Lu, v ; x in [16, 4.18], so that [16, Lemma 4.4] also applies to P
N
x . For N = 0 with d 4,
we have cf., 5.36 X
x
P x, t ≡
X
x
P x, t =
X
x
δ
x,o
δ
t,0
+ Lo, 0, o, 0; x, t ≤ δ
t,0
+ ǫ
2
O β
1 + t
d 2
. 5.63
The factor O β is replaced by Oβ
T
if d ≤ 4. For N ≥ 1, we apply Lemma 5.4 to 5.63 N times. We now relate
P
N
x with P
N
x . Note that, by 5.13–5.14, we have
P
N
x = P
N −1
〈u〉; 2
1
〈u〉
〈w 〉, 2
〈w 〉
x = P
N −1
〈u〉; 2
1
〈u〉
x + P
N −1
〈u〉; 2
1
〈u〉
〈w 〉, 2
1
〈w 〉
x .
5.64 It follows by 5.62 and 5.64 that
P
N
x =
N
X
M =0
N M
P
N +M
x ≤ 2
N N
X
M =0
P
N +M
x . 5.65
where the inequality is due to
N M
≤ 2
N
. By Lemma 5.4, we have, for d 4,
X
x
|x|
q
P
N
t
x ≤ δ
q,0
δ
t,0
δ
N ,0
+ ǫ
2
c β
1 ∨N
σ
q
1 + t
d−q2
N ≥ 0, 5.66
for some c ∞. For d ≤ 4, we can simply replace β
1 ∨N
by β
T
ˆ β
∨N−1
T
and σ
2
by σ
2
T
. Therefore, X
x
|x|
q
P
N
t
x ≤ 2
N N
X
M =0
X
x
|x|
q
P
N +M
t
x ≤ 2
N N
X
M =0
δ
q,0
δ
t,0
δ
N +M ,0
+ ǫ
2
c β
N +M
σ
q
1 + t
d−q2
≤ δ
q,0
δ
t,0
δ
N ,0
+ ǫ
2
2c β
N
1 − cβ
σ
q
1 + t
d−q2
. 5.67
This completes the proof of Lemma 5.7.
5.3 Bound on A
~x
J
In this section, we investigate A ~x
J
. First, in Section 5.3.1, we prove a d-independent diagrammatic bound on A
N
v , ~x
J
; C, where we recall A
N
~x
J
= A
N
o, ~x
J
; {o} in 3.25. Then, in Section 5.3.2,
we prove the bound 5.2 for d 4 and the bound 5.4 for d ≤ 4 simultaneously.
5.3.1 Diagrammatic bound on A
N
v , ~x
J
; C
The main result proved in this section is the following proposition:
Lemma 5.8 Diagrammatic bound on A
N
v , ~x
J
; C. For r
≥ 3, ~x
J
∈ Λ
r −1
, v ∈ Λ and C ⊂ Λ,
A
N
v , ~x
J
; C
5.68
≤
X
I 6=∅,J
1
{v∈C}
P {v −→ ~x
I
} ◦ {v −→ ~x
J \I
} +
X
z 6=v
P v , z; C,
ℓ~x
I
τ~x
J \I
− z N = 0,
X
I 6=∅,J
X
z
P
N
v , z; C τ~x
I
− z + P
N
v , z; C, ℓ~x
I
τ~x
J \I
− z N ≥ 1.
852
To prove Lemma 5.8, we first note that, by 3.16–3.17 and 3.19–3.20, A
N
v , ~x
J
; C =
P E
′
v , ~x
J
; C
N = 0, P
b
N
p
b
N
M
N
v
,b
N
; C
P E
′
b
N
,
~x
J
; ˜ C
N −1
N ≥ 1. 5.69
Thus, we are lead to study P E
′
v , ~x
J
; C
. As a result, Lemma 5.8 is a consequence of the following lemma:
Lemma 5.9. For r ≥ 3, ~x
J
∈ Λ
r −1
, v ∈ Λ and C ⊂ Λ,
P E
′
v , ~x
J
; C
≤ X
I 6=∅,J
1
{v∈C}
P {v −→ ~x
I
} ◦ {v −→ ~x
J \I
} +
X
z 6=v
P
v , z; C, ℓ~x
I
τ~x
J \I
− z
. 5.70
Proof of Lemma 5.8 assuming Lemma 5.9. Since Lemma 5.9 and 5.69 immediately imply 5.68 for N = 0, it thus suffices to prove 5.68 for N
≥ 1.
Substituting 5.70 with v = b
N
, C = ˜
C
N −1
into 5.69 and then using 5.51–5.52, we obtain A
N
v , ~x
J
; C
≤ X
I 6=∅,J
X
b
N
p
b
N
M
N
v
,b
N
; C
1
{b
N
∈˜C
N −1
}
P {b
N
−→ ~x
I
} ◦ {b
N
−→ ~x
J \I
} +
X
z
6=b
N
M
N
v
,b
N
; C
P b
N
, z; ˜ C
N −1
, ℓ~x
I
τ~x
J \I
− z
≤ X
I 6=∅,J
X
z
X
η
X
b
N
P
N −1
v , b
N
; C;
ℓ
η
b
N
p
b
N
δ
b
N
,z
| {z
}
X
P {z −→ ~x
I
} ◦ {z −→ ~x
J \I
}
+ X
η
X
c
X
b
N
b
N
6=z
P
N −1
v , b
N
; C;
ℓ
η
c p
b
N
P b
N
, z; c, ℓ~x
I
| {z
}
Y
τ~x
J \I
− z
,
5.71
where P
η
is the sum over the N − 1
st
admissible lines for P
N −1
v , b
N
; C. Ignoring the restriction
b
N
6= z and using an extension of 5.53, we obtain
Y ≤ P
N
v , z; C, ℓ~x
I
. 5.72
For X , we use 5.36 and 5.39 to obtain X
≤ X
η
X
b
N
P
N −1
v , b
N
; C;
ℓ
η
b
N
p
b
N
P b
N
, z; b
N
≤ X
η
X
y
X
b
N
P
N −1
v , b
N
; C;
ℓ
η
y p
b
N
P b
N
, z; y = P
N
v , z; C. 5.73
Finally, we use the BK inequality to bound P {z −→ ~x
I
} ◦ {z −→ ~x
J \I
} by τ~x
I
− z τ~x
J \I
− z.
This completes the proof. 853
Proof of Lemma 5.9. Recall 5.43. We show below that E
′
v , ~x
J
; C
⊂ [
I 6=∅,J
[
z
n
E v, z; C ∩ {v −→ ~x
I
} ◦ {z −→ ~x
J \I
} o
. 5.74
First, we prove 5.70 assuming 5.74. Substituting 5.74 into PE
′
v , ~x
J
; C, we have
P E
′
v , ~x
J
; C
≤ X
I 6=∅,J
X
z
P
E v, z; C ∩ {v −→ ~x
I
} ◦ {z −→ ~x
J \I
} 5.75
= X
I 6=∅,J
1
{v∈C}
P {v −→ ~x
I
} ◦ {v −→ ~x
J \I
} +
X
z 6=v
P
E v, z; C ∩ {v −→ ~x
I
} ◦ {z −→ ~x
J \I
} .
For the sum over z 6= v, we use the BK inequality to extract Pz −→ ~x
J \I
≡ τ~x
J \I
− z and apply
the following inequality that is a result of an extension of the argument around 5.46:
P Ev, z; C ∩ {v −→ ~x
I
} ≤ P
v , z; C, ℓ~x
I
. 5.76
This completes the proof of 5.70. It remains to prove 5.74. Summarising 4.5–4.9, we can rewrite E
′
v , ~x
J
; C as
E
′
v , ~x
J
; C =
˙ [
j ∈J
n
{v −→ ~x
J
} ∩
v
C
−→ x
1
, . . . , x
j −1
c
∩ E
′
v , x
j
; C
o
∩
∄ pivotal bond b for v −→ x
i
∀i such that v
C
−→ b ˙
∪ ¨
˙ [
∅6=IJ
˙ [
b
n
{v −→ ~x
I
} ∩
v
C
−→ x
1
, . . . , x
j
I
−1 c
∩ E
′
v , b; C in ˜ C
b
v
o
∩ {b is occupied} ∩ b
−→ ~x
J \I
in Λ \ ˜C
b
v
« .
5.77 Ignoring
{v
C
−→ x
1
, . . . , x
j −1
}
c
and
{∄ pivotal bond b for v −→ x
i
∀i such that v
C
−→ b} and using E
′
v , z; C ⊂ E v, z; C, we have
E
′
v , ~x
J
; C
⊂ [
j ∈J
E v, x
j
; C
∩ {v −→ ~x
J
j
} ∪
[
∅6=IJ
[
z
n
E v, z; C ∩ {v −→ ~x
I
} ◦ {z −→ ~x
J \I
} o
. 5.78
Note that the first event on the right-hand side is a subset of the second event, when I = J
j
and
z = x
j
, for which J \ I = { j} and {z −→ ~x
J \I
} = {x
j
−→ x
j
} is the trivial event. This completes the proof of 5.74 and hence of Lemma 5.9.
854
5.3.2 Proof of the bound on A
N
~x
