1 Correction to: J Stat Phys (2017) 168:1180–1190 https://doi.org/10.1007/s10955-017-1854-3

The original publication of the article unfortunately contained a mistake in the first sentence of Theorem 1 and in the second part of the proof of Theorem 1. The corrected statement of Theorem as well as the corrected proof are given below. The full text of the corrected version is available at http://arxiv.org/abs/1705.11150.

The first phrase of Theorem 1 is given as follows.

For any fixed \(q> p > \frac{2}{3} \) there exist \( \lambda _c < \lambda _1 (p ) \le \lambda _2 (p) \) such that the following holds.

The corrected version of Step 2. of the proof of Theorem 1 is given as follows.

Proof of Theorem 1

Step 2. We finally consider the case where \(\lambda _1 \) is sufficiently larger than \(\lambda _c\) and where \( q > p \ge \frac{2}{3}. \) Let Q be the monotone coupling between \( \nu _{\lambda _1}\) and \(\nu _{\lambda _2}\) induced by the construction of Proposition 2. Using this coupling and the fact that \( f(0 ) = 0,\) we obtain thanks to Theorem 2 that

$$\begin{aligned}&\lim _{t \rightarrow \infty } \Delta _{p,q} (\lambda _1, \lambda _2,r, t ) \nonumber \\&\quad = \varrho (\lambda _2) \int _{{\mathcal P}({\mathbb Z})} f(\vert B \cap \Lambda \vert ) \nu _{\lambda _2} (dB) - \varrho (\lambda _1) \int _{{\mathcal P}({\mathbb Z})}f(\vert B \cap \Lambda \vert ) \nu _{\lambda _1}(dB)\nonumber \\&\quad = \varrho (\lambda _1 ) \int \int \left( f(|B_2 \cap \Lambda _r |) - f(\vert B_1 \cap \Lambda _r \vert ) \right) Q(dB_1, dB_2) \nonumber \\&\qquad +\, (\varrho (\lambda _2 ) - \varrho (\lambda _1 ) ) \int _{{\mathcal P}({\mathbb Z})} f(| B \cap \Lambda |) \nu _{\lambda _2} (dB) \nonumber \\&\quad =: T_1 (r) + T_2 (r). \end{aligned}$$
(1.1)

We want to show that this expression is negative for sufficiently large values of \( \lambda _1 \) and r. We put \( \varepsilon := 2-p - q.\) Since by assumption \( q > p \ge \frac{2}{3}, \) we have \( 2 ( 1 - \varepsilon ) > \varepsilon \) (this will be important in (1.3) below).

Then \( f(2) - f(1) = -(1- \varepsilon ) f(1) \) and \( f(2) = \varepsilon f(1).\) Writing for short

$$\begin{aligned} Q (r, n, m ) := Q (\{ (B_1, B_2 ) : |B_1 \cap \Lambda _r | = n, |B_2 \cap \Lambda _r | = m \} ) , \end{aligned}$$

it is clear that

$$\begin{aligned} T_1 (r)= & {} \varrho (\lambda _1) \left[ (f(2) - f(1) ) Q(r, 1, 2) + f(2) Q(r, 0, 2 ) + f(1) Q(r, 0, 1 )\right] \\= & {} \varrho (\lambda _1) f(1) \left[ -(1- \varepsilon ) Q(r, 1, 2 ) + \varepsilon Q(r, 0, 2 ) + Q(r, 0, 1 ) \right] . \end{aligned}$$

Applying the last item of Theorem 2, we have that

$$\begin{aligned} \lim _{r \rightarrow \infty } T_1 (r)= & {} \varrho (\lambda _1) f(1) \Big (-2 (1- \varepsilon ) (\varrho (\lambda _2) - \varrho (\lambda _1 ))\varrho (\lambda _1) \\&+ \;\varepsilon (\varrho (\lambda _2) - \varrho (\lambda _1))^2 +2(\varrho (\lambda _2) - \varrho (\lambda _1 ) ) (1- \varrho (\lambda _2) ) \Big ) \\= & {} \varrho (\lambda _1) f(1) (\varrho (\lambda _2) - \varrho (\lambda _1 ) ) \Big (-2 (1- \varepsilon ) \varrho (\lambda _1) + \varepsilon (\varrho (\lambda _2) - \varrho (\lambda _1 ) ) \\&+\; 2(1- \varrho (\lambda _2) ) \Big ). \end{aligned}$$

Moreover,

$$\begin{aligned} \lim _{r \rightarrow \infty } T_2 (r) = (\varrho ( \lambda _2 ) - \varrho (\lambda _1 ) ) \left[ 2 f(1) \varrho (\lambda _2 ) (1- \varrho ( \lambda _2) ) + f(2) \varrho (\lambda _2)^2 \right] . \end{aligned}$$

Putting these results together, we conclude that

$$\begin{aligned}&\lim _{r \rightarrow \infty } \lim _{t \rightarrow \infty } \Delta _{p,q} (\lambda _1, \lambda _2,r, t ) = (\varrho (\lambda _2 ) - \varrho (\lambda _1 ) ) f(1) \nonumber \\&\qquad \times \,\Big \{-2 (1- \varepsilon ) (\varrho (\lambda _1))^2 + \varepsilon \varrho (\lambda _1) (\varrho (\lambda _2) - \varrho (\lambda _1 ) ) + 2 (1- \varrho (\lambda _2) ) \varrho (\lambda _1) \nonumber \\&\qquad +\, 2 \varrho (\lambda _2 ) (1- \varrho (\lambda _2) ) + \varepsilon \varrho (\lambda _2)^2 \Big \}. \end{aligned}$$
(1.2)

Since \( 2 (1- \varepsilon ) > \varepsilon , \) it is possible to choose \( \delta ^*\) such that for all \( \delta \le \delta ^* , \)

$$\begin{aligned} 2 (1 - \varepsilon ) (1- \delta )^2 - \varepsilon (1- \delta ) \delta - 4 \delta - \varepsilon (1 - \delta )^2 \ge \kappa > 0 , \end{aligned}$$
(1.3)

for some (sufficiently small) \( \kappa > 0.\) Recall that \( \varepsilon = \varepsilon (p).\) Since \( \lim _{\lambda \uparrow \infty } \varrho (\lambda ) = 1,\) we may choose \( \lambda _2 (p ) \) sufficiently large such that \(\varrho (\lambda _1 ) \ge 1 - \delta ^* \) for all \( \lambda _1 \ge \lambda _2 (p).\) As a consequence, for all \( \lambda _2 \ge \lambda _1 \ge \lambda _2(p),\)

$$\begin{aligned} \lim _{r \rightarrow \infty } \lim _{t \rightarrow \infty } \Delta _{p,q} (\lambda _1, \lambda _2,r, t ) \ge \kappa > 0 , \end{aligned}$$
(1.4)

which implies the assertion. \(\square \)