“…Then there exists a positive constant c = c(a, β, T , λ 1 , λ 2) (c = c(a, λ 1 , λ 2 ) when β = 0, respectively) such that for all t ∈ [T , ∞) (t > 0 when β = 0, respectively) and x, y ∈ D with δ D (x)∧δ D (y) ≥ aϕ −1 (t) ≥ 2|x − y|, we have p D (t, x, y) ≥ c /ϕ −d (t).Proof By the same proof as that of [10, Proposition 3.4], we deduce the proposition using the parabolic Harnack inequality(see[15, Theorem 4.12] for β = 0 and[6, Theorem 4.11] for β ∈ (0, ∞]) and Lemma 3.2.…”