O β and λ
ǫ
c
= 1 + O β uniformly in ǫ, we have
ˆ ψ
2 ǫ,2ǫ
0, 0 = 1 − ǫ + λ
ǫ
c
ǫ
1 − ǫ + λ
ǫ
c
ǫ
2
− 1 − ǫ
2
+ λ
ǫ
c
ǫ
2
X
x
Dx
2
| {z
}
2−ǫ+Oβǫλ
ǫ c
ǫ
= 2 − ǫ + Oβ
ǫ.
Combining this with 2.36 yields V
ǫ
= 2 − ǫ + Oβ. 2.41
This establishes the claim on V of Theorem 1.2i. For d ≤ 4, on the other hand, β = β
T
converges to zero as T
↑ ∞, so that V
ǫ
is replaced by 2 − ǫ in Theorem 2.1ii.
2.4 Induction in r
In this section, we prove Theorem 2.1 for ǫ ∈ 0, 1] fixed, assuming 2.30 and Proposition 2.2. The
argument in this section is an adaptation of the argument in [20, Section 2.3], adapted so as to deal with the uniformity in the time discretization. In particular, in this section, we prove Theorem 2.1
for oriented percolation for which ǫ = 1.
For r ≥ 3, we will use the notation
¯t = the second-largest element of {t
1
, . . . , t
r −1
}, t =
min {t
1
, . . . , t
r −1
}. 2.42
Proof of Theorem 2.1i assuming Proposition 2.2. We prove that for d 4 there are positive con-
stants L = L
d and V
ǫ
= V
ǫ
d, L such that for λ = λ
ǫ
c
, L ≥ L
and κ ∈ 0, 1 ∧ ∆ ∧
d −4
2
, we have
ˆ τ
r
~t ~k
p
v
ǫ
σ
2
t
= A
ǫ
A
ǫ
2
V
ǫ
t
r −2
ˆ M
r−1
~tt
~k + O ¯t + 1
−κ
r ≥ 3 2.43
uniformly in t ≥ ¯t and in ~k ∈ R
r−1d
with P
r −1
i=1
|k
i
|
2
bounded, and uniformly in ǫ 0. To prove
Theorem 2.1i, we take t = T and replace ~t by T~t. Since, without loss of generality, we may assume that max
i
t
i
= 1 and t
i
≤ 1, we thus have that T ≥ T ¯t, so that 2.43 indeed proves Theorem 2.1i. We prove 2.43 by induction in r, with the initial case of r = 2 given by Theorem 2.1i:
ˆ τ
t
1
k
p
v
ǫ
σ
2
t
= ˆ τ
t
1
k
p
t
1
t
p
v
ǫ
σ
2
t
1
= A
ǫ
e
− |k|
2
t
1
2d t
+ O t
1
+ 1
−κ
, 2.44
using the facts that |k|
2
is bounded, t
1
≤ t and κ
d −4
2
. The induction will be advanced using 2.30. Let r
≥ 3. By 2.37, ˆ ζ
r
~t
~k is an error term. Thus, we are left to determine the asymptotic behaviour of the first term on the right-hand side of 2.30.
Fix ~k with P
r −1
i=1
|k
i
|
2
bounded. To abbreviate the notation, we write ~k
t
= ~k
p v
ǫ
σ
2
t .
2.45
816
Given 0 ≤ s
≤ t, let t = s
∧ t − s . We show that, for every nonempty subset I ⊂ J
1
,
t
J \I
−s
X
• s
1
=2 ǫ
t
I
−s
X
• s
2
=2 ǫ
ˆ ψ
s
1
,s
2
k
t
J \I
, k
t
I
ˆ τ
r1
~t
J \I
−s
1
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
2
−s
~k
t
I
− V
ǫ
ˆ τ
r1
~t
J \I
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
~k
t
I
≤ Cǫt
r −3
t + 1
−κ
. 2.46
Before establishing 2.46, we first show that it implies 2.43. Since |ˆ
τ
s
k
t
| is uniformly bounded by Theorem 2.1 for r = 2, inserting 2.46 into 2.30 and applying 2.37 gives
ˆ τ
r
~t
~k
t
= V
ǫ
ǫ
t
X
• s
=0
ˆ τ
s
k
t
X
I ⊂J
1
: |I|≥1
ˆ τ
r1
~t
J \I
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
~k
t
I
+ Ot
r −3
ǫ
t
X
• s
=0
t + 1
−κ
+ Ot
r −2−κ
. 2.47
Using the fact that κ 1, the summation in the error term can be seen to be bounded by a multiple
of t
1 −κ
≤ t
1 −κ
. With the induction hypothesis and the identity r
1
+ r
2
= r + 1, 2.47 then implies that
ˆ τ
r
~t
~k
t
= A
ǫ
A
ǫ
2
V
ǫ
t
r −2
ǫ
t
X
• s
=0
ˆ M
1
s t
k X
I ⊂J
1
: |I|≥1
ˆ M
r1−1 ~tJ\I −s0
t
~k
J \I
ˆ M
r2−1 ~tI −s0
t
~k
I
+ Ot
r −2−κ
, 2.48
where the error arising from the error terms in the induction hypothesis again contributes an amount Ot
r −3
ǫ X
• t s
=0
t + 1
−κ
≤ Ot
r −2−κ
. The summation on the right-hand side of 2.48, divided by t, is the Riemann sum approximation to an integral. The error in approximating the integral by
this Riemann sum is O ǫt
−1
. Therefore, using 1.17, we obtain ˆ
τ
r
~t
~k
t
= A
ǫ
A
ǫ
2
V
ǫ
t
r −2
Z
t t
ds ˆ
M
1
s
k X
I ⊂J
1
: |I|≥1
ˆ M
r1−1 ~tJ\I −s0
t
~k
J \I
ˆ M
r2−1 ~tI −s0
t
~k
I
+ Ot
r −2−κ
= A
ǫ
A
ǫ
2
V
ǫ
t
r −2
ˆ M
r−1
~tt
~k + Ot
r −2−κ
. 2.49
Since t ≥ ¯t, it follows that t
r −2−κ
≤ C t
r −2
¯t + 1
−κ
. Thus, it suffices to establish 2.46. To prove 2.46, we write the quantity inside the absolute value signs on the left-hand side as
t
J \I
−s
X
• s
1
=2 ǫ
t
I
−s
X
• s
2
=2 ǫ
ˆ ψ
s
1
,s
2
k
t
J \I
, k
t
I
ˆ τ
r1
~t
J \I
−s
1
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
2
−s
~k
t
I
− V
ǫ
ˆ τ
r1
~t
J \I
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
~k
t
I
= T
1
+ T
2
+ T
3
, 2.50
817
with T
1
= ˆ τ
r1
~t
J \I
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
~k
t
I t
J \I
−s
X
• s
1
=2 ǫ
t
I
−s
X
• s
2
=2 ǫ
ˆ ψ
s
1
,s
2
0, 0 − V
ǫ
, 2.51
T
2
= ˆ τ
r1
~t
J \I
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
~k
t
I t
J \I
−s
X
• s
1
=2 ǫ
t
I
−s
X
• s
2
=2 ǫ
ˆ ψ
s
1
,s
2
k
t
J \I
, k
t
I
− ˆ ψ
s
1
,s
2
0, 0 ,
2.52
T
3
=
t
J \I
−s
X
• s
1
=2 ǫ
t
I
−s
X
• s
2
=2 ǫ
ˆ ψ
s
1
,s
2
k
t
J \I
, k
t
I
× ˆ
τ
r1
~t
J \I
−s
1
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
2
−s
~k
t
I
− ˆ τ
r1
~t
J \I
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
~k
t
I
. 2.53
To complete the proof, it suffices to show that for each nonempty I ⊂ J
1
, the absolute value of each T
i
is bounded above by the right-hand side of 2.46. In the course of the proof, we will make use of some bounds on sums involving b
ǫ
s
1
,s
2
:
Lemma 2.3 Bounds on sums involving b
ǫ
s
1
,s
2
. i Let d
4. For every κ ∈ [0, 1 ∧
d −4
2
, there exists a constant C = Cd,
κ such that the following bounds hold uniformly in ǫ ∈ 0, 1]
∞
X
• s
1
,s
2
=2 ǫ
s
1
∨s
2
≤s
s
i
b
ǫ
s
1
,s
2
+ b
ǫ
s
2
,s
1
≤ Cǫ1 + s
1 −κ
,
∞
X
• s
1
,s
2
=2 ǫ
s
1
∨s
2
≥s
b
ǫ
s
1
,s
2
≤ Cǫ1 + s
−κ
. 2.54
ii Let d ≤ 4 with α ≡ bd −
4 −d
2
0, fix α ∈ 0, α, recall β
T
= β
1
T
−bd
and let ˆ β
T
= β
1
T
−α
. There exists a constant C = Cd,
κ such that the following bound holds uniformly in ǫ ∈ 0, 1] β
T
T log T
X
• s
1
,s
2
=2 ǫ
s
1
∨s
2
2ǫ
δ
s
1
,s
2
+ β
T
b
ǫ
s
1
,s
2
+ b
ǫ
s
2
,s
1
≤ C ˆ β
T
ǫ. 2.55
Proof. i This is straightforward from 2.31, when we pay special attention to the number of powers of
ǫ present in b
ǫ
s
1
,s
2
and use the fact that the power of 1 + s
1
and of 1 + s
2
− s
1
is d − 22 1.
ii We shall only perform the proof for d 4, as the proof for d = 4 is a slight modification of the
argument below. Using 2.31, we can perform the sum to obtain LHS of 2.55
≤ Oβ
T
ǫ
2
X
• 2
ǫs
1
≤T log T
1 + s
1 2−d2
+ O β
2
T
ǫ
3
X
• 2
ǫ≤s
1
s
2
≤T log T
1 + s
1 2−d2
1 + s
2
− s
1 2−d2
≤ Oβ
T
ǫ 1 + T log T
4−d2
1 + β
T
1 + T log T
4−d2
≤ O ˆ β
T
ǫ1 + ˆ β
T
, 2.56
as long as α ∈ 0, α. This proves 2.55.
818
We now resume proving 2.46. By the induction hypothesis and the fact that ¯t
I
i
≤ t, it follows that |ˆ
τ
ri
~t
Ii
~k
I
i
| ≤ Ot
r
i
−2
uniformly in ~t
I
i
and ~k
I
i
. Therefore, it follows from 2.36 and the definition of V
ǫ
in 2.40 that |T
1
| ≤ X
• s
1
≥t
J \I
−s or
s
2
≥t
I
−s
Ot
r −3
b
ǫ
s
1
,s
2
≤ Oǫt
r −3
t + 1
−d−42
, 2.57
where the final bound follows from the second bound in 2.54. Similarly, by 2.36 with q = 2, now using the first bound in 2.54,
|T
2
| ≤
t
J \I
−s
X
• s
1
=2 ǫ
t
I
−s
X
• s
2
=2 ǫ
s
1
|k
t
J \I
|
2
+ s
2
|k
t
I
|
2
Ot
r −3
b
ǫ
s
1
,s
2
≤ Oǫt
r −3
t + 1
−κ
, 2.58
using that t
r −4
t + 1
1 −κ
≤ t
r −3
t + 1
−κ
since t ≥ t
. It remains to prove that |T
3
| ≤ Oǫt
r −3
t + 1
−κ
. 2.59
To begin the proof of 2.59, we note that the domain of summation over s
1
, s
2
in 2.53 is contained in
∪
2 j=0
S
j
~t, where S
~t = [0,
1 2
t
J \I
− s ] × [0,
1 2
t
I
− s ],
S
1
~t = [
1 2
t
J \I
− s , t
J \I
− s ] × [0, t
I
− s ],
S
2
~t = [0, t
J \I
− s ] × [
1 2
t
I
− s , t
I
− s ].
Therefore, |T
3
| is bounded by
2
X
j=0
X
• ~s∈S
j
~t
ˆ ψ
s
1
,s
2
k
t
J \I
, k
t
I
ˆτ
r1
~t
J \I
−s
1
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
2
−s
~k
t
I
− ˆ τ
r1
~t
J \I
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
~k
t
I
. 2.60
The terms with j = 1, 2 in 2.60 can be estimated as in the bound 2.57 on T
1
, after using the triangle inequality and bounding the r
i
-point functions by Ot
r
i
−2
. For the j = 0 term of 2.60, we write
ˆ τ
r1
~t
J \I
−s
1
−s
~k
t
J \I
= ˆ τ
r1
~t
J \I
−s
~k
t
J \I
+ ˆ
τ
r1
~t
J \I
−s
1
−s
~k
t
J \I
− ˆ τ
r1
~t
J \I
−s
~k
t
J \I
, 2.61
ˆ τ
r2
~t
I
−s
2
−s
~k
t
I
= ˆ τ
r2
~t
I
−s
~k
t
I
+ ˆ
τ
r2
~t
I
−s
2
−s
~k
t
I
− ˆ τ
r2
~t
I
−s
~k
t
I
. 2.62
We expand the product of 2.61 and 2.62. This gives four terms, one of which is cancelled by ˆ
τ
r1
~t
J \I
−s
~k
t
J \I
ˆ τ
r2
~t
I
−s
~k
t
I
in 2.60. Three terms remain, each of which contains at least one factor from the second terms in 2.61–2.62. In each term we retain one such factor and bound the
other factor by a power of t, and we estimate ˆ ψ using 2.36. This gives a bound for the j = 0
contribution to 2.60 equal to the sum of X
• s
1
,s
2
∈S ~n
Ot
r
2
−2
b
ǫ
s
1
,s
2
ˆτ
r1
~t
J \I
−s
1
−s
~k
t
J \I
− ˆ τ
r1
~t
J \I
−s
~k
t
J \I
2.63
819
plus a similar term with J \ I and r
1
replaced by I and r
2
, respectively. By the induction hypothesis, the difference of r
1
-point functions in 2.63 is equal to A
ǫ
A
ǫ
2
V
ǫ
t
r
1
−2
f ~t
J \I
− s
1
− s t
− f ~t
J \I
− s t
+ Ot
r
1
−2
t + 1
−κ
2.64 with f ~t = ˆ
M
r1−1
~t
~k
J \I
. Using 1.17, the difference in 2.64 can be seen to be at most Os
1
t
−1
. Therefore, 2.63 is bounded above, using 2.54, by
X
• s
1
,s
2
∈S ~t
Os
1
t
r −4
+ Ot
r −3
t + 1
−κ
b
ǫ
s
1
,s
2
+ b
ǫ
s
2
,s
1
≤ Oǫt
r −3
t + 1
−κ
. 2.65
This establishes 2.59. Summarizing 2.57–2.59 yields 2.46. This completes the proof of Theorem 2.1i assuming
Proposition 2.2i. Proof of Theorem 2.1ii assuming Proposition 2.2. The proof of Theorem 2.1ii is similar, now us-
ing Proposition 2.2ii instead of Proposition 2.2i and Lemma 2.3ii instead of Lemma 2.3i. For d
≤ 4, we will prove that for there are positive constants L = L
d and such that, for λ
T
and µ as
in Theorem 1.1ii, L
1
≥ L , with L
T
defined as in 1.7, and δ ∈ 0, 1 ∧ ∆ ∧ α, we have
ˆ τ
r
~t ~k
p σ
2 T
T
= 2 − ǫT
r −2
ˆ M
r−1
~tT
~k + OT
−µ∧δ
r ≥ 3 2.66
uniformly in T ≥ ¯t, in ~t such that max
r −1
i=1
t
i
≤ T log T , and in ~k ∈ R
r−1d
with P
r −1
i=1
|k
i
|
2
bounded, and uniformly in
ǫ 0. We again prove 2.66 by induction in r, with the initial case of r = 2 given by Theorem 2.1ii.
This part is a straightforward adaptation of the argument in 2.44, and is omitted. We now advance the induction hypothesis. By 2.30 and 2.39,
ˆ τ
r
~t
~k =
t −2ǫ
X
• s
=0
ˆ τ
2
s
k X
∅6=I⊂J
1
t
J \I
−s
X
• s
1
=2 ǫ
t
I
−s
X
• s
2
=2 ǫ
ˆ ψ
s
1
,s
2
k
J \I
, k
I
ˆ τ
r1
~t
J \I
−s
1
−s
~k
J \I
ˆ τ
r2
~t
I
−s
2
−s
~k
I
+ OT
r −2−κ
, 2.67
where, since µ ∈ 0, α − δ due to Theorem 1.1ii, and since κ ∈ 0, α is arbitrary due to Proposi-
tion 2.2ii, we may assume κ ≥ µ without loss of generality.
Below, we will frequently use X
~x
I
τ
|I|+1
~t
I
~x
I
≤ O 1 + ¯t
I |I|−1
, uniformly in
ǫ. 2.68
This is an easy consequence of the already-known results for the 2-point function and certain dia- grammatic constructions introduced in Section 5.1. We will prove 2.68 in the beginning of Sec-
tion 5.3.2.
820
By Proposition 2.2ii and Lemma 2.3ii and using ˆ β
T
= β
1
T
−µ
and 2.68, we can bound
t −2ǫ
X
• s
=0
ˆ τ
2
s
k X
∅6=I⊂J
1
t
J \I
−s
X
• s
1
=2 ǫ
t
I
−s
X
• s
2
=2 ǫ
1
{s
1
,s
2
6=2ǫ,2ǫ}
ˆ ψ
s
1
,s
2
k
J \I
, k
I
ˆ τ
r1
~t
J \I
−s
1
−s
~k
J \I
ˆ τ
r2
~t
I
−s
2
−s
~k
I
2.69
≤ C
ψ
¯t + 1
r −3
t −2ǫ
X
• s
=0 T log T
X
• s
1
,s
2
=2 ǫ
s
1
∨s
2
2ǫ
δ
s
1
,s
2
+ β
T
b
ǫ
s
1
,s
2
+ b
ǫ
s
2
,s
1
≤ OT
r −2−κ
.
Fix ~k with P
r −1
i=1
|k
i
|
2
bounded. To abbreviate the notation, we now write ~k
T
= ~k p
σ
2
T
T . By 2.28 and 1.2,
ˆ ψ
2 ǫ,2ǫ
~k
T
J \I
,~k
T
I
− ˆ ψ
2 ǫ,2ǫ
0, 0 = O ǫ|k|
2
T
−1
, 2.70
and, by 2.28 and the fact that λ
T
= 1 + OT
−µ
, ˆ
ψ
2 ǫ,2ǫ
0, 0 = λ
T
ǫ2 − ǫ = ǫ2 − ǫ + OǫT
−µ
. 2.71
As a result, we obtain that ˆ
τ
r
~t
~k
T
= ǫ2 − ǫ
t −2ǫ
X
• s
=0
ˆ τ
2
s
k
T
X
∅6=I⊂J
1
ˆ τ
r1
~t
J \I
−2ǫ−s
~k
T
J \I
ˆ τ
r2
~t
I
−2ǫ−s
~k
T
I
+ OT
r −2−µ
+ O|k|
2
T
r −3
. 2.72
The remainder of the argument can now be completed as in 2.47–2.49, using the induction hypothesis in 2.66 instead of the one in 2.43.
2.5 The continuum limit
