2. Lower bound for the quenched Lyapunov exponent

The aim of this section is to prove part (1) of Theorem 1.1. The key tool here is Russo’s formula for the independent Bernoulli site percolation on $\mathbb{Z}^d$ . Roughly speaking, by applying Russo’s formula to some events with respect to the quenched travel cost, we derive a lower bound for the differences between quenched Lyapunov exponents. However, Russo’s formula is not directly applicable because the quenched travel cost depends on the states of infinitely many sites. To overcome this problem, we first introduce a modification of the quenched travel cost depending only on the states of finitely many sites.

Let V be a subset of $\mathbb{R}^d$ and let T(V) denote the exit time of the simple random walk from V, that is,

(2.1) \begin{align}T(V)\;:\!=\; \inf\{ k \geq 0\;:\; S_k \not\in V \}.\end{align}

Then, for any $x,y \in \mathbb{Z}^d$ and $N \in \mathbb{N}$ , we define the quenched travel cost $a_N(x,y,\omega)$ from x to y restricted to the simple random walk before exiting $[\!-\!N,N]^d$ as follows:

\begin{align*}a_N(x,y,\omega)\;:\!=\; -\log e_N(x,y,\omega),\end{align*}

where

\begin{align*}e_N(x,y,\omega)\;:\!=\; E^x\Biggl[ \exp\!\Biggl\{ -\sum_{k=0}^{H(y)-1}\omega(S_k) \Biggr\} \mathbf{1}_{\{ H(y)<T([-N,N]^d) \}} \Biggr]\end{align*}

if $x \not= y$ , and $e_N(x,y,\omega)\;:\!=\; 1$ otherwise. Note that $a_N(x,y,\omega)$ depends only on the states of finitely many sites, and the monotone convergence theorem shows that for each $0 \leq r \leq 1$ ,

(2.2) \begin{align}\lim_{N \to \infty} \mathbb{E}_r[a_N(x,y,\omega)]= \mathbb{E}_r[a(x,y,\omega)].\end{align}

We need some preparation before applying Russo’s formula to the restricted quenched travel cost. For any $y \in \mathbb{Z}^d$ , the path measure $\widetilde{P}_{N,\omega}^{0,y}$ is defined by

(2.3) \begin{align}\dfrac{{\text{d}}\widetilde{P}_{N,\omega}^{0,y}}{{\text{d}} P^0}= e_N(0,y,\omega)^{-1} \exp\!\Biggl\{ -\sum_{k=0}^{H(y)-1}\omega(S_k) \Biggr\}\mathbf{1}_{\{ H(y)<T([-N,N]^d) \}},\end{align}

and let $\widetilde{E}_{N,\omega}^{0,y}$ denote the expectation with respect to $\widetilde{P}_{N,\omega}^{0,y}$ . Next, for any $z \in \mathbb{Z}^d$ and $m \in \mathbb{N}$ , let $\ell_z(m)$ be the number of visits to z by the simple random walk up to time $m-1$ , that is,

\begin{align*}\ell_z(m)\;:\!=\; \#\{ 0 \leq k<m\;:\; S_k=z \}.\end{align*}

The following proposition is a consequence of Russo’s formula for the restricted quenched travel cost, which is also useful to derive an upper bound for differences between quenched Lyapunov exponents.

Lemma 2.1. For all $0<r<1$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ , we have

\begin{align*}-\dfrac{{\text{d}}}{{\text{d}} r}\mathbb{E}_r[a_N(0,y,\omega)]&= \sum_{z \in \mathbb{Z}^d \cap [-N,N]^d} \bigl\{ \mathbb{E}_r \bigl[ \mathbf{1}_{\{ \omega(z)=1 \}} \log \widetilde{E}_{N,\omega}^{0,y}\bigl[{\text{e}}^{\ell_z(H(y))}\bigr] \bigr]\\&\quad +\mathbb{E}_r\bigl[ \mathbf{1}_{\{ \omega(z)=0 \}}\bigl( -\log \widetilde{E}_{N,\omega}^{0,y}\bigl[{\text{e}}^{-\ell_z(H(y))}\bigr] \bigr) \bigr] \bigr\}.\end{align*}

Proof. Fix $y \in \mathbb{Z}^d \setminus \{0\}$ and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ . Since the random variable $a_N(0,y,\omega)$ depends only on the states of sites in $[-N,N]^d$ , Russo’s formula (see e.g. [Reference Grimmett7, Theorem 2.32]) gives that for any $0<r<1$ ,

(2.4) \begin{align}-\dfrac{{\text{d}}}{{\text{d}} r}\mathbb{E}_r[a_N(0,y,\omega)]= \sum_{z \in \mathbb{Z}^d \cap [-N,N]^d} \mathbb{E}_r\bigl[ a_N\bigl(0,y,\omega_z^1\bigr)-a_N\bigl(0,y,\omega_z^0\bigr) \bigr],\end{align}

where

\begin{align*}\omega_z^1(x)\;:\!=\;\begin{cases}1 & \text{if $ x=z$,}\\\omega(x) & \text{if $ x \not=z$,}\end{cases}\quad\omega_z^0(x)\;:\!=\;\begin{cases}0 & \text{if $ x=z$,}\\\omega(x) & \text{if $ x \not=z$.}\end{cases}\end{align*}

Note that if $\omega(z)=1$ holds, then

\begin{align*}a_N\bigl(0,y,\omega_z^1\bigr)-a_N\bigl(0,y,\omega_z^0\bigr)= \log \widetilde{E}_{N,\omega}^{0,y}\bigl[ {\text{e}}^{\ell_z(H(y))} \bigr].\end{align*}

Furthermore, in the case where $\omega(z)=0$ ,

\begin{align*}a_N\bigl(0,y,\omega_z^1\bigr)-a_N\bigl(0,y,\omega_z^0\bigr)= -\log \widetilde{E}_{N,\omega}^{0,y}\bigl[ {\text{e}}^{-\ell_z(H(y))} \bigr].\end{align*}

This implies that the right-hand side of (2.4) is equal to

\begin{align*}&\sum_{z \in \mathbb{Z}^d \cap [-N,N]^d}\bigl\{ \mathbb{E}_r \bigl[ \mathbf{1}_{\{ \omega(z)=1 \}} \log \widetilde{E}_{N,\omega}^{0,y}\bigl[{\text{e}}^{\ell_z(H(y))}\bigr] \bigr] +\mathbb{E}_r\bigl[ \mathbf{1}_{\{ \omega(z)=0 \}}\bigl( -\log \widetilde{E}_{N,\omega}^{0,y}\bigl[{\text{e}}^{-\ell_z(H(y))}\bigr] \bigr) \bigr] \bigr\},\end{align*}

and the proposition follows.

We are now in a position to show part (1) of Theorem 1.1.

Proof of part (1) of Theorem 1.1. Let $y \in \mathbb{Z}^d \setminus \{0\}$ and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ . Jensen’s inequality and the fact that $-\log(1-t) \geq t$ holds for $0 \leq t<1$ imply that for any $z \in \mathbb{Z}^d \cap [-N,N]^d$ ,

\begin{align*}\log\widetilde{E}_{N,\omega}^{0,y}\bigl[{\text{e}}^{\ell_z(H(y))}\bigr]&\geq -\log\widetilde{E}_{N,\omega}^{0,y}\bigl[{\text{e}}^{-\ell_z(H(y))}\bigr]\\&\geq -\log\bigl\{ 1-(1-{\text{e}}^{-1})\widetilde{P}_{N,\omega}^{0,y}(\ell_z(H(y)) \geq 1) \bigr\}\\&\geq (1-{\text{e}}^{-1})\widetilde{P}_{N,\omega}^{0,y}(\ell_z(H(y)) \geq 1).\end{align*}

Hence Lemma 2.1 yields that for all $0<r<1$ ,

\begin{align*}-\dfrac{{\text{d}}}{{\text{d}} r}\mathbb{E}_r[a_N(0,y,\omega)]&\geq\sum_{z \in \mathbb{Z}^d \cap [-N,N]^d} \mathbb{E}_r \bigl[ (1-{\text{e}}^{-1})\widetilde{P}_{N,\omega}^{0,y}(\ell_z(H(y)) \geq 1) \bigr]\\&= (1-{\text{e}}^{-1})\, \mathbb{E}_r\Biggl[\widetilde{E}_{N,\omega}^{0,y}\Biggl[ \sum_{z \in \mathbb{Z}^d \cap [-N,N]^d} \mathbf{1}_{\{ \ell_z(H(y)) \geq 1 \}} \Biggr] \Biggr]\\&\geq (1-{\text{e}}^{-1})\|y\|_1.\end{align*}

Here the last inequality follows from the fact that $\widetilde{P}_{N,\omega}^{0,y} \textrm{-} \textrm{a.s.}$ , the simple random walk must visit at least $\|y\|_1$ sites before hitting y. Thus, for all $0<r<1$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

\begin{align*}\dfrac{{\text{d}}}{{\text{d}} r}\bigl\{ \mathbb{E}_r[a_N(0,y,\omega)]+(1-{\text{e}}^{-1})r\|y\|_1 \bigr\} \leq 0,\end{align*}

which implies that the function in the above braces is decreasing in r. Therefore, for all $0<p<q<1$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

\begin{align*}\mathbb{E}_p[a_N(0,y,\omega)]-\mathbb{E}_q[a_N(0,y,\omega)]\geq (1-{\text{e}}^{-1})(q-p)\|y\|_1.\end{align*}

It follows by (2.2) and the definition of the quenched Lyapunov exponent that for all $0<p<q<1$ and $x \in \mathbb{Z}^d \setminus \{0\}$ ,

\begin{align*}\dfrac{\alpha_p(x)-\alpha_q(x)}{\|x\|_1} \geq (1-{\text{e}}^{-1})(q-p).\end{align*}

Note that the right-hand side above does not depend on x, and $\alpha_p(\!\cdot\!)$ and $\alpha_q(\!\cdot\!)$ are norms on $\mathbb{R}^d$ . Consequently, the above inequality can be easily extended to all $x \in \mathbb{R}^d \setminus \{ 0 \}$ , and the proof is complete.

3. Upper bound for the quenched Lyapunov exponent

In this section, for a fixed $q_0 \in (0,1)$ , we prove part (2) of Theorem 1.1. The proof is not so long once a proposition is established (see Proposition 3.1 below).

Proof of part (2) of Theorem 1.1. Considering the proof of part (1) of Theorem 1.1, our task is to show that there exists a constant c (which depends only on d and $q_0$ ) such that for all $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

(3.1) \begin{align}-\dfrac{{\text{d}}}{{\text{d}} r}\mathbb{E}_r[a_N(0,y,\omega)] \leq c\|y\|_1.\end{align}

To this end, note that for each $z \in \mathbb{Z}^d$ and the potential $\omega$ ,

\begin{align*}\dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)}=\begin{cases}\widetilde{E}_{N,\omega}^{0,y}\bigl[{\text{e}}^{\ell_z(H(y))}\bigr] & \text{if $ \omega(z)=1$,}\\[5pt] \widetilde{E}_{N,\omega}^{0,y}\bigl[{\text{e}}^{-\ell_z(H(y))}\bigr] & \text{if $ \omega(z)=0$,}\end{cases}\end{align*}

where $\omega_z$ is the potential which agrees with $\omega$ on all sites except for z, that is, for $x \in \mathbb{Z}^d$ ,

\begin{align*}\omega_z(x)\;:\!=\;\begin{cases}1-\omega(z) & \text{if $ x=z$,}\\\omega(x) & \text{if $ x \not=z$.}\end{cases}\end{align*}

Combining this and Lemma 2.1, we have

(3.2) \begin{align}-\dfrac{{\text{d}}}{{\text{d}} r}\mathbb{E}_r[a_N(0,y,\omega)]\leq \sum_{z \in \mathbb{Z}^d \cap [-N,N]^d}\mathbb{E}_r \biggl[ \biggl| \log\dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)} \biggr| \biggr],\end{align}

and the following proposition is the key to estimating the above sum.

Proposition 3.1. There exists a constant $C_3$ (which depends only on d and $q_0$ ) such that for all $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

\begin{align*}\sum_{z \in \mathbb{Z}^d \cap [-N,N]^d}\mathbb{E}_r \left[ \left| \log\dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)} \right| \right]\leq C_3\,\mathbb{E}_r\bigl[ \widetilde{E}_{N,\omega}^{0,y}[\#\mathcal{A}(0,y)] \bigr],\end{align*}

where $\mathcal{A}(0,y)\;:\!=\; \{ S_k\;:\; 0 \leq k<H(y) \}$ .

Since the proof of the above proposition is fairly long, for now let us complete the proof of part (2) of Theorem 1.1. The same argument as in the proof of [Reference Zerner18, Lemma 3] implies that for any $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

(3.3) \begin{align}\mathbb{E}_r\bigl[ \widetilde{E}_{N,\omega}^{0,y}[\#\mathcal{A}(0,y)] \bigr]\leq \dfrac{1+\log(2d)}{-\log\{ {\text{e}}^{-1}+(1-{\text{e}}^{-1})q_0 \}}\|y\|_1.\end{align}

From (3.2), (3.3), and Proposition 3.1, we conclude that for all $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

\begin{align*}-\dfrac{{\text{d}}}{{\text{d}} r}\mathbb{E}_r[a_N(0,y,\omega)]\leq \dfrac{C_3(1+\log(2d))}{-\log\{ {\text{e}}^{-1}+(1-{\text{e}}^{-1})q_0 \}}\|y\|_1,\end{align*}

and (3.1) follows.

It remains to prove Proposition 3.1. We divide the proof into two cases: Sections 3.1 and 3.2, respectively, treat the high-dimensional case ( $d \geq 3$ ) and the low-dimensional case ( $d=1,2$ ).

3.1. High-dimensional case

This subsection is devoted to the proof of Proposition 3.1 for $d \geq 3$ . First of all, we state the following lemma, which is useful to show Proposition 3.1 not only for $d \geq 3$ but also for $d=1,2$ .

Lemma 3.1. Let $d \geq 1$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ . Moreover, for any $z \in \mathbb{Z}^d$ , define

\begin{align*}\psi_d(z,\omega)\;:\!=\; \Biggl( 1-E^z\Biggl[ \exp\!\Biggl\{ -\sum_{k=1}^{H^+(z)-1}\omega(S_k) \Biggr\} \mathbf{1}_{\{ H^+(z)<\infty \}}\Biggr] \Biggr)^{-1},\end{align*}

where $H^+(z)\;:\!=\; \inf\{ k>0\;:\; S_k=z \}$ . Then, for all $z \in \mathbb{Z}^d \cap [-N,N]^d$ ,

(3.4) \begin{align}\bigg| \log\dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)} \bigg|\leq 4\psi_d(z,\omega)\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y)).\end{align}

Proof. We first treat the case where $z \in \mathbb{Z}^d \cap [\!-\!N,N]^d$ satisfies $\omega(z)=1$ . The definition of $\omega_z$ and the strong Markov property show that

\begin{align*}1 \leq \dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)}\leq 1+\dfrac{e_N(z,y,\omega_z)}{e_N(z,y,\omega)}\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y)).\end{align*}

We use the strong Markov property again to obtain that

\begin{align*} e_N(z,y,\omega_z)& \leq \psi_d(z,\omega)E^z\Biggl[ \exp\!\Biggl\{ -\sum_{k=0}^{H(y)-1}\omega_z(S_k) \Biggr\}\mathbf{1}_{\{ H(y) \leq H^+(z),\,H(y)<T([-N,N]^d) \}} \Biggr]\\& \leq e\psi_d(z,\omega) e_N(z,y,\omega).\end{align*}

With these observations,

\begin{align*}0 \leq \log\dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)}\leq \log\bigl\{ 1+e\psi_d(z,\omega)\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y)) \bigr\}.\end{align*}

Since $t \geq \log(1+t)$ holds for $t \geq 0$ , the right-hand side above is not greater than

\[e\psi_d(z)\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y)),\]

and (3.4) follows.

We next consider the case where $z \in \mathbb{Z}^d \cap [-N,N]^d$ satisfies $\omega(z)=0$ . Then the same argument as in the proof of [Reference Zerner18, Lemma 12] shows that

(3.5) \begin{align}0&\leq -\log\dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)}\notag \\&= a_N(0,y,\omega_z)-a_N(0,y,\omega)\notag \\&\leq \min\bigl\{ -\log\widetilde{P}_{N,\omega}^{0,y}(H(y) \leq H(z)),1+\log\psi_d(z,\omega) \bigr\}.\end{align}

Note that the fact that $-\log t \leq 2(1-t)$ holds for $1/2 \leq t \leq 1$ implies that on the event $\bigl\{ \widetilde{P}_{N,\omega}^{0,y}(H(y) \leq H(z)) \geq 1/2 \bigr\}$ ,

\begin{align*}-\log\widetilde{P}_{N,\omega}^{0,y}(H(y) \leq H(z))\leq 2\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y)).\end{align*}

This, together with $\psi_d(z,\omega) \geq 1$ , shows that the right-hand side of (3.5) is smaller than or equal to

\begin{align*}&2\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y))\mathbf{1}_{\{ \widetilde{P}_{N,\omega}^{0,y}(H(y) \leq H(z)) \geq 1/2 \}}\\&\quad +2(1+\log\psi_d(z,\omega))\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y))\mathbf{1}_{\{ \widetilde{P}_{N,\omega}^{0,y}(H(y) \leq H(z))<1/2 \}}\\&\qquad \leq 2(\psi_d(z,\omega)+\log\psi_d(z,\omega))\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y)).\end{align*}

Therefore (3.4) immediately follows from the fact that $\log t \leq t$ for $t \geq 1$ .

We are now in a position to prove Proposition 3.1 for $d \geq 3$ .

Proof of Proposition 3.1 for ${d \geq 3}$ . Since the simple random walk is transient for $d \geq 3$ , we have $P^0(H^+(0)=\infty)>0$ . Hence, for all $z \in \mathbb{Z}^d$ ,

\begin{align*} \psi_d(z,\omega) \leq P^0(H^+(0)=\infty)^{-1}<\infty.\end{align*}

This together with Lemma 3.1 yields that for all $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

\begin{align*}\sum_{z \in \mathbb{Z}^d \cap [-N,N]^d} \mathbb{E}_r \biggl[ \biggl| \log\dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)} \biggr| \biggr]&\leq \sum_{z \in \mathbb{Z}^d \cap [-N,N]^d}\mathbb{E}_r \bigl[ 4\psi_d(z,\omega)\widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y)) \bigr]\\&\leq 4P^0(H^+(0)=\infty)^{-1} \mathbb{E}_r \bigl[ \widetilde{E}_{N,\omega}^{0,y}[\#\mathcal{A}(0,y)] \bigr].\end{align*}

Thus the proof is complete by taking $C_3\;:\!=\; 4P^0(H^+(0)=\infty)^{-1}$ .

3.2. Low-dimensional case

The aim of this section is to prove Proposition 3.1 for $d=1,2$ . In this case, the strategy taken in the previous subsection does not work because the simple random walk is recurrent. Hence we have to estimate $\psi_d(z,\omega)$ in another way.

Let $p_c=p_c(d) \in (0,1)$ be the critical probability for independent Bernoulli site percolation on $\mathbb{Z}^d$ , and fix $R \in 2\mathbb{N}$ such that

(3.6) \begin{align}q_0^{R^d}+q_0^{R^d-1}(1-q_0)R^d<p_c .\end{align}

(This is possible since $0<q_0<1$ .) We now consider the boxes $\Lambda(v)\;:\!=\; Rv+[-R/2,R/2)^d$ , $v \in \mathbb{Z}^d$ . These boxes form a partition of $\mathbb{R}^d$ , and we let $\overline{z}$ denote the (unique) index v such that $z \in \Lambda(v)$ . Furthermore, a site v is said to be open if $\Lambda(v)$ has at most one site x such that $\omega(x)=1$ , and closed otherwise. Note that under $\mathbb{P}_r$ , the family $(\mathbf{1}_{\{ v \text{ is open} \}})_{v \in \mathbb{Z}^d}$ is the independent Bernoulli site percolation on $\mathbb{Z}^d$ with parameter $r^{R^d}+r^{R^d-1}(1-r)R^d$ . This site percolation induces open clusters, which are connected components of open sites. In particular, for each $v \in \mathbb{Z}^d$ , we let $\mathcal{C}_v$ denote the open cluster containing v.

Although $\psi_d(z,\omega)$ is bounded for $d \geq 3$ , the following lemma says that for $d=1,2$ , $\psi_d(z,\omega)$ is dominated by the size of the open cluster containing $\overline{z}$ .

Lemma 3.2. Let $d=1,2$ . Then there exists a constant $C_4$ (which depends only on d and $q_0$ ) such that for all $z \in \mathbb{Z}^d$ ,

\begin{align*}\psi_d(z,\omega) \leq C_4 (1+\#\mathcal{C}_{\overline{z}}).\end{align*}

Proof. Since the simple random walk is recurrent for $d=1,2$ , we have for any $z \in \mathbb{Z}^d$ ,

\begin{align*}\psi_d(z,\omega)= \Biggl( 1-E^z\Biggl[ \exp\!\Biggl\{ -\sum_{k=1}^{H^+(z)-1}\omega(S_k) \Biggr\} \Biggr] \Biggr)^{-1}.\end{align*}

Let us first treat the case where $\overline{z}$ is closed. Then the box $\Lambda(\overline{z})$ contains at least two sites x with $\omega(x)=1$ . Hence we can find a site $x_0 \in \Lambda(\overline{z})$ such that $x_0 \not= z$ and $\omega(x_0)=1$ . It follows that

\begin{align*}1-E^z\Biggl[ \exp\!\Biggl\{ -\sum_{k=1}^{H^+(z)-1}\omega(S_k) \Biggr\} \Biggr]&\geq 1-P^z(H^+(z) \leq H(x_0))-{\text{e}}^{-1}P^z(H^+(z)>H(x_0))\\& = (1-{\text{e}}^{-1})P^z(H^+(z)>H(x_0))\\&\geq (1-{\text{e}}^{-1})\biggl( \dfrac{1}{2d} \biggr)^{{dR}}.\end{align*}

Thus, since $\#\mathcal{C}_{\overline{z}}=0$ in the case where $\overline{z}$ is closed, we have

\begin{align*}\psi_d(z,\omega) \leq \dfrac{(2d)^{dR}}{1-{\text{e}}^{-1}}= \dfrac{(2d)^{dR}}{1-{\text{e}}^{-1}}(1+\#\mathcal{C}_{\overline{z}}).\end{align*}

We next treat the case where $\overline{z}$ is open. Then the same argument as above does not work since the box $\Lambda(\overline{z})$ may not contain any site x with $x \not= z$ and $\omega(x)=1$ (this situation actually occurs when $\omega(z)=1$ and $\omega(x)=0$ for $x \in \Lambda(\overline{z}) \setminus \{z\}$ ). To overcome this problem, let us introduce the region $\mathcal{O}_z\;:\!=\; \bigcup_{v \in \mathcal{C}_{\overline{z}}} \Lambda(v)$ and the stopping time

\begin{align*}\sigma\;:\!=\; \inf\{ k \geq T(\mathcal{O}_z)\;:\; \omega(S_k)=1 \},\end{align*}

where $T(\mathcal{O}_z)$ is the exit time of the simple random walk from $\mathcal{O}_z$ (see (2.1)). The same computation as in the first case gives

(3.7) \begin{align}1-E^z\Biggl[ \exp\!\Biggl\{ -\sum_{k=1}^{H^+(z)-1}\omega(S_k) \Biggr\} \Biggr]\geq (1-{\text{e}}^{-1})P^z(\sigma<H^+(z)).\end{align}

Noting that the site $v\;:\!=\; \overline{S_{T(\mathcal{O}_z)}}$ is closed and $z \not\in \Lambda(v)$ , the event $\{ \sigma<H^+(z) \}$ occurs if the simple random walk follows a shortest path from v to a site $x \in \Lambda(v)$ with $\omega(x)=1$ after exiting $\mathcal{O}_z$ . Hence the strong Markov property shows that

(3.8) \begin{align}P^z(\sigma<H^+(z))\geq \biggl( \dfrac{1}{2d} \biggr)^{dR}P^z(T(\mathcal{O}_z)<H^+(z)).\end{align}

For the last probability, we use the following estimate on the probability that the simple random walk exits an $\ell^1$ -ball before revisiting its starting point (see e.g. [Reference Lawler11, (1.20), (1.38), and Theorem 1.6.6]): There exists a constant $c \geq 1$ (which depends only on d) such that

(3.9) \begin{align}P^0(\tau<H^+(0))^{-1} \leq c \times\begin{cases}\#\mathcal{O}_z & \text{if $ d=1$,}\\\log(\#\mathcal{O}_z) & \text{if $ d=2$,}\end{cases}\end{align}

where $\tau$ is the exit time for the simple random walk from the $\ell^1$ -ball of radius $\#\mathcal{O}_z$ and center 0, that is,

\begin{align*}\tau\;:\!=\; \inf\{ k \geq 0\;:\; \|S_k\|_1>\#\mathcal{O}_z \}.\end{align*}

Noting that $P^0(\tau<H^+(0)) \leq P^z(T(\mathcal{O}_z)<H^+(z))$ and $\#\mathcal{O}_z=R^d(\#\mathcal{C}_{\overline{z}})$ , we have from (3.7), (3.8), and (3.9),

\begin{align*}\psi_d(z,\omega)\leq \dfrac{c(2d)^{dR}}{1-{\text{e}}^{-1}}(\#\mathcal{O}_z)\leq \dfrac{c(2d)^{dR}R^d}{1-{\text{e}}^{-1}} (1+\#\mathcal{C}_{\overline{z}}).\end{align*}

Consequently, the lemma follows by taking $C_{4}\;:\!=\; c(2d)^{dR}R^d/(1-{\text{e}}^{-1})$ .

We now turn to the proof of Proposition 3.1 for $d=1,2$ .

Proof of Proposition 3.1 for ${d=1,2}$ . Lemmata 3.1 and 3.2 imply that for all $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

\begin{align*}&\sum_{z \in \mathbb{Z}^d \cap [-N,N]^d} \mathbb{E}_r \biggl[ \biggl| \log\dfrac{e_N(0,y,\omega_z)}{e_N(0,y,\omega)} \biggr| \biggr]\\&\quad \leq 4C_{4} \sum_{z \in \mathbb{Z}^d \cap [-N,N]^d} \mathbb{E}_r\bigl[ (1+\#\mathcal{C}_{\overline{z}}) \widetilde{P}_{N,\omega}^{0,y}(H(z)<H(y)) \bigr]\\&\quad = 4C_{4} \Biggl( \mathbb{E}_r\bigl[ \widetilde{E}_{N,\omega}^{0,y}[\#\mathcal{A}(0,y)] \bigr]+\mathbb{E}_r\Biggl[ \widetilde{E}_{N,\omega}^{0,y} \Biggl[ \sum_{z \in \mathcal{A}(0,y)} \#\mathcal{C}_{\overline{z}} \Biggr] \Biggr] \Biggr).\end{align*}

Hence it suffices to prove that there exists a constant $C_5$ (which depends only on d and $q_0$ ) such that for all $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

(3.10) \begin{align}\mathbb{E}_r\Biggl[ \widetilde{E}_{N,\omega}^{0,y} \Biggl[ \sum_{z \in \mathcal{A}(0,y)} \#\mathcal{C}_{\overline{z}} \Biggr] \Biggr]\leq C_{5}\mathbb{E}_r\bigl[ \widetilde{E}_{N,\omega}^{0,y}[\#\mathcal{A}(0,y)] \bigr].\end{align}

To show (3.10), we state some results for lattice animals on $\mathbb{Z}^d$ (which are finite connected subsets of $\mathbb{Z}^d$ ). For $n \geq 1$ , let $\mathbb{A}_n$ denote the set of all lattice animals, of size n, containing 0. Moreover, let $(\widetilde{\Gamma}_v)_{v \in \mathbb{Z}^d}$ be i.i.d. random lattice animals with the common law $\mathbb{P}_{q_0}(\mathcal{C}_0 \in \cdot)$ (we write P for the probability measure governing $(\widetilde{\Gamma}_v)_{v \in \mathbb{Z}^d}$ ). Then, due to (3.6), the following lemma is an immediate consequence of [Reference Fontes and Newman6, (2.6) and (2.7)] and [Reference Scholler14, page 183].

Lemma 3.3. The following results hold.

1. (1) For all $n \geq 1$ and $t \geq 0$ , we have

   \begin{align*}\mathbb{P}_{q_0}\Biggl( \sup_{\Gamma \in \mathbb{A}_n} \dfrac{1}{\#\Gamma}\sum_{v \in \Gamma} \#\mathcal{C}_v \geq t \Biggr)\leq P\Biggl( \sup_{\Gamma \in \bigcup_{m=n}^\infty \mathbb{A}_m} \dfrac{1}{\#\Gamma} \sum_{v \in \Gamma} (\#\widetilde{\Gamma}_v)^2 \geq \dfrac{t}{2} \Biggr).\end{align*}

2. (2) There exist constants $C_6$ and $C_7$ (which depend only on d and $q_0$ ) such that for all $n \geq 1$ ,

   \begin{align*}P\Biggl( \sup_{\Gamma \in \mathbb{A}_n} \dfrac{1}{\#\Gamma} \sum_{v \in \Gamma}(\#\widetilde{\Gamma}_v)^2 \geq C_{6} \Biggr)\leq C_{6}\exp\!\{-C_{9}n^{1/5}\}.\end{align*}

For any lattice animal $\Gamma$ on $\mathbb{Z}^d$ , define

\begin{align*}\mathcal{E}(\Gamma)\;:\!=\; \Biggl\{ \dfrac{1}{\#\Gamma} \sum_{v \in \Gamma} \#\mathcal{C}_v>2C_{6} \Biggr\}.\end{align*}

Moreover, for each $y \in \mathbb{Z}^d \setminus \{0\}$ , set

\begin{align*}\overline{\mathcal{A}}(0,y)\;:\!=\; \{ \overline{z}\;:\; z \in \mathcal{A}(0,y) \},\end{align*}

which is a lattice animal on $\mathbb{Z}^d$ containing 0. Then, for all $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ , the left-hand side of (3.10) is smaller than or equal to

\begin{align*}&R^d\mathbb{E}_r\Biggl[ \widetilde{E}_{N,\omega}^{0,y} \Biggl[ \sum_{v \in \overline{\mathcal{A}}(0,y)} \#\mathcal{C}_v \Biggr] \Biggr]\\&\quad \leq R^d \Biggl\{ 2C_{6}\mathbb{E}_r\bigl[ \widetilde{E}_{N,\omega}^{0,y}[\#\overline{\mathcal{A}}(0,y)] \bigr]+\mathbb{E}_r\Biggl[ \widetilde{E}_{N,\omega}^{0,y} \Biggl[ \Biggl( \sum_{v \in \overline{\mathcal{A}}(0,y)} \#\mathcal{C}_v \Biggr) \mathbf{1}_{\mathcal{E}(\overline{\mathcal{A}}(0,y))} \Biggr] \Biggr] \Biggr\}.\end{align*}

Since $\#\overline{\mathcal{A}}(0,y) \leq \#\mathcal{A}(0,y)$ , our task is now to prove that there exists a constant c (which depends only on d and $q_0$ ) such that for all $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ ,

(3.11) \begin{align}\mathbb{E}_r\Biggl[ \widetilde{E}_{N,\omega}^{0,y} \Biggl[ \Biggl( \sum_{v \in \overline{\mathcal{A}}(0,y)} \#\mathcal{C}_v \Biggr) \mathbf{1}_{\mathcal{E}(\overline{\mathcal{A}}(0,y))} \Biggr] \Biggr]\leq c.\end{align}

Indeed, once (3.11) is proved, the left-hand side of (3.10) is bounded from above by

\begin{align*}(2C_{6}+c)R^d\mathbb{E}_r\bigl[ \widetilde{E}_{N,\omega}^{0,y}[\#\mathcal{A}(0,y)] \bigr],\end{align*}

and (3.10) follows by taking $C_{5}\;:\!=\; (2C_{6}+c)R^d$ .

For any $0<r \leq q_0$ , $y \in \mathbb{Z}^d \setminus \{0\}$ , and $N \in \mathbb{N}$ with $N \geq \|y\|_\infty$ , the left-hand side of (3.11) is equal to

\begin{align*}&\sum_{n=1}^\infty \mathbb{E}_r\Biggl[ \widetilde{E}_{N,\omega}^{0,y} \Biggl[ \Biggl( \sum_{v \in \overline{\mathcal{A}}(0,y)} \#\mathcal{C}_v \Biggr) \mathbf{1}_{\mathcal{E}(\overline{\mathcal{A}}(0,y))} \mathbf{1}_{\{ \#\overline{\mathcal{A}}(0,y)=n \}} \Biggr] \Biggr] \leq \sum_{n=1}^\infty \sum_{v \in [-n,n]^d} \mathbb{E}_r\bigl[ \#\mathcal{C}_v \mathbf{1}_{\bigcup_{\Gamma \in \mathbb{A}_n} \mathcal{E}(\Gamma)} \bigr].\end{align*}

Note that $\mathbb{E}_r\bigl[ \#\mathcal{C}_v \mathbf{1}_{\bigcup_{\Gamma \in \mathbb{A}_n}\mathcal{E}(\Gamma)} \bigr]$ is increasing in r. This combined with the Cauchy–Schwarz inequality implies that for all $n \geq 1$ and $v \in [-n,n]^d$ ,

\begin{align*}\mathbb{E}_r\bigl[ \#\mathcal{C}_v \mathbf{1}_{\bigcup_{\Gamma \in \mathbb{A}_n} \mathcal{E}(\Gamma)} \bigr]&\leq \mathbb{E}_{q_0}\bigl[ \#\mathcal{C}_v \mathbf{1}_{\bigcup_{\Gamma \in \mathbb{A}_n} \mathcal{E}(\Gamma)} \bigr]\\&\leq \mathbb{E}_{q_0}\bigl[ \#\mathcal{C}_0^2 \bigr]^{1/2} \mathbb{P}_{q_0}\Biggl( \sup_{\Gamma \in \mathbb{A}_n} \dfrac{1}{\#\Gamma} \sum_{v' \in \Gamma}\#\mathcal{C}_{v'}>2C_{6} \Biggr)^{1/2}.\end{align*}

Lemma 3.3 says that there exists a constant c ^′ (which depends only on d and $q_0$ ) such that for all $n \geq 1$ ,

\begin{align*}\mathbb{P}_{q_0}\Biggl( \sup_{\Gamma \in \mathbb{A}_n} \dfrac{1}{\#\Gamma} \sum_{v' \in \Gamma}\#\mathcal{C}_{v'}>2C_{6} \Biggr)& \leq P\Biggl( \sup_{\Gamma \in \bigcup_{m=n}^\infty \mathbb{A}_m} \dfrac{1}{\#\Gamma} \sum_{v' \in \Gamma} (\#\widetilde{\Gamma}_{v'})^2>C_{6} \Biggr)\\& \leq \sum_{m=n}^\infty P\Biggl( \sup_{\Gamma \in \mathbb{A}_m} \dfrac{1}{\#\Gamma} \sum_{v \in \Gamma} (\#\widetilde{\Gamma}_v)^2 \geq C_{6} \Biggr)\\& \leq c'\exp\!\biggl\{ -\dfrac{C_{7}}{2}n^{1/5} \biggr\}.\end{align*}

With these observations, we can derive (3.11) as follows:

\begin{align*}\mathbb{E}_r\Biggl[ \widetilde{E}_{N,\omega}^{0,y} \Biggl[ \Biggl( \sum_{v \in \overline{\mathcal{A}}(0,y)} \#\mathcal{C}_v \Biggr) \mathbf{1}_{\mathcal{E}(\overline{\mathcal{A}}(0,y))} \Biggr] \Biggr]& \leq \bigl(c'\mathbb{E}_{q_0}\bigl[ \#\mathcal{C}_0^2 \bigr]\bigr)^{1/2} \sum_{n=1}^\infty (2n+1)^d \exp\!\biggl\{ -\dfrac{C_{7}}{4}n^{1/5} \biggr\} <\infty.\end{align*}

Therefore the proof of Proposition 3.1 is complete for $d=1,2$ .

4. Bounds for the annealed Lyapunov exponent

This section is devoted to the proof of Theorem 1.2. We fix $q_0$ with $0<q_0<1$ . For each $0 \leq r \leq 1$ , the path measure $\widetilde{\mathbb{P}}_r^{0,y}$ is defined by

(4.1) \begin{align}\dfrac{{\text{d}} \widetilde{\mathbb{P}}_r^{0,y}}{{\text{d}} P^0}= \mathbb{E}_r[e(0,y,\omega)]^{-1} \mathbb{E}_r\Biggl[ \exp\!\Biggl\{ -\sum_{k=0}^{H(y)-1}\omega(S_k) \Biggr\}\Biggr] \mathbf{1}_{\{ H(y)<\infty \}},\end{align}

and $\widetilde{\mathbb{E}}_r^{0,y}$ is the expectation with respect to $\widetilde{\mathbb{P}}_r^{0,y}$ . Then the following two lemmata are the key to proving Theorem 1.2.

Lemma 4.1. For all $0<r<1$ and $y \in \mathbb{Z}^d \setminus \{0\}$ , we have

\begin{align*}-\dfrac{{\text{d}}}{{\text{d}} r}b_r(0,y)= \widetilde{\mathbb{E}}_r^{0,y} \!\left[\rule{0pt}{24pt}\right.\! \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y)) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y))}}{r+{\text{e}}^{-\ell_z(H(y))}(1-r)} \!\left.\rule{0pt}{24pt}\right]\! .\end{align*}

Lemma 4.2. If $d \geq 3$ , then there exists a constant $C_8$ (which depends only on d and $q_0$ ) such that for all $0<r \leq q_0$ and $y \in \mathbb{Z}^d \setminus \{0\}$ ,

\begin{align*}\widetilde{\mathbb{E}}_r^{0,y} \!\left[\rule{0pt}{24pt}\right.\! \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y)) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y))}}{r+{\text{e}}^{-\ell_z(H(y))}(1-r)}\!\left.\rule{0pt}{24pt}\right]\!\leq C_{8}\|y\|_1.\end{align*}

Lemma 4.1 gives the derivative of $b_r(0,y)$ at r by using the path measure $\widetilde{\mathbb{P}}_r^{0,y}$ . Moreover, Lemma 4.2 guarantees that in the case $d \geq 3$ , the derivative of $b_r(0,y)$ at r can be bounded from below uniformly in $r \in (0,q_0]$ . Let us show Theorem 1.2 before proving the lemmata above.

Proof of Theorem 1.2. We first prove part (1) of Theorem 1.2, which gives lower bounds for differences between annealed Lyapunov exponents. Note that for each $0<r<1$ , the function $f(t)\;:\!=\; (1-t)/\{ r+t(1-r) \}$ is decreasing in $t \in [0,1]$ . Hence Lemma 4.1 and the fact that $\#\mathcal{A}(0,y) \geq \|y\|_1$ holds $\widetilde{\mathbb{P}}_r^{0,y}$ -a.s. yield that for all $0<r<1$ and $y \in \mathbb{Z}^d \setminus \{0\}$ ,

\begin{align*}-\dfrac{{\text{d}}}{{\text{d}} r}b_r(0,y)&= \widetilde{\mathbb{E}}_r^{0,y}\!\left[\rule{0pt}{24pt}\right.\! \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y)) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y))}}{r+{\text{e}}^{-\ell_z(H(y))}(1-r)} \!\left.\rule{0pt}{24pt}\right]\! \\&\geq \dfrac{1-{\text{e}}^{-1}}{r+{\text{e}}^{-1}(1-r)}\,\widetilde{\mathbb{E}}_r^{0,y}[\#\mathcal{A}(0,y)]\\&\geq (1-{\text{e}}^{-1})\|y\|_1.\end{align*}

Thus, similarly to the proof of part (1) of Theorem 1.1, we have for all $0<p<q<1$

\begin{align*}\inf_{x \in \mathbb{R}^d\setminus\{0\}}\dfrac{\beta_p(x)-\beta_q(x)}{\|x\|_1}\geq (1-{\text{e}}^{-1})(q-p),\end{align*}

which is the desired lower bound for the difference between annealed Lyapunov exponents.

Let us next show part (2) of Theorem 1.2, which gives upper bounds for differences between annealed Lyapunov exponents. Lemma 4.1 and the monotonicity of the function $f(t)=(1-t)/\{ r+t(1-r) \}$ , $0 \leq t \leq 1$ , tell us that for all $0<r \leq q_0$ and $y \in \mathbb{Z}^d \setminus \{0\}$ ,

(4.2) \begin{align}-\dfrac{{\text{d}}}{{\text{d}} r}b_r(0,y)&= \widetilde{\mathbb{E}}_r^{0,y}\!\left[\rule{0pt}{24pt}\right.\! \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y)) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y))}}{r+{\text{e}}^{-\ell_z(H(y))}(1-r)} \!\left.\rule{0pt}{24pt}\right]\!\leq r^{-1}\widetilde{\mathbb{E}}_r^{0,y}[\#\mathcal{A}(0,y)].\end{align}

Moreover, by the same arguments as in [Reference Zerner18, Lemma 3], it holds that for all $0<r<1$ and $y \in \mathbb{Z}^d \setminus \{0\}$ ,

(4.3) \begin{align}\widetilde{\mathbb{E}}_r^{0,y}[\#\mathcal{A}(0,y)]\leq \dfrac{1+\log(2d)}{-\log\{ {\text{e}}^{-1}+(1-{\text{e}}^{-1})r \}}\|y\|_1.\end{align}

Hence (4.2) and (4.3) imply that for all $0<r \leq q_0$ and $y \in \mathbb{Z}^d \setminus \{0\}$ ,

\begin{align*}-\dfrac{{\text{d}}}{{\text{d}} r}b_r(0,y)\leq \dfrac{1+\log(2d)}{-\log\{ {\text{e}}^{-1}+(1-{\text{e}}^{-1})q_0 \}}\,r^{-1}\|y\|_1.\end{align*}

Therefore, by taking

\begin{align*}C_2\;:\!=\; \dfrac{1+\log(2d)}{-\log\{ {\text{e}}^{-1}+(1-{\text{e}}^{-1})q_0 \}},\end{align*}

one sees immediately that for all $0<p<q \leq q_0$ ,

\begin{align*}\sup_{x \in \mathbb{R}^d \setminus \{0\}}\dfrac{\beta_p(x)-\beta_q(x)}{\|x\|_1} \leq C_2(\log q-\log p).\end{align*}

In particular, if $d \geq 3$ , then Lemma 4.2 allows us to obtain the following bound for the derivative of $b_r(0,y)$ at r instead of (4.2): for all $0<r \leq q_0$ and $y \in \mathbb{Z}^d \setminus \{0\}$ ,

\begin{align*}-\dfrac{{\text{d}}}{{\text{d}} r}b_r(0,y)= \widetilde{\mathbb{E}}_r^{0,y}\!\left[\rule{0pt}{24pt}\right.\! \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y)) \geq 1}}\dfrac{1-{\text{e}}^{-\ell_z(H(y))}}{r+{\text{e}}^{-\ell_z(H(y))}(1-r)} \!\left.\rule{0pt}{24pt}\right]\!\leq C_{8}\|y\|_1.\end{align*}

It follows immediately that

\begin{align*}\sup_{x \in \mathbb{R}^d \setminus \{0\}}\dfrac{\beta_p(x)-\beta_q(x)}{\|x\|_1} \leq C_{8}(q-p),\end{align*}

and the proof of part (2) of Theorem 1.2 is complete.

We close this section with the proofs of Lemmata 4.1 and 4.2.

Proof of Lemma 4.1. First of all, we often add $\mathbb{S}$ to the notation of H(y) and $\ell_z(N)$ to clarify the dependence of the trajectory $\mathbb{S}$ of the simple random walk: $H(y)=H(y,\mathbb{S})$ and $\ell_z(N)=\ell_z(N,\mathbb{S})$ . Fix $y \in \mathbb{Z}^d \setminus \{0\}$ , and define for $0<r<1$ and the trajectory $\mathbb{S}$ of the simple random walk,

\begin{align*}\phi(r,\mathbb{S})&\;:\!=\; \mathbb{E}_r\Biggl[ \exp\!\Biggl\{ -\sum_{k=0}^{H(y,\mathbb{S})-1}\omega(S_k) \Biggr\}\Biggr] \mathbf{1}_{\{ H(y,\mathbb{S})<\infty \}} \\& = \prod_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y,\mathbb{S}),\mathbb{S}) \geq 1}} \mathbb{E}_r\bigl[ {\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})\omega(0)} \bigr] \mathbf{1}_{\{ H(y,\mathbb{S})<\infty \}}.\end{align*}

It follows from Fubini’s theorem that $b_r(0,y)=-\log E^0[\phi(r,\mathbb{S})]$ . Hence our task is to show that

(4.4) \begin{align}\dfrac{{\text{d}}}{{\text{d}} r} E^0[\phi(r,\mathbb{S})]= E^0\!\left[\rule{0pt}{24pt}\right.\! \phi(r,\mathbb{S}) \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y,\mathbb{S}),\mathbb{S}) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}}{r+{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}(1-r)} \!\left.\rule{0pt}{24pt}\right]\! .\end{align}

Indeed, once (4.4) is proved, we have

\begin{align*}\dfrac{{\text{d}}}{{\text{d}} r}b_r(0,y)&= -E^0[\phi(r,\mathbb{S})]^{-1} E^0\!\left[\rule{0pt}{24pt}\right.\! \phi(r,\mathbb{S}) \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y,\mathbb{S}),\mathbb{S}) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}}{r+{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}(1-r)} \!\left.\rule{0pt}{24pt}\right]\! \\&= -\widetilde{\mathbb{E}}_r^{0,y}\!\left[\rule{0pt}{24pt}\right.\! \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y)) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y))}}{r+{\text{e}}^{-\ell_z(H(y))}(1-r)} \!\left.\rule{0pt}{24pt}\right]\! ,\end{align*}

which is the desired conclusion.

To prove (4.4), note that for the trajectory $\mathbb{S}$ of the simple random walk,

\begin{align*}\mathbb{E}_r\bigl[ {\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})\omega(0)} \bigr]= r+{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}(1-r).\end{align*}

This tells us that $\phi(r,\mathbb{S})$ is differentiable at $r \in (0,1)$ and

(4.5) \begin{align}\dfrac{{\text{d}}}{{\text{d}} r}\phi(r,\mathbb{S})= \phi(r,\mathbb{S}) \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y,\mathbb{S}),\mathbb{S}) \geq 1}}\dfrac{1-{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}}{r+{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}(1-r)}.\end{align}

Therefore, differentiating under the integral sign (the possibility of this operation will be checked in Appendix A), we have for $0<r<1$

\begin{align*}\dfrac{{\text{d}}}{{\text{d}} r} E^0[\phi(r,\mathbb{S})]&=E^0\biggl[ \dfrac{{\text{d}}}{{\text{d}} r}\phi(r,\mathbb{S}) \biggr]= E^0\!\left[\rule{0pt}{24pt}\right.\! \phi(r,\mathbb{S}) \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y,\mathbb{S}),\mathbb{S}) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}}{r+{\text{e}}^{-\ell_z(H(y,\mathbb{S}),\mathbb{S})}(1-r)} \!\left.\rule{0pt}{24pt}\right]\! ,\end{align*}

and (4.4) follows.

Proof of Lemma 4.2. Let $d \geq 3$ and fix $0<r<1$ and $y \in \mathbb{Z}^d \setminus \{0\}$ . Furthermore, for each $z \in \mathbb{Z}^d$ , set $H_1(z)\;:\!=\; H(z)$ and define inductively

\begin{align*}H_{\ell+1}(z)\;:\!=\; \inf\{ k>H_\ell(z)\;:\; S_k=z \},\quad \ell \geq 1.\end{align*}

Then we have

(4.6) \begin{align}&\widetilde{\mathbb{E}}_r^{0,y}\!\left[\rule{0pt}{24pt}\right.\! \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y)) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y))}}{r+{\text{e}}^{-\ell_z(H(y))}(1-r)} \!\left.\rule{0pt}{24pt}\right]\! \notag \\ &\quad = \sum_{z \in \mathbb{Z}^d \setminus \{y\}} \mathbb{E}_r[e(0,y,\omega)]^{-1} \sum_{\ell=1}^\infty \dfrac{1-{\text{e}}^{-\ell}}{r+{\text{e}}^{-\ell}(1-r)} \,\mathbb{E}_r[\rho_\ell(z,\omega)],\end{align}

where

\begin{align*}\rho_\ell(z,\omega)\;:\!=\; E^0\Biggl[ \exp\!\Biggl\{ -\sum_{k=0}^{H(y)-1}\omega(S_k) \Biggr\} \mathbf{1}_{\{ H_\ell(z)<H(y)<H_{\ell+1}(z) \}} \Biggr].\end{align*}

The strong Markov property gives that

\begin{align*}\rho_\ell(z,\omega)&\leq E^z\left[ \exp\!\left\{ -\sum_{k=0}^{H^+(z)-1}\omega(S_k) \right\} \mathbf{1}_{\{ H^+(z)<\infty \}} \right]^{\ell-1}\\&\quad \times E^0 \left[ \exp\!\left\{ -\sum_{k=0}^{H(y)-1}\omega(S_k) \right\} \mathbf{1}_{\{ H(z)<H(y)<H_2(z) \}} \right]\end{align*}

(see the statement of Lemma 3.1 for the notation $H^+(z)$ ). Hence, in the case where $\omega(z)=0$ , $\rho_\ell(z,\omega)$ is bounded from above by

(4.7) \begin{align}P^0(H^+(0)<\infty)^{\ell-1} \, E^0 \!\left[\rule{0pt}{24pt}\right.\! \exp \!\left\{\rule{0pt}{21pt}\right.\! -\sum_{\substack{0 \leq k<H(y)\\ S_k \not= z}}\omega(S_k)\!\left.\rule{0pt}{21pt}\right\}\! \mathbf{1}_{\{ H(z)<H(y)<H_2(z) \}} \!\left.\rule{0pt}{24pt}\right]\! .\end{align}

On the other hand, in the case where $\omega(z)=1$ , $\rho_\ell(z,\omega)$ is smaller than or equal to

(4.8) \begin{align}{\text{e}}^{-\ell} P^0(H^+(0)<\infty)^{\ell-1} \, E^0\!\left[\rule{0pt}{24pt}\right.\! \exp\!\left\{\rule{0pt}{21pt}\right.\! -\sum_{\substack{0 \leq k<H(y)\\ S_k \not= z}}\omega(S_k) \!\left.\rule{0pt}{21pt}\right\}\! \mathbf{1}_{\{ H(z)<H(y)<H_2(z) \}} \!\left.\rule{0pt}{24pt}\right]\!.\end{align}

Note that both (4.7) and (4.8) are independent of $\omega(z)$ , and the simple random walk visits only one time before hitting y on the event $\{ H(z)<H(y)<H_2(z) \}$ . It follows that

\begin{align*}\mathbb{E}_r[\rho_\ell(z,\omega)]&\leq e\{ r+{\text{e}}^{-\ell}(1-r) \}P^0(H^+(0)<\infty)^{\ell-1}\\&\quad \times \mathbb{E}_r \Biggl[ E^0\Biggl[ \exp\!\Biggl\{ -\sum_{k=0}^{H(y)-1}\omega(S_k) \Biggr\} \mathbf{1}_{\{ H(z)<H(y)<H_2(z) \}} \Biggr] \Biggr].\end{align*}

Therefore (4.3), (4.6) and the fact that $P^0(H^+(0)<\infty)<1$ holds for $d \geq 3$ yield that for all $0<r \leq q_0$ and $y \in \mathbb{Z}^d \setminus \{0\}$ ,

\begin{align*}\widetilde{\mathbb{E}}_r^{0,y}\!\left[\rule{0pt}{24pt}\right.\! \sum_{\substack{z \in \mathbb{Z}^d\\ \ell_z(H(y)) \geq 1}} \dfrac{1-{\text{e}}^{-\ell_z(H(y))}}{r+{\text{e}}^{-\ell_z(H(y))}(1-r)} \!\left.\rule{0pt}{24pt}\right]\! & \leq e\sum_{\ell=1}^\infty P^0(H^+(0)<\infty)^{\ell-1}\,\widetilde{\mathbb{E}}_r^{0,y}[\#\mathcal{A}(0,y)]\\& \leq eP^0(H^+(0)=\infty)^{-1}\dfrac{1+\log(2d)}{-\log\{ {\text{e}}^{-1}+(1-{\text{e}}^{-1})q_0 \}}\|y\|_1,\end{align*}

which is the desired conclusion.

5. Comment on the large deviation principle

In this section we discuss the large deviation principle for the simple random walk in the Bernoulli potential. Let $0 \leq r \leq 1$ and let $\omega$ be the Bernoulli potential with parameter r. Consider the path measures $Q_{n,\omega}^\textrm{qu}$ and $Q_{n,r}^\textrm{an}$ defined as follows:

\begin{align*}\dfrac{{\text{d}} Q_{n,\omega}^{\text{qu}}}{{\text{d}} P^0}= \dfrac{1}{Z_{n,\omega}^{\text{qu}}} \exp\!\left\{ -\sum_{k=0}^{n-1}\omega(S_k) \right\}\end{align*}

and

\begin{align*}\dfrac{{\text{d}} Q_{n,r}^{\text{an}}}{{\text{d}} P^0}= \dfrac{1}{Z_{n,r}^{\text{an}}} \mathbb{E}_r\Biggl[ \exp\!\Biggl\{ -\sum_{k=0}^{n-1}\omega(S_k) \Biggr\} \Biggr],\end{align*}

where $Z_{n,\omega}^{\text{qu}}$ and $Z_{n,r}^{\text{an}}$ are the corresponding normalizing constants. Moreover, for $\lambda \geq 0$ , write $\alpha_r(\lambda,\cdot)$ and $\beta_r(\lambda,\cdot)$ , respectively, for the quenched and annealed Lyapunov exponents for the potential $\omega+\lambda=(\omega(x)+\lambda)_{x \in \mathbb{Z}^d}$ . Note that $\alpha_r(\lambda,x)$ and $\beta_r(\lambda,x)$ are continuous in $(\lambda,x) \in [0,\infty) \times \mathbb{R}^d$ and concave increasing in $\lambda$ (see [Reference Flury4, Theorem A], [Reference Mourrat13, Theorem 1.1], and [Reference Zerner18, Proposition 4]). Then, for $x \in \mathbb{R}^d$ , set

\begin{align*}I_r(x)\;:\!=\; \sup_{\lambda \geq 0}\{ \alpha_r(\lambda,x)-\lambda \}\end{align*}

and

\begin{align*}J_r(x)\;:\!=\; \sup_{\lambda \geq 0}\{ \beta_r(\lambda,x)-\lambda \}.\end{align*}

The following proposition states the large deviation principles for the simple random walk in the Bernoulli potential, which is a direct application of [Reference Flury4, Theorem B], [Reference Mourrat13, Theorem 1.10], and [Reference Zerner18, Theorem 19].

Since exactly the same arguments used in the previous sections work for $\alpha_r(\lambda,\cdot)$ and $\beta_r(\lambda,\cdot)$ , we can replace $\alpha_r(\!\cdot\!)$ and $\beta_r(\!\cdot\!)$ with $\alpha_r(\lambda,\cdot)$ and $\beta_r(\lambda,\cdot)$ in Theorems 1.1 and 1.2, respectively. In particular, the constants $C_{1}$ and $C_2$ can be chosen independently of $\lambda \geq 0$ because for any $a,b \in \mathbb{R}$ with $a<b$ , the function $f(\lambda)\;:\!=\; (\lambda+b)/(\lambda+a)$ is decreasing in $\lambda \geq 0$ , and the factor ${\text{e}}^{-\lambda\ell_z(H(y,\mathbb{S}),\mathbb{S})}$ appears in the denominator and numerator of the expressions above and can be canceled out. This derives the following differences between quenched and annealed rate functions.

Proof. We treat only the upper bound for the quenched case since the same argument works for the other cases. For any $0<r<1$ and $x \in \mathbb{R}^d$ , set

\begin{align*}\lambda_r^{\text{qu}}(x)\;:\!=\; \inf\{ \lambda >0\;:\; \partial_- \alpha_r(\lambda,x) \leq 1 \},\end{align*}

where $\partial_-\alpha_r(\lambda,x)$ is the left derivative of $\alpha_r(\lambda,x)$ with respect to $\lambda$ (note that $\lambda_r^{\text{qu}}(x)$ exists in the case where $\|x\|_1 \leq 1$ although it may be equal to $\infty$ ). Clearly, $\lambda_r^{\text{qu}}(x)$ attains the supremum in the definition of $I_r(x)$ . Hence Theorem 1.1 with $\alpha_r(\!\cdot\!)$ replaced by $\alpha_r(\lambda,\cdot)$ implies that for any $0<p<q<1$ , $t \geq 0$ , and $x \in \mathbb{R}^d$ with $0<\|x\|_1 \leq 1$ ,

\begin{align*}\dfrac{I_p(x)-\{ \alpha_q(\lambda_q^{\text{qu}}(x) \wedge t,x)-(\lambda_q^{\text{qu}}(x) \wedge t) \}}{\|x\|_1} &\geq \dfrac{\alpha_p(\lambda_q^{\text{qu}}(x) \wedge t,x)-\alpha_q(\lambda_q^{\text{qu}}(x) \wedge t,x)}{\|x\|_1}\\&\geq (1-{\text{e}}^{-1})(q-p).\end{align*}

Since $\alpha_q(\lambda,x)$ is continuous in $\lambda$ , letting $t \to \infty$ proves (5.1).