Using that we have a family of functions Fk fulfilling the conditions (F), we want to prove that the operator A, given by (8.1), is localized in a shifted left wedge. This means that we have to prove that the commutator [A, φ0(g)] vanishes if g has support in the corresponding right wedge. To show this, we use Prop. 8.4, and the idea for the proof is as follows. Due to properties (F1) and (F2) (see Prop. 3.7), we have fm,n[A](θ,η) = Fm+n(θ+i0,η +iπ −i0); then, it follows that fm+1,n[A] (θ, ξ +iπ,η) = fm,n+1[A] (θ, ξ,η). In matrix elements between vectors of finite particle number we can compute directlyBg−,ξ+iπ = (Bg+,ξ)∗, using also thatg has compact support. Inserting this into Eq. (8.10), we see that [A, φ0(g)] vanishes if it is possible to shift the integration contour inξ from R toR+iπ.
The fact that we can shift the integration contours depends on the growth behaviour of the analytic functions involved, namely Fk and g, and therefore it depends on the localization regions of A and g. So, first we study the growth behaviour of these functions.
Hence, we define for fixed m, n ∈ N0, f ∈ D(Rm+n), q ∈ N0, and ν ∈ Rq, the function
K(ξ) :=e−iµrsinhξYq
j=1
S(ξ−νj)Z
dmθdnη f(θ,η)Fm+n+1(θ+i0, ξ,η+iπ−i0).
(8.16) Since Fk fulfils the conditions (F), this K is analytic for ξ ∈ S(0, π), with boundary values which are distributions. We also define for fixedg ∈ D(R2) the function,
h(ξ) :=eiµrsinhξg−(ξ). (8.17) Since the support of g is compact, this function is entire analytic.
Now, we want to prove the following lemma:
Lemma 8.5. If the Fk fulfil (F1) and (F6), then there exist c, c0 >0 such that
|K(ξ+iλ)| ≤ c ec0ω(coshξ)
(λ(π−λ))(m+n)/2. (8.18)
Proof. We introduce h := min(λ, π−λ)/(m+n+ 1) and νL := (1,2, . . . , m), νR :=
(n, . . . ,2,1). For fixed ξ, λ,θ,η, we define
G(z) :=e−iµrsinhξFm+n+1(θ+zhνL, ξ+iλ,η+iπ−zhνR). (8.19) We can show that whenz ∈R+i(0,1), then (θ+zhνL, ξ+iλ,η+iπ−zhνR)∈ichG+k, cf. Fig. 8.1. That is, we have to show:
0<Imzh < . . . < mhImz < λ < π−nhImz < . . . < π−hImz < π. (8.20) Obviously, we have Imzh >0,π−hImz < π, Imzh < . . . < mhImzandπ−nhImz <
. . . < π−hImz. It remains to show that mhImz < λ < π−nhImz. By definition h ≤ m+n+1λ , hence mhImz ≤ mm+n+1λ ·1 = m+n+1m λ < λ. By definition we have also h≤ m+n+1π−λ , hence π−nhImz ≥π−m+n+1(π−λ)n Imz ≥π−(π−λ) Imz ≥π−(π−λ) =λ.
8.4. LOCALIZATION IN A LEFT WEDGE
λ
h
Im(θ+zh, ξ+iλ) Imz = 1
Imz = 0
Dt ichG+2
t
Figure 8.1: The computation of the pointwise bound (8.18). Dt refers to the disc introduced in the proof of Prop. C.3
Since the function Fm+n+1 is analytic in ichG+k, we have that the function G is analytic in z ∈ R+i(0,1); for the imaginary part of the argument of Fm+n+1, we can show
dist (hImz, . . . , mhImz, λ, π−nhImz, . . . , π−hImz), ∂ichG+k
≥ 1
√2hImz.
(8.21) The proof of (8.21) works as follows. Note that ∂ichG+k = {λi = λi+1 for some i} ∨ {λ1 = 0} ∨ {λk = π}. We compute (8.21) directly, noting that dist(·) denotes the euclidean distance in Rk. Obviously, we have:
dist (hImz, . . . , mhImz, λ, π−nhImz, . . . , π−hImz),{λ1 = 0}
=p
(hImz−0)2+ (2hImz−λ2)2+. . .≥hImz, (8.22) and,
dist (hImz, . . . , mhImz, λ, π−nhImz, . . . , π−hImz),{λk =π}
=p
. . .+ (π−2hImz−λk−1)2+ (π−hImz−π)2 ≥hImz. (8.23) So, it remains to show
dist (hImz, . . . , mhImz, λ, π−nhImz, . . . , π−hImz),{λi =λi+1 for some i}
≥ 1
√2hImz. (8.24) By direct computation, we find
dist (hImz, . . . , mhImz, λ, π−nhImz, . . . , π−hImz),{λi =λi+1 for some i}
=p
. . .+ (hiImz−λi)2+ (h(i+ 1) Imz−λi)2+. . .. (8.25)
We callµ:=hiImz−λi, h(i+ 1) Imz−λi =µ+hImz,a :=hImz. We compute the This concludes the proof of (8.21).
Then, since the argument of Fm+n+1 is a point in the interior of T(G+k), we can apply condition (F6), which gives, for anyR > 0, a constantscR (dependent onR, but not onξ, λ) and c0 such that
|G(z)| ≤cRec0ω(coshξ)(h|Imz|)−k/2 for all z∈(−R, R) +i(0,1), kθk< R, kηk< R.
(8.27) Following, for example, [BF09, Prop. 4.2], we compute a bound for the boundary distribution We prove (8.29) using induction on `. For ` = 0 it follows directly from (8.27). Now assume that (8.29) is true for `−1 in place of `; we prove it for `: above along this curve, we find:
r.h.s.(8.30) =c`−1R ec0ω(coshξ)h−k/2
Choosing ` > k/2 + 1/4 and inserting (8.29) in (8.28), we find
r.h.s. (8.28)≤c0(`)R ec0ω(coshξ)kg(`)k1h−k/2. (8.32)
8.4. LOCALIZATION IN A LEFT WEDGE where the constantcg,Rmight depend on the test functiong ∈ D(−R, R) and the cutoff R, but not onG (and hence not on ξ, λ,θ,η).
Now, we recall (8.16) and (8.19), and we compute
|K(ξ+iλ)| where we have estimated the factors S(ξ − νj +iλ) by 1, since they are bounded functions on the strip S(0, π).
In order to apply (8.33), we perform in the last line of the equation above a coor-dinate transformation t:C×Rm+n−1 →Cm and u:C×Rm+n−1 →Cn, such that we can rewrite that integral as
|K(ξ+iλ)| ≤ where we used that f has compact support in a ball of radius R, and where the last integral is finite; we call c:= (3(m+n+ 1))k/2R
supp ˜fdρ cf ,ρ,R˜ . This gives the result in Lemma 8.5.
Lemma 8.6. Letm, n∈N0. IfFm+nfulfils (F1) and (F6), then there exists an analytic indicatrix ω0 ≥ω such that for all f ∈ D(Rm+n) and g ∈ Dω0(Wr),
Z
K(ξ+i0)h(ξ)dξ = Z
K(ξ+iπ−i0)h(ξ+iπ)dξ, (8.37) where f, g enter the definitions of K, h given by Eq. (8.16) and Eq. (8.17).
Proof. We set ω0(p) := aω(c0ω(p) + m+n+62 log(1 +p)), with c0 as in Lemma 8.5, and aω0 =aω ≥1. This is a valid indicatrix, see Example 1 in Sec. 2.5. We will show below that
Z
K(ξ+i0)h(ξ)dξ = lim
&0
Z
K(ξ+i)h(ξ+i)dξ, (8.38) and similarly for the upper boundary. We can show that for fixed and with some c >0,
∀λ ∈[, π−] :|h(ξ+iλ)K(ξ+iλ)| ≤ c
(1 + coshξ)(m+n+4)/2. (8.39) Indeed, on the strip λ ∈ [, π − ] we can estimate the denominator in (8.18) by (λ(π −λ))(m+n)/2 ≥ m+n, and we can set c := 1/m+n; moreover, using Lemma 8.5 and Prop. 2.6, we have
|h(ξ+iλ)K(ξ+iλ)| ≤cec0ω(coshξ)e−ω0(coshξ)/aω0
≤cexp
c0ω(coshξ)− aω(c0ω(coshξ) + m+n+42 log(1 + coshξ)) aω0
≤cexp
− m+n+ 4
2 log(1 + coshξ)
=c(1 + coshξ)−(m+n+4)/2, (8.40) where we have setaω =aω0.
So, on the strip λ∈[, π−], where we have shown above that the function h·K is analytic and decays fast in real direction, we can apply Cauchy’s formula and we get,
∀ >0 : Z
dξ K(ξ+i)h(ξ+i) = Z
dξ K(ξ+iπ−i)h(ξ+iπ−i) (8.41) Now, to conclude the result (8.37) from this, it remains to show Eq. (8.38). Namely, we need to show that
lim&0
Z
dξ K(ξ+i) h(ξ)−h(ξ+i)
= 0. (8.42)
We denote K(−`) the `-th antiderivative of K, with
∂
∂ζ
j
K(−`)(iπ2) = 0, 0 ≤ j < `.
We will show
|K(−`)(ξ+iλ)| ≤c ec0ω(coshξ)(1 +|ξ|)`[λ(π−λ)](m+n)/2−`−1/4
, ` <(m+n)/2. (8.43) We prove (8.43) using induction on `. For ` = 0, it follows directly from the bound of Lemma 8.5. Now assume that (8.43) is true for`−1 in place of `, we prove it for `.
8.4. LOCALIZATION IN A LEFT WEDGE
We integrateK(−`+1) along the lines from iπ/2 to ξ+iπ/2 and then from ξ+iπ/2 to ξ+iλ. By induction hypothesis, we find
Z ξ+iλ of ω, i.e. (ω1). This concludes the proof of (8.43).
For` >(m+n)/2, we find by repeated integration, with somec00>0 and for allλ,
|K(−`)(ξ+iλ)| ≤c00(1 +|ξ|)`ec0ω(coshξ). (8.45) Using integration by parts and the bound (8.45), we find
lim&0 if we can show that the integral in the last line is finite. To that end, using the bounds on g− from Prop. 2.6 and the definition of ω0 after Eq. (8.37), we have that for all
where we usedc0 := 1/aω0. Inserting in (8.46), we find Z
dξ (1 +|ξ|)`ec0ω(coshξ) sup
0<λ<π
|h(`+1)(ξ+iλ)|
≤ Z
dξ (1 +|ξ|)`ec0ω(coshξ)e−c0ω(coshξ)(1 + coshξ)`−(m+n)/2−2
= Z
dξ (1 +|ξ|)`(1 + coshξ)`−(m+n)/2−2
. (8.48) This integral is finite if we choose m+n <2` ≤m+n+ 2 (where the left inequality is due to the condition ` >(m+n)/2 before (8.45) and the right inequality is due to the requirement that`−(m+n)/2−2<0).
Using this we can prove wedge-locality of A, as discussed in the beginning of this section.
Proposition 8.7. Let Fk be a sequence of functions fulfilling (F1), (F2), (F5) and (F6), and let A be as in (8.1). Then A isω-local in the wedge Wr0.
Proof. We have that A, given by (8.1), is well-defined by properties (F1), (F5) and we havefm,n[A](θ,η) =Fm+n(θ+i0,η+iπ−i0) by (F2). This is due to an application of Prop. 3.7.
Now, by application of Lemma 4.3(iii), it suffices to show that hψ,[A, φ0(g)]χi = 0 for fixed ψ, χ ∈ Hω,f, and for all g ∈ Dω0(Wr), where the indicatrix ω0 is chosen in a suitable way.
We can assume thatψ, χhave fixed particle number and compact support in rapidity space. This is possible because for anyψ ∈ Hω,f, more specificallyψ ∈ Hω∩ Hn, there existsψn∈ Hn∩ D(Rn), such thatkexp(ω(H/µ))(ψ−ψn)k →0 forn→ ∞. Moreover, letψn→ψ, χn→χfor n→ ∞in the sense of the above norm, thenhψn,[A, φ0(g)]χni converges to hψ,[A, φ0(g)]χi, since [A, φ0(g)] is a well-defined element of Qω.
We use Prop. 8.4, and we consider a summand in Eq. (8.10) with fixed m, n, since we can compare in this equation only terms with the same number of creators z† and annihilatorsz. It suffices to show that if g ∈ Dω0(Wr), for fixed q ∈N0, we have
Z
dmθdnη Z
dξ f(θ,η)
Fm+n+1(θ+i0, ξ+iπ−i0,η+iπ−i0)(Bqg+,ξ)∗
−Fm+n+1(θ+i0, ξ+i0,η+iπ−i0)Bqg−,ξ
= 0. (8.49) We use the definitions (2.100), (8.16) and (8.17), and we rewrite (8.49) as follows:
Z
K(ξ+i0)h(ξ)dξ = Z
K(ξ+iπ−i0)h(ξ+iπ)dξ. (8.50) This is given by Lemma 8.6; moreover, here we used the following fact: By Proposition 2.6 the Fourier transformg−of a functiong ∈ Dω0(Wr) extends to an analytic function on the strip S(0, π) with boundary value g−(ξ +iπ) = g+(ξ). Recalling that the