Proof of Lemma 5.1 Freezing, defreezing and reduction to two key lemmas

Therefore, by choosing p close to 1, the proof of the upper bound in Proposition 3.4 reduces to the proof of the following two lemmas. Lemma 5.1. For all R, α 0, lim sup t, κ,T →∞ κ 2 t log E 1 R, α κ, T ; t ≤ 0. 5.7 Lemma 5.2. For all α 0, lim sup R →∞ lim sup t, κ,T →∞ κ 2 t log E 2 R, α κ, T ; t ≤ 6 α 2 ρ1 − ρ 2 P 3 . 5.8 Lemma 5.1 will be proved in Section 5.1.2, Lemma 5.2 in Sections 5.1.3–5.3.3.

5.1.2 Proof of Lemma 5.1

Proof. Fix R, α 0 arbitrarily. Given a path X , an initial configuration η ∈ Ω and k ∈ N, we first derive an upper bound for E η ‚ exp – α κ Z R κ ds X y ∈Z 3 p 6T 1[ κ] X k,κ s , y ξ s κ y − p 6T 1[ κ]+ s κ X k,κ s , y η y ™Œ , 5.9 where X k,κ s = X k−1R κ +s . 5.10 To this end, we use the independent random walk approximation e ξ of ξ cf. [3], Proposition 1.2.1, to obtain 5.9 ≤ Y y ∈A η E Y ‚ exp – α κ Z R κ ds p 6T 1[ κ] X k,κ s , y + Y s κ − p 6T 1[ κ]+ s κ X k,κ s , y ™Œ , 5.11 where Y is simple random walk on Z 3 with jump rate 1 i.e., with generator 1 6 ∆, E Y is expectation w.r.t. Y starting from 0, and A η = {x ∈ Z 3 : ηx = 1}. 5.12 Observe that the expectation w.r.t. Y of the expression in the exponent is zero. Therefore, a Taylor expansion of the exponential function yields the bound E Y ‚ exp – α κ Z R κ ds p 6T 1[ κ] X k,κ s , y + Y s κ − p 6T 1[ κ]+ s κ X k,κ s , y ™Œ ≤ 1 + ∞ X n=2 n Y l=1 ‚ α κ Z R κ s l −1 ds l X y l ∈Z 3 p sl −sl−1 κ y l −1 , y l × p 6T 1[ κ] X k,κ s l , y + y l + p 6T 1[ κ]+ sl κ X k,κ s l , y Œ , 5.13 2109 where s = 0, y = 0, and the product has to be understood in a noncommutative way. Using the Chapman-Kolmogorov equation and the inequality p t z ≤ p t 0, z ∈ Z 3 , we find that Z R κ s l −1 ds l X y l ∈Z 3 p sl −sl−1 κ y l −1 , y l p 6T 1[ κ] X k,κ s l , y + y l + p 6T 1[ κ]+ sl κ X k,κ s l , y Œ ≤ 2 Z ∞ ds p T + s κ 0 = 2κG T 5.14 with G T 0 = Z ∞ T ds p s 5.15 the cut-off Green function of simple random walk at 0 at time T . Substituting this into the above bound for l = n, n − 1, · · · , 3, computing the resulting geometric series, and using the inequality 1 + x ≤ e x , we obtain 5.13 ≤ exp – C T α 2 κ 2 2 Y l=1 Z R κ s l −1 ds l X y l ∈Z 3 p sl −sl−1 κ y l −1 , y l × p 6T 1[ κ] X k,κ s l , y + y l + p 6T 1[ κ]+ sl κ X k,κ s l , y ™ 5.16 with C T = 1 1 − 2αG T , 5.17 provided that 2 αG T 0 1, which is true for T large enough. Note that C T → 1 as T → ∞. Substituting 5.16 into 5.11, we find that 5.9 ≤ exp – C T α 2 κ 2 X y ∈Z 3 2 Y l=1 Z R κ s l −1 ds l X y l ∈Z 3 p sl −sl−1 κ y l −1 , y l × p 6T 1[ κ] X k,κ s l , y + y l + p 6T 1[ κ]+ sl κ X k,κ s l , y ™ . 5.18 Using once more the Chapman-Kolmogorov equation and p t x, y = p t x − y, we may compute the sums in the exponent, to arrive at 5.9 ≤ exp – C T α 2 κ 2 Z R κ ds 1 Z R κ s 1 ds 2 p 12T 1[ κ]+ s2−s1 κ X k,κ s 2 − X k,κ s 1 + 3p 12T 1[ κ]+ s2+s1 κ X k,κ s 2 − X k,κ s 1 ™ . 5.19 Note that this bound does not depend on the initial configuration η and depends on the process X only via its increments on the time interval [k − 1R κ , kR κ ]. By 5.10, the increments over the time intervals labelled k = 1, 2, · · · , ⌊tR κ ⌋ are independent and identically distributed. Using E ν ρ ,0 = R ν ρ dηE X E η , we can therefore apply the Markov property of the exclusion dynamics 2110 ξ t κ t ≥0 at times R κ , 2R κ , · · · , ⌊tR κ ⌋ − 1R κ to the expectation in the r.h.s. of 5.5, insert the bound 5.19 and afterwards use that X t t ≥0 has independent increments, to arrive at log E 1 R, α t ≤ t R κ log E X ‚ exp – C T α 2 κ 2 Z R κ ds 1 Z R κ s 1 ds 2 p 12T 1[ κ]+ s2−s1 κ X s 2 − X s 1 + 3p 12T 1[ κ]+ s2+s1 κ X s 2 − X s 1 ™Œ . 5.20 Hence, recalling the definition of R κ in 5.3, we obtain lim sup t →∞ κ 2 t log E 1 R, α t ≤ 1 R log E X ‚ exp – C T α 2 R R κ Z R κ ds 1 Z R κ s 1 ds 2 p 12T 1[ κ]+ s2−s1 κ X s 2 − X s 1 + 3p 12T 1[ κ]+ s2+s1 κ X s 2 − X s 1 ™Œ . 5.21 Let b X t = X t + Y t κ , 5.22 and let E b X = E X E Y be the expectation w.r.t. b X starting at 0. Observe that p t+s κ z = E Y p t z + Y s κ . 5.23 We next apply Jensen’s inequality w.r.t. the first integral in the r.h.s. of 5.21, substitute s 2 = s 1 + s, take into account that X has independent increments, and afterwards apply Jensen’s inequality w.r.t. E Y , to arrive at the following upper bound for the expectation in 5.21: E X ‚ exp – C T α 2 R R κ Z R κ ds 1 Z R κ s 1 ds 2 p 12T 1[ κ]+ s2−s1 κ X s 2 − X s 1 + 3p 12T 1[ κ]+ s2+s1 κ X s 2 − X s 1 ™ ≤ 1 R κ Z R κ ds 1 E X ‚ exp – C T α 2 R Z ∞ ds E Y p 12T 1[ κ] X s + Y s κ + 3p 12T 1[ κ]+ 2s1 κ X s + Y s κ ™Œ ≤ 1 R κ Z R κ ds 1 E b X ‚ exp – C T α 2 R Z ∞ ds p 12T 1[ κ] b X s + 3p 12T 1[ κ]+ 2s1 κ b X s ™Œ . 5.24 Applying Lemma 2.6, we can bound the last expression from above by exp – 4C T α 2 R b G 2T 1 − 4C T α 2 R b G 2T ™ , 5.25 where b G 2T 0 is the cut-off at time 2T of the Green function b G at 0 for b X which has generator 1[ κ]∆. Since b G 2T 0 → 1 6 G 12T 0 as κ → ∞, and since the latter converges to zero as T → ∞, a combination of the above estimates with 5.21 gives the claim. 2111

5.1.3 Defreezing