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Proof of Weinberg’s theorem

5.7 Weinberg’s Theorem

5.7.1 Proof of Weinberg’s theorem

The proof of Weinberg’s theorem is obtained in three steps:

(A) Step of induction: We show that if the theorem is true for dimI ≤k wherek≥1 then it is true for dimI =k+ 1.

(B) Convergence for dimI = 1: We prove that the integral over an arbitrary one-dimensional subspace I is absolutely convergent providedDI<0.

(C) Covering the one-dimensional subspace I with a finite number subintervals J each of which contributes a certain asymptotic behaviour of the integral.

(D) The sum over the subintervals constructed in (C) is a function of the class An−1 with αI given by (5.118).

The convergence theorem, i.e., conjecture (a) of the theorem is proven with (A) and (B) which is the simple part of the proof.

Ad (A): Let the theorem be true for any subspaceI Rn with dim(I)≤k.

Now letI be a subspace with dimI =k+ 1. Let S1 and S2 be arbitrary non-trivial disjoint subspaces of I with I = S1 ⊕S2 (there are arbitrary many such splittings of I!). Then necessarily dimS1+ dimS2 = dimI =k+ 1 and because S1 and S2 are both non-trivial they are both of a dimension less thenk. By the induction hypothesis the theorem holds for these subspaces. Thus we apply it twice making use of Fubini’s theorem:

(a1)fS2 converges absolutely ifDS2(f)<0 where DS2(f) = max

S00S2

[α(S00) + dimS00]. (5.122) (b1) IfDS2(f)<0 then fS2 ∈An−k2,k2 = dimS2, with

αS2(S0) = max

Λ(S2)S00=S0[α(S00) + dimS00−dimS0]. (5.123) (a2) IffS2 ∈An−k2 cf. (b1) then fI converges absolutely if

DS1(fS2) = max

S0S1

S2(S0) + dimS0]<0. (5.124) (b2) IffS2 ∈An−k2 and DS1(fS2)<0 then fI ∈An−k−1 with

αI(S) = max

Λ(S1)S0S2(S0) + dimS0−dimS]. (5.125) (a1) and (a2) together mean thatfI is absolutely convergent if

D0I(f) = max{DS2(f), DS1(fS2)}<0 (5.126)

5.7 · Weinberg’s Theorem

and this can be written as

D0I(f) = max

S00

[α(S00) + dimS00] (5.127) where max indicates that S00 has to be a subspace of S2 or such subspaces of Rn that Λ(S2)S00 S1. To show that DI0(f) = DI(f) we have to prove that this means that S00 is running in fact over all subspaces of I =S1⊕S2.

Let S0 I but not a subspace of S2. Then let V~ ∈ S0. Because I = S1⊕S2 there exist vectors~L1∈S1 andL~2∈S2 such thatV~ =L~1+~L2. So we have Λ(S2)V~ =L~1∈S1 and thus Λ(S2)S0 S1. Thus since S00 in max runs over all subspaces ofS2 or such subspaces ofRn for which Λ(S2)S00S1 in factS00 runs over all subspaces of I and thus

DI0(f) = max

S00I[α(S00) + dimS00] =DI(f) (5.128) Thus we have shown claim (a) of the theorem forfI.

To show (b) out of the induction hypothesis we have to combine (b1) and (b2) which imme-diately show thatfI ∈An−k−1. Thus we have only to prove (5.118). From (b1) and (b2) we know that

αI(S) = max

Λ(S1)S0=S

dimS0−dimS+ max

Λ(S2)S00=S0[α(S00) + dimS00−dimS0]

=

= max

Λ(S1)S0=S;Λ(S2)S00=S0[α(S00) + dimS00−dimS].

(5.129)

To complete the induction step we have just to show that we can combine the both conditions forS00 to Λ(I)S00=S.

Thus we have to show the equivalence

Λ(S2)S00=S0∧Λ(S1)S0 =S

⇔S = Λ(S1⊕S2)S00 (5.130) LetS = span{L~1, . . . , ~Lr}thenS = Λ(S1)S0means that in the above maximum we can restrict S0 to such subspaces for which existL~01, . . . , ~L0r ∈S1such thatS0 = span{L~1+L~01, . . . , ~Lr+L~0r}. The same argument givesS00 = span{L~1+~L01+L~001, . . . , ~Lr+L~0r+L~00r} if Λ(S2)S00 =S0 with vectorsL~001, . . . , ~L00r ∈S2. But since S1 and S2 are disjoint subspaces this is nothing else than the statement thatS00 runs over all subspaces with Λ(S1⊕S2)S00=S. This proves (A).

Ad (B):Now we like to show (a) for a one-dimensional subspace. Thus we setI = span{~L} and look on

f~L(P~) = Z

−∞

f(P~ +y~L). (5.131)

Sincef ∈An by definition

f(P~ +Ly)~ ∼=

y→∞O

yα(I)(lny)β(L)~

(5.132) This integral converges absolutely if α(L) + 1~ < 0 if it exists for any finite integral wrt. y.

Now the only non-trivial subset ofI isI itself because dimI = 1 and thus DI(f) =α(I) + 1 which proves part (a) completely.15

15The reader who intends just to understand the convergence conjecture (a) for renormalisation theory may stop here because this part is now completely proved. The rest of the section is devoted to the proof of the asymptotic behaviour for a one dimensional integration space.

Ad (C):The rest of the proof is devoted to show that the integral fI of a functionf ∈An over a one-dimensional subspace I = span{L~} is of classAn−1 providedDI(f)<0.

To show this we use the form (5.115), i.e. we have to study the asymptotic behaviour of fL~(P~) =

Z

−∞

dyf(P~ +y~L) (5.133)

which is the same as that of the functionfI defined by (5.116).

Let span{L~1, . . . , ~Lm} Rn where theL~j forj = 1, . . . , mbuild a linearly independent set of vectors which are also linearly independent fromL, and~ W ⊆Rn compact. Then we have to show that

f~L(P~) ∼=

η1,...,ηm→∞O Ym k=1

ηkα~L(span{~L1,...,~Lk})(lnηk)βL~(L~1,...,~Lk)

!

(5.134) with the asymptotic coefficients

α~L(S) = max

Λ(I)S0=S[α(S0) + dimS0−dimS]. (5.135) For this purpose we try to split the integration range R in (5.133) in regions of definite asymptotic behaviour of the integral. Since we have to find only an upper bound of the integral it is enough to construct a finite coverage which needs not necessarily to be disjoint.

We look on the set of vectors

L~1+u1L, ~~ L2+u2L, . . . , ~~ Lr+ur~L, ~L, ~Lr+1, . . . , ~Lm (5.136) with 0 ≤r≤m and u1, . . . , urR.

Now we have to write out what it means thatf ∈Anfor this set of vectors: There exist num-bersbl(u1, . . . , ur) > (0 ≤l≤r) and M(u1, . . . , ur) >0 such that for all ηl > bl(u1, . . . , ur) andC~ ∈W the function fulfils the inequality:

|f[(L~1+ur~L)η1· · ·ηmη0+· · ·+ (~Lr+urL)η~ r· · ·ηmη0+L~r+1ηr+1· · ·ηm+· · ·+L~mηm+C]~ |

≤M(u1, . . . , ur) Yr k=1

ηkα(span{~L1+u1L,...,~~ Lk+ukL})~ (lnηk)β(L~1+u1~L,...,~Lk+ukL)~ ×

×η0α(span{~L1,...,~Lr,~L})ln(η0)β(L~1+u1~L,...,~Lr+urL,~~ L)×

×ηα(span{r+1 L~1,...,~Lr+1,~L})(lnηr+1)β(~L1+u1L,...,~~ Lr+1+ur+1~L,~L)× · · · ×

×ηmα(span{~L1,...,~Lm,~L})(lnηm)β(L~1+u1~L,...,~Lr+urL,~~ Lr+1,...,~Lm,~L). (5.137) Now each u ∈ [−b0, b0] is contained in a closed interval [u−b−11 (u), u+b−11 (u)]. Because [−b0, b0] is compact in R by the Heine-Borel theorem one can find a finite set of points {Ui}i∈J1 (J1 a finite index set) such that |U1|< b0 and 0< λi ≤b−11 (Ui) such that

[

i∈J1

[Ui−λi, Uii] = [−b0, b0]. (5.138)

5.7 · Weinberg’s Theorem

Next take i∈J1 and the closed interval [−b0(Ui), b0(Ui) which can be covered again due to Heine-Borel with a finite set intervals [Uij −λij, Uijij] (j ∈ J2,J2 finite index set) with 0< λij ≤b−12 (Ui, Uij). This construction we can continue and findm finite sets of points

{Ui1}i1∈J1, {Ui1i2}i1∈J1,i2∈J2, . . . ,{Ui1...im}i1∈J1,...,im∈Jm (5.139) and numbers

0< λi1i2...ir ≤b−1r (i1, . . . , ir) withr≤m (5.140) such that

[

ir∈Jr

[Ui1...ir −λi1...ir, Ui1...iri1...ir]⊆[−b0(i1, . . . , ir−1), b0(i1, . . . , ir−1)]. (5.141) Here we have used the abbreviation

bl(i1, . . . , ir) =bl(Ui1, Ui1i2, . . . , Ui1i2...ir). (5.142) Now for given η1, . . . , ηm >1 we define intervals Ji±1...ir(η) with r ≤m which consist of all y which can be written as

y=Ui1η1· · ·ηm+Ui1i2η2· · ·ηm+· · ·+Ui1...imηm+zηr+1. . . ηm (5.143) where z is running over all values with

b0(i1, . . . , ir)≤ |z|=±z≤ηrλi1···ir. (5.144) For r= 0 we defineJ±(η) to consist of ally with

±y=|y| ≥b0η1· · ·ηm. (5.145) Finally we also defineJi01...im to consist of ally that can be written as

y=Ui1η1· · ·ηm+Ui1i2η2· · ·ηm+· · ·+Ui1...imηm+z with|z|< b0(i1, . . . , im). (5.146) Now we show that any y ∈R is contained in at least one of these intervals J. If y /∈J±(η) then by (5.145) we know that|y| ≤b0η1· · ·ηm and thus because of (5.138) it exists ani1∈J1

such that

y∈[η1· · ·ηm(Ui1 −λi1), η1· · ·ηm(Ui1i1)]. (5.147) So we can write

y=η1· · ·ηmUi1 +y0 with|y0| ≤η1· · ·ηmλi1. (5.148) If ±y0 ≥ η2· · ·ηmb0(i1) then y ∈ Ji±1(η) and we are finished. If on the other hand |y0| ≤ η2· · ·ηmb0(i1) by the same line of arguments using our construction below (5.138) we find a y00 with

y=η1· · ·ηmUi1 +Ui1i2η2· · ·ηm+y00

| {z }

y0

. (5.149)

It may happen that we have to continue this process until the final alternative, which then leads to y∈Ji01···im(η). Thusy must indeed be in one of the sets Ji±1...im(η), J±(η) or Ji01...im. Thus we have found an upper boundary for |f~L(P~)|given by

|fL~(P~)| ≤X

±

Xm r=0

X

i1...im

Z

Ji±1...ir(η)

dy|f(P~ +~Ly|+ X

i1...im

Z

Ji1···im(η)

dy|f(P~ +y~L)|. (5.150)

Ad (D): The final step is to determine the asymptotic behaviour of each term on the right hand side of (5.150) where P~ is given by

P~ =~L1η1· · ·ηm+~L2η2· · ·ηm+· · ·+L~mηm+C, ~~ C∈W (5.151) where W ⊆Rn is a compact region.

(i) Lety∈Ji±1...ir(η)

According to the definition of Ji±1...ir(η) we can combine (5.151) with (5.143) to P~ +~Ly= (L~1+Ui1L)η~ 1· · ·ηm+ (~L2+Ui1i2L)η~ 2· · ·ηm+· · ·+ +(~Lr+Ui1...irL)η~ r· · ·ηm+z~Lηr+1· · ·ηm+L~r+1ηr+1· · ·ηm+· · ·+~Lmηm+C~ =

= (L~1+Ui1~L)η1· · · ηr

|z||z|ηr+1· · ·ηm+ (~L2+Ui1i2L)η~ 2· · · ηr

|z||z|ηr+1· · ·ηm+· · ·+ +(~Lr+Ui1...irL)η~ r· · ·ηm± |z|~Lηr+1· · ·ηm+L~r+1ηr+1· · ·ηm+· · ·+L~mηm+C~

(5.152)

Now we define

α(i1, . . . , il) =α(span{L~1+Ui1L, ~~ L1+Ui1i2~L, . . . , ~Ll+Ui1...il~L}), 1≤l≤m (5.153) and apply (5.137) together with the definition (5.143), (5.144) and (5.140) applied toη0=|z| with which we find that for

ηl > bl(i1, . . . , ir) forl6=r (5.154) the following boundary condition is valid:

|f(P~ +~Ly)| ≤M(i1, . . . , ir1α(i1)(lnη1)β(i1)· · ·ηα(ir−11,...,ir−1)(lnηr−1)β(i1,...,ir−1)×

×ηα(span{r+1 L~1,...,~Lr+1,~L})(lnηr+1)β(~L1,...,~Lr+1,~L)· · ·ηα(span{m L~1,···,~Lm,~L})(lnηm)β(~L1,···,~Lm,~L)×

× ηr

|z|

α(i1,...,ir) ln

ηr

|z|

β(i1,...,ir)

|z|α(span{~L1,...,~Lr,~L})(ln|z|)β(L~1,...,~Lr,~L). (5.155) Now introducing z according to (5.152) as the integration variable we find by using the boundary condition definingJi±1...ir(η) cf. (5.144) we find

Z

Ji±

1...ir(η)

dy|f(P~ +y~L)| ≤M(i1, . . . , irα(i1 1)(lnη1)β(i1)· · ·ηα(ir−11,...,ir−1)(lnηr−1)β(i1,...,ir−1)×

×ηα(span{r+1 L~1,...,~Lr+1,~L})(lnηr+1)β(L~1,...,~Lr+1,~L)· · ·ηmα(span{~L1,···,~Lm,~L})(lnηm)β(~L1,···,~Lm,~L)×

×ηr+1· · ·ηm×

×

Z ηrλi1···ir

b0(i1,...,ir)

d|z||z|α(span{L~1,...,~Lr,~L})(ln|z|)β(L~1,...,~Lr,~L)× ηr

|z|

α(i1,...,ir) ln

ηr

|z|

β(i1,...,ir)

. (5.156) Because for our purposes an upper bound of the logarithmic coefficients β(L, . . . , ~~ Ll) is suf-ficient we assume without loss of generality that these are positive integers. Then we use the

5.7 · Weinberg’s Theorem

We give the elementary proof of this lemma at the end of the section. From (5.157) we have with (5.156) Now we look on the infinite intervalsJ±(η). By definition (5.145) of these intervals we find

Z

where|z| ≤b0(i1, . . . , im). Now the region

R0 ={C~0|C~0 =Lz~ +C,~ |z| ≤b0(i1, . . . , im), ~C ∈W} Rn (5.164) is compact. Because f ∈ An by hypothesis there exist numbers M0(i1, . . . , im) > 0 and b0l(i1, . . . , im)>1 such that

|f[(L~1+Ui1L)η~ 1. . . ηm+· · ·+ (L~m+Ui1...imL)η~ m+C~0]| ≤

≤M0(i1, . . . , im1α(i1)(lnη1)β(i1)· · ·ηmα(i1,...,im)(lnηm)β(i1,...,im) (5.165) forC~0 ∈R0 and ηl> b0l1, . . . , ηm). Thus

Z

Ji1...im(η)

dy|f(P~ +~Ly)| ≤2b0(i1, . . . , im)M0(i1, . . . , im

×η1α(i1)(lnη1)β(i1)· · ·ηα(im 1,...,im)(lnηm)β(i1,...,im)

(5.166)

ifηl≥bl(i1, . . . , im) (1≤l≤m).

Now our proof is finished by inspection of (5.158), (5.165) and (5.166) because of the result of the construction in part (C) cf. (5.150). From these estimates of upper bounds forfL~(P~) we read off that this function is indeed contained in the classAn−1and the asymptotic coefficients are given by

αL~(span{L~1, ~L2, . . . , ~Lr}) = max

i1,...,ir

[α(i1, . . . , ir), α(span{~L1, . . . , ~Lr, ~L) + 1] (5.167) where i1, . . . ,ir are running over the index sets defined in (5.153) and it remains to show that this is the same as given by (5.118) for our case of integrating over a one-dimensional subspace.

By definition (5.153) we can rewrite (5.167) by α~L(span{L~1, ~L2, . . . , ~Lr}) =

= max

u1,...,ur

[α(span{L~1+u1L, . . . , ~~ Lr+ur~L}), α(span{L~1, . . . ~Lr, ~L}) + 1]. (5.168) Here the u1, . . . , ur run over the sets Ui1, . . . , Ui1...ir defined above cf. (5.139) respectively.

This expression has to be compared with (5.118).

So let S = span{~L1, . . . , ~Lr} and S0 Rn such that Λ(span{L~}S0 = S. Then of course S0 could be either S⊕span{~L}or for each~v∈S there must exist a w~ ∈S0 and u∈Rsuch that

~v=w~ +u~L. In other words in this there are u1, . . . , urR such that

S0= span{~L1+u1L,~ · · ·, ~Lr+urL~}. (5.169) In the first case we have dimS0 = dimS+ 1 in the second we have dimS0 = dimS since S0 in (5.118) runs over all subspaces disjoint with spanL~ and this shows immediately the equivalence of (5.168) with (5.118) for the case of integration over the one-dimensional subspace span{~L} and this finally finishes the proof. Q.E.D.

5.7 · Weinberg’s Theorem