Proof of Lemma 8.1. getdoc7e8b. 765KB Jun 04 2011 12:04:34 AM

Remark: The parameters β N , β N 1 are now defined by N , α , α 1 and 11. Indeed our choice of N = N α in the above implies β N = 0 and β N 1 = N log N α 1 − α ∈ – 0, K − η γ log N 2 log N − 3loglog N 2 ™ . Therefore, for r η small enough we will have β N 1 ∈ 0, K − η2 γ . Proof : Given Lemma 8.1, Proposition 8.2 is proved by making trivial changes to the proof of Proposition 9.2 in [7]. We omit the details but the intuition should be clear. By Theorem 1.5, X N converges to a super-Brownian motion with drift θ = K − γβ 1 . By the above remark this quantity is bounded below by η2 0 for a good choice of r. Super-Brownian motion with positive drift and large enough initial mass will continue to grow exponentially up to time T with high probability, and so the same should be true for X N for N large. Finally, Lemma 8.1 bounds X N T − X N T and allows us to make the same conclusion for X N T . ƒ To complete the proof of survival it remains to establish Lemma 8.1.

8.2 Proof of Lemma 8.1.

Choose h : R 2 ∪ ∆ → [0, 1] such that [−K L + 3, K L − 3] 2 ⊆ {h = 1} ⊆ Supph ⊆ [−K L + 2, K L − 2] 2 , |h| Lip ≤ 1. 120 We then define for s ≤ t, x ∈ S N the function Ψs, x := P N t −s hx = E[hB x t −s ] = E[hB x t−s∧T ′ x ]. As in the proof of Lemma 3.2 of [6], if f : S N ∪ ∆ → R and Φs, x = P N t −s f x, s ≤ t we have X N t f = X N P N t f + M N t Φ + Z t d N ,2 s Φ, ξ N s + d N ,3 s Φ, ξ N s ds, 121 where M N t Φ is a square-integrable, mean zero martingale. Apply 28 and 121 to obtain E ” X N t 1 − X N t 1 — ≤ E ” X N t 1 − X N t h — = X N 1 − P N t h + E   Z t 3 X i=2 d N ,i s 1, ξ N s − d N ,i s Ψ, ξ N s ds   . 122 We now state and prove four intermediate results. Lemma 8.3. Let {S n } be a mean zero random walk on Z starting at x ∈ Z under P x and such that E |S 1 | 3 ∞. There is a C 8.3 0 such that if U M = inf{n ≥ 0 : S n ∈ 0, M}, M ∈ N, then P x S U M ≥ M ≤ x + C 8.3 M for all x ∈ Z ∩ [0, M]. 1241 Proof. Let p j = P |S 1 | = j. The inequality in the middle of p. 255 of [14] and the inequality P7 on p. 253 of the same reference imply that for x as in the statement, and some c 0, P x S U M ≥ M − x M ≤ c M h M X s=0 1 + sE |S 1 |1 {|S 1 |s} i = c M h M +1 X j=1 X k ≥ j jkpk i ≤ c M h ∞ X k=1 k 2 k + 1 2 pk i ≤ c M E |S 1 | 3 . The result follows. To use Lemma 3.5 and 61, we need the following. Lemma 8.4. There exists a constant C 8.4 such that |Ψ| 1 2 ≤ p 2 σ, |Ψ| Lip ≤ C 8.4 . Proof : The first statement is easy to verify. Indeed, if s s ′ ≤ t, x ∈ S N , then using |h| Lip ≤ 1, applying the Markov property to B x at time t − s ′ and then Cauchy-Schwarz, Ψs ′ , x − Ψs, x ≤ E • B x t−s ′ ∧T ′ x − B x t−s∧T ′ x 1 {T ′ x t−s ′ } ˜ ≤ sup z ∈I ′ ∩S N E B z s ′ −s∧T ′ z − z 2 1 2 ≤ p 2 σ|s ′ − s| 1 2 . Let us now turn to the proof of the second statement of Lemma 8.4. Let x, x ′ ∈ I ′ ∩ S N be such that |x − x ′ | ≤ 1. In this argument we couple the walk B x ′ with B x by setting B x ′ s := x ′ − x + B x s . We have Ψt − s, x − Ψt − s, x ′ = E hB x s ∧T ′ x − hB x ′ s ∧T ′ x′ ≤ E h 1 {T ′ x ∧T ′ x′ ≥s} |hB x s − hB x ′ s | i + E h 1 {T ′ x s≤T ′ x′ } hB x ′ s + 1 {T ′ x′ s≤T ′ x } hB x s i ≤ |x − x ′ | + E h 1 {T ′ x s≤T ′ x′ } hB x ′ s i + E h 1 {T ′ x′ s≤T ′ x } hB x s i , 123 where we used |h| Lip ≤ 1 to bound the first term in the sum. The other two terms are symmetric, so we may as well only handle the first of the two. On the event {T ′ x s ≤ T ′ x ′ }, since B x ′ T ′ x − B x T ′ x = |x ′ − x| ≤ 1, it is easy to see that B x ′ T ′ x ∈ I ′ , inf z ∈I ′ c B x ′ T ′ x − z ≤ |x − x ′ |. So one of the components, say i, of B x ′ T ′ x is within |x − x ′ | of −K L, K L c . Let us assume without loss of generality by symmetry that B x ′ ,i T ′ x ∈ [K L − |x − x ′ |, K L], and let M u = B x ′ ,i T ′ x +u . From 120, 1242 for hB x ′ s to be positive we need that M hits −∞, K L − 2] before [K L, ∞. Therefore if we set ν := inf{u ≥ 0 : M u ∈ K L − 2, K L } and S n is the embedded discrete time rescaled random walk on Z, then E h 1 {T ′ x s≤T ′ x′ } hB x ′ s i ≤ P M ν ≤ K L − 2|M ≥ K L − |x − x ′ | ≤ P ⌈ p N |x−x ′ |⌉ S U ⌈2 p N ⌉ ≥ ⌈2 p N ⌉ ≤ C 8.3 p N + |x − x ′ | ≤ C 8.3 + 1|x ′ − x|, 124 where we used Lemma 8.3 and then |x ′ − x| ≥ N −12 in the last line. Therefore by 123 and 124 we deduce that for any x, x ′ ∈ S N ∩ I ′ such that |x − x ′ | ≤ 1 we have Ψs, x − Ψs, x ′ ≤ 2C 8.3 + 3|x − x ′ |. Since Ψs, . is supported on I ′ the second assertion of Lemma 8.4 follows. ƒ The next intermediate result we need for proving Lemma 8.1 is a version of Lemma 3.6 for the killed process. Let ˆ H ξ N s −t N , x, t N := ˆ E   ξ N s −t N ˆ B x t N 2 Y i=1 1 − ξ N s −t N ˆ B x+e i t N   − ˆE   1 − ξ N s −t N ˆ B x t N 2 Y i=1 ξ N s −t N ˆ B x+e i t N   . Lemma 8.5. There is a C 8.5 such that for any s ≥ t N and Φ : R 2 → R bounded measurable, E h d N ,3 s Φ, ξ N s | F s −t N i − log N 4 N X x ∈S N Φs − t N , x ˆ H ξ N s −t N , x, t N ≤ C 8.5 ||Φ|| 1 2 X N s −t N 1log N −6 . Proof : This is the same as the proof of Lemma 3.6, only we use 27 in place of 25. ƒ One can similarly obtain the obvious analogue of 55 for the killed process. We leave details to the reader. We will finally need an alternative form of Lemma 3.5. It is a cruder estimate but gives better bounds for test functions which are small in an L 1 sense. Let log + x := log1 ∨ x. Lemma 8.6. Assume f : S N → [0, 1] and | f | 1 = 1 N P x ∈S N f x ∞. There exist constants c 8.6 and C 8.6 , depending on β, such that E[X N t f ] ≤ X N P N t f + C 8.6 expc 8.6 tX N 1log N 3 | f | 1 1 + log + 1 | f | 1 . Proof : Let Φs, x = P N t −s f x and use 28 to get E[X N t f ] ≤ X N P N t f + E   3 X i=2 |D N ,i t Φ|   . 125 1243 Moreover, E   3 X i=2 |D N ,i t Φ|   ≤ E   Z t log N 2 N   X x ∈S N Φs, xlog N 2 + βξ N s x + X e ∈S N p N eξ N s x + e     ≤ 1 + βE –Z t log N 3 € X N s P N t −s f + X N s p N ∗ P N t −s f Š ™ . By A7 of [3], there exists C 126 such that P N t −s f x ≤   X y ∈S N C 126 1 + N t − s −1 f y   ∧ 1 ≤ € C 126 t − s −1 | f | 1 Š ∧ 1. 126 Therefore, E   3 X i=2 |D N ,i t Φ|   ≤ C 126 2 + 2 βlog N 3 Z t € [t − s −1 | f | 1 ] ∧ 1 Š E[X N s 1]ds ≤ C 126 2 + 2β1 + C a log N −16 expc 3.4 tlog N 3 X N 1   Z t | f | 1 ∧t | f | 1 u du + | f | 1   , where we used Proposition 3.4 a at the last line. Lemma 8.6 now easily follows from 125 and the above. ƒ We now turn to the actual proof of Lemma 8.1. By 122, E ” X N t 1 − X N t 1 — ≤ X N 1 − P N t h + E   Z t N ∧t 3 X i=2 d N ,i s 1, ξ N s − d N ,i s Ψ, ξ N s   ds +E   3 X i=2 Z t t N ∧t d N ,i s 1 − Ψ, ξ N s ds   + E   Z t t N ∧t d N ,2 s Ψ, ξ N s − d N ,2 s Ψ, ξ N s ds   +E   Z t t N ∧t d N ,3 s Ψ, ξ N s − d N ,3 s Ψ, ξ N s ds   =: U + U 1 + U 2 + U 3 + U 4 . 127 Below we establish bounds on each of the above terms, and |.| ∞ denotes the L ∞ -norm on R 2 . We claim there is a constant C 129 , depending on T , and a δ 130 0 such that for any t ≤ T , U ≤ X N 1Psup u ≤t |B u | ∞ K − 1L − 3, 128 U 1 ≤ C 129 X N 1log N 3 t N , 129 1244 U 2 ≤ C 129 X N 1 + X N 1 2 ‚ log N −δ 130 + Psup u ≤t |B 0,N u | ∞ K − 1L − 3 Œ , 130 U 3 ≤ C 129 ‚ log N −δ 130 X N 1 + X N 1 2 + Z t E[X N s 1 − X N s 1]ds Œ , 131 U 4 ≤ C 129 ‚ log N −δ 130 X N 1 + X N 1 2 + Z t E[X N s 1 − X N s 1]ds Œ . 132 We start with a bound which will be useful for proving both 128 and 130. We have ≤ 1 − Ψs, y = 1 − P N t −s h y ≤ PT ′ y ≤ t − s + PT ′ y t − s, |B y t −s | ∞ K L − 3 ≤ P sup u ≤t−s |B y u | ∞ K L − 3. 133 Since we assumed suppX N ⊂ I, it follows that for any s ≥ 0 X N € P N s 1 − Ψs, . Š ≤ X N 1Psup u ≤t |B u | ∞ K − 1L − 3 134 Using 134 for s = 0 we obtain U = X N 1 − Ψ0, . ≤ X N 1Psup u ≤t |B u | ∞ K − 1L − 3, which is 128. We next show 129. Using 54 for i = 2, 41 for i = 3, and then Proposition 3.4 a, we obtain |U 1 | ≤ 21 ∨ βE –Z t N ∧t log N 3 + log N X N s 1ds ™ ≤ C 129 X N 1log N 3 t N , and we are done. Here the fact that ξ N ≤ ξ N means the above bounds trivially apply to ξ N as well. Let us turn to the proof of 130. Use 61 for i = 2, Lemma 3.9 for i = 3 to get for suitable δ, η 0 and ¯ K as in 16, |U 2 | ≤ 2β C 15 + ¯ K Z t−t N + E[X N s 1 − Ψs, .]ds +C 3.9 + C 61 ||1 − Ψ|| Lip log N −δ Z t−t N + E[X N s 1]ds + C 3.7 + C 61 ||1 − Ψ|| ∞ t N log N E   Z t ∨t N t N I N η s − t N ds   . 135 1245 Use Lemma 3.5 to bound the first term in 135 : Z t−t N + E[X N s 1 − Ψs, .]ds ≤ Z t−t N + expc 3.5 sX N P N s 1 − Ψs, . + C 3.5 log N −δ 3.5 X N 1 + X N 1 2 ds ≤ Z t−t N + expc 3.5 sX N 1Psup u ≤t |B 0,N u | ∞ K − 1L − 3 +C 3.5 log N −δ 3.5 X N 1 + X N 1 2 ds, where, to obtain the last inequality, we applied 134. Inequality 130 then follows by using the above, Proposition 3.4 a, Corollary 3.11 and Lemma 8.4 in 135. We now turn to the proof of the critical 132, and leave the proof of 131 to the reader, as it uses a similar method. Use the fact that X N s ≤ X N s with Lemmas 3.9 and 8.5 to get |U 4 | ≤ € C 3.9 log N −δ 3.7 + C 8.5 log N −6 Š ||Ψ|| Z t−t N + EX N s 1ds + C 3.7 ||Ψ|| ∞ t N log N Z t−t N + E[ I N η 3.7 s]ds +log N 3 P {0 | e 1 | e 2 } t N Z t−t N + EX N s Ψs, . − X N s Ψs, .ds + log N 4 N E h X x ∈S N Z t−t N + Ψs, x ξ N s xP{0 | e 1 | e 2 } t N − ˆ H N ξ N s , x, t N ds i . Using the above, Lemma 8.4, Proposition 3.4 a and Corollary 3.11 together with X N ≤ X N , we deduce there exists a C 136 and δ 136 0 such that |U 4 | ≤ C 136 – log N −δ 136 X N 1 + X N 1 2 + ¯ K Z t E[X N s 1 − X N s 1]ds + E   log N 4 N X x ∈S N Z t−t N + Ψs, x ˆ H N ξ N s , x, t N − ˆ H N ξ N s , x, t N ds   + E   log N 4 N X x ∈S N Z t−t N + Ψs, x ξ N s xP{0 | e 1 | e 2 } t N − ˆ H N ξ N s , x, t N ds   ™ .136 By 47 in Remark 3.8 with ξ N = ξ N s , along with the fact that X N ≤ X N , the last term in the sum in 1246 136 is E   log N 4 N X x ∈S N Z t−t N + Ψs, x ξ N s xP{0 | e 1 | e 2 } t N − ˆ H N ξ N s , x, t N ds   ≤ C 3.8 log N −δ 3.8 ||Ψ||E h Z t X N s 1ds i + C 3.8 ||Ψ|| t N log N E –Z t I N η 3.7 sds ™ 137 ≤ C 137 X N 1 + X N 1 2 log N −δ 3.8 ∧η 3.7 2 , 138 where we used Proposition 3.4 a, Corollary 3.11 and Lemma 8.4 to get the second inequality above. Let us now turn to bound the next-to-last term in the sum in 136. For δ ≥ 0 we let A δ := {x ∈ S N ∩ I ′ : inf y ∈I ′ c |x − y| ≤ δ}, A δ ′ := {x ∈ S N ∩ I ′ : inf y ∈I ′ c |x − y| δ} For x ∈ A t 1 3 N , it is enough to use the straightforward bound ˆ H N ξ N s , x, t N − ˆ H N ξ N s , x, t N ≤ 2 ˆEξ N s ˆ B x t N + ξ N s ˆ B x+e 1 t N to obtain E     log N 4 N X x ∈A t 1 3 N Z t−t N + Ψs, x ˆ H N ξ N s , x, t N − ˆ H N ξ N s , x, t N ds     ≤ 2E     Z t log N 4 N X x ∈A t 1 3 N X w ∈S N ξ N s w ˆ P ˆ B t N = w − x + ˆPˆB e 1 t N = w − x ds     ≤ 2E – Z t log N 4 N X w ∈A 2t 1 3 N c ξ N s w ˆ P |ˆB t N | ∞ t 1 3 N + P|ˆB e 1 t N | ∞ t 1 3 N + X w ∈A 2t 1 3 N 2 ξ N s w ™ ≤ C 139 E – log N 3 Z t X N s 1N −1 + t N t −23 N + X N s A 2t 1 3 N ds ™ , 139 where we used X N ≤ X N and Chebychev’s inequality to obtain the last line above. Use Lemma 8.6, then the fact that suppX N ⊂ I and the bound 17 to get for any t ≤ T , E – log N 3 Z t X N s A 2t 1 3 N ds ™ ≤ CT log N 3 Z t X N PB … s ∈ A 2t 1 3 N + log N 3 X N 1t 1 3 N [1 + log + t −13 N ] ds ≤ CT X N 1log N 3 Z t t 1 3 N s ∧ PB s K − 1L − 2t 1 3 N ds +tlog N 6 X N 1t 1 3 N 1 + 7 loglog N ≤ C 140 X N 1 log N 3 Z t t 1 3 N s ∧ s ds + log N −16 , 140 1247 for some C 140 depending on T . We then use 140 in 139 to deduce E     log N 4 N X x ∈A t 1 3 N Z t−t N + Ψs, x ˆ H N ξ N s , x, t N − ˆ H N ξ N s , x, t N ds     ≤ C 141 X N 1log N −16 141 For x ∈ A t 1 3 N ′ , we claim that ˆ H N ξ N s , x, t N − ˆ H N ξ N s , x, t N ≤ 2 X i=0 ˆ E – ξ N s ˆ B x+e i t N 1 ¨ sup u ≤t N |ˆB x+e i u − x| t 1 3 N «™ 142 First the obvious coupling shows that for i = 0, 1, 2, ξ N s ˆ B x+e i t N ≤ ξ N s ˆ B x+e i t N . Furthermore, if the left side of 142 is non zero, we must have 0 = ξ N s ˆ B x+e i t N ξ N s ˆ B x+e i t N = 1 for some i ∈ {0, 1, 2}. For this choice of i we must have ˆ B x+e i u ∈ I ′ for some u ≤ t N , and since we supposed x ∈ A t 1 3 N ′ this implies sup u ≤t N |ˆB x+e i u − x| t 1 3 N . As the left side of 142 is at most 1, 142 is proved. Now use 142, then Proposition 3.4a, and finally the weak L 1 -inequality for submartingales to obtain for t ≤ T , E     log N 4 N X x ∈A t 1 3 N ′ Z t−t N + Ψs, x ˆ H N ξ N s , x, t N − ˆ H N ξ N s , x, t N ds     ≤ E     log N 4 N Z t X w ∈S N X x ∈A t 1 3 N ′ 2 X i=0 ξ N s wˆ P ˆ B e i t N = w − x, sup u ≤t N |ˆB e i u | t 1 3 N ds     ≤ C ′ a t X N 1log N 3 2 X i=0 ˆ P sup u ≤t N |ˆB e i u | 2 t 2 3 N ≤ C 143 X N 1log N −3 . 143 We finally combine 136, 137, 141, and 143 to deduce 132. Finally, use 127—132 to conclude E ” X N t 1 − X N t 1 — ≤ C 144 X N 1 + X N 1 2 Psup u ≤t |B 0,N t | ∞ K − 1L − 3 +log N −δ 130 ∧16 + Z t E ” X N s 1 − X N s 1 — ds , 144 and therefore Gronwall’s lemma completes the proof of Lemma 8.1. ƒ

8.3 Proof of coexistence, Theorem 1.2.