5.3.2 Proof of the bound on A
N
~x
J
We prove 5.2 for d 4 and 5.4 for d ≤ 4 simultaneously, using Lemmas 5.3 and 5.7–5.8.
As in Section 2.4, we will frequently use 2.68: X
~x
I
τ
~t
I
~x
I
≤ O 1 + ¯t
I |I|−1
. 5.79
where we recall max
i ∈I
t
i
≤ T log T for d ≤ 4. For simplicity, let I = {1, . . . , i}. Then, 5.79 is an easy consequence of Lemma 5.3:
X
~x
I
τ
~t
I
~x
I
≤ X
~x
Ii
,x
i
τ
~t
Ii
~x
I
i
; ℓx
i
, t
i
≤ · · · ≤ X
x
1
,...,x
i
τ
t
1
x
1
; ℓx
2
, t
2
, · · · , ℓx
i
, t
i
. 5.80
First we prove 5.2, for which d 4, for N ≥ 1. By Lemma 5.8, we have
A
N
~x
J
≡ A
N
o, ~x
J
; {o} ≤
X
I 6=∅,J
X
z
P
N
z τ~x
I
− z + P
N
z; ℓ~x
I
τ~x
J \I
− z. 5.81
Note that the number of lines contained in each diagram for P
N
z at any fixed time between 0 and
t
z
is bounded, say, by L , due to its construction. Therefore, by Lemmas 5.3 and 5.7, we obtain
X
z,x
1
P
N
z, s; ℓx
1
, t
1
≤ L ǫ
2
O β
N
1 + s
d 2
1 + s ∧ t
1
≤ L ǫ
2
O β
N
1 + s
d−22
, 5.82
and further that X
z,x
1
,x
2
P
N
z, s; ℓx
1
, t
1
, ℓx
2
, t
2
≤ L L + 1 ǫ
2
O β
N
1 + s
d−22
1 + s ∨ t
1
∧ t
2
, 5.83
where we note that 1 + s ∨ t
1
∧ t
2
= 1 + min{max{s, t
1
}, t
2
}, so that the order of operations is naturally first ‘
∨’, followed by ‘∧’ and then finally ‘+’. More generally, by denoting the second-largest element of
{s,~t
I
} by ¯s
~t
I
, we have X
z, ~x
I
P
N
z, s; ℓ~x
I
,~t
I
≤ L + |I| − 1
L − 1 ǫ
2
O β
N
1 + s
d−22
1 + ¯ s
~t
I
|I|−1
, 5.84
where the combinatorial factor
L +|I|−1 L −1
is independent of β and N . Substituting this and 5.60
into 5.81 and using 5.79, we obtain that, since d − 22 1,
X
~x
J
A
N
~t
J
~x
J
≤ ǫOβ
N
X
I 6=∅,J
ǫ
X
• s
≤t
J
1 1 + s
d 2
O ¯t
I
− s
|I|−1
O ¯t
J \I
− s
|J\I|−1
+ ǫ
X
• s
≤t
J \I
O1 + ¯ s
~t
I
|I|−1
1 + s
d−22
O ¯t
J \I
− s
|J\I|−1
≤ ǫOβ
N
O 1 + ¯t
|J|−2
, 5.85
855
where ¯t = ¯t
J
. This proves 5.2 for N ≥ 1.
To prove 5.4, for which d ≤ 4, for N ≥ 1, we simply replace Oβ
N
in 5.84 by O β
T
O ˆ β
T
N −1
using Lemma 5.7ii instead of Lemma 5.7i. Then, we use the factor β
T
to control the sums over s
∈ ǫZ
+
in 5.85, as in 5.28. Since t
J \I
≤ T log T , β
T
≡ β
1
T
−bd
and ˆ β
T
≡ β
1
T
−α
with α bd −
4 −d
2
, we have β
T
ǫ X
• s
≤t
J \I
1 + s
−d−22
≤ Oβ
T
1 + t
J \I
4−d2
log1 + t
J \I
δ
d,4
≤ O ˆ β
T
. 5.86
This completes the proof of 5.4 for N ≥ 1.
Next we consider the case of N = 0. Similarly to the above computation, the contribution from the latter sum in 5.68 over z
6= v = o in the current setting equals ǫOβ1 + ¯t
r −3
for d 4 and
ǫO ˆ β
T
1+¯t
r −3
for d ≤ 4. It remains to estimate the contribution from P{o −→ ~x
I
}◦{o −→ ~x
J \I
} in 5.68.
If ǫ is large e.g., ǫ = 1, then we simply use the BK inequality to obtain
P {o −→ ~x
I
} ◦ {o −→ ~x
J \I
} ≤ τ~x
I
τ~x
J \I
. 5.87
Therefore, by 5.79, we have X
~x
J
A
~t
J
~x
J
≤ O 1 + ¯t
r −3
. 5.88
If ǫ ≪ 1, then we should be more careful. Since {o −→ ~x
I
} and {o −→ ~x
J \I
} occur bond-disjointly,
and since there is only one temporal bond growing out of o, there must be a nonempty subset I
′
of I or J
\ I and a spatial bond b with b = o such that {b −→ ~x
I
′
} ◦ {o −→ ~x
J \I
′
} occurs. Then, by the BK inequality and 5.79, we obtain
X
~x
J
P {o −→ ~x
I
} ◦ {o −→ ~x
J \I
} ≤
X
~x
J
X
∅6=I
′
J
X
b=o, ·
spatial
P {b −→ ~x
I
′
} ◦ {o −→ ~x
J \I
′
} ≤
X
~x
J
X
∅6=I
′
J
λǫD
⋆
τ~x
I
′
τ~x
J \I
′
≤ ǫ O 1 + ¯t
I
′
|I
′
|−1
1 + ¯t
J \I
′
|J\I
′
|−1
≤ ǫ O 1 + ¯t
|J|−2
. 5.89
This completes the proof of 5.2 for d 4 and 5.4 for d ≤ 4.
6 Bound on
φy
1
, y
2
±
To prove the bound on ˆ ψ
s
1
,s
2
k
1
, k
2
in Proposition 2.2, we first recall 2.24 and 4.58:
ψy
1
, y
2
= X
v
p
ǫ
v Cy
1
− v, y
2
− v, Cy
1
, y
2
= φy
1
, y
2
+
+ φy
2
, y
1
+
− φy
2
, y
1
−
, 6.1
856
hence ˆ
ψ
s
1
,s
2
k
1
, k
2
= ˆ p
ǫ
k
1
+ k
2
ˆ φ
s
1
−ǫ,s
2
−ǫ
k
1
, k
2
+
+ ˆ φ
s
2
−ǫ,s
1
−ǫ
k
2
, k
1
+
− ˆ φ
s
2
−ǫ,s
1
−ǫ
k
2
, k
1
−
. 6.2 Therefore, to show the bound on ˆ
ψ
s
1
,s
2
k
1
, k
2
, it suffices to investigate φy
Kategori