The rest of the proof

7.4 The rest of the proof

Recall the definition of ν and R, and that our goal is to show that for any B ∈ R, P lim n →∞ U n B = νB = 1. 7.13 Let us fix B ∈ R for the rest of the subsection, and simply write U t instead of U t B. Next, recall the limit in 7.11, but note that the underlying diffusion is only asymptotically Ornstein- Uhlenbeck 9 , that is σ 2 n = 1 − 2 −n , and so the transition kernels q n defined by q n x, d y, k := PY 1 k ∈ d y | Y 1 n = x, k ≥ n, are slightly different from q. Note also the decomposition E € U n+m n | F n Š = 2 n X i=1 2 −n EU i m n | F n = 2 −n 2 n X i=1 q n Y i n , B, n + m n . In addition, recall the following facts. 1. If A n := {suppY n 6⊂ B a n }, then lim n →∞ 1 A n = 0, P-a.s.; 2. m t = ζa t = C 3 + C 4 log t; 3. Lemma 23. From these it follows that the limit lim n →∞ sup x ∈B an q n x , B, n + m n − νB = 0, 7.14 which we will verify below, implies 7.13 with U n replaced by U n+m n . Remark 30 n and N n. Notice that 7.13 must then also hold P-a.s. for U n , and even for U t n with any given sequence t n ↑ ∞ replacing n. Indeed, define the sequence Nn by the equation N n + m N n = t n . Clearly, N n = Θt n , and in particular lim n →∞ N n = ∞. Now, it is easy to see that in the proof of Theorem 16, including the remainder of this paper, all the arguments go through when replacing n by N n, yielding thus 7.13 with U n replaced by U N n+m N n = U t n . In those arguments it never plays any role that n is actually an integer. ⋄ We preferred to provide Remark 30 instead of presenting the proof with N n replacing n every- where, and to avoid notation even more difficult to follow 10 . Motivated by Remark 30, we now show 7.14. To achieve this goal, first recall that on the time interval [l, l + 1, Y = Y 1 corresponds to the Ornstein-Uhlenbeck operator 1 2 σ 2 l ∆ − γx · ∇, 9 Unlike in [4], where σ n ≡ 1. 10 For example, one should replace 2 n with 2 ⌊Nn⌋ or n N n everywhere. 1961 where σ 2 l = 1 − 2 −l , l ∈ N. That is, if σ n · is defined by σ n s := σ n+l for s ∈ [l, l + 1, then, given F n and with a Brownian motion W , one has that Y m n − EY m n | F n = Y m n − e −γm n Y = Z m n σ n se γs−m n dW s = Z m n e γs−m n dW s − Z m n [1 − σ n s]e γs−m n dW s . Fix ε 0. By the Chebyshev inequality and the Itô-isometry, P Z m n [1 − σ n s]e γs−m n dW s ε ≤ ε −2 E   ‚Z m n [1 − σ n s]e γs−m n dW s Œ 2   = ε −2 Z m n [1 − σ n s] 2 e 2 γs−m n ds. Now, [1 − σ n s] 2 ≤ [1 − σ n ] 2 = 1 − p 1 − 2 −n 2 =   2 −n 1 + p 1 − 2 −n   2 ≤ 2 −2n. Hence, P Z m n [1 − σ n s]e γs−m n dW s ε ≤ ε −2 Z m n 2 −2n e 2 γs−m n ds. Since e −m n = e −C 3 n −C 4 , we obtain that ε −2 Z m n 2 −2n e 2 γs−m n ds = ε −2 e −2γC 3 2 −2n n −2γC 4 Z m n e 2 γs ds = ε −2 e −2γC 3 2 −2n n −2γC 4 · e 2 γC 3 n 2 γC 4 − 1 2 γ → 0, as n → ∞. Therefore, lim n →∞ P Z m n [1 − σ n s]e γs−m n dW s ε = 0. 7.15 We have q n x , B, n + m n = PY 1 n+m n ∈ B | Y 1 n = x = P ‚Z m n σ n se γs−m n dW s ∈ B − x e −γm n Œ , and qx , B, m n = P ‚Z m n e γs−m n dW s ∈ B − x e −γm n Œ . 1962 For estimating q n x , B, n + m n let us use the inequality ˙ A ε ⊂ A + b ⊂ A ε , for A ⊂ R d , b ∈ R d , |b| ε, ε 0. So, for any ε 0, q n x , B, n + m n = P ‚Z m n e γs−m n dW s − Z m n [1 − σ n s]e γs−m n dW s ∈ B − x e −γm n Œ = P ‚Z m n e γs−m n dW s ∈ B − x e −γm n + Z m n [1 − σ n s]e γs−m n dW s Œ ≤ P ‚Z m n e γs−m n dW s ∈ B ε − x e −γm n Œ +P Z m n [1 − σ n s]e γs−m n dW s ε = qx , B ε , m n + P Z m [1 − σ n s]e γs−m n dW s ε . Taking lim sup n →∞ sup x ∈B an , the second term vanishes by 7.15 and the first term becomes νB ε by 7.11. The lower estimate is similar: q n x , B, n + m n ≥ P ‚Z m n e γs−m n dW s ∈ ˙B ε − x e −γm n Œ −P Z m n [1 − σ n s]e γs−m n dW s ε = qx , ˙ B ε , m n − P Z m n [1 − σ n s]e γs−m n dW s ε . Taking lim inf n →∞ sup x ∈B an , the second term vanishes by 7.15 and the first term becomes ν ˙ B ε by 7.11. Now 7.14 follows from these limits: lim ε↓0 νB ε = lim ε↓0 ν ˙ B ε = νB. 7.16 To verify 7.16 let ε ↓ 0 and use that, obviously, ν∂ B = 0. Then νB ε ↓ νclB = νB because B ε ↓ clB, and ν˙B ε ↑ ν˙B = νB because ˙B ε ↑ ˙B. The proof of 7.14 and that of Theorem 16 is now completed. ƒ 1963 8 On a possible proof of Conjecture 18 In this section we provide some discussion for the reader familiar with [4] and interested in a possible way of proving Conjecture 18. The main difference relative to the attractive case is that, as we have mentioned earlier, in that case one does not need the spine change of measure from [4]. In the repulsive case however, one cannot bypass the spine change of measure. Essentially, an h-transform transforms the outward Ornstein- Uhlenbeck process into an inward Ornstein-Uhlenbeck process, and in the exponential branching clock setting and with independent particles, this inward Ornstein-Uhlenbeck process becomes the ‘spine.’ A possible way of proving Conjecture 18 would be to try to adapt the spine change of measure to unit time branching and dependent particles. 9 The proof of Lemma 25 and that of 7.12

9.1 Proof of Lemma 25