5.2 Proof of Proposition 3.10
Recall that p
N t
x = N P ˆ B
N ,0 t
= x and set p
N ,z t
x := p
N t
z − x. By assumption Nδ
N
→ ∞ as N
→ ∞ so we can use the local limit theorem for the walk ˆB
N
see e.g., Lemma 7.3 c of [7] to obtain the existence of a constant C
71
such that
1
{|x− y|≤
p
δ
N
}
≤ C
71
δ
N
p
N 2
δ
N
y − x. 71
Therefore the desired result will follow if we can bound E
Z
R
2
δ
N
p
N 2
δ
N
y − xdX
N t
xd X
N t
y
as in the Proposition. By Chapman-Kolmogorov we have δ
N
N X
z ∈S
N
E X
N t
p
N ,z δ
N
2
= E
log N
2
N
2
X
x, y ∈S
N
ξ
N t
xξ
N t
yδ
N
1 N
X
z ∈S
N
p
N δ
N
z − xp
N δ
N
z − y
= E
log N
2
N
2
X
x, y ∈S
N
ξ
N t
xξ
N t
yδ
N
p
N 2
δ
N
y − x
= E
Z
R
2
δ
N
p
N 2
δ
N
y − xdX
N t
xd X
N t
y
. 72
Set φ
z s
= p
N ,z t
−s+δ
N
, which satisfies A
N
φ
z s
+ ˙ φ
z s
= 0, so that, from 28, we deduce that E
X
N t
p
N ,z δ
N
2
≤ 4
E[X
N
φ
z 2
] + 〈Mφ
z
〉
t
+ D
N ,2 t
φ
z 2
+ D
N ,3 t
φ
z 2
.
Using 72 and the above, then again Chapman-Kolmogorov, we obtain the bound: E
Z
R
2
δ
N
p
N 2
δ
N
y − xdX
N t
xd X
N t
y
≤ 4[T + T
1
+ T
2
+ T
3
+ T
4
],
1221
where T
:= δ
N
N X
z ∈S
N
E X
N
p
N ,z t+
δ
N
2
= E Z
R
2
δ
N
p
N 2t+
δ
N
y − xdX
N
xd X
N
y
, T
1
:= δ
N
N X
z ∈S
N
E[ 〈Mφ
z
〉
1,t
] = E
δ
N
log N
2
N Z
t
p
N 2t
−s+δ
N
X
x, y ∈S
N
p
N
y − xξ
N s
x − ξ
N s
y
2
ds
,
T
2
:= δ
N
N X
z ∈S
N
E[ 〈Mφ
z
〉
2,t
], T
3
:= E δ
N
log N
4
N
2
Z
t
Z
t
ds
1
ds
2
X
x
1
,x
2
∈S
N
p
N 2t
−s
1
−s
2
+2δ
N
x
2
− x
1
×
2
Y
i=1
β
N
1 − ξ
N s
i
x
i
f
N 1
x
i
, ξ
N s
i
2
− β
N 1
ξ
N s
i
x
i
f
N
x
i
, ξ
N s
i
2
, T
4
:= E δ
N
log N
8
N
2
Z
t
Z
t
ds
1
ds
2
X
x
1
,x
2
∈S
N
p
N 2t
−s
1
−s
2
+2δ
N
x
2
− x
1
×
2
Y
i=1
ξ
N s
i
x
i
f
N
x
i
, ξ
N s
i
2
− 1 − ξ
N s
i
x
i
f
N 1
x
i
, ξ
N s
i
2
. We will handle each of these terms separately. Using 17, we immediately obtain
T ≤
C
17
δ
N
t + δ
N
X
N
1
2
. 73
We then consider T
1
which will turn out to be the main contribution, although not the most difficult to handle. As we already did before, we will condition back a bit in order to be able to use the voter
estimates. This time however, we will need to condition back by u
N
:=
δ
N
2
∧ log N
−11
. Let ∆
N
s, x := E
X
e ∈S
N
p
N
eξ
N s
x1 − ξ
N s
x + e F
s −u
N
− ˆE
X
e ∈S
N
p
N
eξ
N s
−u
N
ˆ B
x u
N
1 − ξ
N s
−u
N
ˆ B
x+e u
N
. By the obvious analogues of 25 and Lemma 2.2,
log N N
X
x ∈S
N
∆
N
s, x ≤ log N
N X
x ∈S
N
ˆ E
1
E
x,x+e uN
X
y ∈˜B
x uN
ξ
N s
−u
N
y
≤ 4u
N
r
N
exp2u
N
r
N
X
N s
−u
N
1 ≤ C
74
log N
−8
X
N s
−u
N
1. 74
1222
Moreover, from the fact that ξ
N s
x ∈ {0, 1}, X
x,e ∈S
N
p
N
eξ
N s
x − ξ
N s
x + e
2
= 2 X
x,e ∈S
N
p
N
eξ
N s
x1 − ξ
N s
x + e ≤ 2 N
log N X
N s
1. Therefore, using 17,
T
1
≤ C
17
δ
N
log N E
Z
t
t − s + δ
N −1
log N N
2 X
x ∈S
N
,e ∈S
N
p
N
eξ
N s
x1 − ξ
N s
x + eds
≤ 2C
17
δ
N
log N E Z
u
N
t − s + δ
N −1
X
N s
1ds
+2C
17
δ
N
log N E
Z
t ∨u
N
u
N
t − s + δ
N −1
log N N
X
x ∈S
N
ˆ E
h ξ
N s
−u
N
ˆ B
x u
N
1 − ξ
N s
−u
N
ˆ B
x+e u
N
i ds
+2C
17
C
74
δ
N
log N
−7
E
Z
t ∨u
N
u
N
t − s + δ
N −1
X
N s
−u
N
1ds
=: T
1,1
+ T
1,2
+ T
1,3
, 75
where we used 74 for s ≥ u
N
in the last inequality. Then using Proposition 3.4 a, we obtain T
1,1
≤ C
17
C
′
a
log N u
N
X
N
1, and as we will see this term is negligible compared to the bound we will obtain for
T
1,2
. Notice that
ξ
N s
−u
N
ˆ B
x u
N
1 − ξ
N s
−u
N
ˆ B
x+e u
N
≤ X
w ∈S
N
ξ
N s
−u
N
w1
{ˆB
x uN
=w}
1
{x|x+e}
uN
. Thus, by translation invariance of our kernel p
N
, T
1,2
≤ 2C
17
δ
N
log N E
Z
t ∨u
N
u
N
t − s + δ
N −1
X
N s
−u
N
1 X
y ∈S
N
ˆ P
ˆ B
u
N
= y, {0 | e}
u
N
≤ C
76
δ
N
log 1 +
T δ
N
X
N
1, 76
where the above line was obtained using Proposition 3.4 a and 13. Finally, using once again Proposition 3.4 a,
T
1,3
≤ 2C
17
C
74
C
′
a
δ
N
log N
−7
log 1 +
T δ
N
X
N
1, so that this term is also negligible compared to
T
1,2
. From 75 and the above estimates we deduce T
1
≤ C
77
δ
N
log 1 +
T δ
N
X
N
1. 77
We now turn to bound T
2
. By summing over z in 32, and then using 17 and Lemma 5.1 a, we easily see that
T
2
≤ C
5.1
C
17
log N
4
N
2
Z
t
X
N s
1 1
t − s + δ
N
ds 1223
Therefore, using Proposition 3.4 a we obtain T
2
≤ C
78
log N
4
N
2
log 1 +
T δ
N
X
N
1, 78
which is negligible compared to the right-hand side of 77 as N → ∞, by our assumption that
lim inf
N →∞
p N
δ
N
0. We turn to bound
T
4
, which comes from the new drift term. We have T
4
≤ 2 δ
N
N X
z ∈S
N
E
Z
t ∨u
N
u
N
d
N ,3 s
φ
z s
− E[d
N ,3 s
φ
z s
| F
s −u
N
]
ds
2
+2
δ
N
N X
z ∈S
N
E
Z
t ∨u
N
u
N
E[d
N ,3 s
φ
z s
| F
s −u
N
] ds
2
+2E
Z
u
N
Z
s
1
δ
N
log N
8
N
2
X
x
1
∈S
N
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
×
2
Y
i=1
ds
i
h ξ
N s
i
x
i
f
N
x
i
, ξ
N s
i
2
− 1 − ξ
N s
i
x
i
f
N 1
x
i
, ξ
N s
i
2
i =: T
4,1
+ T
4,2
+ T
4,3
. We first handle
T
4,1
. Note that if s
2
− u
N
s
1
, E
2
Y
i=1
d
N ,3 s
i
φ
z s
i
− E h
d
N ,3 s
i
φ
z s
i
| F
s
i
−u
N
i
= 0,
therefore T
4,1
= 4 δ
N
N X
z ∈S
N
E h
Z
t ∨u
N
u
N
Z
s
1
+u
N
s
1
ds
1
ds
2 2
Y
i=1
d
N ,3 s
i
φ
z s
i
− E h
d
N ,3 s
i
φ
z s
i
| F
s
i
−u
N
ii .
We have the evident bound on d
N ,3
: |d
N ,3 s
φ
z s
| ≤ log N
4
N X
x ∈S
N
φ
z s
xΞ
N s
x. 79
where we wrote Ξ
N s
x := ξ
N s
x + P
e ∈S
N
p
N
eξ
N s
x + e. Therefore, using Chapman-Kolmogorov once again, and then 17, we find
T
4,1
≤ 4C
17
δ
N
E Z
t ∨u
N
u
N
Z
s
1
+u
N
s
1
log N
8
N
2
X
x
1
,x
2
2t + δ
N
− s
1
− s
2 −1
×
2
Y
i=1
Ξ
N s
i
x
i
+ E h
Ξ
N s
i
x
i
| F
s
i
−u
N
i ds
1
ds
2
. 80
1224
Using the Markov property and Proposition 3.4 a it comes easily that log N
N X
x
i
∈S
N
E h
Ξ
N s
i
x
i
| F
s
i
−u
N
i ≤ 2C
′
a
X
N s
i
−u
N
1. We may therefore expand the product in 80 and bound each of the terms using the Markov prop-
erty and Proposition 3.4. It follows that there exist constants C
81
, C
′
81
depending on T such that, if t
≥ u
N
, T
4,1
≤ C
81
δ
N
log N
6
X
N
1 + X
N
1
2
Z
t u
N
Z
s
1
+u
N
s
1
ds
1
ds
2
2t + δ
N
− s
1
− s
2 −1
≤ C
81
δ
N
u
N
log N
6
X
N
1 + X
N
1
2
Z
t u
N
ds
1
2t + δ
N
− 2s
1
− u
N −1
≤ C
81
u
2 3
N
log N
6
δ
N
X
N
1 + X
N
1
2
δ
1 3
N
log 2t + 2
δ
N
− 3u
N
2 δ
N
− u
N
≤ C
81
log N
−1
δ
N
X
N
1 + X
N
1
2
δ
1 3
N
log 1 +
2t − u
N
δ
N
≤ C
′
81
log N
−1
δ
N
X
N
1 + X
N
1
2
, 81
where we used that u
N
≤ δ
N
2 ∧ log N
−11
in the third line above, and the assumption δ
N
→ 0 in the last.
We now turn to the more difficult bound on T
4,2
. Recall the notation ˆ H
N
ξ
N s
−u
N
, x, u
N
from Subsec- tion 3.2, and H
N
s, x, u
N
from Subsection 4.1. Using Chapman-Kolmogorov again, we see that T
4,2
= E
2
Z
t u
N
Z
s
1
u
N
ds
1
ds
2
log N
8
δ
N
N
2
X
x
1
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1 2
Y
i=1
H
N
s
i
, x
i
, u
N
. 82
However, by 25, H
N
s, x, u
N
− ˆ H
N
ξ
N s
−u
N
, x, u
N
≤ ˆE
1
{E
x,x+e1,x+e2
}
X
y ∈˜B
x,x+e1 uN
ξ
N s
−u
N
y
,
and Lemma 2.2 thus implies log N
N X
x ∈S
N
H
N
s, x, u
N
− ˆ H
N
ξ
N s
−u
N
, x, u
N
≤ C
83
X
N s
−u
N
1u
N
log N
3
. 83
From Proposition 3.4 a and 50, we see that log N N
−1
X
x ∈S
N
|H
N
s, x, u
N
| ≤ C
′
a
X
N s
−u
N
1.
1225
Using this, 17 and 83, we deduce E
2 Z
t u
N
Z
s
1
u
N
ds
2
ds
1
log N
8
δ
N
N
2
X
x
1
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
|H
N
s
1
, x
1
, u
N
| ×
H
N
s
2
, x
2
, u
N
− ˆ H
N
ξ
N s
2
−u
N
, x
2
, u
N
≤ 2C
17
C
83
C
′
a
log N
9
u
N
δ
N
Z
t u
N
Z
s
1
u
N
ds
2
ds
1
2t + δ
N
− s
1
− s
2 −1
E[X
N s
1
−u
N
1X
N s
2
−u
N
1]. Therefore by Proposition 3.4 and the fact that log N
9
u
N
≤ log N
−2
, the above is bounded by 2C
17
C
83
C
′
a
C
ab
log N
−2
δ
N
X
N
1 + X
N
1
2
. Similarly,
log N N
X
x ∈S
N
| ˆ H
ξ
N s
−u
N
, x, u
N
| ≤ log N
N X
x ∈S
N
ˆ E
h ξ
N s
−u
N
ˆ B
x u
N
+ ξ
N s
−u
N
ˆ B
x+e
1
u
N
i ≤ 2X
N s
−u
N
1, so that, using 17 and 83,
E 2
Z
t u
N
Z
s
1
u
N
ds
1
2ds
1
log N
8
δ
N
N
2
X
x
1
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
| ˆ H
N
ξ
N s
2
−u
N
, x
2
, u
N
| ×
H
N
s
1
, x
1
, u
N
− ˆ H
N
s
1
, x
1
, u
N
≤ 2C
′
a
C
17
C
83
log N
9
u
N
δ
N
Z
t u
N
Z
s
1
u
N
ds
2
ds
1
2t + δ
N
− s
1
− s
2 −1
E[X
N s
1
−u
N
1X
N s
2
−u
N
1] ≤ CT log N
−2
δ
N
X
N
1 + X
N
1
2
. Therefore, we see from 82 and the above bounds, for C
84
= C
84
T , that T
4,2
≤ C
84
X
N
1 + X
N
1
2
log N
−2
δ
N
84 +E
Z
t u
N
Z
s
1
u
N
log N
8
δ
N
N
2
X
x
1
,x
2
∈S
N
p
N 2
δ
N
+t−s
1
−s
2
x
2
− x
1 2
Y
i=1
ds
i
ˆ H
N
ξ
N s
i
−u
N
, x
i
, u
N
. Recall from Subsection 3.2 that ˆ
H
N
ξ
N s
−u
N
, x, u
N
= P
3 i=1
F
N i
s − u
N
, x, u
N
. First, by translation
1226
invariance of p, for i = 1, 2, log N
N X
x
i
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
F
N 1
s
i
− u
N
, x
i
, u
N
= log N
N X
w ∈S
N
ξ
N s
i
−u
N
w X
x
i
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
× ˆ
P ˆ B
u
N
= w − x
i
, {0 | e
1
∼ e
2
}
u
N
− ˆPˆB
e
1
u
N
= w − x
i
, {0 | e
1
∼ e
2
}
u
N
≤ log N
N X
w ∈S
N
ξ
N s
i
−u
N
w ˆ E
1
{0|e
1
∼e
2
}
uN
p
N 2t+
δ
N
−s
1
−s
2
w − ˆB
u
N
− x
3 −i
−p
N 2t+
δ
N
−s
1
−s
2
w − ˆB
e
1
u
N
− x
3 −i
Hence, by Lemma 2.1, for i = 1, 2, log N
N X
x
i
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
F
N 1
s
i
− u
N
, x
i
, u
N
≤ C
2.1
log N N
X
w ∈S
N
ξ
N s
i
−u
N
w ˆ E
ˆ
B
u
N
− ˆB
e
1
u
N
2t + δ
N
− s
1
− s
2 −32
≤ C
85
p u
N
X
N s
i
−u
N
12t + δ
N
− s
1
− s
2 −32
. 85
Furthermore, using again translation invariance of p and 13, log N
N X
x
1
∈S
N
F
N 1
s
1
− u
N
, x
1
, u
N
≤ log N
N X
x
1
∈S
N
ˆ E
h ξ
N s
1
−u
N
ˆ B
x
1
u
N
+ ξ
N s
1
−u
N
ˆ B
x
1
+e
1
u
N
1
{x
1
|x
1
+e
1
∼x
1
+e
2
}
uN
i ≤
log N N
X
w ∈S
N
ξ
N s
1
−u
N
w X
x
1
∈S
N
ˆ P ˆ
B
u
N
= w − x
1
, {0 | e
1
∼ e
2
}
u
N
+ˆ P ˆ
B
e
1
u
N
= w − x
1
, {0 | e
1
∼ e
2
}
u
N
≤ 2C
15
X
N s
1
−u
N
1logN u
N −1
≤ C
86
X
N s
1
−u
N
1log N
−1
. 86
We next handle F
N 2
s − u
N
, x, u
N
and F
N 3
s − u
N
, x, u
N
together. Using translation invariance of p
1227
and 17, log N
N X
x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
|F
N 2
s
2
− u
N
, x
2
, u
N
| + |F
N 3
s
2
− u
N
, x
2
, u
N
| ≤
log N N
X
x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
×9 ˆE h
ξ
N s
2
−u
N
ˆ B
x
2
u
N
+ ξ
N s
2
−u
N
ˆ B
x
2
+e
1
u
N
1
{x
2
|x
2
+e
1
|x
2
+e
2
}
uN
i ≤ 9C
17
2t + δ
N
− s
1
− s
2 −1
log N N
X
w ∈S
N
ξ
N s
2
−u
N
w ×
X
x
2
∈S
N
1
X
i=0
ˆ P ˆ
B
e
i
u
N
= w − x
2
, {0 | e
1
| e
2
}
u
N
≤ C
87
log N
−3
2t + δ
N
− s
1
− s
2 −1
X
N s
2
−u
N
1, 87
where we used 6 in the last inequality above, and have set C
87
:= 9C
17
¯ K, ¯
K as in 16. For i = 1, 2, a similar argument leads to
log N N
X
x
i
∈S
N
|F
N 2
s
i
− u
N
, x
i
, u
N
| + |F
N 3
s
i
− u
N
, x
i
, u
N
| ≤ C
88
log N
−3
X
N s
i
−u
N
1. 88
We now use equation 85, first for i = 2, then for i = 1, and then equation 87 to obtain log N
8
δ
N
N
2
Z
t u
N
Z
s
1
u
N
ds
2
ds
1
E X
x
1
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1 2
Y
i=1 3
X
j=1
F
N j
s
i
− u
N
, x
i
, u
N
≤ log N
6
δ
N
Z
t u
N
Z
s
1
u
N
E log N
N X
x
1
∈S
N
F
N 1
s
1
− u
N
, x
1
, u
N
+
3
X
j=2
F
N j
s
1
− u
N
, x
1
, u
N
×C
85
p u
N
X
N s
2
−u
N
1 2t +
δ
N
− s
1
− s
2 −32
ds
2
ds
1
+log N
6
δ
N
Z
t u
N
Z
s
1
u
N
E log N
N X
x
2
∈S
N
3
X
j=2
F
N j
s
2
− u
N
, x
2
, u
N
C
85
p u
N
X
N s
1
−u
N
1 × 2t + δ
N
− s
1
− s
2 −32
ds
2
ds
1
+C
87
log N
3
δ
N
Z
t u
N
Z
s
1
u
N
E X
N s
2
−u
N
1 log N
N X
x
1
∈S
N
3
X
j=2
F
N j
s
1
− u
N
, x
1
, u
N
× 2t + δ
N
− s
1
− s
2 −1
ds
2
ds
1
.
1228
Using 86, 88 it follows from the above inequality that log N
8
δ
N
N
2
Z
t u
N
Z
s
1
u
N
ds
1
ds
2
E X
x
1
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1 2
Y
i=1 3
X
j=1
F
N j
s
i
− u
N
, x
i
, u
N
≤ C
85
C
86
log N +
2C
88
log N
3
p u
N
log N
6
δ
N
Z
t u
N
Z
s
1
u
N
E h
X
N s
1
−u
N
1X
N s
2
−u
N
1 i
2t + δ
N
− s
1
− s
2 3
2
ds
2
ds
1
+C
87
C
88
δ
N
Z
t u
N
Z
s
1
u
N
E h
X
N s
1
−u
N
1X
N s
2
−u
N
1 i
2t + δ
N
− s
1
− s
2 −1
ds
2
ds
1
. Using Proposition 3.4 and the subsequent 40 in the above, it then follows from 84 that
T
4,2
≤ C
89
X
N
1 + X
N
1
2
δ
N
log N
−2
+ log N
5
p u
N
+ 1
≤ 2C
89
X
N
1 + X
N
1
2
δ
N
. 89
The term T
4,3
is much easier to handle. Indeed, using 17 and the trivial bound
2
Y
i=1
h ξ
N s
i
x
i
f
N
x
i
, ξ
N s
i
2
− 1 − ξ
N s
i
x
i
f
N 1
x
i
, ξ
N s
i
2
i ≤
2
Y
i=1
ξ
N s
i
x
i
+ f
N 1
x, ξ
N s
i
, we find
T
4,3
≤ 8C
17
log N
6
δ
N
Z
u
N
Z
s
1
2t + δ
N
− s
1
− s
2 −1
E h
X
N s
1
1X
N s
2
1 i
ds
2
ds
1
. Using u
N
≤ δ
N
2, we see that Z
u
N
Z
s
1
ds
2
ds
1
2t + δ
N
− s
1
− s
2 −1
≤ u
N
log 2t +
δ
N
2t + δ
N
− u
N
≤ u
N
log 2t +
δ
N
2t + δ
N
. Thus, using 40 and our choice of u
N
, T
4,3
≤ 8C
17
C
ab
log N
6
δ
N
X
N
1 + X
N
1
2
u
N
log 2t + 2
δ
N
2t + δ
N
≤ C
90
log N
−5
δ
N
X
N
1 + X
N
1
2
. 90
Recall T
4
= P
3 i=1
T
4,i
. By grouping 81, 89, 90, we finally obtain T
4
≤ CT δ
N
X
N
1 + X
N
1
2
. 91
We finish by providing an upper bound for T
3
. We give less details, because the method is very similar to the one we used for
T
4
, and the smaller power of log N makes this term easier to handle. As we did for
T
4
, we may bound T
3
by the sum of three terms :
1229
T
3
≤ C δ
N
N X
z ∈S
N
E
Z
t ∨u
N
u
N
d
N ,2 s
φ
z s
− E[d
N ,2 s
φ
z s
| F
s −u
N
]
ds
2
+E
Z
t ∨u
N
u
N
E[d
N ,2 s
φ
z s
| F
s −u
N
]ds
2
+C E
Z
u
N
Z
s
1
log N
4
N
2
δ
N
X
x
1
∈S
N
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1 2
Y
i=1
ds
i
Ξ
N s
i
x
i
=: T
3,1
+ T
3,2
+ T
3,3
, where Ξ
N
is defined after 79, and we used f
N
x, ξ
N s
2
≤ 1, f
N 1
x, ξ
N s
2
≤ f
N 1
x, ξ
N s
to bound the integral on [0, u
N
]
2
. As for
T
4,1
, we have T
3,1
= C δ
N
N X
z ∈S
N
E h
Z
t ∨u
N
u
N
Z
s
1
+u
N
s
1
ds
1
ds
2 2
Y
i=1
d
N ,2 s
φ
z s
i
− E h
d
N ,2 s
φ
z s
i
| F
s
i
−u
N
ii .
We then use the following bound on d
N ,2
compare 79: |d
N ,2 s
φ
z s
| ≤ β log N
2
N X
x ∈S
N
φ
z s
xΞ
N s
x. Now we may reason exactly as for
T
4,1
to obtain compare 81 T
3,1
≤ C
92
log N
−5
δ
N
X
N
1 + X
N
1
2
. 92
Let us now deal with T
3,2
. For s ≥ u
N
, we let H
N
s, x, u
N
:= E
ξ
N s
x f
N
x, ξ
N s
+ 1 − ξ
N s
x f
N 1
x, ξ
N s
| F
s −u
N
,
ˆ H
N
s, x, u
N
:= ˆ E
h ξ
N s
−u
N
ˆ B
x u
N
1 − ξ
N s
−u
N
ˆ B
x+e
1
u
N
+ 1 − ξ
N s
−u
N
ˆ B
x u
N
ξ
N s
−u
N
ˆ B
x+e
1
u
N
i .
As for H, ˆ H, we get by the analogues of 25 and Lemma 2.2 that
log N N
X
x ∈S
N
H
N
s, x, u
N
− ˆ H
N
s, x, u
N
≤ C
93
X
N s
−u
N
1u
N
log N
3
. 93
Argue as when dealing with T
4,2
to get that for some CT 0,
E 2
Z
t u
N
Z
s
1
u
N
ds
1
ds
2
log N
4
δ
N
N
2
X
x
1
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
H
N
s
1
, x
1
, u
N
× H
N
s
2
, x
2
, u
N
− ˆ H
N
s
2
, x
2
, u
N
≤ CT log N
−6
δ
N
X
N
1 + X
N
1
2
, 1230
and E
2 Z
t u
N
Z
s
1
u
N
ds
1
ds
2
log N
4
δ
N
N
2
X
x
1
,x
2
∈S
N
p
N 2t+
δ
N
−s
1
−s
2
x
2
− x
1
ˆ H
N
s
1
, x
1
, u
N
× H
N
s
2
, x
2
, u
N
− ˆ H
N
s
2
, x
2
, u
N
≤ CT log N
−6
δ
N
X
N
1 + X
N
1
2
, and therefore
T
3,2
≤ C
94
X
N
1 + X
N
1
2
log N
−6
δ
N
94 +C
94
E
Z
t u
N
Z
s
1
u
N
log N
4
δ
N
N
2
X
x
1
,x
2
∈S
N
p
N 2
δ
N
+t−s
1
−s
2
x
2
− x
1 2
Y
i=1
ds
i
ˆ H
N
s
i
, x
i
, u
N
. Then,
log N
2
N X
x ∈S
N
ˆ H
N
s, x, u
N
= 2
log N
2
N X
x ∈S
N
ˆ E
h ξ
N s
−u
N
ˆ B
x u
N
1 − ξ
N s
−u
N
ˆ B
x+e
1
u
N
1
{x|x+e
1
}
uN
i
≤ 2 log N
2
N X
x,w ∈S
N
ξ
N s
−u
N
wˆ P ˆ
B
u
N
= w − x, {0 | e
1
}
u
N
≤ 2C
15
log N logN u
N −1
X
N s
−u
N
1, by the definition of C
15
in 15. We deduce from the above, 94 and 17 that T
3,2
≤ C
94
X
N
1 + X
N
1
2
log N
−6
δ
N
+C
95
δ
N
E
Z
t u
N
Z
s
1
u
N
X
N s
1
−u
N
1X
N s
2
−u
N
1ds
1
ds
2
≤ C
′
95
X
N
1 + X
N
1
2
δ
N
, 95
where we used Proposition 3.4 in the last line. Finally, using the same method as for bounding
T
4,3
, T
3,3
≤ C
96
log N
−9
δ
N
X
N
1 + X
N
1
2
. 96
Grouping 92, 95 and 96, we find T
3
≤ C
97
δ
N
X
N
1 + X
N
1
2
. 97
We may now conclude the proof of Proposition 3.10. Indeed, using 73, 77, 78, 91, 97, E
Z Z
δ
N
p
N 2
δ
N
y − xdX
N t
xd X
N t
y
≤ CT X
N
1 + X
N
1
2
δ
N
1 + log 1 +
T δ
N
+ 1
t + δ
N
, which, as explained in the beginning of the proof, is the desired bound.
1231
5.3 Space-time first moment bound : proof of Lemma 3.5
Kategori