Proof of Corollary 3.1.7 Proofs of Theorem 3.1.5, Lemma 3.1.6

it is clear that i X i t t ≥0 is a bounded martingale, and hence there exists a random variable Z such that lim t →∞ E Z − i X i t 2 = 0. 3.6.3 We are therefore done if we can show that lim t →∞ E |X i t − X j t | 2 = 0 ∀i, j ∈ . 3.6.4 We will establish 3.6.4 by showing that each sequence t n → ∞ has a subse- quence t ˜nm such that lim p →∞ E |X i t ˜np − X j t ˜np | 2 = 0 ∀i, j ∈ . 3.6.5 So let us fix a sequence t n → ∞. Lemma 5.1 and Remark 5.2 from chapter 4 of [16] imply the existence of a subsequence t nm and a process ˜ X , such that in the sense of weak convergence on path space X t nm + t t ≥0 ⇒ ˜Xt t ≥0 as m → ∞, 3.6.6 where ˜ X has sample paths in D K [0, ∞ and solves the martingale problem for the operator A. Formula 3.6.3 together with the continuous sample paths of ˜ X implies that for all t ≥ 0 i X i t nm + t − i X i t nm ⇒ 0 as m → ∞, 3.6.7 and therefore i X i t − i X i = 0 a.s. t ≥ 0. 3.6.8 Using the fact that i ˜X i t t ≥0 is a martingale and the fact that ˜ X solves the martingale problem for A, we see that E i ˜X i t − i X i 2 = E i ˜X i t 2 − E i X i 2 = i t E[tr w ˜ X i s]ds. 3.6.9 We use this, together with 3.6.8 and the fact that tr w ≥ 0 to conclude that E[tr w ˜ X i t] = 0 i ∈ , t ≥ 0. 3.6.10 The fact that ˜ X solves the martingale problem for the operator A means that for each f ∈ C 2 K f ˜ X t − f ˜X0 − t A f ˜ X sds 3.6.11 is a martingale. Here A f ˜ X s = i j α a j − i ˜X α j s − ˜X α i s ∂ ∂ x α i f ˜ X s + i αβ w αβ ˜ X i s ∂ 2 ∂ x α i ∂ x β i f ˜ X s, 3.6.12 where the second term is zero by 3.6.10. We therefore see that ˜ X also solves the martingale problem for the operator i j α a j − ix α j s − x α i s ∂ ∂ x α i , 3.6.13 and hence is equal in distribution to a solution of the system of differential equa- tions d ˜ X i t = j a j − i ˜X j t − ˜X i tdt. 3.6.14 By the irreducibility of a this implies that lim t →∞ ˜X i t − ˜X j t = 0 a.s. i, j ∈ . 3.6.15 Thus lim t →∞ lim m →∞ E |X i t nm + t − X j t nm + t| 2 = 0 i, j ∈ . 3.6.16 By a diagonal argument we see that there exist T m such that for all s m ≥ T m lim m →∞ E |X i t nm + s m − X j t nm + t m | 2 = 0 i, j ∈ . 3.6.17 In particular, we can find a further subsequence t ˜nm such that t ˜nm − t nm ≥ T m , and therefore lim m →∞ E |X i t ˜nm − X j t ˜nm | 2 = 0 i, j ∈ . 3.6.18 This shows that 3.6.5 holds.