above construction of the dual we reset locations µR
1
+ e
i
to ∆ if they are outside I
′
. If ξ
N s
∆ = 0, the reconstruction of
ξ
N t
x
i
from the dual variables is again valid as is the reasoning which led to 25. More specifically if z
1
= 1, z
2
, z
3
∈ {0, 1}, x
i
∈ I
′
, for i = 1, 2, 3 and we define E
x
i i
≤3
t
:= ¦
∃i ≤ 3 there is at least one coloured arrow encountered by ˆB
x
i
on [0, t] ©
, then
E
3
Y
i=1
1
{ξ
N t
x
i
=z
i
}
− E
3
Y
i=1
1
{ξ
N
ˆ B
xi t
=z
i
}
≤ E
1
E
x1,x2,x3 t
X
y ∈˜B
x1 t
ξ
N
y
. 27
2.3 Martingale problem
By Proposition 3.1 of [7], we have, for any Φ ∈ C
b
[0, T ] ×S
N
such that
Φ :=
∂ Φ ∂ t
∈ C
b
[0, T ] ×S
N
, X
N t
Φt, .= X Φ0, .+ M
N t
Φ+ D
N ,1 t
Φ+ D
N ,2 t
Φ+ D
N ,3 t
Φ, 28
where the reader should be warned that the terms D
N ,2
and D
N ,3
below do not agree with the corresponding terms in [7] although their sums do agree
D
N ,1 t
Φ = Z
t
X
N s
A
N
Φs, .+
Φs, .ds, A
N
Φs, x := X
y ∈S
N
N p
N
y − x Φs, y − Φs, x ,
D
N ,2 t
Φ := Z
t
log N
2
N X
x ∈S
N
Φs, x
β
N
1 − ξ
N s
x f
N 1
x, ξ
N s
2
− β
N 1
ξ
N s
x f
N
x, ξ
N s
2
ds,
D
N ,3 t
Φ := Z
t
log N
4
N X
x ∈S
N
Φs, x
ξ
N s
x f
N
x, ξ
N s
2
− 1 − ξ
N s
x f
N 1
x, ξ
N s
2
ds,
29 and M
N t
Φ is an F
X
N
t
, L
2
-martingale such that 〈M
N
Φ〉
t
= 〈M
N
Φ〉
1,t
+ 〈M
N
Φ〉
2,t
, 30
with 〈M
N
Φ〉
1,t
:= log N
2
N Z
t
X
x ∈S
N
Φs, x
2
X
y ∈S
N
p
N
y − xξ
N s
y − ξ
N s
x
2
ds, 31
1204
and 〈M
N
Φ〉
2,t
:= log N
2
N Z
t
X
x ∈S
N
Φs, x
2
α
N
− 11 − ξ
N s
x f
N 1
x, ξ
N s
2
+ α
N 1
− 1ξ
N s
x f
N
x, ξ
N s
2
ds. 32 Proposition 3.1 of [7] erroneously had a prefactor of
log N
2
N
2
on the final term. The error is insignifi- cant as the
α
N i
− 1 terms in the above are still small enough to make this term approach 0 in L
1
as N
→ ∞ both here and in [7]. Comparing 28 with the martingale problem MP for the super-Brownian motion limit, we see that
to prove Theorem 1.5 we will need to establish the tightness of the laws of {X
N
} on DR
+
, M
F
R
2
, with all limit points continuous, and if X
N
k
→ X weakly, then as N = N
k
→ ∞,
E h
D
N ,1 t
Φ − Z
t
X
N s
σ
2
2 ∆Φ
s
+ ˙ Φ
s
ds i
→ 0, 33
E h
D
N ,2 t
Φ − γβ − β
1
Z
t
X
N s
Φds i
→ 0, 34
E h
D
N ,3 t
Φ − K Z
t
X
N s
Φds i
→ 0, 35
E h
〈M
N
Φ〉
1,t
− 4πσ
2
Z
t
X
N s
Φ
2
ds i
→ 0, 36
E h
〈M
N
Φ〉
2,t
i → 0.
37 Controlling the five terms in 28 in this manner will also be the main ingredients in establishing
the above tightness. As already noted it is the presence of the higher powers of log N in D
N ,3
which will make 35 particularly tricky.
We now give a brief outline of the proof of this particular result which is established in Section 7 although many of the key ingredients are given in earlier Sections and relies on the moment
estimates in Sections 4 and 5. The remaining convergences also treated in Section 7 are closer to the arguments in [7] although their proofs are also complicated by the greater values of
|α
N i
− 1|. In fact some of the new ingredients introduced here would have significantly simplified some of those
proofs. Although we will be able to drop the time dependence in Φ for the convergence proof itself, it will be important for our moment bounds to keep this dependence.
If D
N ,3 t
Φ = R
t
d
N ,3 s
Φds, then a simple L
2
argument see 67 in Section 5 will allow us to approximate D
N ,3 t
Φ with R
t t
N
Ed
N ,3 s
Φ|F
s −t
N
ds recall that t
N
= log N
−19
. So we need to
1205
estimate Ed
N ,3 s
Φ|F
s −t
N
ds =
log N
4
N X
x ∈S
N
Φs, xEξ
N s
x f
N
x, ξ
N s
2
− 1 − ξ
N s
x f
N 1
x, ξ
N s
2
|F
s −t
N
= log N
4
N X
x ∈S
N
Φs, xEξ
N s
x
2
Y
1
1 − ξ
N s
x + e
i
− 1 − ξ
N s
x
2
Y
1
ξ
N s
x + e
i
|F
s −t
N
, where e
1
, e
2
are chosen independently according to p
N
·. To proceed carefully we will need to use the dual ˜
B of ξ
N
but note that on the short interval [s − t
N
, s] we do not expect to find any red or green arrows since log N
3
t
N
≪ 1, and so we can use the voter dual ˆB and the Markov property to calculate the above conditional expectation see Lemma 3.6. This leads to the estimate here ˆ
E only integrates out the dual and e
1
, e
2
Ed
N ,3 s
Φ|F
s −t
N
≈ log N
4
N X
x ∈S
N
Φs, x ˆ E
ξ
N s
−t
N
ˆ B
x t
N
2
Y
1
1 − ξ
N s
−t
N
ˆ B
x+e
i
t
N
38 − 1 − ξ
N s
−t
N
ˆ B
x t
N
2
Y
1
ξ
N s
−t
N
ˆ B
x+e
i
t
N
. There is no contribution to the above expectation if ˆ
B
x
coalesces with ˆ B
x+e
1
or ˆ B
x+e
2
on [0, t
N
]. On the event where ˆ
B
x+e
1
t
N
= ˆ B
x+e
2
t
N
, the integrand becomes ξ
N s
−t
N
ˆ B
x t
N
1 − ξ
N s
−t
N
ˆ B
x+e
1
t
N
− 1 − ξ
N s
−t
N
ˆ B
x t
N
ξ
N s
−t
N
ˆ B
x+e
1
t
N
39 = ξ
N s
−t
N
ˆ B
x t
N
− ξ
N s
−t
N
ˆ B
x+e
1
t
N
, thanks to an elementary and crucial cancellation. Summing by parts and using the regularity of
Φs, ·, one easily sees that the contribution to Ed
N ,3 s
Φ|F
s −t
N
from this term is negligible in the limit. Consider next the contribution from the remaining event
{x | x + e
1
| x + e
2
} on which there is no coalescing of
{B
x
, ˆ B
x+e
1
, ˆ B
x+e
2
} on [0, t
N
]. Recall that the density of 1’s in ξ
N s
−t
N
is O1 log N
and so we expect the contribution from the second term in 38, requiring 1’s at both distinct sites ˆ
B
x+e
i
t
N
, i = 1, 2, to be negligible in the limit. The same reasoning shows the product in the first term in 38 should be close to 1 and so we expect
Ed
N ,3 s
Φ|F
s −t
N
≈ log N
4
N X
x ∈S
N
Φs, x ˆ E
ξ
N s
−t
N
ˆ B
x t
N
1{x | x + e
1
| x + e
2
} ≈
log N N
X
x ∈S
N
Φs, xξ
N s
−t
N
xlog N
3
ˆ P
{0 | e
1
| e
2
} ≈X
N s
−t
N
Φ
s −t
N
K, where the second line is easy to justify by summing over the values of ˆ
B
x t
N
, and the last line is immediate from 6. Integrating over s we arrive at 35. The key ingredient needed to justify the
above heuristic simplification based on the sparseness of 1’s is Proposition 3.10 whose lengthy proof is in Section 5.
1206
3 Intermediate results for the proof of Theorem 1.5
We use the classical strategy of proving tightness of {X
N
} in the space DR
+
, M
F
R
2
, and then identify the limits. The main difficulty will come from the above drift term D
N ,3 t
Φ. Although it will be convenient to have Theorem 1.5 for a continuous parameter N
≥ 3, it suffices to prove it for an arbitrary sequence approaching infinity and nothing will be lost by considering N
∈ N
≥3
. This condition will be in force thoughout the proof of Theorem 1.5, as will the assumption that
ξ
N
is deterministic and all the conditions of Theorem 1.5.
3.1 Tightness, moment bounds.
