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 where k ≥ 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 provided DI< 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 subspace I R
n with dim(I)≤ k.
Now let I 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 then k. 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 if DS2(f ) < 0 where
DS2(f ) = max
S00S2[α(S
00) + dimS00]. (5.122)
(b1) If DS2(f ) < 0 then fS2 ∈ An−k2, k2 = dimS2, with
αS2(S
0) = max Λ(S2)S00=S0
[α(S00) + dimS00− dimS0]. (5.123)
(a2) If fS2 ∈ An−k2 cf. (b1) then fI converges absolutely if
DS1(fS2) = max
S0S1[αS2(S
0) + dimS0] < 0. (5.124)
(b2) If fS2 ∈ An−k2 and DS1(fS2) < 0 then fI ∈ An−k−1 with
αI(S) = max Λ(S1)S0
[αS2(S
0) + dimS0− dimS]. (5.125)
(a1) and (a2) together mean that fI is absolutely convergent if
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 S
2 or such subspaces of R
n 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 and ~L2∈ S2 such that ~V = ~L1+ ~L2. So we have Λ(S2)~V = ~L1∈ S1 and thus
Λ(S2)S0 S1. Thus since S00 in max∗ runs over all subspaces of S2 or such subspaces ofR
n
for which Λ(S2)S00 S1 in fact S00 runs over all subspaces of I and thus
DI0(f ) = max
S00I[α(S
00) + dimS00] = D
I(f ) (5.128)
Thus we have shown claim (a) of the theorem for fI.
To show (b) out of the induction hypothesis we have to combine (b1) and (b2) which imme- diately show that fI ∈ 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− dim S]. (5.129)
To complete the induction step we have just to show that we can combine the both conditions for S00 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)
Let S = span{~L1, . . . , ~Lr} then S = Λ(S1)S0means that in the above maximum we can restrict
S0 to such subspaces for which exist ~L0
1, . . . , ~L0r ∈ S1such that S0 = span{~L1+~L01, . . . , ~Lr+~L0r}.
The same argument gives S00 = span{~L1+ ~L01+ ~L001, . . . , ~Lr+ ~L0r+ ~L00r} if Λ(S2)S00 = S0 with
vectors ~L00
1, . . . , ~L00r ∈ S2. But since S1 and S2 are disjoint subspaces this is nothing else than
the statement that S00 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 set I = span{~L} and look on f~L( ~P ) = Z ∞ −∞ f ( ~P + y~L). (5.131) Since f ∈ An by definition f ( ~P + ~Ly) ∼= y→∞O yα(I)(ln y)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 of I is I 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 function f ∈ An
over a one-dimensional subspace I = span{~L} is of class An−1 provided DI(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 function fI defined by (5.116).
Let span{~L1, . . . , ~Lm} R
n where the ~L
j for j = 1, . . . , m build a linearly independent set of
vectors which are also linearly independent from ~L, and W ⊆R
n compact. Then we have to
show that f~L( ~P ) ∼= η1,...,ηm→∞ O m Y k=1 ηα~L(span{~L1,...,~Lk}) k (ln ηk)βL~( ~L1,...,~Lk) ! (5.134) with the asymptotic coefficients
α~L(S) = max Λ(I)S0=S[α(S
0) + 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
~
L1+ u1L, ~~ L2+ u2L, . . . , ~~ Lr+ ur~L, ~L, ~Lr+1, . . . , ~Lm (5.136)
with 0 ≤ r ≤ m and u1, . . . , ur ∈R.
Now we have to write out what it means that f ∈ Anfor this set of vectors: There exist num-
bers bl(u1, . . . , ur) > (0 ≤ l ≤ r) and M(u1, . . . , ur) > 0 such that for all ηl > bl(u1, . . . , ur)
and ~C ∈ W the function fulfils the inequality:
|f[(~L1+ ur~L)η1· · · ηmη0+· · · + (~Lr+ urL)η~ r· · · ηmη0+ ~Lr+1ηr+1· · · ηm+· · · + ~Lmηm+ ~C]| ≤ M(u1, . . . , ur) r Y k=1 ηα(span{~L1+u1L,...,~~ Lk+ukL})~ k (ln ηk) β(~L1+u1~L,...,~Lk+ukL)~ × ×ηα(span{~L1,...,~Lr,~L}) 0 ln(η0)β(~L1+u1 ~ L,...,~Lr+urL,~~ L)× ×ηα(span{~L1,...,~Lr+1,~L}) r+1 (ln ηr+1)β(~L1+u1L,...,~~ Lr+1+ur+1~L,~L)× · · · × ×ηα(span{~L1,...,~Lm,~L}) m (ln ηm)β(~L1+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
−1
1 (Ui) such that
[
i∈J1
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, Uij + λij] (j ∈ J2, J2 finite index set) with
0 < λij ≤ b−12 (Ui, Uij). This construction we can continue and find m 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) with r≤ m (5.140) such that [ ir∈Jr
[Ui1...ir − λi1...ir, Ui1...ir + λi1...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 define J±(η) to consist of all y with
±y = |y| ≥ b0η1· · · ηm. (5.145)
Finally we also define J0
i1...im to consist of all y 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 an i1∈ J1
such that y∈ [η1· · · ηm(Ui1 − λi1), η1· · · ηm(Ui1+ λi1)]. (5.147) So we can write y = η1· · · ηmUi1 + y 0 with|y0| ≤ η 1· · · ηmλi1. (5.148) If ±y0 ≥ η
2· · · 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+ y 00 | {z } y0 . (5.149)
It may happen that we have to continue this process until the final alternative, which then leads to y∈ J0
i1···im(η). Thus y must indeed be in one of the sets J
±
i1...im(η), J
±(η) or J0 i1...im.
Thus we have found an upper boundary for |f~L( ~P )| given by |fL~( ~P )| ≤ X ± m X r=0 X i1...im Z J± i1...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+· · · + ~Lmηm+ ~C, ~C∈ W (5.151)
where W ⊆R
n is a compact region.
(i) Let y∈ Ji±1...ir(η)
According to the definition of Ji±1...ir(η) we can combine (5.151) with (5.143) to ~ P + ~Ly = (~L1+ Ui1L)η~ 1· · · ηm+ (~L2+ Ui1i2L)η~ 2· · · ηm+· · · + +(~Lr+ Ui1...irL)η~ r· · · ηm+ z~Lηr+1· · · ηm+ ~Lr+1ηr+1· · · ηm+· · · + ~Lmηm+ ~C = = (~L1+ 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+ ~Lr+1ηr+1· · · ηm+· · · + ~Lmηm+ ~C (5.152) Now we define
α(i1, . . . , il) = α(span{~L1+ 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 to η0=|z|
with which we find that for
ηl > bl(i1, . . . , ir) for l6= r (5.154)
the following boundary condition is valid:
|f( ~P + ~Ly)| ≤ M(i1, . . . , ir)η1α(i1)(ln η1)β(i1)· · · ηα(ir−11,...,ir−1)(ln ηr−1)β(i1,...,ir−1)×
×ηα(span{~L1,...,~Lr+1,~L}) r+1 (ln ηr+1)β(~L1,...,~Lr+1,~L)· · · ηmα(span{~L1,··· ,~Lm,~L})(ln ηm)β(~L1,··· ,~Lm,~L)× × ηr |z| α(i1,...,ir) ln ηr |z| β(i1,...,ir) |z|α(span{~L1,...,~Lr,~L})(ln|z|)β(~L1,...,~Lr,~L). (5.155) Now introducing z according to (5.152) as the integration variable we find by using the boundary condition defining Ji±1...ir(η) cf. (5.144) we find
Z
Ji1...ir± (η)
dy|f( ~P + y~L)| ≤ M(i1, . . . , ir)ηα(i1 1)(ln η1)β(i1)· · · ηα(ir−11,...,ir−1)(ln ηr−1)β(i1,...,ir−1)×
×ηα(span{~L1,...,~Lr+1,~L}) r+1 (ln ηr+1)β(~L1,...,~Lr+1,~L)· · · ηmα(span{~L1,··· ,~Lm,~L})(ln ηm)β(~L1,··· ,~Lm,~L)× ×ηr+1· · · ηm× × Z ηrλi1···ir b0(i1,...,ir) d|z||z|α(span{~L1,...,~Lr,~L})(ln|z|)β(~L1,...,~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
Lemma 1. Let 0 < λ < 1, b > 1, α, α0 ∈R and β, β
0 ∈
N>0. Then the following holds:
Z λη b dz η z αh ln η z iβ zα0ηβ0 ∼= η→∞ ηα(ln η)β+β0+1 for α0+ 1 = α ηα(ln η)β for α0+ 1 < α ηα0+1(ln η)β0 for α0+ 1 > α. (5.157)
We give the elementary proof of this lemma at the end of the section. From (5.157) we have with (5.156)
Z
Ji1...ir± (η)
dy|f( ~P + y~L)| ≤ M(i1, . . . , ir)ηα(i1 1)(ln η1)β(i1)· · · ηα(ir−11,...,ir−1)(ln ηr−1)β(i1,...,ir−1)×
×ηα(span{~L1,...,~Lr+1,~L}) r+1 (ln ηr+1)β(~L1,...,~Lr+1,~L)· · · ηmα(span{~L1,··· ,~Lm,~L})(ln ηm)β(~L1,··· ,~Lm,~L)× × ηα(i1,...,ir)
r (ln ηr)β(i1,...,ir)+β(~L1,...,~Lr,~L)+1 if α(i1, . . . , ir) = α(span{~L1, . . . , ~Lr, ~L}) + 1
ηα(i1,...,ir)
r (ln ηr)β(i1,...,ir) if α(i1, . . . , ir) > α(span{~L1, . . . , ~Lr, ~L}) + 1
ηα(span{~L1,...,~Lr,~L})+1
r (ln ηr)β(~L1,...,~Lr,~L) if α(i1, . . . , ir) < α(span{~L1, . . . , ~Lr, ~L}) + 1
(5.158) whenever ηl > bl(i1, . . . , ir) for l 6= r and ηr > c(i1, . . . , ir) because cf. (5.144) the integral
over the interval Ji±1...ir(η) contributes only if
ηr ≥
b0(i1, . . . , ir)
λi1...ir
≤ b0(i1, . . . , ir)br(i1, . . . , ir). (5.159)
Now we look on the infinite intervals J±(η). By definition (5.145) of these intervals we find Z J±(η) dy|f( ~P + y~L)| = Z ∞ b0η1···ηm dy|f( ~P± y~L)|. (5.160) Substitution of y = zη1· · · ηm and (5.151) for ~P gives
Z J±(η)|f( ~ P + y~L)|dy = η1· · · ηm Z ∞ b0 dy|f(±~Lzη1· · · ηm+ ~L1η1· · · ηm+· · · + ~Lmηm+ ~C)| (5.161) and now we can apply (5.137) for r = 0 and η0 = z:
Z J±(η) dy|f( ~P + y~L)| ≤ Mηα(span{~L1,~L}) 1 (ln η1)β(~L1,~L)· · · ηmα(~L1,...,~Lm,~L)(ln ηm)β(~L1,...,~Lm,~L)× × Z ∞ b0 zα(span{~L})(ln z)β(~L). (5.162) The integral in the last row is a finite constant since by hypothesis α(span{~L}) + 1 < 0. Now we are left with the interval Ji01...im. With its definition (5.146) and (5.151) for ~P we have for these intervals
~
where|z| ≤ b0(i1, . . . , im). Now the region
R0 ={ ~C0| ~C0 = ~Lz + ~C, |z| ≤ b0(i1, . . . , im), ~C ∈ W } R
n (5.164)
is compact. Because f ∈ An by hypothesis there exist numbers M0(i1, . . . , im) > 0 and
b0l(i1, . . . , im) > 1 such that
|f[(~L1+ Ui1L)η~ 1. . . ηm+· · · + (~Lm+ Ui1...imL)η~ m+ ~C
0]| ≤
≤ M0(i1, . . . , im)η1α(i1)(ln η1)β(i1)· · · ηmα(i1,...,im)(ln ηm)β(i1,...,im)
(5.165) for ~C0 ∈ R0 and η l> b0l(η1, . . . , ηm). Thus Z Ji1...im(η) dy|f( ~P + ~Ly)| ≤ 2b0(i1, . . . , im)M0(i1, . . . , im)× ×ηα(i1)
1 (ln η1)β(i1)· · · ηmα(i1,...,im)(ln ηm)β(i1,...,im)
(5.166)
if ηl≥ 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 for fL~( ~P ) we
read off that this function is indeed contained in the class An−1and the asymptotic coefficients
are given by
αL~(span{ ~L1, ~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{ ~L1, ~L2, . . . , ~Lr}) =
= max
u1,...,ur
[α(span{~L1+ u1L, . . . , ~~ Lr+ ur~L}), α(span{~L1, . . . ~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 R
n 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, . . . , ur∈R 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 span~L 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