If $\sqrt{t} \ge a \ge \sqrt{s}$, then

$$ 
\begin{aligned}
& \left( \max(\sqrt{t}-a, 0) \right)^2 - \left( \max(\sqrt{s}-a, 0) \right)^2 \\
&= (\sqrt{t}-a)^2 \le (\sqrt{t}-\sqrt{s})^2 \\
&= \left( \frac{t-s}{\sqrt{t}+\sqrt{s}} \right)^2 \le (t-s) \frac{t+s}{(\sqrt{t}+\sqrt{s})^2} \le t-s.
\end{aligned}
$$

In each case, we obtain the desired inequality.

Now, thanks to Lemma 5.3 applied with

<figure><img src="image_8.png" /></figure>

$$ a = \sqrt{2\lambda/N}, \quad t = \phi_X^2 h(x), \quad s = \frac{1}{N} L_N(x), $$

we have

$$ \left\| \tilde{h} - \frac{1}{N} L_N \right\|_1 \le \sum_{x \in \mathbb{Z}^d} \left| \phi_X^2 h(x) - \frac{1}{N} L_N(x) \right| = \left\| \phi_X^2 h - \frac{1}{N} L_N \right\|_1 \le \Gamma_3, $$

where the last inequality is a consequence of inequalities (5.41) and (5.43). We deal finally with the second term of formula (5.46). We set $a = \sqrt{2\lambda/N}$ and we define

$$ B = \left\{ x \in \mathbb{Z}^d : \frac{1}{N} L_N(x) \ge a^2 \right\}. $$

We have

$$ 
\begin{aligned}
\left\| \frac{1}{N} L_N - \frac{1}{N} L_N \right\|_1 &= \sum_{x \in B} \left| \left( \frac{1}{\sqrt{N}} f_N(x) - a \right)^2 - \frac{1}{N} L_N(x) \right| + \sum_{x \in \mathbb{Z}^d \setminus B} \frac{1}{N} L_N(x) \\
&\le \sum_{x \in B} \left| a^2 - 2a \frac{1}{\sqrt{N}} f_N(x) \right| + a^2 \left| \text{supp } f_N \right| \\
&\le 3a \sum_{x \in B} \left| \frac{1}{\sqrt{N}} f_N(x) \right| + a^2 \left| \text{supp } f_N \right| \\
&\le 3a \left| \text{supp } f_N \right|^{1/2} \left\| \frac{1}{N} L_N \right\|_1^{1/2} + a^2 \left| \text{supp } f_N \right| \\
&\le 3 \sqrt{\frac{2\lambda}{N}} cn^d + \frac{2\lambda}{N} cn^d \le 4 \sqrt{\frac{2c\lambda}{n^2}},
\end{aligned}
$$

where the last inequalities come from the fact that, on the event $\mathcal{A}$, the range of the random walk has cardinality at most $cn^d$, and that $\lambda/n^2 = n^{\rho-2}$ goes to 0 as $n$ goes to $\infty$. Plugging the previous inequalities in inequality (5.46), we obtain that

$$ \left\| \tilde{h} - \frac{1}{N} L_N \right\|_1 \le \Gamma_3 + 4 \sqrt{\frac{2c\lambda}{n^2}}. $$

We conclude that, if $h \in C(g, N, r(N))$ and $f_N \in \mathcal{F}$, then $\tilde{h} \in V(\mathcal{F}, \Gamma_4, N)$, where

$$ \Gamma_4 = \Gamma_3 + 4 \sqrt{\frac{2c\lambda}{n^2}}, $$
and

$$V(\mathcal{F}, \Gamma, N) = \left\{ h \in \ell^1(\mathbb{Z}^d) : \exists \ f \in \mathcal{F} \ \left\| h - \frac{1}{N} f^2 \right\|_1 \le \Gamma \right\}.$$

Together with inequalities (5.44) and (5.45), we obtain finally that

$$E(e^{-|R_N|}; \mathcal{B}) \le 2(\Gamma_2 + 1) \exp \left( \frac{N}{\lambda}\Gamma_3 + \frac{2\Gamma_1}{\lambda} + c' \frac{n^{2d}}{[\Gamma_1]} - \inf \left\{ \left| \left\{ x \in \mathbb{Z}^d : \tilde{h}(x) > 0 \right\} \right| + \frac{N - \Gamma_2}{2} \left( \max \left( \sqrt{\mathcal{E}(\sqrt{\tilde{h}})} - \frac{4\sqrt{\Gamma_3}}{n}, 0 \right) \right)^2 : \tilde{h} \in V(\mathcal{F}, \Gamma_4, N) \right\} \right).$$

We have reached our goal, indeed, this last upper bound depends only on the set $\mathcal{F}$ and it is uniform over $g \in \mathcal{G}(X, M, \eta)$. We report this upper bound in the successive decompositions in sums presented in formulas (5.7) and (5.21). These two formulas imply that

$$E(e^{-|R_N|}; f_N \in \mathcal{F}, \mathcal{A}) \le |\mathcal{X}(N, \delta)| \times |\mathcal{G}(X, M, \eta)| \times \sup E(e^{-|R_N|}; \mathcal{B}),$$

where the supremum is over $X \in \mathcal{X}(N, \delta)$ and $g \in \mathcal{G}(X, M, \eta)$. Using the combinatorial bounds (5.4),(5.5),(5.17), and recalling the definition of $C_0$ given in (5.6), we have, for some constant $c_d$ depending on the dimension only,

$$|\mathcal{X}(N, \delta)| \le \exp(c_d C_0), \quad |\mathcal{G}| \le \left(\frac{\sqrt{N}}{\eta}\right)^{\frac{n^d}{M^d}} c_d C_0.$$

Together with our upper bound on the last expectation, this yields

$$(5.47) \quad \ln E(e^{-|R_N|}; f_N \in \mathcal{F}, \mathcal{A}) \le C_1 - \inf \left\{ \left| \left\{ x \in \mathbb{Z}^d : \tilde{h}(x) > 0 \right\} \right| + \frac{N - \Gamma_2}{2} \left( \max \left( \sqrt{\mathcal{E}(\sqrt{\tilde{h}})} - \frac{4\sqrt{\Gamma_3}}{n}, 0 \right) \right)^2 : \tilde{h} \in V(\mathcal{F}, \Gamma_4, N) \right\},$$

where

$$C_1 = c_d C_0 + \frac{n^d}{M^d} c_d C_0 \ln \left( \frac{\sqrt{N}}{\eta} \right) + \ln (2(\Gamma_2 + 1)) + \frac{N}{\lambda}\Gamma_3 + \frac{2\Gamma_1}{\lambda} + c' \frac{n^{2d}}{[\Gamma_1]}.$$

We compute the asymptotic expansion of these different terms in powers of $n$. Recalling the definitions of $C_0$ and $\Gamma_3$ given in (5.6) and (5.43), we obtain, using (5.28),

$$(5.48) \qquad C_0 \underset{n \to \infty}{\overset{\ln}{\sim}} n^{\alpha 2^*}, \quad \Gamma_3 \underset{n \to \infty}{\overset{\ln}{\sim}} n^{\beta-1} + n^{-4\alpha/d}.$$

Together with the relations (5.30), this yields

$$(5.49) \qquad C_1 \underset{n \to \infty}{\overset{\ln}{\sim}} n^{d(1-\beta)+\alpha 2^*} + n^{d+1+\beta-\rho} + n^{d+2-4\alpha/d-\rho}.$$

In addition, we have

$$\Gamma_4 \underset{n \to \infty}{\overset{\ln}{\sim}} n^{\beta-1} + n^{-4\alpha/d} + n^{\rho/2-1}.$$
and

$$V(\mathcal{F}, \Gamma, N) = \left\{ h \in \ell^1(\mathbb{Z}^d) : \exists \ f \in \mathcal{F} \ \left\| h - \frac{1}{N} f^2 \right\|_1 \le \Gamma \right\}.$$

Together with inequalities (5.44) and (5.45), we obtain finally that

$$E(e^{-|R_N|}; \mathcal{B}) \\ \le 2(\Gamma_2 + 1) \exp \left( \frac{N}{\lambda}\Gamma_3 + \frac{2\Gamma_1}{\lambda} + c' \frac{n^{2d}}{[\Gamma_1]} - \inf \left\{ \left| \left\{ x \in \mathbb{Z}^d : \tilde{h}(x) > 0 \right\} \right| + \frac{N - \Gamma_2}{2} \left( \max \left( \sqrt{\mathcal{E}(\sqrt{\tilde{h}})} - \frac{4\sqrt{\Gamma_3}}{n}, 0 \right) \right)^2 : \tilde{h} \in V(\mathcal{F}, \Gamma_4, N) \right\} \right).$$

We have reached our goal, indeed, this last upper bound depends only on the set $\mathcal{F}$ and it is uniform over $g \in \mathcal{G}(X, M, \eta)$. We report this upper bound in the successive decompositions in sums presented in formulas (5.7) and (5.21). These two formulas imply that

$$E(e^{-|R_N|}; f_N \in \mathcal{F}, \mathcal{A}) \le |\mathcal{X}(N, \delta)| \times |\mathcal{G}(X, M, \eta)| \times \sup E(e^{-|R_N|}; \mathcal{B}),$$

where the supremum is over $X \in \mathcal{X}(N, \delta)$ and $g \in \mathcal{G}(X, M, \eta)$. Using the combinatorial bounds (5.4), (5.5), (5.17), and recalling the definition of $C_0$ given in (5.6), we have, for some constant $c_d$ depending on the dimension only,

$$|\mathcal{X}(N, \delta)| \le \exp(c_d C_0), \quad |\mathcal{G}| \le \left(\frac{\sqrt{N}}{\eta}\right)^{\frac{n^d}{M^d}} c_d C_0.$$

Together with our upper bound on the last expectation, this yields

$$(5.47) \quad \ln E(e^{-|R_N|}; f_N \in \mathcal{F}, \mathcal{A}) \le C_1 - \inf \left\{ \left| \left\{ x \in \mathbb{Z}^d : \tilde{h}(x) > 0 \right\} \right| + \frac{N - \Gamma_2}{2} \left( \max \left( \sqrt{\mathcal{E}(\sqrt{\tilde{h}})} - \frac{4\sqrt{\Gamma_3}}{n}, 0 \right) \right)^2 : \tilde{h} \in V(\mathcal{F}, \Gamma_4, N) \right\},$$

where

$$C_1 = c_d C_0 + \frac{n^d}{M^d} c_d C_0 \ln \left( \frac{\sqrt{N}}{\eta} \right) + \ln (2(\Gamma_2 + 1)) + \frac{N}{\lambda}\Gamma_3 + \frac{2\Gamma_1}{\lambda} + c' \frac{n^{2d}}{[\Gamma_1]}.$$

We compute the asymptotic expansion of these different terms in powers of $n$. Recalling the definitions of $C_0$ and $\Gamma_3$ given in (5.6) and (5.43), we obtain, using (5.28),

$$(5.48) \qquad C_0 \stackrel{\ln}{\sim}_{n \to \infty} n^{\alpha 2^*}, \quad \Gamma_3 \stackrel{\ln}{\sim}_{n \to \infty} n^{\beta-1} + n^{-4\alpha/d}.$$

Together with the relations (5.30), this yields

$$(5.49) \qquad C_1 \stackrel{\ln}{\sim}_{n \to \infty} n^{d(1-\beta)+\alpha 2^*} + n^{d+1+\beta-\rho} + n^{d+2-4\alpha/d-\rho}.$$

In addition, we have

$$\Gamma_4 \stackrel{\ln}{\sim}_{n \to \infty} n^{\beta-1} + n^{-4\alpha/d} + n^{\rho/2-1}.$$
We make next a specific choice for the values of the exponents, which satisfies the constraints stated before. However we do not try to look for the best possible exponents. So, we choose

$$ (5.50) \qquad \alpha = \frac{d}{12}, \quad \beta = \frac{3}{4}, \quad \gamma = \frac{d^2}{12(d-2)}, \quad \rho = \frac{15}{8}. $$

With this choice, we obtain that

$$ C_1 \underset{n \to \infty}{\sim} n^{d-1/8}, \quad \Gamma_2 \underset{n \to \infty}{\sim} n^{d+5/3}, \quad \Gamma_3 \underset{n \to \infty}{\sim} n^{-1/4}, \quad \Gamma_4 \underset{n \to \infty}{\sim} n^{-1/16}. $$

The equivalents above are logarithmic, however, by perturbing slightly the values of the exponents $\alpha, \beta, \gamma, \rho$, we can ensure that, for $n$ large enough,

$$ C_1 \le n^{d-1/8}, \quad \Gamma_2 \le Nn^{-1/4}, \quad 4\sqrt{\Gamma_3} \le n^{-1/8}, \quad \Gamma_4 \le n^{-1/16}. $$

Plugging these inequalities in (5.47), we obtain finally the statement of Theorem 2.2.

# 6. Continuous version of Theorem 2.2

## 6.1. Linear interpolation

We need a version of Theorem 2.2 in which the discrete Dirichlet energy is replaced by the continuous one, so that we can use quantitative versions of the Faber-Krahn inequality. The first step is to transform the local time into a function defined continuously everywhere on $\mathbb{R}^d$. We will state such a version in Theorem 6.2. We start preparing for this result.

Let $f$ be a function defined on $\mathbb{Z}^d$ with values in $\mathbb{R}^+$. We define a function $\tilde{f}$ on $\mathbb{R}^d$ by interpolating linearly $f$ successively in the $d$ directions of the axis. More precisely, we set $f_0 = f$ and for all $(x_1, \dots, x_d) \in \mathbb{Z}^d$,

$$ \forall \alpha \in [0,1] \quad f_1(\alpha x_1 + (1-\alpha)(x_1+1), x_2, \dots, x_d) \\ = \alpha f_0(x_1, v, \dots, x_d) + (1-\alpha)f_0(x_1+1, \dots, x_d). $$

We define iteratively, for $2 \le k \le d$, for all $(x_1, \dots, x_{k-1}) \in \mathbb{R}^{k-1}$, for all $(x_k, \dots, x_d) \in \mathbb{Z}^{d-k+1}$,

$$ \forall \alpha \in [0,1] \quad f_k(x_1, \dots, x_{k-1}, \alpha x_k + (1-\alpha)(x_k+1), x_{k+1}, \dots, x_d) \\ = \alpha f_{k-1}(x_1, \dots, x_{k-1}, \dots, x_d) + (1-\alpha)f_{k-1}(x_1, \dots, x_{k-1}+1, \dots, x_d) $$

and finally $\tilde{f} = f_d$. Let $C$ be the unit cube

$$ C = \{ x = (x_1, \dots, x_d) \in \mathbb{R}^d : 0 \le x_i \le 1, 1 \le i \le d \}. $$

Let us denote by $D^d$ the union of all the lines parallel to the axis which go through the points of $\mathbb{Z}^d$. We make the following observations relating $f$ to $\tilde{f}$.
We make next a specific choice for the values of the exponents, which satisfies the constraints stated before. However we do not try to look for the best possible exponents. So, we choose

$$ (5.50) \qquad \alpha = \frac{d}{12}, \quad \beta = \frac{3}{4}, \quad \gamma = \frac{d^2}{12(d-2)}, \quad \rho = \frac{15}{8}. $$

With this choice, we obtain that

$$ C_1 \stackrel{\ln}{\sim} n^{d-1/8}, \quad \Gamma_2 \stackrel{\ln}{\sim} n^{d+5/3}, \quad \Gamma_3 \stackrel{\ln}{\sim} n^{-1/4}, \quad \Gamma_4 \stackrel{\ln}{\sim} n^{-1/16}. $$

The equivalents above are logarithmic, however, by perturbing slightly the values of the exponents $\alpha, \beta, \gamma, \rho$, we can ensure that, for $n$ large enough,

$$ C_1 \le n^{d-1/8}, \quad \Gamma_2 \le Nn^{-1/4}, \quad 4\sqrt{\Gamma_3} \le n^{-1/8}, \quad \Gamma_4 \le n^{-1/16}. $$

Plugging these inequalities in (5.47), we obtain finally the statement of Theorem 2.2.

# 6. Continuous version of Theorem 2.2

## 6.1. Linear interpolation

We need a version of Theorem 2.2 in which the discrete Dirichlet energy is replaced by the continuous one, so that we can use quantitative versions of the Faber-Krahn inequality. The first step is to transform the local time into a function defined continuously everywhere on $\mathbb{R}^d$. We will state such a version in Theorem 6.2. We start preparing for this result.

Let $f$ be a function defined on $\mathbb{Z}^d$ with values in $\mathbb{R}^+$. We define a function $\tilde{f}$ on $\mathbb{R}^d$ by interpolating linearly $f$ successively in the $d$ directions of the axis. More precisely, we set $f_0 = f$ and for all $(x_1, \dots, x_d) \in \mathbb{Z}^d$,

$$ \forall \alpha \in [0,1] \quad f_1(\alpha x_1 + (1-\alpha)(x_1+1), x_2, \dots, x_d) \\ = \alpha f_0(x_1, v, \dots, x_d) + (1-\alpha)f_0(x_1+1, \dots, x_d). $$

We define iteratively, for $2 \le k \le d$, for all $(x_1, \dots, x_{k-1}) \in \mathbb{R}^{k-1}$, for all $(x_k, \dots, x_d) \in \mathbb{Z}^{d-k+1}$,

$$ \forall \alpha \in [0,1] \quad f_k(x_1, \dots, x_{k-1}, \alpha x_k + (1-\alpha)(x_k+1), x_{k+1}, \dots, x_d) \\ = \alpha f_{k-1}(x_1, \dots, x_{k-1}, \dots, x_d) + (1-\alpha)f_{k-1}(x_1, \dots, x_{k-1}+1, \dots, x_d) $$

and finally $\tilde{f} = f_d$. Let $C$ be the unit cube

$$ C = \{ x = (x_1, \dots, x_d) \in \mathbb{R}^d : 0 \le x_i \le 1, 1 \le i \le d \}. $$

Let us denote by $D^d$ the union of all the lines parallel to the axis which go through the points of $\mathbb{Z}^d$. We make the following observations relating $f$ to $\tilde{f}$.
LEMMA 6.1. — The function $\tilde{f}$ is continuous on $\mathbb{R}^d$ and $C^\infty$ on $\mathbb{R}^d \setminus D^d$. We have

$$ (6.1) \qquad \int_{\mathbb{R}^d} \tilde{f}(x) \, dx = \sum_{x \in \mathbb{Z}^d} f(x), $$

$$ (6.2) \qquad \int_{\mathbb{R}^d} \tilde{f}(x)^2 \, dx \le \sum_{x \in \mathbb{Z}^d} f(x)^2, $$

$$ (6.3) \qquad \int_{\mathbb{R}^d} |\nabla \tilde{f}(x)|^2 \, dx \le d\mathcal{E}(f). $$

*Proof.* — With the help of a standard induction, we get the following formula for $\tilde{f}$:

$$ \forall (y_1, \dots, y_d) \in \mathbb{Z}^d \quad \forall (x_1, \dots, x_d) \in (y_1, \dots, y_d) + C $$

$$ \tilde{f}(x_1, \dots, x_d) = \sum_{\varepsilon_1, \dots, \varepsilon_d \in \{0,1\}} f(y_1 + \varepsilon_1, \dots, y_d + \varepsilon_d) \prod_{1 \le k \le d} \left( (1 - \varepsilon_k)(1 - (x_k - y_k)) + \varepsilon_k(x_k - y_k) \right). $$

Let us compute the integral of $\tilde{f}$ over $\mathbb{R}^d$. We have

$$ \int_C \tilde{f}(x) \, dx = \int_{0 \le x_1 \le 1} \left( \dots \left( \int_{0 \le x_d \le 1} \tilde{f}(x) \, dx_d \right) \dots \right) dx_1. $$

Let us fix $(x_1, \dots, x_{d-1}) \in [0, 1]^{d-1}$ and let us compute

$$ \int_{0 \le x_d \le 1} \tilde{f}(x_1, \dots, x_d) \, dx_d = \int_{0 \le x_d \le 1} \left( (1-x_d) f_{d-1}(x_1, \dots, x_{d-1}, 0) + x_d f_{d-1}(x_1, \dots, x_{d-1}, 1) \right) \, dx_d = \frac{1}{2} \left( f_{d-1}(x_1, \dots, x_{d-1}, 0) + f_{d-1}(x_1, \dots, x_{d-1}, 1) \right). $$

Iterating this computation, we obtain

$$ \int_C \tilde{f}(x) \, dx = \frac{1}{2^d} \sum_{x_1, \dots, x_d \in \{0,1\}} f(x_1, \dots, x_d). $$

Each point of $\mathbb{Z}^d$ belongs to $2^d$ integer translates of $C$, therefore

$$ \int_{\mathbb{R}^d} \tilde{f}(x) \, dx = \sum_{y \in \mathbb{Z}^d} \int_{y+C} \tilde{f}(x) \, dx = \sum_{x \in \mathbb{Z}^d} f(x) $$

which proves (6.1). Inequality (6.2) is proved in the same way, by using $d$ times the convexity of the function $t \to t^2$ in the linear interpolations. Our next goal is to compute the integral of $(\nabla \tilde{f})^2$ over $\mathbb{R}^d$. Let us first compute its partial derivatives. For simplicity, we deal with the derivative with respect to the last variable $x_d$, and we consider only the points $x$ in the unit cube $C$:

$$ \frac{\partial \tilde{f}}{\partial x_d}(x) = \sum_{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\}} D_d f(\varepsilon_1, \dots, \varepsilon_{d-1}) \prod_{k=1}^{d-1} \left( (1 - \varepsilon_k)(1 - x_k) + \varepsilon_k x_k \right), $$
LEMMA 6.1. — The function $\tilde{f}$ is continuous on $\mathbb{R}^d$ and $C^\infty$ on $\mathbb{R}^d \setminus D^d$. We have

$$
(6.1) \qquad \int_{\mathbb{R}^d} \tilde{f}(x) \, dx = \sum_{x \in \mathbb{Z}^d} f(x),
$$

$$
(6.2) \qquad \int_{\mathbb{R}^d} \tilde{f}(x)^2 \, dx \le \sum_{x \in \mathbb{Z}^d} f(x)^2,
$$

$$
(6.3) \qquad \int_{\mathbb{R}^d} |\nabla \tilde{f}(x)|^2 dx \le d\mathcal{E}(f).
$$

Proof. — With the help of a standard induction, we get the following formula
for $\tilde{f}$:

$$
\forall (y_1, \dots, y_d) \in \mathbb{Z}^d \quad \forall (x_1, \dots, x_d) \in (y_1, \dots, y_d) + C
$$

$$
\tilde{f}(x_1, \dots, x_d) =
$$

$$
\sum_{\varepsilon_1, \dots, \varepsilon_d \in \{0,1\}} f(y_1 + \varepsilon_1, \dots, y_d + \varepsilon_d) \prod_{1 \le k \le d} \left( (1 - \varepsilon_k)(1 - (x_k - y_k)) + \varepsilon_k(x_k - y_k) \right).
$$

Let us compute the integral of $\tilde{f}$ over $\mathbb{R}^d$. We have

$$
\int_C \tilde{f}(x) \, dx = \int_{0 \le x_1 \le 1} \left( \dots \left( \int_{0 \le x_d \le 1} \tilde{f}(x) \, dx_d \right) \dots \right) dx_1 .
$$

Let us fix $(x_1, \dots, x_{d-1}) \in [0, 1]^{d-1}$ and let us compute

$$
\int_{0 \le x_d \le 1} \tilde{f}(x_1, \dots, x_d) dx_d =
$$

$$
\int_{0 \le x_d \le 1} \left( (1 - x_d) f_{d-1}(x_1, \dots, x_{d-1}, 0) + x_d f_{d-1}(x_1, \dots, x_{d-1}, 1) \right) dx_d
$$

$$
= \frac{1}{2} \left( f_{d-1}(x_1, \dots, x_{d-1}, 0) + f_{d-1}(x_1, \dots, x_{d-1}, 1) \right).
$$

Iterating this computation, we obtain

$$
\int_C \tilde{f}(x) \, dx = \frac{1}{2^d} \sum_{x_1, \dots, x_d \in \{0,1\}} f(x_1, \dots, x_d).
$$

Each point of $\mathbb{Z}^d$ belongs to $2^d$ integer translates of $C$, therefore

$$
\int_{\mathbb{R}^d} \tilde{f}(x) \, dx = \sum_{y \in \mathbb{Z}^d} \int_{y+C} \tilde{f}(x) \, dx = \sum_{x \in \mathbb{Z}^d} f(x)
$$

which proves (6.1). Inequality (6.2) is proved in the same way, by using $d$ times the
convexity of the function $t \to t^2$ in the linear interpolations. Our next goal is to
compute the integral of $(\nabla \tilde{f})^2$ over $\mathbb{R}^d$. Let us first compute its partial derivatives.
For simplicity, we deal with the derivative with respect to the last variable $x_d$, and
we consider only the points $x$ in the unit cube $C$:

$$
\frac{\partial \tilde{f}}{\partial x_d}(x) = \sum_{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\}} D_d f(\varepsilon_1, \dots, \varepsilon_{d-1}) \prod_{k=1}^{d-1} \left( (1-\varepsilon_k)(1-x_k) + \varepsilon_k x_k \right),
$$
where we define, for $\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0, 1\}$,

$$D_d f(\varepsilon_1, \dots, \varepsilon_{d-1}) = f(\varepsilon_1, \dots, \varepsilon_{d-1}, 1) - f(\varepsilon_1, \dots, \varepsilon_{d-1}, 0).$$

We apply Fubini's theorem to write

$$\int_C \left( \frac{\partial \tilde{f}}{\partial x_d}(x) \right)^2 dx = \sum_{\substack{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\} \\ \varepsilon'_1, \dots, \varepsilon'_{d-1} \in \{0,1\}}} D_d f(\varepsilon_1, \dots, \varepsilon_{d-1}) D_d f(\varepsilon'_1, \dots, \varepsilon'_{d-1}) \\ \prod_{k=1}^{d-1} \left( \int_0^1 \left( (1-\varepsilon_k)(1-x_k) + \varepsilon_k x_k \right) \left( (1-\varepsilon'_k)(1-x_k) + \varepsilon'_k x_k \right) dx_k \right).$$

We compute the value of the integrals. For $\varepsilon, \varepsilon' \in \{0, 1\}$, we have

$$\int_0^1 \left( (1-\varepsilon)(1-x) + \varepsilon x \right) \left( (1-\varepsilon')(1-x) + \varepsilon'x \right) dx = \frac{1}{6} (1 + \delta(\varepsilon, \varepsilon')).$$

where $\delta$ is the Kronecker symbol. Next, we have the bound

$$\int_C \left( \frac{\partial \tilde{f}}{\partial x_d}(x) \right)^2 dx \le \sum_{\substack{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\} \\ \varepsilon'_1, \dots, \varepsilon'_{d-1} \in \{0,1\}}} \left( D_d f(\varepsilon_1, \dots, \varepsilon_{d-1}) \right)^2 \prod_{k=1}^{d-1} \frac{1}{6} (1 + \delta(\varepsilon_k, \varepsilon'_k)) \\ = \frac{1}{2^{d-1}} \sum_{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\}} \left( D_d f(\varepsilon_1, \dots, \varepsilon_{d-1}) \right)^2.$$

Next, we sum the previous inequality over all integer translates of the unit cube $C$. We obtain

$$\int_{\mathbb{R}^d} \left( \frac{\partial \tilde{f}}{\partial x_d}(x) \right)^2 dx = \sum_{y \in \mathbb{Z}^d} \int_{y+C} \left( \frac{\partial \tilde{f}}{\partial x_d}(x) \right)^2 dx \\ \le \sum_{y \in \mathbb{Z}^d} \frac{1}{2^{d-1}} \sum_{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\}} \left( f(y_1 + \varepsilon_1, \dots, y_{d-1} + \varepsilon_{d-1}, y_d + 1) \right. \\ \left. \qquad - f(y_1 + \varepsilon_1, \dots, y_{d-1} + \varepsilon_{d-1}, y_d) \right)^2 \\ = \sum_{y \in \mathbb{Z}^d} \left( f(y_1, \dots, y_{d-1}, y_d + 1) - f(y_1, \dots, y_{d-1}, y_d) \right)^2.$$

Summing finally the integrals associated to each partial derivatives, we conclude that

$$\int_{\mathbb{R}^d} |\nabla \tilde{f}|^2 dx \le d \mathcal{E}(f).$$

Notice that, in the definition (1.6) of $\mathcal{E}$, each edge of the lattice appears twice in the summation. This proves (6.3), which finishes the proof of the lemma. $\square$
where we define, for $\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0, 1\}$,

$$D_d f(\varepsilon_1, \dots, \varepsilon_{d-1}) = f(\varepsilon_1, \dots, \varepsilon_{d-1}, 1) - f(\varepsilon_1, \dots, \varepsilon_{d-1}, 0).$$

We apply Fubini's theorem to write

$$\int_C \left(\frac{\partial \tilde{f}}{\partial x_d}(x)\right)^2 dx = \sum_{\substack{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\} \\ \varepsilon'_1, \dots, \varepsilon'_{d-1} \in \{0,1\}}} D_d f(\varepsilon_1, \dots, \varepsilon_{d-1}) D_d f(\varepsilon'_1, \dots, \varepsilon'_{d-1}) \\ \prod_{k=1}^{d-1} \left( \int_0^1 \left( (1-\varepsilon_k)(1-x_k) + \varepsilon_k x_k \right) \left( (1-\varepsilon'_k)(1-x_k) + \varepsilon'_k x_k \right) dx_k \right).$$

We compute the value of the integrals. For $\varepsilon, \varepsilon' \in \{0, 1\}$, we have

$$\int_0^1 \left( (1-\varepsilon)(1-x) + \varepsilon x \right) \left( (1-\varepsilon')(1-x) + \varepsilon'x \right) dx = \frac{1}{6} (1 + \delta(\varepsilon, \varepsilon')).$$

where $\delta$ is the Kronecker symbol. Next, we have the bound

$$\int_C \left(\frac{\partial \tilde{f}}{\partial x_d}(x)\right)^2 dx \le \sum_{\substack{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\} \\ \varepsilon'_1, \dots, \varepsilon'_{d-1} \in \{0,1\}}} \left(D_d f(\varepsilon_1, \dots, \varepsilon_{d-1})\right)^2 \prod_{k=1}^{d-1} \frac{1}{6} (1 + \delta(\varepsilon_k, \varepsilon'_k)) \\ = \frac{1}{2^{d-1}} \sum_{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\}} \left(D_d f(\varepsilon_1, \dots, \varepsilon_{d-1})\right)^2.$$

Next, we sum the previous inequality over all integer translates of the unit cube $C$. We obtain

$$\int_{\mathbb{R}^d} \left(\frac{\partial \tilde{f}}{\partial x_d}(x)\right)^2 dx = \sum_{y \in \mathbb{Z}^d} \int_{y+C} \left(\frac{\partial \tilde{f}}{\partial x_d}(x)\right)^2 dx \\ \le \sum_{y \in \mathbb{Z}^d} \frac{1}{2^{d-1}} \sum_{\varepsilon_1, \dots, \varepsilon_{d-1} \in \{0,1\}} \left( f(y_1 + \varepsilon_1, \dots, y_{d-1} + \varepsilon_{d-1}, y_d + 1) \right. \\ \left. \qquad - f(y_1 + \varepsilon_1, \dots, y_{d-1} + \varepsilon_{d-1}, y_d) \right)^2 \\ = \sum_{y \in \mathbb{Z}^d} \left( f(y_1, \dots, y_{d-1}, y_d + 1) - f(y_1, \dots, y_{d-1}, y_d) \right)^2.$$

Summing finally the integrals associated to each partial derivatives, we conclude that

$$\int_{\mathbb{R}^d} |\nabla \tilde{f}|^2 dx \le d \mathcal{E}(f).$$

Notice that, in the definition (1.6) of $\mathcal{E}$, each edge of the lattice appears twice in the summation. This proves (6.3), which finishes the proof of the lemma. $\square$
## 6.2. Rescaling

We denote by $\mathbb{Z}_n^d$ the lattice $\mathbb{Z}^d$ rescaled by a factor $n$: $\mathbb{Z}_n^d = \frac{1}{n}\mathbb{Z}^d$. We shall simultaneously rescale the space by a factor $n$ and the values of $L_N$ by a factor $n^d/N = 1/n^2$. Starting with the local time $L_N$, which is a function defined on $\mathbb{Z}^d$, we define the function $\ell_N$ on $\mathbb{R}^d$ by setting

$$ (6.4) \qquad \forall x \in \mathbb{R}^d \quad \ell_N(x) = \frac{n^d}{N} L_N(\lfloor nx \rfloor). $$

For $g$ a continuous function from $\mathbb{R}^d$ to $\mathbb{R}$, we define its support as

$$ \text{supp } g = \{ x \in \mathbb{R}^d : g(x) \neq 0 \}, $$

and we denote by $|\text{supp } g|$ its Lebesgue measure.

Here is a continuous analogue of Theorem 2.2.

**THEOREM 6.2.** — Let us denote by $D^d$ the union of all the lines parallel to the axis which go through the points of $\mathbb{Z}^d$. Let $\mathcal{L}$ be a collection of functions from $\mathbb{R}^d$ to $\mathbb{R}^+$ such that $\|\ell\|_2 = 1$ for $\ell \in \mathcal{L}$. For any $\kappa \ge 1$, we have, for $n$ large enough,

$$ \begin{aligned} E(e^{-|R_N|}; \ell_N \in \mathcal{L}) &\le 3 \exp(-\kappa n^d) + \exp \left( 2n^{d-1/8} - \right. \\ & \qquad \left. n^d \inf \left\{ |\text{supp } g| + \frac{1}{2d}(1-n^{-1/4}) \left( \max \left( \sqrt{\int_{\mathbb{R}^d} |\nabla g|^2 dx} - \frac{\sqrt{d}}{n^{1/8}}, 0 \right) \right)^2 \right. \right. \\ & \qquad \left. \left. : g \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus \frac{1}{n}D^d, \right. \right. \\ & \qquad \left. \left. \exists \ell \in \mathcal{L} \quad \|g^2 - \ell\|_{L^1(\mathbb{R}^d)} \le \frac{3}{n^{1/16}} \right\} \right). \end{aligned} $$

*Proof.* — Let us define formally the operator which transforms the function $f_N$ into $\ell_N$ as in (6.4). To a function $f: \mathbb{Z}^d \to \mathbb{R}^+$, we associate a function $\Phi_n(f): \mathbb{R}^d \to \mathbb{R}^+$ by setting

$$ (6.5) \qquad \forall x \in \mathbb{R}^d \quad \Phi_n(f)(x) = \frac{n^d}{N} \left( f(\lfloor nx \rfloor) \right)^2. $$

With this definition, we check that $\ell_n = \Phi_n(f_N)$. Therefore

$$ E(e^{-|R_N|}; \ell_N \in \mathcal{L}) = E(e^{-|R_N|}; f_N \in \mathcal{F}), $$

where $\mathcal{F}$ is the collection of functions defined by

$$ \mathcal{F} = \Phi_n^{-1}(\mathcal{L}) = \{ f : \mathbb{Z}^d \to \mathbb{R}^+, \Phi_n(f) \in \mathcal{L} \}. $$

We apply the upper bound of Theorem 2.2. For $h$ a function from $\mathbb{Z}^d$ to $\mathbb{R}^+$, we denote its support by

$$ \text{supp } h = \{ x \in \mathbb{Z}^d : h(x) \neq 0 \}. $$
## 6.2. Rescaling

We denote by $\mathbb{Z}_n^d$ the lattice $\mathbb{Z}^d$ rescaled by a factor $n$: $\mathbb{Z}_n^d = \frac{1}{n}\mathbb{Z}^d$. We shall simultaneously rescale the space by a factor $n$ and the values of $L_N$ by a factor $n^d/N = 1/n^2$. Starting with the local time $L_N$, which is a function defined on $\mathbb{Z}^d$, we define the function $\ell_N$ on $\mathbb{R}^d$ by setting

$$ (6.4) \qquad \forall x \in \mathbb{R}^d \quad \ell_N(x) = \frac{n^d}{N} L_N(\lfloor nx \rfloor). $$

For $g$ a continuous function from $\mathbb{R}^d$ to $\mathbb{R}$, we define its support as

$$ \text{supp } g = \{ x \in \mathbb{R}^d : g(x) \neq 0 \}, $$

and we denote by $|\text{supp } g|$ its Lebesgue measure.

Here is a continuous analogue of Theorem 2.2.

**THEOREM 6.2.** — Let us denote by $D^d$ the union of all the lines parallel to the axis which go through the points of $\mathbb{Z}^d$. Let $\mathcal{L}$ be a collection of functions from $\mathbb{R}^d$ to $\mathbb{R}^+$ such that $\|\ell\|_2 = 1$ for $\ell \in \mathcal{L}$. For any $\kappa \ge 1$, we have, for $n$ large enough,

$$ \begin{aligned} E(e^{-|R_N|}; \ell_N \in \mathcal{L}) &\le 3 \exp(-\kappa n^d) + \exp \left( 2n^{d-1/8} - \right. \\ & \qquad \left. n^d \inf \left\{ |\text{supp } g| + \frac{1}{2d}(1-n^{-1/4}) \left( \max \left( \sqrt{\int_{\mathbb{R}^d} |\nabla g|^2 dx} - \frac{\sqrt{d}}{n^{1/8}}, 0 \right) \right)^2 \right. \right. \\ & \qquad \left. \left. : g \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus \frac{1}{n}D^d, \right. \right. \\ & \qquad \left. \left. \exists \ell \in \mathcal{L} \quad \left\| g^2 - \ell \right\|_{L^1(\mathbb{R}^d)} \le \frac{3}{n^{1/16}} \right\} \right). \end{aligned} $$

*Proof.* — Let us define formally the operator which transforms the function $f_N$ into $\ell_N$ as in (6.4). To a function $f: \mathbb{Z}^d \to \mathbb{R}^+$, we associate a function $\Phi_n(f): \mathbb{R}^d \to \mathbb{R}^+$ by setting

$$ (6.5) \qquad \forall x \in \mathbb{R}^d \quad \Phi_n(f)(x) = \frac{n^d}{N} \left( f(\lfloor nx \rfloor) \right)^2. $$

With this definition, we check that $\ell_n = \Phi_n(f_N)$. Therefore

$$ E(e^{-|R_N|}; \ell_N \in \mathcal{L}) = E(e^{-|R_N|}; f_N \in \mathcal{F}), $$

where $\mathcal{F}$ is the collection of functions defined by

$$ \mathcal{F} = \Phi_n^{-1}(\mathcal{L}) = \{ f : \mathbb{Z}^d \to \mathbb{R}^+, \Phi_n(f) \in \mathcal{L} \}. $$

We apply the upper bound of Theorem 2.2. For $h$ a function from $\mathbb{Z}^d$ to $\mathbb{R}^+$, we denote its support by

$$ \text{supp } h = \{ x \in \mathbb{Z}^d : h(x) \neq 0 \}. $$
Let $\kappa \ge 1$. For $n$ large enough, we have

$$
\begin{aligned}
E(e^{-|R_N|}; \ell_N \in \mathcal{L}) &\le \exp(-\kappa n^d) + \\
&\quad \exp \left( n^{d-1/8} - \inf \left\{ \left| \text{supp } h \right| + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\mathcal{E}(\sqrt{h})} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \right. \\
&\qquad \left. \left. : h \in \ell^1(\mathbb{Z}^d), \exists f \in \mathcal{F} \ \left\| h - \frac{1}{N} f^2 \right\|_1 \le \frac{1}{n^{1/16}} \right\} \right).
\end{aligned}
$$

In the infimum appearing above, we can limit ourselves to functions $h$ for which the infimum is of order $n^d$, otherwise the exponential becomes negligible compared to the first term $\exp(-\kappa n^d)$. The relevant functions $h$ should be such that

$$ (6.6) \qquad \left| \text{supp } h \right| \le 2\kappa n^d, \quad \mathcal{E}(\sqrt{h}) \le \frac{3\kappa}{n^2}. $$

We make also the change of function $h \to h^2$. We obtain that, for $n$ large enough,

$$
\begin{aligned}
(6.7) \quad E(e^{-|R_N|}; \ell_N \in \mathcal{L}) &\le 3 \exp(-\kappa n^d) \\
&\quad + \exp \left( n^{d-1/8} - \inf \left\{ \left| \text{supp } h \right| + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\mathcal{E}(h)} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \right. \\
&\qquad \left. \left. : h \in \ell^2(\mathbb{Z}^d), h \ge 0, \left| \text{supp } h \right| \le 2\kappa n^d, \mathcal{E}(h) \le \frac{3\kappa}{n^2}, \right. \right. \\
&\qquad \left. \left. \exists f \in \mathcal{F} \ \left\| h^2 - \frac{1}{N} f^2 \right\|_1 \le \frac{1}{n^{1/16}} \right\} \right).
\end{aligned}
$$

We apply now the linear interpolation procedure described in Section 6.1 to the function $h$ appearing in the above infimum. Since $\|f^2\|_1 = N$ for any $f \in \mathcal{F}$, the constraint on the function $h$ implies that

$$ (6.8) \qquad \left| \|h^2\|_1 - 1 \right| \le \frac{1}{n^{1/16}}. $$

Starting from a function $h$ in $\ell^2(\mathbb{Z}^d)$, we apply the interpolation procedure and we obtain a function $\tilde{h}$ in $L^2(\mathbb{R}^d)$ which satisfies, according to inequalities (6.2) and (6.3),

$$ (6.9) \qquad \|\tilde{h}\|_{L^2(\mathbb{R}^d)} \le \|h\|_{\ell^2(\mathbb{Z}^d)}, \quad \int_{\mathbb{R}^d} |\nabla \tilde{h}|^2 dx \le d \mathcal{E}(h). $$

We wish to obtain an infimum involving only continuous functions, so we have to get completely rid of the discrete function $h$. Therefore we should control $|\text{supp } h|$ and the distance between $\tilde{h}$ and the set $\mathcal{L}$. Let us start with $|\text{supp } h|$. The linear interpolation $\tilde{h}$ of $h$ is non-zero only in the unit cubes having at least one vertex in the support of $h$, therefore

$$
(6.10) \quad \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > 0 \right\} \right| \le |\text{supp } h| + \\
\qquad \left| \left\{ x \in \mathbb{Z}^d \setminus \text{supp } h : \exists y \in \text{supp } h \ |x - y|_{\infty} \le 1 \right\} \right|.
$$
Let $\kappa \ge 1$. For $n$ large enough, we have

$$
\begin{aligned}
E(e^{-|R_N|}; \ell_N \in \mathcal{L}) &\le \exp(-\kappa n^d) + \\
&\quad \exp \left( n^{d-1/8} - \inf \left\{ \left| \text{supp } h \right| + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\mathcal{E}(\sqrt{h})} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \right. \\
&\qquad \left. \left. : h \in \ell^1(\mathbb{Z}^d), \exists f \in \mathcal{F} \ \left\| h - \frac{1}{N} f^2 \right\|_1 \le \frac{1}{n^{1/16}} \right\} \right).
\end{aligned}
$$

In the infimum appearing above, we can limit ourselves to functions $h$ for which the infimum is of order $n^d$, otherwise the exponential becomes negligible compared to the first term $\exp(-\kappa n^d)$. The relevant functions $h$ should be such that

$$ (6.6) \qquad \left| \text{supp } h \right| \le 2\kappa n^d, \quad \mathcal{E}(\sqrt{h}) \le \frac{3\kappa}{n^2}. $$

We make also the change of function $h \to h^2$. We obtain that, for $n$ large enough,

$$
\begin{aligned}
(6.7) \quad E(e^{-|R_N|}; \ell_N \in \mathcal{L}) &\le 3 \exp(-\kappa n^d) \\
&\quad + \exp \left( n^{d-1/8} - \inf \left\{ \left| \text{supp } h \right| + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\mathcal{E}(h)} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \right. \\
&\qquad \left. \left. : h \in \ell^2(\mathbb{Z}^d), h \ge 0, \left| \text{supp } h \right| \le 2\kappa n^d, \mathcal{E}(h) \le \frac{3\kappa}{n^2}, \right. \right. \\
&\qquad \left. \left. \exists f \in \mathcal{F} \ \left\| h^2 - \frac{1}{N} f^2 \right\|_1 \le \frac{1}{n^{1/16}} \right\} \right).
\end{aligned}
$$

We apply now the linear interpolation procedure described in Section 6.1 to the function $h$ appearing in the above infimum. Since $\|f^2\|_1 = N$ for any $f \in \mathcal{F}$, the constraint on the function $h$ implies that

$$ (6.8) \qquad \left| \|h^2\|_1 - 1 \right| \le \frac{1}{n^{1/16}}. $$

Starting from a function $h$ in $\ell^2(\mathbb{Z}^d)$, we apply the interpolation procedure and we obtain a function $\tilde{h}$ in $L^2(\mathbb{R}^d)$ which satisfies, according to inequalities (6.2) and (6.3),

$$ (6.9) \qquad \|\tilde{h}\|_{L^2(\mathbb{R}^d)} \le \|h\|_{\ell^2(\mathbb{Z}^d)}, \quad \int_{\mathbb{R}^d} |\nabla \tilde{h}|^2 dx \le d \mathcal{E}(h). $$

We wish to obtain an infimum involving only continuous functions, so we have to get completely rid of the discrete function $h$. Therefore we should control $|\text{supp } h|$ and the distance between $\tilde{h}$ and the set $\mathcal{L}$. Let us start with $|\text{supp } h|$. The linear interpolation $\tilde{h}$ of $h$ is non-zero only in the unit cubes having at least one vertex in the support of $h$, therefore

$$
(6.10) \quad \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > 0 \right\} \right| \le |\text{supp } h| + \\
\qquad \left| \left\{ x \in \mathbb{Z}^d \setminus \text{supp } h : \exists y \in \text{supp } h \mid x - y \mid_{\infty} \le 1 \right\} \right|.
$$
Notice that we use $|\cdot|$ to denote the Lebesgue measure for a continuous set and the
cardinality for a discrete set. Each point $x$ of $\mathbb{Z}^d$ admits $3^d$ points $y \in \mathbb{Z}^d$ such that
$|x - y|_{\infty} = 1$, therefore

$$
(6.11) \qquad \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > 0 \right\} \right| \le (3^d + 1) \left| \text{supp } h \right|.
$$

This inequality will be useful, however we need a better lower bound for $|\text{supp } h|$, therefore we need a better upper bound on the last term in inequality (6.10). This is a delicate matter. Our strategy is to use the bound on the discrete Dirichlet energy to control this boundary term, and to do so, we truncate the function at a fixed level $\lambda > 0$. Since $\tilde{h}(x)$ is a linear interpolation between the $2^d$ values of $h$ at the vertices of the unit cube containing $x$, we have

$$
(6.12) \quad \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| \le \left| \text{supp } h \right| \\
\qquad + \left| \left\{ x \in \mathbb{Z}^d \setminus \text{supp } h : \exists y \in \text{supp } h \mid x - y \mid_{\infty} \le 1, h(y) \ge \lambda \right\} \right|.
$$

Let $C$ be a unit cube with vertices in $\mathbb{Z}^d$ such that one vertex $x$ of $C$ is not in $\text{supp } h$
and another vertex $y$ of $C$ satisfies $h(y) \ge \lambda$. Then there exist two vertices $x', y'$ of
$C$ which are nearest neighbours and which satisfy

$$
h(x') - h(y') \ge \frac{\lambda}{2^d},
$$

and the contribution to the discrete Dirichlet energy $\mathcal{E}(h)$ of the edges belonging to
the boundary of $C$ is larger or equal than

$$
\frac{1}{d} \left( h(x') - h(y') \right)^2 \geq \frac{1}{d} \frac{\lambda^2}{2^{2d}}.
$$

An edge belongs to at most $2^{d-1}$ unit cubes. The number $M$ of cubes making such a contribution satisfies therefore

$$
M \frac{1}{d} \frac{\lambda^2}{2^{2d}} \frac{1}{2^{d-1}} \le \mathcal{E}(h).
$$

Moreover the last term in inequality (6.12) is bounded from above by $M2^{2d}$. Taking
into account the bound on $\mathcal{E}(h)$ given in (6.6), we conclude that

$$
(6.13) \quad \begin{aligned}[t]
|\text{supp } h| \ge & \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| - M2^{2d} \\
& \ge \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| - \frac{d2^{5d} 3\kappa}{\lambda^2 n^2}.
\end{aligned}
$$

In order to delay the rescaling of the space, we define an intermediate operator
$\Phi$ acting on the functions as follows. To a function $f: \mathbb{Z}^d \to \mathbb{R}^+$, we associate a
function $\Phi(f): \mathbb{R}^d \to \mathbb{R}^+$ by setting

$$
\forall x \in \mathbb{R}^d \quad \Phi(f)(x) = \left( f(\lfloor x \rfloor) \right)^2 .
$$

We rewrite (6.5) with the help of $\Phi$ as follows:

$$
(6.14) \qquad \forall x \in \mathbb{R}^d \qquad \Phi_n(f)(x) = \frac{n^d}{N} \Phi(f)(nx).
$$
Notice that we use $|\cdot|$ to denote the Lebesgue measure for a continuous set and the
cardinality for a discrete set. Each point $x$ of $\mathbb{Z}^d$ admits $3^d$ points $y \in \mathbb{Z}^d$ such that
$|x - y|_{\infty} = 1$, therefore

$$
(6.11) \qquad \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > 0 \right\} \right| \le (3^d + 1) \left| \text{supp } h \right|.
$$

This inequality will be useful, however we need a better lower bound for $|\text{supp } h|$, therefore we need a better upper bound on the last term in inequality (6.10). This is a delicate matter. Our strategy is to use the bound on the discrete Dirichlet energy to control this boundary term, and to do so, we truncate the function at a fixed level $\lambda > 0$. Since $\tilde{h}(x)$ is a linear interpolation between the $2^d$ values of $h$ at the vertices of the unit cube containing $x$, we have

$$
(6.12) \quad \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| \le \left| \text{supp } h \right| \\
\qquad + \left| \left\{ x \in \mathbb{Z}^d \setminus \text{supp } h : \exists y \in \text{supp } h \mid x - y \mid_{\infty} \le 1, h(y) \ge \lambda \right\} \right|.
$$

Let $C$ be a unit cube with vertices in $\mathbb{Z}^d$ such that one vertex $x$ of $C$ is not in $\text{supp } h$
and another vertex $y$ of $C$ satisfies $h(y) \ge \lambda$. Then there exist two vertices $x', y'$ of
$C$ which are nearest neighbours and which satisfy

$$
h(x') - h(y') \ge \frac{\lambda}{2^d},
$$

and the contribution to the discrete Dirichlet energy $\mathcal{E}(h)$ of the edges belonging to
the boundary of $C$ is larger or equal than

$$
\frac{1}{d} \left( h(x') - h(y') \right)^2 \geq \frac{1}{d} \frac{\lambda^2}{2^{2d}}.
$$

An edge belongs to at most $2^{d-1}$ unit cubes. The number $M$ of cubes making such a contribution satisfies therefore

$$
M \frac{1}{d} \frac{\lambda^2}{2^{2d}} \frac{1}{2^{d-1}} \le \mathcal{E}(h).
$$

Moreover the last term in inequality (6.12) is bounded from above by $M2^{2d}$. Taking
into account the bound on $\mathcal{E}(h)$ given in (6.6), we conclude that

$$
(6.13) \quad \begin{aligned}[t]
|\text{supp } h| \ge & \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| - M2^{2d} \\
& \ge \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| - \frac{d2^{5d} 3\kappa}{\lambda^2 n^2}.
\end{aligned}
$$

In order to delay the rescaling of the space, we define an intermediate operator
$\Phi$ acting on the functions as follows. To a function $f: \mathbb{Z}^d \to \mathbb{R}^+$, we associate a
function $\Phi(f): \mathbb{R}^d \to \mathbb{R}^+$ by setting

$$
\forall x \in \mathbb{R}^d \quad \Phi(f)(x) = \left( f(\lfloor x \rfloor) \right)^2 .
$$

We rewrite (6.5) with the help of $\Phi$ as follows:

$$
(6.14) \qquad \forall x \in \mathbb{R}^d \qquad \Phi_n(f)(x) = \frac{n^d}{N} \Phi(f)(nx).
$$
Let now $f$ be an element of $\mathcal{F}$ such that

$$ \left\| h^2 - \frac{1}{N} f^2 \right\|_1 \le \frac{1}{n^{1/16}}. $$

By definition of $\mathcal{F}$, we have that $\Phi_n(f) \in \mathcal{L}$. On the one hand, we have

$$ (6.15) \qquad \left\| \Phi(h) - \frac{1}{N} \Phi(f) \right\|_{L^1(\mathbb{R}^d)} = \left\| h^2 - \frac{1}{N} f^2 \right\|_{\ell^1(\mathbb{Z}^d)} \le \frac{1}{n^{1/16}}. $$

On the other hand, we have

$$ (6.16) \qquad \left\| \tilde{h}^2 - \Phi(h) \right\|_{L^1(\mathbb{R}^d)} = \int_{\mathbb{R}^d} \left| \tilde{h}(x)^2 - h(\lfloor x \rfloor)^2 \right| dx. $$

Let us define

$$ \forall x \in \mathbb{Z}^d \quad D(x) = \{ y \in \mathbb{R}^d : \lfloor y \rfloor = x \}. $$

We can rewrite the previous integral as

$$ \begin{align*} (6.17) \quad \left\| \tilde{h}^2 - \Phi(h) \right\|_{L^1(\mathbb{R}^d)} &= \sum_{x \in \mathbb{Z}^d} \int_{D(x)} \left| \tilde{h}(y)^2 - h(\lfloor y \rfloor)^2 \right| dy \\ &\le \sum_{x \in \mathbb{Z}^d} \sup_{y \in D(x)} \left| \tilde{h}(y)^2 - h(x)^2 \right| \\ &\le \sum_{x \in \mathbb{Z}^d} 2^d \max \left\{ \left| h(y)^2 - h(z)^2 \right| : y, z \in \overline{D(x)} \cap \mathbb{Z}^d, |y-z|=1 \right\} \\ &\le \sum_{x \in \mathbb{Z}^d} 2^d \sum_{\substack{y, z \in \overline{D(x)} \cap \mathbb{Z}^d \\ |y-z|=1}} \left| h(y)^2 - h(z)^2 \right| \\ &\le 2^{2d} \sum_{\substack{y, z \in \mathbb{Z}^d \\ |y-z|=1}} \left| h(y) - h(z) \right| \times \left| h(y) + h(z) \right| \\ &\le 2^{2d} \left( \sum_{\substack{y, z \in \mathbb{Z}^d \\ |y-z|=1}} \left| h(y) - h(z) \right|^2 \right)^{1/2} \left( \sum_{\substack{y, z \in \mathbb{Z}^d \\ |y-z|=1}} \left| h(y) + h(z) \right|^2 \right)^{1/2} \\ &\le 2^{2d} (2d\mathcal{E}(h))^{1/2} 8d \|h\|_{\ell^2(\mathbb{Z}^d)}. \end{align*} $$

Using inequalities (6.6) and (6.8) (remember that we changed $h$ into $h^2$ just afterwards), we obtain that

$$ (6.18) \qquad \left\| \tilde{h}^2 - \Phi(h) \right\|_{L^1(\mathbb{R}^d)} \le 32d^2 2^{2d} \frac{\sqrt{3\kappa}}{n}. $$

Combining this inequality and inequality (6.15), we conclude that, for $n$ large enough,

$$ (6.19) \qquad \left\| \tilde{h}^2 - \frac{1}{N} \Phi(f) \right\|_{L^1(\mathbb{R}^d)} \le 32d^2 2^{2d} \frac{\sqrt{3\kappa}}{n} + \frac{1}{n^{1/16}} \le \frac{2}{n^{1/16}}. $$

In addition, the function $\tilde{h}$ has bounded support and it is continuous on $\mathbb{R}^d$. Let us denote by $D^d$ the union of all the lines parallel to the axis which go through the points of $\mathbb{Z}^d$. The function $\tilde{h}$ is also $C^\infty$ on $\mathbb{R}^d \setminus D^d$. Thus we can take the infimum
Let now $f$ be an element of $\mathcal{F}$ such that

$$ \left\| h^2 - \frac{1}{N} f^2 \right\|_1 \le \frac{1}{n^{1/16}}. $$

By definition of $\mathcal{F}$, we have that $\Phi_n(f) \in \mathcal{L}$. On the one hand, we have

$$ (6.15) \qquad \left\| \Phi(h) - \frac{1}{N} \Phi(f) \right\|_{L^1(\mathbb{R}^d)} = \left\| h^2 - \frac{1}{N} f^2 \right\|_{\ell^1(\mathbb{Z}^d)} \le \frac{1}{n^{1/16}}. $$

On the other hand, we have

$$ (6.16) \qquad \left\| \tilde{h}^2 - \Phi(h) \right\|_{L^1(\mathbb{R}^d)} = \int_{\mathbb{R}^d} \left| \tilde{h}(x)^2 - h(\lfloor x \rfloor)^2 \right| dx. $$

Let us define

$$ \forall x \in \mathbb{Z}^d \quad D(x) = \{ y \in \mathbb{R}^d : \lfloor y \rfloor = x \}. $$

We can rewrite the previous integral as

$$ \begin{align*} (6.17) \quad \left\| \tilde{h}^2 - \Phi(h) \right\|_{L^1(\mathbb{R}^d)} &= \sum_{x \in \mathbb{Z}^d} \int_{D(x)} \left| \tilde{h}(y)^2 - h(\lfloor y \rfloor)^2 \right| dy \\ &\le \sum_{x \in \mathbb{Z}^d} \sup_{y \in D(x)} \left| \tilde{h}(y)^2 - h(x)^2 \right| \\ &\le \sum_{x \in \mathbb{Z}^d} 2^d \max \left\{ \left| h(y)^2 - h(z)^2 \right| : y, z \in \overline{D(x)} \cap \mathbb{Z}^d, |y-z|=1 \right\} \\ &\le \sum_{x \in \mathbb{Z}^d} 2^d \sum_{\substack{y, z \in \overline{D(x)} \cap \mathbb{Z}^d \\ |y-z|=1}} \left| h(y)^2 - h(z)^2 \right| \\ &\le 2^{2d} \sum_{\substack{y, z \in \mathbb{Z}^d \\ |y-z|=1}} \left| h(y) - h(z) \right| \times \left| h(y) + h(z) \right| \\ &\le 2^{2d} \left( \sum_{\substack{y, z \in \mathbb{Z}^d \\ |y-z|=1}} \left| h(y) - h(z) \right|^2 \right)^{1/2} \left( \sum_{\substack{y, z \in \mathbb{Z}^d \\ |y-z|=1}} \left| h(y) + h(z) \right|^2 \right)^{1/2} \\ &\le 2^{2d} (2d\mathcal{E}(h))^{1/2} 8d \|h\|_{\ell^2(\mathbb{Z}^d)}. \end{align*} $$

Using inequalities (6.6) and (6.8) (remember that we changed $h$ into $h^2$ just afterwards), we obtain that

$$ (6.18) \qquad \left\| \tilde{h}^2 - \Phi(h) \right\|_{L^1(\mathbb{R}^d)} \le 32d^2 2^{2d} \frac{\sqrt{3\kappa}}{n}. $$

Combining this inequality and inequality (6.15), we conclude that, for $n$ large enough,

$$ (6.19) \qquad \left\| \tilde{h}^2 - \frac{1}{N} \Phi(f) \right\|_{L^1(\mathbb{R}^d)} \le 32d^2 2^{2d} \frac{\sqrt{3\kappa}}{n} + \frac{1}{n^{1/16}} \le \frac{2}{n^{1/16}}. $$

In addition, the function $\tilde{h}$ has bounded support and it is continuous on $\mathbb{R}^d$. Let us denote by $D^d$ the union of all the lines parallel to the axis which go through the points of $\mathbb{Z}^d$. The function $\tilde{h}$ is also $C^\infty$ on $\mathbb{R}^d \setminus D^d$. Thus we can take the infimum
over the set of functions $\tilde{h}$ having these properties. We are now ready to substitute $\tilde{h}$ to $h$ in inequality (6.7). We bound from below the infimum in the exponential with the help of inequalities (6.9), (6.11), (6.13) and (6.19):

$$
\begin{equation} \tag{6.20}
\begin{aligned}
& E(e^{-|R_N|}; \ell_N \in \mathcal{L}) \le 3 \exp(-\kappa n^d) \\
& \quad + \exp \left( n^{d-1/8} - \inf \left\{ \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| \right. \right. \\
& \qquad \left. - \frac{d^2 5^d}{\lambda^2} \frac{3\kappa}{n^2} + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\frac{1}{d} \int_{\mathbb{R}^d} |\nabla \tilde{h}|^2 dx} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \\
& \qquad \left. \left. : \tilde{h} \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus D^d, \right. \right. \\
& \qquad \left. \left. \left| \text{supp } \tilde{h} \right| \le (3^d + 1) 2\kappa n^d, \exists \, f \in \mathcal{F} \quad \left\| \tilde{h}^2 - \frac{1}{N} \Phi(f) \right\|_{L^1(\mathbb{R}^d)} \le \frac{2}{n^{1/16}} \right\} \right).
\end{aligned}
\end{equation}
$$

We are almost done. We make a change of scale in order to replace $\mathcal{F}$ by $\mathcal{L}$ in the infimum. To the function $\tilde{h}: \mathbb{R}^d \to \mathbb{R}^+$, we associate the function $\tilde{h}_n: \mathbb{R}^d \to \mathbb{R}^+$ obtained by rescaling the space by a factor $n$ and the values by $n^{d/2}$:

$$
\forall x \in \mathbb{R}^d \quad \tilde{h}_n(x) = n^{d/2} \tilde{h}(nx).
$$

We have then

$$
\begin{equation} \tag{6.21}
\begin{aligned}
& \left| \text{supp} \tilde{h}_n \right| = n^d \left| \text{supp} \tilde{h} \right|, \\
& (6.21) \qquad \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| = n^d \left| \left\{ x \in \mathbb{R}^d : \tilde{h}_n(x) > n^{d/2} \lambda \right\} \right|, \\
& \qquad \int_{\mathbb{R}^d} |\nabla \tilde{h}_n(x)|^2 dx = n^d \int_{\mathbb{R}^d} |n \nabla \tilde{h}(nx)|^2 dx = n^2 \int_{\mathbb{R}^d} |\nabla \tilde{h}(x)|^2 dx.
\end{aligned}
\end{equation}
$$

Moreover, using the identity (6.14), we see that

$$
\begin{equation} \tag{6.22}
\begin{aligned}
& \left\| \tilde{h}_n^2 - \Phi_n(f) \right\|_{L^1(\mathbb{R}^d)} \\
& \quad = \int_{\mathbb{R}^d} n^d \left| \tilde{h}(nx)^2 - \frac{1}{N} \Phi(f)(nx) \right| dx = \left\| \tilde{h}^2 - \frac{1}{N} \Phi(f) \right\|_{L^1(\mathbb{R}^d)}.
\end{aligned}
\end{equation}
$$
over the set of functions $\tilde{h}$ having these properties. We are now ready to substitute $\tilde{h}$ to $h$ in inequality (6.7). We bound from below the infimum in the exponential with the help of inequalities (6.9), (6.11), (6.13) and (6.19):

$$
\begin{equation} \tag{6.20}
\begin{aligned}
& E(e^{-|R_N|}; \ell_N \in \mathcal{L}) \le 3 \exp(-\kappa n^d) \\
& \quad + \exp \left( n^{d-1/8} - \inf \left\{ \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| \right. \right. \\
& \qquad \left. - \frac{d^2 5^d}{\lambda^2} \frac{3\kappa}{n^2} + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\frac{1}{d} \int_{\mathbb{R}^d} |\nabla \tilde{h}|^2 dx} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \\
& \qquad \left. \left. : \tilde{h} \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus D^d, \right. \right. \\
& \qquad \left. \left. \left| \text{supp } \tilde{h} \right| \le (3^d + 1) 2\kappa n^d, \exists \, f \in \mathcal{F} \quad \left\| \tilde{h}^2 - \frac{1}{N} \Phi(f) \right\|_{L^1(\mathbb{R}^d)} \le \frac{2}{n^{1/16}} \right\} \right).
\end{aligned}
\end{equation}
$$

We are almost done. We make a change of scale in order to replace $\mathcal{F}$ by $\mathcal{L}$ in the infimum. To the function $\tilde{h}: \mathbb{R}^d \to \mathbb{R}^+$, we associate the function $\tilde{h}_n: \mathbb{R}^d \to \mathbb{R}^+$ obtained by rescaling the space by a factor $n$ and the values by $n^{d/2}$:

$$
\forall x \in \mathbb{R}^d \quad \tilde{h}_n(x) = n^{d/2} \tilde{h}(nx).
$$

We have then

$$
\begin{equation} \tag{6.21}
\begin{aligned}
& \left| \text{supp} \tilde{h}_n \right| = n^d \left| \text{supp} \tilde{h} \right|, \\
& (6.21) \quad \left| \left\{ x \in \mathbb{R}^d : \tilde{h}(x) > \lambda \right\} \right| = n^d \left| \left\{ x \in \mathbb{R}^d : \tilde{h}_n(x) > n^{d/2} \lambda \right\} \right|, \\
& \qquad \int_{\mathbb{R}^d} |\nabla \tilde{h}_n(x)|^2 dx = n^d \int_{\mathbb{R}^d} |n \nabla \tilde{h}(nx)|^2 dx = n^2 \int_{\mathbb{R}^d} |\nabla \tilde{h}(x)|^2 dx.
\end{aligned}
\end{equation}
$$

Moreover, using the identity (6.14), we see that

$$
\begin{equation} \tag{6.22}
\begin{aligned}
& \left\| \tilde{h}_n^2 - \Phi_n(f) \right\|_{L^1(\mathbb{R}^d)} \\
& \quad = \int_{\mathbb{R}^d} n^d \left| \tilde{h}(nx)^2 - \frac{1}{N} \Phi(f)(nx) \right| dx = \left\| \tilde{h}^2 - \frac{1}{N} \Phi(f) \right\|_{L^1(\mathbb{R}^d)}.
\end{aligned}
\end{equation}
$$
Remember that $\mathcal{F} = \Phi_n^{-1}(\mathcal{L})$, thus $\Phi_n(f) \in \mathcal{L}$ whenever $f \in \mathcal{F}$. The identities (6.21) and (6.22) allow to rewrite (6.20) as follows:

$$
\begin{equation} \tag{6.23}
\begin{aligned}
& E(e^{-|R_N|}; \ell_N \in \mathcal{L}) \le 3 \exp(-\kappa n^d) + \\
& \quad \exp \left( n^{d-1/8} - \inf \left\{ n^d \middle| \left\{ x \in \mathbb{R}^d : \tilde{h}_n(x) > n^{d/2} \lambda \right\} \right| \right. \\
& \qquad \left. - \frac{d^2 5^d 3\kappa}{\lambda^2 n^2} + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\frac{1}{dn^2} \int_{\mathbb{R}^d} |\nabla \tilde{h}_n|^2 dx} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \\
& \qquad \left. : \tilde{h}_n \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus \frac{1}{n} D^d, \quad |\text{supp } \tilde{h}_n| \le (3^d + 1) 2\kappa, \right. \\
& \qquad \left. \exists \ell \in \mathcal{L} \quad \left\| \tilde{h}_n^2 - \ell \right\|_{L^1(\mathbb{R}^d)} \le \frac{2}{n^{1/16}} \right\}.
\end{aligned}
\end{equation}
$$

We choose $\lambda = n^{-\frac{d+1}{2}}$ and we set

$$
\forall x \in \mathbb{R}^d \quad g(x) = \max \left( \tilde{h}_n(x) - \frac{1}{\sqrt{n}}, 0 \right).
$$

This function $g$ satisfies

$$
\begin{equation} \tag{6.24}
\begin{aligned}
\left| \left\{ x \in \mathbb{R}^d : \tilde{h}_n(x) > n^{d/2} \lambda \right\} \right| &= \left| \left\{ x \in \mathbb{R}^d : g(x) > 0 \right\} \right|, \\
&\quad 1 + \frac{2}{n^{1/16}} \ge \int_{\mathbb{R}^d} |\tilde{h}_n|^2 dx \ge \int_{\mathbb{R}^d} |g|^2 dx, \\
&\quad \int_{\mathbb{R}^d} |\nabla \tilde{h}_n|^2 dx \ge \int_{\mathbb{R}^d} |\nabla g|^2 dx.
\end{aligned}
\end{equation}
$$

Moreover we have

$$
\begin{align*}
(6.25) \quad \|g^2 - \tilde{h}_n^2\|_{L^1(\mathbb{R}^d)} &= \int_{\mathbb{R}^d} |g(x)^2 - \tilde{h}_n(x)^2| \, dx \\
&\le \frac{1}{\sqrt{n}} \int_{\mathbb{R}^d} |g(x) + \tilde{h}_n(x)| \, dx \\
&\le \frac{1}{\sqrt{n}} \left( \|g\|_{L^2(\mathbb{R}^d)} (|\text{supp } g|)^{1/2} + \|\tilde{h}_n\|_{L^2(\mathbb{R}^d)} (|\text{supp } \tilde{h}_n|)^{1/2} \right) \\
&\le \frac{2}{\sqrt{n}} \|\tilde{h}_n\|_{L^2(\mathbb{R}^d)} (|\text{supp } \tilde{h}_n|)^{1/2} \\
&\le \frac{4}{\sqrt{n}} ((3^d + 1)2\kappa)^{1/2}.
\end{align*}
$$
Remember that $\mathcal{F} = \Phi_n^{-1}(\mathcal{L})$, thus $\Phi_n(f) \in \mathcal{L}$ whenever $f \in \mathcal{F}$. The identities (6.21) and (6.22) allow to rewrite (6.20) as follows:

$$
\begin{equation} \tag{6.23}
\begin{aligned}
& E(e^{-|R_N|}; \ell_N \in \mathcal{L}) \le 3 \exp(-\kappa n^d) + \\
& \quad \exp \left( n^{d-1/8} - \inf \left\{ n^d \middle| \left\{ x \in \mathbb{R}^d : \tilde{h}_n(x) > n^{d/2} \lambda \right\} \right| \right. \\
& \qquad \left. - \frac{d^2 5^d 3\kappa}{\lambda^2 n^2} + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\frac{1}{dn^2} \int_{\mathbb{R}^d} |\nabla \tilde{h}_n|^2 dx} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \\
& \quad \left. \text{: } \tilde{h}_n \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus \frac{1}{n} D^d, \quad |\text{supp } \tilde{h}_n| \le (3^d + 1) 2\kappa, \right. \\
& \qquad \left. \exists \ell \in \mathcal{L} \quad \left\| \tilde{h}_n^2 - \ell \right\|_{L^1(\mathbb{R}^d)} \le \frac{2}{n^{1/16}} \right\}.
\end{aligned}
\end{equation}
$$

We choose $\lambda = n^{-\frac{d+1}{2}}$ and we set

$$
\forall x \in \mathbb{R}^d \quad g(x) = \max \left( \tilde{h}_n(x) - \frac{1}{\sqrt{n}}, 0 \right).
$$

This function $g$ satisfies

$$
\begin{equation} \tag{6.24}
\begin{aligned}
\left| \left\{ x \in \mathbb{R}^d : \tilde{h}_n(x) > n^{d/2} \lambda \right\} \right| &= \left| \left\{ x \in \mathbb{R}^d : g(x) > 0 \right\} \right|, \\
&\quad 1 + \frac{2}{n^{1/16}} \ge \int_{\mathbb{R}^d} |\tilde{h}_n|^2 dx \ge \int_{\mathbb{R}^d} |g|^2 dx, \\
&\quad \qquad \int_{\mathbb{R}^d} |\nabla \tilde{h}_n|^2 dx \ge \int_{\mathbb{R}^d} |\nabla g|^2 dx.
\end{aligned}
\end{equation}
$$

Moreover we have

$$
\begin{align*}
(6.25) \quad \|g^2 - \tilde{h}_n^2\|_{L^1(\mathbb{R}^d)} &= \int_{\mathbb{R}^d} |g(x)^2 - \tilde{h}_n(x)^2| \, dx \\
&\le \frac{1}{\sqrt{n}} \int_{\mathbb{R}^d} |g(x) + \tilde{h}_n(x)| \, dx \\
&\le \frac{1}{\sqrt{n}} \left( \|g\|_{L^2(\mathbb{R}^d)} (|\text{supp } g|)^{1/2} + \|\tilde{h}_n\|_{L^2(\mathbb{R}^d)} (|\text{supp } \tilde{h}_n|)^{1/2} \right) \\
&\le \frac{2}{\sqrt{n}} \|\tilde{h}_n\|_{L^2(\mathbb{R}^d)} (|\text{supp } \tilde{h}_n|)^{1/2} \\
&\le \frac{4}{\sqrt{n}} ((3^d + 1)2\kappa)^{1/2}.
\end{align*}
$$
Plugging the inequalities (6.24) and (6.25) in the infimum of (6.23), we get, for $n$ large enough,

$$
\begin{aligned}
& E(e^{-|R_N|}; \ell_N \in \mathcal{L}) \\
\le & 3 \exp(-\kappa n^d) + \exp \left( n^{d-1/8} - \inf \left\{ n^d \middle| \text{supp } g \middle| - d2^{5d} 3\kappa n^{d-1} \right. \right. \\
& \left. \left. + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\frac{1}{dn^2} \int_{\mathbb{R}^d} |\nabla g|^2 dx} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \right. \\
& \left. \left. : g \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus \frac{1}{n} D^d, \exists \ell \in \mathcal{L} \; \|g^2 - \ell\|_{L^1(\mathbb{R}^d)} \le \frac{3}{n^{1/16}} \right\} \right).
\end{aligned}
$$

For $n$ large enough, this inequality can be rewritten as in the statement of Theorem 6.2. $\square$

# 7. Application of the quantitative Faber–Krahn inequality

## 7.1. Statement

A crucial ingredient to extend Bolthausen’s result to dimensions $d \ge 3$ is the quantitative Faber–Krahn inequality. Currently, the best version of this inequality is due to Brasco, De Philippis and Velichkov [BDPV15]. A weaker version was proved before by Fusco, Maggi and Pratelli [FMP09], and we could very well rely on this weaker version to achieve our goal. Let $G$ be an open subset of $\mathbb{R}^d$ having finite Lebesgue measure. Let $\lambda(G)$ be the first eigenvalue of the Dirichlet–Laplacian of $G$, defined by

$$ \lambda(G) = \inf \left\{ \int_G |\nabla u|^2 dx : \|u\|_{2,G} = 1 \right\}. $$

Recall that our notation differs from Bolthausen’s by a factor 1/2: the notation $\lambda(G)$ in [Bol94] is half of the quantity defined above.

To control the distance of $G$ to a ball, we define the Fraenkel asymmetry

$$ \mathcal{A}(G) = \inf \left\{ \frac{|G\Delta B|}{|B|} : B \text{ ball such that } |B| = |G| \right\}, $$

where $G\Delta B$ is the symmetric difference between the sets $G$ and $B$.

**THEOREM 7.1.** — There exists a positive constant $\sigma$, which depends on the dimension $d$ only, such that, for any an open subset $G$ of $\mathbb{R}^d$ having finite Lebesgue measure, we have, for any $d$ dimensional ball $B$,

$$ |G|^{2/d}\lambda(G) - |B|^{2/d}\lambda(B) \ge \sigma \mathcal{A}(G)^2. $$

This result is proved by Brasco, De Philippis and Velichkov [BDPV15]. With the help of this quantitative Faber–Krahn inequality, we shall prove the crucial Lemma 7.2, which is the $d$ dimensional counterpart of lemma A.1 in Bolthausen’s
Plugging the inequalities (6.24) and (6.25) in the infimum of (6.23), we get, for $n$ large enough,

$$ E(e^{-|R_N|}; \ell_N \in \mathcal{L}) \\ \le 3 \exp(-\kappa n^d) + \exp \left( n^{d-1/8} - \inf \left\{ n^d \middle| \text{supp } g \middle| - d2^{5d} 3\kappa n^{d-1} \right. \right. \\ \left. \left. + \frac{N}{2} (1 - n^{-1/4}) \left( \max \left( \sqrt{\frac{1}{dn^2} \int_{\mathbb{R}^d} |\nabla g|^2 dx} - \frac{1}{n^{9/8}}, 0 \right) \right)^2 \right. \right. \\ \left. \left. : g \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus \frac{1}{n} D^d, \exists \ell \in \mathcal{L} \; \|g^2 - \ell\|_{L^1(\mathbb{R}^d)} \le \frac{3}{n^{1/16}} \right\} \right). $$

For $n$ large enough, this inequality can be rewritten as in the statement of Theorem 6.2. □

# 7. Application of the quantitative Faber–Krahn inequality

## 7.1. Statement

A crucial ingredient to extend Bolthausen’s result to dimensions $d \ge 3$ is the quantitative Faber–Krahn inequality. Currently, the best version of this inequality is due to Brasco, De Philippis and Velichkov [BDPV15]. A weaker version was proved before by Fusco, Maggi and Pratelli [FMP09], and we could very well rely on this weaker version to achieve our goal. Let $G$ be an open subset of $\mathbb{R}^d$ having finite Lebesgue measure. Let $\lambda(G)$ be the first eigenvalue of the Dirichlet–Laplacian of $G$, defined by

$$ \lambda(G) = \inf \left\{ \int_G |\nabla u|^2 dx : \|u\|_{2,G} = 1 \right\}. $$

Recall that our notation differs from Bolthausen’s by a factor 1/2: the notation $\lambda(G)$ in [Bol94] is half of the quantity defined above.

To control the distance of $G$ to a ball, we define the Fraenkel asymmetry

$$ \mathcal{A}(G) = \inf \left\{ \frac{|G\Delta B|}{|B|} : B \text{ ball such that } |B| = |G| \right\}, $$

where $G\Delta B$ is the symmetric difference between the sets $G$ and $B$.

**THEOREM 7.1.** — There exists a positive constant $\sigma$, which depends on the dimension $d$ only, such that, for any an open subset $G$ of $\mathbb{R}^d$ having finite Lebesgue measure, we have, for any $d$ dimensional ball $B$,

$$ |G|^{2/d}\lambda(G) - |B|^{2/d}\lambda(B) \ge \sigma\mathcal{A}(G)^2. $$

This result is proved by Brasco, De Philippis and Velichkov [BDPV15]. With the help of this quantitative Faber–Krahn inequality, we shall prove the crucial Lemma 7.2, which is the $d$ dimensional counterpart of lemma A.1 in Bolthausen’s
paper [Bol94]. Let $\lambda_d$ be the principal eigenvalue of $-\Delta$ in the unit ball $B(0, 1)$ of $\mathbb{R}^d$, let $\omega_d$ be the volume of $B(0, 1)$, and let us define

$$ \rho_d = \left( \frac{\lambda_d}{d^2 \omega_d} \right)^{\frac{1}{d+2}}. $$

Let $G$ be an open subset of $\mathbb{R}^d$ having finite Lebesgue measure and let $r > 0$ be such that $|G| = r^d \omega_d$. The classical Faber-Krahn inequality states that

$$ (7.1) \qquad \lambda(G) \ge \frac{\lambda_d}{r^2}, $$

while the inequality of Theorem 7.1 can be rewritten as

$$ (7.2) \qquad \lambda(G) \ge \frac{1}{r^2} \left( \lambda_d + \frac{\sigma A(G)^2}{(\omega_d)^{2/d}} \right). $$

By the classical Faber-Krahn inequality, we have

$$ |G| + \frac{1}{2d} \lambda(G) \ge r^d \omega_d + \frac{\lambda_d}{2dr^2}. $$

Let us define

$$ (7.3) \qquad \forall r > 0 \quad \psi(r) = r^d \omega_d + \frac{\lambda_d}{2dr^2}. $$

The function $\psi$ admits a unique minimum on $\mathbb{R}^+$ at $r = \rho_d$. The constant $\chi_d$ is defined through the variational formula

$$ \chi_d = \inf \left\{ |G| + \frac{1}{2d} \lambda(G) : G \text{ open subset of } \mathbb{R}^d \right\}. $$

We conclude from the previous inequalities that

$$ \chi_d = \frac{d+2}{2} \left( \frac{\lambda_d}{d^2} \right)^{\frac{d}{d+2}} (\omega_d)^{\frac{2}{d+2}}. $$

Let $\phi$ be the eigenfunction of $-\Delta$ in $B(0, \rho_d)$ with Dirichlet boundary conditions associated to $\lambda_d$ and normalized so that $\|\phi\|_2 = 1$, $\phi \ge 0$. We extend $\phi$ to $\mathbb{R}^d$ by setting it equal to 0 outside $B(0, \rho_d)$. For $x \in \mathbb{R}^d$, we denote by $\phi_x$ the translate of $\phi$ defined by

$$ \forall y \in \mathbb{R}^d \quad \phi_x(y) = \phi(y - x). $$

We state next the counterpart of Lemma 7.2 [Bol94, Lemma A.1 of Bolthausen's paper]. The difference is that we work in dimensions $d \ge 3$ and in the full space rather than in the torus. The spirit of the proof is exactly the same as in Bolthausen's case, the major new input is the quantitative Faber-Krahn inequality. Equipped with this powerful inequality, the proof becomes more transparent than Bolthausen's proof, which contains somehow a two-dimensional version of a quantitative isoperimetric inequality.
paper [Bol94]. Let $\lambda_d$ be the principal eigenvalue of $-\Delta$ in the unit ball $B(0, 1)$ of $\mathbb{R}^d$, let $\omega_d$ be the volume of $B(0, 1)$, and let us define

$$ \rho_d = \left( \frac{\lambda_d}{d^2 \omega_d} \right)^{\frac{1}{d+2}}. $$

Let $G$ be an open subset of $\mathbb{R}^d$ having finite Lebesgue measure and let $r > 0$ be such that $|G| = r^d \omega_d$. The classical Faber-Krahn inequality states that

$$ (7.1) \qquad \lambda(G) \ge \frac{\lambda_d}{r^2}, $$

while the inequality of Theorem 7.1 can be rewritten as

$$ (7.2) \qquad \lambda(G) \ge \frac{1}{r^2} \left( \lambda_d + \frac{\sigma A(G)^2}{(\omega_d)^{2/d}} \right). $$

By the classical Faber-Krahn inequality, we have

$$ |G| + \frac{1}{2d} \lambda(G) \ge r^d \omega_d + \frac{\lambda_d}{2dr^2}. $$

Let us define

$$ (7.3) \qquad \forall r > 0 \quad \psi(r) = r^d \omega_d + \frac{\lambda_d}{2dr^2}. $$

The function $\psi$ admits a unique minimum on $\mathbb{R}^+$ at $r = \rho_d$. The constant $\chi_d$ is defined through the variational formula

$$ \chi_d = \inf \left\{ |G| + \frac{1}{2d} \lambda(G) : G \text{ open subset of } \mathbb{R}^d \right\}. $$

We conclude from the previous inequalities that

$$ \chi_d = \frac{d+2}{2} \left( \frac{\lambda_d}{d^2} \right)^{\frac{d}{d+2}} (\omega_d)^{\frac{2}{d+2}}. $$

Let $\phi$ be the eigenfunction of $-\Delta$ in $B(0, \rho_d)$ with Dirichlet boundary conditions associated to $\lambda_d$ and normalized so that $\|\phi\|_2 = 1$, $\phi \ge 0$. We extend $\phi$ to $\mathbb{R}^d$ by setting it equal to 0 outside $B(0, \rho_d)$. For $x \in \mathbb{R}^d$, we denote by $\phi_x$ the translate of $\phi$ defined by

$$ \forall y \in \mathbb{R}^d \quad \phi_x(y) = \phi(y - x). $$

We state next the counterpart of Lemma 7.2 [Bol94, Lemma A.1 of Bolthausen's paper]. The difference is that we work in dimensions $d \ge 3$ and in the full space rather than in the torus. The spirit of the proof is exactly the same as in Bolthausen's case, the major new input is the quantitative Faber-Krahn inequality. Equipped with this powerful inequality, the proof becomes more transparent than Bolthausen's proof, which contains somehow a two-dimensional version of a quantitative isoperimetric inequality.
LEMMA 7.2. — If $g : \mathbb{R}^d \to \mathbb{R}^+$ is $C^\infty$ and such that $\|g\|_2 = 1$, $g \ge 0$, and if

$$ \varepsilon = \inf_{x \in \mathbb{R}^d} \|g - \phi_x\|_2 > 0 $$

is small enough, then

$$ |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \chi_d + \varepsilon^{48}. $$

*Proof.* — There are many constants involved throughout this proof. So we denote by $c$ a generic positive constant which depends only on the dimension $d$, and we warn that of course the value of $c$ changes from one formula to another! Let $g$ be a function as in the statement of the lemma and let us set

$$ G = \{ x \in \mathbb{R}^d : g(x) > 0 \}. $$

We need only to consider the case where $|G| < \infty$ and the function $g$ is such that

$$ |G| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \le 2\chi_d. $$

The Sobolev inequality implies then that there exists a constant $c_S(d)$ depending on the dimension only such that

$$ (7.4) \qquad \|g\|_4^2 \le c_S(d) \int_{\mathbb{R}^d} \left(g^2 + |\nabla g|^2\right) dx \le c_S(d)(1 + 4d\chi_d). $$

Let $r > 0$ be such that $|G| = r^d \omega_d$. Applying inequality (7.2) to the set $G$, we obtain

$$ (7.5) \qquad \begin{aligned} |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx &\ge r^d \omega_d + \frac{1}{2d} \lambda(G) \\ &\ge r^d \omega_d + \frac{1}{2dr^2} \left( \lambda_d + \frac{\sigma A(G)^2}{(\omega_d)^{2/d}} \right) = \psi(r) + \frac{\sigma A(G)^2}{2dr^2(\omega_d)^{2/d}}, \end{aligned} $$

where $\psi(r)$ is the function defined in (7.3). We recall that the function $\psi$ admits a unique minimum at $r = \rho_d$ and moreover $\psi''(\rho_d) > 0$. Therefore there exist two positive constants $\eta, c$ such that $\eta < \rho_d/2$ and

$$ (7.6) \qquad \forall r \in ]\rho_d - \eta, \rho_d + \eta[ \quad \psi(r) \ge \psi(\rho_d) + c(r - \rho_d)^2. $$

For $\varepsilon_0 > 0$ small enough, we can take $\eta$ sufficiently small to ensure that we have in addition

$$ \forall r \in \mathbb{R}^+ \setminus ]\rho_d - \eta/2, \rho_d + \eta/2[ \quad \psi(r) \ge \psi(\rho_d) + (\varepsilon_0)^{48}. $$

In particular, the inequality stated in the Lemma 7.2 holds if $\varepsilon < \varepsilon_0$ and if $r$ does not belong to the interval $]\rho_d - \eta/2, \rho_d + \eta/2[$. From now onwards, we suppose that $\varepsilon < \varepsilon_0$ and we consider only the cases of sets $G$ such that $\rho_d - \eta/2 < r < \rho_d + \eta/2$. From inequality (7.6), we see that we need only to consider the case where

$$ (7.7) \qquad c(r - \rho_d)^2 \le \varepsilon^{48}. $$

Recalling that $\psi(r) \ge \chi_d$, it follows from (7.5) that if

$$ \frac{\sigma A(G)^2}{8d(\rho_d)^2(\omega_d)^{2/d}} \ge \varepsilon^{48}, $$
LEMMA 7.2. — If $g: \mathbb{R}^d \to \mathbb{R}^+$ is $C^\infty$ and such that $\|g\|_2 = 1$, $g \ge 0$, and if

$$ \varepsilon = \inf_{x \in \mathbb{R}^d} \|g - \phi_x\|_2 > 0 $$

is small enough, then

$$ |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \chi_d + \varepsilon^{48}. $$

*Proof.* — There are many constants involved throughout this proof. So we denote by $c$ a generic positive constant which depends only on the dimension $d$, and we warn that of course the value of $c$ changes from one formula to another! Let $g$ be a function as in the statement of the lemma and let us set

$$ G = \{ x \in \mathbb{R}^d : g(x) > 0 \}. $$

We need only to consider the case where $|G| < \infty$ and the function $g$ is such that

$$ |G| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \le 2\chi_d. $$

The Sobolev inequality implies then that there exists a constant $c_S(d)$ depending on the dimension only such that

$$ (7.4) \qquad \|g\|_4^2 \le c_S(d) \int_{\mathbb{R}^d} \left(g^2 + |\nabla g|^2\right) dx \le c_S(d)(1 + 4d\chi_d). $$

Let $r > 0$ be such that $|G| = r^d \omega_d$. Applying inequality (7.2) to the set $G$, we obtain

$$ (7.5) \qquad \begin{aligned} |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx &\ge r^d \omega_d + \frac{1}{2d} \lambda(G) \\ &\ge r^d \omega_d + \frac{1}{2dr^2} \left( \lambda_d + \frac{\sigma A(G)^2}{(\omega_d)^{2/d}} \right) = \psi(r) + \frac{\sigma A(G)^2}{2dr^2(\omega_d)^{2/d}}, \end{aligned} $$

where $\psi(r)$ is the function defined in (7.3). We recall that the function $\psi$ admits a unique minimum at $r = \rho_d$ and moreover $\psi''(\rho_d) > 0$. Therefore there exist two positive constants $\eta, c$ such that $\eta < \rho_d/2$ and

$$ (7.6) \qquad \forall r \in ]\rho_d - \eta, \rho_d + \eta[ \quad \psi(r) \ge \psi(\rho_d) + c(r - \rho_d)^2. $$

For $\varepsilon_0 > 0$ small enough, we can take $\eta$ sufficiently small to ensure that we have in addition

$$ \forall r \in \mathbb{R}^+ \setminus ]\rho_d - \eta/2, \rho_d + \eta/2[ \quad \psi(r) \ge \psi(\rho_d) + (\varepsilon_0)^{48}. $$

In particular, the inequality stated in the Lemma 7.2 holds if $\varepsilon < \varepsilon_0$ and if $r$ does not belong to the interval $]\rho_d - \eta/2, \rho_d + \eta/2[$. From now onwards, we suppose that $\varepsilon < \varepsilon_0$ and we consider only the cases of sets $G$ such that $\rho_d - \eta/2 < r < \rho_d + \eta/2$. From inequality (7.6), we see that we need only to consider the case where

$$ (7.7) \qquad c(r - \rho_d)^2 \le \varepsilon^{48}. $$

Recalling that $\psi(r) \ge \chi_d$, it follows from (7.5) that if

$$ \frac{\sigma A(G)^2}{8d(\rho_d)^2(\omega_d)^{2/d}} \ge \varepsilon^{48}, $$
then the inequality of the Lemma 7.2 is satisfied. From now onwards, we consider only the cases of sets G satisfying

$$ (7.8) \qquad \mathcal{A}(G) < \sqrt{\frac{1}{\sigma} 8d(\rho_d)^2 (\omega_d)^{2/d} \varepsilon^{48}} = c\varepsilon^{24}. $$

Therefore there exists $x_0 \in \mathbb{R}^d$ and $r_0 > 0$ such that

$$ (7.9) \qquad |G\Delta B(x_0, r_0)| \le c(2\rho_d)^d \omega_d \varepsilon^{24} \le c\varepsilon^{24}. $$

This inequality, together with inequality (7.7), imply that

$$ (7.10) \qquad |r_0 - \rho_d| \le c\varepsilon^{24}. $$

Let $\delta > 0$. We shall compare $\int_{\mathbb{R}^d} |\nabla g|^2 dx$ with $\lambda(B(x_0, r_0 + \delta))$. To this end, we truncate smoothly the function $g$ as follows. Let $h$ be a $C^\infty$ function from $\mathbb{R}^d$ to $[0, 1]$ satisfying

$$ \forall x \in \mathbb{R}^d \quad h(x) = \begin{cases} 1 & \text{if } x \in B(x_0, r_0) \\ 0 & \text{if } x \notin B(x_0, r_0 + \delta) \end{cases}, $$

as well as the following bound on its gradient:

$$ \forall x \in \mathbb{R}^d \quad |\nabla h(x)| \le \frac{c}{\delta}. $$

We set $\tilde{g} = hg$ and we estimate the Dirichlet energy of $\tilde{g}$ as follows. Let

$$ A = B(x_0, r_0 + \delta) \setminus B(x_0, r_0), $$

we have

$$ (7.11) \qquad \int_{\mathbb{R}^d} |\nabla \tilde{g}|^2 dx = \int_{B(x_0, r_0)} |\nabla g|^2 dx + \int_A |\nabla(hg)|^2 dx \\ \le \int_{\mathbb{R}^d} |\nabla g|^2 dx + \int_A |\nabla h|^2 g^2 dx + \int_A 2hg |\nabla h| |\nabla g| dx. $$

Moreover

$$ \int_A |\nabla h|^2 g^2 dx \le \left(\frac{c}{\delta}\right)^2 \int_{A \cap G} g^2 dx. $$

Yet, using (7.9), we have

$$ (7.12) \qquad |A \cap G| \le |G \setminus B(x_0, r_0)| \le c\varepsilon^{24}, $$

and, by Hölder's inequality,

$$ (7.13) \qquad \int_{A \cap G} g^2 dx \le \int_{G \setminus B(x_0, r_0)} g^2 dx \\ \le \left( \int_{\mathbb{R}^d} g^4 dx \right)^{\frac{1}{2}} |G \setminus B(x_0, r_0)|^{\frac{1}{2}} \le \|g\|_4^2 \sqrt{c\varepsilon^{24}}. $$
then the inequality of the Lemma 7.2 is satisfied. From now onwards, we consider only the cases of sets G satisfying

$$ (7.8) \qquad \mathcal{A}(G) < \sqrt{\frac{1}{\sigma} 8d(\rho_d)^2 (\omega_d)^{2/d} \varepsilon^{48}} = c\varepsilon^{24}. $$

Therefore there exists $x_0 \in \mathbb{R}^d$ and $r_0 > 0$ such that

$$ (7.9) \qquad |G\Delta B(x_0, r_0)| \le c(2\rho_d)^d \omega_d \varepsilon^{24} \le c\varepsilon^{24}. $$

This inequality, together with inequality (7.7), imply that

$$ (7.10) \qquad |r_0 - \rho_d| \le c\varepsilon^{24}. $$

Let $\delta > 0$. We shall compare $\int_{\mathbb{R}^d} |\nabla g|^2 dx$ with $\lambda(B(x_0, r_0 + \delta))$. To this end, we truncate smoothly the function $g$ as follows. Let $h$ be a $C^\infty$ function from $\mathbb{R}^d$ to $[0, 1]$ satisfying

$$ \forall x \in \mathbb{R}^d \quad h(x) = \begin{cases} 1 & \text{if } x \in B(x_0, r_0) \\ 0 & \text{if } x \notin B(x_0, r_0 + \delta) \end{cases}, $$

as well as the following bound on its gradient:

$$ \forall x \in \mathbb{R}^d \quad |\nabla h(x)| \le \frac{c}{\delta}. $$

We set $\tilde{g} = hg$ and we estimate the Dirichlet energy of $\tilde{g}$ as follows. Let

$$ A = B(x_0, r_0 + \delta) \setminus B(x_0, r_0), $$

we have

$$ (7.11) \qquad \int_{\mathbb{R}^d} |\nabla \tilde{g}|^2 dx = \int_{B(x_0, r_0)} |\nabla g|^2 dx + \int_A |\nabla(hg)|^2 dx \\ \le \int_{\mathbb{R}^d} |\nabla g|^2 dx + \int_A |\nabla h|^2 g^2 dx + \int_A 2hg |\nabla h| |\nabla g| dx. $$

Moreover

$$ \int_A |\nabla h|^2 g^2 dx \le \left(\frac{c}{\delta}\right)^2 \int_{A \cap G} g^2 dx. $$

Yet, using (7.9), we have

$$ (7.12) \qquad |A \cap G| \le |G \setminus B(x_0, r_0)| \le c\varepsilon^{24}, $$

and, by Hölder's inequality,

$$ (7.13) \qquad \int_{A \cap G} g^2 dx \le \int_{G \setminus B(x_0, r_0)} g^2 dx \\ \le \left( \int_{\mathbb{R}^d} g^4 dx \right)^{\frac{1}{2}} |G \setminus B(x_0, r_0)|^{\frac{1}{2}} \le \|g\|_4^2 \sqrt{c\varepsilon^{24}}. $$
We control the last integral of (7.11) as follows:

$$
\begin{align*}
(7.14) \quad \int_A 2|hg| |\nabla h| |\nabla g| dx &\le 2^{\frac{c}{\delta}} \int_{A \cap G} |g\nabla g| dx \\
&\le 2^{\frac{c}{\delta}} \left( \int_{\mathbb{R}^d} |\nabla g|^2 dx \right)^{\frac{1}{2}} \left( \int_{A \cap G} g^2 dx \right)^{\frac{1}{2}}.
\end{align*}
$$

We plug these inequalities in (7.11). Together with inequality (7.4), we obtain

$$
(7.15) \quad \int_{\mathbb{R}^d} |\nabla \tilde{g}|^2 dx \le \int_{\mathbb{R}^d} |\nabla g|^2 dx + c \frac{\varepsilon^{12}}{\delta^2} + c \frac{\varepsilon^6}{\delta}.
$$

The end of the argument is the same as in [Bol94, Lemma A.1]. We denote by $\hat{\phi}_0$
the normalized eigenfunction in $B(x_0, r_0 + \delta)$, that is,

$$
\forall x \in B(x_0, r_0 + \delta) \hat{\phi}_0(x) = \left( \frac{\rho_d}{r_0 + \delta} \right)^{d/2} \phi \left( \frac{\rho_d}{r_0 + \delta} (x - x_0) \right).
$$

With a change of variables, we obtain that

$$
\int_{\mathbb{R}^d} (\hat{\phi}_0)^2 dx = 1
$$

and

$$
\int_{\mathbb{R}^d} |\nabla \hat{\phi}_0|^2 dx = \left( \frac{\rho_d}{r_0 + \delta} \right)^2 \int_{\mathbb{R}^d} |\nabla \phi|^2 dx.
$$

We have then, using (7.12),

$$
\begin{align*}
(7.16) \quad |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx
  &\ge |B(x_0, r_0)| - c\epsilon^{24} + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla \tilde{g}|^2 dx - c\frac{\epsilon^{12}}{\delta^2} - c\frac{\epsilon^6}{\delta} \\
  &\ge |B(x_0, r_0 + \delta)| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla \hat{\phi}_0|^2 dx - c\delta - c\epsilon^{24} \\
  &\phantom{\ge} + \frac{1}{2d} \int_{\mathbb{R}^d} \left( |\nabla \tilde{g}|^2 - |\nabla \hat{\phi}_0|^2 \right) dx - c\frac{\epsilon^{12}}{\delta^2} - c\frac{\epsilon^6}{\delta}.
\end{align*}
$$

By inequality (7.10), we have

$$
(7.17) \qquad |r_0 + \delta - \rho_d| \le c\epsilon^{24} + \delta.
$$

Thus, for $\epsilon$ and $\delta$ small enough, the value $r_0 + \delta$ belongs to the interval $]\rho_d - \eta, \rho_d + \eta[$ and we can apply (7.6) to get

$$
(7.18) \quad |B(x_0, r_0 + \delta)| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla \hat{\phi}_0|^2 dx \ge \psi(r_0 + \delta) \ge \psi(\rho_d) + c(r_0 + \delta - \rho_d)^2.
$$

To estimate the second integral in (7.16), we proceed as in [Bol94, Lemma A.1].
We denote by $\delta_{12}$ the difference between the first and the second eigenvalues of the
Laplacian in $B(x_0, r_0 + \delta)$. We have

$$
\frac{1}{2d} \int_{\mathbb{R}^d} \left( |\nabla \tilde{g}|^2 - |\nabla \hat{\phi}_0|^2 \right) dx \ge \delta_{12} \|\tilde{g} - \hat{\phi}_0\|_2^2.
$$
We control the last integral of (7.11) as follows:

$$
\begin{align*}
(7.14) \quad \int_A 2|hg| |\nabla h| |\nabla g| dx &\le 2^{\frac{c}{\delta}} \int_{A \cap G} |g\nabla g| dx \\
&\le 2^{\frac{c}{\delta}} \left( \int_{\mathbb{R}^d} |\nabla g|^2 dx \right)^{\frac{1}{2}} \left( \int_{A \cap G} g^2 dx \right)^{\frac{1}{2}}.
\end{align*}
$$

We plug these inequalities in (7.11). Together with inequality (7.4), we obtain

$$
(7.15) \quad \int_{\mathbb{R}^d} |\nabla \tilde{g}|^2 dx \le \int_{\mathbb{R}^d} |\nabla g|^2 dx + c \frac{\varepsilon^{12}}{\delta^2} + c \frac{\varepsilon^6}{\delta}.
$$

The end of the argument is the same as in [Bol94, Lemma A.1]. We denote by $\hat{\phi}_0$
the normalized eigenfunction in $B(x_0, r_0 + \delta)$, that is,

$$
\forall x \in B(x_0, r_0 + \delta) \hat{\phi}_0(x) = \left( \frac{\rho_d}{r_0 + \delta} \right)^{d/2} \phi \left( \frac{\rho_d}{r_0 + \delta} (x - x_0) \right).
$$

With a change of variables, we obtain that

$$
\int_{\mathbb{R}^d} (\hat{\phi}_0)^2 dx = 1
$$

and

$$
\int_{\mathbb{R}^d} |\nabla \hat{\phi}_0|^2 dx = \left( \frac{\rho_d}{r_0 + \delta} \right)^2 \int_{\mathbb{R}^d} |\nabla \phi|^2 dx.
$$

We have then, using (7.12),

$$
\begin{align*}
(7.16) \quad |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx
  &\ge |B(x_0, r_0)| - c\varepsilon^{24} + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla \tilde{g}|^2 dx - c\frac{\varepsilon^{12}}{\delta^2} - c\frac{\varepsilon^6}{\delta} \\
  &\ge |B(x_0, r_0 + \delta)| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla \hat{\phi}_0|^2 dx - c\delta - c\varepsilon^{24} \\
  &\phantom{\ge} + \frac{1}{2d} \int_{\mathbb{R}^d} \left( |\nabla \tilde{g}|^2 - |\nabla \hat{\phi}_0|^2 \right) dx - c\frac{\varepsilon^{12}}{\delta^2} - c\frac{\varepsilon^6}{\delta}.
\end{align*}
$$

By inequality (7.10), we have

$$
(7.17) \qquad |r_0 + \delta - \rho_d| \le c\epsilon^{24} + \delta.
$$

Thus, for $\epsilon$ and $\delta$ small enough, the value $r_0 + \delta$ belongs to the interval $]\rho_d - \eta, \rho_d + \eta[$ and we can apply (7.6) to get

$$
(7.18) \quad |B(x_0, r_0 + \delta)| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla \hat{\phi}_0|^2 dx \ge \psi(r_0 + \delta) \ge \psi(\rho_d) + c(r_0 + \delta - \rho_d)^2.
$$

To estimate the second integral in (7.16), we proceed as in [Bol94, Lemma A.1].
We denote by $\delta_{12}$ the difference between the first and the second eigenvalues of the
Laplacian in $B(x_0, r_0 + \delta)$. We have

$$
\frac{1}{2d} \int_{\mathbb{R}^d} \left( |\nabla \tilde{g}|^2 - |\nabla \hat{\phi}_0|^2 \right) dx \ge \delta_{12} \|\tilde{g} - \hat{\phi}_0\|_2^2.
$$
Again, since $r_0 + \delta$ belongs to the interval $]\rho_d - \eta, \rho_d + \eta[$, which is included in $]\rho_d/2, 3\rho_d/2[$, there exists a constant $c > 0$ depending on the dimension $d$ only such that $\delta_{12} \ge c$. Moreover we have

$$ \varepsilon \le \|g - \phi_{x_0}\|_2 \le \|g - \tilde{g}\|_2 + \|\tilde{g} - \hat{\phi}_0\|_2 + \|\hat{\phi}_0 - \phi_{x_0}\|_2. $$

Now, thanks to inequalities (7.4) and (7.13),

$$ \|g - \tilde{g}\|_2^2 = \int_{\mathbb{R}^d} \left( (1-h)g \right)^2 dx \le \int_{G \setminus B(x_0, r_0)} g^2 dx \le c\varepsilon^{12}. $$

Using the fact that the eigenfunction $\phi$ has bounded support and is Lipschitz, we have

$$ \begin{aligned} \|\hat{\phi}_0 - \phi_{x_0}\|_2^2 &= \int_{\mathbb{R}^d} \left( \left( \frac{\rho_d}{r_0 + \delta} \right)^{d/2} \phi \left( \frac{\rho_d}{r_0 + \delta} (x - x_0) \right) - \phi(x - x_0) \right)^2 dx \\ &\le c \left( \left( \frac{\rho_d}{r_0 + \delta} \right)^{d/2} - 1 \right)^2 + c \left( \left( \frac{\rho_d}{r_0 + \delta} \right) - 1 \right)^2 \\ &\le c (r_0 + \delta - \rho_d)^2. \end{aligned} $$

The previous inequalities yield

$$ \frac{1}{2d} \int_{\mathbb{R}^d} \left( |\nabla \tilde{g}|^2 - |\nabla \hat{\phi}_0|^2 \right) dx \ge c \left( \varepsilon - c\varepsilon^{12} - c(r_0 + \delta - \rho_d)^2 \right)^2. $$

Reporting in the inequality (7.16), and using inequalities (7.17) and (7.18), we get

$$ \begin{aligned} \left| \{ g > 0 \} \right| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx &\ge \psi(\rho_d) + c \left( c\varepsilon - c\varepsilon^{12} - c(c\varepsilon^{24} + \delta)^2 \right)^2 - c\delta - c\varepsilon^{24} - c \frac{\varepsilon^{12}}{\delta^2} - c \frac{\varepsilon^6}{\delta}. \end{aligned} $$

Choosing $\delta = \varepsilon^3$, we conclude that, for $\varepsilon$ small enough,

$$ \left| \{ g > 0 \} \right| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \chi_d + c\varepsilon^2. $$

Therefore the inequality stated in the lemma is satisfied for $\varepsilon$ small enough whenever the conditions (7.7) and (7.8) are fulfilled. $\square$

We shall extend slightly Lemma 7.2, in two ways. First, we will relax the condition $\|g\|_2 = 1$, second, we will consider functions which are not $C^\infty$ on the whole space, but on the complement of a countable union of lines. We could probably reach the Sobolev space $W^{1,2}(\mathbb{R}^d)$, however this won't be necessary for our purpose.

**COROLLARY 7.3.** — Let $D$ be a subset of $\mathbb{R}^d$ which is a countable union of lines. Let $\varepsilon > 0$. If $g : \mathbb{R}^d \to \mathbb{R}^+$ is continuous and $C^\infty$ on $\mathbb{R}^d \setminus D$, and if

$$ 1 - \varepsilon^{49} \le \|g\|_2 \le 1 + \varepsilon^{49}, \quad \forall x \in \mathbb{R}^d \quad \|g - \phi_x\|_2 \ge 2\varepsilon, $$

then, for $\varepsilon$ small enough,

$$ \left| \{ g > 0 \} \right| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \chi_d + \varepsilon^{49}. $$
Again, since $r_0 + \delta$ belongs to the interval $]\rho_d - \eta, \rho_d + \eta[$, which is included in $]\rho_d/2, 3\rho_d/2[$, there exists a constant $c > 0$ depending on the dimension $d$ only such that $\delta_{12} \ge c$. Moreover we have

$$ \varepsilon \le \|g - \phi_{x_0}\|_2 \le \|g - \tilde{g}\|_2 + \|\tilde{g} - \hat{\phi}_0\|_2 + \|\hat{\phi}_0 - \phi_{x_0}\|_2. $$

Now, thanks to inequalities (7.4) and (7.13),

$$ \|g - \tilde{g}\|_2^2 = \int_{\mathbb{R}^d} \left( (1-h)g \right)^2 dx \le \int_{G \setminus B(x_0, r_0)} g^2 dx \le c\varepsilon^{12}. $$

Using the fact that the eigenfunction $\phi$ has bounded support and is Lipschitz, we have

$$ \begin{aligned} \|\hat{\phi}_0 - \phi_{x_0}\|_2^2 &= \int_{\mathbb{R}^d} \left( \left( \frac{\rho_d}{r_0 + \delta} \right)^{d/2} \phi \left( \frac{\rho_d}{r_0 + \delta} (x - x_0) \right) - \phi(x - x_0) \right)^2 dx \\ &\le c \left( \left( \frac{\rho_d}{r_0 + \delta} \right)^{d/2} - 1 \right)^2 + c \left( \left( \frac{\rho_d}{r_0 + \delta} \right) - 1 \right)^2 \\ &\le c (r_0 + \delta - \rho_d)^2. \end{aligned} $$

The previous inequalities yield

$$ \frac{1}{2d} \int_{\mathbb{R}^d} \left( |\nabla \tilde{g}|^2 - |\nabla \hat{\phi}_0|^2 \right) dx \ge c \left( \varepsilon - c\varepsilon^{12} - c(r_0 + \delta - \rho_d)^2 \right)^2. $$

Reporting in the inequality (7.16), and using inequalities (7.17) and (7.18), we get

$$ \begin{aligned} &|\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \\ &\ge \psi(\rho_d) + c \left( c\varepsilon - c\varepsilon^{12} - c(c\varepsilon^{24} + \delta)^2 \right)^2 - c\delta - c\varepsilon^{24} - c \frac{\varepsilon^{12}}{\delta^2} - c \frac{\varepsilon^6}{\delta}. \end{aligned} $$

Choosing $\delta = \varepsilon^3$, we conclude that, for $\varepsilon$ small enough,

$$ |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \chi_d + c\varepsilon^2. $$

Therefore the inequality stated in the lemma is satisfied for $\varepsilon$ small enough whenever the conditions (7.7) and (7.8) are fulfilled. $\square$

We shall extend slightly Lemma 7.2, in two ways. First, we will relax the condition $\|g\|_2 = 1$, second, we will consider functions which are not $C^\infty$ on the whole space, but on the complement of a countable union of lines. We could probably reach the Sobolev space $W^{1,2}(\mathbb{R}^d)$, however this won't be necessary for our purpose.

**COROLLARY 7.3.** — Let $D$ be a subset of $\mathbb{R}^d$ which is a countable union of lines. Let $\varepsilon > 0$. If $g : \mathbb{R}^d \to \mathbb{R}^+$ is continuous and $C^\infty$ on $\mathbb{R}^d \setminus D$, and if

$$ 1 - \varepsilon^{49} \le \|g\|_2 \le 1 + \varepsilon^{49}, \quad \forall x \in \mathbb{R}^d \quad \|g - \phi_x\|_2 \ge 2\varepsilon, $$

then, for $\varepsilon$ small enough,

$$ |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \chi_d + \varepsilon^{49}. $$
Proof. — The condition on the regularity is not problematic, in fact the proof of Lemma 7.2 can be used to deal directly with these functions. The only thing we need to do is to rescale the function $g$ in order to have a function of $L^2$ norm one, for which we can use Lemma 7.2. So let $g$ be a function satisfying the hypothesis of Corollary 7.3 and let us set

$$h = \frac{1}{\|g\|_2} g.$$

We have obviously $\|h\|_2 = 1$. Moreover, for any $x \in \mathbb{R}^d$,

$$2\varepsilon \le \|g - \phi_x\|_2 \le \|g - h\|_2 + \|h - \phi_x\|_2 \\ \le \left|1 - \frac{1}{\|g\|_2}\right| \|g\|_2 + \|h - \phi_x\|_2 \le \varepsilon^{49} + \|h - \phi_x\|_2.$$

We can thus apply the inequality of Lemma 7.2 to $h$. Moreover, we have

$$|\{h > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla h|^2 dx = |\{g > 0\}| + \frac{1}{2d(\|g\|_2)^2} \int_{\mathbb{R}^d} |\nabla g|^2 dx \\ \le \frac{1}{(1 - \varepsilon^{49})^2} \left( |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \right).$$

In the end, we get that, for $\varepsilon$ small enough,

$$|\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge (1 - \varepsilon^{49})^2 (\chi_d + \varepsilon^{48}),$$

and the last term is larger than $\chi_d + \varepsilon^{49}$ for $\varepsilon$ sufficiently small. $\square$

<figure><img src="image_7.png" /></figure>

## 7.2. Application of Faber-Krahn: proof of Theorem 1.1

The time has come to apply the quantitative Faber-Krahn inequality to the random walk in order to prove Theorem 1.1. Note that Theorem 1.1 is the analogue of [Bol94, Proposition 3.1].

We shall apply Theorem 6.2 to the set $\mathcal{L}_n$. To this end, we consider a function $g$ satisfying the constraints of the infimum appearing in Theorem 6.2. So, let $g$ be a continuous function $\mathbb{R}^d \to \mathbb{R}^+$ such that $g$ is $C^\infty$ on $\mathbb{R}^d \setminus \frac{1}{n}D^d$, and there exists $\ell \in \mathcal{L}_n$ satisfying

$$\|g^2 - \ell\|_{L^1(\mathbb{R}^d)} \le \frac{3}{n^{1/16}}.$$

This implies in particular that

$$1 - \frac{3}{n^{1/16}} \le \|g\|_2 \le 1 + \frac{3}{n^{1/16}}.$$
*Proof.* — The condition on the regularity is not problematic, in fact the proof of Lemma 7.2 can be used to deal directly with these functions. The only thing we need to do is to rescale the function *g* in order to have a function of *L*² norm one, for which we can use Lemma 7.2. So let *g* be a function satisfying the hypothesis of Corollary 7.3 and let us set

$$h = \frac{1}{\|g\|_2} g.$$

We have obviously $\|h\|_2 = 1$. Moreover, for any $x \in \mathbb{R}^d$,

$$2\varepsilon \le \|g - \phi_x\|_2 \le \|g - h\|_2 + \|h - \phi_x\|_2 \\ \le \left|1 - \frac{1}{\|g\|_2}\right| \|g\|_2 + \|h - \phi_x\|_2 \le \varepsilon^{49} + \|h - \phi_x\|_2.$$

We can thus apply the inequality of Lemma 7.2 to $h$. Moreover, we have

$$|\{h > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla h|^2 dx = |\{g > 0\}| + \frac{1}{2d(\|g\|_2)^2} \int_{\mathbb{R}^d} |\nabla g|^2 dx \\ \le \frac{1}{(1 - \varepsilon^{49})^2} \left( |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \right).$$

In the end, we get that, for $\varepsilon$ small enough,

$$|\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge (1 - \varepsilon^{49})^2 (\chi_d + \varepsilon^{48}),$$

and the last term is larger than $\chi_d + \varepsilon^{49}$ for $\varepsilon$ sufficiently small. $\square$

<figure><img src="image_9.png" /></figure>

## 7.2. Application of Faber-Krahn: proof of Theorem 1.1

The time has come to apply the quantitative Faber-Krahn inequality to the random walk in order to prove Theorem 1.1. Note that Theorem 1.1 is the analogue of [Bol94, Proposition 3.1].

We shall apply Theorem 6.2 to the set $\mathcal{L}_n$. To this end, we consider a function $g$ satisfying the constraints of the infimum appearing in Theorem 6.2. So, let $g$ be a continuous function $\mathbb{R}^d \to \mathbb{R}^+$ such that $g$ is $C^\infty$ on $\mathbb{R}^d \setminus \frac{1}{n}D^d$, and there exists $\ell \in \mathcal{L}_n$ satisfying

$$\|g^2 - \ell\|_{L^1(\mathbb{R}^d)} \le \frac{3}{n^{1/16}}.$$

This implies in particular that

$$1 - \frac{3}{n^{1/16}} \le \|g\|_2 \le 1 + \frac{3}{n^{1/16}}.$$
Since $\ell$ belongs to $\mathcal{L}_n$, then we have, for any $x \in \mathbb{R}^d$, for $n$ large enough,

$$ 
\begin{aligned}
\frac{1}{n^{1/800}} &\le \|\ell - (\phi_x)^2\|_1 \le \|\ell - g^2\|_1 + \|g^2 - (\phi_x)^2\|_1 \\
&\le \frac{3}{n^{1/16}} + \int_{\mathbb{R}^d} |g - \phi_x| \times |g + \phi_x| dx \\
&\le \frac{3}{n^{1/16}} + \|g - \phi_x\|_2 \times \|g + \phi_x\|_2 \\
&\le \frac{3}{n^{1/16}} + 3\|g - \phi_x\|_2.
\end{aligned}
$$

It follows that, for $n$ large enough,

$$ \forall x \in \mathbb{R}^d \quad \|g - \phi_x\|_2 \ge \frac{1}{4n^{1/800}}. $$

We apply the inequality of Corollary 7.3 to $g$ with $\varepsilon^{49} = 3/n^{1/16}$. For $n$ large enough, we have

$$ (7.19) \qquad |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \chi_d + \frac{1}{n^{1/16}}. $$

In order to exploit the inequality of Theorem 6.2, we will first restrict the set of the functions $g$ which are relevant in the infimum. Let $\kappa \ge 1$ and let $\delta > 0$ be such that

$$ \omega_d \left( \frac{\lambda_d}{2\delta} \right)^{d/2} \ge 2\kappa. $$

Let us set

$$ G = \{ x \in \mathbb{R}^d : g(x) > 0 \}. $$

Let $r > 0$ be such that $|G| = r^d \omega_d$. By the classical Faber-Krahn inequality (7.1), we have

$$ \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \lambda(G) (\|g\|_2)^2 \ge \frac{\lambda_d}{r^2} \left(1 - \frac{3}{n^{1/16}}\right)^2. $$

Suppose that

$$ \int_{\mathbb{R}^d} |\nabla g|^2 dx \le \delta. $$

We would then have, for $n$ large enough,

$$ r^2 \ge \frac{\lambda_d}{2\delta}, \quad |G| = \omega_d r^d \ge \omega_d \left( \frac{\lambda_d}{2\delta} \right)^{d/2} \ge 2\kappa, $$

and for such a function $g$, the functional in the infimum is larger or equal than $2\kappa$. This will also be the case if the Dirichlet energy of $g$ is too large, say larger than $4d\kappa$. So, up to terms which are negligible compared to $\exp(-\kappa n^d)$, we can restrict ourselves to functions $g$ such that

$$ \delta \le \int_{\mathbb{R}^d} |\nabla g|^2 dx \le 4d\kappa. $$
Since $\ell$ belongs to $\mathcal{L}_n$, then we have, for any $x \in \mathbb{R}^d$, for $n$ large enough,

$$ 
\begin{aligned}
\frac{1}{n^{1/800}} &\le \|\ell - (\phi_x)^2\|_1 \le \|\ell - g^2\|_1 + \|g^2 - (\phi_x)^2\|_1 \\
&\le \frac{3}{n^{1/16}} + \int_{\mathbb{R}^d} |g - \phi_x| \times |g + \phi_x| dx \\
&\le \frac{3}{n^{1/16}} + \|g - \phi_x\|_2 \times \|g + \phi_x\|_2 \\
&\le \frac{3}{n^{1/16}} + 3\|g - \phi_x\|_2.
\end{aligned}
$$

It follows that, for $n$ large enough,

$$ \forall x \in \mathbb{R}^d \quad \|g - \phi_x\|_2 \ge \frac{1}{4n^{1/800}}. $$

We apply the inequality of Corollary 7.3 to $g$ with $\varepsilon^{49} = 3/n^{1/16}$. For $n$ large enough, we have

$$ (7.19) \qquad |\{g > 0\}| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \chi_d + \frac{1}{n^{1/16}}. $$

In order to exploit the inequality of Theorem 6.2, we will first restrict the set of the functions $g$ which are relevant in the infimum. Let $\kappa \ge 1$ and let $\delta > 0$ be such that

$$ \omega_d \left( \frac{\lambda_d}{2\delta} \right)^{d/2} \ge 2\kappa. $$

Let us set

$$ G = \{ x \in \mathbb{R}^d : g(x) > 0 \}. $$

Let $r > 0$ be such that $|G| = r^d \omega_d$. By the classical Faber-Krahn inequality (7.1), we have

$$ \int_{\mathbb{R}^d} |\nabla g|^2 dx \ge \lambda(G) (\|g\|_2)^2 \ge \frac{\lambda_d}{r^2} \left(1 - \frac{3}{n^{1/16}}\right)^2. $$

Suppose that

$$ \int_{\mathbb{R}^d} |\nabla g|^2 dx \le \delta. $$

We would then have, for $n$ large enough,

$$ r^2 \ge \frac{\lambda_d}{2\delta}, \quad |G| = \omega_d r^d \ge \omega_d \left( \frac{\lambda_d}{2\delta} \right)^{d/2} \ge 2\kappa, $$

and for such a function $g$, the functional in the infimum is larger or equal than $2\kappa$. This will also be the case if the Dirichlet energy of $g$ is too large, say larger than $4d\kappa$. So, up to terms which are negligible compared to $\exp(-\kappa n^d)$, we can restrict ourselves to functions $g$ such that

$$ \delta \le \int_{\mathbb{R}^d} |\nabla g|^2 dx \le 4d\kappa. $$
We have then, for $n$ large enough,

$$ (7.20) \quad \left( \max \left( \sqrt{\int_{\mathbb{R}^d} |\nabla g|^2 dx} - \frac{\sqrt{d}}{n^{1/8}}, 0 \right) \right)^2 \ge \int_{\mathbb{R}^d} |\nabla g|^2 dx - \frac{2\sqrt{d}}{n^{1/8}} \sqrt{4d\kappa} \\ \ge \int_{\mathbb{R}^d} |\nabla g|^2 dx - \frac{1}{n^{1/9}}. $$

We apply Theorem 6.2 and we use inequality (7.20) to simplify the infimum. We have, for $n$ large enough,

$$ E(e^{-|R_N|}; \ell_N \in \mathcal{L}_n) \le 4 \exp(-\kappa n^d) \\ + \exp \left( 2n^{d-1/8} - n^d (1 - n^{-1/4}) \inf \left\{ |\text{supp } g| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx - \frac{1}{2dn^{1/9}} \right. \right. \\ \left. \left. : g \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus \frac{1}{n} D^d, \exists \ell \in \mathcal{L} \; \|g^2 - \ell\|_{L^1(\mathbb{R}^d)} \le \frac{3}{n^{1/16}} \right\} \right). $$

We can finally use inequality (7.19)! We obtain

$$ E(e^{-|R_N|}; \ell_N \in \mathcal{L}_n) \\ \le 4 \exp(-\kappa n^d) + \exp \left( 2n^{d-1/8} - n^d (1 - n^{-1/4}) \left( \chi_d + \frac{1}{n^{1/16}} - \frac{1}{2dn^{1/9}} \right) \right). $$ 
By choosing $\kappa > \chi_d$, for $n$ large enough, we obtain the statement of Theorem 1.1.

## 8. Filling the ball and proof of Theorem 1.2

We now come to the proof of Theorem 1.2. We will show that with high probability under $\tilde{P}_N$ a ball of approximately the right radius is entirely filled. The analogue of this result in [Bol94, Proposition 4.1], is however only valid for $d=2$, whereas the proof below holds for any $d \ge 2$. Fix $x \in \mathbb{R}^d$, and let

$$ \mathcal{G}_{\text{loc},x} = \left\{ \| \ell_N - (\phi_x)^2 \|_1 \le 1/n^s \right\} $$

with $s = 1/800$ is as in Theorem 1.1 and we recall that the rescaled local time profile $\ell_N$ is defined in (6.4). For $\kappa > 0$ and $x \in \mathbb{R}^d$,

$$ \mathcal{R}_{\kappa, x} = \left\{ \forall z \in B(x, \rho_d(1 - n^{-\kappa})) \cap \mathbb{Z}_n^d, \ell_N(z) > 0 \right\}. $$

This is the event that the Euclidean ball of radius $\rho_d n(1 - n^{-\kappa})$ around $nx$, restricted to the unscaled lattice $\mathbb{Z}^d$, is filled by the walk. The main result of this section, which is the analogue of [Bol94, Proposition 4.1], is the following.

**THEOREM 8.1.** — There exists $\kappa > 0$ such that for all $n$ (or equivalently $N$) large enough,

$$ \frac{1}{Z_N} E \left( e^{-|R_N|}; \mathcal{G}_{\text{loc},x}; (\mathcal{R}_{\kappa,x})^c \right) \le \exp(-n^\kappa). $$
We have then, for $n$ large enough,

$$ (7.20) \quad \left( \max \left( \sqrt{\int_{\mathbb{R}^d} |\nabla g|^2 dx} - \frac{\sqrt{d}}{n^{1/8}}, 0 \right) \right)^2 \ge \int_{\mathbb{R}^d} |\nabla g|^2 dx - \frac{2\sqrt{d}}{n^{1/8}} \sqrt{4d\kappa} \\ \ge \int_{\mathbb{R}^d} |\nabla g|^2 dx - \frac{1}{n^{1/9}}. $$

We apply Theorem 6.2 and we use inequality (7.20) to simplify the infimum. We have, for $n$ large enough,

$$ E(e^{-|R_N|}; \ell_N \in \mathcal{L}_n) \le 4 \exp(-\kappa n^d) \\ + \exp \left( 2n^{d-1/8} - n^d (1 - n^{-1/4}) \inf \left\{ |\text{supp } g| + \frac{1}{2d} \int_{\mathbb{R}^d} |\nabla g|^2 dx - \frac{1}{2dn^{1/9}} \right. \right. \\ \left. \left. : g \text{ is continuous } \mathbb{R}^d \to \mathbb{R}^+ \text{ and } C^\infty \text{ on } \mathbb{R}^d \setminus \frac{1}{n} D^d, \exists \ell \in \mathcal{L} \; \|g^2 - \ell\|_{L^1(\mathbb{R}^d)} \le \frac{3}{n^{1/16}} \right\} \right). $$

We can finally use inequality (7.19)! We obtain

$$ E(e^{-|R_N|}; \ell_N \in \mathcal{L}_n) \\ \le 4 \exp(-\kappa n^d) + \exp \left( 2n^{d-1/8} - n^d (1 - n^{-1/4}) \left( \chi_d + \frac{1}{n^{1/16}} - \frac{1}{2dn^{1/9}} \right) \right). $$

By choosing $\kappa > \chi_d$, for $n$ large enough, we obtain the statement of Theorem 1.1.

## 8. Filling the ball and proof of Theorem 1.2

We now come to the proof of Theorem 1.2. We will show that with high probability under $\tilde{P}_N$ a ball of approximately the right radius is entirely filled. The analogue of this result in [Bol94, Proposition 4.1], is however only valid for $d=2$, whereas the proof below holds for any $d \ge 2$. Fix $x \in \mathbb{R}^d$, and let

$$ \mathcal{G}_{\text{loc},x} = \left\{ \| \ell_N - (\phi_x)^2 \|_1 \le 1/n^s \right\} $$

with $s = 1/800$ is as in Theorem 1.1 and we recall that the rescaled local time profile $\ell_N$ is defined in (6.4). For $\kappa > 0$ and $x \in \mathbb{R}^d$,

$$ \mathcal{R}_{\kappa, x} = \left\{ \forall z \in B(x, \rho_d(1 - n^{-\kappa})) \cap \mathbb{Z}_n^d, \ell_N(z) > 0 \right\}. $$

This is the event that the Euclidean ball of radius $\rho_d n(1 - n^{-\kappa})$ around $nx$, restricted to the unscaled lattice $\mathbb{Z}^d$, is filled by the walk. The main result of this section, which is the analogue of [Bol94, Proposition 4.1], is the following.

**THEOREM 8.1.** — There exists $\kappa > 0$ such that for all $n$ (or equivalently $N$) large enough,

$$ \frac{1}{Z_N} E \left( e^{-|R_N|}; \mathcal{G}_{\text{loc},x}; (\mathcal{R}_{\kappa,x})^c \right) \le \exp(-n^\kappa). $$
## 8.1. Time spent in mesoscopic balls

We recall that constants $C, c > 0$ denote constants depending only on the dimension, whose precise numerical value is allowed to change from line to line. We will use Landau's notations $O(\cdot)$, $o(\cdot)$ where the implicit constants are allowed to depend on the dimension only. The notation $\lesssim, \gtrsim, \asymp$ denote inequalities and equality up to constant respectively: thus $a_n \lesssim b_n$ means $a_n \le Cb_n$ for some constant $C$ (depending only on the dimension). $\kappa$ will be a small parameter eventually chosen in a way that depends only on the dimension, so at the end of the proof we will be able to absorb its value in a generic constant $c > 0$, but we will refrain from doing so during the course of the proof.

Without loss of generality we take $x = 0$ in the rest of the proof of Theorem 8.1; write $G_{loc}$ and $R_\kappa$ for $G_{loc, x}$ and $R_{\kappa, x}$. Fix $z \in B(0, \rho_d n(1 - n^{-\kappa}))$. We let $m = \rho_d n^{1-2\kappa}$ where $\kappa$ is sufficiently small; $m$ is a mesoscopic scale (quite close to $n$) and our first task will be to control the amount of time spent in a ball of that scale around the point $z$. Let

$$B = B(z, m) \cap \mathbb{Z}^d$$

be the discretised ball of radius $m$ around $z$. Note that $B \subset B(0, \rho_d n(1 - n^{-\kappa}/2))$. The idea will be to condition on some information outside $B$ including the local time of the random walk on every site outside of $B$. Let $B^\circ$ denote the ball of radius $m/2$ around $x$, i.e., $B^\circ = B(x, m/2) \cap \mathbb{Z}^d$.

We make the following simple deterministic observations (recall that we expect $L_N(x)$ to be typically of order $n^2$ at any point in the bulk of $R_N$).

**LEMMA 8.2.** — If $G_{loc}$ holds, then necessarily, for some sufficiently small but fixed $\kappa$ and $\delta > 0$ depending only on the dimension $d$, we have

$$L_N(B^\circ) \ge \delta m^d n^{2-2\kappa} \quad (8.1)$$

Furthermore,

$$|R_N| \ge \omega_d(\rho_d n)^d - cn^{d-\varepsilon} \quad (8.2)$$

where $\varepsilon$ depends only on $d$.

*Proof.* — Let $\varphi = \phi^2$. Note that as $z \to \partial B(0, \rho_d)$, for some constant $C > 0$ depending only on the dimension,

$$\varphi(z) \sim Cd(z)^2, \text{ where } d(z) = \text{dist}_{\mathbb{R}^d}(z, \partial B(0, \rho_d)). \quad (8.3)$$

Indeed recall that $\phi$, the first eigenfunction in $B(0, \rho_d)$, has nonzero normal derivative on $\partial B(0, \rho_d)$.

Let $y \in B$. Then $\text{dist}(y, \partial B(0, \rho_d n)) \ge \rho_d n^{1-\kappa}/2$ and so $d(y/n) \ge \rho_d/(2n^\kappa)$. Now suppose for contradiction that $L_N(B^\circ) \le \delta m^d n^{2-2\kappa}$. Then necessarily, by definition of $\ell_N$ as a function on $\mathbb{R}^d$ in (6.4),

$$\int_{(B^\circ/n)} \ell_N(y) dy \le \sum_{y \in \mathbb{Z}^d \cap B^\circ} \frac{1}{n^{d+2}} L_N(y) = \frac{L_N(B^\circ)}{n^{d+2}} \le \delta(m/n)^d n^{-2\kappa} = \delta \rho_d^d n^{-2(d+1)\kappa}$$

whereas

$$\int_{(B^\circ/n)} \varphi(y) dy \ge \omega_d \left(\frac{m/2}{n}\right)^d C \rho_d^2 n^{-2\kappa}/4 = \left(\frac{C \rho_d^{d+2} \omega_d}{2^{2+d}}\right) n^{-2(d+1)\kappa}.$$
## 8.1. Time spent in mesoscopic balls

We recall that constants $C, c > 0$ denote constants depending only on the dimension, whose precise numerical value is allowed to change from line to line. We will use Landau's notations $O(\cdot)$, $o(\cdot)$ where the implicit constants are allowed to depend on the dimension only. The notation $\lesssim, \gtrsim, \asymp$ denote inequalities and equality up to constant respectively: thus $a_n \lesssim b_n$ means $a_n \le Cb_n$ for some constant $C$ (depending only on the dimension). $\kappa$ will be a small parameter eventually chosen in a way that depends only on the dimension, so at the end of the proof we will be able to absorb its value in a generic constant $c > 0$, but we will refrain from doing so during the course of the proof.

Without loss of generality we take $x = 0$ in the rest of the proof of Theorem 8.1; write $G_{loc}$ and $R_\kappa$ for $G_{loc, x}$ and $R_{\kappa, x}$. Fix $z \in B(0, \rho_d n(1 - n^{-\kappa}))$. We let $m = \rho_d n^{1-2\kappa}$ where $\kappa$ is sufficiently small; $m$ is a mesoscopic scale (quite close to $n$) and our first task will be to control the amount of time spent in a ball of that scale around the point $z$. Let

$$B = B(z, m) \cap \mathbb{Z}^d$$

be the discretised ball of radius $m$ around $z$. Note that $B \subset B(0, \rho_d n(1 - n^{-\kappa}/2))$. The idea will be to condition on some information outside $B$ including the local time of the random walk on every site outside of $B$. Let $B^\circ$ denote the ball of radius $m/2$ around $x$, i.e., $B^\circ = B(x, m/2) \cap \mathbb{Z}^d$.

We make the following simple deterministic observations (recall that we expect $L_N(x)$ to be typically of order $n^2$ at any point in the bulk of $R_N$).

**LEMMA 8.2.** — If $G_{loc}$ holds, then necessarily, for some sufficiently small but fixed $\kappa$ and $\delta > 0$ depending only on the dimension $d$, we have

$$L_N(B^\circ) \ge \delta m^d n^{2-2\kappa} \quad (8.1)$$

Furthermore,

$$|R_N| \ge \omega_d(\rho_d n)^d - cn^{d-\epsilon} \quad (8.2)$$

where $\epsilon$ depends only on $d$.

*Proof.* — Let $\varphi = \phi^2$. Note that as $z \to \partial B(0, \rho_d)$, for some constant $C > 0$ depending only on the dimension,

$$\varphi(z) \sim Cd(z)^2, \text{ where } d(z) = \text{dist}_{\mathbb{R}^d}(z, \partial B(0, \rho_d)). \quad (8.3)$$

Indeed recall that $\phi$, the first eigenfunction in $B(0, \rho_d)$, has nonzero normal derivative on $\partial B(0, \rho_d)$.

Let $y \in B$. Then $\text{dist}(y, \partial B(0, \rho_d n)) \ge \rho_d n^{1-\kappa}/2$ and so $d(y/n) \ge \rho_d/(2n^\kappa)$. Now suppose for contradiction that $L_N(B^\circ) \le \delta m^d n^{2-2\kappa}$. Then necessarily, by definition of $\ell_N$ as a function on $\mathbb{R}^d$ in (6.4),

$$\int_{(B^\circ/n)} \ell_N(y) dy \le \sum_{y \in \mathbb{Z}^d \cap B^\circ} \frac{1}{n^{d+2}} L_N(y) = \frac{L_N(B^\circ)}{n^{d+2}} \le \delta(m/n)^d n^{-2\kappa} = \delta \rho_d^d n^{-2(d+1)\kappa}$$

whereas

$$\int_{(B^\circ/n)} \varphi(y) dy \ge \omega_d \left(\frac{m/2}{n}\right)^d C \rho_d^2 n^{-2\kappa}/4 = \left(\frac{C \rho_d^{d+2} \omega_d}{2^{2+d}}\right) n^{-2(d+1)\kappa}.$$
Thus if we take $\delta = C\rho_d^{d+2}\omega_d/2^{3+d}$,

$$ \|\ell_N - \varphi\|_1 \ge \left| \int_{(B^\circ/n)} \ell_N(y) dy - \int_{(B^\circ/n)} \varphi(y) dy \right| \ge \left( \frac{C \rho_d^{d+2} \omega_d}{2^{3+d}} \right) n^{-2(d+1)\kappa}. $$

Hence if $\kappa$ is small enough that $2(d+1)\kappa \le s$ we see that this cannot hold at the same time as $\mathcal{G}_{loc}$. This shows (8.1) with this choice of parameters.

The proof of (8.2) follows a similar argument. □

## 8.2. Bridges in mesoscopic balls

Let us call an interval of time $[t_1, t_2]$ a bridge if $X[t_1, t_2] \subset B$ and $X_{t_1}, X_{t_2} \in B^\circ$. (We warn the reader that this differs from the more standard notion of bridge). We call $a = X_{t_1}$ and $b = X_{t_2}$ its endpoints and $t_2 - t_1$ its length; note that we require that $a, b \in B$ be far away from $\partial B$; in fact we require them to be in $B^\circ$. We will consider bridges of length at least $m^2$ and our first goal will be to show that there are sufficiently many such disjoint bridges.

Let $\mathcal{G}_{br} = \mathcal{G}_{br, \delta}$ be the event that there are at least $a\delta n^{2-2\kappa}m^{d-2}$ bridges of length $m^2$, where $\delta$ is as in Lemma 8.2, and where $a$ is a suitable constant depending also only on the dimension, which will be chosen below.

**LEMMA 8.3.** — We can choose $\kappa$ sufficiently small and $a > 0$, depending only on the dimension, so that

$$ E\left(e^{-|R_N|}; \mathcal{G}_{loc}; (\mathcal{G}_{br})^c\right) \le Z_N \exp\left(-cn^{d-6\kappa}\right), $$
for some constant $c > 0$ depending only on $d$.

*Proof.* — We already know that the walk spends at least $\delta n^{2-2\kappa}m^d$ units of time in $B^\circ$. Roughly speaking, every time the walk is in $B^\circ$ there is a positive probability that during the next $m^2$ units of time, the walk stays in $B$ and its position at the end of this interval is again in $B^\circ$, thereby completing a bridge; independently of the past. So we wish to use standard Chernoff bounds for deviations of binomial random variables. In order to implement this strategy, we must however take care that we are working under the weighted probability measure $\tilde{P}_N$ and not $P$. We will deal with this complication by performing a suitable change of measure (which as it turns out is essentially the same as the one used by Bolthausen in his proof of [Bol94, Proposition 4.1]). The first step will be to work in continuous time rather than discrete time. By Lemma 8.2 (and more precisely (8.2)) we write

$$ E\left(e^{-|R_N|}; \mathcal{G}_{loc}; (\mathcal{G}_{br})^c\right) \le \exp\left(-\omega_d(\rho_d n)^d + cn^{d-\epsilon}\right) P\left(\mathcal{G}_{loc}; \mathcal{G}_{br}^c\right) $$

and we interpret the event in the right hand side of the above inequality as an event for the jump chain of a continuous Markov chain whose jump rates from $x$ to $y$ is $1/(2d)$ if $x$ and $y$ are neighbours and zero else. Let $(\tilde{X}_t, t \ge 0)$ be this process, let $\tilde{P}$ denote its law and let $J(\tilde{X})$ be the jump chain of $\tilde{X}$. Set $\tilde{\mathcal{G}}_{loc} = \{J(\tilde{X}) \in \mathcal{G}_{loc}\}$ and, similarly, $\tilde{\mathcal{G}}_{br} = \{J(\tilde{X}) \in \mathcal{G}_{br}\}$. Then $P(\mathcal{G}_{loc}; (\mathcal{G}_{br})^c) = \tilde{P}(\tilde{\mathcal{G}}_{loc}; (\tilde{\mathcal{G}}_{br})^c)$.

Define a modification $\tilde{\phi}$ of $\phi$ as follows:

$$ \tilde{\phi}(x) = \phi(x) + n^{-\kappa}; \quad x \in B(0, \rho_d) $$
Thus if we take $\delta = C\rho_d^{d+2}\omega_d/2^{3+d}$,

$$ \|\ell_N - \varphi\|_1 \ge \left| \int_{(B^\circ/n)} \ell_N(y) dy - \int_{(B^\circ/n)} \varphi(y) dy \right| \ge \left( \frac{C \rho_d^{d+2} \omega_d}{2^{3+d}} \right) n^{-2(d+1)\kappa}. $$

Hence if $\kappa$ is small enough that $2(d+1)\kappa \le s$ we see that this cannot hold at the same time as $\mathcal{G}_{loc}$. This shows (8.1) with this choice of parameters.

The proof of (8.2) follows a similar argument. $\square$

## 8.2. Bridges in mesoscopic balls

Let us call an interval of time $[t_1, t_2]$ a bridge if $X[t_1, t_2] \subset B$ and $X_{t_1}, X_{t_2} \in B^\circ$. (We warn the reader that this differs from the more standard notion of bridge). We call $a = X_{t_1}$ and $b = X_{t_2}$ its endpoints and $t_2 - t_1$ its length; note that we require that $a, b \in B$ be far away from $\partial B$; in fact we require them to be in $B^\circ$. We will consider bridges of length at least $m^2$ and our first goal will be to show that there are sufficiently many such disjoint bridges.

Let $\mathcal{G}_{br} = \mathcal{G}_{br, \delta}$ be the event that there are at least $a\delta n^{2-2\kappa}m^{d-2}$ bridges of length $m^2$, where $\delta$ is as in Lemma 8.2, and where $a$ is a suitable constant depending also only on the dimension, which will be chosen below.

**LEMMA 8.3.** — We can choose $\kappa$ sufficiently small and $a > 0$, depending only on the dimension, so that

$$ E(e^{-|R_N|}; \mathcal{G}_{loc}; (\mathcal{G}_{br})^c) \le Z_N \exp(-cn^{d-6\kappa}), $$

for some constant $c > 0$ depending only on $d$.

*Proof.* — We already know that the walk spends at least $\delta n^{2-2\kappa}m^d$ units of time in $B^\circ$. Roughly speaking, every time the walk is in $B^\circ$ there is a positive probability that during the next $m^2$ units of time, the walk stays in $B$ and its position at the end of this interval is again in $B^\circ$, thereby completing a bridge; independently of the past. So we wish to use standard Chernoff bounds for deviations of binomial random variables. In order to implement this strategy, we must however take care that we are working under the weighted probability measure $\tilde{P}_N$ and not $P$. We will deal with this complication by performing a suitable change of measure (which as it turns out is essentially the same as the one used by Bolthausen in his proof of [Bol94, Proposition 4.1]). The first step will be to work in continuous time rather than discrete time. By Lemma 8.2 (and more precisely (8.2)) we write

$$ E(e^{-|R_N|}; \mathcal{G}_{loc}; (\mathcal{G}_{br})^c) \le \exp(-\omega_d(\rho_d n)^d + cn^{d-\epsilon}) P(\mathcal{G}_{loc}; \mathcal{G}_{br}^c) $$

and we interpret the event in the right hand side of the above inequality as an event for the jump chain of a continuous Markov chain whose jump rates from $x$ to $y$ is $1/(2d)$ if $x$ and $y$ are neighbours and zero else. Let $(\tilde{X}_t, t \ge 0)$ be this process, let $\tilde{P}$ denote its law and let $J(\tilde{X})$ be the jump chain of $\tilde{X}$. Set $\tilde{\mathcal{G}}_{loc} = \{J(\tilde{X}) \in \mathcal{G}_{loc}\}$ and, similarly, $\tilde{\mathcal{G}}_{br} = \{J(\tilde{X}) \in \mathcal{G}_{br}\}$. Then $P(\mathcal{G}_{loc}; (\mathcal{G}_{br})^c) = \tilde{P}(\tilde{\mathcal{G}}_{loc}; (\tilde{\mathcal{G}}_{br})^c)$.

Define a modification $\tilde{\phi}$ of $\phi$ as follows:

$$ \tilde{\phi}(x) = \phi(x) + n^{-\kappa}; \quad x \in B(0, \rho_d) $$
<figure><img src="image_8.png" /><figcaption>Figure 8.1. Choice of function h: as desired, h decreases from $n^{-\kappa}$ to $n^{-\kappa/2}$ over $[\rho, \rho+1]$. This requires the area above the red function on the right hand side (its derivative) to be less than $n^{-\kappa/2}$. The value of the constant in the right hand side is the normal (radial) derivative of $\phi$. To achieve this, the second derivative will be at most $\lesssim n^{\kappa}$, as claimed in (8.4).</figcaption></figure>

We will also define $\tilde{\phi}$ outside of $B(0, \rho_d)$ so that it is positive everywhere and also in such a way that it is reasonably smooth near the boundary of that ball (indeed, if we do not take a positive function we cannot use it to change the underlying probability measure). Of course we could set $\tilde{\phi}$ to be constant outside of that ball, equal to its value on the boundary, but this turns out to not be sufficiently smooth; in particular the Laplacian on this sphere would be too large.

Instead we define

$$\tilde{\phi}(x) = h(|x|)$$

where

$$h: [\rho_d, \rho_d+1] \to [n^{-\kappa}/2, n^{-\kappa}]$$

is a smooth monotone decreasing convex function such that its derivative at $1 + \rho_d$ is 0 while its derivative at $\rho_d$ is the radial derivative of $\phi$ on $\partial B(0, \rho_d)$, and such that

$$(8.4) \qquad \sup_{x \in [\rho_d, \rho_d+1]} \frac{d^2}{dx^2} h \lesssim n^{\kappa}.$$

It is elementary to check that such a function $h$ exists, so that $\tilde{\phi}$ is well defined (see Figure 8.1 for an illustration of the function $\tilde{\phi}$ and its derivative).

With the help of $\tilde{\phi}$ we can define a new probability measure $\tilde{Q}$ to be the law of the Markov chain whose transition rates are given by

$$q(x, y) = \frac{\tilde{\phi}(y/n)}{\tilde{\phi}(x/n)}$$

whenever $x \neq y$ are neighbours in $Z^d$, and $q(x, y) = 0$ otherwise. Note that since $\tilde{\phi}$ is positive, $\tilde{Q}$ is indeed equivalent to $\tilde{P}$ and furthermore, letting $\tilde{t}_N$ be the first time
<figure><img src="image_10.png" /><figcaption>Figure 8.1. Choice of function h: as desired, h decreases from $n^{-\kappa}$ to $n^{-\kappa/2}$ over $[\rho, \rho+1]$. This requires the area above the red function on the right hand side (its derivative) to be less than $n^{-\kappa/2}$. The value of the constant in the right hand side is the normal (radial) derivative of $\phi$. To achieve this, the second derivative will be at most $\lesssim n^{\kappa}$, as claimed in (8.4).</figcaption></figure>

We will also define $\tilde{\phi}$ outside of $B(0, \rho_d)$ so that it is positive everywhere and also in such a way that it is reasonably smooth near the boundary of that ball (indeed, if we do not take a positive function we cannot use it to change the underlying probability measure). Of course we could set $\tilde{\phi}$ to be constant outside of that ball, equal to its value on the boundary, but this turns out to not be sufficiently smooth; in particular the Laplacian on this sphere would be too large.

Instead we define

$$\tilde{\phi}(x) = h(|x|)$$

where

$$h: [\rho_d, \rho_d+1] \to [n^{-\kappa}/2, n^{-\kappa}]$$

is a smooth monotone decreasing convex function such that its derivative at $1 + \rho_d$ is 0 while its derivative at $\rho_d$ is the radial derivative of $\phi$ on $\partial B(0, \rho_d)$, and such that

$$ (8.4) \qquad \sup_{x \in [\rho_d, \rho_d+1]} \frac{d^2}{dx^2} h \lesssim n^{\kappa}. $$

It is elementary to check that such a function $h$ exists, so that $\tilde{\phi}$ is well defined (see Figure 8.1 for an illustration of the function $\tilde{\phi}$ and its derivative).

With the help of $\tilde{\phi}$ we can define a new probability measure $\tilde{Q}$ to be the law of the Markov chain whose transition rates are given by

$$q(x, y) = \frac{\tilde{\phi}(y/n)}{\tilde{\phi}(x/n)}$$

whenever $x \neq y$ are neighbours in $Z^d$, and $q(x, y) = 0$ otherwise. Note that since $\tilde{\phi}$ is positive, $\tilde{Q}$ is indeed equivalent to $\tilde{P}$ and furthermore, letting $\tilde{t}_N$ be the first time
$\tilde{X}$ has jumped $N$ times,

$$ (8.5) \qquad \left. \frac{d\tilde{P}}{d\tilde{Q}} \right|_{\mathcal{F}_{\tilde{t}_N}} = \frac{\tilde{\phi}(\tilde{X}_0/n)}{\tilde{\phi}(\tilde{X}_{\tilde{t}_N}/n)} \exp \left( \int_0^{\tilde{t}_N} \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}} (\tilde{X}_s/n) ds \right). $$

(See, e.g. [RW00, IV, (22.8)]). Here recall that $\Delta_1$ is the discrete Laplacian, i.e. $2d$ times the generator of the Markov chain $\tilde{X}$ under $\tilde{P}$.

*Step 1.* — We will argue that

$$ (8.6) \qquad E \left( e^{-|R_N|} ; \mathcal{G}_{\text{loc}} ; (\mathcal{G}_{\text{br}})^c \right) \le Z_N e^{-N^{1-5\kappa}} + Z_N e^{3n^{d-e}} \tilde{Q} \left( \tilde{\mathcal{G}}_{\text{loc}} ; (\tilde{\mathcal{G}}_{\text{br}})^c \right). $$

Essentially the proof consists in analysing carefully the integral in (8.5). We follow roughly the arguments [Bol94, see equations (4.3) and (4.4)], with additional details. We start by observing that if $x \in B(0, \rho_d n) \cap \mathbb{Z}^d$ is such that all its $2d$ neighbours are also in $B(0, \rho_d n) \cap \mathbb{Z}^d$, then by a Taylor expansion,

$$ \begin{aligned} \Delta_1 \tilde{\phi}(x/n) &= \Delta_1 \phi(x/n) = \frac{1}{2} \Delta \phi(x/n) + O(n^{-3}) \\ &= -\frac{\lambda}{2n^2} \phi(x/n) + O(n^{-3}) \\ &= -\frac{\lambda}{2n^2} \tilde{\phi}(x/n) + O(n^{-3}) + O(n^{-2-\kappa}). \end{aligned} $$

Assume that $\kappa < 1$ without loss of generality so that $O(n^{-3}) = O(n^{-2-\kappa})$. Hence if furthermore $x \in B(0, \rho_d n(1 - n^{-0.9\kappa}))$,

$$ (8.7) \qquad \left| \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}}(x/n) + \frac{\lambda}{2n^2} \right| \lesssim n^{-2-\kappa+0.9\kappa} \le n^{-2-0.1\kappa} $$

using (8.3).

Furthermore, for any other $x$,

$$ |\Delta_1 \tilde{\phi}(x/n)| \lesssim n^{-2+\kappa} $$

by (8.4) (where the implicit constant depends only on $\tilde{\phi}$ and so on the dimension), so that

$$ (8.8) \qquad \left| \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}}(x/n) + \frac{\lambda}{2n^2} \right| \lesssim n^{-2+2\kappa} $$

for such $x$, and the same remark holds about the implicit constant.

Combining (8.7) and (8.8), it follows that

$$ (8.9) \qquad \int_0^{\tilde{t}_N} \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}} (\tilde{X}_s/n) ds \\ = (-\lambda/2) \frac{\tilde{t}_N}{n^2} + O(\tilde{t}_N n^{-2-0.1\kappa}) + O(n^{-2+2\kappa} \tilde{L}_{\tilde{t}_N} \left( B \left( 0, \rho_d n (1 - n^{-0.9\kappa}) \right)^c \right)), $$

where for a set $A \subset \mathbb{Z}^d$ and a time $t > 0$, $\tilde{L}_t(A) = \int_0^t 1_{\{\tilde{X}_s \in A\}} ds$ is the local time of $\tilde{X}$ (in continuous time) in the set $A$ up to time $t$.
$\tilde{X}$ has jumped $N$ times,

$$ (8.5) \qquad \left. \frac{d\tilde{P}}{d\tilde{Q}} \right|_{\mathcal{F}_{\tilde{t}_N}} = \frac{\tilde{\phi}(\tilde{X}_0/n)}{\tilde{\phi}(\tilde{X}_{\tilde{t}_N}/n)} \exp \left( \int_0^{\tilde{t}_N} \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}} (\tilde{X}_s/n) ds \right). $$

(See, e.g. [RW00, IV, (22.8)]). Here recall that $\Delta_1$ is the discrete Laplacian, i.e. 2d times the generator of the Markov chain $\tilde{X}$ under $\tilde{P}$.

*Step 1. — We will argue that*

$$ (8.6) \qquad E \left( e^{-|R_N|} ; \mathcal{G}_{\text{loc}} ; (\mathcal{G}_{\text{br}})^c \right) \le Z_N e^{-N^{1-5\kappa}} + Z_N e^{3n^{d-e}} \tilde{Q} \left( \tilde{\mathcal{G}}_{\text{loc}} ; (\tilde{\mathcal{G}}_{\text{br}})^c \right). $$

Essentially the proof consists in analysing carefully the integral in (8.5). We follow roughly the arguments [Bol94, see equations (4.3) and (4.4)], with additional details. We start by observing that if $x \in B(0, \rho_d n) \cap \mathbb{Z}^d$ is such that all its 2d neighbours are also in $B(0, \rho_d n) \cap \mathbb{Z}^d$, then by a Taylor expansion,

$$ \begin{aligned} \Delta_1 \tilde{\phi}(x/n) &= \Delta_1 \phi(x/n) = \frac{1}{2} \Delta \phi(x/n) + O(n^{-3}) \\ &= -\frac{\lambda}{2n^2} \phi(x/n) + O(n^{-3}) \\ &= -\frac{\lambda}{2n^2} \tilde{\phi}(x/n) + O(n^{-3}) + O(n^{-2-\kappa}). \end{aligned} $$

Assume that $\kappa < 1$ without loss of generality so that $O(n^{-3}) = O(n^{-2-\kappa})$. Hence if furthermore $x \in B(0, \rho_d n(1 - n^{-0.9\kappa}))$,

$$ (8.7) \qquad \left| \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}}(x/n) + \frac{\lambda}{2n^2} \right| \lesssim n^{-2-\kappa+0.9\kappa} \le n^{-2-0.1\kappa} $$

using (8.3).

Furthermore, for any other $x$,

$$ |\Delta_1 \tilde{\phi}(x/n)| \lesssim n^{-2+\kappa} $$

by (8.4) (where the implicit constant depends only on $\tilde{\phi}$ and so on the dimension), so that

$$ (8.8) \qquad \left| \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}}(x/n) + \frac{\lambda}{2n^2} \right| \lesssim n^{-2+2\kappa} $$

for such $x$, and the same remark holds about the implicit constant.

Combining (8.7) and (8.8), it follows that

$$ (8.9) \qquad \int_0^{\tilde{t}_N} \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}} (\tilde{X}_s/n) ds \\ = (-\lambda/2) \frac{\tilde{t}_N}{n^2} + O(\tilde{t}_N n^{-2-0.1\kappa}) + O(n^{-2+2\kappa} \tilde{L}_{\tilde{t}_N} \left( B \left( 0, \rho_d n (1 - n^{-0.9\kappa}) \right)^c \right)), $$

where for a set $A \subset \mathbb{Z}^d$ and a time $t > 0$, $\tilde{L}_t(A) = \int_0^t 1_{\{\tilde{X}_s \in A\}} ds$ is the local time of $\tilde{X}$ (in continuous time) in the set $A$ up to time $t$.
Set $\tilde{\mathcal{A}} = \tilde{\mathcal{G}}_{loc} \cap (\tilde{\mathcal{G}}_{br})^c$. Set $\tilde{\mathcal{A}}'$ to be the event

$$ \tilde{\mathcal{A}}; |\tilde{t}_N - N| \le N^{1-\kappa}; \tilde{L}_{\tilde{t}_N} \left( B \left( 0, \rho_d n (1 - n^{-0.9\kappa}) \right)^c \right) \le N n^{-2.1\kappa}. $$

Then, still writing $R_N$ for the range of the (continuous time) walk at the time $\tilde{t}_N$ of its Nth jump,

$$ (8.10) \qquad E_{\tilde{P}} \left( e^{-|R_N|}; \tilde{\mathcal{A}} \right) \le E_{\tilde{P}} \left( e^{-|R_N|}; \tilde{\mathcal{A}}' \right) + \tilde{P} \left( \tilde{\mathcal{A}} \setminus \tilde{\mathcal{A}}' \right). $$

We bound separately each of those terms. We start with the first term. On $\mathcal{A}'$, we see that

$$ \int_0^{\tilde{t}_N} \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}} (\tilde{X}_s/n) ds = (-\lambda/2)n^d + O(n^{d-(d+2)\kappa}) + O(n^{d-0.1\kappa}) + O(n^{d-0.1\kappa}) $$

so that if $\kappa < \varepsilon/(d+2)$ (which implies $\kappa < 10\varepsilon$), all error terms are

$$ O(n^{d-(d+2)\kappa}) \le n^{d-\varepsilon} $$

and thus, plugging into (8.5), and using the fact that

$$ \max_{B(0, \rho_d)} \tilde{\phi} \lesssim 1, \quad \min_{B(0, \rho_d)} \tilde{\phi} \gtrsim n^{-\kappa}, $$

as well as the already established lower bound of Proposition 2.1 on $Z_N$,

$$ (8.11) \qquad \begin{aligned} E_{\tilde{P}} (e^{-|R_N|}; \tilde{\mathcal{A}}') & \lesssim n^{\kappa} \exp \left( -\omega_d (\rho_d n)^d - (\lambda/2)n^d + 2n^{d-\varepsilon} \right) \tilde{Q}(\tilde{\mathcal{A}}') \\ & \le Z_N e^{2n^{d-\varepsilon} + O(n^{d-1})} \tilde{Q}(\tilde{\mathcal{A}}') \\ & \le Z_N e^{3n^{d-\varepsilon}} \tilde{Q}(\mathcal{G}_{loc}; (\mathcal{G}_{br})^c). \end{aligned} $$

Let us now deal with the second term in (8.10). We have,

$$ (8.12) \qquad \tilde{P}(\tilde{\mathcal{A}} \setminus \tilde{\mathcal{A}}') \le \tilde{P} \left( |\tilde{t}_N - N| \ge N^{1-\kappa} \right) \\ \qquad \qquad + \tilde{P} \left( \tilde{L}_{\tilde{t}_N} \left( B \left( 0, \rho_d n \left( 1 - n^{-0.9\kappa} \right) \right)^c \right) \ge N n^{-2.1\kappa} ; \tilde{\mathcal{G}}_{loc} \right). $$

Now, standard Chernoff estimates for exponential random variables show that

$$ (8.13) \qquad P \left( |\tilde{t}_N - N| \ge N^{1-\kappa} \right) \le \exp \left( -cN^{1-3\kappa} \right) $$

for some constant $c > 0$. Furthermore, on $\mathcal{G}_{loc}$, deterministically we have

$$ L_N \left( B \left( 0, \rho_d n \left( 1 - n^{-0.9\kappa} \right) \right)^c \right) \le N n^{-2.2\kappa}, $$

since otherwise, reasoning as in Lemma 8.2,

$$ \int_{B(0, \rho_d(1-n^{-0.9\kappa}))^c} \ell_N(y) dy \ge n^{-2.2\kappa} \quad \text{while} \quad \int_{B(0, \rho_d(1-n^{-0.9\kappa}))^c} \phi^2(y) dy \le C \omega_d n^{-2.7\kappa} $$

and we would deduce that for $n$ large enough, $\|\ell_N - \phi^2\| > n^{-2.3\kappa}$ which by definition cannot take place on $\mathcal{G}_{loc}$ if $\kappa \le s/2.3$ (with $s = 1/800$ as in Theorem 1.1. Hence

$$ (8.14) \qquad \tilde{P} \left( \tilde{L}_{\tilde{t}_N} \left( B \left( 0, \rho_d n \left( 1 - n^{-0.9\kappa} \right) \right)^c \right) \ge N n^{-2.1\kappa} ; \tilde{\mathcal{G}}_{\text{loc}} \right) \\ \qquad \qquad \le P \left( \sum_{i=1}^{\lfloor Nn^{-2.2\kappa} \rfloor} X_i \ge N n^{-2.1\kappa} \right) $$
Set $\tilde{\mathcal{A}} = \tilde{\mathcal{G}}_{loc} \cap (\tilde{\mathcal{G}}_{br})^c$. Set $\tilde{\mathcal{A}}'$ to be the event

$$ \tilde{\mathcal{A}}; |\tilde{t}_N - N| \le N^{1-\kappa}; \tilde{L}_{\tilde{t}_N} \left( B \left( 0, \rho_d n (1 - n^{-0.9\kappa}) \right)^c \right) \le N n^{-2.1\kappa}. $$

Then, still writing $R_N$ for the range of the (continuous time) walk at the time $\tilde{t}_N$ of its Nth jump,

$$ (8.10) \qquad E_{\tilde{P}} \left( e^{-|R_N|} ; \tilde{\mathcal{A}} \right) \le E_{\tilde{P}} \left( e^{-|R_N|} ; \tilde{\mathcal{A}}' \right) + \tilde{P} \left( \tilde{\mathcal{A}} \setminus \tilde{\mathcal{A}}' \right). $$

We bound separately each of those terms. We start with the first term. On $\mathcal{A}'$, we see that

$$ \int_0^{\tilde{t}_N} \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}} (\tilde{X}_s/n) ds = (-\lambda/2)n^d + O(n^{d-(d+2)\kappa}) + O(n^{d-0.1\kappa}) + O(n^{d-0.1\kappa}) $$

so that if $\kappa < \varepsilon/(d+2)$ (which implies $\kappa < 10\varepsilon$), all error terms are

$$ O(n^{d-(d+2)\kappa}) \le n^{d-\varepsilon} $$

and thus, plugging into (8.5), and using the fact that

$$ \max_{B(0, \rho_d)} \tilde{\phi} \lesssim 1, \quad \min_{B(0, \rho_d)} \tilde{\phi} \gtrsim n^{-\kappa}, $$

as well as the already established lower bound of Proposition 2.1 on $Z_N$,

$$ \begin{align} E_{\tilde{P}} \left( e^{-|R_N|} ; \tilde{\mathcal{A}}' \right) & \lesssim n^{\kappa} \exp \left( -\omega_d (\rho_d n)^d - (\lambda/2)n^d + 2n^{d-\varepsilon} \right) \tilde{Q} \left( \tilde{\mathcal{A}}' \right) \tag{8.11} \\ & \le Z_N e^{2n^{d-\varepsilon} + O(n^{d-1})} \tilde{Q} \left( \tilde{\mathcal{A}}' \right) \nonumber \\ & \le Z_N e^{3n^{d-\varepsilon}} \tilde{Q} \left( \mathcal{G}_{loc} ; (\mathcal{G}_{br})^c \right). \nonumber \end{align} $$

Let us now deal with the second term in (8.10). We have,

$$ (8.12) \qquad \tilde{P}(\tilde{\mathcal{A}} \setminus \tilde{\mathcal{A}}') \le \tilde{P} \left( |\tilde{t}_N - N| \ge N^{1-\kappa} \right) \\ \qquad \qquad + \tilde{P} \left( \tilde{L}_{\tilde{t}_N} \left( B \left( 0, \rho_d n \left( 1 - n^{-0.9\kappa} \right) \right)^c \right) \ge N n^{-2.1\kappa} ; \tilde{\mathcal{G}}_{loc} \right). $$

Now, standard Chernoff estimates for exponential random variables show that

$$ (8.13) \qquad P \left( |\tilde{t}_N - N| \ge N^{1-\kappa} \right) \le \exp \left( -cN^{1-3\kappa} \right) $$

for some constant $c > 0$. Furthermore, on $\mathcal{G}_{loc}$, deterministically we have

$$ L_N \left( B \left( 0, \rho_d n \left( 1 - n^{-0.9\kappa} \right) \right)^c \right) \le N n^{-2.2\kappa}, $$

since otherwise, reasoning as in Lemma 8.2,

$$ \int_{B(0, \rho_d(1-n^{-0.9\kappa}))^c} \ell_N(y) dy \ge n^{-2.2\kappa} \quad \text{while} \quad \int_{B(0, \rho_d(1-n^{-0.9\kappa}))^c} \phi^2(y) dy \le C \omega_d n^{-2.7\kappa} $$

and we would deduce that for $n$ large enough, $\|\ell_N - \phi^2\| > n^{-2.3\kappa}$ which by definition cannot take place on $\mathcal{G}_{loc}$ if $\kappa \le s/2.3$ (with $s = 1/800$ as in Theorem 1.1. Hence

$$ (8.14) \qquad \tilde{P} \left( \tilde{L}_{\tilde{t}_N} \left( B \left( 0, \rho_d n \left( 1 - n^{-0.9\kappa} \right) \right)^c \right) \ge N n^{-2.1\kappa} ; \tilde{\mathcal{G}}_{loc} \right) \\ \le P \left( \sum_{i=1}^{\lfloor N n^{-2.2\kappa} \rfloor} X_i \ge N n^{-2.1\kappa} \right) $$
where $X_i$ are independent unit exponential random variables. Hence for all $n$ (or equivalently $N$) large enough,

$$ P\left(\sum_{i=1}^{\lfloor Nn^{-2.2\kappa} \rfloor} X_i \ge Nn^{-2.1\kappa}\right) \le P\left(\sum_{i=1}^{\lfloor Nn^{-2.2\kappa} \rfloor} X_i \ge 2Nn^{-2.2\kappa}\right) \\ \le \exp(-cNn^{-2.2\kappa}) \le \exp(-N^{1-3\kappa}), $$

for some constant $c > 0$, by elementary Chernoff estimates for exponential random variables. Hence by (8.14),

$$ \tilde{P}\left(\tilde{L}_{\tilde{t}_N} \left(B\left(0, \rho_d n\left(1-n^{-0.9\kappa}\right)\right)\right)^c \ge Nn^{-2.1\kappa}; \tilde{G}_{loc}\right) \le \exp(-N^{1-3\kappa}). $$

Plugging into (8.12), we deduce that if $\kappa$ is sufficiently small, for all $N$ large enough,

$$ \tilde{P}(A \setminus A') \le \exp(-N^{1-4\kappa}). $$

Since $Z_N = \exp(-O(n^d))$, we deduce that

$$ (8.15) \qquad \tilde{P}(A \setminus A') \le Z_N \exp(-N^{1-5\kappa}). $$

Combining (8.15), (8.11) and (8.10), we obtain

$$ E_{\tilde{P}}\left(e^{-|R_N|}; \mathcal{G}_{loc}; (\mathcal{G}_{br})^c\right) \le Z_N e^{-N^{1-5\kappa}} + Z_N e^{3n^{d-\epsilon}} \tilde{Q}\left(\tilde{G}_{loc}; (\tilde{G}_{br})^c\right), $$

as desired in (8.6).

Step 2. — Now it remains to show that $\tilde{G}_{loc} \cap \tilde{G}_{br}$ is overwhelmingly likely under $\tilde{Q}$. More precisely, we will argue that

$$ (8.16) \qquad \tilde{Q}\left(\left(\tilde{G}_{br}\right)^c; \tilde{G}_{loc}\right) \le \exp\left(-cn^{2-2\kappa}m^{d-2}\right). $$

We claim that every time the walk is in $B^\circ$, there is a probability bounded below by a constant $p$, say, depending only on the dimension, such that under $\tilde{Q}$, the walk will perform a bridge of duration $m^2$ (recall that this means the walk remains in $B$ for the next $m^2$ jumps and ends up in $B^\circ$ again after this time), uniformly over the initial position in $B^\circ$ of the walk. To see this, note that $\min_{B/n} \tilde{\phi} = (1-o(1)) \max_{B/n} \tilde{\phi}$, so that the Radon–Nikodym derivative $d\tilde{Q}/d\tilde{P}$ during an interval of time consisting of the first $m^2$ jumps of the chain (call this time $\tilde{t}_{m^2}$) is at least

$$ (8.17) \qquad \begin{aligned} \frac{d\tilde{Q}}{d\tilde{P}} &\ge \frac{\min_{B/n} \tilde{\phi}}{\max_{B/n} \tilde{\phi}} \exp\left(\tilde{t}_{m^2} \inf_{x \in B} \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}}(x/n)\right) \\ &\ge (1+o(1)) \exp\left(-(\lambda/2+o(1))\frac{\tilde{t}_{m^2}}{n^2}\right). \end{aligned} $$

Now, under $\tilde{P}$, $\tilde{t}_{m^2} \le 2m^2$ with probability at least $3/4$, if $m$ (equivalently $n$ or $N$) is large enough. Moreover, the probability of making a bridge of duration $m^2$, under $\tilde{P}$, is clearly at least $p$ for some constant $p > 0$ depending only on $d$. Since the latter event depends only on the jump chain and the former event depends only
where $X_i$ are independent unit exponential random variables. Hence for all $n$ (or equivalently $N$) large enough,

$$ P\left(\sum_{i=1}^{\lfloor Nn^{-2.2\kappa} \rfloor} X_i \ge Nn^{-2.1\kappa}\right) \le P\left(\sum_{i=1}^{\lfloor Nn^{-2.2\kappa} \rfloor} X_i \ge 2Nn^{-2.2\kappa}\right) \\ \le \exp(-cNn^{-2.2\kappa}) \le \exp(-N^{1-3\kappa}), $$

for some constant $c > 0$, by elementary Chernoff estimates for exponential random variables. Hence by (8.14),

$$ \tilde{P}\left(\tilde{L}_{\tilde{t}_N} \left(B\left(0, \rho_d n\left(1 - n^{-0.9\kappa}\right)\right)\right)^c \ge Nn^{-2.1\kappa}; \tilde{G}_{loc}\right) \le \exp(-N^{1-3\kappa}). $$

Plugging into (8.12), we deduce that if $\kappa$ is sufficiently small, for all $N$ large enough,

$$ \tilde{P}(A \setminus A') \le \exp(-N^{1-4\kappa}). $$

Since $Z_N = \exp(-O(n^d))$, we deduce that

$$ (8.15) \qquad \tilde{P}(A \setminus A') \le Z_N \exp(-N^{1-5\kappa}). $$

Combining (8.15), (8.11) and (8.10), we obtain

$$ E_{\tilde{P}}\left(e^{-|R_N|}; \mathcal{G}_{loc}; (\mathcal{G}_{br})^c\right) \le Z_N e^{-N^{1-5\kappa}} + Z_N e^{3n^{d-\epsilon}} \tilde{Q}\left(\tilde{G}_{loc}; (\tilde{G}_{br})^c\right), $$

as desired in (8.6).

Step 2. — Now it remains to show that $\tilde{G}_{loc} \cap \tilde{G}_{br}$ is overwhelmingly likely under $\tilde{Q}$. More precisely, we will argue that

$$ (8.16) \qquad \tilde{Q}\left(\left(\tilde{G}_{br}\right)^c; \tilde{G}_{loc}\right) \le \exp\left(-cn^{2-2\kappa}m^{d-2}\right). $$

We claim that every time the walk is in $B^\circ$, there is a probability bounded below by a constant $p$, say, depending only on the dimension, such that under $\tilde{Q}$, the walk will perform a bridge of duration $m^2$ (recall that this means the walk remains in $B$ for the next $m^2$ jumps and ends up in $B^\circ$ again after this time), uniformly over the initial position in $B^\circ$ of the walk. To see this, note that $\min_{B/n} \tilde{\phi} = (1-o(1)) \max_{B/n} \tilde{\phi}$, so that the Radon–Nikodym derivative $d\tilde{Q}/d\tilde{P}$ during an interval of time consisting of the first $m^2$ jumps of the chain (call this time $\tilde{t}_{m^2}$) is at least

$$ (8.17) \qquad \begin{aligned} \frac{d\tilde{Q}}{d\tilde{P}} &\ge \frac{\min_{B/n} \tilde{\phi}}{\max_{B/n} \tilde{\phi}} \exp\left(\tilde{t}_{m^2} \inf_{x \in B} \frac{\Delta_1 \tilde{\phi}}{\tilde{\phi}}(x/n)\right) \\ &\ge (1+o(1)) \exp\left(-(\lambda/2+o(1))\frac{\tilde{t}_{m^2}}{n^2}\right). \end{aligned} $$

Now, under $\tilde{P}$, $\tilde{t}_{m^2} \le 2m^2$ with probability at least $3/4$, if $m$ (equivalently $n$ or $N$) is large enough. Moreover, the probability of making a bridge of duration $m^2$, under $\tilde{P}$, is clearly at least $p$ for some constant $p > 0$ depending only on $d$. Since the latter event depends only on the jump chain and the former event depends only
on the time parametrisation, which are independent processes under $\tilde{P}$, we conclude
from (8.17) that for any $a \in B^\circ$,

$$
\begin{align*}
\tilde{Q}_a(X[0, \tilde{t}_{m^2}] \text{ is a bridge}) &\ge E_{\tilde{P}_a} \left( \frac{d\tilde{Q}_a}{d\tilde{P}_a}; X[0, \tilde{t}_{m^2}] \text{ is a bridge} \right) \\
&\ge (1+o(1)) E_{\tilde{P}_a} \left( \exp\left(-(\lambda/2 + o(1)) \frac{\tilde{t}_{m^2}}{n^2}; \tilde{t}_{m^2} \le 2m^2; X[0, \tilde{t}_{m^2}] \text{ is a bridge} \right) \right) \\
&\ge (3/4)(1+o(1)) \tilde{P}_a(X[0, \tilde{t}_{m^2}] \text{ is a bridge}) \ge (3/4)p(1+o(1)).
\end{align*}
$$

Hence $\tilde{Q}_a(X[0, \tilde{t}_{m^2}])$ is a bridge) $\ge p/2$ for $n$ large enough; thereby proving what we desired (with $p$ replaced by $p/2$, a distinction which is of no consequence in the rest of the argument).

Since we know by (8.1) that on $\tilde{\mathcal{G}}_{\text{loc}}$, the total amount of discrete steps in $B^\circ$ (and hence in $B$) is deterministically at least $\delta m^d n^{2-2\kappa}$, and each time the walk is in $B^\circ$ there is a probability $p/2$ (under $\tilde{Q}$) to make a bridge over the next $m^2$ jumps, we deduce that the number of bridges stochastically dominates a binomial random variable of parameters $(\delta m^{d-2} n^{2-2\kappa}, p/2)$, under $\tilde{Q}$. More precisely, let

$$
\#(s, t) := \text{number of jumps by } \tilde{X} \text{ during } (s, t],
$$

and define the sequence of stopping times:

$$
\sigma_1 = \inf \{ t > 0 : \tilde{X}_t \in B^\circ \};
$$

$$
\tau_1 = \inf \{ t > 0 : \#(\sigma_1, t) \ge m^2 \} \wedge \inf \{ t > \sigma_1 : \tilde{X}_t \notin B^\circ \};
$$

then inductively, for $i \ge 2$:

$$
\sigma_i = \inf \{ t > \tau_{i-1} : \tilde{X}_t \in B^\circ \},
$$

$$
\tau_i = \inf \{ t > 0 : \#(\sigma_i, t) \ge m^2 \} \wedge \inf \{ t > \sigma_i : \tilde{X}_t \notin B^\circ \}.
$$

Let

$$
\xi_i = 1_{\{\#(\sigma_i, \tau_i) \ge m^2\}}; \quad i \ge 1
$$

be the Bernoulli variable which is the indicator of the event that the $i$th trial results
in a bridge. Then note that by Lemma 8.2, if $j = \lfloor \delta m^{d-2} n^{2-2\kappa} \rfloor$, then $\tau_j \le \tilde{t}_N$ no
matter what on $\tilde{\mathcal{G}}_{\text{loc}}$. Hence if $S = \sum_{i=1}^j \xi_i$, then

$$
\{S \ge (\delta p/4)m^{d-2}n^{2-2\kappa}\} \subset \tilde{\mathcal{G}}_{br}
$$

where the constant $a > 0$ defining $\mathcal{G}_{br}$ is taken to be $a = p/4$.

Moreover by the Markov property and the above,

$$
\tilde{Q}(\xi_i = 1 | (X_t, t \le \sigma_i)) \ge p/2.
$$

Hence $S$ dominates stochastically a Binomial random variables with parameters $j$
and $p/2$. By standard Chernoff bounds for binomials, for some constant $c > 0$, we
deduce that

$$
\tilde{Q}((\tilde{\mathcal{G}}_{br})^c; \tilde{\mathcal{G}}_{loc}) \le \exp(-cn^{2-2\kappa}m^{d-2}),
$$

which shows (8.16).
on the time parametrisation, which are independent processes under $\tilde{P}$, we conclude
from (8.17) that for any $a \in B^\circ$,

$$
\begin{align*}
\tilde{Q}_a(X[0, \tilde{t}_{m^2}] \text{ is a bridge}) &\ge E_{\tilde{P}_a} \left( \frac{d\tilde{Q}_a}{d\tilde{P}_a}; X[0, \tilde{t}_{m^2}] \text{ is a bridge} \right) \\
&\ge (1+o(1)) E_{\tilde{P}_a} \left( \exp\left(-(\lambda/2 + o(1)) \frac{\tilde{t}_{m^2}}{n^2}; \tilde{t}_{m^2} \le 2m^2; X[0, \tilde{t}_{m^2}] \text{ is a bridge} \right) \right) \\
&\ge (3/4)(1+o(1)) \tilde{P}_a(X[0, \tilde{t}_{m^2}] \text{ is a bridge}) \ge (3/4)p(1+o(1)).
\end{align*}
$$

Hence $\tilde{Q}_a(X[0, \tilde{t}_{m^2}])$ is a bridge) $\ge p/2$ for $n$ large enough; thereby proving what we desired (with $p$ replaced by $p/2$, a distinction which is of no consequence in the rest of the argument).

Since we know by (8.1) that on $\tilde{\mathcal{G}}_{\text{loc}}$, the total amount of discrete steps in $B^\circ$ (and hence in $B$) is deterministically at least $\delta m^d n^{2-2\kappa}$, and each time the walk is in $B^\circ$ there is a probability $p/2$ (under $\tilde{Q}$) to make a bridge over the next $m^2$ jumps, we deduce that the number of bridges stochastically dominates a binomial random variable of parameters $(\delta m^{d-2} n^{2-2\kappa}, p/2)$, under $\tilde{Q}$. More precisely, let

$$
\sharp(s, t) := \text{number of jumps by } \tilde{X} \text{ during } (s, t],
$$

and define the sequence of stopping times:

$$
\begin{align*}
\sigma_1 &= \inf \{ t > 0 : \tilde{X}_t \in B^\circ \}; \\
\tau_1 &= \inf \{ t > 0 : \sharp(\sigma_1, t) \ge m^2 \} \wedge \inf \{ t > \sigma_1 : \tilde{X}_t \notin B^\circ \};
\end{align*}
$$

then inductively, for $i \ge 2$:

$$
\begin{align*}
\sigma_i &= \inf \{ t > \tau_{i-1} : \tilde{X}_t \in B^\circ \}, \\
\tau_i &= \inf \{ t > 0 : \sharp(\sigma_i, t) \ge m^2 \} \wedge \inf \{ t > \sigma_i : \tilde{X}_t \notin B^\circ \}.
\end{align*}
$$

Let

$$
\xi_i = 1_{\{\sharp(\sigma_i, \tau_i) \ge m^2\}}; \quad i \ge 1
$$

be the Bernoulli variable which is the indicator of the event that the $i$th trial results
in a bridge. Then note that by Lemma 8.2, if $j = \lfloor \delta m^{d-2} n^{2-2\kappa} \rfloor$, then $\tau_j \le \tilde{t}_N$ no
matter what on $\tilde{\mathcal{G}}_{\text{loc}}$. Hence if $S = \sum_{i=1}^j \xi_i$, then

$$
\{S \ge (\delta p/4)m^{d-2}n^{2-2\kappa}\} \subset \tilde{\mathcal{G}}_{br}
$$

where the constant $a > 0$ defining $\mathcal{G}_{br}$ is taken to be $a = p/4$.

Moreover by the Markov property and the above,

$$
\tilde{Q}(\xi_i = 1 | (X_t, t \le \sigma_i)) \ge p/2.
$$

Hence $S$ dominates stochastically a Binomial random variables with parameters $j$
and $p/2$. By standard Chernoff bounds for binomials, for some constant $c > 0$, we
deduce that

$$
\tilde{Q}((\tilde{\mathcal{G}}_{br})^c; \tilde{\mathcal{G}}_{loc}) \le \exp(-cn^{2-2\kappa}m^{d-2}),
$$

which shows (8.16).
Plugging (8.16) into (8.6), we deduce

$$E(e^{-|R_N|}; \mathcal{G}_{\text{loc}}; (\mathcal{G}_{br})^c) \le Z_N (e^{-N^{1-5\kappa}} + e^{n^{d-\varepsilon}} e^{-cn^{2-2\kappa}m^{d-2}}).$$

Since $m = \rho_d n^{1-2\kappa}$, $\varepsilon$ is fixed (by Lemma 8.2) in a way that depends only on the dimension, and we are free to choose $\kappa$ as small as we want, we can choose it so that $n^{2-2\kappa}m^{d-2} = n^{d-6\kappa}$ is much greater than $n^{d-\varepsilon}$ (i.e., we assume $6\kappa < \varepsilon$) and then for all $n$ large enough we have

$$E(e^{-|R_N|}; \mathcal{G}_{\text{loc}}; (\mathcal{G}_{br})^c) \le Z_N (e^{-N^{1-5\kappa}} + e^{-cn^{2-2\kappa}m^{d-2}}) \le Z_N e^{-cn^{d-6\kappa}}.$$

This completes the proof of Lemma 8.3. □

### 8.3. Conditioning and summation

Call $P^*$ the conditional probability given the local time at every site in $\mathbb{Z}^d \setminus B$. Let $k \ge 1$, $\mathcal{X} = \{x_1, \dots, x_k\} \subset B$ be a subset of $k$ distinct points in $B$ and let $\mathcal{B}_{\mathcal{X}}$ be the bad event that the range in $B$ avoids exactly those $k$ points, i.e., $R_N \cap B = B \setminus \mathcal{X}$. Note that

$$(8.18) \qquad \{B \not\subset R_N\} \subset (\mathcal{G}_{br})^c \cup \bigcup_{k=1}^{|B|} \bigcup_{\mathcal{X} \subset B; |\mathcal{X}|=k} (\mathcal{B}_{\mathcal{X}} \cap \mathcal{G}_{br}).$$

So we are led to try and analyse expectations of the form

$$\frac{1}{Z_N} E \left[ e^{-|R_N \cap B|^c} E^* \left[ e^{-|R_N \cap B|} 1_{\mathcal{B}_{\mathcal{X}} \cap \mathcal{G}_{br}} \right] \right]$$

where $\mathcal{X} \subset B$ and $|\mathcal{X}| = k \ge 1$ is arbitrary between 1 and $|B|$. The key will be the following estimate:

**LEMMA 8.4.** — For constants $C, c$ depending only on the dimension $d$,

$$E^* \left[ e^{-|R_N \cap B|} 1_{\mathcal{B}_{\mathcal{X}}; \mathcal{G}_{br}} \right] \le C e^{-|B|+k} \exp \left( -c \frac{n^{2-2\kappa}k}{m^2(\log m)^{2+10d}} \right).$$

We defer the proof of Lemma 8.4 and instead show how it implies Theorem 8.1.
Plugging (8.16) into (8.6), we deduce

$$E(e^{-|R_N|}; \mathcal{G}_{\text{loc}}; (\mathcal{G}_{br})^c) \le Z_N (e^{-N^{1-5\kappa}} + e^{n^{d-\epsilon}}e^{-cn^{2-2\kappa}m^{d-2}}).$$

Since $m = \rho_d n^{1-2\kappa}$, $\epsilon$ is fixed (by Lemma 8.2) in a way that depends only on the dimension, and we are free to choose $\kappa$ as small as we want, we can choose it so that $n^{2-2\kappa}m^{d-2} = n^{d-6\kappa}$ is much greater than $n^{d-\epsilon}$ (i.e., we assume $6\kappa < \epsilon$) and then for all $n$ large enough we have

$$E(e^{-|R_N|}; \mathcal{G}_{\text{loc}}; (\mathcal{G}_{br})^c) \le Z_N (e^{-N^{1-5\kappa}} + e^{-cn^{2-2\kappa}m^{d-2}}) \le Z_N e^{-cn^{d-6\kappa}}.$$

This completes the proof of Lemma 8.3. □

## 8.3. Conditioning and summation

Call $P^*$ the conditional probability given the local time at every site in $\mathbb{Z}^d \setminus B$. Let $k \ge 1$, $\mathcal{X} = \{x_1, \dots, x_k\} \subset B$ be a subset of $k$ distinct points in $B$ and let $\mathcal{B}_{\mathcal{X}}$ be the bad event that the range in $B$ avoids exactly those $k$ points, i.e., $R_N \cap B = B \setminus \mathcal{X}$. Note that

$$(8.18) \qquad \{B \not\subset R_N\} \subset (\mathcal{G}_{br})^c \cup \bigcup_{k=1}^{|B|} \bigcup_{\mathcal{X} \subset B; |\mathcal{X}|=k} (\mathcal{B}_{\mathcal{X}} \cap \mathcal{G}_{br}).$$

So we are led to try and analyse expectations of the form

$$\frac{1}{Z_N} E \left[ e^{-|R_N \cap B^c|} E^* \left[ e^{-|R_N \cap B|} 1_{\mathcal{B}_{\mathcal{X}} \cap \mathcal{G}_{br}} \right] \right]$$

where $\mathcal{X} \subset B$ and $|\mathcal{X}| = k \ge 1$ is arbitrary between 1 and $|B|$. The key will be the following estimate:

**LEMMA 8.4.** — For constants $C, c$ depending only on the dimension $d$,

$$E^* \left[ e^{-|R_N \cap B|} 1_{\mathcal{B}_{\mathcal{X}}; \mathcal{G}_{br}} \right] \le C e^{-|B|+k} \exp \left( -c \frac{n^{2-2\kappa}k}{m^2(\log m)^{2+10d}} \right).$$

We defer the proof of Lemma 8.4 and instead show how it implies Theorem 8.1.
*Proof of Theorem 8.1, assuming Lemma 8.4.* — Using (8.18) and Lemma 8.3, we see that since $|R_N \cap B| \le |B|,$

$$
\begin{align*}
& \frac{1}{Z_N} E \left( e^{-R_N} 1_{\mathcal{G}_{loc}} 1_{B \not\subset R_N} \right) \\
& \le e^{-cn^{d-6\kappa}} + \sum_{k=1}^{|B|} \sum_{\mathcal{X} \subset B, |\mathcal{X}|=k} \frac{1}{Z_N} E \left[ e^{-|R_N \cap B^c|} E^* \left[ e^{-|R_N \cap B|} 1_{B_{\mathcal{X}}; \mathcal{G}_{br}} \right] \right] \\
& \le e^{-cn^{d-6\kappa}} + C \sum_{k=1}^{|B|} \sum_{\mathcal{X} \subset B, |\mathcal{X}|=k} \frac{1}{Z_N} E \left[ e^{-|R_N \cap B^c|} e^{-|B|+k} e^{-c \frac{n^{2-2\kappa} k}{m^2 (\log m)^{2+10d}}} \right] \\
& \le e^{-cn^{d-6\kappa}} + C \sum_{k=1}^{|B|} \binom{|B|}{k} e^k e^{-c \frac{n^{2-2\kappa} k}{m^2 (\log m)^{2+10d}}} \frac{1}{Z_N} E \left[ e^{-|R_N|} \right] \\
& \le e^{-cn^{d-6\kappa}} + C \sum_{k=1}^{m^d} \binom{|B|}{k} \exp \left( k \left( 1 - c \frac{n^{2-2\kappa} k}{m^2 (\log m)^{2+10d}} \right) \right).
\end{align*}
$$

Now note that on the one hand, the entropic factor satisfies

$$ \binom{|B|}{k} \le (\omega_d m^d)^k = e^{Ckd \log m}. $$

On the other hand since $m = \rho_d n^{1-2\kappa}$, we have that

$$ \frac{n^{2-2\kappa}}{(m^2(\log m)^{2+10d})} \gtrsim \frac{n^{2\kappa}}{(\log n)^{2+10d}}. $$

This is of course much greater than the exponential factor in the entropic term of $C \log m$, and so altogether the above series is exponentially decaying. Hence we can conclude that

$$ \frac{1}{Z_N} E \left( e^{-R_N} 1_{\mathcal{G}_{loc}} 1_{B \not\subset R_N} \right) \le C \exp(-n^\kappa) $$

where $\kappa > 0$ and $C$ depend only on the dimension. Summing over all $O(n^d)$ possible centres of the ball $B$ and using a union bound we immediately deduce the statement of Theorem 8.1. $\square$

## 8.4. Transition probabilities for bridges

We now start the proof of Lemma 8.4. The idea is to show that for each bridge of duration at least $m^2$ there is a good chance of hitting our $k$ points. For this we will need the following bounds on the heat kernel of bridge; we will now further condition on the endpoints of the bridge. Let $P^{a \to b; \tau}$ denote the law of a bridge of duration $\tau$ starting from $a \in B$ and ending in $b$, i.e., simple random walk starting from $a$, conditioned to be in $b$ at time $\tau$ and to remain in $B$ throughout $[0, \tau]$. Implicit in this notation is the fact that $P^a(X_\tau = b) > 0$, i.e., the parity of $b-a$ is the same as $\tau$.

We start with the following Lemma 8.5.
*Proof of Theorem 8.1, assuming Lemma 8.4.* — Using (8.18) and Lemma 8.3, we see that since $|R_N \cap B| \le |B|,$

$$ 
\begin{align*}
& \frac{1}{Z_N} E \left( e^{-R_N} 1_{\mathcal{G}_{loc}} 1_{B \not\subset R_N} \right) \\
& \le e^{-cn^{d-6\kappa}} + \sum_{k=1}^{|B|} \sum_{\mathcal{X} \subset B, |\mathcal{X}|=k} \frac{1}{Z_N} E \left[ e^{-|R_N \cap B^c|} E^* \left[ e^{-|R_N \cap B|} 1_{B_{\mathcal{X}}; \mathcal{G}_{br}} \right] \right] \\
& \le e^{-cn^{d-6\kappa}} + C \sum_{k=1}^{|B|} \sum_{\mathcal{X} \subset B, |\mathcal{X}|=k} \frac{1}{Z_N} E \left[ e^{-|R_N \cap B^c|} e^{-|B|+k} e^{-c \frac{n^{2-2\kappa} k}{m^2 (\log m)^{2+10d}}} \right] \\
& \le e^{-cn^{d-6\kappa}} + C \sum_{k=1}^{|B|} \binom{|B|}{k} e^k e^{-c \frac{n^{2-2\kappa} k}{m^2 (\log m)^{2+10d}}} \frac{1}{Z_N} E \left[ e^{-|R_N|} \right] \\
& \le e^{-cn^{d-6\kappa}} + C \sum_{k=1}^{m^d} \binom{|B|}{k} \exp \left( k \left( 1 - c \frac{n^{2-2\kappa} k}{m^2 (\log m)^{2+10d}} \right) \right).
\end{align*}
$$

Now note that on the one hand, the entropic factor satisfies

$$ \binom{|B|}{k} \le (\omega_d m^d)^k = e^{Ckd \log m}. $$

On the other hand since $m = \rho_d n^{1-2\kappa}$, we have that

$$ n^{2-2\kappa} / (m^2 (\log m)^{2+10d}) \gtrsim n^{2\kappa} / (\log n)^{2+10d}. $$

This is of course much greater than the exponential factor in the entropic term of $C \log m$, and so altogether the above series is exponentially decaying. Hence we can conclude that

$$ \frac{1}{Z_N} E \left( e^{-R_N} 1_{\mathcal{G}_{loc}} 1_{B \not\subset R_N} \right) \le C \exp(-n^\kappa) $$

where $\kappa > 0$ and $C$ depend only on the dimension. Summing over all $O(n^d)$ possible centres of the ball $B$ and using a union bound we immediately deduce the statement of Theorem 8.1. $\square$

## 8.4. Transition probabilities for bridges

We now start the proof of Lemma 8.4. The idea is to show that for each bridge of duration at least $m^2$ there is a good chance of hitting our $k$ points. For this we will need the following bounds on the heat kernel of bridge; we will now further condition on the endpoints of the bridge. Let $P^{a \to b; \tau}$ denote the law of a bridge of duration $\tau$ starting from $a \in B$ and ending in $b$, i.e., simple random walk starting from $a$, conditioned to be in $b$ at time $\tau$ and to remain in $B$ throughout $[0, \tau]$. Implicit in this notation is the fact that $P^a(X_\tau = b) > 0$, i.e., the parity of $b-a$ is the same as $\tau$.

We start with the following Lemma 8.5.
LEMMA 8.5. — Suppose $\mathrm{dist}(x, \partial B) = r$ with $1 \le r \le m$. Then if $s \asymp m^2$,

$$ (8.19) \qquad P_x(X[0, s] \subset B) \asymp \frac{r}{m}. $$

The right hand side is essentially the familiar gambler's ruin probability in one dimension. Here it is important that we use a curved ball $B$ and not a box (otherwise if $x$ is near a corner the probability would be much smaller). This lemma could be deduced from [LL10, Proposition 6.9.4] but we choose to include a proof in order to make the presentation as self-contained as possible.

*Proof.* — The lower bound follows easily from an eigenvalue estimate and optional stopping: let $\lambda_B^1$ denote the principal eigenvalue of the (discrete) Laplacian $-\Delta_1$ in the ball $B$ with Dirichlet boundary conditions. Let $\phi_B^1$ is the corresponding eigenfunction, normalised so that $m^{-d} \sum_{x \in B} \phi(x) = 1$. Then note that

$$ M_s = \frac{1}{(1 - \lambda_B^1)^s} \phi_B^1(X_s), \quad s = 0, 1, \dots $$

is a nonnegative martingale. Apply the optional stopping theorem at the bounded stopping time $\tau_B \wedge s$ (where $\tau_B$ is the first time the walk leaves $B$) to see that

$$ \phi_B^1(x) = \frac{1}{(1 - \lambda_B^1)^s} E \left( \phi_B^1(X_s) 1_{\tau_B > s} \right) \le \frac{\|\phi_B^1\|_{\infty} P_x(\tau_B > s)}{(1 - \lambda_B^1)^s}. $$

Now, we have already mentioned that $\lambda_B^1 \asymp 1/m^2$ (with implied constants depending as usual only on the dimension) so that when $s \asymp m^2$,

$$ \phi_B^1(x) \lesssim \|\phi_B^1\|_{\infty} P_x(\tau_B > s). $$

Moreover, using [Bol94, Lemma 2.1], and using known properties of the principal eigenfunction in the continuum (namely that there is a radial derivative on the boundary of the ball), we see that

$$ \phi_B^1(x) \lesssim \frac{r}{m} + O(1/m) \lesssim \frac{r}{m}. $$

Since furthermore the principal eigenfunction on the unit ball in the continuum is bounded, using again [Bol94, Lemma 2.1], we deduce the lower bound

$$ (8.20) \qquad P_x(\tau_B > s) \gtrsim \frac{r}{m}, $$

which gives the desired lower bound.

In the other direction, let $\pi$ be the hyperplane tangent to the ball $B$ closest to the point $x$. Let $H$ denote the half space in the complement of $\pi$ that contains $x$, and let $\vec{n}$ be the normal vector to $\pi$. Let $\tau_H = \inf\{t \ge 0 : X_t \notin H\}$, and note that since $B \subset H$,

$$ P(\tau_B > s) \le P(\tau_H > s). $$

We will show that

$$ (8.21) \qquad P(\tau_H > s) \lesssim r/m. $$

Consider first the case where $r = \mathrm{dist}(x, \partial B) \ge \log m$. Then, using a KMT approximation (strong coupling with Brownian motion), see [LL10, Theorem 7.1.1], there
LEMMA 8.5. — Suppose $\mathrm{dist}(x, \partial B) = r$ with $1 \le r \le m$. Then if $s \asymp m^2$,

$$
(8.19) \qquad P_x(X[0, s] \subset B) \asymp \frac{r}{m}.
$$

The right hand side is essentially the familiar gambler's ruin probability in one dimension. Here it is important that we use a curved ball *B* and not a box (otherwise if *x* is near a corner the probability would be much smaller). This lemma could be deduced from [LL10, Proposition 6.9.4] but we choose to include a proof in order to make the presentation as self-contained as possible.

Proof. — The lower bound follows easily from an eigenvalue estimate and optional stopping: let $\lambda_B^1$ denote the principal eigenvalue of the (discrete) Laplacian $-\Delta_1$ in the ball $B$ with Dirichlet boundary conditions. Let $\phi_B^1$ is the corresponding eigenfunction, normalised so that $m^{-d} \sum_{x \in B} \phi(x) = 1$. Then note that

$$
M_s = \frac{1}{(1 - \lambda_B^1)^s} \phi_B^1(X_s), \quad s = 0, 1, \dots
$$

is a nonnegative martingale. Apply the optional stopping theorem at the bounded stopping time $\tau_B \wedge s$ (where $\tau_B$ is the first time the walk leaves $B$) to see that

$$
\phi_B^1(x) = \frac{1}{(1 - \lambda_B^1)^s} E \left( \phi_B^1(X_s) 1_{\tau_B > s} \right) \le \frac{\|\phi_B^1\|_{\infty} P_x(\tau_B > s)}{(1 - \lambda_B^1)^s}.
$$

Now, we have already mentioned that $\lambda_B^1 \asymp 1/m^2$ (with implied constants depending as usual only on the dimension) so that when $s \asymp m^2$,

$$
\phi_B^1(x) \lesssim \|\phi_B^1\|_{\infty} P_x(\tau_B > s).
$$

Moreover, using [Bol94, Lemma 2.1], and using known properties of the principal eigenfunction in the continuum (namely that there is a radial derivative on the boundary of the ball), we see that

$$
\phi_B^1(x) \lesssim \frac{r}{m} + O(1/m) \lesssim \frac{r}{m}.
$$

Since furthermore the principal eigenfunction on the unit ball in the continuum is bounded, using again [Bol94, Lemma 2.1], we deduce the lower bound

$$
(8.20) \qquad P_x(\tau_B > s) \gtrsim \frac{r}{m},
$$

which gives the desired lower bound.

In the other direction, let $\pi$ be the hyperplane tangent to the ball $B$ closest to the point $x$. Let $H$ denote the half space in the complement of $\pi$ that contains $x$, and let $\vec{n}$ be the normal vector to $\pi$. Let $\tau_H = \inf\{t \ge 0 : X_t \notin H\}$, and note that since $B \subset H$,

$$
P(\tau_B > s) \le P(\tau_H > s).
$$

We will show that

$$
(8.21) \qquad P(\tau_H > s) \lesssim r/m.
$$

Consider first the case where $r = \mathrm{dist}(x, \partial B) \ge \log m$. Then, using a KMT approximation (strong coupling with Brownian motion), see [LL10, Theorem 7.1.1], there
is a $d$-dimensional Brownian motion $(W_t^d, t \ge 0)$ such that if $s \asymp m^2$ then for some constant $C > 0$,

$$ \sup_{t \le s} |W_t - X_t| \le C \log m $$

on an event $E_1$ of probability at least $1 - m^{-1}$. Let

$$ \mathrm{osc} = \sup_{0 \le t \le s} |W_t^d - W_{[t]}^d| $$

and note that on an event $E_2$ of probability at least $1 - O(m)e^{-(\log m)^2/2} \ge 1 - 1/m$,

$$ \mathrm{osc} \le \log m. $$

Let $\pi^-$ denote a hyperplane parallel to $\pi$ at distance $C \log m$ from $\pi$ such that $\pi^-$ in the half space which does not contain $B$. Let $T_{\pi^-}$ denote the first (continuous) time when $W^d$ hits $\pi^-$. Then on $E_1 \cap E_2$, $\tau_H > s$ implies $T_{\pi^-} > s$ and hence

$$ P_x(\tau_H > s) \le P_x(T_{\pi^-} > s) + P_x(E_1^c) + P_x(E_2^c) \le 2/m + P_x(T^- > s). $$

Using rotational invariance of Brownian motion and projecting onto $\vec{n}$, letting $W = W^1$ be a standard one-dimensional Brownian motion and using the reflection principle,

$$ P_x(T_{\pi^-} > s) = P_{r+C \log m}(W_t \ge 0 \text{ for all } 0 \le t \le s) \asymp \frac{r+C \log m}{m}. $$

Since we assumed initially that $r \ge \log m$ we see that the right hand side above is $\asymp r/m$ and so this proves the upper bound in this case.

Now suppose that $r \le \log m$. Let $\pi^+$ denote another hyperplane parallel to $\pi$, also at distance $\log m$ from $\pi$ but such that $\pi^+$ is contained in the half space containing $B$ (and in particular, $\pi^+$ intersects $B$). Let $S$ be the slab comprised between $\pi$ and $\pi^+$, and let $\tau_S = \inf\{t \ge 0 : X_t \notin S\}$ denote the first time the walk leaves the slab. Then note that if the walk has left $S$ by time $s/2$, it must do so by hitting $\pi^+$ before $\pi$ and must then avoid $\pi$ for at least $s/2$ units of time. Of course, $s/2 \asymp m^2$ hence using the result in the case already proved that $r \ge \log m$, since $\pi^+$ is at distance $\log m$ from $\pi$,

$$ (8.22) \qquad P_x(\tau_H > s) \lesssim P_x(\tau_S > s/2) + P_x(\tau_{\pi^+} < \tau_H) \frac{\log m}{m}. $$

Let us bound the first term in the right hand side above. Note that the slab $S$ has a width equal to $\log m$ by definition. Hence every $(\log m)^2$ units of time, the walk has a probability bounded below by $p > 0$ to exit $S$. We deduce that

$$ P_x(\tau_S > s/2) \le (1-p)^{\lfloor(s/2)/(\log m)^2\rfloor} $$

and since $s \asymp m^2$ we see that this decays faster than any polynomial in $m$ and hence in particular is $O(1/m)$. Moreover, we claim that

$$ (8.23) \qquad P_x(\tau_{\pi^+} < \tau_H) \asymp \frac{r}{\log m} $$

so that, combining with (8.22), we get $P(\tau_H > s) \lesssim r/m$ as desired. To see (8.23), let $x'$ be the point at which $\pi$ is tangent to $B$ and consider the martingale

$$ M_t' = (X_t - x') \cdot \vec{n}, $$
is a $d$-dimensional Brownian motion $(W_t^d, t \ge 0)$ such that if $s \asymp m^2$ then for some constant $C > 0$,

$$ \sup_{t \le s} |W_t - X_t| \le C \log m $$

on an event $E_1$ of probability at least $1 - m^{-1}$. Let

$$ \mathrm{osc} = \sup_{0 \le t \le s} |W_t^d - W_{[t]}^d| $$

and note that on an event $E_2$ of probability at least $1 - O(m)e^{-(\log m)^2/2} \ge 1 - 1/m$,

$$ \mathrm{osc} \le \log m. $$

Let $\pi^-$ denote a hyperplane parallel to $\pi$ at distance $C \log m$ from $\pi$ such that $\pi^-$ in the half space which does not contain $B$. Let $T_{\pi^-}$ denote the first (continuous) time when $W^d$ hits $\pi^-$. Then on $E_1 \cap E_2$, $\tau_H > s$ implies $T_{\pi^-} > s$ and hence

$$ P_x(\tau_H > s) \le P_x(T_{\pi^-} > s) + P_x(E_1^c) + P_x(E_2^c) \le 2/m + P_x(T^- > s). $$

Using rotational invariance of Brownian motion and projecting onto $\vec{n}$, letting $W = W^1$ be a standard one-dimensional Brownian motion and using the reflection principle,

$$ P_x(T_{\pi^-} > s) = P_{r+C \log m}(W_t \ge 0 \text{ for all } 0 \le t \le s) \asymp \frac{r+C \log m}{m}. $$

Since we assumed initially that $r \ge \log m$ we see that the right hand side above is $\asymp r/m$ and so this proves the upper bound in this case.

Now suppose that $r \le \log m$. Let $\pi^+$ denote another hyperplane parallel to $\pi$, also at distance $\log m$ from $\pi$ but such that $\pi^+$ is contained in the half space containing $B$ (and in particular, $\pi^+$ intersects $B$). Let $S$ be the slab comprised between $\pi$ and $\pi^+$, and let $\tau_S = \inf\{t \ge 0 : X_t \notin S\}$ denote the first time the walk leaves the slab. Then note that if the walk has left $S$ by time $s/2$, it must do so by hitting $\pi^+$ before $\pi$ and must then avoid $\pi$ for at least $s/2$ units of time. Of course, $s/2 \asymp m^2$ hence using the result in the case already proved that $r \ge \log m$, since $\pi^+$ is at distance $\log m$ from $\pi$,

$$ (8.22) \qquad P_x(\tau_H > s) \lesssim P_x(\tau_S > s/2) + P_x(\tau_{\pi^+} < \tau_H) \frac{\log m}{m}. $$

Let us bound the first term in the right hand side above. Note that the slab $S$ has a width equal to $\log m$ by definition. Hence every $(\log m)^2$ units of time, the walk has a probability bounded below by $p > 0$ to exit $S$. We deduce that

$$ P_x(\tau_S > s/2) \le (1-p)^{\lfloor(s/2)/(\log m)^2\rfloor} $$

and since $s \asymp m^2$ we see that this decays faster than any polynomial in $m$ and hence in particular is $O(1/m)$. Moreover, we claim that

$$ (8.23) \qquad P_x(\tau_{\pi^+} < \tau_H) \asymp \frac{r}{\log m} $$

so that, combining with (8.22), we get $P(\tau_H > s) \lesssim r/m$ as desired. To see (8.23), let $x'$ be the point at which $\pi$ is tangent to $B$ and consider the martingale

$$ M_t' = (X_t - x') \cdot \vec{n}, $$
that is to say, the (signed) distance to the plane $\pi$ of the walk $X_t$. Note that this is indeed a martingale since $X$ is a martingale and the projection onto $\vec{n}$ is a linear operation. Apply the optional stopping theorem at the time $\tau_S = \tau_H \wedge \tau_{\pi^+}$. If $\tau_S = \tau_H$ then $|M'_{\tau_S}| \le 1$, whereas if $\tau_S = \tau_{\pi^+}$ then $|M_{\tau_S} - \log m| \le 1$. Consequently,

$$r = O(1)(1 - P_x(\tau_{\pi^+} < \tau_H)) + (\log m + O(1))P_x(\tau_{\pi^+} < \tau_H)$$

from which it immediately follows that

$$P_x(\tau_{\pi^+} < \tau_H) \le \frac{r + O(1)}{\log m + O(1)}$$

which proves (8.23). As explained, the Lemma 8.6 follows. □

We will also need a slight improvement of this estimate where the end point is specified. (This would also follow from [LL10, Proposition 6.9.4.] but as above we prefer to provide our own proof).

LEMMA 8.6. — In the same setting as Lemma 8.5, We have

$$P_x(X[0, t] \subset B; X_t = z) \lesssim t^{-d/2} \frac{\mathrm{dist}(x, \partial B)}{\sqrt{t}} \frac{\mathrm{dist}(z, \partial B)}{\sqrt{t}}.$$

Furthermore if $z \in B^\circ$ and $t \asymp m^2$ is such that $P_x(X_t = z) > 0$, the same inequality holds with $\lesssim$ replaced by $\gtrsim$.

Proof. — This uses a simple time reversal argument as well as the Markov property. Split the interval $[0, t]$ into three intervals of length $t/3$ each (for this argument we can assume without loss of generality that $t/3$ is an integer). Observe that the process $(X_{t-s} : 0 \le s \le t)$ is also by a reversibility a random walk which given $X_t = z$ will be starting from $z$. Hence, using the standard fact that $P_{x'}(X_t = z') \lesssim t^{-d/2}$ for all $t \ge 1$ and $x', z' \in \mathbb{Z}^d$,

$$ 
\begin{align*}
& P_x(X[0, t] \subset B; X_t = z) \\
\le & \sum_{x', z' \in B} P_x(X[0, t/3] \subset B; X_{t/3} = x') P_{x'}(X[0, t/3] \subset B; X_{t/3} = z') \\
& \qquad \times P_{z'}(X_{t/3} = z; X[0, t/3] \subset B) \\
\le & \sum_{x', z' \in B} P_x(X[0, t/3] \subset B; X_{t/3} = x') P_{x'}(X[0, t/3] \subset B; X_{t/3} = z') \\
& \qquad \times P_z(X_{t/3} = z'; X[0, t/3] \subset B) \\
\lesssim & \sum_{x', z' \in B} P_x(X[0, t/3] \subset B; X_{t/3} = x') \times \frac{1}{t^{d/2}} \times P_z(X_{t/3} = z'; X[0, t/3] \subset B) \\
\le & t^{-d/2} P_x(X[0, t/3] \subset B) P_z(X[0, t/3] \subset B) \\
\lesssim & t^{-d/2} \frac{\mathrm{dist}(x, \partial B)}{\sqrt{t}} \frac{\mathrm{dist}(z, \partial B)}{\sqrt{t}}
\end{align*}
$$

by (8.21), as desired. When $z \in B^\circ$ and $t \asymp m^2$, the opposite inequality holds using the lower bound in Lemma 8.5, and the fact that a Brownian bridge from a point in $B(0, 1/2)$ to another point in $B(0, 1/2)$ has a probability bounded below
that is to say, the (signed) distance to the plane $\pi$ of the walk $X_t$. Note that this
is indeed a martingale since $X$ is a martingale and the projection onto $\vec{n}$ is a linear
operation. Apply the optional stopping theorem at the time $\tau_S = \tau_H \wedge \tau_{\pi^+}$. If $\tau_S = \tau_H$
then $|M'_{\tau_S}| \le 1$, whereas if $\tau_S = \tau_{\pi^+}$ then $|M_{\tau_S} - \log m| \le 1$. Consequently,

$$
r = O(1)(1 - P_x(\tau_{\pi^+} < \tau_H)) + (\log m + O(1))P_x(\tau_{\pi^+} < \tau_H)
$$

from which it immediately follows that

$$
P_x(\tau_{\pi^+} < \tau_H) \le \frac{r + O(1)}{\log m + O(1)}
$$

which proves (8.23). As explained, the Lemma 8.6 follows.

<figure><img src="image_11.png" /></figure>

We will also need a slight improvement of this estimate where the end point is
specified. (This would also follow from [LL10, Proposition 6.9.4.] but as above we
prefer to provide our own proof).

LEMMA 8.6. — In the same setting as Lemma 8.5, We have

$$
P_x(X[0, t] \subset B; X_t = z) \lesssim t^{-d/2} \frac{\mathrm{dist}(x, \partial B)}{\sqrt{t}} \frac{\mathrm{dist}(z, \partial B)}{\sqrt{t}}.
$$

Furthermore if $z \in B^\circ$ and $t \asymp m^2$ is such that $P_x(X_t = z) > 0$, the same inequality
holds with $\lesssim$ replaced by $\gtrsim$.

Proof. — This uses a simple time reversal argument as well as the Markov property. Split the interval $[0, t]$ into three intervals of length $t/3$ each (for this argument we can assume without loss of generality that $t/3$ is an integer). Observe that the process $(X_{t-s} : 0 \le s \le t)$ is also by a reversibility a random walk which given $X_t = z$ will be starting from $z$. Hence, using the standard fact that $P_{x'}(X_t = z') \lesssim t^{-d/2}$ for all $t \ge 1$ and $x', z' \in \mathbb{Z}^d$,

$$
\begin{align*}
& P_x(X[0,t] \subset B; X_t = z) \\
\le & \sum_{x', z' \in B} P_x(X[0,t/3] \subset B; X_{t/3} = x') P_{x'}(X[0,t/3] \subset B; X_{t/3} = z') \\
& \qquad \times P_{z'}(X_{t/3} = z; X[0,t/3] \subset B) \\
\le & \sum_{x', z' \in B} P_x(X[0,t/3] \subset B; X_{t/3} = x') P_{x'}(X[0,t/3] \subset B; X_{t/3} = z') \\
& \qquad \times P_z(X_{t/3} = z'; X[0,t/3] \subset B) \\
\lesssim & \sum_{x', z' \in B} P_x(X[0,t/3] \subset B; X_{t/3} = x') \times \frac{1}{t^{d/2}} \times P_z(X_{t/3} = z'; X[0,t/3] \subset B) \\
\le & t^{-d/2} P_x(X[0,t/3] \subset B) P_z(X[0,t/3] \subset B) \\
\lesssim & t^{-d/2} \frac{\mathrm{dist}(x, \partial B)}{\sqrt{t}} \frac{\mathrm{dist}(z, \partial B)}{\sqrt{t}}
\end{align*}
$$

by (8.21), as desired. When $z \in B^\circ$ and $t \asymp m^2$, the opposite inequality holds
using the lower bound in Lemma 8.5, and the fact that a Brownian bridge from a
point in $B(0, 1/2)$ to another point in $B(0, 1/2)$ has a probability bounded below
to stay in $B(0, 1)$ throughout, uniformly in the endpoints of the trajectory within $B(0, 1/2)$. $\square$

The estimate we will rely on is the following:

LEMMA 8.7. — Let $t \ge 2$ and $x \in B$ with $\mathrm{dist}(x, \partial B) = r$ with $1 \le r \le m$. Then if $\tau = m^2$, uniformly in $a, b \in B^\circ$, and $s \in [\tau/4; \tau/2]$,

$$ (8.24) \qquad P^{a \to b; \tau}(X_s = x) \gtrsim \frac{1}{m^d} \left(\frac{r}{m}\right)^2 $$

if the parity of $x - a$ is the same as $s$. Moreover, in that case, uniformly in $a, b \in B^\circ$, and $s \in [\tau/4; \tau/2]$, $1 \le t \le \tau/4$, $1 \le r \le m$, the following holds: let $A = \{x \in B : r \le \mathrm{dist}(x, \partial B) < 2r\}$ be the annulus at distance $r$. Then for any $x \in A$,

$$ (8.25) \qquad P^{a \to b; \tau}(X_{s+t} \in A | X_s = x) \lesssim r^2 (\log t)^{10d}/t. $$

*Proof.* — Observe that, uniformly in $a, b \in B^\circ$,

$$ P_a(X_\tau = b, X[0, \tau] \subset B) \asymp P_a(X_\tau = b) \asymp \frac{1}{m^d} $$

since a Brownian bridge from one point in a ball to another point in a ball has positive probability to stay in that ball throughout. Thus using Lemma 8.6 and reversibility (and noting that the parity of $b - x$ is the same as $\tau - s$ under our assumptions),

$$ \begin{align*} P^{a \to b; \tau}(X_s = x) &\asymp \frac{P_a(X_s = x; X[0, s] \subset B) \times P_x(X_{\tau-s} = b; X[0, \tau-s] \subset B)}{1/m^d} \\ &\gtrsim \frac{P_x(X_s = a; X[0, s] \in B) \times \frac{1}{m^d} \frac{r}{m}}{1/m^d} \\ &\gtrsim \frac{\frac{1}{m^d} \frac{r}{m} \times \frac{1}{m^d} \frac{r}{m}}{1/m^d} = \frac{1}{m^d} \left(\frac{r}{m}\right)^2 \end{align*} $$

which proves (8.24).

For the upper bound (8.25), observe that by the Markov property, if $\tau' = \tau - s$,

$$ P^{a \to b; \tau}(X_{s+t} \in A | X_s = x) = P^{x \to b; \tau'}(X_t \in A) = \frac{P_x(X_t \in A; X[0, \tau'] \subset B; X_{\tau'} = b)}{P_x(X[0, \tau'] \subset B; X_{\tau'} = b)}. $$

Consider first the denominator. Observe that by Lemma 8.6, since $b \in B^\circ$ and $\tau' \asymp m^2$,

$$ P_x(X[0, \tau'] \subset B; X_{\tau'} = b) \asymp \frac{r}{m} \frac{1}{m^d}. $$

Now consider the numerator and suppose we condition on $X_t = y \in A$. Since $s \in [\tau/4, \tau/2]$ we have $\tau' = \tau - s \in [\tau/2, 3\tau/4]$. Hence since we also assume that $t \le \tau/4$ we deduce that $\tau' - t \in [\tau/4, \tau/2]$, and hence $\tau' - t \asymp \tau' \asymp \tau$. From this, it is not hard to see that

$$ P_y(X([0, \tau' - t] \subset B; X_{\tau' - t} = b) \asymp \frac{r}{m} \frac{1}{m^d}. $$
to stay in $B(0, 1)$ throughout, uniformly in the endpoints of the trajectory within $B(0, 1/2)$. $\square$

The estimate we will rely on is the following:

LEMMA 8.7. — Let $t \ge 2$ and $x \in B$ with $\mathrm{dist}(x, \partial B) = r$ with $1 \le r \le m$. Then if $\tau = m^2$, uniformly in $a, b \in B^\circ$, and $s \in [\tau/4; \tau/2]$,

$$ (8.24) \qquad P^{a \to b; \tau}(X_s = x) \gtrsim \frac{1}{m^d} \left(\frac{r}{m}\right)^2 $$

if the parity of $x - a$ is the same as $s$. Moreover, in that case, uniformly in $a, b \in B^\circ$, and $s \in [\tau/4; \tau/2]$, $1 \le t \le \tau/4$, $1 \le r \le m$, the following holds: let $A = \{x \in B : r \le \mathrm{dist}(x, \partial B) < 2r\}$ be the annulus at distance $r$. Then for any $x \in A$,

$$ (8.25) \qquad P^{a \to b; \tau}(X_{s+t} \in A | X_s = x) \lesssim r^2 (\log t)^{10d}/t. $$

*Proof.* — Observe that, uniformly in $a, b \in B^\circ$,

$$ P_a(X_\tau = b, X[0, \tau] \subset B) \asymp P_a(X_\tau = b) \asymp \frac{1}{m^d} $$

since a Brownian bridge from one point in a ball to another point in a ball has positive probability to stay in that ball throughout. Thus using Lemma 8.6 and reversibility (and noting that the parity of $b - x$ is the same as $\tau - s$ under our assumptions),

$$ \begin{align*} P^{a \to b; \tau}(X_s = x) &\asymp \frac{P_a(X_s = x; X[0, s] \subset B) \times P_x(X_{\tau-s} = b; X[0, \tau-s] \subset B)}{1/m^d} \\ &\gtrsim \frac{P_x(X_s = a; X[0, s] \in B) \times \frac{1}{m^d} \frac{r}{m}}{1/m^d} \\ &\gtrsim \frac{\frac{1}{m^d} \frac{r}{m} \times \frac{1}{m^d} \frac{r}{m}}{1/m^d} = \frac{1}{m^d} \left(\frac{r}{m}\right)^2 \end{align*} $$

which proves (8.24).

For the upper bound (8.25), observe that by the Markov property, if $\tau' = \tau - s$,

$$ P^{a \to b; \tau}(X_{s+t} \in A | X_s = x) = P^{x \to b; \tau'}(X_t \in A) = \frac{P_x(X_t \in A; X[0, \tau'] \subset B; X_{\tau'} = b)}{P_x(X[0, \tau'] \subset B; X_{\tau'} = b)}. $$

Consider first the denominator. Observe that by Lemma 8.6, since $b \in B^\circ$ and $\tau' \asymp m^2$,

$$ P_x(X[0, \tau'] \subset B; X_{\tau'} = b) \asymp \frac{r}{m} \frac{1}{m^d}. $$

Now consider the numerator and suppose we condition on $X_t = y \in A$. Since $s \in [\tau/4, \tau/2]$ we have $\tau' = \tau - s \in [\tau/2, 3\tau/4]$. Hence since we also assume that $t \le \tau/4$ we deduce that $\tau' - t \in [\tau/4, \tau/2]$, and hence $\tau' - t \asymp \tau' \asymp \tau$. From this, it is not hard to see that

$$ P_y(X([0, \tau' - t] \subset B; X_{\tau' - t} = b) \asymp \frac{r}{m} \frac{1}{m^d}. $$
Consequently the numerator satisfies

$$
\begin{align*}
P_x(X_t \in A; X[0, \tau'] \subset B; X_{\tau'} = b) &\lesssim \frac{r}{m} \frac{1}{m^d} P_x(X_t \in A; X[0, t] \in B) \\
&\lesssim \frac{r}{m} \frac{1}{m^d} \sum_{y \in A} P_x(X_t = y, X[0, t] \in B).
\end{align*}
$$

Now, if $|y - x| \le t^{1/2}(\log t)^{10}$ then we use Lemma 8.5 to obtain that

$$
P_x(X_t = y, X[0, t] \in B) \lesssim \frac{r^2}{t} t^{-d/2},
$$

and since $\#\{y : |y - x| \le t^{1/2}(\log t)^{10}\} \lesssim t^{d/2}(\log t)^{10d}$, the contribution to the sum
from such points is at most

$$
\sum_{y \in A : |y-x| \le (\log t)^{10}} P_x(X_t = y, X[0,t] \in B) \lesssim \frac{r^2}{t} (\log t)^{10}.
$$

Now if $|y - x| \le t^{1/2}(\log t)^{10}$, then

$$
P_x(X_t = y, X[0, t] \in B) \le P_x(X_t = y)
$$

and so summing over all such y, the contribution is at most

$$
P_x\left(|X_t - x| \ge t^{1/2}(\log t)^{10}\right) \le \exp\left(-\frac{(\log t)^{20}}{2}\right) = o(1/t) \lesssim \frac{r^2}{t}
$$

using elementary Chernoff large deviation bounds for sums of i.i.d. random variables.
Hence

$$
\sum_{y \in A} P_x(X_t = y, X[0, t] \in B) \lesssim \frac{r^2 (\log t)^{10d}}{t}
$$

and so the numerator satisfies

$$
P_x(X_t \in A; X[0, \tau'] \subset B; X_{\tau'} = b) \lesssim \frac{r}{m} \frac{1}{m^d} \frac{r^2 (\log t)^{10d}}{t}.
$$

Combining with the bound on the denominator, we deduce

$$
P^{a \to b; \tau} (X_{s+t} \in A | X_s = x) \le \frac{r^2 (\log t)^{10d}}{t}
$$

as desired.

<figure><img src="image_9.png" /></figure>

8.5. End of proof of Theorem 8.1

We now start the proof of Lemma 8.4. Decompose the ball $B = B(0, m)$ into
dyadic annuli

$$
A_j = \{y \in B : \mathrm{dist}(y, \partial B) \in [2^j; 2^{j+1})\}; \quad 0 \le j \le \log_2(m)
$$

Given our point configuration $\mathcal{X}$ of $k$ disjoint points in $B$, let $\mathcal{X}_j = \mathcal{X} \cap A_j$; let $j$ be
such that $|\mathcal{X}_j|$ is maximal. Then note that

$$
|\mathcal{X}_j| \gtrsim \frac{k}{\log m}.
$$
Consequently the numerator satisfies

$$
\begin{align*}
P_x(X_t \in A; X[0, \tau'] \subset B; X_{\tau'} = b) &\lesssim \frac{r}{m} \frac{1}{m^d} P_x(X_t \in A; X[0, t] \in B) \\
&\lesssim \frac{r}{m} \frac{1}{m^d} \sum_{y \in A} P_x(X_t = y, X[0, t] \in B).
\end{align*}
$$

Now, if $|y - x| \le t^{1/2}(\log t)^{10}$ then we use Lemma 8.5 to obtain that

$$
P_x(X_t = y, X[0, t] \in B) \lesssim \frac{r^2}{t} t^{-d/2},
$$

and since $\#\{y : |y - x| \le t^{1/2}(\log t)^{10}\} \lesssim t^{d/2}(\log t)^{10d}$, the contribution to the sum from such points is at most

$$
\sum_{y \in A: |y-x| \le (\log t)^{10}} P_x(X_t = y, X[0,t] \in B) \lesssim \frac{r^2}{t} (\log t)^{10}.
$$

Now if $|y - x| \le t^{1/2}(\log t)^{10}$, then

$$
P_x(X_t = y, X[0, t] \in B) \le P_x(X_t = y)
$$

and so summing over all such y, the contribution is at most

$$
P_x \left( |X_t - x| \ge t^{1/2} (\log t)^{10} \right) \le \exp \left( -(\log t)^{20}/2 \right) = o(1/t) \lesssim r^2/t
$$

using elementary Chernoff large deviation bounds for sums of i.i.d. random variables.
Hence

$$
\sum_{y \in A} P_x(X_t = y, X[0, t] \in B) \lesssim \frac{r^2 (\log t)^{10d}}{t}
$$

and so the numerator satisfies

$$
P_x(X_t \in A; X[0, \tau'] \subset B; X_{\tau'} = b) \lesssim \frac{r}{m} \frac{1}{m^d} \frac{r^2 (\log t)^{10d}}{t}.
$$

Combining with the bound on the denominator, we deduce

$$
P^{a \to b; \tau} (X_{s+t} \in A | X_s = x) \le \frac{r^2 (\log t)^{10d}}{t}
$$

as desired.

□

8.5. End of proof of Theorem 8.1

We now start the proof of Lemma 8.4. Decompose the ball $B = B(0, m)$ into
dyadic annuli

$$
A_j = \{y \in B : \mathrm{dist}(y, \partial B) \in [2^j; 2^{j+1})\}; \quad 0 \le j \le \log_2(m)
$$

Given our point configuration $\mathcal{X}$ of $k$ disjoint points in $B$, let $\mathcal{X}_j = \mathcal{X} \cap A_j$; let $j$ be
such that $|\mathcal{X}_j|$ is maximal. Then note that

$$
|\mathcal{X}_j| \gtrsim \frac{k}{\log m}.
$$
We will show that any bridge of duration $\tau = m^2$ has a probability bounded below uniformly in its endpoints $a, b$ to visit $\mathcal{X}_j$. Given a bridge $X$ of duration $\tau$, let

$$L(\mathcal{X}_j) = \int_{\tau/4}^{\tau/2} 1_{X_s \in \mathcal{X}_j} ds$$

denote its local time spent in $\mathcal{X}_j$ during its second quarter.

LEMMA 8.8. — Uniformly over $a, b \in B^\circ$,

$$P^{a \to b; \tau}(L(\mathcal{X}_j) > 0) \gtrsim \frac{k}{(\log m)^{2+10d} m^d}$$

*Proof.* — We use the trivial identity for nonnegative random variables:

$$P(X > 0) = \frac{E(X)}{E(X|X > 0)}$$

Hence we need an upper bound on $E^{a \to b; \tau}(L(\mathcal{X}_j)|L(\mathcal{X}_j) > 0)$ and a lower bound on $E^{a \to b; \tau}(L(\mathcal{X}_j))$.

We start by the lower bound. Using (8.24),

$$\begin{aligned} E^{a \to b; \tau}(L(\mathcal{X}_j)) &\gtrsim |\mathcal{X}_j|(\tau/4) \frac{1}{m^d} \left(\frac{r}{m}\right)^2 \\ &\gtrsim \frac{k}{\log m} \frac{1}{m^{d-2}} \left(\frac{r}{m}\right)^2 \end{aligned}$$

where $r = 2^j$. On the other hand, for the upper bound we can apply the Markov property at the first hitting of $\mathcal{X}_j$ and (8.25) to deduce that

$$\begin{aligned} E^{a \to b; \tau}(L(\mathcal{X}_j) | L(\mathcal{X}_j) > 0) &\le \sup_{s \in [\tau/4, \tau/2]} \sup_{x \in \mathcal{X}_j} \sum_{y \in \mathcal{X}_j} \int_0^{\tau/4} P^{a \to b; \tau}(X_{s+t} = y | X_s = x) dt \\ &\lesssim \sup_{s \in [\tau/4, \tau/2]} \sup_{x \in \mathcal{X}_j} \int_0^{\tau/4} P^{a \to b; \tau}(X_{s+t} \in A_j | X_s = x) dt \\ &\lesssim 1 + \int_2^{\tau/4} \frac{r^2 (\log t)^{10d}}{t} dt \lesssim r^2 (\log m)^{1+10d}. \end{aligned}$$

Taking the quotient of these two terms, we deduce

$$P^{a \to b; \tau}(L(\mathcal{X}_j) > 0) \gtrsim \frac{k}{(\log m)^{2+10d} m^d}$$

as desired.

□

We are now able to complete the proof of Lemma 8.4.

*Proof.* — Proof of Lemma 8.4. On $\mathcal{G}_{br}$ there are at least $cn^{2-2\kappa}m^{d-2}$ bridges of length $m^2$, by definition of this event, where $c$ depends only on the dimension. When we condition on the endpoints of the bridges they are independent of each other and of anything else under $P^*$. Hence, using Lemma 8.8, and the inequality
We will show that any bridge of duration $\tau = m^2$ has a probability bounded below uniformly in its endpoints $a, b$ to visit $\mathcal{X}_j$. Given a bridge $X$ of duration $\tau$, let

$$L(\mathcal{X}_j) = \int_{\tau/4}^{\tau/2} 1_{X_s \in \mathcal{X}_j} ds$$

denote its local time spent in $\mathcal{X}_j$ during its second quarter.

LEMMA 8.8. — Uniformly over $a, b \in B^\circ$,

$$P^{a \to b; \tau}(L(\mathcal{X}_j) > 0) \gtrsim \frac{k}{(\log m)^{2+10d} m^d}.$$

*Proof.* — We use the trivial identity for nonnegative random variables:

$$P(X > 0) = \frac{E(X)}{E(X|X > 0)}$$

Hence we need an upper bound on $E^{a \to b; \tau}(L(\mathcal{X}_j)|L(\mathcal{X}_j) > 0)$ and a lower bound on $E^{a \to b; \tau}(L(\mathcal{X}_j))$.

We start by the lower bound. Using (8.24),

$$\begin{aligned} E^{a \to b; \tau}(L(\mathcal{X}_j)) &\gtrsim |\mathcal{X}_j|(\tau/4) \frac{1}{m^d} \left(\frac{r}{m}\right)^2 \\ &\gtrsim \frac{k}{\log m} \frac{1}{m^{d-2}} \left(\frac{r}{m}\right)^2 \end{aligned}$$

where $r = 2^j$. On the other hand, for the upper bound we can apply the Markov property at the first hitting of $\mathcal{X}_j$ and (8.25) to deduce that

$$\begin{aligned} E^{a \to b; \tau}(L(\mathcal{X}_j) | L(\mathcal{X}_j) > 0) &\le \sup_{s \in [\tau/4, \tau/2]} \sup_{x \in \mathcal{X}_j} \sum_{y \in \mathcal{X}_j} \int_0^{\tau/4} P^{a \to b; \tau}(X_{s+t} = y | X_s = x) dt \\ &\lesssim \sup_{s \in [\tau/4, \tau/2]} \sup_{x \in \mathcal{X}_j} \int_0^{\tau/4} P^{a \to b; \tau}(X_{s+t} \in A_j | X_s = x) dt \\ &\lesssim 1 + \int_2^{\tau/4} \frac{r^2 (\log t)^{10d}}{t} dt \lesssim r^2 (\log m)^{1+10d}. \end{aligned}$$

Taking the quotient of these two terms, we deduce

$$P^{a \to b; \tau}(L(\mathcal{X}_j) > 0) \gtrsim \frac{k}{(\log m)^{2+10d} m^d}$$

as desired.

□

We are now able to complete the proof of Lemma 8.4.

*Proof.* — Proof of Lemma 8.4. On $\mathcal{G}_{br}$ there are at least $cn^{2-2\kappa}m^{d-2}$ bridges of length $m^2$, by definition of this event, where $c$ depends only on the dimension. When we condition on the endpoints of the bridges they are independent of each other and of anything else under $P^*$. Hence, using Lemma 8.8, and the inequality
$1 - x \le e^{-x}$ valid for all $x \in \mathbb{R}$,

$$
\begin{aligned}
P^*(\mathcal{G}_{br}; \mathcal{B}_\chi) &\le \left(1 - \frac{k}{(\log m)^{2+10d} m^d}\right)^{cn^{2-2\kappa}m^{d-2}} \\
&\le \exp\left(-c \frac{k}{m^d(\log m)^{2+10d}} n^{2-2\kappa} m^{d-2}\right) \\
&\le \exp\left(-c \frac{n^{2-2\kappa}}{m^2(\log m)^{2+10d}} k\right).
\end{aligned}
$$

Since on this event $|R_N \cap B| = |B| - k$, Lemma 8.4 follows. $\square$

As explained this also finishes the proof of Theorem 8.1.

Appendix A. Functional inequalities on the rescaled lattice

Let $x \in \mathbb{R}^d$ and $r > 0$. The cubic box centered at $x$ of side length $r$ is

$$
\Lambda(x, r) = \{ y = (y_1, \dots, y_d) \in \mathbb{R}^d : -r/2 < y_i - x_i \le r/2, 1 \le i \le d \} .
$$

Let $n \ge 1$. We denote by $\mathbb{Z}_n^d$ the lattice $\mathbb{Z}^d$ rescaled by a factor $n$:

$$
\mathbb{Z}_n^d = \frac{1}{n} \mathbb{Z}^d.
$$

Let $x \in \mathbb{Z}_n^d$ and $r$ a rational number such that $nr$ is an odd integer. We define the discrete box $\Lambda_n(x, r)$ as

$$
\Lambda_n(x, r) = \Lambda(x, r) \cap \mathbb{Z}_n^d.
$$

We rewrite first the general discrete Poincaré–Sobolev inequality proved by Bessemou-
lin–Chatard, Chainais–Hillairet and Filbet (see [BCCHF15, Theorem 3]) in the
particular case of a box and a cubic mesh and for the exponent

$$
2^* = \frac{2d}{d-2}.
$$

THEOREM A.9. — [Discrete Poincaré–Sobolev inequality] Let $f$ be a function from $\Lambda_n(x, r)$ to $\mathbb{R}$. There exists a constant $c_{PS} = c_{PS}(x, r, d)$ which depends on $x, r, d$ such that

$$
\left( \sum_{y \in \Lambda_n(x,r)} \frac{1}{n^d} |f(y)|^{2*} \right)^{\frac{1}{2*}}
\le c_{PS} \left( \sum_{y \in \Lambda_n(x,r)} \frac{1}{n^d} |f(y)|^2 \right)^{\frac{1}{2}} + c_{PS} \left( \sum_{\substack{y,z \in \Lambda_n(x,r) \\ |y-z|=1/n}} \frac{1}{n^{d-2}} (f(y)-f(z))^2 \right)^{\frac{1}{2}}.
$$

We rewrite now the discrete Poincaré–Wirtinger inequality proved by Bessemoulin–
Chatard, Chainais–Hillairet and Filbet (see [BCCHF15, Theorem 5]) in the particular
case of a box and a cubic mesh and for the exponent 2.
$1 - x \le e^{-x}$ valid for all $x \in \mathbb{R}$,

$$ 
\begin{aligned}
P^* (\mathcal{G}_{br}; \mathcal{B}_\chi) &\le (1 - \frac{k}{(\log m)^{2+10d} m^d})^{cn^{2-2\kappa}m^{d-2}} \\
&\le \exp \left( -c \frac{k}{m^d (\log m)^{2+10d}} n^{2-2\kappa} m^{d-2} \right) \\
&\le \exp \left( -c \frac{n^{2-2\kappa}}{m^2 (\log m)^{2+10d}} k \right).
\end{aligned}
$$

Since on this event $|R_N \cap B| = |B| - k$, Lemma 8.4 follows. □

As explained this also finishes the proof of Theorem 8.1.

## Appendix A. Functional inequalities on the rescaled lattice

Let $x \in \mathbb{R}^d$ and $r > 0$. The cubic box centered at $x$ of side length $r$ is

$$ \Lambda(x, r) = \{ y = (y_1, \dots, y_d) \in \mathbb{R}^d : -r/2 < y_i - x_i \le r/2, 1 \le i \le d \}. $$

Let $n \ge 1$. We denote by $\mathbb{Z}_n^d$ the lattice $\mathbb{Z}^d$ rescaled by a factor $n$:

$$ \mathbb{Z}_n^d = \frac{1}{n} \mathbb{Z}^d. $$

Let $x \in \mathbb{Z}_n^d$ and $r$ a rational number such that $nr$ is an odd integer. We define the discrete box $\Lambda_n(x, r)$ as

$$ \Lambda_n(x, r) = \Lambda(x, r) \cap \mathbb{Z}_n^d. $$

We rewrite first the general discrete Poincaré–Sobolev inequality proved by Bessemoulin–Chatard, Chainais–Hillairet and Filbet (see [BCCHF15, Theorem 3]) in the particular case of a box and a cubic mesh and for the exponent

$$ 2^* = \frac{2d}{d-2}. $$

**THEOREM A.9.** — [Discrete Poincaré–Sobolev inequality] Let $f$ be a function from $\Lambda_n(x, r)$ to $\mathbb{R}$. There exists a constant $c_{PS} = c_{PS}(x, r, d)$ which depends on $x, r, d$ such that

$$ \left( \sum_{y \in \Lambda_n(x, r)} \frac{1}{n^d} |f(y)|^{2^*} \right)^{\frac{1}{2^*}} \\ \le c_{PS} \left( \sum_{y \in \Lambda_n(x, r)} \frac{1}{n^d} |f(y)|^2 \right)^{\frac{1}{2}} + c_{PS} \left( \sum_{\substack{y, z \in \Lambda_n(x, r) \\ |y-z|=1/n}} \frac{1}{n^{d-2}} (f(y) - f(z))^2 \right)^{\frac{1}{2}}. $$

We rewrite now the discrete Poincaré–Wirtinger inequality proved by Bessemoulin–Chatard, Chainais–Hillairet and Filbet (see [BCCHF15, Theorem 5]) in the particular case of a box and a cubic mesh and for the exponent 2.
THEOREM A.10. — [Discrete Poincaré–Wirtinger inequality] Let $f$ be a function from $\Lambda_n(x, r)$ to $\mathbb{R}$. There exists a constant $c_{PW} = c_{PW}(x, r, d)$ which depends on $x, r, d$ such that

$$ \left( \sum_{y \in \Lambda_n(x, r)} \frac{1}{n^d} (f(y) - \bar{f})^2 \right)^{\frac{1}{2}} \le c_{PW} \left( \sum_{\substack{y, z \in \Lambda_n(x, r) \\ |y-z|=1/n}} \frac{1}{n^{d-2}} (f(y) - f(z))^2 \right)^{\frac{1}{2}}, $$

where

$$ \bar{f} = \frac{1}{(rn)^d} \sum_{y \in \Lambda_n(x, r)} f(y). $$

In general, the constants $c_{PS}$, $c_{PW}$ depend on both $x \in \mathbb{Z}_n^d$ and $r$. However, the lattices $\mathbb{Z}_n^d$ being invariant under a translation by an element of $\mathbb{Z}_n^d$, these constants are the same for all points $x \in \mathbb{Z}_n^d$. From now onwards, we suppose that $x = 0$. Let us examine the dependence of the constants $c_{PS}$, $c_{PW}$ with respect to $r$. Let $f$ be a function defined on $\Lambda_n(x, r)$ with values in $\mathbb{R}$ and let us set

$$ \forall x \in \Lambda_{nr}(0, 1) \quad g(x) = f(rx). $$

We apply the inequality stated in Theorem A.10 to $g$:

$$ \left( \sum_{y \in \Lambda_{nr}(0,1)} \frac{1}{(nr)^d} (g(y) - \bar{g})^2 \right)^{\frac{1}{2}} \le c_{PW}(0,1,d) \left( \sum_{\substack{y,z \in \Lambda_{nr}(0,1) \\ |y-z|=1/(nr)}} \frac{1}{(nr)^{d-2}} (g(y) - g(z))^2 \right)^{\frac{1}{2}}, $$

and we rewrite everything in terms of the function $f$:

$$ \bar{g} = \frac{1}{(nr)^d} \sum_{y \in \Lambda_{nr}(0,1)} f(ry) = \frac{1}{(nr)^d} \sum_{y \in \Lambda_n(0,r)} f(y) = \bar{f}, $$

$$ \sum_{y \in \Lambda_{nr}(0,1)} (g(y) - \bar{g})^2 = \sum_{y \in \Lambda_n(0,r)} (f(y) - \bar{f})^2, $$

$$ \sum_{\substack{y, z \in \Lambda_{nr}(0,1) \\ |y-z|=1/(nr)}} (g(y) - g(z))^2 = \sum_{\substack{y, z \in \Lambda_n(0,r) \\ |y-z|=1/n}} (f(y) - f(z))^2. $$

We obtain the following inequality for the function $f$:

$$ \left( \sum_{y \in \Lambda_n(0, r)} \frac{1}{n^d} (f(y) - \bar{f})^2 \right)^{\frac{1}{2}} \le c_{PW}(0, 1, d) r \left( \sum_{\substack{y, z \in \Lambda_n(0, r) \\ |y-z|=1/n}} \frac{1}{n^{d-2}} (f(y) - f(z))^2 \right)^{\frac{1}{2}}. $$

We conclude that $c_{PW}(0, r, d) = c_{PW}(0, 1, d)r$.
THEOREM A.10. — [Discrete Poincaré–Wirtinger inequality] Let $f$ be a function from $\Lambda_n(x, r)$ to $\mathbb{R}$. There exists a constant $c_{PW} = c_{PW}(x, r, d)$ which depends on $x, r, d$ such that

$$ \left( \sum_{y \in \Lambda_n(x, r)} \frac{1}{n^d} (f(y) - \bar{f})^2 \right)^{\frac{1}{2}} \le c_{PW} \left( \sum_{\substack{y, z \in \Lambda_n(x, r) \\ |y-z|=1/n}} \frac{1}{n^{d-2}} (f(y) - f(z))^2 \right)^{\frac{1}{2}}, $$

where

$$ \bar{f} = \frac{1}{(rn)^d} \sum_{y \in \Lambda_n(x, r)} f(y). $$

In general, the constants $c_{PS}$, $c_{PW}$ depend on both $x \in \mathbb{Z}_n^d$ and $r$. However, the lattices $\mathbb{Z}_n^d$ being invariant under a translation by an element of $\mathbb{Z}_n^d$, these constants are the same for all points $x \in \mathbb{Z}_n^d$. From now onwards, we suppose that $x = 0$. Let us examine the dependence of the constants $c_{PS}$, $c_{PW}$ with respect to $r$. Let $f$ be a function defined on $\Lambda_n(x, r)$ with values in $\mathbb{R}$ and let us set

$$ \forall x \in \Lambda_{nr}(0, 1) \quad g(x) = f(rx). $$

We apply the inequality stated in Theorem A.10 to $g$:

$$ \left( \sum_{y \in \Lambda_{nr}(0,1)} \frac{1}{(nr)^d} (g(y) - \bar{g})^2 \right)^{\frac{1}{2}} \le c_{PW}(0,1,d) \left( \sum_{\substack{y,z \in \Lambda_{nr}(0,1) \\ |y-z|=1/(nr)}} \frac{1}{(nr)^{d-2}} (g(y) - g(z))^2 \right)^{\frac{1}{2}}, $$

and we rewrite everything in terms of the function $f$:

$$ \bar{g} = \frac{1}{(nr)^d} \sum_{y \in \Lambda_{nr}(0,1)} f(ry) = \frac{1}{(nr)^d} \sum_{y \in \Lambda_n(0,r)} f(y) = \bar{f}, $$

$$ \sum_{y \in \Lambda_{nr}(0,1)} (g(y) - \bar{g})^2 = \sum_{y \in \Lambda_n(0,r)} (f(y) - \bar{f})^2, $$

$$ \sum_{\substack{y, z \in \Lambda_{nr}(0,1) \\ |y-z|=1/(nr)}} (g(y) - g(z))^2 = \sum_{\substack{y, z \in \Lambda_n(0,r) \\ |y-z|=1/n}} (f(y) - f(z))^2. $$

We obtain the following inequality for the function $f$:

$$ \left( \sum_{y \in \Lambda_n(0, r)} \frac{1}{n^d} (f(y) - \bar{f})^2 \right)^{\frac{1}{2}} \le c_{PW}(0, 1, d) r \left( \sum_{\substack{y, z \in \Lambda_n(0, r) \\ |y-z|=1/n}} \frac{1}{n^{d-2}} (f(y) - f(z))^2 \right)^{\frac{1}{2}}. $$

We conclude that $c_{PW}(0, r, d) = c_{PW}(0, 1, d)r$.
## Appendix B. Inequalities on the original lattice

We shall adopt a slightly different viewpoint to apply these inequalities. Instead of rescaling the lattice by a factor $n$, we will consider functions defined on the lattice $\mathbb{Z}^d$ but on boxes of side $n$. We shall deduce the relevant inequalities from the previous ones by a simple change of variables $y \to ny$. More precisely, let $x \in \mathbb{Z}^d$, $n \ge 1$ and let $f$ be a function from $\Lambda(x, n) \cap \mathbb{Z}^d$ to $\mathbb{R}$. Let $\Lambda_n(x/n, 1)$ be the box

$$ \Lambda_n(x/n, 1) = \Lambda(x/n, 1) \cap \mathbb{Z}_n^d. $$

We define a function $f_n$ on $\Lambda_n(x/n, 1)$ by setting

$$ \forall y \in \Lambda_n(x/n, 1) \quad f_n(y) = f(ny). $$

We apply then the inequalities stated in Theorems A.9, A.10 to the function $f_n$ and we rewrite everything in terms of $f$. We first introduce some notation before stating the inequalities. Let $f$ be a function defined on a subset $D$ of $\mathbb{Z}^d$ with values in $\mathbb{R}$. For $p \ge 1$, we define its $p$-norm over $D$

$$ \|f\|_{p,D} = \left( \sum_{y \in D} |f(y)|^p \right)^{\frac{1}{p}} $$

and its Dirichlet energy over $D$

$$ \mathcal{E}(f, D) = \frac{1}{2d} \sum_{\substack{y, z \in D \\ |y-z|=1}} (f(y) - f(z))^2. $$

We recall that the exponent $2^*$ is equal to $2^* = 2d/(d-2)$. The Poincaré–Sobolev inequality stated in Theorem A.9 yields the following inequality in a box of side $n$.

**COROLLARY C.11.** — Let $x \in \mathbb{Z}^d$, $n \ge 1$ and let $f$ be a function from $\Lambda(x, n) \cap \mathbb{Z}^d$ to $\mathbb{R}$. There exists a constant $c_{PS}$ which depends on the dimension $d$ only such that

$$ \|f\|_{2^*, \Lambda(x, n)} \le c_{PS} \left( \frac{1}{n} \|f\|_{2, \Lambda(x, n)} + \sqrt{2d \mathcal{E}(f, \Lambda(x, n))} \right). $$

The Poincaré–Wirtinger inequality stated in Theorem A.10 yields the following inequality in a box of side $n$.

**COROLLARY C.12.** — Let $x \in \mathbb{Z}^d$, $n \ge 1$ and let $f$ be a function from $\Lambda(x, n) \cap \mathbb{Z}^d$ to $\mathbb{R}$. There exists a constant $c_{PW}$ which depends on the dimension $d$ only such that

$$ \|f - \bar{f}\|_{2, \Lambda(x, n)} \le n c_{PW} \sqrt{2d \mathcal{E}(f, \Lambda(x, n))}, $$

where

$$ \bar{f} = \frac{1}{n^d} \sum_{y \in \Lambda(x, n)} f(y). $$

Finally, if we send $n$ to $\infty$ in the inequality of Corollary C.11, we get the following result.
Appendix B. Inequalities on the original lattice

We shall adopt a slightly different viewpoint to apply these inequalities. Instead of rescaling the lattice by a factor $n$, we will consider functions defined on the lattice $\mathbb{Z}^d$ but on boxes of side $n$. We shall deduce the relevant inequalities from the previous ones by a simple change of variables $y \to ny$. More precisely, let $x \in \mathbb{Z}^d$, $n \ge 1$ and let $f$ be a function from $\Lambda(x, n) \cap \mathbb{Z}^d$ to $\mathbb{R}$. Let $\Lambda_n(x/n, 1)$ be the box

$$ \Lambda_n(x/n, 1) = \Lambda(x/n, 1) \cap \mathbb{Z}_n^d. $$

We define a function $f_n$ on $\Lambda_n(x/n, 1)$ by setting

$$ \forall y \in \Lambda_n(x/n, 1) \quad f_n(y) = f(ny). $$

We apply then the inequalities stated in Theorems A.9, A.10 to the function $f_n$ and we rewrite everything in terms of $f$. We first introduce some notation before stating the inequalities. Let $f$ be a function defined on a subset $D$ of $\mathbb{Z}^d$ with values in $\mathbb{R}$. For $p \ge 1$, we define its $p$-norm over $D$

$$ \|f\|_{p,D} = \left( \sum_{y \in D} |f(y)|^p \right)^{\frac{1}{p}} $$

and its Dirichlet energy over $D$

$$ \mathcal{E}(f, D) = \frac{1}{2d} \sum_{\substack{y, z \in D \\ |y-z|=1}} (f(y) - f(z))^2. $$

We recall that the exponent $2^*$ is equal to $2^* = 2d/(d-2)$. The Poincaré–Sobolev inequality stated in Theorem A.9 yields the following inequality in a box of side $n$.

**COROLLARY C.11.** — Let $x \in \mathbb{Z}^d$, $n \ge 1$ and let $f$ be a function from $\Lambda(x, n) \cap \mathbb{Z}^d$ to $\mathbb{R}$. There exists a constant $c_{PS}$ which depends on the dimension $d$ only such that

$$ \|f\|_{2^*, \Lambda(x, n)} \le c_{PS} \left( \frac{1}{n} \|f\|_{2, \Lambda(x, n)} + \sqrt{2d \mathcal{E}(f, \Lambda(x, n))} \right). $$

The Poincaré–Wirtinger inequality stated in Theorem A.10 yields the following inequality in a box of side $n$.

**COROLLARY C.12.** — Let $x \in \mathbb{Z}^d$, $n \ge 1$ and let $f$ be a function from $\Lambda(x, n) \cap \mathbb{Z}^d$ to $\mathbb{R}$. There exists a constant $c_{PW}$ which depends on the dimension $d$ only such that

$$ \|f - \bar{f}\|_{2, \Lambda(x, n)} \le n c_{PW} \sqrt{2d \mathcal{E}(f, \Lambda(x, n))}, $$

where

$$ \bar{f} = \frac{1}{n^d} \sum_{y \in \Lambda(x, n)} f(y). $$

Finally, if we send $n$ to $\infty$ in the inequality of Corollary C.11, we get the following result.
COROLLARY C.13. — Let $f$ be a function defined on $\mathbb{Z}^d$ with values in $\mathbb{R}$ having finite support. There exists a constant $c_{PS}$ which depends on the dimension $d$ only such that

$$ \|f\|_{2^*} \le c_{PS} \sqrt{2d\mathcal{E}(f)}. $$

BIBLIOGRAPHY

[Ant95] Peter Antal, *Enlargement of obstacles for the simple random walk*, Ann. Probab. **23** (1995), no. 3, 1061–1101. ↑7

[BCCHF15] Marianne Bessemoulin-Chatard, Claire Chainais-Hillairet, and Francis Filbet, *On discrete functional inequalities for some finite volume schemes*, IMA J. Numer. Anal. **35** (2015), no. 3, 1125–1149. ↑75

[BDPV15] Lorenzo Brasco, Guido De Philippis, and Bozhidar Velichkov, *Faber–Krahn inequalities in sharp quantitative form*, Duke Math. J. **164** (2015), no. 9, 1777–1831. ↑6, 51

[Bol94] Erwin Bolthausen, *Localization of a two-dimensional random walk with an attractive path interaction*, Ann. Probab. **22** (1994), no. 2, 875–918. ↑1, 3, 4, 6, 7, 8, 9, 21, 51, 52, 55, 57, 59, 61, 63, 69

[BP16] Marek Biskup and Eviatar B. Procaccia, *Shapes of drums with lowest base frequency under non-isotropic perimeter constraints*, https://arxiv.org/abs/1603.03871, 2016. ↑8

[BP18] ————, *Eigenvalue versus perimeter in a shape theorem for self-interacting random walks*, Ann. Appl. Probab. **28** (2018), no. 1, 340–377. ↑8

[BY13] Nathanael Berestycki and Ariel Yadin, *Condensation of random walks and the Wulff crystal*, https://arxiv.org/abs/1305.0139, to appear in *Annales de l'Institut Henri Poincaré (B): Probability and Statistics*, 2013. ↑8

[DFSX18] Jian Ding, Ryoki Fukushima, Rongfeng Sun, and Changji Xu, *Geometry of the random walk range conditioned on survival among Bernoulli obstacles*, https://arxiv.org/abs/1806.08319, 2018. ↑7, 8

[DV75] Monroe D. Donsker and S. R. Srinivasa Varadhan, *Asymptotic evaluation of certain Markov process expectations for large time. I. II*, Commun. Pure Appl. Math. **28** (1975), no. 1-2, 1–47; 279–301. ↑3, 5, 6

[DV79] ————, *On the number of distinct sites visited by a random walk*, Commun. Pure Appl. Math. **32** (1979), no. 6, 721–747. ↑24

[Ell06] Richard S. Ellis, *Entropy, large deviations, and statistical mechanics*, reprint of the 1985 original ed., *Classics in Mathematics*, Springer, 2006. ↑21

[Fan53] Ky Fan, *Minimax theorems*, Proc. Natl. Acad. Sci. USA **39** (1953), 42–47. ↑18

[FMP09] Nicola Fusco, Francesco Maggi, and Aldo Pratelli, *Stability estimates for certain Faber–Krahn, isocapacitary and Cheeger inequalities*, Ann. Sc. Norm. Super. Pisa, Cl. Sci. **8** (2009), no. 1, 51–71. ↑51

[LL10] Gregory F. Lawler and Vlada Limic, *Random walk: a modern introduction*, Cambridge Studies in Advanced Mathematics, vol. 123, Cambridge University Press, 2010. ↑69, 71

[MV16] Chiranjib Mukherjee and S. R. Srinivasa Varadhan, *Brownian occupation measures, compactness and large deviations*, Ann. Probab. **44** (2016), no. 6, 3934–3964. ↑8

[Pov99] Tobias Povel, *Confinement of Brownian motion among Poissonian obstacles in $\mathbb{R}^d$, $d \ge 3$*, Probab. Theory Relat. Fields **114** (1999), no. 2, 177–205. ↑7

[RW00] Chris G. Rogers and David Williams, *Diffusions, Markov processes and martingales: Vol. 2, Itô calculus*, 2nd ed., vol. 2, Cambridge University Press, 2000. ↑63
COROLLARY C.13. — Let $f$ be a function defined on $\mathbb{Z}^d$ with values in $\mathbb{R}$ having finite support. There exists a constant $c_{PS}$ which depends on the dimension $d$ only such that

$$ \|f\|_{2^*} \le c_{PS} \sqrt{2d\mathcal{E}(f)}. $$

BIBLIOGRAPHY

[Ant95] Peter Antal, *Enlargement of obstacles for the simple random walk*, Ann. Probab. **23** (1995), no. 3, 1061–1101. ↑7

[BCCHF15] Marianne Bessemoulin-Chatard, Claire Chainais-Hillairet, and Francis Filbet, *On discrete functional inequalities for some finite volume schemes*, IMA J. Numer. Anal. **35** (2015), no. 3, 1125–1149. ↑75

[BDPV15] Lorenzo Brasco, Guido De Philippis, and Bozhidar Velichkov, *Faber–Krahn inequalities in sharp quantitative form*, Duke Math. J. **164** (2015), no. 9, 1777–1831. ↑6, 51

[Bol94] Erwin Bolthausen, *Localization of a two-dimensional random walk with an attractive path interaction*, Ann. Probab. **22** (1994), no. 2, 875–918. ↑1, 3, 4, 6, 7, 8, 9, 21, 51, 52, 55, 57, 59, 61, 63, 69

[BP16] Marek Biskup and Eviatar B. Procaccia, *Shapes of drums with lowest base frequency under non-isotropic perimeter constraints*, https://arxiv.org/abs/1603.03871, 2016. ↑8

[BP18] ————, *Eigenvalue versus perimeter in a shape theorem for self-interacting random walks*, Ann. Appl. Probab. **28** (2018), no. 1, 340–377. ↑8

[BY13] Nathanael Berestycki and Ariel Yadin, *Condensation of random walks and the Wulff crystal*, https://arxiv.org/abs/1305.0139, to appear in *Annales de l'Institut Henri Poincaré (B): Probability and Statistics*, 2013. ↑8

[DFSX18] Jian Ding, Ryoki Fukushima, Rongfeng Sun, and Changji Xu, *Geometry of the random walk range conditioned on survival among Bernoulli obstacles*, https://arxiv.org/abs/1806.08319, 2018. ↑7, 8

[DV75] Monroe D. Donsker and S. R. Srinivasa Varadhan, *Asymptotic evaluation of certain Markov process expectations for large time. I. II*, Commun. Pure Appl. Math. **28** (1975), no. 1-2, 1–47; 279–301. ↑3, 5, 6

[DV79] ————, *On the number of distinct sites visited by a random walk*, Commun. Pure Appl. Math. **32** (1979), no. 6, 721–747. ↑24

[Ell06] Richard S. Ellis, *Entropy, large deviations, and statistical mechanics*, reprint of the 1985 original ed., *Classics in Mathematics*, Springer, 2006. ↑21

[Fan53] Ky Fan, *Minimax theorems*, Proc. Natl. Acad. Sci. USA **39** (1953), 42–47. ↑18

[FMP09] Nicola Fusco, Francesco Maggi, and Aldo Pratelli, *Stability estimates for certain Faber–Krahn, isocapacitary and Cheeger inequalities*, Ann. Sc. Norm. Super. Pisa, Cl. Sci. **8** (2009), no. 1, 51–71. ↑51

[LL10] Gregory F. Lawler and Vlada Limic, *Random walk: a modern introduction*, Cambridge Studies in Advanced Mathematics, vol. 123, Cambridge University Press, 2010. ↑69, 71

[MV16] Chiranjib Mukherjee and S. R. Srinivasa Varadhan, *Brownian occupation measures, compactness and large deviations*, Ann. Probab. **44** (2016), no. 6, 3934–3964. ↑8

[Pov99] Tobias Povel, *Confinement of Brownian motion among Poissonian obstacles in $\mathbb{R}^d$, $d \ge 3$*, Probab. Theory Relat. Fields **114** (1999), no. 2, 177–205. ↑7

[RW00] Chris G. Rogers and David Williams, *Diffusions, Markov processes and martingales: Vol. 2, Itô calculus*, 2nd ed., vol. 2, Cambridge University Press, 2000. ↑63
[Szn91] Alain-Sol Sznitman, *On the confinement property of two-dimensional Brownian motion among Poissonian obstacles*, Commun. Pure Appl. Math. **44** (1991), no. 8-9, 1137–1170.
↑6, 7

Manuscript received on 11th January 2019,
revised on 10th December 2019,
accepted on 25th February 2020.

Recommended by Editor H. Lacoin.

Published under license CC BY 4.0.

<figure><img src="image_10.png" /></figure>

This journal is a member of Centre Mersenne.

<figure><img src="image_11.png" /></figure>

Nathanaël BERESTYCKI
Universität Wien (Austria)
On leave from the
University of Cambridge, (UK)
nathanael.berestycki@univie.ac.at

Raphaël CERF
Ecole Normale Supérieure, PSL University,
CNRS, DMA, 75005, Paris, (France)
Université Paris-Saclay, CNRS,
Laboratoire de mathématiques d'Orsay,
91405, Orsay, (France)
Raphael.Cerf@ens.fr
[Szn91] Alain-Sol Sznitman, *On the confinement property of two-dimensional Brownian motion among Poissonian obstacles*, Commun. Pure Appl. Math. **44** (1991), no. 8-9, 1137–1170.
↑6, 7

Manuscript received on 11th January 2019,
revised on 10th December 2019,
accepted on 25th February 2020.

Recommended by Editor H. Lacoin.

Published under license CC BY 4.0.

<figure><img src="image_12.png" /></figure>

This journal is a member of Centre Mersenne.

<figure><img src="image_13.png" /></figure>

Nathanaël BERESTYCKI
Universität Wien (Austria)
On leave from the
University of Cambridge, (UK)
nathanael.berestycki@univie.ac.at

Raphaël CERF
Ecole Normale Supérieure, PSL University,
CNRS, DMA, 75005, Paris, (France)
Université Paris-Saclay, CNRS,
Laboratoire de mathématiques d'Orsay,
91405, Orsay, (France)
Raphael.Cerf@ens.fr