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12. That is, for each (f, g) ∈ H, there exists (fn , gn ) ∈ Hn such that lim n→∞ ηn f − fn + ηn g − gn = 0. Step 2: (Exponential tightness) Under the convergence assumptions on {Hn }, exponential tightness of {Xn } holds. 5) (I − αH)f = h. Then a viscosity solution Rα h exists for each h ∈ Dα , and Rα extends continuously to all of C(E). In addition, suppose that {Xn (0)} satisfies a large deviation principle with a good rate function I0 . Then a) {Xn } satisfies the large deviation principle with a good rate function I.

And K ∈ Q, then LIM is just buc-convergence. 7. 32). For m = m(n), let E = E = L2 (O), and let Q be the class of compact sets in E. 10) (πn ρ)(x) = m id +1 m i1 +1 m d i1 y1 = m ... id yd = m ρ(y1 , . . , yd )dyd . . dy1 , x=( id i1 , . . , ) ∈ Λm . 11) n→∞ ρ∈K L2 (O) = 0. 5 are satisfied. 11) and the above definition of KnK , lim sup inf ηn (γn ) − γ n→∞ γ ∈K K γ∈K n n L2 (O) ≤ lim sup ηn (πn (ρ)) − ρ n→∞ ρ∈K L2 (O) = 0. 5. 3. GENERAL STATE SPACES (1) supn fn < ∞, (2) for γn ∈ R|Λm(n) | satisfying ηn (γn ) → ρ in f (ρ).

13) 1 lim sup sup log P (∃t, 0 ≤ t ≤ T, Yn (t) ∈ πm (K1 )|Yn (0) = ρ0 ) ≤ −a. 12 is motivated by a physical context. 14) Em (ρ) ≡ 1 2 |∇m ρ(x)|2 m−d + x∈Λm F (ρ(x))m−d , ρ ∈ En ≡ R|Λm | , x∈Λm where ∇m g(x) is the vector with components m ∇km ρ(x) = (ρ(x + m−1 ek ) − ρ(x − m−1 ek ). 2 Em is the desired Lyapunov function. 15) Hn Em (ρ) ≡ e−nEm An enEm (ρ) n 1 = ∆m ρ − F (ρ), −∆m ρ + F (ρ) m + ∆m ρ − F (ρ) 2L2 (Λm ) 2 md d + ( m2−d + F (ρ(x))m−d ) 2n 2 x∈Λm 1 = − ∆m ρ − F (ρ) 2 2 L2 (Λm ) + 1 1 dm2+d + md 2n 2 F (ρ(x))m−d .