n^{2/d}$ , we have

   \begin{align*} n_1 + n_3 = n \left( 1 - O \left( n^{-1+2/d} \right) \right)\kern-2pt. \end{align*}
   Hence the exponential term in the LCLT for $p_{n_1+n_3}\big(y^{\prime}_{\ell-1},y^{\prime}_\ell\big)$ is
   \begin{align*} \exp \left({-}\frac{|y^{\prime}_{\ell} - y^{\prime}_{\ell-1}|^2}{n} \left( 1 + O \big( n^{-1+2/d} \big) \right) \right) = (1 + o(1)) \, \exp \left({-}\frac{\big|y^{\prime}_{\ell} - y^{\prime}_{\ell-1}\big|^2}{n} \right)\kern-2pt, \end{align*}
   where we used that $A_n = 10 \log \log n$ , and hence $\big|y^{\prime}_{\ell} - y^{\prime}_{\ell-1}\big|^2 \le A_n^2 n = n\, o\big( n^{1-2/d} \big)$ .

2. (ii) If we summed over all $x \in \mathbb{Z}^d$ , we would get exactly $p_{n_1+n_3}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big)$ . Thus the claim amounts to showing that

   (51) \begin{equation} \sum_{x \in \mathbb{Z}^d \setminus G} \sum_{u^{\prime} \in \mathbb{T}_N+xN} p_{n_1} \big(y^{\prime}_{\ell-1}, u^{\prime}\big) \, p_{n_3}\big(u^{\prime}, y^{\prime}_\ell\big) = o(1) p_{n_1+n_3} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{equation}
   First, note that from the LCLT we have
   \begin{align*} p_{n_1+n_3} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) = (1 + o(1)) \overline{p}_{n_1+n_3} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big). \end{align*}
   In order to estimate the left-hand side of (51), using (5), we can estimate the contribution of $\big\{ x \in \mathbb{Z}^d \setminus G\, :\,\max \big\{ \big|y^{\prime}_{\ell-1} - xN\big|, \big|x N - y^{\prime}_{\ell-1}\big| \big\} \gt A_n^2 \sqrt{n} \big\}$ as follows. First, we have
   \begin{align*} p_{n_1+n_3} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) \ge \frac{c}{n^{d/2}} \exp \big({-}C A_n^2 (1+ o(1)) \big) \ge \frac{c}{n^{d/2}} \exp \big({-}C (\log \log n)^2 \big). \end{align*}
   Here we used $\big|y^{\prime}_{\ell} - y^{\prime}_{\ell-1}\big|^2 \le A_n^2 n$ and $n_1+n_3 = n(1-o(1))$ .

On the other hand, note that either $n_1 \ge n/3$ or $n_3 \ge n/3$ . Without loss of generality, assume that $n_3 \ge n/3$ . Then the contribution to the left-hand side of (51), using (5), and by summing in dyadic shells with radii $2^k A_n^2 \sqrt{n}$ , $k = 0, 1, 2, \dots$ , we get the bound

(52) \begin{align}\begin{split} &\sum_{k=0}^\infty C \Big(A_n^2 \sqrt{n}\Big)^d 2^{dk} \, \frac{C}{n_1^{d/2}} \, \exp \Big({-}c 2^{2k} A_n^4 n / n_1 \Big) \, \frac{1}{n_3^{d/2}} \, \exp \Big({-}c 2^{2k} A_n^4 n / n_3 \Big) \\ &\qquad \le \sum_{k=0}^\infty C \, A_n^{2d} \, 2^{dk} \, \frac{1}{\varepsilon_n^{d/2}} \, \exp \Big({-}c 2^{2k} A_n^4 \Big) \frac{1}{n^{d/2}} \, \exp \Big({-}c 2^{2k} A_n^4 \Big) \\ &\qquad \le \frac{C}{n^{d/2}} \, \frac{A_n^{2d}}{\varepsilon_n^{d/2}} \, \sum_{k=0}^\infty \exp \Big({-}c 2^{2k} (\log \log n)^4 + d k \log 2 \Big) \\ &\qquad = \frac{C}{n^{d/2}} \, o \left( \exp ({-}100 (\log \log n)^2 ) \right)\kern-2pt.\end{split} \nonumber \\[-36pt]\end{align}

The above lemma allows us to write

(53) \begin{equation}\begin{split} \sum_{x \in G} I(x) &= \frac{1 + o(1)}{N^d} \, p_n\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) \, \sum_{\substack{n_1 + n_2 + n_3 = n \\ n_1, n_3 \ge \varepsilon_n n \\ n_2 \le N^{2 \delta/d}}} \sum_{z^{\prime} \in \partial B_{0,x}} \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}} \big] \\ &= \frac{(1 + o(1)) \, n}{N^d} \, p_n\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) \, \sum_{n_2 \le N^{2 \delta/d}} \sum_{z^{\prime} \in \partial B_{0,x}} \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}} \big].\end{split}\end{equation}

The next lemma will help us extract the $\textrm{Cap}(K)$ contribution from the right-hand side of (43).

Lemma 4.5. We have

(54) \begin{equation}\begin{split} \sum_{n_2=0}^{N^{2 \delta/d}} \sum_{z^{\prime} \in \partial B_{0,x}} \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}} \big] = \frac{1}{2} \textrm{Cap}(K) \, (1 + o(1)).\end{split}\end{equation}

Proof. Performing the sum over $n_2$ allows us to rewrite the expression in the left-hand side of (54) as

\begin{equation*}\begin{split} &\left( \sum_{z^{\prime} \in \partial B_{0,x}} \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} \lt \xi_{B_{0,x}} \big] \right) - \left( \sum_{z^{\prime} \in \partial B_{0,x}} \textbf{P}_{z^{\prime}} \big[ N^{2 \delta/d} \lt H_{\textbf{K}+xN} \lt \xi_{B_{0,x}} \big] \right) \\ &\quad = \frac{1}{2} \textrm{Cap}(K) +o(1) - \sum_{z^{\prime} \in \partial B_{0,x}} \sum_{\textbf{x} \in \textbf{K}} \textbf{P}_{z^{\prime}} \Big[ N^{2 \delta/d} \lt H_{\textbf{K}+xN} \lt \xi_{B_{0,x}},\, Y_{H_{\textbf{K}+xN}} = \textbf{x} + xN \Big].\end{split}\end{equation*}

Here the $1/2$ before $\textrm{Cap}(K)$ comes from the random walk being lazy; see (4). Using time-reversal for the summand in the last term, we get the expression

(55) \begin{equation}\begin{split} &= \frac{1}{2} \textrm{Cap}(K) +o(1) - \sum_{\textbf{x} \in \textbf{K}} \sum_{z^{\prime} \in \partial B_{0,x}} \textbf{P}_{\textbf{x}+xN} \Big[ N^{2 \delta/d} \lt \xi_{B_{0,x}} \lt H_{\textbf{K}+xN},\, Y_{\xi_{B_{0,x}}} = z^{\prime}\Big] \\ &\qquad = \frac{1}{2} \textrm{Cap}(K) +o(1) - \sum_{\textbf{x} \in \textbf{K}} \textbf{P}_{\textbf{x}+xN} \Big[ N^{2 \delta/d} \lt \xi_{B_{0,x}} \lt H_{\textbf{K}+xN} \Big].\end{split}\end{equation}

The subtracted term in the right-hand side of (55) is at most

\begin{align*} |\textbf{K}| \max_{\textbf{x} \in \textbf{K}} \textbf{P}_{\textbf{x}+xN} \Big[ \xi_{B_{0,x}} \gt N^{2 \delta/d} \Big]. \end{align*}

Since $\zeta \lt \delta/d$ , this expression is o(1).

From the above lemma we get that the main contribution equals

(56) \begin{equation} \sum_{x \in G} I(x) = (1 + o(1)) \frac{n}{N^d} \, \frac{1}{2} \, \textrm{Cap}(K) \, p_n\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{equation}

It remains to estimate all the error terms.

4.2.2. The error terms

Lemma 4.6. We have

\begin{align*} \sum_{x \in G} II(x) = o(1) \frac{n}{N^d} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big). \end{align*}

Proof. We split the estimates according to which condition is violated in the sum. Recall that in the proof of Proposition 2.3 we chose $\varepsilon \gt 0$ such that $-\varepsilon+6\varepsilon^2 \lt 0$ . Here we make the further restriction that $\varepsilon \lt 2\delta/d - 2\zeta$ .

Case 1: $n_2 \gt N^{2\delta/d}$ . We claim that

(57) \begin{equation} \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}}, Y_{H_{\textbf{K}+xN}} = x^{\prime} \big] \le C \exp \big({-}N^{\varepsilon/2} \big) p_{n_2}\big(z^{\prime},x^{\prime}\big).\end{equation}

Since, in every time interval of duration $N^{2\zeta}$ , the walk has a positive chance of exiting the ball $B_{0,x}$ , we have

\begin{equation*}\begin{split} \textbf{P}_{z^{\prime}} \big[ H_{\textbf{K}+xN} = n_2 \lt \xi_{B_{0,x}}, Y_{H_{\textbf{K}+xN}} = x^{\prime} \big] &\le \textbf{P}_{z^{\prime}}\Big[ \xi_{B_{0,x}} \gt N^{2\delta /d}\Big] \le C \exp\bigg({-}c \frac{N^{2\delta /d}}{N^{2\zeta}} \bigg) \\ &\le C \exp ({-}N^{\varepsilon} ).\end{split}\end{equation*}

By (5) on $p_{n_2}$ , and since $\zeta \lt \delta/d$ and $ N^{2\delta /d} \lt n_2 \lt n$ , we have

\begin{equation*}p_{n_2}(z^{\prime},x^{\prime})\ge \frac{c}{n_2^{d/2}}\exp\left({-}C\frac{N^{2\zeta}}{n_2}\right)\ge c\exp \big({-}N^{\varepsilon/2}\big).\end{equation*}

Here we lower-bounded $\exp\left({-}C\frac{N^{2\zeta}}{n_2}\right)$ by c. The claim (57) is proved.

We also have the bound

\begin{equation*} \widetilde{p}_{n_1}^{(x)}\big(y^{\prime}_{\ell-1},z^{\prime}\big) \le p_{n_1}\big(y^{\prime}_{\ell-1},z^{\prime}\big).\end{equation*}

We then get (summing over $z^{\prime}$ and $x^{\prime}$ ) that the contribution to $\sum_{x \in \mathbb{Z}^d} II(x)$ from Case 1 is at most

\begin{equation*}\begin{split} &\sum_{n_1 + n_2 + n_3 = n} \sum_{z^{\prime} \in \mathbb{Z}^d} \sum_{x^{\prime} \in \mathbb{Z}^d} p_{n_1}\big(y^{\prime}_{\ell-1},z^{\prime}\big) \, C \exp \big({-}N^{\varepsilon/2} \big) p_{n_2}(z^{\prime},x^{\prime}) \, p_{n_3} \big(x^{\prime},y^{\prime}_{\ell}\big) \\ &\qquad \le C \exp \big({-}N^{\varepsilon/2} \big) \sum_{n_1 + n_2 + n_3 = n} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) \\ \end{split}\end{equation*}

\begin{equation*}\begin{split} &\le C n^2 \exp \big({-}N^{\varepsilon/2} \big) p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) \\ & = o(1) \frac{n}{N^d} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{split}\end{equation*}

Case 2: $n_2 \le N^{2\delta/d}$ and $n_1 \lt \varepsilon_n n$ . Note that since $n_2 \le N^2 \le \varepsilon_n n$ for large enough N, if we put $n^{\prime}_1 = n_1 + n_2$ and $n^{\prime}_2 = n_3$ , we can upper-bound the contribution of this case by

\begin{equation*} \sum_{\substack{n^{\prime}_1 + n^{\prime}_2 = n \\ n^{\prime}_1 \le 2 \varepsilon_n n}} \sum_{x \in \mathbb{Z}^d} \sum_{x^{\prime} \in \textbf{K}+xN} p_{n^{\prime}_1}\big(y^{\prime}_{\ell-1}, x^{\prime}\big) p_{n^{\prime}_2}\big(x^{\prime}, y^{\prime}_\ell\big).\end{equation*}

Now we can make use of the corollaries stated after the proof of Proposition 1 as follows.

Case 2- (i). $N^{2+6 \varepsilon} \le n^{\prime}_1 \le 2 \varepsilon_n n$ , $\big|y^{\prime}_{\ell-1} - x^{\prime}\big| \le \big(n^{\prime}_1\big)^{\frac{1}{2}+\varepsilon}$ and $\big|x^{\prime} - y^{\prime}_\ell\big| \le \big(n^{\prime}_2\big)^{\frac{1}{2}+\varepsilon}$ . Note that for large enough N we have $n^{\prime}_2 \ge (n - 2 \varepsilon_n n) \ge N^{2+6 \varepsilon}$ . Hence, by Corollary 4.2(i) (with $\varepsilon_n$ there replaced by $2 \varepsilon_n$ ), the contribution of this case is

\begin{equation*}\begin{split} o(1) \frac{n}{N^d} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{split}\end{equation*}

Case 2- (ii). $N^{2+6 \varepsilon} \le n^{\prime}_1 \le 2 \varepsilon_n n$ but either $\big|y^{\prime}_{\ell-1} - x^{\prime}\big| \gt \big(n^{\prime}_1\big)^{\frac{1}{2}+\varepsilon}$ or $\big|x^{\prime} - y^{\prime}_\ell\big| \gt \big(n^{\prime}_2\big)^{\frac{1}{2}+\varepsilon}$ . Again, we have $n^{\prime}_2 \ge N^{2+6 \varepsilon}$ . Hence, neglecting the requirement $n^{\prime}_1 \le 2 \varepsilon_n n$ , Corollary 4.3 immediately implies that the contribution of this case is

\begin{equation*}\begin{split} o(1) \frac{n}{N^d} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{split}\end{equation*}

Case 2- (iii). $n^{\prime}_1 \lt N^{2+6\varepsilon}$ . It follows immediately from Corollary 4.4 that the contribution of this case is

\begin{equation*}\begin{split} o(1) \frac{n}{N^d} p_{n}\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{split}\end{equation*}

Case 3: $n_2 \le N^{2\delta/d}$ and $n_3 \lt \varepsilon_n n$ . By symmetry, this case can be handled very similarly to Case 2.

Lemma 4.7. We have

\begin{align*} \sum_{x \in \mathbb{Z}^d \setminus G} \textbf{P} [ A(x) ] = o(1) \frac{n}{N^d} p_n\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big). \end{align*}

Proof. By the same arguments as in Lemma 4.4(ii), we have

\begin{align*} p_n\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) \ge \frac{C}{n^{d/2}} \exp \left({-}100 (\!\log \log n)^2 \right). \end{align*}

For $x \in \mathbb{Z}^d \setminus G$ , let k be the dyadic scale that satisfies

\begin{align*} 2^k A_n^2 \sqrt{n} \le \big|x^{\prime} - y^{\prime}_{\ell-1}\big| \lt 2^{k+1} A_n^2 \sqrt{n}. \end{align*}

The same bounds hold up to constants for $\big|x^{\prime} - y^{\prime}_{\ell}\big|$ .

Then we have

\begin{equation*}\begin{split} \textbf{P} [ A(x) ] &\le \sum_{1 \le m \le n-1} \sum_{x^{\prime} \in \textbf{K}+xN} p_m\big(y^{\prime}_{\ell-1}, x^{\prime}\big) p_{n-m}\big(x^{\prime},y^{\prime}_\ell\big) \\ &\le C |\textbf{K}| \sum_{1 \le m \le n-1} \frac{1}{m^{d/2}} \frac{1}{(n-m)^{d/2}} \exp \left( -c \frac{2^{2k} A_n^4 n}{m} \right) \exp \left( -c \frac{2^{2k} A_n^4 n}{n-m} \right).\end{split}\end{equation*}

By symmetry of the right-hand side, it is enough to consider the contribution of $1 \le m \le n/2$ , which is bounded by

\begin{equation*}\begin{split} &\frac{C}{n^{d/2}} \exp \left( -c 2^{2k} A_n^4 \right) \sum_{1 \le m \le n/2} \frac{1}{m^{d/2}} \exp \left( -c \frac{2^{2k} A_n^4 n}{m} \right) \\ &\qquad \le \frac{C}{n^{d/2}} \exp \left( -c 2^{2k} A_n^4 \right) \sum_{k^{\prime} = 1}^{\lfloor \log_2 n \rfloor} \sum_{m\, :\,2^{k^{\prime}} \le n/m \lt 2^{k^{\prime}+1}} \frac{2^{k^{\prime} d/2}}{n^{d/2}} \exp \left( -c 2^{2k} A_n^4 2^{k^{\prime}} \right) \\ &\qquad \le \frac{C}{n^{d}} \exp \left( -c 2^{2k} A_n^4 \right) \sum_{k^{\prime} = 1}^{\infty} \frac{n}{2^{k^{\prime}}} \exp \left( -c 2^{2k} A_n^4 2^{k^{\prime}} + k^{\prime} d/2 \log 2 \right) \\ &\qquad \le \frac{C n}{n^{d}} \exp \left( -c 2^{2k} A_n^4 \right).\end{split}\end{equation*}

Now, summing over $x \in \mathbb{Z}^d \setminus G$ , we have that the number of the copies of the torus at dyadic scale $2^k A_n^2 \sqrt{n}$ is at most $C \frac{1}{N^d} \left( 2^k A_n^2 \sqrt{n} \right)^d$ . Hence

\begin{align*}\begin{split} \sum_{x \in \mathbb{Z}^d \setminus G} \textbf{P} [ A(x) ] &\le \frac{C n}{n^{d}} \sum_{k=0}^\infty \frac{1}{N^d} \left( 2^k A_n^2 \sqrt{n} \right)^d \exp \left( -c 2^{2k} A_n^4 \right) \\ &\le \frac{C}{n^{d/2}} \frac{n}{N^d} \sum_{k=0}^\infty \exp \left({-}c 2^{2k} A_n^4 + k d \log 2 + 2d \log A_n \right) \\ &= o(1) \frac{1}{n^{d/2}} \frac{n}{N^d} \exp \left({-}100 (\log \log n)^2\right).\\[-24pt]\end{split}\end{align*}

Lemma 4.8. We have

\begin{align*} \sum_{x_1 \not= x_2 \in \mathbb{Z}^d} \textbf{P} [ A(x_1) \cap A(x_2) ] = o(1) \frac{n}{N^d} p_n\big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big). \end{align*}

Proof. The summand on the left-hand side is bounded above by

\begin{equation*}\begin{split} \textbf{P} [ A(x_1) \cap A(x_2) ] \le \sum_{m_1 + m_2 + m_3 = n} \sum_{\substack{x^{\prime}_1 \in \textbf{K}+x_1N \\ x^{\prime}_2 \in \textbf{K}+x_2N}} &\left[ p_{m_1}\big(y^{\prime}_{\ell-1}, x^{\prime}_1\big) p_{m_2}\big(x^{\prime}_1, x^{\prime}_2\big) p_{m_3}\big(x^{\prime}_2, y^{\prime}_{\ell}\big) \right.\\ &\left. +\, p_{m_1}\big(y^{\prime}_{\ell-1}, x^{\prime}_2\big) p_{m_2}\big(x^{\prime}_2, x^{\prime}_1\big) p_{m_3}\big(x^{\prime}_1, y^{\prime}_{\ell}\big) \right]\kern-2pt.\end{split}\end{equation*}

By symmetry it is enough to consider the first term inside the summation. The estimates are again modelled on the proof of Proposition 2.3.

Case 1: $m_1 + m_2 \ge n/2$ and $\big|x^{\prime}_2 - y^{\prime}_{\ell-1}\big| \le 2 C_1 \sqrt{n} \sqrt{\log n}$ . In this case we can use Corollary 4.1 with $y^{\prime} = y^{\prime}_{\ell-1}$ and $y^{\prime\prime} = x^{\prime}_2$ to perform the summation over $x^{\prime}_1$ and $x_1$ and get the upper bound

(58) \begin{equation} C \frac{n}{N^d} \sum_{m^{\prime}_1 + m^{\prime}_2 = n} \sum_{x_2 \in \mathbb{Z}^d} \sum_{x^{\prime}_2 \in \textbf{K}+x_2N} p_{m^{\prime}_1}\big(y^{\prime}_{\ell-1}, x^{\prime}_2\big) p_{m^{\prime}_2} \big(x^{\prime}_2, y^{\prime}_\ell\big),\end{equation}

where we have written $m^{\prime}_1 = m_1 + m_2$ and $m^{\prime}_2 = m_3$ . Using Corollary 4.1 again, this time with $y^{\prime} = y^{\prime}_{\ell-1}$ and $y^{\prime\prime} = y^{\prime}_\ell$ , yields the upper bound

(59) \begin{equation} C \left( \frac{n}{N^d} \right)^2 p_{n} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big) = o(1) \frac{n}{N^d} p_{n} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{equation}

Case 2: $m_1 + m_2 \ge n/2$ and $2 C_1 \sqrt{n} \sqrt{\log n} \lt \big|x^{\prime}_2 - y^{\prime}_{\ell-1}\big| \le n^{\frac{1}{2}+\varepsilon}$ . We are going to use that $\varepsilon \le 1$ , which we can clearly assume. First sum over all $x^{\prime}_1 \in \mathbb{Z}^d$ to get the upper bound

(60) \begin{equation} C n \sum_{m^{\prime}_1 + m^{\prime}_2 = n} \sum_{x_2 \in \mathbb{Z}^d} \sideset{}{^{\prime}}\sum_{x^{\prime}_2 \in \textbf{K}+x_2N} p_{m^{\prime}_1}\big(y^{\prime}_{\ell-1}, x^{\prime}_2\big) p_{m^{\prime}_2} \big(x^{\prime}_2, y^{\prime}_\ell\big),\end{equation}

where the primed summation denotes the restriction $2 C_1 \sqrt{n} \sqrt{\log n} \lt \big|x^{\prime}_2 - y^{\prime}_{\ell-1}\big| \le n^{\frac{1}{2}+\varepsilon}$ . The choice of $C_1$ (recall (41)) implies that $p_{m^{\prime}_1}$ is $o\big(1/n^{1+3d/2} N^d\big)$ . By the triangle inequality we also have $\big|y^{\prime}_\ell - x^{\prime}_2\big| \gt C_1 \sqrt{n} \sqrt{\log n}$ . Using the LCLT for $p_{m^{\prime}_2}$ we get that

(61) \begin{equation} p_{m^{\prime}_2} \big(x^{\prime}_2, y^{\prime}_\ell\big) \le \frac{C}{\big(m^{\prime}_2\big)^{d/2}} \exp \Big({-}d C_1^2 n \log n / m^{\prime}_2 \Big) \le \frac{C}{n^{d/2}} \exp \Big({-}d C_1^2 \log n \Big) \le C p_{n} \big(y^{\prime}_{\ell-1}, y^{\prime}_\ell\big).\end{equation}

Substituting this bound and $p_{m^{\prime}_1} = o\big(1/n^{1+3d/2} N^d\big)$ into (60), we get

\begin{equation*}\begin{split} &C n\, o(1) \left(\frac{1}{n \cdot n^{3d/2} \cdot N^d} \right) \sum_{m^{\prime}_1 + m^{\prime}_2 = n} \sideset{}{^{\prime}}\sum_{x^{\prime}_2} p_n\big(y^{\prime}_{\ell-1},y^{\prime}_\ell\big) \\ &\qquad \le o(1) \left( \frac{n}{N^d} \right) p_{n}\big(y^{\prime}_{\ell-1},y^{\prime}_\ell\big) \sideset{}{^{\prime}}\sum_{x^{\prime}_2} \frac{1}{(n^{1/2+\varepsilon})^d} \\ &\qquad = o\left( \frac{n}{N^d} \right) p_{n}\big(y^{\prime}_{\ell-1},y^{\prime}_\ell\big).\end{split}\end{equation*}

Case 3: $m_1 + m_2 \ge n/2$ and $\big|x^{\prime}_2 - y^{\prime}_{\ell-1}\big| \gt n^{\frac{1}{2}+\varepsilon}$ . Summing over all $x^{\prime}_1 \in \mathbb{Z}^d$ , we get the transition probability $p_{m_1+m_2}\big(y^{\prime}_{\ell-1}, x^{\prime}_2\big)$ . This is stretched-exponentially small, and hence this case satisfies the required bound.

Case 4: $m_2 + m_3 \ge n/2$ . By symmetry, this case can be handled analogously to Cases 1–3.

4.3. Proof of Proposition 2.1

Proof of Proposition 2.1. We start with the proof of the second claim. We denote the error term in (12) by E, which we claim to satisfy $|E| \le E_1 + E_2 + E_3 + E_4$ , with

\begin{equation*}\begin{split} E_1 &= \textbf{P}_{o} \left[ \left\{\frac{S n}{N^d} \lt \left(\sqrt{\log\log n}\right)^{-1} \right\} \cup \left\{\frac{S n}{N^d} \gt \log N \right\}\right]\kern-2pt, \\ E_2 &= \textbf{P}_{o} \left[ \text{$\exists \ell: \ 1 \le \ell \le \log N\frac{N^d}{n}$ such that $Y_{\ell n} \in \varphi^{-1}( B(\textbf{x}_\textbf{0}, N^{\zeta}) )$} \right]\kern-2pt, \end{split}\end{equation*}

\begin{equation*}\begin{split} E_3 &= \textbf{P}_{o} \left[ \text{$\exists\ t\, :\,T \le t \lt S n$ such that $Y_t \in \varphi^{-1}(\textbf{K})$} \right]\kern-2pt, \\ E_4 &= \textbf{P}_{o} \left[ \text{$\exists \ell: \ 1 \le \ell \le \log N\frac{N^d}{n}$ such that $| Y_{\ell n}-Y_{(\ell-1)n} |\gt f(n)n^{\frac{1}{2}}$} \right]\kern-2pt.\end{split}\end{equation*}

Since $T\leq Sn$ , we have

\begin{align*}\left|\textbf{P}_{o} \left[ Y_t \not\in \varphi^{-1}(\textbf{K}),\, 0 \le t \lt T \right]- \textbf{P}_{o} \left[ Y_t \not\in \varphi^{-1}(\textbf{K}),\, 0 \le t \lt Sn \right]\right|\leq E_3.\end{align*}

By the Markov property, for $\big(\tau, (y_\ell, \textbf{y}_\ell)_{\ell=1}^\tau \big) \in \mathcal{G}_{\zeta,C_1}$ ,

\begin{align*}\begin{split} &\prod_{\ell=1}^\tau \textbf{P}_{y^{\prime}_{\ell-1}} \left[ Y_n = y^{\prime}_{\ell},\, \text{$Y_t \not\in \varphi^{-1}(\textbf{K})$ for $0 \le t \lt n$} \right] \\ & \quad= \textbf{P}_{o}\left[ \begin{array}{l}Y_{n \ell}=y^{\prime}_{\ell}\,\, \textrm{for}\,\, 0 \le \ell \le \tau;\, Y_t \not\in \varphi^{-1}(\textbf{K}) \,\, \textrm{for}\,\, 0 \le t \lt \tau n;\, Y_{n\ell } \not\in \\\varphi^{-1}( B(\textbf{x}_\textbf{0}, N^{\zeta}) ) \, \textrm{for}\, 0 \lt \ell \le \tau \end{array}\right]\kern-2pt.\end{split}\end{align*}

We denote the probability on the right-hand side by $p\big(\tau, (y_\ell, \textbf{y}_\ell)_{\ell=1}^\tau\big)$ . On the event of the right-hand side, since $y_{\tau}\in D^{c}$ , we have $S=\tau$ . We claim that

(62) \begin{equation}\begin{split} &\left| \sum_{\big( \tau, (y_\ell, \textbf{y}_\ell)_{\ell=1}^\tau\big) \in \mathcal{G}_{\zeta,C_1}}p\big(\tau, (y_\ell,\textbf{y}_\ell)_{\ell=1}^\tau\big) - \textbf{P}_{o}\left[\text{$Y_t \not\in \varphi^{-1}(\textbf{K})$ for $0 \le t \lt Sn$}\right]\right|\\ &\quad \leq E_1 + E_2 + E_4.\end{split}\end{equation}

Let $\mathcal{E}_1, \mathcal{E}_2, \mathcal{E}_4$ be the events in the definitions of $E_1, E_2, E_4$ respectively. Let

\begin{align*}A\big(\tau, (y_\ell,\textbf{y}_\ell)_{\ell=1}^{\tau}\big)\,:\!= \left\{ \begin{array}{l}{\ }Y_{n \ell}=y^{\prime}_{\ell}\,\, \textrm{for} \,\, 0 \le \ell \le \tau\,; \,Y_t \not\in \varphi^{-1}(\textbf{K})\,\, \textrm{for} \,\,0 \le t \lt \tau n\,; \,Y_{n\ell } \not\in \\ \varphi^{-1}( B(\textbf{x}_\textbf{0}, N^{\zeta}) )\,\, \textrm{for} \,\, 0 \lt \ell \le \tau\, \end{array}\right\}.\end{align*}

Then we have

\begin{equation*}\begin{split} &\textbf{P}_{o}\left[\left\{\text{$Y_t \not\in \varphi^{-1}(\textbf{K})$ for $0 \le t \lt Sn$}\right\} \cap \mathcal{E}^{c}_1 \cap \mathcal{E}^{c}_2 \cap \mathcal{E}^{c}_4\right] \\ &\quad = \sum_{ \frac{N^d}{n\sqrt{\log \log n}} \le \tau \le \frac{N^d \log N}{n}} \textbf{P}_o \left[\left\{S=\tau\right\} \cap \left\{\text{$Y_t \not\in \varphi^{-1}(\textbf{K})$ for $0 \le t \lt \tau n$}\right\} \cap \mathcal{E}^{c}_2 \cap \mathcal{E}^{c}_4 \right] \\ &\quad = \sum_{\big( \tau, (y_\ell, \textbf{y}_\ell)_{\ell=1}^\tau \big) \in \mathcal{G}_{\zeta,C_1}} \textbf{P}_o \left[ A\big(\tau, (y_\ell,\textbf{y}_\ell)_{\ell=1}^{\tau}\big) \cap \mathcal{E}^{c}_2 \cap \mathcal{E}^{c}_4 \right].\end{split}\end{equation*}

From the above, the claim (62) follows.

The proof of the second claim of Proposition 2.1 follows subject to Lemmas 4.9, 4.11, and 4.12 below, which show that $E_j \to 0$ for $j= 2, 3, 4$ . We have already shown $E_1 \to 0$ in Lemma 2.2.

Similarly, the first claim of the proposition follows from Lemmas 2.2, 4.9, and 4.12.

We bound $E_2$ , $E_3$ , and $E_4$ in the following lemmas.

Proof. Since the number of points in $B(\textbf{x}_\textbf{0}, N^{\zeta})$ is $O\big(N^{d\zeta}\big)$ , and since we are considering times after the first stretch, the random walk is well mixed, so by Lemma 2.1 the probability of visiting any point in the torus is $O\big(1/N^d\big)$ . Using a union bound we have

\begin{align*}E_2\leq C \log N\frac{N^d}{n} N^{d\zeta} \frac{1}{N^d}= C \log N \frac{N^{d \zeta}}{n}.\end{align*}

Since $\zeta \lt \delta/d$ , we have

\begin{align*} E_2 \to 0, \quad \text{as $N\to\infty$}.\\[-36pt]\end{align*}

Before we bound the error term $E_3$ , we first introduce the following lemma. Let $T^{(i)}_0 = \inf\Big\{t\ge 0: Y^{(i)}_t = 0 \Big\}$ , where we recall that $Y^{(i)}$ denotes the ith coordinate of the d-dimensional lazy random walk. We will denote by $t_0$ an instance of $T^{(i)}_0$ .

Lemma 4.10. For all $1 \le i \le d$ , for any integer y such that $-t_0 \le y \lt 0$ and $0 \lt t_0 \le n$ we have

\begin{equation*}\begin{split}\textbf{E}_y \left[Y^{(i)}_n \,|\, T^{(i)}_0 = t_0,\, Y_n^{(i)} \gt 0 \right] &\leq C n^{\frac{1}{2}},\\\textbf{E}_y \left[\Big(Y^{(i)}_n\Big)^2 \,|\, T^{(i)}_0 = t_0,\, Y_n^{(i)} \gt 0 \right] &\leq C n.\end{split}\end{equation*}

Proof. Using the Markov property at time $t_0$ , we get

\begin{align*}\textbf{E}_y\left[Y^{(i)}_n \,|\, T^{(i)}_0=t_0, Y_n^{(i)} \gt 0 \right]&= \textbf{E}_0 \left[Y_{n-t_0}^{(i)} \,|\, Y_{n-t_0}^{(i)} \gt 0 \right]=\frac{\textbf{E}_0\left[Y^{(i)}_{n-t_0}\textbf{1}_{\big\{Y^{(i)}_{n-t_0} \gt 0\big\}}\right]} {\textbf{P}_0\left[Y^{(i)}_{n-t_0}\gt 0\right]} \\&\leq C_0 \left(\textbf{E}_0\left[\Big(Y^{(i)}_{n-t_0}\Big)^2\right]\right)^{\frac{1}{2}} \leq C (n-t_0)^{\frac{1}{2}}\leq C n^{\frac{1}{2}},\end{align*}

where the third step is due to the Cauchy–Schwarz inequality and $\textbf{P}_0\left[Y^{(i)}_{n-t_0}\gt 0\right]\geq c_0$ for some $c_0 \gt 0$ , and the second-to-last step is due to $\textbf{E}_0\left[\Big(Y^{(i)}_{n-t_0}\Big)^2\right] = (n-t_0)/2d$ .

We can similarly bound the conditional expectation of $\Big(Y^{(i)}_n\Big)^2$ as follows:

\begin{align*}&\textbf{E}_y\left[\Big(Y^{(i)}_n\Big)^2| T^{(i)}_0=t_0, Y^{(i)}_n\gt 0\right]=\textbf{E}_0\left[\Big(Y^{(i)}_{n-t_0}\Big)^2|Y^{(i)}_{n-t_0}\gt 0\right] \\&\qquad =\frac{\textbf{E}_0\left[\Big(Y^{(i)}_{n-t_0}\Big)^2\textbf{1}_{\big\{Y^{(i)}_{n-t_0} \gt 0\big\}}\right]} {\textbf{P}_0\left[Y^{(i)}_{n-t_0}\gt 0\right]}\leq C_0 \left(\textbf{E}_0\left[\Big(Y^{(i)}_{n-t_0}\Big)^2\right]\right)\leq C (n-t_0)\leq C n.\end{align*}

Let us consider the lazy random walk $(Y_t)_{t\ge 0}$ in multiples of n steps. Let

\begin{equation*} s_1 = \min \{ \ell n\, :\,\ell n \ge T \} - T,\end{equation*}

and similarly, let

\begin{equation*} s_{r+1} = r n + \min \{ \ell n\, :\,\ell n \ge T \} - T, \quad r \ge 1.\end{equation*}

We let $M_0 = L-Y^{(i)}_{T+s_1}$ and $M_r = L-Y^{(i)}_{T+s_{r+1}}$ for $r\ge 1$ . We have that $(M_r)_{r\ge 0}$ is a martingale. Let $\tilde{S} = \inf \{ r \ge 0: M_r \le 0\}$ , and we are going to bound $\textbf{P} \big[ \tilde{S} \gt N^{\varepsilon_1} \big]$ for some small $\varepsilon_1$ that we will choose in the course of the proof. We are going to adapt an argument in [Reference Levin, Peres and Wilmer10, Proposition 17.19] to this purpose.

Define

\begin{equation*}T_h = \inf\big\{r\geq 0: \text{$M_r\leq 0$ or $M_r\geq h$} \big\},\end{equation*}

where we set $h= \sqrt{n}\sqrt{N^{\varepsilon_1}}$ . Let $(\mathcal{F}_r)_{r \ge 0}$ denote the filtration generated by $(M_r)_{r \ge 0}$ . We have

(63) \begin{equation} \textrm{Var}\big(M_{r+1} \,|\, \mathcal{F}_r\big) = n \sigma^2 \quad \text{for all $r \ge 0$};\end{equation}

here, recall that $\sigma^2$ is the variance of $Y^{(i)}_1$ .

We first estimate $\textbf{E} \big[ M_0 \,|\, \tilde{S} \gt 0 \big]$ . Since $0 \le s_1 \lt n$ , by the same argument as in Lemma 4.10 we have that

\begin{equation*}\begin{split}\textbf{E} \Big[ M_0 \,|\, Y^{(i)}_{T+s_1} \lt L\Big]&= \textbf{E} \Big[ L-Y^{(i)}_{T+s_1} \,|\, L - Y^{(i)}_{T+s_1} \gt 0 \Big]\leq C n^{\frac{1}{2}}.\end{split}\end{equation*}

We first bound $\textbf{P}\big[M_{T_h}\ge h \,|\, M_0\big]$ . Since $\big(M_{r\wedge T_h}\big)$ is bounded, by the optional stopping theorem, we have

\begin{equation*}\begin{split}M_0&= \textbf{E} \big[M_{T_h}|M_0\big]= \textbf{E}\left[M_{T_h}\textbf{1}_{\big\{M_{T_h}\leq 0\big\}}|M_0\right] +\textbf{E}\left[M_{T_h}\textbf{1}_{\big\{M_{T_h}\geq h\big\}}|M_0\right]\\&= -m_{-}(h) + \textbf{E}\left[M_{T_h}\textbf{1}_{\big\{M_{T_h}\geq h\big\}}|M_0\right]\\&\geq -m_{-}(h) + h\textbf{P}\left[M_{T_h}\geq h |M_0 \right],\end{split}\end{equation*}

where we denote $\textbf{E}\Big[M_{T_h}\textbf{1}_{\{M_{T_h}\leq 0\}}|M_0\Big]$ by $-m_{-}(h) \leq 0$ , and the last step is due to $M_{T_h} \textbf{1}_{\{M_{T_h} \ge h\}} \ge h \textbf{1}_{\{M_{T_h} \ge h\}}$ . Hence, we have

\begin{align*}M_0+m_{-}(h) \geq h\textbf{P}\left[M_{T_h}\geq h|M_0\right]\kern-2pt.\end{align*}

We bound $m_{-}(h)$ using Lemma 4.10:

\begin{align*}m_{-}(h)\leq \max_{y\leq L} \textbf{E}_y\left[Y^{(i)}_n-L|Y^{(i)}_n \gt L \right]\leq C n^{\frac{1}{2}}.\end{align*}

Hence, we have

\begin{align*}\textbf{P}[M_{T_h}\geq h \,|\, M_0]\leq \frac{M_0}{h}+\frac{C n^{\frac{1}{2}}}{h}.\end{align*}

We now estimate $\textbf{P} [ T_h \geq r \,|\, M_0 ]$ . Let $G_r = M^2_r-hM_r-\sigma^2 nr$ . The sequence $(G_r)$ is a martingale by (63). We can bound both the ‘overshoot’ above h and the ‘undershoot’ below 0 by Lemma 4.10. For the ‘undershoot’ below 0 we have

\begin{equation*}\begin{split} \textbf{E} \Big[ \big(M_{T_h} - h\big) M_{T_h} \,|\, M_{T_h} \le 0, M_0 \Big] &= \textbf{E} \Big[ M_{T_h}^2 \,|\, M_{T_h} \le 0, M_0 \Big] + \textbf{E} \Big[ -h M_{T_h} \,|\, M_{T_h} \le 0, M_0 \Big] \\ &\le C n + C h n^{1/2}.\end{split}\end{equation*}

For the ‘overshoot’ above h, write $M_{T_h} =: N_{T_h}+h$ ; then we have

\begin{equation*}\big(M_{T_h}-h\big)M_{T_h} = N_{T_h}\big(h+N_{T_h}\big).\end{equation*}

Hence

\begin{equation*}\begin{split} \textbf{E} \Big[ \big(M_{T_h} - h\big) M_{T_h} \,|\, M_{T_h} \ge h, M_0 \Big] &= \textbf{E} \Big[ h N_{T_h} \,|\, N_{T_h} \ge 0, M_0 \Big] + \textbf{E} \Big[ N_{T_h}^2 \,|\, N_{T_h} \ge 0, M_0 \Big]\\ &\le C h n^{1/2} + C n.\end{split}\end{equation*}

For $r \lt T_h$ , we have $(M_r - h) M_r \lt 0$ . Therefore, we have

\begin{align*}\textbf{E}\left[M^2_{r\wedge T_h}-hM_{r\wedge T_h}| M_0\right]&\leq C h n^{1/2} + C n.\end{align*}

Since $(G_{r\wedge T_h})$ is a martingale,

\begin{equation*}\begin{split}-hM_0 \le G_0\le \textbf{E}\big[ G_{r\wedge T_h}| M_0\big]&= \textbf{E}\Big[M^2_{r\wedge T_h}-hM_{r\wedge T_h}| M_0\Big]-\sigma^2 n\textbf{E}[r\wedge T_h| M_0]\\&\le Cn^{\frac{1}{2}}h+Cn-\sigma^2 n \textbf{E}[r\wedge T_h| M_0].\end{split}\end{equation*}

We conclude that

$$\textbf{E}[r\wedge T_h \,|\, M_0 ] \leq \frac{h\Big(M_0+Cn^{\frac{1}{2}}\Big)+Cn}{\sigma^2 n}.$$

Letting $r\to\infty$ , by the monotone convergence theorem,

\begin{equation*}\textbf{E}[T_h \,|\, M_0 ] \leq \frac{h\Big(M_0+Cn^{\frac{1}{2}}\Big)+Cn}{\sigma^2 n},\end{equation*}

where $h= \sqrt{n}\sqrt{N^{\varepsilon_1}}$ . This gives

\begin{equation*}\begin{split}\textbf{P}[T_h\gt N^{\varepsilon_1}|M_0]\le \frac{1}{N^{\varepsilon_1}}\left[\frac{\sqrt{n}\sqrt{N^{\varepsilon_1}}M_0+Cn\sqrt{N^{\varepsilon_1}}+Cn}{\sigma^2 n}\right]\kern-2pt.\end{split}\end{equation*}

Taking expectations of both sides, we have

\begin{equation*}\begin{split}\textbf{P}[T_h\gt N^{\varepsilon_1}]&\le \frac{1}{N^{\varepsilon_1}}\left[\frac{\sqrt{n}\sqrt{N^{\varepsilon_1}}\textbf{E} M_0+Cn\sqrt{N^{\varepsilon_1}}+Cn}{\sigma^2 n}\right]\\&= \frac{\textbf{E} M_0}{\sigma^2\sqrt{n}\sqrt{N^{\varepsilon_1}}}+\frac{C}{\sigma^2\sqrt{N^{\varepsilon_1}}}+\frac{C}{\sigma^2N^{\varepsilon_1}}\le \frac{C}{\sqrt{N^{\varepsilon_1}}}.\end{split}\end{equation*}

Combining the above bounds, we get

\begin{align*}\textbf{P}\big[\tilde{S} \gt N^{\varepsilon_1}\big]&\leq\textbf{P}[ M_{T_h} \ge h ] + \textbf{P}[ T_h \gt N^{\varepsilon_1} ]\le \frac{\textbf{E} [ M_0 ]}{h} +\frac{C n^{\frac{1}{2}}}{h}+ \frac{C}{\sqrt{N^{\varepsilon_1}}}\leq \frac{C}{\sqrt{N^{\varepsilon_1}}}.\end{align*}

We now bound the probability that a copy of $\textbf{K}$ is hit between times T and $s_{N^{\varepsilon_1}}$ .

We first show that the probability that the lazy random walk on the torus is in the ball $B(\textbf{x}_0, N^\zeta)$ at time T goes to 0. Indeed, we have

\begin{align*}&\textbf{P}_{o}\left[Y_{T}\in \varphi^{-1}\left(B(\textbf{x}_0, N^{\zeta})\right)\right]= \sum_{y^{\prime} \in\partial ({-}L,L)^d \cap \varphi^{-1}(B(\textbf{x}_0, N^{\zeta}))} \textbf{P}_{o}\left[Y_{T}=y^{\prime}\right]\\&\qquad \leq C N^{\zeta(d-1)} \frac{L^{d-1}}{N^{d-1}} \frac{C}{L^{d-1}}= C N^{(\zeta-1)(d-1)},\end{align*}

where we have $\zeta\lt\delta/d \lt 1$ , so the last expression goes to 0. Here we used that $\textbf{P}_{o} [ Y_T = y^{\prime} ] \le C/L^{d-1}$ , for example using a half-space Poisson kernel estimate [Reference Lawler and Limic9, Theorem 8.1.2]. As for the number of terms in the sum, we have that there are $C L^{d-1}/N^{d-1}$ copies of the torus within $\ell_{\infty}$ distance $\le N$ of the boundary $\partial ({-}L,L)^d$ . Considering the intersection of the union of balls $\varphi^{-1}(B(\textbf{x}_0, N^{\zeta}))$ and the boundary $\partial ({-}L, L)^d$ , the worst case is that within a single copy of the torus the intersection has size at most $C N^{\zeta(d-1)}$ .

Condition on the location $y^{\prime}$ of the walk at the exit time T. For $y^{\prime}\notin \varphi^{-1}\left(B(\textbf{x}_0, N^{\zeta})\right)$ , we first bound the probability of hitting $\textbf{K}$ between the times between 0 and $s_2$ . After time $s_2$ , the random walk is well mixed, and we can apply a simpler union bound.

We thus have the upper bound

\begin{align*}\sum_{t=0}^{s_2}\sum_{x^{\prime}\in\varphi^{-1}(\textbf{K})}p_t(y^{\prime},x^{\prime})\leq \textbf{P}\left[\max_{0\leq t\leq s_2}|Y^{(i)}_{t}-y^{\prime}| \geq n^{\frac{1}{2}+\varepsilon}\right] +\sum_{\substack{x^{\prime}\in\varphi^{-1}(\textbf{K})\\|x^{\prime}-y^{\prime}|\leq n^{\frac{1}{2}+\varepsilon}}}G(y^{\prime},x^{\prime}).\end{align*}

The first term is stretched-exponentially small by the martingale maximal inequality (3). The Green’s function term is bounded by Lemma 4.1(iii).

After time $s_2$ , by Lemma 2.1, we have that

\begin{align*}\sum_{t=s_2}^{s_{N^{\varepsilon_1}}}\textbf{P}_y\left[Y_t\in\varphi^{-1}(K)\right]\leq n\cdot N^{\varepsilon_1}|\textbf{K}|\frac{C}{N^d}= C\, N^{\delta+\varepsilon_1-d}.\end{align*}

Therefore, combining the above upper bounds, we have the required result:

\begin{align*}E_3\leq C \cdot N^{-\frac{\varepsilon_1}{2}} +C \cdot N^{\delta-d+2 \delta \varepsilon} + C\cdot N^{\delta-d+\varepsilon_1} \to 0,\quad \text{as $N\to\infty$,}\end{align*}

if $\varepsilon$ and $\varepsilon_1$ are sufficiently small.

Proof. By the martingale maximal inequality (3), we have that

\begin{align*}E_4 \le C e^{-cf(n)^2}\frac{N^d \log N}{n}.\end{align*}

Taking, say, $C_1 \gt \sqrt{d/c}$ implies that if $f(n)\geq C_1\sqrt{\log N}$ , we have

\begin{align*}E_4\to 0, \quad\text{as $N\to\infty$}.\\[-36pt]\end{align*}

4.4. Proof of Proposition 2.4

Proof. By the martingale maximal inequality (3) used in the second step, we have

\begin{equation*}\begin{split} \textbf{P}_{y^{\prime}_{\ell-1}}\Big[\big|Y_n-y^{\prime}_{\ell-1}\big|\gt\sqrt{n}\big(10\log\log n\big)\Big] &=\textbf{P}_{0}\Big[|Y_n|\gt\sqrt{n}\big(10\log\log n\big)\Big]\\ &\le \exp\Big({-}c 100\big(\log\log n\big)^2\Big).\end{split}\end{equation*}

Hence we have

\begin{equation*}\begin{split} &\textbf{E} \left[\# \left\{ 1 \le \ell \le \frac{N^d}{n}C_1\log N\, :\,|Y_{n\ell} - Y_{n(\ell-1)}| \gt 10 \sqrt{n} \log\log n \right\} \right] \\ &\qquad \le \frac{N^d}{n} C_1 \log N \exp\Big({-}c\big(\log\log n\big)^2\Big) \le \frac{N^d}{n} C \exp \Big({-}(c/2) \big(\log\log n\big)^2 \Big).\end{split}\end{equation*}

By Markov’s inequality, it follows that

\begin{equation*}\begin{split} &\textbf{P} \left[ \# \left\{ 1 \le \ell \le \frac{N^d}{n}C_1\log N\, :\,|Y_{n\ell} - Y_{n(\ell-1)}| \gt 10 \sqrt{n} \log\log n \right\} \ge \frac{N^d}{n} (\log \log n)^{-1} \right] \\ &\qquad \le \frac{\frac{N^d}{n} C \exp \Big({-}(c/2) \big(\log\log n\big)^2 \Big)}{\frac{N^d}{n} \big(\log \log n\big)^{-1}} \le C \frac{\exp \Big({-}(c/2) \big(\log\log n\big)^2 \Big)}{\big(\log \log n\big)^{-1}} \to 0,\end{split}\end{equation*}

as $N\to\infty$ .